2.4. WAVE EQUATION 77
THEOREM 2 (Solution of wave equation in odd dimensions). Assume n
is an odd integer, n 3, and suppose also g
Cm+1(Rn),
h
Cm(Rn),
for
m =
n+1
2
. Define u by (31). 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.
Proof. 1. Suppose first g 0, so that
(32) u(x, t) =
1
γn
1
t

∂t
n−3
2
(
tn−2H(x;
t)
)
.
Then Lemma 2(i) lets us compute
utt
=
1
γn
1
t

∂t
n−1
2
(
tn−1Ht
)
.
From the calculation in the proof of Theorem 2 in §2.2.2, we see as well that
Ht =
t
n

B(x,t)
Δh dy.
Consequently
utt =
1
nα(n)γn
1
t

∂t
n−1
2
B(x,t)
Δh dy
=
1
nα(n)γn
1
t

∂t
n−3
2
1
t
∂B(x,t)
Δh dS .
On the other hand,
ΔH(x; t) = Δx
∂B(0,t)
h(x + y) dS(y) =
∂B(x,t)
Δh dS.
Consequently (32) and the calculations above imply utt = Δu in
Rn×(0,
∞).
A similar computation works if h 0.
2. We leave it as an exercise to confirm, using Lemma 2(ii)–(iii), that u
takes on the correct initial conditions.
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