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Volume 2010, Article ID 201682,33pages doi:10.1155/2010/201682

Research Article

A Semigroup Approach to the System with Primary and Secondary Failures

Abdukerim Haji

College of Mathematics and System Sciences, Xinjiang University, Urumqi 830046, China

Correspondence should be addressed to Abdukerim Haji,abdukerimhaji63@yahoo.com.cn Received 2 July 2009; Accepted 22 February 2010

Academic Editor: Irena Lasiecka

Copyrightq2010 Abdukerim Haji. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We investigate the solution of a repairable parallel system with primary as well as secondary failures. By using the method of functional analysis, especially, the spectral theory of linear operators and the theory of C0-semigroups, we prove well-posedness of the system and the existence of positive solution of the system. And then we show that the time-dependent solution strongly converges to steady-state solution, thus we obtain the asymptotic stability of the time- dependent solution.

1. Introduction

As science and technology develop, the theory of reliability has infiltrated into the basic sciences, technological sciences, applied sciences, and management sciences. It is well known that repairable parallel systems are the most essential and important systems in reliability theory. In practical applications, repairable parallel systems consisting of three units are often used. Since the strong practical background of such systems, many researchers have studied them extensively under varying assumptions on the failures and repairs; see1–5and their references.

The mathematical model of a repairable parallel system with primary as well as secondary failures was first put forward by Gupta; see1. This system is consisted of three independent identical units, which are connected in parallel. In the system, one of those units operates, the other two act as warm standby. If the operating unit fails, a warm standby unit is instantaneously switched into operation. The operating unit submits primary failures and secondary failures. The primary failures are the result of a deficiency in a unit while it is operating within the design limits. The secondary failures are the result of causes that stem from a unit operating in a conditions that are outside its design limits. Two important types

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of secondary failures are common cause failures and human error failures. A Common cause failure refers to the situation where multiple units fail due to a single cause such as fire, earthquake, flood, explosion, design flaw, and poor maintenance; see2,3. A human error failure implies a failure of the system due to a mistake made by a human caused by such reasons as inadequate training, improper tools, and working in a poor lighting environment;

see4,5. There is one repairman available to repair these units. Once repaired, these units are as good as new. The failure rates of units and system are constant and independent. When the system is operating, the repairman can repair only one unit at a time. If all units fail, the entire system is repaired and checked before beginning further operation of these units.

Unlike4,5, the repair times in this system are arbitrarily distributed.

The parallel repairable system with primary and secondary failures can be described by the following equationssee1:

dp0t

dt −λ2αλc0λh0p0t μp1t 5

i3

0

μixpix, tdx, dp1t

dt λ2αp0t−

μλαλc1λh1

p1t μp2t, dp2t

dt λαp1t−

μλλc2λh2

p2t,

∂pix, t

∂t ∂pix, t

∂x −μixpix, t, i3,4,5.

R

Forx0, the boundary conditions

p30, t λp2t, p40, t 2

i0

λcipit,

p50, t 2

i0

λhipit

BC

are prescribed, and we consider the usual initial condition

p00 c∈C, pi0 bi∈C, i1,2,

pjx,0 fjx, j3,4,5, IC

wherefjx∈L10,∞. The most interesting initial condition is p00 1, pi0 0, i1,2,

pjx,0 0, j 3,4,5. IC0

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Here x, t ∈ 0,∞×0,∞;pitrepresents the probability that the system is in state iat timet,i0,1,2;pjx, trepresents the probability that at timetthe failed system is in state j and has an elapsed repair time ofx,j 3,4,5; λ represents failure rate of an operating unit;λcirepresents common-cause failure rates from stateito state 4,i0,1,2;λhi represents human-error rates from stateito state 5,i0,1,2;αrepresents failure rate of standby unit;

μrepresents constant repair rate if the system is operating;μjxrepresents repair-rate when the failed system is in statejand has an elapsed repair time ofxforj3,4,5 which satisfies μjx≥0 j3,4,5;λci i0,1,2,λhi i0,1,2,λ,μ, andαare positive constants.

In1 the author analyzed the system using supplementary variable technique and obtained various expressions including the system availability, reliability, and mean time of the failure using the Laplace transform. And then he discovered that the time-dependent availability decreases as time increases for exponential repair-time distribution under the following hypotheses.

