Introduction to the Boundary Element Method

Introduction to the Boundary Element Method

Salim Meddahi

University of Oviedo, Spain

University of Trento, Trento April 27 - May 15, 2015

1

Potential theory The Sobolev setting

2

Can we extend the technique to the usual Sobolev setting?

We want generalize the previous strategies in the following situation

I Ω ⊂ R^d(d = 2, 3) bounded polygonal/polyhedral and Lipschitz domain

I Jγ uK := ψ ∈ H

1/2(Γ) and J∂ⁿuK := λ ∈ H

−1/2(Γ)

The first step consists in finding weak versions of the single layer potential Sλ(x) :=

Z

Γ

E(x, y)λ(y)dsy x ∈ Ω ∪ Ω^e and the double layer potential

Dψ(x) :=

Z

Γ

∂n_yE(x, y)ψ(y)dsy x ∈ Ω ∪ Ω^e

3

The Sobolev space H

(Ω) and the trace operator

I The norm in L²(Ω) is denoted k·k0,Ω

I H¹(Ω) is the classical Sobolev space of order one

I The norm in H¹(Ω) is denoted k·k1,Ω I H¹²(Γ) := {ψ ∈ L²(Γ); |ψ|1/2,Γ< ∞} with

|ψ|²1/2,Γ:=

Z

Γ

Z

Γ

|ψ(x) − ψ(y)|²

|x − y|^d dsydsx

I The restriction operator defined for v ∈ C⁰( ¯Ω) by γv := v|Γhas a unique extension to a bounded linear operator

γ : H¹(Ω) → H¹²(Γ)

I For any ψ ∈ H¹²(Γ), there exists Eψ ∈ H¹(Ω) such that γEψ = ψ and kEψk1,Ω≤ Ckψk1/2,Γ ∀ψ ∈ H¹²(Γ)

4

The normal trace operator

I Since the inclusion H¹²(Γ) ⊂ L²(Γ) is dense and continuous, we can consider the Gelfand triple

H¹²(Γ) ⊂ L²(Γ) ⊂ H⁻¹²(Γ) := (H¹²(Γ))⁰

I The duality product will be denoted hλ, ϕiΓand if λ ∈ L²(Γ) hλ, ϕiΓ=

Z

Γ

λ(x)ϕ(x)dsx ∀ϕ ∈ H¹²(Γ)

I The linear operator ∂n: {v ∈ H¹(Ω); ∆v ∈ L²(Ω)} → H⁻¹²(Γ) given by h∂nv, ϕiΓ:=

Z

Ω

∇v · ∇Eϕ + ∆v Eϕ ∀ϕ ∈ H¹²(Γ) is well-defined and continuous

5

A weighted Sobolev space

For d = 2 and d = 3 we consider, W¹(Ω^e) :=n

u : R^d→ R; ρ u ∈ L²(Ω^e), ∇u ∈ L²(Ω^e)^do

where ρ(x) :=







(1 + |x|²)^−1/2 if d = 3 (1 + |x|²)^−1/2log(2 + |x|²)⁻¹ if d = 2.

and introduce W0¹(Ω^e) :=D(Ω^e).

I W¹(Ω^e) is a Hilbert space when endowed with kvk²_W1(Ω^e):= kρvk²_0,Ωe+ k∇vk²_0,Ωe I 1 ∈ W¹(Ω^e) for d = 2 but 1 /∈ W¹(Ω^e) for d = 3

I W¹(Ω^e) ⊂ H_loc¹ (Ω^e) (the weight only affects the behaviour at infinity)

I H¹(Ω^e) ,→ W¹(Ω^e)

I There is a bounded right-inverse for the trace operator γ^e: W¹(Ω^e) → H¹²(Γ)

6

A norm equivalence

Proposition 1.

The norms v 7→ k∇vk0,B^e₁ and v 7→ kvk_W1(B^e₁) are equivalent on W0¹(B1^e).

