0
$\begingroup$

I was wondering how the functional derivative $\frac{ \delta _{ } F}{ \delta _{ } f(x)}$ would read in Fourier/reciprocal space? To give some more detail, I'm thinking of the functional $F=F[f(x)]$ and suppose that $f(x)$ can be written as the Fourier series $f(x) = \sum \limits_{k}^{} \tilde{f}_k e^{i kx}$. Can $\frac{ \delta _{ } F}{ \delta _{ } f(x)}$ be written in terms of $\frac{\partial F}{\partial \tilde{f}_k}$ or something similar?

$\endgroup$

1 Answer 1

1
$\begingroup$

Let $F$ be a functional and $f$ a function in the domain of $F$. Furthermore assume that $f(x)=\sum_k f_k e_k$ where $f_k$ are real or complex numbers and $e_k$ are some basis functions (e.g. $e_k=e^{ik2\pi x/L}$). Then, $$ \frac{\partial F[f]}{\partial f_k} = \frac{\partial F[\small\sum_k f_k e_k]}{\partial f_k} = \int \frac{\delta F[\small\sum_k f_k e_k]}{\delta \small\sum_k f_k e_k(y)} \frac{\partial \small\sum_k f_k e_k(y)}{\partial f_k} dy = \int \frac{\delta F[f]}{\delta f(y)} e_k(y) \, dy . $$ The rightmost integral is an inner product $\langle \delta F[f], e_k \rangle,$ where $\delta F[f](y) = \frac{\delta F[f]}{\delta f(y)},$ so by inversion we can write $$ \frac{\delta F[f]}{\delta f(y)} = \sum_k \frac{\partial F[f]}{\partial f_k} e_k(y). $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .