可修排队系统是日常生活中最常见的排队模型之一. 近年来, 对可修排队系统研究被越来越多的学者所关注, 尤其对M/G/1排队系统开展了大量的研究 ^[1-10]. 具有两种故障状态的M/G/1可修排队系统是指服务台可能发生两种故障状态. 一种是由于服务台寿命终止而引发的故障. 另一种是由于服务员对服务台的操作失误等原因而引起的故障. 我们称第一种情况的故障为正常故障, 第二种故障为异常故障. 2002年, 陈洋和朱翼隽等人^[11]通过运用补充变量方法, 首次建立了描述具有两种故障状态的M/G/1可修排队系统的数学模型, 并运用向量Markov过程理论和Laplace变换等方法, 得到了该模型的稳态可用度、故障频度以及更新频度等相关的排队指标和可靠性指标. 2005年, 陈洋和朱翼隽等人^[12]再次运用补充变量方法建立了描述具有三种状态可修排队系统的数学模型, 并给出了该系统稳态解存在的充分必要条件. 2024年, 周学良和张庆红等人 ^[13]运用$ C_0- $半群理论证明了具有两种故障状态的M/G/1可修排队系统存在唯一的非负时间依赖解。除此之外, 至今还没有发现关于该模型的动态分析方面的其它相关文献. 特别地, 当$ \mu(x)=\mu(\text{常数}), \beta_1(y)=\beta_1(\text{常数}), \beta_2(y)=\beta_2(\text{常数}) $时, 具有两种故障状态的M/G/1可修排队系统称为具有两种故障状态的M/M/1可修排队系统.

本文在文献[11] 和 ^[13]研究结果的基础上, 研究具有两种故障状态的M/M/1可修排队系统时间依赖解的渐近行为. 首先通过运用概率母函数, 证明0是一类具有两种故障状态的M/M/1可修排队系统的主算子及其共轭算子几何重数为1的特征值. 其次结合文献[11]文献[13]的研究结果和文献[14] 中的定理14, 并在一定的假设和约束条件下, 估计出该系统的时间依赖解的渐近性质: 即当时刻$ t $趋向于无穷时, 该系统的时间依赖解强收敛于该系统的稳态解.

根据陈洋与朱翼隽等人 ^[11], 当$ \mu(x)=\mu(\text{常数}), \beta_1(y)=\beta_1(\text{常数}), \beta_2(y)=\beta_2(\text{常数}) $时, 具有两种故障状态的M/M/1可修排队系统可由以下偏微分方程组描述. 这里我们不妨设$ \alpha+\gamma+\lambda+\mu=\psi, $

$ \begin{align} &\frac{dp_{1, 0}(t)}{dt}+\lambda p_{1, 0}(t)= \mu\int^{\infty}_{0}p_{1, 1}(x, t)d x, \end{align} $

(2.1)

$ \begin{align} &\frac{\partial p_{1, 1}(x, t)}{\partial t}+\frac{\partial p_{1, 1}(x, t)}{\partial x}=-\psi p_{1, 1}(x, t)+\beta_1\int^{\infty}_{0}p_{2, 1}(x, y, t)dy \\ &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\, \, \, +\beta_2\int^{\infty}_{0}p_{3, 1}(x, y, t)dy, \end{align} $

(2.2)

$ \begin{align} &\frac{\partial p_{1, n}(x, t)}{\partial t}+\frac{\partial p_{1, n}(x, t)}{\partial x}=-\psi p_{1, n}(x, t)+\beta_1\int^{\infty}_{0}p_{2, n}(x, y, t)dy \\ &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\, \, \, +\beta_2\int^{\infty}_{0}p_{3, n}(x, y, t)dy+\lambda p_{1, n-1}(x, t), \quad n\geqslant 2, \end{align} $

(2.3)

$ \begin{align} &\frac{\partial p_{2, 1}(x, y, t)}{\partial t}+\frac{\partial p_{2, 1}(x, y, t)}{\partial y}=-(\lambda+\beta_1)p_{2, 1}(x, y, t), \end{align} $

(2.4)

$ \begin{align} &\frac{\partial p_{2, n}(x, y, t)}{\partial t}+\frac{\partial p_{2, n}(x, y, t)}{\partial y}=-(\lambda+\beta_1)p_{2, n}(x, y, t)+\lambda p_{2, n-1}(x, y, t), \quad n\geqslant 2, \end{align} $

(2.5)

$ \begin{align} &\frac{\partial p_{3, 1}(x, y, t)}{\partial t}+\frac{\partial p_{3, 1}(x, y, t)}{\partial y}=-(\lambda+\beta_2)p_{3, 1}(x, y, t), \end{align} $

(2.6)

$ \begin{align} &\frac{\partial p_{3, n}(x, y, t)}{\partial t}+\frac{\partial p_{3, n}(x, y, t)}{\partial y}=-(\lambda+\beta_2)p_{3, n}(x, y, t)+\lambda p_{3, n-1}(x, y, t), \quad n\geqslant 2, \end{align} $

(2.7)

$ \begin{align} &p_{1, 1}(0, t)=\lambda p_{1, 0}(t)+\mu\int^{\infty}_{0}p_{1, 2}(x, t)dx, \end{align} $

(2.8)

$ \begin{align} &p_{1, n}(0, t)=\mu\int^{\infty}_{0}p_{1, n+1}(x, t)dx, \quad n\geqslant 2, \end{align} $

(2.9)

$ \begin{align} &p_{2, n}(x, 0, t)=\gamma p_{1, n}(x, t), \quad n\geqslant 1, \end{align} $

(2.10)

$ \begin{align} &p_{3, n}(x, 0, t)=\alpha p_{1, n}(x, t), \quad n\geqslant 1, \end{align} $

(2.11)

$ \begin{align} &p_{1, 0}(0)=1, p_{1, j}(x, 0)=0, p_{i, j}(x, y, 0)=0 i=2, 3, j\ge 1. \end{align} $

(2.12)

其中$ (x, t)\in[0, \infty)\times [0, \infty) $, $ (x, y, t)\in[0, \infty)\times [0, \infty)\times [0, \infty) $. $ p_{1, 0}(t) $表示在时刻$ t $该系统中既没有顾客到达也没有服务的概率; $ p_{1, n}(x, t)dx(n\geqslant 1) $表示在时刻$ t $系统中有$ n $个顾客, 服务台处于正常工作状态并且正在接受服务的顾客剩余服务时间在$ (x, x+dx] $内的概率; $ p_{2, n}(x, y, t)dy(n\geqslant 1) $表示在时刻$ t $系统中有$ n $个顾客, 服务台处于异常故障状态并且正在被维修的服务台已经消耗掉的维修时间在$ (y, y+dy] $内的概率; $ p_{3, n}(x, y, t)dy(n\geqslant 1) $表示在时刻$ t $系统中有$ n $个顾客, 服务台处于正常故障状态并且正在被维修的服务台已经消耗掉的维修时间在$ (y, y+dy] $内的概率. 顾客的输入是属于参数为$ \lambda(\lambda>0) $的Poisson过程, 服务台的寿命是服从参数为$ \alpha $的负指数分布. $ \gamma $是表示由于服务员操作不当等原因而引起故障的失效率. $ \mu $是表示服务台的服务率; $ \beta_1 $是表示服务台处于异常故障状态的修复率; $ \beta_2 $表示服务台处于正常故障状态的修复率.

