Dynamic Analysis of the M/G/1 Stochastic Clearing Queueing Model in a Three-Phase Environment

Abstract

In this paper, we consider the M/G/1 stochastic clearing queueing model in a three-phase environment, which is described by integro-partial differential equations (IPDEs). Our first result is semigroup well-posedness for the dynamic system. Utilizing a $C_0$—semigroup theory, we prove that the system has a unique positive time-dependent solution (TDS) that satisfies the probability condition. As our second result, we prove that the TDS of the system strongly converges to its steady-state solution (SSS) if the service rates of the servers are constants. For this asymptotic behavior, we analyze the spectrum of the system operator associated with the system. Additionally, the stability of the semigroup generated by the system operator is also discussed.

1. Introduction

A stochastic clearing system receives and accumulates random inputs at random time intervals until certain predetermined criteria are met; then, some or all of these inputs are immediately cleared. This type of system has many practical applications in manufacturing, healthcare industries, tourism, transportation, etc. The stochastic clearing systems have a wide range of applications in queuing systems, see, e.g., [1,2,3,4,5].

In this paper, we investigate the M/G/1 stochastic clearing queueing model in a three-phase environment. The system is described as follows:

• There are three phases in this system, phase 1 and phase 3 are working phases, and phase 2 is a deterministic time phase without service.
• The Poisson arrival rate is ${\lambda }_j(j=1,2,3)$ in phase j, the service times in phase $i(i=1,3)$ follow a general distribution and the system residing in phase i follows an exponential distribution with parameter ${\theta }_i$.
• After phase 1, the system will enter phase 2, and if a new customer arrives during phase 2, the customer enters the system with a probability p or leaves the system with probability $1-p$, the customers in phase 2 do not receive service, and they experience a fixed duration d.
• After phase 2, the system will enter phase 3. Once phase 3 is complete, the current customer is forced to leave the system without receiving service, and then the system enters phase 1 and restarts a new service cycle.
• ${\mu }_j(·)(j=1,3)$ is the service rate in phase j and satisfies ${\mu }_j(·)\ge 0,{\int }_0^{\infty }{\mu }_j\left(x\right)dx=\infty$.
• $N_r:=e^{-{\lambda }_{2}pd}{\left({\lambda }_{2}pd\right)}^r{(r!)}^{-1}(r\ge 0)$ denotes the probability of r customers arriving in phase 2.

Additionally, we assume that

$$Q_{j,0}\left(t\right)=Prob\left\{\begin{array}{c}\mathrm{the}\mathrm{server}\mathrm{being}\mathrm{idle}\mathrm{at}\mathrm{time}t\mathrm{and}\mathrm{no}\hfill \\ \mathrm{customer}\mathrm{being}\mathrm{present}\mathrm{in}\mathrm{phase}j,j=1,3\hfill \end{array}\right\},$$

$$V_n\left(t\right)=Prob\left\{\begin{array}{c}\mathrm{the}\mathrm{system}\mathrm{in}\mathrm{phase}2\mathrm{and}\mathrm{there}\mathrm{are}n(n\ge 0)\hfill \\ \mathrm{customers}\mathrm{in}\mathrm{the}\mathrm{system}\mathrm{at}\mathrm{time}t\hfill \end{array}\right\},$$

$$p_{j,n}(x,t)dx=Prob\left\{\begin{array}{c}\mathrm{there}\mathrm{are}n(n\ge 1)\mathrm{customers}\mathrm{in}\mathrm{the}\mathrm{system}\mathrm{at}\mathrm{time}t\hfill \\ \mathrm{in}\mathrm{phase}j(j=1,3)\mathrm{and}\mathrm{the}\mathrm{service}\mathrm{time}\hfill \\ \mathrm{used}\mathrm{for}\mathrm{providing}\mathrm{services}\mathrm{to}\mathrm{customers}\mathrm{is}x\hfill \end{array}\right\}.$$

Then, the system can be described by the following IPDEs:

$$\left\{\begin{array}{c}{\displaystyle \frac{d{Q}_{1,0}\left(t\right)}{dt}=-({\lambda }_1+{\theta }_1)Q_{1,0}\left(t\right)+{\int }_0^{\infty }{p}_{1,1}(x,t){\mu }_1\left(x\right)dx+{\theta }_3\sum _{n=1}^{\infty }{\int }_0^{\infty }{p}_{3,n}(x,t)dx}\hfill \\ {\displaystyle +{\theta }_3{Q}_{3,0}\left(t\right)}\hfill \\ {\displaystyle \frac{d{Q}_{3,0}\left(t\right)}{dt}=-({\lambda }_3+{\theta }_3)Q_{3,0}\left(t\right)+N_0{V}_0\left(t\right)+{\int }_0^{\infty }{p}_{3,1}(x,t){\mu }_3\left(x\right)dx,}\hfill \\ {\displaystyle \frac{d{V}_0\left(t\right)}{dt}=-V_0\left(t\right)+{\theta }_1{Q}_{1,0}\left(t\right),}\hfill \\ {\displaystyle \frac{d{V}_n\left(t\right)}{dt}=-V_n\left(t\right)+{\theta }_1{\int }_0^{\infty }{p}_{1,n}(x,t)dx,n\ge 1,}\hfill \\ {\displaystyle {\partial }_t{p}_{j,1}(x,t)+{\partial }_x{p}_{j,1}(x,t)=-[{\lambda }_j+{\theta }_j+{\mu }_j\left(x\right)]p_{j,1}(x,t),j=1,3,}\hfill \\ {\displaystyle {\partial }_t{p}_{j,n}(x,t)+{\partial }_x{p}_{j,n}(x,t)=-[{\lambda }_j+{\theta }_j+{\mu }_j\left(x\right)]p_{j,n}(x,t)+{\lambda }_j{p}_{j,n-1}(x,t),n\ge 2,}\hfill \end{array}\right.$$

(1)

with the following boundary and initial conditions:

$$\left\{\begin{array}{c}{\displaystyle p_{1,1}(0,t)={\lambda }_1{Q}_{1,0}\left(t\right)+{\int }_0^{\infty }{p}_{1,2}(x,t){\mu }_1\left(x\right)dx,}\hfill \\ {\displaystyle p_{1,n}(0,t)={\int }_0^{\infty }{p}_{1,n+1}(x,t){\mu }_1\left(x\right)dx,n\ge 2,}\hfill \\ {\displaystyle p_{3,1}(0,t)={\lambda }_3{Q}_{3,0}\left(t\right)+N_1{V}_0\left(t\right)+N_0{V}_1\left(t\right)+{\int }_0^{\infty }{p}_{3,2}(x,t){\mu }_3\left(x\right)dx,}\hfill \\ {\displaystyle p_{3,n}(0,t)=\sum _{k=0}^n{N}_k{V}_{n-k}\left(t\right)+{\int }_0^{\infty }{p}_{3,n+1}(x,t){\mu }_3\left(x\right)dx,n\ge 2,}\hfill \\ {\displaystyle V_n\left(0\right)=u_{2,n}\ge 0,n\ge 0;Q_{j,1}\left(0\right)={\overline{u}}_{2,j}\ge 0,p_{j,n}(x,0)=u_{j,n}\left(x\right)\ge 0,n\ge 1.}\hfill \end{array}\right.$$

(2)

Here, $(x,t)\in [0,\infty )\times [0,\infty )$, ${\overline{u}}_{2,1}+{\overline{u}}_{2,3}+{\sum }_{n=0}^{\infty }{u}_{2,n}+{\sum }_{j=1,3}{\sum }_{n=1}^{\infty }{\int }_0^{\infty }{u}_{j,n}\left(x\right)dx=1$.

In [6], some steady-state indices for system (1) and (2) such as the steady-state queue length and the steady-state sojourn time distribution of customers were developed under the following hypothesis:

• ${lim}_{t\to \infty }{Q}_{j,0}\left(t\right)=Q_{j,0},j=1,3$,
• ${lim}_{t\to \infty }{V}_n\left(t\right)=V_n,n\ge 0$,
• ${lim}_{t\to \infty }{p}_{j,n}(·,t)=p_{j,n}(·),j=1,3,n\ge 1$,

which means the following hypotheses from the perspective of PDEs, see, e.g., [7,8]:

Hypothesis 1 (H1).

System (1) and (2) admit a nonnegative TDS.

Hypothesis 2 (H2).

The TDS converges to its SSS.

In this paper, we show that the above hypotheses ($\mathbf{H}\mathbf{1}$) and ($\mathbf{H}\mathbf{2}$) are satisfied, and we discuss the stability of the corresponding semigroup of system (1) and (2). The difficulty in the well-posedness and asymptotic behavior is that the system is composed of an infinite number of IPDEs, with integral boundary conditions. However, the system (1) and (2) we consider in our work are problems in an $L^1$—based nonreflexive Banach space, and proving semigroups in this space is not easy. Hence, we provide the asymptotic behavior of system (1) and (2) when the service rates ${\mu }_1(·)$ and ${\mu }_3(·)$ are constants, which simplifies many calculation processes. If we consider system (1) and (2) only in phase 2 and phase 3, then system (1) and (2) become the M/G/1 queueing model with optional deterministic server vacations, see, e.g., [9,10]. In [9], they proved that hypotheses ($\mathbf{H}\mathbf{1}$) and ($\mathbf{H}\mathbf{2}$) hold true for the optional deterministic server vacations queueing model. According to [11], it can be seen that if the reliable retrial queueing model (see, e.g., [12], p. 9), is extended to the case of the retrial queueing model with server breakdowns (see, e.g., [13]), the steady-state assumption of this server breakdowns queueing model does not hold true. Therefore, it is necessary to investigate whether the hypotheses ($\mathbf{H}\mathbf{1}$) and ($\mathbf{H}\mathbf{2}$) of system (1) and (2) are valid.

In this paper, first of all, we transform system (1) and (2) into a Cauchy problem in a Banach space. Second, by using the semigroup theory, we verify that the system operator of system (1) and (2) generates a positive $C_0$—semigroup of contractions in the space, which is isometric under certain conditions. Thus, we show that the system has a unique positive TDS that satisfies the probability condition, which means ($\mathbf{H}\mathbf{1}$) holds under certain conditions.

To answer ($\mathbf{H}\mathbf{2}$), that is, to obtain the asymptotic properties of the TDS of system (1) and (2), we need to calculate the spectrum of the system operator, see, e.g., [7,11,14,15,16,17]. When the service rates ${\mu }_1(·)$ and ${\mu }_3(·)$ are constants, the M/G/1 stochastic clearing queueing model in a three-phase environment is called the M/M/1 stochastic clearing queueing model in a three-phase environment. In this case, by studying the spectral distribution of a system operator on the imaginary axis, we show that zero is a point spectrum of the operator and its adjoint operator with geometric multiplicity one, and other points on the imaginary axis are not spectrums of the operator. Hence, we obtain that the TDS of system (1) and (2) strongly converges to its non-zero SSS. This implies ($\mathbf{H}\mathbf{2}$) holds under strong convergence.

Furthermore, we consider the spectrum-determined growth condition (SDGC), as well as the stability of the semigroup corresponding to system (1) and (2). We obtain that the SDGC is equal to zero, and the semigroup is not asymptotically stable.

The organization of this paper is as follows. In Section 2, we write system (1) and (2) as a Cauchy problem in a Banach space and present the well-posedness. Based on the spectrum analysis, Section 3 is devoted to the asymptotic behavior of the TDS of Equations (1) and (2). The stability of the corresponding semigroup of system (1) and (2) is further investigated in Section 4, which leads to the SDGC. Section 5 concludes the paper.

2. Well-Posedness of the System

We consider system (1) and (2) in the following state Banach space

$$X=\left\{(V,p_1,p_3)\left|\begin{array}{c}V=(Q_{1,0},Q_{3,0},V_0,V_1,V_2,\cdots )\in l^1,p_j=\left(p_{j,1},p_{j,2},\cdots \right),\hfill \\ p_{j,n}\in L^1[0,\infty ),\parallel (V,p_1,p_3)\parallel =\sum _{j=1,3}|Q_{j,0}|\hfill \\ +\sum _{n=0}^{\infty }|V_n|+\sum _{j=1,3}\sum _{n=1}^{\infty }{\parallel p_{j,n}\parallel }_{L^1[0,\infty )}<\infty \hfill \end{array}\right.\right\}.$$

In the following, we define the system operator and its domain. We define the operator A and its domain by

$$\begin{array}{cc}& A(V,p_1,p_3)=\left(\left(\begin{array}{c}-({\lambda }_1+{\theta }_1)Q_{1,0}+{\theta }_3{Q}_{3,0}\\ -({\lambda }_3+{\theta }_3)Q_{3,0}+N_0{V}_0\\ -V_0\\ -V_1\\ ⋮\end{array}\right),\left(\begin{array}{c}-\frac{d{p}_{1,1}}{dx}\\ -\frac{d{p}_{1,2}}{dx}\\ -\frac{d{p}_{1,3}}{dx}\\ ⋮\end{array}\right),\left(\begin{array}{c}-\frac{d{p}_{3,1}}{dx}\\ -\frac{d{p}_{3,2}}{dx}\\ -\frac{d{p}_{3,3}}{dx}\\ ⋮\end{array}\right)\right),\hfill \end{array}$$

$$D\left(A\right)=\left\{(V,p_1,p_3)\left|\begin{array}{c}\frac{d{p}_{j,n}}{dx}\in L^1[0,\infty ),p_{j,n}\mathrm{are}\mathrm{absolutely}\mathrm{continuous}\hfill \\ \mathrm{functions}\mathrm{and}{p}_1\left(0\right)={\int }_0^{\infty }{\mathrm{\Gamma }}_{1,1}Vdx+{\int }_0^{\infty }{\mathrm{\Gamma }}_{1,2}{p}_1\left(x\right)dx,\hfill \\ p_3\left(0\right)={\int }_0^{\infty }{\mathrm{\Gamma }}_{3,1}Vdx+{\int }_0^{\infty }{\mathrm{\Gamma }}_{3,2}{p}_3\left(x\right)dx+\mathrm{\Gamma }V\hfill \end{array}\right.\right\},$$

where $j=1,3;n\ge 1$, and

$${\mathrm{\Gamma }}_{1,1}=\left(\begin{array}{ccc}{\lambda }_1{e}^{-x}& 0& \cdots \\ 0& 0& \cdots \\ ⋮& ⋮& \ddots \end{array}\right),{\mathrm{\Gamma }}_{3,1}=\left(\begin{array}{cccc}0& {\lambda }_3{e}^{-x}& 0& \cdots \\ 0& 0& 0& \cdots \\ ⋮& ⋮& ⋮& \ddots \end{array}\right),$$

$${\mathrm{\Gamma }}_{j,2}=\left(\begin{array}{ccccc}0& {\mu }_j\left(x\right)& 0& 0& \cdots \\ 0& 0& {\mu }_j\left(x\right)& 0& \cdots \\ 0& 0& 0& {\mu }_j\left(x\right)& \cdots \\ ⋮& ⋮& ⋮& \ddots \end{array}\right),\mathrm{\Gamma }=\left(\begin{array}{ccccccc}0& 0& N_1& N_0& 0& 0& \cdots \\ 0& 0& N_2& N_1& N_0& 0& \cdots \\ 0& 0& N_3& N_2& N_1& N_0& \cdots \\ ⋮& ⋮& ⋮& ⋮& \ddots \end{array}\right).$$

Define the operators U and E and their domains by

$$\begin{array}{cc}U(V,p_1,p_3)\hfill & =\left(\left(\begin{array}{c}0\\ 0\\ 0\\ ⋮\end{array}\right),\left(\begin{array}{c}-{\eta }_1\left(x\right)p_{1,1}\left(x\right)\\ {\lambda }_1{p}_{1,1}\left(x\right)-{\eta }_1\left(x\right)p_{1,2}\left(x\right)\\ {\lambda }_1{p}_{1,2}\left(x\right)-{\eta }_1\left(x\right)p_{1,3}\left(x\right)\\ ⋮\end{array}\right),\left(\begin{array}{c}-{\eta }_3\left(x\right)p_{3,1}\left(x\right)\\ {\lambda }_3{p}_{3,1}\left(x\right)-{\eta }_3\left(x\right)p_{3,2}\left(x\right)\\ {\lambda }_3{p}_{3,2}\left(x\right)-{\eta }_3\left(x\right)p_{3,3}\left(x\right)\\ ⋮\end{array}\right)\right),\hfill \end{array}$$

$$\begin{array}{cc}E(V,p_1,p_3)\hfill & =\left(\left(\begin{array}{c}{\int }_0^{\infty }{p}_{1,1}\left(x\right){\mu }_1\left(x\right)dx+{\theta }_3\sum _{n=1}^{\infty }{\int }_0^{\infty }{p}_{3,n}\left(x\right)dx\\ {\int }_0^{\infty }{p}_{3,1}\left(x\right){\mu }_3\left(x\right)dx\\ 0\\ {\theta }_1{\int }_0^{\infty }{p}_{1,1}\left(x\right)dx\\ {\theta }_1{\int }_0^{\infty }{p}_{1,2}\left(x\right)dx\\ ⋮\end{array}\right),\left(\begin{array}{c}0\\ 0\\ ⋮\end{array}\right),\left(\begin{array}{c}0\\ 0\\ ⋮\end{array}\right)\right)\hfill \end{array},$$

$$D\left(U\right)=X,D\left(E\right)=X,$$

where ${\eta }_j(·)={\lambda }_j+{\theta }_j+{\mu }_j(·),j=1,3$. Then, system (1) and (2) can be written as a Cauchy problem in X:

$$\left\{\begin{array}{c}{\displaystyle \frac{d(V\left(t\right),p_1(·,t),p_3(·,t))}{dt}=(A+U+E)(V\left(t\right),p_1(·,t),p_3(·,t)),t\in (0,\infty ),}\hfill \\ (V\left(0\right),p_1(·,0),p_3(·,0))=\left(u_2,u_1(·),u_3(·)\right),\hfill \\ u_2=({\overline{u}}_{2,1},{\overline{u}}_{2,3},u_{2,0},u_{2,1},\cdots ),u_j(·)=(u_{j,1}(·),u_{j,2}(·),\cdots ),j=1,3.\hfill \end{array}\right.$$

(3)

Now, we study the well-posedness of system (3). Firstly, we prove that the operator $A+U+E$ generates a positive $C_0$-semigroup $e^{(A+U+E)t}$ of contractions on X. Secondly, we show that $A+U+E$ is a conservative operator, from which we obtain that $e^{(A+U+E)t}$ is an isometric semigroup on X. Hence, we present the well-posedness of system (3).

