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

22 2 Helmholtz Equationthe normal vector n can be expressed as (see Figure 2.2)n(x) = − 1 (x − y)εand thusHence, we have∂v κ(x, y) = 1∂n x 4π1ε1 − iκ∥x − y∥eiκ∥x−y∥ .∥x − y∥I 2 = 1 2π π24π 0 − π 21ε1 − iκεeiκε ϕ y + ε(cos ϑ cos ψ, sin ϑ cos ψ, sin ψ) ε 2 cos ψ dψ dϑε= 1 2π π2e iκε (1 − iκε)ϕ y + ε(cos ϑ cos ψ, sin ϑ cos ψ, sin ψ) cos ψ dψ dϑ.4π 0 − π 2The Lebesgue dominated convergence theorem allows us to interchange the limit and theintegration. Moreover, because ϕ ∈ C ∞ 0 (R3 ) andlim e iκε (1 − iκε) = 1,ε→0 +we obtainand finallylim I 2 = 1 2π π2ϕ(y) cos ψ dψ dϑ = ϕ(y)ε→0 + 4π 0 − π 2which was to be proved.I = ⟨∆ṽ κ + κ 2 ṽ κ , ϕ⟩ = limε→0 +I 1 − limε→0 +I 2 = −ϕ(y) = ⟨−δ y , ϕ⟩,2.3 Representation FormulaeThe following two theorems form the basis of the boundary element method, providingformulae for calculating the solution to a boundary value problem in any point of thegiven domain. Firstly, we focus on interior problems, i.e., problems on bounded domains.Theorem 2.6 (Representation Formula for Bounded Domains). Let Ω ⊂ R 3 be a boundedC 1 domain, let v κ denote the fundamental solution for the Helmholtz equation in R 3 andlet n denote the unit outward normal vector to ∂Ω. Then for u ∈ C 2 (Ω) we have therepresentation formula∂uu(x) =∂Ω ∂n (y)v κ(x, y) ds y − u(y) ∂v κ(x, y) ds y∂Ω ∂n y− ∆u(y) + κ 2 u(y) v κ (x, y) dy for x ∈ Ω.Ω