Synchronization for memristive chaotic neural networks using Wirtinger-based multiple integral inequality

Abstract

This paper investigates the synchronization problem for a class of memristive chaotic neural networks with time-varying delays. Based on te Wirtinger-based double integral inequality, two novel inequalities are proposed, which are multiple integral forms of the Wirtinger-based integral inequality. Next, by applying the reciprocally convex combination approach for high order case and a free-matrix-based inequality, novel delay-dependent conditions are established to achieve the synchronization for the memristive chaotic neural networks. The results are based on dividing the bounding of activation function into two subintervals with equal length. Finally, a numerical example is provided to demonstrate the effectiveness of the theoretical results.

Appendices

Appendix I

1.1 Proof of Lemma 1

We utilize the mathematical induction to prove inequality (6). Let $\imath =0,$ inequality (6) changes into the Wirtinger-based integral inequality [29]. That is, inequality (6) holds for $\imath =0.$ Now assume that inequality (6) holds for $\imath =k,$ that is, for any scalar $s\ (a<s<b),$ the following inequality holds

$$\begin{aligned} 0\le\,&\Theta _k(a,s,\vartheta ,X)-\frac{(k+1)!}{(s-a)^{k+1}}\rho _k(a,s,\vartheta )^TX\rho _k(a,s,\vartheta )\nonumber \\&-\frac{(k+3)k!}{(s-a)^{k+1}}\Big [\rho _k(a,s,\vartheta )-\frac{k+2}{s-a}\rho _{k+1}(a,s,\vartheta )\Big ]^T \nonumber \\&\times X\Big [\rho _k(a,s,\vartheta )-\frac{k+2}{s-a}\rho _{k+1}(a,s,\vartheta )\Big ]\nonumber \\ =\,&\Theta _k(a,s,\vartheta ,X)+\varpi (s)^T\Omega (s)\varpi (s), \end{aligned}$$

(21)

where $\varpi (s)=\mathrm{col}\big \{\rho _k(a,s,\vartheta ),\ \rho _{k+1}(a,s,\vartheta )\big \}$ and

$$\begin{aligned} \Omega (s)=\big [\Omega _{ij}(s)\big ]_{2\times 2}=\frac{(k+2)k!}{(s-a)^{k+1}}\bigg [\begin{array}{ll} -2X&\frac{k+3}{s-a}X \\ *&-\frac{(k+2)(k+3)}{(s-a)^{2}}X \end{array}\bigg ]. \end{aligned}$$

Note that matrix $X>0$, thus $\Omega _{11}(s)=-\frac{2(k+2)k!}{(s-a)^{k+1}}X<0$ and $\Omega _{22}(s)-\Omega _{21}(s)\Omega _{11}(s)^{-1}\Omega _{12}(s)=-\frac{(k+3)!}{2(s-a)^{k+3}}X<0$. Applying Schur Complements to $\Omega (s)$ yields $\Omega (s)<0$. By Schur Complements again, (21) is equivalent to the following inequality:

$$\begin{aligned} \bigg [\begin{array}{ll}\Theta _k(a,s,\vartheta ,X) &\varpi (s)^T \\ *& \widetilde{\Omega }(s)\end{array}\bigg ]\ge 0, \end{aligned}$$

(22)

where

$$\begin{aligned} \widetilde{\Omega }(s)=\big [\widetilde{\Omega }_{ij}(s)\big ]_{2\times 2}=-\Omega (s)^{-1}= \frac{(s-a)^{k+1}}{(k+1)!}\bigg [\begin{array}{ll} X^{-1}& \frac{s-a}{k+2}X^{-1} \\ *&\frac{2(s-a)^{2}}{(k+2)(k+3)}X^{-1} \end{array}\bigg ]. \end{aligned}$$

Since $\widetilde{\Omega }_{22}(s)=\frac{2(s-a)^{k+3}}{(k+3)!}X^{-1}>0$ and $\widetilde{\Omega }_{11}(s)-\widetilde{\Omega }_{12}(s)\widetilde{\Omega }_{22}(s)^{-1}\widetilde{\Omega }_{21}(s) =\frac{k+1}{2(k+2)!}X^{-1}>0,$ applying Schur Complements to $\widetilde{\Omega }(s)$ leads to $\widetilde{\Omega }(s)>0$.

