Scientific Journal of Control Engineering June 2014, Volume 4, Issue 3, PP.86-93
Exponential Stability Analysis of the Solution of a Repairable Human-Machine System Dongxu Liu1, Wenyi Si2, Zhe Yin1# 1. Department of Mathematics, College of Science, Yanbian University, Yanji 133002, China 2. Network Publishing Center, Yanbian Education Publishing House, Yanji 133002, China #
Email: yinzhe@ybu.edu.cn
Abstract The repairable Human-Machine System is studied in this paper. By the method of strong continuous semi-group, the paper analyzed the essential spectrum of the system operator before and after perturbation. The results show that the dynamic solution of the system is exponential stability and tends to the steady solution of the system. Keywords: Strictly Dominant Eigenvalue; Essential Spectrum; Disturbance; Exponential Stability
1 INTRODUCTION In [1], the author introduced a supplementary variable and got a failure rate estimation formula of the system; In [2], the authors proved the existence of the solution of the system with Laplace transform; In [3], the authors studied the asymptotic stability and reliability of the system. In [7-8], the authors proved 0 was the growth bound of the system operator, and showed that 0 was also the upper spectral bound of the system operator using the concept of cofinal and relative theory. In this paper, we will discuss the exponential stability of this system. This repairable system consists of a running component and a warm standby component. If the running component fails, the warm standby component will take the replace of it. The failure of the system would be classified as hardware failure, general failure and catastrophic failure. Hardware failure makes some component fail. General failure makes the whole system be in its complete failure mode state. The system and its components can be repaired as soon as they fail. According to [1], the state-space diagram of the system is shown in Figure 1:
c
4
1
4 ( x) c
0
c 0
1
2
1
5
3
2
h
0
3 ( x)
3 ( x)
h
1
2
h
2
FIGURE 1 THIS SYSTEM MODEL CAN BE EXPRESSED BY A GROUP OF INTEGRO-DIFFERENTIAL EQUATIONS:
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2 5 dp0 (t ) ( ) p ( t ) p ( t ) j ( x) p j ( x, t )dx; c0 h0 0 i i dt 0 i 0 j 3 dp1 (t ) p0 (t ) ( 1 c1 h1 ) p1 (t ); dt dp2 (t ) p0 (t ) ( 2 c2 h2 ) p2 (t ); dt p j ( x, t ) p j ( x, t ) j ( x) p j ( x, t ), j 3, 4,5; x t 2 2 p3 (0, t ) [ p1 (t ) p2 (t )], p4 (0, t ) ci pi (t ), p5 (0, t ) hi pi (t ); i 0 i 0 p (0) 1, p (0) p (0) 0, p ( x,0) 0, j 3, 4,5 1 2 j 0
(1.1)
The symbols in the equations are defined as follows: i 0 : The running component and the warm standby component are in the normal mode state. i 1 : The running component fails for hardware failure, and the warm standby component gets into working state. i 2 : the warm standby component fails, and the running component gets into working state.
i 3 : The running component and the warm standby component are both failed because of their hardware failure, and the system is in its complete failure mode state. i 4 : The system is in its complete failure mode state for general failure.
i 5 : The system is in its complete failure mode state for catastrophic failure.
: The constant failure rate of the hardware failure of the running component.
: The constant failure rate of the hardware failure of the warm standby component. c : The constant failure rate when the system transforms from state i to state 4, where i 0,1, 2 ; i
h : The catastrophic failure rate when the system transforms from state i to state 5, where i 0,1, 2 ; i
i : The constant repair rate of the state i .
j ( x) : The repair rate when the failed system is in state j and has an elapsed repair time of x for j 3, 4,5 . pi (t ) : The probability that the system is in state i at time t for i 0,1, 2 .
p j ( x, t ) : The probability density (with respect to repair time) that the failed system is in state j and has an elapsed repair time of x for j 3, 4,5 .
For brevity, let a0 c h , a1 1 c h , a2 2 c h . 0
0
1
1
2
2
Considering the actual background, we suppose that:
0
x
j ( ) d e 0 dx , 0 j ( ) , 0 j ( )d ( j 3, 4,5).
