近几年来, 时间模上动态方程的振荡性理论引起了国内外学术界的广泛兴趣和高度关注^[1-27], 这个新理论是德国的Hilger Stefan博士在他的导师Aulbach Bernd指导下于1990年首次提出来的^[1], 目的有两个:其一是为了统一连续分析和离散分析理论, 将微分方程和差分方程的很多研究统一到一种框架下进行; 其二是弥补了微分方程与差分方程"之间"的不足, 如扩充到了$q$ -差分方程, 这类方程在量子理论方面有非常重要的应用.动态方程的新理论在自动控制技术、生物种群动力学、物理学(特别是核物理)、神经网络和社会科学等诸多领域中均有非常重要的应用, 并能解决许多不同领域里微分方程和差分方程不能解决的实际问题.近年来, 时间模上的理论及时间模上的动态方程研究成果非常丰富^[2-28].本文讨论时间模上如下一类具有阻尼项的二阶非线性中立型变时滞Emden-Fowler型动态方程

$ [A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}+b(t)\phi_{1}(y^{\Delta}(t))+P(t)F(\phi_{2}(x(\delta(t))))=0, t\in{\Bbb T} $

(1.1)

的振荡性, 这里$y(t)=x(t)+B(t)g(x(\tau(t))), \phi_{1}(u)=|u|^{\lambda-1}u, $$ \phi_{2}(u)=|u|^{\beta-1}u, \lambda>0, \beta>0$为实常数; ${\Bbb T}$为任意时间模.

由于我们感兴趣的是方程解的振荡性, 所以假设时间模${\Bbb T}$是无界的, 即$\sup{\Bbb T}=\infty$, 且${\Bbb T}$是实数域${\Bbb R}$上的非空闭子集.设$t_{0}\in{\Bbb T}$且$t_{0}>0$, 我们定义时间模区间$[t_{0}, \infty)_{{\Bbb T}}=[t_{0}, \infty)\cap{\Bbb T}$.方程(1.1)的解是指定义在时间模${\Bbb T}$上满足方程(1.1)的非平凡实值函数$x(t), t\in{\Bbb T}$.方程(1.1)的解$x(t)$称为振荡的, 如果$x(t)$既不最终为正, 也不最终为负.否则, 称为非振荡的.方程(1.1)称为振荡的, 如果它的所有解都是振荡的.我们仅关注方程(1.1)的不最终恒为零的解.为了叙述方便, 考虑如下假设:

(H$_{1}$) $\tau, \delta:{\Bbb T}\rightarrow{\Bbb T}$均为滞量函数, 并且$\tau(t)\leqslant t, \lim\limits_{t\rightarrow\infty}\tau(t)=\infty;$ $\delta(t)\leqslant t, \lim\limits_{t\rightarrow\infty}\delta(t)=\infty$, 且$\tau(\delta(t))=\delta(\tau(t))$;

(H$_{2}$) $\tau$是严格递增的, ${\tilde{\mathbb{T}}}=\tau({\Bbb T})\subseteq{\Bbb T}$是一时间模, $\tau\circ\sigma=\sigma\circ\tau$, 并且$\tau^{\Delta}(t)\geqslant\tau_{0}>0$ (这里$\tau_{0}$是常数), $\delta(t)\geqslant\tau(t)$;

(H$_{3}$) $P(t)\in C_{rd}({\Bbb T}, (0, \infty)), B(t)\in C_{rd}({\Bbb T}, {\Bbb R})$, 且$0\leqslant B(t)\leqslant b_{0}<\infty$ (这里$b_{0}$是常数);

(H$_{4}$) $A(t)\in C_{rd}({\Bbb T}, (0, \infty)), b(t)\in C_{rd}({\Bbb T}, [0, \infty))$, 并且$-b/A\in\Re^{+}$;

(H$_{5}$) $g\in C({\Bbb R}, {\Bbb R}), ug(u)>0 (u\neq0)$, 且存在常数$0<\eta\leqslant1$使得$g(u)/u\leqslant\eta (u\neq0)$;

(H$_{6}$) $F\in C({\Bbb R}, {\Bbb R}), uF(u)>0 (u\neq0)$, 且存在常数$L>0$使得$F(u)/u\geqslant L (u\neq0)$.

同样, 我们将分别在条件

$ \int^{\infty}_{t_{0}}\left[\frac{e_{_{-b/A}}(s, t_{0})}{A(s)}\right]^{1/\lambda}\Delta s=\infty $

($C_1$)

和

$ \int^{\infty}_{t_{0}}\left[\frac{e_{_{-b/A}}(s, t_{0})}{A(s)}\right]^{1/\lambda}\Delta s<\infty $

($C_2$)

成立的情况下建立方程(1.1)的振荡准则.

关于方程(1.1)的特殊情形, 已有一些研究成果.如Saker等^[2]运用Riccati变换技术, 研究了时间模上具阻尼项的二阶非线性动态方程

$ (r(t)x^{\Delta}(t))^{\Delta}+p(t)(x^{\Delta\sigma}(t))^{\gamma}+q(t)f(x(\sigma(t)))=0 (t\in{\Bbb T}) $

的振荡性, 得到了该方程振荡的几个充分条件.

Erbe等^[3]利用时间模上的微积分理论和Riccati变换技术研究了具阻尼项的二阶非线性动态方程

$ [r(t)(x^{\Delta}(t))^{\gamma}]^{\Delta}+p(t)(x^{\Delta\sigma}(t))^{\gamma}+q(t)f(x(\tau(t)))=0 (t\in{\Bbb T}) $

在条件

$ \int^{\infty}_{t_{0}}\left[\frac{1}{R(t)}\int^{t}_{t_{0}}Q(s)\beta^{\gamma}(s)\Delta s\right]^{1/\gamma}\Delta t=\infty $

($C_3$)

成立下得到了该方程的一些振荡准则, 推广并改进了已有的一些结果, 上式中

$ R(t)=e_{_{p/r^{\sigma}}}(t, t_{0})r(t), ~~Q(t)=e_{_{p/r^{\sigma}}}(t, t_{0})q(t), ~~\beta(t)=\int^{\infty}_{\tau(t)}R^{-1/\gamma}(s)\Delta s. $

Chen等^[4]研究了时间模上一类具阻尼项的二阶非线性动态方程

$ [(x^{\Delta}(t))^{\gamma}]^{\Delta}+p(t)(x^{\Delta}(t))^{\gamma}+q(t)f(x^{\sigma}(t))=0 (t\in{\Bbb T}) $

(这里$\gamma$是正奇数之比), 给出了方程振荡的一些充分条件.

张全信等^[5-8]利用时间模上的有关理论及Riccati变换技术, 研究了时间模上具阻尼项的二阶半线性动态方程

$ (a(t)|x^{\Delta}(t)|^{\gamma-1}x^{\Delta}(t))^{\Delta}+p(t)|x^{\Delta}(t)|^{\gamma-1}x^{\Delta}(t)+q(t)|x(\delta(t))|^{\gamma-1}x(\delta(t))=0 (t\in{\Bbb T}) $

(1.2)

的振荡性(这里$\gamma>0$是常数), 得到了方程(1.2)振荡的一些非常有意义的结果, 推广并改进了现有文献中的一些结论.但我们看到, 这些结果是在条件"$\delta(t)$严格递增且$\delta({\Bbb T})={\Bbb T}$"成立的情形下得到的.此条件限制性较强, 通常不容易满足.如取${\Bbb T}=\{2, 4, 6, \cdots\}, \delta(t)=t/2$, 则$\delta$是严格递增函数, 并且满足

$ \delta(t)\leqslant t, ~~\lim\limits_{t\rightarrow\infty}\delta(t)=\infty, $

但$\delta({\Bbb T})=\{1, 2, 3, \cdots\}\neq{\Bbb T}$.这样, 文献[5-8]中的条件$\delta({\Bbb T})={\Bbb T}$就不满足.因此, 文献[5-8]中的结果就不一定成立.

对于中立型动态方程, Sahiner^[10], Wu等^[11], Saker等^[12-14]借助时间模上的有关理论和Riccati变换技术及不等式技巧, 研究了一类二阶非线性变时滞中立型动态方程

$ [r(t)((y(t)+p(t)y(\tau(t)))^{\Delta})^{\gamma}]^{\Delta}+f(t, y(\delta(t)))=0 (t\in{\Bbb T}) $

的振荡性, 得到了该方程振荡的一些充分条件, 推广并改进了一些已知的结果.但对中立项系数函数有限制条件$0\leqslant p(t)<1$ (其它文献也都有这样的限制条件, 如文献[18-24, 26]等)且$\gamma>0$是正奇数之比.

孙一冰等^[9]借助时间模上的微积分理论、Riccati变换技术及不等式技巧, 研究了一类二阶半线性中立型时滞阻尼动态方程

$ (a(t)|z^{\Delta}(t)|^{\gamma-1}z^{\Delta}(t))^{\Delta}+p(t)|z^{\Delta}(t)|^{\gamma-1}z^{\Delta}(t)+q(t)|x(\delta(t))|^{\gamma-1}x(\delta(t))=0 (t\in{\Bbb T}) $

(1.3)

的振荡性, 这里$z(t)=x(t)+r(t)x(\tau(t)), \gamma>0$, 得到了方程(1.3)的一些振荡准则, 部分地改进了文献[5-8]的结果.但对中立项系数函数同样有限制条件$0\leqslant r(t)<1$并且要求"$\tau=\delta, \delta$严格递增且$\tau\circ\sigma=\sigma\circ\tau$".这个条件也是一个限制性较强的条件.

此外, 我们注意到, 对于二阶Euler微分方程

$ (t^{2}x'(t))'+q_{_{0}}x(t)=0 $

(1.4)

来说, 由于$\int^{\infty}_{1}\frac{\ln t}{t^{2}}{\rm d}t<\infty$, 所以文献[3]的条件(C$_{3}$), 文献[6]的条件(4.16), 文献[8]的条件(4.14), 以及文献[9]的条件(3.25)均不满足, 因此这些文献中的定理均不能用于方程(1.4).

显然, 方程(1.1)更具有一般性.当$g(u)=u, F(u)=u$且$\lambda=\beta$时, 方程(1.1)就可简化成方程(1.3)的形式; 当$r(t)\equiv0$时, 方程(1.3)就变成了方程(1.2).因此, 研究方程(1.1)是非常有意义的.本文的目的是利用时间模上动态方程的基本理论和广义的Riccati变换, 结合一些分析技巧, 在较宽松的条件下研究方程(1.1)的振荡性, 改善对方程的这些条件限制(如$\delta({\Bbb T})={\Bbb T}, \tau=\delta, 0\leqslant r(t)<1$等), 得到了该方程的几个新的振荡准则, 这些准则不仅推广和改进了一些已知的结论, 使得文献[2-14]中的许多结果成为我们结果的特例, 而且在时间模上统一了具有阻尼项的二阶非线性中立型变时滞Emden-Fowler型微分方程和差分方程的振荡性质.

本文的整体结构是这样的:第2节, 给出了6个引理, 这些引理对于本文结果的证明起到重要的作用; 第3节, 给出本文的6个主要结果, 是分别在条件$(C_{1})$和$(C_{2})$成立的情况下建立的; 第4节, 给出2个具体例子来说明本文结果的应用.

以下给出几个引理, 其中引理2.1是众所周知的, 为文献[15]中的定理2.33(也是文献[5]中的引理3.1);引理2.2是文献[15]中的定理1.90;引理2.3是文献[18]中的一个引理(也可由数学分析的方法得到); 引理2.4是文献[19]中的引理2.2;引理2.5是文献[20]中的引理2.3.

引理2.  如果$g\in \Re^{+}$, 即$g(t)\in C_{rd}({\Bbb T}, {\Bbb R})$, 并且对于任意的$t\in[t_{0}, +\infty)_{{\Bbb T}}$, 满足$1+\mu(t)g(t)>0$.则初值问题$y^{\triangle}(t)=g(t)y(t), y(t_{0})=y_{0}\in {\Bbb R}$在$[t_{0}, +\infty)_{{\Bbb T}}$上有唯一的正解$e_{g}(t, t_{0})$, 这个"指数函数"有时也记为$e_{g}(\cdot, t_{0})$, 它满足半群性质$e_{g}(a, b)e_{g}(b, c)=e_{g}(a, c)$.

引理2.2  设$x(t)$是$\Delta$可微的且最终为正或最终为负, 则有

$ (x^{\lambda}(t))^{\Delta}=\lambda\int^{1}_{0}\left[hx^{\sigma} +(1-h)x\right]^{\lambda-1}x^{\Delta}(t){\rm d}h. $

(2.1)

引理2.3  设$A>0, B>0, \lambda>0$为常数, 则对$x>0$有$Bx-Ax^{\frac{\lambda+1}{\lambda}}\leqslant\frac{\lambda^{\lambda}B^{\lambda+1}}{(\lambda+1)^{\lambda+1}A^{\lambda}}$.

