An Introduction to the Theory of M-Decompositions

Abstract

We provide a short introduction to the theory of M-decompositions in the framework of steady-state diffusion problems. This theory allows us to systematically devise hybridizable discontinuous Galerkin and mixed methods which can be proven to be superconvergent on unstructured meshes made of elements of a variety of shapes. The main feature of this approach is that it reduces such an effort to the definition, for each element K of the mesh, of the spaces for the flux, V (K), and the scalar variable, W(K), which, roughly speaking, can be decomposed into suitably chosen orthogonal subspaces related to the space traces on ∂K of the scalar unknown, M(∂K). We begin by showing how a simple a priori error analysis motivates the notion of an M-decomposition. We then study the main properties of the M-decompositions and show how to actually construct them. Finally, we provide many examples in the two-dimensional setting. We end by briefly commenting on several extensions including to other equations like the wave equation, the equations of linear elasticity, and the equations of incompressible fluid flow.

Appendix: Proof of the Characterization of M-Decompositions

In this Appendix, we provide a proof Theorem 2.3, as it sheds light on the nature of M-decompositions. We closely follow the proof given in [27], and use the existence of the so-called canonical decomposition of Proposition 2.1.

Step 1.:

We take ${\widetilde {{\boldsymbol V}}}\times {\widetilde W}$ given by the canonical M-decomposition and begin by showing that

$$\displaystyle \begin{aligned} \dim {\widetilde{{\boldsymbol V}}{}^\perp}\cdot{\boldsymbol n}|{}_{\partial K} =\dim {\widetilde{{\boldsymbol V}}{}^\perp} \quad \mbox{ and } \quad \dim {\widetilde W}^\perp|{}_{\partial K} =\dim {\widetilde W}^\perp. \end{aligned}$$

Let us prove the first equality. If $\widetilde {{\boldsymbol v}}^\perp \in {\widetilde {{\boldsymbol V}}{ }^\perp }$ is such that $\widetilde {{\boldsymbol v}}^\perp \cdot {\boldsymbol n}|{ }_{\partial K}=0$, for any w ∈ W, we have that

$$\displaystyle \begin{aligned} 0=\langle w, \widetilde{{\boldsymbol v}}^\perp\cdot{\boldsymbol n}\rangle_{\partial K} =(\nabla w, \widetilde{{\boldsymbol v}}^\perp)_K+(\widetilde{w}^\perp,\nabla\cdot\widetilde{{\boldsymbol v}}^\perp)_K =(\widetilde{w}^\perp,\nabla\cdot\widetilde{{\boldsymbol v}}^\perp)_K \end{aligned}$$

since $\nabla w\subset {\widetilde {{\boldsymbol V}}}$. Since W ⊃∇⋅V , we can take $w:=\nabla \cdot \widetilde {{\boldsymbol v}}^\perp$ and conclude that $\nabla \cdot \widetilde {{\boldsymbol v}}^\perp =0$, which means that $\widetilde {{\boldsymbol v}}^\perp \in {{\boldsymbol V}}_{\mathrm {sbb}}$, which means that $\widetilde {{\boldsymbol v}}^\perp =\boldsymbol {0}$. Thus, the first equity holds.

Now, let us prove the second equality. If $\widetilde {w}^\perp \in {\widetilde W}^\perp$ and is zero on ∂K, then, for any v ∈V , we have

$$\displaystyle \begin{aligned} 0=\langle \widetilde{w}^\perp, \boldsymbol{v}\cdot\boldsymbol{n}\rangle_{\partial K} =(\nabla \widetilde{w}^\perp, \boldsymbol{v})_K+(\widetilde{w}^\perp,\nabla\cdot\boldsymbol{v})_K =(\nabla \widetilde{w}^\perp, \boldsymbol{v})_K \end{aligned}$$

since ${\widetilde W}=\nabla \cdot {{\boldsymbol V}}$. Since V  ⊃∇W, we can now take $\boldsymbol {v}:=\nabla \widetilde {w}^\perp$ and conclude that $\widetilde {w}^\perp$ is a constant on K. As a consequence $\widetilde {w}^\perp =0$, and the second equality follows.

