在全球化趋势下, 对于供应链这个日益复杂的系统, 如何分析和提高其可靠性变得日益迫切, 并受到越来越多的关注^{[1-6, 8]}. Thomas于2002年首次将可靠性工程应用到供应链中, 提出用可靠度来度量供应链系统的可靠性^[3]. Sohn等认为供应链的可靠性就是顾客要求的产品质量可靠性^[4].王建、张文杰从单级供应链可靠性分析出发进行了可靠性的定量分析, 并根据分析结果提出了一些提高供应链可靠性的措施^[5].在文献[6]中作者通过分析供应链系统的状态之间的转移关系, 引入补充变量, 用补充变量法^[7], 建立了供应链系统的可靠性模型, 并对该模型系统解的存在唯一性进行讨论和证明.在文献[8]中当修复率为常数时讨论系统解的渐近性质.本文在文献[6]的基础上当修复率为函数时, 通过研究相应算子的谱的特征得到该系统时间依赖解的渐近行为.

由文献[6]知道, 该供应链系统的数学模型用以下方程组描述:

$ \begin{align} &\frac{dp_{0}(t)}{dt}= -\sum_{i=1}^4\lambda_ip_{0}(t)+\sum_{i=1}^4\int_0^\infty p_i(x, t)\mu_i(x)dx, \end{align}$

(2.1)

$\begin{align} &\frac{\partial{p_{i}(x, t)}}{\partial{t}}+ \frac{\partial{p_{i}(x, t)}}{\partial{x}} =-\mu_{i}(x) p_{i}(x, t), \;\; i=1, 2, 3, 4, \end{align}$

(2.2)

$\begin{align} &p_{i}(0, t)=\lambda_i p_{0}(t), \;\; i=1, 2, 3, 4, \end{align}$

(2.3)

$\begin{align} &p_{0}(0)=1, \;p_{i}(x, 0)=0, \;\; i=1, 2, 3, 4, \end{align}$

(2.4)

其中$(x, t)\in [0, \infty)\times [0, \infty);$$p_{0}(t)$表示在时刻t供应链系统正常运作的概率; $p_{i}(x, t)dx\;(i=1, 2, 3, 4)$表示在时刻t供应链系统处于故障状态$i \;(i=1, 2, 3, 4)$, 在该状态已经驻留了x时间, 在$(x, x+dx]$离开故障状态的概率; $\lambda_i\;(i=1, 2, 3, 4)$是从正常运作状态到状态$i\;(i=1, 2, 3, 4)$的失效率; $\mu_i(x)\;(i=1, 2, 3, 4)$表示供应链系统离开状态$i\, (i=1, 2, 3, 4)$的修复率.

取状态空间为

$X=\left\{p\in \mathbb{R}\times \left({L^{1}[0, \infty)}\right)^4\left|\left\|p \right\| =|p_{0}|+\sum_{i=1}^{4}\|p_{i}\|_{L^{1}[0, \infty)}\right.\right\}.$

显然, X是一个Banach空间^[9].为简单起见, 定义

$\begin{align} &B_i f(x)=-\frac{df(x)}{dx}-\mu_i (x)f(x), \;\;\;f\in W^{1, 1}[0, \infty), \;\; i=1, 2, 3, 4, \nonumber\\ &\phi_i g(x)=\int_0^\infty g(x)\mu_i (x)dx, \;\;\;g\in L^1 [0, \infty), \;\; i=1, 2, 3, 4, \nonumber\\ &\Lambda=\lambda_1+\lambda_2+\lambda_3+\lambda_4, \nonumber \end{align}$

则可以定义算子$A_m$和它的定义域$D(A_m)$为

$\begin{eqnarray*} &&A_m p=\begin{pmatrix} -\Lambda &\phi_1 &\phi_2 &\phi_3 &\phi_4\\ 0 &B_1 &0 &0 &0\\ 0 &0 &B_2 &0 &0 \\ 0 &0 &0 &B_3 &0 \\ 0 &0 &0 &0 &B_4 \end{pmatrix} \begin{pmatrix} p_0\\ p_1(x)\\ p_2(x)\\ p_3(x)\\ p_4(x) \end{pmatrix}, \\ &&D( A_m)=\left\{p\in{X} \left|\begin{array}{l} \frac{dp_{i}(x)}{dx}\in {L^1[0, \infty)}, p_{i}(x)\;\text{是绝对连续函数}, i=1, 2, 3, 4 \end{array} \right. \right\}. \end{eqnarray*}$

