具有优化调整状态的供应链系统算子的性质

Properties of the System Operator for the Supply Chain System with State of Optimal Adjustment

摘要: 研究了一类具有优化调整状态的供应链系统。通过选取空间和定义算子将模型方程转化成抽象柯西问题，运用C₀半群理论，证明了系统算子是稠密的预解正算子，得出了系统算子的共轭算子及其定义域，并证明了系统算子的增长界为0。最后运用预解正算子中共尾的概念及相关理论证明了系统算子的谱上界也是0。

Abstract: The paper presents a supply chain system with state of optimal adjustment. By choosing space and defining operator of this system, we transfer this model into an abstract Cauchy problem. Using C₀ semigroup theory, we first prove the system operator is a densely defined resolvent positive op-erator. We obtain the adjoint operator of the system operator and its domain. Furthermore, we prove that 0 is the growth bound of the system operator. Finally, we show that 0 is also the upper spectral bound of the system operator using the concept of cofinal and relative theory.

1. 引言

在经济全球化迅速发展的今天，供应链得到了越来越多企业的重视，供应链战略成为企业战略的一个主要方面。“真正的合作不是企业与企业之间的合作，而是供应链与供应链之间的合作”，这句话高度概括了供应链的主要性。成功的供应链管理能够协调并整合供应链中所有的活动，最终成为无缝衔接的一体化过程，不仅能满足客户的需要，还能降低成本，实现利润。所以此系统的管理和优化受到越来越多人的关注。辛玉红等人在文献 [1] 中把整个供应链系统作为研究对象，运用增补变量法建立了供应链系统的可靠性模型。因为企业所处的状态不是一成不变的，所以在某个时间段，它会从其中一种状态转移到另一种状态，在此期间造成的错误会随着时间的积累导致整个系统瘫痪。为了减少这类错误的发生，辛玉红等人在文献 [2] 中又对先前的供应链系统进行了评估，优化。文献 [3] 研究了一类具有优化调整状态的供应链系统的稳定性分析。

本文研究了一类边界带积分的供应链系统，在前人的方法及结论的基础上，首先运用预解正算子理论，证明了系统算子是稠密的预解正算子，并讨论了系统算子的共轭算子及其定义域，然后运用相关理论证明了系统算子的增长界为0，最后根据共尾定理证明系统算子的谱上界也是零。

2. 模型介绍

模型假设

1) 供应链系统的故障是状态型的，系统故障后得到及时的修理；

2) 企业可能失效的发生时间服从参数为 ${\mu }_0$ 的指数分布，其分布函数 $F_0\left(x\right)=1-{\text{e}}^{-{\mu }_{0}x}$ ；

3) 若系统进入可能失效状态后，在 $T$ 时间内发生失效，则在失效的瞬间立即进行维修，假设系统的维修时间服从一般分布，其分布函数为 $G_i\left(x\right)=1-{\text{e}}^{-{\displaystyle {\int }_0^x{\mu }_i\left(t\right)\text{d}t}}$ ；

4) 若系统进入可能失效状态后，在 $T$ 时间内未发生失效，此时对系统进行优化调整操作，假设优化调整操作的完成时间服从一般分布，其分布函数记为 $G_N\left(x\right)=1-{\text{e}}^{-{\displaystyle {\int }_0^x{\mu }_N\left(t\right)\text{d}t}}$ 。

根据模型假设，我们可以得到系统的状态转移图，如图1。

其中，图中的各个状态的含义如下：

状态0：企业正常生产状态；

状态1：企业处于故障状态；

状态2：企业处于产品积压状态；

状态3：企业处于原材料供应不足状态；

状态4：企业处于产品积压且原材料不足状态；

状态M：企业处于可能失效状态；

状态N：企业进入优化调整状态。

Figure 1. System state transition diagram

图1. 系统的状态转移图

利用增补变量法，此模型可以由下面的微积分方程组描述：

$\frac{\text{d}{P}_0\left(t\right)}{\text{d}t}=-{\mu }_0{P}_0\left(t\right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\displaystyle {\int }_0^{\infty }{\mu }_i\left(x\right)P_i\left(x,t\right)\text{d}x}}+{\displaystyle {\int }_0^{\infty }{\mu }_N\left(x\right)P_N\left(x,t\right)\text{d}x}$ (1)

$\frac{\partial P_i\left(x,t\right)}{\partial t}+\frac{\partial P_i\left(x,t\right)}{\partial x}=-{\mu }_i\left(x\right)P_i\left(x,t\right),i=1,2,3,4$ (2)

$\frac{\partial P_M\left(x,t\right)}{\partial t}+\frac{\partial P_M\left(x,t\right)}{\partial x}=-\left({\mu }_M\left(x\right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(x\right)}\right)P_M\left(x,t\right)$ (3)

$\frac{\partial P_N\left(x,t\right)}{\partial t}+\frac{\partial P_N\left(x,t\right)}{\partial x}=-{\mu }_N\left(x\right)P_N\left(x,t\right)$ (4)

边值条件为：

$P_i\left(0,t\right)={\displaystyle {\int }_0^{\infty }{\mu }_{M_i}\left(x\right)P_M\left(x,t\right)\text{d}x},i=1,2,3,4$ (5)

$P_M\left(0,t\right)={\mu }_0{P}_0\left(t\right)$ (6)

$P_N\left(0,t\right)={\displaystyle {\int }_0^{\infty }{\mu }_M\left(x\right)P_M\left(x,t\right)\text{d}x}$ (7)

初值条件：

$P_0\left(0\right)=1,P_i\left(x,0\right)=P_M\left(x,0\right)=P_N\left(x,0\right)=\text{0},i=1,2,3,4$ (8)

其中， $\left(x,t\right)\in {\text{R}}_{+}\times {\text{R}}_{+}$ ， $P_0\left(t\right)$ 表示t时刻系统处在状态0的概率； $P_j\left(x,t\right)$ 表示t时刻企业处在状态j，在该状态已经驻留了 $x$ 时间，且在时间段 $\left(x,x+\text{d}x\right)$ 离开状态j的概率， $\left(j=1,2,3,4,M,N\right)$ ； ${\mu }_0$ 表示状态0到状态M的可能失效率；系统进入状态M的瞬间开始计时，以时间段T为临界值， ${\mu }_{M_i}\left(x\right)$ 表示系统在T时间内发生失效且没有立即进行维修，此时系统进入i状态的失效率； ${\mu }_M\left(x\right)$ 表示系统在T时间内未发生失效，在M状态驻留时间T后立即进入N状态的优化率； ${\mu }_i\left(x\right)$ 表示系统从状态i到状态0的修复率； ${\mu }_N\left(x\right)$ 表示系统从状态N到状态0的调整率。且满足

1) ${\mu }_i\left(x\right)<\infty$ ， ${\mu }_M(x)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(x\right)}<\infty$ ， ${\mu }_N\left(x\right)<\infty$

2) ${\displaystyle {\int }_0^{\infty }{\mu }_i\left(x\right)\text{d}x}=\infty$ ， ${\displaystyle {\int }_0^{\infty }\left[{\mu }_M\left(x\right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}\mu {}_{M_i}\left(x\right)}\right]\text{d}x}=\infty$ ， ${\displaystyle {\int }_0^{\infty }{\mu }_N}\left(x\right)\text{d}x=\infty$

取状态空间 $X=\left\{P={\left(P_0,P_1,P_2,P_3,P_4,P_M,P_N\right)}^{\text{T}}|P_0\in \text{R};P_j\in L^1\left({\text{R}}_{+}\right),j=1,2,3,4,M,N\right\}$ 。范数定义为 $\Vert P\Vert =\left|P_0\right|+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\Vert P_i\left(x\right)\Vert }_{L^1\left({\text{R}}_{+}\right)}+{\Vert P_M\left(x\right)\Vert }_{L^1\left({\text{R}}_{+}\right)}}+{\Vert P_N\left(x\right)\Vert }_{L^1\left({\text{R}}_{+}\right)}$ 。显然， $\left(X,\Vert \cdot \Vert \right)$ 是一个Banach空间。

在 $X$ 上定义算子 $A,B$ 及它们的定义域。设

$\begin{array}{l}A=\text{diag}(-{\mu }_0,-\frac{\text{d}}{\text{d}x}-{\mu }_{\text{1}}\left(x\right),-\frac{\text{d}}{\text{d}x}-{\mu }_{\text{2}}\left(x\right),-\frac{\text{d}}{\text{d}x}-{\mu }_{\text{3}}\left(x\right),-\frac{\text{d}}{\text{d}x}-{\mu }_{\text{4}}\left(x\right),\\ -\frac{\text{d}}{\text{d}x}-{\mu }_M\left(x\right)-{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(x\right)},-\frac{\text{d}}{\text{d}x}-{\mu }_N(x))\end{array}$

$\begin{array}{l}D\left(A\right)=\{P\in X|P_j\left(x\right)绝对连续,\frac{\text{d}{P}_j\left(x\right)}{\text{d}x}\in L^1\left({\text{R}}_{+}\right),j=1,2,3,4,M,N,且P_i\left(0,t\right)\end{array}$ $={\displaystyle {\int }_0^{\infty }{\mu }_{M_i}\left(x\right)P_M\left(x\right)\text{d}x},\underset{}{\overset{}{}}{P}_M\left(\text{0}\right)={\mu }_0{P}_0,P_N\left(\text{0}\right)={\displaystyle {\int }_0^{\infty }{\mu }_M\left(x\right)P_M\left(x\right)\text{d}x},i=1,2,3,4\}$

$B=\left(\begin{array}{cc}0& B_1\\ 0& 0\end{array}\right),D\left(B\right)=X$

其中

$B_1=\left({\displaystyle {\int }_0^{\infty }{\mu }_{\text{1}}\left(x\right)\text{d}x},{\displaystyle {\int }_0^{\infty }{\mu }_{\text{2}}\left(x\right)\text{d}x},{\displaystyle {\int }_0^{\infty }{\mu }_{\text{3}}\left(x\right)\text{d}x},{\displaystyle {\int }_0^{\infty }{\mu }_{\text{4}}\left(x\right)\text{d}x},0,{\displaystyle {\int }_0^{\infty }{\mu }_N\left(x\right)\text{d}x}\right)$

所以 $D\left(A+B\right)=D\left(A\right)\cap D\left(B\right)=D\left(A\right)$ 。

则系统方程可转化为Banach空间 $X$ 中的抽象Cauchy问题：

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

3. 算子的性质

在研究系统算子 $A+B$ 的性质前首先给出几个相关定义。

定义1. 算子 $A+B$ 的谱上界 $s\left(A+B\right)$ 定义为 $s\left(A+B\right)=\inf \left\{\omega \in \text{R}|\left(\omega ,\infty \right)\subseteq \rho \left(A+B\right)\right\}$ 。

定义2. 若算子 $A+B$ 是半群 $T\left(t\right)$ 的无穷小生成元，则增长界 $\omega \left(A+B\right)$ 定义为，

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

定义3. 设集合 $C$ 为集合 $E$ 的子集，若对任意的 $f\in E$ ，存在 $g\in C$ ，使得 $f\le g$ ，则称 $C$ 在 $E$ 中共尾。

定理1. $D\left(A+B\right)$ 在 $X$ 中稠密。

证明： $\begin{array}{l}L=\{{\left(P_0,P_1\left(x\right),P_2\left(x\right),P_3\left(x\right),P_4\left(x\right),P_M\left(x\right),P_N\left(x\right)\right)}^{\text{T}}|P_j\left(x\right)\in C_0^{\infty }\left({\text{R}}_{+}\right),且存在常数c_j\\ 使得P_j\left(x\right)=0,x\in \left[0,c_j\right],j=1,2,3,4,M,N\}\end{array}$ 。

根据文献 [4] 知 $L$ 在 $X$ 中稠密。故只需证明 $D\left(A\right)$ 在 $X$ 中稠密即可。

取 $P\in L$ ，则存在 $c_j$ 使得 $P_j\left(x\right)=0,x\in \left[0,c_j\right],j=1,2,3,4,M,N$ 。令 $0<2s<\min \left\{c_i,c_M,c_N,i=1,2,3,4\right\}$ ，

