Try to multiple your convection diffusion équation by r^2*sin(thêta)^2 this way you get rid of singularities

Hope it Helps !

Adam Ayouche Thanks for replying.

I had one doubt when you discretise the equation, and we end up with r^2*sin(theta)^2 on the left-hand side, i.e. coefficient for time derivative and convection terms.

finally we need C(i,j,n+1). so if we write expression for C(i,j,n+1). It should appear as

C(i,j,n+1) = 1/( r^2*sin(theta)^2 ) * func(C(i,j,n))

So here is the bold term causing singularity again.

I think that the left hand side term should not bother you !

you can solve them for example for r = 0 or theta = 0, pi

so you should get func(C(i,j,n)) = 0 (normally a developped r**2*sin(theta)^2*D*laplacian_spherical(C) = 0 )

then for r different than zero and theta different than 0 or pi

you solve it without any singularities problems

Best,

Adam

Adam Ayouche I'm solving an unsteady convection-diffusion equation

Ct + u.delta(C) = D*laplacian_spherical(C) + S

so if we multiply with r**2*sin(theta)^2, the left side will affect right?

Can you multiply by r**2*sin(theta)^2 and develop what you get please ?

Then, maybe try to substitue r = 0 and see what are the terms left and we can see if you are still facing singularities problems...