捕食-被捕食系统的动力性质已经被许多文献[1-7]进行了深入研究,尤其是具有时滞的捕食-被捕食系统的持久性、正平衡点的稳定性和Hopf分支方面.这是因为时滞对捕食-被捕食系统的动力性质会带来很大且复杂的影响,例如,在时滞微分方程中,时滞通过Hopf分支可以引起稳定性的失去,也可以引起振荡和产生周期解.

Beddington^[8]和DeAngelis^[9]最先提出了具有Beddington-DeAngelis功能反应函数的捕食-被捕食系统,具体模型如下

(1.1) $\begin{equation}\left\{\begin{array}{ll} x'(t)=x(t)\left(a-bx(t)-\frac{cy(t)}{m_{1}+m_{2}x(t)+m_{3}y(t)}\right), \\[4mm] y'(t)=y(t)\left(-d+\frac{fx(t)}{m_{1}+m_{2}x(t)+m_{3}y(t)}\right), \end{array}\right.\label{1.02}\end{equation}$

其中$x(t)$表示$t$时间被捕食者的密度, $y(t)$表示$t$时间捕食者的密度, $a$为被捕食者的内禀增长率, $b=a/k, k$为周围环境对被捕食种群最大限度的容纳量, $c$捕食者对被捕食者的功能性反应常数, $d, f$分别为捕食种群的死亡率和转化率.在系统(1.1)中,捕食者以Beddington-DeAngelis功能反应函数$cxy/(m_{1}+m_{2}x(t)+m_{3}y(t))$消耗着被捕食者且生长率为$fxy/(m_{1}+m_{2}x(t)+m_{3}y(t))$.

但是某些环境会限制被捕食者必须具有密度制约,同时也有相当多的证据表明有些捕食者种群由于环境因素可能是密度制约的^[10-11].还有,文献[12]表明捕食者的密度制约性不仅对高密度的捕食还是对低密度的捕食都是非常重要的.因此,在讨论捕食-被捕食系统时仅仅考虑被捕食者的密度水平是不够的,我们需要把捕食者的实际密度水平考虑进去.另一方面,过去的工作用模型模拟不同生命阶段的单种群生长^[13-14],从而生态学家和数学家经常在模型中引入时滞变量.因此,在文献[15]中,用微分方程理论和Lyapunov函数研究了密度制约且具有Beddington-DeAngelis功能反应函数的时滞捕食-被捕食系统

(1.2) $\begin{equation}\left\{\begin{array}{ll} x'(t)=x(t)\left(a-bx(t)-\frac{cy(t)}{m_{1}+m_{2}x(t)+m_{3}y(t)}\right), \\[4mm] y'(t)=y(t)\left(-d-ry(t)+\frac{fx(t-\tau)}{m_{1}+m_{2}x(t-\tau)+m_{3}y(t-\tau)}\right)\end{array}\right.\label{1.1}\end{equation}$

的持久性和正平衡点的局部渐近稳定性和全局渐近稳定性等动力性质.在系统(1.2)中$r$表示捕食者的密度制约率, $\tau$表示捕食者从出生到成熟的时间,且所有参数都是非负的.

在本文中,我们以时滞$\tau$为分支参数去讨论$\tau$对系统(1.2)周期轨的稳定性、方向和周期的影响.在第二部分中我们给出系统的稳定性变换,这为第三部分研究系统在临界点处的分支周期解做好准备,第四部分的数值模拟验证了第二部分和第三部分的结论.

由系统(1.2)可知,除$E_{0}(0, 0)$和$E_{1}\left(\frac{a-E_{1}}{b}, 0\right)$为系统(1.2)的非负平衡点外,满足以下代数方程组

(2.1) $\begin{equation}\left\{\begin{array}{ll} a-bx^{*}-\frac{cy^{*}}{m_{1}+m_{2}x^{*}+m_{3}y^{*}}-E_{1}=0, \\[4mm] -d-py^{*}+\frac{f{\rm e}^{-d_{i}\tau}x^{*}}{m_{1}+m_{2}x^{*}+m_{3}y^{*}}-E_{2}=0\end{array}\right.\label{2.112}\end{equation}$

的$E^{*}(x^{*}, y^{*})$也为系统(1.2)的非负平衡点.根据文献[15],分析方程组(2.1)中方程对应的曲线,我们得到如下结论：如果条件

(2.2) $\begin{equation}(a-E_{1})\left[f{\rm e}^{-d_{i}\tau}-(d+E_{2})m_{2}\right]>bm_{1}(d+E_{2})\label{2.212}\end{equation}$

成立,那么平衡点$E^{*}(x^{*}, y^{*})$是系统(1.2)一个正平衡点.

在这一部分中我们主要给出系统正平衡点$E^{*}(x^{*}, y^{*})$的稳定性和系统的稳定性变换,有些结论在文献[15]的基础之上给出,没有加以证明.首先给出线性化后系统的特征方程,再由特征根的符号判断正平衡点的稳定性以及稳定性变换.

令$x(t)=x^{*}+X(t), ~y(t)=y^{*}+Y(t)$,则系统

(2.3) $\begin{equation}\left\{\begin{array}{ll} x'(t)=x(t)\left(a-bx(t)-\frac{cy(t)}{m_{1}+m_{2}x(t)+m_{3}y(t)}\right)=F\left(x, y, x(t-\tau), y(t-\tau)\right), \\[4mm] y'(t)=y(t)\left(-d-ry(t)+\frac{fx(t-\tau)}{m_{1}+m_{2}x(t-\tau)+m_{3}y(t-\tau)}\right)=G\left(x, y, x(t-\tau), y(t-\tau)\right)\end{array}\right.\label{1.15}\end{equation}$

线性化后是

(2.4) $\begin{equation}\left\{\begin{array}{ll}X'=x^{*}F_{x}X(t)+x^{*}F_{y}Y(t), \\Y'=y^{*}G_{x}X(t-\tau)+y^{*}G_{y_{1}}Y(t-\tau)+y^{*}G_{y_{2}}Y(t), \end{array}\right.\label{2.1}\end{equation}$

其中$F_{x}=\frac{cm_{2}y^{*}}{(m_{1}+m_{2}x^{*}+m_{3}y^{*})^{2}}-b, $$F_{y}=-\frac{c(m_{1}+m_{2}x^{*})}{(m_{1}+m_{2}x^{*}+m_{3}y^{*})^{2}} < 0, $$G_{x}=\frac{f(m_{1}+m_{3}y^{*})}{(m_{1}+m_{2}x^{*}+m_{3}y^{*})^{2}}>0, $$G_{y_{1}}=-\frac{fm_{3}x^{*}}{(m_{1}+m_{2}x^{*}+m_{3}y^{*})^{2}} < 0, $$G_{y_{2}}=-r < 0.$如果下列条件

$\left\{\begin{array}{ll} f>dm_{2} ~\mbox{和} ~(f-dm_{2})\left(\frac{a}{b}-\frac{c}{bm_{3}}-\frac{dm_{3}}{rm_{2}}\right)>dm_{1}\\\mbox{或} \\ am_{3}>c+\frac{bdm_{3}^{2}}{rm_{2}}~\mbox{和}~(f-dm_{2})\left(\frac{a}{b}-\frac{c}{bm_{3}}-\frac{dm_{3}}{rm_{2}}\right)>dm_{1}\end{array}\right.{\rm (H}_{1})$

成立,则$a_{1}=(-F_{x})x^{*}+(-G_{y_{2}})y^{*}>0, ~a_{2}=(-G_{y_{1}})y^{*}>0, ~a_{3}=(-F_{x})(-G_{y_{2}})x^{*}y^{*}>0, $$a_{4}=\left[(-F_{x})(-G_{y_{1}})-(-F_{y})(-G_{x})\right]x^{*}y^{*}>0$.那么系统(2.4)的特征方程为

(2.5) $\begin{equation}\lambda^{2}+a_{1}\lambda+a_{2}\lambda{\rm e}^{-\lambda\tau}+a_{3}+a_{4}{\rm e}^{-\lambda\tau}=0. \label{2.2}\end{equation}$

由文献[15]中的引理4.1,引理4.2,定理4.1和注释4.1,我们有下面结论.

