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)/
Aexact(,
±
t, x) ,
Aexact(,
±
t, x) :=
1
(2π)d/22
ˆ(ζ) γ
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)/
Aexact(,
±
t, x) , uapprox =
±
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,
Aexact(,
±
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
Aexact(,
+
t, x) γ(x−e1t)/2 = C ˆ(ζ) γ
eix.ζ
(
e−it(|e1+ζ|−1)/)−e−itζ1
)
.
The definition of the
Hs(Rd)
norm yields
(1.4.8) Aexact(,
+
t, x) γ(x e1t)/2
Hs(RN )
= ζ
s
ˆ(ζ) γ
(
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) γ
(
e−it(|e1+ζ|−1)/

e−itζ1
)
L2(Rd)
C |t| ζ
s|ζ|2
ˆ(ζ) γ
L2
.
Combining (1.4.8) and (1.4.9) yields the estimate of the proposition.
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