Hypothesis 1. The system has a unique positive time-dependent solutionpx, t.

Hypothesis 2. The time-dependent solutionpx, tconverges to the steady-state solutionpx as time tends to infinity, where

px, t

p0t, p1t, p2t, p3x, t, p4x, t, p5x, t ,

px

p0, p1, p2, p3x, p4x, p5x

. 1.1

The availability and the reliability depend on the time-dependent solution of the system.

In fact, the author used the time-dependent solution in calculating the availability and the reliability. But the author did not discuss the existence of the time-dependent solution and its asymptotic stability, that is, the author did not prove the correctness of the above hypotheses.

It is well known that the above hypotheses do not always hold and it is necessary to prove the correctness. Motivated by this, we will show the well-posedness of the system and study the asymptotic stability of the time-dependent solution in this paper, by using the theory of strongly continuous operator semigroups, from6–8. First, we convert the model of the system into an abstract Cauchy problem in a Banach space. Next, we show that the operator corresponding to this model generates a positive contraction C0-semigroup. Furthermore, we prove that the system is well-posed and there is a positive solution for given initial value.

Finally, we prove that the time-dependent solution converging to its static solution in the sense of the norm through studying the spectrum of the operator and irreducibility of the corresponding semigroup, thus we obtain the asymptotic stability of the time-dependent solution of this system.

In this paper, we require the following assumption for the failure rateμjx.

Assumption 1.1general assumption. The functionμj:R → Ris measurable and bounded such that limx→ ∞μjxexists and

μj : lim

x→ ∞μjx>0, j 3,4,5, μ:min

μ3, μ4, μ5

. 1.2

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2. The Problem as an Abstract Cauchy Problem

In this section, we rewrite the underlying problem as an abstract Cauchy problem on a suitable spaceX, see6, Definition II.6.1, also see 7, Definition II.6.1. As the state space for our problem we choose

X:C3×L10,∞3. 2.1

It is obvious thatXis a Banach space endowed with the norm p:2

i0

pi 5

n3

pn

L10,∞, 2.2

wherep p0, p1, p2, p3x, p4x, p5xtX.

For simplicity, let

a0:λλc0λh0, a1:μλαλc1λh1, a2:μλλc2λh2,

2.3

and we denote byψjthe linear functionals

ψj:L10,∞−→C, f−→ψj

f :

0

μjxfxdx, j 3,4,5. 2.4

Moreover, we define the operatorsDjonW1,10,∞as

Djf :− d

dxfμjf, fW1,10,∞, j 3,4,5, 2.5 respectively. To define the appropriate operator A, DA we introduce a “maximal operator”Am, DAmonXgiven as

Am:

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎝

−a0 μ 0 ψ3 ψ4 ψ5

λ2α −a1 μ 0 0 0 0 λα −a2 0 0 0

0 0 0 D3 0 0

0 0 0 0 D4 0

0 0 0 0 0 D5

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎠ ,

DAm:C3×W1,10,∞3.

2.6

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To model the boundary conditions BC we use an abstract approach as in, for example,9. For this purpose we consider the “boundary space”

∂X:C3, 2.7

and then define “boundary operators”LandΦ. As the operatorLwe take

L:DAm−→∂X,

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎝ p0

p1

p2

p3x p4x p5x

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎠

−→L

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎝ p0

p1

p2

p3x p4x p5x

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎠ :

⎜⎜

p30 p40 p50

⎟⎟

, 2.8

and the operatorΦ∈ LDAm, ∂Xis given by

Φp:

⎜⎜

0 0 λ 0 0 0

λc0 λc1 λc2 0 0 0 λh0 λh1 λh2 0 0 0

⎟⎟

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎝ p0

p1

p2

p3x p4x p5x

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎠

, 2.9

wherep p0, p1, p2, p3x, p4x, p5xtDAm.

The operatorA, DAonXcorresponding to our original problem is then defined as Ap:Amp, DA:

pDAm|Lp Φp

. 2.10

Letpj0 pj0, t,j3,4,5,t≥0, then the conditionLp ΦpinDAis equivalent toBC.

