1 问题的提出

时滞微分方程主要用来描述依赖当前和过去历史状态的动力系统，因此它在物理、信息、化学、工程、经济以及生物数学等领域都有重要应用.由于时滞微分方程在实际中应用如此广泛，所以对时滞微分方程的理论研究就显得十分重要，也是非常有意义的.对时滞微分方程的稳定性理论的研究的转折点可以追溯到1892年，这一年俄国数学家Lyapunov发表了一篇名为《运动的稳定性的一般问题》的论文，该论文给出了研究稳定性的一种很有效的方法，称为Lyapunov第二方法，它至今仍是研究时滞微分方程解的稳定性的主要方法，Lyapunov直接法的关键是构造Lyapunov泛函.至今已经有很多学者在这方面有很好的研究成果，文献[1-16]中有很多的介绍.

特别地, 关于时滞微分方程解的稳定性和有界性的研究现状.

2003年, Sadek^[16]研究了如下三阶时滞微分方程

$ \ddddot 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时它的零解渐近稳定的充分条件, 及p(t)≠0时它的所有解有界的充分条件.

2006年, CemilTunc^[10]研究了一类三阶非线性时滞微分方程

$ \ddddot 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.

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近稳定性

$ \begin{gathered} \ddddot x + 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近稳定的充分条件.

受文献[5]的启发，本文研究了一类三阶时滞微分方程解的稳定性和有界性，给出了其零解渐近稳定和所有解有界的充分性条件.

本文研究三阶时滞微分方程

$ \begin{array}{l} x'''\left( t \right) + g\left( {x\left( t \right),x'\left( t \right)} \right)x''\left( t \right) + f\left( {x'\left( {t - r\left( t \right)} \right)} \right) + \\ h\left( {x\left( t \right)} \right)\varphi \left( {x\left( {t - r\left( t \right)} \right)} \right) = p\left( {t,x\left( t \right),x'\left( t \right),x\left( {t - } \right.} \right.\\ \left. {\left. {\left( t \right)} \right),x'\left( {t - r\left( t \right)} \right),x''\left( t \right)} \right),t \ge 0 \end{array} $

(1)

零解的渐近稳定性和所有解的有界性.其中, f(0)=h(0)=g(0, 0)=φ(0)=0，0≤r(t)≤r, g(x, y), f(x), h(x), φ(x)均为连续函数.

系统(1)等价于系统

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

(2)

2 渐近稳定性

当p(t, x(t), x′(t), x(t－r(t)), x′(t－r(t)), x″(t))=0时，讨论式(1)的渐近稳定性.

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, $

(3)

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^[6]  令V(Φ):C_H→ R为满足局部Lipschitz条件的连续泛函，V (0)=0且满足下列条件:

(1) W₁(|ϕ(0)|)≤V(ϕ)≤W₂($\left\| \phi \right\|$)，W₁(r), W₂(r)为权函数;

(2) ${\dot V_{\left(3 \right)}}$(ϕ)≤0, 当ϕ∈C_H, 则系统(3)零解是一致稳定的，若定义Z ={φ∈C_H: ${\dot V_{\left(3 \right)}}$(φ)=0}, 如果Z的最大不变集Q={0}, 则系统(3)的零解是渐近稳定的.

定理1  若存在正整数a, b, c, d, L, M, B且满足：

(1) 0 < [h(x)φ(x)]′ < ab-c;

(2)$\frac{{f\left(y \right)}}{y}$≥b(y≠0), 对任意的y都有|f′(y)| ≤L;

(3) h(x)φ(x)sgnx>0(x≠0), 对任意的x都有|h(x)| ≤B, |φ′(x)| ≤M;

(4) g′_x(x, y)y≤0且g(x, y)≥a+d;

(5) 0≤r(t)≤σ, r′(t)≤β, 0 < β < 1,

那么当

$ \begin{array}{l} \sigma \le \min \left\{ {\frac{{2c\left( {1 - \beta } \right)}}{{\alpha \left( {L + BM} \right)\left( {1 - \beta } \right) + \left( {a + 1} \right)MB}},} \right.\\ \left. {\frac{{2d\left( {1 - \beta } \right)}}{{\left( {L + MB} \right)\left( {1 - \beta } \right) + \left( {a + 1} \right)L}}} \right\}, \end{array} $

则系(1)的零解是渐近稳定的.

