New stability results for impulsive neural networks with time delays

Abstract

This paper investigates the stability of impulsive neural networks with time delays. Based on a new tool called as uniformly exponentially convergent functions, an improved Razumikhin method leads to new, more permissive stability results. By comparison with the existing results, the rigorous restrictions on impulses, which are presented in the previous Razumikhin stability theorems, are removed. Moreover, the obtained results do not restrict that the time derivative of Lyapunov function is negative definite or positive definite under the Razumikhin condition. The effectiveness of the proposed results is demonstrated by three simple numerical examples.

Appendix

Proof of Lemma 1

Necessity If z(t) is a UECF, there exist two positive constants $M\ge 1$ and $\varepsilon >0$ such that

$$\begin{aligned} z(t+\theta )=z(t)\prod _{t < t_i \le t+\theta } \gamma _i {\text {e}}^{\int _t^{t+\theta }\mu (s){\text {d}}s}\le M z(t){\text {e}}^{-\varepsilon \theta }, \forall t \ge t_0. \end{aligned}$$

(40)

Namely, $\prod _{t<t_i \le t+\theta } \gamma _i {\text {e}}^{\int _t^{t+\theta }\mu (s){\text {d}}s} \le M {\text {e}}^{-\varepsilon \theta }$. Therefore, one can choose $\hat{T}(c)=-\frac{1}{\varepsilon } \ln \dfrac{c}{M}$, $d=M$, and then (7) holds.

Sufficiency For any $t^{**}\ge t^* \ge t_0$, there must exist some integer $l\ge 0$ such that $t^{**}=t^*+lT+\theta$, where $\theta \in [0,T)$. Then, one has

$$\begin{aligned} z(t^{**}) = z(t^*)\prod _{t^* < t_i \le t^{**}}\gamma _i {\text {e}}^{\int _{t^*}^{t^{**}}\mu (s){\text {d}}s}. \end{aligned}$$

(41)

When $l=0$, it follows from (41) that

$$\begin{aligned} z(t^{**})=z(t^*)\prod _{t^* < t_i \le t^*+\theta } \gamma _i {\text {e}}^{\int _{t^*}^{t^*+\theta }\mu (s){\text {d}}s} \le d z(t^*){\text {e}}^{\dfrac{\ln c}{T}(t^{**}-\theta -t^*)} =d z\left( t^*\right) {\text {e}}^{\dfrac{-\theta \ln c }{T}} {\text {e}}^{\dfrac{\ln c}{T}\left( t^{**}-t^*\right) }. \end{aligned}$$

(42)

When $l\ge 1$, one can obtain from (41),

$$\begin{aligned} z(t^{**})&= \,z(t^*) \prod _{j=0}^{l-1}\left( \prod _{t^*+jT< t_i \le t^*+(j+1)T} \gamma _i {\text {e}}^{\int _{t^{*}+jT}^{t^*+(j+1)T}} \right) \prod _{t^*+lT < t_i \le t^{**}} \gamma _i {\text {e}}^{\int _{t+lT}^{t^{**}}\mu (s){\text {d}}s} \nonumber \\&\le \,dz(t^*)c^l =dz(t^*) {\text {e}}^{\dfrac{-\theta \ln c}{T}} {\text {e}}^{\dfrac{\ln c}{T}\left( t^{**}-t^*\right) }. \end{aligned}$$

(43)

Denote $M_0=\dfrac{d}{c}$ and $\varepsilon _0=-\dfrac{\ln c}{T}$. Based on (42) and (43), one obtain $z(t^{**})\le M_0 z(t^*) {\text {e}}^{-\varepsilon _0 (t^{**}-t^*)}$, which implies that z(t) is a UECF.

