FINITE ELEMENT APPROXIMATION
JEONGHUN LEE
1. A short review of approximation theory
In this section we review two topics of basic approximation theory, the
Bramble–Hilbert lemma and the scaling argument.
We introduce multi-index notation in Rn . Let
α pα1 , , αn q,
αi : nonnegative integer, 1 ¤ i ¤ n.
We call α a multi-index and define
|α| α1 αn, α! α1! αn!.
For x px1 , , xn q P Rn and an |α|-differentiable function u on Rn ,
B|α|u .
xα xα1 xαn ,
Dα u α
Bx Bxα
1
n
1
1
n
n
If
on Rn of degree less than or equal to k, then u ° u is a polynomial
α
n
|α|¤k aα x with real numbers aα P R. For x0 P R the Taylor expansion
gives
¸ 1 α
D upx0 qpx x0 qα Op|x x0 |m 1 q.
upxq α!
|α|¤m
We denote α ¤ β for two multi-indices α, β if αi ¤ βi for 1 ¤ i ¤ n.
The subtraction β α is the multi-index that the ith coordinate is βi αi ,
1 ¤ i ¤ n. One can observe that
α β
D x
#
β!
pβ αq! x
β α,
0,
α ¤ β,
otherwise,
and
D α xα
α!.
Lemma 1 (Bramble–Hilbert). Let Ω be a convex domain in Rn with a
Lipschitz boundary. Suppose that u P H k 1 pΩq for which k ¥ 0. Then there
exists C ¡ 0 depending on Ω and k but independent of u such that
inf
in which |u|2k
°
}u p}k ¤ C |u|k
}Dαu}20.
Proof. We define a map Pk : H k pΩq Ñ Pk pΩq by
»
»
Dα Pk upxq dx Dα upxq dx,
1
:
P p q
p Pk Ω
Ω
|α|k
1,
1
Ω
1
|α| ¤ k.
2
JEONGHUN LEE
By induction one can show that Dβ Pk u
u P H k 1 p Ω q,
Pk|β | Dβ u for |β |
¤
k. For
¸
}u Pk u}0 ¤ dπΩ }∇pu Pk uq}0 dπΩ }Dαpu Pk uq}20
1
2
|α|1
,
by the Poincare inequality. From the definition of Pk we can apply the
Poincare inequality for Dα u Dα Pk u for |α| ¤ k. If we use the above
argument inductively, then
¸
}u Pk u}0 ¤ C }Dαpu Pk uq}20 ¤ C 1|u|k
1
2
|α|k
because Dα Pk u 0 for |α| k
1,
1.
Corollary 1. Let Ω € Rn be a bounded domain with a Lipschitz boundary
and Π : H k 1 pΩq Ñ Pk pΩq be a bounded linear projection (as an operator
from H k 1 pΩq to H k pΩq). Then there exists C ¡ 0 depending only on Ω, k
and }Π}H k 1 ÑH k such that
}u Πu}k ¤ C |u|k 1.
Proof. Note that Πp p for p P Pk pΩq. By the triangle inequality and the
boundedness of Π,
}u Πu}k }u p Πpu pq}k ¤ }u p}k }Π}}u p}k 1
¤ }u p}k }Π}p}u p}k |u|k 1q,
for any p P Pk pΩq. The last inequality is due to the fact Dα p 0 for
|α| k 1. Since p is arbitrary, we obtain the conclusion by the Bramble–
Hilbert lemma.
Now we introduce the scaling argument. For simplicity we restrict our
discussion in the two dimensions. Let T̂ be the triangle with vertices v̂0 p0, 0qT , v̂1 p1, 0qT , v̂2 p0, 1qT . Let T be a triangle in R2 with vertices
v0 , v1 , v2 and F : T̂ Ñ T be the affine map such that vi F pv̂i q, k 0, 1, 2.
Denoting the coordinates on T̂ and T by x̂ and x, one can see that F x̂ B x̂ v0 with a 2 2 matrix B pbij q, 1 ¤ i, j ¤ 2. In fact,
pv1 v0|v2 v0q,
with columns vectors v1 v0 and v2 v0 .
