Introduction to the Boundary Element Method

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1 Introduction to the Boundary Element Method Salim Meddahi University of Oviedo, Spain University of Trento, Trento April 27 - May 15,

2 Syllabus The Laplace problem Potential theory: the classical approach Potential theory: the Sobolev setting Direct/indirect boundary integral formulations for the Laplace problem Boundary element method for the Laplacian The Poisson problem: The symmetric FEM-BEM formulation The Johnson-Nédélec FEM-BEM formulation Helmholtz equation Helmholtz equation with damping Helmholtz equation without damping Direct/indirect boundary integral formulations for the Helmholtz problem Boundary element method for Helmholtz problem A computational example in 2d 2

3 Main references Colton, D. and Kress, R., Inverse Acoustic and Electromagnetic Scattering Theory. Second edition, Springer-Verlag, Berlin, Kress, R., Linear Integral Equations. Springer-Verlag, Berlin, McLean, W., Strongly Elliptic Systems and Boundary Integral Equations. Cambridge University Press, Nédélec, J.-C. Acoustic and electromagnetic equations. Integral representations for harmonic problems. Applied Mathematical Sciences, 144. Springer-Verlag, New York, 2001 Sauter, S.A. and Schwab, C., Boundary Element Methods. Springer Series in Computational Mathematics, 39. Springer-Verlag, Berlin, Steinbach, O. Numerical approximation methods for elliptic boundary value problems. Finite and boundary elements. Springer, New York,

4 Prerequisites Distributions, convolution and Fourier transform Sobolev spaces Interpolation error estimates in Sobolev spaces Fredholm alternative 4

5 Potential theory The classical approach 5

6 Notations and conventions Ω R d (d = 2, 3) is bounded open set Ω is connected and has a connected complement Ω e = R d \ Ω := Ω is of class C n is the unit normal vector on pointing towards Ω e Given u : R d R, we introduce u i := u Ω and u e := u Ω e If u i C 0 ( Ω) and u e C 0 ( Ω e ) then, for any x γu(x) = lim Ω y x ui (y), γ e u(x) = lim Ω e y x ue (y) γu (x) := γu(x) γ e u(x) {γu} := γe u(x) + γu(x) 2 If u i C 1 ( Ω) and u e C 1 ( Ω e ) then, for any x nu = γ( u) n, nu (x) := nu(x) e nu(x) e nu = γ e ( u) n { nu} := e nu(x) + nu(x) 2 6

7 Motivation We aim to solve The exterior Dirichlet problem u = 0 in Ω e γ e u = φ on u(x) 0 when x The exterior Neumann problem u = 0 in Ω e e nu = λ on u(x) 0 when x 7

8 Integral representation in Ω The function F : R d \ {0} R defined by 1 1 4π x F (x) := 1 2π log 1 x if d = 3 if d = 2 is a fundamental solution of the Laplacian in d-dimensions, i.e., Theorem 1. R d F (x) ϕ(x) dx = ϕ(0) ϕ D(R d ). If u C 2 ( Ω) and u = 0 in Ω then u(x) = E(x, y) nu(y)ds y ny E(x, y)u(y)ds y, x Ω where E(x, y) := F (x y) 8

9 Integral representation in Ω e Corollary 2 (Mean value theorem). If u C 2 ( B r(x)) is harmonic in B r(x) := {y R d ; 1 u(x) = u(y)dy = B r(x) B r(x) 1 B r(x) x y < r} then, u(y)ds y B r(x) Theorem 3. If u C 2 ( Ω e ), u = 0 in Ω e and lim r max ˆx =1 u(rˆx) = 0 then u(x) = E(x, y) nu(y)ds e y + n e y E(x, y)u(y)ds y, In addition, in the case d = 2, e nu ds = 0. x Ω e As a consequence: u(x) = O(1/ x ) and u(x) = O(1/ x d 1 ) when x 9

10 Notice that and A transmission problem { u(x) x Ω E(x, y) nu(y)ds y ny E(x, y)γu(y)ds y = 0 x Ω e { E(x, y) nu(y)ds e y + ny E(x, y)γ e 0 x Ω u(y)ds y = u(x) x Ω e Hence, adding the two identities we get u(x) = E(x, y) nu (y)ds y ny E(x, y) γu (y)ds y x Ω Ω e We conclude that, given φ, λ C 0 (), the solution of the transmission problem u = 0 in Ω Ω e γu = φ on nu = λ on is given by u(x) = lim max u(rˆx) = 0 r ˆx =1 E(x, y)λ(y)ds y ny E(x, y)φ(y)ds y 10

