含无界时滞的微极流方程组稳态解的稳定性

易卢燕; 刘国威

doi:10.21656/1000-0887.450300

摘要: 利用四种不同的技术结合稳定性理论研究了含无界时滞的微极流方程组稳态解的稳定性.结果表明，当无界时滞函数关于时间连续可微时, 非平凡稳态解具有局部稳定性和平凡稳态解具有渐近稳定性；当无界时滞函数关于时间仅连续时, 非平凡稳态解具有全局稳定性；当无界时滞为比例时滞时, 平凡稳态解具有多项式稳定性.

关键词:

Abstract: The stability of stationary solutions to micropolar fluid equations with unbounded delay was studied through combination of 4 different techniques with the stability theory. The results show that, when the unbounded delay function is continuously differentiable with respect to time, the nontrivial stationary solution will be locally stable and the trivial stationary solution will be asymptotically stable; when the unbounded delay function is only continuous with respect to time, the nontrivial stationary solution will be globally stable; when the unbounded delay is a proportional delay, the trivial stationary solution will be polynomially stable.

Key words:

0. 引言

在流体动力学理论中, 微极流是一类可以检测流体微观结构变化的, 伴有非均匀应力张量的流体, 例如动物的血液、高分子悬浮液、液晶和稀释水溶性聚合物溶液.微极流方程组最早由Eringen于1966年在文献[1]中提出，用来描述微极流流体的运动. 至目前，已有大量文献研究了微极流方程组的各类数学问题, 比如文献[2-4]在不同框架下研究了解的适定性, 文献[5-9]在不同框架下建立了解的长时间渐近行为.特别值得一提的是，文献[5]得到了在二维有界域上初值空间为$ \hat{H} $时稳态解的指数稳定性.

在现实世界中, 时滞效应会自然地出现, 例如在风洞实验中流体的运动, 见文献[10].最近, 含时滞的微极流方程组也吸引了众多学者的关注, 比如文献[11-13]和[14-15]分别研究了含有界时滞和无界时滞的微极流方程组弱解的适定性, 拉回吸引子的存在性和性质.其中文献[11]和[14]还分别建立了在有界时滞时初值空间为$ \hat{H} \times L^2(-h, 0 ; \hat{V}) $，在无界时滞时初值空间为$ {c_\gamma }(\hat H) = \{ {\rm{ \boldsymbol \varPhi}} \in c(( - \infty , 0];\hat H)\mid \mathop {\lim }\limits_{s \to - \infty } {{\rm{e}}^{{\gamma _s}}}{\rm{\boldsymbol \varPhi}}(s) \in \hat H\} (\gamma > 0) $时稳态解的指数稳定性.我们注意到，为了建立稳态解的指数稳定性, 参数γ>0起到了重要的作用.若γ=0，稳态解的指数稳定性是否还存在?若不存在, 那么是否具有一定的稳定性?

本文旨在回答上述问题, 去研究含无界时滞的微极流方程组初值空间为

$c_0(\hat{H})=\left\{\boldsymbol{\varPhi} \in c((-\infty, 0] ; \hat{H}) \mid \mathop {\lim }\limits_{s \rightarrow-\infty} \boldsymbol{\varPhi}(s) \in \hat{H}\right\}$

时稳态解的稳定性.方程组具体形式如下：

$\frac{\partial \boldsymbol{u}}{\partial t}-\left(\nu+\nu_{\mathrm{r}}\right) \Delta \boldsymbol{u}+(\boldsymbol{u} \cdot \nabla) \boldsymbol{u}+\nabla_p-2 \nu_{\mathrm{r}} \nabla^{\perp} \omega=\boldsymbol{f}(x, t)+\boldsymbol{g}\left(t, \boldsymbol{u}_t\right),$

(1)

$\nabla \cdot \boldsymbol{u}=0,$

(2)

$\frac{\partial \omega}{\partial t}-\left(c_{\mathrm{a}}+c_{\mathrm{d}}\right) \Delta \omega+(\boldsymbol{u} \cdot \nabla) \omega+4 \nu_{\mathrm{r}} \omega-2 \nu_{\mathrm{r}} \nabla \times \boldsymbol{u}=\tilde{f}(x, t)+\tilde{g}\left(t, \omega_t\right),$

(3)

其中▽和Δ分别表示梯度算子和Laplace算子,

$\nabla^{\perp} \omega=\left(\frac{\partial \omega}{\partial x_2}, -\frac{\partial \omega}{\partial x_1}\right), \nabla \times \boldsymbol{u}=\frac{\partial u_2}{\partial x_1}-\frac{\partial u_1}{\partial x_2},$

$ (t, x) \in \mathbb{R}_{+} \times \varOmega $, Ω为二维有界区域; ν, ν_r, c_a, c_d表示正的黏性系数, u=(u₁(t, x), u₂(t, x))表示流体的速度, p=p(t, x)表示流体的压力, ω=ω(t, x)表示粒子旋转的角速度, f=(f₁(t, x), f₂(t, x))和$ \tilde{f}=\tilde{f}(t, x) $表示不含时滞的外力项, g(t, u_t)=(g₁(t, u_t), g₂(t, u_t))和$ \tilde{g}\left(t, \omega_t\right) $表示具有某种记忆或遗传特征的含时滞的外力项, 其中u_t和ω_t是定义在(-∞, 0]上的时滞函数，

$\boldsymbol{u}_t=\boldsymbol{u}_t(s)=\boldsymbol{u}(t+s), \omega_t=\omega_t(s)=\omega(t+s), \quad s \in(-\infty, 0] .$

设aΩ为区域Ω的边界且足够光滑, 给方程组(1)—(3)赋予下述初边值条件：

$\boldsymbol{u}(t, x)=0, \omega(t, x)=0, \quad(t, x) \in(0, \infty) \times \partial \varOmega,$

(4)

$(\boldsymbol{u}(s, x), \omega(s, x))=\left(\boldsymbol{\phi}(s, x), \phi_3(s, x)\right), \quad s \in(-\infty, 0], x \in \varOmega .$

(5)

本文通过考虑几类特殊形式的无界时滞函数, 运用四种不同的方法建立了初边值问题(1)—(5)稳态解的稳定性.具体地, 当无界时滞函数关于时间连续可微时, 通过Lyapunov稳定性的定义建立了非平凡稳态解的局部稳定性, 再通过构造Lyapunov泛函的方法建立了平凡稳态解的渐近稳定性.当无界时滞函数关于时间仅连续时, 借助Razumikhin技术建立了非平凡稳态解的全局稳定性.最后, 考虑比例时滞时, 借助比例时滞的特殊性质建立了平凡稳态解的多项式稳定性.

本文具体安排如下: 第1节是预备知识, 在这一部分, 首先通过引入抽象算子将初边值问题(1)—(5)改写为泛函微分方程的初值问题, 并给出其弱解和强解的定义.第2节证明了初边值问题(1)—(5)弱解的适定性、稳态解的存在唯一性及正则性.第3节建立了初边值问题(1)—(5)稳态解的稳定性.

1. 预备知识

本文中用$ \mathbb{R} $和$ \mathbb{R}_{+} $分别表示实数和非负实数的集合. C表示一般的常数, 它在不同的地方可能取不同的值, 如果需要强调依赖于某些量时, 我们会用符号C(·, ·, …). L^p(Ω)和W^{m, p}(Ω) 分别表示经典的Lebesgue空间和Sobolev空间，特别地, H^m(Ω)=W^{m, 2}(Ω).H₀¹(Ω)表示C₀^∞(Ω)在H¹(Ω) 空间中的闭包.在此基础上, 引入向量函数空间：$ v=\left\{\boldsymbol{\varphi} \in c_0^{\infty}(\varOmega) \times c_0^{\infty}(\varOmega) \mid \boldsymbol{\varphi}=\left(\varphi_1, \varphi_2\right), \nabla \cdot \boldsymbol{\varphi}=0\right\} ; $ H=V在(L²(Ω))²中的闭包, 范数为‖·‖_H且对偶空间为H^*=H；V=V在(rowspan=""¹(Ω))²中的闭包, 范数为‖·‖_V且对偶空间为V^*；$ \hat{H}=H \times L^2(\varOmega) $, 范数为$ \|\cdot\| _\hat{H} $, 对偶空间为$ \hat{H}^* $；$ \hat{V}=V \times H_0^1(\varOmega) $, 范数为$ \|\cdot\| _\hat{V} $, 对偶空间为$ \hat{V}^* $. 记(·, ·)为L²(Ω), H或$ \hat{H} $中的内积, 〈·, ·〉为V与V^*或$ \hat{V} \text { 与 } \hat{V}^* $中函数的对偶积.在不会引起混淆的情况下, 范数‖·‖_H和$ \|\cdot\|_{\hat{H}} $可简记为‖·‖. 进一步地，令L^p(I; X)为定义在区间I上, 取值在Banach空间X中的p次可积函数全体, 范数为$ \|\cdot\|{ }_{L^p(I ; X)} $和C(I; X)为定义在区间I上, 取值在Banach空间X中连续函数全体, 范数为‖·‖_{C(I; X)}.

