Integration order

Hi,

I am currently working with THERMO_RICHARDS_MECHANICS process. I’ve been checking different benchmark problems and I have found that all of them have different values for the command “<integration_order></integration_order>”. Could anyone please explain to me how do you choose the value for this integration order?

Also, how does this relate to number of nodes I should have in my mesh?
I am asking because I was generating my mesh in gmsh using triangle elements and assigning them to order 4 and then also using the integration_order command in OGS equal to 4. Are these two related or not? Does it work like that? or am I not understaing correctly?

Thanks for your help,
María :smiley: :smiley:

Hi María,
concerning mesh elements, we use in OGS typically linear elements and quadratic elements only. However, quadratic elements are normally used for a Taylor-Hood mixed element scheme, meaning for TRM you would have linear order for T and H and quadratic order for displacement. For this to work, you need quadratic elements. The order is set for each process variable in the <process_variable> section.
If you generate triangles with order 4, I have no idea how this translates to VTK types as they are typically also linear or quadratic elements, there might be some separate Lagrange element types. Higher order types are needed for high-order methods only. For TRM in OGS you are fine with linear or quadratic order, everything higher is probably not implemented.
The integration order is something different but related. It refers to the number of integration points within a cell. See Gauß-Quadratur – Wikipedia for details. Usually, a higher integration order results in more integration points and more computation effort that is needed, but achieving more accurate results. However, for a polynomial integrand you could achieve complete integration with a relatively low of integration points, which should increase with shape function/element order. See Quadrature schemes and extrapolation for details.

Thanks for your response, now it is clear to me :smiley: