时标上一类分布时滞线性动力方程的全局吸引性

Global Attractivity for a Linear Dynamic Equation with a Distributed Delay on Time Scales

摘要: 本文研究时标上一类具有分布时滞的线性动力方程，利用Lyapunov函数方法，得到了该方程全局吸引性的一个充分条件，并且进一步分析了具有混合时滞的线性动力方程在时标上的全局吸引性。最后文章给出了几个例子予以说明。

Abstract: In this paper, we consider a linear dynamic equation with a distributed delay on time scales. By using Lyapunov function method, we obtain a novel sufficient condition for global attractivity of the linear delayed dynamic equation. And the linear dynamic equation with mixed delays on time scales is also discussed. Some examples are given to illustrate our results.

1. 引言

动力方程模型在从经济到工业各个领域中都发挥了重要作用，比如管道中水的流动变化，一年中鱼群数量的变化，具有有限次出入的交通流量变化等等。这其中有的是连续过程，有的是离散过程，有的兼有连续和离散。连续过程和离散过程在以往要分别利用微分方程和差分方程来进行研究，得出一些相似而又有差异的结果。而如今由Stefan Hilger [1] 创立的时标理论能将这些结果统一起来，并同时揭示了这些结果的差异。而且更有意义的是它不局限于单纯的连续或离散，它能很好地解决诸多兼有连续和离散的模型。这在以前是不可能做到的。时标理论近年来正受到越来越广泛的关注。而对于时标上的动力方程的研究也正吸引了不少眼球，许多前沿的结果被推广到时标上(参看 [1] - [4] )。其中时标上的一阶线性动力方程 $x^{\Delta }\left(t\right)+q\left(t\right)x\left(t\right)=f\left(t\right)$ 得以详细讨论。

实际上不容忽视的是动力系统往往存在时滞现象，而时滞可能给系统带来不确定的影响，需要给予充分的考虑，Yu, J.S.和Cheng, S.S.曾在 [5] 中讨论过一类时滞差分方程 $\Delta x_n+q\left(t\right)x_{n-\tau }=0$ ，2009年周展、李红娟在 [6] 中就时标上一类时滞微分方程 $x^{\Delta }\left(t\right)+q\left(t\right)x\left(t-\tau \right)=0$ 的全局吸引性做了研究，给出了相关条件。

现实中，时滞还可能不仅仅来自于过去某一点，而是可表达为过去某一段时间上状态变量的积分，即分布时滞。许多学者在动力系统研究中逐步开始把分布时滞考虑进来(参看文献 [7] [8] [9] )。本文首次就如下具分布时滞线性动力方程的全局吸引性做了讨论：

$x^{\Delta }\left(t\right)+q\left(t\right){\displaystyle {\int }_0^{\tau }K\left(u\right)x\left(t-u\right)\Delta u}=0,t\in {\left[t_0,+\infty \right)}_T$ (1)

受文献 [9] 和 [10] 的启发，接下来综合考虑离散时滞和分布时滞的双重作用，对如下具混合时滞的线性动力方程的全局吸引性也做了探讨：

$x^{\Delta }\left(t\right)+p\left(t\right)x\left(t-\zeta \right)+q\left(t\right){\displaystyle {\int }_0^{\tau }K\left(u\right)x\left(t-u\right)\Delta u}=0,t\in {\left[t_0,+\infty \right)}_T$ (2)

其中 $T$ 表示某一无界时标， $t\in {\left[t_0,+\infty \right)}_T=\left[t_0,+\infty \right)\cap T$ ， $x\left(t\right)$ 表示 $t$ 时刻的系统状态，函数 $p\left(t\right)$ 和 $q\left(t\right)$ 非负有界并且右密连续， $\bar{p}={\sup }_{t\in {\left[0,+\infty \right)}_T}p\left(t\right)$ ， $\bar{q}={\sup }_{t\in {\left[0,+\infty \right)}_T}q\left(t\right)$ 。 $\zeta$ 和 $\tau$ 为非负常数，并且 $t\pm \zeta \in T$ 。 $\mu \left(t\right)$ 是 $T$ 的距离函数， $\bar{\mu }={\text{sup}}_{t\in T}\mu \left(t\right)$ 。 $K\left(u\right):{\left[0,\tau \right]}_T\to \left[0,+\infty \right)$ 右连续且满足

