Revealing network topology and dynamical parameters in delay-coupled complex network subjected to random noise

Abstract

It is well known that random noise and time delay are two inherent ingredients in complex networks, whose dynamical parameters and topological structures are often unknown or uncertain. This paper will employ the techniques of impulsive control and adaptive control to infer dynamical parameters and network topology in delay-coupled complex network under circumstance noise. By constructing an appropriate adaptive–impulsive control strategy in the response network, the unknown dynamical parameters and topology structure contained in the drive network are to be accurately identified; moreover, these two networks will achieve the global exponential synchronization in mean square. Based on the comparison theorem of impulsive differential equations, the accuracy of the proposed identification strategy is rigorously proved. Finally, two examples with networks of chaotic oscillators are presented to illustrate the application of the suggested strategy. Meanwhile, numerical results indicate that our proposed scheme is robust against the impulsive gain, the update gain and the network topology.

Appendix

The Proof of the Theorem.

Proof

Let $\widetilde{\theta }_i (t)=\overline{\theta }_i (t)-\theta _i$, $\widetilde{b_{ij} }(t)=d_{ij} (t)-b_{ij}$, $e_i (t)=y_i (t)-x_i (t)$, $\left( {i,j=1,2,\ldots M} \right)$. Subtracting the drive network (1) from the response network (2) and using the controller (3) and (4), then the synchronization error between the drive network and the response network can be written as

$$\begin{aligned}&\left\{ {{\begin{array}{l} {\hbox {d}}e_i (t)=\Big [g_i (t,y_i (t)){\widetilde{\theta }_i }(t)+F_i (t,y_i (t),\theta _i )\\ \qquad \qquad \quad -F_i (t,x_i (t),\theta _i )+\sum _{j=1}^M {\widetilde{b}_{ij}}(t)Ay_j (t-\tau (t)) \\ \qquad \qquad \quad \left. +\,\sum _{j=1}^M {b_{ij} Ae_j (t-\tau (t))} -k_i (t)e_i (t)\right] dt\\ \qquad \qquad \quad +\,\sigma _i (e_i (t),e_i (t-\tau (t)),t)dW(t),t\ne t_k \\ {\Delta e_i (t_k )=e_i (t_k )\!-\!e_i (t_k^- )=-d_k e_i (t_k^- ),t=t_k ,k\in Z_+ } \\ \end{array} }} \right. \!,\nonumber \\ \end{aligned}$$

(12)

where $\sigma _i (e_i (t),e_i (t-\tau (t)),t)=\sigma _i (y_i (t),y_i (t-\tau (t)),t)-\sigma _i (x_i (t),x_i (t-\tau (t)),t)$, $\phi _i (s)=({\widetilde{\theta }_i }(s),{\widetilde{b}_{ij} }(s),k_i (s),e_i (s))\in R^{m_1 +M+1+n}$ is the initial value of (12).

Introduce the following V function

$$\begin{aligned} V(t)= & {} \frac{(1-d_k )^{2}}{2}\sum _{i=1}^M {\widetilde{\theta _i }^{T}(t)\widetilde{\theta _i }(t)}\\&+\frac{(1-d_k )^{2}}{2}\sum _{i=1}^M {\sum _{j=1}^M {\widetilde{b_{ij} }} } ^{2}(t)\\&+\frac{(1-d_k )^{2}}{2}\sum _{i=1}^M {\frac{1}{\lambda _i }(k_i (t)-L-\alpha )^{2}}\\&+\frac{1}{2}\sum _{i=1}^M {e_i^T } (t)e_i (t), \end{aligned}$$

where $L=\max \limits _{1\le i\le M} \{L_i \}$, $\alpha =\lambda _{\max } \left( \frac{1}{2}CC^{T}\right)$, $C=B\otimes A$, $\otimes$ is the Kronecker product.

Since the impulsive controller acts only at impulsive instant, and $d_k =0$ for any $t\in (t_{k-1} ,t_k )$, thus the V function can be written as for $t\in (t_{k-1} ,t_k )$

$$\begin{aligned} V(t)= & {} \frac{1}{2}\sum _{i=1}^M {\widetilde{\theta } _i}^{T}(t){\widetilde{\theta }_i }(t) +\frac{1}{2}\sum _{i=1}^M {\sum _{j=1}^M {\widetilde{b}_{ij}}} ^{2}(t)\\&+\frac{1}{2}\sum _{i=1}^M {\frac{1}{\lambda _i }(k_i (t)-L-\alpha )^{2}}\\&+\frac{1}{2}\sum _{i=1}^M {e_i^T } (t)e_i (t). \end{aligned}$$

Applying the $It\hat{o}$ formula to the combined equations of the error dynamical network (12) and the adaptive update laws (5)– (7), then the differential operator $\ell V$can be derived as follows

$$\begin{aligned} \ell V= & {} -L\sum _{i=1}^M {e_i ^{T}(t)e_i (t)} -\alpha \sum _{i=1}^M {e_i ^{T}(t)e_i (t)} \nonumber \\&+\sum _{i=1}^M {e_i ^{T}(t)} [F_i (t,y_i (t),\theta _i )-F_i (t,x_i (t),\theta _i )]\nonumber \\&+\sum _{i=1}^M {\sum _{j=1}^M {e_i ^{T}(t)b_{ij} Ae_j (t-\tau (t))} }\nonumber \\&+\frac{1}{2}\sum _{i=1}^M {trace(\sigma _i^T \sigma _i )}. \end{aligned}$$

(13)

From $(H_1 )$ and $(H_2 )$, one can obtain, respectively, that

$$\begin{aligned} e_i^T (t)[F_i (t,y_i (t),\theta _i )-F_i (t,x_i (t),\theta _i )]\le L_i e_i^T (t)e_i (t) \end{aligned}$$

and

$$\begin{aligned} \frac{1}{2}{\hbox {trace}}(\sigma _i ^{T}\sigma _i )\le & {} pe_i ^{T}(t)e_i (t)+qe_i ^{T}(t-\tau (t))\\&\times e_i (t-\tau (t)). \end{aligned}$$

Then (13) can be rewritten as

$$\begin{aligned} \ell V\le & {} -\alpha \sum _{i=1}^M {e_i^T (t)e_i (t)}\nonumber \\&+ \sum _{i=1}^M {\sum _{j=1}^M {e_i ^{T}(t)b_{ij} Ae_j (t-\tau (t))} } \nonumber \\&+q\sum _{i=1}^M {e_i^T (t-\tau (t))e_i (t-\tau (t))} \nonumber \\&+p\sum _{i=1}^M {e_i^T (t)e_i (t)} . \end{aligned}$$

(14)

Let $e(t)=(e_1 ^{T}(t),e_2 ^{T}(t),\ldots ,e_M ^{T}(t))^{T}\in R^{nM}$, then

$$\begin{aligned} \ell V\le & {} -\alpha e^{T}(t)e(t)+e^{T}(t)Ce(t-\tau (t))\\&+pe^{T}(t)e(t)+qe^{T}(t-\tau (t))e(t-\tau (t)). \end{aligned}$$

It follows from Lemma 1 that

$$\begin{aligned}&e^{T}(t)Ce(t-\tau (t))\le \frac{1}{2}e^{T}(t)CC^{T}e(t)\\&\quad +\frac{1}{2}e^{T}(t-\tau (t))e(t-\tau (t)). \end{aligned}$$

Hence,

$$\begin{aligned} \ell V\le & {} -\alpha e^{T}(t)e(t)+\frac{1}{2}e^{T}(t)CC^{T}e(t)\\&+\,\frac{1}{2}e^{T}(t-\tau (t))e(t-\tau (t))+pe^{T}(t)e(t)\\&+\,qe^{T}(t-\tau (t))e(t-\tau (t))\\\le & {} p\sum _{i=1}^M {e_i^T (t)e_i (t)}\\&+\left( q+\frac{1}{2}\right) \sum _{i=1}^M {e_i^T (t-\tau (t))e_i (t-\tau (t))}\\\le & {} 2pV(t)+(2q+1)V(t-\tau (t)). \end{aligned}$$

According to the $It\hat{o}$ formula, we have

$$\begin{aligned} dV(t)= & {} \ell Vdt\\&+\sum _{i=1}^M {e_i^T (t)} \sigma _i (e_i (t),e_i (t-\tau (t)),t)dW(t)\\\le & {} [2pV(t)+(2q+1)V(t-\tau (t))]dt\\&+\sum _{i=1}^M {e_i^T (t)} \sigma _i (e_i (t),e_i (t-\tau (t)),t)dW(t). \end{aligned}$$

Taking mathematical expectation on both sides of the above inequality, one can obtain that

$$\begin{aligned} \frac{\textit{dEV}(t)}{dt}\le 2pEV(t)+(2q+1)EV(t-\tau (t)). \end{aligned}$$

(15)

When $t=t_k$, $e_i (t_k )=\Delta e_i (t_k )+e_i (t_{_k }^- )=-d_k e_i (t_{_k }^- )+e_i (t_{_k }^- )=(1-d_k )e_i (t_{_k }^- )$.

