在系统可靠性的理论研究和应用中很多因素都会影响其可靠性, 因此有必要对产品的存储可靠性进行研究^[1-2]冷贮备可修复系统作为产品的存储可靠性分析中的最重要的系统, 引起了众多国内外学者的研究兴趣^[3-9]刘宝友^[3]研究了修理人员休假的冷贮备可修复系统, 唐应辉和刘艳^[4]研究了修理工单重休假且无法维修如新的冷贮备可修系统, 苏保河^[5]研究了修理工为系统外客户服务的两部件冷贮备系统, 并利用补充变量法和广义马尔可夫过程得到了系统的一些重要可靠性指标, 刘仁彬等^[6]通过引入维修人员延迟休假的概念, 研究了一个N部件串联可维修系统的模型, 陈永燕和郑海鹰^[7]将修理工延误休假的概念推广到冷贮备可修系统, 建立了相应的数学模型并研究了其可靠性指标, El-Sherbeny^[8]研究了泊松冲击下的两部件冷贮备系统, 得出了该系统的可靠性指标, 马梦饶等^[9]研究了离散时间冷贮备可修重试系统, 利用差分方程迭代解法和母函数法等推导出了系统可用度, 可靠度和首次失效前平均时间等关键可靠性指标. 和(冷)贮备系统类似, 鉴于其广泛的应用场景, 具有关闭规则的可靠性系统一直是研究热点, 已有大量相关文献对其进行探讨. 然而, 从所查阅的文献来看, 目前针对带有关闭规则的可靠性模型的研究范畴较为局限, 主要集中在串联系统^[10-11], 表决系统^[12]以及串-并联系统^[13]. 相比之下, 具有关闭规则的贮备系统目前尚未发现有相关文献展开研究, 但此类系统却有着极为重要的应用价值. 在实际生活中, 带有关闭规则的贮备系统有着诸多典型应用. 以常见的智能电话为例, 其内部通常配备两个处理器, 其中一个处理器处于运行状态, 另一个则处于冷贮备状态, 而电源部分与处理器呈串联关系. 当电源部分出现故障失效时, 会关闭处理器; 但若处理器失效, 电源部分有可能不受影响而继续供电. 一旦某个处理器失效, 备用处理器会立即启动进行替换. 再看汽车的情况, 在汽车运行过程中, 当某一个轮胎失效时, 发动机仍能正常运转; 然而, 一旦发动机失效, 轮胎将无法继续工作. 若发动机未出现故障, 便可以使用备用轮胎替换失效轮胎. 在电子设备领域, 也存在类似情况. 当电源部分失效时, 会关闭功能部分; 而当功能部分失效时, 为方便检修, 电源部分仍能保持运行. 若电源部分未失效, 而功能部分出现故障, 就可以使用备用部件替换失效部件, 并对失效部件立即进行修理. 鉴于上述实际情形, 在文献 [14]中, 作者聚焦于单向关闭两不同部件串联一部件冷贮备的可修系统, 假定两部件串联, 部件1失效, 关闭部件2, 而部件2失效, 部件1却可以继续运行. 与此同时假定部件3冷贮备, 当部件2失效后, 部件3立即替换部件2, 当部件3失效后而部件1没有失效时, 如果部件2已修好立即用部件2替换部件3. 作者运用补充变量法对该系统的可靠性进行分析, 成功得出一系列重要的可靠性指标. 截至目前, 除了这篇文献的研究成果外, 尚未发现有其他关于单向关闭两不同部件串联一部件冷贮备的可修系统的研究成果, 在文献 [14]的基础上, 本文运用泛函分析中的强连续算子半群理论研究单向关闭两不同部件串联一部件冷贮备的三部件可修系统的时间依赖解, 证明此系统非负时间依赖解的存在唯一性.

根据文献 [14], 单向关闭两不同部件串联一部件冷贮备的三部件可修系统由以下概率微积分方程组描述：

$ \begin{align} &{\frac{\mathrm{d}P_0(t)}{\mathrm{ d} t}=-(\lambda_1+\lambda_2)P_0(t)+\int_0^{\infty} \mu_1(y) P_2(t, y) \mathrm{d} y+\int_0^{\infty} \mu_2(y) P_7(t, y) \mathrm{d} y} , \end{align} $

(1.1)

$ \begin{align} &\frac{\mathrm{ d}P_1(t)}{\mathrm{ d} t}=-(\lambda_1+\lambda_2) P_1(t)+\int_0^{\infty} \mu_2(y) P_3(t, y) \mathrm{d} y, \end{align} $

(1.2)

$ \begin{align} &\frac{\partial P_2(t, y)}{\partial t}+\frac{\partial P_2(t, y)}{\partial y}=-\mu_1(y) P_2(t, y), \end{align} $

(1.3)

$ \begin{align} &\frac{\partial P_3(t, y)}{\partial t}+\frac{\partial P_3(t, y)}{\partial y}=-(\lambda_1+\lambda_2+\mu_2(y)) P_3(t, y), \end{align} $

(1.4)

$ \begin{align} &\frac{\partial P_4(t, y)}{\partial t}+\frac{\partial P_4(t, y)}{\partial y}=-\mu_2(y) P_4(t, y)+\lambda_1 P_3(t, y), \end{align} $

(1.5)

$ \begin{align} &\frac{\partial P_5(t, y)}{\partial t}+\frac{\partial P_5(t, y)}{\partial y}=-(\lambda_3+\mu_2(y))P_5(t, y)+\lambda_2 P_3(t, y), \end{align} $

(1.6)

$ \begin{align} &\frac{\partial P_6(t, y)}{\partial t}+\frac{\partial P_6(t, y)}{\partial y}=-\mu_2(y)P_6(t, y)+\lambda_3 P_5(t, y), \end{align} $

(1.7)

$ \begin{align} &\frac{\partial P_7(t, y)}{\partial t}+\frac{\partial P_7(t, y)}{\partial y}=-(\lambda_1+\lambda_2+\mu_2(y)) P_7(t, y), \end{align} $

(1.8)

$ \begin{align} &\frac{\partial P_8(t, y)}{\partial t}+\frac{\partial P_8(t, y)}{\partial y}=-\mu_2(y) P_8(t, y)+\lambda_1 P_7(t, y), \end{align} $

(1.9)

$ \begin{align} &\frac{\partial P_9(t, y)}{\partial t}+\frac{\partial P_9(t, y)}{\partial y}=-(\lambda_3+\mu_2(y)) P_9(t, y)+\lambda_2 P_7(t, y), \end{align} $

(1.10)

$ \begin{align} &\frac{\partial P_{10}(t, y)}{\partial t}+\frac{\partial P_{10}(t, y)}{\partial y}=-\mu_2(y)P_{10}(t, y)+\lambda_3 P_9(t, y), \end{align} $

(1.11)

$ \begin{align} &P_2(t, 0)=\lambda_1\left[P_0(t)+P_1(t)\right]+\int_0^{\infty} \mu_2(y) P_4(t, y) \mathrm{d} y+\int_0^{\infty} \mu_2(y) P_8(t, y) \mathrm{d} y, \end{align} $

(1.12)

$ \begin{align} & P_3(t, 0)=\lambda_2 P_0(t)+\int_0^{\infty} \mu_2(y) P_9(t, y) \mathrm{d} y, \end{align} $

(1.13)

$ \begin{align} & P_4(t, 0)=\int_0^{\infty} \mu_2(y) P_{10}(t, y) \mathrm{d} y, \end{align} $

(1.14)

$ \begin{align} & P_7(t, 0)=\lambda_2 P_1(t)+\int_0^{\infty} \mu_2(y) P_5(t, y) \mathrm{d} y, \end{align} $

(1.15)

$ \begin{align} & P_8(t, 0)=\int_0^{\infty} \mu_2(y) P_6(t, y) \mathrm{d} y, \end{align} $

(1.16)

$ \begin{align} & P_5(t, 0)=P_6(t, 0)=P_9(t, 0)=P_{10}(t, 0)=0, \end{align} $

(1.17)

$ \begin{align} &P_0(0)=1, \quad P_1(0)=P_2(0, y)=\cdots=P_{10}(0, y)=0. \end{align} $

(1.18)

