discretizing surface integral using nodal DG method

Question

I am currently learning nodal DG methods, primarily through the book by Warburton, and am a bit confused on how to handle surface integrals using straight edged elements. On page 187 (and on page 214) of Warburton's book, we have

$$ \int_{\partial E_{k}} \mathbf{n} \cdot f_{h} \ell_{i}^{k}(\mathbf{x}) d\mathbf{x}. $$

Recall that $f_{h} = \sum_{j=1}^{N} f_{j}(x,t)\ell_{j}(\mathbf{x})$ and $\ell_{i}(\mathbf{x})$ is our basis function, and so,

$$ \int_{\partial E_{k}} \mathbf{n} \cdot f_{h} \ell_{i}^{k}(\mathbf{x}) d\mathbf{x} = \int_{\partial E_{k}} \mathbf{n} \cdot \left[ \sum_{j=1}^{N} f_{j}(x,t)\ell_{j}(\mathbf{x}) \right]\ell_{i}^{k}(\mathbf{x}) d\mathbf{x}\\ = \sum_{j=1}^{N} \mathbf{n} \cdot f_{j}(x,t) \int_{\partial E_{k}} \ell{j}(\mathbf{x}) \ell_{i}(\mathbf{x}) d \mathbf{x}\\ = \sum_{j=1}^{N} \mathbf{n} \cdot f_{j}(x,t) M^{k}_{ij}. $$ Now, to discretize the surface mass matrix, we must use a quadrature rule on each face (since I am working in 2D it's a line integral along each edge). With that said, since we have an orthonormal basis, we can compute the mass matrix analytically. For volume integrals, Warburton shows that $M^{k}_{ij} = (\mathcal{V} \mathcal{V}^{T})^{-1} J,$ where $J$ is the metric Jacobian. So intuitively, the same would be for the mass matrix, however, the Vandermonde matrix, $\mathcal{V}$ would be constructed so its on the local face points $(r,s)$, where $(r,s) \in [-1,1]$. Now, looking at their code (https://github.com/tcew/nodal-dg/blob/master/Codes1.1/CFD2D/CurvedEulerRHS2D.m), we see that the surface integral term has no surface mass matrix. There is an inverse mass matrix applied to the numerical flux but it is the one from the time-derivative. I was curious if someone could point out why there is no surface mass matrix in their formulation.

Answer 1

In 2D your volume part consists of a double integral, whereas the surface part is a standard line integral. Generally this is described with tensorial notations.

If you consider the Cartesian case, the volume part results in a mass matrix similar to

$M_{\Omega}\equiv \mathbf{M}_{\text{1}} \otimes \mathbf{M}_{\text{1}}$,

whereas the surface part results in a mass matrix similar to

$M_{\partial \Omega}^{\xi} \equiv \mathbf{M}_{\text{1}} \otimes \mathbf{I}_1\\ M_{\partial \Omega}^{\eta} \equiv \mathbf{I}_1 \otimes \mathbf{M}_{\text{1}}$.

Here $\mathbf{I}_1$ is the identity matrix.

If you invert the volume part and apply it on $\xi$-faces you will get something like

$=\left( \mathbf{M}_{\text{1}} \otimes \mathbf{M}_{\text{1}}\right)^{-1}(\mathbf{M}_{\text{1}} \otimes \mathbf{I}_1) \\ = (\mathbf{M}_{\text{1}} ^{-1} \otimes \mathbf{M}_{\text{1}} ^{-1}) (\mathbf{M}_{\text{1}} \otimes \mathbf{I}_1) \\ = (\mathbf{M}_{\text{1}} ^{-1} \mathbf{M}_{\text{1}} ) \otimes (\mathbf{M}_{\text{1}}^{-1} \mathbf{I}_1)\\ = \mathbf{I}_{\text{1}} \otimes \mathbf{M}_{\text{1}}^{-1}$

or similar on $\eta$-faces

$=\left( \mathbf{M}_{\text{1}} \otimes \mathbf{M}_{\text{1}}\right)^{-1}(\mathbf{I}_{\text{1}} \otimes \mathbf{M}_1) \\ = (\mathbf{M}_{\text{1}} ^{-1} \otimes \mathbf{M}_{\text{1}} ^{-1}) (\mathbf{I}_{\text{1}} \otimes \mathbf{M}_1) \\ = (\mathbf{M}_{\text{1}} ^{-1} \mathbf{I}_{\text{1}} ) \otimes (\mathbf{M}_{\text{1}}^{-1} \mathbf{M}_1)\\ = \mathbf{M}_{\text{1}}^{-1} \otimes \mathbf{I}_{\text{1}}.$

Summerizing: I think you are missing the fact, that both mass matrices (volume and surface) cancel, resulting in a 1D inverse mass matrix. Note that numerical fluxes on $\eta$-faces are projected into the volume in $\xi$-direction, whereas fluxes on $\xi$-faces are projected into the volume in $\eta$-direction.

Regards