引理2.1   如果条件$(H_{1})$和

$a_{2}^{2}-a_{1}^{2}+2a_{3}<0 ~~ \mbox{和} ~~a_{3}^{2}-a_{4}^{2}>0~~\mbox{或}~~(a_{2}^{2}-a_{1}^{2}+2a_{3})^{2}<4(a_{3}^{2}-a_{4}^{2})~~~{\rm (H_{2})}$

成立,则当$\tau>0$时,方程(2.5)所有的特征根都有负实部.

引理2.2    (ⅰ)~若(H$_{1})$和

$a_{3}^{2}-a_{4}^{2}<0 ~~\mbox{或}~~a_{2}^{2}-a_{1}^{2}+2a_{3}>0 ~~\mbox{和}~~(a_{2}^{2}-a_{1}^{2}+2a_{3})^{2}=4(a_{3}^{2}-a_{4}^{2})~~~{\rm (H_{3})}$

成立且$\tau=\tau_{j}^{+}$,则特征方程(2.5)有一对纯虚根$\pm {\rm i}\omega_{+}$.

(ⅱ)~若(H$_{1})$和

$a_{3}^{2}-a_{4}^{2}>0, ~~a_{2}^{2}-a_{1}^{2}+2a_{3}>0 ~~ \mbox{和}~~(a_{2}^{2}-a_{1}^{2}+2a_{3})^{2}>4(a_{3}^{2}-a_{4}^{2})~~~{\rm (H_{4})}$

成立且$\tau=\tau_{j}^{+}$($\tau=\tau_{j}^{-}$),则特征方程(2.5)有一对纯虚根$\pm {\rm i}\omega_{+}$ ($\pm {\rm i}\omega_{-}$).

引理2.2中的(ⅰ)、(ⅱ)两种情形表明当$\tau$取某值时,方程(2.5)的特征根存在纯虚根,其中$\omega_{\pm}^{2}$和$\tau_{j}^{\pm}$分别为

$\omega_{\pm}^{2}=\frac{1}{2}\left(a_{2}^{2}-a_{1}^{2}+2a_{3}\right)\pm\frac{1}{2}\left[(a_{2}^{2}-a_{1}^{2}+2a_{3})^{2}-4(a_{3}^{2}-a_{4}^{2})\right]^{\frac{1}{2}}$

和

(2.6) $\begin{equation}\tau_{j}^{\pm}=\frac{1}{\omega_{\pm}}\arccos\left\{\frac{a_{4}(\omega_{\pm}^{2}-a_{3})-a_{1}a_{2}\omega_{\pm}^{2}}{a_{2}^{2}\omega_{\pm}^{2}+a_{4}^{2}}\right\}+\frac{2j\pi}{\omega_{\pm}}, ~~j=0, 1, 2, \cdot\cdot\cdot.\label{2.3}\end{equation}$

定理2.1在系统(1.2)中, $\tau_{j}^{\pm} (j=0, 1, 2, \cdots)$由(2.6)式所定义,

(ⅰ) 若(H$_{1})$和(H$_{2})$成立,则特征方程(2.5)的根对任意的$\tau\geq0$都有负实部；

(ⅱ) 若(H$_{1})$和(H$_{3})$成立,则当$\tau\in[0, \tau_{0}^{+})$时特征方程(2.5)有负实部,当$\tau=\tau_{0}^{+}$,特征方程(2.5)有一对纯虚根$\pm {\rm i}\omega_{+}$,且当$\tau> \tau_{0}^{+}$,特征方程(2.5)至少有一个具有正实部的特征根;

(ⅲ) 若(H$_{1})$和(H$_{4})$成立,则存在一个正整数$k$使得有$k$次从稳定到不稳定再到稳定的稳定性转化.也就是说,当$\tau\in[0, \tau_{0}^{+}], (\tau_{0}^{-}, \tau_{1}^{+}), \cdots, (\tau_{k-1}^{-}, \tau_{k}^{+}), $特征方程(2.5)的所有根都有负实部,当$\tau\in[\tau_{0}^{+}, \tau_{0}^{-}), [\tau_{1}^{+}, \tau_{1}^{-}), \cdots, [\tau_{k-1}^{+}, \tau_{k-1}^{-})$且$\tau>\tau_{k}^{+}, $特征方程(2.5)至少有一个具有正实部的特征根.

记

$\lambda_{j}=\alpha_{j}(\tau)+{\rm i}\omega_{j}(\tau), ~~~j=0, 1, 2, \cdots $

为方程(2.5)的根满足

$\alpha_{j}(\tau_{j}^{\pm})=0, ~~~~~~~~~\omega_{j}(\tau_{j}^{\pm})=\omega_{\pm}.$

则下面的横截条件成立

$\frac{\rm d}{{\rm d}\tau}{\rm Re}\lambda_{j}(\tau_{j}^{+})>0, ~~~~~~~~~\frac{\rm d}{{\rm d}\tau}{\rm Re}\lambda_{j}(\tau_{j}^{-})<0.$

需要注意的是,由sign$\left\{\frac{{\rm d}{\rm Re}\lambda(\tau)}{{\rm d}\tau}\right\}={\rm sign}\left\{\left(\frac{{\rm d}{\rm Re}\lambda(\tau)}{{\rm d}\tau}\right)^{-1}\right\}$知,为了判断$\frac{{\rm d} {\rm Re}\lambda(\tau)}{{\rm d}\tau}$的符号,只需判断${\rm Re}\left(\frac{{\rm d}\lambda(\tau)}{{\rm d}\tau}\right)^{-1}$的符号即可.将$\lambda_{j}$带入特征方程(2.5)且对$\tau$求导,整理后可得

(2.7) $\begin{equation}{\rm Re}\left(\frac{{\rm d}\lambda}{{\rm d}\tau}\right)^{-1}=\frac{a_{2}^{2}\omega^{4}+2a_{4}^{2}\omega^{2}+a_{4}^{2}(a_{1}^{2}-2a_{3})-a_{2}^{2}a_{3}^{2}}{(a_{2}^{2}\omega^{2}+a_{4}^{2})[a_{1}^{2}\omega^{2}+(a_{3}-\omega^{2})^{2}]}.\label{2.300}\end{equation}$

由定理2.1我们可以直接得出下面的定理.

定理2.2   在系统(1.2)中,时滞$\tau$经过临界值$\tau_{j}^{+}, ~j=0, 1, 2, \cdots, k-1$时,系统(1.2)的内部平衡点$E^{*}(x^{*}, y^{*})$将失去它的稳定性从而出现Hopf分支.

由第二部分的分析和定理2.2可知, $(x^{*}, y^{*})$在$[0, \tau_{0}^{+}]$上是稳定的,在$[\tau_{0}^{+}, \tau_{0}^{-})$上是不稳定的.因此,系统(1.2)在$(x^{*}, y^{*})$处经历了稳定性变化且在$\tau_{0}^{+}$临界点处出现了周期轨道,同样我们对其他临界点$\tau_{0}^{- }, \tau_{1}^{+ }, \tau_{1}^{- }\cdots$处的分支周期解也感兴趣.因此,在这一部分我们将分四大步骤考虑Hopf分支的特性.

首先,通过引入无穷小生成元把时滞系统(1.2)转化为无穷维系统.需要下面两小步.

