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.