°
Define û by u F , so ûpx̂q upF x̂q upxq. Regarding xi 2j 1 bij x̂j
B
v0,i
B xi b .
Bx̂j ij
FINITE ELEMENT APPROXIMATION
Thus
3
Bûpx̂q 2̧ Bupxq Bxi 2̧ Bu pxqb .
Bx̂j i1 Bxi Bx̂j i1 Bxi ij
Similarly,
B2ûpx̂q ¸ b b B2u pxq,
Bx̂iBx̂j 1¤k,l¤2 ik jl Bxk Bxl
and by an induction,
¸
|α|k
|D̂αûpx̂q| ¤ c}B }k
¸
|β |k
|Dβ upxq|,
in which }B } sup1¤i,j ¤2 |bij |. A similar argument gives
¸
|Dβ upxq| ¤ c}B 1}k
¸
|D̂αûpx̂q|.
|α|k
|β |k
1
In general, }B } is not necessarily bounded, so we assume that T satisfies
the minimum angle condition, i.e., the minimum angle of T is greater than
or equal to a certain fixed angle θ0 . Let ρT be the circumference of the
inscribed circle of T and it is known that hT ¤ cρT with a uniform constant
c ¡ 0 if T satisfies the minimum angle condition.
In the definition of B, }B } OphT q. For }B 1 }, consider the inscribed
circle in T . Under B 1 the points that the circle contacting the triangle are
mapped to three points ?
on B T̂ . Since the distance of any two point s on B T̂
is less than or equal to 2 ?
and the distance of any two points on B T is less
than ρT , we have }B 1 } ¤ 2{ρT . Thus
¸
|α|k
|D̂αûpx̂q| ¤ chkT
¸
|Dβ upxq|,
|β |k
¸ α
k
|D upxq| ¤ cρT
|D̂ ûpx̂q|.
|β |k
|α|k
¸
Note that
β
»
»
}v} |vpxq| dx |ûpx̂q|2| detpB q|dx̂ 2|T |}û}20,T̂ ,
T
T̂
in which |T | denotes the area of T .
» ¸
» ¸
2k
2
β 2
|u|k,T |D u| dx ¤ cρT
|D̂αûpx̂q|2dx
2
0,T
2
| |k
»
2k
2c|T |ρ
T β
T
¸
| |k
T̂ α
| |k
T α
|D̂αûpx̂q|2dx̂ 2c|T |ρT2k |û|2k,T̂ .
A similar argument gives
2
|û|2k,T̂ ¤ |Tc | h2k
T |u|k,T .
4
JEONGHUN LEE
As an application of the scaling argument, we consider the approximation
of the linear Lagrange finite element. Suppose that u P H 2 pT q and IT be the
Lagrange interpolation map into P1 pT q. On T̂ , IT̂ is defined similarly. As a
consequence of the Bramble–Hilbert lemma, }ûIT̂ û}1,T̂ ¤ c|û|2,T̂ . From the
{
definition of IT and IT̂ , one can see that Iy
T u IT̂ û, so û IT̂ û u IT u.
By the scaling argument,
a
a
}u IT u}0,T ¤ c |T |}û IT̂ û}0,T̂ ¤ c |T ||û|2,T̂ ¤ ch2T |u|2,T .
For a mesh Th and the corresponding Lagrange interpolation operator Ih ,
we have
}u Ihu}0 ¤ ch2|u|2,
by applying the above argument for each triangle T . Here h maxT PTh hT .
Exercise
Let T be a shape regular triangle and suppose that u P H 1 pT q,
³
E u ds 0 for an edge E of T . Prove the following inequality using the
scaling argument.
}u}0,E ¤ chT }∇u}0,T .
1
2
2. H pdivq and H pcurlq elements
In this section we introduce four families of finite element spaces. Two of
the spaces are H pdivq spaces and the other two are H pcurlq spaces. These
spaces can be understood in a unified framework by the finite element exterior calculus (FEEC). In FEEC, they are nothing but canonical finite
element differential forms of certain form degrees. However, we will not introduce FEEC in this section because learning differential form language is
a heavy load to most numerical analysis researchers. Thus we will describe
those elements and investigate their features using only vector calculus language. Nonetheless the author does emphasize that learning FEEC is highly
recommendable because it gives not only a unified understanding on those
mixed finite element spaces but also a new perspective on the theory of finite
element methods.
