微分方程的振动理论是微分方程定性理论的一个重要分支, 它在物理学、力学、机械振动、自动控制工程、神经网络及生物制药等领域都具有广泛的应用背景.由于科学技术的迅猛发展, 进入二十世纪九十年代, 微分方程的一个新分支---时间轴上动态方程诞生了.它一出现就引起了国内外广大学者们的高度关注, 研究成果非常丰富, 并发表了大量有关时间轴上动态方程理论的研究论文和专著^[1-20].基于这些实际背景, 本文研究时间轴$ {\bf T} $上一类非线性非自治变延迟的二阶阻尼Emden-Fowler型动力系统

(1.1) $ \begin{equation} \left[A(t) \phi_{1}\left(y^{\Delta}(t)\right)\right]^{\Delta}+b(t) \phi_{1}\left(y^{\Delta}(t)\right)+P(t) F\left(\phi_{2}(x(\delta(t)))\right)=0, t \in {\bf T}, t \geq t_{0} \end{equation} $

的振动性, 其中$ 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 $, 并假设下列条件$ \left(\mathrm{H}_{1}\right) $–$ \left(\mathrm{H}_{5}\right) $成立:

$ \left(\mathrm{H}_{1}\right) $$ {\bf T} $是任意时间轴, 且$ \sup {\bf T}=+\infty $, $ {{[{{t}_{0}}, +\infty )}_{{\bf T}}}=[{{t}_{0}}, +\infty )\bigcap {\bf T} $ ($ {{t}_{0}}\in {\bf T} $且$ {{t}_{0}}>0 $).

$ \left(\mathrm{H}_{2}\right) $ 函数$ A(t), B(t), b(t), P(t)\in {{C}_{rd}}({\bf T}, {\bf R}) $, 且$ 0\le B(t)<1 $, $ b(t)\ge 0 $, $ P(t)>0 $.

$ \left(\mathrm{H}_{3}\right) $$ \tau (t), \ \delta (t) $均为$ {\bf T} $到$ {\bf T} $上的延迟函数, $ \tau (t)\le t $, 且$ \lim\limits_{t\to +\infty}\tau (t)=+\infty $; $ \delta (t)\le t $, 且$ \lim\limits_{t\to +\infty}\delta (t)=+\infty $.

$ \left(\mathrm{H}_{4}\right) $ 函数$ g(u), F(u)\in C({\bf R}, {\bf R}) $且$ ug(u)>0 $ ($ u\ne 0 $), $ uF(u)>0 $ ($ u\ne 0 $), 并且存在常数$ 0<\eta \le 1 $和$ L>0 $, 使得当$ u\ne 0 $时$ g(u)/u\le \eta $, $ F(u)/u\ge L $.

$ \left(\mathrm{H}_{5}\right) $$ A(t)>0 $, 且$ -b/A\in {{\Re }^{+}} $.

关于系统(1.1)的解及其振动性定义, 英文文献可参见[1-3, 6, 7, 9, 13, 16], 中文文献可参见[4, 5, 8, 10-12, 14, 15].笔者仅关注系统(1.1)的不最终恒为零的解.系统(1.1)包含了多种类型的方程, 如

(1.2) $ \begin{equation} \left[r(t)\left(x^{\Delta}(t)\right)^{\gamma}\right]^{\Delta}+p(t) x^{\gamma}(t)=0, \end{equation} $

(1.3) $ \begin{equation} \left(a(t) x^{\Delta}(t)\right)^{\Delta}+b(t) x^{\Delta}(t)+p(t) f(x(\delta(t)))=0, \end{equation} $

(1.4) $ \begin{equation} \left(a(t)\left|x^{\Delta}(t)\right|^{\lambda-1} x^{\Delta}(t)\right)^{\Delta}+b(t)\left|x^{\Delta}(t)\right|^{\lambda-1} x^{\Delta}(t)+p(t)|x(\delta(t))|^{\lambda-1} x(\delta(t))=0, \end{equation} $

等.对正则系统即系统满足条件

(1.5) $ \begin{equation} \int_{t_{0}}^{+\infty}\left[A^{-1}(s) e_{-b / A}\left(s, t_{0}\right)\right]^{1 / \lambda} \Delta s=+\infty \end{equation} $

的情形, 其振动结果较多.如对正则系统(1.1), 文献[4]得到了其振动的若干准则, 其中一个基本定理如下:

定理A^[4]  设条件$ \left(\mathrm{H}_{1}\right) $–$ \left(\mathrm{H}_{5}\right) $及(1.5)成立, 若存在函数$ \varphi \in C_{{}}^{1}({\bf T}, (0, +\infty )) $, 使得当$ \lambda \leqslant \beta $时

(1.6) $ \begin{equation} \limsup\limits_{t \rightarrow +\infty} \int_{t_{0}}^{t} \varphi(s)\left\{k^{\beta} \Psi(s)-\frac{[A(\sigma(s))]^{\lambda+1}}{M A^{\lambda}(s)}\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{b(s)}{A(\sigma(s))}\right|^{\lambda+1}\right\} \Delta s=+\infty, \end{equation} $

当$ \lambda>\beta $时

(1.7) $ \begin{equation} \limsup\limits_{t \rightarrow +\infty} \int_{t_{0}}^{t} \varphi(s)\left\{k^{\beta} \Psi(s)-\frac{[A(\sigma(s))]^{\beta+1}}{M A^{\beta}(s)}\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{b(s)}{A(\sigma(s))}\right|^{\beta+1}\right\} \Delta s=+\infty, \end{equation} $

这里常数$ k\in (0, 1) $,

$ M=\left\{ \begin{array}{ll} {{(\lambda +1)}^{\lambda +1}}, &\lambda =\beta, \\ \mbox{正常数}, &\lambda \ne \beta, \\ \end{array} \right. $

而函数$ \Psi (t)=LP(t)\ {{[1-\eta B(\delta (t))]}^{\beta }}{{\left[ \frac{\delta (t)}{\sigma (t)} \right]}^{\beta }} $.则系统(1.1)在$ {{[{{t}_{0}}, +\infty )}_{{\bf T}}} $上是振动的.

