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

摘要: 本文讨论了一类可修系统的指数稳定性，这个系统是冷贮备的而且考虑了修理工带休假的情况。首先选择适当的状态空间，然后定义算子，把系统的模型方程化为Cauchy问题，再用C₀半群理论得到系统的动态解的性质，最后得到这个可修系统的指数稳定性。

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

在我们实际生活中，一个系统性能的好坏往往通过这个系统的稳定性来判断，所以研究系统稳定性是很有必要的。文献 [1] [2] [3] 研究了并联可修系统。文献 [2] 首先得到系统非负解是0本征值对应的非负本向量，又得到系统算子的谱分布，从而证明系统的渐进稳定性。

可靠性理论的研究包含各种模型，其中冷贮备系统是很重要的一类。现在很多文献假设部件出现问题之后可以马上处理，如何讨论系统的可靠性以及稳定性，例如文献 [4] [5]。但是现实生活上当系统出现问题时因为各种客观因素，系统不能立马修复。所以把修理工休假等情况引入可修系统的模型中是有意义的。但是很多系统都只是运用传统方法得到可靠性指标而已，比如文献 [6] [7]。

所以本文运用C₀半群理论来研究修理工带休假的冷贮备可修系统，通过分析系统动态解的性质来讨论系统的指数稳定性。

2. 系统模型

2.1. 基本假设

假设1：系统中有两个不同型部件分别为1和2，再加上一个修理工；

假设2：当部件1和2都处于正常状态时，部件1工作，部件2做冷贮备；

假设3：当部件1处于工作状态时，部件2冷贮备，修理工开始休假；

假设4：当部件出现问题时有以下情况(1) 如果这时修理工正处于休假状态，那么等休假完成后进行维修；(2) 如果这时修理工空闲，那么需要马上维修；(3) 如果这时修理工处于工作状态，那么这个有问题的部件必须等待维修；

假设5：修理工维修完成一个部件后，如果在系统中没有发现其他问题就开始休假，如果休假完成之后发现部件有问题，那么马上维修故障部件；如果系统仍然正常，那么必须待在系统里进入空闲状态，等待出现问题之后马上维修；

假设6：在初始状态时刻部件1和2都是新的，而且部件1比2优先使用和维修，维修后的部件和新的一样；

假设7：文中的随机变量是相互独立的。

2.2. 系统的状态描述如下

状态0：部件1工作，部件2贮备，修理工空闲；

状态1：部件1工作，部件2贮备，修理工休假；

状态2：部件1工作，部件2故障，修理工休假；

状态3：部件1工作，部件2在修理；

状态4：部件1和部件2都待修，修理工休假；

状态5：部件1在修理，部件2待修。

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： $F_1\left(t\right)=1-\exp \left(-\lambda t\right)$， $t\ge 0$

(2) 部件的维修时间分布为： $G\left(t\right)={\displaystyle {\int }_0^{t}g\left(y\right)\text{d}y}=1-\exp \left(-{\displaystyle {\int }_0^t\mu \left(y\right)\text{d}y}\right)$

(3) 部件2贮备时间分布为： $F_2\left(t\right)=1-\exp \left(-\beta t\right)$， $t\ge 0$

(4) 修理工休假时间分布为： $H\left(t\right)={\displaystyle {\int }_0^{t}h\left(x\right)\text{d}x}=1-\exp \left(-{\displaystyle {\int }_0^t\alpha \left(x\right)\text{d}x}\right)$

(5) 修理工空闲时间分布为： $F_3\left(t\right)=1-\exp \left(-\eta t\right)$， $t\ge 0$

设 $E=\left\{0,1,2,3,4,5\right\}$ 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E；Y(t)指t时刻故障时对应的修复时间。

系统方程如下

$\{\begin{array}{l}\left(\frac{\text{d}}{\text{d}t}+\eta +\lambda +\beta \right)P_0\left(t\right)={\displaystyle {\int }_0^{\infty }{P}_3\left(t,y\right)\mu \left(y\right)\text{d}y}\\ \left(\frac{\partial }{\partial t}+\frac{\partial }{\partial x}+\lambda +\beta +\alpha \left(x\right)\right)P_1\left(t,x\right)=0\\ \left(\frac{\partial }{\partial t}+\frac{\partial }{\partial x}+\lambda +\alpha \left(x\right)\right)P_2\left(t,x\right)=\beta P_1\left(t,x\right)\\ \left(\frac{\partial }{\partial t}+\frac{\partial }{\partial y}+\lambda +\mu \left(y\right)\right)P_3\left(t,y\right)=0\\ \left(\frac{\partial }{\partial t}+\frac{\partial }{\partial x}+\alpha \left(x\right)\right)P_4\left(t,x\right)=\lambda P_2\left(t,x\right)\\ \left(\frac{\partial }{\partial t}+\frac{\partial }{\partial y}+\mu \left(y\right)\right)P_5\left(t,y\right)=\lambda P_3\left(t,y\right)\end{array}$ (1)

边界条件为

$\{\begin{array}{l}{P}_1\left(t,0\right)=\eta P_0\left(t\right)\\ P_3\left(t,0\right)=\lambda P_0\left(t\right)+\beta P_0\left(t\right)+{\displaystyle {\int }_0^{\infty }\lambda P_1\left(t,x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }{P}_1\left(t,x\right)\alpha \left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }{P}_2\left(t,x\right)\alpha \left(x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }{P}_5\left(t,y\right)\mu \left(y\right)\text{d}y}\\ P_5\left(t,0\right)={\displaystyle {\int }_0^{\infty }{P}_4\left(t,x\right)\alpha \left(x\right)\text{d}x}\\ P_2\left(t,0\right)=P_4\left(t,0\right)=0\end{array}$ (2)

初始条件为：

$P_0\left(0\right)=1$，其余为0 (3)

再做一些假设来增加(1)的合理性

(1) $0\le \alpha \left(x\right)<\infty$， $0\le \mu \left(y\right)<\infty$， ${\displaystyle {\int }_0^x\alpha \left(\xi \right)\text{d}\xi }<\infty$， ${\displaystyle {\int }_0^x\mu \left(\xi \right)\text{d}\xi }<\infty$。

(2) ${\displaystyle {\int }_0^{\infty }\alpha \left(\xi \right)\text{d}\xi }=\infty$， ${\displaystyle {\int }_0^{\infty }\mu \left(\xi \right)\text{d}\xi }=\infty$。

(3) $0<C_1=\underset{x\to \infty }{\lim }\frac{1}{x}{\displaystyle {\int }_0^x\alpha \left(\xi \right)\text{d}\xi }<\infty$， $0<C_2=\underset{y\to \infty }{\lim }\frac{1}{y}{\displaystyle {\int }_0^y\mu \left(\xi \right)\text{d}\xi }<\infty$。

选择以下状态空间

$X=\left\{P={\left(P_0,P_1\left(x\right),P_2\left(x\right),P_3\left(y\right),P_4\left(x\right),P_5\left(y\right)\right)}^{\text{T}}|P_0\in R,P_i\left(x\right)\in L^1\left(R^{+}\right),P_j\left(y\right)\in L^1\left(R^{+}\right),i=1,2,4;j=3,5\right\}$

范数定义为 $\Vert P\Vert =\left|P_0\right|+\Vert P_1\left(x\right)\Vert +\Vert P_2\left(x\right)\Vert +\Vert P_3\left(y\right)\Vert +\Vert P_4\left(x\right)\Vert +\Vert P_5\left(y\right)\Vert$

很明显 $\left(X,\Vert \cdot \Vert \right)$ 是一个Banach空间。

在X中定义算子A，B及定义域

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

算子A的定义域

$\begin{array}{l}D\left(A\right)=\{P\in X|\frac{\text{d}{P}_i\left(x\right)}{\text{d}x},\frac{\text{d}{P}_j\left(y\right)}{\text{d}y}\in L^1\left(R^{+}\right),P_1\left(0\right)=\eta P_0\left(t\right),P_3\left(t,0\right)=\lambda P_0\left(t\right)+{\displaystyle {\int }_0^{\infty }{P}_2\left(t,x\right)\alpha \left(x\right)\text{d}x},\\ \begin{array}{c}\\ \end{array}{P}_5\left(t,0\right)={\displaystyle {\int }_0^{\infty }{P}_4\left(t,x\right)\alpha \left(x\right)\text{d}x},i=1,2,4;j=3,5\}\end{array}$

