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摘要: 研究了节点状态为高维的多层复杂网络系统的能控性问题. 讨论了节点的异质性、层间耦合和层内耦合对网络能控性的影响. 研究发现当节点状态由同质变为异质、内耦合矩阵由相同变为不同时, 对网络能控性均有影响(网络可以由能控变为不能控, 反之亦然). 对层间耦合模式为驱动响应模式和相互依赖模式, 分别给出了网络系统能控的充分条件或必要条件. 相比于直接应用经典的能控性判据, 这些条件更易于验证, 且驱动响应模式比相互依赖模式实现系统完全能控所需的条件更弱.Abstract: This paper studies the controllability of multi-layer complex network systems with high-dimensional node states. We discuss the influence of node heterogeneity, inter-layer coupling, and intra-layer coupling on the controllability of the network. It is found that when the state of the node changes from homogeneous to heterogeneous or the inner-coupling matrix changes from identical to nonidentical, both have an impact on the controllability of the network (the network can be changed from controllable to uncontrollable, and vice versa). The sufficient or necessary conditions for the controllability of the networked system when the inter-layer coupling mode is the drive response mode and the interdependence mode are given, respectively. These conditions are easier to verify than using the classic controllability criteria directly, and the conditions that the drive response mode to achieve fully controllability are weaker than the conditions for the interdependence mode.
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客观世界中的许多系统都可以用复杂网络进行描述, 例如电力网络、交通网络以及互联网络等. 而对复杂网络研究的主要目的之一是让人们更好地了解网络结构和功能, 控制复杂网络的行为使其为人类服务. 例如, 在电力网络中如何选取合适的变电站作为被控节点来实现对整个电力网络的控制; 在交通网络中, 如何对车辆站点进行合理的控制来避免交通拥堵等. 而在对一个网络进行控制之前, 首先要判断这个网络是否能控. 因此复杂网络的能控性是复杂网络系统分析和综合的基本问题[1]. 人们对能控性最早的认识来源于线性系统理论, 并产生了很多能控性判据[2-7]. 然而, 近些年人们发现, 这些判据很难直接应用于大规模复杂网络, 因此很多学者开展了针对大规模复杂网络的能控性的研究, 包括静态网络[8-18]和时序网络[19-23].
实际复杂系统抽象成网络, 如果按照层数来进行分类, 可以将大规模网络分为单层网络和多层网络. 对于单层网络, Wang等[24-25]和Hao等[26-27]基于PBH (Popov-Belevitch-Hautus)能控性判据, 研究了节点为高维同质的单层网络系统的能控性. 在此基础上, Hao等[28]研究了一些典型结构的同质网络的能控性. 另外, 许多学者在单层网络能控性方面取得了丰硕的成果[29-33].
相比于直接用单层网络描述, 许多实际系统更适合用多层网络来描述, 例如公交−地铁构成的交通网络、互联网−电网构成的网络等[34], 多层网络结构的能控性也已得到较为广泛的研究[35-40]. 当每个节点的维数等于1时, Chapman等[41]提出一个多层网络的分析框架, 将因子网络的能控性和能观性扩展到复合网络上, 并得到系统能控的充要条件. Hao等[42]研究了一类特殊的多层网络的能控性, 并给出了能控的充要条件. Chen等[43]根据张量代数以及多项式控制等理论, 发展了超图网络的能控性, 并推导出类Kalman秩条件来确定实现能控所需的最小控制节点的数量. 当每个节点的维数大于1时, Wu等[44]定义了两类典型的层间耦合模式, 即驱动响应模式和相互依赖模式, 并讨论了其对网络系统平衡稳定性的影响. Wu等[45]研究了多层同质网络的状态能控性, 对层间耦合模式分别为驱动响应模式和相互依赖模式的多层网络, 分别给出了系统能控的一些充分或必要条件, 并揭示了不同的层间耦合模式对网络系统能控性的影响. Jiang等[46]以矩阵方程的形式给出了多层同质网络能控的充要条件, 以及层内网络拓扑为一些特殊结构的网络能控的充分或必要条件.
以上的研究都是假设网络中每个节点具有相同的动力学特性, 这样的网络称为同质网络. 然而, 现实中许多系统, 每个节点的动力学特性并不相同, 这样的网络称为异质网络. 当前, 对于异质网络状态能控性研究得相对较少, Xiang等[47]从图论和代数的角度给出了一类单层异质网络完全能控的充要条件, 并在文献[48]中研究了节点异质性对整个单层网络系统能控性的影响. 在此基础上, 在文献[49]中对节点之间输入输出是一维且每个节点具有能控规范型的单层异质网络, 设计了一种控制输入, 得到网络能控的充要条件, 并研究了网络拓扑可对角化时的网络系统能控的充要条件.
综上所述, 目前关于复杂网络能控性的研究, 主要考虑的是单层或多层同质网络的能控性[24-28, 45-46], 关于单层异质网络的能控性也取得了一定进展[47-49], 但多层异质网络的能控性的研究尚属空白, 并且上述研究都是假设节点之间具有相同的内耦合矩阵. 然而, 这个假设较为理想, 在实际中很难满足. 例如, 考虑层间耦合模式为驱动响应模式(信号只能从驱动层传递到响应层)的两层网络, 驱动层由若干个发电机组成, 响应层由若干个电动机组成, 如果将发电机和电动机视为节点, 发电机和电动机的物理参数视为节点的状态. 显然, 每个节点可能具有不同的状态, 并且节点之间可能具有不同的耦合关系. 因此, 部分工程网络建模为具有不相同内耦合矩阵的多层异质网络更符合实际情况. 另外, 许多对于内耦合矩阵相同的多层同质网络成立的结论对多层异质网络(内耦合矩阵相同或不相同)通常不成立或很难直接推广, 所以对于多层异质网络需要展开新的讨论.
基于以上讨论, 本文研究了节点之间具有不相同的内耦合矩阵的多层异质网络的状态能控性. 主要工作和创新点归纳如下:
1)揭示了节点异质性以及内耦合矩阵的不相同会影响多层网络的能控性;
2)针对层间耦合为驱动响应模式的多层异质网络, 分别给出了内耦合矩阵不相同时系统能控的充分条件以及内耦合矩阵相同时系统能控的必要条件;
3)针对层间耦合为相互依赖模式的多层异质网络, 分别给出了内耦合矩阵不相同时系统能控的充分条件以及内耦合矩阵相同时系统完全能控的一个必要条件.
1. 预备知识
本节给出一些数学记号以及必要的引理. 在本文中, $ {\bf{R}}\,({\bf{C}}) $表示实(复)数域, ${{\bf{R}}^n}\,({\bf{C}^{n}})$表示$ n $元有序实(复)数组构成的$ n $维实(复)向量空间, $ {{\bf{R}}^{n \times m}}\,({{\bf{C}}^{n \times m}}) $表示$ n \times m $的实(复)矩阵的集合, $ {{\boldsymbol{I}}_n} $表示$ n \times n $单位矩阵, ${\text{diag}}\{ {a_1},\cdots,{a_n}\}$表示对角元为${a_1},\cdots,{a_n}$的对角矩阵, $ \sigma ({\boldsymbol{A}}) $为矩阵$ {\boldsymbol{A}} $的所有特征值构成的集合, $ \otimes $表示矩阵的Kronecker积运算, $ \left| {\boldsymbol{A}} \right| $表示方阵$ {\boldsymbol{A}} $的行列式. 如果没有明确指明矩阵的规模, 则认为矩阵之间的代数运算是相容的.
加权有向图$G = (V,E,{\boldsymbol{W}})$由节点集$V = \{ {v_1}, {v_2},\cdots, {v_N}\}$, 边集$ E = \{ ({v_j},{v_i})\} $以及加权邻接矩阵${\boldsymbol{W}} = [{w_i}_j] \in {{\bf{R}}^{N \times N}}$构成. 其中, $ {w_i}_j \ne 0 $表示从节点$ {v_j} $到节点$ {v_i} $存在一条有向边; 否则, $ {w_i}_j = 0 $.
引理 1[49]. 由矩阵对$ ({\boldsymbol{A}},{\boldsymbol{B}}) $描述的线性时不变系统, 下列结论等价:
1) 系统$ ({\boldsymbol{A}},{\boldsymbol{B}}) $是能控的;
2) 能控性矩阵${\boldsymbol{Q}} = [{\boldsymbol{B}},{\boldsymbol{AB}}, {{\boldsymbol{A}}^2}{\boldsymbol{B}},\cdots,{{\boldsymbol{A}}^{n - 1}}{\boldsymbol{B}}]$是行满秩的, 即${\rm{rank}}({\boldsymbol{Q}}) = n$;
3) $ {\boldsymbol{A}} $的任何左特征向量$ {\boldsymbol{\alpha }} $都满足$ {\boldsymbol{\alpha B}} \ne 0 $;
4) ${\rm{rank}}(s{{\boldsymbol{I}}_n} - {\boldsymbol{A}},{\boldsymbol{B}}) = n,\forall s \in {\bf{C}} .$
2. 多层异质网络系统
2.1 多层异质网络模型
一个具有$ M $层且每层$ N $个节点的多层异质网络系统的模型为
$$ \left\{ {\begin{aligned} {\dot{\boldsymbol{x}}}_i^K =\;& {\boldsymbol{A}}_i^K{\boldsymbol{x}}_i^K + \sum\limits_{j = 1}^N {w_{ij}^K{\boldsymbol{H}}_i^K{\boldsymbol{y}}_j^K} \,+ \\ & \sum\limits_{J = 1,J \ne K}^M {\sum\limits_{j = 1}^N {d_{ij}^{KJ}{\boldsymbol{H}}_i^{KJ}{\boldsymbol{y}}_j^J} } + \delta _i^K{\boldsymbol{B}}_i^K{\boldsymbol{u}}_i^K \\ {\boldsymbol{y}}_i^K =\;&{\boldsymbol{C}}_i^K{\boldsymbol{x}}_i^K,i = 1,\cdots,N \end{aligned}} \right. $$ (1) 其中, $ {\boldsymbol{x}}_i^K \in {{\bf{R}}^n} $, $ {\boldsymbol{u}}_i^K \in $$ {{\bf{R}}^p} $, $ {\boldsymbol{y}}_i^K \in {{\bf{R}}^m} $分别表示第$ K $层网络的第$ i $个节点的状态向量、输入向量以及输出向量(上标$ K $表示第$ K $层网络, $ K = 1,\cdots,M $; 下标$ i $表示第$ i $个节点, $ i = 1,\cdots,N $). 本文模型所用的符号与含义如表1所示. $ {\boldsymbol{A}}_i^K \in {{\bf{R}}^{n \times n}} $为第$ K $层节点$ i $的状态矩阵, $ {\boldsymbol{B}}_i^K \in {{\bf{R}}^{n \times p}} $是节点$ i $的输入矩阵, $ {\boldsymbol{C}}_i^K \in {{\bf{R}}^{m \times n}} $为节点$ i $的输出矩阵, $ {\boldsymbol{H}}_i^K \in {{\bf{R}}^{n \times m}} $描述了第$ K $层网络的不同节点各分量之间的耦合方式, 称其为内耦合矩阵; $ {\boldsymbol{H}}_i^{KJ} $表示第$ K $层与第$ J $层的不同节点各分量的耦合方式, 称其为层间内耦合矩阵, 特别地, 当$ K = J $时, $ {\boldsymbol{H}}_i^{KJ} $即为$ {\boldsymbol{H}}_i^K $. 值得注意的是, 每层内的不同节点之间的内耦合矩阵以及层间节点间的内耦合矩阵是不同的, 即, 当$ i \ne j $时, ${\boldsymbol{H}}_i^K \ne {\boldsymbol{H}}_j^K$; 当$K \ne J$ 时, $ {\boldsymbol{H}}_i^K \ne {\boldsymbol{H}}_i^J $ 或${\boldsymbol{H}}_i^K \ne {\boldsymbol{H}}_j^J$, 故称节点间具有不相同的内耦合矩阵. 加权邻接矩阵$ {{\boldsymbol{W}}^K} = [w_{ij}^K] \in {{\bf{R}}^{N \times N}} $, $ {{\boldsymbol{D}}^{KJ}} = [d_{ij}^{KJ}] \in {{\bf{R}}^{N \times N}} $分别表示层内网络的拓扑结构和层间网络拓扑结构, 其中, $ w_{ij}^K \ne 0 $表示存在从节点$ {v_j} $到节点$ {v_i} $的有向边; $ w_{ij}^K = 0 $表示不存在从节点$ {v_j} $到节点$ {v_i} $的有向边; 类似地, 如果存在一条有向边从第$ J $层的节点$ j $指向$ K $层的节点$ i $, 则$ d_{ij}^{KJ} \ne 0 $. 如果第$ K $层的节点 $ i $为受控节点, 则$ \delta _i^K = 1 $, 否则$ \delta _i^K = 0 $. 值得注意的是, 系统(1)中每层内的各个节点以及各层之间的节点具有不同的自身动力学, 故称其为多层异质网络. 而当各层内节点的自身动力学相同以及节点间的内耦合矩阵相同时, 表示多层同质网络模型[46]. 本文中考虑的是节点异质的多层网络的能控性.
