As a reference, I asked the same question (https://physics.stackexchange.com/questions/561449/equations-of-motions-of-mathcall-phi-x-phix) in the physics community but I am interested in the mathematical reason why this produces $1=0$.
Suppose a Lagrangian of this form
$$
L[f,x]=f[x] \tag{1}
$$
where $f:\mathbb{R}\to\mathbb{R}$.
The Euler-Lagrange equations are:
$$\partial_\mu\frac{\partial L}{\partial(\partial_\mu f)}=\frac{\partial L}{\partial f}$$
There are no derivates of $f$, thus the left-most term is equal to zero:
$$\partial_\mu\frac{\partial L}{\partial(\partial_\mu f)}=0$$
Finally, the right-most term is:
$$
\frac{\partial L}{\partial f}=1
$$
Thus, the result is $1=0$. Now, I am just overall skeptical of the whole thing; why is it that the Euler-Lagrangian equation able to produce a contradiction? I suspect that $L[f,x]=f[x]$ violates one of the assumptions used to derive the Euler-Lagrange equations?
Best Answer
A solution $f$ to the E.L equations is an stationary point of the action $ S = \int_a^b L[f]dx = \int_a^b f dx $. The E.L equations in this case admit no solution, as there is no stationary point of this functional.