Q:

Stability analysis of system of nonlinear ODEs

I have the following nonlinear system in polar coordinates:
$$
\begin{align}
\dot{r} &= r^2\sin(1/r)\\
\dot{\phi} &= 1,
\end{align}
$$
which can be thought of as a rotating object with some kind of pull on it depending on the radius. I want to analyze the stability of this system, so first I look at special cases.
I know that when $r^* = \pi^{-1}$, $\sin(1/r^*) = 0$, and so $\dot{r} = 0$, for all time $t$, and the trajectory will just be a circle of radius $r^*$. Additionally, for $r_0 > \pi^{-1}$, we see that $\sin(1/r_0) > 0$ since the term inside the sine is bounded $\pi < 1/r_0 < 0$. Thus, since $\dot{r} \propto r^2$, and $\sin(1/r) > 0$, trajectories with $r_0 > \pi^{-1}$ will all diverge. For trajectories with $r_0 < \pi^{-1}$, we don't have $\sin(1/r_0) > 0$ anymore, and it could be positive or negative (because you are now moving counterclockwise on the unit circle from $\pi$). Thus, there will be transient solutions until the trajectories reach some $r = n\pi^{-1}, n\in\mathbb{N}$ that will make the dynamics zero, at which point the trajectory will be a circle. The way I see it is that it's kind of like a black hole with an event horizon at $r^* = \pi^{-1}$, where all solutions diverge outside that radius, and solutions can't get out if they are inside the radius. 
My problem, however, is regarding the equilibrium point $r = 0$. What kind of equilibrium would it be? I tried taking the Jacobian of the system $\dot{x} = f(x)$, where $x = [r,\phi]^\intercal$, to get 
$$
J = 
\begin{bmatrix}
2r\sin(1/r) - \cos(1/r) & 0\\
0 & 0
\end{bmatrix}
$$
However, it is impossible evaluate this at $r = 0$, since both sine and cosine are oscillatory as $1/r \rightarrow \infty$, so I'm not sure how to assess that. Additionally, since the Jacobian has the form
$$
J = 
\begin{bmatrix}
\alpha & 0\\
0 & 0
\end{bmatrix},
$$
we can look at its eigenvalues which are just $\lambda = 0, \alpha$, with $\alpha$ a number that can be either positive or negative. However, this doesn't make much sense to me because this implies it is a saddle-node, as the solution is
$$
x = 
\begin{bmatrix}
r\\
\phi
\end{bmatrix}
= c_1 
\begin{bmatrix}
0\\
1
\end{bmatrix}
+ c_2e^{\alpha t}
\begin{bmatrix}
1\\
0
\end{bmatrix},
$$
and $v_1 = [0,1]^\intercal$ is an eigenvector with $r = 0$, which is infeasible? Any thoughts are appreciated!

A:

The circles $r = 1/(n\pi)$ for positive integer $n$ are all closed trajectories.  Any solution that starts out near (but not at) $r=0$ will either be on one of these or between two of them (say $r = 1/(n \pi)$ and $r = 1/((n+1)\pi)$, and then it will stay between those two forever, never getting closer to the origin than $1/((n+1)\pi)$ nor farther than $1/(n \pi)$.  Thus the equilibrium point $r=0$ is stable but not asymptotically stable.
The analysis in terms of Jacobians is not applicable: polar coordinates are singular at the origin, and if you convert it to rectangular coordinates the system is singular at the origin: the Jacobian does not exist there.