Integrating (22) from a to b yields

$$\begin{aligned} \Bigg [\begin{array}{ll}\Theta _{k+1}(a,b,\vartheta ,X) &\int ^{b}_{a}\varpi (s)^T\mathrm{d}s \\ *& \int ^{b}_{a}\widetilde{\Omega }(s)\mathrm{d}s\end{array}\Bigg ]\ge 0,\end{aligned}$$

(23)

where $\int ^{b}_{a}\varpi (s)\mathrm{d}s=\mathrm{col}\big \{\rho _{k+1}(a,b,\vartheta ),\ \rho _{k+2}(a,b,\vartheta )\big \}$ and

$$\begin{aligned} \int ^{b}_{a}\widetilde{\Omega }(s)\mathrm{d}s=\frac{(b-a)^{k+2}}{(k+2)!}\bigg [\begin{array}{ll} X^{-1}& \frac{b-a}{k+3}X^{-1} \\ *&\frac{2(b-a)^{2}}{(k+3)(k+4)}X^{-1} \end{array}\bigg ]. \end{aligned}$$

Applying Schur Complements again to (23) yields

$$\begin{aligned} \Theta _{k+1}(a,b,\vartheta ,X)\ge \bigg [\begin{array}{c} \rho _{k+1}(a,b,\vartheta ) \\ \rho _{k+2}(a,b,\vartheta ) \end{array}\bigg ]^T\bigg (\int ^{b}_{a}\widetilde{\Omega }(s)\mathrm{d}s\bigg )^{-1}\bigg [\begin{array}{c} \rho _{k+1}(a,b,\vartheta ) \\ \rho _{k+2}(a,b,\vartheta ) \end{array}\bigg ], \end{aligned}$$

(24)

where

$$\begin{aligned} \bigg (\int ^{b}_{a}\widetilde{\Omega }(s)\mathrm{d}s\bigg )^{-1}=\frac{(k+3)(k+1)!}{(b-a)^{k+2}}\bigg [\begin{array}{ll} 2X& -\frac{k+4}{b-a}X \\ *&\frac{(k+3)(k+4)}{(b-a)^{2}}X \end{array}\bigg ]. \end{aligned}$$

Simple calculating yields that (24) is equivalent to inequality (6) with $\imath =k+1.$ This completes the proof.

Appendix II

1.1 Proof of Lemma 3

Equality (8) comes from Ref. [17], we only need to prove that equality (9) holds for any positive integer i. We utilize the mathematical induction again. Let $i=1,$ (9) changes into the following equality

$$\begin{aligned} \rho _0(a,b,\lambda )=\int ^{b}_{a}\lambda (s)\mathrm{d}s=\int ^{b}_{c}\lambda (s)\mathrm{d}s+\int ^{c}_{a}\lambda (s)\mathrm{d}s=\rho _0(c,b,\lambda )+\rho _0(a,c,\lambda ). \end{aligned}$$

That is, equality (9) holds for $i=1.$ Now assume that inequality (9) holds for $i=k.$ Set $\Gamma (s)=\int ^{s}_{a}\lambda (v)\mathrm{d}v,$ then $\Gamma (s)$ is continuous on [a, b]. Based on the assumption of induction, we have

$$\begin{aligned} \rho _{k}(a,b,\lambda )=&\int ^{b}_{a}\int ^{v_1}_{a}\cdots \int ^{v_{k-1}}_{a}\int ^{v_k}_{a}\lambda (s)\mathrm{d}s\mathrm{d}v_k\mathrm{d}v_{k-1}\ldots \mathrm{d}v_1=\rho _{k-1}(a,b,\Gamma )\\ =\,&\rho _{k-1}(c,b,\Gamma )+\sum ^k_{j=1}\frac{(b-c)^{k-j}}{(k-j)!}\rho _{j-1}(a,c,\Gamma )\\ =&\int ^{b}_{c}\int ^{v_1}_{c}\cdots \int ^{v_{k-1}}_{c}\int ^{v_k}_{a}\lambda (s)\mathrm{d}s\mathrm{d}v_k\mathrm{d}v_{k-1}\ldots \mathrm{d}v_1\\&+\sum ^k_{j=1}\frac{(b-c)^{k-j}}{(k-j)!}\int ^{c}_{a}\int ^{v_1}_{a}\cdots \int ^{v_{j-1}}_{a}\int ^{v_k}_{a}\lambda (s)\mathrm{d}s\mathrm{d}v_k\mathrm{d}v_{j-1}\ldots \mathrm{d}v_1\\ =&\int ^{b}_{c}\int ^{v_1}_{c}\cdots \int ^{v_{k-1}}_{c}\bigg (\int ^{c}_{a}\lambda (s)\mathrm{d}s+\int ^{v_k}_{c}\lambda (s)\mathrm{d}s\bigg )\mathrm{d}v_k\mathrm{d}v_{k-1}\ldots \mathrm{d}v_1 \\&+\sum ^k_{j=1}\frac{(b-c)^{k-j}}{(k-j)!}\rho _{j}(a,c,\lambda )\\ =\,&\rho _{k}(c,b,\lambda ) +\frac{(b-c)^{k}}{k!}\rho _{0}(a,c,\lambda )+\sum ^{k+1}_{l=2}\frac{(b-c)^{k+1-l}}{(k+1-l)!}\rho _{l-1}(a,c,\lambda )\\ =\,&\rho _{k}(c,b,\lambda )+\sum ^{k+1}_{j=1}\frac{(b-c)^{k+1-j}}{(k+1-j)!}\rho _{j-1}(a,c,\lambda ). \end{aligned}$$

That is, inequality (9) holds for $i=k+1.$ This completes the proof.

Appendix III

1.1 Proof of Theorem 1

Consider the following Lyapunov–Krasovskii functional candidate:

$$\begin{aligned} V(e_t,t)=\,&e(t)^T{K}e(t)+\sum ^3_{i=1}V_i(e_t,t), \end{aligned}$$

where

$$\begin{aligned} V_1(e_t,t)=&\int ^t_{t-\tau (t)}\kappa (s)^T\mathcal {Q}\kappa (s)\mathrm{d}s+\bar{\tau }\int ^{t}_{t-\bar{\tau }}\int ^t_{\theta }\kappa (s)^T\mathcal {R}\kappa (s)\mathrm{d}s\mathrm{d}\theta +\int ^{t}_{t-\bar{\tau }}\kappa (s)^T\mathcal {S}\kappa (s)\mathrm{d}s,\\ V_2(e_t,t)=&\int ^{t}_{t-\bar{\tau }}\int ^t_{\theta }\big \{\bar{\tau }\nu (s)^T\mathcal {Z}\nu (s)+\dot{e}(s)^TY\dot{e}(s)\big \}\mathrm{d}s\mathrm{d}\theta ,\\ V_3(e_t,t)=&\sum ^m_{i=1}\frac{\bar{\tau }^i}{i!}\bigg \{\Omega _i(t-\bar{\tau },t,\dot{e}(t),U_i) \\&+\int ^{t}_{t-\bar{\tau }}\int ^{v_i}_{t-\bar{\tau }}\cdots \int ^{v_2}_{t-\bar{\tau }}\int ^{t}_{v_1}\dot{e}(s)^TM_i\dot{e}(s)\mathrm{d}v_1\mathrm{d}v_2\ldots \mathrm{d}v_{i-1}\mathrm{d}v_i\bigg \}, \end{aligned}$$

with $\kappa (s)=\mathrm{col}\big \{e_s,\ g(e_s)\big \}, \nu (t)=\mathrm{col}\big \{e_t,\dot{e}(t)\big \}$ and $\Omega _i(\cdot ,\cdot ,\cdot ,\cdot )(i=1,2,\ldots ,m)$ are defined in Lemma 1.

Calculating the time derivatives of $V(e_t,t)$ along the trajectories of the error system (3), we obtain

$$\begin{aligned} \dot{V}(e_t,t)=2e(t)^T{K}\dot{e}(t)+\sum ^3_{i=1}\dot{V}_i(e_t,t), \end{aligned}$$

(25)

where

$$\begin{aligned} \dot{V}_1(e_t,t)=&\kappa (t)^T\big (\mathcal {Q}+\bar{\tau }^2\mathcal {R}+\mathcal {S}\big )\kappa (t) -[1-\dot{\tau }(t)]\kappa (t-\tau (t))^T\mathcal {Q}\kappa (t-\tau (t))\end{aligned}$$

(26)

$$\begin{aligned}&-\bar{\tau }\int ^{t}_{t-\bar{\tau }}\kappa (s)^T\mathcal {R}\kappa (s)\mathrm{d}s-\kappa (t-\bar{\tau })^T\mathcal {S}\kappa (t-\bar{\tau }),\nonumber \\ \dot{V}_2(e_t,t)=&\bar{\tau }^2\nu (t)^T\mathcal {Z}\nu (t)+\bar{\tau }\dot{e}(t)^TY\dot{e}(t) -\int ^{t}_{t-\bar{\tau }}\big \{\bar{\tau }\nu (s)^T\mathcal {Z}\nu (s)+\dot{e}(s)^TY\dot{e}(s)\big \}\mathrm{d}s,\end{aligned}$$

(27)

$$\begin{aligned} \dot{V}_3(e_t,t)=&\sum ^m_{i=1}\frac{\bar{\tau }^i}{i!}\bigg \{\frac{\bar{\tau }^i}{i!}\dot{e}(t)^T(U_i+M_i)\dot{e}(t) \nonumber \\&-\Omega _{i-1}(t-\bar{\tau },t,\dot{e}(t),U_i)-\Theta _{i-1}(t-\bar{\tau },t,\dot{e}(t),M_i)\bigg \}, \end{aligned}$$

(28)

where $\Theta _i(\cdot ,\cdot ,\cdot ,\cdot )(i=0,1,\ldots ,m-1)$ are defined in Lemma 1.