紧接着, 文献[5]改进了定理A中的条件(1.6)和(1.7), 得到了方程(1.1)的一个较为简洁的振动结果:

定理B^[5]  如果存在函数$ \varphi \in C_{{}}^{1}({\bf T}, (0, +\infty )) $, 使得

(1.8) $ \begin{equation} \limsup\limits_{t \rightarrow +\infty} \int_{t_{0}}^{t} \varphi(s)\left\{\Psi(s)-\frac{A^{\omega_{1}}(s)}{k}\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)} -\frac{b(s)}{A(s)}\right|^{\omega_{2}+1}\right\} \Delta s=+\infty, \end{equation} $

其中常数$ k=\left\{\begin{array}{ll}(\lambda+1)^{\lambda+1}, & \mbox{ 当 } \lambda=\beta \mbox{ 时, } \\ \mbox{ 正常数, } & \mbox{ 当 } \lambda \neq \beta \mbox{ 时 }, \end{array} \quad \omega_{1}=\min \left\{1, \frac{\beta}{\lambda}\right\}, \omega_{2}=\min \{\lambda, \beta\}\right. $, 而函数

(1.9) $ \begin{equation} \begin{array}{ll} { } \Psi(t)=L P(t)[1-\eta B(\delta(t))]^{\beta} \xi^{\beta}\left(t, \delta_{0}\right), \\ { } \xi\left(t, \delta_{0}\right)= \bigg(\int_{\delta_{0}}^{\sigma(t)} \frac{1}{A^{1 / \lambda}(s)} \Delta s \bigg)^{-1} \bigg(\int_{\delta_{0}}^{\delta(t)} \frac{1}{A^{1 / \lambda}(s)} \Delta s\bigg), \end{array} \end{equation} $

其中常数$ \delta_{0}=G\left(T_{0}\right)<\delta(t) $, 而$ {{T}_{0}}\ge {{t}_{0}} $充分大.则方程(1.1)在$ {{[{{t}_{0}}, +\infty )}_{{\bf T}}} $上是振动的.

而对非正则系统即系统满足条件

(1.10) $ \begin{equation} \int_{t_{0}}^{+\infty}\left[A^{-1}(s) e_{-b / A}\left(s, t_{0}\right)\right]^{1 / \lambda} \Delta s<+\infty \end{equation} $

的情形其振动结果相对较少.如Grace等^[6], Hassan^[7]研究了动力系统(1.2)的振动性, 在非正则情形即当

(1.11) $ \begin{equation} \int_{t_{0}}^{+\infty} a^{-1 / \lambda}(s) \Delta s<+\infty \end{equation} $

时其振动的充分条件是

(1.12) $ \begin{equation} \int_{t_{0}}^{+\infty}\left[a^{-1}(t) \int_{t_{0}}^{t} p(s)\left(\int_{s}^{+\infty} a^{-1 / \gamma}(u) \Delta u\right)^{\gamma} \Delta s\right]^{1 / \gamma} \Delta t=+\infty. \end{equation} $

但上述条件(1.12)有点苛刻, 导致许多常见的方程也不适用于该结果.如下列类型的非正则Euler (欧拉)方程

(1.13) $ \begin{equation} \left(t^{2} x^{\prime}(t)\right)^{\prime}+q_{0} x(t)=0, \end{equation} $

因为积分$ \int_{1}^{+\infty}t^{-2}\ln t{\rm d}t $是收敛的, 从而导致条件(1.12)不满足, 因此文献[6, 7]中的结果不能用于方程(1.13).

之后, 张全信等^[8]讨论了更一般的动态方程(1.4)的振动性, 在非正则情形即当

(1.14) $ \begin{equation} \int_{t_{0}}^{+\infty}\left[a^{-1}(s) e_{-b / a}\left(s, t_{0}\right)\right]^{1 / \lambda} \Delta s<+\infty \end{equation} $

时其振动结果如下

定理C^[8]  设函数$ \delta :{\bf T}\to {\bf T} $是严格递增且可微的延迟函数, $ \delta ({\bf T})={\bf T} $, 且(1.14)式成立.如果存在一个正的可微函数$ \varphi :{\bf T}\to {\bf R} $使得

(1.15) $ \begin{equation} \limsup\limits_{t \rightarrow +\infty} \int_{t_{0}}^{t}\left[p(s)-\frac{a(\delta(s))}{(\lambda+1)^{\lambda+1}\left(\delta^{\Delta}(s)\right)^{\lambda}}\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{b(s)}{a(s)}\right|^{\lambda+1}\right] \varphi(s) \Delta s=+\infty \end{equation} $

成立, 并且有

(1.16) $ \begin{equation} \int_{t_{0}}^{+\infty}\left[a^{-1}(t) \int_{t_{0}}^{t} e_{-b / a}(t, \sigma(s)) p(s) \Delta s\right]^{1 / \lambda} \Delta t=+\infty, \end{equation} $

则方程(1.4)的每一个解$ x(t) $在区间$ {{[{{t}_{0}}, \infty )}_{{\bf T}}} $上或者振动或者$ \lim\limits_{t\to\infty}x(t)=0 $.

定理C是文献[8]中的定理4.3, 显然, 定理C的结论具有某种不确定性, 其它文献中的结果(如文献[9]中的定理4.3和定理4.4 (讨论的是方程(1.3)), 文献[10]中的定理4.3和定理4.4, 以及文献[11]等中的相应定理)也是如此, 因此在应用时就会受到制约.紧接着, 作者进一步改进了上述结果^{[12, 13]}, 结论如下:

定理D^[12]  设函数$ \delta :{\bf T}\to {\bf T} $是严格递增且可微的延迟函数, $ \delta ({\bf T})={\bf T} $, 如果(1.14)及(1.15)式成立, 并且有

(1.17) $ \begin{equation} \int_{t_{1}}^{+\infty}\left[a^{-1}(t) \int_{t_{1}}^{t} e_{-b / a}(t, \sigma(s)) \theta^{\lambda}(s) p(s) \Delta s\right]^{1 / \lambda} \Delta t=+\infty, \end{equation} $

这里常数$ {{t}_{1}}\ge {{t}_{0}} $, 函数$ \theta (t)=\int^\infty _t{{a}^{-1/\lambda }}(s)\Delta s $, 则方程(1.4)在区间$ {{[{{t}_{0}}, \infty )}_{{\bf T}}} $上是振动的.

定理D文献[12]中的定理4.3, 其它振动准则(如文献[12]中的定理4.4;文献[13]中的定理4.3及定理4.4等)都是以此为基础得出的.显然, 定理D虽然改进了文献[8-11]中的结果, 但是经计算可知, 定理D也不能用于方程(1.13).最近, 文献[14]研究了方程(1.4)的振动性, 在非正则情形得到了该方程振动的若干新准则, 弥补了以上的缺陷, 但遗憾的是却要求"$ \lambda \ge 1 $为两个正奇数之商", 当$ \lambda $为任意正数时却没有振动性结果.

基于以上研究的不足, 本文将考虑更一般的非正则系统(1.1)的振动性, 即在条件(1.10)成立的前提下, 利用Riccati变换和不等式技巧, 得到了非正则系统(1.1)的一个新型的Kamenev型振动准则和一个新型的Philos型振动准则, 改进了现有文献中对方程的种种限制(如文献[8-10, 12, 13]中的"$ \delta $严格递增、可微且$ \delta ({\bf T})={\bf T} $", 文献[11]中的"$ \tau =\delta $, $ \delta $严格递增且$ \tau \circ \sigma =\sigma \circ \tau $"及文献[4, 15]中的"$ {{A}^{\Delta }}(t)\ge 0 $"等), 改进并拓展了已有的结果.

先给出两个引理:

引理2.1 (Keller连锁法则)^[2]  若$ f:{\bf R}\to {\bf R} $是连续可微的, $ g:{\bf T}\to {\bf R} $是$ \Delta $可导的, 则$ f\circ g:{\bf T}\to {\bf R} $是$ \Delta $可导的, 且

(2.1) $ \begin{equation} (f \circ g)^{\Delta}(t)=\left[\int_{0}^{1} f^{\prime}\left(g(t)+h \mu(t) g^{\Delta}(t)\right) \mathrm{d} h\right] g^{\Delta}(t). \end{equation} $

由此引理, 当函数$ x(t) $最终为正或最终为负且$ \Delta $ -可微时, 则有

(2.2) $ \begin{equation} \left(x^{\alpha}(t)\right)^{\Delta}=\alpha \int_{0}^{1}\left[h x^{\sigma}+(1-h) x\right]^{\alpha-1} x^{\Delta}(t) \mathrm{d} h. \end{equation} $

这也是Keller连锁法则的推论.

引理2.2^[4]  设常数$ A>0, B>0 $, $ \lambda >0 $, 则当$ x>0 $时, 有

$ Bx-A{{x}^{\frac{\lambda +1}{\lambda }}}\le \frac{{{\lambda }^{\lambda }}{{B}^{\lambda +1}}}{{{(\lambda +1)}^{\lambda +1}}{{A}^{\lambda }}}. $

引入记号

$ \theta(t)=\int_{t}^{+\infty}\left[A^{-1}(s) e_{-b / A}\left(s, t_{0}\right)\right]^{1 / \lambda} \Delta s, \quad \pi(t)=\left\{\begin{array}{ll} \bar{k}, & \lambda>\beta, \\ 1, & \lambda=\beta, \\ \bar{k} \theta^{\beta-\lambda}(t), & \lambda<\beta, \end{array}\right. $

其中$ \bar{k}>0 $为常数.

定理2.3  设条件$ \left(\mathrm{H}_{1}\right) $–$ \left(\mathrm{H}_{5}\right) $及(1.10)成立,

$ \bar{Q}(t)=1-\eta B(\delta (\tau (t)))\frac{\theta (\tau (\delta (\tau (t))))}{\theta (\delta (\tau (t)))}>0, $

并且存在函数$ \varphi \in C_{{}}^{1}({\bf T}, (0, +\infty )) $, 使得(1.8)式成立.进一步, 如果存在常数$ m>0 $, 使得

(2.3) $ \begin{equation} \limsup\limits_{t \rightarrow +\infty} \int_{t_{1}}^{t}(t-\sigma(s))^{m}\left[L P(\tau(s)) \bar{Q}^{\beta}(s) \pi(s)-\frac{A(s) \bar{A}^{\lambda+1}(t, s)}{(\lambda+1)^{\lambda+1} e_{-b / A}(s, \tau(s))}\right] \Delta s>0, \end{equation} $

这里常数$ {{t}_{1}}\ge {{t}_{0}} $, 函数

$ \bar{A}(t, s)=\frac{{{\theta }^{\lambda }}(s)b(\tau (s))}{{{\theta }^{\lambda }}(\sigma (s))A(\tau (s))}+\frac{\bar{h}(t, s)}{{{(t-\sigma (s))}^{m}}}, \; \; \bar{h}(t, s)=\left\{ \begin{array}{ll} m{{(t-s)}^{m-1}}, &m\ge 1, \\ m{{(t-\sigma (s))}^{m-1}}, &0<m<1, \\ \end{array} \right. $

则系统(1.1)在$ {{[{{t}_{0}}, +\infty )}_{{\bf T}}} $上是振动的.

证  若不然, 则系统(1.1)在$ {{[{{t}_{0}}, +\infty )}_{{\bf T}}} $上存在一个非振动解$ x(t) $, 不妨设其最终为正, 即$ \exists \ {{t}_{1}}\in {{[{{t}_{0}}, +\infty )}_{{\bf T}}} $, 使得$ x(t)>0 $, $ x(\tau (t))>0 $, $ x(\delta (t))>0 $($ t\in {{[{{t}_{1}}, +\infty )}_{{\bf T}}} $).于是$ y(t)>0 $($ t\in {{[{{t}_{1}}, +\infty )}_{{\bf T}}} $).由系统(1.1)得

(2.4) $ \begin{equation} \left[A(t) \phi_{1}\left(y^{\Delta}(t)\right)\right]^{\Delta}+b(t) \phi_{1}\left(y^{\Delta}(t)\right) \leq-L P(t) x^{\beta}(\delta(t))<0, t \in\left[t_{1}, +\infty\right)_{\mathrm{T}}, \end{equation} $

由文献[4]中的引理6的证明知$ \frac{A(t){{\phi }_{1}}({{y}^{\Delta }}(t))}{{{e}_{-b/A}}(t, {{t}_{0}})} $($ t\in {{[{{t}_{1}}, +\infty )}_{{\bf T}}} $)严格递减且$ {{y}^{\Delta }}(t) $最终定号.因此, 只需要考虑两种情形:

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

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

情形(Ⅰ): $ {{y}^{\Delta }}(t)>0 $, $ t\in {{[{{t}_{1}}, +\infty )}_{{\bf T}}} $.此时完全和文献[5]中定理1的证明相同, 得到一个与条件(1.8)矛盾的结果.

