Math 309C (Badger, Autumn 2010) Harmonic FunctionsSupplemental Notes — November 29, 2010

Abstract: We look at examples of harmonic functions and two of their important properties:the mean value property and the maximum principle.

Contents

1 Introduction 11.1 A Definition and Some Examples . . . . . . . . . . . . . . . . . . . . . . . 11.2 Laplace’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Harmonic Functions as Steady-State Solutions . . . . . . . . . . . . . . . . 21.4 Properties of Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . 3

2 Dirichlet Problem 52.1 Dirichlet Problem on a Disk . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Dirichlet Problem on a Ball . . . . . . . . . . . . . . . . . . . . . . . . . . 102.3 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 Mean Value Property 12

4 Maximum Principle 13

5 Exercises 15

1 Introduction

1.1 A Definition and Some Examples

Let Rn denote n-dimensional Euclidean space, so that R1 is the line, R2 is the plane, etc.In these notes we will typically be interested in the case n ≥ 2. Let Ω ⊂ Rn be a domain1.A function h : Ω → R is harmonic if (1) h has continuous second partial derivatives and(2) the sum of the pure second partial derivatives vanishes, i.e. hx1x1 + · · ·+ hxnxn ≡ 0.

Example 1. Let n = 1 and Ω = (a, b). If h : (a, b) → R is harmonic, then h′′(x) = 0 forall x ∈ (a, b). Integrating twice we find out that

h(x) = h(a) +h(b)− h(a)

b− a(x− a).

The only harmonic functions h in a 1-dimensional interval (a, b) are linear functions.

Example 2. Let n = 2 and Ω = R2 be the whole plane. Define a function h : R2 → R by

h(x, y) = x2 − y2.

Then h is harmonic, because hxx + hyy = 2− 2 = 0. In two or more dimensions there area lot of harmonic functions (not just the linear functions).

Example 3. Let n = 3 and Ω = R3. The polynomial

P (x, y, z) = 8x5 − 40x3y2 + 15xy4 − 40x3z2 + 30xy2z2 + 15xz4

is harmonic in R3. To verify this, we just need to compute:

Pxx = 160x3 − 240xy2 − 240xz2

Pyy = −80x3 + 180xy2 + 60xz2

Pzz = −80x3 + 60xy2 + 180xz2.

Thus Pxx + Pyy + Pzz = 0 and P is harmonic.

Example 4. Let n = 2 and Ω = H := (x1, x2) ∈ R2 : x2 > 0 be the “upper half plane”.It can be shown that the argument function θ(x1, x2) = tan−1(x2/x1) with values between0 and π is harmonic in the upper half plane. (However, it is not harmonic in the plane,because it cannot be defined continuously at the origin.)

1A domain is a connected open set. Open means that around every point X ∈ Ω we can find a ballB(X, r) = Y ∈ Rn : |X − Y | < r with center X and radius r such that B(X, r) is contained in Ω.Connected means any pair of points X and Y in Ω can be joined by a continuous path that lies inside Ω.

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Example 5. Let n = 2 and Ω = (x, y) ∈ R2 : x2 + y2 < 1 and (x, y) 6= (0, 0) be the“punctured unit disk”. The Green function G(x, y) := − log

√x2 + y2 is harmonic in Ω.

(We can’t define G to be continuous and real-valued at (x, y) = (0, 0) since log 0 = ∞.)

Example 6. Let n = 3 and Ω = (x, y, z) ∈ R3 : x2 +y2 +z2 < 1 and (x, y, z) 6= (0, 0, 0)be the “punctured unit ball”. The 3D Green function

G(x, y, z) :=1√

x2 + y2 + z2.

is harmonic in Ω. (Like in the last example, G(x, y, z) cannot be made continuous andreal-valued at the origin.)

1.2 Laplace’s Equation

The partial differential equation ux1x1 + · · · + uxnxn = 0 is called Laplace’s equation.Harmonic functions are the solutions of Laplace’s equation. There is another commonway to write Laplace’s equation. First define a “differential operator” ∆,

∆ =∂2

∂x21

+ · · ·+ ∂2

∂x2n

.

The operator ∆ is called the Laplacian and Laplace’s equation is ∆u = 0.