Theorem 1.

If ${\lambda }_j,{\theta }_j>0$ and $0<{\overline{\mu }}_j={sup}_{x\in [0,\infty )}{\mu }_j\left(x\right)<\infty ,j=1,3$, then $A+U+E$ generates a positive $C_0$-semigroup $e^{(A+U+E)t}$ of contractions on X.

Proof. 

The proof will be finished by four steps.

Step 1. If $\gamma >\mathbb{M}=max\{{\overline{\mu }}_1,{\overline{\mu }}_3\}$ and ${\overline{\mu }}_3\ge N_0$, then ${(\gamma I-A)}^{-1}$ exists and is bounded.

For any $(w,y_1,y_3)\in X$, we consider $(\gamma I-A)(V,p_1,p_3)=(w,y_1,y_3)$ for unknown $(V,p_1,p_3)$, where $w=(z_{1,0},z_{3,0},w_0,w_1,\cdots )$, $y_j=(y_{j,1},y_{j,2},y_{j,3},\cdots ),j=1,3$. This is equivalent to

$$(\gamma +{\lambda }_1+{\theta }_1)Q_{1,0}=z_{1,0}+{\theta }_3{Q}_{3,0},$$

(4a)

$$(\gamma +{\lambda }_3+{\theta }_3)Q_{3,0}=z_{3,0}+N_0{V}_0,$$

(4b)

$$(\gamma +1)V_0=w_0+{\theta }_1{Q}_{1,0},$$

(4c)

$$(\gamma +1)V_n=w_n,n\ge 1,$$

(4d)

$$\frac{d{p}_{j,n}\left(x\right)}{dx}=-\gamma p_{j,n}\left(x\right)+y_{j,n}\left(x\right),j=1,3;n\ge 1,$$

(4e)

$$p_{1,1}\left(0\right)={\lambda }_1{Q}_{1,0}+{\int }_0^{\infty }{p}_{1,2}\left(x\right){\mu }_1\left(x\right)dx$$

(4f)

$$p_{1,n}\left(0\right)={\int }_0^{\infty }{p}_{1,n+1}\left(x\right){\mu }_1\left(x\right)dx,n\ge 2,$$

(4g)

$$p_{3,1}\left(0\right)={\lambda }_3{Q}_{3,0}+N_1{V}_0+N_0{V}_1+{\int }_0^{\infty }{p}_{3,2}\left(x\right){\mu }_3\left(x\right)dx,$$

(4h)

$$p_{3,n}\left(0\right)=\sum _{k=0}^n{N}_k{V}_{n-k}+{\int }_0^{\infty }{p}_{3,n+1}\left(x\right){\mu }_3\left(x\right)dx,n\ge 2.$$

(4i)

Solve system (4e) to obtain

$$p_{1,n}\left(x\right)=a_n{e}^{-\gamma x}+e^{-\gamma x}{\int }_0^x{y}_{1,n}\left(\tau \right)e^{\gamma \tau }d\tau ,n\ge 1,$$

(5a)

$$p_{3,n}\left(x\right)=b_n{e}^{-\gamma x}+e^{-\gamma x}{\int }_0^x{y}_{3,n}\left(\tau \right)e^{\gamma \tau }d\tau ,n\ge 1,$$

(5b)

where $a_n=p_{1,n}\left(0\right),b_n=p_{3,n}\left(0\right)$. Combining the boundary conditions system (4f)–(4i) with system (5a) and (5b), we obtain that $a_n$ and $b_n$ satisfy

$$a_1-a_2{\int }_0^{\infty }{\mu }_1\left(x\right)e^{-\gamma x}dx={\lambda }_1{Q}_{1,0}+{\int }_0^{\infty }{\mu }_1\left(x\right)e^{-\gamma x}{\int }_0^x{y}_{1,2}\left(\tau \right)e^{\gamma \tau }d\tau dx,$$

(6a)

$$a_n-a_{n+1}{\int }_0^{\infty }{\mu }_1\left(x\right)e^{-\gamma x}dx={\int }_0^{\infty }{\mu }_1\left(x\right)e^{-\gamma x}{\int }_0^x{y}_{1,n+1}\left(\tau \right)e^{\gamma \tau }d\tau dx,n\ge 2,$$

(6b)

$$b_1-b_2{\int }_0^{\infty }{\mu }_3\left(x\right)e^{-\gamma x}dx={\lambda }_3{Q}_{3,0}+N_1{V}_0+N_0{V}_1+{\int }_0^{\infty }{\mu }_3\left(x\right)e^{-\gamma x}{\int }_0^x{y}_{3,2}\left(\tau \right)e^{\gamma \tau }d\tau dx,$$

(6c)

$$b_n-b_{n+1}{\int }_0^{\infty }{\mu }_3\left(x\right)e^{-\gamma x}dx=\sum _{k=0}^n{N}_k{V}_{n-k}+{\int }_0^{\infty }{\mu }_3\left(x\right)e^{-\gamma x}{\int }_0^x{y}_{3,n+1}\left(\tau \right)e^{\gamma \tau }d\tau dx,n\ge 2.$$

(6d)

If we define

$$C_j:=\left(\begin{array}{cccc}1& -{\alpha }_j& 0& \cdots \\ 0& 1& -{\alpha }_j& \cdots \\ ⋮& ⋮& ⋮& \ddots \end{array}\right),\overrightarrow{a}=\left(\begin{array}{c}{a}_1\\ a_2\\ ⋮\end{array}\right),\overrightarrow{b}=\left(\begin{array}{c}{b}_1\\ b_2\\ ⋮\end{array}\right),j=1,3,$$

(7)

where ${\alpha }_j={\int }_0^{\infty }{\mu }_j\left(x\right)e^{-\gamma x}dx$, then system (6a)–(7) can be written as

$$C_1\overrightarrow{a}=\left(\begin{array}{c}{\lambda }_1{Q}_{1,0}+{\int }_0^{\infty }{\mu }_1\left(x\right)e^{-\gamma x}{\int }_0^x{y}_{1,2}\left(\tau \right)e^{\gamma \tau }d\tau dx\\ {\int }_0^{\infty }{\mu }_1\left(x\right)e^{-\gamma x}{\int }_0^x{y}_{1,3}\left(\tau \right)e^{\gamma \tau }d\tau dx\\ {\int }_0^{\infty }{\mu }_1\left(x\right)e^{-\gamma x}{\int }_0^x{y}_{1,4}\left(\tau \right)e^{\gamma \tau }d\tau dx\\ ⋮\end{array}\right),$$

(8a)

$$C_3\overrightarrow{b}=\left(\begin{array}{c}{\lambda }_3{Q}_{3,0}+\sum _{k=0}^1{N}_k{V}_{1-k}+{\int }_0^{\infty }{\mu }_3\left(x\right)e^{-\gamma x}{\int }_0^x{y}_{3,2}\left(\tau \right)e^{\gamma \tau }d\tau dx\\ \sum _{k=0}^2{N}_k{V}_{2-k}+{\int }_0^{\infty }{\mu }_3\left(x\right)e^{-\gamma x}{\int }_0^x{y}_{3,3}\left(\tau \right)e^{\gamma \tau }d\tau dx\\ \sum _{k=0}^3{N}_k{V}_{3-k}+{\int }_0^{\infty }{\mu }_3\left(x\right)e^{-\gamma x}{\int }_0^x{y}_{3,4}\left(\tau \right)e^{\gamma \tau }d\tau dx\\ ⋮\end{array}\right).$$

(8b)

The inverse of $C_j$ can be directly calculated as

$$C_j^{-1}=\left(\begin{array}{ccccc}1& {\alpha }_j& {\alpha }_j^2& {\alpha }_j^3& \cdots \\ 0& 1& {\alpha }_j& {\alpha }_j^2& \cdots \\ 0& 0& 1& {\alpha }_j& \cdots \\ ⋮& ⋮& ⋮& ⋮& \ddots \end{array}\right),j=1,3.$$

(9)

By Equations (8a)–(9), it is easy to see that

$$a_1={\lambda }_1{Q}_{1,0}+\sum _{m=1}^{\infty }{\left({\int }_0^{\infty }{\mu }_1\left(x\right)e^{-\gamma x}dx\right)}^{m-1}{\int }_0^{\infty }{\mu }_1\left(x\right)e^{-\gamma x}{\int }_0^x{y}_{1,m+1}\left(\tau \right)e^{\gamma \tau }d\tau dx,$$

(10a)

$$a_n=\sum _{m=1}^{\infty }{\left({\int }_0^{\infty }{\mu }_1\left(x\right)e^{-\gamma x}dx\right)}^{m-1}{\int }_0^{\infty }{\mu }_1\left(x\right)e^{-\gamma x}{\int }_0^x{y}_{1,m+n}\left(\tau \right)e^{\gamma \tau }d\tau dx,n\ge 2,$$

(10b)

$$\begin{array}{cc}{b}_1\hfill & {\displaystyle ={\lambda }_3{Q}_{3,0}+\sum _{m=1}^{\infty }{\left({\int }_0^{\infty }{\mu }_3\left(x\right)e^{-\gamma x}dx\right)}^{m-1}}\hfill \\ \hfill & {\displaystyle \times \left(\sum _{k=0}^m{N}_k{V}_{m-k}+{\int }_0^{\infty }{\mu }_3\left(x\right)e^{-\gamma x}{\int }_0^x{y}_{3,m+1}\left(\tau \right)e^{\gamma \tau }d\tau dx\right),}\hfill \end{array}$$

(10c)

$$\begin{array}{cc}{b}_n\hfill & {\displaystyle =\sum _{m=1}^{\infty }{\left({\int }_0^{\infty }{\mu }_3\left(x\right)e^{-\gamma x}dx\right)}^{m-1}}\hfill \\ \hfill & {\displaystyle \times \left(\sum _{k=0}^{m+n-1}{N}_k{V}_{m+n-1-k}+{\int }_0^{\infty }{\mu }_3\left(x\right)e^{-\gamma x}{\int }_0^x{y}_{3,m+n}\left(\tau \right)e^{\gamma \tau }d\tau dx\right),n\ge 2.}\hfill \end{array}$$

(10d)

For any $\gamma >0$, from system (10a)–(10d) and the Fubini theorem, it is easy to calculate that

$$|a_1|\le {\lambda }_1|Q_{1,0}|+\sum _{m=1}^{\infty }{\left(\frac{{\overline{\mu }}_1}{\gamma }\right)}^m{\parallel y_{1,m+1}\parallel }_{L^1[0,\infty )},$$

(11a)

$${\displaystyle |a_n|\le \sum _{m=1}^{\infty }{\left(\frac{{\overline{\mu }}_1}{\gamma }\right)}^m{\parallel y_{1,m+n}\parallel }_{L^1[0,\infty )},n\ge 2,}$$

(11b)

$${\displaystyle |b_1|\le {\lambda }_3|Q_{3,0}|+\sum _{m=1}^{\infty }{\left(\frac{{\overline{\mu }}_3}{\gamma }\right)}^{m-1}\sum _{k=0}^m{N}_k|V_{m-k}|+\sum _{m=1}^{\infty }{\left(\frac{{\overline{\mu }}_3}{\gamma }\right)}^m{\parallel y_{3,m+1}\parallel }_{L^1[0,\infty )},}$$

(11c)

$$|b_n|\le \sum _{m=1}^{\infty }{\left(\frac{{\overline{\mu }}_3}{\gamma }\right)}^{m-1}\sum _{k=0}^{m+n-1}{N}_k|V_{m+n-1-k}|+\sum _{m=1}^{\infty }{\left(\frac{{\overline{\mu }}_3}{\gamma }\right)}^m{\parallel y_{3,m+n}\parallel }_{L^1[0,\infty )},n\ge 2.$$

(11d)

Thus, by using system (5a) and (5b) and inequalities system (11a)–(11d), we compute for all $\gamma >0$ that

$$\begin{array}{ll}\sum _{n=1}^{\infty }{\parallel p_{1,n}\parallel }_{L^1[0,\infty )}& =\left({\int }_0^{\infty }\left|a_n{e}^{-\gamma x}+e^{-\gamma x}{\int }_0^x{y}_{1,n}\left(\tau \right)e^{\gamma \tau }d\tau \right|dx\right)\\ & \le \sum _{n=1}^{\infty }\left(\frac{1}{\gamma }|a_n|+\frac{1}{\gamma }{\parallel y_{1,n}\parallel }_{L^1[0,\infty )}\right)\\ & \le \frac{1}{\gamma }\left[{\lambda }_1|Q_{1,0}|+\sum _{m=1}^{\infty }{\left(\frac{{\overline{\mu }}_1}{\gamma }\right)}^m{\parallel y_{1,m+1}\parallel }_{L^1[0,\infty )}\right]\\ & +\frac{1}{\gamma }\sum _{n=2}^{\infty }\sum _{m=1}^{\infty }{\left(\frac{{\overline{\mu }}_1}{\gamma }\right)}^m\parallel y_{1,m+n}{\parallel }_{L^1[0,\infty )}+\frac{1}{\gamma }\sum _{n=1}^{\infty }{\parallel y_{1,n}\parallel }_{L^1[0,\infty )}\\ & \le \frac{{\lambda }_1}{\gamma }|Q_{1,0}|+\frac{1}{\gamma -{\overline{\mu }}_1}\sum _{n=1}^{\infty }{\parallel y_{1,n}\parallel }_{L^1[0,\infty )},\end{array}$$

(12a)

$$\begin{array}{l}\sum _{n=1}^{\infty }{\parallel p_{3,n}\parallel }_{L^1[0,\infty )}=\left({\int }_0^{\infty }\left|b_n{e}^{-\gamma x}+e^{-\gamma x}{\int }_0^x{y}_{3,n}\left(\tau \right)e^{\gamma \tau }d\tau \right|dx\right)\\ \le \sum _{n=1}^{\infty }\left(\frac{1}{\gamma }|b_n|+\frac{1}{\gamma }{\parallel y_{3,n}\parallel }_{L^1[0,\infty )}\right)\\ \le \frac{1}{\gamma }\left[{\lambda }_3|Q_{3,0}|+\sum _{m=1}^{\infty }{\left(\frac{{\overline{\mu }}_3}{\gamma }\right)}^{m-1}\sum _{k=0}^m{N}_k|V_{m-k}|+\sum _{m=1}^{\infty }{\left(\frac{{\overline{\mu }}_3}{\gamma }\right)}^m{\parallel y_{3,m+1}\parallel }_{L^1[0,\infty )}\right]\\ +\frac{1}{\gamma }\sum _{n=2}^{\infty }\left[\sum _{m=1}^{\infty }{\left(\frac{{\overline{\mu }}_3}{\gamma }\right)}^{m-1}\sum _{k=0}^{m+n-1}{N}_k|V_{m+n-1-k}|\right.\\ \left.{\displaystyle +\sum _{m=1}^{\infty }{\left(\frac{{\overline{\mu }}_3}{\gamma }\right)}^m{\parallel y_{3,m+n}\parallel }_{L^1[0,\infty )}}\right]+\frac{1}{\gamma }\sum _{n=1}^{\infty }{\parallel y_{3,n}\parallel }_{L^1[0,\infty )}\\ \le \frac{{\lambda }_3}{\gamma }|Q_{3,0}|+\frac{1}{\gamma -{\overline{\mu }}_3}\sum _{n=0}^{\infty }|V_n|+\frac{1}{\gamma -{\overline{\mu }}_3}\sum _{n=1}^{\infty }{\parallel y_{3,n}\parallel }_{L^1[0,\infty )}.\end{array}$$

(12b)

When $\gamma >0$, through simple calculations from system (4a)–(4d), we can conclude that

$$|Q_{1,0}|\le \frac{(\gamma +1)(\gamma +{\lambda }_3+{\theta }_3)|z_{1,0}|+{\theta }_3(\gamma +1)|z_{3,0}|+{\theta }_3{N}_0|w_0|}{(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0},$$

(13a)

$$|Q_{3,0}|\le \frac{{\theta }_1{N}_0|z_{1,0}|+(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)|z_{3,0}|+N_0(\gamma +{\lambda }_1+{\theta }_1)|w_0|}{(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0},$$

(13b)

$$|V_0|\le \frac{{\theta }_1(\gamma +{\lambda }_3+{\theta }_3)|z_{1,0}|+{\theta }_1{\theta }_3|z_{3,0}|+(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)|w_0|}{(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0},$$

(13c)

$$|V_n|=\frac{1}{\gamma +1}|w_n|,n\ge 1.$$

(13d)

Hence, for all $\gamma >\mathbb{M}=max\{{\overline{\mu }}_1,{\overline{\mu }}_3\}$ and ${\overline{\mu }}_3\ge N_0$, using the inequalities system (12a)–(13d), we conclude that

$$\begin{array}{l}\parallel (V,p_1,p_3)\parallel \le |Q_{1,0}|+|Q_{3,0}|+|V_0|+\sum _{n=1}^{\infty }|V_n|+\frac{{\lambda }_1}{\gamma }|Q_{1,0}|+\frac{1}{\gamma -{\overline{\mu }}_1}\sum _{n=1}^{\infty }{\parallel y_{1,n}\parallel }_{L^1[0,\infty )}\\ +\frac{{\lambda }_3}{\gamma }|Q_{3,0}|+\frac{1}{\gamma -{\overline{\mu }}_3}\sum _{n=0}^{\infty }|V_n|+\frac{1}{\gamma -{\overline{\mu }}_3}\sum _{n=1}^{\infty }{\parallel y_{3,n}\parallel }_{L^1[0,\infty )}\\ \le \frac{(\gamma +{\lambda }_1)[(\gamma +1)(\gamma +{\lambda }_3+{\theta }_3)|z_{1,0}|+{\theta }_3(\gamma +1)|z_{3,0}|+{\theta }_3{N}_0|w_0|]}{\gamma [(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0]}\\ +\frac{(\gamma +{\lambda }_3)[{\theta }_1{N}_0|z_{1,0}|+(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)|z_{3,0}|+N_0(\gamma +{\lambda }_1+{\theta }_1)|w_0|]}{\gamma [(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0]}\\ +\frac{(\gamma -{\overline{\mu }}_3+1)[{\theta }_1(\gamma +{\lambda }_3+{\theta }_3)|z_{1,0}|+{\theta }_1{\theta }_3|z_{3,0}|+(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)|w_0|]}{\gamma [(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0]}\\ +\frac{\gamma -{\overline{\mu }}_3+1}{(\gamma +1)(\gamma -{\overline{\mu }}_3)}\sum _{n=1}^{\infty }|w_n|+\frac{1}{\gamma -{\overline{\mu }}_1}\sum _{n=1}^{\infty }\parallel y_{1,n}{\parallel }_{L^1[0,\infty )}+\frac{1}{\gamma -{\overline{\mu }}_3}\sum _{n=1}^{\infty }{\parallel y_{3,n}\parallel }_{L^1[0,\infty )}\\ \le \frac{1}{\gamma }|z_{1,0}|+\frac{1}{\gamma }|z_{3,0}|+\frac{1}{\gamma }|w_0|+\frac{1}{\gamma -{\overline{\mu }}_3}\sum _{n=1}^{\infty }|w_n|+\sum _{j=1,3}\frac{1}{\gamma -{\overline{\mu }}_j}\sum _{n=1}^{\infty }{\parallel y_{j,n}\parallel }_{L^1[0,\infty )}\\ <\frac{1}{\gamma -\mathbb{M}}\parallel (w,y_1,y_3)\parallel .\end{array}$$