下面定义三个抽象算子.第一个算子A定义为: 对任意的v=(u, ω), Ψ=(ψ, ψ₃)∈ $ \hat{V} $ (其中u=(u₁, u₂), ψ=(ψ₁, ψ₂)∈V), 则下式成立

$\begin{aligned} & \langle A \boldsymbol{v}, \boldsymbol{\varPsi}\rangle=\left(\nu+\nu_{\mathrm{r}}\right)(\nabla \boldsymbol{u}, \nabla \boldsymbol{\psi})+\left(c_{\mathrm{a}}+c_{\mathrm{d}}\right)\left(\nabla \omega, \nabla \psi_3\right)= \\ & \quad\left(\nu+\nu_{\mathrm{r}}\right) \sum\limits_{j, k=1}^2 \int_{\varOmega} \frac{\partial u_j}{\partial x_k} \frac{\partial \psi_j}{\partial x_k} \mathrm{~d} x+\left(c_{\mathrm{a}}+c_{\mathrm{d}}\right) \sum\limits_{j=1}^2 \int_{\varOmega} \frac{\partial \omega}{\partial x_j} \frac{\partial \psi_3}{\partial x_j} \mathrm{~d} x . \end{aligned}$

则实际上A=-PΔ(P为Leray投影算子), A的定义域为D(A)= $ \hat{V} $ ∩(H²(Ω))³.第二个算子B(·, ·)定义为: 对于任意u=(u₁, u₂)∈V, Ψ=(ψ₁, ψ₂, ψ₃)∈V, Φ=(Φ₁, Φ₂, Φ₃)∈ $ \hat{V} $, 则下式成立

$\langle B(\boldsymbol{u}, \boldsymbol{\varPsi}), \boldsymbol{\varPhi}\rangle=((\boldsymbol{u} \cdot \nabla) \boldsymbol{\varPsi}, \boldsymbol{\varPhi})=\sum\limits_{k=1}^3 \sum\limits_{j=1}^2 \int_{\varOmega} u_j \frac{\partial \psi_k}{\partial x_j} \phi_k \mathrm{~d} x .$

第三个算子N(·)定义为: 对任意的v=(u, ω)∈ $ \hat{V} $ (其中u=(u₁, u₂)∈V), 则下式成立

$N(\boldsymbol{v})=\left(-2 \nu_{\mathrm{r}} \nabla^{\perp} \omega, 4 \nu_{\mathrm{r}} \omega-2 \nu_{\mathrm{r}} \nabla \times \boldsymbol{u}\right) .$

回顾算子A, B(·, ·), N(·)的相关性质和估计, 有

引理1^{[11, 14]}

① A是从$ \hat{V} $到$ \hat{V}^* $, 或者D(A)到H的线性连续算子, N(·)是从$ \hat{V} $到$ \hat{H} $的线性连续算子, 且满足

$\|N(\boldsymbol{\varPsi})\| \leqslant C\left(\nu_{\mathrm{r}}\right)\|\boldsymbol{\varPsi}\| \hat{v}, \quad \forall \boldsymbol{\varPsi} \in \hat{V},$

(6)

$\delta_1\|\boldsymbol{\varPsi}\|_{\hat{V}}^2 \leqslant\langle A \boldsymbol{\varPsi}, \boldsymbol{\varPsi}\rangle+\langle N(\boldsymbol{\varPsi}), \boldsymbol{\varPsi}\rangle, \quad \forall \boldsymbol{\varPsi} \in \hat{V},$

(7)

其中C(ν_r)>0, δ₁=minν, c_a+c_d.

② B(·, ·)是从V× $ \hat{V} $到$ \hat{V}^* $的双线性连续算子，且满足

$\langle B(\boldsymbol{u}, \boldsymbol{\varPsi}), \boldsymbol{\varPhi}\rangle=-\langle B(\boldsymbol{u}, \boldsymbol{\varPhi}), \boldsymbol{\varPsi}\rangle, \langle B(\boldsymbol{u}, \boldsymbol{\varPsi}), \boldsymbol{\varPsi}\rangle=0, \quad \forall \boldsymbol{u} \in V ; \boldsymbol{\varPsi}, \boldsymbol{\varPhi} \in \hat{V},$

(8)

并存在正常数λ, 使得

$\begin{aligned} |\langle B(\boldsymbol{u}, \boldsymbol{\varPsi}), \boldsymbol{\varPhi}\rangle| \leqslant \lambda\|\boldsymbol{u}\|^{1 / 2}\|\nabla \boldsymbol{u}\|^{1 / 2}\|\boldsymbol{\varPhi}\|^{1 / 2}\|\nabla \boldsymbol{\varPhi}\|^{1 / 2} & \|\nabla \boldsymbol{\varPsi}\| \\ & \forall(\boldsymbol{u}, \boldsymbol{\varPhi}, \boldsymbol{\varPsi}) \in V \times \hat{V} \times \hat{V}, \end{aligned}$

(9)

$\text { | }\langle B(\boldsymbol{u}, \boldsymbol{\varPsi}), \boldsymbol{\varPhi}\rangle \mid \leqslant \lambda\|\boldsymbol{u}\|\|\boldsymbol{\varPsi}\|^{1 / 2}\|A \boldsymbol{\varPsi}\|^{1 / 2}\|\boldsymbol{\varPhi}\|, \quad \forall(\boldsymbol{u}, \boldsymbol{\varPsi}, \boldsymbol{\varPhi}) \in V \times D(A) \times \hat{V} .$

(10)

由上述定义的三个算子可知, 初边值问题(1)—(5)在分布意义下可改写成如下的泛函微分方程的初值问题：

$\begin{cases}\frac{\partial \boldsymbol{v}}{\partial t}+A \boldsymbol{v}+B(\boldsymbol{u}, \boldsymbol{v})+N(\boldsymbol{v})=\boldsymbol{F}(x, t)+\boldsymbol{G}\left(t, \boldsymbol{v}_t\right), & (t, x) \in(0, \infty) \times \varOmega, \\ \boldsymbol{v}_0(s, x)=\boldsymbol{v}(s, x)=\boldsymbol{\varPhi}(s, x), & s \in(-\infty, 0], x \in \varOmega, \end{cases}$

(11)

其中v=(u, ω), Φ=(Φ, Φ₃)，F(x, t)=(f(x, t), $ \tilde{f} $ (x, t)), G(t, v_t)=(g(t, u_t), $ \tilde{g} $ (t, ω_t)).为处理无界时滞项, 本文考虑初值空间为C₀($ \hat{H} $),

$c_0(\hat{H})=\left\{\boldsymbol{\varPhi} \in c((-\infty, 0] ; \hat{H}): \exists \mathop {\lim }\limits_{s \rightarrow-\infty} \boldsymbol{\varPhi}(s) \in \hat{H}\right\} .$

参见文献[16], 该空间为Banach空间, 其范数定义为

$\|\boldsymbol{\varPhi}\|_{c_0(\hat{H})}=\sup _{s \in(-\infty, 0]}\|\boldsymbol{\varPhi}(s)\| .$

初值问题(11)的弱解和强解为

定义1 任意给定T>0和v₀= Φ∈C₀($ \hat{H} $), 若函数

$v \in c((-\infty, T] ; \hat{H}) \cap L^2(0, T ; \hat{V})$

在D′(0, T)的意义下满足

$\begin{aligned} & \frac{\mathrm{d}}{\mathrm{~d} t}(\boldsymbol{v}(t), \boldsymbol{\varPsi})+\langle A \boldsymbol{v}(t), \boldsymbol{\varPsi}\rangle+\langle B(\boldsymbol{u}(t), \boldsymbol{v}(t)), \boldsymbol{\varPsi}\rangle+\langle N(\boldsymbol{v}(t)), \boldsymbol{\varPsi}\rangle= \\ & \quad\langle\boldsymbol{F}(t), \boldsymbol{\varPsi}\rangle+\left(\boldsymbol{G}\left(t, \boldsymbol{v}_t\right), \boldsymbol{\varPsi}\right), \end{aligned}$

则称v为初值问题(11)的弱解. 如果弱解$ \boldsymbol{v} \in c((-\infty, 0] ; \hat{V}) \cap L^2(0, T ; D(A)) $, 则v为初值问题(11)的强解.