Step 2.:

Next, we show that

$$\displaystyle \begin{aligned} \dim {\mbox{tr}}({\widetilde{{\boldsymbol V}}{}^\perp}\times {\widetilde W^\perp})= \dim {\widetilde{{\boldsymbol V}}{}^\perp}\cdot{\boldsymbol n}|{}_{\partial K} + \dim {\widetilde W^\perp}|{}_{\partial K}. \end{aligned}$$

To do that, we only need to show that ${\widetilde {{\boldsymbol V}}{ }^\perp }\cdot {\boldsymbol n}|{ }_{\partial K}\cap {\widetilde W}^\perp |{ }_{\partial K}=\{0\}$. So, if $(\widetilde {{\boldsymbol v}}^\perp , \widetilde {w}^\perp )\in {\widetilde {{\boldsymbol V}}{ }^\perp }\times {\widetilde W^\perp }$ we get that

$$\displaystyle \begin{aligned} \langle \widetilde{w}^\perp, \widetilde{{\boldsymbol v}}^\perp\cdot{\boldsymbol n}\rangle_{\partial K} =(\nabla \widetilde{w}^\perp, \widetilde{{\boldsymbol v}}^\perp)_K+(\widetilde{w}^\perp,\nabla\cdot\widetilde{{\boldsymbol v}}^\perp)_K =0, \end{aligned}$$

because $\nabla \widetilde {w}^\perp \in \nabla W\subset {\widetilde {{\boldsymbol V}}}$ and because $\nabla \cdot \widetilde {{\boldsymbol v}}^\perp \in \nabla \cdot {\widetilde {{\boldsymbol V}}}\subset {\widetilde W}$.

Step 3.:

By the inclusion property (a), the number

$$\displaystyle \begin{aligned}\begin{array}{r*{20}l} I:= &\; \dim M - \dim {\widetilde{{\boldsymbol V}}{}^\perp} -\dim {\widetilde W^\perp} \\ =&\; \dim M -\dim {\widetilde{{\boldsymbol V}}{}^\perp}\cdot{\boldsymbol n}|{}_{\partial K} - \dim {\widetilde W^\perp}|{}_{\partial K}. \end{array}\end{aligned} $$

is always nonnegative and is equal to zero if and only if property (c) holds. Next, we show that I = I _M(V  × W); this is the key computation of the proof. Indeed, we have

$$\displaystyle \begin{aligned}\begin{array}{r*{20}l} I:=&\dim M-\dim {\widetilde{{\boldsymbol V}}{}^\perp} - \dim {\widetilde W}^\perp \\=& \dim M-(\dim{{\boldsymbol V}}- \dim {\widetilde{{\boldsymbol V}}}) -(\dim W-\dim {\widetilde W}) \\ =& \dim M-(\dim{{\boldsymbol V}}- \dim \nabla W- \dim {{\boldsymbol V}}_{\mathrm{sbb}}) -(\dim W-\dim \nabla\cdot {{\boldsymbol V}}) \\ =& \dim M-\left(\dim{{\boldsymbol V}}- \dim \nabla\cdot {{\boldsymbol V}}- \dim {{\boldsymbol V}}_{\mathrm{sbb}}\right) -(\dim W-\dim \nabla W) \\ =& \dim M-(\dim\{{\boldsymbol v}\in{{\boldsymbol V}}: \nabla\cdot {\boldsymbol v}=0\}- \dim\{{\boldsymbol v}\in{{\boldsymbol V}}: \nabla\cdot {\boldsymbol v}=0, {\boldsymbol v}\cdot{\boldsymbol n}|{}_{\partial K}=0\} ) \\&- \dim\{ w\in W: \nabla w=0\} \\ =&\dim M-\dim\{{\boldsymbol v}\cdot{\boldsymbol n}|{}_{\partial K}:\, {\boldsymbol v}\in {{\boldsymbol V}}, \nabla \cdot{\boldsymbol v}=0\} -\dim \{w|{}_{\partial K}:\, w\in W, \nabla w=0\} \\=&I_M({{\boldsymbol V}}\times W). \end{array}\end{aligned} $$

Step 4.:

Now, by the inclusion property (a), we have that

$$\displaystyle \begin{aligned} \{{\boldsymbol v}\cdot{\boldsymbol n}|{}_{\partial K}:\, {\boldsymbol v}\in {{\boldsymbol V}}, \nabla \cdot{\boldsymbol v}=0\} \oplus \{w|{}_{\partial K}:\, w\in W, \nabla w=0\}\subset M, \end{aligned}$$

where the sum is L ²(∂K)-orthogonal since

$$\displaystyle \begin{aligned} \langle{\boldsymbol v}\cdot{\boldsymbol n}, w\rangle_{\partial K}=(\nabla\cdot {\boldsymbol v}, w)_K+({\boldsymbol v}, \nabla w)_K=0 \end{aligned}$$

if ∇⋅v = 0 and ∇w = 0. Finally, since the M-index I _M(V  × W) is zero by property (c), the equality holds. This completes the proof of the characterization Theorem 2.3.