Normal-normal continuous symmetric stress approximation in three-dimensional linear elasticity ^†^†thanks: Supported by ANID-Chile through FONDECYT project 1230013

Carsten Carstensen Department of Mathematics, Humboldt–Universität zu Berlin, Unter den Linden 6, 10099 Berlin, Germany, email: [email protected] Norbert Heuer Facultad de Matemáticas, Pontificia Universidad Católica de Chile, Avenida Vicuña Mackenna 4860, Santiago, Chile, email: [email protected]

Abstract

We present a conforming setting for a mixed formulation of linear elasticity with symmetric stress that has normal-normal continuous components across faces of tetrahedral meshes. We provide a stress element for this formulation with $30$ degrees of freedom that correspond to standard boundary conditions. The resulting scheme converges quasi-optimally and is locking free. Numerical experiments illustrate the performance.

AMS Subject Classification: 65N30, 74G15, 74S05

1 Introduction and model problem

Adams & Cockburn [2] and Arnold et al. [4] presented and analyzed, with slightly different degrees of freedom, the first stable element for conforming stress approximations in 3D linear elasticity on tetrahedral meshes. It provides pointwise symmetric piecewise polynomial approximations of degree at least four with continuous normal components across faces. With its 162 degrees of freedom (plus displacement degrees) this element is prohibitively expensive for practical applications. Modifications and generalizations to arbitrary dimension and higher order have been proposed by Hu & Zhang, Hu [16, 15] and Chen & Huang [9]. The difficulty of constructing conforming elements with few degrees of freedom has been the reason for alternative developments, e.g., of discontinuous and non-conforming methods with weak symmetry. But the quest for cheap stable and conforming stress elements goes on, in particular on tetrahedra which provide important geometric flexibility. Hu and Zhang [17] proposed to use low-degree polynomial tensors enriched with edge and face related bubble polynomials to achieve stability, in this way reducing the degrees of freedom to $48$ . A variation of this approach and extension to arbitrary dimension is given in [18].

In this paper we present a pointwise symmetric tetrahedral element of quadratic polynomials with $30$ degrees of freedom (they can be reduced to $12$ by static condensation) and provide a variational framework that renders the element conforming. In geometrical terms our element leads to linear normal-normal components on faces that are continuous across elements. This approach goes back to the seminal TDNNS —tangential displacement and normal-normal stress continuous— method by Pechstein & Schöberl [21, 22, 23] that in turn transfers the Hellan–Herrmann–Johnson idea from plate bending to linear elasticity, cf. [12, 13, 20]. In [7] we presented a continuous and discrete framework for this setting in the case of plane elasticity with mixed approximation on triangular meshes. Its advantages are

•

a low number of degrees of freedom related to
•

physical degrees of freedom that allow for the
•

treatment of any boundary condition that is physically relevant, and a
•

variational framework that makes the element conforming.

Furthermore, it is

•

provably stable and locking free.

This paper solves the critical open case of three-dimensional elasticity on tetrahedral elements. We stress the fact that all the advantages listed above remain valid in three dimensions. We expect that the existence of a conforming framework facilitates, for instance, the analysis of a posteriori error estimation. Perhaps more important is the fact that, differently from the previously mentioned $H(\mathrm{div})$ -conforming methods, the degrees of freedom of our element avoid vertex values and edge moments of stresses. Such degrees do not relate to boundary conditions in engineering applications and are only well posed when making artificially high regularity assumptions.

In order not to be repetitive, we only briefly recall settings and cite proofs from [7], and also refer to [14] for general results on a hybrid framework. Instead, we focus on the new ingredients required for the three-dimensional setting. In particular, we restrict the analysis to homogeneous Dirichlet boundary conditions. The incorporation and analysis of general boundary conditions are similar to the two-dimensional situation treated in [7].

Model problem. We consider a bounded, polyhedral Lipschitz domain $\Omega\subset\mathbb{R}^{3}$ with boundary $\Gamma:=\partial\Omega$ decomposed into non-intersecting relatively open pieces

\displaystyle\Gamma=\mathrm{closure}({\Gamma_{\mathit{hc}}}\cup{\Gamma_{\mathit{sc}}}\cup{\Gamma_{\mathit{ss}}}\cup{\Gamma_{\mathit{sf}}}).

Boundary conditions are of hard clamped (“ $\mathit{hc}$ ”), soft clamped (“ $\mathit{sc}$ ”), simply supported (“ $\mathit{ss}$ ”) and traction (“ $\mathit{sf}$ ” for stress free) types on the corresponding boundary pieces. Individual sets are either empty or unions of connected sets of positive measure, subject to the validity of Korn’s inequality. For the discretization it has to be assumed that the boundary pieces are conforming with respect to the faces of the tetrahedral elements, that is, interfaces between different boundary pieces do not cut faces.

The model problem of linear elasticity with general boundary conditions reads

\displaystyle\sigma=2\mu\varepsilon(u)+\lambda(\operatorname{div}u)\operatorname{\mathrm{id}}\quad\text{and}\quad-\operatorname{div}\sigma=f\quad\text{in}\ \Omega,

	$\displaystyle u$	$\displaystyle={g_{D}}$		$\displaystyle\text{on}\ {\Gamma_{\mathit{hc}}},\qquad$		$\displaystyle u\cdot n={g_{D,n}},\ \pi_{t}(\sigma n)={g_{N,t}}\quad$		$\displaystyle\text{on}\ {\Gamma_{\mathit{sc}}},$		(1)
	$\displaystyle\pi_{t}u$	$\displaystyle={g_{D,t}},\ n\cdot\sigma n={g_{N,n}}\quad$		$\displaystyle\text{on}\ {\Gamma_{\mathit{ss}}},$		$\displaystyle\sigma n={g_{N}}$		$\displaystyle\text{on}\ {\Gamma_{\mathit{sf}}}.$		(1)

Here, $\varepsilon(u)$ and $\operatorname{\mathrm{id}}$ are the strain and identity tensors, respectively, and $\mu,\lambda>0$ are the Lamé parameters. We assume that $\mu$ is fixed but allow for general $\lambda$ , corresponding to an unrestricted Poisson ratio $\nu\in(0,1/2)$ . For short, we write $\sigma=\mathbb{C}\varepsilon(u)$ with elasticity tensor $\mathbb{C}$ , and let $\mathbb{A}:=\mathbb{C}^{-1}$ be the compliance tensor. Furthermore, $n$ is the exterior unit normal vector on $\Gamma$ , and $\pi_{t}u:=(u-(u\cdot n)n)|_{\Gamma}$ is the tangential trace on $\Gamma$ of vector fields $u$ . Boundary data have to be consistent with the existence of a function $u\in H^{1}(\Omega;\mathbb{R}^{3})$ and tensor $\sigma\in H(\operatorname{div},\Omega;\mathbb{R}^{3\times 3})$ that satisfy (1) as appropriate traces. Note that $f,{g_{D}},{g_{D,t}},{g_{N}},{g_{N,t}}$ are vector-valued and ${g_{D,n}},{g_{N,n}}$ are scalar functions.

As mentioned before, our forthcoming variational formulation and mixed finite element discretization are compatible with the general boundary conditions (1). Only for ease of presentation will we restrict ourselves to homogeneous Dirichlet conditions, thus consider

\displaystyle\sigma=\mathbb{C}\varepsilon(u),\quad-\operatorname{div}\sigma=f\quad\text{in}\ \Omega,\qquad u|_{\Gamma}=0,

(2)

and refer to [7] for details concerning general boundary conditions.

Overview. The remainder is structured as follows. In Section 2 we introduce our notation for spaces and norms. At the center is the definition of skeleton trace spaces of tangential and normal displacements, their characterization (Propositions 1, 2) and density in the space of displacement traces (Proposition 4). The proof of the density result is based on a 3D version of Tartar’s result that says that the set of $C^{\infty}$ -functions with compact support that vanish in a neighborhood of a point $x\in\Omega\subset\mathbb{R}^{2}$ are dense in $H^{1}_{0}(\Omega)$ . This is Proposition 3. Having these results at hand, we follow the canonical procedure from [7] to define the space $H_{nn}(\operatorname{div},\mathcal{T};\SS)$ of symmetric piecewise $H(\mathrm{div})$ -tensors with normal-normal continuous traces on polyhedral meshes and a related displacement trace operator. Proposition 5 summarizes corresponding conformity and norm relations. We note that not all the results on the trace spaces are required to prove Proposition 5. We present them for their general relevance and possible future reference. Section 3 presents the mixed variational formulation of the model problem that defines the stress as an element of $H_{nn}(\operatorname{div},\mathcal{T};\SS)$ . Theorem 6 establishes its Poisson-ratio robust well-posedness. Section 4 is devoted to the discrete setting and analysis. In §4.1 we introduce the stress element and disclose its degrees of freedom. The corresponding conforming piecewise polynomial approximation space $P^{2}_{nn}(\mathcal{T})$ is the subject of §4.2, with approximation properties shown by Proposition 10. The resulting mixed finite element scheme is presented in §4.3, and shown to be locking free with expected convergence orders by Theorem 12. In Section 5 we present a numerical example that illustrates the locking-free convergence of our method.

Throughout, $a\lesssim b$ stands for $a\leq Cb$ with a generic constant $C>0$ that is independent of involved functions and mesh $\mathcal{T}$ . Relation $a\gtrsim b$ means that $b\lesssim a$ .

2 Analytical framework

In this section we introduce and analyze spaces and trace operators that provide the basis of our mixed formulation. The next subsection presents some notation and collects results for trace spaces of vector functions. In §2.2 we define the stress space of pointwise symmetric tensors with continuous normal-normal traces and establish duality relations with a trace space.

2.1 Notation, spaces and trace operators

For sub-domains and surfaces $\omega\subset\overline{\Omega}$ we use standard Lebesgue and Sobolev spaces $L_{q}(\omega;U)$ ( $q>1$ ) and $H^{s}(\omega;U)$ ( $s>0$ ) of $U$ -valued functions on $\omega$ with $U\in\{\mathbb{R},\mathbb{R}^{3},\SS\}$ . Here, $\SS\subset\mathbb{R}^{3\times 3}$ denotes the space of symmetric constant tensors. The generic $L_{2}(\omega;U)$ -inner product and norm are $(\cdot\hskip 1.42262pt,\cdot)_{\omega}$ and $\|\cdot\|_{\omega}=\|\cdot\|_{0,\omega}$ , respectively. For integer $s$ , $\|\cdot\|_{s,\omega}$ denotes the standard norm in $H^{s}(\omega;U)$ , except for $s=1$ and $U=\mathbb{R}^{3}$ in which case $\|\cdot\|_{1,\omega}:=\Bigl{(}\|\cdot\|_{\omega}^{2}+\|\varepsilon(\cdot)\|_{\omega}^{2}\Bigr{)}^{1/2}$ with symmetric gradient $\varepsilon(\cdot):=(\nabla(\cdot)+\nabla(\cdot)^{\top})/2$ . For non-integer $s>0$ , $\|\cdot\|_{\omega,s}$ and $|\cdot|_{\omega,s}$ denote the Sobolev-Slobodeckij norms and seminorms, cf. [1]. Given $s,r\geq 0$ and $\omega\subset\Omega$ , we need the space

H^{s,r}(\operatorname{div},\omega;\SS):=\{\tau\in H^{s}(\omega;\SS);\;\operatorname{div}\tau\in H^{r}(\omega;\mathbb{R}^{3}\}

with row-wise application $\operatorname{div}\tau$ of the divergence operator, and denote $H(\operatorname{div},\omega;\SS):=$ $H^{0,0}(\operatorname{div},\omega;\SS)$ (identifying $H^{0}(\Omega;U)$ with $L_{2}(\Omega;U)$ ). The latter space is provided with the graph norm $\|\cdot\|_{\operatorname{div},\omega}:=\Bigl{(}\|\cdot\|_{\omega}^{2}+\|\operatorname{div}\cdot\|_{\omega}^{2}\Bigr{)}^{1/2}$ . We also need the topological dual $H^{-1}(\omega)$ of $H^{1}_{0}(\omega)$ with norm $\|\cdot\|_{-1,\omega}$ . We generally drop index $\omega$ in the notation of norms and inner products when $\omega=\Omega$ . The trace of a tensor $\tau$ is $\operatorname{tr}(\tau):=\tau:\operatorname{\mathrm{id}}$ with identity tensor $\operatorname{\mathrm{id}}$ and Frobenius product “ $:$ ”; the deviatoric part of $\tau$ reads $\operatorname{dev}(\tau):=\tau-\operatorname{tr}(\tau)\operatorname{\mathrm{id}}/3$ .

