2.4. WAVE EQUATION 79
¯
B x, t) denoting the ball in
Rn+1
with center ¯ x and radius t and

S denoting
n-dimensional surface measure on
¯
B x, t). Now
(36)

¯(¯
B x,t )
¯ g

S =
1
(n + 1)α(n +
1)tn

¯(¯
B x,t )
¯ g

S.
Note that
¯
B (x, t) {yn+1 0} is the graph of the function γ(y) :=
(t2
|y
x|2)1/2
for y B(x, t)
Rn.
Likewise
¯
B (x, t) {yn+1 0}
is the graph of −γ. Thus (36) implies
(37)

¯(¯
B x,t )
¯ g

S =
2
(n + 1)α(n + 1)tn
B(x,t)
g(y)(1 +
|Dγ(y)|2)1/2
dy,
the factor “2” entering because
¯
B x, t) comprises two hemispheres. Note
that
(1+|Dγ(y)|2)1/2
=
t(t2
|y
x|2)−1/2.
Our substituting this into (37)
yields


¯(¯
B x,t )
¯ g

S =
2
(n + 1)α(n + 1)tn−1
B(x,t)
g(y)
(t2 |y x|2)1/2
dy
=
2tα(n)
(n + 1)α(n + 1)

B(x,t)
g(y)
(t2
|y
x|2)1/2
dy.
We insert this formula and the similar one with h in place of g into (35)
and find
u(x, t) =
1
γn+1
2α(n)
(n + 1)α(n + 1)

∂t
1
t

∂t
n−2
2
tn

B(x,t)
g(y)
(t2 |y x|2)1/2
dy
+
1
t

∂t
n−2
2
tn

B(x,t)
h(y)
(t2
|y
x|2)1/2
dy .
Since γn+1 = 1 · 3 · 5 · · · (n 1) and α(n) =
πn/2
Γ(
n+2
2
)
, we may compute
γn = 2 · 4 · · · (n 2) · n.
Hence the resulting representation formula for even n is
(38)















u(x, t) =
1
γn

∂t
1
t

∂t
n−2
2
tn−
B(x,t)
g(y)
(t2 |y x|2)1/2
dy
+
1
t

∂t
n−2
2
tn

B(x,t)
h(y)
(t2 |y x|2)1/2
dy ,
where n is even and γn = 2 · 4 · · · (n 2) · n,
for x
Rn,
t 0.
Since γ2 = 2, this agrees with Poisson’s formula (27) if n = 2.
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