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) +
2A2(t,
x) + · · · ,
A0(t, x) = γ(x e1t)/2,
(1.4.11)
Aj =
1
(2π)−d/2 2
ˆ(ζ) γ
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 Aexact 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

jAj
have terms of size
jCj(2j)!/j!
so they diverge no matter how small is . For a nonanalytic γ, for example
γ C0
∞,
matters are worse still. The series

jAj
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
β ξ
e1
eixξ
.
Choose χ C0
∞(Rξ d)
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)
β
L2(Rξ d)
.
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