The system of integrodifferential equationsRcan be written as the following equation:

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

dp0t

dt dp1t

dt dp2t

dt dp3x, t

dt dp4x, t

dt dp5x, t

dt

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎝

−a0 μ 0 ψ3 ψ4 ψ5

λ2α −a1 μ 0 0 0 0 λα −a2 0 0 0

0 0 0 D3 0 0

0 0 0 0 D4 0

0 0 0 0 0 D5

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎠

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎝ p0t p1t p2t p3x, t p4x, t p5x, t

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎠

. 2.11

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Let pt p0t, p1t, p2t, p3·, t, p4·, t, p5·, ttX, then 2.11 is equivalent to the following operator equation:

dpt

dt Apt, t∈0,∞. 2.12

Thus, the above equationsR,BC, andICcan be equivalently formulated as the abstract Cauchy problem

dpt

dt Apt, t∈0,∞, p0

c, b1, b2, f1, f2, f3

t

X.

ACP

IfAis the generator of a strongly continuous semigroupTtt≥0and the initial value inIC satisfiesp0 c, b1, b2, f1, f2, f3tDA, then the unique solution ofR,BC, andICis given by

pit Ttp0i1, 0≤i≤2,

pjx, t Ttp0j1x, 3≤j≤5. 2.13 For this reason it suffices to studyACP.

3. Boundary Spectrum

In this section we investigate the boundary spectrumσAiRofA. In order to characterise σAby the spectrum of a scalar 3×3-matrix, that is, or on the boundary space∂X, we apply techniques and results from10. We start from the operatorA0, DA0defined by

DA0:

pDAmLp0 ,

A0p:Amp 3.1

We give the the representation of the resolvent of the operatorA0needed below to prove the irreducibility of the semigroup generated by the operatorA.

Lemma 3.1. Let

A:

⎜⎜

−a0 μ 0 λ2α −a1 μ 0 λα −a2

⎟⎟

⎠ 3.2

and setS:{γ∈C|Rγ >−μ} \σA.Then one has

SρA0. 3.3

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Moreover, ifγS,then

R γ, A0

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎝

r1,1 r1,2 r1,3 r1,4 r1,5 r1,6

r2,1 r2,2 r2,3 r2,4 r2,5 r2,6

r3,1 r3,2 r3,3 r3,4 r3,5 r3,6

0 0 0 r4,4 0 0 0 0 0 0 r5,5 0 0 0 0 0 0 r6,6

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎠

, 3.4

where

r1,1

γa1

γa2

μλα γa0

γa1

γa2

μλα γa0

μλγa2

,

r1,2 μ

γa2

γa0

γa1

γa2

μλα γa0

μλγa2

,

r1,3 μ2

γa0

γa1

γa2

μλα γa0

μλγa2

,

r1,4

γa1

γa2

μλα ψ3R

γ, D3

γa0

γa1

γa2

μλα γa0

μλγa2

,

r1,5 μ

γa2

ψ4R γ, D4

γa0

γa1

γa2

μλα γa0

μλγa2

,

r1,6 μ2ψ5R

γ, D5

γa0

γa1

γa2

μλα γa0

μλγa2

,

r2,1 λ2α

γa2

γa0

γa1

γa2

μλα γa0

μλγa2

,

r2,2

γa0

γa2

γa0

γa1

γa2

μλα γa0

μλγa2

,

r2,3 μ

γa0

γa0

γa1

γa2

μλα γa0

μλγa2

,

r2,4 λ2α

γa2

ψ3R γ, D3

γa0

γa1

γa2

μλα γa0

μλγa2

,

r2,5 λa0

γa2

ψ4R γ, D4

γa0

γa1

γa2

μλα γa0

μλγa2

,

r2,6 μ

γa0

ψ5R γ, D5

γa0

γa1

γa2

μλα γa0

μλγa2

,

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r3,1 λ2αλα γa0

γa1

γa2

μλα γa0

μλγa2

,

r3,2

γa0

λα γa0

γa1

γa2

μλα γa0

μλγa2

,

r3,3

γa0

γa1

μλγa0

γa1

γa2

μλα γa0

μλγa2

,

r3,4 λ2αλαψ3R

γ, D3

γa0

γa1

γa2

μλα γa0

μλγa2

,

r3,5 λa0λαψ4R

γ, D4

γa0

γa1

γa2

μλα γa0

μλγa2

,

r3,6

λa0λa1μλψ5R

γ, D5

γa0

γa1

γa2

μλα γa0

μλγa2

, r4,4R

γ, D3

, r5,5R

γ, D4

, r6,6R

γ, D3

.

