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06. Laplace

Laplace in Cartesian Coordinates

L f(x):=⁄{∂}{∂x_i} ⁄{∂}{∂x_i} f

Laplace in Spherical Coordinates

Let x be the vector in old (cartesian) coordinates.

Let x̄(x) be the vector field in the new coordinates, as a function of the old coordinates.

Let f(x̄(x)) be the scalar field to be Laplaced.

Then by the chain rule it follows that:

⁄{∂f}{∂x_i}=⁄{∂f}{∂x̄_k}⋅⁄{∂x̄_k}{∂x_i}

By substitution then follows:

L f=⁄{∂}{∂x_i}(⁄{∂f}{∂x̄_k}⋅⁄{∂x̄_k}{∂x_i})
L f=⁄{∂x̄_l}{∂x_i}⋅⁄{∂}{∂x̄_l}(⁄{∂f}{∂x̄_k}⋅⁄{∂x̄_k}{∂x_i})

By product differentiation it follows that:

L f=⁄{∂x̄_l}{∂x_i}⋅((⁄{∂}{∂x̄_l}⁄{∂f}{∂x̄_k})⋅⁄{∂x̄_k}{∂x_i}+⁄{∂f}{∂x̄_k}⋅⁄{∂}{∂x̄_l}⁄{∂x̄_k}{∂x_i})
L f=⁄{∂x̄_l}{∂x_i}⋅((⁄{∂}{∂x̄_l}⁄{∂f}{∂x̄_k})⋅⁄{∂x̄_k}{∂x_i}+⁄{∂f}{∂x̄_k}⋅0)
L f=⁄{∂x̄_l}{∂x_i}⋅(⁄{∂}{∂x̄_l}⁄{∂f}{∂x̄_k})⋅⁄{∂x̄_k}{∂x_i}
L f=⁄{∂x̄_l}{∂x_i}⋅⁄{∂x̄_k}{∂x_i}⋅⁄{∂}{∂x̄_l}⁄{∂}{∂x̄_k}f

Given the spherical transformations:

x_1:=r⋅(\cos φ)⋅\sin ϑ
x_2:=r⋅(\sin φ)⋅\sin ϑ
x_3:=r⋅\cos ϑ

We arrive at the reverse transformations:

√{x_i⋅x_i}=r
\arctan ⁄{x_2}{x_1}=φ
\arccos ⁄{x_3}{r}=ϑ

And their derivations:

⁄{∂r}{∂x_i}=⁄{x_i}{r}
⁄{∂φ}{∂x_i}=⁄{x_1²}{x_1²+x_2²}⋅⁄{∂}{∂x_i}⁄{x_2}{x_1}
⁄{∂ϑ}{∂x_i}=-(1-⁄{x_3²}{r²})^{-⁄{1}{2}}⋅⁄{∂}{∂x_i}⁄{x_3}{r}

Because:

⁄{∂}{∂a} \arctan a=⁄{1}{1+a²}
⁄{∂}{∂a} \arccos a=⁄{-1}{√{1-a²}}

With the following helpers:

⁄{∂}{∂x_i}⁄{1}{r}=-r^{-2}⋅⁄{∂r}{∂x_i}
⁄{∂}{∂x_i}⁄{1}{r²}=-2⋅r^{-3}⋅⁄{∂r}{∂x_i}

And labelling these as the new coordinates:

x̄:=(r,φ,ϑ)

Author: Danny (remove the ".nospam" to send)

Last modification on: Sat, 04 May 2024 .