When $0<{\tau }(t)<\bar{\tau },$ by utilizing the Jensen integral inequality [11] and reciprocally convex combination [25], we obtain from (11) that

$$\begin{aligned} -\bar{\tau }\int ^{t}_{t-\bar{\tau }}\kappa (s)^T\mathcal {R}\kappa (s)\mathrm{d}s =&-\bar{\tau }\int ^{t}_{t-{\tau }(t)}\kappa (s)^T\mathcal {R}\kappa (s)\mathrm{d}s-\bar{\tau }\int ^{t-{\tau }(t)}_{t-\bar{\tau }}\kappa (s)^T\mathcal {R}\kappa (s)\mathrm{d}s\nonumber \\ \le&-\frac{\bar{\tau }}{\tau (t)}\left( \int ^{t}_{t-{\tau }(t)}\kappa (s)\mathrm{d}s\right) ^T\mathcal {R} \left( \int ^{t}_{t-{\tau }(t)}\kappa (s)\mathrm{d}s\right) \nonumber \\&-\frac{\bar{\tau }}{\bar{\tau }-\tau (t)}\left( \int ^{t-{\tau }(t)}_{t-\bar{\tau }}\kappa (s)\mathrm{d}s\right) ^T\mathcal {R}\left( \int ^{t-{\tau }(t)}_{t-\bar{\tau }}\kappa (s)\mathrm{d}s\right) \nonumber \\ \le&-\tilde{\kappa }(t)^T\bigg [\begin{array}{ll} \mathcal {R} &\mathcal {X} \\ *&\mathcal {R} \end{array}\bigg ]\tilde{\kappa }(t), \end{aligned}$$

(29)

where $\tilde{\kappa }(t)=\mathrm{col}\Big \{\int ^{t}_{t-{\tau }(t)}\kappa (s)\mathrm{d}s,\ \int ^{t-{\tau }(t)}_{t-\bar{\tau }}\kappa (s)\mathrm{d}s\Big \}.$

Inspired by the work of [15], the following zero equalities with any symmetric matrices $D_i(i=1,2)$ are proposed according to the Leibniz-Newton formula:

$$\begin{aligned}&\bar{\tau }e_t^TD_1e_t-\bar{\tau }e_\tau ^TD_1e_\tau -2\bar{\tau }\int ^t_{t-\tau (t)}{e}(s)^TD_1\dot{e}(s)\mathrm{d}s=0,\end{aligned}$$

(30)

$$\begin{aligned}&\bar{\tau }e_\tau ^TD_2e_\tau -\bar{\tau }e_{\bar{\tau }}^TD_2e_{\bar{\tau }} -2\bar{\tau }\int ^{t-\tau (t)}_{t-\bar{\tau }}{e}(s)^TD_2\dot{e}(s)\mathrm{d}s=0. \end{aligned}$$

(31)

By utilizing the Jensen integral inequality [11] and reciprocally convex combination [25], we get from (12) that

$$\begin{aligned}&-\bar{\tau }\int ^{t}_{t-\bar{\tau }}\nu (s)^T\mathcal {Z}\nu (s)\mathrm{d}s-2\bar{\tau }\int ^t_{t-\tau (t)}{e}(s)^TD_1\dot{e}(s)\mathrm{d}s-2\bar{\tau }\int ^{t-\tau (t)}_{t-\bar{\tau }}{e}(s)^TD_2\dot{e}(s)\mathrm{d}s\nonumber \\ =&-\bar{\tau }\int ^{t}_{t-{\tau }(t)}\nu (s)^T(\mathcal {Z}+\mathcal {D}_1)\nu (s)\mathrm{d}s-\bar{\tau }\int ^{t-{\tau }(t)}_{t-\bar{\tau }}\nu (s)^T(\mathcal {Z}+\mathcal {D}_2)\nu (s)\mathrm{d}s\nonumber \\ \le&-\frac{\bar{\tau }}{\tau (t)}\left( \int ^{t}_{t-{\tau }(t)}\nu (s)\mathrm{d}s\right) ^T(\mathcal {Z}+\mathcal {D}_1)\left( \int ^{t}_{t-{\tau }(t)}\nu (s)\mathrm{d}s\right) \nonumber \\&-\frac{\bar{\tau }}{\bar{\tau }-\tau (t)}\left( \int ^{t-{\tau }(t)}_{t-\bar{\tau }}\nu (s)\mathrm{d}s\right) ^T(\mathcal {Z}+\mathcal {D}_2)\left( \int ^{t-{\tau }(t)}_{t-\bar{\tau }}\nu (s)\mathrm{d}s\right) \nonumber \\ \le&-\tilde{\nu }(t)^T\bigg [\begin{array}{ll} \mathcal {Z}+\mathcal {D}_1 &\mathcal {Y} \\ *&\mathcal {Z}+\mathcal {D}_2 \end{array}\bigg ]\tilde{\nu }(t), \end{aligned}$$

(32)

where $\tilde{\nu }(t)=\mathrm{col}\Big \{\int ^{t}_{t-{\tau }(t)}\nu (s)\mathrm{d}s,\ \int ^{t-{\tau }(t)}_{t-\bar{\tau }}\nu (s)\mathrm{d}s\Big \}.$

