研究时标T上具有非正中立项的二阶动力方程

(1.1) $ \begin{equation} \{a(t)\vert [x(t)-p(t)x(\tau (t))]^\Delta \vert ^{\alpha -1}[x(t)-p(t)x(\tau (t))]^\Delta \}^\Delta +q(t)f(x(\delta (t))) = 0, t\in {\rm {\bf T}}, t\ge t_0 >0 \end{equation} $

的动力学性质(振动性).方程(1.1)的解及其振动的定义, 可参见文献[1-15]及其参考文献.假设

(H$ _{1} $)常数$ \alpha >0 $; T为任意时标且$ \sup {\rm {\bf T}} = \infty $; $ [t_0 , \infty )_{\rm {\bf T}} = [t_0 , \infty )\cap {\rm {\bf T}} $为时标区间.

(H$ _{2} $)函数$ a, p, q\in C_{rd} ({\rm {\bf T}}, {\rm {\bf R}} $), 并且$ a(t)>0 $, $ q(t)>0 $, $ 0\le p(t)\le p_0 <1 $(此处$ p_{0} $是常数).

(H$ _{3} $)时滞函数$ \tau \in C_{rd} ({\rm {\bf T}}, {\rm {\bf T}} $), 并且$ 0<\tau (t)\le t $, $ \mathop {\lim }\limits_{t\to \infty } \tau (t) = \infty $.

(H$ _{4} $)时滞函数$ \delta \in C_{rd} ({\rm {\bf T}}, {\rm {\bf T}} $)是$ \Delta - $可微的, 并且$ 0<\delta (t)\le t $, $ \delta ^\Delta (t)>0 $, $ \mathop {\lim }\limits_{t\to \infty } \delta (t) = \infty $.

(H$ _{5} $)函数$ f\in C({\rm {\bf R}}, {\rm {\bf R}} $), 当$ u\ne 0 $时$ uf(u)>0 $, 且存在常数$ k> $0, 使得$ \frac{f(u)}{\vert u\vert ^{\alpha -1}u}\ge k $.

我们考虑正则情形的方程是即方程(1.1)满足条件

(1.2) $ \begin{equation} \int_{{\kern 1pt}{\kern 1pt}{\kern 1pt}{\kern 1pt}t_0 }^{{\kern 1pt}{\kern 1pt}{\kern 1pt}\infty } {a^{-1/\alpha }(u)} \Delta u = \infty \end{equation} $

时的振动准则.

最近, 时标上动力方程的有关理论特别是振动和非振动理论的研究引起了中外学者们的极大兴趣^[1-12], 但我们注意到, 在讨论的各类二阶中立型动力方程中, 大多要求中立项系数函数是非负的^[2-7], 当中立项系数是非正函数时, 研究成果相对较少.如Bohner等^[8]研究了一类具有非正中立项的二阶$ p $-Laplace型动力方程

(1.3) $ \begin{equation} (r(t)\vert z^\Delta (t)\vert ^{p-2}z^\Delta (t))^\Delta +q(t)\vert x(\delta (t))\vert ^{p-2}x(\delta (t)) = 0, t\in {\rm {\bf T}} \end{equation} $

的振动性, 其中$ z(t) = x(t)-a(t)x(\tau (t) $), $ p>1 $, $ 0\le a(t)\le a_0 <1 $, 结果如下.

定理1.1^[8]  设$ \int_{{\kern 1pt}{\kern 1pt}{\kern 1pt}{\kern 1pt}t_0 }^{{\kern 1pt}{\kern 1pt}{\kern 1pt}\infty } {r^{-1/(p-1)}(u)} \Delta u = \infty $, $ \int_{{\kern 1pt}{\kern 1pt}{\kern 1pt}{\kern 1pt}t_0 }^{{\kern 1pt}{\kern 1pt}{\kern 1pt}\infty } {q(u)} \Delta u = \infty $, 如果存在一个函数$ \alpha \in C_{rd} ({\rm {\bf T}}, {\rm {\bf T}} $), 它满足$ \alpha (t)>t $, $ \mathop {\lim }\limits_{t\to \infty } \tau ^{-1}(\delta (\alpha (t))) = \infty $, 并使得一阶动力微分不等式

(1.4) $ \begin{equation} u^\Delta (t)+Q(t)u(\tau ^{-1}(\delta (\alpha (t))))\le 0 \end{equation} $

没有最终正解, 其中函数$ Q(t) = \frac{1}{a_0 }\Big( \frac{1}{r(t)}\int_t^{\alpha (t)} q(s)\Delta s \Big)^{1/(p-1)} $, 则方程(1.3)是振动的.

显然, 定理1.1的条件验证起来有点麻烦, 导致使用不方便.之后, Zhang等^[9]研究了一类具非正中立项的二阶动力方程

(1.5) $ \begin{equation} \{[a(t)(x(t)-p(t)x(\tau (t)))^\Delta ]^\alpha \}^\Delta +q(t)x^\alpha (\delta (t)) = 0, t\in {\rm {\bf T}} \end{equation} $

的振动性, 得到了方程(1.5)振动的充分条件, 结果如下.

