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

57with∂v κ(x, y) = 1 e iκ∥x−y∥(iκ∥x − y∥ − 1)⟨x − y, n(x)⟩.∂n x 4π ∥x − y∥3 The corresponding variational problem reads 1⟨D κ g D , t⟩ ∂Ω =2 I − K∗ κ g N , t∂Ωfor all t ∈ H 1/2 (∂Ω). (4.12)Inserting the approximations of the Cauchy data (4.4), (4.5) into (4.12) we obtain theGalerkin equationsNgj D ⟨D κ ϕ j , ϕ i ⟩ ∂Ω =j=1El=1g N l 12 I − K∗ κψ l , ϕ iUsing Theorem 3.14, the left-hand side of (4.13) yieldsNgj D ⟨D κ ϕ j , ϕ i ⟩ ∂Ω =j=1withC N×N ∋ D κ,h [i, j] := 1 4π− κ24π∂Ω∂Ω∂Ω∂ΩFor the right-hand side of (4.13) we obtainandEl=1g N l 12 ψ l, ϕ iEgl N ⟨K∗ κψ l , ϕ i ⟩ ∂Ω =l=1===El=1El=1g N lg N lEl=1∂Ωg N l∂ΩNgj D D κ,h [i, j]j=1for all i ∈ {1, . . . , N}. (4.13)e iκ∥x−y∥∥x − y∥ ⟨curl ∂Ω ϕ j (y), curl ∂Ω ϕ i (x)⟩ ds y ds xe iκ∥x−y∥∥x − y∥ ϕ j(y)ϕ i (x)⟨n(x), n(y)⟩ ds y ds x .=∂ΩEl=1g N l1ϕ i (x) ds x =2 τ lϕ i (x)τ lEl=1g N l∂v κ∂n x(x, y) ds y ds x12 M h[l, i]1eϕ i (x)4π ∂Ωτ iκ∥x−y∥l∥x − y∥ 3 (iκ∥x − y∥ − 1)⟨x − y, n(x)⟩ ds y ds x1ϕ i (y)4πτ eiκ∥x−y∥l ∂Ω ∥x − y∥ 3 (1 − iκ∥x − y∥)⟨x − y, n(y)⟩ ds y ds xEgl N K κ,h[l, i]l=1(4.14)