Finite-Time Synchronization of Hybrid-Coupled Delayed Dynamic Networks via Aperiodically Intermittent Control

Abstract

In this paper, we investigate finite-time synchronization for a class of delayed dynamic networks with hybrid coupling via aperiodically intermittent control. First, more general models of dynamic networks with transmission delay and self-feedback delay are given. Second, a lemma with a newly added parameter is proposed to ensure finite-time synchronization of dynamic networks via an aperiodically intermittent control scheme, in which the newly added parameter can make the convergence time shorter. Third, by constructing a novel piecewise Lyapunov function and applying linear matrix inequality technique, some sufficient conditions ensuring finite-time synchronization for delayed dynamic networks are obtained. Moreover, the convergence time is affected by some decision parameters besides the newly added parameter, one of which is a maximum uncontrolled ratio generated by the definition of aperiodic intermittent control itself. Finally, a numerical example is presented to verify its validity and rationality.

Appendix

Proof of Lemma 4

Denote $M_0=V^{1-\kappa }(0)+\frac{\varsigma }{\nu _1}$ and $W(t)=\exp \{(1-\kappa )\nu _1t\}V^{1-\kappa }(t),~\text{ for }~t\ge 0$. Let $Q(t)=W(t)-M_0+\frac{\varsigma }{\nu _1}\exp \{(1-\kappa )\nu _1t\}$. It is easy to see that $Q(t)=0,~~\text{ for }~~t=0.$

In the following, we will prove that $Q(t)\le 0,~~\text{ for }~~t\in [0,s_0).$

For $\forall t\in [0,s_0)$, it has

$$\begin{aligned} \begin{aligned} \dot{Q}(t)&\le (1-\kappa )V^{-\kappa }(t)\exp \{(1-\kappa )\nu _1t\}\big (-\varsigma V^\kappa (t)+\nu _{1}V(t)\big )+\nu _1(1-\kappa )\\&\quad \exp \{(1-\kappa )\nu _1t\}V^{1-\kappa }(t)+\varsigma (1-\kappa )\exp \{(1-\kappa )\nu _1t\}\\&=0. \end{aligned} \end{aligned}$$

Hence, $Q(t)\le Q(0)=0,~~\text{ for }~~t\in [0,s_0).$

Let $W_1(t)=\exp \{(1-\kappa )\nu _1t\}\exp \{-(1-\kappa )(\nu _1+\nu _2)(t-s_0)\}V^{1-\kappa }(t)=\exp \{(1-\kappa )\nu _1s_0\}\exp \{-(1-\kappa )\nu _2(t-s_0)\}V^{1-\kappa }(t)$, $Q_1(t)=W_1(t)-M_0+\frac{\varsigma }{\nu _1}\exp \{(1-\kappa )\nu _1t\}\exp \{-(1-\kappa )(\nu _1+\nu _2)(t-s_0)\}=W_1(t)-M_0+\frac{\varsigma }{\nu _1}\exp \{(1-\kappa )\nu _1s_0\}\exp \{-(1-\kappa )\nu _2(t-s_0)\}.$ Next, we will prove that for $t\in [s_0,t_1),~Q_1(t)\le 0.$

For $\forall t\in [s_0,t_1)$, we can obtain

$$\begin{aligned} \begin{aligned} \dot{Q}_1(t)\le&-\frac{\varsigma }{\nu _1}((1-\kappa )\nu _2)\exp \{(1-\kappa )\nu _1s_0\}\exp \{-(1-\kappa )\nu _2(t-s_0)\}\\ \le&0.\\ \end{aligned} \end{aligned}$$

Hence, $Q_1(t)\le Q_1(s_0)=Q(s_0)\le 0.$

Together with $Q_1(t)\le 0$, for $t\in [s_0,t_1)$, we can obtain

$$\begin{aligned} \begin{aligned} W(t)&\le \exp \{(1-\kappa )(\nu _1+\nu _2)(t_1-s_0)\}\big (M_0-\frac{\varsigma }{\nu _1}\exp \{(1-\kappa )\nu _1t\}\\&\quad \times \exp \{-(1-\kappa )(\nu _1+\nu _2)(t_1-s_0)\}\big ). \end{aligned} \end{aligned}$$