We briefly review basics of the H pcurl, Ωq space for Ω € R3 , which is
defined by
H pcurl, Ωq : tτ
For σ, τ
P L2pΩ; R3q | curl τ P L2pΩ; R3qu.
P C 1pΩ; R3q we have an integration by parts formula
»
»
»
curl σ τ dx pσ τ q n ds
σ pcurl τ q dx.
Ω
BΩ
Ω
Based on this integration by parts formula one can define trace of the tangential component of τ P H pcurl, Ωq. One can check that the tangential
trace τ n is in H 1{2 pB Ω; R3 q and the trace map from H pcurl, Ωq to
H 1{2 pB Ω; R3 q is bounded but not surjective.
FINITE ELEMENT APPROXIMATION
5
Exercise Let τ P Pk pTh ; R3 q. Then τ P H pcurlq if and only if the tangential
component of τ on every interior face F is continuous.
Now we describe shape functions and local degrees of freedom of the
Raviart–Thomas–Nedelec (RTN), Brezzi–Douglas–Marini (BDM), the Nedelec H pcurlq elements of the first and second kinds (Ned1 , Ned2 ). The first
two are H pdivq elements and the latter two are H pcurlq elements. In the following description, T denotes a triangle or a tetrahedron and V pT q denotes
the set of shape functions on T . An edge on the triangle T or a face on the
tetrahedron T will be called a facet. We will use E, F to denote an edge
and a facet on B T . In two dimensions, edges and facets are same objects.
Here x will denote a column vector
x
x1
x x2 ,
x1
,
x2
xK x3
x 2
x1
,
in two and three dimensions, and Pk pT q, Hk pT q will denote the space of
polynomials on T of degree ¤ k and the space of homogeneous polynomials
on T of degree k. Throughout the following descriptions, we assume k ¥ 1.
Raviart–Thomas–Nedelec spaces (RTNk )
Shape functions :
(1)
V pT q pPk1 pT qqn
xPk1 pT q pPk1 pT qqn
xHk1 pT q.
Local degrees of freedom :
»
(2)
τ
ÞÝÑ pτ nF qp ds,
p P Pk1 pF q,
(3)
τ
ÞÝÑ
ξ
»F
T
τ ξ dx,
P pPk2pT qqn,
k
¥ 2.
k
¥ 2.
Dimension :
(2D) : k pk 2q,
1
(3D) : k pk 1qpk
2
2 q.
Brezzi–Douglas–Marini spaces (BDMk )
Shape functions :
V pT q pPk pT qqn .
(4)
Local degrees of freedom :
»
(5)
(6)
τ
τ
ÞÝÑ pτ nF qp ds,
p P Pk pF q,
ÞÝÑ
ξ
»
F
T
τ ξ dx,
P Ned1k1pT q,
6
JEONGHUN LEE
Dimension :
(2D) : pk 1qpk 2q,
1
(3D) : pk 1qpk 2qpk
2
3 q.
In the above description, Ned1k1 is the Nedelec H pcurlq element of the first
kind which will be defined below.
In order to describe the Nedelec H pcurlq elements, we first define an
auxiliary polynomial space
Nk pT q : tη
P pHk pT qqn | ηpxq x 0 @x P T u.
Nedelec H pcurlq elements of the first kind (Ned1k )
Shape functions :
(7)
V pT q pPk1 pT qqn
Nk p T q #
x K H k 1 pT q ,
x pHk1 pT qq3 ,
pPk1pT qqn
pPk1pT qqn
p2Dq,
p3Dq.
