Part III, Chapter 14
Traces and interface jumps of finite elements
In this chapter, we start the construction of important subspaces of the broken
finite element spaces introduced in Chapter 13. This construction is done
so that these subspaces embed continuously in a functional space like H 1 ,
H(curl), or H(div). Key structural assumptions making this construction
possible are identified. For instance we establish that a necessary and sufficient
condition for a piecewise smooth function to belong to H 1 (resp., H(curl) or
H(div)) is that the jumps of its trace (resp., tangential or normal component
of its trace) across the mesh interfaces vanish.
Remark 14.1 (Compatible node numbering). There are various compatible numberings. For instance, there are 3 compatible ways of numbering
the vertices of a triangle, 4 for the vertices of a square, 3×4 for the vertices
of a tetrahedron, and 4×6 for the vertices of a cube. All these compatible
numberings lead to equivalent geometric maps.
⊓
⊔
b
Remark 14.2 (Enumeration invariance). The degrees of freedom in Σ
are said to be enumeration-invariant if, for any equivalent geometric maps TK
σi (ψT̃K (v))}i∈N are
and T̃K (see Remark ??), the sets {b
σi (ψTK (v))}i∈N and {b
equal for all v ∈ V (K), where ψTK and ψT̃K are the bounded isomorphisms
in (7.6) corresponding to the maps TK and T̃K , respectively. See Exercise ??
for an example.
⊓
⊔
14.1 H 1 -, H(div)-, and H(curl)-conformity
The spaces H 1 (D), H(div; D), and H(curl; D) are composed of squareintegrable functions whose gradient, divergence, and curl, respectively, is
square-integrable. In this section we assume to be given a piecewise smooth
function on the mesh Th , and we want to find necessary and sufficient conditions for this function to be in H 1 (D), H(div; D), or H(curl; D). We call this
176
Chapter 14. Traces and interface jumps of finite elements
property H 1 -, H(div)-, or H(curl)-conformity. It turns out that the answer
to this question entirely hinges on the continuity properties of the function, or
its normal or tangential component, across the mesh interfaces. The content
of this section is crucial to understand the construction of H 1 (D), H(div; D),
and H(curl; D)-conforming finite elements spaces in Chapters 14 and 15.
14.1.1 Interfaces and jumps
Let Th be a mesh of a domain D ⊂ Rd .
Definition 14.3 (Interfaces, boundary faces). We say that a subset F ⊂
D is an interface if it has positive (d−1)-dimensional measure and if there are
distinct mesh cells Kl , Kr ∈ Th such that F = ∂Kl ∩ ∂Kr . The numbering of
the two mesh cells is arbitrary, but kept fixed once and for all, and we nF to
be the unit normal vector to F pointing from Kl to Kr . We say that a subset
F ⊂ D is a boundary face if it has positive (d−1)-dimensional measure and
if there is a mesh cell K ∈ Th such that F = ∂K ∩ ∂D. nF to be the unit
normal vector to F pointing outward D. Interfaces are collected in the set Fh◦ ,
boundary faces in the set Fh∂ , and we set Fh := Fh◦ ∪ Fh∂ .
Let p ∈ [1, ∞] and consider the functional space
W 1,p (Th ; Rq ) := {v ∈ Lp (D; Rq ) | v|K ∈ W 1,p (K; Rq ), ∀K ∈ Th }.
(14.1)
Functions in W 1,p (Th ; Rq ) are cellwise smooth. Owing to the Trace Theorem B.106, these functions have traces in L1 on both sides of every mesh
interface, but these limits may not necessarily coincide.
Definition 14.4 (Jump across interfaces). Let v ∈ W 1,1 (Th ; Rq ). Let F ∈
Fh◦ be a mesh interface. Let Kl , Kr be the two cells such that F = ∂Kl ∩ ∂Kr .
The jump of v across F is defined to be
[[v]]F (x) = v|Kl (x) − v|Kr (x)
a.e. on F .
F
Kl
Kr
Fig. 14.1. Jump of a function across an interface F = ∂Kl ∩ ∂Kr .
(14.2)
Part III. Finite Element Spaces
177
Figure 14.1 illustrates the notion of jump across an interface. The subscript
F is dropped from the jump notation when the context is unambiguous.
