在过去的几十年中,非线性时滞微分方程得到了广泛的研究.例如,越来越多的学者正在研究捕食-被捕食模型、传染病模型和红血细胞模型等等.而红血细胞模型最初是被Wazewska-Czyzewska和Lasota两位科学家于1976年在文章^[1]中所提出来的.最初的模型刻画如下

(1.1) $ \begin{equation} x' (t) = -ax(t)+be^{-cx(t-\tau)}, \ \ \ \ t\geq 0, \end{equation} $

其中$ x(t) $表示在$ t $时刻红血细胞的数量, $ a>0 $是红血细胞的死亡率, $ b $和$ c $是单位时间内红细胞产量呈正相关常数, $ \tau $是生产红血细胞所需时间.

在现实世界中,由于外部因素的影响,红血细胞的生产通常是不连续的,不连续性是典型的现象.当连续性不满足时,系统的周期解是否存在?连续条件下可以得到的模型稳定性结论在不连续的条件下是否依然得到保证?在此基础上,国内外许多学者对不连续模型得到了一些结论.例如,在文章^[3]中研究了一类具有离散时滞的不连续捕获Lasota-Wazewska模型,运用了常规Lyapunov函数方法得到了周期解的存在性和全局指数稳定性结果.在文章^[32]中,作者调查了一类具有线性捕获的离散Wazewska-Lasota模型.通过构造Lyapunov泛函,证明了该模型的解指数收敛于概周期解.此外,一些生态模型中还出现了具有周期时滞的泛函微分方程,如动物红细胞存活模型^[18-19]和渐近概周期函数^{[17, 20]}.为了克服这一困难,许多经典连续微分方程理论中的结果已被证明在不连续动力系统中无效^[11-16].

然而,现有的研究大多集中在周期解、概周期解、甚至平衡点的存在性和稳定性上,因此需要一些方法来实现时滞动力系统的稳定性.而在现实世界中,分布时滞具有非常重要的生物学意义,是当前研究的热点问题之一.由于模型中分布时滞的存在,相比于其他模型会比较复杂,因此较难处理,这导致在证明全局解的存在性和稳定性时有一定的难度.而同时分布时滞在现实生物模型中又起着不可忽视的作用.例如,在文章^[36]中的捕食-被捕食模型和在文章^[37]中的随机Lotka-Volterra模型中,这些模型中出现的分布时滞对生态系统的作用相比于离散时滞有着更为深远的影响.而目前对于具有混合分布时滞和不连续捕获的模型,其现有的动力学定性及稳定性结果还很少.

基于上述讨论和分析,本文将考虑如下具有不连续捕获的混合时滞Lasote-Wazewska模型

(1.2) $ \begin{eqnarray} x' (t)& = &-a(t)x(t)+\sum\limits_{i = 1}^n b_{i}(t)e^{-c_{i}(t)x(t-\tau_{i}(t))}\\ &&+\sum\limits_{i = 1}^n p_{i}(t)\int^{\infty}_{0}K_{i}(s)e^{-q_{i}(t)x(t-s)}{\rm d}s-d(t)H(x(t)), \end{eqnarray} $

其中$ d(t) $表示在时刻$ t $时的死亡率; $ K_{i}(\cdot): [0, \infty)\rightarrow [0, \infty), i = 1, 2, \cdots, n $表示分布时滞的概率核函数;其他参数的意义与方程(1.1)一致,其中$ a $, $ b_{i} $, $ c_{i} $, $ p_{i} $, $ q_{i} $和$ d $是连续的$ \omega $-周期函数,并且对于$ i = 1, 2, \cdots, n $, $ a $, $ b_{i} $, $ c_{i} $, $ p_{i} $, $ q_{i} $为正数, $ d $为非负数.核函数$ K_{i}:[0, \infty)\rightarrow [0, \infty) $是连续可积的,当$ i = 1, 2, \cdots, n $时,存在$ M_{i}>0 $满足对$ \int^{\infty}_{0}K_{i}(s){\rm d}s\leq M_{i} $. $ H(\cdot) $表示不连续捕获函数,对于不连续捕获函数假设其满足如下条件.

(H1) $ H\geq 0 $是几乎处处有界的且单调非减, $ H $除了可数个不连续点之外在$ {\Bbb R} $上是连续的,此外, $ H $在任意$ {\Bbb R} $中的紧致区间上仅有有限个不连续点,并且$ H(0) = H(0^{+}) = 0 $.

由于方程(1.2)中的右端函数是不连续的,因此有必要给出方程(1.2)解的精确含义,利用Filippov微分包含理论^[23-24]给出方程解在不连续条件下的定义.下面通过一些基本定义和引理来为得到本文的主要结论做一些前期准备.

$ {\Bbb R}^{n}(n\geq 1) $表示$ n $维欧几里得空间,若$ X\subseteq{\Bbb R}^{n} $,则定义

$ P_{0}(X)\triangleq 2^{X} = \{A\subset X:A\neq \varnothing\}, \ \ P(x) = P_{0}\cup\{\varnothing\}, $

$ P_{fc}(X) = \{A\subset X: {\rm (convex) \ nonempty \ and \ closed}\}, $

$ P_{kc}(X) = \{A\subset X: {\rm (convex) \ nonempty \ and \ compact}\}. $

为了研究在不连续定义下的系统的周期解及稳定性问题,需要给出Filippov解的性质,首先考虑如下泛函微分方程

(1.3) $ \begin{equation} \frac{{\rm d}x}{{\rm d}t} = f(t, x_{t}), \end{equation} $

其中$ f(t, x_{t}) $在$ x $处是不连续的.

