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

20 2 Helmholtz Equation∂ΩΩ εεynB ε∂B εFigure 2.2: Illustration for the proof of Theorem 2.5.with⎡⎢det J(r, ϑ, ψ) = det ⎣∂x 1∂r∂x 2∂r∂x 3∂r∂x 1∂ϑ∂x 2∂ϑ∂x 3∂ϑ∂x 1∂ψ∂x 2∂ψ∂x 3∂ψ⎤⎥⎦ = r 2 cos ψ.According to Theorem 2.4 we have ṽ κ ∈ L 1 loc (R3 ) and we can identify ṽ κ with a distributionṽ κ : C0 ∞(R3 ) → C defined as⟨ṽ κ , ϕ⟩ :=ṽ κ (x)ϕ(x) dx =R 3 v κ (x, y)ϕ(x) dx.R 3Theorem 2.5. Let y ∈ R 3 . Then the function ṽ κ : R 3 → C,satisfiesṽ κ (x) := v κ (x, y)∆ṽ κ + κ 2 ṽ κ = −δ yin the distributional sense, i.e.,∆ṽκ + κ 2 ṽ κ , ϕ = ⟨−δ y , ϕ⟩ := −ϕ(y) for all ϕ ∈ C ∞ 0 (R 3 ).Proof. Let y ∈ R 3 and ϕ ∈ C ∞ 0 (R3 ) be chosen arbitrarily. We have to prove that∆ṽκ + κ 2 ṽ κ , ϕ = −ϕ(y).Similarly as in (1.4), we get for the left-hand side∆ṽκ + κ 2 ṽ κ , ϕ = ⟨ṽ κ , ∆ϕ + κ 2 ϕ⟩.