On Generalized Gauss–Radau Projections and Optimal Error Estimates of Upwind-Biased DG Methods for the Linear Advection Equation on Special Simplex Meshes

Abstract

Generalized Gauss–Radau (GGR) projections are global projection operators that are widely used for the error analysis of discontinuous Galerkin (DG) methods with generalized numerical fluxes. In previous work, GGR projections were constructed for Cartesian meshes and analyzed through an algebraic approach. In this paper, we first present an alternative energy approach for analyzing the one-dimensional GGR projection, which does not require assembling and explicitly solving a global system over the entire computational domain as that in the algebraic approach. We then generalize this energy argument to construct a global projection operator on special simplex meshes in multidimensions satisfying the so-called flow condition. With this projection, optimal error estimates are proved for upwind-biased DG methods for the linear advection equation on these meshes, which generalizes the error analysis for the purely upwind case by Cockburn et al. (SIAM J Numer Anal 46(3):1250–1265, 2008) in a time-dependent setting.

Appendices

Appendix A: Proof of Lemma 2.8

Before starting, we first state the following proposition, which can be deduced from Proposition 2.5 through symmetry.

Proposition A.1

Proposition 2.5 holds after replacing (2.34b) with $Z^+_{j-1/2} = z$.

The rest of the section is dedicated to the proof of Lemma 2.8.

Proof of Lemma 2.8

By default, we have $1\le j \le N$ within the proof. From Propositions 2.5 and A.1, it can be seen that the following local projection is well-defined a

$$\begin{aligned} {\left( \Pi u,v\right) }_{I_j} =&\; {\left( u,v\right) }_{I_j}, \quad{} & {} \forall v \in \mathcal {P}_{k-1}(I_j),\quad \forall j = 1, \cdots , N,\end{aligned}$$

(A.1a)

$$\begin{aligned} (\Pi u)^-_{j+\frac{1}{2}} =&\; u^-_{j+\frac{1}{2}}, \quad{} & {} \text {if }\theta _j>\frac{1}{2},\end{aligned}$$

(A.1b)

$$\begin{aligned} (\Pi u)^+_{j-\frac{1}{2}} =&\; u^+_{j-\frac{1}{2}}, \quad{} & {} \text {if }\theta _j<\frac{1}{2}. \end{aligned}$$

(A.1c)

Indeed, it can be equivalently written as

$$\begin{aligned} \Pi u = \left\{ \begin{array}{ll} \Pi _1 u, &{}\text {if }\theta _j>\frac{1}{2},\\ \Pi _0 u, &{}\text {if }\theta _j<\frac{1}{2}. \end{array}\right. \end{aligned}$$

(A.2)

Here $\Pi _1$ and $\Pi _0$ correspond to (2.48) with $\theta \equiv 1$ and $\theta \equiv 0$, respectively. All three projections, $\Pi _1$, $\Pi _0$ and $\Pi$, satisfy (2.49) with $C_\theta = C$ independent of $\mu ^*$ and $\mu _*$.

We denote by $\delta = (\Pi _\theta - \Pi ) u$. Subtracting (A.1) from (2.48), one can see that the difference $\delta$ satisfies the following equations.

$$\begin{aligned} {\left( \delta ,v\right) }_{I_j} =&\; 0, \quad{} & {} v \in \mathcal {P}_{k-1}(I_j),\quad \forall j = 1, \cdots , N,\end{aligned}$$

(A.3a)

$$\begin{aligned} \{\delta \}^{(\theta _j)}_{j+\frac{1}{2}} =&\; {\bar{\eta }}_{j+\frac{1}{2}}, \quad{} & {} \text {if }\theta _j>\frac{1}{2},\end{aligned}$$

(A.3b)

$$\begin{aligned} \{\delta \}^{(\theta _j)}_{j-\frac{1}{2}} =&\; {\bar{\zeta }}_{j-\frac{1}{2}}, \quad{} & {} \text {if }\theta _j<\frac{1}{2}. \end{aligned}$$

(A.3c)

