Local uniqueness in separable variables ODE

Question

I did a test in which I had to decide whether the following IVP have the property of local uniqueness:

$ \begin{cases} x' = |x|^{1/2}\\ x(1) = 0\\ \end{cases} $

$ \begin{cases} x' = |x|^{3/2}\\ x(1) = 0\\ \end{cases} $

The theorem that I use states the following:

Let $a \in C(J_1),g \in C(J_2),t_0 \in J_1,x_0 \in J_2$ and consider the IVP $\begin{cases} x' = a(t)g(x) \\ x(t_0) = x_0 \\ \end{cases} $ then the IVP

1. Has solutions.
2. If $g(x_0) \neq 0$ then verifies the property of local uniqueness.
3. If $g(x_0) = 0$, $G \in C(J_2)$ such that $G(x_0) = 0$,$\exists \delta > 0.\forall x \in ]x_0,x_0+\delta[.G'(x) = \frac{1}{g(x)}$ and $a(t_0) \neq 0$ then the IVP doesn't verify the property of local uniqueness.

My problem is that using this theorem the two system don't verify local uniqueness however, according to the solutions only the first should not verify it.

What am I doing wrong? Can I use any result to discard the second equation for not having local uniqueness?

See also this other question of mine.

Answer 1

For further references I leave some computations (I assume that $x > 0$ and $||$ can be dropped because of the initial condition.

$g(x) = |x|^{1/2} \implies G'(x) = \frac{1}{|x|^{1/2}} \implies G(x) = 2\sqrt{x} \implies G(0) = 2\sqrt(0) = 0$

$g(x) = |x|^{3/2} \implies G'(x) = \frac{1}{|x|^{3/2}} \implies G(x) = \frac{-2}{\sqrt{x}} \implies G(0) = \frac{-2}{\sqrt(0)}!!$

From here, the first IVP doesn't have local uniqueness.

Trying to apply Picard-Lindelof:

First I precise the domain in which the second IVP is defined:

$D = \mathbb{R}^2$ and we have to investigate whether $f:D = \mathbb{R}^2 \to \mathbb{R}$ is continuous and locally lipschitz.

First, $f$ is continuous because is the composition of two continuous functions. Then, a sufficient condition for local lipschitz is that $\frac{df}{dx}:D \to \mathbb{R}$ is continuous. But $\frac{df}{dx}(t,x) = \frac{3x}{2\sqrt{|x|}}$ and the only point that presents difficulties is $x = 0$ but since the limit is $0$ one can extend continuously the function in $x = 0$ as $\frac{df}{dx}(t,0) = 0$.

This ensures local uniqueness.

Please feel free to suggest any improvements to this answer or post your own.