calculus of variations - Reconstruct a functional from Euler-Lagrange equation

Assume a functional $I[u(x)] = \int\limits_{{x_{\min }}}^{{x_{\max }}} {F(x,u,u')dx} $.

Given I have the Euler-Lagrange (EL) equation
$u'(x) = 0$, I need to find the form of $F$ that leads to this EL.

I intuitively tried $F(x,u,u')=u(x)u'(x)$ and this yields
\begin{equation}
\frac{{\partial F}}{{\partial u}} - \frac{d}{{dx}}\left( {\frac{{\partial F}}{{\partial u'}}} \right) = u'(x) - u'(x) = 0

\end{equation}
but I would need $F$ such that
\begin{equation}
\frac{{\partial F}}{{\partial u}} - \frac{d}{{dx}}\left( {\frac{{\partial F}}{{\partial u'}}} \right) = u'(x).
\end{equation}
Does such an $F$ that would lead to EL $u'(x) = 0$ exist?

Answer

I. If $u\in\mathbb{C}$ is complex-valued: Then the Lagrangian $F(x,u,u^{\prime})=i\bar{u}u^{\prime}$ works.

II. If $u\in\mathbb{R}$ is real-valued: Then there is no Lagrangian $F(x,u,u^{\prime},u^{\prime\prime},u^{\prime\prime\prime},\ldots)$.

Sketched proof:

1. $F(x,u,u^{\prime},u^{\prime\prime},\ldots)$ can not depend on higher derivatives $u^{\prime\prime},u^{\prime\prime\prime},\ldots$, in order for the higher-order EL equation to be independent of them.




2. The Lagrangian must be an affine function
   $$F(x,u,u^{\prime})=A(x,u) +u^{\prime}B(x,u) \tag{1}$$
   in $u^{\prime}$ in order for the EL equation to be independent of $u^{\prime\prime}$.




3. It is not hard to see that the EL equation
   $$ \frac{\partial A}{\partial u}- \frac{\partial B}{\partial x}~=~0 \tag{2}$$
   
   for the Lagrangian (1) does not depend on derivatives. Contradiction. $\Box$