广东工业大学学报  2016, Vol. 33Issue (1): 62-66.  DOI: 10.3969/j.issn.1007-7162.2016.01.012.
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引用本文 

黄明辉. 多时滞的非线性微分方程的渐近稳定性[J]. 广东工业大学学报, 2016, 33(1): 62-66. DOI: 10.3969/j.issn.1007-7162.2016.01.012.
Huang Ming-hui. Asymptotic Stability of Nonlinear Differential Equation with Time Delays[J]. Journal of Guangdong University of Technology, 2016, 33(1): 62-66. DOI: 10.3969/j.issn.1007-7162.2016.01.012.

基金项目:

广东省自然科学基金资助项目(S2011010005029)

作者简介:

黄明辉(1988-)，男，助教，硕士研究生，主要研究方向为微分动力系统。

文章历史

收稿日期：2014-09-23
Contents              Abstract             Full text              Figures/Tables              PDF
多时滞的非线性微分方程的渐近稳定性
黄明辉    
广州华夏职业学院 基础部，广东 广州 510935
收稿日期：2014-09-23
基金项目：广东省自然科学基金资助项目(S2011010005029)
作者简介：黄明辉(1988-)，男，助教，硕士研究生，主要研究方向为微分动力系统。
摘要: 以时滞的非线性微分方程为研究对象，利用不动点定理证明了时滞的非线性微分方程的渐近稳定性，并得到了零解渐近稳定的充分条件.
关键词: 时滞    渐近稳定性    不动点    非线性    
Asymptotic Stability of Nonlinear Differential Equation with Time Delays
Huang Ming-hui    
Department of Basic Education, Guangzhou Huaxia Technical College, Guangzhou 510935, China
Abstract: This paper takes the nonlinear delay differential equation as the research subject and uses the fixed point theorem to prove asymptotic stability of nonlinear differential equations with delays. Some sufficient conditions for asymptotic stability of the trivial solution are also established.
Key words: time delays    asymptotic stability    fixed point    nonlinear    
1 问题的提出

近年来，时滞微分方程的研究得到了数学、物理以及化学等多个领域学者的关注[1-2].研究下列多时滞非线性微分方程的渐近稳定性

$ x'\left( t \right) = - \sum\limits_{i = 1}^N {{b_i}\left( t \right)f\left( {x\left( {t - {\tau _i}\left( t \right)} \right)} \right)} , $ (1)

其中bi∈ C(R+, R)和τi∈ C(R+, R+)，f连续可导及满足Lipschitz条件，当t→∞时，t－τi(t)→∞，i=1, 2, …, N.

关于方程(1) 的研究已经取得了很多研究成果[3-8].例如，Jin C H、Luo J W[3]利用Banach不动点定理证明了以下时滞微分方程的渐近稳定性

$ x'\left( t \right) = - \sum\limits_{i = 1}^N {{b_i}\left( t \right)x\left( {t - {\tau _i}\left( t \right)} \right)} , $ (2)

上述方程(2) 是方程(1) 中f(x)=x的特殊情况.当N=1和N=2，τ1=0时，方程(1) 相应地改变为

$ x'\left( t \right) = - b\left( t \right)x\left( {t - \tau \left( t \right)} \right) $ (3)

和

$ x'\left( t \right) = - {b_1}\left( t \right)x\left( t \right) - {b_2}\left( t \right)x\left( {t - \tau \left( t \right)} \right). $ (4)

Yorke J A[4]对方程(3) 证明了：如果存在正数β和q，使得

$ 0 < b\left( t \right) \le \beta ,\tau \left( t \right) \le q,\beta q < \frac{3}{2}, $ (5)

那么方程(3) 的零解一致稳定.Krisztin T[5]将Yorke J A[4]的定理进一步推广为：假设bi:R+→R+连续，bi≤β和τi:R+→[0, qi]连续，i=1, 2, …, N，如果$ \sum\limits_{\mathit{i}{\rm{ = 1}}}^\mathit{n} {{\mathit{\beta }_\mathit{i}}{\mathit{q}_\mathit{i}}} $≤1，则方程(2) 的零解是一致稳定的.如果$ \sum\limits_{\mathit{i}{\rm{ = 1}}}^\mathit{n} {{\mathit{\beta }_\mathit{i}}{\mathit{q}_\mathit{i}}} $＜1，则方程(2) 的零解渐近稳定.