1.3 Harmonic Functions as Steady-State Solutions

One of the ways in which harmonic functions appear is as steady-state (time independent)solutions of the heat equation. Recall that the one-dimensional heat equation whichmodels diffusion of heat in a thin metal rod is the partial differential equation ut = α2uxx.Similarly the two-dimensional heat equation ut = α2(ux1x1+ux2x2) models diffusion of heatin a thin metal plate. Hopefully you can see the general pattern. The three-dimensionalheat equation is ut = α2(ux1x1 +ux2x2 +ux3x3). If we wish to continue, then we can definea four-dimensional “heat” equation ut = α2(ux1x1 + ux2x2 + ux3x3 + ux4x4) and so on . . . .Using the Laplacian, each of the heat equations above may be written more compactly as

ut = α2∆u.

The left hand side involves one derivative with respect to t. The right hand side involvesn pure second partial derivatives in the space variables x1, . . . , xn.

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A solution u(x1, . . . , xn, t) of the heat equation ut = ∆u usually depends on time t.However, in the long run, one expects that solutions of the heat equation will convergeas t → ∞ to a function which does not depend on t. A solution u(x1, . . . , xn) of theheat equation ut = ∆u which does not depend on time t is called a steady-state solution.Notice that if u does not change with time, then the left hand side of the heat equationmust be identically zero, i.e. 0 = ∆u. Thus a steady-state solution of the heat equationis a harmonic function.

1.4 Properties of Harmonic Functions

One of the purposes of these notes is to introduce the reader to 2 important properties ofharmonic functions. They are:

1. The Mean Value Property

2. The Maximum Principle

We will discuss these properties in more detail in §3 and §4 below. For now let’s illustratethese properties in the one-dimensional case. Recall from Example 1 that in an interval(a, b) every harmonic functions h is a linear function,

h(x) = h(a) +h(b)− h(a)

b− a(x− a) for all x ∈ (a, b).

The first property involves the average or mean value of a harmonic function.

Property 1 (Mean Value Property in 1D). Suppose that h : (a, b) → R is harmonic.Then for every subinterval (c, d) ⊂ (a, b), the mean value of h on (c, d) is exactly thevalue of h at the center of (c, d):

1

d− c

∫ d

c

h(x)dx = h

(c + d

2

).

Proof. For brevity let’s write the linear function h as h(x) = mx + b, where m and b areconstants. Let (c, d) ⊂ R be any interval. Then

∫ d

c

h(x)dx =

∫ d

c

(mx + b)dx

=1

2mx2 + bx

∣∣∣∣d

c

=1

2m(d2 − c2) + b(d− c)

=1

2m(d + c)(d− c) + b(d− c).

Dividing this equation through by (d− c) yields

1

d− c

∫ d

c

h(x)dx =m

2(d + c) + b = h

(d + c

2

)

as desired.

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The second property concerns the extreme values of a harmonic function. Very roughlyit says that on an interval (a, b) the only place where a nonconstant harmonic functioncan obtain its maximum value is at one of the end points a or b.

Property 2 (The Maximum Principle in 1D). Suppose that h : (a, b) → R is harmonic.Then either h is constant or h(x) does not obtain its maximum value (or minimum value)at any point x inside the interval (a, b).

Proof. Once again let’s write the linear function h as h(x) = mx + b, where m and b areconstants. There are two cases.

CASE I. Suppose that m = 0. Then h(x) = b is a constant function.

CASE II. Suppose that m 6= 0. Then h′(x) = m 6= 0. That is, the derivativeof h never vanishes. Hence the function h does not have any localmaxima (or local minima) in the interval (a, b). We conclude thatthe global maximum (or global minimum) of h on (a, b) cannotoccur inside the interval (a, b). (So it must occur at an endpoint.)

This completes the proof.

A function f(x) which fails the maximum principle is the following. It is not harmonic.

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2 Dirichlet Problem

The Dirichlet problem is a boundary value problem for Laplace’s equation. Here is it is.Suppose that I give you a continuous function f on the boundary ∂Ω of a domain Ω ⊂ Rn.

The Dirichlet problem is to extend f (which is only defined on the boundary ∂Ω of Ω) toa function u defined inside the domain Ω such that (i) ∆u = 0 in Ω (i.e. u is harmonic)and (ii) “u = f” on ∂Ω (precisely we want limX∈Ω→Q∈∂Ω u(X) = f(Q) for all Q ∈ ∂Ω).The central question of potential theory is:

On what domains Ω ⊂ Rn can we solve the Dirichlet problem?

