34 2. FOUR IMPORTANT LINEAR PDE
as ε 0. Hence our sending ε 0 in (24) yields the formula
(25)
u(x) =
∂U
Φ(y x)
∂u
∂ν
(y) u(y)
∂Φ
∂ν
(y x) dS(y)

U
Φ(y x)Δu(y) dy.
This identity is valid for any point x U and any function u
C2(
¯
U ).
Now formula (25) would permit us to solve for u(x) if we knew the
values of Δu within U and the values of u, ∂u/∂ν along ∂U. However, for
our application to Poisson’s equation with prescribed boundary values for u,
the normal derivative ∂u/∂ν along ∂U is unknown to us. We must therefore
somehow modify (25) to remove this term.
The idea is now to introduce for fixed x a corrector function
φx
=
φx(y),
solving the boundary-value problem
(26)
Δφx = 0 in U
φx
= Φ(y x) on ∂U.
Let us apply Green’s formula once more, to compute
(27)

U
φx(y)Δu(y)
dy =
∂U
u(y)
∂φx
∂ν
(y)
φx(y)
∂u
∂ν
(y) dS(y)
=
∂U
u(y)
∂φx
∂ν
(y) Φ(y x)
∂u
∂ν
(y) dS(y).
We introduce next this
DEFINITION. Green’s function for the region U is
G(x, y) := Φ(y x)
φx(y)
(x, y U, x = y).
Adopting this terminology and adding (27) to (25), we find
(28) u(x) =
∂U
u(y)
∂G
∂ν
(x, y) dS(y)
U
G(x, y)Δu(y) dy (x U),
where
∂G
∂ν
(x, y) = DyG(x, y) · ν(y)
is the outer normal derivative of G with respect to the variable y. Observe
that the term ∂u/∂ν does not appear in equation (28): we introduced the
corrector
φx
precisely to achieve this.
Suppose now u
C2(
¯
U ) solves the boundary-value problem
(29)
−Δu = f in U
u = g on ∂U,
for given continuous functions f, g. Plugging into (28), we obtain
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