research article chaotic tolerant synchronization analysis

Research ArticleChaotic Tolerant Synchronization Analysis withPropagation Delay and Actuator Faults

Zhang Qunli

Department of Mathematics, Heze University, Heze, Shandong 274015, China

Correspondence should be addressed to Zhang Qunli; [email protected]

Received 20 June 2015; Accepted 24 August 2015

Academic Editor: Laura Gardini

Copyright © 2015 Zhang Qunli. This is an open access article distributed under the Creative Commons Attribution License, whichpermits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

The criteria for tolerant synchronizationwith a constant propagation delay and actuator faults are presented by usingmatrix analysistechniques. A new algorithm, which constructs the extended error systems in order to make the conservation of the stability lower,is proposed. Based on proper Lyapunov-Krasovskii functional, the novel delay-dependent fault tolerant synchronization analysesare derived. Finally, numerical examples show the effectiveness of the proposed method.

1. Introduction

The realization of OGY (Ott-Grebogi-Yorker) chaos-controlmethod [1] and PC (Pecora and Carroll) synchronizationmethod [2] has been attracting researchers’ attention sincethe 1990s. Chaos synchronization [3–13] is of great practicalsignificance and has aroused great interest in recent years.

They all focused on the design of synchronization undernormal operating conditions in the above works described.But, in practical chaotic secure communication systems,sensors, actuators, and inner components may inevitablyfail, which can lead to sharp performance decline of chaoticsecure. For the reason, fault tolerant synchronization andcontrol [14–20] of chaotic systems have been the hot topic ofintensive researches recently.

More recent works studied the fault tolerant synchro-nization and control, but they were limited to constructLyapunov-Krasovskii functional

𝑉 (𝑡) = 𝑉1(𝑒 (𝑡)) + 𝑉

2(𝑓 (𝑡)) , (1)

which separates 𝑒(𝑡) and 𝑓(𝑡), where 𝑒(𝑡) is the synchroniza-tion error and 𝑓(𝑡) is the error of fault function. Motivatingthe limitation and the extended transformation [21, 22],

the effective new method, which does the extended transfor-mation for the error system, to consider Lyapunov-Krasovskiifunctional

𝑉 (𝑡) = 𝑉 ((𝑒𝑇(𝑡) , 𝑓𝑇(𝑡))𝑇

) , (2)

which do not separate 𝑒(𝑡) and 𝑓(𝑡), for the chaotic faulttolerant synchronization with a constant propagation delayand actuator faults, makes the conservation descent. Finally,numerical examples are given to verify the above method.

Notations. In this paper, 𝑅, 𝑅𝑛, and 𝑅𝑛×𝑚 denote, respectively,the real number, the real 𝑛-vectors, and the real 𝑛×𝑚matrices.The superscript “𝑇” stands for the transpose of a matrix. Thesymbol 𝑋 > 𝑌 (𝑋 ≥ 𝑌), where𝑋 and𝑌 are symmetricmatri-ces, means that 𝑋-𝑌 is positive definite (positive semidefi-nite). 𝐼 is the identity matrix of appropriate dimensions. “∗”denotes the matrix entries implied by symmetry.

2. Preliminaries and Systems Description

Consider a chaotic master system with the actuator faultsitem 𝑓(𝑡) in the following form:

�� (𝑡) = 𝐴𝑥 (𝑡) + 𝑔 (𝑥 (𝑡)) + ℎ (𝑥 (𝑡 − 𝜃)) + 𝐸𝑓 (𝑡) ,

𝑦 (𝑡) = 𝐶𝑥 (𝑡) + 𝐷𝑓 (𝑡) ,

(3)

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2015, Article ID 785861, 14 pageshttp://dx.doi.org/10.1155/2015/785861

2 Mathematical Problems in Engineering

where 𝑥(𝑡) ∈ 𝑅𝑛 is the measurable state vector. 𝑦(𝑡) ∈ 𝑅𝑝 isthe output vector, and 𝐴,𝐶,𝐷, and 𝐸 are proper dimensionconstant matrices. 𝑔(𝑥(𝑡)), ℎ(𝑥(𝑡 − 𝜃)) are known continuousnonlinear functions. 𝜃 > 0 is the delay.

Assumption 1. There exist the matrices 𝑈1, 𝑈2,𝑀1,𝑀2,

𝑊1,𝑊2, 𝑉1, 𝑉2∈ 𝑅𝑛×𝑛, and the nonlinear functions 𝑔(⋅), and

ℎ(⋅) satisfy

(𝑈1(𝑔 (𝑥) − 𝑔 (𝑦)) − 𝑈

2(𝑥 − 𝑦))

𝑇

⋅ (𝑀1(𝑔 (𝑥) − 𝑔 (𝑦)) −𝑀

2(𝑥 − 𝑦)) ≤ 0,

(4)

(𝑊1(ℎ (𝑥) − ℎ (𝑦)) − 𝑊

2(𝑥 − 𝑦))

𝑇

⋅ (𝑉1(ℎ (𝑥) − ℎ (𝑦)) − 𝑉

2(𝑥 − 𝑦)) ≤ 0

(5)

for all 𝑥, 𝑦 ∈ 𝑅𝑛.The slave system linked with the chaotic master system

(3) is described by

𝑥 (𝑡) = 𝐴𝑥 (𝑡) + 𝑔 (𝑥 (𝑡)) + ℎ (𝑥 (𝑡 − 𝜃)) + 𝐸𝑓

− 𝐿 (𝑦 (𝑡 − 𝜏) − 𝑦 (𝑡 − 𝜏)) ,

𝑦 (𝑡) = 𝐶𝑥 (𝑡) + 𝐷𝑓 (𝑡) ,

(6)

where 𝑥(𝑡) ∈ 𝑅𝑛 is the measurable state vector and 𝜏 > 0 is aconstant propagation delay.

Let 𝑒(𝑡) = 𝑥(𝑡) − 𝑥(𝑡), 𝑓(𝑡) = 𝑓(𝑡) − 𝑓(𝑡), and 𝜑(𝑡) =𝑦(𝑡) − 𝑦(𝑡) = 𝐶𝑒(𝑡) + 𝐷𝑓(𝑡), and then

𝑒 (𝑡) = 𝐴𝑒 (𝑡) + 𝑔 (𝑥 (𝑡)) − 𝑔 (𝑥 (𝑡)) + ℎ (𝑥 (𝑡 − 𝜃))

− ℎ (𝑥 (𝑡 − 𝜃)) + 𝐸𝑓 (𝑡) − 𝐿𝐶𝑒 (𝑡 − 𝜏)

− 𝐿𝐷𝑓 (𝑡 − 𝜏) ,

𝜑 (𝑡) = 𝐶𝑒 (𝑡) + 𝐷𝑓 (𝑡) .

(7)

Lemma 2 (see [23, 24]). For any constant matrix𝑊 ∈ 𝑅𝑛×𝑛,𝑊 > 0, scalar 0 < ℎ(𝑡) < ℎ, and vector function 𝑤(𝑡) :[0, ℎ] → 𝑅

𝑛 such that the integrations concerned are welldefined, and then

(∫

ℎ(𝑡)

0

𝑤 (𝑠) 𝑑𝑠)

𝑇

𝑊(∫

ℎ(𝑡)

0

𝑤 (𝑠) 𝑑𝑠)

≤ ℎ (𝑡) ∫

ℎ(𝑡)

0

𝑤𝑇(𝑠)𝑊𝑤 (𝑠) 𝑑𝑠.

(8)

Lemma 3 (see [25]). Let 𝑄 > 0, 𝐻,𝐹(𝑡), and 𝐸 be real matri-ces of appropriate dimensions, with 𝐹(𝑡) satisfying 𝐹𝑇(𝑡)𝐹(𝑡) ≤𝐼. Then, the following inequalities are equivalent:

(1) 𝑄 +𝐻𝐹(𝑡)𝐸 + 𝐸𝑇𝐹𝑇(𝑡)𝐻𝑇 < 0;(2) there exists a scalar 𝜀 > 0 such that 𝑄 + 𝜀−1𝐻𝐻𝑇 +𝜀𝐸𝑇𝐸 < 0.

