How to find energy functional from Euler Lagrange equation?

For $$f: \mathbb{R} \rightarrow \mathbb{R}$$ a given smooth function, the PDE $$- \nabla \cdot (\frac{\nabla u}{|\nabla u|}) = f(u)$$

Find the functional for which this PDE is the Euler Lagrange equation.

I tried to find the functional from the given PDE. I found the term in my functional associated with L.H.S of the above term. I got

$$E(u) = \int_{\Omega} \frac{1}{2} |\nabla u| dx$$

I am not sure what is the functional term for the given right hand side term $$f(u)$$. Can anyone please help me with this?

• $$E(u) = \int_{\Omega} \left(\frac{1}{2} \lvert \nabla u \rvert - F(u) \right) dx$$ where $F = \int f du$. Jan 20, 2023 at 15:12

Recalling that the functional derivative is given by $$\frac{\delta}{\delta u(x)} \equiv -\mathrm{div}_x\mathrm{grad}_{\nabla u} + \mathrm{grad}_u,$$ one deduces that the energy functional should be $$E(u) = \int_\Omega\left(|\nabla u|+V(u)\right)\mathrm{d}x,$$ where $$V$$ is the potential of $$f$$, i.e. $$f = -\mathrm{grad}_uV$$, such that the kinetic term brings indeed $$-\mathrm{div}_x\mathrm{grad}_{\nabla u}\left(\sqrt{\langle\nabla u,\nabla u\rangle}\right) = -\mathrm{div}_x\left(\frac{\nabla u+\nabla u}{2\sqrt{\langle\nabla u,\nabla u\rangle}}\right) = -\nabla\cdot\left(\frac{\nabla u}{|\nabla u|}\right),$$ while the potential energy leads to $$\mathrm{grad}_uV(u) = -f(u)$$, hence the desired equation of motion.
N.B. : you have a superfluous $$\frac{1}{2}$$ factor; you can convince yourself of that with the more usual relation $$\nabla\sqrt{x^2+y^2} = \frac{(x,y)}{\sqrt{x^2+y^2}}$$.