Heat equation in a sphere surface

Click For Summary
SUMMARY

This discussion focuses on solving the heat equation on the surface of a sphere, specifically addressing the initial temperature distribution defined by azimuth and zenith angles. The solution utilizes spherical harmonics, expressed as T(θ, φ, t) = ∑ e^(-l(l+1)t) ∑ a_{l,m} Y_{l,m}(θ, φ), where Y_{l,m} are spherical functions. The conversation confirms that without additional boundary conditions, the solution evolves from the initial profile and that for large times, the temperature becomes uniform across the surface.

PREREQUISITES
  • Understanding of spherical harmonics and their properties
  • Familiarity with the heat equation and its solutions
  • Knowledge of angular periodicity in spherical coordinates
  • Basic concepts of the spherical Laplace operator (Δ_{S^2})
NEXT STEPS
  • Study the derivation of the heat equation in spherical coordinates
  • Explore the properties of spherical harmonics in greater detail
  • Learn about boundary conditions in partial differential equations
  • Investigate the implications of periodic initial conditions on solutions
USEFUL FOR

Mathematicians, physicists, and engineers interested in heat transfer problems, particularly those involving spherical geometries and initial condition analysis.

Clausius2
Science Advisor
Gold Member
Messages
1,433
Reaction score
7
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?.
 
Physics news on Phys.org
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,
T(\theta, \phi, t=0) = \sum_{l=0}^\infty \sum_{m=-l}^{+l}a_{l,m} Y_{l,m}(\theta, \phi)
The heat equation for one of those spherical functions can be solved easily because they have the property \Delta_{S^2} Y_{l,m}(\theta,\phi)=l(l+1)Y_{l,m}(\theta,\phi) with the spherical Laplace operator \Delta_{S^2} (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) T(\theta,\phi,t)=Y_{l,m}(\theta,\phi)e^{-l(l+1)t}.

Since the heat equation is linear one can solve it separately for each part of the infinite sum, and therefore the solution with general initial conditions is
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).

So the problem is practically solved if you have found the series representation of the initial condition.
Perhaps I have saying something stupid but I think for large times the temperature would be uniform in all over the surface.
Yes, that's true.
 
Last edited:
:eek: Bufff!

I was hoping such an answer given by a mathmatician. I just was referring to the physical problem of the non-existence of apparent boundary conditions. Your answer is very technical, although is always welcomed of course. :biggrin:
 
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 development 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.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
994
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K