Elliptic Partial Differential Equationsweb.cs.elte.hu/~csirik/elliptic.pdfMaxwell immediately realized that by writing his of equations in free space (j D 0, ˆD 0),i.e. div D D 0

Elliptic Partial DifferentialEquations

Csirik Mihály

Budapest2014–2015

Contents

Contents v

1 Problems from classical field theories 11.1 Gravitostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Continuum mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Electrostatics and magnetostatics . . . . . . . . . . . . . . . . . . . . . . 11.4 Time-independent solutions to the wave equation . . . . . . . . . . . . . . 4

2 Analytical techniques I. 72.1 The fundamental solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Green’s representation formula . . . . . . . . . . . . . . . . . . . . . . . . 112.3 Poisson’s equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.4 Linear charge distributions . . . . . . . . . . . . . . . . . . . . . . . . . . 142.5 Basic properties of harmonic functions . . . . . . . . . . . . . . . . . . . . 162.6 The maximum principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.7 Uniqueness of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.8 Laplacian in a box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.9 Green’s functions. Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . 282.10 Green’s functions in one dimension . . . . . . . . . . . . . . . . . . . . . . 312.11 Poisson integral for the half-space . . . . . . . . . . . . . . . . . . . . . . 322.12 Inversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.13 Poisson integral for the ball . . . . . . . . . . . . . . . . . . . . . . . . . . 402.14 The Laplacian in polar coordinates . . . . . . . . . . . . . . . . . . . . . . 442.15 The use of complex analysis I. . . . . . . . . . . . . . . . . . . . . . . . . . 472.16 The single-layer potential . . . . . . . . . . . . . . . . . . . . . . . . . . . 532.17 The double-layer potential . . . . . . . . . . . . . . . . . . . . . . . . . . . 572.18 Boundary integral equations . . . . . . . . . . . . . . . . . . . . . . . . . . 602.19 Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 602.20 The principles of Dirichlet and Thomson . . . . . . . . . . . . . . . . . . . 63

3 Analytical techniques II. 653.1 The Helmholtz equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653.2 The Laplacian in curvilinear coordinates . . . . . . . . . . . . . . . . . . . 653.3 The Laplacian in spherical coordinates . . . . . . . . . . . . . . . . . . . . 683.4 Legendre functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 713.5 Spherical harmonics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 773.6 Green’s function of the second kind for the ball . . . . . . . . . . . . . . . 773.7 The Laplacian in cylindrical coordinates . . . . . . . . . . . . . . . . . . . 783.8 Bessel equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

v

3.9 Poisson integral for the exterior of the ball . . . . . . . . . . . . . . . . . 82

4 Analytical techniques III. 834.1 Boundary integral equations . . . . . . . . . . . . . . . . . . . . . . . . . . 83

5 Maximum principles 855.1 Hopf’s boundary point lemma . . . . . . . . . . . . . . . . . . . . . . . . . 85

6 Classical potential theory 876.1 Potential of a signed measure . . . . . . . . . . . . . . . . . . . . . . . . . 876.2 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 916.3 Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 946.4 Frostman’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 946.5 The Poisson integral revisited . . . . . . . . . . . . . . . . . . . . . . . . . 94

7 Hölder spaces and Schauder estimates 957.1 Hölder spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 957.2 Schauder estimates for the Newton potential . . . . . . . . . . . . . . . . 95

8 Potential estimates 998.1 Riesz Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 998.2 Riesz potential estimates on sets of finite measure . . . . . . . . . . . . . 103

9 Weak formulation in Sobolev spaces 107

10 Topological methods 109

11 Critical point theory 11111.1 Ekeland’s variational principle . . . . . . . . . . . . . . . . . . . . . . . . . 11111.2 Palais–Smale compactness criterion . . . . . . . . . . . . . . . . . . . . . 11411.3 Mountain pass theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11511.4 Linking methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

12 Boundary integral equations 117

13 Calderón–Zygmund theory 11913.1 The case p D 2p D 2p D 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11913.2 The general case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12013.3 The Hörmander cancellation condition . . . . . . . . . . . . . . . . . . . . 122

14 Capacity 123

15 Calculus of variations 12515.1 In Hölder spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12515.2 In Sobolev spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

vi

15.3 Quasiconvexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

16 de Giorgi–Moser–Nash theory 12716.1 Caccioppoli-type estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . 12716.2 Local boundedness – de Giorgi’s method . . . . . . . . . . . . . . . . . . . 129

A Tools from analysis 133A.1 Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133A.2 Critical points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135A.3 Ordinary differential equations . . . . . . . . . . . . . . . . . . . . . . . . 136A.4 Sturm–Liouville theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138A.5 Elementary measure theory . . . . . . . . . . . . . . . . . . . . . . . . . . 139A.6 Borel and Radon measures . . . . . . . . . . . . . . . . . . . . . . . . . . . 140A.7 Harmonic analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142A.8 Convergence of integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143A.9 Partitions of unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144A.10 Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144A.11 Measure theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145A.12 Exterior calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148A.13 Hilbert space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

B Solutions 151

Bibliography 169

vii

1 Problems from classical field theories

§1.1 Gravitostatics

§1.2 Continuum mechanics

§1.3 Electrostatics and magnetostaticsMaxwell’s equations consitute one of the greatest human achievements of the 19th cen-tury. Maxwell’s equations summarize the works of Faraday, Gauss, Weber, Ampère,Gibbs, Heaviside, Hertz and many others. Although this present book restricts itselfto stationary problems, i.e. equilibrium states, we shall still record the complete set ofMaxwell’s equations in both differential and integral forms. They read

div D D �

div B D 0

rot E D �@B

@t

rot H D j C@D

@t

9>>>>>>=>>>>>>;

I@˝

D � dA DZ

˝

� dVI@˝

B � dA D 0I@˙

E � ds D �@

@t

Z˙

B � dAI@˙

H � ds DZ

˙

j � dA C@

@t

Z˙

D � dA

9>>>>>>>>>>=>>>>>>>>>>;(1.3.1)

where

• E D E.t;x;y; z/ is called the electric field, B D B.t;x;y; z/ themagnetic induction,D D D.t;x;y; z/ the displacement field andH D H.t;x;y; z/ themagnetic field. Allthese fields are functions of the the type R4 ! R3, and the time variable t is in thefirst argument.

• � W R4 �! R is the (free) charge density and j W R4 �! R3 is the (free) currentdensity.

• If F W R3 �! R3 is some arbitrary vector field, then the divergence of F is definedas

div F D@Fx@x

C@Fy@y

C@Fz@z;

and the curl of F is defined as

rotF D

@Fz@y

�@Fy@z;@Fx@z

�@Fz@x;@Fy@x

�@Fx@y

!1

2 1. PROBLEMS FROM CLASSICAL FIELD THEORIES

• ˝ � R3 is a domain with closed boundary @˝. ˙ � R3 is a surface with closedboundary @˙ . The integral “O” notation signifies these facts. The volume elementis dV, the surface element is dA and the line element is dmathbfs. All surfaces and contours are assumed to be in time.

Let us elaborate on the connection between the fields E and D, B and H. Relationsbetween them are called material laws. For instance, for vacuum, there holds

D D "0E; H D1�0

B;

where "0 � 8:854�10�12 F=mis the permittivity of free space, and�0 � 4��10�7 Vs=Amis the permeability of free space. Homogeneous material laws may be expressed as

D D "E; H D1�B:

Next, we focus on stationary phenomena. Suppose that all the time derivativesvanish in the differential form of Maxwell’s equations in vacuum.

Electrostatics. First, if j D 0 we are talking about electrostatics,

div D D �

rot E D 0

)(1.3.2)

If the domain˝ � R3 onwhich the above equations are defined is simply connected (hasno holes, see Exercise 1), for instance on the whole space ˝ D R3, then there exists afunction ˚ W ˝ �! R, called the electrostatic potential, such that

E D D˚:

For then the second equation is automatically satisfied, and the first becomes Poisson’sequation

div�"D˚

�D �:

If " is constant throughout space, we have Poisson’s equation

4˚ D divD˚ D �

Transmission conditions. Let ˝� � R3 be an open and bounded set, a perfect con-ductor, called the interior domain, whose boundary @˝ admits a smooth unit normalvector field � W @˝ �! R3 and is locally flat. Let ˝C D R3 X˝� be the correspondingexterior domain. Let us introduce the notations D˙ W ˝˙ �! R3 and E˙ W ˝˙ �! R3.Using the integral form of the electrostatic equations (1.3.2), we shall deduce the so-called transmission conditionswhich describe how the tangential componentEt ofE andnormal component Dn of D changes as we pass through the surface @˝.

1.3. ELECTROSTATICS ANDMAGNETOSTATICS 3

We may arrange matters so that 0 2 @˝, and tangent plane of @˝ at 0 is the xy-plane and the normal �.0/ is parallel to the z-axis. Now let us put a small box B DŒ�L;L � Œ�L;L � Œ�h;h around 0. Hence, for D-field, we have Gauss’ lawI

@BD � dA D

ZB� dV:

As h ! 0, the integral on the left side tends to 4L2.DCn .0/ � D�n .0// and the right side

is the constant 4L2�.0/, where � W @˝ �! R denotes the surface charge distribution.In summary, on the boundary

DCn � D�n D �; (1.3.3)

in particular, if there are no charges on the surface, we get that DCn D D�n , i.e. the

normal component of the displacement field is continuous across the boundary.In contrast, the E-field is always continuous across @˝. In fact, by Faraday’s law of

induction applied to the rectange R D Œ�L;L � f0g � Œ�h;h,IRE � ds D 0;

hence .EC.0/�E�.0//��.0/ D 0 as h ! 0. In general, by letting E˙t D E˙

��, we have

ECt D E�t : (1.3.4)

Furthermore, since the potential ˚ may be expressed as the line integral (cf. Newton–Leibniz formula)

˚.b/ � ˚.a/ DZ

E � ds

for any path joining a and b, we obtain the potential is also continuous across theboundary,

˚� D ˚�:

As an illustration, let the uniform permittivity in ˝� be denoted by "� and in ˝C

by "C. As there are no charges outside the conductor ˝�, i.e. � � 0 on ˝C, we havethat

4˚� D�

"�.in ˝�/

4˚C D 0 .in ˝C/

9=;which is augmented by the transmission conditions

˚� D ˚C .in @˝/"�D�˚� D "�D�˚� .in @˝/

)Magnetostatics. Second, if @j=@t D 0, then the governing equations for magneto-

statics arediv B D 0rot H D 0

)

4 1. PROBLEMS FROM CLASSICAL FIELD THEORIES

Suppose that the material law is given by

B D �0.H C M/;

where the field M describes the a priori known magnetization of the material. Thesecond equation curl H D 0 is satisfied by fields of the form

H D D;

where W R3 �! R is called the magnetostatic potential. Then the first equation andthe material law implies

4 D � div M;

which is Poisson’s equation.

Exercises

Ex. 1 — Prove that the vector field

F.x;y/ D

�y

x2 C y2;

xx2 C y2

!

defined on R2 X f0g has zero curl, but no potential function exists for it.