证明  先定义一个Lyapunov泛函V(x_t, y_t, z_t),

$ \begin{array}{l} V\left( {{x_t},{y_t},{z_t}} \right) = a\int_0^x {h\left( \xi \right)\varphi \left( \xi \right){\rm{d}}\xi } + h\left( x \right)\varphi \left( x \right)y + \\ \frac{1}{2}{\left( {ay + z} \right)^2} + \int_0^y {\frac{{f\left( \eta \right)}}{\eta }{\rm{d}}\eta } + a\int_0^y {\left[ {g\left( {x,\eta } \right) - a} \right]\eta {\rm{d}}\eta } + \\ {u_1}\int_{ - r\left( t \right)}^0 {\int_{t + s}^t {{y^2}\left( \theta \right){\rm{d}}\theta {\rm{d}}s} } + {u_2}\int_{ - r\left( t \right)}^0 {\int_{t + s}^t {{z^2}\left( \theta \right){\rm{d}}\theta {\rm{d}}s} } \ge \\ a\int_0^x {h\left( \xi \right)\varphi \left( \xi \right){\rm{d}}\xi } + h\left( x \right)\varphi \left( x \right)y + \frac{1}{2}{\left( {ay + z} \right)^2} + \frac{1}{2}b{y^2} + \\ a\int_0^y {\left[ {g\left( {x,\eta } \right) - a} \right]\eta {\rm{d}}\eta } + {u_1}\int_{ - r\left( t \right)}^0 {\int_{t + s}^t {{y^2}\left( \theta \right){\rm{d}}\theta {\rm{d}}s} } + \\ {u_2}\int_{ - r\left( t \right)}^0 {\int_{t + s}^t {{z^2}\left( \theta \right){\rm{d}}\theta {\rm{d}}s} } = \frac{1}{{2b{y^2}}}\left[ {4\int_0^x {h\left( \xi \right)\varphi \left( \xi \right)} \left\{ {\int_0^y {\left( {ab - } \right.} } \right.} \right.\\ \left. {\left. {\left. {{{\left[ {h\left( \xi \right)\varphi \left( \xi \right)} \right]}^\prime }} \right)\eta {\rm{d}}\eta } \right\}} \right]{\rm{d}}\xi + \frac{1}{{2b}}{\left[ {by + h\left( x \right)\varphi \left( x \right)} \right]^2} + \\ \frac{1}{2}{\left( {ay + z} \right)^2} + a\int_0^y {\left[ {g\left( {x,\eta } \right) - a} \right]\eta {\rm{d}}\eta } + \\ {u_1}\int_{ - r\left( t \right)}^0 {\int_{t + s}^t {{y^2}\left( \theta \right){\rm{d}}\theta {\rm{d}}s} } + {u_2}\int_{ - r\left( t \right)}^0 {\int_{t + s}^t {{z^2}\left( \theta \right){\rm{d}}\theta {\rm{d}}s} } . \end{array} $

明显地，V(0, 0, 0)=0.

由定理1条件(1)知[h(ξ)φ(ξ)]′ < ab－c，则ab－[h(ξ)φ(ξ)]′>c>0.

又由定理1条件(3)和(4)知h(x)φ(x)sgnx>0, g(x, y)－a>d.

显然，对于以上不等式，存在足够小的正常数D_i(i=1, 2, 3)，使得

$ \begin{array}{l} V\left( {{x_t},{y_t},{z_t}} \right) \ge {D_1}{x^2} + {D_2}{y^2} + {D_3}{z^2} + \\ {u_1}\int_{ - r\left( t \right)}^0 {\int_{t + s}^t {{y^2}\left( \theta \right){\rm{d}}\theta {\rm{d}}s} } + {u_2}\int_{ - r\left( t \right)}^0 {\int_{t + s}^t {{z^2}\left( \theta \right){\rm{d}}\theta {\rm{d}}s} } \ge {D_4}\left( {{x^2} + {y^2} + {z^2}} \right). \end{array} $

因为${u_1}\int_{-r\left(t \right)}^0 {\int_{t + s}^t {{y^2}\left(\theta \right){\rm{d}}\theta {\rm{d}}s} } $和${u_2}\int_{-r\left(t \right)}^0 {\int_{t + s}^t {{z^2}\left(\theta \right){\rm{d}}\theta {\rm{d}}s} } $是非负的，其中D₄=min(D₁, D₂, D₃)，所以不等式

$ V\left( {{x_t},{y_t},{z_t}} \right) \ge {D_4}\left( {{x^2} + {y^2} + {z^2}} \right) $

成立，则V(x_t, y_t, z_t)正定.

