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Journal of Applied Mathematics
Volume 2012, Article ID 890243, 16 pages
http://dx.doi.org/10.1155/2012/890243
Research Article

Further Research on the M/G/1 Retrial Queueing Model with Server Breakdowns

College of Mathematics and Systems Science, Xinjiang University, Urumqi 830046, China

Received 13 March 2012; Accepted 28 June 2012

Academic Editor: Song Cen

Copyright © 2012 Ehmet Kasim and Geni Gupur. 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.

Abstract

We study spectral properties of the operator which corresponds to the M/G/1 retrial queueing model with server breakdowns and obtain that all points on the imaginary axis except zero belong to the resolvent set of the operator and 0 is not an eigenvalue of the operator. Our results show that the time-dependent solution of the model is probably strongly asymptotically stable.

1. Introduction

There has been considerable interest in retrial queueing systems, see Atencia et al. [1, 2], Choi et al. [3], Djellab [4], Gupur [5, 6], Kasim and Gupur [7], Li et al. [8], Li and Wang [9], Wang et al. [10, 11], and Yang and Templeton [12]. Many researchers studied the M/G/1 retrial queueing systems with server breakdowns in the steady-state case, see Atencia et al. [1, 2], Choi et al. [3], Djellab [4], Li and Wang [9], and Wang et al. [10]. Only few researchers studied transient solutions of M/G/1 retrial queueing systems, see Wang et al. [11], Gupur [5, 6], Kasim and Gupur [7]. And Wang et al. [11] studied the transient solution of the M/G/1 retrial queueing system with server failure by using Laplace transform and obtained the expression of the probability-generating function. In other words, they studied the existence of the time-dependent solution of the model. Gupur [5, 6, 13, 14], Gupur et al. [15], and Kasim and Gupur [7] did dynamic analysis for several queueing models including retrial queueing models by using functional analysis and obtained the existence and uniqueness of the time-dependent solution of several queueing models and asymptotic behavior of their time-dependent solutions. In this paper, by using the idea in Gupur [6] and Gupur et al. [15], we study the asymptotic behavior of the time-dependent solution of the M/G/1 retrial queueing system with server breakdowns in which the failure states of the server are absorbing states. This queueing system was studied by Wang et al. [10] in 2001. By using the supplementary variable technique they established the corresponding queueing model and obtained explicit expressions of some reliability indices such as the availability, failure frequency for steady-state cases under the following hypothesis: “the time-dependent solution of the model converges to a steady-state solution." By reading their paper we find that the above hypothesis, in fact, implies the following two hypotheses.

Hypothesis 1. The model has a nonnegative time-dependent solution.

Hypothesis 2. The time-dependent solution of the model converges to a steady-state solution.

In 2010, Gupur [6] studied the above two hypotheses. Firstly, he converted the model into an abstract Cauchy problem by selecting a suitable Banach space and defining the underlying operator which corresponds to the model and its domain. Next, by using the Hille-Yosida theorem and the Phillips theorem he proved that the model has a unique nonnegative time-dependent solution and therefore obtained that Hypothesis 1 holds. Then, when the service completion rate is a constant, he studied the asymptotic behavior of its time-dependent solution. By studying resolvent set of of the adjoint operator of the operator which corresponds to the model (in this case, the M/G/1 retrial queueing model with server breakdowns is called the M/M/1 retrial queueing model with server breakdowns) he obtained the resolvent set of the operator on the imaginary axis: all points on the imaginary axis except zero belong to its resolvent set. And he proved that 0 is not an eigenvalue of the operator. Thus, he suggested that the time-dependent solution of the model probably strongly converges to zero. Until now, any other results about this model have not been found in the literature. In this paper, we try to study the asymptotic behavior of the time-dependent solution of the above model when the service completion rate is a function. Firstly, we convert the model into an abstract Cauchy problem by Gupur [6] and we determine the resolvent set of the adjoint operator of the operator corresponding to the model when the service completion rate satisfies a certain condition. Also, we prove that 0 is not an eigenvalue of the operator. If we can verify 0 is not in the residue spectrum of the operator, then from the above results we conclude that the time-dependent solution of the model strongly converges to zero. Naturally, the results obtained in [6] are now special cases of our results.

According to Wang et al. [10], the M/G/1 retrial queueing system with server breakdowns can be described by the following system of partial differential equations with integral boundary conditions: 𝑑𝑝𝐼,0,0(𝑡)𝑑𝑡=𝜆𝑝𝐼,0,0(𝑡)+0𝜇(𝑥)𝑝𝑤,0,1(𝑥,𝑡)𝑑𝑥,𝑑𝑝𝐼,𝑖,0(𝑡)𝑑𝑡=(𝜆+𝑖𝜃)𝑝𝐼,𝑖,0(𝑡)+0𝜇(𝑥)𝑝𝑤,𝑖,1(𝑥,𝑡)𝑑𝑥,𝑖1,𝜕𝑝𝑤,0,1(𝑥,𝑡)+𝜕𝑡𝜕𝑝𝑤,0,1(𝑥,𝑡)𝜕𝑥=(𝜆+𝛼+𝜇(𝑥))𝑝𝑤,0,1(𝑥,𝑡),𝜕𝑝𝑤,𝑖,1(𝑥,𝑡)+𝜕𝑡𝜕𝑝𝑤,𝑖,1(𝑥,𝑡)𝜕𝑥=(𝜆+𝛼+𝜇(𝑥))𝑝𝑤,𝑖,1(𝑥,𝑡)+𝜆𝑝𝑤,𝑖1,1𝑝(𝑥,𝑡),𝑖1,𝑤,𝑖,1(0,𝑡)=𝜆𝑝𝐼,𝑖,0(𝑡)+(𝑖+1)𝜃𝑝𝐼,𝑖+1,0𝑝(𝑡),𝑖0,𝐼,0,0(0)=1,𝑝𝐼,𝑖,0(0)=0,𝑖1;𝑝𝑤,𝑗,1(𝑥,0)=0,𝑗0.(1.1) Here (𝑥,𝑡)[0,)×[0,);  𝑝𝐼,𝑖,0(𝑡)(𝑖0) represents the probability that the server is idle and there are 𝑖 customers in the retrial group at time 𝑡; 𝑝𝑤,𝑖,1(𝑥,𝑡) represents the joint probability that at time 𝑡 there are 𝑖 customers in the retrial group and the server is up and customer is being served with elapsed service time 𝑥;  𝜆 is the arrival rate of customers; 𝛼 is the server failing rate; 𝜃 is the successive interretrial times of customers; 𝜇(𝑥) is the service completion rate at time 𝑥 satisfying 𝜇(𝑥)0,0𝜇(𝑥)𝑑𝑥=.(1.2) In this paper, we use the notations in Gupur [6]: Γ=𝜆𝜃0000𝜆2𝜃0000𝜆3𝜃0000𝜆4𝜃.(1.3) Take a state space as follows: 𝑝𝑋=𝐼,𝑝𝑤|||||||||||||𝑝𝐼=𝑝𝐼,0,0,𝑝𝐼,1,0,𝑝𝐼,2,0,𝑙1𝑝𝑤=𝑝𝑤,0,1,𝑝𝑤,1,1,𝑝𝑤,2,1,𝐿1[0,)×𝐿1[0,)×𝐿1[𝑝0,)×𝐼,𝑝𝑤=𝑖=0||𝑝𝐼,𝑖,0||+𝑖=0𝑝𝑤,𝑖,1𝐿1[0,)<.(1.4) It is obvious that 𝑋 is a Banach space and also a Banach lattice. In the following we define operators and their domains: 𝐴𝑝𝐼,𝑝𝑤=𝑝𝜆0000(𝜆+𝜃)0000(𝜆+2𝜃)0000(𝜆+3𝜃)𝐼,0,0𝑝𝐼,1,0𝑝𝐼,2,0𝑝𝐼,3,0,𝑑𝑑𝑑𝑥0000𝑑𝑑𝑥0000𝑑𝑑𝑥0000𝑝𝑑𝑥𝑤,0,1𝑝(𝑥)𝑤,1,1(𝑝𝑥)𝑤,2,1𝑝(𝑥)𝑤,3,1,𝑝(𝑥)𝐷(𝐴)=𝐼,𝑝𝑤||||||||||𝑝𝑋𝑤,𝑖,1(𝑥)(𝑖0)areabsolutelycontinuousfunctionsandsatisfy𝑝𝑤(0)=Γ𝑝𝐼,𝑛=0𝑑𝑝𝑤,𝑖,1𝑑𝑥𝐿1[0,),𝑈𝑝<𝐼,𝑝𝑤=𝑝000000000𝐼,0,0𝑝𝐼,1,0𝑝𝐼,2,0,𝑝𝔲00𝜆𝔲00𝜆𝔲𝑤,0,1𝑝(𝑥)𝑤,1,1𝑝(𝑥)𝑤,2,1(𝑥),𝐷(𝑈)=𝑋,(1.5) where 𝔲=(𝜆+𝛼+𝜇(𝑥)). 𝐸𝑝𝐼,𝑝𝑤=0𝜇(𝑥)𝑝𝑤,0,1(𝑥)𝑑𝑥0𝜇(𝑥)𝑝𝑤,1,1(𝑥)𝑑𝑥0𝜇(𝑥)𝑝𝑤,2,1,000(𝑥)𝑑𝑥,𝐷(𝐸)=𝑋.(1.6) Then the above system of (1.1) can be expressed as an abstract Cauchy problem: 𝑑𝑝𝐼,𝑝𝑤(𝑡)𝑝𝑑𝑡=(𝐴+𝑈+𝐸)𝐼,𝑝𝑤𝑝(𝑡),𝑡(0,),𝐼,𝑝𝑤(10,00.0)=(1.7)

Gupur [6] have obtained the following results.

Theorem 1.1. If 𝜇=sup𝑥[0,)𝜇(𝑥)<, then 𝐴+𝑈+𝐸 generates a positive contraction 𝐶0-semi group  𝑇(𝑡). And the system (1.7) has a unique nonnegative time-dependent solution (𝑝𝐼,𝑝𝑤)(𝑥,𝑡)=𝑇(𝑡)(𝑝𝐼,𝑝𝑤)(0) satisfying (𝑝𝐼,𝑝𝑤)(,𝑡)1,𝑡[0,).

2. Asymptotic Behavior of the Time-Dependent Solution of the System (1.7)

In this section, firstly we determine the expression of (𝐴+𝑈+𝐸), the adjoint operator of 𝐴+𝑈+𝐸, next we study the resolvent set of (𝐴+𝑈+𝐸), through which we deduce the resolvent set of 𝐴+𝑈+𝐸 on the imaginary axis. Thirdly, we prove that 0 is not an eigenvalue of 𝐴+𝑈+𝐸. Thus, we state our main results in this paper.

It is easy to see that 𝑋, the dual space of 𝑋, is as follows (see Gupur [6]): 𝑋=𝑞𝐼,𝑞𝑤,|||||||||||||𝑞𝐼=𝑞𝐼,0,0,𝑞𝐼,1,0,𝑞𝐼,2,0,𝑙,𝑞𝑤=𝑞𝑤,0,1,𝑞𝑤,1,1,𝑞𝑤,2,1,𝐿[0,)×𝐿[0,)×𝐿[||𝑞0,)×𝐼,𝑄𝑤||=supsup𝑖0||𝑞𝐼,𝑖,0||,sup𝑖0𝑞𝑤,𝑖,1𝐿[0,)<.(2.1) It is not difficult to verify that 𝑋 is a Banach space.