这里$ (t, y)\in [0, \infty)\times[0, \infty) $; $ P_0(t) $表示在$ t $时刻系统中部件1和2在运行, 部件3冷贮备的概率; $ P_1(t) $表示在$ t $时刻部件1和3在运行, 部件2冷贮备的概率; $ P_2(t, y)\mathrm{d}y $表示在$ t $时刻部件1在修理, 部件2和3冷贮备, 部件1已消耗的修理时间在$ (y, y+\mathrm{d}y] $内的概率; $ P_3(t, y)\mathrm{d}y $表示在$ t $时刻部件1和3在运行, 部件2在修理, 部件2已消耗的修理时间在$ (y, y+\mathrm{d}y] $内的概率; $ P_4(t, y)\mathrm{d}y $表示在$ t $时刻部件2在修理, 部件1待修, 部件3冷贮备, 部件2已消耗的修理时间在$ (y, y+\mathrm{d}y] $内的概率; $ P_5(t, y)\mathrm{d}y $表示在$ t $时刻部件2在修理, 部件3待修, 部件1在运行, 且部件2已消耗的修理时间在$ (y, y+\mathrm{d}y] $内的概率; $ P_6(t, y)\mathrm{d}y $表示在$ t $时刻部件2在修理, 部件1和3待修, 且部件2已消耗的修理时间在$ (y, y+\mathrm{d}y] $内的概率; $ P_7(t, y)\mathrm{d}y $表示在$ t $时刻部件1和2在运行, 部件3在修理, 部件3已消耗的修理时间在$ (y, y+\mathrm{d}y] $内的概率; $ P_8(t, y)\mathrm{d}y $表示在$ t $时刻部件1待修, 部件2冷贮备, 部件3在修理, 部件3已消耗的修理时间在$ (y, y+\mathrm{d}y] $内的概率; $ P_9(t, y)\mathrm{d}y $表示在$ t $时刻部件1在运行, 部件2待修, 部件3在修理, 部件3已消耗的修理时间在$ (y, y+\mathrm{d}y] $内的概率; $ P_{10}(t, y)\mathrm{d}y $表示在$ t $时刻部件1和2待修, 部件3在修理, 部件3已消耗的修理时间在$ (y, y+\mathrm{d}y] $内的概率; $ \lambda_i $是部件$ i(i=1, 2) $的失效率; $ \mu_i(y) $是部件$ i(i=1, 2) $在故障状态, 并部件$ i(i=1, 2) $已消耗的修理时间为$ y $的修复率; $ \lambda_3 $是部件2和3均失效后, 部件1的正常运行率.

为研究单向关闭两不同部件串联一部件冷贮备的三部件可修系统的时间依赖解的存在唯一性, 首先通过选择系统的状态空间, 主算子及其定义域, 将该模型转化成Banach空间中的抽象Cauchy问题. 然后运用泛函分析中的强连续算子半群理论研究此系统的时间依赖解, 证明此系统非负时间依赖解的存在唯一性. 为此, 首先将系统转化为抽象Cauchy问题形式，注意到研究问题的实际意义, 选取状态空间为

$ Y =\left\{P \in R^2 \times \left(L^{1} [0, \infty) \right)^9\left|\|P\|=| P_0 |+|P_1|+\sum\limits_{i=2}^{10}\left\|P_i\right\|_{ L^{1} [0, \infty)}\right.<\infty\right\}, $

设$ P=\left(P_0, P_1, P_2(y), P_3(y), P_4(y), P_5(y), P_6(y), P_7(y), P_8(y), P_9(y), P_{10}(y)\right)^{T}, $并定义算子及其它们的定义域,

$ \begin{align*} & A_1 P=\operatorname{diag}\left(-(\lambda_1+\lambda_2), -(\lambda_1+\lambda_2), -\frac{\text{d}}{\text{d} y}, -\frac{\text{d}}{\text{d} y}, -\frac{\text{d}}{\text{d} y}, -\frac{\text{d}}{\text{d} y}, -\frac{\text{d}}{\text{d} y}, -\frac{\text{d}}{\text{d} y}, -\frac{\text{d}}{\text{d} y}, -\frac{\text{d}}{\text{d} y}, -\frac{\text{d}}{\text{d} y}\right) P, \end{align*} $

$ \begin{equation*} D(A_1)=\left\{P\in X\left| \begin{array}{c} \frac{\text{d} P_{i}(y)}{\text{d} y} \in L^{1}[0, \infty), \; P_{i}(y)(i=2, \cdots, 10)\; \text{是绝对连续函数且} \\ P_2(0)=\lambda_1\left(P_0+P_1\right)+\int_0^{\infty} \mu_2(y) P_4(y) \mathrm{d} y+\int_0^{\infty} \mu_2(y) P_8(y) \mathrm{d} y, \\ P_3(0)=\lambda_2 P_0+\int_0^{\infty} \mu_2(y) P_9(y) \mathrm{d} y, P_4(0)=\int_0^{\infty} \mu_2(y) P_{10}(y) \mathrm{d} y, \\ P_7(0)=\lambda_2 P_1+\int_0^{\infty} \mu_2(y) P_5(y) \mathrm{d} y, P_8(0)=\int_0^{\infty} \mu_2(y) P_6(y) \mathrm{d} y, \\ P_5(0)=P_6(0)=P_9(0)=P_{10}(0)=0 \end{array} \right.\right\}, \end{equation*} $

$ \begin{align} A_2 P=&\bigg( 0, 0, -\mu_1(y) P_2(y), -(\lambda_1+\lambda_2+\mu_2(y)) P_3(y), -\mu_2(y) P_4(y)+\lambda_1 P_3(y), \\ &\quad -\left(\lambda_3+\mu_2(y)\right) P_5(y)+\lambda_2 P_3(y), -\mu_2(y) P_6(y)+\lambda_3 P_5(y), -\left(\lambda_1+\lambda_2+\mu_2(y)\right) P_7(y), \\ &\quad -\mu_2(y) P_8(y)+\lambda_1 P_7(y), -\left(\lambda_3+\mu_2(y)\right) P_9(y)+\lambda_2 P_7(y), -\mu_2(y) P_{10}(y)+\lambda_3 P_9(y)\bigg)^T, \\ A_3 P=&\left(\int_0^{\infty} \mu_1(y) P_2(y) \text{d} y+\int_0^{\infty} \mu_2(y) P_7(y)\text{d}y, \int_0^{\infty} \mu_2(y) P_3(y) \text{d} y, 0, 0, 0, 0, 0, 0, 0, 0, 0\right)^T, \\ D\left(A_2\right)&= D\left(A_3\right)=Y \text {. } \end{align} $

则在选取的状态空间$ Y $中, 把方程组(1.1)-(1.18) 可以描述为Banach空间$ Y $上的抽象Cauchy问题:

$ \begin{align*} &\begin{cases} \frac{\text{d} P(t)}{\text{d} t}=(A_1+A_2+A_3) P(t), \quad \forall t\in[0, \infty), \\ P(0)=(1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)^{T}. \end{cases} \end{align*} $

(2.1)

定理3.1  如果$ \mu_1(y) $和$ \mu_2(y) $满足

$ {\mu}_1=\sup _{y \in [0, \infty)} \mu_1(y)<\infty, \; {\mu}_2=\sup _{y \in [0, \infty)} \mu_2(y)<\infty, $

则$ A_1+A_2+A_3 $生成一个正压缩$ C_0 $－半群$ T(t). $

证   分四步证明此定理. 第一步估计 $ A_1 $ 的豫解式. 第二步验证定义域 $ D(A_1) $ 在 $ X $ 中的稠密性. 第三步指出算子 $ A_2 $ 和 $ A_3 $ 的有界性, 并推出主算子 $ A_1+A_2+A_3 $ 生成一个 $ C_0 $- 半群 $ T(t). $ 最后由主算子的耗散性和 Phillips 定理得到主算子生成的半群 $ T(t) $ 是一个正压缩 $ C_0 $-半群.