For x (0, ) , there exists c j 0 , such that 0 cj
1 x 0 j ( )d ( j 3, 4,5). x
We choose the state space in the following way: X C 3 L1[0, ) L1[0, ) L1[0, ),
For y y0 , y1 , y2 , y3 ( x), y4 ( x), y5 ( x) X , We define its norm as follow: y y0 y1 y2 y3
L1
y4
It is obvious that ( X , ) is a Banach space. We define operator A and B : - 87 http://www.sj-ce.org
L1
y5
L1
.
A diag a0 , a1 , a2 , d 3 ( x), d 4 ( x), d 5 ( x) , dx dx dx 0 1 2 0 3 ( x) dx 0 4 ( x) dx 0 5 ( x) dx 0 0 0 0 0 0 0 0 0 0 B . 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 We choose the domain of the operator A in the following way: 2 2 dp j ( x) 1 D( A) P X L ( R ), j 3, 4,5. p3 (0) p1 p2 , p4 (0) ci pi , p5 (0) hi pi . dx i 0 i 0 Then the system (1.1) can be written as an Abstract Chuchy Problem in Banach space X :
dP(t ) ( A B) P(t ), t 0; dt P(0) (1,0,0,0,0,0).
(1.2)
Here P(t ) ( p0 (t ), p1 (t ), p2 (t ), p3 ( x, t ), p4 ( x, t ), p5 ( x, t )) .
2 STEADY-STATE SOLUTION OF THE SYSTEM First, we’ll discuss the existence of the non-zero solution of ( I A B) P 0 , where P ( p0 , p1 , p2 , p3 , p4 , p5 ) 5 ( a0 ) p0 1 p1 2 p2 0 p j ( x, t ) j ( x)dx 0; j 3 p0 ( a1 ) p1 0; p0 ( a2 ) p2 0; dp j ( x) ( j ( x)) p j ( x), j 3, 4,5; dx p3 (0) p1 p2 ; 2 p4 (0) c pi ; i 0 p (0) 2 p . h i 5 i 0
(2.1)
i
i
Solving (2.1), then we can get that: x
( j ( )) d p j ( x) p j (0)e 0 , j 3, 4,5.
Taking it into the first equation of (2.1), we can obtain that: 5
( a0 ) p0 1 p1 2 p2 p j (0) 0 j ( x)e
j 3
x
( j ( )) d 0
dx 0.
Then we have: 5 ( a ) p p p p j (0) j , dx 0; 0 0 1 1 2 2 j 3 p0 ( a1 ) p1 0; p0 ( a2 ) p2 0; p1 p2 p3 (0) 0; 2 p p (0) 0; ci i 4 i 0 2 p p (0) 0. hi i 5 i 0
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(2.2)
Where j , 0 j ( x)e
x
0 ( j ( )) d
dx, j 3, 4,5 .
The system determinant can be written as D( ) , then
D( )
a0 0
c h
0
0
1 2 3, ( a1 ) 0 0 0 ( a2 ) 0
c h
c h
1
1 0 0
2
1
2
4, 0 0
5, 0 0
0 1 0
0 0 1
Then it can be easily found that if C is an eigenvalue of the operator A B , we have D( ) 0 . Contrary, if there is a C such that D( ) 0 , then the equations have non-zero solution ( p0 , p1 , p2 , p3 ( x), p4 ( x), p5 ( x)) . So ( 3 ( )) d ( 4 ( )) d ( 5 ( )) d 0 , p4 (0)e 0 , p5 (0)e 0 p0 , p1 , p2 , p3 (0)e D( A B ) is a solution of (2.1). x
x
x
In particular, when 0 ,
j , 0 j ( x)e
x
0 ( j ( )) d
dx 1, j 3, 4,5.
So D( ) 0 , From that we can know 0 is an eigenvalue of operator A B and the eigenvector corresponding to eigenvalue 0 is ( p0 , p1 , p2 , p3 ( x), p4 ( x), p5 ( x)) . Here p1 p0 ; a1 p2 p0 ; a 2 x 0 3 ( ) d p ( ) e p0 ; 3 a1 a2 x ( ) d p4 ( x) (c c1 c2 )e 0 4 p0 ; 0 a1 a2 x p ( x) ( h1 h2 )e 0 5 ( ) d p 5 h0 0 a1 a2 Let Q (1,1,1,1,1,1) ,
we
have
P, Q p0 p1 p2 0 p3 ( x)dx 0 p4 ( x)dx 0 p5 ( x)dx 0 .
