70 2. FOUR IMPORTANT LINEAR PDE

b. Spherical means. Now suppose n ≥ 2, m ≥ 2, and u ∈

Cm(Rn

×

[0, ∞)) solves the initial-value problem

(11)

utt − Δu = 0 in

Rn

× (0, ∞)

u = g, ut = h on

Rn

× {t = 0}.

We intend to derive an explicit formula for u in terms of g, h. The plan

will be to study ﬁrst the average of u over certain spheres. These averages,

taken as functions of the time t and the radius r, turn out to solve the

Euler–Poisson–Darboux equation, a PDE which we can for odd n convert

into the ordinary one-dimensional wave equation. Applying d’Alembert’s

formula, or more precisely its variant (10), eventually leads us to a formula

for the solution.

NOTATION. (i) Let x ∈

Rn,

t 0, r 0. Deﬁne

(12) U(x; r, t) := −

∂B(x,r)

u(y, t) dS(y),

the average of u(·,t) over the sphere ∂B(x, r).

(ii) Similarly,

(13)

⎧

⎪

⎪

⎨

⎪

⎪

⎩

G(x; r) := −

∂B(x,r)

g(y) dS(y)

H(x; r) := −

∂B(x,r)

h(y) dS(y).

For ﬁxed x, we hereafter regard U as a function of r and t and discover

a partial diﬀerential equation that U solves:

LEMMA 1 (Euler–Poisson–Darboux equation). Fix x ∈

Rn,

and let u

satisfy (11). Then U ∈

Cm(¯

R

+

× [0, ∞)) and

(14)

Utt − Urr −

n−1

r

Ur = 0 in R+ × (0, ∞)

U = G, Ut = H on R+ × {t = 0}.

The partial diﬀerential equation in (14) is the Euler–Poisson–Darboux

equation. (Note that the term Urr +

n−1

r

Ur is the radial part of the Laplacian

Δ in polar coordinates.)

Proof. 1. As in the proof of Theorem 2 in §2.2.2 we compute for r 0

(15) Ur(x; r, t) =

r

n

−

B(x,r)

Δu(y, t) dy.