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Math Help - Sep of variables in PDE

  1. #1
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    Sep of variables in PDE

    Hi,

    Solve \displaystyle u_t = k \Big( u_{xx} + u_{yy} + u_{zz} \Big) with \displaystyle 0 < x < L, 0 < y < H, 0 < z < W

    IC's: \displaystyle u(x,y,z,0) = f(x,y,z)

    not going to list the BC's here.

    So, I start by letting

    \displaystyle u = \Phi (x,y,z) T(t)

    so \displaystyle  \frac{T'}{k T} = \frac{\Phi_{xx} + \Phi_{yy} + \Phi_{zz}}{\Phi} = - \lambda

    so we have \displaystyle T' + \lambda k T = 0 and \Phi_{xx} + \Phi_{yy} + \Phi_{zz} + \lambda \Phi = 0

    Now I let \Phi = X(x) Y(y) Z(z)

    so X''YZ + XY''Z + XYZ'' + \lambda XYZ = 0 (divide by XYZ)

    but I'm not too sure how correct this is because if I carry on I get

    \displaystyle \frac{X''}{X} + \frac{Y''}{Y} + \frac{Z''}{Z} + \lambda = 0......(1)

    Now Im not sure how to proceed from here....I think I need to somehow add another two separation constants....but I'm not sure how to exactly...

    when I try.... I might get something like

    \displaystyle \frac{X''}{X} = \frac{-Y''}{Y} - \frac{Z''}{Z} - \lambda = - \mu

    so I end up with \displaystyle X'' + \mu X = 0.

    Now the answer in the text book says that I should end up with the three equations

    \displaystyle X'' + \mu X = 0, \displaystyle Y'' + \nu X = 0, and \displaystyle Z'' + (\lambda - \mu - \nu) Z = 0

    But I can't see how to get from ....(1)...to the three equations above. Can anyone please help me out?
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  2. #2
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    Your next equation after (1) works just fine. Just continue in that vein:

    \displaystyle -\frac{Y''}{Y}-\frac{Z''}{Z}=-\mu+\lambda

    \displaystyle \frac{Y''}{Y}+\frac{Z''}{Z}=\mu-\lambda

    \displaystyle \frac{Y''}{Y}=-\frac{Z''}{Z}+\mu-\lambda=-\nu

    Can you finish?
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  3. #3
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    Can you finish?
    Yes I can. Thanks for your help.
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  4. #4
    A Plied Mathematician
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    Great. You're welcome, and have a good one!
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