What Is the Next Step in Simplifying the Van der Pol Oscillator Equation?

[kornelthefirst]

Homework Statement

We have a Van der Pol oscillator with small $\epsilon$ and after writing up a Fourier-series we have to bring it to a simpler form.

Relevant Equations

Equation of motion for the Van der Pol oscillator$$\ddot{x} + \epsilon(x^{2} - 1)\dot{x} + x = 0$$ Fourier-series for the limit cycle(already given) $$x_\epsilon^p(t) = \frac{a_0}{2} + \sum\limits_{k=1}^{\infty } [a_k \cos(k \omega t) + b_k \sin(k \omega t)]$$ Equation we need to arrive to$$\epsilon (x_p^2 - 1)\dot{x_p} = \epsilon a_1\omega[(1-\frac{a_1^2}{4})\sin(\omega t)-\frac{a_1^2}{4}\sin(3 \omega t)]$$

First i looked at the case of $\epsilon = 0$ and came to the conclusion, that this oscillator has a circular limit cycle in a phase space trajectory, when plotted with the axes x and $\dot{x}$.
I have found that $x_0^p (t) = a_1 \cos(t)$ which implies that all other Fourier- coefficients have $\epsilon$ of at least power of 1
The limit cycle is independent of the starting conditions unless $x = 0$ and $\dot{x} = 0$, so we can choose $a_1$ to be > 0 and $b_1 > 0$.
If we put the equation of the Fourier-series back to the equation of motion we get$$\sum\limits_{k=1}^{\infty } [ - a_k \cos(k \omega t) - b_k sin( k \omega t)] + \epsilon (x^2-1)\dot{x} + \frac{a_0}{2}+\sum\limits_{k=1}^{\infty } [a_k \cos(k \omega t) + b_k sin( k \omega t)]$$ so simplified $$\epsilon (x^2-1)\dot{x} + \frac{a_0}{2} = 0$$ I am currently stuck here and can't find the next step. I can only assume it will include trigonometric identities, because i can see some patterns for some.

 

[pasmith]

Homework Statement: We have a Van der Pol oscillator with small $\epsilon$ and after writing up a Fourier-series we have to bring it to a simpler form.

I'm not sure this is the best approach to this problem; could you please post the exact problem statement?

Relevant Equations: Equation of motion for the Van der Pol oscillator$$\ddot{x} + \epsilon(x^{2} - 1)\dot{x} + x = 0$$ Fourier-series for the limit cycle(already given) $$x_\epsilon^p(t) = \frac{a_0}{2} + \sum\limits_{k=1}^{\infty } [a_k \cos(k \omega t) + b_k \sin(k \omega t)]$$ Equation we need to arrive to$$\epsilon (x_p^2 - 1)\dot{x_p} = \epsilon a_1\omega[(1-\frac{a_1^2}{4})\sin(\omega t)-\frac{a_1^2}{4}\sin(3 \omega t)]$$

First i looked at the case of $\epsilon = 0$ and came to the conclusion, that this oscillator has a circular limit cycle in a phase space trajectory, when plotted with the axes x and $\dot{x}$.
I have found that $x_0^p (t) = a_1 \cos(t)$ which implies that all other Fourier- coefficients have $\epsilon$ of at least power of 1
The limit cycle is independent of the starting conditions unless $x = 0$ and $\dot{x} = 0$, so we can choose $a_1$ to be > 0 and $b_1 > 0$.
If we put the equation of the Fourier-series back to the equation of motion we get$$\sum\limits_{k=1}^{\infty } [ - a_k \cos(k \omega t) - b_k sin( k \omega t)] + \epsilon (x^2-1)\dot{x} + \frac{a_0}{2}+\sum\limits_{k=1}^{\infty } [a_k \cos(k \omega t) + b_k sin( k \omega t)]$$

You should have <br /> \ddot x = -\sum_{n=1}^\infty n^2\omega^2 (a_n \cos (n\omega t) + b_n \sin (n\omega t)).

I am currently stuck here and can't find the next step. I can only assume it will include trigonometric identities, because i can see some patterns for some.

I think the idea is that x(t) = a_1 \cos \omega t + \epsilon x_p(t) so that \begin{split}<br /> \ddot x + x &amp;= -\epsilon(x^2 - 1)\dot x \\<br /> (1 - \omega^2) \cos \omega t + \epsilon (\ddot x_p + x_p) &amp;= \epsilon a_1 \omega (a_1^2 \cos^2\omega t - 1)\sin \omega t + O(\epsilon^2)\end{split} subject to \dot x_p(0) = x_p(0) = 0. We do not need a sine term in the leading order solution since that just amounts to a shift in the origin of time, which merely moves us to a different point on the limit cycle. It is not necessary to expand x_p as a fourier series in order to solve this, although expressing the right hand side as a series in \sin n\omega t and knowing \ddot y + y where y = \sin n\omega t or t\cos n\omega t will assist.