# Picard's theorem on Second Order Differential Equation

$$-u'' + L * \sin(u) = f(x)$$
$$u(0) = 1, u(1) = 0$$
I feel puzzled as there is no initial condition for a derivative of $$u$$. Can I proceed without it? How do I manage both initial conditions of $$u$$?
• In general, there is no guarantee that there is a solution. For a given initial condition $(u(0),u'(0))$ the solution at time $t=1$, that is $\phi_1(( u(0),u'(0)) )$ is well defined of course, so you are looking for some $u'(0)$ such that $[\phi_1(( u(0),u'(0)) )]_1 = u(1)$. – copper.hat Mar 26 at 4:20