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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 119, pp. 1--12.\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-2009/119\hfil
 Unbounded upper and lower solution method]
{Unbounded upper and lower solution method for third-order
 boundary-value problems on the half-line}

\author[C. Bai, C. Li\hfil EJDE-2009/119\hfilneg]
{Chuanzhi Bai, Chunhong Li}  % in alphabetical order

\address{Chuanzhi  Bai \newline
Department of Mathematics, Huaiyin Normal University,
 Huaian, Jiangsu 223300, China}
 \email{czbai8@sohu.com}

\address{Chunhong  Li \newline
Department of Mathematics, Huaiyin Normal University, Huaian,
Jiangsu 223300, China.
 \newline Department of Mathematics, Yanbian
University, Yanji, Jilin 133002, China} 
\email{lichshy2006@126.com}

\thanks{Submitted April 4, 2009. Published September 27, 2009.}
\subjclass[2000]{34B15, 34B40} 
\keywords{Upper and lower solutions;
third-order boundary-value problem; \hfill\break\indent
 half-line; Nagumo-condition}

\begin{abstract}
 In this article, we prove the existence of unbounded upper and lower
 solutions of third-order boundary-value problems on the half-line.
 Here the Nagumo conditions play an important role in the nonlinear term
 involved in the second-order derivatives.
\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}

Boundary-value problems on the half-line arise naturally in the
study of radially symmetric solutions of nonlinear elliptic
equations and  various physical phenomena, see
\cite{a2,a3,a4,b1,g4,i1},
such as the theory of drain flows, plasma physics, unsteady flow of gas
through a semi-infinite porous media, in determining the
electrical potential in an isolated neutral atom.

 Recently an increasing interest in studying the existence of solutions
 and positive solutions to boundary-value problems for second-order
differential equations on the half-line is observed;  see for
example \cite{b2,e1,g1,g4,l1,l2,p1,y1,y2}. However,  to the best
knowledge of the authors, no work has been done for the third-order
boundary-value problems on the half-line. It is well known that the
study of third-order boundary-value problems is very important. For
finite interval, there are many results, see \cite{d1,g2,g3}. So it
is necessary to discuss the existence of the three-order
boundary-value problems on the half-line.

In this paper, we are concerned with the existence of solutions for
the following boundary-value problem on the half-line for
the third-order differential equation
\begin{equation}
\begin{gathered}
 u'''(t) + a(t) f(t, u(t), u'(t),
u''(t)) = 0, \quad  t \in (0, +\infty), \\
 u(0) = u'(0) = 0,  \quad \lim_{t \to +\infty} u''(t)  =:
 u''(+ \infty) =  0,
 \end{gathered} \label{e1.1}
\end{equation}
 where $a : (0, + \infty) \to (0, + \infty)$,
$f : [0 + \infty) \times \mathbb{R}^3 \to \mathbb{R}$ are continuous.
By using the upper and
lower solutions method, the authors present sufficient
conditions for the existence of unbounded solutions to  \eqref{e1.1}.

This paper is organized as follows. In section 2, some definitions
and lemmas are given. We establish an upper and lower solution
theory for  \eqref{e1.1} in section 3. Sufficient conditions are given
for the existence of solutions. In section 4,  we give an example
to demonstrate our main results.

\section{Preliminaries}

In this section, we introduce some necessary definitions and
preliminary results that will be used to prove our main results.
Let
\begin{align*}
 E = \big\{&x \in C^2[0, + \infty) :
\sup_{0 \leq t < + \infty} \frac{|x(t)|}{(1+t)^2} < + \infty, \;
\sup_{0 \leq t < + \infty} |x'(t)| < + \infty, \\
&\lim_{t \to + \infty} x''(t) \text{ exists } \big\},
\end{align*}
with the norm $\|x\| = \max \{\|x\|_1, \|x'\|_{\infty},
\|x''\|_{\infty}\}$, where
$\|x\|_1 = \sup_{t \in [0, +\infty)} |\frac{x(t)}{(1+t)^2}|$,
$\|x'\|_{\infty} = \sup_{t \in [0, +\infty)}|x'(t)|$,
$\|x''\|_{\infty} = \sup_{t \in [0, +\infty)}|x''(t)|$.
By  standard arguments, we can prove that $(E, \|\cdot\|)$ is
a Banach space.


\begin{definition} \label{def2.1} \rm
A function $\alpha \in E \cap
C^3(0, +\infty)$ is called a lower solution of  \eqref{e1.1} if
\begin{gather*}
\alpha'''(t) + a(t) f(t, \alpha(t), \alpha'(t),
\alpha''(t)) \geq 0,  \quad  t \in (0, + \infty), \\
\alpha(0) \leq 0, \quad  \alpha'(0) \leq  0,  \quad
\alpha''(+ \infty) \leq 0.
\end{gather*}
Similarly we define an upper solution $\beta \in E \cap C^3(0,
+ \infty)$ of  \eqref{e1.1} by reversing the above inequalities.
\end{definition}

\begin{remark} \label{rmk2.1} \rm
If
\begin{equation}
\alpha'(t) \leq \beta'(t),\quad\text{for every $t \in [0, + \infty)$},
\label{e2.1}
\end{equation}
then by integrating \eqref{e2.1} and using the boundary
conditions of Definition \ref{def2.1}, we can easily obtain that
 $\alpha(t) \leq \beta(t)$  for all $t \in [0, +\infty)$.
\end{remark}