${\displaystyle {\int }_0^{\tau }K\left(u\right)\Delta u}=1$ (3)

利用Lyapunov函数方法和引理1，本文得到方程(1)和(2)全局吸引的两个充分条件即定理1和定理2。

2. 预备知识

首先给出时标上一些基本定义和有关引理(参看文献 [1] 和 [2] )，时标 $T$ 是实数 $R$ 上任一非空子集。对

于任何 $t\in T$ ，定义 $t$ 的前跳算子为 $\sigma \left(t\right)=\left\{s\in T:s>t\right\}$ 。若 $\sigma \left(t\right)>t$ ，称t为右扩散；若 $\sigma \left(t\right)=t$ 且 $t<\text{sup}T$ ，称 $t$ 为右稠密。定义距离函数 $\mu \left(t\right)=\sigma \left(t\right)-t$ 。此外集合 $T^k$ 定义如下：如果 $T$ 的最大点为左扩散点 $T^k=T-\left\{m\right\}$ ；否则， $T^k=T$ 。若函数 $f:T\to R$ 在每个右稠密点处连续，并且在右稠密点处存在左极限，

则称 $f$ 为右稠密连续，简称右密连续。

定义1：设 $f:T\to R$ ，当且仅当对于 $\forall \epsilon >0$ ，存在 $t$ 的一个邻域 $U$ ，使得

$\left|f\left(\sigma \left(t\right)\right)-f\left(s\right)-f^{\Delta }\left(t\right)\left(\sigma \left(t\right)-s\right)\right|\le \epsilon \left|\sigma \left(t\right)-s\right|,\forall s\in U\cap T,t\in T^k$

成立，则称 $f$ 在 $t$ 处delta可导，记为 $f^{\Delta }\left(t\right)$ 。

可以证得：

1) 若 $f$ 在 $t$ 处连续且 $t$ 为右扩散点，则 $f^{\Delta }\left(t\right)=\frac{f\left(\sigma \left(t\right)\right)-f\left(t\right)}{\sigma \left(t\right)-t}$ ；

2) 若 $f$ 在 $t$ 处连续且 $t$ 为右稠密点，则 $f^{\Delta }\left(t\right)={\lim }_{s\to t}\frac{f\left(s\right)-f\left(t\right)}{s-t}$ 。

假设函数 $f,g:T\to R$ 在 $t\in T^k$ 处delta可导，则乘积函数 $fg$ 、和函数 $f\pm g$ 也都在 $t$ 处delta可导，即

${\left(fg\right)}^{\Delta }=f^{\Delta }\left(t\right)f\left(t\right)+f\left(\sigma \left(t\right)\right)g^{\Delta }\left(t\right)=f\left(t\right)g^{\Delta }\left(t\right)+f^{\Delta }\left(t\right)g\left(\sigma \left(t\right)\right)$

${\left(f\left(t\right)\pm g\left(t\right)\right)}^{\Delta }=f^{\Delta }\left(t\right)\pm g^{\Delta }\left(t\right)$

定义2：函数 $F:T^k\to R$ 被称为 $f:T\to R$ 的delta原函数，如果对所有的 $t\in T^k$ 有 $F^{\Delta }\left(t\right)=f\left(t\right)$ 成立，于是可以定义 $f$ 的积分为

${\displaystyle {\int }_0^{t}f\left(s\right)\Delta s}=F\left(t\right)-F\left(a\right),t\in T^k$

${\displaystyle {\int }_t^{\sigma \left(t\right)}f\left(s\right)\Delta s}=f\left(t\right)\mu \left(t\right),t\in T^k$

定义3：对于任一初始值，方程(1)的解 $x\left(t\right)$ 都有 ${\lim }_{t\to +\infty }x\left(t\right)=x_0$ 成立，则 $x_0$ 称为方程(1)的全局吸引子，方程(1)的解 $x\left(t\right)$ 是全局吸引的。

3. 主要结果

首先证明一个时标上的二重变上限积分的求导公式如下：

引理1：设函数 $f:T\to R$ 在 $t\in T^k$ 处delta可导，则

${\left({\displaystyle {\int }_a^t{\displaystyle {\int }_u^{t}f\left(s\right)\Delta s\Delta u}}\right)}^{\Delta }={\displaystyle {\int }_0^{t}f\left(t\right)\Delta u}+{\displaystyle {\int }_t^{\sigma \left(t\right)}f\left(s\right)\Delta s},t\in T^k$

