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: [email protected]
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