定义1.1^[23-24]  通过集值分析,方程(1.3)在Filippov意义下的解$ F: {\Bbb R}^{n}\times{\Bbb C}\rightarrow2^{{\Bbb R}^{n}} $定义如下

$ F(t, x_{t}) = \bigcap\limits_{\delta>0}\bigcap\limits_{\mu({\cal N}) = 0}\overline{co}[f(t, {\mathfrak B}(x_{t}, \delta)\setminus{\cal N})], $

其中$ \mu({\cal N}) $表示$ {\cal N} $的Lebesgue测度集; $ {\mathfrak B}(x_{t}, \delta): = \{\hat{x_{t}}\in {\Bbb C}\big| \|\hat{x_{t}}-x_{t}\|_{{\Bbb C}}<\delta \} $表示以$ x_{t} $为中心以$ \delta $为半径的球.$ \overline{co}[{\Bbb E}] $表示集合$ {\Bbb E} $的凸闭包.如果$ x(t) $在任意$ {\cal I} $区间的紧子集$ [t_{1}, t_{2}] $都是绝对连续的,在非退化区间$ {\cal I}\in {\Bbb R} $上的向量值函数$ x(t) $称为不连续系统的Filippov解,则$ x(t) $需满足如下微分包含

(1.4) $ \begin{equation} \frac{{\rm d}x}{{\rm d}t}\in F(t, x_{t}), \ \ {\rm a.e.} \ \ t\in {\cal I}. \end{equation} $

根据上述定义,方程(1.3)的稳定性问题在Filippov解的意义下可以转化为微分包含(1.4)式的稳定性问题.接下来,我们利用集值映射和微分包含的上述理论,结合Filippov框架,讨论具有右端不连续的混合时滞模型(1.2)的解.

定义1.2^{[24, 27]}  称$ x_{t_{0}} = \varphi:(-\infty, T]\rightarrow{\Bbb R}^{n} $, $ T\in(0, \infty] $是模型(1.2)在$ (-\infty, T] $上的一个状态解解,则需满足

(ⅰ) $ x $在$ (-\infty, T] $上连续且在区间$ [0, T) $的任意的紧子区间是绝对连续;

(ⅱ)存在可测函数$ \gamma:[0, T)\rightarrow{\Bbb R}^{n} $满足$ \gamma(t)\in \overline{co}[H(x(t))] $对几乎处处$ t\in [0, T) $有

(1.5) $ \begin{eqnarray} x' (t)& = &-a(t)x(t)+\sum\limits_{i = 1}^n b_{i}(t)e^{-c_{i}(t)x(t-\tau_{i}(t))}\\ &&+\sum\limits_{i = 1}^n p_{i}(t)\int^{\infty}_{0}K_{i}(s)e^{-q_{i}(t)x(t-s)}{\rm d}s-d(t)\gamma(t). \end{eqnarray} $

由于模型(1.2)是在生物学背景下定义的,故只有正解才是有意义的.因此,给出以下初始条件

(1.6) $ \begin{equation} x_{t_{0}} = \varphi \ \ \mbox{且} \ \ \varphi(0)>0. \end{equation} $

令$ x_{t}(t_{0}, \varphi)(x(t;t_{0}, \varphi)) $是模型(1.2)在$ x_{t_{0}}(t_{0}, \varphi) = \varphi $和$ t_{0}\in {\Bbb R} $定义下的初值.令$ [t_{0}, \eta(\varphi)) $是解$ x_{t}(t_{0}, \varphi) $的一个最大存在区间.

引理1.1^[5]  假设$ \Omega $是Banach空间$ X $的一个紧有界凸子,若存在一个集值映射$ \phi:\Omega\rightarrow P_{kc}(\Omega) $是上半连续的凸紧映射,则$ \phi $在$ \Omega $中具有一个不动点,即存在$ x\in \Omega $有$ x\in \phi(t) $.

定义1.3^[12]  称$ V(x):{\Bbb R}^{n}\rightarrow{\Bbb R}^{n} $是C -正则函数,当且仅当满足

(ⅰ) $ V(x) $在$ {\Bbb R}^{n} $是正则的;

(ⅱ) $ V(x) $是正定的,即对$ x\neq0 $有$ V(x)>0 $和$ V(0) = 0 $;

(ⅲ) $ V(x) $是径向无界,即当$ \|x\|\rightarrow+\infty $时有$ V(x)\rightarrow+\infty $.

定义1.4^[27]  假设$ X, Y $为Hausdorff拓扑空间,并且$ F: X\rightarrow P_{0}(Y) $是严格的,若对于包含$ F(x_{0}) $的任意开集$ U $,总存在一个对应$ x_{0} $的领域$ V $,使$ F(V)\subset U $,则称$ F $在点$ x_{0}\in X $处上半连续(USC).若$ F $在每一点$ x\in X $都是上半连续的,则称$ F $在整个$ X $上是上半连续的.

引理1.2^[5]  若$ V(x):{\Bbb R} ^{n}\rightarrow{\Bbb R} $是C -正则函数,且$ x(t):[0, +\infty)\rightarrow{\Bbb R} ^{n} $在任意紧子区间$ [0, +\infty) $上绝对收敛.则在$ t\in[0, +\infty) $上, $ x(t) $和$ V(x(t)):[0, +\infty)\rightarrow{\Bbb R} $是几乎处处可微的,并有

$ \begin{eqnarray*} \frac{{\rm d}V(x(t))}{{\rm d}t} = \left\langle\varsigma(t), \frac{{\rm d}x(t)}{{\rm d}t}\right\rangle, \quad \forall\varsigma(t)\in \partial V(x(t)). \end{eqnarray*} $

最后,对于有界函数$ g: {\Bbb R}\rightarrow{\Bbb R} $,定义$ g^{+} $和$ g^{-} $为:

$ g^{+} = \sup\limits_{t\in {\Bbb R}} g(t), \qquad g^{-} = \inf\limits_{t\in {\Bbb R}} g(t). $

本节首先基于多值映射的不动点定理和时滞微分包含理论,来研究模型(1.2)的周期解存在性问题,给出定义

$ \begin{eqnarray*} \|x\|_{C} = \max|x_{i}|_{\infty}, \ |x_{i}|_{\infty} = \max\limits_{t\in[0, \omega]}|x_{i}(t)|, \ i = 1, 2, \cdots , n. \end{eqnarray*} $

并给出如下映射$ \phi(x): X\rightarrow2^{X} $:

$ \phi(x) = \int^{t+\omega}_{t} G(t, s){\mathfrak F}(s, x(s)){\rm d}s, $

其中

$ G(t, s) = \frac{e^{\int^{s}_{t}a(u){\rm d}u}}{e^{\int^{t+\omega}_{t}a(u){\rm d}u}-1}, \ G^{+} = \frac{e^{a^{-}}}{e^{a^{-}}-1} , \ G^{-} = \frac{e^{a^{+}}}{e^{a^{+}}-1}, $

$ {\mathfrak F}(s, x(s)) = \sum\limits_{i = 1}^n b_{i}(s)e^{-c_{i}(s)x(s-\tau_{i}(s))}+\sum\limits_{i = 1}^n p_{i}(s)\int^{\infty}_{0}K_{i}(s)e^{-q_{i}(s)x(s-v)}{\rm d}v-d(t)\overline{co}[H(s)]. $

针对不连续捕获策略函数,再给定以下一个基本假设.

(H2)存在非负且单调非减的函数$ W $满足

$ \sup\limits_{\gamma(t)\in \overline{co}[H(x)]}|\gamma(t)|\leq W(t) , $

其中$ W(t) $满足$ \int^{\infty}_{0}\frac{1}{1+r+W(r)}{\rm d}r = \infty $, $ \overline{co}[H(x)] = [\min\{H(x^{-}), H(x^{+})\}, \max\{H(x^{-}), H(x^{+})\}] $.

注2.1  在现有的文献中, $ W(t) $通常是一个线性方程,即当$ W(t) $有界且满足$ a|x|^{\alpha}+b(\alpha\in(0, 1], a, b>0) $时,上述假设$ (H2) $就满足了.而在本文的假设中,系统的Filippov状态解的存在区间可以扩张到$ [0, \infty) $,并且当假设的控制函数$ W(t) $取更一般的情况时,也能得到解的存在性.

命题2.1  称$ x(t) $是模型(1.2)Filippov意义下的$ \omega $周期解当且仅当在初值条件(1.6)下$ x(t) $满足如下积分包含

$ \begin{eqnarray*} x(t)\in\phi(x), \ \ \ t\in[0, \omega]. \end{eqnarray*} $

证  假设$ x(t) $是系统(1.2)的一个$ \omega $ -周期解,根据(1.5)式可知有

$ \begin{eqnarray*} &&\frac{{\rm d}x(t)}{{\rm d}t}\in -a(t)x(t)+\sum\limits_{i = 1}^n b_{i}(t)e^{-c_{i}(t)x(t-\tau_{i}(t))}\\ & &+\sum\limits_{i = 1}^n p_{i}(t)\int^{\infty}_{0}K_{i}(s)e^{-q_{i}(t)x(t-s)}{\rm d}s-d(t)\overline{co}[H(x)] , \end{eqnarray*} $

即对$ t\in[0, \infty) $,有

(2.1) $ \begin{eqnarray} [x(t)e^{\int^{t}_{0}a(s){\rm d}s}]' \in e^{\int^{t}_{0}a(s){\rm d}s}{\mathfrak F}(s, x(s)), \end{eqnarray} $

再根据$ x(t) $的周期性,对(2.1)式在区间上$ [t, t+\omega](0\leq t\leq\omega) $两边同时积分,可得

$ \begin{eqnarray*} x(t)\in\int^{t+\omega}_{t} G(t, s){\mathfrak F}(s, x(s)){\rm d}s, \ \ \ \ t\in[0, \omega]. \end{eqnarray*} $

这就说明了$ \omega $周期解$ x(t) $满足积分包含$ x(t)\in\phi(t) $.

另一方面,假设一个$ \omega $周期解$ x(t) $满足积分包含$ x(t)\in\phi(t) $.则可知对几乎处处$ t\in[0, \infty) $,存在一个可测函数$ \gamma: [0, \infty)\rightarrow{\Bbb R}^{n} $使得$ \gamma(t)\in \overline{co}[H(x)] $,并有

(2.2) $ \begin{eqnarray} x(t) = \int^{t+\omega}_{t} G(t, s){\mathfrak F}(s, \gamma(s)){\rm d}s, \end{eqnarray} $

由于(2.2)式的右端是一个绝对连续函数,对(2.2)式两边同时乘以$ e^{\int^{t}_{0}a(u){\rm d}u} $并对$ t $求导可得

$ \begin{eqnarray*} \frac{{\rm d}x(t)}{{\rm d}t}e^{\int^{t}_{0}a(u){\rm d}u}+x(t)a(t)e^{\int^{t}_{0}a(u){\rm d}u} = (G(t, t+\omega){\mathfrak F}(t+\omega, \gamma)-G(t, t){\mathfrak F}(t, \gamma))e^{\int^{t}_{0}a(u){\rm d}u}. \end{eqnarray*} $

由于$ x(t) $是一个周期函数,故当$ t\in[0, \infty) $时有

$ \begin{eqnarray*} \frac{{\rm d}x(t)}{{\rm d}t} = -a(t)x(t)+{\mathfrak F}(t, \gamma). \end{eqnarray*} $

根据微分包含理论可知, $ x(t) $是系统(1.2)的$ \omega $ -周期解.得证.

