Consensus of Networked Multi-agent Systems with Antagonistic Interactions and Communication Delays

Abstract

A common feature of previous consensus results is the focus on cooperative systems. The consensus of these systems is asymptotically achieved through collaboration, which is characterized by the diffusive coupling and the nonnegative weights among agents.

A common feature of previous consensus results is the focus on cooperative systems [1,2,3]. The consensus of these systems is asymptotically achieved through collaboration, which is characterized by the diffusive coupling [4] and the nonnegative weights among agents [5,6,7,8]. In many real-world cases, however, it is more reasonable to consider that some agents collaborate with each other, while others are competitive. Networks with antagonistic interactions are ubiquitous in real world [9], and it becomes a focus for studying in recent years [10,11,12,13,14,15]. Altafini [10] proved that bipartite consensus can be achieved over networks with antagonistic interactions. Furthermore, in [16], emergent behaviors were investigated over signed random dynamical networks. In [17], flocking behaviors were studied by using results about signed graph. In [18], the leader-following bipartite consensus issue for single-integrator multi-agent systems was investigated, where the signed digraph was considered to be structurally balanced and had a spanning tree.

To achieve the consensus, each node in a network has to transmit its state information to its neighbors via connections. However, because of physical and environmental limitations, communication constraints between connected nodes are unavoidable. As is well-known, the communication delay is one of the most universal communication constraints. Motivated by the aforementioned discussions, we investigate the consensus problem of signed networks with antagonistic interactions and communication delays in this chapter. To the best of our knowledge, only a few results have been done concerning such problem. Due to the difficulty that antagonistic interactions and communication delays need to be simultaneously considered, new techniques are required to deal with this problem. According to matrix theory, Lyapunov theorem, and some other mathematical analysis, we found that bipartite consensus can be achieved for those systems with communication delays. Furthermore, in order to obtain the final bipartite consensus solution, we construct an invariant function to study the relationship of the states of nodes and their initial states. Using some mathematical analysis skills, we provide the bipartite consensus solution with an explicit expression.

6.1 Continuous-Time Multi-agent Consensus

6.1.1 Linear Coupling

In this section, we consider a multi-agent system formed by N linearly coupled identical nodes, where each node’s dynamic is described as follows:

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} \dot{x}_{i}(t)=& \sum_{j=1}^N\mid a_{ij}\mid [\mbox{sgn}(a_{ij})x_{j}(t-\tau_{ij})-x_{i}(t)],~i\in\mathcal{N}, \end{array} \end{aligned} $$

(6.1)

where $x_{i}(t)\in \mathbb {R}^n$ is the state of node i at time t, and τ _ij > 0 denotes the communication delay from ν _j to ν _i for i ≠ j and τ _ii = 0. A = [a _ij]_N×N is the adjacency matrix of the network that is symmetric. Here it is assumed that there is no self-closed loop, which means that a _ii = 0.

Throughout this section, the bipartite consensus of dynamical system (6.1) is said to be realized if lim_t→∞ x _i(t) = α for $i\in \mathcal {V}_{1}$ and lim_t→∞ x _i(t) = −α for $i\in \mathcal {V}_{2}$.

Theorem 6.1

Consider the networked multi-agent system (6.1) with a connected signed graph G(A). The bipartite consensus can be asymptotically reached if G(A) is structurally balanced. If instead G(A) is structurally unbalanced, then lim_t→∞ x(t) = 0.

Proof

We first consider the case that G(A) is structurally balanced. According to Lemma 1.8, one can obtain that $\exists D\in \mathcal {D}$ such that DAD has all nonnegative entries. Let z(t) = Dx(t), we obtain that

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} z_{i}(t)=\sigma_{i}x_{i}(t)~,~i\in\mathcal{N}. \end{array} \end{aligned} $$

(6.2)

Substituting (6.2) into (6.1) results in

$$\displaystyle \begin{aligned} \begin{array}{rcl} \sigma_{i}\dot{z}_{i}(t)=& \sum_{j=1}^N\mid a_{ij}\mid [\mbox{sgn}(a_{ij})\sigma_{j}z_{j}(t-\tau_{ij}) -\sigma_{i}z_{i}(t)]~,i\in\mathcal{N}. \end{array} \end{aligned} $$

Since DAD is a nonnegative matrix, we have σ _isgn(a _ij)σ _j = 1. Using $\sigma _{i}^{2}=1$, one can obtain the following equation:

$$\displaystyle \begin{aligned} \dot{z}_{i}(t)=&\sum_{j=1}^N\mid a_{ij}\mid [\sigma_{i}\mbox{sgn}(a_{ij})\sigma_{j}z_{j}(t-\tau_{ij})-\sigma_{i}^{2}z_{i}(t)]\\ =&\sum_{j=1}^N\mid a_{ij}\mid [z_{j}(t-\tau_{ij})-z_{i}(t)]~,~i\in\mathcal{N}.{} \end{aligned} $$

(6.3)

Following [19], the consensus of networks system (6.3) is asymptotically reached. That is

$$\displaystyle \begin{aligned} \begin{array}{rcl} \lim_{t\rightarrow\infty}z_{i}(t)\rightarrow\alpha,~\forall i\in\mathcal{N}, \end{array} \end{aligned} $$

(6.4)

where $\alpha \in \mathbb {R}^{n}$ is a constant vector.

Hence, we can get that $\lim \limits _{t\rightarrow \infty }x_{i}(t)\rightarrow \sigma _{i}\alpha$ for $i\in \mathcal {N}$. Then, the bipartite consensus of system (6.1) can be reached if G(A) is structurally balanced.

Next, we consider the case that G(A) is structurally unbalanced. Following Lemma 1.10, we can conclude that G(A) contains one or more negative cycles. For the sake of simplicity, let us first consider the simplest case of G(A) with only one negative cycle. Without loss of generality, we assume that (ν ₁, ν ₂) belongs to the negative cycle and a ₁₂ = a ₂₁ = a < 0. According to Lemma 1.10, one can obtain that there is no $D\in \mathcal {D}$ such that DAD is a nonnegative matrix. However, for the subgraph G(B), which denotes the rest part of G(A) reducing the edge (ν ₁, ν ₂), it admits a bipartition of the nodes $\mathcal {V}_{1}$ and $\mathcal {V}_{2}$. Furthermore, one can find that G(B) is connected and matrix B is irreducible. Now, we can make a hypothesis that nodes ν ₁ and ν ₂ simultaneously belong to $\mathcal {V}_{1}$ ( or $\mathcal {V}_{2}$ ) and the rest nodes remain unchanged. Based on this hypothesis, we can choose D ₁ = diag(σ) with σ satisfying σ _i = 1 for $\nu _{i}\in \mathcal {V}_{1}$ and σ _i = −1 for $\nu _{i}\in \mathcal {V}_{2}$. Then $D_{1}AD_{1}=A'=[a^{\prime }_{ij}]_{N\times N}$ has exactly two negative elements, i.e., $a^{\prime }_{12}=a^{\prime }_{21}=a<0$, and the rest elements are nonnegative. The following is a decomposition of the matrix A′:

$$\displaystyle \begin{aligned} \begin{array}{rcl} A'=A_{12}+A_{21}+B'~, \end{array} \end{aligned} $$

(6.5)

where A _ij, i, j ∈{1, 2}, denotes a matrix in which the element lied in the intersection of ith row and jth column is a _ij ≠ 0 and others all are 0. $B'=[b_{ij}^{\prime }]_{N\times N}$ is a nonnegative adjacency matrix. In order to clearly express the matrix B′, we define a function as follows:

$$\displaystyle \begin{aligned}c(i,j)=\left\{ \begin{aligned} &0, & &(i,j)=(1,2)~ or ~(2,1) ;&\\ &1,& &\mbox{otherwise}.& \end{aligned} \right. \end{aligned}$$

Hence, we get $b_{ij}^{\prime }=c(i,j)|a_{ij}|$. It is easy to find that B′ is irreducible. Let $\bar {B}=[\bar {b}_{ij}]_{N\times N}$ be the Laplacian matrix of G(B′), and its elements are defined as : $\bar {b}_{ij}=b_{ij}^{\prime }~(i\neq j)$, $\bar {b}_{ii}=-\sum _{j=1}^{N}b_{ij}^{\prime }$. Therefore, ξ = (1, 1, ⋅⋅⋅, 1)^⊤ is the left eigenvector of $\bar {B}$ corresponding to the zero eigenvalue, i.e., $\xi ^{\top }\bar {B}=\mathbf {0}$ , which implies that

$$\displaystyle \begin{aligned} \begin{array}{rcl} \bar{b}_{ii}=-\sum_{j=1,j\neq i}^{N}\bar{b}_{ji}. \end{array} \end{aligned} $$

(6.6)

Further because $\bar {b}_{ii}=-\sum _{j=1}^{N}b_{ij}^{\prime }$, one can obtain that

$$\displaystyle \begin{aligned} \begin{array}{rcl} \sum_{j=1}^{N}b_{ij}^{\prime}=\sum_{j=1}^{N}b_{ji}^{\prime}~~\mbox{and}~~\sum_{i=1}^{N}b_{ji}^{\prime}=\sum_{i=1}^{N}b_{ij}^{\prime}. \end{array} \end{aligned} $$

(6.7)

Let z(t) = Dx(t), i.e., z _i(t) = σ _i x _i(t) for any $i\in \mathcal {N}$, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} \dot{z}_{i}(t)=\sum_{j=1}^N\mid a_{ij}\mid [\sigma_{i}\sigma_{j}\mbox{sgn}(a_{ij})z_{j}(t-\tau_{ij})-z_{i}(t)]. \end{array} \end{aligned} $$

(6.8)

Consider the following Lyapunov functional for system (6.1):

$$\displaystyle \begin{aligned} \begin{array}{rcl} V(t)=V_{1}(t)+V_{2}(t), \end{array} \end{aligned} $$

(6.9)

where

$$\displaystyle \begin{aligned} \begin{array}{rcl} V_{1}(t)=\frac{1}{2}\sum_{i=1}^{N}x_{i}^{\top}(t)x_{i}(t), \end{array} \end{aligned} $$

(6.10)

and

$$\displaystyle \begin{aligned} \begin{array}{rcl} V_{2}(t)=\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}\int_{t-\tau_{ji}}^{t}|a_{ji}|x_{i}^{\top}(\theta)x_{i}(\theta)d\theta. \end{array} \end{aligned} $$

(6.11)

Calculating the time derivative of V _i(t) (i = 1, 2) along the trajectories of system (6.1), we have

$$\displaystyle \begin{aligned} \dot{V}_{1}(t)=&\sum_{i=1}^{N}x_{i}^{\top}(t)\dot{x}_{i}(t) \\ =&\sum_{i=1}^{N}\sum_{j=1}^{N}|a_{ij}|[x_{i}^{\top}(t)\mbox{sgn}(a_{ij})x_{j}(t-\tau_{ij})-x_{i}^{\top}(t)x_{i}(t)]\\ =&\sum_{i=1}^{N}\sum_{j=1}^{N}|a_{ij}|[\sigma_{i}\sigma_{j}z_{i}^{\top}(t)\mbox{sgn}(a_{ij})z_{j}(t-\tau_{ij})-z_{i}^{\top}(t)z_{i}(t)]\\ =&\sum_{i=1}^{N}\sum_{j=1}^{N}b_{ij}^{\prime}[z_{i}^{\top}(t)z_{j}(t-\tau_{ij})-z_{i}^{\top}(t)z_{i}(t)]+a_{12}[z_{1}^{\top}(t)z_{2}(t-\tau_{12})\\ &+z_{1}^{\top}(t)z_{1}(t)] +a_{21}[z_{2}^{\top}(t)z_{1}(t-\tau_{21})+z_{2}^{\top}(t)z_{2}(t)]\\ =&\sum_{i=1}^{N}\sum_{j=1}^{N}b_{ij}^{\prime}[z_{i}^{\top}(t)z_{j}(t-\tau_{ij})-z_{i}^{\top}(t)z_{i}(t)]+\frac{1}{2}a[2z_{1}^{\top}(t)z_{2}(t-\tau_{12})\\ &+2z_{2}^{\top}(t)z_{1}(t-\tau_{21})+2z_{1}^{\top}(t)z_{1}(t)+2z_{2}^{\top}(t)z_{2}(t)]{}, \end{aligned} $$

