Jost J. Partial Differential Equations

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1. The Laplace Equation as the Prototype of

an Elliptic Partial Diﬀerential Equation of

Second Order

1.1 Harmonic Functions. Representation Formula for

the Solution of the Dirichlet Problem on the Ball

(Existence Techniques 0)

In this section Ω is a bounded domain in R

for which the divergence theorem

holds; this means that for any vector ﬁeld V of class C

(Ω) ∩ C

(

Ω),



div V (x)dx =



∂Ω

V (z) · ν(z)do(z), (1.1.1)

where the dot · denotes the Euclidean product of vectors in R

, ν is the

exterior normal of ∂Ω,anddo(z) is the volume element of ∂Ω. Let us recall

the deﬁnition of the divergence of a vector ﬁeld V =(V

,...,V

):Ω → R

div V (x):=



i=1

∂V

∂x

(x).

In order that (1.1.1) hold, it is, for example, suﬃcient that ∂Ω be of class

Lemma 1.1.1: Let u, v ∈ C

(

Ω). Then we have Green’s 1

formula



v(x)Δu(x)dx +



∇u(x) ·∇v(x)dx =



∂Ω

v(z)

∂u

∂ν

(z)do(z) (1.1.2)

(here, ∇u is the gradient of u), and Green’s 2

formula



{v(x)Δu(x) − u(x)Δv(x)}dx =



∂Ω



v(z)

∂u

∂ν

(z) − u(z)

∂v

∂ν

(z)



do(z).

(1.1.3)

Proof: With V (x)=v(x)∇u(x), (1.1.2) follows from (1.1.1). Interchanging

u and v in (1.1.2) and subtracting the resulting formula from (1.1.2) yields

(1.1.3). 

8 1. The Laplace Equation.

In the sequel we shall employ the following notation:

B(x, r):=



y ∈ R

: |x −y|≤r



(closed ball)

and

B(x, r):=



y ∈ R

: |x −y| <r



(open ball)

for r>0, x ∈ R

Deﬁnition 1.1.1: A function u ∈ C

(Ω) is cal led harmonic (in Ω)if

Δu =0 in Ω.

In Deﬁnition 1.1.1, Ω may be an arbitrary open subset of R

.Webegin

with the following simple observation:

Lemma 1.1.2: The harmonic functions in Ω form a vector space.

Proof: This follows because Δ is a linear diﬀerential operator. 

Examples of harmonic functions:

(1) In R

, all constant functions and, more generally, all aﬃne linear func-

tions are harmonic.

(2) There also exist harmonic polynomials of higher order, e.g.,

u(x)=





−





for x =



,...,x



∈ R

(3) For x, y ∈ R

with x = y, we put

Γ (x, y):=Γ (|x − y|):=



2π

log |x − y| for d =2,

d(2−d)ω

|x − y|

2−d

for d>2,

(1.1.4)

where ω

is the volume of the d-dimensional unit ball B(0, 1) ⊂ R

We have

∂

∂x

Γ (x, y)=

dω



− y



|x − y|

−d

∂

∂x

Γ (x, y)=

dω



|x − y|

− d



− y



− y





|x − y|

−d−2

Thus, as a function of x, Γ is harmonic in R

\{y}.SinceΓ is symmetric

in x and y, it is then also harmonic as a function of y in R

\{x}.The

reason for the choice of the constants employed in (1.1.4) will become

apparent after (1.1.8) below.

1.1 Existence Techniques 0 9

Deﬁnition 1.1.2: Γ from (1.1.4) is called the fundamental solution of the

Laplace equation.

What is the reason for this particular solution Γ of the Laplace equation

in R

\{y}? The answer comes from the rotational symmetry of the Laplace

operator. The equation

Δu =0

is invariant under rotations about an arbitrary center y. (If A ∈ O(d) (or-

thogonal group) and y ∈ R

, then for a harmonic u(x), u(A(x − y)+y)is

likewise harmonic.) Because of this invariance of the operator, one then also

searches for invariant solutions, i.e., solutions of the form

u(x)=ϕ(r)withr = |x − y|.