本文仍沿用文献[13] 中的符号. 记

$ X=\left\{(p_1, p_2, p_3)\left| \begin{array}{ll} p_1\in X_1, p_2\in Y_2, p_3\in Y_3, \\ \|(p_1, p_2, p_3)\|=\|p_1\|_{X_1}+\|p_2\|_{Y_2}+\|p_3\|_{Y_3<\infty} \end{array} \right. \right\}, $

$ X_1=\left\{p_1\left| \begin{array}{ll} p_1=(p_{1, 0}, p_{1, 1}, p_{1, 2}, p_{1, 3}, \cdots)\in R\times L^1[0, \infty)\times L^1[0, \infty)\times \cdots, \\ \|p_1\|=\|p_{1, 0}\|+\sum\limits_{k=i}^\infty\|p_{1, i}\|_{ L^1[0, \infty)}<\infty \end{array} \right. \right\}, $

$ Y_2=\left\{p_2\left| \begin{array}{ll} p_2=(p_{2, 1}, p_{2, 2}, p_{2, 3}, p_{2, 4}, \cdots)\in L^1[0, \infty)\times L^1[0, \infty)\times L^1[0, \infty)\times \cdots, \\ \|p_2\|=\sum\limits_{k=i}^\infty\|p_{2, i}\|_{ L^1[0, \infty)}<\infty \end{array} \right. \right\}, $

$ Y_3=\left\{p_3\left| \begin{array}{ll} p_3=(p_{3, 1}, p_{3, 2}, p_{3, 3}, p_{3, 4}, \cdots)\in L^1[0, \infty)\times L^1[0, \infty)\times L^1[0, \infty)\times \cdots, \\ \|p_3\|=\sum\limits_{k=i}^\infty\|p_{3, i}\|_{ L^1[0, \infty)}<\infty \end{array} \right. \right\}, $

容易验证, $ X $构成一个Banach空间.

为简便起见, 引入以下符号

$ \Psi_{1}= \begin{pmatrix} e^{-x} & 0 & 0 & 0 &\cdots\\ \lambda e^{-x} & 0 & \mu & 0 & \cdots\\ 0 & 0 & 0 & \mu & \cdots\\ 0 & 0 & 0 & 0 & \cdots\\ \vdots & \vdots & \vdots & \vdots & \vdots\\ \end{pmatrix}, \Psi_{2}= \begin{pmatrix} 0&\gamma &0&\cdots\\ 0 &0& \gamma & \cdots\\ 0 & 0 & 0& \cdots\\ 0 & 0 & 0& \cdots\\ \vdots & \vdots & \vdots & \vdots\\ \end{pmatrix}, \Psi_{3}= \begin{pmatrix} 0& \alpha &0&\cdots\\ 0 &0& \alpha & \cdots\\ 0 & 0 & 0& \cdots\\ 0 & 0 & 0& \cdots\\ \vdots & \vdots & \vdots & \vdots\\ \end{pmatrix}, $

下面我们来定义算子及其定义域.

$ A(p_{1}, p_{2}, p_{3})= \left( \begin{array}{ccccccc }\left( \begin{array}{ccccccc} -\lambda&0&0&\cdots\\ 0&-\cfrac{d}{dx}&0&\cdots\\ 0&0&-\cfrac{d}{dx}&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right)\left( \begin{array}{ccccc} p_{1, 0}\\ p_{1, 1}(x)\\ p_{1, 2}(x)\\ p_{1, 3}(x)\\ \vdots \end{array} \right) , \end{array} \right. $

$ \left. \begin{array}{cccccc}\left( \begin{array}{cccccc} -\frac{\partial}{\partial y}&0&0&\cdots\\ 0&-\frac{\partial}{\partial y}&0&\cdots\\ 0&0&-\frac{\partial}{\partial y}&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right) \left( \begin{array}{c} p_{2, 1}(x, y)\\ p_{2, 2}(x, y)\\ p_{2, 3}(x, y)\\ \vdots \end{array} \right), \left( \begin{array}{cccccc} -\frac{\partial}{\partial y}&0&0&\cdots\\ 0&-\frac{\partial}{\partial y}&0&\cdots\\ 0&0&-\frac{\partial}{\partial y}&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right) \left( \begin{array}{c} p_{3, 1}(x, y)\\ p_{3, 2}(x, y)\\ p_{3, 3}(x, y)\\ \vdots \end{array} \right)\end{array} \right), $

$ B(p_{1}, p_{2}, p_{3})= \left( \begin{array}{ccccccc }\left( \begin{array}{ccccccc} 0&0&0&\cdots\\ 0&-\psi&0&\cdots\\ 0&\lambda&-\psi&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right)\left( \begin{array}{ccccc} p_{1, 0}\\ p_{1, 1}(x)\\ p_{1, 2}(x)\\ \vdots \end{array} \right) , \end{array} \right. $

$ \begin{array}{ccccccc }\left( \begin{array}{ccccccc} -(\lambda+\beta_1)&0&0&\cdots\\ 0&-(\lambda+\beta_1)&0&\cdots\\ 0&\lambda &-(\lambda+\beta_1)&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right)\left( \begin{array}{ccccc} p_{2, 1}(x, y)\\ p_{2, 2}(x, y)\\ p_{2, 3}(x, y)\\ \vdots \end{array} \right) , \end{array} $

$ \left. \begin{array}{ccccccc }\left( \begin{array}{ccccccc} -(\lambda+\beta_2)&0&0&\cdots\\ \lambda &-(\lambda+\beta_2)&0&\cdots\\ 0&\lambda&-(\lambda+\beta_2)&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right)\left( \begin{array}{ccccc} p_{3, 1}(x, y)\\ p_{3, 2}(x, y)\\ p_{3, 3}(x, y)\\ \vdots \end{array} \right) \end{array} \right), $

$ E(p_{1}, p_{2}, p_{3})= \left( \begin{array}{ccccccc }\left( \begin{array}{ccccccc} \mu\int^{\infty}_{0}p_{1, 1}(x)dx\\ \beta_1\int^{\infty}_{0}p_{2, 1}(x, y)dy+\beta_2\int^{\infty}_{0}p_{3, 1}(x, y)dy\\ \beta_1\int^{\infty}_{0}p_{2, 2}(x, y)dy+\beta_2\int^{\infty}_{0}p_{3, 2}(x, y)dy\\ \beta_1\int^{\infty}_{0}p_{2, 3}(x, y)dy+\beta_2\int^{\infty}_{0}p_{3, 3}(x, y)dy\\ \vdots \end{array} \right), \left( \begin{array}{ccccc} 0\\ 0\\ 0\\ 0\\ \vdots \end{array} \right) , \left( \begin{array}{ccccc} 0\\ 0\\ 0\\ 0\\ \vdots \end{array} \right) \end{array} \right), $

$ D(A)=\left\{(p_1, p_2, p_3)\in X\left| \begin{array}{lllll} \frac{dp_{1, i}}{dx}\in L^1[0, \infty), i\ge 1, \\ \frac{\partial p_{2, i}}{\partial y}, \frac{\partial p_{3, i}}{\partial y}\in L^1([0, \infty)\times[0, \infty)), i\ge 1, \\ p_{1, k}(x), p_{2, i}(x, y), p_{3, i}(x, y) \mbox{是绝对连续并且满足}\\ p_1(0)=\int_{0}^{\infty}\Psi_1p_1(x)dx, p_2(x, 0)=\int_{0}^{\infty}\Psi_2p_1(x)dx, \\ p_3(x, 0)=\int_{0}^{\infty}\Psi_3p_1(x)dx. \end{array} \right. \right\}, $

$ \qquad D(B)=D(E)=X, $则上述偏微分方程组$ (2.1)-(2.12) $可改写为Banach空间$ X $中的抽象Cauchy问题:

$ \begin{align} &\frac{d(p_1, p_2, p_3)(t)}{dt}=(A+B+E)(p_1, p_2, p_3)(t), \, \, \, \forall t\in (0, \infty), \end{align} $

(2.13)

$ \begin{align} &(p_1, p_2, p_3)(0)=(1, 0, 0, \cdot\cdot\cdot). \end{align} $

(2.14)

2024年, 周学良和张庆红等人 ^[13]证明了以下结果:

定理2.1  $ A+B+E $生成一个正压缩$ C_0 $–半群$ T(t) $. 系统$ (2.13)-(2.14) $存在唯一的、正时间依赖解$ (p_1, p_2, p_3)(t) $并且满足

$ \|\big(p_{1, 0}(t), p_{1, n}(\cdot, t), p_{2, n}(\cdot, \cdot, t), p_{3, n}(\cdot, \cdot, t)\big)\|=1, \quad n\geq1, \forall t\in [0, \infty). $

根据文献[14] 的思想和方法, 不难证明$ X $的共轭空间$ X^* $为

$ X^*=\left\{(q_1^*, q_2^*, q_3^*)\left| \begin{array}{ll} q_1^*\in X_1^*, q_2^*\in Y_2^*, q_3^*\in Y^*_3, \\ \||(q_1^*, q_2^*, q_3^*)\||=\sup\{\|q_1^*\|_{X_1^*}, \|q_2^*\|_{Y_2^*}, \|q_3^*\|_{Y_3^*}\} \end{array} \right. \right\}, $

