2D
Consider \(d = 2\), a triangle with nodes \(\{x_0, x_1, x_2\}\), and let the tangents be given by
\[
\tau_{ij} = x_j - x_i
\]
Following Eggers and Radu (2020), we consider the following basis functions
\[\begin{split}
\begin{aligned}
\varphi_0 &= \lambda_0 \lambda_2 \tau_{02} + \lambda_0 \lambda_1 \tau_{01} \\
\varphi_1 &= \lambda_1 \lambda_0 \tau_{10} + \lambda_1 \lambda_2 \tau_{12} \\
\varphi_2 &= \lambda_2 \lambda_1 \tau_{21} + \lambda_2 \lambda_0 \tau_{20}
\end{aligned}
\end{split}\]
At the cell center, each \(\lambda_i = \frac1{d + 1}\), so we have the following identity:
\[\begin{split}
\begin{aligned}
(\varphi_0 - \varphi_1)(x_c) &= \frac{1}{(d + 1)^2}(\tau_{02} + \tau_{01} - \tau_{10} - \tau_{12}) \\
&= \frac{1}{(d + 1)^2}(3 \tau_{01} - (\tau_{01} + \tau_{12} + \tau_{20}))
= \frac{3}{(d + 1)^2}\tau_{01}
= \frac{1}{d + 1}\tau_{01}
\end{aligned}
\end{split}\]
Let \(\phi_i^j\) denote the basis function for the the degree of freedom at face \(f_j\) and node \(x_i\). Letting \(x^j\) be the node opposite face \(f_j\),
we propose the following basis function
\[
\begin{aligned}
\phi_1^0 &= \pm \frac{1}{d |\Delta|} \left(\lambda_1 \tau_{01} - (\varphi_0 - \varphi_1) \right)
\end{aligned}
\]
More generally, we define
\[
\begin{aligned}
\phi_i^j &= \pm \frac{1}{d |\Delta|} \left(\lambda_1 \tau_{ji} - (\varphi_j - \varphi_i) \right)
\end{aligned}
\]
where the sign is determined by whether the normal of face \(j\) is outward with respect to the element.
For the cell degrees of freedom, we define
\[
\begin{aligned}
\phi_k &= \frac{1}{d |\Delta|} \varphi_k
\end{aligned}
\]
Note that \(\phi_0 + \phi_1 + \phi_2 = 0\), so we can span all three with \(\{\phi_0, \phi_1\}\).
Properties
Degrees of freedom
We now show that these basis functions have the desired properties. First, we have
\[
n \cdot \phi_k = 0
\]
on each face because either \(\phi_k = 0\) or it is tangential to the face. In turn, we have \(\phi_k = 0\) at the nodes.
Second, we have at the cell-centers
\[
\phi_i^j(x_c) = \pm \frac{1}{d |\Delta|} \left(\frac1{d + 1} \tau_{ji} - \frac{1}{d + 1}\tau_{ji} \right) = 0
\]
This means that we can evaluate the degrees of freedom by considering the nodal and cell center values of a function.
The divergence
The divergence of the cell basis functions can be calculated as:
\[\begin{split}
\begin{aligned}
\nabla \cdot \varphi_k
&= \nabla \cdot(\lambda_k \lambda_{k - 1} \tau_{k,k - 1} + \lambda_k \lambda_{k + 1} \tau_{k, k + 1}) \\
&= (\lambda_k (\nabla \lambda_{k - 1}) + (\nabla \lambda_k) \lambda_{k - 1}) \cdot \tau_{k,k - 1} + (\lambda_k (\nabla \lambda_{k + 1}) + (\nabla \lambda_k) \lambda_{k + 1}) \cdot \tau_{k, k + 1} \\
&= \lambda_k - \lambda_{k - 1} + \lambda_k - \lambda_{k + 1} \\
&= 3\lambda_k - (\lambda_{k - 1} + \lambda_k + \lambda_{k + 1}) \\
&= (d + 1) \lambda_k - 1,
\end{aligned}
\end{split}\]
where \(k\) is understood modulo 3. This has mean zero. For the face degrees of freedom, we compute
\[\begin{split}
\begin{aligned}
\nabla \cdot \phi_i^j
&= \pm \frac{1}{d |\Delta|} \left(\nabla\lambda_i \cdot \tau_{ji} - \nabla \cdot(\varphi_j - \varphi_i) \right) \\
&= \pm \frac{1}{d |\Delta|}\left( 1 - (d + 1) (\lambda_j - \lambda_i) \right)
\end{aligned}
\end{split}\]
From which we easily deduce that \(\int_\Delta \nabla \cdot \phi_i^j = \pm \frac1d\).
