TITLE: How can I show that $y'=\sqrt{|y|}$ has infinitely many solutions?
QUESTION [10 upvotes]: Show that the first order differential equation
$y'(x)=\sqrt{|y(x)|}$
with intial value
$y(1/2)= 1/16$
has infinitely many solutions on the interval [−1, 1].
My thought were to show that this equation has two solutions (just by ansatz, or just looking at the interval $]0, 1]$ to get $y'(x) = \sqrt{|y(x)|}$) and deduce from that that there must be infinitely many solutions since the solution set forms a vector space. I've got a nagging feeling this isn't the way to go though. Can anybody give me a hint where to look?

REPLY [15 votes]: This is a classical example of non uniqueness due to the fact that the right hand side of the equation, $\sqrt{|y|}$, is not Lipschitz at $y=0$. You show this by direct computation. The equation is in separeted variables:
$$
\frac{dy}{\sqrt{|y|}}=dt,\quad y(1/2)=1/16.
$$
Integrating, and taking into account the absolute value, we see that
$$
y(t)=\frac{t\,|t|}{4}=\begin{cases}
t^2/4 & t\ge0,\\-t^2/4 & t<0, \end{cases}
$$
is a solution. But for any $\tau\in(-1,0)$, the function
$$
y_\tau(t)=\begin{cases}
t^2/4 & t\ge0,\\0 & \tau<t<0\\-(t-\tau)^2/4 & t<\tau, \end{cases}
$$
is also a solution. Below is the graph of $y_{-1/2}$.

By the way, the set of solutions is not a vector space.