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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

# 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.

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:

$limt→∞p0,v(t)=p0,v,limt→∞pn,v(x,t)=pn,v(x),n≥1,limt→∞p0,b(t)=p0,b,limt→∞pn,b(x,t)=pn,b(x),n≥1.$(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:

$dp0,v(t)dt=−(θ+λ)p0,v(t)+∫0∞μv(x)p1,v(x,t)dx+∫0∞μb(x)p1,b(x,t)dx,∂p1,v(x,t)∂t+∂p1,v(x,t)∂x=−[θ+λ+μv(x)]p1,v(x,t),∂pn,v(x,t)∂t+∂pn,v(x,t)∂x=−[θ+λ+μv(x)]pn,v(x,t)+λpn−1,v(x,t),∀n≥2,dp0,b(t)dt=−λp0,b(t)+θp0,v(t),∂p1,b(x,t)∂t+∂p1,b(x,t)∂x=−[λ+μb(x)]p1,b(x,t),∂pn,b(x,t)∂t+∂pn,b(x,t)∂x=−[λ+μb(x)]pn,b(x,t)+λpn−1,b(x,t),∀n≥2,$(1.1)

with the integral boundary conditions

$p1,v(0,t)=λp0,v(t)+∫0∞μv(x)p2,v(x)dx,pn,v(0,t)=∫0∞μv(x)pn+1,v(x)dx,∀n≥2,p1,b(0,t)=λp0,b(t)+∫0∞μb(x)p2,b(x)dx+θ∫0∞p1,v(x)dx,pn,b(0,t)=∫0∞μb(x)pn+1,b(x)dx+θ∫0∞pn,v(x)dx,∀n≥2.$(1.2)

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

$p0,v(0)=1,p0,b(0)=0,pm,v(x,0)=pm,b(x,0)=0,∀m≥1,$(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

$μv(x)≥0,∫0∞μv(x)dx=∞.$

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

$μb(x)≥0,∫0∞μb(x)dx=∞.$

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.

$X×Y={(pv,pb)|pv∈X,pb∈Y,‖(pv,pb)‖=‖pv‖+‖pb‖},X={pv∈R×L1[0,∞)×⋯|‖pv‖=|p0,v|+∑n=1∞‖pn,v‖L1[0,∞)<∞},Y={pb∈R×L1[0,∞)×⋯|‖pb‖=|p0,b|+∑n=1∞‖pn,b‖L1[0,∞)<∞}.$

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

$Am((p0,vp1,v(x)p2,v(x)p3,v(x)⋮),(p0,bp1,b(x)p2,b(x)p3,b(x)⋮))=((−(θ+λ)ϕv00⋯0ψv00⋯0λψv0⋯00λψv⋯⋮⋮⋮⋮⋱)(p0,vp1,v(x)p2,v(x)p3,v(x)⋮)+(0ϕb0⋯000⋯000⋯000⋯⋮⋮⋮⋱)(p0,bp1,b(x)p2,b(x)p3,b(x)⋮),(−λ000⋯0ψb00⋯0λψb0⋯00λψb⋯⋮⋮⋮⋮⋱)(p0,bp1,b(x)p2,b(x)p3,b(x)⋮)+(θ00⋯000⋯000⋯000⋯⋮⋮⋮⋱)(p0,vp1,v(x)p2,v(x)p3,v(x)⋮)),$

where

$ϕvf=∫0∞μv(x)f(x)dx,ϕbf=∫0∞μb(x)f(x)dx,f∈L1[0,∞),ψvg=−dg(x)dx−(θ+λ+μv(x))g(x),g∈W1,1[0,∞),ψbg=−dg(x)dx−(λ+μb(x))g(x),g∈W1,1[0,∞).D(Am)={(pv,pb)∈X×Y|dpn,vdx∈L1[0,∞),dpn,bdx∈L1[0,∞),pn,v(x)andpn,b(x)(n≥1)are absolutelycontinuous and∑n=1∞‖dpn,vdx‖L1[0,∞)<∞,∑n=1∞‖dpn,bdx‖L1[0,∞)<∞}.$

We choose the boundary space of X × Y

$∂(X×Y)=l1×l1$

and define two boundary operators as

$L:D(Am)→∂(X×Y),Φ:D(Am)→∂(X×Y),L((p0,vp1,v(x)p2,v(x)p3,v(x)⋮),(p0,bp1,b(x)p2,b(x)p3,b(x)⋮))=((p1,v(0)p2,v(0)p3,v(0)p4,v(0)⋮),(p1,b(0)p2,b(0)p3,b(0)p4,b(0)⋮))Φ((p0,vp1,v(x)p2,v(x)p3,v(x)⋮),(p0,bp1,b(x)p2,b(x)p3,b(x)⋮))=((λ0ϕv00⋯000ϕv0⋯0000ϕv⋯⋮⋮⋮⋮⋮⋱)(p0,vp1,v(x)p2,v(x)p3,v(x)⋮),(λ0ϕb00⋯000ϕb0⋯0000ϕb⋯⋮⋮⋮⋮⋮⋱)(p0,bp1,b(x)p2,b(x)p3,b(x)⋮)+(0θF00⋯00θF0⋯000θF⋯⋮⋮⋮⋮⋱)(p0,vp1,v(x)p2,v(x)p3,v(x)⋮)),$

where $\begin{array}{}\mathcal{F}g={\int }_{0}^{\mathrm{\infty }}\end{array}$ 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 $\begin{array}{}\overline{{\mu }_{v}}=\underset{x\in \left[0,\mathrm{\infty }\right)}{sup}{\mu }_{v}\left(x\right)\end{array}$ < ∞ and μb = $\begin{array}{}\underset{x\in \left[0,\mathrm{\infty }\right)}{sup}\end{array}$ 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

$p0,v(t)+∑n=1∞∫0∞pn,v(x,t)dx+p0,b(t)+∑n=1∞∫0∞pn,b(x,t)dx=1,t∈[0,∞).$

## 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 $\begin{array}{}{\int }_{0}^{\mathrm{\infty }}\lambda x{\mu }_{b}\left(x\right){e}^{-{\int }_{0}^{x}{\mu }_{b}\left(\tau \right)d\tau }dx<1,\end{array}$ then 0 is an eigenvalue of A with geometric multiplicity one.