Theorem 2.

The mapping v 7→ k∇vk0,Ω^e+ (3 − d)|R

Γγ^ev| defines a norm on W¹(Ω^e) and W₀¹(Ω^e) that is equivalent to v 7→ kvk_W1(Ω^e).

Proposition 3.

Assume that d = 3. For any λ ∈ H⁻¹²(Γ) and ψ ∈ H¹²(Γ), the problem

Find u ∈ H¹(Ω) × W¹(Ω^e) such thatJγ uK = ψ and Z

Ω∪Ω^e

∇u · ∇v = hλ, γviΓ ∀v ∈ W¹(R³) (1) is well-posed.

7

Canonical splitting of the solution

It is clear that if u solves Problem (1) then:

I u = (uⁱ, u^e) ∈ H¹(Ω) × W¹(Ω^e)

I ∆u = 0 in Ω ∪ Ω^e

I Jγ uK = ψ and J∂ⁿuK = λ By superposition

u = ˜S(J∂ⁿuK) −D(˜ Jγ uK) where

H⁻¹²(Γ) 3 λ → ˜S(λ) ∈ W¹(R³) satisfies Z

R³

∇ ˜S(λ) · ∇v = hλ, γvi ∀v ∈ W¹(R³) and

H¹²(Γ) 3 ψ → ˜D(ψ) ∈ H¹(Ω) × W¹(Ω^e) is such thatJγD(ψ)˜ K = −ψ and Z

Ω∪Ω^e

∇ ˜D(ψ) · ∇v = 0 ∀v ∈ W¹(R³)

Corollary 4.

The operators ˜S : H⁻¹²(Γ) → W¹(R³) and ˜D : H¹²(Γ) → H¹(Ω) × W¹(Ω^e) are linear and bounded.

8

Weak form of the integral representation formula

Given λ ∈ H⁻¹²(Γ) and ψ ∈ H¹²(Γ), we consider the compactly supported distributions γ^tλ and ∂n^tψ given by

hγ^tλ, ϕi = hλ, γϕiΓ and h∂_n^tψ, ϕi = hψ, ∂nϕiΓ ∀ϕ ∈ D(R³)

Theorem 5.

Assume that w ∈ H¹(Ω) × W¹(Ω^e) and ∆w = 0 in Ω ∪ Ω^e then

−∆w = γ^tJ∂ⁿwK − ∂

t

nJγ wK in D⁰(R³).

Moreover, w is given by

w = F ∗ γ^tJ∂ⁿwK − F ∗ ∂

t nJγ wK and it holds that

F ∗ γ^tλ = ˜Sλ ∀λ ∈ H⁻¹²(Γ) and

F ∗ ∂n^tφ = ˜Dφ ∀φ ∈ H¹²(Γ)

We conclude from the theorem that the unique solution u of (1) is given by u = F ∗ γ^tλ − F ∗ ∂_n^tψ

9

Properties of S and D

As a consequence, we have the following result.

Theorem 6.

There holds the mapping properties (i) S : H⁻¹²(Γ) → W¹(R³) (ii) γS : H⁻¹²(Γ) → H¹²(Γ) (iii) ∂nS, ∂n^eS : H⁻¹²(Γ) → H⁻¹²(Γ)

(iv) D : H¹²(Γ) → H¹(Ω) × W¹(Ω^e) (v) γD, γ^eD : H¹²(Γ) → H¹²(Γ) (vi) ∂nD : H¹²(Γ) → H⁻¹²(Γ) and the jump conditions

(vii) Jγ S λK = 0 ∀λ ∈ H⁻¹²(Γ) (viii) Jγ DϕK = −ϕ ∀ϕ ∈ H¹²(Γ)

(ix) J∂ⁿSλK = λ ∀λ ∈ H⁻¹²(Γ) (x) J∂ⁿDϕK = 0 ∀ϕ ∈ H¹²(Γ)

10

Boundary integral operators

Definition 7.

We introduce the boundary integral operators

I V := γS : H⁻¹²(Γ) → H¹²(Γ) (Single-layer operator)

I W := −∂nD : H¹²(Γ) → H⁻¹²(Γ) (Hypersingular operator)

I K := {γD} : H¹²(Γ) → H¹²(Γ) (Double-layer operator)

I K^t:= {∂nS} : H⁻¹²(Γ) → H⁻¹²(Γ) (Dual of the double-layer operator)

With the above definitions, there holds, with I denoting the identity operator, (i) γS = V

(iv) ∂nS =¹₂I + K^t, ∂n^eS = −¹₂I + K^t

(ii) ∂nD = −W

(iii) γD = −¹₂I + K, γ^eD = ¹₂I + K

11

Ellipticity properties

Theorem 8.

The symmetric bilinear form aV(λ, µ) := hµ, V λiΓis bounded on H⁻¹²(Γ) and

∃α > 0; aV(λ, λ) ≥ α kλk²−1/2,Γ

for all λ ∈ H⁻¹²(Γ) if d = 3 and for all λ ∈ H⁻

1 2

0 (Γ) if d = 2.

Theorem 9.

The symmetric bilinear form aW(ψ, φ) := hW ψ, φiΓis bounded on H¹²(Γ) and

∃β > 0; aW(φ, φ) ≥ β kφk²

H¹2(Γ)/P0

∀ϕ ∈ H¹²(Γ)/P0.

Theorem 10.

aW(ψ, φ) = aV(curlΓψ, curlΓφ) ∀ψ, ϕ ∈ H¹²(Γ)

12

Kernels

Proposition 11.

ker(1

2I + K) = P0

Proposition 12.

ker(1

2I + K^t) = span{∂n^eu0} where u0 is the solution of

−∆u0 = 0 in Ω^e (2)

γ^eu = 1 (3)

13

The Calder´ on Projectors

We have shown that the interior and exterior Cauchy data of a harmonic function u in Ω ∪ Ω^e are related by

 γu

∂nu



= 1

2I + H  γu

∂nu



and γ^eu

∂n^eu



= 1

2I − H γ^eu

∂^enu



where

H =−K V

W K^t



It holds that

 1 2I + H

2

= 1 2I + H

 ,  1

2I − H

2

= 1 2I − H



Notice that we have ∂nu = DtN γu with DtN := V⁻¹(1

2I + K) : H¹²(Γ) → H⁻¹²(Γ)

is the Dirichlet-to-Neumann operator also known as the Steklov-Poincar´e map.

Using the second equation of the Calder´on projector we deduce that DtN = W + (1

2I + K^t)V⁻¹(1 2I + K)

which proves that DtN is H^1/2(Γ)/P0-elliptic 14

Direct and indirect boundary integral formulations for the Laplace problem

15

The interior Dirichlet problem

Given g ∈ H¹²(Γ), the

−∆u = 0 in Ω

u = g on Γ (4)

admits a unique weak solution u ∈ H¹(Ω).

I The indirect method: (d = 3) The problem

Find λ ∈ H⁻¹²(Γ); aV(λ, µ) = hµ, giΓ, ∀µ ∈ H⁻¹²(Γ)

has a unique solution and u = Sλ|Ω∈ H¹(Ω) is the unique (weak) solution of (4).

I The direct method: (d = 3) The problem

Find λ ∈ H⁻¹²(Γ); aV(λ, µ) = hµ, (1

2I + K)giΓ, ∀µ ∈ H⁻¹²(Γ) is well-posed and u = (Sλ − Dg)|Ω∈ H¹(Ω) is the unique (weak) solution of (4).

Moreover, λ = ∂nu.

16

The interior Neumann problem

Given g ∈ H⁻₀¹²(Γ), the problem

−∆u = 0 in Ω

∂nu = g on Γ (5)

admits a unique weak solution u ∈ H¹(Ω)/P0.