For
any
P D( A B) , we have ( A B) P, Q 0 . That is ( A B) Q 0 . So 0 is a simple eigenvalue of operator A B . *
Let p0 1 , we normalize P as follows: 1 pˆ 0 N p0 ; pˆ p ; 1 aN 0 1 ˆ p2 a N p0 ; 2 x 3 ( ) d pˆ 3 ( x) ( )e 0 p0 ; N a1 a2 x pˆ ( x) 1 ( c1 c2 )e 0 4 ( ) d p ; c 0 4 N 0 a1 a2 h1 h2 0x 5 ( ) d 1 ˆ p ( x ) ( )e p0 . h 5 N 0 a1 a2 - 89 http://www.sj-ce.org
(2.3)
Where N 1
a1
a2
(
a1
a2
) 0 e
0x 3 ( ) d
dx (c
c
0
1
a1
c
2
a2
) 0 e
0x 4 ( ) d
dx (h 0
h
1
a1
h
2
a2
) 0 e
0x 5 ( ) d
dx.
So we obtain the positive steady-state solution of the system: P* ( pˆ 0 , pˆ1 , pˆ 2 , pˆ 3 ( x), pˆ 4 ( x), pˆ 5 ( x))
(2.4)
3 EXPONENTIAL STABILITY OF THE SYSTEM The asymptotic stability of the system (1.1) has been already obtained in [3]. We will prove that the system has the exponential stability in certain condition.
Theorem 3.1[3] 1) 0 is a simple eigenvalue of operator A B . 2) { C Re 0 or ia, a R, a 0} ( A B) .
Theorem 3.2 Suppose that A is defined as above and there exists a constant c R such that c min{1 , 2 , c , h , c j } i
(i 0,1, 2, j 3, 4,5) . Then, when Re c , we have (A) and ( I A)1
i
1 . Re c
Proof: When Re c , for y y0 , y1 , y2 , y3 ( x), y4 ( x), y5 ( x) X , we consider the equation ( I A) P y , where P p0 , p1 , p2 , p3 ( x), p4 ( x), p5 ( x) :
( a0 ) p0 =y0; ( a1 ) p1 =y1; ( a2 ) p2 =y2; dp j ( x) ( j ( x)) p j ( x) y j ( x), j 3, 4,5; dx p3 (0) ( p1 p2 ); p4 (0) c0 p0 c1 p1 c2 p2 ; p (0) p p p h0 0 h1 1 h2 2 4
If Re c , we have ai ( i 0,1, 2 ). Solving (3.1) we can get: yi pi a ;(i 0,1, 2) i y1 y2 x ( ) d x ( x ) ( ) d 0 e y3 ( )d ; p3 ( x) e a a 1 2 p ( x) c y0 c y1 c y2 e x ( ) d x e ( x ) ( ) d y ( )d ; 0 4 4 a0 a1 a2 h y1 h y2 x ( ) d x ( x ) ( ) d h y0 0 e y5 ( )d ; p5 ( x) e a0 a1 a2 x
0
x
3
3
x
0
1
2
0
0
1
2
0
x
4
x
x
5
It follows that: 2
5
i 0
j 3
4
P pi p j - 90 http://www.sj-ce.org
5
(3.1)
2
yi y y2 ( 1 ) ai a1 a2
i 0
x
dx e 0
0
x
dx e 0
2
i 0
0
2
c yi
i
2
i
h yi
0
a ) i
i 0
i 0
a ) i 0
2
Re ( x )
i
0
0
e
Re x
x
3 ( ) d y ( ) d 3 x
0
x
0 3 ( ) d dx
c yi
2
a i
i 0
4 ( ) d y ( ) d 4
yi y y2 ( 1 ) ai a1 a2
Re ( x )
0
i
h yi
2
a i
i 0
e
e
0
i
Re x
x
0 4 ( ) d dx
Re x
x
0 5 ( ) d dx
0
x
dx e
Re ( x )
0
x
5 ( ) d y ( ) d 5
e (Re c ) x dx y3 ( ) d e (Re c )( x ) dx
0
e (Re c ) x dx y4 ( ) d e (Re c )( x ) dx
0
e (Re c ) x dx y5 ( ) d e (Re c )( x ) dx
0
yi y1 y2 1 1 ai Re c a1 a2 Re c
c yi
2
a i
i 0
i
1
2
h yi
a Re c i
i 0
i
1
5
Re c j 3
yj
1 1 1 (Re c c0 h0 ) y0 (Re c c1 h1 ) y1 Re c a0 Re c a1 5 1 1 1 (Re c c2 h2 ) y2 yj . Re c a2 Re c j 3 1
Then we obtain that: P
1 y . y y Re c Re c 1
2
i 0
5
i
j 3
j
This shows that if Re c 0,( I A)1 : X X is bounded. So ( A) and ( I A)1
1 Re c
.
