24 1. Simple Examples of Propagation Next, analyze the error in (1.4.6). Reintroduce the subscripts. Then extract the rapidly oscillating factor in (1.4.5) to find exact amplitudes, u±(t, x) = ei(x1∓t)/ exact (, t, x) , exact (, t, x) := 1 (2π)d/22 ˆ(ζ) eix.ζ e∓it(|e1+ζ|−1)/ . (1.4.7) Proposition 1.4.2. The exact and approximate solutions of u = 0 with Cauchy data (1.4.4) are given by u = ± ei(x1∓t)/ exact (, t, x) , u approx = ± ei(x1∓t)/ γ(x e1t) 2 , as in (1.4.7) and (1.4.6). The error is O( ) on bounded time intervals. Precisely, there is a constant C 0 so that for all s, , t, exact (, t, x) γ(x e1t) 2 Hs(RN ) C |t| γ Hs+2(Rd) . Proof. It suffices to estimate the error with the plus sign. The definitions yield A+ exact (, t, x) γ(x−e1t)/2 = C ˆ(ζ) eix.ζ ( e−it(|e1+ζ|−1)/)−e−itζ1 ) . The definition of the Hs(Rd) norm yields (1.4.8) A+ exact (, t, x) γ(x e1t)/2 Hs(RN ) = ζ s ˆ(ζ) ( e−it(|e1+ζ|−1)/ e−itζ1 ) L2(RN ) . Taylor expansion yields for |β| 1/2, | e1 + β | = 1 + β1 + r(β) , |r(β)| C |β|2 . Increasing C if needed, the same inequality is true for |β| 1/2 as well. Applied to β = ζ this yields t ( e1 + ζ 1 ) / t ζ1 x1 C |t||ζ|2 so e−it(|e1+ζ|−1)/ e−itζ1 C |t||ζ|2 . Therefore, (1.4.9) ζ s ˆ(ζ) ( e−it(|e1+ζ|−1)/ e−itζ1 ) L2(Rd) C |t| ζ s |ζ|2 ˆ(ζ) L2 . Combining (1.4.8) and (1.4.9) yields the estimate of the proposition.
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