Lemma 4 (see [26–28]). Let𝐴, 𝐿, 𝐸, and 𝐹(𝑡) be real matricesof appropriate dimensions, with 𝐹(𝑡) satisfying 𝐹𝑇(𝑡)𝐹(𝑡) ≤ 𝐼.Then, one has the following:

(1) for any scalar 𝜀 > 0,

𝐿𝐹𝐸 + 𝐸𝑇𝐹𝑇𝐿𝑇< 𝜀−1𝐿𝐿𝑇+ 𝜀𝐸𝑇𝐸; (9)

(2) for any matrix 𝑃 > 0 and scalar 𝜀 > 0, such that 𝜀𝐼 −𝐸𝐹𝐸𝑇> 0,

(𝐴 + 𝐿𝐹𝐸)𝑇𝑃 (𝐴 + 𝐿𝐹𝐸)

≤ 𝐴𝑇𝑃𝐴 + 𝐴

𝑇𝑃𝐸 (𝜀𝐼 − 𝐸

𝑇𝑃𝐸)−1

𝐸𝑇𝑃𝐴 + 𝜀𝐿

𝑇𝐿.

(10)

3. Fault Tolerant Synchronization Analysis

3.1. Fault Tolerant Synchronization Analysis When 𝑓(𝑡) and𝑓(𝑡) Are Derivable on 𝑡. Let 𝑓(𝑡) = −𝐺𝜀(𝑡) = −𝐺𝐶𝑒(𝑡) −𝐺𝐷𝑓(𝑡), and we have


− ℎ (𝑥 (𝑡 − 𝜃)) + 𝐸𝑓 (𝑡) − 𝐿𝐶𝑒 (𝑡) + 𝐿𝐶𝑒 (𝑡)

− 𝐿𝐶𝑒 (𝑡 − 𝜏) − 𝐿𝐷𝑓 (𝑡) + 𝐿𝐷𝑓 (𝑡)

− 𝐿𝐷𝑓 (𝑡 − 𝜏)

= (𝐴 − 𝐿𝐶) 𝑒 (𝑡) + 𝑔 (𝑥 (𝑡)) − 𝑔 (𝑥 (𝑡))

+ ℎ (𝑥 (𝑡 − 𝜃)) − ℎ (𝑥 (𝑡 − 𝜃)) + (𝐸 − 𝐿𝐷)𝑓 (𝑡)

+ 𝐿𝐶∫

𝑡

𝑡−𝜏

𝑒 (𝑠) 𝑑𝑠 + 𝐿𝐷∫

𝑡

𝑡−𝜏

𝑓 (𝑠) 𝑑𝑠.

(11)

Suppose 𝜂(𝑡) = ( 𝑒(𝑡)𝑓(𝑡)), and then

𝑒 (𝑡) = (𝐴 − 𝐿𝐶, 𝐸 − 𝐿𝐷)(

𝑒 (𝑡)

𝑓 (𝑡)

) + 𝑔 (𝑥 (𝑡))

− 𝑔 (𝑥 (𝑡)) + ℎ (𝑥 (𝑡 − 𝜃)) − ℎ (𝑥 (𝑡 − 𝜃))

+ (𝐿𝐶, 𝐿𝐷)(

∫

𝑡

𝑡−𝜏

𝑒 (𝑠) 𝑑𝑠

∫

𝑡

𝑡−𝜏

𝑓 (𝑠) 𝑑𝑠

)

= (𝐴 − 𝐿𝐶, 𝐸 − 𝐿𝐷) 𝜂 (𝑡) + 𝑔 (𝑥 (𝑡))

− 𝑔 (𝑥 (𝑡)) + (𝐿𝐶, 𝐿𝐷)∫

𝑡

𝑡−𝜏

𝜂 (𝑠) 𝑑𝑠,

𝑓 (𝑡) = (−𝐺𝐶, −𝐺𝐷)(

𝑒 (𝑡)

𝑓 (𝑡)

) = (−𝐺𝐶, −𝐺𝐷) 𝜂 (𝑡) ,

Mathematical Problems in Engineering 3

𝜂 (𝑡) = (

𝐴 − 𝐿𝐶 𝐸 − 𝐿𝐷

−𝐺𝐶 −𝐺𝐷)𝜂 (𝑡)

+ (

𝑔 (𝑥 (𝑡)) − 𝑔 (𝑥 (𝑡))

0)

+ (

ℎ (𝑥 (𝑡 − 𝜃)) − ℎ (𝑥 (𝑡 − 𝜃))

0)

+ (

𝐿𝐶 𝐿𝐷

0 0)∫

𝑡

𝑡−𝜏


𝜂 (𝑡) = 𝐵𝜂 (𝑡) + (

𝑝 (𝑡)

0) + (

𝑞 (𝑡)

0) + 𝑅∫

𝑡

𝑡−𝜏


(12)

where

𝐵 = (


−𝐺𝐶 −𝐺𝐷) ,

𝑝 (𝑡) = 𝑔 (𝑥 (𝑡)) − 𝑔 (𝑥 (𝑡)) ,

𝑞 (𝑡) = ℎ (𝑥 (𝑡 − 𝜃)) − ℎ (𝑥 (𝑡 − 𝜃)) ,

𝑅 = (

𝐿𝐶 𝐿𝐷

0 0) .

(13)

From the assumption, we have

(𝑔 (𝑥) − 𝑔 (𝑦))𝑇

𝑈𝑇

1𝑀1(𝑔 (𝑥) − 𝑔 (𝑦))

− (𝑔 (𝑥) − 𝑔 (𝑦))𝑇

𝑈𝑇

1𝑀2(𝑥 − 𝑦)

− (𝑥 − 𝑦)𝑇

𝑈𝑇

2𝑀1(𝑔 (𝑥) − 𝑔 (𝑦))

+ (𝑥 − 𝑦)𝑇

𝑈𝑇

2𝑀2(𝑥 − 𝑦) ≤ 0.

(14)

So, we can get from assumption that

(𝑝 (𝑡))𝑇

𝑈𝑇

1𝑀1𝑝 (𝑡) − (𝑝 (𝑡))

𝑇

𝑈𝑇

1𝑀2𝑒 (𝑡)

− (𝑒 (𝑡))𝑇𝑈𝑇

2𝑀1𝑝 (𝑡) + (𝑒 (𝑡))

𝑇𝑈𝑇

2𝑀2𝑒 (𝑡) ≤ 0,

(𝑒𝑇(𝑡) 𝑓

𝑇(𝑡))(

−𝑈𝑇

2𝑀20

0 0

)(

𝑒 (𝑡)

𝑓 (𝑡)

)

+ (𝑒𝑇(𝑡) 𝑓

𝑇(𝑡))(

𝑈𝑇

2𝑀1𝑅1

0 𝑅2

)(

𝑝 (𝑡)

0)

+ (𝑝𝑇(𝑡) 0)(

𝑈𝑇

1𝑀20

𝑅3𝑅4

)(

𝑒 (𝑡)

𝑓 (𝑡)

)

+ (𝑝𝑇(𝑡) 0)(

−𝑈𝑇

1𝑀1𝑅5

𝑅6

𝑅7

)(

𝑝 (𝑡)

0) ≥ 0,

(15)

in which the proper dimension matrices 𝑅𝑖∈ 𝑅𝑛×𝑛, 𝑖 =

1, 2, . . . , 7, are arbitrary.

