# 一类环上的三自由度耦合van der Pol方程的同伦分析方法The Homotopy Analysis Method for a Class of Three-Degree-of-Freedom Coupled van der Pol Oscillators

• 全文下载: PDF(2436KB)    PP.22-39   DOI: 10.12677/DSC.2020.91003
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In this paper, the approximate expression of periodic solution for a class of 3-DOF coupled van der Pol oscillator ring system is studied by homotopy analysis method. We first give the general equation of the system, and then divide the general equation into four categories: First, all oscillators move synchronously; second, two oscillators move synchronously, while the third vibrator moves in an independent manner (except that it oscillates in the same period as the second); third, the phase difference between adjacent oscillators on the ring is one third-period motion. Fourth, the two oscillators are 1/2 cycle apart, and the third vibrates at twice their frequency. Four different types of van der Pol oscillators are used to illustrate the validity and wide application of the homotopy analysis method.

1. 引言

Figure 1. The ring of three-degree-of-freedom van der Pol oscillators with coupled

2. 同伦分析方法简述

$M\stackrel{¨}{q}+G\stackrel{˙}{q}+Kq=F\left(q\right),$ (2-1)

$N\left[u\left(r,t\right)\right]=M\frac{{\partial }^{2}u\left(r,t\right)}{\partial {t}^{2}}+G\frac{\partial u\left(r,t\right)}{\partial t}+Ku\left(r,t\right)-F\left[u\left(r,t\right)\right],$ (2-2)

$u\left(r,t\right)={\left({x}_{1}\left(t\right),\cdots ,{x}_{n}\left(t\right)\right)}^{\text{T}},$ (2-3)

$\frac{\partial u\left(r,t\right)}{\partial t}={\left(\frac{\text{d}{x}_{1}}{\text{d}t},\cdots ,\frac{\text{d}{x}_{n}}{\text{d}t}\right)}^{\text{T}},$ (2-4)

$\frac{{\partial }^{2}u\left(r,t\right)}{\partial {t}^{2}}={\left(\frac{{\text{d}}^{2}{x}_{1}}{\text{d}{t}^{2}},\cdots ,\frac{{\text{d}}^{2}{x}_{n}}{\text{d}{t}^{2}}\right)}^{\text{T}},$ (2-5)

$\left(1-p\right)\left\{L\left[\Phi \left(r,t,p\right)-{u}_{0}\left(r,t\right)\right]\right\}=phH\left(t\right)N\left[\Phi \left(r,t,p\right)\right],$ (2-6)

${u}_{m}\left(r,t\right)=\frac{1}{m!}{\frac{{\partial }^{m}\Phi \left(r,t,p\right)}{\partial {p}^{m}}|}_{p=0}$ (2-7)

$\Phi \left(r,t,p\right)={u}_{0}\left(r,t\right)+{\sum }_{m=1}^{+\infty }{u}_{m}\left(r,t\right){p}^{m}.$ (2-8)

$u\left(r,t\right)={u}_{0}\left(r,t\right)+{\sum }_{m=1}^{+\infty }{u}_{m}\left(r,t\right)\text{ }\text{ }.$ (2-9)

${u}_{m}=\left\{{u}_{0}\left(r,t\right),{u}_{1}\left(r,t\right),\cdots ,{u}_{m}\left(r,t\right)\right\}.$ (2-10)

$L\left[{u}_{m}\left(r,t\right)-{\chi }_{m}{u}_{m-1}\left(r,t\right)\right]=hH\left(t\right){R}_{m}\left({u}_{m-1},r,t\right),$ (2-11)

${\chi }_{m}=\left\{\begin{array}{l}0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }m\le 1\\ 1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}m>1\end{array}$ (2-12)

${R}_{m}\left({u}_{m-1},r,t\right)=\frac{1}{\left(m-1\right)!}{\frac{{\partial }^{m-1}N\left(\Phi \left(r,t,p\right)\right)}{\partial {p}^{m-1}}|}_{p=0}.$ (2-13)

3. 同伦分析方法的应用

$\begin{array}{l}{\mu }_{1}{\stackrel{¨}{x}}_{1}+{\epsilon }_{1}\left({x}_{1}^{2}-\lambda \right){\stackrel{˙}{x}}_{1}+{\eta }_{1}{x}_{1}={f}_{1}\left({x}_{1},{x}_{2},{x}_{3}\right)\\ {\mu }_{2}{\stackrel{¨}{x}}_{2}+{\epsilon }_{2}\left({x}_{2}^{2}-\lambda \right){\stackrel{˙}{x}}_{2}+{\eta }_{2}{x}_{2}={f}_{2}\left({x}_{1},{x}_{2},{x}_{3}\right)\\ {\mu }_{3}{\stackrel{¨}{x}}_{3}+{\epsilon }_{3}\left({x}_{3}^{2}-\lambda \right){\stackrel{˙}{x}}_{3}+{\eta }_{3}{x}_{3}={f}_{3}\left({x}_{1},{x}_{2},{x}_{3}\right)\end{array}$ (3-1)

