I have the following initial value problem: $$\tag{IVP}\label{IVP} \begin{cases} x^\prime(t) = -\sqrt[3]{x}\\ x(0) = x_0 \end{cases} $$ and I have to show that for every $x_0 \in \mathbb R$ a solution to \eqref{IVP} exists and it is unique.
Existence is not an issue, as the function $f(x) := -\sqrt[3]{x}$ is continuous, hence Peano's Theorem applies. But what about uniqueness?
Here are my considerations. As long as the solution is positive/negative it is decreasing/increasing. Let us suppose $x_0<0$. Then for $t$ small enough it remains negative so that I can divide $$ \frac{x^\prime(t)}{\sqrt[3]{x}} = -1 $$ which by integration yields $$ \int_{x_0}^{x(t)} \frac{dy}{\sqrt[3]{y}} = - t, \qquad \text{i.e. } x(t) = \left(-\frac{2}{3}t + x_0^{2/3}\right)^{3/2} $$ and here comes already a problem: the formula I have found implies $x\ge 0$ (it is a square root!). Where am I mistaken? How can I show uniqueness? Thanks.
The problem is that the definition of $\sqrt[3]{\cdot}$ is ambiguous, if $x>0$ then there is no problem, i.e this is $x^\frac{1}{3}$, but for $x<0$ you can not use such formula so I suppose the definition is: $$\sqrt[3]{x}=sgn(x) |x|^\frac{1}{3}$$ in particular the antiderivative for $x <0$ is not $\frac{3}{2} x^\frac{2}{3}$ but $sgn(x) \frac{3}{2} x^\frac{2}{3}$.
To be more precise as long as $x(t)<0$ with $z(t)=-x(t)>0$ we have: $$-z'(t)=+z(t)^\frac{1}{3}$$ so: $$z(t)=\left(z_0^\frac{2}{3}-\frac{2}{3}t \right)^\frac{3}{2}$$ i.e in terms of $x$: $$x(t)=-\left((-x_0)^\frac{2}{3}-\frac{2}{3}t \right)^\frac{3}{2}$$ which is indeed negative and well defined fot $t<\frac{3}{2} (-x_0)^\frac{2}{3}$.
For uniqueness, the only case remaining is when $x$ touches $0$ (or $x_0=0$). But with $n(t)=x(t)^2 \geq 0$ you have: $$n'(t)=2 x(t) x'(t) = - 2 |x|^\frac{4}{3} \leq 0$$ so $n$ is non increasing. In particular if $x(t^*)=0$ then for all $t \geq t^*$ we have $n(t)=0$ i.e $x(t)=0$.