\begin{definition} \label{def2.2}\rm
 Given a pair of upper and lower
solutions $\beta, \alpha \in E \cap C^3(0, +\infty)$ of \eqref{e1.1}
satisfying $\alpha'(t) \leq \beta'(t)$, $t \in [0, +\infty)$.
A continuous function
$f : [0, + \infty) \times \mathbb{R}^3 \to \mathbb{R}$
is said to satisfy the Nagumo condition with respect to the
pair of functions $\alpha, \beta$, if there exists a nonnegative
function $\phi \in C[0, + \infty)$ and a positive one
$h \in C[0,+ \infty)$ such that
\begin{equation}
|f(t, x, y, z)| \leq \phi(t) h(|z|), \label{e2.2}
\end{equation}
for all $0 \leq t < + \infty$,
$\alpha(t) \leq x \leq \beta(t)$,
$\alpha'(t) \leq y \leq \beta'(t)$, $z \in \mathbb{R}$ and
\begin{equation}
\int_0^{+\infty} \frac{s}{h(s)} ds = + \infty.   \label{e2.3}
\end{equation}
\end{definition}

The above Nagumo conditions provide a priori estimate for the
second-order derivative $u''$ of a class of the
solutions of problem \eqref{e1.1}.

Now we consider the following boundary-value problem for third-order
differential equation on the half-line:
\begin{equation}
\begin{gathered}
u'''(t) + \sigma(t) = 0, \quad  t \in (0, +\infty), \\
u(0) = u'(0) = 0,  \quad  u''(+ \infty) =  0,
\end{gathered} \label{e2.4}
\end{equation}
where $\sigma \in C[0, + \infty)$.


\begin{lemma} \label{lem2.1}
 Let $\sigma \in C[0, + \infty)$
and $\int_0^{\infty} \sigma(t) dt < \infty$.
Then $u \in C^2[0, + \infty) \cap C^3(0, + \infty)$ is a solution
of \eqref{e2.4} if and
only if $u$ is a solution of the following integral equation:
\begin{equation}
u(t) = \int_0^{\infty} G(t, s)
\sigma(s) ds, \quad t \in  [0, + \infty),  \label{e2.5}
\end{equation}
where
\[
 G(t, s) = \begin{cases}
\frac{1}{2} s(2t - s), & 0 \leq s \leq t, \\
\frac{1}{2} t^2, & t \leq s < \infty.
\end{cases}
\]
\end{lemma}

\begin{proof}
 It is easy to show that the general
solution for the equation in boundary-value problem \eqref{e2.4} is
\begin{equation}
u(t) = -\frac{1}{2} \int_0^t (t-s)^2
\sigma(s) ds + A t^2 + B t + C, \quad  t \in [0, + \infty),
\label{e2.6}
\end{equation}
 where $A, B, C$ are constants. By the boundary condition of
\eqref{e2.4}, we get
$A = \frac{1}{2} \int_0^{\infty}
\sigma(s) ds$, $B = C = 0$.
Substituting the expressions of $A, B$ and $C$ into
\eqref{e2.6}, we know that \eqref{e2.5} holds.
\end{proof}

Let $C_l := \{y \in C[0, +\infty) : \lim_{t \to + \infty} y(t)
\text{ exists } \}$. For $y \in C_l$, define $\|y\| := \sup_{t \in
[0, + \infty)} |y(t)|$. Then $C_l$ is a Banach space (see
\cite{a1}).

\begin{lemma}[\cite{c1,m1}] \label{lem2.2}
 Let $M \subset C_l$.
Then $M$ is relatively compact if the following conditions hold:
\begin{itemize}
\item[(a)]  $M$ is bounded in $C_l$;

\item[(b)] the functions belonging to $M$ are locally
equicontinuous on  $[0, + \infty)$;

\item[(c)] the functions from $M$ are equiconvergent; that is,
given $\epsilon > 0$, there corresponds  $T(\epsilon)> 0$ such
that  $ |x(t) - x(+\infty)| < \epsilon$ for all $t > T(\epsilon)$
 and $x \in M$.
\end{itemize}
\end{lemma}

By Lemma \ref{lem2.2}, similar to the proof of \cite[Theorem 2.2]{l2}, we easily
obtain the following result.


\begin{lemma} \label{lem2.3}
Let $M \subset E$. Then $M$ is
relatively compact if the following conditions hold:
\begin{itemize}
\item[(i)]  $M$ is bounded  in $E$;

\item[(ii)] the functions belonging to
$\{y : y = \frac{x}{(1+t)^2}, x \in M  \}$,
$\{z : z = x'(t), x \in M\}$,
 and $\{w : w = x''(t), x \in M\}$
 are locally equicontinuous on  $[0, + \infty)$;

\item[(iii)]  the functions from
$ \{y : y = \frac{x}{(1+t)^2}, x \in M  \}$,
$\{z : z = x'(t), x \in M\}$,
 and $\{w : w = x''(t), x \in M\}$
 are equiconvergent at $+ \infty$.
\end{itemize}
\end{lemma}

\section{Main result}

In this section, we study the existence of solution
to \eqref{e1.1}.

