1 问题的提出

在时滞微分方程解的稳定性和有界性问题的研究中，行之有效的方法仍是Lyapunov第二方法.Lyapunov第二方法，也称Lyapunov直接法，这种方法可以在没有得到方程具体解的情况下，直接确定方程解的稳定性. Lyapunov第二方法的关键在于构造Lyapunov泛函.目前许多学者在研究时滞微分方程解的稳定性时，都是通过构造Lyapunov泛函的方法，并得到了很多有意义的研究成果，如文献[1-14].

特别地，在文献[1-4]分别有如下的研究结果:

2003年，Sadek^[1]研究了如下三阶时滞微分方程

$ \dddot x + a\ddot x + g\left( {\dot x\left( {t - r\left( t \right)} \right)} \right) + f\left( {x\left( {t - r\left( t \right)} \right)} \right) = p\left( t \right), $

得到了当p(t)=0时它的零解渐进稳定的充分条件, 及p(t)≠0时它的所有解有界的充分条件.

2006年，Cemil T^[2]研究了一类三阶非线性时滞微分方程

$ \dddot x + \varphi \left( {x,\dot x} \right)\ddot x + g\left( {\dot x\left( {t - r\left( t \right)} \right)} \right) + f\left( {x\left( {t - r\left( t \right)} \right)} \right) = 0 $

的零解稳定的充分条件，并证明2003年，Sadek得到的定理2(文献[1]中的定理2)在一般情况下是不成立的，并得到了新的定理.

2010年，Afuwape A U和Omeike M O^[3]研究了一类三阶时滞微分方程

$ \begin{gathered} \dddot x + h\left( {\dot x} \right)\ddot x + g\left( {x\left( {t - r\left( t \right)} \right),\dot x\left( {t - r\left( t \right)} \right)} \right) + \hfill \\ f\left( {x\left( {t - r\left( t \right)} \right)} \right) = p\left( {t,x,\dot x,\ddot x} \right), \hfill \\ \end{gathered} $

并得到当p(t, x, ẋ, ẍ)≡0时零解渐进稳定的充分条件，及p(t, x, ẋ, ẍ)≠0时它的所有解有界的充分条件.

2007年，姚洪兴和孟伟业^[4]讨论了如下三阶双滞量时滞微分方程的全局渐进稳定性

$ \begin{gathered} \dddot x\left( t \right) + g\left( {x\left( t \right),\dot x\left( t \right)} \right)\ddot x\left( t \right) + f\left( {\dot x\left( {t - {\tau _1}} \right)} \right) + \hfill \\ h\left( {x\left( t \right)} \right)\varphi \left( {x\left( {t - {\tau _2}} \right)} \right) = 0, \hfill \\ \end{gathered} $

给出了其零解全局渐进稳定的充分条件.

受文献[1-4]的启发，本文将讨论如下三阶非线性双变滞量时滞微分方程零解的全局渐进稳定性和所有解的有界性

$ \begin{gathered} \dddot x + g\left( {x,\dot x} \right)\ddot x + f\left( {\dot x\left( {t - {\tau _1}\left( t \right)} \right)} \right) + h\left( x \right)\varphi \left( {x\left( {t - } \right.} \right. \hfill \\ \left. {\left. {{\tau _2}\left( t \right)} \right)} \right) = p\left( {t,x,\dot x,\ddot x} \right), \hfill \\ \end{gathered} $

(1)

其中x=x(t), ẋ(0)=0, τ₁(t), τ₂(t)≥0, f(0)=h(0)φ(0)=0.本文中所有函数都是连续的，且能保证系统解的唯一性，如果需要，进一步假定它们是连续可微的.

2 全局渐进稳定性

考虑自治RFDE系统

$ \dot x = f\left( {{x_t}} \right),{x_t} = x\left( {t + \theta } \right), - r \le \theta \le 0,t \ge 0, $

(2)

其中f:C_H→Rⁿ是连续泛函，f(0)=0, C_H={φ∈C([-r, 0, Rⁿ): ‖φ ‖≤H}且满足对H₁ < H，若‖φ‖≤H₁，则存在L(H₁)>0，使|f(φ)|≤L(H₁).

引理1^[15]  若存在函数V(φ):C_H→R为连续泛函，V(0)=0，且满足:

(1) V(φ)≥W(|φ(0)|), W(r)为楔函数；

(2) $ {\dot V_{\left( 2 \right)}}\left( \varphi \right) \le 0, \varphi \in {C_H};$

(3)集合$ Z = \left\{ {\varphi \in {C_H}:{{\dot V}_{\left( 2 \right)}}\left( \varphi \right) = 0} \right\}$中最大不变集M={ 0}，并且系统(2)的所有正半轨线都是有界的，则系统(2)的零解是全局渐进稳定的.

