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Laplace Operator

  1. Dec 13, 2011 #1
    1. The problem statement, all variables and given/known data

    The Laplace operator Δ is defined by: Δ= c65dd028c1c9fb80a8288ca893e949da.png

    Show in polar coordinates r and Θ, that the Laplace operator takes the following form:

    http://upload.wikimedia.org/wikipedia/en/math/0/7/a/07a878276cffd0c680f3f827204aba24.png

    2. Relevant equations

    x=rcos(Θ), y=rsin(Θ), r ≥ 0, Θ ∈ [0,2∏]

    3. The attempt at a solution

    It seems simple; convert the x and y in terms of polar coordinates, and differentiate twice. However, when i attempt to differentiate twice, in respect to r and theta, my r disappears and the trig function simply reverts.
     
  2. jcsd
  3. Dec 13, 2011 #2

    I like Serena

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    Can you show your work where you differentiate the first time?
     
  4. Dec 14, 2011 #3
    Sorry for the delayed response, been studying for other finals.

    Here is where i am stuck.

    For the x partial derivative:

    [itex]\partial[/itex] / [itex]\partial[/itex]x= -rsinθ+cosθ

    For the y partial derivative:

    [itex]\partial[/itex] / [itex]\partial[/itex]y= sinθ+rcosθ

    I know this is incorrect because i was informed by my professor that there is a nested chain ruled inside a product rule in the second differentiation (or something similar). d/dr d/dx +d/dθ d/dx is the format for the x partial derivative, and just replacing x with y gives you the format for the y partial derivative.

    Where am i going wrong?
     
  5. Dec 15, 2011 #4

    I like Serena

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    What did you differentiate? That seems to be missing.
    And how did you get this result?


    Yes, the chain rule for multivariable differentiation is:
    [tex]{\partial \over \partial x}f(u(x,y), v(x,y))={\partial f \over \partial u}{\partial u\over \partial x}+{\partial f \over \partial v}{\partial v \over \partial x}[/tex]

    You should do something similar to f(x(r,θ), y(r,θ))...
     
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