Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 4 / 14D / b / Solution

{title2=$u_t=D r^{-1}(ru)_{rr}$}

For a spherically symmetric field $u(r,t)$ in three dimensions,
$$
\nabla^2u=\frac1{r^2}\frac\partial{\partial r}\left(r^2u_r\right)
=\frac1r\frac{\partial^2}{\partial r^2}(ru).
$$
Thus the <heat equation> becomes $u_t=D r^{-1}(ru)_{rr}$. Introducing $w=ru$ reduces its spatial part to the one-dimensional second derivative $w_{rr}$, while regularity at the centre requires $w(0,t)=0$.

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 Incoming links: Radial heat equation in three dimensions