Fast Synchronization of Complex Networks via Aperiodically Intermittent Sliding Mode Control

Abstract

In the literature, a lot of work focused on studying intermittent control problems via feedback control strategy. No study on the intermittent control problems via sliding mode control method has been reported so far. This paper studies the problem of fast synchronization between two complex dynamical networks via aperiodically intermittent control and sliding mode control. In order to achieve fast synchronization of complex dynamical networks by using aperiodically intermittent sliding mode controller, new differential inequalities are derived firstly. After that, some sufficient finite-time synchronization criteria and finite-time achieving slide mode surface are obtained based on finite-time stability theory, aperiodically intermittent sliding mode control technique and constructing Lyapunov function. Finally, an example is provided to verify the effectiveness of the proposed theoretical methods.

Appendix

Proof of Lemma 4

Take $M_0=V^{1-\eta }(0)+\frac{\alpha }{p_1}$ and $W(t)=V^{1-\eta }(t)\text{ exp }\{(1-\eta )p_1t\}$, where $t\ge 0$. Let $Q(t)=W(t)-M_0+\frac{\alpha }{p_1}\text{ exp }\{(1-\eta )p_1t\}$.

It will be proven in the following parts that

$$\begin{aligned} Q(t)\le 0, \end{aligned}$$

(55)

for all $t \in [0,s_0)$. Otherwise, there is a $a_0\in [0,s_0)$, to make that

$$\begin{aligned}&Q(a_0)=0,\quad \dot{Q}(a_0)> 0, \end{aligned}$$

(56)

$$\begin{aligned}&Q(t)\le 0,\quad 0\le t<s_0. \end{aligned}$$

(57)

With Eqs. (55), (56) and (57), one can obtain

$$\begin{aligned} \dot{Q}(a_0)= & {} (1-\eta )V^{-\eta }(a_0)\dot{V}(a_0)e^{(1-\eta )p_1a_0}\nonumber \\&+\,p_1(1-\eta )V^{1-\eta }(a_0)e^{(1-\eta )p_1a_0}+\alpha (1- \eta )e^{(1-\eta )p_1a_0} \nonumber \\\le & {} (1-\eta )V^{-\eta }(a_0)(-\alpha V^\eta (a_0)- p_1 V(a_0))\cdot e^{(1-\eta )p_1a_0}\nonumber \\&+\,p_1(1-\eta )V^{1-\eta }(a_0)e^{(1-\eta )p_1a_0}+\alpha (1-\eta )e^{(1-\eta )p_1a_0} \nonumber \\= & {} -\alpha (1-\eta )e^{(1-\eta )p_1a_0}-p_1(1-\eta )V^{1-\eta }(a_0)e^{(1-\eta )p_1a_0}\nonumber \\&+\,p_1(1-\eta )V^{1-\eta }(a_0)e^{(1-\eta )p_1a_0}+\alpha (1-\eta )e^{(1-\eta )p_1a_0} \nonumber \\= & {} 0, \end{aligned}$$

(58)

which contradicts the second inequality in (56). Hence, (55) holds.

Let $W_1(t)=V^{1-\eta }(t)e^{(1-\eta )p_1t}e^{-(1-\eta )p_1(t-s_0)}e^{-(1-\eta )p_2(t-s_0)}$, and $H(t)=W_1(t)-M_0+\frac{\alpha }{p_1}e^{(1-\eta )p_1t}e^{-(1-\eta )p_1(t-s_0)}e^{-(1-\eta )p_2(t-s_0)},~t\ge s_0$ and prove that

$$\begin{aligned} H(t)\le 0. \end{aligned}$$

(59)

for $t\in [s_0,t_1)$. Otherwise, there exists a $a_1 \in [s_0,t_1)$ such that

$$\begin{aligned}&H(a_1)=0,\quad \dot{H}(a_1)> 0, \end{aligned}$$

(60)

$$\begin{aligned}&H(t)\le 0,\quad s_0\le t<a_1. \end{aligned}$$

(61)

