Global fixed-time synchronization for coupled time-varying delayed neural networks with multi-weights and uncertain couplings via periodically semi-intermittent adaptive control

Abstract

This paper is concerned with the semi-intermittent synchronization in fixed time for hybrid couple neural networks with uncertain multiple weights and discontinuous nonlinearity. Firstly, the principle of the global convergence in fixed time with respect to nonlinear systems with periodically semi-intermittent scheme is developed. Secondly, in order to realize the global synchronization goal in fixed time, novel controllers, which are composed of the semi-intermittent adaptive controller and semi-intermittent feedback controller, are designed. Under Filippov differential inclusion framework, by applying Lyapunov–Krasovskii functional method and inequality analysis technique, the global synchronization conditions in fixed time are achieved in the form of linear matrix inequalities. In addition, the upper bound of the settling time, which is adjusted in advance by choosing the controller parameters, is estimated accurately. Finally, the correctness of the theoretical results and the feasibility of the designed controller are verified by an example.

Appendices

Appendix I

The following corollaries are extended form Hu et al. (2017) and Zheng et al. (2018), respectively.

Corollary 6.1

Suppose $\varpi =1$, assumptions $(A_1)$, $(A_2)$ are satisfied and ${\dot{\tau }}(t)_\iota \le \mu _\iota <1$, $\iota =1,2,\ldots ,p$. If there exists matrix $S\in R^{Nn\times Nn}$, constant $0<\mu <1$, such that the following LMIs hold:

$$\begin{aligned} \Theta =\begin{pmatrix}\Phi _{11}&{}\Phi _{12}\\ \Phi _{21}&{}\Phi _{22} \end{pmatrix}<0, \\ K-{\hat{\omega }}(I_N\otimes B)-{\hat{\omega }}(I_N\otimes A)>0. \end{aligned}$$

Then, the driven system (1) and the respond (3) can be achieved fixed-time synchronization via controller (8). Moreover, the settling time is given by:

$$\begin{aligned} T({{\tilde{e}}}, {\tilde{\eta }})\le T_{max}=\frac{2}{(Nn)^{\frac{1-q}{2}}{\tilde{\rho }}\varpi (q-1)}+(\frac{2}{\rho \varpi (1-\delta )}), \end{aligned}$$

where

$$\begin{aligned} \begin{aligned}&\Phi _{11}=-2(I_{N}\otimes Q)+2{\hat{z}}(I_{N}\otimes A)+2\sum ^{p}_{\iota =1}u_{\iota }(U^\iota \otimes \Gamma ^\iota )^{s}\\&\quad \quad +u_{\iota }(U^\iota \otimes (\sigma HH^T+\sigma ^{-1}MM^T))\\&\quad \quad +2\sum ^{p}_{\iota =1}v_{\iota }({{\mathcal {V}}}^{\iota }\otimes \beta ^\iota )S(\beta ^\iota \otimes {{\mathcal {V}}}^{\iota })^{T}\\&\quad \quad +\sigma \sum ^{p}_{\iota =1}v_{\iota }({{\mathcal {V}}}^{\iota }\otimes {{\tilde{H}}}{{\tilde{H}}}^T)+\sum ^{p}_{\iota =1}\frac{v_{\iota }}{1-\mu }(I_{N}\otimes I_n),\\&\quad \Phi _{12}={\hat{z}}(I_{N}\otimes B),\quad \Phi _{21}={\hat{z}}(I_{N}\otimes B), \\&\quad \Phi _{22}=(I_{N}\otimes I_{n})+\sigma ^{-1}\sum ^{p}_{\iota =1}v_{\iota }({{\mathcal {V}}}^{\iota }\otimes {{\tilde{M}}}^T{{\tilde{M}}})\\&\quad -\sigma ^{-1}\sum ^{p}_{\iota =1}v_{\iota }(I_{N}\otimes I_n),\\&\quad \rho =\text {min}\{\eta _{1}, \eta _{3}, \lambda _1\},\ \ \ {\tilde{\rho }}=\text {min}\{\eta _{2}, \eta _{4}, \lambda _2\},\\&\quad K=\text {diag}(k_1,k_2,\ldots ,k_N)\ \ \ \ \ \eta =[|e_{i}(t)|, |e_{i}(t-\tau )|]^{T}. \end{aligned} \end{aligned}$$

Suppose $p=1$, the driven system becomes the HCNNs with single weight and uncertain couplings.

