在种群动力学中, 对捕食者和食饵之间的动力学行为的研究已经长时期并且将继续成为生态学和数学生态学研究的重要课题.近年的研究表明, 在很多情况下, 特别当捕食者捕食食物时, 就必须得和其他物种进行分享和竞争食物, 因此一个更符合实际的捕食模型应该是具有比率型的模型.关于有比率型种群捕食模型研究已有大量工作, 诸多学者取得了很多成果^[1-11]. 2003年, 谭德君^[12]研究了具有时滞和基于比率的三种群捕食模型

$ \left\{ \begin{array}{lc} \frac{dx_1(t)}{dt}=x_1(t)\left[a_1-a_{11}x_1(t)-\frac{a_{12}x_2(t)}{m_{12}x_2(t)+x_1(t)}-\frac{a_{13}x_3(t)}{m_{13}x_3(t)+x_1(t)}\right], \\ \frac{dx_2(t)}{dt}=x_2(t)\left[-a_2+\frac{a_{21}x_1(t-\tau_1)}{m_{12}x_2(t-\tau_1)+x_1(t-\tau_1)}\right], \\ \frac{dx_3(t)}{dt}=x_3(t)\left[-a_3+\frac{a_{31}x_1(t-\tau_2)}{m_{13}x_3(t-\tau_2)+x_1(t-\tau_2)}\right] \end{array}\right. $

(1.1)

的持续生存和无时滞时的全局渐近稳定性, 其中 $x_i(i=1, 2, 3)$分别表示捕食者, 食饵种群的密度, $a_i, a_{ij}, m_{ij}(i, j=1, 2, 3)$均为正常数, $\tau_i(i=1, 2)$为非负常数.详细的生物学意义可参见文献[13]. 2007年, 黄建科和吴筱宁^[14]进一步研究了模型(1.1) 的全局渐近稳定性, 推广了文献[12]的结果, 同时也研究了该系统的持久性. Song和Zou ^[15]讨论了一类具有比率型的捕食模型的分支问题. Shi和Li ^[16]分析了类具有比率型的捕食模型的全局渐近稳定性. Bai等学者^[17]考虑了一类具有比率型的随机捕食模型的动力学行为.具体相关研究可参考文献[18-22].

我们知道在很多生物现象中, 生物种群的周期性行为经常出现, 因此为了能更清晰全面地理解该捕食系统的动力学行为, 本文将研究系统(1.1) 的周期现象, 具体而言, 本文将以时滞为参数, 考察因时滞的变化导致系统(1.1) 出现Hopf分支, 得到了系统(1.1) 的平凡解稳定的条件, Hopf分支产生的条件及确定Hopf分支方向和分支周期解的稳定性的计算公式.为了研究简化, 假设 $\tau_1=\tau_2=\tau$, 于是得到下面的系统

$ \left\{ \begin{array}{lc} \frac{dx_1(t)}{dt}=x_1(t)\left[a_1-a_{11}x_1(t)-\frac{a_{12}x_2(t)}{m_{12}x_2(t)+x_1(t)}-\frac{a_{13}x_3(t)}{m_{13}x_3(t)+x_1(t)}\right], \\ \frac{dx_2(t)}{dt}=x_2(t)\left[-a_2+\frac{a_{21}x_1(t-\tau)}{m_{12}x_2(t-\tau)+x_1(t-\tau)}\right], \\ \frac{dx_3(t)}{dt}=x_3(t)\left[-a_3+\frac{a_{31}x_1(t-\tau)}{m_{13}x_3(t-\tau)+x_1(t-\tau)}\right], \end{array}\right. $

(1.2)

这里 $x_i(i=1, 2, 3)$分别表示捕食者, 食饵种群的密度, $a_i, a_{ij}, m_{ij}~(i, j=1, 2, 3)$均为正常数, $\tau$为非负常数.

本文作如下安排:第二节研究了(1.2) 式的平凡解的稳定性和Hopf分支的存在性; 第三节讨论了Hopf分支方向及其分支周期解的稳定性的清晰的计算公式; 第四节进行数值模拟验证理论分析的正确性.

由文献[23]知, 系统的持续生存必然蕴含着正平衡点的存在, 我们设系统(1.2) 的正平衡点为 $E^*(x_1^*, x_2^*, x_3^*)$.令 $\bar{x_i}=x_i-x_i^*(i=1, 2, 3)$, 仍记 $\bar{x_i}$分别为 $x_i$, 则(1.2) 式化为

$ \left\{ \begin{array}{lc} \frac{dx_1(t)}{dt}=p_1x_1(t)+p_2x_2(t)+p_3x_3(t)+p_4x_1^2(t)+p_5x_2^2(t)+p_6x_3^2(t)\\ ~~~~~~~~~~~~~+p_7x_1(t)x_2(t)+p_8x_1(t)x_3(t)+p_9x_1^2(t)x_2(t)\\ ~~~~~~~~~~~~~+p_{10}x_2^2(t)x_1(t)+p_{11}x_3^2(t)x_1(t)+\mbox{h.o.t.}, \\ \frac{dx_2(t)}{dt}=q_1x_1(t-\tau)+q_2x_2(t-\tau)+q_3x_1^2(t-\tau)+q_4x_2^2(t-\tau)\\ ~~~~~~~~~~~~~+q_5x_1(t-\tau)x_2(t-\tau)+q_6x_1^3(t-\tau)+q_7x_2^3(t-\tau)+q_8x_1^2(t-\tau)x_2(t-\tau)\\ ~~~~~~~~~~~~~+q_9x_1(t-\tau)x_2^2(t-\tau)+\mbox{h.o.t.}, \\ \frac{dx_3(t)}{dt}=r_1x_1(t-\tau)+r_2x_3(t-\tau)+r_3x_1^2(t-\tau)+r_4x_3^2(t-\tau)\\ ~~~~~~~~~~~~~+r_5x_1(t-\tau)x_3(t-\tau)+r_6x_1^3(t-\tau)+r_7x_3^3(t-\tau)+r_8x_1^2(t-\tau)x_3(t-\tau)\\ ~~~~~~~~~~~~~+r_9x_1(t-\tau)x_3^2(t-\tau)+\mbox{h.o.t.}, \end{array}\right. $

(2.1)

其中 $p_i(i=1, 2, \cdots, 11); q_j, r_j~ (j=1, 2, \cdots, 9)$的表达式参见附录A.系统(2.1) 的线性部分为

$ \left\{ \begin{array}{l} \frac{{d{x_1}(t)}}{{dt}} = {p_1}{x_1}(t) + {p_2}{x_2}(t) + {p_3}{x_3}(t), \\ \frac{{d{x_2}(t)}}{{dt}} = {q_1}{x_1}(t - \tau ) + {q_2}{x_2}(t - \tau ), \\ \frac{{d{x_3}(t)}}{{dt}} = {r_1}{x_1}(t - \tau ) + {r_2}{x_3}(t - \tau ). \end{array} \right. $

(2.2)

(2.2) 式的特征方程为

$ \lambda^3+m_2\lambda^2+m_1\lambda+(n_2\lambda^2+n_1\lambda+n_0)e^{-\lambda\tau}+(s_1\lambda+s_0)e^{-2\lambda\tau}=0, $

(2.3)

其中

$ \begin{eqnarray*} &&m_1=-p_3r_1, m_2=-p_1, n_0=p_3q_2r_1, n_1=p_1(r_2+q_2)-p_2q_2, \\ &&n_2=-(r_2+q_2), s_0=p_2q_1r_2-p_1q_2r_2, s_1=q_2r_2.\end{eqnarray*} $

当 $\tau=0$时, 方程(2.3) 变为

$ \lambda^3+(m_2+n_2)\lambda^2+(m_1+n_1+s_1)\lambda+n_0+s_0=0, $

(2.4)

根据Routh-Hurwitz准则, 如果下列条件

$ (\mbox{H}1)~~n_0+s_0 > 0, (m_2+n_2)(m_1+n_1+s_1) > n_0+s_0 $

成立, 则方程(2.4) 的所有根都具有负实部.方程的两边同时乘以 $e^{\lambda\tau}$得

$ (\lambda^3+m_2\lambda^2+m_1\lambda)e^{\lambda\tau}+(n_2\lambda^2+n_1\lambda+n_0)+(s_1\lambda+s_0)e^{-\lambda\tau}=0, $

(2.5)

若 ${\rm i}\omega$为(2.5) 式的解, 把 $\lambda={\rm i}\omega$代入(2.5) 式并分离实部和虚部得

$ \left\{ \begin{array}{lc} (s_0-m_1\omega+\omega^3)\cos\omega\tau+s_1\omega\sin\omega\tau=(m_2+n_1s_1+n_2s_0)\omega^2-s_0n_0, \\ (m_1\omega+s_1\omega-\omega^2)\cos\omega\tau-(s_0+m_2\omega^2)\omega\sin\omega\tau=-n_1\omega. \end{array}\right. $

(2.6)

于是

$ \begin{eqnarray*} \cos\omega\tau&=&\frac{[(m_2+n_1s_1+n_2s_0)\omega^2-s_0n_0](s_0+m_2\omega^2)-n_1s_1\omega^2} {(s_0-m_1\omega+\omega^3)(s_0+m_2\omega^2)+s_1\omega(m_1\omega+s_1\omega-\omega^2)}, \\ \sin\omega\tau&=&\frac{[(m_2+n_1s_1+n_2s_0)\omega^2-s_0n_0](m_1\omega+s_1\omega-\omega^2)} {(s_0-m_1\omega+\omega^3)(s_0+m_2\omega^2)+s_1\omega(m_1\omega+s_1\omega-\omega^2)}\\ &~&+\frac{n_1\omega(s_0-m_1\omega+\omega^3)} {(s_0-m_1\omega+\omega^3)(s_0+m_2\omega^2)+s_1\omega(m_1\omega+s_1\omega-\omega^2)}. \end{eqnarray*} $

从而由 $\sin^2\omega\tau+\cos^2\omega\tau=1$, 得

$ m_2^2\omega^{10}+k_8\omega^8+k_7\omega^7+k_6\omega^6+k_5\omega^5+k_4\omega^4+k_3\omega^3+k_2\omega^2+k_1\omega+k_0=0, $

(2.7)

其中 $k_i(i=0, 1, 2, \cdots, 8)$的表达式参见附录B.

假设方程(2.7) 有正根, 不失一般性, 假设方程有10个正根, 记为 $\omega_k(k=1, 2, \cdots, 10)$.于是由(2.6) 式得

$ \tau_k^{(j)}=\frac{1}{\omega_k}\left[\arccos\frac{[(m_2+n_1s_1+n_2s_0)\omega_k^2-s_0n_0](s_0+m_2\omega_k^2)-n_1s_1\omega_k^2} {(s_0-m_1\omega_k+\omega_k^3)(s_0+m_2\omega_k^2)+s_1\omega_k(m_1\omega_k+s_1\omega_k-\omega_k^2)}+2j\pi\right], $

(2.8)

这里 $k=1, 2, \cdots, 10;j=0, 1, 2, \cdots.$记

$ {\tau _0} = \tau _{k0}^{(0)} = _{k \in \{ 1, 2, \cdots, 10\} }^{\;\;\;\;\;\;\min }\{ \tau _k^{(0)}\} . $

(2.9)

设 $\lambda(\tau)=\alpha(\tau)+i\omega(\tau)$为(2.5) 式的满足 $\alpha(\tau_k^{(j)})=0, \omega(\tau_k^{(j)})=\omega_k$的根, 其中 $\tau_k^{(j)}$由(2.8) 式定义.把 $\lambda(\tau)$代入(2.5) 式, 两端对 $\tau$求导, 整理得

$ \left[\frac{d\lambda(\tau)}{d\tau}\right]^{-1}= \frac{(3\lambda^2+2m_2\lambda+m_1)e^{\lambda\tau}+(2n_2\lambda+n_1)} {\lambda\left[(n_2\lambda^2+n_1\lambda+n_0)+2(s_1\lambda+s_0)e^{-\lambda\tau}\right]}-\frac{\tau}{\lambda}. $

(2.10)

于是

$ \begin{eqnarray*} \left[\frac{d\mbox{Re}\lambda(\tau)}{d\tau}\right]^{-1}\Big|_{\tau=\tau_k^{(j)}}&=& \mbox{Re}\left\{\frac{(3\lambda^2+2m_2\lambda+m_1)e^{\lambda\tau}+(2n_2\lambda+n_1)} {\lambda\left[(n_2\lambda^2+n_1\lambda+n_0)+2(s_1\lambda+s_0)e^{-\lambda\tau}\right]}\right\}_{\tau=\tau_k^{(j)}}\\ &=& \frac{A_1A_3-A_2A_4}{A_1^2+A_2^2}, \end{eqnarray*} $

这里

$ \begin{eqnarray*} A_1&=&2\left[s_0\sin\omega_k\tau_k^{(j)}-s_1\omega_k\cos\omega_k\tau_k^{(j)}\right], \\ A_2&=&\omega_k\left[n_0-n_2\omega_k^2+2s_0\cos\omega_k\tau_k^{(j)}+2s_1\omega_k\sin\omega_k\tau_k^{(j)}\right], \\ A_3&=&(m_1-3\omega_k^2)\cos\omega_k\tau_k^{(j)}-2m_2\omega_k\sin\omega_k\tau_k^{(j)}+n_1, \\ A_4&=&2n_2\omega_k.\\ \end{eqnarray*} $

为了得到本文的主要结果, 假设

$ (\mbox{H}2)~~\mbox{Re}\left[\frac{d\lambda}{d\tau}\right]\Big|_{\tau=\tau_k^{(j)}}\neq0. $

由上述讨论和Hale ^[24]的第11章的定理1.1可得

定理2.1  对系统(1.2), 假设条件 $(\mbox{H}1)$和 $(\mbox{H}2)$成立, 则

(ⅰ) $\tau\in[0, \tau_0), $其正平衡点是渐近稳定的;

(ⅱ) $\tau > \tau_0, $其正平衡点是不稳定的;

(ⅲ) $\tau=\tau_k^{(j)}$ $(k=1, 2, 3, \cdots, 10; k=1, 2, \cdots)$是~Hopf分支值, 其中 $\tau_k^{(j)}$由(2.8) 式定义.

本节利用中心流形理论和和规范型方法^[25]给出系统(1.2) 的Hopf分支方向, 分支周期解的稳定性等计算公式.

为方便, 令 $t=s\tau, x_i(s\tau)=\tilde{x}_i(s)(i=1, 2, 3), $仍记 $\tilde{x}_i(s)$为 $x_i(s)$. $\tau=\tau_k^{(j)}+\mu, \mu\in{R}, \tau_k^{(j)}$由(2.8) 式定义, 且仍记 $t=s$, 则系统(1.2) 等价于系统

$ \left\{ \begin{array}{ll} \frac{dx_1(t)}{dt}=(\tau_k^{(j)}+\mu)p_1x_1(t)+p_2x_2(t)+p_3x_3(t)+p_4x_1^2(t)+p_5x_2^2(t)+p_6x_3^2(t)\\ ~~~~~~~~~~~~~+p_7x_1(t)x_2(t)+p_8x_1(t)x_3(t)+p_9x_1^2(t)x_2(t)\\ ~~~~~~~~~~~~~+p_{10}x_2^2(t)x_1(t)+p_{11}x_3^2(t)x_1(t)+\mbox{h.o.t.}, \\ \frac{dx_2(t)}{dt}=(\tau_k^{(j)}+\mu)q_1x_1(t-1)+q_2x_2(t-1)+q_3x_1^2(t-1)+q_4x_2^2(t-1)\\ ~~~~~~~~~~~~~+q_5x_1(t-1)x_2(t-1)+q_6x_1^3(t-1)+q_7x_2^3(t-1)+q_8x_1^2(t-1)x_2(t-1)\\ ~~~~~~~~~~~~~+q_9x_1(t-1)x_2^2(t-1)+\mbox{h.o.t.}, \\ \frac{dx_3(t)}{dt}=(\tau_k^{(j)}+\mu)r_1x_1(t-1)+r_2x_3(t-1)+r_3x_1^2(t-1)+r_4x_3^2(t-1)\\ ~~~~~~~~~~~~~+r_5x_1(t-1)x_3(t-1)+r_6x_1^3(t-1)+r_7x_3^3(t-1)+r_8x_1^2(t-1)x_3(t-1)\\ ~~~~~~~~~~~~~+r_9x_1(t-1)x_3^2(t-1)+\mbox{h.o.t.}. \end{array}\right. $

(3.1)

(3.1) 式的线性部分为

$ \left\{ \begin{array}{ll} \frac{dx_1(t)}{dt}=(\tau_k^{(j)}+\mu)p_1x_1(t)+p_2x_2(t)+p_3x_3(t), \\ \frac{dx_2(t)}{dt}=(\tau_k^{(j)}+\mu)q_1x_1(t-1)+q_2x_2(t-1), \\ \frac{dx_3(t)}{dt}=(\tau_k^{(j)}+\mu)r_1x_1(t-1)+r_2x_3(t-1). \end{array}\right. $

(3.2)

(3.1) 的右端的非线性部分为

$ h=(\tau_k+\mu)\left( \begin{array}{c} f_1(t) \\ f_2(t) \\ f_3(t) \\ \end{array} \right), $

(3.3)

其中

$ \begin{eqnarray*} f_1(t)&=&p_4x_1^2(t)+p_5x_2^2(t)+p_6x_3^2(t)+p_7x_1(t)x_2(t)+p_8x_1(t)x_3(t)\\ &~&+p_9x_1^2(t)x_2(t)+p_{10}x_2^2(t)x_1(t)+p_{11}x_3^2(t)x_1(t)+\mbox{h.o.t.}, \\ f_2(t)&=&q_3x_1^2(t-1)+q_4x_2^2(t-1)+q_5x_1(t-1)x_2(t-1)+q_6x_1^3(t-1)\\ &~&+q_7x_2^3(t-1)+q_8x_1^2(t-1)x_2(t-1)+q_9x_1(t-1)x_2^2(t-1)+\mbox{h.o.t.}, \\ f_3(t)&=&r_3x_1^2(t-1)+r_4x_3^2(t-1)+r_5x_1(t-1)x_3(t-1)+r_6x_1^3(t-1) \\ &~&+r_7x_3^3(t-1)+r_8x_1^2(t-1)x_3(t-1)+r_9x_1(t-1)x_3^2(t-1)+\mbox{h.o.t.}. \end{eqnarray*} $

