\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 $$\label{pu10} (r(t)x^\Delta(\rho(t)))^\Delta+p(t)f(x(t))=0,$$ 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 $$\label{jia30} x''(t)+p(t)f(x(t))=0.$$ 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 $$0<\int^\infty_\epsilon\frac{dx}{f(x)},\quad \int^{-\epsilon}_{-\infty}\frac{dx}{f(x)}<\infty, \quad\text{for all }\epsilon>0.$$ 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} :st$, we say$t$is right-scattered, while if$\rho(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$t=t_{i-1} g(\sigma(t))$. First, if$g(t)\le g(\sigma(t))$we have $$\label{2.7} \frac{g(\sigma(t))-g(t)}{g(t)}\geq \int^{g(\sigma(t))}_{g(t)}\frac{1}{v}dv =\ln\frac{g(\sigma(t))}{g(t)}.$$ 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 $$\label{2.88} \frac{g(\sigma(t))-g(t)}{g(t)}\geq \ln\frac{g(\sigma(t))}{g(t)}.$$ 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 $$\label{2.9} \int_{t_{i-1}}^{t_i}\frac{g^{\Delta}(s)}{g(s)}\Delta s \geq \ln\frac{g(\sigma(t))}{g(t)}=\ln\frac{g(t_i)}{g(t_{i-1})}.$$ If the real interval$[t_{i-1},t_i]\subset \mathbb{T}$, then $$\label{2.10} \int^{t_i}_{t_{i-1}}\frac{g^\Delta(s)}{g(s)}\Delta s =\ln\frac{g(t_i)}{g(t_{i-1})}.$$ 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}}$$$\label{2.12} \int^t_{T_0}\frac{g^\Delta(s)}{g(s)}\Delta s \geq \ln\frac{g(t)}{g(T_0)}.$$ \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 $$\label{liu16} 0<\int^\infty_\epsilon\frac{dx}{f(x)},\quad \int^{-\epsilon}_{-\infty}\frac{dx}{f(x)}<\infty,\quad\text{for all }\epsilon>0.$$ 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$t=t_{i-1} x(\sigma(t))$. First, if$x(t)\le x(\sigma(t))$we have $$\label{12.7} \quad\quad\frac{x(\sigma(t))-x(t)}{f(x(\sigma(t)))} \leq \int^{x(\sigma(t))}_{x(t)}\frac{1}{f(v)}dv =F(x(t))-F(x(\sigma(t))),$$ 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 $$\label{12.888} \frac{x(\sigma(t))-x(t)}{f(x(\sigma(t)))}\le F(x(t))-F(x(\sigma(t))).$$ 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 $$\label{12.9} \int_{t_{i-1}}^{t_i}\frac{x^{\Delta}(s)}{f(x(\sigma(s)))}\Delta s \le F(x(t_{i-1}))-F(x(t_i)).$$ If the real interval$[t_{i-1},t_i]\subset \mathbb{T}$, then $$\label{12.10} \begin{split} \int^{t_i}_{t_{i-1}}\frac{x^\Delta(s)}{f(x(\sigma(s)))}\Delta &=\int^{t_i}_{t_{i-1}}\frac{x^\Delta(s)}{f(x(s))}\Delta s\\ &=\int_{x(t_{i-1})}^{x(t_i)}\frac{1}{f(v)}\,dv\\ &= F(x(t_{i-1}))-F(x(t_i)), \end{split}$$ 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$t\in[T,\infty)_{\mathbb{T}}$, $$\label{2.12b} \int^t_{T}\frac{x^\Delta(s)}{f(x(\sigma(s)))}\Delta s \leq F(x(T))-F(x(t))\leq F(x(T)).