Problem: can we apply Picard theorem for $$u' = \sqrt{\lvert u^2 -1 \rvert}$$ if $$ u(\pi / 2)= 0$$ [$u$ is a function of a variable $x$ so $u = u(x)$] My attempt:
Well, what I need to know is whether the function $f(x,u) = \sqrt{\lvert u^2 -1 \rvert} $ is Lipshitz continuous when observing only the second variable: u, on some rectangle $(\pi/2 - \delta,\pi/2 + \delta)\times(1- \epsilon,1 + \epsilon)$.
I'm however uncertaion whether this holds. I would say that if $\epsilon \geq 1$ the function is definitely not Lipshitz contiuous on the second variable, but I'm really not sure how to prove that it is or isn't if $\epsilon > 1$, it seems to me it's supposed to be, but I can't find a way to prove it.