2. 弱解和稳态解的存在唯一性

这一节将首先建立初值问题(11)弱解的整体适定性, 然后建立初值问题(11)稳态解的存在唯一性和正则性.为此, 需要对含无界时滞的外力项G(t, v_t)做如下假设.

假设1 设函数$ \boldsymbol{G}:[0, T] \times c_0(\hat{H}) \rightarrow \boldsymbol{G}(t, \boldsymbol{\xi}) \in \hat{H} $满足

(i) 对任意的$ \boldsymbol{\xi} \in c_0(\hat{H}) \text {, 映射 }[0, T] \ni t \mapsto \boldsymbol{G}(t, \boldsymbol{\xi}) \in \hat{H} $是可测的.

(ii) G(·, 0)=0.

(iii) 存在某个常数L_G>0, 使得对任意的t∈[0, T], ξ, η∈C₀($ \hat{H} $)满足

$\|\boldsymbol{G}(t, \boldsymbol{\xi})-\boldsymbol{G}(t, \boldsymbol{\eta})\| \leqslant L_{\boldsymbol{G}}\|\boldsymbol{\xi}-\boldsymbol{\eta}\|_{c_0(\hat{H})} .$

由(ii)和(iii)可知

$\|\boldsymbol{G}(t, \boldsymbol{\xi})\| \leqslant L_{\boldsymbol{G}}\|\boldsymbol{\xi}\|_{c_0(\hat{H})}, \quad \forall \boldsymbol{\xi} \in c_0(\hat{H}) .$

(12)

注1 参见文献 [16-18], 含无界变量时滞的函数满足上述条件.

初值问题(11)弱解的整体适定性可被建立.

定理1 对于任意$ T>0, \boldsymbol{F}(t, x) \in L^2\left(0, T ; \hat{V}^*\right), \boldsymbol{\varPhi} \in c_0(\hat{rowspan=""}) \text { 和 } \boldsymbol{G}\left(t, \boldsymbol{v}_t\right) $满足假设1, 则初值问题(11)在区间(-∞, T] 上存在唯一弱解v=(u, ω), 且这个弱解关于初值是连续的.若F(t, x)∈L²(0, T; $ \hat{H} $), Φ∈C₀($ \hat{H} $)且Φ(0)∈ $ \hat{V} $, 则上述弱解v=(u, ω)实际上是初值问题(11)的强解.

证明显然，运用经典的Faedo-Galerkin逼近方法和能量估计方法可证明定理1中弱解的存在唯一性以及关于初值的连续性, 具体证明过程类似文献[14]中定理3.1—3.3的证明过程, 仅需做微小的调整.为了不再重复, 这里省去具体证明过程.当$ \boldsymbol{F}(t, x) \in L^2(0, T ; \hat{H}), \boldsymbol{\varPhi} \in c_0(\hat{H}) \text { 且 } \boldsymbol{\varPhi}(0) \in \hat{V} $时, 可得F(t, x)+G(t, v_t)∈L²(0, T; $ \hat{H} $), 由文献[4]中弱解的正则性结果可知弱解实际上是强解.

对应初值问题(11)的稳态方程为

$A \boldsymbol{v}^*+B\left(\boldsymbol{u}^*, \boldsymbol{v}^*\right)+N\left(\boldsymbol{v}^*\right)=\boldsymbol{F}(x)+\boldsymbol{G}\left(\boldsymbol{v}^*\right),$

(13)

这里G(v^*)=G(0, v^*).

初值问题(11)的稳态解为

定义2 如果$ \boldsymbol{v}^*=\left(\boldsymbol{u}^*, \boldsymbol{\omega}^*\right) \in \hat{V} $满足

$\left\langle A \boldsymbol{v}^*, \boldsymbol{\varPsi}\right\rangle+\left\langle B\left(\boldsymbol{u}^*, \boldsymbol{v}^*\right), \boldsymbol{\varPsi}\right\rangle+\left\langle N\left(\boldsymbol{v}^*\right), \boldsymbol{\varPsi}\right\rangle=\langle\boldsymbol{F}, \boldsymbol{\varPsi}\rangle+\left(\boldsymbol{G}\left(\boldsymbol{v}^*\right), \boldsymbol{\varPsi}\right), \quad \forall \boldsymbol{\varPsi} \in \hat{V},$

则称v^*为稳态方程(13)的解, 即为初值问题(11)的稳态解.

由此开始, 记λ₁为算子A=-PΔ(P为Leray投影算子)的第一个特征值. 由Poincaré不等式可知对于任意的Ψ∈ $ \hat{V} $, 有

$\|\boldsymbol{\varPsi}\| \leqslant \lambda_1^{-1 / 2}\|\boldsymbol{\varPsi}\| \hat{\nu} .$

(14)

初值问题(11)稳态解的存在唯一性和正则性可被建立.

定理2 设G满足假设1，且δ₁> λ₁^-1L_G.

(a) 若F∈ $ \hat{V}^* $, 则式(13)至少存在一个解v^*∈ $ \hat{V} $;

(b) 若$ \left(\delta_1-\lambda_1^{-1} L_G\right)^2>\lambda \lambda_1^{-1 / 2}\|\boldsymbol{F}\|_{\hat{v}} . $ (这里λ来自式(10)), 则式(13)的解是唯一的；

(c) 若F∈ $ \hat{H}f $, 则式(13)的解v^*∈D(A).

证明显然，由Lax-Milgram定理和Schauder不动点定理可证明(a)和(b)的结果成立, 具体步骤类似文献[11]中定理2.1的证明过程, 仅需做微小的调整.

在这里, 我们具体给出(c)的证明过程.由$ \boldsymbol{F} \in \hat{H} \hookrightarrow \hat{V}^* $和(a)可知式(13)存在解v^*∈ $ \hat{V} $. 由式(13)、(10)、(6)、假设1、式(14)和事实‖u^*‖≤‖v^*‖, 可得

$\begin{aligned} & \left\|A \boldsymbol{v}^*\right\| \leqslant\left\|B\left(\boldsymbol{u}^*, \boldsymbol{v}^*\right)\right\|+\left\|N\left(\boldsymbol{v}^*\right)\right\|+\left\|\boldsymbol{G}\left(\boldsymbol{v}^*\right)\right\|+\|\boldsymbol{F}\| \leqslant \\ & C\left\|\boldsymbol{v}^*\right\|\left\|\boldsymbol{v}^*\right\|^{1 / 2}\left\|A \boldsymbol{v}^*\right\|^{1 / 2}+C\left(\boldsymbol{\nu}_{\mathrm{r}}\right)\left\|\boldsymbol{v}^*\right\| \hat{v}+\lambda_1^{-1 / 2} L_{\boldsymbol{G}}\left\|\boldsymbol{v}^*\right\| \hat{v}+\|\boldsymbol{F}\| . \end{aligned}$

由Hölder不等式和式(14), 可得

$\left\|A \boldsymbol{v}^*\right\| \leqslant C \lambda_1^{-3 / 2}\left\|\boldsymbol{v}^*\right\| \hat{i} \hat{v}+\frac{1}{2}\left\|A \boldsymbol{v}^*\right\|+C\left(\nu_{\mathrm{r}}\right)\left\|\boldsymbol{v}^*\right\| \hat{v}+\lambda_1^{-1 / 2} L_G\left\|\boldsymbol{v}^*\right\| \hat{v}+\|\boldsymbol{F}\|,$

即

$\left\|A \boldsymbol{v}^*\right\| \leqslant C \lambda_1^{-3 / 2}\left\|\boldsymbol{v}^*\right\| \hat{\hat{v}}+2 C\left(\nu_{\mathrm{r}}\right)\left\|\boldsymbol{v}^*\right\| \hat{\hat{v}}+2 \lambda_1^{-1 / 2} L_G\left\|\boldsymbol{v}^*\right\| \hat{\hat{v}}+2\|\boldsymbol{F}\| .$

(15)

将式(13)与v^*做对偶积, 可得

$\left\langle A v^*, v^*\right\rangle+\left\langle B\left(u^*, v^*\right), v^*\right\rangle+\left\langle N\left(v^*\right), v^*\right\rangle=\left(\boldsymbol{F}, v^*\right)+\left(\boldsymbol{G}\left(v^*\right), v^*\right) .$