$B=\left(\begin{array}{ccccc}0& 0& {\displaystyle {\int }_0^{\infty }\mu \left(y\right)\text{d}y}& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& \lambda & 0& 0& 0\\ 0& 0& \lambda & 0& 0\end{array}\right)$

因此 $D\left(A+B\right)=D\left(A\right)\cap D\left(B\right)=D\left(A\right)$，在 $D\left(A\right)$ 中 $P_i\left(x\right),P_j\left(y\right)$ 是绝对连续的，所以系统方程可以化为Banach空间X中的抽象Cauchy问题

$\{\begin{array}{l}\frac{\text{d}P\left(t\right)}{\text{d}t}=\left(A+B\right)P\left(t\right),t\in \left[0,\infty \right)\\ P\left(0\right)=P_0={\left(1,0,0,0,0,0\right)}^{\text{T}}\end{array}$

3. 系统动态解的存在唯一性

定义1 E的子集C称为在E中共尾，若满足对每个 $f\in E$，存在 $g\in C$，使得 $f\le g$。

定理1 $D\left(A+B\right)$ 在X中稠密。

证明 由文献 [8] 有D(A)在X中是稠密的，又有 $D\left(A+B\right)=D\left(A\right)$，所以有 $D\left(A+B\right)$ 在X中稠密。

定理2 [9] 算子A + B是预解正算子。

定理3 系统算子A + B的对偶算子 ${\left(A+B\right)}^{*}$ 是

$\left(A+B\right)Q^{*}=\{\begin{array}{l}-\left(\eta +\lambda +\beta \right)Q_0+\eta Q_1\left(0\right)+\lambda Q_3\left(0\right)+\beta Q_5\left(0\right)\\ \left(\frac{\text{d}}{\text{d}x}-\lambda -\beta -\alpha \left(x\right)\right)Q_1\left(x\right)+\lambda Q_2\left(x\right)+\beta Q_4\left(x\right)+\alpha \left(x\right)Q_3\left(0\right)\\ \left(\frac{\text{d}}{\text{d}x}-\lambda -\alpha \left(x\right)\right)Q_2\left(x\right)+\alpha \left(x\right)Q_3\left(0\right)+\lambda Q_4\left(x\right)\\ \left(\frac{\text{d}}{\text{d}y}-\lambda -\mu \left(y\right)\right)Q_3\left(y\right)+\mu \left(y\right)Q_0+\lambda Q_5\left(y\right)\\ \left(\frac{\text{d}}{\text{d}x}-\alpha \left(x\right)\right)Q_4\left(x\right)+\alpha \left(x\right)Q_5\left(0\right)\\ \left(\frac{\text{d}}{\text{d}y}-\mu \left(y\right)\right)Q_5\left(y\right)+\mu \left(y\right)Q_3\left(0\right)\end{array}$ (4)

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

证明 任给 $P\in D\left(A+B\right)$ 和 $Q\in X^{*}$，有

$\begin{array}{l}\langle \left(A+B\right)P,Q\rangle \\ =\left[-\left(\eta +\lambda +\beta \right)P_0+{\displaystyle {\int }_0^{\infty }{P}_3\left(y\right)\mu \left(y\right)\text{d}y}\right]Q_0-{\displaystyle {\int }_0^{\infty }\left(\frac{\text{d}}{\text{d}x}+\lambda +\beta +\alpha \left(x\right)\right)P_1\left(x\right)Q_1\left(x\right)\text{d}x}\\ -{\displaystyle {\int }_0^{\infty }\left(\frac{\text{d}}{\text{d}x}+\lambda +\alpha \left(x\right)\right)P_2\left(x\right)Q_2\left(x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }\lambda P_1\left(x\right)Q_2\left(x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }\beta P_1\left(x\right)Q_4\left(x\right)\text{d}x}\\ -{\displaystyle {\int }_0^{\infty }\left(\frac{\text{d}}{\text{d}y}+\lambda +\mu \left(y\right)\right)P_3\left(y\right)Q_3\left(y\right)\text{d}y}-{\displaystyle {\int }_0^{\infty }\left(\frac{\text{d}}{\text{d}x}+\alpha \left(x\right)\right)P_4\left(x\right)Q_4\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }\lambda P_2\left(x\right)Q_4\left(x\right)\text{d}x}-{\displaystyle {\int }_0^{\infty }\left(\frac{\text{d}}{\text{d}y}+\mu \left(y\right)\right)P_5\left(y\right)Q_5\left(y\right)\text{d}y}+{\displaystyle {\int }_0^{\infty }\lambda P_3\left(y\right)Q_5\left(y\right)\text{d}y}\end{array}$