$\begin{array}{l}{\mu }_{1}{\stackrel{¨}{x}}_{1}+{\epsilon }_{1}\left({x}_{1}^{2}-\lambda \right){\stackrel{˙}{x}}_{1}+{\eta }_{1}{x}_{1}={f}_{1}\left({x}_{1},{x}_{2},{x}_{3}\right)\\ {\mu }_{2}{\stackrel{¨}{x}}_{2}+{\epsilon }_{2}\left({x}_{2}^{2}-\lambda \right){\stackrel{˙}{x}}_{2}+{\eta }_{2}{x}_{2}={f}_{2}\left({x}_{1},{x}_{2},{x}_{3}\right)\\ {\mu }_{3}{\stackrel{¨}{x}}_{3}+{\epsilon }_{3}\left({x}_{3}^{2}-\lambda \right){\stackrel{˙}{x}}_{3}+{\eta }_{3}{x}_{3}={f}_{3}\left({x}_{1},{x}_{2},{x}_{3}\right)\end{array}$ (3-2)

$\begin{array}{l}{\mu }_{1}{\omega }^{2}{\stackrel{¨}{u}}_{1}+{\epsilon }_{1}\omega \left({u}_{1}^{2}-\lambda \right){\stackrel{˙}{u}}_{1}+{\eta }_{1}{u}_{1}={f}_{1}\left({u}_{1},{u}_{2},{u}_{3}\right)\\ {\mu }_{2}{\omega }^{2}{\stackrel{¨}{u}}_{2}+{\epsilon }_{2}\omega \left({u}_{2}^{2}-\lambda \right){\stackrel{˙}{u}}_{2}+{\eta }_{2}{u}_{2}={f}_{2}\left({u}_{1},{u}_{2},{u}_{3}\right)\\ {\mu }_{3}{\omega }^{2}{\stackrel{¨}{u}}_{3}+{\epsilon }_{3}\omega \left({u}_{3}^{2}-\lambda \right){\stackrel{˙}{u}}_{3}+{\eta }_{3}{u}_{3}={f}_{3}\left({u}_{1},{u}_{2},{u}_{3}\right)\end{array}$ (3-3)

${u}_{1}\left(0\right)=a,{{u}^{\prime }}_{1}\left(0\right)=b,{u}_{2}\left(0\right)=c,{{u}^{\prime }}_{2}\left(0\right)=d,{u}_{3}\left(0\right)=e,{{u}^{\prime }}_{3}\left(0\right)=0$ (3-4)

$\begin{array}{l}{u}_{1}\left(\tau \right)={\sum }_{k=0}^{+\infty }\left({a}_{1,k}\mathrm{cos}k\tau +{b}_{1,k}\mathrm{sin}k\tau \right)\\ {u}_{2}\left(\tau \right)={\sum }_{k=0}^{+\infty }\left({a}_{2,k}\mathrm{cos}k\tau +{b}_{2,k}\mathrm{sin}k\tau \right)\\ {u}_{3}\left(\tau \right)={\sum }_{k=0}^{+\infty }\left({a}_{3,k}\mathrm{cos}k\tau +{b}_{3,k}\mathrm{sin}k\tau \right)\end{array}$ (3-5)

$\begin{array}{l}{u}_{1,0}\left(\tau \right)={a}_{0}\mathrm{cos}\tau +{b}_{0}\mathrm{sin}\tau ,\\ {u}_{2,0}\left(\tau \right)={c}_{0}\mathrm{cos}\tau +{d}_{0}\mathrm{sin}\tau ,\\ {u}_{3,0}\left(\tau \right)={e}_{0}\mathrm{cos}\tau .\end{array}$ (3-6)

$L\left(\begin{array}{c}{\Phi }_{1}\left(\tau ,p\right)\\ {\Phi }_{2}\left(\tau ,p\right)\\ {\Phi }_{3}\left(\tau ,p\right)\end{array}\right)={\omega }_{0}^{2}\left(\begin{array}{c}\frac{{\partial }^{2}{\Phi }_{1}\left(\tau ,p\right)}{\partial {\tau }^{2}}+{\Phi }_{1}\left(\tau ,p\right)\\ \frac{{\partial }^{2}{\Phi }_{2}\left(\tau ,p\right)}{\partial {\tau }^{2}}+{\Phi }_{2}\left(\tau ,p\right)\\ \frac{{\partial }^{2}{\Phi }_{3}\left(\tau ,p\right)}{\partial {\tau }^{2}}+{\Phi }_{3}\left(\tau ,p\right)\end{array}\right),$ (3-7)

$L\left(\begin{array}{c}{C}_{1}\mathrm{sin}\tau +{C}_{2}\mathrm{cos}\tau \\ {C}_{3}\mathrm{sin}\tau +{C}_{4}\mathrm{cos}\tau \\ {C}_{5}\mathrm{sin}\tau +{C}_{6}\mathrm{cos}\tau \end{array}\right)=0.$ (3-8)

