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.

Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2010 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.