We consider a regular decomposition $\mathcal{T}$ of $\Omega$ into shape-regular tetrahedra and formally denote by $\mathcal{S}:=\{\partial T;\;T\in\mathcal{T}\}$ its skeleton. We need the piecewise constant function $h_{\mathcal{T}}$ with $h_{\mathcal{T}}|_{T}:=\operatorname{diam}(T)$ for $T\in\mathcal{T}$ . The set of faces of $T\in\mathcal{T}$ is $\mathcal{F}(T)$ , and the sets of all (resp. interior) faces is $\mathcal{F}:=\cup_{T\in\mathcal{T}}\mathcal{F}(T)$ (resp. $\mathcal{F}(\Omega)$ ). In the following, $\langle{}\cdot\hskip 1.42262pt,\cdot\rangle_{K}$ denotes the generic $L_{2}$ -duality on $K\in\mathcal{F}\cup\{\Gamma\}$ . Decomposition $\mathcal{T}$ gives rise to product spaces, with corresponding notation where $\Omega$ is replaced with $\mathcal{T}$ , e.g., $H^{s}(\mathcal{T};U):=\Pi_{T\in\mathcal{T}}H^{s}(T;U)$ . Throughout, we identify functions of product spaces with piecewise defined functions. Mesh $\mathcal{T}$ induces the canonical trace operator

\displaystyle\gamma:\;

\displaystyle\left\{\begin{array}[]{cll}H^{1}(\Omega;\mathbb{R}^{3})&\rightarrow&H(\operatorname{div},\mathcal{T};\SS)^{*},\\ v&\mapsto&\langle{}\gamma(v)\hskip 1.42262pt,\tau\rangle_{\mathcal{S}}:=(\varepsilon(v)\hskip 1.42262pt,\tau)+(v\hskip 1.42262pt,\operatorname{div}\tau)_{\mathcal{T}}\end{array}\right.

with support on $\mathcal{S}$ and trace space

H^{1/2}(\mathcal{S};\mathbb{R}^{3}):=\gamma(H^{1}(\Omega;\mathbb{R}^{3})).

Here, $(\cdot\hskip 1.42262pt,\cdot)_{\mathcal{T}}$ is the generic notation for $L_{2}(\mathcal{T};U)$ dualities, $U\in\{\mathbb{R},\mathbb{R}^{3},\SS\}$ . There is an inherent duality between $H^{1/2}(\mathcal{S};\mathbb{R}^{3})$ and $H(\operatorname{div},\mathcal{T};\SS)$ ,

\displaystyle\langle{}\varphi\hskip 1.42262pt,\tau\rangle_{\mathcal{S}}:=\langle{}\gamma(v)\hskip 1.42262pt,\tau\rangle_{\mathcal{S}}\quad(\varphi\in H^{1/2}(\mathcal{S};\mathbb{R}^{3}),\tau\in H(\operatorname{div},\mathcal{T};\SS)),

(3)

where $v$ is any element of the pre-image $\gamma^{-1}(\varphi)\subset H^{1}(\Omega;\mathbb{R}^{3})$ , $\varphi=\gamma(v)$ .

Given the tangential and normal trace operators

	$\displaystyle\gamma_{t}:\;$	$\displaystyle\left\{\begin{array}[]{cll}H^{1}(\Omega;\mathbb{R}^{3})&\rightarrow&H^{1}(\mathcal{T};\mathbb{R}^{3})^{*},\\ v&\mapsto&\langle{}\gamma_{t}(v)\hskip 1.42262pt,z\rangle_{\mathcal{S}}:=(\operatorname{curl}v\hskip 1.42262pt,z)-(v\hskip 1.42262pt,\operatorname{curl}z)_{\mathcal{T}},\end{array}\right.$
	$\displaystyle\gamma_{n}:\;$	$\displaystyle\left\{\begin{array}[]{cll}H^{1}(\Omega;\mathbb{R}^{3})&\rightarrow&H^{1}(\mathcal{T};\mathbb{R}^{3})^{*},\\ v&\mapsto&\langle{}\gamma_{n}(v)\hskip 1.42262pt,z\rangle_{\mathcal{S}}:=(\operatorname{div}v\hskip 1.42262pt,z)-(v\hskip 1.42262pt,\nabla z)_{\mathcal{T}},\end{array}\right.$

we introduce their kernels $H^{1}_{n}(\Omega,\mathcal{S};\mathbb{R}^{3}):=\mathrm{ker}(\gamma_{t})$ and $H^{1}_{t}(\Omega,\mathcal{S};\mathbb{R}^{3}):=\mathrm{ker}(\gamma_{n}).$ These are closed subspaces of $H^{1}(\Omega;\mathbb{R}^{3})$ and give rise to the trace spaces

	$\displaystyle\widetilde{H}^{1/2}_{0,n}(\mathcal{S};\mathbb{R}^{3})$	$\displaystyle:=\gamma(H^{1}_{n}(\Omega,\mathcal{S};\mathbb{R}^{3})\cap H^{1}_{0}(\Omega;\mathbb{R}^{3})),$		(4)
	$\displaystyle\widetilde{H}^{1/2}_{0,t}(\mathcal{S};\mathbb{R}^{3})$	$\displaystyle:=\gamma(H^{1}_{t}(\Omega,\mathcal{S};\mathbb{R}^{3})\cap H^{1}_{0}(\Omega;\mathbb{R}^{3}))$

which consist of normal and tangential vector functions on $\mathcal{S}$ . To be more specific, let us introduce some further notation. For a function $v\in H^{1}(T;U)$ ( $T\in\mathcal{T}$ , $U\in\{\mathbb{R},\mathbb{R}^{3}\}$ ) its traces on the boundary $\partial T$ and faces $F\in\mathcal{F}(T)$ are denoted by $v|_{\partial T}\in H^{1/2}(\partial T;U)$ and $v|_{F}\in H^{1/2}(F;U)$ , respectively. Restrictions of functions $\varphi\in H^{1/2}(\mathcal{S};\mathbb{R}^{3})$ to subsets of $\mathcal{S}$ are defined through their extensions, e.g., $\varphi|_{\Gamma}:=v|_{\Gamma}$ for $v\in H^{1}(\Omega;\mathbb{R}^{3})$ with $\gamma(v)=\varphi$ . We denote by $\widetilde{H}^{1/2}(F)\subset H^{1/2}(F)$ the restriction onto $F$ of the subspace of functions $v\in H^{1/2}(\partial T)$ that vanish on $\partial T\setminus\overline{F}$ , $F\in\mathcal{F}(T)$ , $T\in\mathcal{T}$ .

Proposition 1 ( $\widetilde{H}^{1/2}_{0,n}(\mathcal{S};\mathbb{R}^{3})$ ).

Any function $\varphi\in H^{1/2}(\mathcal{S};\mathbb{R}^{3})$ satisfies $\varphi\in\widetilde{H}^{1/2}_{0,n}(\mathcal{S};\mathbb{R}^{3})$ if and only if $\pi_{t}\varphi|_{F}=0$ , $\varphi\cdot n|_{F}\in\widetilde{H}^{1/2}(F)$ for all $F\in\mathcal{F}(\Omega)$ , and $\varphi|_{\Gamma}=0$ .

Proof.

Given $\varphi\in H^{1/2}(\mathcal{S};\mathbb{R}^{3})$ , $\pi_{t}\varphi|_{F}=0$ for all $F\in\mathcal{F}(\Omega)$ and $\varphi|_{\Gamma}=0$ imply $\varphi\in\widetilde{H}^{1/2}_{0,n}(\mathcal{S};\mathbb{R}^{3})$ . To see the other direction consider $v\in H^{1}(\Omega;\mathbb{R}^{3})$ with $\varphi=\gamma(v)$ . For a tetrahedron $T\in\mathcal{T}$ let $x_{j}$ denote its vertices with opposite faces $F_{j}$ ( $j=1,\ldots,4$ ). The corresponding barycentric coordinates and normal vectors are $\lambda_{j}$ and $n_{j}$ . Assuming that the tangential traces of $v$ on faces $F_{j}$ vanish we have to show that $v\cdot n_{j}|_{F_{j}}\in\widetilde{H}^{1/2}(F_{j})$ ( $j=1,\ldots,4$ ).

It is enough to show that $v\cdot n_{1}|_{F_{1}}\in\widetilde{H}^{1/2}(F_{1})$ . We complement vector $n_{1}$ with two linearly independent vectors $t_{1,1},t_{1,2}$ that are orthogonal to $n_{1}$ . There are constants $c_{j},d_{j},\alpha_{j}\in\mathbb{R}$ , $\alpha_{j}\not=0$ ( $j=2,3,4$ ) such that

n_{1}+c_{j}t_{1,1}+d_{j}t_{1,2}=\alpha_{j}n_{j^{\prime}}\times n_{j^{\prime\prime}},\quad j=2,3,4,

with $j^{\prime},j^{\prime\prime}\in\{2,3,4\}\setminus\{j\}$ , $j^{\prime}<j^{\prime\prime}$ , that is, $(j,j^{\prime},j^{\prime\prime})$ is a permutation of $(2,3,4)$ . In fact, since $\{n_{1},t_{1,2},t_{1,2}\}$ is a basis of $\mathbb{R}^{3}$ , there are numbers $\xi_{j},\eta_{j},\rho_{j}\in\mathbb{R}$ such that $\xi_{j}n_{1}+\eta_{j}t_{1,1}+\rho_{j}t_{1,2}=n_{j^{\prime}}\times n_{j^{\prime\prime}}$ ( $j=2,3,4$ ). Note that $\xi_{j}\not=0$ because $n_{1}\cdot(n_{j^{\prime}}\times n_{j^{\prime\prime}})\not=0$ by the linear independence of $n_{1},n_{j^{\prime}},n_{j^{\prime\prime}}$ . It follows that $c_{j}=\eta_{j}/\xi_{j}$ , $d_{j}=\rho_{j}/\xi_{j}$ and $\alpha_{j}=1/\xi_{j}$ ( $j=2,3,4$ ). We continue to select

g:=\sum_{j=2}^{4}(n_{1}+c_{j}t_{1,1}+d_{j}t_{1,2})\lambda_{j}

and define $w:=g\cdot v|_{T}\in H^{1}(T)$ . Since by assumption the tangential traces of $v$ on $\partial T$ vanish, $w$ satisfies

\displaystyle w|_{F_{1}}=\sum_{j=2}^{4}\lambda_{j}(n_{1}+c_{j}t_{1,1}+d_{j}t_{1,2})\cdot v|_{F_{1}}=\sum_{j=2}^{4}\lambda_{j}n_{1}\cdot v|_{F_{1}}=n_{1}\cdot v|_{F_{1}}

and, because $\lambda_{j}|_{F_{j}}=0$ and $v|_{F_{j}}=(v\cdot n_{j})n_{j}|_{F_{j}}$ ,

\displaystyle w|_{F_{m}}=\sum_{j=2}^{4}\lambda_{j}(n_{1}+c_{j}t_{1,1}+d_{j}t_{1,2})\cdot v|_{F_{m}}=\sum_{j\in\{2,3,4\}\setminus\{m\}}\lambda_{j}\alpha_{j}(v\cdot n_{m})(n_{j^{\prime}}\times n_{j^{\prime\prime}})\cdot n_{m}|_{F_{m}}=0

for $m=2,3,4$ by the orthogonality of $n_{m}$ and $n_{j^{\prime}}\times n_{j^{\prime\prime}}$ ( $m=j^{\prime}$ or $m=j^{\prime\prime}$ in the sum above). We conclude that $w|_{\partial T}\in H^{1/2}(\partial T)$ is a zero extension of $v\cdot n_{1}|_{F_{1}}$ to the other faces of $T$ , that is, $\varphi\cdot n_{1}|_{F_{1}}=v\cdot n_{1}|_{F_{1}}\in\widetilde{H}^{1/2}(F_{1})$ . ∎

There is no corresponding characterization of $\widetilde{H}^{1/2}_{0,t}(\mathcal{S};\mathbb{R}^{3})$ as face-wise “ $\widetilde{H}^{1/2}$ -space”. To see this, consider a tetrahedron $T$ with vertices $x_{i}$ , faces $F_{i}$ opposite to $x_{i}$ , normal vectors $n_{i}$ on $F_{i}$ , and barycentric coordinates $\lambda_{i}$ , $i=1,\ldots,4$ . Let $e_{1,2}\in\mathbb{R}^{3}\setminus\{0\}$ be a vector that is generated by the edge shared by $F_{1}$ and $F_{2}$ . Function $v:=\lambda_{3}\lambda_{4}e_{1,2}$ satisfies $v\in H^{1}(T;\mathbb{R}^{3})$ , $v\cdot n_{i}|_{F_{i}}=0$ ( $i=1,2$ ) and $v|_{F_{i}}=0$ ( $i=3,4$ ), but its tangential component $\pi_{t}v|_{F_{1}}=v|_{F_{1}}$ does not vanish on the shared edge. Therefore, $\pi_{t}v|_{F_{1}}$ cannot be continuously extended by zero onto $F_{2}$ in the trace space on $\partial T$ , which would be an appropriate characterization of the “ $\widetilde{H}^{1/2}(F_{1};\mathbb{R}^{3})$ -property” that corresponds to the scalar case of normal components in $\widetilde{H}^{1/2}(F)$ just considered. However, the edge-normal components of tangential face-traces of functions from $\widetilde{H}^{1/2}_{0,t}(\mathcal{S};\mathbb{R}^{3})$ do have such a property. A proper definition requires more notation.

