在时滞微分方程理论和应用领域中, 稳定性是非常重要的问题.至今研究时滞微分方程有效的方法还是Lyapunov直接法.对于时滞微分方程解的稳定性研究也得到了很多很好的结果^[1-6].

2007年, Zhang和Si^[3]研究了形如

$ x'''(t)+g(x'(t))x''(t)+f(x(t), x'(t))+h(x(t))=0 $

的三阶非线性时滞微分方程解的渐进稳定性.

2010年, Cemil Tunc^[4]研究了形如

$ \begin{eqnarray*}&& x'''(t)+g(x(t), x'(t))x''(t)+f(x(t-r(t)), x'(t-r(t)))+h(x(t-r(t)))\\ &=&p(t, x(t), x'(t), x(t-r(t)), x'(t-r(t)), x''(t))\end{eqnarray*} $

的三阶非线性时滞微分方程，得到了当$p=0$时此方程零解渐进稳定性的充分性判别准则.

本文, 我们将研究如下三阶非线性时滞微分方程零解的渐进稳定性

$ \begin{eqnarray}&& x'''(t)+g_{1}(x(t), x'(t))x''(t)+g_{2}(x(t), x'(t))x'(t)\nonumber\\ &&+f(x(t-r(t)), x'(t-r(t)))+h(x(t-r(t)))\nonumber\\ &=& p(t, x(t), x'(t), x(t-r(t)), x'(t-r(t)), x''(t)). \end{eqnarray} $

(1.1)

显然, (1.1) 式等价于系统

$ \begin{eqnarray}&& x'(t)=y(t), y'(t)=z(t), \nonumber\\ && z'(t)=-g_{1}(x(t), y(t))z(t)-g_{2}(x(t), y(t))y(t)-f(x(t), y(t))+h(x(t))\nonumber\\ && +\int^{t}_{t-r(t)}f_{x}(x(s), y(s))y(s)ds+\int^{t}_{t-r(t)}f_{y}(x(s), y(s))z(s)ds +\int^{t}_{t-r(t)}h'(x(s))y(s)ds\nonumber \\ && +p(t, x(t), y(t), x(t-r(t)), y(t-r(t)), z(t)), \end{eqnarray} $

(1.2)

其中$0\leq r(t)\leq \alpha, \ \alpha$为正常数, $r, g_{1}, g_{2}, f, h, p$均为连续函数, $f(x, 0)=h(0)=0$.导函数$r'(t)=\frac{dr}{dt}, $ $g_{1x}(x, y)=\frac{\partial g_{1}(x, y)}{\partial x}, $ $g_{2x}(x, y)=\frac{\partial g_{2}(x, y)}{\partial x}, $ $f_{x}(x, y)=\frac{\partial f(x, y)}{\partial x}, $ $f_{y}(x, y)=\frac{\partial f(x, y)}{\partial y}, h'(t)=\frac{dh}{dt}$存在且连续.假设$r'(t)\leq \beta, 0<\beta<1, $ $f(x(t-r(t)), y(t-r(t))), $ $h(x(t-r(t)), $ $g_{1}(x(t), y(t)), $ $g_{2}(x(t), y(t)), $ $p(t, x(t), y(t), x(t-r(t)), y(t-r(t)), z(t))$关于$x(t), y(t), x(t-r(t), y(t-r(t))$和$z(t)$满足Lipschitz条件, 则方程有唯一解.最后, 假设所有的解都是实值的.

定理  设存在正常数$a, b, c, \mu, \delta, \lambda_{1}, \lambda_{2}, K$和$L$, 使得下列条件成立:

(1) $g_{1}(x, y)\geq \mu+a, \ g_{2}(x, y)\geq c, \ yg_{1x}(x, y)\leq 0, \ yg_{2x}(x, y)\leq 0;$

(2)  $f(x, y){\rm sgn} y\geq (b+\delta)|y|, \ -K\leq f_{x}(x, y)\leq 0, \ |f_{y}(x, y)|\leq L;$

(3) $0<h'(x)<ab, {\rm sgn}h(x)={\rm sgn} x.$

当$\alpha<\min\bigg\{\frac{2a(\delta+c)}{a^{2}b+aK+aL+2\lambda_{1}}, \frac{\mu}{ab+K+L+2\lambda_{2}}\bigg\}$时, 其中$\lambda_{1}=\frac{a^{2}b+ab+aK+K}{2(1-\beta)}, \lambda_{2}=\frac{aL+L}{2(1-\beta)}, $则方程(1.1) 的零解是渐进稳定的.

