二阶线性脉冲中立型时滞微分方程的稳定性

Stability of Second Order Linear Impulsive Neutral Delay Differential Equations

摘要: 运用特征方程实根的方法研究具有常系数的二阶线性脉冲中立型时滞微分方程解的渐近性形态和稳定性，最后给出实例证明。

Abstract: By using the method of real root of characteristic equation, the asymptotic behavior and stability of solutions of Second Order Linear Impulsive Neutral Delay Differential Equations with constant coefficients are studied. Finally, an example is given to prove the stability.

1. 引言

脉冲时滞微分方程在工程学，医学，金融，力学等方面有着非常广的应用，这就意味着方程的稳定性非常重要。学者在这方面做了大量的研究 [1] [2] [3]。文献 [4] 给出了变系数二阶时滞微分方程解的稳定性的充要条件。文献 [5] 运用特征方程法给出了一阶线性脉冲时滞微分方程解的稳定性，在此基础上，本文将进一步研究脉冲时滞微分方程，将此方法推广到二阶线性脉冲中立型时滞微分方程。

2. 预备知识

论文中，考虑的二阶中立型脉冲时滞微分方程如下：

${\left[x\left(t\right)+cx\left(t-\sigma \right)\right]}^{\prime \text{}\prime }=ax\left(t\right)+bx\left(t-\tau \right),t\ne t_k,t\ge 0,$ (1)

$\Delta x\left(t_k\right)=l_k,k\in z^{+},$ (2)

$x\left(t\right)=\varphi \left(t\right),t\in \left[-h,0\right],$ (3)

其中：1) $\sigma ,\tau \in z^{+}$， $a,b,c\in R$ ；

2) $\Delta x\left(t_k\right)=x\left(t_k^{+}\right)-x\left(t_k^{-}\right)$ ；

3) 脉冲点 $t_k$ 满足 $0<t_1<t_2<\cdots <t_k<t_{k+1}<\cdots ,\underset{t\to \infty }{\lim }{t}_k=\infty$ ；

4) $h=\max \left\{\sigma ,\tau \right\}$。

定义1.1 $x\left(t\right)$ 是方程(1)~(3)的解，如果

1) $x\left(t\right)$ 在 $t\notin z^{+}$ 处连续；

2) $x\left(t\right)+cx\left(t-\sigma \right)$ 在 $t\ne t_k$ 处二阶连续可微；

3) $x\left(t_k^{+}\right)=x\left(t_k\right)$ ；

4) ${\left[x\left(t\right)+cx\left(t-\sigma \right)\right]}^{\prime \text{}\prime }=ax\left(t\right)+bx\left(t-\tau \right)$。

注1.2 设 $x\left(t\right)={\text{e}}^{{\lambda }_{0}t}$ 带入(1)得特征方程

${\lambda }^2\left(1+c{\text{e}}^{-\lambda \sigma }\right)=a+b{\text{e}}^{-\lambda \tau }$, (4)

假设 ${\lambda }_0$ 是方程(4)的实根，令

$y\left(t\right)={\text{e}}^{-{\lambda }_{0}t}x\left(t\right)$, (5)

把(5)分别代入(1)~(3)，得

$\begin{array}{l}{\left[y\left(t\right)+c{\text{e}}^{-{\lambda }_0\sigma }y\left(t-\sigma \right)\right]}^{\prime \text{}\prime }\\ =\left(a-{\lambda }_0^2\right)y\left(t\right)+b{\text{e}}^{-{\lambda }_0\tau }\left(y-\tau \right)-c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }y\left(t-\sigma \right)-2{\lambda }_0{y}^{\prime }\left(t\right)-2c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{y}^{\prime }\left(t-\sigma \right),\end{array}$ (6)

$y\left(t_k\right)-y\left(t_k^{-}\right)=l_k{\text{e}}^{-{\lambda }_0{t}_k},k\in z^{+},$ (7)

$y\left(t\right)={\text{e}}^{-{\lambda }_{0}t}\varphi \left(t\right),t\in \left[-h,0\right],$ (8)

引理1.3 假设 ${\lambda }_0$ 是(4)的实根，并且 $t_k-\sigma$ 不是脉冲点。如果 $y\left(t\right)$ 是(6)~(8)的唯一解，当且仅当

$y\left(t\right)={\text{e}}^{-{\lambda }_{0}t}\varphi \left(t\right),t\in \left[-h,0\right]$, (9)

$\begin{array}{l}y\left(t\right)+c{\text{e}}^{-{\lambda }_0\sigma }y\left(t-\sigma \right)\\ =\varphi \left(0\right)+c\varphi \left(-\sigma \right)+{\displaystyle {\sum }_{i=1}^{\infty }{l}_i{\text{e}}^{-{\lambda }_0{t}_i}}+b{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{-\tau }{\overset{0}{\int }}{\displaystyle \underset{-\tau }{\overset{m}{\int }}{\text{e}}^{-{\lambda }_{0}s}\varphi \left(s\right)\text{d}s\text{d}m}}\\ -c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{-\sigma }{\overset{0}{\int }}{\displaystyle \underset{-\sigma }{\overset{m}{\int }}{\text{e}}^{-{\lambda }_{0}s}\varphi \left(s\right)\text{d}s\text{d}m}}-2c{\lambda }_0{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{-\sigma }{\overset{0}{\int }}{\displaystyle \underset{-\sigma }{\overset{m}{\int }}{\text{e}}^{-{\lambda }_{0}s}{\varphi }^{\prime }\left(s\right)\text{d}s\text{d}m}}\\ +\left(a-{\lambda }_0^2\right){\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}y\left(s\right)\text{d}s\text{d}m}}-2{\lambda }_0{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}{y}^{\prime }\left(s\right)\text{d}s\text{d}m}}+b{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{0}{\overset{t-\tau }{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}y\left(s\right)\text{d}s\text{d}m}}\\ -c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{0}{\overset{t-\sigma }{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}y\left(s\right)\text{d}s\text{d}m}}-2c{\lambda }_0{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{0}{\overset{t-\sigma }{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}{y}^{\prime }}}\left(s\right)\text{d}s\text{d}m,\end{array}$

