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Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo


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Volume 16, Issue 1

Issues

Volume 13 (2015)

Functional analysis method for the M/G/1 queueing model with single working vacation

Ehmet Kasim / Geni Gupur
Published Online: 2018-07-26 | DOI: https://doi.org/10.1515/math-2018-0074

Abstract

In this paper, we study the asymptotic property of underlying operator corresponding to the M/G/1 queueing model with single working vacation, where both service times in a regular busy period and in a working vacation period are function. We obtain that all points on the imaginary axis except zero belong to the resolvent set of the operator and zero is an eigenvalue of both the operator and its adjoint operator with geometric multiplicity one. Therefore, we deduce that the time-dependent solution of the queueing model strongly converges to its steady-state solution. We also study the asymptotic behavior of the time-dependent queueing system’s indices for the model.

Keywords: M/G/1 queueing model with single working vacation; C0−semigroup; Eigenvalue; Resolvent set; Dirichlet operator

MSC 2010: 47D03; 47A10; 60K25

1 Introduction

Queueing system with server working vacations have arisen many researchers’ attention because the working vacation policy is more appropriate to model the real system in which the server has additional task during a vacation[1, 2, 3]. Unlike a classical vacation policy, the working vacation policy requires the server working at a lower rate rather than completely stopping service during a vacation. Therefore, compared with the classical vacation model, there are also customers who leave the system due to the completion of the services during working vacation. In this way, the number of customers in the system may be reduced. For example, an agent in a call center is required to do additional work after speaking with a customer. The agent may provide service to the next customer at a lower rate while performing additional tasks. In 2002, Servi and Finn [1] first introduced the M/M/1 queueing system with multiple working vacation. Since then, many researchers have extended their work to various type of queueing system (see Chandrasekaran et al.[4]). Kim et al.[5] and Wu and Takagi [6] extended Servi and Finn’s [1] M/M/1 queueing system to an M/G/1 queueing system. Xu et al. [7] and Baba [8] studied a batch arrival MX/M/1 queueing with working vacation. Gao and Yao [9] generalized it to an MX/G/1 queueing system. Baba [10] introduced the general input GI/M/1 queueing model with working vacation. Du [11] and Arivudainambi et al. [12] developed retrial queueing model with the concept of working vacation, etc. In 2012, Zhang and Hou [13] established the mathematical model of the M/G/1 queueing system with single working vacation by using the supplementary variable technique and studied the queueing length distribution and service status at the arbitrary epoch in the steady-state case under the following hypothesis:

limtp0,v(t)=p0,v,limtpn,v(x,t)=pn,v(x),n1,limtp0,b(t)=p0,b,limtpn,b(x,t)=pn,b(x),n1.(H)

According to Zhang and Hou [13], the M/G/1 queueing model with single working vacation can be described by the following partial differential equations:

dp0,v(t)dt=(θ+λ)p0,v(t)+0μv(x)p1,v(x,t)dx+0μb(x)p1,b(x,t)dx,p1,v(x,t)t+p1,v(x,t)x=[θ+λ+μv(x)]p1,v(x,t),pn,v(x,t)t+pn,v(x,t)x=[θ+λ+μv(x)]pn,v(x,t)+λpn1,v(x,t),n2,dp0,b(t)dt=λp0,b(t)+θp0,v(t),p1,b(x,t)t+p1,b(x,t)x=[λ+μb(x)]p1,b(x,t),pn,b(x,t)t+pn,b(x,t)x=[λ+μb(x)]pn,b(x,t)+λpn1,b(x,t),n2,(1.1)

with the integral boundary conditions

p1,v(0,t)=λp0,v(t)+0μv(x)p2,v(x)dx,pn,v(0,t)=0μv(x)pn+1,v(x)dx,n2,p1,b(0,t)=λp0,b(t)+0μb(x)p2,b(x)dx+θ0p1,v(x)dx,pn,b(0,t)=0μb(x)pn+1,b(x)dx+θ0pn,v(x)dx,n2.(1.2)

If we assume the system states when there are no customers in the system and the server is in vacation, i.e.,

p0,v(0)=1,p0,b(0)=0,pm,v(x,0)=pm,b(x,0)=0,m1,(1.3)

where (x, t) ∈ [0, ∞) × [0, ∞); p0,v(t) represents the probability that there is no customer in the system and the server is in a working vacation period at time t; pn,v(x, t)dx (n ≥ 1) is the probability that at time t the server is in a working vacation period and there are n customers in the system with elapsed service time of the customer undergoing service lying in (x, x + dx]; p0,b(t) represents the probability that there is no customer in the system and the server is in a regular busy period at time t; pn,b(x, t)dx (n ≥ 1) is the probability that at time t the server is in a regular busy period and there are n customers in the system with elapsed service time of the customer undergoing service lying in (x, x + dx]; λ is the mean arrival rate of customers; θ is the vacation duration rate of the server; μv(x) is the service rate of the server while the server is in a working vacation period and satisfies

μv(x)0,0μv(x)dx=.

μb(x) is the service rate of the server while the server is in a regular busy period and satisfying

μb(x)0,0μb(x)dx=.

In fact, the above hypothesis (H) implies the following two hypotheses in view of partial differential equations:

  • Hypothesis 1

    The model has a unique time-dependent solution.

  • Hypothesis 2

    The time-dependent solution converges to its steady-state solution.

In 2016, Kasim and Gupur [14] did the dynamic analysis for the above model and gave the detailed proof of the hypothesis 1. Moreover, when the service rates in a working vacation period and in a regular busy period are constant, by using the C0− semigroup theory they obtained that the hypothesis 2 also hold. In the general case, the service rates are function, the hypothesis 2 does not always hold, see Gupur [15] and Kasim and Gupur [16], and it is necessary to study the asymptotic behavior of the time-dependent solution of the model. This paper is an effort on this subject.

The rest of this paper is organized as follows. In Section 2 we convert the model into an abstract Cauchy problem. In Section 3, by investigating the spectral properties of the underlying operator we give the main results of this paper. Firstly, we prove that 0 is an eigenvalue of the underlying operator with geometric multiplicity one by using the probability generating function. Next, to obtain the resolvent set of the underlying operator we apply the boundary perturbation method. We obtain that all points on the imaginary axis except zero belong to the resolvent set of the operator. Last, we determine the adjoint operator and verify that 0 is an eigenvalue of the adjoint operator with geometric multiplicity one. Finally, based on these results we present the desired result in this paper: the time-dependent solution of the model strongly converges to its steady-state solution. In addition, the asymptotic behavior of the queueing system’s indices are discussed. A conclusion is given in Section 4. Section 5 provides a detail proof of some lemmas.

2 Abstract Setting for the system

In this section, we reformulate the equation (1.1)-(1.3) as an abstract Cauchy problem. We start by introducing the state space as follows.

X×Y={(pv,pb)|pvX,pbY,(pv,pb)=pv+pb},X={pvR×L1[0,)×|pv=|p0,v|+n=1pn,vL1[0,)<},Y={pbR×L1[0,)×|pb=|p0,b|+n=1pn,bL1[0,)<}.