引理2.4  若$\tau(t)$是严格递增的, ${\tilde{\mathbb{T}}}:=\tau({\Bbb T})\subseteq{\Bbb T}$是一时间模, $\tau(\sigma(t))=\sigma(\tau(t))$.设$x:{\tilde{\mathbb{T}}}\rightarrow{\Bbb R}$, 如果$\tau^{\Delta}(t)$和$x^{\Delta}(\tau(t))$存在($t\in{\Bbb T}^{k}$), 则$(x(\tau(t)))^{\Delta}$存在, 且

$ (x(\tau(t)))^{\Delta}=x^{\Delta}(\tau(t))\tau^{\Delta}(t). $

(2.2)

引理2.5  如果$0<\lambda\leqslant1, X, Y$为非负实数, 则$X^{\lambda}+Y^{\lambda}\geqslant(X+Y)^{\lambda}$.

引理2.6  设条件(H$_{1}$)-(H$_{6}$)和(C$_{1}$)成立, 若$x(t)$是方程(1.1)的一个最终正解, 则存在$t_{1}\in[t_{0}, +\infty)_{{\Bbb T}}$, 使得当$t\in[t_{1}, +\infty)_{{\Bbb T}}$时, 有

$ y(t)>0, y^{\Delta}(t)>0, A(t)\phi_{1}(y^{\Delta}(t))>0, [A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}<0. $

证  因为$x(t)$是方程(1.1)的一个最终正解, 所以$\exists t_{1}\in[t_{0}, +\infty)_{{\Bbb T}}$, 当$t\in[t_{1}, +\infty)_{{\Bbb T}}$时, 有$x(t)>0, x(\tau(t))>0, x(\delta(t))>0$.从而$y(t)>0$.由方程(1.1)得

$ [A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}+b(t)\phi_{1}(y^{\Delta}(t))\leqslant-LP(t)(x(\delta(t)))^{\beta}<0, t\in[t_{1}, \infty)_{{\Bbb T}}, $

(2.3)

注意到引理2.1, 就有

$ \begin{eqnarray}\left[\frac{A(t)\phi_{1}(y^{\Delta}(t))}{e_{_{-b/A}}(t, t_{0})}\right]^{\Delta} &=&\frac{[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}e_{_{-b/A}}(t, t_{0})-A(t)\phi_{1}(y^{\Delta}(t))[e_{_{-b/A}}(t, t_{0})]^{\Delta}} {e_{_{-b/A}}(t, t_{0})e_{_{-b/A}}(\sigma(t), t_{0})} \\ &=&\frac{[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}+b(t)\phi_{1}(y^{\Delta}(t))}{e_{_{-b/A}}(\sigma(t), t_{0})} \leqslant-\frac{LP(t)(x(\delta(t)))^{\beta}}{e_{_{-b/A}}(\sigma(t), t_{0})}<0, \end{eqnarray} $

(2.4)

所以$\frac{A(t)\phi_{1}(y^{\Delta}(t))}{e_{_{-b/A}}(t, t_{0})}$严格递减且最终定号, 由此我们断定$y^{\Delta}(t)>0, t\in[t_{1}, +\infty)_{{\Bbb T}}$.

事实上, 若不然, 则存在$t_{2}\in[t_{1}, +\infty)_{{\Bbb T}}$, 使得$y^{\Delta}(t_{2})<0$, 于是$t\in[t_{2}, +\infty)_{{\Bbb T}}$时, 就有

$ \frac{A(t)\phi_{1}(y^{\Delta}(t))}{e_{_{-b/A}}(t, t_{0})}\leqslant\frac{A(t_{2})\phi_{1}(y^{\Delta}(t_{2}))}{e_{_{-b/A}}(t_{2}, t_{0})}=-M<0, $

这里$M=-\frac{A(t_{2})\phi_{1}(y^{\Delta}(t_{2}))}{e_{_{-b/A}}(t_{2}, t_{0})} =$$\frac{A(t_{2})|y^{\Delta}(t_{2})|^{\lambda-1}[-y^{\Delta}(t_{2})]}{e_{_{-b/A}}(t_{2}, t_{0})}>0$为常数.所以

$ [-y^{\Delta}(t)]^{\lambda}\geqslant M\frac{e_{_{-b/A}}(t, t_{0})}{A(t)}, $

即

$ y^{\Delta}(t)\leqslant-M^{1/\lambda}\left[\frac{e_{_{-b/A}}(t, t_{0})}{A(t)}\right]^{1/\lambda}, $

因此

$ y(t)\leqslant y(t_{2})-M^{1/\lambda}\int^{t}_{t_{2}}\left[\frac{e_{_{-b/A}}(s, t_{0})}{A(s)}\right]^{1/\lambda}\Delta s\rightarrow-\infty(t\rightarrow\infty), $

这与$y(t)>0$矛盾, 故$y^{\Delta}(t)>0, t\in[t_{1}, +\infty)_{{\Bbb T}}$, 进而$A(t)\phi_{1}(y^{\Delta}(t))>0$.

定理3.1  设条件(H$_{1}$)-(H$_{6}$)和(C$_{1}$)成立, 如果存在函数$\varphi\in C^{1}_{rd}([t_{0}, \infty)_{{\Bbb T}}, (0, \infty))$使得当$\lambda\leqslant\beta$时

$ \begin{eqnarray} &&\limsup\limits_{t\rightarrow\infty}\int^{t}_{t_{0}}\varphi(s)\left\{\varsigma\xi(s)- \frac{\lambda^{\lambda}A(\tau(s))}{(\lambda+1)^{\lambda+1}\alpha^{\lambda}_{1}} \left[\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{b(s)}{A(s)}\right|^{\lambda+1}\right.\right. \\ && \left.\left.+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{\tau_{0}b(\tau(s))}{A(\tau(s))}\right|^{\lambda+1} \right]\right\}\Delta s=\infty, \end{eqnarray} $

(3.1)

当$\lambda>\beta$时

$ \begin{eqnarray} &&\limsup\limits_{t\rightarrow\infty}\int^{t}_{t_{0}}\varphi(s)\left\{\varsigma\xi(s)- \frac{\beta^{\beta}A^{\beta/\lambda}(\tau(s))}{(\beta+1)^{\beta+1}\alpha^{\beta}_{2}} \left[\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{b(s)}{A(s)}\right|^{\beta+1}\right.\right. \\ && \left.\left.+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{\tau_{0}b(\tau(s))}{A(\tau(s))}\right|^{\beta+1} \right]\right\}\Delta s=\infty, \end{eqnarray} $

(3.2)

式中函数$\xi(t)=\min\{P(t), P(\tau(t))\}$, 常数$\varsigma=\left\{\begin{array}{ll} L, &0<\lambda\leqslant1, \\ 2^{1-\beta}L, &\lambda>1, \end{array} \right.$ $\alpha_{_{1}}=k^{\frac{\beta-\lambda}{\lambda}}\beta\tau_{_{0}}, $ $\alpha_{_{2}}=\frac{\beta\tau_{_{0}}}{k^{(\lambda-\beta)/\beta\lambda}}$ ($k>0$为某常数), 则方程(1.1)在$[t_{0}, +\infty)_{{\Bbb T}}$上是振荡的.

证  设方程(1.1)在$[t_{0}, +\infty)_{{\Bbb T}}$上有一个非振荡解$x(t)$, 不失一般性, 不妨设$x(t)>0, $ $x(\tau(t))>0, x(\delta(t))>0, t\in[t_{1}, +\infty)_{{\Bbb T}}, t_{1}\in[t_{0}, +\infty)_{{\Bbb T}}$ (当$x(t)$是最终负解时类似地可以证明), 于是$y(t)>0, y^{\Delta}(t)>0, t\in[t_{1}, +\infty)_{{\Bbb T}}$.

当$\lambda>1$时, 由(2.1)式就有

$ [(y(t))^{\lambda}]^{\Delta}\geqslant\lambda\int^{1}_{0}[hy+(1-h)y]^{\lambda-1}y^{\Delta}(t){\rm d}h=\lambda(y(t))^{\lambda-1}y^{\Delta}(t). $

当$0<\lambda\leqslant1$时, 同样由(2.1)式得

$ [(y(t))^{\lambda}]^{\Delta}\geqslant\lambda\int^{1}_{0}[hy^{\sigma}+(1-h)y^{\sigma}]^{\lambda-1}y^{\Delta}(t){\rm d}h =\lambda(y(\sigma(t)))^{\lambda-1}y^{\Delta}(t), $

所以

$ \begin{eqnarray} \left\{\begin{array}{ll} (y^{\lambda}(t))^{\Delta}\geqslant\lambda y^{\lambda-1}(t)y^{\Delta}(t), &\lambda>1, \\ (y^{\lambda}(t))^{\Delta}\geqslant\lambda y^{\lambda-1}(\sigma(t))y^{\Delta}(t), ~&0<\lambda\leqslant1. \end{array} \right. \end{eqnarray} $

(3.3)

利用(2.3)式, 可得

$ [A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}+b(t)\phi_{1}(y^{\Delta}(t))+LP(t)(x(\delta(t)))^{\beta}\leqslant0, $

且有

$ \frac{[A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta}}{\tau^{\Delta}(t)}+ b(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))+LP(\tau(t))(x(\delta(\tau(t))))^{\beta}\leqslant0, $

于是, 综合上两式可得

$ \left\{[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}+b(t)\phi_{1}(y^{\Delta}(t))+LP(t)(x(\delta(t)))^{\beta}\right\} \\ + (b_{0}\eta)^{\beta}\left\{\frac{[A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta}}{\tau^{\Delta}(t)}+ b(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))+LP(\tau(t))(x(\delta(\tau(t))))^{\beta}\right\} \leqslant0. $

一方面, 如果$0<\beta\leqslant1$, 则由上式, 并分别注意到$\tau^{\Delta}(t)\geqslant\tau_{_{0}}>0, \xi(t)$的定义, 引理2.5, 以及$y(t)\leqslant x(t)+b_{0}\eta x(\tau(t))$和$\tau(\delta(t))=\delta(\tau(t)), \delta(t)\geqslant\tau(t)$, 可得

$ \begin{eqnarray} &&[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}+\frac{(b_{0}\eta)^{\beta}}{\tau_{_{0}}}[A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta} \\ && +b(t)\phi_{1}(y^{\Delta}(t))+(b_{0}\eta)^{\beta}b(\tau(t))\phi_{1}(y^{\Delta}(\tau(t))) \\ &\leqslant& -L\xi(t)\left[(x(\delta(t)))^{\beta}+(b_{0}\eta x(\delta(\tau(t))))^{\beta}\right] \\ &\leqslant&-L\xi(t)\left[x(\delta(t))+b_{0}\eta x(\delta(\tau(t)))\right]^{\beta} \\ &\leqslant& -L\xi(t)(y(\delta(t)))^{\beta}. \end{eqnarray} $

(3.4)

又由于$[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}<0$, 并注意到$t\leqslant\sigma(t), \tau(t)\leqslant\tau(\sigma(t))$, 因此有

$ A(t)\phi_{1}(y^{\Delta}(t))\geqslant A(\sigma(t))\phi_{1}(y^{\Delta}(\sigma(t))), A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))\geqslant A(\tau(\sigma(t)))\\ \phi_{1}(y^{\Delta}(\tau(\sigma(t)))), $

将其代入(3.4)式, 注意到$y^{\Delta}(t)>0$, 得

$ \begin{eqnarray} &&[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}+\frac{(b_{0}\eta)^{\beta}}{\tau_{_{0}}}[A(\tau(t)) \phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta} \\ & \leqslant&-L\xi(t)(y(\tau(t)))^{\beta}-b(t)\frac{A(\sigma(t))\phi_{1}(y^{\Delta}(\sigma(t)))}{A(t)} \\ &&-(b_{0}\eta)^{\beta}b(\tau(t))\frac{A(\tau(\sigma(t)))\phi_{1}(y^{\Delta}(\tau(\sigma(t))))}{A(\tau(t))}. \end{eqnarray} $

(3.5)

另一方面, 若$\beta>1$, 注意到不等式$X^{\beta}+Y^{\beta}\geqslant2^{1-\beta}(X+Y)^{\beta}$(这里$X, Y$为非负实数.此不等式可由函数$f(x)=x^{\beta}(\beta>1)$的凹凸性推得), 则(3.5)式应为

$ \begin{eqnarray} &&[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}+\frac{(b_{0}\eta)^{\beta}}{\tau_{_{0}}}[A(\tau(t)) \phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta}\\ & \leqslant&-2^{1-\beta}L\xi(t)(y(\tau(t)))^{\beta}-b(t)\frac{A(\sigma(t))\phi_{1}(y^{\Delta}(\sigma(t)))}{A(t)} \\ &&-(b_{0}\eta)^{\beta}b(\tau(t))\frac{A(\tau(\sigma(t)))\phi_{1}(y^{\Delta}(\tau(\sigma(t))))}{A(\tau(t))}. \end{eqnarray} $

(3.6)

于是(3.5)和(3.6)式可合成一个式子

$ \begin{eqnarray} &&[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}+\frac{(b_{0}\eta)^{\beta}}{\tau_{_{0}}}[A(\tau(t)) \phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta} \\ &\leqslant&-\varsigma\xi(t)(y(\tau(t)))^{\beta}-\frac{b(t)}{A(t)}A(\sigma(t))\phi_{1}(y^{\Delta}(\sigma(t))) \\ && -\frac{(b_{0}\eta)^{\beta}b(\tau(t))}{A(\tau(t))}A(\tau(\sigma(t)))\phi_{1}(y^{\Delta}(\tau(\sigma(t)))). \end{eqnarray} $

(3.7)

下面分$\lambda\leqslant\beta$和$\lambda>\beta$两种情形来证明.

