DOI:10.12677/AAM.2021.108302,PDF,HTML,XML,被引量下载: 382浏览: 612
作者:佘智凤,廖新元^*,金 薇：南华大学，数理学院，湖南 衡阳
关键词:Nicholson果蝇模型；时滞；持久性；全局吸引性；Nicholson’s Blowflies Equation Model；Delay；Permanence；Global Attractivity

文章引用：佘智凤, 廖新元, 金薇. 一类具双时滞的离散Nicholson果蝇模型的持久性和全局吸引性[J]. 应用数学进展, 2021, 10(8): 2888-2896. https://doi.org/10.12677/AAM.2021.108302

1. 引言

由于果蝇的生长周期很短，便于进行生物实验。因此，不论是在生物学方面，还是数学方面都相继出现了很多关于果蝇的讨论。20世纪80年代，Gurney [1] 首次提出了著名的Nicholson果蝇模型，此后，关于Nicholson果蝇模型的研究迅速发展，并已得到了一些很好的研究结果，如Chen [2] 考虑到外界环境周期性的变化，提出时滞周期型Nicholson苍蝇模型，得到了正周期解存在的一些充分条件；Yang [3] 研究了具有多时变时滞的广义Nicholson果蝇模型，利用一些微分不等式，证明了该模型的所有正解都指数收敛于零平衡点；Wang [4] 研究了一类具有多线性收获项的Nicholson果蝇模型正概周期解的存在性和指数收敛性；Berezansky [5] 等提出了一个带收获双时滞的微分方程模型，对于该模型，在文献 [6] [7] [8] 中已经得到了正平衡点的全局吸引性的相关结果，更多研究结果参见文献 [9] [10] [11] [12] [13]。

相对连续模型，离散模型的动力学性态更丰富也更复杂，许多无法求解或理论分析的连续模型往往需要离散模型进行数值模拟。关于离散的Nicholson模型正平衡点的振荡性、全局吸引性和概周期正解的存在性及指数收敛性等动力学性质得到了广泛地讨论 [14] [15] [16]。近年来，由于非自治离散模型能够更准确地刻画系统丰富且复杂的动力学性质，众多学者 [17] [18] [19] 把目光集中在非自治离散模型的研究上。本文基于文献 [5]，研究了一类具有收获项双时滞的离散非自治Nicholson果蝇模型的持久性和全局吸引性。

2. 预备知识

本文考虑下列具有双时滞的Nicholson果蝇离散模型

$x\left(n+1\right)-x\left(n\right)=-\alpha \left(n\right)x\left(n\right)+\beta \left(n\right)x\left(n-\tau \right)e^{-\gamma \left(n\right)x\left(n-\tau \right)}-h\left(n\right)x\left(n-\sigma \right),$(2.1)

初值函数为 $x\left(k\right)=\phi \left(k\right)>0,k\in \left[-{\tau }^{*},0\right],{\tau }^{*}=\max \left\{\tau ,\sigma \right\}$(2.2)

其中， $\alpha \left(n\right)$为第n代成虫平均死亡率， $\tau$为成长期， $\beta \left(n\right)$为第n代按种群数量计算平均产卵率的最大值， $1/\gamma \left(n\right)$为第n代以最大产卵率繁殖的种群数量， $h\left(n\right)$为第n代的收获率， $\sigma$为成熟期。从可持续经济的角度出发，幼体应该不在捕捞范围内，模型(2.1)考虑到捕捞项，引入了一个具时滞的线性函数 $h\left(n\right)x\left(n-\sigma \right)$。

首先，我们假设下列条件(H₁) (H₂)成立

(H₁) $\left\{\alpha \left(n\right)\right\},\left\{\beta \left(n\right)\right\},\left\{\gamma \left(n\right)\right\},\left\{h\left(n\right)\right\}$都是有界非负序列，且