定理1.2^[9]  设条件(H$ _{1} $)–(H$ _{4} $)及(1.2)式成立, $ \alpha \ge 1 $是两个正奇数之商, $ \delta ([t_0 , \infty )_{\rm {\bf T}} ) = [\delta (t_0 ), \infty )_{\rm {\bf T}} $, 如果

(1.6) $ \begin{equation} \mathop {\limsup }\limits_{t\to \infty } \Bigg[ {\Big( {Q(t)+\alpha \int _t^\infty \delta ^\Delta (s)a^{-\frac{1}{\alpha}}(\delta (s))Q^{ {{\alpha +1} \over \alpha }}(\sigma (s))\Delta s } \Big)\Big( {\int_{t_0 }^{\delta (t)} a^{-\frac{1}{\alpha}}(s)\Delta s} \Big)^\alpha } \Bigg]>1, \end{equation} $

其中函数$ Q(t) = \int_t^\infty q(s)\Delta s $, 则方程(1.5)的所有解$ x(t $)或者振动或者满足$ \mathop {\lim }\limits_{t\to \infty } x(t) = 0 $.

显然定理1.2的条件(1.6)较好, 但是由于当$ \alpha $是大于0的任意实数时没有方程(1.5)的振动准则, 因此使用时就会受到许多制约.

本文的目的是利用广义Riccati变换技术, 并结合时标上的有关理论(特别是时标上的微积分理论)和一些经典不等式来研究方程(1.1), 并获得了方程(1.1)的一系列新的振动准则, 改进并丰富了动力方程的有关理论, 并给出了一些有趣的例子来说明本文的结论.

引理2.1  设条件(H$ _{1} $)–(H$ _{5} $)及(1.2)式成立, 函数$ x(t $)是方程(1.1)的最终正解, 令

(2.1) $ \begin{equation} z(t) = x(t)-p(t)x(\tau (t)), \end{equation} $

则$ z(t $)最终必为下列两种情形之一

(A$ _{1}) \quad z(t)> $0, $ z^{\Delta }(t)> $0, $ [a(t)\vert z^\Delta (t)\vert ^{\alpha -1}z^\Delta (t)]^\Delta \le 0 $;

(A$ _{2}) \quad z(t)< $0, $ z^{\Delta }(t)> $0, $ [a(t)\vert z^\Delta (t)\vert ^{\alpha -1}z^\Delta (t)]^\Delta \le 0 $.

证  因为$ x(t $)是方程(1.1)的最终正解, 所以$ \exists {\kern 1pt}{\kern 1pt}t_1 \in [t_0 , \infty )_{\rm {\bf T}} $, 使得$ x(t)>0 $, $ x(\tau (t))>0 $, $ x(\delta (t))>0 $, $ t\in [t_1 , \infty )_{\rm {\bf T}} $.由方程(1.1)得

(2.2) $ \begin{eqnarray} [a(t)\vert z^\Delta (t)\vert ^{\alpha -1}z^\Delta (t)]^\Delta &\le& -kq(t)\left| {x(\delta (t))} \right|^{\alpha -1}x(\delta (t)){}\\ & = &-kq(t)x^\alpha (\delta (t))\le 0, t\in [t_1 , \infty )_{\rm {\bf T}} , \end{eqnarray} $

所以函数$ a(t)\vert z^\Delta (t)\vert ^{\alpha -1}z^\Delta (t $)当$ t\in [t_1 , \infty )_{\rm {\bf T}} $时单调递减且最终定号, 于是可导出$ z^\Delta (t $)也最终定号, 即存在$ t_2 \in [t_1 , \infty )_{\rm {\bf T}} , $使得$ z^{\Delta }(t)> $0或$ z^{\Delta }(t)< $0($ t\in [t_2 , \infty )_{\rm {\bf T}} $).

我们断言: $ z^{\Delta }(t)> $0, $ t\in [t_2 , \infty )_{\rm {\bf T}} $.

若不然, 即有$ z^{\Delta }(t)< $0($ t\in [t_2 , \infty )_{\rm {\bf T}} $).则利用$ a(t)\vert z^\Delta (t)\vert ^{\alpha -1}z^\Delta (t $)的单调性, 当$ t\in [t_2 , \infty )_{\rm {\bf T}} $时, 有

$ a(t)\vert z^\Delta (t)\vert ^{\alpha -1}z^\Delta (t)\le a(t_2 )\vert z^\Delta (t_2 )\vert ^{\alpha -1}z^\Delta (t_2 ) = -M, $

其中常数$ M = -a(t_2 )\vert z^\Delta (t_2 )\vert ^{\alpha -1}z^\Delta (t_2 ) = a(t_2 )\vert z^\Delta (t_2 )\vert ^{\alpha -1}[-z^\Delta (t_2 )]>0 $.于是

$ z^\Delta (t)\le -M^{1/\alpha }a^{-1/\alpha }(t), $

所以

$ z(t)\le z(t_2 )-M^{1/\alpha }\int_{{\kern 1pt}{\kern 1pt}t_2 }^{{\kern 1pt}{\kern 1pt}t} {a^{-1/\alpha }(u)} \Delta u, $

于是, 根据条件(1.2)知

(2.3) $ \begin{equation} \mathop {\lim }\limits_{t\to \infty } z(t) = -\infty . \end{equation} $

如果$ x(t $)有界, 则由$ z(t $)的定义, $ z(t $)也有界, 这与(2.3)式矛盾!

