Solution of ODE is unique without Lipschitz's condition I am facing difficulties trying to show that an ODE has a unique solution. The problem is given by:
$$\begin{cases} \frac{du}{dt} = F(u) & t\in [0,1/2] \\ u(0) = u_0\end{cases}$$
where $F(x):= \begin{cases} x \sin( \frac{1}{x}) & x\in \mathbb{R}\backslash \{0\} \\0 & x=0 \end{cases}$, which does not fulfill the Lipschitz's criterion. We have not introduced any other criteria. I have tried using Banach fixed-point theorem, which has not worked yet. Any hints will be appreciated.
Thanks in advance.
 A: It's a few years late, but note that $F$ vanishes on the countable set accumulating at $0$ given by $\{\frac{1}{k\pi} : k \in \mathbb{Z} \}$, and that away from $0$, $F$ is $C^1$ and thus locally lipschitz, so that the solutions $x_k(t) = \frac{1}{k\pi}$ are all unique in the sense that no other solution $\phi$ can satisfy $\phi(t_0) = \frac{1}{k\pi}$ for some $t_0$, unless $\phi(t) = \frac{1}{k\pi}$ for all $t$. In particular, by the intermediate value theorem, this shows that $\phi(t) = 0$ for all $t$ is the only solution satisfying $\phi(0) = 0$. This gives uniqueness of solutions to the ODE.
This same sort of trick shows $F_\alpha(x) = x^\alpha \sin(\frac{1}{x})$  has a unique solution for all $\alpha \in (0,2]$ even if for such $\alpha$ the function is not locally lipschitz.
I personally find it rather surprising, because the equation $x' = x^\alpha$ does not have a unique solution for $x \in (0,1)$, but after multiplication by $\sin(1/x)$ it does!
It is worth noting as well, that for $\alpha \in (0,2]$ that solutions exist for all time as well, by using comparison theorems and noting $\lim_{x\rightarrow \infty} \sin(\frac{1}{x})x = 1$