According to Eqs. (60) and (61),

$$\begin{aligned} \dot{H}(a_1)= & {} \dot{W}_1(a_1)+\frac{\alpha }{p_1}(1-\eta )p_1e^{(1-\eta )p_1a_1}[e^{-(1-\eta )p_1(a_1-s_0)}e^{-(1-\eta )p_2(a_1-s_0)}]\nonumber \\&+\,\frac{\alpha }{p_1}e^{(1-\eta )p_1a_1 }[-(1-\eta )(p_1+p_2)e^{-(1-\eta )(p_1+p_2)(a_1-s_0)}] \nonumber \\= & {} (1-\eta )V^{-\eta }(a_1)\dot{V}(a_1)e^{(1-\eta )p_1a_1}e^{-(1-\eta )(p_1+p_2)(a_1- s_0)} \nonumber \\&+\,V^{1-\eta }(a_1)[e^{(1- \eta )p_1a_1} e^{-(1-\eta )p_1(a_1-s_0)}e^{- (1-\eta )p_2(a_1-s_0)}]'\nonumber \\&+\,\frac{\alpha }{p_1}(1-\eta )p_1 e^{(1-\eta )p_1a_1} \nonumber \\&e^{-(1-\eta ) (p_1+p_2)(a_1-s_0)} - \frac{\alpha }{p_1}(1-\eta )(p_1+p_2)e^{(1-\eta )p_1a_1} e^{-(1-\eta ) (p_1+p_2)(a_1-s_0)} \nonumber \\\le & {} p_2(1-\eta )V^{1-\eta }(a_1)e^{(1-\eta )p_1a_1}e^{-(1- \eta )(p_1+p_2)(a_1-s_0)}+V^{1-\eta }(a_1)[(1-\eta )p_1\nonumber \\&e^{(1-\eta )p_1a_1}e^{-(1-\eta )(p_1+p_2)(a_1-s_0)}-(1-\eta )(p_1+p_2)\nonumber \\&e^{(1-\eta )p_1a_1} e^{-(1-\eta ) (p_1+p_2)(a_1-s_0)}] +\frac{\alpha }{p_1}\nonumber \\&(1-\eta )p_1e^{(1-\eta )p_1a_1}e^{-(1-\eta ) (p_1 +p_2)(a_1-s_0)}- \frac{\alpha }{p_1}(1-\eta )(p_1+p_2)\nonumber \\&e^{(1-\eta )p_1a_1} e^{-(1-\eta )(p_1+p_2)(a_1-s_0)} \nonumber \\= & {} p_2(1-\eta )V^{1-\eta }(a_1)e^{(1-\eta )p_1a_1} e^{-(1- \eta )(p_1+p_2)(a_1-s_0)}\nonumber \\&-p_2(1-\eta )V^{1-\eta }(a_1) e^{(1 -\eta )p_1a_1}\nonumber \\&e^{-(1-\eta )(p_1+p_2) (a_1-s_0)} \nonumber \\&\quad - \frac{\alpha }{p_1}\cdot p_2e^{(1-\eta )p_1a_1}e^{-(1-\eta )(p_1+p_2)(a_1-s_0)} < 0, \end{aligned}$$

(62)

which contradicts the second inequality in (60). Hence (59) holds. From (59), it is easy to see that

$$\begin{aligned} \begin{aligned} W(t)&\le e^{(1-\eta )(p_1+p_2)(t-s_0)}\Big [M_0-\frac{\alpha }{p_1}e^{(1-\eta )p_1t}e^{-(1-\eta )(p_1+p_2)(t-s_0)}\Big ]. \end{aligned} \end{aligned}$$

(63)

Consequently,

$$\begin{aligned} \begin{aligned} W(t)&< e^{(1-\eta )(p_1+p_2)(t-s_0)}\Big [M_0-\frac{\alpha }{p_1}e^{(1-\eta )p_1t}e^{-(1-\eta )(p_1+p_2)(t-s_0)}\Big ]\\&\le e^{(1-\eta )(p_1+p_2)(t_1-s_0)}\Big [M_0-\frac{\alpha }{p_1}e^{(1-\eta )p_1t}e^{-(1-\eta )(p_1+p_2)(t_1-s_0)}\Big ]. \end{aligned} \end{aligned}$$

for $t\in [\theta T,T)$.