Remark 14.5 (Alternative definition). An alternative definition of the
jump where Kl , Kr play symmetric roles consists of setting [[v]]∗F = v|Kl ⊗
nKl ,F +v|Kr ⊗nKr ,F , where nKi ,F , i ∈ {l, r}, is the unit normal to F pointing
away from Ki , i.e., [[v]]F ⊗ nF = [[v]]∗F . The advantage of (14.2) over this
definition is that the [[v]]F is Rq -valued instead of being Rq×d -valued. Both
definitions are commonly used in the literature.
⊓
⊔
Let us first consider the integrability of the gradient for a scalar-valued,
piecewise smooth function.
Theorem 14.6 (Integrability of ∇). Let v ∈ W 1,p (Th ; Rq ) with p ∈ [1, ∞].
Then, ∇v ∈ Lp (D; Rq ) if and only if [[v]]F = 0 a.e. on all F ∈ Fh◦ .
Proof. It suffices to prove the result for q = 1; the general case is treated
by working componentwise. Let v ∈ W 1,p (Th ) and let C0∞ (D) be the set of
smooth functions compactly supported in D. Then for all Φ ∈ C0∞ (D),
Z
X Z
v∇·Φ dx =
v|K ∇·Φ dx
D
K∈Th
=−
=−
K
X Z
K∈Th
X Z
K∈Th
K
∇(v|K )·Φ dx +
K
∇(v|K )·Φ dx +
X Z
K∈Th
∂K
F ∈Fh◦
F
X Z
v|K nK ·Φ ds
[[v]]F nF ·Φ ds,
where nK is the outward unit normal to K and nF is as in Definition 14.4.
◦
(1)
R If [[v]]F = 0 a.e.
R F ∈ Fh , we infer from the above identity that
P on all
v∇·Φ dx = − K∈Th K Φ·∇(v|K ) dx, which shows that v admits a weak
D
gradient in Lp (D) and (∇v)|K = ∇(v|K ) for all K ∈ Th . Hence, v ∈ W 1,p (D).
(2) Conversely, let v ∈ W 1,p (D). Since the weak derivative satisfies (∇v)|K =
∇(v
infer from the above formula that
R all K ∈ Th (see Exercise 3.2), we ∞
P |K ) for
[[v]]
n
·Φ
ds
=
0,
for
all
Φ
∈
C
F F
0 (D). That [[v]]F = 0 a.e. on all
F ∈Fh◦ F
◦
F ∈ Fh then follows from a density argument after localizing the support of
Φ so that it intersects only one interface.
⊓
⊔
Theorem 14.6 is fundamental for the construction of finite element spaces.
A simple one-dimensional illustration is presented in Figure 14.2.
Let us now consider the integrability of the divergence or the curl of vectorvalued, piecewise smooth functions. Let v ∈ W 1,p (Th ). The jump of the tangential component of v (for d = 3) and the jump of its normal component
across a mesh interface F ∈ Fh◦ , with F = ∂Kl ∩ ∂Kr , are defined to be
[[v×n]]F (x) = (v|Kl ×nF )(x) − (v|Kr ×nF )(x),
[[v·n]]F (x) = (v|Kl ·nF )(x) − (v|Kr ·nF )(x),
178
Chapter 14. Traces and interface jumps of finite elements
Fig. 14.2. One-dimensional example with two piecewise quadratic functions; the
one on the left is not in H 1 , the one on the right is.
for a.e. x ∈ F . Notice that [[v×n]]F = [[v]]F ×nF and [[v·n]]F = [[v]]F ·nF where
[[v]]F is the componentwise jump of v across F from Definition 14.4.
Theorem 14.7 (Integrability of ∇× and ∇·). Let v ∈ W 1,p (Th ) with
p ∈ [1, ∞]. Then,
(i) If d = 3, ∇×v ∈ Lp (D) if and only if [[v×n]]F = 0 a.e. on all F ∈ Fh◦ .
(ii) ∇·v ∈ Lp (D) if and only if [[v·n]]F = 0 a.e. on all F ∈ Fh◦ .
⊓
⊔
Proof. See Exercise 14.1.