(14)

In the third inequality of the above equation, the following inequalities are utilized

$$0<\frac{(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1(\gamma +{\lambda }_3)({\overline{\mu }}_3-N_0)-{\theta }_1{\theta }_3{\overline{\mu }}_3}{\gamma [(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0]}\le \frac{1}{\gamma },$$

(15a)

$$0<\frac{(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{\overline{\mu }}_3}{\gamma [(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0]}\le \frac{1}{\gamma },$$

(15b)

$$\begin{array}{cc}0\hfill & {\displaystyle <\frac{1}{\gamma [(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0]}}\hfill \\ \hfill & {\displaystyle \times \left\{(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)\right.}\hfill \\ \hfill & {\displaystyle \left.-{\theta }_1{\theta }_3{N}_0-({\overline{\mu }}_3-N_0)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)\right\}\le \frac{1}{\gamma }.}\hfill \end{array}$$

(15c)

Therefore, for any $\gamma >\mathbb{M}=max\{{\overline{\mu }}_1,{\overline{\mu }}_3\}$ and ${\overline{\mu }}_3\ge N_0$, the inequality system (14) implies that

$${(\gamma I-A)}^{-1}:X\to D\left(A\right),\parallel {(\gamma I-A)}^{-1}\parallel <\frac{1}{\gamma -\mathbb{M}}.$$

Step 2. $D\left(A\right)$ is dense in X. Since $(V,p_1,p_3)\in X$, for any $\epsilon >0$, $\exists K\in {\mathbb{N}}_{+}$, if $n>K$, then

$$\sum _{n=K+1}^{\infty }|V_n|<\epsilon ,\sum _{n=K+1}^{\infty }{\parallel p_{j,n}\parallel }_{L^1[0,\infty )}<\epsilon ,j=1,3.$$

Due to the set

$$Y_1=\left\{(V,p_1,p_3)\left|\begin{array}{c}V=(Q_{1,0},Q_{3,0},V_0,V_1,\cdots ,V_K,0,\cdots ),Q_{j,0},V_n\in \mathbb{R},0\le n\le K,\hfill \\ p_j=\left(p_{j,1},p_{j,2},\cdots ,p_{j,K},0,\cdots \right),p_{j,n}\in L^1[0,\infty ),1\le n\le K,K\in {\mathbb{N}}_{+}\hfill \end{array}\right.\right\}$$

being dense in X, by the corollary 2.30 of [18], we know that the set

$$Y_2=\left\{(V,p_1,p_3)\left|\begin{array}{c}V=(Q_{1,0},Q_{3,0},V_0,V_1,\cdots ,V_{K_0},0,\cdots ),Q_{j,0},V_n\in \mathbb{R},0\le n\le K_0,\hfill \\ p_j=\left(p_{j,1},p_{j,2},\cdots ,p_{j,K_0},0,\cdots \right),p_{j,n}\in C_0^{\infty }[0,\infty ),\exists c_{j,n}\mathrm{such}\mathrm{that}\hfill \\ p_{j,n}\left(x\right)=0,x\in [0,c_{j,n}],j=1,3;n=1,2,\cdots ,K_0,K_0\in {\mathbb{N}}_{+}\hfill \end{array}\right.\right\}$$

is dense in $Y_1$. Hence, to prove the denseness of $D\left(A\right)$, it is necessary to prove $Y_2\subset \overline{D\left(A\right)}$. Actually, if $Y_2\subset \overline{D\left(A\right)}$, then $\overline{Y_2}\subset \overline{\overline{D\left(A\right)}}=\overline{D\left(A\right)}$. Then, using $\overline{Y_2}=\overline{\overline{Y_2}}=\overline{Y_1}=X$ and $D\left(A\right)\subset X$, we know that $\overline{D\left(A\right)}\subset \overline{X}=X$. Therefore, $\overline{D\left(A\right)}=X$.

To this aim, we take $(V,p_1,p_3)\in Y_2$. Then, for any $x\in [0,2s]$, we have

$$p_{j,n}\left(x\right)=0,j=1,3;n=1,2,\cdots ,K_0,$$

where $0<2s<min\{c_{1,1},c_{1,2},\cdots ,c_{1,K_0},c_{3,1},c_{3,2},\cdots ,c_{3,K_0}\}$. If we define $(V^{\prime },f_1,f_3)$ by

$$\begin{array}{l}{V}^{\prime }=(Q_{1,0}^{\prime },Q_{3,0}^{\prime },V_0^{\prime },V_1^{\prime },\cdots ,V_{K_0}^{\prime },0,\cdots )=(Q_{1,0},Q_{3,0},V_0,V_1,\cdots ,V_{K_0},0,\cdots ),\\ f_j\left(x\right)=\left(f_{j,1}\left(x\right),f_{j,2}\left(x\right),\cdots ,f_{j,K_0}\left(x\right),0,\cdots \right)=\left(p_{j,1}\left(x\right),p_{j,2}\left(x\right),\cdots ,p_{j,K_0}\left(x\right),0,\cdots \right),\\ f_{1,1}\left(0\right)={\lambda }_1{Q}_{1,0}+{\int }_0^{\infty }{p}_{1,2}\left(x\right){\mu }_1\left(x\right)dx,\\ f_{1,n}\left(0\right)={\int }_0^{\infty }{p}_{1,n+1}\left(x\right){\mu }_1\left(x\right)dx,f_{1,K_0}\left(0\right)=0,\\ f_{3,1}\left(0\right)={\lambda }_3{Q}_{3,0}+N_1{V}_0+N_0{V}_1+{\int }_0^{\infty }{p}_{3,2}\left(x\right){\mu }_3\left(x\right)dx,\\ f_{3,n}\left(0\right)=\sum _{k=0}^n{N}_k{V}_{n-k}+{\int }_0^{\infty }{p}_{3,n+1}\left(x\right){\mu }_3\left(x\right)dx,f_{3,K_0}\left(0\right)=\sum _{k=0}^{K_0}{N}_k{V}_{K_0-k},\end{array}$$

(16)

where $j=1,3$, $2\le n\le K_0-1$, and

$$h_{j,n}=\frac{{\int }_0^s{f}_{j,n}\left(0\right){\left(1-\frac{x}{s}\right)}^2{\mu }_j\left(x\right)dx}{{\int }_s^{2s}{(x-s)}^2{(x-2s)}^2{\mu }_j\left(x\right)dx},f_{j,n}\left(x\right)=\left\{\begin{array}{cc}{f}_{j,n}\left(0\right){\left(1-\frac{x}{s}\right)}^2\hfill & x\in [0,s],\hfill \\ -h_{j,n}{(x-s)}^2{(x-2s)}^2\hfill & x\in [s,2s],\hfill \\ p_{j,n}\left(x\right)\hfill & x\in [2s,\infty )\hfill \end{array}\right.$$

for any $n=1,2,\cdots ,K_0$, then, it is not difficult to prove that $(V^{\prime },f_1,f_3)\in D\left(A\right)$. Therefore,

$$\begin{array}{cc}\hfill & {\displaystyle \parallel (V,p_1,p_3)-(V^{\prime },f_1,f_3)\parallel =\sum _{j=1,3}\sum _{n=1}^{K_0}{\int }_0^{\infty }|p_{j,n}\left(x\right)-f_{j,n}\left(x\right)|dx}\hfill \\ \hfill & {\displaystyle =\sum _{j=1,3}\sum _{n=1}^{K_0}|f_{j,n}\left(0\right)|\frac{s}{3}+\sum _{j=1,3}\sum _{n=1}^{K_0}|h_{j,n}|\frac{s^5}{30}\to 0\mathrm{as}s\to 0.}\hfill \end{array}$$

(17)

That is, $Y_2\subset \overline{D\left(A\right)}$. Hence, we have $\overline{D\left(A\right)}=X$.

Therefore, using Step 1, Step 2 and the Hille–Yosida theorem (see, e.g., [14], Thm 1.68), we obtain that the operator A generates a $C_0$—semigroup.

Step 3. The operators U and E are linear bounded. According to definitions of U and E, for any $(V,p_1,p_3)\in X$, through simple calculations, it can be concluded that

$$\begin{array}{ll}\parallel U(V,p_1,p_3)\parallel & \le \sum _{j=1,3}\sum _{n=1}^{\infty }{\int }_0^{\infty }[2{\lambda }_j+{\theta }_j+{\mu }_j\left(x\right)]|p_{j,n}\left(x\right)|dx\\ & \le max\{2{\lambda }_1+{\theta }_1+{\overline{\mu }}_1,2{\lambda }_3+{\theta }_3+{\overline{\mu }}_3\}\parallel (V,p_1,p_3)\parallel ,\end{array}$$

(18a)

$$\begin{array}{cc}\parallel E(V,p_1,p_3)\parallel \hfill & {\displaystyle \le \sum _{j=1,3}{\theta }_j\sum _{n=1}^{\infty }{\int }_0^{\infty }|p_{j,n}\left(x\right)|dx+\sum _{j=1,3}{\int }_0^{\infty }{\mu }_j\left(x\right)|p_{j,1}\left(x\right)|dx}\hfill \\ \hfill & \le max\{{\theta }_1+{\overline{\mu }}_1,{\theta }_3+{\overline{\mu }}_3\}\parallel (V,p_1,p_3)\parallel .\hfill \end{array}$$

(18b)

The inequalities system (18a) and (18b) mean that U and E are bounded. Obviously, U and E are linear operators. Therefore, the proof of this step is complete.

Thus, by the perturbation theorem of a strongly continuous semigroup (see, e.g., [14], Thm 1.80), we deduce that the operator $A+U+E$ generates a $C_0$—semigroup $e^{(A+U+E)t}$.

Step 4. The operator $A+U+E$ is dispersive. For any $(V,p_1,p_3)\in X$, we define $(B,B_1,B_3)$ by

$$B=\left(\frac{{\left[Q_{1,0}\right]}^{+}}{Q_{1,0}},\frac{{\left[Q_{3,0}\right]}^{+}}{Q_{3,0}},\frac{{\left[V_0\right]}^{+}}{V_0},\frac{{\left[V_1\right]}^{+}}{V_1},\cdots \right),B_j=\left(\frac{{\left[p_{j,1}\left(x\right)\right]}^{+}}{p_{j,1}\left(x\right)},\frac{{\left[p_{j,2}\left(x\right)\right]}^{+}}{p_{j,2}\left(x\right)},\cdots \right),j=1,3,$$

where

$${\left[Q_{j,0}\right]}^{+}=\left\{\begin{array}{cc}{Q}_{j,0},\hfill & Q_{j,0}>0,\hfill \\ 0,\hfill & Q_{j,0}\le 0,\hfill \end{array}\right.{\left[V_n\right]}^{+}=\left\{\begin{array}{cc}{V}_n,\hfill & V_n>0,\hfill \\ 0,\hfill & V_n\le 0,\hfill \end{array}\right.n=0,1,2,\cdots$$

$${\left[p_{j,n}\left(x\right)\right]}^{+}=\left\{\begin{array}{cc}{p}_{j,n}\left(x\right),\hfill & p_{j,n}\left(x\right)>0,\hfill \\ 0,\hfill & p_{j,n}\left(x\right)\le 0,\hfill \end{array}\right.n=1,2,\cdots .$$

By simple calculations (for the detailed calculations see the Equation (2.12) of [19]), it can be concluded that

$${\int }_0^{\infty }\frac{d{p}_{j,n}\left(x\right)}{dx}\frac{{\left[p_{j,n}\left(x\right)\right]}^{+}}{p_{j,n}\left(x\right)}dx=-{\left[p_{j,n}\left(0\right)\right]}^{+},j=1,3;n=1,2,\cdots .$$

Therefore, for any ${\lambda }_j,{\theta }_j>0,j=1,3$ and $0<N_0<1$, using the boundary conditions system (4f)–(4i), it is computed that

$$\begin{array}{l}\langle (A+U+E)(V,p_1,p_3),(B,B_1,B_3)\rangle \\ =\left[-({\lambda }_1+{\theta }_1)Q_{1,0}+{\theta }_3{Q}_{3,0}+{\int }_0^{\infty }{p}_{1,1}\left(x\right){\mu }_1\left(x\right)dx+{\theta }_3\sum _{n=1}^{\infty }{\int }_0^{\infty }{p}_{3,n}\left(x\right)dx\right]\frac{{\left[Q_{1,0}\right]}^{+}}{Q_{1,0}}\\ +\left[-({\lambda }_3+{\theta }_3)Q_{3,0}+N_0{V}_0+{\int }_0^{\infty }{p}_{3,1}\left(x\right){\mu }_3\left(x\right)dx\right]\frac{{\left[Q_{3,0}\right]}^{+}}{Q_{3,0}}\\ +(-V_0+{\theta }_1{Q}_{1,0})\frac{{\left[V_0\right]}^{+}}{V_0}+\sum _{n=1}^{\infty }\left[-V_n+{\theta }_1{\int }_0^{\infty }{p}_{1,n}\left(x\right)dx\right]\frac{{\left[V_n\right]}^{+}}{V_n}\\ +\sum _{n=1}^{\infty }{\int }_0^{\infty }\left[-\frac{d{p}_{1,n}\left(x\right)}{dx}-({\lambda }_1+{\theta }_1+{\mu }_1\left(x\right))p_{1,n}\left(x\right)\right]\frac{{\left[p_{1,n}\left(x\right)\right]}^{+}}{p_{1,n}}dx\\ +\sum _{n=2}^{\infty }{\int }_0^{\infty }{\lambda }_1{p}_{1,n-1}\left(x\right)\frac{{\left[p_{1,n}\left(x\right)\right]}^{+}}{p_{1,n}}dx\\ +\sum _{n=1}^{\infty }{\int }_0^{\infty }\left[-\frac{d{p}_{3,n}\left(x\right)}{dx}-({\lambda }_3+{\theta }_3+{\mu }_3\left(x\right))p_{3,n}\left(x\right)\right]\frac{{\left[p_{3,n}\left(x\right)\right]}^{+}}{p_{3,n}}dx\\ +\sum _{n=2}^{\infty }{\int }_0^{\infty }{\lambda }_3{p}_{3,n-1}\left(x\right)\frac{{\left[p_{3,n}\left(x\right)\right]}^{+}}{p_{3,n}}dx\\ \le {\theta }_1\left(\frac{{\left[V_0\right]}^{+}}{V_0}-1\right){\left[Q_{1,0}\right]}^{+}+{\theta }_3\left(\frac{{\left[Q_{1,0}\right]}^{+}}{Q_{1,0}}-1\right)\sum _{n=1}^{\infty }{\int }_0^{\infty }{\left[p_{3,n}\left(x\right)\right]}^{+}dx\\ +{\theta }_3\left(\frac{{\left[Q_{1,0}\right]}^{+}}{Q_{1,0}}-1\right){\left[Q_{3,0}\right]}^{+}+\left(\frac{{\left[Q_{1,0}\right]}^{+}}{Q_{1,0}}-1\right){\int }_0^{\infty }{\left[p_{1,1}\left(x\right)\right]}^{+}{\mu }_1\left(x\right)dx\\ +\left(\frac{{\left[Q_{3,0}\right]}^{+}}{Q_{3,0}}{N}_0-1\right){\left[V_0\right]}^{+}+\left(\frac{{\left[Q_{3,0}\right]}^{+}}{Q_{3,0}}-1\right){\int }_0^{\infty }{\left[p_{3,1}\left(x\right)\right]}^{+}{\mu }_3\left(x\right)dx\\ +{\theta }_1\sum _{n=1}^{\infty }\left(\frac{{\left[V_n\right]}^{+}}{V_n}-1\right){\int }_0^{\infty }{\left[p_{1,n}\left(x\right)\right]}^{+}dx\le 0.\end{array}$$

(19)

The inequality system (19) shows that the operator $A+U+E$ is dispersive.

