Pinning synchronization for directed networks with node balance via adaptive intermittent control

Abstract

In this paper, the problem of synchronization control for directed networks with node balance is investigated. First, a dynamical model of directed network is proposed. Additionally, a new adaptive intermittent scheme is introduced to realize pinning synchronization and some novel criteria are derived by constructing a piecewise auxiliary function and utilizing piecewise analysis method and the theory of series. Based on those criteria, a feasible bound of the rate of control time is given. Finally, some examples with numerical simulations are given to demonstrate the effectiveness of the results derived.

Appendices

Appendix 1

1.1 Proof of Theorem 1

Proof

Construct a piecewise function described by

$$\begin{aligned} W(t)=\left\{ \begin{array}{lllll} \displaystyle \frac{1}{2}e^{-\mu (t-k\hat{T})}\sum _{i=1}^l \frac{1}{\varepsilon _i}\Big (d_i-d_i(t)\Big )^2,\\ \displaystyle \qquad \qquad k\hat{T}\le t\le (k+\delta )\hat{T},\\ \displaystyle \frac{1}{2}e^{2\beta (t-k\hat{T}\!-\!\delta \hat{T})\!-\!\mu \delta \hat{T}}\sum _{i=1}^l \frac{1}{\varepsilon _i}\Big (d_i\!-\!d_i(k\hat{T}\!+\!\delta \hat{T})\Big )^2,\\ \displaystyle \qquad \qquad (k+\delta )\hat{T}<t<(k+1)\hat{T} \end{array} \right. \end{aligned}$$

for $k\in Z^+$, in which $d_i$ is a positive constant to be determined later. It follows from (10) that $W(t)$ is continuous except for $t=(k+1)\hat{T}$ with $k\in Z^+$ and

$$\begin{aligned} W((k+1)\hat{T})=W_+((k+1)\hat{T})=e^{\alpha \hat{T}}W_-((k+1)\hat{T}),\nonumber \\ \end{aligned}$$

(24)

where $\alpha =\mu \delta -2\beta (1-\delta )$ and $\alpha >0$ from condition (2) in Theorem 1, $W_+((k+1)\hat{T})$ and $W_-((k+1)\hat{T})$ denote the right limit and the left limit of $W(t)$ at time $(k+1)\hat{T}$, respectively.

Introduce the following Lyapunov function

$$\begin{aligned} V(t)= U(t)+W(t), \end{aligned}$$

(25)

where

$$\begin{aligned} U(t)=\frac{1}{2}\sum _{i=1}^N\mathbf {e}_{\mathbf{i}}(t)^T{\mathbf {e}}_{\mathbf{i}}(t). \end{aligned}$$

Evidently, $U(t)$ is continuous for all $t\ge 0$, and $V(t)$ is continuous except for $t=(k+1)\hat{T}$ with $k\in Z^+$ and it is right continuous at $t=(k+1)\hat{T}$.

For $k\hat{T}\le t\le (k+\delta )\hat{T}$, the upper right derivative of $V(t)$ with respect to time $t$ along the error system (12) can be calculated as follows:

