2.4. WAVE EQUATION 73 Solution for n = 2. No transformation like (17) works to convert the Euler–Poisson–Darboux equation into the one-dimensional wave equation when n = 2. Instead we will take the initial-value problem (11) for n = 2 and simply regard it as a problem for n = 3, in which the third spatial variable x3 does not appear. Indeed, assuming u ∈ C2(R2 × [0, ∞)) solves (11) for n = 2, let us write (23) ¯(x1,x2,x3,t) u := u(x1,x2,t). Then (11) implies (24) ¯tt u − Δ¯ u = 0 in R3 × (0, ∞) ¯ u = ¯ g, ¯t u = ¯ h on R3 × {t = 0}, for ¯(x1,x2,x3) g := g(x1,x2), ¯(x1,x2,x3) h := h(x1,x2). If we write x = (x1,x2) ∈ R2 and ¯ x = (x1,x2, 0) ∈ R3, then (24) and Kirchhoff’s formula (in the form (21)) imply (25) u(x, t) = ¯(¯ u x, t) = ∂ ∂t t− ∂ ¯(¯ B x,t ) ¯ g d¯ S + t− ∂ ¯(¯ B x,t ) ¯ h d¯ S, where ¯ B (¯ x, t) denotes the ball in R3 with center ¯, x radius t 0 and where d¯ S denotes two-dimensional surface measure on ∂ ¯ B (¯ x, t). We simplify (25) by observing − ∂ ¯(¯ B x,t ) ¯ g d¯ S = 1 4πt2 ∂ ¯(¯ B x,t ) ¯ g d¯ S = 2 4πt2 B(x,t) g(y)(1 + |Dγ(y)|2)1/2 dy, where γ(y) = (t2 − |y − x|2)1/2 for y ∈ B(x, t). The factor “2” enters since ∂ ¯ B (¯ x, t) consists of two hemispheres. Observe that (1 + |Dγ|2)1/2 = t(t2 − |y − x|2)−1/2. Therefore − ∂ ¯(¯ B x,t ) ¯ g d¯ S = 1 2πt B(x,t) g(y) (t2 − |y − x|2)1/2 dy = t 2 − B(x,t) g(y) (t2 − |y − x|2)1/2 dy.

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