It is obvious that X × Y is a Banach space. Define an operator and its domain.

Am((p0,vp1,v(x)p2,v(x)p3,v(x)),(p0,bp1,b(x)p2,b(x)p3,b(x)))=(((θ+λ)ϕv000ψv000λψv000λψv)(p0,vp1,v(x)p2,v(x)p3,v(x))+(0ϕb0000000000)(p0,bp1,b(x)p2,b(x)p3,b(x)),(λ0000ψb000λψb000λψb)(p0,bp1,b(x)p2,b(x)p3,b(x))+(θ00000000000)(p0,vp1,v(x)p2,v(x)p3,v(x))),

where

ϕvf=0μv(x)f(x)dx,ϕbf=0μb(x)f(x)dx,fL1[0,),ψvg=dg(x)dx(θ+λ+μv(x))g(x),gW1,1[0,),ψbg=dg(x)dx(λ+μb(x))g(x),gW1,1[0,).D(Am)={(pv,pb)X×Y|dpn,vdxL1[0,),dpn,bdxL1[0,),pn,v(x)andpn,b(x)(n1)are absolutelycontinuous andn=1dpn,vdxL1[0,)<,n=1dpn,bdxL1[0,)<}.

We choose the boundary space of X × Y

(X×Y)=l1×l1

and define two boundary operators as

L:D(Am)(X×Y),Φ:D(Am)(X×Y),L((p0,vp1,v(x)p2,v(x)p3,v(x)),(p0,bp1,b(x)p2,b(x)p3,b(x)))=((p1,v(0)p2,v(0)p3,v(0)p4,v(0)),(p1,b(0)p2,b(0)p3,b(0)p4,b(0)))Φ((p0,vp1,v(x)p2,v(x)p3,v(x)),(p0,bp1,b(x)p2,b(x)p3,b(x)))=((λ0ϕv00000ϕv00000ϕv)(p0,vp1,v(x)p2,v(x)p3,v(x)),(λ0ϕb00000ϕb00000ϕb)(p0,bp1,b(x)p2,b(x)p3,b(x))+(0θF0000θF0000θF)(p0,vp1,v(x)p2,v(x)p3,v(x))),

where Fg=0 g(x) dx,   gL1[0, ∞).

Now we define operator A and its domain as

A(pv,pb)=Am(pv,pb),D(A)={(pv,pb)D(Am)|L(pv,pb)=Φ(pv,pb)}.

Then the above equations (1.1)-(1.3) can be written as an abstract Cauchy problem:

{d(pv,pb)(t)dt=A(pv,pb)(t),t(0,)(pv,pb)(0)=((10),(00))(2.1)

Kasim and Gupur [14] have obtained the following results.

Theorem 2.1

If μv(x) and μb(x) are measurable functions and satisfy μv¯=supx[0,)μv(x) < ∞ and μb = supx[0,) μb(x) < ∞, then A generates a positive contraction C0-semigroup T(t). The system (2.1) has a unique positive time-dependent solution (pv, pb)(x, t) = T(t)(pv, pb)(0) satisfying

p0,v(t)+n=10pn,v(x,t)dx+p0,b(t)+n=10pn,b(x,t)dx=1,t[0,).

3 Main results

In this section, firstly we prove that 0 is an eigenvalue of A with geometric multiplicity one, next we study the resolvent set of operator A by using the Greiner’s idea [17] and obtain that all points on the imaginary axis except zero belong to the resolvent set of A. Thirdly, we determine the expression of A*, the adjoint operator of A, and verify that 0 is an eigenvalue of the adjoint operator A* with geometric multiplicity one. Thus, we conclude that the time-dependent solution of the system (2.1) strongly converges to its steady-state solution.

Lemma 3.1

If 0λxμb(x)e0xμb(τ)dτdx<1, then 0 is an eigenvalue of A with geometric multiplicity one.

Proof

We consider the equation A(pv, pb) = 0, which is equivalent to

(θ+λ)p0,v=0μv(x)p1,v(x)dx+0μb(x)p1,b(x)dx,dp1,v(x)dx=(θ+λ+μv(x))p1,v(x),dpn,v(x)dx=(θ+λ+μv(x))pn,v(x)+λpn1,v(x),n2,λp0,b=θp0,v,dp1,b(x)dx=(λ+μb(x))p1,b(x),dpn,b(x)dx=(λ+μb(x))pn,b(x)+λpn1,b(x),n2,(3.1)

with the boundary conditions

p1,v(0)=λp0,v+0μv(x)p2,v(x)dx,pn,v(0)=0μv(x)pn+1,v(x)dx,n2,p1,b(0)=λp0,b+0μb(x)p2,b(x)dx+θ0p1,v(x)dx,pn,b(0)=0μb(x)pn+1,b(x)dx+θ0pn,v(x)dx,n2.(3.2)

By solving (3.1) we obtain

pn,v(x)=e(θ+λ)x0xμv(s))dsk=1npk,v(0)(λx)nk(nk)!,n1,(3.3)

pn,b(x)=eλx0μb(s)dsk=1npk,b(0)(λx)nk(nk)!,n1.(3.4)

Define

bk=0(λx)kk!μb(x)eλx0xμb(s)dsdx,ck=0(λx)kk!μv(x)e(θ+λ)x0xμv(s)dsdx,dk=0(λx)kk!e(θ+λ)x0xμv(s)dsdx,k0.

The probability generating functions of these sequences are given by, for |z| < 1

B(z)=i=0bizi=0μb(x)e(λλz)x0xμb(s)dsdx,C(z)=i=0cizi=0μv(x)e(θ+λλz)x0xμv(s)dsdx,D(z)=i=0dizi=0e(θ+λλz)x0xμv(s)dsdx.

From (3.1)-(3.4) we have

p0,v=[1(θ+λ)]p0,v+c0p1,v(0)+b0p1,b(0),p0,b=(1λ)p0,b+θp0,v,p1,v(0)=λp0,v+c0p2,v(0)+c1p1,v(0),p1,b(0)=λp0,b+b0p2,b(0)+b1p1,b(0)+θd0p1,v(0),p2,v(0)=c0p3,v(0)+c1p2,v(0)+c2p1,v(0),p2,b(0)=b0p3,b(0)+b1p2,b(0)+b2p1,b(0)+θd0p2,v(0)+θd1p1,v(0),pn,v(0)=k=0ncnkpk+1,v(0),n2,pn,b(0)=k=0nbnkpk+1,b(0)+k=1nθdnkpk,v(0),n2.(3.5)

Let q0 = (p0,v,p0,b), qi = (pi,v(0), pi,b(0)), i ≥ 1, and define

U0=(1(θ+λ)θ01λ),U1=(λ00λ),E0=(0c00b0),L0=(c000b0),Li=(ciθdi10bi),i1.