$$\begin{aligned}&D^+{V}(t)\nonumber \\&\quad =\displaystyle \sum _{i=1}^N{\mathbf {e}}_{\mathbf{i}}(t)^T\Big [\mathbf {f}({\mathbf {e}}_{\mathbf{i}}(t) +\mathbf {s}(t))-\mathbf {f}(\mathbf {s}(t))\Big ]\nonumber \\&\qquad +\,c\sum _{i=1}^N\sum _{j=1}^N a_{ij}{\mathbf {e}}_{\mathbf{i}}(t)^T{\varvec{\Gamma }} \mathbf{e}_{\mathbf{j}}(t) -\sum _{i=1}^l d_i(t){\mathbf {e}}_{\mathbf{i}}(t)^T\mathbf {e}_{\mathbf{i}}(t)\nonumber \\&\qquad -\,\frac{\mu }{2}\exp {\Big [-\mu \big (t-k\hat{T}\big )\Big ]}\sum _{i=1}^l \frac{1}{\varepsilon _i}\Big (d_i-d_i(t)\Big )^2\nonumber \\&\qquad -\,\exp {\Big [-\mu \big (t-k\hat{T}\big )\Big ]}\sum _{i=1}^l \big (d_i-d_i(t)\big ){\mathbf {e}}_{\mathbf{i}}(t)^T\mathbf {e}_{\mathbf{i}}(t)\nonumber \\&\quad \le \displaystyle \sum _{i=1}^N\theta {\mathbf {e}}_{\mathbf{i}}(t)^T {\mathbf {e}}_{\mathbf{i}}(t)-\exp {(-\mu \delta \hat{T})}\sum _{i=1}^l d_i{\mathbf {e}}_{\mathbf{i}}(t)^T {\mathbf {e}}_{\mathbf{i}}(t)\nonumber \\&\qquad -\,\frac{\mu }{2}\exp {\Big [-\mu \big (t-k\hat{T}\big )\Big ]}\sum _{i=1}^l \frac{1}{\varepsilon _i}\Big (d_i-d_i(t)\Big )^2\nonumber \\&\qquad +\,c\sum _{i=1}^N\sum _{j=1}^N a_{ij}{\mathbf {e}}_{\mathbf{i}}(t)^T{\varvec{\Gamma }} \mathbf{e}_{\mathbf{j}}(t)\nonumber \\&\quad = \displaystyle \sum _{i=1}^N\sum _{k=1}^ne_i^k(t)\theta e_i^k(t) -e^{-\mu \delta \hat{T}}\sum _{i=1}^l\sum _{k=1}^ne_i^k(t)d_ie_i^k(t)\nonumber \\&\qquad +\,c\sum _{i=1}^N\sum _{j=1}^N\sum _{k=1}^n \gamma _k e_i^k(t)\frac{a_{ji}+a_{ij}}{2} e_j^k(t)\nonumber \\&\qquad -\,\frac{\mu }{2}\exp {\Big [-\mu \big (t-k\hat{T}\big )\Big ]}\sum _{i=1}^l \frac{1}{\varepsilon _i}\Big (d_i-d_i(t)\Big )^2\nonumber \\&\quad = \displaystyle \sum _{k=1}^n ({\tilde{\mathbf {e}}^\mathbf{k}}(t))^T\Big [\theta {\mathbf {I}}_{\mathbf{N}}+c\gamma _k\frac{\mathbf {A}^T+\mathbf {A}}{2} - {\hat{\mathbf {D}}} \big ) \Big ]{\tilde{\mathbf {e}}^\mathbf{k}}(t)\nonumber \\&\qquad -\,\frac{\mu }{2}\exp {\Big [-\mu \big (t-k\hat{T}\big )\Big ]}\sum _{i=1}^l \frac{1}{\varepsilon _i}\Big (d_i-d_i(t)\Big )^2\nonumber \\&\quad = \displaystyle -\frac{\mu }{2}\sum _{k=1}^n \left( {\tilde{\mathbf {e}}^\mathbf{k}}(t)\right) ^T{\tilde{\mathbf {e}}^\mathbf{k}}(t)\!+\! \sum _{k=1}^n ({\tilde{\mathbf {e}}^\mathbf{k}}(t))^T(\mathbf {G}\!-\!{\hat{\mathbf {D}}}){\tilde{\mathbf {e}}^\mathbf{k}}(t)\nonumber \\&\qquad -\,\frac{\mu }{2}\exp {\Big [-\mu \big (t-k\hat{T}\big )\Big ]}\sum _{i=1}^l \frac{1}{\varepsilon _i}\Big (d_i-d_i(t)\Big )^2,\nonumber \\ \end{aligned}$$

(26)

where ${\tilde{\mathbf {e}}^\mathbf{k}}=(e_1^k,e_2^k,\ldots ,e_N^k)^T$ for $k=1,2,\ldots ,n$,

$$\begin{aligned} {\hat{\mathbf {D}}}&= \text{ diag }(e^{-\mu \delta \hat{T}}d_1,\ldots ,e^{-\mu \delta \hat{T}}d_l,\underbrace{0,\ldots ,0}_{N-l}),\\ \mathbf {G}&= \left( \theta +\frac{\mu }{2}\right) {\mathbf {I}}_{\mathbf{N}} +c\gamma _k\frac{\mathbf {A}^T+\mathbf {A}}{2}. \end{aligned}$$

Obviously, $\mathbf {G}$ is a real symmetric matrix, by Lemma 1, $\mathbf {G}-{\hat{\mathbf {D}}}<0$ is equivalent to

$$\begin{aligned} \mathbf {G}_l=\left( \left( \theta +\frac{\mu }{2}\right) {\mathbf {I}}_{\mathbf{N}}+c\gamma _k\frac{\mathbf {A}^T+\mathbf {A}}{2}\right) _l<0 \end{aligned}$$

(27)

when $d_i$ ($i=1,2,\ldots ,l$) are sufficiently large.

