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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 125, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/125\hfil Kwong-Wong-type integral equation]
{Kwong-Wong-type integral equation on time scales}

\author[B. Jia \hfil EJDE-2011/125\hfilneg]
{Baoguo Jia} 

\address{Baoguo Jia \newline
School of Mathematics and Computer Science, Zhongshan University
\newline
Guangzhou, 510275, China}
\email{mcsjbg@mail.sysu.edu.cn}

\thanks{Submitted July 21, 2011. Published September 29, 2011.}
\thanks{Supported by grant 10971232 from the National Natural
 Science Foundation of China}
\subjclass[2000]{34K11, 39A10, 39A99}
\keywords{Nonlinear dynamic equation;
integral equation; \hfill\break\indent nonoscillatory solution}

\begin{abstract}
 Consider the second-order nonlinear dynamic equation
 $$
 [r(t)x^\Delta(\rho(t))]^\Delta+p(t)f(x(t))=0,
 $$
 where $p(t)$ is the backward jump operator.
 We obtain a Kwong-Wong-type integral equation, that is:
 If $x(t)$ is a nonoscillatory solution of the above equation
 on $[T_0,\infty)$, then the integral equation
 $$
 \frac{r^\sigma(t)x^\Delta(t)}{f(x^\sigma(t))}
 =P^\sigma(t)+\int^\infty_{\sigma(t)}\frac{r^\sigma(s)
 [\int^1_0f'(x_h(s))dh][x^\Delta(s)]^2}{f(x(s))
 f(x^\sigma(s))}\Delta s
 $$
 is satisfied for $t\geq T_0$, where
 $P^\sigma(t)=\int^\infty_{\sigma(t)}p(s)\Delta s$, and
 $x_h(s)=x(s)+h\mu(s)x^\Delta(s)$.
 As an application, we show that the superlinear dynamic equation
 $$
 [r(t)x^{\Delta}(\rho(t))]^\Delta+p(t)f(x(t))=0,
 $$
 is oscillatory, under certain conditions.
\end{abstract}

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

\section{Introduction}

Consider the second order nonlinear dynamic equation
\begin{equation}\label{pu10}
(r(t)x^\Delta(\rho(t)))^\Delta+p(t)f(x(t))=0,
\end{equation}
where $r(t),p(t)\in C(\mathbb{T},R)$, $ f(x)\in C(R,R)$, $r(t)>0$
and $\int^\infty_{T_0}[r^\sigma(t)]^{-1}\Delta t=\infty$.
We assume that $\lim_{t\to\infty}\int^t_{T_0}p(s)\Delta s$
exists and is finite;
\begin{gather}\label{pu11}
 xf(x)>0, \text{ for } x\neq 0 \text{ and } f'(x)\geq 0;\\
\lim_{x\to\pm\infty}f(x)=\pm\infty. \label{pu12}
\end{gather}

When $\mathbb{T}=\mathbb{R}$, Equation \eqref{pu10} becomes the second-order
nonlinear differential equations
 \begin{equation}\label{jia30}
x''(t)+p(t)f(x(t))=0.
\end{equation}

 Kwong and Wong \cite{kw} proved the following result.

\begin{theorem} \label{jia20}
Suppose that $f$ satisfies \eqref{pu11} and \eqref{pu12} and
$\lim_{t\to\infty}\int^t_{T_0}p(s)\Delta s$ exists and finite.
If $x(t)$ is a nonoscillatory solution of \eqref{jia30}
on $[T_0,\infty)$, then the integral equation
$$
\frac{x'(t)}{f(x(t))}=P(t)+\int^\infty_t\frac{f'(x(s))[x'(s)]^2}{f^2(x(s))}ds
$$
is satisfied for $t\geq T_0$, where $P(t)=\int^\infty_tp(s)\Delta s$.
\end{theorem}

We note that Theorem \ref{jia20} has been used
by Naito \cite{N} in proving results on asymptotic behavior
of nonoscillatory solution of equation \eqref{jia30}.

In this article, we extend  Theorem \ref{jia20} to dynamic equations
on time scales.
As an application, we prove that the superlinear dynamic equation
$$
[r(t)x^{\Delta}(\rho(t))]^\Delta+p(t)f(x(t))=0
$$
is oscillatory, if
$$
\limsup_{t\to\infty}\frac{1}{r^\sigma(t)}
\int^t_{T_0}P^\sigma(s)\Delta s=\infty,
$$
where $f(x)$ satisfies the superlinearity conditions
\begin{equation}
0<\int^\infty_\epsilon\frac{dx}{f(x)},\quad
\int^{-\epsilon}_{-\infty}\frac{dx}{f(x)}<\infty, \quad\text{for
all }\epsilon>0.
\end{equation}

It should be pointed out that our proof of the main
theorem is different from the one in Kwong and Wong for differential
equation in \cite{kw}.

For completeness, we recall some basic results for dynamic
equations and the calculus on time scales;
see \cite{bp1} and \cite{bp2} for elementary results for the
time scale calculus.
Let $\mathbb{T}$ be a time scale (i.e., a closed nonempty subset
of $\mathbb{R}$) with $\sup\mathbb{T}=\infty $. The forward jump
operator is defined by
\begin{equation*}
\sigma(t) = \inf\{s \in \mathbb{T} :s>t \},
\end{equation*}
and the backward jump operator is defined by
\begin{equation*}
\rho(t) = \sup\{s \in \mathbb{T} :s<t \},
\end{equation*}
where $\sup\emptyset = \inf \mathbb{T}$, where $\emptyset$
denotes the empty set. If $\sigma(t)>t$, we say $t$ is
right-scattered, while if $\rho(t)<t$ we say $t$ is left-scattered.
If $\sigma(t)=t$ we say $t$ is right-dense, while if $\rho(t)=t$ and
$t\neq\inf\mathbb{T}$ we say $t$ is left-dense. Given a time scale interval
$[c,d]_{\mathbb{T}}:=\{t\in \mathbb{T}:c\leq t\leq d\}$ in $\mathbb{T}$ the
notation $[c,d]^{\kappa}_{\mathbb{T}}$ denotes the interval $[c,d]_{\mathbb{T}}$ in case
$\rho(d)=d$ and denotes the interval $[c,d)_{\mathbb{T}}$ in case $\rho(d)<d$.
The graininess function $\mu$ for a time scale $\mathbb{T}$ is
defined by $\mu (t)=\sigma (t)-t$, and for any function
$f:\mathbb{T}\to \mathbb{R}$ the notation $f^\sigma (t)$
denotes $f(\sigma (t))$. We say that $x:\mathbb{T}\to\mathbb{R}$ is differentiable
at $t\in\mathbb{T}$ provided
$$
x^{\Delta}(t):=\lim_{s\to t}\frac{x(t)-x(s)}{t-s},
$$
exists when $\sigma(t)=t$ (here by $s\to t$ it is understood that
$s$ approaches $t$ in the time scale)
and when $x$ is continuous at $t$ and $\sigma(t)>t$
$$
x^{\Delta}(t):=\frac{x(\sigma(t))-x(t)}{\mu(t)}.
$$
Note that if $\mathbb{T}=\mathbb{R}$ , then the delta derivative is just the
standard derivative, and when $\mathbb{T}=\mathbb{Z}$ the delta derivative is just
the forward difference operator. Hence our results contain the
discrete and continuous cases as special cases and generalize
these results to arbitrary time scales.