3.5

The resolvent operators of the differential operatorsDj j3,4,5are given by R

γ, Dj

p

x e−γx−0xμjξdξ x

0

eγx0xμjξdξpsds 3.6

forpL10,∞.

Proof. A combination of11, Proposition 2.1and12, Theorem 2.4yields that the resolvent set ofA0satisfies

ρA0S. 3.7

ForγSwe can compute the resolvent ofA0explicitly applying the formula for the inverse of operator matrices; see 12, Theorem 2.4. This leads to the representation 3.4 of the resolvent ofA0.

Clearly, knowing the operator matrix in 3.4, we can directly compute that it represents the resolvent ofA0.

The following consequence is useful to compute the boundary spectrum ofA.

Corollary 3.2. The imaginary axis belongs to the resolvent set ofA0,that is,

iRρA0. 3.8

The eigenvectors in kerγ−Amcan be computed as follows.

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Lemma 3.3. Forγ∈C,one has

p∈ker γAm

⇐⇒ 3.9

p p0, p1, p2, p3·, p4·, p5·tDAm, with 3.10 p0

γa1

γa2

μλα γa0

γa1

γa2

μλα

μλγa2

×5

j3

cj

0

μjxe−γx−x0μjξdξdx,

3.11

p1 λ2α

γa2 5 j3cj

0 μjxe−γx−0xμjξdξdx γa0

γa1

γa2

μλα

μλγa2

, 3.12

p2 λαλ5

j3cj

0 μjxe−γx−0xμjξdξdx γa0

γa1

γa2

μλα

μλγa2

, 3.13

pjx cje−γx−0xμjξdξ, j3,4,5, 3.14

wherec3, c4, c5∈C.

Proof. If forpX,3.11–3.14are fulfilled, then we can easily compute thatp∈kerγ−Am. Conversely, condition3.9gives a system of differential equations. Solving these differential equations, we see that3.11–3.14are indeed satisfied.

The domainDAmof the maximal operatorAmdecomposes, using10, Lemma 1.2, as

DAm DA0⊕ker γAm

. 3.15

Moreover, sinceLis surjective,

L|kerγ−Am: ker γAm

−→∂X 3.16

is invertible for eachγρA0, see10, Lemma 1.2. We denote its inverse by

Dγ: L|kerγ−Am−1:∂X−→ker γAm

3.17

and call it “Dirichlet operator.”

We can give the explicit form ofDγas follows.

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Lemma 3.4. For eachγρA0,the operatorDγhas the form

Dγ

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎝

d1,1 d1,2 d1,3

d2,1 d2,2 d2,3

d3,1 d3,2 d3,3

d4,1 0 0 0 d5,2 0 0 0 d6,3

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎠

, 3.18

where

d1,1

γa1

γa2

μλα 0μ3xe−γx−0xμ3ξdξdx γa0

γa1

γa2

μλα

μλγa2

,

d1,2

γa1

γa2

μλα 0μ4xe−γx−0xμ4ξdξdx γa0

γa1

γa2

μλα

μλγa2

,

d1,3

γa1

γa2

μλα 0μ5xe−γx−0xμ5ξdξdx γa0

γa1

γa2

μλα

μλγa2

,

d2,1 λ2α

γa2

0 μ3xe−γx−0xμ3ξdξdx γa0

γa1

γa2

μλα

μλγa2

,

d2,2 λ2α

γa2

0 μ4xe−γx−0xμ4ξdξdx γa0

γa1

γa2

μλα

μλγa2

,

d2,3 λ2α

γa2

0 μ5xe−γx−0xμ5ξdξdx γa0

γa1

γa2

μλα

μλγa2

,

d3,1 λαλ

0 μ3xe−γx−0xμ3ξdξdx γa0

γa1

γa2

μλα

μλγa2

,

d3,2 λαλ

0 μ4xe−γx−0xμ4ξdξdx γa0

γa1

γa2

μλα

μλγa2

,

d3,3 λαλ

0 μ5xe−γx−0xμ5ξdξdx γa0

γa1

γa2

μλα

μλγa2

, d4,1e−γx−0xμ3ξdξ,

d5,2e−γx−0xμ4ξdξ, d6,3e−γx−0xμ5ξdξ.