因此，泛函V(x_t, y_t, z_t)满足引理1的条件(1).

下证$\frac{{{\rm{d}}{V_{\left({{x_t}, {y_t}, {z_t}} \right)}}}}{{{\rm{d}}t}} \le 0.$

$ \begin{array}{l} \frac{{{\rm{d}}{V_{\left( {{x_t},{y_t},{z_t}} \right)}}}}{{{\rm{d}}t}} = ah\left( x \right)\varphi \left( x \right)y + {\left[ {h\left( x \right)\varphi \left( x \right)} \right]^\prime }{y^2} + \\ h\left( x \right)\varphi \left( x \right)z + \left( {ay + z} \right)\left( {az + z'} \right) + f\left( y \right)z + ayzg\left( {x,y} \right) + \\ ay\int_0^y {{{g'}_x}\left( {x,y} \right)y{\rm{d}}y} - {a^2}yz + {u_1}{z^2}r\left( t \right) + {u_2}{y^2}r\left( t \right) - \\ {u_1}\left( {1 - r'\left( t \right)} \right)\int_{t - r\left( t \right)}^t {{z^2}\left( \theta \right){\rm{d}}\theta } + {u_2}{y^2}r\left( t \right) - {u_2}\left( {1 - } \right.\\ \left. {r'\left( t \right)} \right) = {y^2}{\left[ {h\left( x \right)\varphi \left( x \right)} \right]^\prime } - ayf\left( y \right) + ay\int_0^y {{{g'}_x}\left( {x,} \right.} \\ \left. y \right)y{\rm{d}}y + a{z^2} - {z^2}g\left( {x,y} \right) + \left( {ay + z} \right)\int_{t - r\left( t \right)}^t {f'\left( y \right)z\left( s \right){\rm{d}}s} + \\ \left( {ayh\left( x \right) + zh\left( x \right)} \right)\int_{t - r\left( t \right)}^t {\varphi '\left( x \right)y\left( s \right){\rm{d}}s} + {u_1}{z^2}r\left( t \right) + \\ {u_2}{y^2}r\left( t \right) - {u_1}\left( {1 - r'\left( t \right)} \right)\int_{t - r\left( t \right)}^t {{z^2}\left( \theta \right){\rm{d}}\theta } - {u_2}\left( {1 - } \right.\\ \left. {r'\left( t \right)} \right)\int_{t - r\left( t \right)}^t {{y^2}\left( \theta \right){\rm{d}}\theta } . \end{array} $

由定理1所给的条件可得

$ \begin{array}{l} \left( {ay + z} \right)\int_{t - r\left( t \right)}^t {f'\left( y \right)z{\rm{d}}s} = ay\int_{t - r\left( t \right)}^t {f'\left( y \right)z\left( s \right){\rm{d}}s} + \\ z\int_{t - r\left( t \right)}^t {f'\left( y \right)z\left( s \right){\rm{d}}s} \le \frac{{aL}}{2}{y^2}r\left( t \right) + \frac{{aL}}{2}\int_{t - r\left( t \right)}^t {{z^2}\left( s \right){\rm{d}}s} + \\ \frac{L}{2}{z^2}r\left( t \right) + \frac{L}{2}\int_{t - r\left( t \right)}^t {{z^2}\left( s \right){\rm{d}}s} \le \frac{{aL\rho }}{2}{y^2} + \frac{{aL}}{2}\int_{t - r\left( t \right)}^t {{z^2}\left( s \right){\rm{d}}s} + \\ \frac{{L\rho }}{2}{z^2} + \frac{L}{2}\int_{t - r\left( t \right)}^t {{z^2}\left( s \right){\rm{d}}s} . \end{array} $

