Can we teach Mathematica about functional differentiation?

Question

The key relation for functional differentiation is $$\frac{\delta}{\delta f(y)}f(x)=\delta(x-y), $$ where $\delta(x-y)$ is the Dirac delta function, and the usual properties of differentiation (e.g. linearity, chain rule) still hold.

Another example: $$\frac{\delta}{\delta f(y)}(f(x)f(x))=2 \delta(x-y) f(x).$$

A possible Mathematica version could look like this

FunctionalD[f[x],f[y]]=DiracDelta[x-y]

How would one go about implementing this?

EDIT:

There's also the case of multiple arguments e.g. $$\frac{\delta}{\delta f(y)}(f(x_1)f(x_2))= \delta(x_1-y) f(x_2)+\delta(x_2-y) f(x_1).$$

Answer 1

FunctionalD[expr_, f_[y_]] := 
  With[{xs = DeleteDuplicates[Cases[expr, f[x_ /; FreeQ[x, _f]] :> x, {0, Infinity}]]},
    Total[D[expr, f[#]] DiracDelta[# - y] & /@ xs]
  ]

edit 1: the FreeQ is needed to get the chain rule correct for f[f[x]]

edit 2: the DeleteDuplicates is needed per Stephen Luttrell's comment. I still need to think about Chris' comment.

Answer 2

Mathematica already knows quite a lot about functional derivatives. In particular, you can do variational derivatives. That is, you have to give it the functional and the function (I would strongly suspect that your problem can be written so as to use the VariationalD function). To get started, have a look at the tutorial for the Variational Methods package.

Edit

In case someone really needs the translation from VariationalD to the functional derivative in the question, here it is. Of course, you have to load the package I mentioned above before doing anything else.

Needs["VariationalMethods`"];
FunctionalD[expr_, fn_, var1_, var2_] := 
 VariationalD[(expr /. var1 -> var2)  DiracDelta[var1 - var2], fn, var2]

The first argument is the expression expr to be differentiated, the second is the function fn with respect to which the derivative is taken. As the third and fourth argument, you have to provide the names of the independent variables: var1 refers to expr and var2 to fn.

An example corresponding to the discussion of Jeremy's answer :

FunctionalD[f'[x], f[y], x, y]

(* ==> Derivative[1][DiracDelta][x - y] *)

The only thing I did in FunctionalD is to multiply the given expression expr by a delta function such that the resulting integral underlying VariationalD leaves only expr evaluated at var1. The rest is done by VariationalD, no need to re-invent those manipulations.

Edit 2

I just came across another approach to defining the functional derivative without recourse to the VariationalMethods package. It's directly from the documentation for DiracDelta (under "Applications"):

FunctionalD[functional_, f_[y_]] := 
 Assuming[ Element[y, Reals], 
   Limit[((functional /. f :>  Function[x, f[x] + 
                           ε DiracDelta[x - y]]) - 
      functional)/ε, ε -> 0]]

FunctionalD[f[x]^2, f[y]]

(* ==> 2 DiracDelta[x - y] f[x] *)

FunctionalD[f[x1] f[x2], f[y]]

(* ==> DiracDelta[x2 - y] f[x1] + DiracDelta[x1 - y] f[x2] *)

This satisfies all the requirements of the question without loading a package.

Answer 3

When I did this in FeynCalc some 17 years ago I did not care about DeltaFunction, since it is not really needed, at least not for deriving Feynman rules. You can find a couple of examples here : http://www.feyncalc.org/FeynCalcBook/FunctionalD/

So

FunctionalD[QuantumField[f]^2, QuantumField[f]] 

results in 2 QuantumField[f] The code is free, so you can just download the package and look at FunctionalD.m and friends (like RightPartialD.m, ExpandPartialD.m , etc. )

It is all a bit rusty and should be rewritten, but that takes time.