20 1. Simple Examples of Propagation The analysis of Exercise 1.3.3 does not apply to the fundamental so- lution since the latter does not have finite energy. However it belongs to Cj(R : Hs−j(R)) for all s 1/2 and j ∈ N. The next result provides a good replacement for (1.3.4). Exercise 1.3.4. Suppose that u is the fundamental solution of the Klein– Gordon equation (1.1.6) and that s 1/2. If 0 ≤ χ ∈ C∞(R) is a plateau cutoff supported on the positive half line, that is χ(x) = 0 for x ≤ 0 and χ(x) = 1 for x ≥ 1 , then for all 0 and R 0 there is a δ 0 so that (1.3.5) lim sup t−∞ χ(R + |x| − (1 − δ)t) u(t, x) Hs(Rx) . Hints. Prove that χ u(t) Hs(R) ≤ C u(0) Hs(R) + ut(0) Hs−1(R) with C independent of t and the initial data. Conclude that it suﬃces to prove (1.3.4) with initial data u(0),ut(0) dense in Hs × Hs−1. Take the dense set to be data with Fourier transform in C∞(R). 0 These examples illustrate the important observation that the propaga- tion of singularities in solutions and the propagation of the majority of the energy may be governed by different rules. For the Klein–Gordon equation, both answers can be determined from considerations of group velocities. 1.4. Fourier synthesis and rectilinear propagation For equations with constant coeﬃcients, solutions of the initial value prob- lem are expressed as Fourier integrals. Injecting short wavelength initial data and performing an asymptotic analysis yields the approximations of geometric optics. This is how such approximations were first justified in the nineteenth century. It is also the motivating example for the more gen- eral theory. The short wavelength approximations explain the rectilinear propagation of waves in homogeneous media. This is the first of the three basic physical laws of geometric optics. It explains, among other things, the formation of shadows. The short wavelength solutions are also the building blocks in the analysis of the laws of reflection and refraction presented in §1.6 and §1.7. Consider the initial value problem (1.4.1) 0 = u := ∂2u ∂t2 − d j=1 ∂2u ∂x2 j , u(0,x) = f, ut(0,x) = g.

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