同理

$ \begin{array}{l} \left( {ayh\left( x \right) + zh\left( x \right)} \right)\int_{t - r\left( t \right)}^t {\varphi '\left( x \right)y\left( s \right){\rm{d}}s} = \\ ayh\left( x \right)\int_{t - r\left( t \right)}^t {\varphi '\left( x \right)y\left( s \right){\rm{d}}s} + zh\left( x \right)\int_{t - r\left( t \right)}^t {\varphi '\left( x \right)y\left( s \right){\rm{d}}s} \le \\ aBM\int_{t - r\left( t \right)}^t {\left| y \right|\left| {y\left( s \right)} \right|{\rm{d}}s} + BM\int_{t - r\left( t \right)}^t {\left| z \right|\left| {y\left( s \right)} \right|{\rm{d}}s} \le \\ \frac{{aBM}}{2}{y^2}r\left( t \right) + \frac{{aBM}}{2}\int_{t - r\left( t \right)}^t {{y^2}\left( s \right){\rm{d}}s} + \frac{{BM}}{2}{z^2}r\left( t \right) + \\ \frac{{BM}}{2}\int_{t - r\left( t \right)}^t {{y^2}\left( s \right){\rm{d}}s} \le \frac{{aBM\rho }}{2}{y^2} + \frac{{aBM}}{2}\int_{t - r\left( t \right)}^t {{y^2}\left( s \right){\rm{d}}s} + \\ \frac{{BM\rho }}{2}{z^2} + \frac{{BM}}{2}\int_{t - r\left( t \right)}^t {{y^2}\left( s \right){\rm{d}}s} . \end{array} $

又因为

$ \begin{array}{l} {u_1}{z^2}r\left( t \right) + {u_2}{y^2}r\left( t \right) \le {u_1}{z^2}\rho + {u_2}{y^2}\rho ,\\ - {u_1}\left( {1 - r'\left( t \right)} \right)\int_{t - r\left( t \right)}^t {{z^2}\left( \theta \right){\rm{d}}\theta } - {u_2}\left( {1 - } \right.\\ \left. {r'\left( t \right)} \right)\int_{t - r\left( t \right)}^t {{y^2}\left( \theta \right){\rm{d}}\theta } \le - {u_1}\left( {1 - \beta } \right)\int_{t - r\left( t \right)}^t {{z^2}\left( \theta \right){\rm{d}}\theta } - \\ {u_2}\left( {1 - \beta } \right)\int_{t - r\left( t \right)}^t {{y^2}\left( \theta \right){\rm{d}}\theta } , \end{array} $

$ {y^2}{\left[ {h\left( x \right)\varphi \left( x \right)} \right]^\prime } - ayf\left( y \right) \le \left( {ab - c} \right){y^2} - ab{y^2} - c{y^2}, $

$ a{z^2} - {z^2}g\left( {x,y} \right) \le a{z^2} - {z^2}\left( {a + d} \right) = - d{z^2}, $

则由上面不等式得

$ \begin{array}{l} \frac{{{\rm{d}}{V_{\left( {{x_t},{y_t},{z_t}} \right)}}}}{{{\rm{d}}t}} \le - c{y^2} - d{z^2} + \frac{{aL\rho }}{2}{y^2} + \frac{{aL}}{2}\int_{t - r\left( r \right)}^t {{z^2}\left( s \right){\rm{d}}s} + \\ \frac{{L\rho }}{2}{z^2} + \frac{L}{2}\int_{t - r\left( t \right)}^t {{z^2}\left( s \right){\rm{d}}s} + \frac{{aBM\rho }}{2}{y^2} + \frac{{aBM}}{2}\int_{t - r\left( t \right)}^t {{y^2}\left( s \right){\rm{d}}s} + \\ \frac{{MB\rho }}{2}{z^2} + \frac{{MB}}{2}\int_{t - r\left( t \right)}^t {{y^2}\left( s \right){\rm{d}}s} + {u_1}{z^2}\rho + {u_2}{z^2}\rho - {u_1}\left( {1 - } \right.\\ \left. \beta \right)\int_{t - r\left( t \right)}^t {{z^2}\left( \theta \right){\rm{d}}\theta } - {u_2}\left( {1 - \beta } \right)\int_{t - r\left( t \right)}^t {{y^2}\left( \theta \right){\rm{d}}\theta } = - \left[ {c - } \right.\\ \left. {\left( {\frac{{aL}}{2} + \frac{{aBM}}{2} + {u_2}} \right)\rho } \right]{y^2} - \left[ {d - \left( {\frac{L}{2} + \frac{{MB}}{2} + {u_1}} \right)\rho } \right]{z^2} + \\ \left[ {\frac{{aL}}{2} + \frac{L}{2} - {u_1}\left( {1 - \beta } \right)} \right]\int_{t - r\left( t \right)}^t {{z^2}\left( \theta \right){\rm{d}}\theta } + \left[ {\frac{{aBM}}{2} + \frac{{BM}}{2} - } \right.\\ \left. {{u_2}\left( {1 - \beta } \right)} \right]\int_{t - r\left( t \right)}^t {{y^2}\left( \theta \right){\rm{d}}\theta } . \end{array} $

