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).