$\begin{array}{l}N\left(\begin{array}{c}{\Phi }_{1}\left(\tau ,p\right)\\ {\Phi }_{2}\left(\tau ,p\right)\\ {\Phi }_{3}\left(\tau ,p\right)\end{array}\right)\\ =\left(\begin{array}{c}{\mu }_{1}{\Omega }^{2}\left(p\right)\frac{{\partial }^{2}{\Phi }_{1}\left(\tau ,p\right)}{\partial {\tau }^{2}}+{\epsilon }_{1}\Omega \left(p\right)\left[{\Phi }_{1}^{2}\left(\tau ,p\right)-\lambda \right]\frac{\partial {\Phi }_{1}\left(\tau ,p\right)}{\partial \tau }+{\eta }_{1}{\Phi }_{1}\left(\tau ,p\right)-{f}_{1}\left({\Phi }_{1}\left(\tau ,p\right),{\Phi }_{2}\left(\tau ,p\right),{\Phi }_{3}\left(\tau ,p\right)\right)\\ {\mu }_{2}{\Omega }^{2}\left(p\right)\frac{{\partial }^{2}{\Phi }_{2}\left(\tau ,p\right)}{\partial {\tau }^{2}}+{\epsilon }_{2}\Omega \left(p\right)\left[{\Phi }_{2}^{2}\left(\tau ,p\right)-\lambda \right]\frac{\partial {\Phi }_{2}\left(\tau ,p\right)}{\partial \tau }+{\eta }_{2}{\Phi }_{2}\left(\tau ,p\right)-{f}_{2}\left({\Phi }_{1}\left(\tau ,p\right),{\Phi }_{2}\left(\tau ,p\right),{\Phi }_{3}\left(\tau ,p\right)\right)\\ {\mu }_{3}{\Omega }^{2}\left(p\right)\frac{{\partial }^{2}{\Phi }_{3}\left(\tau ,p\right)}{\partial {\tau }^{2}}+{\epsilon }_{3}\Omega \left(p\right)\left[{\Phi }_{3}^{2}\left(\tau ,p\right)-\lambda \right]\frac{\partial {\Phi }_{3}\left(\tau ,p\right)}{\partial \tau }+{\eta }_{3}{\Phi }_{3}\left(\tau ,p\right)-{f}_{3}\left({\Phi }_{1}\left(\tau ,p\right),{\Phi }_{2}\left(\tau ,p\right),{\Phi }_{3}\left(\tau ,p\right)\right)\end{array}\right)\end{array}$

$\left(1-p\right)L\left(\begin{array}{c}{\Phi }_{1}\left(\tau ,p\right)-{u}_{1,0}\left(\tau \right)\\ {\Phi }_{2}\left(\tau ,p\right)-{u}_{2,0}\left(\tau \right)\\ {\Phi }_{3}\left(\tau ,p\right)-{u}_{3,0}\left(\tau \right)\end{array}\right)=phN\left(\begin{array}{c}{\Phi }_{1}\left(\tau ,p\right)\\ {\Phi }_{2}\left(\tau ,p\right)\\ {\Phi }_{3}\left(\tau ,p\right)\end{array}\right),$ (3-9)

$\begin{array}{l}{\Phi }_{1}\left(0,p\right)=a\left(p\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\frac{\partial {\Phi }_{1}\left(\tau ,p\right)}{\partial \tau }|}_{\tau =0}=b\left(p\right),\\ {\Phi }_{2}\left(0,p\right)=c\left(p\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\frac{\partial {\Phi }_{2}\left(\tau ,p\right)}{\partial \tau }|}_{\tau =0}=d\left(p\right),\\ {\Phi }_{3}\left(0,p\right)=e\left(p\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\frac{\partial {\Phi }_{3}\left(\tau ,p\right)}{\partial \tau }|}_{\tau =0}=0.\end{array}$ (3-10)

${\Phi }_{1}\left(\tau ,0\right)={u}_{1,0}\left(\tau \right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\Phi }_{2}\left(\tau ,0\right)={u}_{2,0}\left(\tau \right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\Phi }_{3}\left(\tau ,0\right)={u}_{3,0}\left(\tau \right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Omega \left(0\right)={\omega }_{0}.$ (3-11)

${\Phi }_{1}\left(\tau ,1\right)={u}_{1}\left(\tau \right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\Phi }_{2}\left(\tau ,1\right)={u}_{2}\left(\tau \right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\Phi }_{3}\left(\tau ,1\right)={u}_{3}\left(\tau \right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Omega \left(1\right)=\omega .$ (3-12)

$\begin{array}{l}{\Phi }_{1}\left(\tau ,p\right)={u}_{1,0}\left(\tau \right)+{\sum }_{m=1}^{\infty }{u}_{1,m}\left(\tau \right){p}^{m},\\ {\Phi }_{2}\left(\tau ,p\right)={u}_{2,0}\left(\tau \right)+{\sum }_{m=1}^{\infty }{u}_{2,m}\left(\tau \right){p}^{m},\\ {\Phi }_{3}\left(\tau ,p\right)={u}_{3,0}\left(\tau \right)+{\sum }_{m=1}^{\infty }{u}_{3,m}\left(\tau \right){p}^{m},\\ \Omega \left(p\right)={\omega }_{0}+{\sum }_{m=1}^{\infty }{\omega }_{m}{p}^{m}\text{}\end{array}$ (3-13)

$\begin{array}{l}{u}_{1,m}\left(\tau \right)=\frac{1}{m!}{\frac{{\partial }^{m}{\Phi }_{1}\left(\tau ,p\right)}{\partial {p}^{m}}|}_{p=0},\\ {u}_{2,m}\left(\tau \right)=\frac{1}{m!}{\frac{{\partial }^{m}{\Phi }_{2}\left(\tau ,p\right)}{\partial {p}^{m}}|}_{p=0},\\ {u}_{3,m}\left(\tau \right)=\frac{1}{m!}{\frac{{\partial }^{m}{\Phi }_{3}\left(\tau ,p\right)}{\partial {p}^{m}}|}_{p=0},\\ {\omega }_{m}=\frac{1}{m!}{\frac{{\partial }^{m}\Omega \left(p\right)}{\partial {p}^{m}}|}_{p=0}.\end{array}$ (3-14)

$\begin{array}{l}{u}_{1}\left(\tau \right)={u}_{1,0}\left(\tau \right)+{\sum }_{m=1}^{\infty }{u}_{1,m}\left(\tau \right),\\ {u}_{2}\left(\tau \right)={u}_{2,0}\left(\tau \right)+{\sum }_{m=1}^{\infty }{u}_{2,m}\left(\tau \right),\\ {u}_{3}\left(\tau \right)={u}_{3,0}\left(\tau \right)+{\sum }_{m=1}^{\infty }{u}_{3,m}\left(\tau \right),\\ \omega ={\omega }_{0}+{\sum }_{m=1}^{\infty }{\omega }_{m}.\end{array}$ (3-15)

$\begin{array}{l}{u}_{1,n}=\left\{{u}_{1,0}\left(\tau \right),{u}_{1,1}\left(\tau \right),\cdots ,{u}_{1,n}\left(\tau \right)\right\},\\ {u}_{2,n}=\left\{{u}_{2,0}\left(\tau \right),{u}_{2,1}\left(\tau \right),\cdots ,{u}_{2,n}\left(\tau \right)\right\},\\ {u}_{3,n}=\left\{{u}_{3,0}\left(\tau \right),{u}_{3,1}\left(\tau \right),\cdots ,{u}_{3,n}\left(\tau \right)\right\},\\ {\omega }_{n}=\left\{{\omega }_{0},{\omega }_{1},\cdots ,{\omega }_{n}\right\}.\end{array}$ (3-16)

$L\left(\begin{array}{c}{u}_{1,m}\left(\tau \right)-{\chi }_{m}{u}_{1,m-1}\left(\tau \right)\\ {u}_{2,m}\left(\tau \right)-{\chi }_{m}{u}_{2,m-1}\left(\tau \right)\\ {u}_{3,m}\left(\tau \right)-{\chi }_{m}{u}_{3,m-1}\left(\tau \right)\end{array}\right)=h\left(\begin{array}{c}{R}_{1,m}\left({u}_{1,m-1},{\omega }_{m-1}\right)\\ {R}_{2,m}\left({u}_{2,m-1},{\omega }_{m-1}\right)\\ {R}_{3,m}\left({u}_{3,m-1},{\omega }_{m-1}\right)\end{array}\right),$ (3-17)

$\left(\begin{array}{c}{R}_{1,m}\left({u}_{1,m-1},{\omega }_{m-1}\right)\\ {R}_{2,m}\left({u}_{2,m-1},{\omega }_{m-1}\right)\\ {R}_{3,m}\left({u}_{3,m-1},{\omega }_{m-1}\right)\end{array}\right)=\frac{1}{\left(m-1\right)!}{\frac{{\partial }^{m-1}N\left(\begin{array}{c}{\Phi }_{1}\left(\tau ,p\right)\\ {\Phi }_{2}\left(\tau ,p\right)\\ {\Phi }_{3}\left(\tau ,p\right)\end{array}\right)}{\partial {p}^{m-1}}|}_{p=0},$ (3-18)

$\begin{array}{l}{u}_{1,m}\left(0\right)={a}_{m},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{{u}^{\prime }}_{1,m}\left(0\right)={b}_{m},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{u}_{2,m}\left(0\right)={c}_{m},\\ {{u}^{\prime }}_{2,m}\left(0\right)={d}_{m},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{u}_{3,m}\left(0\right)={e}_{m},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{{u}^{\prime }}_{3,m}\left(0\right)=0,\left(m\ge 1\right)\end{array}$ (3-19)

$\begin{array}{l}\frac{1}{\pi }{\int }_{0}^{2\pi }{R}_{1,m}\left({u}_{1,m-1},{\omega }_{m-1}\right)\mathrm{cos}\tau \text{d}\tau =0,\\ \frac{1}{\pi }{\int }_{0}^{2\pi }{R}_{1,m}\left({u}_{1,m-1},{\omega }_{m-1}\right)\mathrm{sin}\tau \text{d}\tau =0,\\ \frac{1}{\pi }{\int }_{0}^{2\pi }{R}_{2,m}\left({u}_{2,m-1},{\omega }_{m-1}\right)\mathrm{cos}\tau \text{d}\tau =0,\\ \frac{1}{\pi }{\int }_{0}^{2\pi }{R}_{2,m}\left({u}_{2,m-1},{\omega }_{m-1}\right)\mathrm{sin}\tau \text{d}\tau =0,\\ \frac{1}{\pi }{\int }_{0}^{2\pi }{R}_{3,m}\left({u}_{3,m-1},{\omega }_{m-1}\right)\mathrm{cos}\tau \text{d}\tau =0,\\ \frac{1}{\pi }{\int }_{0}^{2\pi }{R}_{3,m}\left({u}_{3,m-1},{\omega }_{m-1}\right)\mathrm{sin}\tau \text{d}\tau =0,\end{array}$ (3-20)

