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Research article Special Issues

Learning cooperative strategies in StarCraft through role-based monotonic value function factorization

  • StarCraft is a popular real-time strategy game that has been widely used as a research platform for artificial intelligence. Micromanagement refers to the process of making each unit perform appropriate actions separately, depending on the current state in the the multi-agent system comprising all of the units, i.e., the fine-grained control of individual units for common benefit. Therefore, cooperation between different units is crucially important to improve the joint strategy. We have selected multi-agent deep reinforcement learning to tackle the problem of micromanagement. In this paper, we propose a method for learning cooperative strategies in StarCraft based on role-based montonic value function factorization (RoMIX). RoMIX learns roles based on the potential impact of each agent on the multi-agent task; it then represents the action value of a role in a mixed way based on monotonic value function factorization. The final value is calculated by accumulating the action value of all roles. The role-based learning improves the cooperation between agents on the team, allowing them to learn the joint strategy more quickly and efficiently. In addition, RoMIX can also reduce storage resources to a certain extent. Experiments show that RoMIX can not only solve easy tasks, but it can also learn better cooperation strategies for more complex and difficult tasks.

    Citation: Kun Han, Feng Jiang, Haiqi Zhu, Mengxuan Shao, Ruyu Yan. Learning cooperative strategies in StarCraft through role-based monotonic value function factorization[J]. Electronic Research Archive, 2024, 32(2): 779-798. doi: 10.3934/era.2024037

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  • StarCraft is a popular real-time strategy game that has been widely used as a research platform for artificial intelligence. Micromanagement refers to the process of making each unit perform appropriate actions separately, depending on the current state in the the multi-agent system comprising all of the units, i.e., the fine-grained control of individual units for common benefit. Therefore, cooperation between different units is crucially important to improve the joint strategy. We have selected multi-agent deep reinforcement learning to tackle the problem of micromanagement. In this paper, we propose a method for learning cooperative strategies in StarCraft based on role-based montonic value function factorization (RoMIX). RoMIX learns roles based on the potential impact of each agent on the multi-agent task; it then represents the action value of a role in a mixed way based on monotonic value function factorization. The final value is calculated by accumulating the action value of all roles. The role-based learning improves the cooperation between agents on the team, allowing them to learn the joint strategy more quickly and efficiently. In addition, RoMIX can also reduce storage resources to a certain extent. Experiments show that RoMIX can not only solve easy tasks, but it can also learn better cooperation strategies for more complex and difficult tasks.



    The stochastic clearing queueing system has a wide range of applications across various fields such as manufacturing systems, telecommunications, transportation systems, supply chain and inventory management, etc. For example, in manufacturing, stochastic clearing queueing systems can be applied to optimize production lines. For instance, a production line may operate in batches, where all accumulated orders are processed together. The system can be designed to start production only when a certain number of orders are accumulated, or it can be activated randomly when the number of orders is below the threshold to avoid long delays. In telecommunication networks, stochastic clearing queueing systems can be used to manage data packet transmission. Packets arrive randomly and are processed in batches to optimize bandwidth usage and reduce latency. The clearing mechanism helps in managing the queue of packets efficiently.

    Many scholars have previously investigated stochastic clearing queuing systems [1,2,3,4,5]. The M/G/1 stochastic clearing queuing system is a specialized model within the realm of stochastic clearing queues. Here, "M" denotes Markovian arrivals, signifying that the arrival process adheres to a Poisson process, characterized by exponentially distributed inter-arrival times. "G" represents a general service time distribution, implying that service durations can follow any distribution, not limited to exponential ones. Last, "1" indicates that there is only one server in the system. In this system, customers arrive following a Poisson process. A single server attends to them, with service times governed by a general distribution. At random intervals, the entire system is cleared, removing all customers in the queue as well as those being served. The clearance times are random and follow a specific distribution.

    Zhang et al. [6] were the first to examine the M/G/1 stochastic clearing queueing system in a three-phase environment. They formulated the mathematical model of this system using the method of supplementary variables and explored the steady-state solution and several key stationary performance measures under the following hypothesis:

    limtQj,0(t)=Qj,0,j=1,3;limtVn(t)=Vn,n0,limtpj,n(,t)=pj,n(),j=1,3,n1,

    where Qj,0(t)(j=1,3) represents the probability that there are no customers in phase j and the server is idle at time t; Vn(t)(n0) denotes the probability that there are n customers in the system and the system is in phase 2; pj,n(x,t)dx(j=1,3;n1) is the probability that there are n customers in the system (including the one being served) at time t with the server busy serving a customer whose elapsed service time lies in the interval [x,x+dx) in phase j.

    Drawing on the theory of partial differential equations (see [7,8]), we can infer that the aforementioned hypothesis implies the following two hypotheses:

    Hypothesis 1): This queueing system has a nonnegative time-dependent solution.

    Hypothesis 2): The time-dependent solution of this system converges to its nonzero steady-state solution.

    Recently, in our work [9], we established that Hypothesis 1) holds when the service rate of the server is a bounded function, and Hypothesis 2) holds when the service rate is constant. To address whether Hypothesis 2) holds, as demonstrated in [9], we first identify the adjoint operator of the system operator. Subsequently, we analyze the spectrum of this adjoint operator on the imaginary axis. By leveraging the relationship between the spectrum of the operator and its adjoint, we then determine the spectrum of the system operator on the imaginary axis. However, in the aforementioned literature, we did not address whether Hypothesis 2) remains valid when the service rate is a bounded function. In this article, we aim to resolve this outstanding issue.

    In this paper, we employ the boundary perturbation method of Greiner [10] and probability generating functions [6] to investigate the asymptotic behavior of the time-dependent solution for the M/G/1 stochastic clearing queueing system in a three-phase environment. To achieve this, we first select an appropriate Banach space to serve as the state space for the system. We then define the maximal operator and boundary operators for this queueing within this state space. Utilizing these operators, we construct the system operator for the given system.

    Next, by introducing suitable probability generating functions, we demonstrate that 0 is an eigenvalue of the system operator with a geometric multiplicity of 1. We further define an operator with a boundary condition of 0 for this queueing system using the aforementioned maximal and boundary operators. We then identify the resolvent set of this operator and define the Dirichlet operator. Drawing on a result from Haji and Radl [11], we prove that all points on the imaginary axis, except for 0, belong to the resolvent set of the system operator. Additionally, we directly show that 0 is also an eigenvalue of the adjoint operator of the system, with a geometric multiplicity of 1.

    These results collectively indicate that the time-dependent solution of the M/G/1 stochastic clearing queueing system in a three-phase environment strongly converges to its non-zero steady-state solution. In other words, Hypothesis 2) holds true in the sense of strong convergence. In addition, the conclusion of this article includes the main result of [9].

    According to Zhang et al. [6], the mathematical model of the M/G/1 stochastic clearing queueing system operating in a three-phase environment can be described by the following system of integro-partial differential equations:

    dQ1,0(t)dt=(λ1+θ1)Q1,0(t)+θ3Q3,0(t)+0p1,1(x,t)μ1(x)dx+θ3n=10p3,n(x,t)dx, (2.1)
    dQ3,0(t)dt=(λ3+θ3)Q3,0(t)+N0V0(t)+0p3,1(x,t)μ3(x)dx, (2.2)
    dV0(t)dt=V0(t)+θ1Q1,0(t), (2.3)
    dVn(t)dt=Vn(t)+θ10p1,n(x,t)dx,n1, (2.4)
    pj,1(x,t)t+pj,1(x,t)x=[λj+θj+μj(x)]pj,1(x,t),j=1,3, (2.5)
    pj,n(x,t)t+pj,n(x,t)x=[λj+θj+μj(x)]pj,n(x,t)+λjpj,n1(x,t),n2, (2.6)

    with the following boundary and initial conditions

    p1,1(0,t)=λ1Q1,0(t)+0p1,2(x,t)μ1(x)dx, (2.7)
    p1,n(0,t)=0p1,n+1(x,t)μ1(x)dx,n2, (2.8)
    p3,1(0,t)=λ3Q3,0(t)+1k=0NkV1k(t)+0p3,2(x,t)μ3(x)dx, (2.9)
    p3,n(0,t)=nk=0NkVnk(t)+0p3,n+1(x,t)μ3(x)dx,n2, (2.10)
    Vn(0)=un0,n0;Qj,1(0)=¯uj0,pj,n(x,0)=uj,n(x)0,j=1,3;n1. (2.11)

    Here, (x,t)[0,)×[0,), and

    ¯u1+¯u3+n=0un+n=10u1,n(x)dx+n=10u3,n(x)dx=1;

    Qj,0(t)(j=1,3) represents the probability that there are no customers in phase j and the server is idle at time t; Vn(t)(n0) denotes the probability that there are n customers in the system and the system is in phase 2; pj,n(x,t)dx(j=1,3;n1) is the probability that there are n customers in the system (including the one being served) at time t with the server busy serving a customer whose elapsed service time lies in the interval [x,x+dx) in phase j; λj(j=1,2,3) is the arrival rate of customers when the system is in phase j; μj(x)(j=1,3) is the conditional probability (hazard rate) of completing a service during the interval (x,x+dx) with elapsed time x in phase j. It satisfies the conditions:

    μj(x)0,0μj(x)dx=,j=1,3,

    θj(j=1,3) is the residence rate of the system in phase j.

    The system operates in three distinct phases. The first and third phases are working phases, while the second phase is a deterministic time phase during which no service is provided. Upon completion of the first phase, the system transitions into the second phase. If a customer arrives during the second phase, they will either enter the system with probability p or leave without joining the system with probability q=1p. During this second phase, all customers are unable to receive service for a fixed duration d. After the second phase concludes, the system enters the third phase. Once the third phase is completed, the current customer is forced to leave the system without receiving further service, and the system then returns to the first phase to initiate a new service cycle. We assume that Nr=(r!)1eλ2pd(λ2pd)r is the probability of r(r0) arrivals during phase 2.

    In this paper, we present our main result under the following assumption:

    Assumption 2.1. Let λj>0,θj>0,N0(0,1) and μj(x):[0,)[0,) measurable and

    0<infx[0,)μj(x)μj(x)supx[0,)μj(x)<,j=1,3.

    We choose the state space as follows:

    X={(V,p1,p3)|V=(Q1,0,Q3,0,V0,V1,V2)l1p1=(p1,1,p1,2,)L1[0,)×L1[0,)×p3=(p3,1,p3,2,)L1[0,)×L1[0,)×(V,p1,p3)=|Q1,0|+|Q3,0|+n=0|Vn|+n=1p1,nL1[0,)+n=1p3,nL1[0,)<}.

    It is not difficult to verify that X is a Banach space.

    We define the maximal operator of systems (2.1)–(2.11) as follows:

    Am(V,p1,p3)=(A2,1mV+A2,2mp1+A2,3mp3,A1mp1,A3mp3),

    and

    D(Am)={(V,p1,p3)X|dpj,ndxL1[0,),j=1,3,n1,pj,nare absolutely continuous functions andn=1dpj,ndxL1[0,)<},

    where

    A2,1mV=((λ1+θ1)θ30000(λ3+θ3)N000θ101000001000001)(Q1,0Q3,0V0V1V2),A2,2mp1=(φ200000000θ1φ1000θ1φ10)(p1,1p1,2p1,3p1,4),A2,3mp3=(θ3φ1θ3φ1θ3φ1φ300000)(p3,1p3,2p3,3),Ajmpj=(Bj00λjBj00λjBj)(pj,1pj,2pj,3),j=1,3,

    where φ1f(x):=0f(x)dx, φ2f(x):=0μ1(x)f(x)dx, φ3f(x):=0μ3(x)f(x)dx and

    Bjg:=dg(x)dx[λj+θj+μj(x)]g(x),gW1,1[0,).

    In the following, we choose the boundary space X of X and define the boundary operators Ψ and Φ of systems (2.1)–(2.11) as follows:

    X=l1×l1,Ψ:D(Am)X,Φ:D(Am)X,

    and

    Ψ(V,p1,p3)=((p1,1(0)p1,2(0)p1,3(0)),(p3,1(0)p3,2(0)p3,3(0))),Φ(V,p1,p3)=(Φ1,2V+Φ1,1p1,Φ3,2V+Φ3,3p3),

    where

    Φ1,2V=(λ1000000000000000)(Q1,0Q3,0V0V1),Φ1,1=(0φ20000φ20000φ20000)(p1,1p1,2p1,3p1,4),Φ3,2V=(0λ3N1N00000N2N1N0000N3N2N1N000N4N3N2N100N5N4N3N2)(Q1,0Q3,0V0V1V2),Φ3,3p3=(0φ30000φ30000φ30000)(p3,1p3,2p3,3p3,4).

    Now, we define the system operator A of systems (2.1)–(2.11) by

    A(V,p1,p3)=Am(V,p1,p3),D(A)={(V,p1,p3)D(Am)|Ψ(V,p1,p3)=Φ(V,p1,p3)}.

    Consequently, the systems (2.1)–(2.11) can be written as an abstract Cauchy problem in the Banach space X by

    {d(V,p1,p3)(t)dt=A(V,p1,p3)(t),for allt(0,),(V,p1,p3)(0)=((¯u1¯u3u0u1),(u1,1u1,1u1,3u1,4),(u3,1u3,2u3,3u3,4)). (2.12)

    Recently, in our work [9], we have obtained the following results.

    Theorem 2.1. Let λj>0,θj>0,Nr>0, and 0<supx[0,)μj(x)<,j=1,3;r0; then system operator A generates a positive C0-semigroup eAt of contractions on X. Hence, system (2.12) admits a unique positive time-dependent solution (V(t),p1(,t),p3(,t)) that satisfies

    (V(t),p1(,t),p3(,t))=1,for allt[0,).

    Theorem 2.2. Let μj()=μj be a constant and λj,θj,μj>0,Nr>0,j=1,3;r0, then we have the following results:

    1) All points in

    {γC|sup{1|(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0|sup{λ1(θ1+μ1)|(γ+1)(γ+λ3+θ3)||σ1|,λ21|(γ+1)(γ+λ3+θ3)|σ1,λ3θ1N0(θ3+μ3)|σ3|,λ23θ1N0σ3,θ1θ3(1N0)|γ+λ3+θ3||σ3|μ3,λ3μ3θ1(1N0)|γ+λ3+θ3|σ3(|σ3|μ3),λ1θ3(θ1+μ1)|γ+1||σ1|,λ21θ3|γ+1|σ1,θ1θ23(1N0)|σ3|μ3,λ3(θ3+μ3)|(γ+1)(γ+λ1+θ1)||σ3|,λ23|(γ+1)(γ+λ1+θ1)|σ3,λ3μ3θ1θ3(1N0)σ3(|σ3|μ3),λ3θ1N0(θ1+μ1)|σ1|,λ1λ3θ1N0σ1,λ3N0(θ3+μ3)|γ+λ1+θ1||σ3|,λ23N0|γ+λ1+θ1|σ3,θ3(1N0)|(γ+λ1+θ1)(γ+λ3+θ3)||σ3|μ3,λ3μ3(1N0)|(γ+λ1+θ1)(γ+λ3+θ3)|σ3(|σ3|μ3)},θ3|γ+1||σ3|μ3,λ3μ3|γ+1|σ3(|σ3|μ3),θ1+μ1|σ1|,λ1σ1,θ1|σ1|μ1,λ1μ1σ1(|σ1|μ1),θ3+μ3|σ3|,λ3σ3,θ3|σ3|μ3,λ3μ3σ3(|σ3|μ3)}<1,σj>0,|σj|>μj}

    are belong to the resolvent set ρ(A). In particular, all points on the imaginary axis except zero belong to ρ(A), where σj=γ+λj+θj+μj,0<N0<1, and γ is the real part of γ.

    2) If λj<θj+μj, then zero is an eigenvalue of A with geometric multiplicity one.

    3) Zero is an eigenvalue of A with geometric multiplicity one.

    According to Theorem 1.96 of [12], if we can demonstrate that the intersection of the point spectrum of the system operator A and its adjoint operator A with the imaginary axis is a singleton set containing only zero, and that 0 is an eigenvalue of A with geometric multiplicity one, then we can conclude that the time-dependent solution of system (2.12) strongly converges to its steady-state solution.

    In this section, we first employ probability generating functions to demonstrate that 0 is an eigenvalue of the system operator A, with a geometric multiplicity is 1. Subsequently, we apply the boundary perturbation method of Greiner [10] to show that all points on the imaginary axis except for 0 fall on the resolvent set of A. Additionally, we directly prove that 0 is also an eigenvalue of the adjoint operator A, with a geometric multiplicity of 1. Consequently, the primary findings of this paper are articulated by invoking [12, Theorem 1.96].

    Lemma 3.1. Let λj>0,θj>0, Nr(0,1),r0, and 0<μj()<,j=1,3. Then, 0 is an eigenvalue of A with geometric multiplicity one.

    Proof. Consider the equation A(V,p1,p3)=0, which is equivalent to

    (λ1+θ1)Q1,0=θ3n=10p3,n(x)dx+θ3Q3,0+0p1,1(x)μ1(x)dx, (3.1)
    (λ3+θ3)Q3,0=N0V0+0p3,1(x)μ3(x)dx, (3.2)
    V0=θ1Q1,0, (3.3)
    Vn=θ10p1,n(x)dx,n1, (3.4)
    dp1,1(x)dx=[λ1+θ1+μ1(x)]p1,1(x), (3.5)
    dp1,n(x)dx=[λ1+θ1+μ1(x)]p1,n(x)+λ1p1,n1(x),n2, (3.6)
    dp3,1(x)dx=[λ3+θ3+μ3(x)]p3,1(x), (3.7)
    dp3,n(x)dx=[λ3+θ3+μ3(x)]p3,n(x)+λ3p3,n1(x),n2, (3.8)
    p1,1(0)=λ1Q1,0+0p1,2(x)μ1(x)dx, (3.9)
    p1,n(0)=0p1,n+1(x)μ1(x)dx,n2, (3.10)
    p3,1(0)=λ3Q3,0+N1V0+N0V1+0p3,2(x)μ3(x)dx, (3.11)
    p3,n(0)=nk=0NnkVk+0p3,n+1(x)μ3(x)dx,n2. (3.12)

    By solving the Eqs (3.5)–(3.8), we have

    pj,n(x)=e(λj+θj)xx0μj(τ)dτnk=1(λjx)nk(nk)!pj,k(0),j=1,3;n1. (3.13)

    Due to the difficulty in finding the specific expression of Vn,p1,n(x), and p3,n(x) and proving whether 0 is the eigenvalue of the system operator A, we will solve this problem by introducing the probability generating function.

    pj(x,z):=n=1pj,n(x)zn,¯V(z):=n=0Vnzn,pj(z):=n=1pj,nzn,pj,n:=0pj,n(x)dx,j=1,3.