证明：设函数 $F:T^k\to R$ 是 $f$ 的delta原函数，由定义1和2可得

${\displaystyle {\int }_a^t{\displaystyle {\int }_u^{t}f\left(s\right)\Delta s\Delta u}}={\displaystyle {\int }_a^t\left(F\left(t\right)-F\left(u\right)\right)\Delta u},t\in T^k$

和

$\begin{array}{c}{\left({\displaystyle {\int }_a^t{\displaystyle {\int }_u^{t}f\left(s\right)\Delta s\Delta u}}\right)}^{\Delta }={\left({\displaystyle {\int }_a^t\left(F\left(t\right)-F\left(u\right)\right)\Delta u}\right)}^{\Delta }\\ ={\left({\displaystyle {\int }_a^{t}F\left(t\right)\Delta u}\right)}^{\Delta }-{\left({\displaystyle {\int }_a^{t}F\left(u\right)\Delta u}\right)}^{\Delta }\\ ={\left(F\left(t\right)\left(t-a\right)\right)}^{\Delta }-F\left(t\right)\\ =f\left(t\right)\left(t-a\right)+F\left(\sigma \left(t\right)\right)-F\left(t\right)\\ ={\displaystyle {\int }_a^{t}f\left(t\right)\Delta u}+{\displaystyle {\int }_t^{\sigma \left(t\right)}f\left(s\right)\Delta s}\end{array}$

引理得证。

下面给出方程(1)的全局吸引性的一个充分条件。

定理1：如果

$\underset{t\to +\infty }{\lim }{\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}>\left(\tau +\frac{\bar{\mu }}{2}\right){\bar{q}}^2,t\in {\left[t_0,+\infty \right)}_T$

成立，其中函数 $q(\cdot )$ 非负有界并且右稠密连续， $\bar{q}={\sup }_{t\in {\left[0,+\infty \right)}_T}q\left(t\right)$ ，函数 $K\left(u\right):{\left[0,\tau \right]}_T\to \left[0,+\infty \right)$ 右连续且满足 ${\displaystyle {\int }_0^{\tau }K\left(u\right)\Delta u}=1$ ，则方程(1)的解是全局吸引的。

证明：将(1)式变形可得

$\begin{array}{l}{\left(x\left(t\right)-{\displaystyle {\int }_0^{\tau }K\left(u\right)}{\displaystyle {\int }_{t-u}^{t}q\left(s+u\right)x\left(s\right)\Delta s\Delta u}\right)}^{\Delta }\\ =-x\left(t\right){\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u},t\in {\left[t_0,+\infty \right)}_T\end{array}$ (4)

设

$y\left(t\right)=x\left(t\right)-{\displaystyle {\int }_0^{\tau }K\left(u\right)}{\displaystyle {\int }_{t-u}^{t}q\left(s+u\right)x\left(s\right)\Delta s\Delta u}$ (5)

则有

$y^{\Delta }\left(t\right)=-x\left(t\right){\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u},t\in {\left[t_0,+\infty \right)}_T$ (6)

构造Lyapunov函数 $V\left(t\right)$ 如下：

$V\left(t\right)=y^2\left(t\right)+J\left(t\right)$

其中

$J\left(t\right)={\displaystyle {\int }_0^{\tau }\bar{q}K\left(u\right)\left[{\displaystyle {\int }_{t-u}^t\left({\displaystyle {\int }_s^{t}q\left(v+u\right)x^2\left(v\right)\Delta v}\right)\Delta s}\right]\Delta u}$

显然 $V\left(t\right)\ge 0$ ，由(5)和(6)式得

$\begin{array}{c}{\left(y^2\left(t\right)\right)}^{\Delta }=y^{\Delta }\left(t\right)y\left(t\right)+y\left(\sigma \left(t\right)\right)y^{\Delta }\left(t\right)\\ =\mu \left(t\right){\left[y^{\Delta }\left(t\right)\right]}^2+2y^{\Delta }\left(t\right)y\left(t\right)\\ =\mu \left(t\right)x^2\left(t\right){\left({\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}\right)}^2-2x^2\left(t\right){\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}\\ +2x\left(t\right){\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}\cdot {\displaystyle {\int }_0^{\tau }K\left(u\right)}{\displaystyle {\int }_{t-u}^{t}q\left(s+u\right)x\left(s\right)\Delta s\Delta u}\end{array}$