当p(t, x, ẋ, ẍ)≡0时，考虑如下时滞微分方程零解全局渐进稳定的充分条件

$ \begin{gathered} \dddot x + g\left( {x,\dot x} \right)\ddot x + f\left( {\dot x\left( {t - {\tau _1}\left( t \right)} \right)} \right) + h\left( x \right)\varphi \left( {x\left( {t - } \right.} \right. \hfill \\ \left. {\left. {{\tau _2}\left( t \right)} \right)} \right) = 0, \hfill \\ \end{gathered} $

(3)

其中x=x(t), ẋ(0)=0, τ₁(t), τ₂(t)≥0, f(0)=h(0)φ(0)=0.

显然系统(3)等价于系统

$ \left\{ \begin{array}{l} \dot x\left( t \right) = y\left( t \right),\\ y\left( t \right) = z\left( t \right),\\ \dot z\left( t \right) = - g\left( {x\left( t \right),y\left( t \right)} \right)z\left( t \right) - f\left( {y\left( t \right)} \right) - h\left( {x\left( t \right)} \right)\\ \;\;\;\;\;\varphi \left( {x\left( t \right)} \right) + \int_{ - {\tau _1}\left( t \right)}^0 {{{f'}_y}\left( {y\left( {t + s} \right)} \right)z\left( {t + s} \right){\rm{d}}s} + \\ \;\;\;\;\;h\left( {x\left( t \right)} \right) + \int_{ - {\tau _2}\left( t \right)}^0 {{{\varphi '}_x}\left( {x\left( {t + s} \right)} \right)y\left( {t + s} \right){\rm{d}}s} . \end{array} \right. $

(4)

定理1  假设下列条件成立:

(1) 存在正常数a, b, L, M, γ₁, γ₂, β₁, β₂, 使得$M \le a\sqrt {2\left( {1-{\beta _1}} \right)}, L \le a\sqrt {2\left( {1-{\beta _2}} \right)} $，其中0 < β₁ < 1, 0 < β₂ < 1；

(2) 0 < [h(x)φ(x)]′ < ab－a²γ₁－2a²γ₂；

(3) $\frac{{f\left( y \right)}}{y} \ge b\left( {y \ne 0} \right) $，对任意的y，都有|f′(y)|≤L；

(4) g(x, y)≥a+(a²+1)γ₁+γ₂；

(5)|h(x)φ′_x(x)|≤M；

(6) g′_x(x, y)y≤0；

(7) 0≤τ₁(t)≤γ₁,τ′₁(t)≤β₁；0≤τ₂(t)≤γ₂, τ′₂(t)≤β₂.

则系统(4)的零解全局渐进稳定，从而系统(3)的零解全局渐进稳定.

证明  对于系统(4)，构造Lyapunov泛函

$ \begin{array}{l} V\left( {{x_t},{y_t},{z_t}} \right) = a\int_0^x {h\left( x \right)\varphi \left( x \right){\rm{d}}x} + h\left( x \right)\varphi \left( x \right)y + \\ \frac{1}{2}{\left( {ay + z} \right)^2} + \int_0^y {f\left( y \right){\rm{d}}y} + a\int_0^y {\left[ {g\left( {x,y} \right) - a} \right]y{\rm{d}}y} + \\ {a^2}\int_{ - {\tau _1}\left( t \right)}^0 {\int_{t + s}^t {{z^2}\left( u \right){\rm{d}}u{\rm{d}}s} } + {a^2}\int_{ - {\tau _2}\left( t \right)}^0 {\int_{t + s}^t {{y^2}\left( u \right){\rm{d}}u{\rm{d}}s} } . \end{array} $

(5)

由于f(0)=h(0)φ(0)=0，容易得到V(0, 0, 0)=0.

(1) 证明引理1的条件(1).

根据式(5)，将V(x_t, y_t, z_t)改写为

$ \begin{array}{l} V\left( {{x_t},{y_t},{z_t}} \right) = a\int_0^x {h\left( x \right)\varphi \left( x \right){\rm{d}}x} - \frac{1}{{2b}}{h^2}\left( x \right){\varphi ^2}\left( x \right) + \\ \frac{b}{2}{\left[ {y + \frac{{h\left( x \right)\varphi \left( x \right)}}{b}} \right]^2} + \int_0^y {f\left( y \right){\rm{d}}y} - \frac{1}{2}b{y^2} + \frac{1}{2}{\left( {ay + z} \right)^2} + \\ a\int_0^y {\left[ {g\left( {x,y} \right) - a} \right]y{\rm{d}}y} + {a^2}\int_{ - {\tau _1}\left( t \right)}^0 {\int_{t + s}^t {{z^2}\left( u \right){\rm{d}}u{\rm{d}}s} } + \\ {a^2}\int_{ - {\tau _2}\left( t \right)}^0 {\int_{t + s}^t {{y^2}\left( u \right){\rm{d}}u{\rm{d}}s} } . \end{array} $

