DOI:10.12677/AAM.2018.712187,PDF,HTML,XML,被引量下载: 1,409浏览: 6,339
作者:苏彩娴：广东技术师范学院计算机科学学院，广东 广州
关键词:Bogdanov-Takens分岔；规范型；普适开折；Chua’s系统；Bogdanov-Takens Bifurcation；Normal Form；Universal Unfolding；Chua’s System

文章引用：苏彩娴. Chua’s系统的Bogdanov-Takens分岔分析[J]. 应用数学进展, 2018, 7(12): 1600-1606. https://doi.org/10.12677/AAM.2018.712187

1. 引言

1983年，蔡少棠教授首次提出蔡氏电路 [1] ，这个系统拥有复杂的动力学行为，广泛应用于电子学方面 [2] 。本文主要基于参数依赖的中心流形，利用参数依赖的归一化方法 [3] [4] 来分析Chua’s系统的Bogdanov-Takens分岔。文章第二和第三部分分别介绍了对称系统Bogdanov-Takens (BT)分岔规范型和普适开折的计算公式，第四部分计算Chua’s系统相应分岔的规范型和普适开折，并画出它的分岔图。

2. 规范型计算公式

考虑以下 ${\text{Z}}_{\text{2}}$对称系统

$\stackrel{\dot{}}{x}=f\left(x,\alpha \right),$(1)

其中 $x\in R^n\left(n\ge 2\right)$， $\alpha \in R^2$， $f:R^n\times R^2\to R^n$充分光滑。向量场(1)在变换 $x\to -x$下保持不变，当 $\alpha ={\alpha }^{\ast }$时，有平衡点 $x=x^{\ast }=0$，系统的雅可比矩阵 $J\left(x,\alpha \right)$在 $\left(x^{\ast },{\alpha }^{\ast }\right)$处非零并且有二重零特征值，由向量场(1)的对称性，其可展开为如下形式：

$\stackrel{\dot{}}{x}=Ax+\frac{1}{6}C\left(x,x,x\right)+\cdots ,$(2)

其中 $A=J\left(0,{\alpha }^{\ast }\right)$， $C\left(x,z,w\right)={\displaystyle \underset{i,j,k=1}{\overset{n}{\sum }}\frac{{\partial }^{3}f\left(x^{\ast },{\alpha }^{\ast }\right)}{\partial x_i\partial x_j\partial x_k}{y}_i{z}_j{w}_k}$，其余类似。设(1)的规范型和普适开折分别 [5] [6] [7] [8]：

$\{\begin{array}{l}{\stackrel{\dot{}}{x}}_1=x_2,\\ {\stackrel{\dot{}}{x}}_2=c_{30}{x}_1^3+c_{21}{x}_1^2{x}_2,\end{array}$(3)

$\{\begin{array}{l}{\stackrel{\dot{}}{x}}_1=x_2,\\ {\stackrel{\dot{}}{x}}_2={\eta }_1{x}_1+{\eta }_2{x}_2+c_{30}{x}_1^3+c_{21}{x}_1^2{x}_2.\end{array}$(4)

定理1：根据临界中心流形不变性和Fredholm择一性定理，向量场(1)相应于BT分岔规范型(3)，其系数计算公式如下 [9]：

$c_{30}=\frac{1}{6}{p}_2^{{\rm T}}C\left(q_1,q_1,q_1\right),$(5)

$c_{21}=\frac{1}{2}{p}_2^{{\rm T}}C\left(q_1,q_{\text{1}},q_{\text{2}}\right)+\frac{1}{2}{p}_1^{{\rm T}}C\left(q_1,q_1,q_1\right),$(6)

其中 $q_1,q_2$和 $p_1,p_2$分别为A和A^T的广义特征向量，且满足：

$A{q}_1=0,A{q}_2=q_1,A^{{\rm T}}{p}_2=0,A^{{\rm T}}{p}_1=p_2,$(7)

$p_1^{{\rm T}}{q}_1=p_2^{{\rm T}}{q}_2=1,p_1^{{\rm T}}{q}_2=p_2^{{\rm T}}{q}_1=0.$(8)

3. 普适开折的计算公式

考虑参数 $\alpha$在 ${\alpha }^{\ast }$附近扰动，扰动量为 $\Delta =\alpha -{\alpha }^{\ast }$，此时(1)式可展开为如下形式：