由不等式 $2ab\le a^2+b^2$ 和(3)式，上式可以推得

$\begin{array}{c}{\left(y^2\left(t\right)\right)}^{\Delta }\le \bar{\mu }{x}^2\left(t\right){\left({\displaystyle {\int }_0^{\tau }K\left(t\right)\bar{q}\Delta u}\right)}^2-2x^2\left(t\right){\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}\\ +{\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}\cdot {\displaystyle {\int }_0^{\tau }K\left(u\right)\left({\displaystyle {\int }_{t-u}^{t}q\left(s+u\right)\left(x^2\left(t\right)+x^2\left(s\right)\right)\Delta s}\right)\Delta u}\\ \le \bar{\mu }{x}^2\left(t\right){\bar{q}}^2-2x^2\left(t\right){\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}\\ +\bar{q}{\displaystyle {\int }_0^{\tau }K\left(u\right)}\left({\displaystyle {\int }_{t-u}^{t}q\left(s+u\right)\left(x^2\left(t\right)+x^2\left(s\right)\right)\Delta s}\right)\Delta u\end{array}$

由引理1得

$\begin{array}{c}{J}^{\Delta }\left(t\right)={\displaystyle {\int }_0^{\tau }K\left(u\right)}\left({\displaystyle {\int }_{t-u}^t\bar{q}q\left(t+u\right)x^2\left(t\right)\Delta s}\right)\Delta u\\ +{\displaystyle {\int }_0^{\tau }K\left(u\right)}\left({\displaystyle {\int }_t^{\sigma \left(t\right)}\bar{q}q\left(v+u\right)x^2\left(v\right)\Delta v}\right)\Delta u\\ -{\displaystyle {\int }_0^{\tau }K\left(u\right)}\left({\displaystyle {\int }_{t-u}^{\sigma \left(t\right)}\bar{q}q\left(v+u\right)x^2\left(v\right)\Delta v}\right)\Delta u\\ ={\displaystyle {\int }_0^{\tau }K\left(u\right)}\left({\displaystyle {\int }_{t-u}^t\bar{q}q\left(t+u\right)x^2\left(t\right)\Delta s}\right)\Delta u\\ -{\displaystyle {\int }_0^{\tau }K\left(u\right)}\left({\displaystyle {\int }_{t-u}^t\bar{q}q\left(v+u\right)x^2\left(v\right)\Delta v}\right)\Delta u\end{array}$

而 $V^{\Delta }\left(t\right)={\left(y^2\left(t\right)\right)}^{\Delta }+J^{\Delta }\left(t\right)$ ，所以有

$\begin{array}{c}{V}^{\Delta }\left(t\right)\le \bar{\mu }{x}^2\left(t\right){\bar{q}}^2-2x^2\left(t\right){\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}\\ +\bar{q}{\displaystyle {\int }_0^{\tau }K\left(u\right)}\left({\displaystyle {\int }_{t-u}^{t}q\left(s+u\right)x^2\left(t\right)\Delta s}\right)\Delta u\\ +{\displaystyle {\int }_0^{\tau }K\left(u\right)}\left({\displaystyle {\int }_{t-u}^t\bar{q}q\left(t+u\right)x^2\left(t\right)\Delta s}\right)\Delta u\\ =x^2\left(t\right)\left[\bar{\mu }{\bar{q}}^2-2{\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}+\bar{q}{\displaystyle {\int }_0^{\tau }K\left(u\right)}\left({\displaystyle {\int }_{t-u}^{t}q\left(s+u\right)\Delta s}\right)\Delta u\right]\\ +x^2\left(t\right){\displaystyle {\int }_0^{\tau }K\left(u\right)}\left({\displaystyle {\int }_{t-u}^t\bar{q}q\left(t+u\right)\Delta s}\right)\Delta u\\ \le x^2\left(t\right)\left[\bar{\mu }{\bar{q}}^2-2{\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}+2\bar{q}{\displaystyle {\int }_0^{\tau }K\left(u\right)u\Delta u}\right]\\ \le x^2\left(t\right)\left[\bar{\mu }{\bar{q}}^2-2{\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}+2{\bar{q}}^2\tau \right]\end{array}$