那么当 $x\in \left[0,2s\right]$ 时，有 $P_j\left(x\right)=0,j=1,2,3,4,M,N$ 。

根据文献 [5] ，令

$f^s\left(0\right)={\left(P_0,f_1^s\left(0\right),f_2^s\left(0\right),f_3^s\left(0\right),f_4^s\left(0\right),f_M^s\left(0\right),f_N^s\left(0\right)\right)}^{\text{T}}$

$f^s\left(x\right)={\left(P_0,f_1^s\left(x\right),f_2^s\left(x\right),f_3^s\left(x\right),f_4^s\left(x\right),f_M^s\left(x\right),f_N^s\left(x\right)\right)}^{\text{T}}$

其中：

$f_j^s\left(x\right)=\{\begin{array}{l}{f}_j^s\left(0\right){\left(1-\frac{x}{s}\right)}^2,\text{}\text{}\text{}x\in \left[0,s\right]\\ {\mu }_j{\left(x-s\right)}^2{\left(x-2s\right)}^2,x\in \left[s,2s\right]\\ P_j\left(x\right),x\in \left[2s,\infty \right)\end{array}$

此处，

${\mu }_j=\frac{f_j^s\left(0\right){\displaystyle {\int }_0^s{\mu }_j\left(x\right){\left(1-\frac{x}{s}\right)}^2\text{d}x}}{{\displaystyle {\int }_s^{2s}{\mu }_j\left(x\right){\left(x-s\right)}^2{\left(x-2s\right)}^2\text{d}x}}$

易证 $f^s\left(x\right)\in D\left(A\right)$ 且

故 $D\left(A\right)$ 在 $L$ 中稠密，故 $D\left(A\right)$ 在 $X$ 中稠密。又由于 $D\left(B\right)=X$ ，故 $D\left(A+B\right)$ 在 $X$ 中稠密。

定理2. 系统算子 $A+B$ 为预解正算子。

证明：首先证明 $A$ 为预解正算子。对任意的 $Y\in X$ ，考虑算子方程 $\left(rI-A\right)P=Y$ ，即

$\left(r+{\mu }_0\right)P_0=y_0$ (9)

$\frac{\text{d}{P}_i\left(x\right)}{\text{d}x}+\left(r+{\mu }_i\left(x\right)\right)P_i\left(x\right)=y_i\left(x\right),i=1,2,3,4$ (10)

$\frac{\text{d}{P}_M\left(x\right)}{\text{d}x}+\left(r+{\mu }_M\left(x\right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(x\right)}\right)P_M\left(x\right)=y_M\left(x\right)$ (11)

$\frac{\text{d}{P}_N\left(x\right)}{\text{d}x}+\left(r+{\mu }_N\left(x\right)\right)P_N\left(x\right)=y_N\left(x\right)$ (12)

解之得：

$P_0=\frac{y_0}{r+{\mu }_0}$ (13)

$P_i\left(x\right)=P_i\left(0\right){\text{e}}^{-{\displaystyle {\int }_0^x\left[r+{\mu }_i\left(\xi \right)\right]\text{d}\xi }}+{\displaystyle {\int }_0^x{y}_i}\left(s\right){\text{e}}^{-{\displaystyle {\int }_s^x\left[r+{\mu }_i\left(\xi \right)\right]\text{d}\xi }}\text{d}s$ (14)

$P_M\left(x\right)=P_M\left(0\right){\text{e}}^{-{\displaystyle {\int }_0^x\left[r+{\mu }_M\left(\xi \right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(\xi \right)}\right]\text{d}\xi }}+{\displaystyle {\int }_0^x{y}_M}\left(s\right){\text{e}}^{-{\displaystyle {\int }_s^x\left[r+{\mu }_M\left(\xi \right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(\xi \right)}\right]\text{d}\xi }}\text{d}s$ (15)

$P_N\left(x\right)=P_N\left(0\right){\text{e}}^{-{\displaystyle {\int }_0^x\left[r+{\mu }_N\left(\xi \right)\right]\text{d}\xi }}+{\displaystyle {\int }_0^x{y}_N}\left(s\right){\text{e}}^{-{\displaystyle {\int }_s^x\left[r+{\mu }_N\left(\xi \right)\right]\text{d}\xi }}\text{d}s$ (16)

故有，

$\begin{array}{c}\Vert P\Vert =\left|P_0\right|+{\displaystyle \underset{i=1}{\overset{4}{\sum }}\Vert P_i\left(x\right)\Vert }+\Vert P_M\left(x\right)\Vert +\Vert P_N\left(x\right)\Vert \\ =\left|\frac{y_0}{r+{\mu }_0}\right|+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\displaystyle {\int }_0^{\infty }\left|P_i\left(0\right){\text{e}}^{-{\displaystyle {\int }_{\text{0}}^x\left[r+{\mu }_i\left(\xi \right)\right]\text{d}\xi }}+{\displaystyle {\int }_0^x{\text{e}}^{-{\displaystyle {\int }_{\text{s}}^x\left[r+{\mu }_i\left(\xi \right)\right]\text{d}\xi }}{y}_i\left(s\right)\text{d}s}\right|}\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }\left|P_M\left(0\right){\text{e}}^{-{\displaystyle {\int }_{\text{0}}^x\left[r+{\mu }_M\left(\xi \right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(\xi \right)}\right]\text{d}\xi }}+{\displaystyle {\int }_0^x{\text{e}}^{-{\displaystyle {\int }_s^x\left[r+{\mu }_M\left(\xi \right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(\xi \right)}\right]\text{d}\xi }}{y}_M\left(s\right)\text{d}s}\right|}\text{d}x\\ +{\displaystyle {\int }_0^{\infty }\left|P_N\left(0\right){\text{e}}^{-{\displaystyle {\int }_{\text{0}}^x\left[r+{\mu }_N\left(\xi \right)\right]\text{d}\xi }}+{\displaystyle {\int }_0^x{\text{e}}^{-{\displaystyle {\int }_{\text{s}}^x\left[r+{\mu }_N\left(\xi \right)\right]\text{d}\xi }}{y}_N\left(s\right)\text{d}s}\right|}\text{d}x\end{array}$

