An adaptive two-level method for hypersingular integral equations in R 3
Patrick Mund ∗ Ernst P. Stephan ∗
(Received 7 August 2000)
Abstract
In this paper an a posteriori error estimate for hypersingular inte- gral equations is derived by using hierarchical basis techniques. Based on the properties of a two-level additive Schwarz method easily com- putable local error indicators are obtained. An algorithm for adaptive error control which allows anisotropic refinements of the boundary elements is formulated and numerical results are included.
∗
Institut f¨ ur Angewandte Mathematik, Universit¨ at Hannover, Welfengarten 1, 30167 Hannover, Germany. mailto:[email protected]
0
See http://anziamj.austms.org.au/V42/CTAC99/Mund for this article and ancillary
services, c Austral. Mathematical Soc. 2000. Published 27 Nov 2000.
Contents C1020
Contents
1 A stable two-level subspace decomposition C1020
2 An a posteriori error estimate C1026
3 Numerical results C1028
References C1032
1 A stable two-level subspace decomposition
In recent years adaptive hierarchical basis methods in the finite element method (fem) [1, 2, 4] have become increasingly popular. This approach has meanwhile been also applied to the boundary element method (bem) for weakly singular integral equations [8] and to the fem/bem coupling [7].
Here we extend it to hypersingular integral equations on surfaces.
We consider the hypersingular integral equation W v(x) := − 1
4π
∂
∂n
xZ
Γ
v(y) ∂
∂n
y1
|x − y| dσ
y= f (x) , x ∈ Γ (1)
where Γ is an open plane surface and integration has to be understood in the
Hadamard sense. W is a bijective mapping from ˜ H
1/2(Γ) onto H
−1/2(Γ) and
the bilinear form hW u, vi for u, v ∈ ˜ H
1/2(Γ) is symmetric and positive definite (cf. Costabel [3], Stephan [9 ]). Hence the unique solution v
N∈ S
N⊂ ˜ H
1/2(Γ) of the Galerkin scheme
hW v
N, χi = hf, χi ∀χ ∈ S
N(2) converges quasi-optimally towards the exact solution v ∈ ˜ H
1/2(Γ) of (1). For standard Sobolev spaces L
2(Γ) and H
01(Γ) (which is the completion of C
0∞within H
1(Γ)) the space ˜ H
1/2(Γ) is an interpolation space between L
2(Γ) and H
01(Γ), and H
−1/2(Γ) is the dual space of ˜ H
1/2(Γ). The solution u of ( 1) for given f ∈ H
−1/2(Γ) is the jump across Γ of the solution of a Neumann problem for the Laplacian in R
3\¯Γ, cf. [ 9].
In the following we present a two-level method for the h-version of the Galerkin scheme with bilinear continuous elements. For ease of presentation we consider Γ = [ −1, 1]
2.
Let γ
l(0 ≤ l ≤ L) be a uniform partition of Γ into squares of side length h
l= 2
−land let
X
l= {u ∈ C
0(Γ) : u piecewise bilinear w.r.t. γ
land u
|∂Γ= 0 } (0 ≤ l ≤ L) X
l0= {u ∈ X
l: u = 0 on the nodes of γ
l−1} (1 ≤ l ≤ L).
Let b
l,i(0 ≤ i ≤ n
l) be a piecewise bilinear function with value one at
an interior node of γ
l(not belonging to γ
l−1) and with zero value in all
other nodes of γ
l. Let X
L,i0= span {b
L,i}. We have the following two-level
1 A stable two-level subspace decomposition C1022 decomposition of the space X
L:
X
L= X
L−1⊕ X
L,10⊕ · · · ⊕ X
L,n0 L(3) Let
P
(2)L= P
L−1+
nL
X
i=1
P
L,i(4)
be the two-level additive Schwarz operator belonging to the subspace decom- position (3) and the bilinear form hW ·, ·i, i.e.
hW P
L−1φ, ψi = hW φ, ψi ∀ψ ∈ X
L−1, φ ∈ X
L, hW P
L,iφ, ψi = hW φ, ψi ∀ψ ∈ X
L,i0, φ ∈ X
L. Then there holds
Theorem 1 There exist constants c
1, c
2> 0, independent of L, such that c
1hW u, ui ≤ hW P
(2)Lu, ui ≤ c
2hW u, ui ∀u ∈ X
L. (5) The proof of Theorem 1 is based on the following lemmas where always u ∈ X
Larbitrary with
u = u
L−1+
nL
X
i=1
u
L,iwhere u
L−1∈ X
L−1and u
L,i∈ X
L,i0. Let I
L−1u ∈ X
L−1be the bilinear
interpolant of u at the nodes of γ
L−1.
Lemma 2 [6 , Lemma 2.5, Lemma 3.10] There exists a constant c, inde- pendent of L and u, such that
ku − I
L−1uk
L2(Γ)≤ ch
Lkuk
H1(Γ). (6) Furthermore for any s ∈ [0, 1] there exists a constant c = c(s) > 0 such that
kI
L−1uk
H˜s(Γ)≤ ckuk
H˜s(Γ)for any u ∈ X
L. (7)
Lemma 3 [6, Lemma 3.12] Let Γ
Li∈ γ
Lbe a square with vertices x
k(1 ≤ k ≤ 4). Let v, w be bilinear functions on Γ
Liwith v(x
1) = w(x
1) and w(x
2) = w(x
3) = w(x
4) = 0 . Then there holds
kwk
L2(ΓLi)≤ 4
3 kvk
L2(ΓLi).