(6.12)

and

$$\displaystyle \begin{aligned} \dot{V}_{2}(t)=&\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}|a_{ji}|[x_{i}^{\top}(t)x_{i}(t)-x_{i}^{\top}(t-\tau_{ji})x_{i}(t-\tau_{ji})]\\ =&\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}b_{ji}^{\prime}[x_{i}^{\top}(t)x_{i}(t)-x_{i}^{\top}(t-\tau_{ji})x_{i}(t-\tau_{ji})]\\ &+\frac{1}{2}a_{12}[x_{2}^{\top}(t-\tau_{12})x_{2}(t-\tau_{12})-x_{2}^{\top}(t)x_{2}(t)]\\ &+\frac{1}{2}a_{21}[x_{1}^{\top}(t-\tau_{21})x_{1}(t-\tau_{21})-x_{1}^{\top}(t)x_{1}(t)]\\ =&\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}b_{ij}^{\prime}z_{i}^{\top}(t)z_{i}(t) -\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}b_{ij}z_{j}^{\top}(t-\tau_{ij})z_{j}(t-\tau_{ij})\\ &+\frac{1}{2}a[z_{2}^{\top}(t-\tau_{12})z_{2}(t-\tau_{12}+z_{1}^{\top}(t-\tau_{21})z_{1}(t-\tau_{21})\\ &-z_{2}^{\top}(t)z_{2}(t)-z_{1}^{\top}(t)z_{1}(t)]{}. \end{aligned} $$

(6.13)

Using Eqs. (6.12) and (6.13) gives that

$$\displaystyle \begin{aligned} \dot{V}(t)=&-\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}b_{ij}^{\prime}[z_{i}^{\top}(t) z_{i}(t)-2z_{i}^{\top}(t)z_{j}(t-\tau_{ij})+z_{j}^{\top}(t-\tau_{ij})]\\ &+\frac{1}{2}a\{[2z_{1}^{\top}(t)z_{2}(t-\tau_{12}) +z_{1}^{\top}(t)z_{1}(t)+z_{2}^{\top}(t-\tau_{12})z_{2}(t-\tau_{12})] \\ &+[2z_{2}^{\top}(t)z_{1}(t-\tau_{21})]+z_{2}^{\top}(t)z_{2}(t) +z_{1}^{\top}(t-\tau_{21})z_{1}(t-\tau_{21})]\}\\ =&-\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}b_{ij}^{\prime}[z_{i}-z_{j}(t-\tau_{ij})]^{\top} [z_{i}-z_{j}(t-\tau_{ij})]\\ &+\frac{1}{2}a\{[z_{1}(t)+z_{2}(t-\tau_{12})]^{\top}[z_{1}(t)+z_{2}(t-\tau_{12})]\\ &+[z_{2}(t)+z_{1}(t-\tau_{21})]^{\top}[z_{2}(t)+z_{1}(t-\tau_{21})]\}\\ \leq&0. \end{aligned} $$

Hence, V (t) is non-increasing. Referring to the construction of V (t), one has that V (t) ≥ 0, which shows that lim_t→∞ V (t) exists and is finite. Then, we can get the boundedness of x _i(t) for $i\in \mathcal {N}$. Combining with the expression of V (t), further one can easily show the boundedness of $\dot {x}_{i}(t)$ for $i\in \mathcal {N}$ by referring to system (6.1). Thus, $\dot {z}_{i}(t)=\sigma _{i}\dot {x}_{i}(t)$ is bounded, which implies $\ddot {V}(t)$ is also bounded. According to Barbalat’s Lemma ([20]), we can obtain that $\lim _{t\rightarrow \infty }b_{ij}^{\prime }[z_{i}-z_{j}(t-\tau _{ij})]^{\top }[z_{i}-z_{j}(t-\tau _{ij})]=0$, lim_t→∞[z ₁(t) + z ₂(t − τ ₁₂)]^⊤[z ₁ + z ₂(t − τ ₁₂)] = 0, and lim_t→∞[z ₂(t) + z ₁(t − τ ₂₁)]^⊤[z ₂(t) + z ₁(t − τ ₂₁)] = 0, i.e., lim_t→∞[z _i − z _j(t − τ _ij)] = 0, if $b_{ij}^{\prime }>0$,lim_t→∞[z ₁ + z ₂(t − τ ₁₂)] = 0, and lim_t→∞[z ₂(t) + z ₁(t − τ ₂₁)] = 0. Further considering the expression (6.8), we have

$$\displaystyle \begin{aligned} \dot{z}_{1}(t)=&\sum_{j=1}^N\mid a_{1j}\mid [\sigma_{1}\sigma_{j}\mbox{sgn}(a_{1j})z_{j}(t-\tau_{1j})-z_{1}(t)]\\ =&a_{12}[z_{1}(t)+z_{2}(t-\tau_{12})]+\sum_{j=3}^{N}b_{1j}^{\prime}[z_{j}(t-\tau_{1j})-z_{1}(t)]\\ \rightarrow&\mathbf{0},~\mbox{as}~t\rightarrow\infty~, \end{aligned} $$

$$\displaystyle \begin{aligned} \dot{z}_{2}(t)=&\sum_{j=1}^N\mid a_{2j}\mid [\sigma_{2}\sigma_{j}\mbox{sgn}(a_{2j})z_{j}(t-\tau_{2j})-z_{2}(t)]\\ =&a_{21}[z_{2}(t)+z_{1}(t-\tau_{21})]+\sum_{j=3}^{N}b_{2j}^{\prime}[z_{j}(t-\tau_{2j})-z_{2}(t)]\\ \rightarrow&\mathbf{0},~\mbox{as}~t\rightarrow\infty~, \end{aligned} $$

and

$$\displaystyle \begin{aligned} \dot{z}_{i}(t)=\sum_{j=1}^{N}b_{ij}^{\prime}[z_{j}(t-\tau_{ij})-z_{i}(t)]\rightarrow \mathbf{0},~\mbox{as}~t\rightarrow\infty~\mbox{for}~i\geq3. \end{aligned} $$

Therefore, we get that $\lim _{t\rightarrow \infty }\dot {z}_{i}(t)\rightarrow \mathbf {0}$ for any $i\in \mathcal {N}$. Since the adjacency matrix B′ is irreducible, one can obtain that lim_t→∞ z ₁(t) =lim_t→∞ z ₂(t) = ⋅⋅⋅ =lim_t→∞ z _N(t) by referring to $\lim _{t\rightarrow \infty }b_{ij}^{\prime }[z_{i}-z_{j}(t-\tau _{ij})]=\mathbf {0}$. In addition, we can also get lim_t→∞ z ₁(t) = −lim_t→∞ z ₂(t) by referring to lim_t→∞[z ₂(t) + z ₁(t − τ ₂₁)] = 0. Thus, the following continued equality can be concluded: lim_t→∞ z ₁(t) =lim_t→∞ z ₂(t) = ⋅⋅⋅ =lim_t→∞ z _N(t) = 0. Therefore, we obtain that lim_t→∞ x _i(t) =lim_t→∞ σ _i z _i(t) = 0 for $i\in \mathcal {N}$.

Now consider the case of G(A) with m (m ≥ 2) negative cycles. Referring to the above approach, we respectively select m negative cycles as follows: $(\nu _{i_{1}},\nu _{j_{1}}),\cdot \cdot \cdot ,(\nu _{i_{m}},\nu _{j_{m}})$. There exists a diagonal matrix $D_{2}\in \mathcal {D}$ such that

$$\displaystyle \begin{aligned} D_{2}AD_{2}=A_{i_{1}j_{1}}+A_{j_{1}i_{1}}+\cdot\cdot\cdot+A_{i_{m}j_{m}}+A_{j_{m}i_{m}}+B''. \end{aligned} $$

Similarly we can get that lim_t→∞ x _i(t) = 0 for any $i\in \mathcal {N}$.

Remark 6.2

For the structurally unbalanced network, since the number of negative cycles is not the essential attribute of structurally unbalanced network, we just need to think about those structurally unbalanced networks with one negative cycle instead of all structurally unbalanced networks. This consideration reduces the difficulty of the problem.

Remark 6.3

There is a situation that needs to be considered. If a negative edge simultaneously belongs to two or more negative cycles, we should admit that the edge only belongs to one of those cycles and the rest negative cycles should be viewed positive cycles. Then we can still make a hypothesis that two vertices on this negative edge belong to $\mathcal {V}_{1}$ or $\mathcal {V}_{2}$. According to the proof progress of Theorem 6.1, it is obvious that the results of Theorem 6.1 still hold.

According to Theorem 6.1, if G(A) is structurally balanced, the bipartite consensus can be asymptotically reached, and we have lim_t→∞ x _i(t) = α for $i\in \mathcal {V}_{1}$, lim_t→∞ x _i(t) = −α for $i\in \mathcal {V}_{2}$. Calculating the bipartite consensus value of α = (α ₁, α ₂, ⋅⋅⋅, α _n)^⊤ is not an easy task due to the existence of time delays. Here, the value of α = (α ₁, α ₂, ⋅⋅⋅, α _n)^⊤ will be obtained by an exact expression when the initial conditions of system (6.1) are given. We define $\mathbf {1}=(1,1,\cdot \cdot \cdot ,1)_{1\times N}^{\top }$. The initial conditions about system (6.1) are provided as $x_{i}(s)=\sigma _{i}\varphi _{i}(s)\in C([-\tau ,0],\mathbb {R}^{n})$, where τ = max_i,j{τ _ij}. Hence, we have $z_{i}(s)=\varphi _{i}(s)\in C([-\tau ,0],\mathbb {R}^{n})$. Let ξ(t) = (ξ ₁(t), ξ ₂(t), ⋅⋅⋅, ξ _n(t))^⊤, where $\xi _{r}(t)=(1/N)(\sum _{i=1}^{N}z_{ir}(t)+\sum _{i=1}^{N}\sum _{j=1}^{N}|a_{ij}|\int _{t-\tau _{ij}}^{t}z_{jr}(s)ds)$, r ∈{1, 2, ⋅⋅⋅, n}.

Theorem 6.4

Consider a connected signed graph G(A) that is structurally balanced. If$D\in \mathcal {D}$renders DAD nonnegative, then the bipartite solution of (6.1) is$\lim _{t\rightarrow \infty }x(t)=(D\mathbf {1})\otimes [N\xi (0)/\sum _{i=1}^{N}(1+\sum _{j=1}^{N}|a_{ij}|\tau _{ij})]$.

Proof

Referring to the proof of Theorem 6.1, one can obtain that lim_t→∞ z _i(t) = α and lim_t→∞ x(t) = (D 1) ⊗ α. Using (6.3), we can obtain

$$\displaystyle \begin{aligned} \dot{\xi_{r}}(t)=&\frac{1}{N}\sum_{i=1}^{N}\sum_{j=1}^{N}|a_{ij}|[z_{jr}(t-\tau_{ij})-z_{ir}(t)]\\ &+\frac{1}{N}\sum_{i=1}^{N}\sum_{j=1}^{N}|a_{ij}|[z_{jr}(t)-z_{jr}(t-\tau_{ij})]\\ =&-\frac{1}{N}\sum_{i=1}^{N}\sum_{j=1}^{N}|a_{ij}|z_{ir}(t)+\frac{1}{N}\sum_{i=1}^{N}\sum_{j=1}^{N}|a_{ij}|z_{jr}(t)\\ =&0~.{} \end{aligned} $$

(6.14)

Therefore, ξ _r(t) in (6.14) is a constant. That is,

$$\displaystyle \begin{aligned} \xi_{r}(t)=&\xi_{r}(0)\\ =&\frac{1}{N}(\sum_{i=1}^{N}z_{ir}(0)+\sum_{i=1}^{N}\sum_{j=1}^{N}|a_{ij}|\int_{-\tau_{ij}}^{0}\varphi_{jr}(s)ds).{} \end{aligned} $$

(6.15)

Then, we can get

$$\displaystyle \begin{aligned} \xi_{r}(0)=&\lim_{t\rightarrow\infty}\xi_{r}(t)\\ =&\frac{1}{N}(\sum_{i=1}^{N}\alpha_{r}+\sum_{i=1}^{N}\sum_{j=1}^{N}|a_{ij}|\tau_{ij}\alpha_{r})\\ =&\frac{\alpha_{r}}{N}\sum_{i=1}^{N}(1+\sum_{j=1}^{N}|a_{ij}|\tau_{ij}){}~. \end{aligned} $$

(6.16)

Hence, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \alpha=\frac{N\xi(0)}{\sum_{i=1}^{N}(1+\sum_{j=1}^{N}|a_{ij}|\tau_{ij})}{}~, \end{array} \end{aligned} $$

(6.17)

and

$$\displaystyle \begin{aligned} \begin{array}{rcl} \lim_{t\rightarrow\infty}x(t)=(D\mathbf{1})\otimes\frac{N\xi(0)}{\sum_{i=1}^{N}(1+\sum_{j=1}^{N}|a_{ij}|\tau_{ij})}. \end{array} \end{aligned} $$

(6.18)

Remark 6.5

Referring to expression (6.17), for the case of network without communication delays, one can obtain that $\alpha =(1/N)\sum _{i=1}^{N}\varphi _{i}(0)$. This result is consistent with the one obtained in [10]. This shows that our results are more general. Moreover, we conclude that a bipartite consensus solution is not only associated with initial values of x _i(t) but also closely related to communication delays and network structure.

Remark 6.6

According to Theorem 6.4, it is obvious that α ≠ 0 if G(A) is structurally balanced unless ξ _r(0) = 0.

6.1.2 Nonlinear Coupling

In this subsection, we will investigate the multi-agent systems with nonlinear coupling. Consider the following multi-agent systems:

$$\displaystyle \begin{aligned} \dot{x}_{i}(t)=\sum_{j=1}^N\mid a_{ij}\mid \{\mbox{sgn}(a_{ij})h[x_{j}(t-\tau_{ij})]-h[x_{i}(t)]\},~i\in\mathcal{N}, \end{aligned} $$

(6.19)

where $x_{i}(t)\in \mathbb {R}$ is the state of node i. The function $h(\cdot ): \mathbb {R}\rightarrow \mathbb {R}$ is assumed to be odd and strictly monotone increasing, which implies h(0) = 0 and h(−x) = −h(x). Further we assume that h(⋅) is unbounded.

Theorem 6.7

Consider the nonlinear coupled system (6.19) with a connected signed graph G(A). The bipartite consensus can be asymptotically reached if G(A) is structurally balanced. If instead G(A) is structurally unbalanced, then lim_t→∞ x(t) = 0.

Proof

Following Lemma 1.8, if G(A) is structurally balanced, we can obtain that $\exists D\in \mathcal {D}$ such that DAD has all nonnegative entries. Let Z(t) = Dx(t), i.e., z _i(t) = σ _i x _i(t), one can easily get that

$$\displaystyle \begin{aligned} \dot{z}_{i}(t)=\sum_{j=1}^N\mid a_{ij}\mid \{h[z_{j}(t-\tau_{ij})]-h[z_{i}(t)]\}~,~i\in\mathcal{N}~. \end{aligned} $$

(6.20)

Following [19], we obtain that $\lim _{t\rightarrow \infty }z_{i}(t)\rightarrow \beta \in \mathbb {R}$ for any $i\in \mathcal {N}$, which shows that $\lim _{t\rightarrow \infty }x_{i}(t)\rightarrow \sigma _{i}\beta \in \mathbb {R}$ for $i\in \mathcal {N}$. Therefore, the bipartite consensus of system (6.19) can be reached if G(A) is structurally balanced.

Next, we consider the case that G(A) is structurally unbalanced. From Lemma 1.10, it follows that G(A) contains one or more negative cycles. We first consider the case of G(A) with only one negative cycle. The edge (ν ₁, ν ₂) is assumed to be a negative weighted edge belonging to the negative cycle, and a ₁₂ = a ₂₁ = a < 0. Choosing D ₁ = diag(σ) with σ satisfying σ _i = 1 for $\nu _{i}\in \mathcal {V}_{1}$ and σ _i = −1 for $\nu _{i}\in \mathcal {V}_{2}$, one can obtain that D ₁ AD ₁ = A′ has exactly two negative elements, i.e., $a^{\prime }_{12}=a^{\prime }_{21}=a<0$, and the rest elements are nonnegative. The decomposition of the matrix A′ is shown as follows:

$$\displaystyle \begin{aligned} A'=A_{12}+A_{21}+B', \end{aligned} $$

where the definitions of A ₁₂, A ₂₁, and B′ are similar to the proof in Theorem 6.1.

Let Z(t) = Dx(t), i.e., z _i(t) = σ _i x _i(t) for any $i\in \mathcal {N}$, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} \dot{z}_{i}(t)=\sum_{j=1}^N\mid a_{ij}\mid \{\sigma_{i}\sigma_{j}\mbox{sgn}(a_{ij})h[z_{j}(t-\tau_{ij})]-h[z_{i}(t)]\}. \end{array} \end{aligned} $$

(6.21)

Consider the following Lyapunov–Krasovskii functional for system (6.19)

$$\displaystyle \begin{aligned} W(x(t))=W_{1}(x(t))+W_{2}(x(t))~, \end{aligned} $$

where

$$\displaystyle \begin{aligned} W_{1}(x(t))=\sum_{i=1}^{N}\int_{0}^{x_{i}(t)}h(s)ds~, \end{aligned} $$

and

$$\displaystyle \begin{aligned} W_{2}(x(t))=\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}\int_{t-\tau_{ij}}^{t}|a_{ij}|h^{2}[x_{j}(s)]ds~. \end{aligned} $$

Calculating the time derivative of W _i(t)(i = 1, 2) along the trajectories of system (6.19), we have

$$\displaystyle \begin{aligned} \dot{W}_{1}(x(t)) =&\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}b_{ij}^{\prime}\{2h[z_{i}(t)]h[z_{j}(t-\tau_{ij})]-2h^{2}[z_{i}(t)]\}\\ &+a\{h[z_{1}(t)]h[z_{2}(t-\tau_{12})] +h^{2}[z_{1}(t)]\\ &+h[z_{2}(t)]h[z_{1}(t-\tau_{21})]+h^{2}[z_{2}(t)]{}~, \end{aligned} $$

(6.22)

and

$$\displaystyle \begin{aligned} \dot{W}_{2}(x(t)) =&\frac{1}{2}\sum_{i=1}^{N}b_{ij}^{\prime}\sum_{j=1}^{N}h^{2}[z_{i}(t)] -\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}b_{ij}h^{2}[z_{j}(t-\tau_{ij})]\\ &+\frac{1}{2}a\{h^{2}[z_{2}(t-\tau_{12})]-h^{2}[z_{2}(t)]\\ &+h^{2}[z_{1}(t-\tau_{21})]-h^{2}[z_{1}(t)]\}{}~. \end{aligned} $$

(6.23)

Using Eqs. (6.22) and (6.23), we get that

$$\displaystyle \begin{aligned} \dot{W}(x(t))=&-\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}b_{ij}^{\prime}\{h[z_{i}(t)]-h[z_{j}(t-\tau_{ij})]\}^{2}\\ &+\frac{1}{2}a\{h[z_{1}(t)]+h[z_{2}(t-\tau_{12})]\}^{2}\\ &+\frac{1}{2}a\{h[z_{2}(t)]+h[z_{1}(t-\tau_{21})]\}^{2}\\ \leq&0{}~. \end{aligned} $$

(6.24)

Let $S=\{x(t)~|~\dot {W}(x(t))=0\}$. Then it follows from Eq. (6.24) that $S=\{x\in C([t-\tau ,t],\mathbb {R}^{N})~|~b_{ij}^{\prime }\{h[z_{i}(t)]-h[z_{j}(t-\tau _{ij})]\}=0,~h[z_{1}(t)]+h[z_{2}(t-\tau _{12})]=0, ~\mbox{and} ~h[z_{2}(t)]+h[z_{1}(t-\tau _{21})]=0\}$. Combining with the property of h(⋅), we can get that the set S is an invariant set with respect to system (6.21). According to the LaSalle invariance principle [21], one can easily show that x(t) → S as t →∞. Thus, we have lim_t→∞{h[z _i(t)] − h[z _j(t − τ _ij)]} = 0 for $b_{ij}^{\prime }>0$, lim_t→∞{h[z ₁(t)] + h[z ₂(t − τ ₁₂)]} = 0, and lim_t→∞{h[z ₂(t)] + h[z ₁(t − τ ₂₁)]} = 0. Hence, we have $\lim _{t\rightarrow \infty }\dot {z}_{i}(t)=0$. In addition, since h(⋅) is unbounded and strictly increasing with h(0) = 0, we get that lim_t→∞[z _i(t) − z _j(t − τ _ij)] = 0 when $b_{ij}^{\prime }>0$ and lim_t→∞[z ₁(t) − z ₂(t − τ ₁₂)] = 0. According to the fact that B is irreducible, we conclude that z ₁(t) = z ₂(t) = ⋅⋅⋅ = z _N(t) as t →∞. It follows from lim_t→∞[z ₁(t) − z ₂(t − τ ₁₂)] = 0 that z ₁(t) = −z ₂(t) as t →∞. Hence, the following equality can be concluded: lim_t→∞ z ₁(t) =lim_t→∞ z ₂(t) = ⋅⋅⋅ =lim_t→∞ z _N(t) = 0. Therefore, we obtain that lim_t→∞ x _i(t) =lim_t→∞ σ _i z _i(t) = 0 for $i\in \mathcal {N}$.

The result still holds for the case of G(A) with two or more negative cycles. The proof is similar to the case of linear coupling and is omitted for simplicity.

Similar to the case of linear coupling, we can give the bipartite solution of (6.19). The initial conditions about system (6.19) are provided as $x_{i}(s)=\sigma _{i}\psi _{i}(s)\in C([-\tau ,0],\mathbb {R})$. Hence, we have $z_{i}(s)=\psi _{i}(s)\in C([-\tau ,0],\mathbb {R})$. Let $\zeta (0)=(1/N)(\sum _{i=1}^{N}\psi _{i}(0)+\sum _{i=1}^{N}\sum _{j=1}^{N}|a_{ij}|\int _{-\tau _{ij}}^{0}h(\psi _{j}(s))ds)$.

Theorem 6.8

Consider a connected signed graph G(A) that is structurally balanced. If$D\in \mathcal {D}$renders DAD nonnegative, then the bipartite solution of (6.19) is lim_t→∞ x(t) = βD 1 , where$\beta \in \mathbb {R}$meets a relational expression as follows:

$$\displaystyle \begin{aligned} \beta+h(\beta)\sum_{i=1}^{N}\sum_{j=1}^{N}(|a_{ij}|\tau_{ij})-N\times\zeta(0)=0.{} \end{aligned} $$

(6.25)

Proof

Referring to the proof of Theorem 6.4, one can get the bipartite solution of (6.19) similarly. Let

$$\displaystyle \begin{aligned} \zeta(t)=(1/N)(\sum_{i=1}^{N}z_{i}(t)+\sum_{i=1}^{N}\sum_{j=1}^{N}|a_{ij}|\int_{t-\tau_{ij}}^{t}h(z_{j}(s))ds)~. \end{aligned} $$

(6.26)

Combining with (6.20), we have $\dot {\zeta }(t)=0$, which implies ζ(t) = ζ(0). From ζ(0) =lim_t→∞ ζ(t), we can get the expression (6.25).

Remark 6.9

In this theorem, although the value of β cannot be given by an explicit expression, we can get a numerical solution by iterative algorithm from (6.25). In numerical examples, Example 6.10 gives a numerical solution to (6.25) for h(x) = x + 0.5sin(x), which illustrates the computational feasibility.

6.1.3 Numerical Examples

In this subsection, numerical examples will be provided to demonstrate the effectiveness of our theoretical results.

Example 6.10

Now, we will give an example to illustrate the correctness of our main results. Consider the structurally balanced graph of Fig. 6.1.For systems (6.1) and (6.19), all nonzero communication delays are listed as follows: τ ₁₂ = 0.1, τ ₁₃ = 0.3, τ ₂₁ = 0.15, τ ₂₃ = 0.2, τ ₂₄ = 0.1, τ ₃₁ = 0.11, τ ₃₂ = 0.16, τ ₃₅ = 0.23, τ ₄₂ = 0.1, τ ₄₅ = 0.2, τ ₄₇ = 0.12, τ ₅₃ = 0.1, τ ₅₄ = 0.15, τ ₅₆ = 0.24, τ ₆₅ = 0.25, and τ ₇₄ = 0.15, and the initial states are chosen as x ₁(s) = 1, x ₂(s) = 2, x ₃(s) = 3, x ₄(s) = 4, x ₅(s) = 5, x ₆(s) = −3, and x ₇(s) = −5, ∀s ∈ [−0.3, 0]. Let D = diag{−1, 1, 1, 1, 1, 1, 1}. Then, we have z(t) = [−1, 2, 3, 4, 5, −3, −5], ∀s ∈ [−0.3, 0]. Further, we define that h(s) = s + 0.5sin(s). According to Theorem 6.4, one can easily conclude that α = 1.44. Following Theorem 6.8, one can get the numerical solution β = 1.42 by iterative algorithm. Numerical results are depicted in Figs. 6.2 and 6.3, which verify our theoretical results very well.

Fig. 6.1

Structurally balanced signed undirected connectivity graphs with seven nodes

Fig. 6.2

The bipartite consensus on multi-agent system (6.1) with structurally balanced graph and linear coupling in Example 6.10

Fig. 6.3

The bipartite consensus on multi-agent system (6.19) with structurally balanced graph and nonlinear coupling in Example 6.10

Example 6.11

Now let us consider a more general network topology with 100 nodes and signed weight edges. Here two simple signed networks with 100 nodes are constructed, where one is structurally balanced and another one is structurally unbalanced. The network with structurally balanced coupling is constructed as follows: we present 20 identical circular networks with 5 nodes, whose 5 nodes are numbered 1, 2, 3, 4, 5, respectively, and their adjacency matrix is A = [a _ij]_5×5, where a _ij is chosen from (−10, 0) or (0, 10). Now the first two edges of the circular network are defined as negative edges and others are not negative edges, i.e. a ₁₂ < 0, a ₂₃ < 0, and the rest elements are non-negative Here these circular networks are arranged in a sequence. A connected graph with 100 nodes and structurally balanced coupling can be obtained by stochastic interconnections among the 3rd, 4th, and 5th nodes of adjacent circular networks. Similarly the network with structurally unbalanced coupling can be obtained according to the above method when the first three edges of pentagon are defined as negative edges and other steps are the same. All communication delays of systems (6.1) and (6.19) are uniformly distributed in (0, 1). Figure 6.4 shows that the consensus of system (6.1) can be achieved for a _ij ∈ (−10, 10) and τ _ij ∈ (0, 1). Figure 6.5 shows that the consensus of system (6.19) can also be achieved for the above conditions. Throughout this example, the nonlinear function h(x) = x + 0.5sin(x) is not changed. Figures 6.6 and 6.7 show that the bipartite consensus of systems (6.1) and (6.19) can be asymptotically reached, respectively.

