36 2. FOUR IMPORTANT LINEAR PDE

On the other hand, v(z) = Φ(z − x) −

φx(z),

where

φx

is smooth in U. Thus

lim

ε→0

∂B(x,ε)

∂v

∂ν

w dS = lim

ε→0

∂B(x,ε)

∂Φ

∂ν

(x − z)w(z) dS = w(x),

by calculations as in the proof of Theorem 1. Thus the left-hand side of (31)

converges to w(x) as ε → 0. Likewise the right-hand side converges to v(y).

Consequently

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

b. Green’s function for a half-space. In this and the next subsection

we will build Green’s functions for two regions with simple geometry, namely

the half-space R+

n

and the unit ball B(0, 1). Everything depends upon our

explicitly solving the corrector problem (26) in these regions, and this in

turn depends upon some clever geometric reflection tricks.

First let us consider the half-space

R+

n

= {x = (x1,...,xn) ∈

Rn

| xn 0}.

Although this region is unbounded, and so the calculations in the previous

section do not directly apply, we will attempt nevertheless to build Green’s

function using the ideas developed earlier. Later of course we must check

directly that the corresponding representation formula is valid.

DEFINITION. If x = (x1,...,xn−1,xn) ∈ R+,

n

its reflection in the plane

∂R+ n is the point

˜ x = (x1,...,xn−1, −xn).

We will solve problem (26) for the half-space by setting

φx(y)

:= Φ(y − ˜) x = Φ(y1 − x1,...,yn−1 − xn−1,yn + xn) (x, y ∈

R+).n

The idea is that the corrector

φx

is built from Φ by “reflecting the singular-

ity” from x ∈ R+

n

to ˜ x / ∈ R+.

n

We note

φx(y)

= Φ(y − x) if y ∈

∂R+,n

and thus

Δφx

= 0 in

R+n

φx

= Φ(y − x) on

∂R+,n

as required.