Here ${\bar{\eta }}_{j+1/2} = \{u - \Pi _1 u\}^{(\theta _j)}_{j+1/2}$ and ${\bar{\zeta }}_{j-1/2} = \{u - \Pi _0 u\}^{(\theta _j)}_{j-1/2}$. Using the same argument as that in the proof of Lemma 2.1, it suffices to show the solution to (A.3) satisfies

$$\begin{aligned} \Vert \delta \Vert _{L^2\left( \mathcal {T}_h\right) } + h^\frac{1}{2}\Vert \delta \Vert _{L^2\left( \mathcal {E}_h^+\right) } \le \hat{C}_\theta h^\frac{1}{2}\left( \Vert {\bar{\eta }} \Vert _{L^2\left( \mathcal {E}_h^+\right) } +\Vert {\bar{\zeta }} \Vert _{L^2\left( \mathcal {E}_h^+\right) } \right) \end{aligned}$$

(A.4)

with

$$\begin{aligned} \hat{C}_\theta = C\left( 1+\left( \mu ^*+1/2\right) \mu _*^{-1}\right) \left( 1+\left( \mu ^*+1/2\right) \mu _*^{-1/2}\right) \end{aligned}$$

(A.5)

to complete the proof. We now proceed to prove (A.4).

To facilitate the discussion, we introduce the index sets

$$\begin{aligned}&J^{-,+} = \left\{ j: \theta _j<\frac{1}{2}<\theta _{j+1}\right\} ,\quad&J^{+,-} = \left\{ j: \theta _j>\frac{1}{2}>\theta _{j+1}\right\} ;\end{aligned}$$

(A.6a)

$$\begin{aligned}&J^{-,-} = \left\{ j: \theta _j<\frac{1}{2}, \theta _{j+1}<\frac{1}{2}\right\} ,\quad&J^{+,+} = \left\{ j: \theta _j>\frac{1}{2}, \theta _{j+1}>\frac{1}{2}\right\} , \end{aligned}$$

(A.6b)

and $\mathcal {E}_h^{-,+}$, $\mathcal {E}_h^{+,-}$, $\mathcal {E}_h^{-,-}$ and $\mathcal {E}_h^{+,+}$ for corresponding sets of $\{x_{j+1/2}\}$. It can be seen that we are imposing one condition on $\mathcal {E}_h^{-,-}$ and $\mathcal {E}_h^{+,+}$, two conditions on $\mathcal {E}_h^{+,-}$, and no condition on $\mathcal {E}_h^{-,+}$, for each mesh point.

Step 1: Estimate of $\Vert \delta \Vert _{L^2\left( \mathcal {E}_h^{+,-}\right) }$. Note that for $x_{j+\frac{1}{2}}\in \mathcal {E}_h^{+,-}$, we have

$$\begin{aligned} \theta _j \delta _{j+\frac{1}{2}}^- + \tilde{ \theta }_{j} \delta _{j+\frac{1}{2}}^+ = {\bar{\eta }}_{j+\frac{1}{2}},\end{aligned}$$

(A.7a)

$$\begin{aligned} \theta _{j+1} \delta _{j+\frac{1}{2}}^- + \tilde{\theta }_{j+1} \delta _{j+\frac{1}{2}}^+ = {\bar{\zeta }}_{j+\frac{1}{2}}. \end{aligned}$$

(A.7b)

Since $\theta \ne \tilde{\theta }$, we can solve the equation system (A.7) to get

$$\begin{aligned} \delta ^-_{j+\frac{1}{2}} = \frac{\tilde{\theta }_{j+1} {\bar{\eta }}_{j+\frac{1}{2}}- \tilde{\theta }_j {\bar{\zeta }}_{j+\frac{1}{2}} }{\tilde{\theta }_{j+1} -\tilde{\theta }_j}\quad \text {and} \quad \delta ^+_{j+\frac{1}{2}} = \frac{{\theta }_{j+1} {\bar{\eta }}_{j+\frac{1}{2}}- {\theta }_j {\bar{\zeta }}_{j+\frac{1}{2}} }{{\theta }_{j+1} -{\theta }_j}. \end{aligned}$$

(A.8)