$$ \end{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 $$\label{liu117} \frac{r^\sigma(t)x^\Delta(t)}{f(x^\sigma(t))} =P^\sigma(t)+\int^\infty_{\sigma(t)}\frac{r^\sigma(s) [\int^1_0f'(x_h(s))dh][x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s$$ 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 $$\label{p1} \int^\infty_{T_0}\frac{r^{\sigma}(s)[\int^1_0f'(x_h(s))dh] [x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s <\infty,$$ where$x_h(t)=x(t)+h\mu(t)x^\Delta(t)=(1-h)x(t)+hx(\sigma(t))>0. From \eqref{pu10}, it is easy to see that \begin{align*} \Big(\frac{r(t)x^\Delta(\rho(t))}{f(x(t))}\Big)^\Delta &= [r(t)x^{\Delta}(\rho(t))]^\Delta\frac{1}{f(x(t))} +r^{\sigma}(t)x^\Delta(t)\Big(\frac{1}{f(x(t))}\Big)^\Delta\\ &= -p(t)-\frac{r^{\sigma}(t)[\int^1_0f'(x_h(t))dh][x^\Delta(t)]^2} {f(x(t))f(x^\sigma(t))} \end{align*} Integrating fromT_0$to$t$, $$\label{jia1} \begin{split} &\frac{r(t)x^\Delta(\rho(t))}{f(x(t))}-\frac{r(T_0)x^\Delta(\rho(T_0))}{f(x(T_0))}\\ &= -\int^{t}_{T_0}p(s)\Delta s- \int^{t}_{T_0}\frac{r^{\sigma}(s)[\int^1_0f'(x_h(s))dh] [x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s. \end{split}$$ If \eqref{p1} fails to hold; that is, $$\int^\infty_{T_0}\frac{r^{\sigma}(s)[\int^1_0f'(x_h(s))dh] [x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s =\infty.$$ From \eqref{jia1}, we have $$\label{pu3} \lim_{t\to \infty}\frac{r(t)x^\Delta(\rho(t))}{f(x(t))}=-\infty.$$ We can assume that $$\label{pu4} \frac{r(T_0)x^\Delta(\rho(T_0))}{f(x(T_0))} -\int^{t}_{T_0}p(s)\Delta s<-1,$$ 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$\frac{r(T_1)x^\Delta(\rho(T_1))}{f(x(T_1))}<-(2L+1). So we have \begin{align*} \frac{r(T_1)x^\Delta(\rho(T_1))}{f(x(T_1))} -\int^{t}_{T_1}p(s)\Delta s &< -(2L+1)-\Big[\int^{t}_{T_0}p(s)\Delta s -\int^{T_1}_{T_0}p(s)\Delta s\Big]\\ &\leq -(2L+1)-[-2L]=-1. \end{align*} So we can replaceT_0$by$T_1>T_0$such that \eqref{pu4} still holds. From \eqref{jia1} and \eqref{pu4}, we obtain, for$t\geq T_{0}$, $$\label{p2} \frac{r(t)x^\Delta(\rho(t))}{f(x(t))}+ \int^{t}_{T_0}\frac{r^{\sigma}(s)[\int^1_0f'(x_h(s))dh] [x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s<-1.$$ In particular, $$x^\Delta(t)<0,\quad \text{for } t\geq T_0.$$ Therefore,$x(t)$is strictly decreasing. First assume that$\mathbb{T}is an isolated time scale; that is, $$\mathbb{T}=\{t_0,t_1,t_2,\dots\},\quad t_0y(t). Replacing t by \sigma(t) in \eqref{2.20} and using \cite[Theorem1.75]{bp1}, we obtain $$\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))}.$$ Noting that $$\label{p4} \frac{1}{f(x^{\sigma}(t))} =\frac{1}{f(x(t))}+\Big(\frac{1}{f(x(t))}\Big)^{\Delta}\mu(t).$$ Using \eqref{p4} in the left side of \eqref{p3} and noticing that$$ (\frac{1}{f(x(t))})^{\Delta}=-\frac{\int^1_0f'(x_h(t))dh x^\Delta(t)}{f(x(t))f(x^\sigma(t))}, $$from \eqref{p3}, it is easy to see that $$\label{p5} -\frac{r^{\sigma}(t) x^\Delta(t)}{f(x(t))}>y(t).$$ From \eqref{al} and \eqref{p5}, we obtain $$\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}$$ In the isolated time scale case from \eqref{asas} and \eqref{2.17}, we obtain \begin{equation*} \frac{y(\sigma(t))-y(t)}{y(t)(\sigma(t)-t)} >\frac{f(x(\sigma(t)))-f(x(t))}{x(\sigma(t))-x(t)}\cdot\frac{x(t) -x(\sigma(t))}{f(x(\sigma(t)))[\sigma(t)-t]}. \end{equation*} So \begin{equation*} \frac{y(\sigma(t))}{y(t)}>\frac{ f(x(t))}{f(x(\sigma(t)))}; \end{equation*} that is, $$\label{klkl} \frac{y(t_i)}{y(t_{i-1})}>\frac{ f(x(t_{i-1}))}{f(x(t_i))}.