Consensus of Hybrid Multi-agent Systems

A weighted directed graph $\mathscr{G}(\mathscr{A})=(\mathscr{V},\mathscr{E},\mathscr{A})$ of order $n$ consists of a vertex set $\mathscr{V}=\{s_{1},s_{2},\cdots,s_{n}\}$, an edge set $\mathscr{E}=\{e_{ij}=(s_{i},s_{j})\}\subset\mathscr{V}\times\mathscr{V}$ and a nonnegative matrix $\mathscr{A}=[a_{ij}]_{n\times n}$. The neighbor set of the agent $i$ is $\mathscr{N}_{i}=\{j:a_{ij}>0\}$. A directed path between two distinct vertices $s_{i}$ and $s_{j}$ is a finite ordered sequence of distinct edges of $\mathscr{G}$ with the form $(s_{i},s_{k_{1}}),(s_{k_{1}},s_{k_{2}}),\cdots,(s_{k_{l}},s_{j})$. A directed tree is a directed graph, where there exists a vertex called the root such that there exists a unique directed path from this vertex to every other vertex. A directed spanning tree is a directed tree, which consists of all the nodes and some edges in $\mathscr{G}$. If a directed graph has the property that $(s_{i},s_{j})\in\mathscr{E}\Leftrightarrow(s_{j},s_{i})\in\mathscr{E}$, the directed graph is called undirected. An undirected graph is said to be connected if there exists a path between any two distinct vertices of the graph. The degree matrix $\mathscr{D}=[d_{ij}]_{n\times n}$ is a diagonal matrix with $d_{ii}=\sum_{j:s_{j}\in\mathscr{N}_{i}}a_{ij}$ and the Laplacian matrix of the graph is defined as $\mathscr{L}=[l_{ij}]_{n\times n}=\mathscr{D}-\mathscr{A}.$ It is easy to see that $\mathscr{L}\mathbf{1}_{n}=0$.

A nonnegative matrix is said to be a (row) stochastic matrix if all its row sums are $1$. A stochastic matrix $P=[p_{ij}]_{n\times n}$ is called indecomposable and aperiodic (SIA) if $\lim_{k\rightarrow\infty}P^{k}=\mathbf{1}y^{T}$ , where $y$ is some column vector. $\mathscr{G}$ is said the graph associated with $P$ when $(s_{i},s_{j})\in\mathscr{E}$ if and only if $p_{ji}>0$. The following results propose the relationship between a stochastic matrix and its associated graph.

Based on Lemma 1 and Lemma 2, we give the following result which will be used in the next content.

Proof. (Sufficiency) Let $P=I_{n}-H\mathscr{L}$. From the definition of $\mathscr{L}$, we have $P=K+H\mathscr{A}$ where $K=I_{n}-H\mathscr{D}$. It follows from $h_{i}\in\left(0,\frac{1}{d_{ii}}\right)$ that $K$ is a positive diagonal matrix. Consequently, it is easy to obtain that $P$ is a stochastic matrix with positive diagonal entries. Obviously, for all $i,j\in\mathcal{I}_{n},~i\neq j$, the $(i,j)$-th entry of $P$ is positive if and only if $a_{ij}>0$. Then, $\mathscr{G}$ is the graph associated with $P$. Combining Lemma 1 and Lemma 2, we have when graph $\mathscr{G}$ has a spanning tree, $\lim_{k\rightarrow\infty}P^{k}=\mathbf{1}_{n}\nu^{k}$ where $\nu$ is a nonnegative vector. Moreover, $\nu$ satisfies $P^{T}\nu=\nu$ and $\mathbf{1}_{n}^{T}\nu=1$.

