Finite Time Stabilization of Dynamical Networks Under Pinning Event-Triggered Control
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摘要: 该文研究了基于牵制触发控制动态网络的有限时间镇定. 不同于已有结果有限时间事件触发镇定, 考虑到控制成本和控制大规模节点数目的困难性, 提出了牵制自适应事件触发控制保证动态网络的有限时间镇定. 由于动态网络系统存在维数高的问题, 分析牵制事件触发有限时间镇定相当困难. 通过设计恰当的协议, 借助Lyapunov稳定性理论,得到了动态耦合网络有限时间镇定的充分性条件. 最后, 通过数值仿真验证了定理的有效性.Abstract: The finite time stabilization problem of dynamical directed networks under pinning adaptive event-triggered control was addressed. Unlike the existing results as for finite time stabilization with event-triggered protocol, in view of the difficulties of the control cost and the large node number, a novel pinning event-triggered protocol was designed. Given its high dimension, the analysis of the finite time stabilization under the pinning event-triggered control for dynamical networks is challenging. Based on the Lyapunov stability theory and the appropriately designed protocol, sufficient conditions were derived to guarantee the finite time stabilization. Finally, an example was given to demonstrate the effectiveness of the theoretical results.
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0. 引言
由于分布式传感器网络[1]、无人驾驶[2]、水下车辆系统[3]、空中编队问题[2]等在实际生活中的广泛应用,复杂动态网络的研究受到了大量的关注. 另一方面, 在执行任务时, 网络系统还受到执行时间的限制. 现有的方法通过设计反馈控制协议, 以指数式收敛速度, 存在无穷大的停息时间[4]. 相较于渐近稳定、指数稳定, 有限时间镇定要求系统以有限时间收敛到平衡点[5-14]. 因此能够更好地应用于实际问题中.
最近数十年来, 不同的控制协议被设计来实现动态网络的有限时间镇定[10-22]. 如基于连续性状态反馈的非线性系统的有限时间一致[10], 基于滑膜控制实现动态网络的有限时间一致[19]. 文献[20]研究了基于分布式事件触发控制来实现动态网络系统的有限时间一致, 文中给出的停息时间不仅依赖于状态的初值而且还依赖于事件触发的门槛参数. 众所周知, 由于网络中存在参数错误匹配、节点多等问题. 牵制自适应控制能够有效地解决系统参数不确定问题, 实现网络系统的精准控制. 因此, 研究牵制自适应控制有限时间镇定是一项重要的课题.
除了保证网络系统的收敛速率, 在网络系统中降低通信成本是另一项关键的课题. 采样控制在保证网络系统稳定的前提下, 能有效地节约通信成本. 采样控制方案通常分为周期采样和事件触发采样. 周期采样的采样间隔是等长的. 不同于周期采样, 事件触发机制是基于事件而实行网络的采样. 事件触发机制能够有效避免网络的执行器故障问题. 早期的工作如文献[23-25]. 文献[23]考虑了基于事件触发控制的线性系统及非线性系统镇定问题, 文中所设计的协议避免了更多的执行器执行的要求. 为减少触发的频率, 文献[26]设计了基于能量函数的触发策略, 所设计的触发机制能够最大化减少触发次数. 最近关于事件触发控制的研究可见文献[26-29].
基于以上的启发, 本文考虑了牵制触发控制实现动态网络的有限时间一致,主要的贡献概括为: 1) 针对现实生活中动态网络节点数目多的问题, 设计了新颖的牵制自适应触发控制协议. 不同于文献[20], 在大规模动态网络中, 本文所考虑的协议能够更有效地降低控制成本, 解决复杂动态网络系统镇定问题. 2) 对比于文献[20]所设计的牵制自适应控制增益, 面对系统的扰动和不确定, 本文所设计的控制协议具有优越的表现性.
本文主要由以下几个部分组成: 第1节, 给出了基础知识、模型;第2节,给出了主要结果及证明;第3节, 通过数值模拟验证了定理的有效性;最后, 第4节给出了总结和展望.
符号说明: R表示实数集;Rn×n, Cn×n分别表示实矩阵和复矩阵的集合,In∈Rn×n表示单位矩阵.给定向量x=[x1,⋯,xn]T,定义‖和\|\cdot\|表示Euclid范数或诱导矩阵2范数;\mathbf{1}=[1, \cdots, 1]^{\mathrm{T}},同理于\mathbf{0} ; {\lambda}_{\text {min }}(\cdot)和\lambda_{\text {max }}(\cdot)分别表示对称矩阵的最小特征值和最大特征值;定义f(t): g \rightarrow \mathbb{R}, \mathfrak{I}=\left[t_{0}, \infty\right)是个连续函数,对于任意的t \in \mathfrak{I}, f(t)的右上Dini导数定义为D^{+} f(t)=\overline{\lim }_{\varDelta \rightarrow 0^{+}}((f(t+\varDelta)-f(t)) / \varDelta);定义\operatorname{sig}^{\alpha}(r)=\operatorname{sgn}(r)|r|^{\alpha},其中\operatorname{sgn}(\cdot)表示标准的符号函数,即
\operatorname{sgn}(r)=\left\{\begin{array}{lc} 1, & r>0 \\ 0, & r=0, \\ -1, & r<0 \end{array}\right. 1. 基础知识和模型的建立
1.1 代数图论
考虑简单有向图\mathcal{G},定义\mathcal{G}=\{\nu, \boldsymbol{\varepsilon}, \boldsymbol{A}\},其中\nu表示节点集,\varepsilon=\left\{e_{i j} \mid i, j \in \nu, i \neq j\right\}表示边集,\boldsymbol{A}= \left[a_{i j}\right]表示耦合权重矩阵,若i与j相邻,则a_{i j}>0,否则a_{i j}=0.定义图\mathcal{G}的Laplace矩阵\boldsymbol{L}=\left[l_{i j}\right] \in \mathbb{R}^{N \times N}, l_{i j} =-a_{i j}, i \neq j, l_{i i}=-\sum_{j \neq i} l_{i j}.若图\mathcal{G}的任意两个节点存在一条有向路径,则图\mathcal{G}是有向强连通的.定义矩阵\boldsymbol{A}关于零特征值的左特征向量\boldsymbol{\xi}=\left[\xi_{1}, \cdots, \xi_{N}\right](即\boldsymbol{A}^{\mathrm{T}} \boldsymbol{\xi}=\mathbf{0}),且有\max \left\{\xi_{1}, \cdots, \xi_{N}\right\}=1.若\sum_{j=1}^{N} \xi_{j} a_{j i}= \sum_{i=1}^{N} \xi_{i} a_{i j}成立,则该图\mathcal{G}满足细节性平衡条件.
