# Curl and other functions of functions

Where can one find the source code of the Mathematica function Curl? Alternatively, how can one define a function the argument of which is restricted to be a 3D vector function of three variables named x, y, z?

I would define

curl[u_?(VectorQ[#] && Length[#] == 3 &)] := {D[u[], y] -
D[u[], z], D[u[], z] - D[u[], x],
D[u[], x] - D[u[], y]}


EDIT

This function is intended to deal specifically with lists of length 3. It can handle symbolic inputs, but these must be explicitly defined as having 3 components.

curl[{fx[x, y, z], fy[x, y, z], fz[x, y, z]}]

{-Derivative[0, 0, 1][fy][x, y, z] + Derivative[0, 1, 0][fz][x, y, z],
Derivative[0, 0, 1][fx][x, y, z] - Derivative[1, 0, 0][fz][x, y, z],
-Derivative[0, 1, 0][fx][x, y, z] + Derivative[1, 0, 0][fy][x, y, z]}


This is certainly not the only way (and may not be the best way) of defining curl, but matched the OP's request for an implementation that applied to 3-vectors. (I assumed that the originator was aware of the built-in Curl and wanted something different).

• Thank you, I see now how to specify u, thanks to the use of #. – André Bellaïche Aug 26 '16 at 21:15
• Is it possible to modify this definition in order curl works like D ? I mean: With D, you may do some formal calculus. Try D[f[x], x] or D[f[x, y, z], {{x, y, z}, 1}] or D[2 f[x], x] - 2 D[f[x], x]. But if you enter curl[2 f[x]] - 2 curl[f(x]], you don't get 0. Even if you use Expand, Simplify, etc. you are stuck with your input. – André Bellaïche Aug 26 '16 at 22:47
• @AndréBellaïche you would have to start expanding the program by adding definitions for curl. – QuantumDot Aug 27 '16 at 18:31