一类具有连续分布时滞的Nicholson飞蝇模型正周期解的全局指数稳定性

The Global Exponential Stability of the Positive Periodic Solution for a Class of Nicholson’s Blowflies Model with Continuously Distributed Delays

摘要: 该文研究了具有连续分布时滞的Nicholson飞蝇模型，证明了具有连续分布时滞的Nicholson飞蝇模型存在唯一的正周期解以及该正周期解的全局指数稳定性，我们改变了相关文献的模型，改进了相关文献的条件。

Abstract: In this paper, we study the existence of positive periodic solution for Nicholson’s blowflies system with continuously distributed delay. Under appropriate conditions, we obtain that the system has unique positive periodic solution and its global exponential stability.

1. 引言

在生态动力系统中，环境对于人口模型的影响不可忽视，Nicholson飞蝇模型便是考虑了环境因素的模型。在文献 [1] [2] [3] 中，作者研究了具有线性死亡密度的Nicholson飞蝇模型。在文献 [4] [5] [6] 中，作者考虑了具有非线性死亡密度的Nicholson飞蝇模型，为了更精确地描述其发展规律，文献 [7] [8] [9] 研究了具有离散时滞的Nicholson飞蝇模型。随着对此类模型更深入的研究，发现分布时滞更符合真实环境下的模型。在该文中，我们研究一类具有非线性死亡密度连续分布时滞的Nicholson飞蝇模型正周期解的全局指数稳定性。

在文献 [7] 中，作者考虑了离散时滞的Nicholson飞蝇模型的概周期解及其指数稳定性，在文献 [4] 中，作者研究了非线性死亡密度函数为 $a\left(t\right)-b\left(t\right){\text{e}}^{-x}$ 的Nicholson飞蝇模型。该文研究的是非线性死亡密度函数为 $\frac{a\left(t\right)x}{b\left(t\right)+x}$ 具有连续分布时滞的另一类Nicholson飞蝇模型。

该文考虑下面的具有连续分布时滞的Nicholson飞蝇模型

$x^{\prime }\left(t\right)=-\frac{a\left(t\right)x\left(t\right)}{b\left(t\right)+x\left(t\right)}+{\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(t\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(t\right)}{K}_j\left(s\right)x\left(t-s\right){\text{e}}^{-x\left(t-s\right)}\text{d}s}$ (1.1)

其中a，b， ${\beta }_j:R\to \left(0,+\infty \right)$ 是连续的T-周期函数， $K_j(\cdot )>0$， $j=1,2,\cdots ,m$ 为连续时滞核函数。

定义 $g^{+}=\underset{t\in R}{\sup }g\left(t\right)$， $g^{-}=\underset{t\in R}{\inf }g\left(t\right)$， $r=\underset{1\le j\le m}{\max }{\sigma }_j^{+}>0$。

设 $C=C\left(\left[-r,0\right],R\right)$ 是 $\left[-r,0\right]$ 上全体连续函数的集合组成的Banach空间，赋予上确界范数 $\Vert \cdot \Vert$， $C^{+}=C\left(\left[-r,0\right],R^{+}\right)$，定义 $x_t\left(\theta \right)=x\left(t+\theta \right)$， $\theta \in \left[-r,0\right]$，初始条件为 $x_{t_0}=\phi$， $\phi \in C^{+}$， $\phi \left(0\right)>0$。

结合 $\frac{1-x}{{\text{e}}^x}$ 的单调性可知，存在 $k\in \left(0,1\right)$ 使得

$\frac{1-k}{{\text{e}}^k}=\frac{1}{{\text{e}}^2}$ (1.2)

显然

$\underset{x\ge k}{\sup }\left|\frac{1-x}{{\text{e}}^x}\right|=\frac{1}{{\text{e}}^2}$ (1.3)

由 $x{\text{e}}^{-x}$ 在 $\left[0,1\right]$ 上单调递增，在 $\left[1,+\infty \right)$ 上单调递减，则存在唯一的 $\tilde{k}\in \left(1,+\infty \right)$，使得

$k{\text{e}}^{-k}=\tilde{k}{\text{e}}^{-\tilde{k}}$ (1.4)

引理1.1：假设存在 $K\in \left(k,\tilde{k}\right]$ 使得

$\{\begin{array}{l}\underset{t\in R}{\sup }{\displaystyle \underset{j=1}{\overset{m}{\sum }}\frac{{\beta }_j\left(t\right)\left(b\left(t\right)+K\right)}{a\left(t\right)K\text{e}}{\displaystyle {\int }_0^{{\sigma }_j\left(t\right)}{K}_j\left(s\right)\text{d}s}<1,}\\ \underset{t\in R}{\inf }{\displaystyle \underset{j=1}{\overset{m}{\sum }}\frac{{\beta }_j\left(t\right)b\left(t\right)}{a\left(t\right){\text{e}}^k}{\displaystyle {\int }_0^{{\sigma }_j(t)}{K}_j\left(s\right)\text{d}s}>1,}\end{array}$ (1.5)

