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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 53, 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  (login: ftp)}
\thanks{\copyright 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2005/53\hfil Throughout positive solutions]
{Throughout positive solutions of second-order nonlinear
differential equations}
\author[Z. Zhang, C. Wang,  Q. Li, F. Li \hfil EJDE-2005/53\hfilneg]
{Zhenguo Zhang, Chunjiao Wang, Qiaoluan Li, Fang Li}  % in initial order

\address{Zhenguo Zhang \hfill\break
College of Mathematics and Information Science, Hebei normal
University, Heibei,  Shijiazhuang, 050016, China}
\email{Zhangzhg@mail.hebtu.edu.cn}

\address{Chunjiao Wang \hfill\break
College of Mathematics and
Information Science, Hebei normal University,
Heibei,  Shijiazhuang, 050016, China}
\email{ccjjj0601@sina.com.cn}

\address{Qiaoluan Li \hfill\break
College of Mathematics and Information Science, Hebei normal
University, Heibei,  Shijiazhuang, 050016, China}
\email{qll71125@163.com}

\address{Fang Li \hfill\break
College of Mathematics and Information Science, Hebei normal
University, Heibei,  Shijiazhuang, 050016, China}
\email{lifanglucky@126.com}

\date{}
\thanks{Submitted December 13, 2004. Published May 19, 2005.}
\thanks{Supported by the Natural Science Foundation of Hebei
Province and by the Main \hfill\break\indent
Foundation of Hebei Normal University}

\subjclass[2000]{34C10, 34K11}
\keywords{Nonlinear differential equations; neutral term; \hfill\break\indent
eventually positive solution; throughout positive solution}

\begin{abstract}
 In this paper, we consider the second-order nonlinear and the
 nonlinear neutral functional differential equations
 $$\displaylines{
 (a(t)x'(t))'+f(t,x(g(t)))=0,\quad  t\geq t_0\cr
 (a(t)(x(t)-p(t)x(t-\tau))')'+f(t,x(g(t)))=0,\quad t\geq t_0\,.
 }$$
 Using the Banach contraction mapping principle, we obtain the
 existence of throughout positive solutions for the above equations.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{example}[theorem]{Example}

\section{Introduction}

Recently, there has been an increasing interest in the
study of the oscillation and nonoscillation of solutions of
second-order ordinary and delay neutral differential and difference
equations. Also eventually positive solutions and asymptotic
behavior of nonoscillatory solutions have been investigated
widely. Delay differential equations play a very important role in
many practical problems. The papers
\cite{l1,l2,s1,t1,y2,y3,z1} discuss the
oscillation of second order differential and difference equations.
The papers \cite{e1,l3} discuss the oscillation and non-oscillation
criteria for second order differential equations. Of course there
is also the discussion of the existence of eventually positive
solutions, such as \cite{y1,l4,y4,y5}.
 But there are relatively few which
guarantee the existence of throughout positive solutions. The
paper \cite{w1} studies the positive solutions of the following second
order non-neutral ordinary differential equation
$$
y''(t)+F(t,y(t))=0,\quad  t\geq a
$$
where $F: [a,\infty)\times R \to R$ is continuous and nonnegative.
We have studied further and extended the results of Erik
Wahl\'{e}n \cite{w1} to the
self-conjugate and neutral functional differential equations. We
obtain the existence of throughout positive solutions by
introducing a weighted norm (see \cite{e2,w1}) and using the Banach
contraction mapping principle (see \cite{e2}).

 In this paper, we are concerned with existence of
throughout positive solutions for the following self-conjugate
nonlinear differential equations
\begin{gather}
(a(t)x'(t))'+f(t,x(g(t)))=0,\quad  t\geq t_0 \label{e1.1}\\
(a(t)(x(t)-p(t)x(t-\tau))')'+f(t,x(g(t)))=0,\quad  t\geq t_0 \label{e1.2}
\end{gather}
where $a(t)>0$ is continuous; $f(t,x)$ is continuous and satisfies
$f(t,x)x>0$ for $x\neq 0$;   $g(t)$ is continuous, increasing
and satisfies $g(t)\leq t$, $\lim_{t\to\infty} g(t)=\infty$.

\subsection{Definitions}
A solution of differential equation
is said to be oscillatory if it has arbitrarily large zeros;
 otherwise it is said to be non-oscillatory.

A solution of differential equation
is said to be eventually positive solution if there exists some
$T\geq t_0$ such that $x(t)>0$ for all $t\geq T$.

A solution of differential equation is said to be throughout
positive solution if $x(t)>0$ for all $t\geq t_0$.

\subsection*{Related Lemmas}
 To obtain our main results, we need the following lemma.

\begin{lemma} \label{lem1}
Assume $x(t)$ is bounded, $\lim_{t\to \infty}p(t)=p$,
$p\neq \pm 1$,
 $$
 z(t)=x(t)-p(t)x(t-\tau),\quad  \lim_{t\to \infty}z(t)=l,
 $$
 then $\lim_{t\to \infty}x(t) $ exists and
 $\lim_{t\to \infty}x(t) =l/(1-p)$.
 \end{lemma}

\begin{proof}  \textbf{(1)} $p\in(-\infty, -1)$.
Since $x(t)$ is bounded, we get that $\limsup_{t\to
\infty}x(t)=M$ and $\liminf_{t\to\infty}x(t)=m$ exist.
Then there exists a sequence $\{t_n\}$ such that
$\lim_{n\to \infty}x(t_n-\tau)=M$ and
\[
l =\limsup_{n\to \infty}z(t_n)
=\limsup_{n\to \infty}(x(t_n)-p(t_n)x(t_n-\tau))
\geq m-p M \,.
\]
Similarly there exists a sequence $\{t_n'\}$ such that
$\lim_{n\to \infty}x(t_n'-\tau)=m$ and
\[
l =\liminf_{n\to \infty}z(t_n')
=\liminf_{n\to
\infty}(x(t_n')-p(t_n')x(t_n'-\tau))
\leq M-pm \,.
\]
So we have $M-pm \geq m-pM$, that is, $(1+p)M\geq (1+p)m$.  In
view of $1+p<0$, we get $M\leq m$. Hence $M=m$ and
$\lim_{t\to \infty} x(t)$ exists.
By the assumption, we obtain $\lim_{t\to\infty}x(t) =1/(1-p)$.

