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

24 2 Helmholtz Equationand due to properties of uFor the integral I 2 we obtainlim I 1 = 0.ε→0 +I 2 = 1 2π π24π 0 − π 21ε 2 eiκε (1 − iκε)u x + ε(cos ϑ cos ψ, sin ϑ cos ψ, sin ψ) ε 2 cos ψ dψ dϑ= 1 2π π2e iκε (1 − iκε)u x + ε(cos ϑ cos ψ, sin ϑ cos ψ, sin ψ) cos ψ dψ dϑ4π 0 − π 2and using the Lebesgue dominated convergence theorem and qualities of u we haveFinally, letting ε → 0 + we obtain from (2.8)∆u(y) + κ 2 u(y) v κ (x, y) dylim I 2 = u(x).ε→0 +Ω=∂Ω∂u∂n (y)v κ(x, y) ds y − u(y) ∂v κ(x, y) ds y − u(x).∂Ω ∂n ySince the point x ∈ Ω was arbitrary, we have proved Theorem 2.6.When solving a problem of sound scattering or wave propagation described by theHelmholtz equation we are usually interested in the solution in an unbounded domain.The following theorem gives us the representation formula for such domains (see [6]).Theorem 2.7 (Representation Formula for Unbounded Domains). Let Ω ⊂ R 3 be abounded C 1 domain, let v κ denote the fundamental solution for the Helmholtz equation inR 3 and let n denote the unit outward normal vector to ∂Ω. Let us define Ω ext := R 3 \ Ω.Then for u ∈ C 2 (Ω ext ) satisfying∆u + κ 2 u = 0in Ω extand the Sommerfeld radiation condition x ∇u(x), − iκu(x)1∥x∥ = O ∥x∥ 2for ∥x∥ → ∞we have the representation formulau(x) = u(y) ∂v κ(x, y) ds y −∂n y∂Ω∂Ω∂u∂n (y)v κ(x, y) ds y for x ∈ Ω ext .