The beam equation is given by:
$Mu^{(4)}(x)+Nu''(x) = f(x) \; x \in [0,L]$
This is supposed to represent the bending of a beam of length $L$ when a sectional mass density $f(x) \ge 0$ acts on the beam.
I have to show that a candidate to be a minimum of the functional:
$\mathcal{F}[u] = \int_{0}^L \Big(M \frac{u''(x)^2}{2} - N \frac{u'(x)}{2}-f(x)u(x)\Big) dx$
defined on a suitable set, has to verify the beam equation.
I'm looking for hints on where to define this functional and whether I can use any intermediate results. Otherwise, I'll have to reproduce the proof from other variational problems. What surprises me here is that there aren't any boundary conditions.
My approach
We have the functional $\mathcal{F}[u] = \int_{0}^L \Big( M \frac{u''(x)^2}{2} - N \frac{u'(x)^2}{2} - f(x)u(x)\Big) dx$ and we can denote $F(x,u(x),u'(x),u''(x)) = M \frac{u''(x)^2}{2} - N \frac{u'(x)^2}{2} - f(x)u(x)$.
Now we do variations $g(s) = \mathcal{F}[\overline{u}+s\phi]$ where $\overline{u}$ is the (possible) minimum of the functional. Therefore we have that $g'(0) = 0$. Doing the calculations yields: $$0 = \int_{x_0}^{x_1} \frac{dF}{dy} (x,\overline{u}(x),\overline{u}'(x),\overline{u}''(x)) \phi(x) + \frac{dF}{dz} F(x,\overline{u}(x),\overline{u}'(x),\overline{u}''(x)) \phi'(x) + \frac{dF}{dp} F(x,\overline{u}(x),\overline{u}'(x),\overline{u}''(x)) \phi''(x)$$
Giving
$$ F(x,u,u',u'') = \frac 12 M u''-\frac 12 N u'-f(x)u $$
we have according to Euler-Lagrange conditions
$$ F_u-\left(F_{u'}\right)'+\left(F_{u''}\right)'' = M u^{(4)}-f(x)=0 $$
which obeys the beam line equation.