# Functional Derivative of Convolution

How to carry out the following functional derivative?

$$\frac{\delta F}{\delta n(r)}$$ where $$F=\int dr n(r) \int C(|r-r'|) n(r') dr'$$

is it simply: $$2 \int dr' C(|r-r'|) n(r')$$?

-
The answer to the question (v1) is Yes. – Qmechanic Sep 1 '11 at 18:18

Notice that $F$ is essentially a quadratic form; i.e. if it were matrices then you would have (in summation notation): $$F = x_i C_{ij} x_j.$$ Then you would use the fact that $\frac{\partial x_i}{\partial x_j} = \delta_{ij}$ to get \begin{align} \frac{\partial F}{\partial x_k} &= \delta_{ik} A_{ij} x_{j} + x_i A_{ij} \delta_{jk} \ &= 2 A_ik x_k \end{align} if $A_{ij} = A_{ij}$ i.e. it is symmetric.
Here, we use a similar fact: $$\frac{\delta n(x)}{\delta n(y)} = \delta(x-y)$$ where $\delta$ this time is the Dirac distribution. Your "matrix" in the middle is obviously symmetric, so your proposed answer is correct.