$ X_1^*=\left\{q_1^*\left| \begin{array}{ll} q_1^*=(q_{1, 0}^*, q_{1, 1}^*, q_{1, 2}^*, \cdots)\in l^{\infty}\times L^\infty[0, \infty)\times L^\infty[0, \infty)\times \cdots, \\ \|q_1^*\|=\sup\{\|q_{1, 0}^*\|, \sup\limits_{i\geq1}\|q_{1, i}^*\|_{ L^\infty[0, \infty)}\}<\infty \end{array} \right. \right\}, $

$ Y_2^*=\left\{q_2^*\left| \begin{array}{ll} q_2^*=(q_{2, 1}^*, q_{2, 2}^*, q_{2, 3}^*, \cdots)\in L^\infty[0, \infty)\times L^\infty[0, \infty)\times L^\infty[0, \infty)\times \cdots, \\ \|q_2^*\|=\sup\limits_{i\geq 1}\|q_{2, i}^*\|_{ L^\infty([0, \infty)\times[0, \infty))}<\infty \end{array} \right. \right\}, $

$ Y_3^*=\left\{q_3^*\left| \begin{array}{ll} q_3^*=(q_{3, 1}^*, q_{3, 2}^*, q_{3, 3}^*, \cdots)\in L^\infty[0, \infty)\times L^\infty[0, \infty)\times L^\infty[0, \infty)\times \cdots, \\ \|q_3^*\|=\sup\limits_{i\geq 1}\|q_{3, i}^*\|_{ L^\infty([0, \infty)\times[0, \infty))}<\infty \end{array} \right. \right\}, $

显然, $ X^* $构成一个Banach空间.

根据文献[11] 不难证明以下引理结果.

引理2.1  系统达到稳态平衡的充分必要条件是$ \rho=\frac{\lambda}{\mu}(1+\frac{\gamma}{\beta_1}+\frac{\alpha}{\beta_2})<1 $.

证  我们知道要使系统达到稳态平衡状态, 当且仅当$ \lim\limits_{t\rightarrow \infty}p_0(t)>0 $. 即$ 1-\lim\limits_{t\rightarrow \infty}p_0(t)=1-p_0<1 $, 又由文献[11] 中的引理2可知

$ 1-p_0=\bigg\{ \begin{array}{ll} \rho, \rho<1, \\ 1, \rho\geq 1. \end{array} $

因此, 当系统达到稳态平衡的充分必要条件是$ \rho<1 $.

根据文献[15]和[16]的思想和方法, 由共轭算子的定义不难证明以下引理结果:

引理2.2  $ A+B+E $的共轭算子$ (A+B+E)^* $为

$ \begin{align} (A+B+E)^*(q_1^*, q_2^*, q_3^*)=\left(\cal{F}+\cal{G}+\cal{H}+\cal{I}+\cal{J}\right)(q_1^*, q_2^*, q_3^*), \forall (q_1^*, q_2^*, q_3^*)\in D((A+B+E)^*) \end{align} $

(2.15)

其中

$ \mathcal {F}(q_1^*, q_2^*, q_3^*)= \left( \begin{array}{ccccccc }\left( \begin{array}{ccccccc} -\lambda&0&0&\cdots\\ 0&\mathcal {D}_0&0&\cdots\\ 0&0&\mathcal {D}_0&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right)\left( \begin{array}{ccccc} q^{*}_{1, 0}\\ q^{*}_{1, 1}(x)\\ q^{*}_{1, 2}(x)\\ q^{*}_{1, 3}(x)\\ \vdots \end{array} \right) , \end{array} \right. $

$ \begin{array}{ccccccc }\left( \begin{array}{ccccccc} \mathcal {D}_1&0&0&\cdots\\ 0&\mathcal {D}_1&0&\cdots\\ 0&0&\mathcal {D}_1&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right)\left( \begin{array}{ccccc} q^{*}_{2, 1}(x, y)\\ q^{*}_{2, 2}(x, y)\\ q^{*}_{2, 3}(x, y)\\ q^{*}_{2, 4}(x, y)\\ \vdots \end{array} \right) \end{array}, \left. \begin{array}{ccccccc }\left( \begin{array}{ccccccc} \mathcal {D}_2&0&0&\cdots\\ 0&\mathcal {D}_2&0&\cdots\\ 0&0&\mathcal {D}_2&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right)\left( \begin{array}{ccccc} q^{*}_{3, 1}(x, y)\\ q^{*}_{3, 2}(x, y)\\ q^{*}_{3, 3}(x, y)\\ q^{*}_{3, 4}(x, y)\\ \vdots \end{array} \right) \end{array} \right), $

$ \mathcal {G}(q_1^*, q_2^*, q_3^*)= \left( \begin{array}{ccccccc }\left( \begin{array}{ccccccc} 0&0&0&0&\cdots\\ 0&0&\lambda&0&\cdots\\ 0&0&0&\lambda&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right)\left( \begin{array}{ccccc} q^{*}_{1, 0}\\ q^{*}_{1, 1}(x)\\ q^{*}_{1, 2}(x)\\ q^{*}_{1, 3}(x)\\ \vdots \end{array} \right) , \end{array} \right. $

$ \begin{array}{ccccccc }\left( \begin{array}{ccccccc} 0&\lambda&0&0&\cdots\\ 0&0&\lambda&0&\cdots\\ 0&0&0&\lambda&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right)\left( \begin{array}{ccccc} q^{*}_{2, 1}(x, y)\\ q^{*}_{2, 2}(x, y)\\ q^{*}_{2, 3}(x, y)\\ q^{*}_{2, 4}(x, y)\\ \vdots \end{array} \right) \end{array}, \left. \begin{array}{ccccccc }\left( \begin{array}{ccccccc} 0&\lambda&0&0&\cdots\\ 0&0&\lambda&0&\cdots\\ 0&0&0&\lambda&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right)\left( \begin{array}{ccccc} q^{*}_{3, 1}(x, y)\\ q^{*}_{3, 2}(x, y)\\ q^{*}_{3, 3}(x, y)\\ q^{*}_{3, 4}(x, y)\\ \vdots \end{array} \right) \end{array} \right), $

$ \mathcal {H}(q_1^*, q_2^*, q_3^*)= \left( \begin{array}{ccccccc }\left( \begin{array}{ccccccc} 0&\lambda&0&0&\cdots\\ \mu&0&0&0&\cdots\\ 0&\mu&0&0&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right)\left( \begin{array}{ccccc} q^{*}_{1, 0}\\ q^{*}_{1, 1}(0)\\ q^{*}_{1, 2}(0)\\ q^{*}_{1, 3}(0)\\ \vdots \end{array} \right) , \end{array} \right. $

$ \begin{array}{ccccccc }\left( \begin{array}{ccccccc} 0&\beta_1&0&0&\cdots\\ 0&0&\beta_1&0&\cdots\\ 0&0&0&\beta_1&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right)\left( \begin{array}{ccccc} q^{*}_{1, 0}\\ q^{*}_{1, 1}(x)\\ q^{*}_{1, 2}(x)\\ q^{*}_{1, 3}(x)\\ \vdots \end{array} \right) \end{array}, \left. \begin{array}{ccccccc }\left( \begin{array}{ccccccc} 0&\beta_2&0&0&\cdots\\ 0&0&\beta_2&0&\cdots\\ 0&0&0&\beta_2&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right)\left( \begin{array}{ccccc} q^{*}_{1, 0}\\ q^{*}_{1, 1}(x)\\ q^{*}_{1, 2}(x)\\ q^{*}_{1, 3}(x)\\ \vdots \end{array} \right) \end{array} \right), $

$ \mathcal {I}(q_1^*, q_2^*, q_3^*)= \left( \begin{array}{ccccccc }\left( \begin{array}{ccccccc} 0&0&0&\cdots\\ \gamma&0&0&\cdots\\ 0&\gamma&0&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right)\left( \begin{array}{ccccc} q^{*}_{2, 1}(x, 0)\\ q^{*}_{2, 2}(x, 0)\\ q^{*}_{2, 3}(x, 0)\\ q^{*}_{2, 4}(x, 0)\\ \vdots \end{array} \right) , \end{array} \right. $