I The indirect method The problem

Find ψ ∈ H¹²(Γ)/P0; aW(ψ, φ) = hg, φiΓ, ∀µ ∈ H¹²(Γ)/P0

has a unique solution and u = −D(ψ)|Ωis the unique weak solution of (5) in H¹(Ω)/P0.

I The direct method The problem

Find ψ ∈ H¹²(Γ)/P0; aW(ψ, φ) = h(1

2I − K^t)g, φiΓ, ∀µ ∈ H¹²(Γ)/P0

is well-posed. Notice that the right hand-side is compatible since K1 = −¹₂ yields h(1

2I − K^t)g, 1iΓ= hg, (1

2I − K)1iΓ= hg, 1iΓ= 0.

The function given in Ω by u = Sg − Dψ is the unique weak solution of (5) in

H¹(Ω)/P0. ₁₇

The exterior Dirichlet problem

Given g ∈ H¹²(Γ), the problem

−∆u = 0 in Ω

u = g on Γ (6)

admits a unique weak solution u ∈ W¹(Ω^e).

I The indirect method: (d = 3) The problem

Find λ ∈ H⁻¹²(Γ); aV(λ, µ) = hµ, giΓ, ∀µ ∈ H⁻¹²(Γ)

has a unique solution and u = Sλ|Ω^e ∈ W¹(Ω^e) is the unique weak solution of (6).

I The direct method: (d = 3) The problem

Find λ ∈ H⁻¹²(Γ); aV(λ, µ) = −hµ, (1

2I − K)giΓ, ∀µ ∈ H⁻¹²(Γ) is well-posed and u = (−Sλ + Dg)|Ω^e∈ W¹(Ω^e) is the unique weak solution of (6).

Moreover, λ = ∂n^eu.

18

The exterior Neumann problem

Given g ∈ H⁻¹²(Γ), the problem

−∆u = 0 in Ω

∂_n^eu = g on Γ (7)

admits a unique weak solution u ∈ W¹(Ω^e).

I The indirect method:

The problem

Find ψ ∈ H¹²(Γ)/R; aW(ψ, φ) = h¯g, φiΓ, ∀µ ∈ H¹²(Γ)/R, with ¯g = g −^hg,1i_|Γ|^Γ, has a unique solution and u = −Dψ|Ω^e∈ W¹(Ω^e) is the unique weak solution of (7).

I The direct method:

The problem

Find ψ ∈ H¹²(Γ)/R; aW(ψ, φ) = −h(1

2I + K^t)g, φiΓ, ∀φ ∈ H¹²(Γ)/R is well-posed and u = (−Sg + Dψ)|Ω^e∈ W¹(Ω^e) is the unique weak solution of (7).

Moreover, ψ = γ^eu.

19

The BEM for the Laplacian

20

Surface meshes

We assume that Ω is polyhedral and Lipschitz. Hence Γ = ∪^J_j=1Γj, where each Γj is a plane panel. A surface mesh Thof Γ is a decomposition of Γ into finitely disjoint and non-degenerate triangles T ⊂ Γ,

Γ =∪T ∈T_hT .

Let ˆT := {(ˆx1, ˆx2) ∈ R²; 0 < ˆx2< ˆx1< 1} be the reference element. Given T ∈ Th

with vertices P0, P1, P2, the affine mapping FT: ˆT → R³

ˆ

x = (ˆx1, ˆx2) → FT(ˆx) := P0+ ˆx1(P1− P0) + ˆx2(P2− P0) satisfies FT( ˆT ) = T . We denote

BT= DFT= (P1− P0, P2− P0) ∈ R^3×2 Notice that, according to our hypothesis on Th, if GT := B^t_TBT then

detGT > 0 ∀T ∈ Th

21

Shape regular families of triangulations

The diameter of T ∈ This given by

hT= max

x,y∈T|x − y|

and the inner width ρT is the in-circle diameter of T . We set h = max

T ∈T_hhT. Definition 13.

