修理工带休假的两同型部件冷贮备可修系统的指数稳定性

Exponential Stability of a Cold Standby Repairable System with Two Identical Components and a Repairman on Vacation

摘要: 本文假设部件工作时间和修理工空闲时间服从指数分布，部件维修时间和修理工休假时间服从一般分布。把系统方程化为Cauchy问题，通过C₀半群理论得到系统动态解的性质，再分析算子的性质得到算子的本征值，从而得到修理工带休假的两同型部件冷贮备可修系统的指数稳定性。

Abstract: This paper assumes that component working time and repairman idle time obey exponential distribution, component maintenance time and repairman vacation time obey general distribution. The system equation is transformed into the Cauchy problem, the properties of the dynamic solution of the system are obtained by the theory of C₀ semigroups, and then the eigenvalues of the operator are obtained by analyzing the properties of the operator, the exponential stability of the cold standby repairable system with two identical components is obtained.

1. 引言

冷贮备是当主单元出现问题时，贮备单元才开始运行的系统，贮备期间不发生故障，例如文献 [1]。对于冷贮备系统来说，它的性能通常通过系统稳定性来判断例如 [2]。

目前多数文章都假设系统发生问题之后能马上维修来讨论系统的稳定性，例如文献 [3] [4]。但是实际上，当系统故障时可能由于修理工不一定在场，系统需要一段待处理时间。因此将修理工的休假和空闲等状态考虑到可修系统中是有意义的例如 [5]。这类系统大多数采用的是马尔可夫等方法比如文献 [6] [7]。

因此文章在文献 [7] 的基础上讨论修理工带休假的冷贮备可修系统，主要运用C₀半群理论，根据系统动态解的性质和算子性质得到系统的指数稳定性。

2. 系统模型

2.1. 基本假设

回顾文献 [7] 中系统模型：假设部件分别记为M，N，修理工记为V：

M，N的工作时间分布为 $J_1\left(t\right)=1-\exp \left(-\lambda t\right),t\ge 0$ ；

V空闲时间的分布为 $J_2\left(t\right)=1-\exp \left(-\beta t\right),t\ge 0$ ；

M，N修复时间的分布为 $U\left(t\right)={\displaystyle {\int }_0^{t}u\left(y\right)\text{d}y=1-\exp \left(-{\displaystyle {\int }_0^t\mu \left(y\right)\text{d}y}\right)}$ ；

V休假时间的分布为 $H\left(t\right)={\displaystyle {\int }_0^{t}h\left(x\right)\text{d}x}=1-\exp \left(-{\displaystyle {\int }_0^t\alpha \left(x\right)\text{d}x}\right)$。

2.2. 系统的状态

状态0：M工作，N贮备，V空闲。

状态1：M工作，N贮备，V休假。

状态2：M工作，N故障，V休假。

状态3：M在工作，N在修理。

状态4：M，N都待修，V休假。

状态5：M在修理，N待修。

2.3. 系统的符号与方程

X(t)代表t时系统所处状态的一个变量，Y(t)代表t时系统出现问题时所对应的维修时间。

系统数学方程为

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

边界条件如下

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

初始条件为

$P_0\left(0\right)=1$，其余是0 (3)

为了使方程更合理，做以下假设：

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

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

选择合适的状态空间

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

并且定义范数为

$\Vert \stackrel{\to }{P}\Vert =\left|P_0\right|+\Vert P_1\left(x\right)\Vert +\Vert P_{\text{2}}\left(x\right)\Vert +\Vert P_{\text{3}}\left(y\right)\Vert +\Vert P_4\left(x\right)\Vert +\Vert P_5\left(y\right)\Vert$