情形(Ⅱ): $ {{y}^{\Delta }}(t)<0 $, $ t\in {{[{{t}_{1}}, +\infty )}_{{\bf T}}} $.因为函数$ \frac{A(t){{\phi }_{1}}({{y}^{\Delta }}(t))}{{{e}_{-b/A}}(t, {{t}_{0}})}=\frac{A(t){{(-{{y}^{\Delta }}(t))}^{\lambda -1}}{{y}^{\Delta }}(t)}{{{e}_{-b/A}}(t, {{t}_{0}})} $在$ {{[{{t}_{1}}, +\infty )}_{{\bf T}}} $上单调递减, 所以对$ s\in {{[t, +\infty )}_{{\bf T}}} $, 有

(2.5) $ \begin{equation} \frac{A(s)\left(-y^{\Delta}(s)\right)^{\lambda-1} y^{\Delta}(s)}{e_{-b / A}\left(s, t_{0}\right)} \leq \frac{A(t)\left(-y^{\Delta}(t)\right)^{\lambda-1} y^{\Delta}(t)}{e_{-b / A}\left(t, t_{0}\right)} \leq \frac{A(\tau(t))\left(-y^{\Delta}(\tau(t))\right)^{\lambda-1} y^{\Delta}(\tau(t))}{e_{-b / A}\left(\tau(t), t_{0}\right)}, \end{equation} $

即$ {{y}^{\Delta }}(s)\le \frac{{{A}^{1/\lambda }}(t){{y}^{\Delta }}(t)}{{{[{{e}_{-b/A}}(t, {{t}_{0}})]}^{1/\lambda }}}\frac{{{[{{e}_{-b/A}}(s, {{t}_{0}})]}^{1/\lambda }}}{{{A}^{1/\lambda }}(s)} $, 两边积分, 得

$ y(u) \leq y(t)+\frac{A^{1 / \lambda}(t) y^{\Delta}(t)}{\left[e_{-b / A}\left(t, t_{0}\right)\right]^{1 / \lambda}} \int_{t}^{u}\left[\frac{e_{-b / A}\left(s, t_{0}\right)}{A(s)}\right]^{1 / \lambda} \Delta s, $

令$ u\to +\infty $, 并注意到函数$ \theta (t) $的定义, 由上式则可导出

(2.6) $ \begin{equation} y(t)+\frac{A^{1 / \lambda}(t) y^{\Delta}(t)}{\left[e_{-b / A}\left(t, t_{0}\right)\right]^{1 / \lambda}} \theta(t) \geq 0. \end{equation} $

此外, 利用"指数函数$ {{e}_{g}}(., {{t}_{0}}) $"的半群性质^[2]: $ {{e}_{g}}(a, b){{e}_{g}}(b, c)={{e}_{g}}(a, c) $及$ {{e}_{g}}(a, b)=\frac{1}{{{e}_{g}}(b, a)} $, 由(2.5)式还可导出

(2.7) $ \begin{equation} y^{\Delta}(t) \leq \frac{\left[e_{-b / A}(t, \tau(t)) A(\tau(t))\right]^{1 / \lambda}}{A^{1 / \lambda}(t)} y^{\Delta}(\tau(t)). \end{equation} $

作广义Riccati变换

(2.8) $ \begin{equation} \omega(t)=\frac{A(\tau(t)) \phi_{1}\left(y^{\Delta}(\tau(t))\right)}{\phi_{1}(y(t))}=\frac{A(\tau(t))\left(-y^{\Delta}(\tau(t))\right)^{\lambda-1} y^{\Delta}(\tau(t))}{(y(t))^{\lambda}}, t \in\left[t_{1}, +\infty\right)_{\mathrm{T}}, \end{equation} $

明显有$ \omega (t)<0 $($ t\in {{[{{t}_{1}}, +\infty )}_{{\bf T}}} $).由(2.2)式并注意到$ {{y}^{\Delta }}(t)<0 $, 不难推得

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

当$ 0<\lambda \le 1 $时:分别利用(2.9)式、$ {{y}^{\Delta }}(t)<0 $及(2.7)式, 由(2.8)式就可导出

(2.10) $ \begin{eqnarray} \omega^{\Delta}(t)&=&\frac{\left[A(\tau(t)) \phi_{1}\left(y^{\Delta}(\tau(t))\right)\right]^{\Delta}}{y^{\lambda}(\sigma(t))}-\frac{A(\tau(t))\left(-y^{\Delta}(\tau(t))\right)^{\lambda-1} y^{\Delta}(\tau(t))\left(y^{\lambda}(t)\right)^{\Delta}}{y^{\lambda}(t) y^{\lambda}(\sigma(t))} {}\\ &\leq& \frac{\left[A(\tau(t)) \phi_{1}\left(y^{\Delta}(\tau(t))\right)\right]^{\Delta}}{y^{\lambda}(\sigma(t))}-\frac{A(\tau(t))\left(-y^{\Delta}(\tau(t))\right)^{\lambda-1} y^{\Delta}(\tau(t)) \lambda(y(t))^{\lambda-1} y^{\Delta}(t)}{y^{\lambda}(t) y^{\lambda}(\sigma(t))} {}\\ &\leq& \frac{\left[A(\tau(t)) \phi_{1}\left(y^{\Delta}(\tau(t))\right)\right]^{\Delta}}{y^{\lambda}(\sigma(t))} {}\\ &&-\frac{\lambda A(\tau(t))\left(-y^{\Delta}(\tau(t))\right)^{\lambda-1} y^{\Delta}(\tau(t))}{y^{\lambda+1}(t)} \frac{\left[e_{-b / A}(t, \tau(t)) A(\tau(t))\right]^{1 / \lambda}}{A^{1 / \lambda}(t)} y^{\Delta}(\tau(t)) {}\\ &=&\frac{\left[A(\tau(t)) \phi_{1}\left(y^{\Delta}(\tau(t))\right)\right]^{\Delta}}{y^{\lambda}(\sigma(t))}-\frac{\lambda\left[e_{-b / A}(t, \tau(t))\right]^{1 / \lambda}}{A^{1 / \lambda}(t)}(-\omega(t))^{(\lambda+1) / \lambda}. \end{eqnarray} $

当$ \lambda >1 $时:注意到(2.9)式及$ {{y}^{\Delta }}(t)<0 $, 可推得相同的(2.10)式.

利用$ {{\theta }^{\Delta }}(t)=-{{[{{A}^{-1}}(t){{e}_{-b/A}}(t, {{t}_{0}})]}^{1/\lambda }} $及(2.6)式, 可得

(2.11) $ \begin{eqnarray} \left(\frac{y(t)}{\theta(t)}\right)^{\Delta} &=&\frac{y^{\Delta}(t) \theta(t)+y(t)\left[A^{-1}(t) e_{-b / A}\left(t, t_{0}\right)\right]^{1 / \lambda}}{\theta(t) \theta(\sigma(t))} {}\\ & \geq& \frac{y^{\Delta}(t) \theta(t)-\frac{A^{1 / \lambda}(t) y^{\Delta}(t)}{\left[e_{-b / A}\left(t, t_{0}\right)\right]^{1 / \lambda}} \theta(t)\left[A^{-1}(t) e_{-b / A}\left(t, t_{0}\right)\right]^{1 / \lambda}}{\theta(t) \theta(\sigma(t))}=0. \end{eqnarray} $

这样, 由(2.11)式知函数$ \frac{y(t)}{\theta (t)} $单调递增, 因而有

(2.12) $ \begin{equation} y(\tau(t)) \leq \frac{\theta(\tau(t))}{\theta(t)} y(t), \quad \frac{y(t)}{y(\sigma(t))} \leq \frac{\theta(t)}{\theta(\sigma(t))}. \end{equation} $

注意到$ x(\tau (t))\le y(\tau (t)) $及(2.12)式中的第1个不等式, 则有

(2.13) $ \begin{equation} x(t)=y(t)-B(t) g(x(\tau(t))) \geq y(t)-\eta B(t) x(\tau(t)) \geq\left(1-\eta B(t) \frac{\theta(\tau(t))}{\theta(t)}\right) y(t), \end{equation} $

利用(2.13)式, 于是就有

(2.14) $ \begin{eqnarray} \frac{x^{\beta}(\delta(\tau(t)))}{y^{\lambda}(t)}&\geq &\frac{1}{y^{\lambda}(t)}\left(1-\eta B(\delta(\tau(t))) \frac{\theta(\tau(\delta(\tau(t))))}{\theta(\delta(\tau(t)))}\right)^{\beta} y^{\beta}(\delta(\tau(t))){}\\ &\geq&\left(1-\eta B(\delta(\tau(t))) \frac{\theta(\tau(\delta(\tau(t))))}{\theta(\delta(\tau(t)))}\right)^{\beta} y^{\beta-\lambda}(t)=\bar{Q}^{\beta}(t) y^{\beta-\lambda}(t). \end{eqnarray} $

当$ \lambda >\beta $时, 由$ y(t)>0 $, $ {{y}^{\Delta }}(t)<0 $($ t\in {{[{{t}_{1}}, \infty )}_{{\bf T}}} $)可得$ y(t)\le y({{t}_{1}}) $, 所以$ {{y}^{\beta -\lambda }}(t)\ge {{y}^{\beta -\lambda }}({{t}_{1}})=k $.

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

当$ \lambda <\beta $时, 利用$ \frac{A(t){{(-{{y}^{\Delta }}(t))}^{\lambda -1}}{{y}^{\Delta }}(t)}{{{e}_{-b/A}}(t, {{t}_{0}})} $在区间$ {{[{{t}_{1}}, +\infty )}_{{\bf T}}} $上的单调递减性, 对$ s\in {{[{{t}_{1}}, +\infty )}_{{\bf T}}} $, 则有

$ \frac{A(s)\left(-y^{\Delta}(s)\right)^{\lambda-1} y^{\Delta}(s)}{e_{-b / A}\left(s, t_{0}\right)} \leq \frac{A\left(t_{1}\right)\left(-y^{\Delta}\left(t_{1}\right)\right)^{\lambda-1} y^{\Delta}\left(t_{1}\right)}{e_{-b / A}\left(t_{1}, t_{0}\right)}=-M, $

其中$ M=-\frac{A({{t}_{1}}){{(-{{y}^{\Delta }}({{t}_{1}}))}^{\lambda -1}}{{y}^{\Delta }}({{t}_{1}})}{{{e}_{-b/A}}({{t}_{1}}, {{t}_{0}})}>0 $是常数.于是$ {{y}^{\Delta }}(s)\le -{{M}^{1/\lambda }}{{[{{A}^{-1}}(s){{e}_{-b/A}}(s, {{t}_{0}})]}^{1/\lambda }} $, 此式两边积分, 得

$ y(u) \leq y(t)-M^{1 / \lambda} \int_{t}^{u}\left[A^{-1}(s) e_{-b / A}\left(s, t_{0}\right)\right]^{1 / \lambda} \Delta s, $

所以

$ y(t) \geq y(u)+M^{1 / \lambda} \int_{t}^{u}\left[A^{-1}(s) e_{-b / A}\left(s, t_{0}\right)\right]^{1 / \lambda} \Delta s \geq M^{1 / \lambda} \int_{t}^{u}\left[A^{-1}(s) e_{-b / A}\left(s, t_{0}\right)\right]^{1 / \lambda} \Delta s, $

在上式中令$ u\to +\infty $, 则可导出

$ y(t) \geq M^{1 / \lambda} \int_{t}^{+\infty}\left[A^{-1}(s) e_{-b / A}\left(s, t_{0}\right)\right]^{1 / \lambda} \Delta s=M^{1 / \lambda} \theta(t), $

所以$ {{y}^{\beta -\lambda }}(t)\ge k{{\theta }^{\beta -\lambda }}(t) $, 这里$ k={{M}^{(\beta -\lambda )/\lambda }}>0 $是常数.

综上所述, 并注意到(2.14)式和函数$ \pi (t) $的定义, 则有

(2.15) $ \begin{equation} \frac{x^{\beta}(\delta(\tau(t)))}{y^{\lambda}(t)} \geq \bar{Q}^{\beta}(t) \pi(t). \end{equation} $

将(2.4)式代入(2.10)式, 并注意到(2.15)式及(2.12)式中的第2个不等式, 就有

(2.16) $ \begin{eqnarray} \omega^{\Delta}(t)& \leq&-L P(\tau(t)) \frac{x^{\beta}(\delta(\tau(t)))}{y^{\lambda}(\sigma(t))}-\frac{b(\tau(t)) \phi_{1}\left(y^{\Delta}(\tau(t))\right)}{y^{\lambda}(\sigma(t))}{}\\ &&-\frac{\lambda\left[e_{-b / A}(t, \tau(t))\right]^{1 / \lambda}}{A^{1 / \lambda}(t)}(-\omega(t))^{(\lambda+1) / \lambda} {}\\ &\leq&-L P(\tau(t)) \frac{x^{\beta}(\delta(\tau(t)))}{y^{\lambda}(t)}-\frac{y^{\lambda}(t)}{y^{\lambda}(\sigma(t))} \frac{b(\tau(t)) \phi_{1}\left(y^{\Delta}(\tau(t))\right)}{y^{\lambda}(t)}{}\\ &&-\frac{\lambda\left[e_{-b / A}(t, \tau(t))\right]^{1 / \lambda}}{A^{1 / \lambda}(t)}(-\omega(t))^{(\lambda+1) / \lambda} {}\\ &\leq&-L P(\tau(t)) \bar{Q}^{\beta}(t) \pi(t)-\frac{\theta^{\lambda}(t)}{\theta^{\lambda}(\sigma(t))} \frac{b(\tau(t)) \omega(t)}{A(\tau(t))}{}\\ &&-\frac{\lambda\left[e_{-b / A}(t, \tau(t))\right]^{1 / \lambda}(-\omega(t))^{(\lambda+1) / \lambda}}{A^{1 / \lambda}(t)}. \end{eqnarray} $

