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.