Then (3.5) can be written as

q0=q0U0+q1E0,(3.6)

q1=q0U1+q1L1+q2L0,(3.7)

q2=q1L2+q2L1+q3L0,(3.8)

qk=j=1k+1qjLk+1j,k2.(3.9)

Now, we introduce the row-vector generating functions

Π(z)=i=1qizi,L(z)=i=0Lizi,|z|<1.

Hence, from (3.7)-(3.9) we deduce

Π(z)=i=1qizi=q0U1z+i=1zij=1i+1qjLi+1j=q0U1z+1zi=1qizij=0Ljzjq1L0=q0U1z+1zΠ(z)L(z)q1L0,Π(z)=z(q0U1zq1L0)[zIL(z)]1.(3.10)

An easy computation shows that

[zIL(z)]=(zC(z)θzD(z)0zB(z)),

and

[zIL(z)]1=(1zC(z)θzD(z)[zC(z)][zB(z)]01zB(z)).

This together with (3.10) yields

Π(z)=z(λzp0,vc0p1,v(0)zC(z),(λzp0,vc0p1,v(0))θzD(z)+(λp0,bzb0p1,b(0))[zC(z)][zC(z)][zB(z)]).(3.11)

Thus, we have

Π(z)e=z(λzp0,vc0p1,v(0))[θzD(z)+zB(z)]+(λp0,bzb0p1,b(0))[zC(z)][zC(z)][zB(z)],(3.12)

where e = (1, 1)T.

In the following, by using the Rouche’s theorem we conclude that zC(z) has a unique zero point inside unit circle |z| = 1. Let this root be denoted by γ, this must be root of the numerator of the equation (3.12) too. So, substituting z = γ into (3.12) we get

(λγp0,vc0p1,v(0))[θγD(γ)+γB(γ)]=0p1,v(0)=λγc0p0,v.(3.13)

(3.6) and (3.13) give

p1,b(0)=λ(1γ)+θb0p0,v.(3.14)

By using the L’Hospital rule and (3.12)-(3.14), we determine

i=1pi,v(0)=limz1Π(z)e1=limz1(λzp0,vc0p1,v(0))zzC(z)=λ(1γ)θ0eθx0xμv(s)dsdxp0,v,(3.15)

i=1pi,b(0)=limz1Π(z)e2=limz1z{(λzp0,vc0p1,v(0))θzD(z)+(λzp0,bb0p1,b(0))[zC(z)]}[zC(z)][zB(z)]=limz1{λθzD(z)p0,v+(λzp0,vc0p1,v(0))[θD(z)+θzD(z)]+λzp0,b[zc(z)]+(λzp0,bb0p1,b(0))[1C(z)]}×{[1C(z)][zB(z)]+[zC(z)][1B(z)]}1={λθD(1)p0,v+λ(1γ)[θD(1)+θD(1)]p0,v+θ[1c(1)]p0,vλ(1γ)[1C(z)]p0,v}×{[1C(1)][1B(1)]+[1C(1)][1B(1)]}1=(θ+λ)[λ(1γ)+θ]D(1)λ(1γ)[1C(1)][1B(1)]p0,v=(θ+λ)[λ(1γ)+θ]0eθx0xμv(s)dsdxλ(1γ)θ0eθx0xμv(s)dsdx[10μb(x)λxe0xμb(s)dsdx]p0,v,(3.16)

where e1 = (1, 0)T, e2 = (0,1)T.

By combining (3.15) and (3.16) with (3.3) and (3.4), we estimate

n=10pn,v(x)dx=0e(θ+λ)x0xμv(s)dsn=1k=1npk,v(0)(λx)nk(nk)!dx=k=1pk,v(0)0e(θ+λ)x0xμv(s)dsn=1(λx)nn!dx=k=1pk,v(0)0eθx0xμv(s)ds<,(3.17)

n=10pn,b(x)dx=0eλx0xμb(s)dsn=1k=1npk,b(0)(λx)nk(nk)!dx=k=1pk,b(0)0e0xμb(s)dsdx<.(3.18)

(3.17) and (3.18) imply

(pv,pb)=pv+pb<.

Thus, 0 is an eigenvalue of A. Moreover, from (3.5) it easy to see that the eigenvectors corresponding to zero span one dimensional linear space, i.e., the geometric multiplicity of 0 is one. □

In order to obtain the asymptotic behavior of the time-dependent solution of the system (2.1) we need to know the spectrum of A on the imaginary axis (see Theorem 14 in Gupur et al.[18]). For that purpose we use boundary perturbation method, which is developed by the Greiner [17], through which the spectrum of the operator can be deduced by discussing the boundary operator. It is related to the resolvent set of operator A0 and spectrum of Φ Dγ, where Dγ is inverse of L in ker(γIAm). Hence, we first consider the operator

A0(pv,pb)=Am(pv,pb),D(A0)={(pv,pb)D(Am)|L(pv,pb)=0},

and discuss its inverse. For any given (y, z) ∈ X × Y, we consider the equation (γIA0)(pv, pb) = (y, z), i.e.,

(γ+θ+λ)p0,v=y0+0μv(x)p1,v(x)dx+0μb(x)p1,b(x)dx,dp1,v(x)dx=(γ+θ+λ+μv(x))p1,v(x)+y1(x),dpn,v(x)dx=(γ+θ+λ+μv(x))pn,v(x)+λpn1,v(x)+yn(x),n2,(γ+λ)p0,b=z0+θp0,v,dp1,b(x)dx=(γ+λ+μb(x))p1,b(x)+z1(x),dpn,b(x)dx=(γ+λ+μb(x))pn,b(x)+λpn1,b(x)+zn(x),n2,pn,v(0)=0,pn,b(0)=0,n1.(3.19)

By solving (3.19) we have

p0,v=1γ+θ+λy0+1γ+θ+λ0μv(x)p1,v(x)dx+1γ+θ+λ0μb(x)p1,b(x)dx,(3.20)

p1,v(x)=e0x(γ+θ+λ+μv(ξ))dξ0xy1(τ)e0τ(γ+θ+λ+μv(ξ))dξdτ,(3.21)

pn,v(x)=λe0x(γ+θ+λ+μv(ξ))dξ0xpn1,v(τ)e0τ(γ+θ+λ+μv(ξ))dξdτ+e0x(γ+θ+λ+μv(ξ))dξ0xyn(τ)e0τ(γ+θ+λ+μv(ξ))dξdτ,n1,(3.22)

p0,b=1γ+λz0+θγ+λp0,v,(3.23)

p1,b(x)=e0x(γ+λ+μb(ξ))dξ0xz1(τ)e0τ(γ+λ+μb(ξ))dξdτ,(3.24)

pn,b(x)=λe0x(γ+λ+μb(ξ))dξ0xpn1,n(τ)e0τ(γ+λ+μb(ξ))dξdτ+e0x(γ+λ+μb(ξ))dξ0xzn(τ)e0x(γ+λ+μb(ξ))dξ,n1.(3.25)