Ex. 2 — Using Gauss’s law of electrostatics, prove that the radially symmetric elec-trostatic field E of a uniformly charged ball BR with total charge Q is given by

Er.r/ DQ

4�"0

(r=R3; if r < R1=r2; if r > R

Therefore the electric field outside of a uniformly charged ball is indistinguisable fromthat of the point charge placed at the center of the ball. Generalize this result to arbi-trary radially symmetric charge distributions �.r/.

§1.4 Time-independent solutions to the wave equationMaxwell immediately realized that by writing his of equations in free space (j D 0,� D 0), i.e.

div D D 0div B D 0

rot E D �@B

@t

rot B D1c2@E

@t

9>>>>>>=>>>>>>;

1.4. TIME-INDEPENDENT SOLUTIONS TO THEWAVE EQUATION 5

where c D 1=p"0�0, the systemmay be reduced to the electromagnetic wave equations

1c2@2E

@t2� 4E D 0;

1c2@2B

@t2� 4B D 0

9>>=>>;Here, the vector Laplacian is defined as

4F D�4Fx;4Fy;4Fz

�:

By time-independent we mean that the separation

E.t;x;y; z/ D �.t/F.x;y; z/

holds, and similarly for the B-field. Then we have

1c2�00F ��4F D 0;

then for each ˛ 2 fx;y; zg,1c2�00

�D

4F˛F˛

D �k2;

for some k, since the variables on the left and right sides are separated. The left sideresults in a well known differential equation, solved by

�.t/ D C1 sin.ckt/C C2 cos.ckt/;

and the right side is Helmholz’s equation, of the form

4u C k2u D 0: (1.4.1)

Helmholz’s equation is more difficult to handle than Poisson’s equations, but clearlydesires attention to understand electromagnetic waves. As this equation fits into ourobject of study, we will invesigate it thoroughly.

2 Analytical techniques I.

The methods used by Newton, Euler, Laplace, Poisson, Green, and their colleagues arenowadays called analytical solution techniques. Their primary goal was to produce anexplicit representation of the solution in terms of the boundary-, and/or initial values,and other input functions. The theoretical success of these analytical methods is highlylimited: the representation tends to depend on the domain on which the problem is pre-scribed. Engineering problems involve complicated domains with holes, sharp corners,slits, etc., and this makes the representing integral more complicated – if it exists at all.For a general irregular domain, it is not even known a priori that such a representationexists. Worse yet, the analytical tools for handling the often badly behaved integralswere not available at the time.

Note that these explicit solution techniques and the search for Green’s functions isa still active, and current topic. Simple formulas exists for domains possessing somesymmetry, such as the ball and the half-space, and these particular results proved tobe invaluable for the application of classical field theories: electrodynamics, continuummechanics and gravity.

§2.1 The fundamental solutionConsider the problem of finding a function u 2 C2.˝/, where ˝ � Rn is an open set(possibly the whole space), such that 4u D 0, i.e.

@2u@x21

C : : :C@2u@x2n

D 0:

Such a function u is called harmonic on ˝.The first thing to note is that the so-called fundamental solution (also called fun-

damental kernel, or Newton kernel for n � 3 and logarithmic kernel for n D 2) G 2C1.Rn X f0g/ which is defined as

G.x/ WD

8̂̂̂̂ˆ̂

8 2. ANALYTICAL TECHNIQUES I.

is related to the surface area via !n D �n=n. The formulas for these quantities aresometimes written in terms of the � -function,

�n D2�n=2

� .n=2/; !n D

�n=2

� .n=2 C 1/; (2.1.1)

see Exercise 13.The function G.x/ is sometimes called Newton potential for the case n � 3 and

logarithmic potential for n D 2. We won’t use this terminology since we reserve theword “potential” for other uses. We shall use the notation

8y 2 Rn X fxg W Gx.y/ D G.x � y/;

which specifies x 2 Rn as the singularity (or pole) of the fundamental solution . Notethat G satisfies G.�x/ D G.x/. Moreover, G is radial or rotationally symmetric in thesense that it only depends on kxk D �, hence we sometimes write G.�/ D G.x/, whichis of course an abuse of notation. This symmetry will turn out to be crucial, especiallyin spherically symmetric situations in which case we revert to the notation � D kxk.

To get acquinted with the fundamental solution, let us calculate the partial deriva-tives of G for the case n � 3. We have for x ¤ 0,

DkG.x/ D1�n

¡1�i�n

x2i

!� n2xk; Dk`G.x/ D

1�n

¡1�i�n

x2i

!� n2 �1 ¡1�i�n

x2i

!ık` � nxkx`

!:

Or, returning to the more comprehensible vector notation,

DkG.x/ D1�n

xkkxkn

(2.1.2)

Dk`G.x/ D1�n

1kxknC2

�kxk2ık` � nxkx`

�: (2.1.3)

This relation extends to the case n D 2, as the reader may easily check (Exercise 8).

Lemma 2.1.1. The fundamental solution G is harmonic on Rn X f0g, in other words

8x ¤ 0 W 4G.x/ D 0 (2.1.4)

Proof. For arbitrary x ¤ 0, we have from (2.1.3)

8x ¤ 0 W 4G.x/ D¡

1�k�n

DkkG.x/ D1�n

1kxknC2

¡1�k�n

.kxk2 � nx2k/ D 0

Next, we turn to a very interesting observation of Gauss. Consider a fixed sphereSr D S.x; r/ � Rn, and suppose that we can measure the average electrostatic potentialover Sr. How does the measured average electrostatic potential change as we move acharge y 2 Rn around? The surprising answer is given by the following theorem, whichis closely related to Exercise 2, cf. Exercise 12.

2.1. THE FUNDAMENTAL SOLUTION 9

Gauss’ averaging principle 2.1.2. Let S.y; r/ � Rn be an arbitrary sphere. Then

8x 2 Rn W �ZS.y;r/

Gx D

(G.r/; if x 2 B.y; r/Gy.x/; if x 2 B.y; r/c

The proof is postponed to the next section, as it becomes trivial in that context. How-ever, we ask the reader in Exercise 11 to prove it by directly evaluating the integral,just to appreciate the utility of theoretical tools to be developed.

From the “calculus” point of view, for y D 0, Gauss’ averaging principle says

8r > 0 8x 2 R2 WZSrlog k� � xk d� D 2�

8̂ 0 8x 2 Rn WZSr

d�k� � xkn�2

D �n

8 0, the integral is constant as x is movedinside the ball, and on the outside it is proportional to the fundamental solution withsingularity at 0. On the other hand, if x is fixed, the integral grows linearly as 0 < r <kxk, and on the outside with some power-law (quadratically if n D 3).

Exercises

Ex. 3 — Determine the radial (also called rotationally symmetric) solutions of4u D 0on RnXf0g, in other words, harmonic functions of the form u.x/ D R.r/, where r D kxkand R W Œ0;C1/ �! R.

Ex. 4 — Determine the set of numbers ˛, such that 4.kxk˛/ D 0 for all x 2 Rn X f0g.

Ex. 5 — Show that D�G.r/ D G0.r/ on S.0; r/, where � is the outward-point normalon S.0; r/.

Ex. 6 — Prove thatRSrD�G D 1 for all r > 0, whereD� denotes the normal directional

derivative on the sphere Sr. [Hint: Use Exercise 5]

Ex. 7 — Let Q 2 Rn�n be an orthogonal matrix. Prove that 4.u.Qx// D 4u.Qx/.

Ex. 8 — Work out the case n D 2 for (2.1.2).

Ex. 9 — Show that G 2 L1loc.Rn/, DG 2 L1loc.R

n;Rn/ and D2G 2 L1loc.Rn X f0g;Rn�n/.


Ex. 10 — Find the solutions to Laplace’s equation of the form

u.x1; : : : ; xn/ D ˚1.x1/ � : : : � ˚n.xn/

in the box ˝ D .a1; b1/ � : : : � .an; bn/ � Rn with the following boundary condtions.

(a) u � 0 on @˝

(b) D�u � 0 on @˝

* Ex. 11 — Prove relation (2.1.6) in three dimensions by transforming the integral tospherical coordinates.

* Ex. 12 — Let Br D B.0; r/, and prove the following “solid” version of Gauss’ aver-aging principle,

8r > 0 8x 2 Rn WZBr

dyky � xkn�2

D

8̂

2.2. GREEN’S REPRESENTATION FORMULA 11

§2.2 Green’s representation formulaLet u; v 2 C2.˝/ be arbitrary, and let us apply the divergence theorem to the vectorfield vDu, to get Green’s first identityZ

˝

v4u CZ

˝

.Dv;Du/dx DZ

@˝

vD�u;

which bears some resemblance to the integration by parts formula. Upon changing theroles of u and vwe get a similar relation, but the integral of the inner product .Dv;Du/stays the same, hence by subtracting the two equations we arrive at Green’s secondidentity Z

˝

u4v � v4u DZ

@˝

uD�v � vD�u: (2.2.1)

Figure 2.1: Proof of Green’s representation formula.

Green’s representation formula 2.2.1. Let ˝ � Rn be a smooth, bounded domain, u 2C2.˝/ and Gx WD G. � � x/.

8x 2 ˝ W u.x/ DZ

@˝

uD�Gx � GxD�u CZ

˝

Gx4u (2.2.2)

Proof.We apply Green’s second identity for the domain˝ XB.x; r/, where B.x; r/ � ˝and v WD Gx, see Figure 2.1. We haveZ

˝XB.x;r/u4Gx � Gx4u DZ

@˝

uD�Gx � GxD�u �ZS.x;r/

uD�Gx � GxD�u;(2.2.3)


where theminus sign in front of the integral over the sphereS.x; r/ is due to orientation.Using 4Gx D 0 (on ˝ X fxg), the first term in the left side integral vanishes.

The first integral over @˝ is fine. The surface integral over the sphere S.x; r/ isevaluated by exploiting certain properties of the fundamental solution. First,ˇ̌̌̌

ˇZS.x;r/

GxD�u

ˇ̌̌̌ˇ � G.r/

ZS.x;r/

kDuk � Cr�nC2Cn�1 maxS.x;r/

kDuk � Cr �! 0;

when r �! 0, and the generic constant C is independent of r. Second, note that for� 2 S.x; r/, we have

D�Gx.�/ D�x � �r

;DG.� � x/�

D �1

�nrn�1; (2.2.4)

which is precisely the negative reciprocal of the surface area of S.y; r/, so so

�

ZS.x;r/

uD�Gx D1

�nrn�1

ZS.x;r/

u D �ZS.x;r/

u �! u.x/; (2.2.5)

when r �! 0.After rewriting using characteristic functions, the integrand on the left-hand side

of (2.2.3) satisfies j�˝XB.x;r/Gx4uj � CGx and Gx 2 L1loc.˝/, hence Lebesgue’s theoremyields the desired result.

The significance of the representation formula is manyfold and in what follows, wediscuss its various corollaries. For harmonic functions 4u D 0 and the inhomoge-neous problem 4u D f it might give an explicit formula for the solution of the problem,given that the boundary values and normal derivatives ofu on the boundary are known.Note, however in this case the solution will be given in terms of singular integrals dueto the occurance of Gx and D�Gx, which are singular at the point x.