选取X的边界空间$\partial X:=\mathbb{C}^4, $并且定义边界算子$L: D(A_m)\rightarrow \partial X$与$\Phi : D(A_m)\rightarrow \partial X$如下:

$ Lp= \left( \begin{matrix} p_1(0)\\ p_2(0)\\ p_3(0)\\ p_4(0) \end{matrix} \right ), \;\; \Phi p= \left( \begin{matrix} \lambda_1 &0 &0 &0 &0\\ \lambda_2 &0 &0 &0 &0\\ \lambda_3 &0 &0 &0 &0\\ \lambda_4 &0 &0 &0 &0 \end{matrix} \right) \left(\begin{matrix} p_0\\ p_1(x)\\ p_2(x)\\ p_3(x)\\ p_4(x) \end{matrix} \right).$

如果定义算子$(A, D(A))$为$ Ap=A_mp, \;\; D(A)=\left\{p\in D(A_m)\left|Lp=\Phi p\right.\right\}, $那么方程(2.1)-(2.4) 可以描述为Banach空间X上的抽象Cauchy问题:

$\begin{align} \begin{cases} \frac{dp(t)}{dt}=A p(t), \quad t\in [0, \infty), \\ p(0)=(1, 0, 0, 0, 0). \end{cases} \end{align}$

(2.5)

在文献[6]中作者得到了以下结果.

定理2.1  算子$(A, D(A))$生成一个正压缩$C_0$ -半群$T(t)$.系统(2.5) 存在唯一的正时间依赖解$p(x, t)=T(t)p(0), $并且满足

$\|p(\cdot, t)\|=p_0(t)+\sum\limits_{i=1}^4\displaystyle\int_0^\infty p_i(x, t)dx=1, \; \forall t\in [0, \infty).$

引理3.1  0是A的几何重数为1的特征值.

证  讨论方程$Ap=0$, 即

$\begin{align} &\Lambda p_0=\sum_{i=1}^4 \int_0^\infty p_i(x)\mu_i(x) dx, \end{align}$

(3.1)

$\begin{align} & \frac{d{p_{i}(x)}}{dx}=-\mu_i(x) p_i(x), \;\; i=1, 2, 3, 4, \end{align}$

(3.2)

$\begin{align} & p_{i}(0)= \lambda_i p_0, \;\; i=1, 2, 3, 4. \end{align}$

(3.3)

解(3.2) 有

$\begin{align} &p_{i}(x)=a_{i}e^{-\int_0^x \mu_i(\xi)d\xi}, \;\;i=1, 2, 3, 4. \end{align}$

(3.4)

(3.4) 式结合(3.3) 式推出

$\begin{align} &a_i=p_i(0)=\lambda_i p_0, \;\;i=1, 2, 3, 4. \end{align}$

(3.5)

由(3.4) 与(3.5) 式算出

$ \|p\|=|p_0|+\sum_{i=1}^4 \|p_i\|_{L^1[0, \infty)}=\left(1+\sum_{i=1}^4 \lambda_i \int_0^\infty e^{-\int_0^x \mu_i(\xi)d\xi}dx\right)|p_0|<\infty.$

这说明0是A的特征值.由(3.1), (3.4), (3.5) 式知道对应于0的特征向量空间是一维的线性空间, 即0的几何重数为1.证毕.

下面研究A的豫解集.为此首先定义算子$(A_0, D(A_0))$并研究它的豫解集; 其次通过考虑$(\gamma I-A_m)$的核来定义Dirichlet算子$D_\gamma$并推出$\Phi D_\gamma$的表达式; 然后用文献[10]中的结果得到A的豫解集, 从而推出本文的主要结果.