$ \begin{array}{ccccccc }\left( \begin{array}{ccccccc} 0&0&0&\cdots\\ 0&0&0&\cdots\\ 0&0&0&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right)\left( \begin{array}{ccccc} q^{*}_{2, 1}(x, 0)\\ q^{*}_{2, 2}(x, 0)\\ q^{*}_{2, 3}(x, 0)\\ q^{*}_{2, 4}(x, 0)\\ \vdots \end{array} \right) \end{array}, \left. \begin{array}{ccccccc }\left( \begin{array}{ccccccc} 0&0&0&\cdots\\ 0&0&0&\cdots\\ 0&0&0&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right)\left( \begin{array}{ccccc} q^{*}_{3, 1}(x, 0)\\ q^{*}_{3, 2}(x, 0)\\ q^{*}_{3, 3}(x, 0)\\ q^{*}_{3, 4}(x, 0)\\ \vdots \end{array} \right) \end{array} \right), $

$ \mathcal {\mathcal {J}}(q_1^*, q_2^*, q_3^*)= \left( \begin{array}{ccccccc }\left( \begin{array}{ccccccc} 0&0&0&\cdots\\ \alpha&0&0&\cdots\\ 0&\alpha&0&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right)\left( \begin{array}{ccccc} q^{*}_{3, 1}(x, 0)\\ q^{*}_{3, 2}(x, 0)\\ q^{*}_{3, 3}(x, 0)\\ q^{*}_{3, 4}(x, 0)\\ \vdots \end{array} \right) , \end{array} \right. $

$ \begin{array}{ccccccc }\left( \begin{array}{ccccccc} 0&0&0&\cdots\\ 0&0&0&\cdots\\ 0&0&0&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right)\left( \begin{array}{ccccc} q^{*}_{2, 1}(x, 0)\\ q^{*}_{2, 2}(x, 0)\\ q^{*}_{2, 3}(x, 0)\\ q^{*}_{2, 4}(x, 0)\\ \vdots \end{array} \right) \end{array}, \left. \begin{array}{ccccccc }\left( \begin{array}{ccccccc} 0&0&0&\cdots\\ 0&0&0&\cdots\\ 0&0&0&\cdots\\ \vdots&\vdots&\vdots&\vdots \end{array} \right)\left( \begin{array}{ccccc} q^{*}_{3, 1}(x, 0)\\ q^{*}_{3, 2}(x, 0)\\ q^{*}_{3, 3}(x, 0)\\ q^{*}_{3, 4}(x, 0)\\ \vdots \end{array} \right) \end{array} \right), $

这里

$ \mathcal{D}_0=\frac{\mathrm{d}}{\mathrm{d}x}-\psi, \mathcal{D}_1=\frac{\partial}{\partial y}-(\lambda+\beta_1), \mathcal{D}_2=\frac{\partial}{\partial y}-(\lambda+\beta_2), $

$ D((A+B+E)^*)=\left\{(q_1^*, q_2^*, q_3^*)\in X^* \left| \begin{array}{ll} \frac{\mathrm{d}q_{1, n}(x)}{\mathrm{d}x}, \frac{\partial_{2, n}(x, y)}{\partial y} \mbox{和} \frac{\partial_{3, n}(x, y)}{\partial y}\mbox{都存在, 并且}\\ q_{1, n}^*(\infty)=q_{2, n}^*(x, \infty)=q_{3, n}^*(x, \infty)=\mathcal {T} , n\geq 1 \end{array}\right. \right\}, $

这里$ D((A+B+E)^*) $中的$ \mathcal {T} $是与$ n $无关的常数.

引理3.1  若$ \mu >\lambda, (\mu-\lambda)\beta_1\beta_2-\lambda\gamma\beta_2-\lambda\alpha\beta_1>0, $那么0是系统主算子$ A+B+E $的几何重数为1的特征值.

证  考虑方程$ (A+B+E)(p_1, p_2, p_3)=0 $, 这等价于下列微分方程组

$ \begin{align} &\lambda p_{1, 0}= \mu\int^{\infty}_{0}p_{1, 1}(x)dx, \end{align} $

(3.1)

$ \begin{align} &\frac{dp_{1, 1}(x)}{dx}=-\psi p_{1, 1}(x)+\beta_1\int^{\infty}_{0}p_{2, 1}(x, y)dy+\beta_2\int^{\infty}_{0}p_{3, 1}(x, y)dy, \end{align} $

(3.2)

$ \begin{align} &\frac{d p_{1, n}(x)}{dx}=-\psi p_{1, n}(x)+\beta_1\int^{\infty}_{0}p_{2, n}(x, y)dy+\beta_2\int^{\infty}_{0}p_{3, n}(x, y)dy +\lambda p_{1, n-1}(x), \quad n\geqslant 2, \end{align} $

(3.3)

$ \begin{align} &\frac{\partial p_{2, 1}(x, y)}{\partial y}=-(\lambda+\beta_1)p_{2, 1}(x, y), \end{align} $

(3.4)

$ \begin{align} &\frac{\partial p_{2, n}(x, y)}{\partial y}=-(\lambda+\beta_1)p_{2, n}(x, y)+\lambda p_{2, n-1}(x, y), \quad n\geqslant 2, \end{align} $

(3.5)

$ \begin{align} &\frac{\partial p_{3, 1}(x, y)}{\partial y}=-(\lambda+\beta_2)p_{3, 1}(x, y), \end{align} $

(3.6)

$ \begin{align} &\frac{\partial p_{3, n}(x, y)}{\partial y}=-(\lambda+\beta_2)p_{3, n}(x, y)+\lambda p_{3, n-1}(x, y), \quad n\geqslant 2, \end{align} $

(3.7)

$ \begin{align} &p_{1, 1}(0)=\lambda p_{1, 0}+\mu\int^{\infty}_{0}p_{1, 2}(x)dx, \end{align} $

(3.8)

$ \begin{align} &p_{1, n}(0)=\mu\int^{\infty}_{0}p_{1, n+1}(x)dx, \quad n\geqslant 2, \end{align} $

(3.9)

$ \begin{align} &p_{2, n}(x, 0)=\gamma p_{1, n}(x), \quad n\geqslant 1, \end{align} $

(3.10)

$ \begin{align} &p_{3, n}(x, 0)=\alpha p_{1, n}(x), \quad n\geqslant 1, \end{align} $

(3.11)

解(3.4)—(3.7) 式得到

$ \begin{align} p_{2, 1}(x, y)&=a_{2, 1}(x)e^{-(\lambda+\beta_1)y}, \end{align} $

(3.12)

$ \begin{align} p_{2, 2}(x, y)&=a_{2, 2}(x)e^{-(\lambda+\beta_1)y}+\lambda e^{-(\lambda+\beta_1)y}\int^{y}_{0}e^{(\lambda+\beta_1) \tau} p_{2, 1}(x, \tau)d\tau \\ &=a_{2, 2}(x)e^{-(\lambda+\beta_1)y}+\lambda e^{-(\lambda+\beta_1)y}\int^{y}_{0}e^{(\lambda+\beta_1) \tau} a_{2, 1}(x)e^{-(\lambda+\beta_1)\tau}d\tau \\ &=\bigg[a_{2, 2}(x)+a_{2, 1}(x)\lambda y\bigg] e^{-(\lambda+\beta_1)y}, \end{align} $

(3.13)

$ \begin{align} p_{2, n}(x, y)&=a_{2, n}(x)e^{-(\lambda+\beta_1)y}+\lambda e^{-(\lambda+\beta_1)y}\int^{y}_{0}e^{(\lambda+\beta_1) \tau} p_{2, n-1}(x, \tau)d\tau \\ &=\bigg[a_{2, n}(x)+a_{2, n-1}(x)\lambda y+a_{2, n-2}(x)\frac{(\lambda y)^2}{2!}+\cdots+a_{2, 2}(x)\frac{(\lambda y)^{n-2}}{(n-2)!} \\ &\quad +a_{2, 1}(x)\frac{(\lambda y)^{n-1}}{(n-1)!}\bigg] e^{-(\lambda+\beta_1)y}=\sum\limits_{i=1}^n a_{2, i}(x) \frac{(\lambda y)^{n-i}}{(n-i)!} e^{-(\lambda+\beta_1)y}, n\geq1, \end{align} $

