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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?

J. M.'s missing motivation
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André Bellaïche
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1 Answers1

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I would define

curl[u_?(VectorQ[#] && Length[#] == 3 &)] := {D[u[[3]], y] - 
   D[u[[2]], z], D[u[[1]], z] - D[u[[3]], x], 
  D[u[[2]], x] - D[u[[1]], 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).

mikado
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  • 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