Based on Lemma 5, from inequalities (13)–(14) we get that

$$\begin{aligned}&-\int ^{t}_{t-\bar{\tau }}\dot{e}(s)^TY\dot{e}(s)\mathrm{d}s\nonumber \\ =&-\int ^{t}_{t-{\tau }(t)}\dot{e}(s)^TY\dot{e}(s)\mathrm{d}s-\int ^{t-{\tau }(t)}_{t-\bar{\tau }}\dot{e}(s)^TY\dot{e}(s)\mathrm{d}s\nonumber \\ \le&\theta (t)^T\Big \{{\tau }(t)\Big (H_1+\frac{1}{3}H_3\Big )+\mathrm{sym}\big \{H_4\big [\begin{array}{lll} I&-I&0 \end{array}\big ] +H_5\big [\begin{array}{lll} -I&-I&2I \end{array}\big ]\big \}\Big \}\theta (t)\nonumber \\&+\iota (t)^T\Big \{[\bar{\tau }-{\tau }(t)]\Big (E_1+\frac{1}{3}E_3\Big )+\mathrm{sym}\big \{E_4\big [\begin{array}{lll} I&-I&0 \end{array}\big ] +E_5\big [\begin{array}{lll} -I&-I&2I \end{array}\big ]\big \}\Big \}\iota (t). \end{aligned}$$

(33)

where

$$\begin{aligned} \theta (t)=\mathrm{col}\bigg \{e_t,\ e_\tau ,\ \frac{1}{{\tau }(t)}\int ^{t}_{t-{\tau }(t)}e_s\mathrm{d}s\bigg \},\ \iota (t)=\mathrm{col}\bigg \{e_\tau ,\ e_{\bar{\tau }},\ \frac{1}{\bar{\tau }-{\tau }(t)}\int ^{t-{\tau }(t)}_{t-\bar{\tau }}e_s\mathrm{d}s\bigg \}. \end{aligned}$$

Applying Lemma 3 yields

$$\begin{aligned}&\Omega _{i-1}(t-\bar{\tau },t,\dot{e}(t),U_i)=\Omega _{i-1}(t-\bar{\tau },t-\tau (t),\dot{e}(t),U_i) \end{aligned}$$

(34)

$$\begin{aligned}&+\sum ^i_{j=1}\frac{[\bar{\tau }-\tau (t)]^{i-j}}{(i-j)!}\Omega _{j-1}(t-\tau (t),t,\dot{e}(t),U_i), \nonumber \\&\Theta _{i-1}(t-\bar{\tau },t,\dot{e}(t),M_i)=\Theta _{i-1}(t-\tau (t),t,\dot{e}(t),M_i) \nonumber \\&+\sum ^i_{j=1}\frac{[\tau (t)]^{i-j}}{(i-j)!}\Theta _{j-1}(t-\bar{\tau },t-\tau (t),\dot{e}(t),M_i). \end{aligned}$$

(35)

Based on Lemmas 2, 3 and 6, if conditions (15)-(16) hold, then by simple calculating we obtain from inequalities (34)-(35) that

$$\begin{aligned}&\sum ^m_{i=1}\frac{\bar{\tau }^i}{i!}\big \{\Omega _{i-1}(t-\bar{\tau },t,\dot{e}(t),U_i) +\Theta _{i-1}(t-\bar{\tau },t,\dot{e}(t),M_i)\big \}\nonumber \\ =&\sum ^m_{i=1}\frac{\bar{\tau }^i}{i!}\bigg \{\Omega _{i-1}(t-\bar{\tau },t-\tau (t),\dot{e}(t),U_i) +\Omega _{i-1}(t-\tau (t),t,\dot{e}(t),\mathcal {U}_i(t))\nonumber \\&\quad +\Theta _{i-1}(t-\tau (t),t,\dot{e}(t),M_i) +\Theta _{i-1}(t-\bar{\tau },t-\tau (t),\dot{e}(t),\mathcal {M}_i(t))\bigg \}\nonumber \\ \ge&\bigg (\frac{\bar{\tau }}{\bar{\tau }-\tau (t)}\bigg )^i(\psi '_\omega )^T\bigg [\begin{array}{ll} {U}_i &0 \\ 0& \frac{i+2}{i}{U}_i \end{array}\bigg ]\psi '_\omega +\bigg (\frac{\bar{\tau }}{\tau (t)}\bigg )^i(\psi ''_\omega )^T\bigg [\begin{array}{ll} \mathcal {U}_i(t) &0 \\ 0& \frac{i+2}{i}\mathcal {U}_i(t) \end{array}\bigg ]\psi ''_\omega \nonumber \\&+\bigg (\frac{\bar{\tau }}{\tau (t)}\bigg )^i(\hbar '_\xi )^T\bigg [\begin{array}{ll} {M}_i &0 \\ 0& \frac{i+2}{i}{M}_i \end{array}\bigg ]\hbar '_\xi +\bigg (\frac{\bar{\tau }}{\bar{\tau }-\tau (t)}\bigg )^i(\hbar ''_\xi )^T\bigg [\begin{array}{ll} \mathcal {M}_i(t) &0 \\ 0& \frac{i+2}{i}\mathcal {M}_i(t) \end{array}\bigg ]\hbar ''_\xi \nonumber \\ \ge&\big [\begin{array}{ll} \psi '_\omega&\psi ''_\omega \end{array}\big ]^T\mathcal {U}_i(t)\mathrm{col}\big \{\psi '_\omega ,\psi ''_\omega \big \}+\big [\begin{array}{ll} \hbar '_\xi&\hbar ''_\xi \end{array}\big ]^T\mathcal {M}_i(t)\mathrm{col}\big \{\hbar '_\xi ,\hbar ''_\xi \big \}, \end{aligned}$$