\noindent \textbf{(2)} $p\in (-1,0)$.
Similarly, there exists a sequence $\{t_n\}$ such that
$\lim_{n\to \infty}x(t_n)=M$. Then there exists a
sequence $\{t_n'\}$ such that $\lim_{n\to
\infty}x(t_n')=m$ and
\begin{gather*}
l =\limsup_{n\to \infty}z(t_n)
=\limsup_{n\to \infty}(x(t_n)-p(t_n)x(t_n-\tau))
\geq M-pm ,\\
l =\liminf_{n\to \infty}z(t_n')
=\liminf_{n\to \infty}(x(t_n')-p(t_n')x(t_n'-\tau))
\leq m-p M \,.
\end{gather*}
Therefore, $M-pm\leq m-pM$,  that is,  $(1+p)M\leq (1+p)m$.
In view of $1+p>0$, we get $M\leq m$.  Hence $M=m$  and
$\lim_{t\to \infty} x(t)$ exists. By the
assumption, we obtain
$\lim_{t\to \infty}x(t) =1/(1-p)$.

\noindent \textbf{(3)} $p\in [ 0,1)$.
Similarly, there exists a sequence $\{t_n\}$ such that
$\lim_{n\to \infty}x(t_n)=M$. Then there exists a
sequence $\{t_n'\}$ such that $\lim_{n\to \infty}x(t_n')=m$ and
\begin{gather*}
l =\limsup_{n\to \infty}z(t_n)
=\limsup_{n\to \infty}(x(t_n)-p(t_n)x(t_n-\tau))
\geq M(1-p) ,\\
l =\liminf_{n\to \infty}z(t_n')
=\liminf_{n\to \infty}(x(t_n')-p(t_n')x(t_n'-\tau))
\leq m(1-p)\,.
\end{gather*}
Therefore, $M(1-p)\leq m(1-p)$. In view of $1-p>0$ we get $M\leq m$.
Hence $M=m$  and $\lim_{t\to \infty} x(t)$ exists.
By the assumption, we obtain
$\lim_{t\to \infty}x(t) =1/(1-p)$.

\noindent\textbf{(4)} $p\in (1,+\infty)$.
Similarly, there exists a sequence $\{t_n\}$ such that
$\lim_{n\to \infty}x(t_n-\tau)=M$. Then there exists
a sequence $\{t_n'\}$ such that $\lim_{n\to \infty}x(t_n'-\tau)=m$
and
\begin{gather*}
l =\limsup_{n\to \infty}z(t_n)
=\limsup_{n\to \infty}(x(t_n)-p(t_n)x(t_n-\tau))
\leq M(1-p) ,\\
l =\liminf_{n\to \infty}z(t_n')
=\liminf_{n\to \infty}(x(t_n')-p(t_n')x(t_n'-\tau))
\geq m(1-p) \,.
\end{gather*}
Therefore, $M(1-p)\geq m(1-p)$. In view of $1-p<0$ we get $M\leq m$.
Hence $M=m$  and $\lim_{t\to \infty} x(t)$ exists.
By the assumption,
$\lim_{t\to \infty}x(t) =l/(1-p)$
which completes the proof.
\end{proof}

\section{Main Results}

 In this section we give existence theorems of
throughout positive solutions for equations  \eqref{e1.1} and \eqref{e1.2}.
First of all we need the following conditions:

Assume that the nonlinearity $f$ satisfies a
Lipschitz condition
\begin{equation}
|f(t,u)-f(t,v)| \leq k(t) |u-v| ,\quad \mbox{for } 0\leq u,\; v \leq C \mbox{
and } t\geq t_0, \label{e3.1}
\end{equation}
 where the constant $C$ will be specified in the theorems
below, and $k(t)>0$ is a continuous function satisfying
\begin{equation}
\int_{t_0}^{\infty} \frac{s}{\overline{a}(s)} k(s) ds < \infty ,
 \label{e3.2}
\end{equation}
where $\overline{a}(s)=\min\{a(\theta):\min \{t_0-\tau,g(t_0)\}\leq
\theta \leq s\}$.

\begin{theorem} \label{thm3.1}
 For equation \eqref{e1.1}, we define the set
$$
X=\{u\in C^1[t_0,\infty),\; 0\leq u(t)\leq M, \mbox{ for }
t\geq t_0;  u(t)=u(t_0),\mbox{ for } g(t_0)\leq t<t_0 \}.
$$
Assume that for every $u\in X$,
\begin{equation}
\int_{t_0}^{\infty} \frac{ds}{a(s)} \int_s^{\infty} f(\theta, u(g(\theta))) d\theta <
M \,. \label{e3.3}
\end{equation}
 Let conditions \eqref{e3.1} and \eqref{e3.2} hold for $0\leq u,v \leq M$.
Assume further that there exists a positive integer $N>1$
such that $0<\frac{l}{N}<1$, where
$l(N)=\max\{\frac{G(g(t))}{G(t)}$, $t\geq t_0\}$,
$G(t)= \exp (N \int_t^{\infty}\frac{s}{\overline{a}(s)}k(s) ds)$.
Then equation \eqref{e1.1} has a throughout positive solution $x(t)$ on
$[t_0, \infty)$
satisfying $\lim_{t\to \infty} x(t)=M$.
\end{theorem}