Hence, by the above discussions and the Phillips theorem (see, e.g., [14] Thm 1.77, we see that the result of this Theorem holds true. □

Next, we consider the isometry of the semigroup $e^{(A+U+E)t}$. Clearly, the dual space $X^{*}$ of X is

$$X^{*}=\left\{(V^{*},p_1^{*},p_3^{*})\left|\begin{array}{c}{V}^{*}=(Q_{1,0}^{*},Q_{3,0}^{*},V_0^{*},V_1^{*},\cdots )\in l^{\infty },p_j^{*}=\left(p_{j,1}^{*},p_{j,2}^{*},\cdots \right),\hfill \\ p_{j,n}^{*}\in L^{\infty }[0,\infty ),{\parallel (V^{*},p_1^{*},p_3^{*})\parallel }_{*}\hfill \\ =sup\left\{|Q_{j,0}^{*}|,\underset{n\ge 0}{sup}|V_n^{*}|,\underset{n\ge 1}{sup}{\parallel p_{j,n}^{*}\parallel }_{L^{\infty }[0,\infty )}\right\}<\infty \hfill \end{array}\right.\right\}.$$

If we define

$$X_{+}=\left\{(V,p_1,p_3)\in X\left|\begin{array}{c}V=(Q_{1,0},Q_{3,0},V_0,V_1,\cdots )\in l^1,Q_{j,0}\ge 0,V_n\ge 0,n\ge 0,\hfill \\ p_j=\left(p_{j,1},p_{j,2},\cdots \right),0\le p_{j,n}\in L^1[0,\infty ),j=1,3;n\ge 1\hfill \end{array}\right.\right\},$$

then, by Theorem 1, we have $e^{(A+U+E)t}{X}_{+}\subset X_{+}$. For any $(V,p_1,p_3)\in D\left(A\right)\cap X_{+}$, we choose

$$(V^{*},p_1^{*},p_3^{*})=\parallel (V,p_1,p_3)\parallel \left(\left(\begin{array}{c}1\\ ⋮\end{array}\right),\left(\begin{array}{c}1\\ ⋮\end{array}\right),\left(\begin{array}{c}1\\ ⋮\end{array}\right)\right).$$

Then, it is easy to show that $(V^{*},p_1^{*},p_3^{*})\in X^{*}$ and

$$\begin{array}{l}\langle (V,p_1,p_3),(V^{*},p_1^{*},p_3^{*})\rangle \\ =\left(Q_{1,0}+Q_{3,0}+\sum _{n=0}^{\infty }{V}_n+\sum _{j=1,3}\sum _{n=1}^{\infty }{\int }_0^{\infty }{p}_{j,n}\left(x\right)dx\right)\parallel (V,p_1,p_3)\parallel \\ =\parallel (V,p_1,p_3){\parallel }^2.\end{array}$$

(20)

This means that $(V^{*},p_1^{*},p_3^{*})\in Q(V,p_1,p_3)$, where

$$Q(V,p_1,p_3)=\left\{(V^{*},p_1^{*},p_3^{*})\in X^{*}\left|\begin{array}{c}\langle (V,p_1,p_3),(V^{*},p_1^{*},p_3^{*})\rangle ={\parallel (V,p_1,p_3)\parallel }^2\hfill \end{array}\right.\right\}.$$

In addition, for any $(V,p_1,p_3)\in D\left(A\right)$ and $(V^{*},p_1^{*},p_3^{*})\in Q(V,p_1,p_3)$, using the boundary conditions system (4f)–(4i) and ${\sum }_{n=0}^{\infty }{N}_n=1$, we have

$$\begin{array}{l}\langle A(V,p_1,p_3),(V^{*},p_1^{*},p_3^{*})\rangle \\ =\left[-({\lambda }_1+{\theta }_1)Q_{1,0}+{\theta }_3{Q}_{3,0}+{\int }_0^{\infty }{p}_{1,1}\left(x\right){\mu }_1\left(x\right)dx+{\theta }_3\sum _{n=1}^{\infty }{\int }_0^{\infty }{p}_{3,n}\left(x\right)dx\right]\\ \times \parallel (V,p_1,p_3)\parallel \\ +\left[-({\lambda }_3+{\theta }_3)Q_{3,0}+N_0{V}_0+{\int }_0^{\infty }{p}_{3,1}\left(x\right){\mu }_3\left(x\right)dx\right]\parallel (V,p_1,p_3)\parallel \\ +(-V_0+{\theta }_1{Q}_{1,0})\parallel (V,p_1,p_3)\parallel +\sum _{n=1}^{\infty }\left[-V_n+{\theta }_1{\int }_0^{\infty }{p}_{1,n}\left(x\right)dx\right]\parallel (V,p_1,p_3)\parallel \\ +\sum _{n=1}^{\infty }{\int }_0^{\infty }\left[-\frac{d{p}_{1,n}\left(x\right)}{dx}-({\lambda }_1+{\theta }_1+{\mu }_1\left(x\right))p_{1,n}\left(x\right)\right]\parallel (V,p_1,p_3)\parallel dx\\ +\sum _{n=2}^{\infty }{\int }_0^{\infty }{\lambda }_1{p}_{1,n-1}\left(x\right)\parallel (V,p_1,p_3)\parallel dx+\sum _{n=2}^{\infty }{\int }_0^{\infty }{\lambda }_3{p}_{3,n-1}\left(x\right)\parallel (V,p_1,p_3)\parallel dx\\ +\sum _{n=1}^{\infty }{\int }_0^{\infty }\left[-\frac{d{p}_{3,n}\left(x\right)}{dx}-({\lambda }_3+{\theta }_3+{\mu }_3\left(x\right))p_{3,n}\left(x\right)\right]\parallel (V,p_1,p_3)\parallel dx\\ =\left[-{\lambda }_1{Q}_{1,0}+{\int }_0^{\infty }{p}_{1,1}\left(x\right){\mu }_1\left(x\right)dx-{\lambda }_3{Q}_{3,0}+N_0{V}_0+{\int }_0^{\infty }{p}_{3,1}\left(x\right){\mu }_3\left(x\right)dx\right.\\ -\sum _{n=0}^{\infty }{V}_n+\sum _{n=1}^{\infty }\left(p_{1,n}\left(0\right)-{\int }_0^{\infty }{\mu }_1\left(x\right)p_{1,n}\left(x\right)\right)dx\\ \left.+\sum _{n=1}^{\infty }\left(p_{3,n}\left(0\right)-{\int }_0^{\infty }{\mu }_3\left(x\right)p_{3,n}\left(x\right)\right)dx\right]\parallel (V,p_1,p_3)\parallel =0.\end{array}$$

(21)

system (21) means that $A+U+E$ is a conservative operator with respect to the set $Q(·)$.

By Theorem 3.6.1 of [20] (p. 155), we have the following Theorem 2.

Theorem 2.

If ${\lambda }_j,{\theta }_j>0$,$0<{\overline{\mu }}_j<\infty ,j=1,3$ and the initial value $(V\left(0\right),p_1(·,0),p_3(·,0))$ of the system (3) satisfies $(V\left(0\right),p_1(·,0),p_3(·,0))\in D\left(A^2\right)$, then

$$\parallel e^{(A+U+E)t}(V\left(0\right),p_1(·,0),p_3(·,0))\parallel =\parallel (V\left(0\right),p_1(·,0),p_3(·,0))\parallel ,t\in [0,\infty ).$$

Now, using Theorems 1 and 2, we obtain the main result in this section.

Theorem 3.

If ${\lambda }_j,{\theta }_j>0$,$0<{\overline{\mu }}_j<\infty ,j=1,3$ and the initial value $(V\left(0\right),p_1(·,0),p_3(·,0))$ of the system (3) satisfies $(V\left(0\right),p_1(·,0),p_3(·,0))\in D\left(A^2\right)$, then the system (3) admits a unique positive TDS $(V\left(t\right),p_1(·,t),p_3(·,t))$, which satisfies

$$\parallel (V\left(t\right),p_1(·,t),p_3(·,t))\parallel =1,\forall t\in [0,\infty ).$$

Proof. 

It is not difficult to see that $(V\left(0\right),p_1(·,0),p_3(·,0))\in D\left(A^2\right)\cap X_{+}$. Theorem 1 and Theorem 1.81 of [14] mean that system (3) admits a unique positive TDS $(V\left(t\right),p_1(·,t),p_3(·,t))$, which can be described by

$$(V\left(t\right),p_1(·,t),p_3(·,t))=e^{(A+U+E)t}(V\left(0\right),p_1(·,0),p_3(·,0)),t\in [0,\infty ).$$

Due to Theorem 2, it is easy to see that

$$\parallel (V\left(t\right),p_1(·,t),p_3(·,t))\parallel =\parallel (V\left(0\right),p_1(·,0),p_3(·,0))\parallel =1,t\in [0,\infty ).$$

This shows that the physics reflected by $(V(·),p_1(·,·),p_3(·,·))$ is reasonable. □

Theorem 3 means that ($\mathbf{H}\mathbf{1}$) holds under some conditions.

3. Asymptotic Behavior of the TDS of the System

When the service rates ${\mu }_1(·)={\mu }_1$ and ${\mu }_3(·)={\mu }_3$ are constants, the M/G/1 stochastic clearing queueing model in a three-phase environment is called the M/M/1 stochastic clearing queueing model in a three-phase environment. In this section, we discuss the strong convergence of the TDS of system (3) in the above special case. For this aim, first of all, we determine the adjoint operator ${(A+U+E)}^{*}$ of the system operator $A+U+E$. Then, we find the spectrum of ${(A+U+E)}^{*}$ on the imaginary axis. Hence, we provide the spectrum of $A+U+E$ on the imaginary axis by using the relationship between the spectrum of the operator and its adjoint operator. Finally, we derive that zero is a point spectrum of $A+U+E$ and ${(A+U+E)}^{*}$, and the geometric multiplicity of zero is one. Therefore, we obtain that the TDS of system (3) strongly converges to its SSS. That is, ($\mathbf{H}\mathbf{2}$) holds true under strong convergence.

The following Theorem 4 is the main result of this section.

Theorem 4.

If ${\lambda }_j,{\theta }_j,{\mu }_j>0$ and ${\lambda }_j<{\theta }_j+{\mu }_j,j=1,3$, then the TDS of the system (3) strongly converges to its SSS; that is

$$\underset{t\to \infty }{lim}\parallel (V\left(t\right),p_1(·,t),p_3(·,t))-\langle (V^{*},p_1^{*}(·),p_3^{*}(·)),(u_2,u_1(·),u_3(·))\rangle (V,p_1(·),p_3(·))\parallel =0,$$

where $(V,p_1(·),p_3(·))$ and $(V^{*},p_1^{*}(·),p_3^{*}(·))$ are the eigenvectors in Lemmas 3 and 4, respectively.

To prove the above Theorem 4, first we prove the following four lemmas.

Lemma 1.

The adjoint operator ${(A+U+E)}^{*}$ of $(A+U+E)$ is given by

$${(A+U+E)}^{*}(V^{*},p_1^{*},p_3^{*})=(L+J+G+H)(V^{*},p_1^{*},p_3^{*}),(V^{*},p_1^{*},p_3^{*})\in D\left(L\right),$$

where the operators $L,J,G,H$ and its domain are defined by

$$\begin{array}{cc}& L(V^{*},p_1^{*},p_3^{*})\hfill \\ & =\left(\left(\begin{array}{c}-({\lambda }_1+{\theta }_1)Q_{1,0}^{*}+{\theta }_1{V}_0^{*}\\ {\theta }_3{Q}_{1,0}^{*}-({\lambda }_3+{\theta }_3)Q_{3,0}^{*}\\ N_0{Q}_{3,0}^{*}-V_0^{*}\\ -V_1^{*}\\ -V_2^{*}\\ ⋮\end{array}\right),\left(\begin{array}{c}\left(\frac{d}{dx}-{\eta }_1\right)p_{1,1}^{*}\left(x\right)\\ \left(\frac{d}{dx}-{\eta }_1\right)p_{1,2}^{*}\left(x\right)\\ \left(\frac{d}{dx}-{\eta }_1\right)p_{1,3}^{*}\left(x\right)\\ ⋮\end{array}\right),\left(\begin{array}{c}\left(\frac{d}{dx}-{\eta }_3\right)p_{3,1}^{*}\left(x\right)\\ \left(\frac{d}{dx}-{\eta }_3\right)p_{3,2}^{*}\left(x\right)\\ \left(\frac{d}{dx}-{\eta }_3\right)p_{3,3}^{*}\left(x\right)\\ ⋮\end{array}\right)\right),\hfill \end{array}$$

$$D\left(L\right)=\left\{(V^{*},p_1^{*},p_3^{*})\left|\begin{array}{c}\frac{d{p}_{j,n}^{*}}{dx}existand{p}_{j,n}^{*}(\infty )=\alpha ,j=1,3;n\ge 1\hfill \end{array}\right.\right\},$$

$$\begin{array}{cc}J(V^{*},p_1^{*},p_3^{*})\hfill & =\left(\left(\begin{array}{c}0\\ 0\\ ⋮\end{array}\right),\left(\begin{array}{c}{\lambda }_1{p}_{1,2}^{*}\left(x\right)\\ {\lambda }_1{p}_{1,3}^{*}\left(x\right)\\ ⋮\end{array}\right),\left(\begin{array}{c}{\lambda }_3{p}_{3,2}^{*}\left(x\right)\\ {\lambda }_3{p}_{3,3}^{*}\left(x\right)\\ ⋮\end{array}\right)\right),\hfill \end{array}$$

$$\begin{array}{cc}G(V^{*},p_1^{*},p_3^{*})\hfill & =\left(\left(\begin{array}{c}{\lambda }_1{p}_{1,1}^{*}\left(0\right)\\ {\lambda }_3{p}_{3,1}^{*}\left(0\right)\\ \sum _{k=1}^{\infty }{N}_k{p}_{3,k}^{*}\left(0\right)\\ \sum _{k=0}^{\infty }{N}_k{p}_{3,k+1}^{*}\left(0\right)\\ \sum _{k=0}^{\infty }{N}_k{p}_{3,k+2}^{*}\left(0\right)\\ ⋮\end{array}\right),\left(\begin{array}{c}0\\ {\mu }_1{p}_{1,1}^{*}\left(0\right)\\ {\mu }_1{p}_{1,2}^{*}\left(0\right)\\ {\mu }_1{p}_{1,3}^{*}\left(0\right)\\ ⋮\end{array}\right),\left(\begin{array}{c}0\\ {\mu }_3{p}_{3,1}^{*}\left(0\right)\\ {\mu }_3{p}_{3,2}^{*}\left(0\right)\\ {\mu }_3{p}_{3,3}^{*}\left(0\right)\\ ⋮\end{array}\right)\right),\hfill \end{array}$$

$$\begin{array}{cc}H(V^{*},p_1^{*},p_3^{*})\hfill & =\left(\left(\begin{array}{c}0\\ 0\\ 0\\ ⋮\end{array}\right),\left(\begin{array}{c}{\mu }_1{Q}_{1,0}^{*}+{\theta }_1{V}_1^{*}\\ {\theta }_1{V}_2^{*}\\ {\theta }_1{V}_3^{*}\\ ⋮\end{array}\right),\left(\begin{array}{c}{\theta }_3{Q}_{1,0}^{*}+{\mu }_3{Q}_{3,0}^{*}\\ {\theta }_3{Q}_{1,0}^{*}\\ {\theta }_3{Q}_{1,0}^{*}\\ ⋮\end{array}\right)\right),\hfill \end{array}$$

$$D\left(J\right)=X^{*},D\left(G\right)=X^{*},D\left(H\right)=X^{*};$$

here, ${\eta }_j={\lambda }_j+{\theta }_j+{\mu }_j$, and α is a positive constant irrelevant of j and n.

Proof. 

For every $(V,p_1,p_3)\in D\left(A\right)$ and $(V^{*},p_1^{*},p_3^{*})\in D\left(L\right)$, using the boundary conditions system (4f)–(4i) and integration by parts, we have

$$\begin{array}{l}\langle (A+U+E)(V,p_1,p_3),(V^{*},p_1^{*},p_3^{*})\rangle \\ =\left[-({\lambda }_1+{\theta }_1)Q_{1,0}+{\theta }_3\sum _{n=1}^{\infty }{\int }_0^{\infty }{p}_{3,n}\left(x\right)dx+{\theta }_3{Q}_{3,0}+{\mu }_1{\int }_0^{\infty }{p}_{1,1}\left(x\right)dx\right]Q_{1,0}^{*}\\ {\displaystyle +\left[-({\lambda }_3+{\theta }_3)Q_{3,0}+N_0{V}_0+{\mu }_3{\int }_0^{\infty }{p}_{3,1}\left(x\right)dx\right]Q_{3,0}^{*}}\\ {\displaystyle +(-V_0+{\theta }_1{Q}_{1,0})V_0^{*}+\sum _{n=1}^{\infty }\left[-V_n+{\theta }_1{\int }_0^{\infty }{p}_{1,n}\left(x\right)dx\right]V_n^{*}}\hfill \\ +{\int }_0^{\infty }\left[-\frac{d{p}_{1,1}\left(x\right)}{dx}-({\lambda }_1+{\theta }_1+{\mu }_1)p_{1,1}\left(x\right)\right]p_{1,1}^{*}\left(x\right)dx\hfill \\ +\sum _{n=2}^{\infty }{\int }_0^{\infty }\left[-\frac{d{p}_{1,n}\left(x\right)}{dx}-({\lambda }_1+{\theta }_1+{\mu }_1)p_{1,n}\left(x\right)+{\lambda }_1{p}_{1,n-1}\left(x\right)\right]p_{1,n}^{*}\left(x\right)dx\hfill \\ +{\int }_0^{\infty }\left[-\frac{d{p}_{3,1}\left(x\right)}{dx}-({\lambda }_3+{\theta }_3+{\mu }_3)p_{3,1}\left(x\right)\right]p_{3,1}^{*}\left(x\right)dx\hfill \\ +\sum _{n=2}^{\infty }{\int }_0^{\infty }\left[-\frac{d{p}_{3,n}\left(x\right)}{dx}-({\lambda }_3+{\theta }_3+{\mu }_3)p_{3,n}\left(x\right)+{\lambda }_1{p}_{3,n-1}\left(x\right)\right]p_{3,n}^{*}\left(x\right)dx\hfill \\ =-({\lambda }_1+{\theta }_1)Q_{1,0}{Q}_{1,0}^{*}+{\theta }_3{Q}_{1,0}^{*}\sum _{n=1}^{\infty }{\int }_0^{\infty }{p}_{3,n}\left(x\right)dx+{\theta }_3{Q}_{3,0}{Q}_{1,0}^{*}\hfill \\ +{\mu }_1{Q}_{1,0}^{*}{\int }_0^{\infty }{p}_{1,1}\left(x\right)dx-({\lambda }_3+{\theta }_3)Q_{3,0}{Q}_{3,0}^{*}+N_0{V}_0{Q}_{3,0}^{*}\hfill \\ +{\mu }_3{Q}_{3,0}^{*}{\int }_0^{\infty }{p}_{3,1}\left(x\right)dx+{\theta }_1{Q}_{1,0}{V}_0^{*}-\sum _{n=0}^{\infty }{V}_n{V}_n^{*}+{\theta }_1\sum _{n=1}^{\infty }{V}_n^{*}{\int }_0^{\infty }{p}_{1,n}\left(x\right)dx\hfill \\ +{\lambda }_1{Q}_{1,0}{p}_{1,1}^{*}\left(0\right)+{\mu }_1\sum _{n=1}^{\infty }{p}_{1,n}^{*}\left(0\right){\int }_0^{\infty }{p}_{1,n+1}\left(x\right)dx\hfill \\ +\sum _{n=1}^{\infty }{\int }_0^{\infty }{p}_{1,n}\left(x\right)\left[\frac{d{p}_{1,n}^{*}\left(x\right)}{dx}-({\lambda }_1+{\theta }_1+{\mu }_1)p_{1,n}^{*}\left(x\right)\right]dx\hfill \\ +{\lambda }_1\sum _{n=1}^{\infty }{\int }_0^{\infty }{p}_{1,n}\left(x\right)p_{1,n+1}^{*}\left(x\right)dx+{\lambda }_3{Q}_{3,0}{p}_{3,1}^{*}\left(0\right)+V_0\sum _{k=1}^{\infty }{N}_k{p}_{3,k}^{*}\left(0\right)\hfill \\ +\sum _{n=1}^{\infty }\sum _{k=0}^{\infty }{N}_k{V}_n{p}_{3,k+n}^{*}\left(0\right)+{\mu }_3\sum _{n=1}^{\infty }{p}_{3,n}^{*}\left(0\right){\int }_0^{\infty }{p}_{3,n+1}\left(x\right)dx\hfill \\ +\sum _{n=1}^{\infty }{\int }_0^{\infty }{p}_{3,n}\left(x\right)\left[\frac{d{p}_{3,n}^{*}\left(x\right)}{dx}-({\lambda }_3+{\theta }_3+{\mu }_3)p_{3,n}^{*}\left(x\right)\right]dx\hfill \\ +{\lambda }_3\sum _{n=1}^{\infty }{\int }_0^{\infty }{p}_{3,n}\left(x\right)p_{3,n+1}^{*}\left(x\right)dx\hfill \\ =\langle (V,p_1,p_3),(L+J+G+H)(V^{*},p_1^{*},p_3^{*})\rangle .\hfill \end{array}$$

(22)

system (22) means that this Lemma holds true. □

Lemma 2.

