# Not contradiction of Picard-Lindelöf theorem

I have this homework problem:

Consider the following initial value problem: $$\frac{dy}{dt}=6t\sqrt[3]{y^2}$$ $$y(0)=0$$ Demonstrate that this I.V.P has a different solution of $y(t)=0$, $\forall t \in \mathbb{R}$. Explain why this doesn't contradict the Picard-Lindelöf problem.

So I found a solution for this I.V.P:

$$y=t^6$$

Picard-Lindelöf theorem says that:

For a differential equation of the form: $\frac{dy}{dt}=f(t,y)$ with the initial condition $y_0=y(t_0)$, if $f(t,y)$ is Lipschitz continuous, then I.V.P solution exists and it's unique on a open interval which contains $t_0$.

In this case, $f(t,y)$ is $C^1$, and so, Lipschitz continuous. Then Picard-Lindelöf theorem is applicable. Only for $t=0$, this two solutions are equal. So, I can only say that unicity is valid for $t=0$. Can I consider an open interval of the form $]t_0 -\varepsilon, t_0+\varepsilon[=]-\varepsilon,\varepsilon[$, so that Picard-Lindelöf theorem holds?

Note that the function $y=0$ is also a solution to this I.V.P., which at first seems to be a violation of the Picard-Lindelöf Theorem. The problem is that while $f(t,y)=6ty^{2/3}$ is continuously differentiable almost-everywhere, it is not differentiable (in $y$) at 0. In fact, it is not Lipschitz continuous at 0 either, so you cannot apply the Picard-Lindelöf Theorem.
• You're right, I just assumed that $f(t,y)$ was $C^1$ – Élio Pereira Jun 22 '15 at 14:48