\begin{proof} Define a mapping $\mathcal{T}$ on $X$
as follows
\begin{equation}
(\mathcal{T} x)(t)=\begin{cases}
 M-\int_t^{\infty} \frac{ds}{a(s)} \int_s^{\infty}
f(\theta,x(g(\theta)))d \theta & t\geq t_0\\
(\mathcal{T}x)(t_0) & g(t_0)\leq t<t_0\,.
\end{cases} \label{e3.4}
\end{equation}
 From \eqref{e3.3} we have $0\leq (\mathcal{T}x)(t)\leq M$, so
$\mathcal{T}X\subseteq X$.  From the assumption
$G(t)=\exp (N \int_t^{\infty}\frac{s}{\overline{a}(s)} k(s)ds)$,
we introduce the norm $\|\cdot\|$ on $X$,
$\|x\|=\sup_{t\geq t_0}|x(t)|/ G(t)$.
Note that  $X$ is closed with respect to this norm, and therefore
we have a complete metric space.

We now show that $\mathcal{T}$ is a contraction mapping on $X$.
For any $x_1,x_2\in X$, in view of the
assumptions we have
\begin{align*}
 \frac{|(\mathcal{T}x_1)(t)-(\mathcal{T}x_2)(t)|}{G(t)}
&\leq \frac{1}{G(t)} \int_t^{\infty} \frac{d
s}{a(s)}\int_s^{\infty}|f(\theta, x_1(g(\theta)))-f(\theta,
x_2(g(\theta)))| d\theta\\
&\leq \frac{1}{G(t)} \int_t^{\infty} \frac{d
s}{a(s)}\int_s^{\infty}
\frac{G(g(\theta))k(\theta)|x_1(g(\theta))-x_2(g(\theta))|}{G(g(\theta))}
d\theta\\
&\leq \frac{1}{G(t)} \|x_1-x_2\| \int_t^{\infty} \frac{d s}{a(s)}
\int_s^{\infty}G(\theta)k(\theta) \frac{G(g(\theta))}{G(\theta)}d\theta\\
&\leq \frac{l}{G(t)} \|x_1-x_2\| \int_t^{\infty}
\frac{(s-t)G(s)k(s)}{\overline{a}(s)} ds\\
&\leq \frac{l}{G(t)} \|x_1-x_2\| \int_t^{\infty}
\frac{s G(s)k(s)}{\overline{a}(s)} ds\\
&= \frac{l}{G(t)} \|x_1-x_2\|
\int_t^{\infty}(-\frac{1}{N})G'(s) ds\\
&= l\frac{G(t)-1}{N G(t)}\|x_1-x_2\|\\
&\leq \frac{l}{N}\|x_1-x_2\| .
\end{align*}
Since $0<\frac{l}{N}<1$, $\mathcal{T}$ is a contraction mapping on
$X$. Finally we use the Banach fixed point theorem to deduce the
existence of a unique fixed point in $X$,
$$
x(t)=(\mathcal{T}x)(t)=M-\int_t^{\infty}\frac{ds}{a(s)} \int_s^{\infty}
f(\theta,x(g(\theta)))d \theta  .
$$
From \eqref{e3.3} we know that $x(t)>0$ for $t\geq t_0$.
Clearly $x(t)$ satisfies
$$
(a(t)x'(t))'+f(t,x(g(t)))=0,
$$
thus $x(t)$ is a throughout positive solution of \eqref{e1.1} and
$\lim_{t\to\infty}x(t)=M$.
The proof is complete.
\end{proof}

Now we discuss the equation \eqref{e1.2}.

\begin{theorem} \label{thm3.2}
 Assume that   $\lim_{t\to\infty}p(t)=p$, where $p\in[0,1)$ and
 $0<p(t)\leq p$. Define
\begin{align*}
X&=\big\{u\in C^1[t_0,\infty),\; 0\leq u(t)\leq M, \mbox{ for }t\geq t_0;
\; u(t)=u(t_0), \\
&\quad\mbox{for } \min \{g(t_0),t_0-\tau\}\leq t< t_0\big\}.
\end{align*}
Let condition \eqref{e3.1} and \eqref{e3.2} hold for
$0\leq u,v\leq M$, and we replace \eqref{e3.3} by
\begin{equation}
\int_{t_0}^{\infty}\frac{ds}{a(s)} \int_s^{\infty} f(\theta, u(g(\theta))) d\theta
< M(1-p) . \label{e3.5}
\end{equation}
 Assume further there exists a positive
integer $N>1$ such that $0<(p+\frac{1}{N})l<1$, where
$l(N)=\max\{\frac{G(t-\tau)}{G(t)}, \frac{G(g(t))}{G(t)}, t\geq
t_0\}$,
$G(t)= \exp\ (N \int_t^{\infty}\frac{s}{\overline{a}(s)}k(s) ds)$.
Then equation \eqref{e1.2} has a throughout positive
solution $x(t)$ on $[t_0, \infty)$ satisfying
$\lim_{t\to\infty}x(t)=M$.
\end{theorem}