当 $r>0$ 时，

$\begin{array}{c}\left|P\right|\le \frac{1}{r}\left|y_0\right|+{\displaystyle \underset{i=1}{\overset{4}{\sum }}\left({\displaystyle {\int }_0^{\infty }\left|P_i\left(0\right)\right|{\text{e}}^{-rx}\text{d}x+{\displaystyle {\int }_0^{\infty }\left({\displaystyle {\int }_0^x{\text{e}}^{-r\left(x-s\right)}\left|y_i\left(s\right)\right|\text{d}s}\right)\text{d}x}}\right)}\\ +{\displaystyle {\int }_0^{\infty }\left|P_M\left(0\right)\right|{\text{e}}^{-rx}\text{d}x+{\displaystyle {\int }_0^{\infty }\left({\displaystyle {\int }_0^x{\text{e}}^{-r\left(x-s\right)}\left|y_M\left(s\right)\right|\text{d}s}\right)\text{d}x}}\\ +{\displaystyle {\int }_0^{\infty }\left|P_N\left(0\right)\right|{\text{e}}^{-rx}\text{d}x+{\displaystyle {\int }_0^{\infty }\left({\displaystyle {\int }_0^x{\text{e}}^{-r\left(x-s\right)}\left|y_N\left(s\right)\right|\text{d}s}\right)\text{d}x}}\end{array}$

其中，

${\displaystyle {\int }_0^{\infty }\left|P_1\left(0\right)\right|{\text{e}}^{-rx}}\text{d}x={\displaystyle {\int }_0^{\infty }\left|{\displaystyle {\int }_0^{\infty }{P}_M\left(s\right){\mu }_{M_1}\left(s\right)\text{d}s}\right|{\text{e}}^{-rx}}\text{d}x$

考虑 ${\displaystyle {\int }_0^{\infty }\left|{\displaystyle {\int }_0^{\infty }{P}_M\left(s\right){\mu }_{M_1}\left(s\right)\text{d}s}\right|{\text{e}}^{-rx}}\text{d}x$ ，将(15)带入得：

因为

$\frac{\text{d}}{\text{d}s}\left({\text{e}}^{-{\displaystyle {\int }_0^s\left[r+{\mu }_M\left(\xi \right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(\xi \right)}\right]\text{d}\xi }}\right)$ $=-\left(r+{\mu }_M\left(s\right)\right){\text{e}}^{-{\displaystyle {\int }_0^s\left[r+{\mu }_M\left(\xi \right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(\xi \right)}\right]\text{d}\xi }}-{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(s\right){\text{e}}^{-{\displaystyle {\int }_0^s\left[r+{\mu }_M\left(\xi \right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(\xi \right)}\right]\text{d}\xi }}}$

故

$\begin{array}{l}{\displaystyle {\int }_0^{\infty }\left({\displaystyle {\int }_0^{\infty }\left|P_M\left(0\right)\right|{\text{e}}^{-{\displaystyle {\int }_{\text{0}}^s\left[r+{\mu }_M\left(\xi \right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(\xi \right)}\right]\text{d}\xi }}{\mu }_{M_1}\left(s\right)\text{d}s}\right){\text{e}}^{-rx}}\text{d}x\\ \le \left|y_0\right|{\displaystyle {\int }_0^{\infty }\left({\displaystyle {\int }_0^{\infty }{\text{e}}^{-{\displaystyle {\int }_{\text{0}}^s\left[r+{\mu }_M\left(\xi \right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(\xi \right)}\right]\text{d}\xi }}{\mu }_{M_1}\left(s\right)\text{d}s}\right){\text{e}}^{-rx}}\text{d}x\\ =\left|y_0\right|{\displaystyle {\int }_0^{\infty }\left({\displaystyle {\int }_0^{\infty }\frac{\text{d}}{\text{d}s}\left(-{\text{e}}^{-{\displaystyle {\int }_{\text{0}}^s\left[r+{\mu }_M\left(\xi \right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(\xi \right)}\right]\text{d}\xi }}\right)\text{d}s}\right){\text{e}}^{-rx}}\text{d}x\end{array}$

$\begin{array}{l}-\left|y_0\right|{\displaystyle {\int }_0^{\infty }\left({\displaystyle {\int }_{\text{0}}^{\infty }\left(r+{\mu }_M\left(s\right)+{\displaystyle \underset{i=2}{\overset{4}{\sum }}{\mu }_{M_i}\left(s\right)}\right){\text{e}}^{-{\displaystyle {\int }_{\text{0}}^s\left[r+{\mu }_M\left(\xi \right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(\xi \right)}\right]\text{d}\xi }}\text{d}s}\right){\text{e}}^{-rx}}\text{d}x\\ \le \left|y_0\right|{\displaystyle {\int }_0^{\infty }\left({\displaystyle {\int }_0^{\infty }\frac{\text{d}}{\text{d}s}\left(-{\text{e}}^{-{\displaystyle {\int }_{\text{0}}^s\left[r+{\mu }_M\left(\xi \right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(\xi \right)}\right]\text{d}\xi }}\right)\text{d}s}\right){\text{e}}^{-rx}}\text{d}x\\ =\left|y_0\right|{\displaystyle {\int }_0^{\infty }{\text{e}}^{-rx}\text{d}x}=\frac{1}{r}\left|y_0\right|\end{array}$

又因为

$\begin{array}{l}\frac{\text{d}}{\text{d}s}\left({\text{e}}^{-{\displaystyle {\int }_{\tau }^s\left[r+{\mu }_M\left(\xi \right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(\xi \right)}\right]\text{d}\xi }}\right)\\ =-\left(r+{\mu }_M\left(s\right)\right){\text{e}}^{-{\displaystyle {\int }_{\tau }^s\left[r+{\mu }_M\left(\xi \right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(\xi \right)}\right]\text{d}\xi }}-{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(s\right){\text{e}}^{-{\displaystyle {\int }_{\tau }^s\left[r+{\mu }_M\left(\xi \right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(\xi \right)}\right]\text{d}\xi }}}\end{array}$

