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.
Previous Page Next Page