This paper introduces a new family of non-conforming Crouzeix-Raviart finite elements in 3D, providing explicit basis functions, optimal estimates, and analysis of their properties on simplicial meshes.
Contribution
It develops a novel 3D non-conforming finite element family with explicit basis functions and theoretical analysis, extending prior 2D concepts.
Findings
01
Optimal a priori estimates established
02
Explicit local basis functions derived
03
Analysis of linear independence and spanning properties
Abstract
In this paper we will develop a family of non-conforming "Crouzeix-Raviart" type finite elements in three dimensions. They consist of local polynomials of maximal degree p∈N on simplicial finite element meshes while certain jump conditions are imposed across adjacent simplices. We will prove optimal a priori estimates for these finite elements. The characterization of this space via jump conditions is implicit and the derivation of a local basis requires some deeper theoretical tools from orthogonal polynomials on triangles and their representation. We will derive these tools for this purpose. These results allow us to give explicit representations of the local basis functions. Finally we will analyze the linear independence of these sets of functions and discuss the question whether they span the whole non-conforming space.
Tables2
Table 1. Figure 1 : Symmetric orthogonal polynomials on the reference triangle and corresponding tetrahedron-supported non-conforming basis functions.
Table 2. Figure 2 : Orthogonal polynomials of reflection type and corresponding non-conforming basis functions which are supported on two adjacent tetrahedrons.
The common facet is horizontal and the two tetrahedrons are on top of each other.
Equations361
Given f∈L2(Ω) find u∈H01(Ω)a(u,v):=(A∇u,∇v)=(f,v)∀v∈H01(Ω).
Given f∈L2(Ω) find u∈H01(Ω)a(u,v):=(A∇u,∇v)=(f,v)∀v∈H01(Ω).
b_{p,k}^{\operatorname*{sym}}:=\left\{\begin{array}[c]{ll}r_{p,p-2k}&\text{if }p\text{ is even,}\\
r_{p,p-1-2k}&\text{if }p\text{ is odd,}\end{array}\right.\qquad k=0,1,\ldots,d_{\operatorname*{triv}}\left(p\right)-1.
b_{p,k}^{\operatorname*{sym}}:=\left\{\begin{array}[c]{ll}r_{p,p-2k}&\text{if }p\text{ is even,}\\
r_{p,p-1-2k}&\text{if }p\text{ is odd,}\end{array}\right.\qquad k=0,1,\ldots,d_{\operatorname*{triv}}\left(p\right)-1.
B_{p,k}^{\widehat{K},\operatorname*{nc}}\left(\mathbf{N}\right):=\left\{\begin{array}[c]{ll}b_{p,k}^{\operatorname*{sym}}\circ\chi_{T}^{-1}\left(\mathbf{N}\right)&\forall\mathbf{N}\in\widehat{\mathcal{N}}^{p}\text{ s.t. }\mathbf{N}\in T\text{ for some facet }T\subset\partial\widehat{K},\\
0&\forall\mathbf{N}\in\widehat{\mathcal{N}}^{p}\backslash\partial\widehat{K}\end{array}\right.\qquad k=0,1,\ldots,d_{\operatorname*{triv}}\left(p\right)-1.
B_{p,k}^{\widehat{K},\operatorname*{nc}}\left(\mathbf{N}\right):=\left\{\begin{array}[c]{ll}b_{p,k}^{\operatorname*{sym}}\circ\chi_{T}^{-1}\left(\mathbf{N}\right)&\forall\mathbf{N}\in\widehat{\mathcal{N}}^{p}\text{ s.t. }\mathbf{N}\in T\text{ for some facet }T\subset\partial\widehat{K},\\
0&\forall\mathbf{N}\in\widehat{\mathcal{N}}^{p}\backslash\partial\widehat{K}\end{array}\right.\qquad k=0,1,\ldots,d_{\operatorname*{triv}}\left(p\right)-1.
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Full text
A Family of Crouzeix-Raviart Finite Elements in 3D
Patrick Ciarlet, Jr
([email protected]), POEMS
(CNRS/ENSTA ParisTech/INRIA), 828, Boulevard des Maréchaux, 91762
Palaiseau Cedex, France
Charles F. Dunkl
([email protected]), Dept. of Mathematics, University
of Virginia, Charlottesville, Virginia 22904-4137, USA
Stefan A. Sauter
([email protected]), Institut für Mathematik,
Universität Zürich, Winterthurerstr 190, CH-8057 Zürich,
Switzerland
Abstract
In this paper we will develop a family of non-conforming “Crouzeix-Raviart” type finite elements in three dimensions.
They consist of local polynomials of maximal degree p∈N on
simplicial finite element meshes while certain jump conditions are imposed
across adjacent simplices. We will prove optimal a priori estimates for these
finite elements.
The characterization of this space via jump conditions is implicit and the
derivation of a local basis requires some deeper theoretical tools from
orthogonal polynomials on triangles and their representation. We will derive
these tools for this purpose. These results allow us to give explicit
representations of the local basis functions. Finally we will analyze the
linear independence of these sets of functions and discuss the question
whether they span the whole non-conforming space.
For the numerical solution of partial differential equations, Galerkin finite
element methods are among the most popular discretization methods. In the last
decades, non-conforming Galerkin discretizations have become very
attractive where the test and trial spaces are not subspaces of the natural
energy spaces and/or the variational formulation is modified on the discrete
level. These methods have nice properties, e.g. in different parts of the
domain different discretizations can be easily used and glued together or, for
certain classes of problems (Stokes problems, highly indefinite Helmholtz and
Maxwell problems, problems with “locking”,
etc.), the non-conforming discretization enjoys a better stability behavior
compared to the conforming one. One of the first non-conforming finite element
space was the Crouzeix-Raviart element ([8], see
[3] for a survey). It is piecewise affine with respect to
a triangulation of the domain while interelement continuity is required only
at the barycenters of the edges/facets (2D/3D).
In [6], a family of high order non-conforming (intrinsic) finite
elements have been introduced which corresponds to a family of high-order
Crouzeix-Raviart elements in two dimensions. For Poisson’s equation, this
family includes the non-conforming Crouzeix-Raviart element
[8], the Fortin-Soulie element [11], the
Crouzeix-Falk element [7], and the Gauss-Legendre elements
[1], [15] as well as the standard conforming hp-finite elements.
In our paper we will characterize a family of high-order Crouzeix-Raviart type
finite elements in three dimensions, first implicitly by imposing
certain jump conditions at the interelement facets. Then we derive a local
basis for these finite elements. These new finite element spaces are
non-conforming but the (broken version of the) continuous bilinear form can
still be used. Thus, our results also give insights on how far one can go in
the non-conforming direction while keeping the original forms.
The explicit construction of a basis for these new finite element spaces
require some deeper theoretical tools in the field of orthogonal polynomials
on triangles and their representations which we develop here for this purpose.
As a simple model problem for the introduction of our method, we consider
Poisson’s equation but emphasize that this method is applicable also for much
more general (systems of) elliptic equations.
There is a vast literature on various conforming and non-conforming, primal,
dual, mixed formulations of elliptic differential equations and conforming as
well as non-conforming discretization. Our main focus is the characterization
and construction of non-conforming Crouzeix-Raviart type finite elements from
theoretical principles. For this reason, we do not provide an extensive list
of references on the analysis of specific families of finite elements spaces
but refer to the classical monographs [5], [14], and
[2] and the references therein.
The paper is organized as follows.
In Section 2 we introduce our model problem, Poisson’s
equation, the relevant function spaces and standard conditions on its well-posedness.
In Section 3 we briefly recall classical, conforming
hp-finite element spaces and their Lagrange basis.
The new non-conforming finite element spaces are introduced in Section
4. We introduce an appropriate compatibility condition
at the interfaces between elements of the mesh so that the non-conforming
perturbation of the original bilinear form is consistent with the local error
estimates. We will see that this compatibility condition can be inferred from
the proof of the second Strang lemma applied to our setting. The weak
compatibility condition allows to characterize the non-conforming family of
high-order Crouzeix-Raviart type elements in an implicit way. In this
section, we will also present explicit representations of non-conforming basis
functions of general degree p while their derivation and analysis is the
topic of the following sections.
Section 5 is devoted to the explicit construction of a basis
for these new non-conforming finite elements. It requires deeper theoretical
tools from orthogonal polynomials on triangles and their representation which
we will derive for this purpose in this section.
It is by no means obvious whether the constructed set of functions is linearly
independent and span the non-conforming space which was defined implicitly in
Section 4. These questions will be treated in Section
6.
Finally, in Section 7 we summarize the main results and give
some comparison with the two-dimensional case which was developed in
[6].
2 Model Problem
As a model problem we consider the Poisson equation in a bounded Lipschitz
domain Ω⊂Rd with boundary Γ:=∂Ω.
First, we introduce some spaces and sets of functions for the coefficient
functions and solution spaces.
The Euclidean scalar product in Rd is denoted for a,b∈Rd by a⋅b. For s≥0,
1≤p≤∞, let Ws,p(Ω) denote the classical
(real-valued) Sobolev spaces with norm ∥⋅∥Ws,p(Ω). The space W0s,p(Ω) is the closure with respect to the ∥⋅∥Ws,p(Ω) of all C∞(Ω) functions with compact support. As usual we write
Lp(Ω) short for W0,p(Ω). The
scalar product and norm in L2(Ω) are denoted by
(u,v):=∫Ωuv and ∥⋅∥:=(⋅,⋅)1/2. For p=2, we use Hs(Ω), H0s(Ω) as shorthands for
Ws,2(Ω), W0s,2(Ω). The
dual space of H0s(Ω) is denoted by H−s(Ω). We recall that, for positive integers s, the seminorm
∣⋅∣Hs(Ω) in Hs(Ω) which contains only the derivatives of order s is a norm in
H0s(Ω).
We consider the Poisson problem in weak form:
[TABLE]
Throughout the paper we assume that the diffusion matrix A∈L∞(Ω,Rsymd×d) is symmetric and satisfies
[TABLE]
and that there exists a partition P:=(Ωj)j=1J of
Ω into J (possibly curved) polygons (polyhedra for d=3) such
that, for some appropriate r∈N, it holds
[TABLE]
Assumption (2) implies the well-posedness of problem (1)
via the Lax-Milgram lemma.