成立，则系统(1.1)的解 $x\left(t;t_0,\phi \right)$ 在 $\left[t_0,\eta \left(\phi \right)\right)$ 上是有界的，且系统(1.1)是持久的。

证明：令 $x\left(t\right)=x\left(t;t_0,\phi \right)$，因为 $\phi \in C^{+}$，由引理5.2.1 [10] 有 $x_t\left(t_0,\phi \right)\in C^{+}$， $\forall t\in \left[t_0,\eta \left(\phi \right)\right)$。由 $\frac{a\left(t\right)x\left(t\right)}{b\left(t\right)+x\left(t\right)}\le \frac{a\left(t\right)x\left(t\right)}{b\left(t\right)}$，对 $t\in R$， $x\ge 0$，我们有

$\begin{array}{c}{x}^{\prime }\left(t\right)=-\frac{a\left(t\right)x\left(t\right)}{b\left(t\right)+x\left(t\right)}+{\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(t\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(t\right)}{K}_j\left(s\right)x\left(t-s\right){\text{e}}^{-x\left(t-s\right)}\text{d}s}\\ \ge -\frac{a\left(t\right)x\left(t\right)}{b\left(t\right)}+{\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(t\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(t\right)}{K}_j\left(s\right)x\left(t-s\right){\text{e}}^{-x\left(t-s\right)}\text{d}s}\end{array}$，

已知 $x\left(t_0\right)=\phi \left(0\right)>0$，对上式积分有

$\begin{array}{c}x\left(t\right)\ge {\text{e}}^{-{\displaystyle {\int }_{t_o}^t\frac{a\left(s\right)}{b\left(s\right)}\text{d}s}}{\displaystyle \underset{t_0}{\overset{t}{\int }}{\text{e}}^{{\displaystyle {\int }_{t_o}^s\frac{a\left(\tau \right)}{b\left(\tau \right)}\text{d}\tau }}}{\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(s\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(s\right)}{K}_j\left(\tau \right)x\left(s-\tau \right){\text{e}}^{-x\left(s-\tau \right)}\text{d}\tau \text{d}s}+x\left(t_0\right){\text{e}}^{-{\displaystyle {\int }_{t_o}^t\frac{a\left(s\right)}{b\left(s\right)}\text{d}s}}\\ >0,\forall t\in \left[t_0,\eta \left(\phi \right)\right)\end{array}$。

定义

$N\left(t\right)=\max \left\{\gamma :\gamma \le t,x\left(\gamma \right)=\underset{t_0-r\le s\le t}{\max }x\left(s\right)\right\}$， $\forall t\in \left[t_0,\eta \left(\phi \right)\right)$。

下证 $x\left(t\right)$ 在 $\left[t_0,\eta \left(\phi \right)\right)$ 上是有界的。

否则，当 $t\to \eta \left(\phi \right)$ 时， $N\left(t\right)\to \eta \left(\phi \right)$，有 $\underset{t\to \eta \left(\phi \right)}{\lim }x\left(N\left(t\right)\right)=+\infty$。则存在序列 ${\left\{t_n\right\}}_{n=1}^{+\infty }$，使得 $\underset{n\to +\infty }{\lim }{t}_n=\eta \left(\phi \right)$， $\underset{n\to +\infty }{\lim }x\left(N\left(t_n\right)\right)=+\infty$。

另一方面， $x\left(N\left(t\right)\right)=\underset{t_0-r\le s\le t}{\max }x\left(s\right)$，则 $x^{\prime }\left(N\left(t\right)\right)\ge 0,N\left(t\right)>t_0$。

因此，

$0\le x^{\prime }\left(N\left(t\right)\right)=-\frac{a\left(N\left(t\right)\right)x\left(N\left(t\right)\right)}{b\left(N\left(t\right)\right)+x\left(N\left(t\right)\right)}+{\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(N\left(t\right)\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(N\left(t\right)\right)}{K}_j\left(s\right)x\left(N\left(t\right)-s\right){\text{e}}^{-x\left(N\left(t\right)-s\right)}\text{d}s}$，

即

$\frac{a\left(N\left(t\right)\right)x\left(N\left(t\right)\right)}{b\left(N\left(t\right)\right)+x\left(N\left(t\right)\right)}\le {\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(N\left(t\right)\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(N\left(t\right)\right)}{K}_j\left(s\right)x\left(N\left(t\right)-s\right){\text{e}}^{-x\left(N\left(t\right)-s\right)}\text{d}s}$，

由 $\underset{x\ge 0}{\sup }x{\text{e}}^{-x}=\frac{1}{\text{e}}$，有

$\frac{a\left(N\left(t_n\right)\right)x\left(N\left(t_n\right)\right)}{b\left(N\left(t_n\right)\right)+x\left(N\left(t_n\right)\right)}\le {\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(N\left(t_n\right)\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(N\left(t_n\right)\right)}{K}_j\left(s\right)\text{d}s}\frac{1}{\text{e}}$，