对任意的$ \pi \in Y $, 考虑方程$ \left(\gamma I-A_1\right) P=\pi $, 这等价于

$ \begin{align} & \left(\gamma+\lambda_1+\lambda_2\right) P_i=\pi_i, \quad i=0, 1, \end{align} $

(3.1)

$ \begin{align} & \frac{ \rm{d} P_j(y)}{ \rm{d} y}=-\gamma P_j(y)+\pi_j(y), \quad j=2, \ldots, 10, \end{align} $

(3.2)

$ \begin{align} & P_2(0)=\lambda_1\left(P_0+P_1\right)+\int_0^{\infty} \mu_2(y) P_4(y) \rm{d} y+\int_0^{\infty} \mu_2(y) P_8(y) \rm{d} y, \end{align} $

(3.3)

$ \begin{align} & P_3(0)=\lambda_2 P_0+\int_0^{\infty} \mu_2(y) P_9(y) \rm{d} y , \end{align} $

(3.4)

$ \begin{align} & P_4(0)=\int_0^{\infty} \mu_2(y) P_{10}(y) \rm{d} y, \end{align} $

(3.5)

$ \begin{align} & P_7(0)=\lambda_2 P_1+\int_0^{\infty} \mu_2(y) P_5(y) \rm{d} y , \end{align} $

(3.6)

$ \begin{align} & P_8(0)=\int_0^{\infty} \mu_2(y) P_6(y) \rm{d} y , \end{align} $

(3.7)

$ \begin{align} & P_5(0)=P_6(0)=P_9(0)=P_{10}(0)=0. \end{align} $

(3.8)

解$ (3.1) $和$ (3.2) $得

$ \begin{align} & P_i=\frac{1}{\gamma+\lambda_1+\lambda_2} \pi_i, \quad i=0, 1, \end{align} $

(3.9)

$ \begin{align} & P_j(y)=C_j e^{-\gamma y}+e^{-\gamma y} \int_0^y \pi_j(\tau) e^{\gamma \tau} \rm{d} \tau, \quad j=2, \cdots, 10. \end{align} $

(3.10)

(3.10) 结合边界条件(3.3)–(3.8) 推出

$ \begin{align} C_5=&C_6=C_9=C_{10}=0, \end{align} $

(3.11)

$ \begin{align} C_4=&P_4(0)=\int_0^{\infty} \mu_2(y) P_{10}(y) \rm{d} y=\int_0^{\infty} \mu_2(y) e^{-\gamma y} \int_0^y \pi_{10}(\tau) e^{\gamma \tau} \rm{d} \tau \rm{d} y , \end{align} $

(3.12)

$ \begin{align} C_8=&P_8(0)=\int_0^{\infty} \mu_2(y) P_6(y) \rm{d} y=\int_0^{\infty} \mu_2(y) e^{-\gamma y} \int_0^y \pi_6(\tau) e^{\gamma \tau} \rm{d} \tau \rm{d} y , \end{align} $

(3.13)

$ \begin{align} C_2=&P_2(0)=\lambda_1\left(P_0+P_1\right)+\int_0^{\infty} \mu_2(y) P_4(y) \rm{d} y+\int_0^{\infty} \mu_2(y) P_8(y) \rm{d} y \\ =&\frac{\lambda_1}{\gamma+\lambda_1+\lambda_2}\left(\pi_0+\pi_1\right)+\int_0^{\infty} \mu_2(y)\left[C_4 e^{-\gamma y}+e^{-\gamma y} \int_0^y \pi_4(\tau) e^{\gamma \tau} \rm{d} \tau\right] \rm{d} y \\ =&\frac{\lambda_1}{\gamma+\lambda_1+\lambda_2}\left(\pi_0+\pi_1\right)+C_4 \int_0^{\infty} \mu_2(y) e^{-\gamma y} \rm{d} y+\int_0^{\infty} \mu_2(y) e^{-\gamma y} \int_0^y \pi_4(\tau) e^{\gamma \tau} \rm{d} \tau \rm{d} y \\ =&\frac{\lambda_1}{\gamma+\lambda_1+\lambda_2}\left(\pi_0+\pi_{1}\right)+\int_0^{\infty} \mu_2(y) e^{-\gamma y} \int_0^y \pi_{10}(\tau) e^{\gamma{\tau}} \rm{d} \tau \rm{d} y \cdot \int_0^\infty \mu_2(y) e^{-\gamma y} \rm{d} y \\ &+\int_0^{\infty} \mu_2(y) e^{-\gamma y} \int_0^y \pi_4(\tau) e^{\gamma \tau} \rm{d} \tau \rm{d} y , \end{align} $

(3.14)

$ \begin{align} C_3=&P_3(0)=\lambda_2 P_0+\int_0^\infty \mu_2(y) P_9(y) \rm{d} y \\ =&\frac{\lambda_2}{\gamma+\lambda_1+\lambda_2} \pi_0+\int_0^\infty \mu_2(y) e^{-\gamma y} \int_0^y \pi_9(\tau) e^{\gamma \tau} \rm{d} \tau \rm{d} y, \end{align} $

(3.15)

$ \begin{align} C_7=&P_7(0)=\lambda_2 P_1+\int_0^\infty \mu_2(y) P_5(y) \rm{d} y \\ =&\frac{\lambda_2}{\gamma+\lambda_1+\lambda_2} \pi_1+\int_0^\infty \mu_2(y) e^{-\gamma y} \int_0^y \pi_5(\tau) e^{\gamma \tau} \rm{d} \tau \rm{d} y. \end{align} $

(3.16)

当$ \gamma>\mu_1+\mu_2 $时, 由定理条件, (3.12)–(3.16) 和Fubini定理推出

$ \begin{align} \left|C_2\right| \leqslant & \frac{\lambda_1}{\gamma+\lambda_1+\lambda_2}\left(\left|\pi_0\right|+\left|\pi_1\right|\right)+\mu_2^2 \int_0^{\infty} e^{-\gamma y} \int_0^y\left|\pi_{10}(\tau)\right| e^{\gamma x} \rm{d} \tau \rm{d} y \cdot \int_0^{\infty} e^{-\gamma y} \rm{d} y \\ & +\mu_2 \int_0^{\infty} e^{-\gamma y} \int_0^y\left|\pi_4(\tau)\right| e^{\gamma \tau} \rm{d} \tau \rm{d}y \\ =&\frac{\lambda_1}{\gamma+\lambda_1+\lambda_2}\left(\left|{\pi_0}\right|+\left|\pi_1\right|\right) \\ & +\frac{1}{\gamma} \mu_2^2 \int_0^{\infty}\left|\pi_{10}(\tau)\right| e^{\gamma \tau} \int_\tau^{\infty} e^{-\gamma y} \rm{d} y \rm{d}\tau +\mu_2 \int_0^{\infty}\left|\pi_4(\tau)\right| e^{\gamma \tau} \int_\tau^\infty e^{-\gamma y} \rm{d} y \rm{d} \tau \\ =&\frac{\lambda_1}{\gamma+\lambda_1+\lambda_2}(|\pi_0|+|\pi_1|)+\frac{{\mu_2}^2}{\gamma^2}\|\pi_{10}\|_{L^1[0, \infty)}+\frac{{\mu_2}}{\gamma}\|\pi_{4}\|_{L^1[0, \infty)}, \end{align} $

(3.17)

$ \begin{align} \left|C_3\right| \leqslant & \frac{\lambda_2}{\gamma+\lambda_1+\lambda_2}\left|\pi_0\right|+\mu_2 \int_0^{\infty} e^{-\gamma y} \int_0^y\left|\pi_9(\tau)\right| e^{\gamma \tau} \rm{d} \tau \rm{d} y \\ =& \frac{\lambda_2}{\gamma+\lambda_1+\lambda_2}|\pi_0|+\mu_2 \int_0^\infty\left|\pi_9(\tau)\right| e^{\gamma \tau} \int_\tau^\infty e^{-\gamma y} \rm{d} y \rm{d} \tau \\ =& \frac{\lambda_2}{\gamma+\lambda_1+\lambda_2} |\pi_0|+\frac{1}{\gamma}\left\|\pi_9\right\|_{L^1[0, \infty)}, \end{align} $

(3.18)

$ \begin{align} \left|C_4\right| \leqslant & \mu_2 \int_0^{\infty} e^{-\gamma y} \int_0^y\left|\pi_{10}(\tau)\right| e^{\gamma \tau} \rm{d} \tau \rm{d} y=\mu_2 \int_0^{\infty}\left|\pi_{10 }(\tau)\right| e^{\gamma \tau} \int_\tau^{\infty} e^{-\gamma y} \rm{d} y \rm{d} \tau \\ =& \frac{\mu_2}{\gamma} \left\|\pi_{10}\right\|_{L^1[0, \infty)}, \end{align} $

