﻿ 一类修理工带休假的冷贮备可修系统的指数稳定性

# 一类修理工带休假的冷贮备可修系统的指数稳定性Exponential Stability of a Cold Standby Repairable System with Vacation

Abstract: In this paper, the exponential stability of a repairable system is discussed, which is cold standby and the case of repairman with vacation is considered. Firstly, the suitable state space is selected, then the operator is defined, the model equation of the system is transformed into the Cauchy problem, then the properties of the dynamic solution of the system are obtained by using the C0 semigroup theory, and finally the exponential stability of the repairable system is obtained.

1. 引言

2. 系统模型

2.1. 基本假设

2.2. 系统的状态描述如下

2.3. 系统的符号意义与数学方程

(1) 部件1的工作时间分布为： ${F}_{1}\left(t\right)=1-\mathrm{exp}\left(-\lambda t\right)$$t\ge 0$

(2) 部件的维修时间分布为： $G\left(t\right)={\int }_{0}^{t}g\left(y\right)\text{d}y=1-\mathrm{exp}\left(-{\int }_{0}^{t}\mu \left(y\right)\text{d}y\right)$

(3) 部件2贮备时间分布为： ${F}_{2}\left(t\right)=1-\mathrm{exp}\left(-\beta t\right)$$t\ge 0$

(4) 修理工休假时间分布为： $H\left(t\right)={\int }_{0}^{t}h\left(x\right)\text{d}x=1-\mathrm{exp}\left(-{\int }_{0}^{t}\alpha \left(x\right)\text{d}x\right)$

(5) 修理工空闲时间分布为： ${F}_{3}\left(t\right)=1-\mathrm{exp}\left(-\eta t\right)$$t\ge 0$

$E=\left\{0,1,2,3,4,5\right\}$ 是系统的状态集， $i=0,1,2,3$ 时系统正常运行， $j=4,5$ 时系统发生问题。X(t)指t时刻的状态变量，取值于EY(t)指t时刻故障时对应的修复时间。

$\left\{\begin{array}{l}\left(\frac{\text{d}}{\text{d}t}+\eta +\lambda +\beta \right){P}_{0}\left(t\right)={\int }_{0}^{\infty }{P}_{3}\left(t,y\right)\mu \left(y\right)\text{d}y\\ \left(\frac{\partial }{\partial t}+\frac{\partial }{\partial x}+\lambda +\beta +\alpha \left(x\right)\right){P}_{1}\left(t,x\right)=0\\ \left(\frac{\partial }{\partial t}+\frac{\partial }{\partial x}+\lambda +\alpha \left(x\right)\right){P}_{2}\left(t,x\right)=\beta {P}_{1}\left(t,x\right)\\ \left(\frac{\partial }{\partial t}+\frac{\partial }{\partial y}+\lambda +\mu \left(y\right)\right){P}_{3}\left(t,y\right)=0\\ \left(\frac{\partial }{\partial t}+\frac{\partial }{\partial x}+\alpha \left(x\right)\right){P}_{4}\left(t,x\right)=\lambda {P}_{2}\left(t,x\right)\\ \left(\frac{\partial }{\partial t}+\frac{\partial }{\partial y}+\mu \left(y\right)\right){P}_{5}\left(t,y\right)=\lambda {P}_{3}\left(t,y\right)\end{array}$ (1)

$\left\{\begin{array}{l}{P}_{1}\left(t,0\right)=\eta {P}_{0}\left(t\right)\\ {P}_{3}\left(t,0\right)=\lambda {P}_{0}\left(t\right)+\beta {P}_{0}\left(t\right)+{\int }_{0}^{\infty }\lambda {P}_{1}\left(t,x\right)\text{d}x+{\int }_{0}^{\infty }{P}_{1}\left(t,x\right)\alpha \left(x\right)\text{d}x\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\int }_{0}^{\infty }{P}_{2}\left(t,x\right)\alpha \left(x\right)\text{d}x+{\int }_{0}^{\infty }{P}_{5}\left(t,y\right)\mu \left(y\right)\text{d}y\\ {P}_{5}\left(t,0\right)={\int }_{0}^{\infty }{P}_{4}\left(t,x\right)\alpha \left(x\right)\text{d}x\\ {P}_{2}\left(t,0\right)={P}_{4}\left(t,0\right)=0\end{array}$ (2)

