A dynamical system in polar coordinates is:
$$\Theta'=1, r'= r^2\sin(1/r), r>0, r'=0\mbox{ if }r =0.$$
How to show this is stable at origin?
Intuitively, I really can't believe it because I plotted $r^2\sin(1/r)$ and it goes to infinity as $r$ goes to infinity. This means if I start at some where both $r$ and $r'$ are positive. Then the system might blow up rather than stay close to the origin. (For $\Theta$, I think we can just take it as a constant rotating with time pass by)
Consider a circle $r = \frac{1}{2\pi n - \pi/2}$ where $n = 1,2,3,\ldots$
On this circle $$r' = \frac{1}{(2\pi n - \pi/2)^{2}} \sin(2\pi n - \pi/2) = -\frac{1}{(2\pi n - \pi/2)^{2}}.$$
The trajectories passing through this circle all point inward, i.e. trajectories inside this region cannot escape to infinity.
In fact, we have an infinite number of these circles surrounding the origin, getting more and more tightly packed as $n$ increases.
Thus the origin must be Lyapanov stable since trajectories that start "close" to it cannot escape.
(A similar argument can be employed to show that the origin cannot be asymptotically stable).