情形(a)  $\lambda\leqslant\beta$.

定义广义的Riccati变换

$ \begin{eqnarray} W(t)=\varphi(t)\frac{A(t)\phi_{1}(y^{\Delta}(t))}{\phi_{2}(y(\tau(t)))}=\varphi(t)\frac{A(t)(y^{\Delta}(t))^{\lambda}}{(y(\tau(t)))^{\beta}}, ~~t\in[t_{1}, +\infty)_{{\Bbb T}}, \end{eqnarray} $

(3.8)

则$W(t)>0, t\in[t_{1}, +\infty)_{{\Bbb T}}$.

一方面, 若$0<\beta\leqslant1$, 则由(3.3)及(2.2)式, 得

$ [(y(\tau(t)))^{\beta}]^{\Delta}\geqslant\beta(y(\tau(\sigma(t))))^{\beta-1}y^{\Delta}(\tau(t))\tau^{\Delta}(t). $

于是由(3.8)式, 并利用时间模上2个函数积及商的微分运算法则, 我们可得

$ \begin{eqnarray*} W^{\Delta}(t)&=&\varphi(t)\frac{[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}}{\phi_{2}(y(\tau(t)))}+A(\sigma(t))(y^{\Delta}(\sigma(t)))^{\lambda}\\&& \frac{\varphi^{\Delta}(t)(y(\tau(t)))^{\beta}-\varphi(t)[(y(\tau(t)))^{\beta}]^{\Delta}}{(y(\tau(t)))^{\beta}(y(\tau(\sigma(t))))^{\beta}} \\ &\leqslant&\varphi(t)\frac{[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}}{\phi_{2}(y(\tau(t)))}+\frac{\varphi^{\Delta}(t)W(\sigma(t))}{\varphi(\sigma(t))} \\ &&-\varphi(t)\frac{A(\sigma(t))(y^{\Delta}(\sigma(t)))^{\lambda}\beta(y(\tau(\sigma(t))))^{\beta-1}y^{\Delta}(\tau(t))\tau^{\Delta}(t)} {(y(\tau(t)))^{\beta}(y(\tau(\sigma(t))))^{\beta}} \\ &=&\varphi(t)\frac{[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}}{\phi_{2}(y(\tau(t)))}+\frac{\varphi^{\Delta}(t)W(\sigma(t))}{\varphi(\sigma(t))}\\&& -\frac{\beta\tau_{0}\varphi(t)A(\sigma(t))(y^{\Delta}(\sigma(t)))^{\lambda}y^{\Delta}(\tau(t))}{(y(\tau(t)))^{\beta}y(\tau(\sigma(t)))}. \end{eqnarray*} $

由于$A(t)(y^{\Delta}(t))^{\lambda}$是递减的, 因此$A(\tau(t))(y^{\Delta}(\tau(t)))^{\lambda}\geqslant A(\sigma(t))(y^{\Delta}(\sigma(t)))^{\lambda}$, 即

$ y^{\Delta}(\tau(t))\geqslant\left(\frac{A(\sigma(t))}{A(\tau(t))}\right)^{1/\lambda}y^{\Delta}(\sigma(t)). $

注意到$y(\tau(t))\leqslant y(\tau(\sigma(t)))$, 所以

$ \begin{eqnarray} W^{\Delta}(t)&\leqslant& \varphi(t)\frac{[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}}{\phi_{2}(y(\tau(t)))}+\frac{\varphi^{\Delta}(t)W(\sigma(t))}{\varphi(\sigma(t))} \\ &&-\frac{\beta\tau_{0}\varphi(t)(A(\sigma(t)))^{(\lambda+1)/\lambda}(y^{\Delta}(\sigma(t)))^{\lambda+1}} {(A(\tau(t)))^{1/\lambda}(y(\tau(\sigma(t))))^{\beta+1}}\\ %\eqno(3.9) &=&\frac{\varphi(t)[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}}{(y(\tau(t)))^{\beta}}+\frac{\varphi^{\Delta}(t)W(\sigma(t))}{\varphi(\sigma(t))} \end{eqnarray} $

(3.9)

$ -(y(\tau(\sigma(t))))^{\frac{\beta-\lambda}{\lambda}}\frac{\beta\tau_{0}\varphi(t)(W(\sigma(t)))^{(\lambda+1)/\lambda}} {(A(\tau(t)))^{1/\lambda}[\varphi(\sigma(t))]^{(\lambda+1)/\lambda}}. $

(3.10)

由于当$t\in[t_{1}, +\infty)_{{\Bbb T}}$时$y(t)>0, y^{\Delta}(t)>0$, 所以存在常数$k>0$, 使得$y(\tau(\sigma(t)))\geqslant k$ $(t\in[t_{1}, +\infty)_{{\Bbb T}})$.于是上式可化简为

$ \begin{equation} W^{\Delta}(t)\leqslant\frac{\varphi(t)}{(y(\tau(t)))^{\beta}}[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta} +\frac{\varphi^{\Delta}(t)W(\sigma(t))}{\varphi(\sigma(t))} -\frac{\alpha\varphi(t)(W(\sigma(t)))^{(\lambda+1)/\lambda}} {(A(\tau(t)))^{1/\lambda}[\varphi(\sigma(t))]^{(\lambda+1)/\lambda}}, \end{equation} $

(3.11)

其中$\alpha=k^{(\beta-\lambda)/\lambda}\beta\tau_{_{0}}$为常数.

再定义广义的Riccati变换

$ \begin{eqnarray} V(t)=\varphi(t)\frac{A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))}{\phi_{2}(y(\tau(t)))} =\varphi(t)\frac{A(\tau(t))(y^{\Delta}(\tau(t)))^{\lambda}}{(y(\tau(t)))^{\beta}}, t\in[t_{1}, +\infty)_{{\Bbb T}}, \end{eqnarray} $

(3.12)

则$V(t)>0, t\in[t_{1}, +\infty)_{{\Bbb T}}$.类似地, 同样可得

$ \begin{eqnarray} V^{\Delta}(t)&=&\varphi(t)\frac{[A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta}}{\phi_{2}(y(\tau(t)))} \\ &&+A(\tau(\sigma(t)))\phi_{1}(y^{\Delta}(\tau(\sigma(t)))) \frac{\varphi^{\Delta}(t)(y(\tau(t)))^{\beta}-\varphi(t)[(y(\tau(t)))^{\beta}]^{\Delta}}{(y(\tau(t)))^{\beta}(y(\tau(\sigma(t))))^{\beta}} \\ &\leqslant&\varphi(t)\frac{[A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta}}{(y(\tau(t)))^{\beta}} +\frac{\varphi^{\Delta}(t)V(\sigma(t))}{\varphi(\sigma(t))} \\ &&-A(\tau(\sigma(t)))\phi_{1}(y^{\Delta}(\tau(\sigma(t))))\frac{\varphi(t)\beta(y(\tau(\sigma(t))))^{\beta-1}y^{\Delta}(\tau(t))\tau^{\Delta}(t)} {(y(\tau(t)))^{\beta}(y(\tau(\sigma(t))))^{\beta}} \\ &\leqslant&\varphi(t)\frac{[A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta}}{(y(\tau(t)))^{\beta}} +\frac{\varphi^{\Delta}(t)V(\sigma(t))}{\varphi(\sigma(t))} \\ &&-\frac{\beta\tau_{0}\varphi(t)A(\tau(\sigma(t)))(y^{\Delta}(\tau(\sigma(t))))^{\lambda}} {(y(\tau(\sigma(t))))^{\beta+1}}\left(\frac{A(\tau(\sigma(t)))}{A(\tau(t))}\right)^{1/\lambda}y^{\Delta}(\tau(\sigma(t))) \\ &=&\varphi(t)\frac{[A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta}}{(y(\tau(t)))^{\beta}} +\frac{\varphi^{\Delta}(t)V(\sigma(t))}{\varphi(\sigma(t))} \\ &&-\frac{\beta\tau_{0}\varphi(t)[A(\tau(\sigma(t)))]^{(\lambda+1)/\lambda}(y^{\Delta}(\tau(\sigma(t))))^{\lambda+1}} {A^{1/\lambda}(\tau(t))(y(\tau(\sigma(t))))^{\beta+1}}\\ %\eqno(3.13) &=&\varphi(t)\frac{[A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta}}{(y(\tau(t)))^{\beta}} +\frac{\varphi^{\Delta}(t)V(\sigma(t))}{\varphi(\sigma(t))} \\ &&-(y(\tau(\sigma(t))))^{\frac{\beta-\lambda}{\lambda}}\frac{\beta\tau_{0}\varphi(t)[V(\sigma(t))^{(\lambda+1)/\lambda}} {A^{1/\lambda}(\tau(t))[\varphi(\sigma(t))]^{(\lambda+1)/\lambda}} \end{eqnarray} $

(3.13)

$ \leqslant \frac{\varphi(t)}{(y(\tau(t)))^{\beta}}[A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta} +\frac{\varphi^{\Delta}(t)V(\sigma(t))}{\varphi(\sigma(t))} \\ \ \ \ -\frac{\alpha\varphi(t)[V(\sigma(t))^{(\lambda+1)/\lambda}} {A^{1/\lambda}(\tau(t))[\varphi(\sigma(t))]^{(\lambda+1)/\lambda}}. $

(3.14)

另一方面, 若$\beta>1$.由(3.3)式, 得$[(y(\tau(t)))^{\beta}]^{\Delta}\geqslant\beta(y(\tau(t)))^{\beta-1}y^{\Delta}(\tau(t))\tau^{\Delta}(t)$.用与$0<\beta\leqslant1$时同样的方法可推得, 此时(3.11)式及(3.14)式仍然成立.

于是, 综合(3.11)式和(3.14)式, 并分别注意到(3.7)式及$y(\tau(t))\leqslant y(\tau(\sigma(t)))$, 可得

$ \ \ \ \ \ \ W^{\Delta}(t)+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}V^{\Delta}(t) \\ \leqslant \frac{\varphi(t)}{(y(\tau(t)))^{\beta}}\left\{[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta} +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}[A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta}\right\} \\ \ \ \ \ \ \ +\frac{\varphi^{\Delta}(t)W(\sigma(t))}{\varphi(\sigma(t))} -\frac{\alpha\varphi(t)(W(\sigma(t)))^{(\lambda+1)/\lambda}}{A^{1/\lambda}(\tau(t))\varphi^{(\lambda+1)/\lambda}(\sigma(t))} \\ \ \ \ \ \ \ +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left[\frac{\varphi^{\Delta}(t)V(\sigma(t))}{\varphi(\sigma(t))} -\frac{\alpha\varphi(t)[V(\sigma(t))^{(\lambda+1)/\lambda}} {A^{1/\lambda}(\tau(t))\varphi^{(\lambda+1)/\lambda}(\sigma(t))}\right] \\ \leqslant -\varsigma\xi(t)\varphi(t)-\frac{b(t)\varphi(t)}{A(t)}\frac{A(\sigma(t))\phi_{1}(y^{\Delta}(\sigma(t)))}{(y(\tau(t)))^{\beta}} \\ \ \ \ \ \ \ -\frac{(b_{0}\eta)^{\beta}\varphi(t)b(\tau(t))}{A(\tau(t))}\frac{A(\tau(\sigma(t)))\phi_{1}(y^{\Delta}(\tau(\sigma(t))))}{(y(\tau(t)))^{\beta}} \\ \ \ \ \ \ \ +\frac{\varphi^{\Delta}(t)W(\sigma(t))}{\varphi(\sigma(t))} -\frac{\alpha\varphi(t)(W(\sigma(t)))^{(\lambda +1)/\lambda}}{A^{1/\lambda}(\tau(t))\varphi^{(\lambda+1)/\lambda}(\sigma(t))} \\ \ \ \ \ \ \ +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left[\frac{\varphi^{\Delta}(t)V(\sigma(t))}{\varphi(\sigma(t))} -\frac{\alpha\varphi(t)[V(\sigma(t))^{(\lambda+1)/\lambda}} {A^{1/\lambda}(\tau(t))\varphi^{(\lambda+1)/\lambda}(\sigma(t))}\right] \\ \leqslant -\varsigma\xi(t)\varphi(t)-\frac{b(t)\varphi(t)W(\sigma(t))}{A(t)\varphi(\sigma(t))} -\frac{(b_{0}\eta)^{\beta}b(\tau(t))\varphi(t)V(\sigma(t))}{A(\tau(t))\varphi(\sigma(t))} +\frac{\varphi^{\Delta}(t)W(\sigma(t))}{\varphi(\sigma(t))} \\ \ \ \ \ \ \ -\frac{\alpha\varphi(t)(W(\sigma(t)))^{(\lambda+1)/\lambda}}{A^{1/\lambda}(\tau(t))\varphi^{(\lambda+1)/\lambda}(\sigma(t))} +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left[\frac{\varphi^{\Delta}(t)V(\sigma(t))}{\varphi(\sigma(t))} -\frac{\alpha\varphi(t)[V(\sigma(t))^{(\lambda+1)/\lambda}} {A^{1/\lambda}(\tau(t))\varphi^{(\lambda+1)/\lambda}(\sigma(t))}\right] \\ = -\varsigma\xi(t)\varphi(t)+\left(\varphi^{\Delta}(t)-\frac{b(t)\varphi(t)}{A(t)}\right)\frac{W(\sigma(t))}{\varphi(\sigma(t))} -\frac{\alpha\varphi(t)(W(\sigma(t)))^{(\lambda+1)/\lambda}}{A^{1/\lambda}(\tau(t))\varphi^{(\lambda+1)/\lambda}(\sigma(t))} \\ \ \ \ \ \ +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left[\left(\varphi^{\Delta}(t)-\frac{\tau_{0}b(\tau(t))\varphi(t)}{A(\tau(t))}\right)\frac{V(\sigma(t))}{\varphi(\sigma(t))} -\frac{\alpha\varphi(t)(V(\sigma(t)))^{(\lambda+1)/\lambda}}{A^{1/\lambda}(\tau(t))\varphi^{(\lambda+1)/\lambda}(\sigma(t))}\right]. $