\begin{proof} Define a mapping $\mathcal{T}$ on $X$ as follows
$$
(\mathcal{T} x)(t)=\begin{cases}
M(1-p)+p(t)x(t-\tau)-\int_t^{\infty} \frac{ds}{a(s)}
\int_s^{\infty} f(\theta,x(g(\theta)))d \theta \qquad t\geq t_0\\
(\mathcal{T}x)(t_0) \hfill \min \{t_0-\tau,g(t_0)\}\leq t\leq t_0\,.
\end{cases}
$$
For $t\geq t_0$, from \eqref{e3.5} and $p(t)\leq p$, we have
$0\leq(\mathcal{T}x)(t)\leq M(1-p)+pM=M$, so $\mathcal{T}X\subseteq X$.
We introduce the norm $\|\cdot\|$ on $X$,
$\|x\|=\sup_{t\geq t_0} |x(t)|/G(t)$. Now we
 show that $\mathcal{T}$ is a contraction mapping on $X$. For
any $x_1,x_2\in X$, in view of the assumptions we have
\begin{align*}
& \frac{|(\mathcal{T}x_1)(t)-(\mathcal{T}x_2)(t)|}{G(t)}\\
&\leq p(t)\frac{|x_1(t-\tau)-x_2(t-\tau)|}{G(t)}\\
&\quad +\frac{1}{G(t)} \int_t^{\infty} \frac{d
s}{a(s)}\int_s^{\infty}|f(\theta, x_1(g(\theta)))-f(\theta,
x_2(g(\theta)))| d\theta\\
&\leq p(t)\frac{G(t-\tau)}{G(t)}\frac{|x_1(t-\tau)-x_2(t-\tau)|}{G(t-\tau)}\\
&\quad + \frac{1}{G(t)} \int_t^{\infty} \frac{d
s}{a(s)}\int_s^{\infty}|f(\theta, x_1(g(\theta)))-f(\theta,
x_2(g(\theta)))| d\theta\\
&\leq p\ l\ \|x_1-x_2\| +\frac{1}{G(t)} \int_t^{\infty} \frac{d
s}{a(s)}\int_s^{\infty}
\frac{G(g(\theta))k(\theta)|x_1(g(\theta))-x_2(g(\theta))|}{G(g(\theta))}
d\theta\\
&\leq p\ l\ \|x_1-x_2\|+ \frac{1}{G(t)} \|x_1-x_2\|
\int_t^{\infty} \frac{d s}{a(s)}
\int_s^{\infty}G(\theta)k(\theta)\frac{G(g(\theta))}{G(\theta)}d\theta\\
&\leq p\ l\ \|x_1-x_2\|+ \frac{l}{G(t)} \|x_1-x_2\|
\int_t^{\infty}
\frac{(s-t)G(s)k(s)}{\overline{a}(s)} ds\\
&\leq p\ l\ \|x_1-x_2\|+ \frac{l}{G(t)} \|x_1-x_2\|
\int_t^{\infty}
\frac{s G(s)k(s)}{\overline{a}(s)} ds\\
&= p\ l\ \|x_1-x_2\|+ \frac{l}{G(t)} \|x_1-x_2\| \int_t^{\infty}
(-\frac{1}{N})G'(s)ds\\
&\leq \big(p+\frac{G(t)-1}{N G(t)}\big)\ l\ \|x_1-x_2\|\\
&\leq (p+\frac{1}{N})\ l\ \|x_1-x_2\|.
\end{align*}
Since $0<(p+\frac{1}{N}) l <1$,  $\mathcal{T}$ is a contraction
mapping on $X$. Finally we use the Banach fixed point theorem to
deduce the existence of a unique fixed point in $X$
$$
x(t)=(\mathcal{T}x)(t)=M(1-p)+p(t)x(t-\tau)-\int_t^{\infty}\frac{ds}{a(s)}
\int_s^{\infty} f(\theta,x(g(\theta)))d \theta .
$$
 From the condition \eqref{e3.5} and
$p(t)x(t-\tau)\geq 0$ we know that $x(t)>0$ for $t\geq t_0$.
 Clearly $x(t)$ satisfies
$$
(a(t)(x(t)-p(t)x(t-\tau))')'+f(t,x(g(t)))=0,
$$
thus $x(t)$ is a throughout positive solution of \eqref{e1.2}
and
$$
\lim_{t\to\infty}(x(t)-p(t)x(t-\tau))=M(1-p)\,.
$$
In view of the Lemma \ref{lem1},  $\lim _{t\to\infty}x(t)=M$
which completes the proof.
\end{proof}


\begin{theorem} \label{thm3.3}
Assume that
$\lim_{t\to\infty}p(t)=p$ where $ p\in (-1,0)$ and
$p\leq p(t)<0$ and define
\begin{align*}
Y&=\{\ u\in C^1[t_0,\infty),\; 0\leq u(t)\leq M(1-p),\mbox{ for}
 t\geq t_0;\; u(t)=u(t_0),\\
&\quad \mbox{ for } \min\{g(t_0),t_0-\tau\}\leq t<t_0 \}.
\end{align*}
Let conditions \eqref{e3.1} and \eqref{e3.2} hold for $0\leq u,v \leq M(1-p)$.
Assume that for every $u\in Y$,
\begin{equation}
\int_{t_0}^{\infty} \frac{ds}{a(s)} \int_s^{\infty} f(\theta,
u(g(\theta))) d\theta < M(1-p^2) .\label{e3.6}
\end{equation}
Assume further that there exists a positive integer $N>1$ such that
$0<(\frac{1}{N}-p)l <1$, where $l(N)=\max\{\frac{G(t-\tau)}{G(t)},
\frac{G(g(t))}{G(t)}, t\geq t_0\}$,
$G(t)=\exp(N\int_t^{\infty}\frac{s}{\overline{a}(s)}\ k(s) ds)$.
Then equation \eqref{e1.2} has a throughout positive solution $x(t)$
on $[t_0, \infty)$ satisfying
 $\lim_{t\to\infty}x(t)=M$.
\end{theorem}