For a simplex $T\in\mathcal{T}$ with face $F\in\mathcal{F}(T)$ let $n_{\partial F}$ denote the exterior unit normal vector along $\partial F$ that is normal to $n$ (a vector normal to $F$ ). For an edge $e$ of $F$ , $n_{F,e}$ is the restriction of $n_{\partial F}$ to $e$ . We define $\widetilde{H}^{1/2}_{tn}(F;\mathbb{R}^{3})$ as the subspace of functions $\varphi\in H^{1/2}(F;\mathbb{R}^{3})$ such that, for every edge $e$ of $F$ shared with a different face $F^{\prime}\in\mathcal{F}(T)$ , there exists a function $v\in H^{1/2}(\partial T)$ with $v|_{F}=\varphi|_{F}\cdot n_{F,e}$ and $v|_{F^{\prime}}=0$ . In other words, face-wise tangential traces have edge-normal components that, for every edge, can be continuously extended in $H^{1/2}(\partial T;\mathbb{R}^{3})$ by zero across the edge to the neighboring face (suggesting the “ $tn$ ” index notation).

Proposition 2 ( $\widetilde{H}^{1/2}_{0,t}(\mathcal{S};\mathbb{R}^{3})$ ).

Any function $\varphi\in H^{1/2}(\mathcal{S};\mathbb{R}^{3})$ satisfies $\varphi\in\widetilde{H}^{1/2}_{0,t}(\mathcal{S};\mathbb{R}^{3})$ if and only if $\varphi\cdot n|_{F}=0$ , $\varphi|_{F}\in\widetilde{H}^{1/2}_{tn}(F;\mathbb{R}^{3})$ for all $F\in\mathcal{F}(\Omega)$ , and $\varphi|_{\Gamma}=0$ .

Proof.

The proof is similar to the proof of Proposition 1 and we merely discuss the non-trivial implication. Given $\varphi\in\widetilde{H}^{1/2}_{0,t}(\mathcal{S};\mathbb{R}^{3})$ , $v\in H^{1}(\Omega;\mathbb{R}^{3})$ with $\varphi=\gamma(v)$ , and $T\in\mathcal{T}$ , it is immediate that $\varphi\cdot n|_{\partial T}=0$ , and we have to show that $\varphi|_{F}\in\widetilde{H}^{1/2}_{tn}(F;\mathbb{R}^{3})$ for $F\in\mathcal{F}(T)$ . Let $F,F^{\prime}\in\mathcal{F}(T)$ be given with $F\not=F^{\prime}$ and common edge $e$ . For the corresponding normal vectors $n,n^{\prime}$ , there are $c,\alpha\in\mathbb{R}$ such that $n_{F,e}+cn=\alpha n^{\prime}$ . In fact, $n,n^{\prime},n_{F,e}$ are linearly dependent because they are orthogonal to edge $e$ and $n_{F,e}\cdot n^{\prime}\not=0$ . It follows that $w:=(n_{F,e}+cn)\cdot v|_{T}\in H^{1}(T)$ . Furthermore, $w|_{F}=\varphi\cdot n_{F,e}|_{F}$ and $w|_{F^{\prime}}=\alpha n^{\prime}\cdot\varphi|_{F^{\prime}}=0$ , that is, $\varphi|_{F}\in\widetilde{H}^{1/2}_{tn}(F,\mathbb{R}^{3})$ as wanted. ∎

We generalize the concept of “a point is too small” in two dimensions from Tartar [24, §17] to three dimensions. In two dimensions, it says that $C_{0}^{\infty}(\Omega\setminus\{x\})$ for $x\in\Omega\subset\mathbb{R}^{2}$ is still dense in $H^{1}_{0}(\Omega)$ .

Consider a finite set of pairwise distinct, straight lines $L_{1},\ldots,L_{J}$ in $\mathbb{R}^{3}$ , $L_{j}=\{p_{j}+sq_{j};\;s\in\mathbb{R}\}$ with $p_{j},q_{j}\in\mathbb{R}^{3}$ , $|q_{j}|=1$ , and $L_{j}\cap\Omega\not=\emptyset$ , $j=1,\ldots,J$ . (Here, $\Omega\subset\mathbb{R}^{3}$ can be any bounded open set.) The set of lines $L:=L_{1}\cup\ldots\cup L_{J}$ is “too small”, in the following sense.

Proposition 3 (density).

The set $C_{0}^{\infty}(\Omega\setminus L)$ is dense in $H^{1}_{0}(\Omega)$ .

Proof.

Without loss of generality we assume that $\operatorname{diam}(\Omega)\leq 1$ (otherwise we scale $\Omega$ ). Let $f\in H^{1}_{0}(\Omega)$ and $\epsilon>0$ be given. We approximate $f$ in several steps.

1. Since $C^{\infty}_{0}(\Omega)\subset H^{1}_{0}(\Omega)$ is dense we find $f_{0}\in C^{\infty}_{0}(\Omega)$ with $\|f-f_{0}\|_{1}<\epsilon/5$ .

2. Let us assume that $|f_{0}|\leq 1$ in $\Omega$ (for the other case see the end of the proof). We approximate separately the non-negative and non-positive parts $f_{+}:=\max\{0,f_{0}\}$ and $f_{-}:=f_{+}-f_{0}$ of $f_{0}$ . Note that $f_{\pm}\in H^{1}_{0}(\Omega)$ , cf. [10, Theorem 4.4(iii)], and $0\leq f_{\pm}\leq 1$ .

3. The function $\log\log(1/|\cdot|)$ is a standard example of an unbounded $H^{1}$ -function on sufficiently small two-dimensional neighborhoods of the origin. We consider a parameter $0<\delta<1$ to be selected later, and define

\chi(r):=\Bigl{(}\log(-e\log\delta)-\log\log(1/r)\Bigr{)}_{+}\quad\text{for}\quad 0<r<1,

with non-negative part $(\cdot)_{+}$ . It is elementary that $\chi$ is monotonically increasing in $(0,1)$ , $\chi=0$ in $[0,\delta^{e}]$ , $0<\chi<1$ in $(\delta^{e},\delta)$ , $\chi(\delta)=1$ , and $\chi>1$ in $(\delta,1)$ ; $e$ is Euler’s number and $\delta^{e}$ means $\delta$ raised to the power $e$ . We continue to define

g_{j}(x):=\min\{1,\chi(\operatorname{dist}(x,L_{j}))\}\quad(x\in\Omega,j=1,\ldots,J).

Transformation to cylindrical coordinates along $L_{j}$ reveals

\|g_{j}\|_{1,U_{j}}\leq\sqrt{\pi\delta^{2}-2\pi/\log\delta}\quad\text{for}\quad U_{j}:=\{x\in\Omega;\;\operatorname{dist}(x,L_{j})<\delta\}=\{x\in\Omega;\;g_{j}(x)<1\}

( $j=1,\ldots,J$ ) where we used that $\operatorname{diam}(\Omega)\leq 1$ . We denote $U:=U_{1}\cup\ldots\cup U_{J}=\{x\in\Omega;\;\operatorname{dist}(x,L)<\delta\}$ and conclude for $j=1,\ldots,J$ that

\displaystyle\|g_{j}\|_{1,U}^{2}\leq\sum_{k=1}^{J}\|g_{j}\|_{1,U_{k}}^{2}\leq J(\pi\delta^{2}-2\pi/\log\delta)+\sum_{k\not=j}|U_{k}|\to 0\quad\text{as}\ \delta\to 0\ \text{for}.

(5)

4. We define preliminary approximations of $f_{\pm}$ by $g_{\pm}:=\min\{f_{\pm},g_{1},\ldots,g_{J}\}$ on $\Omega$ . They vanish in a neighborhood of $L\cap\Omega$ , $\operatorname{dist}(\operatorname{supp}g_{\pm},L)>0$ , satisfy $g_{\pm}\in H^{1}_{0}(\Omega)$ and

g_{1}(x)=\ldots=g_{J}(x)=1\geq f_{\pm}(x)=g_{\pm}(x)\quad\text{at}\ x\in\Omega\ \text{with}\ \operatorname{dist}(x,L)>\delta.

Thus $\|f_{\pm}-g_{\pm}\|_{1}=\|f_{\pm}-g_{\pm}\|_{1,U}.$ By the triangle inequality and the coarse bounds $0\leq g_{\pm}\leq f_{\pm}+g_{1}+\ldots+g_{J}$ , $|\nabla g_{\pm}|\leq|\nabla f_{\pm}|+|\nabla g_{1}|+\ldots+|\nabla g_{J}|$ (almost everywhere in $U$ ) we find with (5) that

	$\displaystyle\frac{1}{2}\\|f_{\pm}-g_{\pm}\\|_{1,U}^{2}$	$\displaystyle\leq\\|f_{\pm}\\|_{1,U}^{2}+\\|f_{\pm}+g_{1}+\ldots+g_{J}\\|_{U}^{2}+\\|\|\nabla f_{\pm}\|+\|\nabla g_{1}\|+\ldots\|\nabla g_{J}\|\\|_{U}^{2}$
		$\displaystyle\leq\\|f_{\pm}\\|_{1,U}^{2}+(J+1)\bigl{(}\\|f_{\pm}\\|_{1,U}^{2}+\\|g_{1}\\|_{1,U}^{2}+\ldots+\\|g_{J}\\|_{1,U}^{2}\bigr{)}$
		$\displaystyle=(J+2)\\|f_{\pm}\\|_{1,U}^{2}+(J+1)\Bigl{(}\\|g_{1}\\|_{1,U}^{2}+\ldots+\\|g_{J}\\|_{1,U}^{2}\Bigr{)}\to 0\quad(\delta\to 0)$

because $f_{\pm}\in H^{1}(\Omega)$ and $|U|\to 0$ as $\delta\to 0$ . Therefore, $\|f_{\pm}-g_{\pm}\|_{1}<\epsilon/5$ for some fixed, sufficiently small $\delta>0$ . By standard mollification of $g_{\pm}$ we find $\widetilde{g}_{\pm}\in C^{\infty}_{0}(\Omega\setminus L)$ with $\|g_{\pm}-\widetilde{g}_{\pm}\|_{1}\leq\epsilon/5$ .

A combination of steps 1–4 shows that $\widetilde{g}:=\widetilde{g}_{+}-\widetilde{g}_{-}$ satisfies $\widetilde{g}\in C^{\infty}_{0}(\Omega\setminus L)$ and

\|f-\widetilde{g}\|_{1}\leq\|f-f_{0}\|_{1}+\|f_{+}-g_{+}\|_{1}+\|f_{-}-g_{-}\|_{1}+\|g_{+}-\widetilde{g}_{+}\|_{1}+\|g_{-}-\widetilde{g}_{-}\|_{1}\leq\epsilon.

This finishes the proof for the case $\|f_{0}\|_{\infty}\leq 1$ . Otherwise we repeat steps 2 and 4 where $f_{0}$ is replaced with $\widetilde{f}_{0}:=f_{0}/c_{0}$ for $c_{0}:=\|f_{0}\|_{\infty}<\infty$ , and where bound $\epsilon/5$ is reduced to $\epsilon/(5c_{0})$ . In particular, $\widetilde{f}_{0}=f_{+}-f_{-}$ and $\widetilde{g}:=c_{0}(\widetilde{g}_{+}-\widetilde{g}_{-})$ satisfies $\widetilde{g}\in C^{\infty}_{0}(\Omega\setminus L)$ and

\|f-\widetilde{g}\|_{1}\leq\|f-f_{0}\|_{1}+c_{0}\bigl{(}\|f_{+}-g_{+}\|_{1}+\|f_{-}-g_{-}\|_{1}+\|g_{+}-\widetilde{g}_{+}\|_{1}+\|g_{-}-\widetilde{g}_{-}\|_{1}\Bigr{)}\leq\epsilon.

∎

Proposition 4 (inclusion).

The inclusion

\displaystyle\widetilde{H}^{1/2}_{0,n}(\mathcal{S};\mathbb{R}^{3})\oplus\widetilde{H}^{1/2}_{0,t}(\mathcal{S};\mathbb{R}^{3})

\displaystyle\subset H^{1/2}_{0}(\mathcal{S};\mathbb{R}^{3})

is dense. It is strict if there is a face $F\in\mathcal{F}(\Omega)$ that does not touch the boundary. Moreover, every $\tau\in H(\operatorname{div},\mathcal{T};\SS)$ satisfies

\tau\in H(\operatorname{div},\Omega;\SS)\quad\Leftrightarrow\quad\langle{}\rho\hskip 1.42262pt,\tau\rangle_{\mathcal{S}}=0\quad\forall\rho\in\widetilde{H}^{1/2}_{0,n}(\mathcal{S};\mathbb{R}^{3})\oplus\widetilde{H}^{1/2}_{0,t}(\mathcal{S};\mathbb{R}^{3}).

Proof.

The proof is analogous to the proof of [7, Proposition 2, (3)] by replacing density relation [24, Lemma 17.3] with Proposition 3. ∎

2.2 Normal-normal continuous stress space and dual trace operator

Duality (3) and normal trace space (4) provide us with the definition of a space of pointwise symmetric stresses with continuous normal-normal components across $\mathcal{S}$ ,

H_{nn}(\operatorname{div},\mathcal{T};\SS):=\{\tau\in H(\operatorname{div},\mathcal{T};\SS);\;\langle{}\varphi\hskip 1.42262pt,\tau\rangle_{\mathcal{S}}=0\ \forall\varphi\in\widetilde{H}^{1/2}_{0,n}(\mathcal{S};\mathbb{R}^{3})\}.

It is a closed subspace of $H(\operatorname{div},\mathcal{T};\SS)$ . The duality with this constrained space reduces the canonical trace operator $\gamma$ to

\displaystyle{\gamma_{1,\mathit{nn}}}:\;

\displaystyle\left\{\begin{array}[]{cll}H^{1}(\Omega;\mathbb{R}^{3})&\rightarrow&H_{nn}(\operatorname{div},\mathcal{T};\SS)^{*},\\ v&\mapsto&\langle{}{\gamma_{1,\mathit{nn}}}(v)\hskip 1.42262pt,\tau\rangle_{\mathcal{S}}:=(\varepsilon(v)\hskip 1.42262pt,\tau)+(v\hskip 1.42262pt,\operatorname{div}\tau)_{\mathcal{T}}.\end{array}\right.

It delivers tangential traces on interior faces and canonical traces on faces $F\subset\Gamma$ . The resulting trace spaces (the latter one including homogeneous Dirichlet data) are

\displaystyle H^{1/2}_{t}(\mathcal{S};\mathbb{R}^{3})

\displaystyle:={\gamma_{1,\mathit{nn}}}\bigl{(}H^{1}(\Omega;\mathbb{R}^{3})\bigr{)}\quad\text{and}\quad H^{1/2}_{0,t}(\mathcal{S};\mathbb{R}^{3}):={\gamma_{1,\mathit{nn}}}\bigl{(}H^{1}_{0}(\Omega;\mathbb{R}^{3})\bigr{)}.

It remains to specify norms. For an element $\tau$ of the product space $H(\operatorname{div},\mathcal{T};\SS)$ we use the product norm $\|\tau\|_{\operatorname{div},\mathcal{T}}:=\Bigl{(}\sum_{T\in\mathcal{T}}\|\tau\|_{\operatorname{div},T}^{2}\Bigr{)}^{1/2}$ . Trace space $H^{1/2}_{t}(\mathcal{S};\mathbb{R}^{3})$ is endowed with the canonical trace norm

\|\rho\|_{1/2,t,\mathcal{S}}:=\mathrm{inf}\{\|v\|_{1};\;v\in H^{1}(\Omega;\mathbb{R}^{3}),\ {\gamma_{1,\mathit{nn}}}(v)=\rho\},\quad\rho\in H^{1/2}_{t}(\mathcal{S};\mathbb{R}^{3}).

The following conformity and norm relations are essential to develop and analyze our mixed formulation of (2).

Proposition 5 (duality).

(i) Any $\tau\in H_{nn}(\operatorname{div},\mathcal{T};\SS)$ satisfies

\tau\in H(\operatorname{div},\Omega;\SS)\quad\Leftrightarrow\quad\langle{}\rho\hskip 1.42262pt,\tau\rangle_{\mathcal{S}}=0\quad\forall\rho\in H^{1/2}_{0,t}(\mathcal{S};\mathbb{R}^{3}).

(ii) Any $\rho\in H^{1/2}_{t}(\mathcal{S};\mathbb{R}^{3})$ satisfies

\|\rho\|_{1/2,t,\mathcal{S}}=\sup_{0\not=\tau\in H_{nn}(\operatorname{div},\mathcal{T};\SS),\;(\operatorname{tr}(\tau)\hskip 1.42262pt,1)=0}\frac{\langle{}\rho\hskip 1.42262pt,\tau\rangle_{\mathcal{S}}}{\|\tau\|_{\operatorname{div},\mathcal{T}}}.

Proof.

In two dimensions, statement (i) is [7, Lemma 5]. Its proof is based on a decomposition of (a dense subspace of) $H^{1/2}(\mathcal{S};\mathbb{R}^{2})$ into normal and tangential trace spaces. Proposition 4 does provide such a decomposition in three dimensions and allows to apply the same technique as in [7]. For an abstract setting of this result see [14, Lemma 27]. Also statement (ii) is proved analogously to the two-dimensional case [7, Proposition 6], and for an abstract form (without the trace constraint for test functions), see [14, Lemma 26]. ∎

3 Mixed formulation

An application of trace operator $\gamma_{1,nn}$ and the conformity characterization given by Proposition 5(i) lead to the following mixed formulation of (2). Find $\sigma\in H_{nn}(\operatorname{div},\mathcal{T};\SS)$ , $u\in L_{2}(\Omega;\mathbb{R}^{3})$ , and $\eta\in H^{1/2}_{0,t}(\mathcal{S};\mathbb{R}^{3})$ such that


$\displaystyle(\mathbb{A}\sigma\hskip 1.42262pt,{\delta\!\sigma})+(u\hskip 1.42262pt,\operatorname{div}{\delta\!\sigma})_{\mathcal{T}}-\langle{}\eta\hskip 1.42262pt,{\delta\!\sigma}\rangle_{\mathcal{S}}$	$\displaystyle=0,$	(6a)
$\displaystyle(\operatorname{div}\sigma\hskip 1.42262pt,{\delta\!u})_{\mathcal{T}}-\langle{}{\delta\!\eta}\hskip 1.42262pt,\sigma\rangle_{\mathcal{S}}$	$\displaystyle=-(f\hskip 1.42262pt,{\delta\!u})$	(6b)

for any ${\delta\!\sigma}\in H_{nn}(\operatorname{div},\mathcal{T};\SS)$ , ${\delta\!u}\in L_{2}(\Omega;\mathbb{R}^{3})$ , and ${\delta\!\eta}\in H^{1/2}_{0,t}(\mathcal{S};\mathbb{R}^{3})$ .

We show its well-posedness.

Theorem 6 (well-posedness of mixed formulation).

Given $f\in L_{2}(\Omega;\mathbb{R}^{3})$ , problem (6) is well posed with a unique solution $(\sigma,u,\eta)$ . The couple $(u,\sigma)\in H^{1}(\Omega)\times H(\operatorname{div},\Omega;\SS)$ solves (2), $\eta={\gamma_{1,\mathit{nn}}}(u)$ , and

\displaystyle\|\sigma\|_{\operatorname{div},\mathcal{T}}+\|u\|+\|\eta\|_{1/2,t,\mathcal{S}}

\displaystyle\lesssim\|f\|

holds uniformly with respect to $\lambda\in(0,\infty)$ and $\mathcal{T}$ .

Proof.

The proof uses the standard ingredients of mixed formulations, and is analogous to the two-dimensional case studied in [7, Theorem 10]. We recall the main steps and give some more details on the uniform stability.

We need the representation

\displaystyle(\mathbb{A}\tau\hskip 1.42262pt,{\delta\!\tau})=\frac{1}{2\mu}(\operatorname{dev}\tau\hskip 1.42262pt,\operatorname{dev}{\delta\!\tau})+\frac{1}{3(3\lambda+2\mu)}(\operatorname{tr}(\tau)\hskip 1.42262pt,\operatorname{tr}({\delta\!\tau}))\quad\forall\tau,{\delta\!\tau}\in L_{2}(\Omega;\SS),

(7)

see, e.g., [5, (9.1.8)]. The selection of ${\delta\!\sigma}=\operatorname{\mathrm{id}}$ in (6a) and relation $\langle{}\eta\hskip 1.42262pt,\operatorname{\mathrm{id}}\rangle_{\mathcal{S}}=\langle{}\eta\hskip 1.42262pt,n\rangle_{\Gamma}=0$ for $\eta\in H^{1/2}_{0,t}(\mathcal{S};\mathbb{R}^{3})$ show that $(\operatorname{tr}(\sigma)\hskip 1.42262pt,1)=0$ . We conclude that formulation (6) is equivalent to the same system where $H_{nn}(\operatorname{div},\mathcal{T};\SS)$ is replaced with

\widetilde{H}_{nn}(\operatorname{div},\mathcal{T};\SS):=\{\tau\in H_{nn}(\operatorname{div},\mathcal{T};\SS);\;(\operatorname{tr}(\tau)\hskip 1.42262pt,1)=0\}.

We continue to analyze this formulation.

The bilinear and linear forms of (6) are uniformly bounded by the selection of norms. Inf-sup properties

\displaystyle\sup_{0\not=\tau\in H(\operatorname{div},\Omega;\SS)}\frac{(\operatorname{div}\tau\hskip 1.42262pt,{\delta\!u})}{\|\tau\|_{\operatorname{div}}}\gtrsim\|{\delta\!u}\|\quad\forall{\delta\!u}\in L_{2}(\Omega;\mathbb{R}^{3})

and

\displaystyle\sup_{0\not=\tau\in\widetilde{H}_{nn}(\operatorname{div},\mathcal{T};\SS)}\frac{\langle{}{\delta\!\eta}\hskip 1.42262pt,\tau\rangle_{\mathcal{S}}}{\|\tau\|_{\operatorname{div},\mathcal{T}}}\geq\|{\delta\!\eta}\|_{1/2,t,\mathcal{S}}\quad\forall{\delta\!\eta}\in H^{1/2}_{0,t}(\mathcal{S};\mathbb{R}^{3})

hold by surjectivity of $\operatorname{div}:\;H(\operatorname{div},\Omega;\SS)\to L_{2}(\Omega;\mathbb{R}^{3})$ and Proposition 5(ii), respectively. The conformity relation of Proposition 5(i) and [6, Theorem 3.3] then show the required combined inf-sup relation

\displaystyle\sup_{0\not=\tau\in\widetilde{H}_{nn}(\operatorname{div},\mathcal{T};\SS)}\frac{(\operatorname{div}\tau\hskip 1.42262pt,{\delta\!u})_{\mathcal{T}}-\langle{}{\delta\!\eta}\hskip 1.42262pt,\tau\rangle_{\mathcal{S}}}{\|\tau\|_{\operatorname{div},\mathcal{T}}}\gtrsim\|{\delta\!u}\|+\|{\delta\!\eta}\|_{1/2,t,\mathcal{S}}\quad

for any ${\delta\!u}\in L_{2}(\Omega;\mathbb{R}^{3})$ and ${\delta\!\eta}\in H^{1/2}_{0,t}(\mathcal{S};\mathbb{R}^{3})$ .

For the uniform well-posedness of (6) it remains to verify the uniform coercivity of bilinear form $(\mathbb{A}\cdot\hskip 1.42262pt,\cdot)$ on the kernel

	$\displaystyle K_{0}:=$	$\displaystyle\{\tau\in\widetilde{H}_{nn}(\operatorname{div},\mathcal{T};\SS);\;(\operatorname{div}\tau\hskip 1.42262pt,{\delta\!u})_{\mathcal{T}}=0\ \forall{\delta\!u}\in L_{2}(\Omega;\mathbb{R}^{3}),\;\langle{}{\delta\!\eta}\hskip 1.42262pt,\tau\rangle_{\mathcal{S}}=0\ \forall{\delta\!\eta}\in H^{1/2}_{0,t}(\mathcal{S};\mathbb{R}^{3})\}$
	$\displaystyle=$	$\displaystyle\{\tau\in H(\operatorname{div},\Omega;\SS);\;\operatorname{div}\tau=0,\ (\operatorname{tr}(\tau)\hskip 1.42262pt,1)=0\}.$

The latter identity is due to Proposition 5(i). Critical is the bound

\|\operatorname{tr}(\tau)\|\leq C\bigl{(}\|\operatorname{dev}\tau\|+\|\operatorname{div}\tau\|_{-1}\bigr{)}=C\|\operatorname{dev}\tau\|\quad\forall\tau\in K_{0}

for a constant $C>0$ , see [8, Theorem 1]. The uniform coercivity of $(\mathbb{A}\cdot\hskip 1.42262pt,\cdot)$ on $K_{0}$ then follows from (7). The fact that $u\in H^{1}(\Omega;\mathbb{R}^{3})$ and $\sigma\in H(\operatorname{div},\Omega;\SS)$ solve (2) is immediate, and $\eta={\gamma_{1,\mathit{nn}}}(u)$ holds by (6a) and the definition of ${\gamma_{1,\mathit{nn}}}$ . ∎

4 Finite element approximation

Let us introduce some polynomial related notation. For $K\in\mathcal{F}\cup\mathcal{T}$ and $U\in\{\mathbb{R},\mathbb{R}^{3},\SS\}$ we denote by $P^{k}(K;U)$ the space of polynomials of degree $k$ on $K$ with values in $U$ . The piecewise polynomial spaces are $P^{k}(\mathcal{T};U)$ . We drop $U$ when $U=\mathbb{R}$ .

4.1 An $H(\operatorname{div};\SS)$ element

Let $T\subset\mathbb{R}^{3}$ be a tetrahedron with vertices $x_{i}$ , barycentric coordinates $\lambda_{i}$ , faces $F_{i}$ opposite to $x_{i}$ , and unit exterior normal vectors $n_{i}$ on $F_{i}$ (we also employ the generic notation $n$ ), and denote $\widetilde{n}_{i}:=\nabla\lambda_{i}\in\mathrm{span}\{n_{i}\}$ , $i=1,\ldots,4$ . We also need unit vectors $e_{i,j}$ (in a certain direction) generated by the edges $\overline{F}_{i}\cap\overline{F}_{j}$ , $i\not=j=1,\ldots,4$ . All these simplex-related objects are numbered modulus $4$ , e.g., $x_{5}=x_{1}$ . We abbreviate $a\odot b:=(ab^{\top}+ba^{\top})/2\in\SS$ for vectors $a,b\in\mathbb{R}^{3}$ .

The construction of our $H(\operatorname{div},T;\SS)$ -element is based upon the constant tensors

T_{i}:=\frac{e_{i+1,i+2}\odot e_{i+1,i+3}}{n_{i}\cdot e_{i+1,i+2}\;n_{i}\cdot e_{i+1,i+3}}\quad(i=1,\ldots,4),\quad T_{5}:=e_{1,2}\odot e_{3,4},\quad T_{6}:=e_{1,3}\odot e_{2,4}

which from a basis of $\SS$ (as seen below in Lemma 7), and the polynomial tensor spaces

	$\displaystyle P^{0}_{nn}(T;\SS)$	$\displaystyle:=\mathrm{span}\bigl{\{}T_{1},\ldots,T_{4}\bigr{\}},\quad P^{0}_{0}(T;\SS):=\mathrm{span}\bigl{\{}T_{5},T_{6}\bigr{\}},$
	$\displaystyle P^{2,k}(T,\SS)$	$\displaystyle:=\{\tau\in P^{2}(T;\SS);\;n\cdot\tau n\|_{F}\in P^{k}(F),\ F\in\mathcal{F}(T)\},\quad k\in\{0,1\},$
	$\displaystyle P^{2}_{0}(T;\SS)$	$\displaystyle:=\{\tau\in P^{2}(T;\SS);\;n\cdot\tau n\|_{\partial T}=0\}.$

Remark. The construction of $T_{i}\in\SS$ is such that $T_{i}n_{i+1}=0$ , $i=1,\ldots,4$ , that is, the image of mapping $T_{i}:\;\mathbb{R}^{3}\to\mathbb{R}^{3}$ is the plane through the origin that is parallel to face $F_{i+1}$ . Furthermore, $n_{j}\cdot T_{i}n_{j}=0$ for $j=1,\ldots,4$ , $i=5,6$ , and we will see that $n_{j}\cdot T_{i}n_{j}=\delta_{ij}$ for $i,j=1,\ldots,4$ .

Our $H(\operatorname{div},T;\SS)$ finite element on the tetrahedron $T$ is the vector space

\mathbb{X}(T):=\bigl{(}P^{2}(T)P^{0}_{nn}(T;\SS)\cap P^{2,1}(T,\SS)\bigr{)}\oplus P^{0}_{0}(T;\SS)

(8)

and the following degrees of freedom,


$\displaystyle\|F\|\langle{}n\cdot\tau n\hskip 1.42262pt,q\rangle_{F}\quad$	$\displaystyle(q\in P^{1}(F),\ F\in\mathcal{F}(T)),$	(9a)
$\displaystyle(\tau\hskip 1.42262pt,c)_{T}$	$\displaystyle(c\in P^{0}(T;\SS)),$	(9b)
$\displaystyle(\operatorname{div}\tau\hskip 1.42262pt,v)_{T}$	$\displaystyle(v\in P^{1}(T;\mathbb{R}^{3})).$	(9c)

The proof that ( $T$ , $\mathbb{X}(T)$ ,(9)) is a finite element requires some preparation.

Lemma 7.

We have $P^{0}(T;\SS)=P^{0}_{nn}(T;\SS)\oplus P^{0}_{0}(T;\SS).$

Proof.

We start by proving that $\{T_{1},\ldots,T_{6}\}$ is a basis of $P^{0}(T;\SS)$ . Since $\dim P^{0}(T;\SS)=6$ and $T_{i}\in P^{0}(T;\SS)$ , $i=1,\ldots,6$ , it suffices to show that the tensors $T_{i}$ are linearly independent.

For every $i\in\{1,\ldots,4\}$ , at least one of the two vectors $e_{i+1,i+2}$ and $e_{i+1,i+3}$ is tangential to the faces $F_{j}$ for $j\in\{i+1,i+2,i+3\}$ . Furthermore, both vectors are not tangential to $F_{i}$ . Consequently

(n_{i}\cdot e_{i+1,i+2})(n_{i}\cdot e_{i+2,i+3})\not=0\quad\text{and}\quad(n_{j}\cdot e_{i+1,i+2})(n_{j}\cdot e_{i+1,i+3})=0\quad\text{for}\ j\not=i.

Therefore, tensors $T_{i}$ are well defined and satisfy

n_{j}\cdot T_{i}n_{j}=\delta_{ij}\quad\text{for}\ i,j=1,\ldots,4.

On the other hand, for each face $F_{i}$ , one of the vectors $e_{1,2},e_{3,4}$ and one of the vectors $e_{1,3},e_{2,4}$ is tangential to $F_{i}$ . This implies that

n_{j}\cdot T_{i}n_{j}=0\quad\text{for}\ i=5,6,\ j=1,\ldots,4.

It follows that $T_{1},\ldots,T_{6}$ are linearly independent, and thus form a basis of $P^{0}(T;\SS)$ , if $T_{5}$ and $T_{6}$ are linearly independent. The latter is true because

n_{3}\cdot T_{5}n_{2}=n_{3}\cdot e_{1,2}\;e_{3,4}\cdot n_{2}\not=0\quad\text{but}\quad n_{3}\cdot T_{6}n_{2}=n_{3}\cdot e_{1,3}\;e_{2,4}\cdot n_{2}=0.

The splitting of $P^{0}(T;\SS)$ then holds by definition of the two subspaces. ∎

We use a transformation of $T\in\mathcal{T}$ to a reference tetrahedron ${\widehat{T}}$ to show that (9) are degrees of freedom of $\mathbb{X}(T)$ . We consider the tetrahedron ${\widehat{T}}$ with vertices ${\widehat{x}}_{1}=(0,0,0)$ , ${\widehat{x}}_{2}=(1,0,0)$ , ${\widehat{x}}_{3}=(0,1,0)$ , ${\widehat{x}}_{4}=(0,0,1)$ , and let $G_{T}:\;{\widehat{T}}\to T$ denote the affine diffeomorphism

\begin{pmatrix}{\widehat{x}}\\ {\widehat{y}}\\ {\widehat{z}}\end{pmatrix}\mapsto a_{T}+B_{T}\begin{pmatrix}{\widehat{x}}\\ {\widehat{y}}\\ {\widehat{z}}\end{pmatrix}

with $a_{T}\in\mathbb{R}^{3}$ , $B_{T}\in\mathbb{R}^{3\times 3}$ , and set $J_{T}:=\det(B_{T})$ . We use the generic notation ${\widehat{n}}$ for exterior normal vectors on faces ${\widehat{F}}\in\mathcal{F}({\widehat{T}})$ . Tensors ${\widehat{\tau}},\,{\widehat{\sigma}}:\;{\widehat{T}}\to\SS$ and vector functions ${\widehat{v}}:\;{\widehat{T}}\to\mathbb{R}^{3}$ are transformed onto elements $T\in\mathcal{T}$ as $\mathcal{PK}_{{T}}({\widehat{\tau}})=\tau$ , $\mathcal{PK}^{\perp}_{{T}}({\widehat{\sigma}})=\sigma$ and $\mathcal{P}_{T}({\widehat{v}})=v$ , where

J_{T}^{2}\tau\circ G_{T}:=B_{T}{\widehat{\tau}}B_{T}^{\top},\quad\sigma\circ G_{T}:=J_{T}B_{T}^{-\top}{\widehat{\sigma}}B_{T}^{-1},\quad\text{and}\quad v\circ G_{T}:=J_{T}B_{T}^{-\top}{\widehat{v}}\quad\text{on}\ {\widehat{T}}.

These are re-scaled Piola–Kirchhoff and Piola transformations. They conserve the properties of space $\mathbb{X}(T)$ and degrees of freedom (9), as seen next.

Lemma 8.

Let $T\in\mathcal{T}$ and $\tau\in L_{2}(T;\SS)$ be given with $\tau=\mathcal{PK}_{{T}}({\widehat{\tau}})$ .

(i) Property $\tau\in\mathbb{X}(T)$ holds if and only if ${\widehat{\tau}}\in\mathbb{X}({\widehat{T}})$ .

(ii) For $F\in\mathcal{F}(T)$ , $q\in P^{2}(F)$ , $c\in P^{0}(T;\SS)$ , and $v\in P^{1}(T;\mathbb{R}^{3})$ let ${\widehat{F}}\in\mathcal{F}({\widehat{T}})$ , $\widehat{q}\in P^{2}(\widehat{F})$ , $\widehat{c}\in P^{0}({\widehat{T}};\SS)$ , and $\widehat{v}\in P^{1}({\widehat{T}};\mathbb{R}^{3})$ be the transformed objects with $F=G_{T}({\widehat{F}})$ , $q\circ G_{T}=\widehat{q}$ , $c=\mathcal{PK}^{\perp}_{{T}}(\widehat{c})$ , and $v=\mathcal{P}_{T}({\widehat{v}})$ . Then,

\displaystyle|F|\langle{}n\cdot\tau n\hskip 1.42262pt,q\rangle_{F}=|{\widehat{F}}|\langle{}{\widehat{n}}\cdot{\widehat{\tau}}{\widehat{n}}\hskip 1.42262pt,\widehat{q}\rangle_{{\widehat{F}}},\quad(\tau\hskip 1.42262pt,c)_{T}=({\widehat{\tau}}\hskip 1.42262pt,\widehat{c})_{\widehat{T}},\quad(\operatorname{div}\tau\hskip 1.42262pt,v)_{T}=(\widehat{\operatorname{div}}{\widehat{\tau}}\hskip 1.42262pt,{\widehat{v}})_{\widehat{T}}.