则

$\begin{array}{c}\frac{x\left(N\left(t_n\right)\right)}{b\left(N\left(t_n\right)\right)+x\left(N\left(t_n\right)\right)}\le {\displaystyle \underset{j=1}{\overset{m}{\sum }}\frac{{\beta }_j\left(N\left(t_n\right)\right)}{a\left(N\left(t_n\right)\right)\text{e}}}{\displaystyle {\int }_0^{{\sigma }_j\left(N\left(t_n\right)\right)}{K}_j\left(s\right)\text{d}s}\\ \le \underset{t\in R}{\sup }{\displaystyle \underset{j=1}{\overset{m}{\sum }}\frac{{\beta }_j\left(t\right)}{a\left(t\right)\text{e}}{\displaystyle {\int }_0^{{\sigma }_j\left(t\right)}{K}_j\left(s\right)\text{d}s}}\end{array}$，

令 $n\to +\infty$，有

$1\le \underset{t\in R}{\sup }{\displaystyle \underset{j=1}{\overset{m}{\sum }}\frac{{\beta }_j\left(t\right)}{a\left(t\right)\text{e}}{\displaystyle {\int }_0^{{\sigma }_j\left(t\right)}{K}_j\left(s\right)\text{d}s}}$，

与(1.5)矛盾。因此， $x\left(t\right)$ 在 $\left[t_0,\eta \left(\phi \right)\right)$ 上是有界的。由 [11] 中的定理2.3.1知， $\eta \left(\phi \right)=+\infty$。

下证 $x\left(t\right)<K$， $\forall t\in \left[t_1,+\infty \right)$， $t_1>t_0$。否则 $\exists t_2\in \left(t_1,+\infty \right)$，使得 $x\left(t_2\right)=K$， $x\left(t\right)<K$， $\forall t\in \left[t_1,t_2\right)$，

$\begin{array}{c}0\le x^{\prime }\left(t_2\right)=-\frac{a\left(t_2\right)x\left(t_2\right)}{b\left(t_2\right)+x\left(t_2\right)}+{\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(t_2\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(t_2\right)}{K}_j\left(s\right)x\left(t_2-s\right){\text{e}}^{-x\left(t_2-s\right)}\text{d}s}\\ \le -\frac{a\left(t_2\right)K}{b\left(t_2\right)+K}+{\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(t_2\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(t_2\right)}{K}_j\left(s\right)\text{d}s\frac{1}{\text{e}}}\end{array}$

所以，有

$1\le \underset{t\in R}{\sup }{\displaystyle \underset{j=1}{\overset{m}{\sum }}\frac{{\beta }_j\left(t_2\right)\left(b\left(t_2\right)+K\right)}{a\left(t_2\right)K\text{e}}{\displaystyle {\int }_0^{{\sigma }_j\left(t_2\right)}{K}_j\left(s\right)\text{d}s}}\le \underset{t\in R}{\sup }{\displaystyle \underset{j=1}{\overset{m}{\sum }}\frac{{\beta }_j\left(t\right)\left(b\left(t\right)+K\right)}{a\left(t\right)K\text{e}}{\displaystyle {\int }_0^{{\sigma }_j\left(t\right)}{K}_j\left(s\right)\text{d}s}}$，

这与(1.5)矛盾。所以有 $\underset{t\to +\infty }{\lim }\sup x\left(t\right)\le K$。

下证 $c=\underset{t\to +\infty }{\lim }\inf x\left(t\right)>k$，首先证明 $c>0$，否则 $c=0$。

定义

$h\left(t\right)=\max \left\{\gamma :\gamma \le t,x\left(\gamma \right)=\underset{t_0\le s\le t}{\min }x\left(s\right)\right\}$

则有

$\underset{t\to +\infty }{\lim }h\left(t\right)=+\infty$， $\underset{t\to +\infty }{\lim }x\left(h\left(t\right)\right)=0$ (1.6)

由 $h\left(t\right)$ 的定义，有 $x\left(h\left(t\right)\right)=\underset{t_0\le s\le t}{\min }x\left(s\right)$， $x^{\prime }\left(h\left(t\right)\right)\le 0$， $h\left(t\right)>t_0$。因此，

$\begin{array}{c}0\ge x^{\prime }\left(h\left(t\right)\right)=-\frac{a\left(h\left(t\right)\right)x\left(h\left(t\right)\right)}{b\left(h\left(t\right)\right)+x\left(h\left(t\right)\right)}+{\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(h\left(t\right)\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(h\left(t\right)\right)}{K}_j\left(s\right)x\left(h\left(t\right)-s\right){\text{e}}^{-x\left(h\left(t\right)-s\right)}\text{d}s}\\ \ge -\frac{a\left(h\left(t\right)\right)x\left(h\left(t\right)\right)}{b\left(h\left(t\right)\right)}+{\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(h\left(t\right)\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(h\left(t\right)\right)}{K}_j\left(s\right)x\left(h\left(t\right)-s\right){\text{e}}^{-x\left(h\left(t\right)-s\right)}\text{d}s}\end{array}$

