利用数学模型研究传染病传播规律已有几十年的历史,模型及方法也多种多样.而年龄对某些传染病传播的影响不可忽略,因为不同年龄的人对同一种传染病的感染能力及传播能力不同,因而研究年龄结构传染病模型具有重要的实际意义.

年龄结构传染病数学模型已很丰实^[1-9],其中大部分都假设总人口规模不变.利用这种假设研究某些病程较短的疾病(如流感、手足口病等)或研究某些总人口变化不大的国家、地区内的传染病可以,但用于研究病程较长的传染病(如肝炎、艾滋病等)或总人口变化较大的国家、地区内传染病的传播规律就不太恰当.如肝炎潜伏期为$ 1\sim6 $个月,艾滋病潜伏期为$ 0.5\sim20 $年,平均$ 7\sim10 $年.长病程导致人口总数可能发生变化,因而不能忽略人口总数变化这一因素对疾病传播的影响.

已经有一些数学家注意到这类问题并进行了研究,如在文献[10]中Iannelli和Martcheva建立了非线性发展方程的齐次动力系统理论,且研究了总人口规模变化的年龄结构SIR模型.李学志等在文献[11]中运用这一理论研究了总人口规模变化的年龄结构SEIR模型,可以利用其中的条件对一些病程较长、传染过程又可分为潜伏期和染病期两个阶段的传染病(如肝炎、艾滋病等)进行控制.对于不同年龄阶段的人,由于受到年龄或体质的影响,导致有些易感者被感染后进入潜伏期,另外一些直接变成患者,因此对这种SEIR模型进行改进,引入比例进入潜伏或染病群体这一因素进行研究更加符合艾滋病的实际情况,具有重要的理论及实际意义.

把总人口分为易感类、潜伏类、染病类、康复类,分别用$ S(a, t), E(a, t), I(a, t), R(a, t) $表示各类年龄密度函数, $ a $为年龄, $ t $为时间, $ a_\dagger $为人口的最高年龄($ a_\dagger<+\infty $).则比例进入潜伏或染病群体的年龄结构传染病模型为

(2.1) $ \left\{ \begin{array}{ll} \frac{\partial S(a, t)}{\partial a}+ \frac{\partial S(a, t)}{\partial t} = -\lambda(a, t)S(a, t)-\mu(a)S(a, t), \\ \frac{\partial E(a, t)}{\partial a}+ \frac{\partial E(a, t)}{\partial t} = q\lambda(a, t)S(a, t)-[\alpha (a)+\mu(a)]E(a, t), \\ \frac{\partial I(a, t)}{\partial a}+ \frac{\partial I(a, t)}{\partial t} = (1-q)\lambda(a, t)S(a, t)+\alpha (a)E(a, t)-[\gamma (a)+\mu(a)]I(a, t), \\ \frac{\partial R(a, t)}{\partial a}+ \frac{\partial R(a, t)}{\partial t} = \gamma(a)I(a, t)-\mu(a)R(a, t), \end{array} \right. $

其中$ \mu(a) $为年龄依赖自然死亡率, $ [\alpha (a)]^{-1} $为平均潜伏周期, $ [\gamma(a)]^{-1} $为平均染病周期,这是一个没考虑因病死亡,治愈后暂时获得免疫能力的模型.系统(2.1)的边界条件为

(2.2) $ \begin{array}{ll} S(0, t) = \int_0^{a_{†}}\beta(a)P(a, t){\rm d}a, \ E(0, t) = I(0, t) = R(0, t) = 0, \end{array} $

其中$ \beta(a) $为年龄依赖出生率.总人口年龄密度函数$ P(a, t) = S(a, t)+E(a, t)+I(a, t)+R(a, t), $则$ t $时刻的人口总数为

(2.3) $ N(t) = \int_0^{a_{†}}P(a, t){\rm d}a, $

可见人口总数随时间而变化,令感染力函数^[10]为

(2.4) $ \begin{array}{ll} \lambda (a, t) = k(a)\frac{\int_0^{a_{†}}h(\sigma)I(\sigma, t){\rm d}\sigma}{N(t)} = k(a)\frac{H(t)}{N(t)}, \end{array} $

其中$ h(a) $为年龄依赖的染病率, $ k(a) $为年龄依赖的接触率.

把系统(2.1)中的四个方程相加得标准的Lotka-McKendrick人口动力学模型

(2.5) $ \left\{ \begin{array}{ll} \frac{\partial P(a, t)}{\partial a}+ \frac{\partial P(a, t)}{\partial t} = -\mu(a)P(a, t), \\ P(0, t) = \int_0^{a_{†}}\beta(a)P(a, t){\rm d}a. \end{array} \right. $

因系统(2.1)除第四个方程外,其余都不含$ R(a, t) $.则系统(2.1)等价于

(2.6) $ \left\{ \begin{array}{ll} \frac{\partial S(a, t)}{\partial a}+ \frac{\partial S(a, t)}{\partial t} = -\lambda(a, t)S(a, t)-\mu(a)S(a, t), \\ \frac{\partial E(a, t)}{\partial a}+ \frac{\partial E(a, t)}{\partial t} = q\lambda(a, t)S(a, t)-[\alpha (a)+\mu(a)]E(a, t), \\ \frac{\partial I(a, t)}{\partial a}+ \frac{\partial I(a, t)}{\partial t} = (1-q)\lambda(a, t)S(a, t)+\alpha (a)E(a, t)-[\gamma (a)+\mu(a)]I(a, t), \\ \frac{\partial P(a, t)}{\partial a}+ \frac{\partial P(a, t)}{\partial t} = -\mu(a)P(a, t) \end{array}\right. $