    Multiply both sides of Eq (3.4) by zn,n1, then sum n from 0 to positive infinity, and add Eq (3.3) to obtain

    ¯V(z)=V0+n=1Vnzn=V0+θ1n=10p1,n(x)zndx=θ1Q1,0+θ1p1(z). (3.14)

    Multiply both sides of Eqs (3.6) and (3.8) by zn,n2, then sum n from 0 to positive infinity, and using Eqs (3.5) and (3.7), we have

    n=1pj,n(x)znx=n=1[λj+θj+μj(x)]pj,n(x)zn+λjn=1pj,n1(x)zn.

    That is,

    pj(x,z)x=[λj+θj+μj(x)]pj(x,z)+λjzpj(x,z)=[θj+λj(1z)μj(x)]pj(x,z). (3.15)

    By solving the above equation, we obtain

    pj(x,z)=pj(0,z)e[θj+λj(1z)]xx0μj(τ)dτ,j=1,3. (3.16)

    Using the boundary conditions (3.9), (3.10), and Eq (3.16), we have

    p1(0,z)=n=1p1,n(0)zn=p1,1(0)z+n=2p1,n(0)zn=zλ1Q1,0+0p1,2(x)μ1(x)zdx+n=20p1,n+1(x)μ1(x)zndx=zλ1Q1,0+n=10p1,n+1(x)μ1(x)zndx=zλ1Q1,0+1z0n=1p1,n(x)znμ1(x)dx0p1,1(x)μ1(x)dx=zλ1Q1,0+1z0p1(x,z)μ1(x)dx0p1,1(x)μ1(x)dx=zλ1Q1,0+1zp1(0,z)0μ1(x)e[θ1+λ1(1z)]xx0μ1(τ)dτdx0p1,1(x)μ1(x)dx.

    That is,

    (z0μ1(x)e[θ1+λ1(1z)]xx0μ1(τ)dτdx)p1(0,z)=z2λ1Q1,0z0p1,1(x)μ1(x)dx. (3.17)

    By Eqs (3.2), (3.11), (3.12), (3.16) and

    n=0Nnzn=n=0eλ2pd(λ2pdz)nn!=eλ2pd(1z),

    we calculate that

    p3(0,z)=n=1p3,n(0)zn=p3,1(0)z+n=2p3,n(0)zn=zλ3Q3,0+zN1V0+zN0V1+0p3,2(x)zμ3(x)dx+n=2[nk=0NnkVk+0p3,n+1(x)μ3(x)dx]zn=zλ3Q3,0+n=1znnk=0NnkVk+n=10p3,n+1(x)znμ3(x)dx=zλ3Q3,0+¯V(z)eλ2pd(1z)V0N0+1z0n=1p3,n(x)znμ3(x)d0p3,1(x)μ3(x)dx=zλ3Q3,0+¯V(z)eλ2pd(1z)(λ3+θ3)Q3,0+1z0p3(x,z)μ3(x)dx=[θ3+λ3(1z)]Q3,0+¯V(z)eλ2pd(1z)+1zp3(0,z)0μ3(x)e[θ3+λ3(1z)]xx0μ3(τ)dτdx.

    That is,

    [z0μ3(x)e[θ3+λ3(1z)]xx0μ3(τ)dτdx]p3(0,z)=z[θ3+λ3(1z)]Q3,0+z¯V(z)eλ2pd(1z). (3.18)

    Notice that when z(1,1), according to Rouche's theorem, we deduce that there exists a unique solution (for detailed proof, please refer to [6, Lemma 1]) to equation

    z=0μj(x)e[θj+λj(1z)]xx0μj(τ)dτdx,j=1,3,

    we assume that the solution is γj. Then, by the above Eqs (3.17) and (3.18), we obtain γ1 and γ3 that satisfy the following equations:

    γ1λ1Q1,0=0p1,1(x)μ1(x)dx, (3.19)
    Q3,0=¯V(γ3)eλ2pd(1γ3)[θ3+λ3(1γ3)]. (3.20)

    Using the above Eqs (3.16), (3.19), and (3.20), we have

    p1(0,z)=λ1Q1,0z(zγ1)z0μ1(x)e[θ1+λ1(1z)]xx0μ1(τ)dτdx, (3.21)
    p3(0,z)=z[θ3+λ3(1z)]Q3,0+z¯V(z)eλ2pd(1z)z0μ3(x)e[θ3+λ3(1z)]xx0μ3(τ)dτdx. (3.22)

    Hence, by Eqs (3.16), (3.21), (3.22), and (3.15), we obtain

    p1(1)=limz1p1(z)=limz10p1(x,z)dx=limz1λ1Q1,0z(zγ1)0e[θ1+λ1(1z)]xx0μ1(τ)dτdxz0μ1(x)e[θ1+λ1(1z)]xx0μ1(τ)dτdx=λ1Q1,0(1γ1)0eθ1xx0μ1(τ)dτdx10μ1(x)eθ1xx0μ1(τ)dτdx, (3.23)
    p3(1)=limz1p3(z)=limz10p3(x,z)dx=limz1[z(θ3+λ3(1z))Q3,0+z¯V(z)eλ2pd(1z)]z0μ3(x)e[θ3+λ3(1z)]xx0μ3(τ)dτdx×0μ3(x)e[θ3+λ3(1z)]xx0μ3(τ)dτdx=[θ3Q3,0+¯V(1)]0μ3(x)eθ3xx0μ3(τ)dτdx10μ3(x)eθ3xx0μ3(τ)dτdx, (3.24)
    ¯V(1)=limz1¯V(z)=θ1Q1,0+θ1p1(1)=θ1Q1,0+θ1λ1Q1,0(1γ1)0eθ1xx0μ1(τ)dτdx10μ1(x)eθ1xx0μ1(τ)dτdx. (3.25)

    Consequently, using Eqs (3.19), (3.20), (3.23)–(3.25) and the normalizing condition Q1,0+Q3,0+¯V(1)+p1(1)+p3(1)=1, we can obtain the specific expression for Q1,0, which we have omitted here. Notice that 0μj(x)eθjxx0μj(τ)dτdx<1,j=1,3. Hence, the above discussions mean that 0 is an eigenvalue of A.

    In addition, since Eqs (3.1)–(3.4) and (3.13), it is easy to see that the geometric multiplicity of zero is one.

    Next, we define the operator A0 with boundary conditions pj,n(0)=0,j=1,3;n1 for systems (2.1)–(2.11) as follows:

    A0(V,p1,p3)=Am(V,p1,p3),D(A0)={(V,p1,p3)D(Am)|Ψ(V,p1,p3)=0}.

    And find the resolvent set of this operator. For this objective, we consider the equation (γIA0)(V,p1,p3)=(w,y1,y3) for all (V,p1,p3)X, where w=(z1,0,z3,0,w0,w1,) and yj=(yj,1,yj,2,), j=1,3. This equation is equivalent to

    (γ+λ1+θ1)Q1,0=z1,0+θ3n=10p3,n(x)dx+θ3Q3,0+0p1,1(x)μ1(x)dx, (3.26)
    (γ+λ3+θ3)Q3,0=z3,0+N0V0+0p3,1(x)μ3(x)dx, (3.27)
    (γ+1)V0=w0+θ1Q1,0, (3.28)
    (γ+1)Vn=wn+θ10p1,n(x)dx,n1, (3.29)
    dp1,1(x)dx=[γ+λ1+θ1+μ1(x)]p1,1(x)+y1,1(x), (3.30)
    dp1,n(x)dx=[γ+λ1+θ1+μ1(x)]p1,n(x)+λ1p1,n1(x)+y1,n(x),n2, (3.31)
    dp3,1(x)dx=[γ+λ3+θ3+μ3(x)]p3,1(x)+y3,1(x), (3.32)
    dp3,n(x)dx=[γ+λ3+θ3+μ3(x)]p3,n(x)+λ3p3,n1(x)+y3,n(x),n2, (3.33)
    pj,n(0)=0,j=1,3;n1. (3.34)

    Solve the above Eqs (3.26)–(3.33), and using boundary conditions pj,n(0)=0, we have

    Q1,0=(γ+1)(γ+λ3+θ3)(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0z1,0+(γ+1)θ3(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0z3,0+θ3N0(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0w0+(γ+1)(γ+λ3+θ3)(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N00p1,1(x)μ1(x)dx+(γ+1)θ3(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N00p3,1(x)μ3(x)dx+(γ+1)(γ+λ3+θ3)θ3(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0n=10p3,n(x)dx, (3.35)
    Q3,0=θ1N0(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0z1,0+(γ+1)(γ+λ1+θ1)(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0z3,0+N0(γ+λ1+θ1)(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0w0+θ1N0(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N00p1,1(x)μ1(x)dx+(γ+1)(γ+λ1+θ1)(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N00p3,1(x)μ3(x)dx+θ1θ3N0(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0n=10p3,n(x)dx, (3.36)
    V0=θ1(γ+λ3+θ3)(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0z1,0+θ1θ3(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0z3,0+(γ+λ1+θ1)(γ+λ3+θ3)(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0w0+θ1(γ+λ3+θ3)(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N00p1,1(x)μ1(x)dx+θ1θ3(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N00p3,1(x)μ3(x)dx+θ1θ3(γ+λ3+θ3)(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0n=10p3,n(x)dx, (3.37)
    Vn=1γ+1wn+θ1γ+10p1,n(x)dx, (3.38)
    pj,1(x)=e(γ+λj+θj)xx0μj(τ)dτx0yj,1(τ)e(γ+λj+θj)τ+τ0μj(ξ)dξdτ,j=1,3, (3.39)
    pj,n(x)=e(γ+λj+θj)xx0μj(τ)dτx0yj,n(τ)e(γ+λj+θj)τ+τ0μj(ξ)dξdτ+λje(γ+λj+θj)xx0μj(τ)dτx0pj,n1(τ)e(γ+λj+θj)τ+τ0μj(ξ)dξdτ,n1. (3.40)