(3.19)

$ \begin{align} \left|C_7\right| \leqslant & \frac{\lambda_2}{\gamma+\lambda_1+\lambda_2}\left|\pi_1\right|+\mu_2 \int_0^{\infty} e^{-\gamma y} \int_0^y\left|\pi_9(\tau)\right| e^{\gamma \tau} \rm{d} \tau \rm{d} y \\ & =\frac{\lambda_2}{\gamma+\lambda_1+\lambda_2}\left|\pi_1\right|+\frac{\mu_2}{\gamma} \left\|\pi_9\right\| _{L^1[0, \infty)}, \end{align} $

(3.20)

$ \begin{align} \left|C_8\right| \leqslant & \mu_2 \int_0^{\infty} e^{-\gamma y} \int_0^y\left|\pi_6(\tau)\right| e^{\gamma \tau} \rm{d} \tau \rm{d} y=\frac{\mu_2}{\gamma} \left\|\pi_6\right\|_{L^1[0, \infty)}. \end{align} $

(3.21)

由(3.9), (3.10), (3.17)–(3.21) 与Fubini定理得到

$ \begin{align} \|P\|=& \sum\limits_{i=0}^1\left|P_i\right|+\sum\limits_{j=2}^{10}\left\|P_j\right\|_{L^1[0, \infty)}=\sum\limits_{i=0}^1\left|P_i\right|+\sum\limits_{j=2}^{10} \int_0^{\infty}\left|P_j(y)\right| \rm{d} y \\ \leq &\frac{1}{\gamma+\lambda_1+\lambda_2}\left(\left|\pi_0\right|+\left|\pi_1\right|\right)+\sum\limits_{j=2}^4 \int_0^{\infty}\left(\left|C_j\right| e^{-\gamma y}+e^{-\gamma y} \int_0^y\left|\pi_j(\tau)\right| e^{\gamma \tau} \rm{d} \tau\right) \rm{d} y \\ & +\sum\limits_{k=5}^6 \int_0^{\infty} e^{-\gamma y} \int_0^y\left|\pi_k(\tau)\right| e^{\gamma \tau} \rm{d} \tau \rm{d} y+\sum\limits_{l=7}^8 \int_0^\infty \left(\left|C_l\right| e^{-\gamma y}+e^{-\gamma y} \int_0^y\left|\pi_l(\tau)\right| e^{\gamma\tau} \rm{d} \tau\right) \rm{d} y \\ & +\sum\limits_{m=9}^{10} \int_0^{\infty} e^{-\gamma y} \int_0^y\left|\pi_m(\tau)\right| e^{\gamma \tau} \rm{d} \tau \rm{d} y \\ =& \frac{1}{\gamma+\lambda_1+\lambda_2}\left(\left|\pi_0\right|+\left|\pi_1\right|\right)+\sum\limits_{j=2}^4\left(\frac{1}{\gamma}\left|C_j\right|+\int_0^{\infty}\left|\pi_j(\tau)\right| e^{\gamma \tau} \int_{\tau}^{\infty} e^{-\gamma y} \rm{d} y \rm{d} \tau\right) \\ & +\sum\limits_{k=5}^6 \int_0^{\infty}\left|\pi_k(\tau)\right| e^{\gamma \tau} \int_\tau^{\infty} e^{-\gamma y} \rm{d} y \rm{d} \tau +\sum\limits_{l=7}^8\left(\frac{1}{\gamma}\left|C_l\right|+\int_0^{\infty}\left|\pi_l(\tau)\right| e^{\gamma \tau} \int_\tau^\infty e^{-\gamma y} \rm{d} y \rm{d} \tau\right) \\ & +\sum\limits_{m=9}^{10} \int_0^{\infty}\left|\pi_m(\tau)\right| e^{\gamma \tau} \int_\tau^{\infty} e^{-\gamma y} \rm{d} y \rm{d} \tau \\ =& \frac{1}{\gamma+\lambda_1+\lambda_2}\left(\left|\pi_0\right|+\left|\pi_1\right|\right)+\sum\limits_{j=2}^4\left(\frac{1}{\gamma}\left|C_j\right| +\frac{1}{\gamma}\left\|\pi_j\right\|_{L^1[0, \infty)}\right) \\ & +\sum\limits_{k=5}^6 \frac{1}{\gamma}\left\|\pi_k\right\|_{L^1[0, \infty)}+\sum\limits_{l=7}^8\left(\frac{1}{\gamma}\left|C_l\right|+\frac{1}{\gamma}\left\|\pi_l\right\|_{L^1[0, \infty)} \right)+\sum\limits_{m=9}^{10} \frac{1}{\gamma}\left\|\pi_m\right\|_{L^1[0, \infty)} \\& =\frac{1}{\gamma+\lambda_1+\lambda_2}\left(\left|\pi_0\right|+\left|\pi_1\right|\right)+\sum\limits_{i=2, 3, 4, 7, 8 } \frac{1}{\gamma}\left|C_i\right|+\frac{1}{\gamma} \sum\limits_{j=2}^{10}\left\|\pi_j\right\|_{L^1[0, \infty)} \\ \leqslant & \frac{1}{\gamma+\lambda_1+\lambda_2}\left(\left|\pi_0\right|+\left|\pi_1\right|\right) +\frac{\lambda_1}{\gamma\left(\gamma+\lambda_1+\lambda_2\right)}\left(\left|\pi_0\right|+\left|\pi_1\right|\right) \\ &+\frac{\mu_2^2}{\gamma^3}\left\|\pi_{10}\right\|_{L^1[0, \infty)}+\frac{\mu_2}{\gamma^2}\left\|\pi_4 \right\|_{L^1[0, \infty)} \\ &+\frac{\lambda_2}{\gamma\left(\gamma+\lambda_1+\lambda_2\right)}\left|\pi_0\right|+\frac{1}{\gamma^2}\left\|\pi_9\right\|_{L^1[0, \infty)} +\frac{\mu_2}{\gamma^2}\left\|\pi_{10}\right\|_{L^1[0, \infty)} \\ &+\frac{\lambda_2}{\gamma\left(\gamma+\lambda_1+\lambda_2\right)}|\pi_1|+\frac{\mu_2}{\gamma^2}\left\|\pi_{9}\right\|_{L^1[0, \infty)} +\frac{\mu_2}{\gamma^2}\left\|\pi_6\right\|_{L^1[0, \infty)}+\frac{1}{\gamma} \sum\limits_{i=2}^{10}\left\|\pi_i\right\|_{L^1[0, \infty)} \\ =&\left(\frac{1}{\gamma+\lambda_1+\lambda_2}+\frac{\lambda_1+\lambda_2}{\gamma\left(\gamma+\lambda_1+\lambda_2\right)}\right)\left(\left|\pi_0\right|+|\pi_1 |\right) +\frac{1}{\gamma} \sum\limits_{i=2, 3, 5, 7, 8}\left\|\pi_i\right\|_{L^1[0, \infty)} \\ & +\left(\frac{1}{\gamma}+\frac{\mu_2}{\gamma^2}\right) \sum\limits_{j=4, 6, 9}\left\|\pi_j\right\|_{L^1[0, \infty)} +\left(\frac{1}{\gamma}+\frac{\mu_2}{\gamma^2}+\frac{\mu_2^2}{\gamma^3}\right)\left\|\pi_{10}\right\|_{L^1[0, \infty)} \\ =&\frac{1}{\gamma}\left(\left|\pi_0\right|+\left|\pi_1\right|\right)+\frac{1}{\gamma} \sum\limits_{i=2, 3, 5, 7, 8} \|\pi_i\|_{L^1[0, \infty)} \\ &+\frac{\mu_2+\gamma}{\gamma^2} \sum\limits_{j=4, 6, 9}\| \pi_j\|_{L^1[0, \infty)}+\frac{\gamma^2+\mu_2 \gamma+\mu_2^2}{\gamma^3}\| \pi_{10} \|_{L^1[0, \infty)} \\ \leqslant &\frac{1}{\gamma-\mu_2}\|\pi\|. \end{align} $

(3.22)