Lemma 2.1. (𝐴+𝑈+𝐸), the adjoint operator of 𝐴+𝑈+𝐸, is as follows: (𝐴+𝑈+𝐸)𝑞𝐼,𝑞𝑤=𝑞(+𝒢+)𝐼,𝑞𝑤𝑞,𝐼,𝑞𝑤𝐷(),(2.2) where 𝑞𝐼,𝑞𝑤=𝑞𝜆000(𝜆+𝜃)000(𝜆+2𝜃)𝐼,0,0𝑞𝐼,1,0𝑞𝐼,2,0,𝑑0𝑑𝑑𝑥(𝜆+𝛼+𝜇(𝑥))0𝑞𝑑𝑥(𝜆+𝛼+𝜇(𝑥))𝑤,0,1𝑞(𝑥)𝑤,1,1,𝒢𝑞(𝑥)𝐼,𝑞𝑤=000,𝑞𝜇(𝑥)000𝜇(𝑥)000𝜇(𝑥)𝐼,0,0𝑞𝐼,1,0𝑞𝐼,2,0,𝑞𝐼,𝑞𝑤=𝑞𝜆000𝜃𝜆0002𝜃𝜆0003𝜃𝜆𝑤,0,1𝑞(0)𝑤,1,1𝑞(0)𝑤,2,1𝑞(0)𝑤,3,1(,𝑞0)0𝜆00000𝜆00000𝜆00000𝜆𝑤,0,1𝑞(𝑥)𝑤,1,1𝑞(𝑥)𝑤,2,1𝑞(𝑥)𝑤,3,1,𝑞(𝑥)𝐷()=𝐼,𝑞𝑤𝑋||||||𝑞𝑤,𝑖,1(𝑥)areabsolutelycontinuousandsatisfy𝑞𝑤,𝑖,1,()=𝜂,𝑖0𝐷(𝒢)=𝐷()=𝑋.(2.3) Here, 𝜂 in 𝐷() is a nonzero constant which is irrelevant to 𝑖.

Proof. By using integration by parts and the boundary conditions on (𝑝𝐼,𝑝𝑤)𝐷(𝐴), we have, for any (𝑞𝐼,𝑞𝑤)𝐷(), (𝑝𝐴+𝑈+𝐸)𝐼,𝑝𝑤,𝑞𝐼,𝑞𝑤=𝑖=0(𝜆+𝑖𝜃)𝑝𝐼,𝑖,0+0𝜇(𝑥)𝑝𝑤,𝑖,1(𝑞𝑥)𝑑𝑥𝐼,𝑖,0+0𝑑𝑝𝑤,0,1(𝑥)𝑑𝑥(𝜆+𝛼+𝜇(𝑥))𝑝𝑤,0,1𝑞(𝑥)𝑤,0,1+(𝑥)𝑑𝑥𝑖=10𝑑𝑝𝑤,𝑖,1(𝑥)𝑑𝑥(𝜆+𝛼+𝜇(𝑥))𝑝𝑤,𝑖,1(𝑥)+𝜆𝑝𝑤,𝑖1,1(𝑥)×𝑞𝑤,𝑖,1(=𝑥)𝑑𝑥𝑖=0(𝜆+𝑖𝜃)𝑝𝐼,𝑖,0𝑞𝐼,𝑖,0+𝑖=00𝜇(𝑥)𝑝𝑤,𝑖,1(𝑥)𝑞𝐼,𝑖,0+𝑑𝑥𝑖=00𝑑𝑝𝑤,𝑖,1(𝑥)𝑞𝑑𝑥𝑤,𝑖,1(𝑥)𝑑𝑥𝑖=00(𝜆+𝛼+𝜇(𝑥))𝑝𝑤,𝑖,1(𝑥)𝑞𝑤,𝑖,1+(𝑥)𝑑𝑥𝑖=10𝜆𝑝𝑤,𝑖1,1(𝑥)𝑞𝑤,𝑖,1=(𝑥)𝑑𝑥𝑖=0(𝜆+𝑖𝜃)𝑝𝐼,𝑖,0𝑞𝐼,𝑖,0+𝑖=00𝑝𝑤,𝑖,1(𝑥)𝜇(𝑥)𝑞𝐼,𝑖,0+𝑑𝑥𝑖=0𝑝𝑤,𝑖,1(𝑥)𝑞𝑤,𝑖,1||||||+(𝑥)𝑥=𝑥=00𝑝𝑤,𝑖,1(𝑥)𝑑𝑞𝑤,𝑖,1(𝑥)+𝑑𝑥𝑑𝑥𝑖=00𝑝𝑤,𝑖,1(𝑥)(𝜆+𝛼+𝜇(𝑥))𝑞𝑤,𝑖,1+(𝑥)𝑑𝑥𝑖=00𝑝𝑤,𝑖,1(𝑥)𝜆𝑞𝑤,𝑖+1,1(=𝑥)𝑑𝑥𝑖=0(𝜆+𝑖𝜃)𝑝𝐼,𝑖,0𝑞𝐼,𝑖,0+𝑖=00𝑝𝑤,𝑖,1(𝑥)𝜇(𝑥)𝑞𝐼,𝑖,0+𝑑𝑥𝑖=0𝑝𝑤,𝑖,1(0)𝑞𝑤,𝑖,1+(0)𝑖=00𝑝𝑤,𝑖,1(𝑥)𝑑𝑞𝑤,𝑖,1(𝑥)𝑑𝑥(𝜆+𝛼+𝜇(𝑥))𝑞𝑤,𝑖,1+(𝑥)𝑑𝑥𝑖=00𝑝𝑤,𝑖,1(𝑥)𝜆𝑞𝑤,𝑖+1,1=(𝑥)𝑑𝑥𝑖=0(𝜆+𝑖𝜃)𝑝𝐼,𝑖,0𝑞𝐼,𝑖,0+𝑖=00𝑝𝑤,𝑖,1(𝑥)𝜇(𝑥)𝑞𝐼,𝑖,0+𝑑𝑥𝑖=0𝜆𝑝𝐼,𝑖,0+(𝑖+1)𝜃𝑝𝐼,𝑖+1,0𝑞𝑤,𝑖,1+(0)𝑖=00𝑝𝑤,𝑖,1(𝑥)𝑑𝑞𝑤,𝑖,1(𝑥)𝑑𝑥(𝜆+𝛼+𝜇(𝑥))𝑞𝑤,𝑖,1+(𝑥)𝑑𝑥𝑖=00𝑝𝑤,𝑖,1(𝑥)𝜆𝑞𝑤,𝑖+1,1=(𝑥)𝑑𝑥𝑖=0(𝜆+𝑖𝜃)𝑝𝐼,𝑖,0𝑞𝐼,𝑖,0+𝑖=00𝑝𝑤,𝑖,1(𝑥)𝑑𝑞𝑤,𝑖,1(𝑥)𝑑𝑥(𝜆+𝛼+𝜇(𝑥))𝑞𝑤,𝑖,1(𝑥)+𝜆𝑞𝑤,𝑖+1,1(𝑥)+𝜇(𝑥)𝑞𝐼,𝑖,0+𝑑𝑥𝑖=0𝑝𝐼,𝑖,0𝜆𝑞𝑤,𝑖,1(0)+𝑖=0𝑝𝐼,𝑖+1,0(𝑖+1)𝜃𝑞𝑤,𝑖,1=𝑝(0)𝐼,𝑝𝑤,𝑞(+𝒢+)𝐼,𝑞𝑤.(2.4) From the definition of the adjoint operator and (2.4) we know that the assertion of this lemma is true.

