Synchronization of random coupling delayed complex networks with random and adaptive coupling strength

Abstract

Our paper devotes to synchronization issue of random coupling delayed complex networks with adaptive coupling strength as well as random coupling strength via intermittent control. Different from related literature, intermittent control in this paper is aperiodic. Furthermore, we assume coupling delay is a random variable and coupling function is nonlinear. Meanwhile, adaptive and random coupling strength are, respectively, taken into account in our model compared with most existing literature. Some novel sufficient conditions are derived by utilizing Lyapunov method and graph theory. What is more, synchronization of a second-order Kuramoto model, as an application of our theoretical results, is investigated. Meanwhile, some sufficient conditions are given as well. Eventually, some numerical simulations are given to show the effectiveness of results.

Appendix

1.1 Proof of Theorem 1

Proof

To start with, let $V(e,t)=\sum _{k=1}^{N}c_{k}V_{k}(e_{k},t)$, where $V_{k}(e_{k},t)=e_{k}^{\mathrm {T}}e_{k}+e^{-\rho t}\left( a(t)-a_{0}\right) ^{2}$. Considering digraph $({\mathcal {G}},B)$ is strongly connected, in view of Proposition 2.1 in [49], we can easily avail that $c_{k}>0$. And we have

$$\begin{aligned}&{\mathcal {L}}V_{k}(e_{k}(t),t)\\&\quad =2e_{k}^{\mathrm {T}}(t)\Big [{\tilde{f}}_{k}\left( e_{k}(t),e_{k}(t-\tau _{1}(t)),t\right) \\&\qquad +\beta (t)a(t)\sum _{h=1}^{N}a_{kh}{\tilde{H}}_{kh}\left( e_{k}(t-\tau _{2}(t)),e_{h}(t-\tau _{2}(t))\right) \nonumber \\&\qquad +a(t)(1-\beta (t))\sum _{h=1}^{N}a_{kh}{\tilde{H}}_{kh}(e_{k}(t-\tau _{3}(t)),\\&\qquad \times e_{h}(t-\tau _{3}(t)))+u_{k}(t)\Big ]+\left| {\tilde{g}}_{k}(e_{k}(t),t)\right| ^{2}\nonumber \\&\qquad -\rho e^{-\rho t}(a(t)-a_{0})^{2}+ 2e^{-\rho t}(a(t)-a_{0}){\dot{a}}(t), \end{aligned}$$

where the differential operator ${\mathcal {L}}$ can be discovered in [25]. Based on Assumption 2, the following inequalities can be yielded that

$$\begin{aligned}&2e_{k}^{\mathrm {T}}(t){\tilde{f}}_{k}\left( e_{k}(t),e_{k}(t-\tau _{1}(t)),t \right) \nonumber \\&\quad \le (2\varGamma _{k}+\varLambda _{k})|e_{k}(t)|^{2}+\varLambda _{k}|e_{k}(t-\tau _{1}(t))|^{2} \end{aligned}$$

(20)

and

$$\begin{aligned}&2e_{k}^{\mathrm {T}}(t)a(t)\beta (t)\sum _{h=1}^{N}a_{kh}{\tilde{H}}_{kh}\left( e_{k}(t-\tau _{2}(t)),e_{h}(t-\tau _{2}(t))\right) \nonumber \\&\quad \le 2 a(t)\beta (t)\sum _{h=1}^{N}a_{kh}b_{kh}\left[ |e_{k}(t)|^{2}+|e_{k}(t-\tau _{2}(t))|^{2}\right] \nonumber \\&\qquad +\, a(t)\beta (t)\sum _{h=1}^{N}a_{kh}b_{kh}\nonumber \\&\qquad \times \left[ |e_{h}(t-\tau _{2}(t))|^{2}-|e_{k}(t-\tau _{2}(t))|^{2}\right] . \end{aligned}$$

(21)

The following inequality can be obtained by analogical derivation of (21)

$$\begin{aligned}&2 a(t)(1-\beta (t))e_{k}^{\mathrm {T}}(t)\sum _{h=1}^{N}a_{kh}{\tilde{H}}_{kh}\left( e_{k}(t-\tau _{3}(t)),\right. \nonumber \\&\quad \left. e_{h}(t-\tau _{3}(t))\right) \nonumber \\&\quad \le 2 a(t)(1-\beta (t))\sum _{h=1}^{N}a_{kh}b_{kh}\left[ |e_{k}(t)|^{2}\right. \nonumber \\&\qquad \left. +\,|e_{k}(t-\tau _{3}(t))|^{2}\right] \nonumber \\&\qquad +\,a(t)(1-\beta (t))\sum _{h=1}^{N}a_{kh}b_{kh}\left[ |e_{h}(t-\tau _{3}(t))|^{2}\right. \nonumber \\&\qquad \left. -\,|e_{k}(t-\tau _{3}(t))|^{2}\right] . \end{aligned}$$