Local degrees of freedom :
»
(8)
(2D)
(9)
(10)
(3D)
(11)
(12)
τ
ÞÝÑ pτ tE qp ds,
p P Pk1 pE q,
τ
ÞÝÑ
ξ
τ
τ
τ
»E
»
T
»
E
»
F
τ ξ dx,
P pPk2pT qq2,
¥ 2,
ÞÝÑ pτ tE qp dl,
p P Pk1 pE q,
ÞÝÑ pτ nF q ξ ds,
ξ
P pPk2pF qq2,
k
¥ 2,
ÞÝÑ
η
P pPk3pT qq3,
k
¥ 3.
T
τ η dx,
Dimension :
(2D) : k pk 2q,
1
(3D) : k pk 2qpk
2
3 q.
Nedelec H pcurlq elements of the second kind (Ned2k )
Shape functions :
(13)
k
V pT q pPk pT qqn .
FINITE ELEMENT APPROXIMATION
»
7
Local degrees of freedom :
(14)
(2D)
(15)
(16)
(3D)
(17)
τ
ÞÝÑ pτ tE qp ds,
τ
ÞÝÑ
τ
ÞÝÑ pτ tE qp dl,
p P Pk1 pE q,
ÞÝÑ pτ nF q ξ ds,
ξ
τ η dx,
η
τ
(18)
τ
ÞÝÑ
»E
»T
»
E
»
F
T
p P Pk pE q,
τ ξ dx,
ξ
P RTNk1pT q,
k
¥ 2,
P RTNk1pF q,
k
¥ 2,
P RTNk2pT q,
k
¥ 3.
Dimension :
(2D) : pk 1qpk 2q,
1
(3D) : pk 1qpk 2qpk 3q.
2
2.1. Counting DOFs and dimensions of shape functions. In order to
check unisolvency of these shape functions and DOFs, we need to know the
dimensions of Pk pRn q and Hk pRn q for which k ¥ 1, n ¥ 1. It is well-known
in combinatorics that the number of monomials with n indeterminates of
degree k ¥ 1, is equal to
n k1
pnpn k1q!k!1q! .
n1
Furthermore, one can see that dim Pk pRn q
that
R : Hk p R n
dim Hk pRn
1
q
by checking
q Ñ Pk pRnq, pRhqpx1, , xnq hpx1, , xn, 1q,
x1
x1
n
n 1
k
H : Pk pR q Ñ Hk pR
q, pHpqpx1, , xn, xn 1q xn 1p x , , x
,
1
n 1
are their inverse maps. In conclusion,
pn k 1q! , dim P pRnq dim H pRn
dim Hk pRn q k
k
pn 1q!k!
1
n 1
q pnn!k!kq! .
Exercise (Euler’s identity) Let p P Hk pRn q for which k
¥ 1. Then x∇p kp.
xHk pT q Ñ Pk pT q for k ¥ 0.
Exercise Consider the map div : pPk pT qq
Show that this map is surjective. (Hint : Euler’s identity)
n
For a sequence of vector spaces Vi and linear maps di : Vi
the sequence
di 1
di2
di1
di
ÝÑ
Vi1 ÝÑ Vi ÝÑ
Vi 1 ÝÑ Ñ Vi
1,
i P Z,
is called semi-exact at ith stage if di di1 0. In other words, Imdi1 €
Kerdi . Furthermore, if Imdi1 Kerdi , then the sequence is called exact at
8
JEONGHUN LEE
ith stage. The sequence is called semi-exact (exact, resp.) if the sequence is
semi-exact (exact, resp.) at all stages.
The following famous example is the de Rham sequence which is semiexact in general.
P0 pΩq ãÑ C 8 pΩq ÝÑ C 8 pΩ; R3 q ÝÑ C 8 pΩ; R3 q ÝÑ C 8 pΩq ÝÑ 0.
∇
curl
div
The exactness of the de Rham sequence is related to topological quantities
of Ω, called the Betti numbers.
Let T be a triangle and we claim that the follow sequence is exact.