Proof.

A proof of statement (ii) is analogous to that of the two-dimensional case on triangles, see [7, Lemma 13]. Statement (i) is consequence of relations (ii). In fact, the Piola–Kirchhoff transformation $\mathcal{PK}_{{T}}$ and its inverse conserve the space of symmetric polynomial tensor functions of a fixed degree. The conservation of the moments of the normal-normal traces implies that $\mathcal{PK}_{{T}}$ is a bijective mapping between $P^{0}_{nn}({\widehat{T}};\SS)$ and $P^{0}_{nn}(T;\SS)$ , and between $P^{0}_{0}({\widehat{T}};\SS)$ and $P^{0}_{0}(T;\SS)$ . It also means that the normal-normal traces of $\tau$ on faces $F\in\mathcal{F}(T)$ are linear polynomials if and only if this applies to the normal-normal traces of ${\widehat{\tau}}$ on ${\widehat{F}}\in\mathcal{F}({\widehat{T}})$ . This finishes the proof. ∎

Proposition 9 (degrees of freedom).

For a tetrahedron $T\subset\mathbb{R}^{3}$ relations $P^{0}(T;\SS)\subset\mathbb{X}(T)$ , $\dim\mathbb{X}(T)=30$ hold, and (9) are degrees of freedom of $\mathbb{X}(T)$ for a finite element in the sense of Ciarlet.

Proof.

The first relation follows from of Lemma 7 and the fact that $P^{0}_{nn}(T;\SS)$ and $P^{0}_{0}(T;\SS)$ are subsets of $\mathbb{X}(T)$ by construction.

Dimension. By definition of $P^{0}_{nn}(T;\SS)$ ,

P^{2}(T)P^{0}_{nn}(T;\SS)=T_{1}P^{2}(T)\oplus\cdots\oplus T_{4}P^{2}(T)

has dimension $40$ . Since $n_{j}\cdot(T_{i}P^{2}(T))n_{j}=\delta_{ij}P^{2}(T)$ for $i,j=1,\ldots,4$ , the constraint $\bigl{(}T_{i}P^{2}(T)\bigr{)}\cap P^{2,1}(T,\SS)$ requires to select the subspaces

\displaystyle P^{2,1}_{i}(T)

\displaystyle:=\{v\in P^{2}(T);\;v|_{F_{i}}\in P^{1}(F_{i})\}=P^{1}(T)\oplus\mathrm{span}\{\lambda_{i}\lambda_{i+1},\lambda_{i}\lambda_{i+2},\lambda_{i}\lambda_{i+3}\}

for $i=1,\ldots,4$ , each of dimension $7$ . Summing the dimensions gives $4$ times $7$ plus the dimension $2$ of $P^{0}_{0}(T;\SS)$ in (8).

Degrees of freedom. By Lemma 8, (9) are degrees of freedom of $\mathbb{X}(T)$ for a general tetrahedron $T\in\mathcal{T}$ if and only if they are degrees of freedom for the reference tetrahedron $T={\widehat{T}}$ . In the remainder of this proof we therefore assume without loss of generality that $T={\widehat{T}}$ . We need this assumption only in the final step to verify the invertibility of a $12\times 12$ -matrix. The number of functionals in (9) is $30$ : 12 in (9a), 6 in (9b), and 12 in (9c). It is therefore enough to show the linear independence of (9) on $\mathbb{X}(T)$ .

Linear independence. Let $\tau\in\mathbb{X}(T)$ be given with vanishing functionals (9). We have to show that $\tau=0$ . The vanishing of (9a) leads to

\tau\in\oplus_{i=1}^{4}\bigr{(}T_{i}P^{2}(T)\cap P^{2}_{0}(T,\SS)\bigr{)}\oplus P^{0}_{0}(T;\SS).

Since

\displaystyle T_{i}P^{2}(T)\cap P^{2}_{0}(T,\SS)=T_{i}\,\mathrm{span}\{\phi\in P^{2}(T);\;\phi|_{F_{i}}=0\}=T_{i}\,\mathrm{span}\{\lambda_{i},\lambda_{i}\lambda_{i+1},\lambda_{i}\lambda_{i+2},\lambda_{i}\lambda_{i+3}\}

for $i=1,\ldots,4$ we deduce the representation

\displaystyle\tau=\sum_{i=1}^{4}\lambda_{i}\phi_{i}T_{i}+\phi_{5}T_{5}+\phi_{6}T_{6}\quad\text{with}\quad\phi_{i}\in P^{1}(T)\text{ for }i=1,\ldots,4,\ \text{ and }\phi_{5},\phi_{6}\in\mathbb{R}.

The vanishing of degrees of freedom (9b) means

\sum_{i=1}^{4}(\lambda_{i}\phi_{i}\hskip 1.42262pt,1)_{T}T_{i}+(\phi_{5}\hskip 1.42262pt,1)_{T}T_{5}+(\phi_{6}\hskip 1.42262pt,1)_{T}T_{6}=0,

so that, by the linear independence of $T_{1},\ldots,T_{6}$ ,

\displaystyle(\lambda_{i}\phi_{i}\hskip 1.42262pt,1)_{T}=0\quad\text{for }i=1,\ldots,4,\quad\text{and}\quad\phi_{5}=\phi_{6}=0.

We conclude that

\displaystyle\tau=\sum_{i=1}^{4}\lambda_{i}\phi_{i}T_{i}\quad\text{and}\quad(\lambda_{j}\phi_{j}\hskip 1.42262pt,1)_{T}=0\quad\text{for}\ j=1,\ldots,4.

(10)

In the following we represent $\phi_{i}=\sum_{j=1}^{4}\phi_{i}(x_{j})\lambda_{j}$ and calculate $\nabla\phi_{i}=\sum_{j=1}^{4}\phi_{i}(x_{j})\widetilde{n}_{j}$ for $i=1,\ldots,4$ (recall that $\widetilde{n}_{j}=\nabla\lambda_{j}\in\mathrm{span}(n_{j})$ ). Representation (10) and the vanishing of (9c) reveal that

	$\displaystyle 0=\operatorname{div}\tau(x_{k})$	$\displaystyle=\sum_{i=1}^{4}T_{i}\bigl{(}\phi_{i}\nabla\lambda_{i}+\lambda_{i}\nabla\phi_{i}\bigr{)}(x_{k})=\sum_{i=1}^{4}T_{i}\bigl{(}\phi_{i}\widetilde{n}_{i}+\lambda_{i}\sum_{j=1}^{4}\phi_{i}(x_{j})\widetilde{n}_{j}\bigr{)}(x_{k})$
		$\displaystyle=\sum_{i=1}^{4}T_{i}\bigl{(}\phi_{i}(x_{k})\widetilde{n}_{i}+\delta_{ik}\sum_{j=1}^{4}\phi_{i}(x_{j})\widetilde{n}_{j}\bigr{)}\quad\text{for}\ k=1,\ldots,4.$		(11)

These equations form a homogeneous linear system of $12$ equations for the $16$ unknowns $\phi_{i}(x_{j})$ , $i,j=1,\ldots,4$ . Instead of unknowns $\phi_{i}(x_{i})$ we use the scaled unknowns $2\phi_{i}(x_{i})$ for $i=1,\ldots,4$ and order the equations with respect to $k$ and the unknowns with inner loop over $j$ and outer loop over $i$ , namely

2\phi_{1}(x_{1}),\phi_{1}(x_{2}),\phi_{1}(x_{3}),\phi_{1}(x_{4}),\quad\phi_{2}(x_{1}),2\phi_{2}(x_{2}),\phi_{2}(x_{3}),\phi_{2}(x_{4}),\quad\ldots,\phi_{4}(x_{3}),2\phi_{4}(x_{4}).

Denoting $t_{ij}:=T_{i}\widetilde{n}_{j}$ , the corresponding matrix of system (4.1) becomes

\begin{pmatrix}\begin{array}[]{cccc|cccc|cccc|cccc}t_{11}&t_{12}&t_{13}&t_{14}&t_{22}&0&0&0&t_{33}&0&0&0&t_{44}&0&0&0\\ \hline\cr 0&t_{11}&0&0&t_{21}&t_{22}&t_{23}&t_{24}&0&t_{33}&0&0&0&t_{44}&0&0\\ \hline\cr 0&0&t_{11}&0&0&0&t_{22}&0&t_{31}&t_{32}&t_{33}&t_{34}&0&0&t_{44}&0\\ \hline\cr 0&0&0&t_{11}&0&0&0&t_{22}&0&0&0&t_{33}&t_{41}&t_{42}&t_{43}&t_{44}\end{array}\end{pmatrix}.

It remains to incorporate the relations $(\lambda_{i}\phi_{i}\hskip 1.42262pt,1)_{T}=0$ for $i=1,\ldots,4$ from (10). We note that $(\lambda_{i}\hskip 1.42262pt,\lambda_{j})_{T}/(\lambda_{i}\hskip 1.42262pt,\lambda_{i+1})_{T}=\delta_{ij}+1$ , see, e.g., [25, Theorem 3.1], and recall the representation $\phi_{i}=\sum_{j=1}^{4}\phi_{i}(x_{j})\lambda_{j}$ . We find that $(\lambda_{i}\phi_{i}\hskip 1.42262pt,1)_{T}=\sum_{j=1}^{4}\phi_{i}(x_{j})(\lambda_{i}\hskip 1.42262pt,\lambda_{j})_{T}=0$ holds if and only if

\frac{1}{(\lambda_{i}\hskip 1.42262pt,\lambda_{i+1})_{T}}\sum_{j=1}^{4}\phi_{i}(x_{j})(\lambda_{i}\hskip 1.42262pt,\lambda_{j})_{T}=2\phi_{i}(x_{i})+\phi_{i}(x_{i+1})+\phi_{i}(x_{i+2})+\phi_{i}(x_{i+3})=0,\quad i=1,\ldots,4.

We use this relation to eliminate the unknowns $2\phi_{i}(x_{i})$ for $i=1,\ldots,4$ . With the notation $\widetilde{t}_{ij}:=t_{ij}-t_{ii}$ the reduced $12\times 12$ matrix reads

\begin{pmatrix}\begin{array}[]{ccc|ccc|ccc|ccc}\widetilde{t}_{12}&\widetilde{t}_{13}&\widetilde{t}_{14}&t_{22}&0&0&t_{33}&0&0&t_{44}&0&0\\ \hline\cr\rule{0.0pt}{11.19443pt}t_{11}&0&0&\widetilde{t}_{21}&\widetilde{t}_{23}&\widetilde{t}_{24}&0&t_{33}&0&0&t_{44}&0\\ \hline\cr\rule{0.0pt}{11.19443pt}0&t_{11}&0&0&t_{22}&0&\widetilde{t}_{31}&\widetilde{t}_{32}&\widetilde{t}_{34}&0&0&t_{44}\\ \hline\cr\rule{0.0pt}{11.19443pt}0&0&t_{11}&0&0&t_{22}&0&0&t_{33}&\widetilde{t}_{41}&\widetilde{t}_{42}&\widetilde{t}_{43}\end{array}\end{pmatrix}.

(12)

The proof of the lemma is finished by noting that this matrix is invertible. Let us give the details.

