Period of the limit cycle for the system: $\frac{d}{dt}v(t)=-x(t)-\frac{1}{2}v(t)(x(t)^2+v(t)^2-1)$, $\frac{d}{dt}x(t)=v(t)$

Question

Below is a system of a modified Van der Pol oscillator:

$\frac{d}{dt}v(t)=-x(t)-\frac{1}{2}v(t)(x(t)^2+v(t)^2-1)$, $\frac{d}{dt}x(t)=v(t)$

where $x^2+v^2=1$ is the equation of the limit cycle.

How do you calculate the period of the limit cycle? Below are my steps which I think is wrong as I think the period is $2\pi$ from plotting the trajectories.

The gradient of $f(x,v)=x^2+v^2$ is $[2x,2v]^T$

So the tangent vector field along the limit cycle is simply $[-2v,2x]^T$?

I feel this is the step I am getting wrong as this tangent vector field is not specific to the unit circle.

I then solved this new system for $x(0)=1,v(0)=0$:

$\frac{d}{dt}x(t)=-2v(t)$, $\frac{d}{dt}v(t)=2x(t)$

to get:

$v(t)=sin(2t), x(t)=cos(2t)$

but the trajectories of this system has a period of $\pi$ (not $2\pi$).

I feel like I need to give this system a hint that I'm interested in $f(x,v)=1$ but I don't know how.

Thank you in advance for any help.

Answer 1

This not an answer. It is only an help for understanding the issue.

I programmed this system of differential equations, and I obtained these trajectories, with the unit disk as a the unique limit trajectory

Answer 2

So, you clearly see that the vector field for your "modified Van der Pol oscillator" and for basic harmonic oscillator $\ddot{x} = - x$ are the same along $x^2+v^2=1$. And it is very known fact that integral curves of $\ddot{x} = -x$ are concentric circles $x^2+ \dot{x}^2 = C$. So, $x^2+v^2 = 1$ is the integral curve of both systems, and you can use any of these systems to compute "inner" properties of this trajectory, like it's closedness or what period it has (it's important that vector field coincide along it -- a lot of systems might share the same integral curve, but motions along it can be very different). And now to the simpler question: what is the period of trajectory of harmonic oscillator $\ddot{x} = -x$?

It's $2\pi$!