#### Proof

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

$(θ+λ)p0,v=∫0∞μv(x)p1,v(x)dx+∫0∞μb(x)p1,b(x)dx,dp1,v(x)dx=−(θ+λ+μv(x))p1,v(x),dpn,v(x)dx=−(θ+λ+μv(x))pn,v(x)+λpn−1,v(x),n≥2,λp0,b=θp0,v,dp1,b(x)dx=−(λ+μb(x))p1,b(x),dpn,b(x)dx=−(λ+μb(x))pn,b(x)+λpn−1,b(x),n≥2,$(3.1)

with the boundary conditions

$p1,v(0)=λp0,v+∫0∞μv(x)p2,v(x)dx,pn,v(0)=∫0∞μv(x)pn+1,v(x)dx,n≥2,p1,b(0)=λp0,b+∫0∞μb(x)p2,b(x)dx+θ∫0∞p1,v(x)dx,pn,b(0)=∫0∞μb(x)pn+1,b(x)dx+θ∫0∞pn,v(x)dx,n≥2.$(3.2)

By solving (3.1) we obtain

$pn,v(x)=e−(θ+λ)x−∫0xμv(s))ds∑k=1npk,v(0)(λx)n−k(n−k)!,n≥1,$(3.3)

$pn,b(x)=e−λx−∫0∞μb(s)ds∑k=1npk,b(0)(λx)n−k(n−k)!,n≥1.$(3.4)

Define

$bk=∫0∞(λx)kk!μb(x)e−λx−∫0xμb(s)dsdx,ck=∫0∞(λx)kk!μv(x)e−(θ+λ)x−∫0xμv(s)dsdx,dk=∫0∞(λx)kk!e−(θ+λ)x−∫0xμv(s)dsdx,k≥0.$

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

$B(z)=∑i=0∞bizi=∫0∞μb(x)e−(λ−λz)x−∫0xμb(s)dsdx,C(z)=∑i=0∞cizi=∫0∞μv(x)e−(θ+λ−λz)x−∫0xμv(s)dsdx,D(z)=∑i=0∞dizi=∫0∞e−(θ+λ−λz)x−∫0xμv(s)dsdx.$

From (3.1)-(3.4) we have

$p0,v=[1−(θ+λ)]p0,v+c0p1,v(0)+b0p1,b(0),p0,b=(1−λ)p0,b+θp0,v,p1,v(0)=λp0,v+c0p2,v(0)+c1p1,v(0),p1,b(0)=λp0,b+b0p2,b(0)+b1p1,b(0)+θd0p1,v(0),p2,v(0)=c0p3,v(0)+c1p2,v(0)+c2p1,v(0),p2,b(0)=b0p3,b(0)+b1p2,b(0)+b2p1,b(0)+θd0p2,v(0)+θd1p1,v(0),⋯⋯pn,v(0)=∑k=0ncn−kpk+1,v(0),n≥2,pn,b(0)=∑k=0nbn−kpk+1,b(0)+∑k=1nθdn−kpk,v(0),n≥2.$(3.5)

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

$U0=(1−(θ+λ)θ01−λ),U1=(λ00λ),E0=(0c00b0),L0=(c000b0),Li=(ciθdi−10bi),i≥1.$

Then (3.5) can be written as

$q0=q0U0+q1E0,$(3.6)

$q1=q0U1+q1L1+q2L0,$(3.7)

$q2=q1L2+q2L1+q3L0,$(3.8)

$qk=∑j=1k+1qjLk+1−j,k≥2.$(3.9)

Now, we introduce the row-vector generating functions

$Π(z)=∑i=1∞qizi,L(z)=∑i=0∞Lizi,|z|<1.$

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

$Π(z)=∑i=1∞qizi=q0U1z+∑i=1∞zi∑j=1i+1qjLi+1−j=q0U1z+1z∑i=1∞qizi∑j=0∞Ljzj−q1L0=q0U1z+1zΠ(z)L(z)−q1L0,⟹Π(z)=z(q0U1z−q1L0)[zI−L(z)]−1.$(3.10)

An easy computation shows that

$[zI−L(z)]=(z−C(z)−θzD(z)0z−B(z)),$

and

$[zI−L(z)]−1=(1z−C(z)θzD(z)[z−C(z)][z−B(z)]01z−B(z)).$

This together with (3.10) yields

$Π(z)=z(λzp0,v−c0p1,v(0)z−C(z),(λzp0,v−c0p1,v(0))θzD(z)+(λp0,bz−b0p1,b(0))[z−C(z)][z−C(z)][z−B(z)]).$(3.11)

Thus, we have

$Π(z)e=z(λzp0,v−c0p1,v(0))[θzD(z)+z−B(z)]+(λp0,bz−b0p1,b(0))[z−C(z)][z−C(z)][z−B(z)],$(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

$(λγp0,v−c0p1,v(0))[θγD(γ)+γ−B(γ)]=0⟹p1,v(0)=λγc0p0,v.$(3.13)

(3.6) and (3.13) give

$p1,b(0)=λ(1−γ)+θb0p0,v.$(3.14)

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

$∑i=1∞pi,v(0)=limz→1Π(z)e1=limz→1(λzp0,v−c0p1,v(0))zz−C(z)=λ(1−γ)θ∫0∞e−θx−∫0xμv(s)dsdxp0,v,$(3.15)

$∑i=1∞pi,b(0)=limz→1Π(z)e2=limz→1z{(λzp0,v−c0p1,v(0))θzD(z)+(λzp0,b−b0p1,b(0))[z−C(z)]}[z−C(z)][z−B(z)]=limz→1{λθzD(z)p0,v+(λzp0,v−c0p1,v(0))[θD(z)+θzD′(z)]+λzp0,b[z−c(z)]+(λzp0,b−b0p1,b(0))[1−C′(z)]}×{[1−C′(z)][z−B(z)]+[z−C(z)][1−B′(z)]}−1={λθD(1)p0,v+λ(1−γ)[θD(1)+θD′(1)]p0,v+θ[1−c(1)]p0,v−λ(1−γ)[1−C′(z)]p0,v}×{[1−C′(1)][1−B(1)]+[1−C(1)][1−B′(1)]}−1=(θ+λ)[λ(1−γ)+θ]D(1)−λ(1−γ)[1−C(1)][1−B′(1)]p0,v=(θ+λ)[λ(1−γ)+θ]∫0∞e−θx−∫0xμv(s)dsdx−λ(1−γ)θ∫0∞e−θx−∫0xμv(s)dsdx[1−∫0∞μb(x)λxe−∫0xμb(s)dsdx]p0,v,$(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

$∑n=1∞∫0∞pn,v(x)dx=∫0∞e−(θ+λ)x−∫0xμv(s)ds∑n=1∞∑k=1npk,v(0)(λx)n−k(n−k)!dx=∑k=1∞pk,v(0)∫0∞e−(θ+λ)x−∫0xμv(s)ds∑n=1∞(λx)nn!dx=∑k=1∞pk,v(0)∫0∞e−θx−∫0xμv(s)ds<∞,$(3.17)

