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

23In particular, if u satisfies the Helmholtz equationthenu(x) =∂Ω∆u + κ 2 u = 0 in Ω,∂u∂n (y)v κ(x, y) ds y − u(y) ∂v κ(x, y) ds y for x ∈ Ω. (2.7)∂Ω ∂n yProof. The proof is constructed in the same way as in the case of Theorem 2.5. Let x ∈ Ωbe an arbitrary fixed point. Let us choose ε > 0 such thatB ε (x) := {y ∈ R 3 : ∥x − y∥ < ε} ⊂ B ε (x) ⊂ Ωand let us denote Ω ε := Ω \ B ε (x) (similar situation as in Figure 2.2 with points x and yswapped). From Theorem 2.2 and the symmetry of v κ we have∆ y v κ (x, y) + κ 2 v κ (x, y) = 0 for all y ∈ Ω ε .Using the second Green’s identity with functions u and v κ on Ω ε we obtain=0∆u(y) + κ 2 u(y) v κ (x, y) dy − u(y) ∆ y v κ (x, y) + κ 2 v κ (x, y) dyΩ ε Ω ε∂u=∂Ω ∂n (y)v κ(x, y) ds y − u(y) ∂v κ(x, y) ds y(2.8)∂Ω ∂n y+−∂B ε(x)∂u∂n (y)v κ(x, y) ds y =:I 1Because for y ∈ ∂B ε (x) we have ∥x − y∥ = ε andit holds thatn(y) = 1 (x − y),εu(y) ∂v κ(x, y) ds y∂B ε(x) ∂n y =:I 2.v κ (x, y) = 1 e iκε4π ε ,∇ y v κ (x, y) = 1 1 − iκεeiκε4π ε 3 (x − y),∂v κ(x, y) = ⟨∇ y v κ (x, y), n(y)⟩ = 1 1∂n y 4π ε 2 eiκε (1 − iκε)for all y ∈ ∂B ε (x). Using the parametrization as in the proof of Theorem 2.4 (but with xand y swapped) we have for the integral I 1I 1 = 14π 2π0 π2− π 2= 14π εeiκε 2π0e iκεε π2− π 2∂u x + ε(cos ϑ cos ψ, sin ϑ cos ψ, sin ψ) ε 2 cos ψ dψ dϑ∂n∂u x + ε(cos ϑ cos ψ, sin ϑ cos ψ, sin ψ) cos ψ dψ dϑ∂n