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Consensus Problem of Multi-Agent Systems with Time-Delay under Fixed Topology
DOI: 10.12677/AAM.2019.88163, PDF, HTML, XML, 下载: 957  浏览: 1,339

Abstract: This paper studies the consensus problem of second-order multi-agent systems with time-delay un-der fixed topology. The pure imaginary roots of the characteristic equation of the system are solved by Hopf bifurcation method. Further, the necessary and sufficient conditions for the conditions for the consistency of the system are obtained through analysis.

1. 引言

2. 预备知识以及问题描述

${L}_{ij}=\left\{\begin{array}{l}-{a}_{ij},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}j\ne i\\ \underset{k=1,k\ne i}{\overset{n}{\sum }}{a}_{ik},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}j=i\end{array}$

${v}_{i}$ 的邻居集表示为 ${\mathcal{N}}_{i}=\left\{{v}_{j}\in \mathcal{V}:\left({v}_{j},{v}_{i}\right)\in \mathcal{E}\right\}$，进一步， ${\mathcal{N}}_{i}^{-}=\left\{{v}_{j}\in \mathcal{V}|{a}_{ij}<0\right\}$${\mathcal{N}}_{i}^{+}=\left\{{v}_{j}\in \mathcal{V}|{a}_{ij}>0\right\}$ 分别表示 ${v}_{i}$ 的负邻居集和正邻居集。定义 $B:=diag\left\{{b}_{1},{b}_{2},\cdots ,{b}_{n}\right\}$ 为领航者与跟随者之间的邻接矩阵。

${b}_{i}>0$ 表示领航者与第i个智能体有之间连接。否则 ${b}_{i}=0$

$\left\{\begin{array}{l}{\stackrel{˙}{x}}_{i}\left(t\right)={v}_{i}\left(t\right),\\ {\stackrel{˙}{v}}_{i}\left(t\right)={u}_{i}\left(t\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}i\in I\end{array}$ (1)

${\stackrel{˙}{x}}_{0}\left(t\right)={v}_{0}$

${u}_{i}\left(t\right)=\underset{j\in {N}_{i}}{\sum }{a}_{ij}\left[{x}_{j}\left(t-\tau \right)-{x}_{i}\left(t-\tau \right)\right]+{b}_{i}\left[{x}_{0}\left(t-\tau \right)-{x}_{i}\left(t-\tau \right)\right]+k\left({v}_{0}-{v}_{i}\left(t\right)\right)$ (2)

$u\text{}:={\left({u}_{1},{u}_{2},\cdots ,{u}_{n}\right)}^{\text{T}}\in {R}^{n}$$\stackrel{¯}{x}:=x-{x}_{0}$$\stackrel{¯}{v}:=v-{v}_{0}$，且令 $\epsilon \left(t\right):={\left({\stackrel{¯}{x}}^{\text{T}},{\stackrel{¯}{v}}^{\text{T}}\right)}^{\text{T}}$。那么系统(1)在控制协议(2)下可以写成：

$\stackrel{˙}{\epsilon }\left(t\right)=C\epsilon \left(t\right)+E\epsilon \left(t-\tau \right)$ (3)

$C:=\left(\begin{array}{cc}{0}_{n×n}& {I}_{n}\\ {0}_{n×n}& -k{I}_{n}\end{array}\right)$$E:=\left(\begin{array}{cc}{0}_{n×n}& {0}_{n×n}\\ -H& {0}_{n×n}\end{array}\right)$ 其中 $H:=L+B$

(注；经过变量代换，系统(1)在控制协议(2)下的一致性问题，可转化为系统(3)的稳定性问题。)

3. 主要结果

$\begin{array}{c}g\left(s\right)=\mathrm{det}\left(s{I}_{n}-C-E{\text{e}}^{-\tau s}\right)=\mathrm{det}\left(\begin{array}{cc}s{I}_{n}& -{I}_{n}\\ H{\text{e}}^{-\tau s}& \left(s+k\right){I}_{n}\end{array}\right)\\ =|\left({s}^{2}+ks\right){I}_{n}+H{\text{e}}^{-\tau s}|=\underset{i=1}{\overset{n}{\prod }}\left({s}^{2}+ks+{\lambda }_{i}{\text{e}}^{-\tau s}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\lambda }_{i}\in \Lambda \left(H\right)\\ =:\underset{i=1}{\overset{n}{\prod }}{g}_{i}\left(s\right)\end{array}$

$\tau \in \Omega =\left\{\frac{1}{{\omega }_{i1}}\left(2\stackrel{¯}{k}\text{π}+{\theta }_{i1}\right)|i\in I;\stackrel{¯}{k}=0,1,\cdots \right\}$

${\omega }_{i}^{2}=ik{\omega }_{i}+{\lambda }_{i}{\text{e}}^{-i{\omega }_{i}\tau }$

