Analysis of a Staggered Discontinuous Galerkin Method for Linear Elasticity

Abstract

We develop a staggered discontinuous Galerkin method for linear elasticity problems and prove its a priori error estimates. In our variational formulation the symmetry of the stress tensor is imposed weakly via Lagrange multipliers but the resulting numerical stress tensor is strongly symmetric. Optimal a priori error estimates are obtained and the error estimates are robust in nearly incompressible materials. Numerical experiments illustrating our theoretical analysis are included.

Appendix

In this appendix we provide some lemmas which are used in the proof of our main results.

Let ${\mathcal {M}}_h$ be a shape regular triangulation of a bounded domain $\varOmega$.

For simplicity, we use $\Vert \cdot \Vert _r$ for norms instead of $\Vert \cdot \Vert _{r,M}$ when it is clear that functions are clearly defined on $M$.

Lemma 6

Let $T_i, i=1,ldots, d+1$, be subsimplices of $M \in {\mathcal {M}}_h$ obtained by the barycentric subdivision. For a given $k \ge 1$, let

$$\begin{aligned} V_{M,k}&= \{ v \in H^1\left( M; {\mathbb {R}}^d\right) \,:\, v|_{T_i} \in {\mathcal {P}}_k\left( T_i; {\mathbb {R}}^d\right) \text { for } 1 \le i \le d+1, v|_{\partial M} = 0 \}, \end{aligned}$$

(57)

$$\begin{aligned} Q_{M,k}&= \left\{ p \in L^2(M) \,:\, p|_{T_i} \in {\mathcal {P}}_k(T_i), \int _M p \,dx = 0 \right\} . \end{aligned}$$

(58)

Then there exists $\beta >0$ independent of $M \in {\mathcal {M}}_h$ such that

$$\begin{aligned} \inf _{0 \not = p \in Q_{M,k}} \sup _{0 \not = w \in V_{M,k+1}} \frac{ ({\text {div}}\, w , p)}{\Vert w \Vert _1 \Vert p \Vert _0} \ge \beta . \end{aligned}$$

(59)

Proof

Note that (59) is the inf-sup condition of mixed finite element method $(V_{M,k+1}, Q_{M,k})$ for the Stokes equation, so we simply refer to appropriate literature. When $d = 2$ with $k \ge 3$, (59) is obtained from the stability of Scott–Vogelius elements for the Stokes equation [29]. For $d = 2$ with $k = 1,2$, it is reported in [27]. When $d=3$, it is proved in [37]. $\square$

Lemma 7

Suppose $V_{M,k}$ and $Q_{M,k}$ are defined as in (57) and (58). Then there exists $c >0$ which only depends on the shape regularity of ${\mathcal {M}}_h$ and $k\ge 1$ such that, for any given $p \in Q_{M,k}$, there exists $w \in V_{M,k+1}$ satisfying ${\text {div}}\, w = p$ and $\Vert w \Vert _1 \le c \Vert p \Vert _0$.

Proof

Let $M \in {\mathcal {M}}_h$ be fixed. Note that ${\text {div}}\, w \in Q_{M,k}$ for $w \in V_{M,k+1}$, so we can regard ${\text {div}}$ as a linear map from $V_{M, k+1}$ to $Q_{M,k}$. This ${\text {div}}$ is surjective because, if not, then there exists $0 \not = p \in Q_{M,k}$ such that $({\text {div}}\, w, p) = 0$ for all $w \in V_{M,k+1}$, but it contradicts (59). Suppose that $\{ p_i \}_{i=1}^m$ is an orthonormal basis of $Q_{M,k}$ in $L^2(M)$ and define

$$\begin{aligned} \alpha _i := \inf _{w \in V_{M,k+1}, {\text {div}}\, w = p_i} \frac{({\text {div}}\, w, p_i)}{\Vert w \Vert _1}, \qquad \alpha _M := \min _{1 \le i \le m} \alpha _i. \end{aligned}$$

(60)