定义算子$(A_0, D(A_0))$为

$ A_0 p=A_m p, \quad D(A_0)=\{p\in D(A_m)| Lp=0\}, $

那么对任意$y\in X$, 考虑方程$(\gamma I-A_0)p=y, $这等价于

$\begin{align} &[\gamma+\Lambda]p_0=\sum_{i=1}^4\int_0^\infty p_i(x)\mu_i(x)dx + y_0, \end{align}$

(3.6)

$\begin{align} &\frac{dp_i(x)}{dx}=-[\gamma +\mu_i(x)]p_i(x)+y_i(x), \quad i=1, 2, 3, 4, \end{align}$

(3.7)

$\begin{align} &p_i(0)=0, \quad i=1, 2, 3, 4. \end{align}$

(3.8)

解(3.6) 与(3.7) 推出

$\begin{align} &p_0=\frac{1}{\gamma+\Lambda}\left\{y_0+\sum_{i=1}^4\int_0^\infty\mu_i(x) e^{-\gamma x-\int_0^x \mu_i(\xi)d\xi}\int_0^x y_i(\tau)e^{\gamma\tau+\int_0^\tau \mu_i(\xi)d\xi} d\tau dx \right\}, \end{align}$

(3.9)

$\begin{align} &p_i(x)=e^{-\gamma x-\int_0^x \mu_i(\xi)d\xi} \int_0^x y_i(\tau) e^{\gamma \tau+\int_0^\tau \mu_i(\xi)d\xi} d \tau, \quad i=1, 2, 3, 4. \end{align}$

(3.10)

$\forall f\in L^1 [0, \infty)$, 若记

$ E_i f(x)=e^{-\gamma x-\int_0^x \mu_i(\xi)d\xi} \int_0^x f(\tau) e^{\gamma \tau+\int_0^\tau \mu_i(\xi)d\xi} d \tau, \quad i=1, 2, 3, 4, $

那么(3.9) 与(3.10) 式变为

$\begin{align} &p_0=\frac{1}{\gamma+\Lambda}(y_0+\phi_1E_1 y_1(x)+\phi_2E_2 y_2(x)+\phi_3E_3 y_3(x)+\phi_4E_4 y_4(x)), \nonumber \\ &p_i(x)=E_iy_i(x), \quad i=1, 2, 3, 4.\nonumber \end{align}$

即

$(\gamma I-A_0)^{-1}y = \left( \begin{matrix} \frac{1}{\gamma+\Lambda} &\frac{1}{\gamma+\Lambda}\phi_1 E_1 &\frac{1}{\gamma+\Lambda}\phi_2 E_2 &\frac{1}{\gamma+\Lambda}\phi_3 E_3 &\frac{1}{\gamma+\Lambda}\phi_4 E_4\\ 0 &E_1 &0 &0 &0\\ 0 &0 &E_2 &0 &0\\ 0 &0 &0 &E_3 &0\\ 0 &0 &0 &0 &E_4 \end{matrix} \right) \left( \begin{matrix} y_0\\ y_1(x)\\ y_2(x)\\ y_3(x)\\ y_4(x) \end{matrix} \right).$

由上述表达式和豫解集的定义可得以下结论.

引理3.2  设$\mu_i(x):[0, \infty)\rightarrow[0, \infty) (i=1, 2, 3, 4)$是可测函数, 若

$0<\underline{\mu}=\underset{\tiny{\begin {array}{c} x\in[0, \infty) \\ 1\leq i\leq 4 \end{array}}}{\inf} \mu_i(x) \leq \underset{\tiny{\begin {array}{c} x\in[0, \infty) \\ 1\leq i\leq 4 \end{array}}}{\sup} \mu_i(x)=\overline{\mu}<\infty, $

则$ \left\{\gamma \in \mathbb{C}\left | \text{Re}\gamma +\underline{\mu}>0\right.\right\}\subset\rho(A_0).$

证  对任意的$f\in L^1[0, \infty)$用分部积分法估计出

$\begin{align} \int_0^\infty |E_i f(x)|dx &=\int_0^\infty \left| e^{-\gamma x-\int_0^x \mu_i(\xi)d\xi}\int_0^x f(\tau)e^{\gamma \tau+\int_0^\tau\mu_i(\xi)d\xi} d\tau\right|dx \nonumber \\ &\leq \int_0^\infty e^{-\text{Re} \gamma x -\int_0^x\mu_i(\xi)d\xi}\int_0^xe^{\text{Re}\gamma\tau +\int_0^\tau \mu_i(\xi)d\xi}|f(\tau)|d\tau dx \nonumber\\ &=\int_0^\infty \frac{-1}{\text{Re}\gamma+\mu_i(x)}\left(\int_0^x e^{\text{Re} \gamma \tau+\int_0^\tau \mu_i (\xi)d\xi}|f(\tau)|d\tau\right)d e^{-\text{Re}\gamma x-\int_0^x\mu_i(\xi)d\xi} \nonumber\\ &\leq \frac{-1}{\text{Re}\gamma+\underline{\mu}} \left\{ e^{-\text{Re}\gamma x-\int_0^x\mu_i(\xi)d\xi}\times\left.\left(\int_0^x e^{\text{Re}\gamma\tau+\int_0^\tau \mu_i(\xi)d\xi}|f(\tau)|d\tau\right)\right|_{x=0}^{x=\infty}\right. \nonumber\\ &\quad\left.-\int_0^\infty e^{-\text{Re} \gamma x-\int_0^x \mu_i(\xi)d\xi} e^{\text{Re} \gamma x+\int_0^x \mu_i(\xi)d\xi}|f(x)|dx\right\}\nonumber \\ &=\frac{1}{\text{Re}\gamma+\underline{\mu}}\|f\|_{L^1 [0, \infty)}\nonumber\\ &\Rightarrow \nonumber \\ &\|E_i\|\leq\frac{1}{\text{Re}\gamma+\underline{\mu}}, \quad i=1, 2, 3, 4. \end{align}$

(3.11)

对任意的$y\in X$, 由条件$\text{Re}\gamma+\underline{\mu}>0$与(3.11) 式推出

$\begin{align} &\|(\gamma I-A_0)^{-1}y\|\nonumber\\ =&\left\|\left(\frac{1}{\gamma+\Lambda}(y_0+\phi_1E_1y_1+\phi_2E_2y_2+\phi_3E_3y_3+\phi_4E_4y_4), E_1y_1, E_2y_2, E_3y_3, E_4y_4\right)\right\| \nonumber\\ =&\left|\frac{1}{\gamma+\Lambda}\left(y_0+\sum_{i=1}^4\phi_iE_iy_i\right)\right|+\sum_{i=1}^4\|E_iy_i\|_{L^1[0, \infty)}\nonumber\\ \leq& \frac{1}{|\gamma+\Lambda|}|y_0|+\frac{1}{|\gamma+\Lambda|}\sum_{i=1}^4\|\phi_i\|\|E_i\|\|y_i\|_{L^1[0, \infty)} +\sum_{i=1}^4\|E_i\|\|y_i\|_{L^1[0, \infty)}\nonumber\\ \leq& \frac{1}{|\gamma+\Lambda|}|y_0|+\frac{1}{|\gamma+\Lambda|}\sum_{i=1}^4\frac{\underset{x\in[0, \infty)}{\sup}\mu_i(x)}{\text{Re}\gamma+\underline {\mu}}\|y_i\|_{L^1[0, \infty)} +\sum_{i=1}^4\frac{1}{\text{Re}\gamma+\underline {\mu}}\|y_i\|_{L^1[0, \infty)}\nonumber\\ \leq& \sup\left\{\frac{1}{|\gamma+\Lambda|}, \; \frac{1}{|\gamma+\Lambda|}\frac{\overline{\mu}}{\text{Re}\gamma+\underline {\mu}} +\frac{1}{\text{Re}\gamma+\underline {\mu}} \right\}\|y\|<\infty.\nonumber \end{align}$

此式说明引理的结论成立.证毕.

引理3.3  设$\mu_i(x)\; (i=1, 2, 3, 4)$是可测函数, 且

$0<\underline{\mu}=\underset{\tiny{\begin {array}{c} x\in[0, \infty) \\ 1\leq i\leq 4 \end{array}}}{\inf} \mu_i(x) \leq \underset{\tiny{\begin {array}{c} x\in[0, \infty) \\ 1\leq i\leq 4 \end{array}}}{\sup} \mu_i(x)=\overline{\mu}<\infty.$

若$\gamma \in \rho(A_0), $则

$\begin{align} & p\in \ker (\gamma I-A_m)\Longleftrightarrow\nonumber\\ &p_0=\frac{1}{\gamma+\Lambda}\sum_{i=1}^4 a_i \int_0^\infty \mu_i (x) e^{-\gamma x-\int_0^x \mu_i(\tau)d\tau}dx, \end{align}$

(3.12)

$\begin{align} &p_i(x)=a_i e^{-\gamma x-\int_0^x \mu_i(\tau)d\tau}, \;|a_i|<\infty, \; i=1, 2, 3, 4. \end{align}$

(3.13)

证  如果$p\in\ker(\gamma I-A_m), $则$(\gamma I-A_m)p=0, $这等价于

$\begin{align} &(\gamma+\Lambda)p_0=\sum_{i=1}^4 \int_0^\infty \mu_i(x) p_i(x)dx, \end{align}$

(3.14)

$ \begin{align} &\frac{dp_i(x)}{dx}=-[\gamma+\mu_i(x)]p_i(x), i=1, 2, 3, 4.\; \end{align}$

(3.15)

解(3.15) 推出

$\begin{align} p_i(x)=a_ie^{-\gamma x-\int_0^x\mu_i(\tau)d\tau}, \; i=1, 2, 3, 4. \end{align}$

(3.16)

将(3.16) 式代入(3.14) 式算出

$\begin{align} p_0=\frac{1}{\gamma+\Lambda}\sum_{i=1}^4 a_i \int_0^\infty \mu_i(x)e^{-\gamma x-\int_0^x \mu_i(\tau)d\tau}dx. \end{align}$

(3.17)

由于$p\in \ker(\gamma I-A_m), \; p\in D(A_m)$, 所以用嵌入定理^[11]得到

$\begin{align} |a_i| &\leq\sum_{i=1}^4 |a_i|=\sum_{i=1}^4 |p_i(0)|\leq\sum_{i=1}^4\|p_i\|_{L^\infty [0, \infty)}\nonumber\\ &\leq \sum_{i=1}^4 \|p_i\|_{L^1 [0, \infty)}+\sum_{i=1}^4 \left\|\frac{dp_i}{dx}\right\|_{L^1 [0, \infty)}<\infty, \; i=1, 2, 3, 4. \end{align}$

(3.18)

(3.16)-(3.18) 式说明(3.12) 与(3.13) 式成立.

反之, 如果(3.12), (3.13) 式成立, 则有

$\begin{align} \|p_i \|_{L^1 [0, \infty)}& \leq |a_i|\int_0^\infty e^{-\text{Re}\gamma x-\int_0^x\mu_i(\tau)d\tau} dx\nonumber \\ &\leq |a_i|\int_0^\infty e^{-(\text{Re}\gamma +\underline{\mu})x}dx =\frac{|a_i|}{\text{Re}\gamma+\underline{\mu}}, \; i=1, 2, 3, 4\nonumber\\ &\Longrightarrow \nonumber\\ |p_0|+\sum_{i=1}^4\|p_i\|_{L^1[0, \infty)}&\leq\frac{1}{|\gamma+\Lambda|}\sum_{i=1}^4|a_i|\int_0^\infty \mu_i (x)e^{-\text{Re}\gamma x-\int_0^x \mu_i(\tau)d\tau}dx+\sum_{i=1}^4\|p_i\|_{L^1[0, \infty)} \nonumber\\ &\leq \frac{1}{|\gamma+\Lambda|}\sum_{i=1}^4|a_i|\int_0^\infty \mu_i (x)e^{-\int_0^x \mu_i(\tau)d\tau}dx+\sum_{i=1}^4\frac{|a_i|}{\text{Re}\gamma+\underline{\mu}}\nonumber\\ &= \left(\frac{1}{|\gamma+\Lambda|}+\frac{1}{\text{Re}\gamma+\underline{\mu}}\right)\sum_{i=1}^4 |a_i|<\infty. \end{align}$

(3.19)

由(3.13) 式知道

$ \frac{dp_i(x)}{dx}=- a_i[\gamma +\mu_i(x)]e^{-\gamma x-\int_0^x \mu_i(\tau) d\tau}, \;i=1, 2, 3, 4, $

从而有

$\begin{align} \sum_{i=1}^4\left\|\frac{dp_i}{dx}\right\|_{L^1[0, \infty)}&\leq\sum_{i=1}^4 |a_i|\int_0^\infty|\gamma+\mu_i(x)|e^{-\text{Re}\gamma x-\int_0^x\mu_i(\tau)d\tau}dx \nonumber\\ &\leq[\text{Re}\gamma+\overline{\mu}+|\text{Im} \gamma|]\sum_{i=1}^4 |a_i|\int_0^\infty e^{-\text{Re}\gamma x-\int_0^x\mu_i(\tau)d\tau}dx<\infty. \end{align}$

(3.20)

(3.19) 与(3.20) 式表示$p\in\ker (\gamma I-A_m).$证毕.

由于L是满射, 所以

$L|_{\ker(\gamma I-A_m)}:\ker(\gamma I-A_m)\rightarrow \partial X$

可逆.如果$\gamma\in\rho (A_0), $那么定义Dirichlet算子为

$D_\gamma:=\left(L|_{\ker(\gamma I-A_m)}\right)^{-1}: \partial X \rightarrow \ker(\gamma I-A_m).$

由引理3.3知道$D_\gamma$的具体表达式为

$ D_\gamma \left( \begin{matrix} a_1\\ a_2\\ a_3\\ a_4 \end{matrix} \right) = \left( \begin{matrix} \alpha_1 &\alpha_2&\alpha_3 & \alpha_4\\ \delta_1 &0 &0 &0\\ 0 &\delta_2 &0 &0\\ 0 &0 &\delta_3 &0\\ 0 &0 &0 &\delta_4 \end{matrix} \right) \left( \begin{matrix} a_1\\ a_2\\ a_3\\ a_4 \end{matrix} \right), $

其中$\alpha_i=\frac{1}{\gamma+\Lambda}\displaystyle\int_0^\infty \mu_i (x)\delta_i dx, \; \delta_i= e^{-\gamma x-\int_0^x\mu_i(\tau)d\tau}, \;i=1, 2, 3, 4.$

由$D_\gamma$的表达式和$\Phi$的定义推出$\Phi D_\gamma $的表达式

$\begin{align} \Phi D_\gamma \left( \begin{matrix} a_1\\ a_2\\ a_3\\ a_4 \end{matrix} \right) = \left( \begin{matrix} \varepsilon_{11} &\varepsilon_{12} &\varepsilon_{13} &\varepsilon_{14}\\ \varepsilon_{21} &\varepsilon_{22} &\varepsilon_{23} &\varepsilon_{24}\\ \varepsilon_{31} &\varepsilon_{32} &\varepsilon_{33} &\varepsilon_{34}\\ \varepsilon_{41} &\varepsilon_{42} &\varepsilon_{43} &\varepsilon_{44} \end{matrix} \right) \left( \begin{matrix} a_1\\ a_2\\ a_3\\ a_4 \end{matrix} \right), \end{align}$

(3.21)

这里$\varepsilon_{ij}=\frac{\lambda_i}{\gamma+\Lambda}\displaystyle\int_0^\infty \mu_j(x)e^{-\gamma x-\int_0^x \mu_j(\tau)d\tau}dx, \; i, j=1, 2, 3, 4.$

在文献[10]中作者得到以下结果.

引理3.4  设$\gamma\in\rho(A_0)$且存在$\gamma_0\in \mathbb{C}$使得$1\not\in \sigma(\Phi D_{\gamma_0}) $, 则

$\gamma\in\sigma(A)\Longleftrightarrow 1\in\sigma(\Phi D_\gamma). $

结合引理3.4与文献[12]得到如下结论:

引理3.5  设$\mu_i(x)\; (i=1, 2, 3, 4)$是可测函数, 若

$0<\underline{\mu}=\underset{\tiny{\begin {array}{c} x\in[0, \infty) \\ 1\leq i\leq 4 \end{array}}}{\inf} \mu_i(x) \leq \underset{\tiny{\begin {array}{c} x\in[0, \infty) \\ 1\leq i\leq 4 \end{array}}}{\sup} \mu_i(x)=\overline{\mu}<\infty, $

那么在虚轴上除了0外其他所有点都属于A的豫解集.