2.1 Dirichlet Problem on a Disk

We can solve the Dirichlet problem on a disk in the plane, using separation of variables.Let D = (x, y) ∈ R2 : x2+y2 < 1 be the “unit disk”. It will actually be more convenientto work in polar coordinates (r, θ). In polar coordinates, the unit disk is given by

D = (r, θ) : 0 ≤ r < 1 and 0 ≤ θ < 2πNotice that every point in the disk has a unique representation (r, θ) as defined above,except for the origin which can be written as (0, θ) for any choice of angle 0 ≤ θ < 2π.

Suppose that u(r, θ) = R(r)Θ(θ) is a harmonic function in D. We want to find ODEsthat R and Θ satisfy. Since u is harmonic, ∆u = uxx +uyy = 0. But this form of Laplace’sequation is not helpful, since it consists derivatives with respect to x and y, while we wantderivatives with respect to r and θ. Hence we must change of variables.

Recall that x = r cos θ, y = r sin θ, r =√

x2 + y2 and θ = tan−1( yx). Then

∂xr = xr

= cos θ, ∂yr = yr

= sin θ, ∂xθ = − yr2 = −1

rsin θ and ∂yθ = x

r2 =1rcos θ. By the chain rule, it follows that

ux = ur cos θ − uθsin θ

r,

uxx =

(urr cos θ − urθ

sin θ

r

)cos θ + ur

sin2 θ

r

−(

uθr cos θ − uθθsin θ

r

)sin θ

r+ uθ

cos θ sin θ

r2+ uθ

cos θ sin θ

r2,

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and

uy = ur sin θ + uθcos θ

r,

uyy =

(urr sin θ + urθ

cos θ

r

)sin θ + ur

cos2 θ

r

+

(uθr sin θ + uθθ

cos θ

r

)cos θ

r− uθ

cos θ sin θ

r2− uθ

cos θ sin θ

r2,

Adding the formulas for uxx and uyy together yields

uxx +uyy = urr(cos2 θ+sin2 θ)+ur

(sin2 θ

r+

cos2θ

r

)+uθθ

(sin2 θ

r2+

cos2 θ

r2

).

Thus the Laplacian in polar coordinates is ∆u = urr + r−1ur + r−2uθθ.

If u(r, θ) = R(r)Θ(θ) is harmonic, then Laplace’s equation in polar coordinates implies

R′′Θ + r−1R′Θ + r−2RΘ′′ = 0.

Equivalently,

r2R′′

R+ r

R′

R= −Θ′′

Θ.

Since the left hand side only depends on the variable r and the right hand side onlydepends on the variable θ, both sides must be equal to a constant, say λ:

r2R′′

R+ r

R′

R= −Θ′′

Θ= λ.

Therefore, the functions R(r) and Θ(θ) satisfy the ODEs:

r2R′′ + rR′ − λR = 0 and Θ′′ + λΘ = 0.

Now we want to solve this pair of differential equations. To start, let’s focus on thesimpler of the two equations, namely Θ′′ + λΘ = 0. If u is harmonic in D, then u mustbe continuous in the disk. Hence

limθ↑2π

R(1/2)Θ(θ) = limθ↑2π

u(1/2, θ) = u(1/2, 0) = R(1/2)Θ(0).

(See picture.)

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Dividing through by R(1/2) yields

limθ↑2π

Θ(θ) = Θ(0).

That is, Θ(θ) must be periodic of period 2π. Thus we only need to find the solutions Θ(θ)of the differential equation Θ′′ + λΘ = 0 which have period 2π. There are three cases.

CASE I. Suppose that λ = −µ2 < 0. Then we know Θ(θ) = c1eµθ + c2e

−µθ.This function can be periodic only if c1 = c2 = 0. In this caseΘ ≡ 0 and hence u ≡ 0. This case is not interesting.