$0<{\alpha }^l<{\alpha }^u<1,0<{\beta }^l<{\beta }^u,0<{\gamma }^l<{\gamma }^u,0<h^l<h^u<1,{\alpha }^u+h^u<1.$

(H₂) 存在两个正常数 $d_1>d_2>0$，使得

$d_1\ge \frac{{\beta }^u}{{\alpha }^l{\gamma }^{l}e},\frac{1}{{\gamma }^u}\le d_2\le \frac{{\beta }^l}{{\alpha }^u}{d}_1{e}^{-{\gamma }^u{d}_1}-\frac{h^u}{{\alpha }^u}{d}_1.$

其中，对任意有界序列 $\left\{f\left(n\right)\right\}$，定义 $f^l={\inf }_{k\in Z}f\left(k\right),f^u={\sup }_{k\in Z}f\left(k\right)$。

引理 2.1 [20] 若 $A>0,y\left(0\right)>0$，且

1) 若 $y\left(n+1\right)\le Ay\left(n\right)+B\left(n\right),n=1,2,\cdots$，则对任意正整数 $k\le n$有，

$y\left(n\right)\le A^{k}y\left(n-k\right)+{\displaystyle \underset{i=0}{\overset{k-1}{\sum }}{A}^{i}B}\left(n-i-1\right),n=1,2,\cdots .$

特别地，当 $A<1$，B有上确界M时，则

$\underset{n\to \infty }{\lim }\sup y\left(n\right)\le \frac{M}{1-A}.$

2) 若 $y\left(n+1\right)\ge Ay\left(n\right)+B\left(n\right),n=1,2,\cdots$，则对任意正整数 $k\le n$有，

$y\left(n\right)\ge A^{k}y\left(n-k\right)+{\displaystyle \underset{i=0}{\overset{k-1}{\sum }}{A}^{i}B}\left(n-i-1\right),n=1,2,\cdots .$

特别地，当 $A<1$，B有下确界m时，则

$\underset{n\to \infty }{\lim }\sup y\left(n\right)\ge \frac{m}{1-A}.$

引理2.2 若 $x\left(n\right)$是方程(2.1)在初值条件(2.2)下的任意解，则

$\underset{n\to \infty }{\lim }\sup x\left(n\right)\le M,M=\frac{{\beta }^u}{{\alpha }^l{\gamma }^{l}e}.$

证明：由方程(2.1)可得

$\begin{array}{l}x\left(n+1\right)=\left(1-\alpha \left(n\right)\right)x\left(n\right)+\beta \left(n\right)x\left(n-\tau \right)e^{-\gamma \left(n\right)x\left(n-\tau \right)}-h\left(n\right)x\left(n-\sigma \right)\\ \le \left(1-\alpha \left(n\right)\right)x\left(n\right)+{\beta }^{u}x\left(n-\tau \right)e^{-{\gamma }^{l}x\left(n-\tau \right)}.\end{array}$

根据 $\underset{u>0}{\sup }u{e}^{-\gamma u}=\frac{1}{\gamma e}$，可以得到

$x\left(n+1\right)\le \left(1-\partial \left(n\right)\right)x\left(n\right)+\frac{{\beta }^u}{{\gamma }^{l}e}.$(2.3)

对于式(2.3)，由引理2.1可得

$\underset{n\to \infty }{\lim }\sup x\left(n\right)\le \frac{{\beta }^u}{{\alpha }^l{\gamma }^{l}e}\stackrel{def}{=}M.$

证毕。

引理2.3 [16] 令 $S_0=\left\{\phi |d_2<\phi \left(k\right)<d_1,\forall k\in \left[-{\tau }^{*},0\right]\right\},{\tau }^{*}=\text{max}\left\{\tau ,\sigma \right\}$，那么方程(2.1)在初值函数 $\phi \in S_0$下的任意解 $x\left(k\right)$满足 $d_2<x\left(k\right)<d_1,k\in Z^{+}$。