3 Conforming hp-Finite Element Galerkin
Discretization
In this paper we restrict our studies to bounded, polygonal (d=2) or
polyhedral (d=3) Lipschitz domains Ω⊂Rd and regular
finite element meshes G (in the sense of [5])
consisting of (closed) simplices K, where hanging nodes are not allowed. The
local and global mesh width is denoted by hK:=diamK and
h:=maxK∈GhK. The boundary of a simplex K can be split
into (d−1)-dimensional simplices (facets for d=3 and
triangle edges for d=2) which are denoted by T. The set of all facets in
G is called F; the set of facets lying on
∂Ω is denoted by F∂Ω and defines a
triangulation of the surface ∂Ω. The set of facets in Ω
is denoted by FΩ. As a convention we assume that simplices
and facets are closed sets. The interior of a simplex K is denoted by
K∘ and we write T∘ to denote the
(relative) interior of a facet T. The set of all simplex vertices in the
mesh G is denoted by V, those lying on ∂Ω by V∂Ω, and those lying in Ω by
VΩ. Similar the set of simplex edges in G is
denoted by E, those lying on ∂Ω by E∂Ω, and those lying in Ω by EΩ.
We recall the definition of conforming hp-finite element spaces (see, e.g.,
[14]). For p∈N0:={0,1,…}, let
Ppd denote the space of d-variate polynomials of total
degree ≤p. For a connected subset ω⊂Ω, we write
Pdp(ω) for polynomials of degree ≤p
defined on ω. For a connected m-dimensional manifold ω⊂Rd, for which there exists a subset ω^∈Rm along an affine bijection χω:ω^→ω, we set Ppm(ω):={v∘χω−1:v∈Ppm(ω^)}. If the dimension m is clear from the context, we
write Pp(ω) short for Ppm(ω).
The conforming hp-finite element space is given by
[TABLE]
A Lagrange basis for SG,cp can be defined as
follows. Let
[TABLE]
denote the equispaced unisolvent set of nodal points on the d-dimensional
unit simplex
[TABLE]
For a simplex K∈G, let χK:K→K
denote an affine mapping. The set of nodal points is given by
[TABLE]
The Lagrange basis for SG,cp can be indexed
by the nodal points N∈NΩp and is
characterized by
[TABLE]
where δN,N′ is the Kronecker delta.
Definition 1
For all K∈G, T∈FΩ,
E∈EΩ, V∈VΩ, the
conforming spaces SK,cp, ST,cp, SE,cp, SV,cp are
given as the spans of the following basis functions
[TABLE]
The following proposition shows that these spaces give rise to a direct sum
decomposition and that these spaces are locally defined. To be more specific
we first have to introduce some notation.
For any facet T∈FΩ, vertex V∈VΩ, and E∈EΩ we define the sets
[TABLE]
Proposition 2
Let SK,cp,
ST,cp, SE,cp, SV,cp be as in Definition 1. Then the
direct sum decomposition holds
[TABLE]
4 Galerkin Discretization with Non-Conforming Crouzeix-Raviart Finite
Elements
4.1 Non-Conforming Finite Elements with Weak Compatibility Conditions
In this section, we will characterize a class of non-conforming finite element
spaces implicitly by a weak compatibility condition across the facets. For
each facet T∈F, we fix a unit vector nT which is
orthogonal to T. The orientation for the inner facets is arbitrary but fixed
while the orientation for the boundary facets is such that nT
points toward the exterior of Ω. Our non-conforming finite element
spaces will be a subspace of
[TABLE]
and we consider the skeleton T∈F⋃T as a set of measure zero.
For K∈G, we define the restriction operator γK:CG0(Ω)→C0(K) by
[TABLE]
and on the boundary ∂K by continuous extension. For the inner facets
T∈F, let KT1,KT2 be the two simplices which share
T as a common facet with the convention that nT points into
K2. We set ωT:=KT1∪KT2. The jump [⋅]T:CG0(Ω)→C0(T) across T is defined by
[TABLE]
For vector-valued functions, the jump is defined component-wise. The
definition of the non-conforming finite elements involves orthogonal
polynomials on triangles which we introduce first.
Let T denote the (closed) unit simplex in Rd−1, with
vertices 0, (1,0,…,0)⊺, (0,1,0,…,0)⊺, (0,…,0,1)⊺. For n∈N0, the set of orthogonal polynomials on
T is given by
[TABLE]
We lift this space to a facet T∈F by employing an affine
transform χT:T→T
[TABLE]
The orthogonal polynomials on triangles allows us to formulate the
weak compatibility condition which is employed for the definition of
non-conforming finite element spaces:
[TABLE]
We have collected all ingredients for the (implicit) characterization of the
non-conforming Crouzeix-Raviart finite element space.
Definition 3
The non-conforming finite element space SGp with weak compatibility conditions across facets is given by
[TABLE]
The non-conforming Galerkin discretization of (1) for a given
finite element space S which satisfies SG,ncp⊂S⊂SGp reads:
[TABLE]
where
[TABLE]
4.2 Non-Conforming Finite Elements of Crouzeix-Raviart Type in
3D
The definition of the non-conforming space SGp in
(14) is implicit via the weak compatibility condition. In this
section, we will present explicit representations of non-conforming basis
functions of Crouzeix-Raviart type for general polynomial order p. These
functions together with the conforming basis functions span a space
SG,ncp which satisfies the inclusions
SG,cp⊊SG,ncp⊆SGp (cf. Theorem
10). The derivation of the formula and their
algebraic properties will be the topic of the following sections.
We will introduce two types of non-conforming basis functions: those whose
support is one tetrahedron and those whose support consists of two adjacent
tetrahedrons, that is tetrahedrons which have a common facet. For details and
their derivation we refer to Section 5 while here we focus
on the representation formulae.
4.2.1 Non-Conforming Basis Functions Supported on One Tetrahedron
The construction starts by defining symmetric orthogonal polynomialsbp,ksym, 0≤k≤dtriv(p)−1 on the reference triangle T with vertices (0,0)⊺, (1,0)⊺, (0,1)⊺, where
[TABLE]
We define the coefficients
[TABLE]
where pFq denotes the generalized hypergeometric function (cf.
[9, Chap. 16]). The 4F3-sum is understood to terminate at
i to avoid the 0/0 ambiguities in the formal 4F3-series. These
coefficients allow to define the polynomials
[TABLE]
where bp,k, 0≤k≤p, are the basis for the orthogonal polynomials
of degree p on T as defined afterwards in (35). Then,
a basis for the symmetric orthogonal polynomials is given by
[TABLE]
The non-conforming Crouzeix-Raviart basis function Bp,kK,nc∈Pp(K) on the
unit tetrahedron K is characterized by its values at the nodal
points in Np (cf. (5)). For a
facet T⊂∂K, let χT:T→T
denote an affine pullback to the reference triangle. Then Bp,kK,nc∈Pp(K) is uniquely defined by
[TABLE]
Remark 4
In Sec. 5.3, we will prove that the polynomials
bp,ksym are totally symmetric, i.e., invariant under
affine bijections χ:K→K. Thus, any of these
functions can be lifted to the facets of a tetrahedron via affine pullbacks
and the resulting function on the surface is continuous. As a consequence, the
value Bp,kK,nc(N) in
definition (18) is independent of the choice of T also for
nodal points N which belong to different facets.
It will turn out that the value [math] at the inner nodes could be replaced by
other values without changing the arising non-conforming space. Other choices
could be preferable in the context of inverse inequalities and the condition
number of the stiffness matrix. However, we recommend to choose these values
such that the symmetries of Bp,kK,nc are preserved.
Definition 5
The non-conforming tetrahedron-supported basis functions
on the reference element are given by
[TABLE]
with values Bp,kK,nc(N) as in (18). For a simplex K∈G the
corresponding non-conforming basis functions Bp,kK,nc
are given by lifting Bp,kK,nc via an affine
pullback χK from K to K∈G:
[TABLE]
and span the space
[TABLE]
Example 6
The lowest order of p such that dtriv(p)≥1 is p=2. In this case, we get dtriv(p)=1. In Figure 1 the function bp,ksym and corresponding basis functions Bp,kK,nc are depicted for (p,k)∈{(2,0),(3,0),(6,0),(6,1)}.
4.2.2 Non-Conforming Basis Functions Supported on Two Adjacent
Tetrahedrons
The starting point is to define orthogonal polynomials bp,krefl on the reference triangle T which are
mirror symmetric111The superscript “refl” is a shorthand for “reflection” and explained in Section 5.3.1. with
respect to the angular bisector in T through 0 and
linear independent from the fully symmetric functions bp,ksym. We set
[TABLE]
where
[TABLE]
Let K1, K2 denote two tetrahedrons which share a common facet, say
T. The vertex of Ki which is opposite to T is denoted by
Vi. The procedure of lifting the nodal values to the facets of
ωT:=K1∪K2 is analogous as for the basis functions
Bn,kK,nc. However, it is necessary to choose the
pullback χi,T~:T→T~ of a facet
T~⊂∂Ki\T∘ such that the
origin is mapped to Vi.
[TABLE]
Again, the value [math] at the inner nodes of ωT could be replaced by
other values without changing the arising non-conforming space.
Definition 7
The non-conforming facet-oriented basis functions are
given by
[TABLE]
with values Bp,kT,nc(N) as in
(23) and span the space
[TABLE]
The non-conforming finite element space of Crouzeix-Raviart type is given by
[TABLE]
Remark 8
In Sec. 5.3.3, we will show that the polynomials bp,krefl are mirror symmetric with respect to the angular
bisector in T through 0. Thus, any of these functions
can be lifted to the outer facets of two adjacent tetrahedrons via (oriented)
affine pullbacks as employed in (23) and the resulting
function on the surface is continuous. As a consequence, the value
Bp,kT,nc(N) in definition
(23) is independent of the choice of T also for nodal
points N which belong to different facets.
In Theorem 33, we will prove that (26), in fact, is a
direct sum and a basis is given by the functions
[TABLE]
Also we will prove that SG,cp⊊SG,ncp⊆SGp. This
condition implies that the convergence estimates as in Theorem
10 are valid for this space. We restricted the
reflection-type non-conforming basis functions to the lowest order k=0 in
order to keep the functions linearly independent.
Example 9
The lowest order of p such that drefl(p)≥1 is p=1. In this case, we get drefl(p)=1. In Figure 2 the function bp,krefl and corresponding basis functions Bp,kT,nc are depicted for (p,k)∈{(1,0),(2,0),(4,0),(4,1)}.
4.3 Error Analysis
In this subsection we present the error analysis for the Galerkin
discretization (15) with the non-conforming finite element space
SGp and subspaces thereof. The analysis is based on the
second Strang lemma and has been presented for an intrinsic version of
SGp in [6].
For any inner facet T∈F and any v∈SGp,
condition (13) implies ∫T[v]T=0 : hence,
the jump [v]T is always zero-mean valued. Let hT
denote the diameter of T. The combination of a Poincaré inequality with
a trace inequality then yields
[TABLE]
where
[TABLE]
In a similar fashion we obtain for all boundary facets T∈F∂Ω and all u∈SGp the estimate
[TABLE]
We say that the exact solution {u}\in H$${}_{0}^{1}\left(\Omega\right)
is piecewise smooth over the partition P=(Ωj)j=1J, if there exists some positive integer s such that
[TABLE]
We write u\in$$PH^{1+s}(\Omega) and refer for further properties and
generalizations to non-integer values of s, e.g., to [13, Sec.
4.1.9].
For the approximation results, the finite element meshes G are
assumed to be compatible with the partition P in the
following sense: for all K∈G, there exists a single index j
such that \overset{\circ}{{K}}$$\cap\Omega_{j}\neq\emptyset.
The proof that ∣⋅∣Hpw1(Ω) is a norm on SGp is similar as
in [4, Sect. 10.3]: For w∈H01(Ω) this follows from ∣w∣Hpw1(Ω)=∥∇w∥ and a Friedrichs
inequality; for w∈SGp the condition ∥∇Gw∥=0 implies that w∣K
is constant on all simplices K∈G. The combination with ∫Tw=0 for all T∈F∂Ω leads to w∣K=0 for the outmost simplex layer via a Poincaré
inequality, i.e., w∣K=0 for all K∈G
having at least one facet on ∂Ω. This argument can be iterated
step by step over simplex layers towards the interior of Ω to finally
obtain w=0.
Theorem 10
Let Ω⊂Rd be a
bounded, polygonal (d=2) or polyhedral (d=3) Lipschitz domain and let
G be a regular simplicial finite element mesh for Ω. Let
the diffusion matrix A∈L∞(Ω,Rsymd×d) satisfy assumption (2)
and let f∈L2(Ω). As an additional assumption on
the regularity, we require that the exact solution of (1)
satisfies u∈PH1+s(Ω) for some positive integer
s and ∥A∥PWr,∞(Ω)<∞ holds with r:=min{p,s}. Let the
continuous problem (1) be discretized by the non-conforming
Galerkin method (15) with a finite dimensional space S which
satisfies SG,cp⊂S⊂SGp on a compatible mesh G. Then,
(15) has a unique solution which satisfies
[TABLE]
The constant C only depends on amin, amax, ∥A∥PWr,∞(Ω), p, r,
and the shape regularity of the mesh.
Proof. The second Strang lemma (cf. [5, Theo. 4.2.2]) applied to the
non-conforming Galerkin discretization (15) implies the existence
of a unique solution which satisfies the error estimate
[TABLE]
where
[TABLE]
The approximation properties of S are inherited from the approximation
properties of SG,cp in the first infimum
because of the inclusion SG,cp⊂S. For
the second term we obtain
[TABLE]
Note that f∈L2(Ω) implies that div(A∇u)∈L2(Ω) and, in
turn, that the normal jump [A∇u⋅nT]T equals zero and the restriction (A∇u⋅nT)∣T is well
defined for all T∈F. We may apply simplexwise integration by
parts to (29) to obtain
[TABLE]
Let KT be one simplex in ωT. For 1≤i≤d, let
qi∈Pdp−1(KT) denote the best
approximation of wi:=(∑j=1dAi,j∂ju)KT with respect to the H1(KT) norm. Then, qi∣TnT,i∈Pd−1p−1(T) for 1≤i≤d, and the
inclusion S⊂SGp implies
[TABLE]
Standard trace estimates and approximation properties lead to
[TABLE]
where C depends only on p, r, ∥A∥Wr(KT), and the shape regularity of the mesh.The
combination of (30), (31) and (27),(28) along with the shape regularity of the mesh leads to
the consistency estimate
[TABLE]
which completes the proof.
Remark 11
If one chooses in (13) a degree p′<p for the
orthogonality relations in (13), then the order of convergence
behaves like hr′∥e∥H1+r′(Ω), with r′:=min{p′,s},
because the best approximations qi now belong to Pd−1p′−1(T).
5 Explicit Construction of Non-Conforming Crouzeix-Raviart Finite
Elements
5.1 Jacobi Polynomials
Let α,β>−1. The Jacobi polynomialPn(α,β) is a polynomial of degree n such that
[TABLE]
for all polynomials q of degree less than n, and (cf. [9, Table
18.6.1])
[TABLE]
Here the shifted factorial is defined by (a)n:=a(a+1)…(a+n−1) for n>0 and
(a)0:=1. The Jacobi polynomial has an explicit expression
in terms of a terminating Gauss hypergeometric series (see (cf.
[9, 18.5.7]))
[TABLE]
as follows
[TABLE]
5.2 Orthogonal Polynomials on Triangles
Recall that T is the (closed) unit triangle in R2
with vertices A0=(0,0)⊺,
A1=(1,0)⊺, and
A3=(0,1)⊺. An orthogonal
basis for the space Pn,n−1⊥(T)
was introduced in [12] and is given by the functions bn,k, 0≤k≤n,
[TABLE]
where Pk(0,0) are the Legendre polynomials (see
[9, 18.7.9])222The Legendre polynomials with normalization
Pk(0,0)(1)=1 for all k=0,1,… can
be defined [9, Table 18.9.1] via the three-term recursion
from which the well-known relation Pk(0,0)(x)=(−1)kPk(0,0)(x) for all k∈N0 follows.. From (36)
(footnote) it follows that these polynomials satisfy the following symmetry
relation
[TABLE]
By combining (33) - (35), an elementary calculation leads
to333Further special values are
denote the edges of T. For Z∈{I,II,III}, we introduce
the linear restriction operator for the edge EZ by
γZ:C0(T)→C0([0,1]) by
[TABLE]
which allows to define
[TABLE]
Lemma 12
For any Z∈{I,II,III}, each of the systems (bn,kZ)k=0n, form a basis of Pn([0,1]).
Proof. First note that {xj(x−1)n−j:0≤j≤n} is a basis for Pn([0,1]); this follows from expanding the right-hand side of xm=xm(x−(x−1))n−m. Specialize the formula
[9, 18.5.8]
[TABLE]
to m=n−k, α=0,β=2k+1,s=2x−1 to obtain
[TABLE]
The highest index i of xn−i(x−1)i in bn,kI(x) is n−k with coefficient
(n−k)!(2k+2)n−k=0. Thus the
matrix expressing [bn,0I,…,bn,nI] in terms of [(x−1)n,x(x−1)n−1,…,xn] is triangular and
nonsingular; hence {bn,kI:0≤k≤n} is a basis of Pn([0,1]). The
symmetry relation bn,kII=(−1)kbn,kI for 0≤k≤n (cf. (37)) shows
that {bn,kII:0≤k≤n} is also a
basis of Pn([0,1]). Finally
substituting x1=1−x,x2=x in bn,k results in
[TABLE]
and Pn−k(0,2k+1)(1)=1 (from
(32)). Clearly {Pk(0,0)(1−2x):0≤k≤n} is a basis for Pn([0,1]).
Lemma 13
Let v∈Pn([0,1]). Then, there exist unique orthogonal polynomials uZ∈Pn,n−1⊥(T), Z∈{I,II,III}
with v=γZuZ. Thus, the linear
extension operator EZ:Pn([0,1])→Pn,n−1⊥(T) is well defined by EZv:=uZ.
Proof. From Lemma 12 we conclude that γZ
is surjective. Since the polynomial spaces are finite dimensional the
assertion follows from
[TABLE]
The orthogonal polynomials can be lifted to a general triangle T.
Definition 14
Let T denote a triangle and χT an affine pullback to the reference
triangle T. Then, the space of orthogonal polynomials of degree
n on T is
[TABLE]
From the transformation rule for integrals one concludes that for any
u=v∘χT−1∈Pn,n−1⊥(T) and all
q∈Pn−1(T) it holds
[TABLE]
since q∘χT∈Pn−1(T). Here
∣T∣ denotes the area of the triangle T.
5.3 Totally Symmetric Orthogonal Polynomials
In this section, we will decompose the space of orthogonal polynomials
Pn,n−1⊥(T) into three
irreducible modules (see §5.3.1) and thus, obtain a direct sum
decomposition Pn,n−1⊥(T)=Pn,n−1⊥,sym(T)⊕Pn,n−1⊥,refl(T)⊕Pn,n−1⊥,sign(T). We will derive an explicit representation for a basis
of the space of totally symmetric polynomials Pn,n−1⊥,sym(T) in
§5.3.2 and of the space of reflection symmetric polynomials
Pn,n−1⊥,refl(T)
in §5.3.3.
We start by introducing, for functions on triangles, the notation of total
symmetry. For an arbitrary triangle T with vertices A0,
A1, A2, we introduce the set of permutations
Π={(i,j,k):i,j,k∈{0,1,2} pairwise disjoint}. For π=(i,j,k)∈Π, define
the affine mapping χπ:T→T by
[TABLE]
We say a function u, defined on T, has total symmetry if
[TABLE]
The space of totally symmetric orthogonal polynomials is
[TABLE]
The construction of a basis of Pn,n−1⊥,sym(T) requires some algebraic tools which we develop
in the following.
5.3.1 The decomposition of Pn,n−1⊥(T) or Pn([0,1])
into irreducible S3 modules
We use the operator γI (cf. (39)) to set up an action of the symmetric group S3 on
Pn([0,1]) by transferring its
action on Pn,n−1⊥(T) on the
basis {bn,k}. It suffices to work with two generating
reflections. On the triangle χ{0,2,1}(x1,x2)=(x2,x1) and thus bn,k∘χ{0,2,1}=(−1)kbn,k (this follows
from (37)). The action of χ{0,2,1} is
mapped to ∑k=0nαkbn,kI↦∑k=0n(−1)kαkbn,kI, and
denoted by R. For the other generator we use χ{1,0,2}(x1,x2)=(1−x1−x2,x2). Under
γI this corresponds to the map ∑k=0nαkbn,kI(x)↦∑k=0nαkbn,kI(1−x) which is denoted
by M. We will return later to transformation formulae expressing
[TABLE]
in the {bn,k}-basis. Observe that (MR)3=I because χ{1,0,2}∘χ{0,2,1}(x1,x2)=(1−x1−x2,x1) and this mapping is of period 3. It follows that each of
{M,R} and {χ{1,0,2},χ{0,2,1}} generates (an isomorphic copy of)
S3. It is a basic fact that the relations M2=I,R2=I and
(MR)3=I define S3. The representation
theory of S3 informs us that there are three nonisomorphic
irreducible representations:
[TABLE]
(The subscript “refl”* *designates the *reflection representation). *Then the eigenvectors of σ1,σ2
with −1 as eigenvalue are (−1,0)⊺ and (2,−3)⊺ respectively; these two vectors are a basis for
R2. Similarly the eigenvectors of σ1 and σ2
with eigenvalue +1, namely (0,1)⊺, (2,1)⊺, form a basis. Form a direct sum
[TABLE]
where the Ej(triv),Ej(sign),Ej(refl)
are S3-irreducible and realizations of the representations
τtriv,τsign,τrefl respectively. Let dtriv(n),dsign(n),drefl(n) denote the respective multiplicities, so that
dtriv(n)+dsign(n)+2drefl(n)=n+1. The case n even
or odd are handled separately. If n=2m is even then the number of
eigenvectors of R having −1 as eigenvalue equals m (the cardinality of
{1,3,5,…,2m−1}). The same property holds for M since
the eigenvectors of M in the basis {x2m(x−1)2m−j} are explicitly given by {x2m−2ℓ(x−1)2ℓ−x2ℓ(x−1)2m−2ℓ:0≤ℓ≤m}. Each Ej(refl) contains
one (−1)-eigenvector of χ{1,0,2} and
one of χ{0,2,1} and each Ej(sign) consists of one (−1)-eigenvector of χ{0,2,1}. This gives the equation
drefl(n)+dsign(n)=m. Each Ej(refl) contains
one (+1)-eigenvector of χ{1,0,2} and
one of χ{0,2,1} and each Ej(triv) consists of one (+1)-eigenvector of χ{0,2,1}. There are m+1
eigenvectors with eigenvalue 1 of each of χ{1,0,2}
and χ{0,2,1} thus drefl(n)+dtriv(n)=m+1.
If n=2m+1 is odd then the eigenvector multiplicities are m+1 for both
eigenvalues +1,−1. By similar arguments we obtain the equations
drefl(n)+dsign(n)=m+1, drefl(n)+dtriv(n)=m+1. It remains to find one last
relation for both, even and odd cases.
To finish the determination of the multiplicities dtriv(n),dsign(n),drefl(n) it suffices to find
dtriv(n). This is the dimension of the
space of polynomials in Pn,n−1⊥(T) which are invariant under both χ{0,2,1} and
χ{1,0,2}. Since these two group elements generate
S3 this is equivalent to being invariant under each element of
S3 .This property is called totally symmetric. Under
the action of γI this corresponds to the space of
polynomials in Pn([0,1]) which are
invariant under both R and M. We appeal to the classical theory of
symmetric polynomials: suppose S3 acts on polynomials in
(y1,y2,y3) by permutation of coordinates then the
space of symmetric (invariant under the group) polynomials is exactly the
space of polynomials in {e1,e2,e3} the elementary
symmetric polynomials, namely e1=y1+y2+y3, e2=y1y2+y1y3+y2y3, e3=y1y2y3. To apply this we set up
an affine map from T to the triangle in R3 with
vertices (2,−1,−1), (−1,2,−1), (−1,−1,2). The formula for the map is
[TABLE]
The map takes (0,0),(1,0),(0,1)
to the three vertices respectively. The result is
[TABLE]
Thus any totally symmetric polynomial on T is a linear combination
of e2ae3b with uniquely determined coefficients. The number of
linearly independent totally symmetric polynomials in (j=1⨁nPn,n−1⊥(T))⊕P0(T) equals the number of solutions of
0≤2a+3b≤n with a,b=0,1,2,…. As a consequence
dtriv(n)=card{(a,b):2a+3b=n}. This number is the coefficient of
tn in the power series expansion of
[TABLE]
From dtriv(n)=card({0,2,4,…}∩{n,n−3,n−6,…}) we deduce the formula (cf. (16))
[TABLE]
As a consequence: if n=2m then dsign(n)=dtriv(n)−1 and drefl(n)=m+1−dtriv(n); if
n=2m+1 then dsign(n)=dtriv(n) and drefl(n)=m+1−dtriv(n). From this
the following can be derived: dsign(n)=⌊2n−1⌋−⌊3n−1⌋ and drefl(n)=⌊3n+2⌋. Here is a table of values in terms
of nmod6:
[TABLE]
5.3.2 Construction of totally symmetric polynomials
Let M and R denote the linear maps Mp(x1,x2):=p(1−x1−x2,x2) and Rp(x1,x2):=p(x2,x1) respectively. Both are automorphisms of
Pn,n−1⊥(T). Note Mp=p∘χ{1,0,2} and Rp=p∘χ{0,2,1} (cf. Section 5.3.1).
Proposition 15
Suppose 0≤k≤n then
[TABLE]
Proof. The 4F3-sum is understood to terminate at k to avoid the 0/0
ambiguities in the formal 4F3-series. The first formula was shown in
Section 5.3.1. The second formula is a specialization of
transformations in [10, Theorem 1.7(iii)]: this paper used the
shifted Jacobi polynomial Rm(α,β)(s)=(α+1)mm!Pm(α,β)(1−2s). Setting α=β=γ=0 in the
formulas in [10, Theorem 1.7(iii)] results in bn,k=(−1)kk!(n−k)!θn,k and
Mbn,k=k!(n−k)!ϕn,k, where θn,k,
ϕn,k are the polynomials introduced in [10, p.690]. More
precisely, the arguments v1,v2,v3 in θn,k and ϕn,k are specialized to v1=x1,v2=x2 and v3=1−x1−x2.
Proposition 16
The range of I+RM+MR is exactly the subspace {p∈Pn,n−1⊥(T):RMp=p}.
Proof. By direct computation (MR)3=I (cf. Section 5.3.1). This implies (RM)2=MR. If p satisfies RMp=p then
Mp=Rp and p=MRp. Now suppose RMp=p then (I+RM+MR)31p=p; hence p is in the range of I+RM+MR. Conversely suppose
p=(I+RM+MR)p′ for some polynomial p′, then,
RM(I+RM+MR)p′=(RM+(RM)2+I)p′=p.
Let Mi,j(n),Ri,j(n) denote the
matrix entries of M,R with respect to the basis {bn,k:0≤k≤n}, respectively (that is Mbn,k=∑j=0nbn,jMj,k(n)) . Let Si,j(n)
denote the matrix entries of MR+RM+I. Then
[TABLE]
Thus Si,j(n)=2Mi,j(n)+δi,j
if both i,j are even, Si,j(n)=−2Mi,j(n)+δi,j if both i,j are odd , and Si,j(n)=0 if i−j≡1mod2.
Corollary 17
For 0≤k≤2n each polynomial rn,2k:=20≤j≤n/2∑M2j,2k(n)bn,2j+bn,2k is totally
symmetric and for 0≤k≤2n−1 each polynomial rn,2k+1=−20≤j≤(n−1)/2∑M2j+1,2k+1(n)bn,2j+1+bn,2k+1 satisfies Mp=−p=Rp (the sign representation).
Proof. The pattern of zeroes in [Mi,j(n)] shows
that rn,2k=(MR+RM+I)bn,2k∈span{bn,2j} and thus satisfies Rrn,2k=rn,2k; combined with
RMrn,2k=rn,2k this shows rn,2k is totally symmetric. A similar
argument applies to (MR+RM+I)bn,2k+1.
Theorem 18
The functions bn,ksym, 0≤k≤dtriv(n)−1, as in (17) form a
basis for the totally symmetric polynomials in Pn,n−1⊥(T).
Proof. We use the homogeneous form of the bn,m as in [10], that
is, set
[TABLE]
Formally bn,j′(v)=(−1)j(j!(n−j)!)−1θn,j(v) with
θn,j as in [10, p.690]. The expansion of such a
polynomial is a sum of monomials v1n1v2n2v3n3
with ∑i=13ni=n. Symmetrizing the monomial results in the sum of
v1m1v2m2v3m3 where (m1,m2,m3) ranges over all permutations of (n1,n2,n3). The argument is based on the occurrence of certain indices
in bn,m. For a more straightforward approach to the coefficients we use
the following expansions (with ℓ=n−2k,β=2k+1):
[TABLE]
and
[TABLE]
First let n=2m. The highest power of v3 that can occur in
b2m,2m−2k′ is 2k, with corresponding coefficient (2k)!(4m−4k+1)2k∑j=02m−2kcjv2jv12m−j for certain coefficients {cj}. Recall
that dtriv(n) is the number of solutions
(i,j) of the equation 3j+2i=2m (with i,j=0,1,2,…).
The solutions can be listed as (m,0),(m−3,2),(m−6,4)…(m−3ℓ,2ℓ) where
ℓ=dtriv(n)−1. By hypothesis (m−3k,2k) occurs in the list and thus m−3k≥0 and m−k≥2k.
There is only one possible permutation of v1m−kv2m−kv32k
that occurs in b2m,2m−2k′ and its coefficient is (2m−2k)!(2k−2m)m−k3=0. Hence there is a
triangular pattern for the occurrence of v1mv2m, v1m−1v2m−1v32, v1m−2v2m−2v34,…in the
symmetrizations of b2m,2m′, b2m,2m−2′, … with
nonzero numbers on the diagonal and this proves the basis property when n=2m.
