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) +
x) + · · · ,
A0(t, x) = γ(x − e1t)/2,
hj(t, ζ) dζ
hj(t, ∂/i)γ (x − e1t) .
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(
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, ·)
So, for a typical analytic γ, the series
have terms of size
so they diverge no matter how small is . For a nonanalytic γ, for example
γ ∈ C0
matters are worse still. The series
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) =
requires one more idea to handle the contributions from ξ ≈ 0 in
u(t, x) =
β ξ −
Choose χ ∈ C0
with χ = 1 on a neighborhood of ξ = 0. The cutoff
integrand is equal to
ξ − e1/
, ks(ξ) := ξ