(3.15)

应用引理2.3, 于是由(3.15)式, 我们可得

$ \ \ \ \ \ W^{\Delta}(t)+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}V^{\Delta}(t)\\ \leqslant -\varsigma\xi(t)\varphi(t) +\frac{\lambda^{\lambda}A(\tau(t))}{(\lambda+1)^{\lambda+1}[\alpha\varphi(t)]^{\lambda}}\left|\varphi^{\Delta}(t)- \frac{b(t)\varphi(t)}{A(t)}\right|^{\lambda+1} \\ \ \ \ \ \ +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\frac{\lambda^{\lambda}A(\tau(t))}{(\lambda+1)^{\lambda+1}[\alpha\varphi(t)]^{\lambda}}\left|\varphi^{\Delta}(t)- \frac{\tau_{0}b(\tau(t))\varphi(t)}{A(\tau(t))}\right|^{\lambda+1} \\ = \varphi(t)\left\{-\varsigma\xi(t)+\frac{\lambda^{\lambda}A(\tau(t))}{(\lambda+1)^{\lambda+1}\alpha^{\lambda}} \left[\left|\frac{\varphi^{\Delta}(t)}{\varphi(t)}-\frac{b(t)}{A(t)}\right|^{\lambda+1}+ \frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left|\frac{\varphi^{\Delta}(t)}{\varphi(t)}-\frac{\tau_{0}b(\tau(t))}{A(\tau(t))}\right|^{\lambda+1}\right]\right\}. $

两边积分, 得

$ \begin{eqnarray} &&\int^{t}_{t_{1}}\varphi(s)\left\{\varsigma\xi(s)-\frac{\lambda^{\lambda}A(\tau(s))}{(\lambda+1)^{\lambda+1}\alpha^{\lambda}} \left[\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{b(s)}{A(s)} \right|^{\lambda+1}\right.\right. \\ &&\left.\left.+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}} \left|\frac{\varphi^{\Delta}(s)}{\varphi(s)} -\frac{\tau_{0}b(\tau(s))}{A(\tau(s))}\right|^{\lambda+1}\right]\right\}\Delta s \\ & \leqslant&-\int^{t}_{t_{1}}W^{\Delta}(s)\Delta s-\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\int^{t}_{t_{1}}V^{\Delta}(s)\Delta s \\ &=&-W(t)+W(t_{1})-\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}[V(t)-V(t_{1})] \leqslant W(t_{1})+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}V(t_{1}), \end{eqnarray} $

(3.16)

上式取上极限, 得与(3.1)式矛盾.

情形(b)  $\lambda>\beta$.

定义函数$W(t)$如(3.8)式, 则无论$\beta>1$还是$\beta\leqslant1$, (3.9)式都是成立的, 于是由(3.9)式, 得

$ \begin{eqnarray*} W^{\Delta}(t)&\leqslant& \varphi(t)\frac{[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}}{\phi_{2}(y(\tau(t)))}+\frac{\varphi^{\Delta}(t)W(\sigma(t))}{\varphi(\sigma(t))} \\ && -\frac{\beta\tau_{0}\varphi(t)(A(\sigma(t)))^{(\frac{1}{\lambda}-\frac{1}{\beta})}[W(\sigma(t))]^{(\beta+1)/\beta}} {(y^{\Delta}(\sigma(t)))^{(\lambda-\beta)/\beta}\varphi^{(\beta+1)/\beta}(\sigma(t))A^{1/\lambda}(\tau(t))}. \end{eqnarray*} $

利用$A(t)(y^{\Delta}(t))^{\lambda}$的单调减少性, 对$t\in[t_{1}, +\infty)_{{\Bbb T}}$, 有

$ A(\sigma(t))(y^{\Delta}(\sigma(t)))^{\lambda}\leqslant A(t_{1})(y^{\Delta}(t_{1}))^{\lambda}=k, $

即

$ (y^{\Delta}(\sigma(t)))^{(\lambda-\beta)/\beta}\leqslant\frac{k^{(\lambda-\beta)/\beta\lambda}}{A^{(\lambda-\beta)/\beta\lambda}(\sigma(t))}. $

于是

$ \begin{equation} W^{\Delta}(t)\leqslant\varphi(t)\frac{[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}}{\phi_{2}(y(\tau(t)))}+\frac{\varphi^{\Delta}(t)W(\sigma(t))}{\varphi(\sigma(t))} -\frac{\alpha\varphi(t)[W(\sigma(t))]^{(\beta+1)/\beta}}{\varphi^{(\beta+1)/\beta}(\sigma(t))A^{1/\lambda}(\tau(t))}, \end{equation} $

(3.17)

其中$\alpha=\frac{\beta\tau_{0}}{k^{(\lambda-\beta)/\beta\lambda}}$为常数.

我们再定义函数$V(t)$如(3.12)式, 则无论$\beta>1$还是$0<\beta\leqslant1$, (3.13)式都是成立的, 于是由(3.13)式, 按同样类似的方法, 得

$ \begin{equation} V^{\Delta}(t)\leqslant\varphi(t)\frac{[A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta}}{\phi_{2}(y(\tau(t)))}+\frac{\varphi^{\Delta}(t)V(\sigma(t))}{\varphi(\sigma(t))} -\frac{\alpha\varphi(t)[V(\sigma(t))]^{(\beta+1)/\beta}}{\varphi^{(\beta+1)/\beta}(\sigma(t))A^{1/\lambda}(\tau(t))}. \end{equation} $

(3.18)

综合(3.17)和(3.18)两式, 并分别注意到(3.7)式及$y(\tau(t))\leqslant y(\tau(\sigma(t)))$, 利用与情形(a)同样类似的方法, 得

$ \ \ \ \ \ W^{\Delta}(t)+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}V^{\Delta}(t)\\ \leqslant \frac{\varphi(t)}{(y(\tau(t)))^{\beta}}\left\{[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta} +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}[A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta}\right\} +\frac{\varphi^{\Delta}(t)W(\sigma(t))}{\varphi(\sigma(t))} \\ \ \ \ \ \ -\frac{\alpha\varphi(t)[W(\sigma(t))]^{(\beta+1)/\beta}}{\varphi^{(\beta+1)/\beta}(\sigma(t))A^{1/\lambda}(\tau(t))} +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left\{\frac{\varphi^{\Delta}(t)V(\sigma(t))}{\varphi(\sigma(t))} -\frac{\alpha\varphi(t)[V(\sigma(t))]^{(\beta+1)/\beta}}{\varphi^{(\beta+1)/\beta}(\sigma(t))A^{1/\lambda}(\tau(t))}\right\} \\ \leqslant -\varsigma\xi(t)\varphi(t)-\frac{b(t)\varphi(t)}{A(t)}\frac{A(\sigma(t))\phi_{1}(y^{\Delta}(\sigma(t)))}{(y(\tau(t)))^{\beta}} \\ \ \ \ \ \ -\frac{(b_{0}\eta)^{\beta}\varphi(t)b(\tau(t))}{A(\tau(t))}\frac{A(\tau(\sigma(t)))\phi_{1}(y^{\Delta}(\tau(\sigma(t))))}{(y(\tau(t)))^{\beta}} +\frac{\varphi^{\Delta}(t)W(\sigma(t))}{\varphi(\sigma(t))} \\ \ \ \ \ \ - \frac{\alpha\varphi(t)[W(\sigma(t))]^{(\beta+1)/\beta}}{\varphi^{(\beta+1)/\beta}(\sigma(t))A^{1/\lambda}(\tau(t))} +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left\{\frac{\varphi^{\Delta}(t)V(\sigma(t))}{\varphi(\sigma(t))} -\frac{\alpha\varphi(t)[V(\sigma(t))]^{(\beta+1)/\beta}}{\varphi^{(\beta+1)/\beta}(\sigma(t))A^{1/\lambda}(\tau(t))}\right\} \\ \leqslant -\varsigma\xi(t)\varphi(t)-\frac{b(t)\varphi(t)W(\sigma(t))}{A(t)\varphi(\sigma(t))} -\frac{(b_{0}\eta)^{\beta}b(\tau(t))\varphi(t)V(\sigma(t))}{A(\tau(t))\varphi(\sigma(t))} +\frac{\varphi^{\Delta}(t)W(\sigma(t))}{\varphi(\sigma(t))} \\ \ \ \ \ \ -\frac{\alpha\varphi(t)[W(\sigma(t))]^{(\beta+1)/\beta}}{\varphi^{(\beta+1)/\beta}(\sigma(t))A^{1/\lambda}(\tau(t))} +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left\{\frac{\varphi^{\Delta}(t)V(\sigma(t))}{\varphi(\sigma(t))} -\frac{\alpha\varphi(t)[V(\sigma(t))]^{(\beta+1)/\beta}}{\varphi^{(\beta+1)/\beta}(\sigma(t))A^{1/\lambda}(\tau(t))}\right\} \\ = -\varsigma\xi(t)\varphi(t)+\left(\varphi^{\Delta}(t)-\frac{b(t)\varphi(t)}{A(t)}\right)\frac{W(\sigma(t))}{\varphi(\sigma(t))} -\frac{\alpha\varphi(t)[W(\sigma(t))]^{(\beta+1)/\beta}}{\varphi^{(\beta+1)/\beta}(\sigma(t))A^{1/\lambda}(\tau(t))} \\ \ \ \ \ \ +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left[\left(\varphi^{\Delta}(t)-\frac{\tau_{0}b(\tau(t))\varphi(t)}{A(\tau(t))}\right)\frac{V(\sigma(t))}{\varphi(\sigma(t))} -\frac{\alpha\varphi(t)[V(\sigma(t))]^{(\beta+1)/\beta}}{\varphi^{(\beta+1)/\beta}(\sigma(t))A^{1/\lambda}(\tau(t))}\right] $

(3.19)

$ \leqslant -\varsigma\xi(t)\varphi(t)+\frac{\beta^{\beta}\varphi(t)A^{\beta/\lambda}(\tau(t))}{(\beta+1)^{\beta+1}\alpha^{\beta}} \left|\frac{\varphi^{\Delta}(t)}{\varphi(t)}-\frac{b(t)}{A(t)}\right|^{\beta+1} \\ \ \ \ \ \ +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}} \frac{\beta^{\beta}\varphi(t)A^{\beta/\lambda}(\tau(t))}{(\beta+1)^{\beta+1}\alpha^{\beta}} \left|\frac{\varphi^{\Delta}(t)}{\varphi(t)}-\frac{\tau_{0}b(\tau(t))}{A(\tau(t))}\right|^{\beta+1} \\ = -\varsigma\xi(t)\varphi(t) \\ \ \ \ \ \ +\frac{\beta^{\beta}\varphi(t)A^{\beta/\lambda}(\tau(t))}{(\beta+1)^{\beta+1}\alpha^{\beta}} \left[\left|\frac{\varphi^{\Delta}(t)}{\varphi(t)}-\frac{b(t)}{A(t)}\right|^{\beta+1}+ \frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left|\frac{\varphi^{\Delta}(t)}{\varphi(t)}-\frac{\tau_{0}b(\tau(t))}{A(\tau(t))}\right|^{\beta+1}\right]. $

(3.20)

两边积分, 得

$ \begin{eqnarray*} &&\int^{t}_{t_{1}}\varphi(s)\left\{\varsigma\xi(s)-\frac{\beta^{\beta}A^{\beta/\lambda}(\tau(s))}{(\beta+1)^{\beta+1}\alpha^{\beta}} \left[\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)} -\frac{b(s)}{A(s)}\right|^{\beta+1} \right.\right.\\ &&\left.\left. +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}} \left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{\tau_{0}b(\tau(s))}{A(\tau(s))}\right|^{\beta+1}\right]\right\}\Delta s \\ &\leqslant&-\int^{t}_{t_{1}}W^{\Delta}(s)\Delta s-\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\int^{t}_{t_{1}}V^{\Delta}(s)\Delta s \\ &=&-W(t)+W(t_{1})-\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}[V(t)-V(t_{1})] \leqslant W(t_{1})+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}V(t_{1}), \end{eqnarray*} $

上式取上极限, 得与(3.2)式矛盾.

