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\markboth{\hfil A classification scheme for positive solutions \hfil
EJDE--2000/??}
{EJDE--2000/??\hfil  Xianling Fan, Wan-Tong Li, \& Chengkui Zhong \hfil}
\begin{document}
 \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol.~{\bf 2000}(2000), No.~??, pp.~1--14. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)} 
\vspace{\bigskipamount} \\
 %
  A classification scheme for positive solutions of second order nonlinear
iterative differential equations 
\thanks{{\em Mathematics Subject Classifications:} 34K15. \hfil\break\indent
{\em Key words and phrases:}
Nonlinear iterative differential equation, oscillation, \hfil\break\indent 
eventually positive, asymptotic behavior. \hfil\break\indent
\copyright 2000 Southwest Texas State University and University of North Texas. 
 \hfil \break\indent
Submitted December 11, 1999. Published March 31, 2000. \hfil\break\indent
Supported by the NNSF of China and the Foundation for
University Key Teacher by \hfil\break\indent  Ministry Education.} }
\date{}
\author{ Xianling Fan, Wan-Tong Li, \& Chengkui Zhong }
   \maketitle

\begin{abstract}
This article presents a classification scheme for eventually-positive
solutions of second-order nonlinear iterative differential equations, in
terms of their asymptotic magnitudes. Necessary and sufficient conditions
for the existence of solutions are also provided.
\end{abstract}

\newtheorem{theorem}{Theorem} \newtheorem{lemma}{Lemma}

\section{Introduction}

A systematic study of oscillatory properties and asymptotic behavior of
solutions of functional differential equations began with the works 
\cite{ref4, ref11, ref12}. However, a considerable number of papers dealing 
with these problems are from the last two decades. In 1987, the monograph 
\cite{ref5} presented a systematic investigation of the oscillatory properties 
of solutions to ordinary differential equations with deviating arguments.
Recently, Bainov, Markova and Simeonov \cite{ref3} studied the equation 
\begin{equation}
\label{1}(r(t)x'(t))'+f(t,x(t),x(\Delta (t,x(t))))=0 
\end{equation}
with the condition 
$$
\int_0^\infty \frac{ds}{r(s)}=\infty \,. 
$$
They provide a classification scheme for non-oscillatory solutions, and
provide necessary and sufficient conditions for the existence of solutions.
Such schemes are important since further investigations of qualitative
behaviors of solutions can then be reduced to only a number of cases.
However, a more difficult problem \cite{ref9} is to characterize the case
when 
$$
\int_0^\infty \frac{ds}{r(s)}<\infty \,. 
$$

This paper concerns with the general class of second order nonlinear
differential equations 
\begin{equation}
\label{2}(r(t)(x'(t))^\sigma )'+f(t,x(t),x(\Delta
(t,x(t))))=0 
\end{equation}
with the conditions $\int_0^\infty ds/r(s)^{1/\sigma }=\infty $ and $%
\int_0^\infty ds/r(s)^{1/\sigma }<\infty $. We give a classification scheme
for eventually-positive solutions of this equation in terms of their
asymptotic magnitude, and provide necessary and/or sufficient conditions for
the existence of solutions. Our results extend and improve the results in 
\cite{ref3, ref5}.

When $f(t,x(t),x(\Delta (t,x(t))))=f(t,x(t))$, the oscillation and
asymptotic behavior of the solutions of (\ref{2}) have been studied by Li 
\cite{ref6}-\cite{ref10}, Ruan \cite{ref13} and Wong and Agarwal \cite{ref14}.

It is known \cite{ref3} that the differential equation of the from (\ref{1})
with delay depending on the unknown function have been investigated only in
the papers \cite{ref1}, \cite{ref2}.

Let $T\in {\mathbb R}_{+}=[0,\infty )$. Define $T_{-1}=\inf \{\Delta
(t,x):t\geq T,x\in R\}$.

\paragraph{Definition 1.}
The function $x(t)$ is called a solution of the differential equation (\ref
{2}) in the interval $[T,+\infty )$, if $x(t)$ is defined for $t\geq T_{-1}$%
, it is twice differentiable and satisfies (\ref{2}) for $t\geq T$.

\paragraph{Definition 2.}
The solution $x(t)$ of (\ref{2}) is called regular, if it is defined on some
interval $[T_x,\infty )$ and $\sup\{|x(t)|:t\geq T\}>0$ for $t\geq T_x$.

\paragraph{Definition 3.}
The solution $x(t)$ of (\ref{2}) is said to be:

\noindent(i) eventually positive: if there exists $T\geq 0$ such that $%
x(t)>0 $ for all $t\geq T$;

\noindent(ii) eventually negative: if there exists $T\geq 0$ such that $%
x(t)<0$ for all $t\geq T$;

\noindent(iii) non-oscillatory: if it is either eventually positive or
eventually negative;

\noindent (iv) oscillatory: if it is neither eventually positive nor
eventually negative.

Throughout this paper, we assume that the following conditions hold:

\begin{description}
\item  {H1)} $r\in C({\mathbb R}_{+},{\mathbb R}_{+})$ and $r(t)>0,t\in 
{\mathbb R}_{+}$.

\item  {H2)} $f\in C({\mathbb R}_{+}\times R^2,{\mathbb R})$.

