Appendix 1
Proof of Lemma 4
Define \(g(s)=\frac{3s-a-2b}{2(b-a)},\ \tilde{N}=\mathrm{col}\{N_1,\ \ N_2\},\ \zeta (s)=\mathrm{col}\{\chi _3,\ \ g(s)\chi _3\}.\)
It is easy to see that the following inequality holds
$$\begin{aligned} -2\zeta (s)^T\tilde{N}\theta (s)\le \zeta (s)^T\tilde{N}R^{-1}\tilde{N}^T\zeta (s)+\theta (s)^TR\theta (s). \end{aligned}$$
(26)
Integrating (26) on domain \(\mathcal {D}=\{a\le u\le b,\ u\le s\le b\}\) yields
$$\begin{aligned} -2\int ^b_a\int ^b_u\zeta (s)^T\tilde{N}\theta (s)\mathrm{d}s\mathrm{d}u&\le \int ^b_a\int ^b_u\zeta (s)^T\tilde{N}R^{-1}\tilde{N}^T\zeta (s)\mathrm{d}s\mathrm{d}u\\&\quad +\int ^b_a\int ^b_u\theta (s)^TR\theta (s)\mathrm{d}s\mathrm{d}u. \end{aligned}$$
That is
$$\begin{aligned} -2\chi _3^T[N_1\hbar _8+N_2(-\hbar _8+\hbar _9)\chi _3&\le \frac{(b-a)^2}{2}\chi _3^T \Big (N_1R^{-1}N_1^T+\frac{1}{8}N_2(R^{-1}N_2^T\Big )\chi _3\nonumber \\&\quad +\int ^b_a\int ^b_u\theta (s)^TR\theta (s)\mathrm{d}s\mathrm{d}u. \end{aligned}$$
(27)
Rearranging (27) yields (8), and this completes the proof.
Appendix 2
Proof of Theorem 1
Consider the following LKF:
$$\begin{aligned} V(x_t,i)=e_t^TP_{1i}e_t+\sum ^6_{j=1}V_j(e_t,i), \end{aligned}$$
with
$$\begin{aligned} V_1(e_t,i)=&\int ^t_{t-\tau _i(t)}\varepsilon (s)^T{\mathcal {Q}}_i\varepsilon (s)\mathrm{d}s+\int ^t_{t-\bar{\tau }}\int ^t_{\theta }\varepsilon (s)^T{\mathcal {Q}}\varepsilon (s)\mathrm{d}s\mathrm{d}{\theta },\\ V_2(e_t,i)=&\int _{t-{\bar{\tau }_i}}^{t}\varepsilon (s)^T{{\mathcal {R}}_i}\varepsilon (s)\mathrm{d}s+\int ^t_{t-\bar{\tau }}\int ^t_{\theta }\varepsilon (s)^T{\mathcal {R}}\varepsilon (s)\mathrm{d}s\mathrm{d}{\theta },\\ V_{3}(e_t,i)=&\int ^t_{t-\bar{\tau }_i}\int ^t_{\theta }\dot{e}_s^TP_{2i}\dot{e}_s\mathrm{d}s\mathrm{d}{\theta }+\int ^t_{t-\bar{\tau }}\int ^t_{\theta }\dot{e}_s^TS_1\dot{e}_s\mathrm{d}s\mathrm{d}{\theta },\\ V_{4}(e_t,i)=&\int _{t-{\bar{\tau }_i}}^{t}{\int _\theta ^{t} {\int _u ^t {\dot{e}_s^TP_{3i}\dot{e}_s\mathrm{d}s\mathrm{d}u\mathrm{d}\theta }}}+\int _{t-{\bar{\tau }}}^{t}{\int _\theta ^{t} {\int _u ^t {\dot{e}_s^TS_2\dot{e}_s\mathrm{d}s\mathrm{d}u\mathrm{d}\theta }}},\\ V_5(e_t,i)=&\int _{t-{\bar{\tau }_i}}^{t}{\int _{t - \bar{\tau }_i}^\theta {\int _u ^t {\dot{e}_s^TP_{4i}\dot{e}_s\mathrm{d}s\mathrm{d}u}}}\mathrm{d}\theta +\int _{t-{\bar{\tau }}}^{t}{\int _{t - \bar{\tau }}^\theta {\int _u ^t {\dot{e}_s^TS_3\dot{e}_s\mathrm{d}s\mathrm{d}u}}}\mathrm{d}\theta ,\\ V_6(e_t,i)=&\bar{\tau }_i\int ^t_{t-\bar{\tau }_i}\int ^t_{\theta }e_s^TP_{5i}e_s\mathrm{d}s\mathrm{d}{\theta }+\int ^t_{t-\bar{\tau }}\int ^t_{\theta }e_s^TS_4e_s\mathrm{d}s\mathrm{d}{\theta }, \end{aligned}$$
where \(\varepsilon (s)=\mathrm{col}\{e_s,\ f(e_s)\}.\)
Taking the derivative of V(t) along the trajectory of the coupled neural networks (4) and applying Lemma 8 yields
$$\begin{aligned} \pounds V(e_t,i)=2e_t^TP_{1i}\dot{e}_t+e_t^T\sum _{j = 1}^K\pi _{i j}P_{1j}e_t+\sum ^5_{j=1}\pounds V_j(e_t,i), \end{aligned}$$
(28)
where
$$\begin{aligned} \pounds V_1(e_t,i)&=\varepsilon (t)^T({\mathcal {Q}}_i+\bar{\tau }{\mathcal {Q}}) \varepsilon (t) -[1-\dot{\tau }_i(t)]\varepsilon (t-\tau _i(t))^T{\mathcal {Q}}_i \varepsilon (t-\tau _i(t))\nonumber \\&\quad +\sum _{j = 1}^K\pi _{i j}\int ^t_{t-\tau _j(t)}\varepsilon (s)^T{\mathcal {Q}}_j \varepsilon (s)\mathrm{d}s-\int ^t_{t-\bar{\tau }}\varepsilon (s)^T{\mathcal {Q}} \varepsilon (s)\mathrm{d}s, \end{aligned}$$
(29)
$$\begin{aligned} \pounds V_2(e_t,i)&=\varepsilon (t)^T({\mathcal {R}}_i+\bar{\tau }{\mathcal {R}}) \varepsilon (t) -\varepsilon (t-\bar{\tau }_i)^T{\mathcal {R}}_i\varepsilon (t-\bar{ \tau }_i)\nonumber \\&\quad +\sum _{j = 1}^K\pi _{i j}\int ^t_{t-\bar{\tau }_j}\varepsilon (s)^T {\mathcal {R}}_j\varepsilon (s)\mathrm{d}s-\int ^t_{t-\bar{\tau }}\varepsilon (s)^T {\mathcal {R}}\varepsilon (s)\mathrm{d}s, \end{aligned}$$
(30)
$$\begin{aligned} \pounds V_3(e_t,i)=&\dot{e}_t^T\big (\bar{\tau }_iP_{2i}+\bar{\tau }S_1\big ) \dot{e}_t -\int _{t-{\bar{\tau }_i}}^{t}\dot{e}_s^TP_{2i}\dot{e}_s\mathrm{d}s\nonumber \\&\quad +\sum _{j = 1}^K\pi _{i j}\int ^t_{t-\bar{\tau }_j}\int ^t_{\theta } \dot{e}_s^TP_{2j}\dot{e}_s\mathrm{d}s\mathrm{d}{\theta }-\int ^t_{t-\bar{\tau }} \dot{e}_s^TS_1\dot{e}_s\mathrm{d}s, \end{aligned}$$
(31)
$$\begin{aligned} \pounds V_4(e_t,i)=&\frac{1}{2}\dot{e}_t^T\big (\bar{\tau }^2_iP_{3i}+\bar{\tau }^ 2S_2\big )\dot{e}_t\nonumber \\&\quad +\sum _{j = 1}^K\pi _{i j}\int _{t-{\bar{\tau }_j}}^{t}{\int _\theta ^{t} {\int _u ^t {\dot{e}_s^TP_{3j}\dot{e}_s\mathrm{d}s\mathrm{d}u\mathrm{d}\theta }}}\nonumber \\&\quad -\int _{t-{\bar{\tau }_i}}^{t}{\int _\theta ^{t} \dot{e}_s^TP_{3i} \dot{e}_s\mathrm{d}s}\mathrm{d}\theta -\int _{t-{\bar{\tau }}}^{t}{\int _\theta ^{t} {{\dot{e}_s^TS_2\dot{e}_s\mathrm{d}s\mathrm{d}\theta }}}, \end{aligned}$$
(32)
$$\begin{aligned} \pounds V_5(e_t,i)=&\frac{1}{2}\dot{e}_t^T\big (\bar{\tau }^2_iP_{4i}+\bar{\tau }^ 2S_3\big )\dot{e}_t \nonumber \\&\quad +\sum _{j = 1}^K\pi _{i j}\int _{t-{\bar{\tau }_j}}^{t}{\int _{t-{\bar{ \tau }_j}}^\theta {\int _u ^t {\dot{e}_s^TP_{4j}\dot{e}_s\mathrm{d}s\mathrm{d}u\mathrm{d} \theta }}}\nonumber \\&\quad -\int _{t-{\bar{\tau }_i}}^{t}{\int _{t-{\bar{\tau }_i}}^\theta \dot{e}_s^TP _{4i}\dot{e}_s\mathrm{d}s}\mathrm{d}\theta -\int _{t-{\bar{\tau }}}^{t} {\int _{t-\bar{ \tau }}^\theta \dot{e}_s^TS_3\dot{e}_s\mathrm{d}s}\mathrm{d}\theta , \end{aligned}$$
(33)
$$\begin{aligned} \pounds V_6(e_t,i)=&e_t^T\big (\bar{\tau }_i^2P_{5i}+\bar{\tau }S_4\big )e_t -\bar{\tau }_i\int _{t-{\bar{\tau }_i}}^{t}e_s^TP_{5i}e_s\mathrm{d}s\nonumber \\&\quad +\sum _{j = 1}^K\pi _{i j}\bar{\tau }_j\int ^t_{t-\bar{\tau }_j}\int ^t_{\theta } e_s^TP_{5j}e_s\mathrm{d}s\mathrm{d}{\theta }-\int ^t_{t-\bar{\tau }}e_s^TS_4e_s\mathrm{d}s. \end{aligned}$$
(34)
Based on \(\pi _{i j}\ge 0(i\ne j),\pi _{i i}\le 0,\ {\mathcal {Q}}_j>0(i,j=1,2,\ldots ,K),\) applying (14) yields
$$\begin{aligned} \sum _{j = 1}^K\pi _{i j}\int ^t_{t-\tau _j(t)}\varepsilon (s)^T{\mathcal {Q}}_j\varepsilon (s)\mathrm{d}s\le&\sum _{j = 1,j\ne i}^K\pi _{i j}\int ^t_{t-\tau _j(t)}\varepsilon (s)^T{\mathcal {Q}}_j\varepsilon (s)\mathrm{d}s\nonumber \\ \le&\sum _{j = 1,j\ne i}^K\pi _{i j}\int ^t_{t-\bar{\tau }}\varepsilon (s)^T{\mathcal {Q}}_j\varepsilon (s)\mathrm{d}s\nonumber \\ \le&\int ^t_{t-\bar{\tau }}\varepsilon (s)^T{\mathcal {Q}}\varepsilon (s)\mathrm{d}s. \end{aligned}$$
(35)
Similarly, applying (14) yields
$$\begin{aligned} \sum _{j = 1}^K\pi _{i j}\int ^t_{t-\bar{\tau }_j}\varepsilon (s)^T{\mathcal {R}}_j\varepsilon (s)\mathrm{d}s\le&\sum _{j = 1,j\ne i}^K\pi _{i j}\int ^t_{t-\bar{\tau }_j}\varepsilon (s)^T{\mathcal {R}}_j\varepsilon (s)\mathrm{d}s\nonumber \\ \le&\sum _{j = 1,j\ne i}^K\pi _{i j}\int ^t_{t-\bar{\tau }}\varepsilon (s)^T{\mathcal {R}}_j\varepsilon (s)\mathrm{d}s\nonumber \\ \le&\int ^t_{t-\bar{\tau }}\varepsilon (s)^T{\mathcal {R}}\varepsilon (s)\mathrm{d}s. \end{aligned}$$
(36)
Applying Lemma 3 to \(P_{2i}\)-dependent terms gives
$$\begin{aligned} -\int ^{t}_{t-\tau _i(t)}\dot{e}_s^T{P_{2i}}\dot{e}_s\mathrm{d}s&\le \eth _1^T\bar{\Omega }_1(t)\eth _1, \end{aligned}$$
(37)
$$\begin{aligned} -\int ^{t-\tau _i(t)}_{t-{\bar{\tau }_i}}\dot{e}_s^T{P_{2i}}\dot{e}_s\mathrm{d}s&\le \eth _2^T\bar{\Omega }_2(t)\eth _2, \end{aligned}$$
(38)
where \(\eth _1=\mathrm{col}\{e_t\ \ \ e_{\tau _i} \ \ \ y_1\ \ \ y_3\},\ \ \eth _2=\mathrm{col}\{e_{\tau _i}\ \ \ e_{\bar{\tau }_i} \ \ \ y_2\ \ \ y_4\},\) and
$$\begin{aligned} \bar{\Omega }_1(t)&=\tau _i(t)\Big (X_{1i}P_{2i}^{-1}X_{1i}^T+\frac{1}{3}X_{2i} P_{2i}^{-1}X_{2i}^T +\frac{1}{5}X_{3i}P_{2i}^{-1}X_{3i}^T\Big )\\&\quad +\mathrm{sym}\{X_{1i}(\hbar _4-\hbar _5)+X_{2i}( \hbar _4+\hbar _5-2\hbar _6)+X_{3i}(\hbar _4-\hbar _5+6\hbar _6-6\hbar _7)\},\\ \bar{\Omega }_2(t)&=[\bar{\tau }_i-\tau _i(t)]\Big (X_{4i}P_{2i}^{-1}X_{4i}^T +\frac{1}{3}X_{5i}P_{2i}^{-1}X_{5i}^T+\frac{1}{5}X_{6i}P_{2i}^{-1}X_{6i}^T\Big )\\&\quad +\mathrm{sym}\{X_{4i}(\hbar _4-\hbar _5)+X_{5i}( \hbar _4+\hbar _5-2\hbar _6)+X_{6i}(\hbar _4-\hbar _5+6\hbar _6-6\hbar _7)\}. \end{aligned}$$
Similar to (35), changing the order of integral and applying (15) yields