(22)

In view of (10), one can acquire that

$$\begin{aligned}&e^{-\rho t}{\dot{a}}(t)(a(t)-a_{0})\nonumber \\&\quad \le -a(t)b_{k}|e_{k}(t)|^2-a(t)b_{k}|e_{k}(t-\tau _{2}(t))|^{2}\nonumber \\&\qquad -a(t)b_{k}|e_{k}(t-\tau _{3}(t))|^{2}\nonumber \\&\qquad +a_{0}\sum _{k=1}^{N}b_{k}\left[ |e_{k}(t)|^2+|e_{k}(t-\tau _{2}(t))|^{2}\right. \nonumber \\&\qquad \left. +|e_{k}(t-\tau _{3}(t))|^{2}\right] . \end{aligned}$$

(23)

According to (4), when $t\in [t_{m},s_{m})$, one has $e_{k}^{\mathrm {T}}(t)u_{k}(t)=-d_{k}e_{k}^{\mathrm {T}}(t)e_{k}(t)$. In what follows, we can obtain by applying (20)–(23)

$$\begin{aligned}&{\mathbb {E}}{\mathcal {L}}V(e(t),t)\nonumber \\&\quad \le \sum _{k=1}^{N}c_{k}\left( 2\varGamma _{k}+\varLambda _{k}+\delta ^{2}_{k}-2d_{k}\right) {\mathbb {E}}|e_{k}(t)|^{2}\nonumber \\&\qquad +\,\sum _{k=1}^{N}c_{k}\varLambda _{k}{\mathbb {E}}|e_{k}(t-\tau _{1}(t))|^{2} \nonumber \\&\qquad +\,2a_{0}\sum _{k=1}^{N}c_{k}\sum _{h=1}^{N}b_{h}{\mathbb {E}}\left[ |e_{h}(t)|^{2}+|e_{h}(t-\tau _{2}(t))|^{2}\right. \nonumber \\&\qquad \left. +\,|e_{h}(t-\tau _{3}(t))|^{2}\right] \nonumber \\&\qquad +\,{\mathbb {E}}\beta (t)a(t)\sum _{k=1}^{N}c_{k}\sum _{h=1}^{N}a_{kh}b_{kh}\left[ |e_{h}(t-\tau _{2}(t))|^{2}\right. \nonumber \\&\qquad \left. -\,|e_{k}(t-\tau _{2}(t))|^{2}\right] \nonumber \\&\qquad +\,{\mathbb {E}}(1-\beta (t))a(t)\sum _{k=1}^{N}c_{k}\sum _{h=1}^{N}a_{kh}b_{kh}\left[ |e_{h}(t-\tau _{3}(t))|^{2}\right. \nonumber \\&\qquad \left. -\,|e_{k}(t-\tau _{3}(t))|^{2}\right] \nonumber \\&\qquad -\,\sum _{k=1}^{N}c_{k}\rho e^{-\rho t}(a(t)-a_{0})^{2}. \end{aligned}$$

(24)

According to Theorem 2.2 in [49], we can get

$$\begin{aligned}&\sum _{k=1}^{N}\sum _{h=1}^{N}c_{k}a_{kh}b_{kh}\left[ |e_{h}(t-\tau _{2}(t))|^{2}-|e_{k}(t-\tau _{2}(t))|^{2}\right] \\&\quad =\sum _{{\mathcal {Q}}\in {\mathbb {Q}}}W({\mathcal {Q}})\sum _{(s,r)\in E(C_{{\mathcal {Q}}})}\\&\qquad \times \left[ |e_{s}(t-\tau _{2}(t))|^{2}-|e_{r}(t-\tau _{2}(t))|^{2}\right] , \end{aligned}$$

where $C_{{\mathcal {Q}}}$ denotes the directed cycle of ${\mathcal {Q}}$. We can get that

$$\begin{aligned} \sum _{(s,r)\in E(C_{{\mathcal {Q}}})}\left[ |e_{s}(t-\tau _{2}(t))|^{2}-|e_{r}(t-\tau _{2}(t))|^{2}\right] =0. \end{aligned}$$