$∑n=1∞∫0∞pn,b(x)dx=∫0∞e−λx−∫0xμb(s)ds∑n=1∞∑k=1npk,b(0)(λx)n−k(n−k)!dx=∑k=1∞pk,b(0)∫0∞e−∫0xμb(s)dsdx<∞.$(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.,

$(γ+θ+λ)p0,v=y0+∫0∞μv(x)p1,v(x)dx+∫0∞μb(x)p1,b(x)dx,dp1,v(x)dx=−(γ+θ+λ+μv(x))p1,v(x)+y1(x),dpn,v(x)dx=−(γ+θ+λ+μv(x))pn,v(x)+λpn−1,v(x)+yn(x),n≥2,(γ+λ)p0,b=z0+θp0,v,dp1,b(x)dx=−(γ+λ+μb(x))p1,b(x)+z1(x),dpn,b(x)dx=−(γ+λ+μb(x))pn,b(x)+λpn−1,b(x)+zn(x),n≥2,pn,v(0)=0,pn,b(0)=0,n≥1.$(3.19)

By solving (3.19) we have

$p0,v=1γ+θ+λy0+1γ+θ+λ∫0∞μv(x)p1,v(x)dx+1γ+θ+λ∫0∞μb(x)p1,b(x)dx,$(3.20)

$p1,v(x)=e−∫0x(γ+θ+λ+μv(ξ))dξ∫0xy1(τ)e∫0τ(γ+θ+λ+μv(ξ))dξdτ,$(3.21)

$pn,v(x)=λe−∫0x(γ+θ+λ+μv(ξ))dξ∫0xpn−1,v(τ)e∫0τ(γ+θ+λ+μv(ξ))dξdτ+e−∫0x(γ+θ+λ+μv(ξ))dξ∫0xyn(τ)e∫0τ(γ+θ+λ+μv(ξ))dξdτ,n≥1,$(3.22)

$p0,b=1γ+λz0+θγ+λp0,v,$(3.23)

$p1,b(x)=e−∫0x(γ+λ+μb(ξ))dξ∫0xz1(τ)e∫0τ(γ+λ+μb(ξ))dξdτ,$(3.24)

$pn,b(x)=λe−∫0x(γ+λ+μb(ξ))dξ∫0xpn−1,n(τ)e∫0τ(γ+λ+μb(ξ))dξdτ+e−∫0x(γ+λ+μb(ξ))dξ∫0xzn(τ)e∫0x(γ+λ+μb(ξ))dξ,n≥1.$(3.25)

If we introduce the following two operators as

$Evf(x)=e−∫0x(γ+θ+λ+μv(ξ))dξ∫0xf(τ)e∫0τ(γ+θ+λ+μv(ξ))dξdτ,f∈L1[0,∞),Ebf(x)=e−∫0x(γ+λ+μb(ξ))dξ∫0xf(τ)e∫0τ(γ+λ+μb(ξ))dξdτ,f∈L1[0,∞),$

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

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

$pn,v(x)=λEvpn−1,v(x)+Evyn(x)=λEv[λEvpn−2,0(x)+Evyn−1(x)]+Evyn(x)=λ2Ev2pn−2,0(x)+λEv2yn−1(x)+Evyn(x)=λ3Ev3pn−3,0(x)+λ2Ev3yn−2(x)+λEv2yn−1(x)+Evyn(x)=⋯⋯=λn−1Evny1(x)+λn−2Evn−1y2(x)+λn−3Evn−2y3(x)+⋯+λ2Ev3yn−2(x)+λEv2yn−1(x)+Evyn(x)=∑k=0n−1λkEvk+1yn−k(x),n≥2.$(3.27)

Similarly, we have

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

$pn,b(x)=∑k=0n−1λkEbk+1zn−k(x),n≥2.$(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.

$(γI−A0)−1((y0y1y2y3⋮),(z0z1z2z3⋮))=((1γ+θ+λ1γ+θ+λϕvEv000⋯0Ev000⋯0λEv2Ev00⋯0λ2Ev3λEv2Ev0⋯⋮⋮⋮⋮⋮⋱)(y0y1(x)y2(x)y3(x)⋮)+(01γ+θ+λϕbEb00⋯0000⋯0000⋯0000⋯⋮⋮⋮⋱)(z0z1(x)z2(x)z3(x)⋮),(1γ+λθ(γ+λ)(γ+θ+λ)ϕbEb000⋯0Eb000⋯0λEb2Eb00⋯0λ2Eb3λEb2Eb0⋯⋮⋮⋮⋮⋮⋱)(z0z1(x)z2(x)z3(x)⋮)+(θ(γ+λ)(γ+θ+λ)θ(γ+λ)(γ+θ+λ)ϕvEv0⋯000⋯000⋯000⋯⋮⋮⋮⋱)(y0y1(x)y2(x)y3(x)⋮)).$

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

${γ∈C|Reγ+λ>0Reγ+μv_>0Reγ+μb_>0}⊂ρ(A0).$

#### 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

$(pv,pb)∈ker⁡(γI−Am)⟺p0,v=1γ+θ+λa1,v∫0∞μv(x)e−∫0x(γ+θ+λ+μv(ξ))dξdx+1γ+θ+λa1,b∫0∞μb(x)e−∫0x(γ+λ+μb(ξ))dξdx,pn,v(x)=e−∫0x(γ+θ+λ+μv(ξ))dξ∑k=0n−1(λx)kk!an−k,v,n≥1,p0,b=θ(γ+λ)(γ+θ+λ)a1,v∫0∞μv(x)e−∫0x(γ+θ+λ+μv(ξ))dξdx+θ(γ+λ)(γ+θ+λ)a1,b∫0∞μb(x)e−∫0x(γ+λ+μb(ξ))dξdx,pn,b(x)=e−∫0x(γ+λ+μb(ξ))dξ∑k=0n(λx)kk!an−k,b,n≥1,(a1,v,a2,v,a3,v,⋯)∈l1,(a1,b,a2,b,a3,b,⋯)∈l1.$(3.31)

