Solution to $x'=x\sin(\frac{\pi}{x})$ is unique I want to prove that the only solution to the ODE $$x'=\begin{cases} x\sin(\frac{\pi}{x}) \text{ if } x \neq 0 \\ 0 \text{ else}\end{cases}$$ with $x(0) = 0$ is unique (so it is the constant solution $x=0$).
Since $f(x) = x\sin(\frac{\pi}{x})$ is not locally Lipschitz at $0$ we cannot use any general result about uniqueness.
I have tried to prove that if $x$ is a solution then $x^2$ is locally decreasing around $0$ (a trick that has worked other times ) but the oscillations of the term $\sin{\frac{\pi}{x}}$ make impossible reaching conclusions about sign. I have noticed that since $f$ is odd, any solution $x$ must be an even function, this is $x(-t) = x(t)$.
The equation does not seem to be solvable so more direct approach are out of hand I think.
Can you give me any hint?

If $x(t_0) \neq 0$ for some $t_0 \neq 0$. We assume $t_0>0$ and $x(t_0) >0$, the other cases is similar. Let $x_0 \in (0,x(t_0))$ so that
$$ x_0 \sin \left( \frac{\pi}{x_0}\right) = 0.$$
Since $x(0) = 0$ and $x(t_0) > x_0$, by continuity there is $t_1 \in (0,t_0)$ so that $x(t_1) = x_0$. We also choose the smallest such $t_1$, so that $x(t) < x_0$ for all $t<t_1$. Then the initial value problem
$$ \begin{cases} x' = x \sin \left( \frac{\pi}{x}\right), \\ x(t_1) = x_0 \end{cases}$$
has two solutions: $x(t)$ and the constant solution $y(t) = x_0$. These two are different since $x(t) < x_0$ for all $t<t_1$. It contradicts the uniqueness theorem, since $x\sin \left( \frac{\pi}{x}\right) $ is Locally Lipschitz around $(t_1, x_0)$.
