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.
Best Answer
For $x_0 = 0$ there is multiplicity of solutions. Indeed, one solution is $x(t) \equiv 0$, and another one is $$ y(t) := \begin{cases} (-\frac{2}{3} t)^{3/2}, & \text{if}\ t < 0,\\ 0, & \text{if}\ t \geq 0. \end{cases} $$