Computational Electromagnetics; Chapter 3
9
3.2 Finite dimensional subspaces of H(div; Ω)
We consider conforming finite element approximations of the space H(div; Ω)
based on both simplicial and rectangular triangulations Th of a simply-connected
bounded domain Ω ⊂ lRd in case d = 2 and d = 3. Throughout the following
we denote by Nh , Eh , and Fh the sets of vertices, edges, and faces of the triangulation Th . In particular, for D ⊂ Ω, we refer to Nh (D), Eh (D), and Fh (D)
as the sets of vertices, edges, and faces in D.
We further denote by Pk (D) , k ∈ lN0 and P̃k (D) , k ∈ lN0 the set of polynomials of degree ≤ k and the set of homogeneous polynomials of degree k
on D. Moreover, Q`,m,n (D) , `, m, n ∈ lN0 refers to the set of polynomials in
(x1 , x2 , x3 )T ∈ D the maximum degree of which are ` in x1 , m in x2 , and n in
x3 . If ` = m = n = k, we simply write Qk (D) instead of Qk,k,k (D). Obvious
modifications apply in case D ⊂ lR2 .
3.2.1 Conforming elements for H(div; Ω)
Let Th be a triangulation of Ω and
PK := { q = (q1 , ..., qd )T | qi : K → lR , 1 ≤ i ≤ d } ,
Vh (Ω) := { qh : Ω̄ → lR | qh |K ∈ PK , K ∈ Th } .
(3.20)
(3.21)
The following result gives sufficient conditions for Vh (Ω) ⊂ H(div; Ω).
Theorem 3.7 Sufficient conditions for conformity
Let Th be a triangulation of Ω and let PK , K ∈ Th , and Vh (Ω) be given by
(3.20) and (3.21), respectively. Assume that
PK ⊂ H(div; K) , K ∈ Th ,
(3.22)
[n · q|e ] = 0 for all e = Ki ∩ Kj ∈ Eh (Ω) , q ∈ Vh (Ω), d = 2 , (3.23)
[n · q|f ] = 0 for all f = Ki ∩ Kj ∈ Fh (Ω) , q ∈ Vh (Ω), d = 3 , (3.24)
where [n · q|e ] and [n · q|f ] denote the jump of n · q across e and f , i.e.,
[n · q|g ] := n · q|g∩Ki − n · q|g∩Kj , g := e (d = 2) and g := f (d = 3) . (3.25)
Then Vh (Ω) ⊂ H(div; Ω).
Proof: We prove the result in case d = 3. The case d = 2 can be shown in
literally the same way.
Given qh ∈ Vh (Ω), we have to show that divqh is well defined and divqh ∈
L2 (Ω). In other words, we have to find zh ∈ L2 (Ω) such that
Z
Z
qh · grad ϕ dx = −
Ω
zh ϕ dx ,
Ω
ϕ ∈ D(Ω) .
10
Ronald H.W. Hoppe
In view of (3.22), Green’s formula can be applied elementwise:
Z
qh · grad ϕ dx =
X Z
divqh ϕ dx +
K∈Th K
= −
qh · grad ϕ dx =
K∈Th K
Ω
= −
X Z
X Z
n · qh |∂K ϕ dσ =
K∈Th ∂K
X Z
divqh ϕ dx +
K∈Th K
X
Z
[n · qh |f ] ϕ dσ .
f ∈Fh (Ω) f
Taking advantage of (3.24), the assertion follows for zh with zh |K := divqh ,
K ∈ Th .
3.2.2 Raviart-Thomas elements RTk (K)
Let us first consider the case of simplicial triangulations Th of Ω. For K ∈ Th
and k ∈ lN0 we set
(
2
Rk(∂K) := { ϕ ∈ L (∂K) |
ϕ|e ∈ Pk (e) , e ∈ Eh (K) , d = 2
}.