By finite dimensionality of the polynomial spaces $V_{M,k+1}$ and $Q_{M,k}$, there exists $w_i \in V_{M,k+1}$ such that ${\text {div}}\, w_i = p_i$ and $\Vert w_i \Vert _1 = \alpha _i^{-1}$. Suppose that a given $0 \not = p \in Q_{M,k}$ can be expressed as $p = \sum _{i=1}^m c_i p_i$ with coefficients $c_i \in {\mathbb {R}}, 1 \le i \le m$. Then $\Vert p \Vert _0^2 = \sum _{i=1}^m c_i^2$. If we take $w = \sum _{i=1}^m c_i w_i$, then ${\text {div}}\, w = p$ and

$$\begin{aligned} \Vert w \Vert _1 \le \sum _{i=1}^m |c_i| \Vert w_i \Vert _1 = \sum _{i=1}^m \frac{|c_i|}{\alpha _i} \le \left( \sum _{i=1}^m c_i^2 \right) ^{\frac{1}{2}} \left( \sum _{i=1}^m \alpha _i^{-2} \right) ^{\frac{1}{2}} \le \Vert p \Vert _0 \alpha _M^{-1} \sqrt{m}, \end{aligned}$$

by the triangle inequality, the fact $\Vert w_i \Vert _1 = \alpha _i^{-1}$ for $1 \le i \le m$, the Cauchy-Schwarz inequality, and the definition of $\alpha _M$ in (60). To complete proof, note that the $\alpha _i$’s in (60) are independent of the diameter of $M$ by standard scaling argument. By shape regularity assumption of ${\mathcal {M}}_h$ and standard compactness argument (see, e.g., [31], Lemma3.1]), there exists $\alpha >0$ depending only on the shape regularity of ${\mathcal {M}}_h$ and $k$ such that $\alpha \le \alpha _M$ for any $M \in {\mathcal {M}}_h$. The proof is completed. $\square$

Let ${\mathbb {L}} = {\mathbb {V}}$ if $d = 2$ and ${\mathbb {L}} = {\mathbb M}$ if $d = 3$. For given $k \ge 1$, we define $\varXi _{M,0}$ as

$$\begin{aligned} \varXi _{M,0}&= \left\{ \xi \in H^1(M; {\mathbb {L}}) \,:\, \xi |_{T_i} \in {\mathcal {P}}_{k+1}(T_i;{\mathbb {L}}), 1 \le i \le d+1, \; \xi |_{\partial M} = 0 \right\} . \end{aligned}$$

(61)

We also define $S$ and $\chi$ as

$$\begin{aligned} S \begin{pmatrix} \xi _1 \\ \xi _2 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} \xi _1&\xi _2 \end{pmatrix} \quad \text {for } \xi \in \varXi _{M,0}, \quad \chi (r) = \begin{pmatrix} 0 &{} r \\ -r &{} 0 \end{pmatrix} \quad \text {for } r \in {\mathbb {R}}\qquad \text { if } d = 2, \\ S \xi = \frac{1}{2}\left( \xi ^T - ({\text {tr }}\xi ) \varvec{I}\right) \quad \text {for } \xi \in \varXi _{M,0}, \quad \chi \begin{pmatrix} r_1 \\ r_2 \\ r_3 \end{pmatrix} = \begin{pmatrix} 0 &{} -r_3 &{} r_2 \\ r_3 &{} 0 &{} -r_1 \\ -r_2 &{} r_1 &{} 0 \end{pmatrix} \quad \text {if } d = 3, \end{aligned}$$

where $\xi ^T$ is the transpose of $\xi$ and $\varvec{I}$ is the identity $3 \times 3$ matrix. Note that $S$ and $\chi$ are algebraic isomorphisms. One can verify by a direct computation that

$$\begin{aligned} {\text {skw }}{\text {curl }}\xi = \chi {\text {div}}\, S \xi , \qquad \xi \in \varXi _{M,0}. \end{aligned}$$

(62)

Lemma 8

Assume that $\varXi _{M,0}$ is defined as in (61) for some $k \ge 1$. If $\zeta \in \varXi _{M,0}$, then $({\text {curl }}\zeta ) n|_{\partial M} = 0$ for the unit normal vector field $n$ on $\partial M$.

Proof

If $d = 2$, then ${\text {curl }}\zeta$ is the $\pi /2$-rotation of ${\text {grad}}\, \zeta$. Denoting by $t$, the tangent vector obtained by $\pi /2$-rotation of the normal vector $n$ on $\partial M, ({\text {grad}}\, \zeta ) t = 0$ on $\partial M$ because $\zeta$ is vanishing on $\partial M$. Then we can see that $({\text {curl }}\zeta ) n = ({\text {grad}}\, \zeta ) t = 0$ on $\partial M$.

