Invitation to Partial Di erential Equationsstaff.math.su.se/mleites/books/shubin-invitation.pdf · 2017-02-23 · Appendix: Calculus of variations and classical mechanics 36 x2.6

Invitation

to Partial Differential Equations

Mikhail Shubin

Contents

Preface vii

Selected notational conventions ix

Chapter 1. Linear differential operators 1

§1.1. Definition and examples 1

§1.2. The total and the principal symbols 2

§1.3. Change of variables 4

§1.4. The canonical form of second order operators with constantcoefficients 6

§1.5. Characteristics. Ellipticity and hyperbolicity 7

§1.6. Characteristics and the canonical form of 2nd order operatorsand 2nd order equations for n = 2 9

§1.7. The general solution of a homogeneous hyperbolic equationwith constant coefficients for n = 2 11

§1.8. Appendix: Tangent and cotangent vectors 12

§1.9. Problems 14

Chapter 2. One-dimensional wave equation 17

§2.1. Vibrating string equation 17

§2.2. Unbounded string. The Cauchy problem. D’Alembert’sformula 23

§2.3. A semi-bounded string. Reflection of waves from the end ofthe string 27

i

ii Contents

§2.4. A bounded string. Standing waves. The Fourier method(separation of variables method) 29

§2.5. Appendix: Calculus of variations and classical mechanics 36

§2.6. Problems 43

Chapter 3. The Sturm-Liouville problem 47

§3.1. Ordinary differential equations 47

§3.2. Formulation of the problem 48

§3.3. Basic properties of eigenvalues and eigenfunctions 50

§3.4. The short-wave asymptotics 54

§3.5. The Green function and completeness of the system ofeigenfunctions 56

§3.6. Problems 61

Chapter 4. Distributions 63

§4.1. Motivation of the definition. Spaces of test functions. 63

§4.2. Spaces of distributions 69

§4.3. Topology and convergence in the spaces of distributions 74

§4.4. The support of a distribution 77

§4.5. Differentiation of distributions and multiplication by a smoothfunction 81

§4.6. A general notion of the transposed (adjoint) operator. Changeof variables. Homogeneous distributions 96

§4.7. Appendix: Minkowski inequality 100

§4.8. Appendix: Completeness of distribution spaces 103

§4.9. Problems 105

Chapter 5. Convolution and Fourier transform 107

§5.1. Convolution and direct product of regular functions 107

§5.2. Direct product of distributions 109

§5.3. Convolution of distributions 112

§5.4. Other properties of convolution. Support and singular supportof a convolution 116

§5.5. Relation between smoothness of a fundamental solution andthat of solutions of the homogeneous equation 118

Contents iii

§5.6. Solutions with isolated singularities. A removable singularitytheorem for harmonic functions 122

§5.7. Estimates of derivatives for a solution of a hypoellipticequation 124

§5.8. Fourier transform of tempered distributions 127

§5.9. Applying Fourier transform to find fundamental solutions 131

§5.10. Liouville’s theorem 132

§5.11. Problems 134

Chapter 6. Harmonic functions 137

§6.1. Mean-value theorems for harmonic functions 137

§6.2. The maximum principle 139

§6.3. Dirichlet’s boundary-value problem 142

§6.4. Hadamard’s example 143

§6.5. Green’s function: first steps 146

§6.6. Symmetry and regularity of Green’s functions 150

§6.7. Green’s function and Dirichlet’s problem 155

§6.8. Explicit formulas for Green’s functions 159

§6.9. Problems 171

Chapter 7. The heat equation 175

§7.1. Physical meaning of the heat equation 175

§7.2. Boundary value problems for the heat and Laplace equations 177

§7.3. A proof that the limit function is harmonic. 179

§7.4. A solution of the Cauchy problem for the heat equation andPoisson’s integral 180

§7.5. The fundamental solution for the heat operator. Duhamel’sformula 186

§7.6. Estimates of derivatives of a solution of the heat equation 189

§7.7. Holmgren’s principle. The uniqueness of solution of theCauchy problem for the heat equation 190

§7.8. A scheme of solving the first and second initial-boundaryvalue problems by the Fourier method 193

§7.9. Problems 195

iv Contents

Chapter 8. Sobolev spaces. A generalized solution of Dirichlet’sproblem 197

§8.1. Spaces Hs(Ω) 197

§8.2. Spaces Hs(Ω) 203

§8.3. Dirichlet’s integral. The Friedrichs inequality 207

§8.4. Dirichlet’s problem (generalized solutions) 209

§8.5. Problems 216

Chapter 9. The eigenvalues and eigenfunctions of the Laplace operator219

§9.1. Symmetric and self-adjoint operators in Hilbert space 219

§9.2. The Friedrichs extension 224

§9.3. Discreteness of spectrum for the Laplace operator in abounded domain 229

§9.4. Fundamental solution of the Helmholtz operator and theanalyticity of eigenfunctions of the Laplace operator at theinterior points. Bessel’s equation 230

§9.5. Variational principle. The behavior of eigenvalues undervariation of the domain. Estimates of eigenvalues 237

§9.6. Problems 241

Chapter 10. The wave equation 245

§10.1. Physical problems leading to the wave equation 245

§10.2. Plane, spherical and cylindric waves 250

§10.3. The wave equation as a Hamiltonian system 253

§10.4. A spherical wave caused by an instant flash and a solution ofthe Cauchy problem for the 3-dimensional wave equation 259

§10.5. The fundamental solution for the three-dimensional waveoperator and a solution of the non-homogeneous waveequation 265

§10.6. The two-dimensional wave equation (the descent method) 267


Chapter 11. Properties of the potentials and their calculations 273

§11.1. Definitions of potentials 274

§11.2. Functions smooth up to Γ from each side, and theirderivatives 276

Contents v

§11.3. Jumps of potentials 284

§11.4. Calculating potentials 286


Chapter 12. Wave fronts and short-wave asymptotics for hyperbolicequations 291

§12.1. Characteristics as surfaces of jumps 291

§12.2. The Hamilton-Jacobi equation. Wave fronts, bicharacteristicsand rays 297

§12.3. The characteristics of hyperbolic equations 304

§12.4. Rapidly oscillating solutions. The eikonal equation and thetransport equations 306

§12.5. The Cauchy problem with rapidly oscillating initial data 317

Problems 323

Chapter 13. Answers and Hints. Solutions 325

Bibliography 349

Index 351

vi Contents

ABSTRACT

This book contains a course on partial differential equations which was firsttaught twice to students of experimental groups in the Department of Me-chanics and Mathematics of the Moscow State University. All problemspresented here were used in recitations accompanying the lectures. Laterthis course was also taught in Freie University (Berlin) and, in a shortenedversion, in Northeastern University (Boston).

There are about 100 problems in the book. They are not just exercises.Some of them add essential information though require no new ideas to solvethem. Hints and, sometimes, solutions are provided at the end of the book.

Preface vii

I shall be telling this with a sighSomewhere ages and ages hence:

Two roads diverged in a wood, and I –I took the one less travelled by,

And that has made all the difference.

From The Road not Takenby Robert Frost

Preface

It will not be an overstatement to say that everything we see around us(as well as hear, touch, and so on) can be described by partial differentialequations or, in a slightly more restricted way, by equations of mathematicalphysics. Among the processes and phenomena whose adequate models arebased on such equations, we encounter, for example, string vibrations, waveson the surface of water, sound, electromagnetic waves (in particular, light),propagation of heat and diffusion, gravitational attraction of planets andgalaxies, behavior of electrons in atoms and molecules, and also the BigBang that led to the creation of our Universe.

It is no wonder, therefore, that the theory of partial differential equations(PDE) forms today a vast branch of mathematics and mathematical physicsthat uses methods of all the remaining parts of mathematics (from whichthe PDE theory is inseparable). In turn, PDE influence numerous parts ofmathematics, physics, chemistry, biology and other sciences.

A disappointing conclusion is that no book, to say nothing of a textbook,can be sufficiently complete in describing this area. The only possibilityremains: to show the reader pictures at an exhibition — several typicalmethods and problems — chosen by the author’s taste. In doing so, theauthor has to severely restrict the choice of material.

This book contains, practically intact, a transcript of the PDE coursewhich I taught twice to students of experimental groups at the Departmentof Mechanics and Mathematics of the Moscow State University, and later, in

viii Contents

a shortened version, in Northeastern University (Boston). Professor Louhi-waara also taught this course in Freie Universitat (Berlin).

In Moscow, all problems, presented in this book, were used in recitationsaccompanying the lectures. There was one lecture per week during the wholeacademic year (two semesters), recitations were held once a week during thefirst semester and twice a week during the second one.

My intention was to create a course both concise and modern. Thesetwo requirements proved to be contradictory. To make the course concise, Iwas, first, forced to explain some ideas on simplest examples, and, second,to omit many topics which should be included in a modern course but wouldhave expanded the course beyond a reasonable length. With a particularregret I omit the Schrodinger equation and related material; I hoped thegap would be filled by a parallel course of quantum mechanics.

In any modern course of equations of mathematical physics, somethingshould, perhaps, have been said about nonlinear equations but I found itextremely difficult to select this something.

The bibliography at the end of the book contains a number of textbookswhere the reader can find information complementing this course. Naturally,the material of these books overlap, but it seemed difficult to do a moreprecise selection, so I let this selection to be done according to the reader’staste.

The problems in the book are not just exercises. Some of them addessential information but require no new ideas to solve them.

Acknowledgments

I am very grateful to Prof. S. P. Novikov who organized the experimentalgroups and invited me to teach PDE to them. I am also greatly indebtedto professors of the Chair of Differential Equations of the Moscow StateUniversity for many valuable discussions.

I owe a lot to Prof. Louhiwaara who provided unexpectedly generoushelp in editing of an early version of this text, and encouraged me to publishthis book.

I thank Ognen Milatovic for help in verifying problems.

I am also grateful to Prof. D. A. Leites who translated the book intoEnglish.

Selected notational conventions ix

Selected notational conventions

Cross-references inside the book are natural: Theorem 6.2 stands for Theo-rem 2 in Sect. 6, Problem 6.2 is Problem 2 from Sect. 6, etc.

:= means “denotes by definition”.

≡ means “identically equal to”

By Z, Z+, N, R, and C we will denote the sets of all integers, non-negative integers, positive integers, real, and complex numbers respectively.

I denotes the identity operator.

The Heaviside function is θ(z) =

1 for z ≥ 0

0 for z < 0.

Chapter 1

Linear differential

operators

1.1. Definition and examples

We will start by introducing convenient notations for functions of severalvariables and differential operators. By Z,R and C we will denote the setsof integer, real and complex numbers respectively. A multi-index α is ann-tuple α = (α1, . . . , αn), where αj ∈ Z+ (i.e. αj are non-negative integers).If α is a multi-index, then we set |α| = α1 + . . .+αn and α! = α1! . . . αn!; fora vector x = (x1, . . . , xn) ∈ Cn we define xα = xα1

1 . . . xαnn . If Ω is an opensubset of Rn, then C∞(Ω) denotes the space of complex-valued infinitelydifferentiable functions on Ω, and C∞0 (Ω) denotes the subspace of infinitelydifferentiable functions with compact support, i.e. ϕ ∈ C∞0 (Ω) if ϕ ∈ C∞(Ω)and there exists a compact K ⊂ Ω such that ϕ vanishes on Ω \K. (Here Kdepends on ϕ.)

Denote ∂j = ∂∂xj

: C∞(Ω) −→ C∞(Ω);Dj = −i∂j , where i =√−1.

∂ = (∂1, . . . , ∂n);D = (D1, . . . , Dn);

∂α = ∂α11 . . . ∂αnn ;Dα = Dα1

1 . . . Dαnn .

Therefore, ∂α = ∂|α|

∂xα11 ...∂xαnn

is a mixed derivative operator ∂α : C∞(Ω) −→C∞(Ω), Dα = i−|α|∂α.

If f ∈ C∞(Ω), then instead of ∂αf we will sometimes write f (α).

1

2 1. Linear differential operators

Exercise 1.1. Let f ∈ C[x1, . . . , xn], i.e. f is a polynomial of n variablesx1, . . . , xn. Prove the Taylor formula

(1.1) f(x) =∑α

f (α)(x0)α!

(x− x0)α,

where the sum runs over all multi-indices α. (Actually, the sum is finite, i.e.only a finite number of terms may be non-vanishing.)

A linear differential operator is an operator A : C∞(Ω) −→ C∞(Ω) ofthe form

(1.2) A =∑|α|≤m

aα(x)Dα,

where aα(x) ∈ C∞(Ω). Clearly, instead of Dα we may write ∂α, but theexpression in terms of Dα is more convenient as we will see in the future.Here m ∈ Z+, and we will say that A is an operator of order ≤ m. We saythat A is an operator of order m if it can be expressed in the form (1.2) andthere exists a multi-index α such that |α| = m and aα(x) 6≡ 0.

Examples.

1) The Laplace operator or Laplacian ∆ = ∂2

∂x21

+ . . . + ∂2

∂x2n

= −(D21 +

. . .+D2n).

2) The heat operator ∂∂t −∆ (here the number of independent variables

is equal to n+ 1 and they are denoted by t, x1, . . . , xn).

3) The wave operator or d’Alembertian = ∂2

∂t2−∆.

4) The Sturm-Liouville operator L defined for n = 1 and Ω = (a, b) bythe formula

Lu =d

dx

(p(x)

du

dx

)+ q(x)u,

where p, q ∈ C∞((a, b)). Operators in Examples 1)-3) are operators withconstant coefficients. The Sturm-Liouville operator has variable coefficients.

1.2. The total and the principal symbols

The symbol or the total symbol of the operator (1.2) is

(1.3) a(x, ξ) =∑|α|≤m

aα(x)ξα, x ∈ Ω, ξ ∈ Rn,

1.2. The total and the principal symbols 3

and its principal symbol is the function

(1.4) am(x, ξ) =∑|α|=m

aα(x)ξα, x ∈ Ω, ξ ∈ Rn.

The symbol belongs to C∞(Ω)[ξ], i.e., it is a polynomial in ξ1, . . . , ξn withcoefficients in C∞(Ω) and the principal symbol is a homogeneous polynomialof degree m in ξ1, . . . , ξn with coefficients in C∞(Ω).

Examples.

1) The total and the principal symbols of the Laplace operator coincideand are equal to −ξ2 = −(ξ2

1 + . . .+ ξ2n).

2) The total symbol of the heat operator ∂∂t −∆ is a(t, x, τ, ξ) = iτ + ξ2

and its principal symbol is a2(t, x, τ, ξ) = ξ2.

3) The total and the principal symbols of the wave operator coincideand are equal to

a(t, x, τ, ξ) = a2(t, x, τ, ξ) = −τ2 + ξ2.

4) The total symbol of the Sturm-Liouville operator is

a(x, ξ) = −p(x)ξ2 + ip′(x)ξ + q(x);

its principal symbol is a2(x, ξ) = −p(x)ξ2.

The symbol of an operator A is recovered from the operator by theformula

(1.5) a(x, ξ) = e−ix·ξA(eix·ξ),

where A is applied with respect to x. Here we used the notation x · ξ =x1ξ1 + . . .+ xnξn. Formula (1.5) is obtained from the relation

Dαeix·ξ = ξαeix·ξ,

which is easily verified by induction with respect to |α|. Formula (1.5) alsoimplies that the coefficients aα(x) of A are uniquely defined by this operator,because the coefficients of the polynomial a(x, ξ) for each fixed x are uniquelydefined by this polynomial viewed as a function of ξ.

It is convenient to apply A to a more general exponent than eix·ξ, namelyto eiλϕ(x), where ϕ ∈ C∞(Ω) and λ is a parameter. Then we get

Lemma 1.1. If f ∈ C∞(Ω), then e−iλϕ(x)A(f(x)eiλϕ(x)) is a polynomial ofλ of degree ≤ m with coefficients in C∞(Ω) and

(1.6) e−iλϕ(x)A(f(x)eiλϕ(x)) = λmf(x)am(x, ϕx) + λm−1(. . .) + . . . ,


i.e., the highest coefficient of this polynomial (that of λm) is equal to

f(x)am(x, ϕx), where ϕx = ( ∂ϕ∂x1, . . . , ∂ϕ∂xn ) = gradϕ is the gradient of ϕ.

Proof. We have

Dj [f(x)eiλϕ(x)] = λ(∂jϕ)(x)eiλϕ(x) + (Djf)eiλϕ(x),

so that the Lemma holds for the operators of order ≤ 1. In what follows, tofind the derivatives Dαeiλϕ(x) we will have to differentiate products of theform f(x)eiλϕ(x), where f ∈ C∞(Ω). A new factor λ will only appear whenwe differentiate the exponent. Clearly,

Dαeiλϕ(x) = λ|α|ϕαxeiλϕ(x) + λ|α|−1(. . .) + . . . ,

implying the statement of the Lemma.

Corollary 1.2. Let A, B be two linear differential operators in Ω, let k, lbe their orders, ak, bl their principal symbols, C = A B, their composition,ck+l the principal symbol of C. Then

(1.7) ck+l(x, ξ) = ak(x, ξ)bl(x, ξ).

Remark. Clearly, C is a differential operator of order ≤ k + l.

Proof of Corollary 1.2. We have

C(eiλϕ(x)) = A(λlbl(x, ϕx)eiλϕ(x) + λl−1(. . .) + . . .) =

= λk+lak(x, ϕx)bl(x, ϕx)eiλϕ(x) + λk+l−1(. . .) + . . . .

By Lemma 1.1, we obtain

(1.8) ck+l(x, ϕx) = ak(x, ϕx)bl(x, ϕx), for any ϕ ∈ C∞(Ω).

But then, by choosing ϕ(x) = x · ξ, we get ϕx = ξ and (1.8) becomesck+l(x, ξ) = ak(x, ξ)bl(x, ξ), as required.

1.3. Change of variables

Let κ : Ω −→ Ω1 be a diffeomorphism between two open subsets Ω,Ω1 inRn, i.e. κ is a C∞ map having an inverse C∞ map κ−1 : Ω1 −→ Ω. Wewill denote the coordinates in Ω by x and coordinates in Ω1 by y. The mapκ can be defined by a set of functions y1(x1, . . . , xn), . . . , yn(x1, . . . , xn),yj ∈ C∞(Ω), whose values at x are equal to the respective coordinates ofthe point κ(x).

1.3. Change of variables 5

If f ∈ C∞(Ω1), set κ∗f = f κ, i.e., (κ∗f)(x) = f(κ(x)) or (κ∗f)(x) =f(y1(x), . . . , yn(x)). Function κ∗f is obtained from f by a change of vari-ables or by passage to coordinates x1, . . . , xn from coordinates y1, . . . , yn.Recall that the diffeomorphism κ has the inverse κ−1 which is also a diffeo-morphism. Clearly, κ∗ is a linear isomorphism κ∗ : C∞(Ω1) −→ C∞(Ω) and(κ∗)−1 = (κ−1)∗.

Given an operator A : C∞(Ω) −→ C∞(Ω), define the operator A1 :C∞(Ω1) −→ C∞(Ω1) with the help of the commutative diagram

C∞(Ω) A−→ C∞(Ω)xκ∗ xκ∗C∞(Ω1) A1−→ C∞(Ω1)

i.e., by setting A1 = (κ∗)−1Aκ∗ = (κ−1)∗Aκ∗. In other words,

(1.9) A1v(y) = A[v(y(x))]|x=x(y),

where y(x) = κ(x), x(y) = κ−1(y). Therefore, the operator A1 is, actually,just the expression of A in coordinates y.

If A is a linear differential operator, then so is A1, as easily follows fromthe chain rule and (1.9). Let us investigate a relation between the principalsymbols of A and A1. To this end we will use the notions of tangent andcotangent vectors in Ω, as well as tangent and cotangent bundles over Ω (seeAppendix to this Chapter).

Theorem 1.3. The value of the principal symbol am of an operator A atthe cotangent vector (x, ξ) is the same as that of the principal symbol a′m ofA1 at the corresponding vector (y, η), i.e.,

(1.10) am(x, ξ) = a′m(κ(x), (tκ∗)−1ξ).

In other words, the principal symbol is a well-defined function on T ∗Ω(it does not depend on the choice of C∞-coordinates in Ω).

Proof. A cotangent vector at x can be expressed as the gradient ϕx(x)of a function ϕ ∈ C∞(Ω) at x. It is clear from Lemma 1.1 that am(x, ϕx(x))does not depend on the choice of coordinates. But on the other hand, it isclear that this value does not depend on the choice of a function ϕ with aprescribed differential at x. Therefore, the principal symbol is well-definedon T ∗Ω.


1.4. The canonical form of second order operators with

constant coefficients

By a change of variables we can try to transform an operator to a simplerone. Consider a second order operator with constant coefficients of theprincipal part:

A = −n∑

i,k=1

aik∂2

∂xi∂xk+

n∑l=1

bl∂

∂xl+ c,

where aik are real constants such that aik = aki (the latter can always beassumed since ∂2

∂xi∂xk= ∂2

∂xk∂xi). Ignoring the terms of lower order, we will

transform the highest order part of A

A0 = −n∑

i,k=1

aik∂2

∂xi∂xk

to a simpler form with the help of a linear change of variables:

(1.11) yk =n∑l=1

cklxl,

where ckl are real constants. Consider the quadratic form

Q(ξ) =n∑

i,k=1

aikξiξk,

which only differs from the principal symbol of A by a sign.

With the help of a linear change of variables η = Fξ, where F is anonsingular constant matrix, we can transform Q(ξ) to the following form

(1.12) Q(ξ) = (±η21 ± η2

2 ± . . .± η2r )|η=Fξ.

Denote by C the matrix (ckl)nk,l=1 in (1.11). By Theorem 1.3, the operator Ain coordinates y has the form of a second order operator A1 whose principalsymbol is a quadratic form Q1(η) such that

Q(ξ) = Q1(η)|η=(tC)−1ξ,

where tC is the transpose of C. From this and (1.12) it follows that weshould choose C so that (tC)−1 = F or

C = (tF )−1.

1.5. Characteristics. Ellipticity and hyperbolicity 7

Then the principal part of A under the change of variables y = Cx of theform (1.11) will be reduced to the form

(1.13) ± ∂2

∂y21

± ∂2

∂y22

± . . .± ∂2

∂y2r

,

called the canonical form.

Remark. An operator with variable coefficients may be reduced to theform (1.13) at any fixed point by a linear change of variables.

1.5. Characteristics. Ellipticity and hyperbolicity

Let A be anm-th order differential operator, am(x, ξ) its principal symbol. Anonzero cotangent vector (x, ξ) is called a characteristic vector if am(x, ξ) =0. A surface (of codimension 1) in Ω is called characteristic at x0 if itsnormal vector at this point is a characteristic vector. The surface is calleda characteristic if it is characteristic at every point.

If a surface S is defined by the equation ϕ(x) = 0, where ϕ ∈ C1(Ω),ϕx|S 6= 0, then S is characteristic at x0 if am(x0, ϕx(x0)) = 0. The surfaceS is a characteristic if am(x, ϕx(x)) identically vanishes on S.

All level surfaces ϕ = const are characteristics if and only if am(x, ϕx(x)) ≡0.

Theorem 1.3 implies that the notion of a characteristic does not dependon the choice of coordinates in Ω.

Examples.

1) The Laplace operator ∆ has no real characteristic vectors.

2) For the heat operator ∂∂t−∆ the vector (τ, ξ) = (1, 0, . . . , 0) ∈ Rn+1 is

a characteristic one. The surfaces t = const are characteristics. The surfacet = |x|2 (paraboloid) is characteristic at a single point only (the origin).

3) Consider the wave operator ∂2

∂t2−∆. Its characteristic vectors at each

point (t, x) constitute a cone τ2 = |ξ|2. Any cone (t − t0)2 = |x − x0|2 is acharacteristic. In particular, for n = 1 (i.e. for x ∈ R1) the lines x+t = constand x− t = const are characteristics.

Definition.

a) An operator A is called elliptic if am(x, ξ) 6= 0 for all x ∈ Ω and allξ ∈ Rn \ 0, i.e., if A has no non-zero characteristic vectors.


b) An operator A in the space of functions of t, x, where t ∈ R1, x ∈ Rn,is hyperbolic with respect to t if the equation am(t, x, τ, ξ) = 0 consideredas an equation in τ for any fixed t, x, ξ has exactly m distinct real roots ifξ 6= 0. In this case one might say that the characteristics of A are real anddistinct.

Examples.

a) The Laplace operator ∆ is elliptic.

b) The heat operator is neither elliptic nor hyperbolic with respect to t.

c) The wave operator is hyperbolic with respect to t since the equationτ2 = |ξ|2 has two different real roots τ = ±|ξ| if ξ 6= 0.

d) The Sturm-Liouville operator Lu ≡ ddx(p(x) d

dxu)+ q(x)u is elliptic on(a, b) if p(x) 6= 0 for x ∈ (a, b).

By Theorem 1.3, the ellipticity of an operator does not depend on thechoice of coordinates. The hyperbolicity with respect to t does not dependon the choice of coordinates in Rn

x.

Let us see what it means for the surface x1 = const to be characteristic.The coordinates of the normal vector are (1, 0, . . . , 0). Substituting theminto the principal symbol, we get

am(x; 1, 0, . . . , 0) =∑|α|=m

aα(x)(1, 0, . . . , 0)α = a(m,0,...,0)(x),

i.e., the statement “the surface x1 = const is characteristic at x” means thata(m,0,...,0)(x), the coefficient of (1

i )m ∂m

∂xm1, vanishes at x.

All surfaces x1 = const are characteristics if and only if a(m,0,...,0) ≡ 0.

This remark will be used below to reduce second order operators withtwo independent variables to a canonical form.

It is also possible to find characteristics by solving the Hamilton-Jacobiequation am(x, ϕx(x)) = 0. If ϕ is a solution of this equation, then allsurfaces ϕ = const are characteristic. Integration of the Hamilton-Jacobiequation is performed with the help of the Hamiltonian system of ordinarydifferential equations with the Hamiltonian am(x, ξ) (see Chapter 12 or anytextbook on classical mechanics, e.g., Arnold [1]).

1.6. Characteristics and the canonical form (n = 2) 9

1.6. Characteristics and the canonical form of 2nd order

operators and 2nd order equations for n = 2

For n = 2 characteristics are curves and are rather easy to find. For example,consider a 2nd order operator

(1.14) A = a∂2

∂x2+ 2b

∂2

∂x∂y+ c

∂2

∂y2+ . . . ,

where a, b, c are smooth functions of x, y defined in an open set Ω ⊂ R2 anddots stand for terms containing at most 1st order derivatives. Let (x(t), y(t))be a curve in Ω, (dx, dy) its tangent vector, (−dy, dx) the normal vector.The curve is a characteristic if and only if

a(x, y)dy2 − 2b(x, y)dxdy + c(x, y)dx2 = 0

on it.

If a(x, y) 6= 0, then in a neighborhood of the point (x, y) we may assumethat dx 6= 0; hence, x can be taken as a parameter along the characteris-tic which, therefore, is of the form y = y(x). Then the equation for thecharacteristics is of the form

ay′2 − 2by′ + c = 0.

If b2 − ac > 0, then the operator (1.14) is hyperbolic and has 2 families ofreal characteristics found from ordinary differential equations

(1.15) y′ =b+√b2 − aca

,

(1.16) y′ =b−√b2 − aca

.

Observe that in this case two nontangent characteristics pass througheach point (x, y) ∈ Ω. We express these families of characteristics in theform ϕ1(x, y) = C1 and ϕ2(x, y) = C2, where ϕ1, ϕ2 ∈ C∞(Ω). This meansthat ϕ1, ϕ2 are first integrals of equations (1.15) and (1.16), respectively.Suppose that gradϕ1 6= 0 and gradϕ2 6= 0 in Ω. Then gradϕ1 and gradϕ2

are linearly independent, since characteristics of different families are nottangent. Let us introduce new coordinates: ξ = ϕ1(x, y), η = ϕ2(x, y). Inthese coordinates, characteristics are lines ξ = const and η = const. Butthen the coefficients of ∂2

∂ξ2and ∂2

∂η2 vanish identically and A takes the form

(1.17) A = p(ξ, η)∂2

∂ξ∂η+ . . . ,


called the canonical form. Here p(ξ, η) 6= 0 (recall that we have assumeda 6= 0).

If c(x, y) 6= 0 then the reduction to the canonical form (1.17) is similar.It is not necessary to consider these two cases separately, since we haveonly used the existence of integrals ϕ1(x, y), ϕ2(x, y) with the describedproperties. These integrals can be also found when coefficients a and c

vanish (at one point or on an open set). This case can be reduced to one ofthe above cases, e.g. by a rotation of coordinate axes (if a2 + b2 + c2 6= 0).

One often considers differential equations of the form

(1.18) Au = f,

where f is a known function, A a linear differential operator, and u anunknown function. If A is a hyperbolic 2nd order operator with two in-dependent variables, i.e. an operator of the form (1.14 with b2 − ac > 0)(in this case the equation (1.18) is also called hyperbolic), then, introducingthe above coordinates ξ, η and dividing by p(ξ, η), we reduce (1.18) to thecanonical form

∂2u

∂ξ∂η+ . . . = 0,

where dots stand for terms without 2nd order derivatives of u.

Now let b2 − ac ≡ 0 (then the operator (1.14) and the equation (1.18)with this operator are called parabolic). Let us assume that a 6= 0. Then,for characteristics we get the differential equation

(1.19) y′ =b

a.

Let us find characteristics and write them in the form ϕ(x, y) = const, whereϕ is the first integral of (1.19) and gradϕ 6= 0. Choose a function ψ ∈ C∞(Ω)such that gradϕ and gradψ are linearly independent and introduce newcoordinates ξ = ϕ(x, y), η = ψ(x, y). In the new coordinates the operatorA does not have the term ∂2

∂ξ2, since the lines ϕ = const are characteristics.

But then the coefficient of the term ∂2

∂ξ∂η also vanishes, since the principalsymbol should be a quadratic form of rank 1. Thus, the canonical form ofa parabolic operator is

A = p(ξ, η)∂2

∂η2+ . . .

For a parabolic equation (1.18) the canonical form is

∂2

∂η2+ . . . = 0.

1.7. The general solution (n = 2) 11

Observe that if b2− ac = 0 and a2 + b2 + c2 6= 0, then a and c cannot vanishsimultaneously, since then we would also have b = 0. Therefore, we alwayshave either a 6= 0 or c 6= 0 and the above procedure is always applicable.

Finally, consider the case b2−ac < 0. The operator (1.14) and the equa-tion (1.18) are in this case called elliptic ones. Suppose for simplicity thatfunctions a, b, c are real analytic. Then the existence theorem for complexanalytic solutions of the complex equation

y′ =b+√b2 − aca

implies the existence of a local first integral

ϕ(x, y) + iψ(x, y) = C,

where ϕ,ψ are real-valued analytic functions, gradϕ and gradψ are linearlyindependent. It is easy to verify that by introducing new coordinates ξ =ϕ(x, y), η = ψ(x, y), we can reduce A to the canonical form

A = p(ξ, η)(∂2

∂ξ2+

∂2

∂η2

)+ . . . ,

where p(ξ, η) 6= 0. Dividing (1.18) by p, we reduce it to the form

∂2u

∂ξ2+∂2u

∂η2+ . . . = 0

1.7. The general solution of a homogeneous hyperbolic

equation with constant coefficients for n = 2

As follows from Section 1.6, the hyperbolic equation

(1.20) a∂2u

∂x2+ 2b

∂2u

∂x∂y+ c

∂2u

∂y2= 0,

where a, b, c ∈ R and b2− ac > 0, can be reduced by the change of variables

ξ = y − λ1x, η = y − λ2x,

where λ1, λ2 are roots of the quadratic equation aλ2 − 2bλ + c = 0, to theform

∂2u

∂ξ∂η= 0.

Assuming u ∈ C2(Ω), where Ω is a convex open set in R2, we get ∂∂ξ (∂u∂η ) = 0

implying ∂u∂η = F (η) and

u = f(ξ) + g(η),


where f, g are arbitrary functions of class C2. In variables x, y we have

(1.21) u(x, y) = f(y − λ1x) + g(y − λ2x).

It is useful to consider functions u(x, y) of the form (1.21), where f, g are notnecessarily of the class C2 but from a wider class, e.g. f, g ∈ L1

loc(R), i.e.,f, g are locally integrable. Such functions u are called generalized solutionsof equation (1.20). For example let f(ξ) have a jump at ξ0, i.e., has differentlimits as ξ → ξ0+ and as ξ → ξ0−. Then u(x, y) has a discontinuity alongthe line y − λ1x = ξ0.

Note that the lines y − λ1x=const, y − λ2x = const are characteristics.Therefore, in this case discontinuities of solutions are propagating alongcharacteristics. Similar phenomenon occurs in the case of general hyperbolicequations.

Example. Characteristics of the wave equation ∂2u∂t2

= a2 ∂2u∂x2 are x − at =

const, x+at = const. The general solution of this equation can be expressedin the form

u(t, x) = f(x− at) + g(x+ at).

Observe that f(x − at) is the wave running to the right with the speed a

and g(x + at) is the wave running to the left with the same speed a. Thegeneral solution is the sum (or superposition) of two such waves.

1.8. Appendix: Tangent and cotangent vectors

Let Ω be an open set in Rn. The tangent vector to Ω at a point x ∈ Ωis a vector v which is tangent to a parametrized curve x(t) in Ω passingthrough x at t = 0. Here we assume that the function t 7→ x(t) is a C∞

funciton defined for t ∈ (−ε,+ε) with some ε > 0, and takes values in Ω.In coordinates x1, . . . , xn we have x(t) = (x1(t), . . . , xn(t)) and the tangentvector v is given as

v = x(0) = dxdt

∣∣t=0

= (x1(0), . . . , xn(0))

=(dx1

dt

∣∣∣∣t=0

, . . . ,dxndt

∣∣∣∣t=0

)= (v1, . . . , vn).

Interpreting t as the time variable, we can say that the tangent vectorv = x(0) is the velocity vector at the moment t = 0 for the motion given bythe curve x(t).

1.8. Appendix: Tangent and cotangent vectors 13

The real numbers v1, . . . , vn are called the coordinates of the tangent vec-tor x(0). They depend on the choice of the coordinate system (x1, . . . , xn).In fact, they are defined even for any curvilinear coordinate system.

Two tangent vectors at x can be added (by adding the correspondingcoordinates of these vectors), and any tangent vector can be multiplied byany real number (by multiplying each of its coordinates by this number).These operations do not depend on the choice of coordinates in Ω. So alltangent vectors at x ∈ Ω form a well defined vector space (independent ofcoordinates in Ω), which will be denoted TxΩ and called tangent space to Ωat x.

A cotangent vector ξ at x ∈ Ω is a real-valued linear function on thetangent space TxΩ. All cotangent vectors at x form a linear space which isdenoted T ∗xΩ and called cotangent space to Ω at x.

Denote by TΩ the union of all tangent spaces over all points x ∈ Ω, i.e.

TΩ =⋃x

TxΩ ∼= Ω× Rn.

It is called the tangent bundle of Ω. Similarly, we will introduce the cotangentbundle

T ∗Ω =⋃x

T ∗xΩ ∼= Ω× Rn.

For every tangent vector v at x ∈ Ω and any function f ∈ C∞(Ω) wecan define the derivative of f in the direction v: it is

vf =df(x(t))dt

=n∑j=1

∂f(x)∂xj

xj(0) =n∑j=1

vj∂f(x)∂xj

,

where x(t) is a curve in Ω such that x(0) = x and x(0) = v. Sometimes, vf isalso called directional derivative. Taking f to be a coordinate function xj ,we obtain vxj = vj , so v is uniquely defined by the corresponding directionalderivative, and we can identify v with its directional derivative which is alinear function on C∞(Ω).

In particular, take vectors at x ∈ Ω which are tangent to the coordi-nate axes. Then the corresponding directional derivatives will be simply∂∂x1

, . . . , ∂∂xn

, and they form a basis of TxΩ. (For example, ∂∂x1

is tangent tothe curve x(t) = (x1 + t, x2, . . . , xn), and has coordinates (1, 0, . . . , 0). Forthe jth partial derivative all coordinates vanish except jth one which equals1.)


The dual basis in T ∗xΩ consists of linear functions dx1, . . . , dxn on TxΩsuch that dxi( ∂

∂xj) = δij , where δij = 1 for i = j, and δij = 0 for i 6= j.

For any tangent vector x(0) ∈ Tx(0)Ω which is the velocity vector of aC∞ curve x(t) at t = 0, its coordinates in the basis ∂

∂x1, . . . , ∂

∂xnare equal

to dx1dt |t=0, . . . ,

dxndt |t=0. An example of a cotangent vector is given by the

differential df of a function f ∈ C∞(Ω): it is a function on Tx(0)Ω whosevalue at the tangent vector x(0) is equal to df(x(t))

dt |t=0 and whose coordinatesin the basis dx1, . . . , dxn are ∂f

∂x1, . . . , ∂f∂xn , so

df =n∑j=1

∂f

∂xjdxj .

In euclidean coordinates, the differential of a smooth function f can beidentified with its gradient ∇f , which is the tangent vector with the samecoordinates.

Given a diffeomorphism κ : Ω −→ Ω1, we have a natural map κ∗ :TxΩ −→ Tκ(x)Ω1 (the tangent map or the differential of κ). Namely, for atangent vector x(0), which is the velocity vector of a curve x(t) at t = 0, thevector κ∗x(0) is the velocity vector of the curve (κx)(t) = κ(x(t)) at t = 0.The corresponding dual map of spaces of linear functions tκ∗ : T ∗κ(x)Ω1 −→T ∗xΩ is defined by (tκ∗ξ)(x(0)) = ξ(κ∗x(0)), where ξ ∈ T ∗κ(x)Ω.

Choose bases ∂∂xjnj=1, ∂

∂yjnj=1, dyjnj=1, dxjnj=1 in spaces TxΩ,

Tκ(x)Ω1, T ∗κ(x)Ω1, T ∗xΩ respectively. In these bases the matrix of κ∗ is

(κ∗)kl = ∂yk∂xl

. It is called the Jacobi matrix. The matrix of tκ∗ is transposeof κ∗. Since κ is a diffeomorphism, the maps κ∗ and tκ∗ are isomorphismsat each point x ∈ Ω.

It is convenient to define a cotangent vector at x by a pair (x, ξ), whereξ = (ξ1, . . . , ξn), ξ ∈ Rn, is the set of coordinates of this vector in thebasis dxjnj=1 in T ∗xΩ. Under the isomorphism (tκ∗)−1 : T ∗Ω −→ T ∗Ω1

the covector (y, η) corresponds to the covector (x, ξ), where y = κ(x), η =(tκ∗)−1ξ.

1.9. Problems

1.1. Reduce the equations to the canonical form:

a) uxx + 2uxy − 2uxz + 2uyy + 6uzz = 0;

b) uxy − uxz + ux + uy − uz = 0.

1.9. Problems 15

1.2. Reduce the equations to the canonical form:

a) x2uxx + 2xyuxy − 3y2uyy − 2xux + 4yuy + 16x4u = 0;

b) y2uxx + 2xyuxy + 2x2uyy + yuy = 0;

c) uxx − 2uxy + uyy + ux + uy = 0.

1.3. Find the general solution of the equations:

a) x2uxx − y2uyy − 2yuy = 0;

b) x2uxx − 2xyuxy + y2uyy + xux + yuy = 0.

Chapter 2

One-dimensional wave

equation

2.1. Vibrating string equation

Let us derive an equation describing small vibrations of a string. Note rightaway that our derivation will be not mathematical but rather physical ormechanical. However its understanding is essential to grasp the physicalmeaning of, first, the wave equation itself, and second, but not less impor-tantly, the initial and boundary conditions. A knowledge of the derivationand physical meaning helps also to find various mathematical tools in thestudy of this equation (the energy integral, standing waves, etc.). Generally,derivation of equations corresponding to different physical and mechanicalproblems is important for understanding mathematical physics and is essen-tially a part of it.

So, let us derive the equation of small vibrations of a string. We con-sider the vibrations of a taut string such that each point moves in directionperpendicular to the direction of the string in its equilibriuim. We assumehere that all the forces appearing in the string are negligible compared tothe tension directed along the string (we assume the string to be perfectlyflexible, i.e., nonresistant to bending).

First of all, choose the variables describing the form of the string. Sup-pose that in equilibrium the taut string is stretched along the x-axis. For thestart, we will consider the inner points of the string disregarding its ends.

17

18 2. One-dimensional wave equation

x

y

x

u(t;x)

Figure 1. Vibrating string

Suppose that the vibrations keep the string in the plane (x, y) so thateach point of the string is only shifted parallel to y-axis and this shift at thetime t equals u(t, x), see Fig. 1.

Therefore, at a fixed t the graph of u(t, x), as a function of x, is theshape of the string at the moment t, and for a fixed x the function u(t, x)describes the motion of one point of the string. Let us compute the lengthof the part of the string corresponding to the interval (a, b) on the x-axis.It is equal to ∫ b

a

√1 + u2

xdx.

Our main assumption is the negligibility of the additional string lengthen-ing, compared with the equilibrium length. More precisely, we assume thatu2x << 1 and consider u2

x negligible compared to 1. Note that if α = α(t, x) isthe angle between the tangent to the string and the x-axis then tanα = ux,cosα = 1√

1+u2x

, sinα = ux√1+u2

x

; under our assumptions we should consider

cosα ≈ 1 and sinα ≈ ux. If T = T (t, x) is the string’s tension, then itshorizontal component is T cosα ≈ T and the vertical one is T sinα ≈ Tux.

By the Hooke’s law in elasticity theory (which should be considered anassumption in our model) the tension of the string at any point and atany given moment of time is proportional to the relative lengthening of asmall piece of string at this point, i.e. the lengthening (the increase of thelength compared with the equilibrium) divided by the equilibrium length ofthis piece. Since we assume the additional lengthening to be negligible, weshould have T (t, x) = T = const. (Alternatively, we can argue as follows.Since all points of the string move in the vertical direction, along the y-axis,the sum of the horizontal components of all forces acting on the segmentof the string over [a, b] should vanish. This means that T (t, a) = T (t, b)which due to the arbitrariness of a and b yields T (t, x) = T (t), that is, T isindependent of x. Then, the negligibility of the lengthening implies that infact T (t) = T = const.)

2.1. Vibrating string equation 19

x

y

a b

T T

Figure 2. Tension forces in the string

Let us write the equation of motion of the part of the string above [a, b],see Fig. 2. Let ρ(x) be the linear density of the string at x (the ratio of themass of the infinitesimal segment to the length of this segment at x). Thevertical component of the force is

(2.1) (Tux)|x=b − (Tux)|x=a =∫ b

aTuxx(t, x)dx,

assuming the absence of external forces. In the presence of vertical externalforces distributed with density g(t, x) (per unit mass of the string), we shouldadd to (2.1) the following term

(2.2)∫ b

aρ(x)g(t, x)dx.

The vertical component of the momentum of this segment of the string isequal to

(2.3)∫ b

aρ(x)ut(t, x)dx.

Let us use the well-known corollary of the Newton’s Second and ThirdLaws which states that the rate of momentum’s change in time is equal tothe sum of all external forces. (See Appendix to this Chapter.) Then (2.1)– (2.3) give ∫ b

a[ρ(x)utt(t, x)− Tuxx(t, x)− ρ(x)g(t, x)]dx = 0,

or, since a and b are arbitrary,

(2.4) ρ(x)utt − Tuxx − ρ(x)g(t, x) = 0.

In particular, for ρ(x) = ρ = const and g(t, x) ≡ 0 we get

(2.5) utt − a2uxx = 0, where a =√T/ρ.


We would have arrived at the same equation (2.5) if we had applied d’Alembert’sprinciple equating the sum of all the external and inertial forces to zero.

One may arrive at (2.4) via another route by using Lagrange’s equations.(See Appendix to this Chapter for an elementary introduction to the Cal-culus of variations and Lagrange’s equations.) Suppose the external forcesare absent. Clearly, the kinetic energy of the interval (a, b) of the string is

K =12

∫ b

aρ(x)u2

t (t, x)dx.

To calculate the potential energy of the string whose form is that of the graphof v(x), x ∈ (a, b), we should calculate the work necessary to move the stringfrom the equilibrium position to v(x). Suppose this move is defined by the“curve” v(x, α), α ∈ [0, 1], so that v(x, 0) = 0, v(x, 1) = v(x). The force

Tvx(x+ ∆x, α)− Tvx(x, α) =∫ x+∆x

xTvxx(x, α)dx,

acts on the part of the string corresponding to the interval (x, x + ∆x) onthe x-axis.

As α varies from α to α + ∆α, the displacement of the point with theabscissa x equals

v(x, α+ ∆α)− v(x, α) =∫ α+∆α

αvα(x, α)dα.

Therefore, to move the part of the string (x, x+ ∆x) from α to α+ ∆α, thework

−Tvxx(x, α)vα(x, α)∆x ·∆α+ o(∆x ·∆α)

should be performed. Integrating with respect to x and α we see that thetotal work performed over the segment (a, b) is

A = −∫ 1

0

∫ b

aTvxxvαdxdα.

Integrating by parts, we get

−∫ b

aTvxxvαdx = −Tvxvα|ba +

∫ b

aTvxvxαdx = −Tvxvα|ba +

d

dα

∫ b

aT

12v2xdx

and integrating with respect to α from 0 to 1, we get

(2.6) A = −∫ 1

0Tvx(x, α)vα(x, α)|x=b

x=adα+12

∫ b

aTv2

x(x)dx.

2.1. Vibrating string equation 21

Suppose that the ends of the string are fixed at the points 0 and l, i.e., theirdisplacements are equal to zero during the whole process of motion. Thenwe may assume that v(0, α) = v(l, α) = 0 for α ∈ [0, 1], and write

A =12

∫ l

0Tv2

x(x)dx

instead of (2.6) for a = 0, b = l. This clearly implies that the potentialenergy of the string with fixed ends 0, l at time t is equal to

U =12

∫ l

0Tu2

x(t, x)dx.

Now we can write the Lagrangian of the string:

L = K − U =12

∫ l

0ρ(x)u2

t (t, x)dx− 12

∫ l

0Tu2

x(t, x)dx

which is a functional in u and ut (here u(t, x) plays the role of the set ofcoordinates at time t and ut(t, x) the role of the set of velocities). Let uswrite the action

∫ t1t0Ldt and equate the variation of the action with respect to

u to 0. The usual integration by parts under the assumption that δu|t=t0 =δu|t=t1 = 0 gives:

δ

∫ t1

t0

Ldt =∫ t1

t0

∫ l

0ρ(x)ut(t, x)δut(t, x)dxdt−

−∫ t1

t0

∫ l

0Tux(t, x)δux(t, x)dxdt =

= −∫ t1

t0

∫ l

0[ρ(x)utt(t, x)− Tuxx(t, x)]δu(t, x)dxdt,

which, since δu is arbitrary, implies

(2.7) ρ(x)utt(t, x)− Tuxx(t, x) = 0.

This means that we obtained equation (2.4) (with g(t, x) ≡ 0).

Note an important circumstance here: the total energy of the string isequal to

H = K + U =12

∫ l

0ρ(x)u2

t (t, x)dx+12

∫ l

0Tu2

x(t, x)dx.

Let us prove the energy conservation law: the energy of an oscillating stringwith fixed ends is conserved. We will verify this law by using (2.7). We have

dH

dt=∫ l

0ρ(x)ut(t, x)utt(t, x)dx+

∫ l

0Tux(t, x)utx(t, x)dx.


Integrating by parts in the second integral, we get

(2.8)dH

dt=∫ l

0ut(t, x)[ρ(x)utt(t, x)−Tuxx(t, x)]dx+Tux(t, x)ut(t, x)|x=l

x=0.

The last summand in (2.8) vanishes due to the boundary conditions u|x=0 =u|x=l = 0, since then ut|x=0 = d

dt(u|x=0) = 0 and similarly ut|x=l = 0. Thefirst summand vanishes due to (2.7). Thus, dH

dt = 0, yielding the energyconservation law: H(t) = const.

The conservation of energy implies the uniqueness of the solution for thestring equation (2.4) provided ρ(x) > 0 everywhere and the law of motionfor string’s ends is given:

(2.9) u|x=0 = α(t), u|x=l = β(t),

and the initial values (the position and the velocity of the string) are fixed:

(2.10) u|t=0 = ϕ(x), ut|t=0 = ψ(x) for x ∈ [0, 1].

Indeed, if u1, u2 are two solutions of the equation (2.4) satisfying (2.9) and(2.10) then their difference v = u1 − u2 satisfies the homogeneous equation(2.7) and the homogeneous boundary and initial conditions

v|x=0 = v|x=l = 0,

(2.11) v|t=0 = vt|t=0 = 0 for x ∈ [0, l].

But then the energy conservation law for v implies∫ l

0[ρ(x)v2

t (t, x) + Tv2x(t, x)]dx ≡ 0,

yielding vt ≡ 0 and vx ≡ 0, i.e., v = const. By (2.11) we now have v ≡ 0,i.e., u1 ≡ u2, as required.

Finally, note that the derivation of the string equation could be, ofcourse, carried out without the simplifying assumption u2

x << 1. As aresult, we would have obtained a nonlinear equation that hardly admits in-vestigation by simple methods. The equation (2.4) is in fact the linearization(principal linear part) of this nonlinear equation. Properties of the solutionof (2.4) give some idea of the behavior of solutions of the nonlinear equa-tion. Observe, however, that for large deformations this nonlinear equationis also inadequate to describe the physical problem, since the resistance tothe bending and other effects which were unaccounted for, will arise.

2.2. Unbounded string. D’Alembert’s formula 23

2.2. Unbounded string. The Cauchy problem. D’Alembert’s

formula

From the physical viewpoint, an unbounded string is an idealization mean-ing that we consider an inner part of the string assuming the ends to besufficiently far away, so that during the observation time they do not affectthe given part of the string. As we will see, the consideration of an un-bounded string helps in the investigation of a half-bounded string (which isa similar idealization) and of a bounded string.

Passing to the mathematical discussion, consider the one-dimensionalwave equation (2.5) for x ∈ R, t ≥ 0. A natural problem here is the Cauchyproblem, i.e., an initial value problem for (2.5) with the initial conditions

(2.12) u|t=0 = ϕ(x), ut|t=0 = ψ(x).

From the physical viewpoint (2.12) means that the initial position and theinitial velocity of the string are given. We might expect that, like finite-dimensional mechanical problems, this Cauchy problem is well posed, i.e.,there exists a unique solution continuously depending on the initial data ϕand ψ. As we will see shortly, this is really so.

We use the general solution of the equation (2.5):

u(t, x) = f(x− at) + g(x+ at)

found in Section 1.7. From (2.12) we get the system of two equations todetermine two arbitrary functions f and g:

f(x) + g(x) = ϕ(x),(2.13)

−af ′(x) + ag′(x) = ψ(x).(2.14)

The integration of the second equation yields

(2.15) −f(x) + g(x) =1a

∫ x

x0

ψ(ξ)dξ + C.

Now, from (2.13) and (2.15), we get

f(x) =12ϕ(x)− 1

2a

∫ x

x0

ψ(ξ)dξ − C

2,

g(x) =12ϕ(x) +

12a

∫ x

x0

ψ(ξ)dξ +C

2.


Therefore,

u(t, x) =12ϕ(x− at)− 1

2a

∫ x−at

x0

ψ(ξ)dξ +12ϕ(x+ at) +

12a

∫ x+at

x0

ψ(ξ)dξ

or

(2.16) u(t, x) =ϕ(x− at) + ϕ(x+ at)

2+

12a

∫ x+at

x−atψ(ξ)dξ.

Thus, the solution u(t, x) actually exists, is unique and continuously dependson the initial data ϕ,ψ under an appropriate choice of topologies in the setsof initial data ϕ,ψ and functions u(t, x). For example, it is clear that if u1 isa solution corresponding to the initial conditions ϕ1, ψ1 and for a sufficientlysmall δ > 0 we have

(2.17) supx∈R|ϕ1(x)− ϕ(x)| < δ, sup

x∈R|ψ1(x)− ψ(x)| < δ,

then for arbitrarily small ε > 0 we have

(2.18) supx∈R, t∈[0,T ]

|u1(t, x)− u(t, x)| < ε.

(More precisely, for any ε > 0 and T > 0 there exists δ > 0 such that (2.17)implies (2.18).) Therefore, the Cauchy problem is well posed.

The formula (2.16) which determines the solution of the Cauchy problemis called d’Alembert’s formula. Let us make several remarks concerning itsderivation and application.

First, note that this formula makes sense for any locally integrable func-tions ϕ and ψ and in this case it gives a generalized solution of (2.5). Herewe will only consider continuous solutions. Then we can take any continuousfunction for ϕ and any locally integrable one for ψ. It is natural to call thefunctions u(t, x) obtained in this way the generalized solutions of the Cauchyproblem. We will consider generalized solutions together with ordinary oneson equal footing and skip the adjective “generalized”.

Second, it is clear from d’Alembert’s formula that the value of the solu-tion at (t0, x0) only depends on values of ϕ(x) at x = x0± at0 and values ofψ(x) at x ∈ (x0−at0, x0+at0). In any case, it suffices to know ϕ(x) and ψ(x)on [x0−at0, x0 +at0]. The endpoints of the closed interval [x0−at0, x0 +at0]can be determined as intersection points (in (t, x)-space) of the characteris-tics passing through (t0, x0), with the x-axis, see Fig. 3.

The triangle formed by these characteristics and the x-axis is the set ofpoints in the half-plane t ≥ 0 where the value of the solution is completely

2.2. Unbounded string. D’Alembert’s formula 25

x

t

x0 − at0 x0 + at0

t0;x0

Figure 3. Dependence domain

x

t

x−

at=

dx+

at=

cc d

Figure 4. Influence domain

determined by initial values on the segment [x0−at0, x0 +at0]. This triangleis called the dependence domain for the interval [x0 − at0, x0 + at0].

An elementary analysis of the derivation of d’Alembert’s formula showsthat the formula is true for any solution defined in the triangle with thecharacteristics as its sides and an interval [c, d] of the x-axis as its base (i.e.,it is not necessary to require for a solution to be defined everywhere on thehalf-plane t ≥ 0). Indeed, from (2.13) and (2.14) we find the values f(x)and g(x) at x ∈ [c, d] (if the initial conditions are defined on [c, d]). But thisdetermines the values of u(t, x) when x− at ∈ [c, d] and x+ at ∈ [c, d], i.e.,when characteristics through (t, x) intersect the segment [c, d]. We can, ofcourse, assume f(x) and g(x) to be defined on [c, d], i.e., u(t, x) is defined inthe above described triangle which is the dependence domain of the interval[c, d].

The physical meaning of the dependence domain is transparent: it is thecomplement to the set of points (t, x) such that the wave starting the motion

with the velocity a at a point outside [c, d] at time t = 0 cannot reach the

point x by time t.

Further, the values of ϕ and ψ on [c, d] do not affect u(t, x) if x+ at < c

or x − at > d, i.e., if the wave cannot reach the point x during the time tstarting from an inside point of the interval [c, d]. This explains why thedomain bounded by the segment [c, d] and the rays of the lines x + at = c,


c d

u

xu

xu

xu

xu

xu

x

t= 0

t= "

t=l

2a

t=l

2a+ "

t=l

a

t=l

a+ "

Figure 5. Cartoon to Example 2.1

x − at = d belonging to the half-plane t ≥ 0 is called the influence domainof the interval [c, d]. This domain is shaded on Fig. 4. This domain is thecomplement to the set of points (t, x) for which u(t, x) does not depend onϕ(x) and ψ(x) for x ∈ [c, d].

Example 2.1. Let us draw the form of the string at different times if ψ(x) =0, and ϕ(x) has the form shown on the upper-most of the graphs of Fig. 5,i.e., the graph of ϕ(x) has the form of an isosceles triangle with the base[c, d]. Let d− c = 2l. We draw the form of the string at the most interestingtimes to show all of its essentially different forms.

All these pictures together form a cartoon depicting vibrations of thestring such that at the initial moment the string is pulled at the point12(c+ d) whereas the points c and d kept fixed, and then at t = 0 the whole

2.3. A semi-bounded string. Reflection of waves 27

string is released. The formula

u(t, x) =ϕ(x− at) + ϕ(x+ at)

2

makes it clear that the initial perturbation ϕ(x) splits into two identicalwaves one of which runs to the left, and the other to the right with thespeed a.

The dotted lines on Fig. 5 depict divergent semi-waves 12ϕ(x − at) and

12ϕ(x + at) whose sum is u(t, x). We assume ε > 0 to be sufficiently smallcompared to the characteristic time interval l

a (actually it suffices that ε <l

2a).

2.3. A semi-bounded string. Reflection of waves from the

end of the string

Consider the equation (2.5) for x ≥ 0, t ≥ 0. We need to impose a boundarycondition at the end x = 0. For instance, let the motion of the string leftend be given by

(2.19) u|x=0 = α(t), t ≥ 0,

and the initial conditions be only given for x ≥ 0:

(2.20) u|t=0 = ϕ(x), ut|t=0 = ψ(x), x ≥ 0.

The problem with initial and boundary conditions is usually called amixed problem.

Therefore, the problem with conditions (2.19) - (2.20) for a wave equa-tion (2.5) is an example of a mixed problem (sometimes called the firstboundary value problem for the string equation). Let us verify that thisboundary value problem is also well posed and solve it. N

H Let us recall an important notion of convexity.

Definition 2.1. Let A be a subset in Rn. We will say that A is convex (inRn) if for every a, b ∈ A, the open interval connecting a and b, is in fact asubset of A.

For a real-valued function f : (a.b) → R with real a, b, we will say thatthe function f is convex if its subgraph (x, y)|, x ∈ (a, b), y < f(x) isconvex in R2. N

Examples. 1. Any closed cube in Rn is convex.


2. An open ball in Rn with any non-empty subset of the ball boundaryadded, is convex.

3. (A step) A closed cube in Rn with a part of a smaller cube removed,e.g.

[0, 1]n−2 × ( [0, 1]2 \ (1/2, 1]2 ),

is not convex. N

Returning to the boundary value problem above, we can see that thesimplest way to get an analytic solution is as in the Cauchy problem. Thequadrant t ≥ 0, x ≥ 0 is a convex domain, therefore, we can again representu in the form

u(t, x) = f(x− at) + g(x+ at),

where g(x) is now defined and needed only for x ≥ 0 because x+ at ≥ 0 forx ≥ 0, t ≥ 0. (At the same time f(x) is defined, as before, for all x.)

The initial conditions (2.20) define f(x) and g(x) for x ≥ 0 as in the caseof an unbounded string and via the same formula. Therefore, for x − at ≥0 the solution is determined by d’Alembert’s formula which is, of course,already clear. But now we can make use of the boundary condition (2.19)which gives

f(−at) + g(at) = α(t), t ≥ 0,

implying

(2.21) f(z) = α(−za

)− g(−z), z ≤ 0.

The summands in (2.21) for z = x− at give the sum of two waves runningto the right, one of them is α(t − x

a ) and is naturally interpreted as thewave generated by oscillations of the end of the string, the other one is−g(−(x − at)) and is interpreted as the reflection of the wave g(x + at)running to the left from the left end (note that this reflection results in thechange of the sign). If the left end is fixed, i.e., α(t) ≡ 0, only one wave,−g(−(x−at)), remains and if there is no wave running to the left, then onlythe wave α(t− x

a ) induced by the oscillation α(t) of the end is left.

Let us indicate a geometrically more transparent method for solving theproblem with a fixed end. We wish to solve the wave equation (2.5) fort ≥ 0, x ≥ 0, with the initial conditions (2.20) and the boundary condition

u|x=0 = 0, t ≥ 0.

Let us try to find a solution in the form of the right hand side of d’Alembert’sformula (2.16), where ϕ(x), ψ(x) are some extensions of functions ϕ(x), ψ(x)

2.4. A bounded string. Standing waves. The Fourier method 29

onto the whole axis defined for x ≥ 0 by initial conditions (2.20). Theboundary condition yields

0 =ϕ(−at) + ϕ(at)

2+

12a

∫ at

−atψ(ξ)dξ,

implying that we will reach our goal if we extend ϕ and ψ as odd functions,i.e., if we set

ϕ(z) = −ϕ(−z), ψ(z) = −ψ(−z), z ≤ 0.

Thus, we can construct a solution of our problem. The uniqueness of thesolution is not clear from this method; but, luckily, the uniqueness wasproved above.

Example 2.2. Let an end be fixed, let ψ(x) = 0, and let the graph of ϕ(x)again have the form of an isosceles triangle with the base [l, 3l]. Let us drawa cartoon describing the form of the string. Continuing ϕ(x) as an oddfunction, we get the same problem as in Example 2.1. With a dotted line wedepict the graph over the points of the left half-axis (introducing it is just amathematical trick since actually only the half-axis x ≥ 0 exists). We getthe cartoon described on Fig. 6.

Of interest here is the moment t = 2la when near the end the string is

horizontal and so looks not perturbed. At the next moment t = 2la + ε,

however, the perturbation arises due to a non-vanishing initial velocity. Fort > 3l

a + ε we get two waves running to the right, one of which is negativeand is obtained by reflection from the left end of the wave running to theleft.

2.4. A bounded string. Standing waves. The Fourier method

(separation of variables method)

Consider oscillations of a bounded string with fixed ends, i.e., solve the waveequation (2.5) with initial conditions

(2.22) u|t=0 = ϕ(x), ut|t=0 = ψ(x), x ∈ [0, l]

and boundary conditions

(2.23) u|x=0 = u|x=l = 0, t ≥ 0.

This can be done as in the preceding section. We will, however, applyanother method which also works in many other problems. Note that the


u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

t= 0

t= "

t=l

a

t=l

a+ "

t=3l

2a

t=3l

2a+ "

t=2l

a

t=2l

a+ "

t=3l

a+ "

l 3l

−l−3l

Figure 6. Cartoon to Example 2.2

uniqueness of the solution is already proved with the help of the energyconservation law.

We will find standing waves, i.e., the solutions of the wave equation (2.5)defined for x ∈ [0, l], satisfying (2.23) and having the form

u(t, x) = T (t)X(x).


The term “standing wave” is justified since the form of the string does notactually change with the time except for a factor which is constant in x andonly depends on time. Substituting u(t, x) into (2.5), we get

T ′′(t)X(x) = a2T (t)X ′′(x).

Excluding the uninteresting trivial cases when T (t) ≡ 0 or X(x) ≡ 0, wemay divide this equation by a2T (t)X(x) and get

(2.24)T ′′(t)a2T (t)

=X ′′(x)X(x)

The left-hand side of this relation does not depend on x and the right-hand side does not depend on t. Therefore, they are constants, i.e. (2.24) isequivalent to the relations

(2.25) T ′′(t)− λa2T (t) = 0,

X ′′(x)− λX(x) = 0

with the same constant λ. Further, boundary conditions (2.23) imply

X(0) = X(l) = 0.

The boundary value problem

(2.26) X ′′ = λX, X(0) = X(l) = 0

is a particular case of the so-called Sturm-Liouville problem.

We will discuss the general Sturm-Liouville problem later. Now, we solvethe problem (2.26), i.e., find the values λ (eigenvalues) for which the prob-lem (2.26) has nontrivial solutions (eigenfunctions), and find these eigen-functions.

Consider the following possible cases.

a) λ = µ2, µ > 0. Then the general solution of the equation X ′′ = λX

isX(x) = C1 sinhµx+ C2 coshµx.

(Recall that sinhx = 12(ex − e−x) and coshx = 1

2(ex + e−x).)

From X(0) = 0 we get C2 = 0, and from X(l) = 0 we get C1 sinhµl = 0implying C1 = 0, i.e. X(x) ≡ 0, hence, λ > 0 is not an eigenvalue.

b) λ = 0. Then X(x) = C1x+C2 and boundary conditions imply againC1 = C2 = 0 and X(x) ≡ 0.

c) λ = −µ2, µ > 0. Then

X(x) = C1 sinµx+ C2 cosµx;


0 l

X 1(x)

Figure 7. Standing wave 1

0ll=2

X 2(x)


0 ll=3 2l=3

X 3(x)


hence, X(0) = 0 implies C2 = 0 and X(l) = 0 implies C1 sinµl = 0. Assum-ing C1 6= 0, we get sinµl = 0; therefore, the possible values of µ are:

µk =kπ

l, k = 1, 2, . . . ,

and the eigenvalues and eigenfunctions are:

λk = −(kπ

l

)2

, Xk = sinkπx

l, k = 1, 2, . . .

Let us draw graphs of several first eigenfunctions determining the form ofstanding waves (see pictures 7, 8, 9).

It is also easy to find the corresponding functions T (t). Namely, from(2.25) with λ = λk we get

Tk(t) = Ak coskπat

l+Bk sin

kπat

l,

implying the following general form of the standing wave:

uk(t, x) =(Ak cos

kπat

l+Bk sin

kπat

l

)sin

kπx

l, k = 1, 2, . . .


The frequency of the oscillation of each point x corresponding to the solutionuk is equal to

ωk =kπa

l, k = 1, 2, . . . .

These frequencies are called the eigenfrequencies of the string.

Now let us find the general solution of (2.5) with boundary conditions(2.23) as a sum (superposition) of stationary waves, i.e., in the form

(2.27) u(t, x) =∞∑k=1

(Ak cos

kπat

l+Bk sin

kπat

l

)sin

kπx

l

We need to satisfy the initial conditions (2.22). Substituting (2.27) intothese conditions, we get

(2.28) ϕ(x) =∞∑k=1

Ak sinkπx

l, x ∈ [0, l],

(2.29) ψ(x) =∞∑k=1

(kπa

l

)Bk sin

kπx

l, x ∈ [0, l],

i.e., it is necessary to expand both ϕ(x) and ψ(x) into series in eigenfunctionssin

kπx

l: k = 1, 2, . . .

.

First of all, observe that this system is orthogonal on the segment [0, l].This can be directly verified, but we may also refer to the general factof orthogonality of eigenvectors of a symmetric operator corresponding todistinct eigenvalues. For such an operator one should take the operatorL = d2

dx2 with the domain DL consisting of functions v ∈ C2([0, l]) such thatv(0) = v(l) = 0. Integrating by parts, we see that

(Lv,w) = (v, Lw), v, w ∈ DL.

The parentheses denote the inner product in L2([0, l]) defined via

(v1, v2) =∫ l

0v1(x)v2(x)dx.

Thus, the system sin kπxl k=1,2,... is orthogonal in L2([0, l]). We would like

to establish its completeness.

For simplicity, take l = π (the general case reduces to this one byintroducing the variable y = πx

l ) so that the system takes on the formsin kx∞k=1 and is considered on [0, π]. Let f ∈ L2([0, π]). Extend f as an


odd function onto [−π, π] and expand it into the Fourier series in the system1, cos kx, sin kx : k = 1, 2, . . . on the interval [−π, π].

Clearly, this expansion does not contain constants and the terms withcos kx, since the extension is an odd function. Therefore, on [0, π], we getthe expansion in the system sin kx : k = 1, 2, . . . only.

Note, further, that if f is continuous, has a piecewise continuous deriva-tive on [0, π] and f(0) = f(π) = 0, then its 2π-periodic odd extension is alsocontinuous with a piecewise continuous derivative. Therefore, it can be ex-panded into a uniformly convergent series in terms of sin kx : k = 1, 2, . . ..If this 2π-periodic extension is in Ck and its (k + 1)-st derivative is piece-wise continuous, then this expansion can be differentiated k times and theuniform convergence is preserved under this differentiation.

Thus, expansions into series (2.28) and (2.29) exist and imposing somesmoothness and boundary conditions on ϕ(x), ψ(x), we can achieve arbitrar-ily fast convergence of these series. Coefficients of these series are uniquelydefined:

Ak =

∫ l0 ϕ(x) sin kπx

l dx∫ l0 sin2 kπx

l dx=

2l

∫ l

0ϕ(x) sin

kπx

ldx,

Bk =l

kπa

∫ l0 ψ(x) sin kπx

l dx∫ l0 sin2 kπx

l dx=

2kπa

∫ l

0ψ(x) sin

kπx

ldx.

Substituting these values of Ak and Bk into (2.27), we get the solutionof our problem.

The Fourier method can also be used to prove the uniqueness of solution.(More precisely, the completeness of the trigonometric system enables us toreduce it to the uniqueness theorem for ODE.) We will illustrate this belowby an example of a more general problem.

Consider the problem of a string to which a distributed force f(t, x) isapplied:

(2.30) utt = a2uxx + f(t, x), t ≥ 0, x ∈ [0, l].

The ends of the string are assumed to be fixed (i.e., conditions (2.23) aresatisfied) and for t = 0 the initial position and velocity of the string, i.e.,conditions (2.22), are defined as above.

Since the eigenfunctions sin kπxl : k = 1, 2, . . . constitute a complete

orthonormal system on [0, l], any function g(t, x), defined for x ∈ [0, l] andsatisfying reasonable smoothness conditions and boudnary conditions, can


be expanded in terms of this system, with coefficients depending on t. Inparticular, we can write

u(t, x) =∞∑k=1

uk(t) sinkπx

l,

f(t, x) =∞∑k=1

fk(t) sinkπx

l,

where fk(t) are known and uk(t) are to be determined. Substituting thesedecompositions into (2.30), we get

(2.31)∞∑k=1

[uk(t) + ω2kuk(t)− fk(t)] sin

kπx

l= 0,

where ωk = kπal are the eigenfrequencies of the string. It follows from (2.31),

due to the orthogonality of the system of eigenfunctions, that

(2.32) uk(t) + ω2kuk(t) = fk(t), k = 1, 2, . . .

The initial conditions∞∑k=1

uk(0) sinkπx

l= ϕ(x),

∞∑k=1

uk(0) sinkπx

l= ψ(x),

determine uk(0) and uk(0); hence, we can uniquely determine functions uk(t)and, therefore, the solution u(t, x).

In particular, of interest is the case when f(t, x) is a harmonic oscillationin t, e.g.

f(t, x) = g(x) sinωt.

Then, clearly, the right hand sides fk(t) of equations (2.32) are

fk(t) = gk sinωt.

For example, let gk 6= 0. Then for ω 6= ωk (the nonresonance case) theequation (2.32) has an oscillating particular solutions of the form

uk(t) = Uk sinωt, Uk =g2k

ω2k − ω2

.

Therefore, its general solution also has an oscillating form:

uk(t) = Uk sinωt+Ak cosωkt+Bk sinωkt.


For ω = ωk (the resonance case) there is a particular solution of the form

uk(t) = t(Mk cosωt+Nk sinωt),

which the reader can visualize as an oscillation with the frequency ω andan increasing amplitude. The general solution uk(t) and the function u(t, x)are unbounded in this case.

Observe that to the above described problem a more general problemcan be reduced, namely the one with ends that are not fixed but whosemotion is given by

(2.33) u|x=0 = α(t), u|x=l = β(t), t ≥ 0.

Namely, if u0(t, x) is an arbitrary function satisfying (2.33), e.g.

u0(t, x) =l − xl

α(t) +x

lβ(t),

then for v(t, x) = u(t, x)− u0(t, x) we get a problem with fixed ends.

Finally, note that all the above problems are well-posed which is obviousfrom their explicit solutions.

2.5. Appendix: Calculus of variations and classical

mechanics

Variations of functionals. Euler-Lagrange equations. Calculus ofvariations is an analogue of differential calculus, but for (generally non-linear) functions which are defined not on a finite-dimensional space but onan infinite-dimensional space (usually a space of functions). Such functionson an infinite-dimensional space are usually called functionals to distinguishthem from functions on a finite-dimensional space (which can be argumentsof the functionals).

Let us consider a simplest important example of a functional where thevariational calculus can be applied to finding extrema of this functional.It is defined on the space C2([t0, t1]), consisting of real-valued continuousfunctions x = x(t), t ∈ [t0, t1], such that the derivatives x = dx/dt, x =d2x/dt2 exist and are continuous on [a, b], and is given by the formula

S[x] =∫ t1

t0

L(t, x(t), x(t))dt,

where L = L(t, x, v) is a continuous function of real variables t ∈ [a, b],x, v ∈ R, which has continuous partial derivatives ∂L/∂t, ∂L/∂x, ∂L/∂v,

2.5. Appendix: Calculus of variations and mechanics 37

∂2L/∂t∂v, ∂2L/∂x∂v, ∂2L/∂v2. In mechanics and physics, a functional Slike this is usually called action, whereas L is called Lagrangian.

It may be not clear a priori why our requirements include the existenceand continuity of the second derivative x, as well as many partial derivativesof L. Most of these requirements could be avoided or weakened, but thiswould complicate further exposition because x and the partial derivativesexplicitly or implicitly enter into the extremum condition.

Now let us assume that we need to find a minimum of the action S on aset of functions x ∈ C2([t0, t1]), whose values at the ends t0, t1 are fixed. (Inother words, we impose boundary conditions x(t0) = b0, x(t1) = b1 whereb0, b1 are fixed constants.) Let x = x(t) be a minimizer: a function wherethis minimum is attained. Then the following condition should be satisfied:for any function δx ∈ C2([t0, t1]), satisfying the boundary conditions

δx(t0) = δx(t1) = 0,

the function f = f(s) of one variable s ∈ R, defined by

f(s) = S[x+ s δx] =∫ t1

t0

L(t, x(t) + s δx(t), x(t) + s δx(t))dt,

should attain minimum at s = 0. (In this formula δx should be understoodas (d/dt)δx(t).) Therefore, f ′(0) = 0, which can be rewritten as∫ t1

t0

(∂L

∂xδx+

∂L

∂xδx

)dt = 0,

where ∂L/∂x should be understood as ∂L(t, x(t), v)/∂v|v=x(t). Integratingby parts in the second term, and taking into account the boundary conditionsfor δx, we obtain ∫ t1

t0

(∂L

∂x− d

dt

∂L

∂x

)δx dt = 0,

Since δx can be an arbitrary function from C2([t0, t1]), vanishing at the endst0, t1, we easily obtain that the expression in the parentheses should vanishidentically, i.e. we get the following equation

d

dt

∂L

∂x− ∂L

∂x= 0,

identically on the interval [t0, t1]. This is a second-order ordinary differentialequation which is called Euler-Lagrange equation. Its solution satisfying theboundary conditions x(t0) = b0, x(t1) = b1, is called extremal or extremalpoint of the action functional S. All minimizers and maximizers of S shouldbe extremal points. Sometimes, if it is known by some reasons that, say,


the minimum is attained, and there is only one extremal, then this extremalmust coincide with the minimizer.

Let us also denote by δS the following linear functional on the space offunctions δx ∈ C2([t0, t1]), vanishing at the ends of the interval [t0, t1]:

δS[δx] =d

dsS[x+ s δx]

∣∣∣∣s=0

=∫ t1

t0

(∂L

∂xδx+

∂L

∂xδx

)dt

=∫ t1

t0

(∂L

∂x− d

dt

∂L

∂x

)δx dt.

It is the principal linear part of the functional S at the point x, and it iscalled the variation of the functional S. Using the variation, we can rewritethe extremality condition in the form δS = 0 which means that δS[δx] = 0for all variations δx, which vanish at the ends of the interval [t0, t1]. Clearly,this is equivalent to the Euler-Lagrange equation.

We can also rewrite the last equation in the form

δS = δS[δx] =∫ t1

t0

δS

δxδx dt,

whereδS

δx=∂L

∂x− d

dt

∂L

∂xis called the variational derivative of S.

There is a straightforward extension of the arguments given above, to thecase when the action functional S depends upon several functions x1(t), . . . , xn(t)belonging to C2([t0, t1]). In fact, it can be obtained if we just use the sameformulas as above but for a vector-function x instead of a scalar function.(This vector-function can be interpreted as a parametrized curve in Rn, or,alternatively, a trajectory of a point moving in Rn.) Nevertheless, let usreproduce the formulas in detail, leaving the details of the arguments as anexercise for the readers.

The action S is given by

S[x] =∫ t1

t0

L(t, x1(t), . . . , xn(t), x1(t), . . . , xn(t))dt,

where L = L(t, x1, . . . , xn, v1, . . . , vn) is the Lagrangian, which is a scalarreal-valued function. The variations of the given (real-valued) functionsx1(t), . . . , xn(t) are real-valued C2-functions δx1(t), . . . , δxn(t), vanishing atthe ends t0, t1. The extremum conditions are obtained from the equations

∂

∂sjS[x1 + s1δx1, . . . , xn + snδxn]

∣∣∣∣s1=···=sn=0

= 0, j = 1, . . . , n,


which lead to the Euler-Lagrange equations

d

dt

∂L

∂xj− ∂L

∂xj= 0, j = 1, . . . , n.

These equations are equivalent to the vanishing of the variation δS, whichis a linear functional on vector-functions δx = (δx1, . . . , δxn); it is given by


t0

n∑j=1

(∂L

∂xjδxj +

∂L

∂xjδxj

)dt

=∫ t1

t0

n∑j=1

(∂L

∂xj− d

dt

∂L

∂xj

)δxj dt.

This can be rewritten in the form


t0

n∑j=1

δS

δxjδxj dt.

whereδS

δxj=

∂L

∂xj− d

dt

∂L

∂xj

is called the partial variational derivative of S with respect to xj .

Lagrangian approach in classical mechanics. The most important equa-tions of classical mechanics can be presented in the form of the Euler-Lagrange equations. (In mechanics they are usually called Lagrange’s equa-tions.) Take, for example, a point particle of mass m > 0 moving along thex-axis R in a time-independent (or stationary) field of forces F (x) (which isassumed to be a real-valued continuous function on R). Let x(t) denote thecoordinate of the moving particle at the time t. Then by Newton’s SecondLaw, (mass)·(acceleration)=(force), or, more precisely,

mx(t) = F (x(t))

for all t. It is sometimes more convenient to rewrite this equation in theform

dp

dt= F,

where p = mx is called the momentum of the moving particle.

Let us introduce a potential energy of the particle, which is a real-valuedfunction U = U(x) on R, such that F (x) = −U ′(x) for all x. In other words,

U(x) = U(x0)−∫ x

x0

F (s)ds,


which means that to move the particle from x0 to x you need to do workU(x)− U(x0) against the force F . (If U(x) < U(x0), then in fact the forcedoes the work.) The Newton equation rewrites then as

mx = −U ′(x).

Multiplying both parts by x, we can rewrite the result as

d

dt

mx2

2= − d

dtU(x),

where x = x(t). Taking antiderivative, we obtain

mx2

2+ U(x) = E = const,

where the constant E depends on the solution x = x(t) (but is constant alongthe trajectory). This is the energy conservation law for the one-dimensionalmotion. The quantity

K =mx2

2is called kinetic energy of the moving particle, and the quantity

H = K + U = H(x, x)

is called energy, or full energy, or Hamiltonian. The energy conservation lawcan be rewritten in the form H(x, x) = E = const.

Now let us take the Lagrangian

L = L(x, x) = K − U =mx2

2− U(x),

and the corresponding action S. Then it is easy to see that the extremals ofS will be exactly the solutions of the Newton Second Law equation, satis-fying the chosen boundary conditions. In other words, the Euler-Lagrangeequation for this action coincides with the Newton Second Law equation.

Now consider a motion of a particle in Rn. (Note that any motion ofk particles, say, in R3 can be equivalently presented as a motion of oneparticle in R3k, the space whose coordinates consists of all coordinates ofall particles.) The particle coordinates can be presented as a point x ∈Rn, x = (x1, . . . , xn), so the motion of the particle is given by functionsx1(t), . . . , xn(t), or by one vector-function x(t) (with values in Rn). Let usassume that the force is F (x) = (F1(x), . . . , Fn(x)), so that F (x) ∈ Rn forevery x ∈ Rn, and F is continuous. (In case when x represents a system ofparticles, F may include forces of interaction between these particles, which


may depend on the configuration of these particles.) Then by Newton’sSecond Law, the motion satisfies the equations

mj xj = Fj(x), j = 1, . . . , n.

Let us assume also that the field of forces is a potential field, given by apotential energy U = U(x) (which is a scalar, real-valued function on Rn),so that

Fj(x) = − ∂U∂xj

, j = 1, . . . , n,

or, in the vector analysis notation,

F = −∇U = − gradU.

In this case, it is easy to see that the equations of motion coincide with theEuler-Lagrange equations for the action with the Lagrangian L = K − U ,where the kinetic energy K is given by

K =n∑j=1

mj x2j

2.

The Hamiltonian H = H(x, x) = K + U satisfies the energy conservationlaw H(x(t), x(t)) = E = const, where x(t) is the solution of the equationsof motion (or, equivalently, the extremal of the action). This can be verifiedby a straightforward calculation:

ddtH(x(t), x(t)) =

∂H

∂x· x+

∂H

∂x· x

=n∑j=1

(∂H

∂xjxj +

∂H

∂xjxj

)

=n∑j=1

(∂U

∂xjxj +

∂K

∂xj

(− 1mj

∂U

∂xj

))

=n∑j=1

(∂U

∂xjxj +mj xj

(− 1mj

∂U

∂xj

))= 0.

Now let us consider a system of k particles in R3. Assume that n = 3kand write x = (x1, . . . ,xk), where xj ∈ R3, j = 1, . . . , k. Let us view xj asa point in R3 where jth particle is located. For moving particles we assumethat xj = xj(t) is a C2 function of t. Let us introduce the momentum ofjth particle as pj = mjxj , which is a vector in R3. The momentum of thewhole system of such particles is defined as

p = p1 + · · ·+ pk,


which is again a vector in R3. Now assume that all forces acting on theparticles, are split into pairwise interaction forces between the particles andexternal forces, so that the force acting on jth particle is

Fj =∑i 6=j

Fij + Fextj ,

where Fij = Fij(x) ∈ R3, and the summation is taken over i (with j fixed),Fextj = Fext

j (x) ∈ R3. Here Fij is the force acting on the jth particle dueto its interaction with the ith particle, and Fext

j is the external force actingupon the jth particle.

The equations of motion look as follows:

dpjdt

= Fj =∑i 6=j

Fij + Fextj , j = 1, . . . , k.

Now assume that Newton’s Third Law holds for two interacting particles:

Fij(x) = −Fji(x),

for all i, j with i 6= j, and for all x. Then we get

dpdt

= Fext =k∑j=1

Fextj .

In words, the rate of change of the total momentum in time equals the totalexternal force. (The pairwise interaction forces, satisfying Newton’s ThirdLaw, do not count.) Indeed, calculating the derivative of p, we see that allinteraction forces cancel due to Newton’s Third Law.

In particular, in the absence of external forces the conservation of mo-mentum law holds: for any motion of these particles

p(t) = const .

Indeed, from the relations above we immediately see that dp/dt = 0.

Remark. The mechanical system, describing a particle, moving in Rn,is said to have n degrees of freedom. In the main text of this Chapter wediscuss the motion of a string, which naturally has infinitely many degrees offreedom, because its position is described not by a finite set of numbers butby a function (of one variable), or by an infinite set of its Fourier coefficients.

2.6. Problems 43

Figure 10. The ring attached to the end of the string

Figure 11. The end of the rod is elastically attached

2.6. Problems

2.1. a) Write the boundary condition on the left end of the string if a ringwhich can slide without friction along the vertical line is attached to thisend. The ring is assumed to be massless (see Fig. 10);

b) The same as in 2.1 a) but the ring is of mass m;

c) The same as in 2.1 b) but the ring slides with a friction which isproportional to its velocity.

2.2. a) Derive the energy conservation law for a string if at the left end theboundary condition of Problem 2.1 a) is satisfied and the right end is fixed;

b) The same in Problem 2.1 b) (take into account the energy of the ring);

c) The same in Problem 2.1 c) (take into account the loss of energycaused by friction).

2.3. Derive the equation for longitudinal elastic vibrations of a homogeneousrod.

2.4. Write the boundary condition for the left end of the rod if this end isfree.

2.5. Write the boundary condition for the left end of the rod if this end iselastically restrained; see Fig. 11.


2.6. Derive the energy conservation law for a rod with

a) fixed ends;

b) free ends;

c) one end is fixed, and the other restrained elastically.

2.7. Derive an equation for longitudinal vibrations of a conic rod.

2.8. Prove that the energy of an interval of a string between c+at and d−atis a decreasing function of time. Deduce from this the uniqueness of thesolution of the Cauchy problem and an information about the dependenceand influence domains for [c, d].

2.9. Formulate and prove the energy conservation law for an infinite string.

2.10. Draw a cartoon describing oscillations of an infinite string with the

initial values u|t=0 = 0, ut|t=0 = ψ(x), where ψ(x) =

1 for x ∈ [c, d]

0 for x 6∈ [c, d].

2.11. Describe oscillations of an infinite string for t ∈ (−∞,+∞) such thatsome interval (x0 − ε, x0 + ε) of the string is in equilibrium position at alltimes.

2.12. The same as in the above problem but the interval (x0 − ε, x0 + ε) isonly fixed for t ≥ 0.

2.13. Draw a cartoon describing oscillations of a semi-bounded string witha free end (the boundary condition ux|x=0 = 0, see Problem 2.1 a), and theinitial conditions u|t=0 = ϕ(x), ut|t=0 = 0, where the graph ϕ(x) has theform of an isosceles triangle with the base [l, 3l].

2.14. Draw a cartoon describing oscillations of a semi-bounded string witha fixed end and the initial conditions u|t=0 = 0, ut|t=0 = ψ(x), where ψ(x)is the characteristic function of the interval (l, 3l).

2.15. Derive the energy conservation law for a semi-bounded string with afixed end.

2.6. Problems 45

2.16. Given the wave sinω(t + xa ) running for t ≤ 0 along the rod whose

left end is elastically attached, find the reflected wave.

2.17. Given a source of harmonic oscillations with frequency ω runningalong an infinite string at a speed v < a, i.e., u|x=vt = sinωt for t ≥ 0,describe the oscillations of the string to the left and to the right of thesource. Find frequencies of induced oscillations of a fixed point of the stringto the left and to the right of the source and give a physical interpretationof the result (Doppler’s effect).

2.18. Describe and draw standing waves for a string with free ends.

2.19. Prove the existence of an infinite series of standing waves in a rodwith the fixed left end and elastically attached right one (see Problem 2.5).Prove the orthogonality of eigenfunctions. Find the shortwave asymptoticsof eigenvalues.

2.20. The force F = F0 sinωt, where ω > 0, is applied to the right endof the rod for t ≥ 0. Derive a formula for the solution describing forcedoscillations of the rod and find the conditions for a resonance if the left endis

a) fixed;

b) free.

(Comment: the forced oscillations are the oscillations which have the fre-quency which coincides with the frequency of the external force or otherexternal perturbation.)

2.21. The right end of a rod is oscillating via A sinωt for t ≥ 0. Write aformula for the solution describing forced oscillations and find the conditionsfor resonance if the left end is

a) fixed;

b) free;

c) elastically restrained.

Chapter 3

The Sturm-Liouville

problem

3.1. Ordinary differential equations

We will briefly discuss simplest facts on Ordinary Differential Equations(ODE), especially the existence and uniqueness of solutions with appropriateinitial conditions. These results are universally useful in analysis and canbe found in numerous textbooks (e.g. Arnold [2], Ch. 4; Hartman [11], Ch.I; Taylor [30], vol. 1, Ch. 1.2).

