# Functional Derivative Calculation

Given the functional: $$F[\phi] = \int_V \frac{k_B T}{a^3}\phi\ln(\phi) \ ds = \int_V I(\phi) ds$$ I want to find the functional derivative. I believe this would result in: $$\frac{\delta F}{\delta \phi} = \frac{\partial I}{\partial \phi}=\frac{k_B T}{a^3}[\ln(\phi)+1]$$ However, the paper I am following along has only the first term. Is my calculation correct? Note, in this case I set the functional derivative equal to the partial derivative because the functional doesn't contain any higher derivatives - hence those partials vanish.

• Which paper?... Jul 27, 2022 at 14:56
• The last formula you wrote (that is correct) is not the partial derivative of $F$ but the partial derivative of the integrand of F. Notice that in its present form, your question is not about Physics and looks like a homework-like question. Jul 27, 2022 at 14:59
• Oh thanks, that is a typo: by the partial derivative of $F$ with respect to $\phi$, I had meant that of the integrand which I should have defined as something else - maybe $I(\phi)$. And no this is not a homework question. I am merely trying to follow the derivation of the gradient dynamics equations for $h,\psi$ as given in a thesis I was advised to review for my own research. Given the form of the question I asked though, perhaps it would've been more appropriate to post to the math stackexchange Jul 27, 2022 at 16:14
• Never too late to supplant your I(φ) in... Jul 27, 2022 at 20:37

Take a compactly supported smooth function $$\psi$$, then by the definition of the functional derivative: \begin{align*} \int_V\frac{\delta F[\phi]}{\delta\phi}\psi\; ds \stackrel{!}{=}\left[\frac{\mathrm d}{\mathrm d\varepsilon}F[\phi+\varepsilon \psi]\right]_{\varepsilon=0} =\ldots =\int_V\frac{k_\mathrm{B}T}{a^3}(\ln(\phi)+1)\psi\mathrm ds, \end{align*} from which the result follows with the fundamental theorem of the calculus of variations. I think you can fill in the two missing steps, where you just have to put in your expression of the functional, for yourself.
• Yes, I can fill in the missing steps to obtain the RHS of your answer. Thank you for this response! As a generalization, given some functional with an integrated $F[\phi,\nabla\phi]$, or even possibly higher derivatives, how would this process change? Jul 27, 2022 at 16:19
• The functionals considered in physics are often given as $F[\phi]=\int L(\phi,\nabla\phi)$ with $L$ being the Lagrange density. Then you have: $$\frac{\delta F[\phi]}{\delta\phi}=\frac{\partial L}{\partial\phi}-\nabla\frac{\partial L}{\partial\nabla\phi}.$$ Putting this equal to zero is the Euler-Lagrange equation. If you add in higher derivatives, you get the more general Euler-Possion equation. Jul 27, 2022 at 16:33