如果$ x(t $)无界, 则有序列$ \{s_k \} $, $ s_k \in [t_2 , \infty )_{\rm {\bf T}} (k $ = 1, 2, 3, $ \cdots $), 它满足$ \mathop {\lim }\limits_{k\to \infty } s_k = \infty $, 并且使得$ \mathop {\lim }\limits_{k\to \infty } x(s_k ) = \infty $.令

$ x(\bar {s}_k ) = \max \{x(s):t_0 \le s\le s_k \}, $

显然, $ \mathop {\lim }\limits_{k\to \infty } x(\bar {s}_k ) = \infty $.因为$ \mathop {\lim }\limits_{t\to \infty } \tau (t) = \infty $, 所以, 存在正数$ K $, 当$ k>K $时, 有$ \tau (s_k )>t_0 $.注意到$ \tau (t)<t $, 因此当$ k>K $时, 有

$ x(\tau (\bar {s}_k )) = \max \{x(s):t_0 \le s\le \tau (s_k )\}\le \max \{x(s):t_0 \le s\le s_k \} = x(\bar {s}_k ), $

从而

$ z(\bar {s}_k ) = x(\bar {s}_k )-p(\bar {s}_k )x(\tau (\bar {s}_k ))\ge x(\bar {s}_k )[1-p(\bar {s}_k )]>0, $

这与(2.3)式矛盾!

故当$ t\in [t_2 , \infty )_{\rm {\bf T}} $时必有$ z^{\Delta }(t)> $0.于是函数$ z(t $)最终必是情形(A$ _{1} $)或(A$ _{2} $)之一.

引理2.2  设函数$ x(t $)是方程(1.1)的最终正解, 且$ z(t $)为引理2.1中的情形(A$ _{2} $), 则$ \mathop {\lim }\limits_{t\to \infty } x(t) = 0 $.

证  因为$ z(t)< $0, $ z^{\Delta }(t)> $0, 所以$ \mathop {\lim }\limits_{t\to \infty } z(t) = l\le 0 $, 其中$ l $是常数.因此函数$ z(t $)最终是有界的.与引理2.1中的证明类似, 可推得$ x(t $)也是有界的: $ \mathop {\limsup }\limits_{t\to \infty } x(t) = c $, 则$ 0\le c<\infty $.

若$ c>0 $, 则存在数列$ \{s_k \} $, $ s_k \in [t_0 , \infty )_{\rm {\bf T}}\ (k = 1, 2, 3, \cdots $), 满足$ \mathop {\lim }\limits_{k\to \infty } s_k = \infty $, 并使得$ \mathop {\lim }\limits_{k\to \infty } x(s_k ) = c $.于是对$ \varepsilon = \frac{1-p_0 }{2p_0 }c>0 $, 存在正数$ K $, 当$ k>K $时, 就有$ x(\tau (s_k ))<c+\varepsilon $, 故

$ z(s_k ) = x(s_k )-p(s_k )x(\tau (s_k ))\ge x(s_k )-p_0 (c+\varepsilon ), $

进一步, 得

$ 0\ge l = \mathop {\lim }\limits_{k\to \infty } z(s_k )\ge \mathop {\lim }\limits_{k\to \infty } x(s_k )-p_0 (c+\varepsilon ) = c-p_0 (c+\varepsilon ) = \frac{1-p_0 }{2}c, $

矛盾！因此$ c = 0 $, 从而$ \mathop {\lim }\limits_{t\to \infty } x(t) = 0 $.

引理2.3^[3]  若$ x(t $)是$ \Delta $ -可微的且最终为正或最终为负时, 则

$ (x^\lambda (t))^\Delta = \lambda x^\Delta (t)\int_0^1 [hx(\sigma (t))+(1-h)x(t)]^{\lambda -1}{\rm d}h. $

引理2.4^[4]  设$ A>0, B>0 $, $ \lambda >0 $为常数, 则

$ Bx-Ax^{\frac{\lambda +1}{\lambda }}\le \frac{\lambda ^\lambda B^{\lambda +1}}{(\lambda +1)^{\lambda +1}A^\lambda }\ (x>0). $

引理2.5^[4] (Hölder不等式)  设$ a, b\in {\rm {\bf T}} $且$ a<b $, 函数$ f, g:[a, b]\to {\rm {\bf R}} $是rd -连续函数, 则

$ \int {_a^b } \vert f(u)g(u)\vert \Delta u\le \left( {\int {_a^b } \vert f(u)\vert ^p\Delta u} \right)^{\frac{1}{p}}\left( {\int {_a^b } \vert g(u)\vert ^q\Delta u} \right)^{\frac{1}{q}}\ (p>1, \frac{1}{p}+\frac{1}{q} = 1). $

定理3.1  设条件(H$ _{1} $)–(H$ _{5} $)及(1.2)式成立, 若有函数$ \varphi \in C_{rd}^1 ({\rm {\bf T}}, (0, \infty ) $)使得

(3.1) $ \begin{equation} \mathop {\limsup }\limits_{t\to \infty } \int_{t_1 }^t \Bigg\{ k\varphi (s)q(s)-\frac{a(\delta (s))\vert \varphi^\Delta (s)\vert ^{\alpha +1}}{(\alpha +1)^{\alpha +1}[\varphi (s)\delta ^\Delta (s)]^\alpha }\Bigg\}\Delta s = \infty , \end{equation} $

常数$ t_1 \ge t_0 $充分大, 则在区间$ [t_0 , \infty )_{\rm {\bf T}} $上方程(1.1)的所有$ x(t $)或者振动或者$ \mathop {\lim }\limits_{t\to \infty } x(t) = 0 $.

