sandro salsa gianmaria verzini partial differential...

123

Complements and Exercises

Sandro SalsaGianmaria Verzini

Partial Differential Equations in Action

UN

ITEX

TU

NIT

EXT

UNITEXT – La Matematica per il 3+2

Volume 87

More information about this series athttp://www.springer.com/series/5418

Sandro Salsa � Gianmaria Verzini

Partial DifferentialEquations in ActionComplements and Exercises

Sandro Salsa Gianmaria VerziniDipartimento di Matematica Dipartimento di MatematicaPolitecnico di Milano Politecnico di MilanoMilano, Italy Milano, Italy

Translated by Simon G. Chiossi, UFBA – Universidade Federal da Bahia, Salvador (Brazil).

Translation from the Italian language edition: Equazioni a derivate parziali. Complementi ed esercizi,Sandro Salsa e Gianmaria Verzini, © Springer-Verlag Italia, Milano 2005. All rights reserved.

UNITEXT – La Matematica per il 3+2

ISBN 978-3-319-15415-2 ISBN 978-3-319-15416-9 (eBook)DOI 10.1007/978-3-319-15416-9Springer Cham Heidelberg New York Dordrecht London

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Preface

This book is designed for advanced undergraduate students from various disciplines, in-cluding applied mathematics, physics, and engineering. It evolved during the PDE coursesthat both authors have taught during recent decades at the Politecnico di Milano, and con-sists of problems of various types and difficulties.

In the first part of the book, while much emphasis is placed on the most common meth-ods of resolution, such as separation of variables or the method of characteristics, we alsoinvite the student to handle the basic theoretical tools and properties of the solutions to thefundamental equations of mathematical physics.

The second part is slightly more advanced and requires basic tools from functionalanalysis. A small number of exercises aims to familiarize the student with the first ele-ments of the theory of distributions and of the Hilbertian Sobolev spaces. The focus thenswitches to the variational formulation of the most common boundary value problems foruniformly elliptic equations. A substantial number of problems is devoted to the use of theRiesz representation and the Lax-Milgram theorems together with Fredholm alternativeto analyse well posedness or solvability of those problems. Next, a number of problemsaddresses the analysis of weak solutions to initial-boundary value problems for the heator the wave equation.

The text is completed by two short appendixes, the first dealing with Sturm-Liouvilleproblems and Bessel functions and the second listing frequently used formulas.

Each chapter begins with a brief review of the main theoretical concepts and tools thatconstitute necessary prerequisites for a proper understanding. The text Partial DifferentialEquation in Action [18], by S. Salsa, is the natural theoretical reference.

Within each chapter, the problems are divided into two sections. In the first one wepresent detailed solutions and comments to provide the student with a reasonably com-plete guide. In the second section, we propose a set of problems that each student shouldtry to solve by him- or herself. In each case, a solution can be found at the end of the chap-ter. Some problems are proposed as theoretical complements and may prove particularlychallenging; this is especially true of those marked with one or two asterisks.

Milano, January 2015 Sandro SalsaGianmaria Verzini

Contents

1 Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Backgrounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2.1 The method of separation of variables . . . . . . . . . . . . . . . . . . . . . . . . 31.2.2 Use of the maximum principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.2.3 Applying the notion of fundamental solution . . . . . . . . . . . . . . . . . . 251.2.4 Use of Fourier and Laplace transforms . . . . . . . . . . . . . . . . . . . . . . . 371.2.5 Problems in dimension higher than one . . . . . . . . . . . . . . . . . . . . . . . 43

1.3 Further Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501.3.1 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

2 The Laplace Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 812.1 Backgrounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 812.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

2.2.1 General properties of harmonic functions . . . . . . . . . . . . . . . . . . . . . 842.2.2 Boundary-value problems. Solution methods . . . . . . . . . . . . . . . . . . 952.2.3 Potentials and Green functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117


3 First Order Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1493.1 Backgrounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1493.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152

3.2.1 Conservation laws and applications . . . . . . . . . . . . . . . . . . . . . . . . . . 1523.2.2 Characteristics for linear and quasilinear equations . . . . . . . . . . . . . 181


4 Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2154.1 Backgrounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2154.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217

viii Contents

4.2.1 One-dimensional waves and vibrations . . . . . . . . . . . . . . . . . . . . . . . 2174.2.2 Canonical forms. Cauchy and Goursat problems . . . . . . . . . . . . . . . 2384.2.3 Higher-dimensional problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247


5 Functional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2735.1 Backgrounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2735.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278

5.2.1 Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2785.2.2 Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2915.2.3 Sobolev spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298


6 Variational Formulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3336.1 Backgrounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3336.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336

6.2.1 One-dimensional problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3366.2.2 Elliptic problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3466.2.3 Evolution problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366


Appendix A. Sturm-Liouville, Legendre and Bessel Equations . . . . . . . . . . . . . . . . 405A.1 Sturm-Liouville Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405

A.1.1 Regular equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405A.1.2 Legendre’s equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406

A.2 Bessel’s Equation and Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407A.2.1 Bessel functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407A.2.2 Bessel’s equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410

Appendix B. Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413B.1 Gradient, Divergence, Curl, Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413B.2 Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415B.3 Fourier Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416B.4 Laplace Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419

1

Diffusion

1.1 Backgrounds

We shall recall a few notions and results that crop up frequently concerning the diffusionequation

ut �D�u D f

defined on a cylindrical domainQT D �� .0; T /, where� is a domain (connected, opensubset) of Rn. Here u D u.x; t /, and the Laplacian � is taken with respect to the spatialvariables x only.

• Parabolic boundary. The union of the base of QT (i.e. � � ¹0º) and the lateral sur-face ST D @� � Œ0; T � is the parabolic boundary of QT , which we denote by @pQT . Inwell-posed problems for the diffusion equation this is where one must assign the data.

• Maximum principles. Let� be bounded andw 2 C 2;1 .QT /\C�QT

�a sub-solution

(or super-solution), that is

wt �D�w D q � 0 (resp. � 0) in QT :

Then w reaches its maximum value (resp. minimum) on the parabolic boundary @pQTof QT :

maxQT

w D max@pQT

w

(weak maximum principle). In particular, if w is negative (resp. positive) on @pQT , thenit is negative (resp. positive) over all ofQT . If, further,w.x0; t0/ D maxQT

with x0 2 �;then w is constant on � � Œ0; t0� (strong maximum principle).

• Fundamental solution and global Cauchy problem. The function

�D .x; t / D 1

.4�Dt/n=2e�jxj

2=.4Dt/; t > 0;

is called fundamental solution to the diffusion equation; when t > 0 it solves

© Springer International Publishing Switzerland 2015S. Salsa, G. Verzini, Partial Differential Equations in Action. Complements and Exercises,UNITEXT – La Matematica per il 3+2 87, DOI 10.1007/978-3-319-15416-9_1

2 1 Diffusion

ut �D�u D 0 and is the unique function satisfying

limt#0

�D .x; t / D ın .x/ ;Z

Rn

�D .x; t / dx D 1 for any t > 0;

where ın .x/ denotes the n-dimensional Dirac’s delta function.The fundamental solution enables to construct the general solution to the global Cauchy

problem ´ut �D�u D f .x; t / in Rn� .0;1/

u .x; 0/ D g .x/ in Rn;

by means of the formula

u .x; t / DZ

Rn

�D .x � y; t / g .y/ dy CZ t

0

ZRn

� .x � y; t � s/ f .y; s/ dyds:

The latter holds, for example, when jg .x/j � ceAjxj, f is bounded and f , ft , fxj, fxixj

are continuous on Rn � .0;C1/. At a point x0 of continuity of g we have

u .x; t / ! g .x0/ as .y; t / ! .x0; 0/ , t > 0:

• Random walk and fundamental solution (n D 1). Let us consider a particle of unitmass moving along the x-axis as follows.

i) In a time interval � the particle advances by h, starting from x D 0.

ii) It moves leftwards or rightwards with probability p D 1=2, each time independentlyof the previous action.

At time t D N� , i.e. after N iterations, the particle will have reached point x D mh,where N is a natural number andm an integer. The probability p .x; t/ that it will be in xat time t is the solution of the discrete problem

p .x; t C �/ D 1

2p .x � h; t/C 1

2p .x C h; t/ ; (1.1)

with initial conditions

p .0; 0/ D 1 and p .x; 0/ D 0 if x ¤ 0:

Passing to the limit in (1.1) for h; � ! 0, whilst keeping h2=� D 2D D constant andinterpreting p as a probability density, gives the equation pt D Dpxx , and the initialconditions now read

limt#0

p .x; t/ D ı .ı1 D ı/:

We have already noticed that the unique solution with unit mass is the fundamental solu-tion to the diffusion equation:

p .x; t/ D �D .x; t/ :

1.2 Solved Problems 3

1.2 Solved Problems

� 1:2:1 � 1:2:7 W The method of separation of variables.� 1:2:8 � 1:2:11 W Use of the maximum principle.� 1:2:12 � 1:2:18 W Applying the notion of fundamental solution.� 1:2:19 � 1:2:22 W Use of Fourier and Laplace transforms.� 1:2:23 � 1:2:26 W Problems in dimension higher than one.

1.2.1 The method of separation of variables

Problem 1.2.1 (Cauchy-Dirichlet). LetD > 0 be a constant and g 2 C 1 .Œ0; ��/, withg .0/ D g .�/ D 0. Solve, by separating the variables, the problem:8<

:ut .x; t/ �Duxx.x; t/ D 0 0 < x < �; t > 0

u.x; 0/ D g.x/ 0 � x � �

u.0; t/ D u.�; t/ D 0 t > 0:

Discuss uniqueness and continuous dependence on the initial data.

Solution. We start with two preliminary observations. First of all, the choice of Œ0; ��as domain for the space variable is just to keep the formulas simpler. In case the spacevariable x varies between 0 and L > 0 we can use Fourier series on suitable intervals, orreduce to Œ0; �� by the change y D x�=L, v .y; t/ D u .Ly=�; t/, which would give thefollowing problem for v:8<

:vt � D�2

L2 vyy D 0 0 < y < �; t > 0

v.y; 0/ D g.Ly=�/ 0 � y � �

v.0; t/ D v.�; t/ D 0 t > 0:

Notice that the boundary condition is of Dirichlet type and homogeneous. The first step ofthe method consists in seeking non-zero solutions of the form

u.x; t/ D v.x/w.t/:

Substituting into the equation gives

v.x/w0.t/ �Dv00.x/w.t/ D 0:

Dividing by v.x/w.t/ and rearranging terms we find:

1

D

w0.t/w.t/

D v00.x/v.x/

: (1.2)

This is an identity between two members depending on different variables. Consequentlythey must both be equal to some constant � 2 R. Thus we can split (1.2) into the two

4 1 Diffusion

equationsw0.t/ � �Dw.t/ D 0;

solved byw.t/ D Ce�Dt ; C 2 R, (1.3)

andv00.x/ � �v.x/ D 0: (1.4)

The Dirichlet conditions force v.0/w.t/ D v.�/w.t/ D 0 for any t > 0, i.e.

v.0/ D v.�/ D 0: (1.5)

The boundary-value problem (1.4), (1.5) has non-trivial solutions only for special valuesof �, called eigenvalues. The corresponding solutions are said eigenfunctions. We distin-guish three cases.

Case � D �2 > 0. The general integral of (1.4) is

v.x/ D C1e�x C C2e

��x :

By imposing the boundary conditions we find´C1 C C2 D 0

e�� C1 C e�� C2 D 0;

so C1 D C2 D 0. This gives the zero solution only.

Case � D 0. This situations is essentially the same as the above one. From

v.x/ D C1 C C2x;

the Dirichlet constraints immediately force C1 D C2 D 0.

Case � D ��2 < 0. Now we have

v.x/ D C1 cosx C C2 sinx; v.0/ D v.�/ D 0:

From v .0/ D 0 we deduce C1 D 0; from v .�/ D 0 we get

C2 sin� D 0 H) D k positive integer and C2 arbitrary.

Then, the eigenvalues are �k D �k2 and the eigenfunctions vk .x/ D sin kx: Recalling(1.3), we have the infinitely-many solutions

'k .x; t/ D Ce�k2Dt sin kx, k D 1; 2; : : : ,


fulfilling 'k.0/ D 'k.�/ D 0. None of these functions satisfies the condition u .x; 0/ Dg .x/, except when g .x/ D C sinmx and m is an integer. The idea is then to exploit theproblem linearity by assembling the vk into a linear combination, and trying to determinethe coefficients so to satisfy the initial condition. Then our candidate solution has the form:

u .x; t/ D1XkD1

cke�k2Dt sin kx;

and we seek the constants ck by imposing

u .x; 0/ D1XkD1

ck sin kx D g .x/ : (1.6)

Notice that u .x; 0/ is a sines-Fourier series; therefore we extend g on Œ��; �� as an oddfunction and expand it in a sines-Fourier series:

g .x/ D1XkD1

gk sin kx; gk D 2

�

Z �

0

g .x/ sin kx dx:

By comparison with (1.6) we have ck D gk , and thus

u .x; t/ D1XkD1

gke�k2Dt sin kx (1.7)

is the (formal) solution.

• Analysis of (1.7). The function g is C 1 .Œ0; ��/ and vanishes at the endpoints, so itsodd prolongation on Œ��; �� is C 1 .Œ��; ��/. The theory of Fourier series guarantees that1PkD1

jgkj converges. Since ˇgke

�k2Dtˇ

� jgkj ;the function (1.6) converges uniformly on the entire strip Œ0; �� Œ0;1/ and we mayswap the sum with the limit. This ensures that (1.7) is continuous on Œ0; �� Œ0;1/. Onthe other hand if t � t0 > 0 the fast convergence rate of the exponential as k ! 1 allowsto differentiate term-wise (to any order), and in particular

ut �Duxx D1XkD1

gk Œ.uk/t �D .uk/xx� D 0;

so (1.7) solves the differential equation inside the strip.

• Uniqueness and continuous dependence on initial data. The uniqueness of a solu-tion, continuous on Œ0; �� Œ0;1/, and the fact it depends in a continuous manner uponthe initial data both follow from the maximum principle: indeed, if ug is a solution corre-

6 1 Diffusion

sponding to the datum g, we have

maxŒ0;��Œ0;1/

ˇug1

� ug2

ˇ � maxŒ0;��

jg1 � g2j .

Problem 1.2.2 (Cauchy-Neumann). LetD > 0 be a constant and g 2 C 1 .Œ0; ��/ suchthat g0.0/ D g0.�/ D 0. Solve by separation of variables:8<

:ut .x; t/ �Duxx.x; t/ D 0 0 < x < �; t > 0

u.x; 0/ D g.x/ 0 � x � �

ux.0; t/ D 0; ux.�; t/ D 0 t > 0:

Discuss uniqueness and continuous dependence on the initial data.

Solution. Since the Neumann conditions are homogeneous, we proceed by seekingnon-zero solutions of the form

u.x; t/ D v.x/w.t/:

As in the previous problem, we obtain for w the equation

w0.t/ � �Dw.t/ D 0;

with general solutionw.t/ D Ce�Dt ; C 2 R. (1.8)

For v we have the eigenvalue problem´v00.x/ � �v.x/ D 0

v0.0/ D v0.�/ D 0;

where � is a real number. As usual, we must distinguish three cases.

Case � D �2 > 0. The general integral reads


��x :

The Neumann conditions impose v0.0/ D v0.�/ D 0, so´C1 � C2 D 0

e�� C1 � e�� C2 D 0;

and then C1 D C2 D 0 because .e�� C e��/ ¤ 0: The only solution is trivial.

Case � D 0. Fromv.x/ D C1 C C2x;


and the Neumann conditions, we deduce immediately C2 D 0 and C1 arbitrary. Now theeigenfunctions are constant functions.

Case � D ��2 < 0. We have

v.x/ D C1 cosx C C2 sinx; v0.0/ D v0.�/ D 0:

Sincev0.x/ D �C1 sinx C C2 cosx;

from v0 .0/ D 0 we deduce C2 D 0I from v0 .�/ D 0 we infer

C1 sin� D 0 H) D k 2 N, C2 arbitrary:

The eigenvalues are then �k D �k2 and the eigenfunctions vk .x/ D cos kx:Recalling (1.8), we have infinitely-many solutions

'k .x; t/ D Ce�k2Dt cos kx; k 2 N

satisfying '0k.0/ D '0

k.�/ D 0. None fulfils u .x; 0/ D g .x/ except when g .x/ D

C cosmx, m integer. So let us set

u .x; t/ D1XkD0

cke�k2Dt cos kx

as candidate solution (in particular, for k D 0we also obtain the constant solutions of case� D 0). The coefficients ck must be chosen so that

u .x; 0/ D1XkD0

ck cos kx D g.x/: (1.9)

Since u .x; 0/ is a cosines-Fourier series, we extend g as even function on Œ��; �� andexpand it in cosines-Fourier series:

g .x/ D g0

2C

1XkD1

gk cos kx, gk D 2

�

Z �

0

g .x/ cos kx dx:

Notice that g0=2 is the mean value of the datum g on the interval Œ0; ��. After comparisonwith (1.9) we must have c0 D g0=2, ck D gk , giving the (formal) solution

u .x; t/ D g0

2C

1XkD1

gkuk .x; t/ D g0

2C

1XkD1

gke�k2Dt cos kx: (1.10)

• Analysis of (1.10). The function g belongs to C 1 .Œ0; ��/ and has null derivative atthe endpoints, whence its even prolongation on Œ��; �� is in C 1 .Œ��; ��/ : The theory of

8 1 Diffusion

Fourier series guarantees that1PkD1

jgkj converges. As

ˇgke

�k2Dtˇ

� jgkj ;

the function (1.10) converges uniformly on the strip Œ0; �� Œ0;1/ and we may swap thelimit and the sum. This makes sure that (1.10) is continuous on Œ0; �� Œ0;1/. Now letus check the Neumann conditions on the boundary. Fix t0 > 0I for t close to t0 we candifferentiate term by term, so

ux .x; t/ D �1XkD1

kgke�k2Dt sin kx: (1.11)

Since1 ˇkgke

�k2Dtˇ

� 1p2eDt

jgkj ;the series (1.11) converges uniformly on Œ0; �� Œt0;1/ for any t0 > 0. In particular

lim.x;t/!.0;t0/


kgk lim.x;t/!.0;t0/

Œe�k2Dt sin kx� D 0

lim.x;t/!.�;t0/


kgk lim.x;t/!.�;t0/

Œe�k2Dt sin kx� D 0:

The function is therefore C 1 on any strip Œ0; �� Œt0;1/. Similar computations showthat if t � t0 > 0 the fast convergence to zero of the exponential as k ! 1 allows todifferentiate (to any order) each term separately. In particular

ut �Duxx D1XkD1

gk Œ.uk/t �D .uk/xx� D 0;

so (1.10) is indeed a solution on the strip Œ0; �� .0;1/:

• Uniqueness and continuous dependence on the data. We use an energy method. Sup-pose there exist two solutions u and v of the same problem, defined on Œ0; �� Œ0;1/ andC 1 on Œ0; �� .0;1/. Set w D u � v and

E .t/ DZ �

0

w2 .x; t/ dx:

Then E .t/ � 0; E .0/ D limt#0E .t/ D 0, and for t > 0 also

E 0 .t/ D 2

Z �

0

wwt dx D 2D

Z �

0

wwxx dx:

1 Maximise the function f .x/ D xe�x2Dt .


We integrate by parts and recall that wx vanishes at the endpoints:

E 0 .t/ D �2DZ �

0

.wx/2 dx � 0:

Consequently E decreases and therefore E D 0 for any t � 0. As w is continuous,w .x; t/ � 0.

By Bessel’s equality, moreover,

supt>0

ku .�; t /k2L2.0;�/ D supt>0

Z �

0

u2 .x; t/ dx � �

1XkD0

jgkj2 D � kgk2L2.0;�/ ;

showing that the solution depends continuously (in L2) on the initial datum.

Problem 1.2.3 (Stationary state and asymptotic behaviour). Let u be a solution to:8<:ut .x; t/ D Duxx.x; t/ 0 < x < L; t > 0

u.x; 0/ D g.x/ 0 � x � L

ux.0; t/ D ux.L; t/ D 0 t > 0:

a) Interpret the problem supposing u is the concentration (mass per unit of length) ofa substance under the diffusion. Explain heuristically why

u.x; t/ ! U (constant) as t ! C1:

Find the value of U by integrating the equation suitably.

b) Assume g 2 C .Œ0; L�/, u continuous on Œ0; L�� Œ0;1/ and C 1 on Œ0; L�� Œt0;1/

for any t0 > 0. Show that u.x; t/ ! U as t ! C1 in L2.0; L/, i.e.Z L

0

.u.x; t/ � U/2dx ! 0 as t ! 1.

c) Let g 2 C 1 .Œ0; L�/ with g0.0/ D g0.�/ D 0. Using the formula for u of Prob-lem 1.2.2, show that u.x; t/ ! U uniformly on Œ0; L� as t ! C1.

Solution. a) If the equation governs the (one-dimensional) diffusion of a concentratedsubstance then the Neumann boundary conditions tell us that the flow across the endpointsis zero. Thus it is natural to expect that the total mass is preserved, and that the substancewill tend to distribute uniformly, eventually reaching a state of constant density. At thesame time, neglecting the initial condition, the only stationary solutions (independent oft ) of the problem are exactly the constant solutions. Integrating the equation in x overŒ0; L� and using the Neumann conditions givesZ L

0

ut .x; t/ dx DZ L

0

Duxx.x; t/ dx D Dux.L; t/ �Dux.0; t/ D 0;

10 1 Diffusion

whenced

dt

Z L

0

u.x; t/ dx D 0; (1.12)

which is precisely the conservation of the mass. As u is continuous on Œ0; L�� Œ0;1/, wehave Z L

0

u.x; t/ dx !Z L

0

g.x/ dx for t ! 0:

Therefore Z L

0

u.x; t/ dx �Z L

0

g.x/ dx:

Then, if u.x; t/ ! U as t ! C1, it must be

U D 1

L

Z L

0

g.x/ dx: (1.13)

b) We shall prove that, indeed, u converges inL2.0; L/ to the constant U just defined.Set w.x; t/ D u.x; t/ � U . Clearly w.x; 0/ D g .x/ � U , and by (1.13) the function whas zero spatial mean:

R L0w.x; t/ dx � 0, for any t � 0. As in Problem 1.2.2, if

E .t/ DZ L

0

w2.x; t/ dx

it follows that

E 0 .t/ D �2DZ L

0

w2x.x; t/ dx:

At the very end we shall prove that2Z L

0

w2x.x; t/ dx � E .t/

L: (1.14)

So we have an (ordinary) differential inequality:

E 0.t/ � �2DLE.t/;

that is,d

dtlogE .t/ � �2D

L:

Integrating from 0 to t

logE .t/ � logE .0/ � �2DLt

and thusE.t/ � E.0/e� 2D

L t :

2 Poincaré’s inequality.


Therefore as t ! C1, E.t/ tends to 0, that isZ L

0

.u.x; t/ � U/2dx ! 0 as t ! 1;

i.e. u.�; t / tends to U in L2-norm as t ! 1.

c) Take g 2 C 1 .Œ0; L�/ with g0.0/ D g0.L/ D 0. We use the explicit expression ofthe solution obtained in Problem 1.2.2. Passing from Œ0; �� to Œ0; L� we find

u.x; t/ D a0

2C

1XnD1

ane�D�2

nt cos�nx;

where �n D n�=L, an D 2L

R L0g.x/ cos�nx dx. Now if n � 1, t > 0,ˇ

ane�D�2

nt cos�nxˇ

� e�D�21t janj

and since g 2 C 1 .Œ0; L�/, g0.0/ D g0.L/ D 0;

1XnD1

janj D S < 1:

Henceˇˇ1XnD1

ane�D�2

nt cos�nx

ˇˇ � e�D�

21t

1XnD1

janj D e�D�21tS ! 0 as t ! C1:

Therefore, as t ! C1, u.x; t/ tends (exponentially) to a0=2 D U , uniformly in x.

• Proof of (1.14). By the intermediate value theorem, for any t > 0 one can find x .t/such that

w.x .t/ ; t/ D 1

L

Z L

0

w.x; t/ dx D 0:

The fundamental theorem of calculus then implies, for any x 2 Œ0; L�:

w .x; t/ DZ x

x.t/

wx.s; t/ ds

and by Schwarz’s inequality:

jw .x; t/j DˇZ x

x.t/

wx.s; t/ ds

ˇ�Z L

0

jwx.s; t/j ds � pL

Z L

0

w2x.s; t/ ds

!1=2:

12 1 Diffusion

Squaring both sides and integrating over Œ0; L� givesZ L

0

w2.s; t/ ds � L

Z L

0

w2x.s; t/ ds:

Problem 1.2.4 (Cauchy-Neumann; non-homogeneous equation). Solve, using separa-tion of variables, the problem:8<

:ut .x; t/ � uxx.x; t/ D tx 0 < x < �; t > 0

u.x; 0/ D 1 0 � x � �

ux.0; t/ D ux.�; t/ D 0 t > 0:

Solution. This is a non-homogeneous Neumann problem with homogeneous boundaryconditions. In order to be able to separate the variables it is convenient to consider firstthe homogeneous equation, and in particular the associated eigenvalue problem:´

v00.x/ � �v.x/ D 0

v0.0/ D v0.�/ D 0:

In Problem 1.2.2 we found the eigenvalues �k D �k2 and the eigenfunctions vk .x/ Dcos kx. Let us write the candidate solution as

u .x; t/ D1XkD0

ck .t/ cos kx

and impose (recall that v00k

D �k2vk):

ut � uxx D1XkD0

�c0k .t/C k2ck .t/

�cos kx D tx

with

u .x; 0/ D1XkD0

ck .0/ cos kx D 1:

We expand f .x/ D x in cosines-Fourier series:

x D �

2� 4

�

1XkD0

cos Œ.2k C 1/ x�

.2k C 1/2;

uniformly convergent on Œ0; �� :Comparing the last three equations, the coefficients ck .t/must solve the following Cauchy problems:

c00 .t/ D �2t; c0 .0/ D 1I

c02k .t/C 4k2c2k .t/ D 0; c2k .0/ D 0; k � 1Ic02kC1 .t/C .2k C 1/2c2kC1 .t/ D � 4

�1

.2kC1/2 t; c2kC1 .0/ D 0; k � 0:


0

�=2

� 00:5

11:5

22:5

5

10

1

xt

uDu.x;t/

Fig. 1.1 Solution to Problem 1.2.4 for 0 < x < � , 0 < t < 2:5

Solving the ODE problem, we find

c0 .t/ D �4t2 C 1I

c2k .t/ D 0; k � 1Ic2kC1 .t/ D � 4

�.2kC1/4ht C 1

.2kC1/2�e�.2kC1/2t � 1�i ; k � 0:

Hence the (formal) solution reads (Fig. 1.1):

u .x; t/ D �

4t2 C 1C

1XkD0

c2kC1 .t/ cos Œ.2k C 1/ x� : (1.15)

• Analysis of (1.15). Asˇ�e�.2kC1/

2t � 1�

cos .2k C 1/ xˇ

� 2

we deduce that the series

1XkD0

e�.2kC1/2t � 1.2k C 1/6

cos .2k C 1/ x;

the series of first and second partial derivatives in x, given by

�1XkD0

e�.2kC1/2t � 1.2k C 1/5

sin .2k C 1/ x; �1XkD0

e�.2kC1/2t � 1.2k C 1/4

cos .2k C 1/ x;

14 1 Diffusion

and the series of derivatives in t , given by

�1XkD0

e�.2kC1/2t

.2k C 1/4cos .2k C 1/ x;

all converge uniformly on Œ0; �� Œ0;1/: Hence the derivatives can be carried into thesum, making u a C 2 function on Œ0; �� Œ0;1/.

In particular u solves the diffusion equation on .0; �/�.0;1/ and it assumes the givenboundary values in pointwise sense. Within this class of functions, it can be proved thatthe solution is unique and it depends continuously on the data using the energy method(Problem 1.2.2 on page 6).

Problem 1.2.5 (Non-homogeneous Neumann). Solve by separation of variables thefollowing problem:8<

:ut .x; t/ � uxx.x; t/ D 0 0 < x < �; t > 0

u.x; 0/ D 0 0 � x � �

ux.0; t/ D 0; ux.�; t/ D U t > 0:

If U ¤ 0, can there be a stationary solution u1 D u1 .x/‹

Solution. This Neumann problem has non-homogeneous boundary conditions. Let usobserve immediately that U ¤ 0 prevents the existence of stationary solutions u1 Du1 .x/, for otherwise we would have u001 .x/ D 0; u01 .0/ D 0, u01 .�/ D U , which isimpossible.

To separate variables we reduce to homogeneous conditions by setting

w .x; t/ D u .x; t/ � v .x/where vx .0/ D 0, vx .�/ D U . For example we can choose

v .x/ D Ux2

2�:

The function w solves the problem8<:wt .x; t/ � wxx.x; t/ D U=� 0 < x < �; t > 0

w.x; 0/ D �Ux2=2� 0 � x � �

wx.0; t/ D 0; wx.�; t/ D 0 t > 0:

As in Problem 1.2.4, given the homogeneous Neumann conditions, we write

w .x; t/ D c0 .t/

2C

1XkD1

ck .t/ cos kx

so that the Neumann conditions are (formally) satisfied. We have to find ck .t/ so that

wt � wxx D c00 .t/2

C1XkD1

Œc0k .t/C k2ck .t/� cos kx D U

�


and

w .x; 0/ D c0 .0/

2C

1XkD1

ck .0/ cos kx D �Ux2

2�.

Let us expand g .x/ D Ux2

2�in cosines Fourier series:

Ux2

2�D U

2�

´�2

3C 4

1XkD1

.�1/kk2

cos kx

μ;

which is uniformly convergent on Œ0; ��. The comparison of the last three formulas forcesthe coefficients ck .t/ to solve the Cauchy problems:

c00 .t/ D 2U

�; c0 .0/ D �U�

3I

c0k.t/C k2ck .t/ D 0; ck .0/ D 2U

�

.�1/kC1k2

; k � 1:

We find

c0 .t/ D 2U

�t � U�

3; ck .t/ D 2U

�

.�1/kC1k2

e�k2t ; k � 1;

and the solution reads (Fig. 1.2):

u .x; t/ D U

�t C Ux2

2�� U�

6C 2U

�

1XkD1

.�1/kC1k2

e�k2t cos kx: (1.16)

0

�=2

� 0

1

2

3

0

2

4

x t

uDu.x;t/

Fig. 1.2 Solution to Problem 1.2.5 (U D �)

16 1 Diffusion

• Analysis of (1.16). The series is uniformly convergent on Œ0; �� Œ0;1/, so u is con-tinuous there. All derivatives pass into the sum on Œ0; �� Œt0;1/ for any t0 > 0. Thus usolves the diffusion equation on .0; �/ � .0;1/.

Problem 1.2.6 (Mixed Neumann-Robin). A bar made of a homogeneous material, oflengthL, is insulated on its sides and at its left end in x D 0. The other end is subject toa linear Newton’s law of cooling (the heat flow at this end is a multiple of the differencebetween the temperature u and the ambient temperature U ).

a) Write the mathematical model for the evolution of u when t > 0.

b) Solve the problem by separation of variables, assuming that the initial temperatureprofile is continuous in Œ0; L�. Study the behaviour of the solution as t ! C1.

Solution. a) CallD the diffusivity coefficient, > 0 the coefficient appearing in New-ton’s law at x D L, and g the temperature function at time t D 0, with g 2 C.R/ andL-periodic. The ensuing problem is of Neumann-Robin type:8<

ˆ:ut .x; t/ �Duxx.x; t/ D 0 0 < x < L; t > 0

u.x; 0/ D g.x/ 0 � x � L

ux.0; t/ D 0 t > 0

ux.L; t/ D �.u.L; t/ � U/ t > 0:

b) Since the Robin conditions at x D L is not homogeneous, we set z.x; t/ Du.x; t/ � U . The new unknown satisfies the homogeneous problem8<

ˆ:zt .x; t/ �Dzxx.x; t/ D 0 0 < x < L; t > 0

z.x; 0/ D g.x/ � U 0 � x � L

zx.0; t/ D 0 t > 0

zx.L; t/ D �z.L; t/ t > 0:

Seeking solutions of the form z.x; t/ D v.x/w.t/, the equation for w is

w0 .t/ � �Dw .t/ D 0; with general integral w.t/ D Ce�Dt ; C 2 R,

while for v we are led to the eigenvalue problem8<:v00.x/C �2v.x/ D 0

v0.0/ D 0

v0.L/ D �v.L/:(1.17)

We distinguish three cases:

Case � D �2 > 0. We have


��x


and then ´C1 � C2 D 0

.C /e�L C1 � . � /e��L C2 D 0:

Since3 �.C /e�L � . � /e��L� ¤ 0, we find C1 D C2 D 0:

Case � D 0. Fromv.x/ D C1 C C2x;

the Robin conditions force C1 D C2 D 0, hence again the trivial solution.

