Synchronization of fractional-order chaotic systems using unidirectional adaptive full-state linear error feedback coupling

Abstract

Based on the stability theory of fractional-order system, a novel unidirectional adaptive full-state linear error feedback coupling scheme is extended to control and synchronize all of fractional-order differential (FOD) chaotic systems with in-commensurate (and commensurate) orders. The feedback strength is adaptive to an updated law rather than prescribed as a constant. The convergence speed of feedback strength is regulated by a constant. With rigorous linear algebraic theorems and precisely numerical matrix computations, a reasonable interval in which the ultimate final control strength dwells is suggested, and the reliability of synchronization state is guaranteed. It demonstrates that the unidirectional full-state linear feedback coupling scheme can be adopted to control and synchronize FOD chaotic systems directly. Numerical simulations of three representative FOD chaotic systems illustrate the effectiveness of the proposed scheme.

Appendix: The proof of Theorem 3

Proof

A symmetric matrix has exactly $n$ real eigenvalues (counting multiplicities). There is an orthogonal matrix $Q$, such that

$$\begin{aligned} Q^\mathtt T \left( \frac{A + A^\mathtt T }{2} \right) Q = \bar{\varLambda } = \left( \bar{\lambda }_1, \bar{\lambda }_2, \ldots , \bar{\lambda }_n \right) , \end{aligned}$$

where $\bar{\varLambda }$ is the spectrum of the symmetric part of $A$, and $\bar{\lambda }_i \in {\mathbb {R}}$, $i = 1, 2, \ldots , n$.

Due to

$$\begin{aligned} \bar{\lambda }_{\mathrm{max}} \left( \frac{A + A^\mathtt T }{2} \right) = {\text{ max }} \left\{ \bar{\lambda }_1, \bar{\lambda }_2, \ldots , \bar{\lambda }_n \right\} , \end{aligned}$$

and

$$\begin{aligned} \bar{\lambda }_{\mathrm{max}} \left( \frac{A + A^\mathtt T }{2} \right) = \underset{x \ne 0}{\text{ max }} \frac{x^\mathtt T \left( \frac{A + A^\mathtt T }{2}\right) x}{x^\mathtt T x}, \end{aligned}$$

one has

$$\begin{aligned} \bar{\lambda }_{\mathrm{max}} \left( \frac{A + A^\mathtt T }{2} \right) \ge \frac{u^\mathtt T \left( \frac{A + A^\mathtt T }{2}\right) u}{u^\mathtt T u}. \end{aligned}$$

Let $u^\mathtt T = (1, 1, \ldots , 1)$,

$$\begin{aligned} \bar{\lambda }_{\mathrm{max}} \left( \frac{A + A^\mathtt T }{2} \right)\ge & {} u^\mathtt T \left( \frac{A + A^\mathtt T }{2}\right) u \\= & {} \sum _{ij}^n{\left( \frac{a_{ij} + a_{ji}}{2}\right) }. \end{aligned}$$

On the other hand, according to the Gershgorin’s circle theorem, we have

$$\begin{aligned} \bar{\lambda }_{\mathrm{max}} \left( \frac{A + A^\mathtt T }{2} \right) \le \sum _{i \ne j}^n{\left( \frac{\left| a_{ij} + a_{ji}\right| }{2}\right) + a_{ii}}. \end{aligned}$$

Suppose $q_i$ is the $i$th column of $Q$,

$$\begin{aligned} \left\| x \right\| _2^2= & {} x^\mathtt T x = x^\mathtt T Q Q^\mathtt T x = \left( Q^\mathtt T x \right) ^\mathtt T \left( Q^\mathtt T x \right) \\= & {} \sum _{i = 1}^n {\left( q_i^\mathtt T x\right) ^2}. \end{aligned}$$

And,

$$\begin{aligned} x^\mathtt T A x&= x^\mathtt T \left( \frac{A + A^\mathtt T }{2} \right) x = x^\mathtt T \left( Q \bar{\varLambda } Q^\mathtt T \right) x\\&= \left( Q^\mathtt T x \right) ^\mathtt T \bar{\varLambda } \left( Q^\mathtt T x \right) \\&= \sum _{i = 1}^n {\bar{\lambda }_i \left( q_i^\mathtt T x\right) ^2} \le \underset{i}{\text{ max }} (\bar{\lambda }_i) \sum _{i = 1}^n {\left( q_i^\mathtt T x\right) ^2}\\&= \bar{\lambda }_{\mathrm{max}} \left( \frac{A + A^\mathtt T }{2} \right) \left\| x \right\| _2^2. \end{aligned}$$

Next, $\mathfrak {R}(A) = \xi ^\mathtt H \left( (A + A^\mathtt T )/2 \right) \xi = \xi ^\mathtt H \left( Q \bar{\varLambda } Q^\mathtt T \right) \xi$. Denotes that $y = Q \xi = (y_1, y_2, \ldots , y_n)^\mathtt T$. Obviously, $y^\mathtt H y = 1$ and $\mathfrak {R}(A) = \bar{\lambda }_1 \left| y_1\right| ^2 + \bar{\lambda }_2 \left| y_2\right| ^2 + \cdots + \bar{\lambda }_n \left| y_n\right| ^2$.

Therefore,

$$\begin{aligned} \mathfrak {R}(A)\le & {} \bar{\lambda }_{\mathrm{max}} \left( \frac{A + A^\mathtt T }{2} \right) \left( \left| y_1\right| ^2 + \left| y_2\right| ^2 + \cdots + \left| y_n\right| ^2 \right) \\= & {} \bar{\lambda }_{\mathrm{max}} \left( \frac{A + A^\mathtt T }{2} \right) . \end{aligned}$$

Obviously,

$$\begin{aligned} \mathfrak {R}(A) \le \left\| \frac{A + A^\mathtt T }{2}\right\| _2, \end{aligned}$$

because the spectral radius $\rho$ of a symmetric matrix satisfies,

$$\begin{aligned} \bar{\lambda }_{\mathrm{max}} \left( \frac{A + A^\mathtt T }{2} \right) \le \rho \left( \frac{A + A^\mathtt T }{2}\right) \le \left\| \frac{A + A^\mathtt T }{2}\right\| _2 . \end{aligned}$$

This is the end of the proof. $\square$