\begin{theorem} \label{thm3.1}
Assume  that there are
$\alpha, \beta \in  E \cap C^3(0, +\infty)$ lower and upper
solutions of \eqref{e1.1}, respectively, such that \eqref{e2.1} holds.
Let $f : [0, + \infty) \times \mathbb{R}^3 \to \mathbb{R}$
be a continuous function
satisfying the Nagumo condition with respect to the pair of
functions $\alpha, \beta$, and verifying
\begin{equation}
f(t, \alpha(t), y, z) \leq f(t, x, y, z)
\leq f(t, \beta(t), y, z),   \label{e3.1}
\end{equation}
for $(t, x, y, z) \in [0, + \infty) \times [\alpha(t), \beta(t)]
\times \mathbb{R}^2$.  If
\begin{equation}
{\int_0^{+\infty} \max\{s, 1\} a(s)ds <
+ \infty, \quad \int_0^{+\infty} \max \{s, 1\} a(s) \phi(s) ds < +
\infty,} \label{e3.2}
\end{equation}
then \eqref{e1.1} has at least one solution
$u \in E  \cap  C^3(0, + \infty)$ satisfying
\[
\alpha(t) \leq u(t) \leq   \beta(t), \quad
\alpha'(t) \leq u'(t) \leq \beta'(t), \quad
|u''(t)| \leq N \quad \text{for all }
t \in [0, +  \infty),
\]
where $N$ is a constant dependent only on
$\alpha, \beta, a$ and $\phi$.
\end{theorem}

\begin{proof}
 Define  the auxiliary functions
\[
\omega_0(t, x) = \begin{cases}
\alpha(t), &  x < \alpha(t), \\
x(t), &  \alpha(t) \leq x \leq \beta(t), \\
\beta(t), &  x > \beta(t);
\end{cases}
 \quad  \omega_1(t, y) = \begin{cases}
\alpha'(t), &  y < \alpha'(t), \\
 y(t), &  \alpha'(t) \leq y \leq \beta'(t), \\
\beta'(t), &  y > \beta'(t).
\end{cases}
\]
Consider the boundary-value problem
\begin{equation}
\begin{gathered}
u'''(t) + a(t) f^*(t, u(t), u'(t), u''(t)) = 0, \quad  t \in
(0, +\infty), \\
 u(0) = u'(0) = 0,  \quad \lim_{t \to +\infty} u''(t) =  0,
 \end{gathered}   \label{e3.3}
\end{equation}
where
\begin{equation}
f^*(t, x, y, z) =  f(t, \omega_0(t,x), \omega_1(t, y), z)
+ \frac{\omega_1(t, y) - y}{1 +
|\omega_1(t, y) - y|}.  \label{e3.4}
\end{equation}
For each $u \in E$, we have by \eqref{e2.2}, \eqref{e3.2} and
\eqref{e3.4} that
\begin{equation}
\begin{aligned}
&\big|\int_0^{\infty} a(s) f^*(s,u(s), u'(s), u''(s)) ds\big|\\
&\leq \int_0^{\infty} a(s) [\phi(s) h(|u''(s)|) + 1] ds \\
&\leq \int_0^{\infty} a(s)(H_0 \phi(s) + 1)  ds \\
&\leq \int_0^{\infty} \max \{s, 1\} a(s)(H_0 \phi(s) + 1)
ds  < + \infty,
\end{aligned}  \label{e3.5}
\end{equation}
 where $H_0 = \max_{0 \leq t \leq \|u''\|_{\infty}}h(t)$.
 From \eqref{e3.5} and Lemma \ref{lem2.1}, we know that $u$ is a solution
of \eqref{e3.3} if and only if $u$ solves the operator equation
$u = Tu$. Here, the operator $T$ is defined by
\begin{equation}
(Tu)(t) = \int_0^{\infty} G(t, s) a(s)
f^*(s, u(s), u'(s), u''(s)) ds, \quad
u \in E, \; t \in [0, +\infty).   \label{e3.6}
\end{equation}
We claim that $T : E \to E$ is completely continuous.