当$ \gamma>\mu_2 $时, (3.22) 说明, $ \left(\gamma I-A_1\right)^{-1} $存在, 并且

$ (\gamma I-A_1)^{-1}: Y \rightarrow D(A_1), \|(\gamma I-A_1)^{-1}\| \leqslant \frac{1}{\gamma-\mu_2}. $

第二步, 证明$ D(A_1) $在状态空间$ Y $中稠密, 取集合

$ \begin{gathered} \mho=\left\{P=\left(P_0, P_1, P_2(y), \cdots, P_{10}(y)\right) \mid P_i \in R, \;P_j \in C_0^{\infty}[0, \infty), \; i=0, 1, \; j=2, \cdots, 10\right\}, \end{gathered} $

则由文献 [15]知道$ \mho $在状态空间$ Y $中稠密. 若定义集合

$ \begin{align} \Omega=\left\{P=\left(P_0, P_1, P_2(y), \cdots , P_{10}(y)\right) \left| \begin{array}{c} P_i \in C_0^{\infty}[0, \infty), \;\text{且存在}\; \tau_i>0 , \;\text{当} \\ y\in[0, \tau_i]\; \text{时}, \; \text{有}\;P_i(y)=0, i=2, \cdots, 10 \end{array}\right.\right\} \end{align} $

则容易证明$ \Omega $在$ \mho $中稠密. 由$ \Omega, \mho, Y $的关系知道, 为了证明$ D(A_1) $在$ Y $中稠密, 只需证$ D(A_1) $在$ \Omega $中稠密即可.

任取$ P=\left(P_0, P_1, P_2(y), \cdots, P_{10}(y)\right) \in \Omega $, 存在$ \tau_i>0 $, 使得$ y \in\left[0, \tau_i\right] $时, 有$ P_i(y)=0(i=2, \cdots, 10) $. 若定义$ 2 s=\mathop {\operatorname{min}} \limits_{2\leq k \leq 10}\tau_k $, 那么对$ y \in [0, 2s] $都有 $ P_i(y)=0\;(i=2, \cdots, 10) $.

定义

$ \begin{aligned} h^s(0)= & \left(h_0^s, h_1^s, h_2^s(0), h_3^s(0), h_4^s(0), h_5^s(0), h_6^s(0), h_7^s(0), h_8^s(0), h_9^s(0), h_{10}^s(0)\right) \\ = & \left(P_0, P_1, \lambda_1\left(P_0+P_1\right)+\int_{2s}^{\infty} \mu_2(y) P_4(y) \rm{d} y+\int_{2 s}^{\infty} \mu_2(y) P_8(y) \rm{d} y, \right. \\ & \qquad\left.\lambda_2 P_0+\int_{2 s}^{\infty} \mu_2(y) P_9(y) \rm{d} y , \int_{2 s}^{\infty} \mu_2(y) P_{10}(y) \rm{d} y , \right.\\ & \qquad\left.0, 0, \lambda_2 P_1+\int_{2 s}^{\infty} \mu_2(y) P_5(y) \rm{d} y, \int_{2 s}^{\infty} \mu_2(y) P_6(y) \rm{d} y, 0, 0\right)\\ h^s=&\left(h_0^s, h_1^s, h_2^s(y), h_3^s(y), h_4^s(y), h_5^s(y), h_6^s(y), h_7^s(y), h_8^s(y), h_9^s(y), h_{10}^s(y)\right), \end{aligned} $

其中

$ \begin{aligned} & h_i^s(y)= \begin{cases}h_i^s(0)\left(1-\frac{y}{s}\right)^2, & \text { 当 }\; y \in[0, s) , \\ -d_i(y-s)^2(y-2 s)^2, & \text { 当 } \;y \in[s, 2s) , \quad i=2, \cdots, 10, \\ P_i(y) , &\text { 当 } \;y \in[2 s, \infty), \end{cases} \\ &d_2=\frac{\int_0^s \mu_2(y) (h_2^s(0)+h_8^s(0))\left(1-\frac{y}{s}\right)^2 \mathrm{d} y}{2\int_s^{2 s} \mu_2(y)(y-s)^2(y-2 s)^2 \mathrm{d} y}, \;\; d_3=\frac{\int_0^s \mu_2(y) h_9^s(0)\left(1-\frac{y}{s}\right)^2 \mathrm{d} y}{\int_s^{2 s} \mu_2(y)(y-s)^2(y-2 s)^2 \mathrm{d} y}, \\ & d_4=\frac{\int_0^s \mu_2(y) h_{10}^s(0)\left(1-\frac{y}{s}\right)^2 \mathrm{d} y}{\int_s^{2 s} \mu_2(y)(y-s)^2(y-2 s)^2 \mathrm{d} y} , \;\; d_7=\frac{\int_0^s \mu_2(y) h_{5}^s(0)\left(1-\frac{y}{s}\right)^2 \mathrm{d} y}{\int_s^{2 s} \mu_2(y)(y-s)^2(y-2 s)^2 \mathrm{d} y}, \\ &d_8=\frac{\int_0^s \mu_2(y) h_{6}^s(0)\left(1-\frac{y}{s}\right)^2 \mathrm{d} y}{\int_s^{2 s} \mu_2(y)(y-s)^2(y-2 s)^2 \mathrm{d} y}, \;\;d_5=d_6=d_9=d_{10}=0. \end{aligned} $

则, 容易验证$ h^s \in D\left(A_1\right) $, 且

$ \begin{aligned} \left\|P-h^s\right\|= & \sum\limits_{i=2}^{10} \int_0^{\infty}\left|P_i(y)-h_i^s(y)\right| \mathrm{d} y\\ = & \sum\limits_{i=2}^{10} \int_0^s\left|h_i^s(0)\right|\left(1-\frac{y}{s}\right)^2 \mathrm{d} y+\sum\limits_{i=2}^{10} \int_s^{2 s}\left|d_i\right|(y-s)^2(y-2 s)^2 \mathrm{d} y \\ = & \frac{s}{3} \cdot\sum\limits_{i=2}^{10}\left|h_i^s(0)\right|+\frac{s^5}{30}\cdot\sum\limits_{i=2}^{10}\left|d_i\right| \rightarrow 0, \text { 当 }\; s \rightarrow 0 . \end{aligned} $

上式表明$ D\left(A_1\right) $在$ \Omega $中稠密, 即$ D\left(A_1\right) $在$ Y $中稠密. 由前面两个步骤和Hille-Yosida定理即知, 算子$ A_1 $生成某个$ C_0- $半群.

第三步证明其他两个算子$ A_2 $与$ A_3 $是有界线性算子. 对$ \forall P\in Y $, 根据$ A_2, $ $ A_3 $和状态空间$ Y $中范数的定义有

$ \begin{align} \left\|A_2 P\right\| \leqslant & \mu_1\left\|P_2\right\|_{L^1[0, \infty)}+\left( \lambda_1+ \lambda_2+\mu_2\right) \|P_3\|_{L^1[0, \infty)}+\mu_2\| P_4 \|_{L^1[0, \infty)} +\lambda_1\|P_3\|_{L^1[0, \infty)} \\ &+(\lambda_3+\mu_2)\| P_5 \|_{L^1[0, \infty)}+\lambda_2\| P_3 \|_{L^1[0, \infty)} +\mu_2\| P_6 \|_{L^1[0, \infty)}+\lambda_3\| P_5 \|_{L^1[0, \infty)} \\ &+\left( \lambda_1+ \lambda_2+\mu_2\right) \|P_7\|_{L^1[0, \infty)} +\mu_2\| P_8 \|_{L^1[0, \infty)}+\lambda_1\| P_7 \|_{L^1[0, \infty)} \\ &+(\lambda_3+\mu_2)\| P_9 \|_{L^1[0, \infty)}+\lambda_2\| P_7 \|_{L^1[0, \infty)} +\mu_2\| P_{10} \|_{L^1[0, \infty)}+\lambda_3\| P_9 \|_{L^1[0, \infty)} \\ = & \mu_1 \| P_2 \|_{L^1[0, \infty)}+\left( 2\lambda_1+ 2\lambda_2+\mu_2\right) \|P_3\|_{L^1[0, \infty)}+\mu_2\|P_4\|_{L^1[0, \infty)} \\ &+(2\lambda_3+\mu_2)\|P_5\|_{L^1[0, \infty)}+\mu_2 \|P_6\|_{L^1[0, \infty)} +\left( 2\lambda_1+ 2\lambda_2+\mu_2\right) \|P_7\|_{L^1[0, \infty)} \\ &+\mu_2 \|P_8\|_{L^1[0, \infty)}+(2\lambda_3+\mu_2)\|P_9\|_{L^1[0, \infty)}+\mu_2\|P_{10}\|_{L^1[0, \infty)} \\ \leqslant & \left(2 \lambda_1+2 \lambda_2+2 \lambda_3+{\mu}_1+{\mu}_2\right)\|{P}\|, \end{align} $