The Laplace equation then is transformed into the following equation for y

as a function of r,with



denoting a derivative with respect to r,



(r)+

d − 1



(r)=0.

Solutions have to satisfy



(r)=cr

1−d

with constant c. Fixing this constant plus one further additive constant leads

to the fundamental solution Γ (r).

Theorem 1.1.1 (Green representation formula): If u ∈ C

(

Ω),we

have for y ∈ Ω,

u(y)=



∂Ω



u(x)

∂Γ

∂ν

(x, y) −Γ(x, y)

∂u

∂ν

(x)



do(x)+



Γ (x, y)Δu(x)dx

(1.1.5)

(here, the symbol

∂

∂ν

indicates that the derivative is to be taken in the direc-

tion of the exterior normal with respect to the variable x).

Proof: For suﬃciently small ε>0,

B(y, ε) ⊂ Ω,

since Ω is open. We apply (1.1.3) for v(x)=Γ (x, y)andΩ \B(y,ε) (in place

of Ω). Since Γ is harmonic in Ω \{y}, we obtain



Ω\B(y,ε)

Γ (x, y)Δu(x)dx =



∂Ω



Γ (x, y)

∂u

∂ν

(x) − u(x)

∂Γ(x, y)

∂ν



do(x)



∂B(y,ε)



Γ (x, y)

∂u

∂ν

(x) − u(x)

∂Γ(x, y)

∂ν



do(x). (1.1.6)

10 1. The Laplace Equation.

In the second boundary integral, ν denotes the exterior normal of Ω \B(y, ε),

hence the interior normal of B(y,ε).

We now wish to evaluate the limits of the individual integrals in this

formula for ε → 0. Since u ∈ C

(

Ω), Δu is bounded. Since Γ is integrable,

the left-hand side of (1.1.6) thus tends to



Γ (x, y)Δu(x)dx.

On ∂B(y,ε), we have Γ (x, y)=Γ (ε). Thus, for ε → 0,





∂B(y,ε)

Γ (x, y)

∂u

∂ν

(x)do(x)



≤ dω

d−1

Γ (ε)sup

B(y,ε)

|∇u|→0.

Furthermore,

−



∂B(y,ε)

u(x)

∂Γ(x, y)

∂ν

do(x)=

∂

∂ε

Γ (ε)



∂B(y,ε)

u(x)do(x)

(since ν is the interior normal of B(y,ε))

dω

d−1



∂B(y,ε)

u(x)do(x) → u(y).

Altogether, we get (1.1.5). 

Remark: Applying the Green representation formula for a so-called test func-

tion ϕ ∈ C

∞

(Ω),

we obtain

ϕ(y)=



Γ (x, y)Δϕ(x)dx. (1.1.7)

This can be written symbolically as

Γ (x, y)=δ

, (1.1.8)

where Δ

is the Laplace operator with respect to x,andδ

is the Dirac delta

distribution, meaning that for ϕ ∈ C

∞

(Ω),

[ϕ]:=ϕ(y).

In the same manner, ΔΓ ( · ,y) is deﬁned as a distribution, i.e.,

ΔΓ ( · ,y)[ϕ]:=



Γ (x, y)Δϕ(x)dx.

Equation (1.1.8) explains the terminology “fundamental solution” for Γ ,as

well as the choice of constant in its deﬁnition.

∞

(Ω):={f ∈ C

∞

(Ω), supp(f ):={x : f (x) =0} is a compact subset of Ω}.

1.1 Existence Techniques 0 11

Remark: By deﬁnition, a distribution is a linear functional  on C

∞

that is

continuous in the following sense:

Suppose that (ϕ

)

n∈N

⊂ C

∞

(Ω) satisﬁes ϕ

=0onΩ \K for all n and some

ﬁxed compact K ⊂ Ω as well as lim

n→∞

(x) = 0 uniformly in x for all

partial derivatives D

(of arbitrary order). Then

lim

n→∞

[ϕ

]=0

must hold.

