2.4. WAVE EQUATION 73
Solution for n = 2. No transformation like (17) works to convert the
Euler–Poisson–Darboux equation into the one-dimensional wave equation
when n = 2. Instead we will take the initial-value problem (11) for n = 2
and simply regard it as a problem for n = 3, in which the third spatial
variable x3 does not appear.
Indeed, assuming u
C2(R2
× [0, ∞)) solves (11) for n = 2, let us write
(23) ¯(x1,x2,x3,t) u := u(x1,x2,t).
Then (11) implies
(24)
¯tt u Δ¯ u = 0 in
R3
× (0, ∞)
¯ u = ¯ g, ¯t u =
¯
h on
R3
× {t = 0},
for
¯(x1,x2,x3) g := g(x1,x2),
¯(x1,x2,x3)
h := h(x1,x2).
If we write x = (x1,x2)
R2
and ¯ x = (x1,x2, 0)
R3,
then (24) and
Kirchhoff’s formula (in the form (21)) imply
(25)
u(x, t) = ¯(¯ u x, t)
=

∂t
t−

¯(¯
B x,t )
¯ g

S + t−

¯(¯
B x,t )
¯
h

S,
where
¯
B x, t) denotes the ball in
R3
with center ¯, x radius t 0 and where

S denotes two-dimensional surface measure on
¯
B x, t). We simplify (25)
by observing


¯(¯
B x,t )
¯ g

S =
1
4πt2

¯(¯
B x,t )
¯ g

S
=
2
4πt2
B(x,t)
g(y)(1 +
|Dγ(y)|2)1/2
dy,
where γ(y) =
(t2
|y
x|2)1/2
for y B(x, t). The factor “2” enters
since
¯
B x, t) consists of two hemispheres. Observe that (1 +
|Dγ|2)1/2
=
t(t2
|y
x|2)−1/2.
Therefore


¯(¯
B x,t )
¯ g

S =
1
2πt
B(x,t)
g(y)
(t2 |y x|2)1/2
dy
=
t
2

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