$$\begin{aligned}&\sum _{j = 1}^K\pi _{i j}\int ^t_{t-\bar{\tau }_j}\int ^t_{\theta }\dot{e}_s^TP_{2j}\dot{e}_s\mathrm{d}s\mathrm{d}{\theta }\nonumber \\&\quad \le \sum _{j = 1,j\ne i}^K\pi _{i j}\int ^t_{t-\bar{\tau }_j}\int ^t_{\theta }\dot{e}_s^TP_{2j}\dot{e}_s\mathrm{d}s\mathrm{d}{\theta }\nonumber \\&\quad =\sum _{j = 1,j\ne i}^K\pi _{i j}\int ^t_{t-\bar{\tau }_j}\int ^s_{t-\bar{\tau }_j}\dot{e}_s^TP_{2j}\dot{e}_s\mathrm{d}{\theta }\mathrm{d}s\nonumber \\&\quad =\sum _{j = 1,j\ne i}^K\pi _{i j}\int ^t_{t-\bar{\tau }_j}(s-t+\bar{\tau }_j)\dot{e}_s^TP_{2j}\dot{e}_s\mathrm{d}s\nonumber \\&\quad \le \sum _{j = 1,j\ne i}^K\pi _{i j}\bar{\tau }_j\int ^t_{t-\bar{\tau }_j}\dot{e}_s^TP_{2j}\dot{e}_s\mathrm{d}s\nonumber \\&\quad \le \sum _{j = 1,j\ne i}^K\pi _{i j}\bar{\tau }_j\int ^t_{t-\bar{\tau }}\dot{e}_s^TP_{2j}\dot{e}_s\mathrm{d}s\nonumber \\&\quad \le \int ^t_{t-\bar{\tau }}\dot{e}_s^TS_1\dot{e}_s\mathrm{d}s. \end{aligned}$$
(39)
Obviously the following equality holds for any \(t>0\)
$$\begin{aligned}&\int _{t-{\bar{\tau }_i}}^{t}{\int _\theta ^{t} \dot{e}_s^TP_{3i}\dot{e}_s\mathrm{d}s}\mathrm{d}\theta \nonumber \\&\quad =\int _{t-\bar{\tau }_i}^{t-\tau _i(t)}\int _\theta ^{t-\tau _i(t)}{\dot{e}_s^TP_{3i}\dot{e}_s\mathrm{d}s}\mathrm{d}\theta \nonumber \\&\quad +\int _{t-\tau _i(t)}^{t}\int _\theta ^{t}{\dot{e}_s^TP_{3i}\dot{e}_s\mathrm{d}s}\mathrm{d}\theta +[\bar{\tau }_i-\tau _i(t)]\int _{t-\tau _i(t)}^{t}{\dot{e}_s^TP_{3i}\dot{e}_s\mathrm{d}s}. \end{aligned}$$
(40)
Applying Lemma 4 to \(P_{3i}\)-dependent double integral terms gives
$$\begin{aligned} -\int _{t-\tau _i(t)}^{t}\int _\theta ^{t} \dot{e}_s^TP_{3i}\dot{e}_s\mathrm{d}s\mathrm{d}\theta&\le \eth _3^T\bar{\Omega }_3(t)\eth _3, \end{aligned}$$
(41)
$$\begin{aligned} -\int _{t-\bar{\tau }_i}^{t-\tau _i(t)}\int _\theta ^{t-\tau _i(t)} \dot{e}_s^TP_{3i}\dot{e}_s\mathrm{d}s\mathrm{d}\theta&\le \eth _4^T\bar{\Omega }_4(t)\eth _4, \end{aligned}$$
(42)
where
$$\begin{aligned} \eth _3&=\mathrm{col}\{e_t\ \ \ y_1\ \ \ y_3\},\ \ \eth _4=\mathrm{col}\{e_{\tau _i}\ \ \ y_2\ \ \ y_4\},\\ \bar{\Omega }_3(t)&=\frac{\tau _i^2(t)}{2}\Big (X_{7i}P_{3i}^{-1}X_{7i}^T+ \frac{1}{8}X_{8i}P_{3i}^{-1}X_{8i}^T\Big )+\tau _i(t)\mathrm{sym}\Big \{X_{7i}( \hbar _1-\hbar _2)\\&\quad +\frac{1}{2}X_{8i}(\hbar _1+2\hbar _2-3\hbar _3)\Big \},\\ \bar{\Omega }_4(t)&=\frac{[\bar{\tau }_i-\tau _i(t)]^2}{2}\Big (X_{9i}P_{3i}^{-1} X_{9i}^T+\frac{1}{8}X_{10i}P_{3i}^{-1}X_{10i}^T\Big ) +[\bar{\tau }_i-\tau _i(t)] \mathrm{sym}\Big \{X_{9i}(\hbar _1-\hbar _2)\\&\quad +\frac{1}{2}X_{10i}(\hbar _1+2\hbar _2-3 \hbar _3)\Big \}. \end{aligned}$$
Setting \(\vartheta =\frac{\tau _i(t)}{\bar{\tau }_i},\omega =1-\vartheta ,\) when \(0<\tau _i(t)<\bar{\tau }_i,\) applying inequality (7) to \(P_{3i}\)-dependent single integral terms derives
$$\begin{aligned}{}[\bar{\tau }_i-\tau _i(t)]\int ^{t}_{t-\tau _i(t)}{\dot{e}_s^TP_{3i}\dot{e}_s}\mathrm{d}s&\ge \frac{\omega }{\vartheta }\Big \{(e_t-e_{\tau _i})^T{P_{3i}}(e_t-e_{\tau _i})\nonumber \\&\quad +3(e_t+e_{\tau _i}-2y_{1})^T{P_{3i}}(e_t+e_{\tau _i}-2y_{1})\nonumber \\&\quad +5(e_t-e_{\tau _i}+6y_{1}-6y_3)^T{P_{3i}}(e_t-e_{\tau _i}+6y_{1}-6y_3)\Big \}. \end{aligned}$$
(43)
Similar to (35), changing the order of integral and applying (15) yields
$$\begin{aligned}&\sum _{j = 1}^K\pi _{i j}\int _{t-{\bar{\tau }_j}}^{t}{\int _\theta ^{t} {\int _u ^t {\dot{e}_s^TP_{3j}\dot{e}_s\mathrm{d}s\mathrm{d}u\mathrm{d}\theta }}}\nonumber \\&\quad \le \sum _{j = 1,j\ne i}^K\pi _{i j}\int _{t-{\bar{\tau }_j}}^{t}{\int _\theta ^{t} {\int _u ^t {\dot{e}_s^TP_{3j}\dot{e}_s\mathrm{d}s\mathrm{d}u\mathrm{d}\theta }}} \nonumber \\&\quad =\sum _{j = 1,j\ne i}^K\pi _{i j}\int _{t-{\bar{\tau }_j}}^{t}{\int _\theta ^{t} {\int _\theta ^s {\dot{e}_s^TP_{3j}\dot{e}_s\mathrm{d}u\mathrm{d}s\mathrm{d}\theta }}}\nonumber \\&\quad =\sum _{j = 1,j\ne i}^K\pi _{i j}\int _{t-{\bar{\tau }_j}}^{t}{\int _\theta ^{t} {(s-\theta ) {\dot{e}_s^TP_{3j}\dot{e}_s\mathrm{d}s\mathrm{d}\theta }}}\nonumber \\&\quad \le \sum _{j = 1,j\ne i}^K\pi _{i j}\bar{\tau }_j\int _{t-{\bar{\tau }_j}}^{t}{\int _\theta ^{t} {{\dot{e}_s^TP_{3j}\dot{e}_s\mathrm{d}s\mathrm{d}\theta }}}\nonumber \\&\quad \le \sum _{j = 1,j\ne i}^K\pi _{i j}\bar{\tau }_j\int _{t-{\bar{\tau }}}^{t}{\int _\theta ^{t} {{\dot{e}_s^TP_{3j}\dot{e}_s\mathrm{d}s\mathrm{d}\theta }}}\nonumber \\&\quad \le \int _{t-{\bar{\tau }}}^{t}{\int _\theta ^{t} { {\dot{e}_s^TS_2\dot{e}_s\mathrm{d}s\mathrm{d}\theta }}}, \end{aligned}$$
(44)
and
