2.2. LAPLACE’S EQUATION 39
Combining this calculation with estimate (36), we deduce
|u(x)−g(x0)|
2ε,
provided |x
x0|
is sufficiently small.
c. Green’s function for a ball. To construct Green’s function for the
unit ball B(0, 1), we will again employ a kind of reflection, this time through
the sphere ∂B(0, 1).
DEFINITION. If x
Rn
{0}, the point
˜ x =
x
|x|2
is called the point dual to x with respect to ∂B(0, 1). The mapping x ˜ x
is inversion through the unit sphere ∂B(0, 1).
We now employ inversion through the sphere to compute Green’s func-
tion for the unit ball U =
B0(0,
1). Fix x
B0(0,
1). Remember that we
must find a corrector function φx = φx(y) solving
(37)
Δφx
= 0 in
B0(0,
1)
φx
= Φ(y x) on ∂B(0, 1);
then Green’s function will be
(38) G(x, y) = Φ(y x)
φx(y).
The idea now is to “invert the singularity” from x
B0(0,
1) to ˜ x /
B(0, 1). Assume for the moment n 3. Now the mapping y Φ(y ˜) x is
harmonic for y = ˜. x Thus y
|x|2−nΦ(y
˜) x is harmonic for y = ˜, x and so
(39)
φx(y)
:= Φ(|x|(y ˜)) x
is harmonic in U. Furthermore, if y ∂B(0, 1) and x = 0,
|x|2|y

˜|2
x =
|x|2 |y|2

2y · x
|x|2
+
1
|x|2
=
|x|2
2y · x + 1 = |x
y|2.
Thus (|x||y
˜|)−(n−2)
x = |x
y|−(n−2).
Consequently
(40)
φx(y)
= Φ(y x) (y ∂B(0, 1)),
as required.
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