Solving Heat Equation on a Disk and Sphere

Applied Mathematics
Q. Solve the heat equation on a disk, what are the Fourier components? Same
on a sphere.
In order for us to solve heat equation on a sphere, in our it is a disc, having given
boundaries in temperature. And the task will be solving;
And which is having the PDE as;
Since the PDE involves two derivatives in r and two derivatives in but we know
that PDE has to worked out by using two derivatives in r and other two derivatives
in θ and should be under three more conditions.
And in our case, the perfect thermal contact of periodic BCs in θ is given by;
u(r, −π) = u(r, π) for 0 < r < a
∂u / ∂θ(r, −π) = ∂u / ∂θ u(r, π) for 0 < r < a
The three conditions listed are all linear and homogeneous, so we can try
separation of variables.

The three conditions under which this problem can be solved should be all linear
and homogenous. And therefore the first step will be separating the variables
We firstly start separating Ansatz which is;
u(r, θ) = R(r)Θ(θ)
We then separate the PDE;
Let’s remember that a positive λ works better in this task than −λ.
Then we get the two resulting ODEs as:
Or
And

In the above equations we have the periodic boundary conditions as;
Θ(−π) = Θ(π), Θ0(−π) = Θ0(π)
And we already know our eigenvalues and Eigen functions:
λ0 = 0, λn = n^2, n = 1, 2, . . .
Θ0(θ) = 1, Θn(θ) = c1 cos nθ + c2 sin nθ, n = 1, 2, . . .
Therefore, by sing the eigenvalues above, we can get;
Before we go further let’s review the heat equations on a disk;
If;
R(r) = r p, then
R’(r) = pr^ p−1 and R’’(r) = p(p − 1)r^ p−2,
so that the CE equation;
will turn to;
And by assuming that r^p 6= 0 we then get the characteristic
equation as ;
p (p − 1) + p – n^2 = 0
⇐⇒ p^2 = n^2