According to Lumer-phillips theorem in [4], we can obtain the following corollary easily.
Corollary 3.1 Let A and c be defined as above. S (t ) , the C0 −semigroup of contractions generated by operator A , is exponentially stable. That is for any satisfying c 0 , S (t ) et , t 0 . Noting that the operator B is a finite-rank operator, we know B is a compact operator. So from perturbation and compact perturbation theory of semigroup, we can obtain the following result.
Theorem 3.3 Let A and c be defined as above. T (t ) , the C0 −semigroup of contractions generated by operator A B , has the following properties: 1) When C and Re c 0, ( A B) D( ) 0 . 2) Let 0 0 . For any
k { C Re c, D( ) 0} where k 0 and Re ( k 1) Re k , k 1, 2,3,
, N , 0 0 is a strictly dominant eigenvalue of operator A B .
3) Denote P* ( pˆ 0 , pˆ1 , pˆ 2 , pˆ 3 ( x), pˆ 4 ( x), pˆ 5 ( x)) which satisfies P* , Q 1 is the steady-state solution of the system. - 91 http://www.sj-ce.org
For any chosen 0 , which satisfies Re 1 0 , there exists a constant M 0 , such that T (t ) P(0) P(0), Q P* Me(Re 1 )t , t 0 .
Proof: 1) When Re c , according to theorem(3.2), we have ( A) . Then ( I A B) ( I A)( I R( , A) B) .
Since B is a finite-rank operator, we know that R( , A) B is a compact operator. If a compact operator has non-zero spectrum, the spectrum must be the eigenvalue of the operator. So the necessary and sufficient condition of ( A B) is that 1 is not the eigenvalue of R( , A) B . Therefore, When Re c 0 , we must have
( A B) D( ) 0 . 2) When Re c , it is easily found that D( ) is an analytic function and D( ) contains finite zero-points at most. There is no accumulation point in finite region. From theorem 3.1, we know that all spectrum is in the left semiplane and all the points on the imaginary axis are in the resolvent set except 0 . Nothing that 0 is the simple eigenvalue of operator A B , which has positive eigenvector, from the definition of strictly dominant eigenvalue, we can easily know that 0 is the strictly dominant eigenvalue. Let 0 0 . For any
k { C Re c, D( ) 0} , k 0 where k satisfies Re ( k 1) Re k , k 1, 2,3, dominant eigenvalue of operator A B .
, N , we must have k 0 , k 1, 2,3,
, N . And 0 0 is a strictly
3) At last, according to the perturbation theory of semigroup, essential spectrum bound does not change under compact perturbation. So the semigroup T (t ) generated by operator A B and the semigroup S (t ) generated by operator A have the same essential spectrum bound. According to [5] and [6], we know that the essential spectrum bound ( A B) 0 ( A B) . By finite expansion theorem of semigoup, we obtain that for P(0) X , Q (1,1,1,1,1,1) and the chosen 0 which satisfies Re 1 0 , there exists a constant M 0 , such that T (t ) P(0) P(0), Q P* Me(Re )t , t 0 . 1
As a result, in certain condition, the dynamic solution exponentially converges to the steady-state solution and the exponential stability of the system is obtained.
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AUTHORS - 92 http://www.sj-ce.org
1
Dongxu Liu, birthed in Jilin City, Jilin Province, July 14, 1984 and earned Master of Science in Yanbian Univer-
sity in Yanji in 2010, studied in Functional Analysis. Now he works at Yanbian University and his job title is assistant. He is interested in complex system.
- 93 http://www.sj-ce.org