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I 2D Laplacian in polar coordinates

  1. May 1, 2017 #1
    The 2D Laplacian in polar coordinates has the form of
    $$ \frac{1}{r}(ru_r)_r +\frac{1}{r^2}u_{\theta \theta} =0 $$
    By separation of variables, we can write the ## \theta## part as
    $$ \Theta'' (\theta) = \lambda \Theta (\theta)$$

    Now, the book said because we need to satisfy the condition ## \Theta (0)=\Theta (2\pi)##, ##\lambda = n^2## with n =1,2,3...

    I tried this in detailed steps.
    ##\Theta (\theta) = Acos\sqrt{\lambda}\theta + Bcos\sqrt{\lambda}\theta##
    With the BC,
    ## B = Acos(\sqrt{\lambda}2\pi) + Bsin(\sqrt{\lambda}2\pi) ##
    This is just an equality but not an identity. Why can't I conclude that
    ## B = \frac{cos(\sqrt{\lambda}2\pi)}{1-sin(\sqrt{\lambda}2\pi)} A ##
    instead of imposing constraint on ##\lambda##

    Thanks!
     
  2. jcsd
  3. May 1, 2017 #2

    Orodruin

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    Because you also need to satisfy ##\Theta'(0) = \Theta'(2\pi)##. For general lambda that gives you a system of equations that lead to trivial solutions only. Only for a discrete set of lambda do you get non-trivial solutions.

    Note: You are missing the ##\lambda=0## solution.
     
  4. May 2, 2017 #3
    Thank you! Is it also true that ##\Theta^{(n)}(0)=\Theta^{(n)}(2\pi)##?
     
  5. May 2, 2017 #4

    Orodruin

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    That follows from the differential equation if you have the other two.
     
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