$\stackrel{\dot{}}{x}=Ax+\frac{1}{6}C\left(x,x,x\right)+A_1\left(x,\Delta \right)+\frac{1}{6}{C}_1\left(x,x,x,\Delta \right)+\frac{1}{2}{A}_2\left(x,\Delta ,\Delta \right)+\cdots ,$(9)

其中 $A,C$如前所述， $A_1\left(y,\Delta \right)={\displaystyle \underset{i=1}{\overset{n}{\sum }}{\displaystyle \underset{j=1}{\overset{2}{\sum }}\frac{{\partial }^{2}f\left(0,{\alpha }^{\ast }\right)}{\partial x_i\partial {\alpha }_j}{y}_i{\Delta }_j}}$， $C_1,A_2,\cdots$类似。

将原始参数和开折参数的关系表示为 $\Delta =V\left(\eta \right)$，进一步采用如下平方逼近：

${\Delta }_i=V_i\left(\eta \right)=a_i{\eta }_1+b_i{\eta }_2+c_i{\eta }_1^2+d_i{\eta }_1{\eta }_2+e_i{\eta }_2^2+{\rm O}\left({\Vert \eta \Vert }^3\right),i=1,2.$(10)

定理2：设(1)的BT分岔是非退化的，相应于普适开折(4)，根据Fredholm择一定理，比较同调代数方程中 ${\omega }_1^i{\omega }_2^j{\eta }_3^k{\eta }_4^l\left(i+j=1,k+l=1,2\right)$项的系数得到如下线性代数方程 [10]：

$p_2^{{\rm T}}{A}_1\left(q_1,a\right)=1,$(11)

$p_2^{{\rm T}}{A}_1\left(q_1,b\right)=0,$(12)

$p_1^{{\rm T}}{A}_1\left(q_1,a\right)+p_2^{{\rm T}}{A}_1\left(q_2,a\right)=0,$(13)

$p_1^{{\rm T}}{A}_1\left(q_1,b\right)+q_2^{{\rm T}}{A}_1\left(q_2,b\right)=1.$(14)

由(11)~(14)可解得a，b，以下方程可以求得c，d和e：

$p_2^{{\rm T}}{A}_1\left(q_1,c\right)=p_2^{{\rm T}}\left(h_{0110}-A_1\left(h_{1010},a\right)-\frac{1}{2}{A}_2\left(q_1,a,a\right)\right),$(15)

$p_2^{{\rm T}}{A}_1\left(q_1,d\right)=p_2^{{\rm T}}\left(h_{0101}-A_1\left(h_{1001},a\right)-A_1\left(h_{1010},b\right)-A_2\left(q_1,a,b\right)\right),$(16)

$p_2^{{\rm T}}{A}_1\left(q_1,e\right)=-p_2^{{\rm T}}\left(A_1\left(h_{1001},b\right)+\frac{1}{2}{A}_2\left(q_1,b,b\right)\right),$(17)

$\left(p_1^{{\rm T}},p_2^{{\rm T}}\right)\left(\begin{array}{l}{A}_1\left(q_1,c\right)\\ A_1\left(q_2,c\right)\end{array}\right)=p_1^{{\rm T}}{h}_{0110}-\left(p_1^{{\rm T}},p_2^{{\rm T}}\right)\left(\begin{array}{l}{A}_1\left(h_{1010},a\right)\\ A_1\left(h_{0110},a\right)\end{array}\right)-\frac{1}{2}\left(p_1^{{\rm T}},p_2^{{\rm T}}\right)\left(\begin{array}{l}{A}_2\left(q_1,a,a\right)\\ A_2\left(q_2,a,a\right)\end{array}\right),$(18)

$\left(p_1^{{\rm T}},p_2^{{\rm T}}\right)\left(\begin{array}{l}{A}_1\left(q_1,d\right)\\ A_1\left(q_2,d\right)\end{array}\right)=\left(p_1^{{\rm T}},p_2^{{\rm T}}\right)\left(\begin{array}{l}{h}_{0101}\\ h_{0110}\end{array}\right)-\left(p_1^{{\rm T}},p_2^{{\rm T}}\right)\left(\begin{array}{l}{A}_1\left(h_{1001},a\right)+A_1\left(h_{1010},b\right)\\ A_1\left(h_{0101},a\right)+A_1\left(h_{0110},b\right)\end{array}\right)-\left(p_1^{{\rm T}},p_2^{{\rm T}}\right)\left(\begin{array}{l}{A}_2\left(q_1,a,b\right)\\ A_2\left(q_2,a,b\right)\end{array}\right),$(19)