$\begin{array}{l}=\left[-\left(\eta +\lambda +\beta \right)Q_0+\eta Q_1\left(0\right)+\lambda Q_3\left(0\right)+\beta Q_5\left(0\right)\right]P_0\\ +{\displaystyle {\int }_0^{\infty }\left[\left(\frac{\text{d}}{\text{d}x}-\lambda -\beta -\alpha \left(x\right)\right)Q_1\left(x\right)+\lambda Q_2\left(x\right)+\beta Q_4\left(x\right)\right]P_1\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }\left[\left(\frac{\text{d}}{\text{d}x}-\lambda -\alpha \left(x\right)\right)Q_2\left(x\right)+\alpha \left(x\right)Q_3\left(0\right)+\lambda Q_4\left(x\right)\right]P_2\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }\left[\left(\frac{\text{d}}{\text{d}y}-\lambda -\mu \left(y\right)\right)Q_3\left(y\right)+\mu \left(y\right)Q_0+\lambda Q_5\left(y\right)\right]P_3\left(y\right)\text{d}y}\end{array}$

$\begin{array}{l}+{\displaystyle {\int }_0^{\infty }\left[\left(\frac{\text{d}}{\text{d}x}-\alpha \left(x\right)\right)Q_4\left(x\right)+\alpha \left(x\right)Q_5(0)\right]P_4\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }\left[\left(\frac{\text{d}}{\text{d}y}-\mu \left(y\right)\right)Q_5\left(y\right)+\mu \left(y\right)Q_3\left(0\right)\right]P_5\left(y\right)\text{d}y}\\ =\langle P,{\left(A+B\right)}^{*}Q\rangle \end{array}$

定理4 算子A + B生成正压缩 $C_0$ -半群 $T\left(t\right)$。

证明 由文献 [10] 知A + B生成正 $C_0$ 半群 $T\left(t\right)$。

定理5 系统存在唯一的非负时间依赖解 $P\left(t,\cdot \right)$，满足 $\Vert P\left(t,\cdot \right)\Vert =1,\forall t\in \left[0,\infty \right)$。

证明 由定理4可知系统存在这样的解 $P\left(t,\cdot \right)$，它可以表示成 $\Vert P\left(t,\cdot \right)\Vert =T\left(t\right)P_0,\forall t\in \left[0,\infty \right)$。又因为

$P\left(t,\cdot \right)$ 满足方程组(1)，所以有 $\frac{\text{d}\Vert P\left(t,\cdot \right)\Vert }{\text{d}t}=0$，故 $\Vert P\left(t,\cdot \right)\Vert =\Vert T\left(t\right)P_0\Vert =\Vert P_0\Vert =1,\forall t\in \left[0,\infty \right)$。

4. 系统的指数稳定性

定理6 $\left\{r\in C|\mathrm{Re}r>0或r=ia,a\in R\backslash \left\{0\right\}\right\}$ 包含于系统算子A + B的预解集 $\rho \left(A+B\right)$ 中。

证明 对任意 $G=\left\{g_0,g_1\left(x\right),g_2\left(x\right),g_3\left(y\right),g_4\left(x\right),g_5\left(y\right)\right\}$，考虑算子方程 $\left[rI-\left(A+B\right)\right]P=G$，有以下的方程组

