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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?

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

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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 $\nabla\sqrt{x^2+y^2} = \frac{(x,y)}{\sqrt{x^2+y^2}}$.

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