The general heat problem that you will want to solve is of the form
$$
\partial_t f(x,t) + Lf(x,t) = 0,
$$
where ##L## is a SL operator in ##x##. You might also have a number of boundary conditions, but you will need to somehow get rid of those because the boundary conditions of a SL operator need to be homogeneous, more on that later.
The general idea is the following: According to the Sturm-Liouville theorem, a SL operator ##L## has a complete set of linearly independent eigenfunctions ##u_n(x)## such that
$$
Lu_n(x) = \lambda_n u_n(x).
$$
That the set is complete means that any function of ##x## on the domain can be expressed as a linear combination of these functions, i.e., for any function ##h(x)##, we can find constants ##h_n## such that
$$
h(x) = \sum_n h_n u_n(x).
$$
For a fixed ##t##, the ##f(x,t)## in the heat problem is a function of ##x## and can therefore be expanded in terms of the SL eigenfunctions
$$
f(x,t) = \sum_n f_n(t) u_n(x).
$$
Note that the expansion coefficients ##f_n(t)## will generally depend on the time ##t## since the function is going to be different at different times.
We can now insert this expansion into the heat problem and obtain
$$
\partial_t f(x,t) + Lf(x,t) = \sum_n\left[ f'_n(t) u_n(x) + f_n(t) L u_n(x) \right]
= \sum_n [f'_n(t) + \lambda_n f_n(t)] u_n(x) = 0,
$$
where we have used that ##u_n(x)## does not depend on ##t## and that it is an eigenfunction of the SL operator ##L##.
Now, since the functions ##u_n(x)## are linearly independent, each term in the sum must be equal to zero and you obtain
$$
f'_n(t) + \lambda_n f_n(t) = 0
$$
for all ##n##. This is an ordinary differential equation that can be solved by any means that you usually use for those (this one should be particularly familiar).
About the non-zero boundary conditions: You can rewrite your problem by writing the solution as a sum of two functions, one that satisfies the heat equation and homogeneous boundary conditions and one that is static, satisfies the Laplace equation and the inhomogeneous boundary conditions. In this case this is particularly easy because your boundary condition is a constant.
Let us work with a similar example, let us solve the heat problem on a line. The problem is then of the form
$$
\partial_t f - \partial_x^2 f = 0.
$$
Let us also assume boundary conditions ##f(0,t) = f(L,t) = T_0##. The SL operator in this case is just ##L = -\partial_x^2##.
In order to make the boundary conditions homogeneous, we let ##g(x,t) = f(x,t) - T_0##. Inserting this into the heat equation gives the heat equation for ##g(x,t)##, but with homogeneous boundary conditions (##T_0## is just a constant and a constant term does not survive in the heat equation).
Now, to find the eigenfunctions of the SL operator, we need to solve the eigenvalue equation
$$
Lu_n(x) = \lambda_n u_n(x).
$$
Inserting that ##L = -\partial_x^2##, we end up with
$$
u''_n(x) + \lambda_n u_n(x) = 0.
$$
The only solutions that satisfy this equation along with the boundary conditions are the sine functions
$$
u_n(x) = \sin\left(\frac{\pi n x}{L}\right) \equiv \sin(k_n x)
$$
with corresponding eigenvalues ##\lambda_n = k_n^2##.
That any function on the interval ##0 < x < L## can be expanded in these functions is therefore just saying that you can expand the function in a Fourier series
$$
g(x,t) = \sum_n g_n(t) \sin(k_n x).
$$
Inserting this into the heat equation now gives the ODE
$$
g'_n(t) + k_n^2 g_n(t) = 0.
$$
One can go on from here to solve the problem, but I hope you get the general idea. The same procedure should work on your problem, just exchanging the SL operator for the one you have (i.e., Bessel's operator).
dykuma said:
I am curious though, are my initial steps to solving this problem correct? Do they make sense to you?
Some of it, such as the argument for vanishing ##z## and angular dependence makes sense. However, you are not on the right track in terms of constructing the series solution. To do so, you need to follow the prescription above.
Looking at that particular section of Boas, you can solve your problem by replacing the ##z## term with the time-derivative term in the entire section (but remember to shift your solution to get homogeneous boundary conditions!). To be honest, I do not think Boas' treatment is the clearest in terms of what is actually going on mathematically.
