Limit Cycle at $r=1$?

Question

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.

Answer 1

Assume unit speed for the flow system (primes denote differentiation with respect to arc distance). If $\psi$ is angle the streamline makes to radius vector then from the given relations for $r^{'}$ and $\theta^{'}$

$$ \tan \psi = \dfrac{r d\theta}{dr}= \dfrac{r\theta^{'}}{r{'} }=\dfrac {1+\sin \theta \cos \theta \,(r^2-1)}{(r^2-1)\, {\sin ^2\theta}} $$

which has vanishing denominator at $r=1 $ or $ \psi= \pm\pi/2 $ at any $\theta$ in the flow. That is consider $\theta$ as fixed and $r$ as variable. The flow direction does not change, i.e., the sign of slope of the tangent remains same:

$r=1$ represents a circular asymptote of the Limit Cycle. It is also shown in the $(r,\,\tan \psi) $ graph as the green asymptote. The entire flow has either clockwise or anti-clockwise angular velocity on both sides of the circular asymptote.

Answer 2

Hint.

Putting in the form

$$ \dot u_1 = u_2\\ \dot u_2 = -(u_1^2+u_2^2-1)u_2 - u_1 $$

then

$$ \frac 12(u_1^2+u_2^2)' = -u_2^2(u_1^2+u_2^2-1) $$

and the stream plot gives

NOTE

When $u_1^2+u_2^2 = 1$ then $\dot u_1 = u_2$ and also

$$ 2u_1\dot u_1 +2u_2\dot u_2 = 0\Rightarrow \dot u_2 = - u_1 $$

and those points have the dynamics dictated by

$$ \dot u_1 = u_2\\ \dot u_2 = -u_1 $$

which describes concentric circles.

Answer 3

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..