Now let n=2m+1. The highest power of v3 that can occur in
b2m+1,2m−2k′ is 2k+1, with coefficient (2k+1)!(4m−4k+1)2k+1∑j=02m−2kcjv2jv12m−j for certain coefficients {cj}.
The solutions of 3j+2i=2m+1 can be listed as (m−1,1),(m−4,3),(m−7,5)…(m−1−3ℓ,2ℓ+1)
where ℓ=dtriv(n)−1. By hypothesis
(m−1−3k,2k+1) occurs in this list, thus m−k≥2k+1. There
is only one possible permutation of v1m−kv2m−kv32k+1 that
occurs in b2m+1,2m−2k′ and its coefficient is (2m−2k)!(2k−2m)m−k3=0. As above, there is a
triangular pattern for the occurrence of v1mv2mv3,
v1m−1v2m−1v33, v1m−2v2m−2v35,… in
the symmetrizations of b2m+1,2m′, b2m+1,2m−2′,
… with nonzero numbers on the diagonal and this proves the basis
property when n=2m+1.
The totally symmetric orthogonal polynomials can be lifted to a general
triangle T.
Definition 19
Let T denote a triangle. The space of totally symmetric,
orthogonal polynomials of degree n is
[TABLE]
where the lifted symmetric basis functions are given by bn,mT,sym:=bn,msym∘χT−1 for
bn,msym as in Theorem 18 and an affine
pullback χT:T→T.
5.3.3 A Basis for the τrefl component of
Pn,n−1⊥(T)
As explained in Section 5.3.1 the space Pn,n−1⊥(T) can be decomposed into the τtriv-, the τsign- and the
τrefl-component. A basis for the τtriv component are the fully symmetric basis functions
(cf. Section 5.3.2).
Next, we will construct a basis for all of Pn,n−1⊥(T) by extending the totally symmetric one. It is
straightforward to adjoin the dsign(n)
basis, using the same technique as for the fully symmetric ones: the monomials
which appear in p with Rp=−p=Mp must be permutations of v1n1v2n2v3n3 with n1>n2>n3. As in Theorem
18 for n=2m argue on monomials v1m−kv2m−1−kv32k+1 and the polynomials b2m,2m−2k−1′ with 0≤k≤dsign(n)−1=dtriv(n)−2, and for n=2m+1 use the monomials v1m+1−kv2m−kv32k and b2m+1,2m−2k with 0≤k≤dtriv(n)−1=dsign(n)−1.
As we will see when constructing a basis for the non-conforming finite element
space, the τsign component of Pn,n−1⊥(T) is not relevant, in contrast to
the τrefl component. In this section, we will
construct a basis for the τrefl polynomials in
Pn,n−1⊥(T). Each such polynomial
is an eigenvector of RM+MR with eigenvalue −1. We will show that the
polynomials
[TABLE]
are linearly independent (and the same as introduced in (21))
and, subsequently, that the set
[TABLE]
is a basis for the τrefl subspace of Pn,n−1⊥(T). (The upper limit of k is as
in (52) drefl(n)−1 (cf.
(22)).) Note that
[TABLE]
because (RM)2=MR. Thus the calculation of these
polynomials follows directly from the formulae for [Mij]
and [Rij]. The method of proof relies on complex
coordinates for the triangle.
Lemma 20
For k=0,1,2,…
[TABLE]
Proof. Start with the formula (specialized from a formula for Gegenbauer polynomials
[9, 18.5.10])
with t=1/s2; then t−1t=1−s21 and s2k(1−s21)k=(−1)k(1−s2)k. Also 22k(2k)!(21)2k=k!(21)k(21)2k=k!(k+21)k. This proves the first
formula. Set s=v1+v2v1−v2 then 1−s2=(v1+v2)24v1v2 to obtain the second one.
Introduce complex homogeneous coordinates:
[TABLE]
Recall ω=e2πi/3=−21+2i3 and ω2=ω. The inverse relations are
[TABLE]
Suppose f(z,z,t) is a polynomial in z and
zˉ then Rf(z,z,t)=f(z,z,t) and Mf(z,z,t)=f(ωz,ω2z,t). Thus RMf(z,z,t)=f(ω2z,ωz,t) and
MRf(z,z,t)=f(ωz,ω2z,t). The idea is to write bn,2k in terms of z,z,t and apply the projection Π:=31(2I−MR−RM).
To determine linear independence it suffices to consider the terms of highest
degree in z,z thus we set t=v1+v2+v3=0 in the formula
for bn,2k (previously denoted bn,2k′ using the homogeneous
coordinates, see proof of Theorem 18). From formula
(48) and Lemma 20
[TABLE]
The coefficient of (v1v2)k(v1+v2)n−2k in bn,2k′(v1,v2,0) is nonzero, and
this is the term with highest power of v1v2. Thus {bn,2k′(v1,v2,0):0≤k≤3n−2} is a basis for span{(v1v2)k(v1+v2)n−2k:0≤k≤3n−2}. The next step is to show that the projection Π
has trivial kernel. In the complex coordinates v1+v2=−31(z+z−t)=−31(z+z) and v1v2=91(z2−zz+z2)
(discarding terms of lower order in z,z, that is, set t=0).
Proposition 21
If Π∑k=0⌊(n−1)/3⌋ck(z+z)n−2k(z2−zz+z2)k=0 then ck=0 for all
k.
Proof. For any polynomial f(z,z) we have Πf(z,z)=31(2f(z,z)−f(ω2z,ωz)−f(ωz,ω2z)). In particular
[TABLE]
By hypothesis n−3k≥1. Evaluate the expression at z=eπi/6+ε where ε is real and
near [math]. Note eπi/6=21(3+i). Then
[TABLE]
and
[TABLE]
where C=3(n−−3k−1)/2(n−3k)(4−2(−1)n−3k) (binomial theorem). The dominant term in the
right-hand side is (2−(−1)n−3k)3(n−k)/2−1εk. Now suppose Πk=0∑⌊(n−1)/3⌋ck(z+z)n−2k(z2−zz+z2)k=0. Evaluate the polynomial at z=eπi/6+ε. Let ε→0
implying c0=0. Indeed write the expression as
[TABLE]
Since 2−(−1)n−3k≥1 this shows ck=0 for all k.
We have shown:
Proposition 22
Suppose Πk=0∑⌊(n−1)/3⌋ckbn,2k=0 then ck=0 for all k; the cardinality
of the set (52) is drefl(n).
Theorem 23
a.
The polynomials {Πbn,2k:0≤k≤3n−1} are linearly independent.
2. b.
The set {RMΠbn,2k,MRΠbn,2k:0≤k≤3n−1} is linearly independent and defines a basis for the
τrefl component of Pn,n−1⊥(T).
Proof. In general Πzazb=zazb if a−b≡1,2mod3 and Πzazb=0 if a−b≡0mod3. Expand the polynomials wk(z,z):=Π(z+z)n−3k(z3+z3)k by the binomial theorem to obtain
[TABLE]
Then
[TABLE]
Firstly we show that {RMwk,MRwk} is linearly
independent for 0≤k≤3n−1. For each value of
nmod3 we select the highest degree terms from RMwk and
MRwk: (i) n=3m+1, ω2z3m+1+ωz3m+1 and
ωz3m+1+ω2z3m+1, (ii) n=3m+2,ωz3m+2+ω2z3m+2 and ω2z3m+2+ωz3m+2, (iii) n=3m, (n−3k)(ω2z3mz+ωzz3m) and (n−3k)(ωz3mz+ω2zz3m) (by hypothesis n−3k≥1). In each case the two terms are
linearly independent (the determinant of the coefficients is ±(ω−ω2)=∓i3). Secondly
the same argument as in the previous theorem shows that ∑k=0⌊(n−1)/3⌋{ckRMwk+dkMRwk}=0 implies ckRMwk+dkMRwk=0 for
all k. By the first part it follows that ck=0=dk. This completes the
proof.
Remark 24
The basis bn,k for Pn,n−1⊥(T) in (35) is mirror symmetric with respect to the angular bisector
in T through the origin for even k and is mirror skew-symmetric
for odd k. This fact makes the point 0 in T special
compared to the other vertices. As a consequence the functions defined in
Theorem 23.a reflects the special role of 0. Part b
shows that it is possible to define a basis with functions which are either
symmetric with respect to the angle bisector in T through (1,0)⊺ or through (0,1)⊺ by
“rotating” the functions Πbn,2k to
these vertices:
[TABLE]
Since the dimension of E(refl) is
2drefl(n)=2⌊3n+2⌋ is not (always) a multiple of 3, it is, in general, not
possible to define a basis where all three vertices of the triangle are
treated in a symmetric way.
Definition 25
Let
[TABLE]
This space is lifted to a general triangle T by fixing a vertex P
of T and setting
[TABLE]
where the lifting χP,T is an affine pullback χP,T:T→T which maps 0 to
P.
The basis bn,krefl to describe the restrictions of
facet-oriented, non-conforming finite element functions to the facets is
related to a reduced space and defined as in (51) with lifted
versions
[TABLE]
Remark 26
The construction of the spaces \mathbb{P}_{p,p-1}^{\perp,\operatorname*{sym}}\left(T\right)\and Pp,p−1⊥,refl(T) (cf. Definitions 19 and 25)
implies the direct sum decomposition
[TABLE]
It is easy to verify that the basis functions bp,kP,T are
mirror symmetric with respect to the angle bisector in T through
P. However, the space Pn,n−1⊥,refl(T) is independent of the choice of the
vertex P.
In Appendix A we will define further sets of basis functions for the
τrefl component of Pn,n−1⊥(T) – different choices might be preferable for different
kinds of applications.
