2.4. WAVE EQUATION 71

From this equality we deduce limr→0+ Ur(x; r, t) = 0. We next diﬀerentiate

(15), to discover after some computations that

(16) Urr(x; r, t) = −

∂B(x,r)

Δu dS +

1

n

− 1 −

B(x,r)

Δu dy.

Thus limr→0+ Urr(x; r, t) =

1

n

Δu(x, t). Using formula (16), we can similarly

compute Urrr, etc., and so verify that U ∈

Cm(¯

R

+

× [0, ∞)).

2. Continuing the calculation above, we see from (15) that

Ur =

r

n

−

B(x,r)

utt dy by (11)

=

1

nα(n)

1

rn−1

B(x,r)

utt dy.

Thus

rn−1Ur

=

1

nα(n)

B(x,r)

utt dy,

and so

(

rn−1Ur

)

r

=

1

nα(n)

∂B(x,r)

utt dS

=

rn−1

−

∂B(x,r)

utt dS =

rn−1Utt.

c. Solution for n = 3, 2, Kirchhoﬀ’s and Poisson’s formulas. The

plan in the ensuing subsections will be to transform the Euler–Poisson–

Darboux equation (14) into the usual one-dimensional wave equation. As

the full procedure is rather complicated, we pause here to handle the simpler

cases n = 3, 2, in that order.

Solution for n = 3. Let us therefore hereafter take n = 3, and suppose

u ∈

C2(R3

× [0, ∞)) solves the initial-value problem (11). We recall the

deﬁnitions (12), (13) of U, G, H and then set

(17)

˜

U := rU,

(18)

˜

G := rG,

˜

H := rH.

We now assert that

˜

U solves

(19)

⎧

⎨

⎩

˜tt

U −

˜rr

U = 0 in R+ × (0, ∞)

˜

U =

˜

G,

˜t

U =

˜

H on R+ × {t = 0}

˜

U = 0 on {r = 0} × (0, ∞).