${P}_{0}\left(0\right)=1$，其余为0 (3)

(1) $0\le \alpha \left(x\right)<\infty$$0\le \mu \left(y\right)<\infty$${\int }_{0}^{x}\alpha \left(\xi \right)\text{d}\xi <\infty$${\int }_{0}^{x}\mu \left(\xi \right)\text{d}\xi <\infty$

(2) ${\int }_{0}^{\infty }\alpha \left(\xi \right)\text{d}\xi =\infty$${\int }_{0}^{\infty }\mu \left(\xi \right)\text{d}\xi =\infty$

(3) $0<{C}_{1}=\underset{x\to \infty }{\mathrm{lim}}\frac{1}{x}{\int }_{0}^{x}\alpha \left(\xi \right)\text{d}\xi <\infty$$0<{C}_{2}=\underset{y\to \infty }{\mathrm{lim}}\frac{1}{y}{\int }_{0}^{y}\mu \left(\xi \right)\text{d}\xi <\infty$

$X=\left\{P={\left({P}_{0},{P}_{1}\left(x\right),{P}_{2}\left(x\right),{P}_{3}\left(y\right),{P}_{4}\left(x\right),{P}_{5}\left(y\right)\right)}^{\text{T}}|{P}_{0}\in R,{P}_{i}\left(x\right)\in {L}^{1}\left({R}^{+}\right),{P}_{j}\left(y\right)\in {L}^{1}\left({R}^{+}\right),i=1,2,4;j=3,5\right\}$

$A=\left(\begin{array}{cccccc}-\left(\eta +\lambda +\beta \right)& 0& 0& 0& 0& 0\\ 0& -\frac{\text{d}}{\text{d}x}-\left(\lambda +\beta +\alpha \left(x\right)\right)& 0& 0& 0& 0\\ 0& 0& -\frac{\text{d}}{\text{d}x}-\left(\lambda +\alpha \left(x\right)\right)& 0& 0& 0\\ 0& 0& 0& -\frac{\text{d}}{\text{d}y}-\left(\lambda +\mu \left(y\right)\right)& 0& 0\\ 0& 0& 0& 0& -\frac{\text{d}}{\text{d}x}-\alpha \left(x\right)& 0\\ 0& 0& 0& 0& 0& -\frac{\text{d}}{\text{d}y}-\mu \left( y \right)\end{array}\right)$

$\begin{array}{l}D\left(A\right)=\left\{P\in X|\frac{\text{d}{P}_{i}\left(x\right)}{\text{d}x},\frac{\text{d}{P}_{j}\left(y\right)}{\text{d}y}\in {L}^{1}\left({R}^{+}\right),{P}_{1}\left(0\right)=\eta {P}_{0}\left(t\right),{P}_{3}\left(t,0\right)=\lambda {P}_{0}\left(t\right)+{\int }_{0}^{\infty }{P}_{2}\left(t,x\right)\alpha \left(x\right)\text{d}x,\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{c}\text{ }\\ \text{ }\end{array}{P}_{5}\left(t,0\right)={\int }_{0}^{\infty }{P}_{4}\left(t,x\right)\alpha \left(x\right)\text{d}x,i=1,2,4;j=3,5\right\}\end{array}$

$B=\left(\begin{array}{ccccc}0& 0& {\int }_{0}^{\infty }\mu \left(y\right)\text{d}y& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& \lambda & 0& 0& 0\\ 0& 0& \lambda & 0& 0\end{array}\right)$