命题2.2(正持久性)  如果假设条件(H1)–(H2)满足.则系统(1.2)对应初值条件(1.6)的解$ x(t;t_{0}, \varphi) $的最大存在区间为$ (-\infty, T] $, $ T\in(0, \infty] $,并且在$ t\in[t_{0}, T) $上有$ x(t;t_{0}, \varphi)>0 $.同时, $ \lim\limits_{t\rightarrow\infty}\sup x(t;t_{0}, \varphi)\leq (\sum\limits_{i = 1}^n (b^{+}_{i}+q_{i}^{+}))/a^{-} $.

证  为方便起见,分别记$ x(t;t_{0}, \varphi) $和$ \eta(t_{0}, \varphi) $为$ x(t) $和$ \eta(\varphi) $.首先证明

(2.3) $ \begin{equation} x(t)>0 , \ \ \ t\in(t_{0}, \eta({\varphi})). \end{equation} $

通过反证法,假设(2.3)式不成立,则必然存在$ t^{*}\in(t_{0}, \eta(\varphi)) $,满足当$ t\in[t_{0}, t^{*}) $时$ x(t)>0 $且$ x(t^{*}) = 0 $,则根据假设(H1)–(H2)有

$ \begin{eqnarray*} 0&\geq& D^{-}x(t^{*}) \\ & = &-a(t^{*})x(t^{*})+\sum\limits_{i = 1}^n b_{i}(t^{*})e^{-c_{i}(t^{*})x(t^{*}-\tau_{i}(t^{*}))} \\ & &+\sum\limits_{i = 1}^n p_{i}(t^{*})\int^{\infty}_{0}K_{i}(s)e^{-q_{i}(t^{*})x(t^{*}-s)}{\rm d}s-d(t^{*})\gamma(t^{*}) \\ & = &\sum\limits_{i = 1}^n b_{i}(t^{*})e^{-c_{i}(t^{*})x(t^{*}-\tau_{i}(t^{*}))} +\sum\limits_{i = 1}^n p_{i}(t^{*})\int^{\infty}_{0}K_{i}(s)e^{-q_{i}(t^{*})x(t^{*}-s)}{\rm d}s\\ &>&0, \end{eqnarray*} $

矛盾.故有(2.3)式成立,所以当$ t\in[t_{0}, \eta(\varphi)) $时可得

$ x' (t)\leq -a^{-}x(t)+\sum\limits_{i = 1}^{n}(b^{+}_{i}+q_{i}^{+}\int^{\infty}_{0}K_{i}(s){\rm d}s), $

根据计算当$ t\in[t_{0}, \eta(\varphi)) $时可得

(2.4) $ \begin{eqnarray} x(t)\leq e^{-a^{-}(t-t_{0})}x(t_{0})+\frac{\sum\limits_{i = 1}^{n}(b^{+}_{i}+q_{i}^{+}\int^{\infty}_{0}K_{i}(s){\rm d}s)}{a^{-}}(1-e^{-a(t-t_{0})}). \end{eqnarray} $

故根据(2.3)式和延拓定理可知$ \eta(\varphi) = \infty $.

根据$ \eta(\varphi) = \infty $和(2.4)式可知$ \lim\limits_{t\rightarrow\infty}\sup x(t)\leq (\sum\limits_{i = 1}^n (b^{+}_{i}+q_{i}^{+}\int^{\infty}_{0}K_{i}(s){\rm d}s))/a^{-} $.得证.

命题2.3  如果假设条件(H1)–(H2)满足.再给出以下假设

(H3)

$ \sum\limits_{i = 1}^n b_{i}^{-}e^{-c^{+}_{i}R_{1}}+\sum\limits_{i = 1}^n p_{i}^{-}e^{-q^{+}_{i}R_{1}}\int^{\infty}_{0}K_{i}(s){\rm d}s>d^{+}H^{+}, $

则对任意的$ \theta\in (-\infty, 0] $有$ C_{0} = \{\varphi|\varphi\in C, R_{2}<\varphi(\theta)<R_{1}\} $是系统(1.2)的一个正不变集,其中$ R_{1} = (\sum\limits_{i = 1}^n b_{i}^{+}+\sum\limits_{i = 1}^n p_{i}^{+}\int_{0}^{\infty}K_{i}(s){\rm d}s)/a^{-} $,

$ \begin{eqnarray*} R_{2} = (\sum\limits_{i = 1}^n b_{i}^{-}e^{-c^{+}_{i}R_{1}}+\sum\limits_{i = 1}^n p_{i}^{-}e^{-q^{+}_{i}R_{1}}\int^{\infty}_{0}K_{i}(s){\rm d}s-d^{+}H^{+})/a^{+}. \end{eqnarray*} $

证  令$ \varphi\in C_{0} $.根据上述命题2.2可知,在区间$ [t_{0}, \infty) $上, $ x(t;t_{0}, \varphi) $是系统(1.2)满足$ x(t;t_{0}, \varphi)>0(t\geq t_{0}) $的解.为简便起见,记$ x(t;t_{0}, \varphi) $为$ x(t) $.对任意的$ t\in[t_{0}, T) $,要证