另一方面, 记$ h(s)=(t-s) $, 则当$ m\ge 1 $时, $ {{h}^{{{\Delta }_{s}}}}<0 $, 从而$ h(\sigma (s))\le h(s) $, 所以由(2.2)式容易推得

$ \left[(t-s)^{m}\right]^{\Delta_{s}}=\left[h^{m}\right]^{\Delta_{s}} \geq m \int_{0}^{1}[\tau h+(1-\tau) h]^{m-1} h^{\Delta_{s}} {\rm d} \tau=m h^{m-1} h^{\Delta_{s}}=-m(t-s)^{m-1}. $

当$ 0<m<1 $时, 由(2.2)式同理可得$ {{[{{(t-s)}^{m}}]}^{{{\Delta }_{s}}}}\ge -m{{(t-\sigma (s))}^{m-1}} $.因此

$ \left[(t-s)^{m}\right]^{\Delta_{s}} \geq\left\{\begin{array}{ll} -m(t-s)^{m-1}, &m \geq 1 , \\ -m(t-\sigma(s))^{m-1}, & 0<m<1, \end{array}\right. $

即

(2.17) $ \begin{equation} \left[(t-s)^{m}\right]^{\Delta_{s}} \geq-\bar{h}(t, s). \end{equation} $

现将(2.16)式两边同乘以$ {{(t-\sigma (s))}^{m}} $后再积分, 利用时间轴上的分部积分公式

$ \int_{a}^{b} f(\sigma(t)) g^{\Delta}(t) \Delta t=(f g)(b)-(f g)(a)-\int_{a}^{b} f^{\Delta}(t) g(t) \Delta t, $

并注意(2.17)式及引理2.2, 则有

$ \begin{eqnarray*} &&\int_{t_{1}}^{t} L(t-\sigma(s))^{m} P(\tau(s)) \bar{Q}^{\beta}(s) \pi(s) \Delta s\\ & \leq&-\int_{t_{1}}^{t}(t-\sigma(s))^{m} \omega^{\Delta}(s) \Delta s\\ &&+\int_{t_{1}}^{t}(t-\sigma(s))^{m}\left[\frac{\theta^{\lambda}(s) b(\tau(s))}{\theta^{\lambda}(\sigma(s)) A(\tau(s))}(-\omega(s))-\frac{\lambda\left[e_{-b / A}(s, \tau(s))\right]^{1 / \lambda}(-\omega(s))^{(\lambda+1) / \lambda}}{A^{1 / \lambda}(s)}\right] \Delta s \\ &=&-\left[(t-s)^{m} \omega(s)\right]_{t_{1}}^{t}+\int_{t_{1}}^{t} \omega(s)\left[(t-s)^{m}\right]^{\Delta_{s}} \Delta s \\ &&+\int_{t_{1}}^{t}(t-\sigma(s))^{m}\left[\frac{\theta^{\lambda}(s) b(\tau(s))}{\theta^{\lambda}(\sigma(s)) A(\tau(s))}(-\omega(s))-\frac{\lambda\left[e_{-b / A}(s, \tau(s))\right]^{1 / \lambda}(-\omega(s))^{(\lambda+1) / \lambda}}{A^{1 /\lambda}(s)}\right] \Delta s\\ &\leq&\left(t-t_{1}\right)^{m} \omega\left(t_{1}\right)+\int_{t_{1}}^{t}(t-\sigma(s))^{m}\left[\left(\frac{\theta^{\lambda}(s) b(\tau(s))}{\theta^{\lambda}(\sigma(s)) A(\tau(s))}+\frac{\bar{h}(t, s)}{(t-\sigma(s))^{m}}\right)(-\omega(s))\right.\\ &&\left.-\frac{\lambda\left[e_{-b / A}(s, \tau(s))\right]^{1 / \lambda}(-\omega(s))^{(\lambda+1) / \lambda}}{A^{1 / \lambda}(s)}\right]\\ &\leq&\left(t-t_{1}\right)^{m} \omega\left(t_{1}\right)\\ &&+\int_{t_{1}}^{t}(t-\sigma(s))^{m}\left[\frac{A(s)}{(\lambda+1)^{\lambda+1} e_{-b / A}(s, \tau(s))}\left(\frac{\theta^{\lambda}(s) b(\tau(s))}{\theta^{\lambda}(\sigma(s)) A(\tau(s))}+\frac{\bar{h}(t, s)}{(t-\sigma(s))^{m}}\right)^{\lambda+1}\right] \Delta s. \end{eqnarray*} $

注意到函数$ \bar{A}(t, s) $的表达式, 上式整理即得

$ \int_{t_{1}}^{t}(t-\sigma(s))^{m}\left[L P(\tau(s)) \bar{Q}^{\beta}(s) \pi(s)-\frac{A(s) \bar{A}^{\lambda+1}(t, s)}{(\lambda+1)^{\lambda+1} e_{-b / A}(s, \tau(s))}\right] \Delta s \leq\left(t-t_{1}\right)^{m} \omega\left(t_{1}\right), $

这与条件(2.3)矛盾, 证毕.

记$ {{D}_{0}}=\{(t, s):t>s\ge {{t}_{0}}, t, s\in {\bf T}\} $, $ D=\{(t, s):t\ge s\ge {{t}_{0}}, t, s\in {\bf T}\} $, 若二元函数$ H(t, s)\in {{C}_{rd}}(D, {\bf R}) $, 且满足下列条件

(ⅰ) 当$ t\ge {{t}_{0}} $时$ H(t, t)=0 $; 当$ (t, s)\in {{D}_{0}} $时$ H(t, s)>0 $;

(ⅱ) 在$ {{D}_{0}} $上函数$ H(t, s) $对$ s $有连续非正的偏导数, 即$ {{H}^{{{\Delta }_{s}}}}(t, s)\in {{C}_{rd}} $且$ {{H}^{{{\Delta }_{s}}}}(t, s)\le 0 $, 则称函数$ H(t, s) $属于集合$ \Omega $, 记为$ H\in \Omega $.