$ \begin{align} \left\|A_3 P\right\| & =\left|\int_0^{\infty} \mu_1(y) P_2(y) \mathrm{d} y+\int_0^{\infty} \mu_2(y) P_7(y) \mathrm{d} y\right|+\left|\int_0^{\infty} \mu_2(y) P_3(y) \mathrm{d} y\right| \\ & \leqslant {\mu}_1\left\|P_2\right\|_{L^1[0, \infty)}+{\mu}_2\left\|P_7\right\|_{L^1[0, \infty)}+{\mu}_2\left\|P_3\right\|_{L^1[0, \infty)} \\ & \leqslant \left({\mu}_1+{\mu}_2\right)\|P\|. \end{align} $

上面的两个不等式表明算子$ A_2 $和$ A_3 $是有界的, 同时容易推出这两个算子是线性算子. 从而, 由$ C_0 $－半群的扰动理论^[16]即知, 主算子$ A_1+A_2+A_3 $生成－个$ C_0 $-半群$ T(t) $.

第四步证明主算子$ A_1+A_2+A_3 $是dispersive算子. 对$ \forall P \in D\left(A_1\right) $, 定义

$ \alpha(y)=\left(\frac{\left[P_0\right]^{+}}{P_0}, \frac{\left[P_1\right]^{+}}{P_1}, \frac{\left[P_2(y)\right]^{+}}{P_2(y)}, \frac{\left[P_3(y)\right]^+}{P_3(y)}, \cdots, \frac{\left[P_{10}(y)\right]^{+}}{P_{10}(y)}\right), $

其中

$ \begin{array}{l} {\left[P_n\right]^{+}=\left\{\begin{array}{cc} P_n & \text { 当 } P_n>0 \\ 0 & \text { 当 } P_n \leqslant 0 \end{array}, n=0, 1, \right.} \end{array} \; \begin{array}{ll} {\left[P_k(y)\right]^{+}=\left\{\begin{array}{cc} P_k(y) & \text { 当 } P_k(y)>0 \\ 0 & \text { 当 } P_k(y) \leqslant 0 \end{array}, k=2, \ldots, 10, \right.} \end{array} $

若定义 $ B_k^{+}=\left\{y \in[0, \infty) \mid P_k(y)>0\right\}, B_k^{-}=\left\{y \in[0, \infty) \mid P_k(y) \leqslant 0\right\}(k=2, \cdots, 10),$ 则由$ \left[P_k(y)\right]^{+}(k =2, \cdots, 10) $的定义和文献 [17]同样的方法可以推出

$ \begin{align} & \int_0^{\infty} \frac{\mathrm{d} P_k(y)}{\mathrm{d} y} \cdot \frac{\left[P_k(y)\right]^{+}}{P_k(y)} \mathrm{d} y=\int_{B_k^{+}} \frac{\mathrm{d} P_k(y)}{\mathrm{d} y} \cdot \frac{\left[P_k(y)\right]^{+}}{P_k(y)} \mathrm{d} y+\int_{B_k^{-}} \frac{\mathrm{d} P_k(y)}{\mathrm{d} y} \cdot \frac{\left[P_k(y)\right]^{+}}{P_k(y)} \mathrm{d} y \\ & =\int_{B_k^+}\frac{\mathrm{d} P_k(y)}{\mathrm{d} y} \mathrm{d} y=\int_0^{\infty} \frac{\mathrm{d}\left[P_k(y)\right]^{+}}{\mathrm{d} y}\mathrm{d}y=-\left[P_k(0)\right]^{+}, k=2, \ldots, 10 . \end{align} $

(3.23)

那么对上述定义的$ \alpha(y) $和$ P \in D(A_1) $, 由(3.23), 边界条件和不等式

$ \begin{align} &\int_0^{\infty} \mu_n(y) P_k(y) \mathrm{d} y \leqslant \int_0^\infty \mu_n(y) [P_k(y)]^{+} \mathrm{d} y(n=1, 2, k=2, \cdots, 10), \\ &\int_0^{\infty}P_l(y) \frac{[P_m(y)]^+}{P_m(y)} \mathrm{d} y \leq \int_0^{\infty}[P_l(y)]^+\mathrm{d}y \;(l, m=2, \cdots, 10, l\neq m ), \end{align} $

推出

$ \begin{align} \langle\left(A_1+A_3+A_3\right) P, \alpha\left(y\right)\rangle =& \left\{-\left(\lambda_1+\lambda_2\right) P_0+\int_0^{\infty} \mu_1(y) P_2(y) \mathrm{d} y+\int_0^\infty \mu_2(y) P_7(y) \mathrm{d} y\right\} \cdot \frac{\left[P_0\right] ^+}{P_0} \\ & +\left\{-\left(\lambda_1+\lambda_2\right) P_1+\int_0^{\infty} \mu_2(y) P_3(y) \mathrm{d} y\right\} \cdot \frac{\left[P_1\right]^+}{P_1} \\ & +\int_0^{\infty}\left\{-\frac{\mathrm{d} P_2(y)}{\mathrm{d} y}-\mu_1(y) P_2(y)\right\} \cdot \frac{\left[P_2(y)\right]^+}{P_2(y)} \mathrm{d} y \end{align} $