及条件

(2.7) $ \begin{array}{ll} S(0, t) = P(0, t) = \int_0^{a_{†}}\beta(a)P(a, t){\rm d}a, \ E(0, t) = I(0, t) = 0. \end{array} $

假设所有的参数都非负,且

$ \begin{array}{ll} \alpha(a), \beta(a), \gamma (a), h (a), k(a)\in L^\infty(0, a_{†}), \ \mu(a)\in L_{loc}^1(0, a_{†}), \\ \mbox{且} \int_0^{a_{†}}\mu(a){\rm d}a = \infty, \ \mu(a)\ \mbox{在}\ (0, a_{†})\ \mbox{上几乎处处有界}. \end{array} $

令$ \pi(a) = {\rm e}^{-\int_0^a\mu(\tau){\rm d}\tau} $是存活函数,表示从出生后活到年龄$ a $的概率.

令$ {\mathcal {X}} = L^1(0, a_{†})\times L^1(0, a_{†})\times L^1(0, a_{†})\times L^1(0, a_{†}) $.设

$ L^1_\pi = \bigg\{g\in L^1(0, a_{†})|\frac{g(\cdot)}{\pi(\cdot)}\in L^1(0, a_{†})\bigg\}, $

在$ L^1_\pi $上定义范数

$ \|g\|_{L^1_\pi} = \int_0^{a_{†}}\frac{|g(a)|}{\pi(a)}{\rm d}a, $

则得到此范数$ \|\cdot\|_{L^1_\pi} $下的Banach空间$ L^1_\pi $.令

$ X = \bigg\{f\in{{\cal X}}|\frac{f(\cdot)}{\pi(\cdot)}\in{{\cal X}}\bigg\}, $

在$ X $上定义范数

$ \|f\|_X = \sum\limits_{i = 1}^4 \int_0^{a_{†}} \frac{|f_i(a)|}{\pi(a)}{\rm d}a, $

则得在此范数$ \|\cdot\|_X $下的Banach空间$ X $,且

$ L^1_\pi\subset L^1(0, a_{†}), \ \ X\subset{{\cal X}}. $

令锥$ K $:

$ K = \{f\in X | f_i(a)\geq 0\ (i = 1, 2, 3, 4), \ f_1(a)+f_2(a)+f_3(a)\leq f_4(a), \ \ a\in(0, a_\dagger)\}. $

令线性算子$ A : {{\cal D}}(A)\rightarrow {{\cal X}} $,

$ A = \left( \begin{array}{cccc} -\frac{\partial}{\partial a}-\mu(a) & 0 & 0 & 0\\ 0 & -\frac{\partial}{\partial a}-\mu(a)-\alpha(a) & 0 & 0\\ 0 & \alpha(a) & -\frac{\partial}{\partial a}-\mu(a)-\gamma(a) & 0\\ 0 & 0 & 0 & -\frac{\partial}{\partial a}-\mu(a) \end{array}\right). $

$ {\mathcal{D}}(A) = \bigg\{f\in X | \frac{f_i(\cdot)}{\pi(\cdot)} \mbox{绝对连续, $ f_1(0) = f_4(0) = \int_0^{a_{†}}\beta(a)f_4(a){\rm d}a, \ f_2(0) = f_3(0) = 0 \bigg\}.$} $

故算子$ A $在$ X $上生成$ C_0 $-半群$ {\rm e}^{At} $,且$ {\rm e}^{At}(K)\subset K $.由于(2.3)和(2.4)式定义的函数诱导出空间$ X $上的线性泛函

$ \begin{eqnarray*} &&{{\cal H}}:\ X \rightarrow \Re, \ \ {\mathcal {H}}(f): = \int_0^{a_\dagger}h(a)f_3(a){\rm d}a, \\ &&{{\cal N}}:\ X \rightarrow \Re, \ \ {\mathcal {N}}(f): = \int_0^{a_\dagger}f_4(a){\rm d}a. \end{eqnarray*} $

令非线性算子$ {\cal F} $:

$ {{\cal F}}(f)(a) = \left( \begin{array}{c} -k(a)f_1(a) \frac{{{\cal H}}(f)}{{{\cal N}}(f)}\\ qk(a)f_1(a) \frac{{{\cal H}}(f)}{{{\cal N}}(f)}\\ (1-q)k(a)f_1(a) \frac{{{\cal H}}(f)}{{{\cal N}}(f)}\\ 0 \end{array}\right), \ \forall f\neq0, \ f\in K, $

且$ {{\cal F}}(0) = ( 0, 0, 0, 0)^T $.则$ {{\cal F}} $是齐次Lipschitz连续且Frechét可微算子(除零点外).令泛函$ \varphi $:

(3.1) $ \langle \varphi, f \rangle: = {\mathcal {N}}(f) = \int_0^{a_\dagger}f_4(a){\rm d}a. $

因$ \mu(a), \ \pi(a)\ \mbox{在}\ (0, a_{†})\ \mbox{上几乎处处有界} $,故$ \forall f\in X $, $ \langle \varphi, Af \rangle $是$ X $上的有界算子,即$ \varphi\in{{\cal D}}(A^*) $,这里$ A^* $是$ A $的共轭算子.

令$ f = (S, E, I, P)^T $,则系统(2.6)可转化为

(3.2) $ \left\{ \begin{array}{ll} f'(t) = Af+{{\cal F}}(f), \\ f(0) = f_0. \end{array} \right. $

若系统(2.6)存在持续解: $ S(a, t) = {\rm e}^{\lambda t}S^*(a), \ E(a, t) = {\rm e}^{\lambda t}E^*(a), \ I(a, t) = {\rm e}^{\lambda t}I^*(a), \ P(a, t) = {\rm e}^{\lambda t}P^*(a) $,则满足

(3.3) $ \left\{ \begin{array}{l} \lambda S^*(a)+ \frac{{\rm d}S^*(a)}{{\rm d}a} = -k(a)\frac{H^*}{N^*}S^*(a) -\mu(a)S^*(a), \\ \lambda E^*(a)+ \frac{{\rm d}E^*(a)}{{\rm d}a} = qk(a)\frac{H^*}{N^*}S^*(a) -[\mu(a)+\alpha(a)]E^*(a), \\ \lambda I^*(a)+ \frac{{\rm d}I^*(a)}{{\rm d}a} = (1-q)k(a)\frac{H^*}{N^*}S^*(a) +\alpha(a)E^*(a)-[\mu(a)+\gamma(a)]I^*(a), \\ \lambda P^*(a)+ \frac{{\rm d}P^*(a)}{{\rm d}a} = -\mu(a)P^*(a) \end{array}\right. $

与

(3.4) $ \begin{array}{ll} S^*(0) = P^*(0) = \int_0^{a_{†}}\beta(a)P^*(a){\rm d}a, \ E^*(0) = I^*(0) = 0, \\ H^* = \int_0^{a_{†}}h(a)I^*(a){\rm d}a, \ N^* = \int_0^{a_{†}}P^*(a){\rm d}a. \end{array} $

显然系统(3.3)解的存在性和系统(3.2)的持续解的存在性等价.而系统(3.3)最后一个方程与其余方程相独立,其非零解为

(3.5) $ P^*(a) = b_0{\rm e}^{-\lambda^*a}\pi(a). $

则$ \lambda^* $是特征方程

(3.6) $ \int_0^{a_{†}}\beta(a)\pi(a){\rm e}^{-\lambda a}{\rm d}a = 1 $

的唯一实根,因解必须在超平面上,故

$ N^* = \int_0^{a_{†}}P^*(a){\rm d}a = 1, $

则$ b_0 $满足

$ b_0\int_0^{a_{†}}\pi(a){\rm e}^{-\lambda^*a}{\rm d}a = 1, $

解(3.3)式其余方程得:

(1)无病平衡态($ DFE $):

(3.7) $ \begin{array}{l} S^0(a) = P^0(a) = P^*(a), \ E^0(a) = I^0(a) = 0.\ (\mbox{此时}\ H^* = 0 ). \end{array} $

(2)地方病平衡态($ EE $):

(3.8) $ \begin{array}{rl} S^*(a) = & b_0{\rm e}^{-\lambda^*a}\pi(a){\rm e}^{-H^*\int_0^ak(\xi){\rm d}\xi}, \\ E^*(a) = & b_0{\rm e}^{-\lambda^*a}\pi(a)H^* \int_0^aqk(\sigma){\rm e}^{-H^*\int_0^\sigma k(\xi){\rm d}\xi} {\rm e}^{-\int_\sigma^a \alpha(\xi){\rm d}\xi} {\rm d}\sigma, \\ I^*(a) = & b_0{\rm e}^{-\lambda^*a}\pi(a)H^* \int_0^ak(\sigma){\rm e}^{-H^*\int_0^\sigma k(\xi){\rm d}\xi} {\rm e}^{-\int_\sigma^a \gamma(\xi){\rm d}\xi} \\ &\cdot\bigg[q \int_\sigma^a\alpha(\tau){\rm e}^{\int_\sigma^\tau[\gamma(\xi)-\alpha(\xi)]{\rm d}\xi}{\rm d}\tau+(1-q)\bigg]{\rm d}\sigma, \\ P^*(a) = & b_0{\rm e}^{-\lambda^*a}\pi(a). \end{array} $

将$ I^*(a) $代入$ H^* $后两边约分得关于$ H^* $的特征方程

(3.9) $ \begin{array}[b]{rl} & b_0 \int_0^{a_\dagger}h(a)\pi(a){\rm e}^{-\lambda^*a} \int_0^ak(\sigma) {\rm e}^{-H^*\int_0^\sigma k(\xi){\rm d}\xi} \\ &\cdot {\rm e}^{-\int_\sigma^a\gamma(\xi){\rm d}\xi} \bigg[q \int_\sigma^a\alpha(\tau) {\rm e}^{\int_\sigma^\tau[\gamma(\xi)-\alpha(\xi)]{\rm d}\xi}{\rm d}\tau+(1-q) \bigg] {\rm d}\sigma {\rm d}a = 1. \end{array} $

由(3.8)式得地方病平衡态存在的充要条件是方程(3.9)有一正解$ H^* $.令方程(3.9)左边为$ {{\cal R}}(H^*) $,则$ H^*\rightarrow-\infty $时, $ {{\cal R}}(H^*)\rightarrow+\infty $;当$ H^*\rightarrow+\infty $时, $ {{\cal R}}(H^*)\rightarrow0 $,则方程(3.9)有一正解$ H^* $的充要条件是$ {{\cal R}}(0)>1 $.由方程(3.9)得到一个地方病平衡态存在的阈值参数$ {{{\cal R}}_0 = {\cal R}}(0) $,称为基本再生数^[11],即在染病初期所有人都是易感者时,一个病人在其整个染病周期内平均所感染的病人数.

(3.10) $ \begin{array}[b]{rl} {{\cal R}}_0 = & b_0 \int_0^{a_\dagger}h(a)\pi(a){\rm e}^{-\lambda^*a} \int_0^ak(\sigma){\rm e}^{-\int_\sigma^a\gamma(\xi){\rm d}\xi}\\ &\cdot \bigg[q \int_\sigma^a\alpha(\tau){\rm e}^{\int_\sigma^\tau[\gamma(\xi)-\alpha(\xi)]{\rm d}\xi}{\rm d}\tau+(1-q) \bigg] {\rm d}\sigma {\rm d}a. \end{array} $

于是若$ {{\cal R}}_0>1 $,则方程(3.9)有唯一正解$ H^* $,由(3.8)式得到了唯一的地方病平衡态.若$ {{\cal R}}_0\leq1 $,方程(3.9)无正解,则系统(3.3)只含无病平衡态.因此,有

定理3.1   (1)若$ {{\cal R}}_0\leq1 $,则系统(3.3)对应实特征值$ \lambda^* $有唯一的无病平衡态$ (S^0, E^0, $$ I^0, P^0) $;

(2)若$ {{\cal R}}_0>1 $,则系统(3.3)对应实特征值$ \lambda^* $有两个解:无病平衡态$ (S^0, $$ E^0, I^0, P^0) $及地方病平衡态$ (S^*, $$ E^*, I^*, P^*) $.

令系统(2.6)在无病平衡态$ \omega^0(S^0(a), E^0(a), I^0(a), P^0(a)) $处的线性化算子$ A+{{\cal F}}'(\omega^0) $为$ B_0 $,在地方病平衡态$ \omega^*(S^*(a), E^*(a), I^*(a), P^*(a)) $处的线性化算子$ A+{{\cal F}}'(\omega^*) $为$ B $.先研究$ B_0 $的谱,即找$ B_0 $的特征值,令$ v = (s, e, i, p)^T $,即求

(4.1) $ (B_0-\lambda)v = 0 $

的非平凡解.也即

(4.2) $ \left\{ \begin{array}{l} \frac{{\rm d}s(a)}{{\rm d}a} = -\lambda s-\widehat{x}k(a)S^0(a)-\mu(a)s, \\ \frac{{\rm d}e(a)}{{\rm d}a} = -\lambda e+q\widehat{x}k(a)S^0(a) -[\mu(a)+\alpha(a)]e, \\ \frac{{\rm d}i(a)}{{\rm d}a} = -\lambda i+(1-q)\widehat{x}k(a)S^0(a)+\alpha(a)e -[\mu(a)+\gamma(a)]i, \\ \frac{{\rm d}p(a)}{{\rm d}a} = -\lambda p-\mu(a)p, \end{array}\right. $

其边界条件为

(4.3) $ \left\{ \begin{array}{l} s(0) = p(0) = \int_0^{a_\dagger}\beta(a)p(a){\rm d}a, \ e(0) = i(0) = 0, \\ \widehat{x} = \int_0^{a_\dagger}h(a)i(a){\rm d}a, \ \widehat{n} = \int_0^{a_\dagger}p(a){\rm d}a. \end{array}\right. $

记方程(3.6)的实解为$ \lambda^* $,复解为$ \alpha_j(j = 1, 2, \cdot\cdot\cdot) $.对于以下两种情形:

(1)若$ \lambda = \lambda^* $或对某个$ j , \ \lambda = \alpha_j $,系统(4.2)的最后一个方程及条件(4.3)存在非零解$ p $,则系统(4.2)至少有非平凡解$ (p, 0, 0, p) $,因此$ \lambda^* $, $ \alpha_j $是$ B_0 $的特征值.

(2)若$ \lambda\neq\lambda^* $且$ \lambda\neq\alpha_j $,则系统(4.2)的最后一个方程和条件(4.3)无非零解,因此只有平凡解$ p(a) = 0 $,从而$ s(0) = p(0) = 0 $.由系统(4.2)得

(4.4) $ \begin{array}{ll} s(a) = & -\widehat{x}b_0\pi(a) \int_0^ak(\sigma){\rm e}^{\lambda(\sigma-a)} {\rm e}^{-\lambda^*\sigma}{\rm d}\sigma, \\ e(a) = & \widehat{x}b_0\pi(a) \int_0^aqk(\sigma){\rm e}^{\lambda(\sigma-a)}{\rm e}^{-\lambda^*\sigma} {\rm e}^{-\int_\sigma^a\alpha(\xi){\rm d}\xi}{\rm d}\sigma, \\ i(a) = & \widehat{x}b_0\pi(a) \int_0^ak(\sigma){\rm e}^{\lambda(\sigma-a)}{\rm e}^{-\lambda^*\sigma}{\rm e}^{-\int_\sigma^a\gamma(\xi){\rm d}\xi} \\ & \cdot\bigg[q \int_\sigma^a\alpha(\eta){\rm e}^{\int_\sigma^\eta[\gamma(\xi)-\alpha(\xi)]{\rm d}\xi}{\rm d}\eta+(1-q)\bigg]{\rm d}\sigma. \end{array} $

