Fundamental Solutions

Section 1.3: Fundamental Solutions

1.3 Fundamental Solutions

We begin this section by seeking a harmonic function u, that is, ∆u = 0 in R

, which

depends only on r = |x − a| for some xed a ∈ R

. We set v(r) = u(x). This implies

′′

n − 1

′

= 0

and hence

v(r) =







+ c

log r, n = 2,

+ c

2−n

, n ≥ 3,

where c

are constants for i = 1, 2, 3, 4. We are interested in a function with a singularity

such that

∂B

∂v

∂r

dS = 1 for any r > 0.

Hence we set for any xed a ∈ R

Γ(a, x) =

2π

log |a − x| for n = 2

Γ(a, x) =

(2 − n)

|a − x|

2−n

for n ≥ 3.

To summarize, we have that for xed a ∈ R

, Γ(a, x) is harmonic at x 6= a, that is,

∆

Γ(a, x) = 0 for any x 6= a

and has a singularity at x = a. Moreover, it satises

∂B

(a)

∂Γ

∂n

(a, x)dS

= 1 for any r > 0.

Now we prove the Green’s identity.

Theorem 1.17. Suppose Ω is a bounded domain in R

and that u ∈ C

(

Ω)∩ C

(Ω).

Then for any a ∈ Ω there holds

u(a) =

Ω

Γ(a, x)∆u(x)dx −

∂Ω



Γ(a, x)

∂u

∂n

(x) − u(x)

∂I

′

∂n

(a, x)



Remark 1.18.

(i) For any a ∈ Ω, Γ(a, ·) is integrable in Ω although it has a singularity.

(ii) For a /∈

Ω, the expression in the right side gives 0 .

(iii) By letting u = 1 we have

∂Ω

∂Γ

∂n

(a, x)dS

= 1 for any a ∈ Ω.

Chapter 1: Harmonic Functions

Proof. We apply Green’s formula to u and Γ(a, ·) in the domain Ω\B

(a) for small r > 0

and get

Ω\B

(a)

(Γ∆u − u∆Γ)dx =

∂Ω



∂u

∂n

− u

∂Γ

∂n



−

∂B

(a)



∂u

∂n

− u

∂Γ

∂n



Note ∆Γ = 0 in Ω\B

(a). Then we have

Ω

Γ∆udx =

∂Ω



∂u

∂n

− u

∂Γ

∂n



− lim

r→0

∂B

(a)



∂u

∂n

− u

∂Γ

∂n



For n ≥ 3, we get by denition of Γ



∂B

(a)

∂u

∂n



(2 − n)ω

2−n

∂B

(a)

∂u

∂n



≤

n − 2

sup

∂B

(a)

|Du| → 0 as r → 0,

∂B

(a)

∂Γ

∂n

dS =

n−1

∂B

(a)

udS → u(a) as r → 0.

We get the same conclusion for n = 2 in the same way.

Remark 1.19. We may employ the local version of the Green’s identity to get gradient

estimates without using the mean value property. Suppose u ∈ C





is harmonic in

. For any xed 0 < r < R < 1 choose a cuto function φ ∈ C

∞

) such that φ = 1

in B

and 0 ≤ φ ≤ 1. Apply the Green’s formula to u and φΓ(a, ·) in B

(a) for

a ∈ B

and ρ small enough. We proceed as in the proof of Theorem 1.17 and obtain

u(a) = −

r<|x|<R

u(x)∆

(φ(x)Γ(a, x))dx for any a ∈ B

(0).

Hence one may prove (without using the mean value property)

sup

1/2

|u| ≤ c



|u|



1/p

and sup

1/2

|Du| ≤ c max

|u|

where c is a constant depending only on n.

Now we begin to discuss the Green’s functions. Suppose Ω is a bounded domain in

. Let u ∈ C

(

Ω) ∩ C

(Ω). We have by Theorem 1.17 for any x ∈ Ω

u(x) =

Ω

Γ(x, y)∆u(y)dy −

∂Ω



Γ(x, y)

∂u

∂n

(y) − u(y)

∂1

′

∂n

(x, y)



Section 1.3: Fundamental Solutions

If u solves the Dirichlet boundary value problem







∆u = f in Ω

u = φ on ∂Ω

for some f ∈ C(

Ω) and φ ∈ C(∂Ω), then u can be expressed in terms of f and φ, with

one unknown term. We want to eliminate this term by adjusting Γ.

For any xed x ∈ Ω, consider

(

x, y

) = Γ(

x, y

) + Φ(

x, y

)

for some Φ(x, ·) ∈ C

(

Ω) with ∆

Φ(x, y) = 0 in Ω. Then Theorem 1.17 can be expressed

as follows for any x ∈ Ω

u(x) =

Ω

γ(x, y)∆u(y)dy −

∂Ω



γ(x, y)

∂n

(y) − u(y)

vγ

∂n

(x, y)



since the extra Φ(x, ·) is harmonic. Now by choosing Φ appropriately, we are led to the

important concept of Green’s function.