3.19

The operatorΦDγcan be computed explicitly forγρA0.

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Remark 3.5. ForγρA0the operatorΦDγ can be represented by the 3×3-matrix

ΦDγ

⎜⎜

a1,1 a1,2 a1,3

a2,1 a2,2 a2,3

a3,1 a3,2 a3,3

⎟⎟

, 3.20

where

a1,1 λλαλ

0 μ3xe−γx−0xμ3ξdξdx γa0

γa1

γa2

μλα

μλγa2

,

a1,2 λλαλ

0 μ4xe−γx−0xμ4ξdξdx γa0

γa1

γa2

μλα

μλγa2

,

a1,3 λλαλ

0 μ5xe−γx−0xμ5ξdξdx γa0

γa1

γa2

μλα

μλγa2

,

a2,1 λc0

γa1

γa2

μλα

λc1λ2α γa2

λc2λαλγa0

γa1

γa2

μλα

μλγa2

×

0

μ3xe−γx−0xμ3ξdξdx,

a2,2 λc0

γa1

γa2

μλα

λc1λ2α γa2

λc2λαλγa0

γa1

γa2

μλα

μλγa2

×

0

μ4xe−γx−0xμ4ξdξdx,

a2,3 λc0

γa1

γa2

μλα

λc1λ2α γa2

λc2λαλγa0

γa1

γa2

μλα

μλγa2

×

0

μ5xe−γx−0xμ5ξdξdx,

a3,1 λh0

γa1

γa2

μλα

λh1λ2α γa2

λh2λαλγa0

γa1

γa2

μλα

μλγa2

×

0

μ3xe−γx−0xμ3ξdξdx,

a3,2 λh0

γa1

γa2

μλα

λh1λ2α γa2

λh2λαλγa0

γa1

γa2

μλα

μλγa2

×

0

μ4xe−γx−0xμ4ξdξdx,

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a3,3 λh0

γa1

γa2

μλα

λh1λ2α γa2

λh2λαλγa0

γa1

γa2

μλα

μλγa2

×

0

μ5xe−γx−0xμ5ξdξdx.

3.21 The operators Dγ and Φ allow to characterise the spectrum σA and the point spectrumσpAofA. Before doing so we extend the given operators to the productX×∂X as in13, Section 3.

Definition 3.6. iX:X×∂X.

iiA0:A

m 0

−L0

,DA0:DAm× {0}.

iiiX0:X× {0}DAm× {0}DA0. ivB:0 0

Φ0

,DB:DAm×∂X.

vA:A0BA

m 0

Φ−L0

,DA:DAm× {0}.

Remark 3.7. iNote thatρA0ρA0. ForγρA0the resolvent ofA0is

R γ,A0

R

γ, A0

Dγ

0 0

. 3.22

iiThe partA|X0ofAinX0is

DA|X0 DA× {0}, A|X0 A 0

0 0

. 3.23

Hence,A|X0can be identified with the operatorA, DA.

The spectrum ofAcan be characterise by the spectrum of operators on the boundary space∂Xas follows.

Characteristic Equation 3.8.

LetγρA0. Then i

γσpA⇐⇒1∈σp

ΦDγ

. 3.24

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iiIf, in addition, there existsγ0 ∈Csuch that 1/σΦDγ0, then

γσA⇐⇒1∈σ ΦDγ

. 3.25

Proof. Let us first show the equivalence

γσA⇐⇒1∈σ ΦDγ

. 3.26

We can decomposeγ− Aas

γ− Aγ− A0− B

I − BR γ,A0

γ− A0

. 3.27

We conclude from this that the invertibility ofγ− Ais equivalent to the invertibility ofI − BRγ,A0. From

I − BR γ,A0

IdX 0

−ΦR γ, A0

Id∂X−ΦDγ

, 3.28

one can easily see thatI −BRγ,A0is invertible if and only if 1/σΦDγ. This proves3.26.

Since by our assumption 1/σΦDγ0, it follows thatγ0ρA. Therefore,ρAis not empty.