Recall our assumption $0<\mu _* \le \left| \theta _j -1/2 \right| \le \mu ^* <+\infty$ and let us define

$$\begin{aligned} \kappa = \left( \mu ^*+\frac{1}{2}\right) \mu _*^{-1}. \end{aligned}$$

(A.9)

Then it can be estimated that

$$\begin{aligned} \Vert \delta \Vert _{L^2\left( \mathcal {E}_h^{+,-}\right) } \le C\kappa \left( \Vert {\bar{\eta }} \Vert _{L^2\left( \mathcal {E}_h^{+,-}\right) } +\Vert {\bar{\zeta }} \Vert _{L^2\left( \mathcal {E}_h^{+,-}\right) } \right) . \end{aligned}$$

(A.10)

Step 2: Estimate of $\Vert \delta \Vert _{L^2\left( \mathcal {E}_h^{-,+}\right) }$ and $\Vert \left[ \delta \right] \Vert _{L^2(\mathcal {E}_h^{+,+}\cup \mathcal {E}_h^{-,-})}$. From (A.3a), it can be seen that

$$\begin{aligned} \begin{aligned} 0 =&\sum _{j:\theta _j>\frac{1}{2}}{\left( \delta ,\delta _x\right) }_{I_j} - \sum _{j:\theta _j<\frac{1}{2}}{\left( \delta ,\delta _x\right) }_{I_j} \\ =&\frac{1}{2}\sum _{j:\theta _j>\frac{1}{2}}\left( \left| \delta ^-_{j+\frac{1}{2}}\right| ^2 - \left| \delta ^+_{j-\frac{1}{2}}\right| ^2\right) - \frac{1}{2}\sum _{j:\theta _j<\frac{1}{2}}\left( \left| \delta ^-_{j+\frac{1}{2}}\right| ^2 - \left| \delta ^+_{j-\frac{1}{2}}\right| ^2\right) \\ =&\frac{1}{2}\sum _{j\in J^{+,+}}\left( \left| \delta ^-_{j+\frac{1}{2}}\right| ^2 - \left| \delta ^+_{j+\frac{1}{2}}\right| ^2\right) - \frac{1}{2}\sum _{j\in J^{-,-}}\left( \left| \delta ^-_{j+\frac{1}{2}}\right| ^2 - \left| \delta ^+_{j+\frac{1}{2}}\right| ^2\right) \\&- \frac{1}{2}\sum _{j\in J^{-,+}}\left( \left| \delta ^-_{j+\frac{1}{2}}\right| ^2 + \left| \delta ^+_{j+\frac{1}{2}}\right| ^2\right) + \frac{1}{2}\sum _{j\in J^{+,-}}\left( \left| \delta ^-_{j+\frac{1}{2}}\right| ^2 + \left| \delta ^+_{j+\frac{1}{2}}\right| ^2\right) \\ =&- \sum _{j\in J^{+,+}}\{\delta \}^{(1/2)}_{j+\frac{1}{2}} \left[ \delta \right] _{j+\frac{1}{2}}+ \sum _{j\in J^{-,-}}\{\delta \}_{j+\frac{1}{2}}^{(1/2)} \left[ \delta \right] _{j+\frac{1}{2}} - \Vert \delta \Vert _{L^2\left( \mathcal {E}_h^{-,+}\right) }^2+\Vert \delta \Vert _{L^2\left( \mathcal {E}_h^{+,-}\right) }^2. \end{aligned} \end{aligned}$$

(A.11)

Here we have used the identity $|\delta ^+|^2 - |\delta ^-|^2 =2 \{\delta \}^{(1/2)}\left[ \delta \right]$. With (A.3b) and (A.3c), it can be shown that

$$\begin{aligned} \sum _{j\in J^{+,+}}\{\delta \}^{(\theta _j)}_{j+\frac{1}{2}} \left[ \delta \right] _{j+\frac{1}{2}}- \sum _{j\in J^{-,-}}\{\delta \}^{(\theta _j)}_{j+\frac{1}{2}} \left[ \delta \right] _{j+\frac{1}{2}} = \sum _{j\in J^{+,+}}{\bar{\eta }}_{j+\frac{1}{2}} \left[ \delta \right] _{j+\frac{1}{2}}- \sum _{j\in J^{-,-}}{\bar{\zeta }}_{j+\frac{1}{2}} \left[ \delta \right] _{j+\frac{1}{2}}.\nonumber \\ \end{aligned}$$