5.4 Simplex-Supported and Facet-Oriented Non-Conforming Basis
Functions
In this section, we will define non-conforming Crouzeix-Raviart type functions
which are supported either on one single tetrahedron or on two tetrahedrons
which share a common facet. As a prerequisite, we study in
§5.4.1 piecewise orthogonal polynomials on triangle
stars, i.e., on a collection of triangles which share a common vertex and
cover a neighborhood of this vertex (see Notation 27). We will
derive conditions such that these functions are continuous across common edges
and determine the dimension of the resulting space. This allows us to
determine the non-conforming Courzeix-Raviart basis functions which are either
supported on a single tetrahedron (see §5.4.2) or on two adjacent
tetrahedrons (see §5.4.3) by “closing” triangle stars either by a single triangle or
another triangle star.
5.4.1 Orthogonal Polynomials on Triangle
Stars
The construction of the functions Bp,kK,nc and
Bp,kT,nc as in (20) and
(24) requires some results of continuous, piecewise orthogonal
polynomials on triangle stars which we provide in this section.
Notation 27
A subset C⊂Ω is a triangle star if C is
the union of some, say mC≥3, triangles T∈FC⊂F, i.e., C=T∈FC⋃T and there exists some vertex VC∈V such that
[TABLE]
Here, Dk denotes the regular closed k-gon (in R2).
For a triangle star C, we define
[TABLE]
In the next step, we will explicitly characterize the space Pp,p−1⊥(C) by defining a set of basis functions. Set
A:=VC (cf. (58)) and pick an outer
vertex in FC, denote it by A1, and number the
remaining vertices A2,…,AmC in
FC counterclockwise. We use the cyclic numbering convention
AmC+1:=A1 and also for similar quantities.
For 1≤ℓ≤mC, let eℓ:=[A,Aℓ] be the straight line (convex hull) between and
including A, Aℓ. Let Tℓ∈FC
be the triangle with vertices A, Aℓ,
Aℓ+1. Then we choose the affine pullbacks to the reference
element T by
[TABLE]
In this way, the common edges eℓ are parametrized by
χℓ−1(t,0)=χℓ(t,0) if
3≤ℓ≤mC is odd and by χℓ−1(0,t)=χℓ(0,t) if 2≤ℓ≤mC is even. The final
edge e1 is parametrized by χ1(t,0)=χmC(t,0) if mC is even and by χ1(t,0)=χmC(0,t) (with interchanged arguments!)
otherwise. We introduce the set
For a triangle star C, a basis for Pp,p−1⊥(C) is given by bp,kC, k∈Rp,C. Further
[TABLE]
Proof. We show that (bp,kC)k∈Rp,C is a basis of
Pp,p−1⊥(C) and the dimension formula.
Continuity across eℓfor odd 3≤ℓ≤mC.
The definition of the lifted orthogonal polynomials (see (49),
(55), (57)) implies that the continuity across
eℓ for odd 3≤ℓ≤mC is equivalent to
[TABLE]
From Lemma 12 we conclude that the continuity across such
edges is equivalent to
[TABLE]
Continuity across eℓfor even 2≤ℓ≤mC.
Note that χ2(0,t)=χ3(0,t). Taking
into account (49), (55), (57) we see
that the continuity across eℓ is equivalent to
[TABLE]
From Lemma 12 we conclude that the continuity across
eℓ for even 2≤ℓ≤mC is again equivalent to
[TABLE]
Continuity across e1
For even mC the previous argument also applies for the edge
e1 and the functions bp,kC, 0≤k≤p, are
continuous across e1. For odd mC, note that χ1(t,0)=χmC(0,t). Taking into
account (49), (55), (57) we see that
the continuity across e1 is equivalent to
[TABLE]
Using the symmetry relation (37) we conclude that this is
equivalent to
[TABLE]
From Lemma 12 we conclude that this, in turn, is equivalent
to
[TABLE]
From the above reasoning, the continuity of bp,kC across
e1 follows if αn,k(ℓ)=0 for odd
k and all 1≤ℓ≤mC.
The proof of the dimension formula (60) is trivial.
5.4.2 A Basis for the Symmetric Non-Conforming Space
SK,ncp
In this section, we will prove that SK,ncp (cf.
(20)) satisfies
[TABLE]
where SGp is defined in (4) and, moreover, that
the functions Bp,kK,nc, k=0,1,…,dtriv(p)−1, as in (18),
(20) form a basis of SK,ncp.
Let T denote one facet of K and let C:=∂K\T∘. Since C is a triangle star with mC=3, we can apply Lemma
28 to obtain that
[TABLE]
The continuity of bp,2kC implies that the restriction bp,2k∂T:=bp,2kC∂T is continuous.
From (42) we conclude that
[TABLE]
where P2kE is the Legendre polynomial of even degree 2k scaled to
the edge E with endpoint values +1 and symmetry with respect to the
midpoint of E. Hence, we are looking for orthogonal polynomials
Pp,p−1⊥(T) whose traces on ∂T
are linear combination of bp,2k∂T, 0≤k≤⌊2p⌋. From (37) we deduce that they have
total symmetry, i.e., belong to the space Pp,p−1⊥,sym(T) (cf. Definition 19).
For 0≤m≤dtriv(p)−1, let
bp,m∂K,sym:∂K→R be
defined facet-wise for any T⊂∂K by
[TABLE]
Finally, we extend the function bp,m∂K,sym to
the total simplex K by polynomial extension (cf. (18),
(19))
[TABLE]
These functions are the same as those introduced in Definition
5. The above reasoning leads to the following Proposition.
Proposition 29
For a simplex K, the space of non-conforming, simplex-supported
Crouzeix-Raviart finite elements can be chosen as in (20) and the
functions Bp,kK,nc, 0≤k≤dtriv(p)−1 are linearly independent.
5.4.3 A Basis for ST,ncp
Let T∈FΩ be an inner facet and K1,K2∈G such that T=K1∩K2 and ωT=K1∪K2
(cf. (9)) with the convention that the unit normal
nT points into K2. In this section, we will prove that a
space S~T,ncp which satisfies
[TABLE]
can be chosen as S~T,ncp:=ST,ncp (cf. (25)) and, moreover, that the
functions Bp,kT,nc, k=0,1,…,drefl(p)−1, as in (24) form
a basis of ST,ncp.
Let Ci:=(∂Ki)\T∘,
i=1,2, denote the triangle star (cf. Notation 27) formed by the
three remaining triangles of ∂Ki. We conclude from Lemma
28 that a basis for Pp,p−1⊥(Ci) is given by bp,2kCi, 0≤k≤⌊2p⌋ (cf. (59)). Any function u in
STp satisfies
[TABLE]
Since any function in STp is continuous on Ci, we conclude from
Lemma 28 (with mCi=3) that
To identify a space S~T,ncp which satisfies
(67) we consider the jump condition in
(68) restricted to the boundary ∂T. The symmetry of the
functions bp,2k∂T implies that [u]T∈Pp,p−1⊥,sym(T), i.e.,
there is a function q1∈SK1,ncp(see
(20)) such that [u]T=q1∣T and u~, defined by u~∣K1=u1+q1 and u~∣K2=u2, is continuous
across T. On the other hand, all functions u∈STp whose
restrictions u∣ωT are discontinuous can be
found in SK1,ncp⊕SK2,ncp. In view of the direct sum in (67) we may thus
assume that the functions in S~T,ncp are
continuous in ωT.
To finally arrive at a direct decomposition of the space in the right-hand
side of (67) we have to split the spaces
Pp,p−1⊥(Ci) into a direct sum of the
spaces of totally symmetric orthogonal polynomials and the spaces introduced
in Definition 25 and glue them together in a continuous way. We
introduce the functions bp,kCi,sym:=bp,k∂Ki,symCi, 0≤k≤dtriv(p)−1, with bp,k∂Ki,sym as in (65) and define
bp,kCi,refl, 0≤k≤drefl(p)−1, piecewise by bp,kCi,reflT′:=bp,kAi,T′ for T′⊂Ci with bp,kAi,T′ as in (56). The mirror symmetry of
bp,kAi,T′ with respect to the angular bisector in
T′ through Ai implies the continuity of bp,kCi,refl. Hence,
[TABLE]
Since the traces of bp,kCi,sym and bp,kCi,refl at ∂T are continuous and are, from
both sides, the same linear combinations of edge-wise Legendre polynomials of
even degree, the gluing bp,k∂ωT,symCi:=bp,kCi,sym and bp,k∂ωT,reflC˙i:=bp,kCi,refl, i=1,2, defines
continuous functions on ∂ωT. Since the space
ST,ncp must satisfy a direct sum decomposition (cf.
(67)), it suffices to consider the functions
bp,k∂ωT,refl for the definition of
ST,ncp. The resulting non-conforming facet-oriented
space ST,ncp was introduced in Definition
7 and S~T,ncp can be chosen
to be ST,ncp.
Proposition 30
For any u∈ST,ncp, the following
implication holds
[TABLE]
Proof. Assume there exists u∈ST,ncp with u∣T∈ST,ncpT∩Pp,p−1⊥(T). Let K be a simplex
adjacent to T. Then uK=u∣K satisfies uK∣T′∈Pp,p−1⊥(T′) for all T′⊂∂K and, thus, uK∈SK,ncp. Since SK,ncpT′∩ST,ncpT′={0} for T′∈∂K\T∘ we conclude that uK=0.
Note that Definition 7 and Proposition 30
neither imply a priori that the functions
[TABLE]
are linearly independent nor that
[TABLE]
holds. These properties will be proved next. Recall the projection Π=31(2I−MR−RM) from Proposition 21. We
showed (Theorem 23.a) that{bp,krefl:0≤k≤3p−1} is linearly
independent, where bp,krefl:=Πbp,2k.
Additionally Rbp,krefl=bp,krefl
which implies bp,krefl(0,x1)=bp,krefl(x1,0), and the restriction
x1⟼bp,krefl(x1,1−x1) is invariant under x1↦1−x1. For four non-coplanar points
A0,A1,A2,A3 let K denote the tetrahedron with these vertices.