CASE II. Suppose that λ = 0. Then Θ(θ) = c1 + c2θ, which is periodiconly if c2 = 0. Hence Θ(θ) = c1 is constant. When λ = 0,the differential equation for R becomes r2R′′ + rR′ = 0. ThusR(r) = k1 +k2 log r is the general solution of r2R′′+ rR′ = 0. Butsince we want u to be real-valued (finite) at the origin, we cannotallow R(r) to have a log r term, i.e. we need k2 = 0. Therefore,u(r, θ) = R(r)Θ(θ) = k1c1 is a constant function when λ = 0.

CASE III. Suppose that λ = µ2 > 0. Then Θ(θ) = c1 cos µθ + c2 sin µθ andR(r) = k1r

µ + k2r−µ. In order for Θ to be periodic of period

2π, we need µ = n to be a positive integer. In order for u tobe real-valued at the origin, we cannot allow R(r) to have a r−µ

term, i.e. we must take k2 = 0. Therefore, u(r, θ) = R(r)Θ(θ) =k1r

n(c1 cos nθ + c2 sin nθ) for some positive integer n, when λ > 0.

Therefore all harmonic functions in the disk of the form u(r, θ) = R(r)Θ(θ) are eitherconstant or u(r, θ) = rn(c1 cos nθ + c2 sin nθ) for some positive integer n. These are thefundamental solutions of Laplace’s equation in the unit disk.

Now let’s return to solving the Dirichlet problem on the unit disk. Let f(θ) be acontinuous function defined on ∂D = (1, θ) : 0 ≤ θ < 2π. We want to solve the BVP

∆u = 0 in D,u = f on ∂D

Following our solutions of the 1D heat equation and 1D wave equation, we can try to findu which is a (infinite) linear combination of fundamental solutions of Laplace’s equation.Let

u(r, θ) =a0

2+

∞∑n=1

rn(an cos nθ + bn sin nθ).

Then u is harmonic in D. Then u(r, θ) is a harmonic extension of f(θ) to D provided that

u(1, θ) =a0

2+

∞∑n=1

an cos nθ + bn sin θ = f(θ).

Thus we can solve the Dirichlet problem in the unit disk with boundary data f providedthat f(θ) has a Fourier series expansion with L = π. To summarize, we have outlined theproof of the following theorem.

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Theorem 1 (Fourier Series Solution of Dirichlet Problem on D). Let f(θ) be a continuousfunction on ∂D = (1, θ) : 0 ≤ θ < 2π. If f has a uniformly convergent Fourier series(e.g. if f ′(θ) is piecewise continuous), then

u(r, θ) =a0

2+

∞∑n=1

rn(an cos nθ + bn sin nθ) 0 ≤ r < 1, 0 ≤ θ < 2π

is a harmonic extension of f to the unit disk D. The coefficients an and bn are given by

an =1

π

∫ 2π

0

f(t) cos(nt)dt and bn =1

π

∫ 2π

0

f(t) sin(nt)dt.

It is now possible to say that we have solved the Dirichlet problem on the unit disk.But if we are willing to work a little bit more, a nicer solution is waiting to be found.This extra work will involve complex numbers in the intermediate steps, but nothing moreonerous than Euler’s formula einθ = cos nθ + i sin nθ. Let’s proceed.

For each integer n (positive, zero or negative), define a new coefficient cn by

cn :=1

2π

∫ 2π

0

f(t)e−intdt =1

2π

∫ 2π

0

f(t)(cos(nt)− i sin(nt))dt.

Then we can relate the coefficients cn to the coefficients an and bn by

a0 = 2c0, an = cn + c−n and bn = i(cn − c−n) for all n ≥ 1.

Now we can rewrite the Fourier series for f in terms of cn,

f(θ) =a0

2+

∞∑n=1

an cos(nθ) + bn sin(nθ)

= c0 +∞∑

n=1

(cn + c−n) cos(nθ) + i(cn − c−n) sin(nθ)

= c0 +∞∑

n=1

cn(cos(nθ) + i sin(nθ)) + c−n(cos(nθ)− i sin(nθ))

= c0 +∞∑

n=1

cneinθ + c−ne−inθ

= c0 +∞∑

n=1

cneinθ +

−1∑n=−∞

cneinθ

=∞∑

n=−∞cneinθ

In the last line the infinite sum ranges over all integers n (positive, zero and negative).Similarly

u(r, θ) =∞∑

n=−∞cnr|n|einθ where cn =

1

2π

∫ 2π

0

f(t)e−intdt. (?)