证  设函数$ x(t $)为方程(0.1)的一个非振动解, 不妨设$ x(t $)最终为正(当$ x(t $)最终为负时类似可证), 即$ x(t)>0 $, $ x(\tau (t))>0 $, $ x(\delta (t))>0(t\in [t_1 , \infty )_{\rm {\bf T}} $).根据引理2.1, $ z(t $)必为(A$ _{1} $), (A$ _{2} $)两种情形之一.

若属于情形(A$ _{2} $), 则由引理2.2, $ \mathop {\lim }\limits_{t\to \infty } x(t) = 0 $.

若属于情形(A$ _{1} $), 即$ z(t)> $0, $ z^{\Delta }(t)> $0.由$ z(t $)的定义, 可得

(3.2) $ \begin{equation} x(t) = z(t)+p(t)x(\tau (t))\ge z(t). \end{equation} $

由(2.2)式可导出

(3.3) $ \begin{equation} a(\sigma (t))[z^\Delta (\sigma (t))]^\alpha \le a(t)[z^\Delta (t)]^\alpha \le a(\delta (t))[z^\Delta (\delta (t))]^\alpha , \end{equation} $

并且

(3.4) $ \begin{eqnarray} z(t)& = &z(t_1 )+\int_{{\kern 1pt}{\kern 1pt}t_1 }^{{\kern 1pt}{\kern 1pt}t} {\frac{[a(s)(z^\Delta (s))^\alpha ]^{1/\alpha }}{a^{1/\alpha }(s)}} \Delta s{}\\ &\ge &a^{1/\alpha }(t)z^\Delta (t)\int_{{\kern 1pt}{\kern 1pt}t_1 }^{{\kern 1pt}{\kern 1pt}t} {\frac{1}{a^{1/\alpha }(s)}} \Delta s{}\\ & = &\xi (t)z^\Delta (t). \end{eqnarray} $

根据引理2.3, 易推得^[4]

(3.5) $ \begin{eqnarray} [(z(\delta(t)))^\alpha]^{\Delta}\ge \Bigg\{ \begin{array}{ll} \alpha z^\Delta(\delta(t))\delta^\Delta(t)(z(\delta(t)))^{\alpha-1}, &\alpha>1, \\ \alpha z^\Delta(\delta(t))\delta^\Delta(t)(z(\delta(\sigma(t))))^{\alpha-1}, &0<\alpha\le 1. \end{array} \Bigg. \end{eqnarray} $

令

(3.6) $ \begin{equation} W(t) = \varphi (t)\frac{a(t)\vert z^\Delta (t)\vert ^{\alpha -1}z^\Delta (t)}{z^\alpha (\delta (t))} = \varphi (t)\frac{a(t)(z^\Delta (t))^\alpha }{z^\alpha (\delta (t))}, t\in [t_1 , \infty )_{\rm {\bf T}} , \end{equation} $

则$ W(t)>0\ (t\in [t_1 , \infty )_{\rm {\bf T}} $).

当$ \alpha >1 $时, 分别注意到(2.2), (3.5), (3.2)和(3.3)式, 则由(3.6)式可推出

(3.7) $ \begin{eqnarray} W^\Delta (t)& = &\frac{\varphi (t)}{z^\alpha (\delta (t))}[a(t)\vert z^\Delta (t)\vert ^{\alpha -1}z^\Delta (t)]^\Delta {}\\ &&+a(\sigma (t))(z^\Delta (\sigma (t)))^\alpha \frac{\varphi ^\Delta (t)z^\alpha (\delta (t))-\varphi (t)[z^\alpha (\delta (t))]^\Delta }{z^\alpha (\delta (t))z^\alpha (\delta (\sigma (t)))}{}\\ &\le& -k\frac{\varphi (t)q(t)x^\alpha (\delta (t))}{z^\alpha (\delta (t))}+\varphi ^\Delta (t)\frac{a(\sigma (t))(z^\Delta (\sigma (t)))^\alpha }{z^\alpha (\delta (\sigma (t)))}{}\\ &&-a(\sigma (t))(z^\Delta (\sigma (t)))^\alpha \varphi (t)\frac{\alpha z^\Delta (\delta (t))\delta ^\Delta (t)z^{\alpha -1}(\delta (t))}{z^\alpha (\delta (t))z^\alpha (\delta (\sigma (t)))}{}\\ &\le& -k\varphi (t)q(t)+\varphi ^\Delta (t)\frac{a(\sigma (t))(z^\Delta (\sigma (t)))^\alpha }{z^\alpha (\delta (\sigma (t)))}{}\\ &&-\alpha \varphi (t)\delta ^\Delta (t)\frac{a(\sigma (t))(z^\Delta (\sigma (t)))^\alpha }{z^\alpha (\delta (\sigma (t)))}\cdot \frac{z^\Delta (\delta (t))}{z(\delta (t))}. \end{eqnarray} $