$\begin{array}{l}{R}_{1,m}\left({u}_{1,m-1},{\omega }_{m-1}\right)={\sum }_{n=1}^{{\phi }_{1}\left(m\right)}\left[{a}_{m,n}^{1}\mathrm{cos}n\tau +{b}_{m,n}^{1}\mathrm{sin}n\tau \right]\text{ }\text{ },\\ {R}_{2,m}\left({u}_{2,m-1},{\omega }_{m-1}\right)={\sum }_{n=1}^{{\phi }_{2}\left(m\right)}\left[{a}_{m,n}^{2}\mathrm{cos}n\tau +{b}_{m,n}^{2}\mathrm{sin}n\tau \right]\text{ }\text{ },\\ {R}_{3,m}\left({u}_{3,m-1},{\omega }_{m-1}\right)={\sum }_{n=1}^{{\phi }_{2}\left(m\right)}\left[{a}_{m,n}^{3}\mathrm{cos}n\tau +{b}_{m,n}^{3}\mathrm{sin}n\tau \right]\text{ }\text{ }.\end{array}$ (3-21)

$\begin{array}{l}{a}_{m,n}^{1}=\frac{1}{\pi }{\int }_{0}^{2\pi }{R}_{1,m}\left({u}_{1,m-1},{\omega }_{m-1}\right)\mathrm{sin}\left(n\tau \right)\text{d}\tau ,\\ {b}_{m,n}^{1}=\frac{1}{\pi }{\int }_{0}^{2\pi }{R}_{1,m}\left({u}_{1,m-1},{\omega }_{m-1}\right)\mathrm{sin}\left(n\tau \right)\text{d}\tau ,\\ {a}_{m,n}^{2}=\frac{1}{\pi }{\int }_{0}^{2\pi }{R}_{2,m}\left({u}_{2,m-1},{\omega }_{m-1}\right)\mathrm{cos}\left(n\tau \right)\text{d}\tau ,\\ {b}_{m,n}^{2}=\frac{1}{\pi }{\int }_{0}^{2\pi }{R}_{2,m}\left({u}_{2,m-1},{\omega }_{m-1}\right)\mathrm{sin}\left(n\tau \right)\text{d}\tau ,\\ {a}_{m,n}^{3}=\frac{1}{\pi }{\int }_{0}^{2\pi }{R}_{3,m}\left({u}_{3,m-1},{\omega }_{m-1}\right)\mathrm{cos}\left(n\tau \right)\text{d}\tau ,\\ {b}_{m,n}^{3}=\frac{1}{\pi }{\int }_{0}^{2\pi }{R}_{3,m}\left({u}_{3,m-1},{\omega }_{m-1}\right)\mathrm{sin}\left(n\tau \right)\text{d}\tau ,\end{array}$ (3-22)

${\omega }_{-}=\omega +v.$

4. 数值模拟与讨论

4.1. 所有振子都同步运动

$\begin{array}{l}{f}_{1}\left({x}_{1},{x}_{2},{x}_{3}\right)={x}_{1}-{x}_{2}\\ {f}_{2}\left({x}_{1},{x}_{2},{x}_{3}\right)={x}_{2}-{x}_{3}\\ {f}_{3}\left({x}_{1},{x}_{2},{x}_{3}\right)={x}_{3}-{x}_{1}\end{array}$ (4-1)

Figure 2.The eighth-order approximation of versus is obtained for ${\mu }_{i}=1,{\epsilon }_{i}=1,{\eta }_{i}=1\left(i=1,2,3\right)$ and $\lambda =1$

Figure 3. Comparison of the phase curves of the eighth-order approximation with the numerical integration solution is given for ${\mu }_{i}=1,{\epsilon }_{i}=1,{\eta }_{i}=1,\lambda =1\left(i=1,2,3\right),{a}_{0}=2,{b}_{0}=0,{c}_{0}=2,{d}_{0}=2,{e}_{0}=2$ and ${\omega }_{0}=1$

Figure 4. Comparison of the time history responses of the eighth-order approximation with the numerical integration solution is given when ${\mu }_{i}=1,{\epsilon }_{i}=1,{\eta }_{i}=1,\lambda =1\left(i=1,2,3\right),{a}_{0}=2,{b}_{0}=0,{c}_{0}=2,{d}_{0}=2,{e}_{0}=2$ and ${\omega }_{0}=1$