对 $\varphi=(\varphi_1, \varphi_2, \varphi_3)^T\in{C([-1, 0], R^3)}$, 记

$ L_\mu\varphi=B_1\varphi(0)+B_2\varphi(-1), ~~~ h(\mu, \varphi)=(\tau_k+\mu)\left( \begin{array}{c} h_1(\mu, \varphi) \\ h_2(\mu, \varphi) \\ h_3(\mu, \varphi) \\ \end{array} \right), $

这里

$ \begin{eqnarray*} h_1(\mu, \varphi)&=&p_4\varphi_1^2(0)+p_5\varphi_2^2(0)+p_6\varphi_3^2(0)+p_7\varphi_1(0)\varphi_2(0)+p_8\varphi_1(0)\varphi_3(0)\\ &~&+p_9\varphi_1^2(0)\varphi_2(0)+p_{10}\varphi_2^2(0)\varphi_1(0)+p_{11}\varphi_3^2(0)\varphi_1(0)+\mbox{h.o.t.}, \\ h_2(\mu, \varphi)&=&q_3\varphi_1^2(-1)+q_4\varphi_2^2(-1)+q_5\varphi_1(-1)\varphi_2(-1)+q_6\varphi_1^3(-1)\\ &~&+q_7\varphi_2^3(-1)+q_8\varphi_1^2(-1)\varphi_2(-1)+q_9\varphi_1(-1)\varphi_2^2(-1)+\mbox{h.o.t.}, \\ h_3(\mu, \varphi)&=&r_3\varphi_1^2(-1)+r_4\varphi_3^2(-1)+r_5\varphi_1(-1)\varphi_3(-1)+r_6\varphi_1^3(-1) \\ &~&+r_7\varphi_3^3(-1)+r_8\varphi_1^2(-1)x_3(-1)+r_9\varphi_1(-1)\varphi_3^2(-1)+\mbox{h.o.t.}, \\ B_1&=&(\tau_k^{(j)}+\mu)\left( \begin{array}{ccc} p_1&p_2&p_3\\ 0&0&0\\ 0&0&0\\ \end{array} \right), B_2=(\tau_k^{(j)}+\mu)\left( \begin{array}{ccc} 0&0&0\\ q_2&q_2&0\\ r_1&0&r_2\\ \end{array} \right). \end{eqnarray*} $

于是由Riesz表示定理, 存在分量为有界变差函数的二阶矩阵 $\eta(\theta, \mu): [-1, 0]\rightarrow{R^{3^2}}.$使对任意的 $\varphi\in{C([-1, 0], R^3)}$, 有

$ L_\mu\varphi=\int_{-1}^0d\eta(\theta, \mu)\varphi(\theta). $

事实上, 只要取

$ \eta (\theta ,\mu ) = \left\{ {\begin{array}{*{20}{c}} { - {B_2},} & {\theta = - 1,}\\ {0,} & { - 1 < \theta < 0,}\\ {{B_1},} & {\theta = 0} \end{array}} \right. $

即可.对 $\varphi\in{C([-1, 0], R^3)}, $定义

$ A(\mu )\varphi = \left\{ \begin{array}{l} \frac{{d\varphi (\theta )}}{{d\theta }},\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; - 1 \le \theta < 0,\\ \int_{ - 1}^0 \mathit{d} \eta (s,\mu )\varphi (s),\;\;\;\;\;\;\;\;\;\;\;\;\theta = 0, \end{array} \right.\;\;\;\;\;R\varphi = \left\{ \begin{array}{l} 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\; - 1 \le \theta < 0,\\ h(\mu ,\varphi ),\;\;\;\;\;\;\;\;\;\theta = 0. \end{array} \right. $

于是(3.1) 式可写成如下形式

$ \dot{x_t}=A(\mu)x_t+Rx_t, $

(3.4)

其中 $x=(x_1, x_2, x_3)^T.$对 $\alpha\in{C([-1, 0], ({R^3})^*)}, $定义

$ A^*\alpha(s)=\left\{\begin{array}{lc} -\frac{d\alpha(s)}{ds}, &\mbox{$s\in{(0, 1]}$}, \\ \displaystyle\int_{-1}^0d\eta^T(t, 0)\varphi(-t), &\mbox{$s=0$}. \end{array}\right. $

对 $\varphi\in{C([-1, 0], R^3)}$和 $\psi\in{C([0, 1], ({R^3})^*)}$, 定义双线性积

$ \langle \psi ,\varphi \rangle = \bar \psi (0)\varphi (0) - \int_{ - 1}^0 {\int_{\xi = 0}^\theta {{\psi ^T}} (\xi - \theta )d\eta (\theta )\varphi (\xi )d\xi ,} $

这里 $\eta(\theta)=\eta(\theta, 0)$, 则算子 $A=A(0)$与 $A^*$是共轭算子.

由第2节的讨论及变换 $t=s\tau$可知, $ \pm {\rm{i}}{\omega _k}\tau _k^{(j)}$是算子 $A=A(0)$的特征值, 且 $A$的其它的特征值都具有严格负实部, 从而 $ \pm {\rm{i}}{\omega _k}\tau _k^{(j)}$也是算子 $A^*$的特征值, 于是有

引理3.1   $q(\theta ) = {(1, \alpha, \beta )^T}{e^{{\rm{i}}{\omega _k}\tau _k^{(j)}\theta }}$是算子 $A(0)$关于 $\omega_k\tau_k^{(j)}{\rm i}$的特征向量, $q^{*}(s)=D(1, \alpha^*, \beta^*)^{T}e^{i\omega_k\tau_k^{(j)}s}$是算子 $A^*$关于 $-\omega_k\tau_k^{(j)}{\rm i}$的特征向量并且 $\langle {q^*}, q\rangle = 1, \langle {\mathit{q}^*}, \bar q\rangle = 0, $其中

$ \begin{eqnarray*} \alpha&=&-\frac{q_1e^{-{\rm i}\omega_k\tau_k^{(j)}}}{{\rm i}\omega_k-q_2e^{-{\rm i}\omega_k\tau_k^{(j)}}}, \beta=-\frac{r_1}{{\rm i}\omega_k-r_2e^{-{\rm i}\omega_k\tau_k^{(j)}}}, \\ \alpha^*&=&-\frac{p_2}{q_2e^{-{\rm i}\omega_k\tau_k^{(j)}}+{\rm i}\omega_k}, \beta^*=-\frac{p_3}{r_3e^{-{\rm i}\omega_k\tau_k^{(j)}}+{\rm i}\omega_k}, \\ D&=&\frac{1}{1+\bar{\alpha}\alpha^*+\bar{\beta}\beta^* +(\bar{q_1}\alpha^*+r_1\beta^*+\bar{q_2}\bar{\alpha}\alpha^*+r_1+r_2\bar{\beta}\beta^*)e^{{\rm i}\omega_k\tau_k^{(j)}}}. \end{eqnarray*} $

上述引理的证明直接计算即可, 故省略.

令 $z(t)=\langle q^*, x_t\rangle, $其中 $x_t$是方程(3.4) 当 $\mu=0$时的解, 则由文[25]有

$ \frac{dz(t)}{dt}={\rm i}\tau_k^{(j)}\omega_kz(t)+\bar{q^*}(0) \bar{f}(z, \bar{z}), $

(3.5)

其中 $\bar{f}(z, \bar{z})=h[0, W(z, \bar{z})+2\mbox{Re}{\{zq\}}], $ $W(z, \bar{z})=x_t-2\mbox{Re}{\{zq\}}.$而

$ W(z, \bar{z})=W_{20}\frac{z^2}{2}+W_{11}z\bar{z}+W_{02} \frac{\bar{z}^2}{2}+\cdots. $

(3.6)

再把(3.5) 式改写为

$ \frac{dz(t)}{dt}={\rm i}\tau_k^{(j)}\omega_kz(t)+g(z, \bar{z}), $

其中

$ g(z, \bar{z})=g_{20}\frac{z^2}{2}+g_{11}z\bar{z}+g_{02} \frac{\bar{z}^2}{2}+g_{21}\frac{z^2\bar{z}}{2}+\cdots. $

(3.7)

又

$ W^{'}=x_t^{'}+z^{'}q-\bar{z}^{'}\bar{q}. $

(3.8)

将(3.4), (3.5) 式代入(3.8) 式得

$ \begin{array}{l} W' = \left\{ \begin{array}{l} AW - 2{\rm{Re}}\{ {{\bar q}^*}(0)\bar fq(\theta )\}, \;\;\;\;\;\;\;\;\;\;\; - 1 \le \theta < 0, \\ AW - 2{\rm{Re}}\{ {{\bar q}^*}(0)\bar fq(\theta )\} + \bar f, \;\;\;\;\;\;\;\;\;\theta = 0, \end{array} \right.\\ \;\;\;\;\;\mathop = \limits^{{\bf{def}}} AW + H(z, \bar z, \theta ), \end{array} $

(3.9)

