With the curved sides kept at a temperature \(T_0\), and the base at \(T_1\).
We know \(\nabla^2 T =0\), and we can use seperation of variables to give a solution of the form \(T = R(r)\Theta(\theta)Z(z)\).
Then, in cylinderical coordinates,
\[\frac{1}{R}\frac{1}{r} \dv{r} \qty(r \dv{R}{r}) + \frac{1}{\Theta} \frac{1}{r^2} \dv[2]{\Theta}{\theta} + \frac{1}{Z} \dv[2]{Z}{z} = 0\]
We now have
\[\frac{1}{Z} \dv[2]{Z}{z} = k^2\]
implying
\[Z = \exp(\pm kz)\]
also,
\[\frac{1}{R}\frac{1}{r} \dv{r} \qty(r \dv{R}{r}) + \frac{1}{\Theta} \frac{1}{r^2} \dv[2]{\Theta}{\theta} + k^2 = 0\]
which we can multiply by \(r^2\),
\[\frac{r}{R} \dv{r} \qty(r \dv{R}{r}) + \frac{1}{\Theta} \dv[2]{\Theta}{\theta} + k^2 r^2 = 0\]
from which,
\[\frac{1}{\Theta} \dv[2]{\Theta}{\theta} = -n^2\]
implying that
\[\Theta = \{ \cos(n \theta), \sin(n \theta) \}\]
and the periodicity of \(\theta\) will force \(n\) to be a
natural number. Then
\[\frac{r}{R} \dv{r} \qty(r \dv{R}{r}) + (k^2 r^2 - n^2) = 0\]
and letting \(kr = s\),
\[s \dv{s} \qty( s \dv{R}{s} ) + (s^2 - n^2)R = 0\]
which has the form of Bessel’s differential equation, equation (8), and thus the solutions are Bessel functions, \(J_n(s)\), the complete solution is thus
\[J_n (kr) \qty( A \sin(n\theta) + B \cos(n \theta) ) e^{-kz}\]
We can ignore the Bessel functions which are infinite at \(r=0\), as we need a finite solution there, so the functions of the first-kind are the appropriate solutions.
We know that \(T_1 > T_0\), so \(T>T_0\) everywhere, and so \(T_0\) can be taken as a constant.
The boundary condition of the curved surface at \(r=a\) is where \(J_n(ka) = 0\).
We now need to know the zeros of the Bessel functions, and our solution becomes
\[T = T_0 + \sum_{m=0}^{\infty} c_m J_0 \qty(\alpha_{0m}\frac{r}{a}) \exp(-\qty(\frac{\alpha_{0m}z}{a}))\]
The boundary condition at \(z=0\) is that \(T=T_0\), so
\[T_1 - T_0 = \sum_m c_m J_0 \qty( \alpha_{0m} \frac{r}{a})\]
and using the orthogonality condition,
\[\int_0^a (T_1 - T_0) J_0 \qty( \alpha_{0m} \frac{r}{a} ) r \dd{r} = c_m \frac{a^2}{2} J_1^2(\alpha_{0m})\]
and then, from the indefinite integral relationship \(\int x J_0(x) \dd{x} = x J_1(x)\),
\[\begin{split}\begin{aligned}
(T_1-T_0) \frac{a}{\alpha_{0m}} \qty[ r J_1 \qty( \alpha_{0m} \frac{r}{a})]_0^a &= (T_1 - T_0) \frac{a^2}{\alpha_{0m}} J_1 (\alpha_{0m})\\
&= c_m \frac{a^2}{2} J_1^2 (\alpha_{0m})
\end{aligned}\end{split}\]
with
\[c_m = \frac{2}{\alpha_{0m}} \frac{1}{J_1(\alpha_{0m})} (T_1-T_0)\]
and the overarching solution is thus
\[T = T_0 + \sum_m \frac{2 (T_1-T_0)}{\alpha_{0m}J_1(\alpha_{0m})} J_0 \qty( \alpha_{0m} \frac{r}{a}) \exp( - \qty(\frac{\alpha_{0m}z}{a}) )\]