That is,

𝜂𝑇(𝑡) (−𝑈𝑇

2𝑀20

0 0

) 𝜂 (𝑡)

+ 𝜂𝑇(𝑡) (𝑈𝑇

2𝑀1𝑅1

0 𝑅2

)(

𝑝 (𝑡)

0)

+ (

𝑝 (𝑡)

0)

𝑇

(𝑈𝑇

1𝑀20

𝑅3𝑅4

)𝜂 (𝑡)

+ (

𝑝 (𝑡)

0)

𝑇

(−𝑈𝑇

1𝑀1𝑅5

𝑅6

𝑅7

)(

𝑝 (𝑡)

0) ≥ 0,

𝜂𝑇(𝑡) (1

2(𝑆1+ 𝑆𝑇

1)) 𝜂 (𝑡)

+ 𝜂𝑇(𝑡) (1

2(𝑆2+ 𝑆𝑇

3))(

𝑝 (𝑡)

0)

+ (

𝑝 (𝑡)

0)

𝑇

(1

2(𝑆3+ 𝑆𝑇

2)) 𝜂 (𝑡)

+ (

𝑝 (𝑡)

0)

𝑇

(1

2(𝑆4+ 𝑆𝑇

4))(

𝑝 (𝑡)

0) ≥ 0,

(16)

where

𝑆1= (−𝜀𝑈𝑇

2𝑀20

0 0

) ,

𝑆2= (𝜀𝑈𝑇

2𝑀1𝜀𝑅1

0 𝜀𝑅2

) ,


1𝑀20

𝜀𝑅3𝜀𝑅4

) ,


1𝑀1𝜀𝑅5

𝜀𝑅6

𝜀𝑅7

) ,

𝜀 > 0

(17)

is any constant. Consider

𝜂𝑇(𝑡) (𝑆1+ 𝑆𝑇

1) 𝜂 (𝑡) + 𝜂

𝑇(𝑡) (𝑆2+ 𝑆𝑇

3)(

𝑝 (𝑡)

0)

+ (

𝑝 (𝑡)

0)

𝑇

(𝑆3+ 𝑆𝑇

2) 𝜂 (𝑡)

+ (

𝑝 (𝑡)

0)

𝑇

(𝑆4+ 𝑆𝑇

4)(

𝑝 (𝑡)

0) ≥ 0,

𝜉𝑇(𝑡) Λ1𝜉 (𝑡) ≥ 0,

(18)


where

𝜉 (𝑡) = (𝜂𝑇(𝑡) 𝜂𝑇(𝑡) (𝑝

𝑇(𝑡) 0) (𝑞

𝑇(𝑡) 0) ∫

𝑡

𝑡−𝜏

𝜂𝑇(𝑠) 𝑑𝑠 ∫

𝑡

𝑡−𝜃

𝜂𝑇(𝑠) 𝑑𝑠)

𝑇

,

Λ1=

(((((

(

𝑆1+ 𝑆𝑇

10 0 𝑆

2+ 𝑆𝑇

30 0

∗ 0 0 0 0 0

∗ ∗ 𝑆4+ 𝑆𝑇

40 0 0

∗ ∗ ∗ 0 0 0

∗ ∗ ∗ ∗ 0 0

∗ ∗ ∗ ∗ ∗ 0

)))))

)

.

(19)

Imitating the above inference, we obtain from inequality (5)the following:

𝜉𝑇(𝑡) Λ2𝜉 (𝑡) ≥ 0, (20)

where

Λ2= 𝛿

(((((

(

𝑌𝑇

1Δ𝑌10 0 𝑌

𝑇

1Δ𝑌30 𝑌𝑇

1Δ𝑌2

∗ 0 0 0 0 0

∗ ∗ 0 0 0 0

∗ ∗ ∗ 𝑌𝑇

3Δ𝑌30 𝑌𝑇

3Δ𝑌2

∗ ∗ ∗ ∗ 0 0

∗ ∗ ∗ ∗ ∗ 𝑌𝑇

2Δ𝑌2

)))))

)

,

𝛿 > 0,

𝑌1= (

𝐼 0

0 0) ,

𝑌2= (

−𝐼 0

0 0) ,

𝑌3= (

0 0

𝐼 𝑅8

) ,

Δ = (

−𝑊𝑇

2𝑉2− 𝑉𝑇

2𝑊2𝑊𝑇

2𝑉1+ 𝑉𝑇

2𝑊1

∗ −𝑊𝑇

1𝑉1− 𝑉𝑇

1𝑊1

) .

(21)

We choose Lyapunov-Krasovskii functional

𝑉 (𝑡) = 𝜂𝑇(𝑡) 𝑃𝜂 (𝑡) + ∫

0

−𝜏

∫

𝑡

𝑡+𝑟

𝜂𝑇(𝑠) 𝑄 𝜂 (𝑠) 𝑑𝑠 𝑑𝑟

+ ∫

0

−𝜃

∫

𝑡

𝑡+𝑟

𝜂𝑇(𝑠) Θ 𝜂 (𝑠) 𝑑𝑠 𝑑𝑟.

(22)

Differentiating 𝑉(𝑡) with respect to 𝑡, the following result isyielded:

�� (𝑡) = 2𝜂𝑇(𝑡) 𝑃 𝜂 (𝑡) + 𝜏 𝜂

𝑇(𝑡) 𝑄 𝜂 (𝑡)

− ∫

𝑡

𝑡−𝜏

𝜂𝑇(𝑠) 𝑄 𝜂 (𝑠) 𝑑𝑠

= 2𝜂𝑇(𝑡) 𝑃𝐵𝜂 (𝑡) + 2𝜂

𝑇(𝑡) 𝑃(

𝑝 (𝑡)

0)

+ 2𝜂𝑇(𝑡) 𝑃𝑅∫

𝑡

𝑡−𝜏

𝜂 (𝑠) 𝑑𝑠 + 𝜏 𝜂𝑇(𝑡) 𝑄 𝜂 (𝑡)

− ∫

𝑡

𝑡−𝜏

𝜂𝑇(𝑠) 𝑄 𝜂 (𝑠) 𝑑𝑠 + 𝜃 𝜂

𝑇(𝑡) Θ 𝜂 (𝑡)

− ∫

𝑡

𝑡−𝜃

𝜂𝑇(𝑠) Θ 𝜂 (𝑠) 𝑑𝑠.

(23)

From Lemma 2, we have

�� (𝑡) ≤ 𝜂𝑇(𝑡) (𝑃𝐵 + 𝐵

𝑇𝑃) 𝜂 (𝑡) + 2𝜂

𝑇(𝑡) 𝑃(

𝑝 (𝑡)

0)

+ 2𝜂𝑇(𝑡) 𝑃(

𝑞 (𝑡)

0) + 2𝜂

𝑇(𝑡) 𝑃𝑅∫

𝑡

𝑡−𝜏

𝜂 (𝑠) 𝑑𝑠

+ 𝜏 𝜂𝑇(𝑡) 𝑄 𝜂 (𝑡)

−1

𝜏(∫

𝑡

𝑡−𝜏

𝜂 (𝑠) 𝑑𝑠)

𝑇

𝑄(∫

𝑡

𝑡−𝜏


+ 𝜃 𝜂𝑇(𝑡) Θ 𝜂 (𝑡)

−1

𝜃(∫

𝑡

𝑡−𝜏


𝑇

Θ(∫

𝑡

𝑡−𝜏


= 𝜉𝑇(𝑡) Π𝜉 (𝑡) ≥ 0,

(24)


where

Π =

(((((((

(

𝑃𝐵 + 𝐵𝑇𝑃 0 𝑃 𝑃 𝑃𝑅 0

∗ 𝜏𝑄 0 0 0 0

∗ ∗ 0 0 0 0

∗ ∗ ∗ 0 0 0

∗ ∗ ∗ ∗ −1

𝜏𝑄 0

∗ ∗ ∗ ∗ ∗ −1

𝜃Θ

)))))))

)

. (25)

From model (12), we have

0 = 2 (𝑄1𝜂 (𝑡) + 𝑄

2𝜂 (𝑡))𝑇

⋅ (− 𝜂 (𝑡) + 𝐵𝜂 (𝑡) + (

𝑝 (𝑡)

0) + 𝑅∫

𝑡

𝑡−𝜏


= 𝜉𝑇(𝑡) Ξ𝜉 (𝑡) ,

(26)

whereΞ

=

(((((

(

𝑄𝑇

1𝐵 + 𝐵𝑇𝑄1−𝑄𝑇

1+ 𝐵𝑇𝑄𝑇

2𝑄𝑇

1𝑄𝑇

1𝑄𝑇

1𝑅 0

∗ −𝑄𝑇

2− 𝑄2𝑄𝑇

2𝑄𝑇

2𝑄𝑇

2𝑅 0

∗ ∗ 0 0 0 0

∗ ∗ ∗ 0 0 0

∗ ∗ ∗ ∗ 0 0

∗ ∗ ∗ ∗ ∗ 0

)))))

)

.