Corollary 6.2

Suppose $p=1$, assumptions $(A_1)$, $(A_2)$ are satisfied and ${\dot{\tau }}(t)\le \mu <1$. If there exists matrix $S\in R^{Nn\times Nn}$, constant $0<\mu <1$, such that the following LMIs hold:

$$\begin{aligned} \Theta =\begin{pmatrix}\Phi _{11}&{}\Phi _{12}\\ \Phi _{21}&{}\Phi _{22} \end{pmatrix}<0, \\ K-{\hat{\omega }}(I_N\otimes B)-{\hat{\omega }}(I_N\otimes A)>0. \end{aligned}$$

Then, the driven system (1) and the respond system (3) can be achieved fixed-time synchronization via controller (8). Moreover, the settling time is given as follows:

$$\begin{aligned} T({{\tilde{e}}}, {\tilde{\eta }})\le T_{max}=\frac{2}{(Nn)^{\frac{1-q}{2}}{\tilde{\rho }}\varpi (q-1)}+\left( \frac{2}{\rho \varpi (1-\delta )}\right) , \end{aligned}$$

where

$$\begin{aligned} \begin{aligned}&\Phi _{11}=-2(I_{N}\otimes Q)+2{\hat{z}}(I_{N}\otimes A)+2u(U\otimes \Gamma )^{s}\\&\ \ \ \ \ \ \ \ \ \ \ +u(U\otimes (\sigma HH^T+\sigma ^{-1}MM^T))\\&\ \ \ \ \ \ \ \ \ \ \ +2v({{\mathcal {V}}}\otimes \beta )S(\beta \otimes {{\mathcal {V}}})^{T}\\&\ \ \ \ \ \ \ \ \ \ \ +\sigma v({{\mathcal {V}}}\otimes {{\tilde{H}}}{{\tilde{H}}}^T)+\frac{v}{1-\mu }(I_{N}\otimes I_n),\\&\Phi _{12}={\hat{z}}(I_{N}\otimes B),\quad \Phi _{21}={\hat{z}}(I_{N}\otimes B), \\&\Phi _{22}=(I_{N}\otimes I_{n})+\sigma ^{-1}v({{\mathcal {V}}}\otimes {{\tilde{M}}}^T{{\tilde{M}}})-\sigma ^{-1}v(I_{N}\otimes I_n),\\&\rho =\text {min}\{\eta _{1}, \eta _{3}, \lambda _1\},\ \ \ {\tilde{\rho }}=\text {min}\{\eta _{2}, \eta _{4}, \lambda _2\},\\&K=\text {diag}(k_1,k_2,\ldots ,k_N), \eta =[|e_{i}(t)|, |e_{i}(t-\tau )|]^{T}. \end{aligned} \end{aligned}$$

Appendix II

The parameters in the HCNNs (1) and (3) are described as follows:

$$\begin{aligned}&Q= \begin{bmatrix} 0.33 &{} 0 &{} 0\\ 0 &{} 0.21 &{} 0\\ 0 &{} 0 &{} 0.29 \end{bmatrix}, A= \begin{bmatrix} 0.32 &{} 0.03 &{} 1.24\\ 0.03 &{} 1.32 &{} 0.33\\ 1.24 &{} 0.33 &{} 1.02 \end{bmatrix},\\&B= \begin{bmatrix} 0.41 &{} 0.25 &{} 0.53\\ 0.25 &{} 1.14 &{} 1.19\\ 0.53 &{} 1.19 &{} 1.51 \end{bmatrix}, \Gamma ^1= \begin{bmatrix} 1.11 &{} 0 &{} 0\\ 0 &{} 0.14 &{} 0\\ 0 &{} 0 &{} 1.82 \end{bmatrix},\\&\Gamma ^2= \begin{bmatrix} 1.24 &{} 0 &{} 0\\ 0 &{} 1.24 &{} 0\\ 0 &{} 0 &{} 0.71 \end{bmatrix}, \Gamma ^3= \begin{bmatrix} 1.16 &{} 0 &{} 0\\ 0 &{} 0.34 &{} 0\\ 0 &{} 0 &{} 0.13 \end{bmatrix},\\&\beta ^1= \begin{bmatrix} 0.08 &{} 0 &{} 0\\ 0 &{} 1.08 &{} 0\\ 0 &{} 0 &{} 1.06 \end{bmatrix}, \beta ^2= \begin{bmatrix} 0.12 &{} 0 &{} 0\\ 0 &{} 0.34 &{} 0\\ 0 &{} 0 &{} 0.02 \end{bmatrix},\\&\beta ^3= \begin{bmatrix} 1.45 &{} 0 &{} 0\\ 0 &{} 1.51 &{} 0\\ 0 &{} 0 &{} 0.45 \end{bmatrix}, H^1= \begin{bmatrix} 1 &{} 0 &{} 0\\ 0 &{} 3 &{} 0\\ 0 &{} 0 &{} 5 \end{bmatrix},\\&H^2= \begin{bmatrix} 0.45 &{} 0 &{} 0\\ 0 &{} 0.34 &{} 0\\ 0 &{} 0 &{} 0.69 \end{bmatrix}, H^3= \begin{bmatrix} 2.11 &{} 0 &{} 0\\ 0 &{} 1.03 &{} 0\\ 0 &{} 0 &{} 0.27 \end{bmatrix},\\&F^1(t)= \begin{bmatrix} 0.2\sin t &{} 0 &{} 0\\ 0 &{} 0.15 &{} 0\\ 0 &{} 0 &{} 0.61 \end{bmatrix}, M^1= \begin{bmatrix} 0.31 &{} 0 &{} 0\\ 0 &{} 0.62 &{} 0\\ 0 &{} 0 &{} 0.24 \end{bmatrix},\\&F^2(t)= \begin{bmatrix} 0.12 &{} 0 &{} 0\\ 0 &{} 0.2\sin t &{} 0\\ 0 &{} 0 &{} 0.04 \end{bmatrix}, M^2= \begin{bmatrix} 0.5 &{} 0 &{} 0\\ 0 &{} 0.14 &{} 0\\ 0 &{} 0 &{} 0.38 \end{bmatrix},\\&F^3(t)= \begin{bmatrix} 0.04 &{} 0 &{} 0\\ 0 &{} 0.08 &{} 0\\ 0 &{} 0 &{} 0.13\cos t \end{bmatrix}, M^3= \begin{bmatrix} 0.46 &{} 0 &{} 0\\ 0 &{} 0.18 &{} 0\\ 0 &{} 0 &{} 0.32 \end{bmatrix},\\&{{\tilde{H}}}^1= \begin{bmatrix} 3 &{} 0 &{} 0\\ 0 &{} 2 &{} 0\\ 0 &{} 0 &{} 4 \end{bmatrix}, {{\tilde{H}}}^2= \begin{bmatrix} 0.16 &{} 0 &{} 0\\ 0 &{} 0.42 &{} 0\\ 0 &{} 0 &{} 0.28 \end{bmatrix},\\&{{\tilde{H}}}^3= \begin{bmatrix} 1.62 &{} 0 &{} 0\\ 0 &{} 0.28 &{} 0\\ 0 &{} 0 &{} 1.02 \end{bmatrix}, {{\tilde{F}}}^1(t)= \begin{bmatrix} 0.4\sin t &{} 0 &{} 0\\ 0 &{} 0.46 &{} 0\\ 0 &{} 0 &{} 0.24 \end{bmatrix},\\&{{\tilde{F}}}^2(t)= \begin{bmatrix} 0.28 &{} 0 &{} 0\\ 0 &{} 0.4\sin t &{} 0\\ 0 &{} 0 &{} 0.24 \end{bmatrix}, {{\tilde{M}}}^1= \begin{bmatrix} 0.62 &{} 0 &{} 0\\ 0 &{} 0.34 &{} 0\\ 0 &{} 0 &{} 0.18 \end{bmatrix},\\&{{\tilde{F}}}^3(t)= \begin{bmatrix} 0.24 &{} 0 &{} 0\\ 0 &{} 0.18 &{} 0\\ 0 &{} 0 &{} 0.14\cos t \end{bmatrix}, {{\tilde{M}}}^2= \begin{bmatrix} 0.26 &{} 0 &{} 0\\ 0 &{} 0.16 &{} 0\\ 0 &{} 0 &{} 0.28 \end{bmatrix},\\&{{\tilde{M}}}^3= \begin{bmatrix} 0.16 &{} 0 &{} 0\\ 0 &{} 0.24 &{} 0\\ 0 &{} 0 &{} 0.64 \end{bmatrix},\\&u_1=0.81, \ \ u_2=0.62, \ \ \ u_3=0.74,\\&v_1=0.12,\ \ v_2=0.04\ \ \ v_3=0.95. \end{aligned}$$