Lemma 2.2. Assume that there exist two positive constant 𝜇,𝜇 such that 0<𝜇=inf[𝑥0,)𝜇(𝑥)𝜇=sup[𝑥0,)𝜇(𝑥)<.(2.5) If 𝜇<𝛼+𝜇, then ||||||||||||||𝜆𝛾sup||||𝛾+𝜆,sup𝑖1𝜆+𝑖𝜃||||,𝜆𝛾+𝜆+𝑖𝜃𝜇||||𝛾+𝜆Re𝛾+𝜆+𝛼+𝜇+𝜆Re𝛾+𝜆+𝛼+𝜇,sup𝑖1(𝜆+𝑖𝜃)𝜇Re𝛾+𝜆+𝛼+𝜇||||+𝜆𝛾+𝜆+𝑖𝜃Re𝛾+𝜆+𝛼+𝜇<1,Re𝛾+𝜆+𝛼+𝜇>0(2.6) belongs to the resolvent set of (𝐴+𝑈+𝐸). In particular, all points on the imaginary axis except zero belong to the resolvent set of (𝐴+𝑈+𝐸), which implies that all points on the imaginary axis except zero belong to the resolvent set of 𝐴+𝑈+𝐸.

Proof. For any given (𝑦𝐼,𝑦𝑤)𝑋, consider the equation (𝛾𝐼𝒢)(𝑞𝐼,𝑞𝑤)=(𝑦𝐼,𝑦𝑤), that is, (𝛾+𝜆)𝑞𝐼,0,0=𝜆𝑦𝑤,0,1(0),(2.7)(𝛾+𝜆+𝜃)𝑞𝐼,1,0=𝜃𝑦𝑤,0,1(0)+𝜆𝑦𝑤,1,1(0),(2.8)(𝛾+𝜆+𝑖𝜃)𝑞𝐼,𝑖,0=𝑖𝜃𝑦𝑤,𝑖1,1(0)+𝜆𝑦𝑤,𝑖,1(0),𝑖1,(2.9)𝑑𝑞𝑤,𝑖,1(𝑥)𝑑𝑥=(𝛾+𝜆+𝛼+𝜇(𝑥))𝑞𝑤,𝑖,1(𝑥)𝜇(𝑥)𝑞𝐼,𝑖,0𝜆𝑦𝑤,𝑖+1,1𝑞(𝑥),𝑖0,(2.10)𝑤,𝑖,1()=𝜂,𝑖0.(2.11) By solving (2.7)–(2.10), we have 𝑞𝐼,0,0=𝜆𝑦𝛾+𝜆𝑤,0,1𝑞(0),(2.12)𝐼,𝑖,0=𝑖𝜃𝑦𝛾+𝜆+𝑖𝜃𝑤,𝑖1,1𝜆(0)+𝑦𝛾+𝜆+𝑖𝜃𝑤,𝑖,1𝑞(0),𝑖1,(2.13)𝑤,𝑖,1(𝑥)=𝑎𝑖𝑒𝑥0(𝛾+𝜆+𝛼+𝜇(𝜏))𝑑𝜏+𝑒𝑥0(𝛾+𝜆+𝛼+𝜇(𝜏))𝑑𝜏𝑥0𝜇(𝜏)𝑞𝐼,𝑖,0𝜆𝑦𝑤,𝑖+1,1𝑒(𝜏)𝜏0(𝛾+𝜆+𝛼+𝜇(𝑠))𝑑𝑠𝑑𝜏=𝑎𝑖𝑒𝑥0(𝛾+𝜆+𝛼+𝜇(𝜏))𝑑𝜏𝑒𝑥0(𝛾+𝜆+𝛼+𝜇(𝜏))𝑑𝜏𝑥0𝜇(𝜏)𝑞𝐼,𝑖,0+𝜆𝑦𝑤,𝑖+1,1𝑒(𝜏)𝜏0(𝛾+𝜆+𝛼+𝜇(𝑠))𝑑𝑠𝑑𝜏,𝑖0.(2.14) Through multiplying 𝑒𝑥0(𝛾+𝜆+𝛼+𝜇(𝜏))𝑑𝜏 to two sides of (2.14), we obtain 𝑒𝑥0(𝛾+𝜆+𝛼+𝜇(𝜏))𝑑𝜏𝑞𝑤,𝑖,1(𝑥)=𝑎𝑖𝑥0𝜇(𝜏)𝑞𝐼,𝑖,0+𝜆𝑦𝑤,𝑖+1,1𝑒(𝜏)𝜏0(𝛾+𝜆+𝛼+𝜇(𝑠))𝑑𝑠𝑑𝜏,𝑖0.(2.15) By combining (2.11) with (2.15), we deduce 𝑎𝑖=0𝜇(𝜏)𝑞𝐼,𝑖,0+𝜆𝑦𝑤,𝑖+1,1𝑒(𝜏)𝜏0(𝛾+𝜆+𝛼+𝜇(𝑠))𝑑𝑠𝑑𝜏,𝑖0.(2.16) By inserting (2.16) into (2.14), we get 𝑞𝑤,𝑖,1(𝑥)=𝑒𝑥0(𝛾+𝜆+𝛼+𝜇(𝜏))𝑑𝜏𝑥𝜇(𝜏)𝑞𝐼,𝑖,0+𝜆𝑦𝑤,𝑖+1,1𝑒(𝜏)𝜏0(𝛾+𝜆+𝛼+𝜇(𝑠))𝑑𝑠𝑑𝜏=𝑒𝑥0(𝛾+𝜆+𝛼+𝜇(𝜏))𝑑𝜏𝑥𝜇(𝜏)𝑞𝐼,𝑖,0𝑒𝜏0(𝛾+𝜆+𝛼+𝜇(𝑠))𝑑𝑠𝑑𝜏+𝜆𝑒𝑥0(𝛾+𝜆+𝛼+𝜇(𝜏))𝑑𝜏𝑥𝑦𝑤,𝑖+1,1(𝜏)𝑒𝜏0(𝛾+𝜆+𝛼+𝜇(𝑠))𝑑𝑠𝑞𝑑𝜏(2.17)𝑤,𝑖,1𝐿[0,)0𝑒𝑥0(Re𝛾+𝜆+𝛼)𝑑𝜏𝑥||𝑞𝜇(𝜏)𝐼,𝑖,0||𝑒𝜏0(Re𝛾+𝜆+𝛼)𝑑𝑠𝑒𝜏𝑥𝜇(𝑠)𝑑𝑠𝑑𝜏𝑑𝑥+𝜆0𝑒𝑥0(Re𝛾+𝜆+𝛼)𝑑𝜏𝑥||𝑦𝜇(𝜏)𝑤,𝑖+1,1||𝑒(𝜏)𝜏0(Re𝛾+𝜆+𝛼)𝑑𝑠𝑒𝜏𝑥𝜇(𝑠)𝑑𝑠𝑑𝜏𝑑𝑥𝜇Re𝛾+𝜆+𝛼+𝜇||𝑞𝐼,𝑖,0||𝜆+Re𝛾+𝜆+𝛼+𝜇𝑦𝑤,𝑖+1,1𝐿[0,),𝑖0.(2.18) Equation (2.12) gives ||𝑞𝐼,0,0||𝜆||||||𝑦𝛾+𝜆𝑤,0,1||𝜆(0)||||𝑦𝛾+𝜆𝑤,0,1𝐿[0,).