表 1 本文模型中所用的特殊符号Table 1 Special notations used in the model of this paper特殊符号 含义 ${\boldsymbol{A}}_i^K$ 第$ K $层网络的第$ i $个节点的状态矩阵 ${\boldsymbol{B}}_i^K$ 第$ K $层网络的第$ i $个节点的输入矩阵 ${\boldsymbol{C}}_i^K$ 第$ K $层网络的第$ i $个节点的输出矩阵 ${\boldsymbol{H}}_i^K$ 第$ K $层网络的第$ i $个节点与该层其他节点之间的内耦合矩阵 ${\boldsymbol{H}}_i^{KJ}$ 第$ J $层网络的第$ i $个节点与第$ K $层网络的其他节点之间的内耦合矩阵 ${{\boldsymbol{W}}^K}$ 第$ K $层的网络拓扑 ${{\boldsymbol{D}}^{KJ} }$ 第$ J $层到第$ K $层的网络拓扑 ${\boldsymbol{x}}$ 整个网络系统的状态 ${{\boldsymbol{x}}^K}$ 第$ K $层网络的状态 ${\boldsymbol{u}}$ 整个网络系统的输入 ${{\boldsymbol{u}}^K}$ 第$ K $层网络的输入 ${\boldsymbol{ \Phi}}$ 整个网络系统的状态矩阵 ${{\boldsymbol{\Phi}} _{KK} }$ 第$ K $层网络的状态矩阵 ${{\boldsymbol{\Phi}} _{KJ} }$ 第$ J $层到第$ K $层网络的状态矩阵 ${\boldsymbol{\Psi}}$ 整个网络系统的输入矩阵 ${\boldsymbol{ \Xi}}$ 整个网络系统的输出矩阵 ${{\boldsymbol{\Lambda}} ^K}$ 以${\boldsymbol{A}}_1^K,\cdots,{\boldsymbol{A}}_N^K$为对角元的分块对角矩阵 ${\boldsymbol{ \Delta}}$ 对角矩阵${ {\text{diag} } } \{ {{\boldsymbol{\Delta}} ^1},\cdots,{{\boldsymbol{\Delta}} ^M}\}$ ${{\boldsymbol{\Delta}} ^K}$ 对角矩阵${ {\text{diag} } } \{ \delta _1^K,\cdots,\delta _N^K\}$ 令向量${\boldsymbol{x}} = {[{({{\boldsymbol{x}}^1})^{\rm{T}} },\cdots ,{({{\boldsymbol{x}}^M})^{\rm{T}}}]^{\rm{T}}} \in {{\bf{R}}^{MNn}}$表示整个网络系统的状态, 其中, ${{\boldsymbol{x}}^K} = ({({\boldsymbol{x}}_1^K)^{\rm{T}} } ,$ ${({\boldsymbol{x}}_2^K)^{\rm{T}}} ,\cdots , {({\boldsymbol{x}}_N^K)^{\rm{T}} }{)^{\rm{T}} } \in {{\bf{R}}^{Nn}}$表示系统第$ K $层的状态; ${\boldsymbol{u}} = [{({{\boldsymbol{u}}^1})^{\rm{T}}}, \cdots,{({{\boldsymbol{u}}^M})^{\rm{T}}}]^{\rm{T}} \in {{\bf{R}}^{MNp}}$表示整个网络系统的输入, 其中, $ {{\boldsymbol{u}}^K} = {({({\boldsymbol{u}}_1^K)^{\rm{T}}},\cdots,{({\boldsymbol{u}}_N^K)^{\rm{T}} })^{\rm{T}} } \in $ $ {{\bf{R}}^{Np}} $为第$ K $层输入. 令${\boldsymbol{\Phi }} = [{{\boldsymbol{\Phi }}_{KJ}}] \in {{\bf{R}}^{MNn \times MNn}}$; $ K,J = 1,\cdots,M $, 表示整个网络系统的状态矩阵, 当$ K = J $时, ${{\boldsymbol{\Phi }}_{KK}} = {{\boldsymbol{\Lambda }}^K} +{{\boldsymbol{\Gamma }}^K}$, 其中, ${{\boldsymbol{\Lambda }}^K} = {{\text{diag}}} \{ {\boldsymbol{A}}_1^K,\cdots,{\boldsymbol{A}}_N^K\} \in {{\bf{R}}^{Nn \times Nn}}$, ${{\boldsymbol{\Gamma }}^K} = [{\boldsymbol{\Gamma }}_{ij}^K] \in {{\bf{R}}^{Nn \times Nn}}$, ${\boldsymbol{\Gamma }}_{ij}^K = w_{ij}^K{\boldsymbol{H}}_i^K{\boldsymbol{C}}_j^K \in {{\bf{R}}^{n \times n}};$ 当$K \ne J$ 时, ${{\boldsymbol{\Phi }}_{KJ}} =$$[{\boldsymbol{\Phi }}_{ij}^{KJ}] \in {{\bf{R}}^{Nn \times Nn}}$ , 其中, ${\boldsymbol{\Phi }}_{ij}^{KJ} = d_{ij}^{KJ}{\boldsymbol{H}}_i^{KJ}{\boldsymbol{C}}_j^J$. 令$ {\boldsymbol{\Psi }} = $ $ {{\text{diag}}} \{ {{\boldsymbol{\Psi }}^1},\cdots,{{\boldsymbol{\Psi }}^M}\} $表示整个网络系统的输入矩阵, 其中, ${{\boldsymbol{\Psi }}^K} = {{\text{diag}}} \{ \delta _1^K{\boldsymbol{B}}_1^K,\cdots, \delta _N^K{\boldsymbol{B}}_N^K\,\},$ ${\boldsymbol{\Xi }} = {{\text{diag}}} \{ \delta _1^1{\boldsymbol{C}}_1^1,\cdots,\;\delta _N^1{\boldsymbol{C}}_N^1,\;\cdots,\;\delta _1^M{\boldsymbol{C}}_1^M, \cdots, \delta _N^M{\boldsymbol{C}}_N^M\}$表示整个网络系统的输出矩阵; 令${\boldsymbol{\Delta }} = {{\text{diag}}} \{ {{\boldsymbol{\Delta }}^1},\cdots,{{\boldsymbol{\Delta }}^M}\}$, 其中${{\boldsymbol{\Delta }}^K} = {{\text{diag}}} \{ \delta _1^K,\cdots,$ $ \delta _N^K\} $描述了第K层网络中外部控制输入与节点之间的连接关系. 此时系统(1)等价于
$$ \left\{ {\begin{aligned} &{{\dot{\boldsymbol{x}}} = {\boldsymbol{\Phi x}} + {\boldsymbol{\Psi u}}} \\ & {{\boldsymbol{y}} = {\boldsymbol{\Xi x}}} \end{aligned}} \right. $$ (2) 下面介绍两类典型的层间耦合模式: 驱动响应模式和相互依赖模式. 为了便于叙述, 以两层网络为例.
定义1 (驱动响应模式)[44]. 对于一个两层网络, 若存在从第1层内的节点到第2层内的节点的有向边, 而不存在从第2层到第1层的有向边, 则称层间耦合为驱动响应模式, 第1层称为驱动层, 第2层称为响应层.
定义2 (相互依赖模式)[44]. 对于一个两层网络, 若既存在从第1层内的节点到第2层内的节点的有向边, 又存在从第2层到第1层的有向边, 则称层间耦合为相互依赖模式.
层间耦合模式为驱动响应模式和相互依赖模式的两层网络的示意图分别如图1和图2所示. 图1和图2的各层是以有向环形网络为例的, 层内的网络拓扑可以是任意一种结构的网络.
图 1 层间耦合为驱动响应模式的两层网络Fig. 1 Two-layer networks with the inter-layer drive-response couplings
图 2 层间耦合为相互依赖模式的两层网络Fig. 2 Two-layer networks with the inter-layer interdependent couplings层数大于2层时, 驱动响应模式的层间耦合形式为有向链的形式(如图3(a)所示)或为有向支撑树的形式(如图3(b)所示), 或者上述两种情形的组合; 而相互依赖模式的层间耦合形式可以是任意一种复杂的连接形式, 图4给出了3种连接结构的例子.
图 3 层间耦合为驱动响应模式的M层网络结构示意图Fig. 3 The illustration of M-layer networks with the inter-layer drive-response couplings
图 4 层间耦合为相互依赖模式的3层网络结构示意图Fig. 4 The illustration of three-layer networks with the inter-layer interdependent couplings2.2 节点和内耦合矩阵对能控性的影响
例 1. 考虑层间耦合为驱动响应模式的两层网络, 每层有3个节点, 驱动层的网络拓扑是有向链网络, 响应层的网络拓扑是有向星形网络, 如图5所示. 令$ \delta _1^1 = 1,\delta _2^1 = \delta _3^1 = \delta _1^2 = \delta _2^2 = \delta _3^2 = $0, 且
图 5 驱动层为链网络、响应层为星形网络的两层网络Fig. 5 Two-layer networks with the topology of the drive-layer is a chain network, and the response layer is a star network$$ \begin{array}{c}{ {{\boldsymbol{W}}^1} = \left[ {\begin{array}{*{20}{c}} 0&0&0 \\ 1&0&0 \\ 0&1&0 \end{array}} \right],\;\;{{\boldsymbol{W}}^2} = \left[ {\begin{array}{*{20}{c}} 0&0&0 \\ 1&0&0 \\ 1&0&0 \end{array}} \right] } \end{array} $$ $$ \begin{split} &{{\boldsymbol{D}}^{21}} = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right],\;\;{\boldsymbol{B}}_1^1 = \left[ {\begin{array}{*{20}{c}} 1&3 \\ 0&2 \end{array}} \right] \\ &{\boldsymbol{C}}_i^K = \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right],\;\;K = 1,2;\;i = 1,2,3 \end{split} $$ 由式(2)可知
$$ \begin{array}{c}{ {\boldsymbol{\Phi }} = \left[ {\begin{array}{*{20}{c}} {{\boldsymbol{A}}_1^1} & 0 & 0 & 0 & 0 & 0 \\ {{\boldsymbol{H}}_2^1{\boldsymbol{C}}_1^1} & {{\boldsymbol{A}}_2^1} & 0 & 0 & 0 & 0 \\ 0 & {{\boldsymbol{H}}_3^1{\boldsymbol{C}}_2^1} & {{\boldsymbol{A}}_3^1} & 0 & 0 & 0 \\ {{\boldsymbol{H}}_1^{21}{\boldsymbol{C}}_1^1} & 0 & 0 & {{\boldsymbol{A}}_1^2} & 0 & 0 \\ 0 & {{\boldsymbol{H}}_2^{21}{\boldsymbol{C}}_2^1} & 0 & {{\boldsymbol{H}}_2^2{\boldsymbol{C}}_1^2} & {{\boldsymbol{A}}_2^2} & 0 \\ 0 & 0 & {{\boldsymbol{H}}_3^{21}{\boldsymbol{C}}_3^1} & {{\boldsymbol{H}}_3^2{\boldsymbol{C}}_1^2} & 0 & {{\boldsymbol{A}}_3^2} \end{array}} \right] } \end{array} $$ $$ \begin{array}{c}{ {{\boldsymbol{\Psi }} = \left[ {\begin{array}{*{20}{c}} {{\boldsymbol{B}}_1^1}&{}&{}&{} \\ {}&0&{}&{} \\ {}&{}& \ddots &{} \\ {}&{}&{}&0 \end{array}} \right]}} \end{array} $$ 情况1. 每层网络的节点的状态矩阵以及内耦合矩阵都相同, 即
$$ \begin{split} &{\boldsymbol{A}}_i^1 = \left[ {\begin{array}{*{20}{c}} 7&1 \\ 5&2 \end{array}} \right],\;\;{\boldsymbol{A}}_i^2 = \left[ {\begin{array}{*{20}{c}} 1&3 \\ 6&5 \end{array}} \right]\\ &{\boldsymbol{H}}_i^1 = {\boldsymbol{H}}_i^{21} = {\boldsymbol{H}}_i^2 = \left[ {\begin{array}{*{20}{c}} 4&5 \\ 6&7 \end{array}} \right],\;\;i = 1,2,3 \end{split}$$ 易验证, ${\rm{rank}}({\boldsymbol{\Psi }},{\boldsymbol{\Phi \Psi }},{{\boldsymbol{\Phi }}^2}{\boldsymbol{\Psi }},\cdots,{{\boldsymbol{\Phi }}^{11}}{\boldsymbol{\Psi }}) = 12$, 据引理1可知, 系统(2)是能控的.
情况2. 考虑每层网络的节点的内耦合矩阵都与情况1相同, 但节点的状态矩阵由每层内都是相同的(即情况1)改变为每层节点状态矩阵是不相同的(异质的), 取
$$ \begin{split} & {\boldsymbol{A}}_1^1 = \left[ {\begin{array}{*{20}{c}} 3&0 \\ 2&1 \end{array}} \right],\;\;{\boldsymbol{A}}_2^1 = \left[ {\begin{array}{*{20}{c}} 2&0 \\ 0&1 \end{array}} \right],\;\;{\boldsymbol{A}}_3^1 = \left[ {\begin{array}{*{20}{c}} 3&0 \\ 5&1 \end{array}} \right] \\ & {\boldsymbol{A}}_1^2 = \left[ {\begin{array}{*{20}{c}} 2&0 \\ 1&1 \end{array}} \right],\;\;{\boldsymbol{A}}_2^2 = \left[ {\begin{array}{*{20}{c}} 3&2 \\ 0&1 \end{array}} \right],\;\;{\boldsymbol{A}}_3^2 = \left[ {\begin{array}{*{20}{c}} 2&0 \\ 1&5 \end{array}} \right]\end{split} $$ 易验证, ${\rm{rank}}({\boldsymbol{\Psi }},{\boldsymbol{\Phi \Psi }},{{\boldsymbol{\Phi }}^2}{\boldsymbol{\Psi }},\cdots,{{\boldsymbol{\Phi }}^{11}}{\boldsymbol{\Psi }}) = 11 < 12$, 据引理1可知, 系统(2)是不能控的.
例1的情况1和情况2具有相同的网络结构, 当改变节点动态(节点动力学特性)时, 网络由能控变为不能控. 可见, 网络节点的异质性会改变网络的能控性(由能控变为不能控).
情况3. 考虑每层网络的节点的状态矩阵都与情况1相同, 但节点间的内耦合矩阵由每层内都是相同的(即情况1)改变为节点间的内耦合矩阵是不相同的, 取
$$ \begin{split} &{\boldsymbol{H}}_1^1 = \left[ {\begin{array}{*{20}{c}} 2&1 \\ 3&4 \end{array}} \right],\;{\boldsymbol{H}}_2^1 = \left[ {\begin{array}{*{20}{c}} 0&2 \\ 0&0 \end{array}} \right],\;{\boldsymbol{H}}_3^1 = \left[ {\begin{array}{*{20}{c}} 2&1 \\ 3&7 \end{array}} \right]\\ &{\boldsymbol{H}}_1^{21} = \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&3 \end{array}} \right],\;{\boldsymbol{H}}_2^{21} = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 0&3 \end{array}} \right],\;{\boldsymbol{H}}_3^{21} = \left[ {\begin{array}{*{20}{c}} 4&5 \\ 6&7 \end{array}} \right]\\ & {\boldsymbol{H}}_1^2 = \left[ {\begin{array}{*{20}{c}} 0&3 \\ 1&5 \end{array}} \right],\;{\boldsymbol{H}}_2^2 = \left[ {\begin{array}{*{20}{c}} 2&0 \\ 0&3 \end{array}} \right],\;{\boldsymbol{H}}_3^2 = \left[ {\begin{array}{*{20}{c}} 2&0 \\ 6&7 \end{array}} \right] \end{split}$$ 易验证, ${\rm{rank}}({\boldsymbol{\Psi }},{\boldsymbol{\Phi}} {\boldsymbol{\Psi }},{{\boldsymbol{\Phi}} ^2}{\boldsymbol{\Psi}} ,\cdots,{{\boldsymbol{\Phi}} ^{11}}{\boldsymbol{\Psi}} ) = 10{ { \; < \; }}12$, 据引理1可知, 系统(2)是不能控的.
例1中的情况1和情况3具有相同的网络结构, 当改变节点间的内耦合矩阵时, 网络由能控变为不能控. 可见, 内耦合矩阵的不同同样会改变网络的能控性(由能控变为不能控).
例 2. 考虑与例1的层间耦合模式相同的网络, 且$ {{\boldsymbol{W}}^K},{{\boldsymbol{D}}^{KJ}},{\boldsymbol{C}}_i^K,{\boldsymbol{B}}_i^K,\delta _i^K, $$ K,J = 1,2 \ $, $ i = $$1,2, 3$, 与例1中对应量相同.