Proof of Lemma 2

If for any $u\in [t-T,t]$, $y_1(u)\ge \psi (w(u))$ holds. Then, one obtains

$$\begin{aligned} y_1(t) \le y_1(t-T) \prod _{t-T < t_i \le t} \gamma _i {\text {e}}^{\int _{t-T}^t \mu (s){\text {d}}s}. \end{aligned}$$

(44)

If $y_1(u)\ge \psi (w(u))$ is not satisfied for all $u \in [t-T,t]$, there must exist some $t^*= \sup \{u\in [t-T,t]:y_1(u)<\psi (w(u))\}$. When $t^*=t$, one derives

$$\begin{aligned} y_1(t)\le \psi (w(t)) \le \sup _{t-T\le u\le t}\{\psi (w(u))\} \le \sup _{t-T\le u\le t}\{\psi (w(u))\} \chi _z(T). \end{aligned}$$

(45)

When $t^*<t$, one knows that $y_1(\bar{s})\ge \psi (w(\bar{s}))$, $\bar{s} \in [t^*,t]$. Therefore, one otains

$$\begin{aligned} y_1(t)\le y_1(t^{*-})\prod _{t^*\le t_i \le t} \gamma _i {\text {e}}^{\int _{t^*}^t \mu (s){\text {d}}s} \le \psi (w(t^{*-})) \chi _z(t-t^*) \le \sup _{t-T\le u \le t}\psi (w(u)) \chi _z(T). \end{aligned}$$

(46)

Proof of Theorem 1

Let $v(t)=V(t,x(t))$, $w(t)=\sup _{s \in [-\bar{\tau },0]}V(t+s,x(t+s))$, $\psi (w)=q^{-1}(w)$. Then, based on the conditions of Theorem 1 and Lemma 2, one obtains

$$\begin{aligned} v(t)&\le \,\max \left\{ v(t-T)\prod _{t-T < t_i \le t} \gamma _i {\text {e}}^{\int _{t-T}^t \mu (s){\text {d}}s}, \sup _{t-T \le s \le t} \left\{ q^{-1}(w(s)) \chi _z(T) \right\} \right\} \nonumber \\&\le \,\max \left\{ v(t-T) \eta _z(t), \rho \sup _{t-T \le u \le t} \{ w(u) \} \right\} \nonumber \\&\le \,\max \left\{ v(t-T) \eta _z(t), \rho \sup _{-\bar{T} \le \zeta \le 0} \{ y(t+\zeta ) \} \right\} \nonumber \\&\le \,\sup _{-\bar{T} \le \zeta \le 0} \left\{ v(t+\zeta ) \right\} \max \left\{ \eta _z(t) ,\rho \right\} , \end{aligned}$$

(47)

where $\eta _z(t)=\prod _{t-T < t_i \le t} \gamma _i {\text {e}}^{\int _{t-T}^t \mu (s)}{\text {d}}s$, $\bar{T}=T+\bar{\tau }$. Because $T \in \Xi _z$, one can see that there exists some $\varrho \in (0,1)$ such that $\eta _z(t)\le \varrho$, $t\ge t_0+T$. Denote $\bar{\rho }=\max \{ \rho , \varrho \}$, one has

$$\begin{aligned} v(t)\le \bar{\rho } \sup _{-\bar{T}\le \zeta \le 0}v(t+\zeta ). \end{aligned}$$

(48)

By Lemma 3, the above inequality implies that

$$\begin{aligned} v(t) \le \sup _{-\bar{T}\le \zeta \le 0} \{v(t^*+\zeta )\} {\text {e}}^{\dfrac{\ln \bar{\rho }}{\bar{T}} (t-t^*)}, \end{aligned}$$

(49)

where $t^*=t_0+T$. It follows from Condition (i) and inequality (49) that

$$\begin{aligned} \Vert x(t)\Vert \le \left( \dfrac{c_2}{c_1}\right) ^{\dfrac{1}{p}} \Vert x(t^*)\Vert _{\bar{T}} \exp \left( \dfrac{\ln \bar{\rho }}{p \bar{T}} (t-t^*) \right) . \end{aligned}$$

(50)

where $\Vert x(t^*)\Vert _{\bar{T}} = \sup _{-\bar{T}\le \zeta \le 0} \Vert x(t^*+\zeta )\Vert$. This indicates that the neural network (1) is globally exponentially stable.