$\left(p_1^{{\rm T}},p_2^{{\rm T}}\right)\left(\begin{array}{l}{A}_1\left(q_1,e\right)\\ A_1\left(q_2,e\right)\end{array}\right)=p_2^{{\rm T}}{h}_{0101}-\left(p_1^{{\rm T}},p_2^{{\rm T}}\right)\left(\begin{array}{l}{A}_1\left(h_{1001},b\right)\\ A_1\left(h_{0101},b\right)\end{array}\right)-\frac{1}{2}\left(p_1^{{\rm T}},p_2^{{\rm T}}\right)\left(\begin{array}{l}{A}_2\left(q_1,b,b\right)\\ A_2\left(q_2,b,b\right)\end{array}\right),$(20)

其中 $h_{ijkl}\left(i+j=1,k+l=1\right)$是如下奇异线性代数方程的任意解：

$p_2^{{\rm T}}\left(h_{1010},h_{1001}\right)=-p_1^{{\rm T}}\left(A_1\left(q_1,a\right),A_1\left(q_1,b\right)\right),$(21)

$p_2^{{\rm T}}\left(h_{0110},h_{0101}\right)=p_1^{{\rm T}}\left(h_{1010},h_{1001}\right)-p_1^{{\rm T}}\left(A_1\left(q_2,a\right),A_1\left(q_2,b\right)\right).$(22)

通过以上方程计算，并消去 $h_{10kl}\left(k+l=2\right)$，可以确定参数变换 $\Delta =V\left(\eta \right)$，从而求得开折参数 $\eta =V^{-1}\left(\Delta \right)$，且分岔的横截性条件为：

${\left|\frac{\partial \left({\eta }_1,{\eta }_2\right)}{\partial \left({\alpha }_1,{\alpha }_2\right)}\right|}_{\alpha ={\alpha }^{\ast }}={\left|\frac{\partial \left({\eta }_1,{\eta }_2\right)}{\partial \left({\Delta }_1,{\Delta }_2\right)}\right|}_{\Delta }=\frac{1}{\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|}\ne 0.$

如果(1)的双零分岔满足非退化和横截性条件，则当 $\alpha$在 ${\alpha }^{\ast }$附近扰动，普适开折(4)对不同的 $c_{30}\left(\Delta \right)$和 $c_{21}\left(\Delta \right)$，其分岔图和相图的拓扑结构与 $c_{30}=c_{30}\left(0\right)$和 $c_{21}=c_{21}\left(0\right)$时相同 [13] 。

4. Chua’s系统Bogdanov-Takens分岔分析

以下为本文研究的立方非线性Chua’s系统 [11] 。

$\{\begin{array}{l}{\stackrel{\dot{}}{x}}_1=\beta \left(x_2-\gamma x_1-\delta x_1^3\right),\\ {\stackrel{\dot{}}{x}}_2=x_1-x_2+x_3,\\ {\stackrel{\dot{}}{x}}_3=-h{x}_2.\end{array}$(23)

容易验证：当 ${\alpha }^{\ast }=\left(\beta ,0,\delta ,\beta \right)$时，系统(23)有平衡点 $x^{\ast }=0$，它的Jacobi矩阵在平衡点处为 $A=\left(\begin{array}{ccc}0& \beta & 0\\ 1& -1& 1\\ 0& -\beta & 0\end{array}\right)$。令 $\left|A-\lambda I\right|=0$，解得特征值 ${\lambda }_{1,2}=0$， ${\lambda }_3=-1$，根据(7)和(8)得到广义特征向量为：

$q_1=\left(\begin{array}{c}-v\\ 0\\ v\end{array}\right),q_2=\left(\begin{array}{c}-u-\frac{v}{\beta }\\ -\frac{v}{\beta }\\ u\end{array}\right),p_1=\left(\begin{array}{c}\frac{u\beta }{v^2}+\frac{\beta }{v}\\ -\frac{\beta }{v}\\ \frac{u\beta }{v^2}+\frac{\beta }{v}+\frac{1}{v}\end{array}\right),p_2=\left(\begin{array}{c}-\frac{\beta }{v}\\ 0\\ -\frac{\beta }{v}\end{array}\right),$