$$ Let T_0=t_{n_0} and t=t_n, n>n_0, then using \eqref{klkl} we have that$$ \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))}; $$that is, $$\label{p6} \frac{y(t)}{y(t_{n_0})}>\frac{f(x(t_{n_0}))}{f(x(t))},$$ for t>T_0. To obtain \eqref{p6} in the case where \mathbb{T} is the real interval [T_0, \infty), from \eqref{uiop} and \eqref{asas} we have$$ \frac{y'(t)}{y(t)}>\frac{f'(x(t))[-x'(t)]}{f(x(t))}; $$that is,$$ (\ln y(t))'>-(\ln f(x(t)))'. $$Integrating from T_0 to t, we obtain $$\label{vbvb} \frac{y(t)}{y(T_0)}>\frac{f(x(T_0))}{f(x(t))},\quad t>T_0.$$ Using \eqref{p5} again, from \eqref{p6} and \eqref{vbvb}, we obtain \begin{equation*} -\frac{r^{\sigma}(t) x^\Delta(t)}{f(x(t))}>y(t) >\frac{y(T_0)f(x(T_0))}{f(x(t))}. \end{equation*} If we set L_1:=y(T_0)f(x(T_0)), we obtain \begin{equation*} x^\Delta(t)<-\frac{L_1}{r^{\sigma}(t)}. \end{equation*} Integrating from T_0 to t, we obtain \begin{equation*} x(t)-x(T_0)<-\int^t_{T_0}\frac{L_1}{r^{\sigma}(s)}\Delta s \to-\infty,\quad \text{as } t\to\infty. \end{equation*} which contradicts x(t)>0. In \eqref{jia1}, letting t\to\infty and replacing T_0 by \sigma(\tau), denoting $$\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))},$$ we obtain $$\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))}.$$ We claim that \alpha= 0. In the right side of \eqref{p7}, using$$\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 $$\label{p28} \alpha+\int^\infty_{\sigma(\tau)}p(s)\Delta s+\int^\infty_{\tau}\frac{r^{\sigma}(s)[\int^1_0f'(x_h(s))dh][x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s=\frac{r^{\sigma}(\tau)x^\Delta(\tau)}{f(x(\tau))}.$$ From \eqref{p28}, we have $$\label{jia100} \lim_{t\to\infty}\frac{r^{\sigma}(t)x^\Delta(t)}{f(x(t))}=\alpha.$$ 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 $$\label{liu202} x^\Delta(t)\leq \frac{\alpha}{2}\cdot\frac{f(x(t))}{r^{\sigma}(t)}<0.$$ Thus $$\label{j1} \begin{split} M(T_1) &=: \int^\infty_{T_1}\frac{r^{\sigma}(s)[\int^1_0f'(x_h(s))dh] [x^\Delta(s)]^2}{f(x(s))f(x^\sigma(s))}\Delta s\\ &\geq -\frac{\alpha}{2}\int^\infty_{T_1} \frac{[\int^1_0f'(x_h(s))dh][-x^\Delta(s)]}{f(x^\sigma(s))}\Delta s. \end{split}$$ Assume that t=t_{i-1}0, from \eqref{liu10}, we have that there exists T_3 such that for t\geq T_3, $$\label{liu20} \frac{r^\sigma(t)x^\Delta(t)}{f(x^\sigma(t))}\geq \frac{\alpha}{2}.$$ 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)); $$that is, $$\label{jia12} x^\Delta(t)\geq \frac{\alpha}{2r^\sigma(t)}f(x^\sigma(T_3)).