- #1

fahraynk

- 186

- 6

## Homework Statement

$$

U_{tt}=\alpha^2\bigtriangledown^2U$$ in polar coordinates if solution depends only on R, t.

## Homework Equations

## 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 can't 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?