$\left\{\begin{array}{l}\frac{\text{d}P\left(t\right)}{\text{d}t}=\left(A+B\right)P\left(t\right),t\in \left[0,\infty \right)\\ P\left(0\right)={P}_{0}={\left(1,0,0,0,0,0\right)}^{\text{T}}\end{array}$

3. 系统动态解的存在唯一性

$\left(A+B\right){Q}^{*}=\left\{\begin{array}{l}-\left(\eta +\lambda +\beta \right){Q}_{0}+\eta {Q}_{1}\left(0\right)+\lambda {Q}_{3}\left(0\right)+\beta {Q}_{5}\left(0\right)\\ \left(\frac{\text{d}}{\text{d}x}-\lambda -\beta -\alpha \left(x\right)\right){Q}_{1}\left(x\right)+\lambda {Q}_{2}\left(x\right)+\beta {Q}_{4}\left(x\right)+\alpha \left(x\right){Q}_{3}\left(0\right)\\ \left(\frac{\text{d}}{\text{d}x}-\lambda -\alpha \left(x\right)\right){Q}_{2}\left(x\right)+\alpha \left(x\right){Q}_{3}\left(0\right)+\lambda {Q}_{4}\left(x\right)\\ \left(\frac{\text{d}}{\text{d}y}-\lambda -\mu \left(y\right)\right){Q}_{3}\left(y\right)+\mu \left(y\right){Q}_{0}+\lambda {Q}_{5}\left(y\right)\\ \left(\frac{\text{d}}{\text{d}x}-\alpha \left(x\right)\right){Q}_{4}\left(x\right)+\alpha \left(x\right){Q}_{5}\left(0\right)\\ \left(\frac{\text{d}}{\text{d}y}-\mu \left(y\right)\right){Q}_{5}\left(y\right)+\mu \left(y\right){Q}_{3}\left(0\right)\end{array}$ (4)

$D\left({\left(A+B\right)}^{*}\right)=\left\{Q\in {X}^{*}|\frac{\text{d}{Q}_{i}\left(x\right)}{\text{d}x},\frac{\text{d}{Q}_{j}\left(y\right)}{\text{d}y}\in {L}^{\infty }\left({R}^{+}\right),{Q}_{i}\left(x\right),{Q}_{j}\left(y\right)\in {L}^{\infty }\left({R}^{+}\right),i=1,2,4;j=3,5\right\}$

$\begin{array}{l}〈\left(A+B\right)P,Q〉\\ =\left[-\left(\eta +\lambda +\beta \right){P}_{0}+{\int }_{0}^{\infty }{P}_{3}\left(y\right)\mu \left(y\right)\text{d}y\right]{Q}_{0}-{\int }_{0}^{\infty }\left(\frac{\text{d}}{\text{d}x}+\lambda +\beta +\alpha \left(x\right)\right){P}_{1}\left(x\right){Q}_{1}\left(x\right)\text{d}x\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-{\int }_{0}^{\infty }\left(\frac{\text{d}}{\text{d}x}+\lambda +\alpha \left(x\right)\right){P}_{2}\left(x\right){Q}_{2}\left(x\right)\text{d}x+{\int }_{0}^{\infty }\lambda {P}_{1}\left(x\right){Q}_{2}\left(x\right)\text{d}x+{\int }_{0}^{\infty }\beta {P}_{1}\left(x\right){Q}_{4}\left(x\right)\text{d}x\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-{\int }_{0}^{\infty }\left(\frac{\text{d}}{\text{d}y}+\lambda +\mu \left(y\right)\right){P}_{3}\left(y\right){Q}_{3}\left(y\right)\text{d}y-{\int }_{0}^{\infty }\left(\frac{\text{d}}{\text{d}x}+\alpha \left(x\right)\right){P}_{4}\left(x\right){Q}_{4}\left(x\right)\text{d}x\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\int }_{0}^{\infty }\lambda {P}_{2}\left(x\right){Q}_{4}\left(x\right)\text{d}x-{\int }_{0}^{\infty }\left(\frac{\text{d}}{\text{d}y}+\mu \left(y\right)\right){P}_{5}\left(y\right){Q}_{5}\left(y\right)\text{d}y+{\int }_{0}^{\infty }\lambda {P}_{3}\left(y\right){Q}_{5}\left(y\right)\text{d}y\end{array}$