(36)

where $\psi '_\omega =\mathrm{col}\big \{\psi _{i-1}(t-\bar{\tau },t-\tau (t),{e}_t),\ \omega _{i-1}(t-\bar{\tau },t-\tau (t),{e}_t)\big \},\ \psi ''_\omega =\mathrm{col}\big \{\psi _{i-1}(t-\tau (t),t,{e}_t),\ \omega _{i-1}(t-\tau (t),t,{e}_t)\big \},$ $\hbar '_\xi =\mathrm{col}\big \{\hbar _{i-1}(t-\tau (t),t,{e}_t),\ \xi _{i-1}(t-\tau (t),t,{e}_t)\big \},\ \hbar ''_\xi =\mathrm{col}\big \{\hbar _{i-1}(t-\bar{\tau },t-\tau (t),{e}_t),\ \xi _{i-1}(t-\bar{\tau },t-\tau (t),{e}_t)\big \}.$

Based on (3), the following equalities hold for any positive diagonal matrix P:

$$\begin{aligned} 0=\,&2\dot{e}(t)^T{P}[-\dot{e}(t)+\dot{e}(t)]\nonumber \\ =\,&2\dot{e}(t)^T{P}[-\dot{e}(t)-(C-G_1)e(t)+{G_{2}}e(t-{\tau }(t))]\nonumber \\&+2\sum ^n_{i,j=1}\dot{e}_i(t)p_i\big [{a}_{ij}(y_j(t))f_j(y_j(t))-a_{ij}(x_j(t))f_j(x_j(t))\big ]\nonumber \\&+2\sum ^n_{i,j=1}\dot{e}_i(t)p_i\big [{b}_{ij}(y_j(t))f_j(y_j(t-\tau (t)))-{b}_{ij}(x_j(t))f_j(x_j(t-\tau (t)))\big ], \end{aligned}$$

(37)

where $C=\mathrm{diag}\{c_1,c_2,...,c_n\},P=\mathrm{diag}\{p_1,p_2,...,p_n\}.$

According to Lemma 4 and the Cauchy inequality $2\alpha ^T\beta \le \alpha ^TQ\alpha +\beta ^TQ^{-1}\beta ,$ the following inequalities hold for any positive scalars $\varepsilon _1,\varepsilon _2$:

$$\begin{aligned}&2\sum ^n_{i,j=1}\dot{e}_i(t)p_i\big [{a}_{ij}(y_j(t))f_j(y_j(t))-a_{ij}(x_j(t))f_j(x_j(t))\big ]\nonumber \\ \le&\sum ^n_{i,j=1}\Big \{\varepsilon _1^{-1}\dot{e}_i^2(t)p_i^2+\varepsilon _1\big [{a}_{ij}(y_j(t))f_j(y_j(t)) -a_{ij}(x_j(t))f_j(x_j(t))\big ]^2\Big \}\nonumber \\ \le&\sum ^n_{i,j=1}\Big \{\varepsilon _1^{-1}\dot{e}_i^2(t)p_i^2 +\varepsilon _1\hat{a}_{ij}^2\sigma ^2_j\big [y_j(t)-x_j(t)\big ]^2\Big \}\nonumber \\ =\,&n\varepsilon _1^{-1}\dot{e}(t)^T{P}^2\dot{e}(t)+\varepsilon _1e(t)^T\Sigma \hat{A}\Sigma e(t), \nonumber \\&\times 2\sum ^n_{i,j=1}\dot{e}_i(t)p_i\big [{b}_{ij}(y_j(t))f_j(y_j(t-\tau (t)))-{b}_{ij}(x_j(t))f_j(x_j(t-\tau (t)))\big ]\nonumber \\ \le&\sum ^n_{i,j=1}\Big \{\varepsilon _2^{-1}\dot{e}_i^2(t)p_i^2 +\varepsilon _2\big [{b}_{ij}(y_j(t))f_j(y_j(t-\tau (t)))-{b}_{ij}(x_j(t))f_j(x_j(t-\tau (t)))\big ]^2\Big \} \end{aligned}$$