其中

$ \begin{eqnarray}H(z, \bar{z}, \theta)=H_{20}\frac{z^2}{2}+H_{11}z\bar{z}+H_{02} \frac{\bar{z}^2}{2}+\cdots\end{eqnarray} $

(3.10)

且 $W^{'}=W_zz^{'}+W_{\bar{z}}{\bar{z}}^{'}= AW+H(z, \bar{z}, \theta).$将上述相应的级数展式代入, 比较系数得

$ (A-2{\rm i}\tau_k\omega_k)W_{20}=-H_{20}(\theta), $

(3.11)

$ AW_{11}(\theta)=-H_{11}(\theta). $

(3.12)

注意到 $x_t(\theta)=(x_{1t}(\theta), x_{3t}(\theta), x_{3t}(\theta))=W(z, \bar{z}, \theta)+zq(\theta)+\bar{z}\bar{q}(\theta)$和 $q(\theta ) = {(1, \alpha, \beta )^T}{e^{{\rm{i}}\theta {\omega _k}\tau _k^{(j)}}}, $有

$ \begin{array}{*{20}{l}} {\;\;{x_{1t}}(0)\;\;\; = \;\;z + \bar z + W_{20}^{(1)}(0)\frac{{{z^2}}}{2} + W_{11}^{(1)}(0)z\bar z + W_{02}^{(1)}(0)\frac{{{{\bar z}^2}}}{2} + O(|z,\bar z{|^3}),}\\ {\;\;{x_{2t}}(0)\;\;\; = \;\;z\alpha + \bar z\bar \alpha + W_{20}^{(2)}(0)\frac{{{z^2}}}{2} + W_{11}^{(2)}(0)z\bar z + W_{02}^{(2)}(0)\frac{{{{\bar z}^2}}}{2} + O(|z,\bar z{|^3}),}\\ {\;\;{x_{3t}}(0)\;\;\; = \;\;z\beta + \bar z\bar \beta + W_{20}^{(3)}(0)\frac{{{z^2}}}{2} + W_{11}^{(3)}(0)z\bar z + W_{02}^{(3)}(0)\frac{{{{\bar z}^2}}}{2} + O(|z,\bar z){|^3},}\\ {{x_{1t}}( - 1)\;\; = \;\;z{e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + \bar z{e^{{\rm{i}}{\omega _k}\tau _k^{(j)}}} + W_{20}^{(1)}( - 1)\frac{{{z^2}}}{2} + W_{11}^{(1)}( - 1)z\bar z + W_{02}^{(1)}( - 1)\frac{{{{\bar z}^2}}}{2} + O(|z,\bar z{|^3}),}\\ {{x_{2t}}( - 1)\;\; = \;\;z\alpha {e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + \bar z\bar \alpha {e^{{\rm{i}}{\omega _k}\tau _k^{(j)}}} + W_{20}^{(2)}( - 1)\frac{{{z^2}}}{2} + W_{11}^{(2)}( - 1)z\bar z}\\ {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; + W_{02}^{(2)}( - 1)\frac{{{{\bar z}^2}}}{2} + O(|z,\bar z{|^3}),}\\ {{x_{3t}}( - 1)\;\; = \;\;z\beta {e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + \bar z\bar \beta {e^{{\rm{i}}{\omega _k}\tau _k^{(j)}}} + W_{20}^{(3)}( - 1)\frac{{{z^2}}}{2} + W_{11}^{(3)}( - 1)z\bar z + W_{02}^{(3)}( - 1)\frac{{{{\bar z}^2}}}{2}}\\ {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; + O(|z,\bar z{|^3}).} \end{array} $

于是

$ \begin{array}{l} \;\;\;\;g(z,\bar z) = {{\bar q}^*}(0)\bar f(z,\bar z)\\ = \bar D\tau _k^{(j)}(1,\overline {{\alpha ^*}} ,\overline {{\beta ^*}} )\left( \begin{array}{l} {p_4}x_{1t}^2(0) + {p_5}x_{2t}^2(0) + {p_6}x_{3t}^2(0) + {p_7}{x_{1t}}(0){x_{2t}}(0) + {p_8}{x_{1t}}(0){x_{3t}}(0)\\ \;\;\;\; + {p_9}x_{1t}^2(0){x_{2t}}(0) + {p_{10}}x_{2t}^2(0){x_{1t}}(0) + {p_{11}}x_{3t}^2(0){x_{1t}}(0) + {\rm{h}}{\rm{.o}}{\rm{.t}}{\rm{.}}\\ \;\;\;\;\;\;\;\;{q_3}x_{1t}^2( - 1) + {q_4}x_{2t}^2( - 1) + {q_5}{x_{1t}}( - 1){x_{2t}}( - 1) + {q_6}x_{1t}^3( - 1)\\ \;\;\;\;\; + {q_7}x_{2t}^3( - 1) + {q_8}x_{1t}^2( - 1){x_{2t}}( - 1) + {q_9}{x_{1t}}( - 1)x_{2t}^2( - 1) + {\rm{h}}{\rm{.o}}{\rm{.t}}{\rm{.}}\\ \;\;\;\;\;\;\;\;{r_3}x_{1t}^2( - 1) + {r_4}x_{3t}^2( - 1) + {r_5}{x_{1t}}( - 1){x_{3t}}( - 1) + {r_6}x_{1t}^3( - 1)\\ \;\;\;\;\; + {r_7}x_{3t}^3( - 1) + {r_8}x_{1t}^2( - 1){x_{3t}}( - 1) + {r_9}{x_{1t}}( - 1)x_{3t}^2( - 1) + {\rm{h}}{\rm{.o}}{\rm{.t}}{\rm{.}} \end{array} \right)\\ = \bar D\tau _k^{(j)}\left\{ {\left[ {({p_4} + {p_5}{\alpha ^2} + {p_6}{\beta ^2} + {p_7}\alpha + {p_8}\beta ) + {\alpha ^*}({q_3} + {q_4}{\alpha ^2} + {q_5}\alpha ){e^{ - 2{\rm{i}}{\omega _k}\tau _k^{(j)}}}} \right.} \right.\\ \left. {\;\;\;\; + {\beta ^*}({r_3} + {r_4}{\beta ^2} + {q_5}\beta ){e^{ - 2{\rm{i}}{\omega _k}\tau _k^{(j)}}}} \right]{z^2} + \left[ {(2{p_4} + 2{p_5}|\alpha {|^2} + 2{p_6}|\beta {|^2} + 2{p_7}{\rm{Re}}\{ \alpha \} )} \right.\\ \left. {\;\;\;\; + {\alpha ^*}(2{q_3} + 2{q_4}|\alpha {|^2} + 2{q_5}{\rm{Re}}\{ \alpha \} ) + {\beta ^*}(2{r_3} + 2{r_4}|\alpha {|^2} + 2{r_5}{\rm{Re}}\{ \alpha \} )} \right]z\bar z\\ \;\;\;\; + \left[ {({p_4} + {p_5}{{\bar \alpha }^2} + {p_6}{{\bar \beta }^2} + {p_7}\bar \alpha + {p_8}\bar \beta ) + {\alpha ^*}({q_3} + {q_4}{{\bar \alpha }^2} + {q_5}\bar \alpha ){e^{2{\rm{i}}{\omega _k}\tau _k^{(j)}}}} \right.\\ \left. {\;\;\;\; + {\beta ^*}({r_3} + {r_4}{{\bar \beta }^2} + {r_5}\bar \beta ){e^{2{\rm{i}}{\omega _k}\tau _k^{(j)}}}} \right]{{\bar z}^2} + \left[ {{p_4}(W_{20}^{(1)}(0) + 2W_{11}^{(1)}(0)) + {p_5}(W_{20}^{(2)}(0) + 2W_{11}^{(2)}(0))} \right.\\ \;\;\;\; + {p_6}(W_{20}^{(3)}(0) + 2W_{11}^{(3)}(0)) + {p_7}(W_{11}^{(2)}(0) + \alpha W_{11}^{(1)}(0) + \frac{1}{2}W_{20}^{(2)}(0) + \frac{1}{2}\bar \alpha W_{20}^{(1)}(0))\\ \;\;\;\; + {p_8}(\beta W_{11}^{(1)}(0) + W_{11}^{(3)}(0) + \frac{1}{2}\bar \beta W_{20}^{(1)}(0) + \frac{1}{2}W_{20}^{(3)}(0)) + 2{p_9}{\rm{Re}}\{ \alpha \} + {p_{10}}({\alpha ^2} + 2|\alpha {|^2})\\ \;\;\;\; + {p_{11}}({\beta ^2} + 2|\beta {|^2}) + {\alpha ^*}({q_3}(2W_{11}^{(1)}( - 1){e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + W_{20}^{(1)}( - 1){e^{{\rm{i}}{\omega _k}\tau _k^{(j)}}})\\ \;\;\;\; + {q_4}(2\alpha W_{11}^{(2)}( - 1){e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + W_{20}^{(2)}( - 1)\bar \alpha {e^{2{\rm{i}}{\omega _k}\tau _k^{(j)}}}) + {q_5}(W_{11}^{(2)}( - 1){e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}}\\ \;\;\;\; + \frac{1}{2}W_{20}^{(2)}( - 1){e^{{\rm{i}}{\omega _k}\tau _k^{(j)}}} + \alpha W_{11}^{(1)}( - 1){e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}}) + 2{q_6}{e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + 2{q_7}{\beta ^2}\bar \beta {e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}}\\ \;\;\;\; + 2{q_8}{\rm{Re}}\{ \alpha \} {e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + {q_9}({\alpha ^2} + |\alpha {|^2}){e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + {\beta ^*}({q_3}(2W_{11}^{(1)}( - 1){e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}}\\ \;\;\;\; + W_{20}^{(1)}( - 1){e^{{\rm{i}}{\omega _k}\tau _k^{(j)}}}) + {q_4}(2\alpha W_{11}^{(2)}( - 1){e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + W_{20}^{(2)}( - 1)\bar \alpha {e^{2{\rm{i}}{\omega _k}\tau _k^{(j)}}})\\ \;\;\;\; + {q_5}(W_{11}^{(2)}( - 1){e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + \frac{1}{2}W_{20}^{(2)}( - 1){e^{{\rm{i}}{\omega _k}\tau _k^{(j)}}} + \alpha W_{11}^{(1)}( - 1){e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}})\\ \left. {\left. {\;\;\;\; + 2{q_6}{e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + 2{q_7}{\beta ^2}\bar \beta {e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + 2{q_8}{\rm{Re}}\{ \alpha \} {e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + {q_9}({\alpha ^2} + |\alpha {|^2}){e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}})} \right]{z^2}\bar z + \cdots } \right\}. \end{array} $