In addition, according to Lemma 2 and condition (13),

$$\begin{aligned}&\lambda _{\max }\Big (\left( \theta +\frac{\mu }{2}\right) {\mathbf {I}}_{\mathbf{N}} +c\gamma _k\frac{\mathbf {A}^T+\mathbf {A}}{2}\Big )_l\le \theta \\&\quad +\frac{\mu }{2}+c\gamma _k\lambda _{\max }\Big (\frac{\mathbf {A}^T +\mathbf {A}}{2}\Big )_l\le 0, \end{aligned}$$

which implies that $\mathbf {G}_l<0$ and then $\mathbf {G}-{\hat{\mathbf {D}}}<0$, this together with (26), we have

$$\begin{aligned} D^+V(t)\le -\mu V(t) \end{aligned}$$

(28)

for $k\hat{T}\le t\le (k+\delta )\hat{T}$.

Similarly, for $(k+\delta )\hat{T}<t<(k+1)\hat{T}$,

$$\begin{aligned} D^+{V}(t)&\le \displaystyle \sum _{k=1}^n \left( {\tilde{\mathbf {e}}^\mathbf{k}}(t)\right) ^T\Big [\theta {\mathbf {I}}_{\mathbf{N}}+c\gamma _k\frac{\mathbf {A}^T+\mathbf {A}}{2} \Big ]{\tilde{\mathbf {e}}^\mathbf{k}}(t)\\&+\,2\beta W(t). \end{aligned}$$

Since the graph of (3) is strongly connected, then it follows from Ref. [40] that the coupling matrix $\mathbf {A}=(a_{ij})_{N\times N}$ and $\mathbf {A}^T$ are irreducible. According to Lemma 3, the eigenvalues of $\frac{\mathbf {A}+\mathbf {A}^T}{2}$ can be denoted by $0=\hat{\lambda }_1>\hat{\lambda }_2\ge \hat{ \lambda }_3\ge \ldots \ge \hat{ \lambda }_N$. Besides, it is easy to see that $\frac{\mathbf {A}^T+\mathbf {A}}{2}$ is a real symmetric matrix, then there exists a unitary matrix $\mathbf {P}=(P_1,\ldots , P_N)$ satisfying $\mathbf {P}^T\mathbf {P}={\mathbf {I}}_{\mathbf{N}}$ such that

$$\begin{aligned} \frac{\mathbf {A}^T+\mathbf {A}}{2}=\mathbf {P}^T\mathbf {QP}, \end{aligned}$$

where $\mathbf {Q}=\text{ diag }(\hat{\lambda }_1, \hat{\lambda }_2,\ldots ,\hat{\lambda }_N)$. Therefore,

$$\begin{aligned} D^+{V}(t)&\le \displaystyle \sum _{k\!=\!1}^n \left( {\mathbf {P}\tilde{\mathbf{e}}^\mathbf{k}}(t)\right) ^T\Big [\theta {\mathbf {I}}_{\mathbf{N}}\!+\!c\gamma _k\mathbf {Q}\!-\!\beta {\mathbf {I}}_{\mathbf{N}} \Big ]\left( {\mathbf {P}\tilde{\mathbf{e}}^\mathbf{k}}(t)\right) \\&\displaystyle +\,\beta \sum _{k=1}^n \left( {\tilde{\mathbf {e}}^\mathbf{k}}(t)\right) ^T{\tilde{\mathbf {e}}^\mathbf{k}}(t)+2\beta W(t) \\&= \displaystyle \sum _{k=1}^n \sum _{i=1}^N \Big (\theta +c\gamma _k\hat{\lambda }_i-\beta \Big )\big (y_i^k(t)\big )^2\\&\displaystyle +\,\beta \sum _{k=1}^n \left( {\tilde{\mathbf {e}}^\mathbf{k}}(t)\right) ^T{\tilde{\mathbf {e}}^\mathbf{k}}(t)+2\beta W(t)\\&\le \displaystyle 2\beta V(t), \end{aligned}$$

where ${\mathbf {P}}\tilde{\mathbf{e}}^\mathbf{k}(t)={\mathbf {y}}^\mathbf{k}(t)=\left( y_1^k(t),\ldots ,y_N^k(t)\right) ^T$ for $k=1,2,\ldots ,n$.

