2.4. WAVE EQUATION 77 THEOREM 2 (Solution of wave equation in odd dimensions). Assume n is an odd integer, n 3, and suppose also g Cm+1(Rn), h Cm(Rn), for m = n+1 2 . Define u by (31). Then (i) u C2(Rn × [0, ∞)), (ii) utt Δu = 0 in Rn × (0, ∞), and (iii) lim (x,t)→(x0,0) x∈Rn, t0 u(x, t) = g(x0), lim (x,t)→(x0,0) x∈Rn, t0 ut(x, t) = h(x0) for each point x0 Rn. Proof. 1. Suppose first g 0, so that (32) u(x, t) = 1 γn 1 t ∂t n−3 2 ( tn−2H(x t) ) . Then Lemma 2(i) lets us compute utt = 1 γn 1 t ∂t n−1 2 ( tn−1Ht ) . From the calculation in the proof of Theorem 2 in §2.2.2, we see as well that Ht = t n B(x,t) Δh dy. Consequently utt = 1 nα(n)γn 1 t ∂t n−1 2 B(x,t) Δh dy = 1 nα(n)γn 1 t ∂t n−3 2 1 t ∂B(x,t) Δh dS . On the other hand, ΔH(x t) = Δx ∂B(0,t) h(x + y) dS(y) = ∂B(x,t) Δh dS. Consequently (32) and the calculations above imply utt = Δu in Rn×(0, ∞). A similar computation works if h 0. 2. We leave it as an exercise to confirm, using Lemma 2(ii)–(iii), that u takes on the correct initial conditions.
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