$\begin{array}{l}=\left[-\left(\eta +\lambda +\beta \right){Q}_{0}+\eta {Q}_{1}\left(0\right)+\lambda {Q}_{3}\left(0\right)+\beta {Q}_{5}\left(0\right)\right]{P}_{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\int }_{0}^{\infty }\left[\left(\frac{\text{d}}{\text{d}x}-\lambda -\beta -\alpha \left(x\right)\right){Q}_{1}\left(x\right)+\lambda {Q}_{2}\left(x\right)+\beta {Q}_{4}\left(x\right)\right]{P}_{1}\left(x\right)\text{d}x\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\int }_{0}^{\infty }\left[\left(\frac{\text{d}}{\text{d}x}-\lambda -\alpha \left(x\right)\right){Q}_{2}\left(x\right)+\alpha \left(x\right){Q}_{3}\left(0\right)+\lambda {Q}_{4}\left(x\right)\right]{P}_{2}\left(x\right)\text{d}x\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\int }_{0}^{\infty }\left[\left(\frac{\text{d}}{\text{d}y}-\lambda -\mu \left(y\right)\right){Q}_{3}\left(y\right)+\mu \left(y\right){Q}_{0}+\lambda {Q}_{5}\left(y\right)\right]{P}_{3}\left(y\right)\text{d}y\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\int }_{0}^{\infty }\left[\left(\frac{\text{d}}{\text{d}x}-\alpha \left(x\right)\right){Q}_{4}\left(x\right)+\alpha \left(x\right){Q}_{5}\left(0\right)\right]{P}_{4}\left(x\right)\text{d}x\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\int }_{0}^{\infty }\left[\left(\frac{\text{d}}{\text{d}y}-\mu \left(y\right)\right){Q}_{5}\left(y\right)+\mu \left(y\right){Q}_{3}\left(0\right)\right]{P}_{5}\left(y\right)\text{d}y\\ =〈P,{\left(A+B\right)}^{*}Q〉\end{array}$

$P\left(t,\cdot \right)$ 满足方程组(1)，所以有 $\frac{\text{d}‖P\left(t,\cdot \right)‖}{\text{d}t}=0$，故 $‖P\left(t,\cdot \right)‖=‖T\left(t\right){P}_{0}‖=‖{P}_{0}‖=1,\forall t\in \left[0,\infty \right)$

4. 系统的指数稳定性

$\left\{\begin{array}{l}\left(r+\eta +\lambda +\beta \right){P}_{0}={g}_{0}+{\int }_{0}^{\infty }{P}_{3}\left(y\right)\mu \left(y\right)\text{d}y\\ \frac{\text{d}{P}_{1}\left(x\right)}{\text{d}x}+\left(r+\lambda +\beta +\alpha \left(x\right)\right){P}_{1}\left(x\right)={g}_{1}\left(x\right)\\ \frac{\text{d}{P}_{2}\left(x\right)}{dx}+\left(r+\lambda +\alpha \left(x\right)\right){P}_{2}\left(x\right)={g}_{2}\left(x\right)+\beta {P}_{1}\left(x\right)\\ \frac{\text{d}{P}_{3}\left(y\right)}{\text{d}y}+\left(r+\lambda +\mu \left(y\right)\right){P}_{3}\left(y\right)={g}_{3}\left(y\right)\\ \frac{\text{d}{P}_{4}\left(x\right)}{\text{d}x}+\left(r+\alpha \left(x\right)\right){P}_{4}\left(x\right)={g}_{4}\left(x\right)+\lambda {P}_{2}\left(x\right)\\ \frac{\text{d}{P}_{5}\left(y\right)}{\text{d}y}+\left(r+\mu \left(y\right)\right){P}_{5}\left(y\right)={g}_{5}\left(y\right)+\lambda {P}_{3}\left(y\right)\end{array}$ (5)

