具有一个冷储备两单元串联可修复系统的指数性

Exponential Stability of a Repairable System with Cold Storage Units in Series

摘要: 本文论述具有一个冷储备串联系统的指数稳定性，运用随机过程理论和增补变量法建立了一个可修复系统的数学模型，利用C₀半群理论研究了系统算子的指数稳定性。

Abstract: In this paper, the exponential stability of a cold storage series system is discussed. A mathematical model of repairable system is established by using stochastic process theory and supplementary variable method. The exponential stability of system operators is studied by using C₀ semigroup theory.

1. 引言

随着科学技术的发展，系统的可靠性、稳定性分析变得越来越重要。从数学的角度，对系统的稳定性进行定量和定性的分析，从而给出系统性能的判断，无论在实际上还是在理论上都具有很重要的意义。近代以来，可靠性理论得到了系统、规范的发展，显现得日益成熟，特别是一系列太空计划的实施，使得可靠性理论得到了更大推广，目前已经发展为一门独立的工程基础研究。

2. 系统模型的建立 [1] [2] [3] [4] [5]

本文研究的是两个单元串联，且第二个单元具有冷储备单元的系统适定性。系统实际上有三个单元组成，分别记为①，②，③，其中③为冷储备单元，为了对系统建立模型，做出如下几组描述：

1) 初始状态处于良好状态；2） 系统遵循先故障先修理原则；3) 储备系统只有系统②故障才工作，储备期间不发生故障；4) 系统修复后完好如初；5) 系统①的故障率为常数 ${\lambda }_1$，修复率为非常数 $u_1\left(x\right)$，且 ${\lambda }_1>0$， $u_1\left(x\right)>0$ ；6) 系统②，③的故障率均为常数 ${\lambda }_2$，修复率均为 $u_2\left(x\right)$，且有 ${\lambda }_2>0$， $u_2\left(x\right)>0$ ；7) 修复率 $u_i\left(x\right)\left(i=1,2\right)$ 为非负可测函数，且 ${\displaystyle {\int }_0^{\infty }{u}_i\left(x\right)}=\infty \left(i=1,2\right)$ ；8) 三个单元寿命均服从一般分布 $F\left(t\right)=1-{\text{e}}^{\lambda t},t\ge 0,\lambda >0$ ；9) 三个单元修复时间服从一般分布 $G\left(t\right)=1-{\text{e}}^{-{\displaystyle {\int }_0^{\infty }u\left(x\right)}},u\left(x\right)>0,t\ge 0$。

以S(t)表示系统在t时刻所处的状态，则系统在t时刻所处的状态可划分以下几种情况：1) $S\left(t\right)=0$，单元①，②，③均完好，系统正常工作；2) $S\left(t\right)=1$， 1 故障在修，②，③完好，系统停止工作；3) $S\left(t\right)=2$，②故障在修，③开始工作，系统工作；4) $S\left(t\right)=3$， 3 故障在修，②正常工作，系统正常工作；5) $S\left(t\right)=4$，②故障在修，③故障待修，系统停止工作；6) $S\left(t\right)=5$，②故障在修，①故障待修，③停止工作，系统停止工作；7) ③故障在修，①故障待修， 2 停止工作，系统停止工作。

可得系统方程组如下：

$\frac{\text{d}}{\text{d}t}{P}_0\left(t\right)=-\left({\lambda }_1+{\lambda }_2\right)P_0\left(t\right)+{\displaystyle {\int }_0^{\infty }{u}_1\left(x\right)P_1\left(t,x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }{u}_2\left(x\right)\left(P_2\left(t,x\right)+P_3\left(t,x\right)\right)\text{d}x}$

$\frac{\partial }{\partial t}{P}_1\left(t,x\right)+\frac{\partial }{\partial x}{P}_1\left(t,x\right)=-u_1\left(x\right)P_1\left(t,x\right)$

$\frac{\partial }{\partial t}{P}_2\left(t,x\right)+\frac{\partial }{\partial x}{P}_2\left(t,x\right)=-\left({\lambda }_1+{\lambda }_2+u_2\left(x\right)\right)P_2\left(t,x\right)$