(27)

From formulas (18), (20), (24), and (26), we get

�� (𝑡) ≤ 𝜉𝑇(𝑡) (Λ

1+ Λ2+ Π + Ξ) 𝜉 (𝑡)

= 𝜉𝑇(𝑡) Ω𝜉 (𝑡) ,

(28)

whereΩ = Λ + Π + Ξ.Based on the above derivation, we have the following

result.

Theorem 5. The fault tolerant synchronization (3) and (6) isachieved if there exist constants 𝜏 > 0, 𝜃 > 0, 𝛿 > 0, and𝜀 > 0, the positive definite matrices 𝑃 = 𝑃𝑇 > 0, 𝑄 = 𝑄𝑇 > 0,and Θ > 0, and the matrices 𝑄

𝑗,𝑀𝑗, 𝑈𝑗, 𝑗 = 1, 2, 𝐿, 𝐶,𝐷, 𝑅

𝑖,

𝑖 = 1, 2, . . . , 8, such that the matrix Ω < 0.

Remark 6. After the extended transformation 𝑒(𝑡) → 𝜂(𝑡),system (7) is turned into system (12). Based on the Lyapunovfunctional in [15–21], we take Lyapunov functional

𝑉 (𝑡) = 𝜂𝑇(𝑡) 𝑃𝜂 (𝑡) + ∫

0

−𝜏

∫

𝑡

𝑡+𝑟

𝜂𝑇(𝑠) 𝑄 𝜂 (𝑠) 𝑑𝑠 𝑑𝑟

+ ∫

0

−𝜃

∫

𝑡

𝑡+𝑟

𝜂𝑇(𝑠) Θ 𝜂 (𝑠) 𝑑𝑠 𝑑𝑟,

(29)

and we get matrix Ω. The conservation of stability of errorsystem (7) can be decreased by choosing the matrices 𝑅

𝑖, 𝑖 =

1, 2, . . . , 8, and the constant 𝜀 > 0.

Corollary 7. When 𝐷 = 0 in system (3), result similar toTheorem 5 can be obtained.

When ℎ(𝑥(𝑡 − 𝜃)) = 0, inequality (18) is transformed to

𝜎𝑇(𝑡) Λ𝜎 (𝑡) ≥ 0, (30)

where

𝜎 (𝑡)

= (𝜂𝑇(𝑡) 𝜂𝑇(𝑡) (

𝑝 (𝑡)

0)

𝑇

(∫

𝑡

𝑡−𝜏


𝑇

)

𝑇

,

Λ =(

𝑆1+ 𝑆𝑇

10 𝑆2+ 𝑆𝑇

30

∗ 0 0 0

∗ ∗ 𝑆4+ 𝑆𝑇

40

∗ ∗ ∗ 0

),


2𝑀20

0 0

) ,


2𝑀1𝜀𝑅1

0 𝜀𝑅2

) ,


1𝑀20

𝜀𝑅3𝜀𝑅4

) ,


1𝑀1𝜀𝑅5

𝜀𝑅6

𝜀𝑅7

) ,

𝜀 > 0

(31)

is any constant. The proper dimension matrices 𝑅𝑖∈ 𝑅𝑛×𝑛,

𝑖 = 1, 2, . . . , 7, are arbitrary.We choose Lyapunov functional

𝑉 (𝑡) = 𝜂𝑇(𝑡) 𝑃𝜂 (𝑡) + ∫

0

−𝜏

∫

𝑡

𝑡+𝜃

𝜂𝑇(𝑠) 𝑄 𝜂 (𝑠) 𝑑𝑠 𝑑𝜃, (32)

and differentiating 𝑉(𝑡) with respect to 𝑡 and using Lemma 2yield

�� (𝑡) ≤ 𝜎𝑇(𝑡) Π𝜎 (𝑡) , (33)

where

Π =(

(

𝑃𝐵 + 𝐵𝑇𝑃 0 𝑃 𝑃𝑅

∗ 𝜏𝑄 0 0

∗ ∗ 0 0

∗ ∗ ∗ −1

𝜏𝑄

)

)

. (34)



0 = 2 (𝑄1𝜂 (𝑡) + 𝑄

2𝜂 (𝑡))𝑇

⋅ (− 𝜂 (𝑡) + 𝐵𝜂 (𝑡) + (

𝑝 (𝑡)

0) + 𝑅∫

𝑡

𝑡−𝜏


= 𝜎𝑇(𝑡) Ξ𝜎 (𝑡) ,

(35)

where

Ξ =(

𝑄𝑇

1𝐵 + 𝐵𝑇𝑄1−𝑄𝑇

1+ 𝐵𝑇𝑄𝑇

2𝑄𝑇

1𝑄𝑇

1𝑅

∗ −𝑄𝑇

2− 𝑄2𝑄𝑇

2𝑄𝑇

2𝑅

∗ ∗ 0 0

∗ ∗ ∗ 0

). (36)

Based on the above derivation, we have the following result.

Corollary 8. The fault tolerant synchronization (3) and (5) isachieved if there exist constants 𝜏, 𝜀 > 0, the positive definitematrices 𝑃 = 𝑃𝑇 > 0, 𝑄 = 𝑄𝑇 > 0, and the matrices𝑄𝑗,𝑀𝑗, 𝑈𝑗, 𝑗 = 1, 2, 𝑄

𝑗,𝑀𝑗, 𝑈𝑗, 𝑗 = 1, 2, 𝐿, 𝐶,𝐷, 𝑅

𝑖, 𝑖 =

1, 2, . . . , 7, such that matrix Ω < 0.

3.2. Fault Tolerant Synchronization Analysis When 𝑓(𝑡) and𝑓(𝑡) Are Derivable on 𝑥(𝑡). Let 𝑑𝑓(𝑡)/𝑑𝑥(𝑡) = ℏ(𝑥(𝑡)) =𝐻 ⋅ 𝐹(𝑥(𝑡)) ⋅ 𝑁 and 𝑑𝑓(𝑡)/𝑑𝑥(𝑡) = ℏ(𝑥(𝑡)) = 𝐻 ⋅ 𝐹(𝑥(𝑡)) ⋅𝑁, 𝐻,𝑁 be proper dimension matrices, with satisfying𝐹𝑇(𝑥(𝑡))𝐹(𝑥(𝑡)) ≤ 𝐼, where 𝐻,𝑁 are proper dimension

matrices and 𝐼 is identity matrix. Then,


− ℎ (𝑥 (𝑡 − 𝜃)) + 𝐸𝑓 (𝑡) − 𝐿𝐶𝑒 (𝑡) + 𝐿𝐶𝑒 (𝑡)

− 𝐿𝐶𝑒 (𝑡 − 𝜏) − 𝐿𝐷𝑓 (𝑡) + 𝐿𝐷𝑓 (𝑡)

− 𝐿𝐷𝑓 (𝑡 − 𝜏)

= (𝐴 − 𝐿𝐶) 𝑒 (𝑡) + 𝑔 (𝑥 (𝑡)) − 𝑔 (𝑥 (𝑡))

+ (𝐸 − 𝐿𝐷)𝑓 (𝑡) + 𝐿𝐶∫

𝑡

𝑡−𝜏


+ 𝐿𝐷∫

𝑡

𝑡−𝜏

𝑓 (𝑠) 𝑑𝑠.