由(3.3)式, 可得

$ z^\Delta (\delta (t))\ge \left[ {\frac{a(\sigma (t))}{a(\delta (t))}} \right]^{1/\alpha }z^\Delta (\sigma (t)), $

将其代入(3.7)式, 并注意到$ z(\delta (t))\le z(\delta (\sigma (t)) $)及(3.6)式, 则可推出

(3.8) $ \begin{eqnarray} W^\Delta (t)&\le& -k\varphi (t)q(t)+\varphi ^\Delta (t)\frac{a(\sigma (t))(z^\Delta (\sigma (t)))^\alpha }{z^\alpha (\delta (\sigma (t)))}{}\\ & &-\alpha {\kern 1pt}\varphi (t)\delta ^\Delta (t)a(\sigma (t))\frac{(z^\Delta (\sigma (t)))^{\alpha +1}}{z^{\alpha +1}(\delta (\sigma (t)))}\left[ {\frac{a(\sigma (t))}{a(\delta (t))}} \right]^{1/\alpha }{}\\ & = &-k\varphi (t)q(t)+\frac{\varphi ^\Delta (t)}{\varphi (\sigma (t))}W(\sigma (t))-\frac{\alpha {\kern 1pt}\varphi (t)\delta ^\Delta (t)}{a^{1/\alpha }(\delta (t))\varphi ^{(\alpha +1)/\alpha }(\sigma (t))}W^{\frac{\alpha +1}{\alpha }}(\sigma (t)). \end{eqnarray} $

当$ 0<\alpha \le 1 $时, 利用完全相同的方法, 可得到相同的(3.8)式.

现将引理2.4中的不等式用于(3.8)式, 则可导出

$ W^\Delta (t)\le -k\varphi (t)q(t)+\frac{a(\delta (t))\vert \varphi ^\Delta (t)\vert ^{\alpha +1}}{(\alpha +1)^{\alpha +1}[\varphi (t)\delta ^\Delta (t)]^\alpha }, $

于是

(3.9) $ \begin{eqnarray} &&\int_{t_1 }^t \left\{ {k\varphi (s)q(s)-\frac{a(\delta (s))\vert \varphi ^\Delta (s)\vert ^{\alpha +1}}{(\alpha +1)^{\alpha +1}[\varphi (s)\delta ^\Delta (s)]^\alpha }} \right\}\Delta s{}\\ &\le& -\int _{t_1 }^t W^\Delta (s)\Delta s = W(t_1 )-W(t)\le W(t_1 ), \end{eqnarray} $

这与条件(3.1)矛盾.证毕.

若定理3.1中的条件(3.1)不成立, 则有如下定理.

定理3.2  设条件(H$ _{1} $)–(H$ _{5} $)及(1.2)式成立, 如果存在函数$ \varphi \in C_{rd}^1 ({\rm {\bf T}}, (0, \infty ) $)及$ \zeta _1 , $$ \zeta _2 \in C_{rd} ({\rm {\bf T}}, {\rm {\bf R}} $)使得

(3.10) $ \begin{equation} \mathop {\limsup }\limits_{t\to \infty } \int _u^t \varphi (s)q(s)\Delta s\ge \zeta _1 (u), \end{equation} $

(3.11) $ \begin{equation} \mathop {\limsup }\limits_{t\to \infty } \int _u^t \frac{a(\delta (s))\vert \varphi ^\Delta (s)\vert ^{\alpha +1}}{[\varphi (s)\delta ^\Delta (s)]^\alpha }\Delta s\le \zeta _2 (u) \end{equation} $

对一切$ u\ge t_1 \ge t_0 $成立, 并且函数$ \zeta _1 $和$ \zeta _2 $满足

(3.12) $ \begin{equation} \mathop {\liminf}\limits_{t\to \infty } \int _{t_1 }^t \frac{{\kern 1pt}\varphi (s)\delta ^\Delta (s)[\zeta _1^\sigma (s)-\omega \zeta _2^\sigma (s)]_+^{(\alpha +1)/\alpha } }{a^{1/\alpha }(\delta (s))\varphi ^{(\alpha +1)/\alpha }(\sigma (s))}\Delta s = \infty , \end{equation} $

常数$ t_1 \ge t_0 $充分大, $ \omega = \frac{1}{k(\alpha +1)^{\alpha +1}} $, 而函数$ \zeta _+ (t) = \max \{\zeta (t), 0\} $.则在$ [t_0 , \infty )_{\rm {\bf T}} $上方程(1.1)的每一个解$ x(t $)或者振动或者$ \mathop {\lim }\limits_{t\to \infty } x(t) = 0 $.

