﻿ 一类快慢耦合Duffing-van der Pol系统的平衡点分析

# 一类快慢耦合Duffing-van der Pol系统的平衡点分析Equilibrium Analysis of a Class of Fast-Slow Coupled Duffing-van der Pol System

Abstract: This paper studies the equilibrium point problem of a class of fast-slow coupled Duffing-van der Pol system with the coexistence of quintic term and cubic term. Quintic equation of one variable is obtained by solving the equilibrium points of the system. By using the method of reducing the order of n degree equation of one variable, the order of the equilibrium equation is reduced from five to four. When the equilibrium equation is reduced to the quartic, the coefficient of the constant term is calculated to be zero, so it can be directly reduced to the cubic equation. Morigane formula is used to solve the cubic equation. The distribution of the solution of the equilibrium equation on the F-u1 plane is obtained by synthesizing these four cases.

1. 引言

Duffing方程和van der Pol方程都是典型的非线性振动系统，已有不少学者对这两类方程的动力学行为进行研究 [1] - [6]。本文主要研究了一类5次项和3次项共存的快慢耦合Duffing-van der Pol系统的平衡点 [7] [8] 问题。我们在分析平衡点时，会遇到一元n次方程。

2. 数学模型与平衡点方程

$\left\{\begin{array}{l}\stackrel{˙}{x}=y,\\ \stackrel{˙}{y}={u}_{1}x+{u}_{2}y+{x}^{3}+{x}^{5}+a{x}^{2}y+b{x}^{4}y.\end{array}$ (1)

$\left\{\begin{array}{l}\stackrel{˙}{x}=y,\\ \stackrel{˙}{y}={u}_{1}x+{u}_{2}y+{x}^{3}+{x}^{5}+a{x}^{2}y+b{x}^{4}y+f\mathrm{cos}\omega t,\end{array}$ (2)

$\left\{\begin{array}{l}\stackrel{˙}{x}=y,\\ \stackrel{˙}{y}={u}_{1}x+{u}_{2}y+{x}^{3}+{x}^{5}+a{x}^{2}y+b{x}^{4}y+F.\end{array}$ (3)

$\left\{\begin{array}{l}y=0,\\ {u}_{1}x+{x}^{3}+{x}^{5}+F=0.\end{array}$ (4)

${x}^{5}=-{x}^{3}-{u}_{1}x-F.$ (5)

3. 一元 次方程的降次解法分析

${y}^{\left(5\right)}=-{y}^{\left(3\right)}-{u}_{1}{y}^{\left(1\right)}-Fy.$ (6)

$y\left(t\right)={\sum }_{j=1}^{5}{C}_{j}{Y}_{j}\left(t\right),$ (7)

${Y}_{j}\left(t\right)={t}^{j-1}+{\sum }_{m=1}^{\infty }{}^{5}{\int }_{0}^{t}\left(-{y}_{j\left(m-1\right)}^{\left(3\right)}-{u}_{1}{y}_{j\left(m-1\right)}^{\left(1\right)}-F{y}_{j\left(m-1\right)}\right){\left(\text{d}t\right)}^{5}\left(j=1,2,\cdots ,n\right).$ (8)

${\text{e}}^{\xi t}={\sum }_{j=1}^{5}{C}_{j}{Y}_{j}\left(t\right).$ (9)

$t=0$$t=1$$t=2$$t=3$$t=4$ 代入方程9，得到

${C}_{1}=1,{\text{e}}^{\xi }={\sum }_{j=1}^{5}{C}_{j}{Y}_{j}\left(1\right),\cdots ,{\text{e}}^{4\xi }={\sum }_{j=1}^{5}{C}_{j}{Y}_{j}\left(4\right).$ (10)

${Y}_{1}\left(1\right)=1+\left(-F\right)+{\left(-F\right)}^{2}+\cdots +{\left(-F\right)}^{m}+\cdots =1-\frac{F{\mathrm{lim}}_{m\to \infty }\left[1-{\left(-F\right)}^{m}\right]}{1+F}$ (11)

$\begin{array}{c}{Y}_{2}\left(1\right)=1+\left(-{u}_{1}-\frac{1}{2}F\right)+\left(-{u}_{1}-\frac{1}{2}F\right)\left(-F\right)+\cdots +\left(-{u}_{1}-\frac{1}{2}F\right){\left(-F\right)}^{m-1}+\cdots \\ =1-\left(-{u}_{1}-\frac{1}{2}F\right){Y}_{1}\left( 1 \right)\end{array}$