(A.12)

Note that $\{\delta \}_{j+\frac{1}{2}}^{(\theta _j)} = \{\delta \}_{j+\frac{1}{2}}^{(1/2)} - (\theta _j-1/2)\left[ \delta \right] _{j+\frac{1}{2}}$. Combining (A.11) and (A.12) yields

$$\begin{aligned}{} & {} \sum _{j\in J^{+,+}\cup J^{-,-}}\left| \theta _j - \frac{1}{2}\right| \left[ \delta \right] _{j+\frac{1}{2}}^2 + \Vert \delta \Vert _{L^2\left( \mathcal {E}_h^{-,+}\right) }^2-\Vert \delta \Vert _{L^2\left( \mathcal {E}_h^{+,-}\right) }^2\nonumber \\{} & {} \quad = \sum _{j\in J^{-,-}}{\bar{\zeta }}_{j+\frac{1}{2}} \left[ \delta \right] _{j+\frac{1}{2}} - \sum _{j\in J^{+,+}}{\bar{\eta }}_{j+\frac{1}{2}} \left[ \delta \right] _{j+\frac{1}{2}}. \end{aligned}$$

(A.13)

Using the assumption $\left| \theta _j -1/2\right| \ge \mu _*$ and the Cauchy–Schwartz inequality on the right side, we can obtain

$$\begin{aligned} \begin{aligned}&\mu _*\Vert \left[ \delta \right] \Vert _{L^2\left( \mathcal {E}_h^{+,+}\cup \mathcal {E}_h^{-,-}\right) }^2 + \Vert \delta \Vert _{L^2\left( \mathcal {E}_h^{-,+}\right) }^2-\Vert \delta \Vert _{L^2\left( \mathcal {E}_h^{+,-}\right) }^2\\ \le \;&\Vert {\bar{\eta }} \Vert _{L^2\left( \mathcal {E}_h^{+,+}\right) }\Vert \left[ \delta \right] \Vert _{L^2\left( \mathcal {E}_h^{+,+}\right) } + \Vert {\bar{\zeta }} \Vert _{L^2\left( \mathcal {E}_h^{-,-}\right) }\Vert \left[ \delta \right] \Vert _{L^2\left( \mathcal {E}_h^{-,-}\right) }. \end{aligned} \end{aligned}$$

(A.14)

Using the inequality $ab \le \left( a^2 + b^2\right) /2$ on the right side, we can simplify the inequality as

$$\begin{aligned}&\frac{\mu _*}{2}\Vert \left[ \delta \right] \Vert _{L^2\left( \mathcal {E}_h^{+,+}\cup \mathcal {E}_h^{-,-}\right) }^2 + \Vert \delta \Vert _{L^2\left( \mathcal {E}_h^{-,+}\right) }^2 \nonumber \\&\quad \le (2\mu _*)^{-1}\left( \Vert {\bar{\eta }} \Vert _{L^2\left( \mathcal {E}_h^{+,+}\right) }^2 + \Vert {\bar{\zeta }} \Vert _{L^2\left( \mathcal {E}_h^{-,-}\right) }^2\right) + \Vert \delta \Vert _{L^2\left( \mathcal {E}_h^{+,-}\right) }^2. \end{aligned}$$

(A.15)

Taking the square root and using the inequality $(|a|+|b|)/\sqrt{2}\le \sqrt{a^2+b^2}\le |a|+|b|$ yield

$$\begin{aligned}&{\mu _*^{\frac{1}{2}}}\Vert \left[ \delta \right] \Vert _{L^2\left( \mathcal {E}_h^{+,+}\cup \mathcal {E}_h^{-,-}\right) } + \Vert \delta \Vert _{L^2\left( \mathcal {E}_h^{-,+}\right) } \le \nonumber \\ {}&C{\mu _*^{-\frac{1}{2}}}\left( \Vert {\bar{\eta }} \Vert _{L^2\left( \mathcal {E}_h^{+,+}\right) } + \Vert {\bar{\zeta }} \Vert _{L^2\left( \mathcal {E}_h^{-,-}\right) }\right) + C\Vert \delta \Vert _{L^2\left( \mathcal {E}_h^{+,-}\right) }. \end{aligned}$$

