2.4. WAVE EQUATION 71
From this equality we deduce limr→0+ Ur(x; r, t) = 0. We next differentiate
(15), to discover after some computations that
(16) Urr(x; r, t) = −
∂B(x,r)
Δu dS +
1
n
− 1 −
B(x,r)
Δu dy.
Thus limr→0+ Urr(x; r, t) =
1
n
Δu(x, t). Using formula (16), we can similarly
compute Urrr, etc., and so verify that U ∈
Cm(¯
R
+
× [0, ∞)).
2. Continuing the calculation above, we see from (15) that
Ur =
r
n
−
B(x,r)
utt dy by (11)
=
1
nα(n)
1
rn−1
B(x,r)
utt dy.
Thus
rn−1Ur
=
1
nα(n)
B(x,r)
utt dy,
and so
(
rn−1Ur
)
r
=
1
nα(n)
∂B(x,r)
utt dS
=
rn−1
−
∂B(x,r)
utt dS =
rn−1Utt.
c. Solution for n = 3, 2, Kirchhoff’s and Poisson’s formulas. The
plan in the ensuing subsections will be to transform the Euler–Poisson–
Darboux equation (14) into the usual one-dimensional wave equation. As
the full procedure is rather complicated, we pause here to handle the simpler
cases n = 3, 2, in that order.
Solution for n = 3. Let us therefore hereafter take n = 3, and suppose
u ∈
C2(R3
× [0, ∞)) solves the initial-value problem (11). We recall the
definitions (12), (13) of U, G, H and then set
(17)
˜
U := rU,
(18)
˜
G := rG,
˜
H := rH.
We now assert that
˜
U solves
(19)
⎧
⎨
⎩
˜tt
U −
˜rr
U = 0 in R+ × (0, ∞)
˜
U =
˜
G,
˜t
U =
˜
H on R+ × {t = 0}
˜
U = 0 on {r = 0} × (0, ∞).
