2.2. LAPLACE’S EQUATION 35

THEOREM 12 (Representation formula using Green’s function). If

u ∈

C2(

¯

U ) solves problem (29), then

(30) u(x) = −

∂U

g(y)

∂G

∂ν

(x, y) dS(y) +

U

f(y)G(x, y) dy (x ∈ U).

Here we have a formula for the solution of the boundary-value problem

(29), provided we can construct Green’s function G for the given domain U.

This is in general a diﬃcult matter and can be done only when U has simple

geometry. Subsequent subsections identify some special cases for which an

explicit calculation of G is possible.

Interpreting Green’s function. Fix x ∈ U. Then regarding G as a

function of y, we may symbolically write

−ΔG = δx in U

G = 0 on ∂U,

δx denoting the Dirac measure giving unit mass to the point x.

Before moving on to speciﬁc examples, let us record the general assertion

that G is symmetric in the variables x and y:

THEOREM 13 (Symmetry of Green’s function). For all x, y ∈ U, x = y,

we have

G(y, x) = G(x, y).

Proof. Fix x, y ∈ U, x = y. Write

v(z) := G(x, z), w(z) := G(y, z) (z ∈ U).

Then Δv(z) = 0 (z = x), Δw(z) = 0 (z = y) and w = v = 0 on

∂U. Thus our applying Green’s identity on V := U − [B(x, ε) ∪ B(y, ε)] for

suﬃciently small ε 0 yields

(31)

∂B(x,ε)

∂v

∂ν

w −

∂w

∂ν

v dS(z) =

∂B(y,ε)

∂w

∂ν

v −

∂v

∂ν

w dS(z),

ν

denoting the inward pointing unit vector ﬁeld on ∂B(x, ε)∪∂B(y, ε). Now

w is smooth near x, whence

∂B(x,ε)

∂w

∂ν

v dS ≤

Cεn−1

sup

∂B(x,ε)

|v| = o(1) as ε → 0.