# Spherically symmetric Heat Equation

1. Mar 14, 2016

### throneoo

1. The problem statement, all variables and given/known data
A ball of radius a, originally at T0, is immersed to boiling water at T1 at t=0. From t≥0, the surface (of the ball) is kept at T1
Define u(r,t)=R(r)Q(t)=T(r,t)-T1
ΔT=T0-T1<0

r,t≥0

2. Relevant equations
2u=r-2 ∂/∂r ( r2 ∂u/∂r ) =D-1∂u/∂t
D>0
3. The attempt at a solution
Boundary Condition: u(r=a,t)Ξ0
Initial Condition: u(r,t=0)=ΔT*Top(0,a) where Top(0,a)=Θ(r)-Θ(r-a) is the top hat function with end-points 0 and a

This is not a precise formulation as I want u(0,0) to be ΔT and u(a,0)=0. However it conveniently represents u(0≤r<a,0)=ΔT and u(r>a,0)=0

Let u(r,t)=R(r)Q(t)

Separation of variables give
(Rr2)-1d/dr (r2 R')=(D*Q)-1Q'=-μ2 where μ∈R

The choice of the separation constant is such that the boundary condition can be satisfied

rR''+2R'+μ2rR=0
Q'+μ2DQ=0

The radial equation is simplified by the substitution S=rR, reducing it to S''+μ2S=0

Thus S=sin(μr) or cos(μr) and Q=exp(-μ2D*t) and u=SQ/r

Since I expect u to be finite at r=0, S(r=0)=0 is needed to create an indeterminate expression, ruling out the cosine solution.

S(r=a)=sin(μa)=0⇒μna=nπ ; n∈ℕ

So the solution has the form
u(r,t)=r-1∑An sin(μn r)*exp(-μn2Dt)
u(r,0)=r-1∑An sin(μn r)
u(r,0)*r=∑An sin(μn r)-----------------------(@)

Now multiply both sides with sin(μm r) and integrate both sides of (@) w.r.t. r from 0 to a
I found that An=2ΔT/(π*n/a) * (-1)n+1 ;

I'd appreciate it if someone could point out where I made a mistake...

Last edited: Mar 14, 2016
2. Mar 19, 2016