40 1. Simple Examples of Propagation From section 1.6.3, the leading amplitude d0 must be constant on the rays t → (t , x + c t η/|η|). To determine d0, it suﬃces to know the values d0(t, 0+,x ) at the interface. One could choose d0 to guarantee the continuity of u or of ∂1u, but not both. One cannot construct a good approximate solution consisting of just an incident and transmitted wave. Add to the recipe a reflected wave. Seek a reflected wave in x1 ≥ 0 in the form R ∼ ei(xζ+tσ)/ b(, t, x) , b(, t, x) ∼ b0(t, x) + b1(t, x) + · · · . In order that the reflected wave oscillate with the same phase as the incident wave in the interface x1 = 0, one must have ζ = ξ and σ = −1. To satisfy the wave equation in x1 0 requires σ2 = |ζ|2. Together these imply ζ2 1 = ξ2. 1 To have propagation away from the boundary requires ζ1 = −ξ1 so ζ = ˜. Therefore, (1.7.7) R ∼ ei(x ˜ −t)/ b(, t, x) , b(, t, x) ∼ b0(t, x)+ b1(t, x)+ · · · . Summarizing, seek v = I + R in x1 0, T in x1 0. The continuity required at x1 = 0 forces (1.7.8) ei(x ξ −t)/ ( a(, t, 0,x ) + b(, t, 0,x ) ) = ei(x ξ −t)/ d(, t, 0,x ) . The continuity of u and ∂1u hold if and only if at x1 = 0 one has (1.7.9) a + b = d , and iξ1 a + ∂1a − iξ1 b + ∂1b = iη1 d + ∂1d . The first of these relations yields (1.7.10) aj + bj − dj x 1 =0 = 0, j = 0, 1, 2,.... The second relation in (1.7.9) is expanded in powers of . The coeﬃcients of j must match for all all j ≥ −1. The leading order is −1 and yields (1.7.11) ( a0 − b0 − (η1/ξ1)d0 ) x1=0 = 0 . Since a0 is known, the j = 0 equation from (1.7.10) together with (1.7.11) is a system of two linear equations for the two unknown b0,d0, −1 1 1 η1/ξ1 b0 d0 = a0 a0 . Since the matrix is invertible, this determines the values of b0 and d0 at x1 = 0. The amplitude b0 (resp., d0) is constant on rays with velocity ˜ (resp., cη/|η|). Thus the leading amplitudes are determined throughout the half- spaces on which they are defined.

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