(Necessary) From Lemma 1, if $\mathscr{G}$ does not have a spanning tree, the algebraic multiplicity of eigenvalue $\lambda=1$ of $P$ is $m>1$. Then, it is easy to prove that the rank of $\lim_{k\rightarrow\infty}P^{k}$ is greater than 1, which implies that $\lim_{k\rightarrow\infty}P^{k}\neq\mathbf{1}_{n}\nu^{k}$. $\blacksquare$

Suppose that the hybrid multi-agent system consists of continuous-time and discrete-time dynamic agents. The number of agents is $n$, labelled $1$ through $n$, where the number of continuous-time dynamic agents is $m$ $(m<n)$. Without loss of generality, we assume that agent $1$ through agent $m$ are continuous-time dynamic agents. Then, each agent has the dynamics as follows:

where $h$ is the sampling period, $x_{i}\in\mathbb{R}$ and $u_{i}\in\mathbb{R}$ are the position-like and control input of agent $i$, respectively. The initial conditions are $x_{i}(0)=x_{i0}$. Let $x(0)=[x_{10},x_{20},\cdots,x_{n0}]^{T}$.

Each agent is regraded as a vertex in a graph. Each edge $(s_{i},s_{j})\in\mathscr{E}$ corresponds to an available information link from agent $i$ to agent $j$. Moreover, each agent updates its current state based on the information received from its neighbors. In this paper, we suppose that there exists communication behavior in hybrid multi-agent system (1), i.e., there are agent $i$ and agent $j$ which make $a_{ij}>0$.

In this subsection, we assume that all agents communicate with their neighbours and update their control inputs in the sampling time $t_{k}$. Then, the consensus protocol for hybrid multi-agent system (1) is presented as follows.

where $\mathscr{A}=[a_{ij}]_{n\times n}$ is the weighted adjacency matrix associated with the graph $\mathscr{G}$, $h=t_{k+1}-t_{k}>0$ is the sampling period.

Proof. (Sufficiency) Firstly, we will prove that equation (2) holds. From (4), we know that

Therefore, it follows that

Let $x(t_{k})=[x_{1}(t_{k}),x_{2}(t_{k}),...,x_{n}(t_{k})]^{T}$. Then, equation (6) can be written in matrix form as

According to Lemma 3, since $\mathscr{G}$ has a directed spanning tree and $h<\frac{1}{\max_{i\in\mathcal{I}_{n}}\{d_{ii}\}}$, we have $\lim_{k\rightarrow\infty}(I_{n}-h\mathscr{L})^{k}=\mathbf{1}_{n}\nu^{T}_{1}$, where $(I_{n}-h\mathscr{L})^{T}\nu_{1}=\nu_{1}$. Thus, it is easy to obtain

and $\mathscr{L}^{T}\nu_{1}=0$. As a consequence, equation (2) holds. Moreover,

Secondly, we have

From (5), it is easy to know that

when $t\rightarrow\infty$, we have $t_{k}\rightarrow\infty$. Thus,

Moreover,

which implies that equation (3) holds.

Therefore, hybrid multi-agent system (1) with protocol (4) can solve consensus probem with consensus state $\nu^{T}_{1}x(0)$, where $\mathscr{L}^{T}\nu_{1}=0$.

(Necessity) From Lemma 3, we have $\lim_{k\rightarrow\infty}(I_{n}-h\mathscr{L})^{k}\neq\mathbf{1}_{n}\nu^{T}_{1}$ when $\mathscr{G}$ does not have a directed spanning tree. Therefore, equation (2) does not hold, which implies that hybrid multi-agent system (1) can not reach consensus. $\blacksquare$

In this subsection, we still assume that the interaction among agents happens in sampling time $t_{k}$. However, different from Case 1, we assume that each continuous-time dynamic agent can observe its own state in real time. Thus, the consensus protocol for hybrid multi-agent system (1) is presented as follows.

where $\mathscr{A}=[a_{ij}]_{n\times n}$ is the weighted adjacency matrix associated with the graph $\mathscr{G}$, $h=t_{k+1}-t_{k}>0$ is the sampling period.