Case � D ��2 < 0. We have

v.x/ D C1 cosx C C2 sinx:

Asv0.x/ D �C1 sinx C C2 cosx;

v0.0/ D 0 implies C2 D 0, and v0.L/ D �v.L/ implies

sinL D cosL; i.e. tanL D

. (1.18)

Define s D L, so (1.18) reads tan s D L=s. In Fig. 1.3 we see that when s > 0

the graphs of the tangent function y D tan s and the hyperbola y D L=s cross infinitelymany times, for 0 < s1 D 1L < s2 D 2L < : : : , say. Note that

.n � 1/ � < nL < n�;so n n�=L as n ! 1. Therefore also tannL and sinnL tend to zero as n ! 1.The eigenvalues are

�n D �2n D � s2nL2

�2

32�

52�

1L 2L 3Ls

y

Fig. 1.3 Intersections between the graphs of y D tan s and y D L=s

3 In fact, e2�L > . � / = .C / for any ; > 0.

18 1 Diffusion

and the eigenfunctions vn.x/ D cosnx. The candidate solution is

z.x; t/ D1XnD1

ane�D�2

nt cosnx:

The initial condition requires

z .x; 0/ D1XnD1

an cosnx D g .x/ � U; 0 � x � L:

Problem (1.17) is a regular Sturm-Liouville4 problem, and therefore the eigenfunctionsvn .x/ D cosnx satisfyZ L

0

vn .x/ vm .x/ dx D´0 m ¤ nL2

C sin.2�nL/

4�n� ˇn m D n:

If g 2 C .Œ0; L�/ we can write

g .x/ � U D1XnD1

gn cosnx;

with convergence in L2.0; L/, where

gn D 1

ˇn

Z L

0

Œg .x/ � U � cosnx dx:

This gives

z.x; t/ D1XnD1

gne�D�2

nt cosnx (1.19)

and so u .x; t/ D z .x; t/C U:

• Analysis of (1.19). We remind that n n�=L, so that tannL and sinnL tendto zero as n ! 1. Moreover, ˇn ! L=2 if n ! 1. This implies, given the assumptionson g,

jgnj � 1

ˇn

Z L

0

jg .x/ � U j dx � M

and ˇgne

�D�2nt cosnx

ˇ� Me�D�2

nt ; for a suitable constant M:

The series (1.19) is uniformly convergent on Œ0; ��Œt0;1/ and all derivatives can be car-ried under the sum, for any t0 > 0. Thus u solves the diffusion equation on .0; �/�.0;1/.When t ! C1, we have

z .x; t/ ! 0; uniformly on Œ0; �� ;

hence u .x; t/ ! U .

4 Appendix A.


Problem 1.2.7 (Ill posed problem; backward equation). Let u be the solution to theend-value problem:8<

:ut .x; t/ � uxx.x; t/ D 0 0 < x < �; 0 < t < T

u.x; T / D g.x/ 0 � x � �

u.0; t/ D u.�; t/ D 0 0 < t < T:

(1.20)

a) Show that the variable change t D T � s transforms problem .1:20/ for a forwardequation into an initial-value problem for a backward equation.

b) Solve (formally) by separation of variables, specifying which hypotheses on g guar-antee that the expression found is indeed a solution.

c) Show the solution does not depend continuously on the data, by taking gn .x/ D1n

sin nx.

Solution. a) Set t D T � s and v .x; s/ D u .x; T � s/, so that

vs .x; s/ D �ut .T � s/ and v .x; 0/ D u .x; T / ;

resulting in the following problem for v:8<:vs.x; s/C vxx.x; s/ D 0 0 < x < �; 0 < s < T

v.x; 0/ D g.x/ 0 � x � �

v.0; s/ D u.1; s/ D 0 0 < s < T;

which is a backward in time equation with initial datum g.

b) Exactly as in Problem 1.2.1 (page 3) we find the following expression for the solu-tion:

u .x; t/ D1XkD1

ckvk .x; t/ D1XkD1

cke�k2t sin kx:

This time, though, the coefficients ck must be chosen so that u .x; T / D g .x/, in otherwords

ck D gkek2T

by writing

g .x/ D1XkD1

gk sin kx:

This gives

u .x; t/ D1XkD1

ckvk .x; t/ D1XkD1

gkek2.T�t/ sin kx: (1.21)

If we compare to the initial-value situation, now the exponential “fights against” conver-gence, for the exponent is positive for t < T .

20 1 Diffusion

To be able to differentiate inside the sum, thus ensuring that (1.21) really is a solution,we need the coefficients gk to tend to zero very rapidly. For instance, it suffices to have

1XkD1

jgkj ek2T < 1: (1.22)

In particular, (1.22) implies gk D o .k�m/ for any m � 1, i.e. g must have a 2�-periodicodd prolongation at least of class C1 .R/. This is not so surprising if we think that in theforward problem the solution becomes C1 .R/ for t > 0, at least on 0 < x < � , even ifwe start with a non-regular datum at t D 0. The datum has to take into account the strongregularising effect of the diffusion equation.

c) Let us pick

gn .x/ D 1

nsin nx

as final datum. Equation (1.21) gives

un .x; t/ D 1

nen

2.T�t/ sin nx:

Since

jgn .x/j D 1

njsinnxj � 1

n;

gn ! 0 uniformly on R as n ! 1; on the other hand for n D 2mC 1ˇun

��2; 0�ˇ

D 1

nen

2T ! 1as n ! 1. This implies that the solution does not depend on the data in a continuousway, and the problem is ill-posed.

1.2.2 Use of the maximum principle

Problem 1.2.8 (Maximum principle). Let u be a solution to8<:ut .x; t/ � uxx.x; t/ D 0 0 < x < 1; t > 0

u.x; 0/ D sin�x 0 � x � 1

u.0; t/ D 2te1�t , u.1; t/ D 1 � cos�t t > 0;

that is continuousa on the closure of the half-strip S D .0; 1/ � .0;1/.

a) Prove that u is non-negative.

b) Determine an upper bound for u�12; 12

�and u

�12; 3�.

a Note how the boundary values agree on .0; 0/ and .1; 0/.

Solution. a) The parabolic boundary @pS of the strip is the union of the half-linesx D 0, x D 1, t > 0 and the segment 0 � x � 1 on the x-axis (t D 0). By the maximumprinciple u is non-negative on the entire strip provided u � 0 on @pS . On the boundary


1 2 3 4

1

2

Fig. 1.4 Graph of t 7! 2te1�t

half-lines 2te�t � 0, and 1 � cos�x � 0; also the datum sin�x is non-negative. Henceu � 0 on S .

b) By the maximum principle, the value u .1=2; 1=8/ does not exceed the data on theparabolic boundary of the strip S1=8 D .0; 1/ � .0; 1=8/, which is

¹0 � x � 1; t D 0º [²x D 0; 0 � t � 1

8

³[²x D 1; 0 � t � 1

8

³:

The maximum value of the datum and of 1�cos�t is 1. The graph of 2te1�t (see Fig. 1.4)has an absolute maximum 2 at t D 1; on Œ0; 1=8� the maximum is e7=8=4 ' 0:599 72 < 1.We can only say u .1=2; 1=8/ < 1 (the strict inequality holds by the strong maximumprinciple). Similarly, u

�12; 3�< 2. Actually, we can also say that u .x; t/ < 2 on S:

Problem 1.2.9 (Asymptotic behaviour). Let u be a continuous solution on the closureof S D .0; 1/ � .0;1/ to the problem8<

:ut .x; t/ � uxx.x; t/ D 0 0 < x < 1; t > 0

u.x; 0/ D x.1 � x/ 0 � x � 1

u.0; t/ D u.1; t/ D 0 t > 0:

After proving that u is non-negative, determine positive numbers ˛, ˇ, so that

u.x; t/ � w.x; t/ � ˛x.1 � x/e�ˇt :Deduce that u.x; t/ ! 0 uniformly on Œ0; L� as t ! C1.

Solution. We begin by proving that u � 0. For this we recall that by the maximumprinciple is enough to have u � 0 on the parabolic boundary @pS of the strip S . Fort D 0, in fact, u.x; 0/ D x.1 � x/ is nonnegative on Œ0; 1�. Moreover u D 0 along thesides x D 0, x D 1. Hence u � 0. The idea, to havew larger that u, is to use the maximumprinciple for the continuous function v D w � u. More precisely, we look for ˛; ˇ so that

22 1 Diffusion

v � 0 on @pS and v is a super-solution (i.e. vt � vxx � 0):

w.x; 0/ D ˛x.1 � x/wt .x; t/ D �˛ˇx.1 � x/e�ˇtwxx.x; t/ D �2˛e�ˇt ;

so 8<:vt .x; t/ � vxx.x; t/ D ˛.2 � ˇx.1 � x//e�ˇt 0 < x < 1; t > 0

v.x; 0/ D .˛ � 1/x.1 � x/ 0 � x � 1

v.0; t/ D v.1; t/ D 0 t > 0:

Let us find ˇ > 0 so that 2 � ˇx.1 � x/ � 0. As x.1 � x/ � 14

,

2 � ˇx.1 � x/ � 2 � 1

4ˇ

and then is suffices to choose 0 < ˇ � 8. Let us find the sign of v on @pS . Along the sidesof S we have v D 0, while v.x; 0/ D .˛ � 1/x.1 � x/ � 0, for t D 0, provided ˛ � 1.Hence v � 0 on @pS if ˛ � 1. To sum up, for ˛ � 1 and 0 < ˇ � 8, v is a non-negativesuper-solution. We may then use the maximum principle on v, obtaining non-negativity,and so

0 � u.x; t/ � ˛x.1 � x/e�ˇt � ˛

4e�ˇt

because x.1 � x/ � 1=4. As ˇ > 0, e�ˇt ! 0 for t ! 1, so u.x; t/ ! 0 uniformly onŒ0; 1� as t ! C1.

Problem 1.2.10 (Stationary state and asymptotic behaviour). Consider the problem8<:ut .x; t/ � uxx.x; t/ D 1 0 < x < 1; t > 0

u.x; 0/ D 0 0 � x � 1

u.0; t/ D u.1; t/ D 0 t > 0:

a) Determine the stationary solutionus D us .x/ that satisfies the boundary conditions.

b) Show u.x; t/ � us.x/ when t > 0.

c) Find ˇ > 0 so that u.x; t/ � .1 � e�ˇt /us.x/:d) Deduce u.x; t/ tends to us.x/ as t ! C1, uniformly on Œ0; 1�.

e) Double-check the result by solving the problem by separation of variables.

Solution. a) We recall that a stationary solution does not depend on the time variable(ut .x; t/ � 0). Thus, we are asked to find us.x/ D .x/ such that´

� 00.x/ D 1; 0 < x < 1;

.0/ D .1/ D 0:

The solution is the parabola

us.x/ D 1

2x.1 � x/:


b) Set v.x; t/ D us.x/ � u.x; t/: Then

vt .x; t/ � vxx.x; t/ D 0; 0 < x < 1; t > 0:

As u is continuous on the closure of S D Œ0; 1� � .0;1/, to prove the non-negativity ofv we can show that the boundary data are non-negative on the parabolic boundary, by themaximum principle. In fact v.0; t/ D v.1; t/ D 0 for t > 0, and

v.x; 0/ D 1

2x.1 � x/ � 0;

if 0 � x � 1. The maximum principle implies v � 0, i.e. us � u.

c) As for Problem 1.2.9 we setw.x; t/ D u.x; t/�.1�e�ˇt /us.x/ and look for ˇ > 0so that w is a super-solution and non-negative on @pS . Since

@t Œ.1 � e�ˇt /us.x/� D ˇ

2x.1 � x/e�ˇt

@xxŒ.1 � e�ˇt /us.x/� D �1C e�ˇt ;we have 8<

:wt .x; t/ � wxx.x; t/ D e�ˇt .1 � ˇ

2x C ˇ

2x2/ 0 < x < 1; t > 0

w.x; 0/ D 0 0 � x � 1

w.0; t/ D w.1; t/ D 0 t > 0:

So, we have to find ˇ rendering the right-hand side non-negative. After that, the maximumprinciple allows to conclude. Now the right side of the equation is non-negative if

ˇx2 � ˇx C 2 � 0;

so it suffices to demand ˇ � 8.

d) Take ˇ � 8, so that:

.1 � e�ˇt /us.x/ � u.x; t/ � us.x/;

i.e.

0 � us.x/ � u.x; t/ � e�ˇtus.x/:

Then

supx2Œ0;1�

jus.x/ � u.x; t/j � supx2Œ0;1�

e�ˇtus.x/ � 1

8e�ˇt ! 0 as t ! C1:

Therefore u.x; t/ ! 0 uniformly on Œ0; 1� as t ! C1.

e) The solution has the form

w .x; t/ D 1

2x .1 � x/C

1XkD1

cke�k2�2t sin k�x

24 1 Diffusion

where the ck are chosen so that

w .x; 0/ D 1

2x .1 � x/C

1XkD1

ck sin k�x D 0:

Since x.1 � x/ vanishes at x D 0; 1, we have that S D1P1

jckj < 1. As a consequence,ˇw .x; t/ � 1

2x .1 � x/

ˇ� Se��2t ! 0; as t ! 1:

**Problem 1.2.11 (Hopf’s principle). Let u solve

ut .x; t/ � uxx.x; t/ D 0

on the rectangle QT D .0; 1/ � .0; T /, and assume that u is C 1.QT /.

a) Given 0 < t0 � T suppose

u .x; t/ > m; for 0 � x � 1; 0 < t < t0;

and u.0; t0/ D m. Prove that ux.0; t0/ > 0 (it cannot vanish)a.

b) Deduce that the Neumann problem8<:ut .x; t/ � uxx.x; t/ D 0 in QTu.x; 0/ D g.x/ 0 � x � 1

ux.0; t/ D ux.1; t/ D 0 0 < t � T;

has a unique solution u, and

minŒ0;1�

g D minQT

u � maxQT

u D maxŒ0;1�

g:

a Hint. Compare u with z .x; t/ D ex � 1. Note that z.0; t/ D 0, zx .0; t/ D 1 > 0.

Solution. a) First, set v D u � m so that v .0; t0/ D 0 and v > 0 when 0 � x � 1;

0 < t < t0. The idea is to find a function w smaller than v on a neighbourhood of .0; t0/,that vanishes at .0; t0/ and with wx .0; t0/ > 0. Once w is found we can write

v .h; t0/ � v .0; t0/h

>w .h; t0/ � w .0; t0/

h;

and passing to the limit for h ! 0C we obtain

vx .0; t0/ � wx .0; t0/ > 0.

We now construct w. As a neighbourhood of .0; t0/ we choose the rectangle

R D�0;1

2

��t0

2; t0

:

Note that the functionz.x; t/ Dex � 1


is zero at x D 0: On R, moreover,

0 � z � pe � 1 � a; zx .0; t0/ D 1 > 0; zt � zxx D �ex < 0:

Call m0 (> 0) the minimum of v along the sides x D 1=2 and t D t0=2 of @R. Thefunction

w .x/ D m0

az .x/

is a non-negative sub-solution of the diffusion equation; it is smaller than v on the parabolicboundary of R, it vanishes at x D 0 and wx .0; t0/ D m0

a1e> 0. These are the properties

we wanted.

b) We shall prove minŒ0;1� g D minQTu. The claim about the maximum is similar.

Suppose minŒ0;1� g > minQTu, so the maximum principle implies u has minimum onQT

at points on the sides of QT . If either .0; t0/ or .1; t0/, 0 < t0 � T , were the minimumpoint with smallest t -coordinate, by part a) the spatial derivative should be non-zero there;but this would contradict the homogeneous Neumann conditions.

Finally, the uniqueness of the solution comes from the fact that the difference of anytwo solutions with same data would have zero initial datum.

1.2.3 Applying the notion of fundamental solution

Problem 1.2.12 (Non-instantaneous point source). A polluting agent with concentra-tion u D u .x; t/ (mass per unity of length) diffuses, with coefficientD; along a narrowchannel (the x-axis). At x D 0 the quantity q D q.t/ of pollutant (mass per second perunity of length) enters, where

q .t/ D´Q 0 < t < T

0 t > T:

Determine the pollutant concentration at x D 0 and time t , and its asymptotic behaviouras t ! 1.

Solution. Recall that, if t > s, �D.0; t � s/ indicates the concentration at x D 0 andtime t due to a point source with unit intensity placed at x D 0 at time t D s. If the sourceat t D s produces q.s/, the contribution to the concentration at x D 0 and time t is

�D.0; t � s/q.s/ D8<:0 t < s

1p4�D.t � s/q.s/ t > s:

The sum of all contributions as s < t varies gives

u.0; t/ DZ t

0

�D.0; t � s/q.s/ ds:Therefore, for t < T ,

u.0; t/ DZ t

0

Q�D.0; t � s/ ds D Qp4�D

Z t

0

1pt � s ds D Qp

�D

pt ;

26 1 Diffusion

T

Q

rT

�D

t

u

Fig. 1.5 The concentration u .0; t/ in Problem 1.2.12

while, for t > T ,

u.0; t/ DZ T

0

�D.0; t � s/ ds D Qp4�D

Z T

0

1pt � s ds D Qp

�D

hpt � p

t � Ti:

This means that when t ! 1 we have (Fig. 1.5):

u.0; t/ D Qp�D

hpt � p

t � Ti

D Qp�D

Tpt C p

t � T 2QTp�D

1pt:

Problem 1.2.13 (The error function). Find all solutions to ut .x; t/�Duxx.x; t/ D 0

of the form

u.x; t/ D v

�xpt

:

Use the result to recover the function �D.x; t/.