14.2 Unified formalism
Let p ∈ [1, ∞] and let s > p1 if p > 1 or s = 1 if p = 1. Recall that functions
in W s,p (K; Rq ) have traces in L1 (F ; Rq ) for any cell K ∈ Th where F is a
(subset of a) face of K. To allow for a unified treatment of H 1 -, H(curl)-, and
H(div)-conforming subspaces, we introduce the symbol x ∈ {g, c, d} (referring
to the gradient, curl, and divergence operators), and we define the following
x
trace operators γK,F
: W s,p (K; Rq ) −→ L1 (F ; Rt ):
g
γK,F
(v) = v|F
c
γK,F
(v)
d
γK,F (v)
= v|F ×nF
= v|F ·nF
(q = t = 1),
(14.3a)
(q = t = d = 3),
(14.3b)
(q = d, t = 1),
(14.3c)
leading to the following notion of jump:
x
x
[[v]]xF (x) = γK
(v|Kl )(x) − γK
(v|Kr )(x),
r ,F
l ,F
(14.4)
for all v ∈ W s,p (Th ) (defined similarly to (14.1)) and a.e. x ∈ F . Note that
x
[ v|F ≡ 0 ] =⇒ [ γK,F
(v) ≡ 0 ].
We define the functional spaces V
x,p
(D) with
V g,p (D) := {v ∈ Lp (D) | ∇v ∈ Lp (D)},
V
c,p
V
d,p
(14.5)
p
p
(D) := {v ∈ L (D) | ∇×v ∈ L (D)},
p
p
(D) := {v ∈ L (D) | ∇·v ∈ L (D)}.
(14.6a)
(14.6b)
(14.6c)
Part III. Finite Element Spaces
179
The zero-jump conditions from Theorems 14.6 and 14.7 can then be summarized as follows:
∀v ∈ W 1,p (Th ; Rq ),
[ v ∈ V x,p (D) ] ⇐⇒ [ [[v]]xF = 0, ∀F ∈ Fh◦ ].
(14.7)
14.3 Linking jumps and degrees of freedom
In this section we address the crucial question of how the zero-jump condition (14.5) can be enforced by using the degrees of freedom of the underlying
finite elements. We assume that P b (Th ) ⊂ W 1,∞ (Th ; Rq ), so that functions in
P b (Th ) are smooth enough to have a well-defined γ x -trace.
b Pb, Σ)
b be the reference finite element. Let Th be a matching mesh.
Let (K,
b Pb, Σ)
b
We generate on every K ∈ Th a finite element (K, PK , ΣK ) from (K,
−1 b
using Proposition 7.8 and a mapping ψK . Recall that PK = ψK
(P ) and that
ΣK = {σK,i }i∈N with σK,i = σ
bi ◦ψK for all i ∈ N . Our goal is now to generate
on each mesh face F ∈ Fh a (single) face finite element that is compatible
x
with the γK,F
-trace operator and the finite elements (K, PK , ΣK ) for the (one
or two) mesh cells such that F ∈ FK , i.e., F is a face of K. We proceed in two
steps. First, we consider one mesh cell K such that F ∈ FK and construct
a face finite element on F that depends on K. Then, for all F ∈ Fh◦ with
F = ∂Kl ∩ ∂Kr (F is a face of Kl and of Kr since the mesh is matching),
we want that the two face elements generated on F from Kl and Kr are the
same. Each step hinges on a specific assumption illustrated on some examples
in §14.4.
x
x
Definition 14.8 (NK,F , PK,F
). Let K ∈ Th , let F ∈ FK and PK,F
:=
x
x
x
γK,F (PK ). We say that i ∈ NK,F ⊂ N if and only if there is σK,i ∈ L(PK,F
; R)
x
x
x
x
such that σK,i = σK,i
◦ γK,F
. We denote ΣK,F
:= {σK,i
}i∈NK,F .
x
x
Lemma 14.9 (Face unisolvence). (F, PK,F
, ΣK,F
) is a finite element if and
only the following holds for all p ∈ PK :
x
[ σK,i (p) = 0, ∀i ∈ NK,F ] =⇒ [ γK,F
(p) = 0 ].
(14.8)
Proof. The implication is evident. Conversely, we observe
P that the xfunctions
x
(θK,i )
{γK,F
(θK,i )}i∈NK,F are linearly independent. Indeed, if i∈NK,F λi γK,F
x
vanishes identically, using the linearity of σK,j
for all j ∈ NK,F , we infer that
0=
X
i∈NK,F
x
x
λi σK,j
(γK,F
(θK,i )) =
X
λi σK,j (θK,i ) = λj .
i∈NK,F
x
x
As a result, dim(PK,F
) ≥ card(NK,F ) = card(ΣK,F
), and this property combined with (14.8) means that unisolvence holds.