在b(t)≥0以及τ(t)有界的情况下，Yoneyama T[6]也对Yorke J A[4]的结论进一步推广且证明：如果$ \mathit{\lambda = }_{\mathit{t} \ge \text{0}}^{{\rm{sup}}}\int_{\mathit{t}{\rm{ - }}\mathit{\tau }\left( \mathit{t} \right)}^\mathit{t} {\mathit{b}\left( \mathit{s} \right)} < \frac{3}{2}$和$ \mathit{\mu = }_{\mathit{t} \ge {\rm{0}}}^{{\rm{inf}}}\int_{\mathit{t}{\rm{ - }}\mathit{\tau }\left( \mathit{t} \right)}^\mathit{t} {\mathit{b}\left( \mathit{s} \right)} > 0$，则方程(3) 的零解一致渐近稳定.Yoneyama T[7]证明了在t→0，$ \int_{\mathit{t}{\rm{ - }}\mathit{\tau }\left( \mathit{t} \right)}^\mathit{t} {\mathit{b}\left( \mathit{s} \right)} \to 0$的情况下，x′(t)=－b(t)f(x(t－τ(t)))的零解是一致稳定的.Hara T[8]证明了$ _{\mathit{t} \ge \text{0}}^{{\rm{sup}}}\int_{\mathit{t}{\rm{ - }}\mathit{\tau }\left( \mathit{t} \right)}^\mathit{t} {\mathit{b}\left( \mathit{s} \right)} < 1$以及$ \int_{\rm{0}}^\mathit{t} {\mathit{b}\left( \mathit{s} \right) = \infty } $，则方程(3) 的零解是一致稳定且渐近稳定.进一步地，Muroya Y[9]表明$ _{\mathit{t} \ge {\rm{0}}}^{{\rm{inf}}}\int_{\mathit{t}{\rm{ - }}\mathit{\tau }\left( \mathit{t} \right)}^\mathit{t} {\mathit{b}\left( \mathit{s} \right)} > 0$不是方程(3) 零解渐近稳定的必要条件.

20世纪以来，Lyapunov直接法是研究微分方程零解稳定性的主要方法[10-15].但是，仍然存在很多问题并没有得到解决.本文仿用Jin C H和Luo J W[15]的方法，利用Banach不动点建立了新的渐近稳定性条件，并且不要求τi(t)有界，也不要求bi(t)恒正或者恒负.

2 主要结论

设C(S1, S2)表示所有连续函数φ:S1→S2的集合.定义mi(θ)=inf{s－τi(s):s≥θ}，$ {\mathit{\tilde m}}$(θ)=min{mi(θ):1≤θ≤N}，$ {\mathit{\tilde C}}$(θ)=C([$ {\mathit{\tilde m}}$(θ), θ], R)及上确界的符号为·.

对任意(t0, φ)∈ R +× $ {\mathit{\tilde C}}$(t0)，方程(1) 过点(t0, φ)的解是连续函数x:[$ {\mathit{\tilde m}}$(t0), t0+α]→Rn.存在常数α＞0，使得在[t0, t0+α]，x(t)满足方程(1) 且在[$ {\mathit{\tilde m}}$(t0), t0]，x(s)=φ(s).这样的解表示为x(t)=x(t, t0, φ).

定理1  假设存在常数α∈(0, 1) 以及函数τ0∈ C(R+, R+)，当t→∞时，t－τ0(t)→∞，h∈ C ([$ {\mathit{\tilde m}}$(0), ∞], R)，其中m0(θ)=inf{s－τ0(s):s≥θ}，使得当t≥0时，

(1) $ _{\mathit{t} \to \infty }^{\;{\rm{lim}}}{\rm{inf}}\int_{\rm{0}}^\mathit{t} {\mathit{h}\left( \mathit{s} \right){\rm{d}}\mathit{s}} > - \infty $.

(2) f可导，存在正数M1、M2、L，使得| f(x)|≤M1，|f′(x)| ≤M2，|f′(x)| ≤M2，|f′(x)－f′(y)| ≤L |x－y |，x, y∈ C ([$ {\mathit{\tilde m}}$(0), ∞], R)，以及f(0)=0，f′(0)=0.

(3) $ \int_{\mathit{t}{\rm{ - }}{\mathit{\tau }_{\rm{0}}}\left( \mathit{t} \right)}^\mathit{t} {\mathit{|h}\left( \mathit{s} \right){\rm{|d}}\mathit{s}{\rm{ + }}} $

$ \begin{array}{l} \int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}\left| {h\left( s \right)} \right|} \int_{s - {\tau _0}\left( s \right)}^s {\left| {h\left( v \right)} \right|{\rm{d}}v{\rm{d}}s} + \\ \int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}\left| {h\left( {s - {\tau _0}\left( s \right)} \right)\left( {1 - {{\tau '}_0}\left( s \right)} \right)} \right|{\rm{d}}s} + \\ {{\rm{M}}_2}\int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}\sum\limits_{i = 1}^N {\left| {{b_k}\left( s \right)} \right|{\rm{d}}s} } + \\ \left( {M_2^2 + {M_1}L} \right)\sum\limits_{i = 1}^N {\int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}\sum\limits_{j = k}^N {\left| {{b_j}\left( s \right)} \right|} } } \times \\ \int_{s - {\tau _{k - 1}}}^{s - \tau k} {\sum\limits_{i = 1}^N {\left| {{b_i}\left( v \right)} \right|{\rm{d}}v{\rm{d}}s} } \le \alpha , \end{array} $

其中bi(t)在区间[m(0), ∞)连续，i=1, 2, …, N.

方程(1) 的解渐近稳定当且仅当

(4) $ \int_{\rm{0}}^\mathit{t} {\mathit{h}\left( \mathit{s} \right){\rm{d}}\mathit{s}} \to\infty, \mathit{t} \to \infty $.

证明  假设(4) 成立.对任意t0≥0，设$ \mathit{K = }_{\mathit{t} \ge \text{0}}^{{\rm{sup}}}\left\{ {{{\rm{e}}^{-\int_{\rm{0}}^\mathit{t} {\mathit{h}\left( \mathit{s} \right){\rm{d}}\mathit{s}} }}} \right\}$.并定义$ {\mathit{\tilde C}}$(t0)=C([m(t0), t0], R).设φ∈C(t0)是固定函数，并且定义

S={x∈C([m(t0), ∞], R):当t→0, x(t)→0, 对s∈[m(t0), t0], x(s)=φ(s)}，则S是距离为ρ(x, y)=supt≥t0{ |x(t)－y(t)| }的完备度量空间.

