﻿ Leslie型捕食者–食饵系统的弛豫振荡分析 Relaxation Oscillations Analysis in a Predator-Prey System of Leslie Type

Vol. 08  No. 12 ( 2019 ), Article ID: 33352 , 6 pages
10.12677/AAM.2019.812222

Relaxation Oscillations Analysis in a Predator-Prey System of Leslie Type

Chenrong Pan, Songlin Chen

School of Mathematical Science and Engineering, Anhui University of Technology, Ma’anshan Anhui

Received: Nov. 9th, 2019; accepted: Dec. 2nd, 2019; published: Dec. 9th, 2019

ABSTRACT

In this paper, we consider a predator-prey system of Leslie type with functional response. The equilibrium point type and stability of the system are analyzed, and the existence and uniqueness of relaxation oscillation period of the system are proved by using the entry-exit function and geometric singular perturbation theory.

Keywords:Leslie Predator-Prey System, Entry-Exit Function, Geometric Singular Perturbation Theory, Existence and Uniqueness

Leslie型捕食者–食饵系统的弛豫振荡分析

1. 引言

$\left\{\begin{array}{l}\frac{\text{d}x}{\text{d}t}=rx\left(1-\frac{x}{K}\right)-\frac{mxy}{a{x}^{2}+bx+1},\\ \frac{\text{d}y}{\text{d}t}=sy\left(1-\frac{y}{hx}\right).\end{array}$ (1)

$\stackrel{¯}{t}=rt,\stackrel{¯}{x}=\frac{x}{K},\stackrel{¯}{y}=\frac{my}{r},\stackrel{¯}{a}=a{K}^{2},\stackrel{¯}{b}=bK,\epsilon =\frac{s}{r},\beta =\frac{r}{hmK}.$ (2)

$\stackrel{¯}{x},\stackrel{¯}{y},\stackrel{¯}{t},\stackrel{¯}{a},\stackrel{¯}{b}$ 仍记为 $x,y,t,a,b$，则系统(1)可以化为：

$\left\{\begin{array}{l}\frac{\text{d}x}{\text{d}t}=x\left(1-x\right)-\frac{xy}{a{x}^{2}+bx+1},\\ \frac{\text{d}y}{\text{d}t}=\epsilon y\left(1-\frac{\beta y}{x}\right).\end{array}$ (3)

2. 研究背景

2.1. 系统(3)的平衡点分析

1) 如果 $\stackrel{¯}{\Delta }<0$，则系统(3)有一个基本的正平衡点；

2) 如果 $\stackrel{¯}{\Delta }=0$，并且

2.1) ${\Delta }_{1}>0$，则系统(3)有两个不同的正平衡点：一个是基本的正平衡点，另一个是退化的正平衡点；

2.2) ${\Delta }_{1}=0$，则系统(3)有一个正平衡点，并且为退化的正平衡点；

3)如果 $\stackrel{¯}{\Delta }>0$，且 $b<\mathrm{min}\left\{a,\frac{1}{\beta }+1\right\}$，则系统(3)有三个不同的基本的正平衡点。

2.2. 入–出函数

$\left\{\begin{array}{l}\frac{\text{d}x}{\text{d}t}=\epsilon f\left(x,y,\epsilon \right),\\ \frac{\text{d}y}{\text{d}t}=g\left(x,y,\epsilon \right){y}^{2}.\end{array}$ (4)

(5)

$\epsilon =\text{0}$ 时，解从点 $\left({x}_{0},{y}_{0}\right)$ 开始，其中， ${x}_{0}<0,{y}_{0}>0$，解快速地被吸引到x轴，然后沿着x轴向右漂移，最后被x轴排斥。它在x轴上的一点 ${p}_{\epsilon }\left({x}_{0}\right)$ 重新相交于直线 $y={y}_{0}$。当 $\epsilon \to \text{0}$ 时， ${p}_{\epsilon }\left({x}_{0}\right)\to {p}_{0}\left({x}_{0}\right)$。其中 ${p}_{0}\left({x}_{0}\right)$ 由下面的公式决定。

${\int }_{{x}_{0}}^{{p}_{0}\left({x}_{0}\right)}\frac{g\left(x,0,0\right)}{f\left(x,0,0\right)}\text{d}x=0.$ (6)

2.3. 极限系统的动力学

$\left\{\begin{array}{l}\epsilon \frac{\text{d}x}{\text{d}{t}_{1}}=x\left(1-x\right)-\frac{xy}{a{x}^{2}+bx+1},\\ \frac{\text{d}y}{\text{d}{t}_{1}}=y\left(1-\frac{\beta y}{x}\right).\end{array}$ (7)

，当 $b>-2\sqrt{a}$ 时，有 $x\left(a{x}^{2}+bx+1\right)>0$。从而得到：

$\left\{\begin{array}{l}\epsilon \frac{\text{d}x}{\text{d}\tau }={x}^{2}\left(\left(1-x\right)\left(a{x}^{2}+bx+1\right)-y\right),\\ \frac{\text{d}y}{\text{d}\tau }=y\left(a{x}^{2}+bx+1\right)\left(x-\beta y\right).\end{array}$ (8)

$\left\{\begin{array}{l}\frac{\text{d}x}{\text{d}t}={x}^{2}\left(\left(1-x\right)\left(a{x}^{2}+bx+1\right)-y\right),\\ \frac{\text{d}y}{\text{d}t}=\epsilon y\left(a{x}^{2}+bx+1\right)\left(x-\beta y\right).\end{array}$ (9)

$\left\{\begin{array}{l}0={x}^{2}\left(\left(1-x\right)\left(a{x}^{2}+bx+1\right)-y\right),\\ \frac{\text{d}y}{\text{d}\tau }=y\left(a{x}^{2}+bx+1\right)\left(x-\beta y\right).\end{array}$ (10)

$\left\{\begin{array}{l}\frac{\text{d}x}{\text{d}t}={x}^{2}\left(\left(1-x\right)\left(a{x}^{2}+bx+1\right)-y\right),\\ \frac{\text{d}y}{\text{d}t}=\text{0}.\end{array}$ (11)

${C}_{0}=\left\{\left(x,y\right)|y=f\left(x\right)=\left(1-x\right)\left(a{x}^{2}+bx+1\right)\right\}.$ (12)

3. 弛豫振荡

${h}_{1}\left(x,y\right)=\left(1-x\right)\left(a{x}^{2}+bx+1\right)-y$${h}_{2}\left(x,y\right)=y\left(a{x}^{2}+bx+1\right)\left(x-\beta y\right)$，则(9)变为：

$\left\{\begin{array}{l}\frac{\text{d}x}{\text{d}t}={x}^{2}{h}_{1}\left(x,y\right),\\ \frac{\text{d}y}{\text{d}t}=\epsilon {h}_{2}\left(x,y\right).\end{array}$ (13)

(b)当 ${y}_{0}\to 0$ 时，有 ${\int }_{{y}_{0}}^{{y}_{\mathrm{max}}}\frac{{h}_{1}\left(0,y\right)}{{h}_{2}\left(0,y\right)}\text{d}y\to -\infty$。其中， $\left({x}_{\mathrm{max}},{y}_{\mathrm{max}}\right)$ 为一般的折点， ${x}_{\mathrm{max}}=\frac{2\left(a-1\right)}{3a}$${y}_{\mathrm{max}}=f\left({x}_{\mathrm{max}}\right)$

${\int }_{{y}_{0}^{*}}^{{y}_{\mathrm{max}}}\frac{{h}_{1}\left(0,y\right)}{{h}_{2}\left(0,y\right)}\text{d}y=0.$ (14)

$I\left({y}_{0}^{*}\right):={\int }_{{y}_{0}^{*}}^{{y}_{\mathrm{max}}}\frac{{h}_{1}\left(0,y\right)}{{h}_{2}\left(0,y\right)}\text{d}y={\int }_{{y}_{0}^{*}}^{{y}_{\mathrm{max}}}\frac{1-y}{-\beta {y}^{2}}\text{d}y\to -\infty ,{y}_{0}^{*}\to {0}^{+},$ (15)

${I}^{\prime }\left({y}_{0}^{*}\right)=-\frac{1-y}{-\beta {y}^{2}}=\frac{1-y}{\beta {y}^{2}}>0.$ (16)

$I\left({y}_{0}^{*}\right)={\int }_{{y}_{0}^{*}}^{1}\frac{1-y}{-\beta {y}^{2}}\text{d}y+{\int }_{1}^{{y}_{\mathrm{max}}}\frac{1-y}{-\beta {y}^{2}}\text{d}y,$ (17)

$I\left(1\right)={\int }_{1}^{{y}_{\mathrm{max}}}\frac{1-y}{-\beta {y}^{2}}\text{d}y>0.$ (18)

${\int }_{{y}_{0}^{*}}^{{y}_{\mathrm{max}}}\frac{{h}_{1}\left(0,y\right)}{{h}_{2}\left(0,y\right)}\text{d}y=0.$

Relaxation Oscillations Analysis in a Predator-Prey System of Leslie Type[J]. 应用数学进展, 2019, 08(12): 1937-1942. https://doi.org/10.12677/AAM.2019.812222

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