    For convenience, we introduce Ej as follows:

    Ejf(x):=e(γ+λj+θj)xx0μj(τ)dτx0f(τ)e(γ+λj+θj)τ+τ0μj(ξ)dξdτ,j=1,3, (3.41)

    for any fL1[0,). Then, if (γIA0)1 exists, then by Eqs (3.35)–(3.40), we have

    (γIA0)1(w,y1,y3)=((a1,1a1,2a1,300a2,1a2,2a2,300a3,1a3,2a3,3000001γ+1000001γ+1)(z1,0z3,0w0w1w2)+(a1,4φ2E1000a2,4φ2E1000a3,4φ2E1000θ1γ+1φ1E1000θ1γ+1λ1φ1E21θ1γ+1φ1E100θ1γ+1λ21φ1E31θ1γ+1λ1φ1E21θ1γ+1φ1E10)(y1,1y1,2y1,3y1,4)+(a1,5φ3+a1,6φ1a1,6φ1a1,6φ1a2,5φ3+a2,6φ1a2,6φ1a2,6φ1a3,5φ3+a3,6φ1a3,6φ1a3,6φ1000000)(E300λ3E23E30λ23E33λ3E23E3)(y3,1y3,2y3,3),(E100λ1E21E10λ21E31λ1E21E1)(y1,1y1,2y1,3),(E300λ3E23E30λ23E33λ3E23E3)(y3,1y3,2y3,3)), (3.42)

    where

    φ1f(x)=0f(x)dx,φ2f(x)=0μ1(x)f(x)dx,φ3f(x)=0μ3(x)f(x)dx,

    and

    a1,1=a1,4=(γ+1)(γ+λ3+θ3)(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0, (3.43)
    a1,2=a1,5=(γ+1)θ3(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0, (3.44)
    a1,3=θ3N0(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0, (3.45)
    a1,6=(γ+1)(γ+λ3+θ3)θ3(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0, (3.46)
    a2,1=a2,4=θ1N0(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0, (3.47)
    a2,2=a2,5=(γ+1)(γ+λ1+θ1)(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0, (3.48)
    a2,3=N0(γ+λ1+θ1)(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0, (3.49)
    a2,6=θ1θ3N0(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0, (3.50)
    a3,1=θ1(γ+λ3+θ3)(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0, (3.51)
    a3,2=a3,5=θ1θ3(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0, (3.52)
    a3,3=(γ+λ1+θ1)(γ+λ3+θ3)(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0, (3.53)
    a3,4=θ1(γ+λ3+θ3)(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0, (3.54)
    a3,6=θ1θ3(γ+λ3+θ3)(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0. (3.55)

    Lemma 3.2. Let λj>0,θj>0,N0(0,1), and μj(x):[0,)[0,) be measurable and

    0<infx[0,)μj(x)μj(x)supx[0,)μj(x)<,j=1,3.

    Then, S=:{γC|γ+1>0,γ+θj+infx[0,)μj(x)>0}ρ(A0).

    Proof. The proof of this lemma is provided in the appendix.

    Lemma 3.3. Let λj>0,θj>0,N0(0,1), and μj(x):[0,)[0,) be measurable and

    0<infx[0,)μj(x)μj(x)supx[0,)μj(x)<,j=1,3.

    If γS, then (V,p1,p3)ker(γIAm) if and only if

    Q1,0=a1,4p1,1(0)0μ1(x)e(γ+λ1+θ1)xx0μ1(τ)dτdx+a1,5p3,1(0)0μ3(x)e(γ+λ3+θ3)xx0μ3(τ)dτdx+a1,6n=1nk=1p3,nk+1(0)0(λ3x)k1(k1)!e(γ+λ3+θ3)xx0μ3(τ)dτdx, (3.56)
    Q3,0=a2,4p1,1(0)0μ1(x)e(γ+λ1+θ1)xx0μ1(τ)dτdx+a2,5p3,1(0)0μ3(x)e(γ+λ3+θ3)xx0μ3(τ)dτdx+a2,6n=1nk=1p3,nk+1(0)0(λ3x)k1(k1)!e(γ+λ3+θ3)xx0μ3(τ)dτdx, (3.57)
    V0=a3,4p1,1(0)0μ1(x)e(γ+λ1+θ1)xx0μ1(τ)dτdx+a3,5p3,1(0)0μ3(x)e(γ+λ3+θ3)xx0μ3(τ)dτdx+a3,6n=1nk=1p3,nk+1(0)0(λ3x)k1(k1)!e(γ+λ3+θ3)xx0μ3(τ)dτdx, (3.58)
    Vn=θ1γ+1nk=1p1,nk+1(0)0(λ1x)k1(k1)!e(γ+λ1+θ1)xx0μ1(τ)dτdx, (3.59)
    pj,n(x)=e(γ+λj+θj)xx0μj(τ)dτnk=1(λjx)k1(k1)!pj,nk+1(0),j=1,3;n1, (3.60)
    pj=(pj,1,pj,2,)l1,j=1,3, (3.61)

    where the specific expressions for ak,m,k=1,2,3;m=4,5,6 are given in Eqs (3.43)–(3.55).

    Proof. The proof of this lemma is provided in the appendix.

    It is clear that the boundary operator Ψ is surjective. In addition, for γS, we see that the operator

    Ψ|ker(γIAm):ker(γIAm)X,

    is a reversible operator.

    Now, for γS, we define the Dirichlet operator Dγ as follows:

    Dγ:=(Ψ|ker(γIAm))1:Xker(γIAm).

    Then, for any γS, using Lemma 3.3, we obtain the specific expression of the operator Dγ:

    Dγ(p1(0),p3(0))=((a1,4φ2m1,1000a2,4φ2m1,1000a3,4φ2m1,1000θ1γ+1φ1m1,1000θ1γ+1φ1m1,2θ1γ+1φ1m1,100θ1γ+1φ1m1,3θ1γ+1φ1m1,2θ1γ+1φ1m1,10)(p1,1(0)p1,2(0)p1,3(0)p1,4(0))+(a1,5φ3+a1,6φ1a1,6φ1a1,6φ1a2,5φ3+a2,6φ1a2,6φ1a2,6φ1a3,5φ3+a3,6φ1a3,6φ1a3,6φ1000000)(m3,100m3,2m3,10m3,3m3,2m3,1)(p3,1(0)p3,2(0)p3,3(0)),(m1,100m1,2m1,10m1,3m1,2m1,1)(p1,1(0)p1,2(0)p1,3(0)),(m3,100m3,2m3,10m3,3m3,2m3,1)(p3,1(0)p3,2(0)p3,3(0))), (3.62)

    where

    mj,k=(λjx)k1(k1)!e(γ+λj+θj)xx0μj(τ)dτ,j=1,3;k1.

    Finally, by the definitions of operators Φ and Dγ, we calculate that

    ΦDγ(p1(0),p3(0))=((λ1δ100)+(φ2[m1,2p1,1(0)+m1,1p1,2(0)]φ2[m1,3p1,1(0)+m1,2p1,2(0)+m1,1p1,3(0)]φ2[m1,4p1,1(0)+m1,3p1,2(0)+m1,2p1,3(0)+m1,1p1,4(0)]),(λ3δ2+N1δ3+N0θ1γ+1φ1m1,1p1,1(0)N2δ3+N1θ1γ+1φ1m1,1p1,1(0)+N0θ1γ+12k=1φ1m1,kp1,2k+1(0)Υ)+(φ3[m3,2p3,1(0)+m3,1p3,2(0)]φ3[m3,3p3,1(0)+m3,2p3,2(0)+m3,1p3,3(0)]φ3[m3,4p3,1(0)+m3,3p3,2(0)+m3,2p3,3(0)+m3,1p3,4(0)])), (3.63)

    where

    Υ=N3δ3+N2θ1γ+1φ1m1,1p1,1(0)+N1θ1γ+12k=1φ1m1,kp1,2k+1(0)+N0θ1γ+13k=1φ1m1,kp1,3k+1(0),δ1=a1,4φ2m1,1p1,1(0)+a1,5φ3m3,1p3,1(0)+a1,6n=1φ1m3,nk=1p3,k(0),δ2=a2,4φ2m1,1p1,1(0)+a2,5φ3m3,1p3,1(0)+a2,6n=1φ1m3,nk=1p3,k(0),δ3=a3,4φ2m1,1p1,1(0)+a3,5φ3m3,1p3,1(0)+a3,6n=1φ1m3,nk=1p3,k(0).

    Based on a conclusion drawn by Haji and Radl [11], we know that the following result, Lemma 3.4, holds true.

    Lemma 3.4. If γρ(A0) and there exists some γ0 such that 1 is not in the spectrum σ(ΦDγ0) of ΦDγ0, then the conclusion

    γσ(A)if and only if1σ(ΦDγ)

    holds.

    Hence, using Lemma 3.4 and Nagel [13, p. 297], we have the following result:

    Lemma 3.5. Let λj>0,θj>0,Nr(0,1),r0, and μj(x):[0,)[0,) be measurable and

    0<infx[0,)μj(x)μj(x)supx[0,)μj(x)<,j=1,3.

    Then, all points on the imaginary axis, except for 0, fall on the resolvent set ρ(A) of A.

    Proof. If we take γ=ib,bR{0},p1(0)=(p1,1(0),p1,2(0),)l1 and p3(0)=(p3,1(0),p3,2(0),)l1. The Riemann–Lebesgue Lemma states that for any fL1[0,) (i.e., f is an integrable function on [0,), we have

    limb0f(x)cos(bx)dx=0,limb0f(x)sin(bx)dx=0.