On the other hand, together with (55), one can obtain

$$\begin{aligned} \begin{aligned} W(t)&\le M_0-\frac{\alpha }{p_1} e^{(1-\eta )p_1 t}\\&\le M_0-\frac{\alpha }{p_1}e^{(1-\eta )p_1t} e^{-(1-\eta )(p_1+p_2)(t_1-s_0)}\\&\le e^{(1-\eta )(p_1+p_2)(t_1-s_0)}\Big [M_0-\frac{\alpha }{p_1}e^{(1-\eta )p_1t}e^{-(1-\eta )(p_1 +p_2)(t_1-s_0)}\Big ]. \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} W(t)&< e^{(1-\eta )(p_1+p_2)(t_1-s_0)}\Big [M_0- \frac{\alpha }{p_1}e^{(1-\eta )p_1t}e^{-(1 -\eta )(p_1+p_2)(t_1-s_0)}\Big ], \end{aligned} \end{aligned}$$

for all $~t\in [0,t_1)$.

Furthermore, it can be proved that

$$\begin{aligned} \begin{aligned} W(t)&< e^{(1-\eta )(p_1+p_2)(t_1-s_0)}\Big [M_0-\frac{\alpha }{p_1}e^{(1-\eta )p_1t}e^{-(1-\eta )(p_1+p_2)(t_1-s_0)}\Big ], \end{aligned} \end{aligned}$$

(64)

for $t \in [t_1,s_1)$. Otherwise, there is a $a_2 \in [t_1,s_1)$ to make that

$$\begin{aligned} \widetilde{Q}(a_2)=0,\quad \dot{\widetilde{Q}}(a_2)> 0,\quad \widetilde{Q}(a_2)<0,\quad t\in [t_1,a_2), \end{aligned}$$

(65)

where $\widetilde{Q}(t)=W(t)- e^{(1-\eta )(p_1+p_2)(t_1-s_0)}\Big [M_0-\frac{\alpha }{p_1}e^{(1-\eta )p_1t}e^{-(1-\eta )(p_1+p_2)(t_1-s_0)}\Big ]$.

Similar to (58), one has

$$\begin{aligned} \begin{aligned} \dot{\widetilde{Q}}(a_2)=0, \end{aligned} \end{aligned}$$

which contradicts the second inequality in (65). Hence, (64)holds.

With the same approach, one can prove that

$$\begin{aligned} \begin{aligned} W(t)&< e^{(1-\eta )(p_1+p_2)(t_1-s_0)}e^{(1-\eta )(p_1+ p_2)(t-s_1)} \\&\quad \Big [M_0-\frac{\alpha }{p_1}e^{(1-\eta )p_1t}e^{- (1-\eta )(p_1+p_2)(t_1-s_0)}e^{-(1 -\eta )(p_1+p_2)(t-t_1)}\Big ]\\&=e^{(1-\eta )(p_1+p_2)[(t_1-s_0)+(t-s_1)]}\Big [M_0-\frac{\alpha }{p_1} e^{(1-\eta )p_1t}e^{-(1-\eta )(p_1+p_2)[(t_1-s_0)+(t-s_1)]}\Big ]. \end{aligned} \end{aligned}$$

for $t\in [s_1,t_2)$.

With induction method, the following estimation of W(t) for any m can be derived.

For $t_m \le t <s_m$,

$$\begin{aligned} W(t)< e^{(1-\eta )(p_1+p_2)[\sum _{k=1}^m(t_k-s_{k-1})]}\Big [M_0-\frac{\alpha }{p_1}e^{(1-\eta )p_1t}e^{-(1-\eta )(p_1+p_2)[\sum _{k=1}^m(t_k-s_{k-1})]}\Big ],\nonumber \\ \end{aligned}$$

(66)

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

$$\begin{aligned} W(t)< & {} e^{(1-\eta )(p_1+p_2)[\sum _{k=1}^m(t_k-s_{k-1})+(t-s_m)]}\Big [M_0-\frac{\alpha }{p_1} \nonumber \\&e^{(1-\eta )p_1t}e^{-(1-\eta )(p_1+p_2)[\sum _{k=1}^m(t_k-s_{k-1})+(t-s_m)]}\Big ]. \end{aligned}$$

(67)

As there is a nonnegative integer m such that $t_m\le t<t_{m+1}$ for any $t\ge 0$, the following estimation of W(t) for any t can be deduced by Eqs. (66) and (67).

