78 2. FOUR IMPORTANT LINEAR PDE

Remarks. (i) Notice that to compute u(x, t) we need only have information

on g, h and their derivatives on the sphere ∂B(x, t), and not on the entire

ball B(x, t).

(ii) Comparing formula (31) with d’Alembert’s formula (8) (n = 1), we

observe that the latter does not involve the derivatives of g. This suggests

that for n 1, a solution of the wave equation (11) need not for times t 0

be as smooth as its initial value g: irregularities in g may focus at times

t 0, thereby causing u to be less regular. (We will see later in §2.4.3 that

the “energy norm” of u does not deteriorate for t 0.)

(iii) Once again (as in the case n = 1) we see the phenomenon of ﬁnite

propagation speed of the initial disturbance.

(iv) A completely diﬀerent derivation of formula (31) (using the heat

equation!) is in §4.3.3.

e. Solution for even n. Assume now

n is an even integer.

Suppose u is a

Cm

solution of (11), m =

n+2

2

. We want to fashion a repre-

sentation formula like (31) for u. The trick, as above for n = 2, is to note

that

(33) ¯(x1,...,xn+1,t) u := u(x1,...,xn,t)

solves the wave equation in

Rn+1

× (0, ∞), with the initial conditions

¯ u = ¯ g, ¯t u =

¯

h on

Rn+1

× {t = 0},

where

(34)

¯(x1,...,xn+1) g := g(x1,...,xn)

¯(x1,...,xn+1)

h := h(x1,...,xn).

As n + 1 is odd, we may employ (31) (with n + 1 replacing n) to secure

a representation formula for ¯ u in terms of ¯ g,

¯.

h But then (33) and (34) yield

at once a formula for u in terms of g, h. This is again the method of descent.

To carry out the details, let us ﬁx x ∈

Rn,

t 0, and write ¯ x =

(x1,... , xn, 0) ∈

Rn+1.

Then (31), with n + 1 replacing n, gives

(35)

u(x, t) =

1

γn+1

∂

∂t

1

t

∂

∂t

n−2

2

tn−1

−

∂

¯(¯

B x,t )

¯ g

d¯

S

+

1

t

∂

∂t

n−2

2

tn−1

−

∂

¯(¯

B x,t )

¯

h

d¯

S ,