Hence we obtain from6, Proposition IV.2.17that

σA σA, 3.29

sinceAis the part ofAinX0. This showsii.

To proveiobserve first thatAandAhave the same point spectrum, that is,

σpA σpA. 3.30

Suppose now that 1 ∈σpΦDγ. Then there exists 0/f∂Xsuch thatId∂X−ΦDγf 0.

Since 0/Dγf

0

DA, we can compute

γ− A Dγf

0

IdX 0

−ΦR γ, A0

Id∂X−ΦDγ

γAm

Dγf LDγf

IdX 0

−ΦR γ, A0

Id∂X−ΦDγ

0 f

0 Id∂X−ΦDγ

f

0

0

.

3.31

This shows thatγσpA.

(14)

Conversely, if we assume thatγσpA, then there exists 0/fDAmsuch that γ− Af

0

0. From

0 0

γ− A f 0

IdX 0

−ΦR γ, A0

Id∂X−ΦDγ

γAm

f Lf

γAm

f

−ΦR γ, A0

γAm

f

Id∂X−ΦDγ

Lf

3.32

we conclude thatf∈kerγ−Amand thus

0−ΦR γ, A0

γAm

f

Id∂X−ΦDγ

Lf

Id∂X−ΦDγ

Lf. 3.33

It follows from the decomposition3.15thatLf /0 and hence 1∈σpΦDγ.

Using the Characteristic Equation 3.8 we can show that 0 is in the point spectrum ofA.

Lemma 8.8. For the operatorA, DAone has 0σpA.

Proof. By the Characteristic Equation 3.8 it suffices to prove that 1σpΦD0. Since

ΦD0

⎜⎜

b1,1 b1,2 b1,3

b2,1 b2,2 b2,3

b3,1 b3,2 b3,3

⎟⎟

, 3.34

where

b1,1 λλαλa0

a1a2μλα

μλ2αa2

,

b1,2 λλαλa0

a1a2μλα

μλ2αa2

,

b1,3 λλαλa0

a1a2μλα

μλ2αa2

,

b2,1 λc0

a1a2μλα

λc1λ2αa2λc2λαλa0

a1a2μλα

μλ2αa2

,

(15)

b2,2 λc0

a1a2μλα

λc1λ2αa2λc2λαλγa0

γa1

γa2

μλα

μλγa2

,

b2,3 λc0

a1a2μλα

λc1λ2αa2λc2λαλa0

a1a2μλα

μλ2αa2

,

b3,1 λh0

a1a2μλα

λh1λ2αa2λh2λαλa0

a1a2μλα

μλ2αa2

,

b3,2 λh0

a1a2μλα

λh1λ2αa2λh2λαλa0

a1a2μλα

μλ2αa2

,

b3,3 λh0

a1a2μλα

λh1λ2αa2λh2λαλa0

a1a2μλα

μλ2αa2

.

3.35

We can compute thejth column sumj1,2,3of the 3×3-matrixΦD0as follows:

3 i1

ΦD0i,j b1,jb2,jb3,j

λλαλa0

a1a2μλα

μλ2αa2

λc0

a1a2μλα

λc1λ2αa2λc2λαλa0

a1a2μλα

μλ2αa2

λh0

a1a2μλα

λh1λ2αa2λh2λαλa0

a1a2μλα

μλ2αa2

λc0λh0

a1a2μλα

λc1λh1λ2αa2

a0

a1a2μλα

μλ2αa2

λλc2λc2λαλa0

a1a2μλα

μλ2αa2

a0−λ2α

a1a2μλα

a1

μλα

λ2αa2

a0

a1a2μλα

μλ2αa2

a2μ

λαλa0

a1a2μλα

μλ2αa2

a0

a1a2μλα

a1a2λ2α μλαλa0

a1a2μλα

μλ2αa2

a1a2λ2α−μλ2αa2a2λαλa0

a1a2μλα

μλ2αa2

(16)

a2λαλ2α−μλαλa0

a1a2μλα

μλ2αa2

a0

a1a2μλα

μλ2αa2

a0

a1a2μλα

μλ2αa2

1.

