# prove $\partial_x^2f + \partial_y^2 f = \partial_r^2 g+\frac{1}{r}\cdot \partial_r g + \frac{1}{r^2}\cdot \partial_\Theta^2 g$

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 ...

• 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 $\Theta$!). Dec 10 '20 at 0:57

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