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

$L|ker⁡(γI−Am):ker⁡(γI−Am)→∂(X×Y)$

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

$Dγ=(L|ker⁡(γI−Am))−1:∂(X×Y)→ker⁡(γI−Am),$

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

$Dγ((a1,va2,va3,v⋮),(a1,va2,va3,v⋮))=((1γ+θ+λϕvh0000⋯h1100⋯h21h220⋯h31h32h33⋯⋮⋮⋮⋱)(a1,va2,va3,va4,v⋮)+(1γ+θ+λϕbm0000⋯000⋯000⋯000⋯⋮⋮⋮⋱)(a1,ba2,ba3,ba4,1⋮),(θ(γ+λ)(γ+θ+λϕbm0000⋯m1100⋯m21m220⋯m31m32m33⋯⋮⋮⋮⋱)(a1,ba2,ba3,ba4,1⋮)+(θ(γ+λ)(γ+θ+λ)ϕvh0000⋯000⋯000⋯000⋯⋮⋮⋮⋱)(a1,va2,va3,va4,v⋮)),$

where

$h00=e−∫0x(γ+θ+λ+μv(ξ))dξ,h11=e−∫0x(γ+θ+λ+μv(ξ))dξ,h21=λxe−∫0x(γ+θ+λ+μv(ξ))dξ,h22=e−∫0x(γ+θ+λ+μv(ξ))dξ,h31=(λx)22e−∫0x(γ+θ+λ+μv(ξ))dξ,h32=λxe−∫0x(γ+θ+λ+μv(ξ))dξ,h33=e−∫0x(γ+θ+λ+μv(ξ))dξ,⋯hij=(λx)i−j(i−j)!e−∫0x(γ+θ+λ+μv(ξ))dξ,i=1,2,⋯,j=1,2,⋯,i.m00=e−∫0x(γ+λ+μb(ξ))dξ,m11=e−∫0x(γ+λ+μb(ξ))dξ,m21=λxe−∫0x(γ+λ+μb(ξ))dξ,m22=e−∫0x(γ+λ+μb(ξ))dξ,m31=(λx)22e−∫0x(γ+λ+μb(ξ))dξ,m32=λxe−∫0x(γ+λ+μb(ξ))dξ,m33=e−∫0x(γ+λ+μb(ξ))dξ,⋯mij=(λx)i−j(i−j)!e−∫0x(γ+λ+μb(ξ))dξ,i=1,2,⋯,j=1,2,⋯,i.$

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

$ΦDγ((a1,va2,va3,v⋮),(a1,ba2,ba3,b⋮))=((λγ+θ+λϕvh00+ϕvh21ϕvh2200⋯ϕvh31ϕvh32ϕvh330⋯ϕvh41ϕvh42ϕvh43ϕvh44⋯⋮⋮⋮⋮⋱)(a1,va2,va3,v⋮)+(λγ+θ+λϕbm0000⋯000⋯000⋯⋮⋮⋮⋱)(a1,ba2,ba3,b⋮),(λθ(γ+λ)(γ+θ+λ)ϕbm00+ϕbm21ϕbm2200⋯ϕbm31ϕbm32ϕbm330⋯ϕbm41ϕbm42ϕbm43ϕbm44⋯⋮⋮⋮⋮⋱)(a1,ba2,ba3,b⋮)+(λθ(γ+λ)(γ+θ+λ)ϕvh0000⋯000⋯000⋯⋮⋮⋮⋱)(a1,va2,va3,v⋮)+(θFh11000⋯θFh21θFh2200⋯θFh31θFh32θFh330⋯⋮⋮⋮⋮⋱)(a1,va2,va3,v⋮)).$

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

$γ∈σ(A)⟺1∈σ(ΦDγ).$

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

$limb→∞∫0∞f(x)cos⁡βxdx=0,limb→∞∫0∞f(x)sin⁡βxdx=0,f∈L1[0,∞)$

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

$|∫0∞f(x)e−iβxdx|2=|∫0∞f(x)(cos⁡βx−isin⁡βx)dx|2=(∫0∞f(x)cos⁡βxdx)2+(∫0∞f(x)sin⁡βxdx)2<(∫0∞f(x)dx)2,0

In this formula, by replacing f(x) with $\begin{array}{}{\mu }_{v}\left(x\right){e}^{-{\int }_{0}^{x}\left(\theta +\lambda +{\mu }_{v}\left(\xi \right)\right)d\xi },\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\mu }_{b}\left(x\right){e}^{-{\int }_{0}^{x}\left(\lambda +{\mu }_{b}\left(\xi \right)\right)d\xi },\end{array}$ and

$∑j=l+1∞|∫0∞μv(x)(λx)j−l(j−l)!e−∫0x(iβ+θ+λ+μv(ξ))dξdx|≤∑j=l+1∞∫0∞μv(x)(λx)j−1(j−1)!|e−iβx||e−∫0x(θ+λ+μv(ξ))dξ|dx=∑j=l+1∞∫0∞μv(x)(λx)j−1(j−1)!e−∫0x(θ+λ+μv(ξ))dξdx,l≥1,∑j=l+1∞|∫0∞μb(x)(λx)j−l(j−l)!e−∫0x(iβ+λ+μb(ξ))dξdx|≤∑j=l+1∞∫0∞μb(x)(λx)j−1(j−1)!|e−iβx||e−∫0x(λ+μb(ξ))dξ|dx=∑j=l+1∞∫0∞μb(x)(λx)j−1(j−1)!e−∫0x(λ+μb(ξ))dξdx,l≥1,$

and using the fact $\begin{array}{}{\int }_{0}^{\mathrm{\infty }}{\mu }_{b}\left(x\right){e}^{-{\int }_{0}^{x}{\mu }_{b}\left(\xi \right)d\xi }dx=1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\int }_{0}^{\mathrm{\infty }}\left(\theta +{\mu }_{v}\left(x\right)\right){e}^{-{\int }_{0}^{x}\left(\theta +{\mu }_{v}\left(\xi \right)\right)d\xi }dx=1,\end{array}$ we estimate for a⃗v = (a1,v, a2,v, a3,v,⋯) ∈ l1 and a⃗b = (a1,b, a2,b, a3,b,⋯) ∈ l1,

$‖ΦDγ(av→,ab→)‖≤|λγ+θ+λϕvh00+λθ(γ+λ)(γ+θ+λ)ϕvh00+ϕvh21||a1,v|+∑j=3∞|ϕvhj1||a1,v|+∑j=2∞|ϕvhj2||a2,v|+∑j=3∞|ϕvhj3||a3,v|+⋯+|λγ+θ+λϕbm00+λθ(γ+λ)(γ+θ+λ)ϕbm00+ϕbm21||a1,b|+∑j=3∞|ϕbmj1||a1,b|+∑j=2∞|ϕbmj2||a2,b|+∑j=3∞|ϕbmj3||a3,b|+⋯+∑j=1∞|θFhj1||a1,v|+∑j=2∞|θFhj2||a2,v|+∑j=3∞|θFhj3||a3,v|+⋯≤λ|γ+λ||ϕvh00||a1,v|+∑j=2∞|ϕvhj1||a1,v|+∑j=2∞|ϕvhj2||a2,v|+∑j=3∞|ϕvhj3||a3,v|+⋯+λ|γ+λ||ϕbm00||a1,b|+∑j=2∞|ϕbmj1||a1,b|+∑j=2∞|ϕbmj2||a2,b|+∑j=3∞|ϕbmj3||a3,b|+⋯+θ{∑j=1∞|Fhj1||a1,v|+∑j=2∞|Fhj2||a2,v|+∑j=3∞|Fhj3||a3,v|+⋯}≤∑j=1∞|∫0∞μv(x)(λx)j−1(j−1)!e−∫0x(iβ+θ+λ+μv(ξ))dξdx||a1,v|+∑j=2∞|∫0∞μv(x)(λx)j−2(j−2)!e−∫0x(iβ+θ+λ+μv(ξ))dξdx||a2,v|+∑j=3∞|∫0∞μv(x)(λx)j−3(j−3)!e−∫0x(iβ+θ+λ+μv(ξ))dξdx||a3,v|+⋯+∑j=1∞|∫0∞μb(x)(λx)j−1(j−1)!e−∫0x(iβ+λ+μb(ξ))dξdx||a1,b|$

