By Arthur Gill

ISBN-10: 0130392227

ISBN-13: 9780130392220

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22). 25) N β = g. 10 Theorem. 22) belong to the space H1,k−1,κ (G± ). If k ≥ 1, then these functions are the weak solutions of problems (DN± ), respectively. 8. 3 The Direct Method We begin by establishing the dynamic analog of the third Green’s formula (or Somigliana representation formula, as is known in elasticity theory). 27) X ∈ Γ, where f ∈ H1/2,k,κ (Γ). Then for any p ∈ Cκ , u ˆ(x, p) = Lu(x, t) is the weak solution of the problem Bp2 u ˆ(x, p) + (Aˆ u)(x, p) = 0, x ∈ S + , (γ + u ˆ)(x, p) = fˆ(x, p), x ∈ ∂S, where fˆ ∈ H1/2,p (∂S).

Let Hk± be the ranges of the operators T ± : H1/2,k,κ (Γ) → H−1/2,k−1,κ (Γ). We take any g ∈ H−1/2,k+1,κ (Γ) and set f = (T ± )−1 g ∈ H1/2,k,κ (Γ). Since g = T ± (T ± )−1 g, we have H−1/2,k+1,κ (Γ) ⊂ Hk± . It is obvious that for any k ∈ R, the space H−1/2,k+1,κ (Γ) is dense in H−1/2,k−1,κ (Γ); hence, Hk± are dense in H−1/2,k−1,κ (Γ). The case of (T ± )−1 is treated similarly. 1. Their variational versions consist in ﬁnding u ∈ H1,0,κ (G± ) such that, for g prescribed on Γ, ∞ a± (u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0;S ± dt 0 ∞ (g, v ± )0;∂S dt =± ¯ ± ).

Suppose that the ranges of Tp± are not dense in H−1/2,p (∂S). Then there is a nonzero ϕ ∈ H1/2,p (∂S) such that (Tp± f, ϕ)0;∂S = 0 for all f ∈ H1/2,p (∂S). Taking f = ϕ, we see that (Tp± ϕ, ϕ)0;∂S = 0. 8) implies that z = 0; consequently, ϕ = 0. This contradiction completes the proof. At this stage, we can deﬁne operators Tˆ ± and (Tˆ ± )−1 on the elements F (p) = fˆ(· , p) and G(p) = gˆ(· , p) of H1/2,k,κ (∂S) and H−1/2,k,κ (∂S), k ∈ R, respectively, by setting (Tˆ ± F )(p) = (Tp± fˆ)(· , p), (Tˆ ± )−1 G (p) = (Tp± )−1 gˆ (· , p).

### Applied algebra for the computer sciences by Arthur Gill

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