# 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.

• AFAIK, the Picard-Lindelof theorem (which states the existence and uniqueness of ODE's in which $F$ is Lipschitz continuous) is a consequence of the Banach fixed-point theorem. Apr 28 '18 at 17:33
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 $$\lim_{x\rightarrow \infty} \sin(\frac{1}{x})x = 1$$