$\left\{\begin{array}{l}{P}_{1}\left(0\right)=\eta {P}_{0}\\ {P}_{2}\left(0\right)={P}_{4}\left(0\right)=0\\ {P}_{3}\left(0\right)=\lambda {P}_{0}+\beta {P}_{0}+{\int }_{0}^{\infty }\lambda {P}_{1}\left(x\right)\text{d}x+{\int }_{0}^{\infty }{P}_{1}\left(x\right)\alpha \left(x\right)\text{d}x+{\int }_{0}^{\infty }{P}_{2}\left(x\right)\alpha \left(x\right)\text{d}x+{\int }_{0}^{\infty }{P}_{5}\left(y\right)\mu \left(y\right)\text{d}y\\ {P}_{5}\left(0\right)={\int }_{0}^{\infty }{P}_{4}\left(x\right)\alpha \left(x\right)\text{d}x\end{array}$ (6)

$\left\{\begin{array}{l}{P}_{1}\left(x\right)={P}_{1}\left(0\right){\text{e}}^{-{\int }_{0}^{x}\left(r+\lambda +\beta +\alpha \left(\sigma \right)\right)\text{d}\sigma }+{G}_{1}\left(x\right)\\ {P}_{2}\left(x\right)=\beta x{P}_{1}\left(0\right){\text{e}}^{-{\int }_{0}^{x}\left(r+\lambda +\alpha \left(\sigma \right)\right)\text{d}\sigma }+{G}_{2}\left(x\right)\\ {P}_{3}\left(y\right)={P}_{3}\left(0\right){\text{e}}^{-{\int }_{0}^{y}\left(r+\lambda +\mu \left(\sigma \right)\right)\text{d}\sigma }+{G}_{3}\left(y\right)\\ {P}_{4}\left(x\right)=\left(1-{\text{e}}^{-\lambda x}-\lambda x{\text{e}}^{-\lambda x}\right){P}_{1}\left(0\right){\text{e}}^{-{\int }_{0}^{x}\left(r+\alpha \left(\sigma \right)\right)\text{d}\sigma }+{G}_{4}\left(x\right)\\ {P}_{5}\left(y\right)={P}_{5}\left(0\right){\text{e}}^{-{\int }_{0}^{y}\left(r+\mu \left(\sigma \right)\right)\text{d}\sigma }+\left(1-{\text{e}}^{-\lambda y}\right){P}_{3}\left(0\right){\text{e}}^{-{\int }_{0}^{y}\left(r+\mu \left(\sigma \right)\right)\text{d}\sigma }+{G}_{5}\left(y\right)\end{array}$ (7)

$\begin{array}{l}{G}_{1}\left(x\right)={\int }_{0}^{x}{g}_{1}\left(\tau \right){\text{e}}^{-{\int }_{\tau }^{x}\left(r+\lambda +\beta +\alpha \left(\sigma \right)\right)\text{d}\sigma }\text{d}\tau ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{G}_{2}\left(x\right)={\int }_{0}^{x}\left({g}_{2}\left(\tau \right)+\beta {G}_{1}\left(\tau \right)\right){\text{e}}^{-{\int }_{\tau }^{x}\left(r+\lambda +\alpha \left(\sigma \right)\right)\text{d}\sigma }\text{d}\tau \\ {G}_{3}\left(y\right)={\int }_{0}^{y}{g}_{3}\left(\tau \right){\text{e}}^{-{\int }_{\tau }^{y}\left(r+\lambda +\mu \left(\sigma \right)\right)\text{d}\sigma }\text{d}\tau ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{G}_{4}\left(x\right)={\int }_{0}^{x}\left({g}_{4}\left(\tau \right)+\lambda {G}_{2}\left(\tau \right)\right){\text{e}}^{-{\int }_{\tau }^{x}\left(r+\alpha \left(\sigma \right)\right)\text{d}\sigma }\text{d}\tau \\ {G}_{5}\left(y\right)={\int }_{0}^{y}\left({g}_{5}\left(\tau \right)+\lambda {G}_{3}\left(\tau \right)\right){\text{e}}^{-{\int }_{\tau }^{y}\left(r+\mu \left(\sigma \right)\right)\text{d}\sigma }\text{d}\tau \end{array}$

