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COMSOL won't solve the the diffusion equation in axisymmetric using coefficient form PDE

Laurence Hutton-Smith

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So this should be extremely simple. I want to solve the 3D diffusion equation, u_t = nabla^2 u, using the axisymmetric solver. This is the problem description, in COMSOL:

Set up: Axisymmetric problem, Coefficient Form PDE, time dependant

Geometry: Circle radius 3, Circle radius 1

Coefficient Form PDE IC: Initial value = 1/( (4/3)*pi ) in sphere radius 1, zero elsewhere.

Coefficient Form PDE BC: Zero flux on boundary of outer sphere.

Therefore initially the total concentration (or heat) inside is 1 - and should stay at 1 for all time. However as you can see from this file (attached *.mph file here www.dropbox.com/s/fv2feo3gr18ohwe/diffusion_eq.mph?dl=0) the concentration increases and settles at roughly 3 times the initial total concentration. My question is how can I make comsol actually solve the diffusion equation using the Coefficient Form PDE?

I cannot do this in 1D as this is a reduction of my full problem that breaks symmetry in the spherical theta coordinate. My problem is axisymmetric, and takes too long to run in 3D, hence the need for an axisymmetric solution. I also very much need to use the Coefficient Form PDE solver as my full problem requires it.


3 Replies Last Post 27 feb 2017, 14:19 GMT-5
Jeff Hiller COMSOL Employee

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Posted: 8 years ago 27 feb 2017, 10:23 GMT-5
Hello Laurence,
Your model does not set up the cylindrical laplacian, but rather a cross-section laplacian. See why here: www.comsol.se/blogs/guidelines-for-equation-based-modeling-in-axisymmetric-components/
Best,
Jeff
PS: Your post is reminiscent of this old thread:www.comsol.com/community/forums/general/thread/18950
Hello Laurence, Your model does not set up the cylindrical laplacian, but rather a cross-section laplacian. See why here: https://www.comsol.se/blogs/guidelines-for-equation-based-modeling-in-axisymmetric-components/ Best, Jeff PS: Your post is reminiscent of this old thread:https://www.comsol.com/community/forums/general/thread/18950

Laurence Hutton-Smith

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Posted: 8 years ago 27 feb 2017, 10:37 GMT-5
Thank you very much for your reply, I see this now.

Suppose we have:

- vec{n} . nabla u = p u

i.e. neumann BC at the outer circle. Would these also require transformation? And if so what would they be?
Thank you very much for your reply, I see this now. Suppose we have: - vec{n} . nabla u = p u i.e. neumann BC at the outer circle. Would these also require transformation? And if so what would they be?

Jeff Hiller COMSOL Employee

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Posted: 8 years ago 27 feb 2017, 14:19 GMT-5
Hi Laurence,
The in-plane components of the gradient are expressed in the same way in cartesian and in cylindrical coordinates after you replace x with r (I hope this makes sense to you, I don't know how to say it better), so there is no other transformation needed.
Best,
Jeff
Hi Laurence, The in-plane components of the gradient are expressed in the same way in cartesian and in cylindrical coordinates after you replace x with r (I hope this makes sense to you, I don't know how to say it better), so there is no other transformation needed. Best, Jeff

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