情况1. 每层网络的节点的状态矩阵以及内耦合矩阵都相同, 即
$$ \begin{split} &{\boldsymbol{A}}_i^1 = \left[ {\begin{array}{*{20}{c}} 2&1 \\ 0&3 \end{array}} \right],\;\;{\boldsymbol{A}}_i^2 = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 0&1 \end{array}} \right] \\ &{\boldsymbol{H}}_i^1 = {\boldsymbol{H}}_i^{21} = {\boldsymbol{H}}_i^2 = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 0&5 \end{array}} \right],\;\;i = 1,2,3 \end{split}$$ 易验证, ${\rm{rank}}({\boldsymbol{\Psi }},{\boldsymbol{\Phi \Psi }},{{\boldsymbol{\Phi }}^2}{\boldsymbol{\Psi }},\cdots,{{\boldsymbol{\Phi }}^{11}}{\boldsymbol{\Psi }}) = 10{{ \; < \; }}12$, 据引理1可知, 系统(2)是不能控的.
情况2. 考虑每层网络的节点的内耦合矩阵都与情况1相同, 但节点的状态矩阵由每层内都是相同的(即情况1)改变为每层节点状态矩阵是不相同的(异质的), 取
$$ \begin{split} &{\boldsymbol{A}}_1^1 = \left[ {\begin{array}{*{20}{c}} 2&1 \\ 5&3 \end{array}} \right],\;{\boldsymbol{A}}_2^1 = \left[ {\begin{array}{*{20}{c}} 1&3 \\ 2&5 \end{array}} \right],\;{\boldsymbol{A}}_3^1 = \left[ {\begin{array}{*{20}{c}} 2&5 \\ 1&3 \end{array}} \right] \\ &{\boldsymbol{A}}_1^2 = \left[ {\begin{array}{*{20}{c}} 3&2 \\ 4&1 \end{array}} \right],\;{\boldsymbol{A}}_2^2 = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 3&5 \end{array}} \right],\;{\boldsymbol{A}}_3^2 = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 5&1 \end{array}} \right]\end{split} $$ 易验证, ${\rm{rank}}({\boldsymbol{\Psi }},{\boldsymbol{\Phi \Psi }},{{\boldsymbol{\Phi }}^2}{\boldsymbol{\Psi }},\cdots,{{\boldsymbol{\Phi }}^{11}}{\boldsymbol{\Psi }}) = 12$, 据引理1可知, 系统(2)是能控的.
例2的情况1和情况2具有相同的网络结构, 当改变节点动态(节点动力学特性)时, 网络由不能控变为能控. 可见, 网络节点的异质性会改变网络的能控性(由不能控变为能控).
情况3. 考虑每层网络的节点的状态矩阵都与情况1相同, 但节点间的内耦合矩阵由每层内都是相同的(即情况1)改变为节点间的内耦合矩阵是不相同的, 取
$$ \begin{split} &{\boldsymbol{H}}_1^1 = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 4&3 \end{array}} \right],\;{\boldsymbol{H}}_2^1 = \left[ {\begin{array}{*{20}{c}} 2&1 \\ 3&5 \end{array}} \right],\;{\boldsymbol{H}}_3^1 = \left[ {\begin{array}{*{20}{c}} 1&3 \\ 4&5 \end{array}} \right] \\ &{\boldsymbol{H}}_1^{21} = \left[ {\begin{array}{*{20}{c}} 2&7 \\ 0&5 \end{array}} \right],\;{\boldsymbol{H}}_2^{21} = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 3&5 \end{array}} \right],\;{\boldsymbol{H}}_3^{21} = \left[ {\begin{array}{*{20}{c}} 2&3 \\ 4&5 \end{array}} \right] \\ &{\boldsymbol{H}}_1^2 = \left[ {\begin{array}{*{20}{c}} 3&0 \\ 5&1 \end{array}} \right],\;{\boldsymbol{H}}_2^2 = \left[ {\begin{array}{*{20}{c}} 3&2 \\ 1&5 \end{array}} \right],\;{\boldsymbol{H}}_3^2 = \left[ {\begin{array}{*{20}{c}} 4&5 \\ 3&2 \end{array}} \right] \end{split} $$ 易验证, ${\rm{rank}}({\boldsymbol{\Psi }},{\boldsymbol{\Phi \Psi }},{{\boldsymbol{\Phi }}^2}{\boldsymbol{\Psi }},\cdots,{{\boldsymbol{\Phi }}^{11}}{\boldsymbol{\Psi }}) = 12$, 据引理1可知, 系统(2)是能控的.
例2中的情况1和情况3具有相同的网络结构, 当改变节点间的内耦合矩阵时, 网络由不能控变为能控. 可见, 内耦合矩阵的不相同会改变网络的能控性(由不能控变为能控).
上述两个例子揭示了节点的异质性以及内耦合矩阵的不相同对多层网络能控性的影响. 同时也说明多层异质网络与多层同质网络在能控性条件上存在差异, 一些对多层同质网络适用的能控性判据对多层异质网络并不适用, 所以需要对此展开更深入的研究.
3. 能控性条件
本节给出层间耦合模式分别为驱动响应模式和相互依赖模式时, 多层异质网络系统能控的充分或必要条件.
3.1 层间耦合为驱动响应模式
当层间耦合为有向支撑树且层数为M时, 如图3(b)所示, 则系统(2)中的${\boldsymbol{ \Phi}} $和${\boldsymbol{ \Psi }}$具有下述形式:
$$ \begin{split} &{\boldsymbol{\Phi }} = \left[ {\begin{array}{*{20}{c}} {{{\boldsymbol{\Phi }}_{11}}}&{}&{}&{} \\ {{{\boldsymbol{\Phi }}_{21}}}&{{{\boldsymbol{\Phi }}_{22}}}&{}&{} \\ \vdots &{}& \ddots &{} \\ {{{\boldsymbol{\Phi }}_{M1}}}&{}&{}&{{{\boldsymbol{\Phi }}_{MM}}} \end{array}} \right]\\ &{\boldsymbol{\Psi }} = {{\text{diag}}} \{ {{\boldsymbol{\Psi }}^1},{{\boldsymbol{\Psi }}^2},\cdots,{{\boldsymbol{\Psi }}^M}\} \end{split} $$ (3) 其中, $ {{\boldsymbol{\Phi }}_{KJ}} $和$ {{\boldsymbol{\Psi }}^K} $, $ K,J = 1,\cdots,M $按系统(2)的方式进行定义.
为了便于叙述主要的定理, 每层网络的受控节点集记为
$$ {\cal{U}}^{{K}}=\left\{i|{\delta }_{i}^{K}\ne 0,i=1,\cdots,N\right\} $$ 假设$ \sigma ({\boldsymbol{A}}_1^K) = \cdots = \sigma ({\boldsymbol{A}}_N^K) $, 对于任意给定的$ s \in \sigma ({\boldsymbol{A}}_i^K) $以及$ K \in \{ 1,\cdots,M\} $, 定义矩阵集合
$$ \begin{split} &{\cal{M}}^{K}(s)=\\ &\left\{[{({{\boldsymbol{\alpha}} }_{1}^{K})}^{\mathrm{T}},\cdots,{({{\boldsymbol{\alpha}} }_{N}^{K})}^{\mathrm{T}}]\Bigg|\begin{array}{c}{{\boldsymbol{\alpha}} }_{i}^{K}\in {\cal{M}}_{i}^{K1}(s),i\notin {{\cal{U}}}^{K}\\ {{\boldsymbol{\alpha}} }_{i}^{K}\in {\cal{M}}_{i}^{K2}(s),i\in {{\cal{U}}}^{K}\end{array} \right\}\end{split} $$ 其中,
$$ {\cal{M}}_{i}^{K1}(s)=\left\{{{\boldsymbol{\xi}} }_{i}^{K}\in {\bf{C}}^{1\times n}|{{\boldsymbol{\xi}} }_{i}^{K}(s{\boldsymbol{I}}-{{\boldsymbol{A}}}_{i}^{K})=0\right\}$$ 且
$$ \begin{split} {\cal{M}}_{i}^{K2}(s)=\left\{{{\boldsymbol{\xi}} }_{i}^{K}\in {\bf{C}}^{1\times n}|{{\boldsymbol{\xi}} }_{i}^{K}{{\boldsymbol{B}}}_{i}^{K}=0,{{\boldsymbol{\xi}} }_{i}^{K}\in {\cal{M}}_{i}^{K1}(s)\right\}\\ i = 1,\cdots,N \end{split} $$ $$ {\cal{M}}(s) = \left\{ [{{\boldsymbol{m}}}^{1},\cdots,{{\boldsymbol{m}}}^{M}]|{{\boldsymbol{m}}}^{K} \in {\cal{M}}^{K} (s),K = 1,\cdots,M \right\} $$ 定理 1. 假设$\sigma ({\boldsymbol{A}}_1^K) = \cdots = \sigma ({\boldsymbol{A}}_N^K)$, ${\boldsymbol{B}}_i^K \in {{\bf{R}}^{n \times 1}}$, $ {\boldsymbol{C}}_i^K \in {{\bf{R}}^{1 \times n}} $, $ {\boldsymbol{H}}_i^K = {\boldsymbol{H}}_i^{KJ} \in {{\bf{R}}^{n \times 1}} $, 且$ i \ne j $ 时, ${\boldsymbol{H}}_i^K = {\boldsymbol{H}}_i^{KJ} \ne {\boldsymbol{H}}_j^K = {\boldsymbol{H}}_j^{KJ}$, $ i = $$ 1,\cdots,N $; $ K,J = $ $1,2,\cdots, M$. 则网络系统(2)和(3)能控的充分条件如下:
1) $ ({\boldsymbol{A}}_i^K,{\boldsymbol{H}}_i^K) $, $ i = 1,\cdots,N $; $ K = 1,2,\cdots,M $, 是能控的;
2) $ ({\boldsymbol{A}}_i^K,{\boldsymbol{C}}_i^K) $, $ i = 1,\cdots,N $; $ K = 1,2,\cdots,M $, 是能观测的;
3) 对于任意的$ s \in \sigma ({\boldsymbol{A}}_i^K) $以及${\boldsymbol{\alpha }}_i^K \in {\cal{M}}_i^{K1}(s) \cup {\cal{M}}_i^{K2}(s)$, 若$ {\vartheta ^K}{{\boldsymbol{W}}^K} = 0 $, $ K = $$ 1,2,\cdots,M $, 则${\vartheta ^K} = 0$, 其中, $ {\vartheta ^K} = [{\boldsymbol{\alpha }}_1^K{\boldsymbol{H}}_1^K,\cdots, $ $ {\boldsymbol{\alpha }}_N^K{\boldsymbol{H}}_N^K] $;
4) 对于任意的$s \notin \sigma ({\boldsymbol{A}}_i^K)$, ${\rm{rank}}({\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^K}{{\boldsymbol{W}}^K}) =$ $N$; $K = 1,2,\cdots,M$, 其中${{\boldsymbol{\Sigma }}^K} =$$ {{\text{diag}}} \{ \gamma _1^K,\cdots,\gamma _N^K\} $ , $ \gamma _i^K = {\boldsymbol{C}}_i^K{(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K)^{ - 1}}{\boldsymbol{H}}_i^K. $
证明. 假设存在向量${\boldsymbol{\alpha }} = [{{\boldsymbol{\alpha }}^1},\cdots,{{\boldsymbol{\alpha }}^M}] = [ {\boldsymbol{\alpha }}_1^1,\cdots,$ ${\boldsymbol{\alpha }}_N^1,\cdots,{\boldsymbol{\alpha }}_1^M,\cdots,{\boldsymbol{\alpha }}_N^M]$ , 其中, ${\boldsymbol{\alpha }}_i^K \in {{\bf{R}}^{1 \times n}}$, $ i = $$1,\cdots, N$; $ K = 1,2,\cdots,M $, 使得
$$ \left\{ {\begin{aligned} &{{\boldsymbol{\alpha }}(s{\boldsymbol{I}} - {\boldsymbol{\Phi }}) = 0} \\ &{{\boldsymbol{\alpha \Psi }} = 0} \end{aligned}} \right. $$ 等价于
$$ \left\{\begin{split} &{{\boldsymbol{\alpha }}^1}(s{{\boldsymbol{I}}_{Nn}} - {{\boldsymbol{\Phi }}_{11}}) - {{\boldsymbol{\alpha }}^2}{{\boldsymbol{\Phi }}_{21}} - \cdots - {{\boldsymbol{\alpha }}^M}{{\boldsymbol{\Phi }}_{M1}} = 0\\ &{{\boldsymbol{\alpha }}^K}(s{{\boldsymbol{I}}_{Nn}} - {{\boldsymbol{\Phi }}_{KK}}) = 0;K = 2,\cdots,M \\ &{{\boldsymbol{\alpha }}^1}{{\boldsymbol{\Psi }}_1} = 0,\cdots,{{\boldsymbol{\alpha }}^M}{{\boldsymbol{\Psi }}_M} = 0\end{split} \right.$$ (4) 其中, $ {{\boldsymbol{\Phi }}_{KK}},{{\boldsymbol{\Phi }}_{K1}},{{\boldsymbol{\Psi }}_K},K = 1,\cdots,M $, 已在式(2)中定义. 式(4)又等价于
$$ \begin{split} &{\boldsymbol{\alpha }}_i^1(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^1) - \sum\limits_{j = 1}^N {w_{ji}^1{\boldsymbol{\alpha }}_j^1{\boldsymbol{H}}_j^1{\boldsymbol{C}}_i^1} \;- \\ &\qquad\sum\limits_{K = 2}^M {\sum\limits_{j = 1}^N {d_{ji}^{K1}{\boldsymbol{\alpha }}_j^K{\boldsymbol{H}}_j^{K1}{\boldsymbol{C}}_i^1} } = 0 \end{split} $$ (5) $$ \begin{split} &{\boldsymbol{\alpha }}_i^K(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K) - \sum\limits_{j = 1}^N {w_{ji}^K{\boldsymbol{\alpha }}_j^K{\boldsymbol{H}}_j^K{\boldsymbol{C}}_i^K} = 0 \\ &{{\boldsymbol{\alpha}} }_{i}^{K}{{\boldsymbol{B}}}_{i}^{K}=0,\forall i\in {\cal{U}}^{{K}},i=1,\cdots,N;K=2,\cdots,M \end{split}$$ (6) 如果$ s \in \sigma ({\boldsymbol{A}}_i^K) $, 则 ${\rm{rank}}(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K){{ \; < \; }}n$, $K = 1,\cdots, M$, 由式(5)和式(6)可知, 对于$ i = 1,\cdots, N $, 以下两式成立:
$$ \sum\limits_{j = 1}^N {w_{ji}^1{\boldsymbol{\alpha }}_j^1{\boldsymbol{H}}_j^1} + \sum\limits_{K = 2}^M {\sum\limits_{j = 1}^N {d_{ji}^{K1}{\boldsymbol{\alpha }}_j^K{\boldsymbol{H}}_j^{K1}} } = 0$$ (7) $$ \sum\limits_{j = 1}^N {w_{ji}^K{\boldsymbol{\alpha }}_j^K{\boldsymbol{H}}_j^K = 0} ,K = 2,\cdots,M $$ (8) 否则, $ {\boldsymbol{C}}_i^K $可以由$ s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K $的行向量组线性表示, 即
$$ {\rm{rank}}\left[ {\begin{array}{*{20}{c}} {{\boldsymbol{C}}_i^K} \\ {s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K} \end{array}} \right] = {\rm{rank}}(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K)<n $$ 这与$ ({\boldsymbol{A}}_i^K,{\boldsymbol{C}}_i^K) $, $ i = 1,\cdots,N $; $ K = 1,\cdots,M $是能观测的相矛盾.
