Consider the minimization problem $$min\{ \int_{-1}^1 [\frac{1}{2} (u'')^2-fu] > dx| u(-1)=u(1)=u'(1)=u'(-1)=0\}$$ Determine the solution of the boundary value problem which results from the necessary conditions for a minimum if $f$ is constant.
So first lets determine the boundary value problem, I thought that a necessary condition was that the first variation is $0$ for a minimum. This would lead too
$$\frac{d}{d\epsilon}|_{\epsilon=0} \int_{-1}^1 [\frac{1}{2} (u''+\epsilon\varphi'')^2-fu-f\epsilon \varphi] dx = 0$$ With $u(-1)=u(1)=u'(1)=u'(-1)=0$
Which led me too $$\int_{-1}^1 (u''\varphi''-fu)dx = 0$$
I tried to solve this with the product rule for integration but that did not really help me.
Did I get the boundary value problem right? If so, how do I proceed? If not, how do I find the right one?