(2.19) From (2.13) we estimate ||𝑞𝐼,𝑖,0||𝑖𝜃||||||𝑦𝛾+𝜆+𝑖𝜃𝑤,𝑖1,1||+𝜆(0)||||||𝑦𝛾+𝜆+𝑖𝜃𝑤,𝑖,1||(0)𝑖𝜃||||𝑦𝛾+𝜆+𝑖𝜃𝑤,𝑖1,1𝐿[0,)+𝜆||||𝑦𝛾+𝜆+𝑖𝜃𝑤,𝑖,1𝐿[0,)𝜆+𝑖𝜃||||𝛾+𝜆+𝑖𝜃sup𝑖0𝑦𝑤,𝑖,1𝐿[0,),𝑖1.(2.20) By inserting (2.19) and (2.20) into (2.18), we have 𝑞𝑤,0,1𝐿[0,)𝜇Re𝛾+𝜆+𝛼+𝜇𝜆||||𝑦𝛾+𝜆𝑤,0,1𝐿[0,)+𝜆Re𝛾+𝜆+𝛼+𝜇𝑦𝑤,1,1𝐿[0,)𝜆𝜇||||𝛾+𝜆Re𝛾+𝜆+𝛼+𝜇+𝜆Re𝛾+𝜆+𝛼+𝜇×sup𝑖0𝑦𝑤,𝑖,1𝐿[0,),𝑞𝑤,𝑖,1𝐿[0,)𝜇Re𝛾+𝜆+𝛼+𝜇𝜆+𝑖𝜃||||𝛾+𝜆+𝑖𝜃sup𝑖0𝑦𝑤,𝑖,1𝐿[0,)+𝜆Re𝛾+𝜆+𝛼+𝜇sup𝑖0𝑦𝑤,𝑖,1𝐿[0,)=(𝜆+𝑖𝜃)𝜇Re𝛾+𝜆+𝛼+𝜇||||+𝜆𝛾+𝜆+𝑖𝜃Re𝛾+𝜆+𝛼+𝜇×sup𝑖0𝑦𝑤,𝑖,1𝐿[0,),𝑖0.(2.21) By combining (2.19) and (2.20) with (2.21), we derive ||𝑞𝐼,𝑞𝑤||𝜆sup||||𝛾+𝜆,sup𝑖1𝜆+𝑖𝜃||||,𝜆𝛾+𝜆+𝑖𝜃𝜇||||𝛾+𝜆Re𝛾+𝜆+𝛼+𝜇+𝜆Re𝛾+𝜆+𝛼+𝜇,sup𝑖1(𝜆+𝑖𝜃)𝜇Re𝛾+𝜆+𝛼+𝜇||||+𝜆𝛾+𝜆+𝑖𝜃Re𝛾+𝜆+𝛼+𝜇.(2.22) This shows that [𝐼(𝛾𝐼𝒢)1]1 exists and is bounded when 𝛾 belongs to the set ||||||||||||||𝜆𝛾sup||||𝛾+𝜆,sup𝑖1𝜆+𝑖𝜃||||,𝜆𝛾+𝜆+𝑖𝜃𝜇||||𝛾+𝜆Re𝛾+𝜆+𝛼+𝜇+𝜆Re𝛾+𝜆+𝛼+𝜇,sup𝑖1(𝜆+𝑖𝜃)𝜇Re𝛾+𝜆+𝛼+𝜇||||+𝜆𝛾+𝜆+𝑖𝜃Re𝛾+𝜆+𝛼+𝜇<1,Re𝛾+𝜆+𝛼+𝜇>0.(2.23) Through discussing the solution of the equation (𝛾𝐼𝒢)(𝑞𝐼,𝑞𝑤)=(𝑦𝐼,𝑦𝑤) for any given (𝑦𝐼,𝑦𝑤)𝑋, it is not difficult to verify (𝛾𝐼𝒢)1 exists and is bounded when 𝛾 satisfies (2.23). Therefore, by the resolvent equation []𝛾𝐼(+𝒢+)1=(𝛾𝐼𝒢)𝐼(𝛾𝐼𝒢)11=𝐼(𝛾𝐼𝒢)11(𝛾𝐼𝒢)1,(2.24) we know that (𝛾𝐼𝒢)1 exists and is bounded when 𝛾 belongs to the set (2.23), which means that (2.23) belongs to 𝜌(+𝒢+).
In particular, if 𝛾=𝑖𝑎,𝑎{0},𝑖2=1, then all 𝛾’s belong to (2.23). In fact, by using the condition on this lemma, we have 𝜆𝑎2+𝜆2<1,sup𝑘1𝜆+𝑘𝜃𝑎2+(𝜆+𝑘𝜃)2<1,𝜇<𝛼+𝜇𝜆𝜇<𝛼+𝜇𝑎2+𝜆2𝜆𝜇+𝜆𝑎2+𝜆2<𝛼+𝜇𝑎2+𝜆2+𝜆𝑎2+𝜆2𝜆𝜇+𝜆𝑎2+𝜆2<𝜆+𝛼+𝜇𝑎2+𝜆2𝜆𝜇𝑎2+𝜆2𝜆+𝛼+𝜇+𝜆𝜆+𝛼+𝜇<1,𝜇<𝛼+𝜇(𝜆+𝑘𝜃)𝜇<𝛼+𝜇𝑎2+(𝜆+𝑘𝜃)2(𝜆+𝑘𝜃)𝜇+𝜆𝑎2+(𝜆+𝑘𝜃)2<𝛼+𝜇𝑎2+(𝜆+𝑘𝜃)2+𝜆𝑎2+(𝜆+𝑘𝜃)2(𝜆+𝑘𝜃)𝜇+𝜆𝑎2+(𝜆+𝑘𝜃)2<𝜆+𝛼+𝜇𝑎2+(𝜆+𝑘𝜃)2,sup𝑘1(𝜆+𝑘𝜃)𝜇𝑎2+(𝜆+𝑘𝜃)2𝜆+𝛼+𝜇+𝜆𝜆+𝛼+𝜇<1.(2.25) The above inequalities show that all points on the imaginary axis except zero belong to the resolvent set of (𝐴+𝑈+𝐸). From the relation between the spectrum of 𝐴+𝑈+𝐸 and spectrum of (𝐴+𝑈+𝐸) we know that all points on the imaginary axis except zero belong to the resolvent set of 𝐴+𝑈+𝐸.