If ${\lambda }_j,{\theta }_j,{\mu }_j>0,j=1,3$, then all points in the set

$$\mathrm{\Lambda }:=\left\{\gamma \in \mathbb{C}\left|\begin{array}{cc}& sup\left\{\frac{1}{|(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|}sup\left\{\frac{{\lambda }_1({\theta }_1+{\mu }_1)|(\gamma +1)(\gamma +{\lambda }_3+{\theta }_3)|}{|{\sigma }_1|},\right.\right.\hfill \\ & \frac{{\lambda }_1^2|(\gamma +1)(\gamma +{\lambda }_3+{\theta }_3)|}{\Re {\sigma }_1},\frac{{\lambda }_3{\theta }_1{N}_0({\theta }_3+{\mu }_3)}{|{\sigma }_3|},\frac{{\lambda }_3^2{\theta }_1{N}_0}{\Re {\sigma }_3},\frac{{\theta }_1{\theta }_3(1-N_0)|\gamma +{\lambda }_3+{\theta }_3|}{|{\sigma }_3|-{\mu }_3},\hfill \\ & \frac{{\lambda }_3{\mu }_3{\theta }_1(1-N_0)|\gamma +{\lambda }_3+{\theta }_3|}{\Re {\sigma }_3(|{\sigma }_3|-{\mu }_3)},\frac{{\lambda }_1{\theta }_3({\theta }_1+{\mu }_1)|\gamma +1|}{|{\sigma }_1|},\frac{{\lambda }_1^2{\theta }_3|\gamma +1|}{\Re {\sigma }_1},\frac{{\theta }_1{\theta }_3^2(1-N_0)}{|{\sigma }_3|-{\mu }_3},\hfill \\ & \frac{{\lambda }_3({\theta }_3+{\mu }_3)|(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)|}{|{\sigma }_3|},\frac{{\lambda }_3^2|(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)|}{\Re {\sigma }_3},\frac{{\lambda }_3{\mu }_3{\theta }_1{\theta }_3(1-N_0)}{\Re {\sigma }_3(|{\sigma }_3|-{\mu }_3)},\hfill \\ & \frac{{\lambda }_3{\theta }_1{N}_0({\theta }_1+{\mu }_1)}{|{\sigma }_1|},\frac{{\lambda }_1{\lambda }_3{\theta }_1{N}_0}{\Re {\sigma }_1},\frac{{\lambda }_3{N}_0({\theta }_3+{\mu }_3)|\gamma +{\lambda }_1+{\theta }_1|}{|{\sigma }_3|},\frac{{\lambda }_3^2{N}_0|\gamma +{\lambda }_1+{\theta }_1|}{\Re {\sigma }_3},\hfill \\ & \left.\frac{{\theta }_3(1-N_0)|(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)|}{|{\sigma }_3|-{\mu }_3},\frac{{\lambda }_3{\mu }_3(1-N_0)|(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)|}{\Re {\sigma }_3(|{\sigma }_3|-{\mu }_3)}\right\},\hfill \\ & \frac{{\theta }_3}{|\gamma +1||{\sigma }_3|-{\mu }_3},\frac{{\lambda }_3{\mu }_3}{|\gamma +1|\Re {\sigma }_3(|{\sigma }_3|-{\mu }_3)},\frac{{\theta }_1+{\mu }_1}{|{\sigma }_1|},\frac{{\lambda }_1}{\Re {\sigma }_1},\frac{{\theta }_1}{|{\sigma }_1|-{\mu }_1},\frac{{\lambda }_1{\mu }_1}{\Re {\sigma }_1(|{\sigma }_1|-{\mu }_1)},\hfill \\ & \left.\frac{{\theta }_3+{\mu }_3}{|{\sigma }_3|},\frac{{\lambda }_3}{\Re {\sigma }_3},\frac{{\theta }_3}{|{\sigma }_3|-{\mu }_3},\frac{{\lambda }_3{\mu }_3}{\Re {\sigma }_3(|{\sigma }_3|-{\mu }_3)}\right\}<1,\Re {\sigma }_j>0,|{\sigma }_j|>{\mu }_j\hfill \end{array}\right.\right\}$$

belong to the resolvent set $\rho \left({(A+U+E)}^{*}\right)$. In particular, $i\mathbb{R}\setminus \left\{0\right\}\subset \rho \left({(A+U+E)}^{*}\right),i^2=-1$. Hence, $i\mathbb{R}\setminus \left\{0\right\}\subset \rho (A+U+E)$, where ${\sigma }_j=\gamma +{\lambda }_j+{\theta }_j+{\mu }_j,0<N_0<1$, and $\Re \gamma$ is the real part of γ.

Proof. 

For every $(w^{*},y_1^{*},y_3^{*})\in X^{*}$, we consider

$$(\gamma I-L-G)(V^{*},p_1^{*},p_3^{*})=(J+H)(w^{*},y_1^{*},y_3^{*}),$$

where $w^{*}=(z_{1,0}^{*},z_{3,0}^{*},w_0^{*},w_1^{*},\cdots )$, $y_j^{*}=(y_{j,1}^{*},y_{j,2}^{*},y_{j,3}^{*},\cdots ),j=1,3$. That is,

$$(\gamma +{\lambda }_1+{\theta }_1)Q_{1,0}^{*}={\theta }_1{V}_0^{*}+{\lambda }_1{p}_{1,1}^{*}\left(0\right),$$

(23a)

$$(\gamma +{\lambda }_3+{\theta }_3)Q_{3,0}^{*}={\theta }_3{Q}_{1,0}^{*}+{\lambda }_3{p}_{3,1}^{*}\left(0\right),$$

(23b)

$$(\gamma +1)V_0^{*}=N_0{Q}_{3,0}^{*}+\sum _{k=1}^{\infty }{N}_k{p}_{3,k}^{*}\left(0\right),$$

(23c)

$$(\gamma +1)V_n^{*}=\sum _{k=0}^{\infty }{N}_k{p}_{3,k+n}^{*}\left(0\right),n\ge 1,$$

(23d)

$$\frac{d{p}_{1,1}^{*}\left(x\right)}{dx}=(\gamma +{\lambda }_1+{\theta }_1+{\mu }_1)p_{1,1}^{*}\left(x\right)-{\mu }_1{z}_{1,0}^{*}-{\theta }_1{w}_1^{*}-{\lambda }_1{y}_{1,2}^{*}\left(x\right),$$

(23e)

$$\frac{d{p}_{1,n}^{*}\left(x\right)}{dx}=(\gamma +{\lambda }_1+{\theta }_1+{\mu }_1)p_{1,n}^{*}\left(x\right)-{\theta }_1{w}_n^{*}-{\mu }_1{p}_{1,n-1}^{*}\left(0\right)-{\lambda }_1{y}_{1,n+1}^{*}\left(x\right),n\ge 2,$$

(23f)

$$\frac{d{p}_{3,1}^{*}\left(x\right)}{dx}=(\gamma +{\lambda }_3+{\theta }_3+{\mu }_3)p_{3,1}^{*}\left(x\right)-{\theta }_3{z}_{1,0}^{*}-{\mu }_3{z}_{3,0}^{*}-{\lambda }_3{y}_{3,2}^{*}\left(x\right),$$

(23g)

$$\frac{d{p}_{3,n}^{*}\left(x\right)}{dx}=(\gamma +{\lambda }_3+{\theta }_3+{\mu }_3)p_{3,n}^{*}\left(x\right)-{\theta }_3{z}_{1,0}^{*}-{\mu }_3{p}_{3,n-1}^{*}\left(0\right)-{\lambda }_3{y}_{3,n+1}^{*}\left(x\right),n\ge 2,$$

(23h)

$$p_{1,n}^{*}(\infty )=p_{3,n}^{*}(\infty )=\alpha ,n\ge 1.$$

(23i)

Solve the Equations (23e)–(23h) to obtain

$$\begin{array}{c}{\displaystyle p_{1,1}^{*}\left(x\right)=p_{1,1}^{*}\left(0\right)e^{{\sigma }_{1}x}-e^{{\sigma }_{1}x}{\int }_0^x[{\mu }_1{z}_{1,0}^{*}+{\theta }_1{w}_1^{*}+{\lambda }_1{y}_{1,2}^{*}\left(\tau \right)]e^{-{\sigma }_1\tau }d\tau }\hfill \end{array}$$

(24a)

$$\begin{array}{c}{\displaystyle p_{1,n}^{*}\left(x\right)=p_{1,n}^{*}\left(0\right)e^{{\sigma }_{1}x}-e^{{\sigma }_{1}x}{\int }_0^x[{\theta }_1{w}_n^{*}+{\mu }_1{p}_{1,n-1}^{*}\left(0\right)+{\lambda }_1{y}_{1,n+1}^{*}\left(\tau \right)]e^{-{\sigma }_1\tau }d\tau ,n\ge 2,}\hfill \end{array}$$

(24b)

$$\begin{array}{c}{\displaystyle p_{3,1}^{*}\left(x\right)=p_{3,1}^{*}\left(0\right)e^{{\sigma }_{3}x}-e^{{\sigma }_{3}x}{\int }_0^x[{\theta }_3{z}_{1,0}^{*}+{\mu }_3{z}_{3,0}^{*}+{\lambda }_3{y}_{3,2}^{*}\left(\tau \right)]e^{-{\sigma }_3\tau }d\tau }\hfill \end{array}$$

(24c)

$$\begin{array}{c}{\displaystyle p_{3,n}^{*}\left(x\right)=p_{3,n}^{*}\left(0\right)e^{{\sigma }_{3}x}-e^{{\sigma }_{3}x}{\int }_0^x[{\theta }_3{z}_{1,0}^{*}+{\mu }_3{p}_{3,n-1}^{*}\left(0\right)+{\lambda }_3{y}_{3,n+1}^{*}\left(\tau \right)]e^{-{\sigma }_3\tau }d\tau ,n\ge 2,}\hfill \end{array}$$

(24d)

where ${\sigma }_j=\gamma +{\lambda }_j+{\theta }_j+{\mu }_j,j=1,3$. Through multiplying $e^{-{\sigma }_{1}x}$ and $e^{-{\sigma }_{3}x}$ to two sides of system (24a)–(24d), respectively, then taking a limit for these new equations when $x\to \infty$, and using boundary condition system (23i) and $\Re \gamma +{\lambda }_j+{\theta }_j+{\mu }_j=\Re {\sigma }_j>0,j=1,3$, we have

$$p_{1,1}^{*}\left(0\right)=\frac{{\mu }_1}{{\sigma }_1}{z}_{1,0}^{*}+\frac{{\theta }_1}{{\sigma }_1}{w}_1^{*}+{\lambda }_1{\int }_0^{\infty }{y}_{1,2}^{*}\left(x\right)e^{-{\sigma }_{1}x}dx,$$

(25a)

$$\begin{array}{c}{\displaystyle p_{1,n}^{*}\left(0\right)=\frac{{\mu }_1}{{\sigma }_1}{p}_{1,n-1}^{*}\left(0\right)+\frac{{\theta }_1}{{\sigma }_1}{w}_n^{*}+{\lambda }_1{\int }_0^{\infty }{y}_{1,n+1}^{*}\left(x\right)e^{-{\sigma }_{1}x}dx,n\ge 2,}\hfill \end{array}$$

(25b)

$$p_{3,1}^{*}\left(0\right)=\frac{{\theta }_3}{{\sigma }_3}{z}_{1,0}^{*}+\frac{{\mu }_3}{{\sigma }_3}{z}_{3,0}^{*}+{\lambda }_3{\int }_0^{\infty }{y}_{3,2}^{*}\left(x\right)e^{-{\sigma }_{3}x}dx,$$

(25c)

$$\begin{array}{c}{\displaystyle p_{3,n}^{*}\left(0\right)=\frac{{\theta }_3}{{\sigma }_3}{z}_{1,0}^{*}+\frac{{\mu }_3}{{\sigma }_3}{p}_{3,n-1}^{*}\left(0\right)+{\lambda }_3{\int }_0^{\infty }{y}_{3,n+1}^{*}\left(x\right)e^{-{\sigma }_{3}x}dx,n\ge 2.}\hfill \end{array}$$

(25d)

Repeatedly using system (25a)–(25d), it is not difficult to find that

$$\begin{array}{c}{\displaystyle p_{1,n}^{*}\left(0\right)={\left(\frac{{\mu }_1}{{\sigma }_1}\right)}^n{z}_{1,0}^{*}+\frac{{\theta }_1}{{\mu }_1}\sum _{k=1}^n{\left(\frac{{\mu }_1}{{\sigma }_1}\right)}^{n+1-k}{w}_k^{*}+{\lambda }_1\sum _{k=1}^n{\left(\frac{{\mu }_1}{{\sigma }_1}\right)}^{n-k}{\int }_0^{\infty }{y}_{1,k+1}^{*}\left(x\right)e^{-{\sigma }_{1}x}dx,}\hfill \end{array}$$

(26a)

$$\begin{array}{c}{\displaystyle p_{3,n}^{*}\left(0\right)={\left(\frac{{\mu }_3}{{\sigma }_3}\right)}^n{z}_{3,0}^{*}+\frac{{\theta }_3}{{\mu }_3}\sum _{k=1}^n{\left(\frac{{\mu }_3}{{\sigma }_3}\right)}^k{z}_{1,0}^{*}+{\lambda }_3\sum _{k=1}^n{\left(\frac{{\mu }_3}{{\sigma }_3}\right)}^{n-k}{\int }_0^{\infty }{y}_{3,k+1}^{*}\left(x\right)e^{-{\sigma }_{3}x}dx.}\hfill \end{array}$$

(26b)

Thus, for $\Re {\sigma }_j>0,j=1,3$, by Equations (24a)–(24d) and system (26a) and (26b), we obtain the following expressions of $p_{1,n}^{*}\left(x\right)$ and $p_{3,n}^{*}\left(x\right)$ without boundary conditions

$$\begin{array}{ll}\hfill p_{1,1}^{*}\left(x\right)& {\displaystyle =e^{{\sigma }_{1}x}{\int }_x^{\infty }[{\mu }_1{z}_{1,0}^{*}+{\theta }_1{w}_1^{*}+{\lambda }_1{y}_{1,2}^{*}\left(\tau \right)]e^{-{\sigma }_1\tau }d\tau }\hfill \\ & {\displaystyle =\frac{{\mu }_1}{{\sigma }_1}{z}_{1,0}^{*}+\frac{{\theta }_1}{{\sigma }_1}{w}_1^{*}+{\lambda }_1{e}^{{\sigma }_{1}x}{\int }_x^{\infty }{y}_{1,2}^{*}\left(\tau \right)e^{-{\sigma }_1\tau }d\tau ,}\hfill \end{array}$$

(27a)

$$\begin{array}{ll}\hfill p_{1,n}^{*}\left(x\right)& {\displaystyle =e^{{\sigma }_{1}x}{\int }_x^{\infty }[{\theta }_1{w}_n^{*}+{\mu }_1{p}_{1,n-1}^{*}\left(0\right)+{\lambda }_1{y}_{1,n+1}^{*}\left(\tau \right)]e^{-{\sigma }_1\tau }d\tau }\hfill \\ & {\displaystyle =\frac{{\mu }_1}{{\sigma }_1}{p}_{1,n-1}^{*}\left(0\right)+\frac{{\theta }_1}{{\sigma }_1}{w}_n^{*}+{\lambda }_1{e}^{{\sigma }_{1}x}{\int }_x^{\infty }{y}_{1,n+1}^{*}\left(\tau \right)e^{-{\sigma }_1\tau }d\tau }\hfill \\ & {\displaystyle ={\left(\frac{{\mu }_1}{{\sigma }_1}\right)}^n{z}_{1,0}^{*}+\frac{{\theta }_1}{{\mu }_1}\sum _{k=1}^n{\left(\frac{{\mu }_1}{{\sigma }_1}\right)}^{n+1-k}{w}_k^{*}+{\lambda }_1\sum _{k=1}^{n-1}{\left(\frac{{\mu }_1}{{\sigma }_1}\right)}^{n-k}{\int }_0^{\infty }{y}_{1,k+1}^{*}\left(x\right)e^{-{\sigma }_{1}x}dx}\hfill \\ & {\displaystyle +{\lambda }_1{e}^{{\sigma }_{1}x}{\int }_x^{\infty }{y}_{1,n+1}^{*}\left(\tau \right)e^{-{\sigma }_1\tau }d\tau ,n\ge 2,}\hfill \end{array}$$