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Next we define and examine the function

P (r, θ) =∞∑

n=−∞r|n|einθ.

First observe that we can rewrite P (r, θ) as two geometric series:

P (r, θ) =∞∑

n=0

(reiθ)n +∞∑

n=1

(re−iθ)n.

Since |reiθ| = r < 1 and |re−iθ| = r < 1 these series converge and

P (r, θ) =1

1− reiθ+

1

1− re−iθ− 1.

Adding together the three terms and simplifying yields

P (r, θ) =(1− re−θ) + (1− reiθ)− (1− re−θ)(1− reiθ)

(1− reiθ)(1− re−iθ)=

1− r2

1− r(eiθ + e−iθ) + r2.

But eiθ + e−iθ = 2 cos θ so the formula for P (r, θ) simplifies further to

P (r, θ) =1− r2

1− 2r cos θ + r2.

By the law of cosines 1− 2r cos θ + r2 = ‖(1, 0)− (r, θ)‖2. (See picture).

Therefore

P (r, θ) =1− r2

‖(1, 0)− (r, θ)‖2. (??)

where ‖(r, θ)− (1, 0) · ‖ denotes the distance from the point (r, θ) to the point (1, 0).The next step is to combine (?) and (??). First we write u(r, θ) and apply the definition

of the coefficient cn to obtain

u(r, θ) =∞∑

n=−∞cnr

|n|einθ =∞∑

n=−∞

(1

2π

∫ 2π

0

f(t)e−intdt

)r|n|einθ.

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Bringing the infinite sum inside the integral (this step can be made rigorous) yields

u(r, θ) =1

2π

∫ 2π

0

f(t)∞∑

n=−∞r|n|ein(θ−t)dt =

1

2π

∫ 2π

0

P (r, θ − t)f(t)dt.

This gets rid of the infinite sum! Applying the formula (??) we find

u(r, θ) =1

2π

∫ 2π

0

1− r2

‖(1, 0)− (r, θ − t)‖2f(t)dt.

But ‖(1, 0)− (r, θ− t)‖ = ‖(1, t)− (r, θ)‖ (just rotate the picture by t radians). Therefore,a solution of the Dirichlet problem on the unit disk is given by:

Theorem 2 (Poisson Integral Formula on D). Let f(θ) be a continuous function on ∂D.

Then

u(r, θ) =1

2π

∫ 2π

0

1− r2

‖(1, t)− (r, θ)‖2f(t)dt 0 ≤ r < 1, 0 ≤ θ < 2π

is a harmonic extension of f to the unit disk D.

One advantage of the Poisson integral formula (Theorem 2) over the Fourier seriessolution (Theorem 1) is that the former does not involve an infinite sum. A secondadvantage is that the Poisson integral formula generalizes to higher dimensions.

2.2 Dirichlet Problem on a Ball

In this section we record (but do not prove) the following theorem, which is an analogueof the Poisson integral formula in dimension 3. Let B3 = (x, y, z) ∈ R3 : x2+y2+z2 < 1denote the unit ball in three dimensions. Also let S2 = (x, y, z) ∈ R3 : x2 + y2 + z2 = 1denote the unit sphere (i.e. the boundary of the unit ball).

Theorem 3 (Poisson Integral Formula on B3). Let f(Q) be a continuous function on S2.Then

u(X) =1

4π

∫∫

∂S2

1− r2

‖Q−X‖3f(Q)dS for all X ∈ B3

is a harmonic extension of f to the unit ball B3.

Here the double integral represents a surface integral over the unit sphere S2. Recall thatS2 has surface area 4π.

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2.3 Remarks

Remark 1. Theorem 1 states that every continuous function f on the boundary ∂D ofthe unit disk which has a Fourier series expansion can be extended to a harmonic functionu on the unit disk D. But not every continuous function has a Fourier series expansion,so we only solved the Dirichlet problem on D for functions with Fourier series expansions.However, Theorem 2 actually holds for every continuous function f on the boundary ∂Dof the unit disk, even if f does not have a Fourier series expansion. Therefore the Dirichletproblem on the unit disk D always has a solution. (A complete proof of Theorem 2 isusually given in a graduate level complex analysis course.)