\item  {H3)} There exists $T\in {\mathbb R}_{+}$ such that $uf(t,u,v)>0$ for 
$t\geq T$, $uv>0$ and $f(t,u,v)$ is non-decreasing in $u$ and $v$ for each
fixed $t\geq T$.

\item  {H4)} $\Delta \in C({\mathbb R}_{+}\times R,{\mathbb R})$.

\item  {H5)} There exist a function $\Delta _{*}(t)\in C({\mathbb R}_{+},{%
\mathbb 
R})$ and $T\in {\mathbb R}_{+}$ such that $\lim _{t\to \infty }\Delta
_{*}(t)=+\infty $ and $\Delta _{*}(t)\leq \Delta (t,x)$ for $t\geq T$, $x\in 
{\mathbb R}$.

\item  {H6)} There exist a function $\Delta ^{*}(t)\in C({\mathbb R}_{+},{%
\mathbb 
R})$ and $T\in {\mathbb R}_{+}$ such that $\Delta ^{*}(t)$ is a
nondecreasing function for $t\geq T$ and $\Delta (t,x)\leq \Delta
^{*}(t)\leq t$ for $t\geq T,x\in {\mathbb 
R}$.

\item  {H7)} $\sigma $ is a quotient of odd integers.
\end{description}

\noindent For the sake of convenience, we will employ the following notation 
$$
R(t)=\int_t^\infty \frac{ds}{r(s)^{1/\sigma }},\quad R(t,T)=\int_T^t\frac{ds%
}{r(s)^{1/\sigma }},\quad R_0=\int_0^\infty \frac{ds}{r(s)^{1/\sigma}}\,. 
$$

In the following section, we give several preparatory lemmas which will be
used for later results. In Section 3, we will discuss the case $R_0<\infty $. 
The case $R_0=\infty $ will be studied in Section 4.

\section{Preparatory Lemmas}

\begin{lemma}  Suppose $x(t)$ is an eventually-positive solution of (\ref
{2}). Then $x'(t)$ is of constant sign eventually.
\end{lemma}

\paragraph{Proof.}
Assume that there exists $t_0\geq 0$ such that $x(t)>0$, for $t\geq t_0$. It
follows from (H6) that there exists $t_1\geq t_0$ such that $x(\Delta
(t,x(t)))>0$ for $t\geq t_1$. From (H4) and (\ref{2}) we conclude that 
$(r(t)(x'(t))^\sigma )'<0$ for $t\geq t_1$. If $x^{\prime
}(t)$ is not eventually positive, then there exists $t_2\geq t_1$ such that 
$x'(t_2)\leq 0$. Therefore, $r(t_2)(x'(t_2))^\sigma \leq 0$.
  From (\ref{2}), we have 
$$
r(t)(x'(t))^\sigma -r(t_2)(x'(t_2))^\sigma
+\int_{t_2}^tf(s,x(s),x(\Delta (s,x(s))))ds=0. 
$$
Thus 
$$
r(t)(x'(t))^\sigma \leq -\int_{t_2}^tf(s,x(s),x(\Delta
(s,x(s))))ds<0, 
$$
for $t\geq t_2$. This shows that $x'(t)<0$ for $t\geq t_2$. The
proof is complete.\hfill$\diamondsuit $\smallskip

As a consequence, an eventually positive solution $x(t)$ of (\ref{2}) either
satisfies $x(t)>0$ and $x'(t)>0$ for all large $t$, or, $x(t)>0$ and 
$x'(t)<0$ for all large $t$.

\begin{lemma} Suppose that 
\begin{equation}\label{3}
R_0=\int_0^\infty \frac{ds}{r(s)^{1/\sigma }}<\infty \,, 
\end{equation}
and that $x(t)$ is an eventually positive solution of (\ref{2}). Then 
$\lim _{t\to \infty }x(t)$ exists.
\end{lemma}

\paragraph{Proof.}
If not, then we have $\lim _{t\to \infty }x(t)=\infty $ by Lemma 1. On the
other hand, we have noted that $r(t)(x'(t))^\sigma $ is monotone
decreasing eventually. Therefore, there exists $t_1\geq 0$ such that 
$$
r(t)(x'(t))^\sigma \leq r(t_1)(x'(t_1))^\sigma ,\quad 
\mbox{for }t\geq t_1\,. 
$$
Then 
\begin{equation}
\label{4}x'(t)\leq (r(t_1))^{1/\sigma }x'(t_1)\frac
1{r(t)^{1/\sigma }}, 
\end{equation}
for $t\geq t_1$, and after integrating, 
$$
x(t)-x(t_1)\leq (r(t_1))^{1/\sigma }x'(t_1)R(t_1,t), 
$$
for $t\geq t_1$. But this is contrary to the fact that $\lim _{t\to \infty
}x(t)=\infty $ and the assumption that $R_0<\infty $. The proof is complete. 
\hfill$\diamondsuit$\smallskip 

\begin{lemma} Suppose that $R_0<\infty $. Let $x(t)$ be an eventually
positive solution of (\ref{2}). Then there exist $a_1>0,a_2>0$ and 
$T\geq 0$ such that $a_1R(t)\leq x(t)\leq a_2$ for $t\geq T$.  
\end{lemma}