证明：当 $t\in \left[t_k,t_{k+1}\right)$

$\begin{array}{l}y\left(t\right)+c{\text{e}}^{-{\lambda }_0\sigma }y\left(t-\sigma \right)\\ =y\left(t_k\right)+c{\text{e}}^{-{\lambda }_0\sigma }y\left(t_k-\sigma \right)+\left(a-{\lambda }_0^2\right){\displaystyle \underset{t_k}{\overset{t}{\int }}{\displaystyle \underset{t_k}{\overset{m}{\int }}y\left(s\right)\text{d}s\text{d}m}}\\ +b{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{t_k}{\overset{t}{\int }}{\displaystyle \underset{t_k}{\overset{m}{\int }}y\left(s-\tau \right)\text{d}s\text{d}m}}-c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{t_k}{\overset{t}{\int }}{\displaystyle \underset{t_k}{\overset{m}{\int }}y\left(s-\sigma \right)\text{d}s\text{d}m}}\\ -2{\lambda }_0{\displaystyle \underset{t_k}{\overset{t}{\int }}{\displaystyle \underset{t_k}{\overset{m}{\int }}{y}^{\prime }\left(s\right)\text{d}s\text{d}m}}-2c{\lambda }_0{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{t_k}{\overset{t}{\int }}{\displaystyle \underset{t_k}{\overset{m}{\int }}{y}^{\prime }}\left(s-\sigma \right)\text{d}s}\text{d}m,\end{array}$

因为 $t_k-\sigma$ 不是脉冲点，并且根据(7)，得

$\begin{array}{l}y\left(t\right)+c{\text{e}}^{-{\lambda }_0\sigma }y\left(t-\sigma \right)\\ =l_k{\text{e}}^{-{\lambda }_0{t}_k}+y\left(t_k^{-}\right)+c{\text{e}}^{-{\lambda }_0\sigma }y\left(t_k-\sigma \right)+\left(a-{\lambda }_0^2\right){\displaystyle \underset{t_k}{\overset{t}{\int }}{\displaystyle \underset{t_k}{\overset{m}{\int }}y\left(s\right)\text{d}s\text{d}m}}\\ +b{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{t_k}{\overset{t}{\int }}{\displaystyle \underset{t_k}{\overset{m}{\int }}y\left(s-\tau \right)\text{d}s\text{d}m}}-c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{t_k}{\overset{t}{\int }}{\displaystyle \underset{t_k}{\overset{m}{\int }}y\left(s-\sigma \right)\text{d}s\text{d}m}}\\ -2{\lambda }_0{\displaystyle \underset{t_k}{\overset{t}{\int }}{\displaystyle \underset{t_k}{\overset{m}{\int }}{y}^{\prime }\left(s\right)\text{d}s\text{d}m}}-2c{\lambda }_0{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{t_k}{\overset{t}{\int }}{\displaystyle \underset{t_k}{\overset{m}{\int }}{y}^{\prime }}\left(s-\sigma \right)\text{d}s}\text{d}m,\end{array}$ (10)

因为 $k\in \left[0,\infty \right)$，所以

$\begin{array}{l}y\left(t_k^{-}\right)+c{\text{e}}^{-{\lambda }_0\sigma }y\left(t_k-\sigma \right)\\ =l_{k-1}{\text{e}}^{-{\lambda }_0{t}_k}+y\left(t_{k-1}^{-}\right)+c{\text{e}}^{-{\lambda }_0\sigma }y\left(t_{k-1}-\sigma \right)\\ +\left(a-{\lambda }_0^2\right){\displaystyle \underset{t_{k-1}}{\overset{t}{\int }}{\displaystyle \underset{t_{k-1}}{\overset{m}{\int }}y\left(s\right)\text{d}s\text{d}m}}+b{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{t_{k-1}}{\overset{t}{\int }}{\displaystyle \underset{t_{k-1}}{\overset{m}{\int }}y\left(s-\tau \right)\text{d}s\text{d}m}}\\ -c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{t_{k-1}}{\overset{t}{\int }}{\displaystyle \underset{t_{k-1}}{\overset{m}{\int }}y\left(s-\sigma \right)\text{d}s\text{d}m}}-2{\lambda }_0{\displaystyle \underset{t_{k-1}}{\overset{t}{\int }}{\displaystyle \underset{t_{k-1}}{\overset{m}{\int }}{y}^{\prime }\left(s\right)\text{d}s\text{d}m}}\\ -2c{\lambda }_0{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{t_{k-1}}{\overset{t}{\int }}{\displaystyle \underset{t_{k-1}}{\overset{m}{\int }}{y}^{\prime }}\left(s-\sigma \right)\text{d}s}\text{d}m,\end{array}$

$⋮$

$\begin{array}{l}y\left(t_2^{-}\right)+c{\text{e}}^{-{\lambda }_0\sigma }y\left(t_2-\sigma \right)\\ =l_1{\text{e}}^{-{\lambda }_0{t}_k}+y\left(t_1^{-}\right)+c{\text{e}}^{-{\lambda }_0\sigma }y\left(t_1-\sigma \right)\\ +\left(a-{\lambda }_0^2\right){\displaystyle \underset{t_1}{\overset{t}{\int }}{\displaystyle \underset{t_1}{\overset{m}{\int }}y\left(s\right)\text{d}s\text{d}m}}+b{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{t_1}{\overset{t}{\int }}{\displaystyle \underset{t_1}{\overset{m}{\int }}y}\left(s-\tau \right)\text{d}s\text{d}m}\\ -c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{t_1}{\overset{t}{\int }}{\displaystyle \underset{t_1}{\overset{m}{\int }}y}\left(s-\sigma \right)\text{d}s}\text{d}m-2{\lambda }_0{\displaystyle \underset{t_1}{\overset{t}{\int }}{\displaystyle \underset{t_1}{\overset{m}{\int }}{y}^{\prime }\left(s\right)\text{d}s\text{d}m}}\\ -2c{\lambda }_0{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{t_1}{\overset{t}{\int }}{\displaystyle \underset{t_1}{\overset{m}{\int }}{y}^{\prime }}\left(s-\sigma \right)\text{d}s}\text{d}m,\end{array}$