(38)

$$\begin{aligned} \le&\sum ^n_{i,j=1}\Big \{\varepsilon _2^{-1}\dot{e}_i^2(t)p_i^2 +\varepsilon _2\hat{b}_{ij}^2\sigma ^2_j\big [y_j(t-\tau (t))-x_j(t-\tau (t))\big ]^2\Big \}\nonumber \\ =\,&n\varepsilon _2^{-1}\dot{e}(t)^T{P}^2\dot{e}(t)+\varepsilon _2e(t-\tau (t))^T\Sigma \hat{B}\Sigma e(t-\tau (t)), \end{aligned}$$

(39)

where $\hat{A}=\mathrm{diag}\big \{\sum ^n_{i=1}\hat{a}_{i1}^2,\ \sum ^n_{i=1}\hat{a}_{i2}^2,\ \ldots ,\ \sum ^n_{i=1}\hat{a}_{in}^2\big \},\ \hat{B}=\mathrm{diag}\big \{\sum ^n_{i=1}\hat{b}_{i1}^2,\ \sum ^n_{i=1}\hat{b}_{i2}^2,\ \ldots ,\ \sum ^n_{i=1}\hat{b}_{in}^2\big \}.$

Inspired by the work of [16], the interval satisfied by the activation function is divided into the following two subintervals:

Case I:

$$\begin{aligned}&-\sigma _j\le \frac{g_j(\varsigma )-g_j(\zeta )}{\varsigma -\zeta }\le 0,\quad \forall \ \varsigma ,\zeta \in \mathbb {R},\ \varsigma \ne \zeta ,\ j\in \mathbb {N};\end{aligned}$$

(40)

$$\begin{aligned}&-\sigma _j\le \frac{g_j(\varsigma )}{\varsigma }\le 0,\quad \forall \ \varsigma \in \mathbb {R},\ \varsigma \ne 0,\ j\in \mathbb {N}. \end{aligned}$$

(41)

It should be noted that the conditions (40) and (41) are equivalent to the following inequalities respectively:

$$\begin{aligned}&\left[ g_j(\varsigma )-g_j(\zeta )+\sigma _j(\varsigma -\zeta )\right] \cdot \left[ g_j(\varsigma )-g_j(\zeta )\right] \le 0,\quad \forall \ \varsigma ,\zeta \in \mathbb {R},\ \varsigma \ne \zeta ,\ j\in \mathbb {N};\end{aligned}$$

(42)

$$\begin{aligned}&\left[ g_j(\varsigma )+\sigma _j\varsigma \right] g_j(\varsigma )\le 0,\quad \forall \ \varsigma \in \mathbb {R},\ \varsigma \ne 0,\ j\in \mathbb {N}. \end{aligned}$$

(43)

Based on inequalities (42) and (43), the following matrix inequalities hold for any positive diagonal matrices $W_i(i=1,\ldots ,6)$ with compatible dimensions:

$$\begin{aligned} 0\le&-[g(e_t)+\Sigma e_t]^TW_1g(e_t),\end{aligned}$$

(44)

$$\begin{aligned} 0\le&-[g(e_{\tau })+\Sigma e_{\tau }]^TW_2g(e_{\tau }),\end{aligned}$$

(45)

$$\begin{aligned} 0\le&-[g(e_{\bar{\tau }})+\Sigma e_{\bar{\tau }}]^TW_3g(e_{\bar{\tau }}),\end{aligned}$$

(46)

$$\begin{aligned} 0\le&-\big \{g(e_t)-g(e_{\tau })+\Sigma (e_t-e_{\tau })\big \}^TW_4[g(e_t)-g(e_{\tau })],\end{aligned}$$

(47)

$$\begin{aligned} 0\le&-\big \{g(e_{\tau })-g(e_{\bar{\tau }})+\Sigma (e_{\tau }-e_{\bar{\tau }})\big \}^TW_5[g(e_{\tau })-g(e_{\bar{\tau }})],\end{aligned}$$

(48)

$$\begin{aligned} 0\le&-\big \{g(e_t)-g(e_{\bar{\tau }})+\Sigma (e_t-e_{\bar{\tau }})\big \}^TW_6[g(e_t)-g(e_{\bar{\tau }})]. \end{aligned}$$

(49)

Substituting (26)–(36) and (44)–(49) into (25) yields that

$$\begin{aligned} \dot{V}(e_t,t)\le \eth (t)^T\big \{\Xi +\overline{\Xi }+\widetilde{\Xi }+\Xi _1+n\big (\varepsilon _1^{-1} +\varepsilon _2^{-1}\big )\eta _7^T{P}^2\eta _7\big \}\eth (t).\end{aligned}$$

(50)

It is easy to see that inequality (50) holds for ${\tau }(t)=0$ or ${\tau }(t)=\bar{\tau }$ from the Jensen integral inequality [11]. Therefore, inequality (50) holds for any $t>0$ with $0\le {\tau }(t)\le \bar{\tau }$.