Recall that we are considering the reference tetrahedron $T={\widehat{T}}$ with the previously introduced numbering of vertices. In particular,

	$\displaystyle\lambda_{1}(x,y,z)=1-x-y-z,\ \lambda_{2}(x,y,z)=x,\ \lambda_{3}(x,y,z)=y,\ \lambda_{4}=z,$
	$\displaystyle\widetilde{n}_{1}=-(1,1,1)^{\top},\ \widetilde{n}_{2}=(1,0,0)^{\top},\ \widetilde{n}_{3}=(0,1,0)^{\top},\ \widetilde{n}_{4}=(0,0,1)^{\top},$

and selecting the edge vectors

	$\displaystyle e_{1,2}=(0,1,-1)^{\top},\ e_{1,3}=(1,0,-1)^{\top},\ e_{1,4}=(1,-1,0)^{\top},\ e_{2,3}=(0,0,1)^{\top},$
	$\displaystyle e_{2,4}=(0,1,0)^{\top},\ e_{3,4}=(1,0,0)^{\top},\ e_{i,j}=e_{j,i}\ (i>j)$

we calculate

	$\displaystyle T_{1}=\frac{e_{2,3}\odot e_{2,4}}{n_{1}\cdot e_{2,3}\;n_{1}\cdot e_{2,4}}=\begin{pmatrix}0&0&0\\ 0&0&3/2\\ 0&3/2&0\end{pmatrix},$	$\displaystyle T_{2}=\frac{e_{3,4}\odot e_{3,1}}{n_{2}\cdot e_{3,4}\;n_{2}\cdot e_{3,4}}=\begin{pmatrix}1&0&-1/2\\ 0&0&0\\ -1/2&0&0\end{pmatrix},$
	$\displaystyle T_{3}=\frac{e_{4,1}\odot e_{4,2}}{n_{3}\cdot e_{4,1}\;n_{3}\cdot e_{4,2}}=\begin{pmatrix}0&-1/2&0\\ -1/2&1&0\\ 0&0&0\end{pmatrix},$	$\displaystyle T_{4}=\frac{e_{1,2}\odot e_{1,3}}{n_{4}\cdot e_{1,2}\;n_{4}\cdot e_{1,3}}=\begin{pmatrix}0&1/2&-1/2\\ 1/2&0&-1/2\\ -1/2&-1/2&1\end{pmatrix}.$

This gives (note that $t_{i,i+1}=(0,0,0)^{\top}$ )

	$\displaystyle t_{11}=\begin{pmatrix}0\\ -3/2\\ -3/2\end{pmatrix},\ t_{13}=\begin{pmatrix}0\\ 0\\ 3/2\end{pmatrix},\ t_{14}=\begin{pmatrix}0\\ 3/2\\ 0\end{pmatrix},\$	$\displaystyle\widetilde{t}_{12}=-t_{11},\ \widetilde{t}_{13}=\begin{pmatrix}0\\ 3/2\\ 3\end{pmatrix},\ \widetilde{t}_{14}=\begin{pmatrix}0\\ 3\\ 3/2\end{pmatrix},$
	$\displaystyle t_{22}=\begin{pmatrix}1\\ 0\\ -1/2\end{pmatrix},\ t_{21}=\begin{pmatrix}-1/2\\ 0\\ 1/2\end{pmatrix},\ t_{24}=\begin{pmatrix}-1/2\\ 0\\ 0\end{pmatrix},\$	$\displaystyle\widetilde{t}_{21}=\begin{pmatrix}-3/2\\ 0\\ 1\end{pmatrix},\ \widetilde{t}_{23}=-t_{22},\ \widetilde{t}_{24}=\begin{pmatrix}-3/2\\ 0\\ 1/2\end{pmatrix},$
	$\displaystyle t_{33}=\begin{pmatrix}-1/2\\ 1\\ 0\end{pmatrix},\ t_{31}=\begin{pmatrix}1/2\\ -1/2\\ 0\end{pmatrix},\ t_{32}=\begin{pmatrix}0\\ -1/2\\ 0\end{pmatrix},\$	$\displaystyle\widetilde{t}_{31}=\begin{pmatrix}1\\ -3/2\\ 0\end{pmatrix},\ \widetilde{t}_{32}=\begin{pmatrix}1/2\\ -3/2\\ 0\end{pmatrix},\ \widetilde{t}_{34}=-t_{33},$
	$\displaystyle t_{44}=\begin{pmatrix}-1/2\\ -1/2\\ 1\end{pmatrix},\ t_{42}=\begin{pmatrix}0\\ 1/2\\ -1/2\end{pmatrix},\ t_{43}=\begin{pmatrix}1/2\\ 0\\ -1/2\end{pmatrix},\$	$\displaystyle\widetilde{t}_{41}=-t_{44},\ \widetilde{t}_{42}=\begin{pmatrix}1/2\\ 1\\ -3/2\end{pmatrix},\ \widetilde{t}_{43}=\begin{pmatrix}1\\ 1/2\\ -3/2\end{pmatrix},$

and matrix (12) reads

\begin{pmatrix}\begin{array}[]{ccc|ccc|ccc|ccc}0&0&0&1&0&0&-1/2&0&0&-1/2&0&0\\ 3/2&3/2&3&0&0&0&1&0&0&-1/2&0&0\\ 3/2&3&3/2&-1/2&0&0&0&0&0&1&0&0\\ \hline\cr 0&0&0&-3/2&-1&-3/2&0&-1/2&0&0&-1/2&0\\ -3/2&0&0&0&0&0&0&1&0&0&-1/2&0\\ -3/2&0&0&1&1/2&1/2&0&0&0&0&1&0\\ \hline\cr 0&0&0&0&1&0&1&1/2&1/2&0&0&-1/2\\ 0&-3/2&0&0&0&0&-3/2&-3/2&-1&0&0&-1/2\\ 0&-3/2&0&0&-1/2&0&0&0&0&0&0&1\\ \hline\cr 0&0&0&0&0&1&0&0&-1/2&1/2&1/2&1\\ 0&0&-3/2&0&0&0&0&0&1&1/2&1&1/2\\ 0&0&-3/2&0&0&-1/2&0&0&0&-1&-3/2&-3/2\end{array}\end{pmatrix}.

A lengthy calculation shows that its determinant is $2^{-8}3^{4}5^{2}$ . This finishes the proof. ∎

4.2 Approximation space

Elements $\mathbb{X}(T)$ ( $T\in\mathcal{T}$ ) from (8) generate the product space $\mathbb{X}(\mathcal{T})$ that gives rise to the $H_{nn}(\operatorname{div},\mathcal{T};\SS)$ -conforming finite element space

P^{2}_{nn}(\mathcal{T}):=\mathbb{X}(\mathcal{T})\cap H_{nn}(\operatorname{div},\mathcal{T};\SS)=\{\tau\in\mathbb{X}(\mathcal{T});\;[n\cdot\tau n]|_{F}=0\ \forall F\in\mathcal{F}(\Omega)\}.

Here, $[n\cdot\tau n]|_{F}$ denotes the normal-normal jump of $\tau$ across $F$ . Equivalently, this conformity is achieved by the selection of unique degrees of freedom (9a) assigned to interior faces $F\in\mathcal{F}(\Omega)$ . By also assigning volume-related degrees of freedom (9b), (9c) we obtain an interpolation operator

\mathcal{I}^{nn}_{\mathcal{T}}:\;H(\operatorname{div},\Omega;\SS)\cap L_{q}(\Omega;\SS)\to P^{2}_{nn}(\mathcal{T})\quad(q>2).

In the following, $\operatorname{div}_{\mathcal{T}}$ is the piecewise divergence operator and $\Pi^{1}_{\mathcal{T}}$ denotes the $L_{2}$ -projection onto piecewise linear vector-valued polynomials.

Proposition 10 (interpolation).

Operator $\mathcal{I}^{nn}_{\mathcal{T}}:\;H(\operatorname{div},\Omega;\SS)\cap L_{q}(\Omega;\SS)\to P^{2}_{nn}(\mathcal{T})$ is well defined and bounded for $q>2$ . It commutes with the piecewise divergence operator, $\operatorname{div}_{\mathcal{T}}\mathcal{I}^{nn}_{\mathcal{T}}=\Pi^{1}_{\mathcal{T}}\operatorname{div},$ and has the projection property $\mathcal{I}^{nn}_{\mathcal{T}}\tau=\tau$ for $\tau\in P^{2}_{nn}(\mathcal{T})$ . Given $0<s\leq 1$ , $0\leq r\leq 2$ , any $\tau\in H^{s,r}(\operatorname{div},\Omega;\SS)$ satisfies

	$\displaystyle\\|\tau-\mathcal{I}^{nn}_{\mathcal{T}}\tau\\|$	$\displaystyle\lesssim\\|h_{\mathcal{T}}\\|_{\infty}^{s}\\|\tau\\|_{s}+\\|h_{\mathcal{T}}\\|_{\infty}\\|\operatorname{div}\tau\\|\quad$	$\displaystyle\forall\tau\in H^{s,0}(\operatorname{div},\Omega;\SS),$
	$\displaystyle\\|\operatorname{div}(\tau-\mathcal{I}^{nn}_{\mathcal{T}}\tau)\\|_{\mathcal{T}}$	$\displaystyle\lesssim\\|h_{\mathcal{T}}\\|_{\infty}^{r}\\|\operatorname{div}\tau\\|_{r}\quad$	$\displaystyle\forall\tau\in H^{s,r}(\operatorname{div},\Omega;\SS)$

with intrinsic constants independent of $\tau$ and $\mathcal{T}$ .

Proof.

The proof is almost verbatim to the proof of the two-dimensional case [7, Proposition 17]. We only have to verify that the face degrees of freedom are bounded functionals for $\tau\in L_{q}(\Omega;\SS)$ if $q>2$ , and to recall that $H^{s}(\Omega;\SS)\subset L_{q}(\Omega;\SS)$ for $0<s\leq 1$ and $q=6/(3-2s)>2$ by the Sobolev embedding theorem [1, Theorem 7.57]. The former fact can be proved analogously to [3, Lemma 4.7]. Essential step in the proof of that lemma is the continuity of the extension by zero from the Sobolev space $W^{1-1/q^{\prime},q^{\prime}}(E)$ on an edge $E$ of a face $F$ to $W^{1-1/q^{\prime},q^{\prime}}(\partial F)$ where $1/q^{\prime}+1/q=1$ . This zero extension is also continuous as a mapping from $W^{1-1/q^{\prime},q^{\prime}}(F)$ to $W^{1-1/q^{\prime},q^{\prime}}(\partial T)$ where $T\in\mathcal{T}$ and $F\in\mathcal{F}(T)$ , cf. [11, Corollary 1.4.4.5], and thus gives the result. ∎

The next result can be shown analogously to [7, Lemma 20].

Lemma 11.

Any $\tau\in P^{2}_{nn}(\mathcal{T})$ with

\operatorname{div}_{\mathcal{T}}\tau=0,\quad\langle{}{\gamma_{1,\mathit{nn}}}(w)\hskip 1.42262pt,\tau\rangle_{\mathcal{S}}=0\ \forall w\in P^{1}(\mathcal{T};\mathbb{R}^{3})\cap H^{1}_{0}(\Omega;\mathbb{R}^{3}),\quad\text{and}\quad(\operatorname{tr}(\tau)\hskip 1.42262pt,1)=0

satisfies

\|\operatorname{tr}(\tau)\|\leq C\|\operatorname{dev}\tau\|

with a constant $C>0$ that is independent of $\tau$ and $\mathcal{T}$ .

4.3 Finite element scheme and locking-free convergence

Let $S^{1}_{0}(\mathcal{T};\mathbb{R}^{3}):=P^{1}(\mathcal{T};\mathbb{R}^{3})\cap H^{1}_{0}(\Omega;\mathbb{R}^{3})$ be the space of continuous, piecewise linear vector-valued functions that vanish on $\Gamma$ . We select the discrete trace space $S^{1}_{0}(\mathcal{S}):={\gamma_{1,\mathit{nn}}}\bigl{(}S^{1}_{0}(\mathcal{T};\mathbb{R}^{3})\bigr{)}.$ Given $f\in L_{2}(\Omega;\mathbb{R}^{3})$ , our mixed finite element scheme seeks $\sigma_{h}\in P^{2}_{nn}(\mathcal{T})$ , $u_{h}\in P^{1}(\mathcal{T};\mathbb{R}^{3})$ , and $\eta_{h}\in S^{1}_{0}(\mathcal{S})$ such that


$\displaystyle(\mathbb{A}\sigma_{h}\hskip 1.42262pt,{\delta\!\sigma})+(u_{h}\hskip 1.42262pt,\operatorname{div}{\delta\!\sigma})_{\mathcal{T}}-\langle{}\eta_{h}\hskip 1.42262pt,{\delta\!\sigma}\rangle_{\mathcal{S}}$	$\displaystyle=0,$	(13a)
$\displaystyle(\operatorname{div}\sigma_{h}\hskip 1.42262pt,{\delta\!u})_{\mathcal{T}}-\langle{}{\delta\!\eta}\hskip 1.42262pt,\sigma_{h}\rangle_{\mathcal{S}}$	$\displaystyle=-(f\hskip 1.42262pt,{\delta\!u})$	(13b)

for any ${\delta\!\sigma}\in P^{2}_{nn}(\mathcal{T})$ , ${\delta\!u}\in P^{1}(\mathcal{T};\mathbb{R}^{3})$ , and ${\delta\!\eta}\in S^{1}_{0}(\mathcal{S})$ .