于是, 进一步可得下列一般性定理:

定理2.4  $ \left(\mathrm{H}_{1}\right) $–$ \left(\mathrm{H}_{5}\right) $及(1.10)成立, $ \bar{Q}(t)=1-\eta B(\delta (\tau (t)))\frac{\theta (\tau (\delta (\tau (t))))}{\theta (\delta (\tau (t)))}>0 $, 并且存在函数$ \varphi \in C_{{}}^{1}({\bf T}, (0, +\infty )) $, 使得(1.8)式成立.进一步, 如果存在$ H\in \Omega $, 使得

(2.18) $ \begin{equation} \limsup\limits_{t \rightarrow +\infty} \int_{t_{1}}^{t} H(t, \sigma(s))\left[L P(\tau(s)) \bar{Q}^{\beta}(s) \pi(s)-\frac{A(s) \bar{B}^{\lambda+1}(t, s)}{(\lambda+1)^{\lambda+1} e_{-b / A}(s, \tau(s))}\right] \Delta s>0, \end{equation} $

其中$ {{t}_{1}}\ge {{t}_{0}} $是常数, 函数$ \bar{B}(t, s)=\frac{{{\theta }^{\lambda }}(s)b(\tau (s))}{{{\theta }^{\lambda }}(\sigma (s))A(\tau (s))}-\frac{{{H}^{{{\Delta }_{s}}}}(t, s)}{H(t, \sigma (s))} $, 则系统(1.1)在$ {{[{{t}_{0}}, +\infty )}_{{\bf T}}} $上是振动的.

证  同定理2.3的证明, 可得(2.16)式, 即当$ s\ge {{t}_{1}}\ge {{t}_{0}} $时, 有

$ LP(\tau (s)){{\bar{Q}}^{\beta }}(s)\pi (s)\le -{{\omega }^{\Delta }}(s)-\frac{{{\theta }^{\lambda }}(s)}{{{\theta }^{\lambda }}(\sigma (s))}\frac{b(\tau (s))\omega (s)}{A(\tau (s))}-\frac{\lambda {{[{{e}_{-b/A}}(s, \tau (s))]}^{1/\lambda }}{{(-\omega (s))}^{(\lambda +1)/\lambda }}}{{{A}^{1/\lambda }}(s)}, $

类似于定理2.3, 由上式进一步可得

$ \begin{eqnarray*} &&\int_{t_{1}}^{t} L H(t, \sigma(s)) P(\tau(s)) \bar{Q}^{\beta}(s) \pi(s) \Delta s \\ &\leq&-\int_{t_{1}}^{t} \omega^{\Delta}(s) H(t, \sigma(s)) \Delta s \\ &&+\int_{t_{1}}^{t} H(t, \sigma(s))\left[\frac{\theta^{\lambda}(s) b(\tau(s))}{\theta^{\lambda}(\sigma(s)) A(\tau(s))}(-\omega(s))-\frac{\lambda\Big[e_{-b / A}(s, \tau(s))\Big]^{1 / \lambda}(-\omega(s))^{(\lambda+1) / \lambda}}{A^{1 / \lambda}(s)}\right] \Delta s \\ &= &\omega\left(t_{1}\right) H\left(t, t_{1}\right)+\int_{t_{1}}^{t} H(t, \sigma(s))\Big[\left(\frac{\theta^{\lambda}(s) b(\tau(s))}{\theta^{\lambda}(\sigma(s)) A(\tau(s))}-\frac{H^{\Delta_{s}}(t, s)}{H(t, \sigma(s))}\right)(-\omega(s))\\ &&-\frac{\lambda\left[e_{-b / A}(s, \tau(s))\right]^{1 / \lambda}(-\omega(s))^{(\lambda+1) / \lambda}}{A^{1 / \lambda}(s)}\Big] \Delta s \\ &\leq& \omega\left(t_{1}\right) H\left(t, t_{1}\right)\\ &&+\int_{t_{1}}^{t} H(t, \sigma(s))\Big[\frac{A(s)}{(\lambda+1)^{\lambda+1} e_{-b / A}(s, \tau(s))}\left(\frac{\theta^{\lambda}(s) b(\tau(s))}{\theta^{\lambda}(\sigma(s)) A(\tau(s))}-\frac{H^{\Delta_{s}}(t, s)}{H(t, \sigma(s))}\right)^{\lambda+1}\Big] \Delta s. \end{eqnarray*} $

注意到函数$ \bar{B}(t, s) $的定义, 整理上式即得

$ \int_{t_{1}}^{t} H(t, \sigma(s))\left[ LP(\tau (s)){{{\bar{Q}}}^{\beta }}(s)\pi (s)-\frac{A(s){{{\bar{B}}}^{\lambda +1}}(t, s)}{{{(\lambda +1)}^{\lambda +1}}{{e}_{-b/A}}(s, \tau (s))} \right]\Delta s\le \omega ({{t}_{1}})H(t, {{t}_{1}}), $

这与条件(2.18)矛盾.证毕.

注2.1  显然, 本文讨论的方程更具普遍性, 并放宽了对系统(1.1)的条件要求, 如放弃了文献[8-10, 12, 13]中的条件"$ \delta $严格递增、可微且$ \delta ({\bf T})={\bf T} $", 也放弃了文献[11]中的条件"$ \tau =\delta $, $ \delta $严格递增且$ \tau \circ \sigma =\sigma \circ \tau $", 同时也取消了文献[4, 15]中的条件"$ {{A}^{\Delta }}(t)\ge 0 $"等, 因此本文定理的条件是较为宽松的.

例3.1  考虑非正则Euler方程(1.13) (其中$ {{q}_{0}}>0 $是常数):

$ \left(t^{2} x^{\prime}(t)\right)^{\prime}+q_{0} x(t)=0, t \geq 1, $

这相当于系统(1.1)中$ A(t)={{t}^{2}} $, $ b(t)=0 $, $ B(t)\equiv 0 $, $ P(t)={{q}_{0}} $, $ \tau (t)=\delta (t)=t $, $ F(u)=u $, $ \lambda =\beta =1 $, $ {{t}_{0}}=1 $.由于

$ \int_{t_{0}}^{+\infty}\left[A^{-1}(s) e_{-b / A}\left(s, t_{0}\right)\right]^{1 / \lambda} \Delta s=\int_{t_{0}}^{+\infty} s^{-2} {\rm d} s<+\infty, $

所以条件$ \left(\mathrm{H}_{1}\right) $–$ \left(\mathrm{H}_{5}\right) $及(1.10)显然都成立.取$ \varphi (t)=1 $, 因为$ {\bf T}={\bf R} $, 所以

$ \theta(t)=\int_{t}^{+\infty}\left[A^{-1}(s) e_{-b / A}\left(s, t_{0}\right)\right]^{1 / \lambda} \Delta s=\int_{t}^{+\infty} s^{-2} {\rm d} s=\frac{1}{t}, \pi(t)=1, $