Proof. (Sufficiency) Firstly, we will prove that equation (2) holds. Solving hybrid multi-agent system (1) with protocol (8), we have

Accordingly, at time $t_{k+1}$, the states of agents are

Hence, (10) can be rewritten in compact form as

where $x(t_{k})=[x_{1}(t_{k}),x_{2}(t_{k}),...,x_{n}(t_{k})]^{T}$, $H=\operatorname{diag}\left\{\frac{1-e^{-\sum_{j=1}^{n}a_{1j}h}}{\sum_{j=1}^{n}a_{1j}},\dots,\frac{1-e^{-\sum_{j=1}^{n}a_{mj}h}}{\sum_{j=1}^{n}a_{mj}},h,\dots,h\right\}$.

It is easy to know that $\frac{1-e^{-\sum_{j=1}^{n}a_{ij}h}}{\sum_{j=1}^{n}a_{ij}}<\frac{1}{d_{ii}}$, $i\in\mathcal{I}_{m}$. Owing to $h<\frac{1}{\max_{i\in\mathcal{I}_{n}/\mathcal{I}_{m}}\{d_{ii}\}}$, we have $0<h_{i}<\frac{1}{d_{ii}}$ for $H$. From Lemma 3, because $\mathscr{G}$ has a directed spanning tree, $I_{n}-H\mathscr{L}$ is an SIA matrix, i.e., $\lim_{k\rightarrow\infty}(I_{n}-H\mathscr{L})^{k}=\mathbf{1}_{n}\nu^{T}$, where $(I_{n}-H\mathscr{L})^{T}\nu=\nu$. Hence,

which means that

Obviously, equation (2) holds. Moreover, it follows from $(I_{n}-H\mathscr{L})^{T}\nu=\nu$ that $\mathscr{L}^{T}H\nu=0$.

Secondly, we know from (9) that

Since $t_{k}\rightarrow\infty$ when $t\rightarrow\infty$,

Thus, we have $\lim_{t\rightarrow\infty}x_{i}(t)=\nu^{T}x(0),~~i\in\mathcal{I}_{m}.$ Consequently, equation (3) holds.

Therefore, from Definition 1, hybrid multi-agent system (1) with protocol (8) reaches consensus. Moreover, the consensus state is $\nu^{T}x(0)$.

(Necessity) Similar to the proof of necessity in Theorem 1, we know that if the directed communication network $\mathscr{G}$ does not have a directed spanning tree, then hybrid multi-agent system (1) can not achieve consensus. $\blacksquare$

In this subsection, we assume that all agents interact with each other in a gossip-like manner. Some assumptions are given as follows.

(A1) The communication network of hybrid multi-agent system (1) is undirected, i.e., $a_{ij}=a_{ji}$ for all $i,j\in\mathcal{I}_{n}$.

(A2) At time $t_{k}$, two agents $i$ and $j$ ($i<j$) satisfying $(s_{i},s_{j})\in\mathscr{E}$ are chosen with probability $p_{ij}$, where $p_{ij}\in(0,1)$ and $\sum_{(s_{i},s_{j})\in\mathscr{E}}p_{ij}=1$.

When agents $i$ and $j$ are chosen, their interplay follows the below situations, where $h=t_{k+1}-t_{k}>0$ is the sampling period.

• •

  If $i$ and $j$ are continuous-time dynamic agents, i.e., $i,j\in\mathcal{I}_{m}$, they will communicate during $(t_{k},t_{k+1}]$. The control inputs of two agents are

  $$\left\{\begin{aligned} &u_{i}(t)=a_{ij}(x_{j}(t)-x_{i}(t)),~~&&t\in(t_{k},t_{k+1}],~~i\in\mathcal{I}_{m},\\ &u_{j}(t)=a_{ji}(x_{i}(t)-x_{j}(t)),~~&&t\in(t_{k},t_{k+1}],~~j\in\mathcal{I}_{m}.\end{aligned}\right.$$

  (12)

