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),