    This means that as b approaches infinity, the integrals of f(x) with high-frequency cosine or sine functions tend to zero. Using Euler's formula eibx=cos(bx)isin(bx), the integral |0f(x)eibxdx| can be split into real and imaginary parts:

    |0f(x)eibxdx|=|0f(x)cos(bx)dxi0f(x)sin(bx)dx|. (3.64)

    Then, according to the Riemann–Lebesgue Lemma, for sufficiently large |b|, both of integrals of Eq (3.64) tend to zero.

    In addition, for any 0<fL1[0,), we have

    |0f(x)eibxdx|0f(x)dx. (3.65)

    Then, for any 0<fL1[0,) and |b|>M1 (where M1 is some sufficiently large constant), according to Eqs (3.64), (3.65), and the Riemann–Lebesgue Lemma, we can further obtain

    |0f(x)eibxdx|2=(0f(x)cos(bx)dx)2+(0f(x)sin(bx)dx)2<(0f(x)dx)2. (3.66)

    That is, for any 0<fL1[0,) and |b|>M1, we have

    |0f(x)eibxdx|<0f(x)dx. (3.67)

    Furthermore, through tedious calculations, we have found that the following inequality holds true

    |λ1a1,h|<1,|λ3a2,h|<1,|(1N0)a3,h|<1,h=4,5,6, (3.68)

    where the specific expressions for ak,h(k=1,2,3;h=4,5,6) are given in Eqs (3.43)–(3.55) and N0=eλ2pd(<1) is the probability of zero arrivals during phase 2. In fact,

    |(ib+1)(ib+λ1+θ1)(ib+λ3+θ3)θ1θ3N0|2λ21|(ib+1)(ib+λ3+θ3)|2=b6+b4[2λ1θ1+θ21+(λ3+θ3)2+1]+b2{2λ1θ1+θ21+(λ3+θ3)2+(2λ1θ1+θ21)(λ3+θ3)2+2(λ1+θ1+λ3+θ3+1)θ1θ3N0}+2λ1(λ3+θ3)[θ1(λ3+θ3)θ1θ3N0]+[θ1(λ3+θ3)θ1θ3N0]2>0|λ1a1,4|=λ1|(ib+1)(ib+λ3+θ3)||(ib+1)(ib+λ1+θ1)(ib+λ3+θ3)θ1θ3N0|<1, (3.69)
    |(ib+1)(ib+λ1+θ1)(ib+λ3+θ3)θ1θ3N0|2=b6+b4[(λ1+θ1)2+(λ3+θ3)2+1]+b2{(λ1+θ1)2+(λ3+θ3)2+[(λ1+θ1)(λ3+θ3)]2+2(λ1+θ1+λ3+θ3+1)θ1θ3N0}+[(λ1+θ1)(λ3+θ3)θ1θ3N0]2>θ21[b2+(λ3+θ3)2]>[θ1(1N0)]2[b2+(λ3+θ3)2]=|θ1(1N0)(ib+λ3+θ3)|2|(1N0)a3,4|=θ1(1N0)|ib+λ3+θ3||(ib+1)(ib+λ1+θ1)(ib+λ3+θ3)θ1θ3N0|<1. (3.70)

    Using the same method as Eqs (3.69) and (3.70), we can prove that the remaining inequalities in Eq (3.68) also hold true.

    Hence, for any |b|>M1, using Eqs (3.63), (3.67), and (3.68), the formula

    r=0Nr=r=0eλ2pd(λ2pd)rr!=1,0[θj+μj(x)]ex0[θj+μj(τ)]dτdx=1,j=1,3,

    and 1b2+1<1, through tedious calculations, we estimate that

    ΦDib(p1(0),p3(0))λ1|a1,4||φ2m1,1||p1,1(0)|+λ1|a1,5||φ3m3,1||p3,1(0)|+λ1|a1,6|n=1|φ1m3,n|k=1|p3,k(0)|+|φ2m1,2||p1,1(0)|+|φ2m1,1||p1,2(0)|+|φ2m1,3||p1,1(0)|+|φ2m1,2||p1,2(0)|+|φ2m1,1||p1,3(0)|+|φ2m1,4||p1,1(0)|+|φ2m1,3||p1,2(0)|+|φ2m1,2||p1,3(0)|+|φ2m1,1||p1,4(0)|++λ3|a2,4||φ2m1,1||p1,1(0)|+λ3|a2,5||φ3m3,1||p3,1(0)|+λ3|a2,6|n=1|φ1m3,n|k=1|p3,k(0)|+N1|a3,4||φ2m1,1||p1,1(0)|+N1|a3,5||φ3m3,1||p3,1(0)|+N1|a3,6|n=1|φ1m3,n|k=1|p3,k(0)|+N0θ1|ib+1||φ1m1,1||p1,1(0)|+|φ3m3,2||p3,1(0)|+|φ3m3,1||p3,2(0)|+N2|a3,4||φ2m1,1||p1,1(0)|+N2|a3,5||φ3m3,1||p3,1(0)|+N2|a3,6|n=1|φ1m3,n|k=1|p3,k(0)|+N1θ1|ib+1||φ1m1,1||p1,1(0)|+N0θ1|ib+1|2k=1|φ1m1,k||p1,2k+1(0)|+|φ3m3,3||p3,1(0)|+|φ3m3,2||p3,2(0)|+|φ3m3,1||p3,3(0)|+N3|a3,4||φ2m1,1||p1,1(0)|+N3|a3,5||φ3m3,1||p3,1(0)|+N3|a3,6|n=1|φ1m3,n|k=1|p3,k(0)|+N2θ1|ib+1||φ1m1,1||p1,1(0)|+N1θ1|ib+1|2k=1|φ1m1,k||p1,2k+1(0)|+N0θ1|ib+1|3k=1|φ1m1,k||p1,3k+1(0)|+|φ3m3,4||p3,1(0)|+|φ3m3,3||p3,2(0)|+|φ3m3,2||p3,3(0)|+|φ3m3,1||p3,4(0)|+λ1|a1,4||φ2m1,1||p1,1(0)|+λ1|a1,5||φ3m3,1||p3,1(0)|+λ1|a1,6|n=1|φ1m3,n|k=1|p3,k(0)|+n=2|φ2m1,n||p1,1(0)|+n=1|φ2m1,n||p1,2(0)|+n=1|φ2m1,n||p1,3(0)|+n=1|φ2m1,n||p1,4(0)|++λ3|a2,4||φ2m1,1||p1,1(0)|+λ3|a2,5||φ3m3,1||p3,1(0)|+λ3|a2,6|n=1|φ1m3,n|k=1|p3,k(0)|+j=1Nj|a3,4|φ2m1,1||p1,1(0)|+j=1Nj|a3,5||φ3m3,1||p3,1(0)|+j=1Nj|a3,6|n=1|φ1m3,n|k=1|p3,k(0)|+j=0Njθ1b2+1|φ1m1,1||p1,1(0)|+j=0Njθ1b2+1|φ1m1,2||p1,1(0)|+j=0Njθ1b2+1|φ1m1,3||p1,1(0)|+j=0Njθ1b2+1|φ1m1,4||p1,1(0)|++j=0Njθ1b2+1|φ1m1,1||p1,2(0)|+j=0Njθ1b2+1|φ1m1,2||p1,2(0)|+j=0Njθ1b2+1|φ1m1,3||p1,2(0)|+j=0Njθ1b2+1|φ1m1,4||p1,2(0)|++j=0Njθ1b2+1|φ1m1,1||p1,3(0)|+j=0Njθ1b2+1|φ1m1,2||p1,3(0)|+j=0Njθ1b2+1|φ1m1,3||p1,3(0)|+j=0Njθ1b2+1|φ1m1,4||p1,3(0)|++j=0Njθ1b2+1|φ1m1,1||p1,4(0)|+j=0Njθ1b2+1|φ1m1,2||p1,4(0)|+j=0Njθ1b2+1|φ1m1,3||p1,4(0)|+j=0Njθ1b2+1|φ1m1,4||p1,4(0)|++n=2|φ3m3,n||p3,1(0)|+n=1|φ3m3,n||p3,2(0)|+n=1|φ3m3,n||p3,3(0)|+n=1|φ3m3,n||p3,4(0)|+<(θ3n=1|φ1m3,n|+n=1|φ3m3,n|)k=1|p3,k(0)|+(n=1|φ2m1,n|+θ1n=1|φ1m1,n|)k=1|p1,k(0)|=(θ3n=1|0(λ3x)n1(n1)!e(ib+λ3+θ3)xx0μ3(τ)dτdx|+n=1|0μ3(x)(λ3x)n1(n1)!e(ib+λ3+θ3)xx0μ3(τ)dτdx|)k=1|p3,k(0)|+(n=1|0μ1(x)(λ1x)n1(n1)!e(ib+λ1+θ1)xx0μ1(τ)dτdx|+θ1n=1|0(λ1x)n1(n1)!e(ib+λ1+θ1)xx0μ3(τ)dτdx|)k=1|p1,k(0)|(θ3n=10(λ3x)n1(n1)!e(λ3+θ3)xx0μ3(τ)dτdx+n=10μ3(x)(λ3x)n1(n1)!e(λ3+θ3)xx0μ3(τ)dτdx)k=1|p3,k(0)|+(n=10μ1(x)(λ1x)n1(n1)!e(λ1+θ1)xx0μ1(τ)dτdx+θ1n=10(λ1x)n1(n1)!e(λ1+θ1)xx0μ3(τ)dτdx)k=1|p1,k(0)|=(θ30n=1(λ3x)n1(n1)!e(λ3+θ3)xx0μ3(τ)dτdx+0μ3(x)n=1(λ3x)n1(n1)!e(λ3+θ3)xx0μ3(τ)dτdx)k=1|p3,k(0)|+(0μ1(x)n=1(λ1x)n1(n1)!e(λ1+θ1)xx0μ1(τ)dτdx+θ10n=1(λ1x)n1(n1)!e(λ1+θ1)xx0μ3(τ)dτdx)k=1|p1,k(0)|=0[θ3+μ3(x)]ex0[θ3+μ3(τ)]dτdxk=1|p3,k(0)|+0[θ1+μ1(x)]ex0[θ1+μ1(τ)]dτdxk=1|p1,k(0)|=p1(0),p3(0). (3.71)

    That is, ΦDib<1. Inequality (3.71) means that when |b|>M1, the spectral bound r(ΦDib)ΦDib<1. That is, we have 1σ(ΦDib) if |b|>M1. Therefore, by Lemma 3.4, we deduce that

    {ib|b|>M1}ρ(A),{ib|b|<M1}σ(A)iR. (3.72)

    On the other hand, since the semigroup eAt is positive uniformly bounded by Theorem 2.1, by [13, Corollary 2.3, p. 297] we see that σ(A)iR is imaginary additively cyclic, which shows that if ibσ(A)iR, then ibnσ(A)iR for all integer n. Hence, from which, together with Eq (3.72) and Lemma 3.1, we conclude that ibσ(A)iR={0}.

    It is easy to see that the dual space X of X is as follows (see [9]):

    X={(V,p1,p3)|V=(Q1,0,Q3,0,V0,V1,)l,p1=(p1,1,p1,2,p1,3,)L[0,)×L[0,)×L[0,)×,p3=(p3,1,p3,2,p3,3,)L[0,)×L[0,)×L[0,)×,(V,p1,p3)=sup{|Q1,0|,|Q3,0|,supj0|Vj|,supj1p1,jL[0,),supj1p3,jL[0,)}<}.

    Clearly, X also is a Banach space.

    Lemma 3.6. The adjoint operator A of A is as follows:

    A(V,p1,p3)=(((λ1+θ1)0θ100θ3(λ3+θ3)0000N01000001000001)(Q1,0Q3,0V0V1V2)+(λ100000000)(p1,1(0)p1,2(0)p1,3(0))+(0000λ3000N1N2N3N4N0N1N2N30N0N1N2)(p3,1(0)p3,2(0)p3,3(0)p3,4(0)),(μ1(x)00θ1000000θ1000000θ1000000)(Q1,0Q3,0V0V1)+(0000μ1(x)0000μ1(x)0000μ1(x)0)(p1,1(0)p1,2(0)p1,3(0)p1,4(0))+(ζ1λ100ζ1λ100ζ1)(p1,1(x)p1,2(x)p1,3(x)),(θ3μ3(x)00θ3000θ3000θ3000)(Q1,0Q3,0V0V1)+(0000μ3(x)0000μ3(x)0000μ3(x)0)(p3,1(0)p3,2(0)p3,3(0)p3,4(0))+(ζ3λ300ζ3λ300ζ3)(p3,1(x)p3,2(x)p3,3(x))),

    where ζj=ddx[λj+θj+μj(x)],j=1,3 and

    D(A)={(qI,qw)X|pj,n()(n1)are absolutely continuousfunctions and satisfypj,n()=α,αisa nonzero constant which is irrelevant to n }.

    Proof. For any (V,p1,p3)D(A) and (V,p1,p3)D(A), we calculate that

    A(V,p1,p3),(V,p1,p3)=[(λ1+θ1)Q1,0+θ3n=10p3,n(x)dx+θ3Q3,0+0μ1(x)p1,1(x)dx]Q1,0+[(λ3+θ3)Q3,0+N0V0+0μ3(x)p3,1(x)dx]Q3,0+(V0+θ1Q1,0)V0+n=1[Vn+θ10p1,n(x)dx]Vn+0[dp1,1(x)dx(λ1+θ1+μ1(x))p1,1(x)]p1,1(x)dx+n=20[dp1,n(x)dx(λ1+θ1+μ1(x))p1,n(x)+λ1p1,n1(x)]p1,n(x)dx+0[dp3,1(x)dx(λ3+θ3+μ3(x))p3,1(x)]p3,1(x)dx+n=20[dp3,n(x)dx(λ3+θ3+μ3(x))p3,n(x)+λ1p3,n1(x)]p3,n(x)dx=(λ1+θ1)Q1,0Q1,0+θ3n=10p3,nQ1,0(x)dx+θ3Q3,0Q1,0+0μ1(x)p1,1(x)(x)Q1,0dx(λ3+θ3)Q3,0Q3,0+N0V0Q3,0+0μ3(x)p3,1(x)Q3,0dx+θ1Q1,0V0n=0VnVn+θ1n=10p1,n(x)Vndx+n=1p1,n(0)p1,n(0)+n=10p1,n(x)[dp1,n(x)dx(λ1+θ1+μ1(x))p1,n(x)]dx+λ1n=10p1,n(x)p1,n+1(x)dx+n=1p3,n(0)p3,n(0)+n=10p3,n(x)[dp3,n(x)dx(λ3+θ3+μ3(x))p3,n(x)]dx+λ3n=10p3,n(x)p3,n+1(x)dx=(λ1+θ1)Q1,0Q1,0+θ3n=10p3,nQ1,0(x)dx+θ3Q3,0Q1,0+0μ1(x)p1,1(x)Q1,0dx(λ3+θ3)Q3,0Q3,0+N0V0Q3,0+0μ3(x)p3,1(x)Q3,0dx+θ1Q1,0V0n=0VnVn+θ1n=10p1,n(x)Vndx+λ1Q1,0p1,1(0)+n=10μ1(x)p1,n+1(x)p1,n(0)dx+n=10p1,n(x)[dp1,n(x)dx(λ1+θ1+μ1(x))p1,n(x)]dx+λ1n=10p1,n(x)p1,n+1(x)dx+λ3Q3,0p3,1(0)+n=1nk=0NkVnkp3,n(0)+n=10μ3(x)p3,n+1(x)p3,n(0)dx+n=10p3,n(x)[dp3,n(x)dx(λ3+θ3+μ3(x))p3,n(x)]dx+λ3n=10p3,n(x)p3,n+1(x)dx=(V,p1,p3),A(V,p1,p3). (3.73)

    Eq (3.73) means that Lemma 3.6 holds true.

    Lemma 3.7. Let λj>0,θj>0, Nr(0,1),r0, and 0<μj()<,j=1,3. Then, 0 is an eigenvalue of A with geometric multiplicity 1.

    Proof. We consider the equation A(V,p1,p3)=0. That is,

    (λ1+θ1)Q1,0+θ1V0=0, (3.74)
    θ3Q1,0(λ3+θ3)Q3,0+λ3p3,1(0)=0, (3.75)
    N0Q3,0V0+k=1Nkp3,k(0)=0, (3.76)
    Vn+k=0Nkp3,k+n(0)=0,n1, (3.77)
    dp1,1(x)dx[λ1+θ1+μ1(x)]p1,1(x)+λ1p1,2(x)+θ1V1+μ1(x)Q1,0=0, (3.78)
    dp1,n(x)dx[λ1+θ1+μ1(x)]p1,n(x)+λ1p1,n+1(x)+θ1Vn+μ1(x)p1,n1(0)=0,n2, (3.79)
    dp3,1(x)dx[λ3+θ3+μ3(x)]p3,1(x)+λ3p3,2(x)+θ3Q1,0+μ3(x)Q3,0=0, (3.80)
    dp3,n(x)dx[λ3+θ3+μ3(x)]p3,n(x)+λ3p3,n+1(x)+θ3Q1,0+μ3(x)p3,n1(0)=0,n2, (3.81)
    pj,n()=α,j=1,3;n1. (3.82)

    It is easy to see that

    (V,p1,p3)=((αα),(αα),(αα))D(A),

    is a nonzero solution of Eqs (3.74)–(3.82). In addition, Eqs (3.74)–(3.81) are equivalent to

    Q1,0=θ1(λ3+θ3)(λ1+θ1)(λ3+θ3)N0θ1θ3(λ3N0λ3+θ3p3,1(0)+k=1Nkp3,k(0)), (3.83)
    Q3,0=θ1θ3(λ1+θ1)(λ3+θ3)N0θ1θ3(λ3N0λ3+θ3p3,1(0)+k=1Nkp3,k(0)), (3.84)
    V0=(λ1+θ1)(λ3+θ3)(λ1+θ1)(λ3+θ3)N0θ1θ3(λ3N0λ3+θ3p3,1(0)+k=1Nkp3,k(0)), (3.85)
    Vn=k=0Nkp3,k+n(0),n1, (3.86)
    p1,2(x)=1λ1[dp1,1(x)dx(λ1+θ1+μ1(x))p1,1(x)+θ1k=0Nkp3,k+1(0)+θ1(λ3+θ3)μ1(x)(λ1+θ1)(λ3+θ3)N0θ1θ3(λ3N0λ3+θ3p3,1(0)+k=1Nkp3,k(0))], (3.87)
    p1,n+1(x)=1λ1[dp1,n(x)dx(λ1+θ1+μ1(x))p1,n(x)+θ1k=0Nkp3,k+n(0)]μ1(x)λ1p3,n1(0),n2, (3.88)
    p3,2(x)=1λ3[dp3,1(x)dx(λ3+θ3+μ3(x))p3,1(x)+θ1θ3(λ3+θ3)(λ1+θ1)(λ3+θ3)N0θ1θ3(λ3N0λ3+θ3p3,1(0)+k=1Nkp3,k(0))+θ1θ3μ3(x)(λ1+θ1)(λ3+θ3)N0θ1θ3(λ3N0λ3+θ3p3,1(0)+k=1Nkp3,k(0))], (3.89)
    p3,n+1(x)=1λ3[dp3,n(x)dx(λ3+θ3+μ3(x))p3,n(x)+μ3(x)p3,n1(0)+θ1θ3(λ3+θ3)(λ1+θ1)(λ3+θ3)N0θ1θ3(λ3N0λ3+θ3p3,1(0)+k=1Nkp3,k(0))],n2. (3.90)

    Consequently, Eqs (3.83)–(3.90) mean that the geometric multiplicity of 0 is one.