与(3.7) 式比较得

$ \begin{array}{l} {g_{20}} = 2\bar D\tau _k^{(j)}\left[{({p_4} + {p_5}{\alpha ^2} + {p_6}{\beta ^2} + {p_7}\alpha + {p_8}\beta ) + {\alpha ^*}({q_3} + {q_4}{\alpha ^2} + {q_5}\alpha ){e^{-2{\rm{i}}{\omega _k}\tau _k^{(j)}}}} \right.\\ \left. {\;\;\;\;\;\;\;\;\; + {\beta ^*}({r_3} + {r_4}{\beta ^2} + {q_5}\beta ){e^{-2{\rm{i}}{\omega _k}\tau _k^{(j)}}}} \right], \\ {g_{11}} = \bar D\tau _k^{(j)}\left[{(2{p_4} + 2{p_5}|\alpha {|^2} + 2{p_6}|\beta {|^2} + 2{p_7}{\rm{Re}}\{ \alpha \} ) + {\alpha ^*}(2{q_3} + 2{q_4}|\alpha {|^2} + 2{q_5}{\rm{Re}}\{ \alpha \} )} \right.\\ \left. {\;\;\;\;\;\;\;\;\; + {\beta ^*}(2{r_3} + 2{r_4}|\alpha {|^2} + 2{r_5}{\rm{Re}}\{ \alpha \} )} \right], \\ {g_{02}} = 2\bar D\tau _k^{(j)}\left[{({p_4} + {p_5}{{\bar \alpha }^2} + {p_6}{{\bar \beta }^2} + {p_7}\bar \alpha + {p_8}\bar \beta ) + {\alpha ^*}({q_3} + {q_4}{{\bar \alpha }^2} + {q_5}\bar \alpha ){e^{2{\rm{i}}{\omega _k}\tau _k^{(j)}}}} \right.\\ \left. {\;\;\;\;\;\;\;\;\; + {\beta ^*}({r_3} + {r_4}{{\bar \beta }^2} + {r_5}\bar \beta ){e^{2{\rm{i}}{\omega _k}\tau _k^{(j)}}}} \right], \\ {g_{21}} = 2\bar D\tau _k^{(j)}\left[{{p_4}(W_{20}^{(1)}(0) + 2W_{11}^{(1)}(0)) + {p_5}(W_{20}^{(2)}(0) + 2W_{11}^{(2)}(0)) + {p_6}(W_{20}^{(3)}(0)} \right.\\ \;\;\;\;\;\;\;\;\; + 2W_{11}^{(3)}(0)) + {p_7}(W_{11}^{(2)}(0) + \alpha W_{11}^{(1)}(0) + \frac{1}{2}W_{20}^{(2)}(0) + \frac{1}{2}\bar \alpha W_{20}^{(1)}(0))\\ \;\;\;\;\;\;\;\;\; + {p_8}(\beta W_{11}^{(1)}(0) + W_{11}^{(3)}(0) + \frac{1}{2}\bar \beta W_{20}^{(1)}(0) + \frac{1}{2}W_{20}^{(3)}(0)) + 2{p_9}{\rm{Re}}\{ \alpha \} \\ \;\;\;\;\;\;\;\;\; + {p_{10}}({\alpha ^2} + 2|\alpha {|^2}) + {p_{11}}({\beta ^2} + 2|\beta {|^2}) + {\alpha ^*}({q_3}(2W_{11}^{(1)}(-1){e^{-{\rm{i}}{\omega _k}\tau _k^{(j)}}}\\ \;\;\;\;\;\;\;\;\; + W_{20}^{(1)}(-1){e^{{\rm{i}}{\omega _k}\tau _k^{(j)}}}) + {q_4}(2\alpha W_{11}^{(2)}( - 1){e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + W_{20}^{(2)}( - 1)\bar \alpha {e^{2{\rm{i}}{\omega _k}\tau _k^{(j)}}})\\ \;\;\;\;\;\;\;\;\; + {q_5}(W_{11}^{(2)}( - 1){e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + \frac{1}{2}W_{20}^{(2)}( - 1){e^{{\rm{i}}{\omega _k}\tau _k^{(j)}}} + \alpha W_{11}^{(1)}( - 1){e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}})\\ \;\;\;\;\;\;\;\;\; + 2{q_6}{e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + 2{q_7}{\beta ^2}\bar \beta {e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + 2{q_8}{\rm{Re}}\{ \alpha \} {e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + {q_9}({\alpha ^2} + |\alpha {|^2}){e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}}\\ \;\;\;\;\;\;\;\;\; + {\beta ^*}({q_3}(2W_{11}^{(1)}( - 1){e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + W_{20}^{(1)}( - 1){e^{{\rm{i}}{\omega _k}\tau _k^{(j)}}}) + {q_4}(2\alpha W_{11}^{(2)}( - 1){e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}}\\ \;\;\;\;\;\;\;\;\; + W_{20}^{(2)}( - 1)\bar \alpha {e^{2{\rm{i}}{\omega _k}\tau _k^{(j)}}}) + {q_5}(W_{11}^{(2)}( - 1){e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + \frac{1}{2}W_{20}^{(2)}( - 1){e^{{\rm{i}}{\omega _k}\tau _k^{(j)}}}\\ \;\;\;\;\;\;\;\;\; + \alpha W_{11}^{(1)}( - 1){e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}}) + 2{q_6}{e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + 2{q_7}{\beta ^2}\bar \beta {e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}} + 2{q_8}{\rm{Re}}\{ \alpha \} {e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}}\\ \left. {\;\;\;\;\;\;\;\;\; + {q_9}({\alpha ^2} + |\alpha {|^2}){e^{ - {\rm{i}}{\omega _k}\tau _k^{(j)}}})} \right]. \end{array} $

下面来确定 $W_{11}(\theta), W_{20}(\theta).$

由(3.9) 式, 对 $\theta\in[-1, 0)$,

$ \begin{eqnarray}H(z, \bar{z}, \theta)=-\bar{q^*}(0)\bar{f}q(\theta)-q^*(0)\bar{f}\bar{q}(\theta) =-g(z, \bar{z})q(\theta)-\bar{g}(z, \bar{z})\bar{q}(\theta).\end{eqnarray} $

(3.13)

比较(3.10) 和(3.13) 式的同次幂系数, 得

$ H_{20}=-g_{20}q(\theta)-\bar{g}_{02}\bar{q}(\theta), $

(3.14)

$ H_{11}=-g_{11}q(\theta)-\bar{g}_{11}\bar{q}(\theta). $

(3.15)

由(3.10), (3.14) 式及 $A$的定义, 得

$ \dot{W}_{20}(\theta)=2{\rm i}\omega_k\tau_k^{(j)}W_{20}(\theta)+g_{20}q(\theta) +\bar{g_{02}}\bar{q}(\theta). $

(3.16)

注意到 $q(\theta )={{(1, \alpha, \beta )}^{T}}{{e}^{\rm{i}{{\omega }_{k}}\tau _{k}^{(j)}\theta }}$得

$ W_{20}(\theta)=\frac{{\rm i}g_{20}}{\omega_k\tau_k^{(j)}}q(0) e^{{\rm i}\omega_k\tau_k^{(j)}\theta}+\frac{i\bar{g}_{02}}{3\omega_k\tau_k^{(j)}} \bar{q}(0)e^{-i\omega_k\tau_k^{(j)}\theta} +E_1e^{2i\omega_k\tau_k^{(j)}\theta}, $

(3.17)

其中 $E_1=(E_1^{(1)}, E_1^{(2)}, E_1^{(3)})^T\in{R^3}$为常向量.同理, 由(3.10), (3.15) 式及 $A$的定义, 得