$+∑j=2∞|∫0∞μb(x)(λx)j−2(j−2)!e−∫0x(iβ+λ+μb(ξ))dξdx||a2,b|+∑j=3∞|∫0∞μb(x)(λx)j−3(j−3)!e−∫0x(iβ+λ+μb(ξ))dξdx||a3,b|+⋯+θ{∑j=1∞|∫0∞(λx)j−1(j−1)!e−∫0x(iβ+θ+λ+μv(ξ))dξdx||a1,v|+∑j=2∞|∫0∞(λx)j−2(j−2)!e−∫0x(iβ+θ+λ+μv(ξ))dξdx||a2,v|+∑j=3∞|∫0∞(λx)j−3(j−3)!e−∫0x(iβ+θ+λ+μv(ξ))dξdx||a3,v|+⋯}<∫0∞μv(x)∑j=1∞(λx)j−1(j−1)!e−∫0x(θ+λ+μv(ξ))dξdx|a1,v|+∫0∞μv(x)∑j=2∞(λx)j−2(j−2)!e−∫0x(θ+λ+μv(ξ))dξdx|a2,v|+∫0∞μv(x)∑j=3∞(λx)j−3(j−3)!e−∫0x(θ+λ+μv(ξ))dξdx|a3,v|+⋯+∫0∞μb(x)∑j=1∞(λx)j−1(j−1)!e−∫0x(λ+μb(ξ))dξdx|a1,b|+∫0∞μb(x)∑j=2∞(λx)j−2(j−2)!e−∫0x(λ+μb(ξ))dξdx|a2,b|+∫0∞μb(x)∑j=3∞(λx)j−3(j−3)!e−∫0x(λ+μb(ξ))dξdx|a3,b|+⋯+θ{∫0∞∑j=1∞(λx)j−1(j−1)!e−∫0x(θ+λ+μv(ξ))dξ|a1,v|+∫0∞∑j=2∞(λx)j−2(j−2)!e−∫0x(θ+λ+μv(ξ))dξ|a2,v|+∫0∞∑j=3∞(λx)j−3(j−3)!e−∫0x(θ+λ+μv(ξ))dξ|a3,v|+⋯}=∫0∞μv(x)e−∫0x(θ+μv(ξ))dξdx∑n=1∞|an,v|+∫0∞μb(x)e−∫0xμb(ξ)dξdx∑n=1∞|an,b|+θ∫0∞e−∫0x(θ+μv(ξ))dξ∑n=1∞|an,v|=∫0∞(θ+μv(x))e−∫0x(θ+μv(ξ))dξdx∑n=1∞|an,v|+∑n=1∞|an,b|=∑n=1∞|an,v|+∑n=1∞|an,b|=‖(av→,ab→)‖⟹‖ΦDγ‖<1.$(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

$iβ∈σ(A)⇒iβh∈σ(A),all positive integerh.$

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.

$X∗×Y∗={(qv∗,qb∗)|qv∗∈X∗,qb∗∈Y∗,‖|(qv∗,qb∗)‖|=sup{‖|qv∗‖|,‖|qb∗‖|}},$

here

$X∗={qv∗|qv∗(x)=(q0,v∗,q1,v∗(x),q2,v∗(x),q3,v∗(x),⋯),‖|y∗‖|=sup{|q0,v∗|,supn≥1‖qn,v∗‖L∞[0,∞)}<∞},Y∗={qb∗|qb∗(x)=(q0,b∗,q1,b∗(x),q2,b∗,q3,b∗(x),⋯)‖|qb∗‖|=sup{|q0,b∗|,supn≥1‖qn,b∗‖L∞[0,∞)}<∞}.$

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

#### Lemma 3.6

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

$A∗(qv∗,qb∗)=((−(θ+λ)000⋯0ddx−(θ+λ+μv(x))λ0⋯00ddx−(θ+λ+μv(x))λ⋯⋮⋮⋮⋮⋱)(q0,v∗q1,v∗(x)q2,v∗(x)⋮)+(0λ0⋯μv(x)00⋯0μv(x)0⋯00μv(x)⋯⋮⋮⋮⋱)(q0,v∗q1,v∗(0)q2,v∗(0)q3,v∗(0)⋮)+(θ000⋯0θ00⋯00θ0⋯000θ⋯⋮⋮⋮⋮⋱)(q0,b∗q1,b∗(0)q2,b∗(0)q3,b∗(0)⋮),(−λ000⋯0ddx−(λ+μb(x))λ0⋯00ddx−(λ+μb(x))λ⋯⋮⋮⋮⋮⋱)(q0,b∗q1,b∗(x)q2,b∗(x)q3,b∗(x)⋮)+(0λ00⋯0000⋯0μb(x)00⋯00μb(x)0⋯⋮⋮⋮⋮⋱)(q0,b∗q1,b∗(0)q2,b∗(0)q3,b∗(0)⋮)+(000⋯μv(x)00⋯000⋯⋮⋮⋮⋱)(q0,v∗q1,v∗(0)q2,v∗(0)⋮)),D(A∗)={(qv∗,qb∗)∈X∗×Y∗|dqn,v∗(x)dxanddqn,b∗(x)dxexistandqn,v∗(∞)=qn,b∗(∞)=α,n≥1},$

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 $\begin{array}{}\left({q}_{v}^{\ast },{q}_{b}^{\ast }\right)\end{array}$D(A*)