证  证明的前面部分与定理3.1完全相同, 可得(3.8)和(3.9)式.将(3.8)式中的$ t $改成$ s $, 再两边从$ t_{1} $到$ t $积分, 得

$ \int _{t_1 }^t \frac{\alpha {\kern 1pt}\varphi (s)\delta ^\Delta (s)W^{\frac{\alpha +1}{\alpha }}(\sigma (s))}{a^{1/\alpha }(\delta (s))\varphi ^{(\alpha +1)/\alpha }(\sigma (s))}\Delta s+\int _{t_1 }^t k\varphi (s)q(s)\Delta s\le W(t_1 )+\int _{t_1 }^t \frac{\vert \varphi ^\Delta (s)\vert }{\varphi (\sigma (s))}W(\sigma (s))\Delta s, $

注意到条件(3.10), 由上式就可导出

$ \mathop {\liminf}\limits_{t\to \infty } \int _{t_1 }^t \frac{\alpha {\kern 1pt}\varphi (s)\delta ^\Delta (s)W^{\frac{\alpha +1}{\alpha }}(\sigma (s))}{a^{1/\alpha }(\delta (s))\varphi ^{(\alpha +1)/\alpha }(\sigma (s))}\Delta s+k\zeta _1 (t_1 )\le W(t_1 )+\mathop {\liminf}\limits_{t\to \infty } \int _{t_1 }^t \frac{\vert \varphi ^\Delta (s)\vert }{\varphi (\sigma (s))}W(\sigma (s))\Delta s, $

即

(3.13) $ \begin{equation} \mathop {\liminf}\limits_{t\to \infty } \int _{t_1 }^t \frac{\alpha {\kern 1pt}\varphi (s)\delta ^\Delta (s)W^{\frac{\alpha +1}{\alpha }}(\sigma (s))}{a^{1/\alpha }(\delta (s))\varphi ^{(\alpha +1)/\alpha }(\sigma (s))}\Delta s-\mathop {\liminf}\limits_{t\to \infty } \int _{t_1 }^t \frac{\vert \varphi ^\Delta (s)\vert }{\varphi (\sigma (s))}W(\sigma (s))\Delta s\le C_0 , \end{equation} $

($ C_0 = W(t_1 )-k\zeta _1 (t_1 $)为常数).于是, 根据上式, 我们可进一步推出

(3.14) $ \begin{equation} \mathop {\liminf}\limits_{t\to \infty } \int _{t_1 }^t \frac{\varphi (s)\delta ^\Delta (s)W^{(\alpha +1)/\alpha }(\sigma (s))}{a^{1/\alpha }(\delta (s))\varphi ^{(\alpha +1)/\alpha }(\sigma (s))}\Delta s<\infty . \end{equation} $

事实上, 如果(3.14)式不成立, 则存在数列{$ a_{n} $}: $ a_n \in [t_1 , \infty )_{\rm {\bf T}}\ (n = 1, 2, 3, \cdots $), 满足$ \mathop {\lim }\limits_{n\to \infty } a_n = \infty $, 并且使得

(3.15) $ \begin{equation} \mathop {\lim }\limits_{n\to \infty } \int _{t_1 }^{a_n } \frac{\varphi (s)\delta ^\Delta (s)W^{(\alpha +1)/\alpha }(\sigma (s))}{a^{1/\alpha }(\delta (s))\varphi ^{(\alpha +1)/\alpha }(\sigma (s))}\Delta s = \infty , \end{equation} $

结合(3.13)和(3.15)式, 则可推得

(3.16) $ \begin{equation} \mathop {\lim }\limits_{n\to \infty } \int _{t_1 }^{a_n } \frac{\vert \varphi ^\Delta (s)\vert }{\varphi (\sigma (s))}W(\sigma (s))\Delta s = \infty . \end{equation} $

综合(3.13), (3.15)及(3.16)式知, 存在充分大的正数$ N $, 使得当$ n>N $时就有

$ \int _{t_1 }^{a_n } \frac{\varphi (s)\delta ^\Delta (s)W^{(\alpha +1)/\alpha }(\sigma (s))}{a^{1/\alpha }(\delta (s))\varphi ^{(\alpha +1)/\alpha }(\sigma (s))}\Delta s-\int _{t_1 }^{a_n } \frac{\vert \varphi ^\Delta (s)\vert W(\sigma (s))}{\varphi (\sigma (s))}\Delta s<C_0 +1, $

因此, 对$ 0<\varepsilon <1 $和$ n>N $, 由上式可进一步导出

(3.17) $ \begin{equation} \frac{\int _{t_1 }^{a_n } \frac{\vert \varphi ^\Delta (s)\vert W(\sigma (s))}{\varphi (\sigma (s))}\Delta s}{\int _{t_1 }^{a_n } \frac{\varphi (s)\delta ^\Delta (s)W^{(\alpha +1)/\alpha }(\sigma (s))}{a^{1/\alpha }(\delta (s))\varphi ^{(\alpha +1)/\alpha }(\sigma (s))}\Delta s}>1-\varepsilon >0. \end{equation} $

于是, 利用由引理2.5, 可得

(3.18) $ \begin{eqnarray} &&\int _{t_1 }^{a_n } \frac{\vert \varphi ^\Delta (s)\vert W(\sigma (s))}{\varphi (\sigma (s))}\Delta s{}\\ & = &\int _{t_1 }^{a_n } \left[ {\frac{\varphi (s)\delta ^\Delta (s)W^{(\alpha +1)/\alpha }(\sigma (s))}{a^{1/\alpha }(\delta (s))\varphi ^{(\alpha +1)/\alpha }(\sigma (s))}} \right]^{\alpha /(\alpha +1)}\cdot \frac{a^{1/(\alpha +1)}(\delta (s))\vert \varphi ^\Delta (s)\vert }{[\varphi (s)\delta ^\Delta (s)]^{\alpha /(\alpha +1)}}\Delta s{}\\ &\le& \left[ {\int _{t_1 }^{a_n } \frac{\varphi (s)\delta ^\Delta (s)W^{(\alpha +1)/\alpha }(\sigma (s))}{a^{1/\alpha }(\delta (s))\varphi ^{(\alpha +1)/\alpha }(\sigma (s))}\Delta s} \right]^{\alpha /(\alpha +1)}\cdot \left[ {\int _{t_1 }^{a_n } \frac{a(\delta (s))\vert \varphi ^\Delta (s)\vert ^{\alpha +1}}{[\varphi (s)\delta ^\Delta (s)]^\alpha }\Delta s} \right]^{1/(\alpha +1)}.{\qquad} \end{eqnarray} $