由第一积分中值定理，有

$\begin{array}{c}\frac{a\left(h\left(t\right)\right)x\left(h\left(t\right)\right)}{b\left(h\left(t\right)\right)+x\left(h\left(t\right)\right)}\ge {\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(h\left(t\right)\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(h\left(t\right)\right)}{K}_j\left(s\right)x\left(h\left(t\right)-s\right){\text{e}}^{-x\left(h\left(t\right)-s\right)}\text{d}s}\\ ={\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(h\left(t\right)\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(h\left(t\right)\right)}{K}_j\left(s\right)\text{d}s}x\left(h\left(t\right)-{\tau }_j\left(h\left(t\right)\right)\right){\text{e}}^{-x\left(h\left(t\right)-{\tau }_j\left(h\left(t\right)\right)\right)}\end{array}$

结合(1.6)有 $\underset{t\to +\infty }{\lim }x\left(h\left(t\right)-{\tau }_j\left(h\left(t\right)\right)\right)=0$，其中， ${\tau }_j\left(t\right)\in \left[0,{\sigma }_j\left(h\left(t\right)\right)\right]$， $j=1,2,\cdots ,m$。

由系数函数的连续性及周期性，选取序列 ${\left\{t_n\right\}}_{n=1}^{+\infty }$，使得

$\{\begin{array}{l}{t}_n\to +\infty ,x\left(h\left(t_n\right)\right)\to 0,a\left(h\left(t_n\right)\right)\to a^{\ast }\\ b\left(h\left(t_n\right)\right)\to b^{\ast },{\beta }_j\left(h\left(t_n\right)\right)\to {\beta }_j^{\ast },\\ {\sigma }_j\left(h\left(t_n\right)\right)\to {\sigma }_j^{\ast },{\tau }_j\left(h\left(t_n\right)\right)\to {\tau }_j^{\ast }\end{array}$

因此，

$\begin{array}{c}\frac{a\left(h\left(t_n\right)\right)}{b\left(h\left(t_n\right)\right)+x\left(h\left(t_n\right)\right)}\ge {\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(h\left(t_n\right)\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(h\left(t_n\right)\right)}{K}_j\left(s\right)\text{d}s}\frac{x\left(h\left(t_n\right)-{\tau }_j\left(h\left(t_n\right)\right)\right)}{x\left(h\left(t_n\right)\right)}{\text{e}}^{-x\left(h\left(t_n\right)-{\tau }_j\left(h\left(t_n\right)\right)\right)}\\ \ge {\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(h\left(t\right)\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(h\left(t\right)\right)}{K}_j\left(s\right)\text{d}s}{\text{e}}^{-x\left(h\left(t_n\right)-{\tau }_j\left(h\left(t_n\right)\right)\right)}\end{array}$

可得，

$1\ge {\displaystyle \underset{j=1}{\overset{m}{\sum }}\frac{{\beta }_j\left(h\left(t_n\right)\right)\left(b\left(h\left(t_n\right)\right)+x\left(h\left(t_n\right)\right)\right)}{a\left(h\left(t_n\right)\right)}}{\displaystyle {\int }_0^{{\sigma }_j\left(h\left(t_n\right)\right)}{K}_j\left(s\right)\text{d}s}{\text{e}}^{-x\left(h\left(t_n\right)-{\tau }_j\left(h\left(t_n\right)\right)\right)}$

令 $n\to +\infty$，有

$\begin{array}{c}1\ge \underset{n\to +\infty }{\lim }{\displaystyle \underset{j=1}{\overset{m}{\sum }}\frac{{\beta }_j\left(h\left(t_n\right)\right)\left(b\left(h\left(t_n\right)\right)+x\left(h\left(t_n\right)\right)\right)}{a\left(h\left(t_n\right)\right)}}{\displaystyle {\int }_0^{{\sigma }_j\left(h\left(t_n\right)\right)}{K}_j\left(s\right)\text{d}s}{\text{e}}^{-x\left(h\left(t_n\right)-{\tau }_j\left(h\left(t_n\right)\right)\right)}\\ \ge \underset{t\in R}{\inf }{\displaystyle \underset{j=1}{\overset{m}{\sum }}\frac{{\beta }_j\left(t\right)b\left(t\right)}{a\left(t\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(t\right)}{K}_j\left(s\right)\text{d}s}}\end{array}$，

这与(1.5)矛盾，因此， $c>0$。

下证 $c>k$，否则， $c\le k$。

由波动引理( [10] Lemma A.1.)可知，存在数列 ${\left\{t_i\right\}}_{i\ge 1}$，使得 $t_i\to +\infty$， $x\left(t_i\right)\to \underset{t\to +\infty }{\lim }\inf x\left(t\right)$， $x^{\prime }\left(t_i\right)\to 0$，当 $i\to +\infty$ 时。

因为 $\left\{x_{t_i}\right\}$ 是一致有界且等度连续的，由Ascoli-Arzela定理可知，存在 ${\left\{t_i\right\}}_{i\ge 1}$ 的子列，不妨仍记为它本身，使得 $x_{t_i}\to {\phi }^{\ast }$，其中 ${\phi }^{\ast }\in C^{+}$。此外， ${\phi }^{\ast }\left(0\right)=c\le {\phi }^{\ast }\left(\theta \right)\le K$， $\theta \in \left[-r,0\right)$。