注3.1  选取不同的函数$\varphi$ (如$\varphi(t)\equiv1, \varphi(t)=t$等), 就能得到关于方程(1.1)的不同的具体振荡准则.

注3.2  由定理3.1的条件(3.1)和(3.2)式可以看出, 当$\lambda>\beta$时和$\lambda<\beta$时方程(1.1)的振荡准则是不一样的.当$\lambda=\beta$时, 若取$g(u)=u, F(u)=u$, 则可得方程(1.3)的振荡准则, 但本文没有条件"$\tau=\delta$及$0\leqslant r(t)<1$"的限制; 当$\lambda=\beta$时, 若取$B(t)\equiv0$ (即$b_{0}=0, F(u)=u$), 则可得方程(1.2)的振荡准则, 这就是文献[5, 8]中的定理4.1, 但本文没有条件"$\delta({\Bbb T})={\Bbb T}$"的限制.其它相关结果可参看文献[2-4, 10-14, 18-19, 21, 23, 26]及其参考文献.

定理3.2  设(H$_{1}$)-(H$_{6}$)和(C$_{1}$)成立, 如果存在函数$\varphi\in C^{1}_{rd}([t_{0}, \infty)_{{\Bbb T}}, (0, \infty))$及常数$\omega\geqslant1$, 使得当$\lambda\leqslant\beta$时

$ \begin{eqnarray} &&\limsup\limits_{t\rightarrow\infty}\frac{1}{t^{\omega}}\int^{t}_{t_{1}}(t-s)^{\omega}\varphi(s)\left[\varsigma\xi(s)- \frac{\lambda^{\lambda}A(\tau(s))}{(\lambda+1)^{\lambda+1}\alpha^{\lambda}_{1}} \left(\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{b(s)}{A(s)}\right|^{\lambda+1}\right.\right. \\ &&\left.\left.+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{\tau_{0}b(\tau(s))}{A(\tau(s))}\right|^{\lambda+1} \right)\right]\Delta s=\infty, \end{eqnarray} $

(3.21)

当$\lambda>\beta$时

$ \begin{eqnarray} &&\limsup\limits_{t\rightarrow\infty}\frac{1}{t^{\omega}}\int^{t}_{t_{1}}(t-s)^{\omega}\varphi(s)\left[\varsigma\xi(s)- \frac{\beta^{\beta}A^{\beta/\lambda}(\tau(s))}{(\beta+1)^{\beta+1}\alpha^{\beta}_{2}} \left(\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{b(s)}{A(s)}\right|^{\beta+1}\right.\right. \\ && \left.\left.+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{\tau_{0}b(\tau(s))}{A(\tau(s))}\right|^{\beta+1} \right)\right]\Delta s=\infty, \end{eqnarray} $

(3.22)

其中常数$t_{1}\geqslant t_{0}$, 函数$\xi(t)$及常数$\varsigma, \alpha_{1}, \alpha_{2}$的定义如定理3.1, 则方程(1.1)在$[t_{0}, +\infty)_{{\Bbb T}}$上是振荡的.

证  设方程(1.1)在$[t_{0}, +\infty)_{{\Bbb T}}$上有一个非振荡解$x(t)$, 不失一般性, 不妨设$x(t)>0, $ $x(\tau(t))>0, x(\delta(t))>0, t\in[t_{1}, +\infty)_{{\Bbb T}}, t_{1}\in[t_{0}, +\infty)_{{\Bbb T}}$ (当$x(t)$是最终负解时类似地可以证明), 从而$y(t)>0$.

情形(a)  $\lambda\leqslant\beta$.同定理3.1的证明, 可得(3.15)式.将(3.15)式中的$t$改成$s$, 两边再同乘以$(t-s)^{\omega}$并积分, 由时间模上的分部积分公式, 并注意到$[(t-s)^{\omega}]^{\Delta_{s}}\leqslant0$, 得

$ \int^{t}_{t_{1}}(t-s)^{\omega}\varsigma\xi(s)\varphi(s)\Delta s \leqslant -\int^{t}_{t_{1}}(t-s)^{\omega}W^{\Delta}(s)\Delta s -\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\int^{t}_{t_{1}}(t-s)^{\omega}V^{\Delta}(s)\Delta s \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\int^{t}_{t_{1}}(t-s)^{\omega}\left(\varphi^{\Delta}(s)-\frac{b(s)\varphi(s)}{A(s)}\right)\frac{W(\sigma(s))}{\varphi(\sigma(s))}\Delta s \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\int^{t}_{t_{1}}(t-s)^{\omega}\frac{\alpha\varphi(s)(W(\sigma(s)))^{(\lambda+1)/\lambda}} {A^{1/\lambda}(\tau(s))\varphi^{(\lambda+1)/\lambda}(\sigma(s))}\Delta s \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left[\int^{t}_{t_{1}}(t-s)^{\omega}\left(\varphi^{\Delta}(s)-\frac{\tau_{0}b(\tau(s))\varphi(s)}{A(\tau(s))}\right) \frac{V(\sigma(s))}{\varphi(\sigma(s))}\Delta s\right. \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.-\int^{t}_{t_{1}}\frac{\alpha(t-s)^{\omega}\varphi(s)(V(\sigma(s)))^{(\lambda+1)/\lambda}} {A^{1/\lambda}(\tau(s))\varphi^{(\lambda+1)/\lambda}(\sigma(s))}\Delta s\right] \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (t-t_{1})^{\omega}W(t_{1})+\int^{t}_{t_{1}}[(t-s)^{\omega}]^{\Delta_{s}}W(\sigma(s))\Delta s \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left[(t-t_{1})^{\omega}V(t_{1})+\int^{t}_{t_{1}}[(t-s)^{\omega}]^{\Delta_{s}}V(\sigma(s))\Delta s\right] \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\int^{t}_{t_{1}}(t-s)^{\omega}\left(\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{b(s)}{A(s)}\right)\frac{\varphi(s)W(\sigma(s))} {\varphi(\sigma(s))}\Delta s\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\int^{t}_{t_{1}}\frac{\alpha(t-s)^{\omega}\varphi(s)(W(\sigma(s)))^{(\lambda+1)/\lambda}} {A^{1/\lambda}(\tau(s))\varphi^{(\lambda+1)/\lambda}(\sigma(s))}\Delta s \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left[\int^{t}_{t_{1}}(t-s)^{\omega}\left(\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{\tau_{0}b(\tau(s))}{A(\tau(s))}\right) \frac{\varphi(s)V(\sigma(s))}{\varphi(\sigma(s))}\Delta s\right. \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.-\int^{t}_{t_{1}}\frac{\alpha(t-s)^{\omega}\varphi(s)(V(\sigma(s)))^{(\lambda+1)/\lambda}} {A^{1/\lambda}(\tau(s))\varphi^{(\lambda+1)/\lambda}(\sigma(s))}\Delta s\right] \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \leqslant (t-t_{1})^{\omega}W(t_{1})+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}(t-t_{1})^{\omega}V(t_{1}) \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\int^{t}_{t_{1}}\left\{(t-s)^{\omega}\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{b(s)}{A(s)}\right|\frac{\varphi(s)W(\sigma(s))} {\varphi(\sigma(s))} \right.\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.-\frac{\alpha(t-s)^{\omega}\varphi(s)(W(\sigma(s)))^{(\lambda+1)/\lambda}} {A^{1/\lambda}(\tau(s))\varphi^{(\lambda+1)/\lambda}(\sigma(s))}\right\}\Delta s \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\int^{t}_{t_{1}}\left\{(t-s)^{\omega}\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{\tau_{0}b(\tau(s))}{A(\tau(s))}\right| \frac{\varphi(s)V(\sigma(s))}{\varphi(\sigma(s))} \right.\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.-\frac{\alpha(t-s)^{\omega}\varphi(s)(V(\sigma(s)))^{(\lambda+1)/\lambda}} {A^{1/\lambda}(\tau(s))\varphi^{(\lambda+1)/\lambda}(\sigma(s))}\right\}\Delta s. $

利用引理2.3, 于是由上式可得

$ \begin{eqnarray*} &&\int^{t}_{t_{1}}\varsigma(t-s)^{\omega}\xi(s)\varphi(s)\Delta s \leqslant (t-t_{0})^{\omega}W(t_{1}) +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}(t-t_{0})^{\omega}V(t_{1}) \\ &&+\int^{t}_{t_{1}}\frac{\lambda^{\lambda}(t-s)^{\omega}A(\tau(s))\varphi(s)}{(\lambda+1)^{\lambda+1}\alpha^{\lambda}} \left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{b(s)}{A(s)}\right|^{\lambda+1}\Delta s \\ &&+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\int^{t}_{t_{1}}\frac{\lambda^{\lambda}(t-s)^{\omega}A(\tau(s))\varphi(s)}{(\lambda+1)^{\lambda+1}\alpha^{\lambda}} \left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{\tau_{0}b(\tau(s))}{A(\tau(s))}\right|^{\lambda+1}\Delta s \\ &=&(t-t_{0})^{\omega}W(t_{1})+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}(t-t_{0})^{\omega}V(t_{1}) +\int^{t}_{t_{1}}\frac{\lambda^{\lambda}(t-s)^{\omega}\varphi(s)A(\tau(s))}{(\lambda+1)^{\lambda+1}\alpha^{\lambda}} \\ &&\times \left[\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{b(s)}{A(s)}\right|^{\lambda+1} +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)} -\frac{\tau_{0}b(\tau(s))}{A(\tau(s))}\right|^{\lambda+1}\right]\Delta s, \end{eqnarray*} $

由上式进一步可得

$ \begin{eqnarray*} &&\frac{1}{t^{\omega}}\int^{t}_{t_{1}}(t-s)^{\omega}\varphi(s)\left[\varsigma\xi(s)- \frac{\lambda^{\lambda}A(\tau(s))}{(\lambda+1)^{\lambda+1}\alpha^{\lambda}} \left(\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{b(s)}{A(s)}\right|^{\lambda+1}\right.\right. \\ && \left.\left.+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}} \left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{\tau_{0}b(\tau(s))}{A(\tau(s))}\right|^{\lambda+1}\right)\right]\Delta s \\ &\leqslant&\left(1-\frac{t_{0}}{t}\right)^{\omega}W(t_{1})+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left(1-\frac{t_{0}}{t}\right)^{\omega}V(t_{1}) \leqslant W(t_{1})+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}V(t_{1}), \end{eqnarray*} $

上式取上极限, 即得与(3.21)式矛盾.

情形(b)  $\lambda>\beta$.同定理3.1的证明, 可得(3.19)式, 按照与情形(a)同样类似的方法, 可得到一个与(3.22)矛盾的结果.

注3.3  定理3.2将二阶微分方程的Kamenev型振荡准则推广到一类非常广泛的二阶Emden-Fowler型阻尼动态方程(1.1)上.当$\lambda=\beta$时, 若取$g(u)=u, F(u)=u$, 则可得方程(1.3)的振荡准则, 但本文没有条件"$\tau=\delta$且$0\leqslant r(t)<1$"的限制; 当$\lambda=\beta$时, 若取$B(t)\equiv0$(此时$b_{0}=0, F(u)=u$), 则可得方程(1.2)的振荡准则, 这就是文献[5]中的定理4.2, 但本文没有条件"$\delta({\Bbb T})={\Bbb T}$"的限制.