(37)

Suppose 𝜂(𝑡) = ( 𝑒(𝑡)𝑓(𝑡)); we have

𝑒 (𝑡) = (𝐴 − 𝐿𝐶, 𝐸 − 𝐿𝐷)(

𝑒 (𝑡)

𝑓 (𝑡)

) + 𝑔 (𝑥 (𝑡))

− 𝑔 (𝑥 (𝑡)) + ℎ (𝑥 (𝑡 − 𝜃)) − ℎ (𝑥 (𝑡 − 𝜃))

+ (𝐿𝐶, 𝐿𝐷)(

∫

𝑡

𝑡−𝜏


∫

𝑡

𝑡−𝜏

𝑓 (𝑠) 𝑑𝑠

)

= (𝐴 − 𝐿𝐶, 𝐸 − 𝐿𝐷) 𝜂 (𝑡) + 𝑔 (𝑥 (𝑡)) − 𝑔 (𝑥 (𝑡))

+ ℎ (𝑥 (𝑡 − 𝜃)) − ℎ (𝑥 (𝑡 − 𝜃))

+ (𝐿𝐶, 𝐿𝐷)∫

𝑡

𝑡−𝜏

𝜂 (𝑠) 𝑑𝑠.

(38)

According to 𝑑𝑓(𝑡)/𝑑𝑥(𝑡) = ℏ(𝑥(𝑡)) = 𝐻 ⋅ 𝐹(𝑥(𝑡)) ⋅ 𝑁,𝑑𝑓(𝑡)/𝑑𝑥(𝑡) = ℏ(𝑥(𝑡)) = 𝐻 ⋅ 𝐹(𝑥(𝑡)) ⋅ 𝑁, we get

𝑓 (𝑡) = ∫

1

0

ℏ ((1 − 𝜆) 𝑥 (𝑡) + 𝜆𝑥 (𝑡)) 𝑒 (𝑡) 𝑑𝜆

= ∫

1

0

𝐻 ⋅ 𝐹 ((1 − 𝜆) 𝑥 (𝑡) + 𝜆𝑥 (𝑡)) ⋅ 𝑁 ⋅ 𝑒 (𝑡) 𝑑𝜆,

− 𝛿𝑓 (𝑡)

+ 𝛿∫

1

0

𝐻 ⋅ 𝐹 ((1 − 𝜆) 𝑥 (𝑡) + 𝜆𝑥 (𝑡)) ⋅ 𝑁 ⋅ 𝑒 (𝑡) 𝑑𝜆

= 0,

(𝛿∫

1

0

𝐻 ⋅ 𝐹 ((1 − 𝜆) 𝑥 (𝑡) + 𝜆𝑥 (𝑡)) ⋅ 𝑁𝑑𝜆, −𝛿𝐼) 𝜂 (𝑡)

= 0,

(39)

where 𝛿 > 0. Consider

(

𝐼 0

0 0) 𝜂 (𝑡)

= (


𝛿∫

1

0

𝐻 ⋅ 𝐹 ((1 − 𝜆) 𝑥 (𝑡) + 𝜆𝑥 (𝑡)) ⋅ 𝑁𝑑𝜆 −𝛿𝐼

)𝜂 (𝑡)

+ (

𝑔 (𝑥 (𝑡)) − 𝑔 (𝑥 (𝑡))

0) + (

𝐿𝐶 𝐿𝐷

0 0)∫

𝑡

𝑡−𝜏


𝐸1𝜂 (𝑡) = 𝐵𝜂 (𝑡) + (

𝑝 (𝑡)

0) + (

𝑞 (𝑡)

0) + 𝑅∫

𝑡

𝑡−𝜏


(40)

where

𝐸1= (

𝐼 0

0 0) ,

𝐵

= (


𝛿∫

1

0

𝐻 ⋅ 𝐹 ((1 − 𝜆) 𝑥 (𝑡) + 𝜆𝑥 (𝑡)) ⋅ 𝑁𝑑𝜆 −𝛿𝐼

) ,

𝑝 (𝑡) = 𝑔 (𝑥 (𝑡)) − 𝑔 (𝑥 (𝑡)) ,

𝑅 = (

𝐿𝐶 𝐿𝐷

0 0) .

(41)


From the assumption, we have

𝜍𝑇(𝑡) Λ1𝜍 (𝑡) ≥ 0, (42)

where

𝜍 (𝑡) = (𝜂𝑇(𝑡) 𝑒𝑇(𝑡) (𝑝

𝑇(𝑡) 0) (𝑞

𝑇(𝑡) 0) ∫

𝑡

𝑡−𝜏

𝜂𝑇(𝑠) 𝑑𝑠 ∫

𝑡

𝑡−𝜃

𝑒𝑇(𝑠) 𝑑𝑠)

𝑇

,

Λ1=

(((((

(

𝑆1+ 𝑆𝑇

10 0 𝑆

2+ 𝑆𝑇

30 0

∗ 0 0 0 0 0

∗ ∗ 𝑆4+ 𝑆𝑇

40 0 0

∗ ∗ ∗ 0 0 0

∗ ∗ ∗ ∗ 0 0

∗ ∗ ∗ ∗ ∗ 0

)))))

)

,


2𝑀20

0 0

) ,


2𝑀1𝜀𝑅1

0 𝜀𝑅2

) ,


1𝑀20

𝜀𝑅3𝜀𝑅4

) ,


1𝑀1𝜀𝑅5

𝜀𝑅6

𝜀𝑅7

) ,

𝜀 > 0

(43)


𝑖 = 1, 2, . . . , 7, are arbitrary

𝜍𝑇(𝑡) Λ2𝜍 (𝑡) ≥ 0, (44)

where

Λ2

= 𝛿

(((((

(

𝑌𝑇

1Δ𝑌10 0 𝑌

𝑇

1Δ𝑌30 𝑌

𝑇

1Δ𝑌2

∗ 0 0 0 0 0

∗ ∗ 0 0 0 0

∗ ∗ ∗ 𝑌𝑇

3Δ𝑌30 𝑌

𝑇

3Δ𝑌2

∗ ∗ ∗ ∗ 0 0

∗ ∗ ∗ ∗ ∗ −𝑊𝑇

2𝑉2− 𝑉𝑇

2𝑊2

)))))

)

,

𝛿 > 0,

𝑌1= (

𝐼 0

0 0) ,

𝑌2= (

−𝐼 0

0 0) ,

𝑌3= (

0 0

𝐼 𝑅8

) ,

Δ = (

−𝑊𝑇

2𝑉2− 𝑉𝑇

2𝑊2𝑊𝑇

2𝑉1+ 𝑉𝑇

2𝑊1

∗ −𝑊𝑇

1𝑉1− 𝑉𝑇

1𝑊1

) .

