green function, functional derivative

Question

I am trying to find ${\delta F}/{\delta u}$ for the functional:

$F[u]=\int u(x)\int G(x,y)u(y)dy dx $

G is green function for laplace operator.

is there Euler-Lagrange version for double intrgral?

( lagrangian: $L[x,y,u(x),u(y)]$)

Answer 1

I'm tring to find such result too. What about the following?

Write

$ \delta F[u]=\int \frac{\delta F[u]}{\delta u} \phi(x) dx=[\frac{d}{d\epsilon}\int (u(x)+\epsilon \phi(x)) \int G(x,y) (u(y)+\epsilon \phi(y))dy dx]_{\epsilon=0} $

from which:

$= \int \phi(x) \int G(x,y) u(y) dy dx + \int u(x) \int G(x,y) \phi(y) dy dx$

I can switch the variable of the second integral:

$= \int \int G(x,y) u(y) dy \phi(x) dx + \int \int u(y) G(y,x) \phi(x) dy dx$

and if if $G(x,y)=G(y,x)$

$= \int \int G(x,y) u(y) dy \ \phi(x) dx + \int \int u(y) G(x,y) dy \ \phi(x) \ dx$

$= \int \int 2 G(x,y) u(y) dy \ \phi(x) \ dx = \int \frac{\delta F[u]}{\delta u} \phi(x) dx $

from which:

$ \frac{\delta F[u]}{\delta u} = \int 2 G(x,y) u(y) dy $

what do you think?