1.5. A cautionary example in geometric optics 27
The sin t|ξ|/|ξ| factor is |t|. For small, the distance of e1/ to the support
of χ is C/. Therefore,
χ(ξ)
sin t|ξ|
|ξ|
1
ξ e1/ s
L∞(Rd)
ξ
Cs |t|
s
, 0 1 .
It follows that
χ(ξ)
sin t|ξ|
|ξ|
1
ξ e1/ s
ks(ξ e1/)
L2(Rd)
Cs |t|
s
ks
L2(Rd)
,
with s arbitrarily large. The small frequency contribution is negligible in
the limit 0. It is removed with a cutoff as above, and then the analysis
away from ξ = 0 proceeds by decomposition into plane wave as in the case
with ut(0) = 0. It yields left and right moving waves with the same phases
as before.
Exercise 1.4.3. Solve the Cauchy problem for the anisotropic wave equa-
tion, utt = uxx + 4uyy with initial data given by
u (0,x) = γ(x)
eixξ/
, ut(0,x) = 0 , γ
s
Hs(Rd)
.
Find the leading term in the approximate solution to u+. In particular, find
the velocity of propagation as a function of ξ. Discussion. The velocity is
equal to the group velocity from §1.3.
1.5. A cautionary example in geometric optics
A typical science text discussion of a mathematics problem involves simplify-
ing the underlying equations. The usual criterion applied is to ignore terms
which are small compared to other terms in the equation. It is striking that
in many of the problems treated under the rubric of geometric optics, such
an approach can lead to completely inaccurate results. It is an example of
an area where more careful mathematical consideration is not only useful
but necessary.
Consider the initial value problems
∂tu + ∂xu + u = 0 , u
t=0
= a(x) cos(x/)
in the limit 0. The function a is assumed to be smooth and to vanish
rapidly as |x| ∞, so the initial value has the form of a wave packet.
The initial value problem is uniquely solvable, and the solution depends
continuously on the data. The exact solution of the general problem
∂tu + ∂xu + u = 0 , u
t=0
= f(x) ,
is u(t, x) =
e−t
f(x t), so the exact solution u is
u (t, x) =
e−t
a(x t) cos((x t)/) .
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