如果 ${\lim }_{t\to +\infty }{\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}=L>\left(\tau +\frac{\bar{\mu }}{2}\right){\bar{q}}^2$ ，则 $V^{\Delta }\left(t\right)<0$ 当 $t\to +\infty$ 。根据单调有界准则(参看 [11] )，当 $t\to +\infty$ 时 $V\left(t\right)$ 的极限存在，可以记为 ${\lim }_{t\to +\infty }V\left(t\right)=\rho \ge 0$ 。

令，则 $\alpha >0$ ，对充分大的 $t$ ，有 $V^{\Delta }\left(t\right)\le -\alpha x^2\left(t\right)$ 。设 $T^{*}$ 充分大，使得

${\displaystyle {\int }_{T^{*}}^{+\infty }{V}^{\Delta }\left(t\right)\Delta t}\le -{\displaystyle {\int }_{T^{*}}^{+\infty }\alpha x^2\left(t\right)\Delta t}$

于是

${\displaystyle {\int }_{T^{*}}^{+\infty }\alpha x^2\left(t\right)\Delta t}<V\left(T^{*}\right)-\rho <+\infty$

所以有 ${\lim }_{t\to \infty }x\left(t\right)=0$ ，定理得证。

接下来给出方程(2)的全局吸引性的一个充分条件。

定理2：如果对于 $t\in {\left[t_0,+\infty \right)}_T$ 有

$\underset{t\to +\infty }{\lim }{\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}>\left(\bar{\mu }+\frac{\tau +1}{2}\right){\bar{q}}^2+\tau {\bar{q}}^3$

$\underset{t\to +\infty }{\lim }p\left(t+\zeta \right)>\left(\bar{\mu }+\frac{{\zeta }^2}{2}+\zeta +\frac{\tau }{2}\right){\bar{p}}^2$

成立，其中 $p(\cdot ),q(\cdot )$ 为非负有界函数并且右稠密连续， $\bar{p}={\sup }_{t\in {\left[0,+\infty \right)}_T}p\left(t\right)$ ， $\bar{q}={\sup }_{t\in {\left[0,+\infty \right)}_T}q\left(t\right)$ 。 $\zeta$ 和 $\tau$ 为非负常数，并且 $t\pm \zeta \in T$ 。 $\bar{\mu }={\text{sup}}_{t\in T}\mu \left(t\right)$ 。 $K\left(u\right):{\left[0,\tau \right]}_T\to \left[0,+\infty \right)$ 右稠密连续且满足 ${\displaystyle {\int }_0^{\tau }K\left(u\right)\Delta u}=1$ ，

则方程(2)的解是全局吸引的。

证明：(2)式变形可得：

$\begin{array}{l}{\left(x\left(t\right)-{\displaystyle {\int }_{t-\zeta }^{t}p\left(s+\zeta \right)x\left(s\right)\Delta s}-{\displaystyle {\int }_0^{\tau }K\left(u\right)}{\displaystyle {\int }_{t-u}^{t}q\left(s+u\right)x\left(s\right)\Delta s\Delta u}\right)}^{\Delta }\\ =-p\left(t+\zeta \right)x\left(t\right)-x\left(t\right){\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u},t\in {\left[t_0,+\infty \right)}_T\end{array}$ (7)

设

$y\left(t\right)=x\left(t\right)-{\displaystyle {\int }_{t-\zeta }^{t}p\left(s+\zeta \right)x\left(s\right)\Delta s}-{\displaystyle {\int }_0^{\tau }K\left(u\right)}{\displaystyle {\int }_{t-u}^{t}q\left(s+u\right)x\left(s\right)\Delta s\Delta u}$ (8)

则有

$y^{\Delta }\left(t\right)=-p\left(t+\zeta \right)x\left(t\right)-x\left(t\right){\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u},t\in {\left[t_0,+\infty \right)}_T$ (9)

构造Lyapunov函数 $V\left(t\right)$ 如下：

$V\left(t\right)=y^2\left(t\right)+R\left(t\right)+W\left(t\right)$

其中

$R\left(t\right)=\left(\zeta +1\right){\displaystyle {\int }_{t-\zeta }^t{\displaystyle {\int }_s^t{p}^2\left(u+\zeta \right)x^2\left(u\right)\Delta u\Delta s}}$

