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Heat equation in a sphere surface

  1. Jul 25, 2004 #1


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    I was wondering what happens if I want to solve the heat equation in a sphere surface, neglecting its thickness. I have one initial condition for T(t=0), in particular this initial profile can depend on azimuth and zenith angles, it is not uniform. Perhaps I have saying something stupid but I think for large times the temperature would be uniform in all over the surface. The question is I have not any boundary condition, except those of angular periodicity. Or do you think the solution is precisely the initial profile?.
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  3. Jul 25, 2004 #2
    I've figured out a way of solving it, I'm not sure if it works. Anyway here I go:

    The initial temperature distribution can be written as an infinite sum of spherical functions,
    [tex]T(\theta, \phi, t=0) = \sum_{l=0}^\infty \sum_{m=-l}^{+l}a_{l,m} Y_{l,m}(\theta, \phi)[/tex]
    The heat equation for one of those spherical functions can be solved easily because they have the property [tex]\Delta_{S^2} Y_{l,m}(\theta,\phi)=l(l+1)Y_{l,m}(\theta,\phi)[/tex] with the spherical Laplace operator [tex]\Delta_{S^2}[/tex] (it might be l(l-1) instead of l(l+1), I'm not sure anymore).
    The solution to this particular initial condition is (the constants set equal to 1) [tex]T(\theta,\phi,t)=Y_{l,m}(\theta,\phi)e^{-l(l+1)t}[/tex].

    Since the heat equation is linear one can solve it seperately for each part of the infinite sum, and therefore the solution with general initial conditions is
    [tex]T(\theta, \phi, t) = \sum_{l=0}^\infty e^{-l(l+1)t} \sum_{m=-l}^{+l}a_{l,m} Y_{l,m}(\theta, \phi)[/tex].

    So the problem is practically solved if you have found the series representation of the initial condition.
    Yes, that's true.
    Last edited: Jul 26, 2004
  4. Jul 26, 2004 #3


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    :eek: Bufff!

    I was hoping such an answer given by a mathmatician. I just was reffering to the physical problem of the non-existence of apparent boundary conditions. Your answer is very technical, although is always welcomed of course. :biggrin:
  5. Jul 26, 2004 #4
    The only necessary boundary condition is indeed the initial temperature distribution. With that and the condition that no heat leaves or "enters" the thin surface the whole developement in time is determined.

    I'm not sure if the angular periodicity can be called a boundary condition...its more something that comes up when you use spherical coordinates. I guess when the initial condition is periodic then the solution to the heat equation is automatically periodic for all times. I'm sure one can derive this somehow directly from the equation, though I don't know how.
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