Therefore, one has

$$\begin{aligned}&\sum _{k=1}^{N}\sum _{h=1}^{N}c_{k}a_{kh}b_{kh}\\&\quad \left[ |e_{h}(t-\tau _{2}(t))|^{2}-|e_{k}(t-\tau _{2}(t))|^{2}\right] =0. \end{aligned}$$

It is easy to get that

$$\begin{aligned}&\beta (t)a(t)\sum _{k=1}^{N}c_{k}\sum _{h=1}^{N}a_{kh}b_{kh}\\&\quad \times \left[ |e_{h}(t-\tau _{2}(t))|^{2}-|e_{k}(t-\tau _{2}(t))|^{2}\right] =0. \end{aligned}$$

By a similar method, we can obtain

$$\begin{aligned}&a(t)(1-\beta (t))\sum _{k=1}^{N}c_{k}\sum _{h=1}^{N}a_{kh}b_{kh}\\&\quad \times \left[ |e_{h}(t-\tau _{3}(t))|^{2}-|e_{k}(t-\tau _{3}(t))|^{2}\right] =0. \end{aligned}$$

Therefore, one get

$$\begin{aligned}&{\mathbb {E}}{\mathcal {L}}V(e(t),t)\nonumber \\&\quad \le \sum _{k=1}^{N}c_{k}\left( 2\varGamma _{k}+\varLambda _{k}+\delta ^{2}_{k}-2d_{k}+2\frac{{\overline{c}}}{{\underline{c}}}a_{0}Nb_{k}\right) {\mathbb {E}}|e_{k}(t)|^{2}\nonumber \\&\qquad -\sum _{k=1}^{N}c_{k}\rho e^{-\rho t}(a(t)-a_{0})^{2}\nonumber \\&\qquad +\sum _{k=1}^{N}c_{k}\varLambda _{k}{\mathbb {E}}|e_{k}(t-\tau _{1}(t))|^{2}\nonumber \\&\qquad +2\sum _{k=1}^{N}c_{k}\left( \frac{{\overline{c}}}{{\underline{c}}}a_{0}Nb_{k}\right) \nonumber \\&\qquad \times {\mathbb {E}}\left[ |e_{k}(t-\tau _{2}(t))|^{2}+|e_{k}(t-\tau _{3}(t))|^{2}\right] \nonumber \\&\quad \le -\lambda {\mathbb {E}}\left[ \sum _{k=1}^{N}c_{k}\left( V_{k}(e_{k}(t))+e^{-\rho t}(a(t)-a_{0})^{2}\right) \right] \nonumber \\&\qquad +\eta \sum _{k=1}^{N}c_{k}{\mathbb {E}}\left[ |e_{k}(t-\tau _{1}(t))|^{2}\right. \nonumber \\&\qquad \left. +|e_{k}(t-\tau _{2}(t))|^{2}+|e_{k}(t-\tau _{3}(t))|^{2}\right] . \end{aligned}$$

(25)

For any $t\in [s_{m},t_{m+1})$, we can similarly compute the ${\mathcal {L}}V_{k}(e_{k}(t),t)$ as follows:

$$\begin{aligned} {\mathbb {E}}{\mathcal {L}}V_{k}(e_{k}(t),t)\le & {} \left( 2\varGamma _{k}+\varLambda _{k}+\delta ^{2}_{k}\right) {\mathbb {E}}|e_{k}(t)|^{2}\\&+\,\varLambda _{k}{\mathbb {E}}|(e_{k}(t-\tau _{1}(t))|^{2}\\&+\,2{\mathbb {E}}a(t)\beta (t)\sum _{h=1}^{N}a_{kh}b_{kh}\\&\times \,\left[ |e_{k}(t)|^{2}+|e_{k}(t-\tau _{2}(t))|^{2}\right] \\&+\,{\mathbb {E}}a(t)\beta (t)\sum _{h=1}^{N}a_{kh}b_{kh}\\&\times \,\left[ |e_{h}(t-\tau _{2}(t))|^{2}-|e_{k}(t-\tau _{2}(t))|^{2}\right] \\&+\,2{\mathbb {E}}a(t)(1-\beta (t))\sum _{h=1}^{N}a_{kh}b_{kh}\\&\times \,\left[ |e_{k}(t)|^{2}+|e_{k}(t-\tau _{3}(t))|^{2}\right] \\&+\,{\mathbb {E}}a(t)(1-\beta (t))\sum _{h=1}^{N}a_{kh}b_{kh}\\&\times \,\left[ |e_{h}(t-\tau _{3}(t))|^{2}-|e_{k}(t-\tau _{3}(t))|^{2}\right] \\&-\,{\mathbb {E}}a(t)b_{k}\left[ |e_{k}(t)|^{2}+|e_{k}(t-\tau _{2}(t))|^{2}\right. \\&+\left. |e_{k}(t-\tau _{3}(t))|^{2}\right] \\&+\,{\mathbb {E}}a_{0}\sum _{k=1}^{N}b_{k}\left[ |e_{k}(t)|^{2}+|e_{k}(t-\tau _{2}(t))|^{2}\right. \\&+\left. |e_{k}(t-\tau _{3}(t))|^{2}\right] \\&-\,\rho e^{-\rho t}(a(t)-a_{0})^{2}. \end{aligned}$$

