Re: [deal.II] Question about the numbering of DoFs

[yuesu jin]

Dear Jimmy,
  Thank you! Your answer solved my question. I had the wrong impression
that higher-order polynomials will give more base functions. Shape
functions just play the role of interpolation and construct the solution
space. It could be a high-order polynomial but only one base function per
node.  Thank you very much!
Best regards

On Sun, Aug 2, 2020 at 11:50 PM Jimmy Ho  wrote:

> Hi Yuesu,
>
> To be more precise: Yes, you do have two sets of basis functions in each
> element. A quadratic one for interpolating the vector components, and a
> linear one for interpolating the scalar. But when calculating DOFs
> associated with the vector components, you should only count the basis
> functions that interpolate those components, which is one basis function
> per node.
>
> Hope this helps!
>
> Best,
> Jimmy
>
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> .
>


-- 
Yuesu Jin,
Ph.D student,
University of Houston,
College of Natural Sciences and Mathematics,
Department of Earth and Atmospheric Sciences,
Houston, Texas 77204-5008
346-404-2062

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Re: [deal.II] Question about the numbering of DoFs

[Jimmy Ho]

Hi Yuesu,

To be more precise: Yes, you do have two sets of basis functions in each 
element. A quadratic one for interpolating the vector components, and a linear 
one for interpolating the scalar. But when calculating DOFs associated with the 
vector components, you should only count the basis functions that interpolate 
those components, which is one basis function per node.

Hope this helps!

Best,
Jimmy

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Re: [deal.II] Question about the numbering of DoFs

[Jimmy Ho]

Hi Yuesu,

The 2 in the initialization means that the basis functions (hence the finite 
element for the vector part) are quadratic. Which means that each element has 9 
nodes. But you should still only have one basis function associated with each 
node. That's why you have 9*2=18 DOFs associated with the vector problem.

Best, 
Jimmy

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Re: [deal.II] Question about the numbering of DoFs

[yuesu jin]

Dear Jimmy,
 Thank you for your reply. Yes, I can set up only one base function for
each node. But the problem is that the example sets up two base functions
for a two-component vector. FESystems fe_basis(FE_Q (2),dim,FE_Q
(1),1), which the (2) means the vector has order-2 base functions.
Best regards,
Yuesu

On Sun, Aug 2, 2020 at 10:09 PM Jimmy Ho  wrote:

> Hi Yuesu,
>
> When you have a vector-valued finite element, different components of the
> vector are still interpolated using the same basis functions. So you can
> have two DOFs on each node, but there's only one basis function associated
> with this node.
>
> Hope that helps!
>
> Best,
> Jimmy
>
> --
> The deal.II project is located at http://www.dealii.org/
> For mailing list/forum options, see
> https://groups.google.com/d/forum/dealii?hl=en
> ---
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> .
>


-- 
Yuesu Jin,
Ph.D student,
University of Houston,
College of Natural Sciences and Mathematics,
Department of Earth and Atmospheric Sciences,
Houston, Texas 77204-5008
346-404-2062

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[deal.II] Question about the numbering of DoFs

[Jimmy Ho]

Hi Yuesu,

When you have a vector-valued finite element, different components of the 
vector are still interpolated using the same basis functions. So you can have 
two DOFs on each node, but there's only one basis function associated with this 
node.

Hope that helps!

Best,
Jimmy 

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[deal.II] Question about the numbering of DoFs

[yuesu jin]

Dear all,
  I am reading the page :
https://www.dealii.org/current/doxygen/deal.II/classFiniteElement.html.
I feel confused by the numbering of the degrees of freedom. For example:

FESystem

fe_basis(FE_Q
(2), dim,
FE_Q 
(1),1);

As mentioned in Glossary:

Describing "degrees of freedom" in this context requires us to simply
*enumerate* the basis functions of the space *V**h*. For *Q*1 elements this
means simply enumerating the vertices of the mesh in some way, but for
higher elements one also has to enumerate the shape functions that are
associated with edges, faces, or cell interiors of the mesh.


We have a vector u which has 2 components defined on 9 nodes. On each node,
there are 2 base functions, so the degrees of freedom are 9*2*2 = 36. And
we have a scalar p which has 1 component defined on 4 nodes. On each node,
there is one base function, so the degrees of freedom are 4*1*1=4. The
total degrees of freedom should be 36+4=40.

In the webpage, it gives 2*9+4*1=22 degrees of freedom, it equals the base
function with the vector components.
[image: image.png]

u(0) should have two base element [phi(0),0] and [phi(0) 0], u(1) should
have two base elements as well [0 phi(0)] [0 phi(1)], so here should be 4
rather than 2 degrees of freedom.

Best regards,

-- 
Yuesu Jin,
Ph.D student,
University of Houston,
College of Natural Sciences and Mathematics,
Department of Earth and Atmospheric Sciences,
Houston, Texas 77204-5008
346-404-2062

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