将方程(1) 转换为以下形式：

$ \begin{array}{l} x'\left( t \right) = - \sum\limits_{k = 1}^N {{b_k}\left( t \right)f\left( {x\left( {t - {\tau _0}\left( t \right)} \right)} \right)} - \\ \sum\limits_{k = 1}^N {{b_k}\left( t \right)\int_{t - {\tau _0}\left( t \right)}^{t - {\tau _1}\left( t \right)} {f'\left( {x\left( s \right)} \right)x'\left( s \right){\rm{d}}s} } - \\ \sum\limits_{k = 2}^N {{b_k}\left( t \right)\int_{t - {\tau _1}\left( t \right)}^{t - {\tau _2}\left( t \right)} {f'\left( {x\left( s \right)} \right)x'\left( s \right){\rm{d}}s} } - \cdots - \\ {b_N}\left( t \right)\int_{t - {\tau _{N - 1}}\left( t \right)}^{t - {\tau _N}\left( t \right)} {f'\left( {x\left( s \right)} \right)x'\left( s \right){\rm{d}}s} = \\ - \sum\limits_{k = 1}^N {{b_k}\left( t \right)f\left( {x\left( {t - {\tau _0}\left( t \right)} \right)} \right)} - \\ \sum\limits_{k = 1}^N {\sum\limits_{j = k}^N {{b_j}\left( t \right)\int_{t - {\tau _{k - 1}}\left( t \right)}^{t - {\tau _k}\left( t \right)} {\sum\limits_{i = 1}^N {{b_i}\left( s \right){\rm{f'}}\left( {x\left( s \right)} \right)x'\left( s \right){\rm{d}}s} } } } . \end{array} $

两边同时乘以$ {{\rm{e}}^{ - \int_{\rm{0}}^\mathit{t} {\mathit{h}\left( \mathit{s} \right){\rm{d}}\mathit{s}} }}$并从t0到t积分，整理得

$ \begin{array}{l} x\left( t \right) = \varphi \left( {{t_0}} \right){{\rm{e}}^{ - \int_{{t_0}}^t {h\left( u \right){\rm{d}}u} }} + \int_{{t_0}}^t {{{\rm{e}}^{ - \int_{{t_0}}^t {h\left( u \right){\rm{d}}u} }}h\left( s \right)x\left( s \right){\rm{d}}s} - \\ \int_{{t_0}}^t{{\rm{e}}^{ - \int_{{t_0}}^t {h\left( u \right){\rm{d}}u} }}\sum\limits_{k = 1}^N {{b_k}\left( s \right)f\left( {x\left( {s - {\tau _0}\left( s \right)} \right)} \right){\rm{d}}s} + \\ \sum\limits_{k = 1}^N {\int_{{t_0}}^t {{{\rm{e}}^{ - \int_{{t_0}}^t {h\left( u \right){\rm{d}}u} }}} } \sum\limits_{j = k}^N {{b_j}\left( t \right)\int_{s - {\tau _{k - 1}}\left( t \right)}^{s - {\tau _k}\left( t \right)} {f'\left( {x\left( v \right)} \right)} } \times \\ \sum\limits_{i = 1}^N {{b_i}\left( v \right)f\left( {x\left( {v - {\tau _0}\left( v \right)} \right)} \right){\rm{d}}v{\rm{d}}s} . \end{array} $

定义算子P:S→S，当t∈[m(t0), t0]时，(Px)(t)=φ(t)和t≥t0时，

$ \begin{array}{l} \left( {Px} \right)\left( t \right) = \left( {\varphi \left( {{t_0}} \right) - \int_{{t_0} - {\tau _0}\left( {{t_0}} \right)}^t {h\left( s \right)\varphi \left( s \right){\rm{d}}s} } \right){\rm{e}}\\ \int_{t - {\tau _0}\left( t \right)}^t {h\left( s \right)x\left( s \right){\rm{d}}s} - \int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}h\left( s \right)} \times \\ \int_{s - {\tau _0}\left( s \right)}^s {h\left( v \right)x\left( v \right){\rm{d}}v{\rm{d}}s} + \int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}x\left( {s - {\tau _0}\left( s \right)} \right)h\left( {s - } \right.} \\ \left. {{\tau _0}\left( s \right)} \right)\left( {1 - {{\tau '}_0}\left( s \right)} \right)ds - \int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}\sum\limits_{i = 1}^N {{b_k}\left( s \right)} } \times \\ f\left( {x\left( {s - {\tau _0}\left( s \right)} \right)} \right){\rm{d}}s + \sum\limits_{k = 1}^N {\int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} } \sum\limits_{j = k}^N {{b_j}\left( t \right)} \times \\ \int_{s - {\tau _{k - 1}}\left( t \right)}^{s - {\tau _k}\left( t \right)} {f'\left( {x\left( v \right)} \right)} \sum\limits_{i = 1}^N {{b_i}\left( v \right)f\left( {x\left( {v - {\tau _0}\left( v \right)} \right)} \right){\rm{d}}v{\rm{d}}s} . \end{array} $ (6)

