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.