${F}_{1}={\int }_{0}^{\infty }{\text{e}}^{-{\int }_{0}^{x}\left(r+\alpha \left(\sigma \right)\right)\text{d}\sigma }\text{d}x,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{F}_{2}={\int }_{0}^{\infty }{\text{e}}^{-{\int }_{0}^{y}\left(r+\mu \left(\sigma \right)\right)\text{d}\sigma }\text{d}y,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{F}_{3}={\int }_{0}^{\infty }\mu \left(y\right){\text{e}}^{-{\int }_{0}^{y}\left(r+\lambda +\mu \left(\sigma \right)\right)\text{d}\sigma }\text{d}y$

${F}_{4}=\lambda {\int }_{0}^{\infty }\alpha \left(x\right){\text{e}}^{-{\int }_{0}^{x}\left(r+\lambda +\alpha \left(\sigma \right)\right)\text{d}\sigma }\text{d}x,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{F}_{5}={\int }_{0}^{\infty }\alpha \left(x\right){\text{e}}^{-{\int }_{0}^{x}\left(r+\lambda +\alpha \left(\sigma \right)\right)\text{d}\sigma }\text{d}x$

${W}_{0}={g}_{0}+{\int }_{0}^{\infty }\mu \left(y\right){G}_{3}\left(y\right)\text{d}y,\text{\hspace{0.17em}}{W}_{1}=0,\text{\hspace{0.17em}}{W}_{2}={\int }_{0}^{\infty }\alpha \left(x\right){G}_{2}\left(x\right)\text{d}x,\text{\hspace{0.17em}}{W}_{3}={\int }_{0}^{\infty }\alpha \left(x\right){G}_{4}\left(x\right)\text{d}x$

$\left[\begin{array}{cccc}r+\eta +\lambda +\beta & 0& \begin{array}{c}-{F}_{3}\end{array}& 0\\ -\eta & 1& 0& 0\\ -\lambda -\beta & {F}_{5}-{F}_{4}& r{F}_{2}+{F}_{3}& r{F}_{2}-1\\ 0& r{F}_{1}+{F}_{4}+{F}_{5}-1& 0& 1\end{array}\right]\left[\begin{array}{c}{P}_{0}\\ {P}_{1}\left(0\right)\\ {P}_{3}\left(0\right)\\ {P}_{5}\left(0\right)\end{array}\right]=\left[\begin{array}{c}{W}_{0}\\ {W}_{1}\\ {W}_{2}\\ {W}_{3}\end{array}\right]$,

$T=\left[\begin{array}{cccc}r+\eta +\lambda +\beta & 0& \begin{array}{c}-{F}_{3}\end{array}& 0\\ -\eta & 1& 0& 0\\ -\lambda -\beta & {F}_{5}-{F}_{4}& r{F}_{2}+{F}_{3}& r{F}_{2}-1\\ 0& r{F}_{1}+{F}_{4}+{F}_{5}-1& 0& 1\end{array}\right]$