Solution. We just have to substitute the expression into the equation. To simplify thenotation we set

� D xpt

whence@�

@tD � x

2tpt;

@�

@xD 1p

t;

@2�

@x2D 0:

So,

ut .x; t/ D � x

2tptv0.�/; ux .x; t/ D 1p

tv0.�/; uxx.x; t/ D 1

tv00.�/

and then

� x

2tptv0.�/ �D1

tv00.�/ D 0;

or�

2Dv0.�/C v00.�/ D 0:

This is a first order linear ODE for v0, that gives v0.�/ D C exp .��2=4D/, and, by afurther integration,

v.�/ D C1 C C2

Zexp

� �2

4D

!D C1 C C2

Z �p4D

0

e�z2

dz:


�4 �2 2 4

�1

1

t

u

Fig. 1.6 The function erf .x/

Introducing the error function (Fig. 1.6)

erf.x/ D 2p�

Z x

0

e�z2

dz (Gauss’ error function);

we can write

u.x; t/ D C1 C C2 erf

�xp4Dt

:

Choosing C2 D 1�

we find �D .x; t/ D ux.x; t/.

Remark. The error function erf is invariant under the parabolic dilation x 7! �x, t 7! �2t ;solutions of this type are called self-similar and are useful when the domain and the dataare invariant under parabolic dilations (see Exercise 1.3.11 on page 52).

Problem 1.2.14 (Absorbing barriers and Dirichlet conditions; method of images).Consider a one-dimensional symmetric random walk a of a particle of unit mass ini-tially placed at the origin. Let h and � be the space and time steps, and p D p .x; t/ thetransition probabilityb. Suppose to have an absorbing barrier at L D mh > 0, whichmeans that if the particle is at L � h at time t and moves to the right, at time t C � itwill be absorbed and stop at L.

a) Which problem does p D p .x; t/ solve when we pass to the limit h; � ! 0, main-taining h2=� D 1?

b) Find the explicit expression of p.

c) Show that in the limit the particle reaches L in finite time with probability 1. Ob-serve, in particular, that for t > 0Z L

�1p .x; t/ dx < 1:

a [18, Chap. 2, Sect. 4].b The probability that the particle will reach x at time t .

28 1 Diffusion

Solution. a) This being a symmetric walk, the limit density p solves

pt � 1

2pxx D 0

for x < L and t > 0; as the particle starts from the origin, we have

p .x; 0/ D ı.x/

on .�1; L/. To understand what happens at x D L, let us use the absorbing condition. Ifthe particle is located atL�h at time tC� , it can only come fromL�2h, with probability1=2. By the law of total probability

p .L � h; t C �/ D 1

2p .L � 2h; t/ : (1.23)

Taking the limit as h; � ! 0 gives p .L; t/ D 0, which is a homogeneous Dirichlet con-dition.

b) In order to find p .x; t/ we shall use the so-called method of images, which consistsin starting another random walk at 2L, the symmetric point to the origin with respect toL.Then using the linearity of the heat equation, we consider the difference of the fundamentalsolutions with initial conditions ı .x/ and ı .x � 2L/. Let

pA .x; t/ D �1=2 .x; t/ � �1=2 .x � 2L; t/ D �1=2 .x; t/ � �1=2 .2L � x; t/ : (1.24)

Thus pA is the required solution, for

pA .x; 0/ D �1=2 .x; 0/ � �1=2 .2L � x; 0/ D ı.x/

and

pA .L; t/ D �1=2 .L; t/ � �1=2 .L; t/ D 0

when �1 < x < L.

c) We indicate by X .t/ the position at time t and by TL the first instant at which theparticle reaches L. TL is a random variable, defined by

TL D infs

¹X .s/ D Lº :

We claim that the probability of the event ¹TL < 1º is 1. This follows if we show

Prob ¹TL > tº ! 0 as t ! 1.

Now, ¹TL > tº occurs if and only if at time t the particle is inside the interval .�1; L/ ,


(in particular it will not have yet reached L). The probability of being in .�1; L/ is

Prob ¹TL > tº DZ L

�1pA .x; t/ dx D

Z L

�1��1=2 .x; t/ � �1=2 .x � 2L; t/� dx

D 1p2�t

Z L

�1Œe�x2

2t � e� .x�2L/2t

2

�dx D 1p2�t

Z L

�Le�x2

2t dx

D 1p�

Z L=p2t

�L=p2te�y2

dy

(in the last line we substituted x D p2ty). If now t ! 1, Prob ¹TL > tº ! 0.

Problem 1.2.15 (Problems on the half-line; reflection method). Let g W Œ0;C1/ ! Rbe a continuous and bounded function.

a) Find a formula for the solution to8<:ut .x; t/ �Duxx.x; t/ D 0 x > 0; t > 0

u.x; 0/ D g.x/ x > 0

u.0; t/ D 0 t > 0:

Hint. Extend the initial datum, for x < 0, to an odd function and use the formulafor the global Cauchy problem.

b) Find a formula for the solution to8<:ut .x; t/ �Duxx.x; t/ D 0 x > 0; t > 0

u.x; 0/ D g.x/ x > 0

ux.0; t/ D 0 t > 0:

Hint. Extend the initial datum, for x < 0, to an even function and use the formulafor the global Cauchy problem.

c) Show that either formula provides the unique bounded solution to the respectiveproblem.

Solution. a) In the first case we extend g in an odd way:

Qg.x/ D´g.x/ x � 0

�g.�x/ x < 0(odd reflection).

The new function is continuous on R only if g.0/ D 0. Consider the global Cauchy prob-lem ´

ut .x; t/ �Duxx.x; t/ D 0 x 2 R; t > 0

u.x; 0/ D Qg.x/ x 2 R:

30 1 Diffusion

For any x 2 R, t > 0, the solution reads

Qu.x; t/ DZR

�D.x � y; t/ Qg.y/ dy

DZ C1

0

�D.x � y; t/g.y/ dy �Z 0

�1�D.x � y; t/g.�y/ dy

DZ C1

0

�D.x � y; t/g.y/ dy �Z C1

0

�D.x C y; t/g.y/ dy

where in the last term we wrote y instead of �y and swapped endpoints. Let u.x; t/ denotethe restriction of Qu to the first quadrant. The previous computation tells

u.x; t/ DZ C1

0

Œ�D.x � y; t/ � �D.x C y; t/� g.y/ dy: (1.25)

• Analysis of (1.25). Clearly u is bounded and solves the heat equation on the quadrantx > 0; t > 0. Since �D is even in the spatial variable we obtain5

u.0; t/ DZ C1

0

Œ�D.�y; t/ � �D.y; t/� g.y/ dy D 0 for any t > 0:

Hence u fulfils the Dirichlet condition of the half-line x D 0. Write

gC .x/ D´g .x/ x � 0

0 x < 0and g� .x/ D

´0 x � 0

g .�x/ x < 0;

so

u .x; t/ D uC .x; t/ � u� .x; t/

�Z C1

�1�D.x � y; t/gC.y/ dy �

Z C1

�1�D.x � y; t/g�.y/ dy:

Therefore for every x0 > 0, if .x; t/ ! .x0; 0/ we have

uC .x; t/ ! g .x0/ , and u� .x; t/ ! 0;

because g is continuous at x0. Consequently u is continuous on the closed quadrant, ex-cept possibly for the origin, and in particular u .x; 0/ D g .x/, x > 0. Continuity at theorigin holds if and only if g.0/ D 0, because both gC, g� are continuous at x D 0. Inaddition, Qu 2 C1 on the half-plane t > 0 so u 2 C1 when t > 0, x � 0.

b) The strategy is completely similar to the previous one. Now we prolong g evenlyfor x < 0:

Qg.x/ D´g.x/ x � 0

g.�x/ x � 0(even reflection)

5 There are no problems in passing to the limit x ! 0C when t > 0.


and consider the global Cauchy problem with datum Qg. The solution is

Qu.x; t/ DZR

�D.x � y; t/ Qg.y/ dy D

DZ C1

0

�D.x � y; t/g.y/ dy CZ 0

�1�D.x � y; t/g.�y/ dy D

DZ C1

0

�D.x � y; t/g.y/ dy CZ C1

0

�D.x C y; t/g.y/ dy:

If u.x; t/ denotes Qu restricted to the first quadrant, we have

u.x; t/ DZ C1

0

Œ�D.x � y; t/C �D.x C y; t/� g.y/ dy: (1.26)

• Analysis of (1.26). As before u is bounded and solves the heat equation on x > 0; t >0. Note that Qg is continuous also at x D 0, so u equals the Cauchy datum continuouslyon the half-line x � 0. To verify the Neumann condition we have to compute ux.0; t/.Observe

@x�D.x ˙ y; t/ D @x1p4�Dt

exp

�� .x ˙ y/2

4Dt

D �x ˙ y

2Dt�D.x ˙ y; t/

so at x D 0

@x�D.˙y; t/ D y

2Dt�D.y; t/:

For t > 0 we can differentiate the integrand, obtaining

ux.0; t/ DZ C1

0

Œ@x�D.�y; t/C @x�D.y; t/�g.y/ dy

DZ C1

0

y

2DtŒ�D.y; t/ � �D.y; t/�g.y/ dy

D 0:

In this case, too, the regularity of Qu implies u 2 C1 when t > 0, x � 0.c) If there existed distinct bounded and regular solutions for t > 0, x � 0, the reflec-

tion would generate bounded C 2 functions for t > 0, solving the same global Cauchyproblem, thus contradicting the general theory.

Remark. The functions

��D .x; y; t/ D �D.x � y; t/ � �D.x C y; t/

and

�CD .x; y; t/ D �D.x � y; t/C �D.x C y; t/

are called fundamental solutions for the Cauchy-Dirichlet and Cauchy-Neumann prob-lems on the quadrant t > 0, x > 0, respectively (see Figs. 1.7 and 1.8).

32 1 Diffusion

�20

2 12

34

56

�50

0

50

xt

Fig. 1.7 ��1=4

.x; 1; t/ D �1=4 .x � 1; t/ � �1=4 .x C 1; t/

�20

2 12

34

56

0

50

xt

Fig. 1.8 �C1=4

.x; 1; t/ D �1=4 .x � 1; t/C �1=4 .x C 1; t/

Problem 1.2.16 (Reflection method for a finite interval). Adapting the reflectionmethod of Problem 1.2.15, b), deduce a formula for solving8<

:ut �Duxx D 0 0 < x < L; t > 0

ux .x; 0/ D g .x/ 0 � x � L

ux.0; t/ D ux.1; t/ D 0 t > 0

(1.27)

where g is continuous on Œ0; L�.


Solution. With the given homogeneous Neumann conditions we first extend g to.�L;L� in an even way:

Qg.x/ D´g.x/ 0 � x � L

g.�x/ �L < x < 0 (even reflection),

and then prolongeg to R by setting it equal to zero outside .�L;L�; at last, we define

g� .x/ DC1XnD�1

eg .x � 2nL/ :

The function g� is continuous, 2L-periodic and it coincides witheg on .�L;L�: For anygiven x, at most one summand of the series is non-zero. Let us solve the global Cauchyproblem, with datum g�, using

u .x; t/ DZ C1

�1�D .x � y; t/ g� .y/ dy (1.28)

DC1XnD�1

Z C1

�1�D .x � y; t/eg .y � 2nL/ dy:

Sinceeg .y � 2nL/ vanishes outside .2n � 1/L < y � .2nC 1/L, we can write

u .x; t/ DC1XnD�1

Z .2nC1/L

.2n�1/L�D .x � y; t/eg .y � 2nL/ dy

D.y�2nL/�!y

C1XnD�1

Z L

�L�D .x � y � 2nL; t/eg .y/ dy

DC1XnD�1

Z L

0

Œ�D .x � y � 2nL; t/C �D .x C y � 2nL; t/�g .y/ dy

�Z L

0

ND .x; y; t/ g .y/ dy;

where

ND .x; y; t/ DC1XnD�1

Œ�D .x � y � 2nL; t/C �D .x C y � 2nL; t/� :

The restriction of u to Œ0; L� is the solution of (1.27), and we set out to check this fact.

34 1 Diffusion

Initial datum: by (1.28), u certainly solves the diffusion equation; since g� is contin-uous, it is straightforward to see

u .x; t/ ! g� .x0/ D g .x0/ , if .x; t/ ! .x0; 0/

at x0 2 Œ0; L�. Let us compute the flow at x D 0; for t > 0 we can differentiate theintegrand, so it suffices to verify the conditions on the kernel ND:

@xND .0; y; t/ D 1

2Dt

C1XnD�1

Œ.�y � 2nL/�D .y C 2nL; t/

C .y � 2nL/�D .y � 2nL; t/�

D 1

2Dt

C1XnD�1

Œ� .y C 2nL/�D .y C 2nL; t/

C .y C 2nL/�D .y C 2nL; t/�

D 0:

At x D L, changing the summation indices,

@xND .L; y; t/ D 1

2Dt

C1XnD�1

Œ.L � y � 2nL/�D .L � y � 2nL; t/

C .LC y � 2nL/�D .LC y � 2nL; t/�

D 1

2Dt

C1XnD�1

Œ..2nC 1/L � y/ �D .L � y C 2nL; t/

C .y � .2nC 1/L/ �D .LC y � 2nL; t/�D 0:

Therefore the Neumann conditions hold.

Remark. The function

ND D ND .x; y; t/

is called fundamental solution with Neumann conditions for the interval Œ0; L�. Noticethat, since the solution with datum g .x/ D 1 is u .x; t/ D 1, we have

1 DZ L

0

ND.x; y; t/dy:

In particular, if jg .x/j � ", we deduce ju .x; t/j � "; which shows a continuous depen-dence on the initial datum.


Problem 1.2.17 (Duhamel’s principle). Consider the Neumann problem8<:ut .x; t/ �Duxx.x; t/ D f .x; t/ 0 < x < L; t > 0

u.x; 0/ D 0 0 � x � L

ux.0; t/ D ux.L; t/ D 0 t > 0:

(1.29)

Let f be continuous for 0 � x � L; t � 0. Show that if v.x; t I �/, t � � � 0, solves8<:vt .x; t I �/ �Dvxx.x; t I �/ D 0 0 < x < L; t > �

v.x; � I �/ D f .x; �/ 0 � x � L

vx.0; t I �/ D vx.L; t I �/ D 0 t > 0;

(1.30)

then the solution to (1.29) is

u.x; t/ DZ t

0

v.x; t I �/ d�:Find an explicit formula and examine the dependence of u on f .

Solution. As f is continuous for 0 � x � L; t � 0; from the previous problem weknow that the solution to (1.30) is for any � , 0 � � � t;

v .x; t I �/ DZ L

0

ND .x; y; t � �/ f .y; �/ dyand that it is continuous on that set. Then for every 0 < x < L; t > 0:

ut .x; t/ D v .x; t; t/CZ t

0

vt .x; t I �/ d� D f .x; t/CZ t

0

vt .x; t I �/ d�

uxx .x; t/ DZ t

0

vxx.x; t I �/ d�:Therefore

ut �Duxx D f .x; t/ 0 < x < L; t > 0

and moreover u .x; 0/ D 0. We conclude that u solves (1.29). An explicit expression foru is:

u .x; t/ DZ t

0

v.x; t I �/ d� DZ t

0

Z L

0

ND .x; y; t � �/ f .y; �/ dyd�:There is an alternative formula that arises from separation of variables. For v.x; t I �/ wefind

v .x; t I �/ D f0 .�/

2C

1XkD1

fk .�/ e�k2�2.t��/=L cos

�k�

Lx

where

f0 .�/ D 2

L

Z L

0

f .y; �/ dy;

fk .�/ D 2

L

Z L

0

f .y; �/ cos

�k�

Lx

dy;

36 1 Diffusion

and so

u .x; t/ D 1

2

Z t

0

f0 .�/ d� C1XkD1

cos

�k�

Lx

Z t

0

fk .�/ e�k2�2.t��/=Ld�:

• Continuous dependence. SinceZ L

0

ND .x; y; t � �/ dy D 1

for t > � , if jf .y; �/j � " when 0 � � � T; 0 � y � L, we have

ju .x; t/j �Z t

0

Z L

0

ND .x; y; t � �/ jf .y; �/j dyd� � T ";

which shows continuous dependence on finite time intervals.

