(Non) differentiability of a function in a point.

Question

I'm studying the solution of the following boundary value problem:

$ u''=\lvert\sin x\rvert \lvert\cos u\rvert$

$ u(0)=0, u(2015)=7$

(which - unless I have done it wrong - I showed to exist). Clearly the solution is of class $C^2$.

Now, I would like to determine if it is more regular. I suppose it is not because

$f(t.s):= \lvert\sin(t)\rvert\lvert\cos(s)\rvert$ it is not differentiable at $0$.

But how can I prove it?

Thank you :)

EDIT: I replaced $u(1)=1$ with $u(2015)=7$

Answer 1

Hint: If you can show $u([0,1])= [0,\pi/2-\epsilon]$ for some $\epsilon>0,$ then we actually have

$$u''(x) = \sin x\cdot \cos (u(x)), \,\, x\in [0,1].$$