$ W_{11}(\theta)=-\frac{{\rm i}g_{11}}{\omega_k\tau_k^{(j)}}q(0) e^{{\rm i}\omega_k\tau_k^{(j)}\theta}+\frac{i\bar{g}_{11}}{\omega_k\tau_k^{(j)}} \bar{q}(0)e^{-i\omega_k\tau_k^{(j)}\theta}+E_2, $

(3.18)

其中 $E_2=(E_2^{(1)}, E_2^{(2)}, E_1^{(3)})^T\in{R^3}$为常向量.现在只需求(3.17) 式中的 $E_1$, (3.18) 式中的 $E_2$.由(3.11) 和(3.12) 式及 $A$的定义有

$ \int_{-1}^0d\eta(\theta)W_{20}(\theta)= 2{\rm i}\omega_k\tau_k^{(j)}W_{20}(0)-H_{20}(0), $

(3.19)

$ \int_{-1}^0d\eta(\theta)W_{11}(\theta)=-H_{11}(0), $

(3.20)

其中 $\eta(\theta)=\eta(0, \theta)$.由(3.9) 和(3.10) 式得

$ H_{20}(0)=-g_{20}q(0)-\bar{g_{02}}\bar{q}(0)+2\tau_k^{(j)}(H_1, H_2, H_3)^T, $

(3.21)

$ H_{11}(0)=-g_{11}q(0)-\bar{g_{11}}(0)\bar{q}(0)+2\tau_k^{(j)}(P_1, P_2, P_3)^T, $

(3.22)

这里

$ \begin{eqnarray*} H_1&=&p_4+p_5\alpha^2+p_6\beta^2+p_7\alpha+p_8\beta, H_2=q_3+q_4\alpha^2+q_5\alpha)e^{-2{\rm i}\omega_k\tau_k^{(j)}}, \\ H_3&=&(r_3+r_4\beta^2+q_5\beta)e^{-2{\rm i}\omega_k\tau_k^{(j)}}, \\ P_1&=&p_4+p_5|\alpha|^2+p_6|\beta|^2+p_7\mbox{Re}\{\alpha\}, P_2=q_3+q_4|\alpha|^2+q_5\mbox{Re}\{\alpha\}, \\ P_3&=&r_3+r_4|\alpha|^2+r_5\mbox{Re}\{\alpha\}. \end{eqnarray*} $

注意到

$ \left({\rm i}\omega_k\tau_k^{(j)}I-\int_{-1}^0e^{{\rm i}\theta\omega_k\tau_k^{(j)}\theta} d\eta(\theta)\right)q(0)=0, $

(3.23)

$ \left(-{\rm i}\omega_k\tau_k^{(j)}I-\int_{-1}^0e^{-{\rm i}\omega_k\tau_k^{(j)}\theta} d\eta(\theta)\right)\bar{q}(0)=0. $

(3.24)

把(3.17) 和(3.21) 式代入(3.19) 式得

$ \left(2{\rm i}\omega_k\tau_k^{(j)}I-\int_{-1}^0e^{2{\rm i}\omega_k\tau_k^{(j)}\theta} d\eta(\theta)\right)E_1=2\tau_k^{(j)}(H_1, H_2, H_3)^T. $

(3.25)

也就是

$ \left(2{\rm i}\omega_kI-B_1-B_2e^{-2{\rm i}\omega_k\tau_k^{(j)}}\right)E_1=2\tau_k^{(j)}(H_1, H_2, H_3)^T, $

(3.26)

从而

$ \left( {\begin{array}{*{20}{c}} {2{\rm{i}}{\omega _0} - {p_1}} & { - {p_2}} & { - {p_3}}\\ { - {q_1}{e^{ - 2{\rm{i}}{\omega _k}\tau _k^{(j)}}}} & {2{\rm{i}}{\omega _0} - {q_1}{e^{ - 2{\rm{i}}{\omega _k}\tau _k^{(j)}}}} & 0\\ { - {r_1}{e^{ - 2{\rm{i}}{\omega _k}\tau _k^{(j)}}}} & 0 & {2{\rm{i}}{\omega _0} - {r_2}{e^{ - 2{\rm{i}}{\omega _k}\tau _k^{(j)}}}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {E_1^{(1)}}\\ {E_1^{(2)}}\\ {E_1^{(3)}} \end{array}} \right) = 2\left( {\begin{array}{*{20}{c}} {{H_1}}\\ {{H_2}}\\ {{H_3}} \end{array}} \right). $

(3.27)

于是 $E_1^{(1)}=\frac{\Delta_{11}}{\Delta_1}, ~E_1^{(2)}=\frac{\Delta_{12}}{\Delta_1}, ~E_1^{(3)}=\frac{\Delta_{13}}{\Delta_1}, $这里

$ \begin{align} & {{\Delta }_{1}}=\det \left( \begin{matrix} 2\rm{i}{{\omega }_{0}}-{{p}_{1}} & -{{p}_{2}} & -{{p}_{3}} \\ -{{q}_{1}}{{e}^{-2\rm{i}{{\omega }_{k}}\tau _{k}^{(j)}}} & 2\rm{i}{{\omega }_{0}}-{{q}_{1}}{{e}^{-2\rm{i}{{\omega }_{k}}\tau _{k}^{(j)}}} & 0 \\ -{{r}_{1}}{{e}^{-2\rm{i}{{\omega }_{k}}\tau _{k}^{(j)}}} & 0 & 2\rm{i}{{\omega }_{0}}-{{r}_{2}}{{e}^{-2\rm{i}{{\omega }_{k}}\tau _{k}^{(j)}}} \\ \end{matrix} \right), \\ & {{\Delta }_{11}}=2\det \left( \begin{matrix} {{H}_{1}} & -{{p}_{2}} & -{{p}_{3}} \\ {{H}_{2}} & 2\rm{i}{{\omega }_{0}}-{{q}_{1}}{{e}^{-2\rm{i}{{\omega }_{k}}\tau _{k}^{(j)}}} & 0 \\ {{H}_{3}} & 0 & 2\rm{i}{{\omega }_{0}}-{{r}_{2}}{{e}^{-2\rm{i}{{\omega }_{k}}\tau _{k}^{(j)}}} \\ \end{matrix} \right), \\ & {{\Delta }_{12}}=2\det \left( \begin{matrix} 2\rm{i}{{\omega }_{0}}-{{p}_{1}} & {{H}_{1}} & -{{p}_{3}} \\ -{{q}_{1}}{{e}^{-2\rm{i}{{\omega }_{k}}\tau _{k}^{(j)}}} & {{H}_{2}} & 0 \\ -{{r}_{1}}{{e}^{-2\rm{i}{{\omega }_{k}}\tau _{k}^{(j)}}} & {{H}_{3}} & 2\rm{i}{{\omega }_{0}}-{{r}_{2}}{{e}^{-2\rm{i}{{\omega }_{k}}\tau _{k}^{(j)}}} \\ \end{matrix} \right), \\ & {{\Delta }_{13}}=2\det \left( \begin{matrix} 2\rm{i}{{\omega }_{0}}-{{p}_{1}} & -{{p}_{2}} & {{H}_{1}} \\ -{{q}_{1}}{{e}^{-2\rm{i}{{\omega }_{k}}\tau _{k}^{(j)}}} & 2\rm{i}{{\omega }_{0}}-{{q}_{1}}{{e}^{-2\rm{i}{{\omega }_{k}}\tau _{k}^{(j)}}} & {{H}_{2}} \\ -{{r}_{1}}{{e}^{-2\rm{i}{{\omega }_{k}}\tau _{k}^{(j)}}} & 0 & {{H}_{3}} \\ \end{matrix} \right). \\ \end{align} $

类似地, 把(3.18) 和(3.22) 式代入(3.20) 式, 得

$ \left(\int_{-1}^0 d\eta(\theta)\right)E_2=2(P_1, P_2)^T. $

(3.28)

于是

$ (B+C)E_2=2(-P_1, -P_2, -p_3)^T. $

(3.29)

也就是

$ \left( \begin{array}{ccc} p_1& p_2& p_3 \\ q_1& q_2& 0 \\ r_1& 0& r_2 \\ \end{array} \right) \left( \begin{array}{c} E_2^{(1)} \\ E_2^{(2)} \\ E_2^{(3)} \\ \end{array} \right)=2\left( \begin{array}{c} -P_1 \\ -P_2 \\ -P_3 \\ \end{array} \right). $

(3.30)

从而

$ E_2^{(1)}=\frac{\Delta_{21}}{\Delta_2}, ~E_2^{(2)}=\frac{\Delta_{22}}{\Delta_2}, E_2^{(3)}=\frac{\Delta_{23}}{\Delta_2}, $