According to the Schur Complement, $\Xi +\overline{\Xi }+\widetilde{\Xi }+\Xi _1+n\big (\varepsilon _1^{-1}+\varepsilon _2^{-1}\big )\eta _7^T{P}^2\eta _7<0$ is equivalent to inequality (17) with $p=1.$ Therefore when condition (17) is satisfied for $p=1$, from (50) we get that $\dot{V}(e_t,t)<0.$ That is, error system (3) is asymptotically stable.

Case II:

$$\begin{aligned}&0\le \frac{g_j(\varsigma )-g_j(\zeta )}{\varsigma -\zeta }\le \sigma _j,\quad \forall \ \varsigma ,\zeta \in \mathbb {R},\ \varsigma \ne \zeta ,\ j\in \mathbb {N};\end{aligned}$$

(51)

$$\begin{aligned}&0\le \frac{g_j(\varsigma )}{\varsigma }\le \sigma _j ,\quad \forall \ \varsigma \in \mathbb {R},\ \varsigma \ne 0,\ j\in \mathbb {N}. \end{aligned}$$

(52)

It is obvious that the conditions (51) and (52) are equivalent to the following inequalities respectively:

$$\begin{aligned}&\left[ g_j(\varsigma )-g_j(\zeta )\right] \cdot \left[ g_j(\varsigma )-g_j(\zeta )-\sigma _j(\varsigma -\zeta )\right] \le 0,&\forall \ \varsigma ,\zeta \in \mathbb {R},\ \varsigma \ne \zeta ,\ j\in \mathbb {N};\end{aligned}$$

(53)

$$\begin{aligned}&g_j(\varsigma )\left[ g_j(\varsigma )-\sigma _j\varsigma \right] \le 0,&\forall \ \varsigma \in \mathbb {R},\ \varsigma \ne 0,\ j\in \mathbb {N}. \end{aligned}$$

(54)

From (53) and (54), the following matrix inequalities hold for any positive diagonal matrices $L_i(i=1,\ldots ,6)$ with compatible dimensions:

$$\begin{aligned} 0\le&-g(e_t)^TL_1[g(e_t)-\Sigma e_t],\end{aligned}$$

(55)

$$\begin{aligned} 0\le&-g(e_{\tau })^TL_2[g(e_{\tau })-\Sigma e_{\tau }],\end{aligned}$$

(56)

$$\begin{aligned} 0\le&-g(e_{\bar{\tau }})^TL_3[g(e_{\bar{\tau }})-\Sigma e_{\bar{\tau }}],\end{aligned}$$

(57)

$$\begin{aligned} 0\le&-[g(e_t)-g(e_{\tau })]^TL_4\big \{g(e_t)-g(e_{\tau })-\Sigma (e_t-e_{\tau })\big \},\end{aligned}$$

(58)

$$\begin{aligned} 0\le&-[g(e_{\tau })-g(e_{\bar{\tau }})]^TL_5\big \{g(e_{\tau })-g(e_{\bar{\tau }})-\Sigma (e_{\tau }-e_{\bar{\tau }})\big \},\end{aligned}$$

(59)

$$\begin{aligned} 0\le&-[g(e_t)-g(e_{\bar{\tau }})]^TL_6\big \{g(e_t)-g(e_{\bar{\tau }})-\Sigma (e_t-e_{\bar{\tau }})\big \}. \end{aligned}$$

(60)

Substituting (26)–(36) and (55)–(60) into (25) yields that

$$\begin{aligned} \dot{V}(e_t,t)&\le \eth (t)^T\big \{\Xi +\overline{\Xi }+\widetilde{\Xi }+\Xi _2+n\big (\varepsilon _1^{-1} \nonumber \\&\quad +\varepsilon _2^{-1}\big )\eta _7^T{P}^2\eta _7\big \}\eth (t). \end{aligned}$$

(61)

It is easy to see that inequality (61) holds for ${\tau }(t)=0$ or ${\tau }(t)=\bar{\tau }$ from the Jensen integral inequality [11]. Therefore, inequality (61) holds for any $t>0$ with $0\le {\tau }(t)\le \bar{\tau }$.

Again according to the Schur Complement, $\Xi +\overline{\Xi }+\widetilde{\Xi }+\Xi _2+n\big (\varepsilon _1^{-1}+\varepsilon _2^{-1}\big )\eta _7^T{P}^2\eta _7<0$ is equivalent to inequality (17) with $p=2.$ Therefore when condition (17) is satisfied for $p=2,$ we conclude that the drive system (1) and response system (2) are synchronous. This completes the proof of Theorem 1.