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I have a confusion about the following. If we have a function $f:\mathbb{R}^2 \to \mathbb{R}$ of class $C^2$ and we want to calculate the laplacian of this function polar coordinates.

I don't understand what it means. Does it mean we need to calculate :

$$\frac{\partial}{\partial x} (\frac{\partial f(r \cos \theta, r \sin \theta)}{\partial x}) + \frac{\partial}{\partial y} (\frac{\partial f(r \cos \theta, r \sin \theta)}{\partial y})$$

Or does it mean we need to calculate :

$$\frac{\partial ^2 f}{\partial x^2}(r \cos \theta, r \sin \theta) + \frac{\partial ^2 f}{\partial y^2}(r \cos \theta, r \sin \theta)$$

Moreover, I am really confused since I don't understand the fondamental difference between these two expressions.

Thank you !

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  • $\begingroup$ A function $f(r\cos(\theta),r\sin(\theta))$ does not have a partial derivative with respect to $x$. $\endgroup$ Jan 27, 2019 at 9:24

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The basic interpretation is the second one. The Laplacian $\Delta f(x,y)$ is a function of $x$ and $y$, which is defined as $\frac{\partial^2 f}{\partial x^2}(x,y) + \frac{\partial^2 f}{\partial y^2}(x,y)$, and expressing a function in polar coordinates just means substituting $x=r \cos \theta$ and $y = r \sin \theta$. So first compute the second partials with respect to $x$ and $y$, then substitute.

However, the chain rule gives you another way of obtaining the same result, by performing a different sequence of operations. This involves first expressing $f$ in polar coordinates, and then taking partial derivatives of that expression with respect to $r$ and $\theta$. (Not with respect to $x$ and $y$ – that's sort of pointless, because if you wanted to do that, you should have done it directly on $f(x,y)$ instead of changing to polar coordinates to begin with.) How this works is explained in several other answers on this site, for example here or here.

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  • $\begingroup$ Thank you ! So it means that we have : $\Delta f(x,y) = \frac{\partial ^2 f}{\partial x}(x,y) + \frac{\partial^2f}{\partial y}(x,y)$, and this can be expressed in polar coordinates by substituing so we have : $\Delta f(r \cos \theta, r \sin \theta) = \frac{\partial ^2 f}{\partial x}(r \cos \theta,r \sin \theta) + \frac{\partial^2f}{\partial y}(r \cos \theta,r \sin \theta)$ which is the laplacian in polar coordinates. But an other way to expressed this laplacian in polar coordinates is : $\endgroup$ Jan 27, 2019 at 12:13
  • $\begingroup$ $\frac{\partial}{\partial r} (\frac{\partial f(r \cos \theta, r \sin \theta)}{\partial r}) + \frac{\partial}{\partial \theta} (\frac{\partial f(r \cos \theta, r \sin \theta)}{\partial \theta})$ which can be calculated using the chain rule. $\endgroup$ Jan 27, 2019 at 12:13
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    $\begingroup$ @dghkgfzyukz: No, that's not the correct expression for the Laplacian in polar coordinates! See the questions that I linked to. $\endgroup$ Jan 27, 2019 at 12:57
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    $\begingroup$ @dghkgfzyukz: The first comment was OK, but the formula in the second comment was wrong. $\endgroup$ Jan 27, 2019 at 14:18
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    $\begingroup$ Thanks to the link you provided I think I understand now, just need to do some calculations to see is evrything is effectively clear. Once again thank you for your patience and help. $\endgroup$ Jan 27, 2019 at 14:32

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