(45)

We choose Lyapunov function

𝑉 (𝑡) = 𝑒𝑇(𝑡) 𝑃1𝑒 (𝑡) + ∫

0

−𝜏

∫

𝑡

𝑡+𝑟

𝑒𝑇(𝑠) 𝑄 𝑒 (𝑠) 𝑑𝑠 𝑑𝑟

+ ∫

0

−𝜃

∫

𝑡

𝑡+𝑟

𝑒𝑇(𝑠) Θ 𝑒 (𝑠) 𝑑𝑠 𝑑𝑟


= 𝜂𝑇(𝑡) 𝑃𝐸

1𝜂 (𝑡) + ∫

0

−𝜏

∫

𝑡

𝑡+𝜃

𝑒𝑇(𝑠) 𝑄 𝑒 (𝑠) 𝑑𝑠 𝑑𝜃

+ ∫

0

−𝜃

∫

𝑡

𝑡+𝑟

𝑒𝑇(𝑠) Θ 𝑒 (𝑠) 𝑑𝑠 𝑑𝑟,

(46)

where 𝑃 = ( 𝑃1 𝑃20 𝑃3

), and it is easy to know that 𝑃𝐸1= 𝐸𝑇

1𝑃𝑇.

The derivation of 𝑉(𝑡) on 𝑡 is

�� (𝑡) = 2𝜂𝑇(𝑡) 𝑃𝐸

1𝜂 (𝑡) + 𝜏 𝑒

𝑇(𝑡) 𝑄 𝑒 (𝑡)

− ∫

𝑡

𝑡−𝜏

𝑒𝑇(𝑠) 𝑄 𝑒 (𝑠) 𝑑𝑠 + 𝜃 𝑒

𝑇(𝑡) Θ 𝑒 (𝑡)

− ∫

𝑡

𝑡−𝜃

𝑒𝑇(𝑠) Θ 𝑒 (𝑠) 𝑑𝑠

= 2𝜂𝑇(𝑡) 𝑃𝐵𝜂 (𝑡) + 2𝜂

𝑇(𝑡) 𝑃(

𝑝 (𝑡)

0)

+ 2𝜂𝑇(𝑡) 𝑃𝑅∫

𝑡

𝑡−𝜏

𝜂 (𝑠) 𝑑𝑠 + 𝜏 𝑒𝑇(𝑡) 𝑄 𝑒 (𝑡)

− ∫

𝑡

𝑡−𝜏

𝑒𝑇(𝑠) 𝑄 𝑒 (𝑠) 𝑑𝑠 + 𝜃 𝑒

𝑇(𝑡) Θ 𝑒 (𝑡)

− ∫

𝑡

𝑡−𝜃

𝑒𝑇(𝑠) Θ 𝑒 (𝑠) 𝑑𝑠.

(47)

From Lemma 2, we have

�� (𝑡) ≤ 𝜂𝑇(𝑡) (𝑃𝐵 + 𝐵

𝑇𝑃𝑇) 𝜂 (𝑡) + 2𝜂

𝑇(𝑡) 𝑃(

𝑝 (𝑡)

0

)

+ 2𝜂𝑇(𝑡) 𝑃(

𝑞 (𝑡)

0

) + 2𝜂𝑇(𝑡) 𝑃𝑅∫

𝑡

𝑡−𝜏

𝜂 (𝑠) 𝑑𝑠

+ 𝜏 𝑒𝑇(𝑡) 𝑄 𝑒 (𝑡)

−1

𝜏(∫

𝑡

𝑡−𝜏


𝑇

𝑄(∫

𝑡

𝑡−𝜏


+ 𝜃 𝑒𝑇(𝑡) Θ 𝑒 (𝑡)

−1

𝜃(∫

𝑡

𝑡−𝜃


𝑇

Θ(∫

𝑡

𝑡−𝜃

𝑒 (𝑠) 𝑑𝑠)

= 𝜍𝑇(𝑡) Π𝜍 (𝑡) ,

(48)

where

Π =

(((((((

(

𝑃𝐵 + 𝐵𝑇𝑃𝑇0 𝑃 𝑃 𝑃𝑅 0

∗ 𝜏𝑄 0 0 0 0

∗ ∗ 0 0 0 0

∗ ∗ ∗ 0 0 0

∗ ∗ ∗ ∗ −1

𝜏𝑄 0

∗ ∗ ∗ ∗ ∗ −1

𝜃Θ

)))))))

)

. (49)


0 = 2(𝑄1𝑒 (𝑡) + 𝑄

2𝑒 (𝑡) + 𝑄

3∫

𝑡

𝑡−𝜏


𝑇

(− 𝑒 (𝑡)

+ (𝐴 − 𝐿𝐶, 𝐸 − 𝐿𝐷) 𝜂 (𝑡) + 𝑝 (𝑡) + 𝑞 (𝑡)

+ (𝐿𝐶, 𝐿𝐷)∫

𝑡

𝑡−𝜏

𝜂 (𝑠) 𝑑𝑠) = 𝜍𝑇(𝑡) Ξ𝜍 (𝑡) ,

(50)

where

Ξ =

(((((

(

Ξ11Ξ12Ξ13Ξ13Ξ140

∗ Ξ22Ξ23Ξ23Ξ240

∗ ∗ Ξ330 Ξ340

∗ ∗ ∗ Ξ33Ξ340

∗ ∗ ∗ ∗ Ξ440

∗ ∗ ∗ ∗ ∗ 0

)))))

)

,

Ξ11= (𝑄𝑇

1

0

) (𝐴 − 𝐿𝐶, 𝐸 − 𝐿𝐷) + ((𝑄𝑇

1

0

) (𝐴 − 𝐿𝐶, 𝐸 − 𝐿𝐷))

𝑇

,

Ξ12= −(

𝑄𝑇

1

0

) + (𝑄𝑇

2(𝐴 − 𝐿𝐶, 𝐸 − 𝐿𝐷))

𝑇

,

Ξ13= (𝑄𝑇

1

0

) (𝐼, 𝑅8) ,

Ξ14= (𝑄𝑇

1

0

) (𝐿𝐶, 𝐿𝐷) + (𝑄𝑇

3(𝐴 − 𝐿𝐶, 𝐸 − 𝐿𝐷))

𝑇

,

Ξ22= − (𝑄

2+ 𝑄𝑇

2) ,

Ξ23= 𝑄𝑇

2(𝐼, 𝑅9) ,

Ξ24= −𝑄3+ 𝑄𝑇

2(𝐿𝐶, 𝐿𝐷)

𝑇,

Ξ33= 0,

Ξ34= 𝑄𝑇

3(𝐼, 𝑅9) ,

Ξ44= 𝑄𝑇

3(𝐿𝐶, 𝐿𝐷) + (𝑄

𝑇

3(𝐿𝐶, 𝐿𝐷))

𝑇

,

(51)

where the proper dimension matrices 𝑅9∈ 𝑅𝑛×𝑛 is arbitrary.

From formulas (42), (44), (48), and (50), we get

�� (𝑡) ≤ 𝜍𝑇(𝑡) (Λ + Π + Ξ) 𝜍 (𝑡) = 𝜍

𝑇(𝑡) Ω𝜍 (𝑡) , (52)


where

Ω = Λ + Π + Ξ = (

Ω11Ω12

∗ Ω22

) ,

Ω11= (𝑆1+ 𝑆𝑇

1+ Ξ11+ 𝑌𝑇

1Δ𝑌1+ 𝑃𝐵 + 𝐵

𝑇𝑃𝑇

Ξ12

∗ 𝜏𝑄 + Ξ22

) ,

Ω12= (𝑃 + Ξ

13𝑆2+ 𝑆𝑇

3+ Ξ13+ 𝑃 + 𝑌

𝑇

1Δ𝑌3𝑃𝑅 + Ξ

14𝑌𝑇

1Δ𝑌2

Ξ23

Ξ23

Ξ24

0

) ,

Ω22=(

𝑆4+ 𝑆𝑇

40 Ξ

340

∗ 𝑌𝑇

3Δ𝑌3

Ξ34

𝑌𝑇

3Δ𝑌2

∗ ∗ −𝜏−1𝑄 + Ξ

440

∗ ∗ ∗ −𝑊𝑇

2𝑉2− 𝑉𝑇

2𝑊2

),

𝑃𝐵 + 𝐵𝑇𝑃𝑇= (

𝑃1(𝐴 − 𝐿𝐶) 𝑃

1(𝐸 − 𝐿𝐷) − 𝛿𝑃

2

0 −𝛿𝑃3

) + (

𝑃1(𝐴 − 𝐿𝐶) 𝑃

1(𝐸 − 𝐿𝐷) − 𝛿𝑃

2

0 −𝛿𝑃3

)

