A discrete elasticity complex on three-dimensional Alfeld splits

Abstract

We construct conforming finite element elasticity complexes on the Alfeld splits of tetrahedra. The complex consists of vector fields and symmetric tensor fields, interlinked via the linearized deformation operator, the linearized curvature operator, and the divergence operator, respectively. The construction is based on an algebraic machinery that derives the elasticity complex from de Rham complexes, and smoother finite element differential forms.

Appendix A: Supersmoothness

Consider a tetrahedron T and its Alfeld split $\{ T_i\}$ as in the rest of the paper. Proposition 2.1’s items (1) and (2) are a consequence of the following fact proved in [1]: if $v \in C^1(T)$ and $v|_{T_i}$ is in $C^\infty (T_i)$, then v is $C^2$ at the vertices of T. Such serendipitous “supersmoothness” at some points was observed on triangles earlier [20]. In Theorem A.1 below, we establish a supersmoothness result in the same spirit for 1-forms. In fact, the earlier result of Alfeld follows from the theorem, as noted in Corollary A.2. Items (3) and (4) of Proposition 2.1 follow from the arguments below. (The proof will show that the assumption that $v|_{T_i}$ is infinitely smooth can be relaxed, but this generalization is not important for our purposes.)

Theorem A.1

Suppose v is in $C^0(T)^3$, $v_i = v|_{T_i} \in C^\infty (T_i)^{{3}}$, and ${\mathop {\textrm{curl}\,}}v$ is $C^0$ at the vertices $x_i$ of T. Then v is $C^1$ at $x_i$.

Proof

Let $F_{ij} = \partial T_i \cap \partial T_j$ and let $T F_{ij}$ denote the tangent plane of $F_{ij}$. Let $c \in {{\mathbb {V}}}$ and $\tau \in TF_{ij}$. The first observation needed for this proof is that

$$\begin{aligned} c \cdot ({\mathop {\textrm{ grad}\,}}v_i) \tau = c \cdot ({\mathop {\textrm{ grad}\,}}v_j) \tau \quad \text { on } F_{ij}. \end{aligned}$$

(A.1)

This is because the continuity of v requires $(v _i - v_j)\cdot c$ to vanish on $F_{ij}$ for any $c \in {{\mathbb {V}}}$, so its tangential derivatives also vanish on $F_{ij}$.

We claim that at a vertex of T on $F_{ij}$, we also have

$$\begin{aligned} \tau \cdot ({\mathop {\textrm{ grad}\,}}v_i) c = \tau \cdot ({\mathop {\textrm{ grad}\,}}v_j) c. \end{aligned}$$

(A.2)

To show this, consider $x_1$, a common vertex of T and $F_{23}$. Then, since the scalar $\tau \cdot ({\mathop {\textrm{ grad}\,}}v_i)(x_1) \, c$ equals its transpose, we have

$$\begin{aligned} \tau \cdot ({\mathop {\textrm{ grad}\,}}v_2)\, c&= c \cdot ({\mathop {\textrm{ grad}\,}}v_2)' \tau{} & {} \text { at } x_1, \\&= c \cdot \left( ({\mathop {\textrm{ grad}\,}}v_2)' - ({\mathop {\textrm{ grad}\,}}v_2) \right) \tau + c \cdot ({\mathop {\textrm{ grad}\,}}v_2) \tau \\&= c \cdot \left( ({\mathop {\textrm{ grad}\,}}v_2)' - ({\mathop {\textrm{ grad}\,}}v_2) \right) \tau + c \cdot ({\mathop {\textrm{ grad}\,}}v_3) \tau{} & {} \text { by }(\text {A.1}), \\&= c \cdot \left( ({\mathop {\textrm{ grad}\,}}v_3)' - ({\mathop {\textrm{ grad}\,}}v_3) \right) \tau + c \cdot ({\mathop {\textrm{ grad}\,}}v_3) \tau{} & {} \text { as }{\mathop {\textrm{curl}\,}}v\text { is }C^0\text { at }x_1 \\&= \tau \cdot ({\mathop {\textrm{ grad}\,}}v_3)\, c. \end{aligned}$$

This argument can be repeated at other vertices to finish the proof of (A.2).

Now we are ready to show that v is $C^1$ at $x_i$. Let $\tau _i = (x_i - z) / \Vert x_i - z\Vert$ and $\tau _{ij} = (x_i - x_j) / \Vert x_i - x_j \Vert$. Without loss of generality, we focus on one vertex, say $x_1$. At $x_1$,

$$\begin{aligned} c \cdot ({\mathop {\textrm{ grad}\,}}v_2) \tau _0&= c \cdot ({\mathop {\textrm{ grad}\,}}v_3) \tau _0&\tau _0 ({\mathop {\textrm{ grad}\,}}v_2) c&= \tau _0 ({\mathop {\textrm{ grad}\,}}v_3) c \end{aligned}$$

(A.3a)

$$\begin{aligned} c \cdot ({\mathop {\textrm{ grad}\,}}v_2) \tau _{10}&= c \cdot ({\mathop {\textrm{ grad}\,}}v_3) \tau _{10}&\tau _{10} ({\mathop {\textrm{ grad}\,}}v_2) c&= \tau _{10} ({\mathop {\textrm{ grad}\,}}v_3) c. \end{aligned}$$

(A.3b)

The left equalities follow from (A.1) and the right ones from (A.2). Furthermore, at $x_1$ we have

$$\begin{aligned} \tau _{12} \cdot ({\mathop {\textrm{ grad}\,}}v_2 ) \tau _{13}&= \tau _{12} \cdot ({\mathop {\textrm{ grad}\,}}v_0) \tau _{13}{} & {} \text {by } (\text {A.1})\text { applied to }F_{20} \\&= \tau _{12} \cdot ({\mathop {\textrm{ grad}\,}}v_3) \tau _{13}{} & {} \text {by }(\text {A.2}) \text {applied to }F_{03}. \end{aligned}$$

Therefore, $\tau _{12} \cdot ({\mathop {\textrm{ grad}\,}}v_2 )\tau _{13} = \tau _{12} \cdot ({\mathop {\textrm{ grad}\,}}v_3 )\tau _{13}$ at $x_1$. Writing $\tau _{12}$ as a linear combination of $\tau _0, \tau _{10}$ and $\tau _{13}$, and using the equalities of (A.3) in the right panel, we conclude that

$$\begin{aligned} \tau _{13} \cdot ({\mathop {\textrm{ grad}\,}}v_2) \tau _{13} = \tau _{13} \cdot ({\mathop {\textrm{ grad}\,}}v_3) \tau _{13} \qquad \text { at } x_1. \end{aligned}$$

(A.4)

The identities of (A.4) and (A.3) together yield the equality of ${\mathop {\textrm{ grad}\,}}v_2$ and ${\mathop {\textrm{ grad}\,}}v_3$ at $x_1$. Repeating this argument for every pair of $v_i$ meeting at a vertex, the proof is finished. $\square$

Corollary A.2

If $w \in C^1(T)$ and $w|_{T_i}$ is in $C^\infty (T_i)$, then w is $C^2$ at the vertices of T.

Proof

This follows by applying Theorem A.1 with $v = {\mathop {\textrm{ grad}\,}}w$. $\square$