Lemma 2.3. If 0<𝜇=inf[𝑥0,)𝜇(𝑥)𝜇=sup[𝑥0,)𝜇(𝑥)<,(2.26) then 0 is not an eigenvalue of 𝐴+𝑈+𝐸.

Proof. We consider the equation (𝐴+𝑈+𝐸)(𝑝𝐼,𝑝𝑤)=0, which is equivalent to 𝜆𝑝𝐼,0,0=0𝜇(𝑥)𝑝𝑤,0,1(𝑥)𝑑𝑥,(2.27)(𝜆+𝑖𝜃)𝑝𝐼,𝑖,0=0𝜇(𝑥)𝑝𝑤,𝑖,1(𝑥)𝑑𝑥,𝑖1,(2.28)𝑑𝑝𝑤,0,1(𝑥)𝑑𝑥=(𝜆+𝛼+𝜇(𝑥))𝑝𝑤,0,1(𝑥),(2.29)𝑑𝑝𝑤,𝑖,1(𝑥)𝑑𝑥=(𝜆+𝛼+𝜇(𝑥))𝑝𝑤,𝑖,1(𝑥)+𝜆𝑝𝑤,𝑖1,1𝑝(𝑥),𝑖1,(2.30)𝑤,𝑖,1(0)=𝜆𝑝𝐼,𝑖,0+(𝑖+1)𝜃𝑝𝐼,𝑖+1,0,𝑖0.(2.31) It is hard to determine the concrete expressions of all 𝑝𝐼,i,0,𝑝𝑤,𝑖,1(𝑥) and to prove (𝑝𝐼,𝑝𝑤)𝐷(𝐴). In the following we use another method. We introduce the probability-generating functions 𝑄𝐼(𝑧)=𝑖=0𝑝𝐼,𝑖,0𝑧𝑖,    𝑄𝑤(𝑥,𝑧)=𝑖=0𝑝𝑤,𝑖,1(𝑥)𝑧𝑖 for |𝑧|<1. Theorem 1.1 ensures that 𝑄𝐼(𝑧),𝑄𝑤(𝑥,𝑧) are well-defined. By applying the basic knowledge of power series, the Fubini theorem and (2.27)-(2.28), we have 𝜆𝑝𝐼,0,0+𝑖=1(𝜆+𝑖𝜃)𝑝𝐼,𝑖,0𝑧𝑖=0𝜇(𝑥)𝑝𝑤,0,1(𝑥)𝑑𝑥+𝑖=10𝜇(𝑥)𝑝𝑤,𝑖,1(𝑥)𝑧𝑖𝑑𝑥𝜆𝑄𝐼(𝑧)+𝜃𝑧𝑖=0𝑝𝐼,𝑖,0𝑧𝑖=0𝜇(𝑥)𝑖=0𝑝𝑤,𝑖,1(𝑥)𝑧𝑖𝑑𝑥𝜆𝑄𝐼𝑑(𝑧)+𝜃𝑧𝑄𝑑𝑧𝐼(𝑧)=0𝜇(𝑥)𝑄𝑤(𝑥,𝑧)𝑑𝑥.(2.32) By (2.29) and (2.30), we deduce 𝑑𝑝𝑤,0,1(𝑥)+𝑑𝑥𝑖=1𝑑𝑝𝑤,𝑖,1(𝑥)𝑧𝑑𝑥𝑖=(𝜆+𝛼+𝜇(𝑥))𝑝𝑤,0,1(𝑥)(𝜆+𝛼+𝜇(𝑥))𝑖=1𝑝𝑤,𝑖,1(𝑥)𝑧𝑖+𝑖=1𝜆𝑝𝑤,𝑖1,1(𝑥)𝑧𝑖𝜕𝑖=0𝑝𝑤,𝑖,1(𝑥)𝑧𝑖𝜕𝑥=(𝜆+𝛼+𝜇(𝑥))𝑖=0𝑝𝑤,𝑖,1(𝑥)𝑧𝑖+𝜆𝑖=1𝑝𝑤,𝑖1,1(𝑥)𝑧𝑖𝜕𝑄𝑤(𝑥,𝑧)𝜕𝑥=(𝜆+𝛼+𝜇(𝑥))𝑄𝑤(𝑥,𝑧)+𝜆𝑧𝑖=0𝑝𝑤,𝑖,1(𝑥)𝑧𝑖𝜕𝑄𝑤(𝑥,𝑧)=𝜕𝑥(𝜆𝑧𝜆𝛼𝜇(𝑥))𝑄𝑤𝑄(𝑥,𝑧)𝑤(𝑥,𝑧)=𝑄𝑤(0,𝑧)𝑒𝑥0(𝜆𝑧𝜆𝛼𝜇(𝜏))𝑑𝜏.(2.33) Equation (2.31) gives 𝑄𝑤(0,𝑧)=𝑖=0𝑝𝑤,𝑖,1(0)𝑧𝑖=𝜆𝑖=0𝑝𝐼,𝑖,0𝑧𝑖+𝑖=0(𝑖+1)𝜃𝑝𝐼,𝑖+1,0𝑧𝑖=𝜆𝑄𝐼𝑑(𝑧)+𝜃𝑄𝑑𝑧𝐼(𝑧).