For any k such that 0≤k≤3p−1 define a piecewise polynomial
on the faces of K as follows: choose a local (x1,x2)-coordinate system for A0A1A2 so that the respective coordinates
are (0,0),(1,0),(0,1), and
define Qk(0) on the facet equal to bp,krefl. Similarly define Qk(0) on
A0A2A3 and A0A3A1 (with analogously chosen local
(x1,x2)-coordinate systems), by the property
bp,krefl(0,x1)=bp,krefl(x1,0). Qk(0)
is continuous at the edges A0A1, A0A2, and A0A3. The
values at the boundary of the triangle star equal bp,krefl(x1,1−x1); note the symmetry and
thus the orientation of the coordinates on the edges A1A2, A2A3, A3A1 is immaterial. The value of Qk(0)
on the triangle A1A2A3 is taken to be a degree p polynomial,
totally symmetric, with values agreeing with bp,krefl(x1,1−x1) on each edge.
Similarly Qk(1),Qk(2),Qk(3) are defined by taking A1,A2,A3 as the
center of the construction, respectively.
Theorem 31
a) The functions Qk(i), 0≤k≤drefl(p)−1, i=0,1,2,3 are linearly independent.
The proof involves a series of steps. The argument will depend on the values
of the functions on the three rays A0A1, A0A2, A0A3,
each one of them is given coordinates t so that t=0 at A0 and t=1
at the other end-point. For a fixed k let q(t)=bp,krefl(t,0), q(t)=bp,krefl(1−t,0) and q(t)=bp,krefl(t,1−t).
Lemma 32
Suppose 0≤k≤3p−1 and 0≤t≤1 then
q(t)+q(t)+q(t)=0.
Proof. The actions of RM and MR on polynomials f(x1,x2)
are given by MRf(x1,x2)=f(1−x1−x2,x1) and RMf(x1,x2)=f(x2,1−x1−x2). Polynomials of τrefl-type
satisfy f+RMf+MRf=0. Apply this relation to bp,krefl
with x1=t and x2=0 with the result
[TABLE]
The fact that bp,krefl(x1,x2)=bp,krefl(x2,x1) finishes the
proof.
**Proof of Theorem 31. **Consider the contribution of
Qk(1) to the values on the ray A0A1: because
Qk(1) is constructed taking the origin at A1 and
because of the reverse orientation of the ray we see that the value of
Qk(1) is given by q. The value of
Qk(1) on the ray A0A2 is q (by
the symmetry of q the orientation of the ray does not matter).
The other functions are handled similarly, and the contributions to the three
rays are given in this table:
[TABLE]
We use qk,qk,qk to denote the polynomials
corresponding to bp,krefl. Suppose that the linear
combination ∑k=0⌊(p−1)/3⌋∑i=03ck,iQk(i)=0. Evaluate the sum on the
three rays to obtain the equations:
[TABLE]
We used Lemma 32 to eliminate qk from the equations. In
Theorem 23.b we showed the linear independence of {RMbp,krefl,MRbp,krefl:0≤k≤3p−1}, and in Lemma 12 that the
restriction map f↦f(x1,0) is an isomorphism from
the orthogonal polynomials Pp,p−1⊥ to Pp([0,1]). Thus the projection of the set is
also linearly independent, that is, {qk,qk:0≤k≤3p−1} is a linearly independent set of
polynomials on 0≤t≤1. This implies all the coefficients in the above
equations vanish: the qk terms show ck,0=ck,1=ck,2=ck,3 and then the qk-terms show 2ck,0−ck,0=ck,0=0.
To prove (71) it suffices to transfer the statement to
the reference element T. The pullbacks of the restrictions
Bp,mT′,ncT, T′⊂C, are given by
[TABLE]
Since bn,krefl∈Pn,n−1⊥,refl(T) (cf. (85)) it
follows
[TABLE]
6 Properties of Non-Conforming Crouzeix-Raviart Finite
Elements
The non-conforming Crouzeix-Raviart finite element space SG,ncp satisfies SG,cp⊊SG,ncp⊂SGp (cf. Section 4.2). In this section, we will present a basis
for SG,ncp and discuss whether the inclusion
SG,ncp⊂SGp, in fact,
is an equality.
6.1 A Basis for Non-Conforming Crouzeix-Raviart Finite Elements
We have defined conforming and non-conforming sets of functions which are
spanned by functions with local support. In this section, we will investigate
the linear independence of these functions. We introduce the following spaces
[TABLE]
where SK,ncp and ST,ncp are as
in Definitions 5 and 7. For some 0≤k≤drefl(p)−1, we introduce the subspace
Srefl,ncp,k⊂Srefl,ncp by
[TABLE]
Further we will need the conforming finite element space (cf. (4), Def. 1), where the vertex-oriented functions are removed,
i.e.,
[TABLE]
Theorem 33
The sums
[TABLE]
are direct. The sum
[TABLE]
is not direct. The sum
[TABLE]
is direct.
Proof.Part 1. We prove that the sum Ssym,ncp⊕Srefl,ncp is direct.
From Proposition 30 we know that the sum ST,ncpT⊕Pp,p−1⊥(T) is direct. Let ΠT:L2(T)→Pp−1(T) denote the L2(T) orthogonal projection. Since Pp−1(T)
is the orthogonal complement of Pp,p−1⊥(T)
in Pp(T) and since Pp,p−1⊥(T)∩ST,ncpT={0}, the restricted mapping ΠT:ST,ncpT→Pp−1(T) is injective and the functions qp,kT:=ΠT(Bp,kT,ncT), 0≤k≤drefl(p)−1, are linearly independent and
belong to Pp−1(T). We define the functionals
[TABLE]
Next we consider a general linear combination and show that the condition
[TABLE]
implies that all coefficients are zero.
We apply the functionals Jp,kT to (76) and use the
orthogonality between Pp,p−1⊥(T) and
qp,kT to obtain
[TABLE]
For T′=T it holds Jp,kT(Bp,iT′,nc)=0 since Bp,iT′,ncKT is an orthogonal
polynomial. Thus, equation (77) is equivalent to
[TABLE]
The matrix (Jp,kT(Bp,jT,nc))k,j=0drefl(p)−1 is regular
because
[TABLE]
and (Bp,kT,ncT)k are linearly independent. Hence we conclude from (78)
that all coefficients βjT are zero and the condition
(76) reduces to
[TABLE]
The left-hand side is a piecewise continuous function so that the condition is
equivalent to ∑i=0dtriv(p)−1αiKBp,iK,nc=!0 for all
K∈G. Since Bp,iK,nc is a basis for
SK,ncpK we conclude that all
αiK are zero.
**Part 2. **Next we prove that (Ssym,ncp⊕Srefl,ncp,0)∩S~G,cp={0} and we show this by contradiction. Let u∈(Ssym,ncp⊕Srefl,ncp,0)∩S~G,cp which satisfies u=0. We decompose
u=usym+urefl with
usym∈Ssym,ncp
and urefl∈Srefl,ncp. We prove by contradiction that usym∈C0(Ω). Assume that usym∈/C0(Ω). Then, there exists a facet T⊂FΩ such that [usym]T=0. Then, [urefl]T=−[usym]T is a necessary condition for the
continuity of u. However, [usym]T∈Pp,p−1⊥,sym(T) while
[urefl]T∈Pp,p−1⊥,refl(T) and there is a contradiction
because Pp,p−1⊥,sym(T)∩Pp,p−1⊥,refl(T)={0}. Hence, usym∈C0(Ω)
and, in turn, urefl∈C0(Ω).
Since u=0, at least, one of the functions usym and
urefl must be different from the zero function.
**Case a. **We show usym=0 by contradiction: Assume
usym=0. Then, usym∣T=0 for all facets T∈F. (Proof by
contradiction: If usym∣T=0 for some
T∈F, we pick some K∈F which has T as a facet.
Since usym∣K∈SK,ncpK we have usym∣T′=0 for all facets T′ of K and usym∣K=0. Since
usym is continuous in Ω, the restriction usym∣K′ is zero for any K′∈G which shares a facet with K. This argument can be applied
inductively to show that usym=0 in Ω. This is a
contradiction.) We pick a boundary facet T∈F∂Ω.
The condition u∈S~G,cp implies
u=0 on ∂Ω and, in particular, u∣T=usym∣T+urefl∣T=0. We use again the argument
Pp,p−1⊥,sym(T)∩Pp,p−1⊥,refl(T)={0} which implies usym=0 and this is a
contradiction to the assumption usym=0.
**Case b. **From Case a we know that usym=0, i.e.,
urefl=u, and it remains to show urefl=0. The condition urefl∈S~G,cp implies urefl∣∂Ω=0 and urefl(V)=0 for all vertices V∈V.
The proof of Case b is similar than the proof of Case a and we start by
showing for a tetrahedron, say K, with a facet on the boundary that urefl∣K=0 and employ an induction over
adjacent tetrahedrons to prove that urefl=0 on every
tetrahedron in G.
We consider a boundary facet T0∈F∂Ω with
adjacent tetrahedron K⊂G. We denote the three other facets
of K by Ti, 1≤i≤3, and for 0≤i≤3, the vertex of K
which is opposite to Ti by Ai.
**Case b.1. **First we consider the case that there is one and only one
other facet, say, T1 which lies in ∂Ω. Then
urefl∣T=u2∣T+u3∣T for some ui∈STi,ncp,0:=span{Bp,0Ti,nc}, i=2,3. From Theorem
23.b we conclude that the sum ST2,ncp,0T⊕ST3,ncp,0T is direct. The condition urefl∣T=0 then implies u2=u3=0.
Thus, we have proved urefl∣K=0.
**Case b.2. **The case that there are exactly two other facets
which are lying in ∂Ω can be treated in a similar way.
**Case b.3. **Next, we consider the case that Ti∈FΩ for i=1,2,3. Note that urefl∣T=∑i=13ui∣T for some
ui∈STi,ncp,0. On T we choose a local
(x1,x2)-coordinate system such that A1=0, A2=(1,0)⊺,
A3=(0,1)⊺. From (51) and
(53) we conclude that
[TABLE]
This implies u2∣T=RM(u1∣T)=u1∣T∘χ{3,2,1} and u3∣T=MR(u1∣T)=u1∣T∘χ{2,3,1} (cf. (44)) and, in turn, that the restrictions uiE of
ui to the edge Ei=Ti∩T0, 1≤i≤3, are the
“same”, more precisely, the affine pullbacks
of uiE to the interval [0,1] are the same. From Lemma
13, we obtain that
[TABLE]
where χi:T→Ti are affine pullbacks to the
reference triangle such that χi(0)=A0.