Fig. 6.4

The states of system (6.1) with structurally unbalanced graph and linear coupling in Example 6.11

Fig. 6.5

The states of system (6.19) with structurally unbalanced graph and nonlinear coupling in Example 6.11

Fig. 6.6

The bipartite consensus on multi-agent system (6.1) with structurally balanced graph and linear coupling in Example 6.11

Fig. 6.7

The bipartite consensus on multi-agent system (6.19) with structurally balanced graph and nonlinear coupling in Example 6.11

6.2 Discrete-Time Multi-agent Consensus

6.2.1 Distributed Event-Based Bipartite Consensus

Consider a discrete-time multi-agent network with the dynamics described by

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} \displaystyle x_{i}(k+1)=x_{i}(k)+u_{i}(k),~~~i\in\mathcal{N},~ \end{array} \end{aligned} $$

(6.27)

where $x_{i}(k)\in \mathbb {R}$ is the state of the agent i, and u _i(k) is called the consensus protocol.

In this section, we assume that the protocol u _i(k) is based on the event-triggered information transmission. The event-triggered time sequence of the agent i is given by $t^{i}_{1},\,t^{i}_{2},\,\cdots ,\,t^{i}_{l},\,\cdots$. At each triggering time $t^{i}_{l}$, the agent i will transmit the state information to its neighboring agents. Considering time delays during the information transmission, the consensus protocol is proposed as follows:

$$\displaystyle \begin{aligned} u_{i}(k)=\sum_{j\in \mathcal{N}_{i}}|a_{ij}|(sgn(a_{ij})\widehat{x}_{j}(k-\tau_{ij})-\widehat{x}_{i}(k)),~~~~i\in\mathcal{N}, \end{aligned} $$

(6.28)

where τ _ij > 0 denotes the communication delay from agent j to i, $\widehat {x}_{j}(k-\tau _{ij})=x_{j}(k^{j}_{l^{\prime }}),\,k-\tau _{ij}\in [k^{j}_{l^{\prime }}, \, k^{j}_{l^{\prime }+1})$, and $\widehat {x}_{i}(k)=x_{i}(k^{i}_{l}),\,k\in [k_{l}^{i}, \, k_{l+1}^{i})$. It is assumed in this section that τ _ii = 0, i.e., delays exist only in the information that is actually being transmitted between two different agents. The state measurement error of agent i is defined as

$$\displaystyle \begin{aligned} \displaystyle e_{i}(k)=x_{i}(k)-\widehat{x}_{i}(k). \end{aligned} $$

(6.29)

Denote $\tau =\mbox{max}\{\tau _{ij},\,\,\,i,\,j\in \mathcal {N}\}$. The initial conditions associated with (6.27) are given as x _i(s), s = −τ, ⋯ , − 1, 0.

In this subsection, we will give the distributed event-based bipartite consensus criteria for the considered signed network model. We always assume that the network topology of the signed digraph is strongly connected in this section. Let ξ = (ξ ₁, ξ ₂, . . . , ξ _N) be the normalized left eigenvector of matrix |A| with respect to the eigenvalue 1. From Lemma 1.6, we obtain that $\displaystyle \sum _{i=1}^{N} \xi _{i}=1$ and ξ _i > 0.

Theorem 6.12

Consider the multi-agent system (6.27) with arbitrary finite communication delay τ _ijunder control law (6.28). If the first triggering time is$t^{i}_{1} = 0$ , and agent$i,~i\in \mathcal {N}$ , determines the triggering time sequence$\displaystyle t^{i}_{l}|{ }_{l=2}^{\infty }$by

$$\displaystyle \begin{aligned} \inf\left\{k> t^{i}_{l-1}:\,\,e_{i}^{2}(k)>\frac{\sigma a_{ii}^{2}}{4(1-a_{ii})}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|(\widehat{x}_{i}(k)-sgn(a_{ij})\widehat{x}_{j}(k-\tau_{ij}))^{2}\right\}, \end{aligned}$$

where 0 < σ < 1 is a constant. Then, we can obtain the following results:

• (i) System (6.27) can achieve bipartite consensus asymptotically if the signed digraph$\mathcal {G}$is structurally balanced. Moreover, the consensus value of the network is

  $$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{\sum_{i=1}^{N}\xi_{i}d_{i}x_{i}(0) +\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|\sum_{s=-\tau_{ij}}^{-1}d_{j}x_{j}(s)}{1+\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|\tau_{ij}}. \end{array} \end{aligned} $$

• (ii) If the signed digraph$\mathcal {G}$is structurally unbalanced, then the system (6.27) can achieve consensus and the final consensus value is 0, i.e.,$\displaystyle \lim _{k \to +\infty }x_{i}(k)=0,\,\forall i\in \mathcal {N}$.

Proof

We take two steps for the remaining part of the proof.

Step 1: According to Lemma 1.8, if the network structure is balanced, there exists $D=\{d_{1},\,\cdots ,\,d_{N}\}\in \mathcal {D}$, such that DAD is a stochastic matrix. Since DAD has all nonnegative entries, one can get d _i sgn(a _ij)d _j ≥ 0. Denote y _i(k) = d _i x _i(k) and $\widehat {y}_{i}(k)=d_{i}\widehat {x}_{i}(k)$. Note that $\tau _{ii}=0,\,\forall i\in \mathcal {N}$, then we can obtain that

$$\displaystyle \begin{aligned} \displaystyle y_{i}(k+1)=y_{i}(k)+\sum_{j\in \mathcal{N}_{i}}|a_{ij}|(d_{i}sgn(a_{ij})d_{j}\widehat{y}_{j}(k-\tau_{ij})-\widehat{y}_{i}(k)),~~~~i\in\mathcal {N},~ \end{aligned} $$

(6.30)

i.e.,

$$\displaystyle \begin{aligned} \displaystyle y_{i}(k+1)&=y_{i}(k)+\sum_{j\in \mathcal {N}_{i}}|a_{ij}|(\widehat{y}_{j}(k-\tau_{ij})-\widehat{y}_{i}(k))\\ &=y_{i}(k)+\sum_{j=1,\,j\neq i}^{N}|a_{ij}|(\widehat{y}_{j}(k-\tau_{ij})-\widehat{y}_{i}(k)),~~~~i\in\mathcal {N}.{} \end{aligned} $$

(6.31)

Let E _i(k) = d _i e _i(k). Referring to Theorem 1 in [22], the consensus of system (6.31) can be asymptotically reached under the event-triggered condition

$$\displaystyle \begin{aligned} \displaystyle E_{i}^{2}(k)&>\frac{\sigma a_{ii}^{2}}{4(1-a_{ii})}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|(\widehat{y}_{j}(k-\tau_{ij})-\widehat{y}_{i}(k))^{2}\\ &=\frac{\sigma a_{ii}^{2}}{4(1-a_{ii})}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|(\widehat{y}_{j}^{2}(k-\tau_{ij})+\widehat{y}_{i}^{2}(k)-2\widehat{y}_{j}(k-\tau_{ij})\widehat{y}_{i}(k))\\ &=\frac{\sigma a_{ii}^{2}}{4(1-a_{ii})}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|(\widehat{x}_{j}^{2}(k-\tau_{ij})+\widehat{x}_{i}^{2}(k) \\ & \quad -2sgn(a_{ij})\widehat{x}_{j}(k-\tau_{ij})\widehat{x}_{i}(k)),\\ &=\frac{\sigma a_{ii}^{2}}{4(1-a_{ii})}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|(sgn(a_{ij})\widehat{x}_{j}(k-\tau_{ij})-\widehat{x}_{i}(k))^{2}, ~~~~~~i\in\mathcal {N}.~{} \end{aligned} $$

(6.32)

That is, lim_k→+∞ d _i x _i(k) = c, where c is a constant value. Note that $E_{i}^{2}(k)=e_{i}^{2}(k)$. Hence, the event-triggered condition (6.32) can be rewritten as

$$\displaystyle \begin{aligned} \displaystyle e_{i}^{2}(k)>\frac{\sigma a_{ii}^{2}}{4(1-|a_{ii}|)}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|(sgn(a_{ij})\widehat{x}_{j}(k-\tau_{ij})-\widehat{x}_{i}(k))^{2}, ~~~~~~i\in\mathcal{N}.~ \end{aligned} $$

(6.33)

Therefore, under the event-triggered condition (6.33), the bipartite consensus of system (6.27) can be asymptotically reached if $\mathcal {G}$ is structurally balanced.

Next, the bipartite consensus value c of the multi-agent networks is shown below. Let $\eta (k)=\sum _{i=1}^{N}\xi _{i}y_{i}(k)+\sum _{i=1}^{N}\xi _{i}\sum _{j=1,\,j\neq i}^{N}|a_{ij}| \sum _{k-\tau _{ij}}^{k-1}\widehat {y}_{j}(s)$. Substituting (6.31) into η(k + 1), we can calculate the difference of η(k) as follows:

$$\displaystyle \begin{aligned} \begin{array}{rcl} \varDelta\eta(k)& =&\displaystyle \eta(k+1)-\eta(k)\\ & =&\displaystyle \sum_{i=1}^{N}\xi_{i}(y_{i}(k+1)-y_{i}(k))\\ & &\displaystyle +\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}|a_{ij}| (\sum_{k+1-\tau_{ij}}^{k}\widehat{y}_{j}(s)-\sum_{k-\tau_{ij}}^{k-1}\widehat{y}_{j}(s))\\ & =&\displaystyle \sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|(\widehat{y}_{j}(k-\tau_{ij})-\widehat{y}_{i}(k))\\ & &\displaystyle +\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|(\widehat{y}_{j}(k)-\widehat{y}_{j}(k-\tau_{ij})).{} \end{array} \end{aligned} $$

(6.34)

Note that the row sum of matrix |A| is 1 and {ξ ₁, ξ ₂, . . . , ξ _N} is the normalized left eigenvector of matrix |A| with respect to the eigenvalue 1, we have

$$\displaystyle \begin{aligned}\sum_{j=1}^{N}|a_{ij}|=1\,\,\, \mbox{and}\,\,\, \sum_{i=1}^{N}\xi_{i}|a_{ij}|=\xi_{j}.\end{aligned}$$

Hence, we can obtain that

$$\displaystyle \begin{aligned} \varDelta\eta(k) &=-\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|\widehat{y}_{i}(k)+\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|\widehat{y}_{j}(k)\\ &=-\sum_{i=1}^{N}\xi_{i}\widehat{y}_{i}(k)\sum_{j=1,\,j\neq i}^{N}|a_{ij}|+\sum_{i=1}^{N}\xi_{i}\sum_{j=1}^{N}|a_{ij}|\widehat{y}_{j}(k)-\sum_{i=1}^{N}\xi_{i}a_{ii}\widehat{y}_{i}(k)\\ &=-\sum_{i=1}^{N}\xi_{i}(1-a_{ii})\widehat{y}_{i}(k)+\sum_{i=1}^{N}\xi_{i}|a_{ij}|\sum_{j=1}^{N}\widehat{y}_{j}(k)-\sum_{j=1}^{N}\xi_{j}a_{jj}\widehat{y}_{j}(k)\\ &=-\sum_{i=1}^{N}\xi_{i}(1-a_{ii})\widehat{y}_{i}(k)+\sum_{j=1}^{N}\xi_{j}(1-a_{jj})\widehat{y}_{j}(k)\\ &=0. \end{aligned} $$

(6.35)

Due to Δη(k) = 0 for k ≥ 0, it can be easily obtained that η(k) is a constant. That is,

$$\displaystyle \begin{aligned} \eta(k)=\eta(0)&=\sum_{i=1}^{N}\xi_{i}y_{i}(0)+\sum_{i=1}^{N} \xi_{i}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|\sum_{-\tau_{ij}}^{-1}\widehat{y}_{j}(s)\\ &=\sum_{i=1}^{N}\xi_{i}y_{i}(0) +\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|\sum_{-\tau_{ij}}^{-1}y_{j}(s). \end{aligned} $$

Hence,

$$\displaystyle \begin{aligned} \eta(0)=\lim_{k\rightarrow+\infty}\eta(k)=c+\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|\tau_{ij} c. \end{aligned} $$

Therefore, we can conclude that

$$\displaystyle \begin{aligned} c=\frac{\sum_{i=1}^{N}\xi_{i}d_{i}x_{i}(0) +\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|\sum_{s=-\tau_{ij}}^{-1}d_{j}x_{j}(s)}{1+\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|\tau_{ij}}. \end{aligned} $$

(6.36)

This completes the proof of this step.