 \textbf{Step 1:}   $T : E \to E$  is well defined.  For $u \in E$,
 we get by \eqref{e3.5} that
\begin{equation}
\int_1^{\infty} s a(s) (H_0 \phi(s) +1) ds
\leq \int_0^{\infty} \max \{s, 1\} a(s)(H_0 \phi(s) + 1)  ds
< + \infty,  \label{e3.7}
\end{equation}
 which implies
\begin{equation}
\lim_{t \to + \infty} t a(t)(H_0 \phi(t) + 1) = 0.  \label{e3.8}
\end{equation}
 Since
\begin{equation}
\int_t^{\infty} a(s) (H_0 \phi(s) + 1)
ds \leq \int_t^{\infty} s a(s) (H_0 \phi(s) + 1) ds, \quad t \geq
1,  \label{e3.9}
\end{equation}
 by  \eqref{e3.7} and \eqref{e3.9}, we have
\begin{equation}
 \lim_{t \to \infty}
\int_t^{\infty} a(s)(H_0 \phi(s) + 1) ds =  0.  \label{e3.10}
\end{equation}
 By the Lebesgue dominated convergence theorem,
\eqref{e3.8} and \eqref{e3.10}, we obtain
\begin{align*}
&\lim_{t \to +\infty} \frac{|(Tu)(t)|}{(1+t)^2}\\
& \leq \lim_{t \to + \infty}
\int_0^{\infty} \frac{G(t, s)}{(1+t)^2} a(s) (H_0 \phi(s) + 1) ds
\\
&= \lim_{t \to +\infty} \Big[\int_0^t
\frac{\frac{1}{2}s(2t-s)}{(1+t)^2} a(s)(H_0 \phi(s) + 1) ds +
\int_t^{\infty} \frac{\frac{1}{2}t^2}{(1+t)^2} a(s) (H_0 \phi(s) +
1) ds \Big]
\\
&= \lim_{t \to +\infty}
\frac{\int_0^t s a(s) (H_0 \phi(s) + 1) ds + \frac{t^2}{2} a(t)
(H_0 \phi(t) + 1)}{2(1+t)}
\\
&\quad  + \lim_{t \to +\infty}
\frac{t \int_t^{\infty} a(s) (H_0 \phi(s) + 1) ds - \frac{t^2}{2}
a(t) (H_0 \phi(t) + 1)}{2(1+t)}     \quad \text{(L'Hopital's rule)}
\\
&= \lim_{t \to +\infty}
\frac{1}{2}t a(t) (H_0 \phi(t) + 1) +  \frac{1}{4} \lim_{t
\to + \infty} \frac{t}{1+t} t a(t) (H_0 \phi(t) + 1) \\
&\quad +  \lim_{t \to +\infty}
\frac{1}{2} \Big[\int_t^{\infty} a(s) (H_0 \phi(s) + 1) ds + t a(t)
(H_0 \phi(t) + 1)\Big]  \quad  \text{(L'Hopital's rule)}
\\
&\quad - \frac{1}{4} \lim_{t \to +
\infty} \frac{t}{1+t} t a(t) (H_0 \phi(t) + 1) \\
&= \frac{1}{2} \lim_{t \to
+\infty}  \int_t^{\infty} a(s) (H_0 \phi(s) + 1) ds = 0;
\end{align*}
that is,
\begin{equation}
\lim_{t \to +\infty} \frac{(Tu)(t)}{(1+t)^2} = 0,  \label{e3.11}
\end{equation}
which implies
\[
\sup_{0 \leq t < +\infty}
\frac{|(Tu)(t)|}{(1+t)^2} < + \infty.
\]
 By \eqref{e3.5}, we get
\begin{align*}
 \sup_{0 \leq t < +\infty} |(Tu)'(t)|
&= \sup_{0 \leq t < +\infty}
\big|\int_0^{\infty} \frac{\partial G(t, s)}{\partial t} a(s)
f^*(s, u(s), u'(s), u''(s)) ds\big|
\\
&= \sup_{0 \leq t < +\infty}
\big|\int_0^t s a(s) f^*(s, u(s), u'(s), u''(s)) ds \\
&\quad + \int_t^{\infty} t a(s) f^*(s, u(s),
u'(s), u''(s)) ds \big|
\\
&\leq \sup_{0 \leq t <
+\infty} \Big[\int_0^t s a(s) (H_0 \phi(s) + 1) ds +
\int_t^{\infty} t a(s) (H_0 \phi(s) + 1) ds \Big]
\\
&\leq \int_0^{\infty} s a(s) (H_0 \phi(s) + 1) ds \\
&\leq \int_0^{\infty} \max \{s, 1\} a(s) (H_0
\phi(s) + 1) ds < + \infty.
\end{align*}
 From \eqref{e3.10},  we have
\[
\big|\int_t^{\infty} a(s) f^*(s, u(s), u'(s), u''(s)) ds\big|
\leq \int_t^{\infty} a(s)(H_0 \phi(s) + 1)ds \to 0,  \quad   t \to +
\infty.
\]
 Therefore,
\begin{equation}
\lim_{t \to +\infty} (Tu)''(t) = \lim_{t \to +\infty}
\int_t^{\infty} a(s) f^*(s, u(s), u'(s),
u''(s)) ds = 0.  \label{e3.12}
\end{equation}
 So  $T u \in E$.