For $t_m\le t<s_m$, and according to Definition 2, one has

$$\begin{aligned} \begin{aligned} W(t)&< e^{(1-\eta )(p_1+p_2)[\sum _{k=1}^m(t_k-s_{k-1})]}\Big [M_0-\frac{\alpha }{p_1}e^{(1-\eta )p_1t}e^{-(1-\eta )(p_1+p_2)[\sum _{k=1}^m(t_k-s_{k-1})]}\Big ]\\&= e^{(1-\eta )(p_1+p_2)[\sum _{k=1}^m \frac{t_k-s_{k-1}}{t_k-t_{k-1}}\cdot (t_k-t_{k-1})]}\Big [M_0-\frac{\alpha }{p_1} e^{(1-\eta )p_1t}\\&\quad e^{-(1-\eta )(p_1+p_2)[\sum _{k=1}^m \frac{t_k-s_{k-1}}{t_k-t_{k-1}}\cdot (t_k-t_{k-1})]}\Big ]\\&\le e^{(1-\eta )(p_1+p_2)[\Phi \sum _{k=1}^m (t_k-t_{k-1})]}\Big [M_0-\frac{\alpha }{p_1}e^{(1-\eta )p_1t}e^{-(1-\eta )(p_1+p_2)[\Phi \sum _{k=1}^m (t_k-t_{k-1})]}\Big ]\\&= e^{(1-\eta )(p_1+p_2)\Phi t_m}\Big [M_0-\frac{\alpha }{p_1} e^{(1-\eta )p_1t}e^{-(1-\eta )(p_1+p_2)\Phi t_m}\Big ]\\&\le e^{(1-\eta )(p_1+p_2)\Phi t}\Big [M_0-\frac{\alpha }{p_1} e^{(1-\eta )p_1t}e^{-(1-\eta )(p_1+p_2)\Phi t}\Big ]. \end{aligned} \end{aligned}$$

For $s_m\le t<t_{m+1}$, and with Lemma 3, one can obtain

$$\begin{aligned} W(t)&< e^{(1-\eta )(p_1+p_2)[\sum _{k=1}^m(t_k-s_{k-1})+(t-s_m)]}\\&\quad \Big [M_0-\frac{\alpha }{p_1}e^{(1-\eta )p_1t}e^{-(1-\eta )(p_1+p_2)[\sum _{k=1}^m(t_k-s_{k-1})+(t-s_m)]}\Big ]\\&= e^{(1-\eta )(p_1+p_2)[\sum _{k=1}^m\frac{t_k-s_{k-1}}{t_k-t_{k-1}}\cdot (t_k-t_{k-1})+\frac{t-s_m}{t-t_m}\cdot (t-t_m)]} \cdot \\&\quad \Big [M_0-\frac{\alpha }{p_1}e^{(1-\eta )p_1t} e^{-(1-\eta )(p_1+p_2)[\sum _{k=1}^m\frac{t_k-s_{k-1}}{t_k-t_{k-1}}\cdot (t_k-t_{k-1})+\frac{t-s_m}{t-t_m}\cdot (t-t_m)]}\Big ]\\&\le e^{(1-\eta )(p_1+p_2)[ \sum _{k=1}^m \frac{t_k-s_{k-1}}{t_k-t_{k-1}} (t_k-t_{k-1})+\frac{t_{m+1}-s_m}{t_{m+1}-t_m}\cdot (t-t_m)]}\cdot \\&\quad \Big [M_0-\frac{\alpha }{p_1}e^{(1-\eta )p_1t}e^{-(1-\eta )(p_1+p_2)[ \sum _{k=1}^m \frac{t_k-s_{k-1}}{t_k-t_{k-1}}(t_k-t_{k-1})+\frac{t_{m+1}-s_m}{t_{m+1}-t_m}\cdot (t-t_m)]}\Big ]\\&\le e^{(1-\eta )(p_1+p_2)\Phi [ \sum _{k=1}^m (t_k-t_{k-1})+(t-t_m)]} \\&\quad \Big [M_0-\frac{\alpha }{p_1}e^{(1-\eta )p_1t}e^{-(1-\eta )(p_1+p_2)\Phi [ \sum _{k=1}^m (t_k-t_{k-1})+(t-t_m)]}\Big ]\\&= e^{(1-\eta )(p_1+p_2)\Phi t} \Big [M_0-\frac{\alpha }{p_1}e^{(1-\eta )p_1t}e^{-(1-\eta )(p_1+p_2)\Phi t}\Big ]. \end{aligned}$$