显然，(Px)∈C([m(t0), ∞], R).首先证明当t→∞时，式(6) 每一项都是趋向于零的.由(3) 可知，第1项以及第2项都趋于零，因为t→∞时，x(t)→∞.接着证明当t→∞时，最后一项I6趋向于零.由于当t→∞时，x(t)→∞以及t－τi(t)→∞.对任意ε＞0，存在T1, T2＞0，使得s≥T1时，有s－τi(s)≥T2以及v≥T2时，|v－τi(v)|＜ε，对i=1, 2, …, N.因此，当t≥T1，

$ \begin{array}{l} \left| {{I_6}} \right| = \\ \left| {\sum\limits_{k = 1}^N {\int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} } \sum\limits_{j = k}^N {{b_j}\left( t \right)\int_{s - {\tau _{k - 1}}\left( t \right)}^{s - {\tau _k}\left( t \right)} {f'\left( {x\left( v \right)} \right)} } \times } \right.\\ \left. {\sum\limits_{i = 1}^N {{b_i}\left( v \right)f\left( {x\left( {v - {\tau _0}\left( v \right)} \right)} \right){\rm{d}}v{\rm{d}}s} } \right| \le \\ \sum\limits_{k = 1}^N {\int_{{t_0}}^{{T_1}} {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} } \left| {\sum\limits_{j = k}^N {{b_j}\left( t \right)} } \right|\left| {\int_{s - {\tau _{k - 1}}\left( t \right)}^{s - {\tau _k}\left( t \right)} {f'\left( {x\left( v \right)} \right)} \times } \right.\\ \left. {\sum\limits_{i = 1}^N {{b_i}\left( v \right)f\left( {x\left( {v - {\tau _0}\left( v \right)} \right)} \right){\rm{d}}v} } \right|{\rm{d}}s + \end{array} $
$ \begin{array}{l} \sum\limits_{k = 1}^N {\int_{{T_1}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} } \left| {\sum\limits_{j = k}^N {{b_j}\left( t \right)} } \right|\left| {\int_{s - {\tau _{k - 1}}\left( t \right)}^{s - {\tau _k}\left( t \right)} {f'\left( {x\left( v \right)} \right)} \times } \right.\\ \left. {\sum\limits_{i = 1}^N {{b_i}\left( v \right)x\left( {v - {\tau _0}\left( v \right)} \right){\rm{d}}v} } \right|{\rm{d}}s \le \\ M_2^2\sum\limits_{k = 1}^N {\int_{{t_0}}^{{T_1}} {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} } \left| {\sum\limits_{j = k}^N {{b_j}\left( t \right)} } \right| \times \\ \left| {\int_{s - {\tau _{k - 1}}\left( t \right)}^{s - {\tau _k}\left( t \right)} {\sum\limits_{i = 1}^N {{b_i}\left( v \right)x\left( {v - {\tau _0}\left( v \right)} \right){\rm{d}}v} } } \right|{\rm{d}}s + \\ M_2^2\sum\limits_{k = 1}^N {\int_{{T_1}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} } \left| {\sum\limits_{j = k}^N {{b_j}\left( t \right)} } \right| \times \end{array} $
$ \begin{array}{l} \left| {\int_{s - {\tau _{k - 1}}\left( t \right)}^{s - {\tau _k}\left( t \right)} {\sum\limits_{i = 1}^N {{b_i}\left( v \right)x\left( {v - {\tau _0}\left( v \right)} \right){\rm{d}}v} } } \right|{\rm{d}}s \le \\ M_2^2\mathop {\sup }\limits_{\sigma \ge m\left( {{t_0}} \right)} \left| {x\left( \sigma \right)} \right|\sum\limits_{k = 1}^N {\int_{{t_0}}^{{T_1}} {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} } \times \\ \left| {\sum\limits_{j = k}^N {{b_j}\left( t \right)} } \right|\left| {\int_{s - {\tau _{k - 1}}\left( t \right)}^{s - {\tau _k}\left( t \right)} {\sum\limits_{i = 1}^N {\left| {{b_i}\left( v \right)} \right|{\rm{d}}v} } } \right|{\rm{d}}s + \\ M_2^2\varepsilon \sum\limits_{k = 1}^N {\int_{{T_1}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} } \left| {\sum\limits_{j = k}^N {{b_j}\left( t \right)} } \right| \times \\ \left| {\int_{s - {\tau _{k - 1}}\left( t \right)}^{s - {\tau _k}\left( t \right)} {\sum\limits_{i = 1}^N {\left| {{b_i}\left( v \right)} \right|{\rm{d}}v} } } \right|{\rm{d}}s. \end{array} $