3. 持久性和全局吸引性

定理3.1 若条件(H₁)，(H₂)成立，那么具有初值函数 $\phi \in S_0$的系统(2.1)是持久的，即存在常数 $M,m>0$，使系统(2.1)的任意正解 $x\left(n\right)$满足

$m\le \underset{n\to \infty }{\lim }\inf x\left(n\right)\le \underset{n\to \infty }{\lim }\sup x\left(n\right)\le M.$

证明：由引理2.2、引理2.3可以得到

$\underset{n\to \infty }{\lim }\sup x\left(n\right)\le M=\frac{{\beta }^u}{{\alpha }^l{\gamma }^{l}e},\underset{n\to \infty }{\lim }\inf x\left(n\right)>0.$

接下来，只需证明存在正常数m使得 ${\lim }_{n\to \infty }\inf x\left(n\right)\ge m$，

令

$\eta =\underset{n\to \infty }{\lim }\inf x\left(n\right),G=\min \left\{g\left(\eta \right),g\left(M\right)\right\},g\left(x\right)=x{e}^{-{\gamma }^{u}x},$

由方程(2.1)和 $0<{\alpha }^u<1$可以得到

$\eta =\underset{n\to \infty }{\lim }\inf x\left(n\right)\ge \left[\left(\frac{x\left(0\right)+{\beta }^{l}G}{1+h^u}\right){\left(1-{\alpha }^u\right)}^n+\frac{{\beta }^{l}G}{{\alpha }^u\left(1+h^u\right)}\right]=\frac{{\beta }^{l}G}{{\alpha }^u\left(1+h^u\right)},$

如果 $G=g\left(\eta \right)$，那么

$\eta \ge \frac{{\beta }^l}{{\alpha }^u\left(1+h^u\right)}\eta e^{-{\gamma }^u\eta },$

即

$\eta \ge \frac{1}{{\gamma }^u}\ln \frac{{\beta }^l}{{\alpha }^u\left(1+h^u\right)},$

如果 $G=g\left(M\right)$，那么

$\eta \ge \frac{{\beta }^l}{{\alpha }^u\left(1+h^u\right)}M{e}^{-{\gamma }^{u}M}.$

则

$\underset{n\to \infty }{\lim }\inf x\left(n\right)\ge \min \left\{\frac{1}{{\gamma }^u}\ln \frac{{\beta }^l}{{\alpha }^u\left(1+h^u\right)},\frac{{\beta }^l}{{\alpha }^u\left(1+h^u\right)}M{e}^{-{\gamma }^{u}M}\right\}\stackrel{def}{=}m.$

证毕。

定理3.2 若条件(H₁)，(H₂)成立，且 ${\gamma }^{l}m>1$成立，那么具有初值函数 $\phi \in S_0$的系统(2.1)是全局吸引的，

即对系统(2.1)的任意解 $x\left(n\right),y\left(n\right)$有， $\underset{n\to \infty }{\lim }\left(x\left(n\right)-y\left(n\right)\right)=0$。

证明：由定理3.1可得

$m\le \underset{n\to \infty }{\lim }\inf x\left(n\right)\le \underset{n\to \infty }{\lim }\sup x\left(n\right)\le M,m\le \underset{n\to \infty }{\lim }\inf y\left(n\right)\le \underset{n\to \infty }{\lim }\sup y\left(n\right)\le M.$

对任意足够小的正常数 $\epsilon$，存在一个正整数 $n_0$，使得对任意 $n\ge n_0$，

$m\le x\left(n\right),y\left(n\right)\le M,$(3.1)

由均值定理可以得到

$x\left(n\right)e^{-x\left(n\right)}-y\left(n\right)e^{-y\left(n\right)}=\left(1-\theta \left(n\right)\right)e^{-\theta \left(n\right)}\left(x\left(n\right)-y\left(n\right)\right),x\left(n\right)<\theta \left(n\right)<y\left(n\right).$(3.2)