(ⅰ) 将系统(1.2)无量纲化.令$x(t)-x^{*}\rightarrow x(t)$, $y(t)-y^{*}\rightarrow y(t)$且$t\rightarrow \frac{t}{\tau}$,则得到如下形式

(3.1) $\begin{equation}\left( \begin{array}{ccc} x'(t)\\ y'(t)\\ \end{array}\right)=\tau B\left( \begin{array}{ccc} x(t)\\ y(t)\\ \end{array}\right)+ \tau C\left( \begin{array}{ccc} x(t-1)\\ y(t-1)\\ \end{array}\right)+\tau F\left(x, y, x(t-\tau), y(t-\tau)\right), \label{1.2}\end{equation}$

其中

$\begin{array}{l}\;\;\;\;\;B = {\left. {\left( {\begin{array}{*{20}{l}}{\frac{{\partial P}}{{\partial x}}\frac{{\partial P}}{{\partial y}}}\\{\frac{{\partial Q}}{{\partial x}}\frac{{\partial Q}}{{\partial y}}}\end{array}} \right)} \right|_{\left( {{x^*},{y^*}} \right)}}\\\;\;\;\;\;\; = \left( {\begin{array}{*{20}{c}}{a - 2b{x^*} - \frac{{c{y^*}\left( {{m_1} + {m_3}{y^*}} \right)}}{{{{\left( {{m_1} + {m_2}{x^*} + {m_3}{y^*}} \right)}^2}}}}&{\frac{{ - c{x^*}\left( {{m_1} + {m_2}{x^*}} \right)}}{{{{\left( {{m_1} + {m_2}{x^*} + {m_3}{y^*}} \right)}^2}}}}\\0&{ - d - 2r{y^*} + \frac{{f{x^*}}}{{{m_1} + {m_2}{x^*} + {m_3}{y^*}}}}\end{array}} \right)\\C = {\left. {\left( {\begin{array}{*{20}{c}}{\frac{{\partial P}}{{\partial x(t - \tau )}}\frac{{\partial P}}{{\partial y(t - \tau )}}}\\{\frac{{\partial Q}}{{\partial x(t - \tau )}}\frac{{\partial Q}}{{\partial y(t - \tau )}}}\end{array}} \right)} \right|_{\left( {{x^*},{y^*}} \right)}} = \left( {\frac{{f{y^*}\left( {{m_1} + {m_3}{y^*}} \right)}}{{{{\left( {{m_1} + {m_2}{x^*} + {m_3}{y^*}} \right)}^2}}}\frac{{ - f{m_3}{x^*}{y^*}}}{{{{\left( {{m_1} + {m_2}{x^*} + {m_3}{y^*}} \right)}^2}}}} \right)\\\begin{array}{*{20}{c}}{F(x,y,x(t - \tau ),y(t - \tau )) = \left( {\begin{array}{*{20}{c}}{f(x,y,x(t - \tau ),y(t - \tau ))}\\{l(x,y,x(t - \tau ),y(t - \tau ))}\end{array}} \right)}\\{ = \left( {\begin{array}{*{20}{c}}{\frac{1}{2}{f_{20}}{x^2} + {f_{11}}xy + \frac{1}{2}{f_{02}}{y^2} + \frac{1}{6}{f_{30}}{x^3} + \cdots }\\{\frac{1}{2}{l_{02}}{y^2} + \frac{1}{2}{\rm{ }}{{\tilde l}_{20}}{x^2}(t - 1) + {{\tilde l}_{11}}x(t - 1)y(t - 1) + \frac{1}{2}{\rm{ }}{{\tilde l}_{02}}{y^2}(t - 1) + \frac{1}{6}{\rm{ }}{{\tilde l}_{30}}{x^3}(t - 1) + \cdots }\end{array}} \right),}\end{array}\end{array}$

$f_{ij}=\frac{\partial^{i+j}f(x^{*}, y^{*})}{\partial^{i}x~\partial^{j}y}, ~~l_{ij}=\frac{\partial^{i+j}l(x^{*}, y^{*})}{\partial^{i}x~\partial^{j}y}, ~~\widetilde{l}_{ij}=\frac{\partial^{i+j}l(x^{*}, y^{*})}{\partial^{i}x(t-1)~\partial^{j}y(t-1)}, i, j=1, 2, \cdots$,且通过计算我们有$l_{20}=l_{11}=l_{30}=0$.

令$\tau=\tau_{0}+\mu$,因此$\mu=0$是系统(3.1)的Hopf分支值.对$\phi=(\phi_{1}, \phi_{2})^{T}\in C([-1, 0], \mathbb{R}^{2})$,定义

$L_{\mu}(\phi)=(\tau_{0}+\mu)[B(\tau_{0}+\mu)\phi(0)+C(\tau_{0}+\mu)\phi(-1)].$

由Riesz表现定理,存在一个$2\times 2$矩阵$\eta(\theta, \mu), \theta\in [-1, 0]$,且矩阵中的元素是有界变量函数使得

(3.2) $\begin{equation}L_{\mu}(\phi)=\int_{-1}^{0}[{\rm d}\eta(\theta, \mu)]\phi(\theta), ~\phi\in C([-1, 0], \mathbb{R}^{2}). \label{1.3}\end{equation}$

事实上,我们可以选择

$\eta(\theta, \mu)=(\tau_{0}+\mu)B(\tau_{0}+\mu)\delta(\theta)-C(\tau_{0}+\mu)\delta(\theta+1), $

也就是

$\eta(\theta, \mu)= \left\{\begin{array}{ll}(\tau_{0}+\mu)B(\tau_{0}+\mu), &\theta=0, \\0, ~~~~~~~~~~~~~~~~~~~~~~&\theta\in(-1, 0), \\-(\tau_{0}+\mu)N(\tau_{0}+\mu), ~~&\theta=-1.\end{array}\right.$

则满足方程(3.2).

(ⅱ) 引入无穷小生成元.为了把时滞系统(1.2)转化为抽象常微分形式,我们引入无穷小的生成元$A(\mu)$如下.当$\phi\in C([-1, 0], \mathbb{R}^{2})$,我们定义算子$A(\mu)$为

(3.3) $A(\mu) \phi(\theta)=\left\{\begin{array}{ll}{\frac{\mathrm{d} \phi(\theta)}{\mathrm{d} \theta},} & {\theta \in[-1,0)}, \\ {\int_{-1}^{0}[\mathrm{d} \eta(\xi, \mu)] \phi(\xi),} & {\theta=0},\end{array}\right.$

且

(3.4) $\begin{equation}R(\mu)\phi(\theta)= \left\{\begin{array}{ll}0, ~~~~~~&\theta\in[-1, 0), \\F(\mu, \phi), ~~&\theta=0, \end{array}\right.\label{1.16}\end{equation}$

其中$F(\mu, \phi)=$

$ (\tau_{0}+\mu)\left( \begin{array}{ccc} \frac{1}{2}f_{20}\phi_{1}^{2}(0)+f_{11}\phi_{1}(0)\phi_{2}(0)+\frac{1}{2}f_{02}\phi_{2}^{2}(0)+\frac{1}{6}f_{30}\phi_{1}^{3}(0)+\cdots \\[3mm] \frac{1}{2}l_{02}\phi_{2}^{2}(0)+\frac{1}{2}\widetilde{l}_{20}\phi_{1}^{2}(-1)+\widetilde{l}_{11}\phi_{1}(-1)\phi_{2}(-1)+\frac{1}{2}\widetilde{l}_{02}\phi_{2}^{2}(-1)+\frac{1}{6}\widetilde{l}_{30}\phi_{1}^{3}(-1)+\cdots \\ \end{array}\right), $

则系统(1.2)等价于下面的算子方程

(3.5) $\begin{equation}u'_{t}=A(\mu)u_{t}+R(\mu)u_{t}, \label{1.4}\end{equation}$

其中$u(t)=(x(t), y(t))^{T}$和当$\theta\in [-1, 0]$时$u_{t}=u(t+\theta)$.

其次,利用谱分解理论和无穷维系统的中心流行理论求出限制在中心流形上的流所满足的二维常微分方程.我们进行接下来的两小步.

(ⅰ) 引入$A$的伴随算子$A^{*}$.当$\psi\in C^{1}([0, 1], (\mathbb{R}^{2})^{*})$时,定义

$A^{*}\psi(s)= \left\{\begin{array}{ll} -\frac{{\rm d}\psi(s)}{{\rm d}s}, &s\in(0, 1], \\[3mm] \int_{-1}^{0}\psi(-\xi){\rm d}\eta(\xi, 0), ~~&s=0, \end{array}\right.$

和一个双线性型

$\langle\psi(s), \phi(\theta)\rangle=\overline{\psi}(0)\phi(0)-\int_{-1}^{0}\int_{\xi=0}^{\theta}\overline{\psi}(\xi-\theta){\rm d}\eta(\theta)\phi(\xi){\rm d}\xi, $

其中$\eta(\theta)=\eta(\theta, 0)$.则$A(0)$和$A^{*}$是伴随算子.

我们知道$\pm {\rm i}\omega_{0}\tau_{0}$是$A(0)$的特征值,因此也是$A^{*}$的特征值.再则,不难验证向量$q(\theta)=(q_{1}, q_{2})^{T}{\rm e}^{{\rm i}\omega_{0}\tau_{0}\theta}, \theta\in [-1, 0]$和$q^{*}(s)=D(q_{1}^{*}, q_{2}^{*}){\rm e}^{{\rm i}\omega_{0}\tau_{0}s}, s\in [0, 1]$分别为$A(0)$和$A^{*}$对应特征值${\rm i}\omega_{0}\tau_{0}$和$-{\rm i}\omega_{0}\tau_{0}$的特征向量,也就是满足下面等式

$\left({\rm i}\omega_{0}\tau_{0}I-A(0)\right)\left( \begin{array}{ccc} q_{1}\\ q_{2}\\ \end{array}\right){\rm e}^{{\rm i}\omega_{0}\tau_{0}\theta}=\left( \begin{array}{ccc} 0\\ 0\\ \end{array}\right), ~\left(-{\rm i}\omega_{0}\tau_{0}I-A^{*}\right)D\left( \begin{array}{ccc} q_{1}^{*}\\ q_{2}^{*}\\ \end{array}\right)^{T}{\rm e}^{{\rm i}\omega_{0}\tau_{0}s}=\left( \begin{array}{ccc} 0\\ 0\\ \end{array}\right).$

即

(3.6) $\begin{equation}\left\{\begin{array}{ll}\left({\rm i}\omega_{0}\tau_{0}I-\tau_{0}B-\tau_{0}{\rm e}^{-{\rm i}\omega_{0}\tau_{0}}C\right)\left( \begin{array}{ccc} q_{1}\\ q_{2}\\ \end{array}\right)=\left( \begin{array}{ccc} 0\\ 0\\ \end{array}\right), \\[6mm]\left(-{\rm i}\omega_{0}\tau_{0}I-\tau_{0}B^{T}-\tau_{0}{\rm e}^{-{\rm i}\omega_{0}\tau_{0}}C^{T}\right)\left( \begin{array}{ccc} q_{1}^{*}\\ q_{2}^{*}\\ \end{array}\right)=\left( \begin{array}{ccc} 0\\ 0\\ \end{array}\right).\end{array}\right.\label{1.160}\end{equation}$

令$q_{1}=1, ~q_{1}^{*}=1$,由(3.6)式可得

$q_{2}=\frac{(a-2bx^{*})(m_{1}+m_{2}x^{*}+m_{3}y^{*})^{2}-cy^{*}(m_{1}+m_{3}y^{*})}{cx^{*}(m_{1}+m_{2}x^{*})}-\frac{\omega_{0}(m_{1}+m_{2}x^{*}+m_{3}y^{*})^{2}}{cx^{*}(m_{1}+m_{2}x^{*})}{\rm i}, $

$ q^{*}_{2}=\frac{(-a+2bx^{*})(m_{1}+m_{2}x^{*}+m_{3}y^{*})^{2}}{fy^{*}(m_{1}+m_{3}y^{*})}+\frac{c}{f}-\frac{\omega_{0}(m_{1}+m_{2}x^{*}+m_{3}y^{*})^{2}}{fy^{*}(m_{1}+m_{3}y^{*})}{\rm i}. $

近而,由$\langle q^{*}(s), q(\theta)\rangle=1$和$\langle q^{*}(s), \overline{q}(\theta)\rangle=0$,可得

$1/\overline{D}=1+\overline{q_{2}^{*}}q_{2}+\frac{q_{2}^{*}\tau_{0}{\rm e}^{-{\rm i}\omega_{0}\tau_{0}}fy^{*}(m_{1}-q_{2}m_{3}x^{*}+m_{3}y^{*})}{(m_{1}+m_{2}x^{*}+m_{3}y^{*})^{2}}.$

(ⅱ) 得到二维常微分方程.与Hassard^[16]用同样的符号标记,我们先计算刻画在$\mu=0$处中心流形$C_{0}$的坐标.令$u_{t}$是系统(1.2)当$\mu=0$时的解.定义$z(t)=\langle q^{*}u_{t}\rangle$和

(3.7) $\begin{equation}W(t, \theta)=u_{t}(\theta)-2{\rm Re}\{z(t)q(\theta)\}. \label{1.5}\end{equation}$

在中心流形$C_{0}$上我们有

$ W(t, \theta)=W(z(t), \overline{z}(t), \theta), $

其中

$W(z(t), \overline{z}(t), \theta)=W_{20}(\theta)\frac{z^{2}}{2}+W_{11}(\theta)z\overline{z}+W_{02}(\theta)\frac{\overline{z}^{2}}{2}+W_{30}(\theta)\frac{z^{3}}{6}+\cdots, $

$z$和$\overline{z}$分别是中心流形$C_{0}$在方向$q^{*}$和$\overline{q}^{*}$上的局部坐标.需要指出的是如果$u_{t}$是实的则$W$也是实的,我们仅考虑实数解.

对系统(1.2)的解$u_{t}\in C_{0}$,既然$\mu=0$时

(3.8) $\begin{eqnarray}z'(t)&=&{\rm i}\omega_{0}\tau_{0}z(t)+\left\langle q^{*}(\theta), \tau_{0}F(W(z(t), \overline{z}(t), \theta)+2{\rm Re}\{z(t)q(\theta)\})\right\rangle\\&=&{\rm i}\omega_{0}\tau_{0}z+ \overline{q}^{*}(0)\tau_{0} F(W(z(t), \overline{z}(t), 0)+2{\rm Re}\{z(t)q(0)\})\\&=&{\rm i}\omega_{0}\tau_{0}z+\overline{q}^{*}(0)\tau_{0} F_{0}(z, \overline{z}). \label{1.9}\end{eqnarray}$

我们可以重新整理(3.8)式为

(3.9) $\begin{equation}z'(t)={\rm i}\omega_{0}\tau_{0}z(t)+g(z, \overline{z}), \label{1.6}\end{equation}$

其中

(3.10) $\begin{eqnarray}g(z, \overline{z})&=&\overline{q}^{*}(0)\tau_{0} F(W(z(t), \overline{z}(t), 0)+2{\rm Re}\{z(t)q(0)\})\\&=&g_{20}\frac{z^{2}}{2}+g_{11}z\overline{z}+g_{02}\frac{\overline{z}^{2}}{2}+g_{21}\frac{z^{2}\overline{z}}{2}+\cdots.\label{1.11}\end{eqnarray}$

再则,分析二维常微分方程且通过比较系数法求得这个二维常微分方程的系数.我们需要分(ⅰ)和(ⅱ)两小步得到(3.10)式的系数.

