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2.4. WAVE EQUATION 79 ¯ B (¯ x, t) denoting the ball in Rn+1 with center ¯ x and radius t and d¯ S denoting n-dimensional surface measure on ∂ ¯ B (¯ x, t). Now (36) − ∂ ¯(¯ B x,t ) ¯ g d¯ S = 1 (n + 1)α(n + 1)tn ∂ ¯(¯ B x,t ) ¯ g d¯ 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 d¯ 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 d¯ 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.