Thus, one get

$$\begin{aligned}&{\mathbb {E}}{\mathcal {L}}V(e(t),t)\nonumber \\&\quad \le \sum _{k=1}^{N}c_{k}\left( 2\varGamma _{k}+\varLambda _{k}+\delta ^{2}_{k}+\frac{{\bar{c}}}{{\underline{c}}}a_{0}Nb_{k}\right) {\mathbb {E}}|e_{k}(t)|^{2}\nonumber \\&\qquad -\sum _{k=1}^{N}c_{k}\rho e^{-\rho t}(a(t)-a_{0})^{2}\nonumber \\&\qquad +\sum _{k=1}^{N}c_{k}\varLambda _{k}{\mathbb {E}}|e_{k}(t-\tau _{1}(t))|^{2}\nonumber \\&\qquad +\sum _{k=1}^{N}c_{k}\left( \frac{{\bar{c}}}{{\underline{c}}}a_{0}Nb_{k}\right) {\mathbb {E}}\left[ |e_{k}(t-\tau _{2}(t))|^{2}\right. \nonumber \\&\qquad \left. +|e_{k}(t-\tau _{3}(t))|^{2}\right] \nonumber \\&\quad \le {\bar{\lambda }}{\mathbb {E}}\left( \sum _{k=1}^{N}c_{k}V_{k}(e_{k}(t))+e^{-\rho t}(a(t)-a_{0})^{2}\right) \nonumber \\&\qquad +\eta \sum _{k=1}^{N}c_{k}{\mathbb {E}}\left[ |e_{k}(t-\tau _{1}(t))|^{2}\right. \nonumber \\&\qquad \left. +|e_{k}(t-\tau _{2}(t))|^{2}+|e_{k}(t-\tau _{3}(t))|^{2}\right] . \end{aligned}$$

(26)

Then, combining (25) with (26), we can conclude that

$$\begin{aligned}&{\mathbb {E}}{\mathcal {L}}V(e(t),t)\le \left\{ \begin{array}{ll} \displaystyle -\lambda {\mathbb {E}}V(e(t),t)+\eta \sum _{i=1}^{3}{\mathbb {E}}V\left( e(t-\tau _{i}(t)),t-\tau _{i}(t)\right) , &{}\quad t\in [t_{m},s_{m}), \\ \displaystyle {\bar{\lambda }}{\mathbb {E}}V(e(t),t)+\eta \sum _{i=1}^{3}{\mathbb {E}}V(e(t-\tau _{i}(t)),t-\tau _{i}(t)), &{} \quad t\in [s_{m},t_{m+1}). \end{array} \right. \end{aligned}$$

Therefore, one get

$$\begin{aligned}&{\mathbb {E}}{\mathcal {L}}V(e(t),t)\le \left\{ \begin{array}{ll} \displaystyle -\lambda {\mathbb {E}}V(e(t),t)+3\eta \sup _{t-\tau \le s\le t}\{{\mathbb {E}}V{(e(s),s)}\}, &{}\quad t\in [t_{m},s_{m}),\\ \displaystyle {\bar{\lambda }}{\mathbb {E}}V(e(t),t)+3\eta \sup _{t-\tau \le s\le t}\{{\mathbb {E}}V(e(s),s)\}, &{}\quad t\in [s_{m},t_{m+1}). \end{array} \right. \end{aligned}$$