显然若$ \widehat{x}\neq0 $,则系统(4.2)有非平凡解.把$ i(a) $代入$ \widehat{x} $得含$ \lambda $的特征方程

(4.5) $ \begin{array}[b]{rl} F(\lambda) = & b_0 \int_0^{a_{†}}h(a)\pi(a) \int_0^ak(\sigma){\rm e}^{\lambda(\sigma-a)} {\rm e}^{-\lambda^*\sigma}{\rm e}^{-\int_\sigma^a\gamma(\xi){\rm d}\xi} \\ & \cdot \bigg[q \int_\sigma^a\alpha(\eta){\rm e}^{\int_\sigma^\eta[\gamma(\xi)-\alpha(\xi)]{\rm d}\xi}{\rm d}\eta+(1-q)\bigg]{\rm d}\sigma {\rm d}a = 1. \end{array} $

则此方程的解为$ B_0 $的特征值.而方程(4.5)存在唯一的实解$ \tilde{\lambda} $及系列复解$ \tilde{\alpha_j}(j = 1, 2, \cdot\cdot\cdot) $,且$ {\rm Re}\; \tilde{\alpha_j}<\tilde{\lambda} $.若$ \lambda\rightarrow-\infty $,则$ F(\lambda)\rightarrow+\infty $;若$ \lambda\rightarrow+\infty $,则$ F(\lambda)\rightarrow0 $,又$ F(\lambda) $关于$ \lambda $严格递减,且$ F(\lambda^*) = {{\cal R}}_0 $.故若$ {{\cal R}}_0>1 $(或$ {{\cal R}}_0<1 $),则$ \tilde{\lambda}>\lambda^* $(或$ \tilde{\lambda}<\lambda^* $).

而算子$ B_0 $的预解式为紧算子,因此$ B_0 $的谱全由特征值构成.

定理4.1   $ B_0 $的谱为特征值$ \lambda^*, \ \alpha_j, \ \tilde{\lambda}, \ \tilde{\alpha_j}(j = 1, 2, \cdot\cdot\cdot) $,且

(1)若$ {{\cal R}}_0<1 $,则$ {\rm Re}\; \tilde{\alpha_j}<\lambda^*, \ \mbox{且}\ {\rm Re}\; \tilde{\alpha_j}<\tilde{\lambda}<\lambda^*; $

(2)若$ {{\cal R}}_0>1 $,则$ \tilde{\lambda}>\lambda^*. $

下面在$ {{\cal R}}_0>1 $下探讨算子$ B $的谱,即求其特征值,也就是$ (B-\lambda)v = 0 $的非平凡解,即

(4.6) $ \left\{ \begin{array}{rl} \frac{{\rm d}s(a)}{{\rm d}a} = & -\lambda s-H^*k(a)s- k(a)S^*(a) (\widehat{x}-H^*\widehat{n})-\mu(a)s, \\ \frac{{\rm d}e(a)}{{\rm d}a} = &-\lambda e+qH^*k(a)s+ qk(a)S^*(a)(\widehat{x}-H^*\widehat{n}) -[\mu(a)+\alpha(a)]e, \\ \frac{{\rm d}i(a)}{{\rm d}a} = &-\lambda i+\alpha(a)e+(1-q)H^*k(a)s+ (1-q)k(a)S^*(a)(\widehat{x}-H^*\widehat{n})\\& -[\mu(a)+\gamma(a)]i, \\ \frac{{\rm d}p(a)}{{\rm d}a} = &-\lambda p-\mu(a)p \end{array}\right. $

及边界条件(4.3)的非平凡解.对于以下两种情形:

(1)若$ \lambda = \lambda^* $或对某个$ j , \ \lambda = \alpha_j(j = 1, 2, \cdot\cdot\cdot) $,则系统(4.6)的最后一个方程及条件(4.3)存在非零解$ p $,因此系统(4.6)至少有非平凡解$ (p, 0, 0, p) $,则$ \lambda^* $, $ \alpha_j $是$ B $的特征值.

(2)若$ \lambda\neq\lambda^* $且$ \lambda\neq\alpha_j(j = 1, 2, \cdot\cdot\cdot) $,则系统(4.6)的最后一个方程及条件(4.3)无非零解,因此只有$ p(a) = 0 $,则$ s(0) = p(0) = 0, \ \widehat{n} = 0 $.由系统(4.6)得