Note that $Q(t)\le 0$, for $~t\in [0,s_0)$, we have

$$\begin{aligned} \begin{aligned} W(t)&\le \exp \{(1-\kappa )(\nu _1+\nu _2)(t_1-s_0)\}\big (M_0-\frac{\varsigma }{\nu _1}\exp \{(1-\kappa )\nu _1t\}\\&\quad \times \exp \{-(1-\kappa )(\nu _1+\nu _2)(t_1-s_0)\}\big ). \end{aligned} \end{aligned}$$

So, for $t\in [0,t_1)$, it has

$$\begin{aligned} \begin{aligned} W(t)&\le \exp \{(1-\kappa )(\nu _1+\nu _2)(t_1-s_0)\}\big (M_0-\frac{\varsigma }{\nu _1}\exp \{(1-\kappa )\nu _1t\}\\&\quad \times \exp \{-(1-\kappa )(\nu _1+\nu _2)(t_1-s_0)\}\big ), \end{aligned} \end{aligned}$$

Similarly, we can prove that for $t\in [t_1, s_1)$,

$$\begin{aligned} \begin{aligned} W(t)&\le \exp \{(1-\kappa )(\nu _1+\nu _2)(t_1-s_0)\}\big (M_0-\frac{\varsigma }{\nu _1}\exp \{(1-\kappa )\nu _1t\}\\&\quad \times \exp \{-(1-\kappa )(\nu _1+\nu _2)(t_1-s_0)\}\big ). \end{aligned} \end{aligned}$$

Suppose

$$\begin{aligned} \begin{aligned} Q_2(t)&=W(t)-\exp \{(1-\kappa )(\nu _1+\nu _2)(t_1-s_0)\}\big (M_0-\frac{\varsigma }{\nu _1}\exp \{(1-\kappa )\nu _1t\}\\&\quad \times \exp \{-(1-\kappa )(\nu _1+\nu _2)(t_1-s_0)\}\big ). \end{aligned} \end{aligned}$$

It is easy to prove that $Q_2(t)\le 0$, $\forall t\in [t_1, s_1)$.

For any $t\in [s_1, t_2)$, by taking $W_2(t)=W(t)\exp \{-(1-\kappa )(\nu _1+\nu _2)(t_1-s_0)\}\exp \{-(1-\kappa )(\nu _1+\nu _2 )(t-s_1)\}=\exp \{(1-\kappa )$$\nu _1(s_0+s_1-t_1)\}\exp \{-(1-\kappa )\nu _2(t-(s_0+s_1-t_1))\}V^{1-\kappa }(t)$ and $Q_3(t)=W_2(t)-M_0+\frac{\varsigma }{\nu _1}\exp \{(1-\kappa )p_1t\}\exp \{-(1-\kappa )(\nu _1+\nu _2 )(t_1-s_0)\}$$\exp \{-(1-\kappa )(\nu _1+\nu _2 )(t-s_1)\}=W_2(t)-M_0+\frac{\varsigma }{\nu _1}\exp \{(1-\kappa )\nu _1(s_0+s_1-t_1)\}\exp \{-(1-\kappa )\nu _2(t-(s_0+s_1-t_1))\}$, we can verify $Q_3(t)\le Q_3(s_1)\le 0$ similar to the proof of $Q_1(t)\le 0,~t\in [s_0,t_1).$

Therefore,

$$\begin{aligned} \begin{aligned} W_2(t)&\le M_0-\frac{\varsigma }{\nu _1}\exp \{(1-\kappa )\nu _1t\}\exp \{-(1-\kappa )(\nu _1+\nu _2)(t_1-s_0)\}\\&\quad \times \exp \{-(1-\kappa )(\nu _1+\nu _2)(t-s_1)\}, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} W(t)&\le \exp \{(1-\kappa )(\nu _1+\nu _2)[(t_1-s_0)+(t-s_1)]\}\big (M_0-\frac{\varsigma }{\nu _1}\exp \{(1-\kappa )\nu _1t\}\\&\quad \times \exp \{-(1-\kappa )(\nu _1+\nu _2)[(t_1-s_0)+(t-s_1)]\}\big ). \end{aligned} \end{aligned}$$

By induction, for any integer m, we can deduce the following estimation of W(t) for any t.