$\begin{array}{l}y\left(t_1^{-}\right)+c{\text{e}}^{-{\lambda }_0\sigma }y\left(t_1-\sigma \right)\\ =l_0{\text{e}}^{-{\lambda }_0{t}_k}+y\left(t_0^{-}\right)+c{\text{e}}^{-{\lambda }_0\sigma }y\left(t_0-\sigma \right)\\ +\left(a-{\lambda }_0^2\right){\displaystyle \underset{t_0}{\overset{t}{\int }}{\displaystyle \underset{t_0}{\overset{m}{\int }}y\left(s\right)\text{d}s\text{d}m}}+b{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{t_0}{\overset{t}{\int }}{\displaystyle \underset{t_0}{\overset{m}{\int }}y}\left(s-\tau \right)\text{d}s\text{d}m}\\ -c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{t_0}{\overset{t}{\int }}{\displaystyle \underset{t_0}{\overset{m}{\int }}y}\left(s-\sigma \right)\text{d}s}\text{d}m-2{\lambda }_0{\displaystyle \underset{t_0}{\overset{t}{\int }}{\displaystyle \underset{t_0}{\overset{m}{\int }}{y}^{\prime }\left(s\right)\text{d}s\text{d}m}}\\ -2c{\lambda }_0{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{t_0}{\overset{t}{\int }}{\displaystyle \underset{t_0}{\overset{m}{\int }}{y}^{\prime }}\left(s-\sigma \right)\text{d}s}\text{d}m,\end{array}$

根据(9)，递归代回(10)，得

$\begin{array}{l}y\left(t\right)+c{\text{e}}^{-{\lambda }_0\sigma }y\left(t-\sigma \right)\\ =\varphi \left(0\right)+c\varphi \left(-\sigma \right)+{\displaystyle \underset{i=1}{\overset{\infty }{\sum }}{l}_i{\text{e}}^{-{\lambda }_0{t}_i}}+\left(a-{\lambda }_0^2\right){\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}y\left(s\right)\text{d}s\text{d}m}}\\ +b{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}y}\left(s-\tau \right)\text{d}s\text{d}m}-c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}y}\left(s-\sigma \right)\text{d}s}\text{d}m\\ -2{\lambda }_0{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}{y}^{\prime }\left(s\right)\text{d}s\text{d}m}}-2c{\lambda }_0{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}{y}^{\prime }}\left(s-\sigma \right)\text{d}s}\text{d}m,\end{array}$

同时，根据初始条件(8)可得

$\begin{array}{l}y\left(t\right)+c{\text{e}}^{-{\lambda }_0\sigma }y\left(t-\sigma \right)\\ =\varphi \left(0\right)+c\varphi \left(-\sigma \right)+{\displaystyle \underset{i=1}{\overset{\infty }{\sum }}{l}_i{\text{e}}^{-{\lambda }_0{t}_i}}+b{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{-\tau }{\overset{0}{\int }}{\displaystyle \underset{-\tau }{\overset{m}{\int }}{\text{e}}^{-{\lambda }_{0}s}\varphi }\left(s\right)\text{d}s\text{d}m}\\ -c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{-\sigma }{\overset{0}{\int }}{\displaystyle \underset{-\sigma }{\overset{m}{\int }}{\text{e}}^{-{\lambda }_{0}s}\varphi }\left(s\right)\text{d}s\text{d}m}-2c{\lambda }_0{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{-\sigma }{\overset{0}{\int }}{\displaystyle \underset{-\sigma }{\overset{m}{\int }}{\text{e}}^{-{\lambda }_{0}s}{\varphi }^{\prime }}\left(s\right)\text{d}s\text{d}m}\\ +\left(a-{\lambda }_0^2\right){\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}y\left(s\right)\text{d}s\text{d}m}}-2{\lambda }_0{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}{y}^{\prime }\left(s\right)\text{d}s\text{d}m}}+b{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{0}{\overset{t-\tau }{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}y}\left(s\right)\text{d}s\text{d}m}\\ -c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{0}{\overset{t-\sigma }{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}y}\left(s\right)\text{d}s\text{d}m}-2c{\lambda }_0{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{0}{\overset{t-\sigma }{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}{y}^{\prime }}\left(s\right)\text{d}s}\text{d}m,\end{array}$ (11)

对(11)两边同时求二阶导，可推出方程(6)，同时方程(7)显然可得。引理1.3得证。

推论1.4 设 ${\lambda }_0$ 是(4)的实根，当 $k\in z^{+}$ 时， $t_k-\sigma$ 不是脉冲点。那么 $x\left(t\right)$ 是方程(1)~(3)的唯一解，当且仅当

$y\left(t\right)={\text{e}}^{-{\lambda }_{0}t}x\left(t\right),t\in \left[-h,\infty \right]$

是(6)满足初始条件

$y\left(t\right)={\text{e}}^{-{\lambda }_{0}t}x\left(t\right),t\in \left[-h,0\right]$

的解。

定义1.3 [6] 1) 若(1)~(2)的零解是稳定的，则 $\forall \delta ,\tau >0$， $\exists \delta \left(\sigma ,\tau \right)>0$，对任意的起始函数 $\varphi$ 有

$\Vert \varphi \Vert =\underset{-h\le t\le 0}{\max }\left|\varphi \left(t\right)\right|<\delta \left(\sigma ,\tau \right),\Vert {\varphi }^{\prime }\Vert =\underset{-h\le t\le 0}{\max }\left|{\varphi }^{\prime }\left(t\right)\right|<\delta \left(\sigma ,\tau \right)$

且 $x\left(t\right)=x\left(t;0,\varphi \right)$ 是(1)~(2)的解，那么对 $\forall t\in \left[-h,\infty \right)$

$x\left(t;0,\varphi \right)<\epsilon$。

2) 若(1)~(2)的零解是一致稳定，则 $\delta$ 与 $\sigma ,\tau$ 无关；

3) 若(1)~(2)是零解是渐近稳定的，则 $\exists {\delta }_0>0$，对任意的起始函数 $\varphi$ 有 $\left|\varphi \right|<{\delta }_0\left(\sigma ,\tau \right)$， $\left|{\varphi }^{\prime }\right|<{\delta }_0\left(\sigma ,\tau \right)$ 且 $x\left(t\right)=x\left(t;0,\varphi \right)$ 是(1)~(2)的解，那么对 $\forall t\in \left[-h,\infty \right)$

$\underset{t\to \infty }{\lim }x\left(t;0,\varphi \right)=0$。

3. 解的渐近形态

本章节介绍的是满足初始条件(1)~(3)方程解的渐近性。

定理2.1 若根 ${\lambda }_0$ 满足

$\mu \left({\lambda }_0\right)=\left|c\right|{\text{e}}^{-{\lambda }_0\sigma }+{\displaystyle \underset{i=1}{\overset{\infty }{\sum }}\left|l_i\right|}{\text{e}}^{-{\lambda }_0{t}_i}+\frac{1}{2}\left|c\right|{\sigma }^2{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }+\frac{1}{2}\left|b\right|{\tau }^2{\lambda }_0^2{\text{e}}^{-{\lambda }_0\tau }+{\sigma }^2\left|G\left({\lambda }_0\right)\right|<1,$ (12)

则(1)~(3)的解x满足

$\underset{t\to \infty }{\lim }\left[{\text{e}}^{-{\lambda }^{0}t}x\left(t\right)\right]=\frac{L\left({\lambda }_0;\varphi \right)}{1+2\beta \left({\lambda }_0\right)}.$ (13)

其中： $G\left({\lambda }_0\right)=\max \left\{{\lambda }_0,{\lambda }_0{\text{e}}^{-{\lambda }_0\sigma }\right\}$

$\begin{array}{c}L\left({\lambda }_0;\varphi \right)=\varphi \left(0\right)+c\varphi \left(-\sigma \right)+b{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{-\tau }{\overset{0}{\int }}{\displaystyle \underset{-\tau }{\overset{m}{\int }}{\text{e}}^{-{\lambda }_{0}s}\varphi }\left(s\right)\text{d}s\text{d}m}\\ -c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{-\sigma }{\overset{0}{\int }}{\displaystyle \underset{-\sigma }{\overset{m}{\int }}{\text{e}}^{-{\lambda }_{0}s}\varphi }\left(s\right)\text{d}s\text{d}m}-2c{\lambda }_0{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{-\sigma }{\overset{0}{\int }}{\displaystyle \underset{-\sigma }{\overset{m}{\int }}{\text{e}}^{-{\lambda }_{0}s}{\varphi }^{\prime }}\left(s\right)\text{d}s\text{d}m},\end{array}$ (14)

$\beta \left({\lambda }_0\right)={\tau }^{2}b{\text{e}}^{-{\lambda }_0\tau }+c{\text{e}}^{-{\lambda }_0\sigma }\left(2-{\lambda }^2{\sigma }^2\right),$ (15)

证明：由2.1可得 $\frac{1}{2}\left|\beta \left({\lambda }_0\right)\right|+{\displaystyle \underset{i=1}{\overset{\infty }{\sum }}\left|l_i\right|{\text{e}}^{-{\lambda }_0{t}_i}}<\mu \left({\lambda }_0\right)<1$，所以 $0<1+\frac{1}{2}\beta \left({\lambda }_0\right)<2$。求x是(1)~(3)的解等价于求y是(6)~(8)的解。因为 ${\lambda }_0$ 是(4)的实根，则对 $\forall t>0$，有

$\begin{array}{l}y\left(t\right)+c{\text{e}}^{-{\lambda }_0\sigma }y\left(t-\sigma \right)\\ =L\left({\lambda }_0;\varphi \right)+{\displaystyle \underset{i=1}{\overset{\infty }{\sum }}{l}_i{\text{e}}^{-{\lambda }_0{t}_i}}+\left(a-{\lambda }_0^2\right){\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}y\left(s\right)\text{d}s\text{d}m}}\\ -2{\lambda }_0{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}{y}^{\prime }\left(s\right)\text{d}s\text{d}m}}+b{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{0}{\overset{t-\tau }{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}y}\left(s\right)\text{d}s\text{d}m}\\ -c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{0}{\overset{t-\sigma }{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}y}\left(s\right)\text{d}s\text{d}m}-2c{\lambda }_0{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{0}{\overset{t-\sigma }{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}{y}^{\prime }}\left(s\right)\text{d}s}\text{d}m\end{array}$

$\begin{array}{l}=L\left({\lambda }_0;\varphi \right)+{\displaystyle \underset{i=1}{\overset{\infty }{\sum }}{l}_i{\text{e}}^{-{\lambda }_0{t}_i}}+\left(c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }-b{\text{e}}^{-{\lambda }_0\tau }\right){\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}y\left(s\right)\text{d}s\text{d}m}}\\ -2{\lambda }_0{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}{y}^{\prime }\left(s\right)\text{d}s\text{d}m}}+b{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{0}{\overset{t-\tau }{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}y}\left(s\right)\text{d}s\text{d}m}\\ -c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{0}{\overset{t-\sigma }{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}y}\left(s\right)\text{d}s\text{d}m}-2c{\lambda }_0{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{0}{\overset{t-\sigma }{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}{y}^{\prime }}\left(s\right)\text{d}s}\text{d}m\end{array}$

化解整理得

$\begin{array}{l}y\left(t\right)+c{\text{e}}^{-{\lambda }_0\sigma }y\left(t-\sigma \right)\\ =L\left({\lambda }_0;\varphi \right)+{\displaystyle \underset{i=1}{\overset{\infty }{\sum }}{l}_i{\text{e}}^{-{\lambda }_0{t}_i}}+c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{t-\sigma }{\overset{t}{\int }}{\displaystyle \underset{t-\sigma }{\overset{m}{\int }}y\left(s\right)\text{d}s\text{d}m}}-b{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{t-\tau }{\overset{t}{\int }}{\displaystyle \underset{t-\tau }{\overset{m}{\int }}y}\left(s\right)\text{d}s\text{d}m}\\ -2{\lambda }_0{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}{y}^{\prime }}\left(s\right)\text{d}s}\text{d}m-2c{\lambda }_0{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{0}{\overset{t-\sigma }{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}{y}^{\prime }}\left(s\right)\text{d}s}\text{d}m,\end{array}$ (16)

令 $z\left(t\right)=y\left(t\right)-\frac{L\left({\lambda }_0;\varphi \right)}{1+2\beta \left({\lambda }_0\right)}$，