$\{\begin{array}{l}\left(r+\eta +\lambda +\beta \right)P_0=g_0+{\displaystyle {\int }_0^{\infty }{P}_3\left(y\right)\mu \left(y\right)\text{d}y}\\ \frac{\text{d}{P}_1\left(x\right)}{\text{d}x}+\left(r+\lambda +\beta +\alpha \left(x\right)\right)P_1\left(x\right)=g_1\left(x\right)\\ \frac{\text{d}{P}_2\left(x\right)}{dx}+\left(r+\lambda +\alpha \left(x\right)\right)P_2\left(x\right)=g_2\left(x\right)+\beta P_1\left(x\right)\\ \frac{\text{d}{P}_3\left(y\right)}{\text{d}y}+\left(r+\lambda +\mu \left(y\right)\right)P_3\left(y\right)=g_3(y)\\ \frac{\text{d}{P}_4\left(x\right)}{\text{d}x}+\left(r+\alpha \left(x\right)\right)P_4\left(x\right)=g_4\left(x\right)+\lambda P_2\left(x\right)\\ \frac{\text{d}{P}_5\left(y\right)}{\text{d}y}+\left(r+\mu \left(y\right)\right)P_5\left(y\right)=g_5\left(y\right)+\lambda P_3\left(y\right)\end{array}$ (5)

边界条件为

$\{\begin{array}{l}{P}_1\left(0\right)=\eta P_0\\ P_2\left(0\right)=P_4\left(0\right)=0\\ P_3\left(0\right)=\lambda P_0+\beta P_0+{\displaystyle {\int }_0^{\infty }\lambda P_1\left(x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }{P}_1\left(x\right)\alpha \left(x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }{P}_2\left(x\right)\alpha \left(x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }{P}_5\left(y\right)\mu \left(y\right)\text{d}y}\\ P_5\left(0\right)={\displaystyle {\int }_0^{\infty }{P}_4\left(x\right)\alpha \left(x\right)\text{d}x}\end{array}$ (6)

解方程组(5)可得

$\{\begin{array}{l}{P}_1\left(x\right)=P_1\left(0\right){\text{e}}^{-{\displaystyle {\int }_0^x\left(r+\lambda +\beta +\alpha \left(\sigma \right)\right)\text{d}\sigma }}+G_1\left(x\right)\\ P_2\left(x\right)=\beta x{P}_1\left(0\right){\text{e}}^{-{\displaystyle {\int }_0^x\left(r+\lambda +\alpha \left(\sigma \right)\right)\text{d}\sigma }}+G_2\left(x\right)\\ P_3\left(y\right)=P_3\left(0\right){\text{e}}^{-{\displaystyle {\int }_0^y\left(r+\lambda +\mu \left(\sigma \right)\right)\text{d}\sigma }}+G_3\left(y\right)\\ P_4\left(x\right)=\left(1-{\text{e}}^{-\lambda x}-\lambda x{\text{e}}^{-\lambda x}\right)P_1\left(0\right){\text{e}}^{-{\displaystyle {\int }_0^x\left(r+\alpha \left(\sigma \right)\right)\text{d}\sigma }}+G_4\left(x\right)\\ P_5\left(y\right)=P_5\left(0\right){\text{e}}^{-{\displaystyle {\int }_0^y\left(r+\mu \left(\sigma \right)\right)\text{d}\sigma }}+\left(1-{\text{e}}^{-\lambda y}\right)P_3\left(0\right){\text{e}}^{-{\displaystyle {\int }_0^y\left(r+\mu \left(\sigma \right)\right)\text{d}\sigma }}+G_5\left(y\right)\end{array}$ (7)

其中

$\begin{array}{l}{G}_1\left(x\right)={\displaystyle {\int }_0^x{g}_1\left(\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^x\left(r+\lambda +\beta +\alpha \left(\sigma \right)\right)\text{d}\sigma }}\text{d}\tau },G_2\left(x\right)={\displaystyle {\int }_0^x\left(g_2\left(\tau \right)+\beta G_1\left(\tau \right)\right){\text{e}}^{-{\displaystyle {\int }_{\tau }^x\left(r+\lambda +\alpha \left(\sigma \right)\right)\text{d}\sigma }}\text{d}\tau }\\ G_3\left(y\right)={\displaystyle {\int }_0^y{g}_3\left(\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^y\left(r+\lambda +\mu \left(\sigma \right)\right)\text{d}\sigma }}\text{d}\tau },G_4\left(x\right)={\displaystyle {\int }_0^x\left(g_4\left(\tau \right)+\lambda G_2\left(\tau \right)\right){\text{e}}^{-{\displaystyle {\int }_{\tau }^x\left(r+\alpha \left(\sigma \right)\right)\text{d}\sigma }}\text{d}\tau }\\ G_5\left(y\right)={\displaystyle {\int }_0^y\left(g_5\left(\tau \right)+\lambda G_3\left(\tau \right)\right){\text{e}}^{-{\displaystyle {\int }_{\tau }^y\left(r+\mu \left(\sigma \right)\right)\text{d}\sigma }}\text{d}\tau }\end{array}$