On account of (11) and Lemma 1, we can obtain that

$$\begin{aligned}&{\mathbb {E}}V(e(t),t)\le \sup _{-\tau \le s\le 0}\{{\mathbb {E}}V(e(s),s)\} \exp \{-\sigma t\},\nonumber \\&\quad t\ge 0. \end{aligned}$$

(27)

What is more, by virtue of the definition of V(e(t), t), it yields

$$\begin{aligned} V(e(t),t)\ge \min _{k\in {\mathcal {N}}}\left\{ c_{k}\right\} |e(t)|^{2}. \end{aligned}$$

(28)

Consequently, according to (27) and (28), we can deduce

$$\begin{aligned} \limsup _{t\rightarrow \infty }\frac{1}{t}\mathrm {ln}\left( {\mathbb {E}}|e(t)|^{2}\right) \le -\sigma <0. \end{aligned}$$

Hence, in view of Definition 1, systems (1) and (2) with adaptive coupling strength achieve exponential synchronization in mean square. Then, the proof is completed. $\square$

1.2 Proof of Theorem 2

Proof

According to Assumption 2 and (8), error system (5) can be converted to

$$\begin{aligned} \text {d}{e}_{k}(t)= & {} \Big [{\tilde{f}}_{k}(e_{k}(t),e_{k}(t-\tau _{1}(t)),t)\\&+\,\beta (t){\hat{a}}_{0}\sum _{h=1}^{N}a_{kh}{\tilde{H}}_{kh} (e_{k}(t-\tau _{2}(t)),e_{h}(t-\tau _{2}(t))) \\&+\,\beta (t)(a(t)-{\hat{a}}_{0})\sum _{h=1}^{N}a_{kh}{\tilde{H}}_{kh}(e_{k}(t-\tau _{2}(t)),\\&\quad e_{h}(t-\tau _{2}(t)))\\&+\,(1-\beta (t)){\hat{a}}_{0}\sum _{h=1}^{N}a_{kh}{\tilde{H}}_{kh}(e_{k}(t-\tau _{3}(t)),\\&\quad e_{h}(t-\tau _{3}(t)))\\&+\,(1-\beta (t))(a(t)-{\hat{a}}_{0})\sum _{h=1}^{N}a_{kh}{\tilde{H}}_{kh}(e_{k}(t-\tau _{3}(t)),\\&\quad e_{h}(t-\tau _{3}(t)))\\&+\,u_{k}(t)\Big ]\text {d}t+{\tilde{g}}_{k}(e_{k}(t),t)\text {d}B(t). \end{aligned}$$

Next, let $V(e,t)=\sum _{h=1}^{N}c_{k}V_{k}(e_{k},t)$, where $V_{k}(e_{k},t)=|e_{k}|^{p}$, $c_{k}$ stands for the cofactor of the kth diagonal element of Laplacian matrix of digraph $({\mathcal {G}}, B)$, where $B=(a_{kh}b_{kh})_{N\times N}$. What is more, by virtue of Proposition 2.1 in [49], we can discover that $c_{k}>0$.

By using the Young’s inequality,

$$\begin{aligned} |a|^{p} |b|^{q}\le \frac{p}{p+q}|a|^{p+q}+ \frac{q}{p+q}|b|^{p+q} \end{aligned}$$

and Assumption 2, it is easy to see that

$$\begin{aligned}&|e_{k}(t)|^{p-2}e_{k}^{\mathrm {T}}(t){\tilde{f}}_{k}(e_{k}(t),e_{k}(t-\tau _{1}(t)),t)\nonumber \\&\quad \le |e_{k}(t)|^{p-1}(\varGamma _{k}|e_{k}(t)|+\varLambda _{k}|e_{k}(t-\tau _{1}(t))|)\nonumber \\&\quad \le \frac{p\varGamma _{k}+(p-1)\varLambda _{k}}{p}|e_{k}(t)|^{p}+\frac{\varLambda _{k}}{p}|e_{k}(t-\tau _{1}(t))|^{p}. \end{aligned}$$

(29)

Furthermore, similar to (29), it yields

$$\begin{aligned}&|e_{k}(t)|^{p-2}e_{k}^{\mathrm {T}}(t){\tilde{H}}_{kh}(e_{k}(t-\tau _{2}(t)),e_{h}(t-\tau _{2}(t))) \\&\quad \le b_{kh}\left[ \frac{2(p-1)}{p}|e_{k}(t)|^{p}+\frac{1}{p}|e_{k}(t-\tau _{2}(t))|^{p}\right. \\&\qquad \left. +\frac{1}{p}|e_{h}(t-\tau _{2}(t))|^{p}\right] . \end{aligned}$$