I {Th}his shape regular if there exists c1> 0 independent of h such that

T ∈Tmax_h hT

ρT

≤ c1 ∀h.

I {Th}his quasi-uniform if there exists c2> 0 independent of h such that 1 < h

minT ∈T_hhT

≤ c2 ∀h.

I The mesh This said to be conforming if T 6= T⁰∈ Th ⇒ T ∩ T⁰=







∅

common side common vertex

22

Galerkin approximation of the exterior Dirichlet Laplace problem

We recall that the indirect boundary integral approach to solve the exterior Dirichlet Laplace problem (6) is

Find λ ∈ H⁻¹²(Γ); aV(λ, µ) = hµ, giΓ, ∀µ ∈ H⁻¹²(Γ) (8) Given a triangulation Th(not necessarily conforming) we consider the discrete problem

Find λh∈ Yh; aV(λh, µ) = hµ, giΓ, ∀µ ∈ Yh (9) where Yh⊂ H⁻¹²(Γ) is given by

Yh:= {µ ∈ L²(Γ); µ|T ∈ P0, ∀T ∈ Th}

Theorem 14 (C´ea).

There exists a constant C > 0 independent of h such that kλ − λhk−1/2,Γ≤ C inf

µ_h∈Y_hkλ − µhk−1/2,Γ

23

Galerkin approximation of the exterior Neumann Laplace problem

The indirect boundary integral approach to solve the exterior Neumann Laplace problem (7) is

Find ψ ∈ H¹²(Γ); aW(ψ, φ) + h1, ψiΓh1, φiΓ= hφ, ¯giΓ, ∀φ ∈ H¹²(Γ) (10) Given a conforming triangulation Th, we consider the discrete problem

Find ψh∈ Xh; aW(ψh, φ) + h1, ψhiΓh1, φiΓ= hφ, ¯giΓ, ∀φ ∈ Xh (11) where Xh⊂ H¹²(Γ) is given by

Xh:= {φ ∈ C⁰(Γ); φ|T ∈ P1, ∀T ∈ Th}

Theorem 15 (C´ea).

There exists a constant C > 0 independent of h such that kψ − ψhk1/2,Γ≤ C inf

φ_h∈X_hkψ − φhk1/2,Γ

24

Approximation properties

25

Interpolation of operators

We assume here that Γj is one of the plane panels constituting Γ.

Theorem 16.

Assume that T ∈ L(L²(Γj), L²(Γj)) and T ∈ L(L²(Γj), H¹(Γj)). Then T ∈ L(L²(Γj), H^1/2(Γj)) and

kT k_L(L2(Γ_j),H^1/2(Γ_j))≤q

kT kL(L²(Γ_j),L²(Γ_j))

q

kT kL(L²(Γ_j),H¹(Γ_j))

Theorem 17.

Assume that T ∈ L(L²(Γj), H²(Γj) and T ∈ L(H¹(Γj), H²(Γj)). Then T ∈ L(H¹²(Γj), H²(Γj)) and

kT k_L(H1/2(Γ_j),H²(Γ_j))≤q

kT kL(L²(Γ_j),H²(Γ_j))

q

kT kL(H¹(Γ_j),H²(Γ_j))

26

Change of variable formulas

Proposition 18.

kGTk ≤ 2h²T kG⁻¹_T k ≤ 2 π²(hT

ρT

)⁴h⁻²_T

Proposition 19.

Given u : T → C, we consider ˆu(ˆx) := u(FT(ˆx)). Then, v ∈ H^m(T ) ⇔ ˆv ∈ H^m( ˆT ) and, for all 0 ≤ ` ≤ m,

|v|`,T ≤ C1h<sup>1−`</sup>_T |ˆv|_{`, ˆT}

|ˆv|<sub>`, ˆT</sub> ≤ C2h<sup>`−1</sup>_T |v|`,T

with C1 and C2 independent of T .

