# Solving $y'=2\sqrt{|y|}$, $y(0)=0$

Consider the ODE $y'=2\sqrt{|y|}$ where $y\in\Bbb R$. Find all solutions with initial conditions $y(0)=0$.

This is a homework question, so please just small hints...

The most obvious solution is $y(t)=0$. I was thinking that by Picard's Theorem, then this is, locally, the unique solution. But then I realized that $y(t)=t^2$ is another solution, if $t\ge0$ that satisfies the given initial condition. I have two questions:

1. Why doesn't this contradict Picard's Theorem?
2. Would I solve this ODE by, just, assuming $t\ge0$ then integrating both sides, then assuming $t\le0$ and integrating both sides? It seems that this technique should work since it looks like any solution which is defined on a neighborhood of 0 must vanish on some neighborhood of 0 (Picard's theorem seems to guarantee this, but please correct me if I am wrong).

1) It does not contradict the theorem because the function $f(y)=2\sqrt{\lvert y \rvert}$ is not locally Lipschitz. Otherwise, you would have $|f(y)-f(0)|=2\sqrt{\lvert y \rvert}\le L|y|$ for $|y|$ sufficiently small. But this is equivalent to $\lvert y \rvert\ge 4/L^2$ for $|y|$ sufficiently small.
2) What you describe is indeed the best approach. You should only check at the end that it gives you a $C^1$ function since solutions are supposed to be $C^1$.
• Quick question. We solving an ODE, do we just wish to solve it on some open set (open in the functions domain) of the point in question (so in a neighborhood of 0 in this case)? For example, in the above example, $f(t) = t^2$ is a solution for $t \geq 0$. So this solution isn't defined on an open neighborhood of 0, but is defined on a relatively open neighborhood containing 0. – Jonathan Gafar Mar 3 '16 at 21:42
• Yes, that's the notion of solution, plus it should be $C^1$. – John B Mar 3 '16 at 21:49