$\begin{array}{l}{x}_{1}\left(0\right)=2.008600,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{{x}^{\prime }}_{1}\left(0\right)=0,\\ {x}_{2}\left(0\right)=2.008600,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{{x}^{\prime }}_{2}\left(0\right)=0,\\ {x}_{3}\left(0\right)=2.008600,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{{x}^{\prime }}_{3}\left(0\right)=0,\end{array}$

$\begin{array}{c}{x}_{1}\left(\tau \right)=1.881842\mathrm{cos}\stackrel{¯}{\omega }\tau +0.720593\mathrm{sin}\stackrel{¯}{\omega }\tau +0.720593\mathrm{cos}3\stackrel{¯}{\omega }\tau -1.881842\mathrm{sin}3\stackrel{¯}{\omega }\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+0.158456\mathrm{cos}5\stackrel{¯}{\omega }\tau -0.530028\mathrm{sin}5\stackrel{¯}{\omega }\tau -0.176676\mathrm{cos}7\stackrel{¯}{\omega }\tau -0.475368\mathrm{sin}7\stackrel{¯}{\omega }\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-0.020655\mathrm{cos}9\stackrel{¯}{\omega }\tau -0.218080\mathrm{sin}9\stackrel{¯}{\omega }\tau -0.043616\mathrm{cos}11\stackrel{¯}{\omega }\tau +0.103273\mathrm{sin}11\stackrel{¯}{\omega }\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-0.010428\mathrm{cos}13\stackrel{¯}{\omega }\tau -0.000519\mathrm{sin}13\stackrel{¯}{\omega }\tau -0.000074\mathrm{cos}15\stackrel{¯}{\omega }\tau +0.072994\mathrm{sin}15\stackrel{¯}{\omega }\tau ,\end{array}$

$\begin{array}{c}{x}_{2}\left(\tau \right)=1.881842\mathrm{cos}\stackrel{¯}{\omega }\tau +0.720593\mathrm{sin}\stackrel{¯}{\omega }\tau +0.720593\mathrm{cos}3\stackrel{¯}{\omega }\tau -1.881842\mathrm{sin}3\stackrel{¯}{\omega }\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+0.158456\mathrm{cos}5\stackrel{¯}{\omega }\tau -0.530028\mathrm{sin}5\stackrel{¯}{\omega }\tau -0.176676\mathrm{cos}7\stackrel{¯}{\omega }\tau -0.475368\mathrm{sin}7\stackrel{¯}{\omega }\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-0.020655\mathrm{cos}9\stackrel{¯}{\omega }\tau -0.218080\mathrm{sin}9\stackrel{¯}{\omega }\tau -0.043616\mathrm{cos}11\stackrel{¯}{\omega }\tau +0.103273\mathrm{sin}11\stackrel{¯}{\omega }\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-0.010428\mathrm{cos}13\stackrel{¯}{\omega }\tau -0.000519\mathrm{sin}13\stackrel{¯}{\omega }\tau -0.000074\mathrm{cos}15\stackrel{¯}{\omega }\tau +0.072994\mathrm{sin}15\stackrel{¯}{\omega }\tau ,\end{array}$

$\begin{array}{c}{x}_{3}\left(\tau \right)=1.881842\mathrm{cos}\stackrel{¯}{\omega }\tau +0.720593\mathrm{sin}\stackrel{¯}{\omega }\tau +0.720593\mathrm{cos}3\stackrel{¯}{\omega }\tau -1.881842\mathrm{sin}3\stackrel{¯}{\omega }\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+0.158456\mathrm{cos}5\stackrel{¯}{\omega }\tau -0.530028\mathrm{sin}5\stackrel{¯}{\omega }\tau -0.176676\mathrm{cos}7\stackrel{¯}{\omega }\tau -0.475368\mathrm{sin}7\stackrel{¯}{\omega }\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-0.020655\mathrm{cos}9\stackrel{¯}{\omega }\tau -0.218080\mathrm{sin}9\stackrel{¯}{\omega }\tau -0.043616\mathrm{cos}11\stackrel{¯}{\omega }\tau +0.103273\mathrm{sin}11\stackrel{¯}{\omega }\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-0.010428\mathrm{cos}13\stackrel{¯}{\omega }\tau -0.000519\mathrm{sin}13\stackrel{¯}{\omega }\tau -0.000074\mathrm{cos}15\stackrel{¯}{\omega }\tau +0.072994\mathrm{sin}15\stackrel{¯}{\omega }\tau ,\end{array}$

$\omega =0.942953.$

4.2. 两个振子同步运动，而第三个振子以一无关的方式运动(除它与第二个振子有相同周期地振动外)

$\begin{array}{l}{f}_{1}\left({x}_{1},{x}_{2},{x}_{3}\right)={x}_{1}-{x}_{2}\\ {f}_{2}\left({x}_{1},{x}_{2},{x}_{3}\right)={x}_{2}-{x}_{1}\\ {f}_{3}\left({x}_{1},{x}_{2},{x}_{3}\right)={x}_{1}-{x}_{2}-10{x}_{3}\end{array}$ (4-2)

Figure 5. The first-order approximation of ω versus h is obtained for ${\mu }_{i}=1,{\epsilon }_{i}=1,{\eta }_{i}=1\left(i=1,2,3\right)$ and $\lambda =1$