(A.16)

Note that $(\mu _*)^{-\frac{1}{2}}\le 1/2 + 1/(2\mu _*)\le C\left( 1+\kappa \right)$. Combining with (A.10), it can be shown that

$$\begin{aligned} \begin{aligned}&{\mu _*^\frac{1}{2}}\Vert \left[ \delta \right] \Vert _{L^2\left( \mathcal {E}_h^{+,+}\cup \mathcal {E}_h^{-,-}\right) } + \Vert \delta \Vert _{L^2\left( \mathcal {E}_h^{-,+}\right) } \\ \le \;&C(1+\kappa )\left( \Vert {\bar{\eta }} \Vert _{L^2\left( \mathcal {E}_h^{+,+}\cup \mathcal {E}_h^{+,-}\right) } +\Vert {\bar{\zeta }} \Vert _{L^2\left( \mathcal {E}_h^{-,-}\cup \mathcal {E}_h^{+,-}\right) } \right) \\ \le \;&C(1+\kappa )\left( \Vert {\bar{\eta }} \Vert _{L^2\left( \mathcal {E}_h^+\right) } +\Vert {\bar{\zeta }} \Vert _{L^2\left( \mathcal {E}_h^+\right) } \right) . \end{aligned} \end{aligned}$$

(A.17)

Step 3: Estimate of $\Vert \delta \Vert _{L^2\left( \mathcal {E}_h^+\right) }$. From (A.3b) and (A.3c), we have

$$\begin{aligned} \delta ^-_{j+\frac{1}{2}} = {\bar{\eta }}_{j+\frac{1}{2}} - \tilde{\theta }_j\left[ \delta \right] _{j+\frac{1}{2}}, \quad \forall j\in \mathcal {E}_h^{+,+}. \end{aligned}$$

(A.18a)

$$\begin{aligned} \delta ^+_{j+\frac{1}{2}} = {\bar{\zeta }}_{j+\frac{1}{2}} + \theta _j\left[ \delta \right] _{j+\frac{1}{2}}, \quad \forall j\in \mathcal {E}_h^{-,-}. \end{aligned}$$

(A.18b)

With the triangle inequality, the estimate (A.17), and the fact $|\theta _j|, |\tilde{\theta }_j|\le \mu ^* + 1/2 = \kappa \mu _*$, it can be seen that

$$\begin{aligned} \Vert \delta \Vert _{L^2\left( \mathcal {E}_h^{+,+}\cup \mathcal {E}_h^{-,-}\right) } \le \left( 1+ C(1+\kappa )\kappa \mu _*^\frac{1}{2}\right) \left( \Vert {\bar{\eta }} \Vert _{L^2\left( \mathcal {E}_h^+\right) } +\Vert {\bar{\zeta }} \Vert _{L^2\left( \mathcal {E}_h^+\right) } \right) . \end{aligned}$$

(A.19)

Therefore, with (A.10), (A.17), and (A.19), we have

$$\begin{aligned} \Vert \delta \Vert _{L^2\left( \mathcal {E}_h^+\right) }\le & {} \left( 1+ C(1+\kappa )\left( 1+\kappa \mu _*^\frac{1}{2}\right) \right) \left( \Vert {\bar{\eta }} \Vert _{L^2\left( \mathcal {E}_h^+\right) } +\Vert {\bar{\zeta }} \Vert _{L^2\left( \mathcal {E}_h^+\right) } \right) \nonumber \\\le & {} \hat{C}_\theta \left( \Vert {\bar{\eta }} \Vert _{L^2\left( \mathcal {E}_h^+\right) } +\Vert {\bar{\zeta }} \Vert _{L^2\left( \mathcal {E}_h^+\right) } \right) . \end{aligned}$$

(A.20)