于是, 根据式(5)和式(6), 对于$ i = 1,\cdots,N $, 下式成立:
$$ {\boldsymbol{\alpha }}_i^K(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K) = 0,K = 1,\cdots,M $$ (9) 由式(7), 式(8)以及条件3), 可得
$$ {\boldsymbol{\alpha }}_i^K{\boldsymbol{H}}_i^K = 0,i = 1,\cdots,N;K = 1,\cdots,M $$ (10) 从式(9), 式(10)以及条件1)可以推出
$$ {\boldsymbol{\alpha }}_i^K = 0,i = 1,\cdots,N;K = 1,\cdots,M$$ (11) 所以系统是能控的.
如果$ s \notin \sigma ({\boldsymbol{A}}_i^K) \ $ , 则$s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K$, $ i = 1,\cdots,N $; $K = 1,\cdots,M$是可逆的, 由式(5)和式(6)可以推出对于任意的$ i = 1,\cdots,N$, 有
$$ \begin{split} {\boldsymbol{\alpha }}_i^1 =\;&\sum\limits_{j = 1}^N {w_{ji}^1{\boldsymbol{\alpha }}_j^1{\boldsymbol{H}}_j^1{\boldsymbol{C}}_i^1{{(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^1)}^{ - 1}}} \;+ \\ &\sum\limits_{K = 2}^M {\sum\limits_{j = 1}^N {d_{ji}^{K1}{\boldsymbol{\alpha }}_j^K{\boldsymbol{H}}_j^{K1}{\boldsymbol{C}}_i^1{{(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^1)}^{ - 1}}} } \\ {\boldsymbol{\alpha }}_i^K =\;&\sum\limits_{j = 1}^N {w_{ji}^K{\boldsymbol{\alpha }}_j^K{\boldsymbol{H}}_j^K{\boldsymbol{C}}_i^K{{(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K)}^{ - 1}}} \\ &\qquad\qquad\qquad K = 2,\cdots,M \end{split} $$ 令
$$ \left\{\begin{split} & {\boldsymbol{\zeta }}_i^1 = \sum\limits_{j = 1}^N {w_{ji}^1{\boldsymbol{\alpha }}_j^1{\boldsymbol{H}}_j^1} + \sum\limits_{K = 2}^M {\sum\limits_{j = 1}^N {d_{ji}^{K1}{\boldsymbol{\alpha }}_j^K{\boldsymbol{H}}_j^{K1}} } \\ &{\boldsymbol{\zeta }}_i^K = \sum\limits_{j = 1}^N {w_{ji}^K{\boldsymbol{\alpha }}_j^K{\boldsymbol{H}}_j^K} ,K = 2,\cdots,M \end{split}\right. $$ (12) 则
$$ {\boldsymbol{\alpha }}_i^K = {\boldsymbol{\zeta }}_i^K{\boldsymbol{C}}_i^K{(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K)^{ - 1}},K = 1,\cdots,M $$ (13) 将式(13)代入式(12)中, 并由${\boldsymbol{H}}_i^{KJ} = {\boldsymbol{H}}_i^K \in {{\bf{R}}^{n \times 1}}$得到
$$ \begin{split} {\boldsymbol{\zeta }}_i^1 =\;& \sum\limits_{j = 1}^N {w_{ji}^1{\boldsymbol{\zeta }}_j^1{\boldsymbol{C}}_j^1{{(s{\boldsymbol{I}} - {\boldsymbol{A}}_j^1)}^{ - 1}}{\boldsymbol{H}}_j^1} \;+ \\ &\sum\limits_{K = 2}^M {\sum\limits_{j = 1}^N {d_{ji}^{K1}{\boldsymbol{\zeta }}_j^K{\boldsymbol{C}}_j^K{{(s{\boldsymbol{I}} - {\boldsymbol{A}}_j^K)}^{ - 1}}{\boldsymbol{H}}_j^K} } \\ {\boldsymbol{\zeta }}_i^K =\;&\sum\limits_{j = 1}^N {w_{ji}^K{\boldsymbol{\zeta }}_j^K{\boldsymbol{C}}_j^K{{(s{\boldsymbol{I}} - {\boldsymbol{A}}_j^K)}^{ - 1}}{\boldsymbol{H}}_j^K} \\ &\qquad i = 1,\cdots,N;K = 2,\cdots,M \end{split} $$ 令 $\gamma _i^K = {\boldsymbol{C}}_i^K{(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K)^{ - 1}}{\boldsymbol{H}}_i^K ,$ $i = 1,\cdots ,N ;$ $K = 1,$ $2,\cdots,M \;$, 则对于$i = 1,\cdots,N\;$, 有
$$ {\boldsymbol{\zeta }}_i^1 = \sum\limits_{j = 1}^N {w_{ji}^1{\boldsymbol{\zeta }}_j^1\gamma _j^1} + \sum\limits_{K = 2}^M {\sum\limits_{j = 1}^N {d_{ji}^{K1}{\boldsymbol{\zeta }}_j^K\gamma _j^K} } $$ (14) $$ {\boldsymbol{\zeta }}_i^K = \sum\limits_{j = 1}^N {w_{ji}^K{\boldsymbol{\zeta }}_j^K\gamma _j^K} ,K = 2,\cdots,M \quad $$ (15) 令${{\boldsymbol{\Sigma }}^K} = {{\text{diag}}} \{ \gamma _1^K,\cdots,\gamma _N^K\}$ , $ K = 1,\cdots,M $ ; ${\boldsymbol{\zeta }} =$ $[ {{\boldsymbol{\zeta }}^1},\cdots,{{\boldsymbol{\zeta }}^M}] \in {{\bf{R}}^{1 \times MN}}$, 其中$ {{\boldsymbol{\zeta }}^K} = [{\boldsymbol{\zeta }}_1^K,\cdots,{\boldsymbol{\zeta }}_N^K] $, 则将式(14)和式(15)分别写成矩阵形式, 即
$$ \begin{split} &{{\boldsymbol{\zeta }}^1}({\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^1}{{\boldsymbol{W}}^1}) - \sum\limits_{K = 2}^M {{{\boldsymbol{\zeta }}^K}{{\boldsymbol{\Sigma }}^K}{{\boldsymbol{D}}^{K1}}} = 0 \\ &{{\boldsymbol{\zeta }}^K}({\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^K}{{\boldsymbol{W}}^K}) = 0,K = 2,\cdots,M \end{split}$$ 于是,
$$ {\boldsymbol{\zeta }} \left[ {\begin{array}{*{20}{c}} {{\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^1}{{\boldsymbol{W}}^1}}& 0 & \cdots & 0 \\ { - {{\boldsymbol{\Sigma }}^2}{{\boldsymbol{D}}^{21}}}&{{\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^2}{{\boldsymbol{W}}^2}} & \cdots &0 \\ \vdots & \vdots & \ddots & \vdots \\ { - {{\boldsymbol{\Sigma }}^M}{{\boldsymbol{D}}^{M1}}}& 0 & \cdots &{{\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^M}{{\boldsymbol{W}}^M}} \end{array}} \right] = 0 $$ 由于${\rm{rank}}({\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^K}{{\boldsymbol{W}}^K}) = N$, $ K = 1,\cdots,M $, 则
$$\left[ {\begin{array}{*{20}{c}} {{\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^1}{{\boldsymbol{W}}^1}}& 0& \cdots& 0 \\ { - {{\boldsymbol{\Sigma }}^2}{{\boldsymbol{D}}^{21}}} & {\boldsymbol{I}}- {{\boldsymbol{\Sigma }}^2}{{\boldsymbol{W}}^2} & \cdots & 0 \\ \vdots & \vdots &\ddots& \vdots \\ { - {{\boldsymbol{\Sigma }}^M}{{\boldsymbol{D}}^{M1}}}& 0 &\cdots & {{\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^M}{{\boldsymbol{W}}^M}} \end{array}} \right] $$ 是可逆的, 所以$ {\boldsymbol{\zeta }} = [{{\boldsymbol{\zeta }}^1},\cdots,{{\boldsymbol{\zeta }}^M}] = 0 $. 由式(13)可知, $ {\boldsymbol{\alpha }} = [{\boldsymbol{\alpha }}_1^1,\cdots,{\boldsymbol{\alpha }}_N^M] = 0, $ 故系统是能控的.
□ 注 1. 定理1给出了内耦合矩阵不相同、节点异质的多层网络能控的充分条件, 在判断系统是否能控时, 只需要对各层的节点系统以及各层内的网络拓扑分别进行验证, 而无需考虑层间的网络拓扑, 并且参与运算的矩阵的规模远小于整个网络系统矩阵的规模, 所以在计算上是可行的. 例如, 条件3)和条件4)均仅需验证两个行数和列数为$ N $的矩阵的秩. 另外, 该定理对节点和内耦合矩阵均相同的多层网络同样成立.
注 2. 定理1考虑的层间耦合为有向支撑树的多层网络, 对于层间耦合为有向链的多层网络(如图3(a)), 定理1同样成立, 证明方法相同, 仅在${\boldsymbol{\Phi }}$的具体形式上存在差异.
下面给出一个具体例子说明定理1在实际中的应用.
例 3. 考虑层间耦合为驱动响应模式的两层网络, 每层有3个节点. 令$ \delta _1^1 = 1 $, $\delta _2^1 = \delta _3^1 = \delta _1^2 = \delta _2^2 = \delta _3^2 = 0$, 网络拓扑及节点的状态矩阵、节点之间的内耦合矩阵以及输出矩阵的具体取值如下:
$$ \begin{split} &{{\boldsymbol{W}}^1} = {{\boldsymbol{W}}^2} = \left[ {\begin{array}{*{20}{c}} 0&1&0 \\ 1&0&1 \\ 1&1&0 \end{array}} \right],\;{{\boldsymbol{D}}^{21}} = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right] \\ &{\boldsymbol{A}}_1^1{\text{ = }}{({\boldsymbol{A}}_1^2)^{\rm{T}}} = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 1&1 \end{array}} \right],\;{\boldsymbol{A}}_2^1 = {({\boldsymbol{A}}_2^2)^{\rm{T}}} = \left[ {\begin{array}{*{20}{c}} 3&{ - 2} \\ 1&{ - 1} \end{array}} \right] \\ &{\boldsymbol{A}}_3^1 = {({\boldsymbol{A}}_2^2)^{\rm{T}}} = \left[ {\begin{array}{*{20}{c}} 2&1 \\ 1&0 \end{array}} \right],\;{\boldsymbol{H}}_1^1 = \left[ {\begin{array}{*{20}{c}} 1 \\ 2 \end{array}} \right] \end{split}$$ $$ \begin{split} &{\boldsymbol{H}}_2^1 = {\boldsymbol{H}}_2^2 = \left[ {\begin{array}{*{20}{c}} 1 \\ 0 \end{array}} \right],\;{\boldsymbol{H}}_3^1 = {\boldsymbol{H}}_3^2 = \left[ {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right]\\ &{\boldsymbol{C}}_1^1 = {\boldsymbol{C}}_1^2 = \left[ {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right],\;{\boldsymbol{C}}_2^1 = {\boldsymbol{C}}_2^2 = \left[ {\begin{array}{*{20}{c}} 1 \\ 0 \end{array}} \right]\\ &{\boldsymbol{C}}_3^1 = {\boldsymbol{C}}_3^2 = \left[ {\begin{array}{*{20}{c}} 0 \\ 1 \end{array}} \right],\;{\boldsymbol{H}}_1^2 = \left[ {\begin{array}{*{20}{c}} 1 \\ 0 \end{array}} \right],\;{\boldsymbol{B}}_1^1 = \left[ {\begin{array}{*{20}{c}} 1&3 \\ 0&2 \end{array}} \right] \end{split}$$ 通过直接计算可知, ${\rm{rank}}({\boldsymbol{H}}_i^K,{\boldsymbol{A}}_i^K{\boldsymbol{H}}_i^K) = 2,$ ${\rm{ rank}}({({\boldsymbol{C}}_i^K)^{\rm{T}}},{({\boldsymbol{A}}_i^K)^{\rm{T}}}{({\boldsymbol{C}}_i^K)^{\rm{T}}}) = 2, i = 1,2,3;K = 1,2$, 于是定理1的条件1)和条件2)均满足, 另外, $ {{\boldsymbol{W}}^1}, {{\boldsymbol{W}}^2} $均可逆以及 ${\rm{rank}}({\boldsymbol{I}} -{{\boldsymbol{\Sigma }}^K}{{\boldsymbol{W}}^K}) = 3,$$K = 1,2\ $, 因此定理1的条件3)和条件4)同样满足, 所以上述参数描述的多层网络满足定理1的全部条件. 因此, 根据定理1, 可知上述系统是能控的.
定理2给出了节点异质但内耦合矩阵均相同的层间耦合为驱动响应模式的多层网络能控的必要条件.