由第一积分中值定理，有

$x^{\prime }\left(t\right)=-\frac{a\left(t\right)x\left(t\right)}{b\left(t\right)+x\left(t\right)}+{\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(t\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(t\right)}{K}_j\left(s\right)\text{d}s}x\left(t-{\tau }_j\left(t\right)\right){\text{e}}^{-x\left(t-{\tau }_j\left(t\right)\right)}$，

其中， ${\tau }_j\left(t\right)\in \left[0,{\sigma }_j\left(t\right)\right)$， $j=1,2,\cdots ,m$。

由上可推导出， $c\le {\phi }^{\ast }\left(-{\tau }_j^{\ast }\right)\le K$， $j=1,2,\cdots ,m$。

$x^{\prime }\left(t_i\right)=-\frac{a\left(t_i\right)x\left(t_i\right)}{b\left(t_i\right)+x\left(t_i\right)}+{\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(t_i\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(t_i\right)}{K}_j\left(s\right)\text{d}s}x\left(t_i-{\tau }_j\left(t_i\right)\right){\text{e}}^{-x\left(t_i-{\tau }_j\left(t_i\right)\right)}$

当 $t_i\to +\infty$ 时，有

$\begin{array}{c}0=-\frac{a^{\ast }c}{b^{\ast }+c}+{\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j^{\ast }}{\displaystyle {\int }_0^{{\sigma }_j^{\ast }}{K}_j\left(s\right)\text{d}s}{\phi }^{\ast }\left(-{\tau }_j^{\ast }\right){\text{e}}^{-{\phi }^{\ast }\left(-{\tau }_j^{\ast }\right)}\\ \ge -\frac{a^{\ast }c}{b^{\ast }}+{\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j^{\ast }}{\displaystyle {\int }_0^{{\sigma }_j^{\ast }}{K}_j\left(s\right)\text{d}s}c{\text{e}}^{-c}\end{array}$

与(1.5)矛盾。所以 $c=\underset{t\to +\infty }{\lim }\inf x\left(t\right)>k$。综上，引理1.1得证。

引理1.2：假设引理1.1的条件成立，且

$\underset{t\in R}{\sup }\left\{{\displaystyle \underset{j=1}{\overset{m}{\sum }}\frac{{\beta }_j\left(t\right){\left(b\left(t\right)+K\right)}^2}{a\left(t\right)b\left(t\right){\text{e}}^2}{\displaystyle {\int }_0^{{\sigma }_j\left(t\right)}{K}_j\left(s\right){\text{e}}^s\text{d}s}}\right\}<1$ (1.7)

设 $x\left(t\right)=x\left(t;t_0,\phi \right)$， $x^{\ast }\left(t\right)=x\left(t;t_0,{\phi }^{\ast }\right)$，则存在正常数 $\lambda$， $t^{\ast }$ 使得

$\left|x\left(t;t_0,\phi \right)-x\left(t;t_0,{\phi }^{\ast }\right)\right|\le L_{\phi ,{\phi }^{\ast }}{\text{e}}^{-\lambda t}$， $t\ge t^{\ast }$，

其中，

$L_{\phi ,{\phi }^{\ast }}={\text{e}}^{\lambda t^{\ast }}\left(\underset{t\in \left[t_0-r,t^{\ast }\right]}{\max }\left|x\left(t\right)-x^{\ast }\left(t\right)\right|+1\right)$。

证明：令 $y(t)=x(t)-x^{\ast }(t)$，其中 $t\in [t_0-r,+\infty )$，则有

$\begin{array}{c}{y}^{\prime }\left(t\right)=-\left[\frac{a\left(t\right)x\left(t\right)}{b\left(t\right)+x\left(t\right)}-\frac{a\left(t\right)x^{\ast }\left(t\right)}{b\left(t\right)+x^{\ast }\left(t\right)}\right]\\ +{\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(t\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(t\right)}{K}_j\left(s\right)\left[x\left(t-s\right){\text{e}}^{-x\left(t-s\right)}-x^{\ast }\left(t-s\right){\text{e}}^{-x^{\ast }\left(t-s\right)}\right]\text{d}s}\end{array}$

由引理1.1可知，存在 $t_{\phi ,{\phi }^{\ast }}>t_0$， $k\le x\left(t\right),x^{\ast }\left(t\right)\le K$， $\forall t\in \left[t_{\phi ,{\phi }^{\ast }}-r,+\infty \right)$。

考虑Lyapunov函数 $V\left(t\right)=\left|y\left(t\right)\right|{\text{e}}^{\lambda t}$。

则

$\begin{array}{c}{D}^{-}\left(V\left(t\right)\right)=\lambda \left|y\left(t\right)\right|{\text{e}}^{\lambda t}-{\text{e}}^{\lambda t}\frac{a\left(t\right)x\left(t\right)}{\left(b\left(t\right)+x\left(t\right)\right)\left(b\left(t\right)+x^{\ast }\left(t\right)\right)}\left|x\left(t\right)-x^{\ast }\left(t\right)\right|\\ +\mathrm{sgn}\left(x\left(t\right)-x^{\ast }\left(t\right)\right){\text{e}}^{\lambda t}{\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(t\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(t\right)}{K}_j\left(s\right)\left[x\left(t-s\right){\text{e}}^{-x\left(t-s\right)}-x^{\ast }\left(t-s\right){\text{e}}^{-x^{\ast }\left(t-s\right)}\right]\text{d}s}\end{array}$

断言 $V\left(t\right)=\left|y\left(t\right)\right|{\text{e}}^{\lambda t}<L_{\phi ,{\phi }^{\ast }}$。

否则，存在 $t^{\ast }$，使得 $V\left(t^{\ast }\right)=L_{\phi ,{\phi }^{\ast }}$， $V\left(t\right)<L_{\phi ,{\phi }^{\ast }}$， $t\in \left[t_0-r,t^{\ast }\right)$，由条件知，存在 $\lambda \in \left(0,1\right]$， $0<\mu <1$，且 $\lambda <\frac{a\left(t\right)b\left(t\right)}{{\left(b\left(t\right)+K\right)}^2}$，使得

$\underset{t\in R}{\sup }\left\{{\displaystyle \underset{j=1}{\overset{m}{\sum }}\frac{{\beta }_j\left(t\right){\left(b\left(t\right)+K\right)}^2}{\left(a\left(t\right)b\left(t\right)-\lambda {\left(b\left(t\right)+K\right)}^2\right){\text{e}}^2}{\displaystyle {\int }_0^{{\sigma }_j\left(t\right)}{K}_j\left(s\right){\text{e}}^s\text{d}s}}\right\}<\mu <1$

则

$\begin{array}{c}0\le D^{-}\left(V\left(t^{\ast }\right)\right)=\lambda V\left(t^{\ast }\right)-\frac{a\left(t^{\ast }\right)b\left(t^{\ast }\right)}{\left(b\left(t^{\ast }\right)+x\left(t^{\ast }\right)\right)\left(b\left(t^{\ast }\right)+x^{\ast }\left(t^{\ast }\right)\right)}V\left(t^{\ast }\right)\\ +\mathrm{sgn}\left(x\left(t^{\ast }\right)-x^{\ast }\left(t^{\ast }\right)\right){\text{e}}^{\lambda t^{\ast }}{\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(t^{\ast }\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(t^{\ast }\right)}{K}_j\left(s\right)\left[x\left(t^{\ast }-s\right){\text{e}}^{-x\left(t^{\ast }-s\right)}-x^{\ast }\left(t^{\ast }-s\right){\text{e}}^{-x^{\ast }\left(t^{\ast }-s\right)}\right]\text{d}s}\end{array}$

所以，

$\begin{array}{l}\left(\frac{a\left(t^{\ast }\right)b\left(t^{\ast }\right)}{\left(b\left(t^{\ast }\right)+x\left(t^{\ast }\right)\right)\left(b\left(t^{\ast }\right)+x^{\ast }\left(t^{\ast }\right)\right)}-\lambda \right)L_{\phi ,{\phi }^{\ast }}\\ \le {\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(t^{\ast }\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(t^{\ast }\right)}{K}_j\left(s\right){\text{e}}^{\lambda s}V\left(t^{\ast }-s\right)\text{d}s}\frac{1}{{\text{e}}^2}\\ \le {\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(t^{\ast }\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(t^{\ast }\right)}{K}_j\left(s\right){\text{e}}^s\text{d}s}\frac{1}{{\text{e}}^2}{L}_{\phi ,{\phi }^{\ast }}\end{array}$。

因此，有

$1\le \underset{t\in R}{\sup }\left\{{\displaystyle \underset{j=1}{\overset{m}{\sum }}\frac{{\beta }_j\left(t\right){\left(b\left(t\right)+K\right)}^2}{\left(a\left(t\right)b\left(t\right)-\lambda {\left(b\left(t\right)+K\right)}^2\right){\text{e}}^2}{\displaystyle {\int }_0^{{\sigma }_j\left(t\right)}{K}_j\left(s\right){\text{e}}^s\text{d}s}}\right\}$

这与(1.7)矛盾。所以， $\left|y\left(t\right)\right|<L_{\phi ,{\phi }^{\ast }}{\text{e}}^{-\lambda t}$， $t>t^{\ast }$。

2. 周期解及全局指数稳定性

定理2.1：在引理1.2的条件假设下，系统(1.1)存在唯一的正T-周期解，且该正周期解是全局指数稳定的。

证明：令 $x\left(t\right)=x\left(t;t_0,\phi \right)$，由引理1.1可知，存在 $t_{\phi }>t_0$，使得 $k\le x\left(t;t_0,\phi \right)\le K$， $\forall t\ge t_{\phi }$。由(1.1)的系数和时滞的周期性，对任意的自然数q，有

