# Discussion Forum

Note: This discussion is about an older version of the COMSOL Multiphysics® software. The information provided may be out of date.

Discussion Closed This discussion was created more than 6 months ago and has been closed. To start a new discussion with a link back to this one, click here.

## Number of degrees of freedom solved for: 96195

Dear Sirs,

I am trying to understand from where the number 96195 comes from.

In particular, the mesh I am considering is a tetraedral mesh of a rectangular domain in 3D ( width = 5 cm, depth = 1.5 cm, height = 0.35 cm). The mesh, according to the mesh statistics the domain discretization is made of:

Prism elements: 6600
Triangular elements: 264
Edge elements: 276
Vertex elements: 8

I am using nearly-incompressible model for the material, there a mixed formulation is used to solve the problem and the pressure field is computed (which is not specified but since I use quadratic polynomial functions, it could be P1).
According to the formula presented in www.comsol.com/support/knowledgebase/875/ the number of nodes should be 6600*1.4 = 9240 and the DOFs (for the displacement) = 9240*3 = 27720. Adding the DOFf for the pressure (which is P1) we do not get 96195.

As someone an explanation for the number of DOF???

Many thanks,

Paolo

2 Replies Last Post Aug 16, 2012, 4:21 AM EDT
Posted: 5 years ago
Hi

Comsol uses higer order discretization on the elements, so you get more DoFs through tha (check the KB and the doc)t,and you should check how many depedent variables you have, some BC add local dependent variables too

--
Good luck
Ivar

Posted: 5 years ago
Many thanks!

Have a nice day.

Paolo.

Note that while COMSOL employees may participate in the discussion forum, COMSOL® software users who are on-subscription should submit their questions via the Support Center for a more comprehensive response from the Technical Support team.