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
    Commented Oct 31, 2016 at 3:49

2 Answers 2


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.


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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