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[SOLVED] Van der Pol Equation

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
x'' + \varepsilon(x^2-1)x'+x=0\quad \varepsilon\ll 1
$$
Using multiple scale perturbation and letting $f(x,x') = (x^2-1)x'$, we have
$$
x_{0tt} + \varepsilon x_{1tt} + 2\varepsilon x_{0tT} + \cdots + \varepsilon f(x_0,x_{0t}) + \cdots + x_0 + \varepsilon x_1 + \cdots = 0
$$
where $T$ is our slow time and $T = \varepsilon t$.
\begin{alignat}{4}
\text{order } 1: & \ \ x_{0tt} + x_0 & = & 0\\
\text{order } \varepsilon: & \ \ x_{1tt} + x_1 & = & -2x_{0tT} - f(x_0,x_{0t})
\end{alignat}
So $x_0(t,T) = A(T)\cos t + B(T)\sin t = r(T)\cos(t + \phi(T))$.
\begin{alignat}{3}
x_{1tt} + x_1 & = & -2x_{0tT} - f(x_0,x_{0t})\\
& = & 2[r'\sin(t + \phi) + r\phi'\cos(t+\phi)] - f(r\cos(t+\phi),-r\sin(t+\phi))\\
& = & 2r'\sin\theta + 2r\phi'\cos\theta - f(r\cos\theta,-r\sin\theta)
\end{alignat}
where $\theta = t+\phi$.
$$
f(r\cos\theta,-r\sin\theta) = \frac{a_0}{2} + \sum_{n=1}^{\infty}(a_n\cos n\theta + b_n\sin n\theta)
$$
The only resonance terms will occur when $n=1$.
Then
\begin{alignat}{3}
x_{1tt} + x_1 & = & 2r'\sin\theta + 2r\phi'\cos\theta - a_1\cos\theta - b_1\sin\theta - \frac{a_0}{2}-\sum_{n=2}^{\infty}(a_n\cos n\theta + b_n\sin n\theta)\\
& = & \cos\theta(2r\phi'-a_1) + \sin\theta(2r'-b_1) - \text{non-resonance terms}
\end{alignat}
So $2r\phi'-a_1 = 0\Rightarrow \frac{a_1}{2} = r\phi' $ and $2r'-b_1 = 0\Rightarrow \frac{b_1}{2} = r'$.
$$
a_1 = \frac{-1}{\pi}\int_{-\pi}^{\pi}r(r^2\cos^2\theta - 1)\sin\theta\cos\theta d\theta\\
b_1 = \frac{-1}{\pi}\int_{-\pi}^{\pi}r(r^2\cos^2\theta - 1)\sin\theta\sin\theta d\theta
$$
That is,
$$
r' = \frac{-1}{2\pi}\int_{-\pi}^{\pi}[r^3\cos^2\theta\sin^2\theta - r\sin^2\theta] d\theta
$$
and
$$
r\phi' = \frac{-1}{2\pi}\int_{-\pi}^{\pi}[r^3\cos^3\theta\sin\theta - r\sin\theta\cos\theta] d\theta
$$
For $r'$, we can use the orthonormal basis $\left\{\frac{1}{\sqrt{2}},\cos\theta,\cos 2\theta,\ldots,\sin\theta,\ldots\right\}$ to integrate.
\begin{alignat}{3}
-r^3\langle\cos^2\theta\sin^2\theta\rangle + r\langle\sin^2\theta\rangle & = & -\frac{r^3}{2}\langle\sin^2\theta\rangle -\frac{r^3}{2}\langle\cos 2\theta\sin^2\theta\rangle + \frac{r}{2}\\
& = & -\frac{r^3}{4} -\frac{r^3}{2\sqrt{2}}\left\langle\frac{1}{\sqrt{2}}\cos 2\theta\right\rangle - \frac{r^3}{4}\langle\cos^22\theta\rangle + \frac{r}{2}\\
& = & -\frac{r^3}{4} -\frac{r^3}{4}\left\langle \left(\frac{1}{\sqrt{2}}\right)^2 \right\rangle -\frac{r^3}{4\sqrt{2}}\left\langle \frac{1}{\sqrt{2}}\cos 4\theta \right\rangle + \frac{r}{2}\\
&= &-\frac{r^3}{8} + \frac{r}{2}
\end{alignat}
Therefore,
$$
r' = \frac{1}{8}r(r^2-4).
$$
For $r\phi'$, we will use the orthonormal basis.
$$
r\phi' = 0.
$$
\begin{alignat}{3}
\int\frac{8dr}{r(r^2-4)} & = & \int dT\\
\int\phi' & = & 0
\end{alignat}
Thus, $\phi(T) = \phi_0$.
By partial fractions,
$$
\int\left[\frac{1}{r - 2} + \frac{1}{r + 2} - \frac{2}{r}\right]dr = T + k\Rightarrow r(T) = \frac{2}{\sqrt{1 + Ce^{-T}}}
$$
Let $r(0)=r_0$. Then
$$
r(0) = \frac{2}{\sqrt{1 + C}} = r_0\Rightarrow C = \frac{4}{r^2_0} - 1.
$$
$$
r(T) = \frac{2}{\sqrt{1 + \left(\frac{4}{r^2_0} - 1\right)e^{-T}}}
$$
Recall that $x(t,\varepsilon) = r(T)\cos(t+\phi(T)) + \mathcal{O}(\varepsilon)$. Then
$$
x(t,\varepsilon) = \frac{2}{\sqrt{1 + \left(\frac{4}{r^2_0} - 1\right)e^{-T}}}\cos(t + \phi_0) + \mathcal{O}(\varepsilon) =
\frac{2}{\sqrt{1 + \left(\frac{4}{r^2_0} - 1\right)e^{-\varepsilon t}}}\cos(t + \phi_0) + \mathcal{O}(\varepsilon)
$$
We now that that
$$
\lim_{t\to\infty}x(t,\varepsilon) = 2\cos(t + \phi_0)+ \mathcal{O}(\varepsilon)
$$
and
$$
x(t,\varepsilon = 0) = r_0\cos(t+\phi_0).
$$
So the for small epsilon, the period is $2\pi+ \mathcal{O}(\varepsilon)$ and the radius of the limit cycle approaches 2 as $t\to\infty$.
hw5problem1vanderpollimit.jpg
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hi dwsmith, :)

So what is your question?

Kind Regards,
Sudharaka.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Hi dwsmith, :)

So what is your question?

Kind Regards,
Sudharaka.
I had a question as I was typing but I figure it out. But I decided to finish the problem anyways.