明显得到 $\left(\text{X},\Vert \cdot \Vert \right)$ 是一个Banach空间。

在X中分别定义算子A，B如下：

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

并且算子A的定义域为

$\begin{array}{l}D\left(A\right)=\{\stackrel{\to }{P}\in \text{X}|\frac{\text{d}{P}_i\left(x\right)}{\text{d}x},\frac{\text{d}{P}_j\left(y\right)}{\text{d}y}\in {\text{L}}^1\left({\text{R}}^{+}\right),P_1\left(0\right)=\beta P_0\left(t\right),\\ \begin{array}{c}{}_{}{}^{}\\ \end{array}{P}_3\left(t,0\right)=\lambda P_0\left(t\right)+{\displaystyle {\int }_0^{\infty }{P}_2\left(t,x\right)\alpha \left(x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }{P}_5\left(t,y\right)\mu \left(y\right)\text{d}y},i=1,2,4;j=3,5\}\end{array}$

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

所以 $D\left(A+B\right)=D\left(A\right)\cap D\left(B\right)=D\left(A\right)$， $D\left(A\right)$ 中 $P_i\left(x\right),P_j\left(y\right)$ 绝对连续，因此系统方程转化成Banach空间X中的Cauchy问题：

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

3. 系统动态解的性质

定义1如果对于每一个 $f\in \text{E}$，存在 $g\in \text{O}$，使 $f\le g$，则称E的子集O在E中共尾。

定义2 [8] 设X是Bananch空间，X上的有界线性算子半群 $T\left(t\right),\left(t\ge 0\right)$ 称为有界线性算子强连续半群，如果

$\underset{t\to 0}{\lim }T\left(t\right)x=x,对一切x\in \text{X}成立$

X上的有界线性算子强连续半群简称为C₀类半群。

定理1 [8] 设X是Bananch空间， $T\left(t\right)$ 是X上一个C₀半群。则存在常数 $\omega \ge 0$ 和 $M\ge 1$，使得 $\Vert T\left(t\right)\Vert \le M{\text{e}}^{\omega t}$ 对于 $t\ge 0$ 成立。

定理2 [8] (Lumer-Phillips)若A是闭稠定耗散算子，且 $1\in \rho \left(A\right)$，则A生成一个C₀半群 $T\left(t\right)$，并且满足 $\Vert T\left(t\right)\Vert \le 1$。

定理3 $D\left(A+B\right)$ 在X中是稠密的。

证明由 [9] 知， $D\left(A\right)$ 在X中是稠密的。而且因为 $D\left(A+B\right)=D\left(A\right)$，从而 $D\left(A+B\right)$ 在X中稠密。

定理4 [10] 算子 $A+B$ 是预解正算子。

定理5 算子 $A+B$ 的对偶算子 ${\left(A+B\right)}^{\ast }$ 有

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

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

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

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

$\begin{array}{l}-{\displaystyle {\int }_0^{\infty }\left(\frac{\text{d}}{\text{d}y}+\mu \left(y\right)\right)P_5\left(y\right)Q_5\left(y\right)\text{d}y}+{\displaystyle {\int }_0^{\infty }\lambda P_3\left(y\right)Q_5\left(y\right)\text{d}y}\\ =\left[-\left(\lambda +\beta \right)Q_0+\beta Q_1\left(0\right)+\lambda Q_3\left(0\right)\right]P_0\\ +{\displaystyle {\int }_0^{\infty }\left[\left(\frac{\text{d}}{\text{d}x}-\lambda -\alpha \left(x\right)\right)Q_1\left(x\right)+\lambda Q_2\left(x\right)\right]P_1\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }\left[\left(\frac{\text{d}}{\text{d}x}-\lambda -\alpha \left(x\right)\right)Q_2\left(x\right)+\alpha \left(x\right)Q_3\left(0\right)+\lambda Q_4\left(x\right)\right]P_2\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }\left[\left(\frac{\text{d}}{\text{d}y}-\lambda -\mu \left(y\right)\right)Q_3\left(y\right)+\mu \left(y\right)Q_0+\lambda Q_5\left(y\right)\right]P_3\left(y\right)\text{d}y}\end{array}$

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

定理6 $A+B$ 生成正的压缩C₀半群T(t)。

证明

$\text{X}=\left\{\stackrel{\to }{P}\in \text{R}\times {\text{L}}^1\left({\text{R}}^{+}\right)\times {\text{L}}^1\left({\text{R}}^{+}\right)|\Vert \stackrel{\to }{P}\Vert =\left|P_0\right|+{\Vert P_i\left(x\right)\Vert }_{{\text{L}}^1\left({\text{R}}^{+}\right)}+{\Vert P_j\left(y\right)\Vert }_{{\text{L}}^1\left({\text{R}}^{+}\right)},i=1,2,4;j=3,5\right\}$

