Re: [Libmesh-users] How to setup variable length data per element?

[Robert B. <rob...@gm...>]

Does anyone have any advice on this question?  John, maybe?


On 04/30/2013 11:29 AM, Robert Blake wrote:
> I am trying to run a FEM simulation wherein every element has a
> variable-length C data structure defined in an external non-libmesh file
> format.  What is the easiest way to get this data into libmesh and use
> it in parallel?
>
> I see only a couple of ways of doing this:
>
> 1.) I can easily do this using MeshData and foreign_id_for_elem().  This
> works serially, but MeshData is deprecated and doesn't work in parallel.
>
> 2.) I can create my own data structure that indexes off of Elem->id().
> Is this possible?  Based on my cursory glance at the code, I fear that
> Elem->id() will change after partitioning in parallel.  If this happens,
> this technique won't work.  In addition, I'm not sure how this will work
> if I decide to use AMR
>
> 3.) I can give each mesh a unique subdomain_id(), and create my own data
> structure that uses that subdomain_id() as an index.  This should work
> in parallel and for AMR, but it seems like a horrible abuse of the
> subdomain_id() and might cause problems for parts of the code that
> assume n_subdomains << n_elems.
>
> 4.) I can create an ExplicitSystem with a (CONSTANT,MONOMIAL) term, fill
> that data with an integer ID, and then use that ID to index into an
> external data structure of my choosing.  I know that this is now the
> preferred way of storing element data, but
> a.) It seems like a hack to use a floating point field to store integer data
> b.) I'd have to take extra care to ensure the variable is NOT
> interpolated when doing AMR
> c.) I'm still not sure how to populate these data values.  All the code
> I've seen for populating variables for ExplicitSystems uses
> FunctionBase<> which takes point data as it's argument.  I'm not quite
> sure how to easily access an element's data using (x,y,z) coordinates as
> an argument.
>
> Is there something I've missed?  Help!
>
> Rob
>
>