$$\begin{aligned}&\sum _{j = 1}^K\pi _{i j}\int _{t-{\bar{\tau }_j}}^{t}{\int _{t-{\bar{\tau }_j}}^\theta {\int _u ^t {\dot{e}_s^TP_{4j}\dot{e}_s\mathrm{d}s\mathrm{d}u\mathrm{d}\theta }}} \nonumber \\&\quad \le \sum _{j = 1,j\ne i}^K\pi _{i j}\int _{t-{\bar{\tau }_j}}^{t}{\int _{t-{\bar{\tau }_j}}^\theta {\int _u ^t {\dot{e}_s^TP_{4j}\dot{e}_s\mathrm{d}s\mathrm{d}u\mathrm{d}\theta }}}\nonumber \\&\quad =\sum _{j = 1,j\ne i}^K\pi _{i j}\int _{t-{\bar{\tau }_j}}^{t}{\int _{t-{\bar{\tau }_j}}^\theta {\int _{t-{\bar{\tau }_j}} ^s {\dot{e}_s^TP_{4j}\dot{e}_s\mathrm{d}u\mathrm{d}s\mathrm{d}\theta }}}\nonumber \\&\quad =\sum _{j = 1,j\ne i}^K\pi _{i j}\int _{t-{\bar{\tau }_j}}^{t}{\int _{t-{\bar{\tau }_j}}^\theta {(s-t+{\bar{\tau }_j}) {\dot{e}_s^TP_{4j}\dot{e}_s\mathrm{d}s\mathrm{d}\theta }}}\nonumber \\&\quad \le \sum _{j = 1,j\ne i}^K\pi _{i j}{\bar{\tau }_j}\int _{t-{\bar{\tau }_j}}^{t}{\int _{t-{\bar{\tau }_j}}^\theta {{\dot{e}_s^TP_{4j}\dot{e}_s\mathrm{d}s\mathrm{d}\theta }}}\nonumber \\&\quad \le \sum _{j = 1,j\ne i}^K\pi _{i j}{\bar{\tau }_j}\int _{t-{\bar{\tau }}}^{t}{\int _{t-{\bar{\tau }}}^\theta {{\dot{e}_s^TP_{4j}\dot{e}_s\mathrm{d}s\mathrm{d}\theta }}}\nonumber \\&\quad \le \int _{t-{\bar{\tau }}}^{t} {\int _{t-\bar{\tau }}^\theta \dot{e}_s^TS_3\dot{e}_s\mathrm{d}s}\mathrm{d}\theta . \end{aligned}$$
(45)
It is easy to see that the following equality holds for any \(t>0\)
$$\begin{aligned}&\int _{t-\bar{\tau }_i}^{t}\int _{t - \bar{\tau }_i } ^\theta \dot{e}_s^TP_{4i}\dot{e}_s\mathrm{d}s\mathrm{d}\theta \nonumber \\&\quad =\int _{t-\tau _i(t)}^{t}\int _{t-\tau _i(t)}^\theta \dot{e}_s^TP_{4i}\dot{e}_s\mathrm{d}s\mathrm{d}\theta \nonumber \\&\quad +\int _{t-\bar{\tau }_i}^{t-\tau _i(t)}\int _{t-\bar{\tau }_i} ^\theta \dot{e}_s^TP_{4i}\dot{e}_s\mathrm{d}s\mathrm{d}\theta +\tau _i(t)\int _{t - \bar{\tau }_i }^{t-\tau _i(t)} \dot{e}_s^TP_{4i}\dot{e}_s\mathrm{d}s. \end{aligned}$$
(46)
Applying Lemma 5 to \(P_{4i}\)-dependent double integral terms gives
$$\begin{aligned} -\int _{t-\tau _i(t)}^{t}\int _{t-\tau _i(t)}^\theta \dot{e}_s^TP_{4i}\dot{e}_s\mathrm{d}s\mathrm{d}\theta&\le \eth _5^T\bar{\Omega }_5(t)\eth _5, \end{aligned}$$
(47)
$$\begin{aligned} -\int _{t-\bar{\tau }_i}^{t-\tau _i(t)}\int _{t-\bar{\tau }_i} ^\theta \dot{e}_s^TP_{4i}\dot{e}_s\mathrm{d}s\mathrm{d}\theta&\le \eth _6^T\bar{\Omega }_6(t)\eth _6, \end{aligned}$$
(48)
where
$$\begin{aligned} \eth _5&=\mathrm{col}\{e_{\tau _i}\ \ \ y_1\ \ \ y_3\},\ \ \eth _6=\mathrm{col}\{e_{\bar{\tau }_i}\ \ \ y_2\ \ \ y_4\},\\ \bar{\Omega }_5(t)&=\frac{\tau _i^2(t)}{2}\Big (X_{11i}P_{4i}^{-1}X_{11i}^T+ \frac{1}{8}X_{12i}P_{4i}^{-1}X_{12i}^T\Big )\\&\quad +\tau _i(t)\mathrm{sym}\Big \{X_{11i}(-\hbar _1+\hbar _2)+\frac{1}{2}X_{12i}( \hbar _1-4\hbar _2+3\hbar _3)\Big \},\\ \bar{\Omega }_6(t)&=\frac{[\bar{\tau }_i-\tau _i(t)]^2}{2}\Big (X_{13i}P_{4i}^{-1} X_{13i}^T +\frac{1}{8}X_{14i}P_{4i}^{-1}X_{14i}^T\Big )\\&\quad +[\bar{\tau }_i-\tau _i(t)]\mathrm{sym}\Big \{X_{13i}(-\hbar _1+\hbar _2)+ \frac{1}{2}X_{14i}(\hbar _1-4\hbar _2+3\hbar _3)\Big \}. \end{aligned}$$
For \(0<\tau _i(t)<\bar{\tau }_i,\) applying inequality (7) to \(P_{4i}\)-dependent single integral terms derives
$$\begin{aligned} \tau _i(t)\int ^{t-\tau _i(t)}_{t-{\bar{\tau }_i}}\dot{e}_s^TP_{4i}\dot{e}_s\mathrm{d}s&\ge \frac{\vartheta }{\omega }\Big \{(e_{\tau _i}-e_{\bar{\tau }_i})^T{P_{4i}} (e_{\tau _i}-e_{\bar{\tau }_i})\nonumber \\&\quad +3(e_{\tau _i}+e_{\bar{\tau }_i}-2y_{2})^T{P_{4i}}(e_{\tau _i}+e_{\bar{\tau }_i}-2y_{2})\nonumber \\&\quad +5(e_{\tau _i}-e_{\bar{\tau }_i}+6y_{2}-6y_4)^T{P_{4i}}(e_{\tau _i} -e_{\bar{\tau }_i}+6y_{2}-6y_4)\Big \}. \end{aligned}$$
(49)
According to conditions (16), the following inequalities hold for any \(t>0\)
$$\begin{aligned}&\left[ \begin{array}{c} \sqrt{\frac{\omega }{\vartheta }}(e_t-e_{\tau _i})\\ -\sqrt{\frac{\vartheta }{\omega }}(e_{\tau _i}-e_{\bar{\tau }_i})\end{array}\right] ^T\bigg [\begin{array}{cc} P_{3i} &{} Y_{1 i}\\ *&{} P_{4i} \end{array}\bigg ]\left[ \begin{array}{c} \sqrt{\frac{\omega }{\vartheta }}(e_t-e_{\tau _i})\\ -\sqrt{\frac{\vartheta }{\omega }}(e_{\tau _i}-e_{\bar{\tau }_i})\end{array}\right] \ge 0, \\&\left[ \begin{array}{c} \sqrt{\frac{\omega }{\vartheta }}(e_t+e_{\tau _i}-2y_{1})\\ -\sqrt{\frac{\vartheta }{\omega }}(e_{\tau _i}+e_{\bar{\tau }_i}-2y_{2})\end{array}\right] ^T\bigg [\begin{array}{cc} P_{3i} &{} Y_{2i}\\ *&{} P_{4i} \end{array}\bigg ] \left[ \begin{array}{c} \sqrt{\frac{\omega }{\vartheta }}(e_t+e_{\tau _i}-2y_{1})\\ -\sqrt{\frac{\vartheta }{\omega }}(e_{\tau _i}+e_{\bar{\tau }_i}-2y_{2})\end{array}\right] \ge 0,\\&\left[ \begin{array}{c} \sqrt{\frac{\omega }{\vartheta }}(e_t-e_{\tau _i}+6y_{1}-6y_3)\\ -\sqrt{\frac{\vartheta }{\omega }}(e_{\tau _i}-e_{\bar{\tau }_i}+6y_{2}-6y_4)\end{array}\right] ^T\bigg [\begin{array}{cc} P_{3i} &{} Y_{3i}\\ *&{} P_{4i} \end{array}\bigg ] \left[ \begin{array}{c} \sqrt{\frac{\omega }{\vartheta }}(e_t-e_{\tau _i}+6y_{1}-6y_3)\\ -\sqrt{\frac{\vartheta }{\omega }}(e_{\tau _i}-e_{\bar{\tau }_i}+6y_{2}-6y_4)\end{array}\right] \ge 0. \end{aligned}$$