证  设$\overrightarrow{a}=(a_1, a_2, a_3, a_4), \;|a_k|<\infty \;(k=1, 2, 3, 4), \text{ 并且}\;\gamma=ih, \; h\in R\backslash\{0\}. $由Riemann-Lebesgue引理

$\underset{h\to\infty}{\lim}\int_0^\infty f(x)\cos(hx)dx=0, \; \underset{h\to\infty}{\lim}\int_0^\infty f(x)\sin(hx)dx=0, \; f\in L^1 [0, \infty), \;f(x)\geq 0, $

知道存在非负常数$\mathcal {K}>0$使得对一切$|h|>\mathcal {K}$有

$\begin{align} \left|\int_0^\infty f(x) e^{ihx}dx\right|^2&=\left( \int_0^\infty f(x)\sin (hx)dx\right)^2+\left(\int_0^\infty f(x)\cos(hx)dx\right)^2\nonumber\\ &<\left(\int_0^\infty f(x)dx\right)^2 \nonumber\\ &\Longrightarrow\nonumber \\ \left| \int_0^\infty f(x) e^{-ihx}dx\right|&<\int_0^\infty f(x)dx. \end{align}$

(3.22)

从而对$|h|>\mathcal{K}$, 由(3.22) 式和$\displaystyle\int_0^\infty b(x)e^{-\int_0^x b(\tau)d\tau}dx=1$推出

$ \begin{align} \|\Phi D_\gamma \overrightarrow{a}\|&\leq \frac{\Lambda}{|\gamma+\Lambda|}\sum_{i=1}^4\left|\int_0^\infty\mu_i(x)e^{-\gamma x-\int_0^x\mu_i (\tau) d\tau}dx\right||a_i|\nonumber\\ &=\frac{\Lambda}{\sqrt{h^2+\Lambda^2}}\sum_{i=1}^4\left|\int_0^\infty\mu_i(x)e^{-ih x-\int_0^x\mu_i (\tau) d\tau}dx\right||a_i|\nonumber\\ &<\sum_{i=1}^4|a_i|\int_0^\infty \mu_i(x) e^{-\int_0^x\mu_i(\tau) d\tau}dx =\sum_{i=1}^4 |a_i|=\|\overrightarrow{a}\| \nonumber\\ &\Longrightarrow\nonumber\\ \|\Phi D_\gamma \|&<1. \end{align}$

(3.23)

(3.23) 式表明当$|h|>\mathcal{K}$时谱半径$r(\Phi D_\gamma)<\|\Phi D_\gamma\|<1, $这说明$1\not\in \sigma (\Phi D_\gamma).$此结果结合引理3.4知道当$|h|>\mathcal{K}$时有$\gamma\not\in \sigma(A)$, 即

$\begin{align} \{ih\;|\;|h|>\mathcal{K}\}\subset\rho(A), \qquad \{ih\;| \;|h|\leq\mathcal{K}\}\subset \sigma(A)\cap i\mathbb{R}. \end{align}$

(3.24)

另外由定理2.1与文献[12]中的推论2.3知道$\sigma(A)\cap i\mathbb{R}$是虚加法循环.即

$\begin{align} ih\in \sigma(A)\cap i\mathbb{R}\Rightarrow ihk\in \sigma(A)\cap i\mathbb{R}, \quad k \in \mathbb{N}. \end{align}$

(3.25)

从而由(3.24), (3.25) 式与引理3.1推出$\sigma(A)\cap i\mathbb{R}=\{0\}$.证毕.

由文献[13]知道X的共轭空间为

$X^*=\left\{q^*\in \mathbb{R}\times (L^\infty[0, \infty))^4 \left| ||| q^* |||=\sup \left\{|q_0^*|, \underset{1\leq i \leq 4}{\sup}\|q^*_i\|_{L^\infty[0, \infty)}\right\} \right.\right\}.$