Step 2: If the network structure is unbalanced, according to Lemma 1.10, there does not exist $D=\{d_{1},\,\cdots ,\,d_{N}\}\in \mathcal {D}$, such that DAD is a stochastic matrix. For the sake of simplicity, the case of $\mathcal {G}$ with only one negative cycle is studied firstly. Here, we assume that this negative cycle contains an edge $a_{i_{0}j_{0}}<0$. Without loss of generality, we can assume that there exists $B=\{b_{1},\,\cdots ,\,b_{N}\}\in \mathcal {D}$, such that BAB = [b _i a _ij b _j]_N×N is a nonnegative matrix except the element $b_{i_{0}}a_{i_{0}j_{0}}b_{j_{0}}<0$. (If $\mathcal {G}$ contains k (k ≥ 2) negative cycles, there exists $D_{l}\in \mathcal {D}$ such that D _l AD _l has exactly l (1 ≤ l ≤ k) negative elements. The following proof for this case is similar to the case k = 1, and we omit it here due to space limit.) Denoting y _i(k) = b _i x _i(k) and $\widehat {y}_{i}(k)=b_{i}\widehat {x}_{i}(k)$, then we can obtain that

$$\displaystyle \begin{aligned} \displaystyle y_{i}(k+1)=y_{i}(k)+\sum_{j\in \mathcal{N}_{i}}|a_{ij}|(b_{i}sgn(a_{ij})b_{j}\widehat{y}_{j}(k-\tau_{ij})-\widehat{y}_{i}(k)),~~~~i\in\mathcal{N}.~ \end{aligned} $$

(6.37)

Define the matrix W = [w _ij]_N×N as follows: $w_{i_{0}j_{0}}=0$, $w_{ii}=1-\sum _{j=1}^{N}w_{ij}, \forall i\in \mathcal {N}$, and w _ij = b _i a _ij b _j otherwise. Let $E_{i}(k)=y_{i}(k)-\widehat {y}_{i}(k)$. Consider the Lyapunov functional as

$$\displaystyle \begin{aligned} V(k)=V_{1}(k)+V_{2}(k), \end{aligned} $$

(6.38)

where

$$\displaystyle \begin{aligned} V_{1}(k)=\sum_{i=1}^{N}\xi_{i}y_{i}^{2}(k), \end{aligned} $$

(6.39)

and

$$\displaystyle \begin{aligned} V_{2}(k)=\sum_{i=1}^{N}\xi_{i}\sum_{j=1}^{N}|a_{ij}|\sum_{s=k-\tau_{ij}}^{k-1}\widehat{y}_{j}^{2}(s). \end{aligned} $$

(6.40)

Notice that $w_{i_{0}i_{0}}=a_{i_{0}i_{0}}+|a_{i_{0}j_{0}}|$, and difference of V (k) along the solution of (6.37) gives that

$$\displaystyle \begin{aligned} \varDelta V(k) &\leq-\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}\sum_{l>j,\,l\neq i}^{N}w_{ij}w_{il}(\widehat{y}_{j}(k-\tau_{ij}) -\widehat{y}_{l}(k-\tau_{il}))^{2}\\ &\quad -\xi_{i_{0}}|a_{i_{0}j_{0}}|\sum_{l\neq i_{0},\,l\neq j_{0}}^{N}|a_{i_{0}l}|(\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}})-\widehat{y}_{l}(k-\tau_{i_{0}l}))^{2}\\ &\quad -\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|(a_{ii}-\alpha_{i}) (\widehat{y}_{i}(k) -b_{i}sgn(a_{ij})b_{j}\widehat{y}_{j}(k-\tau_{ij}))^{2}\\ &\quad +\sum_{i=1}^{N}\xi_{i}(1-a_{ii})\frac{1}{\alpha_{i}}E_{i}^{2}(k). \end{aligned} $$

(6.41)

Actually,

$$\displaystyle \begin{aligned} \varDelta V(k)&=\varDelta V_{1}(k)+\varDelta V_{2}(k). \end{aligned} $$

(6.42)

Note that for $i\in \mathcal {N}$, τ _ii = 0, and $E_{i}(k)=y_{i}(k)-\widehat {y}_{i}(k)$, it holds that

$$\displaystyle \begin{aligned} \displaystyle y_{i}(k+1)&=y_{i}(k)+\sum_{j\in \mathcal {N}_{i}}|a_{ij}|(b_{i}sgn(a_{ij})b_{j}\widehat{y}_{j}(k-\tau_{ij})-\widehat{y}_{i}(k))\\ &=y_{i}(k)+\sum_{j=1}^{N}|a_{ij}|(b_{i}sgn(a_{ij})b_{j}\widehat{y}_{j}(k-\tau_{ij})-\widehat{y}_{i}(k))\\ &=y_{i}(k)-\widehat{y}_{i}(k)+\sum_{j=1}^{N}|a_{ij}|b_{i}sgn(a_{ij})b_{j}\widehat{y}_{j}(k-\tau_{ij}), \end{aligned} $$

and

$$\displaystyle \begin{aligned} \displaystyle \sum_{i=1}^{N}\xi_{i}y_{i}^{2}(k+1)=&\sum_{i=1}^{N}\xi_{i}[E_{i}(k)+\sum_{j=1}^{N}|a_{ij}|b_{i}sgn(a_{ij})b_{j}\widehat{y}_{j}(k-\tau_{ij})]^{2}\\ =&\sum_{i=1,i\neq i_{0}}^{N}\xi_{i}[E_{i}(k)+\sum_{j=1}^{N}w_{ij}\widehat{y}_{j}(k-\tau_{ij})]^{2}+\xi_{i_{0}}[E_{i_{0}}(k)\\ &+\sum_{j=1}^{N}w_{i_{0}j}\widehat{y}_{j}(k-\tau_{i_{0}j}) -|a_{i_{0}j_{0}}|(\widehat{y}_{i_{0}}(k)+\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}))]^{2}. \end{aligned} $$

Hence, we have

$$\displaystyle \begin{aligned} \varDelta V_{1}(k) =&\sum_{i=1}^{N}\xi_{i}y_{i}^{2}(k+1)-\sum_{i=1}^{N}\xi_{i}y_{i}^{2}(k)\\ =&\sum_{i=1}^{N}\xi_{i}[E_{i}(k)\!+\!\sum_{j=1}^{N}w_{ij}\widehat{y}_{j}(k-\tau_{ij})]^{2}\!+\!\xi_{i_{0}}[a^{2}_{i_{0}j_{0}}(\widehat{y}_{i_{0}}(k)+ \widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}))^{2} \\&-2|a_{i_{0}j_{0}}|(\widehat{y}_{i_{0}}(k) +\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}))(E_{i_{0}}(k)+ \sum_{j=1}^{N}w_{i_{0}j}\widehat{y}_{j}(k-\tau_{i_{0}j}))]\\& -\sum_{i=1}^{N}\xi_{i}y_{i}^{2}(k)\\ =&\sum_{i=1}^{N}\xi_{i}[E_{i}^{2}(k)+\sum_{j=1,\,j\neq i}^{N}w_{ij}^{2}\widehat{y}_{j}^{2}(k-\tau_{ij})+w_{ii}\widehat{y}_{i}^{2}(k) \\&+2\sum_{j=1,\,j\neq i}^{N} \sum_{l>j,\,l\neq i}^{N}w_{ij}w_{il}\cdot \widehat{y}_{j}(k-\tau_{ij})\widehat{y}_{l}(k-\tau_{ij}) \\&+2\sum_{j=1,\,j\neq i}^{N}w_{ij}w_{ii}\widehat{y}_{j}(k-\tau_{ij}) \widehat{y}_{i}(k)+2w_{ii}\widehat{y}_{i}(k)E_{i}(k)\\ &+2\sum_{j=1,\,j\neq i}^{N}w_{ij}\widehat{y}_{j}(k-\tau_{ij})E_{i}(k)-\sum_{i=1}^{N}\xi_{i}[\widehat{y}_{i}^{2}(k)\\ &+E_{i}^{2}(k)+2\widehat{y}_{i}(k)E_{i}(k)]+\xi_{i_{0}}[a^{2}_{i_{0}j_{0}}(\widehat{y}_{i_{0}}(k)+\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}))^{2}\\& -2|a_{i_{0}j_{0}}|(\widehat{y}_{i_{0}}(k) +\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}))(E_{i_{0}}(k)+\sum_{j=1}^{N}w_{i_{0}j}\widehat{y}_{j}(k-\tau_{i_{0}j}))], \end{aligned} $$

(6.43)

and

$$\displaystyle \begin{aligned} \varDelta V_{2}(k)=&\sum_{i=1}^{N}\xi_{i}\sum_{j=1}^{N}|a_{ij}|[\sum_{k+1-\tau_{ij}}^{k}\widehat{y}_{j}^{2}(s)-\sum_{k-\tau_{ij}}^{k-1}\widehat{y}_{j}^{2}(s)]\\ =&\sum_{i=1}^{N}\xi_{i}\sum_{j=1}^{N}|a_{ij}|[\widehat{y}_{j}^{2}(k)-\widehat{y}_{j}^{2}(k-\tau_{ij})]\\ =&\frac{1}{2}\sum_{i=1}^{N}\xi_{i}\sum_{j=1}^{N}\sum_{l=1}^{N}|a_{ij}||a_{il}|[\widehat{y}_{j}^{2}(k)-\widehat{y}_{j}^{2}(k-\tau_{ij}) +\widehat{y}_{l}^{2}(k)-\widehat{y}_{l}^{2}(k-\tau_{il})]\\ =&\sum_{i=1}^{N}\xi_{i}[\sum_{j=1,\,j\neq i}^{N}a_{ij}^{2}(\widehat{y}_{j}^{2}(k)-\widehat{y}_{j}^{2}(k-\tau_{ij}))+\sum_{j=1,\,j\neq i}^{N}\sum_{l>j,\,l\neq i}^{N}|a_{ij}||a_{il}|\\ &(\widehat{y}_{j}^{2}(k)-\widehat{y}_{j}^{2}(k-\tau_{ij})+\widehat{y}_{l}^{2}(k)-\widehat{y}_{l}^{2}(k-\tau_{ij})) +\sum_{j=1,\,j\neq i}^{N}|a_{ij}||a_{ii}|(\widehat{y}_{j}^{2}(k) \\ &-\widehat{y}_{j}^{2}(k-\tau_{ij}))]\\ =&\sum_{i=1}^{N}\xi_{i}[\sum_{j=1,\,j\neq i}^{N}w_{ij}^{2}(\widehat{y}_{j}^{2}(k)-\widehat{y}_{j}^{2}(k-\tau_{ij}))+\sum_{j=1,\,j\neq i}^{N}\sum_{l>j,\,l\neq i}^{N}w_{ij}w_{il}(\widehat{y}_{j}^{2}(k)\\ &-\widehat{y}_{j}^{2}(k-\tau_{ij})+\widehat{y}_{l}^{2}(k)-\widehat{y}_{l}^{2}(k-\tau_{il})) +\sum_{j=1,\,j\neq i}^{N}w_{ij}w_{ii}(\widehat{y}_{j}^{2}(k)\\ &-\widehat{y}_{j}^{2}(k-\tau_{ij}))]\!+\!\xi_{i_{0}}[a^{2}_{i_{0}j_{0}}(\widehat{y}^{2}_{j_{0}}(k)\!-\!\widehat{y}^{2}_{j_{0}}(k-\tau_{i_{0}j_{0}})) \!+\!|a_{i_{0}j_{0}}||a_{i_{0}i_{0}}|(\widehat{y}^{2}_{j_{0}}(k)\\ &-\widehat{y}^{2}_{j_{0}}(k-\tau_{i_{0}j_{0}})) -|a_{i_{0}j_{0}}|\sum_{j=1,\,j\neq i_{0}}^{N}w_{i_{0}j}(\widehat{y}^{2}_{j}(k)-\widehat{y}^{2}_{j}(k-\tau_{i_{0}j})) \\ &+|a_{i_{0}j_{0}}|\sum_{l\neq i_{0},\,l\neq j_{0}}^{N}|a_{i_{0}l}|(\widehat{y}_{j_{0}}^{2}(k)-\widehat{y}_{j_{0}}^{2}(k-\tau_{i_{0}j_{0}})\!+\!\widehat{y}_{l}^{2}(k)-\widehat{y}_{l}^{2}(k-\tau_{i_{0}l}))]. \end{aligned} $$

(6.44)