\begin{proof} Define a mapping $\mathcal{T}$ on $Y$
as follows
$$
(\mathcal{T} x)(t)=\begin{cases}
 M(1-p)+p(t)x(t-\tau)-\int_t^{\infty} \frac{ds}{a(s)}
\int_s^{\infty}
f(\theta,x(g(\theta)))d \theta \qquad t\geq t_0\\
(\mathcal{T}x)(t_0) \hfill \min \{t_0-\tau,g(t_0)\}\leq t\leq t_0\,.
\end{cases}
$$
Since $p(t)<0$, we easily know that
$0\leq (\mathcal{T}x)(t)\leq M(1-p)$. So $\mathcal{T}X\subseteq X$.
We introduce the norm $\|\cdot\|$ on $Y$,
$\|x\|=\sup_{t\geq t_0} |x(t)|/G(t)$.
We now show that $\mathcal{T}$ is a contraction mapping on $Y$.
Similar to the proof  of  Theorem \ref{thm3.2}, for any $x_1,x_2\in Y$, in
view of the assumptions we have
\begin{align*}
 \frac{|(\mathcal{T}x_1)(t)-(\mathcal{T}x_2)(t)|}{G(t)}
&\leq
|p(t)|\frac{G(t-\tau)}{G(t)}\frac{|x_1(t-\tau)-x_2(t-\tau)|}{G(t-\tau)}\\
&\quad + \frac{1}{G(t)} \int_t^{\infty} \frac{d
s}{a(s)}\int_s^{\infty}|f(\theta, x_1(g(\theta)))-f(\theta,
x_2(g(\theta)))| d\theta\\
&\leq |p|\ l\ \|x_1-x_2\|+\frac{l}{N}\|x_1-x_2\|\\
&= (\frac{1}{N}-p)\ l\ \|x_1-x_2\|.
\end{align*}
Since $0<(\frac{1}{N}-p) l <1$,  $\mathcal{T}$ is a contraction
mapping on $Y$. Finally we use the Banach fixed point theorem to
deduce the existence of a unique fixed point in $Y$,
$$
x(t)=(\mathcal{T}x)(t)=M(1-p)+p(t)x(t-\tau)-\int_t^{\infty}\frac{ds}{a(s)}
\int_s^{\infty} f(\theta,x(g(\theta)))d \theta .
$$
Since $x\in Y$ and $p\leq p(t)<0$, we have $p(t)x(t-\tau)\geq pM(1-p)$.
 From the inequality and the condition \eqref{e3.6}, we obtain
 $$
x(t)> M(1-p)+pM(1-p)-M(1-p^2)=0.
$$
Hence $x(t)>0$ for $t\geq t_0$.
Substituting $x(t)$ into \eqref{e1.2}, we know that  $x(t)$ is a
throughout positive solution of equation \eqref{e1.2} and
$$
\lim_{t\to\infty}(x(t)-p(t)x(t-\tau))=M(1-p).
$$
In view of the Lemma \ref{lem1},  $\lim _{t\to\infty}x(t)=M$
which completes the proof.
\end{proof}


\begin{theorem} \label{thm3.4}
Assume that $\lim_{t\to\infty}p(t)=p$ where $ p\in
(-\infty,-1)$ and $ p(t)\leq p$. Define
\begin{align*}
Z&=\big\{\ u\in C^1[t_0,\infty),\; 0\leq u(t)\leq \frac{M(1+|p|)}{|p|},
\mbox{ for }t\geq t_0;\; u(t)=u(t_0),\\
&\quad\mbox{for } g(t_0)\leq t<t_0 \big\}
\end{align*}
where $M$ is a positive constant. Let conditions \eqref{e3.1}
and \eqref{e3.2} hold for $0\leq u,v \leq \frac{M(1+|p|)}{|p|}$.
Assume that for every $u\in Z$,
\begin{equation}
\int_{t_0}^{\infty} \frac{ds}{a(s)} \int_s^{\infty} f(\theta,
u(g(\theta))) d\theta < \frac{M(p^2-1)}{|p|} . \label{e3.7}
\end{equation}
Assume further there exists a positive integer $N>1$ such that
 $0<\frac{1}{|p|}(1+\frac{l}{N})<1$, where $l(N)=\max\{
\frac{G(g(t))}{G(t)}, t\geq t_0\}$,
$G(t)= \exp\ (N \int_t^{\infty} \frac{s}{\overline{a}(s)}\ k(s) ds)$.
Then equation \eqref{e1.2} has a throughout positive solution $x(t)$ on
$[t_0, \infty)$
satisfying $\lim_{t\to\infty}x(t)=M$.
\end{theorem}


