\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 17, pp. 1--8.\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 2004 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE-2004/17\hfil Positive solutions to nonlinear dynamic equations]
{Classification and existence of positive solutions to nonlinear
dynamic equations on time scales}

\author[C. Y. Huang and W.T. Li\hfil EJDE-2004/17\hfilneg]
{Can-Yun Huang \& Wan-Tong Li} % in alphabetical order

\address{Can-Yun Huang \hfill\break
Department of Applied Mathematics,
Lanzhou University of Technology \\
Lanzhou, Gansu 730050, China}

\address{Wan-Tong Li \hfill\break
Department of Mathematics, Lanzhou University\\
Lanzhou, Gansu 730000, China}
\email{wtli@lzu.edu.cn}

\date{}
\thanks{Submitted December 16, 2003. Published February 6, 2004.}
\thanks{Both authors were supported by the Key Research and Development Program for
\hfill\break\indent Outstanding  Groups at Lanzhou University of Technology.
\hfill\break\indent
W.-T. Li was supported by grant 10171040 from the NNSF of China,
and by grant \hfill\break\indent
ZS011-A25-007-Z from the NSF of Gansu Province of China}
\subjclass[2000]{34B15, 39A10}
\keywords{Time scale, nonlinear dynamic equation, positive solution}

\begin{abstract}
 A classification scheme is given for the eventually positive solutions
 to a class of second order nonlinear dynamic equations, in terms of their
 asymptotic magnitudes. Also we provide necessary and/or sufficient conditions
 for the existence of positive solutions.
\end{abstract}

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

\section{Introduction}

The study of dynamic equations on time scales, which has recently received a
lot of attention, was introduced by Stefan Hilger in his Ph.D. thesis in
1988 in order to unify continuous and discrete analysis \cite{3}. Since
Hilger formed the definition of derivatives and integrals on time scales,
several authors have expended on various aspects of the new theory, see the
paper by Agarwal et al. \cite{1} and the book by Bohner and Peterson \cite{2}.

In recent years there has been much research activity concerning the
oscillation and nonoscillation of some different equations on time scales,
we refer the reader to the papers \cite{11, 12, 13}. However, to the best of
our knowledge, there have not appeared any results concerned with asymptotic
behavior and existence of solutions of dynamic equations on time scales. In
this paper, we classify positive solutions of the second
order nonlinear dynamic equation
\begin{equation}
y^{\Delta \Delta }(t)+r(t)f( y^{\sigma}(t)) =0\quad \text{ for }t\in \mathbb{T}
\label{1.1}
\end{equation}
according to the limiting behavior and then provide sufficient and/or necessary
conditions for their existence, where $r\in C_{rd}( [t_0,+\infty )\cap
\mathbb{T},\text{ }[0,\infty )$, $r\not{\equiv }0$ for $t\in \mathbb{T}$,
$t_0>0$ and $f(y)>0$ is nondecreasing for any $y\in \mathbb{R}-\{0\}$.

We note that if $\mathbb{T}=\mathbb{R}$, then \eqref{1.1} becomes the differential
equation
\begin{equation}
y^{\prime \prime }(t)+r(t)f( y(t)) =0\quad \text{ for }t\in \mathbb{R}.
\label{1.2}
\end{equation}
The asymptotic behavior of solutions of (\ref{1.2}) has been studied by
several authors under different conditions, see Naito \cite{9,10}.
If $\mathbb{T}=\mathbb{Z}$, then \eqref{1.1} becomes the difference equation
\begin{equation}
\Delta ^2(y_n)+r_nf( y_{n+1}) =0\quad \text{ for }n\in \mathbb{Z},
\label{1.3}
\end{equation}
which has been discussed in detail by many authors, one can refer to
\cite{4, 8, 5, 6, 7}.

Since we are interested in asymptotic behavior of positive solutions of
\eqref{1.1}, we suppose that the time scale $\mathbb{T}$ is a time scale
interval of the form $[t_0,\infty )\cap \mathbb{T}$ and shall employ
$t\geq t_0$ to denote $[t_0,\infty )\cap \mathbb{T}$ unless otherwise stated.
Here a solution of \eqref{1.1} means that $y\in C_{rd}^2$ and satisfies \eqref{1.1}
for $t\geq t_0$.