由式(7)、(8)、假设1和式(14), 可得

$\begin{gathered} \delta_1\left\|\boldsymbol{v}^*\right\|_{\hat{v}}^2 \leqslant\|\boldsymbol{F}\| \cdot\left\|\boldsymbol{v}^*\right\|+L_{\boldsymbol{G}}\left\|\boldsymbol{v}^*\right\|^2 \leqslant \\ \lambda_1^{-1 / 2}\|\boldsymbol{F}\| \cdot\left\|\boldsymbol{v}^*\right\| \hat{v}+\lambda_1^{-1} L_{\boldsymbol{G}}\left\|\boldsymbol{v}^*\right\|_{\hat{v}}^2, \end{gathered}$

即

$\left\|v^*\right\| \hat{v} \leqslant \frac{\lambda_1^{-1 / 2}\|\boldsymbol{F}\|}{\delta_1-\lambda_1^{-1} L_G}$

(16)

将式(16)代入式(15), 可得

$\left\|A \boldsymbol{v}^*\right\| \leqslant C \lambda_1^{-3 / 2}\left(\frac{\lambda_1^{-1 / 2}\|\boldsymbol{F}\|}{\delta_1-\lambda_1^{-1} L_G}\right)^3+2\left(C\left(\nu_{\mathrm{r}}\right)+\lambda_1^{-1 / 2} L_G\right) \frac{\lambda_1^{-1 / 2}\|\boldsymbol{F}\|}{\delta_1-\lambda_1^{-1} L_G}+2\|\boldsymbol{F}\| .$

上式意味着v^*∈D(A).

3. 稳态解的稳定性

这一节将考虑几种特殊形式的无界时滞函数, 进一步建立初值问题(11)稳态解的稳定性.首先, 当无界时滞函数关于时间连续可微时, 分别应用Lyapunov稳定性定义和构造Lyapunov泛函的方法建立非平凡稳态解的局部稳定性和平凡稳态解的渐近稳定性.其次, 当无界时滞函数关于时间仅连续时, 结合Razumikhin技术建立非平凡稳态解的全局稳定性.最后, 考虑当无界时滞函数为比例时滞时, 得到平凡稳态解的多项式稳定性.

首先, 回顾Lyapunov稳定性的定义.

定义3^[16] 设v(t), v^*分别为初值问题(11)的弱解和稳态解.

(i) 如果对任意的ε>0, 存在δ>0, 当初值Φ∈C₀($ \hat{H} $) 满足$ \boldsymbol{\varPhi} \in c_0(\hat{H}) \text { 满足 }\left\|\boldsymbol{\varPhi}-\boldsymbol{v}^*\right\|_{c_0(\hat{H})} < \delta $时, 有

$\left\|\boldsymbol{v}(t)-\boldsymbol{v}^*\right\| < \varepsilon, \quad \forall t>0,$

则称稳态解v^*是稳定的;

(ii)若存在$ \tilde{\delta}>0 $, 当初值Φ∈C₀($ \hat{H} $)满足$ \left\|\boldsymbol{\varPhi}-\boldsymbol{v}^*\right\| < \tilde{\delta} $时, 有

$\mathop {\lim }\limits_{t \rightarrow \infty}\left\|\boldsymbol{v}(t)-\boldsymbol{v}^*\right\|=0$

则称稳态解v^*是吸引的;

(iii) 若稳态解v^*同时是稳定的和吸引的, 则称v^*是渐近稳定的.

3.1 当无界时滞函数关于时间连续可微时稳态解的稳定性

在本小节中, 考虑无界时滞函数满足下述条件.

条件1令

$\boldsymbol{G}(t, \boldsymbol{\xi})=\hat{\boldsymbol{G}}(\boldsymbol{\xi}(-\rho(t))),$

其中$ (t, \boldsymbol{\xi}) \in \mathbb{R}_{+} \times c_0(\hat{H}), \hat{\boldsymbol{G}}: \hat{H} \rightarrow \hat{H} $为Lipschitz映射, 其Lipschitz常数为$ L_{\hat{G}}, \hat{\boldsymbol{G}}(0)=\mathbf{0} $, 且

$\rho \in C^1\left(\mathbb{R}_{+} ; \mathbb{R}_{+}\right), \rho^*=\sup _{t \geqslant 0} \rho^{\prime}(t) < 1 .$

注2 由注1可知, 满足条件1的G(t, ξ)满足假设1.

根据Lyapunov稳定性的定义可建立非平凡稳态解的局部稳定性.

定理3 设F∈ $ \hat{H} $, 条件1成立, $ \delta_1>\lambda_1^{-1} L_{\hat{G}} $和

$\delta_1>\frac{\lambda \lambda_1^{-1}\|\boldsymbol{F}\|}{\delta_1-\lambda_1^{-1} L_{\hat{\boldsymbol{G}}}}+\frac{\left(2-\rho^*\right) \lambda_1^{-1} L_{\hat{\boldsymbol{G}}}}{2\left(1-\rho^*\right)},$

(17)

则初值问题(11)存在稳态解$ \boldsymbol{v}^* \in D(A) \text {. 当 } \boldsymbol{F}(t) \equiv \boldsymbol{F}, \boldsymbol{\varPhi} \in c_0(\hat{H}) $时, 初值问题(11)存在唯一的弱解v(t), 满足

$\left\|\boldsymbol{v}(t)-\boldsymbol{v}^*\right\|^2 \leqslant C\left(\left\|\boldsymbol{\varPhi}(0)-\boldsymbol{v}^*\right\|^2+\left\|\boldsymbol{\varPhi}-\boldsymbol{v}^*\right\|_{L^2((-\rho(0), 0) ; \hat{\nu})}^2\right), \quad \forall t \geqslant 0 .$

证明显然，由条件1、$ F \in \hat{H}_、\delta_1>\lambda_1^{-1} L_{\hat{G}} $和定理1可知，当$ \boldsymbol{F}(t) \equiv \boldsymbol{F}, \boldsymbol{\varPhi} \in c_0(\hat{H}) $时, 初值问题(11)存在唯一的弱解v=(u, ω). 由定理2可知初值问题(11)存在稳态解v^*∈D(A).

设z(t)=v(t)-v^*, 则有

$\frac{\mathrm{d}}{\mathrm{~d} t} \boldsymbol{z}(t)+A \boldsymbol{z}(t)+B(\boldsymbol{u}(t), \boldsymbol{v}(t))-B\left(\boldsymbol{u}^*, \boldsymbol{v}^*\right)+N(\boldsymbol{z}(t))=\hat{\boldsymbol{G}}(\boldsymbol{v}(t-\rho(t)))-\hat{\boldsymbol{G}}\left(\boldsymbol{v}^*\right) .$

(18)

将式(18)与z(t)做对偶积, 可得

$\begin{aligned} & \frac{1}{2} \frac{\mathrm{~d}}{\mathrm{~d} t}\|z(t)\|^2+\langle A(z(t)), z(t)\rangle+\langle N(z(t)), z(t)\rangle= \\ & \quad\left\langle B\left(u^*, v^*\right)-B(u(t), v(t)), z(t)\right\rangle+\left(\hat{\boldsymbol{G}}(v(t-\rho(t)))-\hat{\boldsymbol{G}}\left(v^*\right), z(t)\right) . \end{aligned}$

(19)

由式(8)、$ \left\|\boldsymbol{u}(t)-\boldsymbol{u}^*\right\| \leqslant\|\boldsymbol{z}(t)\| 、\left\|\nabla\left(\boldsymbol{u}(t)-\boldsymbol{u}^*\right)\right\| \leqslant\|\nabla \boldsymbol{z}(t)\| $、式(9)和(14), 可得

