I am being asked about the following equation:

$-u'' + L * \sin(u) = f(x)$

with initial conditions

$u(0) = 1, u(1) = 0$

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 $u$?

  • 2
    $\begingroup$ This is called a boundary value problem. $\endgroup$ – Simply Beautiful Art Mar 26 at 2:58
  • $\begingroup$ 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 $[\phi_1(( u(0),u'(0)) )]_1 = u(1)$. $\endgroup$ – copper.hat Mar 26 at 4:20

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