2.2. LAPLACE’S EQUATION 23 This reasoning might suggest that the convolution (8) u(x)= Rn Φ(x − y)f(y) dy = − 1 2π R2 log(|x − y|)f(y) dy (n = 2) 1 n(n−2)α(n) Rn f(y) |x−y|n−2 dy (n ≥ 3) will solve Laplace’s equation (1). However, this is wrong. Indeed, as inti- mated by estimate (7), D2Φ(x − y) is not summable near the singularity at y = x, and so naive differentiation through the integral sign is unjustified (and incorrect). We must proceed more carefully in calculating Δu. Let us for simplicity now assume f ∈ Cc 2(Rn) that is, f is twice contin- uously differentiable, with compact support. THEOREM 1 (Solving Poisson’s equation). Define u by (8). Then (i) u ∈ C2(Rn) and (ii) −Δu = f in Rn. We consequently see that (8) provides us with a formula for a solution of Poisson’s equation (2) in Rn. Proof. 1. We have (9) u(x) = Rn Φ(x − y)f(y) dy = Rn Φ(y)f(x − y) dy hence u(x + hei) − u(x) h = Rn Φ(y) f(x + hei − y) − f(x − y) h dy, where h = 0 and ei = (0,..., 1,..., 0), the 1 in the ith-slot. But f(x + hei − y) − f(x − y) h → fxi (x − y) uniformly on Rn as h → 0, and thus uxi (x) = Rn Φ(y)fxi (x − y) dy (i = 1,...,n). Similarly (10) uxixj (x) = Rn Φ(y)fxixj (x − y) dy (i, j = 1,...,n).

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