令

$w\left(n\right)=x\left(n\right)-y\left(n\right)+{\displaystyle \underset{s=n-\tau }{\overset{n-1}{\sum }}\beta \left(s+\tau \right)\left(x\left(s\right)e^{-\gamma \left(s+\tau \right)x\left(s\right)}-y\left(s\right)e^{-\gamma \left(s+\tau \right)y\left(s\right)}\right)}-{\displaystyle \underset{s=n-\sigma }{\overset{n-1}{\sum }}\beta \left(s+\sigma \right)\left(x\left(s\right)-y\left(s\right)\right)},$

则

$\begin{array}{c}w\left(n+1\right)=\left(1-\alpha \left(n\right)\right)\left(x\left(n\right)-y\left(n\right)\right)+\beta \left(n\right)\left(x\left(n-\tau \right)e^{-\gamma \left(n\right)x\left(n-\tau \right)}-y\left(n-\tau \right)e^{-\gamma \left(n\right)y\left(n-\tau \right)}\right)\\ -h\left(n\right)\left(x\left(n-\sigma \right)-y\left(n-\sigma \right)\right)+{\displaystyle \underset{s=n-\tau +1}{\overset{n}{\sum }}\beta \left(s+\tau \right)\left(x\left(s\right)e^{-\gamma \left(s+\tau \right)x\left(s\right)}-y\left(s\right)e^{-\gamma \left(s+\tau \right)y\left(s\right)}\right)}\\ -{\displaystyle \underset{s=n-\sigma +1}{\overset{n}{\sum }}\beta \left(s+\sigma \right)}\left(x\left(s\right)-y\left(s\right)\right),\end{array}$

$\begin{array}{c}\Delta w\left(n\right)=w\left(n+1\right)-w\left(n\right)\\ =-\alpha \left(n\right)\left(x\left(n\right)-y\left(n\right)\right)+\beta \left(n+\tau \right)\left(x\left(n\right)e^{-\gamma \left(n+\tau \right)x\left(n\right)}-y\left(n\right)e^{-\gamma \left(n+\tau \right)y\left(n\right)}\right)\\ -h\left(n+\sigma \right)\left(x\left(n\right)-y\left(n\right)\right).\end{array}$

构造非负函数 $V\left(n\right)=w^2\left(n\right)$，则

(3.3)

由(3.2)、(3.3)可得

$\begin{array}{l}\Delta V\left(n\right)\\ =-\left(\alpha \left(n\right)+h\left(n+\sigma \right)\right)\left(2-\alpha \left(n\right)-h\left(n+\sigma \right)\right){\left(x\left(n\right)-y\left(n\right)\right)}^2\\ +2\left(1-\alpha \left(n\right)-h\left(n+\sigma \right)\right)\beta \left(n+\tau \right)\left(1-\theta \left(n\right)\gamma \left(n+\tau \right)\right)e^{-\theta \left(n\right)\gamma \left(n+\tau \right)}{\left(x\left(n\right)-y\left(n\right)\right)}^2\\ +2\left(\alpha \left(n\right)+h\left(n+\sigma \right)\right)\left(x\left(n\right)-y\left(n\right)\right){\displaystyle \underset{s=n-\tau }{\overset{n-1}{\sum }}\beta \left(s+\tau \right)\left(1-\theta \left(s\right)\gamma \left(s+\tau \right)\right)e^{-\theta \left(s\right)\gamma \left(s+\tau \right)}\left(x\left(s\right)-y\left(s\right)\right)}\\ +{\left(\beta \left(n+\tau \right)\left(1-\theta \left(n\right)\gamma \left(n+\tau \right)\right)e^{-\theta \left(n\right)\gamma \left(n+\tau \right)}\left(x\left(n\right)-y\left(n\right)\right)\right)}^2\\ +2\beta \left(n+\tau \right)\left(1-\theta \left(n\right)\gamma \left(n+\tau \right)\right)e^{-\theta \left(n\right)\gamma \left(n+\tau \right)}\left(x\left(n\right)-y\left(n\right)\right)\\ \cdot {\displaystyle \underset{s=n-\tau }{\overset{n-1}{\sum }}\beta \left(s+\tau \right)\left(1-\theta \left(s\right)\gamma \left(s+\tau \right)\right)e^{-\theta \left(s\right)\gamma \left(s+\tau \right)}\left(x\left(s\right)-y\left(s\right)\right)}\\ +2\left(\alpha \left(n\right)+h\left(n+\sigma \right)\right)\left(x\left(n\right)-y\left(n\right)\right){\displaystyle \underset{s=n-\sigma }{\overset{n-1}{\sum }}h\left(s+\sigma \right)\left(x\left(s\right)-y\left(s\right)\right)}\\ -2\beta \left(n+\tau \right)\left(1-\theta \left(n\right)\gamma \left(n+\tau \right)\right)e^{-\theta \left(n\right)\gamma \left(n+\tau \right)}\left(x\left(n\right)-y\left(n\right)\right)\cdot {\displaystyle \underset{s=n-\sigma }{\overset{n-1}{\sum }}h\left(s+\sigma \right)\left(x\left(s\right)-y\left(s\right)\right).}\end{array}$(3.4)