\section{Lemmas}

The following condition was introduced in \cite{jep}.

\noindent\textbf{Condition (H):}
 We say that $\mathbb{T}$ satisfies Condition
(H), provided one of the following holds:
\begin{itemize}
\item[(1)] There exists a strictly increasing sequence
$\{t_n\}^\infty_{n=0}\subset\mathbb{T}$ with $\lim_{n\to
\infty}t_n=\infty$ and for each $n\geq 0$ either
$\sigma(t_n)=t_{n+1}$ or the real interval $[t_n,t_{n+1}]\subset
\mathbb{T}$; or

\item[(2)] $\mathbb{T}\bigcap\mathbb{R}=[T_0,\infty)$ for some $T_0\in\mathbb{T}$.
\end{itemize}

We say $\mathbb{T}$ is a \textbf{regular time scale} provided it is a time
scale with $\inf{\mathbb{T}}=T_0$, $\sup{\mathbb{T}}=\infty$ and $\mathbb{T}$ is either an
isolated time scale (all points in $\mathbb{T}$ are isolated) or $\mathbb{T}$ is
the real interval $[T_0,\infty)$. Note that in every regular time
scale the backward jump operator is (delta) differentiable (which
is used in the proof of the following theorem).


\begin{remark} \label{rmk2.1} \rm
Time scales that satisfy Condition (H) include most of the
important time scales, such as $\mathbb{R}$, $\mathbb{Z}$, $q^{\mathbb{N}_0}$, harmonic
numbers $\{\sum_{k=1}^n\frac{1}{k},\;  n\in\mathbb{N}\}$, etc.
\end{remark}

We need the following lemmas.

\begin{lemma}\label{liu203}
Assume that $\mathbb{T}$ satisfies Condition (H) and the function
$g(t)>0$ for $t\in [T_0,\infty)$. Then we have for
$t\in [T_0,\infty)_{\mathbb{T}}$,
$$
\int^t_{T_0}\frac{g^\Delta(s)}{g(s)}\Delta s\geq
\ln\frac{g(t)}{g(T_0)}.
$$
\end{lemma}

\begin{proof}
Assume that $t=t_{i-1}<t_i=\sigma(t)$. Then
  \begin{equation}\label{2.6}
  \int^{\sigma(t)}_t\frac{g^\Delta(s)}{g(s)}\Delta s=\frac{g^\Delta(t)\mu(t)}{g(t)}
 =\frac{g(\sigma(t))-g(t)}{g(t)}.
  \end{equation}
We consider the two possible cases: (i) $g(t)\le g(\sigma(t))$ and
(ii) $g(t)> g(\sigma(t))$. First, if $g(t)\le g(\sigma(t))$ we
have
\begin{equation}\label{2.7}
\frac{g(\sigma(t))-g(t)}{g(t)}\geq
\int^{g(\sigma(t))}_{g(t)}\frac{1}{v}dv
 =\ln\frac{g(\sigma(t))}{g(t)}.
\end{equation}
On the other hand, if  $g(t)> g(\sigma(t))$, then
\begin{equation*}
  \frac{g(t)-g(\sigma(t))}{g(t)}
\leq \int_{g(\sigma(t))}^{g(t)}\frac{1}{v}ds
=\ln\frac{g(t)}{g(\sigma(t))},
\end{equation*}
which implies that
\begin{equation}\label{2.88}
 \frac{g(\sigma(t))-g(t)}{g(t)}\geq \ln\frac{g(\sigma(t))}{g(t)}.
\end{equation}
Hence, whenever $t_{i-1}=t<\sigma(t)=t_i$, we have from \eqref{2.6}
and \eqref{2.7} in the first case and \eqref{2.6} and \eqref{2.88}
in the second case, that
\begin{equation}\label{2.9}
 \int_{t_{i-1}}^{t_i}\frac{g^{\Delta}(s)}{g(s)}\Delta s
\geq \ln\frac{g(\sigma(t))}{g(t)}=\ln\frac{g(t_i)}{g(t_{i-1})}.
\end{equation}
If the real interval $[t_{i-1},t_i]\subset \mathbb{T}$, then
  \begin{equation}\label{2.10}
  \int^{t_i}_{t_{i-1}}\frac{g^\Delta(s)}{g(s)}\Delta s
=\ln\frac{g(t_i)}{g(t_{i-1})}.
  \end{equation}
and so \eqref{2.9} also holds in this case.

 Note that since $\mathbb{T}$ satisfies condition (H), we have from \eqref{2.9}, \eqref{2.10} and the additivity of the integral that for $t\in[T_0,\infty)_{\mathbb{T}}$
 \begin{equation}\label{2.12}
 \int^t_{T_0}\frac{g^\Delta(s)}{g(s)}\Delta s
\geq \ln\frac{g(t)}{g(T_0)}.
 \end{equation}
\end{proof}

\begin{lemma}\label{jia99}
Suppose that $\mathbb{T}$ satisfies Condition {\rm (H)}. $x(t)>0$ is a
solution of \eqref{pu10}. $f(x)$ satisfies the superlinearity
conditions
\begin{equation}\label{liu16}
0<\int^\infty_\epsilon\frac{dx}{f(x)},\quad
\int^{-\epsilon}_{-\infty}\frac{dx}{f(x)}<\infty,\quad\text{for
all }\epsilon>0.
\end{equation}
Then
$$
\int^t_T\frac{x^\Delta(s)}{f(x^\sigma(s))}\Delta s
\leq F(x(T))-F(x(t))\leq F(x(T)),
$$
where $F(x)=\int^\infty_x\frac{dv}{f(v)}$.
\end{lemma}

\begin{proof}
Assume that $t=t_{i-1}<t_i=\sigma(t)$. Then
  \begin{equation}\label{12.6}
  \int^{\sigma(t)}_t\frac{x^\Delta(s)}{f(x(\sigma(s)))}\Delta s
=\frac{x^\Delta(t)\mu(t)}{f(x(\sigma(t)))}
 =\frac{x(\sigma(t))-x(t)}{f(x(\sigma(t)))}.
  \end{equation}
We consider the two possible cases: (i) $x(t)\le x(\sigma(t))$ and
(ii) $x(t)> x(\sigma(t))$. First, if $x(t)\le x(\sigma(t))$ we
have
\begin{equation}\label{12.7}
\quad\quad\frac{x(\sigma(t))-x(t)}{f(x(\sigma(t)))}
\leq \int^{x(\sigma(t))}_{x(t)}\frac{1}{f(v)}dv
 =F(x(t))-F(x(\sigma(t))),
\end{equation}
since $f$ is increasing. On the other hand, if  $x(t)>
x(\sigma(t))$, then
\begin{equation*}
\frac{x(t)-x(\sigma(t))}{f(x(\sigma(t)))}
\geq \int_{x(\sigma(t))}^{x(t)}\frac{1}{f(v)}ds=F(x(\sigma(t)))-F(x(t)),
\end{equation*}
which implies that
\begin{equation}\label{12.888}
 \frac{x(\sigma(t))-x(t)}{f(x(\sigma(t)))}\le F(x(t))-F(x(\sigma(t))).
\end{equation}
Hence, whenever $t_{i-1}=t<\sigma(t)=t_i$, we have from \eqref{12.6}
and \eqref{12.7} in the first case and \eqref{12.6} and
\eqref{12.888} in the second case, that
\begin{equation}\label{12.9}
 \int_{t_{i-1}}^{t_i}\frac{x^{\Delta}(s)}{f(x(\sigma(s)))}\Delta s
\le F(x(t_{i-1}))-F(x(t_i)).
\end{equation}
If the real interval $[t_{i-1},t_i]\subset \mathbb{T}$, then
  \begin{equation}\label{12.10}
\begin{split}
\int^{t_i}_{t_{i-1}}\frac{x^\Delta(s)}{f(x(\sigma(s)))}\Delta
&=\int^{t_i}_{t_{i-1}}\frac{x^\Delta(s)}{f(x(s))}\Delta s\\
&=\int_{x(t_{i-1})}^{x(t_i)}\frac{1}{f(v)}\,dv\\
&= F(x(t_{i-1}))-F(x(t_i)),
\end{split}  \end{equation}
and so \eqref{12.9} also holds in this case.

Note that since $\mathbb{T}$ satisfies condition (H), we have from
\eqref{12.9}, \eqref{12.10} and the additivity of the integral
that for $t\in[T,\infty)_{\mathbb{T}}$,
 \begin{equation}\label{2.12b}
 \int^t_{T}\frac{x^\Delta(s)}{f(x(\sigma(s)))}\Delta s
\leq F(x(T))-F(x(t))\leq F(x(T)).
 \end{equation}
\end{proof}

\section{Main results}

\begin{theorem}\label{jia1001}
 Suppose that $\mathbb{T}$ is a regular time scale. $f(x)$ satisfies
\eqref{pu11} and \eqref{pu12}. If $x(t)$ is a nonoscillatory
solution of \eqref{pu10} on $[T_0,\infty)$.
Then the integral equation
\begin{equation}\label{liu117}
\frac{r^\sigma(t)x^\Delta(t)}{f(x^\sigma(t))}
=P^\sigma(t)+\int^\infty_{\sigma(t)}\frac{r^\sigma(s)
[\int^1_0f'(x_h(s))dh][x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s
\end{equation}
is satisfied for $t\geq T_0$, where $P^\sigma(t)
=\int^\infty_{\sigma(t)}p(s)\Delta s$, $x_h(s)
=x(s)+h\mu(s)x^\Delta(s)$.
\end{theorem}

\begin{proof}
Suppose that $x(t)$ is a nonoscillatory solution of \eqref{pu10}
on $[T_0,\infty)$. Without loss of generality,  assume that $x(t)$
is  positive for $t\in [T_0,\infty)$.

 In the first place, we will prove
\begin{equation}\label{p1}
\int^\infty_{T_0}\frac{r^{\sigma}(s)[\int^1_0f'(x_h(s))dh]
[x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s
<\infty,
\end{equation}
where $x_h(t)=x(t)+h\mu(t)x^\Delta(t)=(1-h)x(t)+hx(\sigma(t))>0$.
From \eqref{pu10}, it is easy to see that
\begin{align*}
\Big(\frac{r(t)x^\Delta(\rho(t))}{f(x(t))}\Big)^\Delta
&= [r(t)x^{\Delta}(\rho(t))]^\Delta\frac{1}{f(x(t))}
+r^{\sigma}(t)x^\Delta(t)\Big(\frac{1}{f(x(t))}\Big)^\Delta\\
&= -p(t)-\frac{r^{\sigma}(t)[\int^1_0f'(x_h(t))dh][x^\Delta(t)]^2}
{f(x(t))f(x^\sigma(t))}
\end{align*}
Integrating from $T_0$ to $t$,
\begin{equation}\label{jia1}
\begin{split}
&\frac{r(t)x^\Delta(\rho(t))}{f(x(t))}-\frac{r(T_0)x^\Delta(\rho(T_0))}{f(x(T_0))}\\
&= -\int^{t}_{T_0}p(s)\Delta s-
\int^{t}_{T_0}\frac{r^{\sigma}(s)[\int^1_0f'(x_h(s))dh]
[x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s.
\end{split}
\end{equation}
If \eqref{p1} fails to hold; that is,
\begin{equation}
\int^\infty_{T_0}\frac{r^{\sigma}(s)[\int^1_0f'(x_h(s))dh]
[x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s
=\infty.
\end{equation}
From \eqref{jia1}, we have
\begin{equation}\label{pu3}
\lim_{t\to \infty}\frac{r(t)x^\Delta(\rho(t))}{f(x(t))}=-\infty.
\end{equation}
We can assume that
\begin{equation}\label{pu4}
\frac{r(T_0)x^\Delta(\rho(T_0))}{f(x(T_0))}
-\int^{t}_{T_0}p(s)\Delta s<-1,
\end{equation}
for $t\geq T_0$.  Otherwise let
$L=\sup_{t\geq T_0}|\int^t_{T_0}p(s)\Delta s|$. By \eqref{pu3},
we can take a large $T_1>T_0$ such that
$\frac{r(T_1)x^\Delta(\rho(T_1))}{f(x(T_1))}<-(2L+1)$. So we have
\begin{align*}
\frac{r(T_1)x^\Delta(\rho(T_1))}{f(x(T_1))}
-\int^{t}_{T_1}p(s)\Delta s
&< -(2L+1)-\Big[\int^{t}_{T_0}p(s)\Delta s
 -\int^{T_1}_{T_0}p(s)\Delta s\Big]\\
&\leq -(2L+1)-[-2L]=-1.
\end{align*}
 So we can replace $T_0$ by $T_1>T_0$ such that \eqref{pu4}
 still holds.
From \eqref{jia1} and \eqref{pu4}, we obtain, for $t\geq T_{0}$,
 \begin{equation}\label{p2}
\frac{r(t)x^\Delta(\rho(t))}{f(x(t))}+
\int^{t}_{T_0}\frac{r^{\sigma}(s)[\int^1_0f'(x_h(s))dh]
[x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s<-1.
\end{equation}
In particular,
\begin{equation}
x^\Delta(t)<0,\quad \text{for } t\geq T_0.
\end{equation}
Therefore, $x(t)$ is strictly decreasing.

First assume that $\mathbb{T}$ is an isolated time scale; that is,
$$
\mathbb{T}=\{t_0,t_1,t_2,\dots\},\quad t_0<t_1<t_2\dots,
\lim_{k\to\infty}t_k=\infty.
$$
If $t=t_{i-1}<t_i=\sigma(t)$, then $x(\sigma(t))<x(t)$, so
\begin{equation}\label{2.17}
\begin{split}
 \int^1_0f'(x_h(s))dh
&= \int^1_0f'((1-h)x(s)+h(x(\sigma(s))))dh\\
&=\frac{f((1-h)x(s)+h(x(\sigma(s))))|^1_0}{x(\sigma(s))-x(s)}\\
&=\frac{f(x(\sigma(s)))-f(x(s))}{x(\sigma(s))-x(s)}.
\end{split}\end{equation}
On the other hand if the $\mathbb{T}$ is the real interval $[T_0,\infty)$,
then
\begin{equation}\label{uiop}
 \int^1_0f'(x_h(s))dh=f'(x(s)).
\end{equation}
Let
\begin{equation}\label{al}
 y(t):=1+\int^t_{T_0}\frac{ r^{\sigma}(s)
\int^1_0f'(x_h(s))dh[x^\Delta(s)]^2} {f(x(s))f(x^{\sigma}(s))}\Delta s.
\end{equation}
Hence from \eqref{p2}, we obtain
\begin{equation}\label{2.20}
 -\frac{r(t) x^\Delta(\rho(t))}{f(x(t))}>y(t).
\end{equation}
Replacing $t$ by $\sigma(t)$ in \eqref{2.20} and using
\cite[Theorem1.75]{bp1}, we obtain
 \begin{equation}\label{p3}
-\frac{r^{\sigma}(t) x^\Delta(t)}{f(x^{\sigma}(t))}
>y^{\sigma}(t)
= y(t)+\frac{ r^{\sigma}(t) \int^1_0f'(x_h(t))dh
[x^\Delta(t)]^2\mu(t)} {f(x(t))f(x^{\sigma}(t))}.
\end{equation}
Noting that
\begin{equation} \label{p4}
\frac{1}{f(x^{\sigma}(t))}
=\frac{1}{f(x(t))}+\Big(\frac{1}{f(x(t))}\Big)^{\Delta}\mu(t).
\end{equation}
Using \eqref{p4} in the left side of \eqref{p3} and noticing that
$$
(\frac{1}{f(x(t))})^{\Delta}=-\frac{\int^1_0f'(x_h(t))dh
x^\Delta(t)}{f(x(t))f(x^\sigma(t))},
$$
from \eqref{p3}, it is easy to see  that
\begin{equation}\label{p5}
 -\frac{r^{\sigma}(t) x^\Delta(t)}{f(x(t))}>y(t).
\end{equation}
From \eqref{al} and \eqref{p5}, we obtain
\begin{equation}\label{asas}
\begin{split}
y^\Delta(t)
&= \frac{ r^{\sigma}(t) \int^1_0f'(x_h(t))dh
 [x^\Delta(t)]^2} {f(x(t))f(x^\sigma(t))}\\
&> y(t)\frac{\int^1_0f'(x_h(t))dh[-x^\Delta(t)]}{f(x^{\sigma}(t))}.
\end{split}\end{equation}
In the isolated time scale case from \eqref{asas} and \eqref{2.17},
we obtain
 \begin{equation*}
\frac{y(\sigma(t))-y(t)}{y(t)(\sigma(t)-t)}
>\frac{f(x(\sigma(t)))-f(x(t))}{x(\sigma(t))-x(t)}\cdot\frac{x(t)
-x(\sigma(t))}{f(x(\sigma(t)))[\sigma(t)-t]}.
\end{equation*}
So
\begin{equation*}
\frac{y(\sigma(t))}{y(t)}>\frac{ f(x(t))}{f(x(\sigma(t)))};
\end{equation*}
that is,
\begin{equation}\label{klkl}
\frac{y(t_i)}{y(t_{i-1})}>\frac{ f(x(t_{i-1}))}{f(x(t_i))}.
\end{equation}
Let $T_0=t_{n_0}$ and $t=t_n, n>n_0$, then using \eqref{klkl}
we have that
$$
\frac{y(t_n)}{y(t_{n_0})}=\prod_{k=0}^{n-n_0-1}
\frac{y(t_{n_0+k+1})}{y(t_{{n_0}+k})}
>\prod_{k=0}^{n-n_0-1}\frac{f(x(t_{n_0+k}))}{f(x(t_{{n_0}+k+1}))}
=\frac{f(x(t_{n_0}))}{f(x(t_n))};
$$
that is,
\begin{equation}\label{p6}
\frac{y(t)}{y(t_{n_0})}>\frac{f(x(t_{n_0}))}{f(x(t))},
\end{equation}
for $t>T_0$.  To obtain \eqref{p6} in the case where $\mathbb{T}$
is the real interval $[T_0, \infty)$,  from \eqref{uiop}
and \eqref{asas} we have
$$
\frac{y'(t)}{y(t)}>\frac{f'(x(t))[-x'(t)]}{f(x(t))};
$$
that is,
$$
(\ln y(t))'>-(\ln f(x(t)))'.
$$
Integrating from $T_0$ to $t$, we obtain
\begin{equation}\label{vbvb}
\frac{y(t)}{y(T_0)}>\frac{f(x(T_0))}{f(x(t))},\quad t>T_0.
\end{equation}
Using \eqref{p5} again, from \eqref{p6} and \eqref{vbvb}, we obtain
\begin{equation*}
 -\frac{r^{\sigma}(t) x^\Delta(t)}{f(x(t))}>y(t)
>\frac{y(T_0)f(x(T_0))}{f(x(t))}.
\end{equation*}
If we set $L_1:=y(T_0)f(x(T_0))$, we obtain
\begin{equation*}
 x^\Delta(t)<-\frac{L_1}{r^{\sigma}(t)}.
\end{equation*}
Integrating from $T_0$ to $t$, we obtain
\begin{equation*}
 x(t)-x(T_0)<-\int^t_{T_0}\frac{L_1}{r^{\sigma}(s)}\Delta s
\to-\infty,\quad \text{as } t\to\infty.
\end{equation*}
which contradicts $x(t)>0$.