We may draw the following consequence from the Green representation

formula: If one knows Δu,thenu is completely determined by its values and

those of its normal derivative on ∂Ω. In particular, a harmonic function on Ω

can be reconstructed from its boundary data. One may then ask conversely

whether one can construct a harmonic function for arbitrary given values on

∂Ω for the function and its normal derivative. Even ignoring the issue that

one might have to impose certain regularity conditions like continuity on

such data, we shall ﬁnd that this is not possible in general, but that one can

prescribe essentially only one of these two data. In any case, the divergence

theorem (1.1.1) for V (x)=∇u(x)impliesthatbecauseofΔ = div grad, a

harmonic u has to satisfy



∂Ω

∂u

∂ν

do(x)=



Δu(x)dx =0, (1.1.9)

so that the normal derivative cannot be prescribed completely arbitrarily.

Deﬁnition 1.1.3: A function G(x, y), deﬁned for x, y ∈

Ω, x = y, is called

a Green function for Ω if

(1) G(x, y)=0for x ∈ ∂Ω;

(2) h(x, y):=G(x, y) −Γ (x, y) is harmonic in x ∈ Ω (thus in particular also

at the point x = y).

We now assume that a Green function G(x, y)forΩ exists (which indeed

is true for all Ω under consideration here), and put v(x)=h(x, y) in (1.1.3)

and add the result to (1.1.5), obtaining

u(y)=



∂Ω

u(x)

∂G(x, y)

∂ν

do(x)+



G(x, y)Δu(x)dx. (1.1.10)

Equation (1.1.10) in particular implies that a harmonic u is already deter-

mined by its boundary values u

|∂Ω

This construction now raises the converse question: If we are given func-

tions ϕ : ∂Ω → R, f : Ω → R, can we obtain a solution of the Dirichlet

problem for the Poisson equation

Δu(x)=f(x)forx ∈ Ω,

u(x)=ϕ(x)forx ∈ ∂Ω,

(1.1.11)

12 1. The Laplace Equation.

by the representation formula

u(y)=



∂Ω

ϕ(x)

∂G(x, y)

∂ν

do(x)+



f(x)G(x, y)dx? (1.1.12)

After all, if u is a solution, it does satisfy this formula by (1.1.10).

Essentially, the answer is yes; to make it really work, however, we need

to impose some conditions on ϕ and f . A natural condition should be the

requirement that they be continuous. For ϕ, this condition turns out to be

suﬃcient, provided that the boundary of Ω satisﬁes some mild regularity

requirements. If Ω is a ball, we shall verify this in Theorem 1.1.2 for the case

f = 0, i.e., the Dirichlet problem for harmonic functions. For f, the situation

is slightly more subtle. It turns out that even if f is continuous, the function u

deﬁned by (1.1.12) need not be twice diﬀerentiable, and so one has to exercise

some care in assigning a meaning to the equation Δu = f. We shall return to

this issue in Sections 10.1 and 11.1 below. In particular, we shall show that

if we require a little more about f , namely, that it be H¨older continuous,

then the function u given by (1.1.12) is twice continuously diﬀerentiable and

satisﬁes

Δu = f.

Analogously, if H(x, y)forx, y ∈

Ω, x = y is deﬁned with

∂

∂ν

H(x, y)=

−1

∂Ω

for x ∈ ∂Ω

and a harmonic diﬀerence H(x, y) − Γ (x, y) as before, we obtain

u(y)=

∂Ω



∂Ω

u(x)do(x) −



∂Ω

H(x, y)

∂u

∂ν

(x)do(x)



H(x, y)Δu(x)dx. (1.1.13)

If now u

and u

are two harmonic functions with

∂u

∂ν

∂u

∂ν

on ∂Ω,

applying (1.1.13) to the diﬀerence u = u

− u

yields

(y) − u

(y)=

∂Ω



∂Ω

(x) − u

(x)) do(x). (1.1.14)

Since the right-hand side of (1.1.14) is independent of y, u

− u

must be

constant in Ω. In other words, a solution of the Neumann boundary value

problem

Here, ∂Ω denotes the measure of the boundary ∂Ω of Ω; it is given as



∂Ω

do(x).