将 $2ab\le a^2+b^2$运用到式(3.4)得

$\begin{array}{l}\Delta V\left(n\right)\le -\left(\alpha \left(n\right)+h\left(n+\sigma \right)\right)\left(2-\alpha \left(n\right)-h\left(n+\sigma \right)\right){\left(x\left(n\right)-y\left(n\right)\right)}^2\\ +2\left(1-\alpha \left(n\right)-h\left(n+\sigma \right)\right)\beta \left(n+\tau \right)\left(1-\theta \left(n\right)\gamma \left(n+\tau \right)\right)e^{-\theta \left(n\right)\gamma \left(n+\tau \right)}{\left(x\left(n\right)-y\left(n\right)\right)}^2\\ +\left(\alpha \left(n\right)+h\left(n+\sigma \right)\right){\displaystyle \underset{s=n-\tau }{\overset{n-1}{\sum }}\beta \left(s+\tau \right)\left(1-\theta \left(s\right)\gamma \left(s+\tau \right)\right)e^{-\theta \left(s\right)\gamma \left(s+\tau \right)}{\left(x\left(n\right)-y\left(n\right)\right)}^2}\\ +\left(\alpha \left(n\right)+h\left(n+\sigma \right)\right){\displaystyle \underset{s=n-\tau }{\overset{n-1}{\sum }}\beta \left(s+\tau \right)\left(1-\theta \left(s\right)\gamma \left(s+\tau \right)\right)e^{-\theta \left(s\right)\gamma \left(s+\tau \right)}{\left(x\left(s\right)-y\left(s\right)\right)}^2}\\ +{\left(\beta \left(n+\tau \right)\left(1-\theta \left(n\right)\gamma \left(n+\tau \right)\right)e^{-\theta \left(n\right)\gamma \left(n+\tau \right)}\left(x\left(n\right)-y\left(n\right)\right)\right)}^2\\ +\beta \left(n+\tau \right)\left(1-\theta \left(n\right)\gamma \left(n+\tau \right)\right)e^{-\theta \left(n\right)\gamma \left(n+\tau \right)}\\ \cdot ({\displaystyle \underset{s=n-\tau }{\overset{n-1}{\sum }}\beta \left(s+\tau \right)\left(1-\theta \left(s\right)\gamma \left(s+\tau \right)\right)e^{-\theta \left(s\right)\gamma \left(s+\tau \right)}{\left(x\left(n\right)-y\left(n\right)\right)}^2}\\ +{\displaystyle \underset{s=n-\tau }{\overset{n-1}{\sum }}\beta \left(s+\tau \right)\left(1-\theta \left(s\right)\gamma \left(s+\tau \right)\right)e^{-\theta \left(s\right)\gamma \left(s+\tau \right)}{\left(x\left(s\right)-y\left(s\right)\right)}^2})\\ +\left(\alpha \left(n\right)+h\left(n+\sigma \right)\right){\displaystyle \underset{s=n-\sigma }{\overset{n-1}{\sum }}h\left(s+\sigma \right){\left(x\left(n\right)-y\left(n\right)\right)}^2}\\ +\left(\alpha \left(n\right)+h\left(n+\sigma \right)\right){\displaystyle \underset{s=n-\sigma }{\overset{n-1}{\sum }}h\left(s+\sigma \right){\left(x\left(s\right)-y\left(s\right)\right)}^2}\\ +\beta \left(n+\tau \right)\left(1-\theta \left(n\right)\gamma \left(n+\tau \right)\right)e^{-\theta \left(n\right)\gamma \left(n+\tau \right)}{\displaystyle \underset{s=n-\sigma }{\overset{n-1}{\sum }}h\left(s+\sigma \right){\left(x\left(n\right)-y\left(n\right)\right)}^2}\\ +\beta \left(n+\tau \right)\left(1-\theta \left(n\right)\gamma \left(n+\tau \right)\right)e^{-\theta \left(n\right)\gamma \left(n+\tau \right)}{\displaystyle \underset{s=n-\sigma }{\overset{n-1}{\sum }}h\left(s+\sigma \right){\left(x\left(s\right)-y\left(s\right)\right)}^2}.\end{array}$(3.5)

