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

On Friday, January 31, 2025 at 3:37:30 AM UTC+5:30 Wolfgang Bangerth wrote:

>
> 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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