72 2. FOUR IMPORTANT LINEAR PDE
Indeed
˜tt
U = rUtt
= r Urr +
2
r
Ur by (14), with n = 3
= rUrr + 2Ur = (U + rUr)r =
˜rr.
U
Notice also that
˜
G rr(0) = 0. Applying formula (10) to (19), we find for
0 r t
(20)
˜
U (x; r, t) =
1
2
[
˜
G (r + t)
˜
G (t r)] +
1
2
r+t
−r+t
˜
H (y) dy.
Since (12) implies u(x, t) = limr→0+ U(x; r, t), we conclude from (17), (18),
(20) that
u(x, t) = lim
r→0+
˜
U (x; r, t)
r
= lim
r→0+
˜
G (t + r)
˜
G (t r)
2r
+
1
2r
t+r
t−r
˜
H (y) dy
=
˜
G (t) +
˜
H (t).
Owing then to (13), we deduce
(21) u(x, t) =

∂t
t−
∂B(x,t)
g dS + t−
∂B(x,t)
h dS.
But

∂B(x,t)
g(y) dS(y) =
∂B(0,1)
g(x + tz) dS(z);
and so

∂t

∂B(x,t)
g dS =
∂B(0,1)
Dg(x + tz) · z dS(z)
=
∂B(x,t)
Dg(y) ·
y x
t
dS(y).
Returning to (21), we therefore conclude
(22) u(x, t) =
∂B(x,t)
th(y)+g(y)+Dg(y)·(y −x) dS(y) (x
R3,
t 0).
This is Kirchhoff’s formula for the solution of the initial-value problem (11)
in three dimensions.
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