𝑇

+ 𝛿(

𝑃2𝐻∫

1

0

𝐹 ((1 − 𝜆) 𝑥 (𝑡) + 𝜆𝑥 (𝑡)) 𝑑𝜆𝑁 0

𝑃3𝐻∫

1

0

𝐹 ((1 − 𝜆) 𝑥 (𝑡) + 𝜆𝑥 (𝑡)) 𝑑𝜆𝑁 0

) + 𝛿(

𝑃2𝐻∫

1

0

𝐹 ((1 − 𝜆) 𝑥 (𝑡) + 𝜆𝑥 (𝑡)) 𝑑𝜆𝑁 0

𝑃3𝐻∫

1

0

𝐹 ((1 − 𝜆) 𝑥 (𝑡) + 𝜆𝑥 (𝑡)) 𝑑𝜆𝑁 0

)

𝑇

= 𝑍 + 𝑍𝑇+ 𝛿(

𝑃2𝐻

𝑃3𝐻)∫

1

0

𝐹 ((1 − 𝜆) 𝑥 (𝑡) + 𝜆𝑥 (𝑡)) 𝑑𝜆 (𝑁 0)

+ 𝛿 [(

𝑃2𝐻

𝑃3𝐻)∫

1

0

𝐹 ((1 − 𝜆) 𝑥 (𝑡) + 𝜆𝑥 (𝑡)) 𝑑𝜆 (𝑁 0)]

𝑇

,

(53)

where 𝑍 = ( 𝑃1(𝐴−𝐿𝐶) 𝑃1(𝐸−𝐿𝐷)−𝛿𝑃20 −𝛿𝑃

3

) .

FromLemma 3, we obtain that 𝑆1+𝑆𝑇

1+Ξ11+𝑃𝐵+𝐵

𝑇𝑃𝑇<

0 is equivalent to

𝑆1+ 𝑆𝑇

1+ Ξ11+ 𝑍 + 𝑍

𝑇

+ 𝜀−1𝛿2(

𝑃2𝐻𝐻𝑇𝑃𝑇

2𝑃2𝐻𝐻𝑇𝑃𝑇

3

∗ 𝑃3𝐻𝐻𝑇𝑃𝑇

3

)

+ 𝜀(𝑁𝑇𝑁 0

0 0

) < 0.

(54)

Based on the above derivation, we have the following result.

Theorem 9. The fault tolerant synchronization (3) and (6)is achieved if there exist constants 𝜏, 𝜀 > 0, 𝛿 > 0, thepositive definite matrices 𝑃

1= 𝑃𝑇

1> 0, 𝑄 = 𝑄𝑇 > 0, and

the matrices 𝑄𝑗,𝑀𝑗, 𝑈𝑗, 𝑗 = 1, 2, 𝑄

𝑗,𝑀𝑗, 𝑈𝑗, 𝑗 = 1, 2,

𝐿, 𝐶,𝐷, 𝑅𝑖, 𝑖 = 1, 2, . . . , 9, such that matrix Σ < 0, where

Σ = (

Σ11Ω12

∗ Ω22

) ,

Θ11= 𝑆1+ 𝑆𝑇

1+ Ξ11+ 𝑍 + 𝑍

𝑇

+ 𝜀−1𝛿2(



3


3

)


0 0

) .

(55)

Remark 10. After extending system (37) to singular system(40) by the transformation 𝑒(𝑡) → 𝜂(𝑡), taking propermatrices 𝑅

𝑖, 𝑖 = 1, 2, . . . , 9, and the constant 𝜀 > 0 can


decrease the conservation of stability of error system (37) byconstructing the Lyapunov functional

𝑉 (𝑡) = 𝑒𝑇(𝑡) 𝑃1𝑒 (𝑡) + ∫

0

−𝜏

∫

𝑡

𝑡+𝑟

𝑒𝑇(𝑠) 𝑄 𝑒 (𝑠) 𝑑𝑠 𝑑𝑟

+ ∫

0

−𝜃

∫

𝑡

𝑡+𝑟

𝑒𝑇(𝑠) Θ 𝑒 (𝑠) 𝑑𝑠 𝑑𝑟


1𝜂 (𝑡) + ∫

0

−𝜏

∫

𝑡

𝑡+𝜃


+ ∫

0

−𝜃

∫

𝑡

𝑡+𝑟

𝑒𝑇(𝑠) Θ 𝑒 (𝑠) 𝑑𝑠 𝑑𝑟.

(56)

When ℎ(𝑥(𝑡 − 𝜃)) = 0, inequality (26) is transformed to

𝜔𝑇(𝑡) Λ𝜔 (𝑡) ≥ 0, (57)

where𝜔 (𝑡)

= (𝜂𝑇(𝑡) 𝑒𝑇(𝑡) (

𝑝 (𝑡)

0)

𝑇

(∫

𝑡

𝑡−𝜏

𝑒 (𝑠) 𝑑𝑠)

𝑇

)

𝑇

,

Λ =(

𝑆1+ 𝑆𝑇

10 𝑆2+ 𝑆𝑇

30

∗ 0 0 0

∗ ∗ 𝑆4+ 𝑆𝑇

40

∗ ∗ ∗ 0

),


2𝑀20

0 0

) ,


2𝑀1𝜀𝑅1

0 𝜀𝑅2

) ,


1𝑀20

𝜀𝑅3𝜀𝑅4

) ,


1𝑀1𝜀𝑅5

𝜀𝑅6

𝜀𝑅7

) ,

𝜀 > 0

(58)


𝑖 = 1, 2, . . . , 7, are arbitrary.We choose Lyapunov function

𝑉 (𝑡) = 𝑒𝑇(𝑡) 𝑃1𝑒 (𝑡) + ∫

0

−𝜏

∫

𝑡

𝑡+𝜃



1𝜂 (𝑡) + ∫

0

−𝜏

∫

𝑡

𝑡+𝜃

𝑒𝑇(𝑠) 𝑄 𝑒 (𝑠) 𝑑𝑠 𝑑𝜃,

(59)

where 𝑃 = ( 𝑃1 𝑃20 𝑃3

), and it is easy to know that 𝑃𝐸1= 𝐸𝑇

1𝑃𝑇,

and differentiating 𝑉(𝑡) with respect to 𝑡 and using Lemma 2yield

�� (𝑡) ≤ 𝜔𝑇(𝑡) Π𝜔 (𝑡) , (60)

where

Π =(

(

𝑃𝐵 + 𝐵𝑇𝑃𝑇0 𝑃 𝑃𝑅

∗ 𝜏𝑄 0 0

∗ ∗ 0 0

∗ ∗ ∗ −1

𝜏𝑄

)

)

. (61)


0 = 2(𝑄1𝑒 (𝑡) + 𝑄

2𝑒 (𝑡) + 𝑄

3∫

𝑡

𝑡−𝜏


𝑇

(− 𝑒 (𝑡)

+ (𝐴 − 𝐿𝐶, 𝐸 − 𝐿𝐷) 𝜂 (𝑡) + 𝑝 (𝑡)

+ (𝐿𝐶, 𝐿𝐷)∫

𝑡

𝑡−𝜏

𝜂 (𝑠) 𝑑𝑠) = 𝜎𝑇(𝑡) Ξ𝜎 (𝑡) ,

(62)

where

Ξ =(

Ξ11Ξ12Ξ13Ξ14

∗ Ξ22Ξ23Ξ24

∗ ∗ Ξ33Ξ34

∗ ∗ ∗ Ξ44

),

Ξ11= (𝑄𝑇

1

0

) (𝐴 − 𝐿𝐶, 𝐸 − 𝐿𝐷) + ((𝑄𝑇

1

0

) (𝐴 − 𝐿𝐶, 𝐸 − 𝐿𝐷))

