Looks a bit confusing.
The Laplacian on the unit sphere with metrics
ds^2 = dtheta^2 + sin^2 theta dphi^2
is with s -> sin theta
Laplacian = s^-1 d_theta s d_theta + s^-2 d_phi d_phi
and the general heat equation with production and destruction terms is
d_t F(t,theta,phi) = diffconst Laplacian( F(t,theta,phi) )
- F(t,theta,phi)/lifetime + productionrate
to avoid the sine switch to cylinder coordinates
u = Cos theta, du = -Sqrt(1-u^2) dtheta, d_theta = - Sqrt(-1-u^2) d_u
with
Laplacian = d_u (1-u^2) d_u = d_u d_u - 2 u d_u + 1/(1-u^2) d_phi d_phi
equns=Simplify@
{
D[F[t,u,p],t] == K D[(1-u^2)D[F[t,u,p],u]],u] +
K/(1-u^2) Derivative[0,0,2][F][t,u,phi]
- a F[t,u,p] + b,
F[t,u,2 Pi]==F[t,u,0],
Derivative[F][t,-1.0,p]==0,
Derivative[F][t,1.0,p]==0,
F[0,u,p]==F0[u,p]}
NDSolve[equns,{t, 0.0, tmax}, {u,-1.0, 1.0}, { p, 0.0, 2.0 N[Pi]}]
The polar points u=+-1 are replaced by reflecting boundary conditions.
This guarantees smothness of the solution at the poles and conservation
of probability.
In case the start distribution F0 is rotational symmetric around some
axis you can choose this direction as north and discard the variable phi
completely.
--
Roland Franzius