$\frac{\partial }{\partial t}{P}_3\left(t,x\right)+\frac{\partial }{\partial x}{P}_3\left(t,x\right)=-\left({\lambda }_1+u_2\left(x\right)\right)P_3\left(t,x\right)$

$\frac{\partial }{\partial t}{P}_4\left(t,x\right)+\frac{\partial }{\partial x}{P}_4\left(t,x\right)=-u_2\left(x\right)P_4\left(t,x\right)+{\lambda }_2{P}_2\left(t,x\right)$

$\frac{\partial }{\partial t}{P}_5\left(t,x\right)+\frac{\partial }{\partial x}{P}_5\left(t,x\right)=-u_2\left(x\right)P_5\left(t,x\right)+{\lambda }_1{P}_2\left(t,x\right)$

$\frac{\partial }{\partial t}{P}_6\left(t,x\right)+\frac{\partial }{\partial x}{P}_6\left(t,x\right)=-u_2\left(x\right)P_6\left(t,x\right)+{\lambda }_1{P}_3\left(t,x\right)$

$P_1\left(t,0\right)={\lambda }_1{P}_0\left(t\right)+{\displaystyle {\int }_0^{\infty }\left(P_5\left(t,x\right)+P_6\left(t,x\right)\right)u_2\left(x\right)\text{d}x}$

$P_2\left(t,0\right)={\lambda }_2{P}_0\left(t\right);P_3\left(t,0\right)=P_4\left(t,0\right)=P_5\left(t,0\right)=P_6\left(t,0\right)=0;P_0\left(0\right)=1,P_i\left(0,x\right)=0,\left(i=1,2,3,4,5,6\right)$

将以上问题转化为抽象柯西问题：

取空间 $X=R\times {\left(L^1\left[0,\infty \right)\right)}^2$，对任意 $P=\left(P_0,P_1\left(x\right),\cdots ,P_6\left(x\right)\right)\in X$，

定义范数 $\Vert P\Vert =\Vert P_0\Vert +{\Vert P_1\left(x\right)\Vert }_{L^1\left[0,\infty \right)}+\cdots +{\Vert P_6\left(x\right)\Vert }_{L^1\left[0,\infty \right)}$，则 $\left(X,\Vert .\Vert \right)$ 是一个Banach空间。

引入算子A，B，C及其定义域如下：

$A=\left(\begin{array}{ccc}{A}_1& 0& 0\\ 0& \ddots & 0\\ 0& 0& A_7\end{array}\right)$ 其中， $A_1=-{\lambda }_1-{\lambda }_2$， $A_2=-\frac{\text{d}}{\text{d}x}-u_1\left(x\right)$， $A_3=-\frac{\text{d}}{\text{d}x}-\left({\lambda }_1+{\lambda }_2+u_2(x))\right)$

$A_4=-\frac{\text{d}}{\text{d}x}-\left({\lambda }_1+u_2\left(x\right)\right)$， $A_5=A_6=A_7=-\frac{\text{d}}{\text{d}x}-u_2\left(x\right)$。

$\begin{array}{l}D\left(A\right)=\{P\in X|\frac{\text{d}}{\text{d}x}{p}_i\left(x\right)\in L^1\left[0,\infty \right),p_i\left(x\right)是绝对连续函数\left(i=1,2,3,4,5,6\right)\\ P_1\left(t,0\right)={\lambda }_1{P}_0\left(t\right)+{\displaystyle {\int }_0^{\infty }\left(P_5\left(t,x\right)+P_6\left(t,x\right)\right)u_2\left(x\right)\text{d}x},\\ \begin{array}{c}\\ \end{array}{P}_2\left(t,0\right)={\lambda }_2{P}_0\left(t\right),P_3\left(t,0\right)=P_4\left(t,0\right)=P_5\left(t,0\right)=P_6\left(t,0\right)=0\}\end{array}$

$B=\left[\begin{array}{ccccccc}0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& {\lambda }_2& 0& 0& 0& 0\\ 0& 0& {\lambda }_1& 0& 0& 0& 0\\ 0& 0& 0& {\lambda }_1& 0& 0& 0\end{array}\right]$