记$H\left( x \right)a\int_0^x {h\left( x \right)\varphi \left( x \right){\rm{d}}x{\rm{ - }}\frac{1}{{2b}}} {h^2}\left( x \right){\varphi ^2}\left( x \right) $, 由条件(2)[h(x)φ(x)]′>0知h(x)φ(x)单调增加，即有sgn[h(x)φ(x)]=sgnx.又因为$H' = \frac{1}{b}\left\{ {ab- \left[{h\left( x \right)\varphi \left( x \right)} \right]'} \right\}h\left( x \right)\varphi \left( x \right)$, H(0)=0知，当x≠0时，H(x)>0.由条件(3)$\int_0^y {f\left( y \right){\rm{d}}y = \int_0^y {\frac{{f\left( y \right)}}{y}y{\rm{d}}y \ge \int_0^y {by{\rm{d}}y = \frac{1}{2}b{y^2}} } } $，容易得到

$\int_0^y {f\left( y \right){\rm{d}}y-} \frac{1}{2}b{y^2} \ge 0$.由条件(4)知g(x, y)－a≥0，则$a\int_0^y {\left[{g\left( {x, y} \right)-a} \right]} y{\rm{d}}y \ge 0$>.因此，泛函V(x_t, y_t, z_t)是正定的，满足引理1的条件(1).

(2) 证明$\frac{{{\rm{d}}V}}{{{\rm{d}}t}}\left| {_{\left( 4 \right)} \le 0} \right.$.

用$\frac{{{\rm{d}}V}}{{{\rm{d}}t}}\left| {_{\left( 4 \right)}} \right.$表示泛函V(x_t, y_t, z_t)沿着系统(4)的解求导.由系统(4)和泛函V(x_t, y_t, z_t)可得

$ \begin{array}{l} \frac{{{\rm{d}}V}}{{{\rm{d}}t}}\left| {_{\left( 4 \right)}} \right. = {y^2}{\left[ {h\left( x \right)\varphi \left( x \right)} \right]^\prime } + ay\int_0^y {{{g'}_x}\left( {x,y} \right)y{\rm{d}}y} + \\ a{z^2} - ayf\left( y \right) - g\left( {x,y} \right){z^2} + {a^2}{y^2}{\tau _2}\left( t \right) - {a^2}\left( {1 - } \right.\\ \left. {{{\tau '}_2}\left( t \right)} \right)\int_{ - {\tau _2}\left( t \right)}^0 {{y^2}\left( {t + s} \right){\rm{d}}s} + \left( {ay + z} \right)\int_{ - {\tau _1}\left( t \right)}^0 {{{f'}_y}\left( {y\left( {t + } \right.} \right.} \\ \left. {\left. s \right)} \right)z\left( {t + s} \right){\rm{d}}s + {a^2}{z^2}{\tau _1}\left( t \right) - {a^2}\left( {1 - {{\tau '}_1}\left( t \right)} \right)\int_{ - {\tau _1}\left( t \right)}^0 {{z^2}\left( {t + } \right.} \\ \left. s \right){\rm{d}}s + \left( {ay + z} \right)h\left( x \right)\int_{ - {\tau _2}\left( t \right)}^0 {{{\varphi '}_x}\left( {x\left( {t + s} \right)} \right)y\left( {t + s} \right){\rm{d}}s} . \end{array} $

由条件(3)并利用2uv≤u²+v²可得到

$ \begin{array}{l} \left( {ay + z} \right)\int_{ - {\tau _1}\left( t \right)}^0 {{{f'}_y}\left( {y\left( {t + s} \right)} \right)z\left( {t + s} \right){\rm{d}}s} \le \\ \int_{ - {\tau _1}\left( t \right)}^0 {\left| {ay + z} \right|\left| {{{f'}_y}\left( {y\left( {t + s} \right)} \right)} \right|\left| {z\left( {t + s} \right)} \right|{\rm{d}}s} \le \\ \int_{ - {\tau _1}\left( t \right)}^0 {\left| {ay + z} \right|\left| {Lz\left( {t + s} \right)} \right|{\rm{d}}s} \le \\ \int_{ - {\tau _1}\left( t \right)}^0 {\frac{1}{2}\left[ {{{\left( {ay + z} \right)}^2} + {L^2}{z^2}\left( {t + s} \right)} \right]{\rm{d}}s} \le \\ \int_{ - {\tau _1}\left( t \right)}^0 {\frac{1}{2}\left[ {{a^2}{y^2} + \left( {{a^2}{y^2} + {z^2}} \right) + {z^2}} \right]{\rm{d}}s} + \\ \int_{ - {\tau _1}\left( t \right)}^0 {\frac{1}{2}{L^2}{z^2}\left( {t + s} \right){\rm{d}}s} \le \\ \left( {{a^2}{y^2} + {z^2}} \right){\tau _1}\left( t \right) + \int_{ - {\tau _1}\left( t \right)}^0 {\frac{1}{2}{L^2}{z^2}\left( {t + s} \right){\rm{d}}s} . \end{array} $