In \eqref{jia1}, letting $t\to\infty$ and replacing $T_0$ by
$\sigma(\tau)$, denoting
\begin{equation}\label{liu10} \alpha
=\lim_{t\to\infty}\frac{r(t)x^\Delta(\rho(t))}{f(x(t))}
=\lim_{t\to\infty}\frac{r^{\sigma}(t)x^\Delta(t)}{f(x^{\sigma}(t))},
\end{equation}
we obtain
\begin{equation}\label{p7}
\alpha+\int^\infty_{\sigma(\tau)}p(s)\Delta s+\int^\infty_{\sigma(\tau)}\frac{r^{\sigma}(s)[\int^1_0f'(x_h(s))dh][x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s=\frac{r^{\sigma}(\tau)x^\Delta(\tau)}{f(x^{\sigma}(\tau))}.
\end{equation}

We claim that $\alpha= 0$.
In the right side of \eqref{p7}, using $$\frac{1}{f(x^\sigma(\tau))}=\frac{1}{f(x(\tau))}-\frac{\int^1_0f^{\prime}(x_h(\tau))dhx^\Delta(\tau)}{f(x(\tau))f(x^\sigma(\tau))}\mu(t)$$
and in the second integral term of left side of \eqref{p7}, noticing that
\begin{align*}
&\int^\infty_{\sigma(\tau)}\frac{r^{\sigma}(s)
[\int^1_0f'(x_h(s))dh][x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s\\
&=\int^\infty_{\tau}\frac{r^{\sigma}(s)[\int^1_0f'(x_h(s))dh]
[x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s
-\int^{\sigma(\tau)}_{\tau}\frac{r^{\sigma}(s)
[\int^1_0f'(x_h(s))dh][x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s\\
&=\int^\infty_{\tau}\frac{r^{\sigma}(s)[\int^1_0f'(x_h(s))dh]
[x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s
-\frac{r^{\sigma}(\tau)[\int^1_0f'(x_h(\tau))dh]
[x^\Delta(\tau)]^2}{f(x(\tau))f(x^\sigma(\tau))}\mu(\tau),
\end{align*}
we obtain
\begin{equation}\label{p28}
\alpha+\int^\infty_{\sigma(\tau)}p(s)\Delta s+\int^\infty_{\tau}\frac{r^{\sigma}(s)[\int^1_0f'(x_h(s))dh][x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s=\frac{r^{\sigma}(\tau)x^\Delta(\tau)}{f(x(\tau))}.
\end{equation}
From \eqref{p28}, we have
\begin{equation}\label{jia100}
\lim_{t\to\infty}\frac{r^{\sigma}(t)x^\Delta(t)}{f(x(t))}=\alpha.
\end{equation}

