# 一类食饵具有常数收获率和Holling第II类功能性反应的捕食者–食饵模型的分支分析Bifurcation Analysis of a Predator-Prey Model with Constant Prey Harvesting Rate and Holling Type II Functional Response

DOI: 10.12677/DSC.2019.84029, PDF, HTML, XML, 下载: 328  浏览: 979

Abstract: In this paper, we study the equilibria and the bifurcation of a predator-prey model with constant prey harvesting rate and Holling type II functional response. Firstly, we analyze the existence condition of the equilibria of the system model. Then, we discuss the type and stability of the equilibria, Hopf bifurcation near positive equilibria and we obtain the conditions of Hopf bifurcation. Finally, the numerical simulation of the model is carried out and the relevant conclusions are obtained.

1. 引言

$\left\{\begin{array}{l}\frac{\text{d}x}{\text{d}t}=rx\left(1-\frac{x}{k}\right)-\frac{emxy}{a+x}-h,\\ \frac{\text{d}y}{\text{d}t}=y\left(\frac{mx}{a+x}-d\right).\end{array}$ (1)

$u=x,v=\frac{em}{r}y,\text{d}\tau =r\text{d}t,$

$\left\{\begin{array}{l}\frac{\text{d}u}{\text{d}\tau }=u\left(1-\frac{u}{k}\right)-\frac{uv}{a+u}-{h}_{0},\\ \frac{\text{d}v}{\text{d}\tau }=v\left(\frac{bu}{a+u}-c\right).\end{array}$ (2)

2. 平衡点类型及正平衡点的稳定性

1) 若 $v=0$，则有

${u}^{2}-ku+k{h}_{0}=0.$ (3)

$\Delta ={k}^{2}-4k{h}_{0}=k\left(k-4{h}_{0}\right)$

a) 当 $k=4{h}_{0}$ 时，则系统(2)有唯一边界平衡点 ${E}_{0}=\left(\frac{k}{2},0\right)$

b) 当 $k>4{h}_{0}$ 时，系统(2)有两个边界平衡点 ${E}_{1}=\left(\frac{k+\sqrt{\Delta }}{2},0\right)$${E}_{2}=\left(\frac{k-\sqrt{\Delta }}{2},0\right)$

2) 若 $v\ne 0$，则且当 $b>c$${h}_{0}<-\frac{u\left[ac-\left(b-c\right)k\right]}{k\left(b-c\right)}$ 时，系统(2)存在唯一的正平衡点 ${E}_{3}\left({u}^{*},{v}^{*}\right)$。其中， ${u}^{*}=\frac{ac}{b-c}$

$J=\left(\begin{array}{cc}1-\frac{2u}{k}-\frac{av}{{\left(a+u\right)}^{2}}& -\frac{u}{a+u}\\ \frac{abv}{{\left(a+u\right)}^{2}}& \frac{bu}{a+u}-c\end{array}\right).$ (4)

1) 判断边界平衡点 ${E}_{0}$ 的类型

${E}_{0}=\left(\frac{k}{2},0\right)$ 代入Jacobi矩阵(4)中，可以得到系统(2)在 ${E}_{0}$ 处的Jacobi矩阵为

${J}_{{E}_{0}}=\left(\begin{array}{cc}0& -\frac{k}{k+2a}\\ 0& \frac{bk}{k+2a}-c\end{array}\right).$

2) 判断边界平衡点 ${E}_{1}$${E}_{2}$ 的类型

${E}_{1}=\left(\frac{k+\sqrt{\Delta }}{2},0\right)$代入Jacobi矩阵(4)中，可得到系统(2)在 ${E}_{1}$${E}_{2}$ 处的Jacobi矩阵为

${J}_{{E}_{1}}=\left(\begin{array}{cc}-\frac{\sqrt{\Delta }}{k}& -\frac{k+\sqrt{\Delta }}{2a+k+\sqrt{\Delta }}\\ 0& \frac{b\left(k+\sqrt{\Delta }\right)}{2a+k+\sqrt{\Delta }}-c\end{array}\right),\text{\hspace{0.17em}}{J}_{{E}_{2}}=\left(\begin{array}{cc}\frac{\sqrt{\Delta }}{k}& -\frac{k-\sqrt{\Delta }}{2a+k-\sqrt{\Delta }}\\ 0& \frac{b\left(k-\sqrt{\Delta }\right)}{2a+k-\sqrt{\Delta }}-c\end{array}\right)$

a) 若 $c>\frac{b\left(k+\sqrt{\Delta }\right)}{2a+k+\sqrt{\Delta }}$，那么 ${E}_{1}$ 为稳定结点；若 $c<\frac{b\left(k+\sqrt{\Delta }\right)}{2a+k+\sqrt{\Delta }}$，则 ${E}_{1}$ 为鞍点。

b) 若 $c<\frac{b\left(k-\sqrt{\Delta }\right)}{2a+k-\sqrt{\Delta }}$，那么 ${E}_{2}$ 为不稳定结点；若 $c>\frac{b\left(k-\sqrt{\Delta }\right)}{2a+k-\sqrt{\Delta }}$，则 ${E}_{2}$ 为鞍点。