上式两边($ \alpha +1 $)次方, 利用(3.17)式, 就可推出

(3.19) $ \begin{eqnarray} 0&<&(1-\varepsilon )^\alpha \int _{t_1 }^{a_n } \frac{\vert \varphi ^\Delta (s)\vert W(\sigma (s))}{\varphi (\sigma (s))}\Delta s{}\\ & = &\frac{\left[ {\int {_{t_1 }^{a_n } } \frac{\vert \varphi ^\Delta (s)\vert W(\sigma (s))}{\varphi (\sigma (s))}\Delta s} \right]^{\alpha +1}}{\left[ {\int {_{t_1 }^{a_n } } \frac{\varphi (s)\delta ^\Delta (s)W^{(\alpha +1)/\alpha }(\sigma (s))}{a^{1/\alpha }(\delta (s))\varphi ^{(\alpha +1)/\alpha }(\sigma (s))}\Delta s} \right]^\alpha }\le \int _{t_1 }^{a_n } \frac{a(\delta (s))\vert \varphi ^\Delta (s)\vert ^{\alpha +1}}{[\varphi (s)\delta ^\Delta (s)]^\alpha }\Delta s, \end{eqnarray} $

根据条件(3.11)知, 上式右边的积分是有界的, 从而

$ \int _{t_1 }^{a_n } \frac{\vert \varphi ^\Delta (s)\vert W(\sigma (s))}{\varphi (\sigma (s))}\Delta s<\infty , $

这与(3.16)式矛盾!这就证明(3.14)式.

另一方面, 由(3.9)式, 得

$ \int _u^t k\varphi (s)q(s)\Delta s\le W(u)+\int _u^t \frac{a(\delta (s))\vert \varphi ^\Delta (s)\vert ^{\alpha +1}}{(\alpha +1)^{\alpha +1}[\varphi (s)\delta ^\Delta (s)]^\alpha }\Delta s $

对一切$ t\ge u\ge t_1 \ge t_0 $成立.利用(3.10)和(3.11)式, 上式取上极限, 就有

$ k\zeta _1 (u)\le w(u)+\frac{\zeta _2 (u)}{(\alpha +1)^{\alpha +1}}, $

也就是说

(3.20) $ \begin{equation} \zeta _1 (u)-\omega \zeta _2 (u)\le k^{-1}w(u). \end{equation} $

于是, 分别注意到(3.20)式和(3.14)式, 则有

$ \begin{eqnarray*} &&\mathop {\liminf}\limits_{t\to \infty } \int _{t_1 }^t \frac{{\kern 1pt}\varphi (s)\delta ^\Delta (s)[\zeta _1^\sigma (s)-\omega \zeta _2^\sigma (s)]_+^{(\alpha +1)/\alpha } }{a^{1/\alpha }(\delta (s))\varphi ^{(\alpha +1)/\alpha }(\sigma (s))}\Delta s \\ &\le& \frac{1}{k^{(\alpha +1)/\alpha }}\mathop {\liminf}\limits_{t\to \infty } \int _{t_1 }^t \frac{{\kern 1pt}\varphi (s)\delta ^\Delta (s)W^{(\alpha +1)/\alpha }(s)}{a^{1/\alpha }(\delta (s))\varphi ^{(\alpha +1)/\alpha }(\sigma (s))}\Delta s<\infty , \end{eqnarray*} $

这就与条件(3.12)矛盾!证毕.

例3.1  考虑具非正中立项的二阶动力方程

(3.21) $ \begin{equation} \Big\{ {t\Big| {[ {x(t)-(\frac{1}{2}+\frac{1}{t})x(\frac{t}{2})} ]^\Delta } \Big|^{4/3}[ {x(t)-(\frac{1}{2}+\frac{1}{t})x(\frac{t}{2})} ]^\Delta } \Big\}^\Delta +\frac{1}{t^2}\Big| {x^{4/3}(\frac{t}{2})} \Big|x(\frac{t}{2}) = 0, \;t\in [2, \infty )_{\rm {\bf T}} , \end{equation} $

这相当于方程(1.1)中$ a(t) = t $, $ p(t) = \frac{1}{2}+\frac{1}{t} $, $ q(t) = \frac{1}{t^2} $, $ \tau (t) = \delta (t) = \frac{t}{2} $, $ f(u) = u $且$ \alpha = \frac{7}{3} $的情形.由于

$ \int _{t_0 }^\infty \frac{1}{a^{1/\alpha }(t)}\Delta t = \int_{_2}^\infty \frac{1}{t^{3/7}}\Delta t = \infty , $

