 Solution

Multiply the separated Sturm-Liouville problem by real $R$ and apply integration by parts:The outer endpoint term vanishes because $R(1)=0$. At zero, regularity gives vanishing radial flux. This can be obtained from the equation without assuming a bounded derivative: let $F=r\kappa R^{\prime }$. Since $F^{\prime }=-\nu rR$ and $R$ is bounded, $F$ has a finite limit at zero. If that limit were nonzero, positive bounded thermal diffusivity would make $R^{\prime }$ have fixed sign and magnitude at least a constant times $1/r$ sufficiently near zero, contradicting bounded $R$. Hence $F(0)=0$, and in fact $F(r)=-\nu {\int }_0^{r}sR(s)ds=O(r^2)$. This also forces $RF\to 0$ for bounded $R$.

The boundary terms therefore vanish, provingAs usual for diffusion, $\kappa$ is nonnegative and positive for $0<r\le 1$; the later power-law case may vanish at the centre. The argument applies in that case as well.