同理，根据条件(5)得

$ \begin{array}{l} \left( {ay + z} \right)h\left( x \right)\int_{ - {\tau _2}\left( t \right)}^0 {{{\varphi '}_x}\left( {x\left( {t + s} \right)} \right)y\left( {t + s} \right){\rm{d}}s} \le \\ \left( {{a^2}{y^2} + {z^2}} \right){\tau _2}\left( t \right) + \int_{ - {\tau _2}\left( t \right)}^0 {\frac{1}{2}{M^2}{y^2}\left( {t + s} \right){\rm{d}}s} . \end{array} $

故

$ \begin{array}{l} \frac{{{\rm{d}}V}}{{{\rm{d}}t}}\left| {_{\left( 4 \right)}} \right. \le - \left\{ {ab - {{\left[ {h\left( x \right)\varphi \left( x \right)} \right]}^\prime }} \right\}{y^2} + \\ ay\int_0^y {{{g'}_x}\left( {x,y} \right)ydy} - \left( {g\left( {x,y} \right) - a} \right){z^2} + {a^2}{y^2}{\tau _2}\left( t \right) - \\ {a^2}\left( {1 - {{\tau '}_2}\left( t \right)} \right)\int_{ - {\tau _2}\left( t \right)}^0 {{y^2}\left( {t + s} \right){\rm{d}}s} + \left( {{a^2}{y^2} + {z^2}} \right){\tau _1}\left( t \right) + \\ \int_{ - {\tau _1}\left( t \right)}^0 {\frac{1}{2}{L^2}{z^2}\left( {t + s} \right){\rm{d}}s} + {a^2}{z^2}{\tau _1}\left( t \right) - {a^2}\left( {1 - {{\tau '}_1}\left( t \right)} \right)\int_{ - {\tau _1}\left( t \right)}^0 {{z^2}} \times \\ \left( {t + s} \right){\rm{d}}s + \left( {{a^2}{y^2} + {z^2}} \right){\tau _2}\left( t \right) + \int_{ - {\tau _2}\left( t \right)}^0 {\frac{1}{2}{M^2}{y^2}\left( {t + s} \right){\rm{d}}s} \le \\ - \left\{ {ab - {{\left[ {h\left( x \right)\varphi \left( x \right)} \right]}^\prime } - 2{a^2}{\tau _2}\left( t \right) - {a^2}{\tau _1}\left( t \right)} \right\}{y^2} - \left( {g\left( {x,y} \right) - } \right.\\ \left. {a - {a^2}{\tau _1}\left( t \right) - {\tau _1}\left( t \right) - {\tau _2}\left( t \right)} \right){z^2} + ay\int_0^y {{{g'}_x}\left( {x,y} \right)ydy} - \\ \left[ {{a^2}\left( {1 - {{\tau '}_2}\left( t \right)} \right) - \frac{1}{2}{M^2}} \right]\int_{ - {\tau _2}\left( t \right)}^0 {{y^2}\left( {t + s} \right){\rm{d}}s} - \left[ {{a^2}\left( {1 - {{\tau '}_1}\left( t \right)} \right) - } \right.\\ \left. {\frac{1}{2}{L^2}} \right]\int_{ - {\tau _1}\left( t \right)}^0 {{z^2}\left( {t + s} \right){\rm{d}}s} \le - \left\{ {ab - {{\left[ {h\left( x \right)\varphi \left( x \right)} \right]}^\prime } - 2{a^2}{\gamma _2} - } \right.\\ \left. {{a^2}{\gamma _1}} \right\}{y^2} - \left( {g\left( {x,y} \right) - a - \left( {{a^2} + 1} \right){\gamma _1} - {\gamma _2}} \right){z^2} + ay\int_0^y {{{g'}_x}\left( {x,} \right.} \\ \left. y \right)y{\rm{d}}y - \left[ {{a^2}\left( {1 - {\beta _2}} \right) - \frac{1}{2}{M^2}} \right]\int_{ - {\tau _2}\left( t \right)}^0 {{y^2}\left( {t + s} \right){\rm{d}}s} - \\ \left[ {{a^2}\left( {1 - {\beta _1}} \right) - \frac{1}{2}{L^2}} \right]\int_{ - {\tau _1}\left( t \right)}^0 {{z^2}\left( {t + s} \right){\rm{d}}s} . \end{array} $

由条件(2)知－{ab－ [h(x)φ(x)]′－2a²τ₂(t)－a²τ₁(t)}y²≤0；由条件(4)知－(g(x, y)－a－(a²+ 1)γ₁－γ₂)z²≤0；由条件(6)知$ay\int_0^y {g{'_x}\left( {x, y} \right)} y{\rm{d}}y \le 0$；由条件(1)知${a^2}\left( {1-{\beta _2}} \right)-\frac{1}{2}{M^2} \ge 0, {a^2}\left( {1-{\beta _1}} \right) - \frac{1}{2}{L^2} \ge 0$，故

$ \begin{array}{l} - \left[ {{a^2}\left( {1 - {\beta _2}} \right) - \frac{1}{2}{M^2}} \right]\int_{ - {\tau _2}\left( t \right)}^0 {{y^2}\left( {t + s} \right){\rm{d}}s} \le 0,\\ - \left[ {{a^2}\left( {1 - {\beta _1}} \right) - \frac{1}{2}{L^2}} \right]\int_{ - {\tau _1}\left( t \right)}^0 {{z^2}\left( {t + s} \right){\rm{d}}s} \le 0. \end{array} $

从而有$\frac{{{\rm{d}}V}}{{{\rm{d}}t}}\left| {_{\left( 4 \right)} \le 0} \right.\left( {\frac{{{\rm{d}}V}}{{{\rm{d}}t}}\left| {_{\left( 4 \right)}} \right. = 0\;{\rm{当且仅当}}y = 0, z = 0{\rm{时成立}}} \right)$.引理1的条件(2)得证.