$\begin{aligned} & \mathrm{I}\left\langle B(\boldsymbol{u}(t), \boldsymbol{v}(t))-B\left(\boldsymbol{u}^*, \boldsymbol{v}^*\right), \boldsymbol{z}(t)\right\rangle \mathrm{I}= \\ & \quad\left|\left\langle B(\boldsymbol{u}(t), \boldsymbol{v}(t))-B\left(\boldsymbol{u}(t), \boldsymbol{v}^*\right)+B\left(\boldsymbol{u}(t), \boldsymbol{v}^*\right)-B\left(\boldsymbol{u}^*, \boldsymbol{v}^*\right), \boldsymbol{z}(t)\right\rangle\right|= \\ & \quad\left|\left\langle B(\boldsymbol{u}(t), \boldsymbol{z}(t))+B\left(\boldsymbol{u}(t)-\boldsymbol{u}^*, \boldsymbol{v}^*\right), \boldsymbol{z}(t)\right\rangle\right|=\left|\left\langle B\left(\boldsymbol{u}(t)-\boldsymbol{u}^*, \boldsymbol{v}^*\right), \boldsymbol{z}(t)\right\rangle\right| \leqslant \\ & \quad \lambda\left\|\boldsymbol{u}(t)-\boldsymbol{u}^*\right\|^{1 / 2}\left\|\nabla\left(\boldsymbol{u}(t)-\boldsymbol{u}^*\right)\right\|^{1 / 2}\|\boldsymbol{z}(t)\|^{1 / 2}\|\nabla \boldsymbol{z}(t)\|^{1 / 2}\left\|\nabla \boldsymbol{v}^*\right\| \leqslant \\ & \quad \lambda \lambda_1^{-1 / 2}\|\boldsymbol{z}(t)\|_{\hat{V}}^2 \cdot\left\|\boldsymbol{v}^*\right\| \hat{v} . \end{aligned}$

(20)

将式(16)代入式(20), 可知

$\left|\left\langle B(\boldsymbol{u}(t), \boldsymbol{v}(t))-B\left(\boldsymbol{u}^*, \boldsymbol{v}^*\right), \boldsymbol{z}(t)\right\rangle\right| \leqslant \frac{\lambda \lambda_1^{-1}\|\boldsymbol{F}\|}{\delta_1-\lambda_1^{-1} L_{\hat{\boldsymbol{G}}}}\|\boldsymbol{z}(t)\|_{\hat{\hat{V}}}^2 .$

(21)

由Hölder不等式、式(14)和假设1, 可得

$\begin{aligned} 2 I & \left(\hat{\boldsymbol{G}}(\boldsymbol{v}(t-\rho(t)))-\hat{\boldsymbol{G}}\left(\boldsymbol{v}^*\right), \boldsymbol{z}(t)\right) \mid \leqslant 2\left\|\hat{\boldsymbol{G}}(\boldsymbol{v}(t-\rho(t)))-\hat{\boldsymbol{G}}\left(\boldsymbol{v}^*\right)\right\|\|\boldsymbol{z}(t)\| \leqslant \\ & 2 L_{\hat{\boldsymbol{G}}}\left\|\boldsymbol{v}(t-\rho(t))-\boldsymbol{v}^*\right\|\|\boldsymbol{z}(t)\| \hat{v} \leqslant \\ & \lambda_1^{-1} L_{\hat{\boldsymbol{G}}}\|\boldsymbol{z}(t)\|_{\hat{V}}^2+L_{\hat{\boldsymbol{G}}}\|\boldsymbol{z}(t-\rho(t))\|^2 . \end{aligned}$

(22)

由式(19)、(7)、(21)、(22), 可得

$\begin{aligned} & \frac{\mathrm{d}}{\mathrm{~d} t}\|\boldsymbol{z}(t)\|^2+2 \delta_1\|\boldsymbol{z}(t)\|_{\hat{V}}^2 \leqslant \\ & \frac{2 \lambda \lambda_1^{-1}\|\boldsymbol{F}\|}{\delta_1-\lambda_1^{-1} L_{\hat{\boldsymbol{G}}}}\|\boldsymbol{z}(t)\|_{\hat{V}}^2+\lambda_1^{-1} L_{\hat{\boldsymbol{G}}}\|\boldsymbol{z}(t)\|_{\hat{V}}^2+L_{\hat{\boldsymbol{G}}}\|\boldsymbol{z}(t-\rho(t))\|^2, \end{aligned}$

即

$\frac{\mathrm{d}}{\mathrm{~d} t}\|\boldsymbol{z}(t)\|^2 \leqslant\left(\lambda_1^{-1} L_{\hat{\boldsymbol{G}}}+\frac{2 \lambda \lambda_1^{-1}\|\boldsymbol{F}\|}{\delta_1-\lambda_1^{-1} L_{\hat{\boldsymbol{G}}}}-2 \delta_1\right)\|\boldsymbol{z}(t)\|_{\hat{V}}^2+L_{\hat{\boldsymbol{G}}}\|\boldsymbol{z}(t-\rho(t))\|^2 .$

(23)

由式(23)在[0, t]上积分, 可得

$\begin{aligned} & \|\boldsymbol{z}(t)\|^2 \leqslant\|\boldsymbol{z}(0)\|^2+\left(\lambda_1^{-1} L_{\hat{\boldsymbol{G}}}+\frac{2 \lambda \lambda_1^{-1}\|\boldsymbol{F}\|}{\delta_1-\lambda_1^{-1} L_{\hat{\boldsymbol{G}}}}-2 \delta_1\right) \int_0^t\|\boldsymbol{z}(s)\|_{\hat{V}}^2 \mathrm{~d} s+ \\ & \quad L_{\hat{\boldsymbol{G}}} \int_0^t\|\boldsymbol{z}(s-\rho(s))\|^2 \mathrm{~d} s . \end{aligned}$

(24)

对于含时滞的积分项, 设η=s-ρ(s)=τ(s), 则有

$L_{\hat{\boldsymbol{G}}} \int_0^t\|\boldsymbol{z}(s-\rho(s))\|^2 \mathrm{~d} s \leqslant \frac{L_{\hat{\boldsymbol{G}}}}{1-\rho^*} \int_{-\rho(0)}^t\|\boldsymbol{z}(\boldsymbol{\eta})\|^2 \mathrm{~d} \boldsymbol{\eta} .$

(25)

由式(24)、(25)和式(14), 可得

$\begin{aligned} & \|z(t)\|^2 \leqslant\|z(0)\|^2+\left(\frac{2 \lambda \lambda_1^{-1}\|\boldsymbol{F}\|}{\delta_1-\lambda_1^{-1} L_{\hat{\boldsymbol{G}}}}-2 \delta_1+\frac{\left(2-\rho^*\right) \lambda_1^{-1} L_{\hat{\boldsymbol{G}}}}{1-\rho^*}\right) \int_0^t\|z(s)\|_{\hat{V}}^2 \mathrm{~d} s+ \\ & \quad \frac{\lambda_1^{-1} L_{\hat{\boldsymbol{G}}}}{1-\rho^*} \int_{-\rho(0)}^0\|z(s)\|_{\hat{V}}^2 \mathrm{~d} s . \end{aligned}$

再由式(17)可得

$\|\boldsymbol{z}(t)\|^2 \leqslant\|\boldsymbol{z}(0)\|^2+\frac{\lambda_1^{-1} L_{\hat{\boldsymbol{G}}}}{1-\rho^*} \int_{-\rho(0)}^0\|\boldsymbol{z}(s)\|_{\hat{V}}^2 \mathrm{~d} s$

(26)

由式(26)可知定理3成立.

然后通过构造Lyapunov泛函的方法建立初值问题(11)平凡稳态解的渐近稳定性.

定理4 若F≡0, 条件1成立, $ \boldsymbol{\varPhi} \in c_0(\hat{H}), \delta_1>\lambda_1^{-1} L_{\hat{G}} $, 则初值问题(11)存在唯一平凡稳态解v≡0. 进一步, 若

$\delta_1>\frac{\lambda_1^{-1} L_{\hat{\boldsymbol{G}}}}{\sqrt{1-\rho^*}}$

(27)

则v≡0是渐近稳定的.

证明显然，由上述条件和定理2(a)、2(b)可知，初值问题(11)存在唯一平凡稳态解v≡0. 下面我们分三步证明稳态解v≡0是渐近稳定的.