1.2 模型的建立
考虑如下单个节点的动力系统:
\dot{\boldsymbol{s}}(t)=\boldsymbol{C s}(t)+\boldsymbol{B} \tilde{\boldsymbol{f}}(\boldsymbol{s}(t)), (1) 其中\boldsymbol{s}(t)=\left[s_{1}(t), \cdots, s_{n}(t)\right]^{\mathrm{T}} \in \mathbb{R}^{n}, \boldsymbol{C} \in \mathbb{R}^{n \times n}和\boldsymbol{B} \in \mathbb{R}^{n \times n}均表示系统参数矩阵,\tilde{\boldsymbol{f}}(\boldsymbol{s}(t))=\left(\tilde{f}_{1}(\boldsymbol{s}(t)), \cdots\right.,\left.\tilde{f}_{n}(\boldsymbol{s}(t))\right)^{\mathrm{T}}表示连续可微非线性函数,且满足\tilde{\boldsymbol{f}}(\mathbf{0})=\mathbf{0}. 考虑如下N个耦合节点的非线性动态网络 [30] :
\begin{gather*} \dot{\boldsymbol{x}}_{i}(t)=\boldsymbol{C} \boldsymbol{x}_{i}(t)+\boldsymbol{B} \tilde{\boldsymbol{f}}\left(\boldsymbol{x}_{i}(t)\right)+c \sum\limits_{j=1}^{N} a_{i j} \boldsymbol{\varGamma}\left[\boldsymbol{x}_{j}(t)-\boldsymbol{x}_{i}(t)+\beta \operatorname{sig}^{\alpha}\left(\boldsymbol{x}_{j}(t)-\boldsymbol{x}_{i}(t)\right)\right]+\boldsymbol{u}_{i}(t), \\ i=1, 2, \cdots, N, \end{gather*} (2) 其中\boldsymbol{x}_{i}(t) \in \mathbb{R}^{n}表示第i个节点的状态;c>0表示动态网络的耦合强度;\boldsymbol{u}_{i}(t) \in \mathbb{R}^{n}是第i个节点的控制输人;\boldsymbol{A}=\left[a_{i j}\right] \in \mathbb{R}^{n \times n},若节点i和节点j相连接,则a_{i j}>0,否则a_{i j}=0, \boldsymbol{\varGamma}=\operatorname{diag}\left\{\gamma_{1}, \cdots, \gamma_{n}\right\} \in \mathbb{R}^{n \times n}表示耦合内连接正定矩阵.参数0<\alpha<1和\beta>0.
注1 由于本文研究的是牵制有限时间镇定问题, 对于每个子系统必须破坏系统的Lipschitz连续性. 受文献[30]的启发, 本文设计了牵制触发控制, 实现动态网络的有限时间镇定.
引理1[31] 假定矩阵L=[lij], i, j=1, 2, …, N,定义矩阵A=[aij],若lij=-aij < 0(i≠j), lii=-∑j≠ilij, 则有以下性质成立:
① L的特征值除代数重数为1零特征值之外, 其余实部都为正实部.
② 若L为不可约矩阵, 假定ξ =[ξ1, …, ξN]是关于零特征值的左特征向量, 则ξi>0对所有的i=1, 2, …, N成立.
③ 若L是一个对称矩阵,则
\boldsymbol{\omega}^{\mathrm{T}} \boldsymbol{L} \boldsymbol{\omega}=\frac{1}{2} \sum\limits_{i, j=1}^{N} a_{i j}\left(\omega_{i}-\omega_{j}\right)^{2} 对于任意的向量\boldsymbol{\omega}=\left[\omega_{1}, \cdots, \omega_{N}\right],其中\omega_{i} \in \mathbb{R}对任意的i=1, 2, \cdots, N成立.
引理2[32] 定义x_{1}, \cdots, x_{N} \geqslant 0, 0<r_{1}<1和r_{2}>1,则有\left(\sum_{i=1}^{N} x_{i}\right)^{r_{1}} \leqslant \sum_{i=1}^{N} x_{i}^{r_{1}} \leqslant N^{1-r_{1}}\left(\sum_{i=1}^{N} x_{i}\right)^{r_{1}}和N^{1-r_{2}}\left(\sum_{i=1}^{N} x_{i}\right)^{r_{2}} \leqslant \sum_{i=1}^{N} x_{i}^{r_{2}}.
引理3[33] 若\boldsymbol{A}=\left[a_{i j}\right]是不可约对称矩阵;\operatorname{rank}(\boldsymbol{A})=N-1和\sum_{j=1}^{N} a_{i j}=0对于任意的i=1, 2, \cdots, N成立,则
\boldsymbol{A}_{1}=\left[\begin{array}{cccc} a_{11}-\epsilon & a_{12} & \cdots & a_{1 N} \\ a_{21} & a_{22} & \cdots & a_{2 N} \\ \vdots & \vdots & & \vdots \\ a_{N 1} & a_{N 2} & \cdots & a_{N N} \end{array}\right] 是负定的.
引理4[4] 考虑动力学方程\dot{\boldsymbol{z}}(t)=\boldsymbol{f}(\boldsymbol{z}(t)),其中\boldsymbol{f}(\mathbf{0})=\mathbf{0}和\boldsymbol{z} \in \mathbb{R}^{n}.若存在一个正定有理函数V(\boldsymbol{z}(t))满足\dot{V}(z(t))+c_{1} V^{r}(z(t)) \leqslant 0, z \in \mathbb{R}^{n} \backslash\{0\},其中c_{1}>0和0<r<1,则原点是有限时间稳定的.停息时间的上界T \leqslant \frac{V^{1-r} z(0)}{c_{1}(1-r)}.
定义1 考虑动态网络系统(1)和(2),如果存在控制器ui(t)和依赖于初值的停息时间T>0,使得\lim _{t \rightarrow T}\left\|\boldsymbol{x}_{i}(t)-s(t)\right\|=0和\left\|\boldsymbol{x}_{i}(t)-\boldsymbol{s}(t)\right\|=0对所有的t \geqslant T成立, 则系统(2)可以有限时间镇定到系统(1).
假设1 对于任意的向量\boldsymbol{q}_{1}(t), \boldsymbol{q}_{2}(t) \in \mathbb{R}^{n},非线性函数\boldsymbol{f}(\cdot)满足
\left|\boldsymbol{f}_{i}\left(\boldsymbol{q}_{1}\right)-\boldsymbol{f}_{i}\left(\boldsymbol{q}_{2}\right)\right| \leqslant \sum\limits_{j=1}^{n} h_{i j}\left|\boldsymbol{q}_{1 j}-\boldsymbol{q}_{2 j}\right|, (3) 其中Lipschitz系数hij≥0,对于任意的i, j=1, 2, …, n成立.
2. 主要结果
2.1 基于牵制触发控制的有限时间镇定
本小节中, 我们将研究基于牵制触发控制动态网络的有限时间镇定. 定义系统(1)和(2)的误差状态为ei(t)=xi(t)-s(t). 耦合节点状态xi(t)和目标节点状态s(t)的测量误差如下:
\boldsymbol{e}_{x i}(t)=\boldsymbol{x}_{i}\left(t_{k^{\prime}}^{i}\right)-\boldsymbol{x}_{i}(t), \boldsymbol{e}_{s i}(t)=\boldsymbol{s}\left(t_{k^{\prime}}^{i}\right)-\boldsymbol{s}(t), (4) 其对于任意的t∈[tk'i, tk'+1i)成立. 由此, 我们可得ei(t)=ei(tk'i)-zi(t),其中zi(t)=exi(t)-esi(t).
