2.4. WAVE EQUATION 75
The proof by induction is left as an exercise.
Now assume
n ≥ 3 is an odd integer
and set
n = 2k + 1 (k ≥ 1).
Henceforth suppose u ∈
Ck+1(Rn
× [0, ∞)) solves the initial-value prob-
lem (11). Then the function U defined by (12) is
Ck+1.
NOTATION. We write
(28)
⎧
⎪
⎪
⎨
⎪
⎪
⎩
˜
U (r, t) :=
(
1
r
∂
∂r
)k−1
(r2k−1U(x;
r, t))
˜
G (r) :=
(
1
r
∂
∂r
)k−1
(r2k−1G(x;
r))
˜
H (r) :=
(
1
r
∂
∂r
)k−1
(r2k−1H(x;
r))
(r 0, t ≥ 0).
Then
(29)
˜
U (r, 0) =
˜
G (r),
˜
U t(r, 0) =
˜
H (r).
Next we combine Lemma 1 and the identities provided by Lemma 2 to
demonstrate that the transformation (28) of U into
˜
U in effect converts the
Euler–Poisson–Darboux equation into the wave equation.
LEMMA 3 (
˜
U solves the one-dimensional wave equation). We have
⎧
⎨
⎩
˜tt
U −
˜rr
U = 0 in R+ × (0, ∞)
˜
U =
˜
G,
˜
U
t
=
˜
H on R+ × {t = 0}
˜
U = 0 on {r = 0} × (0, ∞).
Proof. If r 0,
˜rr
U =
∂2
∂r2
1
r
∂
∂r
k−1
(r2k−1U)
=
1
r
∂
∂r
k
(r2kUr)
by Lemma 2(i)
=
1
r
∂
∂r
k−1
[r2k−1Urr
+
2kr2k−2Ur]
=
1
r
∂
∂r
k−1
r2k−1
Urr +
n − 1
r
Ur (n = 2k + 1)
=
1
r
∂
∂r
k−1
(r2k−1Utt)
=
˜
U tt,