\paragraph{Proof.}
By Lemma 2, there exists $t_0\geq 0$ such that $x(t)\leq a_2$ for some
positive number $a_2$. We know that $x'(t)$ is of constant sign
eventually by Lemma 1. If $x'(t)>0$ eventually, then $R(t)\leq x(t)$
eventually because $\lim _{t\to \infty }R(t)=0$. If $x'(t)<0$
eventually, then since $r(t)(x'(t))^\sigma $ is also eventually
decreasing, we may assume that $x'(t)<0$ and $r(t)(x'(t))^%
\sigma $ is monotone decreasing for $t\geq T$. By (4), we have 
$$
x(s)-x(t)\leq (r(T))^{1/\sigma }x'(T)R(t,s),\quad s\geq t\geq T. 
$$
Taking the limit as $s\to \infty $ on both sides of the above inequality, 
$$
x(t)\geq -(r(T))^{1/\sigma }x'(T)R(t), 
$$
for $t\geq T$. The proof is complete. \hfill$\diamondsuit$\smallskip 

Our next result is concerned with necessary conditions for the function $f$
to hold in order that an eventually positive solution of (\ref{2}) exist.

\begin{lemma} Suppose that $R_0<\infty $ and $x(t)$ is an eventually
positive solution of (\ref{2}). Then
$$
\int_0^\infty \frac 1{r(t)^{1/\sigma }}\left( \int_0^tf(s,x(s),x(\Delta
(s,x(s))))ds\right) ^{1/\sigma }dt<\infty . 
$$
\end{lemma}

\paragraph{Proof.}
In view of Lemma 1, we may assume without loss of generality that $x(t)>0$,
and, $x'(t)>0$ or $x'(t)<0$ for $t\geq 0$. From (\ref{2}),
we have 
$$
r(t)(x'(t))^\sigma -r(0)(x'(0))^\sigma
+\int_0^tf(s,x(s),x(\Delta (s,x(s))))ds=0\,. 
$$
Thus, if $x'(t)>0$ for $t\geq 0$, we have 
$$
\displaylines{
\int_0^u\frac 1{r(t)^{1/\sigma }}\left( \int_0^tf(s,x(s),x(\Delta
(s,.x(s))))ds\right) ^{1/\sigma }dt \cr
\leq (r(0))^{1/\sigma }x'(0)
\int_0^u\frac 1{r(t)^{1/\sigma }}dt\,,  
}%
$$
for $u\geq 0$, and 
$$
\int_0^u\frac 1{r(t)^{1/\sigma }}\left( \int_0^tf(s,x(s),x(\Delta
(s,x(s))))ds\right) ^{1/\sigma }dt\leq (r(0))^{1/\sigma }x'(0)R_0
<\infty \,. 
$$
If $x'(t)<0$ for $t\geq 0$, we have 
$$
\int_0^u\frac 1{r(t)^{1/\sigma }}\left( \int_0^tf(s,x(s),x(\Delta
(s,x(s))))ds\right) ^{1/\sigma }dt\leq -\int_0^\infty x'(s)ds\leq
x(0)<\infty \,. 
$$
The proof is complete. \hfill$\diamondsuit$\smallskip

We now consider the case where $R_0=\infty $.

\begin{lemma} Suppose that 
\begin{equation}
\label{5}R_0=\int_0^\infty \frac{ds}{r(s)^{1/\sigma }}=\infty \,. 
\end{equation}
Let $x(t)$ be an eventually positive solution of (\ref{2}). Then
 $x'(t)$ is eventually positive and there exist $c_1>0$, $c_2>0$ and 
 $T\geq 0$ such that $c_1\leq x(t)\leq c_2R(t,T)$ for $t\geq T$. 
\end{lemma}

\paragraph{Proof.}
In view of Lemma 1, $x'(t)$ is of constant sign eventually. If $%
x(t)>0$ and $x'(t)<0$ for $t\geq T$, then we have 
$$
r(t)(x'(t)^\sigma )\leq r(T)(x'(T)^\sigma )<0\,. 
$$
Thus 
$$
x'(t)\leq r(T)^{1/\sigma }x'(T)\frac 1{r(t)^{1/\sigma
}},\quad t\geq T, 
$$
which after integrating yields 
$$
x(t)-x(T)\leq r(T)^{1/\sigma }x'(T)\int_T^t\frac{ds}{r(s)^{1/\sigma
}}. 
$$
The left hand side tends to $-\infty $ in view of (5), which is a
contradiction. Thus $x'(t)$ is eventually positive, and thus $%
x(t)\geq c_1$ eventually for some positive constant $c_1$. Furthermore, the
same reasoning just used also leads to 
$$
x(t)\leq x(T_0)+r(T_0)^{1/\sigma }x'(T_0)\int_{T_0}^t\frac{ds}{%
r(s)^{1/\sigma }}, 
$$
for $t\geq T_0$, where $T_0$ is a number such that $x(t)>0$ and $x^{\prime
}(t)>0$ for $t\geq T_0$. Since $R_0=\infty $, thus there is $c_2>0$ such
that $x(t)\leq c_2R(T,t)$ for all large $t$. The proof is complete.
\hfill$\diamondsuit$