ϕ|f ∈ Pk (f ) , f ∈ Fh (K) , d = 3
Definition 3.8 Raviart-Thomas elements RTk (K)
Let K be a d-simplex. The Raviart-Thomas element RTk (K), k ∈ lN0 , is
defined by
RTk (K) = Pk (K)d + x P̃k (K) .
(3.26)
For q ∈ RTk (K), the degrees of freedom ΣK are given by
Z
q · n pk dσ , pk ∈ Rk (∂K) ,
(3.27)
q · pk−1 dx , pk−1 ∈ Pk−1 (K)d .
(3.28)
∂K
Z
K
We have
(
dim RTk (K) =
1
2
(k + 1)(k + 3) , d = 2 ,
(k + 1)(k + 2)(k + 4) , d = 3
.
(3.29)
@
@
@
I
@
PP
i
@
@
@
@
@
@
?
Fig. 3.1: RT0 (K) (d = 2)
@
@
1@
@
@
@
@
@
@
Fig. 3.2: RT0 (K) (d = 3)
Computational Electromagnetics; Chapter 3
11
@
@
iP
P
1
@
@
u
@
iP
P
1
@
@
@
iP
1
@P
@
@
@
@
I
@
@
@
u @ @
@
?
?
@
I
@
Fig. 3.3: RT1 (K) (d = 2)
Fig. 3.4: RT1 (K) (d = 3)
Lemma 3.9 Range of the discrete divergence operator
For q ∈ RTk (K) we have
div q ∈ Pk (K) ,
n · q|∂K ∈ Rk (∂K) .
(3.30)
(3.31)
Proof: According to (3.27), q can be written as
q = pk + xpk
,
pk ∈ Pk (K)d , pk ∈ P̃k (K) .
(3.32)
Then
div q = div pk + x · grad pk + 3 pk = div pk + (k + 3) pk
(3.33)
which gives (3.30).
Further, for n = (n1 , ..., nd )T we get
d
X
n · q = n · pk + pk
n i xi .
i=1
Since
Pd
i=1
ni xi = const. on e ⊂ ∂K resp. f ⊂ ∂K, this shows (3.31).
Lemma 3.10 Auxiliary result for unisolvence
Let q ∈ Pk (K)d and assume
Z
q · n pk dσ = 0 ,
pk ∈ Rk (∂K) ,
(3.34)
q · pk−1 dx = 0 ,
pk−1 ∈ Pk−1 (K)d
(3.35)
∂K
Z
K
Then
div q = 0 .
Proof: We apply Green’s formula
Z
K
Z
Z
q · grad pk−1 dx +
div q pk−1 dx = −
K
|
{z
=0
}
n · q pk−1 dσ
∂K
|
{z
=0
}
,
pk−1 ∈ Pk−1 (K) .
12
Ronald H.W. Hoppe
Since div q ∈ Pk−1 (K), we may choose pk−1 = div q which gives the assertion.
Theorem 3.11 Unisolvence of the RTk (K) element
The element (K, RTk (K), ΣK ) is unisolvent.
Proof: We have to show that the relations (3.34) and (3.35) imply q = 0.
Due to (3.30) we have n · q ∈ Rk (∂K) and hence, (3.34) implies n · q = 0 on
each edge resp. face.
In the same way as in Lemma 3.10 we deduce div q = 0. Observing Lemma
3.10 and (3.33), we get pk = 0.
Taking advantage of the affine equivalence, we consider the reference tetrahedron (cf. Figure 3.5).
x3
(0, 0, 1)
@
@
@
f4
@
@ (0, 1, 0)
@
x
2
(1, 0, 0)
x1
Fig. 3.5: The reference tetrahedron
In lights of the results obtained so far, we may assume
q̂i = x̂i ψ̂i
,
ψ̂i ∈ Pk−1 (K̂) , 1 ≤ i ≤ 3 .
If we choose p̂k−1 := (ψ̂1 , ψ̂2 , ψ̂3 )T in (3.33), we obtain
3 Z
X
i=1
x̂i ψ̂i2 dx = 0
K̂
whence ψ̂i = 0 , 1 ≤ i ≤ 3, and thus q̂ = 0.
