# Converting Laplacian to polar coordinates

1. Nov 18, 2016

### fahraynk

1. The problem statement, all variables and given/known data
$$U_{tt}=\alpha^2\bigtriangledown^2U$$ in polar coordinates if solution depends only on R, t.

2. Relevant equations

3. The attempt at a solution
So, the books solution is $$U_{tt}=\alpha^2[U_{rr}+\frac{1}{r}U_r]$$. I am getting stuck along the way cant figure out this last step I think. Here is my attempt:

$$\frac{du}{dx}=\frac{du}{dr}\frac{dr}{dx} + \frac{du}{d\Theta}\frac{d\Theta}{dx}$$

$$\frac{d}{dx}(\frac{du}{dr}\frac{dr}{dx} + \frac{du}{d\Theta}\frac{d\Theta}{dx}) = \frac{d^2u}{dr^2}*(\frac{dr}{dx})^2 + \frac{d^2u}{drd\Theta}\frac{dr}{dx}\frac{d\Theta}{dx} + \frac{du}{dr}\frac{d^2r}{dx^2}+\frac{d^2u}{d\Theta^2}(\frac{d\Theta}{dx})^2 + \frac{d^2u}{d\Theta dr}\frac{d\Theta}{dx}\frac{dr}{dx} + \frac{du}{d\Theta}\frac{d^2\Theta}{dx^2}$$

all the derivatives with theta = 0. So it becomes :
$$U_{rr}r_x^2 + U_rr_{xx}$$
$$x=rcos(\Theta)$$
$$y=rsin(\Theta)$$

$$r=\sqrt(x^2+y^2)$$
$$r_x = \frac{x}{sqrt(x^2+y^2)} = cos(\Theta)$$
$$r_{xx} = \frac{y^2}{(x^2+y^2)^{3/2}} = \frac{sin^2(\Theta)}{r}$$

When I plug in, I get :
$$U_{rr}cos^2(\Theta) + \frac{sin^2(\Theta)}{r}U_r$$
the books answer is the same, but without the sine and cos terms. How do I get rid of them otherwise what am I doing wrong?

2. Nov 18, 2016

### Ray Vickson

You forgot to add $\partial^2 u/ \partial y^2$ to get the Laplacian.

3. Nov 18, 2016

### fahraynk

lol ugh thanks