Thus one obtains
$$\begin{aligned}&\frac{\omega }{\vartheta }(e_t-e_{\tau _i})^TP_{3i}(e_t-e_{\tau _i}) +\frac{\vartheta }{\omega }(e_{\tau _i}-e_{\bar{\tau }_i})^TP_{4i}(e_{\tau _i} -e_{\bar{\tau }_i})\nonumber \\&\quad \ge 2( e_t-e_{\tau _i})^TY_{1i}(e_{\tau _i}-e_{\bar{\tau }_i}), \end{aligned}$$
(50)
$$\begin{aligned}&\frac{\omega }{\vartheta }(e_t+e_{\tau _i}-2y_{1})^TP_{3i} (e_t+e_{\tau _i}-2y_{1})+ \frac{\vartheta }{\omega }(e_{\tau _i}+e_{\bar{\tau }_i}-2y_{2})^TP_{4i}(e_{\tau _i}+ e_{\bar{\tau }_i}-2y_{2})\nonumber \\&\quad \ge 2( e_t+e_{\tau _i}-2y_{1})^T Y_{2i}(e_{\tau _i}+ e_{\bar{\tau }_i}-2y_{2}), \end{aligned}$$
(51)
$$\begin{aligned}&\frac{\omega }{\vartheta }(e_t-e_{\tau _i}+6y_{1}-6y_3)^T{P_{3i}}(e_t-e_{\tau _i}+ 6y_{1}-6y_3)\nonumber \\&\quad +\frac{\vartheta }{\omega }(e_{\tau _i}-e_{\bar{\tau }_i}+6y_{2}-6y_4)^ T{P_{4i}}(e_{\tau _i} -e_{\bar{\tau }_i}+6y_{2}-6y_4)\nonumber \\&\quad \ge 2(e_t-e_{\tau _i}+6y_{1}-6y_3)^TY_{3i}(e_{\tau _i}-e_{\bar{\tau }_i}+ 6y_{2}-6y_4). \end{aligned}$$
(52)
When \(0<\tau _i(t)<\bar{\tau }_i,\) applying the well-known Jensen integral inequality [5] yields
$$\begin{aligned} \bar{\tau }_i\int _{t-{\bar{\tau }_i}}^t {e_s^TP_{5i}e_s}\mathrm{d}s&=\bar{\tau }_i\int _{t-\tau _i(t)}^t {e_s^TP_{5i}e_s}\mathrm{d}s+\bar{\tau }_i\int _{t-\bar{\tau }_i}^{t-\tau _i(t)} {e_s^TP_{5i}e_s}\mathrm{d}s\nonumber \\&\ge \frac{\bar{\tau }_i}{\tau _i(t)}\bigg (\int _{t-\tau _i(t)}^t {e_s}\mathrm{d}s\bigg )^TP_{5i}\bigg (\int _{t-\tau _i(t)}^t {e_s}\mathrm{d}s\bigg )\nonumber \\&\quad +\frac{\bar{\tau }_i}{\bar{\tau }_i-\tau _i(t)}\bigg (\int _{t-\bar{\tau }_i}^{t-\tau _i(t)} {e_s}\mathrm{d}s\bigg )^TP_{5i}\bigg (\int _{t-\bar{\tau }_i}^{t-\tau _i(t)} {e_s}\mathrm{d}s\bigg )\nonumber \\&=\big [1+\hslash (t)\big ]\theta _1(t)^TP_{5i}\theta _1(t) +\Big [1+\frac{1}{\hslash (t)}\Big ]\theta _2(t)^TP_{5i}\theta _2(t), \end{aligned}$$
(53)
where \(\hslash (t)=\frac{\bar{\tau }_i-\tau _i(t)}{\tau _i(t)},\theta _1(t)=\int _{t-\tau _i(t)}^t{e_s}\mathrm{d}s,\theta _2(t)=\int _{t-\bar{\tau }_i}^{t-\tau _i(t)}{e_s}\mathrm{d}s.\)
By applying condition (18), we derive
$$\begin{aligned}&\left[ \begin{array}{c} \sqrt{\hslash (t)}\theta _1(t)\\ -\frac{1}{\sqrt{\hslash (t)}}\theta _2(t)\end{array}\right] ^T\bigg [\begin{array}{cc} P_{5i}&{} Y_{4i}\\ * &{}P_{5i}\end{array}\bigg ]\left[ \begin{array}{c} \sqrt{\hslash (t)}\theta _1(t)\\ -\frac{1}{\sqrt{\hslash (t)}}\theta _2(t)\end{array}\right] \ge 0, \end{aligned}$$
which implies
$$\begin{aligned} \hslash (t)\theta _1(t)^TP_{5i}\theta _1(t) +\frac{1}{\hslash (t)}\theta _2(t)^TP_{5i}\theta _2(t)\ge \theta _1(t)^TY_{4i}\theta _2(t)+\theta _2(t)^TY_{4i}^T\theta _1(t). \end{aligned}$$
(54)
Then substituting (54) into (53) gives
$$\begin{aligned}&-\bar{\tau }_i\int _{t-{\bar{\tau }_i}}^t {e_s^TP_{5i}e_s}\mathrm{d}s\nonumber \\&\quad \le -\theta _1(t)^TP_{5i}\theta _1(t)-\theta _2(t)^TP_{5i}\theta _2(t) -\theta _1(t)^TY_{4i}\theta _2(t)-\theta _2(t)^TY_{4i}^T\theta _1(t)\nonumber \\&\quad =-\bigg [\begin{array}{c} \theta _1(t)\\ \theta _2(t)\end{array}\bigg ]^T\bigg [\begin{array}{cc} P_{5i}&{} Y_{4i}\\ * &{}P_{5i}\end{array}\bigg ]\bigg [\begin{array}{c} \theta _1(t)\\ \theta _2(t)\end{array}\bigg ]. \end{aligned}$$
(55)
Note that when \(\tau _i(t)=0\) or \(\tau _i(t)=\bar{\tau }_i,\) we have
$$\begin{aligned} \theta _1(t)=0,\ \ \theta _2(t)=\int _{t-{\bar{\tau }_i}}^t {e_s}\mathrm{d}s, \end{aligned}$$
or
$$\begin{aligned} \theta _1(t)=\int _{t-{\bar{\tau }_i}}^t {e_s}\mathrm{d}s,\ \ \theta _2(t)=0 \end{aligned}$$
, respectively. Thus, inequality (55) still holds according to Jensen integral inequality.
Similar to inequality (39), changing the order of integral gives
$$\begin{aligned} \sum _{j = 1}^K\pi _{i j}\bar{\tau }_j\int ^t_{t-\bar{\tau }_j}\int ^t_{\theta }e_s^TP_{5j}e_s\mathrm{d}s\mathrm{d}{\theta } \le \sum _{j = 1,j\ne i}^K\pi _{i j}\bar{\tau }_j^2\int ^t_{t-\bar{\tau }}e_s^TP_{5j}e_s\mathrm{d}s. \end{aligned}$$
(56)
Considering networks (4), the following zero equality holds for any positive matrix \(Z_{i}=\mathrm{diag}\{z_{11i},\ldots ,z_{1ni},\ldots ,\)\(z_{(N-1)1i},\ldots ,z_{(N-1)ni}\}\)
$$\begin{aligned}&2\dot{e}_t^TZ_{i}\bigg \{-\dot{e}_t-{C}_ie_t+{A}_if(e_t)+{B}_if(e_{\tau _i}) +\zeta \bigg (\int ^{t}_{t-\tau _i(t)}e_s\mathrm{d}s\bigg )\\&\quad +\alpha _i(t){U}_ie_t+\beta _i(t){V}_ie_{\tau _i}+\gamma _i(t){W}_i\int ^{t}_{t-\tau _i(t)}e_s\mathrm{d}s\bigg \}=0. \end{aligned}$$
Based on Assumption 2, taking mathematical expectation on both sides of above equality yields
$$\begin{aligned}&2\dot{e}_t^TZ_{i}\bigg \{-\dot{e}_t-{C}_ie_t+{A}_if(e_t)+{B}_if(e_{\tau _i})\nonumber \\&\quad +\zeta \bigg (\int ^{t}_{t-\tau _i(t)}e_s\mathrm{d}s\bigg )+\bar{\alpha }_i{U}_ie_t+\bar{\beta }_i{V}_ie_{\tau _i}+\bar{\gamma }_i{W}_i\int ^{t}_{t-\tau _i(t)}e_s\mathrm{d}s\bigg \}=0. \end{aligned}$$
(57)
For any positive scalar \(\tilde{\iota }_i,\) applying inequality (19) and Lemma 6 yields
$$\begin{aligned}&2\dot{e}_t^TZ_{i}\zeta \bigg (\int ^{t}_{t-\tau _i(t)}e_s\mathrm{d}s\bigg )\nonumber \\&\quad =2\sum ^{N-1}_{k=1}\sum ^n_{j=1}\big [\dot{x}_{kj}(t)-\dot{x}_{(k+1)j}(t)\big ]z_{kji}\bigg [\bigwedge ^n_{l=1}\varpi _{jl} \int ^{t}_{t-\tau _i(t)}x_{kl}(s)\mathrm{d}s\nonumber \\&\qquad -\bigwedge ^n_{l=1}\varpi _{jl}\int ^{t}_{t-\tau _i(t)}x_{(k+1)l}(s)\mathrm{d}s+\bigvee ^n_{l=1}\varrho _{jl} \int ^{t}_{t-\tau _i(t)}x_{kl}(s)\mathrm{d}s-\bigvee ^n_{l=1}\varrho _{jl}\int ^{t}_{t-\tau _i(t)}x_{(k+1)l}(s)\mathrm{d}s\bigg ]\nonumber \\&\quad \le 2\sum ^{N-1}_{k=1}\sum ^n_{j=1}\big |\dot{x}_{kj}(t)-\dot{x}_{(k+1)j}(t)\big |z_{kji}\sum ^n_{l=1}(|\varpi _{jl}|+|\varrho _{jl}|) \bigg |\ \int ^{t}_{t-\tau _i(t)}\big [x_{kl}(s)-x_{(k+1)l}(s)\big ]\mathrm{d}s\bigg |\nonumber \\&\quad \le 2s_0\jmath _i\sum ^{N-1}_{k=1}\sum ^n_{j=1}\big |\dot{x}_{kj}(t)-\dot{x}_{(k+1)j}(t)\big |\sum ^n_{l=1} \bigg |\ \int ^{t}_{t-\tau _i(t)}\big [x_{kl}(s)-x_{(k+1)l}(s)\big ]\mathrm{d}s\bigg |\nonumber \\&\quad \le s_0\jmath _i\sum ^{N-1}_{k=1}\sum ^n_{j=1}\sum ^n_{l=1}\bigg \{\tilde{\iota }_i\big [\dot{x}_{kj}(t)-\dot{x}_{(k+1)j}(t)\big ]^2+ \tilde{\iota }_i^{-1}\bigg ( \int ^{t}_{t-\tau _i(t)}\big [x_{kl}(s)-x_{(k+1)l}(s)\big ]\mathrm{d}s\bigg )^2\bigg \}\nonumber \\&\quad =ns_0\bigg \{{\iota }_i\dot{e}_t^T\dot{e}_t+\rho _i^{-1}\bigg ( \int ^{t}_{t-\tau _i(t)}e_s\mathrm{d}s\bigg )^T\bigg ( \int ^{t}_{t-\tau _i(t)}e_s\mathrm{d}s\bigg )\bigg \}, \end{aligned}$$
(58)
where \({\iota }_i=\jmath _i\tilde{\iota }_i,\rho _i=\jmath _i^{-1}\tilde{\iota }_i.\)
On the other hand, based on Assumption 1, the following inequalities hold for any \(u,v\in {\mathbb {R}}\) with \(u\ne v\)
$$\begin{aligned}&\left[ \bar{f}_k(u)-\bar{f}_k(v)-l^-_k(u-v)\right] \cdot \left[ \bar{f}_k(u)-\bar{f}_k(v)-{l^+_k}(u-v)\right] \le 0,\ \ \ \ \ \ k\in {\mathbb {N}}. \end{aligned}$$
Thus, for any positive scalar \(t^j_{1ik}(j=1,2,\ldots ,N-1;\ k\in {\mathbb {N}})\) the following inequalities hold
$$\begin{aligned}&-[x_{jk}(t)-x_{j+1,k}(t)]^2t^j_{1ik}l^-_k{l^+_k}-t^j_{1ik}[\bar{f}_k(x_{jk}(t))-\bar{f}_k(x_{j+1,k}(t))]^2\\&\quad +[x_{jk}(t)-x_{j+1,k}(t)]t^j_{1ik}(l^-_k+{l^+_k})[\bar{f}_k(x_{jk}(t))-\bar{f}_k(x_{j+1,k}(t))]\ge 0,\qquad \qquad \forall \ k\in {\mathbb {N}}. \end{aligned}$$
Denoting \(T^j_{1i}=\mathrm{diag}\{t^j_{1i1},t^j_{1i2},\ldots ,t^j_{1in}\},\) the above inequalities are equivalent to the following ones
$$\begin{aligned}&-[x_{j}(t)-x_{j+1}(t)]^TT^j_{1i}\tilde{L}_1[x_{j}(t)-x_{j+1}(t)]+2[x_{j}(t)\\&\qquad -x_{j+1}(t)]^TT^j_{1i}\tilde{L}_2[\bar{f}(x_{j}(t)) -\bar{f}(x_{j+1}(t))]\\&\quad -[\bar{f}(x_{j}(t))-\bar{f}(x_{j+1}(t))]^TT^j_{1i}[\bar{f}(x_{j}(t))-\bar{f}(x_{j+1}(t))]\ge 0,\ \ \ \ \ \forall \ j=1,2,\ldots ,N-1. \end{aligned}$$
Summing both sides of the above inequalities from \(j=1\) to \(N-1\) yields
$$\begin{aligned} -e_t^TT_{1i}L_1e_t+2e_t^TT_{1i}L_2f(e_t)-(Mf(e_t))^TT_{1i}f(e_t)\ge 0, \end{aligned}$$
where \(T_{1i}=\mathrm{diag}\{T^1_{1i},T^2_{1i},\ldots ,T^{N-1}_{1i}\}.\) That is
$$\begin{aligned} 0\le \varepsilon (t)^T\bigg [\begin{array}{cc}-T_{1i}L_1 &{} T_{1i}L_2\\ *&{} -T_{1i}\end{array}\bigg ]\varepsilon (t). \end{aligned}$$
(59)
Similarly the following matrix inequalities hold for any positive diagonal matrix \(T_{2i},T_{3i}\) with compatible dimensions
$$\begin{aligned} 0&\le \varepsilon (t-\tau _i(t))^T\bigg [\begin{array}{cc}-T_{2i}L_1 &{} T_{2i}L_2\\ *&{} -T_{2i}\end{array}\bigg ]\varepsilon (t-\tau _i(t)), \end{aligned}$$
(60)
$$\begin{aligned} 0&\le \varepsilon (t-\bar{\tau }_i)^T\bigg [\begin{array}{cc}-T_{3i}L_1 &{} T_{3i}L_2\\ *&{} -T_{3i}\end{array}\bigg ]\varepsilon (t-\bar{\tau }_i). \end{aligned}$$
(61)
For \(0<\tau _i(t)<\bar{\tau }_i,\) substituting (29)-(61) into (28) and taking mathematical expectation yields
$$\begin{aligned} {\mathbb {E}}\{\pounds V(e_t,i)\} \le {\mathbb {E}}\left\{ \xi _i(t)^T\Big (\Omega _{i}+\Omega _{i}(\tau _i(t))+ns_0\rho _i^{-1}\epsilon _{12}^T\epsilon _{12}\Big )\xi _i(t)\right\} , \end{aligned}$$
(62)
where
$$\begin{aligned} \Omega _{i}(s)&=s\aleph _1^T\Big (X_{1i}P_{2i}^{-1}X_{1i}^T +\frac{1}{3}X_{2i}P_ {2i}^{-1}X_{2i}^T+\frac{1}{5}X_{3i}P_{2i}^{-1}X_{3i}^T\Big )\aleph _1 \\&\quad +(\bar{ \tau }_i-s)\aleph _2^T\Big (X_{4i}P_{2i}^{-1}X_{4i}^T+\frac{1}{3}X_{5i}P_{2i}^{-1} X_{5i}^T\\&\quad +\frac{1}{5}X_{6i}P_{2i}^{-1}X_{6i}^T\Big )\aleph _2+\frac{s^2}{2}\aleph _3^T \Big (X_{7i}P_{3i}^{-1}X_{7i}^T +\frac{1}{8}X_{8i}P_{3i}^{-1}X_{8i}^T\Big ) \aleph _3+s\aleph _3^T\mathrm{sym}\Big \{X_{7i}(\hbar _1-\hbar _2)\\&\quad +\frac{1}{2}X_{8i}(\hbar _1+2\hbar _2-3\hbar _3)\Big \}\aleph _3+\frac{(\bar{\tau }_i -s)^2}{2}\aleph _4^T\Big (X_{9i}P_{3i}^{-1}X_{9i}^T +\frac{1}{8}X_{10i}P_{3i}^{-1} X_{10i}^T\Big )\aleph _4 +(\bar{\tau }_i-s)\aleph _4^T\\&\quad \cdot \mathrm{sym}\Big \{X_{9i}(\hbar _1-\hbar _2)+\frac{1}{2}X_{10i}(\hbar _1+2 \hbar _2-3\hbar _3)\Big \}\aleph _4 \\&\quad +\frac{s^2}{2}\aleph _5^T\Big (X_{11i}P_{4i}^{-1} X_{11i}^T+\frac{1}{8}X_{12i}P_{4i}^{-1}X_{12i}^T\Big )\aleph _5\\&\quad +s\aleph _5^T\mathrm{sym}\Big \{X_{11i}(-\hbar _1+\hbar _2)+\frac{1}{2}X_{12i}( \hbar _1-4\hbar _2+3\hbar _3)\Big \}\aleph _5 +\frac{(\bar{\tau }_i-s)^2}{2}\aleph _6^T \Big (X_{13i}P_{4i}^{-1}X_{13i}^T \\&\quad +\frac{1}{8}X_{14i}P_{4i}^{-1}X_{14i}^T\Big )\aleph _6 +(\bar{\tau }_i-s) \aleph _6^T\mathrm{sym}\Big \{X_{13i}(-\hbar _1+\hbar _2)+\frac{1}{2}X_{14i}( \hbar _1-4\hbar _2+3\hbar _3)\Big \}\aleph _6. \end{aligned}$$
Based on assumptions (13) and Lemmas 3–5, it is easy to see that inequality (62) still holds for \(\tau _i(t)=0\) or \(\tau _i(t)=\bar{\tau }_i.\)
It is easy to see that the coefficient of \(\tau _i{(t)^2}\) in \(\Omega _{i}(\tau _i(t))\) is
$$\begin{aligned} \bar{\aleph }&=\frac{1}{2}\Big \{\aleph _3^T\Big (X_{7i}P_{3i}^{-1}X_{7i}^T +\frac{1}{8}X_{8i}P_{3i}^{-1}X_{8i}^T\Big )\aleph _3+\aleph _4^T\Big (X_{9i}P_{3i}^{-1}X_{9i}^T +\frac{1}{8}X_{10i}P_{3i}^{-1}X_{10i}^T\Big )\aleph _4 \\&\quad +\aleph _5^T\Big (X_{11i}P_{4i}^{-1}X_{11i}^T+\frac{1}{8}X_{12i}P_{4i}^{-1}X_{12i}^T\Big )\aleph _5+\aleph _6^T\Big (X_{13i}P_{4i}^{-1}X_{13i}^T +\frac{1}{8}X_{14i}P_{4i}^{-1}X_{14i}^T\Big )\aleph _6\Big \}. \end{aligned}$$
Since \(\bar{\aleph }\) is nonnegative definite, based on Lemma 7, the following inequality
$$\begin{aligned} \Omega _{i}+\Omega _{i}(\tau _i(t))+ns_0\rho _i^{-1}\epsilon _{12}^T\epsilon _{12}<0, \end{aligned}$$
