70 2. FOUR IMPORTANT LINEAR PDE b. Spherical means. Now suppose n 2, m 2, and u Cm(Rn × [0, ∞)) solves the initial-value problem (11) utt Δu = 0 in Rn × (0, ∞) u = g, ut = h on Rn × {t = 0}. We intend to derive an explicit formula for u in terms of g, h. The plan will be to study first the average of u over certain spheres. These averages, taken as functions of the time t and the radius r, turn out to solve the Euler–Poisson–Darboux equation, a PDE which we can for odd n convert into the ordinary one-dimensional wave equation. Applying d’Alembert’s formula, or more precisely its variant (10), eventually leads us to a formula for the solution. NOTATION. (i) Let x Rn, t 0, r 0. Define (12) U(x r, t) := ∂B(x,r) u(y, t) dS(y), the average of u(·,t) over the sphere ∂B(x, r). (ii) Similarly, (13) G(x r) := ∂B(x,r) g(y) dS(y) H(x r) := ∂B(x,r) h(y) dS(y). For fixed x, we hereafter regard U as a function of r and t and discover a partial differential equation that U solves: LEMMA 1 (Euler–Poisson–Darboux equation). Fix x Rn, and let u satisfy (11). Then U Cm(¯ R + × [0, ∞)) and (14) Utt Urr n−1 r Ur = 0 in R+ × (0, ∞) U = G, Ut = H on R+ × {t = 0}. The partial differential equation in (14) is the Euler–Poisson–Darboux equation. (Note that the term Urr + n−1 r Ur is the radial part of the Laplacian Δ in polar coordinates.) Proof. 1. As in the proof of Theorem 2 in §2.2.2 we compute for r 0 (15) Ur(x r, t) = r n B(x,r) Δu(y, t) dy.
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