定理 2. 假设$ \sigma \,({\boldsymbol{A}}_1^K) = \cdots = \sigma \,({\boldsymbol{A}}_N^K) ,$ $ {\boldsymbol{B}}_i^K \in $ $ {{\bf{R}}^{n \times 1}} $, $ {\boldsymbol{C}}_i^K = {\boldsymbol{C}} \in {{\bf{R}}^{1 \times n}} $, $ {\boldsymbol{H}}_i^K = {\boldsymbol{H}}_i^{KJ} = {\boldsymbol{H}} \in {{\bf{R}}^{n \times 1}} $, $ i = 1,\cdots,N $; $ K,J = 1,2,\cdots,M $, 则网络系统(2)和(3)能控的必要条件如下:
1) 对于任意的$ s \in \sigma ({\boldsymbol{A}}_i^K) $, $ {{m}} \in {\cal{M}}(s) $, 由$ {{m}} \ne 0 $可推出: $ {{m}}{\tilde{\boldsymbol{W}}} \ne 0 $, 其中, ${{m}} = [{{{m}}^1},\cdots,{{{m}}^M}]$,
$$ {\tilde{\boldsymbol{W}}} = \left[ {\begin{array}{*{20}{c}} {{{\boldsymbol{W}}^1}}&0& \cdots &0 \\ {{{\boldsymbol{D}}^{21}}}&{{{\boldsymbol{W}}^2}}&\cdots &0 \\ \vdots &\vdots& \ddots & \vdots \\ {{{\boldsymbol{D}}^{M1}}}&0& \cdots &{{{\boldsymbol{W}}^M}} \end{array}} \right] $$ 2) 对于任意的$ s \notin \sigma (A_i^K) $, ${\rm{rank}}({\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^1}{{\boldsymbol{W}}^1},$ ${{\boldsymbol{\Delta }}^1}{{\boldsymbol{\theta }}^1}) = N,$ ${\rm{rank}}({\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^2}{{\boldsymbol{W}}^2},{{\boldsymbol{\Sigma }}^2}{{\boldsymbol{D}}^{21}},{{\boldsymbol{\Delta }}^2}{{\boldsymbol{\theta }}^2}) = N , \cdots,$ ${\rm{rank}}({\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^M}{{\boldsymbol{W}}^M},{{\boldsymbol{\Sigma }}^M}{{\boldsymbol{D}}^{M1}},{{\boldsymbol{\Delta }}^M}{{\boldsymbol{\theta }}^M}) = N$中至少有一个成立, 其中, ${{\boldsymbol{\Sigma }}^K} = {{\text{diag}}} \{ \gamma _1^K,\cdots,$ $ \gamma _N^K\} $, ${{\boldsymbol{\theta }}^K} = {{\text{diag}}} \{ \eta _1^K,\cdots,\eta _N^K\} ,$ $\eta _i^K = {\boldsymbol{C}}(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K{)^{ - 1}}{\boldsymbol{B}} ,$ $K = 1,$$2,\cdots,M .$
证明. 假设条件1)不成立, 即存在$ {s_0} \in \sigma ({\boldsymbol{A}}_i^K) $, 非零矩阵$ {{m}} = [{({\boldsymbol{\alpha }}_1^1)^{\rm{T}}},\cdots,{({\boldsymbol{\alpha }}_N^1)^{\rm{T}}},\cdots, $ ${({\boldsymbol{\alpha }}_1^M)^{\rm{T}}},\cdots, {({\boldsymbol{\alpha }}_N^M)^{\rm{T}} }] \in {\cal{M}}({s_0})$, 使得${{m}}\tilde{\boldsymbol{ W}} = 0$. 即, 对于$i = 1,\cdots, N$, 下述等式成立:
$$ \begin{split} &\sum\limits_{j = 1}^N {w_{ji}^1{{({\boldsymbol{\alpha }}_j^1)}^{\rm{T}}}} + \sum\limits_{K = 2}^M {\sum\limits_{j = 1}^N {d_{ji}^{K1}{{({\boldsymbol{\alpha }}_j^K)}^{\rm{T}}}} } = 0 \\ &\sum\limits_{j = 1}^N {w_{ji}^K{{({\boldsymbol{\alpha }}_j^K)}^{\rm{T}}}} = 0,K = 2,\cdots,M \end{split}$$ 令${{\boldsymbol{\alpha }}^K} = [{\boldsymbol{\alpha }}_1^K,\cdots,{\boldsymbol{\alpha }}_N^K] ,$ ${\boldsymbol{\alpha }} = [{{\boldsymbol{\alpha }}^1},\cdots,{{\boldsymbol{\alpha }}^M}] ,$ ${{\tilde{\boldsymbol{\Lambda }}}^K} =$ $ {{\text{diag}}} \{ {s_0}{\boldsymbol{I}} - {\boldsymbol{A}}_1^K,\cdots,{s_0}{\boldsymbol{I}} - {\boldsymbol{A}}_N^K\} ,K = 1,\cdots,M $.
由${{m}} \in {\cal{M}}({s_0})$, 因此${{\boldsymbol{\alpha }}^K}{{\tilde{\boldsymbol{\Lambda }}}^K} = 0$, $ K = 1,\cdots, $ $ M $, 从而
$$ {\begin{split} &{\boldsymbol{\alpha }}({s_0}{\boldsymbol{I}} - {\boldsymbol{\Phi }}) = [{{\boldsymbol{\alpha }}^1},\cdots,{{\boldsymbol{\alpha }}^M}]\;\times\\ &\;\;\;\;\;\left[ {\begin{array}{*{20}{c}} { - {{\boldsymbol{W}}^1} \otimes {\boldsymbol{HC}}}&0& \cdots &0\\ { - {{\boldsymbol{D}}^{21}} \otimes {\boldsymbol{HC}}}&{ - {{\boldsymbol{W}}^2} \otimes {\boldsymbol{HC}}}&\cdots&0\\ \vdots &\vdots& \ddots & \vdots \\ { - {{\boldsymbol{D}}^{M1}} \otimes {\boldsymbol{HC}}}&0& \cdots &{ - {{\boldsymbol{W}}^2} \otimes {\boldsymbol{HC}}} \end{array}} \right] =\\ &\;\;\;\;\; - [{{\boldsymbol{\alpha }}^1}({{\boldsymbol{W}}^1} \otimes {\boldsymbol{HC}}) + \sum\limits_{K = 2}^M {{{\boldsymbol{\alpha }}^K}({{\boldsymbol{D}}^{K1}} \otimes {\boldsymbol{HC}})} \\ &{{\boldsymbol{\alpha }}^2}({{\boldsymbol{W}}^2} \otimes {\boldsymbol{HC}}),\cdots,{{\boldsymbol{\alpha }}^M}({{\boldsymbol{W}}^M} \otimes {\boldsymbol{HC}})]=\\ & - \left[ \begin{array}{*{20}{c}} &\left[\left[\sum\limits_{j = 1}^N {w_{j1}^1{\boldsymbol{\alpha }}_j^1} + \sum\limits_{K = 2}^M \sum\limits_{j = 1}^N d_{j1}^{K1}{\boldsymbol{\alpha }}_j^K\right] {\boldsymbol{HC}}\right]^{\rm{T}} \\ & \\ &\left[\left[\sum\limits_{j = 1,j \ne i}^N {w_{j1}^2{\boldsymbol{\alpha }}_j^2} \right] {\boldsymbol{HC}},\cdots,\left[\sum\limits_{j = 1,j \ne i}^N {w_{jN}^2{\boldsymbol{\alpha }}_j^2}\right] {\boldsymbol{HC}}\right]^{\rm{T}}\\ & \vdots \;\;\;\;\;\\ &\left[\left[\sum\limits_{j = 1,j \ne i}^N {w_{j1}^M{\boldsymbol{\alpha }}_j^M} \right] {\boldsymbol{HC}},\cdots,\left[\sum\limits_{j = 1,j \ne i}^N {w_{jN}^M{\boldsymbol{\alpha }}_j^M} \right] {\boldsymbol{HC}}\right]^{\rm{T}} \end{array} \right]^{\rm{T}} = \\ & 0 \end{split}} $$ 对于任意的$K = 1,\cdots,M$, 当$i \in {\cal{U}}$ 时, ${\boldsymbol{\alpha }}_i^K{\boldsymbol{B}}_i^K = 0$, 当$ i\notin {\cal{U}} $时, $ \delta _i^K = 0 $, 于是 $ \delta _i^K{\boldsymbol{\alpha }}_i^K{\boldsymbol{B}}_i^K = 0 $ 对所有的$ i = 1,\cdots,N $ 成立, 从而 ${\boldsymbol{\alpha \Psi }} = [\delta _1^1{\boldsymbol{\alpha }}_1^1{\boldsymbol{B}}_1^1,\cdots, \delta _N^1{\boldsymbol{\alpha }}_N^1{\boldsymbol{B}}_N^1,\cdots,\delta _N^M{\boldsymbol{\alpha }}_N^M{\boldsymbol{B}}_N^M] = 0$, 这与网络系统是能控的相矛盾.
假设条件2)不成立, 则存在$ {s_0} \notin \sigma ({\boldsymbol{A}}_i^K) $, 使得
$$ \begin{split} &{\rm{rank}}({\boldsymbol{I}} - {\boldsymbol{\Sigma }}_0^1{{\boldsymbol{W}}^1},{{\boldsymbol{\Delta }}^1}{\boldsymbol{\theta }}_0^1) < N \\ &{\rm{rank}}({\boldsymbol{I}} - {\boldsymbol{\Sigma }}_0^2{{\boldsymbol{W}}^2},{\boldsymbol{\Sigma }}_0^2{{\boldsymbol{D}}^{21}},{{\boldsymbol{\Delta }}^2}{\boldsymbol{\theta }}_0^2) < N \\ &\qquad\qquad\qquad{\text{ }} \vdots \\ &{\rm{rank}}({\boldsymbol{I}} - {\boldsymbol{\Sigma }}_0^M{{\boldsymbol{W}}^M},{\boldsymbol{\Sigma }}_0^M{{\boldsymbol{D}}^{M1}},{{\boldsymbol{\Delta }}^M}{\boldsymbol{\theta }}_0^M) < N \end{split} $$ 其中, ${\boldsymbol{\Sigma }}_0^K = {{\text{diag}}} \{ \gamma _{10}^K,\cdots,\gamma _{N0}^K\}$, $ {\boldsymbol{\theta }}_0^K = {{\text{diag}}} \{ \eta _{10}^K,\cdots $ , $\eta _{N0}^K\} , \gamma _{i0}^K = {\boldsymbol{C}}{({s_0}{\boldsymbol{I}} - {\boldsymbol{A}}_i^K)^{ - 1}}{\boldsymbol{H}} , \eta _{i0}^K = {\boldsymbol{C}}({s_0}{\boldsymbol{I}} - {\boldsymbol{A}}_i^K{)^{ - 1}}\times $ ${\boldsymbol{B}}_i^K , i = 1,\cdots,N ; K = 1,\cdots,M . $
因此, 存在非零向量 ${{\boldsymbol{\zeta }}^K} = [{\boldsymbol{\zeta }}_1^K,\cdots,{\boldsymbol{\zeta }}_N^K] \in{{\bf{R}}^{1 \times N}},$ $ K = $$ 1,2,\cdots,M $, 使得
$$ \begin{split} &{{\boldsymbol{\zeta }}^1}({\boldsymbol{I}} - {\boldsymbol{\Sigma }}_0^1{{\boldsymbol{W}}^1},{{\boldsymbol{\Delta }}^1}{\boldsymbol{\theta }}_0^1) = 0 \\ &{{\boldsymbol{\zeta }}^2}({\boldsymbol{I}} - {\boldsymbol{\Sigma }}_0^2{{\boldsymbol{W}}^2},{\boldsymbol{\Sigma }}_0^2{{\boldsymbol{D}}^{21}},{{\boldsymbol{\Delta }}^2}{\boldsymbol{\theta }}_0^2) = 0 \\ &\qquad\quad\qquad\qquad \vdots \\ &{{\boldsymbol{\zeta }}^M}({\boldsymbol{I}} - {\boldsymbol{\Sigma }}_0^M{{\boldsymbol{W}}^M},{\boldsymbol{\Sigma }}_0^M{{\boldsymbol{D}}^{M1}},{{\boldsymbol{\Delta }}^M}{\boldsymbol{\theta }}_0^M) = 0 \end{split} $$ 令$ {{\boldsymbol{\alpha }}^K} = [{\boldsymbol{\alpha }}_1^K,\cdots,{\boldsymbol{\alpha }}_N^K], $ $ K = 1,2,\cdots,M, $${\boldsymbol{\alpha }} = [{{\boldsymbol{\alpha }}^1},\cdots,$$ {{\boldsymbol{\alpha }}^M}] $, 其中, $ {\boldsymbol{\alpha }}_i^K = {\boldsymbol{\zeta }}_i^K{\boldsymbol{C}}{(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K)^{ - 1}} $, 由于$ {{\boldsymbol{\zeta }}^1},\cdots, $ $ {{\boldsymbol{\zeta }}^M} $均为非零向量且矩阵$ {\boldsymbol{C}} \ne 0 $, 于是$ {\boldsymbol{\alpha }}$为非零向量. 因此
$$ \begin{split} {\boldsymbol{\alpha \Psi }} =\;& [{\boldsymbol{\zeta }}_{\text{1}}^{\text{1}}{\boldsymbol{C}}{({s_0}{\boldsymbol{I}} - {\boldsymbol{A}}_{\text{1}}^{\text{1}})^{ - 1}},\cdots,{\boldsymbol{\zeta }}_N^M{\boldsymbol{C}}{({s_0}{\boldsymbol{I}} - {\boldsymbol{A}}_N^M)^{ - 1}}]\,\times\\ &{{\text{diag}}} \{ \delta _{\text{1}}^{\text{1}}{{\boldsymbol{B}}_1},\cdots,\delta _N^M{{\boldsymbol{B}}_N}\} =\\ & [{\boldsymbol{\zeta }}_{\text{1}}^{\text{1}},\cdots,{\boldsymbol{\zeta }}_N^M]{{\text{diag}}} \{ \delta _{\text{1}}^{\text{1}}{\boldsymbol{C}}{({s_0}{\boldsymbol{I}} - {\boldsymbol{A}}_{\text{1}}^{\text{1}})^{ - 1}}{{\boldsymbol{B}}_1},\cdots,\\ &\delta _N^M{\boldsymbol{C}}{({s_0}{\boldsymbol{I}} - {\boldsymbol{A}}_N^M)^{ - 1}}{{\boldsymbol{B}}_N}\} =\\ & [{\boldsymbol{\zeta }}_{\text{1}}^{\text{1}},\cdots,{\boldsymbol{\zeta }}_N^M]{{\text{diag}}} \{ \delta _{\text{1}}^{\text{1}}\eta _{10}^1,\cdots,\delta _N^M\eta _{N0}^M\} =\\ & [{{\boldsymbol{\zeta }}^1},\cdots,{{\boldsymbol{\zeta }}^M}]{{\text{diag}}} \{ {{\boldsymbol{\Delta }}^1}{\boldsymbol{\theta }}_0^1,\cdots,{{\boldsymbol{\Delta }}^M}{\boldsymbol{\theta }}_0^M\}=\\ & [{{\boldsymbol{\zeta }}^1}{{\boldsymbol{\Delta }}^1}{\boldsymbol{\theta }}_0^1,\cdots,{{\boldsymbol{\zeta }}^M}{{\boldsymbol{\Delta }}^M}{\boldsymbol{\theta }}_0^M] = 0 \end{split} $$ $$ \begin{split} & {\boldsymbol{\alpha }}({s_0}{\boldsymbol{I}} - {\boldsymbol{\Phi }}) = [{{\boldsymbol{\alpha }}^1},{{\boldsymbol{\alpha }}^2},\cdots,{{\boldsymbol{\alpha }}^M}] \;\times \\ &\left[ {\begin{array}{*{20}{c}} {{{{\tilde{\boldsymbol{\Lambda }}}}^1} - {{\boldsymbol{W}}^1} \otimes {\boldsymbol{HC}}}&0 \\ { - {{\boldsymbol{D}}^{21}} \otimes {\boldsymbol{HC}}}&{{{{\tilde{\boldsymbol{\Lambda }}}}^2} - {{\boldsymbol{W}}^2} \otimes {\boldsymbol{HC}}} \\ \vdots &\vdots \\ { - {{\boldsymbol{D}}^{M1}} \otimes {\boldsymbol{HC}}}&0 \end{array}} \right.\\ &\qquad\qquad\qquad\qquad\;\;\left. {\begin{array}{*{20}{c}} \cdots &0 \\ \cdots &0 \\ \ddots & \vdots \\ \cdots &{{{{\tilde{\boldsymbol{\Lambda }}}}^M} - {{\boldsymbol{W}}^M} \otimes {\boldsymbol{HC}}} \end{array}} \right] = \end{split} $$ $$ \begin{split} & {\left[ \begin{array}{l} ({{\boldsymbol{\zeta }}^1} \otimes {\boldsymbol{C}} - ({{\boldsymbol{\zeta }}^1}{{\text{diag}}} \{ \gamma _{10}^1,\cdots,\gamma _{N0}^1\} {{\boldsymbol{W}}^1}) \otimes {\boldsymbol{C}} \\ - (\sum\limits_{K = 2}^M {({{\boldsymbol{\zeta }}^K}{{\text{diag}}} \{ \gamma _{10}^K,\cdots,\gamma _{N0}^K\} {{\boldsymbol{D}}^{K1}})} ) \otimes {\boldsymbol{C}}{)^{\rm{T}}} \\ {({{\boldsymbol{\zeta }}^2} \otimes {\boldsymbol{C}} - ({{\boldsymbol{\zeta }}^2}{{\text{diag}}} \{ \gamma _{10}^2,\cdots,\gamma _{N0}^2\} {{\boldsymbol{W}}^2}) \otimes {\boldsymbol{C}})^{\rm{T}}} \\ {\text{ }} \qquad\qquad\qquad\qquad \vdots \\ {({{\boldsymbol{\zeta }}^M} \otimes {\boldsymbol{C}} - ({{\boldsymbol{\zeta }}^M}{{\text{diag}}} \{ \gamma _{10}^M,\cdots,\gamma _{N0}^M\} {{\boldsymbol{W}}^M}) \otimes {\boldsymbol{C}})^{\rm{T}} } \end{array} \right]^{\rm{T}}} =\\ & {\left[{\begin{array}{*{20}{c}} \begin{array}{c} (\left\{ {{{\boldsymbol{\zeta }}^1}({\boldsymbol{I}} - {\boldsymbol{\Sigma }}_0^1{{\boldsymbol{W}}^1})} \right\} \otimes {\boldsymbol{C}} -\\ (\sum\limits_{K = 2}^M {({{\boldsymbol{\zeta }}^K}{\boldsymbol{\Sigma }}_0^K{{\boldsymbol{D}}^{K1}})} ) \otimes {\boldsymbol{C}})^{\rm{T}} \\ {\text{ }}\left\{ {{{\boldsymbol{\zeta }}^2}({\boldsymbol{I}} - {\boldsymbol{\Sigma }}_0^2{{\boldsymbol{W}}^2})} \right\} \otimes {\boldsymbol{C}} \\ {\text{ }} \vdots \\ {\text{ }}\left\{ {{{\boldsymbol{\zeta }}^M}({\boldsymbol{I}} - {\boldsymbol{\Sigma }}_0^M{{\boldsymbol{W}}^M})} \right\} \otimes {\boldsymbol{C}} \end{array} \end{array}} \right]^{\rm{T}}} = 0 \end{split} $$ 这与网络系统是能控的相矛盾.