§2.3 Poisson’s equationGreen’s representation formula has a vast number of important consequences, manyof which will be studied in this chapter. In this section, we seek solutions to Poisson’sequation

4u D .2 � n/�n� .in Rn/

where � is some known function. The constant factor .n�2/�n serves later convenience.This type of problem occurs when there are charges present in free space describedby the charge distribution function �, and we want to determine the potential of theresulting electrostatic field. Poisson’s equation is thus the inhomogeneous version ofLaplace’s equation 4u D 0.

First, an easy corollary of Green’s representation formula for compactly supportedsmooth functions.1

1It is always a good idea to get rid of the boundary terms by restricting the support.

2.3. POISSON’S EQUATION 13

Lemma 2.3.1. Let u 2 C2c.Rn/. Then

u.x/ DZ

Gx4u D �1

.n � 2/�n

Z4u.y/dyky � xkn�2

: (2.3.1)

Proof. Fix x 2 Rn, and let ˝ �� supp f. Applying Green’s representation formula tothe function u, we get

u.x/ DZ

@˝

uD�Gx � GxD�u CZ

˝

Gx4u:

The integral over @˝ vanishes because of the choice of ˝.Now we make the bold move of replacing “4u” with .2 � n/�nf on the right side of(2.3.1), and denote the resulting integral with

U�.x/ DZ

�.y/dyky � xkn�2

: (2.3.2)

The functionU� is called theNewton potential (a.k.a. volume potential) of �, and consti-tutes one of the main objects of study of classical potential theory. This type of formulais sometimes referred to as a superposition integral, because it may be viewed as a lim-iting case of discrete charge distributions, see [15, p. 3] for details. We now prove thatthe Newton potential solves Poisson’s equation given that the “input” � is smooth andcompactly supported.

Theorem 2.3.2. If � 2 C2c.Rn/, then U� 2 C2.Rn/ and

4U� D .2 � n/�n� .on Rn/:

Proof.We may write U� as the convolution

U�.x/ DZ�.x � y/

1kykn�2

dy D

� �

1k � kn�2

!.x/;

so Theorem (A.10.1) yields that (since G 2 L1loc.Rn/, see Exercise 9) U� 2 C2.Rn/ and

4U�.x/ D

4� �

1k � kn�2

!.x/ D

Z4�.y/ dy

ky � xkn�2D .2 � n/�n�.x/; (2.3.3)

where the last equality followed from (2.3.1).It is entirely conceivable in practical situations that the charge distribution � is “con-

centrated” on some “thin” subset of Rn. In other words, we wish to determine the po-tential of a charged disk or a spherical shell, or – in the extreme – of a single point. Itis not at all obvious, but nevertheless true, that no such function � 2 C2c.R

n/ exists foreither of the examples. The satisfactory treatment is postponed to a later chapter (see


§6.1), as it requires heavy use of measure theory. However, charge distributions onlines are handled in §2.4 and on surfaces in §2.16.

The (kinetic) energy stored by the field generated by the charge distribution � isdefined as (see [15, p. 34] for a derivation)

E.�/ D12

ZU�.x/�.x/dx:

Under the hypothesis of Theorem 2.3.2, this can be given a simpler form.

Theorem 2.3.3. Let � 2 C2c.Rn/. Then E.�/ may be given by the so-called Dirichlet

integral

E.�/ D12

ZkDU�.x/k2 dx:

§2.4 Linear charge distributionsAn analog of the Newton potential (2.3.2) may for used to determine the potential oflinear charge distributions. Suppose that I � Rn is a curve, possibly infinite. Even if Iis assumed to be finite, no continuous function exists on Rn with compact support I. Toremedy the situation, we formally change “�.x/dx” to “�.�/ d�”, where � is in C2.I/,and is called a linear charge distribution. This results in the formula

U�.x/ DZI

�.�/ d�k� � xkn�2

:

The correct introduction of such a formula is postponed to §6.1.Homogeneous line segment. Weshall calculate the potential of a line segment Œ�a;a�

f0g � f0g � R3, with constant charge density �. By symmetry, we may assume thatx3 D 0. We have

U�.x1;x2; 0/ D �Z a

�a

d�q.� � x1/2 C x22

D�

x2

Z a�a

d�r�� x1x2

�2C 1

:

By the substitution � D .� � x1/=x2, d� D x2 d�, we get

�

Z .a�x1/=x2.�a�x1/=x2

d�p�2 C 1

D �

Z arsinh .a�x1/=x2arsinh .�a�x1/=x2

dt D � log

q.x1 � a/2 C x22 � .x1 � a/q.x1 C a/2 C x22 � .x1 C a/

(2.4.1)via � D sinh t, d� D cosh t dt.

This result can be recast into a simpler form, which we do along the lines of [12].Let xC D xC .a; 0/ and x� D x� .a; 0/, rC D kxCk and r� D kx�k, then with e D .1; 0/,

2.4. LINEAR CHARGE DISTRIBUTIONS 15

we have r2C � .xC; e/2 D r2� � .x�; e/

2, or

r� � .x�; e/rC � .xC; e/

DrC C .xC; e/r� C .x�; e/

:

But this can be rewritten as2

r� � .x�; e/rC � .xC; e/

DrC C r� C .xC � x�; e/rC C r� C .x� � xC; e/

DrC C r� C 2arC C r� � 2a

D

rC C r�2a

C 1

rC C r�2a

� 1:

Note that what we have calculated is really just the expression inside the logarithm in(2.4.1), hence we may write (cf. [9, p. 56])

U�.x1;x2; 0/ D 2� arcthrC C r�

2a: (2.4.2)

By the triangle inequality 2a � rC C r�, so the argument is valid to arcth W R XŒ�1; 1 �! R, except when 2a D rC C r�.

It is now trivial to determine the equipotentials of U�, i.e. surfaces (curves in ourcase, since x3 D 0) of the form fx 2 R3 W U�.x/ D Cg, where C is a constant. Weultimately get from (2.4.2) that

rC C r� D C0;

where C0 D 2a cth C2� > 0, since the case C0 < 0 would yield the empty set. The locus of

points satisfying this relation is some ellipse with foci .�a; 0; 0/ and .a; 0; 0/. Therefore,by symmetry, the three-dimensional equipotentials are ellipsoids. Note that as C �!C1, C0 D 2a, hence the ellipsoid degenerates into the line segment.

Homogeneous circle. Consider a uniformly charged circle I D Sa � f0g � R3, i.e.

I D f.a cos �; a sin �; 0/ 2 R2 W 0 � � < 2�g:

The rotational symmetry along the x3-axis of the I and the constancy of the chargedistribution � implies that the potentialU� is also rotationally symmetric along the x3-axis. Therefore, it is enough to consider points x 2 R � f0g � R. The cosine theoremimplies that for all � 2 Sa, there holds

k� � xk2 D a2 C x21 C x23 � 2.�; x/ D a

2C x21 C x

23 � 2ax1 cos �:

By transforming to polar coordinates,

U�.x1; 0;x3/ D �Z 2�

0

a d�qa2 C x21 C x

23 � 2ax1 cos �

D 2a�Z �

0

d�qa2 C x21 C x

23 � 2ax1 cos �

(2.4.3)

2If a=b D c=d, thenab

Dcd

Da � cb � d

Da C ca C d


where the last equality follows from the fact that cos.� � �/ D cos � for � 2 Œ�; 2�.Now let ' D ��2 , then d� D �2;d' and cos � D 2 sin

2 ' � 1, hence the last integral of(2.4.3) becomes

2Z �

2

0

d'qa2 C x21 C x

23 C 2ax1 � 4ax1 sin

2 '

:

By introducing q2 D a2 C x21 C x23 C 2ax1, p

2 D 4ax1, and k2 D p2=q2

2q

Z �2

0

d'p1 � k2 sin2 '

:

Note that the integral only depends on the single parameter k > 0. This integral is avery famous one, much studied in the eighteenth century, and is called complete ellipticintegral of the first kind. According to the usual convention, it is denoted as K.k/, soour final answer is

U�.x1; 0;x3/ D4�aqK.k/ D

4�aqa2 C x21 C x

23 C 2ax1

K

s4ax1

a2 C x21 C x23 C 2ax1

!:

It is trivial that K.0/ D �=2, it follows that on the x3-axis, the potential assumes thesimple form

U�.0; 0;x3/ D2��aqa2 C x23

:

Exercises

** Ex. 15 — Prove that no function ı W R �! R exists, such that .ı � f/.0/ D f.0/ forall f 2 C.R/.

§2.5 Basic properties of harmonic functionsNext, we prove one of the most important properties of harmonic functions.

Mean value property 2.5.1. Let ˝ � Rn be open and u 2 C2.˝/. For all B.x;R/ � ˝,we have

4u D 0 .on ˝/ H) u.x/ D �ZS.x;R/

u

4u � 0 .on ˝/ H) u.x/ � �ZS.x;R/

u

4u � 0 .on ˝/ H) u.x/ � �ZS.x;R/

u

2.5. BASIC PROPERTIES OF HARMONIC FUNCTIONS 17

Proof. For all R > r > 0, we haveZB.x;r/

4u DZS.x;r/

D�u:

Hence, using the substitution � D r� , d� D rn�1 d� , we haveZS.x;r/

D�u DZS.0;r/

D�u.x C �/d� D rn�1ZS.0;1/

D�u.x C r�/ d�:

But note that since k�k D 1,

Dru.x C r�/ D .Du.x C r�/; �/ D D�u.x C r�/;

we getZS.x;r/

D�u D rn�1ZS.0;1/

Dru.x C r�/ d� D rn�1DrZS.0;1/

u.x C r�/ d� D rn�1A0.r/:

where

A.r/ DZS.0;1/

u.x C r�/ d�; so that A.0/ D �nu.x/; A.R/ D1

Rn�1

ZS.x;R/

u

The three sign conditions 4u D 0, 4u � 0 and 4u � 0 imply, respectively, thatA0.r/ D0, A0.r/ � 0 and A0.r/ � 0. Therefore A.0/ D A.R/, A.0/ � A.R/ and A.0/ � A.R/,which is precisely what we wanted to show.

As a practical application of this theorem, considerGauss’ averaging principle, (2.1.6).In Exercise (11) we asked the reader to prove the theorem in the case n D 3, whichis a moderately difficult calculation. The general case would be much harder to provedirectly, but with the aid of the mean value propery, the proof becomes child’s play. Weonly prove the theorem for n � 3, the case n D 2 is surprisingly technical, see [8, p. 21].The proof below follows that reasoning.