(63)
holds if and only if the following inequalities hold, respectively,
$$\begin{aligned}&\Omega _{i}+\Omega _{li}+\tilde{\Omega }_{li}+ns_0\rho _i^{-1}\epsilon _{12}^T\epsilon _{12}<0,\ \ \ \ \ l=1,2, \end{aligned}$$
(64)
where
$$\begin{aligned} \tilde{\Omega }_{1i}&=\bar{\tau }_i\aleph _1^T\Big (X_{1i}P_{2i}^{-1}X_{1i}^T +\frac{1}{3}X_{2i}P_{2i}^{-1}X_{2i}^T+\frac{1}{5}X_{3i}P_{2i}^{-1}X_{3i}^T\Big )\aleph _1 \\&\quad +\frac{\bar{\tau }_i^2}{2}\aleph _3^T\Big (X_{7i}P_{3i}^{-1}X_{7i}^T + \frac{1}{8}X_{8i}P_{3i}^{-1}X_{8i}^T\Big )\aleph _3\\&\quad +\frac{\bar{\tau }_i^2}{2}\aleph _5^T\Big (X_{11i}P_{4i}^{-1}X_{11i}^T +\frac{1}{8}X_{12i}P_{4i}^{-1}X_{12i}^T\Big )\aleph _5,\\ \tilde{\Omega }_{2i}&=\bar{\tau }_i\aleph _2^T\Big (X_{4i}P_{2i}^{-1}X_{4i}^T+\frac{1}{3}X_{5i}P_{2i}^{-1}X_{5i}^T +\frac{1}{5}X_{6i}P_{2i}^{-1}X_{6i}^T\Big )\aleph _2\\&\quad +\frac{\bar{\tau }_i^2}{2}\aleph _4^T\Big (X_{9i}P_{3i}^{-1}X_{9i}^T +\frac{1}{8}X_{10i}P_{3i}^{-1}X_{10i}^T\Big )\aleph _4\\&\quad +\frac{\bar{\tau }_i^2}{2}\aleph _6^T\Big (X_{13i}P_{4i}^{-1}X_{13i}^T +\frac{1}{8}X_{14i}P_{4i}^{-1}X_{14i}^T\Big )\aleph _6. \end{aligned}$$
Utilizing the well-known Schur complement yields that inequalities (64) are equivalent to inequalities (20). Therefore, if inequalities (20) hold, then we can derive from (62) that \({\mathbb {E}}\{\pounds V(e_t,i)\}<0.\)
Denote \(\upsilon =\min _{i\in {\mathcal {K}},l=1,2}\big \{\lambda _{\min }\big (-\Omega _{i}-\Omega _{li}-\tilde{\Omega }_{li}-ns_0\rho _i^{-1}\epsilon _{12}^T\epsilon _{12}\big )\big \}.\) It follows from inequality (20) that \(\upsilon >0.\) For any \(i\in {\mathcal {K}},\) applying (62) gives
$$\begin{aligned} {\mathbb {E}}\{\pounds V(e_t,i)\} \le -\upsilon {\mathbb {E}}\left\{ ||\xi _i(t)||^2\right\} \le -\upsilon {\mathbb {E}}\left\{ ||e(t)||^2\right\} . \end{aligned}$$
Thus utilizing Itô’s formula yields
$$\begin{aligned} {\mathbb {E}}\{V(e_t,i)\}- {\mathbb {E}}\{V(e_0,\eta _0)\}={\mathbb {E}}\left\{ \int _0^t\pounds V(e_s,\eta (s))\mathrm{d}s\right\} \le -\upsilon {\mathbb {E}}\left\{ \int _0^t||e(s)||^2\mathrm{d}s\right\} . \end{aligned}$$
(65)
On the other hand, for any \(i\in {\mathcal {K}},\) it follows from the definition of \(V(e_t,i)\) that \({\mathbb {E}}\{V(e_t,i)\}\ge 0.\) Applying inequality (65) yields
$$\begin{aligned} {\mathbb {E}}\left\{ \int _0^t||e(s)||^2\mathrm{d}s\right\} \le \frac{1}{\upsilon }{\mathbb {E}}\{V(e_0,\eta _0)\}. \end{aligned}$$
That is, the integral \({\mathbb {E}}\big \{\int _0^{+\infty }||e(s)||^2\mathrm{d}s\big \}\) is convergent. According to Barbalat’s Lemma (see, e.g., [22]), we have
$$\begin{aligned} \lim _{t\rightarrow +\infty }{\mathbb {E}}\left\{ ||e(t)||^2\right\} =0. \end{aligned}$$
Therefore, from Definition 1, networks (4) are globally asymptotically synchronized in mean square. This completes the proof.
Appendix 3
Proof of Theorem 2
Consider the following LKF:
$$\begin{aligned} V(e_t)=e_t^TP_{1}e_t+\sum ^3_{j=1}V_j(e_t), \end{aligned}$$
with
$$\begin{aligned} V_1(e_t)=&\int ^t_{t-\tau (t)}\varepsilon (s)^T{\mathcal {Q}}\varepsilon (s)\mathrm{d}s+\int _{t-{\bar{\tau }}}^{t}\varepsilon (s)^T{{\mathcal {R}}}\varepsilon (s)\mathrm{d}s,\\ V_2(e_t)=&\int ^t_{t-\bar{\tau }}\int ^t_{\theta }\dot{e}_s^TP_{2}\dot{e}_s\mathrm{d}s\mathrm{d}{\theta }+\int _{t-{\bar{\tau }}}^{t}{\int _\theta ^{t} {\int _u ^t {\dot{e}_s^TP_{3}\dot{e}_s\mathrm{d}s\mathrm{d}u\mathrm{d}\theta }}},\\ V_3(e_t)=&\int _{t-{\bar{\tau }}}^{t}{\int _{t - \bar{\tau }}^\theta {\int _u ^t {\dot{e}_s^TP_{4}\dot{e}_s\mathrm{d}s\mathrm{d}u}}}\mathrm{d}\theta +\int ^t_{t-\bar{\tau }}\int ^t_{\theta }{e}_s^TP_{5}{e}_s\mathrm{d}s\mathrm{d}{\theta }. \end{aligned}$$
Following the same line as in Theorem 1, we can easily complete the proof of Theorem 2.
