THE GREEN FUNCTION
CRISTIAN E. GUTIÉRREZ
NOVEMBER 5, 2013
Contents
1. Third Green’s formula
2. The Green function
2.1. Estimates of the Green function near the pole
2.2. Symmetry of the Green function
2.3. The Green function for the ball
2.4. Application 1
2.5. Application 2
References
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1. Third Green’s formula
1
Let n ≥ 3 and Γ(x) =
|x|2−n , where ωn−1 is the surface area of the unit
ωn−1 (2 − n)
sphere in Rn . The third Green formula reads
!
Z
Z
∂Γ(x − y)
∂u
u(y) =
u(x)
− Γ(x − y) (x) dσ(x) +
Γ(x − y) ∆u(x) dx,
∂η(x)
∂η
∂Ω
Ω
where u ∈ C2 (Ω) ∩ C1 (Ω̄), and ∆u ∈ L1 (Ω) and the domain Ω is sufficiently regular.
This formula follows from the 2nd Green formula (1) applied in Ω \ B (y) and
letting → 0. In obtaining the formula the average on the ball B (y) appears and
this is the reason of the value of the constant in the definition of Γ.
2. The Green function
Given y ∈ Ω, define for x , y, φ y (x) = Γ(x − y). The function φ y is continuous in
∂Ω. Consider the Dirichlet problem ∆h = 0 in Ω and h = φ y on ∂Ω, and suppose
that this problem has a solution h y (x). From the maximum principle this solution
is unique.
1
2
C. E. GUTIÉRREZ NOVEMBER 5, 2013
The Green function for the domain Ω and with pole at the point y is defined by
G(x, y) = −h y (x) + Γ(x − y).
With the aid of G we will represent any solution of the Dirichlet problem ∆u = F
in Ω with u = f on ∂Ω. For this we recall the 2nd Green formula:
!
Z
Z
∂v
∂u
(u(x)∆v(x) − v(x)∆u(x)) dx =
(1)
u(x) (x) − v(x) (x) dσ(x).
∂η
∂η
Ω
∂Ω
Applying this formula with v = h y and since h y (x) = Γ(x − y) for x ∈ ∂Ω we get
!
Z
Z
∂h y
∂u
(x) − Γ(x − y) (x) dσ(x)
−
h y (x)∆u(x) dx =
u(x)
∂η
∂η
Ω
∂Ω
!
Z
Z
∂G(x, y)
∂Γ(x − y)
∂u
=−
u(x)
dσ(x) +
u(x)
(x) − Γ(x − y) (x) dσ(x)
∂η
∂η
∂η
∂Ω
∂Ω
Z
Z
∂G(x, y)
=−
u(x)
dσ(x) + u(y) −
Γ(x − y)∆u(x) dx,
∂η
∂Ω
Ω
from the third Green formula. We therefore obtain the following important representation formula valid for y ∈ Ω and u ∈ C2 (Ω) ∩ C1 (Ω̄), with ∆u ∈ L1 (Ω):
Z
Z
∂G(x, y)
(2)
u(y) =
u(x)
dσ(x) +
G(x, y)∆u(x) dx.
∂η(x)
∂Ω
Ω
2.1. Estimates of the Green function near the pole. Since h y (x) is harmonic in Ω,
it follows from the maximum principle that
(3)
m y = min h y (x) ≤ Γ(x − y) − G(x, y) ≤ M y = max h y (x),
x∈∂Ω