$ 令\;{u_1} = \frac{{\left( {a + 1} \right)L}}{{2\left( {1 - \beta } \right)}},{u_2} = \frac{{\left( {a + 1} \right)MB}}{{2\left( {1 - \beta } \right)}},则 $

$ \begin{array}{l} \frac{{{\rm{d}}{V_{\left( {{x_t},{y_t},{z_t}} \right)}}}}{{{\rm{d}}t}} \le - \left[ {c - \left( {\frac{{aL}}{2} + \frac{{aBM}}{2} + {u_2}} \right)\rho } \right]{y^2} - \\ \left[ {d - \left( {\frac{L}{2} + \frac{{MB}}{2} + {u_1}} \right)\rho } \right]{z^2}. \end{array} $

若取

$ \begin{array}{l} \sigma \le \min \left\{ {\frac{{2c\left( {1 - \beta } \right)}}{{a\left( {L + BM} \right)\left( {1 - \beta } \right) + \left( {a + 1} \right)MB}},} \right.\\ \left. {\frac{{2d\left( {1 - \beta } \right)}}{{\left( {L + MB} \right)\left( {1 - \beta } \right) + \left( {a + 1} \right)L}}} \right\},可得 \end{array} $

$ \frac{{{\rm{d}}{V_{\left( {{x_t},{y_t},{z_t}} \right)}}}}{{{\rm{d}}t}} \le - {D_5}{y^2} - {D_6}{z^2} \le 0, $

$ 其中\;{D_5} = \left[ {c - \left( {\frac{{aL}}{2} + \frac{{aBM}}{2} + {u_2}} \right)\rho } \right] \ge 0, $

$ {D_6} = \left[ {d - \left( {\frac{L}{2} + \frac{{MB}}{2} + {u_1}} \right)\rho } \right] \ge 0. $

因此，泛函V(x_t, y_t, z_t)满足引理1的条件(2).

容易证明Z中最大不变集是Q={0}，在这里Z= {Φ∈C_H, V′(Φ)=0 }, 根据

$\frac{{{\rm{d}}V}}{{{\rm{d}}t}}{\rm{ = }}0$和系统(2)，易知x=y=z=0.

因此，引理1的所有条件都满足，所以式(1)的零解是渐近稳定的.

定理得证.

3 有界性

引理2^[7]   Gronwall-Reid-Bellman不等式：若u(t)与α(t)都是[a, b]上的连续函数，β(t)≥0在[a, b]上可积且成立

$ u\left( t \right) \le \alpha \left( t \right) + \int_a^t {\beta \left( s \right)u\left( s \right){\rm{d}}s} ,t \in \left[ {a,b} \right], $

则必有

$ u\left( t \right) \le \alpha \left( t \right) + \int_a^t {\beta \left( s \right)\alpha \left( s \right)\exp \left( {\int_a^t {\beta \left( {{s_1}} \right){\rm{d}}{s_1}} } \right){\rm{d}}s} ; $

若α(t)非减，则成立

$ u\left( t \right) \le \alpha \left( t \right)\exp \left( {\int_a^t {\beta \left( s \right){\rm{d}}s} } \right). $

对于p(t, x(t), x′(t), x(t－r(t)), x′(t－r(t)), x″(t))≠0的情况，有下列结论：

定理2 ^[9]假设定理1的条件成立且连续函数p满足下列条件：

$ \left| {p\left( {t,x\left( t \right),x'\left( t \right),x\left( {t - r\left( t \right)} \right),x'\left( {t - r\left( t \right)} \right)} \right.} \right| \le q\left( t \right), $

其中q(t)∈L¹(0, ∞), L¹(0, ∞)是Lebesgue可积函数空间.