\textbf{Step 2:}  $T : E \to E$ is continuous. For any convergent
sequence $u_n \to u$ in $E$, we have
\[
u_n(t) \to u(t),  \quad  u_n'(t) \to u'(t),  \quad
u_n''(t) \to u''(t), \quad  n \to + \infty, \; t \in [0, +
\infty).
\]
 Now the continuity of $f^*$ implies
\[
  |f^*(s, u_n(s), u_n'(s),
u_n''(s)) - f^*(s, u(s), u'(s),
u''(s))| \to 0, \quad n \to + \infty, \; \forall t \in
 [0, + \infty).
\]
Since $u_n \to u$,  we have
$\sup_{n \in N}\|u_n''\|_{\infty} < + \infty$. Let
\[
  H_p = \max_{0 \leq t \leq \max
\{\|u''\|_{\infty}, \ \sup_{n \in N}
\|u_n''\|_{\infty}\}} h(t).
\]
Then
\begin{equation}
\begin{aligned}
&\int_0^{\infty} s a(s)|f^*(s, u_n(s),
u_n'(s), u_n''(s)) - f^*(s, u(s),
u'(s), u''(s))| ds \\
& \leq 2 \int_0^{\infty} s a(s) (H_p
\phi(s) + 1)ds < + \infty.
\end{aligned} \label{e3.13}
\end{equation}
Hence,  from the Lebesgue dominated convergence
theorem  and \eqref{e3.13}, we have
\begin{equation}
\begin{aligned}
&\|Tu_n - Tu\|_1 \\
&= \sup_{t \in \mathbb{R}^+}  \frac{|(Tu_n)(t) - (Tu)(t)|}{(1+t)^2}
\\
&= \sup_{t \in \mathbb{R}^+} \Big|\int_0^{\infty} \frac{G(t,
s)}{(1+t)^2} a(s) (f^*(s, u_n(s), u_n'(s), u_n''(s)) - f^*(s, u(s),
u'(s), u''(s)) ds\Big|
\\
&\leq \sup_{t \in \mathbb{R}^+}
\Big[\int_0^t \frac{\frac{1}{2}s(2t-s)}{(1+t)^2} a(s) |f^*(s,
u_n(s), u_n'(s), u_n''(s)) - f^*(s, u(s),
u'(s), u''(s))| ds
\\
&\quad + \int_t^{\infty}
\frac{\frac{1}{2}t^2}{(1+t)^2} a(s)|f^*(s, u_n(s),
u_n'(s), u_n''(s)) - f^*(s, u(s),
u'(s), u''(s))| ds\Big]
\\
&\leq \sup_{t \in \mathbb{R}^+}
\Big[\int_0^t s a(s) |f^*(s, u_n(s), u_n'(s),
u_n''(s)) - f^*(s, u(s), u'(s),
u''(s))| ds
\\
&\quad + \int_t^{\infty} \frac{1}{2}
s a(s)|f^*(s, u_n(s), u_n'(s), u_n''(s)) -
f^*(s, u(s), u'(s), u''(s))| ds\Big]
\\
&\leq \int_0^{\infty} s a(s) |f^*(s,
u_n(s), u_n'(s), u_n''(s)) - f^*(s, u(s),
u'(s), u''(s))| ds
\end{aligned}  \label{e3.14}
\end{equation}
which approaches zero as  as  $n \to \infty$.
Also
\begin{equation}
\begin{aligned}
&\|(Tu_n)' - (Tu)'\|_{\infty}\\
&= \sup_{t \in \mathbb{R}^+} |(Tu_n)'(t) -(Tu)'(t)|\\
&= \sup_{t \in \mathbb{R}^+}
\Big|\int_0^t s a(s) (f^*(s, u_n(s), u_n'(s),
u_n''(s)) - f^*(s, u(s), u'(s), u''(s))) ds
\\
&\quad + \int_t^{\infty}
 t a(s)(f^*(s, u_n(s), u_n'(s),
u_n''(s)) - f^*(s, u(s), u'(s),u''(s))) ds\Big|
\\
&\leq \sup_{t \in \mathbb{R}^+}
\Big[\int_0^t s a(s) |f^*(s, u_n(s), u_n'(s),
u_n''(s)) - f^*(s, u(s), u'(s), u''(s))| ds
\\
&\quad + \int_t^{\infty}
 s a(s)|f^*(s, u_n(s), u_n'(s),
u_n''(s)) - f^*(s, u(s), u'(s), u''(s))| ds\Big]
\\
&= \int_0^{\infty} s a(s) |f^*(s,
u_n(s), u_n'(s), u_n''(s)) - f^*(s, u(s),u'(s), u''(s))| ds
\end{aligned}  \label{e3.15}
\end{equation}
which approaches zero as  $n \to \infty$.
 From \eqref{e3.13}, we easily show  that
\begin{equation}
\int_0^{\infty} a(s)|f^*(s, u_n(s),
u_n'(s), u_n''(s)) - f^*(s, u(s),u'(s), u''(s))| ds
< + \infty. \label{e3.16}
\end{equation}
 From the above inequality, we obtain
\begin{equation}
\begin{aligned}
&\|(Tu_n)'' -(Tu)''\|_{\infty}\\
& = \sup_{t \in \mathbb{R}^+}|(Tu_n)''(t) - (Tu)''(t)| \\
&= \sup_{t \in \mathbb{R}^+}
\Big|\int_t^{\infty}  a(s) (f^*(s, u_n(s), u_n'(s),
u_n''(s)) - f^*(s, u(s), u'(s),u''(s))) ds\Big|
\\
&\leq \int_0^{\infty} a(s) |f^*(s,
u_n(s), u_n'(s), u_n''(s)) - f^*(s, u(s),u'(s), u''(s))| ds
\end{aligned}  \label{e3.17}
\end{equation}
which approaches zero as  $n \to \infty$.
 Therefore, by \eqref{e3.14}, \eqref{e3.15} and \eqref{e3.17},
it follows that
 $\|Tu_n - Tu\| \to 0$, as $ n \to +\infty$;
so $T : E \to E$ is continuous.