For each xed x ∈ Ω choose Φ(x, ·) ∈ C

(

Ω) ∩ C

(Ω) such that







∆

Φ(x, y) = 0 for y ∈ Ω

Φ(x, y) = −Γ(x, y) for y ∈ ∂Ω

Denote the resulting γ(x, y) by G(x, y), which is called Green’s function. If such a G

exists, then the solution

to the Dirichlet problem (*) can be expressed as

u(x) =

Ω

G(x, y)f (y)dy +

∂Ω

φ(y)

∂G

∂n

(x, y)dS

Note that Green’s function G(x, y) is dened as a function of y ∈

Ω for each xed x ∈ Ω.

Now we discuss some properties of G as a function of x and y. Our rst observation

is that the Green’s function is unique.

This is proved by the maximum principle since the dierence of two Green’s functions

are harmonic in Ω with zero boundary value. In fact, we have more.

Proposition 1.20. Green’s function G(x, y) is symmetric in Ω × Ω; that is, G(x, y) =

G(y, x) for x 6= y ∈ Ω.

Proof. Pick x

, x

∈ Ω with x

6= x

. Choose r > 0 small such that B

)∩B

) = ∅.

Set G

(y) = G(x

, y) and G

(y) = G(x

, y).

Chapter 1: Harmonic Functions

We apply Green’s formula in Ω \ B

) ∪ B

) and get

Ω\B

)∪B

)

∆G

− G

∆G

) =

∂Ω



∂G

∂n

− G

∂G

∂n



−

∂B

)



∂G

∂n

− G

∂G

∂n



−

∂B

)



∂G

∂n

− G

∂G

∂n



dS.

Since G

is harmonic for y 6= x

, i = 1, 2, and vanishes on ∂Ω, we have

∂B

)



∂G

∂n

− G

∂G

∂n



dS +

∂B

)



∂G

∂n

− G

∂G

∂n



dS = 0.

Note the left side has the same limit as the left side in the following as r → 0:

∂B

)



∂G

∂n

− G

∂Γ

∂n



dS +

∂B

)



∂Γ

∂n

− Γ

∂G

∂n



dS = 0.

Since

∂B

)

∂G

∂n

dS → 0,

∂B

)

∂G

∂n

dS → 0 as r → 0,

∂B

)

∂Γ

∂n

dS → G

∂B

)

∂Γ

∂n

dS → G

) as r → 0,

we obtain G

) − G

) = 0 or equivalently G(x

, x

) = G(x

, x

Proposition 1.21. There holds for x, y ∈ Ω with x 6= y







0 > G(x, y) > Γ(x, y) for n ≥ 3

0 > G(x, y) > Γ(x, y) −

2π

log diam(Ω) for n = 2

Proof. Fix x ∈ Ω and write G( y) = G(x, y). Since lim

y → x

G(y) = −∞ then there exists

an r > 0 such that G(y) < 0 in B

(x). Note that G is harmonic in Ω \B

(x) with G = 0

on ∂Ω and G < 0 on ∂B

(x). The maximum principle implies G(y) < 0 in Ω \B

(x) for

such r > 0. Next, consider the other part of the inequality. Recall the denition of the

Green’s function

G(x, y) = Γ(x, y) + Φ(x, y) where







∆Φ = 0 in Ω,

Φ = −Γ on ∂Ω.

Section 1.3: Fundamental Solutions

For n ≥ 3, we have

Γ(x, y) =

(2 − n)ω

|x − y|

2−n

< 0 for y ∈ ∂Ω,

which implies Φ(x, ·) > 0 on ∂Ω. By the maximum principle, we have Φ > 0 in Ω. For

n = 2 we have

Γ(x, y) =

2π

log |x − y| ≤

2π

log diam(Ω) for y ∈ ∂Ω.

Hence the maximum principle implies Φ > −

2π

log diam(Ω) in Ω.

We may calculate the Green’s functions for some special domains.

Proposition 1.22. The Green’s function for the ball B

(0) is given by

(i)

G(x, y) =

(2 − n)ω

|x − y|

2−n

−



|x|

x −

|x|



2−n

for n ≥ 3,

(ii)

G(x, y) =

2π



log |x − y| − log



|x|

x −

|x|





for n = 2.

Proof. Fix x 6= 0 with |x| < R. Consider X ∈ R

\ B

with X the multiple of x and

|X| · |x| = R

, that is, X =

|x|

x. In other words, X and x are reexive with respect

to the sphere ∂B

. Note the map x 7→ X is conformal; that is, it preserves angles. If

|y| = R, we have by similarity of triangles

|x|

|X|

|y − x|

|y − X|

which implies

|y − x| =

|x|

|y − X| =



|x|

y −

|x|



for any y ∈ ∂B

. (1.5)

Therefore, in order to have zero boundary value, we take for n ≥ 3

G(x, y) =

(2 − n)ω

|x − y|

n−2

−



|x|



n−2

|y − X|

n−2

The case n = 2 is similar.