则(16)等价于

$\begin{array}{l}z\left(t\right)+c{\text{e}}^{-{\lambda }_0\sigma }z\left(t-\sigma \right)\\ ={\displaystyle \underset{i=1}{\overset{\infty }{\sum }}{l}_i{\text{e}}^{-{\lambda }_0{t}_i}}+c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{t-\sigma }{\overset{t}{\int }}{\displaystyle \underset{t-\sigma }{\overset{m}{\int }}z\left(s\right)\text{d}s\text{d}m}}-b{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{t-\tau }{\overset{t}{\int }}{\displaystyle \underset{t-\tau }{\overset{m}{\int }}z}\left(s\right)\text{d}s\text{d}m}\\ -2{\lambda }_0{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}{z}^{\prime }}\left(s\right)\text{d}s}\text{d}m-2c{\lambda }_0{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{0}{\overset{t-\sigma }{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}{z}^{\prime }}\left(s\right)\text{d}s}\text{d}m,\end{array}$ (17)

同时初值条件(8)等价于

$z\left(t\right)={\text{e}}^{-{\lambda }_{0}t}\varphi \left(t\right)-\frac{L\left({\lambda }_0;\varphi \right)}{1+2\beta \left({\lambda }_0\right)},t\in \left[-h,0\right]$ (18)

于是得出(13)的等价方程

$\underset{t\to \infty }{\lim }z\left(t\right)=0,$ (19)

当

$H\left({\lambda }_0;\varphi \right)=\max \left\{1,\underset{t\in \left[-h,0\right]}{\max }\left|{\text{e}}^{-{\lambda }_{0}t}\varphi \left(t\right)-\frac{L\left({\lambda }_0;\varphi \right)}{1+2\beta \left({\lambda }_0\right)}\right|\right\},$ (20)

可得当 $t\in \left[-h,0\right]$ 时， $\left|z\left(t\right)\right|\le H\left({\lambda }_0;\varphi \right)$。

现在将证明当 $t\in \left[-h,\infty \right]$ 时， $\left|z\left(t\right)\right|\le H\left({\lambda }_0;\varphi \right)$。运用反证法，假设 $\exists t^{~}>0$，使得 $\left|z\left(t^{~}\right)\right|>H\left({\lambda }_0;\varphi \right)$，令

$t^{*}=\inf \left\{t^{~}:\left|z\left(t^{~}\right)\right|>H\left({\lambda }_0;\varphi \right)\right\},$

因为 $z\left(t\right)$ 满足右连续，所以当 $t^{*}$ 不是脉冲点时， $\left|z\left(t^{*}\right)\right|=H\left({\lambda }_0;\varphi \right)$，当它是脉冲点时， $\left|z\left(t^{*}\right)\right|\ge H\left({\lambda }_0;\varphi \right)$，也就是说，当 $-h\le t<t^{*}$ 时， $\left|z\left(t\right)\right|\le H\left({\lambda }_0;\varphi \right)$，那么通过(12)和(17)，得

$\begin{array}{c}\left|z\left(t^{*}\right)\right|=|-c{\text{e}}^{-{\lambda }_0\sigma }z\left(t^{*}-\sigma \right){\displaystyle \underset{i=1}{\overset{\infty }{\sum }}{l}_i{\text{e}}^{-{\lambda }_0{t}_i}}\\ +c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{t^{*}-\sigma }{\overset{t^{*}}{\int }}{\displaystyle \underset{t^{*}-\sigma }{\overset{m}{\int }}z\left(s\right)\text{d}s\text{d}m}}-b{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{t^{*}-\tau }{\overset{t^{*}}{\int }}{\displaystyle \underset{t^{*}-\tau }{\overset{m}{\int }}z}\left(s\right)\text{d}s\text{d}m}\\ -2{\lambda }_0{\displaystyle \underset{0}{\overset{t^{*}}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}{z}^{\prime }}\left(s\right)\text{d}s}\text{d}m-2c{\lambda }_0{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{0}{\overset{t^{*}-\sigma }{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}{z}^{\prime }}\left(s\right)\text{d}s}\text{d}m|\end{array}$

$\begin{array}{l}\le \left|c\right|{\text{e}}^{-{\lambda }_0\sigma }\left|z\left(t^{*}-\sigma \right)\right|+{\displaystyle \underset{i=1}{\overset{\infty }{\sum }}\left|l_i\right|{\text{e}}^{-{\lambda }_0{t}_i}}\\ +\left|c\right|{\lambda }_{{}_0}^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{t^{*}-\sigma }{\overset{t^{*}}{\int }}{\displaystyle \underset{t^{*}-\sigma }{\overset{m}{\int }}\left|z\left(s\right)\right|\text{d}s\text{d}m}}-\left|b\right|{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{t^{*}-\tau }{\overset{t^{*}}{\int }}{\displaystyle \underset{t^{*}-\tau }{\overset{m}{\int }}\left|z\left(s\right)\right|\text{d}s\text{d}m}}\\ -2\left|{\lambda }_0\right|{\displaystyle \underset{0}{\overset{t^{*}}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}\left|z^{\prime }\left(s\right)\right|\text{d}s\text{d}m}}-2\left|-c\right|\left|{\lambda }_0\right|{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{t^{*}-\sigma }{\overset{0}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}\left|z^{\prime }\left(s\right)\right|\text{d}s\text{d}m}}\end{array}$

$\begin{array}{l}\le \left|c\right|{\text{e}}^{-{\lambda }_0\sigma }\left|z\left(t^{*}-\sigma \right)\right|+{\displaystyle \underset{i=1}{\overset{\infty }{\sum }}\left|l_i\right|{\text{e}}^{-{\lambda }_0{t}_i}}\\ +\left|c\right|{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{t^{*}-\sigma }{\overset{t^{*}}{\int }}{\displaystyle \underset{t^{*}-\sigma }{\overset{m}{\int }}\left|z\left(s\right)\right|\text{d}s\text{d}m}}+\left|b\right|{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{t^{*}-\tau }{\overset{t^{*}}{\int }}{\displaystyle \underset{t^{*}-\tau }{\overset{m}{\int }}\left|z\left(s\right)\right|\text{d}s\text{d}m}}\\ +2\left|{\lambda }_0\right|{\displaystyle \underset{0}{\overset{t^{*}}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}\left|z^{\prime }\left(s\right)\right|\text{d}s\text{d}m}}+2\left|c\right|\left|{\lambda }_0\right|{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{t^{*}-\sigma }{\overset{0}{\int }}{\displaystyle \underset{0}{\overset{m}{\int }}\left|z^{\prime }\left(s\right)\right|\text{d}s\text{d}m}}\end{array}$

