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 $-(u_{\theta\theta}+u_{rr})$


Not quite.

$$ \frac{\partial}{\partial x}=\frac{\partial}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial}{\partial \theta}\frac{\partial \theta}{\partial x} $$

However, $r=\sqrt{x^2+y^2}$, so $\frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}=\cos (\theta )$. Similarly, $x=r\cos (\theta )$, so $1=\frac{\partial r}{\partial x}\cos (\theta )-r\sin (\theta )\frac{\partial \theta}{\partial x}$, so that $\frac{\partial \theta}{\partial x}=-\frac{\sin (\theta )}{r}$. Plugging this in above, we find

$$ \frac{\partial}{\partial x}=\cos (\theta)\frac{\partial}{\partial r}-\frac{\sin (\theta )}{r}\frac{\partial}{\partial \theta} $$

From here, you have to differentiate this with respect to $x$ to obtain an expression in polar coordiantes for $\frac{\partial ^2}{\partial x^2}$ (you essentially do the same thing I just did, although I'm sure it's a lot more tedious). And then you have to do the same for $y$. It is quite tedious. In the end, you wind up with

$$ \frac{\partial ^2}{\partial x^2}+\frac{\partial ^2}{\partial y^2}=\frac{1}{r}\frac{\partial}{\partial r}\left[ r\frac{\partial}{\partial r}\right] +\frac{1}{r^2}\frac{\partial ^2}{\partial \theta ^2}. $$

I wouldn't worry too much about memorizing the formula. You should know how to derive it if need be and where to look it up if you need to (Wikipedia would surely have it for example).

If you need to see more details about the calculation, let me know.

-Jonny Gleason


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:)

  • 2
    $\begingroup$ This is the way I prefer to do it, but I think you should probably explain what these symbols mean and what formalism you are using here... $\endgroup$ – Zhen Lin May 1 '11 at 18:21

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