Re: Wierdness with the laplacian
- To: mathgroup at smc.vnet.net
- Subject: [mg45161] Re: Wierdness with the laplacian
- From: adam.smith at hillsdale.edu (Adam Smith)
- Date: Fri, 19 Dec 2003 06:57:28 -0500 (EST)
- References: <brpm16$7u4$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
No real weirdness here. The answer is correct. Applying Simplify to
your resulting expression gives what you expect.
<< Calculus`VectorAnalysis`
SetCoordinates[Spherical[r, theta, phi]]
Simplify[Laplacian[Exp[r^2]]]
Out[14]//OutputForm=
2
r 2
2 E (3 + 2 r )
The form you obtained initially is due to the way that Mathematica
represents the Laplacian. Try the following which will produce a
general form of the Laplacian for a function of r,theta and phi
Laplacian[f[r, theta, phi]]
If you compare the resulting output to a typical formula in a
published text you will observe that Mathematica "expands" out all the
terms involving differentian and leaves any terms involving trig
functions of theta unsimplified. The two forms look quite different
but are equivalent.
Adam Smith
Peter Jay Salzman <p at dirac.org> wrote in message news:<brpm16$7u4$1 at smc.vnet.net>...
> Hi all,
>
> When I do:
>
> <<Calculus`VectorAnalysis`
> SetCoordinates[Spherical[r, theta, phi]
> Laplacian[ Exp[r^2] ]
>
> Mathematica gives:
>
> 2 2
> r 2 r 4
> Csc[theta] (6 E r Sin[theta] + 4 E r Sin[theta])
> ------------------------------------------------------
> 2
> r
>
> Which is wrong since Exp[r^2] is not a function of theta or phi.
>
>
> Why is Mathematica giving this result?
>
> Thanks!
> Pete