1.1 Existence Techniques 0 13

Δu(x)=0 forx ∈ Ω,

∂u

∂ν

= g(x)forx ∈ ∂Ω

(1.1.15)

is determined only up to a constant, and, conversely, by (1.1.9), a necessary

condition for the existence of a solution is



∂Ω

g(x)do(x)=0. (1.1.16)

Boundary conditions tend to make the theory of PDEs diﬃcult. Actually, in

many contexts, the Neumann condition is more natural and easier to handle

than the Dirichlet condition, even though we mainly study Dirichlet boundary

conditions in this book as those occur more frequently. – There is in fact

another, even easier, boundary condition, which actually is not a boundary

condition at all, the so-called periodic boundary condition. This means the

following. We consider a domain of the form Ω =(0,L

) ×···×(0,L

) ⊂ R

and require for u :

Ω → R that

u(x

,...,x

i−1

i+1

,...,x

)=u(x

,...,x

i−1

, 0,x

i+1

,...,x

) (1.1.17)

for all x =(x

,...,x

) ∈ Ω, i =1,...,d. This means that u can be period-

ically extended from Ω to all of R

. A reader familiar with basic geometric

concepts will view such a u as a function on the torus obtained by identifying

opposite sides in Ω. More generally, one may then consider solutions of PDEs

on compact manifolds.

Anyway, we now turn to the Dirichlet problem on a ball. As a preparation,

we compute the Green function G for such a ball B(0,R). For y ∈ R

,we

put

¯y :=



|y|

y for y =0,

∞ for y =0.

(¯y is the point obtained from y by reﬂection across ∂B(0,R).) We then put

G(x, y):=



Γ (|x − y|) − Γ



|y|

|x − ¯y|



for y =0,

Γ (|x|) − Γ (R)fory =0.

(1.1.18)

For x = y, G(x, y)isharmonicinx,sincefory ∈

B(0,R), the point ¯y lies

in the exterior of B(0,R). The function G(x, y) has only one singularity in

B(0,R), namely at x = y, and this singularity is the same as that of Γ (x, y).

The formula

G(x, y)=Γ





|x|

+ |y|

− 2x · y





− Γ

⎛

⎜

⎝



|x|

|y|

+ R

− 2x · y



⎞

⎟

⎠

(1.1.19)

14 1. The Laplace Equation.

then shows that for x ∈ ∂B(0,R), i.e., |x| = R,wehaveindeed

G(x, y)=0.

Therefore, the function G(x, y) deﬁned by (1.1.18) is the Green function of

B(0,R).

Equation (1.1.19) also implies the symmetry

G(x, y)=G(y,x). (1.1.20)

Furthermore, since Γ (|x−y|) is monotonic in |x−y|, we conclude from (1.1.19)

that

G(x, y) ≤ 0forx, y ∈ B(0,R). (1.1.21)

Since for x ∈ ∂B(0,R),

|x|

+ |y|

− 2x · y =

|x|

|y|

+ R

− 2x · y,

(1.1.19) furthermore implies for x ∈ ∂B(0,R)that

∂

∂ν

G(x, y)=

∂

∂ |x|

G(x, y)=

dω

|x|

|x − y|

−

dω

|x|

|x − y|

|y|

−|y|

dω

|x − y|

Inserting this result into (1.1.10), we obtain a representation formula for a

harmonic u ∈ C

(B(0,R)) in terms of its boundary values on ∂B(0,R):

u(y)=

−|y|

dω



∂B(0,R)

u(x)

|x − y|

do(x). (1.1.22)

The regularity condition here can be weakened; in fact, we have the following

theorem:

Theorem 1.1.2:

(Poisson representation formula; solution of the

Dirichlet problem on the ball): Let ϕ : ∂B(0,R) → R be continuous.