Remark 2. If the Dirichlet problem on a domain for Ω ⊂ Rn has a solution for anycontinuous boundary data f , we say the domain Ω is regular for the Dirichlet problem.The unit disk D is regular for the Dirichlet problem.

Remark 3. In Theorem 1 and Theorem 2, we only proved that there exists a harmonicextension u of f . We have not yet shown that the function u we obtained is unique.However, this will be a consequence of the maximum principle in §4 below.

Remark 4. We can get a Poisson integral formula on any disk, by doing a change ofvariables (a dilation and translation). Let X0 = (x0, y0) ∈ R2 and write

D = X ∈ R2 : ‖X −X0‖ < R

for the disk with center X and radius R. Also let ∂D = X ∈ R2 : ‖X − X0‖ = Rdenote the boundary of D. If f(Q) is a continuous function on ∂D, then

u(X) =1

2π

∫

∂D

R2 − ‖X −X0‖2

‖Q−X‖2f(Q) dσ

is a harmonic extension of f to D.

Remark 5. A version of the Poisson integral formula for a ball exists in all dimensionsn ≥ 2. We stated the Poisson integral formula for n = 2 and n = 3.

Remark 6. The Dirichlet problem can always be solved on domains with “smooth”boundaries, like the disk or ball. In some cases, the Dirichlet problem can also be solvedon domains with “fractal” boundaries.

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3 Mean Value Property

One of the ways that harmonic functions in Rn generalize linear functions in R is throughthe mean value property. When n = 2, the mean value property says that the averagevalue of a harmonic function over a disk is equal to its value at the center of the disk.(The reader should compare this with the statement for n = 1 in §1.4 above.)

Theorem 4 (2D Mean Value Property). Let Ω ⊂ R2 be a domain and let h : Ω → Rbe a harmonic function. For every disk D = D(X, r) with center X and radius r > 0contained in Ω, the mean value of h over D is exactly the value of h at the center of D:

1

πr2

∫∫

D(X,r)

h(Y )dA = h(X).

For n ≥ 3 the mean value property for harmonic functions is similar, except with disksD(X, r) replaced by balls B(X, r) = Y ∈ Rn : ‖X − Y ‖ < r.

[NOTE: I don’t think we’ll have time to do the proof of the mean value theorem in class.The proof in two dimensions uses Green’s theorem relating area integrals to line integralsover the boundary, which is taught in Math 324.]

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4 Maximum Principle

Another property that harmonic functions satisfy is called the strong maximum principle.It is an important tool, which lets us prove that a solution to the Dirichlet problem mustbe unique.

Theorem 5 (Maximum Principle). Let Ω ⊂ Rn be a bounded domain and let h : Ω → Rbe a harmonic function. Then either

• h is a constant in Ω; or,

• h does not obtain its maximum value (or minimum value) at any point X in Ω.

Proof Sketch. We will only discuss the case n = 2, but the other dimensions are similar.The maximum principle follows from the mean value property for harmonic functions.Here is the main idea. Suppose that X0 ∈ Ω is a point where the function h obtains itsmaximum value in Ω. That means,

h(X0) ≥ h(X) for all X ∈ Ω.

Choose r0 > 0 such that the disk D(X0, r0) centered at X0 of radius r0 is contained in Ω.Then

h(X0) =1

πr20

∫∫

D(X0,r0)

h(X0)dA

by the mean value property. Next we shall rewrite the left hand side by noticing thath(X0) = (πr2

0)−1

∫∫D(X0,r0)

h(X0)dA to obtain

1

πr20

∫∫

D(X0,r0)

h(X0)dA =1

πr20

∫∫

D(X0,r0)

h(X)dA.

Multiply both sides by πr2 and then bring all terms to the left hand side to get∫∫

D(X0,r0)

(h(X0)− h(X))dA = 0.

But the integrand h(X0)−h(X) ≥ 0, since h achieves its maximum value at the point X0.The only way that the integral of a nonnegative continuous function (the left hand side)can equal zero (the right hand side) is if the integrand is zero. Thus, h(X) − h(X0) = 0for all X ∈ D(X0, r). Equivalently, h(X) = h(X0) for all X ∈ D(X0, r) and the functionh is constant on the disk D(X0, r0).