$\begin{array}{l}\le \left|c\right|{\text{e}}^{-{\lambda }_0\sigma }\left|z\left(t^{*}-\sigma \right)\right|+{\displaystyle \underset{i=1}{\overset{\infty }{\sum }}\left|l_i\right|{\text{e}}^{-{\lambda }_0{t}_i}}\\ +\left|c\right|{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{t^{*}-\sigma }{\overset{t^{*}}{\int }}{\displaystyle \underset{t^{*}-\sigma }{\overset{m}{\int }}\left|z\left(s\right)\right|\text{d}s\text{d}m}}+\left|b\right|{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{t^{*}-\tau }{\overset{t^{*}}{\int }}{\displaystyle \underset{t^{*}-\tau }{\overset{m}{\int }}\left|z\left(s\right)\right|\text{d}s\text{d}m}}\\ +2\left|G\left({\lambda }_0\right)\right|{\displaystyle \underset{t^{*}-\sigma }{\overset{t^{*}}{\int }}{\displaystyle \underset{t^{*}-\sigma }{\overset{m}{\int }}\left|z^{\prime }\left(s\right)\right|\text{d}s\text{d}m}}\end{array}$

$\begin{array}{l}\le H\left({\lambda }_0;\varphi \right)\left(\left|c\right|{\text{e}}^{-{\lambda }_0\sigma }+{\displaystyle \underset{i=1}{\overset{\infty }{\sum }}\left|l_i\right|{\text{e}}^{-{\lambda }_0{t}_i}}+\frac{1}{2}\left|c\right|{\sigma }^2{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }+\frac{1}{2}\left|b\right|{\tau }^2{\lambda }_0^2{\text{e}}^{-{\lambda }_0\tau }+{\sigma }^2\left|G\left({\lambda }_0\right)\right|\right)\\ \le H\left({\lambda }_0;\varphi \right)\mu \left({\lambda }_0\right)\\ <H\left({\lambda }_0;\varphi \right),\end{array}$

与假设矛盾，所以当 $t\in \left[-h,\infty \right]$ 时， $\left|z\left(t\right)\right|\le H\left({\lambda }_0;\varphi \right)$ 得证。

同时，对于 $t\ge 0$，有

$\left|z\left(t\right)\right|\le \mu \left({\lambda }_0\right)H\left({\lambda }_0;\varphi \right),$ (21)

得出，对于 $t\ge nh-h\left(n=0,1,\cdots \right)$，有 $\left|z\left(t\right)\right|\le {\left|\mu \left({\lambda }_0\right)\right|}^{n}H\left({\lambda }_0;\varphi \right)$。

根据(12)有， $\underset{n\to \infty }{\lim }{\left|\mu \left({\lambda }_0\right)\right|}^n=0$，所以

$\underset{t\to \infty }{\lim }z\left(t\right)=\underset{t\to \infty }{\lim }\left\{{\text{e}}^{-{\lambda }_{0}t}\varphi \left(t\right)-\frac{L\left({\lambda }_0;\varphi \right)}{1+2\beta \left({\lambda }_0\right)}\right\}=0,$

由此可得

$\underset{t\to \infty }{\lim }\left[{\text{e}}^{-{\lambda }_{0}t}\varphi \left(t\right)\right]=\frac{L\left({\lambda }_0;\varphi \right)}{1+2\beta \left({\lambda }_0\right)}.$

定理2.1得证。

定理2.2若 ${\lambda }_0$ 是(4)的实根且满足(12)，则

$a+b=0,$ (22)

$\left|c\right|+{\displaystyle \underset{i=1}{\overset{\infty }{\sum }}\left|l_i\right|{\text{e}}^{-{\lambda }_0{t}_i}}+\frac{1}{2}\left|b\right|{\tau }^2<1,$ (23)

证明：显然可得。

推论2.3 若(16)~(17)成立，则(1)~(3)的解满足

$\underset{t\to \infty }{\lim }x\left(t\right)=\frac{\varphi \left(0\right)+c\varphi \left(-\sigma \right)+b{\displaystyle \underset{-\tau }{\overset{0}{\int }}{\displaystyle \underset{-\tau }{\overset{m}{\int }}\varphi \left(s\right)\text{d}s\text{d}m}}}{1+{\tau }^{2}b+2c},$ (24)

其中： $0<1+{\tau }^{2}b+2c<2$。

4. 解的稳定性判断

定理3.1 若定理1成立， $\mu \left({\lambda }_0\right)<1$ 成立，

$R\left({\lambda }_0;\varphi \right)=\left\{1,\underset{-h\le t\le 0}{\max }\left|\varphi \left(t\right)\right|,\underset{-h\le t\le 0}{\max }[{\text{e}}^{-{\lambda }_{0}t}\left|\varphi \left(t\right)\right|,\underset{-h\le t\le 0}{\max }\left[{\text{e}}^{-{\lambda }_{0}t}\left|{\varphi }^{\prime }\left(t\right)\right|\right]\right\},$ (25)

则当 $t\in \left[0,\infty \right)$ 时，(1)~(3)的解x满足

$\left|x\left(t\right)\right|\le N\left({\lambda }_0\right)R\left({\lambda }_0;\varphi \right){\text{e}}^{{\lambda }_{0}t}.$ (26)

其中： $N\left({\lambda }_0\right)=\mu \left({\lambda }_0\right)\left(1+\frac{k\left({\lambda }_0\right)}{1+2\beta \left({\lambda }_0\right)}\right)+\left(\frac{k\left({\lambda }_0\right)}{1+2\beta \left({\lambda }_0\right)}\right)$