If we introduce the following two operators as

Evf(x)=e0x(γ+θ+λ+μv(ξ))dξ0xf(τ)e0τ(γ+θ+λ+μv(ξ))dξdτ,fL1[0,),Ebf(x)=e0x(γ+λ+μb(ξ))dξ0xf(τ)e0τ(γ+λ+μb(ξ))dξdτ,fL1[0,),

then (3.21), (3.22), (3.24) and (3.25) imply

p1,v(x)=Evy1(x),(3.26)

pn,v(x)=λEvpn1,v(x)+Evyn(x)=λEv[λEvpn2,0(x)+Evyn1(x)]+Evyn(x)=λ2Ev2pn2,0(x)+λEv2yn1(x)+Evyn(x)=λ3Ev3pn3,0(x)+λ2Ev3yn2(x)+λEv2yn1(x)+Evyn(x)==λn1Evny1(x)+λn2Evn1y2(x)+λn3Evn2y3(x)++λ2Ev3yn2(x)+λEv2yn1(x)+Evyn(x)=k=0n1λkEvk+1ynk(x),n2.(3.27)

Similarly, we have

p1,b(x)=Ebz1(x),(3.28)

pn,b(x)=k=0n1λkEbk+1znk(x),n2.(3.29)

By inserting (3.26) and (3.28) into (3.20), we obtain

p0,v=1γ+θ+λy0+1γ+θ+λϕvEvy1(x)+1γ+θ+λϕbEbz1(x),p0,b=1γ+λz0+θ(γ+λ)(γ+θ+λ)y0+θ(γ+λ)(γ+θ+λ)ϕvEvy1(x)+θ(γ+λ)(γ+θ+λ)ϕbEbz1(x).(3.30)

(3.26)-(3.30) give the expression of (γIA0)−1 as follows if (γIA0)−1 exists.

(γIA0)1((y0y1y2y3),(z0z1z2z3))=((1γ+θ+λ1γ+θ+λϕvEv0000Ev0000λEv2Ev000λ2Ev3λEv2Ev0)(y0y1(x)y2(x)y3(x))+(01γ+θ+λϕbEb00000000000000)(z0z1(x)z2(x)z3(x)),(1γ+λθ(γ+λ)(γ+θ+λ)ϕbEb0000Eb0000λEb2Eb000λ2Eb3λEb2Eb0)(z0z1(x)z2(x)z3(x))+(θ(γ+λ)(γ+θ+λ)θ(γ+λ)(γ+θ+λ)ϕvEv0000000000)(y0y1(x)y2(x)y3(x))).

Therefore, we obtain the following two lemmas and their proof given in the appendix.

Lemma 3.2

If

0<μv_=infx[0,)μv(x)μv¯=supx[0,)μv(x)<,0<μb_=infx[0,)μb(x)μb¯=supx[0,)μb(x)<,

then

{γC|Reγ+λ>0Reγ+μv_>0Reγ+μb_>0}ρ(A0).

Lemma 3.3

Let

0<μv_=infx[0,)μv(x)μv¯=supx[0,)μv(x)<,0<μb_=infx[0,)μb(x)μb¯=supx[0,)μb(x)<.

If γρ(A0), then

(pv,pb)ker(γIAm)p0,v=1γ+θ+λa1,v0μv(x)e0x(γ+θ+λ+μv(ξ))dξdx+1γ+θ+λa1,b0μb(x)e0x(γ+λ+μb(ξ))dξdx,pn,v(x)=e0x(γ+θ+λ+μv(ξ))dξk=0n1(λx)kk!ank,v,n1,p0,b=θ(γ+λ)(γ+θ+λ)a1,v0μv(x)e0x(γ+θ+λ+μv(ξ))dξdx+θ(γ+λ)(γ+θ+λ)a1,b0μb(x)e0x(γ+λ+μb(ξ))dξdx,pn,b(x)=e0x(γ+λ+μb(ξ))dξk=0n(λx)kk!ank,b,n1,(a1,v,a2,v,a3,v,)l1,(a1,b,a2,b,a3,b,)l1.(3.31)

Using the results in Greiner [17], observe that the operator L is surjective. So,

L|ker(γIAm):ker(γIAm)(X×Y)

is invertible if γρ(A0). Its inverse will play an important role in the characterization of the spectrum of A on the imaginary axis and we denote its inverse by

Dγ=(L|ker(γIAm))1:(X×Y)ker(γIAm),

and call it the Dirichlet operator. Furthermore, Lemma 3.3 gives the explicit formula of Dγ for all γρ(A0),

Dγ((a1,va2,va3,v),(a1,va2,va3,v))=((1γ+θ+λϕvh0000h1100h21h220h31h32h33)(a1,va2,va3,va4,v)+(1γ+θ+λϕbm0000000000000)(a1,ba2,ba3,ba4,1),(θ(γ+λ)(γ+θ+λϕbm0000m1100m21m220m31m32m33)(a1,ba2,ba3,ba4,1)+(θ(γ+λ)(γ+θ+λ)ϕvh0000000000000)(a1,va2,va3,va4,v)),

where

h00=e0x(γ+θ+λ+μv(ξ))dξ,h11=e0x(γ+θ+λ+μv(ξ))dξ,h21=λxe0x(γ+θ+λ+μv(ξ))dξ,h22=e0x(γ+θ+λ+μv(ξ))dξ,h31=(λx)22e0x(γ+θ+λ+μv(ξ))dξ,h32=λxe0x(γ+θ+λ+μv(ξ))dξ,h33=e0x(γ+θ+λ+μv(ξ))dξ,hij=(λx)ij(ij)!e0x(γ+θ+λ+μv(ξ))dξ,i=1,2,,j=1,2,,i.m00=e0x(γ+λ+μb(ξ))dξ,m11=e0x(γ+λ+μb(ξ))dξ,m21=λxe0x(γ+λ+μb(ξ))dξ,m22=e0x(γ+λ+μb(ξ))dξ,m31=(λx)22e0x(γ+λ+μb(ξ))dξ,m32=λxe0x(γ+λ+μb(ξ))dξ,m33=e0x(γ+λ+μb(ξ))dξ,mij=(λx)ij(ij)!e0x(γ+λ+μb(ξ))dξ,i=1,2,,j=1,2,,i.

From the expression of Dγ and the definition of Φ, it is easy to determine the explicit form of Φ Dγ as follows.

ΦDγ((a1,va2,va3,v),(a1,ba2,ba3,b))=((λγ+θ+λϕvh00+ϕvh21ϕvh2200ϕvh31ϕvh32ϕvh330ϕvh41ϕvh42ϕvh43ϕvh44)(a1,va2,va3,v)+(λγ+θ+λϕbm0000000000)(a1,ba2,ba3,b),(λθ(γ+λ)(γ+θ+λ)ϕbm00+ϕbm21ϕbm2200ϕbm31ϕbm32ϕbm330ϕbm41ϕbm42ϕbm43ϕbm44)(a1,ba2,ba3,b)+(λθ(γ+λ)(γ+θ+λ)ϕvh0000000000)(a1,va2,va3,v)+(θFh11000θFh21θFh2200θFh31θFh32θFh330)(a1,va2,va3,v)).

Haji and Radl [19] gave the following result, which indicates the relations between the spectrum of A and spectrum of Φ Dγ.

