Solving a time-dependent PDE in cylindrical coordinates

[David Montiel Taboada]

Hello, 

I would like to solve a PDE using cylindrical coordinates (instead of cartesian) but my equations are rotationally symmetric (do not depend on theta). Therefore, I only need to solve the problem in a 2D domain with coordinates r (radius) and z (direction along the axis). 

Can I just do that by employing a cylindrical mesh with dim=2, for example using the function GridGenerator::cylinder, or do I need to pick a specific finite element function and/or modify the weak form of the equations?

I am not sure I understand what the following sentence in the GridGenerator::cylinder function documentation means:

The manifold id for the hull of the cylinder is set to zero, and a CylindricalManifold is attached to it.

Thank you!

David

[Wolfgang Bangerth]

On 5/26/21 10:53 PM, David Montiel Taboada wrote:
>
> I would like to solve a PDE using cylindrical coordinates (instead of
> cartesian) but my equations are rotationally symmetric (do not depend on
> theta). Therefore, I only need to solve the problem in a 2D domain with
> coordinates r (radius) and z (direction along the axis).
>
> Can I just do that by employing a cylindrical mesh with dim=2, for example
> using the function GridGenerator::cylinder, or do I need to pick a
> specific finite element function and/or modify the weak form of the equations?

If you want to solve a rotationally symmetric PDE in a cylinder, then the r-z
domain will just be a rectangle r=0...R, z=0...L. There are functions in
GridGenerator for that.

To describe the weak form of the PDE in that case, you need to use a modified
form of the PDE that accounts for the change of coordinates from x,y,z to r,z.
You might want to search the mailing list archives, where there are several
threads on that topic.

Best
W.


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Wolfgang Bangerth email: bang...@colostate.edu
www: http://www.math.colostate.edu/~bangerth/

[David Montiel Taboada]

Thank you! I will try that.

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