How calculate extremes of the functional?

Question

Is it also here to use the Euler-Lagrange equation? Could someone tell me how it will look like?

$${F}_{u} = \int_{0}^{1} \left( uu' + uu''^{2} + uu'' + u'u'' + 2u'' \right) \mbox{d}x$$

$$u(0) = u'(0) = u(1) = 0$$

$$u'(1) = 1$$

Answer 1

$${L} = f\left(x,u,u',u''\right)$$

$$\frac{ \partial {L} }{ \mbox{d}u } - \frac{ \mbox{d}}{ \mbox{d}x } \left( \frac{ \partial {L} }{ \partial u^{'} } \right) + \frac{ \mbox{d}^{2}}{ \mbox{d}x^{2} }\left( \frac{ \partial {L} }{ \partial u^{''} } \right) = 0$$

But I don't know how calculate this diverates: $$\frac{ \partial {L} }{ \mbox{d}u } =$$

$$\frac{ \partial {L} }{ \mbox{d}u' }= $$

$$\frac{ \mbox{d}}{ \mbox{d}x } \left( \frac{ \partial {L} }{ \partial u^{'} } \right) =$$

$$\frac{ \partial {L} }{ \mbox{d}u'' } = $$

$$\frac{ \mbox{d}^2}{ \mbox{d}x^{2} }\left( \frac{ \partial {L} }{ \partial u^{''} } \right) = $$

Answer 2

Hint: The 1st, 4th & 5th term in OP's Lagrangian are total derivative terms, which don't contribute to the EL equations. The 2nd & 3rd term is $$L~=~u u^{\prime\prime}(u^{\prime\prime}+1),$$ leading to EL equation $$0~=~\frac{\partial L}{\partial u}+ \frac{d^2}{dx^{2}}\frac{\partial L}{\partial u^{\prime\prime}}~=~(3u^{\prime\prime}-2)u^{\prime\prime} +4u^{\prime}u^{\prime\prime\prime} +2u u^{\prime\prime\prime\prime}.$$