Laplace operator on a polar coordinate transformation I have an exercise in my analysis class to do and am entirely lost by the notation of our Prof. Maybe somebody experienced could explain to me what is meant. I will write down first the exercise:
Let $f \in \mathcal{C}^2(\mathbb{R}^2)$ be a function. With the transformation $(x,y) = (r \cos(\varphi), r \sin(\varphi))$ we get a function $F: \mathbb{R}^2 \rightarrow \mathbb{R}$,
$$
F(r(x,y), \varphi(x,y)) := f(r \cos(\varphi), r \sin(\varphi))
$$
This is how the exercise is stated. However, in my opinion, the Prof. uses $\varphi$ once as a function and the other time as an angle. This does not make sense to me.
The goal of the exercise is to compute
$$
\Delta F = F_{rr} + \frac{1}{r} F_r + \frac{1}{r^2} F_{\varphi\varphi}
$$
with
$$
F_{rr} = \frac{\partial^2 F}{\partial r^2} \quad F_{\varphi\varphi} = \frac{\partial^2 F}{\partial \varphi^2}.
$$
Here, $\Delta$ denotes the Laplace operator, defined by $\Delta F = \text{div}(\nabla F)$. It would be enough for me to explain how I can compute the gradient of $F$, because the notation at the very beginning confuses me. Thank you in advance.
 A: Yes $F$ is your function in $(r, \phi)$ coordinates, while $f$ is your function in $(x, y)$ coordinates. A possibly clearer way to define $F$ is
$$F(r, \phi) = f(x(r, \phi), y(r, \phi)).$$
Let $d = \partial$ for ease of typing. Conceptually, you are given an operator, e.g. $\frac{d}{dx}$ that acts on functions $f$ defined in $(x, y)$ coordinates and you want to write down how $\frac{d}{dx}$ changes it's polar coordinate representation $F$. Here, since we need to differentiate with respect to $x$, it is useful to write
$$f(x, y) = F(r(x, y), \phi(x, y)).$$
Naively, we can apply the chain rule to obtain $\frac{df}{dx}$:
$$\frac{df}{dx}(x, y) = \frac{dF}{dr}(r(x,y),\phi(x,y))\frac{dr}{dx}(x, y) + \frac{dF}{d\phi}(r(x, y), \phi(x, y))\frac{d\phi}{dx}(x, y).$$
It is common to supress evaluation and write
$$\frac{df}{dx} = \frac{dF}{dr}\frac{dr}{dx} + \frac{dF}{d\phi}\frac{d\phi}{dx}.$$
But this isn't exactly what we want since $\frac{dr}{dx}$ is an expression involving $x, y$, while we want only expressions involving $r, \phi$. To get these, use the Jacobian matrices of the coordinate transformation maps to get
$$\frac{d(r, \phi)}{d(x, y)} = \left(\frac{d(x, y)}{d(r, \phi)}\right)^{-1}.$$
Even more systematically, using Jacobians,
$$D_{(x, y)}f(x, y) = D_{(r, \phi)}F(r, \phi)\frac{d(r, \phi)}{d(x, y)} = D_{(r, \phi)}F(r, \phi)\left(\frac{d(x, y)}{d(r, \phi)}\right)^{-1}.$$
