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

40 3 Boundary Integral EquationsAccording to Theorem 3.21, the boundary integral equations (3.26) and (3.27) areequivalent to the variational problems 1⟨V κ γ 1,int u, s⟩ ∂Ω =2 I + K κ g D , s for all s ∈ H −1/2 (∂Ω)∂Ωand 12κI − K∗ γ 1,int u, t = ⟨D κ g D , t⟩ ∂Ω for all t ∈ H 1/2 (∂Ω),∂Ωrespectively, where⟨u, v⟩ ∂Ω :=∂Ωu(x)v(x) dsdenotes the duality pairing between H 1/2 (∂Ω) and H −1/2 (∂Ω).The equations (3.26) and (3.27) are not uniquely solvable for all κ ∈ R + . To show this,let us consider the Dirichlet eigenvalue problem for the Laplace equation−∆uλ = λu λ in Ω,γ 0,int (3.28)u λ = 0 on ∂Ω,which can be understood as the Helmholtz boundary value problem with κ 2 = λ and thehomogeneous Dirichlet boundary condition. Let λ ∈ R + be an eigenvalue of (3.28) and u λthe corresponding eigensolution. From the boundary integral equations (3.26) and (3.27)we get(V κ γ 1,int u λ )(x) = 1 2 γ0,int u λ (x) + (K κ γ 0,int u λ )(x) = 0 for x ∈ ∂Ω,12 γ1,int u λ (x) − (K ∗ κγ 1,int u λ )(x) = (D κ γ 0,int u λ )(x) = 0 for x ∈ ∂Ωand hence, if κ 2 coincides with the eigenvalue λ, the operators V κ and (1/2I − K ∗ κ) are notinjective and thus not invertible.On the other hand, for other values of κ both operators are injective. The coercivity ofthe single layer potential operator then ensures a unique solution to the boundary integralequation (3.26) (see Theorem 3.20). The following theorem describes the relation betweenthe weak solution and the solution given by the representation formula (see [12], Theorem7.5).Theorem 3.26. If u ∈ H 1 (Ω, ∆+κ 2 ) is a solution to the interior Dirichlet problem (3.24),then the Neumann trace γ 1,int u satisfies the boundary integral equation (3.26) and u hasthe representation (3.25).Conversely, if γ 1,int u satisfies the boundary integral equation (3.26), then the representationformula (3.25) defines a solution u ∈ H 1 (Ω, ∆+κ 2 ) to the interior Dirichlet problem(3.24).