76 2. FOUR IMPORTANT LINEAR PDE the next-to-last equality holding according to (14). Using Lemma 2(ii) we conclude as well that ˜ U = 0 on {r = 0}. In view of Lemma 3, (29), and formula (10), we conclude for 0 r t that (30) ˜ U (r, t) = 1 2 [ ˜ G (r + t) ˜ G (t r)] + 1 2 t+r t−r ˜ H (y) dy. But recall u(x, t) = limr→0 U(x r, t). Furthermore Lemma 2(ii) asserts ˜ U (r, t) = 1 r ∂r k−1 (r2k−1U(x r, t)) = k−1 j=0 βj krj+1 ∂j ∂rj U(x r, t), and so lim r→0 ˜ U (r, t) β0 kr = lim r→0 U(x r, t) = u(x, t). Thus (30) implies u(x, t) = 1 β0 k lim r→0 ˜ G (t + r) ˜ G (t r) 2r + 1 2r t+r t−r ˜ H (y) dy = 1 β0k [ ˜ G (t) + ˜ H (t)]. Finally then, since n = 2k + 1, (30) and Lemma 2(iii) yield this repre- sentation formula: (31) u(x, t) = 1 γn ∂t 1 t ∂t n−3 2 tn−2− ∂B(x,t) g dS + 1 t ∂t n−3 2 tn−2− ∂B(x,t) h dS where n is odd and γn = 1 · 3 · 5 · · · (n 2), for x Rn, t 0. We note that γ3 = 1, and so (31) agrees for n = 3 with (21) and thus with Kirchhoff’s formula (22). It remains to check that formula (31) really provides a solution of (11).
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