由(4) 可知，存在T3＞T1，当t≥T3时

$ \begin{array}{l} \mathop {\sup }\limits_{\sigma \ge m\left( {{t_0}} \right)} \left| {x\left( \sigma \right)} \right|\sum\limits_{k = 1}^N {\int_{{t_0}}^{{T_1}} {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} } \left| {\sum\limits_{j = k}^N {{b_j}\left( t \right)} } \right| \times \\ \left| {\int_{s - {\tau _{k - 1}}\left( t \right)}^{s - {\tau _k}\left( t \right)} {\sum\limits_{i = 1}^N {\left| {{b_i}\left( v \right)} \right|{\rm{d}}v} } } \right|{\rm{d}}s = \\ \mathop {\sup }\limits_{\sigma \ge m\left( {{t_0}} \right)} \left| {x\left( \sigma \right)} \right|{{\rm{e}}^{ - \int_{{T_1}}^t {h\left( u \right){\rm{d}}u} }}\sum\limits_{k = 1}^N {\int_{{t_0}}^{{T_1}} {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} } \left| {\sum\limits_{j = k}^N {{b_j}\left( t \right)} } \right| \times \\ \left| {\int_{s - {\tau _{k - 1}}\left( t \right)}^{s - {\tau _k}\left( t \right)} {\sum\limits_{i = 1}^N {\left| {{b_i}\left( v \right)} \right|{\rm{d}}v} } } \right|{\rm{d}}s < \varepsilon . \end{array} $

由(3) 可知，|I6|≤2M2L1ε.因此，t→∞时，|I6|→0.类似地，当t→∞时，其余各项趋向于零.从而得到t→∞时，(Px)(t)→0，因此Px∈S.

接下来，证明P是压缩映射.对任意x, y∈S，

$ \begin{array}{l} \left| {\left( {Px} \right)\left( t \right) - \left( {Py} \right)\left( t \right)} \right| \le \\ \int_{t - {\tau _0}\left( t \right)}^t {h\left( s \right)\left| {x\left( s \right) - y\left( s \right)} \right|{\rm{d}}s} + \\ \int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} \left| {h\left( s \right)} \right|\int_{t - {\tau _0}\left( t \right)}^t {h\left( v \right)\left| {x\left( v \right) - } \right.} \\ \left. {y\left( v \right)} \right|{\rm{d}}v{\rm{d}}s + \int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} \left| {x\left( {s - {\tau _0}\left( s \right)} \right) - } \right.\\ \left. {y\left( {s - {\tau _0}\left( s \right)} \right)} \right|\left| {h\left( {s - {\tau _0}\left( s \right)} \right)\left( {1 - {{\tau '}_0}\left( s \right)} \right)} \right|{\rm{d}}s + \end{array} $
$ \begin{array}{l} \int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} \sum\limits_{i = 1}^N {\left| {{b_k}\left( s \right)} \right|\left| {f\left( {x\left( {s - {\tau _0}\left( s \right)} \right)} \right) - } \right.} \\ \left. {f\left( {y\left( {s - {\tau _0}\left( s \right)} \right)} \right)} \right|{\rm{d}}s + \sum\limits_{k = 1}^N {\int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} \times } \\ \sum\limits_{j = k}^N {\left| {{b_j}\left( t \right)} \right|} \int_{s - {\tau _{k - 1}}\left( t \right)}^{s - {\tau _k}\left( t \right)} {\sum\limits_{i = 1}^N {\left| {{b_i}\left( v \right)} \right|\left| {f'\left( {x\left( v \right)} \right)} \right.} } \times \\ \left. {f\left( {x\left( {v - {\tau _0}\left( v \right)} \right)} \right) - f'\left( {y\left( v \right)} \right)f\left( {y\left( {v - {\tau _0}\left( v \right)} \right)} \right)} \right| \times \\ {\rm{d}}v{\rm{d}}s \le \left( {\int_{t - {\tau _0}\left( t \right)}^t {\left| {h\left( s \right)} \right|{\rm{d}}s} + \int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} \left| {h\left( s \right)} \right| \times } \right. \end{array} $
$ \begin{array}{l} \int_{s - {\tau _0}\left( s \right)}^t {\left| {h\left( v \right)} \right|{\rm{d}}v{\rm{d}}s} + \int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} \left| {h\left( {s - {\tau _0}\left( s \right)} \right) \times } \right.\\ \left. {\left( {1 - {{\tau '}_0}\left( s \right)} \right)} \right|{\rm{d}}s + {M_2}\int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} \sum\limits_{i = 1}^N {\left| {{b_k}\left( s \right)} \right|{\rm{d}}s} + \\ \left( {M_2^2 + {M_1}L} \right)\sum\limits_{k = 1}^N {\int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} } \sum\limits_{j = k}^N {\left| {{b_j}\left( s \right)} \right|} \times \\ \int_{s - {\tau _{k - 1}}}^{s - {\tau _k}} {\sum\limits_{i = 1}^N {\left| {{b_i}\left( v \right)} \right|\left. {{\rm{d}}v{\rm{d}}s} \right)\left| {x - y} \right|} } \le \alpha x - y. \end{array} $

根据压缩映射原理，P在S中存在唯一的不动点x，x是方程(1) 的解，即在[m(t0), t0]上且初始函数为φ(s)的解x(t)=x(t, t0, φ)→0，当t→∞时.