The nonlinear activation functions:

$$\begin{aligned}&f_{i1}(t,x_{i1}(t))=0.02x_{i1}(t)+0.015\text {sign}(x_{i1}(t)),\\&f_{i2}(t,x_{i2}(t))=0.08x_{i2}(t)+0.003\text {sign}(x_{i2}(t)),\\&f_{i3}(t,x_{i3}(t))=0.06x_{i3}(t)+0.008\text {sign}(x_{i3}(t)),\\ \end{aligned}$$

have been considered. It can be easily verified that assumptions $A_2$ hold for $z_{1}=0.02$, $\omega _{1}=0.03$, $z_{2}=0.08$, $\omega _{2}=0.006$, $z_{3}=0.06$, $\omega _{3}=0.016$.

The non-delayed and time-varying delayed coupling configuration matrices read as:

$$\begin{aligned}&U^1= \begin{bmatrix} -4 &{} 2 &{} 0 &{} 1 &{} 1\\ 2 &{} 2 &{} -3&{} -1 &{} 0\\ 0 &{} -3 &{} -2 &{} 0 &{} 5\\ 1 &{} -1 &{} 0 &{} -2 &{} 2\\ 1 &{} 0 &{} 5 &{} 2 &{} -8 \end{bmatrix},\\&U^2= \begin{bmatrix} 4 &{} -1 &{} -1 &{} -1 &{} -1\\ -1 &{} 3 &{} 0 &{} 0 &{} -2\\ -1 &{} 0 &{} 2 &{} 1 &{} -2\\ -1 &{} 0 &{} 1 &{} -2 &{} 2 \\ -1&{} -2&{} -2&{} 2 &{} 3 \end{bmatrix},\\&U^3= \begin{bmatrix} -1.4 &{} 0.5 &{} 0.2 &{} 0.3 &{} 0.41\\ 0.5 &{} 1.5 &{} -1.1 &{} -0.2 &{} -0.7\\ 0.2 &{} -1.1 &{} 0.8 &{} 0.1 &{} 0\\ 0.3 &{} 0.2 &{} 0.1 &{} -3.3 &{} 3.1 \\ 0.4&{} -0.7&{} 0&{} 3.1 &{} -3.4 \end{bmatrix}, \end{aligned}$$

and

$$\begin{aligned}&{{\mathcal {V}}}^1= \begin{bmatrix} -2 &{} 0 &{} 1 &{} 1 &{} 0\\ 0 &{} 1 &{} -3 &{} 1 &{} 1\\ 1 &{} -3 &{} 1 &{} 1 &{} 0\\ 1 &{} 1 &{} 1 &{} -1 &{} -2\\ 0 &{} 1 &{} 0 &{}-2 &{} 1 \end{bmatrix},\\&{{\mathcal {V}}}^2= \begin{bmatrix} 1 &{} -1 &{} 0 &{} 0&{} 0\\ -1 &{} 2 &{} 0 &{} 0 &{} -1\\ 0 &{} 0&{} -1 &{} 2 &{} -1\\ 0 &{} 0 &{} 2 &{} -3 &{} 1\\ 0 &{} -1&{} -1 &{} 1 &{} 1 \end{bmatrix},\\&{{\mathcal {V}}}^3= \begin{bmatrix} -1 &{} -1 &{} 0 &{} 0&{} 2\\ -1 &{} 2 &{} 1 &{} -1 &{} -1\\ 0 &{} 1&{} -1 &{} 2 &{} -2\\ 0 &{} -1 &{} 2 &{} -1 &{} 0\\ 2 &{} -1&{} -2 &{} 0 &{} 3 \end{bmatrix}, \end{aligned}$$

respectively.