𝑇

,

Ξ12= −(

𝑄𝑇

1

0

) + (𝑄𝑇

2(𝐴 − 𝐿𝐶, 𝐸 − 𝐿𝐷))

𝑇

,

Ξ13= (𝑄𝑇

1

0

) (𝐼, 𝑅8) ,

Ξ14= (𝑄𝑇

1

0

) (𝐿𝐶, 𝐿𝐷) + (𝑄𝑇

3(𝐴 − 𝐿𝐶, 𝐸 − 𝐿𝐷))

𝑇

,

Ξ22= − (𝑄

2+ 𝑄𝑇

2) ,

Ξ23= 𝑄𝑇

2(𝐼, 𝑅8) ,

Ξ24= −𝑄3+ 𝑄𝑇

2(𝐿𝐶, 𝐿𝐷)

𝑇,

Ξ33= 0,

Ξ34= 𝑄𝑇

3(𝐼, 𝑅8) ,

Ξ44= 𝑄𝑇

3(𝐿𝐶, 𝐿𝐷) + (𝑄

𝑇

3(𝐿𝐶, 𝐿𝐷))

𝑇

,

(63)

where the proper dimension matrices 𝑅8∈ 𝑅𝑛×𝑛 is arbitrary.

Based on the above derivation, we have the followingresult.

Corollary 11. The fault tolerant synchronization (3) and (5)is achieved if there exist constants 𝜏, 𝜀 > 0, 𝛿 > 0, thepositive definite matrices 𝑃

1= 𝑃𝑇

1> 0, 𝑄 = 𝑄𝑇 > 0, and


0 0.5 1 1.5 2 2.5 3 3.5 4

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

4 6 80 12102

Figure 1: The state responses of 𝑥(𝑡) and 𝑥(𝑡) when 𝑓(𝑡) and 𝑓(𝑡) are derivable on 𝑡.

the matrices 𝑄𝑗,𝑀𝑗, 𝑈𝑗, 𝑗 = 1, 2, 𝑄

𝑗,𝑀𝑗, 𝑈𝑗, 𝑗 = 1, 2,

𝐿, 𝐶,𝐷, 𝑅𝑖, 𝑖 = 1, 2, . . . , 8, such that matrix Θ < 0, where

Θ = (

Θ11Ω12

∗ Ω22

) ,

Θ11= 𝑆1+ 𝑆𝑇

1+ Ξ11+ 𝑍 + 𝑍

𝑇

+ 𝜀−1𝛿2(



3


3

)


0 0

) .

(64)

4. Numerical Examples

Example 1. Consider a typical delayed Hopfield neural net-works [29–32] with two neurons

𝐴 = −𝐼,

𝑔 (𝑥 (𝑡)) = (

2.0 −0.1

−5.0 3.0)(

tanh (𝑥1(𝑡))

tanh (𝑥2(𝑡))) ,

ℎ (𝑥 (𝑡 − 𝜃)) = (

0.15 −0.1

−0.2 −2.5)(

tanh (𝑥1(𝑡 − 𝜃))

tanh (𝑥2(𝑡 − 𝜃))

) ,

𝜏 = 0.01,

𝜃 = 1,

𝐿𝐶 = 𝐿𝐷 = −38𝐼,

𝐸 = 2𝐼,

𝑥 (𝑡) = (

𝑥1

𝑥2

) .

(65)

When 𝑓(𝑡) and 𝑓(𝑡) are derivable on 𝑡, we take

𝑓 (𝑡) ={

{

{

(0, 0)𝑇, 0 ≤ 𝑡 ≤ 2

(0, 2 sin 𝑡)𝑇 , 𝑡 ≥ 2,

𝑓 (𝑡) = (

0.2 0

0 0)(

𝑥1(𝑡) − 𝑥

3(𝑡)

𝑥2(𝑡) − 𝑥

4(𝑡)) − 2𝑓 (𝑡) .

(66)

Simulation results are shown in Figure 1.When 𝑓(𝑡) and 𝑓(𝑡) are derivable on 𝑥(𝑡), we take

𝑓 (𝑡)

={

{

{

(0, 0)𝑇, 0 ≤ 𝑡 ≤ 10

(0.2 sin (𝑥1(𝑡)) , 0.4 sin (𝑥

2(𝑡)))𝑇

, 𝑡 ≥ 10.

(67)

Simulation results are shown in Figure 2.

Example 2. Consider chaotic Lu system [31, 32]; that is,

𝐴 = (

−36 36 0

0 20 0

0 0 −3

) ,

𝑔 (𝑥 (𝑡)) = (

0

−𝑥1𝑥3

𝑥1𝑥2

),

(68)

where 𝑥(𝑡) = (𝑥1

𝑥2

𝑥3

). We choose

𝐿𝐶 = (

0 −38 0

0 −38 0

0 0 0

) ,

𝐿𝐷 = (

0 0 0

0 −38 0

0 0 0

) ,


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

4 6 8 10 120 2 4 6 8 10 120 2

Figure 2: The state responses of 𝑥(𝑡) and 𝑥(𝑡) when 𝑓(𝑡) and 𝑓(𝑡) are derivable on 𝑥(𝑡).

2 86 100 4

−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

2 4 6 8 100

0

1

2

3

4

5

6

2 4 6 8 100

−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Figure 3: The state responses of 𝑥(𝑡) and 𝑥(𝑡) when 𝑓(𝑡) and 𝑓(𝑡) are derivable on 𝑡.

𝐸 = (

2 0 0

0 2 0

0 0 2

) ,

𝜏 = 0.01.

(69)

When 𝑓(𝑡) and 𝑓(𝑡) are derivable on 𝑡, we take

𝑓 (𝑡) ={

{

{

(0, 0, 0)𝑇, 0 ≤ 𝑡 ≤ 2

(0, 2 sin 𝑡, 0)𝑇 , 𝑡 ≥ 2,(70)


2 4 6 8 10 12 140

−0.2

0

0.2

0.4

0.6

0.8

−1

0

1

2

3

4

5

6

42 6 8 10 12 14042 6 8 10 12 140

−0.2

0

0.2

0.4

0.6

0.8

Figure 4: The state responses of 𝑥(𝑡) and 𝑥(𝑡) when 𝑓(𝑡) and 𝑓(𝑡) are derivable on 𝑥(𝑡).

𝑓 (𝑡) = (

0.2 0

0 0)(

𝑥1(𝑡) − 𝑥

3(𝑡)

𝑥2(𝑡) − 𝑥

4(𝑡)) − 2𝑓 (𝑡) . (71)

Simulation results are shown in Figure 3.When 𝑓(𝑡) and 𝑓(𝑡) are derivable on 𝑥(𝑡), we take

𝑓 (𝑡)

={

{

{

(0, 0, 0)𝑇, 0 ≤ 𝑡 ≤ 5

(0, 0.4 sin (𝑥1(𝑡)) , 0.4 sin (𝑥

2(𝑡)))𝑇

, 𝑡 ≥ 5.

(72)

Simulation results are shown in Figure 4.

5. Conclusions

In this paper, taking into account the constant propagationdelay and actuator faults and the extended transformation ofthe error systems, the problem of the chaotic fault tolerantsynchronization has been addressed. The conservation ofstability of the error systems is lower than that of otherexisting literatures.

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper.

Acknowledgments

Thiswork is supported by theNational Natural Science Foun-dation of China (11071141), the Natural Science Foundation ofShandong Province of China (ZR2011AL018, ZR2011AQ008),a Project of Shandong Province Higher Educational Sci-ence and Technology Program (J13LI02, J11LA06), and theResearch Fund Project of Heze University (XY10KZ01).

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