(2.34) By combining (2.34) with (2.33) and using (2.32), we calculate 𝜆𝑄𝐼𝑑(𝑧)+𝜃𝑧𝑄𝑑𝑧𝐼(𝑧)=0𝜇(𝑥)𝑄𝑤=(𝑥,𝑧)𝑑𝑥0𝜇(𝑥)𝑄𝑤(0,𝑧)𝑒𝑥0[𝜆𝑧𝜆𝛼𝜇(𝜏)]𝑑𝜏=𝑑𝑥𝜆𝑄𝐼𝑑(𝑧)+𝜃𝑄𝑑𝑧𝐼(𝑧)0𝜇(𝑥)𝑒𝑥0[𝜆𝑧𝜆𝛼𝜇(𝜏)]𝑑𝜏𝑑𝑥𝑑𝑄𝐼(𝑧)𝑄𝐼=𝜆(𝑧)𝜃0𝜇(𝑥)𝑒𝑥0[𝜆𝑧𝜆𝛼𝜇(𝜏)]𝑑𝜏𝑑𝑥1𝑧0𝜇(𝑥)𝑒𝑥0[𝜆𝑧𝜆𝛼𝜇(𝜏)]𝑑𝜏𝑄𝑑𝑥𝑑𝑧𝐼(𝑧)=𝑄𝐼𝜆(0)exp𝜃𝑧00𝜇(𝑥)𝑒𝑥0[𝜆𝑠𝜆𝛼𝜇(𝜏)]𝑑𝜏𝑑𝑥1𝑠0𝜇(𝑥)𝑒𝑥0[𝜆𝑠𝜆𝛼𝜇(𝜏)]𝑑𝜏𝑑𝑥𝑑𝑠=𝑝𝐼,0,0𝜆exp𝜃𝑧00𝜇(𝑥)𝑒𝑥0[𝜆𝑠𝜆𝛼𝜇(𝜏)]𝑑𝜏𝑑𝑥1𝑠0𝜇(𝑥)𝑒𝑥0[𝜆𝑠𝜆𝛼𝜇(𝜏)]𝑑𝜏.𝑑𝑥𝑑𝑠(2.35) This together with (2.34) gives 𝑄𝑤(0,𝑧)=𝜆(𝑧1)𝑧0𝜇(𝑥)𝑒𝑥0[𝜆𝑧𝜆𝛼𝜇(𝜏)]𝑑𝜏𝑄𝑑𝑥𝐼(𝑧).(2.36) From (2.35), we derive 𝑖=0𝑝𝐼,𝑖,0=lim𝑧1𝑄𝐼(𝑧)=lim𝑧1𝑝𝐼,0,0𝜆exp𝜃𝑧00𝜇(𝑥)𝑒𝑥0[𝜆𝑠𝜆𝛼𝜇(𝜏)]𝑑𝜏𝑑𝑥1𝑠0𝜇(𝑥)𝑒𝑥0[𝜆𝑠𝜆𝛼𝜇(𝜏)]𝑑𝜏𝑑𝑥𝑑𝑠=𝑝𝐼,0,0𝜆exp𝜃100𝜇(𝑥)𝑒𝑥0[𝜆𝑠𝜆𝛼𝜇(𝜏)]𝑑𝜏𝑑𝑥1𝑠0𝜇(𝑥)𝑒𝑥0[𝜆𝑠𝜆𝛼𝜇(𝜏)]𝑑𝜏.𝑑𝑥𝑑𝑠(2.37) In the following, by the Rouche theorem we know that 𝑠0𝜇(𝑥)𝑒𝑥0[𝜆𝑠𝜆𝛼𝜇(𝜏)]𝑑𝜏𝑑𝑥 has a unique zero point inside the unit circle |𝑧|=1. And by the residue theorem we determine the above integral.
Let 𝑓(𝑧)=𝑧 and 𝑔(𝑧)=0𝜇(𝑥)𝑒(𝜆+𝛼𝜆𝑧)𝑥𝑥0𝜇(𝜏)𝑑𝜏𝑑𝑥. It is well known that 𝑓(𝑧) and 𝑔(𝑧) are differentiable inside and continuous on the contour |𝑧|=1. And |𝑓(𝑧)|=1 on |𝑧|=1. Since, for (𝜆+𝛼𝜆Re𝑧)>0, ||𝑔||=||||(𝑧)0𝜇(𝑥)𝑒(𝜆+𝛼𝜆𝑧)𝑥𝑥0𝜇(𝜏)𝑑𝜏||||𝑑𝑥0𝑒(𝜆+𝛼𝜆Rez)𝑥𝑑𝑒𝑥0𝜇(𝜏)𝑑𝜏𝑒=(𝜆+𝛼𝜆Re𝑧)𝑥𝑒𝑥0𝜇(𝜏)𝑑𝜏||||||𝑥=𝑥=0+(𝜆+𝛼𝜆Rez)0𝑒(𝜆+𝛼𝜆Re𝑧)𝑥𝑥0𝜇(𝜏)𝑑𝜏𝑑𝑥=1(𝜆+𝛼𝜆Re𝑧)0𝑒(𝜆+𝛼𝜆Re𝑧)𝑥𝑥0𝜇(𝜏)𝑑𝜏||||,𝑑𝑥<1=𝑓(𝑧)(2.38) we have |𝑓(𝑧)|<|𝑔(𝑧)| on |𝑧|=1. Consequently, all conditions of Rouche’s theorem are satisfied. Obviously 𝑓(𝑧)𝑔(𝑧) has only one zero inside the unit circle |𝑧|=1, since 𝑓(𝑧) has one. If we denote this zero by 𝑧0, then it is a simple pole of 0𝜇(𝑥)𝑒(𝜆+𝛼𝜆𝑧)𝑥𝑥0𝜇(𝜏)𝑑𝜏𝑑𝑥1𝑧0𝜇(𝑥)𝑒(𝜆+𝛼𝜆𝑧)𝑥𝑥0𝜇(𝜏)𝑑𝜏.