TITLE: How to transform a Laplacian operator from (x,y) coordinate system to polar system?
QUESTION [8 upvotes]: for any functions in $C_2^2$, we have a $-D^2$ operator $-D^2u=-(u_{xx}+u_{yy})$
However now i need to transform this operator from $(x,y)$ to $(r,\theta)$ for some sphere boundary condition, how am I suppose to do? As we know that $x=r\cos\theta,y=r\sin\theta$.
the form should be something near $-(u_{\theta\theta}+u_{rr})$

REPLY [8 votes]: here is another calculation: $\Delta f=*d*df$.
$df=\partial_r f\, dr + \partial_\theta f\,d\theta$. We have  $*dr=r\,d\theta$ and  $*d\theta=-dr/r$, so $*df=(\partial_r f)\, r\,d\theta - (\partial_\theta f)\,dr/r$, hence $d*df=\partial_r(r(\partial_r f))\,dr\wedge d\theta + (\partial_\theta(\partial_\theta f)/r)\,dr\wedge d\theta$, and finally $*d*df=\partial_r(r(\partial_r f))/r + \partial_\theta(\partial_\theta f)/r^2$.
edit:(on request) I think I won't explain what a differential form is. $*$ denotes the Hodge operator, which is here simply the rotation by $90$ degrees (for $1$-forms) and which maps the area form ($dx\wedge dy = r dr\wedge d\theta$) to $1$. This is probably familiar to anybody who knows what a differential form is, so the answer is not very useful (besides saying - learn differential forms if you don't know them yet:)