(3) 考虑集合$Z = \left\{ {\varphi \in {C_H}:{{\dot V}_{\left( 2 \right)}}\left( \varphi \right) = 0} \right\}$，易得集合Z的最大不变集M={0}.由于$\frac{{{\rm{d}}V}}{{{\rm{d}}t}}\left| {_{\left( 4 \right)}} \right. = 0$当且仅当y=0, z=0时成立，所以K⊂{(x, y, z):x∈ R, y=0, z=0}.设轨线的ω极限集为ω_r，由极限集的性质知，ω_r非空.又有泛函V(x_t, y_t, z_t)存在下界且$\frac{{{\rm{d}}V}}{{{\rm{d}}t}}\left| {_{\left( 4 \right)} \le 0} \right.$，知ω_r位于V(x_t, y_t, z_t)的某个等值曲面V=c上.

下证ω_r只有唯一的点x=0.假设ω_r包含点x′≠0，由引理1的条件(1)可得，c=V(x′)>0，从而ω_r不会包含原点.此外，ω_r是由整轨线构成的，在轨线上有V=c，即有$\frac{{{\rm{d}}V}}{{{\rm{d}}t}} = 0$，所以极限点应该包含在$\frac{{{\rm{d}}V}}{{{\rm{d}}t}} = 0$所确定的点集内，因此有ω_r={0}.即有集合Z的最大不变集M={0}.

再证系统(4)的每一条正半轨线都是有界的.考虑集合

$ D = \left\{ {\left( {x,y,z} \right):V\left( {x,y,z} \right) < L,\left| y \right| < N} \right\}, $

其中L, N是任意给定的正常数.显然D是有界集.要证系统(4)通过D中任意一点的正半轨线不会越出区域D.由于$\frac{{{\rm{d}}V}}{{{\rm{d}}t}} \le 0$，因此若有通过D中任意一点的正半轨线越出区域D，须有穿过区域D的边界∂D上的平面部分，即有这样的瞬间T使|y(T)|=N成立.根据V(x_t, y_t, z_t)的表达式，易知区域D中任意一点都满足$\frac{1}{2}{\left( {ay + z} \right)^2} < L$，即

$ - ay - \sqrt {2L} < z < \sqrt {2L} - ay, $

若y=N，上式右边可得$z < \sqrt {2L}-aN < 0$.若y=－N, 上式左边可得$z > - ay - \sqrt {2L} > 0$，因此可以选择足够大的N，使${\rm{z}}{\mathop{\rm sgn}} {\rm{y}} = \frac{{{\rm{d}}y}}{{{\rm{d}}t}}{\mathop{\rm sgn}} {\rm{y}} < 0$.则当系统(4)的轨线交区域D的边界的平面部分时，轨线应自外向内穿过.即系统(4)的正半轨线有界.

综上，由引理知系统(4)的零解是全局渐进稳定的，从而系统(3)的零解全局渐进稳定.

例题1  考虑三阶非线性时滞微分方程

$ \begin{gathered} \dddot x\left( t \right) + \left( {4\dot x\left( t \right) + 4} \right)\ddot x\left( t \right) + 3\dot x\left( {t - {\tau _1}\left( t \right)} \right) + \hfill \\ \sin \dot x\left( {t - {\tau _1}\left( t \right)} \right) + x\left( {t - {\tau _2}\left( t \right)} \right) = 0, \hfill \\ \end{gathered} $

(6)

系统(6)可以表示为如下系统

$ \left\{ \begin{array}{l} \dot x = y,\\ \dot y = z,\\ \dot z = - \left( {4y + 4} \right)z - \left( {3y + \sin y} \right) - x + \\ \;\;\;\;\int_{ - {\tau _1}\left( t \right)}^0 {\left[ {3 + \cos y\left( {t + s} \right)} \right]z\left( {t + s} \right){\rm{d}}s} + \\ \;\;\;\;\int_{ - {\tau _2}\left( t \right)}^0 {y\left( {t + s} \right){\rm{d}}s} . \end{array} \right. $

(7)

