Solution to $x'=x\sin(\frac{\pi}{x})$ is unique

Solution 1:

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 $(t_1, x_0)$.