$⋮$

$\begin{array}{c}{b}_{0}=2\left[{Y}_{2}\left(1\right){Y}_{3}\left(1\right){Y}_{4}\left(2\right){Y}_{5}\left(3\right)-{Y}_{2}\left(1\right){Y}_{3}\left(1\right){Y}_{5}\left(2\right){Y}_{4}\left(3\right)+{Y}_{2}\left(1\right){Y}_{5}\left(1\right){Y}_{3}\left(2\right){Y}_{4}\left(3\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-{Y}_{2}\left(1\right){Y}_{5}\left(1\right){Y}_{4}\left(2\right){Y}_{3}\left(3\right)-{Y}_{3}\left(1\right){Y}_{4}\left(1\right){Y}_{2}\left(2\right){Y}_{5}\left(3\right)+{Y}_{3}\left(1\right){Y}_{4}\left(1\right){Y}_{5}\left(2\right){Y}_{2}\left(3\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{Y}_{4}\left(1\right){Y}_{5}\left(1\right){Y}_{2}\left(2\right){Y}_{3}\left(3\right)-{Y}_{4}\left(1\right){Y}_{5}\left(1\right){Y}_{3}\left(2\right){Y}_{2}\left(3\right)\right],\end{array}$

$\begin{array}{c}{b}_{1}=\frac{2}{{b}_{0}}\left[{Y}_{3}\left(1\right){Y}_{2}\left(2\right){Y}_{4}\left(3\right){Y}_{5}\left(4\right)-{Y}_{3}\left(1\right){Y}_{2}\left(2\right){Y}_{5}\left(3\right){Y}_{4}\left(4\right)-{Y}_{3}\left(1\right){Y}_{4}\left(2\right){Y}_{2}\left(3\right){Y}_{5}\left(4\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{Y}_{3}\left(1\right){Y}_{4}\left(2\right){Y}_{5}\left(3\right){Y}_{2}\left(4\right)+{Y}_{3}\left(1\right){Y}_{5}\left(2\right){Y}_{2}\left(3\right){Y}_{4}\left(4\right)-{Y}_{3}\left(1\right){Y}_{5}\left(2\right){Y}_{4}\left(3\right){Y}_{2}\left(4\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{Y}_{5}\left(1\right){Y}_{2}\left(2\right){Y}_{3}\left(3\right){Y}_{4}\left(4\right)-{Y}_{5}\left(1\right){Y}_{2}\left(2\right){Y}_{4}\left(3\right){Y}_{3}\left(4\right)-{Y}_{5}\left(1\right){Y}_{3}\left(2\right){Y}_{2}\left(3\right){Y}_{4}\left(4\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{Y}_{5}\left(1\right){Y}_{3}\left(2\right){Y}_{4}\left(3\right){Y}_{2}\left(4\right)+{Y}_{5}\left(1\right){Y}_{4}\left(2\right){Y}_{2}\left(3\right){Y}_{3}\left(4\right)-{Y}_{5}\left(1\right){Y}_{4}\left(2\right){Y}_{3}\left(3\right){Y}_{2}\left(4\right)\right],\end{array}$

$\begin{array}{c}{b}_{2}=\frac{2}{{b}_{0}}\left[{Y}_{2}\left(1\right){Y}_{3}\left(1\right){Y}_{4}\left(3\right){Y}_{5}\left(4\right)-{Y}_{2}\left(1\right){Y}_{3}\left(1\right){Y}_{5}\left(3\right){Y}_{4}\left(4\right)+{Y}_{2}\left(1\right){Y}_{5}\left(1\right){Y}_{3}\left(3\right){Y}_{4}\left(4\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-{Y}_{2}\left(1\right){Y}_{5}\left(1\right){Y}_{4}\left(3\right){Y}_{3}\left(4\right)-{Y}_{3}\left(1\right){Y}_{4}\left(1\right){Y}_{2}\left(3\right){Y}_{5}\left(4\right)+{Y}_{3}\left(1\right){Y}_{4}\left(1\right){Y}_{5}\left(3\right){Y}_{2}\left(4\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{Y}_{4}\left(1\right){Y}_{5}\left(1\right){Y}_{2}\left(3\right){Y}_{3}\left(4\right)-{Y}_{4}\left(1\right){Y}_{5}\left(1\right){Y}_{3}\left(3\right){Y}_{2}\left(4\right)\right],\end{array}$