$$ Integrating \eqref{jia12} from T_3 to t, we obtain x(t)\to+\infty as t\to\infty, which contradicts the boundedness of x(t). If \mathbb{T} is the real interval [T_3,\infty), then from \eqref{liu20} and the monotonicity of f, we have that for t\geq T_3, $$\label{liu15} x'(t)\geq \frac{\alpha}{2r(t)}f(x(t)) \geq \frac{\alpha}{2r(t)}f(x(T_2)).$$ Integrating \eqref{liu15} from T_3 to t, we obtain x(t)\to+\infty as t\to\infty, which also contradicts the boundedness of x(t). Therefore \alpha=0. From \eqref{p7}, we obtain \eqref{liu117}. The proof is complete. \end{proof} \begin{theorem}\label{jia101} Suppose \mathbb{T} is a regular time scale, r(t)>0 with \int^\infty_{T_0}[r^\sigma(t)]^{-1}\Delta t=\infty and suppose that \lim_{t\to\infty}\int^t_{T_0}p(s)\Delta s exists and finite. Let P(t)=\int^\infty_t p(s)\Delta s. f(x) satisfies the superlinearity conditions $$\label{liu18} 0<\int^\infty_\epsilon\frac{dx}{f(x)}, \int^{-\epsilon}_{-\infty}\frac{dx}{f(x)}<\infty,\quad\text{for all }\epsilon>0.$$ Then the superlinear dynamic equation $$\label{liu23} [r(t)x^{\Delta}(\rho(t))]^\Delta+p(t)f(x(t))=0,$$ is oscillatory, if $$\label{liu22} \limsup_{t\to\infty}\frac{1}{r^\sigma(t)} \int^t_{T_0}P^\sigma(s)\Delta s=\infty.$$ \end{theorem} \begin{proof} Suppose that x(t) is a nonoscillatory solution of \eqref{pu10} on [T_0,\infty). Without loss of generality, assume that x(t) is positive for t\in [T_0,\infty). From Theorem \ref{jia1001}, x(t) satisfies the integral equation \eqref{liu117}. Dropping the last integral term in \eqref{liu117}, we have the inequality $$\label{liu19} \frac{r^\sigma(t)x^\Delta(t)}{f(x^\sigma(t))}\geq P^\sigma(t).$$ Dividing \eqref{liu19} by r^\sigma(t) and integrating from T_0 to t and using Lemma \ref{jia99}, we find$$ 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. $$This contradicts \eqref{liu22}, so equation \eqref{liu23} is oscillatory. \end{proof} \section{Examples} \begin{example} \label{examp4.1} \rm Consider the superlinear difference equation $$\label{liu300} \Delta^2x(n-1)+p(n)x^\gamma(n)=0,\quad \gamma>1,$$ where P(n)=\frac{1}{n}+\frac{2(-1)^n}{\sqrt{n}}, $$\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)}}.$$ It is to see that \sum^\infty_{n=1}P(n+1)=\infty. So from Theorem \ref{jia101}, \eqref{liu300} is oscillatory. \end{example} In \cite[Theorem 2.5]{jep1}, we proved that the equation \Delta^2x(n-1)+q(n)x^\gamma(n)=0,\gamma>1, is oscillatory, if \sum^\infty_{n=1}nq(n)=\infty. In the following, we will prove that $$\label{jia1000} \sum^{2k+1}_{n=1}np(n)\to-\infty\quad \text{as}\quad k\to\infty.$$ So \cite[Theorem 2.5]{jep1} would not apply in \eqref{liu300}. We need the following lemmas. The first lemma may be regarded as a discrete version of L'Hopital's rule and can be found in \cite[page 48]{bp1}. \begin{lemma}[Stolz-Ces\'{a}ro Theorem] \label{l3.2} Let \{a_n\}_{n\geq 1} and \{b_n\}_{n\geq 1} be two sequences of real number. Assume b_n is positive, strictly increasing and unbounded and the following limit exists:$$ \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. $$\end{lemma} We will use Lemma \ref{l3.2} to prove the following result. \begin{lemma} For each real number d>0, we have $$\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}.