According to the definition of W(t), one obtains

$$\begin{aligned} \begin{aligned} V^{1-\eta }(t)e^{(1-\eta )p_1 t}&\le e^{(1-\eta )(p_1+p_2)\Phi t}\Big [V^{1-\eta }(0)+\frac{\alpha }{p_1}-\frac{\alpha }{p_1}e^{(1-\eta )p_1 (1-\Phi ) t}e^{-(1-\eta )p_2\Phi t}\Big ]\\&=e^{(1-\eta )(p_1+p_2)\Phi t}\Big [V^{1-\eta }(0)-\frac{\alpha }{p_1}(e^{(1-\eta )p_1(1-\Phi ) t}e^{-(1-\eta )p_2\Phi t}-1)\Big ]. \end{aligned} \end{aligned}$$

The proof is completed. $\square$

Proof of Lemma 5

Take $M_0=V^{1-\eta }(0)$ and $W(t)=V^{1-\eta }(t)+\alpha {(1-\eta )t}$. Let $Q(t)=W(t)-M_0$.

In the following, we will prove that

$$\begin{aligned} \dot{Q}(t)\le 0, \end{aligned}$$

(68)

for all $t \in [0,s_0)$. Otherwise, there is a $a_0\in [0,s_0)$, to make that

$$\begin{aligned}&Q(a_0)=0,\quad \dot{Q}(a_0)> 0, \end{aligned}$$

(69)

$$\begin{aligned}&Q(t)\le 0,\quad 0\le t<s_0. \end{aligned}$$

(70)

For $\forall ~t \in [0,s_0)$, one can obtain

$$\begin{aligned} \dot{Q}(t_0)= & {} (1-\eta )V^{-\eta }(t)\dot{V}(t)+\alpha (1-\eta ) \nonumber \\\le & {} (1-\eta )V^{-\eta }(t)(-\alpha V^\eta (t))+\alpha (1-\eta ) \nonumber \\= & {} -\alpha (1-\eta )+\alpha (1-\eta )=0 \end{aligned}$$

(71)

which contradicts the second inequality in (69). Hence, (68) holds.

Let $H(t)=W(t)-M_0-\alpha (1-\eta )(t-s_0)$. Next, we prove that for $t\in [s_0,t_1)$

$$\begin{aligned} \begin{aligned} H(t)=W(t)-M_0-\alpha (1-\eta )(t-s_0)\le 0. \end{aligned} \end{aligned}$$

(72)

For $\forall ~t \in [s_0,t_1)$, one can get

$$\begin{aligned} \dot{H}(t)= & {} (1-\eta )V^{-\eta }(t_1)\dot{V}(t_1)+\alpha (1-\eta )-\alpha (1-\eta ) \nonumber \\\le & {} \alpha (1-\eta )-\alpha (1-\eta ) \nonumber \\= & {} 0, \end{aligned}$$

(73)

This is to say that H(t) is decreasing in $(s_0,t_1)$. Hence,

$$\begin{aligned} \begin{aligned} H(t)\le H(s_0) =W(s_0)-M_0-\alpha (1-\eta )(s_0-s_0)= W(s_0)-M_0\le 0. \end{aligned} \end{aligned}$$

Hence, Eq. (72) holds.