(2.5) $ \begin{eqnarray} 0<x(t)<R_{1}, \end{eqnarray} $

假设(2.5)式不成立,即存在$ t_{1}\in[t_{0}, T) $对所有的$ t\in(-\infty, t_{1}) $满足

$ \begin{eqnarray*} x(t_{1}) = R_{1}, \ \ \ \ 0<x(t)<R_{1}, \end{eqnarray*} $

则由系统(1.2)和(2.1)可得

$ \begin{eqnarray*} 0&\leq& D_{-}x(t_{1}) \\ & = &-a(t_{1})x(t_{1})+\sum\limits_{i = 1}^n b_{i}(t_{1})e^{-c_{i}(t_{1})x(t_{1}-\tau_{i}(t_{1}))}\\ & &+\sum\limits_{i = 1}^n p_{i}(t_{1})\int^{\infty}_{0}K_{i}(s)e^{-q_{i}(t_{1})x(t_{1}-s)}{\rm d}s-d(t_{1})\gamma(t_{1}) \\ &<& -a^{-}x(t_{1})+\sum\limits_{i = 1}^n b_{i}^{+}+\sum\limits_{i = 1}^n p_{i}^{+}\int^{\infty}_{0}K_{i}(s){\rm d}s \\ & = &-a^{-}R_{1}+\sum\limits_{i = 1}^n b_{i}^{+}+\sum\limits_{i = 1}^n p_{i}^{+}\int^{\infty}_{0}K_{i}(s){\rm d}s \\ & = &0. \end{eqnarray*} $

这就得到矛盾,则可知(2.5)式成立.

接下来证明对任意的$ t\in[t_{0}, T) $有

(2.6) $ \begin{eqnarray} x(t)>R_{2}, \end{eqnarray} $

若不然,则存在$ t_{2}\in(t_{0}, \infty) $对任意的$ t\in(-\infty, t_{2}) $满足

(2.7) $ \begin{eqnarray} x(t_{2}) = R_{2}, \ \ \ \ x(t)>R_{2}, \end{eqnarray} $

根据(2.1)和(2.7)计算可得

$ \begin{eqnarray*} 0&\geq& D^{-}x(t_{2}) \\ & = &-a(t_{2})x(t_{2})+\sum\limits_{i = 1}^n b_{i}(t_{2})e^{-c_{i}(t_{2})x(t_{2}-\tau_{i}(t_{2}))}\\ & &+\sum\limits_{i = 1}^n p_{i}(t_{2})\int^{\infty}_{0}K_{i}(s)e^{-q_{i}(t_{2})x(t_{2}-s)}{\rm d}s-d(t_{2})H(x(t_{2})) \\ &>&-a^{+}R_{2}+\sum\limits_{i = 1}^n b_{i}^{-}e^{-c^{+}_{i}R_{1}}+\sum\limits_{i = 1}^n p_{i}^{-}e^{-q^{+}_{i}R_{1}}\int^{\infty}_{0}K_{i}(s){\rm d}s-d^{+}H^{+} \\ & = &0. \end{eqnarray*} $

则可推出矛盾,故(2.7)式成立.综合(2.5)和(2.7)式,命题得证.

命题2.4  如果假设条件(H1)–(H3)满足,则对每一个$ x\in C_{0} $,集值映射$ \phi(x) $是凸集.

证  对任意$ x\in C_{0} $,设$ u\in \phi(x) $和$ u^{*}\in \phi(x) $.则根据微分包含理论,存在可测函数$ \gamma:[0, T)\rightarrow{\Bbb R}^{n} $满足$ \gamma(t)\in \overline{co}[H(x(t))] $且$ |\gamma(t)|<W(t) $,同样存在可测函数$ \gamma^{*}:[0, T)\rightarrow{\Bbb R}^{n} $满足$ \gamma^{*}(t)\in \overline{co}[H(x(t))] $且$ |\gamma^{*}(t)|<W(t) $,则可得下式

$ u(t) = \int^{t+\omega}_{t} G(t, s){\cal F}(s, x(s)){\rm d}s\in \int^{t+\omega}_{t} G(t, s){\mathfrak F}(s, x(s)){\rm d}s = \phi(x)(t), $

$ u^{*}(t) = \int^{t+\omega}_{t} G(t, s){\cal F}^{*}(s, x(s)){\rm d}s\in \int^{t+\omega}_{t} G(t, s){\mathfrak F}(s, x(s)){\rm d}s = \phi(x)(t), $

其中

$ {\cal F}(s, x(s)) = \sum\limits_{i = 1}^n b_{i}(s)e^{-c_{i}(s)x(s-\tau_{i}(s))}+\sum\limits_{i = 1}^n p_{i}(s)\int^{\infty}_{0}K_{i}(s)e^{-q_{i}(s)x(s-v)}{\rm d}v-d(s)\gamma(s), $

$ {\cal F}^{*}(s, x(s)) = \sum\limits_{i = 1}^n b_{i}(s)e^{-c_{i}(s)x(s-\tau_{i}(s))}+\sum\limits_{i = 1}^n p_{i}(s)\int^{\infty}_{0}K_{i}(s)e^{-q_{i}(s)x(s-v)}{\rm d}v-d(s)\gamma^{*}(s). $

令$ 0\leq\lambda\leq1 $,注意到$ \lambda\gamma(t)+(1-\lambda)\gamma^{*}(t)\in \overline{co}[H(x(t))] $,即有