(19)
P0 pT q ãÑ Pk pT q ÝÑ pPk1 pT qq2
curl
xHk1 pT q ÝÑ Pk1 pT q ÝÑ 0.
div
0
Since the kernel of curl is P0 pT q, the sequence is exact at Pk pT q. By a
consequence of one of the above exercises, div is surjective, so the sequence is
exact at Pk1 pT q, the second stage from the end. Thus we only need to check
the exactness at the third stage, which is equivalent to Im curl Ker div. To
check it, we use dimension counting argument. By the dimension theorem
in linear algebra,
dimppPk1 pT qq2
xHk1 pT qq dimpKer divq
dimpKer divq
Again by the dimension theorem,
dim Pk pT q dimpKer curlq
dimpIm curlq 1
dimpIm divq
dim Pk1 pT q.
dimpIm curlq.
One can check that dimpIm curlq dimpKer divq by counting dimensions
using the above identities. By a similar argument, we can show that
(20)
P0 pT q ãÑ Pk pT q ÝÑ pPk1 pT qq2
∇
P0 pT q ãÑ Pk
(21)
P0 pT q ãÑ Pk
(22)
are exact sequences for all k
xHk1 pT q ÝÑ Pk1 pT q ÝÑ 0,
rot
0
div
0
curl
Pk1 pT q ÝÑ 0,
pT q ÝÑ
pPk pT qq2 ÝÑ
∇
0
2 rot
1 pT q ÝÑ pPk pT qq ÝÑ Pk1 pT q ÝÑ 0,
1
¥ 1 (exercise).
As an application of the above exact sequences we show the unisolvency of
two dimensional RTN elements. One can check that the dimension of V pT q
and the number of DOFs are same (exercise). To prove the unisolvency,
suppose that τ P V pT q and all DOFs of τ given by (2) and (3) vanish.
We first claim that τ |E nE is in Pk1 pE q. To see it, let τ ξ xp
for which ξ P pPk1 pT qq2 and p P Hk1 pT q. It is obvious that ξ |E nE P
Pk1 pE q. For xp, note that x nE is constant on E because E is included in
the line represented by nE x b for some b P R. Thus p|E x nE P Pk1 pE q.
Since the DOFs in (2) are vanishing and τ |E nE P Pk1 pE q, τ nE 0
on E. If k 1, there is no more to prove, so assume that k ¥ 2. By the
integration by parts,
»
T
pdiv τ q2 dx »
T
τ ∇pdiv τ q dx 0,
FINITE ELEMENT APPROXIMATION
9
where the last equality is due to the fact ∇pdiv τ q P Pk2 pT q and (3), and
therefore we have div τ 0.
Let τ ξ xp for which ξ P pPk1 pT qq2 and p P Hk1 pT q. By the Euler’s
identity,
divpxpq divpxqp
x ∇p pk
1qp,
so div τ div ξ pk 1qp. Recall that div ξ P Pk2 pT q and p P Hk1 pT q,
so div τ 0 implies that p 0 and div ξ 0. In other words, τ ξ P
pPk1pT qq2. By the exact sequence (21), there exists q P Pk pT q such that
curl q τ . Now it is enough to show that q 0.
Note that
Bq ,
0 τ |E nE curl q |E nE ∇q |E tE Bt
which implies that q is constant on B T . We may assume that q 0 on B T
because curl q does not change under constant addition. If k 2, there is
no more to prove because q P P2 pT q is vanishing on B T , so q 0. If k ¥ 3,
q has a form of q bT q̃ where bT is the cubic bubble function on T and
q̃ P Pk3 pT q. There exists η P pPk2 pT qq2 such that rot η q̃ by (22). By
the vanishing DOFs of (3), and the integration by parts,
0
»
T
curl q ηdx »
T
q rot η dx »
bT q̃ 2 dx,
T
hence q̃ 0, and thus, q 0.
To check unisolvency for general RTNk , BDMk , Ned1k , Ned2k elements, we
need to know dimensions of the shape functions and the numbers of DOFs.
In the definition of the finite elements, the only unknown quantity is dim Nk .
In two dimensions, dim Nk is nothing but dim Hk1 pR2 q. However, dim Nk
in three dimensions is not obvious and we will show that it is
kpk 2q.