Let

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} \varDelta_{1} & =&\displaystyle \sum_{i=1}^{N}\xi_{i}[\sum_{j=1,\,j\neq i}^{N}w_{ij}^{2}\widehat{y}_{j}^{2}(k-\tau_{ij})+w_{ii}\widehat{y}_{i}^{2}(k) +2\sum_{j=1,\,j\neq i}^{N}\sum_{l>j,\,l\neq i}^{N}w_{ij}w_{il}\\ & &\displaystyle \widehat{y}_{j}(k-\tau_{ij})\widehat{y}_{l}(k-\tau_{ij}) +2\sum_{j=1,\,j\neq i}^{N}w_{ij}w_{ii}\widehat{y}_{j}(k-\tau_{ij})\widehat{y}_{i}(k)-\sum_{i=1}^{N}\xi_{i}\widehat{y}_{i}^{2}(k)\\ & &\displaystyle +\sum_{i=1}^{N}\xi_{i}[\sum_{j=1,\,j\neq i}^{N}w_{ij}^{2} (\widehat{y}_{j}^{2}(k)-\widehat{y}_{j}^{2}(k-\tau_{ij})) +\sum_{j=1,\,j\neq i}^{N}\sum_{l>j,\,l\neq i}^{N}w_{ij}w_{il}(\widehat{y}_{j}^{2}(k)\\ & &\displaystyle -\widehat{y}_{j}^{2}(k-\tau_{ij})\!+\!\widehat{y}_{l}^{2}(k)\!-\!\widehat{y}_{l}^{2}(k-\tau_{il})) \!+\!\sum_{j=1,\,j\neq i}^{N}w_{ij}w_{ii}(\widehat{y}_{j}^{2}(k) -\widehat{y}_{j}^{2}(k-\tau_{ij}))]\\ & =&\displaystyle \sum_{i=1}^{N}\xi_{i}[\sum_{j=1,\,j\neq i}^{N}w_{ij}^{2}\widehat{y}_{j}^{2}(k)+w_{ii}^{2}\widehat{y}_{i}^{2}(k)+\sum_{j=1,\,j\neq i}^{N}\sum_{l>j,\,l\neq i}^{N}w_{ij}w_{il}(\widehat{y}_{j}^{2}(k)+\widehat{y}_{l}^{2}(k)\\ & &\displaystyle -\widehat{y}_{j}^{2}(k-\tau_{ij})-\widehat{y}_{l}^{2}(k-\tau_{il})+2\widehat{y}_{j}(k-\tau_{ij})\widehat{y}_{l}(k-\tau_{ij})) +\sum_{j=1,\,j\neq i}^{N}w_{ij}w_{ii}\\ & &\displaystyle \cdot(\widehat{y}_{j}^{2}(k)-\widehat{y}_{j}^{2}(k-\tau_{ij}) +2\widehat{y}_{j}(k-\tau_{ij})\widehat{y}_{i}(k))]-\sum_{i=1}^{N}\xi_{i}\widehat{y}_{i}^{2}(k)\\ & =&\displaystyle \sum_{i=1}^{N}\xi_{i}[\sum_{j=1}^{N}w_{ij}^{2}\widehat{y}_{j}^{2}(k)+\sum_{j=1,\,j\neq i}^{N}\sum_{l>j,\,l\neq i}^{N}w_{ij}w_{il}(\widehat{y}_{j}^{2}(k)+\widehat{y}_{l}^{2}(k))+\sum_{j< i}^{N}w_{ij}w_{ii}\\ & &\displaystyle \cdot(\widehat{y}_{j}^{2}(k)+\widehat{y}_{i}^{2}(k)) +\sum_{l> i}^{N}w_{il}w_{ii}(\widehat{y}_{l}^{2}(k)+\widehat{y}_{i}^{2}(k))-\widehat{y}_{i}^{2}(k)]\\ & &\displaystyle -\sum_{i=1}^{N}\xi_{i}[\sum_{j=1,\,j\neq i}^{N}\sum_{l>j,\,l\neq i}^{N}\\ & &\displaystyle w_{ij}w_{il}(\widehat{y}_{j}(k-\tau_{ij}) -\widehat{y}_{l}(k-\tau_{il}))^{2}+\sum_{j=1,\,j\neq i}^{N}w_{ij}w_{ii}(\widehat{y}_{i}(k)-\widehat{y}_{j}(k-\tau_{ij}))^{2}]\\ & =&\displaystyle \sum_{i=1}^{N}\xi_{i}[\sum_{j=1}^{N}w_{ij}^{2}\widehat{y}_{j}^{2}(k)+\sum_{j=1}^{N}\sum_{l=1,\,l\neq j}^{N}w_{ij}w_{il}\widehat{y}_{j}^{2}(k) -\widehat{y}_{i}^{2}(k)]\\ & &\displaystyle -\sum_{i=1}^{N}\xi_{i}[\sum_{j=1,\,j\neq i}^{N}\sum_{l>j,\,l\neq i}^{N}\\ & &\displaystyle w_{ij}w_{il}(\widehat{y}_{j}(k-\tau_{ij}) -\widehat{y}_{l}(k-\tau_{il}))^{2}+\sum_{j=1,\,j\neq i}^{N}w_{ij}w_{ii}(\widehat{y}_{i}(k)-\widehat{y}_{j}(k-\tau_{ij}))^{2}]\\ & =&\displaystyle \sum_{i=1}^{N}\xi_{i}[\sum_{j=1}^{N}\sum_{l=1}^{N}w_{ij}w_{il}\widehat{y}_{j}^{2}(k) -\widehat{y}_{i}^{2}(k)]\\ & &\displaystyle -\sum_{i=1}^{N}\xi_{i}[\sum_{j=1,\,j\neq i}^{N}\sum_{l>j,\,l\neq i}^{N}w_{ij}w_{il}(\widehat{y}_{j}(k-\tau_{ij}) \\ & &\displaystyle -\widehat{y}_{l}(k-\tau_{il}))^{2}+\sum_{j=1,\,j\neq i}^{N}w_{ij}w_{ii}(\widehat{y}_{i}(k)-\widehat{y}_{j}(k-\tau_{ij}))^{2}]\\ & =&\displaystyle \sum_{i=1}^{N}\xi_{i}[\sum_{j=1}^{N}w_{ij}\widehat{y}_{j}^{2}(k) -\widehat{y}_{i}^{2}(k)]-\sum_{i=1}^{N}\xi_{i}[\sum_{j=1,\,j\neq i}^{N}\sum_{l>j,\,l\neq i}^{N}w_{ij}w_{il}(\widehat{y}_{j}(k-\tau_{ij}) \\ & &\displaystyle -\widehat{y}_{l}(k-\tau_{il}))^{2} +\sum_{j=1,\,j\neq i}^{N}w_{ij}w_{ii}(\widehat{y}_{i}(k)-\widehat{y}_{j}(k-\tau_{ij}))^{2}], \end{array} \end{aligned} $$

(6.45)

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} \varDelta_{2}& =&\displaystyle \xi_{i_{0}}[a^{2}_{i_{0}j_{0}}(\widehat{y}_{i_{0}}(k)\!+\!\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}))^{2}-2|a_{i_{0}j_{0}}|(\widehat{y}_{i_{0}}(k) +\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}))\sum_{j=1}^{N}w_{i_{0}j}\\ & &\displaystyle \cdot\widehat{y}_{j}(k-\tau_{i_{0}j})]+\xi_{i_{0}}[a^{2}_{i_{0}j_{0}}(\widehat{y}^{2}_{j_{0}}(k)-\widehat{y}^{2}_{j_{0}}(k-\tau_{i_{0}j_{0}})) +|a_{i_{0}j_{0}}||a_{i_{0}i_{0}}|(\widehat{y}^{2}_{j_{0}}(k) \\ & &\displaystyle -\widehat{y}^{2}_{j_{0}}(k-\tau_{i_{0}j_{0}}))-|a_{i_{0}j_{0}}|\sum_{j=1,\,j\neq i_{0}}^{N}w_{i_{0}j}(\widehat{y}^{2}_{j}(k)-\widehat{y}^{2}_{j}(k-\tau_{i_{0}j}))\\ & &\displaystyle +|a_{i_{0}j_{0}}|\sum_{l\neq i_{0},\,l\neq j_{0}}^{N}|a_{i_{0}l}|(\widehat{y}_{j_{0}}^{2}(k)-\widehat{y}_{j_{0}}^{2}(k-\tau_{i_{0}j_{0}})+\widehat{y}_{l}^{2}(k)-\widehat{y}_{l}^{2}(k-\tau_{i_{0}l}))]\\ & =&\displaystyle \xi_{i_{0}}[a^{2}_{i_{0}j_{0}}(\widehat{y}^{2}_{i_{0}}(k)+2\widehat{y}_{i_{0}}(k)\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}))-2|a_{i_{0}j_{0}}|w_{i_{0}i_{0}}\widehat{y}^{2}_{i_{0}}(k) -2|a_{i_{0}j_{0}}|w_{i_{0}i_{0}}\\ & &\displaystyle \cdot\widehat{y}_{i_{0}}(k)\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}) +a^{2}_{i_{0}j_{0}}\widehat{y}^{2}_{j_{0}}(k)+|a_{i_{0}j_{0}}||a_{i_{0}i_{0}}|(\widehat{y}^{2}_{j_{0}}(k)+\widehat{y}^{2}_{i_{0}}(k)\\ & &\displaystyle -\widehat{y}^{2}_{j_{0}}(k-\tau_{i_{0}j_{0}})L-\widehat{y}^{2}_{i_{0}}(k)) -|a_{i_{0}j_{0}}|\sum_{j=1,\,j\neq i_{0}}^{N}w_{i_{0}j}(\widehat{y}^{2}_{j}(k)+\widehat{y}^{2}_{i_{0}}(k))\\ & &\displaystyle -|a_{i_{0}j_{0}}|\sum_{l\neq i_{0},\,l\neq j_{0}}^{N}|a_{i_{0}l}|(\widehat{y}_{j_{0}}^{2}(k)\\ & &\displaystyle +\widehat{y}_{l}^{2}(k))+|a_{i_{0}j_{0}}|\sum_{j=1,\,j\neq i_{0}}^{N}w_{i_{0}j}(-\widehat{y}^{2}_{i_{0}}(k)-\widehat{y}^{2}_{j}(k-\tau_{i_{0}j})\\ & &\displaystyle -2\widehat{y}_{i_{0}}(k)\widehat{y}_{j}(k-\tau_{i_{0}j}))\\ & &\displaystyle +|a_{i_{0}j_{0}}|\sum_{l\neq i_{0},\,l\neq j_{0}}^{N}|a_{i_{0}l}|(-\widehat{y}_{j_{0}}^{2}(k-\tau_{i_{0}j_{0}})\\ & &\displaystyle -\widehat{y}_{l}^{2}(k-\tau_{i_{0}l})-2\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}})\widehat{y}_{l}(k-\tau_{i_{0}j})]\\ & =&\displaystyle \xi_{i_{0}}[a^{2}_{i_{0}j_{0}}\widehat{y}^{2}_{i_{0}}(k)-2|a_{i_{0}j_{0}}|w_{i_{0}i_{0}}\widehat{y}^{2}_{i_{0}}(k)+a^{2}_{i_{0}j_{0}}\widehat{y}^{2}_{j_{0}}(k)+|a_{i_{0}j_{0}}||a_{i_{0}i_{0}}|(\widehat{y}^{2}_{j_{0}}(k) \\ & &\displaystyle +\widehat{y}^{2}_{i_{0}}(k))+|a_{i_{0}j_{0}}||a_{i_{0}i_{0}}|(-\widehat{y}^{2}_{i_{0}}(k)-\widehat{y}^{2}_{j_{0}}(k-\tau_{i_{0}j_{0}})\\ & &\displaystyle -2\widehat{y}_{i_{0}}(k) \widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}))-|a_{i_{0}j_{0}}|\sum_{j=1,\,j\neq i_{0}}^{N}\\ & &\displaystyle w_{i_{0}j}(\widehat{y}^{2}_{j}(k)+\widehat{y}^{2}_{i_{0}}(k))+|a_{i_{0}j_{0}}|\sum_{l\neq i_{0},\,l\neq j_{0}}^{N}|a_{i_{0}l}|(\widehat{y}_{j_{0}}^{2}(k)+\widehat{y}_{l}^{2}(k))\\ & &\displaystyle +|a_{i_{0}j_{0}}|\sum_{j=1,\,j\neq i_{0}}^{N}w_{i_{0}j}\cdot(\widehat{y}_{i_{0}}(k)\\ & &\displaystyle -\widehat{y}_{j}(k-\tau_{i_{0}j}))^{2} -|a_{i_{0}j_{0}}|\sum_{l\neq i_{0},\,l\neq j_{0}}^{N}|a_{i_{0}l}|(\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}})-\widehat{y}_{l}(k-\tau_{i_{0}l}))^{2}]\\ & =&\displaystyle \xi_{i_{0}}[-|a_{i_{0}j_{0}}|\widehat{y}^{2}_{i_{0}}(k)+|a_{i_{0}j_{0}}|\widehat{y}^{2}_{j_{0}}(k)-|a_{i_{0}j_{0}}||a_{i_{0}i_{0}}|(\widehat{y}_{i_{0}}(k)\\ & &\displaystyle \quad +\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}))^{2}+|a_{i_{0}j_{0}}|\\ & &\displaystyle \cdot \sum_{j=1,\,j\neq i_{0}}^{N}w_{i_{0}j}(\widehat{y}_{i_{0}}(k)-\widehat{y}_{j}(k-\tau_{i_{0}j}))^{2} -|a_{i_{0}j_{0}}|\sum_{l\neq i_{0},\,l\neq j_{0}}^{N}|a_{i_{0}l}|(\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}})\\ & &\displaystyle -\widehat{y}_{l}(k-\tau_{i_{0}l}))^{2}], \end{array} \end{aligned} $$

