2.4. WAVE EQUATION 71 From this equality we deduce limr→0+ Ur(x r, t) = 0. We next differentiate (15), to discover after some computations that (16) Urr(x r, t) = − ∂B(x,r) Δu dS + 1 n − 1 − B(x,r) Δu dy. Thus limr→0+ Urr(x r, t) = 1 n Δu(x, t). Using formula (16), we can similarly compute Urrr, etc., and so verify that U ∈ Cm(¯ R + × [0, ∞)). 2. Continuing the calculation above, we see from (15) that Ur = r n − B(x,r) utt dy by (11) = 1 nα(n) 1 rn−1 B(x,r) utt dy. Thus rn−1Ur = 1 nα(n) B(x,r) utt dy, and so ( rn−1Ur ) r = 1 nα(n) ∂B(x,r) utt dS = rn−1 − ∂B(x,r) utt dS = rn−1Utt. c. Solution for n = 3, 2, Kirchhoff’s and Poisson’s formulas. The plan in the ensuing subsections will be to transform the Euler–Poisson– Darboux equation (14) into the usual one-dimensional wave equation. As the full procedure is rather complicated, we pause here to handle the simpler cases n = 3, 2, in that order. Solution for n = 3. Let us therefore hereafter take n = 3, and suppose u ∈ C2(R3 × [0, ∞)) solves the initial-value problem (11). We recall the definitions (12), (13) of U, G, H and then set (17) ˜ U := rU, (18) ˜ G := rG, ˜ H := rH. We now assert that ˜ U solves (19) ⎧ ⎨ ⎩ ˜tt U − ˜rr U = 0 in R+ × (0, ∞) ˜ U = ˜ G, ˜t U = ˜ H on R+ × {t = 0} ˜ U = 0 on {r = 0} × (0, ∞).

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