= Homoclinic basin boundary in a sinusoidal radial flow
Consider
$$
\dot r=r(1-r)(r-2\sin\theta),\qquad\dot\theta=r.
$$
The 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 equation>
$$
\frac{dw}{d\theta}=(1-2\sin\theta)w-1.
$$
Let $A(\theta)=\theta+2\cos\theta$. The solution with $w(\pi)=1$ is
$$
w(\theta)=e^{A(\theta)-A(\pi)}\left[1+\int_\theta^\pi e^{A(\pi)-A(s)}\,ds\right].
$$
At the endpoint, $w'(\pi)=0$ and $w''(\pi)=2$, 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. Also
$$
w(-\pi)=e^{-2\pi}+\int_{-\pi}^{\pi}e^{-\pi-2-s-2\cos s}\,ds<e^{-2\pi}+\int_{-\pi}^{\pi}e^{-\pi-s}\,ds=1.
$$
Every 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.
Back to article page
