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Taking the Laplace transform in $t$ and using the initial condition $u(r,0)=u_0$, the radial heat equation in three dimensions becomesPut $W=rU$ and $q=\sqrt{s}$. ThensoFiniteness at the centre forces $B(s)=0$.

The transformed boundary condition isSincesubstitution and cancellation of the two $u_0/s$ terms givesHenceand the required explicit transform is

For spherical symmetry the angular mode is $\ell =0$. Put $u(t,r)=r\psi (t,r)$. Direct differentiation givesso the radial heat equation in three dimensions becomes the ordinary one-dimensional heat equation $u_t=\kappa u_{rr}$. Since $r_{\pm }>0$, the zero endpoint temperatures become $u(t,r_{\pm })=0$.

Write $L=r_{+}-r_{-}$. The two initial sine modes already satisfy these Dirichlet boundary conditions; their squared wave numbers are $(\pi /L{)}^2$ and $(2\pi /L{)}^2$. Each therefore evolves independently with its corresponding exponential decay. Dividing back by $r$ givesAt $t=0$ the initial profile is recovered, and each sine is zero at both endpoints. The first mode decays four times more slowly in its exponential rate than the second. This is the two-mode instance of the spherical-shell heat equation via radial rescaling; uniqueness follows, for example, by applying the heat equation energy identity to the difference of two solutions.

On a shell $0<r_{-}<r<r_{+}$ with homogeneous Dirichlet boundary conditions, put $L=r_{+}-r_{-}$ and $u=r\psi$. The radial heat equation in three dimensions becomes $u_t=\kappa u_{rr}$, with zero values at both endpoints. Expanding $u(0,r)=r\psi (0,r)$ in a Fourier sine series gives the displayed solution, with $b_n=(2/L){\int }_{r_{-}}^{r_{+}}r\psi (0,r)\sin [n\pi (r-r_{-})/L]dr$. Each radial mode decays at the square of its wave number. The positive inner radius makes division by $r$ harmless throughout the shell.