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关于非自治时滞 Beverton-Holt 模型的注记。

A note on the nonautonomous delay Beverton-Holt model.

机构信息

Department of Mathematics, Xavier University of Louisiana, New Orleans, LA 70125, USA.

出版信息

J Biol Dyn. 2010 Mar;4(2):131-9. doi: 10.1080/17513750902803588.

Abstract

It is well known that the periodic cycle {x(n)} of a periodically forced nonlinear difference equation is attenuant (resonant) if av(x(n)) < av(K(n))(av(x(n)) > av(K(n))),where {K ( n )} is the carrying capacity of the environment and av(t(n)) = (1/p)∑(p−1) (i=0) ti (arithmetic mean of the p-periodic cycle {t ( n )}). In this article, we extend the concept of attenuance and resonance of periodic cycles using the geometric mean for the average of a periodic cycle. We study the properties of the periodically forced nonautonomous delay Beverton-Holt model x(n+1) = r(n)x(n)/1 + (r(n−l) − 1)x(n−k)/K(n−k), n= 0, 1, . . . , where {K ( n )} and {r ( n )} are positive p-periodic sequences; (K ( n )>0, r ( n )>1) as well as k and l are nonnegative integers. We will show that for all positive solutions {x ( n )} of the previous equation lim sup (n→∞) (∏(n−1)(i=0)xi)(1/n) ≤ ((∏(p−1)(i=0)ri)(1/p) − 1)(∏(p−1)(i=0)(ri − 1))(−1/p)(∏(p−1)(i=0)Ki)(1/p). In particular, in the case where {x(n)} is a p-periodic solution of the above equation (assuming that such solution exists) and r ( n )=r>1, the periodic cycle is g-attenuant, that is (∏(p−1)(i=0)x(i))(1/p)<(∏(p−1)(i=0)K(i))(p−1) Surprisingly, the obtained results show that the delays k and l do not play any role.

摘要

众所周知,如果 av(x(n)) < av(K(n))(av(x(n)) > av(K(n))),则周期性受迫非线性差分方程的周期{x(n)}衰减(共振),其中{K(n)}是环境的承载能力,av(t(n)) = (1/p)∑(p−1) (i=0) ti(p 个周期{x(t(n))}的算术平均值)。在本文中,我们使用周期性的几何平均值来扩展周期的衰减和共振的概念。我们研究了具有正 p 周期序列{K(n)}和{r(n)}的非自治时滞 Beverton-Holt 模型 x(n+1) = r(n)x(n)/1 + (r(n−l)−1)x(n−k)/K(n−k),n=0,1,....的周期性受迫的性质,其中 k 和 l 是非负整数。我们将证明,对于前面方程的所有正解{x(n)},lim sup(n→∞) ((∏(n−1)(i=0)xi)(1/n) ≤ ((∏(p−1)(i=0)ri)(1/p)−1)(∏(p−1)(i=0)(ri−1))(−1/p)(∏(p−1)(i=0)Ki)(1/p)。特别是,在{x(n)}是上述方程的 p 周期解的情况下(假设存在这样的解)并且 r(n)=r>1,周期环是 g-衰减的,即 (∏(p−1)(i=0)x(i))(1/p)<(∏(p−1)(i=0)K(i))(p−1)。令人惊讶的是,所得到的结果表明延迟 k 和 l 没有任何作用。

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