$\begin{array}{l}{\left[x\left(t+\left(q+1\right)T\right)\right]}^{\prime }={\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(t\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(t\right)}{K}_j\left(s\right)x\left(t+\left(q+1\right)T-s\right){\text{e}}^{-x\left(t+\left(q+1\right)T-s\right)}\text{d}s}\\ -\frac{a\left(t\right)x\left(t+\left(q+1\right)T\right)}{b\left(t\right)+x\left(t+\left(q+1\right)T\right)},t+\left(q+1\right)T\in \left[t_0,+\infty \right)\end{array}$

这表明 $x\left(t+\left(q+1\right)T\right)$ 在 $t\ge t_0-r-\left(q+1\right)T$ 时为系统(1.1)的解。 $x\left(t+T\right)\left(t\in \left[t_0-r,+\infty \right)\right)$ 是系统(1.1)在初值 $\psi \left(s\right)=x\left(s+t_0+T\right)$， $s\in \left[-r,0\right]$ 下的解。

由引理1.2知，对任意的非负整数h，存在正常数 $Q>t_0$，使得当 $t+qT>Q$ 时，有

$\begin{array}{l}\left|x\left(t+\left(q+1\right)T;t_0,\phi \right)-x\left(t+qT;t_0,\phi \right)\right|\\ =\left|x\left(t+qT;t_0,\psi \right)-x\left(t+qT;t_0,\phi \right)\right|\\ \le L_{\phi ,\psi }{\text{e}}^{-\lambda \left(t+qT\right)}\end{array}$

其中

$L_{\phi ,\psi }={\text{e}}^{\lambda Q}\left(\underset{t\in \left[t_0-r,Q\right]}{\max }\left|x\left(t\right)-x^{\ast }\left(t\right)\right|+1\right)$。

令 $\left[a,b\right]\subset R$ 是R中的任意区间，选择非负整数 $p_0$，使得当 $t\in \left[a,b\right]$ 时，有 $t+p_{0}T\ge Q$，则对 $\forall t\in \left[a,b\right]$ 和 $p>p_0$，有

$x\left(t+pT\right)=x\left(t+p_{0}T\right)+{\displaystyle \underset{q=p_0}{\overset{p-1}{\sum }}\left[x\left(t+\left(q+1\right)T\right)-x\left(t+qT\right)\right]}$。

由 $\left[a,b\right]$ 的任意性可知， ${\left\{x\left(t+pT\right)\right\}}_p$ 在R上内闭一致收敛到 $x^{\ast }\left(t\right)$，且 $k\le x^{\ast }\left(t\right)\le K$， $\forall t\in R$。取极限有

$\begin{array}{c}{x}^{\ast }\left(t+T\right)=\underset{p\to +\infty }{\lim }x\left(\left(t+T\right)+pT\right)\\ =\underset{p+1\to +\infty }{\lim }x\left(t+\left(p+1\right)T\right)\\ =x^{\ast }(t)\end{array}$

所以， $x^{\ast }\left(t\right)$ 是T-周期解。

$\begin{array}{l}x\left(t+pT\right)-x\left(t_0+pT\right)\\ ={\displaystyle {\int }_{t_0}^t\left[-\frac{a\left(s\right)x\left(s+pT\right)}{b\left(s\right)+x\left(s+pT\right)}+{\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(s\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(s\right)}{K}_j\left(u\right)x\left(s+pT-u\right){\text{e}}^{-x\left(s+pT-u\right)}\text{d}u}\right]\text{d}s}\end{array}$

令 $p\to +\infty$，有

$\begin{array}{l}{x}^{\ast }\left(t\right)-x^{\ast }\left(t_0\right)\\ ={\displaystyle {\int }_{t_0}^t\left[-\frac{a\left(s\right)x^{\ast }\left(s\right)}{b\left(s\right)+x^{\ast }\left(s\right)}+{\displaystyle \underset{j=1}{\overset{m}{\sum }}{\beta }_j\left(s\right)}{\displaystyle {\int }_0^{{\sigma }_j\left(s\right)}{K}_j\left(u\right)x^{\ast }\left(s-u\right){\text{e}}^{-x^{\ast }\left(s-u\right)}\text{d}u}\right]\text{d}s}\end{array}$

所以， $x^{\ast }\left(t\right)$ 是系统(1.1)在 $\left[t_0-r,+\infty \right)$ 上的解。类似于引理2.1的证明，可证正周期解 $x^{\ast }\left(t\right)$ 是全局指数稳定的。

3. 举例应用

例3.1：考虑下面的系统

$\begin{array}{c}{x}^{\prime }\left(t\right)=-\frac{\left(16+\frac{1}{100}{\sin }^2\left(t\right)\right)x\left(t\right)}{12+\frac{1}{100}{\sin }^2\left(t\right)+x\left(t\right)}+\frac{80+{\cos }^2\left(t\right)}{100}{\displaystyle {\int }_0^1{\text{e}}^{s}x\left(t-s\right){\text{e}}^{-x\left(t-s\right)}\text{d}s}\\ +\frac{80+{\cos }^2\left(2t\right)}{100}{\displaystyle {\int }_0^1{\text{e}}^{s}x\left(t-s\right){\text{e}}^{-x\left(t-s\right)}\text{d}s}\end{array}$