Lemma 3.4

If γρ(A0) and there exists γ0 ∈ ℂ such that 1 ∉ σ(Φ Dγ0), then

γσ(A)1σ(ΦDγ).

From Lemma 3.4 and Nagel [20], we obtain the resolvent set of A on the imaginary axis.

Lemma 3.5

If

0<μv_=infx[0,)μv(x)μv¯=supx[0,)μv(x)<,0<μb_=infx[0,)μb(x)μb¯=supx[0,)μb(x)<,

then all points on the imaginary axis except zero belong to the resolvent set of A.

Proof

Let γ = iβ, β ∈ ℝ ∖ {0}. The Riemann-Lebesgue lemma

limb0f(x)cosβxdx=0,limb0f(x)sinβxdx=0,fL1[0,)

implies that there exists a positive constant 𝓚 > 0 such that ∀ |β| > 𝓚,

|0f(x)eiβxdx|2=|0f(x)(cosβxisinβx)dx|2=(0f(x)cosβxdx)2+(0f(x)sinβxdx)2<(0f(x)dx)2,0<fL1[0,).

In this formula, by replacing f(x) with μv(x)e0x(θ+λ+μv(ξ))dξ,μb(x)e0x(λ+μb(ξ))dξ, and

j=l+1|0μv(x)(λx)jl(jl)!e0x(iβ+θ+λ+μv(ξ))dξdx|j=l+10μv(x)(λx)j1(j1)!|eiβx||e0x(θ+λ+μv(ξ))dξ|dx=j=l+10μv(x)(λx)j1(j1)!e0x(θ+λ+μv(ξ))dξdx,l1,j=l+1|0μb(x)(λx)jl(jl)!e0x(iβ+λ+μb(ξ))dξdx|j=l+10μb(x)(λx)j1(j1)!|eiβx||e0x(λ+μb(ξ))dξ|dx=j=l+10μb(x)(λx)j1(j1)!e0x(λ+μb(ξ))dξdx,l1,

and using the fact 0μb(x)e0xμb(ξ)dξdx=1,0(θ+μv(x))e0x(θ+μv(ξ))dξdx=1, we estimate for a⃗v = (a1,v, a2,v, a3,v,⋯) ∈ l1 and a⃗b = (a1,b, a2,b, a3,b,⋯) ∈ l1,

ΦDγ(av,ab)|λγ+θ+λϕvh00+λθ(γ+λ)(γ+θ+λ)ϕvh00+ϕvh21||a1,v|+j=3|ϕvhj1||a1,v|+j=2|ϕvhj2||a2,v|+j=3|ϕvhj3||a3,v|++|λγ+θ+λϕbm00+λθ(γ+λ)(γ+θ+λ)ϕbm00+ϕbm21||a1,b|+j=3|ϕbmj1||a1,b|+j=2|ϕbmj2||a2,b|+j=3|ϕbmj3||a3,b|++j=1|θFhj1||a1,v|+j=2|θFhj2||a2,v|+j=3|θFhj3||a3,v|+λ|γ+λ||ϕvh00||a1,v|+j=2|ϕvhj1||a1,v|+j=2|ϕvhj2||a2,v|+j=3|ϕvhj3||a3,v|++λ|γ+λ||ϕbm00||a1,b|+j=2|ϕbmj1||a1,b|+j=2|ϕbmj2||a2,b|+j=3|ϕbmj3||a3,b|++θ{j=1|Fhj1||a1,v|+j=2|Fhj2||a2,v|+j=3|Fhj3||a3,v|+}j=1|0μv(x)(λx)j1(j1)!e0x(iβ+θ+λ+μv(ξ))dξdx||a1,v|+j=2|0μv(x)(λx)j2(j2)!e0x(iβ+θ+λ+μv(ξ))dξdx||a2,v|+j=3|0μv(x)(λx)j3(j3)!e0x(iβ+θ+λ+μv(ξ))dξdx||a3,v|++j=1|0μb(x)(λx)j1(j1)!e0x(iβ+λ+μb(ξ))dξdx||a1,b|

+j=2|0μb(x)(λx)j2(j2)!e0x(iβ+λ+μb(ξ))dξdx||a2,b|+j=3|0μb(x)(λx)j3(j3)!e0x(iβ+λ+μb(ξ))dξdx||a3,b|++θ{j=1|0(λx)j1(j1)!e0x(iβ+θ+λ+μv(ξ))dξdx||a1,v|+j=2|0(λx)j2(j2)!e0x(iβ+θ+λ+μv(ξ))dξdx||a2,v|+j=3|0(λx)j3(j3)!e0x(iβ+θ+λ+μv(ξ))dξdx||a3,v|+}<0μv(x)j=1(λx)j1(j1)!e0x(θ+λ+μv(ξ))dξdx|a1,v|+0μv(x)j=2(λx)j2(j2)!e0x(θ+λ+μv(ξ))dξdx|a2,v|+0μv(x)j=3(λx)j3(j3)!e0x(θ+λ+μv(ξ))dξdx|a3,v|++0μb(x)j=1(λx)j1(j1)!e0x(λ+μb(ξ))dξdx|a1,b|+0μb(x)j=2(λx)j2(j2)!e0x(λ+μb(ξ))dξdx|a2,b|+0μb(x)j=3(λx)j3(j3)!e0x(λ+μb(ξ))dξdx|a3,b|++θ{0j=1(λx)j1(j1)!e0x(θ+λ+μv(ξ))dξ|a1,v|+0j=2(λx)j2(j2)!e0x(θ+λ+μv(ξ))dξ|a2,v|+0j=3(λx)j3(j3)!e0x(θ+λ+μv(ξ))dξ|a3,v|+}=0μv(x)e0x(θ+μv(ξ))dξdxn=1|an,v|+0μb(x)e0xμb(ξ)dξdxn=1|an,b|+θ0e0x(θ+μv(ξ))dξn=1|an,v|=0(θ+μv(x))e0x(θ+μv(ξ))dξdxn=1|an,v|+n=1|an,b|=n=1|an,v|+n=1|an,b|=(av,ab)ΦDγ<1.(3.32)

(3.32) shows that 1 ∉ σ(Φ Dγ) when |β | > 𝓚. This together with Lemma 3.4 give

{iβ||β|>K}ρ(A),{iβ||β|K}σ(A).(3.33)

Theorem 2.1 and Lemma 3.1 ensures that T(t) is a positive contraction C0−semigroup and its spectral bound is zero. By Nagel [20] we know that σ(A) is imaginary additively cyclic (see also Thorem 1.88 in [21]) which states that

iβσ(A)iβhσ(A),all positive integerh.

From which together with (3.33) and Lemma 3.1 it follows that iℝ ∩ σ(A) = {0}. □

A trivial verification shows that X* × Y*, the dual space of X × Y, is as follows.