□ 注 3. 定理2对节点之间的内耦合矩阵相同的多层异质网络给出了能控的必要条件, 但对于内耦合矩阵不相同的网络, 给出的证明方法无法确保定理2中的条件同样成立, 需要考虑内耦合矩阵以及输出矩阵的具体形式.
注 4. 定理2考虑的是层间耦合为有向支撑树的多层网络. 对于层间耦合为有向链的多层网络, 系统能控的必要条件与定理2类似, 只需将描述网络拓扑的矩阵$ {\tilde{\boldsymbol{W}}} $中$ {{\boldsymbol{D}}^{KJ}} $变为有向链的矩阵, 证明方法与定理2的证明方法相同.
3.2 层间耦合为相互依赖模式
本节考虑层间耦合为相互依赖模式的多层网络系统能控的充分或必要条件. 由于层间耦合为相互依赖模式时层间的耦合形式非常复杂, 故在此仅讨论层数是2的网络. 此时系统(2)中的$ {\boldsymbol{\Phi }} $和$ {\boldsymbol{\Psi }} $具有下述形式:
$$ {\boldsymbol{\Phi }} = \left[ {\begin{array}{*{20}{c}} {{{\boldsymbol{\Phi }}_{11}}}&{{{\boldsymbol{\Phi }}_{12}}} \\ {{{\boldsymbol{\Phi }}_{21}}}&{{{\boldsymbol{\Phi }}_{22}}} \end{array}} \right],{\boldsymbol{\Psi }} = {{\text{diag}}} \{ {{\boldsymbol{\Psi }}^1},{{\boldsymbol{\Psi }}^2}\} $$ (16) 其中, $ {{\boldsymbol{\Phi }}_{KJ}} $和$ {{\boldsymbol{\Psi }}^K} $, $ K,J = 1,2 $, 按系统(2)的方式进行定义.
定理3给出了节点异质且内耦合矩阵不相同的层间耦合为相互依赖模式的多层网络能控的充分条件.
定理 3. 假设$ \sigma ({\boldsymbol{A}}_1^K) = \cdots = \sigma ({\boldsymbol{A}}_N^K) $, $ {\boldsymbol{B}}_i^K \in $ $ {{\bf{R}}^{n \times 1}} $, $ {\boldsymbol{C}}_i^K \in {{\bf{R}}^{1 \times n}} $, $ {\boldsymbol{H}}_i^K = {\boldsymbol{H}}_i^{KJ} \in {{\bf{R}}^{n \times 1}} $, 且$ i \ne j $时, $ {\boldsymbol{H}}_i^K = {\boldsymbol{H}}_i^{KJ} \ne {\boldsymbol{H}}_j^K = {\boldsymbol{H}}_j^{KJ} $, $ i = 1,\cdots,N $; $K,J = 1,2$. 则网络系统(2)和(16)能控的充分条件如下:
1) $ ({\boldsymbol{A}}_i^K,{\boldsymbol{H}}_i^K) $, $ i = 1,\cdots,N $; $ K = 1,2 $, 是能控的;
2) $ ({\boldsymbol{A}}_i^K,{\boldsymbol{C}}_i^K) $, $ i = 1,\cdots,N $; $ K = 1,2 $, 是能观测的;
3) 对于任意的$s \;\in\; \sigma\;(\;{\boldsymbol{A}}_i^K\;)$, ${\rm{rank}}\;({{\boldsymbol{W}}^1}) \;= {\rm{rank}}\{ {{\boldsymbol{W}}^2} \;-$${{\boldsymbol{D}}^{21}}{{({{\boldsymbol{W}}^1})}^{ - 1}}{{\boldsymbol{D}}^{12}} \} = N $或 ${\rm{rank}}({{\boldsymbol{W}}^2}) = {\rm{rank}}\{ {{\boldsymbol{W}}^1} - {{\boldsymbol{D}}^{12}}{{({{\boldsymbol{W}}^2})}^{ - 1}}{{\boldsymbol{D}}^{21}} \} = N;$
4) 对于任意的$s \notin \sigma ({\boldsymbol{A}}_i^K)$ , ${\rm{rank}}({\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^1} {{\boldsymbol{W}}^1}) =N$且${\rm{rank}}\{ ({\boldsymbol{I}} -{{\boldsymbol{\Sigma }}^2}{{\boldsymbol{W}}^2}) -{{\boldsymbol{\Sigma }}^2}{{\boldsymbol{D}}^{21}}({\boldsymbol{I}}- {{\boldsymbol{\Sigma }}^1} {{\boldsymbol{W}}^1}{)^{ - 1}}$${{\boldsymbol{\Sigma }}^1}\,\times {{\boldsymbol{D}}^{12}}\} = N,$ 其中, $ {{\boldsymbol{\Sigma }}^K} = {{\text{diag}}} \{ \gamma _1^K,\cdots,\gamma _N^K\} $ , $\gamma _i^K = {\boldsymbol{C}}_i^K{(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K)^{ - 1}}{\boldsymbol{H}}_i^K$.
证明. 假设存在向量$ {\boldsymbol{\alpha }} = [{{\boldsymbol{\alpha }}^1},{{\boldsymbol{\alpha }}^2}] = [{\boldsymbol{\alpha }}_1^1,\cdots, $ $ {\boldsymbol{\alpha }}_N^1,{\boldsymbol{\alpha }}_1^2,\cdots,{\boldsymbol{\alpha }}_N^2] $, 其中, $ {\boldsymbol{\alpha }}_i^K \in {{\bf{R}}^{1 \times n}} $, $ i = 1,\cdots, N; $ $ K = 1,2 $, 使得
$$ \left\{ {\begin{aligned} &{{\boldsymbol{\alpha }}(s{\boldsymbol{I}} - {\boldsymbol{\Phi }}) = 0} \\ & {{\boldsymbol{\alpha \Psi }} = 0} \end{aligned}} \right. $$ 等价于
$$ \left\{\begin{split} &{{\boldsymbol{\alpha }}^1}(s{{\boldsymbol{I}}_{Nn}} - {{\boldsymbol{\Phi }}_{11}}) - {{\boldsymbol{\alpha }}^2}{{\boldsymbol{\Phi }}_{21}} = 0 \\ & - {{\boldsymbol{\alpha }}^1}{{\boldsymbol{\Phi }}_{12}} + {{\boldsymbol{\alpha }}^2}(s{{\boldsymbol{I}}_{Nn}} - {{\boldsymbol{\Phi }}_{22}}) = 0 \\ & {{\boldsymbol{\alpha }}^1}{{\boldsymbol{\Psi }}_1} = 0,{{\boldsymbol{\alpha }}^2}{{\boldsymbol{\Psi }}_2} = 0 \end{split} \right.$$ (17) 其中, $ {{\boldsymbol{\Phi }}_{11}},{{\boldsymbol{\Phi }}_{21}},{{\boldsymbol{\Phi }}_{22}},{{\boldsymbol{\Psi }}_1} $和 $ {{\boldsymbol{\Psi }}_2} $已在式(2)中定义. 式(17)又等价于: 对于$ i = 1,\cdots,N $, 有
$$ \begin{split} &{\boldsymbol{\alpha }}_i^1(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^1) - \sum\limits_{j = 1}^N {w_{ji}^1{\boldsymbol{\alpha }}_j^1{\boldsymbol{H}}_j^1{\boldsymbol{C}}_i^1} \;- \\ &\qquad\sum\limits_{j = 1}^N {d_{ji}^{21}{\boldsymbol{\alpha }}_j^2{\boldsymbol{H}}_j^{21}{\boldsymbol{C}}_i^1} = 0 \end{split} $$ (18) $$ \begin{split} &{\boldsymbol{\alpha }}_i^2(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^2) - \sum\limits_{j = 1}^N {d_{ji}^{12}{\boldsymbol{\alpha }}_j^1{\boldsymbol{H}}_j^{12}{\boldsymbol{C}}_i^2} \;- \\ &\qquad\sum\limits_{j = 1}^N {w_{ji}^2{\boldsymbol{\alpha }}_j^2{\boldsymbol{H}}_j^2{\boldsymbol{C}}_i^2} = 0 \\ &{{\boldsymbol{\alpha}} }_{i}^{K}{{\boldsymbol{B}}}_{i}^{K}=0,\forall i\in {\cal{U}}^{{K}},K=1,2 \end{split} $$ (19) 如果$ s \in \sigma ({\boldsymbol{A}}_i^K) $, 类似于定理1的分析过程可以推出, 对于$ i = 1,\cdots,N $, 有以下两式成立:
$$ \sum\limits_{j = 1}^N {w_{ji}^1{\boldsymbol{\alpha }}_j^1{\boldsymbol{H}}_j^1} + \sum\limits_{j = 1}^N {d_{ji}^{21}{\boldsymbol{\alpha }}_j^2{\boldsymbol{H}}_j^{21}} = 0 $$ (20) $$ \sum\limits_{j = 1}^N {d_{ji}^{12}{\boldsymbol{\alpha }}_j^1{\boldsymbol{H}}_j^{12}} + \sum\limits_{j = 1}^N {w_{ji}^2{\boldsymbol{\alpha }}_j^2{\boldsymbol{H}}_j^2} = 0 $$ (21) 于是, 根据式(18)和式(19), 对于$i = 1,\cdots, N$, 下式成立:
$$ {\boldsymbol{\alpha }}_i^K(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K) = 0,K = 1,2 $$ 令${\tilde{\boldsymbol{W}}} = [{w_{ij}}] = \left[ {\begin{align} {{{\boldsymbol{W}}^1}}\;\;&\;\;{{{\boldsymbol{D}}^{12}}} \\ {{{\boldsymbol{D}}^{21}}}\;\;&\;\;{{{\boldsymbol{W}}^2}} \end{align}} \right] \in {{\bf{R}}^{2N \times 2N}}$, 由分块矩阵的行列式计算公式可知
$$ \left| {{\tilde{\boldsymbol{W}}}} \right| = \left\{ {\begin{array}{*{20}{c}} {\left| {{{\boldsymbol{W}}^1}} \right|\left| {{{\boldsymbol{W}}^2} - {{\boldsymbol{D}}^{21}}{{({{\boldsymbol{W}}^1})}^{ - 1}}{{\boldsymbol{D}}^{12}}} \right|, \; 若\;{{\boldsymbol{W}}^1}\;可逆}\\ {\left| {{{\boldsymbol{W}}^2}} \right|\left| {{{\boldsymbol{W}}^1} - {{\boldsymbol{D}}^{12}}{{({{\boldsymbol{W}}^2})}^{ - 1}}{{\boldsymbol{D}}^{21}}} \right|, \; 若\;{{\boldsymbol{W}}^{\rm{2}}}\;可逆} \end{array}} \right. $$ 由条件3)可知, $| {{\tilde{\boldsymbol{W}}}} | \ne 0$, 因此$ {\tilde{\boldsymbol{W}}} = [{w_{ij}}] \in $ $ {{\bf{R}}^{2N \times 2N}} $是可逆的. 再根据式(20)和式(21)以及$ {\boldsymbol{H}}_i^2 = {\boldsymbol{H}}_i^{21} $, $ {\boldsymbol{H}}_i^1 = {\boldsymbol{H}}_i^{12} $可以推出
$$ {\boldsymbol{\alpha }}_i^1{\boldsymbol{H}}_i^1 = 0,{\boldsymbol{\alpha }}_i^2{\boldsymbol{H}}_i^2 = 0,i = 1,\cdots,N $$ 由 $\left\{ {\begin{aligned} &{{\boldsymbol{\alpha }}_i^K(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K) = 0} \\ &{{\boldsymbol{\alpha }}_i^K{\boldsymbol{H}}_i^K = 0} \end{aligned}} \right.$, $ i = 1,\cdots,N $; $ K = 1,2 $, 以及$ ({\boldsymbol{A}}_i^K,{\boldsymbol{H}}_i^K) $, $ i = 1,\cdots,N $; $ K = 1,2 $ 是能控的, 于是有$ {\boldsymbol{\alpha }} = [{\boldsymbol{\alpha }}_1^1,\cdots,{\boldsymbol{\alpha }}_N^1,{\boldsymbol{\alpha }}_1^2,\cdots,{\boldsymbol{\alpha }}_N^2] = 0 $, 故系统是能控的.