接着，证明方程零解的稳定性.给定ε＞0以及选择δ＞0(δ＜ε)满足$ 2\mathit{\delta K}{\text{e}^{\int_{\rm{0}}^{{\mathit{t}_{\rm{0}}}} {\mathit{h}\left( \mathit{u} \right){\rm{d}}\mathit{u}} }} + \mathit{\alpha \varepsilon < \varepsilon }$.对任意φ＜δ，x(t)=x(t, t0, φ)是方程(1) 的解，则x(t)=(Px)(t).接下来，证明对所有t≥t0有|x(t)|＜ε.在[m(t0), t0]上，|x(t)|＜ε.如果存在t*＞t0使得x(t*)=ε以及当m(t0)≤s＜t*，x(s)＜ε，则根据式(6) 有

$ \begin{array}{l} x\left( {{t^ * }} \right) \le \left\| \varphi \right\|\left( {1 + \int_{{t_0} - {\tau _0}\left( {{t_0}} \right)}^t {h\left( s \right){\rm{d}}s} } \right){{\rm{e}}^{ - \int_{{t_0}}^{{t^ * }} {h\left( u \right){\rm{d}}u} }} + \\ \varepsilon \int_{t - {\tau _0}\left( t \right)}^t {\left| {h\left( s \right)} \right|{\rm{d}}s} + \varepsilon \int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}\left| {h\left( s \right)} \right|} \times \\ \int_{s - {\tau _0}\left( s \right)}^s {\left| {h\left( v \right)} \right|{\rm{d}}v{\rm{d}}s} + \varepsilon \int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} \times \\ \left| {h\left( {s - {\tau _0}\left( s \right)} \right)\left( {1 - {{\tau '}_0}\left( s \right)} \right)} \right|{\rm{d}}s + \\ \varepsilon {M_1}\int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} \sum\limits_{i = 1}^N {\left| {{b_k}\left( s \right)} \right|{\rm{d}}s} + \\ \varepsilon {M_1}{M_2}\sum\limits_{k = 1}^N {\int_{{t_0}}^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}} } \times \\ \sum\limits_{j = k}^N {\left| {{b_j}\left( t \right)} \right|} \int_{s - {\tau _{k - 1}}}^{s - {\tau _k}} {\sum\limits_{i = 1}^N {\left| {{b_i}\left( v \right)} \right|{\rm{d}}v{\rm{d}}s} } \le \\ 2\delta K{{\rm{e}}^{\int_0^{{t_0}} {h\left( u \right)du} }} + \alpha \varepsilon < \varepsilon , \end{array} $ (7)

这与t*的定义不相符.所以，对所有t≥t0有|x(t)|＜ε且方程(1) 的零解是稳定的.这就表明了当条件(4) 成立时，方程(1) 的零解渐近稳定.

相反地，如果(4) 不成立.由1) 可知，存在序列{tn}，当n→∞，tn→∞，即对于l∈ R，有$ \mathop {{\rm{lim}}}\limits_{\mathit{n} \to \infty } \int_{\rm{0}}^{{\mathit{t}_\mathit{n}}} {\mathit{h}\left( \mathit{s} \right){\rm{d}}\mathit{s = l}} $.选择正数J，使得对任意n≥1，满足

$ - J \le \int_0^{{t_n}} {h\left( s \right){\rm{d}}s} \le J. $

设

$ \begin{array}{l} \omega \left( s \right) = \left| {h\left( s \right)} \right|\int_{s - {\tau _0}\left( s \right)}^s {\left| {h\left( v \right)} \right|{\rm{d}}v} + \\ \left| {h\left( {s - {\tau _0}\left( s \right)} \right)\left( {1 - {{\tau '}_0}\left( s \right)} \right)} \right| + {M_1}\left| {\sum\limits_{k = 1}^N {{b_k}\left( s \right)} } \right| + \\ {M_1}{M_2}\sum\limits_{k = 1}^N {\left| {\sum\limits_{j = 1}^N {{b_j}\left( s \right)} } \right|} \left| {\int_{s - {\tau _{k - 1}}\left( s \right)}^{s - {\tau _k}\left( s \right)} {\sum\limits_{i = 1}^N {\left| {{b_i}\left( s \right)} \right|{\rm{d}}v} } } \right|,\\ s \ge 0. \end{array} $

由3) 可知，

$ \int_0^{{t_n}} {{{\rm{e}}^{ - \int_s^{{t_n}} {h\left( u \right){\rm{d}}u} }}\omega \left( s \right)} \le \alpha , $

从而，有

$ \int_0^{{t_n}} {{{\rm{e}}^{\int_0^s {h\left( u \right){\rm{d}}u} }}\omega \left( s \right){\rm{d}}s} \le \alpha {e^{\int_0^{{t_n}} {h\left( u \right){\rm{d}}u} }} \le {{\rm{e}}^J}, $

序列$ \left\{ {\int_{\rm{0}}^{{\mathit{t}_\mathit{n}}} {{{\rm{e}}^{ - \int_{\rm{0}}^\mathit{s} {\mathit{h}\left( \mathit{u} \right){\rm{d}}\mathit{u}} }}\mathit{\omega }\left( \mathit{s} \right)} {\rm{d}}\mathit{s}} \right\}$是有界的，因此存在一个收敛的子序列.假设

$ \mathop {\lim }\limits_{n \to \infty } \int_0^{{t_n}} {{{\rm{e}}^{\int_0^s {h\left( u \right){\rm{d}}u} }}\omega \left( s \right){\rm{d}}s} = \gamma ,\gamma \in {{\bf{R}}^ + }. $

选择充分大的正整数k，使得，对任意n≥k，有

$ \mathop {\lim }\limits_{n \to \infty } \int_{{t_{\bar k}}}^{{t_n}} {{{\rm{e}}^{\int_0^s {h\left( u \right){\rm{d}}u} }}\omega \left( s \right){\rm{d}}s} < \frac{{{\delta _0}}}{{4K}}, $

其中δ0＞0，满足2δ0KeJ+α＜1.