$ \begin{align} & +\int_0^{\infty}\left\{-\frac{\mathrm{d} P_3(y)}{\mathrm{d} y}-\left(\lambda_1+\lambda_2+\mu_2(y)\right) P_3(y)\right\} \cdot \frac{\left[P_3(y)\right]^{+}}{P_3(y)} \mathrm{d} y \\ & +\int_0^{\infty}\left\{-\frac{\mathrm{d} P_4(y)}{\mathrm{d} y}-\mu_2(y) P_4(y)+\lambda_1 P_3(y)\right\} \cdot \frac{\left[P_4(y)\right]^+}{P_4(y)} \mathrm{d} y \\ &+\int_0^{\infty}\left\{-\frac{\mathrm{d} P_5(y)}{\mathrm{d} y} -\left(\lambda_3+\mu_2(y)\right) P_5(y)+\lambda_2 P_3(y)\right\} \cdot \frac{\left[P_5(y)\right]^+}{P_5(y)} \mathrm{d} y \\ & +\int_0^{\infty}\left\{-\frac{\mathrm{d} P_6(y)}{\mathrm{d} y}-\mu_2(y) P_6(y)+\lambda_3 P_5(y)\right\} \cdot \frac{\left[P_6(y)\right]^{+}}{P_6(y)} \mathrm{d} y \\ & +\int_0^{\infty}\left\{-\frac{\mathrm{d} P_7(y)}{\mathrm{d} y}-\left(\lambda_1+\lambda_2+\mu_2(y)\right)P_7(y)\right\} \cdot \frac{\left[P_7(y)\right]^{+}}{P_7(y)} \mathrm{d} y \\ & +\int_0^{\infty}\left\{-\frac{\mathrm{d} P_8(y)}{\mathrm{d} y}-\mu_2(y) P_8(y)+\lambda_1 P_7(y)\right\}\cdot \frac{\left[P_8(y)\right]^{+}}{P_8(y)} \mathrm{d} y \\ &+\int_0^{\infty}\left\{-\frac{\mathrm{d} P_9(y)}{\mathrm{d} y}-\left(\lambda_3+\mu_2(y)\right) P_9(y)+\lambda_2 P_7(y)\right\} \cdot \frac{\left[P_9(y)\right]^+}{P_9(y)} \mathrm{d} y \\ & +\int_0^{\infty}\left\{-\frac{\mathrm{d} P_{10}(y)}{\mathrm{d} y}-\mu_2(y) P_{10}(y)+\lambda_3 P_9(y)\right\} \cdot \frac{\left[P_{10}(y)\right]^+}{P_{10}(y)} \mathrm{d} y \\ =& -\left(\lambda_1+\lambda_2\right)\left[P_0\right]^{+}+\frac{\left[P_0\right]^{+}}{P_0} \int_0^\infty \mu_1(y) P_2(y) \mathrm{d} y+\frac{\left[P_0\right]^{+}}{P_0} \int_0^\infty \mu_2(y) P_7(y) \mathrm{d} y \\ & -\left(\lambda_1+\lambda_2\right)\left[P_1\right]^{+}+\frac{\left[P_1\right]^+}{P_1} \int_0^{\infty} \mu_2(y) P_3(y) \mathrm{d} y-\sum_{k=2}^{10} \int_0^{\infty} \frac{\mathrm{d} P_k(y)}{\mathrm{d} y}\cdot\frac{\left[P_k(y)\right]^{+}}{P_k(y)} \mathrm{d} y \\ & -\int_0^\infty \mu_1(y)\left[P_2(y)\right]^+ \mathrm{d} y-\int_0^{\infty}\left(\lambda_1+\lambda_2+\mu_2(y)\right)\left[P_3(y)\right]^{+} \mathrm{d} y \\ & -\int_0^\infty \mu_2(y)\left[P_4(y)\right]^+\mathrm{d} y+\int_0^{\infty} \lambda_1 P_3(y) \cdot \frac{\left[P_4(y)\right]^+}{P_4(y)} \mathrm{d} y \\ & -\int_0^{\infty}\left(\lambda_3+\mu_2(y)\right)\left[P_5(y)\right]^+\mathrm{d} y+\int_0^{\infty} \lambda_2 P_3(y) \frac{[P_5(y)]^+}{P_5(y)} \mathrm{d} y \\ & -\int_0^{\infty} \mu_2(y)\left[P_6(y)\right]^+\mathrm{d} y+\int_0^{\infty} \lambda_3 P_5(y) \frac{\left[P_6(y)\right]^{+}}{P_6(y)} \mathrm{d} y \\ & -\int_0^\infty\left(\lambda_1+\lambda_2+\mu_2(y)\right)\left[P_7(y)\right]^+ \mathrm{d} y-\int_0^\infty \mu_2(y)\left[P_8(y)\right]^{+} \mathrm{d} y \\ & +\int_0^\infty \lambda_1 P_7(y) \cdot \frac{\left[P_8(y)\right]^{+}}{P_8(y)} \mathrm{d} y-\int_0^{\infty}\left(\lambda_3+\mu_2(y)\right)\left[P_9(y)\right]^{+} \mathrm{d} y \\ & +\int_0^{\infty} \lambda_2 P_7(y) \frac{\left[P_9(y)\right]^+}{P_9(y)} \mathrm{d} y-\int_0^{\infty} \mu_2(y)\left[P_{10}(y)\right]^+ \mathrm{d} y+\int_0^{\infty} \lambda_3 P_9(y) \frac{\left[P_{10}(y)\right]^+}{P_{10}(y)} \mathrm{d} y \\ \leqslant & -\left(\lambda_1+\lambda_2\right)\left(\left[P_0\right]^{+}+\left[P_1\right]^{+}\right) \\ & +\frac{\left[P_0\right]^+}{P_0} \left(\int_0^{\infty} \mu_1(y)\left[P_2(y)\right]^{+} \mathrm{d} y+\int_0^\infty \mu_2(y)\left[P_7(y)\right]^{+} \mathrm{d} y\right) \\ & +\frac{\left[P_1\right]^+}{P_1} \int_0^{\infty} \mu_2(y) [P_3(y)]^{+} \mathrm{d} y+\sum_{k=2}^{10}\left[P_k(0)\right]^{+} -\int_0^{\infty}\mu_1(y)\left[P_2(y)\right]^+ \mathrm{d} y \end{align} $

$ \begin{align} &-\left(\lambda_1+\lambda_2\right) \int_0^{\infty}\left[P_3(y)\right]^{+} \mathrm{d} y-\int_0^{\infty} \mu_2(y)\left[P_3(y)\right]^{+} \mathrm{d} y-\int_0^{\infty} \mu_2(y)\left[P_4(y)\right]^+ \mathrm{d} y \\ &+\lambda_1 \int_0^{\infty}\left[P_3(y)\right]^{+} \mathrm{d} y-\lambda_3 \int_0^{\infty}\left[P_5(y)\right]^+\mathrm{d} y-\int_0^{\infty} \mu_2(y)\left[P_5(y)\right]^+\mathrm{d} y \\ & +\lambda_2 \int_0^{\infty}\left[P_3(y)\right]^+\mathrm{d} y-\int_0^{\infty} \mu_2(y)\left[P_6(y)\right]^+ \mathrm{d} y+\lambda_3 \int_0^{\infty}\left[P_5(y)\right]^{+} \mathrm{d} y \\ & -\left(\lambda_1+\lambda_2\right) \int_0^{\infty}\left[P_7(y)\right]^+ \mathrm{d} y-\int_0^{\infty} \mu_2(y)\left[P_7(y)\right]^+\mathrm{d} y-\int_0^{\infty} \mu_2(y) [P_8(y)]^+ \mathrm{d} y \\ & +\lambda_1 \int_0^{\infty}\left[P_7(y)\right]^+ \mathrm{d} y -\lambda_3 \int_0^{\infty}\left[P_9(y)\right]^{+} \mathrm{d} y-\int_0^{\infty} \mu_2(y)\left[P_9(y)\right]^{+} \mathrm{d} y \\ & +\lambda_2 \int_0^{\infty}\left[P_7(y)\right]^{+} \mathrm{d} y-\int_0^\infty \mu_2(y)\left[P_{10}(y)\right]^{+} \mathrm{d} y+\lambda_3 \int_0^{\infty}\left[P_9(y)\right]^{+} \mathrm{d} y \\ =& -\left(\lambda_1+\lambda_2\right)\left(\left[P_0\right]^{+}+\left[P_1\right]^{+}\right) \\ & +\frac{\left[P_0\right]^{+}}{P_0}\left(\int_0^{\infty} \mu_1(y)\left[P_2(y)\right]^+\mathrm{d} y+\int_0^{\infty} \mu_2(y)\left[P_7(y)\right]^+ \mathrm{d} y\right) \\ & +\frac{\left[P_1\right]^{+}}{P_1} \int_0^{\infty} \mu_2(y)\left[P_3(y)\right]^+ \mathrm{d} y +\lambda_1\left(\left[P_0\right]^{+}+\left[P_1\right]^{+}\right) \\ & +\int_0^\infty \mu_2(y)\left[P_4(y)\right]^+\mathrm{d} y+\int_0^{\infty} \mu_2(y) [P_8(y)]^{+} \mathrm{d} y \\ & +\lambda_2 [P_0]^{+}+\int_0^{\infty} \mu_2(y)\left[P_9(y)\right]^+ \mathrm{d} y+\int_0^{\infty} \mu_2(y)\left[P_{10}(y)\right]^+ \mathrm{d} y \\ & +\lambda_2\left[P_1\right]^{+}+\int_0^{\infty} \mu_2(y)\left[P_5(y)\right]^+ \mathrm{d} y+\int_0^{\infty} \mu_2(y)\left[P_6(y)\right]^+\mathrm{d} y \\ & -\int_0^{\infty} \mu_1(y)\left[P_2(y)\right]^{+} \mathrm{d} y-\int_0^{\infty} \mu_2(y)\left[P_3(y)\right]^+ \mathrm{d} y \\ &-\int_0^{\infty} \mu_2(y)\left[P_4(y)\right]^+ \mathrm{d} y-\int_0^{\infty} \mu_2(y)\left[P_5(y)\right]^{+} \mathrm{d} y \\ & -\int_0^{\infty} \mu_2(y)\left[P_6(y)\right]^{+} \mathrm{d} y-\int_0^{\infty} \mu_2(y)\left[P_7(y)\right]^{+} \mathrm{d} y \\ &-\int_0^{\infty} \mu_2(y)\left[P_8(y)\right]^+ \mathrm{d} y-\int_0^{\infty} \mu_2(y)\left[P_9(y)\right]^+ \mathrm{d} y -\int_0^{\infty} \mu_2(y)\left[P_{10}(y)\right]^{+} \mathrm{d} y \\ =& \left(\frac{\left[P_0\right]^{+}}{P_0}-1\right)\left(\int_0^{\infty} \mu_1(y)\left[P_2(y)\right]^{+} \mathrm{d} y+\int_0^{\infty} \mu_2(y)\left[P_7(y)\right]^{+} \mathrm{d} y\right) \\ &+\left(\frac{\left[P_1\right]^{+}}{P_1}-1\right) \int_0^{\infty} \mu_2(y)\left[P_3(y)\right]^{+} \mathrm{d} y \\ \leqslant& 0. \end{align} $

上式表明算子$ A_1+A_2+A_3 $是dispersive算子. 由这个结论, 第一步, 第二步和Phillips定理推出算子$ A_1+A_2+A_3 $生成一个正压缩$ C_0 $-半群. 由$ C_0 $-半群的唯一性理论^[16]知道, $ A_1+A_2+A_3 $生成的$ C_0 $-半群就是$ T(t) $.

