40 2. FOUR IMPORTANT LINEAR PDE

DEFINITION. Green’s function for the unit ball is

(41) G(x, y) := Φ(y − x) − Φ(|x|(y − ˜)) x (x, y ∈ B(0, 1), x = y).

The same formula is valid for n = 2 as well.

Assume now u solves the boundary-value problem

(42)

Δu = 0 in

B0(0,

1)

u = g on ∂B(0, 1).

Then using (30), we see

(43) u(x) = −

∂B(0,1)

g(y)

∂G

∂ν

(x, y) dS(y).

According to formula (41),

Gyi (x, y) = Φyi (y − x) − Φ(|x|(y − ˜))yi x .

But

Φyi (y − x) =

1

nα(n)

xi − yi

|x − y|n

,

and furthermore

Φ(|x|(y − ˜))yi x . =

−1

nα(n)

yi|x|2

− xi

(|x||y −

˜|)n

x

= −

1

nα(n)

yi|x|2

− xi

|x −

y|n

if y ∈ ∂B(0, 1). Accordingly

∂G

∂ν

(x, y) =

n

i=1

yiGyi (x, y)

=

−1

nα(n)

1

|x − y|n

n

i=1

yi((yi − xi) −

yi|x|2

+ xi)

=

−1

nα(n)

1 −

|x|2

|x − y|n

.

Hence formula (43) yields the representation formula

u(x) =

1 −

|x|2

nα(n)

∂B(0,1)

g(y)

|x −

y|n

dS(y).

Suppose now instead of (42) u solves the boundary-value problem

(44)

Δu = 0 in

B0(0,r)

u = g on ∂B(0,r)