Existence and uniqueness of solutions for continuous and directionally differentiable ODE

Question

Given $f:\mathbb{R}^n \to \mathbb{R}^n$ continuous and directionally differentiable (i.e., such that the directional derivative of $f$ exists for any direction) at a neighborhood $N$ of $x_0\in\mathbb{R}^n$, consider the ODE \begin{align*} \dot{x}=f(x) \end{align*} with initial condition $x(0)=x_0$. By Peano's existence theorem, we know that this Initial Value Problem has at least one solution. Is this solution unique? Directionally differentiable functions are not necessarily locally Lipschitz so Picard's uniqueness theorem does not apply. If uniqueness does not hold, are there any known counterexamples?

Answer 1

No. Consider $x'=g(x,y)$, $y'=h(x,y)$. If we take $g(x,y)=2|x|^{1/2}$ for $y=x^2$ similarly $h=4|x|^{3/2}$ on $y=x^2$, then we can check directly that $x=t^2$, $y=t^4$, $t\ge 0$, and $x=y=0$ are solutions with initial value $(0,0)$, independently of how we define $g,h$ off the parabola.

To define $g,h$ everywhere, we interpolate linearly, starting from the value on the parabola, by letting $g(ta,ta^2)=tg(a,a^2)$, $a\in\mathbb R$, $-1\le t\le 1$, and we also set $g(x,0)=g(0,y)=0$.

So far, $g$ has been defined on the closed set $\{ |y|\ge x^2\}\cup \{ (x,0)\}$, and it is continuous there. To verify continuity on the $y$ axis, note that if $(ta,ta^2)\to (0,b)$ with $b\not= 0$, then $|a|\to\infty$, so $tg(a,a^2)=2t|a|^{1/2}\to 0$ (and also $t|a|^{3/2}\to 0$, which we need in the corresponding step for $h$).

The existence of all directional derivatives at the origin is already guaranteed because the current domain contains a line segment in every direction, and $g$ is linear there. Finally, we extend $g$ to $\{ 0<|y|<x^2\}$. We can make $g$ smooth away from the origin.