On the other hand, together with (68), one can obtain

$$\begin{aligned} \begin{aligned} W(t)\le M_0\le M_0+\alpha (1-\eta )(t_1-s_0), \end{aligned} \end{aligned}$$

for $t\in [0,t_1)$.

Similarly, with the same approach of Eq. (71), one can prove that: for $t\in [t_1,s_1)$,

$$\begin{aligned} \begin{aligned} W(t)\le M_0+\alpha (1-\eta )(t_1-s_0). \end{aligned} \end{aligned}$$

Let $Q_1(t)=W(t)-M_0-\alpha (1-\eta )(t_1-s_0)$, it is obtained that $\dot{Q}_1(t)\le 0$, for $t\in [t_1,s_1)$. Similar to the proof of Eq. (72), one can verify

$$\begin{aligned} \begin{aligned} W(t)\le M_0+\alpha (1-\eta )(t_1-s_0)+\alpha (1-\eta )(t-t_1), \end{aligned} \end{aligned}$$

for $t\in [s_1,t_2)$. Take $H_1(t)=W(t)-M_0+\alpha (1-\eta )(t_1-s_0)-\alpha (1-\eta )(t-t_1)$. Then, one can easy obtain that $\dot{H}_1(t)\le 0$, for $t\in [s_1,t_2)$.

With induction method, one can derive the following estimation of W(t) for any integer m.

For $t_m\le t<s_m$,

$$\begin{aligned} \begin{aligned} W(t)\le M_0+\alpha (1-\eta )\sum _{k=1}^m(t_k-s_{k-1}), \end{aligned} \end{aligned}$$

(74)

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

$$\begin{aligned} \begin{aligned} W(t)\le M_0+\alpha (1-\eta )\left( \sum _{k=1}^m(t_k-s_{k-1})+(t-s_m)\right) . \end{aligned} \end{aligned}$$

(75)

Since for any $t\ge 0$, there exists a nonnegative integer m, such that $t_m\le t<t_{m+1}$ for any $t\ge 0$, the following estimation of W(t) for any t can be deduced by Eqs. (74) and (75).

For $t_m\le t<s_m$, and according to Definition 2, one has

$$\begin{aligned} \begin{aligned} W(t)&\le M_0+\alpha (1-\eta )\sum _{k=1}^m(t_k-s_{k-1})\\&= M_0+\alpha (1-\eta )\sum _{k=1}^m \frac{t_k-s_{k-1}}{t_k-t_{k-1}}\cdot (t_k-t_{k-1})\\&\le M_0+\alpha (1-\eta )\Phi \sum _{k=1}^m (t_k-t_{k-1})\\&=M_0+\alpha (1-\eta )\Phi t_m\\&\le M_0+\alpha (1-\eta )\Phi t. \end{aligned} \end{aligned}$$

For $s_m\le t<t_{m+1}$, and with Lemma 3, one can obtain

$$\begin{aligned} \begin{aligned} W(t)&\le M_0+\alpha (1-\eta )\left( \sum _{k=1}^m(t_k-s_{k-1})+(t-s_m)\right) \\&= M_0+\alpha (1-\eta )\left( \sum _{k=1}^m \frac{t_k-s_{k-1}}{t_k-t_{k-1}} (t_k-t_{k-1})+\frac{t-s_m}{t-t_m}(t-t_m)\right) \\&\le M_0+\alpha (1-\eta )\left( \sum _{k=1}^m \frac{t_k-s_{k-1}}{t_k-t_{k-1}} (t_k-t_{k-1})+\frac{t_{m+1}-s_{m+1}}{t-t_m}(t-t_m)\right) \\&\le M_0+\alpha (1-\eta )\Phi (\sum _{k=1}^m (t_k-t_{k-1})+(t-t_m))\\&=M_0+\alpha (1-\eta )\Phi t \end{aligned} \end{aligned}$$

According to the definition of W(t), one obtains

$$\begin{aligned} \begin{aligned} V^{1-\eta }(t)\le M_0-\alpha (1-\eta )(1-\Phi )t. \end{aligned} \end{aligned}$$

The proof is completed. $\square$