Next, we calculate the normal derivative of Green’s function on the sphere.

Chapter 1: Harmonic Functions

Corollary 1.23. Suppose G is the Green’s function in B

(0). Then there holds

∂G

∂n

(x, y) =

− |x|

R|x − y|

for any x ∈ B

and y ∈ ∂B

Proof. We just consider the case n ≥ 3. Recall with X = R

x/|x|

G(x, y) =

(2 − n)ω

|x − y|

2−n

−



|x|



n−2

|y − X|

2−n

for x ∈ B

, y ∈ ∂B

. Hence we have for such x and y

G(x, y) = −

− y

|x − y|

−



|x|



n−2

− y

|X − y|

− |x|

|x − y|

by (1.5) in the proof of Proposition 1.22. We obtain with n

for |y| = R

∂G

∂n

(x, y) =

i=1

G(x, y) =

− |x|

|x − y|

Denote by K(x, y) the function in Corollary 1.23 for x ∈ Ω, y ∈ ∂Ω. It is called a

Poisson kernel and has the following properties:

(i) K(x, y) is smooth for x 6= y;

(ii) K(x, y) > 0 for |x| < R;

(iii)

|y |=R

K(x, y)dS

= 1 for any |x| < R.

The following result gives the existence of harmonic functions in balls with prescribed

Dirichlet boundary value.

Theorem 1.24 (Poisson Integral Formula). For φ ∈ C(∂B

(0)), the function u dened

u(x) =







∂B

(0)

K(x, y)φ(y)dS

, |x| < R,

φ(x), |x| = R,

satises u ∈ C(Ω) ∩ C

∞

(Ω) and







∆u = 0 in Ω,

u = φ on ∂Ω.

For the proof, see ref.

Section 1.3: Fundamental Solutions

Remark 1.25. In the Poisson integral formula, by letting x = 0, we have

u(0) =

−

∂B

(0)

φ(y)dS

which is the mean value property.

Lemma 1.26 (Harnack’s Inequality). Suppose u is harmonic in B

) and u ≥ 0.

Then there holds



R + r



n−2

R − r

R + r

u(x

) ≤ u(x) ≤



R − r



n−2

R + r

R − r

u(x

)

where r = |x − x

| < R.

Proof. We may assume x

= 0 and u ∈ C(

). Note that u is given by the Poisson

integral formula

u(x) =

∂B

− |x|

|y − x|

u(y)dS

Since R −|x| ≤ |y − x| ≤ R + |x| for |y| = R, we have

R − |x|

R + |x|



R + |x|



n−2

∂B

u(y)dS

≤ u(x) ≤

R + |x|

R − |x|



R − |x|



n−2

∂B

u(y)dS

The mean value property implies

u(0) =

n−1

∂B

u(y)dS

This nishes the proof.

Corollary 1.27. If harmonic function u in R

is bounded above or below, then u ≡

const.

Proof. We assume u ≥ 0 in R

. Take any point x ∈ R

and apply Lemma 1.26 to any

ball B

(0) with R > |x|. We obtain



R + |x|



n−2

R − |x|

R + |x|

u(0) ≤ u(x) ≤



R − |x|



n−2

R + |x|

R − |x|

u(0),

which implies u(x) = u(0) by letting R → +∞.

Next we prove a result concerning the removable singularity.

Chapter 1: Harmonic Functions

Theorem 1.28. Suppose u is harmonic in B

\ {0} and satises

u(x) =







o(log |x|), n = 2,

o(|x|

2−n

), n ≥ 3,

as |x| → 0.

Then u can be dened at 0 so that it is C

and harmonic in B

Proof. Assume u is continuous in 0 < |x| ≤ R. Let v solve







∆v = 0 in B

v = u on ∂B

We will prove u = v in B

\{0}. Set w = v −u in B

\{0} and M

= max

∂B

|w|. We

prove for n ≥ 3. It is obvious that

|w(x)| ≤ M

n−2

|x|

n−2

on ∂B

Note w and

|x|

n−2

are harmonic in B

\ B

. Hence the maximum principle implies

|w(x)| ≤ M

n−2

|x|

n−2

for any x ∈ B

\ B

where M

= max

∂B

|v − u| ≤ max

∂B

|v| + max

∂B

|u| ≤ M + max

∂B

|u| with M =

max

∂B

|u|. Hence we have for each xed x 6= 0

|w(x)| ≤

n−2

|x|

n−2

M +

|x|

n−2

max

∂B

|u| → 0 as r → 0,

that is w = 0 in B

\{0}.