${\text{X}}^{\ast }=\left\{\stackrel{\to }{Q}\in \text{R}\times {\text{L}}^{\infty }\left({\text{R}}^{+}\right)\times {\text{L}}^{\infty }\left({\text{R}}^{+}\right)|\Vert \stackrel{\to }{Q}\Vert =\max \left\{\left|Q_0\right|,{\Vert Q_i\left(x\right)\Vert }_{{\text{L}}^{\infty }\left({\text{R}}^{+}\right)},{\Vert Q_j\left(y\right)\Vert }_{{\text{L}}^{\infty }\left({\text{R}}^{+}\right)}i=1,2,4;j=3,5\right\}\right\}$

${\text{X}}^{\ast }$ 的正锥

${\text{X}}_{+}^{\ast }={\text{X}}^{\ast }\cap \left\{Q=\left(Q_0,Q_1,Q_2,Q_3,Q_4,Q_5\right)|Q_0\ge 0,Q_i\left(x\right),Q_j\left(y\right)\ge 0,i=1,2,4;j=3,5\right\}$

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

$D{\left({\left(A+B\right)}^{\ast }\right)}_{+}={\text{X}}_{+}^{\ast }\cap D\left({\left(A+B\right)}^{\ast }\right)$

从而

$\forall \stackrel{\to }{f}={\left(f_0,f_i\left(x\right),f_j\left(y\right)\right)}^{\text{T}}\in {\text{X}}_{+}^{\ast }$

$\Vert \stackrel{\to }{f}\Vert =\max \left\{\left|f_0\right|,{\Vert f_i\left(x\right)\Vert }_{{\text{L}}^1\left({\text{R}}^{+}\right)},{\Vert f_j\left(y\right)\Vert }_{{\text{L}}^1\left({\text{R}}^{+}\right)}\right\}$

因此

$\Vert \stackrel{\to }{f}\Vert \ge \left|f_0\right|,\Vert \stackrel{\to }{f}\Vert \ge f_i\left(x\right),\Vert \stackrel{\to }{f}\Vert \ge f_j\left(y\right),x\ge 0,y\ge 0$

又因为 $1\left(x\right),1\left(y\right)\in {\text{L}}^{\infty }\left({\text{R}}^{+}\right)$ 且绝对连续，而且 $\frac{\text{d}}{\text{d}x}1\left(x\right),\frac{\text{d}}{\text{d}x}1\left(y\right)\in {\text{L}}^{\infty }(\; R\; +\; )$

对于

$\Vert \stackrel{\to }{g}\Vert =\left(\left|f_0\right|,\Vert \stackrel{\to }{f}\Vert 1\left(x\right),\Vert \stackrel{\to }{f}\Vert 1\left(x\right),\Vert \stackrel{\to }{f}\Vert 1\left(y\right),\Vert \stackrel{\to }{f}\Vert 1\left(x\right),\Vert \stackrel{\to }{f}\Vert 1\left(y\right)\right)\in D{\left({\left(A+B\right)}^{\ast }\right)}_{+}$

$\Vert \stackrel{\to }{f}\Vert -f_0\ge 0,\Vert \stackrel{\to }{f}\Vert 1\left(x\right)-f_i\left(x\right)\ge 0,\Vert \stackrel{\to }{f}\Vert 1\left(y\right)-f_j\left(y\right)\ge 0,x\ge 0,y\ge 0$

有 $\stackrel{\to }{g}-\stackrel{\to }{f}\ge 0$ 也就是 $\stackrel{\to }{f}\le \stackrel{\to }{g}$。因此根据定义1有，在 ${\text{X}}_{+}^{\ast }$ 中任选一个 $\stackrel{\to }{f}$，存在 $\stackrel{\to }{g}\in D{\left({\left(A+B\right)}^{\ast }\right)}_{+}$ 使 $\stackrel{\to }{f}\le \stackrel{\to }{g}$，即 $D{\left({\left(A+B\right)}^{\ast }\right)}_{+}$ 在 ${\text{X}}_{+}^{\ast }$ 中共尾。而且 $A+B$ 是稠定的，再由定理4，因此算子 $A+B$ 生成了正的C₀半群T(t)。