容易证明$X^*$是一个Banach空间^[7].根据文献[8]知道A的共轭算子$A^*$为

$ A^* q^*= (\mathcal {L}+\mathcal{N})q^*, \quad q^*\in D(\mathcal{L}), $

其中

$\begin{align} \mathcal{L}q^*&= \begin{pmatrix} -\Lambda &0 &0 &0 &0\\ 0 &\frac{d}{dx} -\mu_1(x) &0 &0 &0\\ 0 &0 &\frac{d}{dx} -\mu_2(x) &0 &0\\ 0 &0 &0 &\frac{d}{dx} -\mu_3(x) &0\\ 0 &0 &0 &0 &\frac{d}{dx} -\mu_4(x)\\ \end{pmatrix} \begin{pmatrix} q_0^*\\ q_1^*(x)\\ q_2^*(x)\\ q_3^*(x)\\ q_4^*(x) \end{pmatrix}, \nonumber\\ D(\mathcal{L})&=\left\{q^*\in X^*\left | \frac{dq_i^*(x)}{dx} \; \text{存在且}\; q_i^*(\infty)=\alpha, \; i=1, 2, 3, 4\right.\right\}, \nonumber\\ \mathcal{N} q^*&= \begin{pmatrix} 0 &\lambda_1 &\lambda_2 &\lambda_3 &\lambda_4\\ \mu_1(x) &0 &0 &0 &0\\ \mu_2(x) &0 &0 &0 &0\\ \mu_3(x) &0 &0 &0 &0\\ \mu_4(x) &0 &0 &0 &0\\ \end{pmatrix} \begin{pmatrix} q_0^*\\ q_1^*(0)\\ q_2^*(0)\\ q_3^*(0)\\ q_4^*(0) \end{pmatrix}.\nonumber \end{align}$

下面证明0是$A^*$的几何重数为1的特征值.

引理3.6  0是$A^*$的几何重数为1的特征值.

证  考虑方程$A^*q^*=0, $即

$\begin{align} &-\Lambda q_0^*+\sum_{i=1}^4 \lambda_i q_i^*(0)=0, \end{align}$

(3.26)

$\begin{align} &\frac{dq_i^*(x)}{dx}-\mu_i(x)q_i^*(x)+\mu_i(x) q_0^*=0, \;i=1, 2, 3, 4, \end{align}$

(3.27)

$\begin{align} &q_i^*(\infty)=\alpha, \;i=1, 2, 3, 4. \end{align}$

(3.28)

解(3.27) 有

$\begin{align} q_i^*(x)= c_i e^{\int_0^x\mu_i(\tau)d\tau}-e^{\int_0^x\mu_i(\tau)d\tau}\int_0^x\mu_i(\tau) q_0^*e^{-\int_0^\tau\mu_i(\xi)d\xi}d\tau, \; i=1, 2, 3, 4. \end{align}$

(3.29)

(3.29) 式两边同时乘$e^{-\int_0^x\mu_i(\tau)d\tau}$, 并用(3.28) 式推出

$\begin{align} 0=c_i-q_0^* \int_0^\infty \mu_i(\tau) e^{-\int_0^\tau \mu_i(\xi)d\xi}d\tau \Rightarrow c_i=q_0^*. \end{align}$

(3.30)

(3.30) 式代入(3.29) 式可得

$\begin{align} q_i^*(x)= q_0^* e^{\int_0^x\mu_i(\tau)d\tau}\left[1+\int_0^x e^{-\int_0^\tau\mu_i(\xi)d\xi}d\left(-\int_0^\tau \mu_i(\xi)d\xi\right)\right]=q_0^*, ~i=1, 2, 3, 4. \end{align}$

(3.31)

(3.31) 式表明

$||| q* |||=\sup \left\{|q_0^*|, \underset{1\leq i \leq 4}{\sup}\|q^*_i\|_{L^\infty[0, \infty)}\right\}=|q_0^*|<\infty \Rightarrow q^*\in X^*.$

即0是$A^*$的特征值.由(3.31) 式看出对应于0的特征向量空间是1维的.换句话说, 0的几何重数为1.证毕.

结合定理2.1, 引理3.1, 引理3.5, 引理3.6与文献[14]中的定理14推出本文的主要结论.

定理4.1  设$\mu_i(x)\; (i=1, 2, 3, 4)$是可测函数, 且满足

$0<\underline{\mu}=\underset{\tiny{\begin {array}{c} x\in[0, \infty) \\ 1\leq i\leq 4 \end{array}}}{\inf} \mu_i(x) \leq \underset{\tiny{\begin {array}{c} x\in[0, \infty) \\ 1\leq i\leq 4 \end{array}}}{\sup} \mu_i(x)=\overline{\mu}<\infty, $

则系统(2.5) 的时间依赖解强收敛于该系统的稳态解, 即

$\underset{t\to \infty}{\lim}\|p(\cdot, t)-\beta p(\cdot)\|=0, $

其中$p(x)$是引理3.1中的特征向量.