$-{\omega }_{i}^{2}+ik{\omega }_{i}+{\lambda }_{i}\left[\mathrm{cos}\left({\omega }_{i}\tau \right)-i\mathrm{sin}\left({\omega }_{i}\tau \right)\right]=0$(4)

$k{\omega }_{i}-\mathrm{Re}\left({\lambda }_{i}\right)\mathrm{sin}\left({\omega }_{i}\tau \right)+\mathrm{Im}\left({\lambda }_{i}\right)\mathrm{cos}\left({\omega }_{i}\tau \right)=0$(5)

$-{\omega }_{i}^{2}+\mathrm{Re}\left({\lambda }_{i}\right)\mathrm{cos}\left({\omega }_{i}\tau \right)+\mathrm{Im}\left({\lambda }_{i}\right)\mathrm{sin}\left({\omega }_{i}\tau \right)=0$(6)

${\mathrm{sin}}^{2}\left({\omega }_{i}\tau \right)+{\mathrm{cos}}^{2}\left({\omega }_{i}\tau \right)=1$。通过(5)，(6)式解得

$\mathrm{cos}\left({\omega }_{i}\tau \right)=\frac{{\omega }_{i}^{2}\mathrm{Re}\left({\lambda }_{i}\right)-k{\omega }_{i}\mathrm{Im}\left({\lambda }_{i}\right)}{{|{\lambda }_{i}|}^{2}}$(7)

$\mathrm{sin}\left({\omega }_{i}\tau \right)=\frac{k{\omega }_{i}\mathrm{Re}\left({\lambda }_{i}\right)+{\omega }_{i}^{2}\mathrm{Im}\left({\lambda }_{i}\right)}{{|{\lambda }_{i}|}^{2}}$(8)

$\frac{{|{\lambda }_{i}|}^{2}\cdot \left({\omega }_{i}^{4}+{k}^{2}{\omega }_{i}^{2}\right)}{{|{\lambda }_{i}|}^{4}}=1$

${\omega }_{i}^{4}+{k}^{2}{\omega }_{i}^{2}-{|{\lambda }_{i}|}^{2}=0$

${\omega }_{i1}=\sqrt{\frac{-{k}^{2}+\sqrt{{k}^{4}+4{|{\lambda }_{i}|}^{2}}}{2}}$${\omega }_{i2}=-\sqrt{\frac{-{k}^{2}+\sqrt{{k}^{4}+4{|{\lambda }_{i}|}^{2}}}{2}}$

$\begin{array}{l}\tau \in \Omega =\left\{\frac{1}{{\omega }_{i1}}\left(2\stackrel{¯}{k}\text{π}+{\theta }_{i1}\right)|i\in I;\stackrel{¯}{k}=0,1,\cdots \right\}\cup \left\{\frac{1}{{\omega }_{i2}}\left(2\stackrel{¯}{k}\text{π}+{\theta }_{i2}\right)|i\in I;\stackrel{¯}{k}=-1,-2,\cdots \right\}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left\{\frac{1}{{\omega }_{i1}}\left(2\stackrel{¯}{k}\text{π}+{\theta }_{i1}\right)|i\in I;\stackrel{¯}{k}=0,1,\cdots \right\}\cup \left\{\frac{1}{{\omega }_{i2}}\left(2\stackrel{¯}{k}\text{π}-{\theta }_{i2}\right)|i\in I;\stackrel{¯}{k}=1,2,\cdots \right\}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left\{\frac{1}{{\omega }_{i1}}\left(2\stackrel{¯}{k}\text{π}+{\theta }_{i1}\right)|i\in I;\stackrel{¯}{k}=0,1,\cdots \right\}\end{array}$

(充分性)下面验证 $\tau =\frac{1}{{\omega }_{i1}}\left(2\stackrel{¯}{k}\text{π}+{\theta }_{i1}\right)$ 为方程式(4)的根。将它代入(4)式的左边，得

$ik{\omega }_{i}+\left[\mathrm{Re}\left({\lambda }_{i}\right)+i\mathrm{Im}\left({\lambda }_{i}\right)\right]\cdot \left[\frac{{\omega }_{i}^{2}\mathrm{Re}\left({\lambda }_{i}\right)-k{\omega }_{i}\mathrm{Im}\left({\lambda }_{i}\right)}{{|{\lambda }_{i}|}^{2}}-i\frac{{\omega }_{i}^{2}\mathrm{Im}\left({\lambda }_{i}\right)+k{\omega }_{i}\mathrm{Re}\left({\lambda }_{i}\right)}{{|{\lambda }_{i}|}^{2}}\right]=0$

${\mathrm{Re}\left(\frac{\text{d}s}{\text{d}\tau }\right)|}_{\tau \in \Omega }>0$ 恒成立。