\section{The case $R_0<\infty $}

We have shown in the previous section that when $x(t)$ is an eventually
positive solution of (\ref{2}), then $(r(t)(x'(t))^\sigma )'$
is eventually decreasing and $x'(t)$ is eventually of constant sign.
We have also shown that under the assumption that $R_0<\infty$, $x(t)$ must
converge to some (nonnegative) constant. As a consequence, under the
condition $R_0<\infty $, we may now classify an eventually positive solution 
$x(t)$ of (\ref{2}) according to the limits of the sequences $x(t)$ and $%
r(t)(x'(t))^\sigma $. For this purpose, we first denote the set of
eventually-positive solutions of (\ref{2}) by $P$. We then single out
eventually-positive solutions of (\ref{2}) which converge to zero or to
positive constants, and denote the corresponding subsets by $P_0$ and $%
P_\alpha $ respectively. But for any $x(t)$ in $P_\alpha $, since $%
r(t)(x'(t))^\sigma $ either tends to a finite limit or to $-\infty $%
, we can further partition $P_{+}$ into $P_\alpha ^\beta $ and $P_\alpha
^{-\infty }$.

\begin{theorem} Suppose $R_0<\infty $. Then any eventually positive
solutions of (\ref{2}) must belong to one of the following classes:
$$ \displaylines{
P_0=\left\{ x(t)\in P|\lim _{t\to \infty }x(t)=0\right\} , \cr 
P_\alpha ^\beta =\left\{ x(t)\in P|\lim _{t\to \infty }x(t)=\alpha
>0,\quad\lim _{t\to \infty }r(t)(x'(t)^\sigma )=\beta
\right\} ,\cr 
P_\alpha ^{-\infty }=\left\{ x(t)\in P|\lim _{t\to \infty
}x(t)=\alpha >0,\quad\lim _{t\to \infty }r(t)(x'
(t)^\sigma )=-\infty \right\} . 
}$$
\end{theorem}

To justify the above classification scheme, we will derive several existence
theorems.

\begin{theorem} Suppose $R_0<\infty $. Then a necessary and
sufficient condition for (\ref{2}) to have an eventually positive
solution $x(t)$ which belong to $P_\alpha $ is that for some $C>0$,
\begin{equation}
\label{6}\int_0^\infty \left( \frac 1{r(t)}\int_0^tf(s,C,C)ds\right)
^{1/\sigma }dt<\infty \,.
\end{equation}
\end{theorem} 

\paragraph{Proof.}
Let $x(t)$ be any eventually positive solution of (\ref{2}) such that \\ 
$\lim _{t\to \infty }x(t)=c>0$. Thus, in view of (H6), there exist $C_1>0$, 
$C_2>0$ and $T\geq 0$ such that $C_1\leq x(t)\leq C_2$, $C_1\leq x(\Delta
(t,x(t)))\leq C_2$ for $t\geq T$. On the other hand, using Lemma 4 we have 
$$
\int_T^\infty \left( \frac 1{r(t)}\int_0^tf(s,x(s),x(\Delta
(s,x(s))))ds\right) ^{1/\sigma }dt<\infty . 
$$
Since $f(t,u,v)$ is nondecreasing in $u$ and $v$ for each fixed $t$, thus we
have 
$$
\int_T^\infty \left( \frac 1{r(t)}\int_0^tf(s,C_1,C_1)ds\right) ^{1/\sigma
}dt<\infty . 
$$
Conversely, let $a=C/2$. In view of (\ref{6}), we may choose a $T\geq 0$ so
large that 
\begin{equation}
\label{7}\int_T^\infty \left( \frac 1{r(t)}\int_0^tf(s,C,C)ds\right)
^{1/\sigma }dt<a. 
\end{equation}
Define the set 
$$
\Omega =\left\{ 
\begin{array}{l}
x\in C([T_{-1},+\infty ), 
{\mathbb R}) : x(t)=a, \mbox{ for }T_{-1}\leq t<T, \\ \mbox{ and } a\leq
x(t)\leq 2a,\mbox{ for }t\geq T 
\end{array}
\right\} . 
$$
Then $\Omega $ is a bounded, convex and closed subset of $C([T_{-1},+\infty
),{\mathbb R})$. Let us further define an operator $F:\Omega \to
C([T_{-1},+\infty ),{\mathbb R})$ by 
\begin{equation}
\label{8}Fx(t)=\left\{ 
\begin{array}{ll}
a+\int_{t\ }^\infty \left( \frac 1{r(s)}\int_0^sf(u,x(u),x(\Delta
(u,x(u))))du\right) ^{1/\sigma }ds, & t\geq T, \\ 
[3pt] Fx(T), & T_{-1}\leq t\geq T. 
\end{array}
\right. 
\end{equation}
The mapping $F$ have the following properties. $F$ maps $\Omega $ into $%
\Omega $. Indeed, if $x(t)\in \Omega $, then 
$$
\displaylines{
a\leq Fx(t)=a+\int_t^\infty \left( \frac 1{r(s)}\int_0^sf(u,x(u),x(\Delta
(u,x(u))))du\right) ^{1/\sigma }ds  \cr
\leq a+\int_t^\infty \left( \frac 1{r(s)}\int_0^sf(u,C,C)du\right)
^{1/\sigma }ds\leq 2a. 
}%
$$
Next, we show that $F$ is continuous. To see this, let $\epsilon >0$. Choose 
$M\geq T$ so large that 
\begin{equation}
\label{9}\int_t^\infty \left( \frac 1{r(s)}\int_0^sf(u,C,C)du\right)
^{1/\sigma }ds<\frac \epsilon 2\,. 
\end{equation}
Let $\{x^{(n)}\}$ be a sequence in $\Omega $ such that $x^{(n)}\to x $.
Since $\Omega $ is closed, $x\in \Omega $. Furthermore, for any $s\geq t\geq
M$, 
\begin{eqnarray*} 
\lefteqn{ \left| Fx^{(n)}(t)-Fx(t)\right| }\\
&\leq& \int_t^\infty \left( \frac 1{r(s)}\int_0^sf(u,C,C)du\right) ^{1/\sigma
}ds+\int_t^\infty \left( \frac 1{r(s)}\int_0^sf(u,C,C)du\right) ^{1/\sigma
}ds \\
&\leq& 2\int_t^\infty \left( \frac 1{r(s)}\int_0^sf(u,C,C)du\right) ^{1/\sigma
}ds<\epsilon \,. 
\end{eqnarray*}
For $T\leq t\leq s\leq M$, 
\begin{eqnarray*}
\lefteqn{ \left| Fx^{(n)}(t)-Fx(t)\right| }\\
&\leq& \int_M^\infty \left( \frac 1{r(s)}\int_0^sf(u,C,C)du\right) ^{1/\sigma
}ds+\int_M^\infty \left( \frac 1{r(s)}\int_0^sf(u,C,C)du\right) ^{1/\sigma
}ds \\ 
&&+\int_t^M\left( \frac 1{r(s)}\int_0^sf(u,C,C)du\right) ^{1/\sigma
}ds-\int_s^M\left( \frac 1{r(s)}\int_0^sf(u,C,C)du\right) ^{1/\sigma }ds  \\
&\leq& \epsilon +\int_t^s\left( \frac 1{r(s)}\int_0^sf(u,C,C)du\right)
^{1/\sigma }ds \\
&\leq& \epsilon +\max _{T\leq u\leq M}\frac
1{r(u)}\int_0^uf(v,C,C)dv\left| s-t\right| \\
&\leq& \epsilon +C_0\left| s-t\right| <2\epsilon ,\quad\mbox{if}\left| s-t\right|
<\frac \epsilon {C_0}\,, 
\end{eqnarray*}
where $C_0=\max_{T\leq u\leq M} \int_0^uf(v,C,C)\,dv /r(u)$. And for $%
T_{-1}\leq t\leq s<T$, 
$$
| Fx^{(n)}(t)-Fx(t) | =0\,. 
$$
These statements show that $\| Fx^{(v)}-Fx\| $ tends to zero, i.e., $F$ is
continuous.