$$ \end{lemma} \begin{proof} By Taylor's formula, we have $$\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).$$ By \eqref{3.5} and the Stolz-Ces\'{a}ro Theorem (Lemma \ref{l3.2}), it is easy to see that $$\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}$$ Using \eqref{3.5} in \eqref{3.6}, it follows that \eqref{3.3} holds. \end{proof} So given 0<\epsilon<1, for large m, we have the inequality $$\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.$$ Set d=1/2. we have $$\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}.$$ From \eqref{Lynn1} and using Taylor's expansion, we have $$\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}$$ Using the Euler formula \cite[page 205]{kp},$$ C=\lim_{n\to\infty}\Big(\sum^n_{k=1}\frac{1}{k}-\ln n\Big), where C is called Euler constant, we obtain $$\label{jia500} \sum^{2k+1}_{n=1}\frac{1}{n+1}=o(1)+C-1+\ln (2k+2).$$ 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 $$\label{liu303} x''(t)+p(t)x^\gamma(t)=0,\quad \gamma>1,$$ 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 $$\label{jia40} \limsup_{t\to\infty}\int^t_1P(s)ds=\infty.$$ 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 $$\label{liu304} (x(q^{-1}t))^{\Delta\Delta}+p(t)x^\gamma(t)=0,\quad \gamma>1,$$ 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} \begin{thebibliography}{00} \bibitem{bp1} M. Bohner and A. Peterson; \emph{Dynamic Equation on Time Scales: An Introduction with Applications}, Birkh\"{a}user, Boston, 2001. \bibitem{bp2} M. Bohner and A. Peterson, Editors; \emph{Advances in Dynamic Equations on Time Scales}, Birkh\"{a}user, Boston, 2003. \bibitem{ejp} L. Erbe, B. G. Jia and A. Peterson; \emph{Nonoscillation for second order sublinear dynamic equations on time scales}, J. Computational and Applied Mathematics, 232(2009), 594-599. \bibitem{kw} M. K. Kwong and J. S. W. Wong; \emph{An application of integral inequality to second order nonlinear oscillation}, J. Differential Equation \textbf{46} (1982), 63-77. \bibitem{kp} W. Kelly and A. Peterson; \emph{Difference Equation: An Introduction with Applications}, 2nd ed., Harcourt/Academic Press, 2001. \bibitem{N} M. Naito; \emph{Asymptotic behavior of solutions of second order differential equation with integrable coefficients}, Trans. Amer. Maths. Soc. \textbf{282} (1984), 577-588. \bibitem{jep} Baoguo Jia, Lynn Erbe, Allan Peterson; \emph{An Oscillation theorem for second order superlinear dynamic equations on time scales}, Applied Math and Applications, in press. \bibitem{jep1} Baoguo Jia, Lynn Erbe, Allan Peterson; \emph{Kiguradze-type oscillation theorems for second order superlinear dynamic equations on time scales}, Canad. Math. Bulletin, doi:10.4153/CMB-2011-034-4. \end{thebibliography} \end{document}