x∈∂Ω
for x ∈ Ω̄, x , y.
Since G(x, y) = 0 for x ∈ ∂Ω, we have M y = maxx∈∂Ω Γ(x− y) and m y = minx∈∂Ω Γ(x−
y). From the definition of Γ, a simple calculation shows that
My =
1
1
,
ωn−1 (2 − n) maxx∈∂Ω |x − y|n−2
my =
1
1
.
ωn−1 (2 − n) dist(y, ∂Ω)n−2
Then we obtain from (3) that
My
my
G(x, y)
≤
≤1−
,
for x ∈ Ω̄, x , y.
Γ(x − y) Γ(x − y)
Γ(x − y)
!n−2
my
|x − y|
We have 1 −
= 1−
≤ 1, for |x − y| ≤ dist(y, ∂Ω). On the
Γ(x − y)
dist(y, ∂Ω)
!n−2
My
|x − y|
other hand, 1 −
= 1−
for some x0 ∈ ∂Ω. Given 0 < α < 1,
Γ(x − y)
|x0 − y|
1−
THE GREEN FUNCTION November 5, 2013
3
My
|x − y|
≤ α and we obtain 1 −
≥ 1 − αn−2 .
|x0 − y|
Γ(x − y)
Consequently, we get the following estimates for G (since Γ ≤ 0):
if |x − y| ≤ α dist(y, ∂Ω), then
(4)
G(x, y) ≥ Γ(x − y),
for 0 < |x − y| < dist(y, ∂Ω),
and
(5)
G(x, y) ≤ (1 − αn−2 ) Γ(x − y),
for 0 < |x − y| ≤ α dist(y, ∂Ω).
Since G(x, y) = 0 for x ∈ ∂Ω, then from (4) and the maximum principle applied to
G(·, y) − Γ(· − y) in the domain Ω \ Bdist(y,∂Ω) (y) yields that
G(x, y) ≥ Γ(x − y),
for all x ∈ Ω̄, x , y.
In addition, from (5) we get that G(x, y) ≤ 0 in Bαdist(y,∂Ω) (y) and since G(x, y) = 0
on ∂Ω, then once again by the maximum principle
G(x, y) ≤ 0,
for all x ∈ Ω̄, x , y.
2.2. Symmetry of the Green function. We have the following
G(x, y) = G(y, x),
for all x, y ∈ Ω, x , y.
Pick > 0 such that B (x), B (y) ⊂ Ω and B (x) ∩ B (y) = ∅. Applying the second
Green formula in the domain Ω\(B (x)∪B (y)) with u(z) = G(z, x) and v(z) = G(z, y),
and letting → 0, the symmetry follows.
2.3. The Green function for the ball. Let BR (x0 ) be the Euclidean ball in Rn . Let
y ∈ BR (x0 ) and let y∗ on the line joining x0 and y such that
|y∗ − x0 | |y − x0 | = R2 .
(6)
We have y∗ − x0 = λ(y − x0 ) and inserting in (6) yields λ =
R2
and so
|y − x0 |2
R2
y = x0 +
(y − x0 ).
|y − x0 |2
∗
We claim that
(7)
|y∗ − x|
R
=
,
|y − x|
|y − x0 |
for all |x − x0 | = R.
Writing x = x0 + Rw with |w| = 1 it follows expanding the inner products and
using (6) that
|y∗ − x|2
R2
=
|y − x|2
|y − x0 |2
4
C. E. GUTIÉRREZ NOVEMBER 5, 2013
and the claim follows. We then let for 0 < |x − y| ≤ R
!n−2
R
Γ(x − y∗ ).
G(x, y) = Γ(x − y) −
|y − x0 |
We have that G(·, y) is harmonic in BR (x0 ) \ {y}, and G(x, y) = 0 for |x − y| = R. We
also have by calculation