令 $F_1={\displaystyle {\int }_0^{\infty }{\text{e}}^{-{\displaystyle {\int }_0^x\left(r+\alpha \left(\sigma \right)\right)\text{d}\sigma }}\text{d}x},F_2={\displaystyle {\int }_0^{\infty }{\text{e}}^{-{\displaystyle {\int }_0^y\left(r+\mu \left(\sigma \right)\right)\text{d}\sigma }}\text{d}y},F_3={\displaystyle {\int }_0^{\infty }\mu \left(y\right){\text{e}}^{-{\displaystyle {\int }_0^y\left(r+\lambda +\mu \left(\sigma \right)\right)\text{d}\sigma }}\text{d}y}$

$F_4=\lambda {\displaystyle {\int }_0^{\infty }\alpha \left(x\right){\text{e}}^{-{\displaystyle {\int }_0^x\left(r+\lambda +\alpha \left(\sigma \right)\right)\text{d}\sigma }}\text{d}x},F_5={\displaystyle {\int }_0^{\infty }\alpha \left(x\right){\text{e}}^{-{\displaystyle {\int }_0^x\left(r+\lambda +\alpha \left(\sigma \right)\right)\text{d}\sigma }}\text{d}x}$

$W_0=g_0+{\displaystyle {\int }_0^{\infty }\mu \left(y\right)G_3\left(y\right)\text{d}y},W_1=0,W_2={\displaystyle {\int }_0^{\infty }\alpha \left(x\right)G_2\left(x\right)\text{d}x},W_3={\displaystyle {\int }_0^{\infty }\alpha \left(x\right)G_4\left(x\right)\text{d}x}$

将(7)代入 $\left(r+\eta +\lambda +\beta \right)P_0=g_0+{\displaystyle {\int }_0^{\infty }{P}_3\left(y\right)\mu \left(y\right)\text{d}y}$ 及边界条件(6)中，得到矩阵方程：

$\left[\begin{array}{cccc}r+\eta +\lambda +\beta & 0& \begin{array}{c}-F_3\end{array}& 0\\ -\eta & 1& 0& 0\\ -\lambda -\beta & F_5-F_4& r{F}_2+F_3& r{F}_2-1\\ 0& r{F}_1+F_4+F_5-1& 0& 1\end{array}\right]\left[\begin{array}{c}{P}_0\\ P_1\left(0\right)\\ P_3\left(0\right)\\ P_5\left(0\right)\end{array}\right]=\left[\begin{array}{c}{W}_0\\ W_1\\ W_2\\ W_3\end{array}\right]$,

令 $T=\left[\begin{array}{cccc}r+\eta +\lambda +\beta & 0& \begin{array}{c}-F_3\end{array}& 0\\ -\eta & 1& 0& 0\\ -\lambda -\beta & F_5-F_4& r{F}_2+F_3& r{F}_2-1\\ 0& r{F}_1+F_4+F_5-1& 0& 1\end{array}\right]$

根据T的表达式知道矩阵T是不可约的，而且是按列对角占优的， $\det T\ne 0$。所以方程组 $\left[rI-\left(A+B\right)\right]P=G$ 有唯一解，即 $rI-\left(A+B\right)$ 是满射。由 $rI-\left(A+B\right)$ 是闭的，而且 $D\left(A+B\right)$ 在X中稠密，所以由逆算子定理可知 ${\left[rI-\left(A+B\right)\right]}^{-1}$ 存在并且是线性有界的。因此当 $\left\{r\in C|\mathrm{Re}r>0或r=ia,a\in R\backslash \left\{0\right\}\right\}$ 时，有 $r\in \rho \left(A+B\right)$。