(4.7) $ \begin{array}{rl} s(a) = & -b_0\widehat{x}\pi(a){\rm e}^{-H^*\int_0^ak(\tau){\rm d}\tau} \int_0^ak(\sigma){\rm e}^{\lambda(\sigma-a)}{\rm e}^{-\lambda^*\sigma}{\rm d}\sigma, \\ e(a) = & b_0\widehat{x}\pi(a) \int_0^aqk(\sigma){\rm e}^{\lambda(\sigma-a)}{\rm e}^{-\lambda^*\sigma} {\rm e}^{-H^*\int_0^\sigma k(\tau){\rm d}\tau}{\rm e}^{-\int_\sigma^a \alpha(\tau){\rm d}\tau} \\ & \cdot \bigg[1-H^* \int_\sigma^ak(\eta){\rm e}^{\int_\sigma^\eta [\alpha(\tau)-H^*k(\tau)]{\rm d}\tau} {\rm d}\eta\bigg]{\rm d}\sigma, \\ i(a) = & b_0\widehat{x}\pi(a) \int_0^ak(\sigma){\rm e}^{\lambda(\sigma-a)}{\rm e}^{-\lambda^*\sigma} {\rm e}^{-H^*\int_0^\sigma k(\tau){\rm d}\tau}{\rm e}^{-\int_\sigma^a \gamma(\tau){\rm d}\tau} \bigg\{q\int_\sigma^a\alpha(\xi) \\ & \cdot {\rm e}^{\int_\sigma^\xi [\gamma(\tau)-\alpha(\tau)]{\rm d}\tau} \bigg [1-H^* \int_\sigma^\xi k(\eta){\rm e}^{\int_\sigma^\eta [\alpha(\tau)-H^*k(\tau)]{\rm d}\tau} {\rm d}\eta \bigg]{\rm d}\xi\\ & +(1-q) \bigg[1-H^*\int_\sigma^a k(\xi){\rm e}^{\int_\sigma^\xi[\gamma(\tau)-H^*k(\tau)]{\rm d}\tau} {\rm d}\xi \bigg]\bigg\}{\rm d}\sigma. \end{array} $

将$ i(a) $代入$ \widehat{x} $两边约掉$ \widehat{x} $后,得关于$ \lambda $的特征方程

(4.8) $ G(\lambda) = 1, $

其中

(4.9) $ \begin{array}[b]{rl} G(\lambda) = & b_0 \int_0^{a_\dagger}h(a)\pi(a) \int_0^ak(\sigma){\rm e}^{\lambda(\sigma-a)}{\rm e}^{-\lambda^*\sigma} {\rm e}^{-H^*\int_0^\sigma k(\tau){\rm d}\tau}{\rm e}^{-\int_\sigma^a \gamma(\tau){\rm d}\tau} \bigg\{q\int_\sigma^a\alpha(\xi) \\ &\cdot {\rm e}^{\int_\sigma^\xi [\gamma(\tau)-\alpha(\tau)]{\rm d}\tau} \bigg [1-H^* \int_\sigma^\xi k(\eta){\rm e}^{\int_\sigma^\eta [\alpha(\tau)-H^*k(\tau)]{\rm d}\tau} {\rm d}\eta\bigg]{\rm d}\xi\\ & +(1-q)\bigg[1-H^* \int_\sigma^a k(\xi){\rm e}^{\int_\sigma^\xi[\gamma(\tau)-H^*k(\tau)]{\rm d}\tau} {\rm d}\xi \bigg]\bigg\}{\rm d}\sigma {\rm d}a. \end{array} $

为求$ B $的特征值,假设^[4]

(4.10) $ \begin{array}{ll} \frac{E^*(a_\dagger)}{qP^*(a_\dagger)}\leq {\rm e}^{-\int_0^{a_\dagger}\alpha(\tau){\rm d}\tau}, \\ \frac{I^*(a_\dagger)-\int_0^{a_\dagger}\alpha(a)E^*(a) {\rm e}^{-\int_a^{a_\dagger}[\gamma(\tau)+\mu(\tau)+\lambda^*]{\rm d}\tau}{\rm d}a}{(1-q)P^*(a_\dagger)} \leq {\rm e}^{-\int_0^{a_\dagger}\gamma(\tau){\rm d}\tau}. \end{array} $

事实上,条件(4.10)可变形为

$ \begin{array}{ll} H^* \int_0^{a_\dagger} k(\sigma){\rm e}^{\int_0^\sigma [\alpha(\tau)-H^*k(\tau)]{\rm d}\tau} {\rm d}\sigma\leq1, \ H^*\int_0^{a_\dagger} k(\sigma){\rm e}^{\int_0^\sigma [\gamma(\tau)-H^*k(\tau)]{\rm d}\tau} {\rm d}\sigma\leq1. \end{array} $

由(4.10)式第一个不等式得

$ H^*\int_0^{a_\dagger}k(\sigma){\rm e}^{-\int_\sigma^{a_\dagger} \alpha(\tau){\rm d}\tau}{\rm e}^{-H^*\int_0^\sigma k(\tau){\rm d}\tau}{\rm d}\sigma\leq {\rm e}^{-\int_0^{a_\dagger}\alpha(\tau){\rm d}\tau}, $

则

(4.11) $ {\rm e}^{-H^*\int_0^a k(\tau){\rm d}\tau}\geq\int_0^a\alpha(\sigma) {\rm e}^{-\int_\sigma^a \alpha(\tau){\rm d}\tau}{\rm e}^{-H^*\int_0^\sigma k(\tau){\rm d}\tau}{\rm d}\sigma. $

于是(4.9)式第一个中括号内非负.同理由(4.10)式第二个不等式得到(4.9)式第二个中括号内非负.因而若(4.10)式两个条件同时成立,那么$ G(\lambda) $关于$ \lambda $严格递减,且