其中 $u,v\ne 0$为任意非零实数，计算 $C,A_1,C_1,A_2$为：

$\begin{array}{l}C\left(y,z,\omega \right)=\left(\begin{array}{c}-6\beta \delta y_1{z}_1{\omega }_1\\ 0\\ 0\end{array}\right),A_1\left(y,u\right)=\left(\begin{array}{c}-\beta y_1{u}_2+y_2{u}_1\\ 0\\ 0\end{array}\right),\\ C_1\left(y,z,w,u\right)=\left(\begin{array}{c}-6\delta y_1{z}_1{w}_1{u}_1\\ 0\\ 0\end{array}\right),A_2\left(y,u,v\right)=\left(\begin{array}{c}-y_1{u}_1{v}_2-y_1{u}_2{v}_1\\ 0\\ 0\end{array}\right).\end{array}$

由(5)和(6)得到(23)的系数规范型

$\{\begin{array}{l}{c}_{30}=\frac{1}{6}{p}_2^{{\rm T}}\left(q_1,q_1,q_1\right)=-v^2{\beta }^2\delta ,\\ c_{21}=\frac{1}{2}{p}_2^{{\rm T}}\left(q_1,q_1,q_2\right)=3v^2\beta \delta \left(\beta -1\right).\end{array}$(24)

系统(23)满足BT分岔非退化条件 $c_{30}{c}_{21}\ne 0$，从而 $\beta \delta \ne 0$和 $\beta \ne 1$时，然后扰动参数向量 $\left(\beta ,\gamma ,\delta ,h\right)$，由临界值 $\left(\beta ,0,\delta ,\beta \right)$变成 $\left(\beta +{\Delta }_1,{\Delta }_2,\delta ,\beta \right)$由(11)~(14)得到线性项的系数如下：

$a_1=1-\frac{1}{\beta },b_1=1,a_2=-\frac{1}{{\beta }^2},b_2=0,$

将以上系数代入(21)和(22)得

$h_{1010}=\left(\begin{array}{c}\frac{-u-v+\left(C_4-C_5+C_6-C_8\right){\beta }^2}{\beta }\\ -\frac{u}{\beta }\\ -\left(C_4-C_3+C_6-C_8\right)\beta \end{array}\right),$

$h_{1001}=\left(\begin{array}{c}u+\left(C_7-C_8\right)\beta +\left(\frac{1}{2}{C}_1-\frac{1}{2}{C}_2+\frac{1}{2}{C}_3-C_5\right){\beta }^2\\ -u-\left(C_7-C_8\right)\beta +\left(\frac{1}{2}{C}_1-\frac{1}{2}{C}_2+\frac{1}{2}{C}_3-C_5\right){\beta }^2\end{array}\right),$

$h_{0110}=\left(\begin{array}{c}\frac{-2u+2\left(C_4-C_5+C_6-C_8\right)\beta +\left(C_1-C_2+C_3\right){\beta }^2}{2\beta }\\ C_4-C_5+C_6-C_8\\ -\left(C_4-C_5+C_6-C_8\right)\beta \end{array}\right),$

$h_{0101}=\left(\begin{array}{c}\frac{-2v+2\left(C_7-C_8\right)\beta +\left(C_1-C_2+C_3-2C_5+2C_8\right){\beta }^2}{2\beta }\\ \left(\frac{1}{2}{C}_1-\frac{1}{2}{C}_2+\frac{1}{2}{C}_3-C_5\right)\beta \\ -u-\left(C_7-C_8\right)\beta -\left(\frac{1}{2}{C}_1-\frac{1}{2}{C}_2+\frac{1}{2}{C}_3-C_5\right){\beta }^2\end{array}\right).$

这里 $u,v$和 $C_i\left(i=1,2,\cdots ,8\right)$为任意实数，由上面(15)~(20)，得到二次项系数为：

$c_1=\frac{1}{{\beta }^2},d_1=-\frac{2}{\beta },e_1=1,c_2=\frac{2}{{\beta }^3}-\frac{1}{{\beta }^4},d_2=\frac{1}{{\beta }^2}+\frac{1}{{\beta }^3},e_2=0.$

令 ${\eta }_i=f_i{\Delta }_1+g_i{\Delta }_2+h_i{\Delta }_1^2+j_1{\Delta }_1{\Delta }_2+k_i{\Delta }_2^2+{\rm O}\left({\Vert \Delta \Vert }^3\right),i=1,2$，并代入 $\Delta =V\left(\eta \right)$，将以上两式代入(10)式，比较两端系数可以得到开折参数如下：

$\{\begin{array}{l}{\eta }_1=-{\beta }^2{\Delta }_2+{\beta }^4{\Delta }_2^2-\left(\beta -{\beta }^2\right){\Delta }_1{\Delta }_2+{\rm O}\left({\Vert \Delta \Vert }^3\right),\\ {\eta }_2={\Delta }_1-\left(\beta -{\beta }^2\right){\Delta }_2-{\Delta }_1^2-\left(1-2\beta +3{\beta }^2\right){\Delta }_1{\Delta }_2\\ +\left({\beta }^3-2{\beta }^4\right){\Delta }_2^2+{\rm O}\left({\Vert \Delta \Vert }^3\right).\end{array}$(25)

由于 $\frac{1}{\left|\begin{array}{l}\begin{array}{cc}{a}_1& b_1\end{array}\\ \begin{array}{cc}{a}_2& b_2\end{array}\end{array}\right|}=\frac{1}{{\beta }^2}\ne 0$，故横截性满足，从而当 $\beta \delta \ne 0$和 $\beta \ne 1$时， $\beta$和 $\gamma$可以作为系统(23)的分岔参数使系统发生完整双零分岔，本文可以计算2次精确度的开折参数，而文献 [12] 中计算的分岔参数仅是我们计算的线性部分。

最后，取 $\beta =2$， $\delta =\text{1}$，则系统参数变成 $\left(\beta ,\gamma ,\delta ,h\right)=\left(2+{\Delta }_1,{\Delta }_2,1,2\right)$，由上面(24)和(25)得相应系数 $c_{30}=4v^2$， $c_{21}=-6v^2$， ${\eta }_1=-4{\Delta }_2+2{\Delta }_1{\Delta }_2+16{\Delta }_2^2+{\rm O}\left({\Vert \Delta \Vert }^3\right)$， ${\eta }_{\text{2}}={\Delta }_{\text{1}}+\text{2}{\Delta }_2-{\Delta }_1^2-9{\Delta }_1{\Delta }_2-24{\Delta }_2^2+{\rm O}\left({\Vert \Delta \Vert }^3\right)$.根据文献 [13] ，可得分岔曲线如下：

$R=\left\{\left({\eta }_1\left(\Delta \right),{\eta }_2\left(\Delta \right)\right)|{\eta }_1=0,{\eta }_2\ne 0\right\}=\left\{\left({\Delta }_1,{\Delta }_2\right)|{\Delta }_2=0,{\Delta }_1\ne 0\right\},$

$H=\left\{\left({\eta }_1\left(\Delta \right),{\eta }_2\left(\Delta \right)\right)|{\eta }_1=0,{\eta }_2<0\right\}=\left\{\left({\Delta }_1,{\Delta }_2\right)|{\Delta }_2=-\frac{1}{2}{\Delta }_1+\frac{5{\Delta }_1^2}{4}+{\rm O}\left({\Delta }_1^3\right),{\Delta }_1<0\right\},$

$HL=\left\{\left({\eta }_1\left(\Delta \right),{\eta }_2\left(\Delta \right)\right)|{\eta }_2=\frac{c_{21}}{5\left|c_{30}\right|}{\eta }_1+{\rm O}\left({\eta }_1^{3/2}\right),{\eta }_1<0\right\}=\left\{\left({\Delta }_1,{\Delta }_2\right)|{\Delta }_2=-\frac{5}{4}{\Delta }_1+\frac{205{\Delta }_1^2}{8}+{\rm O}\left({\Delta }_1^3\right),{\Delta }_1<0\right\}.$

其分岔图见图1。

5. 结束语

本文利用同调代数方法计算Chua’s系统的规范型和普适开折，计算出来的开折参数精确到2次项，相对于其他研究的计算精确度更高，计算方法也更加简便。最后利用开折参数来分析Chua’s系统的分岔，并画出它的分岔图。

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