构造Lyapunov泛函

$ \begin{array}{l} V\left( {{x_t},{y_t},{z_t}} \right) = 2\int_0^x {\mu {\rm{d}}\mu } + xy + \frac{1}{2}{\left( {2y + z} \right)^2} + \\ \int_0^y {\left( {4\nu + \sin \nu } \right){\rm{d}}\nu } + 2\int_0^y {\left( {4\xi + 2} \right)\xi {\rm{d}}\xi } + \\ 4\int_{ - {\tau _1}\left( t \right)}^0 {\int_{t + s}^t {{z^2}\left( u \right){\rm{d}}u{\rm{d}}s} } + 4\int_{ - {\tau _2}\left( t \right)}^0 {\int_{t + s}^t {{y^2}\left( u \right){\rm{d}}u{\rm{d}}s} } = 2\int_0^x {\mu {\rm{d}}\mu } - \\ \frac{1}{2}{x^2} + \frac{1}{2}{\left( {y + x} \right)^2} + \int_0^y {\left( {4\nu + \sin \nu } \right){\rm{d}}\nu } - \frac{1}{2}{y^2} + \frac{1}{2}\left( {2y + } \right.\\ {\left. z \right)^2} + 2\int_0^y {\left( {4\xi + 2} \right)\xi {\rm{d}}\xi } + 4\int_{ - {\tau _1}\left( t \right)}^0 {\int_{t + s}^t {{z^2}\left( u \right){\rm{d}}u{\rm{d}}s} } + \\ 4\int_{ - {\tau _2}\left( t \right)}^0 {\int_{t + s}^t {{y^2}\left( u \right){\rm{d}}u{\rm{d}}s} } . \end{array} $

有$H\left( x \right) = 2\int_0^x {\mu {\rm{d}}\mu-\frac{1}{2}} {x^2} = {x^2}-\frac{1}{2}{x^2} = \frac{1}{2}{x^2} \ge 0$, 并且$\int_0^y {\frac{{\left( {4v + \sin v} \right)}}{v}} {\rm{d}}v = \int_0^y {\left( {4 + \frac{{\sin v}}{v}} \right)} {\rm{d}}v \ge \int_0^y {y{\rm{d}}y} = \frac{1}{2}{y^2}$, 易得$\int_0^y {\left( {4v + \sin v} \right)} {\rm{d}}v-\frac{1}{2}{y^2} \ge 0$, 故泛函V(x_t, y_t, z_t)是正定的.

$ \begin{array}{l} \frac{{{\rm{d}}V}}{{{\rm{d}}t}}\left| {_{\left( 7 \right)}} \right. = {y^2} + 2{z^2} - 2y\left( {4y + \sin y} \right) - \left( {4y + 4} \right){z^2} + \\ 4{y^2}{\tau _2}\left( t \right) - 4\left( {1 - {{\tau '}_2}\left( t \right)} \right)\int_{ - {\tau _2}\left( t \right)}^0 {{y^2}\left( {t + s} \right){\rm{d}}s} + \left( {2y + } \right.\\ \left. z \right)\int_{ - {\tau _1}\left( t \right)}^0 {\left[ {4 + \cos y\left( {t + s} \right)} \right]z\left( {t + s} \right){\rm{d}}s} + 4{z^2}{\tau _1}\left( t \right) - 4\left( {1 - } \right.\\ \left. {{{\tau '}_1}\left( t \right)} \right)\int_{ - {\tau _1}\left( t \right)}^0 {{z^2}\left( {t + s} \right){\rm{d}}s} + \left( {2y + z} \right)\int_{ - {\tau _2}\left( t \right)}^0 {y\left( {t + s} \right){\rm{d}}s} . \end{array} $

根据定理1, 易得

$ \begin{array}{l} \frac{{{\rm{d}}V}}{{{\rm{d}}t}}\left| {_{\left( 7 \right)}} \right. \le - \left( {1 - 8{\gamma _2} - 4{\gamma _1}} \right){y^2} - \left\{ {4y + 2 - 5{\gamma _1} - } \right.\\ \left. {{\gamma _2}} \right){z^2} - \left[ {4\left( {1 - {\beta ^2}} \right) - \frac{1}{2}{M^2}} \right]\int_{ - {\tau _2}\left( t \right)}^0 {{y^2}\left( {t + s} \right){\rm{d}}s} - \\ \left[ {4\left( {1 - {\beta _1}} \right) - \frac{1}{2}{L^2}} \right]\int_{ - {\tau _1}\left( t \right)}^0 {{z^2}\left( {t + s} \right){\rm{d}}s} . \end{array} $

容易找到正常数L, M, γ₁, γ₂, β₁, β₂, 使得$\frac{{{\rm{d}}V}}{{{\rm{d}}t}}\left| {_{\left( 7 \right)}} \right. \le 0$成立, 并且$\frac{{{\rm{d}}V}}{{{\rm{d}}t}}\left| {_{\left( 7 \right)}} \right.{\rm{ = }}0$当且仅当y=0, z=0时成立.故系统(6)的零解全局渐进稳定.

3 有界性

当p(t, x, ẋ, ẍ)≠0时, 下面讨论系统(1)所有解有界的充分条件.