当

$ \begin{array}{l} \sigma \le \min \left\{ {\frac{{2c\left( {1 - \beta } \right)}}{{a\left( {L + BM} \right)\left( {1 - \beta } \right) + \left( {a + 1} \right)MB}},} \right.\\ \left. {\frac{{2d\left( {1 - \beta } \right)}}{{\left( {L + MB} \right)\left( {1 - \beta } \right) + \left( {a + 1} \right)L}}} \right\} \end{array} $

时，存在一个有限正常数M使系统(2)的解x(t)对所有的t≥t₀满足不等式

$ \left| {x\left( t \right)} \right| \le \sqrt M ,\left| {x'\left( t \right)} \right| \le \sqrt M ,\left| {x''\left( t \right)} \right| \le \sqrt M . $

注：这里用来证明定理2的Lyapunov函数V(x_t, y_t, z_t)与定理1相同.

证明  对于p(t, x(t), x′(t), x(t－r(t)), x′(t－r(t)), x″(t))≠0的情况，有

$ \begin{array}{l} \frac{{{\rm{d}}{V_{\left( {{x_t},{y_t},{z_t}} \right)}}}}{{{\rm{d}}t}} \le - {D_5}{y^2} - {D_6}{z^2} + \left| {\left( {ay + z} \right)p\left( {t,x\left( t \right),} \right.} \right.\\ \left. {\left. {x'\left( t \right),x\left( {t - r\left( t \right)} \right),x'\left( {t - r\left( t \right)} \right),x''\left( t \right)} \right)} \right| \le \left| {\left( {ay + } \right.} \right.\\ \left. {\left. z \right)} \right|\left| {p\left( {t,x\left( t \right),x'\left( t \right),x\left( {t - r\left( t \right)} \right),x'\left( {t - r\left( t \right)} \right),} \right.} \right.\\ \left. {\left. {x''\left( t \right)} \right)} \right| \le \left( {a\left| y \right| + \left| z \right|} \right)\left| {p\left( {t,x\left( t \right),x'\left( t \right),x\left( {t - } \right.} \right.} \right.\\ \left. {\left. {\left. {r\left( t \right)} \right),x'\left( {t - r\left( t \right)} \right),x''\left( t \right)} \right)} \right| \le {D_7}\left( {\left| y \right| + } \right.\\ \left. {\left| z \right|} \right)q\left( t \right). \end{array} $

其中D₇=max{ 1, a }.

由不等式|y|≤1+y²和|z| ≤1+z², 可得

$ \frac{{{\rm{d}}{V_{\left( {{x_t},{y_t},{z_t}} \right)}}}}{{{\rm{d}}t}} \le {D_7}\left( {2 + {y^2} + {z^2}} \right)q\left( t \right). $

由y²+z²≤D₄^－1V(x_t, y_t, z_t)，可以得到

$ \begin{array}{l} \frac{{{\rm{d}}{V_{\left( {{x_t},{y_t},{z_t}} \right)}}}}{{{\rm{d}}t}} \le {D_7}\left( {2 + D_4^{ - 1}V\left( {{x_t},{y_t},{z_t}} \right)} \right)q\left( t \right) = \\ 2{D_7}q\left( t \right) + {D_7}D_4^{ - 1}V\left( {{x_t},{y_t},{z_t}} \right)q\left( t \right). \end{array} $

(4)

对式(4)从0到t进行积分，并利用假设q∈L¹(0, ∞)和Gronwall-Reid-Bellman不等式，可得

$ \begin{array}{l} V\left( {{x_t},{y_t},{z_t}} \right) \le V\left( {{x_0},{y_0},{z_0}} \right) + 2{D_7}A + \\ {D_7}D_4^{ - 1}\int_0^t {V\left( {{x_s},{y_s},{z_s}} \right)q\left( s \right){\rm{d}}s} \le \left( {V\left( {{x_0},{y_0},{z_0}} \right) + } \right.\\ \left. {2{D_7}A} \right)\exp \left( {{D_7}D_4^{ - 1}\int_0^t {q\left( s \right){\rm{d}}s} } \right) \le \left( {V\left( {{x_0},{y_0},{z_0}} \right) + } \right.\\ \left. {2{D_7}A} \right)\exp \left( {{D_7}D_4^{ - 1}A} \right) = {M_1} < \infty , \end{array} $

其中M₁是正常数，M₁=(V(x₀, y₀, z₀)+2D₇A)exp(D₇D₄^－1A)和$A = \int_0^t {q\left(s \right){\rm{d}}s} $.所以

$ {x^2} + {y^2} + {z^2} \le D_4^{ - 1}V\left( {{x_t},{y_t},{z_t}} \right) \le M, $

其中M=D₄^－1M₁.

