2.4. WAVE EQUATION 79 ¯ B x, t) denoting the ball in Rn+1 with center ¯ x and radius t and S denoting n-dimensional surface measure on ¯ B x, t). Now (36) ¯(¯ B x,t ) ¯ g S = 1 (n + 1)α(n + 1)tn ¯(¯ B x,t ) ¯ g S. Note that ¯ B (x, t) {yn+1 0} is the graph of the function γ(y) := (t2 |y x|2)1/2 for y B(x, t) Rn. Likewise ¯ B (x, t) {yn+1 0} is the graph of −γ. Thus (36) implies (37) ¯(¯ B x,t ) ¯ g S = 2 (n + 1)α(n + 1)tn B(x,t) g(y)(1 + |Dγ(y)|2)1/2 dy, the factor “2” entering because ¯ B x, t) comprises two hemispheres. Note that (1+|Dγ(y)|2)1/2 = t(t2 |y x|2)−1/2. Our substituting this into (37) yields ¯(¯ B x,t ) ¯ g S = 2 (n + 1)α(n + 1)tn−1 B(x,t) g(y) (t2 |y x|2)1/2 dy = 2tα(n) (n + 1)α(n + 1) B(x,t) g(y) (t2 |y x|2)1/2 dy. We insert this formula and the similar one with h in place of g into (35) and find u(x, t) = 1 γn+1 2α(n) (n + 1)α(n + 1) ∂t 1 t ∂t n−2 2 tn B(x,t) g(y) (t2 |y x|2)1/2 dy + 1 t ∂t n−2 2 tn B(x,t) h(y) (t2 |y x|2)1/2 dy . Since γn+1 = 1 · 3 · 5 · · · (n 1) and α(n) = πn/2 Γ( n+2 2 ) , we may compute γn = 2 · 4 · · · (n 2) · n. Hence the resulting representation formula for even n is (38) u(x, t) = 1 γn ∂t 1 t ∂t n−2 2 tn− B(x,t) g(y) (t2 |y x|2)1/2 dy + 1 t ∂t n−2 2 tn B(x,t) h(y) (t2 |y x|2)1/2 dy , where n is even and γn = 2 · 4 · · · (n 2) · n, for x Rn, t 0. Since γ2 = 2, this agrees with Poisson’s formula (27) if n = 2.
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