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Given $g(r, \Theta) = f(x(r,\Theta), y(r,\Theta))$ and $x = r \cdot \cos (\Theta)$ and $y = r \cdot \sin(\Theta)$ (so f is undetermined)

First thing I gotta clear. Is my assumption righ: $\partial_r^2 g = 2\,\sin\Theta\,\cos\Theta\,\partial_x\partial_y\,f+\cos^2\Theta\,\partial_x^2\,f+\sin^2\Theta \partial_y^2 \,f$

Second thing: Is it accordingly correct: $\partial_\Theta^2\, g = -2\,r^2\,\cos\Theta\,\sin\Theta\,\partial_x\partial_y\,f+r^2\,\sin^2\Theta\,\partial_x^2\,f+ r^2\,\cos^2\Theta \partial_y^2 \,f$

Seems almost correct 'cos $r^2$ and $\frac{1}{r^2}$ cancel out and factoring out seems probable.

But $+ \frac{1}{r}\,\partial_r\,g = \frac{1}{r} \cdot (\cos\Theta\,\partial_x\,g+\sin\Theta\,\partial_y\,g)$ kinda destroys the entire symmetry ...

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    $\begingroup$ The formula for $\partial_{\Theta}^2 g$ is incorrect, you forgot to apply the product rule after the first derivative ($\sin$ and $\cos$ depend on $\Theta$!). $\endgroup$
    – Jose27
    Dec 10 '20 at 0:57
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Note that this is actually the Laplacian in two dimensions which is expressed in polar and Cartesian coordinates. You can easily find the proof, see this and this for example. In three dimensions we have

$$\Delta f = f_{xx} + f_{yy} + f_{zz} $$ And

$$\Delta f = \frac{1}{r} (rf_r)_{r} + \frac{1}{r^2} f_{\theta \theta} + f_{zz}$$

Which reduces to two dimensions case when $f$ doesn't depend on $z$.

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