(ⅰ) 计算$g_{20}, g_{11}$和$g_{02}$.由(3.5), (3.7)和(3.9)式,我们有

(3.11) $\begin{equation}W'=u_{t}'-z'q-\overline{z}'\overline{q}= \left\{\begin{array}{ll}AW-2{\rm Re}\left\{\overline{q}^{*}(0)\tau_{0}F_{0}q(\theta)\right\}, &\theta\in([-1, 0), \\AW-2{\rm Re}\left\{\overline{q}^{*}(0)\tau_{0}F_{0}q(0)\right\}+\tau_{0}F_{0}, ~~&\theta=0, \end{array}\right.\label{1.12}\end{equation}$

其中

(3.12) $\begin{equation}H(z, \overline{z}, \theta)=H_{20}(\theta)\frac{z^{2}}{2}+H_{11}(\theta)z\overline{z}+H_{02}(\theta)\frac{\overline{z}^{2}}{2}+\cdots.\label{1.7}\end{equation}$

展开上述级数并进行系数比较后可得

(3.13) $\begin{equation}\left\{\begin{array}{ll}-H_{20}(\theta)=(A-2{\rm i}\omega_{0}\tau_{0})W_{20}(\theta), \\-H_{11}(\theta)=AW_{11}(\theta), \\-H_{02}(\theta)=(A+2{\rm i}\omega_{0}\tau_{0})W_{02}(\theta), \\\cdots.\end{array}\right.\label{1.8}\end{equation}$

由(3.8)和(3.9)式,令$F_{0}(z, \overline{z})=F_{z^{2}}\frac{z^{2}}{2}+F_{z\overline{z}}z\overline{z}+F_{\overline{z}^{2}}\frac{\overline{z}^{2}}{2}+F_{\overline{z}z^{2}}\frac{z^{2}\overline{z}}{2}+\cdots$,我们有

$g_{20}=\overline{q}^{*}(0)\tau_{0}F_{z^{2}}, ~~g_{11}=\overline{q}^{*}(0)\tau_{0}F_{z\overline{z}}, ~~g_{02}=\overline{q}^{*}(0)\tau_{0}F_{\overline{z}^{2}}, ~~g_{21}=\overline{q}^{*}(0)\tau_{0}F_{z^{2}\overline{z}}.$

由(3.7)式,我们有$u_{t}(\theta)=W(t, \theta)+z(t)q(\theta)+\overline{z}(t)\overline{q}(\theta)$,近而可得

$\begin{eqnarray*}u(t)&=&W(t, 0)+z(t)q(0)+\overline{z}(t)\overline{q}(0)\\&=&\left( \begin{array}{ccc} 1\\ q_{2}\\ \end{array}\right)z+\left( \begin{array}{ccc} 1\\ \overline{q_{2}}\\ \end{array}\right)\overline{z}+W_{20}(0)\frac{z^{2}}{2}+W_{11}(0)z\overline{z}+W_{02}(0)\frac{\overline{z}^{2}}{2}+\cdots\end{eqnarray*}$

且

$\begin{eqnarray*}u(t-1)&=&W(t, -1)+z(t)q(-1)+\overline{z}(t)\overline{q}(-1)\\&=&{\rm e}^{-{\rm i}\omega_{0}\tau_{0}}\left( \begin{array}{ccc} 1\\ q_{2}\\ \end{array}\right)z+{\rm e}^{{\rm i}\omega_{0}\tau_{0}}\left( \begin{array}{ccc} 1\\ \overline{q_{2}}\\ \end{array}\right)\overline{z}+W_{20}(-1)\frac{z^{2}}{2}+W_{11}(-1)z\overline{z}+W_{02}(-1)\frac{\overline{z}^{2}}{2}+\cdots, \end{eqnarray*}$

其中

$u(t)=\left( \begin{array}{ccc} x(t)\\ y(t)\\ \end{array}\right), u(t-1)=\left( \begin{array}{ccc} x(t-1)\\ y(t-1)\\ \end{array}\right), $

$ W(t, 0)=\left( \begin{array}{ccc} W^{(1)}(t, 0)\\ W^{(2)}(t, 0)\\ \end{array}\right), W(t, -1)=\left( \begin{array}{ccc} W^{(1)}(t, -1)\\ W^{(2)}(t, -1)\\ \end{array}\right). $

因此

(3.14) $\begin{equation}\left\{\begin{array}{ll}x(t)=z+\overline{z}+W^{(1)}(t, 0), \\y(t)=q_{2}z+\overline{q}_{2}\overline{z}+W^{(2)}(t, 0), \\x(t-1)={\rm e}^{-{\rm i}\omega_{0}\tau_{0}}z+{\rm e}^{{\rm i}\omega_{0}\tau_{0}}\overline{z}+W^{(1)}(t, -1), \\y(t-1)={\rm e}^{-{\rm i}\omega_{0}\tau_{0}}q_{2}z+{\rm e}^{{\rm i}\omega_{0}\tau_{0}}\overline{q}_{2}\overline{z}+W^{(2)}(t, -1).\end{array}\right.\label{1.10}\end{equation}$

由(3.14)式和

$\begin{eqnarray*}g(z, \overline{z})&=&\overline{q}^{*}(0)\tau_{0}F_{0}\\&=&\overline{D}\tau_{0}\left( \begin{array}{ccc} 1 &\overline{ q^{*}_{2}} \end{array}\right)\left( \begin{array}{ccc} \frac{1}{2}f_{20}x^{2}+f_{11}xy+\frac{1}{2}f_{02}y^{2}+\cdots \\[3mm] \frac{1}{2}l_{02}y^{2}+\frac{1}{2}\widetilde{l}_{20}x^{2}(t-1)+\widetilde{l}_{11}x(t-1)y(t-1)+\frac{1}{2}\widetilde{l}_{02}y^{2}(t-1)+\cdots \end{array}\right), \end{eqnarray*}$

与(3.10)式比较系数,可得

$\begin{equation}\left\{\begin{array}{ll}g_{20}=\tau_{0}\overline{D}\left[f_{20}+2f_{11}q_{2}+f_{02}q_{2}^{2}+\overline{q^{*}_{2}}q_{2}^{2}\left(l_{02}+\widetilde{l}_{02}{\rm e}^{-2{\rm i}\omega_{0}\tau_{0}}\right)+\overline{q}^{*}_{2}{\rm e}^{-2{\rm i}\omega_{0}\tau_{0}}\left(\widetilde{l}_{20}+2q_{2}\widetilde{l}_{11}\right)\right], \\[2mm]g_{11}=\tau_{0}\overline{D}\left[f_{20}+\overline{q}^{*}_{2}\widetilde{l}_{20}+f_{02}q_{2}\overline{q}_{2}+\left(f_{11}+\overline{q}^{*}_{2}\widetilde{l}_{11}\right)\left(q_{2}+\overline{q}_{2}\right)+\overline{q^{*}_{2}}q_{2}\overline{q}_{2}\left(l_{02}+\widetilde{l}_{02}\right)\right], \\[2mm]g_{02}=\tau_{0}\overline{D}\left[f_{20}+2f_{11}\overline{q}_{2}+f_{02}\overline{q}_{2}^{2}+\overline{q^{*}_{2}}\overline{q}_{2}^{2}\left(l_{02}+\widetilde{l}_{02}{\rm e}^{2{\rm i}\omega_{0}\tau_{0}}\right)+\overline{q}^{*}_{2}{\rm e}^{2{\rm i}\omega_{0}\tau_{0}}\left(\widetilde{l}_{20}+2\overline{q}_{2}\widetilde{l}_{11}\right)\right].\end{array}\right. \label{1.2400}\end{equation}$

因此,我们可以求出$g_{20}, g_{11}$和$g_{02}$.为了得到Hopf分支的特性,我们只需要如下算出$g_{21}$即可.