The time-varying delay

$$\begin{aligned} \begin{aligned} \tau (t)=0.5+0.5cos(t-1), \end{aligned} \end{aligned}$$

has been considered. It is easy to check that $\tau _M=1$ and ${\dot{\tau }}=0.5$. The initial values of x on $[-1,0]$ are selected as:

$$\begin{aligned} \begin{aligned}&{\tilde{\varphi }}_{1}(t)=[2t^{2}+1,2.1cos2t+2,\sin 2t]^T,\\&{\tilde{\varphi }}_{2}(t)=[-2t^{2}+1,2\sin 2t+3,2\cos 2t]^T,\\&{\tilde{\varphi }}_{3}(t)=[0.2\sin 2t+1,0.1e^{3t}+e^{t},2t]^T,\\&{\tilde{\varphi }}_{4}(t)=[2\sin 2t-0.5,e^{t},3\cos 3t]^T\\&{\tilde{\varphi }}_{5}(t)=[2t^{2},2\cos 3t,3e^{2t}]^T. \end{aligned} \end{aligned}$$

The corresponding isolated HCNN of y for the initial conditions:

$$\begin{aligned} {\tilde{\psi }}(t)=[2\cos 3t,2\sin 2t,1.5e^{2t}]^T. \end{aligned}$$

The parameters in the controllers (8) are described as follows: $\eta _1=3.125$, $\eta _2=4.613$, $\eta _3=3.795$, $\eta _4=4.763$. Given the controller gains:

$$\begin{aligned}&K_1= \begin{bmatrix} 1.25 &{} 0 &{} 0\\ 0 &{} 1.05 &{} 0\\ 0 &{} 0 &{} 1.55 \end{bmatrix}, K_2= \begin{bmatrix} 2.05 &{} 0 &{} 0\\ 0 &{} 1.75 &{} 0\\ 0 &{} 0 &{} 1.37 \end{bmatrix},\\&K_3= \begin{bmatrix} 1.36 &{} 0 &{} 0\\ 0 &{} 1.92 &{} 0\\ 0 &{} 0 &{} 1.14 \end{bmatrix}, K_4= \begin{bmatrix} 3.211&{} 0&{} 0\\ 0&{} 4.06 &{} 0\\ 0 &{} 0 &{} 1.02 \end{bmatrix},\\&K_5= \begin{bmatrix} 1.82 &{} 0 &{} 0\\ 0 &{} 1.75 &{} 0\\ 0 &{} 0 &{} 1.32 \end{bmatrix}. \end{aligned}$$

In the simulation, all parameters are selected as $\lambda _1=1.114$, $\lambda _2=4.201$, $T=0.5$, $\theta =0.7$, $m=5$, $\delta =0.5$, $q=1.5$. Thus, by simple calculation, we have:

$$\begin{aligned}&\triangle \Gamma ^1= \begin{bmatrix} 0.062\sin t &{} 0 &{} 0\\ 0 &{} 0.279 &{} 0\\ 0 &{} 0 &{} 0.732 \end{bmatrix},\\&\triangle \Gamma ^2= \begin{bmatrix} 0.024 &{} 0 &{} 0\\ 0 &{} 0.0119\sin t &{} 0\\ 0 &{} 0 &{} 0.0105 \end{bmatrix},\\&\triangle \Gamma ^3= \begin{bmatrix} 0.039 &{} 0 &{} 0\\ 0 &{} 0.015 &{} 0\\ 0 &{} 0 &{} 0.011\cos t \end{bmatrix},\\&\triangle \beta ^1= \begin{bmatrix} 0.744\sin t &{} 0 &{} 0\\ 0 &{} 0.1344 &{} 0\\ 0 &{} 0 &{} 0.1728 \end{bmatrix},\\&\triangle \beta ^2= \begin{bmatrix} 0.0116 &{} 0 &{} 0\\ 0 &{} 0.027 &{} 0\\ 0 &{} 0 &{} 0.0188 \end{bmatrix},\\&\triangle \beta ^3= \begin{bmatrix} 0.062 &{} 0 &{} 0\\ 0 &{} 0.017 &{} 0\\ 0 &{} 0 &{} 0.091\cos t \end{bmatrix}. \end{aligned}$$

By solving the LMIs (11) and (12), we obtain as:

$$\begin{aligned}&S_1=\text {diag}[1.42,1.73,2.14,2.91,1.35,0.14,\\&\ \ \ \ \ 3.25,1.41,0.15,1.03,1.47,1.28,1.39,1.75,1.01],\\&S_2=\text {diag}[1.05,0.43,1.24,1.01,0.52,0.84,1.25,\\&\ \ \ \ \ 2.07,0.45,1.34,1.29,2.14,0.16,0.18,1.37],\\&S_3=\text {diag}[1.47,0.53,1.16,0.25,1.17,0.85,0.36,\\&\ \ \ \ \ 1.49,1.24,0.36,0.19,2.14,1.34,0.17,0.13],\\&S_4=\text {diag}[1.26,0.31,0.14,1.41,1.35,0.14,0.11,\\&\ \ \ \ \ \ 0.41,0.46,1.05,1.37,0.16,0.19,3.18,2.07]\\&S_5=\text {diag}[0.94,1.01,3.17,1.47,0.35,0.04,1.37,\\&\ \ \ \ \ \ 0.55,0.31,0.58,0.14,0.39,1.04,2.45,1.07]. \end{aligned}$$

Appendix III

The parameters in the HCNNs (1) and (3) are described as follows:

$$\begin{aligned}&Q= \begin{bmatrix} 0.91 &{} 0 &{} 0\\ 0 &{} 0.12 &{} 0\\ 0 &{} 0 &{} 0.57 \end{bmatrix}, A= \begin{bmatrix} 0.32 &{} 0.03 &{} 0.74\\ 0.03 &{} 1.32 &{} 0.33\\ 0.74 &{} 0.33 &{} 0.35 \end{bmatrix},\\&B= \begin{bmatrix} 0.34 &{} 0.61 &{} 0.04\\ 0.61 &{} 0.24 &{} 0.14\\ 0.04 &{} 0.14 &{} 1.07 \end{bmatrix}, \Gamma ^1= \begin{bmatrix} 4.01 &{} 0 &{} 0\\ 0 &{} 3.14 &{} 0\\ 0 &{} 0 &{} 4.05 \end{bmatrix},\\&\Gamma ^2= \begin{bmatrix} 2.14 &{} 0 &{} 0\\ 0 &{} 1.71 &{} 0\\ 0 &{} 0 &{} 1.58 \end{bmatrix}, \Gamma ^3= \begin{bmatrix} 2.14 &{} 0 &{} 0\\ 0 &{} 1.41 &{} 0\\ 0 &{} 0 &{} 1.08 \end{bmatrix},\\&\beta ^1= \begin{bmatrix}, 2.14 &{} 0 &{} 0\\ 0 &{} 2.05 &{} 0\\ 0 &{} 0 &{} 2.19 \end{bmatrix}, \beta ^2= \begin{bmatrix}, 1.01 &{} 0 &{} 0\\ 0 &{} 1.27 &{} 0\\ 0 &{} 0 &{} 1.34 \end{bmatrix},\\&\beta ^3= \begin{bmatrix}, 1.81 &{} 0 &{} 0\\ 0 &{} 1.24 &{} 0\\ 0 &{} 0 &{} 1.34 \end{bmatrix}, H^1= \begin{bmatrix} 2 &{} 0 &{} 0\\ 0 &{} 3 &{} 0\\ 0 &{} 0 &{} 1 \end{bmatrix},\\&H^2= \begin{bmatrix} 0.28 &{} 0 &{} 0\\ 0 &{} 0.16 &{} 0\\ 0 &{} 0 &{} 0.34 \end{bmatrix}, H^3= \begin{bmatrix} 1.05 &{} 0 &{} 0\\ 0 &{} 2.31 &{} 0\\ 0 &{} 0 &{} 1.24 \end{bmatrix},\\&F^1(t)= \begin{bmatrix} 0.12\sin t &{} 0 &{} 0\\ 0 &{} 0.11 &{} 0\\ 0 &{} 0 &{} 0.56 \end{bmatrix}, M^1= \begin{bmatrix} 0.16 &{} 0 &{} 0\\ 0 &{} 0.28 &{} 0\\ 0 &{} 0 &{} 0.34 \end{bmatrix},\\&F^2(t)= \begin{bmatrix} 0.24 &{} 0 &{} 0\\ 0 &{} 0.32\sin t &{} 0\\ 0 &{} 0 &{} 0.44 \end{bmatrix}, M^2= \begin{bmatrix} 0.54 &{} 0 &{} 0\\ 0 &{} 0.38 &{} 0\\ 0 &{} 0 &{} 0.26 \end{bmatrix},\\&F^3(t)= \begin{bmatrix} 0.24 &{} 0 &{} 0\\ 0 &{} 0.48 &{} 0\\ 0 &{} 0 &{} 0.22\cos t \end{bmatrix}, M^3= \begin{bmatrix} 0.32 &{} 0 &{} 0\\ 0 &{} 0.24 &{} 0\\ 0 &{} 0 &{} 0.18 \end{bmatrix},\\&{{\tilde{H}}}^1= \begin{bmatrix} 2 &{} 0 &{} 0\\ 0 &{} 1 &{} 0\\ 0 &{} 0 &{} 3 \end{bmatrix}, {{\tilde{H}}}^2= \begin{bmatrix} 0.06 &{} 0 &{} 0\\ 0 &{} 0.32 &{} 0\\ 0 &{} 0 &{} 0.16 \end{bmatrix},\\&{{\tilde{H}}}^3= \begin{bmatrix} 1.32 &{} 0 &{} 0\\ 0 &{} 1.28 &{} 0\\ 0 &{} 0 &{} 0.34 \end{bmatrix}, {{\tilde{F}}}^1(t)= \begin{bmatrix} 0.24\sin t &{} 0 &{} 0\\ 0 &{} 0.34 &{} 0\\ 0 &{} 0 &{} 0.16 \end{bmatrix},\\&{{\tilde{F}}}^2(t)= \begin{bmatrix} 0.14 &{} 0 &{} 0\\ 0 &{} 0.42\sin t &{} 0\\ 0 &{} 0 &{} 0.36 \end{bmatrix},\\&{{\tilde{F}}}^3(t)= \begin{bmatrix} 0.36 &{} 0 &{} 0\\ 0 &{} 0.12 &{} 0\\ 0 &{} 0 &{} 0.24\cos t \end{bmatrix},\\&{{\tilde{M}}}^1= \begin{bmatrix} 0.32 &{} 0 &{} 0\\ 0 &{} 0.26 &{} 0\\ 0 &{} 0 &{} 0.28 \end{bmatrix}, {{\tilde{M}}}^2= \begin{bmatrix} 0.16 &{} 0 &{} 0\\ 0 &{} 0.28 &{} 0\\ 0 &{} 0 &{} 0.34 \end{bmatrix},\\&{{\tilde{M}}}^3= \begin{bmatrix} 0.11 &{} 0 &{} 0\\ 0 &{} 0.23 &{} 0\\ 0 &{} 0 &{} 0.31 \end{bmatrix},\\&u_1=0.82, \ \ u_2=0.61, \ \ \ u_3=0.71,\\&v_1=0.12,\ \ v_2=0.14\ \ \ v_3=0.55. \end{aligned}$$

