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Let $$\Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$$ Consider the polar coordinates with $x=r\cos\theta$ and $y=r\sin\theta$. I want to show that $$\Delta=\frac{\partial^2}{\partial r^2}+\frac1r\frac\partial{\partial r}+\frac1{r^2}\frac{\partial^2}{\partial \theta^2}.$$

I use the chain rule to compute $$\frac{\partial}{\partial x} = \frac{x}{r} \frac{\partial}{\partial r} -\frac{y}{r^2} \frac{\partial}{\partial \theta}$$ and $$\frac{\partial}{\partial y} = \frac{y}{r} \frac{\partial}{\partial r} +\frac{x}{r^2} \frac{\partial}{\partial \theta}$$

So $$\Delta=\left (\frac{x}{r} \frac{\partial}{\partial r} -\frac{y}{r^2} \frac{\partial}{\partial \theta} \right )^2 + \left (\frac{y}{r} \frac{\partial}{\partial r} +\frac{x}{r^2} \frac{\partial}{\partial \theta} \right )^2$$

Expanding, I get $$\Delta=\frac{\partial^2}{\partial r^2}+\frac1{r^2}\frac{\partial^2}{\partial\theta^2}$$

Where does the term $\frac1r\frac\partial{\partial r}$ disappear?

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  • $\begingroup$ You need the second derivatives with respect to $x$, $y$. And note that the square of the partial is not the second partial. $\endgroup$ Nov 2, 2013 at 4:34
  • $\begingroup$ @AndréNicolas Could you please be more specific? I'm not sure where the second derivatives should go. $\endgroup$
    – Mika H.
    Nov 2, 2013 at 4:35
  • $\begingroup$ You need to compute (among other things) $\frac{\partial^2}{\partial x^2}$, so you need to differentiate your expression for $\frac{\partial}{\partial x}$ again with respect to $x$. $\endgroup$ Nov 2, 2013 at 4:38
  • $\begingroup$ @AndréNicolas Could you show, as an example, how to compute $\frac{\partial^2}{\partial x^2}$? I'm not sure how to do that. $\endgroup$
    – Mika H.
    Nov 2, 2013 at 4:42

1 Answer 1

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You have to rewrite everything in terms of $r$ and $\theta$, so where you are leaving $x$ and $y$ behind you have to clear up. Also, these are operators, so you have to have them act on a test function $f$ -- this also makes you be a bit more careful.

What you'll find is that, since $\Delta = \nabla \cdot \nabla$, you need to compute the Jacobian for the transformation, that is \begin{equation} \frac{\partial f}{\partial x} = \frac{\partial f}{\partial \theta} \frac{\partial \theta}{\partial x} + \frac{\partial f}{\partial r} \frac{\partial r}{\partial x} \end{equation} etc. This gets you how to rewrite the $\nabla$ operator in polar coordinates. Then you need to figure out what $\nabla \cdot (\nabla f)$ is to figure out what $\Delta f$ is.

There is also a metric tensor way to rewrite the Laplacian, see under http://mathworld.wolfram.com/Laplacian.html for example.

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  • $\begingroup$ This doesn't help me do the computation I'm stuck with ... $\endgroup$
    – Mika H.
    Nov 2, 2013 at 5:05
  • $\begingroup$ Sure it does. Your calculation is leaving $x$ and $y$ floating around in the partial derivatives. If you do a full change of coordinates, you have to completely eliminate the old coordinates. That's what I told you how to do. $\endgroup$
    – webb
    Nov 4, 2013 at 0:48

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