考虑非自治RFDE系统

$ \dot {\bar x} = F\left( {t,{{\bar x}_t}} \right),{{\bar x}_t} = \bar x\left( {t + \theta } \right), - r \le \theta \le 0,t \ge 0, $

(8)

其中f: R×C→ Rⁿ是连续泛函, 并且f将R×C中的有界集映射为Rⁿ中的有界集.

显然系统(1)等价于系统

$ \left\{ \begin{array}{l} \dot x\left( t \right) = y\left( t \right),\\ \dot y\left( t \right) = z\left( t \right),\\ \dot z\left( t \right) = - g\left( {x\left( t \right),y\left( t \right)} \right)z\left( t \right) - f\left( {y\left( t \right)} \right) - \\ \;\;\;\;h\left( {x\left( t \right)} \right)\varphi \left( {x\left( t \right)} \right) + \int_{ - {\tau _1}\left( t \right)}^0 {{{f'}_y}\left( {y\left( {t + s} \right)} \right)z\left( {t + } \right.} \\ \;\;\;\;\left. s \right){\rm{d}}s + h\left( {x\left( t \right)} \right)\int_{ - {\tau _2}\left( t \right)}^0 {{{\varphi '}_x}\left( {x\left( {t + s} \right)} \right)y\left( {t + } \right.} \\ \;\;\;\;\left. s \right){\rm{d}}s + p\left( {t,x\left( t \right),y\left( t \right),z\left( t \right)} \right). \end{array} \right. $

(9)

定理2  假设定理1中的条件均成立，另外对于所有的t, x, y, z有

$ \left| {p\left( {t,x\left( t \right),y\left( t \right),z\left( t \right)} \right)} \right| \le q\left( t \right), $

其中q(t)< ∞并且q(t)∈L¹(0, ∞)，L¹(0, ∞)是Lebesgue可积函数空间.此时存在一个有限正常数K使得系统(9)的解x(t)对所有的t≥t₀满足

$ \left| {x\left( t \right)} \right| \le K,\left| {x'\left( t \right)} \right| \le K,\left| {x''\left( t \right)} \right| \le K, $

其中x∈C²([t₀－r, t₀], R).

证明  应用定理1的Lyapunov函数V(x_t, y_t, z_t)及定理1的结果，易得$\frac{{{\rm{d}}V}}{{{\rm{d}}t}}\left| {_{\left( 4 \right)}} \right. \le-{K_1}\left( {{y^2} + {z^2}} \right)$，其中K₁为正常数.当p(t, x, ẋ, ẍ)≠0时，显然可以得到V(x_t, y_t, z_t)沿系统(9)的解满足不等式

$ \begin{array}{l} \frac{{{\rm{d}}V}}{{{\rm{d}}t}}\left| {_{\left( 9 \right)}} \right. \le - {K_1}\left( {{y^2} + {z^2}} \right) + \left| {y + z} \right| \times \\ \left| {p\left( {t,x,y,z} \right)} \right| \le - {K_1}\left( {{y^2} + {z^2}} \right) + \left| {y + z} \right|q\left( t \right) \le \\ - {K_1}\left( {{y^2} + {z^2}} \right) + \left( {\left| y \right| + \left| z \right|} \right)q\left( t \right) \le \\ \left( {\left| y \right| + \left| z \right|} \right)q\left( t \right). \end{array} $

由于|y| ≤1+y²和|z|≤1+z²可得$\frac{{{\rm{d}}V}}{{{\rm{d}}t}}\left| {_{\left( 9 \right)}} \right. \le \left( {2 + {y^2} + {z^2}} \right)q\left( t \right)$.

同时根据定理1的结果，存在正常数D，使得V(x_t, y_t, z_t)≥D(x²+y²+z²)，从而有

$ \left( {{x^2} + {y^2} + {z^2}} \right) \le {D^{ - 1}}V\left( {x,y,z} \right). $

则有

$ \begin{array}{l} \frac{{{\rm{d}}V}}{{{\rm{d}}t}}\left| {_{\left( 9 \right)}} \right. \le \left( {2 + {D^{ - 1}}V\left( {x,y,z} \right)} \right)q\left( t \right) = 2q\left( t \right) + \\ {D^{ - 1}}V\left( {x,y,z} \right)q\left( t \right). \end{array} $

(10)

对式(10)在0到t上进行积分，并利用假设条件q(t)∈L¹(0, ∞)和Gronwall-Reid-Bellman不等式，可得

$ \begin{array}{l} V\left( {x,y,z} \right) \le V\left( {{x_0},{y_0},{z_0}} \right) + 2A + {D^{ - 1}}\int_0^1 {\left( {V\left( {{x_s},{y_s},} \right.} \right.} \\ \left. {\left. {{z_s}} \right)} \right)q\left( s \right){\rm{d}}s \le \left( {V\left( {{x_0},{y_0},{z_0}} \right) + 2A} \right)\exp \left( {{D^{ - 1}}\int_0^t {q\left( s \right){\rm{d}}s} } \right) \le \\ \left( {V\left( {{x_0},{y_0},{z_0}} \right) + 2A} \right)\exp \left( {{D^{ - 1}}A} \right) = {K_2} < \infty , \end{array} $