X×Y={(qv,qb)|qvX,qbY,|(qv,qb)|=sup{|qv|,|qb|}},

here

X={qv|qv(x)=(q0,v,q1,v(x),q2,v(x),q3,v(x),),|y|=sup{|q0,v|,supn1qn,vL[0,)}<},Y={qb|qb(x)=(q0,b,q1,b(x),q2,b,q3,b(x),)|qb|=sup{|q0,b|,supn1qn,bL[0,)}<}.

It is evident that X* × Y* is a Banach space.

Lemma 3.6

A*, the adjoint operator of A, is as follows.

A(qv,qb)=(((θ+λ)0000ddx(θ+λ+μv(x))λ000ddx(θ+λ+μv(x))λ)(q0,vq1,v(x)q2,v(x))+(0λ0μv(x)000μv(x)000μv(x))(q0,vq1,v(0)q2,v(0)q3,v(0))+(θ0000θ0000θ0000θ)(q0,bq1,b(0)q2,b(0)q3,b(0)),(λ0000ddx(λ+μb(x))λ000ddx(λ+μb(x))λ)(q0,bq1,b(x)q2,b(x)q3,b(x))+(0λ0000000μb(x)0000μb(x)0)(q0,bq1,b(0)q2,b(0)q3,b(0))+(000μv(x)00000)(q0,vq1,v(0)q2,v(0))),D(A)={(qv,qb)X×Y|dqn,v(x)dxanddqn,b(x)dxexistandqn,v()=qn,b()=α,n1},

here α in D(A*) is a constant which is independent of n.

Proof

By using integration by parts and the boundary conditions on (pv, pb) ∈ D(A), we have, for (qv,qb)D(A*)

A(pv,pb),(qv,qb)={(θ+λ)p0,v+0μv(x)p1,v(x)dx+0μb(x)p1,b(x)dx}q0,v+0{dp1,v(x)dx(λ+μv(x))p1,v(x)}q1,v(x)dx+n=20{dpn,v(x)dx(λ+μv(x))pn,v(x)+λpn1,v(x)}qn,v(x)dx+{λp0,b+θp0,v}q0,b+0{dp1,b(x)dx(λ+μb(x))p1,b(x)}q1,b(x)dx+n=20{dpn,b(x)dx(λ+μb(x))pn,b(x)+λpn1,b(x)}qn,b(x)dx=(θ+λ)p0,vq0,v+0p1,v(x)μv(x)q0,vdx+0p1,b(x)μb(x)q0,vdx+n=10dpn,v(x)dxqn,v(x)dxn=10(λ+μv(x))pn,v(x)qn,v(x)dx+n=20pn1,v(x)λqn,v(x)dx+p0,b(λq0,b)+p0,vθq0,b+n=10dpn,b(x)dxqn,b(x)dxn=10pn,b(x)(λ+μb(x))qn,b(x)dx+n=20pn1,b(x)λqn,b(x)dx=p0,v[(θ+λ)q0,v+θq0,b]+0p1,v(x)μv(x)q0,vdx+0p1,b(x)μb(x)q0,vdx+n=1[pn,v(x)qn,v(x)|x=0x=+0pn,v(x)dqn,v(x)dxdx]+n=10pn,v(x)[(λ+μv(x))qn,v(x)]dx+n=10pn,b(x)[λqn+1,v(x)]dx+p0,b(λq0,b)+n=1[pn,b(x)qn,b(x)(x)|x=0x=+0pn,b(x)dqn,b(x)(x)dx]+n=10pn,b(x)[(λ+μb(x))qn,b(x)(x)]dx+n=10pn,b(x)[λqn+1,b(x)(x)]dx=p0,v[(θ+λ)q0,v+θq0,b]+0p1,v(x)μv(x)q0,vdx+0p1,b(x)μb(x)q0,vdx+n=1pn,v(0)qn,v(0)+n=10pn,v(x)dqn,v(x)dxdx+n=10pn,v(x)[(λ+μv(x))qn,v(x)]dx+n=10pn,v(x)[λqn+1,v(x)]dx+p0,b(λq0,b)+n=1pn,b(0)qn,b(0)+n=10pn,b(x)dqn,b(x)dxdx+n=10pn,b(x)[(λ+μb(x))qn,b(x)]dx+n=10pn,b(x)[λqn+1,b(x)]dx=p0,v[(θ+λ)q0,v+θq0,b]+0p1,v(x)μv(x)q0,vdx+0p1,b(x)μb(x)q0,vdx+p0,vλq1,v(0)+n=10μv(x)pn+1,v(x)qn,v(0)dx+n=10pn,v(x)[dqn,v(x)dx(λ+μv(x))qn,v(x)+λqn+1,v(x)]dx+p0,b[λq0,b+λq1,b(0)]+n=1[0μb(x)pn+1,b(x)dx+θ0pn,v(x)dx]qn,b(0)+n=10pn,b(x)[dqn,b(x)dx(λ+μb(x))qn,b(x)+λqn+1,b(x)]dx=p0,v[(θ+λ)q0,v+λq1,v(0)+θq0,b]+0p1,v(x)[dq1,v(x)dx(λ+μv(x))q1,v(x)+λq2,v(x)+μv(x)q0,v+θq1,b(0)]dx+n=20pn,v(x)[dqn,v(x)dx(λ+μv(x))qn,v(x)+λqn+1,v(x)+μv(x)qn1,v(0)+θqn,b(0)]dx+p0,b[λq0,b+λq1,b(0)]+0p1,b(x)[dq1,b(x)dx(λ+μb(x))q1,b(x)+λq2,b(x)+μb(x)q0,v]dx+n=20pn,b(x)[dqn,b(x)dx(λ+μb(x))qn,b(x)+λqn+1,b(x)+μb(x)qn1,b(0)]dx=(pv,pb),A(qv,qb).

From this together with the definition of adjoint operator the assertion follows. □

From Theorem 2.1, Lemma 3.1 and Arendt et al. [22], we know that 0 is an eigenvalue of A*. Furthermore, we deduce the following result.

Lemma 3.7

If 0λxμb(x)e0xμb(τ)dτdx<1, then 0 is an eigenvalue of A* with geometric multiplicity one.

Now, combining the Theorem 2.1, Lemma 3.1, Lemma 3.5 and Lemma 3.7 with Theorem 14 in Gupur et al. [18] we obtain the following main result.

Theorem 3.8

If

0<μv_=infx[0,)μv(x)μv¯=supx[0,)μv(x)<,0<μb_=infx[0,)μb(x)μb¯=supx[0,)μb(x)<.

then the time-dependent solution of the system (2.1) strongly converges to its steady-state solution, i.e.,

limt(pv,pb)(,t)ω(pv,pb)()=0,

here (pv, pb)(x) is the eigenvector in Lemma 3.1 and ω is decided by the eigenvector in Lemma 3.7 and the initial value (pv, pb)(0).

