In my exercise bundle about Sturm-Liouville problems and solving partial differential equations with the separation method there is an exercise that goes as follows:
Calculate the temperature distribution in a circular disc (infinitely thin) with radius $b$. The temperature on the boundary is equal to $f(\theta) = u(b,\theta)$. The temperature in the disc is finite.
I've solved a few of this kind of exercise, but in those you could find the homogeneous boundary conditions. In this exercise I don't see them. First I thought: in the center the temperature gradient must be zero, because it is in a minimum there. But that can't be true because the temperature on the circle (the boundary) is not constant.
I used the Laplace equation to solve this (heat equation with stationary temperature distribution).
After separation:
$$ r^2\frac{R''}{R}+r\frac{R'}{R}=-\frac{T''}{T}=\lambda, $$
with lambda of course the eigenvalue of the eigenfunctions we are looking for.
Thank you
Kind regards
To continue what I started in the comments.
Here, when you use first boundary condition, you will find that $$ A=A\cos \sqrt{\lambda}2\pi+B\sin\sqrt{\lambda}2\pi. $$ From the secomd boundary condition $$ B\sqrt{\lambda}=-A\sqrt{\lambda}\sin \sqrt{\lambda}2\pi+B\sqrt{\lambda}\cos \sqrt{\lambda}2\pi. $$ This is a homogeneous system of two equations with two unknowns $A,B$. To have a nontrivial solution, the determinant of this system has to be zero. Can you finish?