Implementation
We consider the following spatial basis functions (in order)
\[
\{
\lambda_0, \ \lambda_1, \ \lambda_2, \ \lambda_0 \lambda_1, \lambda_0 \lambda_2, \lambda_1 \lambda_2 \}
\]
Since these are the basis functions used for Lagrange2, we can fetch the inner products from there. After applying a Kronecker product, this local mass matrix is adapted to the vector-valued setting.
As is common in PyGeoN, we then create an array \(\Psi\) whose rows contain the coefficents for each basis function. As an example, let us compute the row for \(\phi_1^0\). We first introduce the helper functions
\[\begin{split}
\begin{aligned}
\psi_0 &=
\begin{bmatrix}
0 \\ 0 \\ 0 \\ \tau_{01} \\ \tau_{02} \\ 0
\end{bmatrix}
&
\psi_1 &=
\begin{bmatrix}
0 \\ 0 \\ 0 \\ \tau_{10} \\ 0 \\ \tau_{12}
\end{bmatrix}
&
\psi_2 &=
\begin{bmatrix}
0 \\ 0 \\ 0 \\ 0 \\ \tau_{20} \\ \tau_{21}
\end{bmatrix}
\end{aligned}
\end{split}\]
These are computed by looping over the edges of the element.
Using these, we can rapidly compute, for example
\[\begin{split}
\Psi_1^0 = \pm \frac1{d |\Delta|}\left(
\begin{bmatrix}
0 \\ \tau_{01} \\ 0 \\ 0 \\ 0 \\ 0
\end{bmatrix}
- (\psi_0 - \psi_1)
\right).\text{ravel}()
\end{split}\]
and for the cell-based degrees of freedom:
\[
\Psi_0 = \frac{1}{d |\Delta|} \psi_0
\]
3D
The three-dimensional generalization of \(\varphi\) is given by
\[
\begin{aligned}
\varphi_k &=
\sum_{i \ne k} \lambda_k \lambda_i \tau_{ki}
\end{aligned}
\]
A similar calculation as in 2D shows us that
\[\begin{split}
\begin{aligned}
(\varphi_0 - \varphi_1)(x_c)
&= \frac{1}{(d + 1)^2}(\tau_{01} + \tau_{02} + \tau_{03} - \tau_{10} - \tau_{12} - \tau_{13}) \\
&= \frac{1}{(d + 1)^2}(4 \tau_{01} - (\tau_{01} + \tau_{12} + \tau_{20}) - (\tau_{01} + \tau_{13} + \tau_{30}))
= \frac{4\tau_{01}}{(d + 1)^2}
= \frac{\tau_{01}}{d + 1}
\end{aligned}
\end{split}\]
so we again propose the basis functions
\[\begin{split}
\begin{aligned}
\phi_i^j &= \pm \frac{1}{d |\Delta|} \left(\lambda_1 \tau_{ji} - (\varphi_j - \varphi_i) \right) \\
\phi_k &= \frac{1}{d |\Delta|} \varphi_k
\end{aligned}
\end{split}\]
The properties concerning the degrees of freedom follow immediately. All we need to double check is the divergence
The divergence
\[\begin{split}
\begin{aligned}
\nabla \cdot \varphi_k &=
\sum_{i \ne k} \nabla \cdot \lambda_k \lambda_i \tau_{ki} \\
&= \sum_{i \ne k} \lambda_k - \lambda_i \\
&= (d + 1) \lambda_k - 1
\end{aligned}
\end{split}\]
this gives us
\[
\begin{aligned}
\nabla \cdot (\varphi_j - \varphi_i) &=
(d + 1) (\lambda_j - \lambda_i)
\end{aligned}
\]
In turn, we have the proper generalization from the 2D case:
\[\begin{split}
\begin{aligned}
\nabla \cdot \phi_i^j
&= \pm \frac{1}{d |\Delta|}\left( 1 - (d + 1) (\lambda_j - \lambda_i) \right) \\
\nabla \cdot \phi_k
&= \frac{1}{d |\Delta|}\left((d + 1) \lambda_k - 1 \right)
\end{aligned}
\end{split}\]