$W\left(t\right)={\displaystyle {\int }_0^{\tau }\left(\bar{q}+1\right)K\left(u\right)}\left[{\displaystyle {\int }_{t-u}^t\left({\displaystyle {\int }_s^t{q}^2\left(v+u\right)x^2\left(v\right)\Delta v}\right)\Delta s}\right]\text{Δ}u$

显然 $V\left(t\right)\ge 0$ ，由(7)和(8)式得

$\begin{array}{c}{\left(y^2\left(t\right)\right)}^{\Delta }=\mu \left(t\right){\left[y^{\Delta }\left(t\right)\right]}^2+2y^{\Delta }\left(t\right)y\left(t\right)\\ =\mu \left(t\right)x^2\left(t\right){\left(p\left(t+\zeta \right)+{\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}\right)}^2-2x^2\left(t\right)p\left(t+\zeta \right)\\ -2x^2\left(t\right){\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}+2x\left(t\right)p\left(t+\zeta \right){\displaystyle {\int }_{t-\zeta }^{t}p\left(s+\zeta \right)x\left(s\right)\Delta s}\\ +2x\left(t\right)p\left(t+\zeta \right){\displaystyle {\int }_0^{\tau }K\left(u\right)}{\displaystyle {\int }_{t-u}^{t}q\left(s+u\right)x\left(s\right)\Delta s\Delta u}\\ +2x\left(t\right){\displaystyle {\int }_{t-\zeta }^{t}p\left(s+\zeta \right)x\left(s\right)\Delta s}\cdot {\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}\\ +2x\left(t\right){\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}\cdot {\displaystyle {\int }_0^{\tau }K\left(u\right)}{\displaystyle {\int }_{t-u}^{t}q\left(s+u\right)x\left(s\right)\Delta s\Delta u}\end{array}$

由不等式 $2ab\le a^2+b^2$ 和(3)式，上式可以推得

$\begin{array}{c}{\left(y^2\left(t\right)\right)}^{\Delta }\le 2\mu \left(t\right)x^2\left(t\right){\left({\displaystyle {\int }_0^{\tau }K\left(t\right)q\left(t+u\right)\Delta u}\right)}^2+2\mu \left(t\right)x^2\left(t\right)p^2\left(t+\zeta \right)\\ -2x^2\left(t\right)p\left(t+\zeta \right)-2x^2\left(t\right){\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}\\ +{\displaystyle {\int }_{t-\zeta }^t\left(p^2\left(t+\zeta \right)x^2\left(t\right)+p^2\left(s+\zeta \right)x^2\left(s\right)\right)\Delta s}\\ +{\displaystyle {\int }_0^{\tau }K\left(u\right)}\left({\displaystyle {\int }_{t-u}^t\left(p^2\left(t+\zeta \right)x^2\left(t\right)+q^2\left(s+u\right)x^2\left(s\right)\right)\Delta s}\right)\Delta u\\ +x^2\left(t\right){\left({\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}\right)}^2+{\left({\displaystyle {\int }_{t-\zeta }^{t}p\left(s+\zeta \right)x\left(s\right)\Delta s}\right)}^2\\ +\bar{q}{\displaystyle {\int }_0^{\tau }K\left(u\right)}{\displaystyle {\int }_{t-u}^{t}q\left(s+u\right)\left(x^2\left(t\right)+x^2\left(s\right)\right)\Delta s\Delta u}\end{array}$

由引理1得

$\begin{array}{c}{R}^{\Delta }\left(t\right)=\left(\zeta +1\right)[{\displaystyle {\int }_{t-\zeta }^t{p}^2\left(t+\zeta \right)x^2\left(t\right)\Delta s}+{\displaystyle {\int }_t^{\sigma \left(t\right)}{p}^2\left(u+\zeta \right)x^2\left(u\right)\Delta u}\\ -{\displaystyle {\int }_{t-\zeta }^{\sigma \left(t\right)}{p}^2\left(u+\zeta \right)x^2\left(u\right)\Delta u}]\\ =\left(\zeta +1\right)\left[\zeta p^2\left(t+\zeta \right)x^2\left(t\right)-{\displaystyle {\int }_{t-\zeta }^t{p}^2\left(u+\zeta \right)x^2\left(u\right)\Delta u}\right]\end{array}$