第一步(构造Lyapunov泛函) 令$ J(\boldsymbol{v}(t)): \mathbb{R}_{+} \times c_0(\hat{H}) \rightarrow \mathbb{R}_{+} $, 定义为

$J(\boldsymbol{v}(t))=\|\boldsymbol{v}(t)\|^2+\frac{C}{1-\rho^*} \int_{t-\rho(t)}^t\|\boldsymbol{v}(s)\|^2 \mathrm{~d} s,$

(28)

其中C>0将在后面确定.由式(28)和其中积分项的正性, 可知

$J(\boldsymbol{v}(t))>\|\boldsymbol{v}(t)\|^2 .$

(29)

另一方面, 由式(28)可知

$\begin{gathered} J(\boldsymbol{v}(0))=\|\boldsymbol{v}(0)\|^2+\frac{C}{1-\rho^*} \int_{-\rho(0)}^0\|\boldsymbol{v}(s)\|^2 \mathrm{~d} s \leqslant \\ \|\boldsymbol{\varPhi}\|_{c_0(\hat{H})}^2+\frac{C}{1-\rho^*} \int_{-\rho(0)}^0\|\boldsymbol{\varPhi}(s)\|^2 \mathrm{~d} s \leqslant \\ \|\boldsymbol{\varPhi}\|_{c_0(\hat{H})}^2+\frac{C \boldsymbol{\rho}(0)}{1-\rho^*}\|\boldsymbol{\varPhi}\|_{c_0(\hat{H})}^2 . \end{gathered}$

因此, 存在常数$ \gamma_1=1+\frac{C \rho(0)}{1-\rho^*}>0 $使得

$J(\boldsymbol{v}(0)) \leqslant \gamma_1\|\boldsymbol{\varPhi}\|_{c_0(\hat{H})}^2 .$

(30)

第二步(稳定性) 由式(28)和条件1, 可得

$\frac{\mathrm{d} J(\boldsymbol{v}(t))}{\mathrm{d} t}=\frac{\mathrm{d}}{\mathrm{~d} t}\|\boldsymbol{v}(t)\|^2+\frac{C}{1-\rho^*}\|\boldsymbol{v}(t)\|^2-\frac{C\left(1-\rho^{\prime}(t)\right)}{1-\rho^*}\|\boldsymbol{v}(t-\rho(t))\|^2 \leqslant \frac{\mathrm{d}}{\mathrm{~d} t}\|\boldsymbol{v}(t)\|^2+\frac{C}{1-\rho^*}\|\boldsymbol{v}(t)\|^2-C\|\boldsymbol{v}(t-\rho(t))\|^2 .$

(31)

由F≡0和条件1可知，初值问题(11)中的第一个方程可写为

$\frac{\mathrm{d} \boldsymbol{v}}{\mathrm{~d} t}=-A \boldsymbol{v}-B(\boldsymbol{u}, \boldsymbol{v})-N(\boldsymbol{v})+\hat{\boldsymbol{G}}(\boldsymbol{v}(t-\rho(t))), \quad \forall t \geqslant 0 .$

将上式与v(t)做对偶积, 利用式(7)、(8)、假设1和Hölder不等式, 可得

$\begin{aligned} & \frac{\mathrm{d}}{\mathrm{~d} t}\|\boldsymbol{v}(t)\|^2=2\langle-A \boldsymbol{v}-B(u, \boldsymbol{v})-N(\boldsymbol{v}), \boldsymbol{v}\rangle+2(\hat{\boldsymbol{G}}(\boldsymbol{v}(t-\rho(t))), \boldsymbol{v}) \leqslant \\ & \quad-2 \delta_1\|\boldsymbol{v}(t)\|_{\hat{V}}^2+2 L_{\hat{\boldsymbol{G}}}\|\boldsymbol{v}(t-\rho(t))\|\|\boldsymbol{v}(t)\| \leqslant \\ & \quad-2 \delta_1\|\boldsymbol{v}(t)\|_{\hat{V}}^2+\frac{L_{\hat{\boldsymbol{G}}}^2}{C}\|\boldsymbol{v}(t)\|^2+C\|\boldsymbol{v}(t-\rho(t))\|^2 . \end{aligned}$

(32)

将式(32)代入式(31), 由式(14)可得

$\begin{aligned} & \frac{\mathrm{d} J(\boldsymbol{v}(t))}{\mathrm{d} t} \leqslant-2 \delta_1\|\boldsymbol{v}(t)\|_{\hat{V}}^2+\frac{L_{\hat{G}}^2}{C}\|\boldsymbol{v}(t)\|^2+\frac{C}{1-\rho^*}\|\boldsymbol{v}(t)\|^2 \leqslant \\ & \quad\left(-2 \delta_1+\lambda_1^{-1}\left(\frac{L_{\hat{G}}^2}{C}+\frac{C}{1-\rho^*}\right)\right)\|\boldsymbol{v}(t)\|_{\hat{V}}^2, \end{aligned}$

(33)

取$ C=L_{\hat{G}} \sqrt{1-\rho^*} $, 由式(27)可知

$\frac{\mathrm{d} J(\boldsymbol{v}(t))}{\mathrm{d} t} \leqslant-2\left(\delta_1-\frac{\lambda_1^{-1} L_{\hat{\boldsymbol{G}}}}{\sqrt{1-\rho^*}}\right)\|\boldsymbol{v}(t)\| \hat{v} < 0 .$

(34)

因此, 由式(34)可知J(v(t))单调递减.再由式(29)、(30)可知

$\|\boldsymbol{v}(t)\|^2 < J(\boldsymbol{v}(t)) < J(\boldsymbol{v}(0)) < \gamma_1\|\boldsymbol{\varPhi}\|_{c_0(\hat{H})}^2,$

(35)

即v≡0是稳定的.

第三步(吸引性) 由式(34)，在[0, t]上进行积分, 可得

$\begin{aligned} J(\boldsymbol{v}(t))-J(\boldsymbol{v}(0)) \leqslant-2\left(\delta_1-\frac{\lambda_1^{-1} L_{\hat{\boldsymbol{G}}}}{\sqrt{1-\rho^*}}\right) \int_0^t\|\boldsymbol{v}(s)\|_{\hat{\boldsymbol{V}}}^2 \mathrm{~d} s \end{aligned}$

由式(35)可知

$\begin{aligned} & \int_0^t\|\boldsymbol{v}(s)\|_{\hat{V}}^2 \mathrm{~d} s \leqslant \frac{1}{2\left(\delta_1-\frac{\lambda_1^{-1} L_{\hat{\boldsymbol{G}}}}{\sqrt{1-\rho^*}}\right)}(J(\boldsymbol{v}(0))-J(\boldsymbol{v}(t))) \leqslant \\ & \frac{1}{2\left(\delta_1-\frac{\lambda_1^{-1} L_{\hat{\boldsymbol{G}}}}{\sqrt{1-\rho^*}}\right)}\left(\gamma_1\|\boldsymbol{\varPhi}\|_{c_0(\hat{H})}^2-\|\boldsymbol{v}(t)\|^2\right) \leqslant \\ & \frac{\gamma_1}{2\left(\delta_1-\frac{\lambda_1^{-1} L_{\hat{\boldsymbol{G}}}}{\sqrt{1-\rho^*}}\right)}\|\boldsymbol{\varPhi}\|_{c_0(\hat{H})}^2 \cdot \end{aligned}$

由上式可知$ \mathop {\lim }\limits_{t \rightarrow \infty}\|\boldsymbol{v}(t)\| \hat{v}=0 $.

综上所述, v≡0是渐近稳定的, 即定理4成立.

3.2 当无界时滞函数仅关于时间连续时稳态解的稳定性

本小节考虑无界时滞函数满足下述条件.

条件2令

$\boldsymbol{G}(t, \boldsymbol{\xi})=\hat{\boldsymbol{G}}(\boldsymbol{\xi}(-\rho(t))), \quad \forall(t, \boldsymbol{\xi}) \in \mathbb{R}_{+} \times c_0(\hat{H}),$

其中$ \hat{\boldsymbol{G}}: \hat{H} \rightarrow \hat{H} $为Lipschitz映射, 其Lipschitz常数为$ L_{\hat{\boldsymbol{G}}}, \hat{\boldsymbol{G}}(0)=0, \rho \in c\left(\mathbb{R}_{+} ; \mathbb{R}_{+}\right) $.

为了建立非平凡稳态解的全局稳定性, 这一小节需要对假设1中做如下修正：

(i)′对任意的ξ∈C₀($ \hat{H} $), 映射$ [0, T] \ni t \mapsto \boldsymbol{G}(t, \boldsymbol{\xi}) \in \hat{H} $是连续的.

注3 由注1可知, 满足条件2的时滞项满足假设1和(i)′.

于是可以建立一个重要的引理.

