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)
