Amit:
I studied tutorial 38 that demonstrates the use of Laplace-Beltrami
operator in deal.ii. In the "Possibilities for extension" section of the
tutorial an example of a 3d surface obtained by transforming a half-
sphere is given. It has also been mentioned that because of the
transformation, the SphericalManifold is no longer attached to the
triangulation. Therefore, irrespective of the degree of the MappingQ
class we use, the mapping will remain bi-linear only.
1. So, is my understanding correct that in this case we are not
calculating Laplace-Beltrami anymore but just a regular Laplacian?
That's only a semantic difference. We generally say that the
Laplace-Beltrami operator is the "Laplace operator on a surface".
Whether that surface is (piecewise) flat or not does not make a difference.
2. I want to use the Laplace-Beltrami operator on surfaces that have
equations like "Ax^2 + B*y^2 + C*z^2 = R^2". Can I simply import a CAD
geometry and use OpenCascade::NormalToMeshProjectionManifold and achieve
this effect?
Yes, that's one way. The other way is of course as in step-38: You could
start from a flat domain and transform it to the shape you want. You
could then also attach a manifold object that describes exactly the kind
of object you have for your domain.
3. The documentation of MappingManifold says that SphericalManifold
cannot be used as for MappingManifold. So I am confused how exactly in
Step-38 in the half-sphere example Laplace-Beltrami is being calculated.
I can see that MappingQ2 is used to construct FEValues object. But
MappingQ2 is not the same as a spherical surface.
I must admit that I don't quite understand the comment in the
documentation. But I also don't quite understand your question: step-38
does not use MappingManifold. Can you elaborate what your question is?
Best
W.
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