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I am trying to convert $\Delta=\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$ to polar coordinates. If anyone has any references on how to do that I would appreciate it.

In my evaluation, I am messing with a lot of partials. Alghough, I am not sure as to if partials are commutative, and google has not been helping me. Enough of a prompt; is this true? $$ \frac{\partial u}{\partial x}\frac{\partial y}{\partial r}=\frac{\partial u}{\partial r}\frac{\partial y}{\partial x} $$ In the previous equation I just switched the denominators.

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    $\begingroup$ If you just want the answer, Wikipedia has it (look at cylindrical and remove the $z$ part). And no, $$\frac{\partial u}{\partial x}\frac{\partial y}{\partial r}\color{red}{\ne}\frac{\partial u}{\partial r}\frac{\partial y}{\partial x}$$ in general. $\endgroup$ – user137731 Oct 31 '16 at 3:49
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First of all: you cannot just switch the denominators like that.

Partial derivatives kind of commute, but not in the sense you are implying. Suppose I have a function $f(x,y)$. Then it is the case that

$$ \frac{\partial}{\partial x} \frac{\partial}{\partial y} f(x,y) = \frac{\partial}{\partial y} \frac{\partial}{\partial x} f(x,y) $$

The partial derivatives commute in this particular case since $x,y$ are independent. That is, $\partial x/\partial y = \partial y/\partial x = 0$.

It is not the case, however, that arbitrary partial derivatives commute. For example, if we were to introduce $r = \sqrt{x^2 + y^2}$, then the partials would not commute.

$$ \frac{\partial}{\partial x} \frac{\partial}{\partial r} f(x,y) \ne \frac{\partial}{\partial r} \frac{\partial}{\partial x} f(x,y) $$

It also isn't the case that partial derivatives commute with functions. For example

$$ \frac{\partial}{\partial x} f(x,y) g(x,y) \ne f(x,y) \frac{\partial}{\partial x} g(x,y) $$

To compute gradients and other vector calculus operations in different coordinate systems, you need to use the multivariate chain rule.

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What you're looking for is the "Polar Laplacian". Googling "Polar Laplacian Derivation" leads to many papers such as this one that go through step-by-step with how to derive it.

If after reading this you're still interested in a challenge, try deriving the Laplacian for Spherical or Cylindrical coordinates.

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