When $s,t\geq M$, by (\ref{9}) we have 
\begin{eqnarray*}
\left| Fx(s)-Fx(t)\right| 
&\leq& \int_t^\infty \left( \frac 1{r(s)}\int_0^sf(u,C,C)du\right) ^{1/\sigma
}ds \\
&&+\int_t^\infty \left( \frac 1{r(s)}\int_0^sf(u,C,C)du\right) ^{1/\sigma}ds
<\epsilon \,, 
\end{eqnarray*}
which holds for any $x\in \Omega $. Therefore, $F\Omega $ is precompact. In
view of Schauder's fixed point theorem, we see that there is an $x^{*}\in
\Omega $ such that F$x^{*}=x^{*}$. It is easy to check that $x^{*}$ is an
eventually positive solution of (\ref{2}). The proof is complete.\hfill$%
\diamondsuit $

\begin{theorem} \label{thm3}
Suppose $R_0<\infty $. A necessary and sufficient
condition for (\ref{2}) to have an eventually-positive solution $x(t)$
which belongs to $P_\alpha ^\beta $ is that (6) holds for some $C>0$ and 
that for some $D>0$,
\begin{equation}
\label{10}\int_0^\infty f(t,D,D)dt<\infty \,.
\end{equation}
\end{theorem} 

\paragraph{Proof.}
If $x(t)$ is an eventually-positive solution in $P_\alpha^\beta $, then, in
view of Theorem 2, we see that (\ref{6}) holds. Furthermore, as in the proof
of Theorem 2, $0<C_1\leq x(t)\leq C_2,C_1\leq x(\Delta (t,x(t)))\leq C_2$
for $t\geq T$. In view of (\ref{2}), we see that 
\begin{eqnarray*}
\int_T^\infty f(s,C_1,C_1)ds & \leq& \int_T^\infty f(s,x(s),x(\Delta
(s,x(s))))ds \\  
& =& r(T)(x'(T))^\sigma -\lim _{t\to \infty }r_m(t)(x'(t))^\sigma 
<\infty\, . 
\end{eqnarray*}
Conversely, in view of (\ref{10}), we can choose a $T\geq 0$ such that 
$$
\int_T^\infty f(t,D,D)dt<\left( \frac D{2R_0}\right) ^\sigma . 
$$
We define the subset $\Omega $ of $C([T_{-1},+\infty ),{\mathbb R})$ as
follows 
$$
\Omega =\left\{ 
\begin{array}{l}
x\in C([T_{-1},+\infty ), 
{\mathbb R}) : x(t)=D/2 \mbox{ for } T_{-1}\leq t<T, \\ \mbox{ and } D/2\leq
x(t)\leq D,\mbox{ for } t\geq T 
\end{array}
\right\} . 
$$
Then $\Omega $ is a bounded, convex and closed subset of $C([T_{-1},+\infty
),{\mathbb R})$. In view of $R_0$ and (\ref{10}), we can further define an
operator $F:\Omega \to C([T_{-1},+\infty ),{\mathbb R})$ as 
$$
Fx(t)=\left\{ 
\begin{array}{l}
D-\int_{t\ }^\infty \left( \frac 1{r(s)}\int_s^\infty f(u,x(u),x(\Delta
(u,x(u))))du\right) ^{1/\sigma }ds \quad t\geq T, \\ 
[3pt] Fx(T) \hfill T_{-1}\leq t\geq T\,. 
\end{array}
\right. 
$$
Then, arguments similar to those in the proof of Theorem 2 show that $F$ has
a fixed point $u$ which satisfies 
$$
r(t)(u'(t))^\sigma =\int_t^\infty f(s,u(s),u(\Delta (s,u(s))))ds,
\quad t\geq T\,. 
$$
Hence $\lim _{t\to \infty }r(t)(u'(t))^\sigma =0$ as required.
Choose a $T\geq 0$ such that 
$$
\int_T^\infty f(t,D,D)dt<\left( \frac D{4R_0}\right) ^\sigma \quad\mbox{and}%
\quad R(t)<\left( \frac D{4R_0}\right) ^\sigma 
$$
for $t\geq T$, and let 
$$
Fx(t)=\left\{ 
\begin{array}{lr}
D-\int_t^\infty \big( \frac 1{r(s)}+\frac 1{r(s)}\int_0^sf(u,x(u),x(\Delta
(u,x(u))))du\big) ^{1/\sigma }ds, \quad t\geq T, &  \\ 
[3pt] Fx(T), \hfill T_{-1}\leq t<T\,. &  
\end{array}
\right. 
$$
Then under the same conditions (\ref{6}) and (\ref{10}), we can shows that 
$F$ has a fixed point $u$ which satisfies $\lim _{t\to \infty }u(t)=D>0 $
and 
$$
r(t)(u'(t))^\sigma =1+\int_t^\infty f(s,u(s),u(\Delta (s,u(s))))ds,
\quad t\geq T\,. 
$$
Therefore, $\lim _{t\to \infty }r(t)(u'(t))^\sigma =1>0$, and the
present proof is complete. \hfill$\diamondsuit$ \smallskip 

In view of Theorem 3, the following result is obvious.

\begin{theorem} \label{thm4}
Suppose $R_0<\infty $. A necessary and sufficient
condition for (\ref{2}) to have an eventually-positive solution $x(t)$
which belongs to $P_\alpha ^{-\infty }$ is that (6) holds for some $C>0$ and 
that for any $D>0$,
\begin{equation} \label{12}
\int_0^\infty f(t,D,D)dt=\infty 
\end{equation}
\end{theorem} 

Our final result concerns with the existence of eventually-positive
solutions in $P_0$ .

\begin{theorem} 
Suppose $R_0<\infty $ and $\sigma =1$. If for some $C>0$,
\begin{equation} \label{13}
\int_0^\infty f(t,CR(t),CR(\Delta _{*}(t)))dt<\infty ,
\end{equation}
then (\ref{2}) has an eventually-positive solution in $P_0$. 
Conversely,  if (\ref{2}) has an eventually-positive solution $x(t)$ such 
that $\lim_{t\to \infty }x(t)=0$ and  \\
$\lim _{t\to \infty}r(t)(x'(t))^\sigma =d\neq 0$, then for some $C>0$,
$$
\int_0^\infty f(t,CR(t),CR(\Delta _{*}(t)))dt<\infty \,. 
$$
\end{theorem}

\paragraph{Proof.}
Suppose (\ref{13}) holds. Then there exists a $T\geq 0$ such that 
$$
\int_t^\infty f(s,CR(s),CR(\Delta _{*}(s)))ds<\frac C2\quad \mbox{for }t\geq
T\,. 
$$
Consider the equation 
\begin{equation}
\label{14}x(t)=\left\{ 
\begin{array}{ll}
R(t)\left( \frac C2+\int_T^tf(s,x(s),x(\Delta (s,x(s))))ds\right) &  \\ 
+\int_t^\infty R(s)f(s,x(s),x(\Delta (s,x(s))))ds & t\geq T, \\ 
[4pt] Fx(T) & T_{-1}\leq t<T\,. 
\end{array}
\right. 
\end{equation}
It is easy to check that a solution of (\ref{14}) must be a solution of (\ref
{2}). We shall show that (\ref{14}) has a positive solution $x(t)$ which
belongs to $P_0$ by means of the method of successive approximations.
Consider the sequence $\{x_k(t)\}$ of successive approximating sequences
defined as follows. 
$$
\displaylines{
x_1(t)=0 \quad\mbox{for } t\geq T_{-1}, \cr 
x_{k+1}(t)=Fx_k(t),\quad \mbox{for }t\geq T_{-1},\; k=1,2,\dots , 
}%
$$
where $F$ is defined by 
$$
Fx(t)=\left\{ 
\begin{array}{ll}
R(t)\left( \frac C2+\int_T^tf(s,x(s),x(\Delta (s,x(s))))ds\right) &  \\ 
+\int_t^\infty R(s)f(s,x(s),x(\Delta (s,x(s))))ds & t\geq T, \\[4pt] 
Fx(T) & T_{-1}\leq t<T\,. 
\end{array}\right. 
$$
In view of (H3), it is easy to see that $0\leq x_k(t)\leq x_{k+1}(t)$ for $%
t\geq T$ and $k=1,2,\dots$. On the other hand, 
$$
x_2(t)=Fx_1(t)=\frac C2R(t)\leq CR(t),\quad t\geq T\,, 
$$
and inductively, 
\begin{eqnarray*}
Fx_k(t)&\leq &\frac C2R(t)+R(t)\int_T^tf(s,CR(s),CR(\Delta ^*(s)))ds \\  
&&+R(t)\int_t^\infty f(s,CR(s),CR(\Delta ^*(s)))ds \\  
&\leq& \frac C2R(t)+R(t)\int_T^\infty f(s,CR(s),CR(\Delta ^*(s)))ds \\  
&\leq& CR(t)\,, 
\end{eqnarray*}
for $k\geq 2$. Therefore, by means of Lebesgue's dominated convergence
theorem, we see that $Tx^*=x^*$. Furthermore, it is clear that $x(t)$
converges to zero as $t\to \infty $.