27

Interpolation error estimates in Y

We introduce for s = 1/2, 1, 2 the Sobolev spaces

H^spw(Γ) := {µ ∈ L²(Γ); v|Γ_j ∈ H^s(Γj), j = 1, · · · , J } Notice that H^1/2(Γ) ,→ H^1/2pw(Γ).

Theorem 20.

Let Πhbe the L²(Γ)-projection onto Yh. If {Th}his shape regular then there exists a constant C > 0 independent of h such that

kµ − Πhµk0,Γ≤ C h^tkµkH^t_pw(Γ) ∀µ ∈ H^tpw(Γ) (t = 1, 1/2)

Theorem 21.

Let Πhbe the L²(Γ)-projection onto Yh. If {Th}his shape regular then there exists a constant C > 0 independent of h such that

kµ − Πhµk₋1

2,Γ≤ C h^3/2kµk_H1

pw(Γ) ∀µ ∈ H¹pw(Γ)

28

Interpolation error estimates in X

We notice that H²_pw(Γ) ∩ H¹(Γ) ,→ C⁰(Γ). Consequently, if This conforming, we can then define the usual Lagrange interpolation operator Ih: H²pw(Γ) ∩ H¹(Γ) → Xh. Theorem 22.

If {Th}his shape regular then there exists a constant C > 0 independent of h such that kφ − Ihφk1/2,Γ≤ C h^3/2kφk_H2

pw(Γ) ∀φ ∈ H²pw(Γ) ∩ H¹(Γ)

29

Inverse estimates

Proposition 23.

Assume that {Th}his shape regular and quasi-uniform. Then, there exists a constant C > 0 independent of h such that,

kµhk1

2,Γ≤ Ch⁻¹²kµhk0,Γ ∀µh∈ Yh

Proposition 24.

Assume that {Th}his shape regular and quasi-uniform. Then, there exists a constant C > 0 independent of h such that,

kµhk0,Γ≤ Ch⁻¹²kµhk₋1

2,Γ ∀µh∈ Yh

30

Error estimates for the exterior Dirichlet Laplace problems

Assume that the solution λ ∈ H⁻¹²(Γ) of problem (6) belongs to H¹pw(Γ). Then, kλ − λhk₋1

2,Γ≤ C1h^3/2kλkH¹_pw(Γ)

with C1> 0 independent of h. Moreover, if {Th}his quasi-uniform, kλ − λhk0,Γ≤ C2h kλk_H1

pw(Γ)

Let us consider the function

˜ u =

Z

Γ

E(x, y)λh(y) x ∈ Ω^e.

As y 7→ E(x, y) ∈ H¹²(Γ), if λ ∈ H¹_pw(Γ) we have that

|u(x) − ˜u(x)| ≤ kE(x, ·)k1

2,Γkλ − λhk₋1

2,Γ≤ C h^3/2kE(x, ·)k1 2,Γkλk_H1

pw(Γ)

for all x ∈ Ω^e.

The last estimate can be improved to

|u(x) − ˜u(x)| = O(h³)

31

Error estimates for the exterior Neumann Laplace/Helmholtz problems

Theorem 26.

Assume that the solution ψ ∈ H¹²(Γ) of problem (7) belongs to H²pw(Γ) ∩ H¹(Γ). Then, kψ − ψhk1

2,Γ≤ Ch^3/2kψkH²_pw(Γ)

with C > 0 independent of h.

Here again, if we consider

˜

u(x) = −Dψh(x) x ∈ Ω^e then

|u(x) − ˜u(x)| ≤ kD1k₋1

2,Γkψ − ψhk1

2,Γ≤ C h^3/2kD1k₋1

2,Γkψk_H2 pw(Γ)

for all x ∈ Ω^e.

32