$C=\left[\begin{array}{ccccccc}0& {\displaystyle {\int }_0^{\infty }{u}_1\left(x\right)P_1\left(t,x\right)\text{d}x}& {\displaystyle {\int }_0^{\infty }{u}_2\left(x\right)P_2\left(t,x\right)\text{d}x}& {\displaystyle {\int }_0^{\infty }{u}_3\left(x\right)P_3\left(t,x\right)\text{d}x}& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\end{array}\right]$

$D\left(B\right)=D\left(C\right)=X$

则上述问题可表示为Banach空间上的抽象柯西问题：

$\{\begin{array}{l}\frac{\text{d}}{\text{d}t}P\left(t\right)=\left(A+B+C\right)P\left(t\right),t\ge 0\\ P\left(0\right)=\left(1,0,0,0,0,0,0\right)\\ P\left(t\right)={\left(P_0\left(t\right),P_1\left(t,x\right),P_2\left(t,x\right),\cdots ,P_6\left(t,x\right)\right)}^{\text{T}}\end{array}$ (2)

3. 可修复系统动态解的存在唯一性 [1] [6] [7]

命题1：当 $\alpha >0$ 时， $\alpha \in \rho \left(A\right)$，并且 $\Vert {\left(\alpha I-A\right)}^{-1}\Vert <\frac{1}{\alpha }$。

命题2：算子A是闭稠定算子。

命题3：A为耗散算子。

命题4：算子 $A+B+C$ 生成正压缩 $C_0$ 半群。

命题5：抽象柯西问题(2)存在唯一的非负解 $P\left(t,x\right)$ 且满足 $\Vert P\left(t,.\right)\Vert =1$ ； $t\ge 0$。

4. 系统的指数稳定性 [1] [6] [8]

命题1:0是算子 $A+B+C$ 的简单本征值。

证：取 $P=\left(P_0,P_1\left(X\right),\cdots ,P_6\left(X\right)\right)$，考虑 $\left(A+B+C\right)P=0$，其解析形式为

$-\left({\lambda }_1+{\lambda }_2\right)+{\displaystyle {\int }_0^{\infty }{u}_1\left(x\right)P_1\left(x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }\left(u_2\left(x\right)P_2\left(x\right)+u_2\left(x\right)P_3\left(x\right)\right)\text{d}x}=0$。

$-\frac{\text{d}}{\text{d}x}{P}_1\left(x\right)-u_1\left(x\right)P_1\left(x\right)=0$， $-\frac{\text{d}}{\text{d}x}{P}_2\left(x\right)-\left({\lambda }_1+{\lambda }_2+u_2\left(x\right)\right)P_2\left(x\right)=0$，

$-\frac{\text{d}}{\text{d}x}{P}_3\left(x\right)-\left({\lambda }_1+u_2\left(x\right)\right)P_3\left(x\right)=0$， $-\frac{\text{d}}{\text{d}x}{P}_4\left(x\right)-u_2\left(x\right)P_4\left(x\right)+{\lambda }_2{P}_2\left(x\right)=0$，

$-\frac{\text{d}}{\text{d}x}{P}_5\left(x\right)-u_2\left(x\right)P_5\left(x\right)+{\lambda }_1{P}_2\left(x\right)=0$， $-\frac{\text{d}}{\text{d}x}{P}_6\left(x\right)-u_2\left(x\right)P_6\left(x\right)+{\lambda }_1{P}_3\left(x\right)=0$

$P_1\left(0\right)={\lambda }_1{P}_0$， $P_2\left(0\right)={\lambda }_2{P}_0$， $P_3\left(0\right)=P_4\left(0\right)=P_5\left(0\right)=P_6\left(0\right)=0$，解得

$P_1\left(x\right)={\lambda }_1{P}_0{\text{e}}^{-{\displaystyle {\int }_0^x{u}_1\left(\xi \right)\text{d}\xi }}$， $P_2\left(x\right)={\lambda }_2{P}_0{\text{e}}^{-{\displaystyle {\int }_0^x\left({\lambda }_1+{\lambda }_2+u_2\left(\xi \right)\right)\text{d}\xi }}$， $P_3\left(x\right)=0$，