This scheme is locking free and converges quasi-optimally.

Theorem 12 (locking-free convergence).

For given $f\in L_{2}(\Omega;\mathbb{R}^{3})$ let $(\sigma,u,\eta)$ denote the solution to (6). Mixed scheme (13) has a unique solution $(\sigma_{h},u_{h},\eta_{h})$ . It satisfies

	$\displaystyle\\|\sigma-\sigma_{h}\\|_{\operatorname{div},\mathcal{T}}+\\|u-u_{h}\\|+\\|\eta-\eta_{h}\\|_{1/2,t,\mathcal{S}}$
	$\displaystyle\lesssim\inf\bigl{\{}\\|\sigma-\tau\\|_{\operatorname{div},\mathcal{T}}+\\|u-v\\|+\\|\eta-\rho\\|_{1/2,t,\mathcal{S}};\;\tau\in P^{2}_{nn}(\mathcal{T}),v\in P^{1}(\mathcal{T};\mathbb{R}^{3}),\rho\in S^{1}_{0}(\mathcal{S})\bigr{\}}$

with a hidden constant that is independent of $\mathcal{T}$ and $\lambda\in(0,\infty)$ .

Let us furthermore assume that $u\in H^{1+s}(\Omega;\mathbb{R}^{3})$ and $f\in H^{r}(\Omega;\mathbb{R}^{3})$ for fixed $0<s\leq 1$ , $s\leq r\leq 2$ , and that there is a regularity shift $0<{s^{\prime}}\leq 1$ with

\|(\operatorname{div}\mathcal{C}\varepsilon)^{-1}g\|_{1+{s^{\prime}}}\leq c\|g\|\quad\forall g\in L_{2}(\Omega;\mathbb{R}^{3})

(inversion subject to the homogeneous Dirichlet condition) for a constant $c>0$ that is independent of $g$ and $\lambda\in(0,\infty)$ . Then the estimates

$\displaystyle\\|\sigma-\sigma_{h}\\|_{\operatorname{div},\mathcal{T}}+\\|u-u_{h}\\|+\\|\eta-\eta_{h}\\|_{1/2,t,\mathcal{S}}$	$\displaystyle\lesssim\\|h_{\mathcal{T}}\\|_{\infty}^{s}(\\|u\\|_{1+s}+\\|\sigma\\|_{s}+\\|f\\|_{s}),$	(14)
$\displaystyle\\|\operatorname{div}(\sigma-\sigma_{h})\\|_{\mathcal{T}}$	$\displaystyle\lesssim\\|h_{\mathcal{T}}\\|_{\infty}^{r}\\|f\\|_{r},$
$\displaystyle\\|u-u_{h}\\|$	$\displaystyle\lesssim\\|h_{\mathcal{T}}\\|_{\infty}^{s+{s^{\prime}}}\bigl{(}\\|u\\|_{1+s}+\\|f\\|_{s}\bigr{)}$

hold with intrinsic constants that are independent of $\mathcal{T}$ and $\lambda\in(0,\infty)$ .

Proof.

The proof is analogous to the proofs of Theorems 21 (quasi-optimal convergence) and 22 (approximation orders) in [7]. To this end we note that, for homogeneous Dirichlet data, the condition $u_{D}\in H^{1+s}(\Omega;\mathbb{R}^{3})$ with $s>0$ for a Dirichlet extension $u_{D}$ in [7, Theorem 21] is obsolete and that the tensors $\sigma_{0}=\sigma_{0,h}\in\SS$ appearing in [7, Theorem 22] vanish, cf. definitions (9) and (28) there. Specifically, Lemma 11 is critical for the robust stability of the discrete scheme and, having established the quasi-optimal convergence, error estimate (14) is consequence of Proposition 10. ∎

5 Numerical experiments

We use $\Omega=(0,1)^{3}$ and consider Dirichlet boundary conditions everywhere given by the manufactured displacement

u(x_{1},x_{2},x_{3})=\begin{pmatrix}\sin(3x_{1})\cos(3x_{2})\cos(3x_{3})\\ \cos(3x_{1})\sin(3x_{2})\cos(3x_{3})\\ \cos(3x_{1})\cos(3x_{3})\sin(3x_{3})\end{pmatrix},

and select $f:=-\operatorname{div}\sigma$ where $\sigma=\mathbb{C}\varepsilon(u)$ . We choose Young’s modulus $E=1$ and different values of $\nu$ . The Lamé parameters are $\mu=\frac{E}{2(1+\nu)}=\frac{1}{2(1+\nu)}$ and $\lambda=\frac{E\nu}{(1+\nu)(1-2\nu)}=\frac{\nu}{(1+\nu)(1-2\nu)}$ .

We use quasi-uniform meshes and plot the individual approximation errors $\|u-u_{h}\|$ (“u”), $\|\varepsilon(u)-\varepsilon(\widetilde{u}_{h})\|$ (“strain”), $\|\sigma-\sigma_{h}\|$ (“sigma”), and $\|\operatorname{div}(\sigma-\sigma_{h})\|_{\mathcal{T}}$ (“div sigma”), normalized by the norm of the exact solution $\bigl{(}\|u\|_{1}^{2}+\|\sigma\|_{\operatorname{div}}^{2}\bigr{)}^{1/2}$ . Here, $\widetilde{u}_{h}\in S^{1}_{0}(\mathcal{T};\mathbb{R}^{3})$ is the piecewise linear interpolation of the continuous, piecewise linear trace approximation $\eta_{h}$ . Note that $\|\varepsilon(u)-\varepsilon(\widetilde{u}_{h})\|$ is, up to the $L_{2}$ part $\|u-\widetilde{u}_{h}\|$ , an upper bound of $\|{\gamma_{1,\mathit{nn}}}(u)-\eta_{h}\|_{1/2,t,\mathcal{S}}$ . The right-hand side functional $(f\hskip 1.42262pt,\cdot)$ is approximated by a 4-point Gauss formula (integrating quadratic polynomials exactly), and the errors are calculated by a 14-point Gauss formula (an overkill to be close to the exact errors), cf. [19]. The non-homogeneous Dirichlet boundary condition is approximated by assigning the corresponding vertex values to trace approximation $\eta_{h}$ .

Figure 1(a) shows the results for $\nu=0.3$ , and confirms the approximation orders with respect to $h:=\|h_{\mathcal{T}}\|_{\infty}$ proved by Theorem 12: $O(h)$ for the stress and trace approximations, and $O(h^{2})$ for the approximations of the displacement and the divergence of the stress.

The sequence of results for $\nu=0.3,0.45,0.49,0.4999$ shown in Figure 1 confirms that our scheme is locking free. (The fact that some relative errors are initially smaller for larger values of $\lambda$ does not contradict this.) Note that $\operatorname{tr}(\sigma)=\operatorname{tr}(\mathbb{C}\varepsilon(u))=9(2\mu+3\lambda)\cos(3x_{1})\cos(3x_{2})\cos(3x_{3})$ and

\operatorname{dev}\sigma=-6\mu\begin{pmatrix}0&\sin(3x_{1})\sin(3x_{2})\cos(3x_{3})&\sin(3x_{1})\cos(3x_{2})\sin(3x_{3})\\ *&0&\cos(3x_{1})\sin(3x_{2})\sin(3x_{3})\\ *&*&0\end{pmatrix},

that is, $\|\operatorname{tr}(\sigma)\|$ and $(\operatorname{tr}(\sigma)\hskip 1.42262pt,1)$ are unbounded as $\nu\to 1/2$ while $\operatorname{dev}\sigma$ is bounded.

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	$\displaystyle\frac{1}{2}\\|f_{\pm}-g_{\pm}\\|_{1,U}^{2}$	$\displaystyle\leq\\|f_{\pm}\\|_{1,U}^{2}+\\|f_{\pm}+g_{1}+\ldots+g_{J}\\|_{U}^{2}+\\|\|\nabla f_{\pm}\|+\|\nabla g_{1}\|+\ldots\|\nabla g_{J}\|\\|_{U}^{2}$
		$\displaystyle\leq\\|f_{\pm}\\|_{1,U}^{2}+(J+1)\bigl{(}\\|f_{\pm}\\|_{1,U}^{2}+\\|g_{1}\\|_{1,U}^{2}+\ldots+\\|g_{J}\\|_{1,U}^{2}\bigr{)}$
		$\displaystyle=(J+2)\\|f_{\pm}\\|_{1,U}^{2}+(J+1)\Bigl{(}\\|g_{1}\\|_{1,U}^{2}+\ldots+\\|g_{J}\\|_{1,U}^{2}\Bigr{)}\to 0\quad(\delta\to 0)$

	$\displaystyle\\|\tau-\mathcal{I}^{nn}_{\mathcal{T}}\tau\\|$	$\displaystyle\lesssim\\|h_{\mathcal{T}}\\|_{\infty}^{s}\\|\tau\\|_{s}+\\|h_{\mathcal{T}}\\|_{\infty}\\|\operatorname{div}\tau\\|\quad$	$\displaystyle\forall\tau\in H^{s,0}(\operatorname{div},\Omega;\SS),$
	$\displaystyle\\|\operatorname{div}(\tau-\mathcal{I}^{nn}_{\mathcal{T}}\tau)\\|_{\mathcal{T}}$	$\displaystyle\lesssim\\|h_{\mathcal{T}}\\|_{\infty}^{r}\\|\operatorname{div}\tau\\|_{r}\quad$	$\displaystyle\forall\tau\in H^{s,r}(\operatorname{div},\Omega;\SS)$

$\displaystyle\\|\sigma-\sigma_{h}\\|_{\operatorname{div},\mathcal{T}}+\\|u-u_{h}\\|+\\|\eta-\eta_{h}\\|_{1/2,t,\mathcal{S}}$	$\displaystyle\lesssim\\|h_{\mathcal{T}}\\|_{\infty}^{s}(\\|u\\|_{1+s}+\\|\sigma\\|_{s}+\\|f\\|_{s}),$	(14)
$\displaystyle\\|\operatorname{div}(\sigma-\sigma_{h})\\|_{\mathcal{T}}$	$\displaystyle\lesssim\\|h_{\mathcal{T}}\\|_{\infty}^{r}\\|f\\|_{r},$
$\displaystyle\\|u-u_{h}\\|$	$\displaystyle\lesssim\\|h_{\mathcal{T}}\\|_{\infty}^{s+{s^{\prime}}}\bigl{(}\\|u\\|_{1+s}+\\|f\\|_{s}\bigr{)}$

Normal-normal continuous symmetric stress approximation in three-dimensional linear elasticity ††thanks: Supported by ANID-Chile through FONDECYT project 1230013

Abstract

1 Introduction and model problem

2 Analytical framework

2.1 Notation, spaces and trace operators

Proposition 1 (H~0,n1/2​(𝒮;ℝ3)\widetilde{H}^{1/2}_{0,n}(\mathcal{S};\mathbb{R}^{3})).

Proof.

Proposition 2 (H~0,t1/2​(𝒮;ℝ3)\widetilde{H}^{1/2}_{0,t}(\mathcal{S};\mathbb{R}^{3})).

Proof.

Proposition 3 (density).

Proof.

Proposition 4 (inclusion).

Proof.

2.2 Normal-normal continuous stress space and dual trace operator

Proposition 5 (duality).

Proof.

3 Mixed formulation

Theorem 6 (well-posedness of mixed formulation).

Proof.

4 Finite element approximation

4.1 An H​(div;SS)H(\operatorname{div};\SS) element

Lemma 7.

Proof.

Lemma 8.

Proof.

Proposition 9 (degrees of freedom).

Proof.

4.2 Approximation space

Proposition 10 (interpolation).

Proof.

Lemma 11.

4.3 Finite element scheme and locking-free convergence

Theorem 12 (locking-free convergence).

Proof.

5 Numerical experiments

References

Normal-normal continuous symmetric stress approximation in three-dimensional linear elasticity ^†^†thanks: Supported by ANID-Chile through FONDECYT project 1230013

Proposition 1 ( $\widetilde{H}^{1/2}_{0,n}(\mathcal{S};\mathbb{R}^{3})$ ).

Proposition 2 ( $\widetilde{H}^{1/2}_{0,t}(\mathcal{S};\mathbb{R}^{3})$ ).

4.1 An $H(\operatorname{div};\SS)$ element