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

d¯

S + t−

∂

¯(¯

B x,t )

¯

h

d¯

S,

where

¯

B (¯ x, t) denotes the ball in

R3

with center ¯, x radius t 0 and where

d¯

S denotes two-dimensional surface measure on ∂

¯

B (¯ x, t). We simplify (25)

by observing

−

∂

¯(¯

B x,t )

¯ g

d¯

S =

1

4πt2

∂

¯(¯

B x,t )

¯ g

d¯

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

d¯

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.