The Boundary Element Method for the Helmholtz Equation ... - FEI VÅ B

32 3 Boundary Integral EquationsBecause the function V κ s satisfies the Helmholtz equation in the weak sense, we actuallyhaveV κ : H −1/2 (∂Ω) → H 1 loc (Ω, ∆ + κ2 )and thus we can compose the Neumann trace operator with the single layer potentialoperator to obtain the linear continuous mappingγ 1 Vκ : H −1/2 (∂Ω) → H −1/2 (∂Ω).Theorem 3.10. For s ∈ H −1/2 (∂Ω) there holdsγ 1,int ( V κ s)(x) = σ(x)s(x) + (K ∗ κs)(x) for x ∈ ∂Ω, (3.6)γ 1,ext ( V κ s)(x) = (σ(x) − 1)s(x) + (K ∗ κs)(x) for x ∈ ∂Ω, (3.7)where Kκ ∗ denotes the adjoint double layer potential operator(Kκs)(x) ∗ ∂v κ:= (x, y)s(y) ds y for x ∈ ∂Ω∂n xand∂Ω∂v κσ(x) := lim(x, y) ds y . (3.8)ε→0 +y∈Ω : ∥x−y∥=ε ∂n yProof. The proof can be found in [9], following Lemma 2.29.Remark 3.11. For the function σ we get1σ(x) = lim⟨x − y, n(y)⟩ eiκ∥x−y∥ 1ε→0 + 4π∥x − y∥ 2 ∥x − y∥ − iκ ds y .For n we can substituteto obtainy∈Ω : ∥x−y∥=ε1σ(x) = limε→0 + 4πn(y) =y∈Ω : ∥x−y∥=ε1 1= limε→0 + 4π eiκε ε 2 − iκ εx − y∥x − y∥e iκ∥x−y∥∥x − y∥ 1∥x − y∥ − iκ ds y1 ds y .y∈Ω : ∥x−y∥=εThus, the function σ depends on the interior angle of Ω in x ∈ ∂Ω. In particular, fordomains with boundary smooth in x ∈ ∂Ω we get σ(x) = 1/2. For Lipschitz domainsthe relation σ(x) = 1/2 is valid almost everywhere and we can simplify (3.6) and (3.7) toobtainγ 1,int ( V κ s)(x) = 1 2 s(x) + (K∗ κs)(x) for x ∈ ∂Ω, (3.9)γ 1,ext ( V κ s)(x) = − 1 2 s(x) + (K∗ κs)(x) for x ∈ ∂Ω. (3.10)