(3.14)

$ \begin{align} p_{3, 1}(x, y)&=a_{3, 1}(x)e^{-(\lambda+\beta_2)y}, \end{align} $

(3.15)

$ \begin{align} p_{3, 2}(x, y)&=a_{3, 2}(x)e^{-(\lambda+\beta_2)y}+\lambda e^{-(\lambda+\beta_2)y}\int^{y}_{0}e^{(\lambda+\beta_2) \tau} p_{3, 1}(x, \tau)d\tau \\ &=a_{3, 2}(x)e^{-(\lambda+\beta_2)y}+\lambda e^{-(\lambda+\beta_2)y}\int^{y}_{0}e^{(\lambda+\beta_2) \tau} a_{3, 1}(x)e^{-(\lambda+\beta_2)\tau}d\tau \\ &=\bigg[a_{3, 2}(x)+a_{3, 1(x)}\lambda y\bigg] e^{-(\lambda+\beta_2)y}, \end{align} $

(3.16)

$ \begin{align} p_{3, n}(x, y)&=a_{3, n}(x)e^{-(\lambda+\beta_2)y}+\lambda e^{-(\lambda+\beta_2)y}\int^{y}_{0}e^{(\lambda+\beta_2) \tau} p_{3, n-1}(x, \tau)d\tau \\ &=\bigg[a_{3, n}(x)+a_{3, n-1}(x)\lambda y+a_{3, n-2}(x)\frac{(\lambda y)^2}{2!}+\cdots+a_{3, 2}(x)\frac{(\lambda y)^{n-2}}{(n-2)!} \\ &\quad +a_{3, 1}(x)\frac{(\lambda y)^{n-1}}{(n-1)!}\bigg] e^{-(\lambda+\beta_1)y}=\sum\limits_{i=1}^n a_{3, i}(x) \frac{(\lambda y)^{n-i}}{(n-i)!} e^{-(\lambda+\beta_2)y}, n\geq1. \end{align} $

(3.17)

解$ (3.2) $和$ (3.3) $式, 并将$ (3.12)-(3.17) $式代入计算得(令$ e^{-\psi x} \int^{x}_{0}e^{\psi \tau}f(\tau)d\tau=Ef(x) $)

$ \begin{align} p_{1, 1}(x)=&a_{1, 1}e^{-\psi x}+e^{-\psi x}\int^{x}_{0}e^{\psi \tau}\bigg[\beta_1\int^{\infty}_{0}p_{2, 1}(\tau, y)dy+\beta_2\int^{\infty}_{0}p_{3, 1}(\tau, y)dy\bigg]d\tau \\ =&a_{1, 1}e^{-\psi x}+E\bigg[\beta_1\int^{\infty}_{0}p_{2, 1}(\tau, y)dy+\beta_2\int^{\infty}_{0}p_{3, 1}(\tau, y) y\bigg], \end{align} $

(3.18)

$ \begin{align} p_{1, 2}(x)=&a_{1, 2}e^{-\psi x}+\lambda e^{-\psi x}\int^{x}_{0}e^{\psi \tau} p_{1, 1}(\tau)d\tau+e^{-\psi x}\int^{x}_{0}e^{\psi \tau}\bigg[\beta_1\int^{\infty}_{0}p_{2, 2}(\tau, y)dy \\ &+\beta_2\int^{\infty}_{0}p_{3, 2}(\tau, y)dy\bigg]d\tau \\ =&a_{1, 2}e^{-\psi x}+a_{1, 1}\lambda x e^{-\psi x}+\lambda E^2\bigg[\beta_1\int^{\infty}_{0}p_{2, 1}(\tau, y)dy+\beta_2\int^{\infty}_{0}p_{3, 1}(\tau, y)dy\bigg] \\ &+E\bigg[\beta_1\int^{\infty}_{0}p_{2, 2}(\tau, y)dy+\beta_2\int^{\infty}_{0}p_{3, 2}(\tau, y)dy\bigg], \end{align} $

(3.19)

$ \begin{align} p_{1, n}(x)=&a_{1, n}e^{-\psi x}+\lambda e^{-\psi x}\int^{x}_{0}e^{\psi \tau} p_{1, n-1}(\tau)d\tau+e^{-\psi x}\int^{x}_{0}e^{\psi \tau}\bigg[\beta_1\int^{\infty}_{0}p_{2, n}(\tau, y)dy \\ &+\beta_2\int^{\infty}_{0}p_{3, n}(\tau, y)dy\bigg]d\tau \\ =&e^{-\psi x}\sum\limits_{i=1}^n a_{1, i} \frac{(\lambda x)^{n-i}}{(n-i)!}+\sum\limits_{i=1}^n \lambda^{i-1}E^i\bigg[\beta_1\int^{\infty}_{0}p_{2, n-i+1}(\tau, y)dy \\ &+\beta_2\int^{\infty}_{0}p_{3, n-i+1}(\tau, y)dy\bigg], n\geqslant 2, \end{align} $

(3.20)

结合$ (3.12)-(3.20) $与$ (3.8)-(3.11) $式得到

$ \begin{align} a_{1, 1}=&p_{1, 1}(0)=\lambda p_{1, 0}+\mu\int^{\infty}_{0}p_{1, 2}(x)dx \\ =&\lambda p_{1, 0}+\mu\int^{\infty}_{0}\bigg\{a_{1, 2}e^{-\psi x}+a_{1, 1}\lambda x e^{-\psi x}+\lambda E^2\bigg[\beta_1\int^{\infty}_{0}p_{2, 1}(\tau, y)dy \\ &+\beta_2\int^{\infty}_{0}p_{3, 1}(\tau, y)dy\bigg]+E\bigg[\beta_1\int^{\infty}_{0}p_{2, 2}(\tau, y)dy+\beta_2\int^{\infty}_{0}p_{3, 2}(\tau, y)dy\bigg]\bigg\}dx, \\ a_{1, n}=&p_{1, n}(0)=\mu\int^{\infty}_{0}p_{1, n+1}(x)dx \end{align} $

(3.21)

$ \begin{align} =&\mu\int^{\infty}_{0}\bigg\{e^{-\psi x}\sum\limits_{i=1}^{n+1} a_{1, i} \frac{(\lambda x)^{n-i+1}}{(n-i+1)!}+\sum\limits_{i=1}^{n+1} \lambda^{i-1}E^i\bigg[\beta_1\int^{\infty}_{0}p_{2, n-i+2}(\tau, y)dy \\ &+\beta_2\int^{\infty}_{0}p_{3, n-i+1}(\tau, y)dy\bigg]\bigg\}dx, \quad n\geqslant 2, \end{align} $

(3.22)

$ \begin{align} a_{2, n}(x)=&p_{2, n}(x, 0)=\gamma p_{1, n}(x) \\ =&\gamma\bigg\{e^{-\psi x}\sum\limits_{i=1}^n a_{1, i} \frac{(\lambda x)^{n-i}}{(n-i)!}+\beta_1e^{-\psi x}\int^{x}_{0}e^{\psi \tau}\int^{\infty}_{0}\sum\limits_{i=1}^n a_{2, i}(\tau) \frac{(\lambda y)^{n-i}}{(n-i)!} e^{-(\lambda+\beta_1)y}dyd\tau \\ &\quad+\beta_2e^{-\psi x}\int^{x}_{0}e^{\psi \tau}\int^{\infty}_{0}\bigg[\sum\limits_{i=1}^n a_{3, i}(\tau) \frac{(\lambda y)^{n-i}}{(n-i)!} e^{-(\lambda+\beta_2)y}\bigg]dy d\tau\bigg\} n\geqslant 1, \end{align} $

(3.23)

$ \begin{align} a_{3, n}(x)=&p_{3, n}(x, 0)=\alpha p_{1, n}(x) \\ =&\alpha\bigg\{e^{-\psi x}\sum\limits_{i=1}^n a_{1, i} \frac{(\lambda x)^{n-i}}{(n-i)!}+\beta_1e^{-\psi x}\int^{x}_{0}e^{\psi \tau}\int^{\infty}_{0}\sum\limits_{i=1}^n a_{2, i}(\tau) \frac{(\lambda y)^{n-i}}{(n-i)!} e^{-(\lambda+\beta_1)y}dyd\tau \\ &\quad+\beta_2e^{-\psi x}\int^{x}_{0}e^{\psi \tau}\int^{\infty}_{0}\bigg[\sum\limits_{i=1}^n a_{3, i}(\tau) \frac{(\lambda y)^{n-i}}{(n-i)!} e^{-(\lambda+\beta_2)y}\bigg]dyd\tau\bigg\} n\geqslant 1, \end{align} $