(ⅱ) 计算$g_{21}$.类似于$g_{20}, g_{11}$和$g_{02}$,我们也有

(3.16) $\begin{eqnarray}g_{21}&=&\tau_{0}\overline{D}\Big[f_{20}\left(2W^{(1)}_{11}(0)+W^{(1)}_{20}(0)\right)+f_{11}\Big(2W^{(2)}_{11}(0)+W^{(2)}_{20}(0)+\overline{q}_{2}W^{(1)}_{20}(0)\\&&+2q_{2}W^{(1)}_{11}(0)\Big)\Big]+\tau_{0}\overline{D}\Big[\left(f_{02}+\overline{q}^{*}_{2}l_{02}\right)\left(\overline{q}_{2}W^{(2)}_{20}(0)+2q_{2}W^{(2)}_{11}(0)\right)\\&&+\overline{q}^{*}_{2}\widetilde{l}_{20}\left(2{\rm e}^{-{\rm i}\omega_{0}\tau_{0}}W^{(1)}_{11}(-1)+{\rm e}^{{\rm i}\omega_{0}\tau_{0}}W^{(1)}_{20}(-1)\right)\Big]+\tau_{0}\overline{D}\overline{q}^{*}_{2}\widetilde{l}_{11}\Big[2{\rm e}^{-{\rm i}\omega_{0}\tau_{0}}W^{(2)}_{11}(-1)\\&&+{\rm e}^{{\rm i}\omega_{0}\tau_{0}}W^{(2)}_{20}(-1)+{\rm e}^{{\rm i}\omega_{0}\tau_{0}}W^{(1)}_{20}(-1)+2q_{2}{\rm e}^{-{\rm i}\omega_{0}\tau_{0}}W^{(1)}_{11}(-1)\Big]\\&&+\tau_{0}\overline{D}\overline{q}^{*}_{2}\widetilde{l}_{02}\Big[\overline{q}_{2}{\rm e}^{{\rm i}\omega_{0}\tau_{0}}W^{(2)}_{20}(-1)+2q_{2}{\rm e}^{-{\rm i}\omega_{0}\tau_{0}}W^{(2)}_{11}(-1)\Big].\label{1.24}\end{eqnarray}$

由(3.16)式,为了计算$g_{21}$,我们需要知道当$\theta\in [-1, 0)$时的$W_{20}(\theta)$和$W_{11}(\theta)$,具体过程如下.

我们讨论(3.11)式中$\theta\in [-1, 0)$的情形.由(3.11)式有

$\begin{eqnarray*}W'&=&AW-2{\rm Re}\left\{\overline{q}^{*}(0)\tau_{0}F_{0}q(\theta)\right\}=AW-\overline{q}^{*}(0)\tau_{0}F_{0}q(\theta)-q^{*}(0)\tau_{0}\overline{F}_{0}\overline{q}(\theta)\\&=&AW-g(z, \overline{z})q(\theta)-\overline{g}(z, \overline{z})\overline{q}(\theta)\end{eqnarray*}$

和

$W'=AW+H(z, \overline{z}, \theta).$

因此

$\begin{eqnarray*}H(z, \overline{z}, \theta)&=&g(z, \overline{z})q(\theta)-\overline{g}(z, \overline{z})\overline{q}(\theta)\\&=&-\left[g_{20}\frac{z^{2}}{2}+g_{11}z\overline{z}+g_{02}\frac{\overline{z}^{2}}{2}\right]q(\theta)-\left[\overline{g}_{20}\frac{\overline{z}^{2}}{2}+\overline{g}_{11}z\overline{z}+\overline{g}_{02}\frac{z^{2}}{2}\right]\overline{q}(\theta)\\&=&H_{20}(\theta)\frac{z^{2}}{2}+H_{11}(\theta)z\overline{z}+H_{02}(\theta)\frac{\overline{z}^{2}}{2}.\end{eqnarray*}$

由上式比较系数可得

(3.17) $\begin{equation}H_{20}(\theta)=-g_{20}q(\theta)-\overline{g}_{02}\overline{q}(\theta)=-(A-2{\rm i}\omega_{0}\tau_{0})W_{20}(\theta)\label{1.13}\end{equation}$

且

(3.18) $\begin{equation}H_{11}(\theta)=-g_{11}q(\theta)-\overline{g}_{11}\overline{q}(\theta)=-AW_{11}(\theta).\label{1.14}\end{equation}$

由(3.17)式得

$AW_{20}(\theta)=2{\rm i}\omega_{0}\tau_{0}W_{20}(\theta)+g_{20}q(\theta)+\overline{g}_{02}\overline{q}(\theta), $

再由(3.3)式,可得

$W'_{20}(\theta)=2{\rm i}\omega_{0}\tau_{0}W_{20}(\theta)+g_{20}q(0){\rm e}^{{\rm i}\omega_{0}\tau_{0}\theta}+\overline{g}_{02}\overline{q}(0){\rm e}^{-{\rm i}\omega_{0}\tau_{0}\theta}.$

求解$W_{20}(\theta)$,得

(3.19) $\begin{equation}W_{20}(\theta)=K_{1}{\rm e}^{2{\rm i}\omega_{0}\tau_{0}\theta}-\frac{g_{20}q(0)}{{\rm i}\omega_{0}\tau_{0}}{\rm e}^{{\rm i}\omega_{0}\tau_{0}\theta}-\frac{\overline{g}_{02}\overline{q}(0)}{3{\rm i}\omega_{0}\tau_{0}}{\rm e}^{-{\rm i}\omega_{0}\tau_{0}\theta}.\label{1.17}\end{equation}$

利用同样的方法,由(3.18)式,我们有

$AW_{11}(\theta)=g_{11}q(\theta)+\overline{g}_{11}\overline{q}(\theta), $

再由(3.3)式,可得

$W'_{11}(\theta)=g_{11}q(0){\rm e}^{{\rm i}\omega_{0}\tau_{0}\theta}+\overline{g}_{11}\overline{q}(0){\rm e}^{-{\rm i}\omega_{0}\tau_{0}\theta}.$

求解$W_{11}(\theta)$,得

(3.20) $\begin{equation}W_{11}(\theta)=K_{2}+\frac{g_{11}q(0)}{{\rm i}\omega_{0}\tau_{0}}{\rm e}^{{\rm i}\omega_{0}\tau_{0}\theta}-\frac{\overline{g}_{11}\overline{q}(0)}{{\rm i}\omega_{0}\tau_{0}}{\rm e}^{-{\rm i}\omega_{0}\tau_{0}\theta}.\label{1.18}\end{equation}$

接下来,为了求$K_{1}$和$K_{2}$,我们考虑(3.11)式中$\theta=0$的情形.