3) 正平衡点 ${E}_{3}$ 的稳定性

${J}_{{E}_{3}}=\left(\begin{array}{cc}{f}_{1}\left({v}^{*}\right)& -\frac{c}{b}\\ \frac{{v}^{*}{\left(b-c\right)}^{2}}{ab}& 0\end{array}\right).$ (5)

$D=\mathrm{det}{J}_{{E}_{3}}=\frac{c{v}^{*}{\left(b-c\right)}^{2}}{a{b}^{2}},T=trace{J}_{{E}_{3}}={f}_{1}\left({v}^{*}\right)=1-\frac{2ac}{k\left(b-c\right)}-\frac{{v}^{*}{\left(b-c\right)}^{2}}{a{b}^{2}}$

$D>0$$T<0$ ；我们可得到系统(2)在正平衡点 ${E}_{3}\left({u}^{*},{v}^{*}\right)$ 处是局部渐近稳定的；

$D>0$$T>0$ ；我们可得到系统(2)在正平衡点 ${E}_{3}\left({u}^{*},{v}^{*}\right)$ 处是不稳定的。

${f}_{1}\left({v}^{*}\right)=0$，既有 $D>0$$T=0$ ；由参考文献 [12]，则Jacobi矩阵 ${J}_{{E}_{3}}$ 的特征方程有一对纯虚的特征根 $±i{\omega }_{0}$ (其中i为虚数单位， ${\omega }_{0}=\frac{b-c}{b}\sqrt{\frac{{v}^{*}c}{a}}>0$ )。

${E}_{3}\left({u}^{*},{v}^{*}\right)$ 代入 ${f}_{1}\left({v}^{*}\right)=0$ 中，解得 ${h}_{0}^{*}=\frac{a{c}^{2}\left[a\left(b+c\right)-k\left(b-c\right)\right]}{{\left(b-c\right)}^{3}k}$

a) 若 ${h}_{0}>{h}_{0}^{*}$，则系统(2)的正平衡点是局部渐进稳定的；

b) 若 $0<{h}_{0}<{h}_{0}^{*}$，则系统(2)的正平衡点是不稳定的；

c) 若 ${h}_{0}={h}_{0}^{*}$，则系统(2)在正平衡点 ${E}_{3}\left({u}^{*},{v}^{*}\right)$ 处出现Hopf分支。

3. Hopf分支

${\lambda }^{2}-\lambda {f}_{1}\left({v}^{*}\right)+\frac{{v}^{*}c{\left(b-c\right)}^{2}}{a{b}^{2}}=0.$ (6)

$k{\left(b-c\right)}^{2}\left(cv+{h}_{0}b\right)+abc\left[ac-k\left(b-c\right)\right]=0.$

$f\left(v\right)=k{\left(b-c\right)}^{2}\left(cv+{h}_{0}b\right)+abc\left[ac-k\left(b-c\right)\right]$，则有 $f\left({v}^{*}\right)=0$${f}^{\prime }\left({v}^{*}\right)>0$

$f\left({v}^{*}\right)=0$ 解得 ${v}^{*}\left({h}_{0}\right)=-\frac{\left\{ac\left[ac-k\left(b-c\right)\right]+{\left(b-c\right)}^{2}{h}_{0}k\right\}b}{{\left(b-c\right)}^{2}kc}$，且有 ${v}^{*}{}^{\prime }\left({h}_{0}\right)=\frac{b}{c}>0$

$p\left({h}_{0}\right)=\frac{{f}_{1}\left({v}^{*}\right)}{2}$。于是有

${p}^{\prime }\left({h}_{0}^{*}\right)<0$ 可知系统(2)满足产生Hopf分支的条件 [13]。因此系统(2)在 ${h}_{0}={h}_{0}^{*}$ 附近可产生Hopf分支。接下来我们来讨论产生的Hopf分支的方向。

${w}_{1}=u-{u}^{*}$${w}_{2}=v-{v}^{*}$。则系统(2)变成以下形式

$\left\{\begin{array}{l}\frac{\text{d}{w}_{1}}{\text{d}\tau }=-\frac{c}{b}{w}_{2}-\frac{{a}^{2}{b}^{3}-k{v}^{*}{\left(b-c\right)}^{3}}{{a}^{2}{b}^{3}k}{w}_{1}^{2}-\frac{{\left(b-c\right)}^{2}}{a{b}^{2}}{w}_{1}{w}_{2}-\frac{{v}^{*}{\left(b-c\right)}^{4}}{{a}^{3}{b}^{4}}{w}_{1}^{3}+\frac{{\left(b-c\right)}^{3}}{{a}^{2}{b}^{3}}{w}_{1}^{2}{w}_{2}+\cdots ,\\ \frac{\text{d}{w}_{2}}{\text{d}\tau }=\frac{{v}^{*}{\left(b-c\right)}^{2}}{ab}{w}_{1}-\frac{{v}^{*}{\left(b-c\right)}^{3}}{{a}^{2}{b}^{2}}{w}_{1}^{2}+\frac{{\left(b-c\right)}^{2}}{ab}{w}_{1}{w}_{2}+\frac{{v}^{*}{\left(b-c\right)}^{4}}{{a}^{3}{b}^{3}}{w}_{1}^{3}-\frac{{\left(b-c\right)}^{3}}{{a}^{2}{b}^{2}}{w}_{1}^{2}{w}_{2}+\cdots .\end{array}$ (7)