(3.24)

由上述微分方程组容易看出直接求出每个$ p_{1, n}(x), p_{2, n}(x, y), p_{3, n}(x, y) $表达式, 并且证明$ p_{1, n}(x), p_{2, n}(x, y), p_{3, n}(x, y)\in D(A+B+E) $非常困难. 因此我们可运用文献[15]和[16]中的思想和方法, 首先引入下列概率母函数,

$ P_1(x, z)=\sum\limits_{n=1}^\infty p_{1, n}(x)z^n, P_2(x, y, z)=\sum\limits_{n=1}^\infty p_{2, n}(x, y)z^n, P_3(x, y, z)=\sum\limits_{n=1}^\infty p_{3, n}(x, y)z^n, |z|<1. $

定理2.1保证$ P_1(x, z) $, $ P_2(x, y, z) $和$ P_3(x, y, z) $都具有意义. 由幂级数的基本知识, Fubini定理以及(3.2) 和(3.3) 式计算得

$ \begin{align} \frac{\partial P_1(x, z)}{\partial x}&=-\psi\sum\limits_{n=1}^\infty p_{1, n}(x)z^n+\beta_1\sum\limits_{n=1}^\infty\int^{\infty}_{0}p_{2, n}(x, y)z^ndy+\beta_2\sum\limits_{n=1}^\infty\int^{\infty}_{0}p_{3, n}(x, y)z^ndy \\ &\quad+\lambda\sum\limits_{n=2}^\infty p_{1, n-1}(x)z^n \\ &=-\psi P_1(x, z)+\beta_1\int^{\infty}_{0}P_2(x, y, z)dy+\beta_2\int^{\infty}_{0}P_3(x, y, z)dy+\lambda zP_1(x, z) \\ &=(\lambda z-\psi)P_1(x, z)+\beta_1\int^{\infty}_{0}P_2(x, y, z)dy+\beta_2\int^{\infty}_{0}P_3(x, y, z)dy, \end{align} $

(3.25)

由(3.4) 和(3.5) 式得到

$ \begin{align} \frac{\partial P_2(x, y, z)}{\partial y}&=-(\lambda+\beta_1)\sum\limits_{n=1}^\infty p_{2, n}(x, y)z^n+\lambda\sum\limits_{n=2}^\infty p_{2, n-1}(x, y)z^n \\ &=-(\lambda+\beta_1)P_2(x, y, z)+\lambda zP_2(x, y, z)=(\lambda z-\lambda-\beta_1)P_2(x, y, z), \\ &\Longrightarrow \\ P_2(x, y, z)&=P_2(x, 0, z)e^{(\lambda z-\lambda-\beta_1)y}, \end{align} $

(3.26)

由(3.6) 和(3.7) 式得到

$ \begin{align} \frac{\partial P_3(x, y, z)}{\partial y}&=-(\lambda+\beta_2)\sum\limits_{n=1}^\infty p_{3, n}(x, y)z^n+\lambda\sum\limits_{n=2}^\infty p_{3, n-1}(x, y)z^n \\ &=-(\lambda+\beta_1)P_3(x, y, z)+\lambda zP_3(x, y, z)=(\lambda z-\lambda-\beta_2)P_3(x, y, z), \\ &\Longrightarrow \\ P_3(x, y, z)&=P_3(x, 0, z)e^{(\lambda z-\lambda-\beta_2)y}, \end{align} $

(3.27)

由$ (3.8)-(3.11) $和(3.1) 式, 计算得到

$ \begin{align} P_1(0, z)=&\sum\limits_{n=1}^\infty P_1(0)z^n=\lambda z p_{1, 0}+ \mu\int^{\infty}_{0}\sum\limits_{n=1}^\infty p_{1, n+1}(x)z^ndx \\ =&\lambda z p_{1, 0}+\frac{\mu}{z}\int^{\infty}_{0}\bigg(\sum\limits_{n=1}^\infty p_{1, n}(x)z^n-p_{1, 1}(x)z\bigg)dx \\ =&\lambda z p_{1, 0}+\frac{\mu}{z}\int^{\infty}_{0}P_1(x, z)dx-\mu\int^{\infty}_{0}p_{1, 1}(x)dx \\ =&\frac{\mu}{z}\int^{\infty}_{0}P_1(x, z)d x+\lambda (z-1) p_{1, 0}, \end{align} $

(3.28)

$ \begin{align} P_2(x, 0, z)=&\sum\limits_{n=1}^\infty P_2(x, 0)z^n=\gamma\sum\limits_{n=1}^\infty p_{1, n}(x)z^n=\gamma P_1(x, z), \end{align} $

(3.29)

$ \begin{align} P_3(x, 0, z)=&\sum\limits_{n=1}^\infty P_3(x, 0)z^n=\alpha\sum\limits_{n=1}^\infty p_{1, n}(x)z^n=\alpha P_1(x, z), \end{align} $

(3.30)

将(3.29) 和(3.30) 分别代入(3.26) 和(3.27) 得到

$ \begin{align} P_2(x, y, z)&=P_2(x, 0, z)e^{(\lambda z-\lambda-\beta_1)y}=\gamma P_1(x, z)e^{(\lambda z-\lambda-\beta_1)y}, \end{align} $

(3.31)

$ \begin{align} P_3(x, y, z)&=P_3(x, 0, z)e^{(\lambda z-\lambda-\beta_2)y}=\alpha P_1(x, z)e^{(\lambda z-\lambda-\beta_2)y}, \end{align} $

(3.32)

将(3.31) 和(3.32) 代入(3.25) 式, 计算得到

$ \begin{align} \frac{\partial P_1(x, z)}{\partial x}&=(\lambda z-\psi)P_1(x, z)+\beta_1\int^{\infty}_{0}P_2(x, y, z)dy+\beta_2\int^{\infty}_{0}P_3(x, y, z)dy \\ &=(\lambda z-\psi)P_1(x, z)+\gamma\beta_1\int^{\infty}_{0}P_1(x, z)e^{(\lambda z-\lambda-\beta_1)y}dy \\ &\quad+\alpha\beta_2\int^{\infty}_{0}P_1(x, z)e^{(\lambda z-\lambda-\beta_2)y}dy \\ &=(\lambda z-\psi)P_1(x, z)-\frac{\beta_1 \gamma}{\lambda z-\lambda-\beta_1} P_1(x, z)-\frac{\beta_2 \alpha}{\lambda z-\lambda-\beta_2} P_1(x, z) \\ &=\bigg[(\lambda z-\psi)-\frac{\beta_1 \gamma}{\lambda z-\lambda-\beta_1}-\frac{\beta_2 \alpha}{\lambda z-\lambda-\beta_2}\bigg]P_1(x, z), \\ &\Longrightarrow \\ P_1(x, z)&=P_1(0, z)e^{[(\lambda z-\psi)-\frac{\beta_1 \gamma}{\lambda z-\lambda-\beta_1}-\frac{\beta_2 \alpha}{\lambda z-\lambda-\beta_2}]x}, \end{align} $

(3.33)

不妨设函数$ \omega(z)=(\lambda z-\psi)-\frac{\beta_1 \gamma}{\lambda z-\lambda-\beta_1}-\frac{\beta_2 \alpha}{\lambda z-\lambda-\beta_2} $, 将(3.33) 式代入(3.28) 式, 计算得到

$ \begin{align} P_1(0, z)=&\frac{\mu}{z}\int^{\infty}_{0}P_1(x, z)dx+\lambda (z-1) p_{1, 0} \\ =&\frac{\mu}{z}\int^{\infty}_{0}P_1(0, z)e^{[(\lambda z-\psi)-\frac{\beta_1 \gamma}{\lambda z-\lambda-\beta_1}-\frac{\beta_2 \alpha}{\lambda z-\lambda-\beta_2}]x}dx+\lambda (z-1) p_{1, 0} \\ =&\frac{\mu}{z}\int^{\infty}_{0}P_1(0, z)e^{\omega(z)x}dx+\lambda (z-1) p_{1, 0} \\ =&\frac{-\mu}{z\omega(z)}P_1(0, z)+\lambda (z-1) p_{1, 0}, \\ \Longrightarrow& \\ \bigg[1+\frac{\mu}{z\omega(z)}\bigg]P_1(0, z)=&\lambda (z-1) p_{1, 0}, \\ \Longrightarrow& \\ P_1(0, z)=&\frac{\lambda(z-1)}{1+\frac{\mu}{z\omega(z)}}p_{1, 0}, \end{align} $

