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,