\section{Some preliminaries}

In this section, we give a short introduction to the time scale calculus.
For the explanation and results we refer to the book by Bohner and Peterson
\cite{2}. \smallskip

\noindent\textbf{Definition}.
The forward jump operator $\sigma
(t)$ and $\rho (t)$ are defined by
\[
\sigma (t):=\inf \{s\in \mathbb{T}:s>t\}\quad\text{and}\quad
\rho (t):=\sup \{s\in \mathbb{T}:s<t\}.
\]
The graininess $\mu $ of the time scale is defined by $\mu (t)=\sigma (t)-t$.
Naturally, $\sigma (t)=t+\mu (t)\geq t$. \smallskip

\noindent\textbf{Definition}.
A point $t\in \mathbb{T}$ is said to
be left-dense if $\rho (t)=t$, right-dense if $\sigma (t)=t$,
left-scattered if $\rho (t)<t$ and right-scattered if $\sigma
(t)>t$.

A function $f:\mathbb{T}\to \mathbb{R}$ is said to be rd-continuous if it
is continuous at each right-dense point and if there exists a finite
left-hand sided limit at all left-dense points. \smallskip

\noindent\textbf{Definition}.
For a function $f:\mathbb{T}\to
\mathbb{R}$, the (delta) derivative is defined by
\[
f^\Delta (t)=\lim_{s\to t}\frac{f( \sigma (t)) -f(s)}{
\sigma (t)-s}=\begin{cases}
\lim_{s\to t}\frac{f( \sigma (t)) -f(s)}{t-s},&\text{if } \mu (t)=0, \\
\frac{f(\sigma (t)) -f(t)}{\sigma (t)-t},&\text{if }\mu (t)>0.
\end{cases}
\]
The function $f$ is said to be differentiable if its derivative exists.

The derivative of $f$ and the jump operator $\sigma $ are related by the
formula
\begin{equation}
f( \sigma (t)) =f^{\sigma}(t)=f(t)+f^\Delta (t)\mu (t)\quad
\text{ for }t\in \mathbb{T}.  \label{2.1}
\end{equation}

\begin{lemma}[\cite{2}] \label{lm1}
The function $f(t)$ is increasing in $t$ if $f^\Delta (t)>0$.
\end{lemma}

 From this lemma, we can say that $f$ is decreasing, nondecreasing and
nonincreasing if $f^\Delta (t)<0$, $f^\Delta (t)\geq 0$ and $f^\Delta
(t)\leq 0$ for $t\in \mathbb{T}$, respectively. \smallskip

\noindent\textbf{Definition} (Antiderivative).
A function
$F:\mathbb{T}\to \mathbb{R}$ is called an antiderivative of
$f:\mathbb{T}\to \mathbb{R}$ provided
\[
F^\Delta (t)=f(t)\quad \text{ for all }t\in \mathbb{T}.
\]

If a function $f:\mathbb{T}\to \mathbb{R}$ is rd-continuous, its
antiderivative exists, denoted by
\[
F(t)=\int_{t_0}^tf(s)\Delta s\quad \text{ for $t_0$ and $t\in \mathbb{T}$},
\]
where
\[
\int_{t_0}^tf(s)\Delta s=\begin{cases}
\int_{t_0}^tf(s)ds,& \text{if }\mathbb{T}=\mathbb{R};\\
\sum_{s\in [t_0,t)}\mu (s)f(s),&\text{if $[t_0,t]$
consists of only isolated points.}
\end{cases}
\]
Obviously, if $f(t)$ is differentiable, its antiderivative exists, and
\begin{equation}
f(t)-f(t_0)=\int_{t_0}^tf^\Delta (s)\Delta s.  \label{2.2}
\end{equation}
For the sake of convenience, we still call the antiderivative $F(t)$ of $
f(t) $ as the integral of $f(t)$, antiderivative calculus as integral
calculus, respectively. Similarly, infinite integral is defined as
\[
\int_{t_0}^\infty f^\Delta (s)\Delta s=\lim_{t\to \infty
}\int_{t_0}^tf^\Delta (s)\Delta s.
\]
If $\mathbb{T}=\mathbb{R}$, then
$\sigma (t)=\rho (t)=t$, $\mu (t)\equiv 0$, $f^\Delta =f'$ and
\[
\int_a^bf(s)\Delta s=\int_a^bf(s)ds.
\]
If $\mathbb{T}=\mathbb{Z}$, then
$\sigma (t)=t+1$, $\rho (t)=t-1$, $\mu (t)\equiv 1$,
$f^\Delta =\Delta f$,  and
\[
\int_a^bf^\Delta (s)\Delta s=\sum_{t=a}^{b-1}f(t)\quad \text{for }a<b.
\]

\section{Main Results}

Let $y(t)$ be a positive solution of \eqref{1.1}. Then from \eqref{1.1} we have
\[
y^{\Delta \Delta }(t)=-r(t)f( y^\sigma (t)) \leq 0,
\]
which, by Lemma \ref{lm1}, implies that $y^\Delta (t)$ is nonincreasing. Thus we
claim that
\[
y^\Delta (t)\geq 0\quad \text{ for }t\geq t_0.
\]
If not, there exists a sufficiently large $t_1\geq t_0$, such that
$y^\Delta(t)<-c$, where $c>0$ is a constant. Hence, for $t>t_1$, we obtain
\[
y(t)-y(t_1)=\int_{t_1}^ty^\Delta (s)\Delta s<\int_{t_1}^t(-c)\Delta
s=-c(t-t_1).
\]
This means that $\lim y(t)=-\infty $, which contradicts $y(t)>0$.

Since $y^\Delta (t)$ is nonincreasing and $y^\Delta (t)\geq 0$ for $t\geq
t_0 $, then there are positive constants $\alpha $ and $\beta $ such that
\[
\alpha \leq y(t)\leq \beta t\quad \text{ for }t\geq t_0.
\]
In view of the above considerations, we may now make the following
classifications. Let $\Omega $ be the set of all non-oscillatory solutions
of \eqref{1.1} and $\Omega ^{+}$ be the subset of $\Omega $ containing those
which are ultimately positive. Then any non-oscillatory solution in
$\Omega^{+}$ must belong to one of the following three sets:
\begin{gather*}
C[\text{max}] =\big\{ y\in \Omega ^{+}:\lim_{t\to +\infty}y^\Delta (t)
=\alpha >0\big\}; \\
C[\text{int}] =\big\{ y\in \Omega ^{+}:{\lim_{t\to +\infty }}y(t)
=\infty \text{ and }{\lim_{t\to +\infty }}y^\Delta (t)=0\big\}; \\
C[\text{min}] =\big\{ y\in \Omega ^{+}:{\lim_{t\to +\infty }}y(t)
=\beta >0\big\} .
\end{gather*}

In the following, we will give several necessary and/or sufficient
conditions for the existence of positive solutions of \eqref{1.1}.

\begin{theorem} \label{thm1}
Equation \eqref{1.1} has a positive solution of class $C[{\rm max}]$
if and only if
\begin{equation}
\int_{t_0}^{+\infty }r(s)f( b\sigma (s)) <\infty \quad \text{for some }
b>0.  \label{3.1}
\end{equation}
\end{theorem}