(27b)

$$\begin{array}{ll}{p}_{3,1}^{*}\left(x\right)& {\displaystyle =e^{{\sigma }_{3}x}{\int }_x^{\infty }[{\theta }_3{z}_{1,0}^{*}+{\mu }_3{z}_{3,0}^{*}+{\lambda }_3{y}_{3,2}^{*}\left(\tau \right)]e^{-{\sigma }_3\tau }d\tau }\hfill \\ & {\displaystyle =\frac{{\mu }_3}{{\sigma }_3}{z}_{3,0}^{*}+\frac{{\theta }_3}{{\sigma }_3}{z}_{1,0}^{*}+{\lambda }_3{e}^{{\sigma }_{3}x}{\int }_x^{\infty }{y}_{3,2}^{*}\left(\tau \right)e^{-{\sigma }_3\tau }d\tau ,}\hfill \end{array}$$

(27c)

$$\begin{array}{ll}{p}_{3,n}^{*}\left(x\right)& {\displaystyle ={\int }_x^{\infty }[{\theta }_3{z}_{1,0}^{*}+{\mu }_3{p}_{3,n-1}^{*}\left(0\right)+{\lambda }_3{y}_{3,n+1}^{*}\left(\tau \right)]e^{-{\sigma }_3\tau }d\tau }\hfill \\ & {\displaystyle =\frac{{\mu }_3}{{\sigma }_3}{p}_{3,n-1}^{*}\left(0\right)+\frac{{\theta }_3}{{\sigma }_3}{z}_{1,0}^{*}+{\lambda }_3{e}^{{\sigma }_{3}x}{\int }_x^{\infty }{y}_{3,n+1}^{*}\left(\tau \right)e^{-{\sigma }_3\tau }d\tau }\hfill \\ & {\displaystyle ={\left(\frac{{\mu }_3}{{\sigma }_3}\right)}^n{z}_{3,0}^{*}+\frac{{\theta }_3}{{\mu }_3}\sum _{k=1}^n{\left(\frac{{\mu }_3}{{\sigma }_3}\right)}^k{z}_{1,0}^{*}+{\lambda }_3\sum _{k=1}^{n-1}{\left(\frac{{\mu }_3}{{\sigma }_3}\right)}^{n-k}{\int }_0^{\infty }{y}_{3,k+1}^{*}\left(x\right)e^{-{\sigma }_{3}x}dx}\hfill \\ & {\displaystyle +{\lambda }_3{e}^{{\sigma }_{3}x}{\int }_x^{\infty }{y}_{3,n+1}^{*}\left(\tau \right)e^{-{\sigma }_3\tau }d\tau ,n\ge 2.}\hfill \end{array}$$

(27d)

Hence, when $|{\sigma }_j|>{\mu }_j$ and $\Re {\sigma }_j>0,j=1,3$, from system (27a)–(27d), we conclude that the following inequalities hold true

$$\begin{array}{ll}\parallel p_{1,1}^{*}{\parallel }_{L^{\infty }[0,\infty )}& {\displaystyle \le \frac{{\mu }_1}{|{\sigma }_1|}|z_{1,0}^{*}|+\frac{{\theta }_1}{|{\sigma }_1|}|w_1^{*}|+\frac{{\lambda }_1}{\Re {\sigma }_1}{\parallel y_{1,2}^{*}\parallel }_{L^{\infty }[0,\infty )}}\hfill \\ & {\displaystyle \le sup\left\{\frac{{\theta }_1+{\mu }_1}{|{\sigma }_1|},\frac{{\lambda }_1}{\Re {\sigma }_1}\right\}\parallel (w^{*},y_1^{*},y_3^{*})\parallel ,}\hfill \end{array}$$

(28a)

$$\begin{array}{ll}\underset{n\ge 2}{sup}{\parallel p_{1,n}^{*}\parallel }_{L^{\infty }[0,\infty )}& {\displaystyle \le \underset{n\ge 2}{sup}\left\{{\left(\frac{{\mu }_1}{|{\sigma }_1|}\right)}^n|z_{1,0}^{*}|+\frac{{\theta }_1}{{\mu }_1}\sum _{k=1}^n{\left(\frac{{\mu }_1}{|{\sigma }_1|}\right)}^{n+1-k}|w_k^{*}|\right.}\hfill \\ & {\displaystyle \left.+\frac{{\lambda }_1}{\Re {\sigma }_1}\sum _{k=1}^{n-1}{\left(\frac{{\mu }_1}{|{\sigma }_1|}\right)}^{n-k}\parallel y_{1,k+1}^{*}{\parallel }_{L^{\infty }[0,\infty )}+\frac{{\lambda }_1}{\Re {\sigma }_1}{\parallel y_{1,n+1}^{*}\parallel }_{L^{\infty }[0,\infty )}\right\}}\hfill \\ & {\displaystyle \le sup\left\{\frac{{\theta }_1}{|{\sigma }_1|-{\mu }_1},\frac{{\lambda }_1{\mu }_1}{(\Re {\sigma }_1)(|{\sigma }_1|-{\mu }_1)}\right\}\parallel (w^{*},y_1^{*},y_3^{*})\parallel ,}\hfill \end{array}$$

(28b)

$$\begin{array}{ll}\parallel p_{3,1}^{*}{\parallel }_{L^{\infty }[0,\infty )}& {\displaystyle \le \frac{{\mu }_3}{|{\sigma }_3|}|z_{3,0}^{*}|+\frac{{\theta }_3}{|{\sigma }_3|}|z_{1,0}^{*}|+\frac{{\lambda }_3}{\Re {\sigma }_3}{\parallel y_{3,2}^{*}\parallel }_{L^{\infty }[0,\infty )}}\hfill \\ & {\displaystyle \le sup\left\{\frac{{\theta }_3+{\mu }_3}{|{\sigma }_3|},\frac{{\lambda }_3}{\Re {\sigma }_3}\right\}\parallel (w^{*},y_1^{*},y_3^{*})\parallel ,}\hfill \end{array}$$

(28c)

$$\begin{array}{ll}\underset{n\ge 2}{sup}{\parallel p_{3,n}^{*}\parallel }_{L^{\infty }[0,\infty )}& {\displaystyle \le \underset{n\ge 2}{sup}\left\{{\left(\frac{{\mu }_3}{|{\sigma }_3|}\right)}^n|z_{3,0}^{*}|+\frac{{\theta }_3}{{\mu }_3}\sum _{k=1}^n{\left(\frac{{\mu }_3}{|{\sigma }_3|}\right)}^k|z_{1,0}^{*}|\right.}\hfill \\ & \left.{\displaystyle +\frac{{\lambda }_3}{\Re {\sigma }_3}\sum _{k=1}^{n-1}{\left(\frac{{\mu }_3}{|{\sigma }_3|}\right)}^{n-k}\parallel y_{3,k+1}^{*}{\parallel }_{L^{\infty }[0,\infty )}+\frac{{\lambda }_3}{\Re {\sigma }_3}{\parallel y_{3,n+1}^{*}\parallel }_{L^{\infty }[0,\infty )}}\right\}\hfill \\ & {\displaystyle \le sup\left\{\frac{{\theta }_3}{|{\sigma }_3|-{\mu }_3},\frac{{\lambda }_3{\mu }_3}{(\Re {\sigma }_3)(|{\sigma }_3|-{\mu }_3)}\right\}\parallel (w^{*},y_1^{*},y_3^{*})\parallel .}\hfill \end{array}$$

(28d)

On the other hand, if $|{\sigma }_j|>{\mu }_j$ and $\Re {\sigma }_j>0,j=1,3$, then, by Equations (23a)–(23d), through simple but tedious calculations, we obtain

$$\begin{array}{ll}\hfill Q_{1,0}^{*}& =\frac{{\lambda }_1(\gamma +1)(\gamma +{\lambda }_3+{\theta }_3)}{(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0}{p}_{1,1}^{*}\left(0\right)\hfill \\ & {\displaystyle +\frac{{\lambda }_3{\theta }_1{N}_0}{(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0}{p}_{3,1}^{*}\left(0\right)}\hfill \\ & {\displaystyle +\frac{{\theta }_1(\gamma +{\lambda }_3+{\theta }_3)}{(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0}\sum _{k=1}^{\infty }{N}_k{p}_{3,k}^{*}\left(0\right),}\hfill \end{array}$$

(29a)

$$\begin{array}{ll}{Q}_{3,0}^{*}& =\frac{{\lambda }_1{\theta }_3(\gamma +1)}{(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0}{p}_{1,1}^{*}\left(0\right)\hfill \\ & {\displaystyle +\frac{{\lambda }_3(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)}{(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0}{p}_{3,1}^{*}\left(0\right)}\hfill \\ & {\displaystyle +\frac{{\theta }_1{\theta }_3}{(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0}\sum _{k=1}^{\infty }{N}_k{p}_{3,k}^{*}\left(0\right),}\hfill \end{array}$$

(29b)

$$\begin{array}{ll}{\displaystyle V_0^{*}}& =\frac{{\lambda }_1{\theta }_3{N}_0}{(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0}{p}_{1,1}^{*}\left(0\right)\hfill \\ & {\displaystyle +\frac{{\lambda }_3{N}_0(\gamma +{\lambda }_1+{\theta }_1)}{(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0}{p}_{3,1}^{*}\left(0\right)}\hfill \\ & {\displaystyle +\frac{(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)}{(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0}\sum _{k=1}^{\infty }{N}_k{p}_{3,k}^{*}\left(0\right),}\hfill \end{array}$$

(29c)

$$\begin{array}{ccc}\hfill V_n^{*}& & =\frac{1}{\gamma +1}\sum _{k=0}^{\infty }{N}_k{p}_{3,k+n}^{*}\left(0\right).\hfill \end{array}$$

(29d)

Then, by system (29a)–(29d) and system (26a) and (26b), it is not difficult to see that $Q_{1,0}^{*},Q_{3,0}^{*}$ and $V_n,n\ge 0$ satisfy the following inequalities

$$\begin{array}{ll}|Q_{1,0}^{*}|& \le \frac{1}{|(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|}\hfill \\ & {\displaystyle \times sup\left\{\frac{{\lambda }_1({\theta }_1+{\mu }_1)|(\gamma +1)(\gamma +{\lambda }_3+{\theta }_3)|}{|{\sigma }_1|},\frac{{\lambda }_1^2|(\gamma +1)(\gamma +{\lambda }_3+{\theta }_3)|}{\Re {\sigma }_1},\right.}\hfill \\ & {\displaystyle \frac{{\lambda }_3{\theta }_1{N}_0({\theta }_3+{\mu }_3)}{|{\sigma }_3|},\frac{{\lambda }_3^2{\theta }_1{N}_0}{\Re {\sigma }_3},\frac{{\theta }_1{\theta }_3(1-N_0)|\gamma +{\lambda }_3+{\theta }_3|}{|{\sigma }_3|-{\mu }_3},}\hfill \\ & {\displaystyle \left.\frac{{\lambda }_3{\mu }_3{\theta }_1(1-N_0)|\gamma +{\lambda }_3+{\theta }_3|}{(\Re {\sigma }_3)(|{\sigma }_3|-{\mu }_3)}\right\}\parallel (w^{*},y_1^{*},y_3^{*})\parallel ,}\hfill \end{array}$$

(30a)

$$\begin{array}{ll}|Q_{3,0}^{*}|& \le \frac{1}{|(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|}sup\left\{\frac{{\lambda }_1{\theta }_3({\theta }_1+{\mu }_1)|\gamma +1|}{|{\sigma }_1|},\right.\hfill \\ & \frac{{\lambda }_1^2{\theta }_3|\gamma +1|}{\Re {\sigma }_1},\frac{{\lambda }_3({\theta }_3+{\mu }_3)|(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)|}{|{\sigma }_3|},\frac{{\theta }_1{\theta }_3^2(1-N_0)}{|{\sigma }_3|-{\mu }_3},\hfill \\ & \left.\frac{{\lambda }_3^2|(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)|}{\Re {\sigma }_3},\frac{{\lambda }_3{\mu }_3{\theta }_1{\theta }_3(1-N_0)}{\Re {\sigma }_3(|{\sigma }_3|-{\mu }_3)}\right\}\parallel (w^{*},y_1^{*},y_3^{*})\parallel ,\hfill \end{array}$$

(30b)

$$\begin{array}{ll}|V_0^{*}|& {\displaystyle \le \frac{1}{|(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|}}\hfill \\ & {\displaystyle \times sup\left\{\frac{{\lambda }_3{\theta }_1{N}_0({\theta }_1+{\mu }_1)}{|{\sigma }_1|},\frac{{\lambda }_1{\lambda }_3{\theta }_1{N}_0}{\Re {\sigma }_1}\right.,}\hfill \\ & {\displaystyle \frac{{\lambda }_3{N}_0({\theta }_3+{\mu }_3)|\gamma +{\lambda }_1+{\theta }_1|}{|{\sigma }_3|},\frac{{\lambda }_3^2{N}_0|\gamma +{\lambda }_1+{\theta }_1|}{\Re {\sigma }_3},}\hfill \\ & {\displaystyle \frac{{\theta }_3(1-N_0)|(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)|}{|{\sigma }_3|-{\mu }_3},}\hfill \\ & {\displaystyle \left.\frac{{\lambda }_3{\mu }_3(1-N_0)|(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)|}{\Re {\sigma }_3(|{\sigma }_3|-{\mu }_3)}\right\}\parallel (w^{*},y_1^{*},y_3^{*})\parallel ,}\hfill \end{array}$$

(30c)

$$\begin{array}{ccc}\hfill \underset{n\ge 1}{sup}|V_n^{*}|& & \le sup\left\{\frac{{\theta }_3}{|\gamma +1||{\sigma }_3|-{\mu }_3},\frac{{\lambda }_3{\mu }_3}{|\gamma +1|\Re {\sigma }_3(|{\sigma }_3|-{\mu }_3)}\right\}\parallel (w^{*},y_1^{*},y_3^{*})\parallel .\hfill \end{array}$$

(30d)

Therefore, when $|{\sigma }_j|>{\mu }_j$ and $\Re {\sigma }_j>0,j=1,3$, from the inequalities system (28a)–(28d) and system (30a)–(30d), it can be computed that

$$\begin{array}{l}{\displaystyle \parallel (V^{*},p_1^{*},p_3^{*}){\parallel }_{*}=sup\{|Q_{1,0}^{*}|,|Q_{3,0}^{*}|,\underset{k\ge 0}{sup}|V_k^{*}|,\underset{k\ge 1}{sup}\parallel p_{1,k}{\parallel }_{L^{\infty }[0,\infty )},\underset{k\ge 1}{sup}\parallel p_{3,k}{\parallel }_{L^{\infty }[0,\infty )}\}}\hfill \\ {\displaystyle \le sup\left\{\frac{1}{|(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|}\right.}\hfill \\ \times sup\left\{\frac{{\lambda }_1({\theta }_1+{\mu }_1)|(\gamma +1)(\gamma +{\lambda }_3+{\theta }_3)|}{|{\sigma }_1|},\frac{{\lambda }_1^2|(\gamma +1)(\gamma +{\lambda }_3+{\theta }_3)|}{\Re {\sigma }_1},\frac{{\lambda }_3^2{\theta }_1{N}_0}{\Re {\sigma }_3}\right.,\hfill \\ \frac{{\lambda }_3{\theta }_1{N}_0({\theta }_3+{\mu }_3)}{|{\sigma }_3|},\frac{{\theta }_1{\theta }_3(1-N_0)|\gamma +{\lambda }_3+{\theta }_3|}{|{\sigma }_3|-{\mu }_3},\frac{{\lambda }_3{\mu }_3{\theta }_1(1-N_0)|\gamma +{\lambda }_3+{\theta }_3|}{\Re {\sigma }_3(|{\sigma }_3|-{\mu }_3)},\hfill \\ \frac{{\lambda }_1{\theta }_3({\theta }_1+{\mu }_1)|\gamma +1|}{|{\sigma }_1|},\frac{{\lambda }_1^2{\theta }_3|\gamma +1|}{\Re {\sigma }_1},\frac{{\lambda }_3({\theta }_3+{\mu }_3)|(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)|}{|{\sigma }_3|},\hfill \\ \frac{{\lambda }_3^2|(\gamma +1)(\gamma +{\lambda }_1+{\theta }_1)|}{\Re {\sigma }_3},\frac{{\theta }_1{\theta }_3^2(1-N_0)}{|{\sigma }_3|-{\mu }_3},\frac{{\lambda }_3{\mu }_3{\theta }_1{\theta }_3(1-N_0)}{\Re {\sigma }_3(|{\sigma }_3|-{\mu }_3)},\frac{{\lambda }_3{\theta }_1{N}_0({\theta }_1+{\mu }_1)}{|{\sigma }_1|},\hfill \\ \frac{{\lambda }_1{\lambda }_3{\theta }_1{N}_0}{\Re {\sigma }_1},\frac{{\lambda }_3{N}_0({\theta }_3+{\mu }_3)|\gamma +{\lambda }_1+{\theta }_1|}{|{\sigma }_3|},\frac{{\lambda }_3^2{N}_0|\gamma +{\lambda }_1+{\theta }_1|}{\Re {\sigma }_3},\hfill \\ \frac{{\theta }_3(1-N_0)|(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)|}{|{\sigma }_3|-{\mu }_3},\hfill \\ \left.\frac{{\lambda }_3{\mu }_3(1-N_0)|(\gamma +{\lambda }_1+{\theta }_1)(\gamma +{\lambda }_3+{\theta }_3)|}{\Re {\sigma }_3(|{\sigma }_3|-{\mu }_3)}\right\},\hfill \\ \frac{{\theta }_3}{|\gamma +1||{\sigma }_3|-{\mu }_3},\frac{{\lambda }_3{\mu }_3}{|\gamma +1|\Re {\sigma }_3(|{\sigma }_3|-{\mu }_3)},\frac{{\theta }_1+{\mu }_1}{|{\sigma }_1|},\frac{{\lambda }_1}{\Re {\sigma }_1},\frac{{\theta }_1}{|{\sigma }_1|-{\mu }_1},\hfill \\ \left.\frac{{\lambda }_1{\mu }_1}{\Re {\sigma }_1(|{\sigma }_1|-{\mu }_1)},\frac{{\theta }_3+{\mu }_3}{|{\sigma }_3|},\frac{{\lambda }_3}{\Re {\sigma }_3},\frac{{\theta }_3}{|{\sigma }_3|-{\mu }_3},\frac{{\lambda }_3{\mu }_3}{\Re {\sigma }_3(|{\sigma }_3|-{\mu }_3)}\right\}\parallel (w^{*},y_1^{*},y_3^{*})\parallel .\hfill \end{array}$$

(31)

The inequality system (31) means that if $\gamma \in \mathrm{\Lambda }$, then the operator ${[I-{(\gamma I-L-G)}^{-1}(J+H)]}^{-1}$ exists and is bounded. Now, we consider $(\gamma I-L-G)(V^{*},p_1^{*},p_3^{*})=(w^{*},y_1^{*},y_3^{*}),(w^{*},y_1^{*},y_3^{*})\in X^{*}$. Then, it is not difficult to show that the operator ${(\gamma I-L-G)}^{-1}$ also exists and is bounded. By the relationship

$$\begin{array}{c}{\displaystyle {[\gamma I-(L+G+J+H)]}^{-1}={\left\{(\gamma I-L-G)[I-{(\gamma I-L-G)}^{-1}(J+H)]\right\}}^{-1}}\hfill \\ {\displaystyle ={[I-{(\gamma I-L-G)}^{-1}(J+H)]}^{-1}{(\gamma I-L-G)}^{-1},}\hfill \end{array}$$

(32)

we obtain that if $\gamma \in \mathrm{\Lambda }$, then the operator ${(\gamma I-L-G-J-H)}^{-1}$ also exists and is bounded. Therefore, all points in the set $\mathrm{\Lambda }$ belong to the resolvent set $\rho (L+J+G+H)$.