Gauss’ averaging principle 2.5.2. For n D 3, we have

8r > 0 8x 2 Rn WZSr


D �n

8


which is precisely what we wanted to show.If, on the other hand x 2 Br, consider the function u W Rn �! R defined via the

instruction

8x 2 Rn W u.x/ D �ZSr


:

If we manage to prove that u is locally bounded, rotationally symmetric and harmonicon Br, then it follows from the solution of Exercise 3, that u is a constant function and

u.0/ D �ZSr

d�k�kn�2

D1

rn�2:

Local boundedness is trivial, as the integrand is continuous on every compact K �Br. Rotational symmetry is also easy: letQ 2 Rn�n orthogonal, then by the substitution� D Q�,

u.Qx/ D �ZSr

d�k� � Qxkn�2

D1

jSrj

ZQ�1.Sr/

d�kQ� � Qxkn�2

D �

ZSr


D u.x/:

It remains to show thatu is harmonic. Formally, the integrand is harmonicwith respectto x, see (2.1.4). All it remains is to justify the differentiation under the integral sign,for which we apply Theorem (A.8.3). For the first derivatives,

Dxk1

k� � xkn�2D .n � 2/

�k � xkk� � xkn

which is continuous on Br � Sr. Applying the theorem again,

Dxkxk1

k� � xkn�2D .n � 2/

k� � xk2 � n.�k � xk/2

k� � xknC2

which is still continuous on Br � Sr.Let x 2 Srx 2 Srx 2 Sr. First, if fxkg � Br and xk �! x, then u.xk/ D �nr. If, on the other hand

fxkg � Bcr, then we need to take the limit in the integral

u.xk/ D �ZSr

d�k� � xkkn�2

:

Wemay choose a subsequence of fxkg, denoted with the same symbol, such that kxkk &kxk D r. Then k� � xkk � k� � xk for all � 2 Sr and all k � 1. Hence

1k� � xkkn�2

�1

k� � xkn�2

which provides an integrable majorant for Lebesgue’s dominated convergence theo-rem. Hence the claim follows.

2.5. BASIC PROPERTIES OF HARMONIC FUNCTIONS 19

Harnack inequality 2.5.3.

Flux. The notion of the flux of a vector field occurs in the foundation of field theories.For example, Gauss’ law of electrostatics states that the flux of the electric field E overa closed surface � (sometimes called a Gaussian surface) is proportional to the signedsum of charges contained in volume enclosed by � . Therefore, we ought to investigatethe flux of harmonic functions.

Definition 2.5.4. Let U � Rn be open and let ˙ � U be a smooth .n � 1/-dimensionalorientable surface. The flux˚ of a harmonic function u W U �! R, through � is definedas

˚ D

Z�

D�u:

The following lemma shows that the flux does not depend on the particular choice of� as long as it encloses a fixed compact set. In other words, the flux through a closedsurface is well-defined.

Lemma 2.5.5. Let U � Rn be open, K � U compact and u W U X K �! R harmonic. IfV1 and V2 are open sets, such that

K � Vi � Vi � U .for i D 1; 2/;

then Z@V1

D�u DZ

@V2D�u:

Proof. Let W be open, such that K � W � W � Vi, and apply the divergence theoremto the function Du on the open set Vi X W,

0 DZViXW

4u DZ

@ViD�u �

Z@WD�u

For example, if u is harmonic in the ball Br, then its flux through any S�, where0 < � < r, is zero. Interesting things will happen if u has a singularity inside Br, seeExercise 16.

Exercises

Ex. 16 — Letu be harmonic in the open setUwith boundary� , which is an orientableC1-hypersurface. Prove that Z

�

D�u D 0:

Does this mean that the flux of a harmonic function is always zero?


Ex. 17 — Calculate the flux of the fundamental solution Gx through the sphere Sr fordifferent values of x 2 Rn. You should get

ZSrD�Gx.�/ d� D

8̂

2.6. THEMAXIMUM PRINCIPLE 21

Proposition 2.6.2. A harmonic function u W ˝ �! R cannot have strict, local maximaor minima in the domain ˝.

In the case of Poisson’s equation 4u D f however, the sign of the sum of the eigenvaluesof D2u.a/ is not enough to decide whether nC D n or n� D n, therefore the secondderivative test cannot be used for determining the type of a critical point if n � 2. Evenworse, calculating the set of critical points of a function can be highly nontrivial.

Thus, we retreat to the problem of localization of the maximum points, and thequantitative estimation of the maximum values in bounded domains. Weierstrass’ the-orem implies that the maximum is always attained in such a domain for a continuousfunction.

Strict maximum principle 2.6.3. Let ˝ � Rn be bounded domain, u W ˝ �! R a C2-function that is continuous on ˝. Then,

4u > 0 .on ˝/ H) max˝

u D max@˝

u:

Moreover, the maximum over ˝ is attained in @˝.

Proof. The inequalitymax

˝

u � max@˝

u

is automatic. Assume for contradiction that

max˝

u > max@˝

u: (2.6.1)

By Weierstrass’ theorem, the maximum over˝ is attained at some point a 2 ˝, that is

u.a/ D max˝

u:

Moreover, we have a 2 ˝ due to (2.6.1). By part (2) of the second derivative test, a isa critical point and nC D 0. Therefore 4u.a/ D trD2u.a/ � 0, a contradiction.

As for the “moreover” part of the theorem, a 2 ˝ would imply (2.6.1), which wasjust shown to be impossible.Note that a dual result holds for the minimum, automatically giving rise to a “strictminimum principle”. Manufacturing such versions are left to the reader.

Maximum principle 2.6.4. Let ˝ � Rn be bounded domain, u W ˝ �! R a C2-functionthat is continuous on ˝. Then,

4u � 0 .on ˝/ H) max˝

u D max@˝

u:


Proof. The theorem is reduced to strict maximum principle as follows. Let " > 0, and

u".x/ D u.x/C "ex1 :

Then,4u".x/ D 4u.x/C "ex1 � "ex1 > 0

so the strict maximum principle is applicable, giving

max˝

u" D max@˝

u":

As " & 0, u" �! u uniformly on ˝, hence the proof is finished.

Strong maximum principle 2.6.5. Let ˝ � Rn be bounded domain, u W ˝ �! R aC2-function that is continuous on ˝. If u is harmonic in ˝, and the maximum in ˝ isattained at an interior point, then u is constant.

Proof. Suppose that the maximum is attained at a 2 ˝, i.e.

u.a/ D max˝

u:

Now consider the setA D fx 2 ˝ W u.x/ D u.a/g:

Since u is continuous, A is closed. Also, A is open, because if x 2 A and B.x;R/ � ˝,then by the mean value property, for all 0 � r < R,

u.a/ D u.x/ D �ZB.x;r/

u � maxB.x;r/

u � max˝

u D u.a/:

We claim that u � u.a/ on B.x;R/. In fact, 0 � u.a/ � u 2 C.B.x;R// and thus

80 � r < R WZB.x;r/

u.a/ � u.y/ dy D 0;

from which u.a/ � u � 0 on B.x;R/ follows. In summary, B.x;R/ � A.Thus, A � ˝ is open and closed. The open set ˝ was assumed to be connected3,

therefore A D ˝, since A ¤ ;.

Comparision principle 2.6.6. Let˝ � Rn be a bounded domain, andu; v 2 C2.˝/\C.˝/be harmonic. Then

(1) u � v .in @˝/ H) u � v .in ˝/

(2) juj � v .in @˝/ H) juj � v .in ˝/3We remind the reader that a topological space X is connected, iff the only sets that are open and

closed are X and ;. Equivalently, X cannot be partitioned into two nonempty closed sets.

2.7. UNIQUENESS OF SOLUTIONS 23

Proof. For (1), the function u � v is harmonic on ˝, it follows from the maximum prin-ciple that

max˝

.u � v/ D max@˝

.u � v/ � 0;

hence u � v on˝. Finally, for (2), the function �u� v is also harmonic on˝, and hence

max˝

.�u � v/ D max@˝

.�u � v/ � 0;

so �u � v on ˝. In summary ˙u � v on ˝.

§2.7 Uniqueness of solutions

A boundary value problem for Poisson’s equation on a domain ˝ consists of finding afunction, such that 4u D f and u satisfies certain constraints on the boundary @˝.In other words, from the vector space of functions P � C2.˝/ with 4u D f on ˝,a particular subset P1 � P gets selected which satisfies the given constraints, com-monly referred to as boundary conditions. The question of existence, i.e. whether P1is nonempty, is lot more difficult and will occupy our attention in a number of sections.Uniqueness however, i.e. whether P1 consists of at most one function is usually moresimpler to settle.

Boundary conditions are prescribed on a part � � @˝ of the boundary, which maycoincide with thewhole of @˝. First, let us consider the case when � is at not at infinity.

(i) Dirichlet. The values u.x/ D g.x/ are prescribed for all x 2 � .

(ii) Neumann. The outward-pointing normal directional derivatives D�u.x/ D h.x/are prescribed for all x 2 � .

(iii) Robin. A linear combination of the values and the normal derivatives ˛u.x/ CˇD�u.x/ D g.x/ are prescribed for all x 2 � .

When � D @˝, it is customary to add the adjective “pure” to the type of the boundarycondition, whereas if disjoint parts �1 and �2 prescribe different types of boundaryconditions, the adjective “mixed” is used.

As the boundary conditions studied here are linear, i.e. any linear combination ofsolutions is also a solution, in particular satisfies the boundary conditions, the followingreduction is used in various places. If u1 and u2 both solves the problem, then u1 � u2does also, but with homogeneous boundary conditions, i.e. with g � 0 and/or h � 0.Hence, for uniqueness, it suffices to prove that the homogeneous problem is only solvedby the function u � 0, except for the pure Neumann case, where the only solutions areof the form u � c, where c 2 R is arbitrary.


Uniqueness theorem 2.7.1. Let˝ � Rn be a bounded domain, f 2 C.˝/ and g 2 C.@˝/.Then there are at most one function u 2 C2.˝/ \ C.˝/ solving

4u D f .in ˝/u D g .in @˝/

)

Proof. Suppose that u1 and u2 both solves the problem. Then the function u D u1 � u2satisfies

4u D 0 .in ˝/u D 0 .in @˝/

)The comparision principle implies (with the choice v D 0) that u D u1 � u2 D 0 on˝.

§2.8 Laplacian in a boxIn this section we solve Laplace’s equation 4u D 0 and more generally, Helmoltz’sequation 4u C �u D 0 in the nondegenerate box

˝ D .0;a1/ � : : : .0;an/ � Rn

with various boundary conditions described in the previous section.Pure homogeneous Dirichlet boundary. We first consider the boundary value prob-

lem4u C �u D 0 .on ˝/

u D 0 .on @˝/

)for some � 2 R. We look for separated a solution u of the form

u.x1; : : : ;xn/ D ˚1.x1/ � : : : � ˚n.xn/;

and then apply the uniqueness theorem to obtain that in fact we have completely solvedthe boundary value problem. Now Helmholtz’s equation reads

˚ 001 .x1/ � : : : � ˚n.xn/C : : :C ˚1.x1/ � : : : � ˚00n .xn/C �˚1.x1/ � : : : � ˚n.xn/ D 0: (2.8.1)

Suppose that ˚1.x1/ � : : : � ˚n.xn/ ¤ 0 for all x 2 T. Then,

˚ 001 .x1/˚1.x1/

C : : :C˚ 00n .xn/˚n.xn/

C � D 0:

Putting everything to the right side except for the quotient depending on x1, we obtainthat in fact

˚ 001 .x1/˚1.x1/

D Cx2;:::;xn;�:

2.8. LAPLACIAN IN A BOX 25

Now recall the boundary conditions, we have ˚.0/ D ˚.a1/ D 0. This is a regularSturm–Liouville problem, which admits a family of solutions

8k1 2 N W Cx2;:::;xn;� D ��2k21a21

; ˚1;k1.x1/ D sin��k1a1

x1�:

Continuing thisway, and noting that the constantsC� are actually independent of x1; : : : ;xn,we get

8k 2 Nn W �k D �2 k21a21

C : : :Ck2na2n

!;

and

8k 2 Nn W 'k.x1; : : : ; xn/ D sin��k1a1

x1�

� : : : � sin��knan

xn�

which in fact satisfies (2.8.1), as the reader may easily verify.Pure Neumann boundary. The boundary value problem now reads

4u C �u D 0 .on ˝/D�u D 0 .on @˝/

)

for some � 2 R. The above technique is applicable and along similar lines we get

8k 2 Nn W �k D �2 k21a21

C : : :Ck2na2n

!;

and

8k 2 Nn W k.x1; : : : ;xn/ D cos��k1a1

x1�

� : : : � cos��knan

xn�:

Fourier’s method. Originally, J.-B. J. Fourier was interested in solving the heatequation and thus developed the series method now bearing his name. His crucial ob-servation was that the set f'kgk2Nn is orthogonal in the sense that

8k; l 2 Nn W k ¤ ` H)Z

˝

'k.x/'`.x/dx D 0:

In fact, by Green’s indentityZ˝

'k'` D �1�k

Z˝

.4'k/'` D1�k

Z˝

.D'k;D'`/ D �1�k

Z˝

'k4'` D�`

�k

Z˝

'k'`;

wherewehave exploited the homogeneousDirichlet boundary conditionmultiple times.Therefore if k ¤ `, then �k ¤ �` so the integral must be zero. In modern terms, wesay that f'kg � L2.˝/ is orthogonal with respect to the usual L2.˝/ inner product. Itimmediately follows that f'kg is linearly independent.


Moreover, we can rescale each 'k so that the resulting set fe'kg becomes orthonor-mal. The square of the L2.˝/ norm of uk isZ

˝

'2k D

n£iD1

Z ai0

sin2��kiai

xi�dxi D

n£iD1

ai�

Z �0

sin2.kiyi/dyi

D

n£iD1

ai2�

Z �0

1 � cos.2kiyi/ dyi Da1 � : : : � an

2n

Hence, by setting

e'k.x/ Ds

2n

a1 � : : : � an'k.x/;

the set fe'kg � L2.˝/ is orthonormal, i.e. .e'k;e'`/L2.˝/ D ık`. Let us put the whole thinginto the proper context. The negative Lapacian �4 W L2.˝/ �! L2.˝/ is defined tohave the domain

dom.�4/ D fu 2 C2.˝/ \ C.˝/ W uˇ̌@˝

D 0g:

The set fe'kg is complete in dom.�4/ � L2.˝/ in the sense that its orthogonal comple-ment satisfies fe'kg? D f0g, in other words if a function in dom.�4/ is orthogonal toevery element of fe'kg, then it must be zero. Elementary Hilbert space theory tells usthat using a complete orthonormal basis such as fe'kg every element u 2 dom.�4/ canbe uniquely written as

u D¡k2Nn

.u;e'k/L2.˝/e'k D ¡k2Nn

bu.k/e'k .in L2.˝// (2.8.2)where we introduced the shorthand notationbu.k/ D .u;e'k/L2.˝/ for the Fourier coeffi-cents. We remind the reader that innocent-looking “.in L2.˝//” indicates that

u �¡

k2Nn

bu.k/e'k

L2.˝/

�! 0 .N �! 1/:

As the series in (2.8.2) is a trigonometric series, classical convergence theorems ofFourier analysis may be applied to improve the L2.˝/-convergence to pointwise oreven uniform convergence. For example, if u is Hölder continuous, then the Fourierseries converges uniformly.

The Fourier method can be used to solve the inhomogeneous boundary value prob-lem

�4u D f .on ˝/u D 0 .on @˝/

)for f Hölder continuous. Note the minus sign in front of the Laplacian: f'kg are in factthe eigenvectors of negative Laplacian �4, which has positive eigenvalues f�kg. Thisis only a matter of convenience.

2.8. LAPLACIAN IN A BOX 27

Suppose that the Fourier series of f in the orthonormal basis fe'kg isf D

¡k2Nn

bf.k/e'k; (2.8.3)where againbf.k/ D .f;e'k/L2.˝/. The first observation is that if the series on the rightside of (2.8.2) converges uniformly (u is also Hölder continuous, see regularity), thenwe may interchange the Laplacian and the summation to get

f D �4u D¡k2Nn

bu.k/.�4e'k/ D ¡k2Nn

bu.k/�ke'k: (2.8.4)The key point is that Fourier series are unique, therefore the Fourier coefficents in(2.8.3) and (2.8.4) must be equal,

bf.k/ Dbu.k/�k; or bu.k/ Dbf.k/�k

:

In summary, the solution may be given by a uniformly converging Fourier series

u D¡k2Nn

bf.k/�ke'k:

From the computational point of view, determining the solution amounts to finding theFourier coeffcientsbfk of the input f.

Mixed boundary value problems.

Exercises

Ex. 19 — Consider the boundary value problem

�4u D f .on ˝/u D 0 .on @˝/

)

where ˝ D .0;a1/ � .0;a2/ � R2 and

(i) f.x1;x2/ D sin�

�a1x1�sin�

�a2x2�

(ii) f.x1;x2/ D sin�2�a1x1�sin�3�a2x2�

(iii) f D 1

Ex. 20 — Develop a systematic treatment of the pureNeumann boundary value prob-lem using Fourier’s method.


Ex. 21 — Prove that for any f 2 L2.˝/, we have

kukL2.˝/ � �� n¡

iD1

1a2i

�kfkL2.˝/

§2.9 Green’s functions. SymmetryThe behavior of a harmonic function u on ˝ is thus completely determined by its be-havior near the boundary @˝, i.e.

8x 2 ˝ W u.x/ DZ

@˝

GxD�u � uD�Gx

Unfortunately, both uˇ̌@˝

and D�uˇ̌@˝

is required, but only one is available in concretesituations. Green’s representation formula is thus asking toomuch information. Canweperhaps eliminate either D�u or u? Note also that the partial derivative D�u actuallyrequires information from an open neighborhood U � ˝ of @˝.

Elimination of the normal derivative D�u results in the solution of the Dirichletboundary value problem, i.e. u

ˇ̌@˝

is known, and conversely, elimination of uˇ̌@˝

resultsin the solution of the (pure) Neumann boundary value problem, i.e. D�u

ˇ̌@˝

is known.Suppose that the boundary @˝ is partitioned into two nonempty subsets �0 and

�1. When on �0 Dirichlet-, and on �1 Neumann boundary data is prescribed, we aretalking about a mixed boundary value problem. Mixed boundary value problems arevery important in the applications, but are notoriously difficult to solve analytically.

Green’s function (of the first kind). We derive a representation formula for thesolution of the Dirichlet boundary value problem

4u D f .on ˝/u D g .on @˝/

)(2.9.1)

Suppose that for all x 2 ˝ we can find a function vx, such that vx solves the followingauxiliary problem,

4vx D 0 .on ˝/vx D �Gx .on @˝/

)Applying Green’s second identity (2.2.1) to vx and u, we get the relationZ

˝

vx4u DZ

@˝

�GxD�u � uD�vx;

and by adding this to Green’s representation formula

u.x/ DZ

@˝

GxD�u � uD�Gx �Z

˝

Gx4u;

2.9. GREEN’S FUNCTIONS. SYMMETRY 29

we see that GxD�u cancels (this is why vx is sometimes called a “corrector” function)and we are left with

u.x/ D �Z

@˝

u.D�Gx C D�vx/ �Z

˝

.Gx C vx/4u

By introducing the so-called Green’s function

Gx WD Gx C vx (2.9.2)

we get the representation formula for the solution of the Dirichlet problem (2.9.1),

u.x/ D �Z

@˝

uD�Gx �Z

˝

Gx4u

D �

Z@˝

gD�Gx �Z

˝

Gxf(2.9.3)

Observe that since the integral involving the normal derivativeD�u disappears, hencethe representation is entirely in terms of the Dirichlet boundary data g. The possiblysingular integral kernel Gx is called a Green’s function (of the first kind). It satsfies

4Gx D 0 .on ˝ X fxg/Gx D 0 .on @˝/

Gx D Gx C vx .on ˝/

9>=>; (2.9.4)for all x 2 ˝. Its existence and possible calculation are no simple matters, and in thefollowing sections, we discuss a few simple cases.

Green’s function of the second kind. Let us consider the pure Neumann boundaryproblem

4u D f .on ˝/D�u D h .on @˝/

)(2.9.5)

Before going into the details of describing Green’s functions of the second kind, let usmake some observations. If u 2 C2.˝/ \ C1.˝/ is a solution, then we necessarily haveZ

˝

f DZ

˝

�Du DZ

@˝

D�u DZ

@˝

h: (2.9.6)

Furthermore, if the problem admits a solution u 2 C2.˝/ \ C1.˝/, then u C c is asolution too, for arbitraty c 2 R. No need to worry however, we will show later that noother solutions exists. This is an important characteristic of pure Neumann boundaryvalue problems: a solution must contain an undetermined constant.

To derive a formula for the problem (2.9.5) analogous to (2.9.3), we might try to dosame trick as before. Similarly, suppose that vx solves

4vx D 0 .on ˝ X fxg/D�vx C D�Gx D c .on @˝/

)


for some c 2 R. Note that we have relaxed the requirement that D�vx should cancelD�Gx. Then Green’s second identity applied to u and vx,Z

˝

vx4u DZ

@˝

vxD�u � uD�vx

and Green’s representation formula

u.x/ DZ

@˝

GxD�u � uD�Gx �Z

˝

Gx4u;

can be added to yield

u.x/ DZ

@˝

.Gx C vx/D�u � cZ

@˝

u �Z

˝

.Gx C vx/4u: (2.9.7)

Using the notation Kx D Gx C vx for the Green’s function of the second kind, we getthe representation formula

u.x/ DZ

@˝

KxD�u � cZ

@˝

u �Z

˝

Kx4u

D

Z@˝

Kxh � cZ

@˝

u �Z

˝

Kxf(2.9.8)

We would like to stress that although the occurance of the Green’s functions of thesecond kind is less common in the literature, they are equally important in the appli-cations. The reason for this is the fact that they are notoriously difficult to calculate,even for simple domains, see Section 3.6. In summary a Green’s function of the secondkind should solve

4Kx D 0 .on ˝ X fxg/D�Kx D c .on @˝/

)(2.9.9)

for all x 2 ˝ and some c 2 R.A very important remark is in order. There are various heuristic arguments (“deriva-

tions”) for obtaining various Green’s functions mostly cooked up by physicists and godforbid, electrical engineers. These result in perfectly valid representiation formulas.However, if some input is given, we cannot simply put them in a representation for-mula, we need to prove that it holds true under various technical hypotheses on thedata. The formulas sketched above are merely a schematic description of how solu-tions should look like.

The notation Gx.y/ is meant to express the fact that x 2 ˝ is fixed point at whichwe want to express u.x/. However G .x;y/ D Gx.y/ is more common, because of thefollowing surprising result.

Symmetry of Green’s function 2.9.1. LetGx be aGreen’s function satsfying (2.9.9). Then

8x;y 2 ˝ W x ¤ y H) G .x;y/ D G .y;x/:

2.10. GREEN’S FUNCTIONS IN ONE DIMENSION 31

Proof. Let x;y 2 ˝ be distinct points. Let v.z/ D G .x; z/ and w.z/ D G .z;y/, thenby hypothesis 4v D 0 on ˝ X fxg, 4w D 0 on ˝ X fyg and v D w D 0 on @˝. Green’ssecond identity (2.2.1) may be applied to the functions v andw, on the punctured domain˝ X .B.x; r/ [ B.y; r//, for small " > 0, to getZ

S.x;r/vD�w � wD�v D

ZS.y;r/

wD�v � vD�w:

Wewish to prove that as r & 0, the left side converges to v and the right side convergesto w. We only handle left side, as the other is completely similar. First,D�w is boundedon B.x; r/ if r > 0 is small enough, henceˇ̌̌Z

S.x;r/vD�w

ˇ̌̌� C

ˇ̌̌ZS.x;"/

Gx C vxˇ̌̌

� C�nrn�1.r�nC2 C 1/ �! 0;

and�

ZS.x;r/

wD�v D �ZS.x;r/

wD�Gx �ZS.x;r/

wD�vx �! w.x/;

where the first integral is handled using similar reasoning as (2.2.5), and the secondintegral vanishes because w is bounded near x and vx is harmonic.

§2.10 Green’s functions in one dimensionThis section is concerned with the n D 1 special case of the Green’s function technique.We warn the reader that the triviality and simplicity of this special case is misleading.This section serves as a “proof-of-concept”, it may be safely skipped for more interest-ing material.

Let us first determine the Green’s function for the one dimensional domain .a; b/ �R with Dirichlet boundary conditions. In our case, the “corrector” function vx shouldsatisfy

v00x.y/ D 0 .y 2 .a; b/ X fxg/

vx.a/ D �Gx.a/ D 12.x � a/

vx.b/ D �Gx.b/ D 12.b � x/

9>=>;for all x 2 .a; b/. The first equation says that vx should be linear, so the unique solutionis

vx.y/ D12a C b � 2xb � a

.y � a/Cx � a

2:

Since Gx D vx C Gx, we have

Gx.y/ D12a C b � 2xb � a

.y � a/Cx � a

2�

jy � xj2

D1

b � a

(.b � x/.y � a/ if a < y < x.x � a/.b � y/ if x < y < b


To specialize (2.9.3) to our case, we need to evaluate the normal derivative D�Gx at aand b, which is simply

D�Gx.a/ D �b � xb � a

; D�Gx.b/ Da � xb � a

;

therefore the representation formula for the boundary value problem

8y 2 .a; b/ W u00.y/ D fu.a/ D cu.b/ D d

9>=>;becomes

u.x/ Db � xb � a

c Cx � ab � a

d Cb � xb � a

Z xa.y � a/f.y/dy C

x � ab � a

Z bx.b � y/f.y/dy:

We leave it as exercise to prove that u defined so solves the problem.

§2.11 Poisson integral for the half-spaceFinding Green’s functions often involves the trickery called method of image charges.The bounded harmonic term vx in (2.9.2) must compensate the fundamental solution Gxon the boundary. More precisely, if wemanage to find a harmonic vx that is proportionalto Gx on the boundary, then we have a Green’s function by replacing vx with a suitableconstant multiple of it.

This compensation, or “corrector”, is easiest to find in the case of the half-space. Bytranslation and rotation, we may assume that the normal of the bounding hyperplaneis parallel to the xn-axis, so let

H D fx 2 Rn W xn > 0g;

so @H D fx 2 Rn W xn D 0g. Let x be an arbitrary point, and let us denote the reflectionof x onto @H by x0. We claim that Gx and Gx0 are equal on the hyperplane @H. First ofall, since

x0 D x � 2xn�;

where � D .0; : : : ; 0; 1/, we easily have kxk D kx0k, and if � 2 @H, that is �n D 0, then

k� � x0k2 D k�k2 C kxk2 � 2.x0; �/ D k�k2 C kxk2 � 2.x; �/C 4xn�nD k� � xk2

Then, n � 3n � 3n � 3, we have

8� 2 @H WGx0.�/Gx.�/

D

k� � xkk� � x0k

!n�2D 1:

2.11. POISSON INTEGRAL FOR THE HALF-SPACE 33

In summary, Gx � Gx0 D 0 on @H. It is easy to see that this is true for the case n D 2n D 2n D 2.Therefore the Green’s function for the half-space is given by

8z 2 H W Gx.z/ D G .x; z/ D Gx.z/ � Gx0.z/

D

8̂̂̂̂


solution u, i.e. a harmonic function that coincides with g on @H. In other words, can gbe extended “harmonically” to the whole half-space H?

Let

8x 2 H 8� 2 @H W P.x; �/ D2xn/�n

1k� � xkn

denote the Poisson kernel for the half-space. The following crucial properties of thePoisson kernel shall be used, which are called approximation of unity in harmonic anal-ysis, because the family fP.x; � / W x 2 Hg provides an approximation to the ı-distribution in a well-defined way. Intuitively, P.x; � / “blows up” as x approaches @˝,but in certain, controlled way.

(a) Positivity. P.x; �/ � 0

(b) Normalization. Putting the harmonic function u � 1 in (2.11.1), we find

8x 2 H WZ

@HP.x; �/ d� D 1: (2.11.2)

(c) Vanishing away from the singularity. For all ı > 0, � 2 @H and fxkg �B.�; ı/ with xk �! �, there holdsZ

@HXB.�;ı/P.xk; �/ d� �! 0; as k �! 1: (2.11.3)

In fact, by writing the sequence fxkg � B.�; ı/ as xk D �k C tk�, wheref�kg � B.�; ı/, �k �! � and tk & 0 with tk � ı after possibly dropping afinite number of terms, we have the following estimate using the triangleinequality

k� � �k � k� � �kk C k�k � �k � k� � xkk C tk C ı � 3k� � xkk;

if ı � k� � xkk. Hence,Z@HXB.�;ı/

2tk�n

1k� � xkkn

d� �2tk3n�n

ZHXB.�;ı/

1k� � �kn

d�;

which tends to zero as k �! 0 for all fixed ı > 0, since the last integralis convergent.

The following theorem says that for bounded and continuous boundary function g, thePoisson integral provides us with a solution.


Theorem 2.11.1. Suppose that g 2 Cb.@H/, and let

8x 2 H W u.x/ DZ

@HP.x; �/g.�/ d�: (2.11.4)

Then,

(1) u 2 C1b .H/,

(2) 4u D 0 on H,

(3) If fxkg � H, � 2 @H and xk �! � , then u.xk/ �! g.�/ as k �! 1

Proof. Introduce the notation x D .x;xn/, then the Poisson kernel takes the form

P.x; �/ D2xn�n

1�k� � xk2 C x2n

�n=2To prove (1), let fxkg � H, x 2 H, such that xk �! x. There exists a closed cylinder

Z D Br � Œa; b � H, such that fxkg � Z. Then, using the triangle ineqality,

k� � xkk2 � .k�k � r/2

the following pointwise estimate holds for the integrand

jP.xk; �/g.�/j �2�n

b�k� � xkk2 C a2

�n=2kgk1 � C�.k�k � r/2 C a2

�n=2 ;which is an integrable majorant independent of k. Lebesgue’s dominated convergencetheorem implies that u.xk/ �! u.x/ as k �! 1, and the trivial estimate of the Poissonintegral implies that ju.x/j < C1.

For (2), note that x 7! G .�; x/ is harmonic for all � ¤ x. Using the symmetry ofGreen’s function (see (2.9.1), or directly), we may express the Poisson kernel as

x 7! P.x; �/ D x 7! D�.� 7! G .x; �// D x 7! D�.� 7! G .�; x//:

Hence,

4.x 7! P.x; �// D 4.x 7! D�.� 7! G .�; x/// D D�.� 7! 4.x 7! G .�; x/// D 0:

Finally, for (3), let � 2 @H, by the continuity of g,

8" > 0 9ı > 0 8� 2 @H W k� � �k < ı H) jg.�/ � g.�/j < ":

Then, for fxkg � H \ B.�; ı=2/, we have

ju.xk/ � g.�/j �

Z@H\B.�;ı/

C

Z@HXB.�;ı/

!P.xk; �/jg.�/ � g.�/j d�;


using the normalization property of the Poisson kernel and the triangle inequality. Thefirst integral is � ", by continuity of g and normalization. The second integral may beestimated trivially from above by

2kgk1Z

@HXB.�;ı/P.xk; �/ d�

which tends to 0 as k �! 0, for all fixed ı > 0, because of property (c) of the Poissonkernel.

Green’s function of the second kind. Let us now investigate the pureNeumann case.The normal derivative of the fundamental solution Gx reads

8� 2 @H W D��Gx.�/ D�DGx.�/;��

�D

1�n

k� � xk�n.� � x;��/;

where we have used (2.1.2). Hence

8� 2 @H W D��Gx.�/C D��Gx0.�/ D1�n

k� � xk�n.� � x � x0;��/ D 0;

because .�; �/ D 0 and .x C x0; �/ D xn C xn � 2xn D 0. Therefore the Green’s functionof the second kind for the half-space is given by

8� 2 @H W K�;x.�/ D Gx.�/C Gx0.�/ D

8̂̂


Exercises

Ex. 22 — Prove by direct calculation that the potential u � c on the half-space, if it isknown that it assumes the constant value c 2 R on the hyperplane @H D f.x; �/ D 0g �Rn, where

(a) n D 2

(b) n D 3

(c) n arbitrary [Hint: Evaluate the resulting integral using the coarea for-mula to arrive at some known improper integral.]

Note that these results provide direct proofs of the fact that the Poisson kernel hasintegral 1.

Ex. 23 — Calculate the gradient of (2.11.1) rigorously. What is the physical relevanceof this?

Ex. 24 — Determine the Green’s functions for the following domains.

(a) The quadrant fx1 > 0; x2 > 0g � R2.

(b) The octant fx1 > 0g \ fx2 > 0g \ fx3 > 0g � R3.

Ex. 25 — Calculate the potentialu in the half-plane fx2 > 0g for the followingDirichletboundary data.

(a) u.x1; 0/ D �Œ�a;a.x1/, where � is the characteristic function. Determinethe equipotentials, i.e. curves on which the potential u is constant.

(b) u.x1; 0/ D x1�Œ�a;a.x1/.