Then u, deﬁned by

u(y):=



−|y|

dω



∂B(0,R)

ϕ(x)

|x−y|

do(x) for y ∈

B(0,R),

ϕ(y) for y ∈ ∂B(0,R),

(1.1.23)

is harmonic in the open ball

B(0,R) and continuous in the closed ball B(0,R).

1.1 Existence Techniques 0 15

Proof: Since G is harmonic in y, so is the kernel of the Poisson representation

formula

K(x, y):=

∂G

∂ν

(x, y)=

−|y|

dω

|x − y|

−d

Thus u is harmonic as well.

It remains only to show continuity of u on ∂B(0,R). We ﬁrst insert the

harmonic function u ≡ 1 in (1.1.22), yielding



∂B(0,R)

K(x, y)do(x) = 1 for all y ∈

B(0,R). (1.1.24)

We now consider y

∈ ∂B(0,R). Since y is continuous, for every ε>0there

exists δ>0with

|ϕ(y) − ϕ(y

)| <

for |y − y

| < 2δ. (1.1.25)

With

μ := sup

y∈∂B(0,R)

|ϕ(y)|,

by (1.1.23), (1.1.24) we have for |y − y

| <δthat



u(y) − u(y

)





∂B(0,R)

K(x, y)(ϕ(x) − ϕ(y

)) do(x)



≤



|x−y

|≤2δ

K(x, y) |ϕ(x) − ϕ(y

)|do(x)



|x−y

|>2δ

K(x, y) |ϕ(x) − ϕ(y

)|do(x)

≤

+2μ



−|y|



d−1

. (1.1.26)

(For estimating the second integral, note that because of |y −y

| <δ,for

|x − y

| > 2δ also |x − y|≥δ.) Since |y

| = R, for suﬃciently small |y − y

then also the second term on the right-hand side of (1.1.26) becomes smaller

than ε/2, and we see that u is continuous at y

. 

Corollary 1.1.1: For ϕ ∈ C

(∂B(0,R)), there exists a unique solution u ∈

(

B(0,R)) ∩C

(B(0,R)) of the Dirichlet problem

Δu(x)=0 for x ∈

B(0,R),

u(x)=ϕ(x) for x ∈ ∂B(0,R).

16 1. The Laplace Equation

Proof: Theorem 1.1.2 shows the existence. Uniqueness follows from (1.1.10);

however, in (1.1.10) we have assumed u ∈ C

(B(0,R)), while more generally,

here we consider continuous boundary values. This diﬃculty is easily over-

come: Since u is harmonic in

B(0,R), it is of class C

B(0,R), for example

by Corollary 1.1.2 below. Consequently, for |y| <r<R, applying (1.1.22)

with r in place of R,weget

u(y)=

−|y|

dω



∂B(0,r)

u(x)

|x − y|

do(x),

and since u is continuous in B(0,R), we may let r tend to R in order to get

the representation formula in its full generality. 

Corollary 1.1.2: Any harmonic function u : Ω → R is real analytic in Ω.

Proof: Let z ∈ Ω and choose R such that B(z, R) ⊂ Ω. Then by (1.1.22), for

y ∈

B(z, R),

u(y)=

−|y − z|

dω



∂B(z,R)

u(x)

|x − y|

do(x),

which is a real analytic function of y ∈

B(z, R). 

1.2 Mean Value Properties of Harmonic Functions.

Subharmonic Functions. The Maximum Principle

Theorem 1.2.1 (Mean value formulae): A continuous u : Ω → R is

harmonic if and only if for any ball B(x

,r) ⊂ Ω,

u(x

)=S(u, x

,r):=

dω

d−1



∂B(x

,r)

u(x)do(x) (spherical mean),

(1.2.1)

or equivalently, if for any such ball

u(x

)=K(u, x

,r):=



B(x

,r)

u(x)dx (ball mean). (1.2.2)

Proof: “⇒”:

Let u be harmonic. Then (1.2.1) follows from Poisson’s formula (1.1.22) (since

we have written (1.1.22) only for the ball B(0,R), take the harmonic function

v(x):=u(x + x

) and apply the formula at the point x = 0). Alternatively,