取$ {{\delta }_{0}}=G({{T}_{0}})=1/2<\delta (t) $, 则

$ \begin{eqnarray*} \xi\left(t, \delta_{0}\right)&=&\left(\int_{\delta_{0}}^{\sigma(t)} \frac{1}{A^{1 / \lambda}(s)} \Delta s\right)^{-1} \left(\int_{\delta_{0}}^{\delta(t)} \frac{1}{A^{1 / \lambda}(s)} \Delta s\right)\\ &=&\left(\int_{1 / 2}^{t} \frac{1}{s^{2}} \Delta s\right)^{-1}\left(\int_{1 / 2}^{t} \frac{1}{s^{2}} \Delta s\right)=1, \end{eqnarray*} $

于是$ \Psi (t)={{q}_{0}} $, 且有

$ \limsup\limits_{t \rightarrow +\infty} \int_{t_{0}}^{t} \varphi(s)\left\{\Psi(s)-\frac{A^{\omega_1}(s)}{k}\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{b(s)}{A(s)}\right|^{\omega_{2}+1}\right\} \Delta s=\limsup\limits_{t \rightarrow +\infty} \int_{1}^{t} q_{0} {\rm d} s=+\infty. $

再取$ H(t, s)=\frac{{{(t-s)}^{2}}}{s} $, 则当$ {{q}_{0}}>1/4 $时

$ \begin{eqnarray*} &&\limsup\limits_{t \rightarrow +\infty} \int_{t_{1}}^{t} H(t, \sigma(s))\left[L P(\tau(s)) \bar{Q}^{\beta}(s) \pi(s)-\frac{A(s) \bar{B}^{\lambda+1}(t, s)}{(\lambda+1)^{\lambda+1} e_{-b / A}(s, \tau(s))}\right] \Delta s \\ &=&\limsup\limits_{t \rightarrow +\infty} \int_{1}^{t} \frac{(t-s)^{2}}{s}\left[q_{0}-\frac{s^{2}}{4}\left(\frac{2}{t-s}+\frac{1}{s}\right)^{2}\right] \mathrm{d} s>0, \end{eqnarray*} $

最后一个不等式的成立可参见文献[16].所以由定理2.4知, 当$ {{q}_{0}}>1/4 $时二阶Euler方程(1.13)是振动的.

例3.2  考虑二阶动态方程

(3.1) $ \begin{equation} \left[t^{2} x^{\Delta}(t)\right]^{\Delta}+\frac{2 t-1}{t} F(x(\frac{t}{3}))=0, t \in T=3^{z}, t \geq 3, \end{equation} $

显然这是二阶3 -差分方程, 其中$ F(u)=u\left[5+\ln ^{3}\left(1+u^{2}\right)\right] $.令$ A(t)={{t}^{2}} $, $ B(t)\equiv 0 $, $ b(t)\equiv 0 $, $ P(t)=\frac{2t-1}{t} $, $ \delta (t)=\frac{t}{3} $, $ \lambda =\beta =1 $, $ {{t}_{0}}=3 $.则显然有$ {{e}_{-b/A}}(t, {{t}_{0}})=1 $, $ \sigma (t)=3t $, $ \frac{F(u)}{u}=5+{{\ln }^{3}}(1+{{u}^{2}})\ge 5=L $, 且

$ \zeta(t)=L P(t)[1-\eta B(\delta(t))]^{\beta} \xi^{\beta}\left(t, \delta_{0}\right)\!=\!5 \frac{2 t\!-\!1}{t}, $

$ \int_{t_{0}}^{+\infty}\left[A^{-1}(s) e_{-b / A}\left(s, t_{0}\right)\right]^{1 / \lambda} \Delta s\!=\!\int_{3}^{+\infty} s^{-2} \Delta s\!<\!+\infty. $

取$ \varphi (t)=1 $, 则容易计算出

$ \limsup\limits_{t \rightarrow +\infty} \int_{t_{0}}^{t} \varphi(s)\left\{\Psi(s)-\frac{A^{\omega_{1}}(s)}{k}\left|\frac{\varphi^{\Delta}(s)}{\varphi(s)}-\frac{b(s)}{A(s)}\right|^{\omega_{2}+1}\right\} \Delta s =5 \limsup\limits_{t \rightarrow \infty} \int_{3}^{t} \frac{2 s-1}{s} \Delta s=+\infty. $

另一方面, 令$ m=2 $, 则有$ \bar{Q}(t)=1 $, $ \pi (t)=1 $, 且

$ \theta(t)=\int_{t}^{\infty}\left[A^{-1}(s) e_{-b / A}\left(s, t_{0}\right)\right]^{1 / \lambda} \Delta s=\int_{t}^{\infty} s^{-2} \Delta s=\lim _{u \rightarrow \infty} \frac{u^{-1}-t^{-1}}{t^{-1}-1}=\frac{1}{t-1}, $

因此$ \bar{A}(t, s)=\frac{2(t-s)}{{{(t-3s)}^{2}}} $, 并且

$ \begin{eqnarray*} &&\limsup\limits_{t \rightarrow +\infty} \int_{t_{2}}^{t}(t-\sigma(s))^{m}\left[L P(\tau(s)) \bar{Q}^{\beta}(s) \pi(s)-\frac{A(s) \bar{A}^{\lambda+1}(t, s)}{(\lambda+1)^{\lambda+1} e_{-b / A}(s, \tau(s))}\right] \Delta s \\ &=&\limsup\limits_{t \rightarrow +\infty} \int_{1}^{t}(t-3 s)^{2}\left[\frac{2 s-1}{s}-\frac{s^{2}(s-t)^{2}}{(t-3 s)^{4}}\right] \Delta s>0 . \end{eqnarray*} $

所以由定理2.3知, 方程(3.1)是振动的.

注3.1  若用文献[8-11]中的定理来判别方程(1.13)或方程(3.1)的振动性, 只能得到"方程(1.13)和(3.1)的每一个解$ x(t) $或者振动或者满足$ \lim\limits_{x\to +\infty }x(t)=0 $", 显然这个结论具有不确定性, 不能令人满意的; 而文献[6, 7, 12, 13]中的相关定理的条件不满足(这是因为$ \int^{+\infty}_1{{t}^{-2}}\ln t{\rm d}t<+\infty $或$ \int^{+\infty }_3{{t}^{-2}}\ln t\Delta t<+\infty $所致).因而不能用于方程(1.13)和(3.1).因此本文定理改进并拓展了已有的结果.