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Is it possible to compute

$\Delta F=\nabla^2 F$

for $F=F(x^1,x^2,…,x^D)$ with $D$ finite?

I seem to be only able to specify spherical coorindates, for example, if I want to leave standard Cartesian coordinates.

The function I'm actually using looks like this

$F(x^1,x^2,…,x^D)=f^1(x^1)\times f^2(x^2) \times … \times f^D(x^D)$

(separation of variables). Is there a way mathematica can do this?

Phibert
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  • There is in version 9 a new function Laplacian, however in earlier versions on could do it this way Using D to find a symbolic derivative. – Artes Jun 25 '14 at 14:42
  • But using version 9, I don't know how to tell mathematica which are the variables when using this Laplacian function. It seems to be default at Cartesian coordinates and works for spherical, cylindrical, etc. But not for how ever many dimensions you like. – Phibert Jun 25 '14 at 14:48
  • Since what you're asking for is "calculate the Laplacian in $d$ dimensions in a specific coordinate system", you might want to say that in the question – acl Jun 25 '14 at 20:39

1 Answers1

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Using the Laplacian operator of v9 you can extract the list of variables and insert it into the second slot as follows:

ql = Laplacian[#, List @@ #] &

so eg

ql[f[x, y, z, p, q]]

or even

ql[f @@ (x[#] & /@ Range[10])]

works. This assumes Cartesian coordinates.

No doubt there are better methods.

acl
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  • Does it matter what coordinates I'm using? I'm actually using $D$ dimensional space-time i.e. coordinates $t,r,angles$ where there can be $D-2$ angles. This will still work? – Phibert Jun 25 '14 at 15:53
  • If you evaluate what I gave you'll see it assumes Cartesian coordinates. Have you looked at the third argument of Laplacian in the docs? Although personally I'd just implement it by hand (the Laplacian for d-dimensional cylindrical coordinates is not hard to calculate and you can just modify my code above to implement that instead of the Cartesian one) – acl Jun 25 '14 at 16:03