\begin{proof}
Let $y(t)\in C[{\rm max}]$ of \eqref{1.1}, then
\[
{\lim_{t\to +\infty }}y^\Delta (t)=\alpha >0\quad \text{for }t\geq t_0.
\]
Hence there exist a sufficiently large $t_1$ such that
\[
\frac 12\alpha <y^\Delta (t)<\frac 32\alpha \quad \text{for }t\geq t_1,
\]
so that
\[
\frac 12\alpha t<y(t)<\frac 32\alpha t\quad \text{for }t>t_2>t_1.
\]
Set $b=\alpha/2$. Then the nondecreasing property of $f$ implies
\begin{equation}
f( y(t)) \geq f(bt)\quad \text{and}\quad
f( y^\sigma (t)) \geq f( b\sigma (t)).  \label{3.2}
\end{equation}
Integrating both sides of \eqref{1.1} from $t_1$ to $t$, we see
\[
y^\Delta (t_1)-y^\Delta (t)=\int_{t_1}^tr(s)f( y^\sigma (s)) \Delta s.
\]
Taking limits on both sides of the above equality, we get
\[
{\lim_{t\to \infty }}\int_{t_2}^tr(s)f( y^\sigma (s)) \Delta s
=y^\Delta (t_1)-\alpha ,
\]
which implies
\begin{equation}
\int_{t_1}^\infty r(s)f( y^\sigma (s)) \Delta s<\infty . \label{3.3}
\end{equation}
 From (\ref{3.2}) and (\ref{3.3}), it follows that
\[
\int_{t_2}^\infty r(s)f( b\sigma (s)) \Delta s<\infty .
\]
Conversely, assume that (\ref{3.1}) holds. Then there exists a large number $
T$ such that
\begin{equation}
\int_t^\infty r(s)( f(b\sigma (s))) \Delta s<\frac b2\quad \text{ for }
t\geq T.  \label{3.4}
\end{equation}
Consider the sequence $\{x_n\}_0^\infty $ defined by
$x_0(t)=b/2$ for $t\geq T$
and
\begin{equation}
x_{n+1}(t)=\frac b2+\frac 1t\int_T^t\int_{\tau}^{\infty}
r(s)f( \sigma (s)x_n^\sigma (s)) \Delta s\Delta \tau \label{3.5}
\end{equation}
for $t\geq T$, $n=0,1,2,\cdots $. In view of (\ref{3.4}), the sequence
$\{x_n(t)\}_0^\infty $ is well defined. In fact,
\begin{align*}
x_1(t) &=\frac b2+\frac 1t\int_T^t\int_\tau ^\infty r(s)f( \frac b2\sigma (s))
\Delta s\Delta \tau \\
&\leq \frac b2+\frac{t-T}t\int_T ^\infty r(s)f(b\sigma (s)) \Delta s\Delta \tau \\
&\leq \frac b2+\int_{T}^\infty r(s)f( b\sigma (s)) \Delta s<\frac b2+\frac b2=b
\end{align*}
and $x_1(t)\geq x_0(t)$  for $t\geq T$.
By induction and the nondecreasing property of $f$, we have
\begin{equation}
x_{n+1}(t)\geq x_n(t)\quad \text{for }t\geq T,\; n=0,\; 1,2,\cdots .  \label{3.6}
\end{equation}

Next, we prove $\left\{ x_n(t)\right\} _0^\infty $ is bounded for $t\geq T$.
Since
\[
x_0(t)=\frac b2<b\quad \text{and}\quad x_1(t)<b,
\]
so if we assume $x_n(t)<b$ for $t\geq T$, then
$\sigma (s)x_n^\sigma (s)<b\sigma (s)$ and
\begin{equation}
\begin{aligned}
x_{n+1}(t) &=\frac b2+\frac 1t\int_T^t\int_\tau ^\infty r(s)
f( \sigma (s)x_n^\sigma (s)) \Delta s\Delta \tau  \\
&\leq \frac b2+\frac 1t\int_T^t\int_\tau ^\infty r(s)f( b\sigma (s))
 \Delta s\Delta \tau   \\
&\leq \frac b2+\int_{T}^\infty r(s)f( b\sigma (s)) \Delta s<b\quad
\text{ for }t\geq T,
\end{aligned} \label{3.7}
\end{equation}
which, by induction implies $\{x_n(t)\}_0^\infty $ is bounded for $t\geq T$.
In view of (\ref{3.6}), we know $\{x_n(t)\}_0^\infty $ is pointwise
convergent to some function $x^{*}(t)$. By means of the Lebesgue's dominated
convergence theorem, see \cite{0, 101}, we obtain
\[
x^{*}(t)=\frac b2+\frac 1t\int_T^t\int_\tau ^\infty r(s)f( \sigma (s)x^\sigma (s))
\Delta s\Delta \tau \quad \text{ for }t\geq T
\]
and $\frac b2\leq x^{*}(t)<b$.
Setting $y(t)=tx^{*}(t)$, i.e.,
\[
y(t)=\frac b2t+\int_T^t\int_\tau ^\infty r(s)f(
y^\sigma (s)) \Delta s\Delta \tau \text{ for }t\geq T,
\]
obviously, $y(t)$ is a solution of (\ref{3.1}) and belongs to $C[{\rm max}]$.
The proof is complete.
\end{proof}

\begin{theorem}\label{thm2}
Equation \eqref{1.1} has a positive solution of class $C[{\rm min}]$
if and only if
\begin{equation}
\int_{t_0}^{+\infty }\int_\tau ^{+\infty \text{ }}r(s)f(d)\Delta
s\Delta \tau <\infty \quad \text{ for some }d>0.  \label{4.1}
\end{equation}
\end{theorem}