和

$\begin{array}{c}{W}^{\text{Δ}}\left(t\right)=\left(\bar{q}+1\right)[{\displaystyle {\int }_0^{\tau }K\left(u\right)}\left({\displaystyle {\int }_{t-u}^t{q}^2\left(t+u\right)x^2\left(t\right)\Delta s}\right)\text{Δ}u+{\displaystyle {\int }_0^{\tau }K\left(u\right)}\left({\displaystyle {\int }_t^{\sigma \left(t\right)}{q}^2\left(v+u\right)x^2\left(v\right)\Delta v}\right)\text{Δ}u\\ -{\displaystyle {\int }_0^{\tau }K\left(u\right)}\left({\displaystyle {\int }_{t-u}^{\sigma \left(t\right)}{q}^2\left(v+u\right)x^2\left(v\right)\Delta v}\right)\text{Δ}u]\\ =\left(\bar{q}+1\right)\left[{\displaystyle {\int }_0^{\tau }K\left(u\right)q^2\left(t+u\right)x^2\left(t\right)u\text{Δ}u}-{\displaystyle {\int }_0^{\tau }K\left(u\right)}\left({\displaystyle {\int }_{t-u}^t{q}^2\left(v+u\right)x^2\left(v\right)\Delta v}\right)\text{Δ}u\right]\end{array}$

同时利用不等式 ${\left(A_1+A_2+\cdots +A_n\right)}^2\le n\left(A_1^2+A_2^2+\cdots +A_n^2\right)$ 有

$\begin{array}{c}{V}^{\Delta }\left(t\right)\le 2\mu \left(t\right)x^2\left(t\right){\left({\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}\right)}^2+2\mu \left(t\right)x^2\left(t\right)p^2\left(t+\zeta \right)\\ -2x^2\left(t\right)p\left(t+\zeta \right)-2x^2\left(t\right){\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}\\ +\left({\zeta }^2+2\zeta \right)p^2\left(t+\zeta \right)x^2\left(t\right)+{\displaystyle {\int }_0^{\tau }K\left(u\right)}\left({\displaystyle {\int }_{t-u}^t{p}^2\left(t+\zeta \right)x^2\left(t\right)\Delta s}\right)\Delta u\\ +x^2\left(t\right){\left({\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}\right)}^2+\bar{q}{\displaystyle {\int }_0^{\tau }K\left(u\right)}\left({\displaystyle {\int }_{t-u}^t{q}^2\left(s+u\right)x^2\left(t\right)\Delta s}\right)\Delta u\\ +\left(\bar{q}+1\right){\displaystyle {\int }_0^{\tau }K\left(u\right)q^2\left(t+u\right)x^2\left(t\right)u\Delta u}\\ \le x^2\left(t\right)[2\bar{\mu }{\bar{q}}^2+2\bar{\mu }{\bar{p}}^2-2p\left(t+\zeta \right)-2{\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}+\left({\zeta }^2+2\zeta \right){\bar{p}}^2\\ +{\bar{p}}^2{\displaystyle {\int }_0^{\tau }K\left(u\right)u\Delta u}+{\bar{q}}^2+{\bar{q}}^3{\displaystyle {\int }_0^{\tau }K\left(u\right)u\Delta u}+\left({\bar{q}}^2+{\bar{q}}^3\right){\displaystyle {\int }_0^{\tau }K\left(u\right)u\Delta u}]\\ \le x^2\left(t\right)\left[\left(2\bar{\mu }+{\zeta }^2+2\zeta \right){\bar{p}}^2+{\bar{p}}^2{\displaystyle {\int }_0^{\tau }K\left(u\right)u\Delta u}-2p\left(t+\zeta \right)\right]\\ +x^2\left(t\right)\left[-2{\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}+2\bar{\mu }{\bar{q}}^2+{\bar{q}}^2+\left(2{\bar{q}}^3+{\bar{q}}^2\right){\displaystyle {\int }_0^{\tau }K\left(u\right)u\Delta u}\right]\end{array}$

如果 ${\lim }_{t\to +\infty }{\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}=M>\left(\bar{\mu }+\frac{\tau +1}{2}\right){\bar{q}}^2+\tau {\bar{q}}^3$ ，而且 ${\lim }_{t\to +\infty }p\left(t+\zeta \right)=N>\left(\bar{\mu }+\frac{{\zeta }^2}{2}+\zeta +\frac{\tau }{2}\right){\bar{p}}^2$ 则 $V^{\Delta }\left(t\right)<0$ 当 $t\to +\infty$ 。根据单调有界准则(参看 [11] )，当 $t\to +\infty$ 时 $V\left(t\right)$ 的极限存在，可以记为 ${\lim }_{t\to +\infty }V\left(t\right)=\rho \ge 0$ 。

