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