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Partial Differentiation Problem

  1. Apr 18, 2006 #1
    Hi to all,

    I have been given the following problem as an assignment.

    [tex] \frac{\partial ^2 \phi}{\partial \rho^2} + \frac{1}{\rho}\frac{\partial \phi}{\partial \rho} + \frac{1}{\rho^2}\frac{\partial \phi}{\partial \chi^2} + \frac{\partial ^2 \phi}{\partial Z^2}+B^2\phi = 0 [/tex]

    Here is my attempt to the problem:
    Assuming [tex] \phi = S(\rho,\chi)Z(z) [/tex]

    [tex] \frac{1}{S(\rho,\chi)}\frac{\partial ^2 S(\rho,\chi)}{\partial \rho^2} + \frac{1}{S(\rho,\chi) \rho }\frac{\partial S(\rho,\chi)}{\partial \rho} + \frac{1}{S(\rho,\chi) \rho^2}\frac{\partial S(\rho,\chi)}{\partial \chi^2} + \frac {1}{Z}\frac{\partial ^2 Z}{\partial Z^2} + B^2 = 0 [/tex]

    Separating the variables we get

    [tex] \frac{\partial ^2 z}{\partial Z^2} + B^2 Z = 0 [/tex]

    [tex] \frac{1}{S(\rho,\chi)}\frac{\partial ^2 S(\rho,\chi)}{\partial \rho^2} + \frac{1}{S(\rho,\chi) \rho }\frac{\partial S(\rho,\chi)}{\partial \rho} + \frac{1}{S(\rho,\chi) \rho^2}\frac{\partial S(\rho,\chi)}{\partial \chi^2} + B^2 = 0 [/tex]

    Assuming [tex] S(\rho, \chi) = \rho(\rho) \chi(\chi) [/tex]

    [tex] \frac{1}{\rho}\frac{\partial ^2 \rho}{\partial \rho^2} + \frac{1}{\rho^2 }\frac{\partial \rho}{\partial \rho} + \frac{1}{\rho^2 \chi}\frac{\partial ^2 \chi}{\partial \chi^2} + B^2 = 0 [/tex]

    How can I solve this last PDE?

    Thank you in advance
     
    Last edited: Apr 18, 2006
  2. jcsd
  3. Apr 18, 2006 #2
    Okay, I haven't actually tried the problem, but you could try first to classify the pde (hyperbolic, elliptic, parabolic). In this case, the class of the pde depends on your choice of [tex]\rho[/tex]. Have you tried it?
     
  4. Apr 18, 2006 #3

    Dr Transport

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    if you multiply by [tex] \rho^{2} [/tex] you'll be able to separate the variables completely.
     
  5. Apr 19, 2006 #4
    [Hyperreality] Okay, I haven't actually tried the problem, but you could try first to classify the pde (hyperbolic, elliptic, parabolic). In this case, the class of the pde depends on your choice of [tex]\rho[/tex]
    . Have you tried it? [/Q]

    I don't know as to which class the PDE falls to. Let me state the whole problem so that it becomes clear to everyone.
    Solve the following

    [tex] \frac{\partial ^2 \phi}{\partial \rho^2} + \frac{1}{\rho}\frac{\partial \phi}{\partial \rho} + \frac{1}{\rho^2}\frac{\partial ^2 \phi}{\partial \chi^2} + \frac{\partial ^2 \phi}{\partial Z^2}+B^2\phi = 0 [/tex]

    Where: 0 < [tex] \rho [/tex] < R, 0 < [tex] \chi [/tex] < [tex] \pi [/tex], [tex] -\frac{H}{2} < z < \frac{H}{2} [/tex]

    The Boundary Conditions are

    [tex] \phi(R,\chi,z) = 0 [/tex]
    [tex] \phi(\rho,0,z) = \phi(\rho,\pi,z) = 0 [/tex]
    [tex] \phi(\rho,\chi, \pm \frac{H}{2}) = 0 [/tex]


    [Dr Transport] if you multiply by [tex] \rho^{2} [/tex] you'll be able to separate the variables completely. [/Q]

    As you suggested to multiply the last PDE by [tex] \rho^{2} [/tex], when separating the PDE's involving both [tex] \rho [/tex] and [tex] \chi [/tex] I still have [tex] \rho^{2} [/tex] on one of the PDE's involving [tex] \chi [/tex]. See below

    [tex] \frac{1}{\chi}\frac{\partial ^2 \chi}{\partial \chi^2} + B^2 \rho^2 = 0 [/tex]

    And

    [tex] \rho \frac{\partial ^2 \rho}{\partial \rho^2} + \frac{\partial \rho}{\partial \rho} + B^2 \rho^2 = 0 [/tex]
     
    Last edited: Apr 19, 2006
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