\textbf{Step 3:}   $T : E \to E$ is compact. Let $A$ be any bounded
subset of $E$, then for  $u \in A$, let
$H_q = \sup_{0 \leq t \leq \|u''\|_{\infty},  u \in A} h(t) < + \infty$,
similar to the proof of \eqref{e3.14}, \eqref{e3.15} and \eqref{e3.17},
 by \eqref{e2.2} and \eqref{e3.2} one has
\begin{align*}
\|Tu\|_1
&= \sup_{t \in \mathbb{R}^+} \frac{|(Tu)(t)|}{(1+t)^2}\\
& \leq \sup_{t \in \mathbb{R}^+}
\int_0^{\infty} \frac{G(t, s)}{(1+t)^2} a(s) |f^*(s, u(s),
u'(s), u''(s))| ds
\\
&\leq \int_0^{\infty} s a(s) |f^*(s,
u(s), u'(s), u''(s))| ds \\
&\leq \int_0^{\infty} s a(s) (H_q \phi(s) + 1) ds < + \infty,
\end{align*}
\begin{align*}
\|(Tu)'\|_{\infty}
&\leq \int_0^{\infty} s a(s) |f^*(s, u(s), u'(s),u''(s))| ds \\
&\leq \int_0^{\infty} s a(s) (H_q \phi(s)
+ 1) ds < + \infty,
\end{align*}
 and
\begin{align*}
\|(Tu)''\|_{\infty}
&\leq \int_0^{\infty}  a(s) |f^*(s, u(s), u'(s), u''(s))| ds\\
&\leq \int_0^{\infty}  a(s) (H_q \phi(s) +1) ds < + \infty,
\end{align*}
which implies that $\|Tu\| < + \infty$. Thus  $TA$ is
uniformly  bounded. Meanwhile, for any $B > 0$, if
$t_1, t_2 \in [0, B]$, we have
\begin{align*}
&\big|\frac{(Tu)(t_1)}{(1+t_1)^2}  -
\frac{(Tu)(t_2)}{(1+t_2)^2}\big|\\
&= \big|\int_0^{\infty} \Big(\frac{G(t_1, s)}{(1+t_1)^2} - \frac{G(t_2,
 s)}{(1+t_2)^2}\Big) a(s) f^*(s, u(s), u'(s), u''(s)) ds\big|
 \\
&\leq  \int_0^{\infty} \big|\frac{G(t_1, s)}{(1+t_1)^2} - \frac{G(t_2,
 s)}{(1+t_2)^2}\big| a(s) (H_q \phi(s) + 1)ds
\end{align*}
which approaches zero  as $t_1 \to t_2$.
Also
\begin{align*}
&|(Tu)'(t_1) - (Tu)'(t_2)| \\
& =  \Big|\int_0^{t_1} s a(s) f^*(s,
u(s), u'(s), u''(s)) ds  +
\int_{t_1}^{\infty} t_1 a(s) f^*(s, u(s), u'(s), u''(s)) ds
\\
& - \int_0^{t_2} s a(s) f^*(s,
u(s), u'(s), u''(s)) ds  -
\int_{t_2}^{\infty} t_2 a(s) f^*(s, u(s), u'(s),u''(s)) ds \Big|
\\
&\leq \Big|\int_{t_1}^{t_2} s a(s)
f^*(s, u(s), u'(s), u''(s)) ds \Big|
+\Big|t_1 \int_{t_1}^{t_2} a(s) f^*(s, u(s), u'(s),u''(s)) ds \Big|
\\
&\quad + \big|(t_2 - t_1)\int_{t_2}^{\infty} a(s) f^*(s, u(s), u'(s),
u''(s)) ds \big|
\\
&\leq \int_{t_1}^{t_2} s a(s) (H_q \phi(s) + 1)ds
+ t_1 \int_{t_1}^{t_2} a(s) (H_q \phi(s) + 1) ds
\\
&\quad + |t_2 - t_1| \int_{t_2}^{\infty} a(s) (H_q \phi(s) + 1) ds
\end{align*}
which approaches zero as $t_1 \to t_2$.
Also
\begin{align*}
|(Tu)''(t_1) - (Tu)''(t_2)|
&= \big|\int_{t_1}^{t_2}  a(s) f^*(s,
u(s), u'(s), u''(s)) ds\big| \\
&\leq \int_{t_1}^{t_2}  a(s) (H_q \phi(s) + 1) ds
\end{align*}
which approaches zero as  $t_1 \to t_2$.
 As a result,  $TA$ is equicontinuous.  From \eqref{e3.11}, we
get
\[
\Big|\frac{(Tu)(t)}{(1+t)^2}  - \lim
_{t \to + \infty} \frac{(Tu)(t)}{(1+t)^2} \Big|
=\Big|\frac{(Tu)(t)}{(1+t)^2}\Big| \to 0, \quad
\text{as $t \to + \infty$}.
\]
Since
\begin{align*}
 \Big|\int_t^{\infty} t a(s) f^*(s,
u(s), u'(s), u''(s)) ds\Big|
&\leq \int_t^{\infty} t a(s) (H_0 \phi(s) + 1) ds\\
&\leq \int_t^{\infty} s a(s) (H_0 \phi(s) + 1) ds
\end{align*}
which approaches zero as $ t \to + \infty$,
we have
\begin{align*}
\lim_{t \to + \infty} (T u)'(t)
&= \lim_{t \to + \infty} \Big[\int_0^t s
a(s) f^*(s, u(s), u'(s), u''(s))) ds\\
&\quad + \int_t^{\infty}
 t a(s) f^*(s, u(s), u'(s),u''(s))) ds\Big] \\
&= \int_0^{\infty} s a(s) f^*(s, u(s),u'(s), u''(s))) ds.
\end{align*}
Thus,
\begin{align*}
&|(Tu)'(t) - \lim _{t \to + \infty} (T u)'(t)|\\
&= \Big| \int_t^{\infty}  t a(s) f^*(s, u(s), u'(s), u''(s))) ds
- \int_t^{\infty}  s a(s) f^*(s, u(s), u'(s),u''(s))) ds \Big|
\end{align*}
which approaches zero as $t \to + \infty$.
 Moreover,  by \eqref{e3.12}, we have
\begin{align*}
\big|(Tu)''(t) - \lim_{t \to + \infty} (T u)''(t)\big|
&= |(Tu)''(t)| \\
&= \Big|\int_t^{\infty}  a(s) f^*(s,
u(s), u'(s), u''(s))) ds \Big|
\end{align*}
which approaches zero  as $t \to + \infty$.
 That is, $TA$ is equiconvergent at infinity. Then $TA$
is relatively compact. Hence, $T : E \to E$ is completely
continuous.
\end{proof}

By the Schauder fixed point theorem, we can easily obtain that $T$
has at least one fixed point $u \in E$. Thus $u$ is a solution of
\eqref{e3.3}.

Next, we show that $u$ satisfies the inequalities
\[
\alpha(t) \leq u(t) \leq \beta(t),  \quad
\alpha'(t) \leq u'(t) \leq  \beta'(t),
\quad \forall t \in \mathbb{R}^+,
\]
which implies that $u$ is a solution of \eqref{e1.1}.  First,
we show that $u'(t) \leq \beta'(t)$ for all
$t \in [0, + \infty)$.  Suppose not, then
\[
\sup_{0 \leq t < + \infty} (u'(t) - \beta'(t)) > 0.
\]
Since $\lim_{t \to + \infty}(u''(t) -\beta''(t)) < 0$,
there are two cases.

\textbf{Case 1.}  There exists a $t_0 \in (0, \infty)$ such that
\[
u'(t_0) - \beta'(t_0)
  = \sup_{t \in \mathbb{R}^+} (u'(t) - \beta'(t)) > 0.
\]
So we have $u''(t_0) =\beta''(t_0)$ and
\begin{equation}
u'''(t_0) \leq \beta'''(t_0). \label{e3.18}
\end{equation}
By \eqref{e3.1}, \eqref{e3.3} and \eqref{e3.4},  we get
\begin{align*}
u'''(t_0) &= - a(t_0)
\Big[f(t_0, \omega_0(t_0, u), \omega_1(t_0, u'),
u''(t_0)) + \frac{\omega_1(t_0, u') -
u'(t_0)}{1 + |\omega_1(t_0, u') - u'(t_0)|} \Big]
\\
&= - a(t_0) \Big[f(t_0,
 \omega_0(t_0, u(t_0)), \beta'(t_0)),
 \beta''(t_0)) + \frac{ \beta'(t_0) -
 u'(t_0)}{1 + |\beta'(t_0) - u'(t_0)|} \Big]
\\
&\geq - a(t_0)f(t_0, \beta(t_0),
\beta'(t_0)), \beta''(t_0)) + a(t_0) \frac{
u'(t_0) - \beta'(t_0)}{1 + |u'(t_0) -
\beta'(t_0)|}
\\
&> - a(t_0) f(t_0, \beta(t_0),
\beta'(t_0)), \beta''(t_0)) \geq
\beta'''(t_0),
\end{align*}
which is a contradiction.

\textbf{Case 2.}
 $ u'(0) - \beta'(0) = \lim _{t \to 0^+} (u'(t) - \beta'(t))
= \sup _{t \in \mathbb{R}^+} (u'(t) - \beta'(t)) > 0$.
 By the boundary condition, we have the
contradiction
$u'(0) - \beta'(0) \leq 0$.
 Consequently, $u'(t) \leq \beta'(t)$
holds for all $t \in [0, +\infty)$.
Using an analogous technique,
we prove that $\alpha'(t) \leq u'(t)$,  for all $t
\in [0, +\infty)$. By integration,
\[
 \alpha(t) \leq u(t) \leq \beta(t), \quad \forall t
\in [0, + \infty).
\]
Let $\sigma > 0$ and choose
\begin{equation}
r \geq \max \big\{\sup_{t \in
[\sigma, + \infty)} \frac{\beta'(t) -
\alpha'(0)}{t}, \ \sup_{t \in [\sigma, + \infty)}
 \frac{\beta'(0) - \alpha'(t)}{t}\big\}  \label{e3.19}
\end{equation}
and $N > r$, such that
\begin{equation}
\int_r^N \frac{s}{h(s)} ds \geq  m
\Big(\sup_{0 \leq t < + \infty} \beta'(t) - \inf
_{0 \leq t \leq + \infty} \alpha'(t)\Big),
\label{e3.20}
\end{equation}
where  $ m = \sup_{t \in [0, + \infty)} a(t) \phi(t) < +\infty$.

\begin{remark} \label{rmk3.1} \rm
  By condition \eqref{e3.2}, it is easy to
know that $\int_0^{+\infty} a(s) \phi(s) ds < + \infty$. Thus we
have $m < + \infty$.
\end{remark}