\begin{proof} Define a mapping $\mathcal{T}$ on $Z$
as follows
$$
(\mathcal{T} x)(t)=\left\{\begin{array}{l}
 \frac{1}{-p(t+\tau)}\big[
M(1-p)-x(t+\tau)-\int_{t+\tau}^{\infty} \frac{ds}{a(s)}
\int_s^{\infty} f(\theta,x(g(\theta)))d \theta \big] \; t\geq t_0\\
(\mathcal{T}x)(t_0) \hfill g(t_0)\leq t\leq t_0.
\end{array}\right.
$$
 From \eqref{e3.7}, we have
 $0\leq (\mathcal{T}x)(t)\leq \frac{M(1+|p|)}{|p|}$. So
  $\mathcal{T}Z\subseteq Z$.
We introduce the norm $\|\cdot\|$ on $Z$,
$\|x\|=\sup_{t\geq t_0} |x(t)|/G(t)$.
We now show that $\mathcal{T}$ is a contraction mapping on $Z$.
For any $x_1,x_2\in Z$, in view of the assumptions we have
\begin{align*}
& \frac{|(\mathcal{T}x_1)(t)-(\mathcal{T}x_2)(t)|}{G(t)}\\
&\leq \frac{-1}{G(t+\tau)p(t+\tau)}|x_1(t+\tau)-x_2(t+\tau)|\\
&\quad + \frac{-1}{G(t)p(t+\tau)}\int_{t+\tau}^{\infty} \frac{d s}{a(s)}
\int_s^{\infty}|f(\theta, x_1(g(\theta)))-f(\theta, x_2(g(\theta)))| d\theta \\
&\leq  \frac{1}{|p|}\|x_1-x_2\| +\frac{1}{G(t)|p|}
\int_{t+\tau}^{\infty} \frac{d s}{a(s)}\int_s^{\infty}
\frac{G(g(\theta))k(\theta)|x_1(g(\theta))-x_2(g(\theta))|}{G(g(\theta))}
d\theta\\
&\leq \frac{1}{|p|}\|x_1-x_2\|+ \frac{1}{G(t)|p|} \|x_1-x_2\|
\int_{t+\tau}^{\infty} \frac{d s}{a(s)}
\int_s^{\infty}G(\theta)k(\theta)\frac{G(g(\theta))}{G(\theta)}d\theta\\
&\leq \frac{1}{|p|}\|x_1-x_2\|+ \frac{l}{G(t)|p|} \|x_1-x_2\|
\int_{t+\tau}^{\infty}
\frac{(s-t-\tau)G(s)k(s)}{\overline{a}(s)} ds\\
&\leq \frac{1}{|p|}\|x_1-x_2\|+ \frac{l}{G(t)|p|} \|x_1-x_2\|
\int_{t+\tau}^{\infty}
\frac{s G(s)k(s)}{\overline{a}(s)} ds\\
&= \frac{1}{|p|}\|x_1-x_2\|+ \frac{l}{G(t)|p|} \|x_1-x_2\|
\int_{t+\tau}^{\infty}
(-\frac{1}{N})G'(s)ds\\
&\leq \frac{1}{|p|}\left(1+l\ \frac{G(t+\tau)-1}{N G(t)}\right)\|x_1-x_2\|\\
&\leq \frac{1}{|p|}(1+\frac{l}{N})\|x_1-x_2\|.
\end{align*}
Since $0<\frac{1}{|p|}(1+\frac{l}{N})<1$,  $\mathcal{T}$ is a
contraction mapping on $Z$. Finally we use the Banach fixed point
theorem to deduce the existence of a unique fixed point in $Z$,
\begin{align*}
x(t)&=(\mathcal{T}x)(t)\\
&=\frac{1}{-p(t+\tau)}\Big[M(1-p)-x(t+\tau)-\int_{t+\tau}^{\infty}
\frac{ds}{a(s)}\int_s^{\infty}
f(\theta,x(g(\theta)))d\theta \Big].
\end{align*}
 Since $x\in Z$, we have
$x(t+\tau)\leq \frac{M(1+|p|)}{|p|}$. From the inequality and the
condition \eqref{e3.7}, we obtain
$$
x(t)>\frac{1}{-p(t+\tau)}\big[M(1-p)-\frac{M(1+|p|)}{|p|}-\frac{M(p^2-1)}{|p|}
\big]=0.
$$
Hence $x(t)>0$ for $t\geq t_0$. Substituting $x(t)$ into
\eqref{e1.2}, we know that $x(t)$ is a throughout positive
solution of  \eqref{e1.2} and
$$
\lim_{t\to\infty}(x(t)-p(t)x(t-\tau))=M(1-p).
$$
In view of Lemma \ref{lem1}, we have $\lim _{t\to\infty}x(t)=M$.
The proof is complete.
\end{proof}


\begin{theorem} \label{thm3.5}
 Assume that $\lim_{t\to\infty}p(t)=p$ where $ p\in (1,+\infty)$
and $ p(t)\geq p$. Define
\begin{align*}
\Omega&=\big\{u\in C^1[t_0,\infty),\;0\leq u(t)\leq \frac{M(1+p)}{p},
\mbox{ for }t\geq t_0;\;u(t)=u(t_0), \\
&\quad \mbox{for }  g(t_0)\leq t<t_0\big\}
\end{align*}
where $M$ is a positive constant. Let conditions \eqref{e3.1} and \eqref{e3.2}
hold for $0\leq u,v\leq \frac{M(1+p)}{p}$.
We assume that for every $u\in \Omega$
\begin{equation}
\int_{t_0}^{\infty} \frac{ds}{a(s)} \int_s^{\infty} f(\theta,
u(g(\theta))) d\theta \leq \frac{p-1}{p}M . \label{e3.8}
\end{equation}
Assume further that there exists a positive integer $N>1$ such that
 $0<\frac{1}{p}(1+\frac{l}{N})<1$, where
$l(N)=\max\{\frac{G(g(t))}{G(t)}, t\geq t_0\}$,
$G(t)= \exp(N \int_t^{\infty}\frac{s}{\overline{a}(s)}\ k(s) ds)$.
Then \eqref{e1.2} has a throughout positive solution~ $x(t)$ on
$[t_0, \infty)$ satisfying
 $\lim_{t\to\infty}x(t)=M$.
\end{theorem}