In the following, by applying the Theorem 3.8 we briefly discuss the queueing system’s indices. It is easily seen that the time-dependent queueing size at the departure point converges to a positive number, i.e.,

limtπj(t)=limt{K00μv(x)pj+1,v(x,t)dx+K00μb(x)pj+1,b(x,t)dx}={K00μv(x)pj+1,v(x)dx+K00μb(x)pj+1,b(x)dx}=πj,j0.

and the time-dependent queueing length L(t) converges to the steady-state queueing length L, that is,

limtL(t)=limt{n=1n0pn,v(x,t)dx+n=1n0pn,b(x,t)dx}=n=1n0pn,v(x)dx+n=1n0pn,b(x)dx=L.

From this we can obtain that other queuing indices Lq(t), W(t) and Wq(t) also converge to a positive number Lq, W and Wq respectively.

4 Conclusion

In this paper, we study an M/G/1 queueing model with single working vacation, in which the service time is generally distributed. The system is described by infinite number of partial differential equations with integral boundary conditions which we have converted into an abstract Cauchy problem in the Banach space. Then, by investigating the spectrum of the operator on the imaginary axis, which corresponds to the M/G/1 queueing model with single working vacation, we proved that the time-dependent solution of the model strongly converges to its steady-state solution. In other words, we verified that the hypothesis 2 holds in the sense of strong convergence.

In this paper and our previous paper, we only studied spectra of the operator on the right half complex plane and imaginary axis, which corresponds to the M/G/1 queueing model with single working vacation, so it is worth studying spectra of the operator on the left half complex plane.

5 Appendix

Proof of Lemma 3.2

For any fL1[0, ∞), by using integration by parts, we have

EvfL1[0,)=0|Evf(x)|dx=0e0x(γ+θ+λ+μv(ξ))dξ0xf(τ)e0τ(γ+θ+λ+μv(ξ))dξdτdx0e0x(Reγ+θ+λ+μv(ξ))dξ0x|f(τ)|e0τ(Reγ+θ+λ+μv(ξ))dξdτdx=01Reγ+θ+λ+μv(x)0x|f(τ)|e0τ(Reγ+θ+λ+μv(ξ))dξdτde0x(Reγ+θ+λ+μv(ξ))dξ01Reγ+θ+λ+μv_0x|f(τ)|e0τ(Reγ+θ+λ+μv(ξ))dξdτde0x(Reγ+θ+λ+μv(ξ))dξ=1Reγ+θ+λ+μv_{e0x(Reγ+λ+μv(ξ))dξ0x|f(τ)|e0τ(Reγ+θ+λ+μv(ξ))dξdτ|x=0x=0|f(x)|e0x(Reγ+λ+μv(ξ))dξe0x(Reγ+θ+λ+μv(ξ))dξdx}=1Reγ+θ+λ+μv_fL1[0,)Ev1Reγ+θ+λ+μv_.(A.1)

Ebf=0|Ebf(x)|dx=0e0x(γ+λ+μb(ξ))dξ0xf(τ)e0τ(γ+λ+μb(ξ))dξdτdx1Reγ+λ+μb_fL1[0,)Eb1Reγ+λ+μb_.(A.2)

From (A.1) and (A.2) together with condition of this lemma and using ∥ϕv∥ ≤ μv, ∥ϕb∥ ≤ μb we deduce, for any (y, z) ∈ X × Y

(γIA0)1(y,z)=|1γ+θ+λy0+1γ+θ+λϕvEvy1+1γ+θ+λϕbEbz1|+Evy1L1[0,)+λEv2y1+Evy2L1[0,)+λ2Ev3y1+λEv2y2+Evy3L1[0,)+k=03λkEvk+1y4kL1[0,)++k=0n1λkEvk+1ynkL1[0,)++|1γ+λz0+θ(γ+λ)(γ+θ+λ)ϕbEbz1+θ(γ+λ)(γ+θ+λ)y0+θ(γ+λ)(γ+θ+λ)ϕvEvy1|+Ebz1L1[0,)+λEb2z1+Ebz2L1[0,)+λ2Eb3z1+λEb2z2+Ebz3L1[0,)+k=03λkEbk+1z4kL1[0,)++k=0n1λkEbk+1znkL1[0,)+1|γ+θ+λ||y0|+1|γ+θ+λ|ϕvEvy1L1[0,)+1|γ+θ+λ|ϕbEbz1L1[0,)+n=1k=0n1λkEvk+1ynkL1[0,)+1|γ+λ|z0+θ|γ+λ||γ+θ+λ|ϕbEbz1L1[0,)+θ|γ+λ||γ+θ+λ||y0|+θ|γ+λ||γ+θ+λ|ϕvEvy1L1[0,)+n=1k=0n1λkEbk+1znkL1[0,)1|γ+θ+λ||y0|+μv¯|γ+θ+λ|(Reγ+θ+λ+μv_)y1L1[0,)+μb¯|γ+θ+λ|(Reγ+λ+μb_)z1L1[0,)+1Reγ+θ+λ+μv_k=0λReγ+θ+λ+μv_kn=1ynL1[0,)+1|γ+λ|z0+μb¯|γ+λ|(Reγ+λ+μb_)z1L1[0,)+1|γ+λ||y0|+μv¯|γ+λ|(Reγ+θ+λ+μv_)y1L1[0,)+1Reγ+λ+μb_k=0λReγ+λ+μb_kn=1znL1[0,)sup{1|γ+λ|,μv¯|γ+λ|(Reγ+θ+λ+μv_)+1Reγ+θ+μv_}|y0|+n=1ynL1[0,)+sup{1|γ+λ|,μb¯|γ+λ|(Reγ+λ+μb_)+1Reγ+μb_}|z0|+n=1znL1[0,)sup{1|γ+λ|,μv¯|γ+λ|(Reγ+θ+λ+μv_)+1Reγ+θ+μv_,+μb¯|γ+λ|(Reγ+λ+μb_)+1Reγ+μb_}(y,z)<.

This shows that the result of this lemma is right. □

Proof of Lemma 3.3

If (pv, pb) ∈ ker(γIAm), then (γIAm)(pv, pb) = 0, which is equivalent to

(γ+θ+λ)p0,v=0μv(x)p1,v(x)dx+0μv(x)p1,b(x)dx,(A.3)

dp1,v(x)dx=(γ+θ+λ+μv(x))p1,v(x),(A.4)

dpn,v(x)dx=(γ+θ+λ+μv(x))pn,v(x)+λpn1,v(x),n2,(A.5)

(γ+λ)p0,b=θp0,v,(A.6)

dp1,b(x)dx=(γ+λ+μv(x))p1,b(x),(A.7)

dpn,b(x)dx=(γ+λ+μv(x))pn,b(x)+λpn1,b(x),n2.(A.8)

By solving (A.4), (A.5) and (A.7), (A.8), we have

p1,v(x)=a1,ve0x(γ+θ+λ+μv(ξ))dξ,pn,v(x)=an,ve0x(γ+θ+λ+μv(ξ))dξ(A.9)

+λe0x(γ+θ+λ+μv(ξ))dξ0xpn1,v(τ)e0τ(γ+θ+λ+μv(τ))dξdτ,n2,(A.10)

p1,b(x)=a1,be0x(γ+λ+μb(ξ))dξ,pn,b(x)=an,be0x(γ+λ+μb(ξ))dξ(A.11)