⊓
⊔
180
Chapter 14. Traces and interface jumps of finite elements
x
x
Assumption 14.10 (Face element). The triple (F, PK,F
, ΣK,F
) is a finite
element for every K ∈ Th and every F ∈ FK .
Assumption 14.11 (Face matching). For all F ∈ Fh◦ with F = ∂Kl ∩∂Kr ,
x
x
PK
= PK
and there is a bijective map χlr : NKl ,F −→ NKr ,F such that
r ,F
l ,F
x
x
x
x
σKl ,i = σKr ,χlr (i) for all i ∈ NKl ,F , i.e., ΣK
= ΣK
.
r ,F
l ,F
Lemma 14.12 (Zero γ x -jump). Under Assumptions 14.10 and 14.11 the
following holds for all vh ∈ P b (Th ) and all F ∈ Fh◦ :
[ σKl ,i (vh|Kl ) − σKr ,χlr (i) (vh|Kr ) = 0, ∀i ∈ NKl ,F ] ⇐⇒ [ [[vh ]]xF = 0 ], (14.9)
x
x
Proof. Set vl := γK
(vh|Kl ) and vr := γK
(vh|Kr ) so that [[vh ]]xF = vl − vr .
r ,F
l ,F
x
x
x
Note that vl ∈ γKl ,F (PKl ) = PKl ,F . Similarly, vr ∈ PK
, which owing to
r ,F
x
x
x
Assumption 14.11 also means that vr ∈ PKl ,F . Since (F, PK
, ΣK
) is a
l ,F
l ,F
x
finite element owing to Assumption 14.10, we infer that [[vh ]]F = 0 if and only
x
x
x
x
(vh|Kl )) =
(vl − vr ) = 0 for all i ∈ NKl ,F . Now, σK
(vl ) = σK
(γK
if σK
l ,i
l ,i
l ,F
l ,i
σKl ,i (vh|Kl ), and Assumption 14.11 implies that
x
x
x
x
σK
(vr ) = σK
(vr ) = σK
(γK
(vh|Kr )) = σKr ,χlr (i) (vh|Kr ).
r ,F
l ,i
r ,χlr (i)
r ,χlr (i)
x
(vl − vr ) = σKl ,i (vh|Kl ) − σKr ,χlr (i) (vh|Kr ).
Hence, σK
l ,i
⊓
⊔
x
Remark 14.13 (Kernels). Definition 14.8 implies that ker(γK,F
|PK ) ⊂
∩i∈NK,F ker(σK,i ), whereas (14.8) means that the other inclusion holds. Hence,
x
Assumption 14.10 holds if and only if ∩i∈NK,F ker(σK,i ) = ker(γK,F
⊓
⊔
|PK ).
14.4 Examples
We now present various examples of finite elements satisfying the assumptions
of §14.3. For brevity, we focus on matching simplicial meshes; all the examples
can be adapted to matching Cartesian meshes.
14.4.1 Lagrange elements
The Lagrange finite elements from §6.3 can be used to build discrete functions
g
with integrable gradient. Let the γ g -trace be defined by (14.3a) and let ψK
be
the pullback by the geometric map TK . We do not necessarily consider affine
−1
(F );
meshes. Let F ∈ Fh and let K ∈ Th be such that F ∈ FK . Set Fb := TK
b since the mesh is matching. Let NK,F be the collection of the
Fb is a face of K
indices of the Lagrange nodes in F , i.e., aK,i ∈ F for all i ∈ NK,F . Note that
g
card(NK,F ) = Pk,d−1 , see, e.g., Table 6.1. Let us set σK,i
(φ) = φ(aK,i ) for all
g
g
g
g
i ∈ NK,F and all φ ∈ γK,F (PK ) =: PK,F . Let us set ΣK,F = {σK,i
}i∈NK,F .
Part III. Finite Element Spaces
181
g
g
Lemma 14.14 (Verification of Assumption 14.10). The triple (F, PK,F
, ΣK,F
)
is a (Lagrange) finite element, i.e., Assumption 14.10 holds.
Proof. We use Lemma 14.9 and verify that (14.8) holds. Let p ∈ PK be
such that σK,i (p) = p(aK,i ) = 0 for all i ∈ NK,F . Recalling that p = pb ◦
−1
TK
with pb ∈ Pk,d and that aK,i = TK (b
ai ), we infer that pb(b
ai ) = 0 for all
i ∈ NK,F . Moreover, {b
ai }i∈NK,F are the reference Lagrange nodes located in
−1
b follows from the fact that the mesh is
Fb = TK
(F ); that Fb is a face of K
matching. We infer from the proof of Proposition 6.12 that pb|Fb = 0. Since
g
g
−1
γK,F
(p) = pb|Fb ◦ TK|F
, we conclude that γK,F
(p) = 0.
⊓
⊔
Lemma 14.15 (Verification of Assumption 14.11).
tion 14.11 holds.
Then, Assump-
−1
Proof. Let us consider F ∈ Fh◦ with F = ∂Kl ∩ ∂Kr , and set Fbl = TK
(F )
l
−1
b
and Fr = TKr (F ). It follows from the assumption that we are working with a
b By construction, the mapping
matching mesh that Fbl and Fbr are faces of K.
−1
b
b
Trl := TKl ◦ TKr |Fbr : Fr −→ Fl is bijective. Even when the mappings TKl and
TKr are nonlinear, the mapping Trl turns out to be affine, see Exercise 14.6.
−1
−1
Then, the mapping Srl = TK
◦ TKr ,F = TFb−1 ◦ TK
◦ TKr ◦ TFbr is affine
l ,F
l
l
d−1
b
and bijective from S
onto itself, see Figure 14.3. This implies in particular
that Pk,d−1 is invariant by Srl , i.e., Pk,d−1 ◦ Srl = Pk,d−1 . Using this property
g
−1
proved above, we
together with the generic identity PK,F
= Pk,d−1 ◦ TK,F
infer that
g
g
−1
−1
−1
PK
= Pk,d−1 ◦ TK
= (Pk,d−1 ◦ Srl ) ◦ TK
= Pk,d−1 ◦ TK
= PK
,
r ,F
r ,F
r ,F
l ,F
l ,F
g
g
g
g
which establishes that PK
= PK
. Let us now show that ΣK
= ΣK
.
r ,F
r ,F
l ,F
l ,F
⊓
⊔
T Fl
Sbd−1
TFbl
TFbr
T Fr
T Kl
Fbl
b
K
Fbr
Kl
F
Kr
T Kr
Fig. 14.3. Two-dimensional example (d = 2): geometric maps associated with an
interface F , the reference faces Fbl and Fbr , and the unit simplex Sbd−1 .
182
Chapter 14. Traces and interface jumps of finite elements
Fig. 14.4. 2D example (left) and counter-example (right) for Assumption 14.11 on
the P2 Lagrange element (the triangles Kl and Kr are drawn slightly apart).
14.4.2 Canonical hybrid elements
Consider affine meshes and assume d = 3 (the case d = 2 is similar, but
g
simpler). We consider again the γ g -trace defined by (14.3a), and ψK
is again
the pullback by the geometric map TK .
−1
Let F ∈ Fh and let K ∈ Th be such that F ∈ FK . Set Fb = TK
(F ).
Recall the affine bijective mappings TFb : Sbd−1 −→ Fb and TK,F := TK ◦ TFb :
Sbd−1 −→ F , and use a similar notation for the edges of F . Recall that
g
−1
PK,F
= Pk,d−1 ◦ TK,F
. The set NK,F now collects the indices of all the
degrees of freedom involving an integral over F of a product with a function from the fixed basis {ζm }m∈{1: nssh } of Pk−3,2 (see (6.11c)), an integral
over one of the edges E of F of a product with a function from the fixed
basis {µm }m∈{1: nesh } of Pk−2,1 (see (6.11b)), or the evaluation at one of the
vertices of F (see R(6.11a)). Observe that card(NK,F ) = dim(Pk,d−1 ). Let us
g
g
−1
)φ ds for some mi ∈ {1: nssh }, or σK,i
(φ) =
(φ) = |F1 | F (ζmi ◦ TK,F
set σK,i
R
−1
1
′
e
′
|E| E (µmi ◦ TK,E )φ ds for some edge E of F and some mi ∈ {1: nsh }, or
g
σK,i (φ) = φ(zji ) for some vertex zji of F . The fact that Assumption 14.10
holds results from Lemma 14.9 and the proof of Proposition 6.16. The resulting face element is a 2D canonical hybrid element.
Lemma 14.16 (Verification of Assumption 14.11). Assume that the chosen bases {ζm }m∈{1: nssh } and {µm }m∈{1: nesh } are invariant by permutation of
the vertices of the unit simplices Sb2 and Sb1 , respectively, i.e., for any affine
bijective mapping S : Sb2 −→ Sb2 based on a vertex permutation of Sb2 , there
exists a permutation χS of the set {1: nssh } such that ζm ◦ S = ζχS (m) for
all m ∈ {1: nssh }, and similarly for the edge basis functions. Then, Assumption (14.11) holds.
g
g
Proof. We have already seen that PK
= PK
. Consider a degree of freer ,F
l ,F
g
dom in Kl , say σK
,
that
is
an
integral
over
the
face F of the product with
l ,i
the basis function ζmi , mi ∈ {1: nssh }. Let Srl be the affine bijective mapping
from Sbd−1 onto itself introduced in the proof of Lemma 14.15; note thatSrl
is based on a vertex permutation. We have
Z
Z
1
1
g
−1
−1
σK
(φ)
=
(ζ
◦
T
((ζm ◦ Srl ) ◦ TK
)φ
ds
=
)φ ds,
m
Kl ,F
r ,F
l ,i
|F | F
|F | F
g
for all φ ∈ PK
. The permutation invariance implies that there is χrl (m) such
l ,F
g
that ζm ◦ Srl = ζχrl (m) , so that σK
is one of the degrees of freedom in Kr
l ,i
Part III. Finite Element Spaces
183
attached to F . The reasoning is similar for the edges, and it is straightforward
for the vertices.
⊓
⊔
14.4.3 Raviart–Thomas elements
Consider affine meshes. The Raviart–Thomas elements from §10.3 can be used
to build discrete functions with integrable divergence. We consider the γ d trace defined by (14.3c) and the contravariant Piola transformation such that
Let F ∈ Fh and let K ∈ Th be such that F ∈ FK . Consider the reference
−1
b The degrees of freedom attached to F in the Raviart–
face Fb = TK
(F ) of K.
Thomas finite element generated in K are integrals over F of the component
−1
1
),
JT (b
nFb ◦ TK|F
along the unit normal vector to F given by nsK,F = kJT n
b kℓ2 K
K bF
see §10.4. Since this vector may be different from nF , we consider the family
composed of the 2d+1 reference simplices with all the possible orientations of
the faces (2 possible orientations per face) and we pick in this family the unique
b Fbd defining the
b Fb0 , . . . n
reference finite element such that the unit vectors n
reference degrees of freedom are mapped to nF0 , . . . nFd by the contravariant
Piola transformation, i.e., nsK,i F = nFi , i ∈ {0: d}.
Recall the affine bijective mappings TFb : Sbd−1 −→ Fb and TK,F := TK ◦
TFb : Sbd−1 −→ F . Let NK,F be the collection of the degrees of freedom of
R
−1
) ds for some mi ∈ {1: nssh } where
the form σK,i (v) := F (v·nF )(ζmi ◦ TK,F
{ζm }m∈{1: nssh } is the fixed basis of Pk,d−1 . Note that card(NK,F ) = nssh =
R
−1
d
)φ ds for all i ∈ NK,F
dim(Pk,d−1 ). Finally, let us set σK,i
(φ) = F (ζmi ◦ TK,F
d
d
d
RTk,d ), and ΣK,F = {σK,i }i∈NK,F .
and all φ ∈ γK,F (RT
d
d
RTk,d ) =
Lemma 14.17 ([Verification of Assumption 14.11). PK,F
:= γK,F
(RT
−1
d
d
Pk,d−1 ◦ TK,F , and (F, PK,F
, ΣK,F
) is a (modal) finite element, i.e., Assumption 14.10 holds.
