76 2. FOUR IMPORTANT LINEAR PDE
the next-to-last equality holding according to (14). Using Lemma 2(ii) we
conclude as well that
˜
U = 0 on {r = 0}.
In view of Lemma 3, (29), and formula (10), we conclude for 0 r t
that
(30)
˜
U (r, t) =
1
2
[
˜
G (r + t)
˜
G (t r)] +
1
2
t+r
t−r
˜
H (y) dy.
But recall u(x, t) = limr→0 U(x; r, t). Furthermore Lemma 2(ii) asserts
˜
U (r, t) =
1
r

∂r
k−1
(r2k−1U(x;
r, t))
=
k−1
j=0
βj
krj+1
∂j
∂rj
U(x; r, t),
and so
lim
r→0
˜
U (r, t)
β0 kr
= lim
r→0
U(x; r, t) = u(x, t).
Thus (30) implies
u(x, t) =
1
β0
k
lim
r→0
˜
G (t + r)
˜
G (t r)
2r
+
1
2r
t+r
t−r
˜
H (y) dy
=
1
β0k
[
˜
G (t) +
˜
H (t)].
Finally then, since n = 2k + 1, (30) and Lemma 2(iii) yield this repre-
sentation formula:
(31)















u(x, t) =
1
γn

∂t
1
t

∂t
n−3
2
tn−2−
∂B(x,t)
g dS
+
1
t

∂t
n−3
2
tn−2−
∂B(x,t)
h dS
where n is odd and γn = 1 · 3 · 5 · · · (n 2),
for x Rn, t 0.
We note that γ3 = 1, and so (31) agrees for n = 3 with (21) and thus
with Kirchhoff’s formula (22).
It remains to check that formula (31) really provides a solution of (11).
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