由(3.11)式,可得

$W'=AW+\tau_{0}F_{0}-2{\rm Re}\left\{\overline{q}^{*}(0)\tau_{0}F_{0}q(0)\right\}=AW+H(z, \overline{z}, 0).$

因此

$\begin{eqnarray*}H(z, \overline{z}, 0)&=&-2{\rm Re}\left\{\overline{q}^{*}(0)\tau_{0}F_{0}q(0)\right\}+\tau_{0}F_{0}=-g(z, \overline{z})q(\theta)-\overline{g}(z, \overline{z})\overline{q}(\theta)+\tau_{0}F_{0}\\&=&-\left[g_{20}\frac{z^{2}}{2}+g_{11}z\overline{z}+g_{02}\frac{\overline{z}^{2}}{2}\right]q(0)-\left[\overline{g}_{20}\frac{\overline{z}^{2}}{2}+\overline{g}_{11}z\overline{z}+\overline{g}_{02}\frac{z^{2}}{2}\right]\overline{q}(0)\\&&+\left( \begin{array}{ccc} \frac{1}{2}f_{20}x^{2}+f_{11}xy+\frac{1}{2}f_{02}y^{2}+\cdots \\[3mm] \frac{1}{2}l_{02}y^{2}+\frac{1}{2}\widetilde{l}_{20}x^{2}(t-1)+\widetilde{l}_{11}x(t-1)y(t-1)+\frac{1}{2}\widetilde{l}_{02}y^{2}(t-1)+\cdots \end{array}\right)\\&=&H_{20}(0)\frac{z^{2}}{2}+H_{11}(0)z\overline{z}+H_{02}(0)\frac{\overline{z}^{2}}{2}.\end{eqnarray*}$

将(3.14)式代入上面等式中,比较$\frac{z^{2}}{2}$和$z\overline{z}$的系数可得

(3.21) $\begin{equation}H_{20}(0)=-g_{20}q(0)-\overline{g}_{02}\overline{q}(0)+\left( \begin{array}{ccc} f_{20}+2q_{2}f_{11}+f_{02}q^{2}_{2}\\ l_{02}q^{2}_{2}+\widetilde{l}_{20}{\rm e}^{-2{\rm i}\omega_{0}\tau_{0}}+2\widetilde{l}_{11}q_{2}{\rm e}^{-2{\rm i}\omega_{0}\tau_{0}}+\widetilde{l}_{02}q^{2}_{2}{\rm e}^{-2{\rm i}\omega_{0}\tau_{0}} \\ \end{array}\right) \label{1.19}\end{equation}$

且

(3.22) $\begin{equation}H_{11}(0)=-g_{11}q(0)-\overline{g}_{11}\overline{q}(0)+\left( \begin{array}{ccc} f_{20}+f_{11}(q_{2}+\overline{q}_{2})+f_{02}q_{2}\overline{q}_{2} \\ l_{02}q_{2}\overline{q}_{2}+\widetilde{l}_{20}+\widetilde{l}_{11}(q_{2}+\overline{q}_{2})+\widetilde{l}_{02}q_{2}\overline{q}_{2} \\ \end{array}\right).\label{1.20}\end{equation}$

由$AW_{20}(0)=2{\rm i}\omega_{0}\tau_{0}W_{20}(0)-H_{20}(0)$,我们有

$\begin{equation}\tau_{0}BW_{20}(0)+\tau_{0}CW_{20}(-1)=2{\rm i}\omega_{0}\tau_{0}W_{20}(0)-H_{20}(0).\label{1.21}\end{equation}$

把(3.19)和(3.21)式代入(3.23)式中,有

$\begin{eqnarray*}&&\tau_{0}BK_{1}-\frac{g_{20}}{{\rm i}\omega_{0}\tau_{0}}\tau_{0}Bq(0)-\frac{\overline{g}_{02}}{3{\rm i}\omega_{0}\tau_{0}}\tau_{0}B\overline{q}(0)+{\rm e}^{-2{\rm i}\omega_{0}\tau_{0}}\tau_{0}CK_{1}-\frac{g_{20}{\rm e}^{-{\rm i}\omega_{0}\tau_{0}}}{{\rm i}\omega_{0}\tau_{0}}\tau_{0}Cq(0)\\&&-\frac{\overline{g}_{02}{\rm e}^{{\rm i}\omega_{0}\tau_{0}}}{3{\rm i}\omega_{0}\tau_{0}}\tau_{0}C\overline{q}(0)-2{\rm i}\omega_{0}\tau_{0}\left(K_{1}-\frac{g_{20}}{{\rm i}\omega_{0}\tau_{0}}q(0)-\frac{\overline{g}_{02}}{3{\rm i}\omega_{0}\tau_{0}}\overline{q}(0)\right)\\&=&g_{20}q(0)+\overline{g}_{02}\overline{q}(0)-\left( \begin{array}{ccc} f_{20}+2q_{2}f_{11}+f_{02}q^{2}_{2} \\ l_{02}q^{2}_{2}+\widetilde{l}_{20}{\rm e}^{-2{\rm i}\omega_{0}\tau_{0}}+2\widetilde{l}_{11}q_{2}{\rm e}^{-2{\rm i}\omega_{0}\tau_{0}}+\widetilde{l}_{02}q^{2}_{2}{\rm e}^{-2{\rm i}\omega_{0}\tau_{0}} \end{array}\right).\end{eqnarray*}$

由

$\begin{equation}\left(\tau_{0}B+{\rm e}^{-{\rm i}\omega_{0}\tau_{0}}\tau_{0}C\right)q(0)={\rm i}\omega_{0}\tau_{0}q(0), ~~~~\left(\tau_{0}B+{\rm e}^{{\rm i}\omega_{0}\tau_{0}}\tau_{0}C\right)\overline{q}(0)=-{\rm i}\omega_{0}\tau_{0}\overline{q}(0), \label{1.23}\end{equation}$

可得

$\left(2{\rm i}\omega_{0}\tau_{0}I-\tau_{0}B-{\rm e}^{-2{\rm i}\omega_{0}\tau_{0}}\tau_{0}C\right)K_{1}=\left(\begin{array}{ccc} f_{20}+2q_{2}f_{11}+f_{02}q^{2}_{2}\\ l_{02}q^{2}_{2}+\widetilde{l}_{20}{\rm e}^{-2{\rm i}\omega_{0}\tau_{0}}+2\widetilde{l}_{11}q_{2}{\rm e}^{-2{\rm i}\omega_{0}\tau_{0}}+\widetilde{l}_{02}q^{2}_{2}{\rm e}^{-2{\rm i}\omega_{0}\tau_{0}} \end{array}\right), $

因此

$K_{1}=\tau_{0}\left(2{\rm i}\omega_{0}I-B-{\rm e}^{-2{\rm i}\omega_{0}\tau_{0}}C\right)^{-1}\left(\begin{array}{ccc} f_{20}+2q_{2}f_{11}+f_{02}q^{2}_{2} \\ l_{02}q^{2}_{2}+\widetilde{l}_{20}{\rm e}^{-2{\rm i}\omega_{0}\tau_{0}}+2\widetilde{l}_{11}q_{2}{\rm e}^{-2{\rm i}\omega_{0}\tau_{0}}+\widetilde{l}_{02}q^{2}_{2}{\rm e}^{-2{\rm i}\omega_{0}\tau_{0}} \end{array}\right).$

由$AW_{11}(0)=-H_{11}(0)$,我们有

(3.25) $\begin{equation}\tau_{0}BW_{11}(0)+\tau_{0}CW_{11}(-1)=-H_{11}(0). \label{1.22}\end{equation}$

把(3.20)和(3.22)式代入(3.25)式中可得

$\begin{eqnarray*}&&\tau_{0}B\left(K_{2}+\frac{g_{11}}{{\rm i}\omega_{0}\tau_{0}}q(0)-\frac{\overline{g}_{11}}{{\rm i}\omega_{0}\tau_{0}}\overline{q}(0)\right)+\tau_{0}C\left(K_{2}+{\rm e}^{-{\rm i}\omega_{0}\tau_{0}}\frac{g_{11}}{{\rm i}\omega_{0}\tau_{0}}q(0)-{\rm e}^{{\rm i}\omega_{0}\tau_{0}}\frac{\overline{g}_{11}}{{\rm i}\omega_{0}\tau_{0}}\overline{q}(0)\right)\\&&=g_{11}q(0)+\overline{g}_{11}\overline{q}(0)-\left( \begin{array}{ccc} f_{20}+f_{11}(q_{2}+\overline{q}_{2})+f_{02}q_{2}\overline{q}_{2}\\ l_{02}q_{2}\overline{q}_{2}+\widetilde{l}_{20}+\widetilde{l}_{11}(q_{2}+\overline{q}_{2})+\widetilde{l}_{02}q_{2}\overline{q}_{2} \end{array}\right).\end{eqnarray*}$