$ \begin{eqnarray*} &&\lambda{\cal F}(s, x(s))+(1-\lambda){\cal F}^{*}(s, x(s))\nonumber\\ & = &\lambda\Big[\sum\limits_{i = 1}^n b_{i}(s)e^{-c_{i}(s)x(s-\tau_{i}(s))}+\sum\limits_{i = 1}^n p_{i}(s)\int^{\infty}_{0}K_{i}(s)e^{-q_{i}(s)x(s-v)}{\rm d}v-d(s)\gamma(s)\Big] \nonumber\\ & &+(1-\lambda)\Big[\sum\limits_{i = 1}^n b_{i}(s)e^{-c_{i}(s)x(s-\tau_{i}(s))}+\sum\limits_{i = 1}^n p_{i}(s)\int^{\infty}_{0}K_{i}(s)e^{-q_{i}(s)x(s-v)}{\rm d}v-d(s)\gamma^{*}(s)\Big] \nonumber\\ &&\in\lambda\Big[\sum\limits_{i = 1}^n b_{i}(s)e^{-c_{i}(s)x(s-\tau_{i}(s))}+\sum\limits_{i = 1}^n p_{i}(s)\int^{\infty}_{0}K_{i}(s)e^{-q_{i}(s)x(s-v)}{\rm d}v-d(s)\overline{co}[H(x(s))]\Big] \nonumber\\ & &+(1-\lambda)\Big[\sum\limits_{i = 1}^n b_{i}(s)e^{-c_{i}(s)x(s-\tau_{i}(s))}+\sum\limits_{i = 1}^n p_{i}(s)\int^{\infty}_{0}K_{i}(s)e^{-q_{i}(s)x(s-v)}{\rm d}v-d(s)\overline{co}[H(x(s))]\Big] \nonumber\\ & = &\sum\limits_{i = 1}^n b_{i}(s)e^{-c_{i}(s)x(s-\tau_{i}(s))}+\sum\limits_{i = 1}^n p_{i}(s)\int^{\infty}_{0}K_{i}(v)e^{-q_{i}(s)x(s-v)}{\rm d}v-d(s)\overline{co}[H(x(s))]\nonumber\\ & = &{\mathfrak F}(s, x(s)), \end{eqnarray*} $

因此对所有的$ t\in[0, \omega] $有

$ \begin{eqnarray*} \lambda u(t)+(1-\lambda )u^{*}(t)& = &\int^{t+\omega}_{t} G(t, s)\big(\lambda{\cal F}(s, x(s))+(1-\lambda){\cal F}^{*}(s, x(s))\big){\rm d}s \nonumber\\ &\in& \int^{t+\omega}_{t} G(t, s){\mathfrak F}(s, x(s)){\rm d}s = \phi(t). \end{eqnarray*} $

这就意味着在$ C_{0} $上对每一个$ x\in C_{0} $, $ \phi(x) $是凸集.得证.

命题2.5  假设条件(H1)–(H3)满足,则集值映射$ \phi: C_{0}\rightarrow P_{kc}(C_{0}) $是一个紧映射.

证  根据Ascoli-Arzela定理,我们只需证明$ \phi(C_{0}) $是一致有界且等度连续的即可.

首先证明$ \phi(C_{0}) $是一致有界的.令任意$ y\in C_{0} $和$ u\in\phi(y) $.则存在一个可测函数$ \gamma:[0, T)\rightarrow{\Bbb R}^{n} $满足$ \gamma(t)\in \overline{co}[H(x(t))] $.显然,对任意的$ x\in C_{0} $,根据假设条件(H2)和(2.5)式可知对几乎处处$ t\in[0, T) $有$ \gamma(t)\leq W(t) $且

$ \begin{eqnarray*} u(t) = \int^{t+\omega}_{t}G(t, s){\mathfrak F}(s, \gamma){\rm d}s>0, \end{eqnarray*} $

因此

$ \begin{eqnarray*} |u(t)|_{\infty}&\leq & G^{+}\int^{t+\omega}_{t}{\mathfrak F}(s, \gamma){\rm d}s \\ &\leq & G^{+}\omega\Big[\sum\limits^{n}_{i = 1}\big(b^{+}e^{-c^{-}R_{2}}+p^{+}e^{-q^{-}R_{2}}\int^{\infty}_{0}K_{i}(s){\rm d}s\big)+d^{+}(\max\limits_{0\leq s\leq t}\{W(s)\}+\|\psi\|_{\infty})\Big] \end{eqnarray*} $

其中$ \|\psi\|_{\infty} = \mathop{\rm ess \sup}\limits_{s\in(-\infty, 0]}|\psi(s)| $,即对任意的$ x\in C_{0} $可得

$ \begin{eqnarray*} \|u\|_{C_{0}}\leq G^{+}\omega\Big[\sum\limits^{n}_{i = 1}\big(b^{+}e^{-c^{-}R_{2}}+p^{+}e^{-q^{-}R_{2}}\int^{\infty}_{0}K_{i}(s){\rm d}s\big)+d^{+}(\max\limits_{0\leq s\leq t}\{W(s)\}+\|\psi\|_{\infty})\Big]. \end{eqnarray*} $

则可知对任意的$ x\in C_{0} $来说, $ \phi(C_{0}) $是一致有界的.

然后我们只需证明$ \phi(C_{0}) $是等度连续的.对任意$ u\in \phi(C_{0}) $,假设$ t_{1}, t_{2}\in[0, \omega] $,有

$ \begin{eqnarray*} |u(t_{1})-u(t_{2})|& = &|\int^{t_{1}+\omega}_{t_{1}}G(t_{1}, s){\mathfrak F}(s, \gamma){\rm d}s-\int^{t_{2}+\omega}_{t_{2}}G(t_{2}, s){\mathfrak F}(s, \gamma){\rm d}s|\\ &\leq&|\int^{t_{1}+\omega}_{t_{1}}G(t_{1}, s){\mathfrak F}(s, \gamma){\rm d}s-\int^{t_{1}+\omega}_{t_{1}}G(t_{2}, s){\mathfrak F}(s, \gamma){\rm d}s|\\ &&+|\int^{t_{1}+\omega}_{t_{1}}G(t_{2}, s){\mathfrak F}(s, \gamma){\rm d}s-\int^{t_{2}+\omega}_{t_{2}}G(t_{2}, s){\mathfrak F}(s, \gamma){\rm d}s|\\ &\leq&|\int^{t_{1}+\omega}_{t_{1}}[G(t_{1}, s)-G(t_{2}, s)]{\mathfrak F}(s, \gamma){\rm d}s| \\ &&+|\int^{t_{1}+\omega}_{t_{2}+\omega}G(t_{2}, s){\mathfrak F}(s, \gamma){\rm d}s|+|\int^{t_{1}}_{t_{2}}G(t_{2}, s){\mathfrak F}(s, \gamma){\rm d}s|\\ &\leq&\max\limits_{t\leq s\leq t+\omega}\{|G(t_{1}, s)-G(t_{2}, s)|\}\int^{\omega}_{0}{\mathfrak F}(s, \gamma){\rm d}s\\ &&+G^{+}\int^{t_{1}+\omega}_{t_{2}+\omega}|{\mathfrak F}(s, \gamma)|{\rm d}s+G^{+}\int^{t_{1}}_{t_{2}}|{\mathfrak F}(s, \gamma)|{\rm d}s. \end{eqnarray*} $

同时根据假设(H1)和(H2),对任意的$ x\in C_{0} $,可得

$ \begin{eqnarray*} |{\mathfrak F}(s, \gamma)|&\leq&\sum\limits_{i = 1}^n |b_{i}(s)e^{-c_{i}(s)x(s-\tau_{i}(s))}|+\sum\limits_{i = 1}^n |p_{i}(s)\int^{\infty}_{0}K_{i}(s)e^{-q_{i}(s)x(s-v)}{\rm d}v|+|d(t)\gamma(t)| \\ &\leq&\sum\limits_{i = 1}^n |b_{i}^{+}e^{-c_{i}^{-}R_{2}}|+\sum\limits_{i = 1}^n |p_{i}^{+}e^{-q_{i}^{-}R_{2}}\int^{\infty}_{0}K_{i}(s){\rm d}s|+d^{+}\max\limits_{0\leq s\leq t}\{W(s)\}\triangleq \Re. \end{eqnarray*} $

这就蕴含着

$ \begin{eqnarray*} |u(t_{1})-u(t_{2})|\leq\max\limits_{t\leq s\leq t+\omega}\{|G(t_{1}, s)-G(t_{2}, s)|\}\omega\Re+2G^{+}\Re|t_{1}-t_{2}|. \end{eqnarray*} $

当$ t_{1}\rightarrow t_{2} $时可知不等式的右端趋于0.因此可得当$ t_{1}\rightarrow t_{2} $时$ \|u(t_{1})-u(t_{2})\|\rightarrow 0 $.这就证明了$ \phi(C_{0}) $是等度连续的.命题得证.

命题2.6  假设条件(H1)–(H3)满足,则集值映射$ \phi: C_{0}\rightarrow P_{kc}(C_{0}) $是上半连续(USC)映射.

证  只需证明$ \phi $是闭图像即可.令$ {\Bbb F}(t, x(t)) = -a(t)x(t)+{\mathfrak F}(t, x(t)) $,令$ |||{\Bbb F}(t, x(t))||| = \sup\{|u|:u\in{\Bbb F}(t, x(t))\} $, $ L^{1}([0, \omega], {\Bbb R}^{n}) $表示一个Banach空间,空间内所有函数$ u:[0, \omega]\rightarrow{\Bbb R}^{n} $都是勒贝格可积的.定义算子$ {\Bbb F}:C_{0}\rightarrow L^{1}([0, \omega], {\Bbb R}^{n}) $容易得到对每一个固定的$ x\in C_{0} $来说, $ {\Bbb F}(x) $是非空的.

考虑线性连续算子$ \hbar:L^{1}([0, \omega], {\Bbb R}^{n})\rightarrow C([0, \omega], {\Bbb R}^{n}) $, $ \hbar u(t) = \int^{t+\omega}_{t}G(t, s)u(s){\rm d}s, $$ t\in[0, \omega]. $因此根据定义1.4可知$ \phi = L\circ{\Bbb F} $是一个闭图像算子.这就意味着$ \phi $是一个上半连续(USC)算子.得证.

根据第二章的引理1.1,结合本小节的命题2.1–2.6可知,集值映射$ \phi: C_{0}\rightarrow P_{kc}(C_{0}) $有至少一个不动点$ x^{*}(t)\in C_{0} $满足$ x^{*}(t)\in \phi(x^{*})(t) $.因此,根据Kakutani's不动点理论,可得到如下的关于系统(1.2)的周期解存在性定理.

定理2.1  假设(H1)–(H3)条件满足,则具有不连续捕获系统(1.2)具有至少一个$ \omega $ -周期解.

本节在$ \omega $ -周期解存在性基础上研究模型(1.2)周期解的全局指数稳定性.

定理3.1  如果假设条件(H1)–(H3)成立.再给出以下假设

(H4)

$ R_{0} = -a^{-}+\sum\limits_{i = 1}^n b_{i}^{+}c_{i}^{+}+\sum\limits_{i = 1}^n p_{i}^{+}q_{i}^{+}\int^{\infty}_{0}K_{i}(s){\rm d}s<0, $

则对满足初值条件$ \varphi, \varphi^{*}\in C_{0} $的解,当$ t\geq t_{0} $时,存在一个正常数$ \varepsilon>0 $使得

$ \begin{eqnarray*} |x(t;t_{0}, \varphi)-x(t;t_{0}, \varphi^{*})|<(R_{1}-R_{2})e^{-\varepsilon(t-t_{0})}. \end{eqnarray*} $

也就是说系统(1.2)的解是全局指数稳定的.

