I was reading up in the help about the Laplacian operator. and tried to understand the one example I saw:
Laplacian[Sin[r^2],{r,\[Theta]},"Polar"]
results in:
$4 Cos[r^2]-4 r^2 Sin[r^2]$
which is not what I expected. However,
Laplacian[Sin[r^2],{r,\[Theta]}]
does, ${2 Cos[r^2] - 4 r^2 Sin[r^2]}$
What does adding "Polar" to the Laplacian instruction do? Where does the "4" come from (instead of 2)?
thanks
Laplacian[f[r, \[Theta]], {r, \[Theta]}, "Polar"]
vs.Laplacian[f[r, \[Theta]], {r, \[Theta]}]
to see where the difference comes from. It becomes even clearer when you assume only a radially changing function like in your example:Laplacian[f[r], {r, \[Theta]}, "Polar"]
vs.Laplacian[f[r], {r, \[Theta]}]
. Essentially, the"Polar"
version has an additionalf'[r]/r
term that the cartesian version doesn't have. $\endgroup$ – Thies Heidecke Apr 24 '19 at 15:35