This implies that the functions ui at A0 have the same
value (say w0) and, from the condition urefl(A0)=3w0=0, we conclude that ui(A0)=0. The values of ui at the vertex Ai of K (which is opposite to Ti) also coincide and we denote this
value by v0. Since urefl∣T=0
it holds urefl(Ai)=2w0+v0=0. From w0=0 we conclude that also v0=0. Let χi,T0:T→T0 denote an affine pullback with the property
χi,T0(0)=Ai. Hence,
[TABLE]
with values zero at the vertices of T^. Note that
[TABLE]
The vertex properties (81) along the definition of bp,krefl (cf. (51)) imply that
[TABLE]
Since cp=0 for p≥1 we conclude that ui=0 holds.
Relation (80) implies ui∣T0=0 and
thus ui=0. From urefl∣T=∑i=13ui∣T we deduce that urefl∣K=0.
The Cases b.1-.3 allow to proceed with the same induction argument as for Case
a and urefl=0 follows by induction.
**Part 3. **An inspection of Part 2 shows that, for the proof of Case a,
it was never used that the vertex-oriented basis functions have been removed
from SG,cp and Case a holds verbatim for
SG,cp. This implies that the first sum in
(73) is direct.
Part 4. The fact that the sum SG,cp+Srefl,ncp is not direct
is postponed to Proposition 34.
Proposition 34
For any vertex V∈VΩ it
holds Bp,VG∈Ssym,ncp⊕Srefl,ncp,0⊕S~G,cp.
Proof. We will show the stronger statement Bp,VG∈Srefl,ncp,0⊕S~G,cp. It suffices to construct a continuous
function uV∈Srefl,ncp
which coincides with Bp,VG at all vertices
V′∈V and vanishes at ∂Ω; then,
Bp,VG−uV∈S~G,cp and the assertion follows. Recall the known values of
bp,0refl at the vertices of the reference triangle and
the definition of cp as in (82). Let K∈G be a
tetrahedron with V as a vertex. The facets of K are denoted by
Ti, 0≤i≤3, and the vertex which is opposite to Ti is
denoted by Ai. As a convention we assume that A0=V. For every Ti, 1≤i≤3, we define the function
uTi∈STi,ncp by setting (cf.
(56))
[TABLE]
where χAi,T0:T→T0 is an affine
pullback which satisfies χAi,T0(0)=Ai. (It is easy to see that the definition of
uTi is independent of the side of Ti, where the tetrahedron K
is located.) From (51) and (53) we conclude that
∑i=13uTiT0=0 holds. We proceed in
the same way for all tetrahedrons K∈GV (cf.
(9)). This implies that
[TABLE]
vanishes at Ω\ωV∘ (cf.
(9)). By construction the function u~V is continuous. At V, the function uTi has
the value (cf. (82))
[TABLE]
so that u~V(V)=Ccp, where C
is the number of terms in the sum (83). Since cp>0 for all
p≥1, the function uV:=Ccp1u~V is well
defined and has the desired properties.
Remark 35
We have seen that the extension of the basis functions of SG,cp by the basis functions of Srefl,ncp leads to linearly depending functions. On the other
hand, if the basis functions of the subspace Srefl,ncp,0 are added and the vertex-oriented basis functions
in SG,cp are simply removed, one arrives at
a set a linear independent functions which span a larger space than
SG,cp. Note that Srefl,ncp,0=Srefl,ncp
for p=1,2,3.
One could add more basis functions from Srefl,ncp but then has to remove further basis functions from
S~G,cp or formulate side constraints
in order to obtain a set of linearly independent functions.
We finish this section by an example which shows that there exist meshes with
fairly special topology, where the inclusion
[TABLE]
is strict. We emphasize that the left-hand side in (84), for
p≥4, defines a larger space than the space in (75) since it
contains all non-conforming functions of reflection type.
Example 36
Let us consider the octahedron Ω with vertices A±:=(0,0,±1)⊺ and A1:=(1,0,0)⊺, A2:=(0,1,0)⊺, A3:=(−1,0,0)⊺,
A4:=(0,−1,0)⊺. Ω is subdivided
into a mesh G:={Ki:1≤i≤8} consisting of
eight congruent tetrahedrons sharing the origin 0 as a common
vertex. The six vertices at ∂Ω have the special topological
property that each one belongs to exactly four surface facets.
Note that the space defined by the left-hand side of (84) does not
contain functions whose restriction to a surface facet, say T, belongs to
the τsign component of Pn,n−1⊥(T). Hence, the inclusion in (84) is strict if we
identify a function in SGp whose restriction to some surface
facet is an orthogonal polynomial of “sign
type”. Let q=0 be a polynomial which belongs
to the τsign component of Pn,n−1⊥(T) on the reference element. Denote the (eight) facet on
∂Ω with the vertices A±, Ai,
Ai+1 by Ti± for 1≤i≤4 (with cyclic numbering
convention) and choose affine pullbacks χ±,i:T→Ti± as χ±,i(x):=A±+x1(Ai−A±)+x2(Ai+1−A±). Then, it is easy to verify (use
Lemma 28 with even mC) that the function q:∂Ω→R, defined by q∣Ti±:=q∘χ±,i−1 is continuous on ∂Ω. Hence
the “finite element extension” to the
interior of Ω via
[TABLE]
defines a function in SGp which is not in the space defined
by the left-hand side of (84).
We state in passing that the space SGp does not contain any
function whose restriction to a boundary facet, say T, belongs to the
τsign component of Pp,p−1⊥(T) if there exists at least one surface vertex which belongs to an
odd number of surface facets. In this sense, the topological situation
considered in this example is fairly special.
7 Conclusion
In this article we developed explicit representation of a local basis for
non-conforming finite elements of the Crouzeix-Raviart type. As a model
problem we have considered Poisson-type equations in three-dimensional
domains; however, this approach is by no means limited to this model problem.
Using theoretical conditions in the spirit of the second Strang lemma, we have
derived conforming and non-conforming finite element spaces of arbitrary
order. For these spaces, we also derived sets of local basis functions. To the
best of our knowledge, such explicit representation for general polynomial
order p are not available in the existing literature. The derivation
requires some deeper tools from orthogonal polynomials of triangles, in
particular, the splitting of these polynomials into three irreducible
irreducible S3 modules.
Based on these orthogonal polynomials, simplex- and facet-oriented
non-conforming basis functions are defined. There are two types of
non-conforming basis functions: those whose supports consist of one
tetrahedron and those whose supports consist of two adjacent tetrahedrons. The
first type can be simply added to the conforming hp basis functions. It is
important to note that the span of the functions of the second type contains
also conforming functions and one has to remove some conforming functions in
order to obtain a linearly independent set of functions. We have proposed a
non-conforming space which consists of a) all basis functions of the first
type and b) a reduced set of basis functions of the second type and c) of the
conforming basis functions without the vertex-oriented ones. This leads to a
set of linearly independent functions and is in analogy to the well known
lowest order Crouzeix-Raviart element.
It is interesting to compare these results with high-order Crouzeix-Raviart
finite elements for the two-dimensional case which have been presented in
[6]. Facets T of tetrahedrons in 3D correspond to edges E of
triangles in 2D. As a consequence the dimension of the space of orthogonal
polynomials Pp,p−1⊥(E) equals one. For
even degree p, one has only non-conforming basis functions of
“symmetric” type (which are supported on a
single triangle) and for odd degree p, one has only non-conforming basis
functions of “reflection” type (which are
supported on two adjacent triangles). It turns out that adding the non
conforming symmetric basis function to the conforming hp finite element
space leads to a set of linearly independent functions which is the analogue
of the first sum in (73). If the non-conforming basis functions
of reflection type are added, the set of vertex-oriented conforming basis
functions have to be removed from the conforming space. This is in analogy to
the properties (74) and (75).
Future research is devoted on numerical experiments and the application of
these functions to system of equations as, e.g., Stokes equation and the
Lamé system.
Acknowledgement This work was supported in part by ENSTA,
Paris, through a visit of S.A. Sauter during his sabbatical. This support is
gratefully acknowledged.
Appendix A Alternative Sets of “Reflection-type” Basis Functions
In this Appendix we define further sets of basis functions for the
τrefl component of Pn,n−1⊥(T) – different choices might be preferable for different
kinds of applications. All these sets have in common that two vertices of
T are special – any basis function is symmetric/skew symmetric
with respect to the angular bisector of one of these two vertices.
Remark 37
The functions bn,2k can be characterized as the range of I+R. We
project these functions onto τrefl, that is, the space
E(refl):={p:RMp+MRp=−p}.
Let
[TABLE]
The range of both is E(refl). We will show
that {T1bn,2k,T2bn,2k,0≤k≤(n−2)/3} is a basis for E(refl).
Previously we showed {RMqk,MRqk} is a basis, where
qk=(2I−MR−RM)bn,2k=(T1+T2)bn,2k
(cf. (51). Observe that
[TABLE]
holds, so the basis is made up out of linear combinations of {T1bn,2k,T2bn,2k,0≤k≤(n−1)/3}.
These can be written as elements of the range of T1(I+R)
and T2(I+R). Different linear combinations will behave
differently under the reflections R,M,RMR (that is (x,y)→(y,x),(1−x−y,y),(x,1−x−y) respectively). After some computations
we find
[TABLE]
Any two of these types can be used in producing bases from the bn,2k.
Also each pair (first two, second two, third two) are orthogonal to each
other. Note R fixes (0,0) and reflects in the line x=y,
M fixes (0,1), reflects in 2x+y=1, and RMR fixes
(1,0), reflects in x+2y=1.
If we allow for a complex valued basis, the three vertices of T
can be treated more equally as can be seen from the following remark.
Remark 38
The basis functions can be complexified: set ω=e2πi/3; any polynomial in E(refl) can be expressed as p=p1+p2 such that
MRp=ωp1+ω2p2 (consequently RMp=ω2p1+ωp2 ), then
[TABLE]
These lead to another basis built up from the bn,2k. Let
[TABLE]
Applying these operators to bn,2k produces a basis {S1bn,2k,S2bn,2k:0≤k≤(n−1)/3}
satisfying
[TABLE]
This is a basis which behaves similarly at each vertex.
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