(2) ${\stackrel{¯}{g}}_{i}\left(s,\tau \right)$ 在点 $\left(i{\omega }_{0},{\tau }_{0}\right)$ 处连续。(3) $\frac{\partial {\stackrel{¯}{g}}_{i}}{\partial s}$$\frac{\partial {\stackrel{¯}{g}}_{i}}{\partial \tau }$ 是连续的，且 ${\frac{\partial {\stackrel{¯}{g}}_{i}}{\partial s}|}_{\left(i{\omega }_{0},{\tau }_{0}\right)}\ne 0$。那么，在 ${g}_{i}\left(s\right)=0$ 关于 $\tau$ 求导，得

$2s\frac{\text{d}s}{\text{d}\tau }+k\frac{\text{d}s}{\text{d}\tau }+{\lambda }_{i}\left(-s-\tau \frac{\text{d}s}{\text{d}\tau }\right){\text{e}}^{-\tau s}=0$

$\frac{\text{d}s}{\text{d}\tau }=\frac{{\lambda }_{i}s{\text{e}}^{-\tau s}}{2s+k-{\lambda }_{i}\tau {\text{e}}^{-\tau s}}$

$\mathcal{X}=\mathrm{Re}\left({\lambda }_{i}\right)\mathrm{sin}\left({\omega }_{ij}\right)-\mathrm{Im}\left({\lambda }_{i}\right)\mathrm{cos}\left({\omega }_{ij}\right)$$\mathcal{Y}=\mathrm{Re}\left({\lambda }_{i}\right)\mathrm{cos}\left({\omega }_{ij}\right)+\mathrm{Im}\left({\lambda }_{i}\right)\mathrm{sin}\left({\omega }_{ij}\right)$，那么

${\frac{\text{d}s}{\text{d}\tau }|}_{s=i{\omega }_{ij}}=\frac{{\omega }_{ij}\left(\mathcal{X}+i\mathcal{Y}\right)}{k-\tau \mathcal{Y}+i\left(2{\omega }_{ij}+\tau \mathcal{X}\right)}$

${\mathrm{Re}\left(\frac{\text{d}s}{\text{d}\tau }\right)|}_{\tau \in \Omega }=\frac{{\omega }_{ij}\left[\mathcal{X}\left(k-\tau \mathcal{Y}\right)+\mathcal{Y}\left(2{\omega }_{ij}+\mathcal{X}\tau \right)\right]}{{\left(k-\tau \mathcal{Y}\right)}^{2}+{\left(2{\omega }_{ij}+\mathcal{X}\tau \right)}^{2}}$

$h\left({\omega }_{ij},\tau \right)=k-\tau \mathcal{Y}+i\left(2{\omega }_{ij}+\tau \mathcal{X}\right)$，利用(7)，(8)，可以得到

$\begin{array}{l}{|h\left({\omega }_{ij},\tau \right)|}^{2}{\mathrm{Re}\left(\frac{\text{d}s}{\text{d}\tau }\right)|}_{\tau \in \Omega }\\ ={\omega }_{ij}\left[\mathcal{X}\left(k-\tau \mathcal{Y}\right)+\mathcal{Y}\left(2{\omega }_{ij}+\mathcal{X}\tau \right)\right]={\omega }_{ij}\left[\mathcal{X}k+2\mathcal{Y}{\omega }_{ij}\right]\\ =\left[{\omega }_{ii}k\mathrm{Re}\left({\lambda }_{i}\right)+2{\omega }_{ij}^{2}\mathrm{Im}\left({\lambda }_{i}\right)\right]\mathrm{sin}\left({\omega }_{ij}\right)+\left[{\omega }_{ij}k\left(-\mathrm{Im}\left({\lambda }_{i}\right)\right)+2{\omega }_{ij}^{2}\mathrm{Re}\left({\lambda }_{i}\right)\right]\mathrm{cos}\left({\omega }_{ij}\tau \right)\\ =\left[2{\omega }_{ij}^{4}+{k}^{2}{\omega }_{ij}^{2}\right]\left[{\mathrm{Re}}^{2}\left({\lambda }_{i}\right)+{\mathrm{Im}}^{2}\left({\lambda }_{i}\right)\right]/{|{\lambda }_{i}|}^{2}=2{\omega }_{ij}^{4}+{k}^{2}{\omega }_{ij}^{2}\end{array}$

$2{\omega }_{ij}^{4}+{k}^{2}{\omega }_{ij}^{2}>0$，显然， ${\mathrm{Re}\left(\frac{\text{d}s}{\text{d}\tau }\right)|}_{\tau \in \Omega }>0$ 成立。

$\tau <{\tau }^{\ast }=\underset{i\in I}{\mathrm{min}}\left\{\frac{{\theta }_{i1}}{{\omega }_{i1}}\right\}$

4. 结论

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