We first note that the map pHk pR3 qq3 Ñ Hk 1 pR3 q defined by
τ ÞÝÑ x τ,
τ P pHk pR3 qq3 ,
is surjective because any p P Hk 1 pR3 q can be written as p x1 p1 x2 p2
x3 p3 for some pi P Hk pR3 q, i 1, 2, 3. Since Nk is the kernel of the above
dim Nk
map, the dimension theorem in linear algebra gives
dimpHk pR3 qq3
dim Nk
By a direct computation we have dim Nk
dim Hk
k pk
Lemma 2. Let T be a tetrahedron in R3 . Then
1
2 q.
pR 3 q .
pHpT qq3 Nk pT q ` ∇Hk 1pT q.
Proof. It is obvious that Nk pT q ∇Hk 1 pT q € pHpT qq3 . To see the opposite inclusion, it is enough to show that Nk pT q X ∇Hk 1 pT q t0u because dim ∇Hk 1 pT q dim Hk 1 pT q and dim Nk pT q dim Hk 1 pT q (23)
10
JEONGHUN LEE
dimpHpT qq3 . Suppose that τ P Nk pT q and τ ∇p for some p P Hk 1 pT q.
Then 0 x τ x ∇p pk 1qp by Euler’s identity, thus τ ∇p 0. 2.2. Proof of unisolvency. We will prove unisolvency of the four families
of finite elements only for the three dimensions. Two dimensional cases are
left as exercises.
Let T be a tetrahedron T with vertices v0 , v1 , v2 , v3 . We set Fi as the
face which does not meet vertex vi , and λi as the barycentric coordinate,
i.e., the linear polynomial on T attaining 1 at vi and 0 at all other vertices
for i 0, 1, 2, 3. From this definition one can see that ∇λi is a vector which
is normal to the face Fi .
Lemma 3. For τ
τ
P pPk pT qq3 there exist pi, qi P Pk pT q, i 1, 2, 3 such that
3̧
q1∇λ2 ∇λ3
pi ∇λi
q2 ∇λ3 ∇λ1
q3 ∇λ1 ∇λ2 .
i 1
Proof. Observe that
t∇λ1, ∇λ2, ∇λ3u,
t∇λ1 ∇λ2, ∇λ2 ∇λ3, ∇λ3 ∇λ1u,
are bases of R3 . The polynomials pi , qi ’s are obtained by rewriting τ with
these bases.
Lemma 4. Suppose that τ P pPk pT qq3 and τ n
for some p1 , p2 , p3 P Pk1 pT q, τ has a form of
τ
p1λ1p∇λ2 ∇λ3q
p2 λ2 p∇λ3 ∇λ1 q
Proof. By the previous lemma,
τ
q1∇λ2 ∇λ3
q2 ∇λ3 ∇λ1
0 on F1, F2, F3.
Then,
p3 λ3 p∇λ1 ∇λ2 q.
q3 ∇λ1 ∇λ2 .
On F1 , since nF1 is parallel to ∇λ1 ,
0 τ |F1 nF1
q1nF p∇λ2 ∇λ3q.
The quantity nF p∇λ2 ∇λ3 q  0 is non-vanishing because it is the
determinant of three vectors nF , ∇λ2 , ∇λ3 . Thus q1 0 on F1 and q1 p1 λ1 for some p1 P Pk1 pT q. A similar argument works for F2 and F3 , and
1
1
1
the conclusion follows.
Raviart–Thomas–Nedelec spaces (RTNk )
Suppose that τ P V pT q of RTNk and all DOFs in are vanishing. Vanishing
DOFs of normal components leads to τ n 0 on B T . By the integration
by parts, and the vanishing interior DOFs, we have
»
T
pdiv τ q
2
dx »
T
τ ∇pdiv τ q dx 0,
so div τ 0. In the definition of V pT q, τ
p P Hk1 pT q. Then p 0 because
0 div τ
div τ0
divpxpq div τ0
pk
τ0
xp for τ0
P pPk1pT qq3,
2qp P Pk2 pT q ` Hk1 pT q.
FINITE ELEMENT APPROXIMATION
11
If k 1, proof is complete because τ P pP0 pT qq3 and τ n 0 on B T implies
τ 0. Hence we assume that k ¥ 2. By Lemma 4,
p1λ1p∇λ2 ∇λ3q p2λ2p∇λ3 ∇λ1q p3λ3p∇λ1 ∇λ2q.
for some p1 , p2 , p3 P Pk2 pT q. Take ξ p1 ∇λ1 p2 ∇λ1 p3 ∇λ3 P pPk2 pT qq3 .