引理2 设G满足假设1和(i)′, $ F \in \hat{H}, v^* \in D(A) $是初值问题(11)的稳态解, 其对于任意的Ψ=(ψ, ψ₃)∈C₀($ \hat{H} $), Ψ(0)=(ψ(0), ψ₃(0))∈ $ \hat{V} $ (其中, ψ=(ψ₁, ψ₂)∈C₀(H), ψ(0)=(ψ₁(0), ψ₂(0))∈V), 且Ψ≠v^*,

$\left\|\boldsymbol{\varPsi}-\boldsymbol{v}^*\right\|_{c_0(\hat{H})}=\left\|\boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right\|$

(36)

中的函数Ψ满足

$\begin{gathered} -\left(A\left(\boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right), \boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right)-\left(B(\boldsymbol{\psi}(0), \boldsymbol{\varPsi}(0))-B\left(\boldsymbol{u}^*, \boldsymbol{v}^*\right), \boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right)- \\ \left(N\left(\boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right), \boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right)+\left(\boldsymbol{G}(t, \boldsymbol{\varPsi})-\boldsymbol{G}\left(t, \boldsymbol{v}^*\right), \boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right) < 0 . \end{gathered}$

(37)

那么对任意$ \boldsymbol{\varPhi}=\left(\boldsymbol{\phi}, \boldsymbol{\phi}_3\right) \in c_0(\hat{H}), \boldsymbol{\varPhi}(0)=\left(\boldsymbol{\varPhi}(0), \phi_3(0)\right) \in \hat{V} $ (其中$ \boldsymbol{\phi}=\left(\phi_1, \phi_2\right) \in c_0(H), \boldsymbol{\phi}(0)=\left(\phi_1(0), \right.\left.\phi_2(0)\right) \in V $), 初值问题(11)的解v(t)满足

$\left\|\boldsymbol{v}(t)-\boldsymbol{v}^*\right\| \leqslant\left\|\boldsymbol{\varPhi}-\boldsymbol{v}^*\right\|_{c_0(\hat{H})}, \quad \forall t \geqslant 0 .$

(38)

证明显然，当Φ=v^*时, 结论显然成立.当Φ≠v^*时, 下面用反证法证明引理2的结论. 假设存在初值Φ满足Φ∈C₀($ \hat{H} $), Φ(0)∈ $ \hat{V} $, 且Φ≠v^*使得式(38)不成立.则存在t>0使得

$\left\|\boldsymbol{v}(t)-\boldsymbol{v}^*\right\|>\left\|\boldsymbol{\varPhi}-\boldsymbol{v}^*\right\|_{c_0(\hat{H})} .$

记

$\boldsymbol{\sigma}=\inf \left\{t>0:\left\|\boldsymbol{v}(t)-\boldsymbol{v}^*\right\|>\left\|\boldsymbol{\varPhi}-\boldsymbol{v}^*\right\|_{c_0(\hat{H})}\right\} .$

则对任意0≤s≤σ, 有

$\left\|\boldsymbol{v}(s)-\boldsymbol{v}^*\right\| \leqslant\left\|\boldsymbol{v}(\boldsymbol{\sigma})-\boldsymbol{v}^*\right\|=\left\|\boldsymbol{\varPhi}-\boldsymbol{v}^*\right\|_{c_0(\hat{H})}$

(39)

和存在子列$ \left\{t_k\right\}_{k \geqslant 1} \subset(\sigma, \infty), t_k \downarrow \sigma(k \rightarrow \infty) $使得

$\left\|\boldsymbol{v}\left(t_k\right)-\boldsymbol{v}^*\right\|>\left\|\boldsymbol{v}(\boldsymbol{\sigma})-\boldsymbol{v}^*\right\| .$

(40)

由式(39)可知

$\left\|\boldsymbol{v}_\sigma-\boldsymbol{v}^*\right\| c_0(\hat{H})=\sup _{\theta \leqslant 0}\left\|\boldsymbol{v}(\sigma+\theta)-\boldsymbol{v}^*\right\|=\left\|\boldsymbol{v}(\sigma)-\boldsymbol{v}^*\right\| .$

(41)

由式(36)、(37)、(41)可知, 取Ψ=v_σ时有

$\begin{array}{r} -\left(A\left(v(\sigma)-v^*\right), v(\sigma)-v^*\right)-\left(B(u(\sigma), v(\sigma))-B\left(u^*, v^*\right), v(\sigma)-v^*\right)- \\ \quad\left(N\left(v(\sigma)-v^*\right), v(\sigma)-v^*\right)+\left(G\left(\sigma, v_\sigma(\cdot)\right)-G\left(t, v^*\right), v(\sigma)-v^*\right) < 0 . \end{array}$

由算子A, B, N, G的连续性可知, 存在$ \epsilon>0 $, 对任意的t∈[σ, σ+ $ \epsilon $], 有

$\begin{gathered} -\left(A\left(v(t)-v^*\right), v(t)-v^*\right)-\left(B(u(t), v(t))-B\left(u^*, v^*\right), v(t)-v^*\right)- \\ \left(N\left(v(t)-v^*\right), v(t)-v^*\right)+\left(G\left(t, v_t\right)-G\left(t, v^*\right), v(t)-v^*\right) < 0 . \end{gathered}$

记z(t)=v(t)-v^*, 由式(19)可得, 对任意t∈[σ, σ+ $ \epsilon $], 下式成立：

$\begin{aligned} & \frac{1}{2} \frac{\mathrm{~d}}{\mathrm{~d} t}\|z(t)\|^2=-(A z(t), z(t))-\left(B(\boldsymbol{u}(t), \boldsymbol{v}(t))-B\left(\boldsymbol{u}^*, \boldsymbol{v}^*\right), z(t)\right)- \\ & \quad(N(z(t)), z(t))+\left(\boldsymbol{G}\left(t, \boldsymbol{v}_t\right)-\boldsymbol{G}\left(t, \boldsymbol{v}^*\right), z(t)\right) < 0 \end{aligned}$

取$ t_{k(\epsilon)} \in(\sigma, \sigma+\epsilon] $, 并对上式在$ \left[\sigma, t_{k(\epsilon)}\right] $上积分, 可知

$\begin{aligned} & \left\|z\left(t_{k(\epsilon)}\right)\right\|^2-\|z(\sigma)\|^2 \leqslant \\ & \quad 2 \int_\sigma^{t_{k(\epsilon)}}\left[-(A z(t), z(t))-\left(B(\boldsymbol{u}(t), \boldsymbol{v}(t))-B\left(\boldsymbol{u}^*, \boldsymbol{v}^*\right), z(t)\right)-(N(z(t)), z(t))\right] \mathrm{d} t+ \\ & \quad 2 \int_\sigma^{t_{k(\epsilon)}}\left(\boldsymbol{G}\left(t, \boldsymbol{v}_t\right)-\boldsymbol{G}\left(t, \boldsymbol{v}^*\right), z(t)\right) \mathrm{d} t < 0, \end{aligned}$

即$ \left\|z\left(t_{k(\epsilon)}\right)\right\| < \|z(\sigma)\| $, 这与式(40)矛盾, 所以式(38)成立.

利用上述引理, 可得非平凡稳态解的全局稳定性.

定理5 假设条件2成立, $ \boldsymbol{F} \in \hat{H}, \delta_1 $满足$ \delta_1>\lambda_1^{-1} L_{\hat{G}} $,

$\delta_1>\lambda_1^{-1} L_{\hat{\boldsymbol{G}}}+\frac{\lambda \cdot \lambda_1^{-1}\|\boldsymbol{F}\|}{\delta_1-\lambda_1^{-1} L_{\hat{\boldsymbol{G}}}} .$

(42)

则对任意的$ \boldsymbol{\varPhi} \in c_0(\hat{H}), \boldsymbol{\varPhi}(0) \in \hat{V} $, 初值问题(11)存在稳态解v^*∈D(A)和强解v(t)，且满足

$\left\|\boldsymbol{v}(t)-\boldsymbol{v}^*\right\| \leqslant\left\|\boldsymbol{\varPhi}-\boldsymbol{v}^*\right\|_{c_0(\hat{H})}, \quad \forall t \geqslant 0 .$

(43)

证明显然，由条件2、$ F \in \hat{H}_{\checkmark} \delta_1>\lambda_1^{-1} L_{\hat{G}} $、定理1和定理2可得初值问题(11)存在稳态解v^*∈D(A)和强解v(t).下证式(43)成立. 设$ \boldsymbol{\varPsi}=\left(\boldsymbol{\psi}, \psi_3\right) \in c_0(\hat{H}), \boldsymbol{\varPsi}(0)=\left(\boldsymbol{\psi}(0), \psi_3(0)\right) \in \hat{V}\left(\text { 其中 } \boldsymbol{\psi}=\left(\psi_1, \psi_2\right) \in\right.\left.c_0(H), \boldsymbol{\psi}(0)=\left(\psi_1(0), \psi_2(0)\right) \in V\right) \text {, 且 } \boldsymbol{\varPsi} \neq \boldsymbol{v}^* $满足