动态耦合网络的节点状态xi(t)触发仅在时刻tk'i. 因此, 检测器需要检测目标节点的测量值s(tk'i)来确定是否更新控制器的节点i的状态. 考虑如下的事件触发机制:
\left\|{\boldsymbol{z}}_{i}(t)\right\| \leqslant \sigma_{i}\left\|\boldsymbol{e}_{i}(t)\right\|, \quad t \in\left[t_{k^{\prime}}^{i}, t_{k^{\prime}+1}^{i}\right), (5) 其中\sigma_{i} \in \mathbb{R}表示第i个节点触发门槛参数和\sigma_{i}>0, i=1, \cdots, N.定义\sigma=\max \left(\sigma_{1}, \cdots, \sigma_{N}\right).
在实际应用中,动态网络存在大量的节点. 因此, 控制所有的节点实现系统的稳定是难以实现的. 牵制控制能有效地减少动态网络中控制器的数目. 为实现动态网络的有限时间镇定, 减少控制成本, 我们设计如下的牵制控制方案:
\boldsymbol{u}_{i}(t)=-c {\hat{k}_{i}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)-d_{1} \boldsymbol{\varGamma}_{\mathrm{sig}^{\alpha}}\left(\boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)\right), \quad i=1, \cdots, l, (6) 其中d1>0. 考虑如下的更新定律:
\dot{\hat{k}}_{i}(t)=\operatorname{proj}\left(\hat{k}_{i}(t)\right)= \begin{cases}0, & (1-\sigma) \hat{k}_{i}(t) \geqslant \bar{k}_{1} \\ \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}\left(t_{k^{\prime}}^{i}\right) \boldsymbol{\varGamma} \boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right), & (1-\sigma) \hat{k}_{i}(t)<\bar{k}_{1}, \end{cases} (7) 其中\hat{k}_{i}(t)表示自适应控制增益,且初值满足4 \bar{k}_{1} \sigma^{2} \leqslant \hat{k}_{i}(0) \leqslant \bar{k}_{1} /(1-\sigma) ; \operatorname{proj}(\cdot)为投影算子.
注2 本文采用了牵制自适应触发控制实现有限耦合网络的有限时间镇定. 相较于文献[20]所有节点控制,牵制控制能够有效减少控制成本, 便于大规模网络实现控制. 自适应控制具有抗干扰等优点, 因此, 对比于文献[20]的反馈触发控制, 耦合网络在噪声环境下,具有优越的表现性.
基于式(1)、(2)和以上分析, 可得如下误差系统:
\left\{\begin{array}{r} \dot{\boldsymbol{e}}_{i}(t)=\boldsymbol{C} \boldsymbol{e}_{i}(t)+\boldsymbol{B} \boldsymbol{f}\left(\boldsymbol{e}_{i}(t)\right)+c \sum\limits_{j=1}^{N} a_{i j} \boldsymbol{\varGamma}\left[\boldsymbol{e}_{j}(t)-\boldsymbol{e}_{i}(t)+\beta \operatorname{sig}^{\alpha}\left(\boldsymbol{e}_{j}(t)-\boldsymbol{e}_{i}(t)\right)\right]+\boldsymbol{u}_{i}(t), \\ i=1, \cdots, l, \\ \dot{\boldsymbol{e}}_{i}(t)=\boldsymbol{C} \boldsymbol{e}_{i}(t)+\boldsymbol{B} \boldsymbol{f}\left(\boldsymbol{e}_{i}(t)\right)+c \sum\limits_{j=1}^{N} a_{i j} \boldsymbol{\varGamma}\left[\boldsymbol{e}_{j}(t)-\boldsymbol{e}_{i}(t)+\beta \operatorname{sig}^{\alpha}\left(\boldsymbol{e}_{j}(t)-\boldsymbol{e}_{i}(t)\right)\right], \;\;\;\;\;\;\;\;\;\;\;\\ i=l+1, 2, \cdots, N . \end{array}\right. (8) 引理6 若动态网络(2)是强连通和细节性平衡的,则
2 \sum\limits_{i=1}^{N} \sum\limits_{j=1}^{N} \xi_{i} a_{i j} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \operatorname{sig}^{\alpha}\left(\boldsymbol{e}_{j}(t)-\boldsymbol{e}_{i}(t)\right)=-\sum\limits_{i=1}^{N} \sum\limits_{j=1}^{N} \sum\limits_{k=1}^{n} \xi_{i} a_{i j} \boldsymbol{\gamma}_{k}\left|\boldsymbol{e}_{j k}(t)-\boldsymbol{e}_{i k}(t)\right|^{\alpha+1}. 证明 因为
\begin{align*} & 2 \sum\limits_{i=1}^{N} \sum\limits_{j=1}^{N} \xi_{i} a_{i j} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \operatorname{sig}^{\alpha}\left(\boldsymbol{e}_{j}(t)-\boldsymbol{e}_{i}(t)\right)= \\ & \quad \sum\limits_{i=1}^{N} \sum\limits_{j=1}^{N} \xi_{i} a_{i j} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \operatorname{sig}^{\alpha}\left(\boldsymbol{e}_{j}(t)-\boldsymbol{e}_{i}(t)\right)+\sum\limits_{i=1}^{N} \sum\limits_{j=1}^{N} \xi_{j} a_{j i} \boldsymbol{e}_{j}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \operatorname{sig}^{\alpha}\left(\boldsymbol{e}_{i}(t)-\boldsymbol{e}_{j}(t)\right) . \end{align*} (9) 由网络是细节性平衡的和式(9),可得
\begin{aligned} 2 \sum\limits_{i=1}^N & \sum\limits_{j=1}^N \xi_i a_{i j} \boldsymbol{e}_i^{\mathrm{T}}(t) \boldsymbol{\varGamma} \mathrm{sig}^\alpha\left(\boldsymbol{e}_j(t)-\boldsymbol{e}_i(t)\right)= \\ & \sum\limits_{i=1}^N \sum\limits_{j=1}^N \xi_i a_{i j}\left(\boldsymbol{e}_i(t)-\boldsymbol{e}_j(t)\right)^{\mathrm{T}} \boldsymbol{\varGamma}_{\operatorname{sig}^\alpha}\left(\boldsymbol{e}_j(t)-\boldsymbol{e}_i(t)\right)=\\ &-\sum\limits_{i=1}^N \sum\limits_{j=1}^N \sum\limits_{k=1}^n \xi_i a_{i j} \gamma_k\left|\boldsymbol{e}_{j k}(t)-\boldsymbol{e}_{i k}(t)\right|^{\alpha+1}. \end{aligned} 注3 细节性平衡条件处理有向网络下含有符号函数的耦合状态时,具有重要作用, 具有不可替代性.
定理1 假设耦合网络(2)是强连通和细节性平衡的. 若假设1成立, 且参数0<\sigma<\phi, \phi= \min \left\{\frac{\xi_{\text {min }}}{N n^{(1-\mu) /(2 \mu+2)}}, \frac{1}{2}\right\}和\xi_{\text {min }}=\min \left\{\xi_{1}, \xi_{2}, \cdots, \xi_{N}\right\},则耦合网络系统(2)在控制方案(6)下以有限时间T镇定到目标状态s(t).其中停息时间T=\frac{4 V_{0}^{(1-\alpha) / 2}(0)}{\rho_{1}(1-\alpha)}+\frac{V(0)}{\rho_{0} \kappa_{m}+\rho_{1} \kappa_{m}^{(1+\alpha) / 2}}.