Suppose that $\alpha<0$. Then from \eqref{jia100} there exists
a large $T_1$ such that for $t>T_1$,
we have
$$
\frac{r^{\sigma}(t)x^\Delta (t)}{f(x(t))}\leq \frac{\alpha}{2}.
$$
So
\begin{equation}\label{liu202}
x^\Delta(t)\leq \frac{\alpha}{2}\cdot\frac{f(x(t))}{r^{\sigma}(t)}<0.
\end{equation}
Thus
 \begin{equation}\label{j1}
\begin{split}
M(T_1)
&=: \int^\infty_{T_1}\frac{r^{\sigma}(s)[\int^1_0f'(x_h(s))dh]
 [x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s\\
&\geq  -\frac{\alpha}{2}\int^\infty_{T_1}
 \frac{[\int^1_0f'(x_h(s))dh][-x^\Delta(s)]}{f(x^\sigma(s))}\Delta s.
\end{split}\end{equation}
Assume that $t=t_{i-1}<t_i=\sigma(t)$. From \cite[Theorem 1.75]{bp1},
\eqref{2.17} and $x^\Delta(t)<0$, we have
\begin{equation}\label{y1}
\begin{split}
\int^{\sigma(t)}_t\frac{[\int^1_0f'(x_h(s))dh]
 [-x^\Delta(s)]}{f(x^\sigma(s))}\Delta s
&= \frac{[\int^1_0f'(x_h(t))dh][-x^\Delta(t)]
 (\sigma(t)-t)}{f(x^\sigma(t))}\\
&= \frac{f(x(t))-f(x^\sigma(t))}{f(x^\sigma(t))}\\
&\geq \int^{f(x(t))}_{f(x^\sigma(t))}\frac{1}{v}dv\\
&= \ln\frac{f(x(t))}{f(x^\sigma(t))}\\
&= \ln\frac{f(x(t_{i-1}))}{f(x(t_i))}.
\end{split}
\end{equation}
In the isolated time scale, from \eqref{y1}, we obtain
\begin{equation}\label{liu200}
\int^t_{T_1}\frac{[\int^1_0f'(x_h(s))dh]
 [-x^\Delta(s)]}{f(x^\sigma(s))}\Delta s
\geq  \ln\frac{f(x(T_1))}{f(x(t))}.
\end{equation}
In the case where $\mathbb{T}$ is the real interval $[T_0,\infty)$,
from \eqref{uiop}, we have
\begin{equation}\label{liu201}
\begin{split}
\int^t_{T_1}\frac{[\int^1_0f'(x_h(s))dh]
 [-x^\Delta(s)]}{f(x^\sigma(s))}\Delta s
 &= \int^t_{T_1}\frac{[\int^1_0f'(x_h(s))dh][-x'(s)]}{f(x(s))}ds\\
 &= \ln\frac{f(x(T_1))}{f(x(t))}.
\end{split}
\end{equation}
From \eqref{j1}, \eqref{liu200}, \eqref{liu201} and the additivity
of the integral, it is easy to see that
\[
M(T_1)\geq  -\frac{\alpha}{2}\lim_{t\to\infty}\int^t_{T_1}
\frac{[\int^1_0f'(x_h(s))dh][-x^\Delta(s)]}{f(x^\sigma(s))}\Delta s
\geq  -\frac{\alpha}{2}\lim_{t\to\infty} \ln\frac{f(x(T_1))}{f(x(t))}.
\]
 So there exists a large $T_2$ such that for $t\geq T_2$, we have that
$$
\ln\frac{f(x(T_1))}{f(x(t))}\leq -\frac{2M(T_1)}{\alpha}+1.
$$
Thus for $t\geq T_2$,
$$
f(x(t))\geq {f(x(T_1))}\exp\Big(\frac{2M(T_1)}{\alpha}-1\Big).
$$
By \eqref{liu202} and noticing that $\alpha<0$, we have that for
$t\geq T_2$
\begin{equation}\label{y2}
x^\Delta(t)\leq \frac{\alpha}{2}\cdot\frac{1}{r^{\sigma}(t)}
\cdot f(x(T_1))\exp{\Big(\frac{2M(T_1)}{\alpha}-1\Big)}.
\end{equation}
Integrating \eqref{y2} from $T_2$ to $t$, we obtain $x(t)\to-\infty$
as $t\to\infty$, which is  a contradiction.

If $\alpha>0$, from \eqref{liu10}, we have that there exists $T_3$
such that for $t\geq T_3$,
\begin{equation}\label{liu20}
\frac{r^\sigma(t)x^\Delta(t)}{f(x^\sigma(t))}\geq \frac{\alpha}{2}.
\end{equation}
Therefore, from Lemma \ref{liu203}, we have
\begin{align*}
\int^\infty_{T_3}\frac{r^\sigma(s)[\int^1_0f'(x_h(s))dh]
 [x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s
&\geq  \frac{\alpha}{2}\int^\infty_{T_3}
 \frac{[\int^1_0f'(x_h(s))dh][x^\Delta(s)]}{f(x(s))}\Delta s\\
&= \frac{\alpha}{2}\lim_{t\to\infty}\int^t_{T_3}
 \frac{[f(x(s))]^\Delta}{f(x(s))}\Delta s\\
&\geq  \frac{\alpha}{2}\lim_{t\to\infty}\ln\frac{f(x(t))}{f(x(T_3))}.
\end{align*}
Due to condition \eqref{pu11} and \eqref{pu12}, it is easy to
know that $x(t)$ is bounded.

On the other hand, from \eqref{liu20} and the monotonicity of $f$,
we obtain that in the isolated time scale case
\begin{gather*}
r^\sigma(T_3)x^\Delta(T_3)\geq \frac{\alpha}{2}f(x^\sigma(T_3)),\\
r^{\sigma^2}(T_3)x^\Delta(\sigma(T_3))\geq \frac{\alpha}{2}
 f(x^{\sigma^2}(T_3))\geq \frac{\alpha}{2}f(x^\sigma(T_3)).
\end{gather*}
By induction, it is easy to get that for $t\geq T_3$,
$$
r^\sigma(t)x^\Delta(t)\geq \frac{\alpha}{2}f(x^\sigma(T_3));
$$
that is,
\begin{equation}\label{jia12}
x^\Delta(t)\geq \frac{\alpha}{2r^\sigma(t)}f(x^\sigma(T_3)).
\end{equation}
Integrating \eqref{jia12} from $T_3$ to $t$, we obtain
$x(t)\to+\infty$ as $t\to\infty$, which contradicts the
boundedness of $x(t)$.

If $\mathbb{T}$ is the real interval $[T_3,\infty)$, then from \eqref{liu20}
and the monotonicity of $f$, we have that for $t\geq T_3$,
\begin{equation}\label{liu15}
x'(t)\geq \frac{\alpha}{2r(t)}f(x(t))
\geq \frac{\alpha}{2r(t)}f(x(T_2)).
\end{equation}
Integrating \eqref{liu15} from $T_3$ to $t$, we obtain
$x(t)\to+\infty$ as $t\to\infty$, which also contradicts
the boundedness of $x(t)$. Therefore $\alpha=0$.