$\left\|\boldsymbol{\varPsi}-\boldsymbol{v}^*\right\|_{c_0(\hat{H})}=\left\|\boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right\| .$

(44)

由式(7)、(10)、(44)和假设1, 可得

$\begin{aligned} & -\left(A\left(\boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right), \boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right)-\left(B(\boldsymbol{\psi}(0), \boldsymbol{\varPsi}(0))-B\left(\boldsymbol{u}^*, \boldsymbol{v}^*\right), \boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right)- \\ & \quad\left(N\left(\boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right), \boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right)+\left(\boldsymbol{G}(t, \boldsymbol{\varPsi})-\boldsymbol{G}\left(t, \boldsymbol{v}^*\right), \boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right) \leqslant \\ & \quad-\delta_1\left\|\boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right\|_{\hat{v}}^2-\left(B\left(\boldsymbol{\psi}(0)-\boldsymbol{u}^*, \boldsymbol{v}^*\right), \boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right)+ \\ & \quad L_{\hat{G}}\left\|\boldsymbol{\varPsi}-\boldsymbol{v}^*\right\|_{c_0(\hat{H})}\left\|\boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right\| \leqslant \\ & \quad-\delta_1\left\|\boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right\|_{\hat{v}}^2+\left|\left(B\left(\boldsymbol{\psi}(0)-\boldsymbol{u}^*, \boldsymbol{v}^*\right), \boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right)\right|+ \\ & \quad \lambda_1^{-1} L_{\hat{G}}\left\|\boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right\|_{\hat{v}}^2 . \end{aligned}$

(45)

由式(24)、(42)和式(45), 可得

$\begin{aligned} - & \left(A\left(\boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right), \boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right)-\left(B(\boldsymbol{\psi}(0), \boldsymbol{\varPsi}(0))-B\left(\boldsymbol{u}^*, \boldsymbol{v}^*\right), \boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right)- \\ & \left(N\left(\boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right), \boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right)+\left(\boldsymbol{G}(t, \boldsymbol{\varPsi})-\boldsymbol{G}\left(t, \boldsymbol{v}^*\right), \boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right) \leqslant \\ & \left(-\delta_1+\lambda_1^{-1} L_{\hat{G}}+\frac{\lambda \cdot \lambda_1^{-1}\|\boldsymbol{F}\|}{\delta_1-\lambda_1^{-1} L_{\hat{G}}}\right)\left\|\boldsymbol{\varPsi}(0)-\boldsymbol{v}^*\right\|_{\hat{v}}^2 < 0 . \end{aligned}$

故式(37)成立.进而, 由引理2可知定理5成立.

3.3 当无界时滞函数为比例时滞时稳态解的多项式稳定性

本小节考虑无界时滞函数满足以下条件.

条件3令

$\boldsymbol{G}(t, \boldsymbol{\xi})=\hat{\boldsymbol{G}}(\boldsymbol{\xi}(-(1-\gamma) t)), \quad(t, \boldsymbol{\xi}) \in \mathbb{R}_{+} \times c_0(\hat{H}),$

其中$ \gamma \in(0, 1), \hat{\boldsymbol{G}}: \hat{H} \rightarrow \hat{H} $为Lipschitz映射, 其Lipschitz常数为$ L_{\hat{\boldsymbol{G}}} \text { 且 } \hat{\boldsymbol{G}}(0)=\mathbf{0} $.

注4 由注1可知，满足条件3的时滞项也满足假设1.

首先回顾一个重要结论.

引理3(文献[19]中引理3.5) 考虑下述的Pantograph方程的初值问题:

$\left\{\begin{array}{l} x^{\prime}(t)=a x(t)+b x(\gamma t), \quad \forall t \geqslant 0, \gamma \in(0, 1), \\ x(0)=x_0 . \end{array}\right.$

(46)

设a < 0, b∈ $ \mathbb{R} $, x是初值问题(46)的解, 则存在C=C(a, b, γ)>0使得

$|x(t)| \leqslant C|x(0)|(1+t)^\mu, \quad t \geqslant 0,$

其中μ∈ $ \mathbb{R} $满足

$0=a+|b| \gamma^\mu .$

利用上述引理, 可得平凡稳态解的多项式稳定性.

定理6 假设条件3成立, $ \boldsymbol{F} \equiv \mathbf{0}, \delta_1>\lambda_1^{-1} L_{\hat{G}} $, 则对任意的$ \boldsymbol{\varPhi} \in c_0(\hat{H}) $, 初值问题(11)存在唯一的平凡稳态解且存在$ C=C\left(\delta_1, L_{\hat{G}}, \gamma\right)>0, \mu < 0 $使得

$\|\boldsymbol{v}(t)\|^2 < C\|\boldsymbol{\varPhi}(0)\|^2(1+t)^\mu, \quad \forall t \geqslant 0 .$

证明显然，由$ \boldsymbol{F} \equiv \mathbf{0} 、\delta_1>\lambda_1^{-1} L_{\hat{\boldsymbol{G}}} $、条件3及定理2(a)、2(b)可知，初值问题(11)存在唯一的平凡稳态解.初值问题(11)中的第一个方程与v(t)做对偶积, 可得

$\begin{aligned} & \frac{1}{2} \frac{\mathrm{~d}}{\mathrm{~d} t}\|\boldsymbol{v}(t)\|^2+\langle A \boldsymbol{v}(t), \boldsymbol{v}(t)\rangle+\langle B(\boldsymbol{u}(t), \boldsymbol{v}(t)), \boldsymbol{v}(t)\rangle+\langle N(\boldsymbol{v}(t)), \boldsymbol{v}(t)\rangle= \\ & \quad(\hat{\boldsymbol{G}}(\boldsymbol{v}(\gamma t)), \boldsymbol{v}(t)), \end{aligned}$

(47)

由式(7)、(8)和条件3, 可知

$\frac{\mathrm{d}}{\mathrm{~d} t}\|\boldsymbol{v}(t)\|^2+2 \delta_1\|\boldsymbol{v}(t)\|_{\hat{V}}^2 \leqslant 2 L_{\hat{G}}\|\boldsymbol{v}(\gamma t)\|\|\boldsymbol{v}(t)\| \leqslant L_{\hat{G}}\|\boldsymbol{v}(\gamma t)\|^2+L_{\hat{G}}\|\boldsymbol{v}(t)\|^2 .$

由式(14)可得

$\frac{\mathrm{d}}{\mathrm{~d} t}\|\boldsymbol{v}(t)\|^2+2 \lambda_1 \delta_1\|\boldsymbol{v}(t)\|^2 \leqslant L_{\hat{\boldsymbol{G}}}\|\boldsymbol{v}(\gamma t)\|^2+L_{\hat{\boldsymbol{G}}}\|\boldsymbol{v}(t)\|^2 .$

整理得

$\frac{\mathrm{d}}{\mathrm{~d} t}\|\boldsymbol{v}(t)\|^2 \leqslant\left(-2 \lambda_1 \delta_1+L_{\hat{\boldsymbol{G}}}\right)\|\boldsymbol{v}(t)\|^2+L_{\hat{\boldsymbol{G}}}\|\boldsymbol{v}(\gamma t)\|^2 .$

(48)

由$ \delta_1>\lambda_1^{-1} L_{\hat{G}} $, 可知

$-2 \lambda_1 \delta_1+L_{\hat{G}} < 0$

由引理3和式(48)可知，存在$ C=C\left(\delta_1, L_{\hat{G}}, \gamma\right)>0, \mu \in \mathbb{R} $，使得

$\|\boldsymbol{v}(t)\|^2 \leqslant C\|\boldsymbol{\varPhi}(0)\|^2(1+t)^\mu,$

其中μ满足

$-2 \delta_1 \lambda_1+L_{\hat{G}}+L_{\hat{G}} \gamma^\mu=0$

解之, 得

$\mu=\log _\gamma\left(\frac{2 \delta_1 \lambda_1}{L_{\hat{G}}}-1\right) .$

(49)

由于$ \delta_1 \lambda_1>L_{\hat{\boldsymbol{G}}} $, 所以有

由式(49)和γ∈(0, 1), 可得μ < 0.

致谢: 本文作者衷心感谢重庆师范大学研究生科研创新项目(YKC24005)对本文的资助.

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