$〈A(pv,pb),(qv∗,qb∗)〉={−(θ+λ)p0,v+∫0∞μv(x)p1,v(x)dx+∫0∞μb(x)p1,b(x)dx}q0,v∗+∫0∞{−dp1,v(x)dx−(λ+μv(x))p1,v(x)}q1,v∗(x)dx+∑n=2∞∫0∞{−dpn,v(x)dx−(λ+μv(x))pn,v(x)+λpn−1,v(x)}qn,v∗(x)dx+{−λp0,b+θp0,v}q0,b∗+∫0∞{−dp1,b(x)dx−(λ+μb(x))p1,b(x)}q1,b∗(x)dx+∑n=2∞∫0∞{−dpn,b(x)dx−(λ+μb(x))pn,b(x)+λpn−1,b(x)}qn,b∗(x)dx=−(θ+λ)p0,vq0,v∗+∫0∞p1,v(x)μv(x)q0,v∗dx+∫0∞p1,b(x)μb(x)q0,v∗dx+∑n=1∞∫0∞−dpn,v(x)dxqn,v∗(x)dx−∑n=1∞∫0∞(λ+μv(x))pn,v(x)qn,v∗(x)dx+∑n=2∞∫0∞pn−1,v(x)λqn,v∗(x)dx+p0,b(−λq0,b∗)+p0,vθq0,b∗+∑n=1∞∫0∞−dpn,b(x)dxqn,b∗(x)dx−∑n=1∞∫0∞pn,b(x)(λ+μb(x))qn,b∗(x)dx+∑n=2∞∫0∞pn−1,b(x)λqn,b∗(x)dx=p0,v[−(θ+λ)q0,v∗+θq0,b∗]+∫0∞p1,v(x)μv(x)q0,v∗dx+∫0∞p1,b(x)μb(x)q0,v∗dx+∑n=1∞[−pn,v(x)qn,v∗(x)|x=0x=∞+∫0∞pn,v(x)dqn,v∗(x)dxdx]+∑n=1∞∫0∞pn,v(x)[−(λ+μv(x))qn,v∗(x)]dx+∑n=1∞∫0∞pn,b(x)[λqn+1,v∗(x)]dx+p0,b(−λq0,b∗)+∑n=1∞[−pn,b(x)qn,b∗(x)(x)|x=0x=∞+∫0∞pn,b(x)dqn,b∗(x)(x)dx]+∑n=1∞∫0∞pn,b(x)[−(λ+μb(x))qn,b∗(x)(x)]dx+∑n=1∞∫0∞pn,b(x)[λqn+1,b∗(x)(x)]dx=p0,v[−(θ+λ)q0,v∗+θq0,b∗]+∫0∞p1,v(x)μv(x)q0,v∗dx+∫0∞p1,b(x)μb(x)q0,v∗dx+∑n=1∞pn,v(0)qn,v∗(0)+∑n=1∞∫0∞pn,v(x)dqn,v∗(x)dxdx+∑n=1∞∫0∞pn,v(x)[−(λ+μv(x))qn,v∗(x)]dx+∑n=1∞∫0∞pn,v(x)[λqn+1,v∗(x)]dx+p0,b(−λq0,b∗)+∑n=1∞pn,b(0)qn,b∗(0)+∑n=1∞∫0∞pn,b(x)dqn,b∗(x)dxdx+∑n=1∞∫0∞pn,b(x)[−(λ+μb(x))qn,b∗(x)]dx+∑n=1∞∫0∞pn,b(x)[λqn+1,b∗(x)]dx=p0,v[−(θ+λ)q0,v∗+θq0,b∗]+∫0∞p1,v(x)μv(x)q0,v∗dx+∫0∞p1,b(x)μb(x)q0,v∗dx+p0,vλq1,v∗(0)+∑n=1∞∫0∞μv(x)pn+1,v(x)qn,v∗(0)dx+∑n=1∞∫0∞pn,v(x)[dqn,v∗(x)dx−(λ+μv(x))qn,v∗(x)+λqn+1,v∗(x)]dx+p0,b[−λq0,b∗+λq1,b∗(0)]+∑n=1∞[∫0∞μb(x)pn+1,b(x)dx+θ∫0∞pn,v(x)dx]qn,b∗(0)+∑n=1∞∫0∞pn,b(x)[dqn,b∗(x)dx−(λ+μb(x))qn,b∗(x)+λqn+1,b∗(x)]dx=p0,v[−(θ+λ)q0,v∗+λq1,v∗(0)+θq0,b∗]+∫0∞p1,v(x)[dq1,v∗(x)dx−(λ+μv(x))q1,v∗(x)+λq2,v∗(x)+μv(x)q0,v∗+θq1,b∗(0)]dx+∑n=2∞∫0∞pn,v(x)[dqn,v∗(x)dx−(λ+μv(x))qn,v∗(x)+λqn+1,v∗(x)+μv(x)qn−1,v∗(0)+θqn,b∗(0)]dx+p0,b[−λq0,b∗+λq1,b∗(0)]+∫0∞p1,b(x)[dq1,b∗(x)dx−(λ+μb(x))q1,b∗(x)+λq2,b∗(x)+μb(x)q0,v∗]dx+∑n=2∞∫0∞pn,b(x)[dqn,b∗(x)dx−(λ+μb(x))qn,b∗(x)+λqn+1,b∗(x)+μb(x)qn−1,b∗(0)]dx=〈(pv,pb),A∗(qv∗,qb∗)〉.$

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 $\begin{array}{}{\int }_{0}^{\mathrm{\infty }}\lambda x{\mu }_{b}\left(x\right){e}^{-{\int }_{0}^{x}{\mu }_{b}\left(\tau \right)d\tau }dx<1,\end{array}$ 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.,

$limt→∞πj(t)=limt→∞{K0∫0∞μv(x)pj+1,v(x,t)dx+K0∫0∞μb(x)pj+1,b(x,t)dx}={K0∫0∞μv(x)pj+1,v(x)dx+K0∫0∞μb(x)pj+1,b(x)dx}=πj,j≥0.$

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

$limt→∞L(t)=limt→∞{∑n=1∞n∫0∞pn,v(x,t)dx+∑n=1∞n∫0∞pn,b(x,t)dx}=∑n=1∞n∫0∞pn,v(x)dx+∑n=1∞n∫0∞pn,b(x)dx=L.$

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

$∥Evf∥L1[0,∞)=∫0∞|Evf(x)|dx=∫0∞e−∫0x(γ+θ+λ+μv(ξ))dξ∫0xf(τ)e∫0τ(γ+θ+λ+μv(ξ))dξdτdx≤∫0∞e−∫0x(Reγ+θ+λ+μv(ξ))dξ∫0x|f(τ)|e∫0τ(Reγ+θ+λ+μv(ξ))dξdτdx=∫0∞−1Reγ+θ+λ+μv(x)∫0x|f(τ)|e∫0τ(Reγ+θ+λ+μv(ξ))dξdτde−∫0x(Reγ+θ+λ+μv(ξ))dξ≤∫0∞−1Reγ+θ+λ+μv_∫0x|f(τ)|e∫0τ(Reγ+θ+λ+μv(ξ))dξdτde−∫0x(Reγ+θ+λ+μv(ξ))dξ=−1Reγ+θ+λ+μv_{e−∫0x(Reγ+λ+μv(ξ))dξ∫0x|f(τ)|e∫0τ(Reγ+θ+λ+μv(ξ))dξdτ|x=0x=∞−∫0∞|f(x)|e∫0x(Reγ+λ+μv(ξ))dξe−∫0x(Reγ+θ+λ+μv(ξ))dξdx}=1Reγ+θ+λ+μv_∥f∥L1[0,∞)⟹∥Ev∥≤1Reγ+θ+λ+μv_.$(A.1)

