Optimal control theory: Unclear dependence of control input on Lagrangian

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Here I do not understand the sentence

Note that even if $L$ does not depend on $u$, the cost $J$ depends on the control $u(\cdot)$ through $x(\cdot)$ which is the trajectory that this control generates.

How can an input $u$ generate a trajectory when it is not present in $L$? It seems that $u$ generates $x$ just by the fact that it is not present somewhere else!

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Q: How can an input u generate a trajectory when it is not present in L ?

A: Simply by being present in the differential equation that defines the trajectory (along with the initial conditions).

Example:

$$\dot{x} = u(t), \; x(0) = x_0$$

The solution to this is

$$x(u(t), t) = x_0 + \int_0^t u(\tau) \text{d}\tau$$

You could choose the Lagrangian to not explicitly depend on $u$

$$L = x^2,$$

but it would still do since $x$ does. So does the cost $J$.