I have to prove the following:
Let $f: \mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R}$ defined by $f(t,x) = x^\frac{2}{3}, \forall (t,x) \in \mathbb{R} \times \mathbb{R}$. Let $x:[-1,0]\rightarrow \mathbb{R}, x(t) = 0, \forall t \in [-1,0]$ be a soluction of the IVP that follows: \begin{equation*} x' = f(t,x), x(a) = \epsilon \end{equation*} Show that there are at least two maximal/saturated solutions of the same IVP problem that extend $x$.
What I did so far: Basically I have only "translated" the information I am given but i dont know how to prove that there are two solutions that extend $x$. Any help would be really apreciatted.
Thanks for your time guys.
The trick as usual is that you do not have uniqueness at $x=0$ you can branch in and out. You get a general solution $$ x(t)=\begin{cases} \sqrt{27}(x-x_0)^3,&x<x_0,\\ 0,&x_0\le x\le x_1,\\ \sqrt{27}(x-x_0)^3,&x_1<x.\\ \end{cases} $$ Obviously, $x_0\le -1$ and $0\le x_1$. That a non-zero value is prescribed fixes $x_1$ (assuming $a$ and $ϵ$ positive).