\begin{proof}
Let $y(t)\in C[{\rm min}]$ for \eqref{1.1}, then
\[
{\lim_{t\to +\infty }}y(t)=\beta >0\quad\text{and}\quad
\lim_{t\to +\infty }y^\Delta (t)=0\quad \text{for }t\geq t_0.
\]
Hence there exists a sufficient large $t_1$ such that
\[
\frac 12\beta <y(t)<\frac 32\beta \quad \text{for }t\geq t_1.
\]
Set $d=\beta/2 $, then the nondecreasing property implies
\[
f( y(t)) >f(d)\quad \text{and}\quad
f( y^\sigma (t))>f( d) \quad \text{for }t>t_1.
\]
Integrating both sides of \eqref{1.1} from $t$ to $\infty $ for $t>t_1$, we
obtain
\begin{equation}
\beta -y(t)=\int_t^\infty \int_\tau ^\infty r(s)f(y^\sigma (s))
 \Delta s\Delta \tau ,  \label{4.2}
\end{equation}
which implies
\[
\int_{t_1}^\infty \int_\tau ^\infty r(s)f(d)\Delta
s\Delta \tau <\infty .
\]
i.e., (\ref{4.1}) holds.
The rest of the proof is similar to that of Theorem \ref{thm1}, we omit it here.
\end{proof}

\begin{theorem} \label{thm3}
If \eqref{1.1} has a positive solution in $C[{\rm int}]$,
then
\begin{gather}
\int_{t_0}^\infty r(s)f( a) \Delta s<\infty \quad\text{for some }a>0\,,
\label{5.1}\\
\int_{t_0}^\infty \int_\tau ^\infty r(s)f( b\sigma
(s)) \Delta s\Delta \tau =\infty \quad\text{for every }b>0.  \label{5.2}
\end{gather}
\end{theorem}

\begin{proof}
Let $y\in C[{\rm int}]$, be a solution of \eqref{1.1}, then
${\lim_{t\to +\infty }}y(t)=\infty $ and ${\lim_{t\to +\infty }}y^\Delta (t)=0$.
Hence there exist two positive
constants $a$, $b$ and a sufficient large $t_1>t_0$ such that $a<y(t)<bt$
for $t>t_1$, which, in view of the nondecreasing property of $f$, implies
that
\begin{equation}
\begin{gathered}
f( y(t)) \geq f(a)\quad \text{and}\quad f( y^\sigma (t)) \leq f( a) ,  \\
f( y(t)) \leq f(bt)\quad \text{and}\quad f( y^\sigma (t)) \leq f( b\sigma (t))
\quad \text{for }t>t_1.
\end{gathered}\label{5.3}
\end{equation}
 From equation \eqref{1.1}, we have
\begin{equation}
y^\Delta (t)+\int_{t_1}^tr(s)f( y^\sigma (s)) \Delta
s=y^\Delta (t_1)\text{ for }t>t_1.  \label{5.4}
\end{equation}
In view of ${\lim_{t\to +\infty }}y^\Delta (t)=0$, (\ref{5.4}) yields
\[
\int_{t_1}^\infty r(s)f( y^\sigma (s)) \Delta
s=y^\Delta (t_1),
\]
and so
$\int_{t_1}^\infty r(s)f(a)\Delta s<\infty$,
which implies (\ref{5.1}) holds.

Furthermore, since ${\lim_{t\to +\infty }}y^\Delta (t)=0$,
we obtain
\begin{equation}
\int_s^\infty r(s)f( y^\sigma (s)) \Delta s=y^\Delta (s)\quad
\text{for }s>t_1.  \label{5.5}
\end{equation}
Integrating on both sides of (\ref{5.5}) from $t_1$ to $t$, we obtain
\begin{align*}
y(t)-y(t_1) &=\int_{t_1}^t\int_\tau ^\infty r(s)f(
y^\sigma (s)) \Delta s\Delta \tau \\
&\leq \int_{t_1}^t\int_\tau ^\infty r(s)f( b\sigma
(s)) \Delta s\Delta \tau \text{ for }t>t_1.
\end{align*}
Hence, (\ref{5.3}) and ${\lim_{t\to +\infty }}y(t)=\infty $
imply
\[
\int_{t_1}^\infty \int_\tau ^\infty r(s)f( b\sigma
(s)) \Delta s\Delta \tau =\infty .
\]
The proof is complete.
\end{proof}