Proof. Assumption 14.10 follows from Lemma 14.9 and Lemma 10.12. More−1
d
d
RTk,d ) ⊂ Pk,d−1 ◦ TK,F
over, PK,F
:= γK,F
(RT
owing to Lemma 10.7. Equality
then follows from a dimension argument.
⊓
⊔
Lemma 14.18 (Permutation invariance). Assume that the chosen basis
{ζm }m∈{1: nssh } of Pk,d−1 is invariant by vertex permutations of Sbd−1 . Then,
Assumption 14.11 holds.
−1
Proof. Let F ∈ Fh◦ with F = ∂Kl ∩ ∂Kr , and set Fbl = TK
(F ) and Fbr =
l
−1
−1
TKr (F ). Recall the mapping Srl = TKl ,F ◦ TKr ,F , see Figure 14.3. Then we
−1
−1
−1
d
observe that PK
= Pk,d−1 ◦TK
= (Pk,d−1 ◦Srl )◦TK
= Pk,d−1 ◦TK
=
l ,F
r ,F
r ,F
l ,F
d
PKr ,F , as in the proof of Lemma 14.15. Moreover, letting χrl be the index
d
permutation associated with the mapping Srl , any degree of freedom σK
in
l ,i
d
d
ΣKl ,F is also in ΣKr ,F since
184
Chapter 14. Traces and interface jumps of finite elements
d
σK
(φ)
l ,i
=
=
Z
Z
−1
TK
)φ ds
l ,F
Z
F
(ζmi ◦
F
−1
(ζχrl (mi ) ◦ TK
)φ ds.
r ,F
=
F
−1
)φ ds
(ζmi ◦ Srl ◦ TK
r ,F
⊓
⊔
14.4.4 Nédélec elements
Consider affine meshes. The Nédélec elements from §11.3 can be used to build
discrete functions with integrable curl. Assume that d = 3 (the construction is
analogous, but simpler, for d = 2). We consider the γ c -trace defined by (14.3b)
c
and the covariant Piola transformation ψK
(v) = JT
K (v ◦ TK ).
Let F ∈ Fh and let K ∈ Th be such that F ∈ FK . Consider the reference
−1
b The degrees of freedom attached to F in the Nédélec
face Fb = TK
(F ) of K.
finite element generated in K are integrals over the three edges E ⊂ ∂F of the
b = T −1 (E)) and
component along the tangent vectors tK,E = JK tbEb (with E
K
integrals over F of the component along the tangent vectors tK,F,j = JK tbFb,j ,
j ∈ {1, 2}, see §11.4. Let us now fix all the tangent vectors on all the faces of
K once and for all. This means that for any mesh face F ∈ Fh , we fix the two
face vectors tF,j , j ∈ {1, 2}, and for any mesh edge E ∈ Eh , we fix the edge
vector tE . We now pick among the 26 ×(12×22 ) possible reference elements
the unique one such that tE = JK tbEb for all the edges E and tF,j = JK tbFb,j
for all the faces.
Let us now verify that Assumption 14.10 holds. Let F ∈ Fh and let K ∈
Th be such that F ∈ FK . Let E be any of the three edges of F . Recall
the affine bijective maps TFb (resp., TEb ) from the unit simplex Sb2 of R2 to
b and let us set TK,F := TK ◦ T b : Sb2 −→ F
Fb (resp., Sb1 of R1 to E),
F
(resp., TK,E := TK ◦ TEb : Sb1 −→ E). Let NK,F be the collection of the
R
−1
) ds
degrees of freedom that of the form σK,i (v) := |F1 | F (v·tF,j )(ζmi ◦ TK,F
for some j ∈ {1, 2} and mi ∈ {1: nssh }, where {ζm }m∈{1: nssh } is the fixed basis
R
−1
1
(v·tE )(µm′i ◦ TK,E
) dl for some
of Pk−1,2 , or of the form σK,i (v) := |E|
E
′
e
edge E of F and mi ∈ {1: nsh }, where {µm }m∈{1: nesh } is the fixed basis of
Pk,1 . Note that card(NK,F ) = 2× dim(Pk−1,2
1)(k +
R ) + 3 dim(Pk,1 ) = (k + −1
c
Nk,2 ). Let us set σK,i
) ds
3) = dim(N
(φ) = |F1 | F ((nF ×φ)·tF,j )(ζmi ◦ TK,F
R
−1
1
c
or σK,i (φ) = |E| E ((nF ×φ)·tE )(µm′i ◦ TK,E ) dl for all i ∈ NK,F and all
c
c
c
Nk,d ), and ΣK,F
φ ∈ γK,F
(N
= {σK,i
}i∈NK,F . Since (nF ×(h×nF ))·t = h·t for
3
c
c
any h ∈ R and any t tangent vector on F , σK,i = σK,i
◦γK,F
for all i ∈ NK,F .
