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.
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