From \eqref{p7}, we obtain \eqref{liu117}. The proof is complete.
\end{proof}

\begin{theorem}\label{jia101}
Suppose  $\mathbb{T}$ is a regular time scale, $r(t)>0$ with
$\int^\infty_{T_0}[r^\sigma(t)]^{-1}\Delta t=\infty$ and suppose
that $\lim_{t\to\infty}\int^t_{T_0}p(s)\Delta s$ exists and finite.
Let $P(t)=\int^\infty_t p(s)\Delta s$. $f(x)$ satisfies the
superlinearity conditions
\begin{equation}\label{liu18}
0<\int^\infty_\epsilon\frac{dx}{f(x)},
\int^{-\epsilon}_{-\infty}\frac{dx}{f(x)}<\infty,\quad\text{for
all }\epsilon>0.
\end{equation}
Then the superlinear dynamic equation
\begin{equation}\label{liu23}
[r(t)x^{\Delta}(\rho(t))]^\Delta+p(t)f(x(t))=0,
\end{equation}
is oscillatory, if
\begin{equation}\label{liu22}
\limsup_{t\to\infty}\frac{1}{r^\sigma(t)}
\int^t_{T_0}P^\sigma(s)\Delta s=\infty.
\end{equation}
\end{theorem}

\begin{proof}
Suppose that $x(t)$ is a nonoscillatory solution of \eqref{pu10}
on $[T_0,\infty)$. Without loss of generality,  assume that $x(t)$
is  positive for $t\in [T_0,\infty)$. From Theorem \ref{jia1001},
$x(t)$ satisfies the integral equation \eqref{liu117}. Dropping
the last integral term in \eqref{liu117}, we have the inequality
\begin{equation}\label{liu19}
\frac{r^\sigma(t)x^\Delta(t)}{f(x^\sigma(t))}\geq P^\sigma(t).
\end{equation}
Dividing \eqref{liu19} by $r^\sigma(t)$ and integrating from
$T_0$ to t and using Lemma \ref{jia99}, we find
$$
F(x(T_0))\geq \int^t_{T_0}\frac{x^\Delta(s)}{f(x^\sigma(s))}\Delta s
\geq \frac{1}{r^\sigma(t)}\int^t_{T_0}P^\sigma(s)\Delta s.
$$
This contradicts \eqref{liu22}, so equation \eqref{liu23} is
oscillatory.
\end{proof}

\section{Examples}

\begin{example} \label{examp4.1} \rm
 Consider the superlinear difference equation
\begin{equation}\label{liu300}
\Delta^2x(n-1)+p(n)x^\gamma(n)=0,\quad \gamma>1,
\end{equation}
where $P(n)=\frac{1}{n}+\frac{2(-1)^n}{\sqrt{n}}$,
\begin{equation}\label{Lynn1}
 p(n)=P(n)-P(n+1)=\frac{1}{n(n+1)}
+\frac{2(-1)^n(\sqrt{n+1}+\sqrt{n})}{\sqrt{n(n+1)}}.
 \end{equation}
It is to see that $\sum^\infty_{n=1}P(n+1)=\infty$.
So from Theorem \ref{jia101}, \eqref{liu300} is oscillatory.
\end{example}

In \cite[Theorem 2.5]{jep1}, we proved that the equation
$\Delta^2x(n-1)+q(n)x^\gamma(n)=0,\gamma>1$, is oscillatory,
if $\sum^\infty_{n=1}nq(n)=\infty$.
In the following, we will prove that
 \begin{equation}\label{jia1000}
 \sum^{2k+1}_{n=1}np(n)\to-\infty\quad \text{as}\quad k\to\infty.
 \end{equation}
So \cite[Theorem 2.5]{jep1} would not apply in \eqref{liu300}.

 We need the following lemmas.
 The first lemma may be regarded as a discrete version
of L'Hopital's rule and can be found in \cite[page 48]{bp1}.

 \begin{lemma}[Stolz-Ces\'{a}ro Theorem] \label{l3.2}
Let $\{a_n\}_{n\geq 1}$ and $\{b_n\}_{n\geq 1}$ be two sequences
of real number. Assume  $b_n$ is positive, strictly increasing and
unbounded and the following limit exists:
 $$
 \lim_{n\to \infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=l.
 $$
 Then
 $$
 \lim_{n\to \infty}\frac{a_n}{b_n}=l.
 $$
 \end{lemma}

 We will use Lemma \ref{l3.2} to prove the following result.

 \begin{lemma}
For each real number $d>0$, we have
  \begin{equation}\label{3.3}
 \lim_{m\to \infty}\frac{\sum^m_{i=1} i^d-\frac{m^{d+1}}{d+1}}{m^{d}}=\frac{1}{2}.
 \end{equation}
  \end{lemma}

\begin{proof}
  By Taylor's formula, we have
 \begin{equation}\label{3.5}
 \big(1+\frac{1}{m}\big)^d=1+\frac{d}{m}+\frac{d(d-1)}{2m^2}
+o\big(\frac{1}{m^2}\big).
 \end{equation}
By \eqref{3.5} and the Stolz-Ces\'{a}ro Theorem
(Lemma \ref{l3.2}), it is easy to see that
\begin{equation}\label{3.6}
\begin{split}
\lim_{m\to\infty}\frac{\sum^m_{i=1}i^d-\frac{m^{d+1}}{d+1}}{m^{d}}
 &= \lim_{m\to\infty}\frac{(m+1)^d-\frac{(m+1)^{d+1}}{d+1}
 +\frac{m^{d+1}}{d+1}}{(m+1)^{d}-m^{d}}\\
&= \lim_{m\to\infty}\frac{(1+\frac{1}{m})^{d}
 -\frac{m+1}{d+1}(1+\frac{1}{m})^{d}
 +\frac{m}{d+1}}{(1+\frac{1}{m})^{d}-1}.
\end{split} \end{equation}
Using \eqref{3.5} in \eqref{3.6}, it follows that \eqref{3.3} holds.
 \end{proof}

