1.4. Fourier synthesis and rectilinear propagation 23 The approximation is a wave packet with envelope A and wavelength . The wave packet translates rigidly with velocity equal to e1. The waveform γ is arbitrary. The approximate solution resembles the columnated light from a flashlight. If the support of γ is small, the approximate solution resembles a light ray. The amplitude A satisfies the transport equation ∂A ∂t + ∂A ∂x1 = 0, so it is constant on the rays x = x + te1. The construction of a family of short wavelength approximate solutions of D’Alembert’s wave equations requires only the solution of a simple transport equation. The dispersion relation of the family of plane waves, ei(xξ+τt) = ei(xξ−|ξ|t), is τ = −|ξ|. The velocity of transport, v = (1, 0,..., 0), is the group velocity v = −∇ξτ(ξ) = ξ/|ξ| at ξ = (1, 0,..., 0). For the opposite choice of sign, the dispersion relation is τ = |ξ|, the group velocity is −e1, and the rays are the lines x = x − te1. Had we taken data with oscillatory factor eixξ/, then the approximate solution would have been the sum of two wave packets with group velocities ±ξ/|ξ|, 1 2 ei(xξ−t|ξ|)/ γ x − t ξ |ξ| + ei(xξ+t|ξ|)/ γ x + t ξ |ξ| . The approximate solution (1.4.6) is a function H(x − te1) with H(x) = eix1/ h(x). When h has compact support, or more generally tends to zero as |x| → ∞, the approximate solution is localized and has velocity equal to e1. The next result shows that when d 1, no exact solution can have this form. In particular the distribution δ(x −e1t) that is the most intuitive notion of a light ray is not a solution of the wave equation or Maxwell’s equation. Proposition 1.4.1. If d 1, s ∈ R, K ∈ Hs(Rd) and u = K(x − e1t) satisfies u = 0, then K = 0. Exercise 1.4.1. Prove Proposition 1.4.1. Hint. Prove the following lemma. Lemma. If k ≤ d, s ∈ R, and w ∈ Hs(Rd) satisfies 0 = ∑ d k ∂2w/∂2xj, then w = 0.

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