2.2. LAPLACE’S EQUATION 25 (Remember from §A.3 that a slash through an integral denotes an average.) 4. Combining now (11)–(15) and letting ε → 0, we find −Δu(x) = f(x), as asserted. Theorem 1 is in fact valid under far less stringent smoothness require- ments for f: see Gilbarg–Trudinger [G-T]. Interpretation of fundamental solution. We sometimes write −ΔΦ = δ0 in Rn, δ0 denoting the Dirac measure on Rn giving unit mass to the point 0. Adopt- ing this notation, we may formally compute −Δu(x) = Rn −ΔxΦ(x − y)f(y) dy = Rn δxf(y) dy = f(x) (x ∈ Rn), in accordance with Theorem 1. 2.2.2. Mean-value formulas. Consider now an open set U ⊂ Rn and suppose u is a harmonic function within U. We next derive the important mean-value formulas, which declare that u(x) equals both the average of u over the sphere ∂B(x, r) and the average of u over the entire ball B(x, r), provided B(x, r) ⊂ U. These implicit formulas involving u generate a remarkable number of consequences, as we will momentarily see. THEOREM 2 (Mean-value formulas for Laplace’s equation). If u ∈ C2(U) is harmonic, then (16) u(x) = − ∂B(x,r) u dS = − B(x,r) u dy for each ball B(x, r) ⊂ U. Proof. 1. Set φ(r) := − ∂B(x,r) u(y) dS(y) = − ∂B(0,1) u(x + rz) dS(z). Then φ (r) = − ∂B(0,1) Du(x + rz) · z dS(z),

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