𝑑𝑥(2.39) By the residue theorem, we calculate 100𝜇(𝑥)𝑒(𝜆+𝛼𝜆𝑠)𝑥𝑥0𝜇(𝜏)𝑑𝜏𝑑𝑥1𝑠0𝜇(𝑥)𝑒(𝜆+𝛼𝜆𝑠)𝑥𝑥0𝜇(𝜏)𝑑𝜏𝑑𝑥𝑑𝑠=lim𝑧𝑧00𝜇(𝑥)𝑒(𝜆+𝛼𝜆𝑧)𝑥𝑥0𝜇(𝜏)𝑑𝜏𝑑𝑥110𝜆𝑥𝜇(𝑥)𝑒(𝜆+𝛼𝜆𝑧)𝑥𝑥0𝜇(𝜏)𝑑𝜏=𝑑𝑥0𝜇(𝑥)𝑒(𝜆+𝛼𝜆𝑧0)𝑥𝑥0𝜇(𝜏)𝑑𝜏𝑑𝑥110𝜆𝑥𝜇(𝑥)𝑒(𝜆+𝛼𝜆𝑧0)𝑥𝑥0𝜇(𝜏)𝑑𝜏.𝑑𝑥(2.40) Thus, we obtain 𝑖=0𝑝𝐼,𝑖,0=𝑝𝐼,0,0𝜆exp𝜃0𝜇(𝑥)𝑒(𝜆+𝛼𝜆𝑧0)𝑥𝑥0𝜇(𝜏)𝑑𝜏𝑑𝑥110𝜆𝑥𝜇(𝑥)𝑒(𝜆+𝛼𝜆𝑧0)𝑥𝑥0𝜇(𝜏)𝑑𝜏𝑑𝑥<.(2.41) From (2.36), we derive lim𝑧1𝑄𝑤(0,𝑧)=lim𝑧1𝜆(𝑧1)𝑧0𝜇(𝑥)𝑒𝑥0[𝜆𝑧𝜆𝛼𝜇(𝜏)]𝑑𝜏𝑄𝑑𝑥𝐼(𝑧)=lim𝑧1𝜆(𝑧1)𝑧0𝜇(𝑥)𝑒𝑥0[𝜆𝑧𝜆𝛼𝜇(𝜏)]𝑑𝜏𝑑𝑥lim𝑧1𝑄𝐼=(𝑧)𝜆(11)10𝜇(𝑥)𝑒𝑥0[𝛼+𝜇(𝜏)]𝑑𝜏𝑄𝑑𝑥𝐼(1)=0.(2.42) By combining (2.33) with (2.42), we estimate 𝑖=0𝑝𝑤,𝑖,1(𝑥)=lim𝑧1𝑄𝑤(𝑥,𝑧)=lim𝑧1𝑄𝑤(0,𝑧)𝑒𝑥0(𝜆𝑧𝜆𝛼𝜇(𝜏))𝑑𝜏=lim𝑧1𝑄𝑤(0,𝑧)𝑒𝑥0(𝛼+𝜇(𝜏))𝑑𝜏=0.(2.43) Since Theorem 1.1 shows that each 𝑝𝑤,𝑖,1(𝑥) is nonnegative, (2.43) implies 𝑝𝑤,𝑖,1(𝑥)=0 for all 𝑖0. This together with (2.27) and (2.28) gives 𝑝𝐼,𝑖,0=0 for all 𝑖0. In other words, (𝐴+𝑈+𝐸)(𝑝𝐼,𝑝𝑤)=0 has only zero solution; that is, 0 is not an eigenvalue of 𝐴+𝑈+𝐸.

Lemma 2.3 shows that the system (1.7) does not have nonzero steady-state solution.

If we can prove that 0 is not in the residue spectrum of 𝐴+𝑈+𝐸, then by Theorem 1.1, Lemmas 2.2 and 2.3, and ABLV Theorem (see [16] or [17]), we deduce that the time-dependent solution of system (1.7) is strongly asymptotically stable. This result is quite different from other queueing models, see Gupur [1315] and Kasim and Gupur [7].

Acknowledgments

The authors’ research work is supported by Science Foundation of Xinjiang University (no: XY110101) and the Natural Science Foundation of Xinjiang (no: 2012211A023).

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