(c) If a; b > 0, let

u.x1; 0/ D

(�a; if x1 < 0b; if x1 > 0

Determine the equipotentials, and Du.

(d) If a > 0, let

u.x1; 0/ D

8̂ a

* Ex. 26 — Calculate the potential u in the half-space having the following Dirichletboundary data.


(a) In R2, u.x1; 0/ D sin!x1, where ! > 0.

(b) In R2, u.x1; 0/ D x1=.x21 C 1/.

(c) In R3, u.x1;x2; 0/ D cos x1 cos x2.

(d) In R2, u.x1; 0/ D 1=.x21 C a2/.

Ex. 27 — Prove that if g 2 L2.R/, then

u.x1;x2/ DZ C1

�1

.F g/.�/e2�ix1�e�2�x2j�j d�

solves Laplace’s equation on the upper half plane with u D g on fx2 D 0g.

(a) Deduce that ju.x1;x2/j � 1p2�x2 kgkL2 .

Ex. 28 — Consider Laplace’s equation on the semi-infinite strip ˝ � R2, i.e.

˝ D f.x1;x2/ 2 R2 W 0 < x1 < �; x2 > 0g:

Solve the following problem:

4u D 0 .on ˝/u D 0 .on f0; �g � .0;C1//u D g .on .0; �/ � f0g/

9>=>;where g is a continuous function on Œ0; �, vanishing at the endpoints and having theFourier series expansion

g.t/ D1¡

kD1

bg.k/ sinkt; where bg.k/ D 2�

Z �0

g.s/ sinkt ds:

(1) Prove that the functions

uk.x1;x2/ D e�kx2 sinkx1

solve the above problem with the boundary data set to g.t/ D sinkt.Note that in this casebg.k/ D 1, while all the other Fourier coefficentsare zero.

(2) Prove that fukg are orthogonal with respect to the L2.˝/ inner product.

(3) By writing the solution u in terms of fukg, deduce that

u.x1;x2/ D1¡

kD1

bg.k/uk.x1;x2/:(4) * Sum the series!

2.12. INVERSION 39

§2.12 InversionThat trick which led to the Green’s function for the half-space works for the ball, too.Reflection onto a ball ismain subject of study in inversive geometry. Itworks as follows.

Let x 2 B WD B.0; 1/, and consider its inversion (a.k.a Kelvin transform), x� ontothe sphere S WD S.0; 1/, via

8x ¤ 0 W x� Dx

kxk2:

The inversion is conveniently extended to the one-point (a.k.a Alexandroff-) compacti-fication of Rn by the instructions

0� WD 1; .1/� D 0:

The inversion is easily extended to arbitrary spheres by means of dilation and trans-lation (Exercise 29), the reason for choosing the origin-centered unit sphere is the im-mense simplification of formulas. We record the following properties of the inversion,their proof is in Exercise 30.

(1) kx�kkxk D 1

(2) .x�/� D x

(3) x 2 S () x� D x

(4) The following symmetry property holds true:

8x1;x2 2 B W kx1kkx�1 � x2k D kx2kkx1 � x�2k (2.12.1)

Let˝ � Rn be an open set with y … ˝, and let u W ˝ �! R. Define the inversion of˝ with respect to the sphere S.y; r/ via ˝� D fx� W x 2 ˝g, and the Kelvin transformof u, u� W ˝� �! R via

8x 2 ˝� W u�.x/ Drn�2

kx � ykn�2u.x�/

Theorem 2.12.1. A function u is harmonic if and only if u� is harmonic.

Proof. A direct calculation which is made simpler by the formalism of the Laplace–Beltrami operator, see Exercise 49.

Exercises

Ex. 29 — Extend the results of this section to the arbitrary sphere S.y; r/.

* Ex. 30 — Prove properties (1)–(4) of inversion. [Hint: for (2.12.1), expand kx�1 �x2k2.]


** Ex. 31 — Prove Therorem (2.12.1). [Hint: Suppose that B D B.0; 1/ and calculate4u.x�/.]

Ex. 32 — Consider the maps '.x/ D x=kxk and .x/ D x=kxk2. Prove that D' isnever invertible while D is always invertible.

§2.13 Poisson integral for the ballIn this section, we calculate the Green’s function of the ballB D B.0; 1/ � Rn and obtaina specialized form of Green’s representation formula. This result will have enormoustheoretical importance and we shall study it further in the next section.

Let us begin the derivation by finding a connection betweenGx� andGx, where x ¤ 0(equivalently, x� ¤ 1), and the inversion is taken to be with respect to S D S.0; 1/. Let� D kxk from now on in this section. Also, let � 2 S be arbitrary, i.e. �� D � , then forn D 2,

Gx.�/ � Gx�.�/ D12�

logk�� xkk� � x�k

D12�

logk�k

kxkD

12�

log1�; (2.13.1)

where we have used (2.12.1). Therefore, the difference of Gx.�/ and Gx�.�/ is constant(in �), for all � 2 Sr.

As for, n � 3, the situation is similar, although this time the ratio ofGx.�/ andGx�.�/is constant,

Gx.�/Gx�.�/

D

k� � x�kk�� xk

!n�2D

k�k

kxk

!n�2D

1�n�2

: (2.13.2)

In other words Gx.�/ is a constant (in its argument, �) multiple of Gx�.�/ for all � 2 S.By dilation x 7! x=r, we obtain similar formulas for a general ballB D B.0; r/, r > 0.

The corrector function Gx� is bounded on B, hence Green’s function for the ball,

8z 2 Br W Gx.z/ WD

8̂̂

2.13. POISSON INTEGRAL FOR THE BALL 41

Figure 2.2: For n � 3, the values of Gx and Gx� are proportional on the sphere Sr.

(2.12.1), we have for � 2 S,

��nDGx.�/ D� � x

k� � xkn�

1�n�2

� � x�

kz � x�kn

D1

k� � xkn

.� � x/ �

1�n�2

k�� xkn

k� � x�kn

� �

1�2x

!!

D1

k� � xkn

.� � x/ �

�n

�n�2

� �

1�2x

!!

D1

k� � xkn

.� � x/ � �2

� �

1�2x

!!

D1 � �2

k� � xkn�

hence, by dilation x 7! x=r,

D�Gx.�/ D

DGx.�/;

�

k�k

!D

�1�nr

r2 � �2

k� � xkn: (2.13.3)

In summary, we have obtained the Poisson integral for the ball for a harmonic functionu on Br,

8x 2 Br W u.x/ Dr2 � �2

�nr

ZSr

u.�/ d�k� � xkn

; (2.13.4)

where � D kxk. As in the case of the half-space, this formula represents the pointwisevalues of u in terms of its values on the boundary. Similarly, we define the Poisson


kernel for the ball as

8x 2 Br 8� 2 Sr W P.x; �/ Dr2 � �2

�nr1

k� � xkn:

Analogously to the case of the half-space, we have the following properties.

(a) Positivity. P.x; �/ � 0

(b) Normalization. Putting the harmonic function u � 1 in (2.13.4), we find

8x 2 Br WZSrP.x; �/ d� D 1: (2.13.5)

(c) Vanishing away from the singularity. For all ı > 0, � 2 Sr and fxkg �B.�; ı/ with �k �! �, there holdsZ

SrXB.�;ı/P.xk; �/ d� �! 0; as k �! 1:

In fact, we may assume that r D 1, and write xk D �k�k, where �k �! 1,j1 � �kj � ı and k�kk D 1. By using the triangle inequality

k� � �kk � k� � xkk C j1 � �kj � 2k� � xkk;

if k� � xkk � ı. We haveZk�kD1; k��k�ı

1 � �2k�n

1k� � xkkn

d� �1 � �2k2n�n

Zk�kD1; k��k�ı

1k� � �kkn

d�

which clearly tends to zero as k �! 1 for any fixed ı > 0, since the lastintegral is convergent.

Given that the boundary function is continuous (hence bounded, because Sr is compact),the Dirichlet boundary value problem is solved by the Poisson integral.

Theorem 2.13.1. Suppose that g 2 C.Sr/, and let

8x 2 Br W u.x/ DZSrP.x; �/g.�/ d�: (2.13.6)

Then,

(1) u 2 C1b .Br/,

(2) 4u D 0 on Br,

(3) If fxkg � Br, � 2 Sr and xk �! � , then u.xk/ �! g.�/ as k �! 1

2.13. POISSON INTEGRAL FOR THE BALL 43

Proof. The proof is completely analogous to that of Theorem (2.11.1), only the techni-calities have to be checked.

Let us write the case n D 2, explicitly out in polar coordinates,

80 � � < r80 � ˛ � 2� W u.�; ˛/ D12�

Z 2�0

r2 � �2

r2 � 2�r cos.� � ˛/C �2u.r; �/ d�:

(2.13.7)Note that the factor of 1=r disappeared in front of the integral because of the changeof variables. In both the general and the two dimensional case Poisson integrals canbe thought of as a convolution (on the boundary). In the case n D 2, the 2�-periodicPoisson kernel is particularly nice in form, i.e.

P�.ˇ/ Dr2 � �2

r2 � 2�r cosˇ C �2; (2.13.8)

so that

u.�; ˛/ D12�

Z 2�0

P�.˛ � �/u.r; �/ d� D .P� � u.r; � //.�/:

Various properties of this very important integral kernel are explored in the exercises.

Exercises

Ex. 33 — For simplicity, consider the case r D 1 of Poisson kernel (2.13.8) in n D 2,i.e.

P�.˛/ D1 � �2

1 � 2� cos˛ C �2

(a) Prove using complex arithmetic that if ! D ei˛, then

P�.˛/ D Re

1 C �!1 � �!

!:

(b) Find a series expansion using (a).

(c) From (b), deduce that P� has integral 2� .

(d) Show that P�.˛/ > 0 for all 0 � � < 1 and 0 � ˛ < 2� .

(e) Show that P�.˛/ � P�.ı/ if 0 < ı � j˛j � � .

(f) Show that for all 0 � � � 1 and ı > 0, there exists a constant Cı > 0such that the following estimate holds

P�.ı/ � Cı.1 � �2/:

Ex. 34 — Prove that if g 2 Lp.Sr/, then u� �! g in Lp.Sr/ as � �! r, where u�.�/ Du.��/ for � 2 Sr and 0 � � < 1.


§2.14 The Laplacian in polar coordinatesThese successes with the ball may encourage us to investigate Laplace’s equation on adisk and ball directly. It is easy to see that the Laplacian is invariant under rotations(and translations, too), see Exercise 7. Note, however, that a transformation to polar orspherical coordinates is not a mere rotation (it is nonlinear), so we should expect thatLaplace’s equation becomes quite different in form in the new coordinate systems.

The case of arbitrary curvilinear coordinates is treated in §3.2. The separated so-lution of the Laplacian in spherical coordinates introduce special functions hence arepostponted to §3.3.