在状态空间$ Y $中定义集合

$ \Theta=\left\{P \in Y \mid P=\left(P_0, P_1, P_2(y), \cdots, P_{10}(y)\right), P_0 \geqslant 0, P_1 \geqslant 0, P_k(y) \geqslant 0, k=2, \cdots, 10\right\}, $

则所定义的集合$ \Theta $是$ Y $中的锥. 由$ Y $的定义易证它的共轭空间

$ Y^*=\left\{P^* \in R^2 \times \left(L^{\infty}[0, \infty)\right)^9\left|\left|\left|\left|P^*\right|\right|\right|=\sup \left\{\left|P_0^*\right|, \left|P_1^*\right|, \sup _{2 \leqslant k \leqslant 10}\left\|P_k\right\|_{L^{\infty}[0, \infty)}\right\}<\infty\right.\right\}, $

是一个Banach空间^[18]. 对$ P \in D\left(A_1\right) \cap \Theta, \text{取} P^*=\|P\|(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) $, 同时利用边界条件推出

$ \begin{align} & \langle\left(A_1+A_2+A_3\right) P, P^*\rangle \\ =&\left\{-\left(\lambda_1+\lambda_2\right) P_0+\int_0^{\infty} \mu_1(y) P_2(y) \mathrm{d} y+\int_0^{\infty} \mu_2(y) P_7(y) \mathrm{d} y\right\}\|P\| \\ & +\left\{-\left(\lambda_1+\lambda_2\right) P_1+\int_0^{\infty} \mu_2(y) P_3(y) \mathrm{d} y\right\}\|P\| \\ & +\int_0^{\infty}\left\{-\frac{\mathrm{d} P_2(y)}{\mathrm{d} y}-\mu_1(y) P_2(y)\right\}\|P\| \mathrm{d} y \\ &+\int_0^{\infty}\left\{-\frac{\mathrm{d} P_3(y)}{\mathrm{d} y}-\left(\lambda_1+\lambda_2+\mu_2(y)\right) P_3(y)\right\}\|P\| \mathrm{d} y \\ & +\int_0^{\infty}\left\{-\frac{\mathrm{d} P_4(y)}{\mathrm{d} y}-\mu_2(y) P_4(y)+\lambda_1 P_3(y)\right\}\|P\| \mathrm{d} y \\ &+\int_0^{\infty}\left\{-\frac{\mathrm{d} P_5 (y)}{\mathrm{d} y}-\left(\lambda_3+\mu_2(y)\right) P_5(y)+\lambda_2 P_3(y)\right\}\|P\| \mathrm{d} y \\ &+\int_0^{\infty}\left\{-\frac{\mathrm{d} P_6(y)}{\mathrm{d} y}-\mu_2(y)P_6(y)+\lambda_3 P_5(y)\right\}\|P\|\mathrm{d} y \\ &+\int_0^{\infty}\left\{-\frac{\mathrm{d} P_7(y)}{\mathrm{d} y}-\left(\lambda_1+\lambda_2+\mu_2(y)\right)P_7(y)\right\}\|P\| \mathrm{d} y \\ & +\int_0^{\infty}\left\{-\frac{\mathrm{d} P_8(y)}{\mathrm{d} y}-\mu_2(y) P_8(y)+\lambda_1 P_7(y)\right\}\|P\| \mathrm{d} y \\ &+\int_0^{\infty}\left\{-\frac{\mathrm{d} P_9(y)}{\mathrm{d} y}-\left(\lambda_3+\mu_2(y)\right) P_9(y)+\lambda_2 P_7(y)\right\}\|P\| \mathrm{d} y \\ & +\int_0^{\infty}\left\{-\frac{\mathrm{d} P_{10}(y)}{\mathrm{d} y}-\mu_2(y) P_{10}(y)+\lambda_3 P_9(y)\right\} \|P\| \mathrm{d} y \\ =&-\left(\lambda_1+\lambda_2\right)\|{P}\|\left(P_0+P_1\right)+\|P\| \sum\limits_{i=2}^{10} P_i(0)-\|P\| \sum\limits_{\substack{i=4 \\ i\neq7}}^{10} \int_0^{\infty} \mu_2(y) P_i(y) \mathrm{d} y \\ =&-\left(\lambda_1+\lambda_2\right)\|P\|\left(P_0+P_1\right)+\lambda_1\|P\|\left(P_0+P_1\right)+\|P\| \int_0^{\infty} \mu_2(y) P_4(y) \mathrm{d} y \\ & +\|P\| \int_0^\infty \mu_2(y) P_8(y) \mathrm{d} y +\lambda_2\|P\| P_0+\|P\|\int_0 ^{\infty}\mu_2(y) P_9 (y) \mathrm{d} y \\ &+\left\| P\right\|\int_0^{\infty} \mu_2(y) P_{10}(y) \mathrm{d} y+\lambda_2\left\| P\right\|P_1+\left\| P\right\| \int_0^{\infty} \mu_2(y) P_5(y) \mathrm{d} y \\ & +\|P\| \int_0^{\infty} \mu_2(y) P_6(y) \mathrm{d} y-\|P\| \sum\limits_{\substack{i=4 \\ i \neq 7}}^{10} \int_0^{\infty} \mu_2(y) P_i(y) \mathrm{d} y \\ & =0 \text {. } \end{align} $

上式表示算子$ A_1+A_2+A_3 $是保守算子. 由于初值$ P(0) \in D\left(A^2_1\right) \cap\Theta $, 因比可以用Fatforini定理^[19], 并得到如下结论：

定理3.2   $ T(t) $对于三部件系统(2.1) 的初值$ P(0) $是等距算子, 即

$ \begin{align} \|T(t)P(0)\|=\|P(0)\|(\forall t\in [0, \infty)). \end{align} $

(3.24)

综合定理3.1和定理3.2得到本文的主要结果：

定理3.3  如果$ \mu_1(y) $和$ \mu_2(y) $满足

$ {\mu}_1=\sup _{y \in[0, \infty)} \mu_1(y)<\infty, \quad {\mu}_2=\sup _{y \in[0, \infty)} \mu_2(y)<\infty, $

则系统(2.1) 存在唯一的非负时间依赖解$ P(t, y) $, 并满足$ \|P(t, \cdot)\|=1, \;\;\forall t\in [0, \infty). $

证   由于$ P(0) \in D\left({A_1}^2\right) \cap\Theta $, 所以由文献 [15]中的定理11与本文中的定理3.1知道, 所讨论的三部件系统(2.1) 存在唯一的正时间依赖解$ P(t, y) $, 并其可表示为

$ P(t, y)=T(t)P(0), \;\; \forall t \in[0, \infty), $

上式结合(3.24) 式得到

$ \begin{align} \|P(t, \cdot)\|=\|T(t)(1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)\|=\|(1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)\|=1, \;\;\forall t\in [0, \infty). \end{align} $

本文运用泛函分析中的强连续算子半群理论, 研究了一个由单向关闭的两不同部件串联与一部件冷贮备构成的三部件可修系统的时间依赖解的存在唯一性. 具体而言, 首先通过引入状态空间, 算子及其定义域, 将该系统的模型转化为Banach空间中的抽象Cauchy问题. 随后, 证明了该系统模型非负时间依赖解的存在唯一性, 为后续研究时间依赖解的渐近行为奠定了基础. 因此, 可以借鉴文献 [20]中的思想方法, 进一步探讨该系统时间依赖解的渐近行为.