Then the V function can be rewritten as

$$\begin{aligned} V(t_k )= & {} \frac{(1-d_k )^{2}}{2}\sum _{i=1}^M {\widetilde{\theta }_i }^{T}(t_{_k }^- ){\widetilde{\theta }_i} (t_{_k }^- )\\&+\,\frac{(1-d_k )^{2}}{2}\sum _{i=1}^M {\sum _{j=1}^M {\widetilde{b}_{ij}}} ^{2}(t_{_k }^- ) \\&+\,\frac{(1-d_k )^{2}}{2}\sum _{i=1}^M {\frac{1}{\lambda _i }(k_i (t_{_k }^- )-L-\alpha )^{2}} \\&+\,\frac{(1-d_k )^{2}}{2}\sum _{i=1}^M {e_i^T } (t_{_k }^- )e_i (t_{_k }^- ) \\= & {} (1-d_k )^{2}V(t_{_k }^- ). \end{aligned}$$

In view of (8), we have $V(t_k )\le \beta V(t_{_k }^- )$, hence

$$\begin{aligned} EV(t_k )\le \beta EV(t_{_k }^- ). \end{aligned}$$

(16)

For any $\delta >0$, let $\upsilon (t)$ be a unique solution of the following impulsive delay system

$$\begin{aligned}&{}\left\{ {{\begin{array}{l} \dot{\upsilon }(t)=2p\upsilon (t)+(2q+1)\upsilon (t-\tau (t))+\delta ,t\ne t_k \\ {\upsilon (t_k )=\beta \upsilon (t_{_k }^- ),t=t_k ,k\in Z_+ } \\ {\upsilon (s)=E\left\| {\phi (s)} \right\| ^{2},s\in [-\bar{\tau },0]} \\ \end{array} }} \right. \!, \end{aligned}$$

(17)

where $\phi (s)=(\phi _1^T (s),\phi _2^T (s),\ldots ,\phi _M^T (s))^{T}$ is the initial value of (12).

Since $EV(s)\le E\left\| {\phi (s)} \right\| ^{2}=\upsilon (s)$ for $s\in [-\bar{\tau },0]$, it follows from Lemma 2 (i.e., the comparison theorem of impulsive differential equations) that

$$\begin{aligned} \upsilon (t)\ge EV(t)\ge 0,\forall t\ge 0. \end{aligned}$$

(18)

Considering the formula for the variation of parameters [41], we have

$$\begin{aligned} \upsilon (t)= & {} P(t,0)\upsilon (0)\nonumber \\&+\int _0^t {P(t,s)} [(2q+1)\upsilon (s-\tau (s))+\delta ]ds,\nonumber \\ \end{aligned}$$

(19)

where $P(t,s),\, t,s\ge 0$ is the Cauchy matrix of linear system

$$\begin{aligned} \left\{ {{\begin{array}{l} {\dot{\eta }(t)=2p\eta (t),t\ne t_k } \\ {\eta (t_k )=\beta \eta (t_{_k }^- ),t=t_k } \\ \end{array} }} \right. . \end{aligned}$$

In view of the representation of the Cauchy matrix, we can get the following estimation

$$\begin{aligned} P(t,s)= & {} e^{2p(t-s)}\prod _{s\le t_k \le t} \beta \le e^{2p(t-s)}\beta ^{\frac{t-s}{\zeta }-1}\nonumber \\= & {} \beta ^{-1}e^{\left( 2p+\frac{In\beta }{\zeta }\right) (t-s)}=\beta ^{-1}e^{-a(t-s)}, \end{aligned}$$

(20)

where $a=-(2p+\frac{In\beta }{\zeta })$.

Let $\gamma =\beta ^{-1}\sup \limits _{-\bar{\tau }\le s\le 0} \{E\left\| {\phi (s)} \right\| ^{2}\}$, then it can be derived from (19) and (20)

$$\begin{aligned}&\upsilon (t)\le \gamma e^{-at}+\int _0^t \beta ^{-1}e^{-a(t-s)}\nonumber \\&\quad \times [(2q+1)\upsilon (s-\tau (s))+\delta ] {\hbox {d}}s. \end{aligned}$$

(21)

Now we introduce the following function

$$\begin{aligned} h(\mu )=\mu -a+(2q+1)\beta ^{-1}e^{\mu \bar{\tau }}. \end{aligned}$$

It follows from (9) that $h(0)=-a+(2q+1)\beta ^{-1}<0$. Since $h(+\infty )=+\infty$ and $h^{{\prime }}(\mu )>0$, there exists a unique $\lambda >0$ such that

$$\begin{aligned} \lambda -a+(2q+1)\beta ^{-1}e^{\lambda \bar{\tau }}=0. \end{aligned}$$