$P_4\left(x\right)={\lambda }_2^2{P}_0{\displaystyle {\int }_0^x{\text{e}}^{-{\displaystyle {\int }_0^t\left({\lambda }_1+{\lambda }_2+u_2\left(\xi \right)\right)\text{d}\xi }}{\text{e}}^{-{\displaystyle {\int }_t^x{u}_2\left(\xi \right)\text{d}\xi }}\text{d}t}$， $P_5\left(x\right)={\lambda }_1{\lambda }_2{P}_0{\displaystyle {\int }_0^x{\text{e}}^{-{\displaystyle {\int }_0^t\left({\lambda }_1+{\lambda }_2+u_2\left(\xi \right)\right)\text{d}\xi }}{\text{e}}^{-{\displaystyle {\int }_t^x{u}_2\left(\xi \right)\text{d}\xi }}\text{d}t}$， $P_6\left(x\right)=0$，

$\begin{array}{c}{\displaystyle {\int }_0^{\infty }{u}_1\left(x\right)P_1\left(x\right)\text{d}x}={\displaystyle {\int }_0^{\infty }{u}_1\left(x\right){\lambda }_1{P}_0{\text{e}}^{-{\displaystyle {\int }_0^x{u}_1\left(\xi \right)\text{d}\xi }}\text{d}x}\\ ={\lambda }_1{P}_0{\displaystyle {\int }_0^{\infty }{u}_1\left(x\right){\text{e}}^{-{\displaystyle {\int }_0^x{u}_1\left(\xi \right)\text{d}\xi }}\text{d}x}\\ ={\lambda }_1{P}_0\left[-{\displaystyle {\int }_0^{\infty }\text{d}\left({\text{e}}^{-{\displaystyle {\int }_0^x{u}_1\left(\xi \right)\text{d}\xi }}\right)}\right]={\lambda }_1{P}_0\end{array}$

同理得， ${\displaystyle {\int }_0^{\infty }\left(u_2\left(x\right)P_2\left(x\right)+u_2\left(x\right)P_3\left(x\right)\right)\text{d}x}={\displaystyle {\int }_0^{\infty }{u}_2\left(x\right){\lambda }_2{P}_0{\text{e}}^{-{\displaystyle {\int }_0^x\left({\lambda }_1+{\lambda }_2+u_2\left(\xi \right)\right)\text{d}\xi }}\text{d}x}=\frac{{\lambda }_2{P}_0}{{\lambda }_1+{\lambda }_2}$

代入 $\left(\frac{{\lambda }_2}{{\lambda }_1+{\lambda }_2}+{\lambda }_1-\left({\lambda }_1+{\lambda }_2\right)\right)P_0=0=\left(\frac{{\lambda }_2}{{\lambda }_1+{\lambda }_2}-{\lambda }_2\right)P_0=\left(\frac{{\lambda }_2\left(1-\left({\lambda }_1+{\lambda }_2\right)\right)}{{\lambda }_1+{\lambda }_2}\right)P_0\ge -\frac{{\lambda }_2}{2}{P}_0$，得 $P_0\ge 0$

易知， $P_i\left(x\right)\in L^1\left[0,\infty \right),i=1,2,\cdots ,6$， $P={\left(P_0,P_1,\cdots ,P_6\right)}^{\text{T}}$，0对应于算子 $A+B+C$ 的本征向量，取 $Q=\left(1,1\left(x\right),\cdots ,1\left(x\right)\right)$，则 $\langle P,Q\rangle =P_0+{\displaystyle {\int }_0^{\infty }{P}_1\left(x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }{P}_2\left(x\right)\text{d}x}+\cdots +{\displaystyle {\int }_0^{\infty }{P}_6\left(x\right)\text{d}x}>0$，且对任意 $P\in D\left(A+B+C\right)$，有 $\langle \left(A+B+C\right)P,Q\rangle =0$，即 $\left(A+B+C\right)\ast Q=0$，即0是算子 $A+B+C$ 的简单本征值。

命题2：若存在正数a使得 $a=\min \left\{C_i,{\lambda }_1,{\lambda }_2\right\}$，则当 $\mathrm{Re}\gamma >-C$ 时， $\gamma \in \rho \left(A\right)$，且 $\Vert {\left(\gamma I-A\right)}^{-1}\Vert \le \frac{1}{\mathrm{Re}\gamma +C}$