$\left\{\begin{array}{l}\left(r+\eta +\lambda +\beta \right){P}_{0}-{\int }_{0}^{\infty }{P}_{3}\left(y\right)\mu \left(y\right)\text{d}y=0\\ \frac{\text{d}{P}_{1}\left(x\right)}{\text{d}x}+\left(r+\lambda +\beta \right){P}_{1}\left(x\right)=0\\ \frac{\text{d}{P}_{2}\left(x\right)}{\text{d}x}+\left(r+\lambda +\alpha \left(x\right)\right){P}_{2}\left(x\right)-\beta {P}_{1}\left(x\right)=0\\ \frac{\text{d}{P}_{3}\left(y\right)}{\text{d}y}+\left(r+\lambda +\mu \left(y\right)\right){P}_{3}\left(y\right)=0\\ \frac{\text{d}{P}_{4}\left(x\right)}{\text{d}x}+\left(r+\alpha \left(x\right)\right){P}_{4}\left(x\right)-\lambda {P}_{2}\left(x\right)=0\\ \frac{\text{d}{P}_{5}\left(y\right)}{\text{d}y}+\left(r+\mu \left(y\right)\right){P}_{5}\left(y\right)-\lambda {P}_{3}\left(y\right)=0\\ {P}_{1}\left(0\right)=\eta {P}_{0}\\ {P}_{3}\left(0\right)=\lambda {P}_{0}+{\int }_{0}^{\infty }{P}_{2}\left(0\right)\alpha \left(x\right)\text{d}x\\ {P}_{5}\left(0\right)={\int }_{0}^{\infty }{P}_{4}\left(0\right)\alpha \left(x\right)\text{d}x\\ {P}_{2}\left(0\right)={P}_{4}\left(0\right)=0\end{array}$ (8)

$\begin{array}{l}{P}_{1}\left(x\right)={P}_{1}\left(0\right){\text{e}}^{-\left(r+\lambda +\beta \right)x}\\ {P}_{2}\left(x\right)=\beta {P}_{1}\left(0\right){\int }_{0}^{x}{\text{e}}^{-{\int }_{0}^{\tau }\left(r+\lambda +\beta \right)\text{d}\xi }\cdot {\text{e}}^{-{\int }_{\tau }^{x}\left(r+\lambda +\alpha \left(\xi \right)\right)\text{d}\xi }\text{d}\tau \\ {P}_{3}\left(y\right)={P}_{3}\left(0\right){\text{e}}^{-{\int }_{0}^{y}\left(r+\lambda +\mu \left(\xi \right)\right)\text{d}\xi }\\ {P}_{5}\left(y\right)={P}_{5}\left(0\right){\text{e}}^{-{\int }_{0}^{y}\left(r+\mu \left(\xi \right)\right)\text{d}\xi }+\left(1-{\text{e}}^{-\lambda y}\right){P}_{3}\left(0\right){\text{e}}^{-{\int }_{0}^{y}\left(r+\mu \left(\xi \right)\right)\text{d}\xi }\end{array}$ (9)

$\left\{\begin{array}{l}\left(r+\eta +\lambda +\beta \right){P}_{0}-{\int }_{0}^{\infty }{\text{e}}^{-{\int }_{0}^{y}\left(r+\lambda +\mu \left(\xi \right)\right)\text{d}\xi }\text{d}y{P}_{3}\left(0\right)=0\\ \eta {P}_{0}-{P}_{1}\left(0\right)=0\\ {P}_{2}\left(0\right)=0\\ \lambda {P}_{0}+{\int }_{0}^{\infty }\alpha \left(x\right)\text{d}x{P}_{2}\left(0\right)-{P}_{3}\left(0\right)=0\\ {P}_{4}\left(0\right)=0\\ {\int }_{0}^{\infty }\alpha \left(x\right)\text{d}x{P}_{4}\left(0\right)-{P}_{5}\left(0\right)=0\end{array}$ (10)

$D\left(r\right)=|\begin{array}{cccccc}r+\eta +\lambda +\beta & 0& 0& {\int }_{0}^{\infty }{\text{e}}^{-{\int }_{0}^{y}\left(r+\lambda +\mu \left(\xi \right)\right)\text{d}\xi }\text{d}y& 0& 0\\ \eta & -1& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\\ \lambda & 0& {\int }_{0}^{\infty }\alpha \left(x\right)\text{d}x& -1& 0& 0\\ 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& {\int }_{0}^{\infty }\alpha \left(x\right)\text{d}x& -1\end{array}|$