依据 ${\gamma }^{l}m>1,{\alpha }^u+h^u<1,\underset{x\in \left[1,+\infty \right)}{\max }\left(1-x\right)e^{-x}<0$，不等式(3.5)可写成

$\begin{array}{l}\Delta V\left(n\right)\le -\left(\alpha \left(n\right)+h\left(n+\sigma \right)\right)\left(2-\alpha \left(n\right)-h\left(n+\sigma \right)\right){\left(x\left(n\right)-y\left(n\right)\right)}^2\\ \le -\left({\alpha }^u+h^u\right)\left(2-{\alpha }^u-h^u\right){\left(x\left(n\right)-y\left(n\right)\right)}^2\\ \le -\delta {\left(x\left(n\right)-y\left(n\right)\right)}^2.\end{array}$

取一个足够小的正常数 $\delta >0$使得

$\left({\alpha }^u+h^u\right)\left(2-{\alpha }^u-h^u\right)\le \delta ,$

则

$\Delta V\left(n\right)\le -\delta {\left(x\left(n\right)-y\left(n\right)\right)}^2.$(3.6)

将不等式(3.6)左右两边分别从 $n_0+\tau$累加到n可得

${\displaystyle \underset{s=n_0+\tau }{\overset{n}{\sum }}\left(V\left(s+1\right)-V\left(s\right)\right)}\le -\delta {\displaystyle \underset{s=n_0+\tau }{\overset{n}{\sum }}{\left(x\left(s\right)-x\left(s\right)\right)}^2},$

即

$V\left(n+1\right)+\delta {\displaystyle \underset{s=n_0+\tau }{\overset{n}{\sum }}{\left(x\left(s\right)-x\left(s\right)\right)}^2}\le V\left(n_0+\tau \right),$

则

${\displaystyle \underset{s=n_0+\tau }{\overset{n}{\sum }}{\left(x\left(s\right)-x\left(s\right)\right)}^2}\le \frac{V\left(n_0+\tau \right)}{\delta }.$

由(3.1)可知 $V\left(n_0+\tau \right)$是有界的，因此

${\displaystyle \underset{s=n_0+\tau }{\overset{n}{\sum }}{\left(x\left(s\right)-x\left(s\right)\right)}^2}\le \frac{V\left(n_0+\tau \right)}{\delta }<+\infty ,$

