# Can a functional derivative be calculated if we have a function of more than one variable?

Can a functional derivative be calculated if we have a function of more than one variable?

The functional derivative of, for example, $F[b(x)]=e^{\int_0^{x'} dx a(x,y) b(x)}$ is

$$\frac{\delta F[b(x)]}{\delta b(z)} = a(z,y) e^{\int_0^{x'} dx a(x,y) b(x)}$$

But what about if there it's a functional of more than one variable? - $F[b(x,k)]=e^{\int_0^{x'} \int_0^p dx dk a(x,y,k) b(x,k) }$? Can we write the following?

$$\frac{\delta F[b(x,k)]}{\delta b(z,k)} = a(z,y,k)e^{\int_0^{x'} \int_0^p dx dk a(x,y,k) b(x,k )}$$

Edit: I just want to note that I've been told that you cannot do this by a lecturer. Therefore, is the answer is yes (or no!), please give some details! (I wasn't convinced that he was correct).

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Yes we can – user346 Feb 25 '11 at 14:57
@Deepak thanks! Please see my edit. – Jane Feb 25 '11 at 20:25
for functional differentiation I suggest you check out Peskin and Schroeder, Chap. 9. That should clarify any doubts you have. At the level of a physicist I'd say this is nothing more than the chain rule, though mathematicians would be concerned with such things as measures, compact support etc. – user346 Feb 26 '11 at 2:40
1) The answer to the specific question (v2) Can we write the following? is, strictly speaking No, because the variable $k$ appears on the right-hand side both as an integration variable and also outside the integral. This is inconsistent. However, there is an easy fix, as explained in Roy Simpson's answer. 2) Secondly, I would probably write $F[b]$ instead of $F[b(x,k)]$, as it may give the false impression that the functional $F$ actually depends on a specific $x$ and $k$. – Qmechanic May 2 '11 at 13:04

Jane,

The problem here might be with the exact form of this expression, which seems to have free and bound variables mixed up. Here is an alternative (the problem is with the k).

$$\frac{\delta F[b(x,k)]}{\delta b(z,k')} = a(z,y,k')e^{\int_0^{x'} \int_0^p dx dk a(x,y,k) b(x,k )}$$

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@Roy Simpson, thanks, but is there a reason why I can't set $k=k'$ and/or $z=x$ to find $\delta F[b(x,k)]/\delta b(x,k) = a(x,y,k)e^{\int_0^x' \int_0^p dx dk a(x,y,k) b(x,k)}$? – Jane Feb 28 '11 at 18:30
Or did you just mean that the $x$ and $k$ in $b(x,k)$ are specific values, i.e. $x_1$ and $k_1$, and are not the integration variable? – Jane Mar 1 '11 at 14:56
@Jane, that is essentially all that I meant here. The k is inside the integral under dk, but also outside. There may be different notational conventions in this area, but your example expression seemed to mix up (x to z, but k to k). So your lecturer might have just reacted to an expression that wasnt presented in a notation that made sense, as has Kostya above. – Roy Simpson Mar 1 '11 at 15:23
@Roy Simpson Thanks. Actually the example the lecturer gave included a sum $\delta F[b_l(x,k)]/ \delta b_m(z,k′)=a_m(z,k′)Exp[\sum_l \int^x_0 \int^p_0 dx dk a_l(x,k)b_l(x,k)]$ where $F[b_l(x,k)] = Exp[\sum_l \int^x_0 \int^p_0 dx dk a_l(x,k)b_l(x,k)]$. He said this was wrong, though I don't understand why. – Jane Mar 2 '11 at 17:53
This expression has another problem of a similar sort. Note that x is the limit of the integral, the integration variable and the value in b on the LHS. I would have to study it more to see whether that was the real problem here. – Roy Simpson Mar 2 '11 at 18:05

Kostya's response is almost, but not quite correct. It confuses the Gateaux (or Frechet) derivative with the functional derivative. Let F be a functional, acting on a space M of functions defined on a space X. Then the Gateaux derivative of F at the function f is the functional DF[f] such that, to "first order", we have

F[f + $\delta$f] - F[f] = DF[f] [$\delta$f]

To define the functional derivative, we assume that DF[f] can be written as an integral operator:

DF[f][$\delta$f] = $\int$ ${{\delta F} \over {\delta f(x)}} \delta f(x) dx$

In other words, the functional derivative of F at f is the function (not functional!) of x such that:

F[f + $\delta f$] - F[f] = $\int$ ${{\delta F} \over {\delta f(x)}} \delta f(x) dx$

(Since this is a function, I actually prefer to write it as ${{\delta F} \over {\delta f}}(x)$, but, although this makes more sense, it's in violent conflict with the standard notation.)

This definition of the functional derivative always works, as opposed to definitions using multiples of delta functions (which often don't even belong to the set of functions in question!).

Note that since the functional derivative depends the space of functions, not the points of the underlying space, it doesn't make sense to take a partial functional derivative with respect to a particular coordinate direction. In fact, we may be doing this on a manifold, which has no chosen set of coordinates!

However, it is possible to have a functional defined on a product of spaces of functions. For example, with M as above, define F[f,g] = $\int f(x)g(x)^2 dx$. Then it does make sense to define partial functional derivatives with respect to the different spaces of functions. In our example,

F[f, g+$\delta$g] - F[f, g] = $\int f(x)(g(x)+\delta g(x))^2 dx$ - $\int f(x)g(x)^2 dx$

= $\int 2f(x) g(x) \delta g(x) dx$ = $\int {{\partial F} \over {\partial g(x)}} \delta g(x) dx$

That is, at (f, g), the partial functional derivative of F with respect to the second slot is the function:

${{\partial F} \over {\partial g(x)}} = 2f(x)\cdot g(x)$

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I will not do all the math for you -- just the general idea.

Given a functional $F[g(x_1,x_2,...,x_n)]$. It takes a function of $n$ variables ( I will not write them further) and returns a number. Now you slightly change the argument $F[g+\delta g]$.

Usually the value of this functional for small $\delta g$ is approximately:
$F[g+\delta g] \simeq F[g]+G[g]\delta g$

And $G$ is called the functional derivative $G=\frac{\delta F}{\delta g}$ -- compare this to usual deriavative $f(x+\epsilon)\simeq f(x)+\frac{df}{dx}\epsilon$. So generally you can take a functional derivative over function of any number of variables. An that is the general answer to your general question.

Concerning your example -- there is something strange happening. The expression $\frac{\delta F[b(x,k)]}{\delta b(y,k)}$ just doesn't make any sense to me. Try going step by step in a way I explained.

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Thanks. Why doesn't $\frac{\delta F[b(x,k)]}{\delta b(y,k)}$ make sense? – Jane Feb 28 '11 at 19:56