由(3.24)式,可得

$\left(\tau_{0}B+\tau_{0}C\right)K_{2}=-\left( \begin{array}{ccc} f_{20}+f_{11}(q_{2}+\overline{q}_{2})+f_{02}q_{2}\overline{q}_{2}\\ l_{02}q_{2}\overline{q}_{2}+\widetilde{l}_{20}+\widetilde{l}_{11}(q_{2}+\overline{q}_{2})+\widetilde{l}_{02}q_{2}\overline{q}_{2} \end{array}\right), $

因此

$K_{2}=-\tau_{0}\left(B+C\right)^{-1}\left( \begin{array}{ccc} f_{20}+f_{11}(q_{2}+\overline{q}_{2})+f_{02}q_{2}\overline{q}_{2} \\ l_{02}q_{2}\overline{q}_{2}+\widetilde{l}_{20}+\widetilde{l}_{11}(q_{2}+\overline{q}_{2})+\widetilde{l}_{02}q_{2}\overline{q}_{2} \end{array}\right).$

由$K_{1}$和$K_{2}$,我们可以得到$W_{11}(\theta)$和$W_{20}(\theta)$.则$g_{21}$就可以确定.

最后,得到hopf分支关于稳定性、方向和周期的特性.将有下面的(ⅰ)和(ⅱ)两小步去完成.

(ⅰ) 计算参数$C_{1}(0), \mu_{2}, T_{2}, \beta_{2}$.基于以上分析,我们可以看出(3.15)和(3.16)式中的$g_{ij}$都是由系统(1.2)中的参数和时滞所确定的.因此,我们可以计算下列特征参数

(3.26) $\begin{aligned} C_{1}(0)=& \frac{\mathrm{i}}{2 \omega_{0} \tau_{0}}\left(g_{20} g_{11}-2\left|g_{11}\right|^{2}-\frac{1}{3}\left|g_{02}\right|^{2}\right)+\frac{g_{21}}{2}, \quad \mu_{2}=-\frac{\operatorname{Re}\left\{C_{1}(0)\right\}}{\operatorname{Re}\left\{\lambda^{\prime}\left(\tau_{0}\right)\right\}} \\ & T_{2}=-\frac{\operatorname{Im}\left\{C_{1}(0)\right\}+\mu_{2} \operatorname{Im}\left\{\lambda^{\prime}\left(\tau_{0}\right)\right\}}{\omega_{0} \tau_{0}}, \quad \beta_{2}=2 \operatorname{Re}\left\{C_{1}(0)\right\} \end{aligned}$

(ⅱ) 因此,我们可以计算出参数$C_{1}(0), \mu_{2}, T_{2}$和$\beta_{2}$,它们可以确定在临界点$\tau_{0}$处分支周期解的特性且有如下结论.

定理3.1   对系统(1.2)来说,下列结论成立

(1)  如果$\mu_{2}>0 (< 0)$,则Hopf分支是向前的(向后的).

(2)  如果$\beta_{2} < 0 (>0)$在中心流形上的分支周期解是稳定的(不稳定的).

(3)  如果$T_{2}>0 (< 0)$,周期是递增的(递减的).

在这一部分中,我们将用数值模拟表明系统(1.2)有趣的动力行为.从下面例子可以看出,随着时滞$\tau$的变化,系统(1.2)正平衡点的稳定性也会发生改变,并会有Hopf分支出现,这也是对我们前面第二部分和第三部分结论的一个数值验证.

例4.1   令$a=0.4, ~b=0.0015,~c=1.32, ~d=0.37,~r=0.02, ~f=3, m_{1}=10, m_{2}=1, $$m_{3}=1, $则系统(1.2)就变成

(4.1) $\begin{equation}\left\{\begin{array}{ll} x'(t)=x(t)\left(0.4-0.0015x(t)-\frac{1.32y(t)}{10+x(t)+y(t)}\right), \\[4mm] y'(t)=y(t)\left(-0.37-0.02y(t)+\frac{3x(t-\tau)}{10+x(t-\tau)+y(t-\tau)}\right).\end{array}\right.\label{4.1}\end{equation}$

通过数值计算,我们有$E^{*}=(2.9689, 5.5490)$且满足引理2.2中条件(H$_{1})$和(H$_{3})$.由(2.6)式、(2.7)式以及第三节中的公式,计算可得

$\tau_{0}^{+}=1.9823, \ {\rm Re}\{\lambda'(\tau_{0}^{+})\}=16.3571>0, \g_{20}=-0.0639+0.0467{\rm i}, \ g_{11}=0.0914+0.0392{\rm i}, $

$g_{02}=-0.0779-0.3521{\rm i}, \ g_{21}=-0.0631+0.1296{\rm i}, \ C_{1}(0)=-0.033+0.0085{\rm i}, $

再由公式(3.26)可知$\mu_{2}>0, \beta_{2}>0$.因此,数值模拟结果图 1和图 2表明当$\tau < \tau_{0}^{+}$时正平衡点$E^{*}$是稳定的,当$\tau$穿过临界点$\tau>\tau_{0}^{+}$时, $E^{*}$将失去它的稳定性,就出现了Hopf分支.由$\mu_{2}>0$和$\beta_{2}>0$得Hopf分支的方向是$\tau>\tau_{0}^{+}$,且从$E^{*}$点出现的分支周期解在$\tau_{0}^{+}$处是稳定的.

本文研究了密度制约且具有Beddington-DeAngelis功能反应函数的时滞捕食-被捕食系统,在文献[15]研究持久性和正平衡点的局部渐近稳定性和全局渐近稳定性等动力性质的基础之上,对系统(1.2)的Hopf分支性质进行讨论,是对文献[15]在研究内容上的一个推广.

在前期准备工作中得到:条件(H$_{1})$和(H$_{2})$成立时,对任意的$\tau\geq 0$,系统(1.2)的正平衡点是稳定的;条件(H$_{1})$和(H$_{3})$成立时,当$\tau\in[0, \tau_{0}^{+})$,系统(1.2)的正平衡点是稳定的,当$\tau>\tau_{0}^{+}$,系统(1.2)的正平衡点不稳定, $\tau=\tau_{0}^{+}$是临界点;条件(H$_{1})$和(H$_{4})$成立时,当$\tau\in[0, \tau_{0}^{+}], (\tau_{0}^{-}, \tau_{1}^{+}), \cdots, (\tau_{k-1}^{-}, \tau_{k}^{+}), $系统(1.2)的正平衡点是稳定的,当$\tau\in[\tau_{0}^{+}, \tau_{0}^{-}), [\tau_{1}^{+}, \tau_{1}^{-}), \cdots, [\tau_{k-1}^{+}, \tau_{k-1}^{-})$且$~\tau>\tau_{k}^{+}, $系统(1.2)的正平衡点不稳定,出现$k$次稳定性转化现象:稳定$\rightarrow$不稳定$\rightarrow$稳定$\cdot\cdot\cdot$.在Hopf分支方面,我们得到Hopf分支的特性指标表达式并且可以直接判断Hopf分支方向、稳定性以及周期情况.在数值模拟这部分对系统稳定性变换和Hopf分支得到的结论进行了验证.