Let $x(t)$ be an eventually positive solution of (\ref{2}) such that $%
x(t)\to 0$ and $r(t)(x'(t))^\sigma \to $ $d<0$ (the proof of the
case $d>0$ being similar). Then there exist $C_1>0$, $C_2>0$ and $T\geq 0$
such that $-C_1<r(t)(x'(t))^\sigma <-C_2,$ for $t\geq T$. Hence, 
$$
-C_1^{1/\sigma }\frac 1{r(t)^{1/\sigma }}<x'(t)<-C_2^{1/\sigma
}\frac 1{r(t)^{1/\sigma }}, 
$$
and, after integrating, 
$$
-C_1^{1/\sigma }R(s,t)<x(s)-x(t)<-C_2^{1/\sigma }R(s,t), 
$$
for $s>t\geq T$. Let $s\to \infty $, then $-C_1^{1/\sigma
}R(t)<-x(t)<-C_2^{1/\sigma }R(t)$. That is, $C_2^{1/\sigma
}R(t)<x(t)<C_1^{1/\sigma }R(t)$. On the other hand, by (\ref{2}), 
$$
r(t)(x'(t))^\sigma =r(T)(x'(T))^\sigma
+\int_T^tf(s,x(s),x(\Delta (s,x(s))))ds,\quad t\geq T\,.  
$$
Since lim$_{t\to \infty }r(t)(x'(t))^\sigma =d<0$, we have 
$$
\int_T^\infty f(s,x(s),x(\Delta (s,x(s))))ds=r(T)(x'(T))^\sigma
-d<\infty \,. 
$$
Thus, 
$$
\int_T^\infty f(s,C_1^{1/\sigma }R(s),C_1^{1/\sigma }R(\Delta
_{*}(s)))ds\leq \int_T^\infty f(s,x(s),x(\Delta (s,x(s))))ds<\infty \,. 
$$
The proof is complete.\hfill$\diamondsuit $

\section{The case $R_0=\infty $}

In this section, we assume that $R_0=\infty $. Let $P$ denotes the set of
all eventually-positive solutions of (\ref{2}). Recall that if $x(t)$
belongs to $P$, then $r(t)(x'(t))^\sigma $ is eventually
decreasing. Furthermore, in view of Lemma 5, we see that $x'(t)$,
and hence $r(t)(x'(t))^\sigma $, are eventually positive. Hence $%
x(t)$ either tends to a positive constant or to positive infinity, and $%
r(t)(x'(t))^\sigma $ tends to a nonnegative constant. Note that if $%
x(t)$ tends to a positive constant, then $r(t)(x'(t))^\sigma $ must
tend to zero. Otherwise $r(t)(x'(t))^\sigma \geq d>0$ for $t$
larger than or equal to $T$, so that 
$$
x'(t)\geq d^{1/\sigma }\frac 1{r^{1/\sigma }(t)}, 
$$
and 
$$
x(t)\geq x(T)d^{1/\sigma }\int_T^t\frac 1{r^{1/\sigma }(s)}ds\to \infty \,,
\mbox{ as } t\to \infty\,,  
$$
which is a contradiction.

\begin{theorem} \label{thm6} Suppose that $R_0=\infty $. 
Then any eventually-positive solution $x(t)$ of (\ref{2}) must belong to 
one of the following  classes: 
$$ \displaylines{
P_\alpha ^0=\left\{ x(t)\in P|\lim _{t\to \infty }x(t)\in (0,\infty
),\quad \lim _{t\to \infty }r(t)(x'(t))^\sigma =0\right\},  \cr
P_\infty ^0=\left\{ x(t)\in P|\lim _{t\to \infty }x(t)=+\infty ,
\quad\lim _{t\to \infty }r(t)(x'(t))^\sigma =0\right\} ,  \cr
P_\infty ^\beta =\left\{ x(t)\in P|\lim _{t\to \infty }x(t)=+\infty ,
\quad\lim _{t\to \infty }r(t)(x'(t))^\sigma =\beta \neq
0\right\} . 
}$$
\end{theorem}

In order to justify our classification scheme, we present the following two
results.

