Why $L(x,u,p)=\sqrt{\frac{1+p^2}{u}}$ does not depend on $x$, even when $u=u(x)$?

Question

I'm reading an article which writes that a Lagrangian:

$L(x,u,p)=\sqrt{\frac{1+p^2}{u}}$ does not depend on $x$, even when $u$ is function of $x$.

So how does it not depend on $x$? Is it some increase in abstraction (thinking that it could work for any kind of similar $u$ so the specific nature of $x$ doesn't matter)?

http://www-users.math.umn.edu/~olver/ln_/cv.pdf, p. 15-16.

Answer 1

Formally, your Lagrangian does not depend on $x$. It is only by solving the variational problem that you get the dependence of $u$ on $x$ required to minimize the action along all possible trajectories $u(x)$.