evaluation FE function outside domain

[Mathieu]

Hello everyone,

I was wondering if deal.II has built-in functions to evaluate continuous finite element functions (FEQ) at points outside the mesh?

E.g.: given a point P in 2d/3d, which is not contained in the 2d/3d mesh but close to the boundary, how could one evaluate (extrapolate) the FE function at P? Or is it not a good idea to extrapolate shape functions?

Thank you

Math

[Bruno Turcksin]

Math,

The basis functions are not defined outside of an element, so you cannot extrapolate them outside of the mesh. What are you trying to achieve?

Best,

Bruno

[Mathieu]

Hi Bruno, 

for a post-processing step, I have to evaluate my temperature field at some points outside the mesh close to its boundary. 

I know that shape functions are not defined outside the mesh, that's why I wanted to extrapolate them. 

My idea was to use the gradient of the temperature field to come up with an extrapolation. 

But I do not know if that's reasonable. 

Best,

Math

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[Bruno Turcksin]

Math,

It really depends on your problem. If outside of your simulation domain, you have a domain that has the same properties and the solution changes slowly that's probably a good approximation. If outside of your domain, the solution changes quickly that's probably a bad approximation. If your solution has a shock, your approximation will be totally off. If you are solving the heat equation, the approximation is probably fine.

Best,

Bruno

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[Wolfgang Bangerth]

On 6/6/23 14:24, Mathieu wrote:
>
> for a post-processing step, I have to evaluate my temperature field at some
> points outside the mesh close to its boundary.
>
> I know that shape functions are not defined outside the mesh, that's why I
> wanted to extrapolate them.
> My idea was to use the gradient of the temperature field to come up with an
> extrapolation.
> But I do not know if that's reasonable.

Like Bruno already said in other words, you are starting at the wrong end of
the question. In a first step, define mathematically *what* it is you want to
do (and justify why this is the right choice). Then, in a second step, one can
think about *how* to implement this.

Best
W.

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Wolfgang Bangerth email: bang...@colostate.edu
www: http://www.math.colostate.edu/~bangerth/

[Mathieu]

"Like Bruno already said in other words, you are starting at the wrong end of
the question. In a first step, define mathematically *what* it is you want to
do (and justify why this is the right choice). Then, in a second step, one can
think about *how* to implement this."

I need this function (triangulation) for post-processing, but also during my assemble routine.

Basically, the values of this function at some points are assembled into my linear system.

But it may happen that I have to evaluate the function at a point which is not contained in its triangulation,

since I do not know the bounds a priori. 

Currently, my program is aborted if I would need to evaluate at a point outside the triangulation. 

I want to circumvent this by extrapolating the function somehow (bilinear, ...).

The function should be increasing in x-and y-direction and is quite smooth at the boundary.

Any suggestions?

Thank you,

Math

[Wolfgang Bangerth]

On 6/7/23 10:57, Mathieu wrote:
> "Like Bruno already said in other words, you are starting at the wrong end of
> the question. In a first step, define mathematically *what* it is you want to
> do (and justify why this is the right choice). Then, in a second step, one can
> think about *how* to implement this."
>

> [...]

> I want to circumvent this by extrapolating the function somehow (bilinear, ...).

You'll need to say what this "somehow" is supposed to be. I can think of a
half-dozen ways of extrapolating functions, but like Bruno already outlined,
which one is right depends on what you want to do.

In other words, come up with a *mathematical definition* of the operation you
want to do, and we can help with how that can be implemented.

[Mathieu]

A bilinear extrapolation seems to be a suitable first try since the shape functions are bilinear. 

Going this way, I would compute the gradient of the function at a point at the boundary to come up with the extrapolated function value. 

Call the point outside the grid P, I would project P to the closest at the boundary of the triangulation and evaluate the gradient there. 

So, given P, how can I find the closest point at the boundary? 

Or would you choose a different point at the boundary? 

-Math

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[Wolfgang Bangerth]

On 6/7/23 11:42, Mathieu wrote:
> A bilinear extrapolation seems to be a suitable first try since the shape
> functions are bilinear.
> Going this way, I would compute the gradient of the function at a point at the
> boundary to come up with the extrapolated function value.
> Call the point outside the grid P, I would project P to the closest at the
> boundary of the triangulation and evaluate the gradient there.
>
> So, given P, how can I find the closest point at the boundary?
> Or would you choose a different point at the boundary?

This is feasible, though finding the closest vertex is an expensive operation.
For this to work, you will have to find which cell is closest to the point in
question, for which I would use
GridTools::find_closest_vertex()
and
GridTools::find_cells_adjacent_to_vertex()

[Mathieu]

" This is feasible, though finding the closest vertex is an expensive operation.
For this to work, you will have to find which cell is closest to the point in
question, for which I would use
   GridTools::find_closest_vertex()
and
   GridTools::find_cells_adjacent_to_vertex() "

I was not aware that GridTools::find_closest_vertex() works if the point

is outside the domain.

Say, P=(Px,Py) is the point outside the domain,

T=(Tx, Tx) the closest vertex at the boundary as computed above,

and grad_T the gradient of the finite element function at T.

Then I would compute the function value at P as

f(P) = f(T) + grad_T[0]*(Px-Tx) + grad_T[1]*(Py-Ty).

Is that reasonable?

Thanks,

Math

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[Wolfgang Bangerth]

On 6/9/23 08:27, Mathieu wrote:
>
> Say, P=(Px,Py) is the point outside the domain,
> T=(Tx, Tx) the closest vertex at the boundary as computed above,
> and grad_T the gradient of the finite element function at T.
> Then I would compute the function value at P as
> f(P) = f(T) + grad_T[0]*(Px-Tx) + grad_T[1]*(Py-Ty).
> Is that reasonable?

Yes, that is just a linear extrapolation and that's reasonable if your
solution is smooth and if P is close to T.

If T is a vertex, then the solution generally has a kink there and the
gradient can have several values depending on which adjacent cell you take it
on. Whether that matters to you is a separate question -- it may be sufficient
to simply take the gradient from one of the adjacent cells.

[Mathieu]

" If T is a vertex, then the solution generally has a kink there and the
gradient can have several values depending on which adjacent cell you take it
on. "

To have a unique gradient, it would be better to not search for the closest vertex,

but rather to the closest point at the boundary, which may be a point at the line connecting two vertices.

But there is no straightforward way to obtain this point, right?

At least, I did not find a functionality in the GridTools namespace, which is probably the right place to look for.

-Math

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[Wolfgang Bangerth]

On 6/9/23 09:24, Mathieu wrote:
>
> To have a unique gradient, it would be better to not search for the closest
> vertex,
> but rather to the closest point at the boundary, which may be a point at the
> line connecting two vertices.

Perhaps, but it is not clear to me that this would actually be more accurate.

> But there is no straightforward way to obtain this point, right?
> At least, I did not find a functionality in the GridTools namespace, which is
> probably the right place to look for.

Correct.

[Mathieu]

"

Say, P=(Px,Py) is the point outside the domain,

T=(Tx, Tx) the closest vertex at the boundary as computed above,

and grad_T the gradient of the finite element function at T.

Then I would compute the function value at P as

f(P) = f(T) + grad_T[0]*(Px-Tx) + grad_T[1]*(Py-Ty). "

Let me clarify one last thing:

The gradient (deriv. with regards to x and y) of the above extrapolation is given by

[ grad_T[0]; grad_T[1] ].

Correct?

-Math

 

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