- #1

BiGyElLoWhAt

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## Homework Statement

As written: "Calculate ##\frac{\delta H[f]}{\delta f(z)} \ \text{where} \ H=\int G(x,y)f(y)dy##"

and ##\frac{\delta H[f]}{\delta f(z)}## is the functional derivative of H[f] with respect to f(y) at z (I think that's what it is).

## Homework Equations

...

## The Attempt at a Solution

So if I think what's happening here is we're concerned with how the value of H[f] changes with f(y) when y = z.

This gives rise to (changing notation slightly) ##\frac{dF[f]}{d f(y)} = \lim_{\epsilon\ \to\ 0} \frac{F[f(y) + \epsilon \delta (x-y)] - F[f(y)] }{\epsilon}## where ##\delta(x-y)## is the dirac delta function centered at y= x, (I believe this is so, please correct me, my book is not explicit) so this gives us our F[f + epsilon] - F[f] when y = x and F[f]-F[f]=0 when y doesn't equal x. If this is right, going through the example:

##\frac{d H[f]}{d f(z)} ##

##=lim_{\epsilon\ \to\ 0}\frac{\int[G(x,y)(f(z) + \epsilon \delta (x-y))] - G(x,y)f(z)]}{\epsilon }##

##=\left \{

\begin{array}{lr}

-\int G(x,y)dy\ \text{if y=z} \\

0\ otherwise\\

\end{array} \right. ##

I tried to show my train of thought, please correct any misconceptions I have about this. I would also appreciate it if someone would check my answers. Apparently my book doesn't have the answer's to it's exercises.