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I am studying about the functional derivative and found some specific examples and formulas for the energy and potential functionals used in physics. https://en.wikipedia.org/wiki/Functional_derivative#Thomas%E2%80%93Fermi_kinetic_energy_functional

These formulas, such as Coulomb potential energy functional, have the form as $F[\rho(r)] = \int f(\rho(r))dr$; that is, this is a local functional with respect to $r$.

For such functional derivative of $F$, is it enough to simply differentiate $f(\rho(r))$ in the integral?

My assumption may be different for more general functionals, but for such a simple local form of the functional, is this basically the right understanding?

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Yes, this is a simple case. The variational differential is $$ \delta F[\rho] = \int f'(\rho(r)) \, \delta\rho(r) \, dr. $$

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  • $\begingroup$ Thank you for the quick reply. I understood! $\endgroup$
    – neco
    Sep 9, 2021 at 12:15

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