In my lecture we introduced the Picard-Lindelöf Theorem the following way.
Theorem (Picard-Lindelöf): Let $f(t,y)$ be continuous on a cylinder
$$D = \{(t,y) \in \mathbb{R} \times \mathbb{R}^d | |t-t_0| \leq a, |y-u_0| \leq b\}$$ Let $f$ be bounded and satisfy a Lipschitz condition in its second argument. Then the IVP $$u' = f(t,u)\hspace{15pt}u(t_0) = u_0$$ is uniquely solvable on the interval $I = [t_0 - T, t_0 + T]$ for some $T> 0$
Now I'm looking at the following IVP:
$$u'(t) = u(t)^\frac{1}{4}\hspace{5pt}\forall t \geq 0\hspace{30pt}u(0) = 0$$
I'm asked to give to different solutions to the IVP and show that $f(x) = x^\frac{1}{4}$ is Lipschitz-continuous on every interval $[\epsilon, \infty)$ with $\epsilon > 0$, but not on $[0, \infty)$. I do understand the exercise and was able to solve it. I should take home, that the existence of multiple solution is possible, since the Lipschitz-condition in the Picard-Lindelöf theorem is not fulfilled.
Question: Let's look over the fact that $f$ is not Lipschitz continuous in $0$. How would I apply the Picard-Lindelöf theorem for an IVP, where $f$ is only defined for $t \geq t_0$, since I can't define a cylinder $D$ as in the theorem?