*Problem 1.2.18. Let g be boundeda (jg .x/j � M for any x 2 R) and

u .x; t/ DZ

R�D .x � y; t/ g .y/ dy:

Show that if g is continuous on x0; then u .x; t/ ! g .x0/ for .x; t/ ! .x0; 0/.

a Actually it is enough to have constants C , A such that jg .x/j � CeAx2

.

Solution. We recallR

R �D .x � y; t/ dy D 1, for any t > 0, x 2 R. Therefore wemay write

u .x; t/ � g .x0/ DZ

R�D .x � y; t/ Œg .y/ � g .x0/�dy: (1.31)

Given " > 0, let ı" be chosen so that if jy � x0j � 2ı" then

jg .y/ � g .x0/j < ":

Now write:ZR�D .x � y; t/ Œg .y/ � g .x0/�dy D

Z¹jy�x0j�2ı"º

� � � dy CZ¹jy�x0j>2ı"º

� � � dy:

Then: ˇZ¹jy�x0j�2ı"º

� � � dyˇ

�Z¹jy�x0j�2ı"º

�D .x � y; t/ jg .y/ � g .x0/j„ ƒ‚ …�"

dy � ":


For the second integral, if jx � x0j � ı" and jy � x0j > 2ı", it follows jx � yj > ı"; so,jx � x0j � ı" implies:ˇZ

¹jy�x0j>2ı"º� � � dy

ˇ�Z¹jy�xj>ı"º

�D .x � y; t/ jg .y/ � g .x0/j„ ƒ‚ …�2M

dy

� 2Mp4�Dt

Z¹jy�xj>ı"º

e�.x�y/2

4Dt dy D�y � x D z

p4�Dt

�� 2Mp

�

Z C1ı"p

4�Dt

e�z2

dz ! 0 for t # 0:

In conclusion, if jx � x0j � ı" and t > 0 is small enough,

ju .x; t/ � g .x0/j � 2"

as claimed.

1.2.4 Use of Fourier and Laplace transforms

Problem 1.2.19 (Fourier transform and fundamental solution). Using the Fouriertransform with respect to x recover the formula for the solution to the global Cauchyproblem ´

ut �Duxx D f .x; t/ �1 < x < 1; t > 0

u .x; 0/ D g .x/ �1 < x < 1:

Solution. Define bu .�; t/ D RR u .x; t/ e

�ix�d�, the partial Fourier transform of u.Thenbu solves the Cauchy problem (formally)´but CD�2bu D bf .�; t/ �1 < � < 1; t > 0bu .�; 0/ Dbg .�/ �1 < � < 1;

where bf denotes the partial Fourier transform of f . We find

bu .�; t/ Dbg .�/ e�D�2t CZ t

0

e�D�2.t�s/bf .�; s/ ds.

We remind that the inverse transform of the exponential e�D�2t is �D .x; t/, and the in-verse transform of a product is the convolution of the inverse transforms. This yields (atleast formally)

u .x; t/ DZ

R�D .x � y; t/ g .y/ dy C

Z t

0

ZR�D .x � y; t � s/ f .y; s/ ds.

38 1 Diffusion

Problem 1.2.20 (Dirichlet conditions on the half-line). a) Using the Fourier sinetransform find a formula for the bounded solution to8<

:ut .x; t/ � uxx.x; t/ D 0 x > 0; t > 0

u.x; 0/ D 0 x � 0

u.0; t/ D g .t/ t > 0;

(1.32)

where g is continuous and bounded. Show that this is the only solution with thegiven properties.

b) Prove that without the condition that u is bounded, problem (1.32) does not have,in general, a unique solution. Hint. Use the functions w1 .x; t/ D ex cos .2t C x/

and w2 .x; t/ D e�x cos .2t � x/.

Solution. a) The Fourier sine transform in x is defined as:

S .u/ .�; t/ D U .�; t/ D 2

�

Z 1

0

u .x; t/ sin .�x/ dx;

with inverse formula

u .x; t/ DZ 1

0

U .�; t/ sin .�x/ d�:

Notice that U is an odd function in �. Assuming that both u and ux vanish at infinity wehave

S .uxx/ .�; t/ D 2

��u .0; t/ � �2U .�; t/ ,

and therefore U solves the problem´Ut .�; t/C �2U .�; t/ D 2

��g .t/ � > 0; t > 0

U.�; 0/ D 0 � � 0:

We have

U .�; t/ D 2

��

Z t

0

e��2.t�s/g .s/ ds:

Anti-transforming gives:

u .x; t/ DZ 1

0

U .�; t/ sin.�x/d� D 1

�

Z t

0

g .s/

Z 1

0

2�e��2.t�s/ sin.�x/d�

�ds

D �Z t

0

g .s/

�.t � s/²h

sin.�x/e��2.t�s/d�

i10

� xZ 1

0

e��2.t�s/ cos.�x/d�

³ds

D x

�

Z t

0

g .s/

t � sZ 1

0

e��2.t�s/ cos.�x/d�

�ds:


Observe that Z 1

0

e�a�2

cos.�x/d� D 1

2

Z 1

�1e�a�

2Ci�xd� Dr�

4ae�

x2

4a :

Substituting a D t � s finally gives (at least formally)

u .x; t/ D x

2p�

Z t

0

g .s/

.t � s/3=2 e� x2

4.t�s/ ds: (1.33)

To show uniqueness, let u1; u2 be bounded solutions, so that also v D u1 � u2 is abounded solution to (1.33) as well, with g.t/ � 0. But then its odd prolongation (v.x; t/ D�v.�x; t/ on x < 0) is regular, bounded and it solves the global Cauchy problem for theheat equation with null initial value. The general theory says v must vanish identically,i.e. u1 and u2 coincide.

b) The functions w1; w2 are solutions of wt � wxx D 0 (on the whole .x; t/-plane);additionally

w1 .x; 0/ D ex cos x; w1 .0; t/ D cos 2t

w2 .x; 0/ D e�x cos x; w2 .0; t/ D cos 2t:

We will modify these functions in order to have zero initial value and the same Dirichletdatum on x D 0, t > 0. To this end we recall from Problem 1.2.15 (page 29) that6

v1.x; t/ DZ C1

0

��.x; y; t/ ey cos y dy (1.34)

v2 .x; t/ DZ C1

0

��.x; y; t/ e�y cos y dy (1.35)

solve vt � vxx D 0 on the quadrant x > 0; t > 0, with vanishing lateral datum and initialdatum

v1 .x; 0/ D ex cos x; v2 .x; 0/ D e�x cos x;

respectively. Thenu1 D w1 � v1; u2 D w2 � v2

have zero initial value, Dirichlet value on x D 0 equal to cos 2t and it is not hard to checkthat they are different (for example at

��2; �2

�). The problem therefore has no unique so-

lution. Note that w1 (hence u1) is unbounded on the quadrant.

Problem 1.2.21 (Linear reaction coefficient). Use Fourier transforms to solve´ut D uxx C xu x 2 R; t > 0

u .x; 0/ D g .x/ x 2 R

where g is continuous and L2 on R. Study the effect of the reaction term by choosingg .x/ D ı .x/. Does anything change if the reaction terms is �xu‹

6 See the remark at the end of Problem 1.2.15.

40 1 Diffusion

Solution. Let bu .�; t/ DZ

Ru .x; t/ e�ix�d�

be the partial Fourier transform of u. Since the transform of xu .x; t/ is ibu� .�; t/,bu sat-isfies (formally) the Cauchy problem´but � ibu� D ��2bu �1 < � < 1; t > 0bu .�; 0/ Dbg .�/ �1 < � < 1:

The equation is linear, non-homogeneous, of order one and can be solved by the methodof characteristics, as described in Chap. 3. The characteristic curves have parametric equa-tions

t D t .�/ ; � D � .�/ ; z D z .�/

and solve (d=d� D P) 8<:

Pt D 1; t .0/ D 0

P� D �i; � .0/ D s

Pz D ��2z; z .0/ Dbg .s/ :The first two give t D � , � D s � i� , the third

z .�; s/ Dbg .s/ e�s2��is�2C �3

3 :

Eliminating the parameters s; � we find

bu .�; t/ Dbg .� C i t/ e��2t�3i�t2C 7

3 t3

and then

u .x; t/ D 1

2�e

73 t

3

ZRei�.x�3t2/e��

2tbg .� C i t/ d�.

Now recallbg .� C i t/ is the transform of

extg .x/ ;

while e��2t that of

�1 .x; t/ D 1p4�t

e�x2

4t :

Hence

u .x; t/ DZ

R�1�x � 3t2 � y; t� eytC 7

3 t3

g .y/ dy.

Choosing g .x/ D ı .x/ gives (Fig. 1.9)

u .x; t/ D e73 t

3

�1�x � 3t2; t� D � � � D 1p

4�texp

²1

12t3 � .x � 6t2/x

4t

³:


�20

24

68

0:20:4

0:60:8

11:2

0

10

xt

Fig. 1.9 Effect of the reaction term xu: the solution of Problem 1.2.21 for t > 0:01

What we see is that for t ! 1 the solution diverges to C1 at any point x. The reac-tion term cancels out the damping effect of the diffusion. Had there been �x instead of xnothing would have changed: setting y D �x would reduce to the above case.

Problem 1.2.22. The initial temperature of a homogeneous bar of length L and smallcross-section is zero. Determine a formula for the temperature knowing that

u .0C; t / D ı .t/ ; u .L; t/ D 0; t > 0,

where ı .t/ is the Dirac distribution centred at the origin. Use the Laplace transform.

Solution. We assume that u admits Laplace transform with respect to t

L .u/ .x; �/ D U .x; �/ DZ 1

0

e��tu .x; t/ dt

defined on the half-plane Re � > ˛, ˛ > 0 a suitable constant. Recalling that

L .ut / .x; �/ D �U .x; �/ � u .x; 0/ D �U .x; �/ ;

we deduce that U solves7

�U � Uxx D 0; 0 < x < L

U .0; �/ D 1; U .L; �/ D 0:

7 Set D D 1 for simplicity.

42 1 Diffusion

Solving the ODE gives:

U .x; �/ D C1ep�x C C2e

�p�x

with p� D

pj�jei arg �

2 :

The boundary conditions force

C1 C C2 D 1; C1ep�L C C2e

�p�L D 0;

whence

C2 D ep�L

2 sinh�p

�L� and C1 D � e�

p�L

2 sinh�p

�L� :

Therefore

U .x; �/ D ep�.L�x/ � e

p�.x�L/

ep�L � e�p�L

Dsinh

hp�.L � x/

isinh

�p�L� :

To anti-transform U we recall8 that the inverse transform of .a; �/ D e�ap� is

‰ .a; t/ D a

2p�t�3=2e�a2=4t :

So let us try to write U as a sum of terms e�ap� :

ep�.L�x/ � e

p�.x�L/

ep�L � e�p�L D e�x

p� � e�.2L�x/

p�

1 � e�2p�L

Dhe�x

p� � e�.2L�x/

p�i 1XnD0

e�2nLp�

D1XnD0

e�.2nLCx/p� �

1XnD1

e�.2nL�x/p�

D1XnD0

.2nLC x; �/ �1XnD1

.2nL � x; �/ :

Thus, noticing that ‰ is odd with respect to a:

u .x; t/ D1X

nD�1‰ .2nLC x; t/ D 1

2p�t3=2

1XnD�1

.2nLC x/e�.2nLCx/2=4t :

8 Appendix B.


1.2.5 Problems in dimension higher than one

Problem 1.2.23 (Dirichlet on the rectangle). The surfaces of a thin rectangular plateof length a and width b are thermally insulated while the four edges are kept at zerotemperature. Determine the evolution of the temperature, knowing its initial value.

Solution. The problem is essentially two-dimensional, so that we can suppose u Du .x; y; t/. Then (for simplicity we fix the diffusion coefficientD D 1) we have the prob-lem

ut � .uxx C uyy/ D 0; 0 < x < a; 0 < y < b; t > 0

with initial conditionu .x; y; 0/ D g .x; y/

and Dirichlet conditions´u .0; y; t/ D 0; u .a; y; t/ D 0; 0 < y < b; t > 0

u .x; 0; t/ D 0; u .x; b; t/ D 0; 0 < x < a; t > 0:

We can separate the variables. First we look for non-zero solutions of the form

u .x; y; t/ D v .x; y/ z .t/

satisfying the Dirichlet conditions. Substituting in the equation and separating variablesgives:

vxx C vyy

vD z0

zD �:

For z there are no problems; we have

z .t/ D ce�t :

For v we obtain the eigenvalue problem

vxx C vyy D �v (1.36)

on the rectangle .0; a/ � .0; b/, with´v .0; y/ D 0; v .a; y/ D 0; 0 � y � b

v .x; 0/ D 0; v .x; b/ D 0; 0 � x � a:(1.37)

We separate again variables, setting

v .x; y/ D X .x/ Y .y/ :

Substituting in (1.36) and separating, we find

Y 00 .y/Y .y/

� � D �X00 .x/X .x/

D

44 1 Diffusion

with constant. Set D �� :

We have the following eigenvalue problems for X and Y :´X 00 C X D 0 in .0; a/

X .0/ D X .a/ D 0;

´Y 00 C Y D 0 in .0; b/

Y .0/ D Y .b/ D 0:

We have already solved these problems in previous exercises. The eigenvalues and thecorresponding eigenfunctions are:

Xm .x/ D Am sin�m�xa

�; m D m2�2

a2 ; m D 1; 2; : : :

Yn .x/ D Bn sin�n�yb

�; n D n2�2

b2 ; n D 1; 2; : : : :

As� D �. C /;

we conclude that the eigenvalues for problem (1.36), (1.37) are

�mn D ��2�m2

a2C n2

b2

; m; n D 1; 2; : : :

with eigenfunctions

vmn .x; y/ D Cmn sin�m�x

a

�sin�n�yb

�; m; n D 1; 2; : : : :

Summarising, we have the solutions

umn .x; y; t/ D Cmne��2

�m2

a2 Cn2

b2

�t sin

�m�xa

�sin�n�yb

�;

which vanish on the boundary of the rectangle. To match the initial condition we superposethe functions umn:

u .x; y; t/ D1X

m;nD1Cmne

��2�

m2

a2 Cn2

b2

�t sin

�m�xa

�sin�n�yb

�(1.38)

and impose1X

m;nD1Cmn sin

�m�xa

�sin�n�yb

�D g .x; y/ :

If we assume g can be expanded in double sines-Fourier series, it suffices that the Cmnequal the corresponding Fourier coefficients of g:

Cmn D 4

ab

Z a

0

Z b

0

sin�m�x

a

�sin�n�yb

�g .x; y/ dxdy.


As usual, if g is smooth enough, for instance of classC 1 on the closed rectangle, the seriesis uniformly convergent, and the fast convergence to zero of the exponentials ensures that(1.38) solve the problem.

Problem 1.2.24 (Fourier transform on the half-plane). Let g D g .x; y/ W R �Œ0;C1/ ! R be continuous and bounded. Using Fourier transforms solve the fol-lowing Dirichlet problem on S D R � .0;C1/ � .0;C1/:8<

:ut .x; y; t/ ��u.x; y; t/ D 0 x 2 R; y > 0; t > 0

u.x; y; 0/ D g.x; y/ x 2 R; y > 0

u.x; 0; t/ D 0 x 2 R; t > 0:

Solution. Denote bybu .�; y; t/ D RR u .x; y; t/ e

�ix�d� the partial Fourier transformof u in x. Since the transform of uxx .x; y; t/ is ��2bu .�; y; t/,bu .�; �; �/ satisfies (formally)the Cauchy problem on the quadrant y > 0, t > 0;8<

:but �buyy C �2bu D 0 y > 0; t > 0bu .�; y; 0/ Dbg .�; y/ y > 0bu .�; 0; t/ D 0 t > 0;

where � 2 R. We eliminate the reaction term by setting v .�; y; t/ D e�2tbu .�; y; t/; the

function v solves vt � vyy D 0 with the same initial and boundary data. The reflectionmethod used in Problem 1.2.15 on page 29 gives:

v .�; y; t/ DZ C1

0

Œ�1.y � z; t/ � �1.y C z; t/�bg.�; z/ dzand then

u .�; y; t/ D e��2t

Z C1

0

Œ�1.y � z; t/ � �1.y C z; t/�bg.�; z/ dzwhere �1 .y; t/ is the fundamental solution for the operator @t � @yy . Note the inverse

transform of e��2tbg.�; y/ isZ C1

�1�1 .x � w; t/ g .w; y/ dw

and that

�1 .x; t/ �1 .y; t/ D 1

2p�te�x2

4t1

2p�te�

y2

4t

D 1

4�te�

x2Cy2

4t D �1.x; y; t/

46 1 Diffusion

where �1.x; y; t/ is the fundamental solution for the operator @t � .@xx C @yy/. The finalformula reads:

u .x; y; t/ DZ C1

�1

Z C1

0

Œ�1.x � w; y � z; t/ � �1.x � w; y C z; t/�g.w; z/ dzdw:

As g is continuous and bounded, the study of the solution goes as in Problem 1.2.15. Inparticular, for any x0 2 R and y0 > 0, if .x; y; t/ ! .x0; y0; 0/

u .x; y; t/ ! g .x0; y0/ ;

so u is continuous on the closure of S except possibly for the half-plane y D 0. Continuityalong y D 0 holds precisely if g.x; 0/ D 0.

