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)/

dζ . (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

)

dζ .

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.