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,

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