Problem 1.2.25 (Dirichlet in the ball). Let BR be a ball of radius R in R3, made of ahomogeneous material and at temperature U > 0 (constant) at time t D 0. Describehow the temperature of BR evolves at any of its points, in case the temperature on thesurface is kept constant and equal 0. Verify that the temperature at the centre of the balltends to zero exponentially as t ! C1.

Solution. The problem is invariant under rotations, so the temperature depends onlyupon time and the distance from the centre, which we take to be the origin: u D u .r; t/

with r D jxj. The equation for u is9:

ut ��u D ut � .urr C 2

rur / D 0; 0 < r < R; t > 0

with conditions

u .r; 0/ D U , 0 � r < R

u .R; t/ D 0, ju .0; t/j < 1; t > 0.

Using the identity

urr C 2

rur D 1

r.ru/rr (1.39)

we can set v D ru and write for v a one-dimensional problem:8<:vt � vrr D 0 0 < r < R; t > 0

v .r; 0/ D rU 0 � r < R

v .R; t/ D v .0; t/ D 0 t > 0:

Recalling Problem 1.2.1 on page 3, we have

v .r; t/ D1XkD1

ck exp

�k

2�2

R2t

�sin

k�r

R

9 For the expression of the Laplace operator � in spherical coordinates see Appendix B.


where

ck D 2U

R

Z R

0

r sink�r

Rdr D 2RU

�k.�1/kC1 :

Finally:

u .r; t/ D 2RU

�r

1XkD1

.�1/kC1k

exp

�k

2�2

R2t

�sin

k�r

R.

The analysis follows the lines of the one carried out for Problem 1.2.1. In particular, bythe maximum principle, u .r; t/ � 0; moreover, if t > 0, we can take the limit as r ! 0

and find the temperature in the middle:

0 � u .0; t/ D 2U

1XkD1

.�1/kC1 exp

�k

2�2

R2t

�� 2U exp

��

2

R2t

�;

since the series is alternating with decreasing terms. We deduce that u .0; t/ ! 0 expo-nentially as t ! C1.

Problem 1.2.26 (Dirichlet in a cylinder). Determine the temperature u inside the cylin-der

C D ®.x; y; z/ W r2 � x2 C y2 < R2; 0 < z < b

¯;

knowing that the surface is kept at temperature u D 0, and the initial temperature isg D g .r; z/.

Solution. The problem to solve has axial symmetry and cylindrical coordinates are themost natural ones. Hence let u D u .r; z; t/; in these coordinates we have10

�u D urr C 1

rur C uzz :

We have to find a bounded u such that:8<:ut �D.urr C 1

rur C uzz/ D 0 0 < r < R; 0 < z < b; t > 0

u.r; z; 0/ D g.r; z/ 0 < r < R; 0 < z < b;

u.r; z; t/ D 0 if r D R; z D 0; or z D b; t > 0:

We seek solutions u .r; z; t/ D v .r; z/w .t/, that satisfy the Dirichlet conditions. Substi-tuting into the differential equation, with the usual procedure we are led to the followingequation for w,

w0 .t/ D D�w .t/ ;

whence w .t/ D ce�Dt , and to the eigenvalue problem´vrr C 1

rvr C vzz D �v 0 < r < R; 0 < z < b;

v.r; z/ D 0 if r D R; z D 0 or z D b;(1.40)

10 Appendix B.

48 1 Diffusion

where v must additionally be bounded. We solve also this problem by separation of vari-ables, setting v .r; z/ D h .r/Z .z/. Substituting gives:�

h00 .r/C 1rh0 .r/ � �h .r/�h .r/

D �Z00 .z/Z .z/

D (constant)

from which the eigenvalue problem:

Z00 C Z D 0; Z .0/ D Z .b/ D 0 (1.41)

arises. Setting D �� , we find for h the equation

h00 .r/C 1

rh0 .r/C h .r/ D 0, h .R/ D 0; h bounded. (1.42)

Problem (1.41) is solved by

Zm .z/ D Am sin�m�zb

�; m D m2�2

b2; m D 1; 2; : : : :

Problem (1.42) has non-trivial solutions only if D 2 > 0. This is clear by multiplyingby rh and integrating by parts on .0;R/; this gives, in fact:

0 DZ R

0

.rhh00 C h0h/dr C

Z R

0

h2dr D Œrhh0�R0 �Z R

0

r.h0/2dr C

Z R

0

h2dr

and then

DZ R

0

r.h0/2dr=Z R

0

h2dr > 0:

The equation of problem (1.42) is thus a Bessel equation of order zero,11 and its onlybounded solutions are of the type

h .r/ D J0 .r/

where D 2, > 0, and

J0 .s/ D1XkD0

.�1/k.2kkŠ/2

s2k D 1 � s2

4C s4

32� � � �

is the Bessel function of order zero.For the condition h .R/ D 0 to hold we must require

J0 .R/ D 0:

Since J0 has infinitely many simple zeroes s1 < s2 < � � � < sn < � � � in .0;1/, we obtaininfinitely many eigenvalues

n D 2n D s2n=R2;

11 Appendix A.


10 20 30 40 50

�0:5

0:5

1

s

J0.s/

Fig. 1.10 Graph of the Bessel function s 7! J0.s/

with eigenfunctionshn .r/ D J0

�snrR

�.

Recalling that � D � � , we get 12-many eigenvalues

�nm D � s2nR2

� m2�2

b2

for problem (1.40), with eigenfunctions

vmn .r; t/ D cmn sin�m�zb

�J0

�snrR

�m; n D 1; 2; : : :

Now since w .t/ D ce�Dt , we can construct a candidate solution as

u .r; z; t/ D1X

m;nD1cmn sin

�m�zb

�J0

�snrR

�exp

��D

s2nR2

C m2�2

b2

�t

:

The coefficients cmn should be chosen so that

1Xm;nD1

cmn sin�m�zb

�J0

�snrR

�D g .r; z/ . (1.43)

To find them, we remind that the functions

'n .r/ D J0

�snrR

�have the following orthogonality properties12Z R

0

r'n .r/ 'm .r/ dr D´0 m ¤ nR2

2J1 .sn/

2 m D n:

12 Appendix A.

50 1 Diffusion

Let us multiply (1.43) by

rJ0 .skr=R/ sin .j�z=b/

and integrate over .0;R/� .0; b/. Keeping in mind of the orthogonality of Bessel’s func-tions and the trigonometric functions we obtain:

cjk D 4

bR2J1 .sk/2

Z R

0

Z b

0

r sin

�j�z

b

J0

�skrR

�g .r; z/ drdz:

The formal expression thus found is a solution provided g is smooth enough. Uniquenesscan be proved using the energy method.

1.3 Further Exercises

1.3.1. Let u denote the temperature of a homogeneous bar of density � and constant (small)cross-section, placed along the interval 0 � x � L: Set u .x; 0/ D g .x/. Suppose:

i) The lateral surface is not thermally insulated, and exchanges heat with the ambient. The latter isat temperature � , and obeys Newton’s law of cooling.

ii) The bar ends are insulated.

iii) The bar is heated by an electric current of intensity I .

Write down the mathematical model describing the evolution of u for t > 0.

1.3.2. A pipe of constant cross-section S and length L is filled with a homogeneous substanceof constant porosity ˛ (the ratio of volume of the pores over the total volume). Inside the pipe is agas, of concentration u D u .x; t/ ; 0 < x < L, that diffuses under Nerst’s law:

Q D �Dux

where Q D Q.x; t/ is the quantity of gas crossing from left to right the surface element S per unitof time, at the point x and at time t . The walls are leakproof. Write the mathematical model for uwhen t > 0, in the following cases: u .x; 0/ D g .x/ and:

a) Beginning at t D 0, at x D 0 the gas concentration is c D c .t/ ; while at x D L there is no flux.

b) A constant gaseous inflow c0 is maintained at x D 0, while a porous membrane diaphragm atx D L lets the gas seep in accordance with Newton’s law. There is no gas in the ambient.

1.3.3. LetD > 0 be a constant and g 2 C 1 .Œ0; ��/. Solve by separation of variables the follow-ing mixed problem: 8<

:ut .x; t/ �Duxx.x; t/ D 0 0 < x < �; t > 0

u.x; 0/ D 2 sin .3x=2/ 0 � x � �

u.0; t/ D ux.�; t/ D 0 t > 0:

(1.44)

Find a solution formula with a general initial profile u .x; 0/ D g .x/ :

1.3 Further Exercises 51

1.3.4. Let D > 0; h > 0 be constants, g 2 C 1 .Œ0; ��/. Solve by separation of variables themixed problem: 8<

ˆ:ut .x; t/ �Duxx.x; t/ D 0 0 < x < L; t > 0

u.x; 0/ D U 0 � x � L

u.0; t/ D 0 t > 0

ux.L; t/C hu.L; t/ D 0 t > 0:

1.3.5. Let u be a solution tout D Duxx C bux C cu:

a) Determine h and k so that the function

v .x; t/ D u .x; t/ ehxCkt

solves vt �Dvxx D 0.

b) Write the formula for the global Cauchy problem with initial datum u .x; 0/ D u0 .x/.

1.3.6. Solve the following Dirichlet problem using separation of variables:8<:ut D uxx CmuC sin 2�x C 2 sin 3�x 0 < x < 1; t > 0

u .x; 0/ D 0 0 < x < 1

u .0; t/ D u .1; t/ D 0 t > 0:

1.3.7. (Periodicity at one end) Write the formal solution to the problem:´ut D Duxx 0 < x < 1;�1 < t < �1u .0; t/ D 0; u .1; t/ D p .t/ �1 < t < �1

where p is of class C 1 .R/ and T -periodic: p .t C T / D p .t/ : Study the case D D 1; p .t/ Dcos 2t .

1.3.8. Let g be bounded (jg .x/j � M for any x 2 R) and

u .x; t/ DZ

R�D .x � y; t/ g .y/ dy:

Discusslim

.x;t/!.x0;0/u .x; t/

in the case g has a jump discontinuity at x0:

1.3.9. Solve the global Cauchy problem with the characteristic function of the interval .0; 1/ asinitial datum. Study the limit of the solution as .x; t/ tends to .x0; 0/.

1.3.10. Adapting the reflection method used in Problem 1.2.15 on page 29, find a formula for thesolution to 8<

:ut �Duxx D 0 0 < x < L; t > 0

u .x; 0/ D g .x/ 0 � x � L

u.0; t/ D u.L; t/ D 0 t > 0

(1.45)

with g continuous on Œ0; L�, g .0/ D g .L/ D 0.

52 1 Diffusion

1.3.11. Find the bounded solution of the problem

ut � uxx D 0 on the quadrant x > 0; t > 0;

satisfying the following conditions:

a) u .0; t/ D 0, u .x; 0/ D U .b) u .0; t/ D U , u .x; 0/ D 0.c) u .0; t/ D 0,

u .x; 0/ D´0 0 < x < L

1 x > L:

1.3.12. Referring to Problem 1.2.14 on page 27:

a) compute the probability that the particle reaches L before the instant t , i.e.13

F .L; t/ D Prob ¹TL � tº :b) Determine the probability that the particle is absorbed at x D L in the time interval .t; t C dt/.

1.3.13. (Reflecting barriers and Neumann conditions) Refer to the random walk of Problem 1.2.14(page 27) and consider the situation where a reflecting barrier is placed atL D mh: then the particle,at L � h

2 at time t and moving rightwards, is reflected and returns to L � h2 at time t C � .

a) Which problem does the transition probability p D p .x; t/ solve, when passing to the limith; � ! 0, with h2=� D 2D?

b) Find the explicit expression for p.

1.3.14. (Duhamel’s principle) Consider the Dirichlet problem8<:ut .x; t/ �Duxx.x; t/ D f .x; t/ 0 < x < L; t > 0

u.x; 0/ D 0 0 � x � L

u.0; t/ D u.L; t/ D 0 t > 0:

(1.46)

Let f be continuous for 0 � x � L; t � 0. Show that if v.x; t I �/, t � � � 0, is the solution to8<:vt .x; t I �/ �Dvxx.x; t I �/ D 0 0 < x < L; t > �

v.x; � I �/ D f .x; �/ 0 � x � L

v.0; t I �/ D v.L; t I �/ D 0 t > 0

(1.47)

then (1.29) is solved by

u.x; t/ DZ t

0v.x; t I �/ d�:

Determine an explicit formula for u.

1.3.15. A round wire with constant cross-section S and length L, (centred at the origin) withgiven initial temperature, is heated by a sine-wave current of intensity I .t/. Write the mathematicalmodel for the temperature u and find the expression of u.

13 F .L; t/ D Prob ¹TL � tº is called (probability) distribution for the variable TL: The derivativeFt is the (associated probability) density.


1.3.16. (Neumann condition on the half-line; Fourier cosine transform) a) Taking into accountProblem 1.2.20 on page 38, find a formula for a bounded solution to8<

:ut .x; t/ � uxx.x; t/ D 0 x > 0; t > 0

u.x; 0/ D 0 x � 0

ux.0; t/ D g .t/ t > 0;

(1.48)

where g is continuous and bounded. Prove that there is only one bounded solution.b) Prove that dropping the assumption on boundedness problem (1.48) does not, in general, guar-

antee uniqueness of the solution; you can use

w1 .x; t/ D ex sin .2t C x/ and w2 .x; t/ D �e�x sin .2t � x/ :

1.3.17. (Drift variable) Using Fourier transforms solve:´ut D uxx C xux x 2 R; t > 0

u .x; 0/ D g .x/ x 2 R;

where g is continuous and L2 on R. Examine the effect of the transport term when g .x/ Dı .x � x0/ (use the method of characteristics, Chap. 3).

1.3.18. The initial temperature of a semi-infinite homogeneous bar with small cross-section iszero. Determine a formula for the temperature knowing that

u .0C; t / D ı .t/ ; u .1; t / D 0;

where ı .t/ denotes Dirac’s distribution centred at the origin. Use the Laplace transform.

1.3.19. (Neumann on the ball) Let BR be a ball of radius R in R3 made of a homogeneous ma-terial. Describe how the temperature of BR evolves at any of its points, in case the heat quantity q(constant) flows across the surface and the initial temperature is qr , with r being the distance of thepoint from the centre.

1.3.20. (Fourier transform and fundamental solution) Using Fourier transforms in x, recover theformula for the global Cauchy problem in dimension n:´

ut �D�u D f .x; t / x 2 Rn; t > 0

u .x; 0/ D g .x/ x 2 Rn:

Assuming g and f are in L2 .Rn/ and L2�RnC1

�respectively, explain in which sense the initial

condition is attained.

1.3.21. Determine the temperature u D u .x; y; z; t/ in the region between the parallel planesz D 0 and z D 1, knowing that u D 0 on the planes, and initially

u.x; y; z; 0/ D g .x; y; z/ :

1.3.22. (Maximum principle) Consider in RnC1 the cylinder

QT D � � .0; T / ;

54 1 Diffusion

where � � Rn is a bounded domain with parabolic boundary

@pQT D .@� � Œ0; T �/ [ .� � ¹0º/ :

Leta D a.x; t /; b D b.x; t / 2 Rn; c D c.x; t /

be continuous on QT and such thata.x; t / � a0 > 0:

Consider a function u 2 C 2;1.QT / \ C.QT / satisfying

Lu D ut � a�uC b � ruC cu � 0 (resp. � 0) on QT :

a) Show that if c.x; t / � 0 and u has a positive maximum (or negative minimum), then this valueis taken on the parabolic boundary:

maxQT

u D M > 0 ) max@pQT

u D M:

(resp. minQT

u D m < 0 H) min@pQTu D m.)

Deduce that if u � 0 on @pQT (or u � 0) then u � 0 in QT (resp. u � 0).b) Deduce that if g is continuous on @pQT , the Dirichlet problem´

Lu D 0 in QTu D g on @pQT

has a unique solution in C 2;1.QT / \ C.QT /.c) Show that part b) holds even when the assumption c .x; t / � 0 is replaced by the more general

requirement jc.x; t/j � M .

