﻿ 具有一个冷储备两单元串联可修复系统的指数性

# 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

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

2. 系统模型的建立 [1] [2] [3] [4] [5]

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 ${u}_{i}\left(x\right)\text{\hspace{0.17em}}\left(i=1,2\right)$ 为非负可测函数，且 ${\int }_{0}^{\infty }{u}_{i}\left(x\right)=\infty \text{\hspace{0.17em}}\left(i=1,2\right)$ ；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 $G\left(t\right)=1-{\text{e}}^{-{\int }_{0}^{\infty }u\left(x\right)},u\left(x\right)>0,t\ge 0$

$\frac{\text{d}}{\text{d}t}{P}_{0}\left(t\right)=-\left({\lambda }_{1}+{\lambda }_{2}\right){P}_{0}\left(t\right)+{\int }_{0}^{\infty }{u}_{1}\left(x\right){P}_{1}\left(t,x\right)\text{d}x+{\int }_{0}^{\infty }{u}_{2}\left(x\right)\left({P}_{2}\left(t,x\right)+{P}_{3}\left(t,x\right)\right)\text{d}x$

$\frac{\partial }{\partial t}{P}_{1}\left(t,x\right)+\frac{\partial }{\partial x}{P}_{1}\left(t,x\right)=-{u}_{1}\left(x\right){P}_{1}\left(t,x\right)$

$\frac{\partial }{\partial t}{P}_{2}\left(t,x\right)+\frac{\partial }{\partial x}{P}_{2}\left(t,x\right)=-\left({\lambda }_{1}+{\lambda }_{2}+{u}_{2}\left(x\right)\right){P}_{2}\left(t,x\right)$

$\frac{\partial }{\partial t}{P}_{3}\left(t,x\right)+\frac{\partial }{\partial x}{P}_{3}\left(t,x\right)=-\left({\lambda }_{1}+{u}_{2}\left(x\right)\right){P}_{3}\left(t,x\right)$

$\frac{\partial }{\partial t}{P}_{4}\left(t,x\right)+\frac{\partial }{\partial x}{P}_{4}\left(t,x\right)=-{u}_{2}\left(x\right){P}_{4}\left(t,x\right)+{\lambda }_{2}{P}_{2}\left(t,x\right)$

$\frac{\partial }{\partial t}{P}_{5}\left(t,x\right)+\frac{\partial }{\partial x}{P}_{5}\left(t,x\right)=-{u}_{2}\left(x\right){P}_{5}\left(t,x\right)+{\lambda }_{1}{P}_{2}\left(t,x\right)$

$\frac{\partial }{\partial t}{P}_{6}\left(t,x\right)+\frac{\partial }{\partial x}{P}_{6}\left(t,x\right)=-{u}_{2}\left(x\right){P}_{6}\left(t,x\right)+{\lambda }_{1}{P}_{3}\left(t,x\right)$

${P}_{1}\left(t,0\right)={\lambda }_{1}{P}_{0}\left(t\right)+{\int }_{0}^{\infty }\left({P}_{5}\left(t,x\right)+{P}_{6}\left(t,x\right)\right){u}_{2}\left(x\right)\text{d}x$

${P}_{2}\left(t,0\right)={\lambda }_{2}{P}_{0}\left(t\right);\text{\hspace{0.17em}}{P}_{3}\left(t,0\right)={P}_{4}\left(t,0\right)={P}_{5}\left(t,0\right)={P}_{6}\left(t,0\right)=0;\text{\hspace{0.17em}}{P}_{0}\left(0\right)=1,{P}_{i}\left(0,x\right)=0,\left(i=1,2,3,4,5,6\right)$

$A=\left(\begin{array}{ccc}{A}_{1}& 0& 0\\ 0& \ddots & 0\\ 0& 0& {A}_{7}\end{array}\right)$ 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}\left(x\right)\right)\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)$