$ \begin{array}{ll} G(\lambda^*)\leq & b_0 \int_0^{a_\dagger}h(a)\pi(a){\rm e}^{-\lambda^*a} \int_0^ak(\sigma) {\rm e}^{-H^*\int_0^\sigma k(\xi){\rm d}\xi} \\ & \cdot {\rm e}^{-\int_\sigma^a\gamma(\xi){\rm d}\xi}[q \int_\sigma^a\alpha(\tau) {\rm e}^{\int_\sigma^\tau[\gamma(\xi)-\alpha(\xi)]{\rm d}\xi}{\rm d}\tau+(1-q)] {\rm d}\sigma {\rm d}a = 1, \end{array} $

这里用到(3.9)式,则(4.8)式有唯一的实解$ \widehat{\lambda} $和复解$ \widehat{\alpha}_j(j = 1, 2, \cdot\cdot\cdot) $,且

$ {\rm Re}\; \widehat{\alpha}_j<\widehat{\lambda}<\lambda^* . $

显然算子$ B $的预解式为紧算子,则$ B $的谱仅由特征值构成.因此,有

定理4.2  算子$ B $的谱为特征值$ \lambda^*, \alpha_j, \widehat{\lambda}, \widehat{\alpha}_j $,且若(4.10)式成立,则有$ {\rm Re}\, \alpha_j<\lambda^* $且$ {\rm Re}\, \widehat{\alpha}_j<\widehat{\lambda}<\lambda^*. $

探讨无病平衡态及地方病平衡态的局部稳定性.定理4.1和定理4.2得出算子$ B_0 $及$ B $的谱,也即算子$ S_0 $及$ S $的谱,由文献[11]中的定理1.1及定理1.2得,要证明平衡态的稳定性,即证算子$ S_{0\varphi} $及$ S_\varphi $生成的半群最终紧的.只要证$ S_\varphi $生成最终紧半群即可,因$ S_{0\varphi} $生成的半群为$ S_\varphi $生成的半群的特殊情形.

由文献[11]中的(2.12)及(2.14)式,得$ S_\varphi = A+{\mathcal{F}}'(\omega^*)-\langle\varphi, B\rangle\omega^* $,这里$ \omega^* $对应于地方病平衡态($ EE $).记$ S_\varphi $为$ S_\varphi = {{\cal A}}+T $,其中

$ \begin{array}{l} {{\cal A}} = \left( \begin{array}{cccc} -\frac{\partial}{\partial a}-\mu(a)-H^*k(a) & 0 & 0 & 0\\ qH^*k(a) & -\frac{\partial}{\partial a}-\mu(a)-\alpha(a) & 0 & 0\\ (1-q)H^*k(a) & \alpha(a) & -\frac{\partial}{\partial a}-\mu(a)-\gamma(a) & 0\\ 0 & 0 & 0 & -\frac{\partial}{\partial a}-\mu(a) \end{array}\right) \end{array} $

与$ A $有同样的定义域.在无病平衡态处$ H^* = 0 $, $ {{\cal A}} $与$ A $相同.令$ v = (s, e, i, p)^T $

$ (Tv)(a) = \left( \begin{array}{c} -k(a)S^*(a) (\widehat{x}-H^*\widehat{n})-\langle\varphi, Bv\rangle S^*(a)\\ qk(a)S^*(a) (\widehat{x}-H^*\widehat{n})-\langle\varphi, Bv\rangle E^*(a)\\ (1-q)k(a)S^*(a) (\widehat{x}-H^*\widehat{n})-\langle\varphi, Bv\rangle I^*(a)\\ -\langle\varphi, Bv\rangle P^*(a) \end{array}\right), $

这里$ T $是Volterra积分算子, $ T $为有界线性算子且有限维,故为紧算子.因而要证明$ S_\varphi $最终生成紧半群,只需证$ {{\cal A}} $最终生成紧半群.为此用特征线法讨论系统

$ \left\{ \begin{array}{ll} \frac{\partial f_1(a, t)}{\partial a}+ \frac{\partial f_1(a, t)}{\partial t} = -k(a)H^*f_1(a, t)-\mu(a)f_1(a, t), \\ \frac{\partial f_2(a, t)}{\partial a}+ \frac{\partial f_2(a, t)}{\partial t} = qk(a)H^*f_1(a, t)-[\alpha (a)+\mu(a)]f_2(a, t), \\ \frac{\partial f_3(a, t)}{\partial a}+ \frac{\partial f_3(a, t)}{\partial t} = (1-q)k(a)H^*f_1(a, t)+\alpha(a) f_2(a, t)-[\gamma(a)+\mu(a)]f_3(a, t), \\ \frac{\partial f_4(a, t)}{\partial a}+ \frac{\partial f_4(a, t)}{\partial t} = -\mu(a)f_4(a, t) \end{array}\right. $

及条件$ f_1(0, t) = f_4(0, t) = \int_0^{a_{†}}\beta(a)f_4(a, t){\rm d}a, \ f_2(0, t) = f_3(0, t) = 0 $的解.