因此，当t≥t₀时，不等式|x| ≤$\sqrt M $, |y| ≤ $\sqrt M $, |z| ≤ $\sqrt M $成立.

即当t≥t₀时，|x(t)| ≤ $\sqrt M $, |x′(t) |≤ $\sqrt M $, |x″(t)| ≤ $\sqrt M $成立.

证毕.

例题1  考虑三阶时滞微分方程

$ \begin{array}{l} x'''\left( t \right) + \left( {9 + {{\left( {x'\left( t \right)} \right)}^2}} \right)x''\left( t \right) + 4x'\left( {t - r\left( t \right)} \right) + 2{\rm{arctg}}x\left( {t - r\left( t \right)} \right) = \\ \frac{1}{{1 + {t^2} + {x^2}\left( t \right) + {x^2}\left( {t - r\left( t \right)} \right) + {{x''}^2}\left( t \right) + {{x'}^2}\left( {t - r\left( t \right)} \right)}}. \end{array} $

(5)

方程(5)可表示为如下系统

$ \left\{ \begin{array}{l} x' = y,\\ y' = z,\\ z' = - 2{\rm{arctg}}x\left( t \right) - 4y\left( t \right) - \left( {9 + {y^2}\left( t \right)} \right)z\left( t \right) + \\ \;\;\;\;4\int_{t - r\left( t \right)}^t {z\left( s \right){\rm{d}}s} + 2\int_{t - r\left( t \right)}^t {\frac{{y\left( s \right)}}{{1 + {x^2}\left( s \right)}}{\rm{d}}s + p(t,} \\ x\left( t \right),x'\left( t \right),x\left( {t - r\left( t \right)} \right),x'\left( {t - r\left( t \right)} \right),x\left( t \right)). \end{array} \right. $

令

$ \begin{array}{l} p\left( {t,x\left( t \right),x'\left( t \right),x\left( {t - r\left( t \right)} \right),x'\left( {t - r\left( t \right)} \right),x''\left( t \right)} \right) = \\ \frac{1}{{1 + {t^2} + {x^2}\left( t \right) + {x^2}\left( {t - r\left( t \right)} \right) + {{x''}^2}\left( t \right) + {{x'}^2}\left( {t - r\left( t \right)} \right)}}, \end{array} $

显然方程(5)是式(1)的特殊形式，可知

$ \begin{array}{l} 0 < {\left[ {h\left( x \right)\varphi \left( x \right)} \right]^\prime } = \frac{2}{{{{\left( {1 + x} \right)}^2}}} \le 2 = 3 - 1,a = 1,\\ b = 3,c = 1,\frac{{f\left( y \right)}}{y} = 4 \ge 3 = b,\left( {y \ne 0} \right),\left| {f'\left( y \right)} \right| = \\ 4 \le 5 = L, \end{array} $

$ \begin{array}{l} {\mathop{\rm sgn}} \arctan x = {\rm{sgn}}x,\varphi '\left( x \right) = \frac{2}{{1 + {x^2}}} \le 2 = M,g\left( {x,y} \right) = \\ 9 + {y^2} \ge 9 = 1 + 8,d = 8,{{g'}_x}\left( {x,y} \right)y = 0 \le 0. \end{array} $

由于

$ \begin{array}{l} p\left( {t,x\left( t \right),x'\left( t \right),x\left( {t - r\left( t \right)} \right),x'\left( {t - r\left( t \right)} \right),x''\left( t \right)} \right) = \\ \frac{1}{{1 + {t^2} + {x^2}\left( t \right) + {x^2}\left( {t - r\left( t \right)} \right) + {{x''}^2}\left( t \right) + {{x'}^2}\left( {t - r\left( t \right)} \right)}} \le \\ \frac{1}{{1 + {t^2}}}, \end{array} $

则

$ \int_0^\infty {q\left( s \right){\rm{d}}s} = \int_0^\infty {\frac{1}{{1 + {s^2}}}{\rm{d}}s} = \frac{{\rm{ \mathsf{ π} }}}{2} \le \infty , $

其中q∈L¹(0, +∞).

因此定理1和定理2的所有条件都成立，当p≡0时方程(5)的零解渐近稳定，当p≠0时方程(5)所有解有界.