2.4. WAVE EQUATION 79
¯
B (¯ x, t) denoting the ball in
Rn+1
with center ¯ x and radius t and
d¯
S denoting
n-dimensional surface measure on ∂
¯
B (¯ x, t). Now
(36) −
∂
¯(¯
B x,t )
¯ g
d¯
S =
1
(n + 1)α(n +
1)tn
∂
¯(¯
B x,t )
¯ g
d¯
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
d¯
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
d¯
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.
