![Figure 7. The DPG method performs better than the DG method under both h and p refinements for the Peterson example [17]. Both plots are in a log-log scale.](/figures/figure-7-the-dpg-method-performs-better-than-the-dg-method-2sfzsglz.webp)
Q2. What is the result of the rewriting of the left hand side?
When rewriting the left hand side as a sum over the mesh edges, the authors observe that all contributions due to edges interior to Λ` cancel out.
Q3. How can the authors get a low-order generalization of the DPG method to the case?
by simply modifying their test space Vp(K) to align with ΠRTh ~β within each element K, the authors obtain a low-order generalization of the DPG method to the case of variable ~β with minimal modifications.
Q4. What is the first in the finite element family of methods?
The authors believe their method is the first in the finite element (not just DG) family of methods for the transport equation which has provably optimal convergence rates on very general meshes.
Q5. how can the outflow variable be interpreted?
the identification of inflow and outflow faces can only be as accurate as the round-off errors permit, i.e., the strict inequalities (9) can only be implemented up to round-off.
Q6. How can the identification of inflow and outflow faces be done?
the identification of inflow and outflow faces can only be as accurate as the round-off errors permit, i.e., the strict inequalities (9) can only be implemented up to round-off.
Q7. What is the first error estimate of the theorem?
By the inequality of arithmetic and geometric means, this implies|||φ− φh|||21 h , 1 β ,K ≤ 3h−1β,K‖φ−ΠMφ‖21 β ,∂outK∪∂inK ,which gives the first error estimate of the theorem.
Q8. What is the norm on the test space?
The functional |||(u, φ)|||h is a norm on the space X = L2(Ω)×M , where M = {µ : µ|F ∈ L2(F ) for all mesh edges F and µ|∂inΩ = 0}.
Q9. What is the spectral method for the composite method?
The authors put together the spectral method on each element to get the following composite method on the mesh Th: Find (uh, φh) ∈ Xh def = Wh ×Mh satisfyingah( (uh, φh), vh) = (f, vh)Ω − 〈~β · ~n g, vh〉∂inΩ, (25) for all vh ∈ Vh, whereah( (uh, φh), vh) =∑K∈Th[ 〈φh, vh〉∂outK − 〈φh, vh〉∂inK\\∂inΩ − (uh, ~β · ~∇ vh)K ]
Q10. How does the process of splitting the mesh into layers S yield a flux degree of freedom?
the process of splitting the mesh into layers S` (see § 3.4) yields an enumeration of flux degrees of freedom that makes the matrix for φh’s triangular, and hence suitable for fast solution by backsubstitution.
Q11. What is the ole r in the case of a general tetra?
The ole r as the authors re tricted ourselves to th two-dimensional case in Section 3 is that Lemma 3.1 does not hold for general tetrahedral meshes.