Additionally, if we take $\gamma =ib,i^2=-1,b\in \mathbb{R}\setminus \left\{0\right\}$, then, for any ${\lambda }_j,{\theta }_j,{\mu }_j>0,j=1,3$ and $0<N_0<1$, through simple but tedious calculations, we obtain the following inequalities:

$$\begin{array}{c}{\displaystyle \frac{{\lambda }_j}{|ib+{\lambda }_j+{\theta }_j+{\mu }_j|-{\mu }_j}=\frac{{\lambda }_j}{\sqrt{b^2+{({\lambda }_j+{\theta }_3+{\mu }_j)}^2}-{\mu }_j}<1,}\hfill \end{array}$$

(33a)

$$\begin{array}{c}{\displaystyle \frac{{\theta }_j}{|ib+{\lambda }_j+{\theta }_j+{\mu }_j|-{\mu }_j}=\frac{{\theta }_j}{\sqrt{b^2+{({\lambda }_j+{\theta }_j+{\mu }_j)}^2}-{\mu }_j}<1,}\hfill \end{array}$$

(33b)

$$\begin{array}{c}{\displaystyle \frac{{\theta }_j+{\mu }_j}{|ib+{\lambda }_j+{\theta }_j+{\mu }_j|}=\frac{{\theta }_j+{\mu }_j}{\sqrt{b^2+{({\lambda }_j+{\theta }_j+{\mu }_j)}^2}}<1,}\hfill \end{array}$$

(33c)

$$\begin{array}{l}{\displaystyle |(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0{|}^2-{\lambda }_1^2{|(ib+1)(ib+{\lambda }_3+{\theta }_3)|}^2}\hfill \\ {\displaystyle =b^6+b^4[2{\lambda }_1{\theta }_1+{\theta }_1^2+{({\lambda }_3+{\theta }_3)}^2+1]+b^2\{2{\lambda }_1{\theta }_1+{\theta }_1^2+{({\lambda }_3+{\theta }_3)}^2}\hfill \\ {\displaystyle +(2{\lambda }_1{\theta }_1+{\theta }_1^2){({\lambda }_3+{\theta }_3)}^2+2({\lambda }_1+{\theta }_1+{\lambda }_3+{\theta }_3+1){\theta }_1{\theta }_3{N}_0\}}\hfill \\ {\displaystyle +2{\lambda }_1({\lambda }_3+{\theta }_3)[{\theta }_1({\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0]+{[{\theta }_1({\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0]}^2>0}\hfill \\ {\displaystyle \Rightarrow \frac{{\lambda }_1|(ib+1)(ib+{\lambda }_3+{\theta }_3)|}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|}<1}\hfill \\ {\displaystyle \Rightarrow \frac{{\lambda }_1({\theta }_1+{\mu }_1)|(ib+1)(ib+{\lambda }_3+{\theta }_3)|}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0||ib+{\lambda }_1+{\theta }_1+{\mu }_1|}<1,}\hfill \end{array}$$

(33d)

$$\begin{array}{l}{\displaystyle \frac{{\lambda }_1^2|(ib+1)(ib+{\lambda }_3+{\theta }_3)|}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|({\lambda }_1+{\theta }_1+{\mu }_1)}<1,}\hfill \end{array}$$

(33e)

$$\begin{array}{l}\frac{{\lambda }_3{\theta }_1{N}_0}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|}<1\\ \Rightarrow \\ \frac{{\lambda }_3{\theta }_1{N}_0({\theta }_3+{\mu }_3)}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0||ib+{\lambda }_3+{\theta }_3+{\mu }_3|}<1,\end{array}$$

(33f)

$$\begin{array}{c}{\displaystyle \frac{{\lambda }_3^2{\theta }_1{N}_0}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|({\lambda }_3+{\theta }_3+{\mu }_3)}<1,}\hfill \end{array}$$

(33g)

$$\begin{array}{c}{\displaystyle \frac{{\lambda }_3{\theta }_1{N}_0({\theta }_1+{\mu }_1)}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0||ib+{\lambda }_1+{\theta }_1+{\mu }_1|}<1,}\hfill \end{array}$$

(33h)

$$\begin{array}{c}{\displaystyle \frac{{\lambda }_1{\lambda }_3{\theta }_1{N}_0}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0||ib+{\lambda }_1+{\theta }_1+{\mu }_1|}<1,}\hfill \end{array}$$

(33i)

$$\begin{array}{l}\frac{{\theta }_1(1-N_0)|ib+{\lambda }_3+{\theta }_3|}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|}<1\\ \Rightarrow \\ \frac{{\theta }_1{\theta }_3(1-N_0)|ib+{\lambda }_3+{\theta }_3|}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|(|ib+{\lambda }_3+{\theta }_3+{\mu }_3|-{\mu }_3)}\\ <1,\end{array}$$

(33j)

$$\begin{array}{c}{\displaystyle \frac{{\lambda }_3{\mu }_3{\theta }_1(1-N_0)|ib+{\lambda }_3+{\theta }_3|({\lambda }_3+{\theta }_3+{\mu }_3)(|ib+{\lambda }_3+{\theta }_3+{\mu }_3|-{\mu }_3)}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|}<1,}\hfill \end{array}$$

(33k)

$$\begin{array}{l}\frac{{\lambda }_1{\theta }_3|ib+1|}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|}<1\hfill \\ {\displaystyle \Rightarrow \frac{{\lambda }_1{\theta }_3({\theta }_1+{\mu }_1)|ib+1|}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|({\lambda }_1+{\theta }_1+{\mu }_1)}<1,}\hfill \end{array}$$

(33l)

$$\begin{array}{ccc}\hfill & & {\displaystyle \frac{{\lambda }_1^2|(ib+1)(ib+{\lambda }_3+{\theta }_3)|}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|({\lambda }_1+{\theta }_1+{\mu }_1)}<1,}\hfill \end{array}$$

(33m)

$${\displaystyle \frac{{\lambda }_3({\theta }_3+{\mu }_3)|(ib+1)(ib+{\lambda }_1+{\theta }_1)|}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0||ib+{\lambda }_3+{\theta }_3+{\mu }_3|}<1,}$$

(33n)

$${\displaystyle \frac{{\lambda }_3^2|(ib+1)(ib+{\lambda }_1+{\theta }_1)|}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|({\lambda }_3+{\theta }_3+{\mu }_3)}<1,}$$

(33o)

$${\displaystyle \frac{{\theta }_1{\theta }_3^2(1-N_0)}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|(|ib+{\lambda }_3+{\theta }_3+{\mu }_3|-{\mu }_3)}<1,}$$

(33p)

$${\displaystyle \frac{{\lambda }_3{\mu }_3{\theta }_1{\theta }_3(1-N_0)({\lambda }_3+{\theta }_3+{\mu }_3)(|ib+{\lambda }_3+{\theta }_3+{\mu }_3|-{\mu }_3)}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|}<1,}$$

(33q)

$${\displaystyle \frac{{\lambda }_3{N}_0({\theta }_3+{\mu }_3)|ib+{\lambda }_1+{\theta }_1|}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0||ib+{\lambda }_3+{\theta }_3+{\mu }_3|}<1,}$$

(33r)

$${\displaystyle \frac{{\lambda }_3^2{N}_0|ib+{\lambda }_1+{\theta }_1|}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|({\lambda }_3+{\theta }_3+{\mu }_3)}<1,}$$

(33s)

$${\displaystyle \frac{{\theta }_3(1-N_0)|(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)|}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|(|ib+{\lambda }_3+{\theta }_3+{\mu }_3|-{\mu }_3)}<1,}$$

(33t)

$${\displaystyle \frac{{\lambda }_3{\mu }_3(1-N_0)|(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)({\lambda }_3+{\theta }_3+{\mu }_3)(|ib+{\lambda }_3+{\theta }_3+{\mu }_3|-{\mu }_3)|}{|(ib+1)(ib+{\lambda }_1+{\theta }_1)(ib+{\lambda }_3+{\theta }_3)-{\theta }_1{\theta }_3{N}_0|}<1.}$$

(33u)

The inequalities system (33a)–(33u) imply that $i\mathbb{R}\setminus \left\{0\right\}\subset \rho \left({(A+U+E)}^{*}\right)$. Then, by the relationship between the spectrum $\sigma (A+U+E)$ and $\sigma \left({(A+U+E)}^{*}\right)$, we obtain that $i\mathbb{R}\setminus \left\{0\right\}\subset \rho (A+U+E)$. □

Let ${\sigma }_p(A+U+E)$ denote the point spectrum of $A+U+E$.

Lemma 3.

If ${\lambda }_j,{\theta }_j,{\mu }_j>0$ and ${\lambda }_j<{\theta }_j+{\mu }_j,j=1,3$, then $0\in {\sigma }_p(A+U+E)$, and the geometric multiplicity of zero is one.

Proof. 

We consider $(A+U+E)(V,p_1,p_3)=0$, which is equivalent to

$${\displaystyle ({\lambda }_1+{\theta }_1)Q_{1,0}={\theta }_3\sum _{n=1}^{\infty }{\int }_0^{\infty }{p}_{3,n}\left(x\right)dx+{\theta }_3{Q}_{3,0}+{\mu }_1{\int }_0^{\infty }{p}_{1,1}\left(x\right)dx,}$$

(34a)

$${\displaystyle ({\lambda }_3+{\theta }_3)Q_{3,0}=N_0{V}_0+{\mu }_3{\int }_0^{\infty }{p}_{3,1}\left(x\right)dx,}$$

(34b)

$$V_0={\theta }_1{Q}_{1,0},V_n={\theta }_1{\int }_0^{\infty }{p}_{1,n}\left(x\right)dx,n\ge 1,$$

(34c)

$$\frac{d{p}_{j,1}\left(x\right)}{dx}=-({\lambda }_j+{\theta }_j+{\mu }_j)p_{j,1}\left(x\right),j=1,3,$$

(34d)

$$\frac{d{p}_{j,n}\left(x\right)}{dx}=-({\lambda }_j+{\theta }_j+{\mu }_j)p_{j,n}\left(x\right)+{\lambda }_j{p}_{j,n-1}\left(x\right),n\ge 2;j=1,3,$$

(34e)

$$p_{1,1}\left(0\right)={\lambda }_1{Q}_{1,0}+{\mu }_1{\int }_0^{\infty }{p}_{1,2}\left(x\right)dx,$$

(34f)

$$p_{1,n}\left(0\right)={\mu }_1{\int }_0^{\infty }{p}_{1,n+1}\left(x\right)dx,n\ge 2,$$

(34g)

$$p_{3,1}\left(0\right)={\lambda }_3{Q}_{3,0}+N_1{V}_0+N_0{V}_1+{\mu }_3{\int }_0^{\infty }{p}_{3,2}\left(x\right)dx,$$

(34h)

$$p_{3,n}\left(0\right)=\sum _{k=0}^n{N}_{n-k}{V}_k+{\mu }_3{\int }_0^{\infty }{p}_{3,n+1}\left(x\right)dx,n\ge 2.$$

(34i)

Solving Equations (34d) and (34e), we obtain

$${\displaystyle p_{j,n}\left(x\right)=e^{-({\lambda }_j+{\theta }_j+{\mu }_j)x}\sum _{k=1}^n\frac{{\left({\lambda }_{j}x\right)}^{n-k}}{(n-k)!}{p}_{j,k}\left(0\right),j=1,3;n\ge 1.}$$

(35)

In the following, by introducing the probability generating functions (PGFs), see, e.g., [6], we show that $0\in {\sigma }_p(A+U+E)$. For any conplex number z such that $|z|<1$, we define the PGFs by

$$\overline{V}\left(z\right):=\sum _{n=0}^{\infty }{V}_n{z}^n,p_j(x,z):=\sum _{n=1}^{\infty }{p}_{j,n}\left(x\right)z^n,j=1,3.$$

Then, Theorem 3 means that $\overline{V}\left(z\right)$ and $p_j(x,z)$ are well-posed. From Equations (34b)–(34e), it is easy to compute that

$$\overline{V}\left(z\right)=V_0+\sum _{n=1}^{\infty }{V}_n{z}^n={\theta }_1{Q}_{1,0}+{\theta }_1\sum _{n=1}^{\infty }{\int }_0^{\infty }{p}_{1,n}\left(x\right)z^{n}dx,$$

(36a)

$$\begin{array}{ll}& \frac{\partial {\sum }_{n=1}^{\infty }{p}_{j,n}\left(x\right)z^n}{\partial x}=-\sum _{n=1}^{\infty }({\lambda }_j+{\theta }_j+{\mu }_j)p_{j,n}\left(x\right)z^n+{\lambda }_j\sum _{n=1}^{\infty }{p}_{j,n-1}\left(x\right)z^n\\ \Rightarrow & p_j(x,z)=p_j(0,z)e^{-[{\theta }_j+{\lambda }_j(1-z)+{\mu }_j]x},j=1,3.\end{array}$$

(36b)

From the boundary conditions system (34f) and (34g) and system (36b), we have

$$\begin{array}{c}{\displaystyle p_1(0,z)=p_{1,1}\left(0\right)z+\sum _{n=2}^{\infty }{p}_{1,n}\left(0\right)z^n}\hfill \\ {\displaystyle =z{\lambda }_1{Q}_{1,0}+{\mu }_1{\int }_0^{\infty }{p}_{1,2}\left(x\right)zdx+{\mu }_1\sum _{n=2}^{\infty }{\int }_0^{\infty }{p}_{1,n+1}\left(x\right)z^{n}dx}\hfill \\ {\displaystyle =z{\lambda }_1{Q}_{1,0}+\frac{{\mu }_1}{z[{\theta }_1+{\lambda }_1(1-z)+{\mu }_1]}{p}_1(0,z)-{\mu }_1{\int }_0^{\infty }{p}_{1,1}\left(x\right)dx.}\hfill \end{array}$$

That is,

$$\left(z-\frac{{\mu }_1}{{\theta }_1+{\lambda }_1(1-z)+{\mu }_1}\right)p_1(0,z)=z^2{\lambda }_1{Q}_{1,0}-z{\mu }_1{\int }_0^{\infty }{p}_{1,1}\left(x\right)dx.$$

(37)

By Equations (34b) and (34c), (34h)–(34i), and (36b) and

$$\sum _{n=0}^{\infty }{N}_n{z}^n=\sum _{n=0}^{\infty }{e}^{-{\lambda }_{2}pd}\frac{{\left({\lambda }_{2}pdz\right)}^n}{n!}=e^{-{\lambda }_{2}pd(1-z)},$$

we compute that

$$\begin{array}{c}{\displaystyle p_3(0,z)=p_{3,1}\left(0\right)z+\sum _{n=2}^{\infty }{p}_{3,n}\left(0\right)z^n=z{\lambda }_3{Q}_{3,0}+z{N}_1{V}_0+z{N}_0{V}_1+{\mu }_3{\int }_0^{\infty }{p}_{3,2}\left(x\right)zdx}\hfill \\ {\displaystyle +\sum _{n=2}^{\infty }\left[\sum _{k=0}^n{N}_{n-k}{V}_k+{\mu }_3{\int }_0^{\infty }{p}_{3,n+1}\left(x\right)dx\right]z^n}\hfill \\ {\displaystyle =-[{\theta }_3+{\lambda }_3(1-z)]Q_{3,0}+\overline{V}\left(z\right)e^{-{\lambda }_{2}pd(1-z)}+\frac{{\mu }_3}{z[{\theta }_3+{\lambda }_3(1-z)+{\mu }_3]}{p}_3(0,z).}\hfill \end{array}$$

This is equivalent to

$$\left(z-\frac{{\mu }_3}{{\theta }_3+{\lambda }_3(1-z)+{\mu }_3}\right)p_3(0,z)=-z[{\theta }_3+{\lambda }_3(1-z)]Q_{3,0}+z\overline{V}\left(z\right)e^{-{\lambda }_{2}pd(1-z)}.$$

(38)

It is not difficult to show that if ${\lambda }_j<{\theta }_j+{\mu }_j,j=1,3$, then

$$0<{\gamma }_j:=\frac{{\lambda }_j+{\theta }_j+{\mu }_j-\sqrt{{({\lambda }_j+{\theta }_j+{\mu }_j)}^2-4{\lambda }_j{\mu }_j}}{2{\lambda }_j}<1,$$

and ${\gamma }_j$ is the unique solution of the function $z=\frac{{\mu }_j}{{\theta }_j+{\lambda }_j(1-z)+{\mu }_j}$. Hence, using system (37) and (38), we obtain

$$Q_{1,0}=\frac{{\mu }_1}{{\gamma }_1{\lambda }_1}{\int }_0^{\infty }{p}_{1,1}\left(x\right)dx,Q_{3,0}=\frac{\overline{V}\left({\gamma }_3\right)e^{-{\lambda }_{2}pd(1-{\gamma }_3)}}{{\theta }_3+{\lambda }_3(1-{\gamma }_3)}.$$