• •

  If $i$ is continuous-time dynamic agent and $j$ is discrete-time dynamic agent, i.e., $i\in\mathcal{I}_{m},~j\in\mathcal{I}_{n}/\mathcal{I}_{m}$, the control inputs of two agents are

  $$\left\{\begin{aligned} &u_{i}(t)=a_{ij}(x_{j}(t_{k})-x_{i}(t)),~&&i\in\mathcal{I}_{m},\\ &u_{j}(t_{k+1})=x_{j}(t_{k})+ha_{ji}(x_{i}(t_{k})-x_{j}(t_{k})),~&&j\in\mathcal{I}_{n}/\mathcal{I}_{m}.\end{aligned}\right.$$

  (13)

• •

  If $i$ and $j$ are discrete-time dynamic agents, i.e., $i,j\in\mathcal{I}_{n}/\mathcal{I}_{m}$, the control inputs of two agents are

  $$\left\{\begin{aligned} &u_{i}(t_{k+1})=x_{i}(t_{k})+ha_{ij}(x_{j}(t)-x_{i}(t_{k})),~&&i\in\mathcal{I}_{n}/\mathcal{I}_{m},\\ &u_{j}(t_{k+1})=x_{j}(t_{k})+ha_{ji}(x_{i}(t_{k})-x_{j}(t_{k})),~&&j\in\mathcal{I}_{n}/\mathcal{I}_{m}.\end{aligned}\right.$$

  (14)

For each $r\in\mathcal{I}_{n}/\{i,j\}$, it keeps static, i.e.,

Proof. (Sufficiency) It suffices to prove that

and

hold for any initial states.

Firstly, we will show that equation (16) holds. From (12)–(15), if agents $i$ and $j$ are selected to interplay at time $t_{k}$, the states of all agents at time $t_{k+1}$ are three cases:

Note that (18) can be rewritten in the following compact form as

where

According to (A2), it is not hard to know that $\{x({t_{k}})\}$ is a stochastic linear system

Therefore, it follows that

Due to $0<h<\frac{1}{\max_{i,j\in\mathcal{I}_{n}}\{a_{ij}\}}$, we have $\frac{1-e^{-2a_{ij}h}}{2}\in(0,\frac{1}{2})$, $1-e^{-a_{ij}h}\in(0,1)$ and $ha_{ij}\in(0,1)$. Thus, $\Phi_{ij}$ is a stochastic matrix with positive diagonal entries. Moreover, the $(i,j)$-th and $(j,i)$-th entries of $\Phi_{ij}$ are positive, while all other non-diagonal entries are zeros. Hence, noticing that

we know that $\mathbb{E}(\Phi_{k})$ is a stochastic matrix satisfying

• •

  all diagonal entries are positive;

• •

  the $(i,j)$-th and $(j,i)$-th entries are positive if and only if $(s_{i},s_{j})\in\mathscr{E}$.

Consequently, $\mathscr{G}$ is the graph associated with $\mathbb{E}(\Phi_{k})$. Since $\mathscr{G}$ is connected, combining Lemma 1 and Lemma 2, we have

From (20), we have

which implies that equation (16) holds.

Secondly, at time $t\in(t_{k},t_{k+1}]$, the state of each $i\in\mathcal{I}_{m}$ follows the below three scenarios:

• •

  if $i$ is selected to communicate with its  continuous-time dynamical neighbour $j\in\mathcal{I}_{m}$,

  $$x_{i}(t)=\frac{1+e^{-2a_{ij}(t-t_{k})}}{2}x_{i}(t_{k})+\frac{1-e^{-2a_{ij}(t-t_{k})}}{2}x_{j}(t_{k});$$

• •

  if $i$ is selected to communicate with its  discrete-time dynamical neighbour $j\in\mathcal{I}_{n}/\mathcal{I}_{m}$,

  $$x_{i}(t)=e^{-a_{ij}(t-t_{k})}x_{i}(t_{k})+(1-e^{-a_{ij}(t-t_{k})})x_{j}(t_{k});$$

• •

  if $i$ is not selected, $x_{i}(t)=x_{i}(t_{k}).$