Figure 6. Comparison of the phase curves of the first-order approximation with the numerical integration solution is given for ${\mu }_{i}=1,{\epsilon }_{i}=1,{\eta }_{i}=1,\lambda =1\left(i=1,2,3\right),{a}_{0}=2,{b}_{0}=0,{c}_{0}=2,{d}_{0}=2,{e}_{0}=2$ and ${\omega }_{0}=1$

Figure 7. Comparison of the time history responses of the first-order approximation with the numerical integration solution is given when ${\mu }_{i}=1,{\epsilon }_{i}=1,{\eta }_{i}=1,\lambda =1\left(i=1,2,3\right),{a}_{0}=2,{b}_{0}=0,{c}_{0}=2,{d}_{0}=2,{e}_{0}=2$ and ${\omega }_{0}=1$

$\begin{array}{c}{x}_{1}\left(\tau \right)=1.844822\mathrm{cos}\stackrel{¯}{\omega }\tau -0.118627\mathrm{sin}\stackrel{¯}{\omega }\tau -0.273745\mathrm{cos}3\stackrel{¯}{\omega }\tau -1.844822\mathrm{sin}3\stackrel{¯}{\omega }\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+0.453207\mathrm{cos}5\stackrel{¯}{\omega }\tau -0.120863\mathrm{sin}5\stackrel{¯}{\omega }\tau -1.836718\mathrm{cos}7\stackrel{¯}{\omega }\tau -0.135962\mathrm{sin}7\stackrel{¯}{\omega }\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-0.047096\mathrm{cos}9\stackrel{¯}{\omega }\tau -0.125599\mathrm{sin}9\stackrel{¯}{\omega }\tau -0.144917\mathrm{cos}11\stackrel{¯}{\omega }\tau +0.235481\mathrm{sin}11\stackrel{¯}{\omega }\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+0.500366\mathrm{cos}13\stackrel{¯}{\omega }\tau -0.003442\mathrm{sin}13\stackrel{¯}{\omega }\tau -0.021553\mathrm{cos}15\stackrel{¯}{\omega }\tau -0.035026\mathrm{sin}15\stackrel{¯}{\omega }\tau ,\end{array}$

$\begin{array}{c}{x}_{2}\left(\tau \right)=1.844822\mathrm{cos}\stackrel{¯}{\omega }\tau -0.118627\mathrm{sin}\stackrel{¯}{\omega }\tau -0.273745\mathrm{cos}3\stackrel{¯}{\omega }\tau -1.844822\mathrm{sin}3\stackrel{¯}{\omega }\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+0.453207\mathrm{cos}5\stackrel{¯}{\omega }\tau -0.120863\mathrm{sin}5\stackrel{¯}{\omega }\tau -1.836718\mathrm{cos}7\stackrel{¯}{\omega }\tau -0.135962\mathrm{sin}7\stackrel{¯}{\omega }\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-0.047096\mathrm{cos}9\stackrel{¯}{\omega }\tau -0.125599\mathrm{sin}9\stackrel{¯}{\omega }\tau -0.144917\mathrm{cos}11\stackrel{¯}{\omega }\tau +0.235481\mathrm{sin}11\stackrel{¯}{\omega }\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+0.500366\mathrm{cos}13\stackrel{¯}{\omega }\tau -0.003442\mathrm{sin}13\stackrel{¯}{\omega }\tau -0.021553\mathrm{cos}15\stackrel{¯}{\omega }\tau -0.035026\mathrm{sin}15\stackrel{¯}{\omega }\tau ,\end{array}$

$\begin{array}{c}{x}_{3}\left(\tau \right)=1.124328\mathrm{cos}\stackrel{¯}{\omega }\tau +0.573061\mathrm{sin}\stackrel{¯}{\omega }\tau +0.134443\mathrm{cos}3\stackrel{¯}{\omega }\tau -1.131251\mathrm{sin}3\stackrel{¯}{\omega }\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+0.359854\mathrm{cos}5\stackrel{¯}{\omega }\tau -0.670447\mathrm{sin}5\stackrel{¯}{\omega }\tau -0.001711\mathrm{cos}7\stackrel{¯}{\omega }\tau -0.223488\mathrm{sin}7\stackrel{¯}{\omega }\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+0.271384\mathrm{cos}9\stackrel{¯}{\omega }\tau +0.348288\mathrm{sin}9\stackrel{¯}{\omega }\tau +0.054345\mathrm{cos}11\stackrel{¯}{\omega }\tau -0.059881\mathrm{sin}11\stackrel{¯}{\omega }\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-0.171917\mathrm{cos}13\stackrel{¯}{\omega }\tau -0.066372\mathrm{sin}13\stackrel{¯}{\omega }\tau +0.167520\mathrm{cos}15\stackrel{¯}{\omega }\tau -0.089623\mathrm{sin}15\stackrel{¯}{\omega }\tau ,\end{array}$

