FEA Theory and Application with Ansys

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45) is not invertible. See [287, 288, 141]. 45), we decompose the right-hand side into a constant and a zero-mean part: g = g0 + (g − g0 ), g0 = (1/|∂D|) ∂D g dσ. We also decompose the unknown density φ into a constant and a zero-mean part: φ = φ0 + (φ − φ0 ), φ0 = (1/|∂D|) ∂D φ dσ. Since KD (1) = 1/2, the constant part of the unknown density φ is simply given by φ0 = (1/(λ − (1/2))) g0 . 45) can be written as ∗ (λI − KD )(φ − φ0 ) = g − 1 ∗ g0 (λI − KD )(1) ∈ L20 (∂D) , λ − 12 which is well behaved.

L2 (∂D) Now assume that |k − 1| is small or, equivalently, λ is large. Then φ L2 (∂D) ∗ Since KD φ ≤ 1 λφ λ L2 (∂D) L2 (∂D) ≤C φ φ ≤ 1 ∗ (λI − KD )φ λ L2 (∂D) L2 (∂D) ≤ L2 (∂D) + 1 ∗ K φ λ D L2 (∂D) . for some C then, if λ > 2C, we get 2 ∗ )φ (λI − KD λ L2 (∂D) . 44) depends continuously on k, by a compactness argument, the proof is complete. ✷ We will also prove the following theorem due to Verchota [298]. 24 Let D be a bounded Lipschitz domain in Rd . Then the single layer potential SD maps L2 (∂D) into W12 (∂D) boundedly and KD : W12 (∂D) → W12 (∂D) is a bounded operator.

Let Ω be a bounded Lipschitz domain in Rd , d = 2, 3, containing an inclusion D. Assume that the conductivity of the background Ω \ D is A and that of D is A. The conductivity profile of the body Ω is then given by χ(Ω \ D)A + χ(D)A. 87) ν · A∇u =g, ⎪ ⎪ ∂Ω ⎪ ⎪ ⎪ ⎪ ⎪ u(x) dσ(x) = 0 . ⎩ ∂Ω Introduce A A H A (x) := −SΩ (g)(x) + DΩ (f )(x), x ∈ Ω, f := u|∂Ω ∈ W 12 (∂Ω) . 86). 88). 89) where the pair (φ, ψ) is the unique solution in L20 (∂D) × L2 (∂D) to the system of integral equations ⎧ ⎨S A ψ − S A φ = H A D D on ∂D .

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