显然, 条件(H$ _{1} $)–(H$ _{5} $)及(1.2)式都满足.现在定理3.1中取$ \varphi (t) = t $, 则

$ \begin{eqnarray*} &&\mathop {\limsup }\limits_{t\to \infty } \int _{t_0 }^t \left\{ {k\varphi (s)q(s)-\frac{a(\delta (s))\vert \varphi ^\Delta (s)\vert ^{\alpha +1}}{(\alpha +1)^{\alpha +1}[\varphi (s)\delta ^\Delta (s)]^\alpha }} \right\}\Delta s\\ & = &\mathop {\limsup }\limits_{t\to \infty } \int_{2}^t \left\{ {\frac{1}{s}-\left( {\frac{3}{11}} \right)^{11/3}\frac{2^{4/3}}{s^{4/3}}} \right\}\Delta s = \infty , \end{eqnarray*} $

根据定理3.1, 方程(3.21)的每一个解$ x(t $)或者振动或者满足$ \mathop {\lim }\limits_{t\to \infty } x(t) = 0 $.

注3.1  由例3.1看出, 本文定理的条件验证起来非常简单且方便.另一方面, 由于方程()是具有非正中立项的, 因此文献[1-7, 10-12]中的定理都不能用于方程(3.21).

例3.2  考虑具负中立项的二阶动力方程

(3.22) $ \begin{equation} [t^{\frac{2}{5}}(x(t)-\frac{1}{2}x(\frac{t}{2}))^\Delta ]^\Delta +t^{-\frac{11}{10}}x(\frac{t}{2}) = 0, t\in {\rm {\bf T}} = 2^{\rm {\bf Z}}, t\ge t_0 = 2, \end{equation} $

这是具非正中立项的二阶非线性的- 2差分方程, 相当于方程(1.1)中$ \tau (t) = \delta (t) = \frac{t}{2} $, $ a(t) = t^{\frac{2}{5}} $, $ p(t) = -\frac{1}{2} $, $ q(t) = t^{-\frac{11}{10}} $, $ \alpha = 1 $, $ k = 1 $.可以验证

$ \int_{t_0 }^t \frac{1}{a^{1/\alpha }(s)}\Delta s = \int_2^t s^{-\frac{2}{5}}\Delta s = \frac{t^{3/5}-2^{3/5}}{2^{3/5}-1}\to \infty (t\to \infty ). $

若取$ \varphi (t) = 1 $, 则当$ t\ge 2 $时, 有

$ \begin{eqnarray*} \mathop {\limsup }\limits_{t\to \infty } \int _u^t \varphi (s)q(s)\Delta s& = &\mathop {\limsup }\limits_{t\to \infty } \int _u^t s^{-\frac{11}{10}}\Delta s \\ & = &\mathop {\limsup }\limits_{t\to \infty } \frac{t^{-1/10}-u^{-1/10}}{u^{-1/10}-1}\\ & = &\frac{1}{u^{1/10}-1} \ge \frac{1}{u^{1/10}} = \zeta _1 (u), \end{eqnarray*} $

$ \mathop {\limsup }\limits_{t\to \infty } \int _{u}^t \frac{a(\delta (s))\vert \varphi ^\Delta (s)\vert ^{\alpha +1}}{[\varphi (s)\delta ^\Delta (s)]^\alpha }\Delta s = 0 = \zeta _2 (u), $

且有

$ \begin{eqnarray*} &&\mathop {\liminf}\limits_{t\to \infty } \int _{t_1 }^t \frac{{\kern 1pt}\varphi (s)\delta ^\Delta (s)[\zeta _1^\sigma (s)-\omega \zeta _2^\sigma (s)]_+^{(\alpha +1)/\alpha } }{a^{1/\alpha }(\delta (s))\varphi ^{(\alpha +1)/\alpha }(\sigma (s))}\Delta s\\ & = &\mathop {\liminf}\limits_{t\to \infty } \int _2^t [\frac{1}{(2s)^{1/10}}]^2\frac{1}{s^{2/5}}\Delta s = \frac{1}{2^{1/5}}\mathop {\liminf}\limits_{t\to \infty } \int_2^t \frac{1}{s^{3/5}}\Delta s = \infty , \end{eqnarray*} $

以上三式说明, 条件(3.10)–(3.12)均满足, 因此, 由定理3.2, 方程(3.22)的每一个解$ x(t $)或者振动或者满足$ \mathop {\lim }\limits_{t\to \infty } x(t) = 0 $.

注3.2  因为(取$ \varphi (t) = 1 $)

$ \mathop {\limsup }\limits_{t\to \infty } \int _{t_1 }^t \left\{ k\varphi (s)q(s)-\frac{a(\delta (s))\vert \varphi^\Delta (s)\vert ^{\alpha +1}}{(\alpha +1)^{\alpha +1}[\varphi (s)\delta ^\Delta (s)]^\alpha } \right\}\Delta s = \mathop {\limsup }\limits_{t\to \infty } \int _2^t \frac{1}{s^{11/10}}\Delta s<\infty , $

所以定理3.1的条件(3.1)不满足, 因此定理3.1不能用于方程(3.1).此外, 由于方程(3.22)是具有非正中立项的, 因此, 现有文献如^[1-12]中定理都不能应用于方程(3.22).