定理7 $r=0$ 是算子A + B的几何重数为1的本征值的充要条件是 $D\left(r\right)=0$。

证明 设 $r\in C$ 是A + B的一个本征值，P是一个本征向量，即 $\left[rI-\left(A+B\right)\right]P=0$。从而这个问题相当于讨论以下方程组非零解的存在性

$\{\begin{array}{l}\left(r+\eta +\lambda +\beta \right)P_0-{\displaystyle {\int }_0^{\infty }{P}_3\left(y\right)\mu \left(y\right)\text{d}y}=0\\ \frac{\text{d}{P}_1\left(x\right)}{\text{d}x}+\left(r+\lambda +\beta \right)P_1\left(x\right)=0\\ \frac{\text{d}{P}_2\left(x\right)}{\text{d}x}+\left(r+\lambda +\alpha \left(x\right)\right)P_2\left(x\right)-\beta P_1\left(x\right)=0\\ \frac{\text{d}{P}_3\left(y\right)}{\text{d}y}+\left(r+\lambda +\mu \left(y\right)\right)P_3\left(y\right)=0\\ \frac{\text{d}{P}_4\left(x\right)}{\text{d}x}+\left(r+\alpha \left(x\right)\right)P_4\left(x\right)-\lambda P_2\left(x\right)=0\\ \frac{\text{d}{P}_5\left(y\right)}{\text{d}y}+\left(r+\mu \left(y\right)\right)P_5\left(y\right)-\lambda P_3\left(y\right)=0\\ P_1\left(0\right)=\eta P_0\\ P_3\left(0\right)=\lambda P_0+{\displaystyle {\int }_0^{\infty }{P}_2\left(0\right)\alpha \left(x\right)\text{d}x}\\ P_5\left(0\right)={\displaystyle {\int }_0^{\infty }{P}_4\left(0\right)\alpha \left(x\right)\text{d}x}\\ P_2\left(0\right)=P_4\left(0\right)=0\end{array}$ (8)

解微分方程组得到

$\begin{array}{l}{P}_1\left(x\right)=P_1\left(0\right){\text{e}}^{-\left(r+\lambda +\beta \right)x}\\ P_2\left(x\right)=\beta P_1\left(0\right){\displaystyle {\int }_0^x{\text{e}}^{-{\displaystyle {\int }_0^{\tau }\left(r+\lambda +\beta \right)\text{d}\xi }}\cdot {\text{e}}^{-{\displaystyle {\int }_{\tau }^x\left(r+\lambda +\alpha \left(\xi \right)\right)\text{d}\xi }}\text{d}\tau }\\ P_3\left(y\right)=P_3\left(0\right){\text{e}}^{-{\displaystyle {\int }_0^y\left(r+\lambda +\mu \left(\xi \right)\right)\text{d}\xi }}\\ P_5\left(y\right)=P_5\left(0\right){\text{e}}^{-{\displaystyle {\int }_0^y\left(r+\mu \left(\xi \right)\right)\text{d}\xi }}+\left(1-{\text{e}}^{-\lambda y}\right)P_3\left(0\right){\text{e}}^{-{\displaystyle {\int }_0^y\left(r+\mu \left(\xi \right)\right)\text{d}\xi }}\end{array}$ (9)

于是得到如下代数方程组

$\{\begin{array}{l}\left(r+\eta +\lambda +\beta \right)P_0-{\displaystyle {\int }_0^{\infty }{\text{e}}^{-{\displaystyle {\int }_0^y\left(r+\lambda +\mu \left(\xi \right)\right)\text{d}\xi }}\text{d}y}{P}_3\left(0\right)=0\\ \eta P_0-P_1\left(0\right)=0\\ P_2\left(0\right)=0\\ \lambda P_0+{\displaystyle {\int }_0^{\infty }\alpha }\left(x\right)\text{d}x{P}_2\left(0\right)-P_3\left(0\right)=0\\ P_4\left(0\right)=0\\ {\displaystyle {\int }_0^{\infty }\alpha }\left(x\right)\text{d}x{P}_4\left(0\right)-P_5\left(0\right)=0\end{array}$ (10)

