= Solution
Substitute $T=e^{-\nu t}R(r)$ into the axisymmetric <heat equation> and cancel the exponential:
$$
-\nu R=\frac1r(r\kappa R')'.
$$
Thus the radial <ordinary differential equation> and its <Sturm-Liouville form> are
$$
\boxed{(r\kappa R')'+\nu rR=0,\qquad -(pR')'=\nu wR,\quad p(r)=r\kappa(r),\quad w(r)=r}.
$$
The weight in this <Sturm-Liouville problem> is therefore $\boxed{w=r}$. The outer <Dirichlet boundary condition> is $R(1)=0$, and the physical solution is regular at the centre. The factor $r$ in the <weighted inner product> is also the radial part of the polar area element.
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