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Converting Laplacian to polar coordinates

  1. Nov 18, 2016 #1
    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. jcsd
  3. Nov 18, 2016 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    You forgot to add ##\partial^2 u/ \partial y^2## to get the Laplacian.
     
  4. Nov 18, 2016 #3
    lol ugh thanks
     
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