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

42 3 Boundary Integral EquationsSimilarly, applying the Neumann trace operator we get the Fredholm integral equation ofthe first kind(D κ γ 0,int u)(x) = 1 2 g N(x) − (K ∗ κg N )(x) for x ∈ ∂Ω. (3.35)Alternatively, we may consider the variational problems 1κ2 I + K γ 0,int u, s = ⟨V κ g N , s⟩ ∂Ω for all s ∈ H −1/2 (∂Ω),∂Ωand⟨D κ γ 0,int u, t⟩ ∂Ω = 12 I − K∗ κ g N , t∂Ωfor all t ∈ H 1/2 (∂Ω).corresponding to (3.34) and (3.35), respectively.To study the solvability of (3.34) and (3.35) let us first consider the Neumann eigenvalueproblem for the Laplace equation−∆uµ = µu µ in Ω,γ 1,int u µ = 0on ∂Ω.(3.36)Let µ ∈ R + be an eigenvalue of (3.36) with the corresponding eigensolution u µ . From(3.34) and (3.35) we obtain for κ 2 = µ12 γ0,int u µ (x) + (K κ γ 0,int u µ )(x) = (V κ γ 1,int u µ )(x) = 0 for x ∈ ∂Ω,(D κ γ 0,int u µ )(x) = 1 2 γ1,int u µ (x) − (K ∗ κγ 1,int u µ )(x) = 0 for x ∈ ∂Ω,implying that the function γ 0,int u µ belongs to the kernel of the boundary integral operatorsD κ and (1/2I + K κ ) and thus these operators are not invertible.For values of κ 2 not coinciding with eigenvalues of (3.36) both operators are injective.Since the hypersingular operator D κ is coercive, the boundary integral equation (3.35) isuniquely solvable.Theorem 3.27. If u ∈ H 1 (Ω, ∆ + κ 2 ) is a solution to the interior Neumann problem(3.32), then the Dirichlet trace γ 0,int u satisfies the boundary integral equation (3.35) and uhas the representation (3.33).Conversely, if γ 0,int u satisfies the boundary integral equation (3.35), then the representationformula (3.33) defines a solution u ∈ H 1 (Ω, ∆ + κ 2 ) to the interior Neumannproblem (3.32).