Therefore, when $t\in [t_{m},s_{m})$, the following inequality can be deduced

$$\begin{aligned}&{\mathbb {E}}{\mathcal {L}}V_{k}(e_{k}(t),t)\\&\quad \le \left[ p\varGamma _{k}+(p-1)\varLambda _{k}+{\hat{a}}_{0}\sum _{h=1}^{N}2a_{kh}b_{kh}(p-1)\right. \\&\qquad \left. -pd_{k}+\frac{p(p-1)}{2}\delta _{k}^{2}\right] {\mathbb {E}}|e_{k}(t)|^{p}\\&\qquad +\varLambda _{k}{\mathbb {E}}|e_{k}(t-\tau _{1}(t))|^{p}\\&\qquad +2\beta _{0}{\hat{a}}_{0}\sum _{h=1}^{N}a_{kh}b_{kh}{\mathbb {E}}|e_{k}(t-\tau _{2}(t))|^{p}\\&\qquad +\beta _{0}{\hat{a}}_{0}\sum _{h=1}^{N}a_{kh}b_{kh}\left[ {\mathbb {E}}|e_{h}(t-\tau _{2}(t))|^{p}\right. \\&\qquad \left. -{\mathbb {E}}|e_{k}(t-\tau _{2}(t))|^{p}\right] \\&\qquad +2(1-\beta _{0}){\hat{a}}_{0}\sum _{h=1}^{N}a_{kh}b_{kh}{\mathbb {E}}|e_{k}(t-\tau _{3}(t))|^{p}\\&\qquad +(1-\beta _{0}){\hat{a}}_{0}\sum _{h=1}^{N}a_{kh}b_{kh}\left[ {\mathbb {E}}|e_{h}(t-\tau _{3}(t))|^{p}\right. \\&\qquad \left. -{\mathbb {E}}|e_{k}(t-\tau _{3}(t))|^{p}\right] . \end{aligned}$$

In virtue of Theorem 2.2 in [49], we can see that

$$\begin{aligned}&\beta _{0}{\hat{a}}_{0}\sum _{k=1}^{N}\sum _{h=1}^{N}c_{k}a_{kh}b_{kh}\left[ {\mathbb {E}}|e_{h}(t-\tau _{2}(t))|^{p}\right. \\&\quad \left. -{\mathbb {E}}|e_{k}(t-\tau _{2}(t))|^{p}\right] =0,\\&(1-\beta _{0}){\hat{a}}_{0}\sum _{k=1}^{N}\sum _{h=1}^{N}c_{k}a_{kh}b_{kh}\left[ {\mathbb {E}}|e_{h}(t-\tau _{3}(t))|^{p}\right. \\&\quad \left. -{\mathbb {E}}|e_{k}(t-\tau _{3}(t))|^{p}\right] =0. \end{aligned}$$

Besides, one can deduce as follows:

$$\begin{aligned}&{\mathbb {E}}{\mathcal {L}}V(e(t),t)\nonumber \\&\quad \le \sum _{k=1}^{N}c_{k}\left[ p\varGamma _{k}+(p-1)\varLambda _{k}\right. \nonumber \\&\qquad +{\hat{a}}_{0}\sum _{h=1}^{N}2a_{kh}b_{kh}(p-1)-pd_{k}\nonumber \\&\qquad \left. +\frac{p(p-1)}{2}\delta _{k}^{2}\right] {\mathbb {E}}|e_{k}(t)|^{2} \nonumber \\&\qquad +\sum _{k=1}^{N}c_{k}\varLambda _{k}{\mathbb {E}}|e_{k}(t-\tau _{1}(t))|^{p}\nonumber \\&\qquad +\sum _{k=1}^{N}2c_{k}\beta _{0}{\hat{a}}_{0}\sum _{h=1}^{N}a_{kh}b_{kh}{\mathbb {E}}|e_{k}(t-\tau _{2}(t))|^{p}\nonumber \\&\qquad +\sum _{k=1}^{N}2c_{k}(1-\beta _{0}){\hat{a}}_{0}\sum _{h=1}^{N}a_{kh}b_{kh}{\mathbb {E}}|e_{k}(t-\tau _{3}(t))|^{p}. \end{aligned}$$