故(10)有非零解的充分必要条件是系数行列式为零。计算(10)的系数行列式

$D\left(r\right)=\left|\begin{array}{cccccc}r+\eta +\lambda +\beta & 0& 0& {\displaystyle {\int }_0^{\infty }{\text{e}}^{-{\displaystyle {\int }_0^y\left(r+\lambda +\mu \left(\xi \right)\right)\text{d}\xi }}\text{d}y}& 0& 0\\ \eta & -1& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\\ \lambda & 0& {\displaystyle {\int }_0^{\infty }\alpha \left(x\right)\text{d}x}& -1& 0& 0\\ 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& {\displaystyle {\int }_0^{\infty }\alpha \left(x\right)\text{d}x}& -1\end{array}\right|$

当 $D\left(r\right)=0$ 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

$\left(p_0,p_1\left(0\right),p_2\left(0\right),p_3\left(0\right),p_4\left(0\right),p_5\left(0\right)\right):\{\begin{array}{l}{P}_1\left(0\right)=\eta P_0\\ P_2\left(0\right)=0\\ P_3\left(0\right)=\lambda P_0\\ P_4\left(0\right)=0\\ P_5\left(0\right)=0\end{array}$ (11)

所以 $\left(p_0,p_1\left(0\right),p_2\left(0\right),p_3\left(0\right),p_4\left(0\right),p_5\left(0\right)\right)\in D\left(A+B\right)$ 是方程(1)的一个解，从而(10)给出相应于r的A + B的本征函数。令 $\hat{P}=\frac{P}{\Vert P\Vert },\hat{P}=\left({\hat{P}}_1\left(x\right),{\hat{P}}_2\left(x\right),{\hat{P}}_3\left(y\right),{\hat{P}}_4\left(x\right),{\hat{P}}_5\left(y\right)\right)$ 是A + B的非负本征向量，即非负稳定解。其中这个向量是本征值0所对应的。因此0在X中的几何重数是1。

定理8 ${\left(A+B\right)}^{*}$ 的代数重数为1的本征值是0，且 ${\left(1,1,1,1,1,1\right)}^{\text{T}}$ 为0的对应的特征向量。

证明 考虑方程 ${\left(A+B\right)}^{*}Q=0$。由定理3有

$\{\begin{array}{l}\left(\eta +\lambda +\beta \right)Q_0=\eta Q_1\left(0\right)+\beta Q_3\left(0\right)\\ \frac{\text{d}}{\text{d}x}{Q}_1\left(x\right)=\left(\lambda +\beta \right)Q_1\left(x\right)-\beta Q_2\left(x\right)\\ \frac{\text{d}}{\text{d}x}{Q}_2\left(x\right)=\left(\lambda +\alpha \left(x\right)\right)Q_2\left(x\right)-\alpha \left(x\right)Q_3\left(0\right)-\lambda Q_4\left(x\right)\\ \frac{\text{d}}{\text{d}y}{Q}_3\left(y\right)=\left(\lambda +\mu \left(y\right)\right)Q_3\left(y\right)-\mu \left(y\right)Q_0-\lambda Q_5\left(y\right)\\ \frac{\text{d}}{\text{d}x}{Q}_4\left(x\right)=\alpha \left(x\right)Q_4\left(x\right)-\alpha \left(x\right)Q_5\left(0\right)\\ \frac{\text{d}}{\text{d}y}{Q}_5\left(y\right)=\mu \left(y\right)Q_5\left(y\right)-\mu \left(y\right)Q_3\left(0\right)\end{array}$ (12)

解上述方程组有 $Q_0=Q_1\left(x\right)=Q_2\left(x\right)=Q_3\left(y\right)=Q_4\left(x\right)=Q_5\left(y\right)$。

显然 $Q\in D\left({\left(A+B\right)}^{*}\right)$，所以0是 ${\left(A+B\right)}^{*}$ 的本征值，而且几何重数是1，把 $Q=\left(1,1,1,1,1,1\right)$ 代入方程组(12)成立，因此 $Q=\left(1,1,1,1,1,1\right)$ 是0的对应特征向量。

定理9 系统(1)的时间依赖解是强收敛于稳定解的，即 ${\lim }_{t\to \infty }P\left(t,\cdot \right)=\hat{P}$，并且有 $\Vert P\left(t,\cdot \right)-\hat{P}\Vert \le M{\text{e}}^{-\lambda t}$,这里的M是某一个合适的常数。