\begin{proof} Define a mapping $\mathcal{T}$ on
$\Omega$ as follows
$$
(\mathcal{T} x)(t)=\begin{cases}
 \frac{1}{p(t+\tau)}\big[
M(p-1)+x(t+\tau)+\int_{t+\tau}^{\infty} \frac{ds}{a(s)}
\int_s^{\infty}f(\theta,x(g(\theta)))d \theta \big] \quad t\geq t_0\\
(\mathcal{T}x)(t_0) \hfill g(t_0)\leq t\leq t_0\,.
\end{cases}
$$
 From \eqref{e3.8}, we have $0\leq (\mathcal{T}x)(t)\leq \frac{p+1}{p}M$.
So $\mathcal{T}\Omega \subseteq \Omega$.
We introduce the norm $\|\cdot\|$ on $\Omega$,
$\|x\|=\sup_{t\geq t_0} |x(t)|/G(t)$.
We now show that $\mathcal{T}$ is a contraction mapping on
$\Omega$. Similar to the proof of Theorem \ref{thm3.4}, for any $x_1,x_2\in
\Omega$, in view of the assumptions we have
\begin{align*}
& \frac{|(\mathcal{T}x_1)(t)-(\mathcal{T}x_2)(t)|}{G(t)}\\
&\leq \frac{1}{G(t+\tau)p(t+\tau)}|x_1(t+\tau)-x_2(t+\tau)|\\
&\quad + \frac{1}{G(t)p(t+\tau)}\int_{t+\tau}^{\infty} \frac{d
s}{a(s)}\int_s^{\infty}|f(\theta, x_1(g(\theta)))-f(\theta,
x_2(g(\theta)))| d\theta \\
&\leq \frac{1}{p}(1+\frac{l}{N})\|x_1-x_2\|.
\end{align*}
Since $0<\frac{1}{p}(1+\frac{l}{N})<1$,  $\mathcal{T}$ is a
contraction mapping on $\Omega$. Finally we use the Banach fixed
point theorem to deduce the existence of a unique fixed point in
$\Omega$,
\begin{align*}
x(t)&=(\mathcal{T}x)(t)\\
&=\frac{1}{p(t+\tau)}\big[M(p-1)+x(t+\tau)+\int_{t+\tau}^{\infty}\frac{ds}{a(s)} \int_s^{\infty}
f(\theta,x(g(\theta)))d \theta \big].
\end{align*}
 Because $p>1$, that is $M(p-1)>0$, and all the other terms which are in
the expression of $x(t)$ are nonnegative, we easily know that $x(t)>0$ for
 $t\geq t_0$. Substituting $x(t)$ into \eqref{e1.2}, we know that $x(t)$ is a
throughout positive solution of equation \eqref{e1.2} and
$$
\lim_{t\to\infty}(x(t)-p(t)x(t-\tau))=M(1-p).
$$
In view of the Lemma \ref{lem1} we have $\lim _{t\to\infty}x(t)=M$.
The proof is complete.
\end{proof}


\section{Examples}

\begin{example} \label{ex1} \rm
 Consider the second order self-conjugate differential equation
\begin{equation}
(tx'(t))'+\frac{4(t-1)^6}{t^6(t-2)^3}\quad
x^3(t-1)=0,\quad t\geq t_0=6. \label{e4.1}
\end{equation}
In our notation,
$a(t)=t$, $\overline{a}(s)=5$, $g(t)=t-1$,
$f(t,u)=\dfrac{4(t-1)^6}{t^6(t-2)^3}~u^3$.
We choose
$M=1$, $k(t)=\dfrac{12(t-1)^6}{t^6(t-2)^3}$, $N=3$.
 We know that for any $0\leq u,v\leq 1$,
\begin{eqnarray*}
|f(t,u)-f(t,v)|&=&|~\frac{4(t-1)^6}{t^6(t-2)^3}~(u^3-v^3)| \leq
\frac{12(t-1)^6}{t^6(t-2)^3}~|u-v|.
\end{eqnarray*}
For any $u,v\in X$
$$
\int_{t_0}^{\infty}\frac{s}{\overline{a}(s)}~k(s)~ds=\frac{1}{5}\int_6^{\infty}
\frac{12(s-1)^6}{s^5(s-2)^3}~ds< \infty$$
\begin{align*}
\int_{t_0}^{\infty}\frac{d
s}{a(s)}\int_s^{\infty}f(\theta,u(g(\theta)))\,d\theta
&=\int_6^{\infty}\frac{4}{s}~ds\int_s^{\infty}\frac{(\theta-1)^6
(u(\theta-1))^3}{\theta^6 (\theta-2)^3}\,d\theta\\
&\leq \int_6^{\infty}\frac{4}{s}\,ds
\int_s^{\infty}\frac{d\theta}{(\theta-2)^3}\\
&= \frac{1}{4}+\frac{1}{2}\ln\frac{4}{6}
\leq \frac{1}{4} < 1\,,
\end{align*}
 $$
l=\exp\Big(N \int_{t_0-1}^{t_0}\frac{s}{t_0-1} \frac{12(s-1)^6}{s^6(s-2)^3} ds
\Big)
 =\exp\Big(3 \int_{5}^6\frac{s}{5}\frac{12(s-1)^6}{s^6(s-2)^3} ds\Big)<3.
$$
Thus the conditions in Theorem \ref{thm3.1} are satisfied. So
\eqref{e4.1} has a throughout positive solution $x(t)$ on $[t_0,\infty)$
and $\lim_{t\to \infty}x(t)=1$. In fact,
$x(t)=1-\frac{1}{t^2}$ is such a solution.
\end{example}