也就是说

${\displaystyle \underset{s=n_0+\tau }{\overset{+\infty }{\sum }}{\left(x\left(s\right)-x\left(s\right)\right)}^2}\le \frac{V\left(n_0+\tau \right)}{\delta }<+\infty .$

显而易见， $\underset{n\to \infty }{\lim }{\left(x\left(n\right)-y\left(n\right)\right)}^2=0$，即 $\underset{n\to \infty }{\lim }\left(x\left(n\right)-y\left(n\right)\right)=0$。

证毕。

4. 数值模拟

考虑下列方程

$\begin{array}{l}x\left(n+1\right)-x\left(n\right)=-\left(0.5+0.01\sin \left(\sqrt{2}n\right)\right)x\left(n\right)+\left(2.79+0.01\sin \left(\sqrt{3}n\right)\right)x\left(n-1\right)\\ \cdot e^{-\left(0.82+0.01\sin n)x\left(n-1\right)\right)}-\left(0.2+0.01\sin \sqrt{2}n\right)x\left(n-1\right)\end{array}$(4.1)

初始条件为

$x\left(-1\right)=1.4,x\left(0\right)=1.6$

其中，

$\begin{array}{l}{\alpha }^u=0.51,{\alpha }^l=0.49,{\beta }^u=2.8,{\beta }^l=2.78,{\gamma }^u=0.83,{\gamma }^l=0.81,h^u=0.21,\\ h^l=0.19,M=2.25,m=1.56,{\alpha }^u+h^u<1,{\gamma }^{l}m\approx 1.26>1.\end{array}$

$\begin{array}{l}x\left(n+1\right)-x\left(n\right)=-\left(0.5+0.02\sin \left(\sqrt{2}n\right)\right)x\left(n\right)+\left(2.79+0.02\sin \left(\sqrt{3}n\right)\right)x\left(n-1\right)\\ \cdot e^{-\left(0.82+0.02\sin n)x\left(n-1\right)\right)}-\left(0.2+0.02\sin \sqrt{2}n\right)x\left(n-1\right)\end{array}$(4.2)

初始条件为

$x\left(-1\right)=1.3,x\left(0\right)=1.4$

$\begin{array}{l}x\left(n+1\right)-x\left(n\right)=-\left(0.51+0.02\sin \left(\sqrt{2}n\right)\right)x\left(n\right)+\left(2.80+0.02\sin \left(\sqrt{3}n\right)\right)x\left(n-1\right)\\ \cdot e^{-\left(0.80+0.02\sin n)x\left(n-1\right)\right)}-\left(0.21+0.02\sin \sqrt{2}n\right)x\left(n-1\right)\end{array}$(4.3)

初始条件为

$x\left(-1\right)=1.5,x\left(0\right)=1.4$

$\begin{array}{l}x\left(n+1\right)-x\left(n\right)=-\left(0.51+0.01\sin \left(\sqrt{2}n\right)\right)x\left(n\right)+\left(2.80+0.02\sin \left(\sqrt{3}n\right)\right)x\left(n-1\right)\\ \cdot e^{-\left(0.82+0.01\sin n)x\left(n-1\right)\right)}-\left(0.21+0.02\sin \sqrt{2}n\right)x\left(n-1\right)\end{array}$(4.4)

初始条件为

$x\left(-1\right)=1.3,x\left(0\right)=1.5$

显然，上述系统均满足条件(H₁)，(H₂) (H₃)，依据定理3.1和定理3.2，系统(4.1)、(4.2)、(4.3)、(4.4)的解 $x\left(n\right)$是持久且全局吸引的的(如图1)。

5. 结论

本文考虑了具有时滞和收获项的离散非自治Nicholson果蝇模型。对于该模型，通过构造适当的Lyapunov函数，我们得到了它的持久性和全局吸引性存在的充分条件，并通过Matlab数值模拟充分验证了结果。

NOTES

^*通讯作者。

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