 So given $0<\epsilon<1$, for large $m$, we have the inequality
\begin{equation}\label{3.10}
 \frac{m^{d+1}}{d+1}+\frac{m^d}{2}-\epsilon m^d<\sum^m_{i=1}i^d
< \frac{m^{d+1}}{d+1}+\frac{m^d}{2}+\epsilon m^d.
 \end{equation}
Set $d=1/2$. we have
 \begin{equation}\label{3.15}
 \frac{2}{3}m^{3/2}+\frac{1}{2}m^{1/2}
-\epsilon m^{1/2}<\sum^m_{i=1}i^{1/2}< \frac{2}{3}m^{3/2}
+\frac{1}{2}m^{1/2}+\epsilon m^{1/2}.
 \end{equation}
From \eqref{Lynn1} and using Taylor's expansion, we have
 \begin{equation}\label{jia305}
\begin{split}
np(n)&= \frac{1}{n+1}+2(-1)^n\sqrt{n}
\big[1+\big(1+\frac{1}{n}\big)^{-1/2}\big]\\
 &= \frac{1}{n+1}+2(-1)^n\sqrt{n}\big[2-\frac{1}{2n}
+O\big(\frac{1}{n^2}\big)\big]\\
 &= \frac{1}{n+1}+4(-1)^n\sqrt{n}
-\frac{(-1)^n}{\sqrt{n}}+O\big(\frac{1}{n^{3/2}}\big).
\end{split}
\end{equation}
Using the Euler formula \cite[page 205]{kp},
$$
C=\lim_{n\to\infty}\Big(\sum^n_{k=1}\frac{1}{k}-\ln n\Big),
$$
where $C$ is called Euler constant,
we obtain
\begin{equation}\label{jia500}
\sum^{2k+1}_{n=1}\frac{1}{n+1}=o(1)+C-1+\ln (2k+2).
\end{equation}
From \eqref{3.15} and using Taylor's expansion, we have
\begin{align}
&\sum^{2k+1}_{n=1}(-1)^n\sqrt{n} \nonumber \\
&= -\sum^{2k+1}_{n=1}\sqrt{n}+2\sqrt{2}\sum^{k}_{n=1}\sqrt{n} \nonumber\\
&< -\big[\frac{2}{3}(2k+1)^{3/2}+(\frac{1}{2} -\epsilon)(2k+1)^{1/2}\big]
+2\sqrt{2}\big[\frac{2}{3}k^{3/2}+(\frac{1}{2}+\epsilon)k^{1/2}\big]
 \nonumber\\
&=\frac{4\sqrt{2}}{3}k^{3/2}
\big[1-\big(1+\frac{1}{2k}\big)^{3/2}\big]
+\sqrt{2}k^{1/2}\big[1-\frac{1}{2}\big(1+\frac{1}{2k}\big)^{1/2}
\big] \nonumber\\
&\quad +\epsilon \sqrt{2} k^{1/2}\big[2+\big(1+\frac{1}{2k}
 \big)^{1/2}\big] \nonumber\\
&=\frac{4\sqrt{2}}{3}k^{3/2}
 \big[-\frac{3}{4k}+O\big(\frac{1}{k^2}\big)\big]
+\sqrt{2}k^{1/2}\big[\frac{1}{2}+O(\frac{1}{k})\big] \nonumber \\
&\quad +\epsilon \sqrt{2} k^{1/2}[3+O(\frac{1}{k})] \label{jia306}\\
&=-(\frac{1}{2}-3\epsilon)\sqrt{2}k^{1/2}+O\big(\frac{1}{k^{1/2}}\big).
\nonumber
\end{align}
Take $\epsilon<1/6$. From \eqref{jia305}, \eqref{jia500},
\eqref{jia306} and noticing that
$\sum^\infty_{n=1}\frac{(-1)^n}{\sqrt{n}}$ and
$\sum^\infty_{n=1}O\big(\frac{1}{n^{3/2}}\big)$ are convergent,
we obtain that
$$
\sum^{2k+1}_{n=1}np(n)\to-\infty, \quad \text{as } k\to\infty.
$$
So we complete the proof of \eqref{jia1000}.

\begin{example} \label{examp4.2} \rm
Consider the differential equation
\begin{equation}\label{liu303}
x''(t)+p(t)x^\gamma(t)=0,\quad \gamma>1,
\end{equation}
where $P(t)=(t^{1/4}\sin\sqrt{t})'$.
So we have $\int^t_1P(s)ds=t^{1/4}\sin\sqrt{t}-\sin1$ and
\begin{equation}\label{jia40}
\limsup_{t\to\infty}\int^t_1P(s)ds=\infty.
\end{equation}
It is easy to see that
\begin{align*}
p(t)&=-(P(t))'=\frac{3}{16}t^{-7/4}\sin\sqrt{t}
-\frac{1}{8}t^{-5/4}\cos\sqrt{t}\\
&\quad +\frac{1}{8}t^{-5/4}\cos\sqrt{t}
 -\frac{1}{4}t^{-3/4}\sin\sqrt{t}.
\end{align*}
When $\beta<-\frac{1}{2}$, $\int^\infty_1t^\beta\sin\sqrt{t}dt$
and $\int^\infty_1t^\beta\cos\sqrt{t}dt$ are convergent,
we have that $\lim_{t\to\infty}\int^t_1p(s)ds$ exists and finite.
From \eqref{jia40} and Theorem \ref{jia101}, \eqref{liu303}
is oscillatory.
\end{example}

\begin{example} \label{examp4.3} \rm
Consider the q-difference equation
\begin{equation}\label{liu304}
(x(q^{-1}t))^{\Delta\Delta}+p(t)x^\gamma(t)=0,\quad \gamma>1,
\end{equation}
where $t=q^n$, $n\in \mathbb{N}_0$, $P(t)=\frac{1+2(-1)^n}{t}$. It is easy
to see that
$$
p(t)=-P^\Delta(t)=\big[1+\frac{2(-1)^n(1+q)}{q-1}\big]\frac{1}{qt^2}.
$$
We have that $\int^\infty_1p(t)\Delta t$ is convergent and
$\int^\infty_1P^\sigma(t)\Delta t=\infty$. From Theorem
\ref{jia101}, \eqref{liu304} is oscillatory.
\end{example}

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\end{thebibliography}

\end{document}