证：当 $\mathrm{Re}\gamma >-C$ 时，对任意 $\eta =\left({\eta }_0,{\eta }_1\left(x\right),\cdots ,{\eta }_6\left(x\right)\right)\in X$，考虑 $\left(\gamma I-A\right)P=\eta$，

而 $P=\left(P_0,P_1\left(x\right),\cdots ,P_6\left(x\right)\right)$，则有 $\left(\gamma +{\lambda }_1+{\lambda }_2\right)P_0={\eta }_0$，

$\frac{\text{d}}{\text{d}x}{P}_1\left(x\right)+\left(\gamma +u_1\left(x\right)\right)P_1\left(x\right)={\eta }_1(x)$

$\frac{\text{d}}{\text{d}x}{P}_2\left(x\right)+\left(\gamma +{\lambda }_1+{\lambda }_2+u_2\left(x\right)\right)P_2\left(x\right)={\eta }_2(x)$

$\frac{\text{d}}{\text{d}x}{P}_3\left(x\right)+\left(\gamma +{\lambda }_1+u_2\left(x\right)\right)P_3\left(x\right)={\eta }_3(x)$

$\frac{\text{d}}{\text{d}x}{P}_4\left(x\right)+\left(\gamma +u_2\left(x\right)\right)P_4\left(x\right)={\eta }_4(x)$

$\frac{\text{d}}{\text{d}x}{P}_5\left(x\right)+\left(\gamma +u_2\left(x\right)\right)P_5\left(x\right)={\eta }_5(x)$

$\frac{\text{d}}{\text{d}x}{P}_6\left(x\right)+\left(\gamma +u_2\left(x\right)\right)P_6\left(x\right)={\eta }_6\left(x\right)$ ； $P_1\left(0\right)={\lambda }_1{P}_0$， $P_2\left(0\right)={\lambda }_2{P}_0$， $P_3\left(0\right)=P_4\left(0\right)=P_5\left(0\right)=P_6\left(0\right)=0$。

当 $\mathrm{Re}\gamma >-C$ 时，有 $\gamma \ne -\left({\lambda }_1+{\lambda }_2\right)$，解上述方程组得 $P_0=\frac{{\eta }_0}{\gamma +{\lambda }_1+{\lambda }_2}$

$P_1\left(x\right)=\frac{{\lambda }_1{\eta }_0}{\gamma +{\lambda }_1+{\lambda }_2}{\text{e}}^{-\gamma x-{\displaystyle {\int }_0^x{u}_1\left(\xi \right)\text{d}\xi }}+{\displaystyle {\int }_0^x{\text{e}}^{-\gamma \left(x-t\right)-{\displaystyle {\int }_t^x{u}_1\left(\xi \right)\text{d}\xi }}{\eta }_1\left(t\right)\text{d}t}$

$P_2\left(x\right)=\frac{{\lambda }_2{\eta }_0}{\gamma +{\lambda }_1+{\lambda }_2}{\text{e}}^{-\left(\gamma +{\lambda }_1+{\lambda }_2\right)x-{\displaystyle {\int }_0^x{u}_2\left(\xi \right)\text{d}\xi }}{\displaystyle {\int }_0^x{\text{e}}^{-\left(\gamma +{\lambda }_1+{\lambda }_2\right)\left(x-t\right)-{\displaystyle {\int }_t^x{u}_2\left(\xi \right)\text{d}\xi }}{\eta }_2\left(t\right)\text{d}t}$

$P_3\left(x\right)={\displaystyle {\int }_0^x{\text{e}}^{-\left(\gamma +{\lambda }_1\right)\left(x-t\right)-{\displaystyle {\int }_t^x{u}_2\left(\xi \right)\text{d}\xi }}{\eta }_3\left(t\right)\text{d}t}$ ； $P_4\left(x\right)={\displaystyle {\int }_0^x{\text{e}}^{-\gamma \left(x-t\right)-{\displaystyle {\int }_t^x{u}_2\left(\xi \right)\text{d}\xi }}{\eta }_4\left(t\right)\text{d}t}$