如果$ s \notin \sigma ({\boldsymbol{A}}_i^K) $ , 则$ s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K $, $ i = 1,\cdots,N $ ; $K = 1, 2$ 是可逆的, 由式(18)和式(19)可以推出对于任意的$ i = 1,\cdots,N $, 有
$$ \begin{split} {\boldsymbol{\alpha }}_i^1 =\;&\sum\limits_{j = 1}^N {w_{ji}^1{\boldsymbol{\alpha }}_j^1{\boldsymbol{H}}_j^1{\boldsymbol{C}}_i^1{{(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^1)}^{ - 1}}} \;+ \\ &\sum\limits_{j = 1}^N {d_{ji}^{21}{\boldsymbol{\alpha }}_j^2{\boldsymbol{H}}_j^{21}{\boldsymbol{C}}_i^1{{(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^1)}^{ - 1}}} \\ {\boldsymbol{\alpha }}_i^2 = \;&\sum\limits_{j = 1}^N {d_{ji}^{12}{\boldsymbol{\alpha }}_j^1{\boldsymbol{H}}_j^{12}{\boldsymbol{C}}_i^2{{(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^2)}^{ - 1}}}\; + \\ &\sum\limits_{j = 1}^N {w_{ji}^2{\boldsymbol{\alpha }}_j^2{\boldsymbol{H}}_j^2{\boldsymbol{C}}_i^2{{(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^2)}^{ - 1}}} \end{split} $$ 令
$$ {\boldsymbol{\zeta }}_i^1 = \sum\limits_{j = 1}^N {w_{ji}^1{\boldsymbol{\alpha }}_j^1{\boldsymbol{H}}_j^1} + \sum\limits_{j = 1}^N {d_{ji}^{21}{\boldsymbol{\alpha }}_j^2{\boldsymbol{H}}_j^{21}} $$ (22) $$ {\boldsymbol{\zeta }}_i^2 = \sum\limits_{j = 1}^N {d_{ji}^{12}{\boldsymbol{\alpha }}_j^1{\boldsymbol{H}}_j^{12}} + \sum\limits_{j = 1}^N {w_{ji}^2{\boldsymbol{\alpha }}_j^2{\boldsymbol{H}}_j^2} $$ (23) 则
$$ \left\{\begin{split} &{\boldsymbol{\alpha }}_i^1 = {\boldsymbol{\zeta }}_i^1{\boldsymbol{C}}_i^1{(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^1)^{ - 1}} \\ &{\boldsymbol{\alpha }}_i^2 = {\boldsymbol{\zeta }}_i^2{\boldsymbol{C}}_i^2{(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^2)^{ - 1}} \end{split} \right.$$ (24) 将式(24)分别代入式(22)和式(23)中, 并由$ {\boldsymbol{H}}_i^{21} = $${\boldsymbol{ H}}_i^2 $, $ {\boldsymbol{H}}_i^{12} = {\boldsymbol{H}}_i^1 $可知, 对于$ i = 1,\cdots,N $, 有
$$ \begin{split} {\boldsymbol{\zeta }}_i^1 = \;&\sum\limits_{j = 1}^N {w_{ji}^1{\boldsymbol{\zeta }}_j^1{\boldsymbol{C}}_j^1{{(s{\boldsymbol{I}} - {\boldsymbol{A}}_j^1)}^{ - 1}}{\boldsymbol{H}}_j^1} \;+ \\ &\sum\limits_{j = 1}^N {d_{ji}^{21}{\boldsymbol{\zeta }}_j^2{\boldsymbol{C}}_j^2{{(s{\boldsymbol{I}} - {\boldsymbol{A}}_j^2)}^{ - 1}}{\boldsymbol{H}}_j^2} \\ {\boldsymbol{\zeta }}_i^2 = \;&\sum\limits_{j = 1}^N {d_{ji}^{12}{\boldsymbol{\zeta }}_j^1{\boldsymbol{C}}_j^1{{(s{\boldsymbol{I}} - {\boldsymbol{A}}_j^1)}^{ - 1}}{\boldsymbol{H}}_j^1} \;+ \\ &\sum\limits_{j = 1}^N {w_{ji}^2{\boldsymbol{\zeta }}_j^2{\boldsymbol{C}}_j^2{{(s{\boldsymbol{I}} - {\boldsymbol{A}}_j^2)}^{ - 1}}{\boldsymbol{H}}_j^2} \end{split} $$ 令 ${\boldsymbol{\gamma }}_i^K = {\boldsymbol{C}}_i^K{(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K)^{ - 1}}{\boldsymbol{H}}_i^K$, $i = 1,\cdots,N$; $K = 1,$ $ 2 $, 则对于$ i = 1,\cdots,N $, 有
$$ {\boldsymbol{\zeta }}_i^1 = \sum\limits_{j = 1}^N {w_{ji}^1{\boldsymbol{\zeta }}_j^1{\boldsymbol{\gamma }}_j^1} + \sum\limits_{j = 1}^N {d_{ji}^{21}{\boldsymbol{\zeta }}_j^2{\boldsymbol{\gamma }}_j^2} $$ (25) $$ {\boldsymbol{\zeta }}_i^2 = \sum\limits_{j = 1}^N {d_{ji}^{12}{\boldsymbol{\zeta }}_j^1{\boldsymbol{\gamma }}_j^1} + \sum\limits_{j = 1}^N {w_{ji}^2{\boldsymbol{\zeta }}_j^2{\boldsymbol{\gamma }}_j^2} $$ (26) 令$ {{\boldsymbol{\Sigma }}^K} = {{\text{diag}}} \{ \gamma _1^K,\cdots,\gamma _N^K\} $ , $ K = 1,2 $ , ${\boldsymbol{\zeta }} = [{{\boldsymbol{\zeta }}^1}, {{\boldsymbol{\zeta }}^2}] \in$$ {{\bf{R}}^{1 \times 2N}} $, 其中, ${{\boldsymbol{\zeta }}^1}= [{\boldsymbol{\zeta }}_1^1,\cdots,{\boldsymbol{\zeta }}_N^1]$, ${{\boldsymbol{\zeta }}^2} = [{\boldsymbol{\zeta }}_1^2,\cdots, {\boldsymbol{\zeta }}_N^2]$. 将式(25)和式(26)分别写成矩阵形式, 即
$$ \begin{split} &{{\boldsymbol{\zeta }}^1}({\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^1}{{\boldsymbol{W}}^1}) - {{\boldsymbol{\zeta }}^2}{{\boldsymbol{\Sigma }}^2}{{\boldsymbol{D}}^{21}} = 0 \\ & - {{\boldsymbol{\zeta }}^1}{{\boldsymbol{\Sigma }}^1}{{\boldsymbol{D}}^{12}} + {{\boldsymbol{\zeta }}^2}({\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^2}{{\boldsymbol{W}}^2}) = 0 \end{split} $$ 于是,
$$ \left[ {{{\boldsymbol{\zeta }}^1},{{\boldsymbol{\zeta }}^2}} \right]\left[ {\begin{array}{*{20}{c}} {{\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^1}{{\boldsymbol{W}}^1}}&{ - {{\boldsymbol{\Sigma }}^1}{{\boldsymbol{D}}^{12}}} \\ { - {{\boldsymbol{\Sigma }}^2}{{\boldsymbol{D}}^{21}}}&{{\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^2}{{\boldsymbol{W}}^2}} \end{array}} \right] = 0 $$ 根据分块矩阵的行列式计算公式, 则
$$\begin{split} &\left| {\begin{array}{*{20}{c}} {{\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^1}{{\boldsymbol{W}}^1}}&{ - {{\boldsymbol{\Sigma }}^1}{{\boldsymbol{D}}^{12}}} \\ { - {{\boldsymbol{\Sigma }}^2}{{\boldsymbol{D}}^{21}}}&{{\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^2}{{\boldsymbol{W}}^2}} \end{array}} \right| = \left| {{\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^1}{{\boldsymbol{W}}^1}} \right| \times \\ & \quad \left| {({\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^2}{{\boldsymbol{W}}^2}) - {{\boldsymbol{\Sigma }}^2}{{\boldsymbol{D}}^{21}}{{({\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^1}{{\boldsymbol{W}}^1})}^{ - 1}}{{\boldsymbol{\Sigma }}^1}{{\boldsymbol{D}}^{12}}} \right| \end{split} $$ 由条件4)可知
$$ \left| {\begin{array}{*{20}{c}} {{\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^1}{{\boldsymbol{W}}^1}}&{ - {{\boldsymbol{\Sigma }}^1}{{\boldsymbol{D}}^{12}}} \\ { - {{\boldsymbol{\Sigma }}^2}{{\boldsymbol{D}}^{21}}}&{{\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^2}{{\boldsymbol{W}}^2}} \end{array}} \right| \ne 0 $$ 因此, $\left[ {\begin{aligned}{{\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^1}{{\boldsymbol{W}}^1}}\;\;\;\;{ - {{\boldsymbol{\Sigma }}^1}{{\boldsymbol{D}}^{12}}} \\\;\; { - {{\boldsymbol{\Sigma }}^2}{{\boldsymbol{D}}^{21}}}\;\;{{\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^2}{{\boldsymbol{W}}^2}} \end{aligned}} \right]$是可逆的, 所以${\boldsymbol{\zeta }} = $ $ [{{\boldsymbol{\zeta }}^1},{{\boldsymbol{\zeta }}^2}] = 0 $. 由式(24)可知, $ {\boldsymbol{\alpha }} = [{\boldsymbol{\alpha }}_1^1,\cdots,{\boldsymbol{\alpha }}_N^1, $ $ {\boldsymbol{\alpha }}_1^2,\cdots,{\boldsymbol{\alpha }}_N^2] = 0 $, 故系统是能控的.
□ 注 5. 与定理1类似, 定理3的条件验证起来是容易的, 尤其是当网络的规模较大时, 验证每个条件需要计算的矩阵的规模远小于整个网络系统的规模. 与定理1所讨论的层间耦合为驱动响应模式不同的是, 这里的条件3)需要综合考虑层内的网络拓扑以及层间的网络拓扑, 即定理1的条件3)无需考虑层间网络拓扑$ {{\boldsymbol{D}}^{21}} $, 而定理3的条件3)既需要考虑层间网络拓扑$ {{\boldsymbol{W}}^1} $, $ {{\boldsymbol{W}}^2} $, 又需要考虑层内网络拓扑${{\boldsymbol{D}}^{12}} ,$ $ {{\boldsymbol{D}}^{21}} $. 这表明层间耦合为相互依赖模式相比于驱动响应模式使系统完全能控所需要的条件更强.
定理4给出了节点异质且内耦合矩阵相同的层间耦合为相互依赖模式的多层网络能控的必要条件.
定理 4. 假设$ \sigma ({\boldsymbol{A}}_1^K) = \cdots = \sigma ({\boldsymbol{A}}_N^K) $, $ {\boldsymbol{H}}_i^K = $ $ {\boldsymbol{H}}_i^{KJ} = {\boldsymbol{H}} \in {{\bf{R}}^{n \times 1}} $, $ {\boldsymbol{C}}_i^K = {\boldsymbol{C}} \in {{\bf{R}}^{1 \times n}} $, $ {\boldsymbol{B}}_i^K \in {{\bf{R}}^{n \times 1}} $, $ i = 1,\cdots,N $; $ K,J = 1,2 $. 则网络系统(2)和(16)能控的必要条件如下:
1) 对于任意的$ s \in \sigma ({\boldsymbol{A}}_i^K) $, $ {{m}} \in {\cal{M}}(s) $, 由$ {{m}} \ne 0 $ 可推出 ${{m}}\tilde W\; \ne \;0,$ 其中, ${{m}} = [{{{m}}^1},\;\;{{{m}}^2}]$, ${\tilde{\boldsymbol{W}}}= \left[ {\begin{aligned} {{{\boldsymbol{W}}^1}}\;\;{{{\boldsymbol{D}}^{12}}} \\ {{{\boldsymbol{D}}^{21}}}\;\;{{{\boldsymbol{W}}^2}} \end{aligned}} \right].$
2) 对于任意的$ s \notin \sigma ({\boldsymbol{A}}_i^K) $, ${\rm{rank}}({\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^1} {{\boldsymbol{W}}^1}, {{\boldsymbol{\Sigma }}^1}{{\boldsymbol{D}}^{12}},{{\boldsymbol{\Delta }}^1}{{\boldsymbol{\theta }}^1}) = N$或${\rm{rank}}({\boldsymbol{I}} - {{\boldsymbol{\Sigma }}^2}{{\boldsymbol{W}}^2},{{\boldsymbol{\Sigma }}^2} {{\boldsymbol{D}}^{21}}, {{\boldsymbol{\Delta }}^2}\,\times {{\boldsymbol{\theta }}^2}) = N,$ 其中, ${{\boldsymbol{\Sigma }}^K} = {{\text{diag}}} \,\{ \gamma _1^K,\,\gamma _2^K,\,\cdots,\gamma _N^K\}$, ${{\boldsymbol{\theta }}^K}\, = \,{{\text{diag}}}\, \{ \eta _1^K,\,\cdots,\eta _N^K\},$ ${\boldsymbol{\eta }}_i^K = {\boldsymbol{C}}{(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K)^{ - 1}}{\boldsymbol{B}} ,$ ${\boldsymbol{\gamma }}_i^K = {\boldsymbol{C}}(s{\boldsymbol{I}} - {\boldsymbol{A}}_i^K)^{ - 1}{\boldsymbol{H}};$ $ K = 1,2. $
证明. 假设条件1)不成立, 即存在${s_0} \in \sigma ({\boldsymbol{A}}_i^K)$以及非零矩阵$ {{m}} = [{({\boldsymbol{\alpha }}_1^1)^{\rm{T}}},\cdots,{({\boldsymbol{\alpha }}_N^1)^{\rm{T}}}, $ ${({\boldsymbol{\alpha }}_1^2)^{\rm{T}}},\cdots, {({\boldsymbol{\alpha }}_N^2)^{\rm{T}}}] \in$$ {\cal{M}}({s_0}) $, 使得${{m}}\tilde {\boldsymbol{W}} = 0$. 即, 对于$i = 1,\cdots, N$, 以下两式成立:
$$ \begin{split} &\sum\limits_{j = 1}^N {w_{ji}^1{{({\boldsymbol{\alpha }}_j^1)}^{\rm{T}} }} + \sum\limits_{j = 1}^N {d_{ji}^{21}{{({\boldsymbol{\alpha }}_j^2)}^{\rm{T}} }} = 0 \\ & \sum\limits_{j = 1}^N {d_{ji}^{12}{{({\boldsymbol{\alpha }}_j^1)}^{\rm{T}}}} + \sum\limits_{j = 1}^N {w_{ji}^2{{({\boldsymbol{\alpha }}_j^2)}^{\rm{T}} }} = 0 \end{split} $$ 令${\boldsymbol{\alpha }} = [{{\boldsymbol{\alpha }}^1},{{\boldsymbol{\alpha }}^2}]$, 其中, ${{\boldsymbol{\alpha }}^K}=[{\boldsymbol{\alpha }}_1^K,\cdots,{\boldsymbol{\alpha }}_N^K]$, $K = 1, 2 $. 由 $ {{m}} \in {\cal{M}}({s_0}) $, 则 $ {{\boldsymbol{\alpha }}^K}{{\tilde{\boldsymbol{\Lambda }}}^K} = 0 $, 从而, 有
$$ \begin{split} &{\boldsymbol{\alpha }}({s_0}{\boldsymbol{I}} - {\boldsymbol{\Phi }}) =\\ &\;\;\;\; [{{\boldsymbol{\alpha }}^1},{{\boldsymbol{\alpha }}^2}]\left[ {\begin{array}{*{20}{c}} { - {{\boldsymbol{W}}^1} \otimes {\boldsymbol{HC}}}&\qquad { - {{\boldsymbol{D}}^{12}} \otimes {\boldsymbol{HC}}} \\ { - {{\boldsymbol{D}}^{21}} \otimes {\boldsymbol{HC}}}&\,\qquad{ - {{\boldsymbol{W}}^2} \otimes {\boldsymbol{HC}}} \end{array}} \right] = \\ & \;\;\;\; - [{{\boldsymbol{\alpha }}^1}({{\boldsymbol{W}}^1} \otimes {\boldsymbol{HC}}) + {{\boldsymbol{\alpha }}^2}({{\boldsymbol{D}}^{21}} \otimes {\boldsymbol{HC}})\\ &{{\boldsymbol{\alpha }}^1}({{\boldsymbol{D}}^{12}} \otimes {\boldsymbol{HC}}) + {{\boldsymbol{\alpha }}^2}({{\boldsymbol{W}}^2} \otimes {\boldsymbol{HC}})] = \\ &\;\; - \left[ {\begin{array}{*{20}{c}} \left(\displaystyle\sum\limits_{j = 1}^N {w_{j1}^1{\boldsymbol{\alpha }}_j^1} + \sum\limits_{j = 1}^N d_{j1}^{21}{\boldsymbol{\alpha }}_j^2\right)^{\rm{T}} {{({\boldsymbol{HC}})}^{\rm{T}} } \\ \vdots \\ {{{\left(\displaystyle\sum\limits_{j = 1}^N {d_{jN}^{12}{\boldsymbol{\alpha }}_j^1} + \sum\limits_{j = 1,j \ne N}^N {w_{jN}^2{\boldsymbol{\alpha }}_j^2} \right)}^{\rm{T}} } {{({\boldsymbol{HC}})}^{\rm{T}} }} \end{array}} \right]^{\rm{T}} = 0 \end{split} $$ 当$ i\in {\cal{U}} $时, $ {\boldsymbol{\alpha }}_i^K{\boldsymbol{B}}_i^K = 0,K = 1,2 $, 当$ i\notin {\cal{U}} $时, $\delta _i^K = 0$, $ K = 1,2 $ , 于是 $ \delta _i^K{\boldsymbol{\alpha }}_i^K{\boldsymbol{B}}_i^K = 0 $ 对所有的$K = 1,2$, $ i = 1,\cdots,N $ 成立, 从而${\boldsymbol{\alpha \Psi}} = [\delta _1^1{\boldsymbol{\alpha }}_1^1{\boldsymbol{B}}_1^1,\cdots, $$\delta _N^1{\boldsymbol{\alpha }}_N^1{\boldsymbol{B}}_N^1, \delta _2^2{\boldsymbol{\alpha }}_2^2{\boldsymbol{B}}_2^2,\cdots,\delta _N^2{\boldsymbol{\alpha }}_N^2{\boldsymbol{B}}_N^2] = 0$, 这与网络系统是能控的相矛盾.