τ
By the vanishing interior DOFs assumption and direct computations,
0
therefore pi
»
T
τ ξ dx » 3̧
T
0 for i 1, 2, 3.
p2i λi
dxp∇λ1 ∇λ2 q ∇λ3 ,
i 1
Brezzi–Douglas–Marini spaces (BDMk )
Suppose that τ P V pT q and all DOFs of τ vanish. By a similar argument
in the RTNk proof, we have τ n 0 on B T and div τ 0. For p P Hk pT q,
0
»
T
div τ p dx »
T
τ ∇p dx.
Combining it with the vanishing interior DOFs assumption, we obtain
»
(24)
T
τ ξ dx 0,
ξ
P pPk1pT qq3,
because
pPk1pT qq3 pPk2pT qq3 ` Nk1 ` ∇Hk pT q
Nedk1 ` ∇Hk pT q.
pby(23)q
By Lemma 4,
p1λ1p∇λ2 ∇λ3q p2λ2p∇λ3 ∇λ1q p3λ3p∇λ1 ∇λ2q.
for some p1 , p2 , p3 P Pk1 pT q. If we take ξ p1 ∇λ1 p2 ∇λ1 p3 ∇λ3 P
pPk2pT qq3 and use (24), then we can derive p1 p2 p3 0.
Lemma 5. For ξ P pHk pT qq3 , the followings hold.
(25)
curl x ξ x div ξ pk 2qξ,
(26)
x curl ξ ∇x ξ pk 1qξ.
τ
Proof. The proof is left as an exercise.
Lemma 6. The following decompositions hold.
(27)
(28)
pHk pT qq3 ∇Hk 1pT q ` x pHk1pT qq3,
pHk pT qq3 curlpHk 1pT qq3 ` xHk1.
Proof. We first prove that
pHk pT qq3 € curlpHk 1pT qq3 xHk1,
pHk pT qq3 € ∇Hk 1pT q x pHk1pT qq3.
12
JEONGHUN LEE
P pHk pT qq3. By (25) and (25),
2qξ curlpx ξ q x div ξ
P curlpHk 1pT qq3
1qξ x curl ξ ∇px ξ q
P x pHk1pT qq3
To see it, let ξ
pk
pk
xHk1 pT q,
∇Hk
1
pT q .
The opposite direction of inclusion is trivial, so it is enough to show that
curlpHk
pT qq3 X xHk1pT q t0u,
x pHk1 pT qq3 X ∇Hk 1 pT q t0u.
Suppose that ξ P pHk1 pT qq3 X ∇Hk 1 pT q, i.e., ξ ∇p x η for some
p P Hk 1 pT q and η P pHk1 pT qq3 . Then, by Euler’s identity,
0 x px η q x ξ x ∇p pk 1qp,
thus ξ ∇p 0. To complete proof, suppose that ξ P curlpHk 1 pT qq3 X
xHk1 pT q, i.e., ξ curl η xp for some η P pHk 1 pT qq3 and p P Hk1 pT q.
By (27), we may assume that η x η̃ for η̃ P pHk pT qq3 because curl ∇ 0.
1
Then
0 x pxpq x ξ
x pcurl ηq x pcurlpx η̃q x div η̃q pk 2qx η̃,
where the last equality is due to (25), therefore η x η̃ 0, and ξ curl η 0.
Corollary 2. The following holds.
Nk pT q x pHk1 pT qq3 .
Proof. In the definition of Nk pT q, it is obvious that x pHk1 pT qq3
Recall the decompositions
€ Nk p T q .
pHk pT qq3 Nk pT q ` ∇Hk 1pT q,
pHk pT qq3 x pHk1pT qq3 ` ∇Hk 1pT q,
in (23) and (27). These direct sums means that Nk pT q and x pHk1 pT qq3
are vector spaces of same dimensions, so they are same.
(To be appended)
Last updated : 2013, October 15.