We have to minimize:
$ \phi = \int_{0}^1 (u'^2 + u'^4 -u'x -u)dx$
with boundary condition u(1) = 1, and we have to show that "it is indeed a minima".
My approach:
$\frac{d}{dx} (\frac{dL}{du'}) = 2u''+12u'^2*u''-1$
Hence:
$2u''+12u'^2*u''-1+1 = 0$
$2u''+12u'^2*u'' = 0$
$2u''(1+6u'^2) = 0$
Hence:
1) $u'' = 0 \to u = c_1x+c_2$,
2) $1+6u'^2 = 0 \to u' = \pm \sqrt{-\frac{1}{6}}$.
This seems not entirely correct and I can't determine if I have a minima at all – where did I go wrong?
Best Answer
Note by integration by part
$$\int_0^1 u' x \ dx = u(1) \cdot 1 - u(0) \cdot 0 - \int_0^1 u\ dx= 1- \int_0^1 u\ dx, $$
so $\phi$ can be simplified as
$$ \phi = \int_0^1 (u'^2 + u'^4 )dx -1 = \int_0^1 \left( u'^2 + \frac 12\right)^2 dx - \frac 54.$$
From here it is quite obvious that $\phi$ is minimized when $u$ is constantly one.
Remark When one uses E-L equations and are asked to find global minimum, it is always tricky... One has to know that global minima EXISTS, and in general this is a nontrivial analysis problem. We are lucky this time that $\phi$ has a nice expression.