定理7 系统存在非负时间解 $P\left(t,\cdot \right)$，而且是唯一的，并且满足 $\Vert P\left(t,\cdot \right)\Vert =1,\forall t\in \left[0,\infty \right)$。

证明 根据定理6可知，系统存在时间依赖解 $P\left(t,\cdot \right)$，而且唯一非负，还可以表示成

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

$\Vert P\left(t,\cdot \right)\Vert =\Vert T\left(t\right)P_0\Vert =\Vert P_0\Vert =1,\forall t\in \left[0,\infty \right)$。

4. 算子的性质

定义3 如果算子 $A+B$ 是半群 $T\left(t\right)$ 的无穷小生成元，那么增长界

$\omega \left(A+B\right)=\inf \left\{\omega \in \text{R}|\exists M\ge 1,s.t.\forall t\ge 0,\Vert T\left(t\right)\Vert \le M{\text{e}}^{\omega t}\right\}$

定义4 算子 $A+B$ 的谱上界 $s\left(A+B\right)=\inf \left\{\omega \in \text{R}|\left(\omega \text{,}\infty \right)\subseteq \rho \left(A+B\right)\right\}$

定理8 算子 $A+B$ 的增长界 $\omega \left(A+B\right)=0$。

证明

$\{\begin{array}{l}\frac{\text{d}}{\text{d}t}{P}_0\left(t\right)=-\left(\lambda +\beta \right)P_0\left(t\right)+{\displaystyle {\int }_0^{\infty }{P}_3\left(t,y\right)\mu \left(y\right)\text{d}y}\\ \frac{\partial }{\partial t}{P}_1\left(t,x\right)=-\left(\frac{\partial }{\partial x}+\lambda +\alpha \left(x\right)\right)P_1\left(t,x\right)\\ \frac{\partial }{\partial t}{P}_2\left(t,x\right)=-\left(\frac{\partial }{\partial x}+\lambda +\alpha \left(x\right)\right)P_2\left(t,x\right)+\lambda P_1\left(t,x\right)\\ \frac{\partial }{\partial t}{P}_3\left(t,y\right)=-\left(\frac{\partial }{\partial y}+\lambda +\mu \left(y\right)\right)P_3\left(t,y\right)\\ \frac{\partial }{\partial t}{P}_4\left(t,x\right)=-\left(\frac{\partial }{\partial x}+\alpha \left(x\right)\right)P_4\left(t,x\right)+\lambda P_2\left(t,x\right)\\ \frac{\partial }{\partial t}{P}_5\left(t,y\right)=-\left(\frac{\partial }{\partial y}+\mu \left(y\right)\right)P_5\left(t,y\right)+\lambda P_3\left(t,y\right)\end{array}$ (4)

把(4)从0到 $+\infty$ 积分，代入(2)，(3)，同时把每个式子左右分别相加有 $\frac{\text{d}\Vert \stackrel{\to }{P}\Vert }{\text{d}t}=0$。

所以方程组对应的半群是非扩张的，由初始条件 $P_0\left(0\right)=1$ 可得 $\Vert T\left(t\right)\Vert =1$。因此半群的增长界 $\omega \left(A+B\right)=0$。

定理9 算子 $A+B$ 的谱上界 $s\left(A+B\right)=0$。

证明由定理6和 [11] 中的定理2.2，有 $s\left(A+B\right)=\omega \left(A+B\right)=0$。

5. 系统的指数稳定性

定理10 $\left\{r\in \text{C}|\mathrm{Re}r>0或r=ia,a\in \text{R}\left\{\text{0}\right\}\right\}$ 包含于算子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 $\left(rI-\left(A+B\right)\right)P=G$，得到方程组