$\begin{array}{c}{b}_{3}=\frac{2}{{b}_{0}}\left[{Y}_{2}\left(1\right){Y}_{3}\left(1\right){Y}_{4}\left(2\right){Y}_{5}\left(4\right)-{Y}_{2}\left(1\right){Y}_{3}\left(1\right){Y}_{5}\left(2\right){Y}_{4}\left(4\right)+{Y}_{2}\left(1\right){Y}_{5}\left(1\right){Y}_{3}\left(2\right){Y}_{4}\left(4\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-{Y}_{2}\left(1\right){Y}_{5}\left(1\right){Y}_{4}\left(2\right){Y}_{3}\left(4\right)-{Y}_{3}\left(1\right){Y}_{4}\left(1\right){Y}_{2}\left(2\right){Y}_{5}\left(4\right)+{Y}_{3}\left(1\right){Y}_{4}\left(1\right){Y}_{5}\left(2\right){Y}_{2}\left(4\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{Y}_{4}\left(1\right){Y}_{5}\left(1\right){Y}_{2}\left(2\right){Y}_{3}\left(4\right)-{Y}_{4}\left(1\right){Y}_{5}\left(1\right){Y}_{3}\left(2\right){Y}_{2}\left(4\right)\right],\end{array}$

$c=0.$

${z}^{4}+{b}_{3}{z}^{3}+{b}_{2}{z}^{2}+{b}_{1}z+c=0.$ (12)

${z}^{3}+{b}_{3}{z}^{2}+{b}_{2}z+{b}_{1}=0.$ (13)

${z}_{1}$${z}_{2}$${z}_{3}$${z}_{4}$ 是方程(12)的4个根，那么 $\mathrm{ln}{z}_{1}$$\mathrm{ln}{z}_{2}$$\mathrm{ln}{z}_{3}$$\mathrm{ln}{z}_{4}$ 就是方程(5)的4个根， $-\left(\mathrm{ln}{z}_{1}+\mathrm{ln}{z}_{2}+\mathrm{ln}{z}_{3}+\mathrm{ln}{z}_{4}\right)$ 是第5个根。

4. 平衡点分析

$\begin{array}{l}A={b}_{3}^{2}-3{b}_{2},\\ B={b}_{2}{b}_{3}-9{b}_{1},\\ C={b}_{2}^{2}-3{b}_{1}{b}_{3},\\ \text{Δ}={B}^{2}-4AC.\end{array}$ (14)

4.1. 情形①： $A=B=0$

$A=B=0$，即

${b}_{3}^{2}-3{b}_{2}=0.$ (15)

${b}_{2}{b}_{3}-9{b}_{1}=0.$ (16)

Figure 1. The range of z on the F-u1 plane under case ①: (a) No condition z > 0; (b) Have condition z > 0

4.2. 情形②： $\text{Δ}>0$

$\text{Δ}>0$，即

${\left({b}_{2}{b}_{3}-9{b}_{1}\right)}^{2}-4\left({b}_{3}^{2}-3{b}_{2}\right)\left({b}_{2}^{2}-3{b}_{1}{b}_{3}\right)>0$ (17)

Figure 2. Δ = 0 on the F-u1 plane

Figure 3. The range of z on the F-u1 plane under case ②

4.3. 情形③： $\text{Δ}=0$

$\text{Δ}=0$，即

${\left({b}_{2}{b}_{3}-9{b}_{1}\right)}^{2}-4\left({b}_{3}^{2}-3{b}_{2}\right)\left({b}_{2}^{2}-3{b}_{1}{b}_{3}\right)=0$ (18)

Figure 4. The range of z on the F-u1 plane under case ③

4.4. 情形④： $\text{Δ}<0$

$\text{Δ}<0$，即

${\left({b}_{2}{b}_{3}-9{b}_{1}\right)}^{2}-4\left({b}_{3}^{2}-3{b}_{2}\right)\left({b}_{2}^{2}-3{b}_{1}{b}_{3}\right)<0$ (19)

Figure 5. The range of z on the F-u1 plane under case ④

4.5. 解个数分析

Figure 6. Number distribution of x

5. 结论

NOTES

*通讯作者。

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