$\begin{array}{l}D\left(A\right)=\left\{P\in X|\frac{\text{d}}{\text{d}x}{p}_{i}\left(x\right)\in {L}^{1}\left[0,\infty \right),{p}_{i}\left(x\right)是绝对连续函数\left(i=1,2,3,4,5,6\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{P}_{1}\left(t,0\right)={\lambda }_{1}{P}_{0}\left(t\right)+{\int }_{0}^{\infty }\left({P}_{5}\left(t,x\right)+{P}_{6}\left(t,x\right)\right){u}_{2}\left(x\right)\text{d}x,\\ \begin{array}{c}\text{ }\\ \text{ }\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{P}_{2}\left(t,0\right)={\lambda }_{2}{P}_{0}\left(t\right),{P}_{3}\left(t,0\right)={P}_{4}\left(t,0\right)={P}_{5}\left(t,0\right)={P}_{6}\left(t,0\right)=0\right\}\end{array}$

$B=\left[\begin{array}{ccccccc}0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& {\lambda }_{2}& 0& 0& 0& 0\\ 0& 0& {\lambda }_{1}& 0& 0& 0& 0\\ 0& 0& 0& {\lambda }_{1}& 0& 0& 0\end{array}\right]$

$C=\left[\begin{array}{ccccccc}0& {\int }_{0}^{\infty }{u}_{1}\left(x\right){P}_{1}\left(t,x\right)\text{d}x& {\int }_{0}^{\infty }{u}_{2}\left(x\right){P}_{2}\left(t,x\right)\text{d}x& {\int }_{0}^{\infty }{u}_{3}\left(x\right){P}_{3}\left(t,x\right)\text{d}x& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\end{array}\right]$

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

$\left\{\begin{array}{l}\frac{\text{d}}{\text{d}t}P\left(t\right)=\left(A+B+C\right)P\left(t\right),t\ge 0\\ P\left(0\right)=\left(1,0,0,0,0,0,0\right)\\ P\left(t\right)={\left({P}_{0}\left(t\right),{P}_{1}\left(t,x\right),{P}_{2}\left(t,x\right),\cdots ,{P}_{6}\left(t,x\right)\right)}^{\text{T}}\end{array}$ (2)

3. 可修复系统动态解的存在唯一性 [1] [6] [7]

4. 系统的指数稳定性 [1] [6] [8]

$-\left({\lambda }_{1}+{\lambda }_{2}\right)+{\int }_{0}^{\infty }{u}_{1}\left(x\right){P}_{1}\left(x\right)\text{d}x+{\int }_{0}^{\infty }\left({u}_{2}\left(x\right){P}_{2}\left(x\right)+{u}_{2}\left(x\right){P}_{3}\left(x\right)\right)\text{d}x=0$

$-\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}}^{-{\int }_{0}^{x}{u}_{1}\left(\xi \right)\text{d}\xi }$${P}_{2}\left(x\right)={\lambda }_{2}{P}_{0}{\text{e}}^{-{\int }_{0}^{x}\left({\lambda }_{1}+{\lambda }_{2}+{u}_{2}\left(\xi \right)\right)\text{d}\xi }$${P}_{3}\left(x\right)=0$

${P}_{4}\left(x\right)={\lambda }_{2}^{2}{P}_{0}{\int }_{0}^{x}{\text{e}}^{-{\int }_{0}^{t}\left({\lambda }_{1}+{\lambda }_{2}+{u}_{2}\left(\xi \right)\right)\text{d}\xi }{\text{e}}^{-{\int }_{t}^{x}{u}_{2}\left(\xi \right)\text{d}\xi }\text{d}t$${P}_{5}\left(x\right)={\lambda }_{1}{\lambda }_{2}{P}_{0}{\int }_{0}^{x}{\text{e}}^{-{\int }_{0}^{t}\left({\lambda }_{1}+{\lambda }_{2}+{u}_{2}\left(\xi \right)\right)\text{d}\xi }{\text{e}}^{-{\int }_{t}^{x}{u}_{2}\left(\xi \right)\text{d}\xi }\text{d}t$${P}_{6}\left(x\right)=0$

$\begin{array}{c}{\int }_{0}^{\infty }{u}_{1}\left(x\right){P}_{1}\left(x\right)\text{d}x={\int }_{0}^{\infty }{u}_{1}\left(x\right){\lambda }_{1}{P}_{0}{\text{e}}^{-{\int }_{0}^{x}{u}_{1}\left(\xi \right)\text{d}\xi }\text{d}x\\ ={\lambda }_{1}{P}_{0}{\int }_{0}^{\infty }{u}_{1}\left(x\right){\text{e}}^{-{\int }_{0}^{x}{u}_{1}\left(\xi \right)\text{d}\xi }\text{d}x\\ ={\lambda }_{1}{P}_{0}\left[-{\int }_{0}^{\infty }\text{d}\left({\text{e}}^{-{\int }_{0}^{x}{u}_{1}\left(\xi \right)\text{d}\xi }\right)\right]={\lambda }_{1}{P}_{0}\end{array}$