First, let us deal with polar coordinates. For .r; �/ 7! .x;y/, the transformationrules are

x D r cos �y D r sin �

)while the inverse .x;y/ 7! .r; �/ are

r Dpx2 C y2

� D arctan�yx

�9=;Let u W R2 �! R, and introduce Qu W Œ0;C1/ � Œ0; 2� �! R via

Qu.r; �/ D u.r cos �; r sin �/;

in other words,u.x;y/ D Qu.r; �/ D Qu

�px2 C y2; arctan

�yx

��:

Note the following

Dxr Dxp

x2 C y2D cos �; Dyr D

ypx2 C y2

D sin �;

Dx� D �y

x2 C y2D �

sin �

r; Dy� D

xx2 C y2

Dcos �

r;

Dxxr Dy2

.x2 C y2/3=2D

sin2 �

r; Dyyr D

x2

.x2 C y2/3=2D

cos2 �

r

Dxx� D2xy

x2 C y2D 2 cos � sin �; Dyy� D �

2xyx2 C y2

D �2 cos � sin �

Before calculating the Laplacian, let us derive the gradient of u, Du in polar coordi-nates. First Du.x;y/ D .Dxu;Dyu/, so by the chain rule

Dxu.x;y/ D Dx Qu.r; �/ D Dr QuDxr C D� QuDx� D cos �Dr Qu �sin �

rD� Qu

Dyu.x;y/ D Dy Qu.r; �/ D Dr QuDyr C D� QuDy� D sin �Dr Qu Ccos �

rD� Qu

2.14. THE LAPLACIAN IN POLAR COORDINATES 45

which certainly does not equal .Dr Qu;D� Qu/ as one might naively think. Physicists andengineers have a suggesive way of writing this by introducing the notations

er.r; �/ D�cos �sin �

�; e�.r; �/ D

�� sin �cos �

�which form a basis of the tangent space at the point .r; �/. Then they write

Du.x;y/ D Dr Qu er C1rD� Qu e� : (2.14.1)

Note very carefully the factor of 1=r, and also the fact that contrary to the standardbasis vectors ex and ey accuring in the Cartesian expression of the gradient, the tangentvectors er and e� depend on the point .r; �/. Anotherway of remembering the 1=r factoris that Dr Qu has units length

�1 but D� Qu is unitless, hence by muliplying by 1=r we canadd the two quantities.

The Laplacian is the divergence of the gradient, so forgetting the trigonometricexpression we calculated as an illustration, we obtain

4u D Dx.Dr QuDxr C D� QuDx�/C Dy.Dr QuDyr C D� QuDy�/

D Drr Qu.Dxr/2 C(((((((D�r QuDx�Dxr C Dr QuDxxr C(((((

((Dr� QuDxrDx� C D�� Qu.Dx�/2 C��D� QuDxx�

C Drr Qu.Dyr/2 C(((((((D�r QuDy�Dyr C Dr QuDyyr C(((((

((Dr� QuDyrDy� C D�� Qu.Dy�/2 C��D� QuDyy�

Nowwemake a couple of simple observations. First, themixed derivativeDr� Qu vanish,because its cofficient is

2Dx�Dxr C 2Dy�Dyr D 0:

Also, D� Qu cancel for similar reasons. What remains simplifies to

4u D Drr Qu C1rDr Qu C

1r2D�� Qu;

which is often written in the product form

4u D1rDr.rDr Qu/C

1r2D�� Qu: (2.14.2)

What’s the use of this complicated expression? First, it becomes easy to separate vari-ables in polar coordinates, see Exercise (36).

Exercises


Ex. 35 — Find the solution to the following boundary value problems on the annulus:

(a)

8̂

2.15. THE USE OF COMPLEX ANALYSIS I. 47

Ex. 38 — Consider the Laplace–Dirichlet boundary value problem on the unit diskB1 � R2,

4u D 0 .on B1/u D g .on S1/

)for some continuous function g on S1 given by its Fourier series

g.�/ Da02

C

1¡kD1

.ak cos k� C bk sink�/:

(1) Prove that the function

u.r; �/ Da02

C

1¡kD1

rk.ak cos k� C bk sink�/

solves the problem.

(2) Prove that the Dirichlet integral satisfies

D.u/ DZB1

kDbuk2 D Z 2�0

Z 10

h.Dru/2 C

1r2.D�u/2

ir dr d� D �

1¡kD1

k.a2k C b2k/

wherebu.x;y/ D u.r; �/.(3) Prove that if g.�/ D sin k� , then D.u/ D �k.

(4) Now consider the boundary function

g.�/ D1¡

kD1

sin.kŠ�/k2

:

Prove that g is continuous, but D.u/ D C1.

§2.15 The use of complex analysis I.In Exercise 36, we have witnessed an interesting family of harmonic functions on theunit disk B1 � R2, i.e.

uk.r; �/ D rk.ak cos k� C bk sink�/:

The occurence of the trigonometric functions suggest that we should work with com-plex exponantials as per Euler’s formula

eik� D cos k� C i sin k�:


In fact uk.r; �/ may be written as the real part of ckrkeik� , where ck D ak � ibk. Thatfunction happens to be complex analytic.

Indeed, it is a well-known fact from complex analysis that the real and imaginaryparts of a complex analytic function are harmonic. We quickly remind the reader thata complex analytic function f W C �! C may be viewed as a functionef W R2 �! R2,where

f.x C iy/ D u.x;y/C iv.x;y/; ef.x;y/ D �u.x;y/; v.x;y/�:Complex analyticity means thatef is differentiable (as an R2 �! R2 function), and thatderivative corresponds to a complex number. Since the derivative at a fixed point isa linear map R2 �! R2, in other words a (real) matrix R2�2, and since we know fromalgebra that there is an injective morphism R2�2 �! C given by�

a b�b a

�7! u C iv;

we have that the derivative matrix Def.x0;y0/ has to be of this particular form, because�Dxu Dyu

�Dxv Dyv

�.x0;y0/

must correspont to a complex number. We get theCauchy–Riemann partial differentialequations,

Dxu D DyvDyu D �Dxv

)Then it is obvious that both u, the real part of f and v, the imaginary part of f areharmonic,

Dxxu C Dyyu D Dxyv � Dyxv D 0Dxxv C Dyyv D �Dxyu C Dyxu D 0

by the fact that a complex analytic function is infinitely differentiable, so Young’s the-orem applies for the mixed derivative.

For simply connected domains˝ the converse also holds: for every harmonic func-tion u W ˝ �! R there exists a so-called harmonic conjugate v W ˝ �! R, which isharmonic and f D u C iv is complex analytic.

Let us now investigate the case of the open diskBr � C. Ifu W Br �! R is harmonic,then it is the real part of some analytic function f W Br �! C. For that analytic function,we have the Taylor series expansion

8z 2 Br W f.z/ D1¡

kD0

ckzk:


This can be rewritten using complex exponentials for all 0 � � < r and 0 � � < 2� as

f.�ei�/ D1¡

kD0

ck�keik� :

Hence u D Re f is given by

u.�; �/ D12

1¡kD0

ck�keik� C12

1¡kD0

ck�ke�ik�

D12

1¡kD0

ck�keik� C12

0¡kD�1

c�k��keik� D1¡

kD�1

dk�jkjeik� ;

(2.15.1)

where we set

dk D

8̂ 0

D

8̂ 0

Dirichlet boundary conditions. We first impose a continuous boundary data g.�/ on@Br D Sr, i.e. let

u.r; �/ D g.�/:

Then, we have the following condition on the coefficients dk,

g.�/ D1¡

kD�1

dkrjkjeik� ; (2.15.2)

in other words the dk’s are the Fourier coefficients of the known boundary data g.�/.But we shouldn’t stop here. By multiplying (2.15.2) by e�i`� and integrating with re-spect to � , we get

2�dkrjkj DZ 2�

0g.�/e�ik� d�;

because the integration and the summation can be interchanged due to the uniformconvergence and Z 2�

0eik�e�i`� d� D 2�ık`:

Writing this expression for the coefficients dk back into (2.15.1), we get

u.�; �/ D12�

1¡kD�1

��r

�jkjeik�

Z 2�0

g.˛/e�ik˛ d˛ D12�

Z 2�0

g.˛/1¡

kD�1

��r

�jkjeik.��˛/ d˛:


The sum inside can be evaluated explicitly. To this end, let ! D ei.��˛/. Then

1¡kD�1

��r

�jkj!k D

1¡kD0

��r

�k!k C

1¡kD1

��r

�k!k D

rr � �!

C�!

r � �!

Dr2 � �2

r2 � �2 � 2�rRe!D

r2 � �2

r2 � 2�r cos.� � ˛/ � �2:

In summary, we have recovered thePoisson integral formula for theDirichlet boundaryvalue problem for the disk,

u.�; �/ D12�

Z 2�0

r2 � �2

r2 � 2�r cos.� � ˛/ � �2g.˛/ d˛:

For simplicity, we consider the unit disk, i.e. r D 1, the general case follows bydilation. We impose a continuous boundary data g.�/ on @B1 D S1 and prove that theDirichlet problem

4u D 0 .on B1/u D g .on S1/

)(2.15.3)

admits a unique solution.

Theorem 2.15.1. Let g.�/ be continuous function on S1. Then there exists a uniquecontinuous function u on B1, such that u satisfies (2.15.3). Moreover, u is given by thePoisson integral formula

u.�; �/ D

8


may restrict ourselves to the boundary point � D 0. That is, we need to check that forall " > 0 there exists some ı > 0 and ı0 > 0 such that

j� j < ı and 1 � ı0 < � < 1 H) ju.�; �/ � g.0/j < ":

Showing this amounts to estimating the Poisson integral, because

ju.�; �/ � g.0/j �12�

Z 2�0

P�.� � ˛/jg.˛/ � g.0/jd˛

D12�

Zj˛j 0, such that

j˛j < ı H) jg.˛/ � g.0/j <"

2:

Moreover, from part (c) (and (d)) of Exercise 33, we may estimate the integral averageover Œ0; 2� of P� by 1. Therefore, the first integral is small, i.e. lesser than "2 .

The second integral is also small, since g is bounded and the Poisson kernel is smallnear the boundary and away from its singularity, see part (e) and (f) of Exercise 33.More precisely, if j� j < ı2 and j˛j � ı, then j��˛j �

ı2 . Therefore,P�.��˛/ � Cı.1��

2/.LetM D max jg.˛/ � g.0/j, and choose ı0 so that

1 � ı0 < � < 1 H) P�.� � ˛/ �"

2M:

Putting it all together, the second integral can be estimated by "2 . Hence the claimfollows.

Neumann boundary conditions. If, on the other hand the normal derivative is pre-scribed on the boundary, i.e.

D�u.r; �/ D h.�/;

we may do as follows. First of all, from the divergence theorem, we deduce thatZSrh D 0 (2.15.4)

necessarily. Also, note that D� D D�, so from (2.15.1), we have

h.�/ D1¡

kD�1

jkjdkrjkj�1eik� ;

hence along similar lines as before

2�jkjrjkj�1dk DZ 2�

0h.�/e�ik� d�

52 2. ANALYT

Documents

Elliptic Partial Differential Equationsweb.cs.elte.hu/~csirik/elliptic.pdfMaxwell immediately realized that by writing his of equations in free space (j D 0, ˆD 0),i.e. div D D 0