When $d = 3$, let $\zeta _i, i=1,2,3$, be the $i$-th row of $\zeta$. It suffices to show that ${\text {curl }}\zeta _i \cdot n \equiv 0$ on $\partial M$. By the Stokes’ theorem, for any $\phi \in C^1(\overline{M})$,

$$\begin{aligned} 0&= \int _M {\text {div}}\, ({\text {curl }}\zeta _i) \phi \,dx = \int _{\partial M} ({\text {curl }}\zeta _i \cdot n) \phi \,ds - \int _M ({\text {curl }}\zeta _i) \cdot {\text {grad}}\, \phi \,dx \\&= \int _{\partial M} ({\text {curl }}\zeta _i \cdot n) \phi \,ds + \int _{\partial M} (\zeta _i \times n) \cdot {\text {grad}}\, \phi \,ds - \int _M \zeta _i \cdot {\text {curl }}({\text {grad}}\, \phi ) \,dx \\&= \int _{\partial M} ({\text {curl }}\zeta _i \cdot n) \phi \,ds + \int _{\partial M} (\zeta _i \times n) \cdot {\text {grad}}\, \phi \,ds. \end{aligned}$$

Since $\zeta _i \equiv 0$ on $\partial M, \int _{\partial M} (\zeta _i \times n) \cdot {\text {grad}}\, \phi \,ds = 0$ and then ${\text {curl }}\zeta _i \cdot n \equiv 0$ on $\partial M$ because $\phi \in C^1(\overline{M})$ is arbitrary. $\square$

Lemma 9

For a given $k \ge 1$, define $\varGamma _{M,0}$ as

$$\begin{aligned} \varGamma _{M,0} = \left\{ \varvec{\eta } \in L^2(M; {\mathbb K}) \,:\, \varvec{\eta }|_{T_i} \in {\mathcal {P}}_k(T_i; {\mathbb K}), 1 \le i \le d+1, \int _M \varvec{\eta }\,dx = 0 \right\} , \end{aligned}$$

and recall the definition of $\Sigma _{M,0}$ in (21) with same $k$. For any $\varvec{\eta } \in \varGamma _{M,0}$ there is $\varvec{\tau } \in \Sigma _{M,0}$ such that ${\text {div}}\, \varvec{\tau } = 0, {\text {skw }}\varvec{\tau } = \varvec{\eta }$, and $\Vert \varvec{\tau } \Vert _0 \le c \Vert \varvec{\eta } \Vert _0$ with $c>0$ independent of $M \in {\mathcal {M}}_h$.

Proof

We only prove the three dimensional case because the two dimensional one can be proved similarly. For the given $k$, we consider $V_{M,k+1}, Q_{M,k}$ as in (57), (58), and $\varXi _{M,0}$ as in (61). By the definition of $\chi$, note that $\chi ^{-1} \varvec{\eta } \in L^2(M; {\mathbb {V}})$ for $\varvec{\eta } \in \varGamma _{M,0}$ and each component of $\chi ^{-1} \varvec{\eta }$ is in $Q_{M,k}$. Applying Lemma 7 to each component of $\chi ^{-1} \varvec{\eta }$, one can find $\varvec{\xi } \in H^1(M; {\mathbb M})$ such that ${\text {div}}\, \varvec{\xi } = \chi ^{-1} \varvec{\eta }, \Vert \varvec{\xi } \Vert _1 \le c \Vert \varvec{\eta } \Vert _0$, and each row can be identified with an element in $V_{M,k+1}$. If we set $\varvec{\tau } = {\text {curl }}S^{-1} \varvec{\xi }$, then ${\text {div}}\, \varvec{\tau } = 0$ and also $\varvec{\tau } \in \Sigma _{M,0}$ by Lemma 8 because $S^{-1} \varvec{\xi } \in \varXi _{M,0}$. Furthermore, by (62) and the assumptions on $\varvec{\xi }$,

$$\begin{aligned} {\text {skw }}\varvec{\tau }&= {\text {skw }}{\text {curl }}S^{-1} \varvec{\xi } = \chi {\text {div}}\, \varvec{\xi } = \varvec{\eta }, \\ \Vert \varvec{\tau } \Vert _0&= \Vert {\text {curl }}S^{-1} \varvec{\xi } \Vert _0 \le c \Vert \varvec{\xi } \Vert _1 \le c \Vert \varvec{\eta } \Vert _0. \end{aligned}$$

The proof is completed. $\square$