证明 考虑如下的Lyapunov函数:
V(t)=V_{0}(t)+V_{1}(t), (10) V_{0}(t)=\sum\limits_{i=1}^{N} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{e}_{i}(t), V_{1}(t)=c \sum\limits_{i=1}^{l}\left(\hat{k}_{i}(t)-\bar{k}_{1}\right)^{2} . (11) 沿着误差系统(8)V(t)的导数有
\begin{align*} \dot{V}(t)&=2 \sum\limits_{i=1}^{N} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \dot{\boldsymbol{e}}_{i}(t)+2 c \sum\limits_{i=1}^{l}\left(\hat{k}_{i}(t)-\bar{k}_{1}\right) \dot{\hat{k}}_{i}(t)= \\ & 2 \sum\limits_{i=1}^{N} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t)\left[\boldsymbol{C} \boldsymbol{e}_{i}(t)+\boldsymbol{B} \boldsymbol{f}\left(\boldsymbol{e}_{i}(t)\right)+c \sum\limits_{j=1}^{N} a_{i j} \boldsymbol{\varGamma}\left(\boldsymbol{e}_{j}(t)-\boldsymbol{e}_{i}(t)+\beta \operatorname{sig}^{\alpha}\left(\boldsymbol{e}_{j}(t)-\boldsymbol{e}_{i}(t)\right)\right)\right]+ \\ & 2 \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t)\left[-c \hat{k}_{i}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)-d_{1} \boldsymbol{\varGamma} \operatorname{sig}^{\alpha}\left(\boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)\right)\right]+2 c \sum\limits_{i=1}^{l}\left(\hat{k}_{i}(t)-\bar{k}_{1}\right) \dot{\hat{k}}_{i}(t)= \\ & 2 \sum\limits_{i=1}^{N} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t)\left[\boldsymbol{C} \boldsymbol{e}_{i}(t)+\boldsymbol{B} \boldsymbol{f}\left(\boldsymbol{e}_{i}(t)\right)+c \sum\limits_{j=1}^{N} a_{i j} \boldsymbol{\varGamma} \boldsymbol{e}_{j}(t)\right]-2 d_{1} \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \mathrm{sig}^{\alpha}\left(\boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)\right)- \\ & 2 c \sum\limits_{i=1}^{l} \hat{k}_{i}(t) \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)+2 c \beta \sum\limits_{i=1}^{N} \sum\limits_{j=1}^{N} \xi_{i} a_{i j} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \mathrm{sig}^{\alpha}\left(\boldsymbol{e}_{j}(t)-\boldsymbol{e}_{i}(t)\right)+ \\ & 2 c \sum\limits_{i=1}^{l}\left(\hat{k}_{i}(t)-\bar{k}_{1}\right) \dot{\hat{k}}_{i}(t) . \end{align*} (12) 基于假设1, 可得
2 \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{B} \boldsymbol{f}\left(\boldsymbol{e}_{i}(t)\right) \leqslant \boldsymbol{e}_{i}^{\mathrm{T}}(t)\left(\boldsymbol{B} \boldsymbol{B}^{\mathrm{T}}+\boldsymbol{L}_{1}^{\mathrm{T}} \boldsymbol{L}_{1}\right) \boldsymbol{e}_{i}(t), (13) 其中L1=[hij]n×n.
基于更新律(7), 则有以下情形.
情形A (1-\sigma) \hat{k}_{i}(t) \geqslant \bar{k}_{1}和\dot{\hat{k}}_{i}(t)=0
2 c \sum\limits_{i=1}^{l}\left(\hat{k}_{i}(t)-\bar{k}_{1}\right) \dot{\hat{k}}_{i}(t)-2 c \sum\limits_{i=1}^{l} \hat{k}_{i}(t) \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right) \leqslant-c \bar{k}_{1} \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t). (14) 情形B (1-\sigma) \hat{k}_{i}(t)<\bar{k}_{1}
2 c \sum\limits_{i=1}^{l}\left(\hat{k}_{i}(t)-\bar{k}_{1}\right) \dot{\hat{k}}_{i}(t)-2 c \sum\limits_{i=1}^{l} \hat{k}_{i}(t) \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right) \leqslant-c \bar{k}_{1} \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t). (15) 式(14)和(15)的具体推导见文后附录.
定义\boldsymbol{P}=\left[p_{i j}\right]_{N \times N}=\boldsymbol{\varXi} \tilde{\boldsymbol{A}}+\tilde{\boldsymbol{A}}^{\mathrm{T}} \boldsymbol{\boldsymbol { \varXi }},其中\tilde{\boldsymbol{A}}=\left[\tilde{a}_{i j}\right], \tilde{\boldsymbol{A}}=\boldsymbol{A}-\operatorname{diag}\overbrace{\left(\frac{\bar{k}_{1}}{2}, \frac{\bar{k}_{1}}{2}\right.}^{l}, \cdots, 0),我们有p_{i j}=\tilde{a}_{i j} \xi_{j}+\tilde{a}_{j i} \xi_{i}.因此,我们有\left|p_{i i}\right|>\sum_{j \neq i}^{N}\left|p_{i j}\right|, i=1, \cdots, l和\left|p_{i i}\right|=\sum_{j \neq i}^{N}\left|p_{i j}\right|, i=l+1, \cdots, N.由引理3得\lambda_{\text {max }}(\boldsymbol{P})< 0.基于以上分析,可得
2 c \sum\limits_{i=1}^{N} \sum\limits_{j=1}^{N} \xi_{i} \tilde{a}_{i j} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{j}(t)=c \sum\limits_{k=1}^{n} \gamma_{k}\left(\boldsymbol{e}^{k}(t)\right)^{\mathrm{T}}\left(\boldsymbol{\varXi} \tilde{\boldsymbol{A}}+\tilde{\boldsymbol{A}}^{\mathrm{T}} \boldsymbol{\varXi}\right) \boldsymbol{e}^{k}(t) \leqslant \begin{align*} & c {\lambda}_{\text {max }}\left(\boldsymbol{\varXi} \tilde{\boldsymbol{A}}+\tilde{\boldsymbol{A}}^{\mathrm{T}} \boldsymbol{\varXi}\right) \gamma_{\text {min }} \sum\limits_{i=1}^{N} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{e}_{i}(t) \leqslant \\ & c {\lambda}_{\text {max }}\left(\boldsymbol{\varXi} \tilde{\boldsymbol{A}}+\tilde{\boldsymbol{A}}^{\mathrm{T}} \boldsymbol{\varXi}\right) \gamma_{\text {min }} \sum\limits_{i=1}^{N} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{e}_{i}(t)= \\ & c \lambda_{\text {max }}(\boldsymbol{P}) \gamma_{\text {min }} V_{0}(t), \end{align*} (16) 其中\boldsymbol{e}^{k}(t)=\left(\boldsymbol{e}_{1 k}, \cdots, \boldsymbol{e}_{N k}\right), \gamma_{\text {min }}=\min \left\{\gamma_{1}, \gamma_{2}, \cdots, \gamma_{n}\right\}和\max _{1 \leqslant i \leqslant N} \xi_{i}=1.由引理2和引理5,可得
