- #1
Bishamonten
- 17
- 1
Homework Statement
"The functional ## J[f] = \int [f(y)]^pφ(y)\, dy ## has a functional derivative with respect to ## f(x) ## given by:
$$ \frac {δJ[f]} {δf(x)} = \lim_{ε \rightarrow 0} \frac 1 ε \left[ \int[f(y) + εδ(y-x)]^pφ(y)\, dy - \int [f(y)]^pφ(y)\, dy\right] $$
$$ = p[f(x)]^{p-1}φ(x) $$
Homework Equations
Limit definition of a functional derivative:
$$ \frac {δF} {δf(x)} = \lim_{ε \rightarrow 0} \frac {F[f(x') + εδ(x-x')] - F[f(x')]} ε$$
The Attempt at a Solution
From line 1:
$$ = \lim_{ε \rightarrow 0} \frac 1 ε \left[ \int[f(y) + εδ(y-x)]^pφ(y)\, dy - \int [f(y)]^pφ(y)\, dy\right] $$
$$ = \lim_{ε \rightarrow 0} \frac 1 ε \left[ \int \sum_{k=0}^p
\begin{pmatrix}
p \\
k \\
\end{pmatrix}
[f(y)]^{p-k}[εδ(y-x)]^kφ(y)\, dy - \int [f(y)]^pφ(y)\, dy\right] $$
$$ \lim_{ε \rightarrow 0} \frac 1 ε \left[ \int \sum_{k=0}^p \frac {p!} {k!(p-k)!} [f(y)]^{p-k}[εδ(y-x)]^kφ(y)\, dy - \int [f(y)]^pφ(y)\, dy\right] $$
This comes straight out of Lancaster and Blundell's "Quantum Field Theory for the Gifted Amateur", and is in the first chapter where they are introducing Lagrangians, Fermat's principle of least time, and Hamilton's principle of least action. This is simply them demonstrating functional derivatives, only, I wanted to see how they arrived from the limit definition of the functional ## J[f] ## to ## p[f(x)]^{p-1}φ(x) ##.
I wanted to evaluate the limit myself, through first evaluating the integral. I saw the power of the sum expression and thought ok, maybe I should introduce the binomial theorem, but I feel like I'm only making a mess of things and am not arriving any closer at ## p[f(x)]^{p-1}φ(x) ##, may someone let me know if I'm going at this in the right direction, and if not, any pointers, or if I'm simply making this more complicated than it should be?