# Understanding functional derivative

1. Oct 25, 2017

### Bishamonten

1. The problem statement, all variables and given/known data

"The functional $J[f] = \int [f(y)]^pφ(y)\, dy$ has a functional derivative with respect to $f(x)$ given by:

$$\frac {δJ[f]} {δf(x)} = \lim_{ε \rightarrow 0} \frac 1 ε \left[ \int[f(y) + εδ(y-x)]^pφ(y)\, dy - \int [f(y)]^pφ(y)\, dy\right]$$

$$= p[f(x)]^{p-1}φ(x)$$

2. Relevant equations

Limit definition of a functional derivative:

$$\frac {δF} {δf(x)} = \lim_{ε \rightarrow 0} \frac {F[f(x') + εδ(x-x')] - F[f(x')]} ε$$

3. 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?

2. Oct 25, 2017

### Orodruin

Staff Emeritus
Which terms in your binomial expansion survive the limit $\varepsilon \to 0$?

3. Oct 25, 2017

### Bishamonten

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Oooh, ok I don't know why I didn't put it in its expanded form earlier. Maybe it was my lack of confidence talking Thank yo so much for responding!

So when I expand the term out, the 1st term in the series matches $-\int [f(y)]^pφ(y)dy$, except it's positive, thus cancelling. Every other term in the series besides the 2nd term $\frac {p!} {(p-1)!} [f(y)]^{p-1}δ(y-x)φ(y)$ zeroes out when evaluating the limit because they will all continue to have the kth power of epsilon multiplying it, whereas the 2nd term has it's epsilon cancel the one distributed, and of course $\frac {p!} {(p-1)!}$ is just p.

So here's what I got so far, I am a bit uncomfortable with the dirac delta function argument I make in this last step however, as you'll see:

$$p\int[f(y)]^{p-1}φ(y)δ(y-x)dy$$

$$[f(y)]^{p-1}φ(y) ≡ g(y)$$

$$= p\int g(y)δ(y-x)dy = g(x) = p[f(x)]^{p-1}φ(x)$$

Is that a proper use of the indefinite integral of this delta function?

4. Oct 25, 2017

### Orodruin

Staff Emeritus
That is the definition of how the delta function works, so yes, it is perfectly fine.

5. Oct 25, 2017

### Bishamonten

Thank you very much for the help. We touched up on the dirac delta function during the last part of our differential equations class, but looks like I need to review how it works a bit. I remember that this was the definition of how it works, but the intuition for why it works that way didn't survive me unfortunately.

Also, I wish to ask another question while I'm at it, related to tensors, covariant and contravariant vectors. May I post it here, or should I start a new thread?

6. Oct 25, 2017

### Orodruin

Staff Emeritus
It works like that because it is defined to work like that. For more in depth on what it really is you would need to study distributions (it really is a distribution and not a function), in particular in terms of derivatives and how it relates to differential equations.

You should start a new thread. If it is a problem like this one, post it in the appropriate homework forum. If it is a more general question, you can post it in one of the technical forums.

7. Oct 25, 2017

### Bishamonten

Once again, thank you for all your concise help Orodruin, I've no doubt I'll be needing it again soon enough

8. Oct 25, 2017

### Bishamonten

Sorry, post made by mistake