\begin{align*} & -2 d_{1} \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \operatorname{sig}^{\alpha}\left(\boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)\right)= \\ & \quad-2 d_{1} \sum\limits_{i=1}^{l} \sum\limits_{k=1}^{n} \gamma_{k} \xi_{i} \boldsymbol{e}_{i k}(t) \operatorname{sig}^{\alpha}\left(\boldsymbol{e}_{i k}(t)+{\boldsymbol{z}}_{i k}(t)\right) \leqslant \\ & \quad-2 d_{1} \sum\limits_{i=1}^{l} \sum\limits_{k=1}^{n} \gamma_{k} \xi_{i} \boldsymbol{e}_{i k}(t) \operatorname{sig}^{\alpha}\left(\boldsymbol{e}_{i k}(t)\right)+2 d_{1} \sum\limits_{i=1}^{l} \sum\limits_{k=1}^{n} \gamma_{k} \xi_{i}\left|\boldsymbol{e}_{i k}(t)\right|\left|{\boldsymbol{z}}_{i k}(t)\right|^{\alpha} \leqslant \\ & \quad-2 d_{1} \sum\limits_{i=1}^{l} \sum\limits_{k=1}^{n} \gamma_{k} \xi_{i}\left|\boldsymbol{e}_{i k}(t)\right|^{\alpha+1}+2 d_{1} \sum\limits_{i=1}^{l} \sum\limits_{k=1}^{n} \gamma_{k} \xi_{i}\left(\frac{1}{\alpha+1}\left|\boldsymbol{e}_{i k}(t)\right|^{\alpha+1}+\frac{\alpha}{\alpha+1}\left|{\boldsymbol{z}}_{i k}(t)\right|^{\alpha+1}\right)= \\ & \quad-\frac{2 d_{1} \alpha}{\alpha+1} \sum\limits_{i=1}^{l} \sum\limits_{k=1}^{n} \gamma_{k} \xi_{i}\left|\boldsymbol{e}_{i k}(t)\right|^{\alpha+1}+\frac{2 d_{1} \alpha}{\alpha+1} \sum\limits_{i=1}^{l} \sum\limits_{k=1}^{n} \gamma_{k} \xi_{i}\left|{\boldsymbol{z}}_{i k}(t)\right|^{\alpha+1}, \end{align*} (17) \begin{align*} & \sum\limits_{i=1}^{l} \sum\limits_{k=1}^{n} \xi_{i} \gamma_{k}\left|{\boldsymbol{z}}_{i k}(t)\right|^{\alpha+1} \leqslant n^{(1-\alpha) / 2} \sum\limits_{i=1}^{l} \xi_{i}\left(\sum\limits_{k=1}^{n} \gamma_{k}^{2 /(\alpha+1)}\left|{\boldsymbol{z}}_{i k}(t)\right|^{2}\right)^{(\alpha+1) / 2} \leqslant \\ & \sigma^{\alpha+1} n^{(1-\alpha) / 2} \sum\limits_{i=1}^{l} \sum\limits_{k=1}^{n} \xi_{i} \gamma_{k}\left|\boldsymbol{e}_{i k}(t)\right|^{\alpha+1} . \tag{18} \end{align*} (18) 上述不等式使用了事件触发条件|zik(t)|≤σ |eik(t)|和引理2.
通过引理2、引理6、式(17)和(18),则有
\begin{align*} & 2 c \beta \sum\limits_{i=1}^{N} \sum\limits_{j=1}^{N} \xi_{i} a_{i j} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma}_{\mathrm{sig}^{\alpha}}\left(\boldsymbol{e}_{j}(t)-\boldsymbol{e}_{i}(t)\right)-2 d_{1} \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma}_{\operatorname{sig}^{\alpha}}\left(\boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)\right) \leqslant \\ & \quad-c \beta\left(\sum\limits_{k=1}^{n} \sum\limits_{i=1}^{N} \sum\limits_{j=1}^{N} \xi_{i} \gamma_{k} a_{i j}\left|\boldsymbol{e}_{j k}(t)-\boldsymbol{e}_{i k}(t)\right|^{1+\alpha}+\right. \\ & \left.\quad \frac{2 d_{1}}{c \boldsymbol{\beta}} \frac{\alpha}{\alpha+1}\left(1-\boldsymbol{\sigma}^{\alpha+1} n^{(1-\alpha) / 2}\right) \sum\limits_{i=1}^{l} \sum\limits_{k=1}^{n} \xi_{i} \gamma_{k}\left|\boldsymbol{e}_{i k}(t)\right|^{1+\alpha}\right) \leqslant \\ & \quad-c \boldsymbol{\beta} \gamma_{\text {min }}\left(\sum _ { k = 1 } ^ { n } \left(\sum\limits_{i=1}^{N} \sum\limits_{j=1}^{N}\left(\xi_{i} a_{i j}\right)^{2 /(1+\alpha)\left|\boldsymbol{e}_{j k}(t)-\boldsymbol{e}_{i k}(t)\right|^{2}+}\right.\right. \\ & \left.\left.\quad \sum\limits_{i=1}^{l}\left(\frac{\bar{d}_{1} \xi_{i}}{c \boldsymbol{\beta}}\right)^{2 /(1+\alpha)}\left|\boldsymbol{e}_{i k}(t)\right|^{2}\right)\right)^{(1+\alpha) / 2}= \\ & \quad-c \boldsymbol{\beta} \gamma_{\min }\left(\boldsymbol{e}^{\mathrm{T}}(t)\left((\overline{\boldsymbol{L}}+\overline{\boldsymbol{D}}) \otimes \boldsymbol{I}_{n}\right) \boldsymbol{e}(t)\right)^{(1+\alpha) / 2} \leqslant \\ & \quad-c \boldsymbol{\beta} \gamma_{\min } \lambda_{\min }^{(1+\alpha) / 2}(\boldsymbol{Q})\left(\sum\limits_{i=1}^{N} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{e}_{i}(t)\right)^{(1+\alpha) / 2} \leqslant \\ & \quad-c \boldsymbol{\beta} \gamma_{\min } \lambda_{\min }^{(1+\alpha) / 2}(\boldsymbol{Q}) V_{0}^{(1+\alpha) / 2}(t), \end{align*} (19) 其中\overline{\boldsymbol{A}}=\left[\bar{a}_{i j}\right]_{N \times N}, \bar{a}_{i j}=2\left(\xi_{i} a_{i j}\right)^{2 /(1+\alpha)}和\bar{d}_{1}=\frac{2 d_{1} \alpha}{\alpha+1}\left(1-\sigma^{\alpha+1} n^{(1-\alpha) / 2}\right).基于引理1,定义\boldsymbol{Q}=\overline{\boldsymbol{L}}+\overline{\boldsymbol{D}},这里\overline{\boldsymbol{L}}= \left[\bar{I}_{i j}\right] \in \mathbb{R}^{N \times N}, \bar{I}_{i j}=-\bar{a}_{i j},其中i \neq j, \bar{I}_{i i}=\sum\limits_{j \neq i}^{N} \bar{a}_{i j},和\overline{\boldsymbol{D}}=\operatorname{diag}(\overbrace{\left(\frac{\bar{d}_{1} \xi_{1}}{c \beta}\right)^{2 /(1+\alpha)}}^{l}, \left(\frac{\bar{d}_{1} \xi_{2}}{c \beta}\right)^{2 /(1+\alpha)}, \cdots, 0, 0) \in \mathbb{R}^{N \times N}是个对角矩阵.由引理3可知,矩阵\boldsymbol{Q}是正定的.
基于式(12)—(19), 可得
\dot{V}(t) \leqslant \sum\limits_{i=1}^{N} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t)\left[\boldsymbol{C}^{\mathrm{T}}+\boldsymbol{C}+\boldsymbol{L}_{1} \boldsymbol{L}_{1}^{\mathrm{T}}+\boldsymbol{B} \boldsymbol{B}^{\mathrm{T}}+c \gamma_{\min } \lambda_{\max }(\boldsymbol{P}) \boldsymbol{I}_{n}\right] \boldsymbol{e}_{i}(t)-\\ \;\;\;\;\;\;\;\;c \beta \gamma_{\min } \lambda_{\min }^{(1+\alpha) / 2}(\boldsymbol{Q}) V_{0}^{(1+\alpha) / 2}(t) . (20) 选取矩阵\boldsymbol{P}中参数k \bar{I}_{1}充分大,则存在正数\rho_{0}>0满足
-\rho_{0}=\lambda_{\max }\left(\boldsymbol{C}^{\mathrm{T}}+\boldsymbol{C}+\boldsymbol{L}_{1} \boldsymbol{L}_{1}^{\mathrm{T}}+\boldsymbol{B} \boldsymbol{B}^{\mathrm{T}}\right)+c \gamma_{\min } \lambda_{\max }(\boldsymbol{P}) . 由此, 下列不等式成立:
\dot{V}(t) \leqslant-\rho_{0} V_{0}(t)-\rho_{1} V_{0}^{(1+\alpha) / 2}(t), (21) 其中\rho_{1}=c \beta \gamma_{\text {min }} \lambda_{\text {min }}^{(1+\alpha) / 2}(\boldsymbol{Q})>0.因此,我们有下式成立:
\begin{align*} \dot{V}_{0}(t) & =\dot{V}(t)-\dot{V}_{1}(t) \leqslant \\ & -\rho_{0} V_{0}(t)-\rho_{1} V_{0}^{(1+\alpha) / 2}(t)+2 c \sum\limits_{i=1}^{l}\left(\bar{k}_{1}-\hat{k}_{i}(t)\right) \dot{\hat{k}}_{i}(t) . \end{align*} (22) 基于更新律(7), 考虑以下情形:
① 若(1-\sigma) \hat{k}_{i}(t) \geqslant \bar{k}_{1},则对所有的i=1, 2, \cdots, l,有\dot{\hat{k}}_{i}(t)=0.
② 若(1-\sigma) \hat{k}_{i}(t)<\bar{k}_{1},对i=1, 2, \cdots, l,由更新定律(7), \boldsymbol{e}_{i}(t)=\boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)-{\boldsymbol{z}}_{i}(t)和\hat{k}_{i}(t)>0,可得
\begin{align*} & 2 \sum\limits_{i=1}^{l}\left(\bar{k}_{1}-\hat{k}_{i}(t)\right) \dot{\hat{k}}_{i}(t)=2 c \sum\limits_{i=1}^{l}\left(\bar{k}_{1}-\hat{k}_{i}(t)\right) \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}\left(t_{k^{\prime}}^{i}\right) \boldsymbol{\varGamma} \boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)= \\ & \quad 2 c \sum\limits_{i=1}^{l}\left(\bar{k}_{1}-\hat{k}_{i}(t)\right) \xi_{i}\left(\boldsymbol{e}_{i}(t)+\boldsymbol{z}_{i}(t)\right)^{\mathrm{T}} \boldsymbol{\varGamma}\left(\boldsymbol{e}_{i}(t)+\boldsymbol{z}_{i}(t)\right)= \\ & \quad 2 c \sum\limits_{i=1}^{l}\left(\bar{k}_{1}-\hat{k}_{i}(t)\right) \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t)+4 c \sum\limits_{i=1}^{l}\left(\bar{k}_{1}-\hat{k}_{i}(t)\right) \xi_{i} \boldsymbol{e}_{i}(t)^{\mathrm{T}} \boldsymbol{\varGamma} \boldsymbol{z}_{i}(t)+ \\ & \quad 2 c \sum\limits_{i=1}^{l}\left(\bar{k}_{1}-\hat{k}_{i}(t)\right) \xi_{i} {\boldsymbol{z}}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{z}_{i}(t) \text {. } \end{align*} (23) 基于杨氏不等式(Young’s inequality), 由式(23)进一步可得
\begin{align*} & 2 \sum\limits_{i=1}^{l}\left(\bar{k}_{1}-\hat{k}_{i}(t)\right) \dot{\hat{k}}_{i}(t) \leqslant \\ & \quad 2 c \sum\limits_{i=1}^{l}\left(\bar{k}_{1}-\hat{k}_{i}(t)\right) \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t)+2 c \sum\limits_{i=1}^{l}\left(\bar{k}_{1}-\hat{k}_{i}(t)\right) \xi_{i} \boldsymbol{z}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} {\boldsymbol{z}}_{i}(t)+ \\ & \quad c \sum\limits_{i=1}^{l}\left(\bar{k}_{1}+\hat{k}_{i}(t)\right) \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t)+4 c \sum\limits_{i=1}^{l}\left(\bar{k}_{1}+\hat{k}_{i}(t)\right) \xi_{i} {\boldsymbol{z}}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} {\boldsymbol{z}}_{i}(t) \leqslant \\ & \quad\left(2+2 \sigma^{2}\right) c \bar{k}_{1} \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t)+\mu_{1} \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t)= \\ & \quad \mu \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{e}_{i}(t), \end{align*} (24) 其中\mu=\frac{-6 \sigma^{3}+10 \sigma^{2}-3 \sigma+4}{1-\sigma} \bar{k}_{1} c \gamma_{\text {max }}, \gamma_{\text {max }}=\max \left\{\gamma_{1}, \cdots, \gamma_{n}\right\}和\mu_{1}=\left(1+\frac{1}{1-\sigma}\right)\left(1+4 \sigma^{2}\right) c \bar{k}_{1}.
综合情形①和②,我们可得到下列不等式:
\begin{align*} \dot{V}_{0}(t) & \leqslant-\rho_{0} V_{0}(t)-\rho_{1} V_{0}^{(1+\alpha) / 2}(t)+\mu \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{e}_{i}(t) \leqslant \\ & -\frac{\rho_{1}}{2} V_{0}^{(1+\alpha) / 2}(t)+\left[\left(\mu-\rho_{0}\right) V_{0}^{(1-\alpha) / 2}(t)-\frac{\rho_{1}}{2}\right] V_{0}^{(1+\alpha) / 2}(t) . \end{align*} (25) 定义紧集:
\bar{\varOmega}=\left\{\boldsymbol{e}(0) \mid V_{0}(\boldsymbol{e}(0)) \leqslant \kappa_{m}\right\}, 其中κm=(ρ1/(2μ-2ρ0))2/(1-α). 基于不等式(25), 我们讨论下列情形:
情形A1 V_{0}(\boldsymbol{e}(0)) \leqslant \kappa_{m}.由式(25)可得\dot{V}_{0}(t) \leqslant-\left(\rho_{1} / 2\right) V_{0}^{(1+\alpha) / 2}(t).进一步,由引理4可知,\boldsymbol{e}(t)以有限时间收玫到\mathbf{0},停息时间T_{1}=\frac{4 V_{0}^{(1-\alpha) / 2}(0)}{\rho_{1}(1-\alpha)}.