(3.34)

$ \begin{align} \frac{d\omega(z)}{dz}=&\omega^{\prime}(z)=\lambda+\frac{\lambda\gamma\beta_1}{(\lambda z-\lambda-\beta_1)^2}+\frac{\lambda\alpha\beta_2}{(\lambda z-\lambda-\beta_2)^2}, \end{align} $

(3.35)

$ \begin{align} \omega^{\prime}(1)=&\lambda+\frac{\lambda\gamma\beta_1}{(-\beta_1)^2}+\frac{\lambda\alpha\beta_2}{(-\beta_2)^2} =\frac{\lambda(\beta_1\beta_2+\gamma\beta_2+\alpha\beta_1)}{\beta_1\beta_2}, \end{align} $

(3.36)

$ \begin{align} \omega(1)=&(\lambda-\psi)-\frac{\beta_1 \gamma}{-\beta_1}-\frac{\beta_2 \alpha}{-\beta_2}=(\lambda-\psi)+\gamma+\alpha=-\mu, \end{align} $

(3.37)

由(3.34) 式和罗必达法则计算得

$ \begin{align} \lim\limits_{z\rightarrow 1}P_1(0, z)=&\lim\limits_{z\rightarrow 1}\frac{\lambda(z-1)}{1+\frac{\mu}{z\omega(z)}}p_{1, 0} =\lim\limits_{z\rightarrow 1}\frac{\lambda}{\frac{-\mu[\omega(z)+z\omega^{\prime}(z)]}{(z\omega(z))^2}}p_{1, 0} =\lim\limits_{z\rightarrow 1}\frac{-\lambda (z\omega(z))^2}{\mu[\omega(z)+z\omega^{\prime}(z)]}p_{1, 0}\\ =&\frac{\lambda\mu\beta_1\beta_2}{(\mu-\lambda)\beta_1\beta_2-\lambda\gamma\beta_2-\lambda\alpha\beta_1}p_{1, 0}<\infty, \end{align} $

(3.38)

联立(3.33) 和(3.38) 式得到

$ \begin{align} \sum\limits_{n=1}^\infty p_{1, n}(x)=&\lim\limits_{z\rightarrow 1}P_1(x, z)=\lim\limits_{z\rightarrow 1}P_1(0, z)e^{[(\lambda z-\psi)-\frac{\beta_1 \gamma}{\lambda z-\lambda-\beta_1}-\frac{\beta_2 \alpha}{\lambda z-\lambda-\beta_2}]x}, \\ =&\frac{(\lambda\mu\beta_1\beta_2)p_{1, 0}}{(\mu-\lambda)\beta_1\beta_2-\lambda\gamma\beta_2-\lambda\alpha\beta_1}e^{-\mu x}, \end{align} $

(3.39)

$ \begin{align} \Longrightarrow &\\ \sum\limits_{n=1}^\infty\int^{\infty}_{0}p_{1, n}(x)dx=&\frac{1}{\mu}\times \frac{(\lambda\mu\beta_1\beta_2)p_{1, 0}}{(\mu-\lambda)\beta_1\beta_2-\lambda\gamma\beta_2-\lambda\alpha\beta_1}\\ =&\frac{(\lambda\beta_1\beta_2)p_{1, 0}}{(\mu-\lambda)\beta_1\beta_2-\lambda\gamma\beta_2-\lambda\alpha\beta_1}<\infty, \end{align} $

(3.40)

联合(3.26), (3.27) 和(3.39) 式计算得

$ \begin{align} \sum\limits_{n=1}^\infty p_{2, n}(x, y)=&\lim\limits_{z\rightarrow 1}P_2(x, y, z)=\lim\limits_{z\rightarrow 1}P_2(x, 0, z)e^{(\lambda z-\lambda-\beta_1)y}=\lim\limits_{z\rightarrow 1}\gamma P_1(x, z)e^{(\lambda z-\lambda-\beta_1)y}\\ =&\frac{(\gamma\lambda\mu\beta_1\beta_2)p_{1, 0}}{(\mu-\lambda)\beta_1\beta_2-\lambda\gamma\beta_2-\lambda\alpha\beta_1}e^{-\beta_1 y}, \end{align} $

(3.41)

$ \begin{align} \Longrightarrow & \\ \sum\limits_{n=1}^\infty\int^{\infty}_{0}p_{2, n}(x, y)dy=&\frac{1}{\beta_1}\times\frac{(\gamma\lambda\mu\beta_1\beta_2)p_{1, 0}}{(\mu-\lambda)\beta_1\beta_2-\lambda\gamma\beta_2-\lambda\alpha\beta_1}\\ =&\frac{(\gamma\lambda\mu\beta_2)p_{1, 0}}{(\mu-\lambda)\beta_1\beta_2-\lambda\gamma\beta_2-\lambda\alpha\beta_1}<\infty, \end{align} $

(3.42)

$ \begin{align} \sum\limits_{n=1}^\infty p_{3, n}(x, y)=&\lim\limits_{z\rightarrow 1}P_3(x, y, z)=\lim\limits_{z\rightarrow 1}P_3(x, 0, z)e^{(\lambda z-\lambda-\beta_2)y}=\lim\limits_{z\rightarrow 1}\alpha P_1(x, z)e^{(\lambda z-\lambda-\beta_2)y}\\ =&\frac{(\alpha\lambda\mu\beta_1\beta_2)p_{1, 0}}{(\mu-\lambda)\beta_1\beta_2-\lambda\gamma\beta_2-\lambda\alpha\beta_1}e^{-\beta_2 y}, \end{align} $

(3.43)

$ \begin{align} \Longrightarrow & \\ \sum\limits_{n=1}^\infty\int^{\infty}_{0}p_{3, n}(x, y)dy=&\frac{1}{\beta_2}\times\frac{(\alpha\lambda\mu\beta_1\beta_2)p_{1, 0}}{(\mu-\lambda)\beta_1\beta_2-\lambda\gamma\beta_2-\lambda\alpha\beta_1}\\ =&\frac{(\alpha\lambda\mu\beta_1)p_{1, 0}}{(\mu-\lambda)\beta_1\beta_2-\lambda\gamma\beta_2-\lambda\alpha\beta_1}<\infty, \end{align} $

(3.44)

由(3.40), (3.42) 和(3.44) 式估计出

$ \begin{align} \|(p_1, p_2, p_3)\|=\|p_1\|+\|p_2\|+\|p_3\|=\frac{\lambda(\beta_1\beta_2+\gamma\mu\beta_2+\alpha\mu\beta_1)p_{1, 0}}{(\mu-\lambda)\beta_1\beta_2-\lambda\gamma\beta_2-\lambda\alpha\beta_1}<\infty. \end{align} $

(3.45)

(3.45) 式说明$ \, (p_1, p_2, p_3)\in X $, 即0是$ A+B+E $的特征值. 此外, $ (3.1), (3.12)-(3.24) $容易看出对应于主算子特征值0的特征向量子空间的维数是1, 即特征值0的几何重数为1.

注  在一个排队系统中, 因为$ \mu $表示服务率, $ \lambda $表示顾客到达率, 所以可推出$ \mu>\lambda $. 否则, 随着时间的延长, 排队系统的队长将达到无限长.

根据引理2.1 $ \rho=\frac{\lambda}{\mu}(1+\frac{\gamma}{\beta_1}+\frac{\alpha}{\beta_2})<1 $可推得

$ \begin{align} &\lambda(1+\frac{\gamma}{\beta_1}+\frac{\alpha}{\beta_2})<\mu \\ &\Longrightarrow \lambda\beta_1\beta_2+\lambda\gamma\beta_2+\lambda\alpha\beta_1<\mu\beta_1\beta_2\\ &\Longrightarrow \mu\beta_1\beta_2-\lambda\beta_1\beta_2-\lambda\gamma\beta_2-\lambda\alpha\beta_1>0\\ &\Longrightarrow (\mu-\lambda)\beta_1\beta_2-\lambda\gamma\beta_2-\lambda\alpha\beta_1>0 \end{align} $

上式说明引理的条件具有实际的物理背景, 符合实际应用的合理性. 引理3.1证毕.