$k\left({\lambda }_0\right)=1+\left|c\right|+\frac{1}{2}\left|b\right|{\tau }^2{\text{e}}^{-{\lambda }_0\tau }+\frac{1}{2}\left|c\right|{\lambda }^2{\sigma }^2{\text{e}}^{-{\lambda }_0\sigma }+\left|c\right|\left|{\lambda }_0\right|{\sigma }^2{\text{e}}^{-{\lambda }_0\sigma }$。

同时，(1)~(3)的解x是

1) 稳定的。若 ${\lambda }_0=0$ 时，(2.9)~(2.10)成立；

2) 渐近稳定的。若 ${\lambda }_0<0$ ；

3) 不稳定的。若 ${\lambda }_0>0$。

证明：假设x是(1)~(3)的解，对于 $\forall t\ge -h$

$z\left(t\right)=y\left(t\right)-\frac{L\left({\lambda }_0;\varphi \right)}{1+2\beta \left({\lambda }_0\right)},y\left(t\right)={\text{e}}^{-{\lambda }_{0}t}x\left(t\right),$

根据(21)，得

$\begin{array}{c}\left|y\left(t\right)\right|\le \left|z\left(t\right)\right|+\left|\frac{L\left({\lambda }_0;\varphi \right)}{1+2\beta \left({\lambda }_0\right)}\right|\\ \le \mu \left({\lambda }_0\right)H\left({\lambda }_0;\varphi \right)+\frac{\left|L\left({\lambda }_0;\varphi \right)\right|}{1+2\beta \left({\lambda }_0\right)},\end{array}$

再根据(12)，(13)可得

$\begin{array}{c}\left|L\left({\lambda }_0;\varphi \right)\right|\le \left|\varphi \left(0\right)\right|+\left|c\right|\varphi \left(-\sigma \right)+\left|b\right|{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{-\tau }{\overset{0}{\int }}{\displaystyle \underset{-\tau }{\overset{m}{\int }}{\text{e}}^{-{\lambda }_{0}s}\varphi }\left(s\right)\text{d}s\text{d}m}\\ +\left|c\right|{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{-\sigma }{\overset{0}{\int }}{\displaystyle \underset{-\sigma }{\overset{m}{\int }}{\text{e}}^{-{\lambda }_{0}s}\varphi }\left(s\right)\text{d}s\text{d}m}+2\left|c\right|\left|{\lambda }_0\right|{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{-\sigma }{\overset{0}{\int }}{\displaystyle \underset{-\sigma }{\overset{m}{\int }}{\text{e}}^{-{\lambda }_{0}s}{\varphi }^{\prime }}\left(s\right)\text{d}s\text{d}m}\\ \le \left(1+\left|c\right|+\frac{1}{2}\left|b\right|{\tau }^2{\text{e}}^{-{\lambda }_0\tau }+\frac{1}{2}\left|c\right|{\lambda }_0^2{\sigma }^2{\text{e}}^{-{\lambda }_0\sigma }+\left|c\right|{\lambda }_0{\sigma }^2{\text{e}}^{-{\lambda }_0\sigma }\right)R\left({\lambda }_0;\varphi \right)\\ \le N\left({\lambda }_0\right)R\left({\lambda }_0;\varphi \right).\end{array}$

另一方面，根据定理2.1

$\begin{array}{c}H\left({\lambda }_0;\varphi \right)\le \max \left\{1,R\left({\lambda }_0;\varphi \right)+\frac{L\left({\lambda }_0;\varphi \right)}{1+2\beta \left({\lambda }_0\right)}\right\}\\ \le R\left({\lambda }_0;\varphi \right)+\frac{L\left({\lambda }_0;\varphi \right)}{1+2\beta \left({\lambda }_0\right)}\\ \le R\left({\lambda }_0;\varphi \right)+\frac{k\left({\lambda }_0\right)R\left({\lambda }_0;\varphi \right)}{1+2\beta \left({\lambda }_0\right)}\\ =\left(1+\frac{k\left({\lambda }_0\right)}{1+2\beta \left({\lambda }_0\right)}\right)R\left({\lambda }_0;\varphi \right),\end{array}$

于是

$\begin{array}{c}y\left(t\right)\le \mu \left({\lambda }_0\right)H\left({\lambda }_0;\varphi \right)+\frac{L\left({\lambda }_0;\varphi \right)}{1+2\beta \left({\lambda }_0\right)}\\ \le \mu \left({\lambda }_0\right)\left(1+\frac{k\left({\lambda }_0\right)}{1+2\beta \left({\lambda }_0\right)}\right)R\left({\lambda }_0;\varphi \right)+\frac{k\left({\lambda }_0\right)R\left({\lambda }_0;\varphi \right)}{1+2\beta \left({\lambda }_0\right)}\\ \le \left\{\mu \left({\lambda }_0\right)\left(1+\frac{k\left({\lambda }_0\right)}{1+2\beta \left({\lambda }_0\right)}\right)+\frac{k\left({\lambda }_0\right)}{1+2\beta \left({\lambda }_0\right)}\right\}R\left({\lambda }_0;\varphi \right)\\ =N\left({\lambda }_0\right)R\left({\lambda }_0;\varphi \right),\end{array}$

所以，当 $t\ge 0$ 时

$\left|x\left(t\right)\right|\le N\left({\lambda }_0\right)R\left({\lambda }_0;\varphi \right){\mathrm{e}}^{{\lambda }_{0}t},$

综上所述，定理3.1第一部分得证。

现在将证明定理3.1中的三条稳定性。设 $\varphi \in C\left(\left[-h,0\right],R\right)$，x是(1.1)~(1.3)的解

1) 当 ${\lambda }_0=0$ 时，对 $\forall t\ge 0$

$\left|x\left(t\right)\right|\le N\left({\lambda }_0\right)R\left({\lambda }_0;\varphi \right),$

因为 $\Vert \varphi \Vert =\underset{-h\le t\le 0}{\max }\left|\varphi \left(t\right)\right|\le R\left({\lambda }_0;\varphi \right)$，则对 $\forall \epsilon >0$， $\exists \delta =\frac{\epsilon }{N\left({\lambda }_0\right)}$，使 $R\left({\lambda }_0;\varphi \right)<\delta$，则

$\left|x\left(t\right)\right|\le N\left({\lambda }_0\right)R\left({\lambda }_0;\varphi \right)\le N\left({\lambda }_0\right)\delta =\epsilon ,$