Definition 3.12 The Raviart-Thomas finite element space RTk (Ω; Th )
The Raviart-Thomas finite element space RTk (Ω; Th ) is given by
RTk (Ω; Th ) := { q ∈ L2 (Ω)d | q|K ∈ RTk (K) , K ∈ Th } .
It is a finite dimensional subspace of H(div; Ω).
(3.36)
Computational Electromagnetics; Chapter 3
13
3.2.3 Raviart-Thomas elements RT[k] (K)
Let Th be a rectangular triangulation of Ω. In much the same way as before
we define
(
2
Qk (∂K) := {q ∈ L (K) |
q|e ∈ Pk (e)2 , e ∈ Eh (K) , d = 2
q|f ∈ Qk (f ) , f ∈ Fh (K) , d = 3
)
. (3.37)
Moreover, we set
(
Ψk (K) :=
Qk−1,k (K) × Qk,k−1 (K) , d = 2 ,
. (3.38)
Qk−1,k,k (K) × Qk,k−1,k (K) × Qk,k,k−1 (K) , d = 3
Definition 3.13 Raviart-Thomas elements RT[k] (K)
Let K be a d-rectangle. The Raviart-Thomas element RT[k] (K) , k ∈ lN0 , is
defined by
RT[k] (K) = Qk (K)d + x Qk (K) .
(3.39)
For q ∈ RT[k] (K), the degrees of freedom ΣK are given by
Z
q · n dσ , pk ∈ Qk (∂K) ,
(3.40)
∂K
Z
q · pk dx , pk ∈ Ψk (K) .
(3.41)
K
We have
(
dim RT[k] (K) =
2 (k + 1)(k + 2) , d = 2 ,
3 (k + 1)2 (k + 2) , d = 3
.
(3.42)
Theorem 3.14 Unisolvence of the RT[k] (K) element
The element (K, RT[k] (K), ΣK ) is unisolvent.
Proof: The proof is left as an exercise.
•
Definition 3.15 The Raviart-Thomas finite element space RT[k] (Ω; Th )
The Raviart-Thomas finite element space RTk (Ω; Th ) is given by
RT[k] (Ω; Th ) := { q ∈ L2 (Ω)d | q|K ∈ RT[k] (K) , K ∈ Th } .
It is a finite dimensional subspace of H(div; Ω).
(3.43)
14
Ronald H.W. Hoppe
3.3 Boundary element approximation of the boundary integral formulation of Maxwell’s exterior domain problem
We remind that the variational formulation of the boundary integral approach
to the exterior domain problem for the time-harmonic Maxwell equations is as
follows:
Find j ∈ H−1/2 (divΓ ; Γ) such that for all q ∈ H−1/2 (divΓ ; Γ):
Z
Z
G(x, y) j(y) dσ(y) ∧ nx (x)] · q(x) dσ(x) −
[nx (x) ∧
(3.44)
Γ
Γ
1 Z Z
− 2
G(x, y) divΓ j(y) divΓ q(x) dσ(y) dσ(x) =
k
Γ Γ
1 Z
=
π t (Einc )(x) · q(x) dσ(x) .
iωµ0
Γ
We assume that Th is a geometrically conforming simplicial triangulation of Γ
and approximate H−1/2 (divΓ ; Γ) by the Raviart-Thomas finite element space
RTk (Γ, Th ), k ∈ lN0 . The Galerkin approximation of (3.44) then reads as follows:
Find jh ∈ RTk (Γ, Th ) such that for all qh ∈ RTk (Γ, Th ):
Z
Z
G(x, y) jh (y) dσ(y) ∧ nx (x)] · qh (x) dσ(x) − (3.45)
[nx (x) ∧
Γ
Γ
1 Z Z
− 2
G(x, y) divΓ jh (y) divΓ qh (x) dσ(y) dσ(x) =
k
Γ Γ
1 Z
=
π t (Einc )(x) · qh (x) dσ(x) .
iωµ0
Γ