(6.46)

and

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} \varDelta_{3}& =&\displaystyle \sum_{i=1}^{N}\xi_{i}[2w_{ii}\widehat{y}_{i}(k)E_{i}(k)+2\sum_{j=1,\,j\neq i}^{N}w_{ij}\widehat{y}_{j}(k-\tau_{ij})E_{i}(k)]-2\sum_{i=1}^{N}\xi_{i}\widehat{y}_{i}(k)\\& &\displaystyle E_{i}(k) -2|a_{i_{0}j_{0}}|(\widehat{y}_{i_{0}}(k) +\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}))E_{i_{0}}(k)\\ & =&\displaystyle 2\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}w_{ij}E_{i}(k)(\widehat{y}_{j}(k-\tau_{ij})-\widehat{y}_{i}(k))-2\xi_{i_{0}}|a_{i_{0}j_{0}}|E_{i_{0}}(k)(\widehat{y}_{i_{0}}(k) \\ & &\displaystyle +\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}))\\ & \leq&\displaystyle \sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}w_{ij}[\frac{1}{\alpha_{i}}E_{i}^{2}(k)+\alpha_{i}(\widehat{y}_{j}(k-\tau_{ij})-\widehat{y}_{i}(k))^{2}]+\xi_{i_{0}}|a_{i_{0}j_{0}}|\\& &\displaystyle [\frac{1}{\alpha_{i_{0}}}E_{i_{0}}^{2}(k) +\alpha_{i_{0}}(\widehat{y}_{i_{0}}(k) +\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}))^{2}]. \end{array} \end{aligned} $$

(6.47)

Substituting (6.45), (6.46), and (6.47) into (6.42), we can obtain that

$$\displaystyle \begin{aligned} \begin{array}{rcl} \varDelta V(k) & =&\displaystyle \varDelta_{1}+\varDelta_{2}+\varDelta_{3}\\ & =&\displaystyle \sum_{i=1}^{N}\xi_{i}[\sum_{j=1}^{N}w_{ij}\widehat{y}_{j}^{2}(k) -\widehat{y}_{i}^{2}(k)]+\xi_{i_{0}}[-|a_{i_{0}j_{0}}|\widehat{y}^{2}_{i_{0}}(k)+|a_{i_{0}j_{0}}|\widehat{y}^{2}_{j_{0}}(k)] \\ & &\displaystyle -\sum_{i=1}^{N}\xi_{i}[\sum_{j=1,\,j\neq i}^{N}\sum_{l>j,\,l\neq i}^{N}w_{ij}w_{il}(\widehat{y}_{j}(k-\tau_{ij}) -\widehat{y}_{l}(k-\tau_{il}))^{2} \\& &\displaystyle +\sum_{j=1,\,j\neq i}^{N}w_{ij}w_{ii}(\widehat{y}_{i}(k)-\widehat{y}_{j}(k-\tau_{ij}))^{2}]+ \xi_{i_{0}}[-|a_{i_{0}j_{0}}||a_{i_{0}i_{0}}|(\widehat{y}_{i_{0}}(k)\\& &\displaystyle +\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}))^{2} +|a_{i_{0}j_{0}}|\sum_{j=1,\,j\neq i_{0}}^{N}w_{i_{0}j}(\widehat{y}_{i_{0}}(k)-\widehat{y}_{j}(k-\tau_{i_{0}j}))^{2} \\& &\displaystyle -|a_{i_{0}j_{0}}|\sum_{l\neq i_{0},\,l\neq j_{0}}^{N}|a_{i_{0}l}|(\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}) -\widehat{y}_{l}(k-\tau_{i_{0}l}))^{2}]\\& &\displaystyle +2\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}w_{ij}E_{i}(k)(\widehat{y}_{j}(k-\tau_{ij})-\widehat{y}_{i}(k))\\& &\displaystyle -2\xi_{i_{0}}|a_{i_{0}j_{0}}|E_{i_{0}}(k)(\widehat{y}_{i_{0}}(k) +\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}))\\ & =&\displaystyle \sum_{i=1}^{N}\xi_{i}\sum_{j=1}^{N}|a_{ij}|\widehat{y}_{j}^{2}(k) -\sum_{i=1}^{N}\xi_{i}\widehat{y}_{i}^{2}(k)-\sum_{i=1}^{N}\xi_{i}[\sum_{j=1,\,j\neq i}^{N}\sum_{l>j,\,l\neq i}^{N}w_{ij}w_{il}\cdot\\& &\displaystyle (\widehat{y}_{j}(k-\tau_{ij}) -\widehat{y}_{l}(k-\tau_{il}))^{2}+\sum_{j=1,\,j\neq i}^{N}w_{ij}a_{ii}(\widehat{y}_{i}(k)-\widehat{y}_{j}(k-\tau_{ij}))^{2}] \\& &\displaystyle -\xi_{i_{0}}[|a_{i_{0}j_{0}}||a_{i_{0}i_{0}}|(\widehat{y}_{i_{0}}(k) +\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}))^{2}\\& &\displaystyle +|a_{i_{0}j_{0}}|\sum_{l\neq i_{0},\,l\neq j_{0}}^{N}|a_{i_{0}l}|(\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}})-\widehat{y}_{l}(k-\tau_{i_{0}l}))^{2}] \\& &\displaystyle +2\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}w_{ij}E_{i}(k)(\widehat{y}_{j}(k-\tau_{ij})-\widehat{y}_{i}(k))\\& &\displaystyle -2\xi_{i_{0}}|a_{i_{0}j_{0}}|E_{i_{0}}(k)(\widehat{y}_{i_{0}}(k) +\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}))\\ & \leq&\displaystyle -\sum_{i=1}^{N}\xi_{i}[\sum_{j=1,\,j\neq i}^{N}\sum_{l>j,\,l\neq i}^{N}w_{ij}w_{il}(\widehat{y}_{j}(k-\tau_{ij}) -\widehat{y}_{l}(k-\tau_{il}))^{2}\\ & &\displaystyle +\sum_{j=1,\,j\neq i}^{N}w_{ij}(a_{ii}-\alpha_{i})(\widehat{y}_{i}(k)-\widehat{y}_{j}(k-\tau_{ij}))^{2}]\\ & &\displaystyle -\xi_{i_{0}}[|a_{i_{0}j_{0}}|(a_{i_{0}i_{0}}-\alpha_{i_{0}})(\widehat{y}_{i_{0}}(k) +\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}))^{2}\\ & &\displaystyle +|a_{i_{0}j_{0}}|\sum_{l\neq i_{0},\,l\neq j_{0}}^{N}|a_{i_{0}l}|(\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}}) -\widehat{y}_{l}(k-\tau_{i_{0}l}))^{2}]\\ & &\displaystyle +\xi_{i_{0}}|a_{i_{0}j_{0}}|\frac{1}{\alpha_{i_{0}}}E_{i_{0}}^{2}(k)+\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}w_{ij}\frac{1}{\alpha_{i}}E_{i}^{2}(k)\\ & =&\displaystyle -\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}\sum_{l>j,\,l\neq i}^{N}w_{ij}w_{il}(\widehat{y}_{j}(k-\tau_{ij}) -\widehat{y}_{l}(k-\tau_{il}))^{2}\\& &\displaystyle -\xi_{i_{0}}|a_{i_{0}j_{0}}|\sum_{l\neq i_{0},\,l\neq j_{0}}^{N}|a_{i_{0}l}|(\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}})-\widehat{y}_{l}(k-\tau_{i_{0}l}))^{2} \\& &\displaystyle -\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|(a_{ii}-\alpha_{i})(\widehat{y}_{i}(k) -b_{i}sgn(a_{ij})b_{j}\widehat{y}_{j}(k-\tau_{ij}))^{2}\\& &\displaystyle +\sum_{i=1}^{N}\xi_{i}(1-a_{ii})\frac{1}{\alpha_{i}}E_{i}^{2}(k), \end{array} \end{aligned} $$

(6.48)

which implies (6.41) holds.

Let $f(\alpha _{i})=\frac {\alpha _{i}(a_{ii}-\alpha _{i})}{1-a_{ii}}$. We aim to reduce the number of event-triggering time instants as much as possible when the parameter α _i is chosen. That is to say, the event-triggered condition needs to be more difficult to be satisfied when we select the parameter α _i. To realize this objective, we choose $\alpha _{i}=\frac {a_{ii}}{2}$ such that f(α _i) can be maximized. Note that the event-triggered condition (6.32) can be rewritten as follows for $i\in \mathcal {N}$:

$$\displaystyle \begin{aligned} \displaystyle E_{i}^{2}(k)=e_{i}^{2}(k)&>\frac{\sigma a_{ii}^{2}}{4(1-a_{ii})}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|(sgn(a_{ij})\widehat{x}_{j}(k-\tau_{ij})-\widehat{x}_{i}(k))^{2}\\ &=\frac{\sigma a_{ii}^{2}}{4(1-a_{ii})}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|(b_{i}sgn(a_{ij})b_{j}\widehat{y}_{j}(k-\tau_{ij})-\widehat{y}_{i}(k))^{2}. \end{aligned} $$

(6.49)

Under the event-triggered condition (6.49), we have that

$$\displaystyle \begin{aligned} & \displaystyle -\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|(a_{ii}-\alpha_{i})(\widehat{y}_{i}(k)-b_{i}sgn(a_{ij})b_{j}\widehat{y}_{j}(k-\tau_{ij}))^{2}\\&~~~~~+\sum_{i=1}^{N}\xi_{i}(1-a_{ii})\frac{1}{\alpha_{i}}E_{i}^{2}(k)\\ &\leq- \sum_{i=1}^{N}\xi_{i}(a_{ii}-\frac{a_{ii}}{2})\frac{4(1-a_{ii})}{\sigma a_{ii}^{2}}e_{i}^{2}(k)+\sum_{i=1}^{N}\xi_{i}(1-a_{ii})\frac{2}{a_{ii}}E_{i}^{2}(k)\\ &=- \sum_{i=1}^{N}\xi_{i}\frac{2(1-a_{ii})}{a_{ii}}(\frac{1}{\sigma}-1)E_{i}^{2}(k). \end{aligned} $$

(6.50)

Note that 0 < σ < 1. Hence, under the trigger condition (6.49), it holds that for ∀k ≥ 0,

$$\displaystyle \begin{aligned} \varDelta V(k)\leq&-\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}\sum_{l>j,\,l\neq i}^{N}w_{ij}w_{il}(\widehat{y}_{j}(k-\tau_{ij}) -\widehat{y}_{l}(k-\tau_{il})^{2}\\ &-\xi_{i_{0}}|a_{i_{0}j_{0}}|\cdot \sum_{l\neq i_{0},\,l\neq j_{0}}^{N}|a_{i_{0}l}|(\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}})-\widehat{y}_{l}(k-\tau_{i_{0}l}))^{2}\\ &-\sum_{i=1}^{N}\xi_{i}\frac{2(1-a_{ii})}{a_{ii}}\cdot(\frac{1}{\sigma}-1)E_{i}^{2}(k)\\ \leq& 0. \end{aligned} $$

(6.51)

According to LaSalle’s invariance principle, all the agents in the network will converge to the maximal positively invariant set of Φ = {θ ∈ Υ _−τ, x(k + θ) ∈ X : ΔV (k) = 0} asymptotically. Note that ΔV (k) = 0 if and only if e _i(k) = 0,

$$\displaystyle \begin{aligned} &-\sum_{i=1}^{N}\xi_{i}\sum_{j=1,\,j\neq i}^{N}\sum_{l>j,\,l\neq i}^{N}w_{ij}w_{il}(\widehat{y}_{j}(k-\tau_{ij}) -\widehat{y}_{l}(k-\tau_{il})^{2}-\xi_{i_{0}}|a_{i_{0}j_{0}}|\sum_{l\neq i_{0},\,l\neq j_{0}}^{N}\\&|a_{i_{0}l}|(\widehat{y}_{j_{0}}(k-\tau_{i_{0}j_{0}})-\widehat{y}_{l}(k-\tau_{i_{0}l}))^{2}=0, \end{aligned} $$