假设条件2)不成立, 则存在${s_0} \notin \sigma ({\boldsymbol{A}}_i^K)$, 使得
$$ \begin{split} &{\rm{rank}}({\boldsymbol{I}} - {\boldsymbol{\Sigma }}_0^1{{\boldsymbol{W}}^1},{\boldsymbol{\Sigma }}_0^1{{\boldsymbol{D}}^{12}},{{\boldsymbol{\Delta }}^1}{\boldsymbol{\theta }}_0^1) < N \\ &{\rm{rank}}({\boldsymbol{I}} - {\boldsymbol{\Sigma }}_0^2{{\boldsymbol{W}}^2},{\boldsymbol{\Sigma }}_0^2{{\boldsymbol{D}}^{21}},{{\boldsymbol{\Delta }}^2}{\boldsymbol{\theta }}_0^2) < N \end{split} $$ 其中, $ {\boldsymbol{\Sigma }}_0^K = {{\text{diag}}} \{ \gamma _{10}^K,\cdots,\gamma _{N0}^K\} $, $ {\boldsymbol{\theta }}_0^K = {{\text{diag}}} \{ \eta _{10}^K,\cdots, $ $ \eta _{N0}^K\} ,$ $\gamma _{i0}^K = {\boldsymbol{C}}{({s_0}{\boldsymbol{I}} - {\boldsymbol{A}}_i^K)^{ - 1}}{\boldsymbol{H}}$, $\eta _{i0}^K = {\boldsymbol{C}}({s_0}{\boldsymbol{I}} -$${\boldsymbol{A}}_i^K{)^{ - 1}} \times {\boldsymbol{B}}_i^K$, $i = 1,\cdots,N ;$ $K = 1,2 .$
因此, 存在非零向量 ${{\boldsymbol{\zeta }}^1} = [{\boldsymbol{\zeta }}_1^1,\cdots,{\boldsymbol{\zeta }}_N^1] \in {{\bf{R}}^{1 \times N}},$ ${{\boldsymbol{\zeta }}^2} = [{\boldsymbol{\zeta }}_1^2,\cdots,{\boldsymbol{\zeta }}_N^2] \in {{\bf{R}}^{1 \times N}},$ 使得
$$ \begin{split} &{{\boldsymbol{\zeta }}^1}({\boldsymbol{I}} - {\boldsymbol{\Sigma }}_0^1{{\boldsymbol{W}}^1},{\boldsymbol{\Sigma }}_0^1{{\boldsymbol{D}}^{12}},{{\boldsymbol{\Delta }}^1}{\boldsymbol{\theta }}_0^1) = 0 \\ &{{\boldsymbol{\zeta }}^2}({\boldsymbol{I}} - {\boldsymbol{\Sigma }}_0^2{{\boldsymbol{W}}^2},{\boldsymbol{\Sigma }}_0^2{{\boldsymbol{D}}^{21}},{{\boldsymbol{\Delta }}^2}{\boldsymbol{\theta }}_0^2) = 0 \end{split} $$ 令 ${{\boldsymbol{\alpha }}^1} = [{\boldsymbol{\alpha }}_1^1,\cdots,{\boldsymbol{\alpha }}_N^1]$ , ${{\boldsymbol{\alpha }}^2} = [{\boldsymbol{\alpha }}_1^2,\cdots,{\boldsymbol{\alpha }}_N^2]$ , $ {\boldsymbol{\alpha }} = [{{\boldsymbol{\alpha }}^1}, $ $ {{\boldsymbol{\alpha }}^2}] $, 其中, $ {\boldsymbol{\alpha }}_i^K = {\boldsymbol{\zeta }}_i^K{\boldsymbol{C}}{({s_0}{\boldsymbol{I}} - {\boldsymbol{A}}_i^K)^{ - 1}} $, 由$ {{\boldsymbol{\zeta }}^1} $和$ {{\boldsymbol{\zeta }}^2} $均为非零向量, 可得${\boldsymbol{\alpha}}$为非零向量. 因此
$$ \begin{split} & {\boldsymbol{\alpha }}({s_0}{\boldsymbol{I}} - {\boldsymbol{\Phi }}) =\\ &[{{\boldsymbol{\alpha }}^1},{{\boldsymbol{\alpha }}^2}] \left[ {\begin{array}{*{20}{c}} {{{{\tilde{\boldsymbol{\Lambda }}}}^1} - {{\boldsymbol{W}}^1} \otimes {\boldsymbol{HC}}}\;\;\quad{ - {{\boldsymbol{D}}^{12}} \otimes {\boldsymbol{HC}}} \\ \;\;\;{ - {{\boldsymbol{D}}^{21}} \otimes {\boldsymbol{HC}}}\qquad{{{{\tilde{\boldsymbol{\Lambda }}}}^2} - {{\boldsymbol{W}}^2} \otimes {\boldsymbol{HC}}} \end{array}} \right ] =\\ &{\left[ {\begin{array}{*{20}{c}} \begin{array}{l} {(\{ {{\boldsymbol{\zeta }}^1}({\boldsymbol{I}} - {\boldsymbol{\Sigma }}_0^1{{\boldsymbol{W}}^1}) - {{\boldsymbol{\zeta }}^2}{\boldsymbol{\Sigma }}_0^2{{\boldsymbol{D}}^{21}}\} \otimes {\boldsymbol{C}})^{\rm{T}}} \\ {(\{ - {{\boldsymbol{\zeta }}^1}{\boldsymbol{\Sigma }}_0^1{{\boldsymbol{D}}^{12}} + {{\boldsymbol{\zeta }}^2}({\boldsymbol{I}} - {\boldsymbol{\Sigma }}_0^2{{\boldsymbol{W}}^2})\} \otimes {\boldsymbol{C}})^{\rm{T}} } \end{array} \end{array}} \right]^{\rm{T}}} = 0\end{split} $$ 类似于定理2的证明过程, 容易验证
$$ {\boldsymbol{\alpha \Psi }} = 0$$ 这与网络系统是能控的相矛盾.
□ 注 6. 定理4中所提供的方法同样仅对内耦合矩阵相同的多层异质网络适用. 相比于层间耦合为驱动响应模式, 层间耦合为相互依赖模式的多层网络能控的必要条件相对复杂. 例如定理2的条件1)中只需对分块下三角矩阵进行验证, 计算量相对较小, 而定理4的条件1)中的矩阵并非分块下三角矩阵, 计算量相对较大. 但相对于整个网络系统的规模, 需要计算的矩阵的规模仍较小. 另外, 若将整个网络系统看成具有$ MN $个节点的单层大规模网络, 则无法区分不同的层间耦合模式带来的能控性的条件上的差异.
注 7. 本文所得结论还存在一定的局限性. 定理1和定理2对层数大于2的网络成立, 但定理3和定理4仅对层数为2的网络成立, 对于层数大于2的网络, 层间耦合为相互依赖模式的网络需要考虑层间的连接方式以及与整个系统的网络拓扑对应的分块矩阵的逆是否存在的问题, 因此, 不易推广. 另外, 本文定理中节点状态矩阵不属于完全异质的, 而是加上了一个限制条件, 即, 节点的状态矩阵都有相同的特征值集合; 此外, 在定理1和定理3中假设部分节点之间的内耦合矩阵是相同的.
4. 结束语
本文考虑了节点异质且内耦合矩阵不相同的多层网络系统的能控性问题, 揭示了节点的异质性以及内耦合矩阵对网络系统能控性的影响: 网络节点由同质变为异质时, 网络既可由能控变为不能控, 又可由不能控变为能控; 同样, 当网络内耦合矩阵由相同变为不同时, 网络也是既可由能控变为不能控, 又可由不能控变为能控. 本文给出了层间耦合为驱动响应模式和相互依赖模式时, 多层网络系统能控的充分和必要条件, 这些条件对于多层异质网络更容易验证, 且可以清晰地表明多层网络的各组成部分如何具体地影响整个网络系统能控性. 同时, 本文还发现层间耦合为驱动响应模式实现系统完全能控所需的条件弱于相互依赖模式, 即, 在某些情况下无需考虑层间的网络拓扑结构, 更容易判断系统的能控性. 本文的研究结果可为现实的一般网络控制问题提供参考.
本文对节点异质的多层网络能控性问题仅为初步探讨, 定理中节点状态矩阵不属于完全异质并存在一定的限制条件, 对于相互依赖模式, 仅考虑了2层的网络结构. 而对于更一般的情况, 还需进一步的研究.
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图 1 层间耦合为驱动响应模式的两层网络
Fig. 1 Two-layer networks with the inter-layer drive-response couplings
图 2 层间耦合为相互依赖模式的两层网络
Fig. 2 Two-layer networks with the inter-layer interdependent couplings
图 3 层间耦合为驱动响应模式的M层网络结构示意图
Fig. 3 The illustration of M-layer networks with the inter-layer drive-response couplings
图 4 层间耦合为相互依赖模式的3层网络结构示意图
Fig. 4 The illustration of three-layer networks with the inter-layer interdependent couplings
图 5 驱动层为链网络、响应层为星形网络的两层网络
Fig. 5 Two-layer networks with the topology of the drive-layer is a chain network, and the response layer is a star network
表 1 本文模型中所用的特殊符号
Table 1 Special notations used in the model of this paper
特殊符号 含义 ${\boldsymbol{A}}_i^K$ 第$ K $层网络的第$ i $个节点的状态矩阵 ${\boldsymbol{B}}_i^K$ 第$ K $层网络的第$ i $个节点的输入矩阵 ${\boldsymbol{C}}_i^K$ 第$ K $层网络的第$ i $个节点的输出矩阵 ${\boldsymbol{H}}_i^K$ 第$ K $层网络的第$ i $个节点与该层其他节点之间的内耦合矩阵 ${\boldsymbol{H}}_i^{KJ}$ 第$ J $层网络的第$ i $个节点与第$ K $层网络的其他节点之间的内耦合矩阵 ${{\boldsymbol{W}}^K}$ 第$ K $层的网络拓扑 ${{\boldsymbol{D}}^{KJ} }$ 第$ J $层到第$ K $层的网络拓扑 ${\boldsymbol{x}}$ 整个网络系统的状态 ${{\boldsymbol{x}}^K}$ 第$ K $层网络的状态 ${\boldsymbol{u}}$ 整个网络系统的输入 ${{\boldsymbol{u}}^K}$ 第$ K $层网络的输入 ${\boldsymbol{ \Phi}}$ 整个网络系统的状态矩阵 ${{\boldsymbol{\Phi}} _{KK} }$ 第$ K $层网络的状态矩阵 ${{\boldsymbol{\Phi}} _{KJ} }$ 第$ J $层到第$ K $层网络的状态矩阵 ${\boldsymbol{\Psi}}$ 整个网络系统的输入矩阵 ${\boldsymbol{ \Xi}}$ 整个网络系统的输出矩阵 ${{\boldsymbol{\Lambda}} ^K}$ 以${\boldsymbol{A}}_1^K,\cdots,{\boldsymbol{A}}_N^K$为对角元的分块对角矩阵 ${\boldsymbol{ \Delta}}$ 对角矩阵${ {\text{diag} } } \{ {{\boldsymbol{\Delta}} ^1},\cdots,{{\boldsymbol{\Delta}} ^M}\}$ ${{\boldsymbol{\Delta}} ^K}$ 对角矩阵${ {\text{diag} } } \{ \delta _1^K,\cdots,\delta _N^K\}$ 
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