证  先定义一个Lyapunov函数

$ \begin{eqnarray}V&\;=&\;V(x_{t}, y_{t}, z_{t})=a\int^{x}_{0}h(\xi)d\xi+h(x)y+\frac{1}{2}(ay+z)^{2}+ a\int^{y}_{0}[g_{1}(x, \eta)-a]\eta d\eta\nonumber\\ &&+\int^{y}_{0}g_{2}(x, \eta)\eta d\eta +\int^{y}_{0}f(x, \eta)d\eta+ \lambda_{1}\int^{0}_{-r(t)}\int^{t}_{t+s}y^{2}(\theta)d\theta ds \nonumber\\&&+ \lambda_{2}\int^{0}_{-r(t)}\int^{t}_{t+s}z^{2}(\theta)d\theta ds. \end{eqnarray} $

(2.1)

显然, $V(0, 0, 0)=0, $并且(2.1) 式可以改写为

$ \begin{eqnarray}&& V =a\int^{x}_{0}h(\xi)d\xi-\frac{1}{2b}h(x)^{2}+\frac{b}{2}\bigg[y+\frac{h(x)}{b}\bigg]^{2} +\frac{1}{2}(ay+z)^{2}\nonumber\\ &&+ a\int^{y}_{0}\bigg[g_{1}(x, \eta)+\frac{1}{a}g_{2}(x, \eta)-a\bigg]\eta d\eta\nonumber\\ && +\int^{y}_{0}f(x, \eta)d\eta-\frac{b}{2}y^{2}+ \lambda_{1}\int^{0}_{-r(t)}\int^{t}_{t+s}y^{2}(\theta)d\theta ds + \lambda_{2}\int^{0}_{-r(t)}\int^{t}_{t+s}z^{2}(\theta)d\theta ds.\end{eqnarray} $

(2.2)

由条件$0<h'(x)<ab, $ ${\hbox{sgn}}h(x)={\hbox{sgn}}x$和$g_{1}(x, y)\geq \mu+a$可得

$ \begin{eqnarray*} a\int^{x}_{0}h(\xi)d\xi-\frac{1}{2b}h(x)^{2} &\;=&\; a\int^{x}_{0}h(\xi)d\xi-\frac{1}{b}\int^{x}_{0}h(\xi)h'(\xi)d\xi\\ &\;=&\;\int^{x}_{0}[a-b^{-1}h'(\xi)]h(\xi)d\xi =b^{-1}\int^{x}_{0}[ab-h'(\xi)]h(\xi)d\xi>0 \end{eqnarray*} $

和

$ a\int^{y}_{0}\bigg[g_{1}(x, \eta)+\frac{1}{a}g_{2}(x, \eta)-a\bigg] \eta d\eta\geq \frac{a\mu+c}{2}y^{2}. $

另一方面, 由条件$f(x, y){\hbox{sgn}}y\geq (b+\delta)|y|, $可得

当$y>0$时, $f(x, y)\geq(b+\delta)y, $ $f(x, y)y\geq(b+\delta)y^{2}.$

当$y<0$时, $f(x, y)\leq(b+\delta)y, $ $f(x, y)y\geq(b+\delta)y^{2}.$

从而必有

$ \int^{y}_{0}f(x, \eta)d\eta-\frac{b}{2}y^{2} =\int^{y}_{0}[f(x, \eta)-b\eta]d\eta\geq 0. $

综上述

$ \begin{eqnarray*} V&\;\geq&\; b^{-1}\int^{x}_{0}[ab-h'(\xi)]h(\xi)d\xi+\frac{a\mu+c}{2}y^{2}+\frac{1}{2}(ay+z)^{2}\\ && +\lambda_{1}\int^{0}_{-r(t)}\int^{t}_{t+s}y^{2}(\theta)d\theta ds + \lambda_{2}\int^{0}_{-r(t)}\int^{t}_{t+s}z^{2}(\theta)d\theta ds.\end{eqnarray*} $

显然对于这个不等式, 存在足够小的正常数$D_{i}~ (i=1, 2, 3), $使得

$ \begin{eqnarray} V&\;=&\;D_{1}x^{2}+D_{2}y^{2}+D_{3}z^{2}+\lambda_{1}\int^{0}_{-r(t)}\int^{t}_{t+s}y^{2}(\theta)d\theta ds + \lambda_{2}\int^{0}_{-r(t)}\int^{t}_{t+s}z^{2}(\theta)d\theta ds\nonumber\\ &\geq&\; D_{4}(x^{2}+y^{2}+z^{2})+\lambda_{1}\int^{0}_{-r(t)}\int^{t}_{t+s}y^{2}(\theta)d\theta ds + \lambda_{2}\int^{0}_{-r(t)}\int^{t}_{t+s}z^{2}(\theta)d\theta ds\nonumber\\ &\geq&\;D_{4}(x^{2}+y^{2}+z^{2}), \end{eqnarray} $

(2.3)

其中$D_{4}=\min\{D_{1}, D_{2}, D_{3}\}, $从而$x^{2}+y^{2}+z^{2}\leq D_{4}^{-1}V(x_{t}, y_{t}, z_{t}), \ y^{2}+z^{2}\leq D_{4}^{-1}V(x_{t}, y_{t}, z_{t}).$

接下来, 证明存在一个连续函数$u(x)\geq 0$且$u(|\varphi(0)|)\geq 0$, 使得$u(|\varphi(0)|)\leq V(\varphi).$

事实上, $V$沿系统(1.2) 的导数为

$ \frac{dV}{dt}\;=\;-af(x, y)y+h'(x)y^{2}-[g_{1x}(x, y)-a]z^{2} +ay\int^{y}_{0}g_{1x}(x, \eta)\eta d\eta+\\\;\;\;\;y\int^{y}_{0}g_{2x}(x, \eta)\eta d\eta\nonumber\\ \;\;\;\;+y\int^{y}_{0}f_{x}(x, \eta)d\eta-ag_{2}(x, y)y^{2} +(ay+z)\int^{t}_{t-r(t)}h'(x(s))y(s)ds\nonumber\\ \;\;\;\;+(ay+z)\int^{t}_{t-r(t)}f_{x}(x(s), y(s))y(s)ds \;\;\;\;+(ay+z)\int^{t}_{t-r(t)}f_{y}(x(s), y(s))z(s)ds+\\\;\;\;\;\lambda_{1}r(t)y^{2}\nonumber\\ \;\;\;\;-\lambda_{1}(1-r'(t))\int^{t}_{t-r(t)}y^{2}(s)ds+\lambda_{2}r(t)z^{2} -\lambda_{2}(1-r'(t))\int^{t}_{t-r(t)}z^{2}(s)ds. $

(2.4)

根据定理假设和等式$2|st|=s^{2}+t^{2}, $可得

$ \begin{eqnarray}\frac{dV}{dt}&\;\leq&\; [a(\delta+c)-(\frac{a^{2}b}{2}+\frac{aK}{2}+\frac{aL}{2}+\lambda_{1})]y^{2} -[\mu-(\frac{ab}{2}+\frac{K}{2}+\frac{L}{2}+\lambda_{2})]z^{2}\nonumber\\ && +[\frac{a^{2}b}{2}+\frac{ab}{2}+\frac{aK}{2}+\frac{K}{2}-\lambda_{1}(1-\beta)] \int^{t}_{t-r(t)}y^{2}(s)ds\nonumber\\ &&+[\frac{aL}{2}+\frac{L}{2}-\lambda_{2}(1-\beta)]\int^{t}_{t-r(t)}z^{2}(s)ds, \end{eqnarray} $

(2.5)

从而$\frac{dV}{dt}\leq 0.$易证集合$Z=\{\varphi\in C:\dot{V}(\varphi)=0\}$的最大不变集$Q=\{0\}.$于是方程(1.1) 满足$\frac{dV(x_{t}, y_{t}, z_{t})}{dt}=0$的唯一解是$x=y=z=0.$因此, 方程(1.1) 的零解是渐进稳定的(根据见参考文献[5]).证毕.

下面给出以下的一个三阶非线性时滞微分方程来阐述主要结论的具体应用.例如三阶非线性时滞微分方程零解的渐进稳定性

$ \begin{eqnarray}&& x'''(t)+[1+e^{-x(t)x'(t)}+\frac{1}{3+\cos(x'(t))}]x''(t)+[3+e^{-x(t)x'(t)}]x'(t)\nonumber\\ && +x'(t-r)+\frac{1}{10}x(t-r(t))=0.\end{eqnarray} $

(2.6)

显然, (1.1) 式等价于系统

$ \begin{eqnarray*}&& x'(t)=y(t), y'(t)=z(t), \\ && z'(t)=-[1+e^{-x(t)x'(t)}+\frac{1}{3+\cos(x'(t))}]z(t)-[3+e^{-x(t)x'(t)}]y(t)-x'(t-r)\\ &&+\frac{1}{10}x(t-r(t))+\int^{t}_{t-r(t)}\frac{4-y^{2}(s)-y(s)}{(4+y^{2}(s))^{2}}z(s)ds +\frac{1}{10}\int^{t}_{t-r(t)}y(s)ds, \end{eqnarray*} $

其中

$ g_{1}(x, y)=2+\sin(x)+\frac{1}{3+\cos(y)}, g_{2}(x, y)=3+\frac{1}{1+x^{2}+y^{2}}, f(x, y)=y, h(x)=\frac{1}{10}x. $

显然

(1) $g_{1}(x, y)=1+e^{-xy}+\frac{1}{3+\cos(y)}\geq1+\frac{1}{4}:= \mu+a,   yg_{1x}(x, y)=-y^{2}e^{-xy}\leq 0, $

$ g_{2}(x, y)=3+e^{-xy}\geq3:= c, yg_{2x}(x, y)=-y^{2}e^{-xy}\leq 0; $

(2) $f(x, y)$sgn$y=y$ sgn$y=|y|=(\frac{1}{2}+\frac{1}{2})|y|:=(b+\delta)|y|, $

$ -1:=-K< f_{x}(x, y)=0, |f_{y}(x, y)|=1:=L; $

(3)  $h(x)=\frac{1}{10}x, ~~h(0)=0, ~~0<h'(x)=\frac{1}{10}<ab=\frac{1}{8}, $ sgn$h(x)=$sgn$x.$

即定理的所有条件被满足, 因此方程(2.6) 的零解是渐近稳定的.