$\omega =1.208175.$

4.3. 环上相邻的振子之间彼此都相位差1/3周期的运动

$\begin{array}{l}{f}_{1}\left({x}_{1},{x}_{2},{x}_{3}\right)=\sqrt{3}/9{\left({x}_{1}+2{x}_{2}\right)}^{3}\\ {f}_{2}\left({x}_{1},{x}_{2},{x}_{3}\right)=\sqrt{3}/9{\left({x}_{2}+2{x}_{3}\right)}^{3}\\ {f}_{3}\left({x}_{1},{x}_{2},{x}_{3}\right)=\sqrt{3}/9{\left({x}_{3}+2{x}_{1}\right)}^{3}\end{array}$ (4-3)

Figure 8. Comparison of the phase curves of the first-order approximation with the numerical integration solution is given for ${\mu }_{i}=1,{\epsilon }_{i}=1,{\eta }_{i}=4,\lambda =1\left(i=1,2,3\right),{a}_{0}=-\sqrt{10}/5,{b}_{0}=\sqrt{30}/5,{c}_{0}=-\sqrt{10}/5,{d}_{0}=-\sqrt{30}/5,{e}_{0}=2\sqrt{10}/5$ and ${\omega }_{0}=2$

Figure 9. Comparison of the time history responses of the first-order approximation with the numerical integration solution is given for ${\mu }_{i}=1,{\epsilon }_{i}=1,{\eta }_{i}=4,\lambda =1\left(i=1,2,3\right),{a}_{0}=-\sqrt{10}/5,{b}_{0}=\sqrt{30}/5,{c}_{0}=-\sqrt{10}/5,{d}_{0}=-\sqrt{30}/5,{e}_{0}=2\sqrt{10}/5$ and ${\omega }_{0}=2$

Figure 10. Comparison of time history diagrams of ${x}_{1},{x}_{2}$ and ${x}_{3}$ in the same condition

${x}_{1}\left(0\right)=-0.632455,{{x}^{\prime }}_{1}\left(0\right)=1.095445,{x}_{2}\left(0\right)=-0.632455,{{x}^{\prime }}_{2}\left(0\right)=-1.095445,{x}_{3}\left(0\right)=1.264911,{{x}^{\prime }}_{3}\left(0\right)=0$

${x}_{1}\left(\tau \right)=-0.649572\mathrm{cos}\tau +1.089516\mathrm{sin}\tau +0.017116\mathrm{cos}3\tau +0.001976\mathrm{sin}3\tau ,$

${x}_{2}\left(\tau \right)=-0.615339\mathrm{cos}\tau -1.101374\mathrm{sin}\tau -0.017116\mathrm{cos}3\tau +0.001976\mathrm{sin}3\tau ,$

${x}_{3}\left(\tau \right)=1.264911\mathrm{cos}\tau +0.094868\mathrm{sin}\tau -0.031623\mathrm{sin}3\tau ,$

$\omega =2.$

4.4. 两个振子相位差1/2周期，而第三个振子2倍于它们的频率振动

$\begin{array}{l}{f}_{1}\left({x}_{1},{x}_{2},{x}_{3}\right)=\left(1-{x}_{1}^{2}\right){{x}^{\prime }}_{2}\\ {f}_{2}\left({x}_{1},{x}_{2},{x}_{3}\right)=\left(1-{x}_{2}^{2}\right){{x}^{\prime }}_{1}\\ {f}_{3}\left({x}_{1},{x}_{2},{x}_{3}\right)={x}_{1}^{3}{{x}^{\prime }}_{2}^{3}\end{array}$ (4-4)

Figure 11. Comparison of the phase curves of the first-order approximation with the numerical integration solution is given for ${\mu }_{i}=1,{\epsilon }_{i}=1,{\eta }_{i}=1,\lambda =1\left(i=1,2\right),{\mu }_{3}=1,{\epsilon }_{3}=1/16,{\eta }_{3}=4,{a}_{0}=1,{b}_{0}=0,{c}_{0}=-1,{d}_{0}=0,{e}_{0}=1,{\omega }_{0}=1$

Figure 12. Comparison of the time history responses of the first-order approximation with the numerical integration solution is given for ${\mu }_{i}=1,{\epsilon }_{i}=1,{\eta }_{i}=1,\lambda =1\left(i=1,2\right),{\mu }_{3}=1,{\epsilon }_{3}=1/16,{\eta }_{3}=4,{a}_{0}=1,{b}_{0}=0,{c}_{0}=-1,{d}_{0}=0,{e}_{0}=1$ and ${\omega }_{0}=1$

Figure 13. Comparison of time history diagrams of ${x}_{1},{x}_{2}$ and ${x}_{3}$ in the same condition

${x}_{1}\left(0\right)=1,{{x}^{\prime }}_{1}\left(0\right)=0,{x}_{2}\left(0\right)=-1,{{x}^{\prime }}_{2}\left(0\right)=0,{x}_{3}\left(0\right)=1,{{x}^{\prime }}_{3}\left(0\right)=0$

${x}_{1}\left(\tau \right)=\mathrm{cos}\tau ,$

${x}_{2}\left(\tau \right)=-\mathrm{cos}\tau ,$

${x}_{3}\left(\tau \right)=\mathrm{cos}2\tau ,$

$\omega =1.$

5. 结论

NOTES

*通讯作者。

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