其中

$ \begin{align} & {{\Delta }_{2}}=\det \left( \begin{matrix} {{p}_{1}} & {{p}_{2}} & {{p}_{3}} \\ {{q}_{1}} & {{q}_{2}} & 0 \\ {{r}_{1}} & 0 & {{r}_{2}} \\ \end{matrix} \right), ~~~{{\Delta }_{21}}=2\det \left( \begin{matrix} -{{P}_{1}} & {{p}_{2}} & {{p}_{3}} \\ -{{P}_{2}} & {{q}_{2}} & 0 \\ -{{P}_{3}} & 0 & {{r}_{2}} \\ \end{matrix} \right), \\ & {{\Delta }_{22}}=2\det \left( \begin{matrix} {{p}_{1}} & -{{P}_{1}} & {{p}_{3}} \\ {{q}_{1}} & -{{P}_{2}} & 0 \\ {{r}_{1}} & -{{P}_{3}} & {{r}_{2}} \\ \end{matrix} \right), ~~~{{\Delta }_{23}}=2\det \left( \begin{matrix} {{p}_{1}} & {{p}_{2}} & -{{P}_{1}} \\ {{q}_{1}} & {{q}_{2}} & -{{P}_{2}} \\ {{r}_{1}} & 0 & -{{P}_{3}} \\ \end{matrix} \right). \\ \end{align} $

根据(3.17) 和(3.18) 式, 可以计算 $g_{21}$的值, 于是可以计算如下数值

$ \begin{eqnarray*} c_1(0)&=&\frac{{\rm i}}{2\omega_k\tau_k^{(j)}}\left(g_{20}g_{11}-2|g_{11}|^2-\frac{|g_{02}|^2}{3}\right)+\frac{g_{21}}{2}, \\ \mu_2&=&-\frac{\mbox{Re}\{c_1(0)\}}{\mbox{Re}\{\lambda^{'}(\tau_k^{(j)})\} }, \\ \beta_2&=&2\mbox{Re}{(c_1(0))}, \\ T_2&=&-\frac{\mbox{Im}{\{c_1(0)\}}+\mu_2\mbox{Im}\{\lambda^{'}(\tau_k^{(j)})\}}{\omega_k\tau_k^{(j)}}. \end{eqnarray*} $

定理3.1  假设条件 $(\mbox{H}1)$和 $(\mbox{H}2)$成立, 则 $\mu=0$是系统(1.2) 的Hopf分支值.分支方向由 $\mu$确定, 如果 $\mu > 0$ $(\mu < 0)$, 则系统(1.2) 有超临界分支(次临界分支); 分支周期解稳定性由 $\beta_2$确定, 如果 $\beta_2 > 0$ $(\beta_2 < 0)$, 分支周期解是稳定的(不稳定的); 分支周期解的周期由 $T_2$确定, 如果 $T_2 > 0$ $(T_2 < 0)$, 分支周期解的周期是增加的(减小的).

在这一节, 取不同的时滞 $\tau$对系统(1.2) 进行数值模拟, 考虑下列系统

$ \left\{ \begin{array}{lc} \frac{dx_1(t)}{dt}=x_1(t)\left[0.5-0.2x_1(t)-\frac{3x_2(t)}{12x_2(t)+x_1(t)}-\frac{0.3x_3(t)}{10x_3(t)+x_1(t)}\right], \\ \frac{dx_2(t)}{dt}=x_2(t)\left[-0.2+\frac{0.82x_1(t-\tau_1)}{12x_2(t-\tau_1)+x_1(t-\tau_1)}\right], \\ \frac{dx_3(t)}{dt}=x_3(t)\left[-0.2+\frac{0.5x_1(t-\tau_2)}{10x_3(t-\tau_2)+x_1(t-\tau_2)}\right]. \end{array}\right. $

(4.1)

显然, 系统(4.1) 有正平衡点 $E^*(1.465, 0.375, 0.22)$并且满足定理2.1的条件, 经计算得 $\tau_0=8.5, \lambda^{'}(\tau_0)=0.1275-0.7182{\rm i}, $从而, $c_1(0)=-12.0332-21.2541{\rm i}, \mu_2 > 0, \beta_2 < 0, T_2 > 0.$故得到当 $\tau < \tau_0=8.5$时, 正平衡点 $E^*(1, 0.5)$是渐近稳定的, 当 $\tau$通过临界值 $\tau_0=8.5$时, 正平衡点 $E^*(1, 0.5)$失去稳定性, Hopf分支产生, 由 $\mu_2 > 0, \beta_2 < 0$知Hopf分支为超临界的, 分支方向为 $\tau > \tau_0$且从 $E^*(1.465, 0.375, 0.22)$附近产生的Hopf分支周期解是稳定的, 其数值模拟见图 1-2.

$ \begin{align} & {{p}_{1}}=\left[\frac{{{a}_{12}}x_{2}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{2}}}+\frac{{{a}_{13}}x_{3}^{*}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{2}}}-{{a}_{11}} \right]x_{1}^{*}, \\ & {{p}_{2}}=\left[\frac{{{a}_{12}}{{m}_{12}}x_{2}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{2}}}-\frac{{{a}_{12}}}{{{m}_{12}}x_{2}^{*}+x_{1}^{*}} \right]x_{1}^{*}, \\ & {{p}_{3}}=\left[\frac{{{a}_{13}}{{m}_{13}}x_{3}^{*}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{2}}}+\frac{{{a}_{13}}}{{{m}_{13}}x_{3}^{*}+x_{1}^{*}} \right]x_{1}^{*}, \\ & {{p}_{4}}=\left[\frac{{{a}_{12}}x_{2}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{2}}}+\frac{{{a}_{13}}x_{3}^{*}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{2}}}-{{a}_{11}} \right]\left[\frac{{{a}_{13}}x_{3}^{*}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{3}}}-\frac{{{a}_{12}}x_{2}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{2}}} \right], \\ & {{p}_{5}}=\left[\frac{{{a}_{12}}{{m}_{12}}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{2}}}-\frac{{{a}_{12}}}{m_{12}^{2}x_{2}^{*}+x_{1}^{*}} \right]x_{1}^{*}, {{p}_{6}}=\left[\frac{{{a}_{13}}{{m}_{13}}}{{{({{m}_{13}}x_{2}^{*}+x_{1}^{*})}^{2}}}+\frac{{{a}_{13}}}{{{(m_{13}^{2}x_{3}^{*}+x_{1}^{*})}^{3}}} \right]x_{1}^{*}, \\ & {{p}_{7}}=\left[\frac{{{a}_{12}}-2{{a}_{12}}x_{2}^{*}}{{{({{m}_{13}}x_{2}^{*}+x_{1}^{*})}^{2}}} \right]x_{1}^{*}+\left[\frac{{{a}_{12}}{{m}_{12}}x_{2}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{2}}}-\frac{{{a}_{12}}}{{{m}_{12}}x_{2}^{*}+x_{1}^{*}} \right], \\ & {{p}_{8}}=\left[\frac{{{a}_{13}}{{m}_{13}}x_{3}^{*}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{2}}}+\frac{{{a}_{13}}}{{{m}_{13}}x_{3}^{*}+x_{1}^{*}} \right]\left[\frac{{{a}_{13}}-2{{a}_{12}}{{m}_{13}}x_{3}^{*}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{3}}} \right]x_{1}^{*}, \\ & {{p}_{9}}=\left[\frac{3{{a}_{12}}{{m}_{12}}x_{2}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{3}}}-\frac{{{a}_{12}}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{2}}} \right]\left[\frac{{{a}_{12}}-2{{a}_{12}}x_{2}^{*}}{{{({{m}_{13}}x_{2}^{*}+x_{1}^{*})}^{2}}} \right]x_{1}^{*}, \\ & {{p}_{10}}=\left[\frac{3{{a}_{12}}m_{12}^{2}x_{2}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{3}}}-\frac{2{{a}_{12}}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{2}}} \right]\left[\frac{{{a}_{12}}{{m}_{12}}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{2}}}-\frac{{{a}_{12}}}{m_{12}^{2}x_{2}^{*}+x_{1}^{*}} \right]x_{1}^{*}, \\ & {{p}_{11}}=\left[\frac{3{{a}_{13}}m_{13}^{2}x_{3}^{*}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{4}}}-\frac{2{{a}_{13}}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{3}}} \right]\left[\frac{{{a}_{13}}{{m}_{13}}x_{3}^{*}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{2}}}+\frac{{{a}_{13}}}{{{m}_{13}}x_{3}^{*}+x_{1}^{*}} \right]x_{1}^{*}, \\ & {{q}_{1}}=\left[\frac{{{a}_{21}}}{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}-\frac{{{a}_{21}}x_{1}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{2}}} \right]x_{2}^{*}, {{q}_{2}}=-\left[\frac{{{a}_{21}}{{m}_{12}}x_{1}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{2}}} \right]x_{2}^{*}, \\ & {{q}_{3}}=\left[\frac{{{a}_{21}}m_{12}^{2}x_{1}^{*}x_{2}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{3}}}-\frac{{{a}_{21}}m12x_{1}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{2}}} \right], \\ & {{q}_{4}}=\left[\frac{2{{a}_{21}}{{m}_{12}}x_{1}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{3}}}-\frac{{{a}_{21}}}{{{({{m}_{12}}x_{3}^{*}+x_{1}^{*})}^{2}}} \right]\left[\frac{{{a}_{21}}}{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}-\frac{{{a}_{21}}x_{1}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{2}}}-{{a}_{11}} \right]x_{2}^{*}, \\ & {{q}_{5}}=\left[\frac{2{{a}_{21}}{{m}_{12}}x_{1}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{3}}}-\frac{{{a}_{21}}{{m}_{12}}}{{{(m_{12}^{2}x_{2}^{*}+x_{1}^{*})}^{2}}} \right]\left[\frac{{{a}_{21}}}{{{m}_{12}}x_{2}^{*}+x_{1}^{*}}-\frac{{{a}_{21}}x_{1}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{2}}} \right]x_{2}^{*}, \\ & {{q}_{6}}=\left[\frac{{{a}_{21}}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{3}}} \right]x_{2}^{*}, {{q}_{7}}=\left[\frac{{{a}_{21}}m_{12}^{3}x_{1}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{4}}} \right]x_{2}^{*}, \\ & {{q}_{8}}=\left[\frac{3{{a}_{21}}{{m}_{12}}x_{1}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{4}}}+\frac{2{{a}_{21}}{{m}_{12}}x_{1}^{*}}{{{(m_{12}^{2}x_{2}^{*}+x_{1}^{*})}^{3}}} \right]\left[\frac{{{a}_{21}}}{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}-\frac{{{a}_{21}}x_{1}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{2}}}-{{a}_{11}} \right]x_{2}^{*}, \\ & {{q}_{9}}=\left[\frac{3{{a}_{21}}m_{12}^{2}x_{1}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{4}}}+\frac{{{a}_{21}}m_{12}^{2}x_{1}^{*}}{{{(m_{12}^{2}x_{2}^{*}+x_{1}^{*})}^{3}}} \right]\left[\frac{2{{a}_{21}}{{m}_{12}}x_{1}^{*}}{{{({{m}_{12}}x_{2}^{*}+x_{1}^{*})}^{3}}}-\frac{{{a}_{21}}}{{{({{m}_{12}}x_{3}^{*}+x_{1}^{*})}^{2}}} \right]x_{2}^{*}, \\ & {{r}_{1}}=\left[\frac{{{a}_{31}}}{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}-\frac{{{a}_{31}}x_{1}^{*}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{2}}} \right]x_{2}^{*}, {{r}_{2}}=-\left[\frac{{{a}_{31}}{{m}_{13}}x_{1}^{*}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{2}}} \right]x_{3}^{*}, \\ & {{r}_{3}}=\left[\frac{{{a}_{31}}m_{13}^{2}x_{1}^{*}x_{3}^{*}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{3}}}-\frac{{{a}_{31}}m13x_{1}^{*}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{2}}} \right], \\ & {{r}_{4}}=\left[\frac{2{{a}_{31}}{{m}_{13}}x_{1}^{*}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{3}}}-\frac{{{a}_{31}}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{2}}} \right]\left[\frac{{{a}_{21}}}{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}-\frac{{{a}_{31}}x_{1}^{*}}{{{({{m}_{12}}x_{3}^{*}+x_{1}^{*})}^{2}}} \right]x_{2}^{*}, \\ & {{r}_{5}}=\left[\frac{2{{a}_{31}}{{m}_{13}}x_{1}^{*}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{3}}}-\frac{{{a}_{31}}{{m}_{13}}}{{{(m_{13}^{2}x_{2}^{*}+x_{1}^{*})}^{2}}} \right]\left[\frac{{{a}_{31}}}{{{m}_{13}}x_{3}^{*}+x_{1}^{*}}-\frac{{{a}_{31}}x_{1}^{*}}{{{({{m}_{13}}x_{2}^{*}+x_{1}^{*})}^{2}}} \right]x_{2}^{*}, \\ & {{r}_{6}}=\left[\frac{{{a}_{31}}}{{{({{m}_{13}}x_{2}^{*}+x_{1}^{*})}^{3}}} \right]x_{3}^{*}, {{r}_{7}}=\left[\frac{{{a}_{31}}m_{13}^{3}x_{1}^{*}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{4}}} \right]x_{2}^{*}, \\ & {{r}_{8}}=\left[\frac{3{{a}_{31}}{{m}_{13}}x_{1}^{*}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{4}}}+\frac{2{{a}_{31}}{{m}_{13}}x_{1}^{*}}{{{(m_{13}^{2}x_{3}^{*}+x_{1}^{*})}^{3}}} \right]\left[\frac{{{a}_{31}}}{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}-\frac{{{a}_{31}}x_{1}^{*}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{2}}} \right]x_{3}^{*}, \\ & {{r}_{9}}=\left[\frac{3{{a}_{31}}m_{13}^{2}x_{1}^{*}}{{{({{m}_{13}}x_{2}^{*}+x_{1}^{*})}^{4}}}+\frac{{{a}_{31}}m_{13}^{2}x_{1}^{*}}{{{(m_{13}^{2}x_{3}^{*}+x_{1}^{*})}^{3}}} \right]\left[\frac{2{{a}_{31}}{{m}_{13}}x_{1}^{*}}{{{({{m}_{13}}x_{2}^{*}+x_{1}^{*})}^{3}}}-\frac{{{a}_{31}}}{{{({{m}_{13}}x_{3}^{*}+x_{1}^{*})}^{2}}} \right]x_{3}^{*}. \\ \end{align} $

$ \begin{align} & {{k}_{0}}=s_{0}^{4}, {{k}_{1}}=2{{m}_{1}}s_{0}^{3}, \\ & {{k}_{2}}={{[{{n}_{1}}{{s}_{0}}-{{s}_{0}}{{n}_{0}}({{m}_{1}}+{{s}_{1}})]}^{2}}-2s_{0}^{2}{{n}_{0}}[{{s}_{0}}({{m}_{2}}+{{n}_{1}}{{s}_{1}}+{{n}_{2}}{{s}_{0}}) \\ & \ \ \ \ \ \ \ -{{m}_{2}}{{s}_{0}}{{n}_{0}}-{{n}_{1}}{{s}_{1}}]-2s_{0}^{2}(m-2{{s}_{0}}-{{s}_{1}}{{m}_{1}}+s_{1}^{2}), \\ & {{k}_{3}}=-2({{m}_{1}}{{n}_{1}}+{{s}_{0}}{{n}_{0}})[{{n}_{1}}{{s}_{0}}-{{s}_{0}}{{n}_{0}}({{m}_{1}}+{{s}_{1}})]+2{{m}_{1}}{{s}_{0}}(m-2{{s}_{0}}-{{s}_{1}}{{m}_{1}}+s_{1}^{2}), \\ & {{k}_{4}}=2({{m}_{1}}+{{s}_{1}})({{m}_{2}}+{{n}_{1}}{{s}_{1}}+{{n}_{2}}{{s}_{0}})[{{n}_{1}}{{s}_{0}}-{{s}_{0}}{{n}_{0}}({{m}_{1}}+{{s}_{1}})] \\ & ~~\ \ \ \ \ \ \ -2s_{0}^{2}{{n}_{0}}{{m}_{2}}({{m}_{2}}+{{n}_{1}}{{s}_{1}}+{{n}_{2}}{{s}_{0}})+2{{m}_{1}}{{s}_{0}}({{s}_{0}}-{{s}_{1}}-{{m}_{1}}{{m}_{2}}), \\ & {{k}_{5}}=-2({{m}_{1}}+{{s}_{1}})({{m}_{2}}+{{n}_{1}}{{s}_{1}}+{{n}_{2}}{{s}_{0}})({{m}_{1}}{{n}_{1}}+{{s}_{0}}{{n}_{0}})-2s_{0}^{2}{{m}_{2}} \\ & \ \ \ \ \ \ \ \ -2({{s}_{0}}-{{s}_{1}}-{{m}_{1}}{{m}_{2}})({{m}_{2}}{{s}_{0}}-{{s}_{1}}{{m}_{1}}+s_{1}^{2}), \\ & {{k}_{6}}={{s}_{0}}({{m}_{2}}+{{n}_{1}}{{s}_{1}}+{{n}_{2}}{{s}_{0}})-{{m}_{2}}{{s}_{0}}{{n}_{0}}-{{n}_{1}}{{s}_{1}} \\ & \ \ \ \ \ \ \ \ -2({{n}_{1}}-{{m}_{2}}-{{n}_{1}}{{s}_{1}}-{{n}_{2}}{{s}_{0}})({{m}_{1}}{{n}_{1}}+{{s}_{0}}{{n}_{0}})-({{s}_{0}}-{{s}_{1}}-{{m}_{1}}{{m}_{2}}), \\ & {{k}_{7}}=2{{m}_{2}}({{m}_{2}}{{s}_{0}}-{{m}_{1}}{{s}_{1}}+s_{1}^{2})-2({{n}_{1}}-{{m}_{2}}-{{n}_{1}}{{s}_{1}}-{{n}_{2}}{{s}_{0}}) \\ & \ \ \ \ \ \ \ \ \times ({{m}_{1}}+{{s}_{1}})({{m}_{2}}+{{n}_{1}}{{s}_{1}}+{{n}_{2}}{{s}_{0}}), \\ & {{k}_{8}}=2{{m}_{2}}({{s}_{0}}-{{s}_{1}}-{{m}_{1}}{{m}_{2}})-{{m}_{2}}{{({{m}_{2}}+{{n}_{1}}{{s}_{1}}+n-2{{s}_{0}})}^{2}} \\ & \ \ \ \ \ \ \ \ -{{({{n}_{1}}-{{m}_{2}}-{{n}_{1}}{{s}_{1}}-{{n}_{2}}{{s}_{0}})}^{2}}. \\ \end{align} $