(6.52)

and

$$\displaystyle \begin{aligned} \sum_{i=1}^{N}\xi_{i}\frac{a_{ii}}{2}\sum_{j=1,\,j\neq i}^{N}a_{ij}(b_{i}sgn(a_{ij})b_{j}\widehat{y}_{j}(k-\tau_{ij})-\widehat{y}_{i}(k))^{2}=0. \end{aligned} $$

(6.53)

Hence, ΔV (k) = 0 if and only if e _i(k) = 0, and

$$\displaystyle \begin{aligned} \widehat{y}_{j}(k-\tau_{ij})=sgn(a_{ij})\widehat{y}_{i}(k),\,\,\forall i,\,j\in\mathcal{N}_{i}, \end{aligned} $$

(6.54)

or

$$\displaystyle \begin{aligned} \widehat{y}_{j}(k-\tau_{ij})=\widehat{y}_{i}(k),\,if\, \,w_{ij}>0, \end{aligned} $$

(6.55)

$$\displaystyle \begin{aligned} \widehat{y}_{j_{0}}(k-\tau_{ij})=-\widehat{y}_{i_{0}}(k). \end{aligned} $$

(6.56)

Substituting (6.54) into (6.37) yields that

$$\displaystyle \begin{aligned} y_{i}(k+1)=y_{i}(k),\,\,\forall i\in\mathcal{N}. \end{aligned} $$

(6.57)

Hence, we have

$$\displaystyle \begin{aligned} y_{i}(k)=\widehat{y}_{i}(k)=\widehat{y}_{j}(k-\tau_{ij})=y_{j}(k-\tau_{ij})=y_{j}(k),\,\,\forall w_{ij}>0. \end{aligned} $$

(6.58)

It follows that $\mathcal {G}$ is strongly connected, and it implies

$$\displaystyle \begin{aligned} y_{i}(k)=y_{j}(k),\,\,k\geq-\tau_{ij},\,\forall i,\,j\in\mathcal {N}, \end{aligned} $$

(6.59)

and

$$\displaystyle \begin{aligned} y_{i_{0}}(k)=-y_{j_{0}}(k),\,\,k\geq-\tau_{ij}. \end{aligned} $$

(6.60)

Hence, ΔV (k) = 0 if and only if $y_{i}(k)=0,~~\forall i\in \mathcal {N}$. By LaSalle’s invariance principle, we have

$$\displaystyle \begin{aligned}\lim_{k\rightarrow+\infty}y_{1}(k) = \lim_{k\rightarrow+\infty}y_{2}(k) =\cdots=\lim_{k\rightarrow+\infty}y_{N}(k)=0,\end{aligned}$$

which implies that

$$\displaystyle \begin{aligned}\lim_{k\rightarrow+\infty}x_{1}(k) = \lim_{k\rightarrow+\infty}x_{2}(k) =\cdots=\lim_{k\rightarrow+\infty}x_{N}(k)=0.\end{aligned}$$

Remark 6.13

It can be observed that only the received neighboring states are used in the trigger condition (6.30). Hence, the event-based protocol proposed in this section is distributed. Zeno behavior is defined as an infinite number of triggering occurring in a finite-time interval, which should be avoided in the event-based consensus protocol. Nevertheless, the Zeno behavior can always be excluded in discrete-time multi-agent system since the maximum triggering number is the length of the finite-time interval.

6.2.2 Self-triggered Approach

In Theorem 6.12, we have proved that the proposed event-based protocol is effective to realize the bipartite consensus of the network model. However, the triggering condition needs to be continuously verified for each agent. In this section, we aim to solve this difficult problem by designing a self-triggered algorithm, i.e., the next update time is precomputed based on predictions using the received data. Under the proposed self-triggered algorithm, the signal remains unchanged until next triggering time of multi-agent networks. The appropriate equation for obtaining the triggering time guarantees desired levels of performance. Hence, self-triggered communication schemes for multi-agent networks can effectively reduce the communication costs.

Different from the event-triggered communication strategy, for self-triggered algorithm, the agent i will predict next triggering time instant $t^{i}_{l+1}$ according to the information at time $t^{i}_{l}$. Next, we will give an algorithm to determine the time instant $t^{i}_{l+1}$.

Denote

$$\displaystyle \begin{aligned} l(k-\tau_{ij})&=\arg \mbox{max}_{l\in \mathbb{N}}\{t^{j}_{l}|t^{j}_{l}\leq k-\tau_{ij}\}. \end{aligned} $$

(6.61)

Let

$$\displaystyle \begin{aligned} p_{i}(k)&=\sum_{j=1,\,j\neq i}^{N}|a_{ij}|(sgn(a_{ij})\widehat{x}_{j}(t^{j}_{l(k-\tau_{ij})})-x_{i}(t^{i}_{l})), \end{aligned} $$

(6.62)

and

$$\displaystyle \begin{aligned} q_{i}(k)&=\frac{\sigma a_{ii}^{2}}{4(1-a_{ii})}\sum_{j=1,\,j\neq i}^{N}|a_{ij}|(\widehat{x}_{i}(t^{i}_{l})-sgn(a_{ij})\widehat{x}_{j}(t^{j}_{l(k-\tau_{ij})}))^{2}. \end{aligned} $$

(6.63)

For $k\in [t^{i}_{l},t^{i}_{l+1})$, recall that $e^{2}_{i}(k)=(x_{i}(k)-x_{i}(t^{i}_{l}))^{2}$. For the positive integer m, we have

$$\displaystyle \begin{aligned} \displaystyle x_{i}(t^{i}_{l}+m)=&x_{i}(t^{i}_{l}+m-1)+\sum_{j\in \mathcal{N}_{i}}|a_{ij}|(sgn(a_{ij})\widehat{x}_{j}(t^{i}_{l}+m-1-\tau_{ij})\\ &-\widehat{x}_{i}(t^{i}_{l}+m-1))\\ =&x_{i}(t^{i}_{l}+m-2)+\sum_{j\in \mathcal {N}_{i}}|a_{ij}|(sgn(a_{ij})\widehat{x}_{j}(t^{i}_{l}+m-2-\tau_{ij})\\ &-\widehat{x}_{i}(t^{i}_{l}+m-2))+\sum_{j\in \mathcal{N}_{i}}|a_{ij}|(sgn(a_{ij})\widehat{x}_{j}(t^{i}_{l}+m-1-\tau_{ij})\\ &-\widehat{x}_{i}(t^{i}_{l}+m-1))\\ =&\cdots\cdots\\ =&x_{i}(t^{i}_{l})+\sum_{j\in \mathcal{N}_{i}}|a_{ij}|(sgn(a_{ij})\widehat{x}_{j}(t^{i}_{l}-\tau_{ij})-x_{i}(t^{i}_{l}))+\cdots+\\ &+\sum_{j\in \mathcal{N}_{i}}|a_{ij}|(sgn(a_{ij})\widehat{x}_{j}(t^{i}_{l}+m-1-\tau_{ij})-\widehat{x}_{i}(t^{i}_{l}+m-1)).{} \end{aligned} $$

(6.64)

To propose the self-triggered algorithm to find $t^{i}_{l+1}$, set Σ = 0 and $s=t^{i}_{l}$. The following two cases are considered:

Case 1::

For k > s, if agent i does not receive the renewed information from its neighbors, it follows from (6.64) that

$$\displaystyle \begin{aligned} \displaystyle (x_{i}(k)-x_{i}(s))^{2}=[\varSigma+p_{i}(s)(k-s)]^{2}, ~~~~~~i\in\mathcal {N}.~ \end{aligned} $$

(6.65)

Solving the inequality [Σ + p _i(s)(k − s)]² − q _i(s) > 0, we can obtain that the minimum $k=\omega ^{i}_{l}$ satisfying the above inequality. Hence, according to the event-triggered condition (6.33), the event-triggered time instant is is $t^{i}_{l+1}=\omega ^{i}_{l}$ in this case.

Case 2::

If agent i firstly receives the renewed information from some of its neighbors at time $k^{0}<\omega ^{i}_{l}$, it follows from (6.64) that

$$\displaystyle \begin{aligned} \displaystyle (x_{i}(k^{0})-x_{i}(s))^{2}=[p_{i}(s)(k^{0}-s)]^{2}, ~~~~~~i\in\mathcal {N}.~ \end{aligned} $$

(6.66)

Set Σ = 0 + p _i(s)(k ⁰ − s). According to the event-triggered condition (6.33), we should update s = k ⁰ and then go back to Case 1.

Based on the above discussions, an efficient algorithm to find $t^{i}_{l+1},\,\forall i\in \mathcal {N}$ can be summarized as follows.

Algorithm 6.1 Self-triggered algorithm for system (6.27)

Step 1. :

For each agent $i\in \mathcal {N}$, set Σ = 0 and $s=t^{i}_{l}$.

Step 2. :

Solving the inequality [Σ + p _i(s)(k − s)]² − q _i(s) > 0, we can obtain the minimum $k=\omega ^{i}_{l}$ such that the inequality holds.

Step 3. :

For k ≥ s, if agent i does not receive the renewed information from its neighbors until $k=\omega ^{i}_{l}$, then set $t_{l+1}^{i}=\omega ^{i}_{l}$ and stop the algorithm.

Step 4. :

If agent i firstly receives the renewed information from some of its neighbors at time $k^{0}<\omega ^{i}_{l}$, set Σ = Σ + p _i(s)(k ⁰ − s). Update s = k ⁰ and go to Step 2.

According to the above analysis, the following Theorem 6.14 can be obtained.

Theorem 6.14

Consider the multi-agent system (6.27) with arbitrary finite communication delay τ _ijunder control law (6.28). If the first triggering time$t^{i}_{1} = 0$ , agent$i,~i\in \mathcal {N}$ , determines the triggering time sequence$\displaystyle t^{i}_{l}|{ }_{l=2}^{\infty }$by self-triggered algorithm6.1 . Then, we can obtain the following results:

• (i) System (6.27) can achieve bipartite consensus asymptotically if signed digraph$\mathcal {G}$is structurally balanced.

• (ii) If signed digraph$\mathcal {G}$is structurally unbalanced, then the system (6.27) can achieve consensus and the final consensus value is 0, i.e.,$\displaystyle \lim _{k \to +\infty }x_{i}(k)=0,\,\forall i\in \mathcal {N}$.

6.2.3 Numerical Example

Example 6.15

Consider a signed multi-agent network with structurally balanced topology and structurally unbalanced topology, respectively (see Fig. 6.8). Set σ = 0.9 in event-triggered condition (6.30) and the distinct communication delays are as follows:

Fig. 6.8

Network topology in Example 6.15. (a) structurally balanced. (b) Structurally unbalanced

According to Theorem 6.12, one can easily conclude that under the proposed event-triggered condition (6.30),

• (i) the system with communication delays will achieve bipartite consensus when the network topology is shown in Fig. 6.8a;

• (ii) the states of all the agents will converge to zero when the network topology is shown in Fig. 6.8b.

The evolvement of the agents under the event-triggered condition (6.30) is shown in Figs. 6.9 and 6.10, respectively. The numerical results in Fig. 6.9 show that the individual state of the multi-agent system converges to the bipartite constant limit that has the same modulus and different signs. The numerical results in Fig. 6.10 show that the individual state of the multi-agent system converges to zero. Figures 6.9 and 6.10 agree well with the proposed theoretical result.

Fig. 6.9

The states of multi-agent system (6.27) associated with signed diagraph with balanced structure in Fig. 6.8a

Fig. 6.10

The states of multi-agent system (6.27) associated with signed digraph with unbalanced structure in Fig. 6.8b

The individual event time instants corresponding to Figs. 6.9 and 6.10 under the proposed event-triggered protocol are shown in Figs. 6.11 and 6.12, respectively. Table 6.1 illustrates the event-triggering frequency under two different network topologies. One can conclude from the simulation example that the event-based strategy in this chapter can significantly decrease the information transmission during the bipartite consensus process of the signed network model with distinct communication delays.

Fig. 6.11

Event-trigger times associated with signed digraph in Fig. 6.8a

Fig. 6.12

Event-trigger times associated with signed digraph in Fig. 6.8b

Table 6.1 The total number of triggering over the total number of iterations in simulation under structurally balanced topology and structurally unbalanced topology, respectively

6.3 Summary

In this chapter, the bipartite consensus of continuous-time and discrete-time multi-agent system was studied. For the continuous-time model, according to Perron–Frobenius theorem and some other mathematical analysis, it was found that the bipartite consensus can be asymptotically reached if the strongly connected signed digraph $\mathcal {G}$ is structurally balanced. For the discrete-time model, communication delays and event-based strategy were considered simultaneously. It is shown that under the proposed event-triggered condition the bipartite consensus can be asymptotically achieved if the network topology is structurally balanced, and all the agents converge to zero if the signed digraph is structurally unbalanced. Numerical examples were provided to demonstrate the effectiveness of our derived results.