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1/2 For a test function v T ∈ HT (Γ ), we then obtain the variational problem S Lame uT , v T Γ = − S Lame gn, v T Γ , which is uniquely solvable due to the mapping properties of the Steklov– Poincar´e operator. Note that one may also consider mixed boundary value problems with sliding boundary conditions only on a part ΓS , but standard Dirichlet or Neumann boundary conditions elsewhere. However, to ensure uniqueness, one needs to assume Dirichlet boundary conditions somewhere for each component.

57), u ¯ = γ0ext ue (x) = γ0int ui (x) , αi γ1int ui (x) = αe γ1ext ue (x) for x ∈ Γ, we obtain a coupled Steklov–Poincar´e operator equation to find u ¯ ∈ H 1/2 (Γ ) such that ¯)(x) + αe (S ext u ¯)(x) = αi (S int u (S int γ0int up )(x) − γ1int up (x) + αe 1 I − K (V −1 u0 )(x) 2 is satisfied for x ∈ Γ . 59) Γ S int γ0int up − γ1int up + αe 1 I − K V −1 u0 , v 2 Γ is satisfied for all v ∈ H 1/2 (Γ ). 59) finally follows from the ellipticity estimates for the interior and exterior Steklov–Poincar´e operators S int and S ext .

48) describes the correct solution for any scaling parameter. 1), we now consider an inhomogeneous Poisson equation with some given right hand side. The Dirichlet boundary value problem for the Poisson equation reads −Δu(x) = f (x) for x ∈ Ω, γ0int u(x) = g(x) for x ∈ Γ. 3), we then obtain the representation formula u∗ (x, y)t(y)dsy − u(x) = Γ int u∗ (x, y)g(y)ds + γ1,y y Γ u∗ (x, y)f (y)dy Ω for x ∈ Ω, where t = γ1int u is the yet unknown Neumann datum. 55) where u∗ (x, y)f (y)dy (N0 f )(x) = for x ∈ Γ Ω is the Newton potential entering the right hand side.

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