The nonlinear activation functions:

$$\begin{aligned}&f_{i1}(t,x_{i1}(t))=0.02x_{i1}(t)+0.015\text {sign}(x_{i1}(t)),\\&f_{i2}(t,x_{i2}(t))=0.08x_{i2}(t)+0.003\text {sign}(x_{i2}(t)),\\&f_{i3}(t,x_{i3}(t))=0.06x_{i3}(t)+0.008\text {sign}(x_{i3}(t)), \end{aligned}$$

have been considered. It can be easily verified that assumptions $A_2$ hold for $z_{1}=0.02$, $\omega _{1}=0.03$, $z_{2}=0.08$, $\omega _{2}=0.006$, $z_{3}=0.06$, $\omega _{3}=0.016$.

The non-delayed and time-varying delayed coupling configuration matrices read as:

$$\begin{aligned}&U^1= \begin{bmatrix} 2 &{} 0 &{} 1 \\ 0 &{} -2&{} 2\\ 1 &{} 2 &{} -3 \end{bmatrix}, U^2= \begin{bmatrix} -1 &{} -1 &{} 2\\ -1 &{} 2 &{} -1 \\ 2 &{} -1 &{} -1 \end{bmatrix}, \end{aligned}$$

$$\begin{aligned}&U^3= \begin{bmatrix} 0.2 &{} 0.3 &{} 0.4\\ 0.3 &{} 1.5 &{} -1.8\\ 0.4 &{} -1.8 &{} 1.4 \end{bmatrix},\\ \end{aligned}$$

and

$$\begin{aligned}&{{\mathcal {V}}}^1= \begin{bmatrix} 1 &{} 1 &{} -2\\ 1 &{} -1 &{} 0 \\ -2 &{} 0 &{} 2 \end{bmatrix}, {{\mathcal {V}}}^2= \begin{bmatrix} 1 &{} -1 &{} 0 \\ -1 &{} 2 &{} -1 \\ 0 &{} -1&{} 1 \end{bmatrix},\\&{{\mathcal {V}}}^3= \begin{bmatrix} -1 &{} -1 &{} 0 \\ -1 &{} 2 &{} -1\\ 0 &{} -1&{} 1 \end{bmatrix},\\ \end{aligned}$$