故

$\begin{array}{l}{\displaystyle {\int }_0^{\infty }\left({\displaystyle {\int }_0^{\infty }\left({\displaystyle {\int }_0^s{\text{e}}^{-{\displaystyle {\int }_{\tau }^s\left[r+{\mu }_M\left(\xi \right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(\xi \right)}\right]\text{d}\xi }}}\left|y_M\left(\tau \right)\right|\text{d}\tau \right){\mu }_{M_1}\left(s\right)\text{d}s}\right){\text{e}}^{-rx}}\text{d}x\\ ={\displaystyle {\int }_0^{\infty }\left({\displaystyle {\int }_0^{\infty }\left|y_M\left(\tau \right)\right|\left({\displaystyle {\int }_{\tau }^{\infty }{\text{e}}^{-{\displaystyle {\int }_{\tau }^s\left[r+{\mu }_M\left(\xi \right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(\xi \right)}\right]\text{d}\xi }}}{\mu }_{M_1}\left(s\right)\text{d}s\right)\text{d}\tau }\right){\text{e}}^{-rx}}\text{d}x\\ ={\displaystyle {\int }_0^{\infty }\left({\displaystyle {\int }_0^{\infty }\left|y_M\left(\tau \right)\right|\left({\displaystyle {\int }_{\tau }^{\infty }\frac{\text{d}}{\text{d}s}\left(-{\text{e}}^{-{\displaystyle {\int }_{\tau }^s\left[r+{\mu }_M\left(\xi \right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(\xi \right)}\right]\text{d}\xi }}\right)\text{d}s}\right)\text{d}\tau }\right){\text{e}}^{-rx}}\text{d}x\end{array}$

考虑

${\displaystyle {\int }_0^{\infty }\left({\displaystyle {\int }_0^x{\text{e}}^{-r\left(x-s\right)}\left|y_1\left(s\right)\right|\text{d}s}\right)\text{d}x}={\displaystyle {\int }_0^{\infty }\left|y_1\left(s\right)\right|\left({\displaystyle {\int }_s^{\infty }{\text{e}}^{-r\left(x-s\right)}\text{d}x}\right)\text{d}s=\frac{1}{r}\Vert y_1\left(s\right)\Vert }$

所以

$\Vert P_1\left(x\right)\Vert \le \frac{1}{r}\left[\left|y_0\right|+\Vert y_1\left(s\right)\Vert +\Vert y_M\left(s\right)\Vert \right]$

同理

$\Vert P_i\left(x\right)\Vert \le \frac{1}{r}\left[\left|y_0\right|+\Vert y_i\left(s\right)\Vert +\Vert y_M\left(s\right)\Vert \right]$

$\Vert P_M\left(x\right)\Vert \le \frac{1}{r}\left[\left|y_0\right|+\Vert y_M\left(s\right)\Vert \right]$

$\Vert P_N\left(x\right)\Vert \le \frac{1}{r}\left[\left|y_0\right|+\Vert y_M\left(s\right)\Vert +\Vert y_N\left(s\right)\Vert \right]$

故整理得

$\begin{array}{c}\Vert P\Vert =\left|P_0\right|+{\displaystyle \underset{i=1}{\overset{4}{\sum }}\Vert P_i\left(x\right)\Vert }+\Vert P_M\left(x\right)\Vert +\Vert P_N\left(x\right)\Vert \\ \le \frac{7}{r}\left|y_0\right|+\frac{1}{r}\Vert y_1\left(x\right)\Vert +\frac{1}{r}\Vert y_2\left(x\right)\Vert +\frac{1}{r}\Vert y_3\left(x\right)\Vert +\frac{1}{r}\Vert y_4\left(x\right)\Vert +\frac{6}{r}\Vert y_M\left(x\right)\Vert +\frac{1}{r}\Vert y_N\left(x\right)\Vert \\ \le \frac{7}{r}\Vert Y\Vert \end{array}$

故当 $r>0$ 时， ${\left(rI-A\right)}^{-1}$ 存在，且 $\Vert {\left(rI-A\right)}^{-1}\Vert \le \frac{7}{r}$ 。由 $Y$ 的任意性可知， $rI-A$ 为满射。所以 $\left(0,\infty \right)\subset \rho \left(A\right)$ ， $A$ 为预解正算子。再由 $B$ 的表达式知， $B$ 为正算子，且

$B\le \sup \left\{{\displaystyle {\int }_0^{\infty }{\mu }_{F_i}\left(x\right)\text{d}x,{\displaystyle {\int }_0^{\infty }{\mu }_N\left(x\right)\text{d}x}},i=1,2,3,4\right\}=G$

故当 $r>7G$ 时，

$\Vert {\left(rI-A\right)}^{-1}B\Vert \le \Vert {\left(rI-A\right)}^{-1}\Vert \Vert B\Vert \le \frac{7M}{r}<1$

由文献 [6] 定理“设 $A\in C^{n\times n}$ ，若 $\Vert A\Vert \le 1$ ，则 $I-A$ 为非奇异矩阵，且 $\Vert {\left(I-A\right)}^{-1}\Vert \le {\left(1-\Vert A\Vert \right)}^{-1}$ ”，知 $I-{\left(rI-A\right)}^{-1}B$ 可逆，且 $\Vert {\left[I-{\left(rI-A\right)}^{-1}B\right]}^{-1}\Vert \le {\left[1-\Vert {\left(rI-A\right)}^{-1}B\Vert \right]}^{-1}$ ，即 ${\left[I-{\left(rI-A\right)}^{-1}B\right]}^{-1}$ 有界，其中 $r>7G$ 。

因

${\left[I-{\left(rI-A\right)}^{-1}B\right]}^{-1}={\displaystyle \underset{k=0}{\overset{\infty }{\sum }}{\left[{\left(rI-A\right)}^{-1}B\right]}^k}$

且 $A$ 为预解正算子， $B$ 为正算子，故 ${\left[I-{\left(rI-A\right)}^{-1}B\right]}^{-1}$ 也为正算子。而

$\begin{array}{c}{\left[rI-\left(A+B\right)\right]}^{-1}={\left[\left(rI-A\right)-\left(rI-A\right){\left(rI-A\right)}^{-1}B\right]}^{-1}\\ ={\left\{\left(rI-A\right)\left[I-{\left(rI-A\right)}^{-1}B\right]\right\}}^{-1}\\ ={\left[I-{\left(rI-A\right)}^{-1}B\right]}^{-1}{\left(rI-A\right)}^{-1}\end{array}$

