Understanding Laplace-Beltrami Operator on a 3D surface

[Amit Singh]

Dear all

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?

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?

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.

- Amit

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[Wolfgang Bangerth]

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.

[Amit Singh]

Thank you for clarifying the meaning of Laplace-Beltrami operator and the suggestion about making the mesh and manifold.

About MappingManifold and SphericalManifold: I misunderstood the note to mean that SphericalManifold is not usable by ANY of the Mapping classes. Now it is clear that the comment is specifically for the MappingManifold class.

I was unclear about the interaction between Mapping class and Manifold objects. But then I read the geometry paper which has a figure of a MappingQ2 cell imposed on a part of a sphere. So my current understanding is that the MappingQ2 object uses SphericalManifold to identify the location of the extra nodes (extra with respect to the bi-linear quad element) but it interpolates them using a polynomial only.

Thank you for your time and help.

-Amit