26 1. Simple Examples of Propagation Here, hj(t, ζ) is a polynomial in t, ζ. Injecting in the formula for Aexact(, t, x) yields an expansion Aexact(, t, x) A0(t, x) + A1(t, x) + 2 A2(t, x) + · · · , A0(t, x) = γ(x e1t)/2, (1.4.11) Aj = 1 (2π)−d/2 2 ˆ(ζ) ei(xζ−tζ1) hj(t, ζ) = 1 2 hj(t, ∂/i)γ (x e1t) . (1.4.12) The series is asymptotic as 0 in the sense of Taylor series. For any s, N, truncating the series after N terms yields an approximate amplitude which differs from Aexact by O( N+1 ) in Hs uniformly on compact time intervals. Exercise 1.4.2. Compute the precise form of the first corrector a1. Formula (1.4.11) implies that if the Cauchy data are supported in a set O, then the amplitudes Aj are all supported in the tube of rays (1.4.13) T := (t, x) : x = x + te1, x O . Warning 1. Though the Aj are supported in this tube, it is not true, when d 2, that A exact is supported in the tube. If it were, then u would be supported in the tube. When d 2, the function u = 0 is the only solution of D’Alembert’s equation with support in a tube of rays with compact cross section (see Exercise 5.2.9). Warning 2. By a closer inspection of (1.4.11) or by the analysis after Exercise 5.2.9, one can show that Aj(t, ·) L∞ (j!)−1 |β|≤2j ∂βγ||L∞. So, for a typical analytic γ, the series j Aj have terms of size j Cj(2j)!/j! so they diverge no matter how small is . For a nonanalytic γ, for example γ C∞, 0 matters are worse still. The series j Aj is a divergent series that gives an accurate asymptotic expansion as 0. It is a nonconvergent Taylor expansion of Aexact(, t, x). To analyze the oscillatory initial value problem with u(0) = 0, ut(0) = β(x) eix1/ requires one more idea to handle the contributions from ξ 0 in the expression u(t, x) = (2π)−d/2 sin t|ξ| |ξ| ˆ ξ e1 eixξ . Choose χ C∞(Rd) 0 ξ with χ = 1 on a neighborhood of ξ = 0. The cutoff integrand is equal to χ(ξ) sin t|ξ| |ξ| 1 ξ e1/ s ks(ξ −e1/) eixξ , ks(ξ) := ξ s ˆ(ξ) L2(Rd) ξ .
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