(39)

In addition, from system (37) and (38), it is easy to obtain that

$${\displaystyle p_1(0,z)=\frac{{\lambda }_1{Q}_{1,0}z(z-{\gamma }_1)}{z-\frac{{\mu }_1}{{\theta }_1+{\lambda }_1(1-z)+{\mu }_1}}=\frac{{\lambda }_1{Q}_{1,0}z(z-{\gamma }_1)[{\theta }_1+{\lambda }_1(1-z)+{\mu }_1]}{z[{\theta }_1+{\lambda }_1(1-z)+{\mu }_1]-{\mu }_1},}$$

(40a)

$${\displaystyle p_3(0,z)=\frac{-z[{\theta }_3+{\lambda }_3(1-z)]Q_{3,0}+z\overline{V}\left(z\right)e^{{\lambda }_{2}pd(1-z)}}{z-\frac{{\mu }_3}{{\theta }_3+{\lambda }_3(1-z)+{\mu }_3}}.}$$

(40b)

Taking z on both sides of system (40a) and (40b) tends towards the limit of one, and it can be obtained that

$$p_1(0,1)=\frac{{\lambda }_1{Q}_{1,0}(1-{\gamma }_1)({\theta }_1+{\mu }_1)}{{\theta }_1},$$

(41a)

$$p_3(0,1)=\frac{[-{\theta }_3{Q}_{3,0}+\overline{V}\left(1\right)]({\theta }_3+{\mu }_3)}{{\theta }_3}.$$

(41b)

Then, using system (36b), (41a), and (39), we obtain

$${\displaystyle \sum _{n=1}^{\infty }{\int }_0^{\infty }{p}_{1,n}\left(x\right)dx=\underset{z\to 1}{lim}{\int }_0^{\infty }{p}_1(0,z)e^{-[{\theta }_1+{\lambda }_1(1-z)+{\mu }_1]x}dx=\frac{{\lambda }_1{Q}_{1,0}(1-{\gamma }_1)}{{\theta }_1},}$$

(42a)

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{n=1}^{\infty }{\int }_0^{\infty }{p}_{3,n}\left(x\right)dx=\underset{z\to 1}{lim}{\int }_0^{\infty }{p}_3(0,z)e^{-[{\theta }_3+{\lambda }_3(1-z)+{\mu }_3]x}dx}\hfill \\ \hfill & {\displaystyle =\frac{-{\theta }_3{Q}_{3,0}+{\theta }_1{Q}_{1,0}+{\lambda }_1{Q}_{1,0}(1-{\gamma }_1)}{{\theta }_3}.}\hfill \end{array}$$

(42b)

Therefore, using system (34c), (39), (42a), and (42b) and by simple calculations, it can be obtained that

$$\begin{array}{l}{Q}_{1,0}+Q_{3,0}+\sum _{n=0}^{\infty }{V}_n+\sum _{n=1}^{\infty }{\int }_0^{\infty }{p}_{1,n}\left(x\right)dx+\sum _{n=1}^{\infty }{\int }_0^{\infty }{p}_{3,n}\left(x\right)dx\hfill \\ =\frac{(1+{\theta }_1){\theta }_1{\theta }_3+{\theta }_1^2+{\lambda }_1(1-{\gamma }_1)({\theta }_1+{\theta }_3)}{{\theta }_1{\theta }_3}{Q}_{1,0}.\hfill \end{array}$$

(43)

system (43) means that $0\in {\sigma }_p(A+U+E)$. Additionally, from system (34a)–(34c) and system (34f)–(35), it is not difficult to show that the geometric multiplicity of zero is one. □

Lemma 4.

If ${\lambda }_j,{\theta }_j,{\mu }_j>0,j=1,3$, then $0\in {\sigma }_p\left({(A+U+E)}^{*}\right)$, and the geometric multiplicity of zero is one.

Proof. 

We consider ${(A+U+E)}^{*}(V^{*},p_1^{*},p_3^{*})=0$; that is,

$$-({\lambda }_1+{\theta }_1)Q_{1,0}^{*}+{\theta }_1{V}_0^{*}+{\lambda }_1{p}_{1,1}^{*}\left(0\right)=0,$$

(44a)

$${\theta }_3{Q}_{1,0}^{*}-({\lambda }_3+{\theta }_3)Q_{3,0}^{*}+{\lambda }_3{p}_{3,1}^{*}\left(0\right)=0,$$

(44b)

$$N_0{Q}_{3,0}^{*}-V_0^{*}+\sum _{k=1}^{\infty }{N}_k{p}_{3,k}^{*}\left(0\right)=0,$$

(44c)

$$-V_n^{*}+\sum _{k=0}^{\infty }{N}_k{p}_{3,k+n}^{*}\left(0\right)=0,n\ge 1,$$

(44d)

$$\frac{d{p}_{1,1}^{*}\left(x\right)}{dx}-({\lambda }_1+{\theta }_1+{\mu }_1)p_{1,1}^{*}\left(x\right)+{\lambda }_1{p}_{1,2}^{*}\left(x\right)+{\theta }_1{V}_1^{*}+{\mu }_1{Q}_{1,0}^{*}=0,$$

(44e)

$$\frac{d{p}_{1,n}^{*}\left(x\right)}{dx}-({\lambda }_1+{\theta }_1+{\mu }_1)p_{1,n}^{*}\left(x\right)+{\lambda }_1{p}_{1,n+1}^{*}\left(x\right)+{\theta }_1{V}_n^{*}+{\mu }_1{p}_{1,n-1}^{*}\left(0\right)=0,n\ge 2,$$

(44f)

$$\frac{d{p}_{3,1}^{*}\left(x\right)}{dx}-({\lambda }_3+{\theta }_3+{\mu }_3)p_{3,1}^{*}\left(x\right)+{\lambda }_3{p}_{3,2}^{*}\left(x\right)+{\theta }_3{Q}_{1,0}^{*}+{\mu }_3{Q}_{3,0}^{*}=0,$$

(44g)

$$\frac{d{p}_{3,n}^{*}\left(x\right)}{dx}-({\lambda }_3+{\theta }_3+{\mu }_3)p_{3,n}^{*}\left(x\right)+{\lambda }_3{p}_{3,n+1}^{*}\left(x\right)+{\theta }_3{Q}_{1,0}^{*}+{\mu }_3\left(x\right)p_{3,n-1}^{*}\left(0\right)=0,n\ge 2.$$

(44h)

It is not difficult to see that

$$(V^{*},p_1^{*},p_3^{*})=\left(\left(\begin{array}{c}\alpha \\ ⋮\end{array}\right),\left(\begin{array}{c}\alpha \\ ⋮\end{array}\right),\left(\begin{array}{c}\alpha \\ ⋮\end{array}\right)\right)\in D\left({(A+U+E)}^{*}\right)$$

is a positive eigenvalue of the system of Equations (44a)–(44h). Moreover, by system (44a)–(44h), it is not difficult to calculate that

$$\begin{array}{ll}{Q}_{1,0}^{*}& {\displaystyle =\frac{{\lambda }_1({\lambda }_3+{\theta }_3)}{({\lambda }_1+{\theta }_1)({\lambda }_3+{\theta }_3)-N_0{\theta }_1{\theta }_3}{p}_{1,1}^{*}\left(0\right)+\frac{{\lambda }_3{\theta }_1{N}_0}{({\lambda }_1+{\theta }_1)({\lambda }_3+{\theta }_3)-N_0{\theta }_1{\theta }_3}{p}_{3,1}^{*}\left(0\right)}\hfill \\ & +\frac{{\theta }_1({\lambda }_3+{\theta }_3)}{({\lambda }_1+{\theta }_1)({\lambda }_3+{\theta }_3)-N_0{\theta }_1{\theta }_3}\sum _{k=1}^{\infty }{N}_k{p}_{3,k}^{*}\left(0\right),\end{array}$$

(45a)

$$\begin{array}{ll}{Q}_{3,0}^{*}& {\displaystyle =\frac{{\lambda }_1{\theta }_3}{({\lambda }_1+{\theta }_1)({\lambda }_3+{\theta }_3)-N_0{\theta }_1{\theta }_3}{p}_{1,1}^{*}\left(0\right)+\frac{{\lambda }_3({\lambda }_1+{\theta }_1)}{({\lambda }_1+{\theta }_1)({\lambda }_3+{\theta }_3)-N_0{\theta }_1{\theta }_3}{p}_{3,1}^{*}\left(0\right)}\\ & +\frac{{\theta }_1{\theta }_3}{({\lambda }_1+{\theta }_1)({\lambda }_3+{\theta }_3)-N_0{\theta }_1{\theta }_3}\sum _{k=1}^{\infty }{N}_k{p}_{3,k}^{*}\left(0\right),\end{array}$$

(45b)

$$\begin{array}{ll}{V}_0^{*}& =\frac{{\lambda }_1{\theta }_3{N}_0}{({\lambda }_1+{\theta }_1)({\lambda }_3+{\theta }_3)-N_0{\theta }_1{\theta }_3}{p}_{1,1}^{*}\left(0\right)+\frac{{\lambda }_3({\lambda }_1+{\theta }_1)N_0}{({\lambda }_1+{\theta }_1)({\lambda }_3+{\theta }_3)-N_0{\theta }_1{\theta }_3}{p}_{3,1}^{*}\left(0\right)\\ & +\frac{({\lambda }_1+{\theta }_1)({\lambda }_3+{\theta }_3)}{({\lambda }_1+{\theta }_1)({\lambda }_3+{\theta }_3)-N_0{\theta }_1{\theta }_3}\sum _{k=1}^{\infty }{N}_k{p}_{3,k}^{*}\left(0\right),\end{array}$$

(45c)

$$V_n^{*}=\sum _{k=0}^{\infty }{N}_k{p}_{3,k+n}^{*}\left(0\right),n\ge 1,$$

(45d)

$$p_{1,2}^{*}\left(x\right)=-\frac{1}{{\lambda }_1}\left[\frac{d{p}_{1,1}^{*}\left(x\right)}{dx}-({\lambda }_1+{\theta }_1+{\mu }_1)p_{1,1}^{*}\left(x\right)+{\theta }_1{V}_1^{*}+{\mu }_1{Q}_{1,0}^{*}\right],$$

(45e)

$$p_{1,n+1}^{*}\left(x\right)=-\frac{1}{{\lambda }_1}\left[\frac{d{p}_{1,n}^{*}\left(x\right)}{dx}-({\lambda }_1+{\theta }_1+{\mu }_1)p_{1,n}^{*}\left(x\right)+{\theta }_1{V}_n^{*}+{\mu }_1{p}_{1,n-1}^{*}\left(0\right)\right],n\ge 2,$$

(45f)

$$p_{3,2}^{*}\left(x\right)=-\frac{1}{{\lambda }_3}\left[\frac{d{p}_{3,1}^{*}\left(x\right)}{dx}-({\lambda }_3+{\theta }_3+{\mu }_3)p_{3,1}^{*}\left(x\right)+{\theta }_3{Q}_{1,0}^{*}+{\mu }_3{Q}_{3,0}^{*}\right],$$

(45g)

$$p_{3,n+1}^{*}\left(x\right)=-\frac{1}{{\lambda }_3}\left[\frac{d{p}_{3,n}^{*}\left(x\right)}{dx}-({\lambda }_3+{\theta }_3+{\mu }_3)p_{3,n}^{*}\left(x\right)+{\theta }_3{Q}_{1,0}^{*}+{\mu }_3{p}_{3,n-1}^{*}\left(0\right)\right],n\ge 2.$$

(45h)

Equations (45a)–(45h) mean that the geometric multiplicity of zero is one. □

Proof of Theorem 4.

Theorem 3 means that $A+U+E$ generates a uniformly bounded $C_0$—semigroup on the Banach space X. Moreover, by Lemmas 2, 3, and 4 it is easy to see that ${\sigma }_p(A+U+E)\cap i\mathbb{R}={\sigma }_p\left({(A+U+E)}^{*}\right)\cap i\mathbb{R}=\left\{0\right\}$, $\{\gamma \in \mathbb{C}\mid \gamma =ib,b\ne 0,b\in \mathbb{R}\}\subset \rho (A+U+E)$, and $0\in {\sigma }_p\left({(A+U+E)}^{*}\right)$, and the geometric multiplicity of zero is one. Therefore, by theorem 1.96 of [14], we obtain that the TDS of system (3) strongly converges to its SSS; that is

$$\underset{t\to \infty }{lim}\parallel (V\left(t\right),p_1(·,t),p_3(·,t))-\langle (V^{*},p_1^{*}(·),p_3^{*}(·)),(u_2,u_1(·),u_3(·))\rangle (V,p_1(·),p_3(·))\parallel =0,$$

where $(V,p_1(·),p_3(·))$ and $(V^{*},p_1^{*}(·),p_3^{*}(·))$ are the eigenvectors corresponding to the zero in Lemmas 3 and 4, respectively. □

4. Stability of the Semigroup

In this section, when the service rates ${\mu }_1(·)$ and ${\mu }_3(·)$ are constants, based on the preceding discussion, we study the stability of the $C_0$-semigroup $e^{\mathcal{A}t}$ generated by $\mathcal{A}:=A+U+E$, from which we derive that the SDGC holds true.

Theorem 5.

The SDGC holds for the semigroup $e^{\mathcal{A}t}$, and ${\omega }_0\left(\mathcal{A}\right)=s\left(\mathcal{A}\right)=0$, where ${\omega }_0\left(\mathcal{A}\right)=\{\omega \mid \mathit{there}\mathit{exists}\mathit{an}\mathbb{M}\mathit{such}\mathit{that}\parallel e^{\mathcal{A}t}\parallel \le \mathbb{M}{e}^{\omega t}\}$ is the growth bound of the $C_0$—semigroup $e^{\mathcal{A}t}$, and $s\left(\mathcal{A}\right)=\{\Re \gamma \mid \gamma \in \sigma (\mathcal{A}\left)\right\}$ is the spectral bound of $\mathcal{A}$. Furthermore, $e^{\mathcal{A}t}$ is not asymptotically stable.

Proof. 

Obviously, X is a Banach lattice, and the semigroup $e^{\mathcal{A}}$ is positive on the Banach lattice X by Theorem 1. For $(V,p_1,p_3)$, $(w,y_1,y_3)\in X_{+}$, where $w=(z_{1,0},z_{3,0},w_0,w_1,\cdots )$, $y_j=(y_{j,1},y_{j,2},y_{j,3},\cdots ),j=1,3$, it is easy to obtain that

$$\begin{array}{cc}& \parallel (V,p_1,p_3)+(w,y_1,y_3){\parallel }_{+}\hfill \\ \hfill & {\displaystyle =|Q_{1,0}+z_{1,0}|+|Q_{3,0}+z_{3,0}|+\sum _{n=0}^{\infty }|V_n+w_n|+\sum _{j=1,3}\sum _{n=1}^{\infty }{\int }_0^{\infty }|p_{j,n}\left(x\right)+y_{j,n}\left(x\right)|dx}\hfill \\ \hfill & {\displaystyle =Q_{1,0}+Q_{3,0}+\sum _{n=0}^{\infty }{V}_n+\sum _{j=1,3}\sum _{n=1}^{\infty }{\int }_0^{\infty }{p}_{j,n}\left(x\right)dx}\hfill \\ \hfill & {\displaystyle +z_{1,0}+z_{3,0}+\sum _{n=0}^{\infty }{w}_n+\sum _{j=1,3}\sum _{n=1}^{\infty }{\int }_0^{\infty }{y}_{j,n}\left(x\right)dx}\hfill \\ \hfill & =\parallel (V,p_1,p_3){\parallel }_{+}+{\parallel (w,y_1,y_3)\parallel }_{+},\hfill \end{array}$$

where $X_{+}$ is introduced in Section 2, and ${\parallel ·\parallel }_{+}$ is the norm on $X_{+}$. This shows that X is an AL-space. By Lemmas 2 and 3, we see that $0\in {\sigma }_p\left(\mathcal{A}\right)$, and $\sigma \left(\mathcal{A}\right)\subset \{\gamma \in \mathbb{C}\mid \Re \gamma <0\}\cup \left\{0\right\}$. Then, we have $s\left(\mathcal{A}\right)\in \sigma \left(\mathcal{A}\right)$ and $s\left(\mathcal{A}\right)=0$ according to corollary 12.9 of [21] (p. 188). Thus, using Theorem 12.17 of [21] (p. 193) to obtain the SDGC holds, and ${\omega }_0\left(\mathcal{A}\right)=s\left(\mathcal{A}\right)=0$. This implies that $e^{\mathcal{A}t}$ is not exponentially stable, see, e.g., [22] (Thm VI.1.14, p. 357). That is, there are no constants $K>0$ and $\epsilon >0$ such that

$$\parallel e^{\mathcal{A}t}\parallel \le K{e}^{-\epsilon t},\forall t\ge 0.$$

Furthermore, since $e^{\mathcal{A}t}$ is uniformly bounded (Thm 1), Lemmas 2 and 3 show that $\sigma \left(\mathcal{A}\right)\cap i\mathbb{R}\subset {\sigma }_p\left(\mathcal{A}\right)$, and the asymptotic stability then follows from Theorem 3.26 of [23] (p. 130). This is different from the recent result [11]. □

5. Conclusions

In this paper, we investigate the well-posedness and asymptotic behavior of the M/G/1 stochastic clearing queueing model in a three-phase environment. In the natural state space $L^1$, it is not easy to prove semigroup property. Based on semigroup theory and spectrum analysis on the imaginary axis, we show that this queueing model admits a unique positive TDS. When the service rates of servers are constants, we prove that the TDS of the queueing model is strongly convergent to its nonzero SSS. However, we did not answer whether the general situation holds true, and the solution exponentially converges to its SSS. Moreover, exponential convergence of the TDS of Equations (1) and (2) depends on the growth bound and essential growth bound of the corresponding semigroup, see, e.g., [7,15,16,17,24]. For this aim, we need to know the spectrum of the system operator on the left half of the complex plane. This is work to complete in the future.

Additionally, we show that the SDGC is equal to zero, and the semigroup generated by the system operator is not asymptotically stable. However, we did not study the compactness, reducibility, and ergodicity of the corresponding semigroup of the system. There have been many studies on these asymptotic behavior of semigroups, such as corresponding queueing models, reliable models, and population models, see, e.g., [11,14,16,17,24]. These are also future research topics.