$D\left(r\right)=0$ 时， $r=0$ 是A + B的本征值。反过来若 $r=0$ 使得 $D\left(r\right)=0$，则方程(10)有非零解

$\left({p}_{0},{p}_{1}\left(0\right),{p}_{2}\left(0\right),{p}_{3}\left(0\right),{p}_{4}\left(0\right),{p}_{5}\left(0\right)\right):\left\{\begin{array}{l}{P}_{1}\left(0\right)=\eta {P}_{0}\\ {P}_{2}\left(0\right)=0\\ {P}_{3}\left(0\right)=\lambda {P}_{0}\\ {P}_{4}\left(0\right)=0\\ {P}_{5}\left(0\right)=0\end{array}$ (11)

$\left\{\begin{array}{l}\left(\eta +\lambda +\beta \right){Q}_{0}=\eta {Q}_{1}\left(0\right)+\beta {Q}_{3}\left(0\right)\\ \frac{\text{d}}{\text{d}x}{Q}_{1}\left(x\right)=\left(\lambda +\beta \right){Q}_{1}\left(x\right)-\beta {Q}_{2}\left(x\right)\\ \frac{\text{d}}{\text{d}x}{Q}_{2}\left(x\right)=\left(\lambda +\alpha \left(x\right)\right){Q}_{2}\left(x\right)-\alpha \left(x\right){Q}_{3}\left(0\right)-\lambda {Q}_{4}\left(x\right)\\ \frac{\text{d}}{\text{d}y}{Q}_{3}\left(y\right)=\left(\lambda +\mu \left(y\right)\right){Q}_{3}\left(y\right)-\mu \left(y\right){Q}_{0}-\lambda {Q}_{5}\left(y\right)\\ \frac{\text{d}}{\text{d}x}{Q}_{4}\left(x\right)=\alpha \left(x\right){Q}_{4}\left(x\right)-\alpha \left(x\right){Q}_{5}\left(0\right)\\ \frac{\text{d}}{\text{d}y}{Q}_{5}\left(y\right)=\mu \left(y\right){Q}_{5}\left(y\right)-\mu \left(y\right){Q}_{3}\left(0\right)\end{array}$ (12)

[1] 曹晋华, 程侃.可靠性数学引论[M]. 北京: 科学出版社, 1986.

[2] 郭卫华, 徐厚宝, 朱广田. 两部件并联维修系统解的渐进稳定性[J]. 系统工程理论与实践, 2006, 26(12): 62-68.

[3] 郭卫华, 党艳霞, 高超. 两不同部件并联可修复系统指数稳定性分析[J]. 2009(2): 108-114.

[4] 张欣. 电站两辅助设备冷贮备系统的可靠性分析[J]. 系统工程理论与实践, 2013, 33(10): 2615-2622.

[5] 任寒景, 张玉峰, 张欣. 含同原因故障和一冷储备部件的系统稳定性[J]. 应用泛函分析学报, 2016, 18(1): 41-49.

[6] 杨会崇. 修理工带休假的两同型部件冷贮备可修系统[J]. 数学理论与应用, 2009, 29(1): 23-28.

[7] 陈永燕, 郑海鹰. 冷贮备可修系统的一个新模型及其可靠性分析[J]. 浙江大学学报(理学版), 2011, 38(3): 274-278.

[8] 郭卫华. 两相同部件温储备可修的人机系统解的性质分析[J]. 数学的实践与认识, 2003, 33(7): 88-95.

[9] ZHANG, X. (2015) Reliability Analysis of a Cold Standby Repairable System with Repairman Extra Work. Journal of Systems Science and Complexity, 28, 1015-1032.
https://doi.org/10.1007/s11424-015-4081-5

[10] 霍慧霞, 原文志. 一类冷贮备可修系统的指数稳定性[J]. 数学的实践与认识, 2018, 48(1): 185-191.

Top