80 2. FOUR IMPORTANT LINEAR PDE
THEOREM 3 (Solution of wave equation in even dimensions). Assume n
is an even integer, n 2, and suppose also g
Cm+1(Rn),
h
Cm(Rn),
for m =
n+2
2
. Define u by (38). Then
(i) u
C2(Rn
× [0, ∞)),
(ii) utt Δu = 0 in
Rn
× (0, ∞),
and
(iii) lim
(x,t)→(x0,0)
x∈Rn, t0
u(x, t) =
g(x0),
lim
(x,t)→(x0,0)
x∈Rn, t0
ut(x, t) =
h(x0)
for each point
x0

Rn.
This follows from Theorem 2. Observe, in contrast to formula (31), that
to compute u(x, t) for even n we need information on u = g, ut = h on all
of B(x, t) and not just on ∂B(x, t).
Huygens’ principle. Comparing (31) and (38), we observe that if n is odd
and n 3, the data g and h at a given point x
Rn
affect the solution u only
on the boundary {(y, t) | t 0, |x y| = t} of the cone C = {(y, t) | t 0,
|x−y| t}. On the other hand, if n is even, the data g and h affect u within
all of C. In other words, a “disturbance” originating at x propagates along
a sharp wavefront in odd dimensions, but in even dimensions it continues
to have effects even after the leading edge of the wavefront passes. This is
Huygens’ principle.
2.4.2. Nonhomogeneous problem.
We next investigate the initial-value problem for the nonhomogeneous
wave equation
(39)
utt Δu = f in
Rn
× (0, ∞)
u = 0, ut = 0 on
Rn
× {t = 0}.
Motivated by Duhamel’s principle (introduced earlier in §2.3.1), we define
u = u(x, t; s) to be the solution of
(40s)
utt(·; s) Δu(·; s) = 0 in
Rn
× (s, ∞)
u(·; s) = 0, ut(·; s) = f(·,s) on
Rn
× {t = s}.
Now set
(41) u(x, t) :=
t
0
u(x, t; s) ds (x
Rn,t
0).
Duhamel’s principle asserts this is a solution of
(42)
utt Δu = f in
Rn
× (0, ∞)
u = 0, ut = 0 on
Rn
× {t = 0}.
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