respectively. The time-varying delay

$$\begin{aligned} \begin{aligned} \tau (t)=0.5+0.4cos(t-1), \end{aligned} \end{aligned}$$

has been considered. It is easy to check that $\tau _M=1$ and ${\dot{\tau }}=0.4$. The initial values of x on $[-1,0]$ are selected as:

$$\begin{aligned} \begin{aligned}&{\tilde{\varphi }}_{1}(t)=[0.2\sin 2t+1,0.1e^{3t}+e^{t},2t]^T,\\&{\tilde{\varphi }}_{2}(t)=[2\sin 2t-0.5,e^{t},3\cos 3t]^T,\\&{\tilde{\varphi }}_{3}(t)=[2t^{2},2\cos 3t,3e^{2t}]^T. \end{aligned} \end{aligned}$$

The corresponding isolated HCNN of y(t) for the initial conditions

$$\begin{aligned} {\tilde{\psi }}(t)=[2\cos 3t,2\sin 2t,1.5e^{2t}]^T. \end{aligned}$$

The parameters in the controllers (28) are described as follows: $\eta _1=2.256$, $\eta _2=3.249$, $\eta _3=2.316$, $\eta _4=3.001$ and $\varepsilon _i=0.12$. Given the controller gains:

$$\begin{aligned}&K_1= \begin{bmatrix} 1.05 &{} 0 &{} 0\\ 0 &{} 1.15 &{} 0\\ 0 &{} 0 &{} 1.51 \end{bmatrix},\\&K_2= \begin{bmatrix} 1.05 &{} 0 &{} 0\\ 0 &{} 1.31 &{} 0\\ 0 &{} 0 &{} 0.52 \end{bmatrix},\\&K_3= \begin{bmatrix} 1.13 &{} 0 &{} 0\\ 0 &{} 1.04 &{} 0\\ 0 &{} 0 &{} 1.26 \end{bmatrix}. \end{aligned}$$

In the simulation, all parameters are selected as $T=0.4$, $\theta =0.6$, $m=5$, $\delta =1.5$, $q=0.5$. Thus, by simple calculation, we have

$$\begin{aligned}&\triangle \Gamma ^1= \begin{bmatrix} 0.0384\sin t &{} 0 &{} 0\\ 0 &{} 0.0924 &{} 0\\ 0 &{} 0 &{} 0.1904 \end{bmatrix},\\&\triangle \Gamma ^2= \begin{bmatrix} 0.0363 &{} 0 &{} 0\\ 0 &{} 0.019\sin t &{} 0\\ 0 &{} 0 &{} 0.035 \end{bmatrix},\\&\triangle \Gamma ^3= \begin{bmatrix} 0.08 &{} 0 &{} 0\\ 0 &{} 0.266 &{} 0\\ 0 &{} 0 &{} 0.049\cos t \end{bmatrix}, \end{aligned}$$

$$\begin{aligned}&\triangle \beta ^1= \begin{bmatrix} 0.0013 &{} 0 &{} 0\\ 0 &{} 0.038\sin t &{} 0\\ 0 &{} 0 &{} 0.019 \end{bmatrix},\\&\triangle \beta ^2= \begin{bmatrix} 0.1536 &{} 0 &{} 0\\ 0 &{} 0.0884 &{} 0\\ 0 &{} 0 &{} 0.1344 \end{bmatrix},\\&\triangle \beta ^3= \begin{bmatrix} 0.052 &{} 0 &{} 0\\ 0 &{} 0.035 &{} 0\\ 0 &{} 0 &{} 0.025\cos t \end{bmatrix}, \end{aligned}$$

By solving the LMIs (29) and (30), we obtain that

$$\begin{aligned}&S_1=\text {diag}[1.417,0.053,1.106,0.215,\\&\ \ \ \ \ \ \ \ \ \ \ 1.107,0.815,0.306,1.014,1.348],\\&S_2=\text {diag}[1.106,0.501,0.254,1.106,\\&\ \ \ \ \ \ \ \ \ \ \ \ 1.109,0.368,0.456,0.528,0.327]\\&S_3=\text {diag}[0.201,0.617,1.017,0.528,\\&\ \ \ \ \ \ \ \ \ \ \ \ \ 0.169,1.204,0.168,0.429,0.558]. \end{aligned}$$