现在考察方程(6) 的解x(t)=x(t, tk, φ)，其中φ(tk)=δ0以及s≤tk时，|φ(s)|≤δ0.与方程(7) 相类似的结论表示为t≥tk时，|x(t)|≤1.选择φ使得

$ \varphi \left( {{t_{\bar k}}} \right) - \int_{{t_{\bar k}} - {\tau _0}\left( {{t_{\bar k}}} \right)}^{{t_{\bar k}}} {h\left( s \right)\varphi \left( s \right)ds} \ge \frac{1}{2}{\delta _0}. $

由式(6) 以及x(t)=(Px)(t)，当n≥tk时，

$ \begin{array}{l} \left| {x\left( {{t_n}} \right) - \int_{{t_n} - {\tau _0}\left( {{t_n}} \right)}^{{t_n}} {h\left( s \right)x\left( s \right){\rm{d}}s} } \right| \ge \\ \frac{1}{2}{\delta _0}{{\rm{e}}^{ - \int_{{t_{\bar k}}}^{{t_n}} {h\left( u \right){\rm{d}}u} }} - \int_{{t_{\bar k}}}^{{t_n}} {{{\rm{e}}^{\int_s^{{t_n}} {h\left( u \right){\rm{d}}u} }}\omega \left( s \right){\rm{d}}s} = \\ \frac{1}{2}{\delta _0}{{\rm{e}}^{ - \int_{{t_{\bar k}}}^{{t_n}} {h\left( u \right){\rm{d}}u} }} - {{\rm{e}}^{ - \int_0^{{t_n}} {h\left( u \right){\rm{d}}u} }}\int_{{t_{\bar k}}}^{{t_n}} {{{\rm{e}}^{\int_0^s {h\left( u \right){\rm{d}}u} }}\omega \left( s \right){\rm{d}}s} = \\ {{\rm{e}}^{ - \int_{{t_{\bar k}}}^{{t_n}} {h\left( u \right){\rm{d}}u} }}\left( {\frac{1}{2}{\delta _0} - {{\rm{e}}^{ - \int_0^{{t_{\bar k}}} {h\left( u \right){\rm{d}}u} }}\int_{{t_{\bar k}}}^{{t_n}} {{{\rm{e}}^{\int_0^s {h\left( u \right){\rm{d}}u} }}\omega \left( s \right){\rm{d}}s} } \right) \ge \\ {{\rm{e}}^{ - \int_{{t_{\bar k}}}^{{t_n}} {h\left( u \right){\rm{d}}u} }}\left( {\frac{1}{2}{\delta _0} - K\int_{{t_{\bar k}}}^{{t_n}} {{{\rm{e}}^{\int_0^s {h\left( u \right){\rm{d}}u} }}\omega \left( s \right){\rm{d}}s} } \right) \ge \\ \frac{1}{4}{\delta _0}{{\rm{e}}^{ - \int_{{t_{\bar k}}}^{{t_n}} {h\left( u \right){\rm{d}}u} }} \ge \frac{1}{4}{\delta _0}{{\rm{e}}^{ - 2J}} > 0. \end{array} $ (8)

另一方面，假设方程(1) 的零解渐近稳定，则t→∞时，x(t)=x(t, tk, φ)→0.因为当n→∞时，tn－τ(tn)→∞以及3) 的成立.因此，

$ x\left( {{t_n}} \right) - \int_{{t_n} - {\tau _0}\left( {{t_n}} \right)}^{{t_n}} {h\left( s \right)x\left( s \right){\rm{d}}s} \to 0,n \to \infty , $

这与式(8) 相矛盾，所以(4) 是方程(1) 的零解渐近稳定的必要条件.证明完毕.

3 算例

考察以下标量方程

$ \begin{array}{l} x'\left( t \right) = - {b_1}\left( t \right)f\left( {x\left( {t - {\tau _1}\left( t \right)} \right)} \right) - {b_2}\left( t \right)f\left( {x\left( {t - } \right.} \right.\\ \left. {\left. {{\tau _2}\left( t \right)} \right)} \right), \end{array} $ (9)

其中τ1(t)=0.15t，τ2(t)=0.28t，$ {\mathit{b}_{\rm{1}}}\left( \mathit{t} \right) = \frac{{1.55}}{{0.85\mathit{t} \text{+} {\rm{1}}}}, {\mathit{b}_2}\left( \mathit{t} \right) = \frac{{0.45}}{{0.72\mathit{t} \text{+} {\rm{1}}}}, $以及

$ f\left( x \right) = \left\{ \begin{array}{l} \gamma {{\rm{e}}^{ - \frac{1}{{{x^2}}}}},x = 0,\\ 0,\;\;\;\;x = 0, \end{array} \right. $

γ为充分小的正数.