\begin{example} \rm
Consider the second-order neutral differential
equation
\begin{equation}
(x(t)-\frac{1}{2}x(t-1))''+\frac{2(t-1)^3-t^3}{(t-1)^3(t-2)^3}x^3(t-1)=0,\quad
t\geq t_0=13 .\label{e4.2}
\end{equation}
 Here $a(t)=1$, $\overline{a}(s)=1$,
$p(t)=\dfrac{1}{2}$, $g(t)=t-1$,
$f(t,u)=\dfrac{[2(t-1)^3-t^3]u^3}{(t-1)^3(t-2)^3}$.
We choose
$M=1$, $k(t)=\dfrac{3[2(t-1)^3-t^3]}{(t-1)^3(t-2)^3}$, $N=4$.
 It is easy to show that for any $0\leq u,v\leq 1$,
$$
|f(t,u)-f(t,v)|=|\frac{2(t-1)^3-t^3}{(t-1)^3(t-2)^3}(u^3-v^3)|\leq
 \frac{3[2(t-1)^3-t^3]}{(t-1)^3(t-2)^3}|u-v|.
$$
 For any $u,v\in X$
$$
\int_{t_0}^{\infty}\frac{s}{\overline{a}(s)}k(s)\,ds
=\int_{13}^{\infty}3s \frac{2(s-1)^3-s^3}{(s-1)^3(s-2)^3}\,ds< \infty\,,
$$
\begin{align*}
\int_{t_0}^{\infty}\frac{d
s}{a(s)}\int_s^{\infty}f(\theta,u(g(\theta)))d\theta
&=\int_{13}^{\infty}\int_s^{\infty}\frac{2(\theta-1)^3-\theta^3}{(\theta-1)^3
(\theta-2)^3}(u(\theta-1))^3d\theta ds\\
&=\int_{13}^{\infty}
(\theta-t)\frac{2(\theta-1)^3-\theta^3}{(\theta-1)^3(\theta-2)^3}
(u(\theta-1))^3d\theta\\
&\leq \int_{13}^{\infty}\frac{2\theta}{(\theta-2)^3}~d\theta\\
&= \frac{24}{121}< \frac{1}{2}\,,
\end{align*}
$$
(p+\frac{1}{N})l=(\frac{1}{2}+\frac{1}{4})\ \exp\ \big(4\int_{12}^{13}
s\frac{3[2(s-1)^3-s^3]}{(s-1)^3(s-2)^3}\ ds\big)<1.
$$
Thus the conditions in Theorem \ref{thm3.2} are satisfied. So  \eqref{e4.2} has a
throughout positive solution $x(t)$ on $[t_0,\infty)$ and
$\lim_{t\to \infty}x(t)=1$. In fact,
$x(t)=1-\frac{1}{t}$ is such a solution.
\end{example}


\begin{example} \rm
Consider the second-order self-conjugate neutral differential equation
\begin{equation}
\big[\frac{t^3(t-1)}{4((t-1)^3+2t^3)}(x(t)+2x(t-1))'\big]'
+\frac{(t-1)^3}{t^3(t-2)^3}~x^3(t-1)=0,\ t\geq t_0=9. \label{e4.3}
\end{equation}
In our notation,
$p(t)=-2$, $g(t)=t-1$, $\tau=1$, $a(t)=\dfrac{t^3(t-1)}{4((t-1)^3+2t^3)}$,
$\overline{a}(s)=\frac{896}{1367}$, $f(t,u)=\dfrac{(t-1)^3}{t^3(t-2)^3}~u^3$.
We choose that $M=1$,
$k(t)=\dfrac{27(t-1)^3}{4t^3(t-2)^3}$, $N=3$.
 Here we define
$Z=\{u\in C^1[t_0,\infty): 0\leq u(t)\leq \frac{3}{2},\; t\geq t_0 \}$.
 It is easy to show that for any $0\leq u,v\leq \frac{3}{2}$,
$$
|f(t,u)-f(t,v)|=|\frac{(t-1)^3}{t^3(t-2)^3}(u^3-v^3)|\leq
 \frac{27(t-1)^3}{4t^3(t-2)^3} |u-v|.
$$
For any $u,v\in Z$,
$$
\int_{t_0}^{\infty}\frac{s}{\overline{a}(s)}k(s)\,ds=\frac{1367}{896}
 \int_9^{\infty}
\frac{27(s-1)^3}{4s^2(s-2)^3}~ds< \infty\,,
$$
\begin{align*}
&\int_{t_0}^{\infty}\frac{ds}{a(s)}\int_s^{\infty}f(\theta,u(g(\theta)))d\theta\\
&\leq \frac{4((t_0-1)^3+2t_0^3)}{t_0^3(t_0-1)}\int_9^{\infty}ds\int_s^{\infty}
\frac {(\theta-1)^3(u(\theta-1))^3}{\theta^3(\theta-2)^3}~d\theta\\
&\leq \frac{12}{t_0-1}\int_9^{\infty}ds\int_s^{\infty}
\frac{d\theta}{(\theta-2)^3}\\
&= \frac{3}{28} < \frac{3}{2}\,,
\end{align*}
$$
\frac{1}{|p|}(1+\frac{l}{N})=\frac{1}{2}\Big[1+\frac{1}{3}
\exp \Big(3\int_{8}^9 s\frac{4((9-2)^3+2(9-1)^3)}{(9-1)^3(9-2)}
\frac{27(s-1)^3}{4s^2(s-2)^3}~ds\Big)\Big]<1.
$$
Thus the conditions in Theorem \ref{thm3.4} are satisfied. So  \eqref{e4.3}
has a throughout positive solution $x(t)$ on $[t_0,\infty)$ and
$\lim_{t\to \infty}x(t)=1$. In fact,
$x(t)=1-\frac{1}{t^2}$ is such a solution.
\end{example}


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\end{document}