For $t_m\le t<s_m$, we can get

$$\begin{aligned} \begin{aligned} W(t)&\le \exp \big \{(1-\kappa )(\nu _1+\nu _2)\sum \limits _{k=1}^{m}(t_k-s_{k-1})\big \}\big (M_0-\frac{\varsigma }{\nu _1}\exp \{(1-\kappa )\nu _1t\}\\&\quad \times \exp \big \{-(1-\kappa )(\nu _1+\nu _2)\sum \limits _{k=1}^{m}(t_k-s_{k-1})\big \}\big ), \end{aligned} \end{aligned}$$

and for $s_m\le t<t_{m+1}$, we can get

$$\begin{aligned} \begin{aligned} W(t)&\le \exp \big \{(1-\kappa )(\nu _1+\nu _2)\sum \limits _{k=1}^{m}[(t_k-s_{k-1})+(t-s_m)]\big \}\big (M_0-\frac{\varsigma }{\nu _1}\\&\quad \times \exp \{(1-\kappa )\nu _1t\}\exp \big \{-(1-\kappa )(\nu _1+\nu _2)\sum \limits _{k=1}^{m}[(t_k-s_{k-1})+(t-s_m)]\big \}\big ). \end{aligned} \end{aligned}$$

For $t_m\le t<s_m$, by applying Definition 2, it has

$$\begin{aligned} \begin{aligned} W(t)&\le \exp \big \{(1-\kappa )(\nu _1+\nu _2)\sum \limits _{k=1}^{m}\frac{t_k-s_{k-1}}{t_k-t_{k-1}}(t_k-t_{k-1})\big \}\big (M_0-\frac{\varsigma }{\nu _1}\\&\quad \times \exp \{(1-\kappa )\nu _1t\}\exp \big \{-(1-\kappa )(\nu _1+\nu _2)\sum \limits _{k=1}^{m}\frac{t_k-s_{k-1}}{t_k-t_{k-1}}(t_k-t_{k-1})\big \}\big )\\&\le \exp \{(1-\kappa )(\nu _1+\nu _2)\Psi t\}\big (M_0-\frac{\varsigma }{\nu _1}\exp \{(1-\kappa )\nu _1t\}\\&\quad \times \exp \{-(1-\kappa )(\nu _1+\nu _2)\Psi t\}\big ). \end{aligned} \end{aligned}$$

For $s_m\le t<t_{m+1}$, by applying Definition 2 and Lemma 1, it has

$$\begin{aligned} \begin{aligned} W(t)&\le \exp \big \{(1-\kappa )(\nu _1+\nu _2)\sum \limits _{k=1}^{m}\big [\frac{t_k-s_{k-1}}{t_k-t_{k-1}}(t_k-t_{k-1})+\frac{t_{m+1}-s_m}{t_{m+1}-t_m}(t-t_m)\big ]\big \}\\&\quad \times \big (M_0-\frac{\varsigma }{\nu _1}\exp \{(1-\kappa )\nu _1t\}\exp \big \{-(1-\kappa )(\nu _1+\nu _2)\Psi \sum \limits _{k=1}^{m}\big ((t_k-t_{k-1})\\&\quad +(t-t_m)\big )\big \}\big )\\&\le \exp \{(1-\kappa )(\nu _1+\nu _2)\Psi t \}\big (M_0-\frac{\varsigma }{\nu _1}\exp \{(1-\kappa )\nu _1t\}\exp \{-(1-\kappa )(\nu _1\\&\quad +\nu _2)\Psi t\}\big ). \end{aligned} \end{aligned}$$

From the definition of W(t), we can obtain

$$\begin{aligned} \begin{aligned} V^{1-\kappa }(t)\exp \{(1-\kappa )\nu _1t\}&\le \exp \{(1-\kappa )(\nu _1+\nu _2)\Psi t \}\big (M_0-\frac{\varsigma }{\nu _1}\exp \{(1-\kappa )\nu _1t\}\\&\quad \times \exp \{-(1-\kappa )(\nu _1+\nu _2)\Psi t\}\big ). \end{aligned} \end{aligned}$$

With Lemma 3, the settling time $T^*$ can be obtained in the following form $V^{1-\kappa }(0)+\frac{\varsigma }{\nu _1}=\frac{\varsigma }{\nu _1}\exp \{(1-\kappa )\nu _1T^*\}\exp \{-(1-\kappa )(\nu _1+\nu _2)\Psi T^*\}$.

From (10), one can obtain that

$$\begin{aligned} T^*=\frac{\ln (1+\frac{\nu _1}{\varsigma }V^{1-\kappa }(0))}{(1-\kappa )\big ( \nu _1-(\nu _1+\nu _2)\Psi \big )}. \end{aligned}$$

The proof is completed. $\square$