3.36 This shows thatΦD0 is column stochastic, its transposeΦD0tis row stochastic, and hence 1 ∈ σpΦD0t. Since σpΦD0 σpΦD0t, also 1 ∈ σpΦD0holds. Therefore, by the Characteristic Equation 3.8 we conclude that 0∈σpA.

Indeed, 0 is even the only spectral value ofAon the imaginary axis.

Lemma 8.9. UnderAssumption 1.1, the spectrumσAofAsatisfies

σAiR{0}. 3.37

Proof. For anya∈R,a /0,Ψ ψ0, ψ1, ψ2, ψ3x, ψ4x, ψ5x∈X,we consider the resolvent equation

aiId−AP Ψ, 3.38 whereP p0, p1, p2, p3x, p4x, p5x. This equation is equivalent to the following system of equations:

ai−a0p0μp15

j3

0

μjxpjxdxψ0, 3.39

−λ2αp0 ai−a1p1μp2ψ1, 3.40

−λαp1 ai−a2p2ψ2, 3.41 dpjx

dx

aiμjx

pjx ψjx, j3,4,5, 3.42

p30 λp2, t >0, 3.43

p40 2

i0

λcipi, 3.44

p50 2

i0

λhipi. 3.45

Solving3.42–3.45, we get

pjx pj0e−aix−0xμjξdξe−aix−0xμjξdξ x

0

ψjueaiu0uμjξdξdu. 3.46

(17)

Since

0

pjx dx pj0

0

e−aix−0xμjξdξdx

0

e0xμjξdξ x

0

ψju e0uμjξdξdu

dx,

0

e0xμjξdξ x

0

ψju e0uμjξdξdu

dx

0

ψju e0uμjξdξ

u

e0xμjξdξdx

du.

3.47

ByAssumption 1.1, we have

ulim→∞e0uμjξdξ

u

e0xμjξdξdx lim

u→∞

u ex0μjξdξdx e0uμjξdξ lim

u→∞

e0uμjξdξdx μjue0uμjξdξ lim

u→∞

1 μju

<∞.

3.48

It follows thatpjx∈L10,∞,j 3,4,5.Let

Ij

0

μjxe−aix−0xμjξdξdx,

Kj

0

μjxe−aix−0xμjξdξ x

0

ψjueaiu0uμjξdξdu

dx.

3.49

Then

Ij

0

μjxe0xμjξdξdx1, Kj

0

μjxe0xμjξdξ x

0

ψju e0uμjξdξdu

dx.

3.50

Since

0

μjxpjxdxpj0IjKj, 3.51

henceμjxpjx∈L10,∞,j3,4,5.

(18)

Substitutingpjxinto3.39–3.41we get the following system of equations:

aia0λc0I4λh0I5p0

μλc1λh1I5

p1

−λI3λc2I4λh2I5p2ψ0K3K4K5,

−λ−2αp0 aia1p1μp2ψ1,

−λ−αp1 aia2p2ψ2.

3.52

The matrix of the coefficient of the above system is denoted by

D

⎜⎜

aia0λc0I4λh0I5 −μ−λc1λh1I5 −λI3λc2I4λh2I5

−λ−2α aia1 −μ

0 −λ−α aia2

⎟⎟

. 3.53

Since

|aia0λc0I4λh0I5| ≥ |aia0| −λc0|I4| −λh0|I5|

>|aia0| −λc0λh0

> a0λc0λh0 λ2α, −μ−λc1λh1I5 |−λ−α| ≤μλc1|I4|λh1|I5|λα

< μλc1λh1λαa1

<|aia1|,

|−λI3λc2I4λh2I5| −μ ≤λ|I3|λc2|I4|λh2|I5|μ

< λλc2λh2μa2

<|aia2|.

3.54

This shows that the matrixDis a diagonally dominant matrix, it follows that the determinant of the matrixD is not equal to 0.Therefore, system3.52has a unique solutionp0, p1, p2. Combining this with 3.46we obtain that the equation aiId−AP Ψ has exactly one solutionp0, p1, p2, p3x, p4x, p5x∈DA,this yieldsaiρA.

4. Well-Posedness of the System

The main gaol in this section is to prove the well-posedness of the system. In order to prove this, we will need some lemmas.

Lemma 8.1. A:DARAXis a closed linear operator andDAis dense inX.

Odkazy

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