${\omega }_{0}=\frac{b-c}{b}\sqrt{\frac{{v}^{*}c}{a}}$，再作变换：

$\xi =-\frac{{v}^{*}{\left(b-c\right)}^{2}}{ab}{w}_{1},\eta =-{\omega }_{0}{w}_{2},$

$\left\{\begin{array}{l}\frac{\text{d}\xi }{\text{d}\tau }=-{\omega }_{0}\eta +\frac{{a}^{2}{b}^{3}{\xi }^{2}-k{v}^{*}{\left(b-c\right)}^{3}{\xi }^{2}}{a{b}^{2}{v}^{*}k{\left(b-c\right)}^{2}}+\frac{{\left(b-c\right)}^{2}\xi \eta }{a{b}^{2}{\omega }_{0}}+\frac{{\xi }^{3}}{a{b}^{2}{v}^{*}}+\frac{\left(b-c\right){\xi }^{2}\eta }{a{b}^{2}{v}^{*}{\omega }_{0}}+\cdots ,\\ \frac{\text{d}\eta }{\text{d}\tau }={\omega }_{0}\xi +\frac{{\omega }_{0}{\xi }^{2}}{{v}^{*}\left(b-c\right)}-\frac{\xi \eta }{{v}^{*}}+\frac{{\omega }_{0}{\xi }^{3}}{{v}^{*}{}^{2}{\left(b-c\right)}^{2}}-\frac{{\xi }^{2}\eta }{{v}^{*}{}^{2}\left(b-c\right)}+\cdots .\end{array}$ (8)

$\frac{\text{d}z}{\text{d}\tau }=i{\omega }_{0}z+\underset{2\le k+l\le \text{3}}{\sum }{g}_{kl}{z}^{k}{\stackrel{¯}{z}}^{l},$

${g}_{20}=\frac{1}{4}\left[\frac{ab}{{v}^{*}k{\left(b-c\right)}^{2}}-\frac{b-c}{a{b}^{2}}-\frac{1}{{v}^{*}}\right]+\frac{1}{4}i\left[\frac{{\omega }_{0}}{{v}^{*}\left(b-c\right)}-\frac{{\left(b-c\right)}^{2}}{a{b}^{2}{\omega }_{0}}\right],$

${g}_{11}=\frac{1}{2}\left[\frac{ab}{{v}^{*}k{\left(b-c\right)}^{2}}-\frac{b-c}{a{b}^{2}}\right]+\frac{1}{2}i\frac{{\omega }_{0}}{{v}^{*}\left(b-c\right)},$

${g}_{21}=\frac{1}{8{v}^{*}}\left[\frac{3}{a{b}^{2}{v}^{*}}-\frac{1}{{v}^{*}\left(b-c\right)}\right]+\frac{1}{8{v}^{*}}i\left[\frac{3{w}_{0}}{{v}^{*}{\left(b-c\right)}^{2}}-\frac{b-c}{a{b}^{2}{\omega }_{0}}\right].$

${l}_{1}\left(0\right)=\frac{1}{2{\omega }_{0}^{2}}\mathrm{Re}\left(i{g}_{20}{g}_{11}+{\omega }_{0}{g}_{21}\right),$

${l}_{1}\left(0\right)=\frac{1}{8{\omega }_{0}}\left\{\frac{1}{2a{b}^{2}{\omega }_{0}^{2}}\left[\frac{ab{v}^{*}+5}{{v}^{*}{}^{2}k}-\frac{{\left(b-c\right)}^{3}}{a{b}^{2}}\right]-\frac{ab}{{v}^{*}{}^{2}{\left(b-c\right)}^{3}k}\right\}.$

a) 当 ${l}_{1}\left(0\right)<0$ 时，模型(1)在  附近可以产生超临界的Hopf分支；

b) 当 ${l}_{1}\left(0\right)>0$ 时，模型(1)在 ${E}_{3}$ 附近可以产生次临界的Hopf分支。

4. 仿真算例

(a) (b)

Figure 1. (a) When $a=1.8,b=3.2,c=1.2,k=3.6,{h}_{0}=0.066$, there is a stable limit cycle near ${E}_{3}$ ; (b) When $a=1.9,b=2.3,c=0.5,k=1.5,{h}_{0}=0.14$, there is a unstable limit cycle near ${E}_{3}$

5. 结论

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