引理3.2  如果$ \mu >\lambda, (\mu-\lambda)\beta_1\beta_2-\lambda\gamma\beta_2-\lambda\alpha\beta_1>0, $那么0是共轭算子$ (A+B+E)^* $的几何重数为1的特征值.

证  讨论方程$ (A+B+E)^*(q_1^*, q_2^*, q_3^*)=0, $即

$ \begin{align} &-\lambda q^*_{1, 0}+\lambda q_{1, 1}^*(0)=0, \end{align} $

(3.46)

$ \begin{align} &\frac{d q_{1, 1}^*(x)}{d x}=\psi q_{1, 1}^*(x)-\mu q^*_{1, 0}-\gamma q^*_{2, 1}(x, 0)-\alpha q^*_{3, 1}(x, 0)-\lambda q^*_{1, 2}(x), \end{align} $

(3.47)

$ \begin{align} &\frac{d q_{1, n}^*(x)}{d x}=\psi q_{1, n}^*(x)-\mu q^*_{1, n-1}(0)-\gamma q^*_{2, n}(x, 0)-\alpha q^*_{3, n}(x, 0)-\lambda q^*_{1, n+1}(x), n\geqslant 2, \end{align} $

(3.48)

$ \begin{align} &\frac{d q_{2, n}^*(x, y)}{d y}=(\overset{\thicksim }{\gamma}+\lambda+\beta_1)q_{2, n}^*(x, y)-\beta_1 q^*_{1, n}(x)-\lambda q^*_{2, n+1}(x, y), \, \, n\geqslant 1, \end{align} $

(3.49)

$ \begin{align} &\frac{d q_{3, n}^*(x, y)}{d y}=(\overset{\thicksim }{\gamma}+\lambda+\beta_2)q_{3, n}^*(x, y)-\beta_2 q^*_{1, n}(x)-\lambda q^*_{3, n+1}(x, y), \, \, n\geqslant 1, \end{align} $

(3.50)

$ \begin{align} &q_{1, n}^*(\infty)=q_{2, n}^*(x, \infty)=q_{3, n}^*(x, \infty)=\mathcal {T}, \, \, n\geqslant 1. \end{align} $

(3.51)

容易验证

$ (q_1^*, q_2^*, q_3^*)= \left( \begin{array}{ccccccc }\left( \begin{array}{ccccccc} \mathcal {\mathcal {T}}\\ \mathcal {\mathcal {T}}\\ \mathcal {\mathcal {T}}\\ \vdots\\ \end{array} \right), \left( \begin{array}{ccccc} \mathcal {\mathcal {T}}\\ \mathcal {\mathcal {T}}\\ \mathcal {\mathcal {T}}\\ \vdots\\ \end{array} \right) , \left( \begin{array}{ccccc} \mathcal {\mathcal {T}}\\ \mathcal {\mathcal {T}}\\ \mathcal {\mathcal {T}}\\ \vdots\\ \end{array} \right) \end{array} \right)\in D((A+B+E)^*) $

是方程组$ (3.46)-(3.51) $的一个非零解(注意到$ \mathcal {\mathcal {T}}\neq0 $), 即0是该系统共轭算子$ (A+B+E)^* $的特征值. 此外, $ (3.46)-(3.50) $等价于

$ \begin{align} q^*_{1, 0}=&q_{1, 1}^*(0), \end{align} $

(3.52)

$ \begin{align} q^*_{1, 2}(x)=&\frac{1}{\lambda}\bigg[-\frac{d q_{1, 1}^*(x)}{d x}+\psi q_{1, 1}^*(x)-\mu q^*_{1, 0}-\gamma q^*_{2, 1}(x, 0)-\alpha q^*_{3, 1}(x, 0)\bigg] , \end{align} $

(3.53)

$ \begin{align} q^*_{1, n+1}(x)=&\frac{1}{\lambda}\bigg[-\frac{d q_{1, n}^*(x)}{d x}+\psi q_{1, n}^*(x)-\mu q^*_{1, n-1}(0)-\gamma q^*_{2, n}(x, 0)\\ &-\alpha q^*_{3, n}(x, 0)\bigg], n\geqslant 2, \end{align} $

(3.54)

$ \begin{align} q^*_{2, n+1}(x, y)=&\frac{1}{\lambda}\bigg[-\frac{d q_{2, n}^*(x, y)}{d y}+(\overset{\thicksim }{\gamma}+\lambda+\beta_1)q_{2, n}^*(x, y)-\beta_1 q^*_{1, n}(x)\bigg], \, \, n\geqslant 1, \end{align} $

(3.55)

$ \begin{align} q^*_{3, n+1}(x, y)=&\frac{1}{\lambda}\bigg[-\frac{d q_{3, n}^*(x, y)}{d y}+(\overset{\thicksim }{\gamma}+\lambda+\beta_2)q_{3, n}^*(x, y)-\beta_2 q^*_{1, n}(x)\bigg], \, \, n\geqslant 1, \end{align} $

(3.56)

由$ (3.52)-(3.56) $可知对应于特征值0的特征向量生成一维的线性子空间, 所以特征值0的几何重数为1.

由文献[14] 中的定理14可以看出, 系统$ (2.13)-(2.14) $的时间依赖解的渐进性质是由主算子$ (A+B+E) $在虚轴上的谱分布决定. 如果我们能得到在虚轴上除了0点外其它所有点都属于该系统主算子$ (A+B+E) $的豫解集, 那么由此结果并结合定理2.1, 引理2.2, 引理3.1, 引理3.2和文献[14] 中的定理14得到系统$ (2.13)-(2.14) $的时间依赖解的渐近行为: 即系统$ (2.13)-(2.14) $的时间依赖解强收敛于该系统的稳态解.

定理4.1  若$ \mu >\lambda, (\mu-\lambda)\beta_1\beta_2-\lambda\gamma\beta_2-\lambda\alpha\beta_1>0, $则当时刻$ t $趋向于无穷时, 系统(2.13)-(2.14) 的时间依赖解强收敛于其稳态解, 即

$ \underset{t\rightarrow \infty}{\mathrm{lim}}\|(p_1, p_2, p_3)(\cdot, \cdot, t)-\mathcal {K}(p_1, p_2, p_3)(\cdot)\|=0. $

其中$ (p_1, p_2, p_3) $是在引理3.1中的特征向量, 并且$ \mathcal {K} $是由引理3.2中的特征向量和初值$ (p_1, p_2, p_3)(0) $确定.

本文运用文献[15]和[16]中的研究思想和方法, 并在文献[11]和[13]的研究基础上, 当$ \mu(x)=\mu(\text{常数}), \beta_1(y)=\beta_1(\text{常数}), \beta_2(y)=\beta_2(\text{常数}) $时, 对具有两种故障状态的M/M/1可修排队系统时间依赖解的渐近性质进行了研究. 首先是通过运用概率母函数, 证明了0是该系统的主算子及其共轭算子几何重数为1的特征值. 其次是在一定的假设和约束条件下, 并结合定理2.1, 引理2.2, 引理3.1, 引理3.2和文献[14] 中的定理14可推出该系统$ (2.13)-(2.14) $的时间依赖解的渐近性质. 根据文献[14] 定理14可以看出, 系统$ (2.13)-(2.14) $的时间依赖解的渐进性质是由主算子$ (A+B+E) $在虚轴上的谱分布决定. 因此, 我们非常有必要进一步研究在虚轴上除了0点外其它所有点是否都属于该模型主算子的豫解集. 根据查阅相关文献所知, 至今还没有发现这方面的研究结果. 接下来可以利用文献[7], [9], [17], [18]中的方法和理论, 运用边界扰动思想把原来的系统进行简化, 然后讨论相应的边界算子得到所需要的主算子的豫解集. 从而再结合相关结果得到该系统$ (2.13)-(2.14) $的时间依赖解强收敛于其稳态解, 这也是我们下一步即将开展的工作.