How can I show that the ode $$ \dot x(t) = t^2x^3 \sin(\frac{1}{t^3x}), x(0)=x_0 $$ doesn't have a unique solution on $[-1,1]$ ?
It seems $x\equiv 0$ is a solution.
Now clearly, for $t,x\neq 0$, $g(t,x)=t^2x^3 \sin(\frac{1}{t^3x})$ and $g(t,0)=0=g(0,x)$ is continuous on $(-1,1)\times \mathbb{R}$, $x\mapsto g(t,x)$ is continuously differentiable, however $g$ is not locally Lipschitz that is for any compact $K\subset (-1,1)\times \mathbb{R}$, we cannot find a constant $L_K$ such that : $$ |g(t,x)-g(t,y)|\le L_K|x-y| $$ for all $(t,x), (t,y) \in K$.
I think that for a suitable $x_0$, the solution explodes in finite time but I am struggling to prove it.
Here is the phase portrait.
we clearly see a blow-up in finite time