由上式可知，

$a\left(t\right)=16+\frac{1}{100}{\sin }^2\left(t\right)$， $b\left(t\right)=12+\frac{1}{100}{\sin }^2\left(t\right)$， ${\beta }_1\left(t\right)=\frac{80+{\cos }^2\left(t\right)}{100}$， ${\beta }_2\left(t\right)=\frac{80+{\cos }^2\left(2t\right)}{100}$， ${\sigma }_1\left(t\right)={\sigma }_2\left(t\right)=1$， $K_1\left(s\right)=K_2\left(s\right)=e^s$。

则 $r=1$， $a^{+}=16.01$， $a^{-}=16$， $b^{+}=12.01$， $b^{-}=12$， ${\beta }_1^{+}={\beta }_2^{+}=\frac{4}{5}$， ${\beta }_1^{-}={\beta }_2^{-}=\frac{81}{100}$。

取 $K=0.9$， $\lambda =0.4$， $\mu =0.94$，由计算可知， $k\approx 0.72154$， $\tilde{k}\approx 1.34228$。

$\begin{array}{l}\underset{t\in R}{\sup }{\displaystyle \underset{j=1}{\overset{m}{\sum }}\frac{{\beta }_j\left(t\right)\left(b\left(t\right)+K\right)}{a\left(t\right)K\text{e}}{\displaystyle {\int }_0^{{\sigma }_j\left(t\right)}{K}_j\left(s\right)\text{d}s}}\\ \text{=}\underset{t\in R}{\sup }\left\{\frac{\frac{80+{\cos }^2\left(t\right)}{100}\left(12+\frac{1}{100}{\sin }^2\left(t\right)+\frac{9}{10}\right)}{16+\frac{1}{100}{\sin }^2\left(t\right)\frac{9}{10}\text{e}}{\displaystyle {\int }_0^1{\text{e}}^s\text{d}s}+\frac{\frac{80+{\cos }^2\left(2t\right)}{100}\left(12+\frac{1}{100}{\sin }^2\left(t\right)+\frac{9}{10}\right)}{16+\frac{1}{100}{\sin }^2\left(t\right)\frac{9}{10}\text{e}}{\displaystyle {\int }_0^1{\text{e}}^s\text{d}s}\right\}\\ <0.54<1\end{array}$

$\begin{array}{l}\underset{t\in R}{\inf }{\displaystyle \underset{j=1}{\overset{m}{\sum }}\frac{{\beta }_j\left(t\right)b\left(t\right)}{a\left(t\right){\text{e}}^k}{\displaystyle {\int }_0^{{\sigma }_j\left(t\right)}{K}_j\left(s\right)\text{d}s}}\\ =\underset{t\in R}{\inf }\left\{\frac{\frac{80+{\cos }^2\left(t\right)}{100}\left(12+\frac{1}{100}{\sin }^2\left(t\right)\right)}{\left(16+\frac{1}{100}{\sin }^2\left(t\right)\right){\text{e}}^k}{\displaystyle {\int }_0^1{\text{e}}^s\text{d}s}+\frac{\frac{80+{\cos }^2\left(2t\right)}{100}\left(12+\frac{1}{100}{\sin }^2\left(t\right)\right)}{\left(16+\frac{1}{100}{\sin }^2\left(t\right)\right){\text{e}}^k}{\displaystyle {\int }_0^1{\text{e}}^s\text{d}s}\right\}\\ >1.07>1\end{array}$

$\begin{array}{l}\underset{t\in R}{\sup }\left\{{\displaystyle \underset{j=1}{\overset{m}{\sum }}\frac{{\beta }_j\left(t\right){\left(b\left(t\right)+K\right)}^2}{a\left(t\right)b\left(t\right){\text{e}}^2}{\displaystyle {\int }_0^{{\sigma }_j\left(t\right)}{K}_j\left(s\right){\text{e}}^s\text{d}s}}\right\}\\ =\underset{t\in R}{\sup }\left\{\frac{\left(\frac{80+{\cos }^2\left(t\right)}{100}+\frac{80+{\cos }^2\left(2t\right)}{100}\right){\left(12+\frac{1}{100}{\sin }^2\left(t\right)+\frac{9}{10}\right)}^2}{\left(\left(16+\frac{1}{100}{\sin }^2\left(t\right)\right)\left(12+\frac{1}{100}{\sin }^2\left(t\right)\right)-\frac{2}{5}{\left(12+\frac{1}{100}{\sin }^2\left(t\right)+\frac{9}{10}\right)}^2\right){\text{e}}^2}{\displaystyle {\int }_0^1{\text{e}}^{2s}\text{d}s}\right\}\\ <0.94<1\end{array}$

综上，满足定理2.1的全部条件，所以上述系统存在唯一的全局指数稳定的正2π-周期解。

基金项目

国家自然科学基金资助项目(No. 11571088，No. 11471109，No.11526111)，浙江省自然科学项目(No. LY14A010024)，湖南省教育厅项目(No. 14A098)。