$∥Ebf∥=∫0∞|Ebf(x)|dx=∫0∞e−∫0x(γ+λ+μb(ξ))dξ∫0xf(τ)e∫0τ(γ+λ+μb(ξ))dξdτdx≤1Reγ+λ+μb_∥f∥L1[0,∞)⟹∥Eb∥≤1Reγ+λ+μb_.$(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

$(γI−A0)−1(y,z)=|1γ+θ+λy0+1γ+θ+λϕvEvy1+1γ+θ+λϕbEbz1|+∥Evy1∥L1[0,∞)+λEv2y1+Evy2L1[0,∞)+λ2Ev3y1+λEv2y2+Evy3L1[0,∞)+∑k=03λkEvk+1y4−kL1[0,∞)+⋯+∑k=0n−1λkEvk+1yn−kL1[0,∞)+⋯+|1γ+λz0+θ(γ+λ)(γ+θ+λ)ϕbEbz1+θ(γ+λ)(γ+θ+λ)y0+θ(γ+λ)(γ+θ+λ)ϕvEvy1|+∥Ebz1∥L1[0,∞)+λEb2z1+Ebz2L1[0,∞)+λ2Eb3z1+λEb2z2+Ebz3L1[0,∞)+∑k=03λkEbk+1z4−kL1[0,∞)+⋯+∑k=0n−1λkEbk+1zn−kL1[0,∞)+⋯≤1|γ+θ+λ||y0|+1|γ+θ+λ|∥ϕv∥∥Ev∥∥y1∥L1[0,∞)+1|γ+θ+λ|∥ϕb∥∥Eb∥∥z1∥L1[0,∞)+∑n=1∞∑k=0n−1λk∥Ev∥k+1∥yn−k∥L1[0,∞)+1|γ+λ|z0+θ|γ+λ||γ+θ+λ|∥ϕb∥∥Eb∥∥z1∥L1[0,∞)+θ|γ+λ||γ+θ+λ||y0|+θ|γ+λ||γ+θ+λ|∥ϕv∥∥Ev∥∥y1∥L1[0,∞)+∑n=1∞∑k=0n−1λk∥Eb∥k+1∥zn−k∥L1[0,∞)≤1|γ+θ+λ||y0|+μv¯|γ+θ+λ|(Reγ+θ+λ+μv_)∥y1∥L1[0,∞)+μb¯|γ+θ+λ|(Reγ+λ+μb_)∥z1∥L1[0,∞)+1Reγ+θ+λ+μv_∑k=0∞λReγ+θ+λ+μv_k∑n=1∞∥yn∥L1[0,∞)+1|γ+λ|z0+μb¯|γ+λ|(Reγ+λ+μb_)∥z1∥L1[0,∞)+1|γ+λ||y0|+μv¯|γ+λ|(Reγ+θ+λ+μv_)∥y1∥L1[0,∞)+1Reγ+λ+μb_∑k=0∞λReγ+λ+μb_k∑n=1∞∥zn∥L1[0,∞)≤sup{1|γ+λ|,μv¯|γ+λ|(Reγ+θ+λ+μv_)+1Reγ+θ+μv_}|y0|+∑n=1∞∥yn∥L1[0,∞)+sup{1|γ+λ|,μb¯|γ+λ|(Reγ+λ+μb_)+1Reγ+μb_}|z0|+∑n=1∞∥zn∥L1[0,∞)≤sup{1|γ+λ|,μv¯|γ+λ|(Reγ+θ+λ+μv_)+1Reγ+θ+μv_,+μb¯|γ+λ|(Reγ+λ+μb_)+1Reγ+μb_}∥(y,z)∥<∞.$

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

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

$dp1,v(x)dx=−(γ+θ+λ+μv(x))p1,v(x),$(A.4)

$dpn,v(x)dx=−(γ+θ+λ+μv(x))pn,v(x)+λpn−1,v(x),n≥2,$(A.5)

$(γ+λ)p0,b=θp0,v,$(A.6)

$dp1,b(x)dx=−(γ+λ+μv(x))p1,b(x),$(A.7)

$dpn,b(x)dx=−(γ+λ+μv(x))pn,b(x)+λpn−1,b(x),n≥2.$(A.8)

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

$p1,v(x)=a1,ve−∫0x(γ+θ+λ+μv(ξ))dξ,pn,v(x)=an,ve−∫0x(γ+θ+λ+μv(ξ))dξ$(A.9)

$+λe−∫0x(γ+θ+λ+μv(ξ))dξ∫0xpn−1,v(τ)e∫0τ(γ+θ+λ+μv(τ))dξdτ,n≥2,$(A.10)

$p1,b(x)=a1,be−∫0x(γ+λ+μb(ξ))dξ,pn,b(x)=an,be−∫0x(γ+λ+μb(ξ))dξ$(A.11)

$+λe−∫0x(γ+λ+μb(ξ))dξ∫0xpn−1,b(τ)e∫0τ(γ+λ+μb(ξ))dξdτ,n≥2.$(A.12)

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

$p2,v(x)=a2,ve−∫0x(γ+θ+λ+μv(ξ))dξ+λe−∫0x(γ+θ+λ+μv(ξ))dξ∫0xa1,vdτ=e−∫0x(γ+θ+λ+μv(ξ))dξ[a2,v+λxa1,v],p3,v(x)=a3,ve−∫0x(γ+θ+λ+μv(ξ))dξ+λe−∫0x(γ+θ+λ+μv(ξ))dξ∫0x[a2,v+λτa1,v]dτ=e−∫0x(γ+θ+λ+μv(ξ))dξa3,v+λxa2,v+(λx)22a1,v,⋯⋯pn,v(x)=e−∫0x(γ+θ+λ+μv(ξ))dξan,v+λxan−1,v+(λx)22an−2,v+⋯+(λx)nn!a1,v=e−∫0x(γ+θ+λ+μv(ξ))dξ∑k=0n−1(λx)kk!an−k,v,n≥1.$(A.13)