记$ {{\cal B}}(t) = \int_0^{a_{†}}\beta(a)f_4(a, t){\rm d}a $,为$ t $时刻出生的婴儿函数,则$ {{\cal A}} $生成的半群为

$ ({\rm e}^{{{\cal A}}t}w_0)(a) = \left( \begin{array}{c} {{\cal B}}(t-a)\pi(a){\rm e}^{-H^*\int_0^ak(\tau){\rm d}\tau}\\ H^*{\mathcal{B}}(t-a)\pi(a) \int_0^aqk(\sigma){\rm e}^{-\int_\sigma^a \alpha(\tau){\rm d}\tau}{\rm e}^{-H^*\int_0^\sigma k(\tau){\rm d}\tau}{\rm d}\sigma\\ H^*{\mathcal{B}}(t-a)\pi(a) \int_0^ak(\sigma){\rm e}^{-H^*\int_0^\sigma k(\xi){\rm d}\xi} {\rm e}^{-\int_\sigma^a \gamma(\xi){\rm d}\xi}\\ \cdot \bigg[q\int_\sigma^a\alpha(\tau){\rm e}^{\int_\sigma^\tau [\gamma(\xi)-\alpha(\xi)]{\rm d}\xi}{\rm d}\tau+(1-q)\bigg]{\rm d}\sigma\\ {{\cal B}}(t-a)\pi(a) \end{array}\right). $

则^[11]

(5.1) $ {{\cal B}}(t) = F(t)+\int_0^t{{\cal K}}(s){\mathcal {B}}(t-s){\rm d}s, \ \ t\geq0, $

这里

$ F(t) = \int_0^{a_\dagger}{{\cal K}}(s+t)\frac{f_4(s)}{\pi(s)}{\rm d}s, $

及$ {{\cal K}}(a) = \beta(a)\pi(a) $.而当$ a\geq a_\dagger $时, $ {{\cal K}}(a) = 0 $.因此若$ t\geq a_\dagger $,则$ F(t) = 0 $.且

$ |F(t)|\leq\overline{\beta}\|f_4\|_{L_\pi^1}. $

这里$ \beta(a)\leq \overline{\beta} = \|\beta\|_{L^\infty} $.

由Gronwall引理, (5.1)式转化为

$ |{{\cal B}}(t)|\leq {\rm e}^{\overline{\beta}t}\overline{\beta}\|f_4\|_{L_\pi^1}. $

由Gronwall不等式得

$ |{{\cal B}}(t+h)-{{\cal B}}(t)|\leq {\rm e}^{\overline{\beta}t}\overline{\beta}{\mathcal {O}}(h)\|f_4\|_{L_\pi^1}+Ch{\rm e}^{\overline{\beta}t}\|f_4\|_{L_\pi^1} = {\mathcal{O}}_1(h){\rm e}^{\overline{\beta}t}\|f_4\|_{L_\pi^1}, $

其中$ C $是一个只与$ \overline{\beta} $有关而与$ t $和$ h $无关的常数, $ {\mathcal{O}}(h) $与$ f_4 $无关仅与$ h $有关

${\mathcal{O}}(h) = \int_0^{a_\dagger}|{{\cal K}}(s+h)-{{\cal K}}(s)|{\rm d}s\rightarrow0, \ \ h\rightarrow0. $

故若$ t>2a_\dagger $,则

(5.2) $ \int_0^{a_\dagger}|{\mathcal{B}}(t+h-a)-{\mathcal{B}}(t-a)|{\rm d}a\leq{\mathcal {O}}_2(h)\|w^0\|_X, $

这里$ {{\cal O}}_2(h) = a_\dagger{\mathcal{O}}_1(h){\rm e}^{\overline{\beta}t} $.

再证若$ t>2a_\dagger $, $ {\rm e}^{{{\cal A}}t} $是紧半群,即证$ X $中的任意有界初值集经半群$ {\rm e}^{{{\cal A}}t} $作用后的列向量函数族在$ X $中是紧的,也即证所有列向量函数与$ \pi(a) $的商组成的集合在$ {{\cal X}} $里列紧,只需证明这些列向量的每个分量与$ \pi(a) $的商构成的集合在$ L^1 $空间里列紧.由$ L^1 $空间里的Frećhet-Kolmogorov紧性判据^[11]得,对分量$ x_i(a) $,若$ h\rightarrow0 $, $ \int_0^{a_\dagger}|x_i(a)-x_i(a-h)|{\rm d}a\rightarrow0 $$ (i = 1, 2, 3, 4) $.对于$ X $中的有界初值集,考虑充分小的$ h>0 $,则第一分量估算为

(5.3) $ \begin{array}[b]{rl} & \int_0^{a_\dagger}|{\mathcal{B}}(t-a){\rm e}^{-H^*\int_0^a k(\tau){\rm d}\tau}-{\mathcal{B}}(t+h-a){\rm e}^{-H^*\int_0^{a-h}k(\tau){\rm d}\tau}|{\rm d}a\\ \leq& \int_0^{a_\dagger}|{\mathcal{B}}(t-a)-{\mathcal{B}}(t+h-a)|{\rm d}a+H^* \overline{k}h \int_0^{a_\dagger}{\mathcal{B}}(t-a){\rm d}a\\ \leq&{\mathcal {O}}_3(h)\|w^0\|_X. \end{array} $

若$ h\rightarrow0 $, $ {{\cal O}}_3(h)\rightarrow0 $,且和$ w^0 $无关.

同样可估算第二、第三、第四个分量,由Fre$ \acute{c} $het-Komogorov紧性判据得:若$ t>2a_\dagger $,则$ {\cal A} $生成的半群是紧的.因此由文献[11]得:

定理5.1   (1)若$ {{\cal R}}_0<1 $,则无病平衡态局部渐近稳定;

(2)若$ {{\cal R}}_0>1 $,则无病平衡态不稳定,且若条件(4.10)成立,则地方病平衡态局部渐近稳定.