 Homoclinic basin boundary in a sinusoidal radial flow

ConsiderThe circle $r=1$ is a periodic orbit. Writing $r=1+\rho$ gives $\dot{\rho }=-(1-2\sin \theta )\rho +O({\rho }^2)$, while $\theta =t+{\theta }_0$ on the circle. Its normal Floquet multiplier is therefore $\exp [-{\int }_0^{2\pi }(1-2\sin \theta )d\theta ]=e^{-2\pi }<1$.

Inside the circle, use the increasing angle as independent variable and set $w=(1-r{)}^{-1}$. Direct differentiation gives the linear ordinary differential equationLet $A(\theta )=\theta +2\cos \theta$. The solution with $w(\pi )=1$ isAt the endpoint, $w^{\prime }(\pi )=0$ and $w^{\prime \prime }(\pi )=2$, so $w=1+(\theta -\pi {)}^2+O((\theta -\pi {)}^3)>1$ immediately before $\pi$. A crossing of $w=1$ has slope $-2\sin \theta$. Thus there is no crossing on $(0,\pi )$ compatible with this terminal behavior. AlsoEvery crossing on $(-\pi ,0)$ has positive slope, giving a unique ${\theta }_b\in (-\pi ,0)$ where $w=1$. Hence $r=1-1/w>0$ for ${\theta }_b<\theta <\pi$ and tends to zero at both ends. At the incoming endpoint $r\sim (\pi -\theta {)}^2$, so $dt=d\theta /r$ has a divergent integral. At the outgoing endpoint the zero is simple and the time integral diverges logarithmically. The curve is a homoclinic orbit to the origin.

Uniqueness for the angular scalar equation prevents positive-radius trajectories from crossing this loop. Inside it they approach the origin; outside it they approach the attracting circle. The origin therefore attracts an open set, but is not Lyapunov stable: points arbitrarily close to its outgoing branch follow trajectories to a fixed positive radius before returning or escaping. Attraction and stability are distinct properties.