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Example text

38) and the following representation holds. 41). Proof. 13) that x ∈ Ω. 38 2 Transmission Problem 1 1 I − KΩ (H|∂Ω ) = −(SΩ g)|∂Ω + I + KΩ ((SD φ)|∂Ω ) 2 2 on ∂Ω . 49) Since by Green’s formula U = −SΩ (g) + DΩ (U |∂Ω ) in Ω, we have 1 I − KΩ (U |∂Ω ) = −(SΩ g)|∂Ω . 33) that − 1 I − KΩ ((ND φ)|∂Ω ) = (SD φ)|∂Ω . 15 that 1 I − KΩ f = 0 , f ∈ L2 (∂Ω) ⇒ f = constant. 51), we conclude that 1 I − KΩ 2 H|∂Ω − U |∂Ω + 1 I + KΩ ((ND φ)|∂Ω ) 2 =0. Therefore, we have H|∂Ω − U |∂Ω + 1 I + KΩ ((ND φ)|∂Ω ) = C (constant).

M. Recall that the quadratic form QD (u) is defined by QD (u) := u, u D . 34) IRd s=l m (kl − 1)QBl (f + Φ) + QIRd (Φ) . 30), we get (kl − 1) ∂f ∂Φ = ∂ν (l) ∂ν (l) + − kl ∂Φ ∂ν (l) on ∂Bl , − l = 1, . . , d .

We claim that φ = (k − 1) ∂u ∂ν . 42) that ∂u ∂ν = − ∂ ∂H + SD φ ∂ν ∂ν = − 1 ∂H 1 ∗ + (− I + KD φ. 47) by Green’s formula. Let g ∈ L20 (∂Ω) and U (y) := N (x, y)g(x) dσ(x) . 38) and the following representation holds. 41). Proof. 13) that x ∈ Ω. 38 2 Transmission Problem 1 1 I − KΩ (H|∂Ω ) = −(SΩ g)|∂Ω + I + KΩ ((SD φ)|∂Ω ) 2 2 on ∂Ω . 49) Since by Green’s formula U = −SΩ (g) + DΩ (U |∂Ω ) in Ω, we have 1 I − KΩ (U |∂Ω ) = −(SΩ g)|∂Ω . 33) that − 1 I − KΩ ((ND φ)|∂Ω ) = (SD φ)|∂Ω . 15 that 1 I − KΩ f = 0 , f ∈ L2 (∂Ω) ⇒ f = constant.

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Advanced engineering mathematics with Matlab by Harman et al.


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