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

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

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

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

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

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

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

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

### [deal.II] Question about the numbering of DoFs

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] Question about the numbering of DoFs

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