所以当 $r>7G$ 时， ${\left[rI-\left(A+B\right)\right]}^{-1}$ 为正算子，即 $A+B$ 为预解正算子。

定理3. 算子 $A+B$ 的对偶算子 ${\left(A+B\right)}^{\ast }$ ， $Q=(Q_0,Q_1\left(x\right),Q_2\left(x\right),Q_3\left(x\right),Q_4\left(x\right),Q_M\left(x\right),Q_N(x))$

${\left(A+B\right)}^{\ast }Q=\left[\begin{array}{c}{\mu }_0{Q}_M\left(0\right)-{\mu }_0{Q}_0\\ \left(\frac{\text{d}}{\text{d}x}-{\mu }_{\text{1}}\left(x\right)\right)Q_1\left(x\right)+{\mu }_{\text{1}}\left(x\right)Q_0\\ \left(\frac{\text{d}}{\text{d}x}-{\mu }_{\text{2}}\left(x\right)\right)Q_2\left(x\right)+{\mu }_{\text{2}}\left(x\right)Q_0\\ \left(\frac{\text{d}}{\text{d}x}-{\mu }_{\text{3}}\left(x\right)\right)Q_3\left(x\right)+{\mu }_{\text{3}}\left(x\right)Q_0\\ \left(\frac{\text{d}}{\text{d}x}-{\mu }_{\text{4}}\left(x\right)\right)Q_4\left(x\right)+{\mu }_{\text{4}}\left(x\right)Q_0\\ \left(\frac{\text{d}}{\text{d}x}-{\mu }_M\left(x\right)-{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(x\right)}\right)Q_M\left(x\right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(x\right)Q_i\left(0\right)+{\mu }_M\left(x\right)Q_N\left(0\right)}\\ \left(\frac{\text{d}}{\text{d}x}-{\mu }_N\left(x\right)\right)Q_N\left(x\right)+{\mu }_N\left(x\right)Q_0\end{array}\right]$

$D{\left(A+B\right)}^{\ast }=\left\{Q\in X^{\ast }|Q_j\left(x\right)绝对连续且Q_j\left(x\right),\frac{\text{d}}{\text{d}x}{Q}_j\left(x\right)\in L^{\infty }\left({\text{R}}_{+}\right),j=1,2,3,4,M,N\right\}$

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

$\begin{array}{c}\langle \left(A+B\right)P,Q\rangle =\left\{-{\mu }_0{P}_0+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\displaystyle {\int }_0^{\infty }{\mu }_i\left(x\right)P_i\left(x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }{\mu }_N\left(x\right)P_N\left(x\right)\text{d}x}}\right\}{Q}_0\\ +{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\displaystyle {\int }_0^{\infty }\left\{-\frac{\text{d}{P}_i\left(x\right)}{\text{d}x}-{\mu }_i\left(x\right)P_i\left(x\right)\right\}}{Q}_i\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }\left\{-\frac{\text{d}{P}_M\left(x\right)}{\text{d}x}-\left({\mu }_M\left(x\right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(x\right)}\right)P_M\left(x\right)\right\}{Q}_M\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }\left\{-\frac{\text{d}{P}_N\left(x\right)}{\text{d}x}-{\mu }_N\left(x\right)P_N\left(x\right)\right\}{Q}_N\left(x\right)\text{d}x}\end{array}$

$\begin{array}{l}=-{\mu }_0{P}_0{Q}_0+{\displaystyle {\int }_0^{\infty }{\mu }_{\text{1}}\left(x\right)P_1\left(x\right)}\text{d}x{Q}_0+{\displaystyle {\int }_0^{\infty }{\mu }_{\text{2}}\left(x\right)P_2\left(x\right)}\text{d}x{Q}_0\\ +{\displaystyle {\int }_0^{\infty }{\mu }_3\left(x\right)P_3\left(x\right)\text{d}x{Q}_0}+{\displaystyle {\int }_0^{\infty }{\mu }_4\left(x\right)P_4\left(x\right)\text{d}x{Q}_0}+{\displaystyle {\int }_0^{\infty }{\mu }_N\left(x\right)P_N\left(x\right)\text{d}x{Q}_0}\\ +{\displaystyle {\int }_0^{\infty }-\frac{\text{d}{P}_1\left(x\right)}{\text{d}x}{Q}_1\left(x\right)\text{d}x}-{\displaystyle {\int }_0^{\infty }{\mu }_{\text{1}}\left(x\right)P_1\left(x\right)Q_1\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }-\frac{\text{d}{P}_2\left(x\right)}{\text{d}x}{Q}_2\left(x\right)\text{d}x}-{\displaystyle {\int }_0^{\infty }{\mu }_{\text{2}}\left(x\right)P_2\left(x\right)Q_2\left(x\right)\text{d}x}\end{array}$

$\begin{array}{l}+{\displaystyle {\int }_0^{\infty }-\frac{\text{d}{P}_3\left(x\right)}{\text{d}x}{Q}_3\left(x\right)\text{d}x}-{\displaystyle {\int }_0^{\infty }{\mu }_{\text{3}}\left(x\right)P_3\left(x\right)Q_3\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }-\frac{\text{d}{P}_4\left(x\right)}{\text{d}x}{Q}_4\left(x\right)\text{d}x}-{\displaystyle {\int }_0^{\infty }{\mu }_{\text{4}}\left(x\right)P_4\left(x\right)Q_4\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }-\frac{\text{d}{P}_M\left(x\right)}{\text{d}x}{Q}_M\left(x\right)\text{d}x}-{\displaystyle {\int }_0^{\infty }\left[{\mu }_M\left(x\right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(x\right)}\right]P_M\left(x\right)Q_M\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }-\frac{\text{d}{P}_N\left(x\right)}{\text{d}x}{Q}_N\left(x\right)\text{d}x}-{\displaystyle {\int }_0^{\infty }{\mu }_N\left(x\right)P_N\left(x\right)Q_N\left(x\right)\text{d}x}\end{array}$

其中

$\begin{array}{c}{\displaystyle {\int }_0^{\infty }-\frac{\text{d}{P}_i\left(x\right)}{\text{d}x}{Q}_i\left(x\right)\text{d}x}=-{Q_i\left(x\right)P_i\left(x\right)|}_0^{\infty }+{\displaystyle {\int }_0^{\infty }\frac{\text{d}{Q}_i\left(x\right)}{\text{d}x}{P}_i\left(x\right)\text{d}x}\\ =Q_i\left(0\right)P_i\left(0\right)+{\displaystyle {\int }_0^{\infty }\frac{\text{d}{Q}_i\left(x\right)}{\text{d}x}{P}_i\left(x\right)\text{d}x}\\ ={\displaystyle {\int }_0^{\infty }{\mu }_{M_i}\left(x\right)P_M\left(x\right)\text{d}x{Q}_i\left(0\right)}+{\displaystyle {\int }_0^{\infty }\frac{\text{d}{Q}_i\left(x\right)}{\text{d}x}{P}_i\left(x\right)\text{d}x},i=1,2,3,4\end{array}$

同理

${\displaystyle {\int }_0^{\infty }-\frac{\text{d}{P}_M\left(x\right)}{\text{d}x}{Q}_M\left(x\right)\text{d}x}={\mu }_0{P}_0{Q}_M\left(0\right)+{\displaystyle {\int }_0^{\infty }\frac{\text{d}{Q}_M\left(x\right)}{\text{d}x}{P}_M\left(x\right)\text{d}x}$

${\displaystyle {\int }_0^{\infty }-\frac{\text{d}{P}_N\left(x\right)}{\text{d}x}{Q}_N\left(x\right)\text{d}x}={\displaystyle {\int }_0^{\infty }{\mu }_M\left(x\right)P_M\left(x\right)\text{d}x{Q}_N\left(0\right)}+{\displaystyle {\int }_0^{\infty }\frac{\text{d}{Q}_N\left(x\right)}{\text{d}x}{P}_N\left(x\right)\text{d}x}$

所以

$\begin{array}{l}\langle \left(A+B\right)P,Q\rangle =-{\mu }_0{Q}_0{P}_0+{\displaystyle {\int }_0^{\infty }{\mu }_{\text{1}}\left(x\right)Q_0{P}_1\left(x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }{\mu }_{\text{2}}\left(x\right)Q_0{P}_2\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }{\mu }_{\text{3}}\left(x\right)Q_0{P}_3\left(x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }{\mu }_{\text{4}}\left(x\right)Q_0{P}_4\left(x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }{\mu }_N\left(x\right)Q_0{P}_N\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }{\mu }_{M_1}\left(x\right)Q_1\left(0\right)P_M\left(x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }\frac{\text{d}{Q}_1\left(x\right)}{\text{d}x}{P}_1\left(x\right)\text{d}x}-{\displaystyle {\int }_0^{\infty }{\mu }_{\text{1}}\left(x\right)Q_1\left(x\right)P_1\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }{\mu }_{M_2}\left(x\right)Q_2\left(0\right)P_M\left(x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }\frac{\text{d}{Q}_2\left(x\right)}{\text{d}x}{P}_2\left(x\right)\text{d}x}-{\displaystyle {\int }_0^{\infty }{\mu }_{\text{2}}\left(x\right)Q_2\left(x\right)P_2\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }{\mu }_{M_3}\left(x\right)Q_3\left(0\right)P_M\left(x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }\frac{\text{d}{Q}_3\left(x\right)}{\text{d}x}{P}_3\left(x\right)\text{d}x}-{\displaystyle {\int }_0^{\infty }{\mu }_{\text{3}}\left(x\right)Q_3\left(x\right)P_3\left(x\right)\text{d}x}\end{array}$

$\begin{array}{l}+{\displaystyle {\int }_0^{\infty }{\mu }_{M_4}\left(x\right)Q_4\left(0\right)P_M\left(x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }\frac{\text{d}{Q}_4\left(x\right)}{\text{d}x}{P}_4\left(x\right)\text{d}x}-{\displaystyle {\int }_0^{\infty }{\mu }_{\text{4}}\left(x\right)Q_4\left(x\right)P_4\left(x\right)\text{d}x}\\ +{\mu }_0{Q}_M\left(0\right)P_0+{\displaystyle {\int }_0^{\infty }\frac{\text{d}{Q}_M\left(x\right)}{\text{d}x}{P}_M\left(x\right)\text{d}x}-{\displaystyle {\int }_0^{\infty }\left[{\mu }_M\left(x\right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(x\right)}\right]Q_M\left(x\right)P_M\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{\infty }{\mu }_M\left(x\right)Q_N\left(0\right)P_M\left(x\right)\text{d}x}+{\displaystyle {\int }_0^{\infty }\frac{\text{d}{Q}_N\left(x\right)}{\text{d}x}{P}_N\left(x\right)\text{d}x}-{\displaystyle {\int }_0^{\infty }{\mu }_N\left(x\right)Q_N\left(x\right)P_N\left(x\right)\text{d}x}\\ =\left({\mu }_0{Q}_M\left(0\right)-{\mu }_0{Q}_0\right)P_0+{\displaystyle {\int }_0^{\infty }{P}_1\left(x\right)}\left(\frac{\text{d}{Q}_1\left(x\right)}{\text{d}x}-{\mu }_{\text{1}}\left(x\right)Q_1\left(x\right)+{\mu }_{\text{1}}\left(x\right)Q_0\right)\text{d}x\\ +{\displaystyle {\int }_0^{\infty }{P}_2\left(x\right)}\left(\frac{\text{d}{Q}_2\left(x\right)}{\text{d}x}-{\mu }_{\text{2}}\left(x\right)Q_2\left(x\right)+{\mu }_{\text{2}}\left(x\right)Q_0\right)\text{d}x\end{array}$