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Next choose a point X1 inside D(X0, r0) which is also near the boundary of the disk.Then we can find r1 > 0 such that the disk D(X1, r1) with center at X1 and radius r1 > 0is contained inside Ω and so that D(X1, r1) also contains points of Ω outside D(X0, r0).(See picture.)

Since X1 ∈ D(X0, r0) we know h(X1) = h(X0) and h obtains its maximum value at X1.Now the same argument used above shows that h(X) = h(X1) for all X ∈ D(X1, r1).Thus, h achieves its maximum value and is constant on the union of the two disksD(X0, r0) and D(X1, r1) (the area shaded green in the picture above). Now pick a pointX2 in either D(X0, r1) or D(X1, r1) which is near the boundary of the union of the disksand continue the argument... Each time we pick a new point, we can find a small diskaround that point where h is constant and obtains its maximum value. The area insideΩ where we know h achieves its maximum grows at each step and will eventually fill outthe entire domain Ω. (This can be made precise using a type of math called topology.)Here is what we proved. If there exists even a single point X0 ∈ Ω such that h obtains itsmaximum value at X0, then h(X) = h(X0) for all X0 ∈ Ω and h is a constant function.Therefore, if h is a nonconstant harmonic function in a domain Ω, then h cannot obtainits maximum value inside Ω.

The Dirichlet problem on a bounded domain is always unique.

Theorem 6 (Uniqueness). Let Ω ⊂ Rn be a bounded domain, and let f be continuous onthe boundary ∂Ω of Ω. If the Dirichlet problem

∆u = 0 in Ω and u = f on ∂Ω

has a solution u, then u is the unique solution.

Proof. Let Ω ⊂ Rn be a bounded domain, let f be continuous on the boundary ∂Ω, andsuppose that u1 and u2 are two solutions to the Dirichlet problem with boundary data f .Then the function v = u1−u2 is harmonic inside Ω, since ∆v = ∆u1−∆u2 = 0−0 = 0 inΩ, and v = 0 on ∂Ω, since v = u1 − u2 = f − f = 0 on ∂Ω. Since v is harmonic in Ω, themaximum value of v in Ω must occur at the boundary of Ω by the maximum principle.But 0 is the only value of v on the boundary, so the maximum value of v in Ω must be 0.Similarly, the minimum value of v in Ω is 0 (apply the maximum principle to −v). Thus,since the maximum and the minimum value of v in Ω are both zero, v must be identicallyzero in Ω. Recalling that v = u1−u2 we conclude that u1 = u2 in Ω. Therefore, solutionsto the Dirichlet problem with continuous boundary data are unique.

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5 Exercises

Exercise 1. Prove that the polynomial x3 − 3xy2 is harmonic in R2.

Exercise 2. Find a harmonic polynomial in R2 of degree 4.

Exercise 3. Find a harmonic polynomial in R2 of degree 5.

Exercise 4. Show that the collection H(Rn) of harmonic functions in Rn is a vector spaceover R. (You need to check closure under addition and closure under scalar multiplication.)

Exercise 5. Show that if h(x, y) is a harmonic function in R2 with continuous partialderivatives of all orders, then the partial derivatives hx and hy of h are also harmonic.(Hint: To show hx is harmonic, compute ∆hx. Proceed similarly for hy.)

Exercise 6. Show the polynomial x2(y − z) + y2(z − x) + z2(x− y) is harmonic in R3.

Exercise 7. Show that the polynomial x21 + x2

2 − x23 − x2

4 is harmonic in R4.

Exercise 8. Show that the “4D Green function”

G(x, y, z, w) =1

x2 + y2 + z2 + w2

is harmonic in R4\(0, 0, 0, 0) (i.e. verify ∆G(x, y, z, w) = 0 when (x, y, z, w) 6= (0, 0, 0, 0)).

Exercise 9. Let f(x, y) and g(x, y) be twice differentiable functions in R2 Prove that

∆(fg) = ∆(f) + 2∇f · ∇g + ∆(g)

where ∆f is the Laplacian of f and ∇f = (fx, fy) is its gradient.

Exercise 10. Suppose f(x, y) and g(x, y) are harmonic functions in R2. Prove that theirproduct fg is harmonic in R2 if and only if ∇f · ∇g = 0. (Use the result of Exercise 9.)

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