Convert Laplace's equation into polar coordinates

Question

When I read Ahlfors's book, I came across this problem the$$u(x,y)$$is a harmonic function and satisfies Laplace's equation $$\Delta u＝\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}＝0$$ He converted Laplace's equations into polar coordinates $$r\frac{\partial(r\frac{\partial u}{\partial r})}{\partial r}+\frac{\partial^2 u}{\partial \theta}＝0$$But I don't know how he changed

Answer 1

The general formula for the Laplacian with a diagonal metrics in $\mathbb{R}^n$ with scale factors $h_k(x_1,\dots,x_n), k=1,\dots, n$ derives from metric invariant form of the local scalar square of the gradient

$$\nabla f . \nabla g = \sum_k (\frac{1}{h_k} \partial_k f) (\frac{1}{h_k} \partial_k g)$$ integrated over all of $\mathbb{R}^n$ with volume measure $\prod_k h_k dx_k$

On the class of functions with zero boundary value integral over a large sphere one has by partial integration over all coordinates the definition of the Laplacian

$$\int \prod_m h_m dx_m \sum_k (\frac{1}{h_k} \partial_k f) (\frac{1}{h_k} \partial_k g) = -\int \prod_m h_m dx_m \ \ f \ \ \left(\sum_k (\frac{1}{h_k} \partial_k \left( (\frac{\prod_m h_m}{h_k} \partial_k g)\right)\right)=-\int \left(\prod_m h_m dx_m \right)\ \ f\ \ \Delta g $$

For polar coordinates in $\mathbb R^2$ with $h_r=1, \ h_\phi= r $ the resulting Laplacian is

$$\Delta = \frac{1}{r}\partial_r r \partial r + \frac{1}{r^2} \partial_{\phi,\phi}$$

For the Laplace equation one power of r can be discarded.

Answer 2

I think in calculus this formula is derived like this

Second

Third