其中K₂是正整数，K₂=(V(x₀, y₀, z₀)+2A)×exp(D^－1A)，并且$A = \int_0^t {q\left( s \right){\rm{d}}s} $, 则

$ \left( {{x^2} + {y^2} + {z^2}} \right) \le {D^{ - 1}}V\left( {x,y,z} \right) \le {D^{ - 1}}{K_2}. $

令K=D^－1K₂，对于任意的t≥t₀有

$ \left| {x\left( t \right)} \right| \le K,\left| {y\left( t \right)} \right| \le K,\left| {z\left( t \right)} \right| \le K, $

即对于任意的t≥t₀有

$ \left| {x\left( t \right)} \right| \le K,\left| {x'\left( t \right)} \right| \le K,\left| {x''\left( t \right)} \right| \le K, $

则定理2得证.

例题2  考虑三阶非线性时滞微分方程

$ \begin{gathered} \dddot x\left( t \right) + \left( {4\dot x\left( t \right) + 4} \right)\ddot x\left( t \right) + 3\dot x\left( {t - {\tau _1}\left( t \right)} \right) + \sin \dot x\left( {t - } \right. \hfill \\ \left. {{\tau _1}\left( t \right)} \right) + x\left( {t - {\tau _2}\left( t \right)} \right) = \frac{2}{{1 + {t^2} + {x^2}\left( t \right) + {{\dot x}^2}\left( t \right) + {{\ddot x}^2}\left( t \right)}}, \hfill \\ \end{gathered} $

(11)

系统(11)可以表示为如下系统

$ \left\{ \begin{array}{l} \dot x = y,\\ \dot y = z,\\ \dot z = - \left( {4y + 4} \right)z - \left( {3y + \sin y} \right) - x + \\ \;\;\;\;\int_{ - {\tau _1}\left( t \right)}^0 {\left[ {3 + \cos y\left( {t + s} \right)} \right]z\left( {t + s} \right){\rm{d}}s} + \\ \;\;\;\;\int_{ - {\tau _2}\left( t \right)}^0 {y\left( {t + s} \right){\rm{d}}s} + \frac{2}{{1 + {t^2} + {x^2}\left( t \right) + {y^2}\left( t \right) + {z^2}\left( t \right)}}. \end{array} \right. $

(12)

易知

$ \frac{2}{{1 + {t^2} + {x^2}\left( t \right) + {y^2}\left( t \right) + {z^2}\left( t \right)}} \le \frac{2}{{1 + {t^2}}} = q\left( t \right) $

对于所有的t, x, y, z均成立，并且有

$ \int_0^\infty {q\left( s \right){\rm{d}}s} = \int_0^\infty {\frac{2}{{1 + {s^2}}}{\rm{d}}s} = {\rm{ \mathsf{ π} }} < \infty ,q\left( t \right) \in {L^1}\left( {0,\infty } \right). $

易得V(x_t, y_t, z_t)沿系统(12)的解满足以下不等式

$ \begin{array}{l} \frac{{{\rm{d}}V}}{{{\rm{d}}t}}\left| {_{\left( {12} \right)}} \right. \le - {K_1}\left( {{y^2} + {z^2}} \right) + \left| {y + z} \right| \times \\ \left| {p\left( {t,x,y,z} \right)} \right| \le - {K_1}\left( {{y^2} + {z^2}} \right) + \frac{{\left| {y + z} \right|}}{{1 + {t^2}}} \le - {K_1}\left( {{y^2} + } \right.\\ \left. {{z^2}} \right) + \frac{{\left| y \right| + \left| z \right|}}{{1 + {t^2}}} \le \frac{{\left| y \right| + \left| z \right|}}{{1 + {t^2}}} \le \frac{{2 + {y^2} + {z^2}}}{{1 + {t^2}}} \le \frac{4}{{1 + {t^2}}} + \\ \frac{{2{D^{ - 1}}}}{{1 + {t^2}}}V\left( {{x_t},{y_t},{z_t}} \right). \end{array} $

(13)

其中K₁, D为正常数.对式(13)在0到t上进行积分，并利用假设条件q(t)∈L¹(0, ∞)和Gron wall-Reid-Bellman不等式，可得

$ V\left( {x,y,z} \right) \le \left( {V\left( {{x_0},{y_0},{z_0}} \right) + 2A} \right)\exp \left( {{D^{ - 1}}A} \right) = {K_2} < \infty . $

其中K₂是正整数, K₂=(V(x₀, y₀, z₀)+2A)×exp(D^－1A), 并且$A = \int_0^t {q\left( s \right){\rm{d}}s} $, 则

$ \left( {{x^2} + {y^2} + {z^2}} \right) \le {D^{ - 1}}V\left( {x,y,z} \right) \le {D^{ - 1}}{K_2}. $

令K=D^－1K₂, 对于任意的t≥t₀,|x(t)|≤K,|x′(t)|≤K,|x″(t)|≤K均成立.故系统(11)的所有解有界.