1.3.23. Provide an explicit formula for the solution to global Cauchy problem (in R3):´ut .x; t / D a.t/�u.x; t /C b.t/ � ru.x; t /C c.t/u.x; t / x 2 R3; t > 0

u.x; 0/ D g.x/ x 2 R3;

where a, b, c and f are continuous and a.t/ � a0 > 0.

1.3.24. Answer the following questions.

a) Let u solve ut � uxx D �1, in 0 < x < 1; t > 0, with

u .x; 0/ D 0 and u .0; t/ D u .1; t/ D sin�t:

Can there exist a point x0, 0 < x0 < 1 such that u .x0; 1/ D 1‹

b) Establish whether there exists a solution to8<:ut C uxx D 0 �1 < x < 1; 0 < t < Tu .x; 0/ D jxj �1 < x < 1u .0; t/ D u .1; t/ D 0 0 < t < T:


c) Verify that the function

u .x; t/ D @x�1 .x; t/

solves ´ut � uxx D 0 x 2 R; t > 0

u.x; 0/ D 0 x 2 R

and that u .x; t/ ! 0 if t ! 0, for any x given. Does this contradict the uniqueness of thesolution to the global Cauchy problem?

d) Let u D u .x; t/ be a continuous solution to the Robin problem8<:ut � uxx D 0 0 < x < 1; 0 < t < T

u .x; 0/ D sin�x 0 � x � 1

�ux .0; t/ D ux .1; t/ D �hu, h > 0 0 � t � T:

Show that u cannot have a negative minimum. What is the maximum of u?

1.3.25. (Evolution of a chemical solution) Consider a tube of lengthL and constant cross-sectionA, where x is the symmetry axis. The tube contains a saline solution of concentration c. Let A besmall enough so that we can assume that the concentration c depends only on x and t , so that thediffusion of salt can be thought of as one-dimensional, along x. Let also the fluid speed be negligible.

From the left end of the pipe, at x D 0, a solution of constant concentration C0 enters at a rateof R0 cm3/s, while at the other end x D L the solution is removed at the same speed.

Using Fick’s law show that c solves a diffusion Neumann-Robin problem. Then find the explicitsolution and verify that, for t ! C1, c .x; t/ tends to a steady state.

1.3.26. (Random particle subject to an elastic force) The 1-dimensional motion of a random par-ticle is subject to the following rules. Let N be a natural integer.

1. At each time step � the particle takes one step of h units of length, starting from x D 0:

2. If the particle is at the point mh, �N � m � N , it moves to the right, or to the left, withprobability

p D 1

2

�1 � m

N

�or q D 1

2

�1C m

N

�independently of the previous step.

Prove that if h2=� D 2D and N� D > 0, when h; � ! 0 and N ! 1, the limit transitionprobability p .x; t/ is a solution of the equation

pt D Dpxx C 1

.xp/x :

Give a formula for the solution with initial data p .x; 0/ D ı .x/ when D D D 1.

1.3.27. Solve the following initial-value/Neumann problem in the unit ball B1 D ¹x 2 R3 Wjxj < 1º: 8<

:ut D �u x 2 B1; t > 0u .x; 0/ D jxj x 2 B1u� .� ; t / D 1 � 2 @B1; t > 0:

56 1 Diffusion

1.3.28. Solve the following non-homogeneous initial-value/Dirichlet problem in B1 (u Du .r; t/ ; r D jxj): 8<

:ut �

�urr C 2

rur

D qe�t 0 < r < 1; t > 0

u .r; 0/ D U 0 � r � 1

u .1; t/ D 0 t > 0:

.

1.3.29. (An . . . invasion problem) A population of density P D P .x; y; t/ and total mass M .t/

is initially (t D 0) concentrated at a single point on the plane (say, the origin .0; 0/). It grows linearlyat a rate a > 0 and spreads with diffusion constant D.

a) Write the problem governing the evolution of P , then solve it.b) Determine the evolution of the mass M .t/ D R

R2 P .x; y; t/ dxdy.c) Let BR be the circle centred at .0; 0/ with radius R. Determine R D R .t/ so thatZ

R2nBR.t/

P .x; y; t/ dxdy D M .0/ :

d) Call metropolitan area the region BR.t/ and rural area the region R2nBR.t/. Determine thevelocity of the metropolitan advancing front.

1.3.1 Solutions

Solution 1.3.1. The temperature solves the following Neumann problem:8<ˆ:ut D �

�cvuxx � ˇ

�cv.u � �/C

I2R

�cv0 < x < L; t > 0

ux .0; t/ D ux .L; t/ D 0 t > 0

u .x; 0/ D g .x/ 0 � x � L;

where the term I2Rc is due to the heat produced by the current and � ˇ

c .u � �/ comes from theheat exchange with the surrounding ambient (Newton’s law).

Solution 1.3.2. As the pipe is insulated, in either case the concentration of the gas satisfies thediffusion equation

ut D auxx (1.49)

in 0 < x < L, t > 0, with a D D=˛. In fact, the gas found at time t between x and x C�x isZ xCx

x˛u .x; t/ dx;

so the conservation of mass gives:Z xCx

x˛ut .x; t/ dx D �D Œux .x; t/ � ux .x C�x; t/�

where we kept Nerst’s law into account. Dividing by �x and passing to the limit as �x ! 0 gives(1.49). Let us consider the ends conditions.


a) We have a Dirichlet condition at x D 0 Wu .0; t/ D c .t/

and a homogeneous Neumann condition at x D L Wux .L; t/ D 0:

b) Here we have a non-homogeneous Neumann condition at x D 0 W�Dux .0; t/ D c0

and a Robin condition at x D L WDux .L; t/ D �hu .L; t/ :

Solution 1.3.3. The solution is (Fig. 1.11)

u .x; t/ D 2e� 9D4

2t sin

�3

2x

: (1.50)

In fact, writing u .x; t/ D v .x/w .t/ leads to the equations

w00 � �Dw D 0

solved byw.t/ D Ce�Dt ; C 2 R,

and the eigenvalue problemv00.x/ � �v.x/ D 0

with mixed conditionsv.0/ D v0.�/ D 0:

0

�=3

2�=3

� 02

46

8

�2

0

2

x t

Fig. 1.11 u .x; t/ D 2e�t=4 sin�32x�

58 1 Diffusion

In case � D 2 � 0 we have only the zero solution. If � D �2 < 0 the eigenvalues are

�k D ��.2k C 1/

2

2k D 0; 1; : : :

with eigenfunctions

vk .x/ D sin

�.2k C 1/

2x

:

Therefore we have infinitely many solutions

uk .x; t/ D ck sin

�.2k C 1/

2x

e��

.2kC1/2

�2Dt;

fulfilling the mixed conditions at the endpoints. For c1 D 2, u1 satisfies the initial condition as well.This gives (1.50).

Taking u .x; 0/ D g .x/ as initial datum produces as (formal) solution:

u .x; t/ D1XkD0

ck sin

�.2k C 1/

2x

e��

.2kC1/2

�2Dt

where

ck D 2

�

Z �

0g .x/ sin

�.2k C 1/

2x

dx

are the Fourier coefficients of g with respect to the family ¹vkº :

Solution 1.3.4. The solution is

u .x; t/ D U

1XkD1

cke��2

kDt sinkx

where k are the positive solutions to

h tanL D �and

ck D coskL � 1˛kk

(1.51)

with

˛k D L

2� sin .2kL/

4k. (1.52)

In fact, by setting u .x; t/ D v .x/w .t/ we are led to

w00 � �Dw D 0

solved byw.t/ D Ce�Dt ; C 2 R,

and to the Sturm-Liouville problem

v00.x/ � �v.x/ D 0

with mixed conditionsv.0/ D v0.L/C hv .L/ D 0:


If � D 2 � 0 the only solution is the trivial one, If � D �2 < 0 we obtain

v .x/ D C1 cosx C C2 sinx;

and sincev0 .x/ D �C1 sinx C C2 cosx;

the mixed conditions read ´C1 D 0

. cosLC h sinL/C2 D 0.

The eigenvalues are therefore the positive solutions k , k � 1, to the equation

h tanL D �while the eigenfunctions are

vk .x/ D sinkx.

Equations (1.51) and (1.52) follow from the fact thatZ L

0sinkx dx D coskL � 1

kand

Z L

0sin2 kx dx D ˛k .

Solution 1.3.5. a) We have

vt D Œut C ku�ehxCkt

vx D Œux C hu�ehxCkt vxx D Œuxx C 2hux C h2u�ehxCkt

and hence, by ut D Duxx C bux C cu;

vt �Dvxx D ehxCkt Œut �Duxx � 2Dhux C .k �Dh2/u� DD ehxCkt Œ.b � 2Dh/ux C .k �Dh2 C c/u�:

If we chooseh D b

2Dk D b2

4D� c

the function v solves the heat equation vt �Dvxx D 0.

b) The formula is (Fig. 1.12)

u .x; t/ D e

�c� b2

4D

�tZ

Re

b2D .y�x/�D .y � x; t/ u0 .y/ dy:

Solution 1.3.6. Using the previous exercise we may simplify, and set

w .x; t/ D u .x; t/ e�mt :The function w satisfies

wt �wxx D e�mt Œsin 2�x C 2 sin 3�x�

with homogeneous Dirichlet conditions. Define w .x; t/ D v .x/ z .t/. The eigenfunctions associ-ated to the Dirichlet problem are

vk .x/ D sin k�x; k D 1; 2; : : :

with eigenvalues �k D �k2�2. The right-hand side of the differential equation has the form

e�mt Œv2 .x/C v3 .x/�;

60 1 Diffusion

�20

24

68

2

4

6

0

10

20

x

t

Fig. 1.12 Exercise 1.3.5: the function .x; t/ 7! 30e2x�t�1=4 .x; t/ (t > 0:6)

whence the candidate solution will have the form

w .x; t/ D c2 .t/ sin 2�x C 2c3 .t/ sin 3�x;

due to the homogeneous Dirichlet conditions. The cj .t/, j D 1; 2, are determined imposingcj .0/ D 0 and

wt � wxx D Œc0j .t/C j 2�2cj .t/� sin j�x D e�mt sin j�x;

i.e.c0j .t/C j 2�2cj .t/ D e�mt ; cj .0/ D 0:

Then, if m ¤ j 2�2, we find

cj .t/ D 1

j 2�2 �m�e�mt � e�j2�2t

�;

while, if m D j 2�2;

cj .t/ D te�j2�2t :

The solution is therefore (Fig. 1.13)

u .x; t/ D emt Œc2 .t/ sin 2�x C 2c3 .t/ sin 3�x�:

Solution 1.3.7. Let us expand p in Fourier series; it is convenient to use the complex form:

p .t/ DC1XnD�1

pn exp

�2�int

T

where

pn D 1

T

Z T

0p .t/ exp

�2�int

T

dt:

We recall that since p is real-valued, we have

p�n D pn:


00:2

0:40:6

0:81 0

0:1

0:2

0:3

0

5 � 10�2

xt

Fig. 1.13 The solution to Exercise 1.3.6; m D 0:5; 0 < t < 0:3. Reaction, source and diffusionbalance each other

It is reasonable to assume u is T -periodic in t , so we set

u .x; t/ DC1XnD�1

pnun .x/ exp

�2�int

T

and we set out to find the coefficients un from:

ut �Duxx DC1XnD�1

�2�in

T

un .x/ �Du00n .x/

�exp

�2�int

T

D 0;

whence

u00n .x/ � 2�in

DTun .x/ D 0; n D 0;˙1;˙2; : : : (1.53)

andun .0/ D 0; un .1/ D 1: (1.54)

As u is real, too, we haveu�n .x/ D un .x/

so it suffices to determine un for n D 0; 1; 2; : : : With n D 0, we have

u0 .x/ D x:

With n > 0, the general solution of (1.53) is:

un .x/ D an exp ¹cn .1C i/ xº C bn exp ¹�cn .1C i/ xº ; cn Dr�n

DT:

Equations (1.54) hold if:

an C bn D 0; an exp ¹cn .1C i/º C bn exp ¹�cn .1C i/º D 1

which give, after elementary manipulations:

an D �bn D 1

2i sin Œcn .1C i/�:

62 1 Diffusion

00:2

0:40:6

0:81 0

12

34

56

�2

0

2

xt

uDu.x;t/

Fig. 1.14 The solution to Exercise 1.3.7

Finally, returning to real functions:

u .x; t/ D p0x C1XnD1

Re

²pn

sin Œcn .1C i/� x

sin Œcn .1C i/�exp

��2�int

T

³:

In case

p .t/ D cos 2t D eit C e�it2

we have T D �; cn D pn;

p1 D p�1 D 1

2, pn D 0 if n ¤ ˙1

so

a1 D �b1 D 1

4i sin .1C i/, an D bn D 0 if n ¤ 1:

Then:

u1 .x/ D 1

4i sin .1C i/

he.1Ci/x � e�.1Ci/x

iD sin Œ.1C i/ x�

2 sin .1C i/

and (see Fig. 1.14)

u .x; t/ D u1 .x/ e�2it C u1 .x/e

2it D 2Rehu1 .x/ e

�2it i D Re

sin Œ.1C i/ x�

sin .1C i/exp .�2it/

�:

Solution 1.3.8. Up to translating the x-axis we may assume x0 D 0. Let us set lC D g .0C/ andl� D g .0�/, and introduce the extensions

gC .x/ D´g .x/ x > 0

lC x � 0;g� .x/ D

´l� x > 0

g .x/ x � 0:

The maps gC and g� are continuous on R. Observe that we can write

u .x; t/ DZ 0

�1�D .x � y; t/ g .y/ dx C

Z C1

0�D .x � y; t/ g .y/ dy � uC .x; t/C u� .x; t/


and

uC .x; t/ DZ

R�D .x � y; t/ gC .y/ dy � lC

Z C1

0�D .x � y; t/ dy

u� .x; t/ DZ

R�D .x � y; t/ g� .y/ dy � l�

Z 0

�1�D .x � y; t/ dy:

From Problem 1.2.18 (page 36) we know that if .x; t/ ! .0; 0/ then:ZR�D .x � y; t/ gC .y/ dy ! lC;

ZR�D .x � y; t/ g� .y/ dy ! l�:

Thus, it is enough to examine the limits ofZ C1

0�D .x � y; t/ dy and

Z 0

�1�D .x � y; t/ dy:

We haveZ C1

0�D .x � y; t/ dy D 1p

4D�t

Z C1

0e�

.x�y/2

4Dt dy D 1p�

Z C1

� xp4Dt

e�z2

dz

Z 0

�1�D .x � y; t/ dy D 1p

�

Z � xp4Dt

�1e�z2

dz

from which we deduce the following facts:a) As the limit of x=

pt for .x; t/ ! .0; 0/ does not exist, neither does the limit of uC. Similarly,

the limits of u� and u do not exist.b) If x D o .t/ (this implies, in particular, that .x; t/ ! .0; 0/ tangentially along the t -axis) then

uC ! lC=2, u� ! l�=2 and u ! �lC C l�

�=2:

c) Take t D o .jxj/ (implying, in particular, .x; t/ ! .0; 0/ tangentially to the x-axis); if x > 0,then uC ! lC and u� ! 0, whereas if x < 0; then uC ! 0 and u� ! l�. In the former caseu ! lC, in the latter u ! l�.


u .x; t/ DZ 1

0�D .x � y; t/ dy:

Set .x; t/ ! .x0; 0/. From the previous exercise we infer:

jx0j > 1 H) u .x; t/ ! 0;

0 < x0 < 1 H) u .x; t/ ! 1;

and if x0 D 0 or x0 D 1, the limit does not exist.

Solution 1.3.10. With the given homogeneous Dirichlet conditions we extend g to Œ�L;L� to anodd map

Qg.x/ D´g.x/ 0 � x � L

�g.�x/ �L � x < 0(odd reflection).

