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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