Step 4: Estimate of $\Vert \delta \Vert _{L^2\left( \mathcal {T}_h\right) }$. We can deduce from Propositions 2.5 and A.1 that

$$\begin{aligned} \Vert \delta \Vert _{L^2\left( I_j\right) }\le C h_j^{\frac{1}{2}}\left| \delta ^-_{j+\frac{1}{2}}\right| , \quad \forall j: \theta _j >\frac{1}{2}.\end{aligned}$$

(A.21a)

$$\begin{aligned} \Vert \delta \Vert _{L^2\left( I_j\right) }\le C h_j^{\frac{1}{2}}\left| \delta ^+_{j-\frac{1}{2}}\right| , \quad \forall j: \theta _j <\frac{1}{2}. \end{aligned}$$

(A.21b)

Taking the square and summing over all mesh cells yield

$$\begin{aligned} \Vert \delta \Vert _{L^2\left( \mathcal {T}_h\right) }^2 = \sum _{j = 1}^N \Vert \delta \Vert _{L^2\left( I_j\right) }^2 \le C h \Vert \delta \Vert _{L^2\left( \mathcal {E}_h^+\right) }^2. \end{aligned}$$

(A.22)

Apply the estimate in (A.20) and take the square root. One can obtain

$$\begin{aligned} \Vert \delta \Vert _{L^2\left( \mathcal {T}_h\right) } \le \hat{C}_\theta h^\frac{1}{2}\left( \Vert {\bar{\eta }} \Vert _{L^2\left( \mathcal {E}_h^+\right) } +\Vert {\bar{\zeta }} \Vert _{L^2\left( \mathcal {E}_h^+\right) } \right) . \end{aligned}$$

(A.23)

Finally, the estimate (A.4) can be obtained by combining (A.20) and (A.23).

Appendix B: Proof of Lemma 3.2

Proof

Two-dimensional case: It is known that the shape-regularity condition (3.2) is equivalent to the following minimal angle condition in 2D (also known as the Zlámal’s condition [8, Exercise 3.1.3]):

$$\begin{aligned} \text {There exists a constant } \alpha _0 > 0, \text { such that } \alpha _K \ge \alpha _0 \text { for all } K \in \mathcal {T}_h, \end{aligned}$$

(B.1)

where $\alpha _K$ is the minimum angle of K.

Fig. 4

Triangular elements for the proof of Lemma 3.2 in 2D

Now we consider the triangular elements $K = \triangle ABC$ in Fig. 4. Let $e_K^+ = {AB}$ be the outflow edge. Suppose $\pmb {\beta }$ starts at C. Due to the flow condition (3.3a), we must have the extension of $\pmb {\beta }$ intersect the line segment AB at some point E. Furthermore, we set O to be the foot of the altitude from C to AB. Then we will have either $|OA|\ge |OE|$ or $|OB|\ge |OE|$. Without loss of generality, we assume $|OA|\ge |OE|$, which implies $\angle OCE\le \angle OCA$. As a result, we have

$$\begin{aligned} \pmb {\beta }\cdot {\pmb {n}_{e_K^+}} = |\pmb {\beta }|\cos \angle OCE \ge |\pmb {\beta }|\cos \angle OCA =|\pmb {\beta }|\sin \angle OAC \ge |\pmb {\beta }|\sin \alpha _0>0. \end{aligned}$$

(B.2)

Here we have used the Zlámal’s condition (B.1). Hence the transversality condition (3.4) holds with $\gamma = \sin \alpha _0$.

Three-dimensional case: In 3D, it is proven in [3] that the shape-regularity condition (3.2) is equivalent to the following minimal angle condition.

$$\begin{aligned} \begin{aligned}&\text {There exists a constant } \alpha _0 > 0, \text { such that for any simplex }K \in \mathcal {T}_h, \text { any dihedral}\\&\text {angle } \alpha , \text { and any solid angle }\alpha \text { of }K,\text { we have } \alpha \ge \alpha _0. \end{aligned} \end{aligned}$$

(B.3)