$\begin{array}{l}+{\displaystyle {\int }_0^{\infty }{P}_3\left(x\right)}\left(\frac{\text{d}{Q}_3\left(x\right)}{\text{d}x}-{\mu }_{\text{3}}\left(x\right)Q_3\left(x\right)+{\mu }_{\text{3}}\left(x\right)Q_0\right)\text{d}x\\ +{\displaystyle {\int }_0^{\infty }{P}_4\left(x\right)}\left(\frac{\text{d}{Q}_4\left(x\right)}{\text{d}x}-{\mu }_{\text{4}}\left(x\right)Q_4\left(x\right)+{\mu }_{\text{4}}\left(x\right)Q_0\right)\text{d}x\\ +{\displaystyle {\int }_0^{\infty }{P}_M\left(x\right)}[\frac{\text{d}{Q}_M\left(x\right)}{\text{d}x}-\left({\mu }_M\left(x\right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(x\right)}\right)Q_M\left(x\right)\\ +{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(x\right)Q_i\left(0\right)+{\mu }_M\left(x\right)Q_N\left(0\right)}]\text{d}x\\ +{\displaystyle {\int }_0^{\infty }{P}_N\left(x\right)}\left(\frac{\text{d}{Q}_N\left(x\right)}{\text{d}x}-{\mu }_N\left(x\right)Q_N\left(x\right)+{\mu }_N\left(x\right)Q_0\right)\text{d}x\end{array}$

显然

$D{\left(A+B\right)}^{\ast }=\left\{Q\in X^{\ast }|Q_j\left(x\right)绝对连续且Q_j\left(x\right),\frac{\text{d}}{\text{d}x}{Q}_j\left(x\right)\in L^{\infty }\left({\text{R}}_{+}\right),j=1,2,3,4,M,N\right\}$

定理4. $\omega \left(A+B\right)=0$

证明：将系统状态方程组(1)-(4)左右两边分别对 $x$ 从0到 $\infty$ 积分并整理得：

$\frac{\text{d}{P}_0\left(t\right)}{\text{d}t}=-{\mu }_0{P}_0\left(t\right)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\displaystyle {\int }_0^{\infty }{\mu }_i}}\left(x\right)P_i\left(x,t\right)\text{d}x+{\displaystyle {\int }_0^{\infty }{\mu }_N}\left(x\right)P_N\left(x,t\right)\text{d}x$

$\frac{\text{d}{P}_i\left(x,t\right)}{\text{d}t}=P_i\left(0,t\right)-{\displaystyle {\int }_0^{\infty }{\mu }_i\left(x\right)P_i\left(x,t\right)\text{d}x},i=1,2,3,4$

$\frac{\text{d}{P}_M\left(x,t\right)}{\text{d}t}=P_M\left(0,t\right)-{\displaystyle {\int }_0^{\infty }\left({\mu }_M(x)+{\displaystyle \underset{i=1}{\overset{4}{\sum }}{\mu }_{M_i}\left(x\right)}\right)P_M\left(x,t\right)\text{d}x}$

$\frac{\text{d}{P}_N\left(x,t\right)}{\text{d}t}=P_N\left(0,t\right)-{\displaystyle {\int }_0^{\infty }{\mu }_N\left(x\right)P_N\left(x,t\right)\text{d}x}$

将边值条件(5)-(7)代入上面方程组，并将等式左右相加，可得 $\frac{\text{d}\Vert P\Vert }{\text{d}t}=0$ 。故方程组所对应的半群是非扩张半群。根据初值条件(8)式，知 $\Vert T\left(t\right)\Vert =1$ 。故半群的增长界 $\omega \left(A+B\right)=0$ 。

定理5. $s\left(A+B\right)=0$

证明：由前文知，

$X=\left\{P|P_0\in \text{R};P_i\left(x\right),P_M\left(x\right),P_N\left(x\right)\in L^1\left({\text{R}}_{+}\right),\Vert P\Vert =\left|P_0\right|+{\displaystyle \underset{i=1}{\overset{4}{\sum }}\Vert P_i\left(x\right)\Vert }+\Vert P_M\left(x\right)\Vert +\Vert P_N\left(x\right)\Vert \right\}$

$X^{\ast }=\left\{Q|Q_0\in \text{R};Q_j\left(x\right)\in L^{\infty }\left({\text{R}}_{+}\right),\Vert Q\Vert =\max \left\{\left|Q_0\right|,\Vert Q_j\left(x\right)\Vert ,j=1,2,3,4,M,N\right\}\right\}$

$X_{+}^{\ast }=\left\{Q\in X^{\ast }|Q_0\ge \text{0};Q_j\left(x\right)\ge 0,j=1,2,3,4,M,N\right\}$

任取

$f={\left(f_0,f_1\left(x\right),f_2\left(x\right),f_3\left(x\right),f_4\left(x\right),f_M\left(x\right),f_N\left(x\right)\right)}^{{\rm T}}\in X_{+}^{\ast }$

则

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

因此

$\Vert f\Vert \ge \left|f_0\right|,\Vert f\Vert \ge f_i\left(x\right),\Vert f\Vert \ge f_M\left(x\right),\Vert f\Vert \ge f_N\left(x\right),x\ge 0$

又因为 $1\left(x\right)\in L^{\infty }\left({\text{R}}_{+}\right)$ ，且 $1\left(x\right)$ 绝对连续，且 $\frac{\text{d}}{\text{d}x}1\left(x\right)\in L^{\infty }\left({\text{R}}_{+}\right)$ ，故存在

$g={\left(\Vert f\Vert ,\Vert f\Vert 1\left(x\right),\Vert f\Vert 1\left(x\right),\Vert f\Vert 1\left(x\right),\Vert f\Vert 1\left(x\right),\Vert f\Vert 1\left(x\right),\Vert f\Vert 1\left(x\right)\right)}^{{\rm T}}\in D{\left({\left(A+B\right)}^{\ast }\right)}_{+}$

故 $g-f\ge 0$ ，即 $f\le g$ 。所以对任意 $f\in X_{+}^{\ast }$ ，存在 $g\in D{\left({\left(A+B\right)}^{\ast }\right)}_{+}$ ，使得 $f\le g$ 。所以 $D{\left({\left(A+B\right)}^{\ast }\right)}_{+}$ 在 $X_{+}^{\ast }$ 中共尾。又因为 $A+B$ 为稠密的预解正算子，根据 [7] 中定理2.2知， $s\left(A+B\right)=\omega \left(A+B\right)=0$ 。

4. 结论

具有优化调整状态的供应链系统是一个边界带有积分的复杂的数学模型。本文运用补充变量法建立了此模型的微积分方程组，通过对系统的合理假设及相关理论得到了抽象Cauchy问题，并对系统算子的性质进行了深入的研究，这为优化调整状态的供应链系统的后续研究奠定了基础。