$P_5\left(x\right)={\displaystyle {\int }_0^x{\text{e}}^{-\gamma \left(x-t\right)-{\displaystyle {\int }_t^x{u}_2\left(\xi \right)\text{d}\xi }}{\eta }_5\left(t\right)\text{d}t}$ ； $P_6\left(x\right)={\displaystyle {\int }_0^x{\text{e}}^{-\gamma \left(x-t\right)-{\displaystyle {\int }_t^x{u}_2\left(\xi \right)\text{d}\xi }}{\eta }_6\left(t\right)\text{d}t}$

由于

$\begin{array}{c}\Vert P\Vert =\left|P_0\right|+{\displaystyle {\sum }_{i=1}^6\Vert P_i\Vert }\\ \le \left|\frac{{\eta }_0}{\gamma +{\lambda }_1+{\lambda }_2}\right|+\left|\frac{{\lambda }_1{\eta }_0}{\gamma +{\lambda }_1+{\lambda }_2}\right|{\displaystyle {\int }_0^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)x}\text{d}x}+{\displaystyle {\int }_0^{\infty }\left|{\eta }_1\left(t\right)\right|\text{d}t}{\displaystyle {\int }_t^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)\left(x-t\right)}\text{d}x}\\ +\left|\frac{{\lambda }_2{\eta }_0}{\gamma +{\lambda }_1+{\lambda }_2}\right|{\displaystyle {\int }_0^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)x}\text{d}x}+{\displaystyle {\int }_0^{\infty }\left|{\eta }_2\left(t\right)\right|\text{d}t}{\displaystyle {\int }_t^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)\left(x-t\right)}\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }\left|{\eta }_3\left(t\right)\right|\text{d}t}{\displaystyle {\int }_t^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)\left(x-t\right)}\text{d}x}+{\displaystyle {\int }_0^{\infty }\left|{\eta }_4\left(t\right)\right|\text{d}t}{\displaystyle {\int }_t^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)\left(x-t\right)}\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }\left|{\eta }_5\left(t\right)\right|\text{d}t}{\displaystyle {\int }_t^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)\left(x-t\right)}\text{d}x}+{\displaystyle {\int }_0^{\infty }\left|{\eta }_6\left(t\right)\right|\text{d}t}{\displaystyle {\int }_t^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)\left(x-t\right)}\text{d}x}\end{array}$

$\begin{array}{l}=\left|\frac{{\eta }_0}{\gamma +{\lambda }_1+{\lambda }_2}\right|+\left|\frac{\left({\lambda }_1+{\lambda }_2\right){\eta }_0}{\gamma +{\lambda }_1+{\lambda }_2}\right|{\displaystyle {\int }_0^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)x}\text{d}x}+{\displaystyle {\sum }_{i=1}^6\Vert {\eta }_i\Vert {\displaystyle {\int }_t^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)\left(x-t\right)}\text{d}x}}\\ \le \frac{1}{\mathrm{Re}\gamma +C}\left(\Vert {\eta }_0\Vert +{\displaystyle {\sum }_{i=1}^6\Vert {\eta }_i\Vert }\right)=\frac{1}{\mathrm{Re}\gamma +C}\Vert \eta \Vert \end{array}$

得 $P\le \frac{1}{\mathrm{Re}\gamma +C}\Vert \eta \Vert$。这表明当 $\mathrm{Re}\gamma +C>0$ 时， ${\left(\gamma I-A\right)}^{-1}:X\to x$ 是有界的，所以 $\gamma \in \rho \left(A\right)$，并且 $\Vert {\left(\gamma I-A\right)}^{-1}\Vert \le \frac{1}{\mathrm{Re}\gamma +C}$。由Lumer-Phillips半群生成定理得以下推论：

推论：算子A生成的压缩半群S(t)是指数衰减的，即对任意 $0<\omega <\text{C}$ 有 $\Vert S\left(t\right)\Vert \le {\text{e}}^{-\omega t}$ ； $t\ge 0$