情形B1 V_{0}(\boldsymbol{e}(0))>\kappa_{m}, \boldsymbol{e}(t)以有限时间到达区域\bar{\varOmega}.否则,V_{0}(t)>\kappa_{m}对任意时刻t成立.基于不等式(21),我们可得
\begin{align*} & V(0) \geqslant V(0)-V(t)=-\int_{0}^{t} \dot{V}(s) \mathrm{d} s \geqslant \\ & \quad \int_{0}^{t} \rho_{0} V_{0}(s)+\rho_{1} V_{0}^{(1+\alpha) / 2}(s) \mathrm{d} s>t\left(\rho_{0} \boldsymbol{\kappa}_{m}+\rho_{1} \kappa_{m}^{(1+\alpha) / 2}\right) . \end{align*} (26) 由于V(\boldsymbol{e}(0))是有界的,因此在t \rightarrow \infty时同式(26)矛盾.因此V(0)是V(t)的上界,可得T_{2}= \frac{V(0)}{\rho_{0} \kappa_{m}+\rho_{1} \kappa_{m}^{(1+\alpha) / 2}}和\boldsymbol{e}(t)在T_{2}内收敛到\bar{\varOmega}.结合以上分析,我们可得\boldsymbol{e}(t)最终在有限时间T=T_{1}+T_{2}收敛到 0.
注4 基于以上分析, 耦合网络(2)被镇定在系统(1)时, 停息时间的估计小于等于T. 由参数T1和μ可推得触发门槛常数对停息时间的影响具有重要作用. 换言之, 参数σ越大, 停息时间越大.
2.2 Zeno行为的排除
定理2 在定理1的假设下, 可以合理的排除Zeno行为, 换言之, tk'+1i-tk'i有一个正的下界.
证明 基于不等式(21), 我们可得到V(e(t))≤V(e(0)),由此ei(t)和ki(t)是有界的. 由触发条件(5), ‖zi(t)‖≤σi‖ei(t)‖,对t∈[tk'i, tk'+1i)成立. 分析下界内部时间如下, 基于ei(t)=ei(tk'i)-zi(t),考虑受牵制节点i的系统, 则有下式:
\begin{align*} & \left\|\dot{\boldsymbol{z}}_{i}(t)\right\|=\left\|\dot{\boldsymbol{e}}_{i}(t)\right\|= \\ & \| \boldsymbol{C e}_{i}(t)+\boldsymbol{B}_{1} \boldsymbol{f}\left(\boldsymbol{e}_{i}(t)\right)+c \sum\limits_{j=1}^{N} a_{i j}\left[\boldsymbol{\varGamma} \boldsymbol{e}_{j}(t)+\beta \operatorname{sig}^{\alpha}\left(\boldsymbol{e}_{j}(t)-\boldsymbol{e}_{i}(t)\right)\right]+ \\ & \quad k_{i}(t) \boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)+k_{1} \operatorname{sig}^{\mu}\left(\boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)\right) \| \leqslant \\ & \quad\left(\|\boldsymbol{C}\|+h\left\|\boldsymbol{B}_{1}\right\|+c\left\|a_{i i}\right\|\|\boldsymbol{\varGamma}\|\right)\left\|\boldsymbol{e}_{i}(t)\right\|+c \sum\limits_{j \neq i}^{N} a_{i j} \boldsymbol{\varGamma}\left[\left\|\boldsymbol{e}_{j}(t)\right\|+\right. \\ & \quad \left.\beta\left\|\boldsymbol{e}_{j}(t)-\boldsymbol{e}_{i}(t)\right\|^{\alpha}\right]+\left\|k_{i}(t) \boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)+k_{1} \operatorname{sig}^{\mu}\left(\boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)\right)\right\| \leqslant \\ & \quad \left\|\boldsymbol{G}_{i}\right\|\left\|\boldsymbol{z}_{i}(t)\right\|+\boldsymbol{\phi}_{k}{ }^{i}, \end{align*} (27) 其中h=\max h_{k, j}(k, j=1, 2, \cdots, n), \left\|\boldsymbol{G}_{i}\right\|=\|\boldsymbol{C}\|+h\left\|\boldsymbol{B}_{1}\right\|+c\left\|a_{i i}\right\|\|\boldsymbol{\varGamma}\|, \boldsymbol{\phi}_{k}{ }^{i}=c\left\|\sum_{j \neq i}^{N} a_{i j} \boldsymbol{\varGamma}\boldsymbol{e}_{j}(t)\right\|+ \beta\left\|\left(\boldsymbol{e}_{j}(t)-\boldsymbol{e}_{i}(t)\right)^{\alpha}\right\|+\left\|k_{i}(t) \boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)+k_{1} \operatorname{sig}^{\mu}\left(\boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)\right)\right\|+\left\|\boldsymbol{G}_{i}\right\|\left\|\boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)\right\|.
进一步, 可得
\left\|{\boldsymbol{z}}_{i}(t)\right\| \leqslant \frac{\phi_{k}^{i}}{\left\|\boldsymbol{G}_{i}\right\|}\left(\boldsymbol{e}^{\left\|\boldsymbol{G}_{i}\right\|}\left(t-t_{k^{\prime}}^{i}\right)-1\right) (28) 成立. 由触发机制(5), 下一个触发时刻tk'+1i满足
\left\|{\boldsymbol{z}}_{i}\left(t_{k^{\prime}+1}^{i}\right)\right\|>\sigma_{i}\left\|\boldsymbol{e}_{i}\left(t_{k^{\prime}+1}^{i}\right)\right\| . (29) 由不等式(29)可知, 存在一个常数M1>0满足‖zi(tk'+1i)‖>M1>σi‖ei(tk'+1i)‖. 同式(28), 我们可得下式:
\frac{\phi_{k}^{i}}{\left\|\boldsymbol{G}_{i}\right\|}\left(\boldsymbol{e}^{\left\|\boldsymbol{G}_{i}\right\|\left(t_{k^{\prime}+1}^i-t_{k^{\prime}}^i\right)}-1\right)>M_{1}>0 进一步, 可得
t_{k^{\prime}+1}^{i}-t_{k^{\prime}}^{i}>\frac{\ln \left(M_{1}\left\|\boldsymbol{G}_{i}\right\| /\left(\phi_{k}^{i}\right)+1\right)}{\left\|\boldsymbol{G}_{i}\right\|}>0 . 证毕.
