By Samuel Zaidman
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To conclude this section, we make one more remark. 18. We leave the details to the reader. 54 3. 3. 1. Problem Formulation. , let h ∈ C 1 (∂D) and ∂D be given by ∂D = x ˜:x ˜ = x + h(x)ν(x), x ∈ ∂D . 34) ⎧ ∆u + ω 2 u = 0 ⎪ ⎪ ⎪ ⎪ ⎪ ω2 ⎪ ⎪ u =0 ∆u + ⎪ ⎪ k ⎪ ⎨ u |+ − u |− = 0 ⎪ ∂u ∂u ⎪ ⎪ −k =0 ⎪ ⎪ ⎪ ∂ν + ∂ν − ⎪ ⎪ ⎪ ⎪ ⎩ ∂u = 0 ∂ν in Ω \ D , in D , on ∂D , on ∂D , on ∂Ω. 34) to the calculation of the asymptotic expressions of the characteristic values of the operator-valued function A (ω) given by ⎛ 1 ω I − KΩ ⎜ 2 ⎜ ω ω → A (ω) := ⎜ DΩ ⎜ ⎝ ∂ Dω ∂ν Ω ⎞ ω −SD ω SD 1 ω ∗ I + (KD ) 2 0 ω √ k −SD ω √ 1 −k(− I + (KDk )∗ ) 2 ⎟ ⎟ ⎟.
A0 (ω)−1 Anp (ω)ω n . 10). 2. Leading-Order Terms. As a simplest case, let us now ﬁnd the leadingorder term in the asymptotic expansion of µj − µj as → 0. 11) Recalling that 0 SB 2 A0 (ω)−1 A1 (ω)ωdω. 14). It is now ⎛ 1 ω −1 ) 0 ( I − KΩ ⎜ −1 2 A0 (ω) = ⎝ ω 1 ω −1 0 −1 −CDΩ ( I − KΩ ) [·](z) ϕe (SB ) 2 where C := cap(∂B). 31) that (A0 )(ω)−1 A1 (ω) ω (A0 )(ω)−1 A1 (ω) ω (A0 )(ω) −1 A1 (ω) ω = 0, 11 21 22 (A0 )(ω)−1 A1 (ω) ω 12 easy to see that ⎞ ⎟ ⎠, = NΩω (x, z) 0 −1 ω = (SB ) [∇DΩ [·](z) · x], ω = −CDΩ [NΩω (·, z)](z)ϕe · dσ(y) ∂B √ −1ω Cϕe · dσ(y).
3. Radiation Condition. Let us formulate the radiation conditions for the elastic waves when Im ω ≥ 0 and ω = 0. 49) kT = ω ω =√ cT µ and kL = ω ω = √ . 50) ( + kT2 )u(p) = 0, ∇ × u(p) = 0, ( 2 )u(s) = 0, + kL ∇ · u(s) = 0. 21) for solutions of the Helmholtz equation by requiring that √ ∂r u(p) (x) − −1kT u(p) (x) = o(r −1 ), as r = |x| → +∞. 51). By a straightforward calculation, one can see that the single- and double-layer potentials satisfy the radiation condition. We refer to [1, 159] for details.