(22)

From (9), it is obvious that $a\beta -(2q+1)>0$, therefore,

$$\begin{aligned} \upsilon (t)= & {} E\left\| {\phi (t)} \right\| ^{2}\le \gamma <\gamma e^{-\lambda t}\\&+\frac{\delta }{a\beta -(2q+1)}, \qquad \forall t\in [-\bar{\tau },0]. \end{aligned}$$

In the following, we shall prove that

$$\begin{aligned} \upsilon (t)<\gamma e^{-\lambda t}+\frac{\delta }{a\beta -(2q+1)}, \forall t\ge 0. \end{aligned}$$

(23)

If this is not true, there will exist a smallest nonnegative $t^{*}$ such that

$$\begin{aligned} \upsilon (t)<\gamma e^{-\lambda t}+\frac{\delta }{a\beta -(2q+1)}, 0\le t<t^{*} \end{aligned}$$

(24)

and

$$\begin{aligned} \upsilon (t^{*})\ge \gamma e^{-\lambda t^{*}}+\frac{\delta }{a\beta -(2q+1)}. \end{aligned}$$

(25)

Combining (21) with (24), we have

$$\begin{aligned}&\upsilon (t^{*})\le \gamma e^{-at^{*}}\nonumber \\&\quad +\int _0^{t^{*}} \beta ^{-1}e^{-a(t^{*}-s)}[(2q+1)\upsilon (s-\tau (s))+\delta ] {\hbox {d}}s\nonumber \\&\quad <\gamma e^{-at^{*}}+\int _0^{t^{*}} \beta ^{-1}e^{-a(t^{*}-s)}\Bigg [(2q+1)\nonumber \\&\qquad \qquad \left. \left( \gamma e^{-\lambda (s-\tau (s))}+\frac{\delta }{a\beta -(2q+1)}\right) +\delta \right] {\hbox {d}}s \nonumber \\&\quad <e^{-at^{*}}\left\{ \gamma +\int _0^{t^{*}} \beta ^{-1}e^{as}\left[ (2q+1)\left( \gamma e^{-\lambda (s-\tau (s))}\right. \right. \right. \nonumber \\&\qquad \qquad \qquad \left. \left. \left. +\frac{\delta }{a\beta -(2q+1)}\right) +\delta \right] ds\right. \nonumber \\&\qquad \left. +\frac{\delta }{a\beta -(2q+1)}\right\} . \end{aligned}$$

(26)

In view of (22) and (26), we have

$$\begin{aligned}&\upsilon (t^{*})\,{<}\,e^{-at^{*}}\left\{ \gamma +\int _0^{t^{*}} {\beta ^{-1}e^{as}(2q+1)\gamma e^{-\lambda s}e^{\lambda \tau (s)}{\hbox {d}}s} \right. \\&\qquad +\int _0^{t^{*}} {\beta ^{-1}e^{as}(2q+1)\frac{\delta }{a\beta -(2q+1)}{\hbox {d}}s} \\&\qquad \left. +\int _0^{t^{*}} {\beta ^{-1}e^{as}\delta } ds+\frac{\delta }{a\beta -(2q+1)}\right\} \\&\quad =\gamma e^{-\lambda t^{*}}+\frac{\delta }{a\beta -(2q+1)}. \end{aligned}$$

Obviously, the above inequality contradicts (25). Thus, the inequality (23) holds. Letting $\delta \rightarrow 0$, we can get from (18) that

$$\begin{aligned} EV(t)\le \upsilon (t)<\gamma e^{-\lambda t}, \,t\ge 0. \end{aligned}$$

Therefore, we can derive that $E\left\| {e_i (t)} \right\| ^{2}<2\gamma e^{-\lambda t}\left( {i=1,2,\ldots ,M} \right)$ from the V function. From the Definition, we know that the solution $e_i (t)=0$ of (12) is exponentially stable in mean square, which means that the response network (2) can exponentially synchronize the drive network (1) in mean square. Moreover, from $(H_3 )$, Equations (7) and (12) one can obtain that ${\widetilde{\theta }_i}(t)=0, {\widetilde{b}_{ij} }(t)=0$ are also exponentially stable in mean square, and the feedback strength $k_i(t)$ convergences to some constants. Thus, the unknown dynamical parameters and topology structure contained in the drive network (1) are to be accurately inferred during the process of synchronization between the drive network (1) and the response network (2). This completes the proof of the Theorem. $\square$