□ 3. 数值模拟
考虑混沌细胞神经网络[34]
\dot{\boldsymbol{s}}(t)=\boldsymbol{C s}(t)+\boldsymbol{B} \tilde{\boldsymbol{f}}(\boldsymbol{s}(t)), (30) 其中\boldsymbol{C}=\operatorname{diag}\{-1, -1, -1\},
\boldsymbol{B}=\left[\begin{array}{ccc} 1.16 & -1.5 & -1.5 \\ -1.5 & 1.16 & -2.0 \\ -1.2 & 2.0 & 1.16 \end{array}\right], 其中\boldsymbol{f}_{i}(\boldsymbol{s}(t))=0.5(|\boldsymbol{s}(t)+1|-|\boldsymbol{s}(t)-1|), \boldsymbol{s}(t)是单节点的状态轨迹.初值\boldsymbol{s}(0)=(0.1, 0.2, 0.3)^{\mathrm{T}}时,系统(30)有个混沌吸引子,如图 1(a)所示.系统(30)的平衡点为
\begin{aligned} & \boldsymbol{x}_{1}^{*}=(-1.22259936, -0.655802861, 0.697535771)^{\mathrm{T}}, \\ & \boldsymbol{x}_{2}^{*}=(0, 0, 0)^{\mathrm{T}}, \boldsymbol{x}_{3}^{*}=(1.22259936, 0.655802861, -0.697535771)^{\mathrm{T}} . \end{aligned} 本例中,我们选择\boldsymbol{x}_{3}^{*}作为目标状态.考虑系统(1)和(2).本文考虑了30个节点的耦合网络(30个系统(2)),其中内连接矩阵\boldsymbol{\varGamma}=\boldsymbol{I}_{3}.假定网络是强连通的和细节性平衡的.选取参数d_{1}=1, c=70, \bar{k}_{1}=1, \beta= 0.5, \alpha=0.5和\hat{k}_{i}(0)=0.2.通过计算,选取参数\sigma=0.2.基于文献[33],我们选取受牵制的节点5, 10, 11, 12,和27实现耦合网络的有限时间镇定.基于定理1,在控制方案(6)下,系统(2)被有限时间镇定到目标节点s(t),如图 1(b).由图 1(b)可得出停息时间约在2 s处.图 2(a)刻画了自适应耦合强度,图 2(b)刻画了受牵制节点的触发次数.
4. 结论
本文研究了基于牵制触发控制动态耦合网络的有限时间镇定. 对比与有限时间触发控制, 对于复杂的动态网络, 牵制触发控制能最大化减少控制的成本, 更有效地实现网络的镇定. 为实现动态网络的有限时间镇定, 新颖的牵制触发控制方案被设计. 此外, 本文借助Lyapunov稳定理论给出了耦合动态网络有限时间镇定的充分条件. 最后数值仿真验证了定理的合理性.
附录
式(14)的证明:
\begin{align*} & -2 c \sum\limits_{i=1}^{l} \hat{k}_{i}(t) \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)+2 c \sum\limits_{i=1}^{l}\left(\hat{k}_{i}(t)-\bar{k}_{1}\right) \dot{k}_{i}(t)= \\ & \quad-c \bar{k}_{1} \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t)+c \bar{k}_{1} \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t)-2 c \sum\limits_{i=1}^{l} \hat{k}_{i}(t) \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma}\left(\boldsymbol{e}_{i}(t)+\boldsymbol{z}_{i}(t)\right)= \\ & \quad-c \bar{k}_{1} \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t)+\sum\limits_{i=1}^{l}\left(c \bar{k}_{1}-2 c \hat{k}_{i}(t)\right) \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t)-2 c \sum\limits_{i=1}^{l} \hat{k}_{i}(t) \dot{\boldsymbol{\xi}}_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} {\boldsymbol{z}}_{i}(t) \leqslant \\ & \quad-c \bar{k}_{1} \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t)+c \sum\limits_{i=1}^{l}\left(\bar{k}_{1}-2 \hat{k}_{i}(t)\right) \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t)+ \\ & \quad c \sum\limits_{i=1}^{l} \hat{k}_{i}(t) \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t)+c \sum\limits_{i=1}^{l} \hat{k}_{i}(t) \xi_{i} {\boldsymbol{z}}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{z}_{i}(t) \leqslant \\ & \quad-c \bar{k}_{1} \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t)+c \sum\limits_{i=1}^{l}\left(\bar{k}_{1}+\hat{k}_{i}(t)\left(\boldsymbol{\sigma}^{2}-1\right)\right) \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t) \leqslant \\ & \quad-c \bar{k}_{1} \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t) \text {. } \end{align*} (A1) 式(15)的证明:
\begin{align*} & -2 c \sum\limits_{i=1}^{l} \hat{k}_{i}(t) \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)+2 c \sum\limits_{i=1}^{l}\left(\hat{k}_{i}(t)-\bar{k}_{1}\right) \dot{k}_{i}(t)= \\ &\quad -2 c \sum\limits_{i=1}^{l} \hat{k}_{i}(t) \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}\left(t_{k^{\prime}}^{i}\right)+2 c \sum\limits_{i=1}^{l}\left(\hat{k}_{i}(t)-\bar{k}_{1}\right) \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}\left(t_{k^{\prime}}^{i}\right) \boldsymbol{\varGamma} \boldsymbol{e}\left(t_{i}\left(t_{k^{\prime}}\right)=\right. \\ &\quad -2 c \sum\limits_{i=1}^{l} \hat{k}_{i}(t) \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t)-2 c \sum\limits_{i=1}^{l} \hat{k}_{i}(t) \dot{\xi}_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{z}_{i}(t)+ \\ &\quad 2 c \sum\limits_{i=1}^{l}\left(\hat{k}_{i}(t)-\bar{k}_{1}\right) \xi_{i}\left(\boldsymbol{e}_{i}(t)+{\boldsymbol{z}}_{i}(t)\right)^{\mathrm{T}} \boldsymbol{\varGamma}\left(\boldsymbol{e}_{i}(t)+{\boldsymbol{z}}_{i}(t)\right)= \\ & \quad-2 c \bar{k}_{1} \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t)+2 c \sum\limits_{i=1}^{l}\left(\hat{k}_{i}(t)-2 \bar{k}_{1}\right) \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{z}_{i}(t)+ \\ & \quad2 c \sum\limits_{i=1}^{l}\left(\hat{k}_{i}(t)-\bar{k}_{1}\right) \xi_{i} {\boldsymbol{z}}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{z}_{i}(t) \leqslant \\ &\quad -2 c \bar{k}_{1} \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t)+2 c \sum\limits_{i=1}^{l}\left(\hat{k}_{i}(t)-\bar{k}_{1}\right) \xi_{i} {\boldsymbol{z}}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} {\boldsymbol{z}}_{i}(t)+ \\ & \quad2 c \sum\limits_{i=1}^{l}\left(2 \bar{k}_{1}-\hat{k}_{i}(t)\right) \xi_{i}\left(\frac{1}{4} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t)+{\boldsymbol{z}}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{z}_{i}(t)\right) \leqslant \\ & \quad-c \bar{k}_{1} \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t)-2 c\left(\frac{\hat{k}_{i}(0)}{4}-\bar{k}_{1} \sigma^{2}\right) \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t) \leqslant \\ & \quad-c \bar{k}_{1} \sum\limits_{i=1}^{l} \xi_{i} \boldsymbol{e}_{i}^{\mathrm{T}}(t) \boldsymbol{\varGamma} \boldsymbol{e}_{i}(t), \end{align*} (A2) 最后一步由式(7)\hat{k}_{i}(0)>4 \bar{k}_{1} \sigma^{2}可得.
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