11 The single and double layer potentials Given φ C 0 () we introduce the single layer potential Sφ(x) := E(x, y)φ(y) ds y and the double layer potential Dφ(x) := ny E(x, y)φ(y) ds y x Ω Ω e x Ω Ω e Since E C (R d \ {x}) and ye = 0 in R d \ {x}, Sφ, Dφ belong to C (Ω Ω e ) and (Sφ) = (Dφ) = 0 in Ω Ω e Moreover, (Sφ)(x) = 1 ( φ) log(1/ x ) + O(1/ x ) 2π (d = 2) as x (Dφ)(x) = O(1/ x ) (Sφ)(x) = O(1/ x ) (d = 3) as x (Dφ)(x) = O(1/ x 2 ) 11

12 Boundary integral operators The operators C 0 () φ V φ(x) = E(x, y)φ(y)ds y, x C 0 () φ Kφ(x) = ny E(x, y)φ(y)ds y, x and C 0 () φ K t φ(x) = nx E(x, y)φ(y)ds y, x exist as improper integrals. Theorem 4 (Jump properties). It holds that, for any φ C 0 (), γsφ = 0 nsφ = φ γdφ = φ ndφ = 0 12

13 It can be shown that γdφ = nsφ = Boundary integral operators ( 12 ) ( ) I + K φ γ e 1 Dφ = 2 I + K φ ( ) 1 2 I + Kt φ nsφ e = ( 12 ) I + Kt φ Moreover, Kφ, ψ = φ, K t ψ φ, ψ C 1 () The remaining boundary integral equation W φ = ndφ = ndφ e has formally the hyper-singular kernel nx ny E(x, y) = 1 ( ) n(x) n(y) n(x) (x y)n(y) (x y) d ω d x y d x y d+2 W should be interpreted as a distribution represented by the Hadamard finite-part corresponding to this kernel 13

14 Positiveness of V and W Theorem 5. Suppose that φ, ψ C 0 (). If d = 3 then V φ, ψ := Sφ Sψ Ω Ω e and V φ, φ = 0 implies φ = 0. When d = 2, we have the same result provided φ = 0. Theorem 6. If d = 3, for every φ, ψ C 0 (), W ψ, φ = Dψ Dφ Ω Ω e and W φ, φ = 0 implies φ = constant. Moreover, if φ, ψ C 1 (), W ψ, φ = curl φ, V curl ψ 14

15 Reduction of an interior/exterior Laplace problem to a boundary integral equation 15

16 The interior Dirichlet problem u = 0 in Ω u = g on The indirect method u = Sλ in Ω where λ : R (with λ = 0 if d = 2) solves V λ = g on The direct method u = S nu Dγu in Ω and then we have to solve for λ := nu : R the boundary integral equation ( ) 1 V λ = 2 I + K g 16

17 The interior Neumann problem u = 0 in Ω nu = g on g = 0 The indirect method u = Dφ where φ : R, φ = 0 solves W φ = g in Ω on The direct method u = S nu Dγu in Ω and then we have to solve for φ := γu : R the boundary integral equation ( ) 1 W φ = 2 I Kt g Notice that K1 = 1/2. 17

18 The exterior Dirichlet problem u = 0 in Ω e u = g on, u(x) 0 as x The indirect method u = Sλ in Ω e where λ : R (with λ = 0 if d = 0) solves V λ = g on The direct method u = S e nu + Dγ e u in Ω e and then we have to solve for λ := nu e : R the boundary integral equation ( ) 1 V λ = 2 I K g 18

19 The exterior Neumann problem u = 0 in Ω e e nu = g on, u(x) 0 as x The indirect method u = Dφ where φ : R, φ = 0 solves W φ = g in Ω e on The direct method u = S nu + Dγu in Ω e and then we have to solve for φ := γu : R the boundary integral equation ( ) 1 W φ = 2 I + Kt g 19

The Factorization Method for Inverse Scattering Problems Part I

The Factorization Method for Inverse Scattering Problems Part I Andreas Kirsch Madrid 2011 Department of Mathematics KIT University of the State of Baden-Württemberg and National Large-scale Research Center