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

$pn,b(x)=e−∫0x(γ+λ+μb(ξ))dξ∑k=0n−1(λx)kk!an−k,b,n≥1.$(A.14)

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

$p0,v=1γ+θ+λa1,v∫0∞μv(x)e−∫0x(γ+θ+λ+μv(ξ))dξdx+1γ+θ+λa1,b∫0∞μb(x)e−∫0x(γ+λ+μb(ξ))dξdx,p0,b=θ(γ+λ)(γ+θ+λ)a1,v∫0∞μv(x)e−∫0x(γ+θ+λ+μv(ξ))dξdx$(A.15)

$+θ(γ+λ)(γ+θ+λ)a1,b∫0∞μb(x)e−∫0x(γ+λ+μb(ξ))dξdx.$(A.16)

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

$∑n=1∞|an,v|=∑n=1∞|pn,v(0)|≤∑n=1∞∥pn,v∥L∞[0,∞)≤∑n=1∞∥pn,v∥L1[0,∞)+∑n=1∞dpn,vdxL1[0,∞)<∞,∑n=1∞|an,b|=∑n=1∞|pn,b(0)|≤∑n=1∞∥pn,b∥L∞[0,∞)≤∑n=1∞∥pn,b∥L1[0,∞)+∑n=1∞dpn,bdxL1[0,∞)<∞.$

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

Conversely, if (2.55) is right, then by using $\begin{array}{}{\int }_{0}^{\mathrm{\infty }}{x}^{k}{e}^{-Mx}dx=\frac{k!}{{M}^{k+1}},\phantom{\rule{1em}{0ex}}k\ge 1,\phantom{\rule{thickmathspace}{0ex}}M>0,\end{array}$ integration by parts and the Fubini theorem we estimate

$∥pn,v∥L1[0,∞)=∫0∞e−∫0x(γ+θ+λ+μv(ξ))dξ∑k=0n−1(λx)kk!an−k,vdx≤∑k=0n−1|an−k,0|λkk!∫0∞xke−Reγ+θ+λ+μv_xdx=∑k=0n−1λk(Reγ+θ+λ+μv_)k+1|an−k,v|⟹|p0,v|+∑n=1∞∥pn,v∥L1[0,∞)=μv¯|γ+θ+λ|(Reγ+θ+λ+μv_)|a1,v|+μb¯|γ+θ+λ|(Reγ+λ+μb_)|a1,b|+1Reγ+θ+μv_∑n=1∞|an,v|<∞.$(A.17)

Similarly, we get

$|p0,b|+∑n=1∞∥pn,b∥L1[0,∞)=μb¯|γ+θ|(Reγ+θ+λ+μb_)|a1,v|+μb¯|γ+λ|(Reγ+λ+μb_)|a1,b|+1Reγ+θ+μb_∑n=1∞|an,b|<∞.$(A.18)

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

$|p0,v|+|p0,b|+∑n=1∞∥pn,v∥L1[0,∞)+∑n=1∞∥pn,b∥L1[0,∞)<∞.$

Since

$dp1,v(x)dx=−(γ+θ+λ+μv(x))p1,vdpn,v(x)dx=−(γ+θ+λ+μv(x))pn,v(x)+λpn−1,v(x),n≥2,dp1,b(x)dx=−(γ+λ+μb(x))p1,b(x),dpn,b(x)dx=−(γ+λ+μb(x))pn,b(x)+λpn−1,b(x),n≥2.$

It is immediately obtained

$∑n=1∞dpn,vdxL1[0,∞)≤(Reγ+θ+λ+μv¯)∑n=0∞∥pn,v∥L1[0,∞)+λ∑n=2∞∥pn−1,v∥L1[0,∞)≤Reγ+θ+λ+μv¯Reγ+θ+μv_+λReγ+θ+μv_∑n=1∞|an,v|<∞,$(A.19)

$∑n=1∞dpn,bdxL1[0,∞)≤(Reγ+λ+μb¯)∑n=1∞∥pn,b∥L1[0,∞)+λ∑n=2∞∥pn−1,b∥L1[0,∞)≤Reγ+λ+μb¯Reγ+μb_+λReγ+μb_∑n=0∞|an,b|<∞.$(A.20)

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

#### Proof of Lemma 3.7

We consider the equation $\begin{array}{}{A}^{\ast }\left({q}_{v}^{\ast },{q}_{b}^{\ast }\right)=0,\end{array}$ which is equivalent to

$−(θ+λ)q0,v∗+λq1,v∗(0)+θq0,b∗=0,dq1,v∗(x)dx−(λ+μv(x))q1,v∗(x)+λq2,v∗(x)+μv(x)q0,v∗+θq1,b∗(0)=0,dqn,v∗(x)dx−(λ+μv(x))qn,v∗(x)+λqn+1,v∗(x)+μv(x)qn−1,v∗(0)+θqn,b∗(0)=0,n≥2,−λq0,b∗+λq1,b∗(0)=0,dq1,b∗(x)dx−(λ+μb(x))q1,b∗(x)+λq2,b∗(x)+μb(x)q0,v∗=0,dqn,b∗(x)dx−(λ+μb(x))qn,b∗(x)+λqn+1,b∗(x)+μb(x)qn−1,b∗=0,n≥2,qn,v∗(∞)=qn,b∗(∞)=α,n≥1.$(A.21)

It is easy to see that

$(qv∗,qb∗)=αα⋮,αα⋮∈D(A∗)$

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

$q0,v∗=λθ+λq1,v∗(0)+θθ+λq0,b∗,q2,v∗(x)=1λ{−dq1,v∗(x)dx+(λ+μv(x))q1,v∗(x)−μv(x)q0,v∗−θq1,b∗(0)},qn+1,0∗(x)=1λ{−dqn,v∗(x)dx+(λ+μv(x))qn,v∗(x)−μv(x)qn−1,v∗(0)−θqn,b∗(0)},n≥2,q0,b∗=q1,b∗(0),q2,b∗(x)=1λ{−dq1,b∗(x)dx+(λ+μb(x))q1,b∗(x)−μb(x)q0,v∗},qn+1,b∗(x)=1λ{−dqn,b∗(x)dx+(λ+μb(x))qn,b∗(x)−μb(x)qn−1,b∗},n≥2.$(A.22)

(A.22) show that we can determine each $\begin{array}{}{q}_{n,v}^{\ast }\end{array}$ (x) and $\begin{array}{}{q}_{n,b}^{\ast }\end{array}$ (x) for all n ≥ 1 if $\begin{array}{}{q}_{1,v}^{\ast }\end{array}$ (x) and $\begin{array}{}{q}_{1,b}^{\ast }\end{array}$ (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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Accepted: 2018-05-24

Published Online: 2018-07-26

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 767–791, ISSN (Online) 2391-5455,

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