\begin{theorem} \label{thm7} 
Suppose that $R_0=\infty $. A necessary and sufficient
condition for (\ref{2}) to have an eventually-positive solution $x(t)$
which belongs to $P_\alpha ^0$ is that for some $C>0$, 
\begin{equation} \label{15}
\int_0^\infty \left( \frac 1{r(t)}\int_t^\infty f(s,C,C)ds\right)
^{1/\sigma }dt<\infty \,. 
\end{equation}
\end{theorem}

\paragraph{Proof.}
Let $x(t)$ be an eventually-positive solution of (\ref{2}) which belong to $%
P_\alpha ^0$, i.e., $\lim _{t\to \infty}x(t)=\alpha >0$ and $\lim _{t\to
\infty }r(t)(x'(t))^\sigma =0$. Then there exist two positive
constants $C_1$, $C_2$ and $T\geq 0$ such that $C_1\leq x(t)\leq C_2$, $%
C_1\leq x(\Delta (t,x(t))\leq C_2 $ for $t\geq T$. On the other hand, in
view of (\ref{2}) we have 
$$
r(t)(x'(t))^\sigma =\int_t^\infty f(s,x(s),x(\Delta (s,x(s))))ds\,, 
$$
for $t\geq T$. After integrating, we see that 
\begin{eqnarray*}
\lefteqn{ \int_T^\infty \left( \frac 1{r(t)}\int_t^\infty f(s,C,C)ds\right) 
^{1/\sigma
}dt }\\ 
&\leq& \int_0^\infty \left( \frac 1{r(t)}\int_t^\infty f(s,x(s),x(\Delta
(s,x(s))))ds\right) ^{1/\sigma }dt \\
&\leq& \alpha -x(T). 
\end{eqnarray*}
The proof of the converse is similar to that of Theorem 1 and hence is
sketched. In view of (\ref{15}), we may choose a $T\geq 0$ so large that 
\begin{equation}
\label{16}\int_T^\infty \left( \frac 1{r(t)}\int_t^\infty f(s,C,C)ds\right)
^{1/\sigma }<\frac C2\,. 
\end{equation}
Define a bounded, convex, and closed subset $\Omega $ of $C([T_{-1},\infty ),%
{\mathbb R})$ and an operator $F:\Omega \to \Omega $ as 
$$
\Omega =\left\{ 
\begin{array}{l}
x\in C([T_{-1},+\infty ), 
{\mathbb R}) : x(t)=\frac C2 \mbox{ for }T_{-1}\leq t<T, \\ \mbox{ and }
\frac C2\leq x(t)\leq C,\mbox{ for } t\geq T\,, 
\end{array}
\right\} 
$$
and 
$$
Fx(t)=\left\{ 
\begin{array}{l}
\frac C2+\int_t^\infty \left( \frac 1{r(s)}\int_s^\infty f(u,x(u),x(\Delta
(u,x(u))))du\right) ^{1/\sigma }ds\quad t\geq T, \\ 
Fx(T) \hfill T_{-1}\leq t<T\,, 
\end{array}
\right. 
$$
respectively. As in the proof of Theorem 3, we prove that $F$ maps $\Omega $
into $\Omega $, that $F$ is continuous, and that F$\Omega $ is precompact.
The fixed point $x^*(t)$ of $F$ will converge to $C/2$ and satisfies (\ref{2}%
). The proof is complete. \hfill$\diamondsuit$\smallskip

We remark that Theorem 7 extends Theorem 6 of Bainov, Markova and Simeonov 
\cite{ref3}. The proof of the following result is again similar to that of
Theorem 3 and hence is omitted.

\begin{theorem} \label{thm8} Suppose $R_0=\infty $. If  for a positive 
constant $C$, 
\begin{equation}
\label{17}\int_0^\infty f(t,CR(t,0),CR(\Delta ^*(t),0))dt<\infty  
\end{equation}
then (\ref{2}) has a solution in $P_\infty ^\beta $. 
Conversely, if (\ref{2}) has a solution $x(t)$ in $P_\infty ^\beta $, then
for some positive constant $C$,
$$
\int_0^\infty f(t,CR(t,0),CR(\Delta _{*}(t),0))dt<\infty \,.$$
\end{theorem}

We remark that our Theorem 8 extends Theorem 5 of Bainov, Markova and
Simeonov \cite{ref3}. In view of Theorems 7 and 8, the following result is
clear.

\begin{theorem} \label{thm9} Suppose $R_0=\infty $. If for any positive
constant $C$ and for some positive constant $D$ such that
$$ \displaylines{
\int_0^\infty \left( \frac 1{r(t)}\int_t^\infty f(s,C,C)ds\right) ^{1/\sigma
}dt=\infty \,,\cr 
\int_0^\infty f(t,DR(t,0),DR(\Delta ^*(t),0))dt<\infty \,,
}$$
then (\ref{2}) has a solution in $P_\infty ^0$. 
\end{theorem}

We remark that our Theorem 9 extends Theorem 7 in \cite{ref3}, and that
several oscillation statements for (\ref{2}) can be proven. Since the method
is similar to that of \cite{ref3}, we omit them here.

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\smallskip

\noindent{\sc Xianling Fan,  Wan-Tong Li, \& Chengkui Zhong} \\ 
Department of Mathematics, Lanzhou University\\ 
Lanzhou, Gansu, 730000, People's Republic of China \\ 
e-mail: liwt@gsut.edu.cn

\end{document}