Fig. 5

Tetrahedral elements for the proof of Lemma 3.2 in 3D

Now let us consider the tetrahedrons $K = ABCD$ in Fig. 5. We assume $e_K^+ = \triangle ABC$ to be the outflow face. Suppose $\pmb {\beta }$ starts at D. Due to the flow condition (3.3a), to ensure a unique outflow face, we need that the extension of $\pmb {\beta }$ intersect $\triangle ABC$ within the triangle at some point E. Furthermore, we set O to be the foot of the altitude from D to $\triangle ABC$. We connect OE and extend it until it intersects the edge of $\triangle ABC$ at some point F (so that $|OF|\ge |OE|$). Then we must have

$$\begin{aligned} \cos \angle ODE \ge \cos \angle ODF = \sin \angle OFD. \end{aligned}$$

(B.4)

Without loss of generality, we assume that F is on the edge AB. Note we have either $|OA| \ge |OF|$ or $|OB| \ge |OF|$. We only consider the case $|OA|\ge |OF|$ and the other case can be proved similarly. When $|OA|\ge |OF|$, we have

$$\begin{aligned} \sin \angle OFD \ge \sin \angle OAD = \frac{|OD|}{|DA|}. \end{aligned}$$

(B.5)

Then we set G to be the foot of the altitude from D to AB on $\triangle ABD$. In can be seen that

$$\begin{aligned} \frac{|OD|}{|DA|} = \frac{|OD|}{|DG|}\cdot \frac{|DG|}{|DA|} = \sin \angle OGD \cdot \sin \angle GAD \ge \sin ^2 \alpha _0. \end{aligned}$$

(B.6)

Here we have used the fact that $\angle OGD$ is the dihedral angle between the plane ABC and the plane ABD and $\angle GAD$ is a solid angle in $\triangle ABD$, which are both greater than or equal to $\alpha _0$ according to the minimal angle condition (B.3). Combining (B.4), (B.5), and (B.6), we get

$$\begin{aligned} \pmb {\beta }\cdot {\pmb {n}_{e_K^+}} = |\pmb {\beta }| \cos \angle ODE \ge |\pmb {\beta }| \sin ^2 \alpha _0>0. \end{aligned}$$

(B.7)

Hence the transversality condition (3.4) holds with $\gamma = \sin ^2 \alpha _0$.

Appendix C: Dimension Count in the Proof of Lemma 3.4

Proposition C.1

The finite-dimensional linear system determined by (3.23) is square.

Proof

On each mesh cell $K\in \mathcal {T}_h$, the degrees of freedom of the unknown $\delta$ is $\dim (\mathcal {P}_{k}(K))$, the number of equations associated with (3.23a) is $\dim (\mathcal {P}_{k-1}(K))$, and the number of equations associated with (3.23b) is $\dim (\mathcal {P}_{k}(e_{K}^+))$. Since

$$\begin{aligned} \dim \left( \mathcal {P}_{k}(K)\right)= & {} \left( {\begin{array}{c}k+d\\ d\end{array}}\right) , \end{aligned}$$

(C.8a)

$$\begin{aligned} \dim \left( \mathcal {P}_{k-1}(K)\right) = \left( {\begin{array}{c}k-1+d\\ d\end{array}}\right) , \ \end{aligned}$$

(C.8b)

$$\begin{aligned} \dim \left( \mathcal {P}_{k}\left( e_{K}^+\right) \right)= & {} \left( {\begin{array}{c}k+d-1\\ d-1\end{array}}\right) , \end{aligned}$$

(C.8c)

and

$$\begin{aligned} \left( {\begin{array}{c}k+d\\ d\end{array}}\right) = \left( {\begin{array}{c}k-1+d\\ d\end{array}}\right) + \left( {\begin{array}{c}k+d-1\\ d-1\end{array}}\right) , \end{aligned}$$

(C.9)

we know that $\dim (\mathcal {P}_{k}(K)) = \dim (\mathcal {P}_{k-1}(K)) + \dim (\mathcal {P}_{k}(e_{K}^+))$ — the degrees of freedom equals to the number of equations on each mesh cell. As a result, the global system (3.23) is square with $|\mathcal {T}_h|\left( {\begin{array}{c}k+d\\ d\end{array}}\right)$ unknowns, where $|\mathcal {T}_h|$ is the number of mesh cells.