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 !
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.