Limit Cycle at $r=1$? Consider the non-linear ODE
$$u''+(u^2+u'^2-1)u'+u=0.$$ Transforming this to polar coordinates:
$$r'=-(r^2-1)r\sin^2(\theta),$$
$$\theta'=-\sin(\theta)\cos(\theta)(r^2-1)-1.$$

If we consider an annulus (trapping region), $\frac{1}{2}<x^2+y^2<2,$ how do we deal with the $\sin^2(\theta)$ term?

We can use the fact that $$\sin^2(\theta)=\frac{1-\cos(2\theta)}{2}.$$
If we take $\cos(2\theta)=-1\implies r'>0$ for $r<\frac{1}{2}$ and $r'<0$ for $r>2$.
But, if we take $\cos(2\theta)=1\implies r'=0$ for $r<\frac{1}{2}$ and $r'=0$ for $r>2$.
Ideally $r'$ should point towards the annular region for both cases. I can't see an error in my logic.

Consider
$$
\frac{d}{dt}|r-1|={\rm sign}(r-1)r'=-|r-1|r(r+1)\sin^2θ.
$$
This is always non-positive for $r>0$. And as no solution can stay at $θ=k\pi$, the radius will always move towards $r=1$ over finite time steps..