所以(1)~(3)的零解是稳定的。

2) 当 ${\lambda }_0<0$ 时，对 $\forall t\ge 0$，

$\left|x\left(t\right)\right|\le N\left({\lambda }_0\right)R\left({\lambda }_0;\varphi \right),$

即

$\underset{t\to \infty }{\lim }x\left(t\right)=0,$

所以(1.1)~(1.3)的零解是渐近稳定的。

3) 当 ${\lambda }_0>0$ 时，我们运用反证法。假设 ${\lambda }_0>0$ 时，(1)~(3)的零解是稳定的。则 $\exists \delta >0$， $\Vert \varphi \Vert <\delta$ 即当 $t\ge -h$ 时，

$\left|x\left(t\right)\right|<1,$ (27)

设 ${\varphi }_0\left(t\right)={\text{e}}^{{\lambda }_{0}t},t\in \left[-h_0,0\right]$，则

$\begin{array}{c}L\left({\lambda }_0;{\varphi }_0\right)\le {\varphi }_0\left(0\right)+c{\varphi }_0\left(-\sigma \right)+b{\text{e}}^{-{\lambda }_0\tau }{\displaystyle \underset{-\tau }{\overset{0}{\int }}{\displaystyle \underset{-\tau }{\overset{m}{\int }}{\text{e}}^{-{\lambda }_{0}s}{\varphi }_0}\left(s\right)\text{d}s\text{d}m}\\ +c{\lambda }_0^2{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{-\sigma }{\overset{0}{\int }}{\displaystyle \underset{-\sigma }{\overset{m}{\int }}{\text{e}}^{-{\lambda }_{0}s}\varphi }\left(s\right)\text{d}s\text{d}m}+2c{\lambda }_0{\text{e}}^{-{\lambda }_0\sigma }{\displaystyle \underset{-\sigma }{\overset{0}{\int }}{\displaystyle \underset{-\sigma }{\overset{m}{\int }}{\text{e}}^{-{\lambda }_{0}s}{{\varphi }^{\prime }}_0}\left(s\right)\text{d}s\text{d}m}\\ \le 1+\frac{1}{2}b{\tau }^2{\text{e}}^{-{\lambda }_0\tau }+c{\text{e}}^{-{\lambda }_0\sigma }\left(1+\frac{1}{2}{\lambda }_0^2{\sigma }^2\right)\\ \le 1+2\beta \left({\lambda }_0\right)>0,\end{array}$ (28)

设 $\exists {\delta }_0>0$，满足 $0<{\delta }_0<\delta$，则

$\varphi =\frac{{\delta }_0}{\Vert {\varphi }_0\Vert }{\varphi }_0\in C\left(\left[-h,0\right],R\right),$

其中 ${\varphi }_0={\delta }_0<\delta$，于是(1)~(3)满足(27)。

另一方面，根据(13)有

$\underset{t\to \infty }{\lim }\left[{\text{e}}^{-{\lambda }_{0}t}x\left(t\right)\right]=\frac{L\left({\lambda }_0;\varphi \right)}{1+\frac{1}{2}\beta \left({\lambda }_0\right)}=\frac{\frac{{\delta }_0}{\Vert \varphi \Vert }L\left({\lambda }_0;\varphi \right)}{1+\frac{1}{2}\beta \left({\lambda }_0\right)}=\frac{{\delta }_0}{\Vert {\varphi }_0\Vert }>0,$

但是由(27)可知，当 ${\lambda }_0>0$， $\underset{t\to \infty }{\lim }\left[{\text{e}}^{-{\lambda }_{0}t}x\left(t\right)\right]=0$，矛盾。所以当 ${\lambda }_0>0$ 时，(1.1)~(1.3)的零解是稳定的。综上所述定理3.1得证。

5. 实例验证

${\left[x\left(t\right)-\frac{1}{3\text{e}}x\left(t-\frac{1}{2}\right)\right]}^{\prime \text{}\prime }=\frac{1}{3}x\left(t\right)-\frac{1}{3}x\left(t\right)-\frac{1}{\text{e}}x\left(t-\frac{1}{5}\right),t\ne k$ (29)

$\Delta x\left(t_k\right)={\left(-\frac{1}{\text{e}}\right)}^k,k\in z^{+}$

$x\left(t\right)=\varphi \left(t\right),\frac{1}{2}\le t\le 0$

其中 $\varphi \in \left(\left[-\frac{1}{2},0\right],R\right)$。

解：(29)的特征方程

${\lambda }^2\left(1-\frac{1}{3\text{e}}{\text{e}}^{-\frac{\lambda }{2}}\right)=\frac{1}{3}-\frac{1}{\text{e}}{\text{e}}^{{}^{-\frac{\lambda }{5}}}$

由图1可知特征方程的根有两个，解得 ${\lambda }_1\approx -4.25$， ${\lambda }_2\approx 0.07$。

Figure 1. (29) graph of characteristic equation

图1. (29)特征方程的曲线图

1) 当 ${\lambda }_1\approx -4.25$ 时， ${\displaystyle \underset{i=1}{\overset{\infty }{\sum }}\left|l_i\right|{\text{e}}^{-{\lambda }_1{t}_i}\to \infty }$，则 $\mu \left({\lambda }_1\right)\to \infty$，所以方程(29)不适用于定义3.1。

2) 当 ${\lambda }_2\approx 0.07$

$\mu \left({\lambda }_2\right)=\left(-0.07\right)=\frac{1}{3}{\text{e}}^{-1.035}+{\displaystyle \underset{i=1}{\overset{\infty }{\sum }}{\text{e}}^{-1.07i}+\frac{0.0049}{12}{\text{e}}^{-1.35}}+\frac{0.049}{100}{\text{e}}^{-1.012}+\frac{0.049}{4}\approx 0.6419<1$

满足定理3.1的条件，所以方程(2.9)的零解是渐近稳定的。

6. 总结

文章研究了二阶线性脉冲中立型时滞微分方程解的渐近形态和稳定性，采用的是特征方程实根的方法，通过特征方程的根与零的关系来判断方程零解的稳定性，最后通过一个实例来验证结果.

NOTES

^*通讯作者。