\begin{theorem} \label{thm4}
Equation \eqref{1.1} has a positive solution in $C[{\rm int}]$
provided that
\begin{gather}
\int_{t_0}^\infty r(s)f( a\sigma (s)) \Delta s<\infty \quad
\text{ for some }a>0  \label{5.6} \\
\int_{t_0}^\infty \int_\tau ^\infty r(s)f(b)\Delta
s\Delta \tau =\infty \quad \text{ for every }b>0. \label{5.7}
\end{gather}
\end{theorem}

\begin{proof}
In view of (\ref{5.6}) and (\ref{5.7}), there exist two
positive constants $a, b$ and a sufficiently large $t_1$ such that
\begin{equation}
\frac bt<a, \quad \frac bt+\frac 1t\int_{t_1}^t\int_\tau ^\infty r(s)
f( a\sigma (s)) \Delta s\Delta \tau <a\quad \text{for }t\geq t_1.  \label{5.8}
\end{equation}
Consider the sequence $\left\{ x_n(t)\right\} _0^\infty $ defined by
$x_0(t)=0$ and
\[
x_{n+1}(t)=Px_n(t)=\frac bt+\frac 1t\int_{t_1}^t\int_\tau ^\infty r(s)
f( \sigma (s)x_n^\sigma (s)) \Delta
s\Delta \tau \quad \text{for }t\geq t_1,n=0,1,\cdots .
\]
It is easy to see that $\left\{ x_n(t)\right\} _0^\infty $ is well defined.
In fact,
\[
x_1(t)=\frac bt<a,\text{ }x_1^\sigma (s)<a\quad \text{for }t\geq t_1
\]
and
\begin{align*}
x_2(t) &=\frac bt+\frac 1t\int_{t_1}^t\int_\tau ^\infty r(s)
f( \sigma (s)x_1^\sigma (s)) \Delta s\Delta \tau \\
&\leq \frac bt+\frac 1t\int_{t_1}^t\int_\tau ^\infty r(s)f( a\sigma (s))
 \Delta s\Delta \tau <a\quad \text{for }t\geq t_1.
\end{align*}
Furthermore, if we assume that $x_n(t)<a$ for $t\geq t_1$, then
$x_n^\sigma (t)=x_n( \sigma (t)) <a$ and
\begin{align*}
x_{n+1}(t) &=\frac bt+\frac 1t\int_{t_1}^t\int_\tau ^\infty r(s)
f( \sigma (s)x_n^\sigma (s)) \Delta s\Delta \tau  \\
&\leq \frac bt+\frac 1t\int_{t_1}^t\int_\tau ^\infty r(s)f( a\sigma (s))
\Delta s\Delta \tau <a\quad \text{for }t\geq t_1,
\end{align*}
which, by induction, shows that $\{ x_n(t)\} _0^\infty $ is
bounded, i.e.,
\begin{equation}
0\leq x_n(t)<a\quad \text{for }t\geq t_1,\; n=0,1,\cdots .
\label{5.10}
\end{equation}
In view of $x_0(t)\leq x_1(t)$ and the nondecreasing property of $f$, we
have
\begin{equation}
x_{n+1}(t)\geq x_n(t)\quad \text{for }t\geq t_1,\; n=0,1,\cdots . \label{5.11}
\end{equation}
Hence, Lebesgue's dominated convergence theorem \cite{0, 101} implies that
\[
x^{*}(t)=\frac at+\frac 1t\int_{t_1}^t\int_\tau ^\infty r(s)
f( \sigma (s)x^{*}(\sigma (s))) \Delta s\Delta \tau \quad\text{ for }t\geq t_1.
\]
Set $y(t)=tx^{*}(t)$. Then
\[
y(t)=a+\int_{t_1}^t\int_\tau ^\infty r(s)f(
y^\sigma (s)) \Delta s\Delta \tau \quad \text{for }t\geq t_1.
\]
It is easily verified that $y(t)$ is a solution of \eqref{1.1} and belongs
to $C[$int$]$. The proof is complete.
\end{proof}

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