证  令$ \varphi, \varphi^{*}\in C_{0} $,为方便起见,分别把$ x(t;t_{0}, \varphi) $和$ x(t;t_{0}, \varphi^{*}) $记作$ x(t) $和$ x^{*}(t) $,根据上一小节的命题2.3可知对任意的$ t\in[t_{0}, T) $有

$ \begin{eqnarray*} R_{2}<x(t), x^{*}(t)<R_{1}. \end{eqnarray*} $

根据假设条件(H4)可知必存在一个足够小的$ \varepsilon>0 $使得

$ \begin{eqnarray*} -(a^{-}-\varepsilon)+\sum\limits_{i = 1}^n b_{i}^{+}c_{i}^{+}e^{\varepsilon \tau}+\sum\limits_{i = 1}^n p_{i}^{+}q_{i}^{+}\int^{\infty}_{0}K_{i}(s)e^{\varepsilon s}{\rm d}s<0. \end{eqnarray*} $

对$ t\geq t_{0} $,再考虑下述Lyapunov函数

$ \begin{eqnarray*} V(t) = |x(t)-x^{*}(t)|e^{\varepsilon t}. \end{eqnarray*} $

显然, $ V(t) $是绝对连续的,根据求导的链式法则,对任意的$ t\geq t_{0} $可得

$ \begin{eqnarray*} V(t) = |x(t)-x^{*}(t)|e^{\varepsilon t}<(R_{1}-R_{2})e^{\varepsilon t_{0}}, \end{eqnarray*} $

用反正法证明上式成立,如不然,假设存在$ t_{3}>t_{0} $,对任意的$ t\in[t_{0}, t_{3}) $满足

$ \begin{eqnarray*} V(t_{3}) = (R_{1}-R_{2})e^{\varepsilon t_{0}}, \ \ \ \ V(t)<(R_{1}-R_{2})e^{\varepsilon t_{0}}. \end{eqnarray*} $

则计算可得

$ \begin{eqnarray*} 0&\leq &D_{-}V(t_{3}) \\ & = &\varepsilon e^{\varepsilon t_{3}}|x(t_{3})-x^{*}(t_{3})| +e^{\varepsilon t_{3}}{\rm sgn}\{x(t_{3})-x^{*}(t_{3})\}\times\Big[-a(t_{3})(x(t_{3})-x^{*}(t_{3})) \\ & &+\sum\limits_{i = 1}^n p_{i}(t_{3})\Big(\int^{\infty}_{0}K_{i}(s)e^{-q_{i}(t_{3})x(t_{3}-s)}{\rm d}s-\int^{\infty}_{0}K_{i}(s)e^{-q_{i}(t_{3})x^{*}(t_{3}-s)}{\rm d}s\Big) \\ &&+\sum\limits_{i = 1}^n b_{i}(t_{3})\big(e^{-c_{i}(t_{3})x(t_{3}-\tau_{i}(t_{3}))}-e^{-c_{i}(t_{3})x^{*}(t_{3}-\tau_{i}(t_{3}))}\big) -d(t_{3})(\gamma(t_{3})-\gamma^{*}(t_{3}))\Big] \\ &\leq &-(a^{-}-\varepsilon)|x(t_{3})-x^{*}(t_{3})|e^{\varepsilon t_{3}} \\ & &+\sum\limits_{i = 1}^n p_{i}^{+}e^{\varepsilon t_{3}}\int^{\infty}_{0}K_{i}(s)|e^{-q_{i}(t_{3})x(t_{3}-s)}-e^{-q_{i}(t_{3})x^{*}(t_{3}-s)}|{\rm d}s \\ &&+\sum\limits_{i = 1}^n b_{i}^{+}e^{\varepsilon t_{3}}\big(e^{-c_{i}(t_{3})x(t_{3}-\tau_{i}(t_{3}))}-e^{-c_{i}(t_{3})x^{*}(t_{3}-\tau_{i}(t_{3}))}\big) -d^{-}e^{\varepsilon t_{3}}|\gamma(t_{3})-\gamma^{*}(t_{3})| \\ &\leq &-(a^{-}-\varepsilon)|x(t_{3})-x^{*}(t_{3})|e^{\varepsilon t_{3}}\\ & &+\sum\limits_{i = 1}^n b_{i}^{+}c_{i}^{+}|x(t_{3}-\tau_{i}(t_{3}))-x^{*}(t_{3}-\tau_{i}(t_{3}))|e^{\varepsilon(t_{3}-\tau_{i}(t_{3}))}e^{\tau_{i}(t_{3})} \\ &&+\sum\limits_{i = 1}^n p_{i}^{+}q_{i}^{+}\int^{\infty}_{0}K_{i}(s)|x(t_{3}-s)-x^{*}(t_{3}-s)|e^{\varepsilon(t_{3}-s)}e^{\varepsilon s}{\rm d}s \\ &\leq &-(a^{-}-\varepsilon)(R_{1}-R_{2})e^{\varepsilon t_{0}}+\sum\limits_{i = 1}^n b_{i}^{+}c_{i}^{+}fe^{\varepsilon \tau}(R_{1}-R_{2})e^{\varepsilon t_{0}}\\ & &+\sum\limits_{i = 1}^n p_{i}^{+}q_{i}^{+}\int^{\infty}_{0}K_{i}(s)e^{\varepsilon s}{\rm d}s(R_{1}-R_{2})e^{\varepsilon t_{0}} \\ & = & \big(-(a^{-}-\varepsilon)+\sum\limits_{i = 1}^n b_{i}^{+}c_{i}^{+}e^{\varepsilon \tau}+\sum\limits_{i = 1}^n p_{i}^{+}q_{i}^{+}\int^{\infty}_{0}K_{i}(s)e^{\varepsilon s}{\rm d}s\big)(R_{1}-R_{2})e^{\varepsilon t_{0}} \\ &<&0, \end{eqnarray*} $

其中$ t>t_{0} $,这与$ \varepsilon $的选择产生矛盾.故定理得证.