    Finally, using Lemmas 3.1, 3.5, and 3.7 and Theorem 1.96 of [12], we obtain the following desired result.

    Theorem 3.1. Let λj>0,θj>0,Nr(0,1),r0, and μj(x):[0,)[0,) be measurable and

    0<infx[0,)μj(x)μj(x)supx[0,)μj(x)<,j=1,3.

    Then, the time-dependent solution of system (2.12) strongly converges to its steady-state solution, that is,

    limt(V(t),p1(,t),p3(,t))(V,p1(),p3()),(V,p1,p3)(0))(V,p1(),p3())=0, (3.91)

    where (V,p1(),p3()) and (V,p1(),p3()) are the eigenvectors in Lemmas 3.1 and 3.7, respectively, and (V,p1,p3)(0)) is the initial value of system (2.12).

    Proof. Theorem 2.1 establishes that the operator A generates a uniformly bounded C0semigroup on the Banach space X. Furthermore, leveraging Lemmas 3.5, 3.1, and 3.7, we can readily deduce that

    σp(A)iR=σp(A)iR={0},

    and that the set {γCγ=ib,b0,bR} is a subset of the resolvent set ρ(A). Additionally, zero is an eigenvalue of A with geometric multiplicity one.

    Consequently, invoking Theorem 1.96 from [12], we conclude that the time-dependent solution of the system (2.12) converges strongly to its steady-state solution. Specifically, the limit Eq (3.91) holds true.

    In this paper, we investigate the asymptotic behavior of the M/G/1 stochastic clearing queueing model in a three-phase environment, specifically when the service rate of the server is a bounded function. By employing probability generating functions and the boundary perturbation method of Greiner, we demonstrate that all points on the imaginary axis, except for 0, fall within the resolvent set of the system operator. Additionally, we highlight that 0 is an eigenvalue of the adjoint operator of the system operator, with a geometric multiplicity of 1. This finding implies that the time-dependent solution of the system strongly converges to its steady-state solution.

    This theoretical result provides a solid foundation for understanding the long-term behavior of the system. However, the implications of strong convergence for practical system performance metrics, such as queue-length distributions, transient behavior, and the rate of convergence, are equally important for real-world applications. Strong convergence implies that over time, the system's transient behavior will closely approximate its steady-state behavior. This means that for sufficiently large times, the queue-length distribution and other performance metrics will be well-represented by their steady-state counterparts. In practical settings, this suggests that after an initial transient period, the system will exhibit stable performance characteristics that can be effectively estimated using steady-state analysis.

    However, for the more general case of the service rate function, where 0μj(), we have not yet determined whether the above results still hold. Furthermore, it remains unclear whether the time-dependent solution exponentially converges to its steady-state solution. To address the exponential stability of this system, we need to investigate the spectrum of the system operator on the left half of the complex plane, as discussed in [14,15]. These topics are among our future research endeavors.

    The approach outlined in this paper is specifically tailored for queuing systems that are characterized through partial differential equations [16]. It is not applicable to the queuing systems discussed in [17,18]. There have been extensive studies on the asymptotic behavior of semigroups (see [7,8,12,19]). which are also of significant interest for our future research.

    The authors declare that they have not used Artificial Intelligence (AI) tools in the creation of this article.

    We are grateful to the anonymous referees, who read carefully the manuscript and made valuable comments and suggestions. This work was supported by the Natural Science Foundation of Xinjiang Uygur Autonomous Region (No: 2024D01C229) and the National Natural Science Foundation of China (No: 12301150).

    The authors declare there is no conflict of interest.

    The proof of Lemma 3.2. For all fL1[0,), we compute

    0|Ejf(x)|dx=0|e(γ+λj+θj)xx0μj(τ)dτx0f(τ)e(γ+λj+θj)τ+τ0μj(ξ)dξdτ|dx0e(γ+λj+θj)xx0μj(τ)dτx0|f(τ)|e(γ+λj+θj)τ+τ0μj(ξ)dξdτdx=01γ+λj+θj+μj(x)x0|f(τ)|e(γ+λj+θj)τ+τ0μj(ξ)dξdτde(γ+λj+θj)xx0μj(τ)dτ=1γ+λj+θj+μj(x)e(γ+λj+θj)xx0μj(τ)dτx0|f(τ)|e(γ+λj+θj)τ+τ0μj(ξ)dξdτ|x=x=0+01γ+λj+θj+μj(x)e(γ+λj+θj)xx0μj(τ)dτe(γ+λj+θj)x+x0μj(τ)dτ|f(x)|dx=limx1γ+λj+θj+μj(x)e(γ+λj+θj)xx0μj(τ)dτx0|f(τ)|e(γ+λj+θj)τ+τ0μj(ξ)dξdτ+01γ+λj+θj+μj(x)|f(x)|dx=01γ+λj+θj+μj(x)|f(x)|dx1γ+λj+θj+infx[0,)μj(x)fL1[0,),j=1,3.

    This means that

    Ej1γ+λj+θj+infx[0,)μj(x),j=1,3. (A.1)

    Then, for any (w,y1,y3)X, using the inequalities φ11, φ2supx[0,)μ1(x), φ3supx[0,)μ3(x) and Eq (3.56), we estimate

    (γIA0)1(w,y1,y3)=(a1,1+a2,1+a3,1)z1,0+(a1,2+a2,2+a3,2)z3,0+(a1,3+a2,3+a3,3)w0+1γ+1n=1wn+(a1,4+a2,4+a3,4)φ2E1y1,1+θ1γ+1[φ1E1y1,1+(λ1φ1E21y1,1+φ1E1y1,2)+(λ21φ1E31y1,1+λ1φ1E21y1,2+φ1E1y1,3)+(λ31φ1E41y1,1+λ21φ1E31y1,2+λ1φ1E21y1,3+φ1E1y1,4)+]+(a1,5+a2,5+a3,5)φ3E3y3,1+(a1,6+a2,6+a3,6)[φ1E3y3,1+(λ3φ1E23y3,1+φ1E3y3,2)+(λ23φ1E33y3,1+λ3φ1E23y3,2+φ1E3y3,3)+(λ33φ1E43y3,1+λ23φ1E33y3,2+λ3φ1E23y3,3+φ1E3y3,4)+]+(E1y1,1+λ1E21y1,1+E1y1,2+λ21E31y1,1+λ1E21y1,2+E1y1,3+λ31E41y1,1+λ21E31y1,2+λ1E21y1,3+)+(E3y3,1+λ3E23y3,1+E3y3,2+λ23E33y3,1+λ3E23y3,2+E3y3,3+λ33E43y3,1+λ23E33y3,2+λ3E23y3,3+)|a1,1+a2,1+a3,1||z1,0|+|a1,2+a2,2+a3,2||z3,0|+|a1,3+a2,3+a3,3||w0|+1|γ+1|n=1|wn|+|a1,4+a2,4+a3,4|φ2E1y1,1L1[0,)+θ1|γ+1|[φ1E1y1,1L1[0,)+λ1φ1E12y1,1L1[0,)+φ1E1y1,2L1[0,)+λ21φ1E13y1,1L1[0,)+λ1φ1E12y1,2L1[0,)+φ1E1y1,3L1[0,)+λ31φ1E14y1,1L1[0,)+λ21φ1E13y1,2L1[0,)+λ1φ1E12y1,3L1[0,)+φ1E1y1,4L1[0,)+]+|a1,5+a2,5+a3,5|φ3E3y3,1L1[0,)+|a1,6+a2,6+a3,6|[φ1E3y3,1L1[0,)+λ3φ1E32y3,1L1[0,)+φ1E3y3,2L1[0,)+λ23φ1E33y3,1L1[0,)+λ3φ1E32y3,2L1[0,)+φ1E3y3,3L1[0,)+λ33φ1E34y3,1L1[0,)+λ23φ1E33y3,2L1[0,)+λ3φ1E32y3,3L1[0,)+φ1E3y3,4L1[0,)+]+E1y1,1L1[0,)+λ1E12y1,1L1[0,)+E1y1,2L1[0,)+λ21E13y1,1L1[0,)+λ1E12y1,2L1[0,)+E1y1,3L1[0,)+λ31E14y1,1L1[0,)+λ21E13y1,2L1[0,)+λ1E12y1,3L1[0,)++E3y3,1L1[0,)+λ3E32y3,1L1[0,)+E3y3,2L1[0,)+λ23E33y3,1L1[0,)+λ3E32y3,2L1[0,)+E3y3,3L1[0,)+λ33E34y3,1L1[0,)+λ23E33y3,2L1[0,)+λ3E32y3,3L1[0,)+=|a1,1+a2,1+a3,1||z1,0|+|a1,2+a2,2+a3,2||z3,0|+|a1,3+a2,3+a3,3||w0|+1|γ+1|n=1|wn|+|a1,4+a2,4+a3,4|φ2E1y1,1L1[0,)+θ1|γ+1|φ1k=1λk11E1kn=1y1,nL1[0,)+|a1,5+a2,5+a3,5|φ3E3y3,1L1[0,)+|a1,6+a2,6+a3,6|φ1k=1λk13E3kn=1y3,nL1[0,)+k=1λk11E1kn=1y1,nL1[0,)+k=1λk13E3kn=1y3,nL1[0,)|a1,1+a2,1+a3,1||z1,0|+|a1,2+a2,2+a3,2||z3,0|+|a1,3+a2,3+a3,3||w0|+1γ+1n=1|wn|+|a1,4+a2,4+a3,4|supx[0,)μ1(x)γ+λ1+θ1+infx[0,)μ1(x)y1,1L1[0,)+θ1γ+11γ+λ1+θ1+infx[0,)μ1(x)×k=1(λ1γ+λ1+θ1+infx[0,)μ1(x))k1n=1y1,nL1[0,)+|a1,5+a2,5+a3,5|supx[0,)μ3(x)γ+λ3+θ3+infx[0,)μ3(x)y3,1L1[0,)+|a1,6+a2,6+a3,6|1γ+λ3+θ3+infx[0,)μ3(x)×k=1(λ3γ+λ3+θ3+infx[0,)μ3(x))k1n=1y3,nL1[0,)+1γ+λ1+θ1+infx[0,)μ1(x)×k=1(λ1γ+λ1+θ1+infx[0,)μ1(x))k1n=1y1,nL1[0,)+1γ+λ3+θ3+infx[0,)μ3(x)×k=1(λ3γ+λ3+θ3+infx[0,)μ3(x))k1n=1y3,nL1[0,)sup{|a1,1+a2,1+a3,1|+|a1,2+a2,2+a3,2|+|a1,3+a2,3+a3,3|+1γ+1,|a1,4+a2,4+a3,4|supx[0,)μ1(x)γ+λ1+θ1+infx[0,)μ1(x)+θ1γ+11γ+θ1+infx[0,)μ1(x)+1γ+θ1+infx[0,)μ1(x),|a1,5+a2,5+a3,5|supx[0,)μ3(x)γ+λ3+θ3+infx[0,)μ3(x)+|a1,6+a2,6+a3,6|1γ+θ3+infx[0,)μ3(x)+1γ+θ3+infx[0,)μ3(x)}(w,y1,y3)=sup{|(γ+1)(γ+λ3+θ3)+θ1N0+θ1(γ+λ3+θ3)||(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0|+|(γ+1)θ3+(γ+1)(γ+λ1+θ1)+θ1θ3||(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0|+|θ3N0+(γ+λ1+θ1)N0+(γ+λ1+θ1)(γ+λ3+θ3)||(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0|+1γ+1,|(γ+1)(γ+λ3+θ3)+θ1N0+θ1(γ+λ3+θ3)||(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0|supx[0,)μ1(x)γ+λ1+θ1+infx[0,)μ1(x)+θ1γ+11γ+θ1+infx[0,)μ1(x)+1γ+θ1+infx[0,)μ1(x),|(γ+1)θ3+(γ+1)(γ+λ1+θ1)+θ1θ3||(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0|supx[0,)μ3(x)γ+λ3+θ3+infx[0,)μ3(x)+|(γ+1)(γ+λ3+θ3)θ3+θ1θ3N0+θ1θ3(γ+λ3+θ3)||(γ+1)(γ+λ1+θ1)(γ+λ3+θ3)θ1θ3N0|1γ+θ3+infx[0,)μ3(x)+1γ+θ3+infx[0,)μ3(x)}(w,y1,y3). (A.2)