定理3.3  设(H$_{1}$)-(H$_{6}$)和(C$_{1}$)成立, 如果存在函数$\varphi\in C^{1}_{rd}([t_{0}, \infty)_{{\Bbb T}}, (0, \infty))$及常数$\omega\geqslant1$, 使得当$\lambda\leqslant\beta$时

$ \begin{eqnarray} &&\limsup\limits_{t\rightarrow\infty}\frac{1}{t^{\omega}}\int^{t}_{t_{1}}\left\{\varsigma(t-s)^{\omega}\xi(s)\varphi(s)- \frac{\lambda^{\lambda}A(\tau(s))\varphi^{\lambda+1}(\sigma(s))}{(\lambda+1)^{\lambda+1}[\alpha_{1}(t-s)^{\omega}\varphi(s)]^{\lambda}}\right.\\ &&\times \left.\left[|h_{1}(t, s)|^{\lambda+1}+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}|h_{2}(t, s)|^{\lambda+1}\right]\right\}\Delta s=\infty, \end{eqnarray} $

(3.23)

当$\lambda>\beta$时

$ \begin{eqnarray} &&\limsup\limits_{t\rightarrow\infty}\frac{1}{t^{\omega}}\int^{t}_{t_{1}}\left\{\varsigma(t-s)^{\omega}\xi(s)\varphi(s)- \frac{\beta^{\beta}A^{\beta/\lambda}(\tau(s))\varphi^{\beta+1}(\sigma(s))}{(\beta+1)^{\beta+1}[\alpha_{2}(t-s)^{\omega}\varphi(s)]^{\beta}}\right.\\ &&\times \left.\left[|h_{1}(t, s)|^{\beta+1}+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}|h_{2}(t, s)|^{\beta+1}\right]\right\}\Delta s=\infty, \end{eqnarray} $

(3.24)

其中常数$t_{1}\geqslant t_{0}$, 函数$\xi(t)$及常数$\varsigma, \alpha_{1}, \alpha_{2}$的定义如定理3.1,

$ h_{1}(t, s)=(t-s)^{\omega}\left(\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{b(s)}{A(s)}\right)\frac{\varphi(s)}{\varphi(\sigma(s))} -\omega(t-\sigma(s))^{\omega-1}, \\ h_{2}(t, s)=(t-s)^{\omega}\left(\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{\tau_{0}b(\tau(s))}{A(\tau(s))}\right) \frac{\varphi(s)}{\varphi(\sigma(s))}-\omega(t-\sigma(s))^{\omega-1}, $

则方程(1.1)在$[t_{0}, +\infty)_{{\Bbb T}}$上是振荡的.

证  设方程(1.1)在$[t_{0}, +\infty)_{{\Bbb T}}$上有一个非振荡解$x(t)$, 不失一般性, 不妨设$x(t)>0, x(\tau(t))>0, x(\delta(t))>0, t\in[t_{1}, +\infty)_{{\Bbb T}}, t_{1}\in[t_{0}, +\infty)_{{\Bbb T}}$ (当$x(t)$是最终负解时类似地可以证明), 从而$y(t)>0$.

由文献[22]中定理2.9的证明, 有

$ \begin{eqnarray} [(t-s)^{\omega}]^{\Delta_{s}}\leqslant-\omega(t-\sigma(s))^{\omega-1}, \end{eqnarray} $

(3.25)

这里$t\geqslant\sigma(s), \omega\geqslant1$.

情形(a)  当$\lambda\leqslant\beta$时.同定理3.1的证明, 可得(3.15)式.将(3.15)式中的$t$改成$s$, 两边再同乘以$(t-s)^{\omega}$并积分, 由时间模上的分部积分公式, 并注意到(3.25)式, 可得

$ \begin{eqnarray*} &&\int^{t}_{t_{1}}(t-s)^{\omega}\varsigma\xi(s)\varphi(s)\Delta s\\ &\leqslant& -\int^{t}_{t_{1}}(t-s)^{\omega}W^{\Delta}(s)\Delta s -\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\int^{t}_{t_{1}}(t-s)^{\omega}V^{\Delta}(s)\Delta s \\ &&+\int^{t}_{t_{1}}(t-s)^{\omega}\left(\varphi^{\Delta}(s)-\frac{b(s)\varphi(s)}{A(s)}\right)\frac{W(\sigma(s))}{\varphi(\sigma(s))}\Delta s \\ &&-\int^{t}_{t_{1}}(t-s)^{\omega}\frac{\alpha\varphi(s)(W(\sigma(s)))^{(\lambda+1)/\lambda}} {A^{1/\lambda}(\tau(s))\varphi^{(\lambda+1)/\lambda}(\sigma(s))}\Delta s \\ &&+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left[\int^{t}_{t_{1}}(t-s)^{\omega}\left(\varphi^{\Delta}(s)-\frac{\tau_{0}b(\tau(s))\varphi(s)}{A(\tau(s))}\right) \frac{V(\sigma(s))}{\varphi(\sigma(s))}\Delta s\right. \\ &&\left.-\int^{t}_{t_{1}}\frac{\alpha(t-s)^{\omega}\varphi(s)(V(\sigma(s)))^{(\lambda+1)/\lambda}} {A^{1/\lambda}(\tau(s))\varphi^{(\lambda+1)/\lambda}(\sigma(s))}\Delta s\right] \\ &\leqslant& (t-t_{1})^{\omega}W(t_{1})-\int^{t}_{t_{1}}\omega(t-\sigma(s))^{\omega-1}W(\sigma(s))\Delta s \\ &&+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left[(t-t_{1})^{\omega}V(t_{1})-\int^{t}_{t_{1}}\omega(t-\sigma(s))^{\omega-1}V(\sigma(s))\Delta s\right] \\ &&+\int^{t}_{t_{1}}(t-s)^{\omega}\left(\varphi^{\Delta}(s)-\frac{b(s)\varphi(s)}{A(s)}\right)\frac{W(\sigma(s))}{\varphi(\sigma(s))}\Delta s \\ && -\int^{t}_{t_{1}}(t-s)^{\omega}\frac{\alpha\varphi(s)(W(\sigma(s)))^{(\lambda+1)/\lambda}} {A^{1/\lambda}(\tau(s))\varphi^{(\lambda+1)/\lambda}(\sigma(s))}\Delta s \\ &&+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left[\int^{t}_{t_{1}}(t-s)^{\omega}\left(\varphi^{\Delta}(s)-\frac{\tau_{0}b(\tau(s))\varphi(s)}{A(\tau(s))}\right) \frac{V(\sigma(s))}{\varphi(\sigma(s))}\Delta s\right. \\ &&\left.-\int^{t}_{t_{1}}\frac{\alpha(t-s)^{\omega}\varphi(s)(V(\sigma(s)))^{(\lambda+1)/\lambda}} {A^{1/\lambda}(\tau(s))\varphi^{(\lambda+1)/\lambda}(\sigma(s))}\Delta s\right] \\ &=&(t-t_{1})^{\omega}W(t_{1})+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}(t-t_{1})^{\omega}V(t_{1}) \\ &&+\int^{t}_{t_{1}}\left\{h_{1}(t, s)W(\sigma(s))-\frac{\alpha(t-s)^{\omega}\varphi(s)(W(\sigma(s)))^{(\lambda+1)/\lambda}} {A^{1/\lambda}(\tau(s))\varphi^{(\lambda+1)/\lambda}(\sigma(s))}\right\}\Delta s \\ &&+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\int^{t}_{t_{1}}\left\{h_{2}(t, s)V(\sigma(s))-\frac{\alpha(t-s)^{\omega}\varphi(s)(V(\sigma(s)))^{(\lambda+1)/\lambda}} {A^{1/\lambda}(\tau(s))\varphi^{(\lambda+1)/\lambda}(\sigma(s))}\right\}\Delta s. \end{eqnarray*} $

将引理2.3应用于上式, 可得

$ \begin{eqnarray*} \int^{t}_{t_{1}}(t-s)^{\omega}\varsigma\xi(s)\varphi(s)\Delta s &\leqslant& (t-t_{1})^{\omega}W(t_{1}) +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}(t-t_{1})^{\omega}V(t_{1}) \\ &&+\int^{t}_{t_{1}}\frac{\lambda^{\lambda}A(\tau(s))[\varphi(\sigma(s))|h_{1}(t, s)|]^{\lambda+1}} {(\lambda+1)^{\lambda+1}[\alpha(t-s)^{\omega}\varphi(s)]^{\lambda}}\Delta s \\ && +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\int^{t}_{t_{1}}\frac{\lambda^{\lambda}A(\tau(s))[\varphi(\sigma(s))|h_{2}(t, s)|]^{\lambda+1}} {(\lambda+1)^{\lambda+1}[\alpha(t-s)^{\omega}\varphi(s)]^{\lambda}}\Delta s, \end{eqnarray*} $

所以

$ \begin{eqnarray*} &&\frac{1}{t^{\omega}}\int^{t}_{t_{1}}\left\{\varsigma(t-s)^{\omega}\xi(s)\varphi(s)- \frac{\lambda^{\lambda}A(\tau(s))\varphi^{\lambda+1}(\sigma(s))}{(\lambda+1)^{\lambda+1}[\alpha(t-s)^{\omega}\varphi(s)]^{\lambda}}\right. \\ &&\times\left.\left[|h_{1}(t, s)|^{\lambda+1}+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}|h_{2}(t, s)|^{\lambda+1}\right]\right\}\Delta s \\ &\leqslant& \left(1-\frac{t_{1}}{t}\right)^{\omega}W(t_{1})+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left(1-\frac{t_{1}}{t}\right)^{\omega}V(t_{1}) \leqslant W(t_{1})+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}V(t_{1}), \end{eqnarray*} $

上式取$t\rightarrow\infty$的上极限, 就得到与(3.23)式矛盾.

情形(b)  $\lambda>\beta$.同定理3.1的证明, 可得(3.19)式, 按照与情形(a)同样类似的方法, 可得到一个与(3.24)式矛盾的结果.定理证毕.

定理3.4  设条件(H$_{1}$)-(H$_{6}$)和(C$_{2}$)成立, 并且存在函数$\varphi\in C^{1}_{rd}([t_{0}, \infty)_{{\Bbb T}}, (0, \infty))$使得当$\lambda\leqslant\beta$时(3.1)式成立, 当$\lambda>\beta$时(3.2)式成立.如果

$ \begin{eqnarray} && \limsup\limits_{t\rightarrow\infty}\int^{t}_{t_{1}}\theta^{\lambda}(\sigma(s))\left\{\varsigma\xi(s)\pi(s)- \frac{A(s)}{(\lambda+1)^{\lambda+1}}\left[\left(\frac{b(s)}{A(s)}+\frac{\overline{\theta}(s)}{\theta^{\lambda}(\sigma(s))}\right)^{\lambda+1}\right.\right. \\ && \left.\left.+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left(\frac{\tau_{0}b(\tau(s))}{A(\tau(s))} +\frac{\overline{\theta}(s)}{\theta^{\lambda}(\sigma(s))}\right)^{\lambda+1}\right]\right\}\Delta s=\infty, \end{eqnarray} $

(3.26)

其中常数$t_{1}\geqslant t_{0}$, 函数$\theta(t)=\int^{\infty}_{t}A^{-1/\lambda}(s)\Delta s$,

$ \overline{\theta}(t)=\left\{\begin{array}{ll} \lambda\theta^{\lambda-1}(t)A^{-1/\lambda}(t), &\lambda>1, \\ \lambda\theta^{\lambda-1}(\sigma(t))A^{-1/\lambda}(t), ~&0<\lambda\leqslant1, \end{array} \right.~~ \pi(t)=\left\{\begin{array}{ll} k, &\lambda>\beta, \\ 1, &\lambda=\beta, \\ k\theta^{\beta-\lambda}(t), ~&\lambda<\beta \end{array} \right. $

($k>0$为某常数), 则方程(1.1)在$[t_{0}, +\infty)_{{\Bbb T}}$上是振荡的.

证  设方程(1.1)在$[t_{0}, +\infty)_{{\Bbb T}}$上有一个非振荡解$x(t)$, 不失一般性, 不妨设$x(t)>0, $ $x(\tau(t))>0, x(\delta(t))>0, t\in[t_{1}, +\infty)_{{\Bbb T}}, t_{1}\in[t_{0}, +\infty)_{{\Bbb T}}$ (当$x(t)$是最终负解时类似地可以证明), 于是, 由引理2.6的证明过程(并注意到(2.3)式)知, 当$t\in[t_{1}, +\infty)_{{\Bbb T}}$时$y(t)>0, $ $[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}<0$且$y^{\Delta}(t)$或者最终为正或者最终为负.因此我们只需要考虑下列两种情形:

(ⅰ) $y^{\Delta}(t)>0, t\in[t_{1}, +\infty)_{{\Bbb T}}$;

(ⅱ) $y^{\Delta}(t)<0, t\in[t_{1}, +\infty)_{{\Bbb T}}$.

情形(ⅰ)  $y^{\Delta}(t)>0, t\in[t_{1}, +\infty)_{{\Bbb T}}$.此时引理2.6成立, 于是同定理3.1的证明, 知方程(1.1)在$[t_{0}, +\infty)_{{\Bbb T}}$上是振荡的.

情形(ⅱ)  $y^{\Delta}(t)<0(t\in[t_{1}, +\infty)_{{\Bbb T}})$.令

$ \begin{eqnarray} v(t)=\frac{A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))}{\phi_{1}(y(t))}=\frac{A(\tau(t))(-y^{\Delta}(\tau(t)))^{\lambda-1}y^{\Delta}(\tau(t))} {(y(t))^{\lambda}}, ~~t\in[t_{1}, +\infty)_{{\Bbb T}}, \end{eqnarray} $

(3.27)

则$v(t)<0, t\in[t_{1}, +\infty)_{{\Bbb T}}$.由于$A(t)(-y^{\Delta}(t))^{\lambda-1}y^{\Delta}(t)$是单调减少的, 所以当$s\in[t, +\infty)_{{\Bbb T}}$时, 有

$ A(s)(-y^{\Delta}(s))^{\lambda-1}y^{\Delta}(s)\leqslant A(t)(-y^{\Delta}(t))^{\lambda-1}y^{\Delta}(t)\leqslant A(\tau(t))(-y^{\Delta}(\tau(t)))^{\lambda-1}y^{\Delta}(\tau(t)), $

即$y^{\Delta}(s)\leqslant\frac{A^{1/\lambda}(\tau(t))y^{\Delta}(\tau(t))}{A^{1/\lambda}(s)}$, 进一步可得$y(u)\leqslant y(t)+A^{1/\lambda}(\tau(t))y^{\Delta}(\tau(t))\int^{u}_{t}A^{-1/\lambda}(s)\Delta s$, 令$u\rightarrow\infty$, 则有$y(t)+A^{1/\lambda}(\tau(t))y^{\Delta}(\tau(t))\theta(t)\geqslant0$, 从而$-1\leqslant\frac{A^{1/\lambda}(\tau(t))y^{\Delta}(\tau(t))}{y(t)}\theta(t)\leqslant0, $于是, 利用(3.27)式, 就有

$ \begin{eqnarray} -1\leqslant v(t)\theta^{\lambda}(t)\leqslant0. \end{eqnarray} $

(3.28)

另一方面, 由于$y^{\Delta}(t)<0$, 于是由(2.1)式, 有

$ \left\{\begin{array}{ll} (y^{\lambda}(t))^{\Delta}\leqslant\lambda y^{\lambda-1}(\sigma(t))y^{\Delta}(t), ~&\lambda>1, \\ (y^{\lambda}(t))^{\Delta}\leqslant\lambda y^{\lambda-1}(t)y^{\Delta}(t), ~&0<\lambda\leqslant1. \end{array} \right. $

现对(3.27)式求$\Delta$-导数, 注意到上式及$y^{\Delta}(t)<0$, 当$0<\lambda\leqslant1$时, 得

$ v^{\Delta}(t)=\frac{[A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta}}{y^{\lambda}(\sigma(t))} -\frac{A(\tau(t))(-y^{\Delta}(\tau(t)))^{\lambda-1}y^{\Delta}(\tau(t))(y^{\lambda}(t))^{\Delta}}{y^{\lambda}(t)y^{\lambda}(\sigma(t))} \\ \ \ \ \ \ \ \ \ \ \leqslant\frac{[A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta}}{y^{\lambda}(\sigma(t))}- \frac{A(\tau(t))(-y^{\Delta}(\tau(t)))^{\lambda-1}y^{\Delta}(\tau(t))\lambda y^{\lambda-1}(t)y^{\Delta}(t)}{y^{\lambda}(t)y^{\lambda}(\sigma(t))} \\ \ \ \ \ \ \ \ \ \ \leqslant \frac{[A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta}}{y^{\lambda}(t)}-\frac{\lambda A(\tau(t)) (-y^{\Delta}(\tau(t)))^{\lambda-1}y^{\Delta}(\tau(t))}{y^{\lambda+1}(t)}\frac{A^{1/\lambda}(\tau(t))}{A^{1/\lambda}(t)}y^{\Delta}(\tau(t)) \\ \ \ \ \ \ \ \ \ \ = \frac{[A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta}}{y^{\lambda}(t)}- \frac{\lambda(-v(t))^{(\lambda+1)/\lambda}}{A^{1/\lambda}(t)}. $

(3.29)

与前面一样, 当$\lambda>1$时(3.29)式也是成立的.再令

$ \begin{equation} w(t)=\frac{A(t)\phi_{1}(y^{\Delta}(t))}{\phi_{1}(y(t))}=\frac{A(t)(-y^{\Delta}(t))^{\lambda-1}y^{\Delta}(t)} {(y(t))^{\lambda}}, ~~t\in[t_{1}, +\infty)_{{\Bbb T}}, \end{equation} $

(3.30)

类似地, 同样有$w(t)\leqslant0, t\in[t_{1}, +\infty)_{{\Bbb T}}$, 并且

$ \begin{equation} -1\leqslant w(t)\theta^{\lambda}(t)\leqslant0. \end{equation} $

(3.31)

$ \begin{equation} w^{\Delta}(t)\leqslant\frac{[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}}{y^{\lambda}(t)}- \frac{\lambda(-w(t))^{(\lambda+1)/\lambda}}{A^{1/\lambda}(t)}. \end{equation} $

(3.32)

由于(3.4)式仍然成立, 类似于(3.7)式, 并注意到$y^{\Delta}(t)<0$, 可得

$ \begin{eqnarray} &&[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}+\frac{(b_{0}\eta)^{\beta}}{\tau_{_{0}}}[A(\tau(t)) \phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta} \\ & \leqslant&-\varsigma\xi(t)(y(t))^{\beta}-b(t)\phi_{1}(y^{\Delta}(t))-(b_{0}\eta)^{\beta}b(\tau(t))\phi_{1}(y^{\Delta}(\tau(t))) \\ &\leqslant&-\varsigma\xi(t)(y(t))^{\beta}-\frac{b(t)\phi_{1}(y(t))}{A(t)}w(t)- \frac{(b_{0}\eta)^{\beta}b(\tau(t))\phi_{1}(y(t))}{A(\tau(t))}v(t). \end{eqnarray} $

(3.33)

于是, 综合(3.32)和(3.29)式, 并注意到(3.33)式, 可得

$ w^{\Delta}(t)+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}v^{\Delta}(t) \leqslant \frac{[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}}{y^{\lambda}(t)}- \frac{\lambda(-w(t))^{(\lambda+1)/\lambda}}{A^{1/\lambda}(t)} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left\{\frac{[A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta}}{y^{\lambda}(t)}- \frac{\lambda(-v(t))^{(\lambda+1)/\lambda}}{A^{1/\lambda}(t)}\right\} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{y^{\lambda}(t)}\left\{[A(t)\phi_{1}(y^{\Delta}(t))]^{\Delta}+ \frac{(b_{0}\eta)^{\beta}}{\tau_{0}}[A(\tau(t))\phi_{1}(y^{\Delta}(\tau(t)))]^{\Delta}\right\} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\frac{\lambda}{A^{1/\lambda}(t)}\left\{(-w(t))^{\frac{\lambda+1}{\lambda}}+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}(-v(t))^{\frac{\lambda+1}{\lambda}}\right\} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \leqslant -\varsigma\xi(t)(y(t))^{\beta-\lambda}-\frac{b(t)}{A(t)}w(t)-\frac{(b_{0}\eta)^{\beta}b(\tau(t))}{A(\tau(t))}v(t) \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\frac{\lambda}{A^{1/\lambda}(t)}\left\{(-w(t))^{\frac{\lambda+1}{\lambda}}+ \frac{(b_{0}\eta)^{\beta}}{\tau_{0}}(-v(t))^{\frac{\lambda+1}{\lambda}}\right\}. $

(3.34)

当$\lambda>\beta$时, 由$y(t)>0, y^{\Delta}(t)<0(t\in[t_{1}, +\infty)_{{\Bbb T}})$知, $y(t)\leqslant y(t_{1})$, 即$y^{\beta-\lambda}(t)\geqslant y^{\beta-\lambda}(t_{1})=k$.

当$\lambda=\beta$时, $y^{\beta-\lambda}(t)=1$.

当$\lambda<\beta$时, 再次利用$A(t)(-y^{\Delta}(t))^{\lambda-1}y^{\Delta}(t)$的单调减少性, 得当$s\in[t_{1}, +\infty)_{{\Bbb T}}$时, 有

$ A(s)(-y^{\Delta}(s))^{\lambda-1}y^{\Delta}(s)\leqslant A(t_{1})(-y^{\Delta}(t_{1}))^{\lambda-1}y^{\Delta}(t_{1})=-M, $

其中$M=-A(t_{1})(-y^{\Delta}(t_{1}))^{\lambda-1}y^{\Delta}(t_{1})>0$为常数, 于是$A(s)(-y^{\Delta}(s))^{\lambda}\geqslant M$即$y^{\Delta}(s)\leqslant-M^{1/\lambda}A^{-1/\lambda}$, 进一步就有$y(u)\leqslant y(t)-M^{1/\lambda}\int^{u}_{t}A^{-1/\lambda}(s)\Delta s$, 即

$ y(t)\geqslant y(u)+M^{1/\lambda}\int^{u}_{t}A^{-1/\lambda}(s)\Delta s\geqslant M^{1/\lambda}\int^{u}_{t}A^{-1/\lambda}(s)\Delta s, $

在上式中令$u\rightarrow\infty$, 得$y(t)\geqslant M^{1/\lambda}\int^{\infty}_{t}A^{-1/\lambda}(s)\Delta s=M^{1/\lambda}\theta(t)$, 即$y^{\beta-\lambda}(t)\geqslant k\theta^{\beta-\lambda}(t)$, 这里$k=M^{(\beta-\lambda)/\lambda}>0$是常数.

综合上述3种情形及函数$\pi(t)$的定义, 根据(3.34)式, 有

$ \begin{eqnarray*} w^{\Delta}(t)+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}v^{\Delta}(t) &\leqslant&-\varsigma\xi(t)\pi(t)-\frac{b(t)}{A(t)}w(t) -\frac{(b_{0}\eta)^{\beta}b(\tau(t))}{A(\tau(t))}v(t) \\ &&- \frac{\lambda}{A^{1/\lambda}(t)}\left\{(-w(t))^{\frac{\lambda+1}{\lambda}}+ \frac{(b_{0}\eta)^{\beta}}{\tau_{0}}(-v(t))^{\frac{\lambda+1}{\lambda}}\right\}, \end{eqnarray*} $

上式两边同乘$\theta^{\lambda}(\sigma(t))$后再积分, 由分部积分公式, 并注意到

$ [\theta^{\lambda}(t)]^{\Delta}\geqslant\left\{\begin{array}{ll} -\lambda\theta^{\lambda-1}(t)A^{-1/\lambda}, &\lambda>1, \\ -\lambda\theta^{\lambda-1}(\sigma(t))A^{-1/\lambda}, ~&0<\lambda\leqslant1. \end{array} \right. $

即$[\theta^{\lambda}(t)]^{\Delta}\geqslant-\overline{\theta}(t)$以及引理2.3, 得

$ \ \ \ \ \int^{t}_{t_{1}}\varsigma\xi(s)\pi(s)\theta^{\lambda}(\sigma(s))\Delta s\\ \leqslant -\int^{t}_{t_{1}}\theta^{\lambda}(\sigma(s))w^{\Delta}(s)\Delta s -\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\int^{t}_{t_{1}}\theta^{\lambda}(\sigma(s))v^{\Delta}(s)\Delta s \\ \ \ \ \ -\int^{t}_{t_{1}}\frac{\theta^{\lambda}(\sigma(s))b(s)}{A(s)}w(s)\Delta s -(b_{0}\eta)^{\beta}\int^{t}_{t_{1}}\frac{\theta^{\lambda}(\sigma(s))b(\tau(s))}{A(\tau(s))}v(s)\Delta s \\ \ \ \ \ -\int^{t}_{t_{1}}\frac{\lambda\theta^{\lambda}(\sigma(s))}{A^{1/\lambda}(s)}\left\{(-w(s))^{\frac{\lambda+1}{\lambda}}+ \frac{(b_{0}\eta)^{\beta}}{\tau_{0}}(-v(s))^{\frac{\lambda+1}{\lambda}}\right\}\Delta s \\ = \theta^{\lambda}(t_{1})w(t_{1})-\theta^{\lambda}(t)w(t)+\int^{t}_{t_{1}}[\theta^{\lambda}(t)]^{\Delta}w(s)\Delta s \\ \ \ \ \ +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left\{\theta^{\lambda}(t_{1})v(t_{1})-\theta^{\lambda}(t)v(t) +\int^{t}_{t_{1}}[\theta^{\lambda}(t)]^{\Delta}v(s)\Delta s\right\} \\ \ \ \ \ -\int^{t}_{t_{1}}\frac{\theta^{\lambda}(\sigma(s))b(s)}{A(s)}w(s)\Delta s-(b_{0}\eta)^{\beta}\int^{t}_{t_{1}}\frac{\theta^{\lambda}(\sigma(s))b(\tau(s))}{A(\tau(s))}v(s)\Delta s \\ \ \ \ \ -\int^{t}_{t_{1}}\frac{\lambda\theta^{\lambda}(\sigma(s))}{A^{1/\lambda}(s)}\left\{(-w(s))^{\frac{\lambda+1}{\lambda}}+ \frac{(b_{0}\eta)^{\beta}}{\tau_{0}}(-v(s))^{\frac{\lambda+1}{\lambda}}\right\}\Delta s \\ = \theta^{\lambda}(t_{1})w(t_{1})-\theta^{\lambda}(t)w(t)+ \frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left[\theta^{\lambda}(t_{1})v(t_{1})-\theta^{\lambda}(t)v(t)\right] \\ \ \ \ \ +\int^{t}_{t_{1}}\left[\left(\frac{b(s)}{A(s)}\theta^{\lambda}(\sigma(s))+\overline{\theta}(s)\right)(-w(s)) -\frac{\lambda\theta^{\lambda}(\sigma(s))}{A^{1/\lambda}(s)}(-w(s))^{(\lambda+1)/\lambda}\right]\Delta s \\ \ \ \ \ +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\int^{t}_{t_{1}}\left[\left(\frac{\tau_{0}b(\tau(s))}{A(\tau(s))}\theta^{\lambda}(\sigma(s)) +\overline{\theta}(s)\right)(-v(s))-\frac{\lambda\theta^{\lambda}(\sigma(s))}{A^{1/\lambda}(s)}(-v(s))^{(\lambda+1)/\lambda}\right]\Delta s \\ \leqslant \theta^{\lambda}(t_{1})w(t_{1})-\theta^{\lambda}(t)w(t)+ \frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left[\theta^{\lambda}(t_{1})v(t_{1})-\theta^{\lambda}(t)v(t)\right] \\ \ \ \ \ +\int^{t}_{t_{1}}\frac{A(s)\left(\frac{b(s)\theta^{\lambda}(\sigma(s))}{A(s)} +\overline{\theta}(s)\right)^{\lambda+1}}{(\lambda+1)^{\lambda+1}\theta^{\lambda^{2}}(\sigma(s))}\Delta s+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\int^{t}_{t_{1}}\frac{A(s)\left(\frac{\tau_{0}b(\tau(s))\theta^{\lambda}(\sigma(s))}{A(\tau(s))} +\overline{\theta}(s)\right)^{\lambda+1}}{(\lambda+1)^{\lambda+1}\theta^{\lambda^{2}}(\sigma(s))}\Delta s \\ = \theta^{\lambda}(t_{1})w(t_{1})-\theta^{\lambda}(t)w(t)+ \frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left[\theta^{\lambda}(t_{1})v(t_{1})-\theta^{\lambda}(t)v(t)\right] \\ \ \ \ \ +\int^{t}_{t_{1}}\frac{A(s)\theta^{\lambda}(\sigma(s))}{(\lambda+1)^{\lambda+1}}\left[\left(\frac{b(s)}{A(s)} +\frac{\overline{\theta}(s)}{\theta^{\lambda}(\sigma(s))}\right)^{\lambda+1}+ \frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left(\frac{\tau_{0}b(\tau(s))}{A(\tau(s))}+ \frac{\overline{\theta}(s)}{\theta^{\lambda}(\sigma(s))}\right)^{\lambda+1}\right]\Delta s, $

将(3.28)和(3.31)式用于上式, 得

$ \begin{eqnarray*} &&\int^{t}_{t_{1}}\theta^{\lambda}(\sigma(s))\left\{\varsigma\xi(s)\pi(s)- \frac{A(s)}{(\lambda+1)^{\lambda+1}}\left[\left(\frac{b(s)}{A(s)}+\frac{\overline{\theta}(s)}{\theta^{\lambda}(\sigma(s))}\right)^{\lambda+1} \right.\right. \\ &&\left.\left. +\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left(\frac{\tau_{0}b(\tau(s))}{A(\tau(s))} +\frac{\overline{\theta}(s)}{\theta^{\lambda}(\sigma(s))}\right)^{\lambda+1}\right]\right\}\Delta s \\ & \leqslant& \theta^{\lambda}(t_{1})w(t_{1})+1+\frac{(b_{0}\eta)^{\beta}}{\tau_{0}}\left[\theta^{\lambda}(t_{1})v(t_{1})+1\right], \end{eqnarray*} $

这与(3.26)式矛盾.

注3.4  从条件(3.26)的结构来看, 本文定理3.4的条件(3.26)比文献[5]中定理4.3的条件(4.14)式及文献[8]中定理3.5的条件(3.25)更为合理, 而且也更容易验证.

运用与定理3.4相同的证明思路和方法, 并结合定理3.2和定理3.3, 容易得到下面结果.

定理3.5  设条件(H$_{1}$)-(H$_{6}$), (C$_{2}$)及(3.26)式成立, 若存在函数$\varphi\in C^{1}_{rd}([t_{0}, \infty)_{{\Bbb T}}, (0, \infty))$及常数$\omega\geqslant1$, 使得当$\lambda\leqslant\beta$时(3.21)式成立, 当$\lambda>\beta$时(3.22)式成立, 则方程(1.1)在$[t_{0}, +\infty)_{{\Bbb T}}$上是振荡的.

定理3.6  设条件(H$_{1}$)-(H$_{6}$), (C$_{2}$)及(3.26)式成立, 若存在函数$\varphi\in C^{1}_{rd}([t_{0}, \infty)_{{\Bbb T}}, (0, \infty))$及常数$\omega\geqslant1$, 使得当$\lambda\leqslant\beta$时(3.23)式成立, 当$\lambda>\beta$时(3.24)式成立, 则方程(1.1)在$[t_{0}, +\infty)_{{\Bbb T}}$上是振荡的.

注3.5  本文结果给出了时间模上一类非常广泛的具有阻尼项的二阶非线性中立型变时滞Emden-Fowler型泛函动态方程(1.1)振荡的几个充分条件, 推广并改进了现有文献中的结果, 并使得现有文献中的一些结果成为我们结果的特例.这些结果即使当${\Bbb T}={\Bbb R}$和${\Bbb T}={\Bbb N}$时也是新的, 并且在时间模上统一了相应的时滞微分方程和时滞差分方程振荡的有关结论.

例1  考虑二阶微分方程

$ (t^{2}x'(t))'+q_{_{0}}x(t)=0, t\geqslant1, $

($E_1$)

其中常数$q_{0}>0$.显然, 这是著名的二阶Euler微分方程.令$A(t)=t^{2}, b(t)=0, B(t)=0, $ $P(t)=q_{0}, \tau(t)=\delta(t)=t, F(u)=u, \lambda=\beta=1$, 则$\varsigma=1, \tau_{0}=1, b_{0}=0$.显然满足条件(H$_{1}$)-(H$_{6}$)及(C$_{2}$).由于${\Bbb T}={\Bbb R}$, 因此

$ \xi(t)=\min\{P(t), P(\tau(t))\}=q_{0}, ~~ \theta(t)=\int^{\infty}_{t}A^{-1/\lambda}(s)\Delta s=\int^{\infty}_{t}s^{-2}{\rm d}s=\frac{1}{t}, \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \overline{\theta}(t)=\lambda\theta^{\lambda-1}(\sigma(t))A^{-1/\lambda}(t)=t^{-2}, ~~ \pi(t)=1. $

取$\varphi(t)=t$, 并记(3.1)式左边为$L_{1}$, 记(3.26)式左边为$L_{2}$, 则

$ L_{1}=\limsup\limits_{t\rightarrow\infty}\int^{t}_{1}s\left\{q_{0}-\frac{s^{2}}{2^{2}}\left[\left|\frac{1}{s}-0\right|^{2}+0\right]\right\}{\rm d}s =\left(q_{0}-\frac{1}{4}\right)\limsup\limits_{t\rightarrow\infty}\int^{t}_{1}s{\rm d}s, $

并且

$ L_{2}=\limsup\limits_{t\rightarrow\infty}\int^{t}_{1}\frac{1}{s}\left\{q_{0}- \frac{s^{2}}{2^{2}}\left[\left(0+\frac{s^{-2}}{1/s}\right)^{2}+0\right]\right\}{\rm d}s =\left(q_{0}-\frac{1}{4}\right)\limsup\limits_{t\rightarrow\infty}\int^{t}_{1}\frac{1}{s}{\rm d}s, $

所以当$q_{0}>\frac{1}{4}$时, 条件(3.1)和(3.26)都满足, 即定理3.4的条件全部满足, 因此由定理3.4知, 当$q_{0}>\frac{1}{4}$时方程$(E_{1})$是振荡的.显然, 这与众所周知的结果是一致的.

注4.1  我们注意到, 若将文献[5]或[7]的定理4.3以及文献[9]的定理3.5用于方程$(E_{1})$, 只能得到"方程$(E_{1})$的每一个解或者振荡或者收敛于零", 不能判定方程的振荡性.另外, 由于$\int^{\infty}_{1}\frac{\ln t}{t^{2}}{\rm d}t<\infty$, 文献[3, 6, 8]中有关定理的条件不满足, 因而不能用于方程$(E_{1})$.其它文献如[2, 4, 10-14, 18-24, 26]中的定理也不能用于方程$(E_{1})$.

例2  考虑时间模${\Bbb T}$上具阻尼项的二阶变时滞泛函动态方程

$ \left\{t^{\frac{8}{9}}|y^{\Delta}(t)|^{\frac{1}{3}}y^{\Delta}(t)\right\}^{\Delta}+\frac{1}{t^{2}}|y^{\Delta}(t)|^{\frac{1}{3}}y^{\Delta}(t) +\frac{1}{t}f(|x(\frac{t}{3})|^{\frac{3}{2}}x(\frac{t}{3}))=0, ~~t\in{\Bbb T}, t\geqslant3, $

(4.1)

这里$y(t)=x(t)+(9-\cos t)g(x(\frac{t}{3}))$.这是方程(1.1)中$\lambda=\frac{4}{3}, \beta=\frac{5}{2}, A(t)=t^{\frac{8}{9}}, $ $B(t)=9-\cos t, b(t)=\frac{1}{t^{2}}, $$P(t)=\frac{1}{t}, \tau(t)=\delta(t)=\frac{t}{3}$的情形.若取$g(u)=\frac{u}{\sqrt{1+\sin^{4}(u+5)}}, $ $f(u)=u[3^{\frac{5}{2}}+\ln^{\frac{5}{2}}(1+u^{2})]$, 并取时间模${\Bbb T}=3^{{\Bbb Z}}$, 则此时动态方程(4.1)即为二阶时滞3 -差分方程.由于当$u\neq0$时, 有

$ \frac{g(u)}{u}=\frac{1}{\sqrt{1+\sin^{4}(u+5)}}\leqslant1=\eta, ~~ \frac{f(u)}{u}=3^{\frac{5}{2}}+ \ln^{\frac{5}{2}}(1+u^{2})\geqslant3^{\frac{5}{2}}=L, \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0<B(t)=9-\cos t\leqslant10=b_{0}. $

当$t\geqslant3$时, $1-\mu(t)\frac{b(t)}{A(t)}=1-t\cdot\frac{t^{-2}}{t^{8/9}}=1-\frac{1}{t^{17/9}}>0$, 即$-b/A\in\Re^{+}$.进一步, 由文献[25]中的引理2得, 当$t\geqslant3$时, 有

$ \begin{eqnarray*} e_{_{-b/A}}(t, t_{0})&\geqslant& 1-\int^{t}_{3}\frac{b(s)}{A(s)}\Delta s=1-\int^{t}_{3}s^{-26/9}\Delta s>1-\int^{t}_{3}s^{-8/3}\Delta s \\ &=&1-\frac{t^{-5/3}-3^{-5/3}}{3^{-5/3}-1} =\frac{t^{-5/3}+1-2\cdot3^{-5/3}}{1-3^{-5/3}}\\ &\geqslant& t^{-5/3}+1-2\cdot3^{-5/3}>1-2\cdot3^{-5/3}>\frac{1}{2}, \end{eqnarray*} $

从而

$ \begin{eqnarray*} \int^{t}_{t_{0}}\left[\frac{e_{_{-b/A}}(s, t_{0})}{A(s)}\right]^{1/\lambda}\Delta s &=& \int^{t}_{3}s^{-\frac{2}{3}}[e_{_{-b/A}}(t, t_{0})]^{\frac{3}{4}}\Delta s \geqslant 2^{-\frac{3}{4}}\int^{t}_{3}s^{-\frac{2}{3}}\Delta s \\ &=&2^{-\frac{3}{4}}\frac{t^{1/3}-3^{1/3}}{3^{1/3}-1}\rightarrow\infty~~(t\rightarrow\infty), \end{eqnarray*} $

所以条件(H$_{1}$)-(H$_{6}$)及(C$_{1}$)均满足.为了简单并考虑到$\lambda<\beta$, 现在定理3.1中取$\varphi(t)=1$, 则

$ \xi(t)=\min\{P(t), P(\tau(t))\}=\frac{1}{t}, $

同样, 记(3.1)式左边为$L_{1}$, 则

$ L_{1}=\limsup\limits_{t\rightarrow\infty}\int^{t}_{3}\left\{\frac{3^{5/2}}{2^{3/2}}\frac{1}{s}-\frac{4^{4/3}3^{1/9}[1+10^{5/2}3^{26/9}]}{7^{7/3}\alpha^{4/3}} \frac{1}{s^{158/9}}\right\}\Delta s=\infty, $

所以(3.1)式成立, 因此定理3.1的条件均满足, 于是由定理3.1知, 此时方程(4.1)是振荡的.

注4.2  由于方程(4.1)中的$\lambda\neq\beta$且中立项的系数函数$B(t)>1$, 因此最近文献(如[2-14, 18-24, 26-27]等)中的结果都不能判定方程(4.1)的振荡性