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

边界条件为

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

解方程组(5)可得

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

其中

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

令

$F_1={\displaystyle {\int }_0^{\infty }{\text{e}}^{-{\displaystyle {\int }_0^x\left(r+\alpha \left(\sigma \right)\right)\text{d}\sigma }}}\text{d}x$

$F_2={\displaystyle {\int }_0^{\infty }{\text{e}}^{-{\displaystyle {\int }_0^y\left(r+\mu \left(\sigma \right)\right)\text{d}\sigma }}}\text{d}y$

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

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

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

$W_0=g_0+{\displaystyle {\int }_0^{\infty }\mu \left(y\right)G_3\left(y\right)\text{d}y}$

$W_1=0$

$W_2={\displaystyle {\int }_0^{\infty }\mu \left(y\right)G_5\left(y\right)\text{d}y}$

$W_3={\displaystyle {\int }_0^{\infty }\alpha \left(x\right)G_4\left(x\right)\text{d}x}$

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

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

令

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

下证矩阵K是非奇异的。

由于

$\begin{array}{l}\left|-\beta \right|+\left|-\lambda \right|=\lambda +\beta <r+\lambda +\beta \\ \left|F_5-F_4\right|+\left|r{F}_1+F_4+F_5-1\right|=1-r{F}_1={\displaystyle {\int }_0^{\infty }\alpha \left(x\right){\text{e}}^{-{\displaystyle {\int }_0^x\left(r+\alpha \left(\sigma \right)\right)\text{d}\sigma }}\text{d}x}<1\\ \left|-F_3\right|=F_3<r{F}_2+F_3\\ \left|r{F}_2-1\right|=1-r{F}_2={\displaystyle {\int }_0^{\infty }\mu \left(y\right){\text{e}}^{-{\displaystyle {\int }_0^y\left(r+\mu \left(\sigma \right)\right)\text{d}\sigma }}\text{d}y}<1\end{array}$

所以矩阵K不可约且按列对角占优， $\det K\ne 0$。从而得到方程组 $\left(rI-\left(A+B\right)\right)P=G$ 有唯一的解，说明 $rI-\left(A+B\right)$ 是满射。又 $rI-\left(A+B\right)$ 是闭的，且有定理3，所以根据逆算子定理， ${\left[rI-\left(A+B\right)\right]}^{-1}$ 是存在的而且是线性有界的。因此 $\left\{r\in \text{C}|\mathrm{Re}r>0或r=ia,a\in \text{R}\left\{\text{0}\right\}\right\}$ 时， $r\in \rho \left(A+B\right)$。

定理11 0是 $A+B$ 的本征值，而且几何重数为1。

证明 K的行列式

$\begin{array}{c}\left|K\right|=\left|\begin{array}{cccc}r+\lambda +\beta & 0& -F_3& 0\\ -\beta & 1& 0& 0\\ -\lambda & F_5-F_4& r{F}_2+F_3& r{F}_2-1\\ 0& r{F}_1+F_4+F_5-1& 0& 1\end{array}\right|\\ =r\left|\begin{array}{cccc}1& F_1& F_2& F_2\\ -\beta & 1& 0& 0\\ -\lambda & F_5-F_4& r{F}_2+F_3& r{F}_2-1\\ 0& r{F}_1+F_4+F_5-1& 0& 1\end{array}\right|\\ =r\beta [r^2{F}_1{F}_2\left(F_1-F_2\right)+F_3\left(F_1{F}_4+F_1{F}_5-F_2{F}_4-F_2{F}_5\right)+F_2\left(F_3+F_4+F_5\right)\\ +r\left(F_1^2{F}_3+F_1{F}_2{F}_4+F_1{F}_2{F}_5+F_2^2-F_1{F}_2{F}_3-F_2^2{F}_4-F_2^2{F}_5\right)]+r\left[\left(r+\lambda \right)F_2+F_3\right]\end{array}$

令 $\left|K\right|=0$，则上式中 $r=0$ 或者

$\begin{array}{l}\beta [r^2{F}_1{F}_2\left(F_1-F_2\right)+F_3\left(F_1{F}_4+F_1{F}_5-F_2{F}_4-F_2{F}_5\right)+F_2\left(F_3+F_4+F_5\right)\\ +r\left(F_1^2{F}_3+F_1{F}_2{F}_4+F_1{F}_2{F}_5+F_2^2-F_1{F}_2{F}_3-F_2^2{F}_4-F_2^2{F}_5\right)]+r\left[\left(r+\lambda \right)F_2+F_3\right]=0\end{array}$

因此0是系统算子的本征值， $P=\left(P_0,P_1\left(x\right),P_2\left(x\right),P_3\left(y\right),P_4\left(x\right),P_5\left(y\right)\right)$ 是对应本向量。令

$\hat{P}=\frac{P}{\Vert P\Vert },\hat{P}=({\hat{P}}_1\left(x\right),{\hat{P}}_2\left(x\right),{\hat{P}}_3\left(y\right),{\hat{P}}_4\left(x\right),{\hat{P}}_5(\; y\; ))$

是A + B本征值所对应的非负的本征向量，而且是系统的非负稳态解，显然0在X中的几何重数为1。

定理12 A + B的本征值是0，代数重数是1，而且 ${\left(1,1,1,1,1,1\right)}^{\text{T}}$ 是0对应的特征向量。

证明 用反证法：想要得到0的代数重数是1，要证明0的代数指数为1。假定0在X中的代数指数是2，那么存在Z使 $\left(A+B\right)Z=\hat{P}$。因此讨论 ${\left(A+B\right)}^{\ast }Q=0$。根据定理5有

$\{\begin{array}{l}-\left(\lambda +\beta \right)Q_0+\beta Q_1\left(0\right)+\lambda Q_3\left(0\right)=0\\ \left(\frac{\text{d}}{\text{d}x}-\lambda -\alpha \left(x\right)\right)Q_1\left(x\right)+\lambda Q_2\left(x\right)+\alpha \left(x\right)Q_3\left(0\right)=0\\ \left(\frac{\text{d}}{\text{d}x}-\lambda -\alpha \left(x\right)\right)Q_2\left(x\right)+\alpha \left(x\right)Q_3\left(0\right)+\lambda Q_4\left(x\right)=0\\ \left(\frac{\text{d}}{\text{d}y}-\lambda -\mu \left(y\right)\right)Q_3\left(y\right)+\mu \left(y\right)Q_0+\lambda Q_5\left(y\right)=0\\ \left(\frac{\text{d}}{\text{d}x}-\alpha \left(x\right)\right)Q_4\left(x\right)+\alpha \left(x\right)Q_5\left(0\right)=0\\ \left(\frac{\text{d}}{\text{d}y}-\mu \left(y\right)\right)Q_5\left(y\right)+\mu \left(y\right)Q_3\left(0\right)=0\end{array}$ (8)

解上述方程组有 $Q_0=Q_1\left(x\right)=Q_2\left(x\right)=Q_3\left(y\right)=Q_4\left(x\right)=Q_5\left(y\right)$，容易验证 $Q\in D\left({\left(A+B\right)}^{\ast }\right)$，而且0为 ${\left(A+B\right)}^{\ast }$ 的本征值，令 $Q^{\ast }=\left(1,1,1,1,1,1\right)$，有 $Q^{\ast }$ 为0对应的本向量，并且几何重数是1，即 ${\left(A+B\right)}^{\ast }{Q}^{\ast }=0$， $\langle \hat{P},Q^{\ast }\rangle =\langle \left(A+B\right)Z,Q^{\ast }\rangle =\langle Z,{\left(A+B\right)}^{\ast }{Q}^{\ast }\rangle =0$，但是 $\langle \hat{P},Q^{\ast }\rangle =1$，存在矛盾，因此0是本征值，而且代数重数为1。

定理13 系统(1)的时间依赖解是有强收敛性的，并且收敛于稳态解，也就是 $\underset{t\to \infty }{\lim }P\left(t,\cdot \right)=\hat{P}$，而且 $\Vert P\left(t,\cdot \right)-\hat{P}\Vert \le M{\text{e}}^{-\lambda t}$，其中M为某个适当的常数。