+λe0x(γ+λ+μb(ξ))dξ0xpn1,b(τ)e0τ(γ+λ+μb(ξ))dξdτ,n2.(A.12)

By using (A.9) and (A.10) repeatedly, we obtain

p2,v(x)=a2,ve0x(γ+θ+λ+μv(ξ))dξ+λe0x(γ+θ+λ+μv(ξ))dξ0xa1,vdτ=e0x(γ+θ+λ+μv(ξ))dξ[a2,v+λxa1,v],p3,v(x)=a3,ve0x(γ+θ+λ+μv(ξ))dξ+λe0x(γ+θ+λ+μv(ξ))dξ0x[a2,v+λτa1,v]dτ=e0x(γ+θ+λ+μv(ξ))dξa3,v+λxa2,v+(λx)22a1,v,pn,v(x)=e0x(γ+θ+λ+μv(ξ))dξan,v+λxan1,v+(λx)22an2,v++(λx)nn!a1,v=e0x(γ+θ+λ+μv(ξ))dξk=0n1(λx)kk!ank,v,n1.(A.13)

Similarly, by applying (A.11) and (A.12) repeatedly, we deduce

pn,b(x)=e0x(γ+λ+μb(ξ))dξk=0n1(λx)kk!ank,b,n1.(A.14)

Through inserting (A.9) and (A.11) into (A.3), we derive

p0,v=1γ+θ+λa1,v0μv(x)e0x(γ+θ+λ+μv(ξ))dξdx+1γ+θ+λa1,b0μb(x)e0x(γ+λ+μb(ξ))dξdx,p0,b=θ(γ+λ)(γ+θ+λ)a1,v0μv(x)e0x(γ+θ+λ+μv(ξ))dξdx(A.15)

+θ(γ+λ)(γ+θ+λ)a1,b0μb(x)e0x(γ+λ+μb(ξ))dξdx.(A.16)

Since (pv, pb) ∈ ker(γID(Am)), (pv, pb) ∈ D(Am) implies by the imbedding theorem in Adams [23],

n=1|an,v|=n=1|pn,v(0)|n=1pn,vL[0,)n=1pn,vL1[0,)+n=1dpn,vdxL1[0,)<,n=1|an,b|=n=1|pn,b(0)|n=1pn,bL[0,)n=1pn,bL1[0,)+n=1dpn,bdxL1[0,)<.

from which together with (A.13)-(A.16) we know that (2.55) holds.

Conversely, if (2.55) is right, then by using 0xkeMxdx=k!Mk+1,k1,M>0, integration by parts and the Fubini theorem we estimate

pn,vL1[0,)=0e0x(γ+θ+λ+μv(ξ))dξk=0n1(λx)kk!ank,vdxk=0n1|ank,0|λkk!0xkeReγ+θ+λ+μv_xdx=k=0n1λk(Reγ+θ+λ+μv_)k+1|ank,v||p0,v|+n=1pn,vL1[0,)=μv¯|γ+θ+λ|(Reγ+θ+λ+μv_)|a1,v|+μb¯|γ+θ+λ|(Reγ+λ+μb_)|a1,b|+1Reγ+θ+μv_n=1|an,v|<.(A.17)

Similarly, we get

|p0,b|+n=1pn,bL1[0,)=μb¯|γ+θ|(Reγ+θ+λ+μb_)|a1,v|+μb¯|γ+λ|(Reγ+λ+μb_)|a1,b|+1Reγ+θ+μb_n=1|an,b|<.(A.18)

(A.17) and (A.18) gives

|p0,v|+|p0,b|+n=1pn,vL1[0,)+n=1pn,bL1[0,)<.

Since

dp1,v(x)dx=(γ+θ+λ+μv(x))p1,vdpn,v(x)dx=(γ+θ+λ+μv(x))pn,v(x)+λpn1,v(x),n2,dp1,b(x)dx=(γ+λ+μb(x))p1,b(x),dpn,b(x)dx=(γ+λ+μb(x))pn,b(x)+λpn1,b(x),n2.

It is immediately obtained

n=1dpn,vdxL1[0,)(Reγ+θ+λ+μv¯)n=0pn,vL1[0,)+λn=2pn1,vL1[0,)Reγ+θ+λ+μv¯Reγ+θ+μv_+λReγ+θ+μv_n=1|an,v|<,(A.19)

n=1dpn,bdxL1[0,)(Reγ+λ+μb¯)n=1pn,bL1[0,)+λn=2pn1,bL1[0,)Reγ+λ+μb¯Reγ+μb_+λReγ+μb_n=0|an,b|<.(A.20)

(A.17)–(A.20) show that (pv, pb) ∈ ker(γIAm). □

Proof of Lemma 3.7

We consider the equation A(qv,qb)=0, which is equivalent to

(θ+λ)q0,v+λq1,v(0)+θq0,b=0,dq1,v(x)dx(λ+μv(x))q1,v(x)+λq2,v(x)+μv(x)q0,v+θq1,b(0)=0,dqn,v(x)dx(λ+μv(x))qn,v(x)+λqn+1,v(x)+μv(x)qn1,v(0)+θqn,b(0)=0,n2,λq0,b+λq1,b(0)=0,dq1,b(x)dx(λ+μb(x))q1,b(x)+λq2,b(x)+μb(x)q0,v=0,dqn,b(x)dx(λ+μb(x))qn,b(x)+λqn+1,b(x)+μb(x)qn1,b=0,n2,qn,v()=qn,b()=α,n1.(A.21)

It is easy to see that

(qv,qb)=αα,ααD(A)

is a solution of (A.21). In addition, (A.21) is equivalent to

q0,v=λθ+λq1,v(0)+θθ+λq0,b,q2,v(x)=1λ{dq1,v(x)dx+(λ+μv(x))q1,v(x)μv(x)q0,vθq1,b(0)},qn+1,0(x)=1λ{dqn,v(x)dx+(λ+μv(x))qn,v(x)μv(x)qn1,v(0)θqn,b(0)},n2,q0,b=q1,b(0),q2,b(x)=1λ{dq1,b(x)dx+(λ+μb(x))q1,b(x)μb(x)q0,v},qn+1,b(x)=1λ{dqn,b(x)dx+(λ+μb(x))qn,b(x)μb(x)qn1,b},n2.(A.22)

(A.22) show that we can determine each qn,v (x) and qn,b (x) for all n ≥ 1 if q1,v (x) and q1,b (x) are given. That is to say, geometric multiplicity of zero is one. □

Acknowledgement

The authors would like to express their sincere thanks to the anonymous referees and associated editor for his/her careful reading of the manuscript. The author’ research work was supported by the National Natural Science Foundation of China (No:11371303) and Natural Science Foundation of Xinjiang University(No:BS130104).

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About the article

Received: 2017-10-21

Accepted: 2018-05-24

Published Online: 2018-07-26


Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 767–791, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2018-0074.

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© 2018 Kasim and Gupur, published by De Gruyter. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License. BY-NC-ND 4.0

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