c
c
Lemma 14.19 (Face element). (F, PK,F
, ΣK,F
) is a (Nédélec) finite ele−T
−1
c
Nk,2 ◦TK,F )×nF , i.e., Assumption 14.10 holds.
ment and PK,F = JK,F (N
Proof. Assumption 14.10 results from Lemma 14.9 and Lemma 11.11. That
−1
c
c
Nk,3 ) ⊂ J−T
Nk,2 ◦ TK,F
PK,F
:= γK,F
(N
)×nF can be shown as in the proof
K,F (N
of Lemma 11.7, and equality results from a dimension argument.
⊓
⊔
Part III. Finite Element Spaces
185
Lemma 14.20 (Permutation invariance). Assume that the chosen basis
{ζm }m∈{1: nssh } and {µm }m∈{1: nesh } are invariant by vertex permutations of Sb2
and Sb1 , respectively. Then, Assumption 14.11 holds.
⊓
⊔
Proof. Similar to that of Lemma 14.18.
Exercises
Exercise 14.1 (H(div)- and H(curl)-conformity). Prove Theorem 14.7.
(Hint: use Theorem B.113 for the divergence and Corollary B.115 for the
curl.)
Exercise 14.2 (Orientation). Assume that the tetrahedron from Figure 3.9,
say Km , has geometric nodes such that (j(i, m))i∈{1: 4} = (42, 67, 35, 89). Determine the type of the corresponding reference tetrahedron.
Exercise 14.3 (NK,F ). Prove that, under Assumption 14.10, i ∈ NK,F if and
x
only if γK,F
(θK,i ) 6= 0.
b Assume
Exercise 14.4 (Reference face element). Let Fb be a face of K.
bnx : PbFxb :=
that there is a subset NFb ⊂ N and that there are linear forms σ
for all n ∈ N b . Assume that the
bn = σ
bnx ◦ γ x
γ x (Pb ) −→ R such that σ
b F
b
K,
b F
b
K,
F
b x := {b
b x ) with Σ
σnx }n∈NFb is a finite element. Let F = TK (Fb )
triple (F, PbFxb , Σ
b
b
F
F
x
be a face of K. Assume that there is a transformation ψK,F : PK,F
−→ PbFxb
x
x
such that ψK,F ◦ γK,F = γK,
b F
b ◦ ψK . Verify that the face element generated
x bx
b
on F from (F, P , Σ ) using the transformation ψK,F is the finite element
b
F
b
F
x
x
, ΣK,F
) from Assumption 14.10 with NK,F = NFb .
(F, PK,F
Exercise 14.5 (Permutation invariance). Let Sb1 = [0, 1] and consider the
bases B1 = {µ1 (s) = 1 − s, µ2 (s) = s} and B2 = {µ1 (s) = 1, µ2 (s) = s}. Are
these bases invariant by permutation of the vertices of Sb1 ?
Exercise 14.6 (Affine map between faces). Let F ∈ Fh◦ with F = ∂Kl ∩
−1
−1
(F ). Prove that the mapping Trl =
∂Kr , and set Fbl = TK
(F ) and Fbr = TK
r
l
−1
b
TKl ◦TKr |Fbr is affine. (Hint: let {λi }i∈{0: d}
b
be the barycentric coordinates of K,
assume without loss of generality that
T Kl
Kl
b0 = 0}, let {b
b ∩ {λ
b
Fbl = ∂ K
zi }i∈{1: d}
x
Fbl
be the vertices of Fbl , and let Tlrλ :
Kr
F
b
K
Fbl −→ Fbr be defined such that Tlrλ (b
x) = yb Fbr
P
b x)Tlr (b
b ∈ Fbl . Note
zi ) for all x
TK−1r
i∈{1: d} λi (b
λ
that Tlr is affine and bijective, see the figure.)