A. Let us consider an ODE with an initial condition

(3.1)dy

dx= f(x, y), y(x0) = y0.

Here we assume that an open interval I ⊂ R1 and an open subset V ⊂ Rn

are given, x0 ∈ I, y0 ∈ V , f is a continuous function on I × V with thevalues in Rn. A function y = y(x) on an open interval J ⊂ I with values inV is called a solution of the problem (3.1) if y(x) satisfies the equation forall x ∈ J and the initial condition at x = x0.

Assume also that f satisfies the Lipschitz condition in y, that is

(3.2) |f(x, y1)− f(x, y2)| ≤ L|y1 − y2|,

where x ∈ I and y1, y2 ∈ V , L is a non-negative constant.

Theorem 3.1. (A.L.Cauchy) Under the conditions above, the equation (3.1)has a unique solution y = y(x) with x ∈ J , where J is in a sufficiently smallinterval J ⊂ I, containing x0.

47

48 3. The Sturm-Liouville problem

We will skip the proof which can be found, e.g., in the textbooks quotedabove at the beginning of this Section.

Note that a higher-order ODE problem

y(n)(t) = f(t, y, y′, . . . , y(n−1))

can be reduced to an equivalent first-order system by introducing a systemof new variables u = (u0, u1, . . . , un−1) with

u0 = y, uj = y(j), 1 ≤ j ≤ n− 1,

which leads to

du

dt= (u1, . . . , un−1, f(t, u0, . . . , un−1)).

It is easy to see that the last system of equations with appropriate boundaryconditions is equivalent to the original initial-value problem.

B. It is also important to extend the solution (say, y = y(x)) to themaximal interval J of the independent variable x, where the solution stillexists. More precisely, assume that J = (c, d) ⊂ R, x0 ∈ J , so that thegraph of the solution y = y(x) comes to a singularity (in particular, goes toinfinity), or does not have any finite limits as x→ c+ or x→ d−. Then weshould declare that y(x) has a singular point at this x = c or x = d.

The singular points are an important topic in mathematics, because find-ing and investigating them may be sufficient for a complete understandingof the equations under the study.

3.2. Formulation of the problem

Now we will start a study of a seemingly special PDE which appears as amodel for a hyperbolic second-order equation with two independent vari-ables, t and x, so that in a physical presentation t means “time” and x is a“space” variable. We will start by simplifying the equation by a change ofvariables.

Let us consider the equation

(3.3) ρ(x)utt = (p(x)ux)x − q(x)u,

which is more general than the equation derived in Chapter 2 for oscil-lations of a non-homogeneous string. For the future calculations to bevalid, we will assume that ρ ∈ C1([0, l]), ρ > 0 on [a, b], p ∈ C1([a, b])

3.2. Formulation of the problem 49

and q ∈ C([a, b]). Let us try to simplify the equation by a change of vari-ables u(t, x) = z(x)v(t, x), where z(x) is a known function. This will allowus to obtain an equation of the form (3.3) with ρ(x) ≡ 1.

Indeed, we have

utt = zvtt, ux = zvx + zxv, uxx = zvxx + 2zxvx + zxxv.

Substituting this into (3.3) and dividing by zρ, we get

vtt =p

ρvxx +

(2pρ· zxz

+pxρ

)vx +B(x)v.

This equation is of the form (3.3) with ρ(x) ≡ 1 whenever(p

ρ

)x

=2pρ· zxz

+pxρ.

Performing the differentiation in the left hand side we see that this can beequivalently rewritten in the form

zxz

= −ρx2ρ

and we may take, for example,

z(x) = ρ(x)−1/2.

Since we always assume ρ(x) > 0, such a substitution is possible. Thus,consider the equation

utt = (p(x)ux)x − q(x)u.

Solving it by separation of variables on the closed interval [0, l], and assumingthat the ends are fixed, we get the following eigenvalue problem:

(3.4) −(p(x)X ′(x))′ + q(x)X(x) = λX(x),

with the boundary conditions

(3.5) X(0) = X(l) = 0.

Generalizing it, we can take the same differential operator while replac-ing the boundary conditions by more general conditions:

αX ′(0) + βX(0) = 0, γX ′(l) + δX(l) = 0,

where α, β, γ, δ are real numbers, α2 + β2 6= 0, γ2 + δ2 6= 0. This eigenvalueproblem is called the Sturm-Liouville problem. Considering this problemwe usually suppose that

p ∈ C1([0, l]), q ∈ C([0, l]) and p(x) 6= 0 for x ∈ [0, l] (ellipticity).


For simplicity, we assume that p(x) ≡ 1 and the boundary conditions are asin (3.5). The general case is treated similarly. Thus, consider the problem

(3.6) −X ′′(x) + q(x)X(x) = λX(x),

with the boundary conditions (3.5) and assume also that

(3.7) q(x) ≥ 0.

This is not a restriction since we can achieve (3.7) by adding the sameconstant to q and to λ.

3.3. Basic properties of eigenvalues and eigenfunctions

We will start with the following Oscillation Theorem by C.Sturm (1836):

Theorem 3.2. Let y, z, q1, q2 ∈ C2([a, b]) be real-valued functions, such thatz(a) = z(b) = 0, z 6≡ 0, y and z satisfy the equations

(3.8) y′′(x) + q1(x)y(x) = 0,

(3.9) z′′(x) + q2(x)z(x) = 0.

Assume also that

(3.10) q1(x) ≥ q2(x), for all x ∈ [a, b].

Then, either there exists x0 ∈ (a, b) such that y(x0) = 0, or q1(x) ≡ q2(x)and y(x) ≡ cz(x) on [a, b], where c is a constant.

Proof. (a) By a null-point of a real-valued function z = z(x) ∈ C2([a, b]),satisfying (3.9), we will call a point x0 ∈ [a, b], such that z(x0) = 0, z 6≡ 0.We will start by proving, that such null-points are simple (non-degenerate),which means that if z = z(·) is a solution of (3.9), x0 ∈ [a, b], z(x0) = 0 andz(x) 6≡ 0, then z′(x0) 6= 0.

Indeed, if we assume that z′(x0) = 0, then z(x0) = z′(x0) = 0, whichimplies that z(x) ≡ 0 by the existence and uniqueness theorem for the linearsecond order ODE in C2([a, b]) (see Theorem 3.1 and further commentsthere). Moreover, the same arguments, as above, present all solutions of(3.9) as a 2-dimensional linear subspace (in C2([a, b])), because taking theinitial conditions z(·) 7→ (z(c), z′(c)) maps the space of all solutions of(3.9) to R2 by a linear isomorphism for any fixed c ∈ [a, b].

It follows that the set of all null-points of the solution z(x), which satisfythe condition z(x0) = 0 with a fixed x0 ∈ [a, b], is discrete, hence finite,because the interval [a, b] is compact.

3.3. Eigenvalues and eigenfunctions 51

In particular, all null-points of z(x) are isolated.

Note also that for any null-point x0 ∈ [a, b], the set of solutions z(x)satisfying z(x0) = 0, is a 1-dimensional subspace in C2([a, b]).

The set z of all null-points of the function z(x), belonging to [a, b], isdiscrete, hence finite, because the interval [a, b] is compact.

(b) Let us consider the Wronskian

(3.11) W (x) = y(x)z′(x)− y′(x)z(x), x ∈ [a, b].

Differentiating this identity with respect to x, we obtain

(3.12) W ′(x) = y(x)z′′(x)−y′′(x)z(x) = (q1(x)−q2(x))y(x)z(x), x ∈ [a, b].

Obviously, it suffices to consider the case when a and b are neighboringnull-points of the function z(·) on [a, b]. Then, without loss of generality, wemay assume that z(x) > 0 for all x ∈ (a, b). It follows that z′(a) > 0 andz′(b) < 0.

Now, looking at y(·), we see two possibilities. The first one is that thereexists x1 ∈ (a, b) such that y(x1) = 0, which is enough for the validity ofone of the statements of this theorem. The alternative is that y(x) 6= 0 forall x ∈ (a, b). Then we can assume that y(x) > 0 for all x ∈ (a, b), which wewill assume from this moment to the end of the proof. In this case we alsohave y(x) ≥ 0) for all x ∈ [a, b].

Integrating (3.9) over [a, b], we get W (b)−W (a) ≥ 0. Here the equalityholds if and only if q1(x) ≡ q2(x) on [a, b].

On the other hand, by the arguments above, y(a) ≥ 0 and y(b) ≥ 0.Then we have

W (b)−W (a) = y(b)z′(b)− y(a)z′(a) ≤ 0,

with equality if and only if y(a) = y(b) = 0.

Comparing the obtained inequalities, we see that under our assumptionsW (b)−W (a) = 0, implying q1(x) ≡ q2(x) and y(a) = y(b) = 0. Since z(a) =z(b) = 0, the solutions y(x) and z(x) must be proportional, i.e. y(x) =Cz(x), where C is a constant. (Indeed, taking C such that y′(a) = Cz′(a),we obtain y(x) ≡ Cz(x) due to the uniqueness theorem for the second orderODE.)

Corollary 3.3. Let us assume that q1(x) = q2(x) for all x ∈ [a, b] (that is,the equations (3.8) and (3.9) coincide).Then W (x) = c = const for any pairy, z of solutions (with the constant c depending of the choice of the pair).


Sturm-Liouville’s operators

1. Notations. Let us introduce the Sturm-Liouville operators of theform L = − d2

dx2 + q(x). Let us define L as a linear operator with thedomain DL consisting of all v ∈ C2([0, l]) satisfying the boundary conditionsv(0) = v(l) = 0. The Sturm-Liouville problem is to find eigenvalues andeigenfunctions of this operator.

Let us recall that an eigenvalue λ and an eigenvector v of the operator Lare defined by the relation Lv = λv, where it is assumed that λ ∈ C, v ∈ DL

and v 6= 0.

2. Symmetry. The operator L is symmetric, i.e.

(Lv1, v2) = (v1, Lv2) for v1, v2 ∈ DL,

where (·, ·) is the scalar product (sometimes also called the inner product)

(u, v) =∫ l

0u(x)v(x)dx.

Indeed,

(Lv1, v2)− (v1, Lv2) =∫ l

0[−v′′1 v2 + v1v

′′2 ]dx =

=∫ l

0

d

dx[−v′1v2 + v1v

′2]dx = [−v′1v2 + v1v

′2]|l0 = 0.

This implies that all the eigenvalues are real and the eigenfunctions corre-sponding to different eigenvalues are orthogonal.

3. Simplicity. All eigenvalues are simple, i.e. all eigenspaces are 1-dimensional. This was actually established above in the proof of Theorem3.2. In short, any two solutions of the equation y′′ + p(x)y′ + q(x)y = 0,that vanish at some point x0, are proportional to each other (and each ofthem is proportional, due to the uniqueness theorem, to the solution withthe initial conditions y(x0) = 0, y′(x0) = 1).

4. Enumeration of the eigenvalues. For the convenience of notations wewill use a complex parameter k =

√λ, so that the main equation takes the

form

(3.13) −X ′′(x) + q(x)X(x) = k2X(x),

and we will use the Dirichlet boundary conditions X(0) = X(l) = 0 whenindicated. Denote by ψ = ψ(x, k) the solution of (3.13) with the initialconditions

ψ(0, k) = 0, ψ′x(0, k) = k

3.3. Eigenvalues and eigenfunctions 53

(if q(x) ≡ 0, then ψ(x, k) = sin kx). Since the eigenvalues of L are real, wewill be mostly interested in the case when k ≥ 0 if k2 ≥ 0, and we can specifyk = ±i|k| when necessary. The positive eigenvalues of the Sturm-Liouvilleoperator L have the form λ = k2, where k is such that ψ(l, k) = 0.

The Sturm theorem implies that the number of the null-points of ψ(x, k)belonging to any fixed segment [0, a], where a ≤ l, is an increasing functionof k. Indeed, if 0 = x1(k) < x2(k) < · · · < xn(k) are all such zeros of ψ(·, k)(xn(k) ≤ a), and k′ > k, then every open interval (xj(k), xj+1(k)) containsat least one zero of ψ(·, k′). Recalling that x1(k) = 0 for all k, we see thatthe total number of zeros of ψ(·, k′) on [0, a] is greater or equal than n, whichproves the claim.

Therefore, as k grows, all zeros of ψ(x, k) move to the left. In fact, it iseasy to see that this motion is continuous, even smooth. Indeed, the zerosxj(k) satisfy the equation ψ(xj(k), k) = 0. Note that the function ψ =ψ(x, k) has continuous partial derivatives in (x, k) and ∂ψ

∂x (xj(k), k) 6= 0, asobserved above. By the Implicit Function Theorem, the function xj = xj(k)is continuous and has continuous derivative x′j(k), which can be obtained ifwe differentiate the equation ψ(xj(k), k) = 0 with respect to k, arriving to

x′j(k) = −∂ψ∂k∂ψ∂x

∣∣∣∣∣x=xj(k)

.

The eigenvalues correspond to the values k for which a new zero appearsat l. Since the number of these zeros is finite for any k, this implies that theeigenvalues form a discrete sequence

λ1 < λ2 < λ3 < . . . ,

which is either finite or tends to infinity. The eigenfunction Xn(x) corre-sponding to the eigenvalue λn, has exactly n− 1 zeros on the open interval(0, l).

It is easy to see that the number of eigenvalues is actually infinite. In-deed, Sturm’s theorem implies that the number of zeros of ψ(x, k) on (0, l)is not less that the number of zeros on (0, l) for the corresponding solutionof the equation

−y′′ +My = k2y, where M = supx∈[0,l]

q(x).

But this solution is c sin√k2 −Mx with c 6= 0, and the number of its zeros

on (0, l) tends to infinity as k → +∞. Thus, we have proved


Theorem 3.4. The eigenvalues are simple (non-multiple) and they form anincreasing sequence

λ1 < λ2 < λ3 < . . .

tending to +∞. If q1 ≥ 0, then λ1 > 0. The eigenfunctions Xn(x), cor-responding to different eigenvalues λn, are orthogonal. The eigenfunctionXn(x) has exactly n− 1 zeros on the open interval (0, l).

Remark 3.5. To prove the simplicity of any eigenvalue λ of the Sturm-Liouville operator L = − d2

dx2 + q(x), with q ∈ C([0, l]) and the boundaryconditions y(0) = y(l) = 0, it suffices to consider the case λ = 0. Otherwisewe can replace q and λ by q−λ and 0 respectively, which shifts every eigen-value by subtracting λ and preserves the multiplicities and eigenfunctionsof every eigenvalue. So, let us assume λ = 0.

Let us take the linear space M consisting of all functions y ∈ C2([0, l]),which satisfy the equation −y′′ = qy. (Not all of them are the eigenfunctionsbecause we ignored the boundary conditions.) Clearly, taking the initialvalues of the solutions (or the map y 7→ By = y(0), y′(0)), we obtain alinear isomorphism B : M → R2. In particular, dimM = 2.

3.4. The short-wave asymptotics

Let us describe the asymptotic behavior of large eigenvalues and the corre-sponding eigenfunctions. The asymptotic formulas obtained are often calledthe high-frequency asymptotics or short-wave asymptotics, meaning that theycorrespond to high frequencies and therefore to short wavelengths in thenonstationary (i.e., time-dependent) problem.

The behavior of eigenvalues is easily described with the help of oscillationtheorems. Namely, the eigenvalues of the operator L = − d2

dx2 + q(x) arecontained between the eigenvalues of L1 = − d2

dx2 and the eigenvalues ofL2 = − d2

dx2 +M , where M = maxx∈[0,l]

q(x). Since the eigenvalues of L1 and L2

are equal to (πnl )2 and (πnl )2 +M respectively (with n = 1, 2, . . .), we get(πnl

)2≤ λn ≤

(πnl

)2+M

In particular, this implies the asymptotic formula

λn =(πnl

)2(

1 +O

(1n2

)),

3.4. The short-wave asymptotics 55

or, putting kn =√λn:

(3.14) kn =πn

l

(1 +O

(1n2

))=πn

l+O

(1n

).

Now, let us find the asymptotics of the eigenfunctions Xn(x). The idea isthat for large k the term k2X in (3.13) plays a greater role than q(x)X.Therefore, let us solve the equation

(3.15) ψ′′ + k2ψ = q(x)ψ

with the initial conditions

(3.16) ψ(0) = 0, ψ′(0) = k,

taking q(x)ψ as the right hand side. We get an integral equation for ψ =ψ(x) = ψ(x, k) which can be solved by successive approximations. Let uswrite

ψ(x) = C1(x) cos kx+ C2(x) sin kx,

then the equations for C1(x) and C2(x), obtained by variation of parameters,are

C ′1(x) cos kx+ C ′2(x) sin kx = 0,(3.17)

−kC ′1(x) sin kx+ kC ′2(x) cos kx = q(x)ψ.(3.18)

Solving these equations, we find C ′1(x) and C ′2(x). After integration thisdetermines C1(x) and C2(x) up to arbitrary additive constants which arefixed by the initial values

(3.19) C1(0) = 0, C2(0) = 1,

obtained from (3.16). Then

C1(x) = −∫ x

0q(τ)ψ(τ)

sin kτk

dτ,

C2(x) = 1 +∫ x

0q(τ)ψ(τ)

cos kτk

dτ,

implying an integral equation (of so called Volterra type):

(3.20) ψ(x) = sin kx+1k

∫ x

0sin k(x− τ)q(τ)ψ(τ)dτ.


Clearly, the solution ψ of this equation satisfies (3.15) and (3.16) so it isequivalent to (3.15) with initial conditions (3.16). To solve (3.20) by suc-cessive approximations method, consider the integral operator A defined bythe formula

Aψ(x) =∫ x

0sin k(x− τ)q(τ)ψ(τ)dτ.

The equation (3.20) becomes(I − 1

kA

)ψ = sin kx,

and its solution may be expressed in the form

(3.21) ψ =(I − 1

kA

)−1

sin kx =∞∑n=0

1knAn(sin kx),

provided the series in the right hand side of (3.21) converges and A can beapplied termwise.

It can be proved that the series (3.21) indeed converges uniformly inx ∈ [0, l] for all k but, since we are only interested in large k, we can restrictourselves to an obvious remark that it converges uniformly on [0, l] for largek because A is norm bounded uniformly in k as an operator in the Banachspace C([0, l]) (with sup norm), and the norm of sin kx in C([0, l]) is notgreater than 1. In particular, we get

(3.22) ψ(x, k) = sin kx+O

(1k

)as k → +∞.

This gives the short-wave asymptotics of the eigenfunctions Xn. Indeed, itis clear from (3.14) and (3.22) that

Xn(x) = ψ(x, kn) = sin(knx) +O

(1kn

)= sin

(πn

l+O

(1n

))+O

(1n

),

hence

Xn(x) = sinπnx

l+O

(1n

)as n→ +∞.

3.5. The Green function and completeness of the system of

eigenfunctions

Consider the Schrodinger operator L = − d2

dx2 + q(x) as a linear operatormapping its domain

DL = v(x) ∈ C2([0, l]) : v(0) = v(l) = 0

3.5. The Green function and eigenfunctions 57

into C([0, l]). We will see that for q ≥ 0 this operator is invertible and theinverse operator L−1 can be expressed as an integral operator

(3.23) L−1f(x) =∫ l

0G(x, y)f(y)dy,

where G ∈ C([0, l]× [0, l]). The function G(x, y) is called the Green functionof L. In general, if an operator is expressed in the form of the right handside on (3.23), then G(x, y) is called its kernel or Schwartz’ kernel (in honorof L. Schwartz). Therefore, L−1 has a continuous Schwartz kernel equal toG(x, y). Clearly, the function G(x, y) is uniquely defined by (3.23).

To find L−1f , we have to solve the equation

(3.24) −v′′(x) + q(x)v(x) = f(x)

with the boundary conditions

(3.25) v(0) = v(l) = 0.

This is performed by the variation of parameters. It is convenient to usetwo solutions y1(x), y2(x) of the homogeneous equation

−y′′(x) + q(x)y(x) = 0

satisfying the boundary conditions

(3.26) y1(0) = 0, y′1(0) = 1;

and

(3.27) y2(l) = 0, y′2(l) = −1.

Clearly, the solutions y1(x) and y2(x) are linearly independent since 0 is notan eigenvalue of L because q(x) ≥ 0. Now, by the variation of parameters,writing

v(x) = C1(x)y1(x) + C2(x)y2(x),

we arrive to the equations

C ′1(x)y1(x) + C ′2(x)y2(x) = 0(3.28)

C ′1(x)y′1(x) + C ′2(x)y′2(x) = −f(x).(3.29)

The determinant of this system of linear equations for C ′1(x) and C ′2(x) isthe Wronskian

W (x) = y1(x)y′2(x)− y′1(x)y2(x).


It is well-known (and is easy to verify by differentiation of W (x) – see calcu-lation in the proof of Theorem 3.2) that W (x)=const and W (x) = 0 if andonly if y1 and y2 are linearly dependent. Therefore, in our case

W (x) = W = const 6= 0.

Solving the system (3.28), (3.29) by Cramer’s rule, we get

C ′1(x) =1Wf(x)y2(x), C ′2(x) = − 1

Wf(x)y1(x).

Taking into account the boundary condition (3.25) for v and the boundaryconditions (3.26), (3.27) for y1, y2, we see that C1(x) and C2(x) should bechosen so that

C1(l) = 0, C2(0) = 0.

This implies

C1(x) = − 1W

∫ l

xf(ξ)y2(ξ)dξ, C2(x) = − 1

W

∫ x

0f(ξ)y1(ξ)dξ.

Therefore, v exists, is unique and is determined by the formula

v(x) = − 1W

∫ x

0y1(ξ)y2(x)f(ξ)dξ − 1

W

∫ l

xy1(x)y2(ξ)f(ξ)dξ,

which can be expressed in the form

(3.30) v(x) =∫ l

0G(x, ξ)f(ξ)dξ,

where

(3.31) G(x, ξ) = − 1W

[θ(x− ξ)y1(ξ)y2(x) + θ(ξ − x)y1(x)y2(ξ)],

where θ(z) is the Heaviside function (θ(z) = 1 for z ≥ 0 and θ(z) = 0 forz < 0).

Note that G(x, ξ) is continuous and symmetric, i.e.,

(3.32) G(x, ξ) = G(ξ, x).

(The latter property is clear without explicit computation. Indeed, L−1

should be symmetric due to symmetry of L, and the symmetry of L−1 isequivalent to that of G(x, ξ).)

Consider now the operator G in L2([0, l]) with the kernel G(x, ξ)

Gf(x) =∫ l

0G(x, ξ)f(ξ)dξ.

3.5. The Green function and eigenfunctions 59

Since the kernel is continuous, this is a symmetric compact operator. Bythe Hilbert-Schmidt theorem it has a complete orthogonal system of eigen-functions Xn(x) with real eigenvalues µn, where n = 1, 2, . . . , and µn → 0as n → ∞. (For the proofs of these facts from functional analysis see e.g.Sect. VI.5, especially Theorem VI.16, in Reed and Simon [22].) We have

(3.33) GXn = µnXn.

If Xn are continuous, then, applying L to (3.33), we see that

Xn = µnLXn.

This implies that µn 6= 0 and λn = 1/µn is an eigenvalue of L. The continuityof Xn follows easily from (3.33), provided µn 6= 0. In general, if f ∈ L2([0, l]),then Gf ∈ C([0, l]) because by the Cauchy–Schwarz inequality

|(Gf)(x′)− (Gf)(x′′)| =∣∣∣∣∫ l

0[G(x′, ξ)−G(x′′, ξ)]f(ξ)dξ

∣∣∣∣ ≤≤ sup

ξ∈[0,l]|G(x′, ξ)−G(x′′, ξ)| ·

∫ l

0|f(ξ)|dξ ≤

≤ supξ∈[0,l]

|G(x′, ξ)−G(x′′, ξ)| ·√l

(∫ l

0|f(ξ)|2dξ

)1/2

,

and G(x, ξ) is uniformly continuous on [0, l]× [0, l]. It remains to prove thatG cannot have zero as an eigenvalue on L2([0, l]). But if u ∈ L2([0, l]) andGu = 0, then u is orthogonal to the image of G, since then

(3.34) (u,Gf) = (Gu, f) = 0.

But all the functions v ∈ DL can be represented in the form Gf since thenv = G(Lv) by construction of G. Since DL is dense in L2([0, l]), we see that(3.34) implies u = 0.

Thus, the set of eigenfunctions of G coincides exactly with the set ofeigenfunctions of L. In particular, we have proved the completeness of thesystem of eigenfunctions of L in L2([0, l]).

The following properties of the Green function are easily deducible withthe help of (3.31):

a) G(x, ξ) has continuous derivatives up to the second order for x 6= ξ

and satisfies (also for x 6= ξ) the equation LxG(x, ξ) = 0 in x, where Lx =− d2

dx2 + q(x).


b) G(x, ξ) is continuous everywhere, whereas its derivative G′x has leftand right limits at x = ξ and the jump is equal to −1:

G′x(ξ + 0, ξ)−G′x(ξ − 0, ξ) = −1.

c) the boundary conditions G|x=0 = G|x=l = 0 are satisfied.

These properties may be used in order to find G without variation ofparameters. It is easy to verify that G(x, ξ) is uniquely defined by theseconditions.

Conditions a) and b) can be easily expressed with the help of the Diracδ-function. Later on we will do it more carefully, but now we just give someheuristic arguments (on the “physical” level of rigor). It is convenient toapply formula (3.23) which expresses L−1 in terms of its kernel from thevery beginning. Let us formally apply L to (3.23) and denote

δ(x, ξ) = LxG(x, ξ).

Then

(3.35) f(x) =∫ l

0δ(x, ξ)f(ξ)dξ.

It is clear that δ(x, ξ) should vanish for ξ 6= x and at the same time∫ l

0δ(x, ξ)dξ = 1.

Taking f with a small support and shifting the support we easily obtainfrom (3.35) that δ(x + z, ξ + z) = δ(x, ξ). Therefore, δ(x, ξ) should onlydepend on ξ−x, i.e., δ(x, ξ) = δ(ξ−x). We see that δ(x) vanishes for x 6= 0and at the same time ∫

δ(x)dx = 1.

There is no locally integrable function with this property; nevertheless, itis convenient to make use of the symbol δ(x) if it is a part of an integrand,keeping in mind that ∫

δ(x)f(x)dx = f(0).

The “function” δ(x) is called Dirac’s δ-function or the Dirac measure. Inthe theory of distributions (or generalized functions) δ(x) is understood asa linear functional on smooth functions whose value at a function f(x) isequal to f(0).

The δ-function can be interpreted as a point mass, or a point charge, ora point load. For example, let a force f be applied to a nonhomogeneous

3.6. Problems 61

string. Then when oscillations subside (e.g. due to friction), we see that thestationary form of the string v(x) should satisfy (3.24), (3.25) and, therefore,is expressed by formula (3.30). If the load is concentrated near a point ξand the total load equals 1, i.e.

∫f(x)dx = 1, then the form of the string

is precisely the graph of G(x, ξ). This statement can easily be made into amore precise one by passing to the limit when a so-called δ-like sequence ofloads fn is chosen, e.g. such that fn(x) ≥ 0; fn(x) = 0, for |x− ξ| ≥ 1/n and∫fn(x)dx = 1. We skip the details leaving them as an easy exercise for the

reader.

3.6. Problems

3.1. Derive an integral equation for a function Φ(x) satisfying

−Φ′′ + q(x)Φ = k2Φ, Φ(0) = 1, Φ′(0) = 0,

with q ∈ G([0, l]) and deduce from this integral equation the asymptotics ofΦ(x) and Φ′(x) as k → +∞.

3.2. Using the result of Problem 3.1, prove the existence of an infinite num-ber of eigenvalues for the following Sturm-Liouville problem:

−X ′′ + q(x)X = λX, X ′(0) = X(l) = 0.

Prove the symmetry of the corresponding operator and the orthogonality ofeigenfunctions.

3.3. Write explicitly the Green function of the operator L = − d2

dx2 with theboundary conditions v(0) = v(l) = 0. Give a physical interpretation.

3.4. Write explicitly the Green function of the operator L = − d2

dx2 + 1 withthe boundary conditions v′(0) = v′(l) = 0. Give a physical interpretation.

3.5. With the help of the Green function, prove the completeness of thesystem of eigenfunctions for the operator L = − d2

dx2 +q(x) with the boundaryconditions v′(0) = v′(l) = 0 corresponding to free ends.

3.6. Prove that if q(x) ≥ 0, then the Green function G(x, ξ) of L = − d2

dx2 +q(x) with the boundary conditions v(0) = v(l) = 0 is positive for x 6= 0, land ξ 6= 0, l.


3.7. Prove that under the conditions of the previous problem, the Greenfunction G(x, ξ) is a positive kernel, i.e. the matrix (G(xi, xj))ni,j=1 is positivedefinite for any finite set of distinct points x1, . . . , xn ∈ (0, l).

3.8. Expand the Green function G(x, ξ) of the Sturm-Liouville problem intothe series with respect to eigenfunctions Xn(x) and express in terms of theGreen function the following sums:

a)∞∑n=1

1λn,

b)∗∞∑n=1

1λ2n

, where λn (n = 1, 2, . . .) is the set of eigenvalues of this prob-

lem. In particular, calculate∞∑n=1

1n2 and

∞∑n=1

1n4 .

Chapter 4

Distributions

4.1. Motivation of the definition. Spaces of test functions.

H In Analysis and Mathematical Physics one often meets technical diffi-culties caused by non-differentiability of certain functions. The theory ofdistributions (generalized functions) makes it possible to overcome thesedifficulties. The Dirac δ-function, which naturally appears in many situ-ations (one of them was mentioned above, in Section 3.5), is an exampleof a distribution. In the theory of distributions a number of notions andtheorems of analysis acquire simpler formulations and are free of unnaturaland irrelevant restrictions.

The origin of the notion of distribution or generalized function may beexplained as follows. Let f(x) be a physical quantity (such as temperature,pressure, etc.) which is a function of x ∈ Rn. If we measure this quantityat a point x0 with the help of some device (thermometer, pressure gauge,etc.), then we actually measure a certain average of f(x) over a neighbor-hood of x0, that is an integral of the form

∫f(x)ϕ(x)dx, where ϕ(x) is a

function characterizing the measuring gadget and smeared somehow over aneighborhood of x0.

The idea is to completely abandon the function f(x) and consider insteada linear functional by assigning to each test function ϕ the number

(4.1) 〈f, ϕ〉 =∫f(x)ϕ(x)dx.

Considering arbitrary linear functionals (not necessarily of the form (4.1))we come to the notion of a distribution or generalized function. N

63

64 4. Distributions

H Let us introduce now the needed spaces of test functions. Let Ω bean open subset of Rn. Let E(Ω) = C∞(Ω) be the space of smooth (infinitelydifferentiable) functions on Ω; let D(Ω) = C∞0 (Ω) be the space of smoothfunctions with compact support in Ω, i.e., functions ϕ ∈ C∞(Ω) such thatthere exists a compact K ⊂ Ω with the property ϕ|Ω\K = 0.

In general, the support of ϕ ∈ C(Ω) (denoted by suppϕ) is the closure(in Ω) of the set x ∈ Ω : ϕ(x) 6= 0. Therefore, suppϕ is the minimalclosed set F ⊂ Ω such that ϕ|Ω\F = 0 or, equivalently, the complement tothe maximal open set G ⊂ Ω such that ϕ|G = 0. Therefore, D(Ω) consistsof ϕ ∈ C∞(Ω) such that suppϕ is compact in Ω.

For a compact subset K ⊂ Rn denote by D(K) = C∞0 (K) the space offunctions ϕ ∈ C∞(Rn) such that suppϕ ⊂ K.

Clearly, D(Ω) is a union of all D(K) over compacts K ⊂ Ω.

Finally, as a space of test functions we will also use S(Rn), L. Schwartz’sspace, consisting of functions ϕ(x) ∈ C∞(Rn) such that sup

x∈Rn|xα∂βϕ(x)| <

+∞ for any multiindices α and β.

Theory of distributions makes use of topology in the spaces of test func-tions. We will introduce it with the help of a system of seminorms. First,we will give general definitions.

A seminorm in a linear space E is a map p : E −→ [0,+∞) such that

a) p(x+ y) ≤ p(x) + p(y), x, y ∈ E;

b) p(λx) = |λ|p(x), x ∈ E, λ ∈ C.

The property p(x) = 0 does not necessarily imply x = 0 (if this implica-tion holds, then the seminorm is called a norm).

Let a system of seminorms pjj∈J , where J is a set of indices, be definedon E. For any ε > 0 and j ∈ J , set

Uj,ε = x : x ∈ E, pj(x) < ε

and introduce the topology in E whose basis of neighborhoods of zero con-sists of all finite intersections of the sets Uj,ε (a basis of neighborhoods ofany other point x0 is obtained by the shift by x0, i.e., consists of finite in-tersections of the sets x0 + Uj,ε). Therefore, G ⊂ E is open if and onlyif together with each point x0 it also contains a finite intersection of setsx0 + Uj,ε. Clearly, lim

n→+∞xn = x if and only if lim

n→+∞pj(xn − x) = 0 for any

j ∈ J . N

4.1. Motivation. Spaces of test functions 65

H To avoid perpetual reference to finite intersections, we may assumethat for any j1, j2 ∈ J there exists j3 ∈ J such that pj1(ϕ) ≤ pj3(ϕ) andpj2(ϕ) ≤ pj3(ϕ) for any ϕ ∈ E (in this case Uj3,ε ⊂ (Uj1,ε

⋂Uj2,ε)). If this

does not hold, we can add (to the existing system of seminorms) additionalseminorms of the form

pj1...jk(ϕ) = maxpj1(ϕ), . . . , pjk(ϕ).

This does not affect the topology defined by seminorms in E and we willalways assume that this is already done. In other words, we will assumethat the basis of neighborhoods of zero in E consists of all sets Uj,ε.

The same topology in E can be defined by a different set of seminorms.Namely, assume that we have another system of seminorms p′j′j′∈J ′ , withthe corresponding neighborhoods of 0 denoted U ′j′,ε. Clearly, the systemsof seminorms pjj∈J , p′j′j′∈J ′ define the same topology if and only if thefollowing two conditions are satisfied:

• for every j′ ∈ J ′ and ε′ > 0, there exist j ∈ J and ε > 0 such thatUj,ε ⊂ U ′j′,ε′ ;• for every j ∈ J and ε > 0, there exist j′ ∈ J ′ and ε′ > 0 such that

U ′j′,ε′ ⊂ Uj,ε.It is easy to see that these conditions are satisfied if and only if

• for every j′ ∈ J ′ there exist j ∈ J and C = Cj′,j > 0, such thatp′j′(x) ≤ Cpj(x), x ∈ E;

• for every j ∈ J there exist j′ ∈ J ′ and C ′ = C ′j,j′ > 0, such thatpj(x) ≤ C ′p′j′(x), x ∈ E.

If these conditions are satisfied for two systems of seminorms, pj, p′j′,then we will say that the systems are equivalent. Since we are only interestedin translation invariant topologies on vector spaces E, we can freely replacea system of seminorms by any equivalent system.

If f is a linear functional on E, then it is continuous if and only if thereexist j ∈ J and C > 0 such that

(4.2) |〈f, ϕ〉| ≤ Cpj(ϕ), ϕ ∈ E,

where 〈f, ϕ〉 denotes the value of f on ϕ.

Indeed, the continuity of f is equivalent to the existence of a set Uj,εsuch that |〈f, ϕ〉| ≤ 1 for ϕ ∈ Uj,ε and the latter is equivalent to (4.2) withC = 1/ε.

66 4. Distributions

Note that the seminorm pj itself, as a function on E, is continuous on E.Indeed, this is obvious at the point ϕ = 0 because the set ϕ| pj(ϕ) < ε isa neighborhood of 0 for any ε > 0. Now the continuity of pj at an arbitrarypoint ϕ0 ∈ E follows, because the triangle inequality a) above implies

p(ϕ0)− p(ϕ) ≤ pj(ϕ0 + ϕ) ≤ p(ϕ0) + p(ϕ),

so the continuity at ϕ0 follows from the continuity at 0.

We will usually suppose that the following separability condition is sat-isfied:

(4.3) if pj(ϕ) = 0 for all j ∈ J, then ϕ = 0.

This implies that any two distinct points x, y ∈ E have nonintersectingneighborhoods (the Hausdorff property of the topology). Indeed, there existsj ∈ J such that pj(x−y) > 0. But then the neighborhoods x+Uj,ε, y+Uj,εdo not intersect for ε < 1

2pj(x− y), because if

z = x+ t = y + s ∈ (x+ Uj,ε) ∩ (y + Uj,ε),

thenpj(x− y) = pj(s− t) ≤ pj(s) + pj(t) ≤ 2ε < pj(x− y).

Now, consider the case which is most important for us when J is a countable(infinite) set. We may assume that J = N. Introduce a metric on E by theformula

(4.4) ρ(x, y) =∞∑k=1

12k

pk(x− y)1 + pk(x− y)

.

The symmetry and the triangle inequality are easily checked. Due to sepa-rability condition (4.3), ρ(x, y) = 0 implies x = y.

Let us prove that the topology defined by the metric coincides with thetopology defined above by seminorms. Since the metric (4.4) is invariantwith respect to the translations (i.e., ρ(x+z, y+z) = ρ(x, y)), it is sufficientto verify that each neighborhood Uj,ε contains a ball Bδ(0) = x : ρ(x, 0) <δ of radius δ > 0 with the center at the origin, and, conversely, each ballBδ(0) contains a set of the form Uj,ε. (Note that in the future we will alsowrite B(δ, 0) instead of Bδ(0).)

First, take an arbitrary δ > 0. Let us prove that there exist j, ε suchthat Uj,ε ⊂ Bδ(0). If j is such that p1(x) ≤ pj(x), . . . , pN (x) ≤ pj(x), then

ρ(x, 0) ≤N∑k=1

pk(x)2k

+∞∑

k=N+1

12k

pk(x)1 + pk(x)

≤ pj(x) +1

2N.

4.1. Motivation. Spaces of test functions 67

Therefore, if ε < δ/2 and 1/2N < δ/2, then Uj,ε ⊂ Bδ(0).

Conversely, given j and ε, we deduce the existence of δ > 0 such thatBδ(0) ⊂ Uj,ε from the obvious inequality

12j

pj(x)1 + pj(x)

≤ ρ(x, 0).

Namely, take δ so that δ <12j

ε

1 + ε. Then, since f(t) = 1

2jt

1+t is strictly

increasing for t > 0, the inequality ρ(x, 0) < δ implies pj(x) < ε. N

Definition 4.1. A countably normed space is a vector space endowed witha countable number of seminorms satisfying the separability condition.

Acting by induction, we can always replace a countable system of semi-norms in E by an increasing system of seminorms p1 ≤ p2 ≤ p3 ≤ . . . , without changing topology. (Here the inequality p ≤ p′ between two semi-norms means that p(x) ≤ p′(x) for every x ∈ E.) Indeed, we can al-ways replace p2(x) by p′2(x) = maxp1(x), p2(x), x ∈ E, then p3(x) byp′3(x) = maxp1(x), p2(x), p3(x), etc.

Since the topology in a countably normed space E is defined by a metric,the notion of completeness of E is well defined with respect to this metric(which is chosen as in (4.4)). Note, however, that completeness for generalmetric spaces is not determined by the topology (that is, for two metricsdefining the same topology, one may be complete, while another is not). Sofor the moment, we only allow choosing the metric as in (4.4), and try toformulate the notion of completeness in the language of topology. Let usrecall that a metric space E with a metric ρ is called complete if every Cauchysequence xj | j = 1, 2, . . . ⊂ E converges in E (with respect to the metricρ). Here a sequence xj | j = 1, 2, . . . is called a Cauchy sequence (withrespect to ρ) if ρ(xj , xk) → 0, as j, k → +∞. But due to the translationinvariance of the metric (and topology), this condition can be also formulatedas follows:

(4.5) xj − xk → 0 as j, k → +∞,

or, in other words, for every neighborhood U of 0 in E there exists m =m(U) > 0, such that for every j, k, which are both greater than m, we havexj−xk ∈ U . Clearly, the last condition only depends upon the topology in E.But the notion of convergence also depends on the topology only. Therefore,the completeness property of a countably normed space is independent ofthe choice of a system of seminorms in the same equivalence class.

68 4. Distributions

Now we can also formulate the notion of completeness in terms of semi-norms. Using (4.5) and the definition of topology in terms of seminorms,we see that a sequence xj | j = 1, 2, . . . is a Cauchy sequence (in terms oftopology or metric) if and only if

(4.6) pl(xj − xk)→ 0, as j, k → +∞,for every seminorm pl. So to check the completeness of a countably normedspace it suffices to check convergence of the sequences satisfying (4.6), whichis sometimes more convenient.

Definition 4.2. Frechet space is a topological vector space whose topologyis defined by a structure of a complete countably normed topological vec-tor space. In other words, it is a complete topological vector space witha countable system of seminorms defined up to the equivalence explainedabove. Equivalently, we can also say that a Frechet space is a topologi-cal vector space with a selected equivalence class of countable systems ofseminorms.

Since the topology in a countably normed space E can be determinedby a metric ρ(x, y), the continuity of functions on E and, in general, of anymaps of E into a metric space, can be determined in terms of sequences. Forinstance, a linear functional f on E is continuous if and only if lim

k→+∞ϕk = 0

implies limk→+∞

〈f, ϕk〉 = 0.

Definition 4.3. The space of continuous linear functionals in E is calledthe dual space to E and is denoted by E′.

Now, let us turn D(K), E(Ω) and S(Rn) into countably normed spaces.

1) Space D(K). Denote

pm(ϕ) =∑|α|≤m

supx∈K|∂αϕ(x)|, ϕ ∈ D(K).

The seminorms pm(ϕ), m = 1, 2, . . . , define a structure of a countablynormed space on D(K). Clearly, the convergence ϕp → ϕ in the topol-ogy of D(K) means that if α is any multiindex, then ∂αϕp(x) → ∂αϕ(x)uniformly on K.

Let us show that D(K) is in fact a Frechet space. The only non-obviousthing to check is its completeness. Let ϕj | j = 1, 2, . . . be a Cauchysequence in D(K). This means that for any multiindex α the sequence∂αϕj is a Cauchy sequence in C0(K), the space of all continuous functions

4.2. Spaces of distributions 69

on Rn supported in K (i.e. vanishing in Rn \ K), with the sup norm p0.Then there exists ψα ∈ C0(K) such that ∂αϕj → ψα in C0(K) as j → +∞(i.e. converges uniformly on Rn). It remains to prove that in fact ψα = ∂αψ0

for all α. But this follows from the standard analysis results on possibility todifferentiate a uniformly convergent sequence, if the sequence of derivativesalso converges uniformly.

2) Space E(Ω). Let Kl ⊂ Ω, l = 1, 2, . . ., be a sequence of compacts suchthat K1 ⊂ K2 ⊂ K3 ⊂ . . . and for every x ∈ Ω there exists l such that xbelongs to the interior of Kl (i.e. K contains a neighborhood of x). Forinstance, we may set

Kl = x : x ∈ Ω, |x| ≤ l, ρ(x, ∂Ω) ≥ 1/l,

where ∂Ω is the boundary of Ω (i.e., ∂Ω = Ω\Ω) and ρ is the usual Euclideandistance in Rn. Set

pl(ϕ) =∑|α|≤l

supx∈Kl

|∂αϕ(x)|, ϕ ∈ E(Ω).

These seminorms turn E(Ω) into a countably normed space. Clearly, theconvergence ϕp → ϕ in the topology of E(Ω) means that ∂αϕp(x)→ ∂αϕ(x)uniformly on any compact K ⊂ Ω and for any fixed multiindex α. Inparticular, this implies that the topology defined in this way does not dependon the choice of a system of compacts Kl described above.

3) Space S(Rn). Define the system of seminorms

pk(ϕ) =∑

|α+β|≤k

supx∈Rn

|xα∂βϕ(x)|,

and the convergence ϕp → ϕ in the topology of S(Rn) means that if A isa differential operator with polynomial coefficients on Rn, then Aϕp(x) →Aϕ(x) uniformly on Rn.

Both E(Ω) and S(Rn) are also Frechet spaces. Their completeness canbe established by the same arguments as the ones used for the space D(K)above.

4.2. Spaces of distributions

The above procedure of transition from E to its dual E′ makes it possibleto define E ′(Ω) and S′(Rn) as dual spaces of E(Ω) and S(Rn), respectively.We have not yet introduced topology in D(Ω) (the natural topology in D(Ω)is not easy to define) but we will not need it. Instead we define D′(Ω) as

70 4. Distributions

the space of linear functionals f on D(Ω) such that f |D(K) is continuous onD(K) for any compact K ⊂ Ω. This continuity can be also defined in termsof sequences (rather than seminorms), because the topology in D(K) can bedefined by a metric.

It is also convenient to introduce the convergent sequences in D(Ω),saying that a sequence ϕkk=1,2,..., ϕk ∈ D(Ω), converges to ϕ ∈ D(Ω) ifthe following two conditions are satisfied:

(a) there exists a compact K ⊂ Ω such that ϕ ∈ D(K) and ϕk ∈ D(K)for all k;

(b) ϕk → ϕ in D(K), or, in other words, for every fixed multiindex α,∂αϕk → ∂αϕ uniformly on K (or on Ω).

Using this definition, we can easily see that a linear functional f on D(Ω)is a distribution if and only if 〈f, ϕk〉 → 0 for every sequence ϕk, such thatϕk → 0 in D(Ω).

Definition 4.4.

(1) The elements of D′(Ω) are called distributions in Ω.

(2) The elements of E ′(Ω) are called distributions with compact supportin Ω. (The notion of support of a distribution that justifies theabove term will be defined later.)

(3) The elements of S′(Rn) are called tempered distributions on Rn.

Example 4.1. The “regular” distributions on Ω. Let L1loc(Ω) be the space of

(complex-valued) functions on Ω which are Lebesgue integrable with respectto the Lebesgue measure on any compact K ⊂ Ω. To any f ∈ L1

loc(Ω), assignthe functional on D(Ω) (denoted by the same letter), setting

(4.7) 〈f, ϕ〉 =∫

Ωf(x)ϕ(x)dx,

where ϕ ∈ D(Ω). Clearly, we get a distribution on Ω (called regular). Animportant fact is that if two functions f1, f2 ∈ L1

loc(Ω) determine the samedistribution, they coincide almost everywhere. This is a corollary of thefollowing lemma.

Lemma 4.5. Let f ∈ L1loc(Ω) and

∫Ω f(x)ϕ(x)dx = 0 for any ϕ ∈ D(Ω).

Then f(x) = 0 for almost all x.


Proof of this lemma requires the existence of an ample set of functions inD(Ω). We do not know yet whether there exist such nontrivial (not identi-cally vanishing) functions. Let us construct a supply of functions belongingto D(Rn).

First, let ψ(t) = θ(t)e−1/t, where t ∈ R and θ(t) is the Heaviside function.Clearly, ψ(t) ∈ C∞(R). Therefore, if we consider the function ϕ(x) =ψ(1 − |x|2) in Rn, we get a function ϕ(x) ∈ C∞0 (Rn) such that ϕ(x) = 0for |x| ≥ 1 and ϕ(x) > 0 for |x| < 1. It is convenient to normalize ϕ

considering instead the function ϕ1(x) = c−1ϕ(x), where c =∫ϕ(x)dx.

Now, set ϕε(x) = ε−nϕ1(x/ε). Then ϕε(x) = 0 for |x| ≥ ε and ϕε(x) > 0for |x| < ε. Besides,

∫ϕε(x)dx = 1.

Now, introduce an important averaging (or mollifying) operation: forany f(x) ∈ L1

loc(Ω) take a family of convolutions

(4.8) fε(x) =∫f(x− y)ϕε(y)dy =

∫f(y)ϕε(x− y)dy

defined for x ∈ Ωε = x : x ∈ Ω, ρ(x, ∂Ω) > ε. It is clear from thedominated convergence theorem that the last integral in (4.8) is a C∞-function and

∂αfε(x) =∫f(y)(∂αϕε)(x− y)dy,

hence fε ∈ C∞(Ωε). Note also the following properties of averaging:

a) If f(x) = 0 outside of a compact K ⊂ Ω, then fε(x) = 0 outside of theε-neighborhood of K. In particular, in this case fε ∈ D(Ω) for a sufficientlysmall ε > 0.

b) If f(x) = 1 for x ∈ Ω2ε, then fε(x) = 1 for x ∈ Ω3ε. In particular,if f is the characteristic function of Ω2ε, then fε ∈ D(Ω) and fε(x) = 1 forx ∈ Ω3ε.

c) If f ∈ C(Ω), then fε(x) → f(x) uniformly on any fixed compactK ⊂ Ω as ε→ 0+.

Indeed,

fε(x)− f(x) =∫

[f(x− y)− f(x)]ϕε(y)dy,

implying

|fε(x)− f(x)| ≤ sup|y|≤ε|f(x− y)− f(x)|

so that the uniform continuity of f on the ε-neighborhood of K implies ourstatement.

72 4. Distributions

d) If f ∈ Lploc(Ω), where p ≥ 1, then fε → f with respect to the norm inLp(K) for any fixed compact K ⊂ Ω as ε→ 0+.

Since the values fε(x) at x ∈ K only depend on the values of f onthe ε-neighborhood of the compact K, we may assume that f(x) = 0 forx ∈ Ω\K1, where K1 is a compact in Ω. Let the norm in Lp(Ω) be denotedby ‖u‖p, i.e.

‖u‖p =(∫

Ω|u(x)|pdx

)1/p

.

Due to Minkowski’s inequality (see Appendix to this chapter: Sect. 4.7)

‖fε‖p =∥∥∥∥∫ f(· − y)ϕε(y)dy

∥∥∥∥p

≤∫‖f(· − y)‖p|ϕε(y)|dy = ‖f‖p.

Due to c), the property d) holds for f ∈ C(Ω). But if f ∈ Lp(Ω) and f = 0outside K1, then we can approximate f with respect to the norm in Lp(Ω)first, by step functions, and then by continuous functions on Ω which vanishoutside a compact K2 ⊂ Ω. Let g be a continuous function on Ω whichvanishes outside K2 and such that ‖f − g‖p < δ. Then

lim supε→0+

‖fε−f‖p ≤ lim supε→0+

‖fε−gε‖p+lim supε→0+

‖gε−g‖p+lim supε→0+

‖g−f‖p =

lim supε→0+

‖(f − g)ε‖p + ‖g − f‖p ≤ 2‖f − g‖p ≤ 2δ,

which due to the arbitrariness of δ > 0 implies lim supε→0+

‖fε − f‖p = 0, as

required.

Proof of Lemma 4.5. Let f ∈ L1loc(Ω) and

∫Ω f(x)ϕ(x)dx = 0 for every

ϕ ∈ D(Ω). This implies fε(x) = 0 for all x ∈ Ωε. Therefore, the statementof the lemma follows from the property d).

Lemma 4.5 allows us to identify the functions f ∈ L1loc(Ω) with the

distributions that they define by formula (4.7). Note that the expression

(4.9) 〈f, ϕ〉 =∫f(x)ϕ(x)dx

is also often used for distributions f (in this case the left-hand side of (4.9)is the definition for the right hand side; when the right-hand side makessense, this definition is consistent).


Example 4.2. “Regular” distributions in E ′(Ω) and S′(Rn). If f ∈ L1comp(Ω),

i.e., f ∈ L1(Rn) and f(x) = 0 outside a compact K ⊂ Ω, then we may con-struct via the standard formula (4.9) a functional on E(Ω) which is a dis-tribution with compact support. We will identify f with the correspondingelement of E ′(Ω).

If f ∈ L1loc(Rn) and

(4.10)∫|f(x)|(1 + |x|)−N < +∞

for some N > 0, then (4.9) defines a functional on S(Rn) which is a tempereddistribution. In particular, (4.10) holds if f is measurable and

|f(x)| ≤ C(1 + |x|)N−n−1.

Example 4.3. Dirac’s δ-function. This is the functional defined as follows:

(4.11) 〈δ(x− x0), ϕ(x)〉 = ϕ(x0).

If x0 ∈ Ω, then δ(x − x0) ∈ D′(Ω) and, moreover, δ(x − x0) ∈ E ′(Ω).Obviously, δ(x− x0) ∈ S′(Rn) for any x0 ∈ Rn. Instead of (4.11) one oftenwrites, in accordance with the above-mentioned convention,∫

δ(x− x0)ϕ(x)dx = ϕ(x0),

though, clearly, δ(x−x0) is not a regular distribution. Indeed, if we assumethat δ(x−x0) is regular, then Lemma 4.5 would imply that δ(x−x0) vanishesalmost everywhere in Rn\x0, hence, it vanishes almost everywhere in Rn,and so vanishes as a distribution in Rn.

Example 4.4. Let L be a linear differential operator in Rn, Γ a smoothcompact hypersurface in Rn, dS the area element of Γ. Then the formula

(4.12) ϕ 7→∫

Γ(Lϕ)|ΓdS

defines a nonregular distribution with compact support in Rn. As Γ we maytake a compact submanifold (perhaps with a boundary) of any codimensionin Rn. In this case as dS we may take any density on Γ (or a differentialform of the maximal order if the orientation of Γ is fixed). In particular, theDirac δ-function appears as a particular case when Γ is a point.

74 4. Distributions

4.3. Topology and convergence in the spaces of distributions

Let F be one of the spaces of test functions D(Ω), E(Ω) or S(Rn), and letF ′ be the corresponding dual space of distributions (continuous linear func-tionals on F). We will consider the space F ′ with so-called weak topology,i.e., the topology defined by the seminorms

pϕ(f) = |〈f, ϕ〉|, ϕ ∈ F , f ∈ F ′.

In particular, the convergence fk → f in this topology means that

〈fk, ϕ〉 → 〈f, ϕ〉 for any ϕ ∈ F .

In this case we will also use the usual notation limk→∞

fk = f .

Up to now we assumed that the limit f is itself a distribution. Now letus simply assume that fk| k = 1, 2, . . . is a sequence of distributions (fromF ′), such that for any ϕ ∈ F there exists a limit

〈f, ϕ〉 := limk→+∞

〈fk, ϕ〉.

Will the limit f be always a distribution too? It is clear that the limit is alinear functional with respect to ϕ. But will it be continuous? The answer isaffirmative, and the corresponding property is called completeness (or weakcompleteness) of the distribution space. It is a particular case of a generalresult which holds for any Frechet space (see Appendix to this chapter: Sect.4.8).

Example 4.5. δ-like sequence. Consider the above constructed functionsϕε(x) ∈ D(Rn). Recall that they satisfy:

a) ϕε(x) ≥ 0 for all x;

b)∫ϕε(x)dx = 1;

c) ϕε(x) = 0 for |x| ≥ ε.Let us prove that the properties a)–c) imply

(4.13) limε→0+

ϕε(x) = δ(x)

in D′(Rn). Indeed, this means that

(4.14) limε→0+

∫ϕε(x)ψ(x)dx = ψ(0) for any ψ ∈ D(Rn).

With the help of b) we may rewrite (4.14) in the form of the relation

limε→0+

∫ϕε(x)[ψ(x)− ψ(0)]dx = 0,

4.3. Topology and convergence 75

which is obviously true due to the estimate∣∣∣∣∫ ϕε(x)[ψ(x)− ψ(0)]dx∣∣∣∣ ≤ sup

|x|≤ε|ψ(x)− ψ(0)|.

Note that (4.13) holds not only in D′(Rn) but also in S′(Rn), E ′(Rn) andeven in E ′(Ω) if 0 ∈ Ω.

Sequences of “regular” functions converging to the δ-function are calledδ-like sequences. Properties a)–c) can be considerably weakened retainingδ-likeness of the sequence. Thus, in the theory of Fourier series one provesthe δ-likeness (e.g. in D′((−π, π))) of the sequence of Dirichlet kernels

Dk(t) =1

2πsin(k + 1

2)tsin t

2

,

defined by the condition that 〈Dk, ϕ〉 is the sum of the k first terms of theFourier series of ϕ at t = 0. Similarly, a δ-like sequence in D′((−π, π)) isconstituted by the Fejer kernel

Fk(t) =1

2πksin2 kt

2

sin2 t2

,

defined from the condition that 〈Fk, ϕ〉 is the arithmetic mean of the k firstpartial sums of the Fourier series. Later on we will encounter a number ofother important examples of δ-like sequences. Note that the δ-likeness isessentially always proved in the same way as for ϕε(x).

Example 4.6. Distributions 1x±i0 and v.p. 1

x . The “regular” functions 1x+iε ,

and 1x−iε (here i =

√−1) of x ∈ R1 for ε > 0 determine tempered distribu-

tions (elements of S′(R1)). It turns out that there exist limits in S′(R1)

1x+ i0

= limε→0+

1x+ iε

,1

x− i0 = limε→0+

1x− iε

The left hand sides here are, by definition, the limits on the right hand sides.

Let us prove, e.g. the existence of the limit 1x+i0 in S′(R1).

We must prove that if ϕ ∈ S(R1) then

(4.15) limε→0+

∫ ∞−∞

ϕ(x)x+ iε

dx

exists and is a continuous linear functional of ϕ ∈ S(R1). The simplest wayto do this is to integrate by parts. Namely, we have

1x+ iε

=d

dxln(x+ iε),

76 4. Distributions

where we take an arbitrary branch of ln(x+ iε) (but keeping continuity forall x ∈ R, ε > 0). Integrating by parts, we get

(4.16)∫ ∞−∞

ϕ(x)x+ iε

dx = −∫ ∞−∞

ϕ′(x) ln(x+ iε)dx.

Since ln(x+ iε) = ln |x+ iε|+ iarg (x+ iε), it is clear that, by the dominatedconvergence theorem, the limit of the right-hand side of (4.16), as ε→ 0+,is equal to(4.17)

−∫ ∞−∞

(ln |x|)ϕ′(x)dx− iπ∫ 0

−∞ϕ′(x)dx = −

∫ ∞−∞

(ln |x|)ϕ′(x)dx− πiϕ(0).

Since ln |x| ∈ L1loc(R) and ln |x| grows at infinity not faster than |x|ε for any

ε > 0, it is obvious that the right hand side of (4.17) is finite and defines acontinuous functional of ϕ ∈ S(Rn). This functional is denoted by 1

x+i0 .

Let us study in detail the first summand in (4.17). We have

−∫ ∞−∞

(ln |x|)ϕ′(x)dx = limε→0+

[−∫|x|≥ε

ln |x| · ϕ′(x)dx

]=

(4.18) = limε→0+

[ln |x| · ϕ(x)|ε−ε +

∫|x|≥ε

1xϕ(x)dx

]=

= limε→0+

∫|x|≥ε

1xϕ(x)dx,

since limε→0+

ln ε[ϕ(ε) − ϕ(−ε)] = limε→0+

O(ε) ln ε = 0. In particular, we have

proved that the last limit in (4.18) exists and defines a functional belongingto S′(R1). This functional is denoted v.p. 1

x (the letters v.p. stand for thefirst letters of French words “valeur principale” meaning the principal value).Therefore, by definition,

(4.19) 〈v.p.1x, ϕ〉 = lim

ε→0+

∫|x|≥ε

1xϕ(x)dx.

Moreover, we may now rewrite (4.17) in the form

(4.20)1

x+ i0= v.p.

1x− πiδ(x).

Similarly, we get

(4.21)1

x− i0 = v.p.1x

+ πiδ(x).

4.4. The support of a distribution 77

Formulas (4.20) and (4.21) are called Sokhotsky’s formulas. They, in par-ticular, imply

1x+ i0

− 1x− i0 = −2πiδ(x).

The existence of limits in (4.15) and (4.19) may be also proved by expandingϕ(x) in the sum of a function equal to ϕ(0) in a neighborhood of 0 and afunction which vanishes at 0. For each summand, the existence of limits iseasy to verify. In the same way, one gets the Sokhotsky formulas, too.

Distributions 1x±i0 and v.p. 1

x are distinct “regularizations” of the nonin-tegrable function 1

x , i.e., they allow us to make sense of the divergent integral∫∞−∞

1xϕ(x)dx. We see that there is an ambiguity: various distributions cor-

respond to the nonintegrable function 1x . The regularization procedure is

important if we wish to use 1x as a distribution (e.g. to differentiate it).

This procedure (or a similar one) is applicable to many other non-integrable functions.

4.4. The support of a distribution

Let Ω1,Ω2 be two open subsets of Rn and Ω1 ⊂ Ω2. Then D(Ω1) ⊂ D(Ω2).Therefore, if f ∈ D′(Ω2) we may restrict f to D(Ω1) and get a distributionon Ω1 denoted by f |Ω1 . It is important that the operation of restriction hasthe following properties:

a) Let Ωj : j ∈ J be a cover of an open set Ω by open sets Ωj , j ∈ J ,i.e., Ω =

⋃j∈J Ωj . If f ∈ D′(Ω) and f |Ωj = 0 for j ∈ J , then f = 0.

b) Let again Ω =⋃j∈J Ωj and let fj ∈ D′(Ωj) a set of distributions such

that fk|Ωk∩Ωl = fl|Ωk∩Ωl for any k, l ∈ J . Then there exists a distributionf ∈ D′(Ω) such that f |Ωj = fj for any j ∈ J .

Properties a) and b) mean that distributions constitute a sheaf.

Let us prove the property a), which is important for us. To do this, letus use a partition of unity, i.e. a family of functions ϕjj∈J , such that

1) ϕj ∈ C∞(Ω), suppϕj ⊂ Ωj ;

2) The family ϕjj∈J is locally finite, i.e., any point x0 ∈ Ω has aneighborhood in which only a finite number of functions of this family,ϕj1 , . . . , ϕjk , do not vanish;

3)∑

j∈J ϕj ≡ 1.

78 4. Distributions

The existence of a partition of unity is proved in courses of geometry andanalysis, see e.g. Theorem 10.8 in Rudin [23] or Theorem 1.11 in Warner[32].

If ϕ ∈ D(Ω), then property 1) implies ϕjϕ ∈ D(Ωj) and properties 2)and 3) imply

ϕ =∑j∈J

ϕjϕ,

where the sum locally contains only a finite number of summands which arenot identically zero, since the compact suppϕ intersects, due to property 2),only a finite number of supports suppϕj . If f |Ωj = 0 for all j, then

〈f, ϕ〉 =∑j∈J〈f, ϕjϕ〉 = 0,

implying f = 0, due to the arbitrariness of ϕ. This proves property a).

Property b) may be proved by constructing f ∈ D′(Ω) from distributionsfj ∈ D′(Ωj) with the help of the formula

〈f, ϕ〉 =∑j∈J〈fj , ϕjϕ〉.

If ϕ ∈ D(Ωk), where k is fixed, then

〈f, ϕ〉 =∑j∈J〈fj , ϕjϕ〉 =

∑j∈J〈fk, ϕjϕ〉 = 〈fk,

∑j∈J

ϕjϕ〉 = 〈fk, ϕ〉

since supp(ϕjϕ) ⊂ suppϕj ∩ suppϕ ⊂ Ωj ∩ Ωk. We have therefore verifiedthat f |Ωk = fk, proving property b).

Property a) allows us to introduce for f ∈ D′(Ω) a well-defined maximalopen subset Ω′ ⊂ Ω such that f |Ω′ = 0 (here Ω′ is the union of all opensets Ω1 ⊂ Ω such that f |Ω1 = 0). The closed subset F = Ω\Ω′ is called thesupport of the distribution f (denoted by supp f). Therefore, supp f is thesmallest closed subset F ⊂ Ω such that f |Ω\F = 0.

Example 4.7. supp δ(x) = 0. The support of the distribution fromExample 4.4 belongs to Γ. In general, if supp f ⊂ F , then f is said to besupported on F . Therefore, a distribution of the form (4.12) is supported onΓ.

Now, let us define supports of distributions from E ′(Ω) and S′(Rn). Thisis easy if we note that there are canonical embeddings

E ′(Ω) ⊂ D′(Ω),

4.4. The support of a distribution 79

S′(Rn) ⊂ D′(Rn).

They are constructed with the help of embeddings

D(Ω) ⊂ E(Ω),

D(Rn) ⊂ S(Rn),

which enable us to define maps

(4.22) E ′(Ω) −→ D′(Ω)

and

(4.23) S′(Rn) −→ D′(Rn)

by restricting functionals onto the smaller subspace.

The continuity of these functionals on D(Ω) and D(Rn) follows from thecontinuity of embeddings

D(K) ⊂ E(Ω) for a compact K in Ω

D(K) ⊂ S(Rn) for a compact K in Rn.

Finally, the dual maps (4.22), (4.23) are embeddings, because D(Ω) is densein E(Ω) and D(Rn) is dense in S(Rn), therefore any continuous linear func-tional on E(Ω) (resp. S(Rn)) is uniquely determined by its values on a densesubset D(Ω) (resp. D(Rn)).

In what follows, we will identify distributions from E ′(Ω) and S′(Rn)with their images in D′(Ω) and D′(Rn), respectively.

Proposition 4.6. Let f ∈ D′(Ω). Then f ∈ E ′(Ω) if and only if supp f is

a compact subset of Ω, or, in other words, if f is supported on a compact

subset of Ω.

Proof. Let f ∈ E ′(Ω). Then f is continuous with respect to a seminorm inE(Ω), i.e.,

(4.24) |〈f, ϕ〉| ≤ C∑|α|≤l

supx∈K|∂αϕ(x)|,

where the compact K ⊂ Ω and numbers C, l do not depend on ϕ. But thisimplies that 〈f, ϕ〉 = 0 for any ϕ ∈ D(Ω\K), so f |Ω\K = 0, i.e., supp f ⊂ K.

Conversely, let f ∈ D′(Ω) and supp f ⊂ K1, where K1 is a compact in Ω.Clearly, if K ⊂ Ω contains a neighborhood of K1, then f is determined by itsrestriction onto D(K) and (4.24) holds for ϕ ∈ D(K); hence, for ϕ ∈ D(Ω)(since 〈f, ϕ〉 does not depend on ϕ|Ω\K).

80 4. Distributions

Therefore, f can be continuously extended to a functional f ∈ E ′(Ω).Note that this extension essentially reduces to the following two steps: 1)decompose ϕ ∈ E(Ω) into a sum ϕ = ϕ1 + ϕ2, where ϕ1 ∈ D(Ω) and ϕ2 = 0in a neighborhood of supp f , and 2) set 〈f, ϕ〉 = 〈f, ϕ1〉. Proposition 4.6 isproved.

The following important theorem describes distributions supported at apoint.

Theorem 4.7. Let f ∈ E ′(Rn), supp f = 0. Then f has the form

(4.25) 〈f, ϕ〉 =∑|α|≤l

cα(∂αϕ)(0),

where l and constants cα do not depend on ϕ.

Proof. To any ϕ ∈ E(Rn) assign its l-jet at 0, i.e., the set of derivatives

jl0(ϕ) = (∂αϕ)(0)|α|≤l.Choose l so as to satisfy (4.24), where K is a compact which is a neighbor-hood of 0. Let us verify that 〈f, ϕ〉 actually depends only on jl0(ϕ). Thenthe statement of the theorem becomes obvious since it reduces to the de-scription of linear functionals on a finite dimensional vector space of l-jetsat 0 and these functionals are defined as in the right-hand side of (4.25).Thus, it remains to verify that jl0(ϕ) = 0 implies 〈f, ϕ〉 = 0.

Let χ ∈ D(Rn) is chosen so that χ(x) = 1 if |x| ≤ 1/2 and χ(x) = 0 if|x| ≥ 1. Set χε(x) = χ(xε ), ε > 0. Clearly, supp f = 0 implies 〈f, ϕ〉 =〈f, χεϕ〉 for any ε > 0. We will see below that jl0(ϕ) = 0 implies

(4.26) supx∈Rn

|∂α[χε(x)ϕ(x)]| → 0 for |α| ≤ l as ε→ 0 + .

Therefore〈f, ϕ〉 = 〈f, χεϕ〉 = lim

ε→0+〈f, χεϕ〉 = 0

due to (4.24). It remains to prove (4.26). But by the Leibniz rule

∂α[χε(x)ϕ(x)] =∑

α′+α′′=α

cα′α′′ [∂α′χε(x)][∂α

′′ϕ(x)] =

=∑

α′+α′′=α

cα′α′′ε−|α′|

[(∂α

′χ)(xε

)][∂α

′′ϕ(x)].

It follows from Taylor’s formula that

∂α′′ϕ(x) = O(|x|l+1−|α′′|) as x→ 0.

4.5. Differentiation and multiplication by a smooth function 81

Note that |x| ≤ ε on suppχ(xε ). Therefore, as ε→ 0+,

supx∈Rn

ε−|α′|∣∣∣(∂α′χ)

(xε

)∣∣∣ |∂α′′ϕ(x)| = O(ε−|α′|+l+1−|α′′|) = O(εl+1−|α|)→ 0,

implying (4.26).

4.5. Differentiation of distributions and multiplication by a

smooth function

Differentiation of distributions should be a natural extension of differentia-tion of smooth functions. To define it note that if u ∈ C1(Ω) and ϕ ∈ D(Ω),then

(4.27)∫

∂u

∂xjϕdx = −

∫u∂ϕ

∂xjdx

or

(4.28)⟨∂u

∂xj, ϕ

⟩= −

⟨u,∂ϕ

∂xj

⟩.

Therefore, it is natural to define ∂u∂xj

for u ∈ D′(Ω) by the formula (4.28)(which we will sometimes write in the form (4.27) in accordance with theabove-mentioned convention). Namely, define the value of the functional∂u∂xj

at ϕ ∈ D(Ω) (i.e., the left-hand side of (4.27)) as the right-hand side

which makes sense because ∂ϕ∂xj∈ D(Ω). Since ∂

∂xjcontinuously maps D(K)

to D(K) for any compact K ⊂ Ω, we see that ∂u∂xj∈ D′(Ω). We can now

define higher derivatives

∂α : D′(Ω) −→ D′(Ω),

where α = (α1, . . . , αn) is a multiindex, as products of operators ∂∂xj

, namely,

by setting ∂α = ( ∂∂x1

)α1 . . . ( ∂∂xn

)αn . Another (equivalent) way to do this isto write

(4.29) 〈∂αu, ϕ〉 = (−1)|α|〈u, ∂αϕ〉,which enables us to immediately compute derivatives of an arbitrary order.The successive application of (4.28) shows that the result obtained from(4.29) is equal to that obtained by applying ∂α considered as the compositionof operators ∂

∂xj. Since ∂α continuously maps E(Ω) to E(Ω) and S(Rn) to

S(Rn), it is clear that ∂α maps E ′(Ω) to E ′(Ω) and S′(Rn) to S′(Rn).

It is also clear from (4.29) that ∂α is a continuous operator in D′(Ω),E ′(Ω) and S′(Rn). Therefore, we may say that ∂α is the extension by con-tinuity of the usual differentiation onto the space of distributions. Such an

82 4. Distributions

extension is unique, because, as we will see in the next section, D(Ω) is densein D′(Ω).

Finally, it is clear that

supp(∂αu) ⊂ suppu, for all u ∈ D′(Ω).

Example 4.8. The derivative of the Heaviside function θ(x). The distribu-tional derivative of θ(x) can be calculated as follows:

〈θ′, ϕ〉 = −〈θ, ϕ′〉 = −∫ ∞

0ϕ′(x)dx = ϕ(0), ϕ ∈ D(R1),

i.e.,θ′(x) = δ(x).

Example 4.9. The derivative ∂αδ(x) of δ(x) ∈ D′(Rn). By definition,

〈∂αδ(x), ϕ(x)〉 = (−1)|α|〈δ(x), ∂αϕ(x)〉 = (−1)|α|ϕ(α)(0).

Therefore, ∂αδ(x) assigns the number (−1)|α|ϕ(α)(0) to the function ϕ(x).Theorem 4.7 may now be formulated as follows: any distribution supportedat 0 is a finite linearbination of derivatives of the δ-function.

Example 4.10. The fundamental solution of the Laplace operator. Considerthe Laplace operator ∆ in Rn and try to find a distribution u ∈ D′(Rn) suchthat

∆u = δ(x).

Such u is called a fundamental solution for the operator ∆. It is not unique:we can add to it any distributional solution of the Laplace equation ∆v = 0(in this case v will be actually a C∞ and even real-analytic function, aswe will see later). Fundamental solutions play an important role in whatfollows. We will try to select a special (and the most important) fundamentalsolution, using symmetry considerations. (It will be later referred to as thefundamental solution.) The operator ∆ commutes with rotations of Rn

(or, equivalently, preserves its form under an orthogonal transformation ofcoordinates). Therefore, by rotating a solution u we again get a solutionof the same equation (it is easy to justify the meaning of the process, but,anyway, we now reason heuristically). But then we may average with respectto all rotations. It is also natural to suppose that u has no singularities atx 6= 0. Therefore, we will seek a distribution u(x) of the form f(r), wherer = |x|, for x 6= 0. Let us calculate ∆f(r). We have

∂

∂xjf(r) = f ′(r)

∂r

∂xj= f ′(r)

∂

∂xj

(√x2

1 + . . .+ x2n

)= f ′(r)

xjr


implying∂2

∂x2j

f(r) = f ′′(r)x2j

r2+ f ′(r)

[1r−x2j

r3

].

Summing over j = 1, . . . , n we get

∆f(r) = f ′′(r) +n− 1r

f ′(r).

Let us solve the equation

f ′′(r) +n− 1r

f ′(r) = 0.

Denoting f ′(r) = g(r), we get for g(r) a separable ODE

g′(r) +n− 1r

g(r) = 0,

which yields g(r) = Cr−(n−1), where C is a constant. Integrating, we get

f(r) =

C1r

−(n−2) + C2 for n 6= 2;

C1 ln r + C2 for n = 2.

Since constants C2 are solutions of the homogeneous equation ∆u = 0, wemay take

f(r) = Cr2−n for n 6= 2,

f(r) = C ln r for n = 2.

We will only consider the case n ≥ 2 (the case n = 1 is elementary). andwill be later analyzed in a far more general form). For convenience of thefuture calculations introduce

(4.30) un(x) =1

2− nr2−n for n ≥ 3,

(4.31) u2(x) = ln r.

Then un(x) is locally integrable in Rn, and hence may be considered as adistribution. The factor 1

2−n is introduced so that un satisfies ∂un∂r = r1−n

for all n ≥ 2.

Now it is convenient to make a break in our exposition to include someintegral formulas for the fundamental solutions (4.30), (4.31) and relatedfunctions.

84 4. Distributions

Some Integral Formulas

We will use several formulas relating integrals over a bounded domain(open set) Ω with some integrals over its boundary. The first formula of thiskind is the Fundamental Theorem of Calculus∫ b

af ′(x)dx = f(b)− f(a), f ∈ C1([a, b]),

which is due to Newton and Leibniz. All other formulas of the describedtype (due to Gauss, Ostrogradsky, Green, Stokes and others) follow fromthis one, most often through integration by parts.

We will proceed through the following Green formula:

(4.32)∫

Ω(u∆v − v∆u)dx =

∫∂Ω

(u∂v

∂n− v ∂u

∂n

)dS,

where the notations and conditions will be explained now.

Notations and Assumptions

• Ω is a bounded domain (bounded open subset) in Rn.

• The closure of Ω is the smallest closed set Ω such that Ω ⊂ Ω ⊂ Rn.

• The boundary ∂Ω of Ω in Rn is Ω \ Ω.

It seems natural that ∂Ω should be called the boundary of Ω. However,this would allow too much freedom in the global structure of Ω and ∂Ω. Toavoid these problems we will later introduce some additional “smoothness”restrictions.

• For the closure Ω of Ω in Rn we have Ω = Ω ∪ ∂Ω, but it is possiblethat Ω ∩ ∂Ω 6= ∅. (For example, consider a non-empty closed straight lineinterval J ⊂ Ω and then replace Ω by Ω \ J .)

• dS is the area element on the boundary.

• Let k be fixed, k ∈ 1, 2, . . . , or k = ∞. (Sometimes it is alsoconvenient to use k = 0, which means that only continuity of the integrand isrequired.) The space Ck(Ω) is defined as the space of all functions u ∈ C(Ω)such that any derivative ∂αu with |α| ≤ k exists and is continuous in Ω. Thespace C∞(Ω) can be defined as the intersection of all spaces Ck(Ω). Thisintersection is a Frechet space.

• n = n(x) is an outgoing unit vector field which is based at everypoint x ∈ ∂Ω, tangent to Rn and orthogonal to ∂Ω at all points x ∈ ∂Ω.

• Let us also assume that n(x) is continuous with respect to x. It iseasy to see from the Implicit Function Theorem that locally the problem has


at most one solution. This means that the solution exists and is unique ina small neighborhood of any outgoing unit vector field which is orthogonalto ∂Ω and Rn (at the points x where a local continuous outgoing normalvector exists).

• Now we will formulate conditions which guarantee that the closureΩ of Ω in Rn and the boundary ∂Ω ⊂ Rn are sufficiently regular at a pointx. For simplicity it suffices that Ω is a finite union of simplest cells such asformulated above.

Definition 4.8 (Smoothness of the boundary). Assume that U ⊂ Rn isopen and bounded in Rn, k ∈ 1, 2, . . . . We will say that the boundary∂Ω is Ck (or ∂Ω ∈ C∞ ) if for every x ∈ ∂Ω, there exist r > 0 and aCk function h : Rn−1 → R such that after relabeling and reorienting thecoordinates axes, if necessary, we have

(4.33) U ∩B(x, r) = x ∈ B(x, r) |xn > h(x1, . . . , xn−1).

(See Fig. on page 626 in Evans.)

The outward unit normal vector field

Definition 4.9. (i) If ∂U is C1, then along ∂U we can define the outwardunit normal vector field

ν = (ν1, . . . , νn).

This field is normal at any point x ∈ ∂U if ν(x) = ν = (ν1, . . . νn).

(ii) Let u ∈ C1(∂Ω). We define

∂u

∂ν= ν ·Du,

which is called the (outward) normal derivative of u. (Here D = ∂/∂x.)

We will sometimes need to change coordinates near a point x ∈ ∂Ω so asto “flatten out” the boundary. To be more specific, fix x ∈ ∂Ω, and chooser, h, etc. as above. Define then Φ1(x), . . . ,Φn(x) by the formulas

yi = xi = Φi(x) (i = 1, . . . , n− 1)

yn = xn − h(x1, . . . , xn−1) = Φn(x),

and write

y = Φ(x).

86 4. Distributions

Similarly, we define Ψ1(x), . . . ,Ψn(x) from the formulasyi = xi = Ψi(x) (i = 1, . . . , n− 1)

yn = xn + h(x1, . . . , xn−1) = Ψn(x),

and writey = Ψ(x).

Then Φ = Ψ−1, and the mapping x 7→ Φ(x) = y “flattens out ∂Ω” near x.Observe also that det Φ = det Ψ = 1.

Remark 4.10. In particular, considering the geometric features of thecurves in the picture (4.33), [page 626 in EVANS] it is not difficult to con-clude that the curves can not blow up in a finite time or in any boundeddomain.

Remark 4.11. A perturbation of liquid in a picture like [EVANS] maycontain waves, bulges, ripples, swirls and many other forms of the flowwhich can hardly be reproduced in computer pictures (especially for 3 ormore space coordinates). This makes the problem of describing such flowsextraordinary challenging.

H Let us discuss what kind of regularity we need to require for theboundary to deserve to be called “sufficiently regular”. Unfortunately, thereis no easy answer to this question: the answer is dictated by the goal. Morespecifically, suppose that we need to make sense of the formula (4.32), soas to give sufficient conditions for it to hold. Obviously, it is reasonable torequest that we have a continuous unit normal vector and the area element(at least almost everywhere). To make sense of the left hand side, it isnatural to require that u, v ∈ C2(Ω) and ∂Ω ∈ C1, which will guaranteethat the integrals on the both sides make sense. In fact, if u, v ∈ C2(Ω) and∂Ω ∈ C1, then Green’s formula (4.32) can be indeed verified.

Additionally, if we need a C2 change of variables, which moves theboundary (say, if we would like to straighten the C1 boundary locally),then the second order derivatives of the equation of the boundary (i.e. ofthe function h in Definition 4.8) will show up in the transformed operator∆ and will be out of control. Therefore, to guarantee the continuity of thecoefficients of the transformed equation, it is natural to assume ∂Ω ∈ C2. N

Other integral formulas

Here we will establish more elementary integral formulas (Gauss, Green,Ostrogradsky, Stokes, . . . ). We will start with the deduction of Green’s


formula (4.32), from simpler integral identities, because the intermediateformulas have at least an equal importance.

The divergence formula, alternatively called Gauss-Ostrogradsky for-mula, holds:

If ∂Ω ∈ C1and F is a C1 vector field in a neighborhood of Ω in Rn, then

(4.34)∫

Ωdiv F dx =

∫∂Ω

F · n dS,

where

div F = ∇ · F =n∑j=1

∂Fj∂xj

.

Here F1, . . . , Fn are the coordinates of the vector F. (For the proof see, forexample, Rudin [23], Theorem 10.51.)

For our purposes it suffices to have this formula for the bounded do-mains Ω with smooth boundary ∂Ω (say, class Ck, with k ≥ 1), that is, theboundary which is a compact closed hypersurface (of class Ck) in Rn. Moregenerally, we can allow the domains with a piecewise Ck boundary. Butthe analysis of “piecewise smooth” situations is very difficult compared withthe “smooth” case. So we will not discuss the piecewise smooth objects inthis book.

H Let us return to the deduction of Green’s formula (4.32). TakingF = u∇v, we find for u ∈ C1(Ω) and v ∈ C2(Ω), we easily see that

div(u∇v) = u∆v +∇u · ∇v,hence (4.34) takes the form

(4.35)∫

Ω(u∆v +∇u · ∇v)dx =

∫∂Ωu∂v

∂ndS.

Assuming now that u, v ∈ C2(Ω) and subtracting from this formula the oneobtained by transposition u and v, we obtain (4.32).

The Green formula (4.32) is also a particular case of the Stokes formulafor general differential forms (see e.g. [23], Theorem 10.33):

(4.36)∫

Ωdω =

∫∂Ωω,

where ω is an (n− 1)-form on Ω, ω ∈ C1, and dω is its exterior differential.Take the differential form

ω =n∑j=1

(−1)j−1

(u∂v

∂xj− v ∂u

∂xj

)dx1 ∧ . . . ∧ dxj ∧ . . . ∧ dxn,

88 4. Distributions

where dxj means that dxj is omitted. Clearly,

dω = (u∆v − v∆u)dx1 ∧ . . . ∧ dxn,so the left-hand side of (4.36) coincides with the left-hand side of (4.32).The right-hand side of (4.36) in this case may be easily reduced to the right-hand side of (4.32) and we leave the corresponding details to the reader asan exercise. N

H Let us return to the calculation of ∆un in D′(Rn). For any ϕ ∈ D(Rn),

〈∆un, ϕ〉 = 〈un,∆ϕ〉 =∫

Rnun(x)∆ϕ(x)dx = lim

ε→0+

∫|x|≥ε

un(x)∆ϕ(x)dx.

We now use Green’s formula (4.32) with u = un, v = ϕ and

Ω = x : ε ≤ |x| ≤ R, where R is so large that ϕ(x) = 0 for |x| ≥ R − 1.Since ∆un(x) = 0 for x 6= 0, we get∫

|x|≥εun(x)∆ϕ(x)dx =

∫|x|=ε

(un∂ϕ

∂n− ϕ∂un

∂n

)dS.

Clearly,∂ϕ

∂n= −∂ϕ

∂r,∂un∂n

= −∂un∂r

= −r1−n,

implying

(4.37)∫|x|≥ε

un(x)∆ϕ(x)dx = −fn(ε)∫|x|=ε

∂ϕ

∂rdS + ε1−n

∫|x|=ε

ϕ(x)dS,

where fn(ε) = 12−nε

2−n for n ≥ 3 and fn(ε) = ln ε for n = 2. Since the areaof a sphere of radius ε in Rn is equal to σn−1ε

n−1 (here σn−1 is the area of asphere of radius 1), the first summand in the right hand side of (4.37) tendsto 0 as ε → 0+, and the second summand tends to σn−1ϕ(0). Finally, weobtain

〈∆un, ϕ〉 = σn−1ϕ(0),

or∆un = σn−1δ(x).

Now, setting

(4.38) En(x) =1

(2− n)σn−1r2−n, for n ≥ 3,

(4.39) E2(x) =1

2πln r,

we clearly get∆En(x) = δ(x),


i.e., En(x) is a fundamental solution of the Laplace operator in Rn. It willbe later referred to as the fundamental solution of the Laplace operator. Itwill be clear later that for n ≥ 3 it is the only fundamental solution suchthat it is a function of r = |x| and, besides, tends to 0 as |x| → ∞. Forn = 2 there is no obvious way to choose the additive constant.

Remark 4.12. The area σn−1 of the unit sphere in Rn. Let us computeσn−1. To do this consider Gaussian integrals

I1 =∫ ∞−∞

e−x2dx,

In =∫

Rne−|x|

2dx =

∫Rne−(x2

1+...+x2n)dx1 . . . dxn = In1 .

Expressing In in polar coordinates, we get

In =∫ ∞

0e−r

2σn−1r

n−1dr.

Setting r2 = t, we get

In = σn−1

∫ ∞0

e−ttn−1

2 d(√t) = σn−1 ·

12

∫ ∞0

tn2−1e−tdt =

σn−1

2Γ(n

2

),

where Γ is the Euler Gamma-function. This implies σn−1 = 2InΓ(n

2) . At the

same time, for n = 2, we have

I2 = 2π∫ ∞

0e−r

2rdr = −πe−r2 |+∞0 = π,

therefore I1 =√π and In = πn/2. Thus,

(4.40) σn−1 =2πn/2

Γ(n2 ).

Note that this formula can also be rewritten without Γ-function, since thenumbers Γ(n2 ) are easy to compute. Indeed, the functional equation

Γ(s + 1) = sΓ(s) enables us to express Γ(n2 ) in terms of Γ(1) for even n

and in terms of Γ(12) for odd n. But for n = 1 we deduce from (4.40) that

Γ(12) =

√π. (Clearly, σ0 = 2. However, we could also use the formula

σ2 = 4π taking n = 3 and using the relation Γ(32) = 1

2Γ(12).)

For n = 2, we deduce from (4.40) that Γ(1) = 1, which is also easy toverify without this formula. Therefore, for n = 2k + 2 we get

σ2k+1 =2πk+1

k!,

90 4. Distributions

and for n = 2k + 1 we get

σ2k =2 · (2π)k

(2k − 1)!!, where (2k − 1)!! = 1 · 3 · 5 · . . . · (2k − 1).

Remark 4.13. A physical meaning of the fundamental solution of the Laplaceoperator. For an appropriate choice of units, the potential u(x) of the elec-trostatic field of a system of charges distributed with density ρ(x) in R3

satisfies the Poisson equation

(4.41) ∆u = ρ.

In particular, for a point charge at zero we have ρ(x) = δ(x); hence, u(x)is a fundamental solution for ∆. Only potentials decaying at infinity havephysical meaning. In what follows, we will prove Liouville’s theorem, whichimplies in particular that a solution u(x) of the Laplace equation ∆u = 0defined on Rn tending to 0 as |x| → +∞, is identically zero. Therefore,there is a unique fundamental solution tending to 0 as |x| → +∞, namelyE3(x) = − 1

4πr . It defines the potential of the point unit charge situated at 0.Besides, the potential of an arbitrary distribution of charges should, clearly,by the superposition principle, be determined by the formula

(4.42) u(x) =∫E3(x− y)ρ(y)dy.

Applying formally the Laplace operator, we get

∆u(x) =∫δ(x− y)ρ(y)dy = ρ(x),

i.e., Poisson’s equation (4.41) holds. These arguments may serve to deducePoisson’s equation if, from the very beginning, the potential is defined by(4.42). Its justification may be obtained with the help of the convolution ofdistributions introduced below.

Let us indicate a meaning of the fundamental solution E2(x) in R2. InR3 consider an infinite uniformly charged string (with the constant lineardensity of the charge equal to 1) placed along x3-axis. From symmetryconsiderations, it is clear that its potential u(x) does not depend on x3 andonly depends on r =

√x2

1 + x22. Let x = (x1, x2). Poisson’s equation for u

takes the form ∆u(x) = δ(x) implying u(x) = 12π ln r+C. Therefore, in this

case the potential is of the form E2(x1, x2) + C.

Note, however, that the potential of the string is impossible to computewith the help of (4.42) which, in this case, is identically equal to +∞. Whatis the meaning of this potential? This is easy to understand if we recall that


the observable physical quantity is not the potential but the electrostaticfield

E = −gradu.

This field can be computed via the Coulomb inverse square law, and theintegral that defines it converges (at the points outside the string). Thepotential u, which can be recovered from E up to an additive constant,equals E2(x1, x2) + C.

This potential can be defined by another method when we first assumethat the string is of finite length 2l and subtract from the potential of finitestring a constant depending on l (this does not affect E!) and then passto the limit as l → +∞. It is easy to see that it is possible to choose theconstants in such a way that the limit of the potentials of the finite stringsreally exists. Then this limit must be equal to E2(x, y) + C by the abovereasoning. We actually subtract an infinite constant from the potential ofthe string of the form (4.42), but this does not affect E. This procedure iscalled in physics renormalization of charge and has its analogues in quantumelectrodynamics.

Let us prove the possibility of renormalizing the charge. Let us writethe potential of a segment of the string x3 ∈ [−l, l]:

ul(x1, x2, x3) = − 14π

∫ l

−l

dt√x2

1 + x22 + (t− x3)2

.

Since√x2

1 + x22 + (t− x3)2 ∼ |t| as t→∞, it is natural to consider instead

of ul the function

vl(x1, x2, x3) = − 14π

∫ l

−l

[1√

x21 + x2

2 + (t− x3)2− 1√

1 + t2

]dt,

that differs from ul by a constant depending on l:

vl = ul +1

4π

∫ l

−l

dt√1 + t2

.

The integrand in the formula for vl is

1√x2

1 + x22 + (t− x3)2

− 1√1 + t2

=√

1 + t2 −√x2

1 + x22 + (t− x3)2√

x21 + x2

2 + (t− x3)2 ·√

1 + t2=

=1 + 2tx3 − x2

1 − x22 − x2

3√x2

1 + x22 + (t− x3)2 ·

√1 + t2

(√x2

1 + x22 + (t− x3)2 +

√1 + t2

) ,which is asymptotically O

(1t2

)as t → +∞, so that there exists a limit

of the integral as l → +∞. Clearly, the integral itself and its limit may

92 4. Distributions

be computed explicitly but they are of no importance to us, since we cancompute the limit up to a constant by another method.

• Multiplication by a smooth function.

This operation is introduced similarly to differentiation. Namely, if f ∈L1

loc(Ω), a ∈ C∞(Ω), ϕ ∈ D(Ω), then, clearly,

(4.43) 〈af, ϕ〉 = 〈f, aϕ〉.

This formula may serve as a definition of af when f ∈ D′(Ω), a ∈ C∞(Ω). Itis easy to see that then we again get a distribution af ∈ D′(Ω). If f ∈ E ′(Ω)then af ∈ E ′(Ω) and supp(af) ⊂ supp f .

Now let f ∈ S′(Rn). Then we can use (4.43) for ϕ ∈ S(Rn) only ifaϕ ∈ S(Rn). Moreover, if we want af ∈ S′(Rn), it is necessary that theoperator mapping ϕ to aϕ be a continuous operator from S(Rn) to S(Rn).To this end, it suffices, e.g. that a ∈ C∞(Rn) satisfies

|∂αa(x)| ≤ Cα(1 + |x|)mα , x ∈ Rn,

for every multiindex α with some constants Cα and mα.

In particular, the multiplication by a polynomial maps S′(Rn) into S′(Rn).

A function f ∈ L1loc(Ω) can be multiplied by any continuous function

a(x). For a distribution f , conditions on a(x) justifying the existence of afcan also be weakened. Namely, let e.g. f ∈ D′(Ω) be such that for someinteger m ≥ 0 and every compact K ⊂ Ω we have

(4.44) |〈f, ϕ〉| ≤ CK∑|α|≤m

supx∈K|∂αϕ(x)|, ϕ ∈ D(K).

Then we say that f is a distribution of finite order (not exceeding m). Denoteby D′m(Ω) the set of such distributions. In general case the estimate (4.44)only holds for a constant m depending on K, and now we require m to beindependent of K. Clearly, if f ∈ E ′(Ω), then the order of f is finite. Iff ∈ L1

loc(Ω), then f ∈ D′0(Ω).

If f ∈ D′m(Ω), then we may extend f by continuity to a linear functionalon the space Dm(Ω) consisting of Cm-functions with compact support inΩ. Clearly, Dm(Ω) =

⋃K Dm(K), where K is a compact in Ω,Dm(K) is

the subset of Dm(Ω) consisting of functions with the support belonging toK. Introducing in Dm(K) the norm equal to the right-hand side of (4.44)without CK , we see that Dm(K) becomes a Banach space, and f ∈ D′m(Ω)can be extended to a continuous linear functional on Dm(K). Therefore, it


defines a linear functional on Dm(Ω). It is continuous in the sense that itsrestriction on Dm(K) is continuous for any compact K ⊂ Ω.

For instance, it is obvious that δ(x− x0) ∈ D′0(Ω) for x0 ∈ Ω, so that δ-function defines a linear functional onD0(Ω). It actually defines a continuouslinear functional on C(Ω).

In general, if f ∈ D′m(Ω) and supp f ⊂ K, then f can be extended to acontinuous linear functional on Cm(Ω) if the topology in Cm(Ω) is defined byseminorms given by the right-hand side of (4.44) with any compact K ⊂ Ω.

Now, let f ∈ D′m(Ω) and a ∈ Cm(Ω). Then (4.43) makes sense forϕ ∈ D(Ω), since then aϕ ∈ Dm(Ω). Therefore, af is defined for f ∈ D′m(Ω)and a ∈ Cm(Ω).

Example 4.11. a(x)δ(x) = a(0)δ(x) for a ∈ C(Rn).

Example 4.12. Let n = 1. Let us compute a(x)δ′(x) for a ∈ C∞(R1). Wehave

〈aδ′, ϕ〉 = 〈δ′, aϕ〉 = −〈δ, (aϕ)′〉 = −a′(0)ϕ(0)− a(0)ϕ′(0)

implyinga(x)δ′(x) = a(0)δ′(x)− a′(0)δ(x).

Note, in particular, that a(0) = 0 does not necessarily imply a(x)δ′(x) = 0.

Example 4.13. Leibniz rule.

Let f ∈ D′(Ω), a ∈ C∞(Ω). Let us prove that

(4.45)∂

∂xj(af) =

∂a

∂xj· f + a

∂f

∂xj.

Indeed, if ϕ ∈ D(Ω) then⟨∂

∂xj(af), ϕ

⟩= −

⟨af,

∂ϕ

∂xj

⟩= −

⟨f, a

∂ϕ

∂xj

⟩=

= −⟨f,

∂

∂xj(aϕ)

⟩+⟨f,

∂a

∂xjϕ

⟩=⟨∂f

∂xj, aϕ

⟩+⟨f,

∂a

∂xjϕ

⟩=

=⟨a∂f

∂xj+

∂a

∂xjf, ϕ

⟩,

as required.

By (4.45) we may differentiate the product of a distribution by a smoothfunction as the product of the regular functions. It follows that the Leibnizrule for higher order derivatives also holds.

94 4. Distributions

Example 4.14. A fundamental solution for an ordinary (one-dimensional)differential operator.

On R1 consider a differential operator

L =dm

dxm+ am−1

dm−1

dxm−1+ . . .+ a1

d

dx+ a0,

where the aj are constants. Let us find its fundamental solution E(x), i.e.,a solution of Lu = δ(x). Clearly, E(x) is determined up to a solution of thehomogeneous equation Lu = 0. Besides LE(x) = 0 should hold for x 6= 0.Therefore, it is natural to seek E(x) in the form

E(x) =

y1(x), for x < 0

y2(x), for x > 0,

where y1, y2 are solutions of Ly = 0. Moreover, subtracting y1(x), we mayassume that

(4.46) E(x) = θ(x)y(x),

where y(x) is a solution of Ly = 0. Now, let us find LE(x). Clearly,

(θ(x)y(x))′ = y(0)δ(x) + θ(x)y′(x),

and the higher order differentiations give rise to derivatives of δ(x). If wewish these derivatives to vanish and the δ-function to appear only at thevery last step, we should assume

y(0) = y′(0) = . . . = y(m−2)(0) = 0, y(m−1)(0) = 1.

Such a solution y(x) exists and is unique. For such a choice we get

(θ(x)y(x))′ = θ(x)y′(x)

(θ(x)y(x))′′ = (θ(x)y′(x))′ = θ(x)y′′(x)

· · · · · · · · · · · · · · ·(θ(x)y(x))(m−1) = θ(x)y(m−1)(x)

(θ(x)y(x))(m) = y(m−1)(0)δ(x) + θ(x)y(m)(x) = δ(x) + θ(x)y(m)(x).

Therefore,L(θ(x)y(x)) = θ(x)Ly(x) + δ(x) = δ(x),

as required.

Why should the fundamental solution be a regular C∞-function for x 6=0? This follows from the following


Theorem 4.14. Let u ∈ D′((a, b)) be a solution of the differential equation

(4.47) u(m) + am−1(x)u(m−1) + . . .+ a0(x)u = f(x),

where aj(x), f(x) are C∞-functions on (a, b). Then u ∈ C∞((a, b)).

Proof. Subtracting a smooth particular solution of (4.47) which exists, asis well-known, we see that everything is reduced to the case when f ≡ 0.Further, for f ≡ 0, equation (4.47) is easily reduced to the system of theform

(4.48) v′ = A(x)v,

where v is a vector-valued distribution (i.e. a vector whose components aredistributions) and A(x) a matrix with entries from C∞((a, b)). Let Ψ(x) bean invertible matrix of class C∞ satisfying

Ψ′(x) = A(x)Ψ(x),

i.e. Ψ(x) is a matrix whose columns constitute a fundamental system ofsolutions of (4.48). Set v = Ψ(x)w, i.e. denote w = Ψ−1(x)v. Then w is avector-valued distribution on (a, b). Substituting v = Ψ(x)w in (4.48), weget w′ = 0. Now, it remains to prove the following

Lemma 4.15. Let u ∈ D′((a, b)) and u′ = 0. Then u = const.

Proof. The condition u′ = 0 means that 〈u, ϕ′〉 = 0 for any ϕ ∈ D((a, b)).But, clearly, ψ ∈ D((a, b)) can be represented as ψ = ϕ′, where ϕ ∈ D((a, b)),if and only if

(4.49) 〈1, ψ〉 =∫ b

aψ(x)dx = 0.

Indeed, if ψ = ϕ′ then (4.49) holds due to the main theorem of calculus.Conversely, if (4.49) holds, then we may set ϕ(x) =

∫ xa ψ(t)dt and, clearly,

ϕ ∈ D((a, b)) and ϕ′ = ψ.

Consider the map I : D((a, b)) −→ C sending ψ to 〈1, ψ〉. Since 〈u, ψ〉 =0 for ψ ∈ KerI, it is clear that 〈u, ψ〉 only depends on Iψ; but since thisdependence is linear, then 〈u, ψ〉 = C〈1, ψ〉, where C is a constant. But thismeans that u = C.

96 4. Distributions

4.6. A general notion of the transposed (adjoint) operator.

Change of variables. Homogeneous distributions

The differentiation and multiplication by a smooth function are particularcases of the following general construction. Let Ω1,Ω2 be two domains in Rn

and L : C∞(Ω2) −→ C∞(Ω1) a linear operator. Let tL : D(Ω1) −→ D(Ω2)be a linear operator such that

(4.50) 〈Lf, ϕ〉 = 〈f, tLϕ〉, ϕ ∈ D(Ω1),

for f ∈ C∞(Ω2). The operator tL is called the transposed or adjoint or dualof L. Suppose that, for any compact K1 ⊂ Ω1, there exists a compactK2 ⊂ Ω2 such that tLD(K1) ⊂ D(K2) and tL : D(K1) −→ D(K2) is acontinuous map. Then formula (4.50) enables us to extend L to a linearoperator L : D′(Ω2) −→ D′(Ω1).

If Ω1 = Ω2 = Rn and the operator tL can be extended to a continuouslinear operator tL : S(Rn) −→ S(Rn), then it is clear that L determines alinear map L : S′(Rn) −→ S′(Rn).

Note that L thus constructed is always weakly continuous. This is ob-vious from (4.50), since |〈Lf, ϕ〉| is a seminorm of the general form for Lfand |〈f,t Lϕ〉| is a seminorm for f .

Examples of transposed operators.

a) If L = ∂∂xj

then tL = − ∂∂xj

.

b) If L = a(x) (the left multiplication by a(x) ∈ C∞(Ω)), then tL = L =a(x).

In particular, the operators ∂∂xj

and a(x) are continuous on D′(Ω) with

respect to the weak topology. Moreover, ∂∂xj

is continuous on S′(Rn).

c) A diffeomorphism χ : Ω1 −→ Ω2 naturally defines a linear map χ∗ :C∞(Ω2) −→ C∞(Ω1) by the formula

(χ∗f)(x) = f(χ(x)).

Clearly, the same formula defines a linear map χ∗ : L1loc(Ω2) −→ L1

loc(Ω1).Besides, χ∗ transforms D(Ω2) to D(Ω1). We wish to extend χ∗ to a contin-uous linear operator

χ∗ : D′(Ω2) −→ D′(Ω1)

To do so, we find the transposed operator. Let ϕ ∈ D(Ω1). Then

4.6. Transposed operator. Change of variables 97

〈χ∗f, ϕ〉 =∫

Ω1

f(χ(x))ϕ(x)dx =∫

Ω2

f(z)ϕ(χ−1(z))∣∣∣∣∂χ−1(z)

∂z

∣∣∣∣ dz=⟨f(z),

∣∣∣∣∂χ−1(z)∂z

∣∣∣∣ · ϕ(χ−1(z))⟩,

where χ−1 is the inverse of χ and∂χ−1(z)∂z

its Jacobian.

It is clear now that the transposed operator is of the form

(tχ∗ϕ)(z) =∣∣∣∣∂χ−1(z)

∂z

∣∣∣∣ · ϕ(χ−1(z)).

Since |∂χ−1(z)∂z | ∈ C∞(Ω2), it is clear that tχ∗ maps D(Ω1) to D(Ω2) and

defines a continuous linear map of D(K1) to D(χ(K1)). Therefore, thegeneral scheme determines a continuous linear map χ∗ : D′(Ω2) −→ D′(Ω1).If Ω1 = Ω2 = Rn and χ is an affine transformation (or at least coincides withan affine transformation outside a compact set), then tχ∗ continuously mapsS(Rn) to S(Rn), and in this case |∂χ−1(z)

∂z | 6= 0 everywhere (it is constanteverywhere if χ is an affine map, and a constant near infinity in the generalcase). So a continuous linear map χ∗ : S′(Rn) −→ S′(Rn) is well defined.

Examples.

a) Translation operator f(x) 7→ f(x − t) with t ∈ Rn can be extended toa continuous linear operator from D′(Ω) to D′(t + Ω) and from S′(Rn) toS′(Rn). In particular, the notation δ(x − x0) used above agrees with thedefinition of the translation by x0.

b) The homothety f(x) 7→ f(tx), where t ∈ R\0, is defined on D′(Rn) andmaps D′(Rn) to D′(Rn) and S′(Rn) to S′(Rn). For instance, let us calculateδ(tx). We have∫

δ(tx)ϕ(x)dx =∫δ(z)ϕ

(zt

)|t|−ndz = |t|−nϕ(0) = |t|−n

∫δ(z)ϕ(z)dz.

Therefore,

δ(tx) = |t|−nδ(x), t ∈ R\0.

Definition 4.16. A distribution f ∈ S′(Rn) is called homogeneous of orderm ∈ R if

(4.51) f(tx) = tmf(x), t > 0.

98 4. Distributions

It is easily checked that the requirement f ∈ S′(Rn) can be actuallyreplaced by f ∈ D′(Rn) (then (4.51) implies f ∈ S′(Rn)).

Sometimes, one says “homogeneous of degree m” instead of “homoge-neous of order m”.

Examples.

a) If f ∈ L1loc(Rn) and for any t > 0, (4.51) holds almost everywhere, then

f is a homogeneous distribution of order m.

b) δ-function δ(x) is homogeneous of order −n.

It is easy to verify that ∂∂xj

f(tx) = t(∂f∂xj

)(tx). Therefore, if f is

homogeneous of order m, then ∂f∂xj

is homogeneous of order m − 1. Forinstance, ∂αδ(x) is homogeneous of order −n− |α|.

Homogeneity considerations allow us to examine without computing thestructure of the fundamental solution for the Laplace operator and its pow-ers. Namely, the order of homogeneity of the distribution r2−n for n ≥ 3is 2 − n, and for r 6= 0 this distribution satisfies the equation ∆u = 0.Therefore, ∆r2−n is supported at 0 and its order of homogeneity is −n.But, since the order of homogeneity of ∂αδ(x) is −n − |α|, it is clear that∆r2−n = Cδ(x).

For n = 2, we have ∆(ln r) = 0 for r 6= 0, and ∂∂xj

(ln r) is homogeneous oforder −1. Therefore, ∆(ln r) is supported at 0 and its order of homogeneityis −2, so ∆(ln r) = Cδ(x).

It is not clear a priori why the following does not hold: ∆r2−n = 0 inD′(Rn). Later on we will see, however, that if u ∈ D′(Ω) and ∆u = 0,then u ∈ C∞(Ω) (and, moreover, u is real analytic in Ω). Since r2−n has asingularity at 0, it follows that ∆r2−n 6= 0 implying ∆r2−n = Cδ(x), whereC 6= 0. Similarly, for n = 2 we have ∆(ln r) = Cδ(x) for C 6= 0.

Finally, note that a rotation operator is defined in D′(Rn) and in S′(Rn).This enables us to give a precise meaning to our arguments on averaging onwhich we based our conclusion about the existence of a spherically symmetricfundamental solution for ∆.

Example 4.15. Fundamental solutions for powers of the Laplace operator.

The spherical symmetry and homogeneity considerations make it clearthat in order to find the fundamental solution of ∆m in Rn, it is natural toconsider the function r2m−n. Then we get ∆m−1r2m−n = C1r

2−n. Can it

4.6. Transposed operator. Change of variables 99

be C1 = 0? It turns out that if 2m − n 6∈ 2Z+ (i.e., 2m − n is not an evennonnegative integer) or, equivalently, r2m−n 6∈ C∞(Rn), then C1 6= 0 and,therefore, ∆mr2m−n = Cδ(x), where C 6= 0. Indeed, for r 6= 0 we have, forany α ∈ R:

∆rα =(d2

dr2+n− 1r

d

dr

)rα = [α(α−1)+α(n−1)]rα−2 = α(α+n−2)rα−2.

Therefore,

∆r2k−n = (2k − n)(2k − 2)r2k−n−2

and consecutive applications of ∆ to r2m−n yield a factor of the form (2m−2)(2m−n), then (2m−4)(2m−n−2), etc., implying C1 6= 0 for 2m−n 6∈ 2Z+.

Therefore, for 2m − n 6∈ 2Z+ the fundamental solution Emn (x) of ∆m isof the form

Emn (x) = Cm,nr2m−n.

Further, let 2m−n ∈ 2Z+ so that r2m−n is a polynomial in x. Then considerthe function r2m−n ln r. We get

∆m−1(r2m−n ln r) =

C1 ln r for n = 2

C2r2−n for n ≥ 4

(C2 depends on n), where C1 6= 0 and C2 6= 0. Consider for the sake ofdefiniteness the case n ≥ 4. Since

∆m−1f(r) =(d2

dr2+n− 1r

d

dr

)m−1

f(r)

and ddr ln r = 1

r , it follows that

∆m−1(r2m−n ln r) = ∆m−1(r2m−n) · ln r + C2r2−n = C2r

2−n

because r2m−n is a polynomial in x of degree 2m − n. Here C2 6= 0 sincer2m−n ln r 6∈ C∞(Rn) and the equation ∆u = 0 has no solutions with singu-larities. (Indeed, if it had turned out that C2 = 0, then for some k we wouldhave had that u = ∆k−1(r2m−n ln r) had a singularity at 0 and satisfied∆u = 0). Note though that C2 may be determined directly. Therefore, for2m− n ∈ 2Z+,

Emn (x) = Cm,nr2m−n ln r.

100 4. Distributions

4.7. Appendix: Minkowski inequality

Let (M,µ) be a space with a positive measure. Here M is a set where aσ-algebra of subsets is chosen. (We will call a subset S ⊂ M measurable ifit belongs to the σ-algebra.) The measure µ is assumed to be defined on theσ-algebra and take values in [0,+∞]. We will also assume that µ is σ-finite,i.e. there exist measurable sets M1 ⊂ M2 ⊂ . . . , such that µ(Mj) is finitefor every j, and M is the union of all Mj ’s.

Then the function spaces Lp(M,µ), 1 ≤ p < +∞, are well defined, withthe norm

‖f‖p =(∫

M|f(x)|pµ(dx)

)1/p

, 1 ≤ p < +∞.

The Minkowski inequality in its simplest form is the triangle inequalityfor the norm ‖ · ‖p:

‖f + g‖p ≤ ‖f‖p + ‖g‖p, f, g ∈ Lp(M,µ).

By induction, this easily extends to several functions:

‖f1 + f2 + · · ·+ fk‖p ≤ ‖f1‖p + ‖f2‖p + · · ·+ ‖fk‖p,

where fj ∈ Lp(M,µ), j = 1, 2, . . . , k. Multiplying each fj by aj > 0, weobtain

(4.52) ‖a1f1 + a2f2 + · · ·+ akfk‖p ≤ a1‖f1‖p + a2‖f2‖p + · · ·+ ak‖fk‖p.

A natural generalization of this inequality can be obtained if we replacesummation by integration over a positive measure. More precisely, let (N, ν)be another space with a positive measure ν, and f = f(x, y) be a measurablefunction on (M×N,µ×ν), where µ×ν is the product measure defined by theproperty (µ×ν)(A×B) = µ(A)ν(B) for any measurable sets A ⊂M,B ⊂ Nwith finite µ(A) and ν(B). Then we have

Theorem 4.17 (Minkowski’s inequality). Under the conditions above

(4.53)∥∥∥∥∫

Nf(·, y)ν(dy)

∥∥∥∥p

≤∫N‖f(·, y)‖pν(dy),

or, equivalently,(4.54)(∫

M

∣∣∣∣∫Nf(x, y)ν(dy)

∣∣∣∣p µ(dx))1/p

≤∫N

(∫M|f(x, y)|pµ(dx)

)1/p

ν(dy).

4.7. Appendix: Minkowski inequality 101

The inequality (4.52) is a particular case of (4.53). To see this it sufficesto choose N to be a finite set of k points, and take measures of these pointsto be a1, . . . , ak.

Vice versa, if we know that (4.52) holds, then (4.53) becomes heuristi-cally obvious. For example, if we could present the integrals in (4.53) aslimits of the corresponding Riemann sums, then (4.53) would follow from(4.52). But this is not so easy to do rigorously even if the functions aresufficiently regular. The easiest way to give a rigorous proof, is by a directuse of the following very important

Theorem 4.18 (Holder’s inequality). Let f ∈ Lp(M,µ), g ∈ Lq(M,µ),where 1 < p, q < +∞, and

(4.55)1p

+1q

= 1.

Then fg ∈ L1(M,µ) and

(4.56)∣∣∣∣∫Mf(x)g(x)µ(dx)

∣∣∣∣ ≤ ‖f‖p‖g‖q.Proof. Let us start with the following elementary inequality:

(4.57) ab ≤ ap

p+bq

q,

where a ≥ 0, b ≥ 0, and p, q are as in Lemma above. It is enough toconsider the case a > 0, b > 0. Consider the curve t = sp−1, s ≥ 0, in thepositive quadrant of the real plane R2

s,t, and two disjoint bounded domainsG1 and G2, so that each of them is bounded by this curve together withthe coordinate axes and straight line intervals as follows (see fig. ??): theboundary of G1 consists of two straight line intervals (s, 0) : 0 ≤ s ≤ a,(a, t) : 0 ≤ t ≤ ap−1, and the curvilinear part (s, t) : t = sp−1, 0 ≤ s ≤a; the boundary of G2 also consists of two straight line intervals (0, t) :0 ≤ t ≤ b, (s, b) : 0 ≤ s ≤ b1/(p−1), and generally different curvilinearpiece of the same curve, namely, (s, t) : s = t1/(p−1), 0 ≤ t ≤ b.

It is easy to see that the areas of G1, G2 are ap/p and bq/q respectively,and their union contains the rectangle [0, a]× [0, b] up to a set of measure 0.Therefore, (4.57) follows by comparison of the areas.

To prove (4.56) note first that replacing f , g by |f |, |g|, we may assumethat f ≥ 0 and g ≥ 0. Dividing both parts of (4.56) by ‖f‖p‖g‖q, wesee that it suffices to prove it for normalized functions f, g, i.e. such that


‖f‖p = ‖g‖q = 1. To this end apply (4.57) with a = f(x), b = g(x) to get

f(x)g(x) ≤ f(x)p

p+g(x)q

q.

Integrating this over M with respect to µ and using (4.55), we come to thedesired result.

Proof of Theorem 4.17. To prove (4.54) let us make some simplifyingassumptions. Without loss of generality we can assume that f is positive(otherwise, replace it by |f |, which does not change the right hand side, andcan only increase the left hand side).

We can assume that both sides of (4.54) are finite. If the right hand sideis infinite, then the inequality is obvious. If the left hand side is infinite,we can appropriately truncate f (using σ-finiteness of the measures), andthen deduce the proof in the general case in the limit from a monotoneapproximation.

Introducing a function

h(x) :=∫Nf(x, y)ν(dy),

which is measurable on M by Fubini’s theorem, we can rewrite the pth powerof the left hand side of (4.54) as∫

M h(x)pµ(dx) =∫M

(∫Nf(x, y)ν(dy)

)h(x)p−1µ(dx)

=∫N

(∫Mf(x, y)h(x)p−1µ(dx)

)ν(dy)

Now applying Holder’s inequality (4.56) to the internal integral in the righthand side, and taking into account that q = p/(p− 1), we obtain∫Mf(x, y)h(x)p−1µ(dx) ≤

(∫Mf(x, y)pµ(dx)

)1/p(∫Mh(x)pµ(dx)

)(p−1)/p

.

Using this to estimate the right hand side of the previous identity, we obtain∫Mh(x)pµ(dx) ≤

∫N

(∫Mf(x, y)pµ(dx)

)1/p

ν(dy)(∫

Mh(x)pµ(dx)

)(p−1)/p

.

Now dividing both sides by the last factor in the right hand side, we obtainthe desired inequality (4.54). This division is possible if we assume that thelast factor does not vanish. But in opposite case f = 0 almost everywhereand the statement is trivially true.

4.8. Appendix: Completeness of distribution spaces 103

4.8. Appendix: Completeness of distribution spaces

Let E be a Frechet space with the topology given by seminorms pj | j =1, 2, . . . , such that

p1(x) ≤ p2(x) ≤ . . . , for all x ∈ E.Let E′ denote the dual space of all linear continuous functionals f : E → R.Let us say that a set F ⊂ E′ is bounded if there exist an integer j ≥ 1 anda positive constant C such that

(4.58) |〈f, ϕ〉| ≤ Cpj(ϕ), for all f ∈ F,ϕ ∈ E.Note that for F consisting of just one element f , this is always true, beingequivalent to the continuity of f – see (4.2). So the requirement (4.58) meansequicontinuity, or uniform continuity of all functionals f ∈ F .

Let us also say that F is weakly bounded if for every ϕ ∈ E there existsa positive Cϕ such that

(4.59) |〈f, ϕ〉| ≤ Cϕ, for all f ∈ F.The following theorem is a simplest version of the Banach-Steinhaus theo-rem, which is also called the uniform boundedness principle.

Theorem 4.19. If under the assumptions described above, F is weaklybounded, then it is bounded.

Completeness of E plays important role in the proof, which relies on thefollowing

Theorem 4.20. (Baire’s Theorem) A non-empty complete metric space cannot be presented as at most countable union of nowhere dense sets.

Here nowhere dense set S in a metric space M is a set whose closure doesnot have interior points. So the theorem can be equivalently reformulatedas follows: a non-empty complete metric space can not be presented as atmost countable union of closed sets without interior points.

Proof of Theorem 4.20. Let M be a non-empty complete metric spacewith the metric ρ, Sj | j = 1, 2, . . . be a sequence of nowhere dense subsetsof M . We will produce a point x ∈M which does not belong to any Sj .

Passing to the closures we can assume that Sj is closed for every j.Now, clearly S1 6= M because otherwise S1 would be open in M , so it wouldnot be nowhere dense. Hence there exists a closed ball B1 = B(x1, r1) =


x ∈ M | ρ(x, x1) ≤ r1, such that B1 ∩ S1 = ∅. By induction we canconstruct a sequence of closed balls Bj = B(xj , rj), j = 1, 2, . . . , such thatBj+1 ⊂ Bj , rj+1 ≤ rj/2 and Bj ∩ Sj = ∅ for all j. It follows that Bjdoes not intersect with S1 ∪ S2 ∪ · · · ∪ Sj . On the other hand, the sequenceof centers xj | j = 1, 2, . . . is a Cauchy sequence because by the triangleinequality, for m < n,

ρ(xm, xn) ≤ ρ(xm, xm+1)+· · ·+ρ(xn−1, xn) ≤ rm+rm+1+· · · ≤ 2rm ≤ 2−m+2r1.

Therefore, due to completeness of M , xm → x in M , and x belongs to allballs Bj . It follows that x 6∈ Sj , for all j.

Proof of Theorem 4.19. Let us take for any j = 1, 2, . . . ,

Sj = ϕ|ϕ ∈ E, supf∈F|〈f, ϕ〉| ≤ jpj(ϕ)

Then Sj ⊂ Sj+1 for all j, each set Sj is closed, and E is the union of allSj , j = 1, 2, . . . . Therefore, by Baire’s Theorem, there exists j such that Sjhas a non-empty interior. Taking a fundamental system of neighborhoodsof any point ϕ0 ∈ E, consisting of sets

Bj(ϕ0, ε) = ϕ0 + ϕ| pj(ϕ) < ε,

where ε > 0, j = 1, 2, . . . , we see that we can find ϕ0, j and ε, such thatBj(ϕ0, ε) ⊂ Sj . Then for any f ∈ F and any ϕ ∈ Bj(0, ε)

|〈f, ϕ〉| ≤ |〈f, ϕ+ϕ0〉|+ |〈f, ϕ0〉| ≤ jpj(ϕ+ϕ0)+C1 ≤ jpj(ϕ)+C2 ≤ jε+C2,

where C1, C2 > 0 are independent of ϕ and f (but may depend on ϕ0, j andε). We proved this estimate for ϕ ∈ E such that pj(ϕ) < ε. By scaling wededuce that

|〈f, ϕ〉| ≤ (j + ε−1C2)pj(ϕ)

for all f ∈ F and ϕ ∈ E, which implies the desired result.

Corollary 4.21 (Completeness of the dual space). Let E be a Frechet space,fk ∈ E′| k = 1, 2, . . . be a sequence of continuous linear functionals suchthat for every ϕ ∈ E there exists limit

〈f, ϕ〉 := limk→+∞

〈fk, ϕ〉.

Then the limit functional f is also continuous, i.e. f ∈ E′.

4.9. Problems 105

Proof. Clearly, the set F = fk| k = 1, 2, . . . , is weakly bounded. There-fore, by Theorem 4.19 there exists a seminorm pj , which is one of the semi-norms defining topology in E, and a constant C > 0, such that

|〈fk, ϕ〉| ≤ Cpj(ϕ), ϕ ∈ E, k = 1, 2, . . . .

Taking limit as k → +∞, we obtain

|〈f, ϕ〉| ≤ Cpj(ϕ), ϕ ∈ E,with the same C and j. This implies continuity of f .

Applying this corollary Theorem 4.19 to the distribution spaces D′(Ω),E ′(Ω) and S′(Rn), we obtain their completeness. (In case of D′(Ω) we shouldapply the corollary to E = D(K)).

4.9. Problems

4.1. Find limε→0+

χε(x) in D′(R1), where χε(x) = 1/√ε for x ∈ (−ε, ε) and

χε(x) = 0 for x 6∈ (−ε, ε).


ψε(x) in D′(R1), where ψε(x) = 1/ε for x ∈ (−ε, ε) and

ψε(x) = 0 for x 6∈ (−ε, ε).


1x sin x

ε in D′(R1).

4.4. Prove that if f ∈ C∞(R), then 1x(f(x)− f(0)) ∈ C∞(R).

4.5. Consider the canonical projection p : R1 −→ S1 = R/2πZ. It inducesan isomorphism p∗ : C∞(S1) −→ C∞2π(R), where C∞2π(R) is the set of all 2π-periodic C∞-functions on R. Introduce a natural Frechet topology in thespace C∞(S1) and define the set of distributions on S1 as the dual spaceD′(S1) = (C∞(S1))′. Identify each function f ∈ L1(S1) (or, equivalently,locally integrable 2π-periodic function f on R) with a distribution on S1 bythe formula

〈f, ϕ〉 =∫ 2π

0f(x)ϕ(x)dx.

Prove that p∗ can be extended by continuity to an isomorphism

D′(S1) −→ D′2π(R),

where D′2π(R) is the set of all 2π-periodic distributions.


4.6. Prove that the trigonometric series∑k∈Z

fkeikx

converges in D′(R1) if and only if there exist constants C and N such that|fk| ≤ C(1+ |k|)N . Prove that if f is the sum of this series, then f ∈ D′2π(R)and the series above is in fact its Fourier series, that is

fk =1

2π〈f, e−ikx〉,

where the angle brackets denote the natural pairing between D′(S1) andC∞(S1).

4.7. Prove that any 2π-periodic distribution (or, which is the same, distri-bution on S1) is equal to the sum of its Fourier series.

4.8. Find the Fourier series expansion of the δ-function on S1 or, equiva-lently, of the distribution

∞∑k=−∞

δ(x+ 2kπ) ∈ D′2π(R1).

4.9. Use the result of the above problem to prove the following Poissonsummation formula:

∞∑k=−∞

f(2kπ) =1

2π

∞∑k=−∞

f(k),

where f ∈ S(R1), and f is the Fourier transform of f .

4.10. Find all u ∈ D′(R), such that:

a) xu(x) = 1;

b) xu′(x) = 1;

c) xu′(x) = δ(x).

4.11. Verify that −eikr

4πr, where r = |x|, is the fundamental solution of the

Helmholtz operator ∆ + k2 in R3.

Chapter 5

Convolution and

Fourier transform

5.1. Convolution and direct product of regular functions

The convolution of locally integrable functions f and g on Rn is the integral

(5.1) (f ∗ g)(x) =∫f(x− y)g(y)dy,

over Rn. Obviously, the integral (5.1) is not always defined. Let us considerit when f, g ∈ C(Rn) and one of these functions has a compact support (inthis case (5.1) makes sense and, clearly, f ∗ g ∈ C(Rn)). The change ofvariables x− y = z turns (5.1) into

(f ∗ g)(x) =∫f(z)g(x− z)dz,

i.e., the convolution is commutative:

f ∗ g = g ∗ f.

Further, let f ∈ C1(Rn). Then it is clear that (5.1) may be differentiatedunder the integral sign, and we get

∂

∂xj(f ∗ g) =

∂f

∂xj∗ g.

In general, if f ∈ Cm(Rn), then

(5.2) ∂α(f ∗ g) = (∂αf) ∗ g for |α| ≤ m.

107

108 5. Convolution and Fourier transform

The convolution is associative, i.e.,

f ∗ (g ∗ h) = (f ∗ g) ∗ h

if f, g, h ∈ C(Rn) and two of the three functions have compact supports.This can be verified by the change of variables but a little later we willprove this in another way.

Let ϕ ∈ C(Rn) and ϕ have a compact support. Then

〈f ∗ g, ϕ〉 =∫f(x− y)g(y)ϕ(x)dydx

may be transformed by the change x = y + z to the form

(5.3) 〈f ∗ g, ϕ〉 =∫f(z)g(y)ϕ(z + y)dydz.

By Fubini’s theorem, this makes obvious the already proved commutativityof the convolution. Further, this also implies

〈(f ∗ g) ∗ h, ϕ〉 =∫f(x)g(y)h(z)ϕ(x+ y + z)dxdydz = 〈f ∗ (g ∗ h), ϕ〉,

proving the associativity of the convolution.

Note the important role played in (5.3) by the function f(x)g(y) ∈C(R2n) called the direct or tensor product of functions f and g and denotedby f ⊗ g or just f(x)g(y) if we wish to indicate the arguments.

Suppose that f and g have compact supports. Then in (5.3) we may setϕ(x) = e−iξ·x, where ξ ∈ Rn. We get

f ∗ g(ξ) = f(ξ) · g(ξ),

where the tilde denotes the Fourier transform:

f(ξ) =∫f(x)e−iξ·xdx.

Therefore, the Fourier transform turns the convolution into the usual prod-uct of Fourier transforms.

Let us analyze the support of a convolution. Let A,B ⊂ Rn. Set

A+B = x+ y : x ∈ A, y ∈ B

(sometimes A + B is called the arithmetic sum of subsets A and B). As-suming that A is closed and B is compact in Rn, we easily see that A + B

is also closed in Rn. Taking A = supp f , B = supp g, we obtain

(5.4) supp(f ∗ g) ⊂ supp f + supp g.

5.2. Direct product of distributions 109

Indeed, it follows from (5.3) that if suppϕ⋂

(supp f + supp g) = ∅, then〈f ∗ g, ϕ〉 = 0 yielding (5.4).

5.2. Direct product of distributions

Let Ω1 ⊂ Rn1 , Ω2 ⊂ Rn2 , f1 ∈ D′(Ω1), f2 ⊂ D′(Ω2). Let us define the director tensor product f1 ⊗ f2 ∈ D′(Ω1 × Ω2) of distributions f1 and f2. Insteadof f1 ⊗ f2, we will often write also f1(x)f2(y), where x ∈ Ω1, y ∈ Ω2.

Let ϕ(x, y) ∈ D(Ω1 × Ω2). For fj ∈ L1loc(Ωj) the Fubini theorem yields

(5.5) 〈f1 ⊗ f2, ϕ〉 = 〈f1(x), 〈f2(y), ϕ(x, y)〉〉 = 〈f2(y), 〈f1(x), ϕ(x, y)〉〉.

This formula should be considered as a basis for definition of f1 ⊗ f2 if f1

and f2 are distributions. Let us show that it makes sense. First, note thatany compact K in Ω1 ×Ω2 is contained in a compact of the form K1 ×K2,where Kj is a compact in Ωj , j = 1, 2. In particular, suppϕ ⊂ K1 ×K2 forsome K1,K2. Further, ϕ(x, y) can be now considered as a C∞-function ofx with values in D(K2) (the C∞-function with values in D(K2) means thatall the derivatives are limits of their difference quotients in the topology ofD(K2)). Therefore, 〈f2(y), ϕ(x, y)〉 ∈ D(K1), since f2 is a continuous linearfunctional on D(K2). This implies that the functional

〈F,ϕ〉 = 〈f1(x), 〈f2(y), ϕ(x, y)〉〉

is well defined. It is easy to verify that F ∈ D′(Ω1 × Ω2) and suppF ⊂supp f1× supp f2. We can also construct the functional G ∈ D′(Ω1×Ω2) bysetting

〈G,ϕ〉 = 〈f2(y), 〈f1(x), ϕ(x, y)〉〉.Let us verify that F = G. Note that if ϕ = ϕ1 ⊗ ϕ2, where ϕj ∈ D(Ωj) i.e.,ϕ(x, y) = ϕ1(x)ϕ2(y), then, clearly,

〈F,ϕ〉 = 〈G,ϕ〉 = 〈f1, ϕ1〉〈f2, ϕ2〉.

Therefore, obviously, to prove F = G it suffices to prove the following

Lemma 5.1. In D(Ω1 × Ω2), finite linear combinations of the functionsϕ1 ⊗ ϕ2, where ϕj ∈ D(Ωj) for j = 1, 2, are dense. More precisely, if Kj

is a compact in Ωj, j = 1, 2, and Kj is a compact in Ωj, such that Kj iscontained in the interior of Kj, then any function ϕ ∈ D(K1 ×K2) can beapproximated, as close as we like, in D(K1×K2) by finite sums of functionsof the form ϕ1 ⊗ ϕ2, where ϕj ∈ D(Kj).


Proof. Consider the cube Qd in Rn1+n2 :

Qd =

(x, y) : |xj | ≤d

2, |yl| ≤

d

2, j = 1, . . . , n1; l = 1, . . . , n2

,

i.e., the cube with the center at the origin with edges of length d parallel tothe coordinate axes. Take d large enough for the compact set K1 ×K2 tobelong to the interior of Qd. In L2(Qd), the exponentials

exp(

2πidk · z

): k ∈ Zn, n = n1 + n2; z ∈ Rn

considered as functions of z, constitute a complete orthonormal system. Afunction ϕ = ϕ(x, y) = ϕ(z) can be expanded into the Fourier series

(5.6) ϕ(z) =∑k

ϕk exp(

2πidk · z

),

where

ϕk =1dn

∫Qd

ϕ(z) exp(−2πi

dk · z

)dz.

The Fourier series (5.6) converges absolutely and uniformly inQd to ϕ(z) andit can be term-wise differentiated any number of times. Indeed, integratingby parts shows that the absolute values of

(1 + |k|2)Nϕk =1dn

∫Qd

ϕ(z)(

1− d2

4π2∆z

)Nexp

(−2πi

dk · z

)dz =

=1dn

∫Qd

[(1− d2

4π2∆z

)Nϕ(z)

]exp

(−2πi

dk · z

)dz

are bounded by a constant CN independent of k, i.e.,

|ϕk| ≤ CN (1 + |k|2)−N .

This implies the convergence of (5.6) and the possibility of its termwisedifferentiation.

Therefore, (5.6) converges in the topology of C∞(Qd). Now, let χj ∈D(Kj), where χj = 1 in a neighborhood of Kj . We get

ϕ(x, y) = χ1(x)χ2(y)ϕ(x, y) =

=∑k1,k2

ϕkχ1(x) exp(

2πidk1 · x

)χ2(y) exp

(2πidk2 · y

),

where k1 ∈ Zn1 , k2 ∈ Zn2 , k = (k1, k2) and the series converges in thetopology D(K1 × K2). But then the partial sums of this series give therequired approximation for ϕ(x, y).

5.2. Direct product of distributions 111

Now we are able to introduce the following

Definition 5.2. If fj ∈ D′(Ωj), j = 1, 2, then the direct or tensor productof f1 and f2 is the distribution on Ω1×Ω2 defined via (5.5) and denoted byf1 ⊗ f2 or f1(x)f2(y) or f1(x)⊗ f2(y).

Clearly,supp(f1 ⊗ f2) ⊂ supp f1 × supp f2,

and if fj ∈ S′(Rnj ) then f1 ⊗ f2 ∈ S′(Rn1+n2). Further, it is easy to verifythat

∂

∂xj(f1 ⊗ f2) =

∂f1

∂xj⊗ f2,

∂

∂yl(f1 ⊗ f2) = f1 ⊗

∂f2

∂yl.

Example 5.1. δ(x)δ(y) = δ(x, y).

Example 5.2. Let t ∈ R1, x ∈ Rn−1, α(x) ∈ C(Rn−1). The distributionδ(t)⊗ α(x) ∈ D′(Rn

t,x) is called the single layer with density α on the planet = 0. Its physical meaning is as follows: it describes the charge supported onthe plane t = 0 and smeared over the plane with the density α(x). Clearly,

(5.7) 〈δ(t)⊗ α(x), ϕ(t, x)〉 =∫

Rn−1x

α(x)ϕ(0, x)dx.

For β(x) ∈ C(Rn−1), the distribution δ′(t) ⊗ β(x) is called the doublelayer with density β on the plane t = 0. Its physical meaning is as follows:it describes the distribution of dipoles smeared with the density β over theplane t = 0 and oriented along the t-axis.

Recall that a dipole in electrostatics is a system of two very large chargesof the same absolute value and of opposite signs placed very close to eachother. The product of charges by the distance between them should be adefinite finite number called the dipole moment. Clearly, in D′(Rn), thefollowing limit relation holds

δ′(t)⊗ β(x) = limε→+0

[δ(t+ ε)

2ε⊗ β(x)− δ(t− ε)

2ε⊗ β(x)

],

implying the discussed above interpretation of the double layer. It is alsoclear that

(5.8) 〈δ′(t)⊗ β(x), ϕ(t, x)〉 = −∫

Rn−1x

∂ϕ

∂t(0, x)β(x)dx.

The formulas similar to (5.7), (5.8) can be used to define the single anddouble layers on any smooth surface Γ of codimension 1 in Rn. Namely, if


α, β ∈ C(Γ) then we can define the distributions αδΓ and ∂∂n(βδΓ) by the

formulas

〈αδΓ, ϕ〉 =∫

Γα(x)ϕ(x)dS,⟨

∂

∂n(βδΓ), ϕ

⟩= −

∫Γβ(x)

∂ϕ

∂ndS,

where dS is the area element of Γ, n being the external normal to Γ. Note,however, that, locally, a diffeomorphism reduces these distributions to theabove direct products δ(t)⊗ α(x) and δ′(t)⊗ β(x).

Note another important property of the direct product: it is associativei.e., if fj ∈ D′(Ωj), j = 1, 2, 3, then

(f1 ⊗ f2)⊗ f3 = f1 ⊗ (f2 ⊗ f3) in D′(Ω1 × Ω2 × Ω3).

The proof follows from the fact that finite linear combinations of functionsof the form ϕ1 ⊗ ϕ2 ⊗ ϕ3 with ϕj ∈ D(Ωj), j = 1, 2, 3, are dense in D(Ω1 ×Ω2 × Ω3) in the same sense as in Lemma 5.1.

5.3. Convolution of distributions

It is clear from (5.3) that it is natural to try to define the convolution ofdistributions f, g ∈ D′(Rn) with the help of the formula

(5.9) 〈f ∗ g, ϕ〉 = 〈f(x)g(y), ϕ(x+ y)〉, ϕ ∈ D(Rn).

This formula (and the convolution itself) does not always make sense, sinceϕ(x + y) 6∈ D(R2n) for ϕ 6= 0. Nevertheless, sometimes the right-hand sideof (5.9) can be naturally defined. Let for example f ∈ E ′(Rn). Let us set

(5.10) 〈f(x)g(y), ϕ(x+ y)〉 = 〈f(x), 〈g(y), ϕ(x+ y)〉〉,which makes sense since 〈g(y), ϕ(x + y)〉 ∈ E(Rn

x). We may proceed theother way around, setting

(5.11) 〈f(x)g(y), ϕ(x+ y)〉 = 〈g(y), 〈f(x), ϕ(x+ y)〉〉,which also makes sense, since 〈f(x), ϕ(x+ y)〉 ∈ D(Rn

y ).

Let us show that both methods give the same answer. To this end, takea function χ ∈ D(Rn) such that χ(x) = 1 in a neighborhood of supp f .Then f = χf and we have

〈f(x), 〈g(y), ϕ(x+ y)〉〉 = 〈χ(x)f(x), 〈g(y), ϕ(x+ y)〉〉 =

= 〈f(x), χ(x)〈g(y), ϕ(x+ y)〉〉 = 〈f(x), 〈g(y), χ(x)ϕ(x+ y)〉〉,

5.3. Convolution of distributions 113

and, similarly,

〈g(y), 〈f(x), ϕ(x+ y)〉〉 = 〈g(y), 〈f(x), χ(x)ϕ(x+ y)〉〉.

The right-hand sides of these equations are equal, due to the property ofthe direct product (5.5) proved above, since χ(x)ϕ(x+ y) ∈ D(R2n). There-fore, their left-hand sides coincide. Now we may introduce the followingdefinition.

Definition. The convolution of two distributions f, g ∈ D′(Rn) one of whichhas a compact support is the distribution f ∗ g ∈ D′(Rn) defined via (5.9)understood in the sense of (5.10) or (5.11).

It follows from the arguments above that the convolution is commutative,i.e.,

f ∗ g = g ∗ f, f ∈ E ′(Rn), g ∈ D′(Rn);

besides, it is associative, i.e.

(f ∗ g) ∗ h = f ∗ (g ∗ h),

if two of the three distributions have compact support. This is easy to verifyfrom the associativity of the direct product and the equation

〈(f ∗ g) ∗ h, ϕ〉 = 〈f(x)g(y)h(z), ϕ(x+ y + z)〉,

because the same holds if (f ∗ g) ∗ h is replaced by f ∗ (g ∗ h).

The rule (5.2) of differentiating the convolution applies also to distribu-tions. Indeed,

〈∂α(f ∗ g), ϕ〉 = (−1)|α|〈f ∗ g, ∂αϕ〉 = (−1)|α|〈f(x)g(y), (∂αϕ)(x+ y)〉,

but (∂αϕ)(x+ y) can be represented in the form ∂αxϕ(x+ y) or ∂αy ϕ(x+ y)and then, moving ∂α backwards, we see that (5.2) holds:

∂α(f ∗ g) = (∂αf) ∗ g = f ∗ (∂αg).

Example 5.3. δ(x) ∗ f(x) = f(x), f ∈ D′(Rn).

From the algebraic viewpoint, we see that the addition and convolutionendow E ′(Rn) with the structure of an associative and commutative algebra(over C) with the unit δ(x), and D′(Rn) is a two-sided module over thisalgebra.

H Now we will introduce one of the key notions in partial differential equa-tions (with constant coefficients): fundamental solution.


Definition 5.3. Let p(D) be a linear differential operator with constantcoefficients in Rn. A distribution E(x) ∈ D′(Rn) is its fundamental solutionif

p(D)E(x) = δ(x).

Example 5.4. A solution of nonhomogeneous equation with constant coef-ficients.

Proposition 5.4. Let p(D) be a linear differential operator with constantcoefficients in Rn, f ∈ E ′(Rn). Then the equation

p(D)u = f

has a particular solutionu = E ∗ f.

Proof. Indeed,

p(D)(E ∗ f) = [p(D)E ] ∗ f = δ ∗ f = f,

as required.

Example 5.5. Using the above example we can find a particular solution ofa nonhomogeneous ordinary differential equation with constant coefficientswithout variation of parameters. Indeed, let n = 1 and p(D) = L, where

L =dm

dxm+ am−1

dm−1

dxm−1+ . . . a1

d

dx+ a0,

and all coefficients aj are constants. In this case, we may take E(x) =θ(x)y(x), where y is the solution of the equation Ly = 0 satisfying theinitial conditions

y(0) = y′(0) = . . . = y(m−2)(0) = 0, y(m−1)(0) = 1.

(see (4.46)). This implies that if f ∈ C(R1) has a compact support, then

u1(x) =∫E(x− t)f(t)dt =

∫ x

−∞y(x− t)f(t)dt

is a particular solution of the equation Lu = f . Note that since y(x − t)satisfies, as a function of x, the equation Ly(x − t) = 0 for any t ∈ R, itfollows that

u2(x) =∫ 0

−∞y(x− t)f(t)dt

satisfies the equation Lu2 = 0. Therefore, together with u1(x), the function

(5.12) u(x) =∫ x

0y(x− t)f(t)dt

5.3. Convolution of distributions 115

is also a particular solution of Lu = f . Formula (5.12) gives a solutionof Lu = f for an arbitrary f ∈ C(R1) (not necessarily with a compactsupport) since the equality Lu = f at x depends only on the behavior off in a neighborhood of x, and the solution u itself given by this formula isdetermined in a neighborhood of x by the values of f on a finite segmentonly.

An example: a particular solution of the equation u′′ + u = f is givenby the formula

u(x) =∫ x

0sin(x− t)f(t)dt.

Remark 5.5. If the potential in the right hand side of an integral like theones taken above, is multiplied by a constant (for example, changes sign),then a particular solution can be obtained by the same procedure.

Example 5.6. Potentials. If En is the fundamental solution of the Laplaceoperator ∆ in Rn defined by (4.38), (4.39), then u = −En∗f with f ∈ E ′(Rn)is called a potential and satisfies the Poisson equation ∆u = −f .

In particular, if f = ρ, where ρ is piecewise continuous, then for n = 3we get the Newtonian potential

u(x) =1

4π

∫ρ(y)|x− y|dy,

and for n = 2, the logarithmic potential

u(x) = − 12π

∫ρ(y) ln |x− y|dy.

In these examples, the convolution En ∗ f is locally integrable for any n. Asa distribution, it coincides with a locally integrable function

u(x) = −∫En(x− y)ρ(y)dy,

since En(x− y)ρ(y) is locally integrable in R2nx,y and, by the Fubini theorem,

we can make the change of variables in the integral 〈u, ϕ〉 which leads us tothe definition of the convolution of distributions.

Let us give other examples of potentials. The convolution

(5.13) u = −En ∗ αδΓ,

where Γ is a smooth compact surface of codimension 1 in Rn and α ∈ C(Γ),is a simple layer potential and may be interpreted as a potential of a systemof charges smeared over Γ with the density α.


The convolution

(5.14) u = −En ∗∂

∂n(βδΓ)

is called a double layer potential and denotes the potential of a system ofdipoles smeared over Γ and oriented along the exterior normal with thedensity of the dipole moment equal to β(x).

Note that outside Γ both potentials (5.13) and (5.14) are C∞-functionsgiven by the formula

u(x) = −∫

ΓEn(x− y)α(y)dSy,

u(x) =∫∂En(x− y)

∂nyβ(y)dSy.

The behavior of the potentials near Γ is of interest, and we will later discussit in detail.

5.4. Other properties of convolution. Support and singular

support of a convolution

First, let us study the convolution of a smooth function and a distribution.

Proposition 5.6. Let f ∈ E ′(Rn) and g ∈ C∞(Rn), or f ∈ D′(Rn) andg ∈ C∞0 (Rn). Then f ∗ g ∈ C∞(Rn) and

(f ∗ g)(x) = 〈f(y), g(x− y)〉 = 〈f(x− y), g(y)〉

Proof. First, observe that g(x − y) is an infinitely differentiable functionof x with values in C∞(Rn

y ). If, moreover, g ∈ C∞0 (Rn), then the supportof g(x − y) as of a function of y, belongs to a compact set if x runs over acompact set. Therefore, 〈f(y), g(x− y)〉 ∈ C∞(Rn). The equation

〈f(x− y), g(y)〉 = 〈f(y), g(x− y)〉follows since, by definition, the change of variables in distributions is per-formed as if we deal with an ordinary integral. It remains to verify that ifϕ ∈ D(Rn), then

〈f ∗ g, ϕ〉 = 〈〈f(y), g(x− y)〉, ϕ(x)〉or

〈f(x), 〈g(y), ϕ(x+ y)〉〉 = 〈〈f(y), g(x− y)〉, ϕ(x)〉.Since g ∈ C∞(Rn), this means that⟨

f(x),∫g(y)ϕ(x+ y)dy

⟩=∫〈f(y), g(x− y)〉ϕ(x)dx,

5.4. Support of a convolution 117

or ⟨f(y),

∫g(x− y)ϕ(x)dx

⟩=∫〈f(y), g(x− y)〉ϕ(x)dx,

i.e., we are to prove that the continuous linear functional f may be insertedunder the integral sign. This is clear, since the integral (Riemann) sums ofthe integral ∫

g(x− y)ϕ(x)dx,

considered as functions of y, tend to the integral in the topology in C∞(Rny ),

because derivatives ∂αy of the Riemann integral sums are Riemann integralsums themselves for the integrand in which g(x−y) is replaced with ∂αy g(x−y). Proposition 5.6 is proved.

Proposition 5.7. Let fk → f in D′(Rn) and g ∈ E ′(Rn), or fk → f inE ′(Rn) and g ∈ D′(Rn). Then fk ∗ g → f ∗ g in D′(Rn).

Proof. The proposition follows immediately from

〈fk ∗ g, ϕ〉 = 〈fk(x), 〈g(y), ϕ(x+ y)〉〉.

Corollary 5.8. C∞0 (Rn) is dense in D′(Rn) and in E ′(Rn).

Proof. If ϕk ∈ C∞0 (Rn), ϕk(x) → δ(x) in E ′(Rn) as k → +∞ and f ∈D′(Rn) then ϕk ∗ f ∈ C∞(Rn) and lim

k→+∞ϕk ∗ f = δ ∗ f = f in D′(Rn).

Therefore, C∞(Rn) is dense in D′(Rn). But, clearly, C∞0 (Rn) is dense inC∞(Rn). Therefore, C∞0 (Rn) is dense in D′(Rn).

It is easy to see that if fl ∈ E ′(Rn) and suppfl ⊂ K, whereK is a compactin Rn (independent of l), then fl → f in D′(Rn) if and only if fl → f inE ′(Rn). But then it follows from the above argument that C∞0 (Rn) is densein E ′(Rn).

A similar argument shows that C∞0 (Ω) is dense in E ′(Ω) and D′(Ω) and,besides, C∞0 (Rn) is dense in S′(Rn).

These facts also say that in order to verify any distributional formulawhose both sides continuously depend, for example, on f ∈ D′(Rn), it suf-fices to do so for f ∈ C∞0 (Rn).

Proposition 5.9. Let f ∈ E ′(Rn), g ∈ D′(Rn). Then

supp(f ∗ g) ⊂ supp f + supp g.


Proof. We have to prove that if ϕ ∈ D(Rn) and suppϕ ∩(supp f+supp g) =∅, then 〈f∗g, ϕ〉 = 0. But this is clear from (5.9) since in this case suppϕ(x+y) does not intersect with supp[f(x)g(y)].

Definition 5.10. The singular support of a distribution f ∈ D′(Ω) is thesmallest closed subset F ⊂ Ω such that f |Ω\F ∈ C∞(Ω\F ), and is denotedsing supp f .

Proposition 5.11. Let f ∈ E ′(Rn), g ∈ D′(Rn). Then

(5.15) sing supp(f ∗ g) ⊂ sing supp f + sing supp g.

Proof. Using cut-off functions we can, for any ε > 0, represent any distri-bution h ∈ D′(Rn) as a sum h = hε + h

′ε, where supphε is contained in

the ε-neighborhood of sing supph and h′ε ∈ C∞(Rn). Let us choose such

representation for f and g:

f = fε + f′ε, g = gε + g

′ε.

Thenf ∗ g = fε ∗ gε + fε ∗ g

′ε + f

′ε ∗ gε + f

′ε ∗ g

′ε.

In this sum the three last summands belong to C∞(Rn) by Proposition 5.6.Therefore,

sing supp(f ∗ g) ⊂ sing supp(fε ∗ gε) ⊂ supp(fε ∗ gε) ⊂ supp fε + supp gε.

The latter set is contained in the 2ε-neighborhood of sing supp f+sing supp g.Since ε is arbitrary, this implies (5.15), as required.

5.5. Relation between smoothness of a fundamental solution

and that of solutions of the homogeneous equation

We will first prove the following simple but important

Lemma 5.12. Let p(D) be a linear differential operator with constant co-efficients, E = E(x) its fundamental solution, u ∈ E ′(Rn) and p(D)u = f .Then u = E ∗ f .

Proof. We have

E ∗ f = E ∗ p(D)u = p(D)E ∗ u = δ ∗ u = u,

as required.

5.5. Smoothness of solutions 119

Remark 5.13. The requirement u ∈ E ′(Rn) is essential here (the assertionis clearly wrong without it). Namely, it allows us to move p(D) from u to Ein the calculation given above.

Theorem 5.14. Assume that E is a fundamental solution of p(D), andE ∈ C∞(Rn\0). If u ∈ D′(Ω) is a solution of p(D)u = f with f ∈ C∞(Ω),then u ∈ C∞(Ω).

Proof. Note that the statement of the theorem is local, i.e., it suffices toprove it in a neighborhood of any point x0 ∈ Ω. Let χ ∈ D(Ω), χ(x) = 1 ina neighborhood of x0. Consider the distribution χu and apply p(D) to it.Denoting p(D)(χu) by f1, we see that f1 = f in a neighborhood of x0 and,in particular, x0 6∈ sing supp f1. By Lemma 5.12, we have

(5.16) χu = E ∗ f1.

We now use Proposition 5.11. Since sing supp E = 0, we get

sing supp(χu) ⊂ sing supp f1.

Therefore, x0 6∈ sing supp(χu) and x0 6∈ sing suppu, i.e., u is a C∞-functionin a neighborhood of x0, as required.

A similar fact is true for the analyticity of solutions.

Recall that a function u = u(x) defined in an open set Ω ⊂ Rn, withvalues in R or C, is called real analytic if for every y ∈ Ω there exists aneighborhood Uy of y in Ω, such that u|Uy = u|Uy(x) can be presented in Uyas a sum of a convergent power series

u(x) =∑α

cα(x− y)α, x ∈ Uy,

with the sum over all multiindices α.

Consider first the case of the homogeneous equation p(D)u = 0.

Theorem 5.15. Assume that a fundamental solution E(x) of p(D) is realanalytic in Rn\0, u ∈ D′(Ω) and p(D)u = 0. Then u is a real analyticfunction in Ω.

Proof. We know already that u ∈ C∞(Ω). Using the same argument asin the proof of Theorem 5.14, consider (5.16), where now f1 ∈ D(Ω) andf1(x) = 0 in a neighborhood of x0. Then for x close to x0 we have

(5.17) u(x) =∫|y−x0|≥ε>0

E(x− y)f1(y)dy.


Note that the analyticity of E(x) for x 6= 0 is equivalent to the possibilityto extend E(x) to a holomorphic function in a complex neighborhood of theset Rn\0 in Cn, i.e., in an open set G ⊂ Cn such that G ∩ Rn = Rn\0.

Recall, that a complex-valued function ϕ = ϕ(z) defined for z ∈ G,where G is an open subset in Cn, is called holomorphic or complex analyticor even simply analytic, if one of the following two equivalent conditions isfulfilled:

a) ϕ(z) can be expanded in a power series in z − z0 in a neighborhoodof each point z0 ∈ G, i.e.,

ϕ(z) =∑α

cα(z − z0)α,

where z is close to z0, cα ∈ C and α runs over all multiindices;

b) ϕ(z) is continuous inG and differentiable with respect to each complexvariable zj (i.e., is holomorphic with respect to each zj).

Proof can be found on the first pages of any textbook on functions ofseveral complex variables (e.g. Gunning and Rossi [10], Ch.1, Theorem A2,or Hormander [13], Ch.2).

Note that in (5.17) the variable y varies over a compact K ⊂ Rn. Fixy0 6= x0. Then for |z−x0| < δ , |w−y0| < δ with z, w ∈ Cn and δ sufficientlysmall, we can define E(z −w) as a holomorphic function in z and w. Then,clearly, ∫

|y−y0|≤δE(z − y)f1(y)dy

is well defined for |z−x0| < δ and holomorphic in z since we can differentiatewith respect to z under the integral sign. But using a partition of unity, wecan express the integral in (5.17) as a finite sum of integrals of this form.Therefore, we can define u(z) for complex z, which are close to x0, so thatu(z) is analytic in z. In particular, u(x) can be expanded in a Taylor seriesin a neighborhood of x0, i.e., u(x) is real analytic.

The case of a nonhomogeneous equation reduces to the above due to theCauchy-Kovalevsky theorem which we will give in a form convenient for us.

Theorem 5.16. (Cauchy-Kovalevsky) Consider the equation

(5.18)∂mu

∂xmn+

∑|α|≤m,αn<m

aα(x)Dαu(x) = f(x),

5.5. Smoothness of solutions 121

where aα(x), f(x) are real analytic in a neighborhood of 0 ∈ Rn. Considerthe Cauchy problem for this equation, i.e., the initial value problem with theinitial conditions

(5.19) u|xn=0 = ϕ0(x′), . . . ,∂m−1u

∂xm−1n

∣∣∣∣xn=0

= ϕm−1(x′),

where x′ ∈ Rn−1, ϕj(x′) are real analytic functions in a neighborhood of0 ∈ Rn−1. The problem (5.18) − (5.19) has a unique solution, which is realanalytic in a sufficiently small ball |x| ≤ ε in Rn.

We will not prove this theorem (the reader can find the proof e.g. inPetrovskii [21], Ch.1, Hormander [12], Theorem 5.1.2 or [14], v.1, Theorem9.4.5). Note only that one of the ways to prove it is to seek u(x) as a powerseries whose coefficients are uniquely determined by (5.18), (5.19), and thenestimate the coefficients, proving the convergence of this series.

Note that if P = p(x,D) is a linear differential operator of order ≤ m

with analytic coefficients in a neighborhood of 0 whose principal symbolpm(x, ξ) is such that pm(0, ξ) 6≡ 0 (i.e., one of the highest coefficients doesnot vanish at 0), then by rotating the coordinate axes, we can reduce Pu = f

to the form (5.18). Namely, we should choose coordinate axes so that theplane xn = 0 were noncharacteristic at 0. Therefore, the equation Pu = f ,where f is an analytic function in a neighborhood of 0, always has a solutionu(x) analytic in a neighborhood of 0. In particular, this always holds foroperators with constant coefficients. It is clear now that Theorem 5.15implies

Theorem 5.17. Let a fundamental solution E(x) of p(D) be real analyticfor x 6= 0. If u ∈ D′(Ω), p(D)u = f and f is real analytic in Ω, then so isu.

Corollary 5.18. If u ∈ D′(Ω), ∆mu = f , where f is real analytic in Ω,then so is u.

Remark 5.19. Any differential operator p(D) with constant coefficientshas a fundamental solution. It is clear from Theorem 5.14 that the followingconditions on p(D) are equivalent:

a) Operator p(D) has a fundamental solution E(x) ∈ C∞(Rn\0);b) if u ∈ D′(Ω) and p(D)u = 0, then u ∈ C∞(Ω).


The operators p(D) with this property are called hypoelliptic. It can beproved that p(D) is hypoelliptic if and only if

∂p(ξ)∂ξj

/p(ξ)→ 0 as |ξ| → ∞, j = 1, . . . , n.

Further, by Theorem 5.15, the following two conditions are equivalent:

c) p(D) has a fundamental solution E(x) which is real analytic for x 6= 0.

d) If u ∈ D′(Ω) and p(D)u = 0, then u is real analytic in Ω.

It can be proved that c) and d) are equivalent to the ellipticity of p(D)in the sense of Section 1.5.

Proofs of all statements formulated in this remark, can be found, forexample, in Shilov [27], Ch.III, or Hormander [14], Theorem 11.1.1.

5.6. Solutions with isolated singularities. A removable

singularity theorem for harmonic functions

Consider a hypoelliptic operator p(D) i.e., an operator p(D) with a funda-mental solution E(x) ∈ C∞(Rn\0). The fundamental solution E(x) itselfsatisfies, for x 6= 0, the equation p(D)u = 0 and at x = 0 has a singularity.The same applies to ∂αE(x). It turns out that if the singularity of u is notworse than that of a power of |x|−1, this example actually exhausts all thepossibilities.

Theorem 5.20. Let u(x) ∈ C∞(Ω\0), where Ω is a domain in Rn, 0 ∈ Ω,and let u(x) be a solution of the equation p(D)u = 0 in Ω\0, where p(D) isa hypoelliptic operator. Let E(x) be a fundamental solution of p(D). Assumealso that in a neighborhood of 0 we have

(5.20) |u(x)| ≤ C|x|−N , x 6= 0,

for some C and N . Then u(x) can be represented in Ω\0 as

(5.21) u(x) =∑|α|≤p

cα∂αE(x) + u0(x),

where cα ∈ C, u0(x) ∈ C∞(Ω) and p(D)u0 = 0.

In the proof we will need the following

Lemma 5.21. If u(x) ∈ C∞(Ω\0) satisfies (5.20) and (5.21) then thereexists u ∈ D′(Ω), such that u|Ω\0 = u.

5.6. Solutions with isolated singularities 123

Proof. We would like to give a meaning to the integral∫u(x)ϕ(x)dx, where

ϕ(x) ∈ D(Ω), in such a way that its value on ϕ ∈ D(Ω\0) would be natural(for such ϕ this is the usual integral of a continuous function with a compactsupport). To this end subtract from ϕ its Taylor polynomial at 0, and notethat

ϕ(x)−∑

|α|≤N−1

ϕ(α)(0)α!

xα = O(|x|N ) as |x| → 0.

Let χ ∈ D(Ω), χ(x) = 1 in a neighborhood of 0. Set

〈u, ϕ〉 =∫

Ωu(x)

[ϕ(x)− χ(x)

∑|α|≤N−1

ϕ(α)(0)α!

xα]dx,

where the integral is already defined since the integrand is bounded. Clearly,if ϕ ∈ D(Ω\0), then 〈u, ϕ〉 =

∫u(x)ϕ(x)dx, because ϕ(α)(0) = 0 for any

α. It is also easy to see that u ∈ D′(Ω), thus proving the Lemma.

Proof of Theorem 5.20. Clearly, p(D)u is a distribution with the supportat 0, hence, due to Theorem 4.7 (see also Example 4.8),

p(D)u(x) =∑|α|≤p

cα∂αδ(x).

But setting

u0(x) = u(x)−∑|α|≤p

cα∂αE(x),

we, clearly, get u0 ∈ D′(Ω) and p(D)u0 = 0 implying u0 ∈ C∞(Ω).

In particular, for the Laplace operator and n ≥ 3, a solution of theequation ∆u = 0 (called a harmonic function) defined in Ω\0 and sat-isfying (5.20) is a sum of homogeneous functions of homogeneity orders2 − n, 2 − n − 1, . . . , 2 − n − p and a harmonic function in Ω. For n = 2such a solution is the sum of c0 ln |x|, homogeneous functions of orders−1,−2, . . . ,−p, and a harmonic function in Ω. In particular, this impliesthe following removable singularity theorem for harmonic functions:

Theorem 5.22. (On removable singularity) Let u(x) be a harmonic func-tion in Ω\0 such that in a neighborhood of 0

lim|x|→0

[|x|n−2 · u(x)] = 0, for n ≥ 3,


lim|x|→0

[(ln |x|)−1u(x)] = 0, for n = 2.

Then u(x) can be extended to a harmonic function in Ω.

5.7. Estimates of derivatives for a solution of a hypoelliptic

equation

Recall that an operator p(D) with constant coefficients in Rn is called hy-poelliptic if it has a fundamental solution E(x) ∈ D′(Rn)

⋂C∞(Rn\0) or,

equivalently, any solution u ∈ D′(Ω) of the equation p(D)u = 0 actuallybelongs to C∞(Ω).

It is clear from the calculation of the fundamental solution of the Lapla-cian ∆, given above, that ∆ is hypoelliptic. Another example, to be dis-cussed below in Section 7, is the heat operator.

Proposition 5.23. Let Ω1,Ω2 be two bounded open sets in Rn such thatΩ2 ⊂ Ω1 (here “bar” means the closure in Rn), and p(D) is a hypoellipticoperator in Rn. Then solutions u ∈ C∞(Ω1) of the equation p(D)u = 0satisfy

(5.22) supx∈Ω2

|∂αu(x)| ≤ Cα∫

Ω1

|u(x)|dx,

where α is an arbitrary multiindex, and constants Cα do not depend on thechoice of u (but may depend on the choice of p, Ω1 and Ω2, as well as ofα). In particular, estimates

(5.23) supx∈Ω2

|∂αu(x)| ≤ C ′α supx∈Ω1

|u(x)|

hold, where C ′α are other constants, also independent of u.

Proof. It suffices to consider the case when the right hand side of (5.22) isfinite (otherwise the inequality is obvious).

We will use the closed graph theorem in the following form:

If E1, E2 are two Banach spaces and a linear operator A : E1 → E2 is

everywhere defined and has a closed graph (i.e., the set (x,Ax) : x ∈ E1 isclosed in E1 × E2 or, in other words, xn → x and Axn → y imply Ax = y),then this operator is continuous (or, which is the same, bounded). (For theproof, based on the Baire Theorem 4.20, see e.g. Reed and Simon [22],Theorem III.12.)

As E1, we take the subspace of L1(Ω1) consisting of the solutions of theequation p(D)u = 0 (the operator is applied in the distributional sense).

5.7. Estimates of derivatives of a solution 125

It is important that E1 is a closed subspace of L1(Ω1). In fact, if un → u

in L1(Ω1) and p(D)un = 0, then un → u weakly in D′(Ω1), i.e., 〈un, ϕ〉 →〈u, ϕ〉 for any ϕ ∈ D(Ω1). But then 〈un, p(−D)ϕ〉 → 〈u, p(−D)ϕ〉 for anyϕ ∈ D(Ω1) since in this case p(−D)ϕ ∈ D(Ω1). The latter limit meansthat p(D)un → p(D)u weakly in D′(Ω1) implying p(D)u = 0, as required.Therefore, E1 is a Banach space. Notice that, actually, E1 ⊂ C∞(Ω1) byhypoellipticity of p(D).

As E2, we take the Banach space of functions u ∈ C∞(Ω2) such thatp(D)u = 0 and all derivatives of u of order no greater than k (here k ∈ Z+

is arbitrary) are bounded. In E2, introduce the usual norm

‖u‖(k) =∑|α|≤k

supx∈Ω2

|∂αu(x)|.

If un → u with respect to ‖·‖(k) and p(D)un = 0, then clearly p(D)u = 0 (inD′(Ω2)) and, therefore, u ∈ C∞(Ω2). Therefore, E2 is a Banach space (it isa closed subspace of a Banach space of functions u ∈ Ck(Ω2) with boundedderivatives of order up to k included).

Now, consider a linear operator A : E1 → E2 that maps u into u|Ω2 . Toverify that the graph of A is closed, we need to prove that if un → u in E1

and Aun → v in E2, then v = Au, i.e., v = u|Ω2 . But un → u in E1 impliesthat un → u weakly in D′(Ω1) and therefore, clearly, un|Ω2 → u|Ω2 weaklyin D′(Ω2). But on the other hand, we know that un|Ω2 → v in E2 (hence,weakly in D′(Ω2)). Since one sequence cannot have two different weak limitsin D′(Ω2), it follows that v = u|Ω2 , as required.

By the closed graph theorem A is bounded. But this means that

‖u‖(k) ≤ Ck∫

Ω1

|u(x)|dx,

which, since k is arbitrary, implies (5.22). Applying an obvious inequality∫Ω1

|u(x)|dx ≤ C supx∈Ω1

|u(x)|,

we also get (5.23).

This proof is somewhat ineffective, in the sense that it does not allow toget any information about the constants Ck, because the constants whichappear due to the application of the closed graph theorem, are out of ourcontrol. But it is possible to get effective constants in a weaker result, whereevery higher derivative ∂αu is estimated through L1-norms of the derivativesof order ≤ deg p. We will sketch now how this can be done.


We will argue as in the proof of Theorem 5.14. Let us choose a functionχ ∈ D(Ω1), such that χ = 1 in a neighborhood of Ω2. Then we obtain forany u ∈ C∞(Ω1),

χu = E ∗ f, where f = p(D)(χu).

Now note that if additionally p(D)u = 0 in Ω1, then f = 0 in a neighborhoodU of Ω2, which implies that for x ∈ Ω2 we can write

u(x) =∫

Ω1\UE(x− y)f(y)dy,

where the integrand is in fact a C∞ function of (x, y) for all x ∈ U , andthe integration is taken over a compact set which is independent of x ∈ Ω2.Therefore we can write the derivatives ∂αu(x) for x ∈ Ω2 by differentiatingunder the integral sign, i.e.

∂αu(x) =∫

Ω1\U∂αx E(x− y)f(y)dy.

it follows that

|∂αu(x)| ≤ C∑

|α|≤deg p

∫Ω1

|∂αu(y)|dy,

for every x ∈ Ω2.

Proposition 5.23 implies

Corollary 5.24. If uj | j = 1, 2, . . . is a sequence of distributional (henceC∞) solutions of the equation p(D)uj = 0 on Ω1, p(D) is hypoelliptic, anduj → u, in L1(Ω1), then uj → u in C∞(Ω1).

Making the arguments given here more precise, it is easy to see that ifuj → u in D′(Ω) and p(D)uj = 0, then uj → u in the topology of C∞(Ω1).In particular, it follows that for solutions of p(D)u = 0, the convergence inL1(Ω1) implies the uniform convergence of derivatives of any order on Ω2.

More precise estimates can be obtained in concrete situations (e.g. forthe Laplacian in a ball).

Proposition 5.23 also makes it possible, for instance, to get informationon the growth or decay of derivatives of a solution of the equation p(D)u = 0from similar data concerning the solution itself. We will see an importantexample of such an argument in Section 7.6.

5.8. Fourier transform of tempered distributions 127

5.8. Fourier transform of tempered distributions

Denote by F the Fourier transform given by

(F f)(ξ) = f(ξ) =∫f(x)e−ix·ξdx.

It is easy to see that it is well defined on functions f ∈ S(Rn) (f = f(x)),because of their fast decay (faster than any polynomial) as |x| → ∞. Let usdescribe properties of f assuming that f ∈ S(Rn). It is obvious that Ff isbounded. Differentiating f (with respect to ξ), we come to integrals of thesame type, so it follows that F maps S(Rn) to C∞(Rn), and every derivativeof Ff is bounded. Multiplying Ff(ξ) by ξα (where α is a multiindex) andintegrating by parts, we get

ξαf(ξ) =∫f(x)(−Dx)αe−ix·ξdx =

∫(Dαf)(x)e−ix·ξdx.

This is also an integral of the same type (the Fourier transform of a functionDαf ∈ S(Rn)). So we see that f remains bounded after differentiation of anyorder and multiplication by any polynomial. This means that Ff ∈ S(Rn

ξ ),so F is a linear map of S(Rn

x) to S(Rnξ ).

It easily follows from the argument above that F : S(Rnx) → S(Rn

ξ ) iscontinuous if the spaces S(Rn

x) and S(Rnξ ) are considered with their Frechet

topologies (introduced in the previous chapter). Moreover, it is known fromFourier analysis that in fact the map

F : S(Rnx) −→ S(Rn

ξ )

is a topological isomorphism (a continuous linear operator with continuousinverse) and the inverse operator is given by the formula

(5.24) (F−1g)(x) = (2π)−n∫g(ξ)eix·ξdξ.

(See, for example, Hormander [12], Corollary 1.7.1, Hormander [14], Theo-rem 7.1.5, or Schwartz [25], Ch.5.)

Clearly, if f ∈ S(Rn), then

(5.25) 〈Ff, ϕ〉 = 〈f, Fϕ〉, ϕ ∈ S(Rn).

In other words, tF = F . Therefore, using (5.25) we may extend F to acontinuous map F : S′(Rn) −→ S′(Rn). Clearly, the Fourier transformF gives a topological isomorphism of S′(Rn) onto S′(Rn), and the inverseoperator F−1 is also equal to t(F−1), where F−1 is given by (5.24), i.e.,

(5.26) 〈F−1f, ϕ〉 = 〈f, F−1ϕ〉, f ∈ S′(Rn), ϕ ∈ S(Rn).


Thus we can define the Fourier transform of any function f(x) such that|f(x)| ≤ C(1+ |x|)N . The Fourier transform is in general not a function buta distribution from S′(Rn).

Note that since S(Rn) is dense in S′(Rn), then F on S′(Rn) is the exten-sion by continuity of the Fourier transform defined on S(Rn). This remarkmakes it possible to extend by continuity certain facts from S(Rn) to S′(Rn)without proving them separately. For instance, we do not need to verify for-mula (5.26), since it holds for f ∈ S(Rn), ϕ ∈ S(Rn) and both sides arecontinuous with respect to f ∈ S′(Rn) in the weak topology of S′(Rn).

As in the case of f ∈ S(Rn), the Fourier transform of a distributionf ∈ S′(Rn) will be denoted by f(ξ).

Let us find the Fourier transform of Dαf . If f ∈ S(Rn), then integrationby parts yields

Dαf(ξ) =∫e−ix·ξDαf(x)dx =

∫ [(−D)αe−ix·ξ

]f(x)dx =

=∫ξαe−ix·ξf(x)dx = ξαf(ξ).

Therefore

(5.27) Dαf(ξ) = ξαf(ξ)

implying

(5.28) p(D)f(ξ) = p(ξ)f(ξ)

for any polynomial p(ξ). Thus, a Fourier transform converts p(D) into themultiplication by the symbol p(ξ).

Note that formula (5.28) holds by continuity for f ∈ S′(Rn).

Example 5.7. Let us find the Fourier transform of the δ-function. Forϕ ∈ S(Rn) we have

〈δ(ξ), ϕ(ξ)〉 =⟨δ(x),

∫ϕ(ξ)e−iξ·xdξ

⟩=∫ϕ(ξ)dξ = 〈1, ϕ(ξ)〉

implying

(5.29) δ(ξ) = 1.

Let us also find the Fourier transform of 1. We have

〈F (1), ϕ(ξ)〉 =∫ϕ(ξ)e−iξ·xdξdx = (2π)nϕ(0)

5.8. Fourier transform of tempered distributions 129

by the inversion formula (5.24). Therefore,

F (1)(ξ) = (2π)nδ(ξ).

Example 5.8. Let us find the Fourier transform of the Heaviside functionθ(x). It is most convenient to represent θ(x) as the limit of functions whichdecay at infinity, so that their Fourier transforms can be understood in theordinary sense. This can be done, for instance, as follows: clearly,

θ(x) = limε→0+

θ(x)e−εx in S′(R1).

We have∫θ(x)e−εxe−iξ·xdx =

∫ ∞0

e−x(iξ+ε)dx =1

iξ + ε= − i

ξ − iε .

Passing to the limit as ε→ +0, we get

θ(ξ) = − i

ξ − i0 .

Note that ddxθ(x) = δ(x) and iDx(θ(x)) = δ(x). This agrees with (5.27) and

(5.29) due to the obvious identity iξ(− iξ−i0) = 1.

We can try to use the last observation for an alternative way to calculateθ(ξ). Namely, since θ′(x) = δ(x), it follows that iξ · θ(ξ) = 1. A distributiong(ξ) ∈ S′(R1) satisfying ξg(ξ) = 1 is not, however, uniquely defined, though,clearly, it should be equal to 1/ξ for ξ 6= 0. It is easy to see that all theambiguity reduces to adding Cδ(ξ), where C is an arbitrary constant. Thisconstant can be determined by applying the definition of the Fourier trans-form of distributions to a particular function ϕ(ξ) ∈ S(R1). It is simpler,however, to use the direct method explained above.

Proposition 5.25. Let f ∈ E ′(Rn). Then f(ξ) ∈ C∞(Rn) and

(5.30) f(ξ) = 〈f, e−iξ·x〉.

There exists a constant N > 0 such that

(5.31) |∂αξ f(ξ)| ≤ Cα(1 + |ξ|)N

for any multiindex α.

Proof. Let us choose constants N , C and a compact K ⊂ Rn so that

(5.32) |〈f, ϕ〉| ≤ C∑|α|≤N

supx∈K|∂αϕ(x)|, ϕ ∈ C∞(Rn).


Set g(ξ) = 〈f, e−iξ·x〉. Since e−iξ·x is an infinitely differentiable function inξ with values in C∞(Rn), then, clearly, g(ξ) ∈ C∞(Rn

ξ ) and

∂αξ g(ξ) = 〈f, (−ix)αe−iξ·x〉,implying that (5.31) holds for g(ξ). It remains to verify that g(ξ) = f(ξ),i.e., that⟨

f(x),∫ϕ(ξ)e−iξ·xdξ

⟩=∫ϕ(ξ)〈f(x), e−iξ·x〉dξ, ϕ ∈ S(Rn

ξ ).

But this follows from the convergence of∫ϕ(ξ)e−iξ·xdξ in the topology of

C∞(Rnx) which means that this integral and integrals obtained from it by

taking derivatives of any order converge uniformly with respect to x ∈ K(where K is a compact set in Rn).

Remark 5.26. The condition f ∈ C∞(Rn) and the estimate (5.31) arenecessary but not sufficient for a distribution f to have a compact support.For instance, if f ∈ E ′(Rn), then, clearly, f(ξ) may be defined via (5.30) forξ ∈ Cn and we get an entire (i.e., homomorphic everywhere in Cn) functionsatisfying

|f(ξ)| ≤ C(1 + |ξ|)NeA|Im ξ|.

It turns out that this condition on f is already necessary and sufficient forf ∈ E ′(Rn). This fact is a variant of the Paley-Wiener theorem and its proofmay be found in Hormander [12], Theorem 1.7.7, or [14], Theorem 7.3.1.

Proposition 5.27. Let f ∈ E ′(Rn), g ∈ S′(Rn). Then

(5.33) f ∗ g(ξ) = f(ξ)g(ξ).

Proof. Note that the right-hand side of (5.33) makes sense due to Proposi-tion 5.25. It is easy also to verify that f ∗ g ∈ S′(Rn) so that the left-handside of (5.33) makes sense. Indeed, if ϕ ∈ S(Rn), then 〈f(y), ϕ(x + y)〉 ∈S(Rn

x) due to (5.32). Therefore, for ϕ ∈ S(Rn) the equality 〈f ∗ g, ϕ〉 =〈g(x), 〈f(y), ϕ(x+ y)〉〉 makes sense and we clearly get f ∗ g ∈ S′(Rn).

Now let us verify (5.33). For ϕ ∈ S(Rn) we get

〈f ∗ g(ξ), ϕ(ξ)〉 =⟨f ∗ g,

∫ϕ(ξ)e−iξ·xdξ

⟩=⟨g(x),

⟨f(y),

∫ϕ(ξ)e−iξ·(x+y)dξ

⟩⟩=⟨g(x),

∫ϕ(ξ)e−iξ·x

⟨f(y), e−iξ·y

⟩dξ

⟩=⟨g(x),

∫ϕ(ξ)f(ξ)e−iξ·xdξ

⟩= 〈g(ξ), f(ξ)ϕ(ξ)〉 = 〈f(ξ)g(ξ), ϕ(ξ)〉,

5.9. Applying Fourier transform 131

as required.

5.9. Applying Fourier transform to find fundamental

solutions

Let p(D) be a differential operator with constant coefficients, E ∈ S′(Rn) itsfundamental solution. The Fourier transform E(ξ) ∈ S′(Rn) satisfies

(5.34) p(ξ)E(ξ) = 1.

Clearly, E(ξ) = 1/p(ξ) on the open set ξ : p(ξ) 6= 0. If p(ξ) 6= 0 everywherein Rn, then we should just take E(ξ) = 1/p(ξ) and apply the inverse Fouriertransform. (It can be proved, though non-trivially, that in this case 1/p(ξ)is estimated by (1 + |ξ|)N with some N > 0, hence 1/p(ξ) ∈ S′(Rn

ξ ) – see,e.g., Hormander [14], Example A.2.7 in Appendix A2, vol. 2.) When p(ξ)has zeros we should regularize 1/p(ξ) in neighborhoods of zeros of p(ξ) so asto get a distribution E(ξ) equal to 1/p(ξ) for p(ξ) 6= 0 and satisfying (5.34).For instance, for n = 1 and P (ξ) = ξ we can set E(ξ) = v.p.1

ξ or E(ξ) = 1ξ±i0

or, in general, E(ξ) = v.p.1ξ + Cδ(ξ), where C is an arbitrary constant.

In the general case, we must somehow give meaning to integrals of theform ∫

Rn

ϕ(ξ)p(ξ)

dξ, ϕ ∈ S(Rn)

to get a distribution with needed properties. It can often be done, e.g. byentering the complex domain (the construction of the distribution 1

ξ±i0 is anexample). Sometimes 1/p(ξ) is locally integrable. Then it is again naturalto assume E(ξ) = 1/p(ξ). For instance, for p(D) = ∆, i.e., p(ξ) = −ξ2, wecan set E(ξ) = −1/ξ2 for n ≥ 3.

Since, as is easy to verify, the Fourier transform and its inverse preservethe spherical symmetry and transform a distribution of homogeneity degreeα into a distribution of homogeneity degree−α−n, then, clearly, F−1( 1

|ξ|2 ) =C|x|2−n, where C 6= 0. Therefore, we get the fundamental solution En(x)described above.

For n = 2 the function 1|ξ|2 is not integrable in a neighborhood of 0. Here

a regularization is required. For instance, we can set

〈E2(ξ), ϕ(ξ)〉 = −∫

R2

ϕ(ξ)− χ(ξ)ϕ(0)|ξ|2 dξ,

where χ(ξ) ∈ C∞0 (R2), χ(ξ) = 1 in a neighborhood of 0. It can be verifiedthat E2(x) = 1

2π ln |x|+ C, where C depends on the choice of χ(ξ).


5.10. Liouville’s theorem

Theorem 5.28. Assume that the symbol p(ξ) of an operator p(D) vanishes(for ξ ∈ Rn) only at ξ = 0. If u ∈ S′(Rn) and p(D)u = 0, then u(x) isa polynomial in x. In particular, if p(D)u = 0, where u(x) is a boundedmeasurable function on Rn, then u = const.

Remarks.

a) The inclusion u ∈ S′(Rn) holds, for instance, if u ∈ C(Rn) and|u(x)| ≤ C(1 + |x|)N , x ∈ Rn, for some C and N .

b) If p(ξ0) = 0 for some ξ0 6= 0, then p(D)u = 0 has a bounded noncon-stant solution eiξ0·x. Therefore, p(ξ) 6= 0 for ξ 6= 0 is a necessary conditionfor the validity of the theorem.

c) If p(ξ) 6= 0 for all ξ ∈ Rn, then p(D)u = 0 for u ∈ S′(Rn) implies, aswe will see from the proof of the theorem, that u ≡ 0.

Proof of Theorem 5.28. Applying the Fourier transform we get

p(ξ)u(ξ) = 0.

Since p(ξ) 6= 0 for ξ 6= 0, then, clearly, the support of u(ξ) is equal to0 (provided u 6≡ 0). In more detail: if ϕ(ξ) ∈ D(Rn\0), then ϕ(ξ)

p(ξ) ∈D(Rn\0) implying

〈u(ξ), ϕ(ξ)〉 =⟨u(ξ), p(ξ)

ϕ(ξ)p(ξ)

⟩=⟨p(ξ)u(ξ),

ϕ(ξ)p(ξ)

⟩= 0.

Thus, supp u(ξ) ⊂ 0. Therefore,

u(ξ) =∑|α|≤p

aαδ(α)(ξ), aα ∈ C,

implying

u(x) =∑|α|≤p

cαxα, cα ∈ C,

as required.

Example. If u(x) is a harmonic function everywhere on Rn and |u(x)| ≤M , then u = const (this statement is often called Liouville’s theorem forharmonic functions). If u(x) is a harmonic function in Rn and |u(x)| ≤C(1 + |x|)N , then u(x) is a polynomial. The same holds for solutions of∆mu = 0.

Later we will prove a more precise statement:

5.10. Liouville’s theorem 133

If u is harmonic and bounded below (or above), e.g., u(x) ≥ −C, x ∈ Rn,then u = const (see e.g. Petrovskii [21]).

This statement fails already for the equation ∆2u = 0 satisfied by thenonnegative polynomial u(x) = |x|2.

Let us give one more example of application of Liouville’s theorem. Letfor n ≥ 3 a function u(x) be defined and harmonic in the exterior of a ball,i.e., for |x| > R and u(x) → 0 as |x| → +∞. We wish to describe in detailthe behavior of u(x) as |x| → ∞. To this end we may proceed, for example,as follows. Take a function χ(x) ∈ C∞(Rn) such that

χ(x) =

0 for |x| ≤ R+ 1,

1 for |x| ≥ R+ 2.

Now, consider the following function u(x) ∈ C∞(Rn):

u(x) =

χ(x)u(x) for |x| > R,

0 for |x| ≤ R.Clearly, ∆u(x) = f(x) ∈ C∞0 (Rn). Let us verify that

(5.35) u = En ∗ f,where En is the standard fundamental solution of the Laplace operator.

Clearly, (En ∗f)(x)→ 0 as |x| → ∞ and, by assumption, the same holdsfor u(x) (and hence, for u(x), since u(x) = u(x) for |x| > R+ 2). Moreover,

∆(En ∗ f) = ∆En ∗ f = f,

so ∆(u − En ∗ f) = 0 everywhere on Rn. But, by Liouville’s theorem, thisimplies that u− En ∗ f = 0, as required.

Formula (5.35) can be rewritten in the form

u(x) =∫|y|≤R+2

En(x− y)f(y)dy, |x| > R+ 2.

The explicit form of En now implies

|u(x)| ≤ C

|x|n−2, |x| > R+ 2.

The above argument is also applicable in a more general situation. Forharmonic functions a more detailed information on the behavior of u(x) as|x| → ∞ can be also obtained with the help of the Kelvin transformation

u(x) 7→ v(x) = |x|2−nu(

x

|x|2).


It is easy to verify that the Kelvin transformation preserves harmonicity.For n = 2 the Kelvin transformation is the change of variables induced bythe inversion x 7→ x/|x|2.

If u(x) is defined for large |x|, then v(x) is defined in a punctured neigh-borhood of 0, i.e., in Ω\0, where Ω is a neighborhood of 0. If u(x)→ 0 as|x| → +∞, then v(x)|x|n−2 → 0 as |x| → 0, and by the removable singularitytheorem, v(x) can be extended to a function which is harmonic everywherein Ω. (We will denote the extension by the same letter v). Observe thatKelvin transformation is its own inverse (or, as they say, is an involution),i.e., we may express u in terms of v via the same formula u(y) = |y|2−nv( y

|y|2 ).Expanding v(x) into Taylor series at x = 0 we see that u(y) has a convergentexpansion into a series of the form

u(y) = |y|2−n∑α

cαyα

|y|2|α|

in a neighborhood of infinity. In particular,

u(y) = c0|y|2−n +O(|y|1−n) as |y| → +∞.

5.11. Problems

5.1. Find the Fourier transforms of the following distributions:

a) 1x+i0 ;

b) δ(r − r0), r = |x|, x ∈ R3;

c) sin(r0|x|)|x| , x ∈ R3;

d) xkθ(x), k ∈ Z+.

5.2. Find the inverse Fourier transforms of the following distributions:

a) 1/|ξ|2, ξ ∈ Rn, n ≥ 3;

b) 1|ξ|2+k2 , ξ ∈ R3, k > 0;

c) 1|ξ|2−k2±i0 , ξ ∈ R3, k > 0.

5.3. Using the result of Problem 5.2 b), find the fundamental solution forthe operator −∆ + k2 in R3.

5.11. Problems 135

5.4. We will say that the convolution f ∗g of two distributions f, g ∈ D′(Rn)is defined if for any sequence χk ∈ D(Rn) such that χk(x) = 1 for |x| ≤k, k = 1, 2, . . . the limit f ∗ g = lim

k→∞f ∗ (χkg) exists in D′(Rn) and does not

depend on the choice of the sequence χk. Prove that if f, g ∈ D′(R) and thesupports of f and g belong to [0,+∞), then f ∗ g is defined, and additionand convolution define on the set D′(R+) of distributions with supports in[0,+∞) the structure of an associative commutative ring with unit.

5.5. Prove that δ(k)(x) ∗ f(x) = f (k)(x) for f ∈ D′(R1).

5.6. Find the limits limt→−∞

eixt

x+i0 and limt→+∞

eixt

x+i0 in D′(R1).

5.7. For f ∈ L1loc(R+) define an integration operator setting If =

∫ x0 f(ξ)dξ.

Prove that If = f ∗ θ and I can be extended by continuity to D′(R+).

5.8. Using the associativity of the convolution, prove that

Imf =[

1(m− 1)!

xm−1θ(x)]∗ f,

where m = 1, 2, . . ., for f ∈ D′(R+).

5.9. Set xλ+ = θ(x)xλ for Re λ > −1, so that xλ+ ∈ L1loc(R+) and, in partic-

ular, xλ+ ∈ D′(R+). Define the fractional integral Iλ by the formula

Iλf =xλ−1

+

Γ(λ)∗ f, Re λ > 0.

Prove that IλIµ = Iλ+µ, Re λ > 0, Re µ > 0.

5.10. For any λ ∈ C define the fractional integral Iλ by

Iλf =(d

dx

)kIλ+kf for f ∈ D′(R+),

where k ∈ Z+ is such that Re λ+ k > 0. Prove that Iλ is well defined, i.e.,the result does not depend on the choice of k ∈ Z+. Prove that IλIµ = Iλ+µ

for any λ, µ ∈ C. Prove that I0f = f and I−kf = f (k) for k ∈ Z+ and anyf ∈ D′(R+). So I−λf for λ > 0 can be considered a fractional derivative (oforder λ) of the distribution f .

Chapter 6

Harmonic functions

6.1. Mean-value theorems for harmonic functions

Harmonic function in an open set Ω ⊂ Rn (n ≥ 1) is a (real-valued orcomplex-valued) function u ∈ C2(Ω) satisfying the Laplace equation

(6.1) ∆u = 0,

where ∆ is the standard Laplacian on Rn. As we already know from Ch. 5,any harmonic function is in fact in C∞(Ω), and even real-analytic in Ω.This is true even if we a priori assume only that u ∈ D′(Ω), i.e. if u is adistribution in Ω.

For n = 1 the harmonicity of a function on an interval (a, b) means thatthis function is linear. Though in this case all statements, discussed here,are also true, they are usually trivial, so we can concentrate on the casen ≥ 2.

In this Chapter we will provide a sample of classical properties of har-monic functions and related topics. All proofs are relatively elementary: wedo not use advanced functional analysis, such as Sobolev spaces (they willbe discussed later in this book). Among other things, we illustrate, how alocal equation (6.1) may imply a range of global properties of solutions.

We will start with a property of harmonic functions which is called meanvalue theorem, or Gauss’s law of the arithmetic mean.

Theorem 6.1. (Mean Value Theorem for spheres) Let B = Br(x) be anopen ball in Rn with the center at x and the radius r > 0, S = Sr(x) isits boundary (the sphere with the same center and radius), B is the closure

137

138 6. Harmonic functions

of B (so B = B ∪ S). Let u be a continuous function on B such that itsrestriction to B is harmonic. Then the average of u over S equals u(x) (thevalue of u at the center). In other words,

(6.2) u(x) =1

voln−1(S)

∫Su(y)dSy =

1σn−1rn−1

∫Sr(x)

u(y)dSy,

where dSy means the standard area element on the sphere S, σn−1 is the(n− 1)-volume of the unit sphere.

Proof. Let us rewrite the average of u in (6.2) as an average over the unitsphere. To this end, denote z = r−1(y−x) and make the change of variablesy 7→ z in the last integral. We get then

1σn−1rn−1

∫Sr(x)

u(y)dSy =1

σn−1

∫|z|=1

u(x+ rz)dSz.

Now, if we replace r by a variable ρ, 0 ≤ ρ ≤ r, then the integral becomes afunction I = I(ρ) on [0, r], which is continuous due to the uniform continuityof u on B. Clearly, I(0) = u(x). Our theorem will be proved if we showthat I(ρ) = const, or, equivalently, that I ′(ρ) ≡ 0 on [0, r). To this end,let us calculate the derivative I ′(ρ) by differentiating under the integral sign(which is clearly possible):

I ′(ρ) =1

σn−1

∫|z|=1

d

dρu(x+ ρz)dSz =

1σn−1

∫|z|=1

∂u

∂nρ(x+ ρz)dSz,

where nρ is the outgoing unit normal to the sphere Sρ(x), at the pointx+ ρz. We will see that the last integral vanishes, if we apply the followingcorollary of Green’s formula (4.32), which is valid for any bounded domainwith a C2 boundary

(6.3)∫

Ω∆u dx =

∫∂Ω

∂u

∂ndS

(take v ≡ 1 in (4.32)). For our purposes we should take Ω = Bρ(x). Since uis harmonic, we come to the desired conclusion that I ′(ρ) = 0.

The sphere can be replaced by ball in Theorem 6.1:

Theorem 6.2. (Mean Value Theorem for balls) In the same notation asin Theorem 6.1, let u be a continuous function on B = Br(x) such that itsrestriction to B is harmonic. Then the average of u over B equals u(x). Inother words,

(6.4) u(x) =1

vol(B)

∫Bu(y)dy =

n

σn−1rn

∫Br(x)

u(y)dy,

6.2. Maximum principle 139

where dy means the standard volume element (Lebesgue measure) on Rn.

Proof. Let ρ denote the distance from x, considered as a function on Br(x),as in the proof of Theorem 6.2. Since | grad ρ| = 1, we can replace integrationover Br(x) by two subsequent integrations: first integrate over the spheresSρ(x) for every ρ ∈ [0, r], then integrate with respect to the radii ρ (noadditional Jacobian type factor is needed). Using Theorem 6.1, we obtain:∫

Br(x)u(y)dy =

∫ r

0

∫Sρ(x)

u(y)dSρ(y)dρ =∫ r

0u(x)voln−1(Sρ(x))dρ

= u(x) vol(Br(x)),

which is what we need.

Remark 6.3. Proof of Theorem 6.1 shows a way how a local equationcan imply a global property (in this case, the mean value property (6.2)). Itsuggests that we can try to connect the desired global property with a trivialone by a path (in our case, the path is parametrized by the radius ρ ∈ [0, r])and expand the validity of the desired property along this path, startingfrom a trivial case (ρ = 0 in our case), relying on the local information whenmoving by an infinitesimal step with respect to the parameter.

The proof of Theorem 6.2 does not belong to this type, because it simplydeduces a global result from another global one.

6.2. The maximum principle

In this section we will often assume for simplicity of formulations, that Ωis an open connected subset in Rn. (In case if Ω is not connected, we canthen apply the results on each connected component separately.) Recall thatconnectedness of an open set Ω ⊂ Rn can be defined by one of the followingequivalent conditions:

(i) Ω can not be presented as a union of two disjoint non-empty opensubsets.

(ii) If F ⊂ Ω is non-empty open and at the same time closed (in Ω),then F = Ω.

(iii) Every two points x, y ∈ Ω can be connected by a continuous pathin Ω (that is, there exists a continuous map γ : [0, 1] → Ω, suchthat γ(0) = x and γ(1) = y).

(For the proofs see e.g. Armstrong [1], Sect. 3.5.)


The maximum principle for harmonic functions can be formulated asfollows:

Theorem 6.4. (The Maximum Principle for Harmonic Functions) Let u bea real-valued harmonic function in a connected open set Ω ⊂ Rn. Then u

can not have a global maximum in Ω unless it is constant. In other words,if x0 ∈ Ω and u(x0) ≥ u(x) for all x ∈ Ω, then u(x) = u(x0) for all x ∈ Ω.

Proof. Let x0 ∈ Ω be a point of global maximum for u, B = Br(x0) is aball in Rn (with the center at x0), such that B ⊂ Ω (this is true if the radiusr is sufficiently small). Now, applying (6.4) with x = x0, we obtain

0 = u(x0)− 1vol(B)

∫Bu(y)dy =

1vol(B)

∫B

(u(x0)− u(y))dy.

Since u(x0) − u(y) ≥ 0 for all y ∈ Ω, we conclude that u(y) = u(x0) for ally ∈ B.

Let us consider the set

M = y ∈ Ω|u(y) = u(x0).As we just observed, for every point y ∈ M there exist a ball B centeredat y, such that B ⊂ M . This means that M is open. But it is obviouslyclosed too. It is non-empty because it contains x0. Since Ω is connected,we conclude that M = Ω which means that u(y) = u(x0) = const for ally ∈ Ω.

Corollary 6.5. Let u be a real-valued harmonic function in a connectedopen set Ω ⊂ Rn. Then u can not have a global minimum in Ω unless it isconstant.

Proof. Applying Theorem 6.4 to −u, we obtain the desired result.

Corollary 6.6. Let u be a complex-valued harmonic function in a connectedopen set Ω ⊂ Rn. Then |u| can not have a global maximum in Ω unlessu = const.

Proof. Assume that x0 ∈ Ω and |u(x0)| ≥ |u(x)| for all x ∈ Ω. If |u(x0)| =0, then u ≡ 0 and the result is obvious. Otherwise, present u(x0) =|u(x0)|eiϕ0 , 0 ≤ ϕ0 < 2π, and consider a new function u = e−iϕ0u, which isalso harmonic. Denoting the real and imaginary parts of u by v, w, so thatu = v + iw, we see that v and w are real-valued harmonic functions on Ω,and for all x ∈ Ω

(6.5) v(x0) = |u(x0)| ≥ |u(x)| = |u(x)| =√v2(x) + w2(x) ≥ v(x).

6.2. Maximum principle 141

Applying Theorem 6.4 to v, we see that v(x) = v(x0) = const for all x ∈ Ω,and all the inequalities in (6.5) become equalities. In particular, we havev(x0) =

√v2(x0) + w2(x), hence w ≡ 0, which implies the desired result.

For bounded domains Ω we can take into account what happens nearthe boundary to arrive to other important facts.

Theorem 6.7. 1) Let u be a real-valued harmonic function in an openbounded Ω ⊂ Rn, and let u can be extended to a continuous function (stilldenoted by u) on the closure Ω. Then for any x ∈ Ω

(6.6) min∂Ω

u ≤ u(x) ≤ max∂Ω

u,

where ∂Ω = Ω \ Ω is the boundary of Ω.

2) Let u be a complex-valued harmonic function on Ω, which can be extendedto a continuous function on Ω. Then for any x ∈ Ω

(6.7) |u(x)| ≤ max∂Ω|u|.

3) The inequalities in (6.6) and (6.7) become strict for x in those connectedcomponents of Ω where u is not constant.

Proof. Since Ω is compact, u should attain its maximum and minimumsomewhere on Ω. But due to Theorem 6.4 and Corollary 6.5 this mayhappen only on the boundary or in the connected components of Ω where uis constant. The desired results for a real-valued u immediately follow. Forcomplex-valued u the same argument works if we use Corollary 6.6 insteadof Theorem 6.4 and Corollary 6.5.

Corollary 6.8. Let Ω be an open bounded set in Rn, u, v be harmonicfunctions, which can be extended to continuous functions on Ω, and suchthat u = v on ∂Ω. Then u ≡ v everywhere in Ω.

Proof. Taking w = u − v, we see that w is harmonic and w = 0 on ∂Ω.Now it follows from (6.7) that w ≡ 0 in Ω.

This corollary means that any harmonic function u in Ω, which can beextended to a continuous function on Ω, is uniquely defined by its restrictionto the boundary ∂Ω. This fact can be reformulated as uniqueness of solutionfor the following boundary-value problem for the Laplace equation, which is


called Dirichlet’s problem:

(6.8)

∆u(x) = 0, x ∈ Ω;u|∂Ω = ϕ(x), x ∈ ∂Ω.

Here, for the start, let us assume that ϕ ∈ C(∂Ω), and we are interestedin a classical solution u, that is u ∈ C(Ω), u is harmonic in Ω. We justproved that for any bounded Ω and any continuous ϕ, there may be at mostone solution of this problem. It is natural to ask whether such solutionexists for a fixed bounded Ω but with any given ϕ. This proves to be trueif the boundary ∂Ω is smooth or piecewise smooth. But in general case thecomplete answer, characterizing open sets Ω for which the Dirichlet problemis solvable for any continuous ϕ, is highly non-trivial. This answer was givenby N. Wiener, and it is formulated in terms of capacity, the notion whichappeared in electrostatics and was brought by Wiener to mathematics inthe 1920’s. The whole story is beyond the scope of this course and can befound in the books on Potential theory (see e.g. Wermer [33]).

However, it occurs that in some weaker sense the problem is alwayssolvable. The appropriate language and necessary technique (functionalanalysis, Sobolev spaces) will be discussed in Chapter 8. N

6.3. Dirichlet’s boundary-value problem

To require the existence and uniqueness of solution of a mathematical prob-lem which appears as a model of a real-life situation, is quite natural. (Forexample, if it is not unique, then we should additionally investigate whichone corresponds to reality.) But this is not enough. Another reasonablerequirement is well-posedness of the problem which means that the solutioncontinuously depends on the data of the problem, where the continuity isunderstood with respect to some natural topologies in the spaces of dataand solutions.

The well-posedness of the Dirichlet problem in the sup-norm immedi-ately follows from the maximum principle. More precisely, we have

Corollary 6.9. Let Ω be an open bounded domain in Rn. Let u1, u2 beharmonic functions in Ω, which can be extended to continuous functions onΩ, ϕj = uj |∂Ω, j = 1, 2. Then

(6.9) maxΩ|u1 − u2| ≤ max

∂Ω|ϕ1 − ϕ2|.

Proof. The result immediately follows from (6.7) in Theorem 6.7.

6.4. Hadamard’s example 143

Now assume that uk| k = 1, 2, . . . is a sequence of harmonic functionsin Ω, continuously extendable to Ω, whose restrictions on ∂Ω (call them ϕk)converge uniformly to ϕ on ∂Ω as k →∞. Then ϕk is a Cauchy sequencein the Banach space C(∂Ω) with the sup-norm. It follows from (6.9) thatuk is a Cauchy sequence in C(Ω). Hence, uk → u ∈ C(Ω) with respectto this norm. It is easy to see that the limit function u is also harmonic.Indeed, the uniform convergence in Ω obviously implies convergence uk → u

in distributions (that is, in D′(Ω)), hence ∆uk → ∆u and ∆u = 0 in Ω, sou is harmonic. N

H The Laplace equation and the Dirichlet problem appear in electro-statics as the equation and boundary-value problem for a potential in adomain Ω which is free of electric charges. It is also natural to consider amodification which allows charges, including point charges and also the onesdistributed in some volume, or on a surface. To this end we should replacethe Laplace equation by the Poisson equation ∆u = f (see (4.41)), where fis an appropriate distribution. Allowing charges inside Ω, we can considerDirichlet’s problem for the Poisson equation:

(6.10)

∆u(x) = f(x), x ∈ Ω;u|∂Ω = ϕ(x), x ∈ ∂Ω.

To make it precise, we can assume that u is a distribution in Ω, such that itis in fact a continuous function in a neighborhood U of the boundary ∂Ω inΩ. We can also take f ∈ D′(Ω), and ϕ ∈ C(∂Ω). Then we can extend theresult of Corollary 6.8 to the problem (6.10):

Corollary 6.10. Assume that u, v ∈ D′(Ω) are solutions of the boundary-value problem (6.10) with the same f and ϕ. Then u = v in Ω.

Proof. Clearly, w = u− v is a distribution in Ω, satisfying ∆w = 0. There-fore, by Theorem 5.14, w is a harmonic function in Ω. By our assumptions,w is continuous up to the boundary and w = 0 on ∂Ω. Hence, w ≡ 0 by themaximum principle.

6.4. Hadamard’s example

Now I will explain an example, due to Jacques Hadamard, which showsthat there exist uniquely solvable problems which are not well-posed withrespect to natural norms, i.e. small perturbations of the data may lead touncontrollably large perturbations of the solution.


Let us consider the initial-value problem for the Laplacian in a two-dimensional strip ΠT = (x, y) ∈ R2|x ∈ R, 0 ≤ y < T:

(6.11)

∆u = 0 in ΠT ,

u(x, 0) = ϕ(x), ∂u∂y (x, 0) = ψ(x) for all x ∈ R,

where we assume that u ∈ C2(ΠT ), ϕ ∈ C2(R) and ψ ∈ C1(R).

PICTURE BELOW HADAMARD (6.12)

A solution of this problem may not exist (see Problem 6.5), but if itexists, then it is unique. To prove the uniqueness we will need the following

Lemma 6.11. Let Ω = Rn with the coordinates (x′, xn), where x′ ∈ Rn−1,xn ∈ R. Assume that a function u ∈ C1(Ω) is harmonic in the open sets

Ω± = x ∈ Ω| ± xn > 0,which are obtained as two subsets of Ω lying above and below the hyperplanexn = 0 respectively. Then u can be extended from u ∈ L1

loc(Ω+ ∪ Ω−) to a

harmonic function in Ω.

Proof. It suffices to prove that ∆u = 0 in the sense of distributions, i.e.that for any test function Φ ∈ C∞0 (Ω) we have

(6.12)∫

Ωu∆Φ dx = 0.

Note that it is sufficient to prove this equality locally, in a small neighbor-hood of any chosen point x0 ∈ Ω, that is for the test functions Φ supportedin such a neighborhood. Since this is obvious for the points x0 which donot belong to the separating hyperplane, we can assume that x0 = (x′0, 0),and that Ω is a small ball centered at such a point. Then, due to Green’sformula (4.32) we can write:∫

Ω−

u∆Φ dx =∫

Ω∩xn=0

(u∂Φ∂xn

− Φ∂u

∂xn

)dx′,

where we used that ∆u = 0 in Ω−. Similarly,∫Ω+

u∆Φ dx = −∫

Ω∩xn=0

(u∂Φ∂xn

− Φ∂u

∂xn

)dx′,

where the minus sign in the right hand side appears due to the fact thatthe direction of the xn-axis is for Ω− and incoming for Ω+. We also usedthe continuity of u and ∂u/∂xn which implies that these quantities are thesame from both sides of xn = 0. Adding these equalities leads to (6.12),which ends the proof.

6.4. Hadamard’s example 145

Now let us turn to the proof of the uniqueness in (6.11). For two solutionsu1, u2, their difference u = u1 − u2 is a solution of (6.11) with ϕ ≡ ψ ≡ 0.Extend u by reflection with respect to the line y = 0, as an odd functionwith respect to y, to a function u on a doubled strip

ΠT = (x, y) ∈ R2|x ∈ R, y ∈ (−T, T ).

Namely, define u(x, y) = −u(x,−y) for −T < y < 0. Due to the initialconditions on u we clearly have u ∈ C1(ΠT ). Also, u is harmonic in thedivided strip ΠT \ (x, 0)|x ∈ R, i.e. everywhere in ΠT except at the x-axis. It follows from Lemma 6.11 that u is harmonic in ΠT . Therefore,it is real-analytic in ΠT by Theorem 5.17 (or Corollary 5.18). But it isalso easy to check that the equation ∆u = 0 and the initial conditionsu|y=0 = ∂u/∂y|y=0 = 0 imply by induction in |α|, that ∂αu|y=0 = 0 for every2-multiindex α. Hence, by the Taylor expansion of u at the point x = y = 0,we conclude that u ≡ 0 in ΠT , which proves the desired uniqueness.

Remark 6.12. Uniqueness of Cm-solution of the initial-value problem forany order m equation with real-analytic coefficients and the initial condi-tions on any non-characteristic hypersurface, is the content of a Holmgrentheorem (see e.g. John [15], Sect.3.5). This theorem is deduced, by a du-ality argument, from the existence of solution for an adjoint problem inreal-analytics functions, which is the content of the Cauchy-Kovalevsky the-orem (see Theorem 5.17). In the next chapter we will explain Holmgren’smethod in detail but for the heat equation.

Now we can proceed to the Hadamard example . Let us try to find theexponential solutions of the Laplace equation ∆u(x, y) = 0, that is, solutionsof the form

u(x, y) = ekx+ly,

where k, l ∈ C. Substituting this function into the Laplace equation, weimmediately see that it is a solution if and only if k2 + l2 = 0. For examplewe can take k = iω, l = ω for a real ω, to get

u(x, y) = eiωxeωy.

We can also take the real part of this solution which is

u(x, y) = uω(x, y) = eωy cosωx.


Let us try to understand whether we can use the Ck-norms on functionsϕ ∈ Ck(R):

(6.13) ‖ϕ‖(k) =∑|α|≤k

supx∈R|∂αϕ(x)|,

for the initial data of the problem (6.11), and investigate whether a value ofthe solution u at a point can be done small if the Ck-norms of the initial dataare small. Equivalently, we can ask whether a small deviation of ϕ,ψ fromthe equilibrium values ϕ ≡ ψ ≡ 0 necessarily leads to small perturbation ofthe solution u. More precisely, fixing an arbitrarily large k and choosing,for example, a point (0, σ) ∈ ΠT , where σ > 0, we would like to know theanswer to the following question: is it true, that for any ε > 0 there existsδ > 0 such that the inequality ‖ϕ‖(k) +‖ψ‖(k) < δ implies that |u(0, σ)| < ε?

The answer to this question is emphatically negative. Indeed, for u = uωwith ω > 1, we obtain ϕ(x) = cosωx, ψ(x) = −ω sinωx, and

‖ϕ‖(k) + ‖ψ‖(k) ≤ 2(k + 1)ωk+1.

At the same time u(0, σ) = eωσ, which is much bigger if ω is large. In par-ticular, taking instead the solution uω = e−ωσ/2uω with the correspondinginitial data

ϕω(x) = e−ωσ/2 cosωx, ψω(x) = −e−ωσ/2ω sinωx,

we see that u(0, σ) = eωσ/2 exponentially grows as ω → +∞, whereas

‖ϕω‖(k) + ‖ψω‖(k) ≤ 2(k + 1)ωk+1e−ωσ/2,

which exponentially decays as ω → +∞. In particular, now the negativeanswer to the question becomes obvious: the value of u at a point can notbe controlled by the Ck-norms of the initial data.

As we have seen in Chapter 2, the situation is quite opposite for thewave equation (also in dimension 2): there the values of u are estimated bythe sup-norms of the initial data. In fact, this holds for hyperbolic equationsin higher dimensions too. We will see this for the wave equation in Chapter10.

6.5. Green’s function: first steps

In Sect. 3.5 we introduced Green’s function of the Sturm-Liouville operatoron [0, l] with vanishing boundary conditions at the ends 0 and l. Here wewill define a similar object for the Dirichlet problem (6.8) in a bounded openset Ω ⊂ Rn. Let us start with the fundamental solution En = En(x) of the

6.5. Green’s function: first steps 147

Laplacian ∆ in Rn. Then En = En(x) is a distribution in Rn, which has apoint singularity at 0 ∈ Rn and satisfies the equation ∆En(x) = δ(x). Sucha solution is defined up to adding a constant. To choose a definite copy incase n ≥ 3, we can first require that En(x)→ 0 as |x| → ∞. Then it is easyto show that En(x) is given by (4.38).

In case n = 2 the fundamental solution can not decay at infinity, butwe can select a radially symmetric solution by (4.39), which looks like anarbitrary choice of the constant, but note that ∇E2(x) does not depend uponthis choice.

Note that for any n ≥ 2 the fundamental solution En(x) is harmonic forall x 6= 0.

In this section we will define, study and apply Green’s function G =G(x, y), where x, y ∈ Ω, x 6= y. More precisely, we will assume G to bea distribution on Ω × Ω, which is defined by a locally integrable functionG = G(x, y). For every fixed y ∈ Ω, it should be harmonic with respect tox ∈ Ω if x 6= y, and we also impose the boundary condition G(x, y) = 0 if xor y is in the boundary of Ω.

As usual, two such functions are identified if they coincide almost every-where in Ω×Ω, or, equivalently, if the corresponding distributions coincidein Ω× Ω. (For simplicity we will simply say that G is locally integrable.)

Summarizing, we get

Definition 6.13. The Green function G = G(x, y) of the Laplacian ∆ in Ωwith the Dirichlet boundary condition, is a locally integrable solution of theequation

(6.14) ∆xG(x, y) = δ(x− y),

for any fixed y ∈ Ω and any x ∈ Ω, x 6= y, satisfying the boundary condition

(6.15) G(x, y) = 0, y ∈ Ω, x ∈ Ω, x 6= y, and y ∈ ∂Ω or x ∈ ∂Ω.

Define also Ωy = (z, y) ∈ Ω × Ω| z ∈ Ω for every fixed y ∈ Ω, ∆x is theLaplacian with respect to x in Ωy.

Comments to Definition 6.13.

1) The equation (6.14) with the Dirichlet boundary condition corre-sponds to a physical requirement that the resulting solution represents theelectrostatic Coulomb interaction between electric charges, so that the charge


located at y ∈ Ω, has the potential G(·, y) with the fixed y. Due to the sym-metry principle, which we will prove later, G(x, y) = G(y, x) for all x, y ∈ Ω,hence the same will hold for any charge located at x and tested at y.

2) The Dirichlet boundary condition means that G(x, y) = 0 when x ory is at the boundary of Ω.

3) In more detail, we will assume that Ω and ∂Ω are in C2(Ω) and∂C2(Ω) respectively. The Green function G will have a singularity on thediagonal

diag(Ω× Ω) = (x, x)|x ∈ Ω ⊂ Ω× Ω,

and it is in C2 outside the diagonal. (Sometimes we will need C3 instead ofC2.) The singularities of G(x, y) near the diagonal x = y are similar to thesimplest singularity in case Ω = Rn.

The simplest singularity with Ω = Rn has an unbounded Ω, but it isuseful to keep its structure in mind. In this case, normalizing it to havecharge 1 and singularity at x = y, we have, by definition,

(6.16) G(x, y) = G0(x, y) = En(x− y), x, y ∈ Rn,

where En = En(x) is the standard fundamental solution of the Laplacian ∆in Rn, x, y ∈ Rn, n ≥ 2 (see (4.38), (4.39)).

In an analogy with the case n = 1, we will define Green’s function of theLaplacian in Ω ⊂ Rn for n ≥ 2 and

G = G(x, y), x, y ∈ Ω, by the formula

(6.17) G(x, y) := En(x− y) + g(x, y),

where g = g(x, y) is called the remainder, which is expected to satisfy thefollowing properties:

(6.18) Gu(x) =∫

ΩG(x, y)u(y)dy, u ∈ C∞0 (Ω), x ∈ Ω.

boundary condition in a bounded open set Ω ⊂ Rn and with a sufficientlysmooth boundary condition (C2). In this case, the solution of the Dirichletproblem is a solution G(x, y) (with x, y ∈ Ω, x 6= y) of the following systemof equations:

We will always assume that for the chosen open bounded set Ω, theGreen function G = G(x, y), x, y ∈ Ω, x 6= y, exists.

Now take such a Green function G = G(x, y) for the Laplacian ∆ ona bounded open set ω, with x, y ∈ Ω, G ∈ D′(Ω × Ω), so that (6.10) with

6.5. Green’s function: first steps 149

f(x) = δy(x) = δ(x−y) and ϕ ≡ 0 . Here y ∈ Ω is considered as a parameter.So we will have then

(6.19)

∆x,yG(x, y) = δ(x− y), x, y ∈ Ω;∆x,yG(x, y)|x∈∂Ω = 0.

As follows from Corollary 6.10, if for some y ∈ Ω such a distribution G(·, y)exists, then it is unique. Concerning the existence, we have

Lemma 6.14. Assume that the Dirichlet problem (6.8) has a solution forany ϕ ∈ C(∂Ω). Then Green’s function G = G(·, y) exists for all y ∈ Ω.

Proof. Let us start with a particular solution u(x) of the equation ∆u(x) =δ(x−y), which is u(x) = En(x−y), where En(x) is the fundamental solutionof the Laplacian ∆, given by (4.38) and (4.39). Then the remainder

(6.20) g(x, y) := G(x, y)− En(x− y)

should be harmonic with respect to x and satisfy the boundary conditiong(x, y) = −En(x− y) for x ∈ ∂Ω. This means that g(·, y) is a solution of theDirichlet problem (6.8) with the boundary condition ϕ(x) = −En(x−y), forany fixed y ∈ Ω. Due to our assumption about the solvability of Dirichlet’sproblem in Ω, such a function g exists, which ends the proof.

Remark 6.15. In fact, for the existence of Green’s function G(x, y) we needthe existence of a solution of Dirichlet’s problem (6.8) not for all ϕ ∈ C(∂Ω),but only for specific functions of the form ϕy = −En(·−y)|∂Ω, y ∈ Ω. In thefuture, when we mention Green’s function, then we assume that it exists inthe sense of this remark.

Let us consider the set L of all ϕ ∈ C(∂Ω), such that Dirichlet’s problem(6.8) has a solution u = uϕ ∈ C(Ω). Clearly, L is a linear subspace inC(∂Ω). Moreover, L is closed in the Banach space C(∂Ω). The linearoperator S : L → C(Ω), ϕ 7→ uϕ, is bounded. (It has norm 1 due tothe maximum principle.) The existence of Green’s function is equivalent toinclusion ϕy| y ∈ Ω ⊂ L, which is equivalent to saying that the closedlinear span of the set of all ϕy’s is a subset of L.

We will later refer to S as the solving operator for Dirichlet’s problem(6.8).

Let us investigate the behavior of G(x, y) as a function of (x, y) on Ω×Ω.Since we know En explicitly, it is sufficient to investigate the behavior of theremainder g(x, y) in (6.20).


Lemma 6.16. Let G and g be the corresponding Green function and theremainder (as in (6.20)). Then

(a) G(x, y) can be extended to a continuous function of (x, y) on the setΩ× Ω \ diag(Ω× Ω) ⊂ Rn × Rn, where diag(Ω× Ω) = (x, x)|x ∈ Ω.

(b) The function x 7→ G(x, y) is harmonic in Ω \ y for every fixedy ∈ Ω.

(c) g(x, y) can be extended to a continuous function of (x, y) on the setΩ× Ω \ diag(Ω× Ω) ⊂ Rn × Rn.

(d) The function x 7→ g(x, y) is harmonic in Ω \ y for every fixedy ∈ Ω.

Proof. The result follows if we note that the function x, y 7→ En(x− y) onΩ×Ω \ diag(Ω×Ω), together with the functions G, g, posses the continuityand harmonicity properties indicated in the formulation of the Lemma forG(x, y) and g(x, y). It remains to note that all the properties are linear.

Corollary 6.17. Under the same conditions

G = G(x, y) ∈ C(Ω× Ω \ diag(Ω× Ω)).

6.6. Symmetry and regularity of Green’s functions

Up to this moment we did not impose any smoothness restrictions uponΩ, except that it is an open bounded subset in Rn. But such restrictionsare needed if we want to know more about the behavior of the solutionsof Dirichlet’s problem near the boundary, aside of just being continuous.We will introduce such restrictions in the future. Now we will start from aconditional result.

Proposition 6.18. Let Ω be a bounded domain, ∂Ω ∈ C2. Assume thatGreen’s function G = G(x, y) for Dirichlet’s problem (for the Laplacian ∆in Ω) exists and, besides,

(6.21) G(·, y) ∈ C2(Ω \ y) for every y ∈ Ω,

or, equivalently,

(6.22) g(·, y) ∈ C2(Ω \ y) for every y ∈ Ω,

where g is the remainder, g(x, y) = G(x, y) − En(x − y). Then both G andg are symmetric, that is, G(x, y) = G(y, x) for all x, y ∈ Ω, x 6= y, andg(x, y) = g(y, x) for all x, y ∈ Ω, x 6= y.

6.6. Symmetry and regularity of Green’s function 151

Proof. Let us recall Green’s formula

(6.23)∫

Ω(u∆v − v∆u)dx =

∫∂Ω

(u∂v

∂n− v ∂u

∂n

)dS,

where u, v ∈ C2(Ω) (see (4.32) and the discussion which follows it). Letus try to apply (6.23) to the functions u = G(·, y) and v = G(·, z), wherey, z ∈ Ω, y 6= z. Due to the boundary conditions on G, we have

u|∂Ω = v|∂Ω = 0, hence the right hand side vanishes (for any fixed y, z ∈ Ω).We also have ∆u = δ(· − y), ∆v = δ(· − z), so the left hand side equalsu(z)− v(y) = G(z, y)−G(y, z), if the integrals there are understood in thesense of distributions. Therefore, if Green’s formula can be extended to suchu, v, then we will get the desired symmetry of Green’s function.

Note, that the only reason that (6.23) can not be directly applied to ouru and v is that they have singularities at the points y and z respectively. Soa natural way is to approximate u, v by functions uε, vε ∈ C2(Ω), where ε ∈(0, ε0), so that all the terms in (6.23) are limits, as ε ↓ 0, of the correspondingterms in this formula with u, v replaced by uε, vε . To this end, let us takean arbitrary function χ ∈ C∞(Rn), such that χ = 0 in B1/2(0), χ = 1 inRn \ B1(0), 0 ≤ χ(x) ≤ 1 for all x ∈ Rn. Then take χε(x) = χ(ε−1x).Finally, define uε(x) = χε(x− y)u(x) and vε(x) = χε(x− z)u(x).

We will choose ε0 > 0 so that the closed balls Bε0(y) and Bε0(z) aredisjoint subsets in Ω (that is ε0 < |y − z|/2, and ε0 is strictly less than thedistances from y and z to ∂Ω).

Clearly, |uε(x)| ≤ |u(x)| and |vε(x)| ≤ |v(x)| for all x ∈ Ω. Also, uε = u

in Ω \Bε(y), and vε = v in Ω \Bε(z). In particular, these equalities hold ina neighborhood of ∂Ω.

Note that u and v are locally integrable near their singularities, henceintegrable in Ω. By the Lebesgue dominated convergence theorem uε → u,vε → v, as ε ↓ 0, in L1(Ω), hence, in D′(Ω). It follows that ∆uε → ∆u and∆vε → ∆v in D′(Ω).

Since uε = u ∈ C∞ in a neighborhood of supp ∆v (which is a subset ofBε0(z)), we clearly have∫

Ωuε∆vεdx = 〈∆vε, uε〉 = 〈∆vε, u〉 → 〈∆v, u〉 = 〈δ(·−z), u〉 = u(z) = G(z, y).

Here the angle brackets 〈·, ·〉 stand for the duality between E ′(Ω1) andC∞(Ω1), where Ω1 is an appropriate open subset in Ω.


Similarly, we obtain that∫Ωvε∆uεdx = 〈∆uε, vε〉 = 〈∆uε, v〉 → 〈∆u, v〉 = 〈δ(·−y), v〉 = v(y) = G(y, z).

Now Green’s formula (6.23) with u = G(·, y), v = G(·, z) (which isequivalent to the symmetry of Green’s function) can be obtained if we replaceu, v by uε, vε respectively, and then take limit as ε ↓ 0.

Remark 6.19. The symmetry of Green’s function can be interpreted as theself-adjointness of an operator in L2(Ω) which is inverse to the Laplacianwith Dirichlet’s boundary condition. The boundary condition defines anappropriate domain of the Laplacian, and Green’s function becomes theSchwartz kernel of this operator. We will discuss this approach in laterchapters. It requires a powerful technique of the Sobolev spaces, which willbe discussed in Chapter 8. It is noticeable that this technique does notrequire any restrictions on Ω except its boundedness.

H Now we will discuss how to relieve Proposition 6.18 of the redundantregularity requirements. This becomes possible if we introduce more delicatemeasure of smoothness than simply Ck: Holder spaces, which allow a precisedescriptions of the smoothness of solutions of the Dirichlet problem for thePoisson equation (6.10).

First, for functions u ∈ Ck(Ω), k = 0, 1, 2, . . . , introduce the norm

(6.24) ‖u‖(k) =∑|α|≤k

supx∈Ω|∂αu(x)|,

defined whenever u ∈ Ck(Ω) and the right hand side makes sense (i.e. if itis finite). (We already used this norm on R – see (6.13).) With this norm,Ck(Ω) becomes a Banach space.

Now for any γ, 0 < γ < 1, and any function u ∈ C(Ω) introduce aseminorm

|u|(γ) = supx,y∈Ω, x 6=y

|u(x)− u(y)||x− y|γ .

If this seminorm is finite, then we will say that this function is a Holderfunction with the exponent γ. (If the same is true with γ = 1, then onesays that u is a Lipschitz function, but we will rarely use Lipschitz functionsin this book.) We will denote the space of all Holder functions with theexponent γ by Cγ(Ω).

6.6. Symmetry and regularity of Green’s function 153

Note that replacing Ω by Ω may change the seminorm |u|(γ), thoughthis does not happen in case of a good boundary (say, of class C1 and non-selfintersecting).

Similarly, for any integer k ≥ 0 and γ ∈ (0, 1) define the space Ck,γ(Ω),consisting of functions u on Ω, such that u ∈ Ck(Ω) and ∂αu ∈ Cγ(Ω) forall multiindices α with |α| = k. It is a Banach space with the norm

‖u‖k,γ = ‖u‖(k) +∑

α: |α|=k

|∂αu|(γ).

We can also use two versions of the Holder spaces: Ck,γ(Ω) and Ck,γ(Ω),where Ω means the closure of an open set Ω ⊂ Rn, k is a non-negativeinteger, γ ∈ (0, 1). But for practical purposes only the first version is needed.(For example, these two versions coincide for bounded domains Ω with asufficiently smooth boundary.)

The spaces Ck,γ can be used to measure the smoothness of the boundaryas well. Namely, we will say that Ω has a Ck,γ boundary and write ∂Ω ∈ Ck,γif ∂Ω ∈ Ck and the functions h(x′), which enter into local representationsof the boundary as the graph xn = h(x′) (see Definition 4.8), are in fact inCk,γ on every compact subset of the domain where they are defined.

If k ≥ 1, then it is easy to see that the space Ck,γ(Ω) and Ck,γ smooth-ness of ∂Ω are invariant under Ck,γ diffeomorphisms. (This follows from thestatements formulated in Problem 6.6.)

In fact, on the Ck,γ boundary ∂Ω itself, for k ≥ 1, there exists a structureof a Ck,γ manifold, which is given by the sets of local coordinates related toeach other in the usual way, by Ck,γ diffeomorphisms between two domainsin Rn−1, which are images of the intersection of the two coordinate neigh-borhoods in Ω. Such local coordinates can be taken to be x′ for the point(x′, h(x′)) in the local graph presentation of ∂Ω.

Let us return to (6.10) which is the Dirichlet problem for the Poissonequation.

Here is a remarkable theorem by J. Schauder, which gives a precisesolvability and regularity, and also shows that the Holder spaces Ck,γ workgood for the problem (6.10).

Theorem 6.20. (a) (Existence and uniqueness) Assume that k is an integer,k ≥ 0, γ ∈ (0, 1), ∂Ω ∈ Ck+2,γ, f ∈ Ck,γ(Ω) and ϕ ∈ Ck+2,γ(∂Ω). Then asolution u ∈ C2(Ω) ∩ C(Ω) of the boundary-value problem (6.10) exists andis unique.


(b) (Regularity) In fact, the unique solution u belongs to Ck+2,γ(Ω).

(c ) (Isomorphism Operator) The operator

(6.25) A : Ck+2,γ(Ω)→ Ck,γ(Ω)× Ck+2,γ(∂Ω), Au = ∆u, u|∂Ω,

is a linear topological isomorphism (that is, a bounded linear operator, whichhas a bounded inverse).

Corollary 6.21. Assume that ∂Ω ∈ C∞, f ∈ C∞(Ω) and ϕ ∈ C∞(∂Ω).Then the problem (6.10) has a solution u ∈ C∞(Ω). A solution u of thisproblem, such that u ∈ C2(Ω) ∩ C(Ω), is unique and therefore in fact wehave u ∈ C∞(Ω). Moreover, the operator

(6.26) A : C∞(Ω)→ C∞(Ω)× C∞(∂Ω), Au = ∆u, u|∂Ω,

is a linear topological isomorphism of Frechet spaces.

Proof. The statement immediately follows from Theorem 6.20 if we notethat for any fixed γ ∈ (0, 1), the space C∞(Ω) is the intersection of all spacesCk,γ(Ω), k = 0, 1, 2, . . . , and the Frechet topology in C∞(Ω) can be definedby norms ‖ · ‖k,γ .

Remarks. 1. I will not provide a proof of Theorem 6.20. The interestedreader can find complete proofs in the monographs by Ladyzhenskaya andUral’tseva [17], Ch. III; Gilbarg and Trudinger [8], Ch. 6. In both books asubstantial extension of this result to general second-order elliptic operatorsis proved. (It is also due to Schauder.)

2. All three statements of Theorem 6.20 fail in the usual classes Ck(Ω)even locally. Namely, the equation ∆u = f with f ∈ C(Ω) may fail to havea solution u ∈ C2(Ω1) where Ω1 is a fixed subdomain of Ω. Also f ∈ C(Ω)does not even imply u ∈ C2(Ω). However, it is possible to deduce a weakerexistence and (separately) regularity statements in terms of the usual Ck

spaces.

Here is the regularity result:

Corollary 6.22. Assume that k ≥ 1 is an integer, Ω is a bounded domainin Rn, such that ∂Ω ∈ Ck+2. Let u ∈ C2(Ω) ∩ C(Ω) be a solution of (6.10)with f = ∆u ∈ Ck(Ω) and ϕ = u|∂Ω ∈ Ck+2(∂Ω). Then u ∈ Ck+1(Ω).

In particular, if ∂Ω ∈ Ck+2, u ∈ C(Ω) is harmonic in Ω, and ϕ = u|∂Ω ∈Ck+2(∂Ω), then u ∈ Ck+1(Ω).

6.7. Green’s Function and Dirichlet’s Problem 155

Proof. Choose γ ∈ (0, 1), and note that the given conditions imply the in-clusions ∂Ω ∈ Ck+1,γ , f ∈ Ck−1,γ(Ω), and ϕ ∈ Ck+2,γ(∂Ω). Then Theorem6.20 implies that u ∈ Ck+1,γ(Ω), hence u ∈ Ck+1(Ω).

About the existence result see Problem 6.8. N

H Now let us return to the existence, smoothness and symmetry ofGreen’s function.

Theorem 6.23. Assume that Ω is a bounded domain in Rn, such that

∂Ω ∈ C2,γ, where γ ∈ (0, 1). Then the following statements hold:

(a) There exists Green’s function G = G(x, y) of the operator ∆ withDirichlet’s boundary conditions in Ω (see (6.19));

(b) For the remainder g(x, y) = G(x, y)−En(x−y) we have g ∈ C2,γ(Ω×Ω \ diag(Ω× Ω)), and g(·, y) ∈ C2,γ(Ω \ y) for any y ∈ Ω;

(c) Green’s function is symmetric, i.e. G(x, y) = G(y, x) for all x, y ∈ Ω,x 6= y.

Proof. Due to Theorem 6.20, for any fixed y ∈ Ω, we can find g = g(x, y)such that g(·, y) ∈ C2,γ(Ω \ y), by solving Dirichlet’s problem (6.8) withϕ(x) = −En(x − y). Therefore, Green’s function exists, and (a), (b) aresatisfied. Now (c) follows by a straightforward application of Proposition6.18.

Corollary 6.24. Assume ∂Ω ∈ C2,γ . Then g(x, y) can be extended to acontinuous function in Ω× Ω \ diag(Ω× Ω).

Proof. The statement immediately follows from Lemma 6.16, and the sym-metry of Green’s function and the remainder g(x, y).

6.7. Green’s function and Dirichlet’s problem

We will start by deducing an integral formula for the solution u of the non-homogeneous Dirichlet problem (6.10) through the right hand side f andthe boundary data ϕ, assuming that we know Green’s function G = G(x, y)of Ω. (In particular, we assume that Green’s function exists.)

Let us also assume that Ω is a bounded domain in Rn with a piecewisesmooth boundary. (More precisely, it suffices to assume that ∂Ω is piecewiseC2.) This will guarantee that the Green-Stokes type formulas work for Ω.Specifically, we need the formula (6.23) to hold for all functions u, v ∈ C2(Ω).We will also require that G ∈ C2 on (x, y) ∈ (Ω× Ω) ∪ (Ω× Ω)|x 6= y.


All conditions above (both on Ω and on Green’s function) will be satisfiedif we impose a slightly stronger Holder regularity conditions on the boundary∂Ω and the given data:

∂Ω ∈ C2,γ , f ∈ Cγ(Ω), ϕ ∈ C2,γ(∂Ω).

(Here 0 < γ < 1; see a discussion about necessary definitions and factsabove, before Theorem 6.20.) By Schauder’s Theorem 6.20 the conditionsabove imply that u ∈ C2,γ(Ω).

Starting with the general Green’s formula (6.23), take an arbitrary u =u(x), u ∈ C2(Ω), and v = v(·) = v(x) = G(·, y) = v(x, y), where the notationG(·, y) means that G(x, y) is considered as a function (or distribution) ofx ∈ Ω for a fixed y ∈ Ω (that is, y is considered to be a parameter).

Note, however, that the function v = G(·, y) has a singularity at x = y,so that v is not in C2 (with respect to x) in any neighborhood of y. But(6.23) still holds if v has a point singularity at y ∈ Ω, as is clear from thearguments given above in the proof of Proposition 6.18, which are nothingelse but just an appropriate approximation of v by smooth functions.

To substitute the functions u and v into (6.23), note that

∆v = ∆xG(x, y) = δ(x−y),∫

Ωu(x) ∆xG(x, y)dx =

∫Ωu(x)δ(x−y)dx = u(y).

So the left hand side of (6.23) has the form

u(y)−∫

ΩG(x, y)∆u(x)dx,

whereas the right hand side is

−∫∂Ω

∂G(x, y)∂nx

u(x)dSx,

because G(x, y) = 0 if x ∈ ∂Ω. Equality of these two expressions, afterinterchanging of x and y, together with the symmetry of Green’s functionimply the following Proposition:

Proposition 6.25. Let us assume that Ω is a bounded open set with apiecewise smooth (C2) boundary ∂Ω in Rn, and the Dirichlet Problem forthe Laplacian in Ω has Green’s function G = G(x, y), which is defined forall (x, y) ∈ (Ω × Ω) ∪ (Ω × Ω). Then for every u ∈ C2(Ω) the followingintegral representation holds:

(6.27) u(x) =∫

ΩG(x, y)∆u(y)dy −

∫∂Ω

∂G(x, y)∂ny

u(y)dSy, x ∈ Ω.

6.7. Green’s Function and Dirichlet’s Problem 157

In other words, if u ∈ C2(Ω) is the (unique) solution of the Dirichletproblem (6.10), that is ∆u = f in Ω and u|∂Ω = ϕ on ∂Ω, then

(6.28) u(x) =∫

ΩG(x, y)f(y)dy −

∫∂Ω

∂G(x, y)∂ny

ϕ(y)dSy, x ∈ Ω.

This Proposition gives an explicit presentation of the solution for theDirichlet problem in terms of Green’s function. Two important particularcases are obtained when ϕ ≡ 0 or f ≡ 0: if ϕ ≡ 0, then

(6.29) u(x) =∫

ΩG(x, y)f(y)dy, x ∈ Ω;

if f ≡ 0, then

(6.30) u(x) = −∫∂Ω

∂G(x, y)∂ny

ϕ(y)dSy, x ∈ Ω.

Corollary 6.26. Let u ∈ C2(Ω) satisfy the same conditions as in (6.30)above. In particular, it suffices for Ω to be bounded, for ∂Ω to be in C2 andfor u to be harmonic in Ω. Then u satisfies (6.30).

Remark 6.27. Here are more details about the physical meaning of theformulas (6.28)–(6.30) in electrostatics. (About the simplest notions of elec-trostatics see physics textbooks, e.g. [7], Vol. 2, Chapters 4–8, and alsoChapters 4 and 5 here.)

The situation to describe is a unit charge sitting in a body Ω, which isbounded by a thin closed conductive shell ∂Ω. Let us explain what thesewords actually mean.

Consider Ω to be a closed vessel bounded by a thin shell ∂Ω, madeof a conductive material (e.g. a metal), which means a material wherethe electrons can move freely. Equivalently, we may say that there are noforces acting on the electrons, hence the potential is locally constant on theboundary ∂Ω, i.e. it is constant on each of the connected components of∂Ω. By adding an appropriate solution of the Dirichlet problem in Ω, wecan force these constants to vanish. Then all connected components of theboundary ∂Ω are grounded, i.e. connected with the Earth by wires. It isassumed that the Earth has a potential which vanishes on its surface, hencethe boundary condition u|∂Ω = 0 is satisfied. For simplicity, assume thatthe interior of the vessel is made of a dielectric (that is non-conducting)material.

Now let us consider Green’s function G(x, y), x, y ∈ Ω, x 6= y. It isinterpreted as a static (i.e. time-independent) potential of a unit charge


located at y ∈ Ω, whereas the potential itself is u = u(x) = G(·, y). We areinterested in the electric field in the vessel Ω, which has no charges inside Ω(except at y). The boundary condition G(x, y) = 0 for x ∈ ∂Ω, requires Ωto be a closed vessel bounded by a thin conductive shell ∂Ω.

Green’s function G(x, y) replaces the fundamental solution En(x − y),when the whole space Rn is replaced by the domain Ω.

For a system of charges the potentials add up by the superposition prin-ciple, and for a distributed charge the potential of this charge becomes anintegral, as in (6.28)–(6.30).

Note that for n ≥ 3 all potentials with compactly supported chargestend to 0 as |x| → ∞. This reflects a physically natural requirement thatthe potential u should be 0 at infinity (up to an additive constant, whichcan be chosen so that, indeed, u(∞) = 0).

By the superposition principle, the integral in (6.29) is the potential uof the system of charges which are distributed in Ω with the density f(y)(or f(y)dy). Assuming that u|∂Ω = 0 we obtain the (unique) solution of theboundary-value problem

(6.31)

∆u(x) = f(x), x ∈ Ω;u|∂Ω = 0.

If the boundary is not connected, then the condition, that the boundaryis conductive, means that the potential u is constant on each of the connectedcomponents of the boundary. This is a weaker condition compared with thevanishing condition for Green’s function G(x, y) = 0, x ∈ ∂Ω.

Physically, in the same electrostatic model, the property of the boundaryto be conductive is equivalent to one that no energy is needed to transporta small charge along the boundary. (The latter statement follows from thefact that the energy needed to transport a small charge along a curve is equalto the difference between the potentials of the ends of the curve, where thevalue of the potential at the end of the curve should be subtracted from thevalue at the beginning of the curve.)

Remark 6.28. Turning back to the integral formulas (6.28)–(6.30), we seethat the solution of the Dirichlet boundary-value problem for the Poissonequation (6.31), under the smoothness requirements described above, isgiven by (6.29), if such a solution exists. Moreover, it suffices to requiref ∈ C(Ω), ∂Ω ∈ C2 and u ∈ C2(Ω) to assure the possibility of substitutionof u to (6.31), and directly check both conditions. (Even this is often not

6.8. Explicit formulas for Green’s functions 159

needed; the conditions are satisfied automatically, if all data are sufficientlysmooth, e.g. f ∈ C1(∂Ω), ∂Ω ∈ C2 and u ∈ C3(Ω).) But it may happenthat such solution does not always exist.

Similar statements are true for the boundary-value problem (6.8) andthe integral representation (6.30).

As to the more general problem (6.10), its solution can be presented asa sum u = uf + uϕ, where uf , uϕ are solutions of (6.10) with ϕ = 0 andf = 0 respectively. So to solve the boundary-value problem (6.10) we needto add corresponding solutions of (6.29) and (6.30).

Remark 6.29. If Green’s function is known, the solution u of the boundary-value problem (6.31) is found by a simple integration from the right hand sidef , according to (6.29). But to find solution of (6.8), given Green’s functionand the boundary data ϕ, we still need to calculate the normal derivative∂G/∂ny and then integrate over the boundary. This significant differenceappears because the solution operator needs to produce a function of nvariables from a function of n−1 variables. Here one degree of freedom getsconsumed by the requirement for the solution u to be a harmonic functionin Ω.

Note also that the right hand side of the formula (6.30) is nothing elsebut the double layer potential (see Ch. 5).

6.8. Explicit formulas for Green’s functions

The cases, where Green’s function can be found explicitly and as a finiteexpression, are rare. Nevertheless, they are worth learning because theyshow what can you expect in the general case. We will discuss method ofimages (sometimes called method of reflections).

6.8.1. Method of images (generalities). This method allows to findexplicitly Green’s function and formulas for solving the Dirichlet problemfor the Laplace operator in some important examples.

Let us first explain the idea of the method for domains in Rn where n ≥ 3.It is convenient to consider some unbounded domains too. In this case, thecondition G(x, y)|x∈∂Ω = 0 may be not sufficient to select a unique Greenfunction, and we will impose additionally the following natural condition:

Decay Assumption. For every fixed y ∈ Ω,

(6.32) G(x, y)→ 0 as |x| → ∞, x ∈ Ω.


Lemma 6.30. If n ≥ 3, then there is at most one Green’s function G =G(x, y), existing for a given domain Ω ⊂ Rn (possibly unbounded) and sat-isfying (6.32).

Proof. Let BR(0) be the open ball with the radius R > 0 and center at 0in Rn. Denote ΩR = Ω∩BR(0) which is a bounded open set in Rn. (It mayhappen that ΩR is empty.) Clearly, union of all ΩR with R > 0 is Ω.

Now, let u = u(x) = G1(x, y) − G2(x, y), where G1, G2 are two Greenfunctions (of ∆ in Ω), both satisfying (6.32), y ∈ Ω is fixed. Then u isharmonic in Ω, and u|∂Ω = 0. If Ω was bounded, then we would be ableto conclude that u ≡ 0 due to the maximum principle (6.7). Unfortunately,this argument does not work for unbounded Ω. To modify it, note first, that∂ΩR ⊂ (∂Ω ∩ BR(0)) ∪ (Ω ∩ ∂BR(0)), hence,

max∂ΩR|u| ≤ max

∂Ω∩BR(0)|u|+ max

Ω∩∂BR(0)|u| = max

∂BR(0)|u|,

because u|∂Ω = 0. It follows, again by the maximum principle, that

maxΩR|u| ≤ max

Ω∩∂BR(0)|u|.

Due to (6.32), the right hand side tends to 0 as R → ∞, whereas theleft hand side increases as R increases. It follows that u ≡ 0 in ΩR forsufficiently large R, which implies that u ≡ 0 in Ω.

Main idea

Consider the fundamental solution En = En(z) of the Laplacian ∆ in Rn

(see (4.38),(4.39)), n ≥ 3. Due to the equation ∆zE(z) = δ(z) and Liouville’sTheorem 5.28, this is the only fundamental solution with En(z) → 0 as|z| → ∞. It is clear from the definition of Green’s function, that for Ω = Rn

we have

G(x, y) = GRn(x, y) = En(x− y).

The idea of the method of images is to construct Green’s function of adomain Ω ⊂ Rn by taking the fundamental solution of the Laplacian in Rn

and adding a correction term (6.20) which has the form of the potentialof a point charged particle or a system of such particles. These additionalcharged particles must be located outside Ω (to make the additional sum ofthe potentials harmonic in Ω and continuous in Ω), and also assure that thetotal potential of all charges vanishes on the boundary of Ω. Thus, the totalpotential will be Green’s function of Ω. N


We will start from the case when the correcting term is the potentialof just one charged particle. Let the original charged particle be located aty ∈ Ω and carry a unit charge q0 = 1, so it creates an electric field with thepotential

u0 = u0(·, y) = En(· − y) = En(x− y), x, y ∈ Rn, x 6= y.

Let us denote by y∗ the location of the correcting charge, which meansy∗ = y∗(y) ∈ Rn \Ω, and let us denote the corresponding correcting chargeby q∗ = q∗(y) ∈ R. The total potential of the system of these two chargedparticles is

(6.33) u = u(x) = u(x, y) = En(x− y) + q∗(y)En(x− y∗),

where y is considered a parameter (a fixed vector which can be chosen ar-bitrary). For Green’s function of Ω to coincide with the right hand side of(6.33), we need to require that u|x∈∂Ω = 0, for all x ∈ ∂Ω, y ∈ Ω, x 6= y. Inother words,

(6.34) En(x− y) + q∗(y)En(x− y∗(y)) = 0, x ∈ ∂Ω, y ∈ Ω.

In this case, Green’s function of Ω is

(6.35) G(x, y) = En(x− y) + q∗(y)En(x− y∗(y)),

where x, y ∈ Ω, x 6= y, y∗ ∈ Rn \Ω.

Remark 6.31. In case of unbounded domain the above discussed condi-tions on Ω should be supplemented by the condition (6.32) which requiresvanishing of u at infinity. We will abbreviate it as u(∞, y) = 0 (for ally ∈ Ω).

Note that q∗(y) < 0 for all y ∈ Ω, due to (6.34).

Remark 6.32. Since En(x−y) is actually well-defined and real-analytic forall x 6= y, the sum in (6.35) is naturally defined for all x, y, y∗ ∈ Rn, suchthat x 6= y, x 6= y∗, whatever is q∗ ∈ R (which can be also variable). Inparticular, we can take charges, which are arbitrarily located with respectto Ω. This means physically that we actually consider the charges at y, y∗

to be much larger by absolute value than the charge at x, which is thereforeconsidered as a “small, weakly charged test particle”. The influence of thelatter upon the charges at y, y∗ is considered negligible and therefore ignored.Also, the diameters of all particles are considered negligible if compared withthe distances between them, so they are thought to be point particles.


In (classical) gravitation the equations are the same as in electrostatics,except all charges are positive (they are masses) and they only attract eachother (there is no repulsion), this would mean that we have 3 “point par-ticles”: two heavy stars which are considered “attached” to their places inthe sky, or, more precisely, moving very slowly, compared with the motionof the third particle which is a light planet, located at x and moving underthe influence of the stars gravitation attraction. The problem of describingthe motion of such a system of three gravitating point masses, is referred to,in the language of Celestial Mechanics, as one of the simplified (restricted)versions of so-called classical (as opposed to quantum) three-body problem.

Description of admissible domains

Let us attempt to calculate Green’s function by finding y∗ = y∗(y) andq∗ = q∗(y), which are both functions on Ω with values in Rn \Ω and R \0,respectively, and satisfy (6.34). We will see that such functions exist onlyfor very special domains Ω, and we will completely describe such domains.

The explicit form of En allows to rewrite (6.34) in the form

(6.36) |x− y|2−n + q∗|x− y∗|2−n = 0, x ∈ ∂Ω,

or, equivalently (for x 6= y),

(6.37)|x− y∗|2|x− y|2 = (−q∗)2/(n−2), x ∈ ∂Ω, y ∈ Ω,

which implies, in particular, that the left-hand side is independent of x. Letus denote

(6.38) Q∗ = Q∗(q∗) = (−q∗)−2/(n−2), Q∗ > 0,

which is equivalent to

(6.39) q∗ = −(Q∗)−(n−2)/2.

Now we can rewrite (6.37) as

(6.40) |x− y|2 −Q∗|x− y∗|2 = 0, x ∈ ∂Ω, y ∈ Ω.

Proposition 6.33. Under the conditions above, let u = u(x) = u(x, y) =u(x, y, x∗, y∗, q∗) be a two-charges potential (as in (6.33) ) such that Q∗ > 0,Q∗ 6= 1. Then the zero nodal domain u−1(0) is a geometric sphere.

It is a hyperplane in case Q∗ = 1.


Proof. It follows that for every fixed y ∈ Ω, the set of all x, satisfying (6.40),is a quadric, which depends upon y ∈ Ω. It is easy to see by looking at theprincipal part of the quadric equation (second degree terms with respect tox), that this quadric is a sphere or a hyperplane. Let us recall that underour assumptions Q∗ > 0. We will prove below that it is a sphere if Q∗ 6= 1and a hyperplane if Q∗ = 1. We will determine the precise parameters ofthe quadric later.

Remark 6.34. We can simplify the equation of the quadric (6.40) (or(6.37)). Let us choose y ∈ Ω, and find y∗ = y∗(y) and q∗ = q∗(y). (Westill have to specify how to do this explicitly.) Then y∗ ∈ Rn \Ω. Takex0 to be the intersection point of ∂Ω with the straight line segment [y, y∗],connecting y and y∗. This segment intersects ∂Ω by exactly one intersectionpoint, because the points y, y∗ are located on the different sides of ∂Ω whichis a sphere or a hyperplane. Then the right hand side of (6.37) is equal toits value for x = x0, which means that

(6.41)|x− y∗||x− y| =

|x0 − y∗||x0 − y|

, x ∈ ∂Ω, y ∈ Ω.

Here we will usually take y ∈ Rn \0, but sometimes it is better to use one-point compactification Sn = Sn∪∞, extending y∗ = y∗(y) to a continuousmap Sn → Sn.

It is interesting that (6.41) does not contain dimension n.

H Denoting the standard scalar product of vectors x, y ∈ Rn by x · y, weobtain

0 =Q∗(|x|2 − 2x · y∗ + |y∗|2)− (|x|2 − 2x · y + |y|2)

=(Q∗ − 1)|x|2 − 2x · (Q∗y∗ − y) + (Q∗|y∗|2 − |y|2).

If Q∗ 6= 1, then completing the square in the right hand side gives

0 =(Q∗ − 1)[|x|2 − 2x · (Q∗y∗ − y)

Q∗ − 1+|Q∗y∗ − y|2(Q∗ − 1)2

]− |Q

∗y∗ − y|2 − (Q∗|y∗|2 − |y|2)(Q∗ − 1)Q∗ − 1

=(Q∗ − 1)∣∣∣∣x− Q∗y∗ − y

Q∗ − 1

∣∣∣∣2 − Q∗

Q∗ − 1|y∗ − y|2,

or, equivalently, ∣∣∣∣x− Q∗y∗ − yQ∗ − 1

∣∣∣∣2 =Q∗

(Q∗ − 1)2|y∗ − y|2.(6.42)


This means that our quadric is a sphere SR(c) with the center c ∈ Rn andthe radius R > 0, given by the formulas

(6.43) c =Q∗y∗ − yQ∗ − 1

, R =(Q∗)1/2

|Q∗ − 1| |y∗ − y|.

H Similarly to the relations deduced above, we can also obtain relations

|(Q∗ − 1)x− (Q∗y∗ − y)|2 = Q∗|y∗ − y|2,(6.44)

|Q∗(x− y∗)− (x− y)|2 = Q∗|y∗ − y|2.(6.45)

Their advantage is that they also hold for Q∗ = 1, in which case the spherebecomes hyperplane. N

Solving the first equation in (6.43) with respect to y∗ and y, we get

(6.46) y∗ =1Q∗

y +(

1− 1Q∗

)c, y = Q∗y∗ + (1−Q∗)c.

It follows that c, y and y∗ are on the same line in Rn. If Q∗ − 1 > 0 (resp.Q∗ − 1 < 0) , then they follow in the order c, y∗, y (resp. c, y, y∗).

Substituting the expression for y∗ from (6.46) into (6.43), and then solv-ing the resulting equation with respect to Q∗, we easily find that

(6.47) Q∗ =|y − c|2R2

.

Now using the first equation of (6.43), we obtain

(6.48) y∗ − c =R2

|y − c| ·y − c|y − c| ,

which implies that y∗ − c and y − c are proportional and

|y − c| · |y∗ − c| = R2.

This means that the transformation from y to y∗ is the standard inversionwith respect to the sphere SR(c).

Corollary 6.35. The variables y∗, Q∗ can be real-analytically expressed interms of 2n + 1 geometric quantities c,R, y, except on a finite number ofalgebraic submanifolds.

The formulas above also imply that the geometric parameters c,R, yuniquely define all other parameters y∗ (inverse to y with respect to thesphere SR(c)), Q∗ (see (6.47)), q∗ (see (6.39)). By translation we can achievec = 0 (translating all other parameters respectively). N


Remark 6.36. If (6.48) is considered as a map

∗ : Rn \c → Rn \c, y 7→ y∗,

then it is involutive or an involution, which means that ∗2 = Id (the identitymap), with SR(c) being its fixed point set.

The particular involution ∗ : y 7→ y∗ of Rn \c given by (6.48), is calledreflection with respect to the sphere SR(c).

Remark 6.37. In the limit R → +∞, with the appropriate choice of thecenter c = c(R), we can obtain a limit relation SR(c(R))→ S∞, where S∞ isa hyperplane in Rn and the convergence is in C2-topology of hypersurfaces(defined by C2-topologies on functions representing S∞ as graphs of C2-functions in local coordinates). Then, for any fixed y we get the reflectiony∗ = y∗(R), such that y∗(R) → y∗∞, where y∗∞ is the mirror image of ywith respect to the limit hyperplane S∞. (See more about this below in thisSection.)

H To understand the general cause of involutivity, we can consider generalcharged particles y, q, where y ∈ Rn, q ∈ R \0, and q is the chargelocated at the point y. Let y1, q1 = y∗, q∗ be the “conjugate” pair (toy, q) with respect to a fixed sphere SR(c). This means that y1 6= y, and

q

|x− y|n−2= − q1

|x− y1|n−2, x ∈ SR(c).

Clearly, the nodal hypersurface for this system depends only upon the ratioq1/q (i.e. does not change under the transformations (q, q1) 7→ (tq, tq1),where t ∈ R \0. Previously we only considered the “normalized” pairsy, q, satisfying the condition q = 1. Assuming that we also have a chargedparticle y2, q2, which is conjugate to y1, q1, then we also have y2 6= y1,and

q1

|x− y1|n−2= − q2

|x− y2|n−2, x ∈ SR(c).

It follows thatq

|x− y|n−2=

q2

|x− y2|n−2, x ∈ SR(c).

Using (6.48) (implying the uniqueness of y∗ for fixed c,R, y), we see thaty2 = y, which means (y∗)∗ = y for all y, i.e. involutivity of the map ∗. N

We will summarize the arguments above in the following theorem.


Theorem 6.38. Let Ω be a bounded domain with a C∞ boundary ∂Ω inRn, n ≥ 3. Assume that for every y ∈ Ω and every q > 0 there exist a pointy∗ ∈ Rn \Ω and a number q∗ < 0, such that the potential u of the system oftwo charges 1, q∗ at the points y, y∗ respectively, has the nodal hypersurface

u−1(0) = x ∈ Rn : u(x) = 0,

which coincides with ∂Ω. Then the following statements hold:

(a) Ω is a ball BR(c) with the center c ∈ Ω and radius R > 0.

(b) With Q∗ = (−q∗)−2/(n−2) the quantities Q∗, R, c, q∗, y∗ are uniqueand given by the formulas (6.43), (6.33).

(c) The identities (6.37), (6.40), (6.41), (6.44), (6.45), (6.46), (6.47),(6.48) hold.

Making the calculations in the opposite direction, we can conclude thatthe following proposition holds:

Proposition 6.39. For every sphere SR(c) and every point y ∈ Rn \SR(c)(n ≥ 3), there exist a unique point y∗ ∈ Rn and a unique number q∗ ∈R \0, such that the potential u of the system of two charged point particlesat y, y∗ with the charges 1, q∗ respectively, has the nodal hypersurface whichcoincides with the sphere SR(c).

Proof. There are two excluding possibilities:

(a) ∂Ω ⊂ SR(c);

(b) ∃ y ∈ (Rn \∂Ω) ∩ SR(c).

It is easy to see that (a) implies the equality ∂Ω = SR(c), and (b) contradictsthe conditions of Proposition 6.39.

• Green’s function for a half-space, n ≥ 2.

Now we lift the restriction n ≥ 3 and assume n ≥ 2. The Laplacian ∆2

does not have a fundamental solution E2(·) satisfying E2(x)→ 0 as |x| → ∞.Also, there is no natural fundamental solution for ∆2. The general one hasthe form E2 +C, where C is an arbitrary constant. Fixing C, we choose E2,and this choice does not make any difference. In this section we will chosethe constant so that E2(x) = (2π)−1 ln |x|, as in (4.39).

Let L be a hyperplane in M = Rn, n ≥ 3, y ∈ M \ L. Then M \ L =M+ ∪M−, where M± are the disjoint connected components of M \ L, i.e.M+ and M− are open half-spaces of M , defined by L. Assume that y ∈M−.


Let us denote Ω = M−. Then Rn \Ω = M+. So we can define y∗, theconjugate point to y. It is easy to see that y∗ is the mirror image of y in themirror L.

We can choose orthogonal coordinates x1, . . . , xn in M , such that inthese coordinates L = x ∈ Rn |xn = 0, M+ = x ∈M |xn > 0. Then fory = (y1, . . . , yn) we have

(6.49) y∗ = (y1, . . . , yn−1,−yn).

Now it is easy to check that the function

(6.50) G(x, y) = En(x− y)− En(x− y∗)

satisfies all conditions required from Green’s function (including the DecayAssumption (6.32), which is required from all unbounded domains).

In the notations, introduced earlier in this Chapter, the reflected chargeis q∗ = −1, y∗ is given by (6.49), that is, y∗ is the reflected in the mirror(which is L) unit charge at y, which also changed sign at reflection.

H The desired result also follows from symmetry considerations, moreprecisely, from the uniqueness of Green’s function combined with the invari-ance of the problem with respect to all translations parallel to the mirror L,rotations with a fixed point c ∈ L and mirror reflections with respect to themirror L. N

• Green’s function in 2 dimensions

Even though the fundamental solution E2(z) of the Laplacian in R2 (thatis, for n = 2) behaves differently at infinity (for example, it does not go to 0when |z| → ∞), almost nothing changes for other objects, e.g. for derivativesof Green’s functions. In particular, a Green function of the half-plane

Ω = x = (x1, x2) : x2 < 0 ⊂ R2,

can be taken in the form

G(x, y) = E2(x− y) + g(x, y),

where E2(x) = 12π log |x|, g = g(x, y) is the compensating reflected term,

g(x, y) = −E2(x− y∗), x ∈ Ω,

with

y∗ = (y1, y2)∗ = (y1,−y2).


In particular, G(x, y) generally stays unbounded if x→∞ and |y| is bounded.We also see from (6.40) that

(6.51) G(x, y) = E2(x− y)− E2(x− y∗).

6.8.2. Green’s function of a ball. Consider the open ball Ω = BR(0)with the radius R > 0, centered at the origin 0 ∈ Rn, n ≥ 2. Let us fix y ∈ Ω,and take the fundamental solution En = En(x) of the Laplace operator. Nowchoose a smooth function g in (Ω× Ω) ∪ (Ω× Ω),

(6.52) g = g(·, y) = g(x, y) = q∗(y)En(x− y∗(y)),

the other notations the same as in (6.33). We would like the sum

(6.53) G(x, y) = En(x− y) + g(x, y) = En(x− y) + q∗(y)En(x− y∗(y)),

to be Green’s function of Dirichlet’s problem for the ball Ω. Simplifyingthe sufficient conditions on Ω, we see that it is sufficient to determine thecoefficients q∗, y∗ in (6.52) by substitution the expression (6.53) into theconditions for Green’s functions:

(a) g ∈ C2(Ω× Ω);

(b) ∆xG(·, y) = δ(·, y) for all (x, y) ∈ (Ω× Ω) ∪ (Ω× Ω),

∆yG(x, ·) = δ(x, ·) for all (x, y) ∈ (Ω× Ω) ∪ (Ω× Ω);

(c) G(x, y) = 0 if (x, y) ∈ (∂Ω× Ω) ∪ (Ω× ∂Ω).

Note that due to the location and local structure of singularities, wehave, for G defined on Ω× Ω,

sing supp G = diag(Ω× Ω).

Moreover, y ∈ Ω implies y∗ ∈ Rn \Ω, so all interior singularities have thesame form as En(x− y). More precisely, G(x, y)− En(x− y) ∈ C∞(Ω× Ω).We also established above that one corrective term is sufficient only if Ω is aball or a half space, in which case the formulae (6.39), (6.42), (6.43), (6.46),(6.47), (6.48) provide effective relations between q∗, Q∗, y, y∗, c and R.

Using the arguments on Green’s function in a smooth bounded domainabove, we see that a solution g = g(x, y) of the Laplace equations ∆xg =∆yg = 0 with any prescribed Dirichlet boundary condition g(x, y) for allx ∈ ∂Ω, y ∈ Ω (see also (6.30)) will necessarily coincide with the originalsolution u. This also could be seen from the Maximum Principle. Hence, itis enough to investigate the hypothetic solution from the formula (6.30).

Note first that for any y ∈ Ω we have q∗ ∈ (−∞,−1) (or, equivalently,Q∗ ∈ (0, 1), see (6.38) and (6.39)). Due to the above mentioned uniqueness of


Green’s function, this leads to the same Green function and other parametersas above.

The argument above makes it natural to try test functions of the form(6.35) and adjust them to become Green’s functions. Fixing y∗ = y∗(y), wesee, as above, that y∗ must be a conjugate point to y for every y ∈ Ω. Italso follows that if we choose y∗ to be a conjugate point to y (for every fixedy ∈ Ω), then, after some natural analytic continuations, the set of all y’sbecomes a sphere, and so does the set of all y∗’s.

Now we can choose the set of common zeros of all functions (6.35) asa sphere BR(0), and arrive to a sphere which depend upon the choices ofc ∈ Rn, R > 0 and q∗ < −1.

Straightforward construction, starting from the sphere, described above,may be done as follows.

A. Given a sphere SR(0), where R > 0 is its radius, 0 is the center of thesphere (the origin of Rn), define Ω = BR(0) which is the open ball with thesame center and radius. Introduce the natural inclusion map BR(0) ⊂ Rn

(naturally with respect to the canonical analytic inclusion).

B. Adding a point ∞ to Rn, introduce topology on the union Sn =Rn ∪∞, so that it naturally extends the inclusion Rn ⊂ Sn = Rn ∪∞ to thehomeomorphism of the union to the standard sphere Sn.

C. Define the involutions *: Sn → Sn, y 7→ y∗, by

y∗ =R2y

|y|2 ,

on Rn \0 and extended to the points 0 and ∞ by continuity.

D. Using the charge 1 at a point y ∈ Ω = BR(0) and the involution ∗,introduced above (the reflection with respect to the sphere SR(0)), definethe reflected charge q∗ at y∗ for every y ∈ Sn and adjust its size so that thetotal potential of two charges at y, y∗ vanishes on the sphere.

E. For Green’s functionG = G(x, y) of Ω, we haveG = En(x−y)+g(x, y),where g ∈ C∞((Ω × Ω) ∪ (Ω × Ω)), ∆xg = ∆yg = 0 on Ω × Ω. Besides,G(x, y) = 0 if x ∈ ∂Ω or y ∈ ∂Ω.

Remark 6.40. In fact, the condition E is sufficient for the uniqueness ofGreen’s function of a given domain Ω. But the introductory geometricterminology considerably simplifies exposition. Indeed, assuming n ≥ 3,y ∈ Ω = BR(0), we can take the reflection y∗ = R2|y|−2y, and then writeG = G(x, y) by the formula above, finding g(x, y) as a one-charge potential


q∗(y)En(x− y∗(y)) with the coefficient q∗(y) to be found from the Dirichletboundary condition G(x, y) = 0 for all x ∈ ∂Ω (or all y ∈ ∂Ω).

In these notations we obtain

Proposition 6.41. Let us consider the set of all conjugate pairs y, y∗where y ∈ Ω = BR(0), and y∗ = R2|y|−2y, that is, y∗ is conjugate to y,y∗ ∈ Rn \Ω. Now take G = G(x, y) = En(x − y) + g(x, y), where g ∈C∞(Ω × (Rn \Ω)) satisfies the equations and boundary conditions requiredfrom Green’s functions, that is ∆xg = 0,∆yg = 0 on Ω×Ω, and G(x, y) = 0if x = 0 or y = 0.

Then G is Green’s function of the Laplacian in Ω.

Proof. Immediately follows from the equations and boundary conditionsimposed on G, g.

H Now we can establish the geometric description of the electric fields inthe ball Ω = BR(0), when the equilibrium of the main point charge (locatedat x ∈ Ω) is given by exactly one additional charge which is located outsidethe ball. According to our present notations, we can take c = 0 and y ∈ Ωarbitrary. This means that the following relations hold:

(i) 0 is the center of the ball Ω,

(ii) R > 0 is the radius of Ω,

(iii) q∗ and Q∗ are related by (6.38) and (6.39),

(iv) Q∗ is expressed through R by (6.47),

(v) y∗ is related with y by (6.48): y∗ = R2|y|−2y.

It is easy to see that all other quantities like q∗, Q∗, etc. can be easilyexpressed through the geometric parameters c,R. This allows to restrictconsideration of the problems relating any quantities like this, to elementaryproblems relating c,R, y. N

H Let us write down explicit formulas for Green’s function of the ballΩ = BR(0). According to the arguments above, it is reasonable to try

(6.54) G(x, y) = En(x− y) + q∗(y)En(x−R2|y|−2y), x ∈ Ω, y ∈ Ω,

where q∗(·) is found by properties of G which we have not used in relationto q∗. In particular, we can use the symmetry of Green’s function, i.e. therelation G(x, y) = G(y, x) for all x, y ∈ Ω. This means

q∗(y)En(x− y) = q∗(x)En(y − x), x, y ∈ Ω, x 6= y.

6.9. Problems 171

Since En(−z) = En(z) for all z ∈ Rn \0, we can cancel the factor En andconclude that q∗(x) = q∗(y) for all x, y ∈ Ω. Therefore, we conclude thatq∗(·) = const. Denote this constant simply q∗. Then, we can rewrite theequality (6.34) as

En(x− y) + q∗En(x−R2|y|−2y) = 0,

for every x ∈ SR(0) = ∂Ω, with a constant q∗. (Compare (6.34).) N

6.9. Problems

6.1. Prove the statements inverse to the ones of Theorems 6.1, 6.2: if acontinuous function u in an open set Ω ⊂ Rn has mean value property forall balls B with B ⊂ Ω or for all spheres which are boundaries of suchballs (that is, either (6.2) holds for such spheres or (6.4) holds for all suchballs), then u is harmonic in Ω. Moreover, it suffices to take balls or spheres,centered at every point x with sufficiently small radii r ∈ (0, ε(x)), whereε(x) > 0 continuously depends upon x ∈ Ω.

6.2. Let u be a real-valued C2 function on an open ball B = Br(x) ⊂ Rn,which extends to a continuous function on the closure B, and ∆u ≥ 0everywhere in B. Prove that

(6.55) u(x) ≤ 1voln−1(S)

∫Su(y)dSy =

1σn−1rn−1

∫Sr(x)

u(y)dSy,

and

(6.56) u(x) ≤ 1vol(B)

∫Bu(y)dy =

n

σn−1rn

∫Br(x)

u(y)dy,

where the notations are the same as in Section 6.1.

6.3. Prove the inverse statement to the one in the previous problem: ifu ∈ C2(Ω) is real-valued and for all balls B with B ⊂ Ω (or for all sphereswhich are boundaries of such balls), (13.2) (respectively, (13.1)) holds, then∆u ≥ 0 in Ω. Moreover, it suffices to take balls or spheres, centered atevery point x with sufficiently small radii r ∈ (0, ε(x)), where ε(x) > 0continuously depends upon x ∈ Ω.

Remark 6.42. Let a real-valued function u ∈ C2(Ω) be given. It is calledsubharmonic if ∆u ≥ 0 in Ω. Note, however, that the general notion of sub-harmonicity does not require that u ∈ C2 and allows more general functionsu (or even distributions, in which case ∆u ≥ 0 means that ∆u is a positiveRadon measure).


6.4. Let u be a real-valued subharmonic function in a connected open setΩ ⊂ Rn. Prove that u can not have local maxima in Ω, unless it is constant.In other words, if x0 ∈ Ω and u(x0) ≥ u(x) for all x in a neighborhood ofx0, then u(x) = u(x0) for all x ∈ Ω.

6.5. Consider the initial value problem (6.11) with ϕ ∈ C∞0 (Rn) and ψ ≡ 0.Assume that it has a solution u. Prove that in this case ϕ ≡ 0 on R andu ≡ 0 on ΠT .

6.6. Let U, V,W be open sets in Rm,Rn,Rp respectively, and g : U → V ,f : V → W be maps which are locally Ck,γ (that is, they belong to Ck,γ

on any relatively compact subdomain of of the corresponding domain ofdefinition).

(a) Assume that k ≥ 1. Then prove that the composition f g : U →W

is also locally Ck,γ .

(b) Prove that the statement (a) above does not hold for k = 0.

(c) Prove that if k ≥ 1, Ω is a domain in Rn with ∂Ω ∈ Ck,γ , U is aneighborhood of Ω in Rn, and ψ : U → Rn is a Ck,γ diffeomorphism of Uonto a domain V ⊂ Rn, then the image Ω′ = ψ(Ω) is a domain with a Ck,γ

boundary, Ω′ ⊂ V .

6.7. (a) Prove that if Ω is a bounded domain in Rn and ∂Ω ∈ Ck,γ withk ≥ 1, then ∂Ω is a Ck,γ manifold, dimR ∂Ω = n− 1.

(b) Prove that for any non-negative integers k, k′ and numbers

γ, γ′ ∈ (0, 1), such that k+γ ≥ k′+γ′, k ≥ 1, the class Ck′,γ′ is locally invari-

ant under Ck,γ diffeomorphisms. In particular, function spaces Ck′,γ′(∂Ω)

are well defined.

(c) Under the same assumptions as in (a) and (b), for a real-valuedfunction ϕ : ∂Ω → R, the inclusion ϕ ∈ Ck

′,γ′(∂Ω) is equivalent to theexistence of an extension ϕ ∈ Ck′,γ′(Ω), such that ϕ|∂Ω = ϕ.

6.8. Assume that k ≥ 1 is an integer, Ω is a bounded domain in Rn, suchthat ∂Ω ∈ Ck+2, f ∈ Ck(Ω) and ϕ ∈ Ck+2(∂Ω). Using Schauder’s Theorem6.20, prove that the problem (6.10) has a solution u ∈ Ck+1(Ω).

6.9. Assume that ∂Ω ∈ C2, and there exists Green’s function G = G(x, y) ofΩ, such that G ∈ C2 for x 6= y, x, y ∈ Ω. Consider an integral representationof a function u ∈ C2(Ω) given by the formula (6.28), with f ∈ C(Ω) and

6.9. Problems 173

ϕ ∈ C(∂Ω). Prove that the pair f, ϕ, representing a fixed function u ∈C2(Ω) by (6.28), is unique.

Hint. The function f can be recovered from u by the formula f = ∆u,whereas at any regular point of the boundary x ∈ ∂Ω, ϕ is the jump of theoutgoing normal derivative of u on ∂Ω:

(6.57) ϕ(x) = limε→0+

(∂u

∂n(x+ εn)− ∂u

∂n(x− εn)

).

6.10. Provide detailed proofs of identities (6.44) and (6.45).

6.11. Provide detailed proofs of identities (6.46), (6.47) and (6.48).

6.12. Deduce the expression for Green’s function in a half-space of Rn,n ≥ 3, from symmetry considerations.

Hint. Use the uniqueness of Green’s function combined with the invarianceof the problem with respect to all translations parallel to the mirror L,rotations with a fixed point c ∈ L and mirror reflections with the mirror L.

Chapter 7

The heat equation

7.1. Physical meaning of the heat equation

The heat equation is

(7.1)∂u

∂t= a2∆u,

where u = u(t, x), t ∈ R, x ∈ Rn, a > 0, and ∆ is the Laplace operator withrespect to x. This equation is an example of a parabolic type equation. Itdescribes, for instance, the temperature distribution of a homogeneous andisotropic medium (in which case u(t, x) is the temperature at the point xat the time t). This equation is also satisfied by the density of a diffusingmatter, e.g. the density of Brownian particles when there are sufficientlymany of them so that we can talk about their density and consider theevolution of this density as a continuous process. This is a reason why theequation (7.1) is often called the diffusion equation.

Deriving of the heat equation is based on a consideration of the energybalance. More precisely, we will deal with the transmission of the heatenergy, which is the kinetic energy of the atoms and molecules chaotic motionin the gases and liquids or the energy of the atoms vibrations in metals,alloys and other solid materials. Instead of “heat energy” we will use theword “heat”, which is shorter and historically customary in the books onthermodynamics.

H The heat equation for temperature u(t, x) is derived from the followingnatural empirically verified physical assumptions:

175

176 7. The heat equation

1) The heat Q needed to warm up a part of matter with the mass mfrom the temperature u1 to u2 is proportional to m and u2 − u1:

Q = cm(u2 − u1).

The coefficient c is called the specific heat (per unit of mass and unit oftemperature).

2) Fourier’s law: The quantity of heat ∆Q passing through a smallsurface element with the area S during the time ∆t, is proportional to S,∆t, and ∂u

∂n (the rate of growth of u in the direction n of the unit normalvector to the surface). More precisely,

∆Q = −kS ∂u∂n

∆t,

where k > 0 is the heat conductivity coefficient characterizing heat propertiesof the media; the minus sign indicates that heat is moving in the directionopposite to that of the growth of the temperature. N

One should keep in mind that both hypotheses are only approximations,and so is the equation (7.1) which they imply. As we will see below, aconsequence of equation (7.1) is an infinite velocity for propagation of heat,which is physically absurd. In the majority of applied problems, however,these hypotheses and equation (7.1) are sufficiently justified. A more precisemodel of heat distribution should take into account molecular structure ofthe matter leading to problems of statistical physics far more difficult thanthe solution of the model equation (7.1).

Note, however, that, regardless of its physical meaning, equation (7.1)and its analogues play an important role in mathematics. They are used,e.g., in the study of elliptic equations.

H Let us derive the heat equation (for n = 3). We use the energyconservation law (heat conservation in our case) in a volume Ω ⊂ R3. Letus assume that Ω has a smooth boundary ∂Ω. The rate of change of theheat energy of the matter in Ω is obviously equal to

dQ

dt=

d

dt

∫Ωcρu dV,

where ρ is the volume density of the matter (mass per unit volume), dV isthe volume element in R3. Assuming that c = const and ρ = const, we get

dQ

dt= cρ

∫Ω

∂u

∂tdV.

7.2. Boundary value problems 177

Now, let P be the rate of the heat escape through ∂Ω. Clearly,

P = −∫∂Ωk∂u

∂ndS,

where n is the outward unit normal to ∂Ω , dS the area element of ∂Ω. Herein the right hand side we have the flow of the vector −k gradu through ∂Ω.By the divergence theorem (4.34) we get

P = −∫

Ωdiv(k grad u)dV

or, for k = const,

P = −k∫

Ω∆u dV.

The heat energy conservation law in the absence of heat sources means that

dQ

dt= −P.

Substituting the above expressions for dQdt and P , we get∫

Ωcρ∂u

∂tdV =

∫Ωk∆u dV,

which, since Ω is arbitrary, results in (7.1) with a2 = kcρ . N

7.2. Boundary value problems for the heat and Laplace

equations

A physical interpretation of the heat equation suggests natural setting forboundary value problems for this equation. The Cauchy problem (sometimesalso called the initial value problem) is the simplest one:

(7.2)

∂u∂t = a2∆u, t > 0, x ∈ Rn,

u|t=0 = ϕ(x) x ∈ Rn

Its physical meaning is to determine the temperature of the medium at allmoments t > 0 provided the temperature distribution is known at t = 0.

Sometimes we are only interested by the temperature in a domain Ω ⊂Rn regardless of what takes place outside of it. Then, on ∂Ω, additional con-ditions should be imposed: e.g. the temperature or the heat flow should begiven. Thus we come to the two following initial-boundary value problems.


The first initial-boundary value problem (or Cauchy-Dirichlet’s initial-boundary value problem):

(7.3)

∂u∂t = a2∆u, t > 0, x ∈ Ω,u|t=0 = ϕ(x), x ∈ Ω,u|∂Ω = α(t, x), x ∈ ∂Ω, t > 0.

The second boundary value problem (or Cauchy-Neumann’s boundaryvalue problem) is obtained by replacing the latter of conditions (7.3) by

∂u

∂n

∣∣∣∣∂Ω

= β(t, x), x ∈ ∂Ω, t > 0,

where n is the outward unit normal to ∂Ω.

For time-independent boundary conditions it is often interesting to de-termine temperature behavior as t → +∞. Suppose that a limit v(x) ofthe solution u(t, x) of the heat equation exists as t→ +∞ (we say that thesolution “stabilizes” or “tends to an equilibrium”). Then it is natural toexpect that v(x) satisfies the Laplace equation ∆v(x) = 0, i.e., v(x) is aharmonic in x, at least because the only possible finite limit of ∂u(t, x)/∂tcan be 0. It is very easy to make the meaning of this statement precise butwe will not do so at the moment in order not to divert our attention.

The first and second initial-boundary value problems for the heat equa-tion turn in the limit into the following boundary value problems for theLaplace equation:

Dirichlet’s problem

(7.4)

∆v(x) = 0, x ∈ Ω;v|∂Ω = α(x), x ∈ ∂Ω.

Neumann’s problem:

(7.5)

∆v(x) = 0, x ∈ Ω;∂v∂n

∣∣∂Ω

= β(x), x ∈ ∂Ω.

In this model Dirichlet’s problem means the following: find the steadystate temperature inside Ω when the temperature α(x) is maintained on theboundary. The meaning of Neumann’s problem is to find the steady statetemperature inside Ω if the heat flow β(x) on the boundary is given.

We already mentioned the Dirichlet problem in the previous chapter – seeCorollary 6.8, where the uniqueness of the classical solution was established.Clearly, the solution of Neumann’s problem is not unique: any constant can

7.3. A proof that the limit function is harmonic. 179

always be added to the solution. Further, it is clear that the steady statedistribution of heat can exist in Ω only if the total heat flux through theboundary vanishes, i.e.,

(7.6)∫∂Ωβ(x)dS = 0.

(This also follows from the formula (6.3), which implies that if v is a solutionof (7.5) then (7.6) holds.)

We will see later that the above problems for the heat and Laplaceequations are well-posed in the sense that they have a unique solution (upto adding a constant in case of the Neumann problem) in natural classesof functions, and, besides, the solutions depend continuously on initial andboundary values for an appropriate choice of spaces.

Note that the Dirichlet and Neumann problems for the Laplace equationhave many physical interpretations apart from the above ones. For instance,a small vertical deflection of a membrane u(t, x1, x2) and a small displace-ment u(t, x1, x2, x3) of an elastic solid body satisfy the wave equation

∂2u

∂t2= a2∆u.

In the time-independent case we come again to the Laplace equation andthen the Dirichlet problem. For instance, for a membrane this is the problemof recovering the membrane form from that of its boundary. The boundarycondition of Neumann’s problem is interpreted as prescribing the boundaryvalue of the vertical component of the force acting on the membrane along itsboundary. This topic is related with the famous question by Mark Kac “CanOne Hear the Shape of a Drum?” (1968), which led to a wide developmentof Spectral Geometry.

7.3. A proof that the limit function is harmonic.

There are several ways to justify the harmonicity of the limit function. Wewill give one of them.

Proposition 7.1. Let a bounded solution u(t, x) of the heat equation bedefined and bounded for t > 0, x ∈ Ω (for an open subset Ω ⊂ Rn). Supposethat for almost all x ∈ Ω there exists a limit

(7.7) v(x) = limt→+∞

u(t, x).

Then v(x) is a harmonic function in Ω.


Remark 7.2. The solution u(t, x) may be a priori considered as a boundedmeasurable function (here the equation (7.1) should be understood in thedistributional sense). However we will see below that the heat operatorhas a fundamental solution which is infinitely differentiable for |t| + |x| 6=0 and, therefore, actually, u(t, x) ∈ C∞(R+ × Ω), where R+ = t : t >0. Moreover, under the assumptions of Proposition 7.1, the limit in (7.7)is actually uniform together with any finite number of derivatives on anycompact set K ⊂ Ω, i.e., it is the limit in the topology of C∞(Ω).

Proof of Proposition 7.1. Clearly,

limτ→+∞

u(t+ τ, x) = v(x)

for all t > 0 and almost all x ∈ Ω. If ϕ(t, x) ∈ D(R+ × Ω), then thedominated convergence theorem implies

(7.8) limτ→+∞

∫u(t+ τ, x)ϕ(t, x)dtdx =

∫v(x)ϕ(t, x)dtdx,

or, in other words,

limτ→+∞

u(t+ τ, x) = v(x) in D′(R+ × Ω)

(here v(x) should be understood as 1 ⊗ v, i.e., as the right-hand side of(7.8)). Applying the heat operator ∂

∂t − a2∆x to the equality above, we caninterchange this operator with the passage to the limit (this can be done inthe space of distributions; for example, in our case everything is reduced toreplacing ϕ(t, x) in (7.8) by another function (− ∂

∂t−a2∆x)ϕ(t, x) ∈ D(R+×Ω)). Then ( ∂∂t − a2∆x)v(x) = 0, i.e. v(x) is a distributional solution of∆v(x) = 0. But, as we know, v(x) is then a regular harmonic function, asrequired.

7.4. A solution of the Cauchy problem for the heat equation

and Poisson’s integral

Let us investigate the Cauchy problem (7.2). First, assume that ϕ(x) ∈S(Rn) and u(t, x) ∈ S(Rn

x) for any t ≥ 0, so that u(t, x) is a continuousfunction of t for t ≥ 0 with values in S(Rn

x) and, moreover, C∞-function oft for t > 0 with values in S(Rn

x).

As we will see below, a solution u(t, x) with the described propertiesexists (and is unique in a very broad class of solutions). For the time being,note that we may perform the Fourier transform in x and since the Fourier

7.4. Poisson’s integral 181

transform is a topological isomorphism F : S(Rnx) −→ S(Rn

ξ ), it commuteswith ∂

∂t . Thus, let

u(t, ξ) =∫e−ix·ξu(t, x)dx

(hereafter, integration is performed over the whole Rn unless otherwise in-dicated).

The standard integration by parts gives∫e−ix·ξ

(∂u

∂t− a2∆xu

)dx =

[∂

∂t+ a2|ξ|2

] ∫e−ix·ξu(t, x)dx =

∂u

∂t+a2|ξ|2u.

Therefore (7.1) is equivalent to

(7.9)∂u(t, ξ)∂t

+ a2|ξ|2u(t, ξ) = 0.

The initial condition takes the form

(7.10) u|t=0 = ϕ(ξ), where ϕ = Fϕ.

Equation (7.9) is an ordinary differential equation with the independentvariable t and with a parameter ξ. It is easy to solve it, and we get with(7.10):

u(t, ξ) = e−ta2|ξ|2ϕ(ξ).

From this formula we see that u(t, ξ) is actually a (continuous for t ≥ 0 andC∞ for t > 0) function of t with values in S(Rn

ξ ). Therefore, applying theinverse Fourier transform, we get a solution u(t, x) of the Cauchy problemsatisfying the above conditions.

Let us derive an explicit formula. We have

u(t, x) = (2π)−n∫eix·ξu(t, ξ)dξ = (2π)−n

∫eix·ξ−ta

2|ξ|2ϕ(ξ)dξ

= (2π)−n∫ei(x−y)·ξ−ta2|ξ|2ϕ(y)dydξ.

Interchanging the order of integration (by Fubini’s theorem) we get

u(t, x) =∫

Γ(t, x− y)ϕ(y)dy,

where

(7.11) Γ(t, x) = (2π)−n∫eix·ξ−ta

2|ξ|2dξ

(we have essentially repeated the argument proving that the Fourier trans-form turns multiplication into convolution). Let us express Γ(t, x) explicitly,


computing the integral in (7.11). For this, complete the square in the expo-nent:

ix · ξ − ta2|ξ|2 = −ta2ξ · ξ + ix · ξ = −ta2

(ξ − ix

2ta2

)·(ξ − ix

2ta2

)− |x|

2

4ta2.

The change of coordinates ξ − ix2ta2 = η, i.e., the shift of the contour of

integration over each ξj , yields:

Γ(t, x) = (2π)−ne−|x|2

4ta2

∫e−ta

2|η|2dη = (2π)−ne−|x|2

4ta2 (a√t)−n

∫e−|ξ|

2dξ

(we have also made the change of variables ξ = a√tη).

Using the formula ∫e−|ξ|

2dξ = πn/2,

we finally get

Γ(t, x) = (2a√πt)−n exp

(− |x|

2

4ta2

).

The solution of the Cauchy problem is expressed in the form

(7.12) u(t, x) =1

(2a√πt)n

∫e−|x−y|2

4a2t ϕ(y)dy,

called Poisson’s integral.

Let us analyze this formula. First, observe that the formula makes sensefor a larger class of initial functions ϕ(x) than the class S(Rn) from whichwe have started. For instance, it is clear that if ϕ(x) is continuous and

(7.13) |ϕ(x)| ≤ Ceb|x|2 , b > 0,

then the integral (7.12) converges if 14a2t

> b, i.e., for 0 < t < 14a2b

. Here theintegral itself and integrals obtained from it after differentiation with respectto t or x any number of times uniformly converge for t ∈ [A,B], |x| ≤ R,where 0 < A < B < 1

4a2b. Therefore, we may differentiate any number of

times under the integral sign. The function u(t, x) is a solution of the heatequation (7.1) for the indicated values of t. Indeed, it suffices to show thatΓ(t, x) is a solution of (7.1) for t > 0. This can be directly verified, but toavoid this calculation we may also note that, as we already know,

0 =(∂

∂t− a2∆x

)u(t, x) =

∫ (∂

∂t− a2∆x

)Γ(t, x− y)ϕ(y)dy

for any ϕ ∈ S(Rn), implying(∂

∂t− a2∆x

)Γ(t, x− y) = 0, t > 0.


It is easy to verify that if ϕ is continuous and satisfies (7.13) and u is givenby (7.12), then u|t=0 = ϕ(x) in the sense that

(7.14) limt→0+

u(t, x) = ϕ(x), for any x ∈ Rn.

We will even prove this in two ways.

1st method. Note that Γ(t, x) = ε−nf(xε ), where ε = 2a√t and

f(x) = π−n/2 exp(−|x|2), so that f(x) ≥ 0,∫f(x)dx = 1 and ε → 0+

as t→ 0+. So we deal with a classical δ-like family of positive functions (inparticular, Γ(t, x)→ δ(x) in D′(Rn) as t→ 0+). Now, (7.14) is verified viathe standard scheme of the proof of δ-likeness, see Example 4.5. We skipthe corresponding details; the reader should consider this a recommendedexercise.

2nd method. Suppose we wish to verify (7.14) for x = x0. Let usdecompose ϕ(x) into the sum ϕ(x) = ϕ0(x) + ϕ1(x), where ϕ0(x) is con-tinuous, coincides with ϕ(x) for |x− x0| ≤ 1 and vanishes for |x− x0| ≥ 2.Accordingly, ϕ1(x) is continuous, vanishes for |x − x0| ≤ 1 and satisfies(7.13). Poisson’s integral (7.12) for ϕ splits into the sum of similar integralsfor ϕ0 and ϕ1 (we denote them by u0(t, x) and u1(t, x)). It suffices to verify(7.14) for u0(t, x) and u1(t, x) separately.

First, let us deal with u1(t, x). We have

u1(t, x0) =∫|y−x0|≥1

Γ(t, x0 − y)ϕ1(y)dy.

Clearly, the integrand tends to zero as t → 0+ (since limt→+0

Γ(t, x) = 0

for x 6= 0) and is majorized for 0 < t ≤ B < 14a2b

, by an independent of tfunction of the form C1 exp(−ε|y|2), where ε > 0, C1 > 0. By the dominatedconvergence theorem lim

t→+0u1(t, x0) = 0.

It remains to consider the case of an initial function with a compact sup-port. Here the argument of Example 4.5 will do without any modifications.There is, however, another method based on approximation by smooth func-tions with a subsequent passage to the limit. It is based on the followingimportant

Lemma 7.3. (The maximum principle) Let a function u(t, x) be definedby Poisson’s integral (7.12). Then

(7.15) infx∈Rn

ϕ(x) ≤ u(t, x) ≤ supx∈Rn

ϕ(x).


If one of these inequalities turns into the equality for some t > 0 and x ∈ Rn,then u = const.

Proof. We have

u(t, x) =∫

Γ(t, x− y)ϕ(y)dy ≤ supx∈Rn

ϕ(x) ·∫

Γ(t, x− y)dy = supx∈Rn

ϕ(x).

The second inequality in (7.15) is proved similarly. Since Γ(t, x) > 0 forall t > 0, x ∈ Rn, then the equality is only possible if ϕ(x) = const almosteverywhere, and then u = const as required.

Remark 7.4. As we will see later, a solution u(t, x) of the Cauchy problem(7.2), satisfying the growth restriction of type (7.13) with respect to x uni-formly in t, is unique. It will follow that the solution u(t, x) in fact can berepresented by the Poisson integral, hence satisfies the maximum principle(7.15).

Let us finish the proof of (7.14). Let ϕ be a continuous function witha compact support; ϕk, k = 1, 2, . . . , a sequence of functions from C∞0 (Rn)such that

supx∈Rn

|ϕ(x)− ϕk(x)| → 0 for k → +∞.

(It is possible to obtain a sequence ϕk, for example, via mollifying, cf. Section4.2.) Let

uk(t, x) =∫

Γ(t, x− y)ϕk(y)dy.

Since ϕk ∈ S(Rn), it follows that limt→+0

uk(t, x) = ϕk(x) in the topology of

S(Rn) (in particular, uniformly on Rn). But by Lemma 7.3

supt>0, x∈Rn

|uk(t, x)− u(t, x)| ≤ supx∈Rn

|ϕk(x)− ϕ(x)|;

this, clearly, implies limt→+0

u(t, x) = ϕ(x). Indeed, given ε > 0, we may

choose k such that supx∈Rn

|ϕk(x) − ϕ(x)| < ε3 , and then t0 > 0 such that

supx∈Rn

|uk(t, x)− ϕk(x)| < ε3 for 0 < t < t0. This clearly implies

|u(t, x)− ϕ(x)| ≤ |u(t, x)− uk(t, x)|+ |uk(t, x)− ϕk(x)|+ |ϕk(x)− ϕ(x)| ≤

2 supx∈Rn

|ϕk(x)− ϕ(x)|+ |uk(t, x)− ϕk(x)| < ε for 0 < t < t0,

as required.


Observe also that Poisson’s integral often makes sense for distributionsϕ as well. For instance, if ϕ ∈ S′(Rn), then Poisson’s integral makes thefollowing natural sense:

u(t, x) = 〈ϕ(y),Γ(t, x− y)〉,

since Γ(t, x − y) ∈ S(Rny ) for any t > 0, x ∈ Rn. It is also easily verified

that u(t, x) is a solution of (7.1). The relation limt→+0

u(t, ·) = ϕ(·) now holds

in the sense of weak convergence in S′(Rn). Indeed, it is easy to verify thatif ψ(x) ∈ S(Rn), then∫

u(t, x)ψ(x)dx =∫〈ϕ(·),Γ(t, x−·)〉ψ(x)dx =

⟨ϕ(·),

∫Γ(t, x− ·)ψ(x)dx

⟩= 〈ϕ(·), v(t, ·)〉,

where v(t, x) is the Poisson integral corresponding to the initial function ψ(x)(we justify the possibility to interchange the integration with respect to x

and application of the functional ϕ by the convergence of the integral in thetopology of S(Rn

y ), cf. an identical argument in the proof of Proposition 5.6).

Since v(t, ·)→ ψ(·) as t→ +0 in the topology of S(Rn), it follows that

limt→+0

∫u(t, x)ψ(x)dx = 〈ϕ,ψ〉,

implying that

limt→+0

u(t, ·) = ϕ(·) in S′(Rn).

We have already noted above that limt→+0

Γ(t, x) = δ(x) in S′(Rn). This

relation is often written in a shorter form

Γ|t=0 = δ(x)

and is a particular case of the statement just proved.

Finally, let (7.13) be true for any b > 0 (with a constant C > 0 dependingon b). Then the Poisson integral and, therefore, solution of the Cauchy prob-lem are defined for any t > 0. This is the case e.g. if |ϕ(x)| ≤ C exp(b0|x|2−ε)for some ε > 0, b0 > 0, and, in particular, if ϕ(x) is bounded.

If ϕ ∈ Lp(Rn) for p ∈ [1,∞), then Poisson’s integral converges byHolder’s inequality. In this case lim

t→+0u(t, ·) = ϕ(·) with respect to the

norm Lp(Rn) (the proof is similar to the argument proving the convergenceof mollifiers with respect to the Lp-norm, see Section 4.2).


It is easy to verify that if ϕ(x) satisfies (7.13), then the function u(t, x)determined by Poisson’s integral satisfies a similar estimate:

|u(t, x)| ≤ C1eb1|x|2 , 0 ≤ t ≤ B <

14a2b

,

where the constants C1, b1 depend on ϕ and B. It turns out that a solutionsatisfying such an estimate is unique; we will prove this later. At the sametime it turns out that the estimate

|u(t, x)| ≤ Ceb|x|2+ε

does not guarantee the uniqueness for any ε > 0. (For the proof see e.g.John [15], Ch.7.)

The uniqueness of solution makes it natural to apply Poisson’s integralin order to find a meaningful solution (as a rule, solutions of fast growth haveno physical meaning). For instance, the uniqueness of a solution takes placein the class of bounded functions. By the maximum principle (Lemma 7.3),the Cauchy problem is well-posed in the class of bounded functions (if ϕis perturbed by adding a uniformly small function, then the same happenswith u(t, x)).

In conclusion notice that the fact that Γ(t, x) > 0 for all t > 0, x ∈ Rn

implies that “the speed of the heat propagation is infinite” (since the initialperturbation is supported at the origin!). But Γ(t, x) decreases very fastas |x| → +∞, which is practically equivalent to a finite speed of the heatpropagation.

7.5. The fundamental solution for the heat operator.

Duhamel’s formula

Theorem 7.5. The following locally integrable function

E(t, x) = θ(t)Γ(t, x) = (2a√πt)−nθ(t) exp

(− |x|

2

4a2t

)is a fundamental solution for ∂

∂t − a2∆x in Rn+1. Here θ(t) is the Heavisidefunction.

Proof. First, let us give a heuristic explanation why E(t, x) is a fundamentalsolution. We saw that Γ(t, x) is a continuous function in t with values inS′(Rn

x) for t ≥ 0. The function E(t, x) = θ(t)Γ(t, x) is not continuous asa function of t with values in S′(Rn

x), but has a jump at t = 0 equal to

7.5. The fundamental solution. Duhamel’s formula 187

δ(x). Therefore, applying ∂∂t , we get δ(t) ⊗ δ(x) + f(t, x), where f(t, x) is

continuous in t in the similar sense. But then(∂

∂t− a2∆x

)E(t, x) = δ(t)⊗ δ(x) = δ(t, x),

since the summand which is continuous with respect to t, vanishes because( ∂∂t − a2∆x)E(t, x) = 0 for |t| + |x| 6= 0. This important argument makes itpossible to write down the fundamental solution in all cases when we canwrite the solution of the Cauchy problem in a Poisson integral-like form. Wewill not bother with its justification, however, since it is simpler to verifydirectly that E(t, x) is a fundamental solution.

We begin with the verification of local integrability of E(t, x). Clearly,E(t, x) ∈ C∞(Rn+1\0). So we only have to verify local integrability in aneighborhood of the origin. Since

exp(−1τ

)≤ Cατα for τ > 0, for any α ≥ 0,

we have

(7.16) |E(t, x)| ≤ Cαt−n/2(

t

|x|2)α

= Cαt−n/2+α|x|−2α.

Let α ∈ (n2 −1, n2 ) so that −n

2 +α > −1 and −2α > −n. It is clear from(7.16) that E(t, x) is majorized by an integrable function in a neighborhoodof the origin; therefore, it is locally integrable itself.

Now, let f(t, x) ∈ C∞0 (Rn+1). We have⟨(∂

∂t− a2∆x

)E(t, x), f(t, x)

⟩=⟨E(t, x),

(− ∂

∂t− a2∆x

)f(t, x)

⟩=

(7.17) = limε→+0

∫t≥εE(t, x)

(− ∂

∂t− a2∆x

)f(t, x)dtdx.

Let us move ∂∂t and ∆x in the integral back onto E(t, x). Since ( ∂∂t −

a2∆x)E(t, x) = 0 for t > 0, the result only contains a boundary integralover the plane t = ε obtained after integrating by parts with respect to t:∫

E(ε, x)f(ε, x)dx =∫

Γ(ε, x)f(ε, x)dx.

The latter integral is uε(ε, 0), where uε(t, x) is Poisson’s integral with theinitial function ϕε(x) = f(ε, x). Consider also Poisson’s integral u(t, x) withthe initial function ϕ(x) = f(0, x). By the maximum principle

|u(ε, 0)− uε(ε, 0)| ≤ supx|ϕε(x)− ϕ(x)| = sup

x|f(ε, x)− f(0, x)|.


This, clearly, implies that

u(ε, 0)− uε(ε, 0)→ 0 for ε→ +0.

But, as we have already seen,

u(ε, 0)→ ϕ(0) = f(0, 0) for ε→ +0,

implying limε→+0

uε(ε, 0) = f(0, 0) which, in turn, yields

limε→+0

∫t≥εE(t, x)

(− ∂

∂t− a2∆x

)f(t, x)dtdx = f(0, 0).

This, together with (7.17), proves the theorem.

Corollary 7.6. If u(t, x) ∈ D′(Ω), where Ω is a domain in Rn+1, and

( ∂∂t − a2∆x)u = 0, then u ∈ C∞(Ω).

Therefore, the heat operator is an example of a hypoelliptic but notelliptic operator. Note that the heat operator has C∞ but non-analyticsolutions (e.g. E(t, x) in a neighborhood of (0, x0), where x0 ∈ Rn\0).

The knowledge of the fundamental solution makes it possible to solve thenon-homogeneous equation. For instance, a solution u(t, x) of the problem

(7.18)

∂u∂t − a2∆xu = f(t, x), t > 0, x ∈ Rn;u|t=0 = ϕ(x), x ∈ Rn;

can be written in the form

u(t, x) =∫τ≥0E(t− τ, x− y)f(τ, y)dτdy +

∫Γ(t, x− y)ϕ(y)dy =

(7.19) =∫ t

0dτ

∫Rn

Γ(t− τ, x− y)f(τ, y)dy +∫

Γ(t, x− y)ϕ(y)dy

under appropriate assumptions concerning f(t, x) and ϕ(x) which we willnot formulate exactly. The first integral in (7.19) is of interest. Its innerintegral

v(τ, t, x) =∫

Γ(t− τ, x− y)f(τ, y)dy

is a solution of the Cauchy problem

(7.20)

( ∂∂t − a2∆x)v(τ, t, x) = 0, t > τ, x ∈ Rn,

v|t=τ = f(τ, x).

7.6. Derivatives of solution of heat equation 189

The Duhamel formula is a representation of the solution of the Cauchyproblem (7.18) with zero initial function ϕ in the form

(7.21) u(t, x) =∫ t

0v(τ, t, x)dτ,

where v(τ, t, x) is the solution of the Cauchy problem (7.20) (for a homoge-neous equation!).

A similar representation is possible for any evolution equation. In par-ticular, precisely this representation will do for operators of the form ∂

∂t −A(x,Dx), where A(x,Dx) is a linear differential operator with respect tox with variable coefficients. Needless to say, in this case a2∆x should bereplaced by A(x,Dx) in (7.20). The formal verification is very simple: byapplying ∂

∂t −A(x,Dx) to (7.21) we get in the right hand side

v(t, t, x) +∫ t

0

(∂

∂t−A(x,Dx)

)v(τ, t, x)dτ = v(t, t, x) = f(t, x),

as required.

7.6. Estimates of derivatives of a solution of the heat

equation

Since the heat operator is hypoelliptic, we can use Proposition 5.23 to esti-mate derivatives of a solution of the heat equation through the solution itselfon a slightly larger domain. We will only need it on an infinite horizontalstrip.

Lemma 7.7. Let us assume that a solution u(t, x) of the heat equation (7.1)is defined in the strip t, x : t ∈ (a, b), x ∈ Rn and satisfies the estimate

(7.22) |u(t, x)| ≤ Ced|x|p ,

where p ≥ 0, C > 0, d ∈ R. Then in a somewhat smaller strip

t, x : t ∈ (a′, b′), x ∈ Rn, where a < a′ < b′ < b,

similar estimates hold:

(7.23) |∂αu(t, x)| ≤ Cαed′|x|p ,

where the constant p is the same as in (7.22); d′ = 0 if d = 0 and d′ = d+ ε

with an arbitrary ε > 0 if d 6= 0; Cα depend upon the (n + 1)-dimensionalmulti-index α and also upon ε, a′ and b′.


Proof. Let us use Proposition 5.23 with

Ω1 = t, x : t ∈ (a, b), |x| < 1, Ω2 = t, x : t ∈ (a′, b′), |x| < 1/2,and estimate (5.23) applied to the solution uy(t, x) = u(t, x+ y) of the heatequation depending on y ∈ Rn as on a parameter. We get

|∂αu(t, x)| ≤ Cα sup|x′−x|<1

|u(t, x′)| ≤ C ′α sup|x′−x|<1

ed|x′|p ≤ C ′ed′|x|p

for t ∈ (a′, b′), as required.

7.7. Holmgren’s principle. The uniqueness of solution of the

Cauchy problem for the heat equation

Holmgren’s principle is, roughly speaking, the fact that the uniqueness ofthe solution of the abstract Cauchy problem for the equation

(7.24)du

dt= A(t)u, t > 0,

with the initial condition u(0) = u0, follows from the existence of a solutionof the adjoint Cauchy problem directed opposite to the time axis:

(7.25)

dvdt = −A′(t)v, t < t0,

v|t=t0 = ψ.

Here u = u(t), v = v(t) are functions of t with values in vector spaces E andE′ respectively, with a pairing 〈·, ·〉, i.e., a bilinear form 〈·, ·〉 : E×E′ −→ C.This pairing should be non-degenerate, in the sense that if u ∈ E and〈u, ψ〉 = 0 for all ψ ∈ E′, then u = 0. Further, A(t) and A′(t) are linearoperators in E and E′ respectively, which are assumed to be adjoint to eachother with respect to the given pairing, i.e.

〈Au, v〉 = 〈u,A′v〉, u ∈ E, v ∈ E′.

To define the derivatives dudt and dv

dt , we need some topology in E andE′ but we will not give a precise meaning to these derivatives. Instead, wewill describe the scheme of the proof of the uniqueness of the solution of theCauchy problem for the equation (7.24). This scheme will serve as a basisof such a proof in any concrete situation. Therefore, we postpone attachinga precise meaning to the subsequent arguments until a concrete situationarises.

So, let u(t) be a solution of (7.24) defined for 0 < t < T and having zeroinitial value u(0) = 0. We want to prove that u(t) ≡ 0. To this end consider

7.7. Holmgren’s principle. The uniqueness of solution 191

for some t0 ∈ (0, T ) a solution v(t) of the adjoint Cauchy problem (7.25).We have

d

dt〈u(t), v(t)〉 =

⟨du(t)dt

, v(t)⟩

+⟨u(t),

dv(t)dt

⟩= 〈A(t)u(t), v(t)〉 − 〈u(t), A′(t)v(t)〉 = 0, 0 < t < t0.

This main calculation, which needs subsequent justifying, explains why theadjoint Cauchy problem appears. The formulas yield:

〈u(t), v(t)〉 = const = 〈u(t0), ψ〉 = 〈u(0), v(0)〉 = 0

(passing to the limit as t → t0 and t → 0 should be also justified). Now, ifthe problem (7.25) is solvable (and the solution has properties which enableus to perform all the preceding calculations) for a class of initial values ψsuch that 〈u, ψ〉 = 0 for all ψ implies u = 0 then, clearly, u(t0) = 0, but thenu(t) ≡ 0, since t0 is arbitrary.

Now, armed with the abstract scheme of Holmgren’s principle we willprove the uniqueness of the solution of the Cauchy problem for the heatequation.

Theorem 7.8 (A.N.Tikhonov, 1935). Let a function u(t, x) be continuousin the strip [0, T ) × Rn, satisfy the heat equation (7.1) , and satisfy theestimate

(7.26) |u(t, x)| ≤ Ceb|x|2

in the open strip (0, T )× Rn. If u(0, x) = 0 for all x ∈ Rn, then u(t, x) ≡ 0in the whole strip.

Proof. Consider the adjoint Cauchy problemdv(t,x)dt = −a2∆xv(t, x), t < t0,

v|t=t0 = ψ(x),

where ψ ∈ D(Rn). Let us express the solution of this problem via Poisson’sintegral:

v(t, x) =∫

Γ(t0 − t, x− y)ψ(y)dy.

Since ψ has a compact support, this solution satisfies

|v(t, x)| ≤ Cε supy∈suppψ

exp(− |x− y|2

4a2(t0 − t)

)≤ C ′ε exp

(− c|x|2

4a2(t0 − t)

),

for t < t0 − ε, where ε > 0, c > 0, |x| ≥ R, and R is sufficiently large.


The same estimate holds also for any derivative ∂αv(t, x), where α is a(n+ 1)-dimensional multi-index. Let t0 > 0 be so small that c

4a2t0> b, i.e.,

t0 <c

4a2b. Then the integral

〈u(t, x), v(t, x)〉 =∫u(t, x)v(t, x)dx

makes sense and converges uniformly for 0 ≤ t ≤ t0 − ε. As Lemma 7.7shows, the integrals obtained by replacing u or v by their derivatives withrespect to t or x also converge uniformly for 0 < ε ≤ t ≤ t0− ε, where ε > 0is arbitrary. Therefore, the function χ(t) = 〈u(t, x), v(t, x)〉 is continuous att ∈ [0, t0), vanishes at t = 0 and its derivative with respect to t at t ∈ (0, t0)is

dχ(t)dt

=∫∂(u(t, x)v(t, x))

∂tdx =

∫ (∂u

∂t· v + u · ∂v

∂t

)dx =

= a2

∫[(∆xu)v − u(∆xv)]dx.

The latter integral vanishes, since, by Green’s formula (4.32), it may beexpressed in the form

limR→∞

∫|x|≤R

[(∆xu)v − u(∆xv)]dx =

= limR→∞

∫|x|=R

(∂u

∂n· v − u · ∂v

∂n

)dS = 0,

because ∂u∂n ·v and u· ∂v∂n tend to 0 exponentially fast as |x| → +∞. Therefore,

χ(t) = const = 0.

Now, let us verify that∫u(t0, x)ψ(x)dx = lim

t→t0−0

∫u(t, x)v(t, x)dx.

(Actually, the right hand side equals limt→t0−0

χ(t) = 0.) We have∫u(t, x)v(t, x)dx =

∫u(t, x)Γ(t0 − t, x− y)ψ(y)dydx =

=∫ (∫

Γ(t0 − t, x− y)u(t, x)dx)ψ(y)dy →

∫u(t0, y)ψ(y)dy, as t→ t0 − 0,

since the argument we used in the consideration of Poisson’s integral showsthat the convergence

limt→t0−0

∫Γ(t0 − t, x− y)u(t, x)dx = u(t0, y)

7.8. Fourier method 193

is uniform on any compact in Rny . Thus, clearly,∫

u(t0, x)ψ(x)dx = 0, 0 ≤ t0 ≤ δ, ψ ∈ D(Rn),

if δ < c4a2b

. Therefore, u(t, x) ≡ 0 for 0 ≤ t ≤ δ.But now we may pass to defining initial conditions for t = δ and prove

that u(t, x) = 0 already for 0 ≤ t ≤ 2δ, etc. Finally, we get u(t, x) ≡ 0 onthe whole strip, as required.

Thus, in the class of functions satisfying (7.26), a solution of the Cauchyproblem for the heat equation is unique.

Corollary 7.9. Let u = u(t, x) be a solution of the Cauchy problem (7.2) in[0, T )×Rn and u satisfies the growth restriction (7.26). Then the maximumprinciple (7.15) holds for all t ∈ [0, T ) and x ∈ Rn.

Proof. Let us prove the first inequality in (7.15). (The second one is easilyreduced to the first if we replace u by −u and ϕ by −ϕ.) Note that the resultis obvious if inf

x∈Rnϕ(x) = −∞. So from now on assume that the infimum is

finite, that is, ϕ is bounded below. But then, adding a constant, we canreduce the general statement to the case when the infimum of ϕ is 0, inparticular, ϕ ≥ 0, and we have to prove that u ≥ 0 everywhere. Due to theuniqueness theorem, u(t, x) can be presented by Poisson’s integral (7.12) forsmall t, 0 < t < 1

4a2b. Then, clearly, u(t, x) ≥ 0 for such t, due to positivity

of the Poisson kernel Γ. But then, moving by small steps (e.g. 1/(8a2b)) int, we obtain that u ≥ 0 for all t ∈ [0, T ).

7.8. A scheme of solving the first and second

initial-boundary value problems by the Fourier method

Let us try to solve the first initial-boundary value problem with zero bound-ary conditions

(7.27)

∂u∂t = a2u, t > 0, x ∈ Ω,u|x∈∂Ω = 0, t > 0, x ∈ ∂Ω,u|t=0 = ϕ(x), x ∈ Ω

by the Fourier method. Here we assume Ω to be a bounded domain in Rn.If the function u(t, x) = ψ(t)v(x) satisfies the first two conditions in (7.27),we formally get

(7.28)ψ′(t)a2ψ(t)

=∆v(x)v(x)

= −λ, v|∂Ω = 0,


leading to the eigenvalue problem for the operator −∆ with zero boundaryconditions on ∂Ω:

(7.29)

−∆v = λv, x ∈ Ω,v|∂Ω = 0.

For n = 1 this problem becomes the simplest Sturm-Liouville problemconsidered above (with the simplest boundary conditions: v should vanishat the endpoints of the segment). We will prove below that problem (7.29)possesses properties very similar to those of the Sturm-Liouville problem.In particular, it has a complete orthogonal system of eigenfunctions vk(x),k = 1, 2, . . ., with eigenvalues λk, such that λk > 0 and λk → +∞ ask → +∞.

It follows from (7.28) that ψ(t) = Ce−λa2t which makes it clear that it

is natural to seek u(t, x) as a sum of the series

u(t, x) =∞∑k=1

Cke−λka2tvk(x).

The coefficients Ck are chosen from the initial condition u|t=0 = ϕ(x) andare coefficients of the expansion of ϕ(x) in terms of the orthogonal systemvk(x)∞k=1.

The second initial-boundary value problem∂u∂t = a2∆u, t > 0, x ∈ Ω,∂u∂n |x∈∂Ω = 0, t > 0, x ∈ ∂Ω,u|t=0 = ϕ(x), x ∈ Ω

is solved similarly. It leads to the eigenvalue problem

(7.30)

−∆v = λv, x ∈ Ω∂v∂n |∂Ω = 0

with the same properties as the problem (7.29), the only exception beingthat λ = 0 is an eigenvalue of (7.30).

The same approach, as for n = 1, yields solutions of more general prob-lems, for example,

∂u∂t = a2∆u+ f(t, x), t > 0, x ∈ Ω,u|x∈∂Ω = α(t, x), t > 0, x ∈ ∂Ω,u|t=0 = ϕ(x), x ∈ Ω.

7.9. Problems 195

For this, subtracting an auxiliary function, we make α(t, x) ≡ 0 and thenseek the solution u(t, x) in the form of a series

u(t, x) =∞∑k=1

ψk(t)vk(x),

which leads to first order ordinary differential equations for ψk(t).

The uniqueness of the solution of these problems may be proved byHolmgren’s principle or derived from the maximum principle: if u(t, x) is a

function continuous in the cylinder [0, T ]× Ω and satisfies the heat equation

on (0, T )× Ω, then

(7.31) u(t, x) ≤ max

supx∈Ω

u(0, x), supx∈∂Ω,t∈[0,T ]

u(t, x)

;

and if the equality takes place then u = const.

We will not prove this physically natural principle. Its proof is quiteelementary and may be found, for instance, in Petrovskii [21].

In what follows, we give the proof of existence of a complete orthogonalsystem of eigenfunctions for the problem (7.29). The boundary conditionin (7.29) will be understood in a generalized sense. The Dirichlet problemwill also be solved in a similar sense. All this requires introducing Sobolevspaces whose theory we are about to present.

7.9. Problems

7.1. Given a rod of length l with a heat-insulated surface (the lateral sidesand ends). For t = 0, the temperature of the left half of the rod is u1 andthat of the right half is u2. Find the rod temperature for t > 0 by theFourier method. Find the behavior of the temperature as t → +∞. Esti-mate the relaxation time needed to decrease by half the difference betweenthe maximal and minimal temperatures of the rod; in particular, find thedependence of this time on the parameters of the rod: length, thickness,density, specific heat and thermal conductivity.

7.2. The rod is heated by a constant electric current. Zero temperature ismaintained at the ends and the side surface is insulated. Assuming that thetemperature in any cross-section of the rod is a constant, find the distribu-tion of the temperature in the rod from an initial distribution and describethe behavior of the temperature as t→ +∞.


7.3. Let a bounded solution u(t, x) of the heat equation ut = a2uxx (x ∈ R1)be determined for t > 0 and satisfy the initial condition u|t=0 = ϕ(x), whereϕ ∈ C(R1), lim

x→+∞ϕ(x) = b, lim

x→−∞ϕ(x) = c. Describe the behavior of u(t, x)

as t→ +∞.

7.4. Prove the maximum principle for the solutions of the heat equation inthe formulation given at the end of this section (see (7.31)).

Chapter 8

Sobolev spaces.

A generalized solution

of Dirichlet’s problem

8.1. Spaces Hs(Ω)

Let s ∈ Z+, Ω an open subset in Rn.

Definition 8.1. The Sobolev space Hs(Ω) consists of u ∈ L2(Ω) such that∂αu ∈ L2(Ω) for |α| ≤ s, where α is a multiindex, and the derivative isunderstood in the distributional sense.

In Hs(Ω), introduce the inner product

(u, v)s =∑|α|≤s

(∂αu, ∂αv),

where (·, ·) is the inner product in L2(Ω), and also the corresponding norm

(8.1) ‖u‖s = (u, u)12s =

∑|α|≤s

∫Ω|∂αu(x)|2dx

12

.

Clearly, Hs(Ω) ⊂ Hs′(Ω) for s ≥ s′. Furthermore, H0(Ω) = L2(Ω). Inparticular, each space Hs(Ω) is embedded into L2(Ω).

197

198 8. Sobolev spaces. Dirichlet’s problem

Proposition 8.2. The inner product (u, v)s defines in Hs(Ω) the struc-ture of a separable Hilbert space (i.e. a Hilbert space which admits at mostcountable orthonormal basis).

Proof. Only the completeness and separability are not obvious. Letus prove the completeness. Let a sequence uk∞k=1 be a Cauchy sequencein Hs(Ω). The definition of ‖ · ‖s implies that all the sequences ∂αuk∞k=1

(for |α| ≤ s) are Cauchy sequences in L2(Ω). But then, by the completenessof L2(Ω), they converge, i.e., lim

k→∞∂αuk = vα with respect to the norm in

L2(Ω). In particular, uk → v0 in L2(Ω). But then ∂αuk → ∂αv0 in D′(Ω)and since, on the other hand, ∂αuk → vα in D′(Ω), it follows that vα = ∂αv0.Therefore, v0 ∈ Hs(Ω) and ∂αuk → ∂αv0 in L2(Ω) for |α| ≤ s. But thismeans that uk → v0 in Hs(Ω).

Let us prove separability. The map u 7→ ∂αu|α|≤s defines an isometricembedding of Hs(Ω) into the direct sum of several copies of L2(Ω) as aclosed subspace. Now, the separability of Hs(Ω) follows from that of L2(Ω).

Let us consider the case Ω = Rn in more detail. The space Hs(Rn) canbe described in terms of the Fourier transform (understood in the sense ofdistributions, in S′(Rn)).

Recall that, by Plancherel’s theorem, the Fourier transformation

F : u(x) 7→ u(ξ) =∫e−ix·ξu(x)dx

determines an isometry of L2(Rnx) onto L2(Rn

ξ ), where L2(Rnξ ) is defined by

the measure (2π)−ndξ. Let us explain the meaning of this statement and itsinterpretation from distributional viewpoint. We know from Analysis thatthe Parseval identity

(8.2)∫|u(x)|2dx = (2π)−n

∫|u(ξ)|2dξ

holds for u ∈ S(Rn). The identity (8.2) for u ∈ S(Rn) follows easily, forinstance, from the inversion formula (5.24). Indeed,∫

|u(ξ)|2dξ =∫ (∫

e−ix·ξu(x)dx)(∫

eiy·ξu(y)dy)dξ

=∫ei(y−x)·ξu(x)u(y)dxdydξ =

∫u(y)

[∫ei(y−x)·ξu(x)dxdξ

]dy

= (2π)n∫|u(y)|2dy.

8.1. Spaces Hs(Ω) 199

Since F defines an isomorphism of S(Rnx) onto S(Rn

ξ ) and S(Rn) is densein L2(Rn), then F can be extended to an isometry F : L2(Rn

x) −→ L2(Rnξ ).

Since the convergence in L2(Rn) implies a (weak) convergence in S′(Rn),the map F : L2(Rn

x) −→ L2(Rnξ ) is continuous in restrictions of the weak

topologies of S′(Rnx) and S′(Rn

ξ ). Therefore, this F is, actually, a restrictionof the generalized Fourier transform F : S′(Rn

x) −→ S′(Rnξ ).

Now, note that the Fourier transform maps Dαu(x) to ξαu(ξ). There-fore, u ∈ Hs(Rn) if and only if ξαu(ξ) ∈ L2(Rn) for |α| ≤ s. But insteadof this, we may write

∑|α|≤s |ξα|2|u(ξ)|2 ∈ L1(Rn). Due to the obvious

estimatesC−1(1 + |ξ|2)s ≤

∑|α|≤s

|ξα|2 ≤ C(1 + |ξ|2)s,

where C > 0 does not depend on ξ, we have u ∈ Hs(Rn) if and only if(1 + |ξ|2)s/2u(ξ) ∈ L2(Rn) and the norm (8.1) for Ω = Rn is equivalent tothe norm

(8.3) ‖u‖s =[∫

(1 + |ξ|2)s|u(ξ)|2dξ] 1

2

.

We will denote the norm (8.3) by the same symbol as the norm (8.1); thereshould not be any misunderstanding. Thus, we have proved

Proposition 8.3. The space Hs(Rn) consists of u ∈ S′(Rn) such that (1 +|ξ|2)s/2u(ξ) ∈ L2(Rn). It can be equipped with the norm (8.3) which isequivalent to the norm (8.1).

This proposition may serve as a basis for definition of Hs(Rn) for anyreal s. Namely, for any s ∈ R we may construct the space of u ∈ S′(Rn),such that u(ξ) ∈ L2

loc(Rn) and (1 + |ξ|2)s/2u(ξ) ∈ L2(Rn), with the norm(8.3). We again get a separable Hilbert space denoted also, as for s ∈ Z+,by Hs(Rn).

In what follows we will need the Banach space Ckb (Rn) (here k ∈ Z+)consisting of functions u ∈ Ck(Rn), whose derivatives ∂αu(x), |α| ≤ k, arebounded for all x ∈ Rn. The norm in Ckb (Rn) is defined by the formula

‖u‖(k) =∑|α|≤k

supx∈Rn

|∂αu(x)|.

Theorem 8.4. (Sobolev’s embedding theorem) For s > n2 + k, we have a

continuous embedding

(8.4) Hs(Rn) ⊂ Ckb (Rn).


Let us clarify the formulation. At the first glance, it seems that it ismeaningless, since a function u ∈ Hs(Rn) can be arbitrarily modified on anyset of measure 0 without affecting the corresponding distribution (and theelement of Hs(Rn)). Changing u at all points with rational coordinates wemay make it discontinuous everywhere. Therefore, we should more preciselydescribe the embedding (8.4), as follows: if u ∈ Hs(Rn), then there existsa unique function u1 ∈ Ckb (Rn) which coincides with the initial function u

almost everywhere (and, for brevity we write u instead of u1). In otherwords, Hs(Rn) is a subspace of Ckb (Rn) provided both are considered assubspaces of S′(Rn) or D′(Rn).

Observe that the uniqueness of a continuous representative is obvious,since in any neighborhood of x0 ∈ Rn there exist points of any set of to-tal measure, so that any modification of a continuous function on a set ofmeasure 0 leads to a function which is discontinuous at all points where themodification took place.

Proof of Theorem 8.4. First, let us prove the estimate

(8.5) ‖u‖(k) ≤ C‖u‖s, u ∈ S(Rn),

where C is a constant independent of u. The most convenient way is toapply the Fourier transform. We have

∂αu(x) =1

(2π)n

∫(iξ)αeix·ξu(ξ)dξ,

implying

|∂αu(x)| ≤ 1(2π)n

∫|ξαu(ξ)|dξ;

hence,

‖u‖(k) ≤ C∫

(1 + |ξ|2)k/2|u(ξ)|dξ.

Dividing and multiplying the integrand by (1 + |ξ|2)s/2, we have by theCauchy-Schwarz inequality:

‖u‖(k) ≤ C∫

(1 + |ξ|2)(k−s)/2(1 + |ξ|2)s/2|u(ξ)|dξ ≤

≤ C(∫

(1 + |ξ|2)(k−s)dξ

) 12

·(∫

(1 + |ξ|2)s/2|u(ξ)|2dξ) 1

2

.

The integral in the first factor in the right hand side converges, since 2(k −s) < −n and, therefore, the integrand decays faster than |ξ|−n−ε as |ξ| →+∞ for a sufficiently small ε > 0. The second factor is equal to ‖u‖s.Therefore, for s > k + n

2 we get (8.5).


Now, note that S(Rn) is dense in Hs(Rn) for any s ∈ R. Indeed, in-troduce the operator Λs which acts on u = u(x) by multiplying the Fouriertransform u(ξ) by (1 + |ξ|2)s/2. This operator is a unitary map of Hs(Rn)onto L2(Rn) mapping S(Rn) isomorphically onto S(Rn). Therefore, S(Rn)is dense in Hs(Rn), because S(Rn) is dense in L2(Rn).

Now, let v ∈ Hs(Rn), um ∈ S(Rn), um → v in Hs(Rn). But (8.5) impliesthat the sequence um is a Cauchy sequence with respect to the norm ‖ · ‖(k).Therefore, um → v1 in Ckb (Rn). But then v and v1 coincide as distributionsand, therefore, almost everywhere. By continuity, (8.5) holds for all u ∈Hs(Rn) proving the continuity of the embedding (8.4).

In particular, Theorem 8.4 shows that it makes sense to talk about valuesof functions u ∈ Hs(Rn) at a point for s > n

2 .

Also of interest is the question of a meaning of restriction (or trace) of afunction from a Sobolev space on a submanifold of an arbitrary dimension.We will not discuss this problem in full generality, however, and only touchthe most important case of codimension l. For simplicity, consider only thecase of a hyperplane xn = 0 in Rn. A point x ∈ Rn will be expressed in theform x = (x′, xn), where x′ ∈ Rn−1.

Theorem 8.5 (Sobolev’s theorem on restrictions or traces). If s > 1/2, thenthe restriction operator S(Rn) → S(Rn−1), u 7→ u|xn=0, can be extended toa linear continuous map

Hs(Rn) −→ Hs− 12 (Rn−1).

Proof. Let us express the restriction operator in terms of the Fouriertransform (for u ∈ S(Rn)):

u(x′, 0) =1

(2π)n

∫eix′·ξ′ u(ξ′, ξn)dξ′dξn.

Denoting by F ′ the Fourier transform with respect to x′, we get

F ′u(x′, 0) =1

2π

∫u(ξ′, ξn)dξn.

For brevity, set v(x′) = u(x′, 0), v(ξ′) = F ′v(x′), so that

(8.6) v(ξ′) =1

2π

∫u(ξ′, ξn)dξn.

The needed statement takes on the form of the estimate

(8.7) ‖v‖′s− 1

2

≤ C‖u‖s,


where ‖·‖′s− 1

2

is the norm in Hs− 12 (Rn−1), u ∈ S(Rn) and C does not depend

on u. Let us prove this estimate. We have

(8.8)(‖v‖′

s− 12

)2=∫

(1 + |ξ′|2)s−12 |v(ξ′)|2dξ′.

From (8.6) we deduce for any s > 0:

(8.9) |v(ξ′)|2 =1

4π2

[∫(1 + |ξ|2)−s/2(1 + |ξ|2)s/2u(ξ)dξn

]2

≤ 14π2

∫(1 + |ξ|2)−sdξn ·

∫(1 + |ξ|2)s|u(ξ)|2dξn.

Let us estimate the first factor. We have∫(1 + |ξ|2)−sdξn =

∫(1 + |ξ′|2 + |ξn|2)−sdξn

= (1 + |ξ′|2)−s∫ 1 +

∣∣∣∣∣ ξn√1 + |ξ′|2

∣∣∣∣∣2−s dξn

= (1 + |ξ′|2)−s+12

∫ ∞−∞

(1 + |t|2)−sdt = C(1 + |ξ′|2)−s+12 .

Here we used the fact that s > 12 , which ensures the convergence of the last

integral. Therefore, (8.9) is expressed in the form

|v(ξ′)|2 ≤ C(1 + |ξ′|2)−s+12

∫(1 + |ξ|2)s|u(ξ)|2dξn.

By applying this inequality to estimate |v(ξ′)|2 in (8.8), we get the desiredestimate (8.7).

Remark. Since s > 1/2, the restriction map Hs(Rn) → Hs− 12 (Rn−1),

u 7→ u|xn=0, is actually surjective (see e.g. Hormander [12], Ch.II), sothe statement of the theorem is the exact one (the index s − 1

2 cannot beincreased).

Theorem 8.5 means that if u ∈ Hs(Rn), then the trace u|xn=0 of u onthe hyperplane xn = 0 is well-defined and is an element of Hs− 1

2 (Rn−1). Itis obtained as follows: represent u in the form of the limit u = lim

m→∞um of

functions um ∈ S(Rn) with respect to the norm ‖ · ‖s . Then the restrictionsum|xn=0 have a limit as m → ∞ with respect to the norm ‖ · ‖′

s− 12

in

the space Hs− 12 (Rn−1), and this limit does not depend on the choice of

an approximating sequence.


If we wish to restrict ourselves to integer values of s we may use the factthat Hs− 1

2 (Rn−1) ⊂ Hs−1(Rn−1). Therefore, the trace u|xn=0 of u ∈ Hs(Rn)belongs to Hs−1(Rn−1).

The space Hs(M), where M is a smooth compact manifold (withoutboundary), can also be defined. Namely, we will write u ∈ Hs(M) if for anycoordinate diffeomorphism κ : U −→ Ω (here U is an open subset of M , Ω anopen subset of Rn) and any ϕ ∈ C∞0 (Ω) we have ϕ(uκ−1) ∈ Hs(Rn). Hereuκ−1 is a distribution u pulled back to Ω via κ−1. It is multiplied by ϕ to geta distribution with support in Ω. Therefore, ϕ(u κ−1) ∈ E ′(Ω) ⊂ E ′(Rn);in particular, ϕ(u κ−1) ∈ S′(Rn), and it makes sense to talk about theinclusion ϕ(u κ−1) ∈ Hs(Rn). The localization helps us to prove thefollowing analogue of Theorem 8.5:

if Ω is a bounded domain with smooth boundary ∂Ω, then for s > 12 the

restriction map u 7→ u|∂Ω can be extended from C∞(Ω) to a linear continuousoperator Hs(Ω) −→ Hs− 1

2 (∂Ω).

8.2. Spaces Hs(Ω)

Definition 8.6. The space Hs(Ω) is the closure of D(Ω) in Hs(Ω) (weassume that s ∈ Z+).

Therefore, Hs(Ω) is a closed subspace in Hs(Ω). Hence, it is a separableHilbert space itself.

Clearly, H0(Ω) = H0(Ω) = L2(Ω). But already H1(Ω) does not neces-sarily coincide with H1(Ω). Indeed, if Ω is a bounded domain with a smoothboundary, then, as had been noted above, a function u ∈ H1(Ω) possessesa trace u|∂Ω ∈ L2(∂Ω). However, if u ∈ H1(Ω), then u|∂Ω = 0. Therefore,if, e.g. u ∈ C∞(Ω) and u|∂Ω 6= 0, then u 6∈ H1(Ω).

The space Hs(Ω) is isometrically embedded into Hs(Rn) as well. Thisembedding extends the embedding D(Ω) ⊂ D(Rn). Such an extension isdefined by continuity, since the norm ‖·‖s, considered on D(Ω), is equivalentto the norm in Hs(Rn).

The following theorem on compact embeddings is important.

Theorem 8.7. Let Ω be a bounded domain in Rn, s, s′ ∈ Z+, s > s′. Thenthe embedding Hs(Ω) ⊂ Hs′(Ω) is a compact operator.

H Recall, that a linear operator A : E1 −→ E2, where E1, E2 are Ba-nach spaces, is called compact if the image of the unit ball B1 = x : x ∈


E1, ‖x‖E1 ≤ 1 is a precompact subset in E2. In turn, a set Q in a metricspace E is called precompact (in E) if its closure is compact or, equivalently,if from any sequence of points of Q we can choose a subsequence convergingto a point of E. If E is a complete metric space, then a set Q ⊂ E is pre-compact if and only if it is totally bounded, i.e. for any ε > 0 it has a finiteε-net: a finite set x1, . . . , xN ∈ E, such that

Q ⊂N⋃j=1

B(xj , ε)

(here B(x, r) means open ball in E of radius r centered at x), or, in otherwords, for any point x ∈ Q there exists k ∈ 1, . . . , N such that ρ(x, xk) <ε, where ρ is the metric in E.

Let K be a compact subset in Rn (i.e., K is closed and bounded). Con-sider the space C(K) of continuous complex-valued functions on K. LetF ⊂ C(K). The Arzela-Ascoli theorem gives a criterion of precompactnessof F :

F is precompact if and only if it is uniformly bounded and equicontinu-ous.

Here the uniform boundedness of the set F means the existence of aconstant M > 0, such that supx∈K |f(x)| ≤ M for any f ∈ F ; the equicon-tinuity means that for any ε > 0 there exists δ > 0, such that |x′ − x′′| < δ

with x′, x′′ ∈ K implies |f(x′)− f(x′′)| < ε for any f ∈ F .

Proofs of the above general facts on compactness and the Arzela-Ascolitheorem may be found in any textbook on functional analysis, see, e.g.,Kolmogorov and Fomin [16], Chapter II, Sect. 6 and 7.

Note the obvious corollary of general facts on compactness: for a setQ ⊂ E, where E is a complete metric space, to be precompact it sufficesthat for any ε > 0 there exists a precompact set Qε ⊂ E, such that Qbelongs to the ε-neighborhood of Qε. Indeed, a finite ε/2-net of Qε/2 is anε-net for Q. N

Proof of Theorem 8.7. We will use the mollifying operation (see Proofof Lemma 4.5). We have introduced there a family ϕε ∈ D(Rn), such thatsupp ϕε ⊂ x : |x| ≤ ε, ϕε ≥ 0 and

∫ϕε(x)dx = 1. Then from f ∈ L1

loc(Rn)we may form a mollified family:

fε(x) =∫f(x− y)ϕε(y)dy =

∫f(y)ϕε(x− y)dy.


In Section 4.2 we have formulated and proved a number of important prop-erties of mollifying. Now we will make use of these properties.

First, observe that a set F is precompact in Hs′(Ω) if and only if all thesets Fα = ∂αf : f ∈ F, where α is a multi-index such that |α| ≤ s′, areprecompact in L2(Ω). Further, f ∈ Hs(Ω) clearly implies ∂αf ∈ Hs−|α|(Ω),where |α| ≤ s. Since s > s′, it is clear that all sets Fα belong to a boundedsubset of H1(Ω). Therefore, it suffices to verify the compactness of theembedding H1(Ω) ⊂ L2(Ω).

Let F = u : u ∈ H1(Ω), ‖u‖1 ≤ 1. We should verify that F isprecompact in L2(Ω) (or, which is the same, in L2(Rn)). The idea of thisverification is as follows: consider the set

Fε = fε(x), f ∈ F

obtained by ε-mollifying of all functions from F . Then prove that for a smallε > 0 the set F belongs to an arbitrarily small neighborhood of Fε (withrespect to the norm in L2(Rn)) and Fε is precompact in C(Ω) for a fixedε (hence, precompact in L2(Ω)). The remark before the proof shows thenthat F is precompact in L2(Ω).

Observe also that definition of the distributional derivative easily implies:(∂f

∂xj

)ε

(x) =∫∂f(y)∂yj

ϕε(x− y)dy = −∫f(y)

∂

∂yjϕε(x− y)dy

=∫f(y)

∂ϕε(x− y)∂xj

dy =∂fε(x)∂xj

, for f ∈ H1(Ω).

In other words, distributional differentiation commutes with mollifying. Clearly,we have (by the Cauchy-Schwarz inequality)

|fε(x)| ≤ Cε∫|f(y)|dy ≤ C ′ε

(∫|f(y)|2dy

) 12

, f ∈ F .

Therefore, Fε is uniformly bounded (on Rn) for a fixed ε > 0. Further,by similar reasoning the derivatives ∂fε

∂xj=(∂f∂xj

)ε

are uniformly boundedfor f ∈ F and a fixed ε > 0. Therefore, Fε is equicontinuous. By theArzela-Ascoli theorem, Fε is precompact in C(Rn) (we may assume thatFε ⊂ C(K), where K is a compact neighborhood of Ω in Rn). It followsthat Fε is precompact in L2(Rn) and Fε|Ω = fε|Ω, fε ∈ Fε is precompactin L2(Ω).

It remains to verify that for any δ > 0 there exists ε > 0, such that Fbelongs to the δ-neighborhood of Fε. To this end, let us estimate the norm


of f − fε in L2(Rn). (This norm will be, for the time being, denoted by just‖ · ‖.) First, assume that f ∈ D(Ω). We have

f(x)−fε(x) =∫

[f(x)−f(x−y)]ϕε(y)dy =∫dy

∫ 1

0dt

d

dt[f(x)−f(x−ty)]ϕε(y)

=∫ 1

0dt

∫dy

n∑j=1

yj∂f

∂xj(x− ty)ϕε(y).

Using Minkowski’s inequality (4.53) with p = 2 and the Cauchy-Schwarzinequality for finite sums, we obtain

‖f − fε‖ ≤∫ 1

0dt

∫dy

n∑j=1

|yj |∥∥∥∥ ∂f∂xj (· − ty)

∥∥∥∥ϕε(y)

=∫ 1

0dt

∫dy

n∑j=1

|yj |∥∥∥∥∂f(·)∂xj

∥∥∥∥ϕε(y) =n∑j=1

∥∥∥∥ ∂f∂xj∥∥∥∥∫ ϕε(y)|yj |dy

≤ δ1

n∑j=1

∥∥∥∥ ∂f∂xj∥∥∥∥ ≤ δ1

√n‖f‖1,

where δ1 = supx∈suppϕε

∑nj=1 |xj |. Clearly, δ1 ≤ Cε, hence δ1 → 0 as ε → 0.

Since f ∈ F means ‖f‖1 ≤ 1, it follows that ‖f − fε‖ ≤ δ = δ1√n for

f ∈ F , f ∈ D(Ω).

In the resulting inequality

(8.10) ‖f − fε‖ ≤ C1ε‖f‖1

(which we proved for all f ∈ D(Ω)) we can pass to the limit along a sequencefk ∈ D(Ω), converging to f ∈ H1(Ω) in the H1-norm ‖ · ‖1, to concludethat the inequality holds for all f ∈ H1(Ω). To make it more precise, weneed the estimate ‖fε‖ ≤ ‖f‖ which can be proved as follows:

‖fε‖ =∥∥∥∥∫ f(x− y)ϕε(y)dy

∥∥∥∥ ≤ ∫ ‖f(· − y)‖ϕε(y)dy = ‖f‖,

where we again used Minkowski’s inequality (4.53). If fk ∈ D(Ω), k =1, 2, . . . , are chosen so that ‖f − fk‖1 ≤ 1/k, hence ‖f − fk‖ ≤ 1/k, then‖fε − (fk)ε‖ = ‖(f − fk)ε‖ ≤ 1/k and, therefore,

‖f−fε‖ ≤ ‖f−fk‖+‖fk−(fk)ε‖+‖fε−(fk)ε‖ ≤ C1ε‖fk‖1+2k≤ C1ε‖f‖1+

C1 + 2k

.

8.3. Dirichlet’s integral. The Friedrichs inequality 207

In the resulting inequality we can pass to the limit as k → +∞, to concludethat (8.10) holds for f ∈ H1(Ω) with the same constant C1. It follows thatF lies in C1ε-neighborhood of Fε which ends the proof of Theorem 8.7.

On a compact manifold M without boundary one can prove compactnessof the embedding Hs(M) ⊂ Hs′(M) for any s, s′ ∈ R such that s > s′. Theembedding Hs(Rn) ⊂ Hs′(Rn) is never compact whatever s and s′.

8.3. Dirichlet’s integral. The Friedrichs inequality

Dirichlet’s integral is

D(u) =∫

Ω

n∑j=1

(∂u

∂xj

)2

dx, where u ∈ H1(Ω).

One of the possible physical interpretations of this integral is the potentialenergy for small vibrations of a string (for n = 1), or a membrane (for n = 2),where u(x) is the displacement (assumed to be small) from the equilibrium.(In case of the membrane we should assume that the membrane points onlymove orthogonally to the coordinate plane. Besides, we will assume thatthe boundary of the membrane is fixed, i.e. it can not move.)

Taking any displacement function u, let us try to understand when u

represents an equilibrium. In case of mechanical systems with finitely manydegrees of freedom (see Sect.2.5) the equilibriums are exactly the critical(or stationary) points of the potential energy. It is natural to expect thatthe same is true for systems with infinitely many degrees of freedom, suchas a string or a membrane. So the equilibrium property means vanishingof the variation of the Dirichlet integral under infinitesimal deformationspreserving u|∂Ω, i.e., the Euler equation for the functional D(u). We will seethat it is just the Laplace equation. For the string this had been actuallyverified in Section 2.1, where string’s Lagrangian was given. Let us verifythis in the multidimensional case.

H Let us take, for simplicity, u ∈ C∞(Ω), δu ∈ C∞(Ω) and δu|∂Ω = 0,where Ω is assumed to be bounded with a smooth boundary. Let us writeexplicitly the condition for D(u) to be stationary, δD(u) = 0, where

δD(u) =d

dtD(u+ t δu)

∣∣∣∣t=0

.


(See Sect. 2.5.) We have

δD(u) = 2n∑j=1

∫Ω

∂u

∂xj

∂

∂xj(δu)dx

= −2∫

Ω

n∑j=1

∂2u

∂x2j

δudx = −2∫

Ω∆u(x) · δu(x)dx.

(The integral over ∂Ω appearing due to integration by parts vanishes sinceδu|∂Ω = 0.) Thus, Dirichlet’s integral for u ∈ C∞(Ω) and fixed u|∂Ω isstationary if and only if ∆u = 0. Therefore, the Dirichlet problem forthe Laplace equation can be considered as a minimization problem for theDirichlet integral in the class of functions such that u|∂Ω = ϕ, where ϕ is afixed function on ∂Ω.

The Laplace equation ∆u(x) = 0 is also easy to derive for u ∈ H1(Ω),if u is a stationary point of Dirichlet’s integral for permissible variationsδu ∈ C∞0 (Ω). The same argument fits, but integration by parts should bereplaced by the definition of derivative.

Note that, in fact, the stationary points of Dirichlet’s integral are thepoints of minima (this is clear from what will follow). Therefore, Dirichlet’sproblem may be considered as a minimization problem for Dirichlet’s integralin the class of functions with fixed boundary values. N

Proposition 8.8. Let Ω be a bounded domain in Rn. Then there exists aconstant C > 0, such that

(8.11)∫

Ω|u(x)|2dx ≤ C

∫Ω

n∑j=1

∣∣∣∣∂u(x)∂xj

∣∣∣∣2 dx, u ∈ H1(Ω).

This inequality is called the Friedrichs inequality.

Proof. Both parts of (8.11) continuously depend on u ∈ H1(Ω) in thenorm of H1(Ω). Since H1(Ω) is the closure of D(Ω) in H1(Ω), it is clearthat it suffices to prove (8.11) for u ∈ D(Ω) with C independent of u.

Let Ω be contained in the strip x : 0 < xn < R (this can always beachieved by taking a sufficiently large R and shifting Ω in Rn with the helpof a translation that does not affect (8.11)). For u ∈ D(Ω) we have

u(x) = u(x′, xn) =∫ xn

0

∂u

∂xn(x′, t)dt.

8.4. Dirichlet’s problem (generalized solutions) 209

By the Cauchy–Schwarz inequality, this implies

|u(x)|2 ≤ |xn| ·∫ xn

0

∣∣∣∣ ∂u∂xn (x′, t)∣∣∣∣2 dt ≤ R ∫ R

0

∣∣∣∣∂u(x)∂xn

∣∣∣∣2 dxn.Integrating this inequality over Ω, we get∫

Ω|u(x)|2dx ≤ R2

∫Ω

∣∣∣∣ ∂u∂xn∣∣∣∣2 dx.

In particular, this implies that (8.11) holds for u ∈ D(Ω) with a constantC = R2.

The Friedrichs inequality implies that, for a bounded open set Ω, thenorm ‖·‖1 in H1(Ω) is equivalent to the norm defined by Dirichlet’s integral,i.e. u 7→ D(u)

12 . Indeed,

‖u‖21 = ‖u‖2 +D(u), u ∈ H1(Ω),

where ‖ · ‖ is the norm in L2(Ω). The Friedrichs inequality means that‖u‖2 ≤ CD(u), for all u ∈ H1(Ω), implying

D(u) ≤ ‖u‖21 ≤ C1D(u), u ∈ H1(Ω),

as required.

Together with Dirichlet’s integral, we will use the corresponding innerproduct

[u, v] =∫

Ω

n∑j=1

∂u

∂xj

∂v

∂xjdx, u, v ∈ H1(Ω).

On H1(Ω), it defines a Hilbert space structure, since H1(Ω) is a closedsubspace in H1(Ω) and ‖u‖1 is equivalent in H1(Ω) to the norm [u, u]

12 =

D(u)12 .

8.4. Dirichlet’s problem (generalized solutions)

Consider two generalized settings of Dirichlet’s boundary value problem forthe Laplace operator. In both cases, the existence and uniqueness of asolution easily follows, after the above preliminary argument, from generaltheorems of functional analysis.

1. The first of the settings contains, as a classical analogue, the followingproblem (the boundary value problem with zero boundary condition for thePoisson equation):

(8.12)

∆u(x) = f(x), x ∈ Ω,u|∂Ω = 0.


Instead of u|∂Ω = 0, write

(8.13) u ∈ H1(Ω).

Now let us try to make sense of ∆u = f . First, let u ∈ C2(Ω), u|∂Ω = 0. Ifv ∈ D(Ω) then, integrating by parts, we get∫

Ω(∆u(x))v(x)dx = −

∫Ω

n∑j=1

∂u

∂xj

∂v

∂xjdx = −[u, v].

If ∆u = f this means that

(8.14) [u, v] = −(f, v),

where (·, ·) is the inner product in L2(Ω). Now note that if both parts of(8.14) are considered as functionals of v, then these functionals are linearand continuous with respect to the norm ‖ · ‖1 in H1(Ω). Therefore, (8.14)holds when v belongs to the closure of D(Ω) in H1(Ω), i.e., when v ∈ H1(Ω).

Conversely, let Ω be bounded and have a smooth boundary,

u ∈ C2(Ω)⋂H1(Ω), f ∈ L2(Ω) and (8.14) holds for any v ∈ H1(Ω). Let us

prove that u is a solution of (8.12).

First, observe that u ∈ H1(Ω) implies u|∂Ω = 0. This follows from thefact that, due to Theorem 8.5, the map u → u|∂Ω extends to a continuousmap H1(Ω) −→ L2(∂Ω) which maps all functions from D(Ω) and, therefore,all functions from H1(Ω) to zero. Note that here we only use, actually, thefact that

u ∈ C2(Ω) ∩ H1(Ω) =⇒ u|∂Ω = 0.

This fact can be obtained directly, avoiding Theorem 8.5 and its generaliza-tion to manifolds. Let us sketch the corresponding argument, omitting thedetails.

Using a partition of unity and locally rectifying the boundary with thehelp of a diffeomorphism, we can, by the same arguments as those used inthe proof of the Friedrichs inequality, get∫

∂Ω|u(x)|2dSx ≤ CD(u)

for u ∈ C1(Ω) with the support in a small neighborhood of a fixed point ofthe boundary. Therefore, if u|∂Ω 6= 0, then it is impossible to approximateu by functions from D(Ω) with respect to the norm in H1(Ω). Thus, ifu ∈ C2(Ω)

⋂H1(Ω), then u|∂Ω = 0.


Conversely, if u ∈ C1(Ω), then

u|∂Ω = 0 =⇒ u ∈ H1(Ω).

This may be proved as follows. The ε-mollified characteristic function ofΩ2ε, where

Ωδ = x : x ∈ Ω, ρ(x, ∂Ω) ≥ δ,is a function χε ∈ D(Ω), such that χε(x) = 1 for x ∈ Ω3ε and |∂αχε(x)| ≤Cαε

−|α| (the constant Cα does not depend on ε). Now, if u ∈ C1(Ω), thenχεu ∈ H1(Ω) and u|∂Ω = 0 implies that if x ∈ Ω\Ω3ε, then |u(x)| ≤ Cε,where C hereafter depends on u but does not depend on ε. This enables usto estimate

‖u− χεu‖21 = ‖u− χεu‖2 +D(u− χεu),

where ‖ · ‖ is the norm in L2(Ω) and D the Dirichlet integral. Since theLebesgue measure of Ω\Ω3ε does not exceed Cε, we have

‖u− χεu‖2 ≤ Cε supΩ\Ω3ε

|u− χεu|2 ≤ C1ε3

and

D(u− χεu) ≤ mes(Ω\Ω3ε) · supΩ\Ω3ε

n∑j=1

∣∣∣∣ ∂∂xj (u− χεu)∣∣∣∣2

≤ Cε supΩ\Ω3ε

n∑j=1

[∣∣∣∣ ∂u∂xj∣∣∣∣2 |(1− χε)|2 + |u|2

∣∣∣∣ ∂∂xj (1− χε)∣∣∣∣2]≤ C1ε,

where C1 does not depend upon ε (but depends upon u). The product of uand ∂

∂xj(1−χε) is bounded because in the 3ε-neighborhood of the boundary

we have u = O(ε) whereas the derivative is O(ε−1).

So, it is possible to approximate u with respect to the norm ‖ · ‖1 byfunctions from C1(Ω) whose supports are compact in Ω. In turn, they canbe easily approximated with respect to the norm ‖ · ‖1 by functions fromD(Ω) with the help of the mollifying.

Thus, supposing u ∈ C2(Ω) for a bounded open set Ω with a smoothboundary we get u|∂Ω = 0 if and only if u ∈ H1(Ω). This justifies thereplacement of the boundary conditions (8.12) with (8.13).

Further, (8.14) holds for u ∈ H1(Ω) and any v ∈ H1(Ω) if and onlyif ∆u = f , where ∆ is understood in the distributional sense. Indeed, if(8.14) holds, then, taking v ∈ D(Ω), we obtain by moving the derivativesfrom u to v (by use of the definition of the distributional derivatives) that


(u,∆v) = (f, v). that ∆u = f , because v can be arbitrary function fromD(Ω).

Conversely, if ∆u = f , u ∈ H1(Ω) and f ∈ L2(Ω), then (8.14) holds forv ∈ D(Ω), hence, by continuity, for any v ∈ H1(Ω).

All this justifies the following generalized setting of the problem (8.12):

• Let Ω be an arbitrary bounded open set in Rn, f ∈ L2(Ω). Findu ∈ H1(Ω), such that (8.14) holds for any v ∈ H1(Ω) (or, which isthe same, for any v ∈ D(Ω)).

In this case u is called a generalized solution of the problem (8.12). Wehave seen above that if Ω is a bounded open set with a smooth boundaryand u ∈ C2(Ω), then u is a generalized solution of (8.12) if and only if it isa solution of this problem in the classical sense.

Theorem 8.9. A generalized solution of (8.12) exists and is unique.

Proof. Rewrite (8.14) in the form

[v, u] = −(v, f)

and consider the right hand side as a functional in v for v ∈ H1(Ω). Thisis a linear continuous functional. Its linearity is obvious and its continuityfollows from the Cauchy–Schwarz inequality:

|(v, f)| ≤ ‖f‖ · ‖v‖ ≤ ‖f‖ · ‖v‖1,where ‖ · ‖ is the norm in L2(Ω).

By the Riesz representation theorem we may (uniquely) write this func-tional in the form [v, u], where u is a fixed element of H1(Ω). This provesthe unique solvability of (8.12) (in the generalized setting).

The generalized setting of the Dirichlet problem leaves two importantquestions open:

1) the exact description of the class of all functions u which appear asthe solutions when f runs through all possible functions from L2(Ω),

2) the existence of a more regular solution for a more regular f .

Observe that local regularity problems are easily solved by the remarkthat if E(x) is a fundamental solution of the Laplace operator in Rn, thenthe convolution

u0(x) = (E ∗ f)(x) =∫

ΩE(x− y)f(y)dy


(defined for almost all x) is a solution of ∆u0 = f . Therefore, ∆(u−u0) = 0,i.e., u − u0 is an analytic function in Ω. Therefore, the regularity of ucoincides locally with that of u0 and, therefore, can be easily described (e.g.if f ∈ C∞(Ω), then u ∈ C∞(Ω), as proved above).

The question of regularity near the boundary is more complicated. More-over, for non-smooth boundary the answer is not completely clear even now.However, if ∂Ω is sufficiently smooth (say, of class C∞) there are exact(though not easy to prove) theorems describing the regularity of u in termsof regularity of f . We will formulate one of such theorems without proof.

Theorem 8.10. Let the boundary of Ω be of class C∞. If f ∈ Hs(Ω),s ∈ Z+, then u ∈ Hs+2(Ω) ∩ H1(Ω).

(For the proof see e.g. Gilbarg and Trudinger [8], Sect. 8.4.)

Conversely, it is clear that if u ∈ Hs+2(Ω), then ∆u = f ∈ Hs(Ω).Therefore,

∆ : Hs+2(Ω) ∩ H1(Ω) −→ Hs(Ω), s ∈ Z+,

is an isomorphism. In particular, in this case all generalized solutions (forf ∈ L2(Ω)) lie in H2(Ω)

⋂H1(Ω).

Another version of an exact description of the regularity, which we al-ready discussed above, is Schauder’s Theorem 6.20 in Holder functionalspaces. There are numerous other descriptions, depending on the choiceof functional spaces. Most of them are useful in miscellaneous problems ofmathematical physics.

Note that the last statement fails in the usual classes Ck(Ω) even locally.Namely, f ∈ C(Ω) does not even imply u ∈ C2(Ω).

2. Let us pass to the generalized setting of the usual Dirichlet problemfor the Laplace equation:

(8.15)

∆u = 0u|∂Ω = ϕ

where Ω is a bounded open set in Rn. First, there arises a question ofinterpretation of the boundary condition. We will do it as by assuming thatthere exists a function v ∈ H1(Ω), such that v|∂Ω = ϕ, and formulatingthe boundary condition for u as follows: u − v ∈ H1(Ω). This impliesu ∈ H1(Ω). Introducing v saves us from describing the smoothness of ϕ(which requires using Sobolev spaces with non-integer indices on ∂Ω if Ω


has a smooth boundary). Besides, it becomes possible to consider the caseof domains with non-smooth boundaries.

Let us give the generalized formulation of Dirichlet’s problem (8.15):

• Given a bounded domain Ω ⊂ Rn and v ∈ H1(Ω), find a function

u ∈ H1(Ω), such that ∆u = 0 and u− v ∈ H1(Ω).

At the first sight this problem seems to be reducible to the above oneif we seek u − v ∈ H1(Ω) and put f = ∆(u − v) = −∆v. But we have noreason to think that ∆v ∈ L2(Ω), and we did not consider a more generalsetting. Therefore, we will consider this problem independently of the firstproblem, especially because its solution is as simple as the solution of thefirst problem.

The reasoning above makes it clear that if Ω has a smooth boundary,v ∈ C(Ω), v|∂Ω = ϕ, u ∈ C2(Ω) and u is a generalized solution of (8.15),then u is a classical solution of this problem.

Theorem 8.11. A generalized solution u of Dirichlet’s problem exists andis unique for any bounded domain Ω ⊂ Rn and any v ∈ H1(Ω). Thissolution has strictly and globally minimal Dirichlet integral D(u) among allu ∈ H1(Ω) such that u− v ∈ H1(Ω).

Conversely, if u is a critical (stationary) point of the Dirichlet integral inthe class of all u ∈ H1(Ω) such that u−v ∈ H1(Ω), then the Dirichlet integralactually reaches the strict and global minimum at u and u is a generalizedsolution of Dirichlet’s problem.

Proof. Let u ∈ H1(Ω). The equation ∆u = 0 can be expressed in theform

(u,∆w) = 0, w ∈ D(Ω),

or, which is the same, in the form

[u,w] = 0, w ∈ D(Ω).

Since [u,w] is a continuous functional in w on H1(Ω), then, by continuity,we get

(8.16) [u,w] = 0, w ∈ H1(Ω).

This condition is equivalent to the Laplace equation ∆u = 0.

Now we can invoke a geometric intepretation of the problem: we haveto find u ∈ H1(Ω), which is orthogonal to H1(Ω) with respect to [·, ·] andsuch that v − u ∈ H1(Ω). Imagining for a moment that H1(Ω) is a Hilbert


space with respect to the inner product [·, ·], we see that u is the componentin the decomposition v = (v−u) +u in the direct orthogonal decomposition

H1(Ω) = H1(Ω) + (H1(Ω))⊥.

It is well known from Functional Analysis that such a decomposition exists,is unique (in any Hilbert space and for any closed subspace of this Hilbertspace), and besides, the perpendicular (to H1(Ω)) component u has thestrictly minimal norm compared with all other decompositions v = (v −w) +w with v−w ∈ H1(Ω). Vice versa, in this case, if u is stationary, thenit is orthogonal to H1(Ω).

However, the “inner product” [·, ·] defines a Hilbert space structure onH1(Ω), but not on H1(Ω) (it is not strictly positive; for example, [1, 1] = 0whereas 1 6= 0). Therefore, the above argument should be made moreprecise. Denote z = v − u ∈ H1(Ω) so that v = u + z. It follows from(8.16) that

(8.17) [w, v] = [w, z], w ∈ H1(Ω).

Now it is easy to prove the existence of solution. Namely, let us consider[w, v] as a linear continuous functional in w on H1(Ω). By the Riesz theoremit can be uniquely represented in the form [w, z], where z is a fixed elementof H1(Ω). It remains to set u = v − z. The definition of z implies (8.17)which in turn implies (8.16). The condition u− v ∈ H1(Ω) is obvious.

The uniqueness of the solution is clear, since if u1, u2 are two solutions,then u = u1 − u2 ∈ H1(Ω) satisfies (8.16), and this yields u = 0.

Let us now prove that the Dirichlet integral is minimal on the solutionu in the class of all u1 ∈ H1(Ω) such that u1 − v ∈ H1(Ω). Set z = u1 − u;hence, u1 = u+ z. Clearly, z ∈ H1(Ω) and (8.16) yields

D(u1) = [u1, u1] = [u, u] + [z, z] = D(u) +D(z) ≥ D(u),

where the equality is only attained for D(z) = 0, i.e., for z = 0 or, which isthe same, for u = u1.

Finally, let us verify that if Dirichlet’s integral is stationary on u ∈ H1(Ω)in the class of all u1 ∈ H1(Ω) such that u1 − v ∈ H1(Ω), then u is ageneralized solution of the Dirichlet problem (hence the Dirichlet integralreaches the strict global minimum at u). The stationarity property of theDirichlet integral at u means that

d

dtD(u+ tz)|t=0 = 0, z ∈ H1(Ω).


But[d

dtD(u+ tz)

]∣∣∣∣t=0

=(d

dt[u+ tz, u+ tz]

)∣∣∣∣t=0

= [u, z] + [z, u] = 2 Re[u, z].

The stationarity property gives Re[u, z] = 0 for any z ∈ H1(Ω). Replacingz by iz, we get [u, z] = 0 for any z ∈ H1(Ω) yielding (8.16). Theorem 8.11is proved.

Finally, we will give without proof a precise description of regularityof solutions from the data on ϕ in the boundary value problem (8.15) fora domain Ω with a smooth boundary. It was already mentioned that ifu ∈ Hs(Ω), s ∈ Z+, s ≥ 1, then u|∂Ω ∈ Hs− 1

2 (∂Ω). It turns out that theconverse also holds: if ϕ ∈ Hs− 1

2 (∂Ω), where s ∈ Z+, s ≥ 1, then thereexists a unique solution u ∈ Hs(Ω) of the problem (8.15) (see e.g. Lions andMagenes [20]).

Similarly, if ϕ ∈ Cm+γ(∂Ω), where m ∈ Z+, γ ∈ (0, 1), then there existsa (unique) solution u ∈ Cm+γ(Ω).

Combining the described regularity properties of solutions of problems(8.12) and (8.15), it is easy to prove that the map

u 7→ ∆u, u|Γcan be extended to a topological isomorphism

Hs(Ω) −→ Hs−2(Ω)⊕Hs− 12 (∂Ω), s ∈ Z+, s ≥ 2,

and also to a topological isomorphism

Cm+2+γ(Ω) −→ Cm+γ(Ω)⊕ Cm+2+γ(∂Ω), m ∈ Z+, γ ∈ (0, 1).

(See e.g. the above quoted books Gilbarg and Trudinger [8], Ladyzhenskayaand Ural’tseva [17].)

Similar theorems can be proved also for general elliptic equations withgeneral boundary conditions satisfying an algebraic condition called the el-lipticity condition (or coercitivity, or Shapiro–Lopatinsky condition). For thegeneral theory of elliptic boundary value problems, see e.g. books by Lionsand Magenes [20], Hormander [12], Ch.X, or [14], vol.3, Ch.20.

8.5. Problems

8.1. Using the Fourier method, solve the following Dirichlet problem: finda function u = u(x, y) such that ∆u = 0 in the disc r ≤ a, where r =√x2 + y2, u|r=a = (x4 − y4)|r=a.

8.5. Problems 217

8.2. By the Fourier method solve the Dirichlet problem in the disk r ≤ a

for a smooth boundary function f(ϕ), where ϕ is the polar angle, i.e., find afunction u = u(x, y) such that ∆u = 0 in the disc r ≤ a and u|r=a = f(ϕ).Justify the solution. Derive for this case Poisson’s formula:

u(r, ϕ) =1

2π

∫ 2π

0

(a2 − r2)f(α)dαa2 − 2ar cos(ϕ− α) + r2

.

8.3. Using the positivity of Poisson’s kernel

K(r, ϕ, α) =a2 − r2

2π[a2 − 2ar cos(ϕ− α) + r2],

prove the maximum principle for the solutions of Dirichlet’s problem ob-tained in the preceding problem. Derive from the maximum principle thesolvability of Dirichlet’s problem for any continuous function f(ϕ).

8.4. Find a function u = u(x, y) such that ∆u = x2 − y2 for r ≤ a andu|r=a = 0.

8.5. In the ring 0 < a ≤ r ≤ b < +∞, find a function u = u(x, y) such that∆u = 0, u|r=a = 1, ∂u

∂r

∣∣r=b

= (cosϕ)2.

8.6. By the Fourier method solve the Dirichlet problem for the Laplaceequation in the rectangle 0 ≤ x ≤ a, 0 ≤ y ≤ b with the boundary conditionsof the form

u|x=0 = ϕ0(y), u|x=a = ϕ1(y), (0 ≤ y ≤ b);u|y=0 = ψ0(x), u|y=b = ψ1(x), (0 ≤ x ≤ a),

where (the continuity condition for the boundary function)

ϕ0(0) = ψ0(0), ϕ0(b) = ψ1(0), ϕ1(0) = ψ0(a), ϕ1(b) = ψ1(a).

Describe a way for solving the general problem, and then solve the problemin the following particular case:

ϕ0(y) = Ay(b− y), ψ0(x) = B sinπx

a, ϕ1(y) = ψ1(x) = 0.

8.7. In the half-plane (x, y) : y ≥ 0 solve Dirichlet’s problem for theLaplace equation in the class of bounded continous functions. For this pur-pose use the Fourier transform with respect to x to derive the followingformula for the solutions which are in L2(R) with respect to x:

u(x, y) =1π

∫ ∞−∞

y

y2 + (x− z)2ϕ(z)dz,


where ϕ(x) = u(x, 0). Then investigate the case of a general bounded con-tinuous function ϕ.

8.8. Investigate for which n, s and α (α > −n) the function u(x) = |x|αbelongs to Hs(B1), where B1 is the open ball of radius 1 with the center atthe origin of Rn.

8.9. Does the function u(x) ≡ 1, where x ∈ (−1, 1), belong to H1((−1, 1))?

8.10. Prove that the inclusion Hs(Rn) ⊂ L2(Rn), s > 0, is not a compactoperator.

Chapter 9

The eigenvalues and

eigenfunctions of the

Laplace operator

9.1. Symmetric and self-adjoint operators in Hilbert space

Let H be a Hilbert space. Recall that a linear operator in H is a collectionof two objects:

a) a linear (but not necessarily closed) subspace D(A) ⊂ H called thedomain of the operator A;

b) a linear map A : D(A) −→ H.

In this case we will write, by an abuse of notations, that a linear operatorA : H −→ H is defined, though, strictly speaking, there is no map H −→ H,because in general A may not be defined on the whole H.

It is possible to add linear operators and take their products (composi-tions). Namely, given two linear operators A1, A2 in H, we set

D(A1 +A2) = D(A1) ∩D(A2),

(A1 +A2)x = A1x+A2x for x ∈ D(A1 +A2).

Further, set

D(A1A2) = x : x ∈ D(A2), A2x ∈ D(A1),

(A1A2)x = A1(A2x) for x ∈ D(A1A2).

219

220 9. The eigenvalues and eigenfunctions

Define alsoKer A = x : x ∈ D(A), Ax = 0,Im A = x : x = Ay, y ∈ D(A).

If Ker A = 0, then the inverse operatorA−1 is defined. Namely, setD(A−1) =Im A and A−1x = y for x = Ay.

If there are two linear operators A,B in H, such that D(A) ⊂ D(B) andAx = Bx for x ∈ D(A), then we say that B is an extension of A and writeA ⊂ B.

An operator A is called symmetric if

(Ax, y) = (x,Ay), x, y ∈ D(A).

In what follows, we will also introduce the notion of a self-adjoint operator(for unbounded operators symmetry and self-adjointness do not coincide!)

Let A be a linear operator such that D(A) is dense in H. Then theadjoint operator A∗ is defined via the identity

(9.1) (Ax, y) = (x,A∗y), x ∈ D(A), y ∈ D(A∗).

More precisely, let D(A∗) consist of y ∈ H such that there exists z ∈ Hsatisfying

(Ax, y) = (x, z) for all x ∈ D(A).

Since D(A) is dense in H, we see that z is uniquely defined and we setA∗y = z.

If the domain of A is dense and A is symmetric, then, clearly, A ⊂ A∗.An operator A is self-adjoint if A∗ = A (it is understood that the domain

of A is dense and D(A∗) = D(A)).

Clearly, any self-adjoint operator is symmetric. The converse is notalways true.

Define the graph of A as the set

GA = x,Ax, x ∈ D(A) ⊂ H ×H.

An operator is closed if its graph is closed in H × H, i.e. if xn ∈ D(A),xn → x and Axn → y (the convergence is understood with respect to thenorm in H) yield x ∈ D(A) and Ax = y.

It is easy to see that every bounded everywhere defined linear operatoris closed; more generally, a bounded linear operator A with a closed domainD(A) is closed. But generally closed operators can be unbounded.

9.1. Symmetric and self-adjoint operators 221

Any self-adjoint operator is closed. A more general fact is true: theoperator A∗ is always closed for any A.

Indeed, if yn ∈ D(A∗), yn → y, and A∗yn → z then passing to the limitas n→∞ in the identity

(Ax, yn) = (x,A∗yn), x ∈ D(A),

we get(Ax, y) = (x, z), x ∈ D(A),

yielding y ∈ D(A∗) and A∗y = z.

If D(A) is a closed subspace in H (in particular, if D(A) = H, i.e., A isdefined everywhere) then the closedness of A is equivalent to its bounded-ness, due to the closed graph theorem (which we already used in Sect. 7.6;see also Reed and Simon [22], Theorem III.12). In particular, a self-adjointeverywhere defined operator A is necessarily bounded.

Let us indicate a way of constructing self-adjoint operators.

Proposition 9.1. Let A be a self-adjoint operator in H and Ker A = 0.Then A−1 is self-adjoint.

Proof. Denote by E⊥ the orthogonal complement to E ⊂ H, namely

E⊥ = y : y ∈ H, (x, y) = 0 for all x ∈ E.Given a linear operator A with D(A) dense in H, we easily derive from thedefinition of A∗ that

(9.2) KerA∗ = (Im A)⊥.

Indeed, the equality (Ax, y) = 0 for all x ∈ D(A) means, due to (9.1),exactly that y ∈ D(A∗) and A∗y = 0.

Note that E⊥ = 0 if and only if the linear span of E is dense in H.

Thus if A is self-adjoint and KerA = 0, then from (9.2) it follows thatD(A−1) = ImA is dense in H.

Therefore, A−1 and (A−1)∗ are defined. It remains to verify that (A−1)∗ =A−1.

We have y ∈ D((A−1)∗) and (A−1)∗y = z if and only if

(9.3) (A−1x, y) = (x, z), x ∈ ImA.

Setting x = Af , f ∈ D(A), we see that (9.3) is equivalent to

(f, y) = (Af, z), f ∈ D(A),


yielding z ∈ D(A∗) and A∗z = y. But, since A∗ = A, this means thatz ∈ D(A) and Az = y which may be also rewritten as z = A−1y. We haveproved that y ∈ D((A−1)∗) with (A−1)∗y = z are equivalent to y ∈ D((A−1))with A−1y = z. Therefore, (A−1)∗ = A−1 as required.

Corollary 9.2. If B is a bounded, everywhere defined symmetric operatorwith KerB = 0, then A = B−1 is self-adjoint.

Elementary Spectral Theory

The importance of self-adjoint operators follows from the spectral theo-rem. Before we formulate it, let us give an example of a self-adjoint operator.

Let M be a space with measure dµ, i.e., a set with a σ-algebra of itssubsets and a countable-additive measure (with values in [0,+∞]) on it.Construct in a usual way the space L2(M,dµ) consisting of classes of mea-surable functions on M , such that the square of their absolute value isintegrable. (Two functions are in the same class if they coincide almosteverywhere.) Then L2(M,dµ) is a Hilbert space. Let a(m) be a real-valuedmeasurable and almost everywhere finite function on M . The operator Aof multiplication by a(m) is defined by the formula Af(m) = a(m)f(m),where

f ∈ D(A) = f : f ∈ L2(M,dµ), af ∈ L2(M,dµ).This operator is always self-adjoint. Indeed, its symmetry is obvious and wehave only to verify that D(A∗) = D(A).

Let u ∈ D(A∗), A∗u = v. Let us check that u ∈ D(A) and a(m)u(m) =v(m) for almost all m. Consider the set

MN = m : m ∈M, |a(m)| ≤ N,and let χN (m) = 1 for m ∈MN and χN (m) = 0 for m 6∈MN . The identityA∗u = v means that∫

Ma(m)f(m)u(m)dµ =

∫Mf(m)v(m)dµ, for any f ∈ D(A).

Note that if g ∈ L2(M,dµ), then χNg ∈ D(A), implying∫MN

a(m)g(m)u(m)dµ =∫MN

g(m)v(m)dµ, for any g ∈ L2(MN , dµ).

But it is clear that(au)|MN

∈ L2(MN , dµ)

since a is bounded on MN . Therefore, the arbitrariness of g implies au|MN=

v|MN(almost everywhere). Therefore, if M =

⋃∞N=1MN , then au|cM = v|cM .

9.1. Symmetric and self-adjoint operators 223

Since a(m) is almost everywhere finite, then the measure of M\M is equal to0. Hence, a(m)u(m) = v(m) almost everywhere. In particular, a(m)u(m) ∈L2(M,dµ), i.e., u ∈ D(A) and au = v, as required.

An important example: consider the Hilbert space of sequences

l2 = x = x1, x2, . . . : xj ∈ C,∞∑j=1

|xj |2 < +∞.

For any sequence of real numbers λ1, λ2, . . . , consider the operator A in l2

which is defined as follows:

D(A) = x1, x2, . . . : xj ∈ C,∞∑j=1

|λj |2|xj |2 < +∞,

Ax1, x2, . . . = λ1x1, λ2x2, . . ..In this example M is a countable set and the measure of each point is equalto 1.

Operators A1 : H1 −→ H1 and A2 : H2 −→ H2 are called unitarilyequivalent if there exists a unitary (i.e., invertible and inner product pre-serving) operator U : H1 −→ H2 such that U−1A2U = A1. (Recall that theequality of operators by definition implies the coincidence of their domains.)

In other words, in these notations the unitary equivalence of A1 andA2 means the existence of a unitary operator U : H1 → H2 such that thediagram

H1A1−→ H1yU yU

H2A2−→ H2

is commutative.

The following theorem is one of the most important results in functionalanalysis.

The Spectral Theorem. Any self-adjoint operator in a Hilbert space isunitarily equivalent to an operator of multiplication by a measurable real-valued function in a space L2(M,dµ).

Proofs can be found , e.g., in Reed and Simon [22], Chapters VII andVIII, or in Berezin and Shubin [4], Appendix 1.

Note that A is unitarily equivalent to the above described operator withreal eigenvalues in l2 if and only ifH is separable and there is an orthonormal


basis ϕ1, ϕ2, . . . in H consisting of eigenvectors of A with real eigenvaluesλ1, λ2, . . .. In this case

D(A) =

ϕ =∞∑j=1

cjϕj :∞∑j=1

|cj |2 < +∞,∞∑j=1

|λj |2|cj |2 < +∞

.

If |λj | → +∞ as j → +∞, then we say that the spectrum of A is discrete.If KerA = 0, then the spectrum of A is discrete if and only if A−1 iseverywhere defined and compact. This follows from the Hilbert-Schmidttheorem (already used in Ch. 3), which states that a compact self-adjointoperator with real eigenvalues has an orthogonal basis of eigenvectors withreal eigenvalues µ1, µ2, . . ., where µj → 0 as j → +∞.

For a self-adjoint operator A the discreteness of spectrum is equivalent tothe requirement that there exists λ0 ∈ R such that the operator (A−λ0I)−1

exists, is everywhere defined and compact. In this case, the discreteness ofspectrum of A is equivalent to the compactness of (A− λ0I)−1.

Important remark: In the subsection above we mainly considered operatorswith purely discrete spectrum. But a self-adjoint operator A may have nonon-vanishing eigenvectors. (See e.g. the multiplication by the functiona(x) = x, x ∈ [0, 1], in the Hilbert space L2([0, 1]).)

9.2. The Friedrichs extension

Let us give an example of a symmetric but not self-adjoint operator. Let ∆be the Laplace operator, Ω a bounded domain in Rn. In L2(Ω), consider anoperator A0 with a domain D(A0) = D(Ω) mapping ϕ ∈ D(Ω) to (−∆ϕ).An elementary integration by parts shows that A0 is symmetric, i.e.,

(A0ϕ,ψ) = (ϕ,A0ψ), ϕ, ψ ∈ D(Ω),

where parentheses denote the inner product in L2(Ω). This operator isnot even closed. Indeed, it is easy to see that if ϕ ∈ C2(Ω) and suppϕis a compact subset in Ω, then there exists a sequence ϕk ∈ D(Ω) (easilyobtained, e.g. by mollyifying) such that ϕk → ϕ and ∆ϕk → ∆ϕ withrespect to the norm in L2(Ω). Taking ϕ 6∈ D(Ω), we see that A0 is notclosed. However, it is closable, in the sense that we may define a closedoperator A0 whose graph is the closure of the graph of A0 (in L2(Ω)×L2(Ω)).This means that if ϕ ∈ L2(Ω) and there exists a sequence ϕk ∈ D(Ω) suchthat ϕk → ϕ and ∆ϕk → f in L2(Ω), then, by definition, ϕ ∈ D(A0) and

9.2. The Friedrichs extension 225

A0ϕ = f . Clearly, in this case we will again have

(A0ϕ,ψ) = (ϕ,A0ψ), ψ ∈ D(Ω).

This, in particular, implies that A0ϕ is well defined (does not depend on thechoice of the sequence ϕk) and A0 ⊂ A∗0.

Incidentally, it is easy to see that (A0)∗ = A∗0 (passing to the limit inthe identity defining A∗0).

It is interesting to understand what A∗0 is. By definition, D(A∗0) consistsof u ∈ L2(Ω) such that there exists v ∈ L2(Ω) such that

(v, ψ) = (u, (−∆)ψ), ψ ∈ D(Ω).

But this means that −∆u = v in the distributional sense. Therefore

(9.4) D(A∗0) = u : u ∈ L2(Ω), ∆u ∈ L2(Ω).It is not trivial to describe D(A0). However, it is easy to see that D(A0) ⊂H1(Ω). Indeed, (A0u, u) = D(u) (the Dirichlet integral of u) for u ∈ D(Ω).Therefore, if ϕk ∈ D(Ω), k = 1, 2, . . . , ϕk → ϕ and A0ϕk → A0ϕ in L2(Ω),then (A0(ϕk − ϕl), ϕk − ϕl)→ 0 as k, l → +∞ implying ϕk → ϕ in H1(Ω);hence, ϕ ∈ H1(Ω).

So, we have proved that D(A0) ⊂ H1(Ω). At the same time, it followsfrom (9.4) that D(A∗0) ⊃ H2(Ω), implying A∗0 6= A0. For example, if Ω hasa smooth boundary and u ∈ C2(Ω), then u ∈ H2(Ω), hence u ∈ D(A∗0); butif we also assume u|∂Ω 6= 0, then u 6∈ H1(Ω); hence, u 6∈ D(A0). For thegeneral bounded Ω it is easy to see that u ≡ 1 is in H2(Ω), hence in D(A∗0),but not in H1(Ω) due to Friedrichs’ inequality (8.11), hence not in D(A0).

Therefore, A0 is symmetric and closed but not self-adjoint.

There exists a natural construction of a self-adjoint extension of anysemibounded symmetric operator called the Friedrichs extension. We haveactually applied this construction to get a generalized solution of the problem∆u = f , u|∂Ω = 0. Now we will describe the Friedrichs extension in a moregeneral abstract setting.

A symmetric operator A0 : H −→ H is semibounded below if there existsa constant C such that

(9.5) (A0ϕ,ϕ) ≥ −C(ϕ,ϕ), ϕ ∈ D(A0).

Adding (C+ε)I, where ε > 0, to A0, we get a new operator (denote it againby A0) such that

(9.6) (A0ϕ,ϕ) ≥ ε(ϕ,ϕ), ϕ ∈ D(A0).


We will assume from the very beginning that this stronger estimate (9.6)holds. From the point of view of the eigenvalue problem the addition of(C + ε)I is of no importance, because it only shifts eigenvalues withoutaffecting eigenfunctions. Note, that the Friedrichs inequality shows that inour main example above the estimate (9.6) is already satisfied.

Set[u, v] = (A0u, v), u, v ∈ D(A0).

This inner product defines a pre-Hilbert structure on D(A0) (the positivedefiniteness follows from (9.6)). Denote by ‖ · ‖1 the corresponding norm(i.e. ‖u‖1 = [u, u]1/2). Then the convergence with respect to ‖ · ‖1 impliesconvergence with respect to the norm ‖ · ‖ in H. In particular, if a sequenceϕk ∈ D(A0) is a Cauchy sequence with respect to ‖ · ‖1, then it is a Cauchysequence with respect to ‖ · ‖; hence, ϕk converges in H.

Denote by H1 the completion of D(A0) with respect to ‖ · ‖1 (in theexample, H1 = H1(Ω)). The inequality (9.6) and the above arguments showthe existence of a natural linear map j : H1 → H. Namely, for any g ∈ H1,its image g∗ = jg ∈ H is the limit in H of a sequence ϕk, ϕk ∈ D(A0),converging to g in H1. Let us prove that this map is injective.

Indeed, assume that g∗ = 0. This means that for the sequence ϕkabove we have ϕk → g in H1 and ϕk → 0 in H. Then by continuity of theinner products,

‖g‖2 = limk,l→∞

[ϕk, ϕl] = limk,l→∞

(A0ϕk, ϕl) = limk→∞

liml→∞

(A0ϕk, ϕl)

= 0,

hence g = 0 which proves the injectivity of j.

Therefore, there is a natural imbedding H1 ⊂ H, so we will identifythe elements of H1 with the corresponding elements of H. Note, that (9.6)implies that

(9.7) ‖u‖2 ≤ ε−1‖u‖21, u ∈ H1,

which means that the norm of the imbedding operator is not greater thanε−1/2.

Now setD(A) = H1 ∩D(A∗0)

andA = A∗0|D(A).

We get an operator A : H −→ H called the Friedrichs extension of A0.

9.2. The Friedrichs extension 227

If for only a weaker estimate (9.5) is satisfied for A0, then the construc-tion of its Friedrichs extension should be modified as follows: we should takethe Friedrichs extension of A0 + (C + ε)I and subtract (C + ε)I from theresult. N

Theorem 9.3. 1. The Friedrichs extension of a semibounded operator A0

is a self-adjoint operator A : H −→ H. It is again semibounded below andhas the same lower bound as A0. For example, if A0 satisfies (9.6), then

(9.8) (Au, u) ≥ ε(u, u), u ∈ D(A),

where ε > 0 is the same constant as in the similar estimate (9.6) for A0.

2. Assuming (9.6) is satisfied, the operator A−1 is everywhere definedand bounded linear operator in H, with the operator norm ‖A−1‖ ≤ ε−1.Moreover, A−1 maps H to H1 as a bounded linear operator with the samenorm not greater than ε−1/2, that is

(9.9) ‖A−1f‖1 ≤ ε−1/2‖f‖, f ∈ H,where ‖ · ‖ and ‖ · ‖1 are the norms in H and H1 respectively.

Proof. 1. Let us verify that A−1 exists, is everywhere defined, boundedand symmetric. This implies that A is self-adjoint (see Corollary 9.2).

First, verify that

(9.10) (Au, v) = [u, v], u, v ∈ D(A).

Indeed, this is true for u, v ∈ D(A0). By continuity this is also true foru ∈ D(A0), v ∈ H1.

Now, make use of the identity

(A0u, v) = (u,A∗0v), u ∈ D(A0), v ∈ D(A∗0).

In particular, this is true for u ∈ D(A0), v ∈ D(A). Then A0u = Au,A∗0v = Av, and therefore, we have

(Au, v) = (u,Av) = [u, v], u ∈ D(A0), v ∈ D(A).

Taking here 2nd and 3rd terms, we may pass to the limit with respect to u(if the limit is taken in H1). We get

(9.11) (u,Av) = [u, v], u ∈ H1, v ∈ D(A).

In particular, this holds for u ∈ D(A), v ∈ D(A). Interchanging u and v, weget (9.10), implying

(9.12) (Au, v) = (u,Av) = [u, v], u, v ∈ D(A).


Therefore, A is symmetric and

(9.13) (Au, u) = [u, u] ≥ ε(u, u), u ∈ D(A),

where ε > 0 is the same as in (9.6).

It follows from (9.13) that KerA = 0 and A−1 is well-defined. Thesymmetry of A implies that of A−1. It follows from (9.13) that A−1 isbounded. Indeed, if u ∈ D(A), then

‖u‖2 ≤ ε−1(Au, u) ≤ ε−1‖u‖ · ‖Au‖,implying

‖u‖ ≤ ε−1‖Au‖, u ∈ D(A).

Setting v = Au, we get

‖A−1v‖ ≤ ε−1‖v‖, v ∈ D(A−1),

which means the boundedness of A−1 together with the desired estimate of‖A−1‖ in H.

Let us prove that D(A−1) = H, i.e. ImA = H. We need to prove thatif f ∈ H, then there exists u ∈ D(A), such that Au = f . But this meansthat u ∈ H1 ∩D(A∗0) and A∗0u = f , i.e.,

(9.14) (A0ϕ, u) = (ϕ, f), ϕ ∈ D(A0).

Taking (9.12) into account, we may rewrite this in the form

(9.15) [ϕ, u] = (ϕ, f), ϕ ∈ D(A0).

Now, by the Riesz theorem about the structure of of continuous linear func-tionals on a Hilbert space, for any f ∈ H there exists u ∈ H1 such that(9.15) holds. It remains to verify that u ∈ D(A∗0). But this is clear because,due to (9.11), we may rewrite (9.15) in the form (9.14).

2. Clearly, A−1 maps H to H1 because D(A) ⊂ H1. Returning to (9.13)and using (9.7), we obtain for any u ∈ D(A)

‖u‖21 = (Au, u) ≤ ‖Au‖ ‖u‖ ≤ ε1/2‖Au‖ ‖u‖1,which immediately implies (9.9) if we take f = Au.

Now, let us pass to the model example.

Let A0 be the operator (−∆) on D(Ω), where Ω is a bounded domainin Rn. As we have already seen, the adjoint operator A∗0 maps u to (−∆u)if u ∈ L2(Ω) and ∆u ∈ L2(Ω) (here ∆ is understood in the distributionalsense). Due to the Friedrichs inequality, A0 is positive, i.e., (9.6) holds. Let

9.3. Discreteness of spectrum in a bounded domain 229

us construct the Friedrichs extension of A. This is an operator in L2(Ω),such that

A0 ⊂ A ⊂ A∗0and

D(A) = H1(Ω) ∩D(A∗0),

i.e.D(A) = u : u ∈ H1(Ω), ∆u ∈ L2(Ω).

By Theorem 9.3, it follows that A is self-adjoint and positive. It is usuallycalled the self-adjoint operator in L2(Ω) defined by (−∆) and the Dirichletboundary condition u|∂Ω = 0. The operator A−1 is everywhere defined andbounded.

9.3. Discreteness of spectrum for the Laplace operator in a

bounded domain

Let Ω be a bounded open subset in Rn.

Theorem 9.4. 1) The spectrum of the operator A defined in L2(Ω) by (−∆)and the Dirichlet boundary condition, is discrete. More precisely, there existsan orthonormal basis of eigenvectors ψj, j = 1, 2, . . . , such that ψj ∈ H1(Ω)and (−∆)ψj = λjψj in L2(Ω), where λj → +∞ as j → +∞.

2) All eigenvalues of A are strictly positive.

Proof. Applying Theorem 9.3 with H = L2(Ω) and A0 = −∆ in L2(Ω)with the domain D(A0) = D(Ω), we obtain A as the Friedrichs extensionof the operator A0. In this construction we get H1 = H1(Ω), and A hasa bounded, everywhere defined inverse operator A−1 in L2(Ω), which is ac-tually a composition of a bounded operator from L2(Ω) to H1(Ω) and theimbedding operator of H1(Ω) into L2(Ω). Since the imbedding is a com-pact operator due to Theorem 8.7, the operator A−1 : L2(Ω) → L2(Ω) isa compact self-adjoint operator. By the Hilbert-Schmidt Theorem (see e.g.Reed-Simon [22], Theorem VI.16) it has an orthonormal basis of eigenfunc-tions in L2(Ω), with real eigenvalues µj , j = 1, 2 . . . , such that µj → 0 asj →∞.

Clearly, 0 can not be an eigenvalue of A−1. Therefore, the eigenvalues λjof A are λj = µ−1

j , with the same eigenfunctions. Therefore, λj → +∞ asj →∞, and the eigenvalues λj are all isolated and have finite multiplicities.


9.4. Fundamental solution of the Helmholtz operator and

the analyticity of eigenfunctions of the Laplace operator

at the interior points. Bessel’s equation

We would like to prove analyticity of eigenfunctions of the Laplace operatorin the interior points of the domain. Since eigenvalues are positive, it sufficesto prove analyticity of any solution u ∈ D′(Ω) of the equation

(9.16) ∆u+ k2u = 0, k > 0,

called the Helmholtz equation. First, let us find a fundamental solution ofthe Helmholtz operator ∆ + k2 in an explicit or asymptotic form. Denotethis fundamental solution by E(k)

n (x). If it turns out that E(k)n (x) is analytic

for x 6= 0, hence any solution u ∈ D′(Ω) of the equation (9.16) is analytic inΩ; cf. Theorem 5.15.

Since ∆ +k2 commutes with rotations, it is natural to seek a sphericallysymmetric fundamental solution, that is one which is invariant under rota-tions. Using operation of averaging over all rotations, we can construct aspherically symmetric fundamental solution starting from an arbitrary one.(The averaging should be done over the group SO(n) of all orientation pre-serving rotations in Rn, with respect to the Haar measure on this group,that is, the invariant measure on SO(n) – see e.g. Rudin [24], Ch. 5.)

For a function on Rn, the spherical symmetry means that this functionhas the form E(k)

n (x) = f(r), where r = |x|, for x 6= 0. Assuming f tobe smooth (C2) for r 6= 0, we easily obtain that f(r) satisfies the ordinarydifferential equation

(9.17) f ′′(r) +n− 1r

f ′(r) + k2f(r) = 0

(see how to compute ∆f(r) in Example 4.10). Clearly, to get a fundamentalsolution we should find a solution f(r) with a singularity as r → 0+ which issimilar to the fundamental solution En(x) = E(0)

n (x) of the Laplace operator.For instance, if it turns out that we have found a solution f(r) of the equation(9.17) such that

f(r) = En(r)g(r), g ∈ C2([0,+∞)), g(0) = 1,

then the verbatim repetition of the verification that ∆En(x) = δ(x) yields(∆ + k2)f(|x|) = δ(x). We will explain the precise meaning of these wordslater in this Section.

9.4. Fundamental solution of the Helmholtz operator 231

The required smoothness of f can be easily deduced from the smoothnessof the fundamental solution of the Helmholtz operator, which holds becausethis operator is elliptic. However, since we did not prove the desired smooth-ness, we can avoid using it, instead deducing the smoothness from the directconstruction by separation of variables, which we sketch below. N

H We will reduce (9.17) to the Bessel equation. This may be done evenfor a more general equation

(9.18) f ′′(r) +α

rf ′(r) +

(k2 +

β

r2

)f(r) = 0,

where α, β, k are arbitrary real (or complex) constants. First, let us rewrite(9.18) in the form

r2f ′′(r) + αrf ′(r) + (β + k2r2)f(r) = 0.

Here the first three terms of the left hand side above constitute Euler’soperator

L = r2 d2

dr2+ αr

d

dr+ β,

applied to f . The Euler operator commutes with the homotheties of thereal line (i.e. with the transformations x 7→ λx, with λ ∈ R \ 0, acting uponfunctions on R by the change of variables), and rκ for any κ ∈ C are itseigenfunctions. (It is also useful to notice that the change of variable r = et

transforms Euler’s operator into an operator commuting with all translationswith respect to t, i.e., into an operator with constant coefficients.)

We can start with the following change of the unknown function in (9.18):

f(r) = rκz(r),

which leads to the equation for z(r):

(9.19) z′′(r) +α+ 2κr

z′(r) +[k2 +

κ(κ+ α− 1) + β

r2

]z(r) = 0.

Here the parameter κ is at our disposal. The first idea is to set κ = −α/2in order to cancel z′(r). Then we get

(9.20) z′′(r) +[k2 +

c

r2

]z(r) = 0.

It is easy to derive from (9.20) that for k 6= 0 the solutions z(r) of thisequation behave, for r → +∞, as solutions of the equation

u′′(r) + k2u(r) = 0.


More precisely, let, for example, k > 0 (this case is most important to us).Then there exist solutions z1(r) and z2(r) of (9.20) whose asymptotics asr → +∞ are

(9.21) z1(r) = sin kr +O

(1r

), z2(r) = cos kr +O

(1r

).

This can be proved by passing to an integral equation as we did when seek-ing asymptotics of eigenfunctions and eigenvalues of the Sturm-Liouvilleoperator. Namely, let us rewrite (9.20) in the form

z′′(r) + k2z(r) = − c

r2z(r)

and seek z(r) in the form

z(r) = C1(r) cos kr + C2(r) sin kr.

The variation of parameters method for functions Cj(r) yields that we cantake C1(r), C2(r) satisfying

C ′1(r) cos kr + C ′2(r) sin kr = 0,−kC ′1(r) sin kr + kC ′2(r) cos kr = − c

r2z(r),

which imply

C1(r) = A+∫ r

r0

c

ρ2

sin kρk

z(ρ)dρ,

C2(r) = B −∫ r

r0

c

ρ2

cos kρk

z(ρ)dρ.

We wish that C1(r)→ A and C2(r)→ B as r → +∞. But then it is naturalto take r0 = +∞ and write

(9.22) z(r) = A cos kr +B sin kr +c

k

∫ ∞r

1ρ2

sin k(r − ρ)z(ρ)dρ,

which is an integral equation for z(r). Let us rewrite it in the form

(I − T )z = A cos kr +B sin kr,

where T is an integral operator mapping z(r) to the last summand in(9.22). Consider this equation in the Banach space Cb([r0,∞)) of continu-ous bounded functions on [r0,∞), with the norm ‖z‖ = supr≥r0 |z(r)|. It isclear that

‖Tz‖ ≤ |c||k|

∫ ∞r0

1ρ2dρ‖z‖ =

c

|k|r0‖z‖.

If r0 is sufficiently large, then ‖T‖ < 1, so the operator I − T is invertible,because

(I − T )−1 = I + T + T 2 + . . . ,


where the series converges with respect to the operator norm in Cb([r0,∞).Therefore, the integral equation (9.22) has a unique solution in Cb([r0,∞)).But its continuous bounded solutions are solutions of (9.20) and if z(r) issuch a solution, then

|z(r)−A cos kr −B sin kr| ≤ C∫ ∞r

1ρ2dρ =

C

r,

i.e.

(9.23) z(r) = A cos kr +B sin kr +O

(1r

)as r → +∞.

Thus, for any A and B, the equation (9.20) has a solution with the as-ymptotic (9.23). In particular, there are solutions z1(r) and z2(r) withasymptotics (9.21). Clearly, these solutions are linearly independent. Theasymptotic (9.23) may also be expressed in the form

z(r) = A sin k(r − r0) +O

(1r

),

which makes it clear that if z(r) is real (this is so, for instance, if all constantsare real), then z(r) has infinitely many zeros with approximately the samebehavior as that of zeros of sin k(r − r0), as r → +∞.

Bessel equation and general cylindric functions.

Now, return to (9.19) and choose the parameter κ so as to get α+2κ = 1.Then we get the equation for z(r):

(9.24) z′′(r) +1rz′(r) +

[k2 − ν2

r2

]z(r) = 0,

where ν2 = κ2 − β. (In particular, we can take ν = ±κ in the case β = 0,which was our starting point; generally, ν may be not real.) Put r = k−1x

or x = kr. Introducing x as a new independent variable, we get instead of(9.24) the following equation for y(x) = z(k−1x) (or z(x) = y(kx)):

(9.25) y′′(x) +1xy′(x) +

(1− ν2

x2

)y(x) = 0.

(One should not confuse the real variable x > 0 here with the x ∈ Rn usedearlier.)

The equation (9.25) is called the Bessel equation and its solutions arecalled cylindric functions. N


H Cylindric functions appear when we try to apply the separation ofvariables (or the Fourier method) to find eigenfunctions of the Laplace op-erator in a disc and also solutions of the Dirichlet problem in a cylinder(over a disc) of finite or infinite height (this accounts for the term “cylindricfunctions”).

Eliminating, as above, y′(x) in (9.25), we see that the asymptotic of anycylindric function as x→ +∞ has the form

(9.26) y(x) =A√x

sin(x− x0) +O

(1x

).

Incidentally, substituting all parameters, we see that (9.19) in our case (forα = 1, κ = −1/2, k = 1, β = −ν2) takes the form

z′′(r) +

[1 +

14 − ν2

r2

]z(r) = 0,

which makes it clear that for ν = ±12 the Bessel equation is explicitly solvable

and the solutions are of the form coinciding with the first term in (9.26),i.e.,

y(x) =1√x

(A cosx+B sinx).

In particular, for n = 3 we can solve explicitly the equation (9.17) that arosefrom the Helmholtz equation. Namely, setting α = 2, β = 0, κ = −1, wesee that (9.19) takes the form z′′ + k2z = 0, implying that for n = 3 thespherically symmetric solutions of the Helmholtz equation are of the form

f(x) =1r

(A cos kr +B sin kr), r = |x|.

We get the fundamental solution if A = 14π . In particular, functions

−eikr

4πr, −e

−ikr

4πr, −cos kr

4πr,

are fundamental solutions, and the function u = sin krr is a solution of the

homogeneous equation

(∆ + k2)u(x) = 0, x ∈ R3.

Let us return to the general equation (9.25). We should describe somehowthe behavior of its solutions as x → 0+. The best method for this is totry to find the solution in the form of a series in x. Since ν2

x2 plays a moreimportant role than 1, as x → 0+, we may expect that y(x) behaves atzero as a solution of Euler’s equation obtained from (9.25), if 1 is removedfrom the coefficient by y(x). But solutions of Euler’s equation are linear


combinations of functions xσ and, perhaps, xσ lnx for appropriate σ. In ourcase, substituting xσ into Euler’s equation, we get σ = ±ν. Therefore, it isnatural to expect that we may find a pair of solutions of (9.25) that behaveas x±ν as x→ +0. This is, however, true up to some corrections only.

For the time being, let us seek solutions y(x) of (9.25) in the form

(9.27) y(x) = xσ(a0 + a1x+ a2x2 + . . .) = a0x

σ + a1xσ+1 + . . . .

Substituting this series in (9.25) and equating coefficients of all powers of xto zero, we get

a0σ(σ − 1) + a0σ − a0ν2 = 0

a1(σ + 1) + a1(σ + 1)− a1ν2 = 0,

· · · · · · · · · · · · · · · · · · · · · · · · · · ·ak(σ + k)(σ + k − 1) + ak(σ + k) + ak−2 − akν2 = 0, k = 2, 3, . . .

or

(9.28) a0(σ2 − ν2) = 0,

(9.29) a1[(σ + 1)2 − ν2] = 0,

(9.30) ak[(σ + k)2 − ν2] + ak−2 = 0, k = 2, 3, . . .

Without loss of generality, we may assume that a0 6= 0 (for a0 = 0 wemay replace σ by σ + k, where k is the subscript number of the first non-zero coefficient ak, and then reenumerate the coefficients). But then (9.28)implies σ = ±ν. If ν is neither integer nor half-integer, then (9.29) and(9.30) imply

a1 = a3 = . . . = 0,

ak = − ak−2

(σ + k)2 − ν2= − ak−2

(σ + ν + k)(σ − ν + k)= − ak−2

k(k + 2σ).

If k = 2m, then this yields

a2m = −a2m−21

22m(m+ σ)= . . . =

(−1)ma0

22mm!(m+ σ)(m+ σ − 1) . . . (σ + 1).

Set a0 = 12σΓ(σ+1) , where Γ denotes the Euler Γ-function. Then, using the

identity sΓ(s) = Γ(s+ 1), we get

a2m =(−1)m

22m+σm!Γ(m+ σ + 1).


The corresponding series for σ = ν is denoted by Jν(x) and is called theBessel or cylindric function of the 1st kind. Thus,

(9.31) Jν(x) =∞∑k=0

(−1)k1

k!Γ(k + ν + 1)

(x2

)2k+ν.

it is easy to see that this series converges for all complex x 6= 0.

All these calculations make sense also for an integer or half-integer σ = ν,if Re ν ≥ 0. We could assume that Re ν ≥ 0 without loss of generality,because only ν2 enters the equation. But it makes sense to consider ν

with Re ν < 0 in the formulas for the solutions, to get the second solutionsupplementing the one given by (9.31). (In other words, we could takeσ = −ν, returning to (9.27).) For ν = −1,−2, . . ., the series (9.31) maybe defined by continuity, since Γ(z) has no zeros but only poles at z =0,−1,−2, . . ..

So, the Bessel equation (9.25) for Re ν ≥ 0 always has a solution of theform Jν(x) = xνgν(x), where gν(x) is an entire analytic function of x, suchthat gν(0) = 1

Γ(ν+1) 6= 0.

We will neither seek the second solution (in the cases when we did notdo this yet) nor investigate other properties of cylindric functions. We willonly remark that in reference books and textbooks on special functions,properties of cylindric functions are discussed with the same minute details,as the properties of trigonometric functions are discussed in textbooks ontrigonometry. Cylindric functions are tabulated and are a part of standardmathematics software, such as Maple, Mathematica and MATLAB. Thesame applies to many other special functions. Many of them are, as cylin-dric functions, solutions of certain second order linear ordinary differentialequations.

For more details about the cylindric functions we refer the reader to thebooks by Lebedev [18], Whittaker and Watson [34]. N

H Let us return to Helmholtz’s equation. We have seen that (9.17)reduces to Bessel’s equation. Indeed, it is of the form (9.18) with α = n−1,β = 0. Let us pass to (9.19) and set κ = 1−α

2 = 2−n2 . Then we get (9.24),

where

ν2 =n− 2

2

[n− 2 +

2− n2

]=(n− 2

2

)2

,

9.5. Variational principle 237

implying ν = ±n−22 . Therefore, in particular, we get a solution of (9.17) of

the formf(r) = r

2−n2 Jn−2

2(kr).

Actually, this solution does not have any singularities (it expands in a seriesin powers of r2).

We may also consider the solution

f(r) = r2−n

2 J 2−n2

(kr).

If n is odd then, multiplying this function by a constant, we may get afundamental solution for Helmholtz’s operator (this solution is, actually, anelementary function). If n is even, then, together with Jn−2

2(x), we should

use the second solution of Bessel’s equation (it is called a cylindric functionof the second kind or Neumann’s function and its expansion contains lnx asx→ 0+). It can also be expressed in terms of Jn−2

2(x) via a quadrature with

the help of a well known Wronskian trick (which gives for the second solutiona 1st order linear differential equation). In any case, we see that there existsa fundamental solution E(k)

n (r), analytic in r for r 6= 0. This implies that alleigenfunctions of the Laplace operator are analytic at interior points of thedomain.

Note, however, that there is a general theorem which guarantees the an-alyticity of any solution of an elliptic equation with analytic coefficients. So-lutions of an elliptic equation with smooth (C∞) coefficients are also smooth.

9.5. Variational principle. The behavior of eigenvalues under

variation of the domain. Estimates of eigenvalues

Let A be a self-adjoint semibounded below operator with discrete spectrumin a Hilbert space H. Let ϕ1, ϕ2, . . . be a complete orthonormal systemof eigenvectors of A with the eigenvalues λ1, λ2, . . ., i.e., Aϕj = λjϕj forj = 1, 2, . . .. The discreteness of the spectrum means that λj → +∞ asj → +∞. Therefore, we may assume that eigenvalues increase with j, i.e.,

λ1 ≤ λ2 ≤ λ3 ≤ . . . ,and we will do this in what follows. Instead of the set of eigenvalues, it isconvenient to consider a (non-strictly) increasing function N(λ) defined forλ ∈ R as the number of eigenvalues not exceeding λ. This function takesonly non-negative integer values, is constant between eigenvalues, and ateigenvalues its jumps are equal to the multiplicities of these eigenvalues. Itis continuous from the right, i.e. N(λ) = N(λ+ 0).


The eigenvalues λj are easily recovered from N(λ). Namely, if λ is areal number, then it is an eigenvalue of multiplicity N(λ + 0) − N(λ − 0)(if N(λ + 0) − N(λ − 0) = 0, then λ is not an eigenvalue). If N(λ − 0) <j ≤ N(λ + 0) and j is a nonnegative integer, then λj = λ. In short, λj isuniquely defined from the condition N(λj − 0) < j ≤ N(λj + 0).

The following proposition is important.

Proposition 9.5. (Glazman’s lemma)

(9.32) N(λ) = supL ⊂ D(A)

(Au, u) ≤ λ(u, u)

u ∈ L

dimL,

where L denotes a finite dimensional linear subspace of D(A) such that(Au, u) ≤ λ(u, u) for all u ∈ L. (Under the “sup” sign stands the indicationthat supremum is taken over all L such that (Au, u) ≤ λ(u, u) for all u ∈ L.)

If A is the Friedrichs extension of A0 and λ is not an eigenvalue thenL ⊂ D(A) may be replaced by a stronger condition L ⊂ D(A0).

Proof. Let Lλ be a finite-dimensional subspace in H spanned by alleigenvectors ϕj with eigenvalues λj ≤ λ. Clearly, Lλ ⊂ D(A) and A − λIhas only non-positive eigenvalues on Lλ; therefore, it is non-positive itself,i.e., (Au, u) ≤ λ(u, u), u ∈ Lλ. In particular, we may take L = Lλ in theright-hand side of (9.32). But dimLλ = N(λ), implying that the left handside of (9.32) is not greater than the right hand side.

Let us verify that the left-hand side of (9.32) is not less than the right-hand side. Let L ⊂ D(A) and (Au, u) ≤ λ(u, u), u ∈ L. We wish toprove that dimL ≤ N(λ). Fix such a subspace L and set Mλ = L⊥λ (theorthogonal complement to Lλ). Clearly, if u ∈Mλ, then u can be expandedwith respect to the eigenfunctions ϕj with the eigenvalues λj > λ. Therefore,if u ∈ D(A)

⋂Mλ and u 6= 0, then (Au, u) > λ(u, u) implying L ∩Mλ = 0.

Now, consider the projection Eλ : H −→ H onto Lλ parallel to Mλ, i.e.,if u = v+w, where v ∈ Lλ, w ∈Mλ, then Eλu = v. Clearly, Ker Eλ = Mλ.Now, consider the operator Eλ|L : L −→ Lλ. Since (KerEλ) ∩ L = 0, thenEλ|L is injective. Therefore, dimL ≤ dimLλ = N(λ), as required.

It remains to verify the last statement of the proposition. Without lossof generality, we may assume that (Au, u) ≥ (u, u), for u ∈ D(A0). Let usmake use of the Hilbert space H1, the completion of D(A0) with respect tothe norm defined by the inner product [u, v] = (A0u, v), where u, v ∈ D(A0).

9.5. Variational principle 239

Instead of (Au, u) in (9.32), we may write [u, u]. Let ‖·‖1 be the norm in H1.Let L be any finite dimensional subspace of D(A). In particular, L ⊂ H1.In L, choose an orthonormal (with respect to the inner product in H) basise1, . . . , ep and let vectors f1, . . . , fp ∈ D(A0) be such that ‖ej − fj‖1 < δ,where δ > 0 is sufficiently small. Let L1 be a subspace spanned by f1, . . . , fp.

It is easy to see that dimL1 = dimL = p for a small δ. Indeed, if∑pj=1 cjfj = 0 with cj ∈ C, then by taking the inner product with fk, we

getp∑j=1

cj(fj , fk) = 0, k = 1, . . . , p.

But (fj , fk) = δjk + εjk, where εjk → 0 as δ → 0. Therefore, the matrix(δjk + εjk), being close to the identity matrix, is invertible for a small δ.Therefore, cj = 0 and f1, . . . , fp are linearly independent, i.e., dimL1 = p.

Further, if [u, u] ≤ λ(u, u) for all u ∈ L, then [v, v] ≤ (λ+ε)(v, v), v ∈ L1,with ε = ε(δ), such that ε(δ)→ 0 as δ → 0. Indeed, if v =

∑pj=1 cjfj , then

u =∑p

j=1 cjej is close to v with respect to ‖ · ‖1, implying the requiredstatement.

Now, set

N1(λ) = supL1 ⊂ D(A0)

(Au, u) ≤ λ(u, u)

u ∈ L1

dimL1.

Then from the above argument, it follows that N(λ − ε) ≤ N1(λ) ≤ N(λ)for any λ ∈ R and ε > 0. If λ is not an eigenvalue, then N(λ) is continuousat λ implying N1(λ) = N(λ).

Remark. Obviously, N(λ) is recovered from N1(λ) (namely, N(λ) =N1(λ+ 0)).

Another description of the eigenvalues in terms of the quadratic form ofA is sometimes more convenient. Namely, the following statement holds.

Proposition 9.6. (Courant’s variational principle)

(9.33) λ1 = minϕ∈D(A)\0

(Aϕ,ϕ)(ϕ,ϕ)

,

(9.34) λj+1 = supL⊂D(A), dimL=j

infϕ∈(L⊥\0)∩D(A)

(Aϕ,ϕ)(ϕ,ϕ)

, j = 1, 2, . . . .


If A is the Friedrichs extension of A0, then, by replacing min by inf in (9.33),we may write ϕ ∈ D(A0)\0 and L ⊂ D(A0), instead of ϕ ∈ D(A)\0and L ⊂ D(A), respectively.

Proof. We will use the same notations as in the proof of Proposition9.5. Let us prove (9.33) first. If ϕ =

∑∞j=1 cjϕj for ϕ ∈ D(A) (i.e., if∑∞

j=1 |cj |2λ2j < +∞), then (Aϕ,ϕ) =

∑∞j=1 λj |cj |2 and (ϕ,ϕ) =

∑∞j=1 |cj |2.

Therefore,

(Aϕ,ϕ) =∞∑j=1

λj |cj |2 ≥ λ1

∞∑j=1

|cj |2 = λ1(ϕ,ϕ),

and the equality is attained at ϕ = ϕ1. This proves (9.33).

Let us prove (9.34). Let Φj be the subspace spanned by ϕ1, . . . , ϕj . TakeL = Φj . We see that the right-hand side of (9.34) is not smaller than theleft hand side.

Now, we should verify that the right hand side is not greater than theleft hand side, i.e., if L ⊂ D(A) and dimL = j, then there exists a non-zero vector ϕ ∈ L⊥ ∩D(A) such that (Aϕ,ϕ) ≤ λj+1(ϕ,ϕ). But, as in theproof of Proposition 9.5, it is easy to verify that Φj+1

⋂L⊥ 6= 0 (if we had

Φj+1⋂L⊥ = 0, then the orthogonal projection onto L would be an injective

map of Φj+1 in L, contradicting to the fact that dim Φj+1 = j + 1 > j =dimL). Therefore, we may take ϕ ∈ Φj+1 ∩ L⊥, ϕ 6= 0, and then it is clearthat (Aϕ,ϕ) ≤ λj+1(ϕ,ϕ).

The last statement of Proposition 9.6 is proved as the last part of Propo-sition 9.5.

Let us return to the eigenvalues of (−∆) in a bounded domain Ω ⊂ Rn

(with the Dirichlet boundary conditions). Denote them by λj(Ω), j =1, 2, . . ., and their distribution function, N(λ), by NΩ(λ). In this caseD(A0) = D(Ω). Now let Ω1 and Ω2 be two bounded domains such thatΩ1 ⊂ Ω2. Then D(Ω1) ⊂ D(Ω2), and Proposition 9.5 implies that

(9.35) NΩ1(λ) ≤ NΩ2(λ);

therefore,λj(Ω1) ≥ λj(Ω2), j = 1, 2, . . . .

The eigenfunctions and eigenvalues may be found explicitly if Ω is a rectan-gular parallelepiped. Suppose Ω is of the form

Ω = (0, a1)× . . .× (0, an).

9.6. Problems 241

It is easy to verify that the eigenfunctions can be taken in the form

ϕk1,...,kn = Ck1,...,kn sink1πx1

a1. . . sin

knπxnan

,

where k1, . . . , kn are (strictly) positive integers, Ck1,...,kn are normalizingconstants. The eigenvalues are of the form

λk1,...,kn =(k1π

a1

)2

+ . . .+(knπ

an

)2

.

The function NΩ(λ) in our case is equal to the number of points of theform (k1πa1

, . . . , knπan ) ∈ Rn belonging to the closed ball of radius√λ with the

center at the origin. If ki are allowed to be any integers, we see that points(k1πa1

, . . . , knπan ) run over a lattice in Rn obtained by obvious homotheties ofan integer lattice. It is easy to see that NΩ(λ) for large λ has a two-sidedestimate in terms of the volume of a ball of radius

√λ, i.e.,

(9.36) C−1λn/2 ≤ NΩ(λ) ≤ Cλn/2,where C > 0.

Now note that for any non-empty bounded open set Ω ⊂ Rn we may takecubes Q1, Q2 such that Q1 ⊂ Ω ⊂ Q2. Then NQ1(λ) ≤ NΩ(λ) ≤ NQ2(λ). Itfollows that (9.36) holds for any such Ω.

Making these arguments more precise, it is possible to prove the followingasymptotic formula by H.Weyl:

NΩ(λ) ∼ (2π)−nVn ·mes(Ω) · λn/2, as λ→ +∞,where Vn is the volume of the unit ball in Rn and mes(Ω) is the Lebesguemeasure of Ω, see e.g. Courant, Hilbert [5], v.1.

9.6. Problems

9.1. Let Ω be a bounded domain in Rn with a smooth boundary. Let L bea self-adjoint operator in L2(Ω) determined by the Laplace operator in Ωwith the Dirichlet boundary conditions. The Green function of the Laplaceoperator in Ω is the Schwartz kernel of L−1, i.e., a distribution G(x, y),x, y ∈ Ω, such that

L−1f(x) =∫

ΩG(x, y)f(y)dy.

Prove that∆xG(x, y) = δ(x− y),

G|x∈∂Ω = 0,


and G is uniquely defined by these conditions. Give a physical interpretationof the Green function.

9.2. Find out what kind of singularity the Green function G(x, y) has atx = y ∈ Ω.

9.3. Prove that the Green function is symmetric, i.e.,

G(x, y) = G(y, x), x, y ∈ Ω.

9.4. Prove that the solution of the problem ∆u = f , u|∂Ω = ϕ in the domainΩ is expressed via the Green function by the formula

u(x) =∫

ΩG(x, y)f(y)dy −

∫∂Ωϕ(y)

∂G(x, y)∂ny

dSy,

where ny is the exterior normal to the boundary at y and dSy is the areaelement of the boundary at y.

9.5. The Green function of the Laplace operator in an unbounded domainΩ ⊂ Rn is a function G(x, y), x, y ∈ Ω, such that ∆xG(x, y) = δ(x − y),G|x∈∂Ω = 0 and G(x, y) → 0 as |x| → +∞ (for every fixed y ∈ Ω) if n ≥ 3;G(x, y) is bounded for |x| → +∞ for every fixed y ∈ Ω if n = 2.

Find the Green function for the half-space xn ≥ 0.

9.6. Using the result of the above problem, write a formula for the solution ofDirichlet’s problem in the half-space xn ≥ 0 for n ≥ 3. Solve this Dirichlet’sproblem also with the help of the Fourier transform with respect to x′ =(x1, . . . , xn−1) and compare the results.

9.7. Find the Green function of a disc in R2 and a ball in Rn, n ≥ 3.

9.8. Write a formula for the solution of the Dirichlet problem in a disc anda ball. Obtain in this way the Poisson formula for n = 2 from Problem 8.2.

9.9. Find the Green function of a half-ball.

9.10. Find the Green function for a quarter of R3, i.e., for (x1, x2, x3) :x2 > 0, x3 > 0 ⊂ R3.

9.6. Problems 243

9.11. The Bessel function Jν(x) is defined for x > 0 and any ν ∈ C, by theseries

Jν(x) =∞∑k=0

(−1)k

k!Γ(k + ν + 1)

(x2

)2k+ν.

Prove that if n ∈ Z+, then J−n(x) = (−1)nJn(x).

9.12. Prove that if n ∈ Z+, then the Bessel equation

y′′ +1xy′ +

(1− n2

x2

)y = 0

has for a solution not only Jn(x) but also a function defined by the series

(lnx) · x−n∞∑k=0

ckxk +

∞∑k=0

dkxk, c0 6= 0. NOT CORRECT

9.13. Prove that the Laurent expansion of ex2

(t− 1t) with respect to t (for

t 6= 0,∞) has the form

ex2

(t− 1t) =

∞∑−∞

Jn(x)tn.

9.14. Prove that the expansion of eix sinϕ into the Fourier series of ϕ is ofthe form

eix sinϕ = J0(x) + 2∞∑n=1

[J2n(x) cos 2nϕ+ iJ2n−1(x) sin(2n− 1)ϕ].

9.15. Find a complete orthonormal system of the eigenvalues and eigen-functions of the Laplace operator in a rectangle.

9.16. Let k ∈ Z+ and αk,1 < αk,2 < . . . be all the zeros of Jk(x) for x > 0.Prove that ∫ R

0Jk

(αk,nR

r)Jk

(αk,mR

r)rdr = 0, for m 6= n.

9.17. Using the result of Problem 9.16, find a complete orthonormal systemof eigenfunctions of the Laplace operator in a disc.

9.18. Describe a way to use the Fourier method to solve the Dirichlet prob-lem in a straight cylinder (of finite height) with a disc as the base.

Chapter 10

The wave equation

10.1. Physical problems leading to the wave equation

There are many physical problems leading to the wave equation

(10.1) utt = a2∆u,

where u = u(t, x), t ∈ R, x ∈ Rn, ∆ is the Laplacian with respect to x. Thefollowing more general non-homogeneous equation is also often encountered:

(10.2) utt = a2∆u+ f(t, x).

We have already seen that small oscillations of the string satisfy (10.1) (forn = 1), and in the presence of external forces they satisfy (10.2). It canbe shown that small oscillations of a membrane satisfy similar equations,if we denote by u = u(t, x), for x ∈ R2, the vertical displacement of themembrane from the (horizontal) equilibrium position. Similarly, under smalloscillations of a gas (acoustic wave) its parameters (e.g. pressure, density,displacement of gas particles) satisfy (10.1) with n = 3.

H One of the most important fields, where equations of the form (10.1)and (10.2) play a leading role, is classical electrodynamics. Before explainingthis in detail, let us recall some notations from vector analysis. (See e.g. Ch.10 in [23].)

Let F = F(x) = (F1(x), F2(x), F3(x)) be a vector field defined in anopen set Ω ⊂ R3. Here x = (x1, x2, x3) ∈ Ω, and we will assume that F issufficiently smooth, so that all derivatives of the components Fj , which weneed, are continuous. It is usually enough to assume that F ∈ C2(Ω) (which

245

246 10. The wave equation

means that every component Fj is in C2(Ω)). If f = f(x) is a sufficientlysmooth function on Ω, then its gradient

grad f = ∇f =(∂f

∂x1,∂f

∂x2,∂f

∂x3

)is an example of a vector field in Ω. Here it is convenient to consider ∇ asa vector

∇ =(

∂

∂x1,∂

∂x2,∂

∂x3

)whose components are operators acting, say, in C∞(Ω), or, as operatorsfrom Ck+1(Ω) to Ck(Ω), k = 0, 1, . . . .

The divergence of a vector field F is a scalar function

div F =∂F1

∂x1+∂F2

∂x2+∂F3

∂x3.

For a vector field F in Ω its curl is another vector field

curl F = ∇× F =(∂F3

∂x2− ∂F2

∂x3,∂F1

∂x3− ∂F3

∂x1,∂F2

∂x1− ∂F1

∂x2

).

Here the cross “×” in ∇×F stands for the cross product of two 3-componentvectors with the product of ∂/∂xj and Fk interpreted as the derivative∂Fk/∂xj .

The following identities, which hold for every smooth scalar function f

and vector field F, are easy to verify:

(10.3) div grad f = ∇ · ∇f = ∆f,

(10.4) curl grad f = ∇×∇f = 0,

(10.5) div curl F = 0,

(10.6) curl(curl F) = ∇× (∇× F) = grad div F−∆F,

where in the last term the Laplacian ∆ is applied to every component of F.

Vector analysis is easier to understand if we use the language of differen-tial forms. To this end, with each vector field F we associate two differentialforms

λF = F1dx1 + F2dx2 + F3dx3,

10.1. Physical problems 247

and

ωF = F1 dx2 ∧ dx3 + F2 dx3 ∧ dx1 + F3 dx1 ∧ dx2.

It is easy to see that dλF = ωcurl F, which can serve as a definition of curl F.Also,

dωF = div F dx1 ∧ dx2 ∧ dx3.

In particular, the identities (10.4) and (10.5) immediately follow from thewell known relation d2 = 0 for the de Rham differential d on differentialforms.

In what follows we will deal with functions and vector fields whichadditionally depend upon the time variable t. In this case all operationsgrad,div, curl should be applied with respect to the space variables x (forevery fixed t). N

H The equations of classical electrodynamics (Maxwell’s equations) are

(M.1) div E =ρ

ε0

(M.2) curl E = −∂B∂t

(M.3) div B = 0

(M.4) c2 curl B =jε0

+∂E∂t

Here E, B are vectors of electric and magnetic fields, respectively (three-dimensional vectors depending on t and x ∈ R3), ρ a scalar function of tand x (density of the electric charge), j a 3-dimensional vector of density ofthe electric current, also depending on t and x (if charges with density ρ ata given point and time move with velocity v, then j = ρv). Numbers c, ε0

are universal constants depending on the choice of units: the speed of lightand the dielectric permeability of vacuum.

For applications of the Maxwell equations it is necessary to complementthem with the formula for the Lorentz force acting on a moving charge. Thisforce is of the form

(M.5) F = q(E + v ×B),

where q is the charge, v its velocity. Formula (M.5) may be used for ex-perimental measurement of E and B (or, say, to define them as physicalquantities).


The discussion of experimental facts on which equations (M.1)-(M.5) arebased, may be found in textbooks on physics (see e.g. Feynmann, Leighton,Sands [7], vol. 2, Ch. 18).

Let us rewrite the Maxwell equations, introducing scalar and vectorpotentials ϕ and A. For simplicity assume that the fields E and B aredefined and sufficiently smooth in the whole space R4

t,x. By the Poincarelemma, it follows from (M.3) that B = curl A, where A is a vector functionof t and x determined up to a field A0 such that curl A0 = 0, hence A0 =gradψ, where ψ is a scalar function. Substituting B = curl A into (M.2),we get curl(E + ∂A

∂t ) = 0. Hence, using the Poincare lemma again, we seethat E + ∂A

∂t = − gradϕ, where ϕ is a scalar function. Therefore, instead of(M.2) and (M.3), we may write

(M.6) B = curl A

(M.7) E = − gradϕ− ∂A∂t

The vector-function A is called the vector potential, and the scalar functionϕ is called the scalar potential. Note that equations (M.6) and (M.7) donot uniquely determine potentials A and ϕ. Without affecting E and B, wemay replace A and ϕ by

(M.8) A′ = A + gradψ, ϕ′ = ϕ− ∂ψ

∂t

(this transformation of potentials is called a gauge transformation). Trans-formations (M.8) may be used to get simpler equations for potentials. Forinstance, using (M.8) we may get any value of

div A′ = div A + div gradψ = div A + ∆ψ,

since solving Poisson’s equation ∆ψ = α(t, x) we may get any value of ∆ψ,hence, of div A.

Let us derive equations for A and ϕ using two remaining Maxwell’sequations. Substituting the expression for E in terms of the potentials into(M.1), we get

(M.9) −∆ϕ− ∂

∂tdiv A =

ρ

ε0.

Substituting E and B in (M.4), we get

c2 curl curl A− ∂

∂t

(− gradϕ− ∂A

∂t

)=

jε0.

10.1. Physical problems 249

Using (10.6), we obtain

(M.10) − c2∆A + c2 grad(div A) +∂

∂tgradϕ+

∂2A∂t2

=jε0.

Now, choose div A with the help of the gauge transformation such that

(M.11) div A = − 1c2

∂ϕ

∂t.

More precisely, let first some potentials ϕ′ and A′ be given. We want to findthe function ψ in the gauge transformation (M.8) so that ϕ and A satisfy(M.11). This yields for ψ an equation of the form

(M.12)1c2

∂2ψ

∂t2−∆ψ = θ(t, x),

where θ(t, x) is a known function. This is a non-homogeneous wave equation.Suppose that we have solved this equation and, finding ψ, got ϕ and Asatisfying (M.11). Then the second and the third terms in (M.10) canceland we get

(M.13)∂2A∂t2− c2∆A =

jε0,

and (M.9) takes the form

(M.14)∂2ϕ

∂t2− c2∆ϕ =

c2ρ

ε0.

Therefore, we may assume that ϕ and A satisfy non-homogeneous waveequations (M.13), (M.14), which, together with (M.11), are equivalent tothe Maxwell equations. The gauge condition (M.11) is called the Lorentzgauge condition.

Denote by the wave operator (or d’Alembertian)

(M.15) =∂2

∂t2− c2∆

and let −1 be the convolution with a fundamental solution of this operator.Then −1 commutes with differentiations. Suppose −1j and −1ρ makesense. Then the potentials A = −1 j

ε0and ϕ = −1 c2ρ

ε0satisfy (M.13) and

(M.14).

Will the Lorentz gauge condition be satisfied for this particular choiceof the potentials? Substituting the found A and ϕ into (M.11) we see thatthis condition is satisfied if and only if

(M.16)∂ρ

∂t+ div j = 0.


But this condition means the conservation of charge. Therefore, if the con-servation of charge is observed, then a solution of the Maxwell equationsmay be found if we can solve a non-homogeneous wave equation.

Reducing Maxwell’s equations to the form (M.13), (M.14) shows thatMaxwell’s equations are invariant with respect to Lorentz transformations(linear transformations of the space-time R4

t,x preserving the Minkowskimetric c2dt2 − dx2 − dy2 − dz2). Moreover, they are invariant under thetransformations from the Poincare group, which is generated by the Lorentztransformations and translations.

In the empty space (in the absence of currents and charges) the potentialsϕ and A satisfy the wave equation of the form (10.1) for n = 3 and a = c.This implies that all components of E and B satisfy the same equation. N

10.2. Plane, spherical and cylindric waves

A plane wave is a solution of (10.1) which, for a fixed t, is constant on eachhyperplane which is parallel to a given one. By a rotation in x-space wemay make this family of planes to be of the form x1 = const. Then a planewave is a solution of (10.1) depending only on t and x1. But then it is asolution of the one-dimensional wave equation utt = a2ux1x1 , i.e., is of theform

u(t, x) = f(x1 − at) + g(x1 + at),

where f, g are arbitrary functions. Before the rotation, each of the sum-mands had clearly been of the form f(r · x − at), where r ∈ Rn, |r| = 1.Another form of expression: f(k · x − ωt), where k ∈ Rn, but may havean arbitrary length. Such a form is convenient to make the argument off dimensionless, and is most often used in physics and mechanics. If thedimension of t is [second] and that of x is [meter], then that of ω is [sec−1]and that of k is [m−1].

Substituting f(k · x − ωt) to (10.1), we see that ω2 = a2|k|2 or a =ω|k| . Clearly, the speed of the plane wave front (the plane where f takes aprescribed value) is equal to a and the equation of such a front is k ·x−ωt =const, where k and ω satisfy ω2 − a2|k|2 = 0. An important example of aplane wave is ei(k·x−ωt). (Up to now we only considered real solutions inthis section, and we will do this in the future, but we write this exponentassuming that either real or imaginary part is considered. They will alsobe solutions of the wave equation.) In this case each fixed point x oscillates

10.2. Plane, spherical and cylindric waves 251

sine-like with frequency ω and for a fixed t the wave depends sine-like onk · x, so that it varies sine-like in the direction of k with the speed |k|.

The vector k is often called wave vector. The relation ω2 = a2|k|2 is thedispersion law for the considered waves (in physics more general dispersionlaws of the form ω = ω(k) are encountered). Many solutions of the waveequation may be actually obtained as a superposition of plane waves withdifferent k. We will, however, obtain the most important types of such wavesdirectly.

In what follows, we will assume that n = 3, and study spherical waves,solutions of (10.1) depending only on t and r, where r = |x|. Thus, letu = u(t, r), where r = |x|. Then (10.1) can be rewritten in the form

(10.7)1a2utt = urr +

2rur.

Let us multiply (10.7) by r and use the identity

rurr + 2ur =∂2

∂r2(ru).

Then, clearly, (10.7) becomes

1a2

(ru)tt = (ru)rr,

implying ru(t, r) = f(r − at) + g(r + at) and

(10.8) u(t, r) =f(r − at)

r+g(r + at)

r.

This is the general form of spherical waves. The wave r−1f(r− at) divergesfrom the origin 0 ∈ R3, as from its source, and the wave r−1g(r + at), con-versely, converges to the origin (coming from infinity). In electrodynamicsone usually considers the wave coming out of a source only, discarding thesecond summand in (10.8) by physical considerations. In this case the func-tion f(r) characterizes properties of the source. The front of the emanatingspherical wave is the sphere r− at = const. It is clear that the speed of thefront is a, as before.

Let us pass to cylindric waves – solutions of (10.1) depending only on t

and the distance to the x3-axis. A cylindric wave only depends, actually, ont, x1, x2 (and even only on t and ρ =

√x2

1 + x22) so that it is a solution of

the wave equation (10.1) with n = 2. It is convenient, however, to considerit as a solution of the wave equation with n = 3, but independent of x3.

One of the methods to construct cylindric waves is as follows: take asuperposition of identical spherical waves with sources at all points of the


x3 axis (or converging to all points of the x3 axis). Let e3 be the unit vectorof the x3-axis. Then we get a cylindric wave, by setting

(10.9) v(t, x) =∫ ∞−∞

f(|x− ze3| − at)|x− ze3|

dz +∫ ∞−∞

g(|x− ze3|+ at)|x− ze3|

dz.

Set

r = |x− ze3| =√ρ2 + (x3 − z)2, ρ =

√x2

1 + x22.

Clearly, v(t, x) does not depend on x3 and only depends on t and ρ.Therefore, we may assume that x3 = 0 in (10.9). Then the integrands areeven functions of z, and it suffices to calculate the integrals over (0,+∞).For the variable of integration take r instead of z, so that

dr =z

rdz, dz =

r

zdr =

r√r2 − ρ2

dr.

Since r runs from ρ to ∞ (for z ∈ [0,∞)), we get

v(t, ρ) = 2∫ ∞ρ

f(r − at)√r2 − ρ2

dr + 2∫ ∞ρ

g(r + at)√r2 − ρ2

dr

or

(10.10) v(t, ρ) = 2∫ ∞ρ−at

f(ξ)dξ√(ξ + at)2 − ρ2

+ 2∫ ∞ρ+at

g(ξ)dξ√(ξ − at)2 − ρ2

.

All this makes sense when the integrals converge. For instance, if f, g arecontinuous functions of ξ, then for the convergence it suffices that∫ ∞

M

|f(ξ)|dξ|ξ| < +∞,

∫ ∞M

|g(ξ)|dξ|ξ| < +∞, for M > 0.

It can be shown that (10.10) is the general form of the cylindric waves.

Let us briefly describe another method of constructing cylindric waves.Seeking a cylindric wave of the form v(t, ρ) = eiωtf(ρ), we easily see thatf(ρ) should satisfy the equation

f ′′ +1ρf ′ + k2f = 0, k =

ω

a,

which reduces to the Bessel equation with ν = 0 (see Section 9). Thenwe can take a superposition of such waves (they are called monochromaticwaves) by integrating over ω.

10.3. The wave equation as a Hamiltonian system 253

10.3. The wave equation as a Hamiltonian system

In Section 2.1 we have already discussed for n = 1 the possibility of ex-pressing the wave equation as a Lagrange equation of a system with aninfinite number of degrees of freedom. Now, let us briefly perform the samefor n = 3, introducing the corresponding quanitites by analogy with theone-dimensional case. Besides, we will discuss a possibility to pass to theHamiltonian formalism. For simplicity, we will only consider real-valuedfunctions u(t, x), which are smooth functions of t with values in S(R3

x), i.e.,we will analyze the equation (10.1) in the class of rapidly decaying functionswith respect to x. Here temporarily we will denote by S(R3

x) the class ofreal-valued functions satisfying the same restrictions as in the definition ofS(R3) which was previously formulated. We prefer to do this rather thancomplicate notations, and we hope that this will not lead to a confusion. Allother functions in this section will be also assumed real-valued.

Let us introduce terminology similar to one used in classical mechanics.

The space M = S(R3x) will be referred to as the configuration space.

An element u = u(x) ∈ S(R3x) may be considered as a “set of coordinates”

of the three-dimensional system, where x is the “label” of the coordinateand u(x) is the coordinate itself with the index x. The tangent bundle onM is the direct product TM = S(R3

x) × S(R3x). As usual, the set of pairs

(u0, v) ∈ TM for a fixed u0 ∈ M is denoted by TMu0 and called thetangent space at u0 (this space is canonically isomorphic to M as in the casewhen M is a finite dimensional vector space). Elements of TM are calledtangent vectors.

A path in M is a function u(t, x) , t ∈ (a, b), infinitely differentiable withrespect to t with values in S(R3

x). The velocity of u(t, x) at t0 is the tangentvector u(t0, x), u(t0, x), where u = ∂u

∂t .

The kinetic energy is the function K on the tangent bundle introducedby the formula

K(u, v) =12

∫R3

v2(x)dx.

Given a path u(t, x) in M then, taking for each t a tangent vector u, uand the value of the kinetic energy at it, we get the function of time

K(u) =12

∫R3

[u(t, x)]2dx,

which we will call the kinetic energy along the path u.


The potential energy is the following function on M :

U(u) =a2

2

∫R3

|ux(x)|2dx,

where ux = gradu(x) =(∂u∂x1

, ∂u∂x2, ∂u∂x3

).

If there is a path in M , then the potential energy along this path is afunction of time. With the help of the canonical projection TM −→M , thefunction U is lifted to a function on TM , again denoted by U . Therefore,U(u, v) = U(u).

The Lagrangian or the Lagrange function is the function L = K − U onTM . The action along the path u(t, x), t ∈ [t0, t1], is the integral along thispath:

S = S[u] =∫ t1

t0

Ldt,

where

L = L(u(t, x), u(t, x)) =12

∫R3

[u(t, x)]2dx− a2

2

∫R3

|ux(x)|2dx.

The wave equation (10.1) may be expressed in the form δS = 0, where δSis the variation (or the differential) of the functional S taken along pathsu(t, x) with fixed source u(t0, x) and the target u(t1, x). Indeed, if δu(t, x)is an admissible variation of the path, i.e., a smooth function of t ∈ [t0, t1]with values in S(R3

x), such that δu(t0, x) = δu(t1, x) = 0, then for such avariation of the path, δS takes the form

δS =d

dεS[u+ ε(δu)]|ε=0 =

∫ t1

t0

∫R3

u(δu)dxdt− a2

∫ t1

t0

∫R3

ux · δuxdxdt

= −∫ ∫

∂2u

∂t2δudxdt+ a2

∫ ∫(∆u)δudxdt =

∫ ∫(−u)δudxdt,

where = ∂2

∂t2− a2∆x is the d’Alembertian.

It is clear now that δS = 0 is equivalent to u = 0, i.e., to the waveequation (10.1).

To pass to the Hamiltonian formalism, we should now introduce thecotangent bundle T ∗M . But in our case each TMu0 is endowed with theinner product determined by the quadratic form 2K which makes it naturalto set T ∗Mu0 = TMu0 and T ∗M = TM .

The Hamiltonian or energy is the following function on TM :

(10.11) H = H(u, v) = K + U =12

∫R3

v2(x)dx+a2

2

∫R3

|ux(x)|2dx.


Given a path u(t, x), we see that H(u, u) is a function of time along thispath.

Proposition 10.1. (The energy conservation law) Energy is constant alongany path u(t, x) satisfying u = 0.

Proof. Along the path u(t, x), we have

dH

dt=∫

R3

uudx+ a2

∫R3

ux · uxdx =∫

R3

uudx− a2

∫R3

∆u · udx

=∫

R3

( u) · udx = 0,

as required.

Corollary 10.2. For any ϕ,ψ ∈ S(R3x) there exists at most one path u(t, x)

satisfying u = 0 and initial conditions

u(t0, x) = ϕ(x), u(t0, x) = ψ(x).

Proof. It suffices to verify that if ϕ ≡ ψ ≡ 0, then u ≡ 0. But thisfollows immediately from H = const = 0 along this path.

The Hamiltonian expression of equations uses also the symplectic formon T ∗M . To explain and motivate further calculations, let us start witha toy model case of a particle which moves along the line R in a potentialfield given by a potential energy U = U(x) (see Appendix to Chapter 2).Here M = R, TM = R × R, T ∗M = R × R, the equation of motion ismx = −U ′(x), and the enery is H(x, x) = 1

2mx2 + U(x). Introducing the

momentum p = mx and using q instead of x (as physicists do) we can rewritethe Hamiltonian in the form

(10.12) H(q, p) =p2

2m+ U(q)

and then the equation of motion is equivalent to a Hamiltonian system

(10.13)

q = ∂H

∂p

p = −∂H∂q .

Let us agree that when we talk about a point in coordinates q, p, thenit is understood that we mean a point in T ∗M , and we will simply writeq, p ∈ T ∗M .

Another important object (in this toy model) is the symplectic 2-formω = dp ∧ dq on T ∗M . It can be also considered as a skew-symmetric formon each tangent space Tq,p(T ∗M), which in this case can identified with the


set of 4-tuples q, p,Xq, Xp, where all entries are reals, and Xq, Xp arethe natural coordinates along the fiber of T (T ∗M) over the point q, p:

ω(q, p,Xq, Xp, q, p, Yq, Yp) = XpYq −XqYp.

Now with any vector field X on T ∗M we can associate a 1-form αX on T ∗Mgiven by the relation

(10.14) αX(Y ) = ω(Y,X).

Note that the form ω is non-degenerate in the following sense: if X is atangent vector to T ∗M at z ∈ T ∗M , such that ω(X,Y ) = 0 for all Y ∈Tz(T ∗M), then X = 0. Therefore, the vector field X is uniquely defined byits 1-form αX . So we will denote by Xα the vector field corresponding to any1-form α on T ∗M , i.e. the relations α = αX and X = Xα are equivalent.

Let us try to find X = XdH where H is a smooth function on T ∗M (aHamiltonian). By definition we should have for any vector field Y on T ∗M

∂H

∂qYq +

∂H

∂pYp = XqYp −XpYq,

hence

Xq =∂H

∂p, Xp = −∂H

∂q,

so XdH is the vector field corresponding to the Hamiltonian system (10.13).Note that this argument works for any smooth Hamiltonian H, not only forthe specific one given by (10.12).

The arguments and calculations above can be easily extended to the caseof motion in M = Rn. In this case we should allow x, p, q to be vectors inRn, x = (x1, . . . , xn), p = (p1, . . . , pn), q = (q1, . . . , qn), the potential energyU = U(x) be a smooth scalar (real-valued) function on Rn, then take theHamiltonian

(10.15) H(q, p) =n∑j=1

p2j

2mj+ U(q),

where q = x. Here H(q, p) should be considered as a function on T ∗M =R2n. Now the equations of motion

mj xj = − ∂U∂xj

, j = 1, . . . , n,

again can be rewritten as a Hamiltonian system (10.13), where ∂H/∂p and∂H/∂q should be understood as gradients with respect to variables p and q


respectively, for example,

∂H

∂p=(∂H

∂p1, . . . ,

∂H

∂pn

).

Instead of the form dp ∧ dq we should take the 2-form

(10.16) ω =n∑j=1

dpj ∧ dqj .

which is usually called canonical symplectic form on T ∗M . It is non-degeneratein the sense explained above. Therefore, (10.14) again defines a one-one cor-respondence between vector-fields and 1-forms on T ∗M . Then again it iseasy to check that the 1-form dH corresponds to the vector field whichgenerates the Hamiltonian system (10.13).

The reader may wonder why do we need fancy notations like M , T ∗M forusual vector spaces Rn, R2n. The point is that in fact M can be an arbitrarysmooth manifold. In fact, choosing local coordinates x = (x1, . . . , xn) in anopen set G ⊂ M , we can define natural coordinates q, p in the cotangentbundle T ∗M , so that q, p represents the cotangent vector pdq = p1dq1 + · · ·+pndqn at q = x. Then the 1-form pdq is well defined on M , i.e. does notdepend on the choice of local coordinates. (We leave the proof to the readeras an easy exercise.) Therefore, the 2-form ω = d(pdq) is also well definedon M .

In even more general context, a 2-form ω on a manifold Z is calledsymplectic if it is closed and non-degenerate. Then the same machineryworks. So, starting with a Hamiltonian H (a smooth function on Z), we cantake the 1-form dH and the corresponding vector field XH . The system ofODE

(10.17) z = XH(z)

is called a Hamiltonian system with the Hamiltonian H. Locally, it is nothingnew: by the Darboux theorem we can choose local coordinates so that ω isgiven by (10.16) in these coordinates. Therefore, the Hamiltonian system(10.17) has the canonical form (10.13). More details on symplectic structure,Hamiltonian systems and related topics can be found in [3].

Now let us make a leap to infinite-dimensional spaces, namely, return toM = S(R3

x).

The symplectic form should be defined on each tangent space to T ∗Mat a fixed point. In our case this tangent space is naturally identified with


T ∗M = TM itself, so we have T (T ∗M) = T ∗M × T ∗M = TM × TM . Anelement of T (T ∗M) should be expressed as a four-tuple u, v, δu, δv con-sisting of functions belonging to S(Rn). The symplectic form itself dependsonly on δu, δv. For brevity let us set δu = α , δv = β. Then define thesymplectic form to be(10.18)

[u, v, α, β, u, v, α1, β1] = [α, β, α1, β1] =∫

R3

(αβ1 − βα1)dx.

(This is an analogue of the canonical 2-form ω =∑n

j=1 dpj ∧ dqj , wheresummation is replaced by integration, though we use brackets instead of ωto simplify notations.)

Choosing an arbitrary u, v ∈ T ∗M , let us take variation of H =H(u, v) (which replaces the differential in the infinite-dimensional case) inthe direction of a vector u, v, δu, δv ∈ T (T ∗M). In our case u, v, δu, δv arearbitrary functions from S(R3

x). We obtain, differentiating the expression ofH in (10.11) and integrating by parts,

δH(δu, δv) =∫

R3

(a2 δH

δux· δux + v δv

)dx =

∫R3

(−a2∆u δu+v δv)dx =∫

R3

(α δu+β δv)dx,

where α = −a2∆u and β = v. Now, if we take another vector Y =u, v, α1, β1 ∈ T (T ∗M) (with the same u, v), and denote by XH Hamil-tonian vector field with the Hamiltonian H, then we should have [Y,XH ] =δH(Y ), which gives∫

R3

(α1β − β1α)dx =∫

R3

(α1v + β1a2∆u)dx =

∫R3

(α1Xα + β1Xβ)dx,

where u, v,Xα, Xβ is the vector of the Hamiltonian vector field on T ∗M .Since α1 and β1 are arbitrary, we should have Xα = v, Xβ = a2∆u, and thecorresponding differential equation on T ∗M has the form

u = v

v = a2∆u,

which is equivalent to the wave equation (10.1) for u. So we establishedthat the wave equation can be rewritten as a Hamiltonian system with theHamiltonian (10.11) and the symplectic form given by (10.18).

Later we will use the fact that the flow of our Hamiltonian system pre-serves the symplectic form (or, as one says, the transformations in this floware canonical). This is true for general Hamiltonian systems (see e.g. [3]) but

10.4. Spherical waves and the Cauchy problem 259

we will not prove this. Instead we will directly establish the correspondinganalytical fact for the wave equation.

Proposition 10.3. Let u(t, x) and v(t, x) be two paths. Consider the fol-lowing function of t:

[u, v] =∫

R3

(uv − uv)dx.

If u = v = 0, then [u, v] = const.

Proof. We have

d

dt[u, v] =

∫R3

d

dt(uv−uv)dx =

∫R3

(uv−uv)dx = a2

∫R3

(u·∆v−∆u·v)dx = 0,

as required.

10.4. A spherical wave caused by an instant flash and a

solution of the Cauchy problem for the 3-dimensional

wave equation

Let us return to spherical waves and consider the diverging wave

(10.19)f(r − at)

r, r = |x|.

The function f(−at) characterizes the intensity of the source (located atx = 0) at the moment t. It is interesting to take the wave emitted by aninstant flash at the source. This corresponds to f(ξ) = δ(ξ). We get thewave

(10.20)δ(r − at)

r=δ(r − at)

at, r = |x|,

whose meaning is to be clarified.

It is possible to make sense to (10.20) in many equivalent ways. We willconsider it as a distribution with respect to x depending on t as a parameter.

First, consider the wave (10.19) and set f = fk, where fk(ξ) ∈ C∞0 (R1),fk(ξ) ≥ 0, fk(ξ) = 0 for |ξ| ≥ 1/k and

∫fk(ξ)dξ = 1, so that fk(ξ) → δ(ξ)

as k → +∞. Now, set

(10.21) δ(r − at) = limk→∞

fk(r − at),

if this limit exists in D′(R3) (or, which is the same, in E ′(R3), since supportsof all functions fk(r − at) belong to a fixed compact provided t belongs toa fixed finite interval of R). Clearly, this limit exists for t < 0 and is 0, i.e.,


δ(r− at) = 0 for t < 0. We will see that the limit exists also for t > 0. Thenthis limit is the distribution δ(r − at) ∈ E ′(R3) and, obviously,

supp δ(r − at) = x : |x| = at.

Since 1/r = 1/|x| is a C∞-function in a neighborhood of supp δ(r − at) fort 6= 0, then the distribution δ(r−at)

r is well-defined and

(10.22)δ(r − at)

r= lim

k→∞

fk(r − at)r

, t 6= 0.

Let us prove the existence of the limit (10.21) for t > 0 and compute it. Letϕ ∈ D(R3). Let us express the integral 〈fk(r − at), ϕ〉 in polar coordinates

〈fk(r − at), ϕ〉 =∫

R3

fk(|x| − at)ϕ(x)dx =

=∫ ∞

0

(∫|x|=r

fk(r − at)ϕ(x)dSr

)dr =

=∫ ∞

0fk(r − at)

(∫|x|=r

ϕ(x)dSr

)dr,

where dSr is the area element of the sphere of radius r. Clearly,∫|x|=r ϕ(x)dSr

is an infinitely differentiable function of r for r > 0. Since limk→∞

fk(ξ) = δ(ξ),

we obviously get

limk→∞〈fk(r − at), ϕ〉 =

∫|x|=at

ϕ(x)dSat

and, therefore, the limit of (10.21) exists and

〈δ(r − at), ϕ〉 =∫|x|=at

ϕ(x)dSat.

Passing to the integration over the unit sphere we get

〈δ(r − at), ϕ〉 = a2t2∫|x′|=1

ϕ((at)x′)dS1,

implying

(10.23)⟨δ(r − at)

r, ϕ

⟩= at

∫|x′|=1

ϕ((at)x′)dS1.

It is worthwhile to study the dependence on the parameter t. It is clear from(10.23) that

(10.24) limt→0

δ(r − at)r

= 0


and, by continuity, we set δ(r−at)r |t=0 = 0. Finally, it is obvious from (10.23)

that δ(r−at)r is infinitely differentiable with respect to t for t > 0 (in the weak

topology). Derivatives with respect to t are easy to compute. For instance,

(10.25)⟨∂

∂t

δ(r − at)r

, ϕ

⟩=

d

dt

⟨δ(r − at)

r, ϕ

⟩= a

∫|x′|=1

ϕ((at)x′)dS1 + a2t3∑j=1

∫|x′|=1

xj∂ϕ

∂xj((at)x′)dS1.

This, in particular, implies that

limt→+0

∂

∂t

δ(r − at)r

= 4πaδ(x).

(the first of integrals in (10.25) tends to 4πaϕ(0) and the second one to 0).Finally, by (10.22) it is clear that

δ(r − at)

r= 0, t > 0.

Therefore, the distribution δ(r−at)r is a solution of the wave equation for t > 0


(10.26)δ(r − at)

r

∣∣∣∣t=0

= 0,∂

∂t

δ(r − at)r

∣∣∣∣t=0

= 4πaδ(x).

The converging wave δ(r+at)r is similarly constructed. It is equal to 0 for

t > 0 and for t < 0 satisfies the wave equation with the initial conditions

(10.27)δ(r + at)

r

∣∣∣∣t=0

= 0,∂

∂t

δ(r + at)r

∣∣∣∣t=0

= −4πaδ(x),

(this is clear, for instance, because δ(r+at)r is obtained from δ(r−at)

r by re-placing t with −t). Besides, we can perform translations with respect to thespatial and time variables and consider waves

δ(|x− x0| − a(t− t0))|x− x0|

,δ(|x− x0|+ a(t− t0))

|x− x0|,

which are also solutions of the wave equation with initial values (for t = t0)obtained by replacing δ(x) with δ(x− x0) in (10.26) and (10.27).

Now, we want to apply the invariance of the symplectic form (Proposi-tion 10.3) to an arbitrary solution u(t, x) of the equation u = 0 and to theconverging wave

v(t, x) =δ(|x− x0|+ a(t− t0))

|x− x0|.


We will assume that u ∈ C1([0, t0] × R3x) and u = 0 holds (where is

understood in the distributional sense). Then we can define

[u, v] =∫

R3

(uv − uv)dx =⟨∂

∂tv(t, x), u(t, x)

⟩−⟨v(t, x),

∂

∂tu(t, x)

⟩,

where 〈·, ·〉 denotes the value of the distribution of the variable x at a testfunction (t is considered as a parameter) which makes sense for u ∈ C1

(not only for u ∈ C∞0 ) due to explicit formulas (10.23), (10.25). Now, weuse [u, v] = const, which holds for u ∈ C2 and is proved similar to Propo-sition 10.3 or by approximating u and v by smooth solutions of the waveequation with compact (with respect to x) supports (this can be performed,for instance, with the help of mollifying). Actually, we only have to ver-ify that d

dt [u, v] = [u, v] + [u, v], which is clear, for example, from explicitformulas (10.23), (10.25) which define the functionals v and v.

Now, let us explicitly write the relation

(10.28) [u, v]|t=0 = [u, v]|t=t0 .

Set u|t=0 = ϕ(x), ∂u∂t

∣∣t=0

= ψ(x). Then

[u, v]|t=0 =∂

∂t

⟨δ(|x− x0|+ a(t− t0))

|x− x0|, ϕ(x)

⟩∣∣∣∣t=0

−⟨δ(|x− x0|+ a(t− t0))

|x− x0|, ψ(x)

⟩∣∣∣∣t=0

= − ∂

∂t0

⟨δ(|x− x0| − at0)|x− x0|

, ϕ(x)⟩−⟨δ(|x− x0| − at0)|x− x0|

, ψ(x)⟩

= − 1at0

∫|x−x0|=at0

ψ(x)dSat0 −∂

∂t0

(1at0

∫|x−x0|=at0

ϕ(x)dSat0

).

Further, by (10.27), we have

[u, v]|t=t0 = −4πa〈δ(x− x0), u(t0, x)〉 = −4πau(t0, x0).

Therefore, (10.28) can be expressed in the form

−4πau(t0, x0) = − 1at0

∫|x−x0|=at0

ψ(x)dSat0−∂

∂t0

(1at0

∫|x−x0|=at0

ϕ(x)dSat0

).

Replacing t0, x0, x by t, x, y we get(10.29)

u(t, x) =∂

∂t

(1

4πa2t

∫|y−x|=at

ϕ(y)dSat

)+

14πa2t

∫|y−x|=at

ψ(y)dSat,

where dSat is the area element of the sphere y : |y − x| = at.


We obtained the formula for the solution of the Cauchy problem

(10.30) u = 0, u|t=0 = ϕ(x),∂u

∂t

∣∣∣∣t=0

= ψ(x).

Formula (10.29) is called the Kirchhoff formula. Let us derive some corol-laries of it.

1) The solution of the Cauchy problem is unique and continuously de-pends on initial data in appropriate topology (e.g. if ψ continuously varies inC(R3) and ϕ continuously varies in C1(R3), then u(t, x) continuously variesin C(R3) for each t > 0).

2) The value u(t, x) only depends on the initial data near the spherey : |y − x| = at.

Suppose that for t = 0 the initial value differs from 0 in a small neigh-borhood of one point x0 only. But then at t the perturbation differs from 0in a small neighborhood of the sphere x : |x− x0| = at only. This meansthat a perturbation spreads at a speed a and vanishes after a short time ateach observation point. Therefore, the spreading wave has a leading edgeand a trailing one.

A strictly localized initial state is observed later at another point as aphenomenon, which is also strictly localized. This phenomenon is called theHuygens principle.

Since Huygens’ principle holds for the wave equation for n = 3, we maytransmit information with the help of sound or light.

As we will see later, the Huygens principle does not hold for n = 2(once started, oscillations never cease, e.g. waves on the water). Therefore,the residents of the Flatland (the flat 2-dimensional world described byE.A.Abbott in 1884) would find it very difficult to pass information throughtheir space (the plane).

We have not yet proved the existence of the solution for the Cauchyproblem (10.30). The simplest way to do this is to take the function u(t, x)constructed by the Kirchhoff formula (10.29) and verify that it is a solutionof (10.30). The fulfillment of u = 0 may be deduced from the fact thatthe right-hand side of (10.29) is the sum of two convolutions ϕ ∗ ∂

∂t1

4πδ(r−at)

r

and ψ ∗ 14π

δ(r−at)r (taken with respect to x ∈ R3 with t as a parameter) of ϕ

and ψ with distributions of x depending smoothly on t and satisfying in anatural sense the wave equation (for t > 0).


We will use more convenient arguments. In order not to bother withdistributions depending on a parameter, it is convenient to consider δ(r−at)

r

as a distribution on R4t,x by setting⟨

δ(r − at)r

, ϕ(t, x)⟩t,x

=∫ ∞

0

⟨δ(r − at)

r, ϕ(t, x)

⟩x

dt =∫ ∞

0

1atdt

∫|x|=at

ϕ(t, x)dSat

=∫ ∞

0dt at

∫|x′|=1

ϕ(t, atx′)dS1, ϕ ∈ D(R4).

This formula implies in particular, that the functional on D(R4) determinedby the left hand side is a distribution which we will again denote by δ(r−at)

r .Therefore, we may assume that δ(r−at)

r ∈ D′(R4).

Clearly, supp δ(r−at)r = K+, where K+ is the upper sheet of the light

cone (t, x) : |x|2 − a2t2 = 0; namely,

K+ = (t, x) : |x| = at = (t, x) : |x|2 − a2t2 = 0, t ≥ 0.

It is also easy to verify that δ(r−at)r = 0 for |t|+ |x| 6= 0. Indeed, δ(r−at)r = 0

outside K+; therefore, it suffices to check that δ(r−at)r = 0 for t > 0. But

this is clear, for example, from (10.22) which is true also when the limit isunderstood in D′((t, x) : t > ε > 0) for any ε > 0.

It is easy to prove that the right-hand side of (10.29) can be written inthe form

(10.31) u = [ψ(x)⊗ δ(t)] ∗ 14π

δ(r − at)r

+ [ϕ(x)⊗ δ(t)] ∗ ∂∂t

14π

δ(r − at)r

and then u = 0 obviously holds due to the properties of convolutions.

The initial conditions in (10.30) may be also deduced from (10.31); wewill, however, deduce them directly from the structure of formula (10.29).Note that Kirchhoff’s formula is of the form

u = uψ +∂

∂tuϕ,

where

uψ(t, x) =1

4πa2t

∫|y−x|=at

ψ(y)dSat,

and uϕ is a similar expression obtained by replacing ψ by ϕ. This structureof Kirchhoff’s formula is not accidental. Indeed, suppose we have provedthat uψ is a solution of the Cauchy problem

(10.32) uψ = 0, uψ|t=0 = 0,∂uψ∂t

∣∣∣∣t=0

= ψ.

10.5. The fundamental solution for the three-dimensional operator 265

Let us prove then that vϕ = ∂∂tuϕ is a solution of the Cauchy problem

(10.33) vϕ = 0, vϕ|t=0 = ϕ,∂vϕ∂t

∣∣∣∣t=0

= 0.

The equation vϕ = 0 obviously holds and so does vϕ|t=0 = ϕ due tothe second equation in (10.32). It remains to verify the second equality in(10.33). Assuming that uϕ ∈ C2 for t ≥ 0, we get

∂vϕ∂t

∣∣∣∣t=0

=∂2uϕ∂t2

∣∣∣∣t=0

= a2∆uϕ|t=0 = a2∆(uϕ|t=0) = 0,

as required. The condition uϕ ∈ C2(t ≥ 0) holds, for example, for ϕ ∈C2(R3), as is clear for example from the expression

uψ(t, x) =t

4π

∫|y′|=1

ψ(x+ aty′)dS1.

This expression makes (10.32) obvious for ψ ∈ C1(R3) also.

Thus, if ψ ∈ C1(R3), ϕ ∈ C2(R3) then Kirchhoff’s formula (10.29) givesa solution of the Cauchy problem (10.30) (the equation u = 0 holds in thedistributional sense for t > 0). If we additionally assume that ψ ∈ C2(R3)and ϕ ∈ C3(R3), then u ∈ C2(R3) and u = 0 holds in the classical sense.Thus the Cauchy problem (10.30) is uniquely solvable. The Kirchhoff’sformula makes it also clear that it is well-posed.

10.5. The fundamental solution for the three-dimensional

wave operator and a solution of the non-homogeneous

wave equation

Set

E3(t, x) =1

4πaδ(|x| − at)|x| ∈ D′(R4).

Theorem 10.4. The distribution E3(t, x) satisfies E3(t, x) = δ(t, x), i.e.,it is a fundamental solution for the d’Alembert operator.

Proof. The statement actually follows from the fact that

E3(t, x) = 0 for |t|+ |x| 6= 0, E3(t, x) = 0 for t < 0,E3(+0, x) = 0, ∂E3

∂t (+0, x) = δ(x).

The function E3(t, x) may be considered as a smooth function of t (for t 6= 0)with values in D′(R3) and for t = 0 it is continuous while ∂E3

∂t has a jumpequal to δ(x). Therefore, applying = ∂2

∂t2−a2∆x to E3, we get δ(t)⊗δ(x) =

δ(t, x).


It is difficult to make the above arguments entirely rigorous (if we try todo this, we face the necessity of a cumbersome struggle with the topologyin D′(R3)). We may, however, consider all the above as a heuristic hint anddirectly verify that E3 = δ. This can be done similarly to the proof for theheat equation (see the proof of Theorem 7.5) and it is left to the reader asan exercise.

Now, using E3(t, x) we can solve the non-homogeneous wave equationu = f by writing

(10.34) u = E3 ∗ f,

if the convolution in the right-hand side makes sense. The convolution(10.34) is called the retarded potential. In a more detailed expression,the retarded potential is of the form

u(t, x) =1

4πa

∫ ∞0

dτ

∫|y|=aτ

f(t− τ, x− y)|y| dSaτ(10.35)

=1

4πa2

∫ ∞0

dτ

τ

∫|y|=aτ

f(t− τ, x− y)dSaτ

(here the integration in the second integrals is done over y).

With the help of E3(t, x), we may obtain the Kirchhoff formula (10.29)for the solution of the Cauchy problem (10.30) in a different way. Namely,let a solution u(t, x) of the Cauchy problem (10.30) be given for t > 0.Extend it by zero onto the half-space t : t < 0 and consider the obtaineddistribution u ∈ D′(R4). Clearly,

(10.36) u = δ′(t)⊗ ϕ(x) + δ(t)⊗ ψ(x).

A solution of this equation vanishing for t < 0 may be found in the form ofa retarded potential (10.34) taking f equal to the right-hand side of (10.36).Clearly, we again get the Kirchhoff formula.

Note also that the retarded potential u(t, x) obtained via (10.35) dependsonly on values f(t′, x′) for t′ ≤ t and |x′−x| = a(t−t′), i.e., on values of f atpoints of the lower part of the light cone with the vertex at (t, x) (the word“lower” here is understood with the reference to the direction of the t-axis).This is what the term “retarded potential” really means. This property ofthe retarded potential is ensured by the fact that supp E3 ⊂ K+.

A fundamental solution for with this property is unique. Even astronger result holds:

10.6. The two-dimensional wave equation (the descent method) 267

Theorem 10.5. There exists exactly one fundamental solution for withthe support belonging to the half-space (t, x) : t ≥ 0, namely, E3(t, x).

Proof If there were two such solutions, then their difference u(t, x) ∈D′(R4) would had satisfied the wave equation u = 0 and would had van-ished for t < 0. If u were a smooth function, then the uniqueness of thesolution of the Cauchy problem would imply u ≡ 0 (Cauchy initial valuesvanish for t = t0 < 0). We can, however, make u smooth by mollifying.

Let ϕε ∈ D(R4), suppϕε ⊂ (t, x) : |t|+ |x| ≤ ε, ϕε ≥ 0 and∫ϕεdtdx =

1, so that ϕε → δ(t, x) in D′(R4) as ε → +0. Then limε→+0(u ∗ ϕε) = u inD′(R4) (see Proposition 5.7). But u ∗ ϕε ∈ C∞(R4) and

(u ∗ ϕε) = (u) ∗ ϕε = 0,

and also

supp(u ∗ ϕε) ⊂ suppu+ suppϕε ⊂ (t, x) : t ≥ −ε.Therefore, due to the uniqueness of the smooth solution of the Cauchy prob-lem for the wave equation, u ∗ ϕε ≡ 0 for any ε > 0, implying u ≡ 0, asrequired.

The importance of Theorem 10.5 stems from the fact that among all thefundamental solutions it chooses the only one which satisfies the causalityprinciple , which claims that it is impossible to transmit information tothe “past”. Therefore, in electrodynamics, to solve equations of the formu = f , satisfied by the scalar and vector potentials, one uses just thisfundamental solution. It seems that the Nature itself has chosen the solutionsatisfying the causality principle among all the solutions of the equation u = f (within the limits of the current accuracy of experiments).

10.6. The two-dimensional wave equation (the descent

method)

Let us solve the Cauchy problem for the equation

(10.37)∂2u

∂t2= a2

(∂2u

∂x21

+∂2u

∂x22

), u = u(t, x1, x2),


(10.38) u|t=0 = ϕ(x1, x2),∂u

∂t

∣∣∣∣t=0

= ψ(x1, x2).

The idea of the solution (the descent method) is very simple: introduce anadditional variable x3 and find the solution of the Cauchy problem for the


three-dimensional wave equation utt = a2∆u (where ∆ is the Laplacian withrespect to x1, x2, x3), with the initial conditions (10.38), which are indepen-dent of x3. Then the solution u(t, x1, x2, x3) does not actually depend on x3,since the function uz(t, x1, x2, x3) = u(t, x1, x2, x3 + z) is a solution of thesame equation utt = a2∆u for any z with the same initial conditions (10.38);therefore, the uniqueness theorem for the solution of the Cauchy problemfor the three-dimensional wave equation implies that uz does not dependon z, i.e., u does not depend on x3. Therefore the solution of the Cauchyproblem (10.37)–(10.38) does exist (e.g. for any ϕ ∈ C2(R2), ψ ∈ C1(R2)).It is unique due to the uniqueness theorem concerning the three-dimensionalcase since the solution of (10.37)–(10.38) can also be regarded as a solutionof the three-dimensional Cauchy problem.

Now, let us express u(t, x1, x2) via the Kirchhoff formula

u(t, x) =∂

∂t

(1

4πa2t

∫|y−x|=at

ϕ(y1, y2)dSat

)(10.39)

+1

4πa2t

∫|y−x|=at

ψ(y1, y2)dSat,

where y = (y1, y2, y3), dSat is the area element of the sphere y : y ∈R3, |y − x| = at and x = (x1, x2) = (x1, x2, 0) (we identify the points ofR2 with the points of R3 whose third coordinate is 0; this coordinate mighthave any other value since it does not affect anything). Let us rewrite thesecond summand in formula (10.39), which we will denote uψ:

(10.40) uψ =1

4πa2t

∫|y−x|=at

ψ(y1, y2)dSat

(the first summand in (10.39) is of the form ∂∂tuϕ).

Consider the sphere in R3y over which we integrate in (10.40). This is

the sphere with the center at x of radius at (see Fig. 1). We are to integrate

y1

y2

y3

x

Figure 1. To the descent method

10.6. The two-dimensional wave equation (the descent method) 269

a function, which is independent upon y3, over the sphere. This actuallymeans that we integrate twice over the projection of the sphere onto theplane y3 = 0. Let dy1dy2 be the Lebesgue measure of this plane, dSat thearea element of the sphere at y whose projection is equal to dy1dy2. Clearly,dy1dy2 = | cosα(y)|dSat, where α(y) is the angle between the normal tothe sphere and the y3-axis. But the normal is proportional to the vectory − x = (y1 − x1, y2 − x2, y3) of length at. Let us integrate over the upperhalf-sphere (y3 > 0) and then double the result. Then |y − x| = at implies

y3 =√a2t2 − (y1 − x1)2 − (y2 − x2)2,

cosα(y) =y3

at=

1at

√a2t2 − (y1 − x1)2 − (y2 − x2)2,

dSat =dy1dy2

cosα(y)=

atdy1dy2√a2t2 − (y1 − x1)2 − (y2 − x2)2

.

Therefore, we can rewrite (10.40) in the form

uψ(t, x) =1

2πa

∫|y−x|≤at

ψ(y)dy√a2t2 − |y − x|2

,

where x = (x1, x2), y = (y1, y2), dy = dy1dy2. The solution of the Cauchyproblem (10.37)–(10.38) is given by the formula(10.41)

u(t, x) =∂

∂t

(1

2πa

∫|y−x|≤at

ϕ(y)dy√a2t2 − |y − x|2

)+

12πa

∫|y−x|≤at

ψ(y)dy√a2t2 − |y − x|2

known as Poisson’s formula.

It is clear from Poisson’s formula that the value of the solution u(t, x) atx for n = 2 depends upon initial values ϕ(y) and ψ(y) in the disc y : |y−x| ≤at and not only on its boundary (recall that for n = 3 it was sufficient toknow initial values on the sphere y : |y − x| = at). In particular, if thesupports on ϕ,ψ are near origin then the solution in a neighborhood of x isalways non-zero after a certain moment. Therefore, a localized perturbationis not seen as a localized one from another point, i.e., the wave does not gotracelessly, but leaves an aftereffect, though decreasing with time as 1

t , as isclear from (10.41). In other words, the Huygens principle fails for n = 2.

The method of descent enables us also to derive d’Alembert’s formulafrom Poisson’s formula, the former determining the solution of the Cauchyproblem for the one-dimensional wave equation. However, we have alreadyobtained this formula by another method.


Let us also find a fundamental solution for the two-dimensional waveoperator. As in the three-dimensional case, we need to solve the Cauchyproblem with the initial conditions

u|t=0 = 0, ut|t=0 = δ(x).

But it is clear from Poisson’s formula that such a solution E2(t, x) has theform

E2(t, x) =θ(at− |x|)

2πa√a2t2 − |x|2

, x ∈ R2.

Now, it is easy to verify directly that E2(t, x) is locally integrable and isa fundamental solution for the two-dimensional wave operator. The latterfact is verified in exactly the same way as in the three-dimensional case andwe leave this to the reader as an exercise.

Finally, d’Alembert’s formula makes it clear that a fundamental solutionfor the one-dimensional wave operator ∂2

∂t2− a2 ∂2

∂x2 is of the form

E1(t, x) =12aθ(at− |x|).

10.7. Problems

10.1. By separation of variables find cylindric waves in R3.

10.2. Solve the “explosion-of-a-ball problem” in the three-dimensional space:find u = u(t, x), x ∈ R3, if

utt = a2∆u, u|t=0 = ϕ, ut|t=0 = 0,

where ϕ is the characteristic function of the ball x : |x| ≤ R. Draw acartoon describing the behavior of u(t, x) as a function of t and |x|.

10.3. The same as in the above problem but with different initial conditions:u|t=0 = 0, ut|t=0 = ψ, where ψ is the characteristic function of the ballx : |x| ≤ R.

10.4. Using the result of Problem 5.1 c), write, with the help of the Fouriertransform, a formula for the solution of the Cauchy problem for the three-dimensional wave equation. (Derive once more Kirchhoff’s formula in thisway.)

10.5. With the help of the Fourier transform with respect to x, solve theCauchy problem for the wave equation

utt = a2∆u, u = u(t, x), x ∈ Rn.

10.7. Problems 271

Write the fundamental solution for the n-dimensional d’Alembertian =∂2

∂t2− a2∆ in the form of an integral and prove that its singularities belong

to the light cone (t, x) : |x|2 = a2t2.

10.6. Describe location of the singularities of the fundamental solution forthe operator∂2

∂t2− ∂2

∂x2 − 4 ∂2

∂y2.

10.7. Let u(t, x) be a solution of the Cauchy problem of the equation utt =∆u, x = (x1, x2) ∈ R2 with the initial conditions u|t=0 = ϕ, ut|t=0 = ψ.

a) Let ϕ,ψ be known in the rectangle x1 ∈ [0, a], x2 ∈ [0, b]. Where canu be determined? Draw the domain in R3

t,x1,x2in which u can be determined

(for t > 0).

b) Let the supports of ϕ,ψ belong to the rectangle x1 ∈ [0, a], x2 ∈ [0, b].Where is u 6= 0 for t > 0? Draw this domain in R3

t,x1,x2.

10.8. Solve Problem 10.7 for x ∈ R3 with the rectangle replaced by arectangular parallelepiped. Instead of the domain in R4

t,x, draw its sectionby the hyperplane t = 1.

Chapter 11

Properties of the

potentials and their

calculations

We have already met the potentials in this book. In Remark 4.13 we dis-cussed the physical meaning of the fundamental solution of the Laplaceoperator in R3 and R2. It is explained there that the fundamental solutionE3(x) of the Laplace operator in R3 has the meaning of the potential of apoint charge, and the fundamental solution E2(x) of the Laplace operatorin R2 has the meaning of the potential of a thin uniformly charged string.It is impossible to define the potential of the string as an ordinary integral(it diverges). We indicated two methods for defining this potential: its re-covery from the strength of the field (which is minus the gradient of thepotential), and making the divergent integral meaningful by renormaliza-tion of the charge. Both methods define the potential of the string up to aconstant, which suffices since in the long run we are only interested in thestrength of the field which is physically observable, whereas the potential isan auxiliary mathematical object (at least in the classical electrodynamicsand gravitation theory).

In Example 5.2 we introduced distributions, called single and doublelayers on the plane t = 0 and on an arbitrary surface, and explained theirphysical meaning. In Example 5.6 we introduced the potentials of the simple

273

274 11. Potentials and their calculations

and double layers as convolutions of the fundamental solution En(x) withthe single and double layers on the surface.

In Section 10.1 the role of the scalar and vector potentials in electrody-namics was clarified.

Here we will discuss in detail properties of potentials (and also howto define and calculate them). However, before we start getting acquaintedwith the details, we advise the reader to reread the above indicated passages.

Potentials will be understood with sufficient versatility. When the pre-liminary definitions that we begin with become unsuitable, we will modifythem at our convenience trying all the time to preserve physical observables.

11.1. Definitions of potentials

Let En(x) be the standard fundamental solution for ∆ in Rn;

En(x) =1

(2− n)σn−1|x|2−n, n ≥ 3,

where σn−1 is the area of the unit sphere in Rn;

E2(x) =1

2πln |x|.

In particular, the case n = 3 is the most important one:

E3(x) = − 14π|x| .

E3(x) is the potential of a point charge, which is equal to +1 for a propersystem of units, and E2(x) is the potential of an infinite uniformly chargedstring (see e.g. Feynman, Leighton, Sands [7], vol. 2, ch. 4).

All potentials have the form

(11.1) −En ∗ ffor different distributions f ∈ E ′(Rn) and, therefore, for n = 3 and 2 havethe meaning of potentials of some distributed systems of charges. Everypotential u of the form (11.1) satisfies in D′(Rn) the Poisson equation

∆u = −f.

Now we will describe the potentials which we need, turn by turn.

a) The volume potential of charges distributed with density ρ(x) is theintegral

(11.2) u(x) = −∫

RnEn(x− y)ρ(y)dy.

11.1. Definitions of potentials 275

We will always suppose that ρ(y) is a locally integrable function with com-pact support and ρ(y) is piecewise smooth, i.e., ρ is a C∞-function outsideof a finitely many smooth hypersurfaces in Rn on which it may have jumps.All the derivatives ∂αρ are supposed to satisfy the same condition (∂αρ istaken here outside of the hypersurfaces). An example of an admissible func-tion ρ(y) is the characteristic function of an arbitrary bounded domain withsmooth boundary.

The most important property of the volume potential is that it satisfiesthe equation

∆u = −ρunderstood in the distributional sense; this follows from the equalities u =−En ∗ ρ and ∆En(x) = δ(x).

b) The single layer potential is the integral

(11.3) u(x) = −∫

ΓEn(x− y)σ(y)dSy,

where Γ is a smooth hypersurface in Rn, σ is a C∞ function on Γ and dSythe area element on Γ. For the time being, we assume that σ has a compactsupport (in the sequel, we will encounter examples when this does not hold).This potential satisfies

∆u = −σδΓ,

where σδΓ is the distribution with the support on Γ defined in Example 5.2.

c) The double layer potential is the integral

(11.4) u(x) =∫

Γ

∂En(x− y)∂ny

α(y)dSy,

where ny is the outward (or in some other way chosen) normal to Γ at y,and the remaining notations are the same as in the preceding case. Thepotential of the double layer satisfies

∆u = − ∂

∂n(α(y)δΓ),

where ∂∂n(αδΓ) — the “double layer” — is the distribution defined in Ex-

ample 5.2. In this case we assume for the time being that α has a compactsupport and that α ∈ C∞(Γ).

It is easy to see that the potentials (11.2)–(11.4) are defined and infin-itely differentiable on Rn\Γ (where in case of the volume potential (11.2), Γis understood as the union of surfaces where ρ or its derivatives have jumps).


We will understand these potentials everywhere in Rn as distributions ob-tained as convolutions (11.1), where f for the volume potential is the sameas ρ in (11.2); for the case of the single layer potential f = σδΓ, where

〈σδΓ, ϕ〉 =∫

Γσ(x)ϕ(x)dSx, ϕ ∈ D(Rn),

whereas for the case of the double layer potential

f =∂

∂n(αδΓ),

where ⟨∂

∂n(αδΓ), ϕ

⟩= −

∫Γα(x)

∂ϕ(x)∂nx

dSx.

Clearly, the convolutions u = −En ∗ f in all the above cases coincide forx ∈ Rn\Γ with the expressions given by the integrals (11.2)–(11.4).

The physical meaning of the potentials (11.2)–(11.4) was described inChapter 5.

11.2. Functions smooth up to Γ from each side, and their

derivatives

Let Ω be a domain in Rn,Γ a smooth hypersurface in Ω, i.e., a closed inΩ submanifold of codimension 1, and u ∈ C∞(Ω\Γ). We will say that u issmooth up to Γ from each side if all the derivatives ∂αu are continuous upto Γ from each side of the hypersurface.

More precisely, if x0 ∈ Γ, then there should exist a neighborhood U ofx0 in Ω such that U = U+ ∪ ΓU ∪ U−, where ΓU = Γ ∩ U , U+ and U−

are open and the restrictions u|U+ and u|U− can be extended to functionsu+ ∈ C∞(U+) and u− ∈ C∞(U−).

In what follows we will often be interested in local questions only, whereone should set from the beginning U = Ω and the above decomposition isof the form Ω = Ω+ ∪ Γ ∪ Ω−, where u+ ∈ C∞(Ω+) and u− ∈ C∞(Ω−).

Our aim is to prove that all the potentials are smooth up to Γ from eachside (this in turn enables us to prove theorems about jumps of the potentials,and calculate the potentials). To this end we will first prove several auxiliarylemmas about functions which are smooth up to Γ from each side.

If u ∈ C∞(Ω\Γ) is smooth up to Γ from each side, then it uniquelydetermines a distribution [u] ∈ D′(Ω), such that [u] ∈ L1

loc(Ω) and [u]|Ω\Γ =u. Our immediate goal is to learn how to apply to [u] differential operators

11.2. Functions smooth up to the boundary 277

with smooth coefficients. Let us agree to write [u] only when u is smoothup to Γ from each side.

In a neighborhood U of every point x0 ∈ Γ we can construct a diffeo-morphism of this neighborhood onto a domain in Rn, that transforms U ∩Γinto a part of the plane x : xn = 0. Here x = (x1, . . . , xn−1, xn) = (x′, xn),where x′ = (x1, . . . , xn−1) ∈ Rn−1. This diffeomorphism induces a change ofvariables in any differential operator transforming it into another differentialoperator of the same order (see Section 1.3). Thus, we will first consider asufficiently small neighborhood U of 0 ∈ Rn and assume that the coordinatesin U are selected so that

Γ = U ∩ x : xn = 0,

U± = x : x ∈ U, x = (x′, xn), ±xn > 0,and the normal n to Γ is of the form n = (0, . . . , 0, 1), i.e., ∂u

∂n = ∂u∂xn

.

For simplicity of notations we will now temporarily assume that U = Ω,so U ∩ Γ = Γ.

Denote by ψk the jump of the k-th normal derivative of u on Γ, i.e.

ψk =∂ku+

∂xkn

∣∣∣∣Γ

− ∂ku−

∂xkn

∣∣∣∣Γ

∈ C∞(Γ),

where k = 0, 1, 2, . . .. Note that the jumps of all the other derivatives areexpressed via ψk by the formulas

(∂αu+)|Γ − (∂αu−)|Γ = ∂α′ψαn ,

if α = (α′, αn), where α′ is an (n− 1)-dimensional multi-index.

Lemma 11.1. The derivatives ∂[u]∂xn

, ∂2[u]∂x2n

and ∂2[u]∂xn∂xj

(j = 1, 2 . . . , n− 1) areexpressed by the formulas

(11.5)∂[u]∂xn

=[∂u

∂xn

]+ ψ0(x′)⊗ δ(xn),

(11.6)∂[u]∂xj

=[∂u

∂xj

], j = 1, . . . , n− 1,

(11.7)∂2[u]∂x2

n

=[∂2u

∂x2n

]+ ψ1(x′)⊗ δ(xn) + ψ0(x′)⊗ δ′(xn),

(11.8)∂2[u]∂xn∂xj

=[

∂2u

∂xn∂xj

]+∂ψ0

∂xj⊗ δ(xn).


Proof. For any ϕ ∈ D(Ω) we have⟨∂[u]∂xn

, ϕ

⟩=−

⟨[u],

∂ϕ

∂xn

⟩= −

∫xn≥0

[u]∂ϕ

∂xndx−

∫xn≤0

[u]∂ϕ

∂xndx

=−∫dx′[∫ ∞

0[u]

∂ϕ

∂xndxn −

∫ 0

−∞[u]

∂ϕ

∂xndxn

]=∫ [

∂u

∂xn

]ϕ(x)dx+

∫u+(x′, 0)ϕ(x′, 0)dx′ −

∫u−(x′, 0)ϕ(x′, 0)dx′

=⟨[

∂u

∂xn

], ϕ

⟩+∫ψ0(x′)ϕ(x′, 0)dx′,

which proves (11.5). Integrating by parts with respect to xj easily leadsto (11.6). The formula (11.7) is obtained if we twice apply (11.5), whereas(11.8) follows from (11.5) and (11.6).

Now, let a ∈ C∞(Ω), a = a(x′, xn). We need to learn how to multiplydistributions encountered in the right-hand sides of (11.5)–(11.8) by smoothfunctions. To this end, first, expand a(x′, xn) by the Taylor formula in xnnear xn = 0. We have

(11.9) a(x′, xn) =N−1∑k=0

ak(x′)xkn + xNn a(N)(x′, xn),

where ak ∈ C∞(Γ), a(N) ∈ C∞(V ) (here V is a neighborhood of Γ), and

ak(x′) =1k!

∂ka

∂xkn

∣∣∣∣xn=0

,

a(N)(x′, xn) =

1(N − 1)!

∫ 1

0

∂Na

∂xNn(x′, txn)(1− t)N−1dt.

Lemma 11.2. If ψ ∈ C∞(Γ), then

(11.10) a(x)(ψ(x′)⊗ δ(xn)) = [a0(x′)ψ(x′)]⊗ δ(xn),

(11.11) a(x)(ψ(x′)⊗δ′(xn)) = [a0(x′)ψ(x′)]⊗δ′(xn)− [a1(x′)ψ(x′)]⊗δ(xn).

Proof. Clearly,

ak(x′)xkn(ψ(x′)⊗ f(xn)) = [ak(x′)ψ(x′)]⊗ [xknf(xn)]

for any distribution f ∈ D′(R). Further,

a(N)(x′, xn)xNn (ψ(x′)⊗ f(xn)) = a(N)(x

′, xn)(ψ(x′)⊗ xNn f(xn)).

The direct calculation shows that

xpnδ(k)(xn) = 0 for p > k;


xnδ′(xn) = −δ(xn),

which with a(x′, xn) in the form (11.9) immediately yields (11.10) and(11.11).

Lemma 11.3. For any set of functions ψ0, ψ1, . . . , ψN ∈ C∞(Γ), there existsa function u ∈ C∞(Ω\Γ) smooth up to Γ from each side, such that ψk is thejump of ∂ku

∂xknon Γ for all k = 0, 1, . . . , N .

Proof. We may, for example, set

u(x) =N∑k=0

1k!ψk(x′)xknθ(xn),

where θ(t) is the Heaviside function.

Lemma 11.4. Let a distribution f ∈ D′(Ω) be of the form

(11.12) f(x) = [f0(x)] + b0(x′)⊗ δ(xn) + b1(x′)⊗ δ′(xn),

where b0, b1 ∈ C∞(Γ), and f0 is smooth up to Γ from each side. Let A bea 2nd order linear differential operator such that Γ is non-characteristic forA at every point. Then for any integer N ≥ 0 there exists a function u on aneighborhood Ω′ of hypersurface Γ in Ω such that u is smooth up to Γ fromeach side, and

(11.13) A[u]− f ∈ CN (Ω′).

Proof. By Lemma 11.3 it is enough to select jumps ψ0, ψ1, . . ., ψN+2 ∈C∞(Γ) of normal derivatives of u. It is clear from Lemmas 11.1 and 11.2that the distribution A[u] = f always has the form (11.12) (but perhapswith other f0, b0, b1). We need to select the jumps ψ0, ψ1, . . . , ψN+2 in sucha way that

f(x) = [f0(x)] + b0(x′)⊗ δ(xn) + b1(x′)⊗ δ′(xn)

with the same functions b0, b1 as in (11.12), and also the jumps of ∂kf0∂xkn

and ∂k f0∂xkn

on Γ coincide for k = 0, 1, . . . , N . (Then we will obviously have

f − f ∈ CN (Ω), as required.)

First, note that the non-characteristicity of Γ for A means that A canbe expressed in the form

(11.14) A = a(x)∂2

∂x2n

+A′,


where a ∈ C∞(Ω), a(x′, 0) 6= 0, A′ does not contain ∂2

∂x2n

. Since a(x) 6= 0 ina neighborhood of Γ we can divide (11.13) by a(x) and reduce the problemto the case when a(x) = 1. Therefore, in what follows we assume that

(11.15) A =∂2

∂x2n

+A′,

where A′ is a 2nd order differential operator that does not contain ∂2

∂x2n

.

Let us start with selecting a function v0 such that

A[v0] = [f0(x)] + b0(x′)⊗ δ(xn) + b1(x′)⊗ δ′(xn),

where f0 is any function (smooth up to Γ from each side), b0 ∈ C∞(Γ).Lemmas 11.1 and 11.2 show that it suffices to take ψ(0)

0 (x′) = b1(x′), whereψ

(0)0 is the jump of v0 on Γ. Now, (11.13) may be rewritten in the form

(11.16) A[u− v0] = [f1] + b2(x′)⊗ δ(xn)

with some f1 and b2, where b2 ∈ C∞(Γ).

Now, select a function v1 such that

A[v1] = [f1] + b2(x′)⊗ δ(xn),

where b2 is the same as in (11.16) and f1 is arbitrary. If ψ(1)0 and ψ

(1)1 are

jumps of v1 and ∂v1∂xn

on Γ, then it suffices to take

ψ(1)0 = 0, ψ

(1)1 (x′) = b2(x′).

Setting u2 = u− v0 − v1 we see that the condition (11.13) for u reduces to

A[u2]− [f2] ∈ CN (Ω′),

where f2 is a function that is smooth up to Γ from each side. We will provethe possibility to select such u2 by induction with respect to N , startingwith N = −1 and defining C−1(Ω′) as the set of all functions in Ω′ that aresmooth up to Γ from each side.

For N = −1 everything reduces to the simple condition u2 ∈ C1(Ω′),i.e., to the absence of jumps of u2 and ∂u2

∂xnon Γ).

Now suppose that we have already selected a function uk+2 such that

(11.17) A[uk+2]− [f2] ∈ Ck−1(Ω′),

where k is a non-negative integer (as we have just seen this is possible fork = 0). Then, as an induction step, we need to select a function uk+3 suchthat

(11.18) A[uk+3]− [f2] ∈ Ck(Ω′).


To this end set uk+3 = uk+2 + vk+2, where vk+2 ∈ Ck+1(Ω′) (we do this inorder to preserve (11.17) when we replace uk+2 by uk+3). Clearly, A′[uk+2] ∈Ck(Ω′) and, therefore, (11.18) reduces to

∂2[vk+2]∂x2

n

− [fk] ∈ Ck(Ω′),

where fk = −(A[uk+2]− [f2]) ∈ Ck−1(Ω′). This can be obtained by setting,for example,

vk+2(x) =1

(k + 2)!xk+2n

[∂kf+

k

∂xkn

∣∣∣∣∣Γ

− ∂kf−k∂xkn

∣∣∣∣∣Γ

]θ(xn)

in order to make the jumps of the k-th derivatives of the functions ∂2[vk+2]∂x2n

and fk on Γ coincide. Thus, we have proved that the induction with respectto N can be applied. This completes the proof.

Global version. Fermi coordinates.

Now we will briefly describe a global version of the results above. Letus introduce appropriate notations.

Let Γ be a closed hypersurface in Rn, that is a (n − 1)-dimensionalcompact closed submanifold of Rn without boundary. Let us take a functions : Rn → R, defined as follows: s(x) = dist(x,Γ) if x is outside of the(closed, compact) body B bounded by Γ (in particular, Γ = ∂B), and s(x) =−dist(x,Γ) if x is inside this body, s = 0 on Γ. Clearly, s < 0 on the interiorof B, s > 0 on Rn \ B. It is easy to see that s ∈ C∞ in a sufficiently smallneighborhood of Γ. Indeed, let n = n(z) denote the outward unit normalvector to Γ at the point z ∈ Γ. Consider the map

(11.19) Γ× R→ Rn, (z, s) 7→ z + sn.

Its differential is obviously bijective at the points (z, 0) ∈ Γ×R. Therefore,by the inverse function theorem, the map (11.19) restricts to a diffeomor-phism of Γ × (−ε, ε) onto an open subset Γε of Rn, provided ε > 0 issufficiently small.

In Γε, it is convenient to use Fermi coordinates which have the form(y′, yn), where y′ = (y1, . . . , yn−1) is a set of local coordinates on Γ, andyn = s ∈ (−ε, ε). If the (n − 1)-tuple y′ represents a point z ∈ Γ, then, bydefinition, y = (y′, s) are coordinates of the point z + sn ∈ Γε. If we havea second order differential operator A such that Γ is non-characterisic forA at every point, then, as in (11.14), we can write in any Fermi coordinate


system (y′, yn):

(11.20) A = b(y)∂2

∂y2n

+B′,

where b ∈ C∞(Γε), b(y′, 0) 6= 0, B′ does not contain ∂2

∂y2n. Now, if we pass

from one Fermi coordinate system to another one y = (y′, yn), then thefirst term in the right hand side of (11.20) will not change its form, exceptchange of variables in the scalar function b. Therefore, the function b is awell defined scalar function on Γε. Then in any equation of the form Au = f

we can divide by b on a small neighborhood of Γ = Γ × 0. For example,we can replace ε by a smaller value and assume that b(y) 6= 0 for all y ∈ Γε.

For the Laplacian ∆ in Rn and Fermi coordinates y with respect to anyhypersurface Γ, the formula (11.20) simplifies: then b(y) = 1 for all y, sothat we have

(11.21) ∆ =∂2

∂y2n

+B′,

and, moreover, B′ does not contain any second derivatives of the form ∂2

∂yj∂xn

with j ≤ n. Both statements follow from a simple geometric fact (a gener-alization of a Gauss lemma), that in Fermi coordinates ∂

∂ynis the gradient

of s = yn, or, equivalently, that the invariantly defined tangent vector field∂∂yn

is orthogonal to the level sets of the signed distance function s = yn.(See the proof in a much more general context in Gray [9], Chapter 2; Rn

can be replaced by an arbitrary Riemannian manifold M , and Γ by a closedsubmanifold of an arbitrary dimension of M . As an exercise, the readercan find an elementary proof based on the fact that for small ε the straightline interval z + tsn| 0 ≤ t ≤ 1 is the shortest way between the levelhypersurfaces yn = 0 and yn = s , where s ∈ [0, ε] is arbitrarily fixed.)

For more information and details about Fermi coordinates see Gray [9],Chapter 2.

To understand how the statement about the second derivatives followsfrom the above geometric fact, let us recall that in any local coordinatesy = (y1, . . . , yn) the Laplacian can be written in the form of the Laplace-Beltrami operator, which is given by

(11.22) ∆u =1√g

∑1≤i,j≤n

∂

∂yi

(gij√g∂u

∂yj

).

Here (gij) is the inverse matrix for (gij), where gij = gij(y) is the symmetricpositive definite matrix of the standard Riemannian metric on Rn (i.e. ds2 =


dx21 + · · ·+ dx2

n) in coordinates y, that is

gij(x) =⟨∂

∂yi,∂

∂yj

⟩x

,

where the tangent vectors ∂∂yj

are considered as vectors in Rn, 〈·, ·〉 is thestandard scalar product in Rn. (See, for example, Taylor [30], Vol. 1,Sect. 2.4 for the details about the Laplace-Beltramy operator.)

Note that the vector ∂∂yn

has length one, hence gnn = 1. The absence of

the mixed second derivatives ∂2

∂yj∂xn, j < n, is equivalent to gjn = 0, which

implies that for the inverse matrix (gij) we also have gnn = 1 and gjn = 0 ifj < n. This implies the desired absence of the mixed derivatives.

Now we can globalize the arguments given above in the proofs of Lemmas11.1-11.4 (the local case), except we need to replace ψ(x′) ⊗ δ(xn) by ψδΓ

where ψ ∈ C∞(Γ), and ψ(x′)⊗δ′(xn) by− ∂∂n(ψδΓ). This leads, in particular,

to the following global version of Lemma 11.4:

Lemma 11.5. Let Γ be a closed compact hypersurface in Rn, and a distri-bution f ∈ D′(Rn) be of the form

(11.23) f(x) = [f0(x)] + b0(x)δΓ +∂

∂n(b1(x)δΓ),

where b0, b1 ∈ C∞(Γ), f0 is locally integrable and smooth up to Γ from eachside. Let A be a 2nd order linear differential operator in Rn, such that Γis non-characteristic for A at every point. Then for any integer N ≥ 0there exists an integrable compactly supported function u in Rn such that uis smooth up to Γ from each side, and

(11.24) A[u]− f ∈ CN (Rn).

We leave the details of the proof as an exercise for the reader. Note onlythat if we found u satisfying (11.24) but without compact support, then wecan replace u by χu where χ ∈ C∞0 (Rn), χ = 1 in a neighborhood of Γ.Then χu will satisfy all the requirements of the Lemma.

Remark 11.6. Lemma 11.5 is a simplified version of a more general result.It is not necessary to require compactness of Γ. For example, the proof worksif Γ is a hyperplane or an infinite cylinder with a compact base. In fact,it suffices that Γ is a closed submanifold in Rn, though the proof requiresa minor modification: we need to glue local solutions into a global one bypartition of unity, and this should be done separately on every induction stepfrom the proof of Lemma 11.4. Moreover, the result holds in an even more


general setting, where Rn is replaced by any general Riemannian manifold(for example, any open set in Rn), and Γ by its closed submanifold.

11.3. Jumps of potentials

Now we are able to prove the main theorem on jumps of potentials.

Theorem 11.7. 1) Let u be one of the potentials (11.2) − (11.4). Then u

is smooth up to Γ from each side.

2) If u is a volume potential, then u and ∂u∂n do not have jumps on Γ, so

u ∈ C1(Rn).

3) If u is a single layer potential, then u does not have any jump on Γ,so u ∈ C(Rn), whereas the jump of ∂u

∂n on Γ is (−σ), where σ is the densityof the charge on Γ in the integral (11.3) defining u.

4) If u is a double layer potential, then the jump of u on Γ is equal to(−α), where α is the density (of dipole moment) on Γ in the integral (11.4)defining u.

Here the direction of the jumps is naturally determined by the normalvector field n.

Proof. I will first explain the main idea of the proof. As we have seenin the previous section, the differentiation (of order 1 or 2) of a functionu which is smooth up to Γ from each side, leads to another function ofthe same kind but with additional singular terms: terms containing thesurface δ-function δΓ and its normal derivative, with coefficients which aresums of jumps of u and its normal derivative on Γ, taken with coefficientsfrom C∞(Γ). For our proof, we need, vice versa, to find some jumps bythe information from the right-hand side of the equation ∆u = −f , whichis satisfied for all potentials with an explicitly given distribution f whichis supported on Γ and has exactly the structure which we have described.Now, from the results of the previous section, we can produce a solution u

of ∆u = −f1, where f1 differs from f by an arbitrarily smooth function.Then ∆(u − u) = f1 − f can be done arbitrarily smooth, and so u − u canbe done arbitrarily smooth too. It will follow that u is smooth up to Γ fromeach side, and the jumps can be recovered from f .

Now let us give more detail for each of the statements 1)-4) above.

1) By Lemma 11.5, for each of the potentials u we may find a compactlysupported integrable function uN smooth up to Γ from each side, such that

11.3. Jumps of potentials 285

∆[uN ] + f ∈ CN (Rn) (here f is the same as in (11.1)). Therefore, ∆(u −uN ) = fN ∈ CN (Rn).

Let us deduce from this that u − uN ∈ CN (Rn). Without loss of gen-erality we may assume that uN and fN have compact supports. But then∆(u− uN − En ∗ fN ) = 0, which implies that u− uN − En ∗ fN ∈ C∞(Rn).At the same time, clearly, En ∗ fN ∈ CN (Rn), because expressing this con-volution as an integral ∫

En(y)fN (x− y)dy,

we can differentiate, by dominated convergence theorem, under the integralN times with respect to x. Therefore, u− uN ∈ CN (Rn). This implies thatthe derivatives ∂αu are continuous for |α| ≤ N up to Γ from each side. SinceN is arbitrary, this proves the first statement.

2) Let u be a volume potential. Clearly, u ∈ C∞(Rn\Γ). In a neighbor-hood of every point of Γ we can rectify Γ and use the equation Au = −f ,obtained by the change of variables from the equation ∆u = −f . Since ∆ isan elliptic operator, A is also elliptic. Therefore, Γ is non-characteristic.But then the inclusion f ∈ L1

loc(Rn) and Lemmas 11.1 and 11.2 implyu ∈ C1(Rn). (If we assume opposite, then either u or ∂u

∂n has a nontriv-ial jump on Γ, and the formulas (11.5)–(11.8),(11.10), and (11.11) wouldclearly lead to the occurrence of a non-vanishing singular distribution oftype ψ0(x′)⊗ δ(xn) or ψ0(x′)⊗ δ′(xn) in the right hand side of the equationAu = −f , whereas f is locally integrable due to our assumptions in the caseof the volume potential.)

3) In a neighborhood of a point x0 ∈ Γ it is convenient to use Fermicoordinates y = (y′, yn). In particular, ∂u∂n = ∂u

∂ynat the boundary points. Let

u be a simple layer potential, defined by (11.3), with the density σ ∈ C∞0 (Γ).Then the Poisson equation ∆u = −σδΓ implies that u itself can not have ajump, because any non-vanishing jump ψ0 of u would lead to the appearanceof the term −ψ0(y′)⊗δ′(yn) in f = −∆u, which would contradict the Poissonequation. Therefore, u is continuous. On the other hand, it is clear from theform of the Laplacian in coordinates y given by (11.21) and from Lemma11.1 that the jump of ∂u

∂n should be −σ for u to satisfy the Poisson equation.

4) The same argument works for the double layer potential. Namely, ajump ψ1 of u leads to the term ∂

∂n(ψ1)δΓ in f = −∆u, which implies thedesired statement.


Remark. In a similar way we may find the jumps of any derivative ofthe potentials.

11.4. Calculating potentials

We will give here simplest examples of calculation of potentials. The explicitcalculation becomes possible in these examples with the help of:

a) symmetry considerations;

b) the Poisson equation satisfied by the potential;

c) theorems on jumps;

d) asymptotics at infinity.

As a rule, these considerations even give a surplus information whichenables us to verify the result.

Example 11.1. The potential of a uniformly charged sphere. Let Γ be thesphere of radius R in R3 with the center at the origin. Suppose a charge isuniformly (with the surface density σ0) distributed over the sphere. Denotethe total charge of the sphere by Q , i.e., Q = 4πR2σ0. Let us find thepotential of the sphere and its field. Denote the desired potential by u.

First, symmetry considerations show that u is invariant under rotations,hence u(x) = f(r), where r = |x|, i.e., u only depends on the distance tothe center of the sphere. Now use the fact that u is a harmonic functionoutside Γ. Since any spherically symmetric harmonic function is of the formC1 + C2r

−1, we have

f(r) =

A+Br−1, r < R;

C +Dr−1, r > R.

Here A, B, C, D are real constants to be determined. First, B = 0 sinceu is harmonic in a neighborhood of 0. This, in particular, implies that u =const for |x| < R and, therefore, E = − gradu = 0 for |x| < R, i.e., the fieldvanishes inside a uniformly charged sphere.

To find the three remaining constants, we need three more conditionsfor u. They can be obtained in different ways. Let us indicate the simplestof them.

1. The continuity of u(x) on Γ implies A = C +DR−1.

2. Considering the representation of u(x) as an integral, we see thatu(x)→ 0 as |x| → ∞, hence C = 0.

11.4. Calculating potentials 287

3. Consider the integral that defines u(x), in somewhat more detail.We have

u(x) =1

4π

∫|y|=R

σ0dSy|x− y| =

σ0

4π

∫|y|=R

dSy|x|

1| x|x| −

y|x| |

=σ0

4π|x|

∫|y|=R

dSy

(1 +O

(1|x|

))=

Q

4π|x|

(1 +O

(1|x|

))=

Q

4πr

(1 +O

(1r

)),

as r → ∞. Hence D = Q4π . From 1 − 3 we deduce that B = C = 0,

D = Q4π , A = Q

4πR . Thus,

u(x) =

Q

4πR for |x| < RQ

4πr for |x| > R.

This means that outside the sphere the potential is the same as if the wholecharge were concentrated at the center. Therefore the homogeneous sphereattracts the charges situated out of it as if the whole charge were supported inits center. This fact was first proved by Newton who considered gravitationalforces. (He actually proved this theorem in a different way: by consideringdirectly the integral that defines the attraction force.)

We will indicate other arguments that may be used to get equations onA, C, D. We can use them to verify the result.

4. The jump of the normal derivative of u on the sphere should beequal to −σ0, i.e., Dr−2|r=R = σ0 or D = σ0R

2 = Q4πR2 · R2 = Q

4π whichagrees with the value of D found above.

5. The value of u(x) at the center of the sphere is equal to

A =1

4π

∫|y|=R

σ0dSyR

=Q

4πR· σ04πR2 =

Q

4πR,

which agrees with the value of A found above.

Using the obtained result, we can immediately find the attracting force(and the potential) of the uniformly charged ball by summing the potentials(or attracting forces) of the spherical layers constituting the ball. In partic-ular, the ball attracts a charge outside of it as if the total charge of the ballwere concentrated at its center.

Example 11.2. The potential of the double layer of the sphere.

Consider the sphere of radius R over which dipoles with the density ofthe dipole moment α0 are uniformly distributed. The potential of the double


layer here is the integral (11.4), where

Γ = x : x ∈ R3, |x| = R, α(y) = α0 = const .

Let us calculate this potential. This can be done in the same way as in thepreceding example. We may immediately note, however, that u(x) = 0 for|x| > R since u(x) is the limit of the potentials of simple layers of a pair ofconcentric spheres uniformly charged by equal charges with opposite signs.In the limit, these spheres tend to each other, and charges grow in inverseproportion to the distance between the spheres. Even before the passage tothe limit, the potential of this pair of spheres vanishes outside the largestof them. Therefore, u(x) = 0 for |x| > R. It is also clear that u(x) = constfor |x| < R, and the theorem on jumps implies (when the exterior normal ischosen) that u(x) = α0 for |x| < R.

Example 11.3. The potential of a uniformly charged plane.

Let us define and calculate the potential of a uniformly charged planewith the surface density of the charges σ0. Let x3 = 0 be the equation of theplane in R3. Observe that we can not use the definition of the potential by(11.3) because the integral diverges. Therefore, it is necessary to introducefirst a new definition of the potential. This may be done in several ways.We will indicate two of them.

The first is to preserve the main equation

(11.25) ∆u = −σ0δΓ = −σ0δ(x3)

and as many symmetry properties of the potential as possible. (Physicistsoften determine a value of an observable quantity from symmetry consider-ations without bothering much with the definitions.) If the integral (11.3)were convergent in our case, then (11.25) would be satisfied and the followingsymmetry conditions would hold:

a) u is not changed by the translations along our plane, i.e., u = u(x3);

b) u is invariant under the reflections with respect to our plane, i.e.,u(−x3) = u(x3).

Now, try to calculate u(x3) using (11.25) and the above symmetry prop-erties. It turns out that u(x3) is uniquely defined up to a constant. There-fore, we can define the potential of the plane to be a distribution u satisfying(11.25), a) and b).

11.5. Problems 289

From (11.25) it is clear that u(x3) is linear in x3 separately for x3 > 0and x3 < 0, i.e.,

u(x3) =

A+Bx3, x3 > 0;

C +Dx3, x3 < 0.

Equation (11.25) means, besides, that the theorem on jumps holds, i.e., uis continuous on Γ (implying A = C) and the jump of ∂u

∂x3on Γ is equal

to −σ0 (implying B − D = −σ0). Finally, the fact that u(x3) is even(condition b)) once again yields A = C and, besides, implies D = −B.Hence, B = −D = −σ0

2 and

u = u(x3) = A− σ0

2|x3|.

Thus we have found u up to an irrelevant additive constant.

Now, recall that our main goal is to find E = − gradu. We see that Eis everywhere perpendicular to the plane and if the direction from the planeis taken as the positive one, then the magnitude of the field is equal to σ0

2 .

Here is another method to define u. The field E can be found by theCoulomb law (E is given, as is easy to see, by a converging integral!) andthen u is found from the equation E = − gradu. Since the symmetry prop-erties clearly hold and div E = σ0δ(x3) leads to (11.25), we can apply thesame method to find u. Therefore we have explicitly found the field strengthE given by a relatively complicated integral.

11.5. Problems

11.1. Calculate the potential of a uniformly charged circle on the plane.

11.2. Calculate the potential of a uniformly charged disc on the plane.

11.3. Define and calculate the potential of the uniformly charged surface ofa straight cylinder in R3.

11.4. Find the potential of the double layer of the circle in R2 and the surfaceof a straight cylinder in R3 with a constant density of the dipole moment.

11.5. Find the potential of a uniformly charged sphere in Rn for n ≥ 3.

11.6. A point charge Q is placed at the distance d from the center of agrounded conducting sphere of radius R (R < d). Find the total chargeinduced on the sphere.


11.7. Under the conditions of the preceding problem find the distribution ofthe induced charge over the surface of the sphere.

11.8. A thin elliptic wire is charged with the total charge Q. How will thecharge distribute along the wire?

Chapter 12

Wave fronts and

short-wave asymptotics

for hyperbolic

equations

12.1. Characteristics as surfaces of jumps

The simplest situation in partial differential equations is when the equationor the boundary value problem can be explicitly solved. However, this case isexceptionally rare. You can more often meet the situation when only a partof the equation is known, or a part of the equation is known approximately.Then it is also very useful to represent the answer in an asymptotic form,when the solution is written in the form of an infinite expansion, and a finitepart of the expansion is conjectured to be an approximate answer. This ideavery often leads to a dramatic success. We will consider and explore severalexamples in this chapter.

We will start with a partial differential equation of the form Au = f ,where A is a linear differential operator with smooth coefficients in an openset Ω ⊂ Rn, f is a known function (or distribution) in Ω, u is an unknownfunction or distribution in a possibly smaller open set.

Let Γ be a closed smooth hypersurface in Ω, i.e., a closed submanifold inΩ of codimension 1. Let S be a closed hypersurface of codimension 1 in Γ.

291

292 12. Wave fronts and short-wave asymptotics

In this way, Γ will be closed in Ω, and

Ω ⊃ Γ ⊃ S,

dim Γ = n− 1,dimS = n− 2.

Let a linear differential operator A of degree m (see Chapter 1),

(12.1) A = a(x,D) =∑|α|≤m

aα(x)Dα,

be given in Ω, with aα ∈ C∞(Ω), a(x, ξ) be the total symbol of A (we usuallyconsider (12.1) to be the definition of both A and a(x,D)). Recall that

(12.2) a(x, ξ) =∑|α|≤m

aα(x)ξα,

so that a(x,D) is obtained from a(x, ξ) if all the ξα are written on the right(as in (12.2)) and then replaced by Dα. The principal symbol am(x, ξ) of Ais of the form

am(x, ξ) =∑|α|=m

aα(x)ξα.

A hypersurface S is called a characteristic of A if for any point x ∈ Sthe unit normal nx to S at x is not 0 but satisfies

(12.3) am(x, nx) = 0.

For any characteristic it follows that the direction of the cotangent vector(x, nx) ∈ T ∗xΩ vanishes. The vanishing of a corresponding vector directionis independent of the choice of local coordinates near x. (See Section 1.3.)

Unfortunately, definitions of characteristic are ambiguous: there aremany (not necessarily equivalent) versions. However, there are many im-portant common features, which allow to create a common use of them.

Let us take now a real-valued function u ∈ C∞(Ω\S) which is infinitelydifferentiable up to S from each side (cf. definition in Section 11.2). Usingthe same notation as in Section 11.2, we locally have Ω = Ω+ ∪ S ∪ Ω−,and there exist u+ ∈ C∞(Ω+) and u− ∈ C∞(Ω−) such that u|Ω+ = u+,u|Ω− = u−.

We can consider u as a locally integrable function on Ω, and we willdenote by u the corresponding distribution too. However, due to the implicitfunction theorem, we will actually have u ∈ C∞(Γ) under the assumptionsabove.

12.1. Characteristics as surfaces of jumps 293

Note that it may happen that actually there is no discontinuity, i.e.,u can be extended to a function from C∞(Ω). This can be considered aregular case, whereas in the opposite (singular) case we will write that S isthe surface of jumps for u. This means that, first, u is infinitely differentiableup to S from each side and, secondly, for each point x0 ∈ S there exists amulti-index α such that ∂αu is discontinuous at x0, i.e., ∂αu does not extendto a function on Ω continuous at x0. In other words, in this case

(∂αu+)(x0) 6= (∂αu−)(x0).

Note that the same inequality also holds on S for x close to x0 (with thesame α).

Theorem 12.1. If S is a surface of jumps for u ∈ L1loc(Ω), and u is a solu-

tion of the equation Au = f in Ω, with f ∈ C∞(Ω), then S is a characteristicof A.

Proof. Since the theorem is local and the notion of a characteristic isinvariant under diffeomorphisms, we may assume that S is of the form

S = x : x ∈ Ω, xn = 0,

where x = (x′, xn) = (x1, . . . , xn−1, xn). Since the normal to S is of the form(0, 0, . . . , 1), the characteristic property of S means that the coefficient of∂m

∂xmnin A vanishes on S.

We will prove the theorem arguing by contradiction. Let S be non-characteristic at some point. This means that the coefficient of ∂m

∂xmnis non-

zero at this point; hence, near it. Then we may assume that it is non-zeroeverywhere in Ω. Dividing the equation Au = f by this coefficient, we mayassume that the equation is of the form

(12.4)∂mu

∂xmn+

m∑j=1

aj(x,D′)∂m−ju

∂xm−jn

= f,

where aj(x,D′) is a differential operator in Ω without any derivatives withrespect to xn (its order is ≤ j though this is of no importance now).

Let u be a C∞-function up to S from each side, Ω± = Ω∩x : ±xn > 0and u± ∈ C∞(Ω±) be such that u|Ω± = u±. We want to prove that u canbe extended to a function from C∞(Ω), i.e., the jumps of all the derivativesactually vanish. Denote these jumps by ϕα, i.e.,

ϕα = (∂αu+)|S − (∂αu−)|S ∈ C∞(S).


As we have seen in Section 11.2, a special role is played by the jumps of thenormal derivatives

(12.5) ψk =∂ku+

∂xkn

∣∣∣∣S

− ∂ku−

∂xkn

∣∣∣∣S

∈ C∞(S).

All the jumps ϕα are expressed in terms of the jumps of the normal deriva-tives ψk via the already mentioned formula

ϕα = ∂α′

x′ψαn , α = (α′, αn).

Therefore, it suffices to prove that ψk = 0 for k = 0, 1, 2, . . . .

Let us find Au in D′(Ω). By Lemma 11.1 we have

∂u

∂xn= ψ0(x′)⊗ δ(xn) +

[∂u

∂xn

],

∂2u

∂x2n

= ψ0(x′)⊗ δ′(xn) + ψ1(x′)⊗ δ(xn) +[∂2u

∂x2n

].

Further calculations of the same type easily yield

(12.6)∂ku

∂xkn=

k−1∑l=0

ψk−l−1(x′)⊗ δ(l)(xn) +[∂ku

∂xkn

].

Now, if α′ is an (n− 1)-dimensional multi-index and ∂α′

= ∂α′

x′, then

(12.7) ∂α′ ∂ku

∂xkn=

k−1∑l=0

[∂α′ψk−l−1(x′)]⊗ δ(l)(xn) +

[∂α′ ∂ku

∂xkn

].

Now, let a ∈ C∞(Ω). Using the same arguments as in the proof of Lemma 11.2,we easily obtain

(12.8) a(x′, xn)(ψ(x′)⊗ δ(l)(xn)) =l∑

q=0

aq(x′)⊗ δ(q)(xn),

where aq ∈ C∞(S).

Formulas (12.6)–(12.8) show that if Au = f outside S, then (12.4) maybe rewritten in the form

(12.9) ψ0(x′)⊗ δ(m−1)(xn) +m−1∑l=1

al(x′)⊗ δ(m−1−l)(xn) = g(x),

where g ∈ L1loc(Ω), al(x′) ∈ C∞(S) and ψ0 is the jump of u on S (see formula

(12.5) for k = 0). It is clear from (12.9) that g ≡ 0 since the left-hand sideof (12.9) is supported on S. Further, after applying the distribution at theleft hand side of (12.9) to a test function ϕ that is equal to ϕ1(x′)xm−1

n in

12.1. Characteristics as surfaces of jumps 295

a neighborhood of S (here ϕ1 ∈ C∞0 (S)), we see that all the terms but thefirst vanish, and the first one is equal to (−1)m−1(m− 1)!

∫ψ0(x′)ϕ1(x′)dx′.

Since ϕ1 is arbitrary, it is clear that ψ0(x′) ≡ 0, i.e., u cannot have a jumpon S.

One similarly proves that the derivatives up to order m− 1 cannot havejumps on S. Namely, let p be the smallest of all k such that ψk 6≡ 0.Therefore,

(12.10) ψ0 ≡ ψ1 ≡ . . . ≡ ψp−1 ≡ 0, ψp 6≡ 0.

We wish to distinguish the most singular term in the left-hand side of (12.4).This is easy to do starting from the formulas (12.6)–(12.8). Namely, calcu-lating the normal derivatives ∂ku

∂xknfor k ≤ p, we get functions from L1

loc(Ω).For k = p+ 1 we get

∂p+1u

∂xp+1n

= ψp(x′)⊗ δ(xn) +[∂p+1u

∂xp+1n

]and for p ≤ m− 1 equation (12.4) takes on the form

ψp(x′)⊗ δ(m−p−1)(xn) +m−p−1∑l=1

al(x′)⊗ δ(m−p−1−l)(x′) = g(x),

where al(x′) ∈ C∞(S), g ∈ L1loc(Ω). As above, this implies ψp ≡ 0, contra-

dicting the hypothesis. Thus, ψp = 0 for p ≤ m− 1.

Let us prove that ψm = 0. This is immediately clear from (12.4) sinceall the derivatives ∂αu, where α = (α′, αn) and αn ≤ m− 1, are continuousin Ω. Therefore, the normal derivative of order m has no jumps either.

Now, consider the case when p > m, where p is the number of the firstdiscontinuous normal derivative; see (12.10). This case can be easily reducedto the case p = m. Indeed, applying ∂p−m

∂xp−mnto (12.4), we see that u satisfies

an equation of the same form as (12.4) but of order p instead of m. Thisimplies that the case p > m is impossible. This concludes the proof.

Theorem 12.1 implies that solutions of elliptic equations with smooth(C∞) coefficients and the right hand side cannot have jumps of the structuredescribed above. It can be even proved that they are always C∞-functions.(The reader can find proofs in more advanced textbooks – see, for example,Taylor [30], vol. 2, Sect. 7.4, Hormander [12], Ch. X, or [14], vol. I, Corollary8.3.2.) The examples given below show that non-elliptic equations may havediscontinuous solutions.


Example 12.1. The equation utt = a2uxx has a solution θ(x − at) with ajump along the characteristic line x = at. A jump of derivatives of arbitrarilylarge order along the characteristic x = at can be arranged by consideringthe solution

u(t, x) = (x− at)Nθ(x− at), where N ∈ Z+.

Similar discontinuities can be obviously constructed along any characteristic(all the characteristics here can be written in one of the forms x− at = c1,x+ at = c2).

Example 12.2. Let us consider a more intricate case of the three-dimensionalwave equation

utt = a2∆u, u = u(t, x), x ∈ R3.

The simplest example of a discontinuous solution of this equation is thespherical wave u(t, x) = θ(|x|−at)

|x| . This solution, clearly, has a jump onthe upper sheet of the light cone (t, x) : |x| = at, t > 0. The spherex : |x| = at ∈ R3 is naturally called the wave front. The wave front movesat a speed a. In geometrical optics, a ray is a parametrically defined (withthe parameter t) curve which is normal to the wave front at any momentof time. In our example the rays are straight lines x = a~e t, where t > 0,~e ∈ R3, |~e| = 1.

Sometimes any curve normal to all the wave fronts (without an explicitlygiven parameter) is also called a ray.

In what follows we will introduce more terminology of the geometricoptics in connection with the Hamilton-Jacobi equation.

Example 12.3. Consider again the three-dimensional wave equation utt =a2∆u, but this time try to construct a solution which is discontinuous ona given smooth compact surface Γ ⊂ R3 at the initial moment. Let, forinstance, Γ = ∂Ω, where Ω is a bounded domain in R3, and the initialconditions are of the form

u|t=0 = χΩ, ut|t=0 = 0,

where χΩ is the characteristic function of Ω.

Let us try to find out how the jumps of u behave as t grows by lookingat the Kirchhoff formula which we will use to construct u. In the Kirchhoffformula we are to integrate χΩ over the sphere of radius at with the centerat x, divide the result by 4πa2t and then take the derivative with respect tot. It is easy to see that this u(t, x) is a smooth function of t, x for t, x such

12.2. The Hamilton-Jacobi equation 297

that the sphere of radius at with the center at x is not tangent to Γ (i.e.this sphere either does not intersect Γ or intersects it transversally.

Therefore, the jumps may occur only at the points where this sphere istangent to Γ. For small t, the set of all such x forms exactly the set of thepoints x which are situated at the distance at from Γ (“wave front”). Forsmall t, the wave front can be constructed as follows: draw all the normalsto Γ (“rays”) and mark on them points at the distance at on both sides. Itis easy to see that on the wave front thus constructed, there is actually ajump of u. For large t normals may begin to intersect. An envelope of thefamily of rays appears and is called a caustic. The singularity of the solutionon the caustic has a much more sophisticated structure and we will not dealwith it.

Even this example shows how objects of geometric optics appear in thewave theory. In what follows we will consider this problem in somewhatmore detail.

12.2. The Hamilton-Jacobi equation. Wave fronts,

bicharacteristics and rays

The Hamilton-Jacobi equation is an equation of the form

(12.11) H

(x,∂S

∂x

)= 0,

where H = H(x, ξ) is a given real-valued function of variables x ∈ Rn

and ξ ∈ Rn (we will usually assume that H ∈ C2), and S = S(x) is anunknown (also real-valued) function with the gradient ∂S

∂x . The Hamilton-Jacobi equation plays an important role in mechanics and physics.

Recall briefly a method for integrating (12.11), more precisely, a connec-tion of this equation with the Hamiltonian system

(12.12)

x = Hξ(x, ξ),

ξ = −Hx(x, ξ),

where the Hamiltonian H = H(x, ξ) is the same function as in (12.11), Hξ =∂H∂ξ = ( ∂H∂ξ1 , . . . ,

∂H∂ξn

), Hx = ∂H∂x = ( ∂H∂x1

, . . . , ∂H∂xn ). (We already discussedHamiltonian systems in Sect. 10.3; here we use x, ξ instead of q, p in (10.13).)

Solutions of this system (curves (x(t), ξ(t))) are called the Hamiltoniancurves (or, more exactly, Hamiltonian curves of the Hamiltonian H(x, ξ))


. Vector field (Hξ,−Hx) in R2nx,ξ defining the system (12.12) is called a

Hamiltonian vector field with the Hamiltonian H.

As was mentioned in Sect. 10.3, any Hamiltonian vector field is welldefined on T ∗Rn

x. This means that the system (12.12) retains the sameform for any choice of curvilinear coordinates in Rn

x assuming that ξ isa cotangent vector expressed in the coordinate system corresponding to agiven coordinate system in Rn

x (i.e., considering ξ1, . . . , ξn as coordinates ofa cotangent vector with respect to the basis dx1, . . . , dxn, see Section 1.1).The reader can find more detail on this in Arnold [3].

Proposition 12.2. The Hamiltonian H(x, ξ) is a first integral of the system(12.12), i.e., H(x(t), ξ(t)) = const along any Hamiltonian curve (x(t), ξ(t)).

Proof. We have

d

dtH(x(t), ξ(t)) =

∂H

∂xx+

∂H

∂ξξ = HxHξ +Hξ(−Hx) = 0.

(Here and below for the sake of simplicity we omit the dot standing for theinner product of two vectors.) This immediately implies the desired result.

In classical mechanics, Proposition 12.2 has the meaning of the en-ergy conservation law. In fact, we already proved it for particular infinite-dimensional linear Hamiltonian system, which is equivalent to the waveequation (see Proposition 10.1).

The Hamiltonian curves with H(x(t), ξ(t)) ≡ 0 are of particular impor-tance for us. If H is the principal symbol of a differential operator, then theyare called bicharacteristics (of the operator or of its symbol). By some abuseof terminology, we will call such a curve bicharacteristic for any H (even if His not a principal symbol of any differential operator). As Proposition 12.2says, if (x(t), ξ(t)) is a Hamiltonian curve defined on an interval of the t-axisand H(x(t0), ξ(t0)) = 0 for some t0, then this curve is a bicharacteristic.

Now, consider the graph of the gradient of S(x), i.e., the n-dimensionalsubmanifold in R2n

x,ξ of the form ΓS = (x, Sx(x)), x ∈ Rn, where Sx = ∂S∂x .

Let us formulate the main statement necessary to integrate the Hamiltonian-Jacobi equation.

Proposition 12.3. If S is a C2 solution of the Hamilton-Jacobi equation(12.11), then the Hamiltonian field is tangent to ΓS at its points.


Proof. The statement means that if (x(t), ξ(t)) is a bicharacteristic and(x0, ξ0) = (x(t0), ξ(t0)) ∈ ΓS , then

d

dt(x(t), Sx(x(t)))|t=t0 =

d

dt(x(t), ξ(t))|t=t0 .

Or, in a shorter form,

d

dtSx(x(t))|t=t0 = ξ(t0).

Butd

dtSx(x(t))|t=t0 = Sxx(x(t0))x(t0) = Sxx(x0)Hξ(x0, ξ0),

where Sxx = ‖ ∂2S∂xi∂xj

‖ni,j=1 is conisidered as a n × n symmetric matrix. On

the other hand, ξ(t0) = −Hx(x0, ξ0), so we need to prove that

(SxxHξ +Hx)|ΓS = 0.

But this immediately follows by differentiation of the Hamilton-Jacobi equa-tion (12.11) with respect to x.

Proposition 12.3 directly implies

Theorem 12.4. ΓS is invariant with respect to the Hamiltonian flow (i.e.,with respect to the motion along the bicharacteristics).

This means that if (x(t), ξ(t)) is a bicharacteristic defined for t ∈ (a, b)and ξ(t0) = Sx(x(t0)) for some t0 ∈ (a, b), then Sx(x(t)) = ξ(t) for allt ∈ (a, b).

Using Theorem 12.4, we can construct a manifold ΓS if it is known overa submanifold M of codimension 1 in Rn

x. If we also know S over M , thenS is recovered by simple integration over ΓS (note, that if ΓS is found, thenS is recovered from a value at a single point). It should be noted, however,that all these problems are easy to study locally whereas we are rarelysuccessful in recovering S globally. We more often succeed in extendingΓS to a manifold which is not the graph of a gradient any longer but isa so-called Lagrangian manifold; this often suffices to solve the majority ofproblems of mathematical physics (the ambiguity of projecting a Lagrangianmanifold onto Rn

x causes the appearance of caustics). The detailed analysisof these questions, however, is beyond the scope of this textbook.

Let us indicate how to construct (locally) S if it itself and its gradient areknown over an initial manifold M of codimension 1 in Rn

x. If the bicharac-teristic L = (x(t), ξ(t)) begins at (x0, ξ0), where x0 ∈M and ξ0 ∈ Sx(x0),


and terminates at (x1, ξ1), then, clearly,

(12.13) S(x1)− S(x0) =∫LSxdx =

∫Lξdx.

The projections of bicharacteristics to Rnx are called rays. We will only con-

sider bicharacteristics starting over M at the points of the form (x, Sx(x)),where x ∈M . Take two rays x1(t) and x2(t) corresponding to these bichar-acteristics. They may intersect at some point x3 ∈ Rn

x and then (12.13) givestwo (generally, distinct) values for S(x) at x3. For this reason the Cauchyproblem for the Hamilton-Jacobi equation (finding S from the values of S|Mand Sx|M ) is rarely globally solvable.

The Cauchy problem is, however, often solvable locally. For its localsolvability it suffices, for example, that rays of the described type thatstart at the points from M , be transversal to M at the initial moment.Then we can consider the map f : M × (−ε, ε) −→ Rn

x, which to a pairx0 ∈ M, t ∈ (−ε, ε) assigns a point x(t) of the ray starting at x0 (i.e.,x(0) = x0) which is the projection of the bicharacteristic (x(t), ξ(t)) suchthat ξ(0) = Sx(x0). (We assume that this map is well defined, i.e. thecorresponding bicharacteristics are defined for t ∈ (−ε, ε).) By the implicitfunction theorem, f determines a diffeomorphism of U × (−ε, ε) (where Uis a small neighborhood of x0 in M and ε is sufficiently small) with anopen set in Rn

x if and only if x(0) is not tangent to M (this is a necessaryand sufficient condition for the derivative map of f to be an isomorphism(Tx0M × R −→ Tx0Rn

x). In this case locally the rays do not intersect andthe Cauchy problem is locally solvable.

Let us give an exact formulation of the Cauchy problem. Given a sub-manifold M ⊂ Rn

x and S0 ∈ C∞(M), let ΓS |M be a submanifold in (T ∗Rnx)|M

which projects to the graph of the gradient of S0 under the natural projec-tion onto T ∗M and belongs to the zero level surface of H(x, ξ). The Cauchyproblem is then to find S such that S|M = S0 and ΓS |M coincides with thesubmanifold thus denoted above and given over M . If the above transver-sality condition holds at all points of M , the Cauchy problem has a solutionin a neighborhood of M .

The level surfaces of S are usually called wave fronts. We will see belowwhy the use of this term here agrees with that in Example 12.2.

Example 12.4. Consider a Hamilton-Jacobi equation of the form

(12.14) |Sx(x)|2 =1a2.


The Hamiltonian in this case is

H(x, ξ) = |ξ|2 − 1a2

and the Hamiltonian equations areξ = 0,

x = 2ξ,

which implies that the Hamiltonian curves are given byξ = ξ0,

x = 2ξ0t+ x0.

In particular, the rays are straight lines in all directions. If such a ray is aprojection of a bicharacteristic belonging to the graph ΓS of the gradient ofthe solution S(x) (due to Theorem 12.4 this means that Sx(x0) = ξ0), then

Sx(x(t)) = ξ(t) = ξ0 =x

2,

along the whole ray, and the ray x(t) is orthogonal to all the wave frontsS(x) = const which it intersects. N

Note that in general rays are not necessarily orthogonal to wave fronts.

The function S(x) = a−1|x| in Rn\0 is one of the solutions of (12.14).Its level lines, i.e., wave fronts, are spheres x : S(x) = t, and they coincidewith the wave fronts of Example 12.2. The rays in this case are straight lineswhich also coincide with the rays, which we mentioned in Example 12.2. N

H Now consider the following particular case of the Hamilton-Jacobiequation:

(12.15)∂S

∂t+A

(x,∂S

∂x

)= 0,

where S = S(t, x); A = A(x, ξ); t ∈ R1; x, ξ ∈ Rn; A ∈ C2 is assumed to bereal-valued.

This is a Hamilton-Jacobi equation in Rn+1t,x with the Hamiltonian

H(t, x, τ, ξ) = τ +A(x, ξ).


The Hamiltonian system that defines the Hamiltonian curves (t(s), x(s), τ(s), ξ(s))is

t = 1,

x = Aξ(x, ξ),

τ = 0,

ξ = −Ax(x, ξ),

(the “dot” stands for the derivative with respect to the parameter s). Itfollows that τ(s) = τ0, t(s) = s+ t0 and (x(s), ξ(s)) is a Hamiltonian curveof the Hamiltonian A(x, ξ). Since a shift of the parameter of a Hamiltoniancurve does not affect anything, we may assume that t0 = 0, i.e. t = s, and re-gard x and ξ as functions of t. Further, when interested in bicharacteristics,we may arbitrarily set x(0) = x0, ξ(0) = ξ0, and then τ0 is uniquely deter-mined: τ0 = −A(x0, ξ0). Therefore arbitrary Hamiltonian curves (x(t), ξ(t))of the Hamiltonian A(x, ξ) are in one-to-one correspondence with bicharac-teristics for H(t, x, τ, ξ) starting at t = 0.

The rays corresponding to the described above bicharacteristics are ofthe form (t, x(t)). In particular, they are transversal to all the planes t =const.

We may set the Cauchy problem for the equation (12.15) imposing theinitial condition

(12.16) S|t=0 = S0(x).

This implies Sx|t=0 = ∂S0∂x , and from (12.15) we find ∂S

∂t

∣∣t=0

. Therefore, thegraph ΓS of the gradient of S is known over the plane M = (t, x) : t =0. Since the rays are transversal to M at the initial moment, ΓS may beextended to define first ΓS and then S over a neighborhood of M .

Namely, in accordance with (12.13), we obtain

S(t, x) = S0(x0) +∫ t

0[τ0dt+ ξ(t)x(t)dt],

where (x(t), ξ(t)) is a Hamiltonian curve of the Hamiltonian A(x, ξ) suchthat x(0) = x0 and x(t) = x, i.e., the points x0 and x are connected by aray (clearly, x0 should be chosen so that the ray with x0 as the source hitsx at the time t); besides,

τ0 = −A(x0, ξ0) = −A(x0,

∂S0

∂x(x0)

)= −A(x(t), ξ(t))


(this means that the corresponding bicharacteristic of H(t, x, τ, ξ) belongsto ΓS). Thus,

(12.17) S(t, x) = S0(x0) +∫ x

x0

(ξdx−Adt),

where the integral is taken over a Hamiltonian curve of the Hamiltonian A

such that its projection joins x0 and x. Instead of (12.17) we can also write

(12.18) S(t, x) = S0(x0) +∫ x

x0

Ldt,

where L = ξx−A is called the Lagrangian corresponding to the HamiltonianA and the integral is taken again over the curve. Formulas (12.17) and(12.18) show that S is similar to the action of classical mechanics (ξ ∈ Rn

plays the role of momentum). Therefore, the wave fronts are level surfacesof the action.

We will not go any deeper into the Hamilton-Jacobi equation theoryand its relations with mechanics and geometric optics. Details can be foundin textbooks on mechanics (see e.g. Arnold [3]). Note, in particular, thatwe have proved here the uniqueness of a solution for the Cauchy problemand gave a method of its construction but did not verify its existence, i.e.,we did not verify that this method actually produces a solution. This isreally so in the domain where the above system of rays starting at t = 0 isdiffeomorphic to the above system of parallel segments (in particular, thisis true in a small neighborhood of the plane (t, x) : t = 0), but we skipthe verification of this simple fact. N

H Let us give one further remark on the Hamilton-Jacobi equation of theform (12.15). Suppose we are interested not in the solution itself but only inthe wave fronts (t, x) : S(t, x) = c, where c is a constant. Fix one value ofthe constant and consider the system of surfaces x : S(t, x) = c belongingto Rn

x and depending on t as on a parameter. If ∂S∂t 6= 0 somewhere, then we

may (locally) solve the equation S(t, x) = c for t and express this system ofsurfaces in the form x : S(x) = t, i.e., in the form of level surfaces of S.

How to write an equation for S? Differentiating the identity S(S(x), x) ≡c with respect to x we get

∂S

∂t· Sx + Sx = 0.


Substituting Sx = −Sx · ∂S∂t into (12.15), we get

(12.19)∂S

∂t+A

(x,−Sx(x)

∂S

∂t

)= 0.

Consider the case most often encountered and the most important for partialdifferential equations: the case when A(x, ξ) is homogeneous of 1st order inξ, i.e.,

A(x, λξ) = λA(x, ξ), x ∈ Rn, ξ ∈ Rn, λ > 0.

We may assume that ∂S∂t < 0 (otherwise replace S by −S which does not

affect the wave fronts). Dividing (12.19) by (−∂S∂t ) we get

(12.20) A(x, Sx(x)) = 1.