Lemma 4 [5] Let {Γ
i, i = 1, . . . , N} be a finite covering of Γ with rectan- gles Γ
iand covering constant σ ∈ N, i.e. we can colour {Γ
i, i = 1, . . . , N}
by at most σ different colours such that subdomains with same colour are disjoint. Let φ = P
Ni=1
φ
i∈ ˜ H
s(Γ) for s ∈ R with φ
i∈ ˜ H
s(Γ
i). Then there holds
kφk
2H˜s(Γ)≤ σ X
Ni=1
kφ
ik
2H˜s(Γi).
1 A stable two-level subspace decomposition C1024 Proof: (of Theorem 1 ) We show that there exist constants c
1, c
2> 0 such that
1
c
2kuk
2H˜1/2(Γ)≤ ku
L−1k
2H˜1/2(Γ)+
nL
X
i=1
ku
L,ik
2H˜1/2(Γ)≤ 1
c
1kuk
2H˜1/2(Γ). (8)
The left inequality follows directly from Lemma 4 with c
2= 9 since to any x ∈ Γ there belong at most 8 different b
L,i’s with x ∈ suppb
L,i(1 ≤ i ≤ n
L) and since
ku
L,ik
H˜1/2(Γ)= ku
L,ik
H˜1/2(supp
bL,i).
It remains to show the right inequality in (8 ). Since u
L−1= I
L−1u Lemma 2 implies
ku
L−1k
H˜1/2(Γ)≤ ckuk
H˜1/2(Γ). (9) Let {x
i}
ni=1Ldenote the set of nodes in γ
Lwhich do not belong to γ
L−1. We decompose the index set {1, 2, . . . , n
L} = M
1∪ M
2∪ M
3into 3 disjoint sets M
ksuch that two indices i, j ∈ {1, 2, . . . , n
L} belong to the same set M
kif
|x
i− x
j| is an integer multiple of 2h
L. The sets M
kare uniquely determined (up to permutation) (cf. Fig. 1 ). For k ∈ {1, 2, 3} the nodes {x
i}
i∈Mkare just the nodes of the coarse grid γ
L−1shifted by h
Lin the x
1- and/or x
2-direction.
There holds for k ∈ {1, 2, 3}
meas(supp b
L,i∩ supp b
L,j) = 0 ∀i, j ∈ M
k, i 6= j
1 1
1
1
1
1 3
3
3 3
3 3
2 2
2 2
Figure 1: Decomposition of the nodes γ
Linto disjoint subsets: Number k ∈ {1, 2, 3} of the node x means x ∈ M
k. The nodes of the coarse mesh γ
L−1are marked with •.
2 An a posteriori error estimate C1026 with the Lebesgue-measure meas( ·). Hence
X
i∈Mk
ku
L,ik
20= X
i∈Mk
u
L,i2
0
≤ 16
9 k˜uk
20(10)
where ˜ u = u − u
L−1and k · k
s= k · k
H˜s(Γ), s ∈ IR.
Here the last inequality follows from Lemma 3 since w = P
i∈Mk
u
L,iand ˜ u coincide at one node of each element in γ
Land w vanishes in all the other nodes. With the standard inverse inequality for finite elements and (10) and (6) we have
nL
X
i=1
ku
L,ik
21/2≤ ch
−1LP
nLi=1
ku
L,ik
20= ch
−1LX
3 k=1X
i∈Mk
ku
L,ik
20≤ c
0h
−1Lk˜uk
20≤ c
0h
−1Lku − I
L−1uk
20≤ ch
Lkuk
21≤ c kuk
21/2.
Together with (9) this implies the right inequality in (8). Thus the proof is complete due to the equivalence of the norms kuk
1/2and hW u, ui
1/2. ♠
2 An a posteriori error estimate
Next we need the saturation assumption: (A
h) There exist constants k
0∈ N and 0 < ρ < 1 such that
kv − v
k+1k
1/2≤ ρkv − v
kk
1/2∀k ≥ k
0where v
k+1denotes the Galerkin solution on the level k + 1 and v the exact solution of (1).
Theorem 5 Suppose (A
h) holds. Then there exist constants c
1, c
2> 0 such that there holds for k ≥ k
0c
1 nLX
j=1
η
2L,j≤ kv − v
L−1k
21/2= c
2 nLX
j=1
η
L,j2(11)
with
η
L,j= |hf − W v
L−1, b
L,ji|
hW b
L−1, b
L,ji
1/2(j = 1, 2, . . . , n
L) . (12)
Proof: The saturation assumption (A
h) yields the equivalence of norms kv
L− v
L−1k
1/2∼ kv − v
L−1k
1/2.
Due to Theorem 1 we have
c
1kv
L− v
L−1k
21/2≤ kP
L−1(v
L− v
L−1) k
21/2+
nL
X
i=1