  Finally, we show that $|u''(t)| < N$ for $t \in [0, + \infty)$.
If $|u''(t)| \leq r$, for every $t \in [0, + \infty)$
then we have $|u''(t)| < N$.
If $u''(t) > r$, for all $t \in [0, + \infty)$, then
for any $R \geq \sigma$,  by \eqref{e3.19} we have
\[
\frac{\beta'(R) - \alpha'(0)}{R} \geq \frac{u'(R) -
u'(0)}{R} = \frac{\int_0^R u''(s) ds}{R}
> r \geq \frac{\beta'(R) -
\alpha'(0)}{R},
\]
   which is a contradiction.
If $u''(t) < - r$,  for every $t \in [0, + \infty)$, a similar
contradiction can be obtained. So, there exists
$t_0 \in [0, + \infty)$ such that
$|u''(t_0)| \leq r$.  Hence, there exists
$[t_1, t_2] \subset [0, + \infty)$ such that $|u''(t_1)| = r$,
$|u''(t)| > r$, $t \in (t_1, t_2]$  or
$|u''(t_2)| = r$, $|u''(t)| > r$, $t \in
[t_1, t_2)$.  Without loss of generality, we suppose that
$u''(t_1) = r$, $u''(t) > r$, $t \in
(t_1, t_2]$. Then, by a convenient change of variable and applying
assumptions \eqref{e2.2} and \eqref{e3.20}, we have
\begin{align*}
\int_{u''(t_1)}^{u''(t_2)}\frac{s}{h(s)} ds
&= \int_{t_1}^{t_2} \frac{u''(t)}{h(u''(t))} u'''(t) dt \\
&=  \int_{t_1}^{t_2} \frac{- a(t)
f(t, u(t), u'(t), u''(t)) u''(t)}{h(u''(t))} dt \\
&\leq \int_{t_1}^{t_2}  a(t) \phi(t) u''(t) dt  \\
&\leq  m \int_{t_1}^{t_2} u''(t) dt
= m (u'(t_2) - u'(t_1))\\
& \leq  m \Big(\sup_{t \in
[0, + \infty)} \beta'(t) - \inf_{t \in [0, + \infty)}
\alpha'(t)\Big)\\
&\leq \int_r^N \frac{s}{h(s)} ds,
\end{align*}
 which concludes that $u''(t_2) \leq N$. Since $t_2$
can be arbitrarily as long as $u''(t) > r$ we can conclude
that, for every $t \in [0, + \infty)$ such that $u''(t) > r$,
we have $u''(t) \leq N$. By a similar way, we can  also obtain
that if $u'(t_1) = - r$, $u'(t) < - r$, $t \in (t_1, t_2]$,
then $u'(t) > - N$, $t \in [0, + \infty)$. Hence,
\[
u'''(t) = - a(t) f^*(t, u(t),
u'(t), u''(t)) = - a(t) f(t, u(t), u'(t), u''(t));
\]
that is, $u$ is a solution of  \eqref{e1.1}.

\section{An example}

In this section, we give an example to illustrate our main result.
Consider the boundary-value problem
\begin{equation}
\begin{gathered}
 u'''(t) + e^{-\gamma t} (t^3 + u^3(t))(1 - u'(t))
 (1 + \arctan ((u''(t))^2) = 0, \quad  t \in (0, +\infty), \\
 u(0) = u'(0) = 0,  \quad    u''(+ \infty) =  0,
 \end{gathered} \label{e4.1}
\end{equation}
 where $\gamma$ is a positive constant.
Set
\[
 a(t) = e^{-\gamma t}, \quad
f(t, x, y, z) = (t^3+x^3)(1-y)(1+\arctan (z^2)).
\]
According to Definition  \ref{def2.1}, it is easy to check that
$\alpha(t) = - t$, $\beta(t) = t$ are a pair of lower and upper
solutions of \eqref{e4.1}. Moreover, we have $\alpha, \beta \in E$,
$\alpha(t) \leq \beta(t)$, $t \in [0, + \infty)$.

Obviously, $f$ is continuous on $[0, + \infty) \times \mathbb{R}^3$ and
increasing in $x$ when $\alpha(t) \leq x(t) \leq \beta(t)$,
$t \in [0, + \infty)$. Meanwhile, when
$0 \leq t < + \infty$, $- t \leq x \leq t$,
$-1 \leq y \leq 1$, it holds
\[
 |f(t, x, y, z)| \leq \phi(t) h(|z|),
\]
where $\phi(t) = 4(1+t^3)$ and $h(z) = 1+z^2$. Since
\[
\int_0^{+\infty} \frac{s}{h(s)} ds =
\int_0^{+\infty} \frac{s}{1+s^2} ds = + \infty,
\]
 $f$ satisfies the Nagumo condition
with respect to $-t, t$. Furthermore,  we have
\[
\int_0^{+\infty} \max \{s, 1\} a(s) ds
= \int_0^1 e^{- \gamma s} ds + \int_1^{+\infty} s e^{- \gamma s}
ds < + \infty
\]
 and
\[
\int_0^{+\infty} \max \{s, 1\} a(s)
\phi(s) = \int_0^1 4(1+s^3)e^{- \gamma s} ds + \int_1^{+\infty}
4s(1+s^3)e^{- \gamma s} ds < + \infty;
\]
 that is, \eqref{e3.2} holds. Therefore, by Theorem \ref{thm3.1}, there
exists at least one solution $u(t)$ for  \eqref{e4.1} such that
\[
- t \leq u(t) \leq t, \quad
-1 \leq u'(t) \leq 1, \quad
t \in [0, + \infty).
\]


\subsection*{Acknowledgements}
The authors want to thank the anonymous the referees for their
comments and suggestions.  This research was supported by grants
10771212 from the National Natural Science Foundation of China, and
06KJB110010 from the Natural Science Foundation of Jiangsu Education
Office.

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\end{document}