72 2. FOUR IMPORTANT LINEAR PDE Indeed ˜tt U = rUtt = r Urr + 2 r Ur by (14), with n = 3 = rUrr + 2Ur = (U + rUr)r = ˜rr. U Notice also that ˜ G rr(0) = 0. Applying formula (10) to (19), we find for 0 r t (20) ˜ U (x r, t) = 1 2 [ ˜ G (r + t) ˜ G (t r)] + 1 2 r+t −r+t ˜ H (y) dy. Since (12) implies u(x, t) = limr→0+ U(x r, t), we conclude from (17), (18), (20) that u(x, t) = lim r→0+ ˜ U (x r, t) r = lim r→0+ ˜ G (t + r) ˜ G (t r) 2r + 1 2r t+r t−r ˜ H (y) dy = ˜ G (t) + ˜ H (t). Owing then to (13), we deduce (21) u(x, t) = ∂t t− ∂B(x,t) g dS + t− ∂B(x,t) h dS. But ∂B(x,t) g(y) dS(y) = ∂B(0,1) g(x + tz) dS(z) and so ∂t ∂B(x,t) g dS = ∂B(0,1) Dg(x + tz) · z dS(z) = ∂B(x,t) Dg(y) · y x t dS(y). Returning to (21), we therefore conclude (22) u(x, t) = ∂B(x,t) th(y)+g(y)+Dg(y)·(y −x) dS(y) (x R3, t 0). This is Kirchhoff’s formula for the solution of the initial-value problem (11) in three dimensions.
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