We could get the same result just assuming that

(12.21) S(t, x) = S(x)− t+ c.

Then (12.20) is immediately obtained without requiring A(x, ξ) to be homo-geneous. The role of homogeneity is in the fact that under the assumptionof homogeneity, (12.15) is actually just an equation for the direction of thenormal vector to the wave fronts and does not depend in any way on pa-rameterization of the wave fronts and the length of the normal vector.

Thus, assuming that S(t, x) is of the form (12.21) or A(x, ξ) is homo-geneous of order 1 in ξ, we see that the wave fronts are determined bythe equation S(x) = t, where S(x) is a solution of the time independentHamilton-Jacobi equation (12.20). N

12.3. The characteristics of hyperbolic equations

We have not yet discussed the existence of characteristics. Clearly, the ellip-tic operator has no characteristics. We will see that a hyperbolic operatorhas plenty of characteristics.

LetA = a(t, x,Dt, Dx) =

∑|α|≤m

aα(t, x)Dα0t Dα′

x ,

where t ∈ R, x ∈ Rn; α = (α0, α′) is a (n + 1)-dimensional multi-index;

aα(t, x) ∈ C∞(Ω), Ω is a domain in Rn+1.

Let us recall what the hyperbolicity of A with respect to a distinguishedvariable t means (see Section 1.1). Consider the principal symbol

(12.22) am(t, x, τ, ξ) =∑|α|=m

aα(t, x)τα0ξα′.

12.3. The characteristics of hyperbolic equations 305

The hyperbolicity of A means that the equation

(12.23) am(t, x, τ, ξ) = 0,

regarded as an equation with respect to τ has exactly m real distinct rootsfor any (t, x) ∈ Ω, ξ ∈ Rn\0. Denote the roots by τ1(t, x, ξ), . . . , τm(t, x, ξ).Clearly, am is a polynomial of degree m in τ and the hyperbolicity impliesthat its highest coefficient a(m,0,...,0) (the coefficient by τm) does not vanishfor (t, x) ∈ Ω (this coefficient does not depend on ξ due to (12.22). If weare only interested in the equation Au = f , then we can divide by thiscoefficient to make the highest coefficient identically 1. Then all coefficientsof the principal symbol am will become real. So we can (and will) assumethat a(m,0,...,0) ≡ 1 from the very beginning.

Since the roots of am with respect to τ are simple, we have

∂am∂τ

∣∣∣∣τ=τj

6= 0.

By the implicit function theorem, we can solve (12.23) for τ in a neighbor-hood of τj and obtain τj(t, x, ξ) as a locally smooth function of (t, x, ξ) ∈Ω × (Rn\0), i.e., in a sufficiently small neighborhood of any fixed point(t0, x0, ξ0) ∈ Ω× (Rn\0).

The last statement can be made global due to hyperbolicity. Indeed, wecan enumerate the roots so that

τ1(t, x, ξ) < τ2(t, x, ξ) < · · · < τm(t, x, ξ).

Then every τj will be a continuous (hence, smooth) function of t, x, ξ onΩ× (Rn\0) because when we continuously move t, x, ξ (keeping ξ 6= 0), theroots also move continuously and can not collide (a collision would contradicthyperbolicity), hence their order will be preserved for any such motion.

Further, by homogeneity of am, i.e., since

am(t, x, sτ, sξ) = smam(t, x, τ, ξ),

it is clear that the set of roots τ1(t, x, sξ), . . . , τm(t, x, sξ) coincides with thesetsτ1(t, x, ξ), . . . , sτm(t, x, ξ). Since the order of the roots τj is preserved underthe homothety, we see that the functions τj(t, x, ξ) are smooth, real-valuedand homogeneous of order 1 in ξ:

τj(t, x, sξ) = sτj(t, x, ξ), s > 0.


In particular, if we know them for |ξ| = 1, then they can be extended on allvalues of ξ 6= 0.

Suppose the surface (t, x) : S(t, x) = 0 is a characteristic. Then thevector (τ, ξ) = (∂S∂t , Sx) on this surface satisfies (12.23). But then it is clearthat for some j we locally have

(12.24)∂S

∂t= τj

(t, x,

∂S

∂x

)(again on the surface S = 0). In particular, for S we may take any solutionof the Hamilton-Jacobi equation (12.24). We only need for the surface S = 0to be non-singular, i.e., S = 0 should imply grad S 6= 0. Such a functionS may be locally obtained, for example, by solving the Cauchy problem forthe equation (12.24) with the initial condition

S|t=t0 = S0(x),

where gradxS0(x) 6= 0. For instance, we may construct a characteristicpassing through (t0, x0) taking S0(x) = ξ · (x− x0), where ξ ∈ Rn\0.

In fact, taking S0(x) to be arbitrary, so that gradS0(x) 6= 0 on x :S0(x) = 0, and choosing turn by turn all τj , j = 1, . . . ,m, we will findm different characteristics, whose intersection with the plane t = t0 is thesame non-singular hypersurface in this plane. This means that the wavefront, preassigned for t = t0, may propagate in m different directions.

We may verify that the obtained characteristics, whose intersection withthe plane t = t0 is the given non-singular hypersurface, do not actuallydepend on the arbitrariness in the choice of the initial function S0 if theinitial surface x : S0(x) = 0 is fixed. We skip a simple verification ofthis fact; it follows, for example, from the study of the above method ofconstructing solutions for the Hamilton-Jacobi equation. For n = 1 such averification is performed in Chapter 1, where we proved that in R2 exactlytwo characteristics of any second-order hyperbolic equation pass througheach point.

12.4. Rapidly oscillating solutions. The eikonal equation and

the transport equations

Now we wish to understand how the wave optics turns into the geometricoptics for very short waves (or, which is the same, for high frequencies).

12.4. The eikonal and transport equations 307

Recall that the plane wave with frequency ω and wave vector k is of theform

u(t, x) = ei(ωt−k·x) = eiω(t− kω·x).

The length of the vector kω is equal to 1

a , where a is the speed of the wavepropagation. For the three-dimensional wave equation, a is a constant, i.e.,it does not depend on ω, whereas ω may be viewed as arbitrary, in particular,arbitrarily large. Therefore, for any vector k0 of length 1

a and any ω thereexists a plane wave of the form

(12.25) u(t, x) = eiω(t−k0·x).

We may assume now k0 to be fixed, and let ω tend to +∞. This is the limittransition to geometric optics for the case of a plane wave.

The argument of the exponent, ϕ = ω(t − k0 · x), is called the phase ofthe plane wave (12.25). Let us fix ω and k0 and consider constant phasesurfaces ϕ = const, which are called wave fronts. For each t a wave front isthe plane

x : k0 · x = t+ c ⊂ R3.

A wave front moves with the speed a in the direction of the vector k0.Therefore, the straight lines in the direction k0 are naturally called rays.

Now consider a general differential operator

A = a(x,D) =∑|α|≤m

aα(x)Dα, x ∈ Ω ⊂ Rn

and try to find a solution of the equation Au = 0, by analogy with the planewave (12.25), in the form

(12.26) u(x) = eiλS(x),

where λ is a large parameter, S(x) a real-valued function called phase.

Calculating Au, we see that

(12.27) Au = λmam(x, Sx(x))eiλS(x) +O(λm−1),

where am(x, ξ) is the principal symbol ofA (see Section 1.2). For the momentwe will ignore the terms denoted in (12.27) by O(λm−1) (they are of the formof eiλS(x) times a polynomial of λ of degree not higher than m− 1). But itis clear from (12.27) that if we want Au = 0 to be satisfied for arbitrarilylarge values of λ, then S(x) should satisfy the equation

(12.28) am

(x,∂S

∂x

)= 0


which is called the eikonal equation in geometric optics. It is a Hamilton-Jacobi equation if the principal symbol am(x, ξ) is real-valued (only thiscase will be considered in what follows). The equation (12.28) means thefulfillment of Au = 0 in the first approximation. More precisely, if u(x) isof the form (12.26), then (12.28) is equivalent to

Au = O(λm−1) as λ→∞.We see that the search for wave fronts (constant phase surfaces) reduces inthis case to solving the Hamilton-Jacobi equation.

Suppose that the solvability conditions for the Cauchy problem holdfor the Hamilton-Jacobi equation (12.28) (this is true, for example, if A ishyperbolic with respect to a variable denoted by t). Then we may constructwave fronts by methods of geometric optics proceeding from their dispositionon the initial manifold (at t = 0 in the hyperbolic case).

Now let us try to satisfy Au = 0 with a better precision. It turns outthat it is a good idea to seek u in the form

(12.29) u = eiλS(x)b(x, λ),

where

b(x, λ) =∞∑j=0

bj(x)λ−j

is a formal power series (in λ) called the amplitude. Therefore u = u(x, λ)is also a formal series.

Let us explain the meaning of Au. Clearly,

(12.30) A(f(x)eiλS(x)) = eiλS(x)m∑l=0

λm−l(Alf),

where Al are linear differential operators. Applying A to the k-th term of(12.29) we get

(12.31) A(eiλS(x)bk(x)λ−k) = eiλS(x)m∑l=0

λm−l−k(Albk).

Set

(12.32) Au = eiλS(x)∞∑j=0

vj(x)λm−j

which is the formal series obtained by applying A termwise to (12.29) andgrouping together all the terms with the same power of λ (as is clear from(12.31), there is only finite number of terms with any given power of λ). We


call u an asymptotic or rapidly oscillating solution if Au = 0, i.e., if all thecoefficients vj in (12.32) vanish.

Let u be an asymptotic solution of the form (12.29). Consider a finitesegment of the series (12.29):

uN (x, λ) = eiλS(x)N∑j=0

bj(x)λ−j .

Then uN is a (regular) function of x depending on λ as on a parameter. Theseries u − uN begins with λ−N−1. Therefore, the series A(u − uN ) beginswith λ−N−1+m. But Au = 0, therefore, A(u− uN ) = −AuN , implying

AuN = O(λ−N−1+m).

Thus, if we know an asymptotic solution u(x, λ), then its finite segmentsuN (x, λ) for large λ are “near-solutions” which become ever more accurateas N increases.

Similar arguments are applicable to a non-homogeneous equation Au =f and to boundary-value problems for this equation. Let a boundary-valueproblem for Au = f has a unique solution continuously depending on f

(this is often possible to deduce, for example, from energy estimates). Then,having found uN such that AuN = fN differs but little from f (for large λ),we see that since A(u − uN ) = f − fN , the approximate solution differsonly a bit from the exact solution u (we assume that uN satisfies the sameboundary conditions as u). The “exact solution”, however, does not havegreater physical meaning than the approximate one since in physics theequations themselves are usually approximate. Therefore, in the sequel wewill speak about approximate solutions and not bother with “exact” one.

Note, however, that an approximate solution satisfies the equation withgood accuracy only for large λ (for “high frequencies” of “in short-waverange”).

How to find asymptotic solutions? Note first that we may assumeb0(x) 6≡ 0 (if b0(x) ≡ 0, then we could factor out λ and reenumerate thecoefficients bj). It may happen that b0(x) 6≡ 0 but b0|U ≡ 0 for some opensubset U ⊂ Rn. In this case we should consider U and its complement sep-arately. With this remark in mind we will assume that b0|U 6≡ 0 for anyU ⊂ Ω. But then the highest term in the series Au is

(12.33) λmam(x, Sx(x))b0(x)eiλS(x).


If we wish it to be 0, then the phase S(x) should satisfy the Hamilton-Jacobiequation (12.28) and if this equation is satisfied, then (12.33) vanishes what-ever b0(x). Thus, let S(x) be a solution of the Hamilton-Jacobi equation.

Now, let us try to find bj(x) so that the succeeding terms would vanish.For this, let us write the coefficients vj in the series Au (formula (12.32))more explicitly. In the above notation, we clearly have

(12.34) vj(x) =∑

l + k = j

0 ≤ l ≤ m

k = 0, 1, 2, . . .

(Albk), j = 0, 1, 2, . . . ,

where Al are linear differential operators depending on S and defined by(12.30). The sum in (12.34) runs over l ∈ 0, 1, . . . ,m and integers k ≥ 0such that l+ k = j; therefore, the sum is finite. For j = 0 the sum containsone summand: A0b0 = am(x, Sx(x))b0(x). This makes it clear that A0 = 0due to the choice of S(x). Therefore, the sum over l in (12.34) runs actuallyfrom 1 to m. We have

v1 = A1b0,

v2 = A1b1 +A2b0,

v3 = A1b2 +A2b1 +A3b0,

. . . . . . . . . . . . . . . . . . . . . . . . .

Therefore, the conditions vj = 0, j = 0, 1, 2, . . . yield, in addition to thealready written Hamilton-Jacobi equation, also linear differential equationsfor functions bj . These equations may be written in the form

A1b0 = 0,A1b1 = −A2b0,

A1b2 = −A2b1 −A3b0,

. . . . . . . . . . . . . . . . . . . . . .

i.e., all of them are of the form

(12.35) A1bj = fj ,

where fj is expressed in terms of b0, . . . , bj−1.

Equations (12.35) are called transport equations. They show the im-portance of the role played by A1.

Let us find A1. We assume that S(x) satisfies the eikonal equation(12.28). Then A1 is determined from the formula

A(f(x)eiλS(x)) = eiλS(x)λm−1(A1f) +O(λm−2).


The direct computation, using the Leibniz product differentiation rule, showsthat for |α| = m we have

Dα(f(x)eiλS(x)) = λm(Sx(x))αf(x)eiλS(x)+λm−1n∑j=1

αj(Sx(x))α−ej [Djf(x)]eiλS(x)+

λm−1cα(x)f(x)eiλS(x) +O(λm−2),

where Dj = i−1 ∂∂xj

, ej is the multi-index with only the jth non-zero entrywhich is equal to 1, cα is a C∞ function which depends upon the first andsecond derivatives of S. Since

αjξα−ej =

∂

∂ξjξα,

it follows that

A(f(x)eiλS(x)) = λm−1

n∑j=1

∂am∂ξj

∣∣∣∣ξ=Sx(x)

(Djf(x)) + c(x)f(x)

eiλS(x)+O(λm−2),

implying

A1 =n∑j=1

∂am∂ξj

∣∣∣∣ξ=Sx(x)

Dj + c(x) = i−1L1 + c(x),

where L1 is the derivative along the vector field

V(x) =∂am∂ξ

∣∣∣∣ξ=Sx(x)

· ∂∂x

What is the geometric meaning of V(x)? Let us write the Hamiltoniansystem with the Hamiltonian am(x, ξ):

x = ∂am∂ξ

ξ = −∂am∂x

and take a bicharacteristic belonging to the graph of the gradient S, i.e.,a bicharacteristic (x(t), ξ(t)) such that ξ(t) = Sx(x(t)). Then V(x(t)) isthe tangent vector to the ray x(t) corresponding to such a bicharacteristic.Therefore, the transport equations (12.35) are of the form

1i

d

dtbj(x(t)) + c(x(t))bj(x(t)) = fj(x(t)),

i.e., are ordinary linear differential equations along rays. Therefore, thevalue bj(x(t)) along the ray x(t) is uniquely determined by bj(x(0)).


Example 12.5. Consider the wave equation in a non-homogeneous medium

(12.36)∂2u

∂t2− c2(x)∆u = 0, x ∈ R3,

where c(x) is a smooth positive function of x. Due to the general recipe, weseek a rapidly oscillating solution in the form

(12.37) u(t, x, λ) = eiλS(x)∞∑j=0

bj(t, x)λ−j .

The principal symbol of the corresponding wave operator is

H(t, x, τ, ξ) = −τ2 + c2(x)|ξ|2,so the phase S(x, t) should satisfy the following eikonal equation:(

∂S

∂t

)2

− c2(x)(∂S

∂x

)2

= 0.

The equations of Hamiltonian curves aret = −2τ

x = 2c2(x)ξ

τ = 0

ξ = −2c(x)|ξ|2 · ∂c∂ximplying τ = τ0, t = −2τ0σ + t0, where σ is the parameter along the curve.Since we are only interested in bicharacteristics, we should also assume thatτ2 = c2(x)|ξ|2 or τ = ∓c(x)|ξ|. From this we find for |ξ| 6= 0:

dx

dt=x

t=

2c2(x)ξ−2τ0

= ±c(x)ξ

|ξ| .

In particular, the absolute value of the ray’s speed is c(x). This easily impliesthat c(x) is the speed of the wave front propagation at each point. N

The first and the last equations of bicharacteristics also imply

dξ

dt=−2c(x)|ξ|2cx(x)

−2τ0= ∓|ξ| · cx(x).

Thus, x(t) and ξ(t) satisfy

(12.38)

dxdt = ±c(x) ξ

|ξ| ,dξdt = ∓|ξ| · cx(x).

This is a Hamiltonian system with the Hamiltonian H(x, ξ) = ±c(x)|ξ|. Inparticular, c(x)|ξ| = H0 = const along the trajectories. The choice of the


constant H0 (determined by the initial conditions) does not affect the raysx(t) because (x(t), λξ(t)) is a solution of (12.38) for any λ > 0 as soon as(x(t), ξ(t)) is. Using this fact, let us try to exclude ξ(t). Replacing |ξ| byH0/c(x) we derive from (12.38) that

(12.39)

dxdt = ±H−1

0 c2(x)ξdξdt = ∓H0

cx(x)c(x)

Now, substituting the expression

(12.40) ξ = ± H0

c2(x)dx

dt

found from the first equation into the second one, we get

(12.41)d

dt

(1

c2(x)dx

dt

)+cx(x)c(x)

= 0.

Thus, we have shown that every ray x(t) satisfies the 2nd order differentialequation (12.41). This is a non-linear differential equation (more precisely,a system of n such equations). It can be resolved with respect to d2x

dt2(its

coefficient is equal to 1c2(x)

and we have assumed that c(x) > 0). Therefore,the initial conditions x(0) = x0, x′(0) = v0 (the “prime” stands for thederivative with respect to t) uniquely define x(t).

Not every solution x(t) of this equation, however, determines a ray, sincethe passage from (12.38) to (12.41) is not exactly an equivalence. Let usconsider this in detail.

Let x(t) be a solution of (12.41). Determine ξ(t) via (12.40) choosing(so far, arbitrarily) a positive constant H0. Then (12.41) is equivalent tothe second of equations (12.39), whereas the first of equations (12.39) meanssimply that ξ(t) is obtained via (12.40). Therefore, if (x(t), ξ(t)) is a solutionof (12.39) for some H0 > 0, then x(t) is a solution of (12.41) and, conversely,if x(t) is a solution of (12.41) and H0 is an arbitrary positive constant, thenthere exists a unique function ξ(t), such that (x(t), ξ(t)) is a solution of(12.39).

How to pass from (12.39) to (12.38)? If (x(t), ξ(t)) is a solution of (12.39)and ξ(t) 6= 0, then clearly (12.38) holds if and only if c(x(t))|ξ(t)| = H0. Notethat for this it suffices that c(x(0))|ξ(0)| = H0. Indeed, it suffices to verifythat the vector field (

±H−10 c2(x)ξ, ∓H0

cx(x)c(x)

)


is tangent to the manifold

M = (x, ξ) : c(x)|ξ| = H0 ⊂ R2nx,ξ.

Take (x0, ξ0) ∈M and a curve (x(t), ξ(t)) which is a solution of (12.39) suchthat (x(0), ξ(0)) = (x0, ξ0). In particular, the velocity vector (x(0), ξ(0))coincides with the vector field at (x0, ξ0). We are to verify that (x(0), ξ(0))is tangent to M at (x0, ξ0), i.e., that

d

dt(c2(x(t))|ξ(t)|2)|t=0 = 0.

But

d

dt(c2(x(t))|ξ(t)|2)|t=0 = 2c(x0)[cx(x0) · x′(0)]|ξ0|2 + 2c2(x0)ξ0 · ξ′(0)

= 2c(x0)|ξ0|2cx(x0) · [±H−10 c2(x0)ξ0] + 2c2(x0)ξ0 · [∓H0c

−1(x0)cx(x0)]

= 2c(x0)[ξ0 · cx(x0)][±H−10 c2(x0)|ξ0|2 ∓H0] = 0,

because c(x0)|ξ0| = H0 due to our assumption that (x0, ξ0) ∈M .

Thus, if c(x(0))|ξ(0)| = H0, then a solution (x(t), ξ(t)) of (12.39) is alsoa solution of (12.38); hence, x(t) is a ray.

Now note that (12.40) implies that for each fixed t the condition c(x(t))|ξ(t)| =H0 is equivalent to |x′(t)| = c(x(t)). In particular, if |x′(0)| = c(x(0)) andx(t) is a solution of (12.41), then x(t) is a ray and |x′(t)| = c(x(t)) for all t.Thus, the rays x(t) are distinguished among the solutions of (12.41) by thefact that their initial values x(0) = x0, x′(0) = v0 satisfy v0 = c(x0).

Let us verify “Fermat’s principle”, which means that that the rays areextrema of the functional T = T (γ) defining time for our wave (e.g. light)to pass along the curve γ, i.e.,

T (γ) =∫γ

ds

c,

where ds is the arc-length element of the curve γ and c = c(x) is considered asa function on γ. The extremum is understood with respect to all variationspreserving the endpoints, i.e., in the class of the paths γ that connect twofixed points. Denote these points x0 and x1. Since the extremality of a curvedoes not depend on a choice of a parameter, it is convenient to assume thatthe parameter runs over [0, 1]. Thus, let γ = x(τ) : τ ∈ [0, 1]. Takevariations of γ preserving the conditions x(0) = x0, x(1) = x1. A curve isan extremum if it satisfies the Euler–Lagrange equations (see Sect. 2.5) for


the functional

T (x(τ)) =∫ 1

0

|x(τ)|dτc(x(τ))

,

where the “dot” denotes the derivative with respect to τ . The Euler-Lagrange equations

d

dτ

(∂L

∂x

)− ∂L

∂x= 0

for the Lagrangian L = |x|c(x) can be rewritten as

(12.42)d

dτ

(1

c(x(τ))1|x(τ)|

dx(τ)dτ

)+|x(τ)|c2(x(τ))

cx(x(τ)) = 0.

Clearly, the extremality of a curve is independent upon a choice of theparameter τ and its range, and so the Euler-Lagrange equations hold forevery such choice, or for none. So the parameter τ may run over any fixedsegment [a, b]). Let now x(t) be a ray with the parameter t (the sameas in (12.38)). We know that |x(t)| = c(x(t)) along the ray. But thensetting τ = t, we see that the Lagrange equation (12.42) turns into (12.41)derived above for rays. Thus, the rays satisfy the Fermat principle. Thisagrees, in particular, with the fact that in a homogeneous medium (i.e., forc(x) = c = const) the rays are straight lines.

Let us write down the transport equations in the described situation. Forthis purpose, substitute a series u(t, x, λ) of the form (12.37) into equation(12.36). We will assume that S(t, x) satisfies the eikonal equation. Thenthe series ∂2u

∂t2− c2(x)∆u begins with λ1. We have


ut(t, x, λ) = iλSteiλS

∞∑j=0

bjλ−j + eiλS

∞∑j=0

∂bj∂tλ−j ;

utt(t, x, λ) = −λ2S2t eiλS

∞∑j=0

bjλ−j + iλStte

iλS∞∑j=0

bjλ−j +

2iλSteiλS∞∑j=0

∂bj∂tλ−j + eiλS

∞∑j=0

∂2bj∂t2

λ−j ;

ux(t, x, λ) = iλSxeiλS

∞∑j=0

bjλ−j + eiλS

∞∑j=0

∂bj∂x

λ−j ;

∆u(t, x, λ) = −λ2S2xeiλS

∞∑j=0

bjλ−j + iλ∆SeiλS

∞∑j=0

bjλ−j +

2iλeiλS∞∑j=0

Sx ·∂bj∂x

λ−j + eiλS∞∑j=0

(∆bj)λ−j .

(Here S2x means Sx · Sx = |Sx|2.)

Now, compute utt − c2(x)∆u. The principal terms (with λ2) in theexpressions for utt and c2(x)∆u cancel due to the eikonal equation. Equatingthe coefficients of all the powers of λ to zero and dividing by 2i, we get

(12.43)

St∂b0∂t − c2(x)Sx · ∂b0∂x + 1

2 [Stt − c2(x)∆S]b0 = 0;St

∂b1∂t − c2(x)Sx · ∂b1∂x + 1

2 [Stt − c2(x)∆S]b1+(2i)−1[∂

2b0∂t2− c2(x)∆b0] = 0;

St∂b2∂t − c2(x)Sx · ∂b2∂x + 1

2 [Stt − c2(x)∆S]b2+(2i)−1[∂

2b1∂t2− c2(x)∆b1] = 0;

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .St

∂bj∂t − c2(x)Sx · ∂bj∂x + 1

2 [Stt − c2(x)∆S]bj+

(2i)−1[∂2bj−1

∂t2− c2(x)∆bj−1] = 0.

These are the transport equations. The Hamiltonian equations imply thatthe vector field

(St,−c2(x)Sx) = (τ,−c2(x)ξ)

is tangent to the rays, i.e., the equations (12.43) are of the form

d

dσbj(t(σ), x(σ)) + a(σ)bj(t(σ), x(σ)) + f(σ) = 0,

12.5. The Cauchy problem with rapidly oscillating initial data 317

where a(σ) only depends on S and f(σ) depends on S, b0, . . . , bj−1 while(t(σ), x(σ)) is a ray with the parameter σ considered at the beginning of thediscussion of this example. N

H Let us indicate another method of analyzing this example and exam-ples of a similar structure. We may seek solutions of (12.36) in the form

(12.44) u(t, x, ω) = eiωtv(x, ω),

where ω is a large parameter and v is a formal series of the form

v(x, ω) = eiωS(x)∞∑k=0

vk(x)ω−k.

Equation (12.36) for a function u(t, x, ω) of the form (12.44) turns into anequation of the Helmholtz type for v(x, ω)

[∆ + ω2c−2(x)]v(x, ω) = 0.

For S(x) we get the eikonal equation

S2x(x) =

1c2(x)

.

We could get the same result assuming initially S(t, x) = t− S(x). We willnot write now the transport equations for functions vk(x). The structureof these equations is close to that of non-stationary transport equations(12.43).

Remark. Transport equations describe transport of energy, polariza-tion and other important physical phenomena in physical problems. We willnot dwell, however, on these questions which rather belong to physics.

12.5. The Cauchy problem with rapidly oscillating initial

data

Consider the equation Au = 0, where A = a(t, x,Dt, Dx), t ∈ R, x ∈ Rn,and A is hyperbolic with respect to t. Let us consider a formal series

(12.45) u(t, x, λ) = eiλS(t,x)∞∑j=0

bj(t, x)λ−j

and try to find out which initial data at t = 0 enable us to recover this seriesif it is an asymptotic solution. We will only consider the phase functionsS(t, x) such that gradx S(t, x) 6= 0. The function S(t, x) should satisfy theeikonal equation

(12.46) am(t, x, St, Sx) = 0,


where am(t, x, τ, ξ) is the principal symbol of A (which can be assumed realwithout loss of generality). But as we have seen above in Section 12.3, thisis equivalent to the fulfillment of one of m equations

(12.47)∂S

∂t= τl(t, x, Sx)

where τl(t, x, ξ), for l = 1, 2, . . . ,m, is the complete set of roots of the equa-tion am(t, x, τ, ξ) = 0, chosen, say, in the increasing order. Setting the initialcondition

(12.48) S|t=0 = S0(x),

where S0(x) is such that gradxS0(x) 6= 0, we may construct exactly m differ-ent solutions of the eikonal equation (12.46) satisfying this initial condition,namely, solutions of each of the equation (12.47) with the same initial con-dition (12.48) (such a solution is unique as soon as we have choosen the rootτl). Solutions of equations (12.47) exist in a small neighborhood of the planet = 0 determined as indicated in Section 12.2. In the same neighborhood,the functions bj(t, x) are uniquely defined from the transport equations pro-vided the initial values are given:

bj |t=0 = bj,0(x), j = 0, 1, 2, . . . .

Let us now formulate the Cauchy problem for the equation Au = 0 byimposing the initial conditions of the form

(12.49)

u|t=0 = eiλS0(x)∑∞

j=0 c(0)j (x)λ−j ,

ut|t=0 = λeiλS0(x)∑∞

j=0 c(1)j (x)λ−j ,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .∂m−1u∂tm−1

∣∣∣t=0

= λm−1eiλS0(x)∑∞

j=0 c(m−1)j (x)λ−j .

Solving this problem would allow to find with arbitrarily good accuracy(for large λ) solutions of the equation Au = 0 with rapidly oscillating initialvalues that are finite segments of the series on the right-hand sides of (12.49).

A solution of this problem is not just a series of the form (12.45): theremay be no such a series. We should take a finite sum of these series withdifferent phases S(t, x). Thus, consider a sum

(12.50) u(t, x, λ) =m∑r=0

ur(t, x, λ),

where

ur(t, x, λ) = eiλSr(t,x)

∞∑j=0

b(r)j (t, x)λ−j .


Sum (12.50) is called a solution of the equation Au = 0 if each ur(t, x, λ)is a solution of this equation. Since all the series ∂kur

∂tk

∣∣∣t=0

are of the form

λkeiλS0(x)∞∑j=0

cj(x)λ−j ,

the initial values u|t=0, ut|t=0, . . . ,∂m−1u∂tm−1

∣∣∣t=0

also are of the same form.

Therefore, it makes sense to say that u(t, x, λ) satisfies the initial condi-tions (12.49).

For instance, we may consider a particular case when the series in (12.49)contain one term each:

u|t=0 = eiλS0(x)ϕ0(x),. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .∂m−1u∂tm−1 |t=0 = λm−1eiλS0(x)ϕm−1(x).

Cutting off the series for u at a sufficiently remote term, we see that anyasymptotic solution provides (regular) functions u(t, x, λ), such that

Au = O(λ−N ),u|t=0 = eiλS0(x)ϕ0(x) +O(λ−N ),. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .∂m−1u∂tm−1

∣∣∣t=0

= λm−1eiλS0(x)ϕm−1(x) +O(λ−N ),

where N may be taken arbitrarily large.

Theorem 12.5. The above Cauchy problem for the hyperbolic equationAu = 0 with initial conditions (12.49) is uniquely solvable in a small neigh-borhood of the initial plane t = 0.

Proof Each of the phases Sr(t, x) should satisfy the eikonal equationand the initial condition S|t=0 = S0(x). There exist, as we have seen,exactly m such phases S1, S2, . . . , Sm. Now the transport equations forthe amplitudes b(r)j (t, x) show that it suffices to set b

(r)j

∣∣∣t=0

= bj,r(x) to

uniquely define all functions b(r)j (t, x). Thus we need to show that the initialconditions (12.49) uniquely determine bj,r(x). Substitute u =

∑mr=1 u

r intothe initial conditions (12.49) and write the obtained equations for b(r)j (t, x)equating the coefficients of all the powers of λ. The first of equations (12.49)gives

m∑r=1

b(r)j

∣∣∣t=0

= c(0)j , j = 0, 1, . . . .


The second of equations (12.49) becomesm∑r=1

(i∂Sr

∂tb(r)j

)∣∣∣∣t=0

+m∑r=1

∂b(r)j−1

∂t

∣∣∣∣∣t=0

= c(1)j , j = 0, 1, . . . .

In general, the equation which determines ∂ku∂tk

∣∣∣t=0

becomes

(12.51)m∑r=1

[(i∂Sr

∂t

)kb(r)j

]∣∣∣∣∣t=0

= f(k)j (x), k = 0, 1, . . . ,m− 1; j = 0, 1, . . . ,

where f (k)j (x) only depends on b

(r)0 , . . . , b

(r)j−1. Now, note that

∂Sr

∂t

∣∣∣∣t=0

= τr

(0, x,

∂S0

∂x

),

where τr(0, x, ξ) 6= τl(0, x, ξ) if r 6= l and ξ 6= 0. Therefore, if ∂S0∂x 6= 0 (which

we assume to be the case in the considered problem), then the system (12.51)for a fixed j is a system of linear equations with respect to b

(r)j

∣∣∣t=0

= bj,r(x)with the determinant equal to the Vandermond determinant∣∣∣∣∣∣∣∣∣

1 . . . 1iτ1 . . . iτm. . . . . . . . .

(iτ1)m−1 . . . (iτm)m−1

∣∣∣∣∣∣∣∣∣ ,where τr = τr(0, x, ∂S0

∂x ). Since, due to hyperbolicity, τj 6= τk for j 6= k, thedeterminant does not vanish anywhere (recall that we assumed ∂S0

∂x 6= 0).The right-hand sides of (12.51) depend only on b

(r)0 , . . . , b(r)j−1. There-

fore, the sequence of the systems of linear equations (12.51) enables us touniquely determine all functions b(r)j by induction. Namely, having found

b0,r(x) for r = 1, . . . ,m from (12.51) with j = 0, we may determine b(r)0 (t, x)from the transport equations with the initial conditions b(r)0 |t=0 = b0,r(x).If b0, . . . , bj−1 are determined, then the functions b(r)j |t=0, r = 1, . . . ,m,are recovered from (12.51), and then the transport equations enable us todetermine b(r)j (t, x).

Let us briefly indicate the connection of Theorem 12.5 with the behaviorof the singularities for solutions of the Cauchy problem. It is well knownthat the singularities of f(x) are connected with the behavior of its Fouriertransform f(ξ) at infinity. For instance, the inversion formula

(12.52) f(x) = (2π)−n∫f(ξ)eix·ξdξ,


implies that if |f(ξ)| ≤ C(1 + |ξ|)−N , then f ∈ CN−n−1(Rn). Therefore,to find the singularities of a solution u(t, x) it suffices to solve the Cauchyproblem up to functions whose Fourier transform with respect to x decreasessufficiently rapidly as |ξ| → ∞.

Formula (12.52) can be regarded as a representation of f(x) in the formof a linear combination of exponents eiξ·x. Since the problem is a linear one,to find solutions of the problem with one of the initial condition equal tof(x), it suffices to consider a solution with the initial data f(x) replaced byeix·ξ and then take the same linear combination of solutions.

If f ∈ E ′(Rn) and supp f ⊂ K, where K is a compact in Rn, it is moreconvenient to somewhat modify the procedure. Namely, let ϕ ∈ C∞0 (Rn),ϕ = 1 in a neighborhood of K. Then

(12.53) f(x) = ϕ(x)f(x) = (2π)−n∫f(ξ)ϕ(x)eix·ξdξ

and it suffices to solve the initial problem by replacing f(x) by ϕ(x)eix·ξ.For instance, if f(x) = δ(x), ϕ ∈ C∞0 (Rn), ϕ(0) = 1, then (12.53) holds withf(ξ) ≡ 1; therefore, having solved the Cauchy problem

(12.54)

Au = 0,

u|t=0 = 0,

ut|t=0 = 0,

. . . . . . . . . . . . . . . . . . . . . .∂m−1u∂tm−1

∣∣∣t=0

= ϕ(x)eix·ξ,

and having denoted the solution by uξ(t, x), we can write

E(t, x) = (2π)−n∫uξ(t, x)dξ

(if the integral exists in some sense, e.g. converges for any fixed t in thetopology of D′(Rn

x)), and then E(t, x) is a solution of the Cauchy problem

(12.55)

AE = 0,

E|t=0 = 0,

Et|t=0 = 0,

. . . . . . . . . . . . . . . . .∂m−1E∂tm−1

∣∣∣t=0

= δ(x).

If we only wish to trace the singularities of E(x), we should investigate thebehavior of uξ(x) as |ξ| → +∞. We may fix the direction of ξ introducing


a large parameter λ = |ξ| and setting η = ξ|ξ| . Then the Cauchy problem

(12.54) takes on the form of the Cauchy problem with rapidly oscillatinginitial data. Namely, in (12.49) we should set

S0(x) = η · x,c

(0)j (x) = c

(1)j (x) = . . . = c

(m−2)j (x) = 0, for j = 0, 1, . . . ;

c(m−1)j (x) = 0, for j 6= m− 1,c

(m−1)m−1 (x) = ϕ(x).

We may find a solution uξ(t, x) = uη(t, x, λ) which satisfies (12.54) withaccuracy up to O(λ−N ) for arbitrarily large N . Then (12.55) are satisfiedup to functions of class CN−n−1(Rn). It can be proved that singularities ofthe genuine solution of (12.55) are the same as for the obtained approximatesolution, i.e., the genuine solution differs from the approximate one by anarbitrarily smooth function (for large values of N).

Where are the singularities of E(t, x)? To find out, we should considerthe integral

EN (t, x) = (2π)−n∫uξ,N (t, x)dξ,

where uξ,N is the sum of the first N + 1 terms of the series which representsthe asymptotic solution uξ. All these terms are of the form

λk∫eiλS

r(t,x,η)ϕ(t, x, η)dξ,

where λ = |ξ|, η = ξ/|ξ|, Sr(t, x, η) is one of the phase functions satisfy-ing the eikonal equation with the initial condition Sr|t=0 = S0(x) = η · x;ϕ(t, x, η) is infinitely differentiable with respect to all variables; ϕ is de-fined for small t and has the (t, x)-support belonging to an arbitrarily smallneighborhood of the set of rays (t, x(t)) with the source in (0, 0). The laststatement is deduced from the form of the transport equations. Thereforewe see that the support of EN (t, x) belongs to an arbitrarily small neighboor-hood of the set of rays passing through (0, 0). Taking into account the factthat the singularities of E and EN coincide (with any accuracy), we see thatthe distribution E(t, x) is infinitely differentiable outside of the set of rayswith (0, 0) as the source. Here we should take all the rays corresponding toall the functions Sr(t, x, η) for each η and each r = 1, . . . ,m. We get the setof m curved cones formed by all rays starting from (0, 0) (for each η ∈ Rn,|η| = 1, and each r = 1, . . . ,m we get a ray smoothly depending on η).

Problems 323

A more detailed description is left to the reader as a useful exercise whichshould be worked out for 2nd order equations, for example, for the equation(12.36) describing light propagation in a non-homogeneous medium.

Problems

12.1. For the equation utt = a2∆u, where u = u(t, x), x ∈ R2, find thecharacteristics whose intersection with the plane t = 0 is

a) the line η · x = 0, where η ∈ R2\0);

b) the circle x : |x| = R.

12.2. Describe the relation between the rays and characteristics of the equa-tion utt = c2(x)uxx.

12.3. For the Cauchy problemutt = x2uxx

u|t=0 = 0

ut|t=0 = eiλx

write and solve the eikonal equation and the first transport equation. Usethis to find the function u(t, x, λ), such that for x 6= 0, we have as λ→∞:

utt − x2uxx = O(λ−1)

u|t=0 = O(λ−2)

ut|t=0 = eiλx +O(λ−1).

Chapter 13

Answers and Hints.

Solutions

1.1. a) upp + uqq + urr = 0; the change of variables is:p = x,

q = −x+ y,

r = x− 12y + 1

2z.

(The reader should keep in mind that a change of variables leading to thecanonical form is not unique; the above one is one of them, and may verywell happen that the reader will find another one).

b) upp − uqq + 2up = 0; the change of variables is:p = x+ y,

q = x− y,r = y + z.

1.2. a) uzw − 1wuz − 1

4zuw − 1z2u = 0; the change of variables is:z = yx−3,

w = xy.

b) uzz + uww + 1z+wuz + 1

2wuw = 0; the change of variables is:z = y2 − x2,

w = x2.

325

326 13. Answers and Hints. Solutions

c) uww + 2uz + uw = 0; the change of variables is:z = x+ y,

w = x.

1.3. a) u(x, y) = f(xy ) +√∣∣∣xy ∣∣∣g(xy), where f , g are arbitrary functions of

one independent variable.

(Here and in b) below the form of the general solution is not unique.)

Hint. The change of variablesz = ln |xy|,w = ln |y/x|,

reduces the equation to the form 2uzw +uz = 0, i.e., ∂∂z (2uw +u) = 0 which

implies 2uw + u = F (w) and u(z, w) = a(w) + exp(−w/2)b(z).

b) u(x, y) = ln | yx |f(xy) + g(xy), where f, g are arbitrary functions ofone independent variable.

Hint. The change of variables:z = ln |xy|,w = ln |y/x|,

reduces the equation to the form uww = 0, hence u(z, w) = wϕ(z) + ψ(z).

2.1. a) ux|x=0 = 0

Hint. Write the dynamical Newton equation of the motion of the ring.

b) (mutt − Tux)|x=0 = 0.

Hint. See Hint to a).

c) (mutt + αut − Tux)|x=0 = 0.

2.2. a) E(t) = 12

∫ l0 [ρu2

t (t, x) + Tu2x(t, x)]dx = E0 = const .

b), c) Let

E(t) =12mu2

t |x=0 +∫ l

0[ρu2

t (t, x) + Tu2x(t, x)]dx.

Then E(t) ≡ E0 = const in case b) and E(t)−E(0) = Afr = −∫ t

0 αu2t |x=0dt

in case c) (here Afr is the work of the friction force).

2.3. ρutt = Euxx or utt = a2uxx with a =√E/ρ. Here u = u(t, x)

is a longitudinal displacement of a point with position x in equilibrium,ρ the volume density of the material of the rod, E the Young modulus

13. Answers and Hints. Solutions 327

(called also the modilus of elongation) which describes the force arising inthe deformed material by the formula F = ES∆l

l (Hook’s law), S the cross-section where the force is measured, ∆l/l the deformation of the small pieceof material around the point of measurement (∆l/l is often called the relativelengthening, here l is the length of the piece in equilibrium position and ∆lis the increment of this length caused by the force).

Hint. Prove that the relative lengthening of the rod at the point withposition x in equilibrium is ux = ux(t, x) so the force F = ESux(t, x) isacting on the left part of the rod in the corresponding cross-section at thepoint x. Consider the motion of the part [x, x + ∆x] of the rod, writeNewton’s dynamical equation for this part and then take the limit as ∆x→0.

2.4. ux|x=0 = 0.

2.5. (ESux − ku)|x=0 = 0. Here k is the rigidity coefficient of the spring,i.e. the force per unit lengthening of the spring.

2.6. a), b)

E(t) ≡ 12

∫ l

0[ρu2

t (t, x) + Eu2x(t, x)]dx = E0 = const .

c) E(t) ≡ 12ku

2|x=0 + 12

∫ l0 [ρSu2

t (t, x) + ESu2x(t, x)]dx = E0 = const.

2.7. ρS(x)utt = ∂∂x(ES(x)∂u∂x), where S(x) is the area of the cross-section

of the rod at the point with position x in equilibrium (the x-axis should bedirected along the axis of the rod).

2.9. E(t) ≡ 12

∫∞−∞[ρu2

t (t, x) + Tu2x(t, x)]dx = E0 = const (perhaps +∞ for

all t)

Hint. Using Problem 2.8 prove that E(t′) ≤ E(t) for t′ ≥ t, and thenreverse the direction of the time-variable.

2.10. See Fig. 1.

2.11. u ≡ 0

Hint. The Cauchy initial conditions are u|x=x0 ≡ 0, ∂u∂x∣∣x=x0

≡ 0.

2.12. u =

f(x− at) for x > x0 + ε, where f(x) = 0 for x < x0 + ε,

g(x+ at) for x < x0 − ε, where g(x) = 0 for x > x0 − ε.2.13. See Fig. 2.

2.14. See Fig. 3.


c dx

x

x

x

x

t= 0

t= "< l=2a; l= d− c

t= l=2a

t= l=2a+ "

t l=2a

Figure 1

Hint. Use the formula12a

∫ x+at

x−atψ(s)ds = Ψ(x+ at)−Ψ(x− at), where Ψ(z) =

12a

∫ z

0ψ(s)ds.

The graphs of Ψ(x+ at) and Ψ(x− at) are drawn on Fig. 3 by dashed lines.

2.15. E(t) ≡ 12

∫∞0 (ρu2

t (t, x) + Tu2x(t, x))dx = E0 = const.

Hint. See Hint to Problem 2.9.

2.16. The reflected wave is(k2

E2S2+ω2

a2

)−1 2ωkaES

[cosω

(t− x

a

)− exp

k

ES(x− at)

]+(ω2

a2− k2

E2S2

)sinω

(t− x

a

),

with notations as in the answer to Problem 2.3.

2.17. On the left u = sinω x+ata+v with the frequency ωl = ωa

a+v < ω.

On the right u = sinω x−atv−a with the frequency ωr = ωaa−v > ω.

2.18. The standing waves are X0(x)(at+b) and Xn(x)(an cosωnt+bn sinωnt)for n = 1, 2, . . . with Xn(x) = cos nπxl and ωn = πna

l .

The graphs of the first Xn(x), n = 1, 2, 3, see on Fig. 4.

2.19. The boundary conditions are u|x=0 = 0 and (ESux+ku)|x=l = 0. Thestanding waves are Xp(x)(ap cosωpt + bp sinωpt), where Xp(x) = sin γpx

l ,ωp = γpa/l, and the γp are the solutions of the equation

cot γ = − kl

ESγ


x

x

x

x

x

x

x

x

x

x

x

x

x

x

t= 0

t= "< l=2a

t= l=2a

t= l=2a+ "

t= l=a

t= l=a+ "

t= 3l=2a

t= 3l=2a+ "

t= 2l=a

t= 2l=a+ "

t= 5l=2a

t= 5l=2a+ "

t= 3l=a+ "

t> 3l=a

l 3l

Figure 2

such that γp ∈ (pπ, (p + 1)π) for p = 0, 1, 2, . . . . The system Xp : p =0, 1 . . . is orthogonal in L2([0, l]), and every function Xp is an eigenfunc-tion of the operator L = −d2/dx2 with the boundary conditions X(0) =0, X ′(0) + αX(0) = 0 for α = k/(ES). The corresponding eigenvalue isλp = (γp/l)2 and its shortwave asymptotic is

λp =[π(2p+ 1)

2l

]2(1 +O

(1p2

)).


x

x

x

x

x

x

x

x

2l

t= 0

t= "< l=a

t= l=a

t= l=a+ "

t= 2l=a

t= 2l=a+ "

t= 3l=a

t> 3l=a

−l l 3l

Figure 3

2.20. a) The resonance frequencies are ωk = (2k+1)πa2l , k = 0, 1, 2, . . .; the

condition for a resonance is ω = ωk for some k. If ω 6= ωk then

u(t, x) = F0aESω cos ω

al sin ω

ax · sinωt+∑∞k=0(−1)k 2F0ωa2

ωklES(ω2−ω2k)

sinωkt · sin (2k+1)πx2l .


0 xl

0 xl

0 xl

0 xl

X 0 ≡ 1

X 1(x)= cosπx

l

X 2(x)= cos2πx

l

X 3(x)= cos3πx

l

Figure 4

If ω = ωk for some k, then

u(t, x) = (−1)k F0a2

ESlω2 [32 sinωt · sin ω

ax− xωa sinωt · cos ωax−

ωt cosωt · sin ωax]+∑

p≥0,p 6=k(−1)p 2F0ωa2

ωplES(ω2−ω2p)

sinωpt · sin (2p+1)πx2l .

Hint. Seek a particular solution in the form u0 = X(x) sinωt satisfyingthe boundary conditions (but not the initial conditions), then find v = u−u0

by the Fourier method in case ω 6= ωk. In case of a resonance ω = ωk, takethe limit of the non-resonance formula for u as ω → ωk.

b) Resonance frequencies is ωk = kπal , k = 1, 2, . . ., the condition for a

resonance is ω = ωk for some k. If ω 6= ωk, then

u(t, x) = − F0aESω sin ωl

a

sinωt · cos ωax+ F0a2tESlω+∑∞

k=0(−1)k 2F0ωa2

ωklES(ω2−ω2k)

sinωkt · cos kπxl .


u(t, x) = (−1)k F0a2

ESl [ 32ω2 sinωt · sin ω

ax− 1ω (xa sinωt · cos ωkxa +

t cosωt · sin ωax)] + F0a2t

ESlω+∑p≥1, p 6=k(−1)p 2F0ωa2

ωplES(ω2−ω2p)

sinωpt · cos pπxl .

Hint. See Hint to a).


2.21. a) The resonance frequencies are ωk = kπal for k = 1, 2, . . .; the

condition for a resonance is ω = ωk for some k. If ω 6= ωk, then

u(t, x) = A sin ωxa

sin ωla

sinωt+∑∞k=1(−1)k+1 2ωAa

l(ω2−ω2k)

sinωkt · sin kπxl .


u(t, x) = (−1)k Aal [ 12ω sinωt · sin ω

ax+ xa sinωt · cos ωxa + t cosωt · sin ω

ax]+∑p≥1, p 6=k(−1)p+1 2ωAa

l(ω2−ω2p)

sinωpt · sin pπxl .

Hint. See Hint to 2.20 a).

b) The resonance frequencies are ωk = (2k+1)πa2l , k = 0, 1, 2, . . ., the con-

dition for a resonance is ω = ωk for some k. If ω 6= ωk, then

u(t, x) = A cos ωxa

cos ωla

sinωt+∑∞k=0(−1)k 2ωAa

l(ω2−ω2k)

sinωkt · cos (2k+1)πa2l .


u(t, x) = (−1)k Aal [ 12ω sinωt · cos ωax+ x

a sinωt · sin ωxa − t cosωt · cos ωxa ]+∑

p≥0, p 6=k(−1)p 2ωAa

l(ω2−ω2p)

sinωpt · cos (2p+1)πa2l .


c) The resonance frequencies are ωp = γpa/l, where γp are as in answerto 2.19, p = 1, 2, . . . The condition for a resonance is ω = ωp for some p. Ifω 6= ωp, then with γ = ωl/a:

u(t, x) = A(cos ωla + kaESω sin ωl

a )−1(cos ωxa + kaESω sin ωx

a ) sinωt+∑∞p=0

4Aγγp sin γp(2γp−sin(2γp))(γ2−γ2

p)sinωpt · (cos γpxl + kl

ESγpsin γpx

l ).

If ω = ωp for some p, then

u(t, x) = 2γA2γ−sin 2γ [xl sinωt · cos(γ(xl − 1)) + at

l cosωt · sin(γ(xl − 1))]+Aγ(1−cos 2γ) sinωt·sin(γ(x

l−1))

(2γ−sin 2γ)2− A sinωt·sin(γ(x

l+1))

2γ−sin 2γ +∑n≥0, n 6=p

4Aγγn sin γn(2γn−sin(2γn))(γ2−γ2

n)sinωnt · (cos γnxl + kl

ESγnsin γnx

l ).



Figure 5

3.1. Φ(x) = cos kx+ 1k

∫ x0 sin k(x− t)q(t)Φ(t)dt;

Φ(x) = cos kx+O( 1k ) =

cos kx+ 1k

∫ x0 sin k(x− t) cos kt q(t)dt+O( 1

k2 );Φ′(x) = −k sin kx+O(1) =−k sin kx+

∫ x0 cos k(x− t) cos kt q(t)dt+O( 1

k ).

3.2. Hint. The values of k satisfying Φ(l) = Φ(l, k) = 0 can be found byimplicit function theorem near k such that cos kl = 0. Differentiation withrespect to k of the integral equation of Problem 3.1 gives an integral equa-tion for ∂Φ/∂k providing information needed to apply the implicit functiontheorem.

3.3. G(x, ξ) = 1l [θ(ξ − x)x(l − ξ) + θ(x− ξ)(l − x)ξ].

A physical interpretation: G(x, ξ) is the form of a string which is pulledat point ξ by the vertical point force F = T , where T is the tension force ofthe string.

3.4. G(x, ξ) = 1sinh l [θ(ξ−x) coshx · cosh(l− ξ) + θ(x− ξ) cosh(l−x) · cosh ξ].

A physical interpretation: G(x, ξ) is the form of a string which is pulledat a point ξ by the vertical point force F = T , has free ends (ends which areallowed to move freely in the vertical direction) and is subject to the actionof an elastically returning uniformly distributed force; see Fig. 5, where thereturning force is realized by the little springs attached to the points of thestring and to a rigid massive body.

3.6. Hint. A solution y = y(x) 6≡ 0 of the equation −y′′ + q(x)y = 0 withq ≥ 0 has at most one root.

3.7. Hint. Use the inequality (Gϕ,ϕ) ≥ 0 for G = L−1; take

ϕ(x) =n∑j=1

cjϕε(x− xj)


with a δ-like family ϕε.

3.8. a)∞∑j=1

1λj

=∫ l

0 G(x, x)dx.

Hint. Use the expansion

G(x, ξ) =∞∑j=1

Xj(x)Xj(ξ)λj

which can be obtained, e.g. as the orthogonal expansion of G(·, ξ) withrespect to Xj(x), j = 1, 2, . . ..

b)∑∞

j=11λ2j

=∫ l

0

∫ l0 |G(x, ξ)|2dxdξ.

Hint. This is the Parseval identity for the orthogonal expansion ofG = G(x, ξ) with respect to the system

Xj(x)Xk(ξ) : j, k = 1, 2, . . ..

In the particular case L = −d2/dx2 on [0, π] with boundary conditionsX(0) = X(π) = 0 we obtain λn = n2, and taking into account the answerto Problem 3.3, we get

∞∑n=1

1n2

=π2

6,

∞∑n=1

1n4

=π4

90.

4.1. 0.

4.2. 2δ(x).

4.3. πδ(x).

4.4. Hint. f(x)− f(0) =∫ 1

0ddtf(tx)dt = x

∫ 10 f′(tx)dt.

4.5. Hint. Write explicit formulas for the map D′(S1) → D′2π(R) and itsinverse.

4.6. Hint. Use the continuity of a distribution from D′(S1) with respect toa seminorm in C∞(S1).

4.7. Hint. Use duality considerations.

4.8.∑

k∈Z δ(x+ 2kπ) = 12π

∑m∈Z e

imx.

4.9. Hint. Multiply the above formula in Answer to 4.9 by f(x) andintegrate.

4.10. a) u(x) = v.p. 1x + Cδ(x) = 1

x+i0 + C1δ(x) = 1x−i0 + C2δ(x).

b) u(x) = ln |x|+ Cθ(x) + C1.


c) u(x) = −δ(x) + Cθ(x) + C1.

5.1. a) −2πiθ(ξ).

b) 4πr0sin r0|ξ||ξ| .

Hint. Use the spherical symmetry and take

ξ = |ξ|(1, 0, 0) = (ξ1, 0, 0).

c) 2π2

ρ0δ(|ξ| − ρ0).

Hint. Use the Answer to b).

d) (−i)k+1

(ξ−i0)k+1 .

5.2. a) 1(n−2)σn−1|x|n−2 .

Hint. It should be a spherically symmetric fundamental solution for(−∆).

b) 14π|x|e

−k|x|.

Hint. Using the spherical symmetry take ξ = (ξ1, 0, 0). Use the regular-ization with multiplication by exp(−ε|ξ|). Calculate the arising 1-dimensionalintegral using residues.

c) 14π|x|e

∓ik|x|.

Hint. Calculate F−1( 1|ξ|2−k2±iε) as in b) and pass to the limit as ε→ +0.

5.3. 14π|x|e

−k|x|.

5.6. 0 as t→ +∞ and −2πiδ(x) as t→ −∞.

6.1. Prove the statements inverse to the ones of Theorems 6.1, 6.2: if acontinuous function u in an open set Ω ⊂ Rn has mean value property forall balls B with B ⊂ Ω or for all spheres which are boundaries of suchballs (that is, either (6.2) holds for such spheres or (6.4) holds for all suchballs), then u is harmonic in Ω. Moreover, it suffices to take balls or spheres,centered at every point x with sufficiently small radii r ∈ (0, ε(x)), whereε(x) > 0 continuously depends upon x ∈ Ω.

6.2. Let u be a real-valued C2 function on an open ball B = Br(x) ⊂ Rn,which extends to a continuous function on the closure B, and ∆u ≥ 0everywhere in B. Prove that

(13.1) u(x) ≤ 1voln−1(S)

∫Su(y)dSy =

1σn−1rn−1

∫Sr(x)

u(y)dSy,


and

(13.2) u(x) ≤ 1vol(B)

∫Bu(y)dy =

n

σn−1rn

∫Br(x)

u(y)dy,

where the notations are the same as in Section 6.1.

6.3. Prove the inverse statement to the one in the previous problem: ifu ∈ C2(Ω) is real-valued and for all balls B with B ⊂ Ω (or for all sphereswhich are boundaries of such balls), (13.2) (respectively, (13.1)) holds, then∆u ≥ 0 in Ω. Moreover, it suffices to take balls or spheres, centered atevery point x with sufficiently small radii r ∈ (0, ε(x)), where ε(x) > 0continuously depends upon x ∈ Ω.

Remark 13.1. Let a real-valued function u ∈ C2(Ω) be given. It is calledsubharmonic, subharmonic function if ∆u ≥ 0 in Ω. Note, however, that thegeneral notion of subharmonicity does not require that u ∈ C2 and allowsmore general functions u (or even distributions, in which case ∆u ≥ 0 meansthat ∆u is a positive Radon measure).

6.4. Let u be a real-valued subharmonic function in a connected open setΩ ⊂ Rn. Prove that u can not have local maxima in Ω, unless it is constant.In other words, if x0 ∈ Ω and u(x0) ≥ u(x) for all x in a neighborhood ofx0, then u(x) = u(x0) for all x ∈ Ω.

6.5. Consider the initial value problem (6.11) with ϕ ∈ C∞0 (Rn) and ψ ≡ 0.Assume that it has a solution u. Prove that in this case ϕ ≡ 0 on R andu ≡ 0 on ΠT .

6.6. Let U, V,W be open sets in Rm,Rn,Rp respectively, and g : U → V ,f : V → W be maps which are locally Ck,γ (that is, they belong to Ck,γ

on any relatively compact subdomain of of the corresponding domain ofdefinition).

(a) Assume that k ≥ 1. Then prove that the composition f g : U →W

is also locally Ck,γ .

(b) Prove that the statement (a) above does not hold for k = 0.

(c) Prove that if k ≥ 1, Ω is a domain in Rn with ∂Ω ∈ Ck,γ , U is aneighborhood of Ω in Rn, and ψ : U → Rn is a Ck,γ diffeomorphism of Uonto a domain V ⊂ Rn, then the image Ω′ = ψ(Ω) is a domain with a Ck,γ

boundary, Ω′ ⊂ V .


6.7. (a) Prove that if Ω is a bounded domain in Rn and ∂Ω ∈ Ck,γ withk ≥ 1, then ∂Ω is a Ck,γ manifold, dimR ∂Ω = n− 1.

(b) Prove that for any non-negative integers k, k′ and numbers

γ, γ′ ∈ (0, 1), such that k+γ ≥ k′+γ′, k ≥ 1, the class Ck′,γ′ is locally invari-

ant under Ck,γ diffeomorphisms. In particular, function spaces Ck′,γ′(∂Ω)

are well defined.

(c) Under the same assumptions as in (a) and (b), for a real-valuedfunction ϕ : ∂Ω → R, the inclusion ϕ ∈ Ck

′,γ′(∂Ω) is equivalent to theexistence of an extension ϕ ∈ Ck′,γ′(Ω), such that ϕ|∂Ω = ϕ.

6.8. Assume that k ≥ 1 is an integer, Ω is a bounded domain in Rn, suchthat ∂Ω ∈ Ck+2, f ∈ Ck(Ω) and ϕ ∈ Ck+2(∂Ω). Using Schauder’s Theorem6.20, prove that the problem (6.10) has a solution u ∈ Ck+1(Ω).

6.9. Assume that ∂Ω ∈ C2, and there exists Green’s function G = G(x, y) ofΩ, such that G ∈ C2 for x 6= y, x, y ∈ Ω. Consider an integral representationof a function u ∈ C2(Ω) given by the formula (6.28), with f ∈ C(Ω) andϕ ∈ C(∂Ω). Prove that the pair f, ϕ, representing a fixed function u ∈C2(Ω) by (6.28), is unique.

Hint. The function f can be recovered from u by the formula f = ∆u,whereas at any regular point of the boundary x ∈ ∂Ω, ϕ is the jump of theoutward normal derivative of u on ∂Ω:

(13.3) ϕ(x) = limε→0+

(∂u

∂n(x+ εn)− ∂u

∂n(x− εn)

).

6.10. Provide detailed proofs of identities (6.44) and (6.45).

6.11. Provide detailed proofs of identities (6.46), (6.47) and (6.48).

7.1. u(t, x) = u1+u22 +

∑∞p=0

2(u1−u2)(−1)p

(2p+1)π exp[−( (2p+1)πal )2t] cos (2p+1)πx

l .

Relaxation time is tr ' l2

4π2a2 ' l2

40a2 ' l2cp40k , where k is the thermal

conductivity, c the specific heat (per units of mass and temperature), ρ thedensity (mass per volume), l the length of the rod.

Hint. The first term in the sum above is much greater than the otherterms at times comparable with the relaxation time.


7.2. Let I be the current, R the electric resistance of the rod, V the volumeof the rod, k the thermal conductivity,

cp =2l

∫ l

0

[ϕ(x)− I2R

2V kx(l − x)

]sin

pπx

ldx, where ϕ = u|t=0.

Then

u(t, x) =I2R

2V kx(l − x) +

∞∑p=1

cp exp[−(pπa

l

)2t

]sin

pπx

l

and

limt→+∞

u(t, x) =I2R

2V kx(l − x).

Hint. The equation describing the process is

ut = a2uxx +I2R

cρV

7.3. limt→+∞

u(t, x) = b+c2 .

Hint. Consider Poisson’s integral giving an explicit formula for u(t, x).

8.1. u(x, y) = a2(x2 − y2).

Hint. The Fourier method gives in the polar coordinates

u(r, ϕ) = a0 +∞∑k=1

rk(ak cos kϕ+ bk sin kϕ).

8.4. u(t, x) = 112(x4 − y4)− a2

12 (x2 − y2).

Hint. Find a particular solution up of ∆u = x2 − y2 and then seekv = u− up.8.5. In the polar coordinates

u(r, ϕ) = 1 +b

2lnr

a+

b3

4(a4 + b4)(r2 − a4r−2) cos 2ϕ.

Hint. The Fourier method gives in the polar coordinates

u(r, ϕ) = A0 +B0 ln r+∞∑k=1

[(Akrk +Bkr−k) cos kϕ+(Ckrk +Dkr

−k) sin kϕ].

8.6. The answer to the last question is

u(x, y) = Bsinh π(b−y)

a

sinh πba

sinπx

a+


+∞∑p=0

8Ab(2p+ 1)2π2

sinh (2p+1)π(a−x)b

sinh (2p+1)πab

sin(2p+ 1)πy

b.

Hint. The general scheme is as follows.

Step 1. Reduce the problem to the case

ϕ0(0) = ϕ0(a) = ϕ1(0) = ψ0(0) = ψ0(b) = ψ1(0) = ψ1(b) = 0

by subtracting a function of the form A0 +A1x+A2y +A3xy.

Step 2. Seek u in the form u = u1 + u2 with u1, u2 such that ∆u1 =∆u2 = 0; u1 satisfies the Dirichlet boundary conditions with ψ0 = ψ1 = 0and the same ϕ0, ϕ1 as for u; u2 satisfies the Dirichlet boundary conditionswith ϕ0 = ϕ1 = 0 and the same ψ0, ψ1 as for u.

Then

u1(x, y) =∑∞

k=1(Ckekπxb +Dke

− kπxb ) sin kπy

b =∑∞k=1(Ak sinh kπx

b +Bk sinh kπ(a−x)b ) sin kπy

b ,

u2(x, y) =∑∞

k=1(A′k sinh kπy

a +B′k sinh kπ(b−y)

a ) sin kπxa ,

whereAk sinh kπa

b = 2b

∫ b0 ϕ1(y) sin kπy

b dy,

Bk sinh kπab = 2

b

∫ b0 ϕ0(y) sin kπy

b dy,

A′k sinh kπb

a = 2a

∫ a0 ψ1(x) sin kπx

a dx,

B′k sinh kπb

a = 2a

∫ a0 ψ0(x) sin kπx

a dx.

8.7. Hint. If u(ξ, y) =∫e−iξxu(x, y)dx, then −ξ2u+ ∂2u

∂y2= 0 and u(ξ, y) =

C1(ξ) exp(−|ξ|y) + C2(ξ) exp(|ξ|y). The second summand should vanish al-most everywhere if we want u to be bounded. Hence, u(ξ, y) = ϕ(ξ) exp(−|ξ|y)and u(x, y) = 1

2π

∫e−|ξ|y+i(x−z)ξϕ(z)dzdξ. Now, change the order of inte-

gration.

8.8. α ∈ 2Z+ or α− s > −n/2.

8.9. No.

Hint. Use Friedrichs’ inequality.

8.10. Hint. Work with the Fourier transform. Show that the question canbe reduced to the case of a multiplication operator in L2(Rn).

9.1. Physical interpretation: G(x, y) is the potential at x of a unit pointcharge located at the point y ∈ Ω inside a conducting grounded surface ∂Ω.

9.2. The same as for En(x− y), where En is the fundamental solution of ∆.


9.3. Hint. Use the fact that L−1 is self-adjoint.

9.4. Hint. Apply Green’s formula (4.32) with v(x) = G(x, y) (consider y asa parameter) and Ω replaced by Ω\B(y, ε), where B(y, ε) = x : |x−y| ≤ ε.Then pass to the limit as ε→ +0.

9.5. G(x, y) =

1

4πln

(x1 − y1)2 + (x2 − y2)2

(x1 − y1)2 + (x2 + y2)2for n = 2;

1(2−n)σn−1

(1

|x−y|n−2 − 1|x−y|n−2

)for n ≥ 3,

where y = (y1, . . . , yn−1,−yn) for y = (y1, . . . , yn−1, yn).

Hint. Seek the Green function in the form

G(x, y) = En(x− y) + cEn(x− y),

where y 6∈ Ω and c is a constant.

9.6.

u(x) =∫

Rn−1

2xnϕ(y′)dy′

σn−1|x−(y′,0)|n =

2σn−1

∫Rn−1

xnϕ(y1,...,yn−1)dy1...yn−1

[(x1−y1)2+...+(xn−1−yn−1)2+y2n]n2.

Hint. Use the formula in Problem 9.4.

9.7. Let the disc or the ball be x : |x| < R. Then

G(x, y) =

12π ln R|x−y|

|y||x−y| for n = 2,1

(2−n)σn−1|x−y|n−2 − Rn−2

|y|n−2(2−n)σn−1|x−y|n−2 for n ≥ 3,

where y = R2y|y|2 is the point which is obtained from y by the inversion with

respect to the circle (sphere) x : |x| = R.Hint. If n = 2, seek G(x, y) in the form

E2(x− y)− E2(x− y)− c(y).

If n ≥ 3, seek G(x, y) in the form En(x− y)− c(y)En(x− y), where (in both

cases) y = R2y|y|2 . Show geometrically that if |x| = R, then

|x− y||x− y| =

|y|R

does

not depend on x.

9.8.

u(x) =1

σn−1R

∫|y|=R

R2 − |x|2|x− y|n f(y)dSy.

Here the disc (or the ball) is Ω = x : |x| < R, f = u|∂Ω.


Hint. Use the formula from Problem 8.4. When calculating ∂∂ny

G(x, y),use that |y| = R implies

∂∂ny|x− y| = − ∂

∂ny|x− y| = −(∇y|x− y|, y|y|) = −( y−x

|y−x| ,y|y|) = (x,y)−R2

R|x−y| .

9.9. For the half-ball x : |x| ≤ R, xn ≥ 0 the answer is

G(x, y) = G0(x, y)−G0(x, y),

whereG0 is the Green function of the ball x : |x| < R and y = (y1, . . . , yn−1,−yn)for y = (y1, . . . , yn−1, yn).

9.10.G(x, y) = − 1

4π√

(x1−y1)2+(x2−y2)2+(x3−y3)2+

1

4π√

(x1−y1)2+(x2+y2)2+(x3−y3)2+

1

4π√

(x1−y1)2+(x2−y2)2+(x3+y3)2−

1

4π√

(x1−y1)2+(x2+y2)2+(x3+y3)2.

Hint. G(x, y) = E3(x − y) − E3(x − T2y) − E3(x − T3y) + E3(x − T2T3y),where T2(y1, y2, y3) = (y1,−y2, y3), T3(y1, y2, y3) = (y1, y2,−y3),

9.11. Hint. Use the fact that Γ(z) has simple poles for z = 0,−1,−2, . . .

9.12. Hint. Use the Liuoville formula for the Wronskian of two linearlyindependent solutions of the Bessel equation.

9.13. Hint. Expand the left-hand side in power series in t and x, and usethe expansion of Jn(x) given in Problem 8.11.

9.14. Hint. Take t = eiϕ in the formula of Problem 8.13 and use the resultof Problem 8.11.

9.15. For the domain x = (x1, . . . , xn) : 0 < xj < aj , j = 1, . . . , n theeigenfunctions are

∏nj=1

sin kjπxjaj

for kj = 1, 2, . . . and the eigenvalues are

λk1...kn = −[(k1πa1)2 + . . .+ (knπan )2].

9.16. Hint. Use the fact that Jk(αk,nx) for all n = 1, 2, . . . , are eigenfunc-tions of the same operator L = d2

dx2 + 1xddx − k2

x2 (on the space of functionsvanishing at x = 1). This operator is symmetric in the space L2((0, 1), xdx)consisting of the functions on (0, 1) which are square integrable with respectto the measure xdx.

9.17. For the disc x : |x| < R, in the polar coordinates the desired systemis

Jk

(αk,nR

r)eikϕ, for k ∈ Z+, n = 1, 2, . . . ,


where αk,n are defined in Problem 8.16.

Hint. Write ∆ in the polar coordinates and expand the eigenfunctioninto Fourier series u(r, ϕ) =

∑kXk(r)eikϕ. Then each term Xk(r)eikϕ will

be an eigenfunction and Xk(r) satisfies the Bessel equation. Using regularityat 0, show that Xk(r) = cJk(αr).

9.18. Hint. For the cylinder Ω = (x, y, z) : x2 + y2 < R2, 0 < z < Hreduce the equation to the case ∆u = f, u|∂Ω = 0, then expand u and f inthe eigenfunctions of ∆ in Ω which are of the form

eikϕJk

(αk,n

r

R

)sin

mπz

H: k ∈ Z+, n = 1, 2, . . . , m = 1, 2, . . .

.

10.1. u(t, x) = eiωtJ0(ωa ρ), ρ =√x2

1 + x22. (The axis of the cylinder is

x : x1 = x2 = 0).

10.2. See Fig. 6.

u(t, r) =

1 if r ≤ R− at;12− at

2rif r + at > R and −R < r − at < R;

0 if r − at ≥ R.

Hint. Introduce a new function v = ru instead of u. (Here r = |x|.)Thenv(t, r) = 1

2 [ϕ(r+ at) + ϕ(r− at)], where ϕ is the extension of rϕ to R as anodd function.

10.3. See Fig. 7.

u(t, r) =

t if r + at ≤ R,

14ar [R2 − (r − at)2] if r + at > R and |r − at| < R,

0 if |r − at| ≥ R.

Hint. Introduce a new function v = ru with r = |x|. Then

v(t, r) =12a

∫ r+at

r−atψ(s)ds = Ψ(r + at)−Ψ(r − at),

where Ψ(r) = 12a

∫ r0 ψ(s)ds and ψ is the extension of rψ to R as an odd

function.


0

u

0

u

0

u

0

u

Rr= jxj

t= 0

t= "< R =a

t= R =a

t> R =a

Figure 6

10.5. Hint. Consider the Cauchy problem with initial conditions u|t=0 = 0and ut|t=0 = ψ(x). Then

u(t, x) = (2π)−n∫

12ia|ξ|e

i[(x−y)·ξ+at|ξ|]ψ(y)dydξ−(2π)−n

∫1

2ia|ξ|ei[(x−y)·ξ−at|ξ|]ψ(y)dydξ.


0 jxjt= 0

t= "< R =2a

t= R =2a

t= R =2a+ "

t= R =a

t> R =a

Figure 7

Consider, e.g. the first summand. Multiplying the integrand by a C∞-function χ(ξ) that is equal to 1 for |ξ| > 1 and vanishes for |ξ| < 1/2 (thisdoes not affect the singularities of u), try to find a differential operatorL =

∑nj=1 cj(t, x, y, ξ)

∂∂ξj

such that Lei[(x−y)·ξ+at|ξ|] = ei[(x−y)·ξ+at|ξ|].

Prove that such an operator L exists if and only if |x− y|2 6= a2t2 (herex, y, t are considered as parameters) and integrate by parts N times thusmoving L from the exponential to other terms.

10.6. (t, x, y) : t ≥ 0, t2 = x2 + y2/4.Hint. Reduce the operator to the canonical form.

10.7. a) (t, x, y) : 0 ≤ t ≤ min(a2 ,b2), x ∈ [t, a− t], y ∈ [t, b− t]


t

x1

x2

a

b

Figure 8

Figure 9

Figure 10

(See Fig. 8: the lid of a coffin.)

b) (t, x, y) : t ≥ 0, dist((x, y),Π) ≤ t, where Π = [0, a] × [0, b]. (SeeFig. 9: an infinitely high cup of a stadium.)

10.8. a) Let (x, y, z) : x ∈ [0, a], y ∈ [0, b], z ∈ [0, c] be the given paral-lelepiped.

Then the answer is(t, x, y, z) : 0 ≤ t ≤ min

(a

2,b

2,c

2

), t ≤ x ≤ a− t, t ≤ y ≤ b− t, t ≤ z ≤ c− t

The section by the hyperplane t = 1 is a smaller parallelepiped, or a rectan-gle, or an interval, or the empty set (Fig. 10).

b) (t, x, y, z) : dist((x, y, z),Π) ≤ t, where Π is the given parallelepiped.The section by the hyperplane t = 1 is plotted on Fig. 11.

11.1.

u(x) =

− Q

2π lnR if |x| ≤ R,

− Q2π ln r if |x| ≥ R,


Figure 11

where r = |x|, the circle is x : |x| = R, Q is the total charge of the circle,i.e.,Q =

∫|y|=R σ(y)dsy, where σ is the density that defines the potential.

11.2.

u(x) =

− Qr2

4πR2 + Q4π −

Q2π lnR if r ≤ R;

− Q2π ln r if r ≥ R.

Here Q is the total charge of the disc x : |x| ≤ R, r = |x|.

11.3. The same answer as in Problem 10.1 with r =√x2 + y2 in the (x, y, z)-

space, where the z-axis is the axis of the cylinder, R is the radius of itscross-section by the xy-plane.

11.4.

u(x) =

α0, r < R,

0, r > R,

where α0 is the density of the dipole moment, r is the distance to the centerof the circle or to the axis of the cylinder, R is the radius of the circle or ofthe cross-section of the cylinder.

11.5.

u(x) =

Q

(n−2)σn−1Rn−2 if |x| ≤ R,Q

(n−2)σn−1|x|n−2 if |x| > R,

where the sphere of radius R is taken with the center at the origin, Q is thetotal charge of the sphere.

11.6. The induced charge is q = −QRd .

Hint. The potential vanishes on the surface of the sphere, hence insideof the sphere. Calculate the potential in the center assuming that we knowthe distribution of the charge on the surface.


11.7. σ(x) =Q(R2 − d2)4πR|x− y|3 , where y is the point at which the initial charge

Q is situated.

Hint. Let y be the point obtained by reflecting y with respect to thesphere (if the sphere is x : |x| = R, then y = R2y

|y|2 ). Place the charge q at

the point y; then the potential u(x) = q4π|x−y| + Q

4π|x−y| coincides with thepotential of all real charges (Q and induced ones) outside the sphere becauseof the uniqueness of the solution of the Dirichlet problem in the exterior.Then

σ(x) = −∂u(x)∂r

, where∂u

∂r=(∇u, x|x|

).

11.8. The charge will be distributed uniformly over the whole wire.

12.1. a) at± x · η|η| = 0;

b) at± (|x| −R) = 0.

12.2. Rays and characteristics are the same curves in the (t, x)-plane.

Hint. Use the description of rays as solutions of (12.39) satisfying|x(t)| = c(x(t)).

12.3. u(t, x, λ) = iλ−1

2x [exp(iλxe−t + 32 t)− exp(iλxe−t − 3

2 t)].

Bibliography

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1978.

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Index

Ck-boundary, 82

Ck(Ω), 82

Ck,γ , 150

Ck,γ boundary, 150

Ckb (Rn), 179

Hs(Rn), 179

L2(M,dµ), 202

S(Rn), 62

T ∗M , 234

N, positive integers, ix

Z+, non-negative integers, ix

δ-function, 58

δ-like sequence, 58, 72

Hs(Ω), 183

D(Ω) = C∞0 (Ω), 62

E(Ω) = C∞(Ω), 62

Hamiltonian, 40

Cauchy-Dirichlet’s boundary value

problem, 158

singular support of a distribution, 115

acoustic wave, 225

action, 36, 282

action along the path, 234

adjoint Cauchy problem, 170

adjoint operator, 200

asymptotic solution, 288

Bessel equation, 213

bicharacteristic, 278

boundary in Ck, 82

boundary value problem,

Cauchy-Dirichlet’s, 158

boundary value problem,

Cauchy-Neumann’s, 158

boundary value problem, first, 158

boundary value problem, second, 158

boundary, piecewise Ck, 83

boundary, piecewise smooth, 83

bundle, cotangent, 13

bundle, tangent, 13

canonical form of 2nd order operators, 6

canonical form of a 2nd order operator, 10,

11

Cauchy problem, 23, 157, 280

Cauchy problem for the Hamilton-Jacobi

equation, 279

Cauchy sequence, 65, 66

Cauchy-Neumann’s boundary value

problem, 158

causality principle, 247

caustic, 276

characteristic hypersurface, 272

characteristic vector, 7

closed graph theorem, 122

compact operator, 183

completeness of distribution space, 72

configuration space, 233

conservation of momentum, 42

convergent sequences in D(Ω), 68

convolution, 105

convolution of distributions, 110

convolution of two distributions, 132

351

352 Index

cotangent bundle, 13, 234

cotangent vector, 13

countably normed space, 65

Courant’s variational principle, 219

cylindric function of the second kind, 217

cylindric functions, 213

cylindric wave, 231

d’Alembert’s formula, 24

d’Alembert’s principle, 20

d’Alembertian, 2, 229, 234

D’Alemberts’ formula, 23

degrees of freedom, 42

dependence domain, 25

derivative, directional, 13

derivative, fractional, 133

descent method, 247

dielectric permeability, 227

differential of a map, 14

differential operator, linear, 2

differential operator, order of, 2

diffusion equation, 155

dipole, 109

dipole moment, 109

Dirac’s δ-function, 58

Dirac’s measure, 58

direct product of distributions, 107, 109

direct product of functions, 106

directional derivative, 13

Dirichlet kernel, 73

Dirichlet’s integral, 187

Dirichlet’s problem, 140, 158

dispersion law, 231

distribution, 61

distribution with compact support, 68

distribution, homogeneous, 95

distribution, regular, 68

distribution, tempered, 68

distributions, 68

divergence formula, 83

domain of the operator, 199

domain, dependence, 25

domain, influence, 26

Doppler’s effect, 44

double layer, 109

double layer potential, 113, 255

dual space, 66

Duhamel’s formula, 168

eigenfrequencies, 32

eigenfunctions, 31

eigenvalues, 31

eikonal equation, 287

energy, 40, 234

energy conservation law, 21, 39, 40, 235

energy, full, 40

energy, kinetic, 39, 233

energy, kinetic along the path, 233

energy, potential, 39, 40, 234

equation, diffusion, 155

equation, Hamilton-Jacobi, 276

equation, parabolic, 155

equations, Lagrange’s, 20, 39

equations, Maxwell, 227

equations, transport, 290, 295

equicontinuity, 184

equivalent systems of seminorms, 63

Euler-Lagrange equation, 37

Euler-Lagrange equations, 38

extremal, 37

extremal point, 37

Fejer kernel, 73

Fermat’s principle, 293

Fermi coordinates, 261

first boundary value problem, 27, 158

forced oscillations, 45

formula, d’Alembert’s, 24

formula, Duhamel’s, 168

formula, Kirchhoff, 243

formula, Poisson’s, 249

formulas, Sokhotsky, 75

Fourier transform, 106

Fourier transformation, 106

Fourier’s law, 156

Frechet space, 66

fractional derivative, 133

fractional integral, 133

freedom, degrees of, 42

Friedrichs extension of an operator, 205,

206

Friedrichs inequality, 188

function, Bessel of the 1st kind, 215

function, cylindric of the 1st kind, 215

function, cylindric of the second kind, 217

function, Green, 54

function, harmonic, 121

function, Lagrange, 234

function, Neumann’s, 217

function, subharmonic, 153

functional, 36

functions smooth up to GGG from each side,

256

Index 353

fundamental solution of Laplace operator,

80

gauge transformation, 228

gauge, Lorentz, 229

Gauss’s law of the arithmetic mean, 135

Gauss-Ostrogradsky formula, 83

Gaussian integrals, 86

generalized function, 61

generalized solution, 12

generalized solution of Dirichlet’s problem,

192

generalized solutions of the Cauchy

problem, 24

Glazman’s lemma, 218

gradient, 14

graph of the operator, 200

Green function, 54

Green function of Laplacian, 142

Holder function, 150

Holder’s inequality, 98

Hadamard’s example, 144

Hamilton-Jacobi equation, 276

Hamiltonian, 234, 277

Hamiltonian curves, 277

Hamiltonian system, 235, 277

Hamiltonian vector field, 277

harmonic function, 121, 135

Hausdorff property, 64

heat conductivity, 156

heat equation, 155

heat operator, 2

Heaviside function, ix

Helmholtz operator, 103

high-frequency asymptotics, 52

Holmgren’s principle, 170

homogeneous distribution, 95

homothety, 94

Hook’s law, 305

Hooke’s law, 18

Huygens principle, 243

hyperbolicity of an operator, 284

hypoelliptic operator, 119, 121

identity, Parseval, 312

inequality, Friedrichs, 188

inequality, Holder’s, 98

influence domain, 26

initial value problem, 23

integral, Dirichlet, 187

integral, fractional, 133

integral, Poisson’s, 162

inverse operator, 200

Jacobi matrix, 14

Kelvin transformation, 131

kernel, Dirichlet, 73

kernel, Fejer, 73

kinetic energy, 39, 233

kinetic energy along the path, 233

Kirchhoff formula, 243

L. Schwartz’s space, 62

Lagrange function, 234

Lagrange’s equations, 20, 39

Lagrangian, 36, 234

Lagrangian corresponding to the

Hamiltonian, 282

Lagrangian manifold, 279

Laplace operator, 2

Laplacian, 2

law of dispersion, 231

law, conservation of energy, 235

law, Fourier, 156

law, Hook’s, 305

linear operator in Hilbert space, 199

Liouville’s theorem for harmonic functions,

130

logarithmic potential, 113

Lorentz gauge, 229

map, tangent , 14

maximum principle, 163, 174

maximum principle for harmonic

functions, 138

Maxwell equations, 227

mean value theorem for harmonic

functions, 135

method of descent, 247

mixed problem, 27

momentum, 39, 41

momentum, conservation of, 42

momentum, total, 41

monochromatic wave, 232

multi-index, 1

Neumann’s function, 217

Neumann’s problem, 158

Newton’s Second Law, 39

Newton’s Third Law, 41

norm, 62

354 Index

operator semibounded below, 205

operator, adjoint, 93, 200

operator, closed, 200

operator, compact, 183

operator, dual, 93

operator, elliptic, 7, 11

operator, extension of, 200

operator, heat, 2

operator, Helmholtz, 103

operator, hyperbolic, 8, 9

operator, hypoelliptic, 119, 121

operator, inverse, 200

operator, Laplace, 2

operator, of translation, 94

operator, one-dimensional Schrodinger, 49

operator, parabolic, 10

operator, Schrodinger, 49

operator, self-adjoint, 200

operator, Sturm-Liouville, 2, 49

operator, symmetric, 200

operator, transposed, 93

operator, wave, 2, 229

order of a distribution, 89

order of differential operator, 2

Parseval identity, 312

partition of unity, 75

path, 233

phase, 286

piecewise Ck boundary, 83

piecewise smooth boundary, 83

plane wave, 230

Poisson’s equation, 87

Poisson’s formula, 249

Poisson’s integral, 162

potential, 113

potential energy, 39, 40, 234

potential of a uniformly charged sphere,

266

potential of the double layer of the sphere,

267

potential, double layer, 255

potential, retarded, 246

potential, single layer, 255

precompact, 183

principle of maximum, 174

principle, causality, 247

principle, d’Alembert’s, 20

principle, Holmgren’s, 170

principle, Huygens, 243

principle, maximum, 163

problem, adjoint Cauchy, 170

problem, Cauchy, 23

problem, first boundary value, 27

problem, initial value, 23

problem, mixed, 27

problem, Sturm-Liouville, 31, 48, 50

rapidly oscillating solution, 288

ray, 275, 279

rays, 286

real analytic function, 117

regular distribution, 68

removable singularity theorem, 121

renormalization, 88

retarded potential, 246

Schrodinger operator, 49

Schrodinger operator, one-dimensional, 49

Schwartz’ kernel, 54

seminorm, 62

seminorms, equivalent systems of, 63

sheaf, 75

short-wave asymptotics, 52

simple layer potential, 113

single layer, 109

single layer potential, 255

Sobolev space, 177

Sobolev’s embedding theorem, 179

Sobolev’s theorem on restrictions, 181

Sokhotsky’s formulas, 75

solution asymptotic, 288

solution rapidly oscillating, 288

solution, generalized, 12

space, countably normed, 65

space, dual, 66

space, Frechet, 66

space, Sobolev, 177

specific heat, 156

spectral theorem, 203

speed of light, 227

spherical waves, 231

standing waves, 30

string, unbounded, 23

Sturm-Liouville operator, 2, 49

Sturm-Liouville problem, 31, 48, 50

subharmonic function, 153

superposition, 12, 32

support, 62

support of the distribution, 76

surface of jumps, 272

surface, characteristic, 7

surface, characteristic at a point, 7

Index 355

symbol, 2

symbol, principal, 3

symbol, total, 2

tangent bundle, 13, 233

tangent map, 14

tangent space, 233

tangent vector, 12

tangent vectors, 233

tempered distribution, 68

tensor product of distributions, 107, 109

tensor product of functions, 106

theorem, Liouville, 130

theorem, mean value, for harmonic

functions, 135

theorem, Sobolev’s on embedding, 179

theorem, Sobolev’s on restrictions, 181

theorem, spectral, 203

theorem, Tikhonov, 171

Tikhonov’s theorem, 171

transformation, Kelvin, 131

transport equations, 290, 295

uniform boundedness, 184

unitarily equivalent operators, 203

variation of a functional, 37

variational derivative, 37

variational derivative, partial, 38

variational principle, Courant’s, 219

vector, characteristic, 7

vector, cotangent, 13

vector, tangent, 12

vector, wave, 231

velocity, 233

volume potential, 254

wave fronts, 286

wave equation, the nonresonance case, 35

wave equation, the resonance case, 35

wave front, 275

wave fronts, 280

wave operator, 2, 229

wave vector, 231

wave, cylindric, 231

wave, monochromatic, 232

waves standing, 30

waves, spherical, 231

weak topology, 72

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