显然|f(x)|≤γ.因为x≠0时，

$ \begin{array}{l} f'\left( x \right) = \frac{{2\gamma }}{{{x^3}}}{{\rm{e}}^{ - \frac{1}{{{x^2}}}}} \le 3\frac{{\sqrt 6 }}{2}\gamma {{\rm{e}}^{ - \frac{3}{2}}},\\ f''\left( x \right) = \frac{\gamma }{{{x^6}}}{{\rm{e}}^{ - \frac{1}{{{x^2}}}}}\left( {4 - 6{x^2}} \right) \le 8\gamma {{\rm{e}}^{ - 2}}. \end{array} $

所以，$ {\mathit{M}_{\rm{1}}} = \mathit{\gamma }{\rm{, }}{\mathit{M}_2} = 3\frac{{\sqrt 6 }}{2}\mathit{\gamma }{{\rm{e}}^{{\rm{ - }}\frac{3}{2}}}{\rm{, }}\mathit{L = }{\rm{8}}\mathit{\gamma }{{\rm{e}}^{{\rm{ - 2}}}}$.

选择τ0(t)=τ1(t)=0.15t和h(t)= $ \frac{{2.3}}{{\mathit{t}{\rm{ + 1}}}}$，则

$ \begin{array}{l} \int_{t - {\tau _0}\left( t \right)}^{{t_n}} {\left| {h\left( s \right)} \right|{\rm{d}}s} = \int_{0.85t}^t {\frac{{2.3}}{{s + 1}}{\rm{d}}s} = \\ 2.3\ln \frac{{t + 1}}{{0.85t + 1}} < 0.3738,\\ \int_0^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}\left| {h\left( s \right)} \right|} \int_{s - {\tau _0}\left( s \right)}^t {\left| {h\left( v \right)} \right|{\rm{d}}v{\rm{d}}s} < 0.3738,\\ \int_0^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}\left| {h\left( {s - {\tau _0}\left( s \right)} \right)\left( {1 - {{\tau '}_0}\left( s \right)} \right){\rm{d}}s} \right|} = \\ \int_0^t {\frac{{0.405}}{{0.85s + 1}}} {{\rm{e}}^{ - \int_s^t {\frac{{2.3}}{{u + 1}}{\rm{d}}u} }}{\rm{d}}s \le \frac{{0.405}}{{2.3 \times 0.85}} < 0.2072. \end{array} $

以及

$ \begin{array}{l} \int_0^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}\left( {\left| {{b_1}\left( s \right) + {b_2}\left( s \right)} \right|} \right){\rm{d}}s} = \\ \int_0^t {{{\rm{e}}^{ - \int_s^t {\frac{{2.3}}{{u + 1}}{\rm{d}}u} }}\left( {\frac{{1.55}}{{0.85s + 1}} + \frac{{0.45}}{{0.72s + 1}}} \right){\rm{d}}s} = \\ \int_0^t {{{\rm{e}}^{ - \int_s^t {\frac{{2.3}}{{u + 1}}{\rm{d}}u} }}\left( {\frac{{1.4985s + 2}}{{\left( {0.72s + 1} \right)\left( {0.85s + 1} \right)}}} \right){\rm{d}}s} \le \\ \frac{{1.4985}}{{0.72}}\int_0^t {{{\rm{e}}^{ - \int_s^t {\frac{{2.3}}{{u + 1}}{\rm{d}}u} }} \times \frac{2}{{\left( {0.85s + 1} \right)}}{\rm{d}}s} \le \\ \frac{{1.4985}}{{0.72 \times 0.85 \times 2.3}} < 1.0646. \end{array} $
$ \begin{array}{l} \int_{t - {\tau _1}\left( t \right)}^{t - {\tau _2}\left( t \right)} {\left( {\left| {{b_1}\left( s \right) + {b_2}\left( s \right)} \right|} \right){\rm{d}}s} = \int_0^t {\left( {\frac{{1.55}}{{0.85s + 1}} + } \right.} \\ \left. {\frac{{0.45}}{{0.72s + 1}}} \right){\rm{d}}s < 0.4065, \end{array} $
$ \begin{array}{l} \int_0^t {{{\rm{e}}^{ - \int_s^t {h\left( u \right){\rm{d}}u} }}\left| {{b_2}\left( s \right)} \right|\left| {\int_{s - {\tau _1}\left( s \right)}^{s - {\tau _2}\left( s \right)} {\left( {\left| {{b_1}\left( v \right) + {b_2}\left( v \right)} \right|} \right){\rm{d}}v} } \right|{\rm{d}}s} \le \\ 0.4065\int_0^t {{{\rm{e}}^{ - \int_s^t {\frac{{2.3}}{{u + 1}}{\rm{d}}u} }} \times \frac{{0.45}}{{0.72s + 1}}{\rm{d}}s} \le \\ \left( {\frac{{1.55}}{{0.85}} + \frac{{0.45}}{{0.72}}} \right)\ln \frac{{0.85}}{{0.72}} < 0.1105, \end{array} $

令α=0.373 8+0.373 8+0.207 2+

$ \begin{array}{l} {M_2}1.0646 + \left( {M_2^2 + {M_1}L} \right)0.1105 = 0.9548 + \\ 1.06463\frac{{\sqrt 6 }}{2}\gamma {{\rm{e}}^{ - \frac{3}{2}}} + 0.1105\left( {\frac{3}{2}{\gamma ^2}{{\rm{e}}^{ - 3}} + 8{\gamma ^2}{{\rm{e}}^{ - 2}}} \right). \end{array} $

因为γ是足够小的正数，可以选取充分小的γ，使得α＜1，由定理1可知，方程(9) 的零解渐近稳定.

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