$P=\left({P}_{0},{P}_{1}\left(x\right),\cdots ,{P}_{6}\left(x\right)\right)$，则有 $\left(\gamma +{\lambda }_{1}+{\lambda }_{2}\right){P}_{0}={\eta }_{0}$

$\frac{\text{d}}{\text{d}x}{P}_{1}\left(x\right)+\left(\gamma +{u}_{1}\left(x\right)\right){P}_{1}\left(x\right)={\eta }_{1}\left(x\right)$

$\frac{\text{d}}{\text{d}x}{P}_{2}\left(x\right)+\left(\gamma +{\lambda }_{1}+{\lambda }_{2}+{u}_{2}\left(x\right)\right){P}_{2}\left(x\right)={\eta }_{2}\left(x\right)$

$\frac{\text{d}}{\text{d}x}{P}_{3}\left(x\right)+\left(\gamma +{\lambda }_{1}+{u}_{2}\left(x\right)\right){P}_{3}\left(x\right)={\eta }_{3}\left(x\right)$

$\frac{\text{d}}{\text{d}x}{P}_{4}\left(x\right)+\left(\gamma +{u}_{2}\left(x\right)\right){P}_{4}\left(x\right)={\eta }_{4}\left(x\right)$

$\frac{\text{d}}{\text{d}x}{P}_{5}\left(x\right)+\left(\gamma +{u}_{2}\left(x\right)\right){P}_{5}\left(x\right)={\eta }_{5}\left(x\right)$

$\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$

$\mathrm{Re}\gamma >-C$ 时，有 $\gamma \ne -\left({\lambda }_{1}+{\lambda }_{2}\right)$，解上述方程组得 ${P}_{0}=\frac{{\eta }_{0}}{\gamma +{\lambda }_{1}+{\lambda }_{2}}$

${P}_{1}\left(x\right)=\frac{{\lambda }_{1}{\eta }_{0}}{\gamma +{\lambda }_{1}+{\lambda }_{2}}{\text{e}}^{-\gamma x-{\int }_{0}^{x}{u}_{1}\left(\xi \right)\text{d}\xi }+{\int }_{0}^{x}{\text{e}}^{-\gamma \left(x-t\right)-{\int }_{t}^{x}{u}_{1}\left(\xi \right)\text{d}\xi }{\eta }_{1}\left(t\right)\text{d}t$

${P}_{2}\left(x\right)=\frac{{\lambda }_{2}{\eta }_{0}}{\gamma +{\lambda }_{1}+{\lambda }_{2}}{\text{e}}^{-\left(\gamma +{\lambda }_{1}+{\lambda }_{2}\right)x-{\int }_{0}^{x}{u}_{2}\left(\xi \right)\text{d}\xi }{\int }_{0}^{x}{\text{e}}^{-\left(\gamma +{\lambda }_{1}+{\lambda }_{2}\right)\left(x-t\right)-{\int }_{t}^{x}{u}_{2}\left(\xi \right)\text{d}\xi }{\eta }_{2}\left(t\right)\text{d}t$

${P}_{3}\left(x\right)={\int }_{0}^{x}{\text{e}}^{-\left(\gamma +{\lambda }_{1}\right)\left(x-t\right)-{\int }_{t}^{x}{u}_{2}\left(\xi \right)\text{d}\xi }{\eta }_{3}\left(t\right)\text{d}t$${P}_{4}\left(x\right)={\int }_{0}^{x}{\text{e}}^{-\gamma \left(x-t\right)-{\int }_{t}^{x}{u}_{2}\left(\xi \right)\text{d}\xi }{\eta }_{4}\left(t\right)\text{d}t$

${P}_{5}\left(x\right)={\int }_{0}^{x}{\text{e}}^{-\gamma \left(x-t\right)-{\int }_{t}^{x}{u}_{2}\left(\xi \right)\text{d}\xi }{\eta }_{5}\left(t\right)\text{d}t$${P}_{6}\left(x\right)={\int }_{0}^{x}{\text{e}}^{-\gamma \left(x-t\right)-{\int }_{t}^{x}{u}_{2}\left(\xi \right)\text{d}\xi }{\eta }_{6}\left(t\right)\text{d}t$

$\begin{array}{c}‖P‖=|{P}_{0}|+{\sum }_{i=1}^{6}‖{P}_{i}‖\\ \le |\frac{{\eta }_{0}}{\gamma +{\lambda }_{1}+{\lambda }_{2}}|+|\frac{{\lambda }_{1}{\eta }_{0}}{\gamma +{\lambda }_{1}+{\lambda }_{2}}|{\int }_{0}^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)x}\text{d}x+{\int }_{0}^{\infty }|{\eta }_{1}\left(t\right)|\text{d}t{\int }_{t}^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)\left(x-t\right)}\text{d}x\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+|\frac{{\lambda }_{2}{\eta }_{0}}{\gamma +{\lambda }_{1}+{\lambda }_{2}}|{\int }_{0}^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)x}\text{d}x+{\int }_{0}^{\infty }|{\eta }_{2}\left(t\right)|\text{d}t{\int }_{t}^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)\left(x-t\right)}\text{d}x\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\int }_{0}^{\infty }|{\eta }_{3}\left(t\right)|\text{d}t{\int }_{t}^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)\left(x-t\right)}\text{d}x+{\int }_{0}^{\infty }|{\eta }_{4}\left(t\right)|\text{d}t{\int }_{t}^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)\left(x-t\right)}\text{d}x\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\int }_{0}^{\infty }|{\eta }_{5}\left(t\right)|\text{d}t{\int }_{t}^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)\left(x-t\right)}\text{d}x+{\int }_{0}^{\infty }|{\eta }_{6}\left(t\right)|\text{d}t{\int }_{t}^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)\left(x-t\right)}\text{d}x\end{array}$

$\begin{array}{l}=|\frac{{\eta }_{0}}{\gamma +{\lambda }_{1}+{\lambda }_{2}}|+|\frac{\left({\lambda }_{1}+{\lambda }_{2}\right){\eta }_{0}}{\gamma +{\lambda }_{1}+{\lambda }_{2}}|{\int }_{0}^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)x}\text{d}x+{\sum }_{i=1}^{6}‖{\eta }_{i}‖{\int }_{t}^{\infty }{\text{e}}^{-\left(\mathrm{Re}\gamma +C\right)\left(x-t\right)}\text{d}x\\ \le \frac{1}{\mathrm{Re}\gamma +C}\left(‖{\eta }_{0}‖+{\sum }_{i=1}^{6}‖{\eta }_{i}‖\right)=\frac{1}{\mathrm{Re}\gamma +C}‖\eta ‖\end{array}$

$P\le \frac{1}{\mathrm{Re}\gamma +C}‖\eta ‖$。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 $‖{\left(\gamma I-A\right)}^{-1}‖\le \frac{1}{\mathrm{Re}\gamma +C}$。由Lumer-Phillips半群生成定理得以下推论：

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 $\stackrel{^}{P}=\left({\stackrel{^}{P}}_{0},{\stackrel{^}{P}}_{1}\left(x\right),\cdots ,{\stackrel{^}{P}}_{6}\left(x\right)\right)$ 是系统的稳态解，满足 $〈\stackrel{^}{P},Q〉=1$，设 $\mathrm{Re}{\gamma }_{1}<\omega <{\gamma }_{0}$，那么对任意的 $P\in X$$Q=\left(1,1,1,1,1,1,1\right)$$‖T\left(t\right)P-〈P,Q〉\stackrel{^}{P}‖\le {\text{e}}^{-\omega t}‖P‖$$t\ge 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 $\stackrel{^}{P}=\left({\stackrel{^}{P}}_{0},{\stackrel{^}{P}}_{1}\left(x\right),\cdots ,{\stackrel{^}{P}}_{6}\left(x\right)\right)$ 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$$Q=\left(1,1,1,1,1,1,1\right)$$‖T\left(t\right)P-〈P,Q〉\stackrel{^}{P}‖\le {\text{e}}^{-\omega t}‖P‖$$t\ge 0$

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