Hence, for $(k+\delta )\hat{T}< t<(k+1)\hat{T}$,

$$\begin{aligned} D^+V(t)\le 2\beta V(t). \end{aligned}$$

(29)

In the following, we will prove that

$$\begin{aligned} \lim _{t\rightarrow +\infty }U(t)=0. \end{aligned}$$

By virtue of (24), (28) and (29), we can derive that

$$\begin{aligned} V_-((k+1)\hat{T})&\le \displaystyle e^{2\beta (1-\delta )\hat{T}}V(k\hat{T}+\delta \hat{T})\\&\le \displaystyle e^{-\alpha \hat{T}}V_+(k\hat{T})\\&= \displaystyle e^{-\alpha \hat{T}}U(k\hat{T})+W_-(k\hat{T})\\&= \displaystyle e^{-\alpha \hat{T}}V_-(k\hat{T})+(1-e^{-\alpha \hat{T}})W_-(k\hat{T}), \end{aligned}$$

which implies that

$$\begin{aligned}&V_-((k+1)\hat{T})-V_-(k\hat{T})\nonumber \\&\quad \le \displaystyle (e^{-\alpha \hat{T}}-1)V_-(k\hat{T})+(1-e^{-\alpha \hat{T}})W_-(k\hat{T})\nonumber \\&\quad =\displaystyle (e^{-\alpha \hat{T}}-1)U(k\hat{T}), \end{aligned}$$

(30)

and then

$$\begin{aligned} V_-((k+1)\hat{T})-V(0)\le (e^{-\alpha \hat{T}}-1)\sum _{i=0}^k U(i\hat{T}), \end{aligned}$$

it shows that

$$\begin{aligned} \sum _{i=0}^{\infty } U(i\hat{T})\le \frac{V(0)}{1-e^{-\alpha \hat{ T}}}, \end{aligned}$$

(31)

therefore, by the theory of series,

$$\begin{aligned} \lim _{i\rightarrow +\infty }U(i\hat{T})=0. \end{aligned}$$

(32)

In addition, for $k\hat{T}\le t< (k+1)\hat{T}$, in view of the nonnegativity of $d_i(t)$, it is easy to estimate that

$$\begin{aligned} D^+{U}(t)&\le \displaystyle \sum _{i=1}^N{\mathbf {e}}_{\mathbf{i}}(t)^T\Big [\mathbf {f}({\mathbf {e}}_{\mathbf{i}}(t) +\mathbf {s}(t))-\mathbf {f}(\mathbf {s}(t))\Big ]\\&\displaystyle +\,c\sum _{i=1}^N\sum _{j=1}^N a_{ji} {\mathbf {e}}_{\mathbf{i}}(t)^T{\varvec{\Gamma }} \mathbf{e}_{\mathbf{j}}(t)\\&\le \displaystyle \sum _{k=1}^n ({\tilde{\mathbf {e}}^\mathbf{k}}(t))^T\Big [\theta {\mathbf {I}}_{\mathbf{N}} +c\gamma _k\frac{\mathbf {A}^T+\mathbf {A}}{2} \Big ]{\tilde{\mathbf {e}}^\mathbf{k}}(t)\\&\le \displaystyle 2\hat{\lambda }U(t), \end{aligned}$$

where $\hat{\lambda }=\max \,\{\hat{\lambda }_i,\ i=1,2,\ldots ,n\}$, $\hat{\lambda }_i$ is the largest eigenvalue of the matrix $\theta {\mathbf {I}}_{\mathbf{N}}+c\gamma _i\frac{\mathbf {A}^T+\mathbf {A}}{2}$. In view of this, we have

$$\begin{aligned} U(t)\le U(k\hat{T})e^{2\hat{\lambda }(t-k\hat{T})},\quad k\hat{T}\le t< (k+1)\hat{T}. \end{aligned}$$

(33)

Evidently, $k\rightarrow \infty$ when $t\rightarrow \infty$, this combines with (32) and (33), we obtain

$$\begin{aligned} \lim _{t\rightarrow +\infty }U(t)=0. \end{aligned}$$

Therefore, the asymptotical synchronization of the controlled network (6) is realized, and the proof of Theorem 1 is completed.$\square$

Appendix 2

1.1 Proof of Corollary 1

Proof

From inequality (14),

$$\begin{aligned} \theta +c\gamma \lambda _{l+1}<0. \end{aligned}$$

Denote $\mu =-2(\theta +c\gamma \lambda _{l+1})$ and $\beta =\theta$, then condition (13) holds and

$$\begin{aligned} \mu \delta -2(1-\delta )\beta =-2(\theta +c\gamma \delta \lambda _{l+1})>0. \end{aligned}$$

It follows from Theorem 1 that the network (6) is globally asymptotically synchronized.$\square$