    Inequality (A.2) means that the result of Lemma 3.2 holds true.

    The proof of Lemma 3.3. We assume that (V,p1,p3)ker(γIAm), then (γIAm)(V,p1,p3)=0. That is,

    (γ+λ1+θ1)Q1,0=θ3n=10p3,n(x)dx+θ3Q3,0+0p1,1(x)μ1(x)dx, (A.3)
    (γ+λ3+θ3)Q3,0=N0V0+0p3,1(x)μ3(x)dx, (A.4)
    (γ+1)V0=θ1Q1,0, (A.5)
    (γ+1)Vn=θ10p1,n(x)dx,n1, (A.6)
    dp1,1(x)dx=[γ+λ1+θ1+μ1(x)]p1,1(x), (A.7)
    dp1,n(x)dx=[γ+λ1+θ1+μ1(x)]p1,n(x)+λ1p1,n1(x),n2, (A.8)
    dp3,1(x)dx=[γ+λ3+θ3+μ3(x)]p3,1(x), (A.9)
    dp3,n(x)dx=[γ+λ3+θ3+μ3(x)]p3,n(x)+λ3p3,n1(x),n2. (A.10)

    By solving Eqs (A.7)–(A.10), we have

    pj,n(x)=e(γ+λj+θj)xx0μi(τ)dτnk=1(λjx)k1(k1)!pj,nk+1(0),j=1,3;n1. (A.11)

    By Eqs (A.3)–(A.6), we have

    Q1,0=θ3γ+λ1+θ1n=10p3,n(x)dx+θ3γ+λ1+θ1Q3,0+1γ+λ1+θ10p1,1(x)μ1(x)dx, (A.12)
    Q3,0=N0γ+λ3+θ3V0+1γ+λ3+θ30p3,1(x)μ3(x)dx, (A.13)
    V0=θ1γ+1Q1,0, (A.14)
    Vn=1γ+10p1,n(x)dx,n1. (A.15)

    Then, from Eqs (A.11)–(A.15), by a simple calculation, we can obtain that Eqs (3.56)–(3.59) in Lemma 3.3 hold true.

    Moreover, since (V,p1,p3)ker(γIAm), using the Sobolev imbedding theorem [20, Theorem 4.12], we estimate

    n=1|pj,n(0)|n=1pj,nL[0,)n=1(pj,nL1[0,)+dpj,ndxL1[0,))<,j=1,3. (A.16)

    Hence, we conclude that Eqs (3.56)–(3.61) in Lemma 3.3 hold true.

    On the other hand, if the Eqs (3.56)–(3.61) holds true, then for any positive constant M0 and k1, using the formula

    0eM0xxkdx=k!Mk+10,

    we estimate

    pj,nL1[0,)=0|e(γ+λj+θj)xx0μj(τ)dτnk=1(λjx)k1(k1)!pj,nk+1(0)|dxnk=1λk1j(k1)!|pj,nk+1(0)|0xk1e[γ+λj+θj+infx[0,)μj(x)]xdxnk=1λk1j(k1)!|pj,nk+1(0)|(k1)!(γ+λj+θj+infx[0,)μj(x))k=nk=1λk1j(γ+λj+θj+infx[0,)μj(x))k|pj,nk+1(0)|. (A.17)

    To sum n from 1 to positive infinity on both sides of the above inequality and using the condition γ+θj+infx[0,)μj(x)>0, we have

    n=1pj,nL1[0,)=n=1nk=1λk1j(γ+λj+θj+infx[0,)μj(x))k|pj,nk+1(0)|=k=1λk1j(γ+λj+θj+infx[0,)μj(x))kn=1|pj,n(0)|=1γ+θj+infx[0,)μj(x)n=1|pj,n(0)|<,j=1,3. (A.18)

    By taking the derivative of x on both sides of Eq (A.18), we can obtain

    dpj,1(x)dx=[γ+λj+θj+μj(x)]pj,1(0)e(γ+λj+θj)xx0μj(τ)dτ=[γ+λj+θj+μj(x)]pj,1(x),j=1,3, (A.19)
    dpj,n(x)dx=[γ+λj+θj+μj(x)]pj,n(x)+λjpj,n1(x),j=1,3;n1. (A.20)

    Combining the Eqs (A.18)–(A.20), we derive

    n=1dpj,ndxL1[0,)[|γ|+2λj+θj+supx[0,)μj(x)]n=1pj,nL1[0,). (A.21)

    Therefore, Eqs (A.18)–(A.21) imply that γρ(A0) and

    (V,p1,p3)ker(γIAm).
    Table 1.  List of notations.
    Symbol First occurrence & Description
    Qj,0(t) Page 2: The probability that there are no customers in phase j(j=1,3) and the server is idle at time t.
    Vn(t) Page 2: The probability that there are n(n0) customers in the system and the system is in phase 2.
    pj,n(x,t)dx Page 2: The probability that there are n(n1) customers in the system at time t with the server busy serving a customer whose elapsed service time lies in the interval [x,x+dx) in phase j(j=1,3).
    λj Page 4: The arrival rate of customers when the system is in phase j(j=1,2,3).
    μj(x) Page 4: The conditional probability of completing a service during the interval (x,x+dx) with elapsed time x in phase j(j=1,3).
    θj Page 4: The residence rate of the system in phase j(j=1,3).
    Nr Page 5: The probability of r(r0) arrivals during phase 2.
    X Page 5: The state space of systems (2.1)–(2.11).
    X Page 24: The dual space of X.
    X Page 6: The boundary space of X.
    Am Page 5: The maximal operator of systems (2.1)–(2.11).
    D(Am) Page 5: The domain of operator Am.
    Φ,Ψ Page 6: The boundary operators of systems (2.1)–(2.2).
    A Page 7: The system operator of systems (2.1)–(2.11).
    A Page 8: The adjoint operator of A.
    A0 Page 12: The system operator with zero boundary conditions for systems (2.1)–(2.11).
    ρ(A) Page 8: The resolvent set of A.
    σ(A) Page 18: The spectrum set of A.
    σp(A) Page 29: The point spectrum set of A.
    γ Page 8: The real part of γ.
    Dγ Page 16: The Dirichlet operator.

     | Show Table
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