注意到B和C是有限秩算子，所以B和C是紧算子，由算子半群的扰动定理以及算子半群的紧扰动得以下结果：

命题3：算子 $A+B+C$ 生成的压缩 $C_0$ 半群T(t)有以下性质：

1) 对任意 $\gamma \in C$， $\mathrm{Re}\gamma +C>0$， $\gamma \in \sigma \left(A\right)$ 的充要条件是 $D\left(r\right)=0$ ；

2) 设 ${\gamma }_0=0$，对任意 ${\gamma }_k\in \sigma \left(A\right)\cap \left\{\gamma \in C|\mathrm{Re}\gamma \ge -C,D\left(r\right)=0\right\}$， ${\gamma }_k\ne {\gamma }_0$， $k=0,1,2,\cdots ,6$，其中 ${\gamma }_k$ 按严格实部递减排序， $\mathrm{Re}{\gamma }_{\left(k+1\right)}\le \mathrm{Re}{\gamma }_k$， $k=1,2,\cdots ,6$，即 ${\gamma }_0=0$ 是严格占优本征值。

3) 设 $\hat{P}=\left({\hat{P}}_0,{\hat{P}}_1\left(x\right),\cdots ,{\hat{P}}_6\left(x\right)\right)$ 是系统的稳态解，满足 $\langle \hat{P},Q\rangle =1$，设 $\mathrm{Re}{\gamma }_1<\omega <{\gamma }_0$，那么对任意的 $P\in X$， $Q=\left(1,1,1,1,1,1,1\right)$ 有 $\Vert T\left(t\right)P-\langle P,Q\rangle \hat{P}\Vert \le {\text{e}}^{-\omega t}\Vert P\Vert$， $t\ge 0$。

证：1) 当 $\mathrm{Re}\gamma >-C$ 时，由前一命题， $\gamma \in \rho \left(A\right)$，则 $\left(\gamma I-\left(A+B+C\right)\right)=\left(\gamma -A\right)\left(I-R\left(\gamma ,A\right)\left(B+C\right)\right)$，注意到B和C是有限秩算子，那么 $R\left(\gamma ,A\right)\left(B+C\right)$ 是紧算子，从而 $\gamma \in \rho \left(A+B+C\right)$ 充要条件是：1是 $R\left(\gamma ,A\right)\left(B+C\right)$ 的本征值，因此对 $\mathrm{Re}\gamma +C>0$， $\gamma \in \rho \left(A\right)$ 的充要条件是 $D\left(r\right)=0$。

2) 当 $\mathrm{Re}\gamma >-C$ 时， $D\left(r\right)$ 是解析函数，至多有可数个零点，设 ${\gamma }_0=0$，对其他本征值按实部递减排序 $\mathrm{Re}{\gamma }_{\left(k+1\right)}\le \mathrm{Re}{\gamma }_k$， $k=0,1,2,\cdots ,6$，则 ${\gamma }_k\in \sigma \left(A\right)\cap \left\{\gamma \in C|\mathrm{Re}\gamma \ge -C,D\left(r\right)=0\right\}$， ${\gamma }_k\ne {\gamma }_0$，由本征值离散及本节命题1得 $\mathrm{Re}{\gamma }_k\le \mathrm{Re}{\gamma }_0$， $k=1,2,\cdots ,6$，因 ${\gamma }_0$ 对应的本征函数是正的，所以 ${\gamma }_0=0$ 是严格占优本征值。

3) 由半群扰动定理，紧扰动不改变半群的本质谱界，算子 $A+B+C$ 生成的半群T(t)与算子A生成的半群S(t)有同样本质谱界，即T(t)本质谱界 $\omega \left(A+B+C\right)\le {\omega }_0\left(A\right)$。设 $\hat{P}=\left({\hat{P}}_0,{\hat{P}}_1\left(x\right),\cdots ,{\hat{P}}_6\left(x\right)\right)$ 是系统稳态解， $\mathrm{Re}{\gamma }_1<\omega <{\gamma }_0$，则由算子半群展开定理可得对任意 $P\in X$， $Q=\left(1,1,1,1,1,1,1\right)$ 有 $\Vert T\left(t\right)P-\langle P,Q\rangle \hat{P}\Vert \le {\text{e}}^{-\omega t}\Vert P\Vert$， $t\ge 0$。

综上，在一定条件下，系统的动态解以指数形式收敛于系统的稳态解。