令，则 $\alpha >0$ ，对充分大的 $t$ ，有 $V^{\Delta }\left(t\right)\le -\alpha x^2\left(t\right)$ 。设 $T^{*}$ 充分大，使得

${\displaystyle {\int }_{T^{*}}^{+\infty }{V}^{\Delta }\left(t\right)\Delta t}\le -{\displaystyle {\int }_{T^{*}}^{+\infty }\alpha x^2\left(t\right)\Delta t}$

于是

${\displaystyle {\int }_{T^{*}}^{+\infty }\alpha x^2\left(t\right)\Delta t}<V\left(T^{*}\right)-\rho <+\infty$

所以有 ${\lim }_{t\to \infty }x\left(t\right)=0$ ，定理得证。

4. 一些例子

在这一部分，文章给出两个例子以验证结论的有效性，并展示了在时标上讨论动力方程的优势。

例1 对方程(1)，设 $\tau =1$ ， $q\left(t\right)=\frac{1}{3}-\frac{1}{t}$ ， $K\left(u\right)=1$ ，

1) 考虑 $T=R$ ，则原方程为

$x^{\prime }\left(t\right)+\left(\frac{1}{3}-\frac{1}{t}\right){\displaystyle {\int }_0^{1}x\left(t-u\right)\text{d}u}=0$ (10)

显然有 ${\lim }_{t\to +\infty }{\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}={\lim }_{t\to +\infty }{\displaystyle {\int }_0^1\left(\frac{1}{3}-\frac{1}{t+u}\right)\text{d}u}=\frac{1}{3}>\left(\tau +\frac{\bar{\mu }}{2}\right){\bar{q}}^2$ ，根据定理1，(10)的解是全局吸引的。

2) 考虑 $T=Z$ ，在原方程为

$x^{\Delta }\left(t\right)+\left(\frac{1}{3}-\frac{1}{t}\right){\displaystyle {\int }_0^{1}x\left(t-u\right)\text{d}}u=0$ (11)

显然有 ${\lim }_{t\to +\infty }{\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}={\lim }_{t\to +\infty }\left(\frac{1}{3}-\frac{1}{t+u}\right)=\frac{1}{3}>\left(\tau +\frac{\bar{\mu }}{2}\right){\bar{q}}^2$ ，根据定理1，(11)的解是全局吸引的。

例2 对于方程(2)，考虑 $T=\left[2k,2k+1\right]$ ， $k=0,1,2,\cdots$ ，设 $\tau =1$ ， $\zeta =2$ ， $K\left(u\right)=1$ ， $p\left(t\right)=\frac{1}{6}-\frac{1}{t^2}$ ， $q\left(t\right)=\frac{1}{5}-\frac{1}{t}$ ，则原方程为

$x^{\Delta }\left(t\right)+\left(\frac{1}{6}-\frac{1}{t^2}\right)x\left(t-\zeta \right)+\left(\frac{1}{5}-\frac{1}{t}\right){\displaystyle {\int }_0^{1}x\left(t-u\right)\text{d}u}=0$ (12)

于是有

$\underset{t\to +\infty }{\lim }{\displaystyle {\int }_0^{\tau }K\left(u\right)q\left(t+u\right)\Delta u}=\{\begin{array}{l}\underset{t\to +\infty }{\lim }q\left(t\right)=\frac{1}{5}t\in \left[2k+1,2k+2\right)\\ \underset{t\to +\infty }{\lim }{\displaystyle {\int }_t^{t+1}q\left(s\right)\text{d}s}=\frac{1}{5}t\in \left[2k,2k+1\right)\end{array}$

可见 $\frac{1}{5}>\left(\bar{\mu }+\frac{\tau +1}{2}\right)\bar{q}+\tau {\bar{q}}^3$ ，且 ${\lim }_{t\to +\infty }p\left(t+\zeta \right)=\frac{1}{6}>\left(\bar{\mu }+\frac{{\zeta }^2}{2}+\zeta +\frac{\tau }{2}\right){\bar{p}}^2$ ，由定理2，可知(12)是全局吸引的。

基金项目

本文由福建省中青年教师教育科研项目基金资助(JA15649)。