(30)

Hence, it yields by (30)

$$\begin{aligned} {\mathbb {E}}{\mathcal {L}}V(e(t),t)\le & {} -\varepsilon _{1}{\mathbb {E}}V(e(t),t)\\&+\,\varLambda {\mathbb {E}}V(e(t-\tau _{1}(t)),t-\tau _{1}(t)) \\&+\,{\bar{b}}{\mathbb {E}}V(e(t-\tau _{2}(t)),t-\tau _{2}(t))\\&+\,{\hat{b}}{\mathbb {E}}V(e(t-\tau _{3}(t)),t-\tau _{3}(t)). \end{aligned}$$

Similarly, when $t\in [s_{m},t_{m+1})$, one has

$$\begin{aligned}&{\mathbb {E}}{\mathcal {L}}V(e(t),t)\nonumber \\&\quad \le \sum _{k=1}^{N}c_{k}\left[ p\varGamma _{k}+(p-1)\varLambda _{k}\right. \nonumber \\&\qquad +{\hat{a}}_{0}\sum _{h=1}^{N}2(p-1)a_{kh}b_{kh}\nonumber \\&\qquad \left. +\frac{p(p-1)}{2}\delta _{k}^{2}\right] {\mathbb {E}}|e_{k}(t)|^{2} \nonumber \\&\qquad +\sum _{k=1}^{N}c_{k}\varLambda _{k}{\mathbb {E}}|e_{k}(t-\tau _{1}(t))|^{p}\nonumber \\&\qquad +\sum _{k=1}^{N}2c_{k}\beta _{0}{\hat{a}}_{0}\sum _{h=1}^{N}a_{kh}b_{kh}{\mathbb {E}}|e_{k}(t-\tau _{2}(t))|^{p}\nonumber \\&\qquad +\sum _{k=1}^{N}2c_{k}(1-\beta _{0}){\hat{a}}_{0}\sum _{h=1}^{N}a_{kh}b_{kh}{\mathbb {E}}|e_{k}(t-\tau _{3}(t))|^{p}\nonumber \\&\quad \le \varepsilon _{2}{\mathbb {E}}V(e(t),t)+\varLambda {\mathbb {E}}V(e(t-\tau _{1}(t)),t-\tau _{1}(t)) \nonumber \\&\qquad +{\bar{b}}{\mathbb {E}}V(e(t-\tau _{2}(t)),t-\tau _{2}(t))\nonumber \\&\qquad +{\hat{b}}{\mathbb {E}}V(e(t-\tau _{3}(t)),t-\tau _{3}(t)). \end{aligned}$$

(31)

Consequently, combining (30) with (31), we have

$$\begin{aligned}&{\mathbb {E}}{\mathcal {L}}V(e(t),t)\\&\quad \le \left\{ \begin{array}{ll} \displaystyle -\varepsilon _{1}{\mathbb {E}}V(e(t),t)+(\varLambda +{\bar{b}}+{\hat{b}})\sup _{-\tau \le s\le 0}\{{\mathbb {E}}V{(e(s),s)}\}, &{}\quad t\in [t_{m},s_{m}),\\ \displaystyle \varepsilon _{2}{\mathbb {E}}V(e(t),t)+(\varLambda +{\bar{b}}+{\hat{b}})\sup _{-\tau \le s\le 0}\{{\mathbb {E}}V(e(s),s)\}, &{}\quad t\in [s_{m},t_{m+1}). \end{array} \right. \end{aligned}$$

In view of (12) and Lemma 1, we can obtain that

$$\begin{aligned} {\mathbb {E}}V(e(t),t)\le \sup _{-\tau \le s\le 0}\{{\mathbb {E}}V(e(s),s)\} \exp \{-\sigma t\}, t\ge 0. \end{aligned}$$

Moreover, we obtain

$$\begin{aligned} \min _{k\in {\mathcal {N}}}\left\{ c_{k}\right\} |e(t)|^{p}\le V(e(t),t). \end{aligned}$$

Thus, it is clear that

$$\begin{aligned} \limsup _{t\rightarrow \infty }\frac{1}{t}\mathrm {ln}({\mathbb {E}}|e(t)|^{p})\le -\sigma <0, \end{aligned}$$

which indicates systems (1) and (2) with random coupling strength reach pth moment exponential synchronization by virtue of Definition 1. Then, the proof is completed. $\square$