!−2 


1
R

 (x − x0 )
Dx G(x, y) =
|x − y|−n 1 −
ωn−1
|y − x0 | 
and since the normal η =
x − x0
we obtain
|x − x0 |

!2 

|y − x0 | 
1
∂
−n
 ,
G(x, y) =
|x − y| |x − x0 | 1 −

ωn−1
R
∂η
and for |x − x0 | = R we get the Poisson kernel
∂
1 R2 − |y − x0 |2
G(x, y) = P(x, y) =
.
ωn−1 R |x − y|n
∂η
(8)
Since the function one is harmonic in Ω and equals one on ∂Ω, then from the
uniqueness of the solution to the Dirichlet problem and (2) we get that
Z
∂G(x, y)
dσ(x) = 1,
Ω ∂η(x)
in particular,
Z
P(x, y) dσ(x) = 1.
|x−x0 |=R
We now solve the Dirichlet problem in the ball BR (x0 ). Let f ∈ C(∂BR (x0 )) and
let
Z
(9)
u(y) =
f (x) P(x, y) dσ(x),
for |y − x0 | < R,
|x−x0 |=R
and u(y) = f (y) for y ∈ ∂BR (x0 ). Since y is in the interior of BR (x0 ) and |x − x0 | = R,
the function P(x, y) is C∞ in a neighborhood of the point y and all its derivatives
are bounded in this neighborhood uniformly for |x−x0 | = R. We can then infinitely
differentiate under the integral sign in (9). By direct calculation, ∆ y P(x, y) = 0 for
each y ∈ Rn \ {x} and only when |x − x0 | = R. We then obtain that ∆u = 0 in BR (x0 ).
It remains to show that u(z) → f (y) when z → y, z ∈ BR (x0 ) for each y ∈ ∂BR (x0 ).
Indeed, we write
Z
Z
Z
u(z)− f (y) =
f (x) − f (y) P(x, z) dσ(x) =
+
= I+II.
|x−x0 |=R
|x−x0 |=R,|x−y|≤δ
|x−x0 |=R,|x−y|>δ
THE GREEN FUNCTION November 5, 2013
5
We have |I| ≤ max|x−y|≤δ | f (x) − f (y)|. If |z − y| < δ/2, then |x − z| ≥ |x − y| − |y −
z| ≥ δ − δ/2 = δ/2. Hence on the region of integration in II, we get P(x, z) ≤
Cn (2/δ)n R2 − |z − x0 |2 = Cn (2/δ)n (R − |z − x0 |)(R + |z − x0 |) ≤ Cn δ−n 2R(R − |z − x0 |)
for |z − y| < δ/2. Then |II| ≤ Cδ max∂BR (x0 ) | f (x)| (R − |z − x0 |) for |z − y| ≤ δ/2. If
z → y, then R − |z − x0 | → 0 and so II → 0.
2.4. Application 1. Since the Dirichlet problem is solvable in a ball, we obtain the
following.
Theorem 1. If u ∈ C(Ω) satisfies the mean value property on any ball B b Ω, then u is
harmonic in Ω.
Proof. Fix a ball B b Ω and let v be harmonic in B such that v = u on ∂B. Then
v satisfies the mean value property on any ball B0 b B and so v − u also satisfies
the same property. Since the mean value property is enough to prove the strong
maximum principle, we obtain that v − u = 0 in B and we are done.
2.5. Application 2. A function u ∈ C(Ω) is subharmonic in Ω if
?
u(y) ≤
u(x) dσ(x)
∂Br (y)
for each ball BR (y) b Ω. Also u is superharmonic in Ω if −u is subharmonic in Ω.
Theorem 2. If u ∈ C2 (Ω) and ∆u ≥ 0 in Ω, then u is subharmonic in Ω. Conversely, if
u ∈ C2 (Ω) is subharmonic in Ω, then ∆u ≥ 0 in Ω.
Proof. Let G(x, y) be the Green function of the ball Br (x0 ). From (2)
Z
Z
∂G(x, x0 )
u(x0 ) =
u(x)
dσ(x) +
G(x, x0 )∆u(x) dx.
∂η(x)
Br (x0 )
∂Br (x0 )
Also from (8)
Z
∂G(x, x0 )
u(x)
dσ(x) =
u(x)P(x, x0 ) dσ(x)
∂η(x)
∂Br (x0 )
∂Br (x0 )
Z
?
1
=
u(x) dσ(x) =
u(x) dσ(x).
ωn−1 Rn−1 ∂Br (x0 )
∂Br (x0 )
Z
Then
(10)
?
u(x0 ) −
∂Br (x0 )
Z
u(x) dσ(x) =
G(x, x0 )∆u(x) dx.
Br (x0 )
Since G ≤ 0 in Br (x0 ), the first part of the theorem follows.
6
C. E. GUTIÉRREZ NOVEMBER 5, 2013
On the other hand, if u is subharmonic, then the left hand side of (10) is less
than or equal to zero and so
Z
G(x, x0 )∆u(x) dx ≤ 0.
Br (x0 )
Since G(·, x0 ) is integrable in Br (x0 ), we then can write
Z
Z
∆u(x0 )
G(x, x0 ) dx ≤
G(x, x0 ) (∆u(x0 ) − ∆u(x)) dx.
Br (x0 )
Since G ≤ 0, we obtain
R
∆u(x0 ) ≥
Br (x0 )
Br (x0 )
G(x, x0 ) (∆u(x0 ) − ∆u(x)) dx
R
→ 0, as r → 0,
G(x,
x
)
dx
0
B (x )
r
0
and we are done.
References
Department of Mathematics, Temple University, Philadelphia, PA 19122
E-mail address: [email protected]