Integral Equations

Theorem 36 The operators I − K 0 and I − K 1 have trivial nullspacesN(I − K 0 ) = {0}, N(I − K 1 ) = {0}, (297)The nullspaces of operators I + K 0 and I + K 1 have dimension one andwithN(I + K 0 ) = span {1}, N(I + K 1 ) = span {ψ 0 } (298)∫Γψ 0 dl y ≠ 0. (299)Proof. Let φ be a solution to the homogeneous integral equation φ − K 0 φ = 0 and define adouble-layer potential∫∂Φ(x, y)v(x) =ϕ(y)dl y , x ∈ D (300)∂n yΓwith a continuous density φ. Then we have, for the limiting values on Γ,∫∂Φ(x, y)v ± (x) =ϕ(y)dl y ± 1 ϕ(x) (301)∂n y 2which yieldsΓ2v − (x) = K 0 ϕ(x) − ϕ = 0. (302)From the uniqueness of the interior Dirichlet problem (Theorem 33) it follows that v = 0 in thewhole domain D. Now we will apply the contunuity of the normal derivative of the double-layerpotential (300) over a smooth contour Γ,∂v(y)∂n y∣ ∣∣∣∣+− ∂v(y)∂n y∣ ∣∣∣∣−= 0, y ∈ Γ, (303)where the internal and extrenal limiting values with respect to Γ are defined as followsand∣∂v(y) ∣∣∣∣− ∂v(y)= lim∂n y y→Γ, y∈D ∂n y∣∂v(y) ∣∣∣∣+ ∂v(y)= lim∂n y y→Γ, y∈R 2 \ ¯D ∂n yNote that (303) can be written asFrom (304)–(306) it follows that= limh→0n y · grad v(y − hn y ) (304)= limh→0n y · grad v(y + hn y ). (305)lim n y · [grad v(y + hn y ) − grad v(y − hn y )] = 0. (306)h→0∂v(y)∂n y∣ ∣∣∣∣+− ∂v(y)∂n y∣ ∣∣∣∣−= 0, y ∈ Γ. (307)Using the uniqueness for the exterior Neumann problem and the fact that v(y) → 0, |y| → ∞,we find that v(y) = 0, y ∈ R 2 \ ¯D. Hence from (301) we deduce ϕ = v + − v − = 0 on Γ. ThusN(I − K 0 ) = {0} and, by the Fredholm alternative, N(I − K 1 ) = {0}.57