Then we extendeg to R by setting it equal zero outside Œ�L;L�, and define

g� .x/ DC1XnD�1

eg .x � 2nL/ :

64 1 Diffusion

The map g� is continuous (for g .0/ D g .L/ D 0), periodic of period 2L and it coincides witheg onŒ�L;L�: Note that, for any x, only one summand survives. Let us solve the global Cauchy problemusing the formula

u .x; t/ DZ C1

�1�D .x � y; t/ g� .y/ dy (1.55)

DC1XnD�1

Z C1

�1�D .x � y; t/eg .y � 2nL/ dy:

We remind thateg .y � 2nL/ is zero outside the interval

.2n � 1/L � y � .2nC 1/L;

so we can write

u .x; t/ DC1XnD�1

Z .2nC1/L

.2n�1/L�D .x � y; t/eg .y � 2nL/ dy

D.y�2nL/�!y

C1XnD�1

Z L

�L�D .x � y � 2nL; t/eg .y/ dy

DC1XnD�1

Z L

0Œ�D .x � y � 2nL; t/ � �D .x C y � 2nL; t/�g .y/ dy

�Z L

0GD .x; y; t/ g .y/ dy

where

GD .x; y; t/ DC1XnD�1

Œ�D .x � y � 2nL; t/ � �D .x C y � 2nL; t/� :

The restriction of u to Œ0; L� is a solution of (1.27). Now let us examine the initial condition. From(1.28), u is certainly a solution of the diffusion equation; since g� is continuous, we have

u .x; t/ ! g� .x0/ D g .x0/ , if .x; t/ ! .x0; 0/

if x0 2 Œ0; L�. Let us verify the Dirichlet data at the endpoints. When t > 0 we can take the limitsas x ! 0C; L� inside the integral, so it will be enough to consider the Dirichlet conditions on thekernel GD ; after an easy change of index in the sum, we find:

GD .0; y; t/ DC1XnD�1

�D .y C 2nL; t/ � �D .y � 2nL; t/

DC1XnD�1

Œ�D .y C 2nL; t/ � �D .y C 2nL; t/�

D 0:


At x D L

DC1XnD�1

Œ�D .L � y � 2nL; t/ � �D .LC y � 2nL; t/�

DC1XnD�1

Œ�D .L � y � 2nL; t/ � �D .�LC y C 2nL; t/�

D 0:

This shows that the Dirichlet conditions hold and, by maximum principle, u is the unique solutionof problem (1.27).

Remark. The function GD D GD .x; y; t/ is called fundamental solution of the heat equation withDirichlet conditions for the interval Œ0; L�.

Solution 1.3.11. The problems of parts a) and b) are invariant under parabolic dilations, so it isreasonable to seek solutions of the form given in Problem 1.2.13 (page 26):

u .x; t/ D C1 C C2 erf

�x

2pt

.

Since erf .0/ D 0, erf .C1/ D 1, it is easy to check that the solution (the only bounded one) toproblem a) is

u .x; t/ D U erf

�x

2pt

:

Analogously, the only bounded solution to problem b) reads14:

u .x; t/ D U

1 � erf

�x

2pt

�:

For the solution to c), we use (1.25) in Problem 1.2.15 (page 29), which tells

u.x; t/ DZ C1

LŒ�1.x � y; t/ � �1.x C y; t/� dy

D 1p�

Z C1.L�x/

2p

t

e�z2

dz �Z C1

.LCx/

2p

t

e�z2

dy

�D

D 1

2

erf

.LC x/

2pt

� erf.L � x/2pt

�:

The graph of u is shown in Fig. 1.15.

Solution 1.3.12. a) From Problem 1.2.14 on page 27 the probability equals

F .L; t/ D 1 �Z L

�1pA .x; t/ dx D 1 � 1p

2�t

Z L

�Le� x2

2t dx

D 2p2�t

Z C1

Le�x2

2t dx DxDp2ty

2p�

Z C1Lp2t

e�y2

2 dy:

b) The required probability coincides with the probability that the particle first reaches x D L

at a time between t and t C dt: The latter event coincides with ¹t < TL � t C dtº, so if dt 0,

P ¹t < TL � t C dtº D F .L; t C dt/ � F .L; t/ ' Ft .L; t/dt:

14 One can also use the formula of Problem 1.2.20 (page 38).

66 1 Diffusion

0

0:5

1

1:5

20:5

11:5

22:5

3

0

0:5

1

x t

uDu.x;t/

Fig. 1.15 The function 12

�erf 1Cx

2pt

� erf 1�x2pt

�, 0 < x < 2, 0:05 < t < 3

Consequently

P ¹t < TL � t C dtº D Lp2�t3=2

e�L2

2t dt:

Remark. It might be instructive to see another method as well. Let us interpretF as the rate at whicha unit quantity of heat, initially concentrated at the origin, will propagate in presence of a heat sinkplaced at x D L: In this case the interpretation is simple: F is nothing but the outgoing heat flowper unit of time at the point L: Fourier’s law prescribes

F .L; t/ D �12pAx .L; t/ D Lp

2�t3=2e�L2

2t ;

in agreement with the result previously found.

Solution 1.3.13. a) The transition probability p D p .x; t/ solves

pt � 1

2pxx D 0

in .�1; L/ � .0;1/, and p .x; 0/ D ı in .�1; L/. Moreover, we must haveZ L

�1p .x; t/ dx D 1: (1.56)

To see what happens at x D L, let us use the reflection principle. By the latter, a particle has thesame probability of being at L � h=2, at time t C � , and at L � 3h=2. By the theorem of totalprobability we can write:

p

�L � 1

2h; t C �

D 1

2p

�L � 3

2h; t

C 1

2p

�L � 1

2h; t

: (1.57)

Now, since

p

�L � 1

2h; t C �

D p

�L � 1

2h; t

C pt

�L � 1

2h; t

� C o .�/


and

p

�L � 3

2h; t

D p

�L � 1

2h; t

� px

�L � 1

2h; t

hC o .h/ ;

substituting into (1.57) gives, after a few simplifications,

pt

�L � 1

2h; t

� C o .�/ D �px

�L � 1

2h; t

hC o .h/ :

Let us divide by h and take the limit h ! 0. Since h2=� D 1 we have �h

! 0, and then

px .L; t/ D 0 t > 0: (1.58)

This is a homogeneous Neumann condition. To sum up p solves, besides (1.56),8<:pt � 1

2pxx D 0 on � 1 < x < L; t > 0

p .x; 0/ D ı.x/ �1 < x < L

px .L; t/ D 0 t > 0:

b) We will use the method of images and place a second walk starting at 2L; the symmetricpoint to the origin with respect to L. By virtue of the heat equation’s linearity, we consider a linearcombination of the two fundamental solutions

pR .x; t/ D �D .x; t/C �D .x � 2L; t/ D �D .x; t/C �D .2L � x; t/ : (1.59)

The function pR thus defined is precisely the solution needed, because for �1 < x < L

pR .x; 0/ D �D .x; 0/C �D .2L � x; 0/ D ı.x/

and

pRx .x; t/ D 1p4�Dt

²� x

2Dte� x2

4Dt C 2L � x2Dt

e�.2L�x/2

4Dt

³;

sopx .L; t/ D 0

(which could have been inferred from the symmetry of pR, without computations). Finally,Z L

�1pR .x; t/ dx D

Z L

�1¹�D .x; t/C �D .2L � x; t/º dx D

letting 2L � x D z in the last integrand,

DZ L

�1�D .x; t/ dx C

Z C1

L�D .z; t/ dz D 1:

Hence also (1.56) holds.

Solution 1.3.14. As f is continuous for 0 � x � L; t � 0; from Exercise 1.3.10 we deduce thatthe solution to problem (1.30) is, for any � , 0 � � � t;

v .x; t I �/ DZ L

0GD .x; y; t � �/ f .y; �/ dy:

The latter is continuous on the same set. We also have, for any 0 < x < L; t > 0:

ut .x; t/ D v .x; t; t/CZ t

0vt .x; t I �/ d� D f .x; t/C

Z t

0vt .x; t I �/ d�

uxx .x; t/ DZ t

0vxx.x; t I �/ d�:

68 1 Diffusion

Hence

ut �Duxx D f .x; t/ 0 < x < L; t > 0

and

u .x; 0/ D 0:

We conclude that u solves (1.29). An explicit formula is:

u .x; t/ DZ t

0v.x; t I �/ d� D

Z t

0

Z L

0GD .x; y; t � �/ f .y; �/ dyd�:

An alternative expression comes from separating the variables. In fact, for v.x; t I �/ we find

v .x; t I �/ D1XkD1

fk .�/ e�k2�2.t��/=L sin

�k�

Lx

where

fk .�/ D 2

L

Z L

0f .y; �/ sin

�k�

Lx

dy

and then

u .x; t/ D1XkD1

sin

�k�

Lx

Z t

0fk .�/ e

�k2�2.t��/=Ld�:

Solution 1.3.15. Let R D L=2� be the radius and use coordinates � and t , then set u D u .�; t/.The temperature u varies with continuity, whence u .�; t/ is periodic in � , of period 2� . We isolate a(cylindrical) portion of the wire V; between � and �Cd� of length ds D Rd� . A current I D I .t/

generates a source of intensity I2, where depends on the physical characteristics of the wire(electric resistance, density, diffusivity). More precisely, I2Rd� represents the quantity of heatgenerated per unit of time in the infinitesimal line element between � and �Cd�: The conservationof energy tells

d

dt

ZVcv�u dv D

Z@V

�� duds

� � � d� C I2R d� (1.60)

where � is the unit tangent vector to the wire and � the unit normal to the boundary of the cylinder.Then:

d

dt

ZVcv�udv D cv�SR

Z �Cd�

�ut�� 0; t

�d� 0: (1.61)

Moreover, rrom � � � D 0 on the boundary of the wire d� D S at � , � C d� , and

du=ds D u�d�=ds D u�=R;

it follows that Z@V

� �

Ru�� d� D ��S

RŒu� .� C d�; t/ � u� .�; t/�: (1.62)

Comparing eqs. (1.60), (1.61) and (1.62), dividing by d� and passing to the limit as d� ! 0 pro-duces the equation

ut D Ku�� C f .t/ ; in 0 < � < 2�; t > 0


with

K D �

cv�R2; f .t/ D I2

cv�S:

Moreover, ´u .�; 0/ D g .�/ 0 � � � 2�

u .0; t/ D u .2�; t/ ; u� .0; t/ D u� .2�; t/ t > 0;

where the initial datum g has period 2� .When searching for solutions of the type v .�/w .t/ we are lead to the following eigenvalue

problem:v00 C �v D 0; v .0/ D v .2�/ , v0 .0/ D v0 .2�/ ;

whose solutions are

vn .�/ D A cos n� C B sin n�; n D 0; 1; 2 : : : :

Therefore we look for a solution like:

u .�; t/ D1XnD0

ŒAn .t/ cosn� C Bn .t/ sinn��

and we impose1XkD0

hA0n .t/C n2KAn .t/

icosn� C

hB 0n .t/C n2Bn .t/

isinn� D f .t/

withAn .0/ D an; Bn .0/ D bn; n D 0; 1; 2 : : :

where an and bn are the Fourier coefficients of g. Necessarily

A00 .t/ D f .t/ , A0 .0/ D a0

2

and when n > 0,A0n .t/C n2KAn .t/ D 0, An .0/ D an

B 0n .t/C n2KBn .t/ D 0, Bn .0/ D bn:

This gives:

A0 .t/ D a0

2CZ t

0f .s/ ds, An .t/ D ane

�n2Kt , Bn .t/ D bne�n2Kt :

Solution 1.3.16. a) We use the Fourier cosine transform in x, defined by

C .u/ .�; t/ D U .�; t/ D 2

�

Z 1

0u .x; t/ cos .�x/ dx;

whose inverse formula is given by

u .x; t/ DZ 1

0U .�; t/ cos .�x/ d�:

Notice thatU is even in � . For functions vanishing with their first x-derivatives, as x goes to infinity,we have

C .uxx/ .�; t/ D � 2

�ux .0; t/ � �2U .�; t /:

Then we look for U solving´Ut .�; t/C �2U .�; t/ D � 2

� g .t/ � > 0; t > 0

U.�; 0/ D 0 � � 0:

70 1 Diffusion

Hence

U .�; t/ D � 2

�

Z t

0e��2.t�s/g .s/ ds:

Transforming back:

u .x; t/ DZ 1

0U .�; t/ cos.�x/d� D � 2

�

Z t

0g .s/

Z 1

0e��2.t�s/ cos.�x/d�

�ds

D � 1p�

Z t

0

g .s/pt � s e

� x2

4.t�s/ ds

D �2Z t

0� .x; t � s/ g .s/ ds

where

� .x; t/ D 1p4�t

exp

�x

2

4t

!is the fundamental solution for the operator @t � @xx .

b) The proof of non-uniqueness is carried out, with minimal variations, as in Problem 1.2.20(page 38). The details are left to the reader.

Solution 1.3.17. Let us denote with

bu .�; t/ DZ

Ru .x; t/ e�ix�dx

the partial Fourier transform of u. Recalling that the transform of xux is ��bu��bu, thenbu (formally)satisfies the Cauchy problem´but C �bu� D �.�2 C 1/bu �1 < � < 1; t > 0bu .�; 0/ Dbg .�/ �1 < � < 1:

The differential equation is linear, non-homogeneous and of order one. The characteristic curves(see Chap. 3) have parametric equations

t D t .�/ ; � D � .�/ ; z D z .�/

and solve (d=d� D P) 8<:

Pt D 1; t .0/ D 0

P� D �; � .0/ D s

Pz D �.�2 C 1/z; z .0/ Dbg .s/ :From the first two we get t D � , � D se� , while the third one gives

z .�; s/ Dbg .s/ exp

��12s2e2� C 1

2s2 � �

:

Eliminating the parameters s; � , we find

bu .�; t/ Dbg ��e�t � exp

� �

2

2C �2

2e�2t � t

!;


and so

u .x; t/ D 1

2�e�t

ZRbg ��e�t � exp

´�1 � e�2t

2�2 C i�x

μd�.

We observe thatbg ��e�t � is the transform of etg�xet

�, while if we set

a .t/ D 1 � e�2t2

,

then e��2a.t/ is the transform15 of

�1 .x; a.t// D 1p4�a.t/

e� x2

4a.t/ :

Therefore

u .x; t/ DZ

R�1 .y; a.t// g

�et .x � y�/dy D e�t

ZR�1�x � e�ty; a .t/�g .y/ dy:

In caseg .x/ D ı .x � x0/ ;

we find (Fig. 1.16)

u .x; t/ D e�tq2��1 � e�2t � exp

´��x � e�tx0

�22�1 � e�2t �

μ: (1.63)

When x0 ¤ 0 the Gaussian curve shifts (to the right if x0 > 0; to the left if negative) withspeed decreasing exponentially in time. When x0 D 0 there is no shift; the damping effect remainsexponential by virtue of the term e�t .

�10

1

2

3 0:20:4

0:60:8

1 1:21:4

0

0:5

xt

Fig. 1.16 Exponential transport and damping for the Gaussian curve in (1.63)

15 Appendix B.

72 1 Diffusion

Solution 1.3.18. Supposing u admits a Laplace transform in t , define

L .u/ .x; �/ D U .x; �/ DZ 1

0e��tu .x; t/ dt:

Then recalling thatL .ut / .x; �/ D �U .x; �/ � u .x; 0/ D �U .x; �/ ;

we see that U solves16

�U � Uxx D 0; x > 0

U .0; �/ D 1; U .1; �/ D 0:

Solving the ODE we findU .x; �/ D C1e

p�x C C2e

�p�x ;where p

� Dp

j�jei arg �2 :

Now, imposing the boundary conditions gives

C1 D 0, C2 D 1;

henceU .x; �/ D e�

p�x :

Finally, anti-transforming17 produces

u .x; t/ D x

2p�t3=2

e�x2

4t :

Solution 1.3.19. The problem is rotation-invariant and thus the temperature depends only on timeand the distance from the centre, which we take to be the origin. Thus u D u .r; t/, with r D jxj.The equation for u reads18:

ut ��u D ut � 1

r.ru/rr D 0; 0 < r < R; t > 0

with ´u .r; 0/ D qr 0 � r < R

ur .R; t/ D q, ju .0; t/j < 1 t > 0.

As in Problem 1.2.25 (page 46) we may set v D ru and write the one-dimensional mixed Dirichlet-Robin problem: 8<

:vt � vrr D 0 0 < r < R; t > 0

v .r; 0/ D qr2 0 � r < R

v .0; t/ D 0, Rvr .R; t/ D v .R; t/CR2q t > 0:

Let us also definew .r; t/ D v .r; t/ � qr2;

16 We set D D 1 for simplicity.17 Appendix B.18 Recall identity (1.39).


so that 8<:wt � wrr D 2q 0 < r < R; t > 0

w .r; 0/ D 0 0 � r < R

w .0; t/ D 0, Rwr .R; t/ D w .R; t/ t > 0:

Remembering the solution to Exercise 1.3.4, the problem has as eigenvalues the numbers k=R;k � 1, i.e. the positive solutions of tan D . The eigenfunctions are

wk .r/ D sinkr

R;

so the solution is

w .r; t/ D1XkD1

ck .t/ sinkr

R

where the ck are chosen so that (recall that w00k

C �2k

R2wk D 0)

wt � wrr D1XkD1

"c0k .t/C 2

k

R2ck .t/

#sin

kr

RD 2q:

Since

2q D1XkD1

qk sinkr

R

where

qk D 2qcosk˛kk

; ˛k D 1

2� sink

4k;

we must have, also by the zero initial condition:

c0k .t/C 2k

R2ck .t/ D qk , ck .0/ D 0, k D 1; 2; : : : :

This gives

ck .t/ D qkR2

2k

1 � e�

�2k

R2 t

!and finally

u .r; t/ D qr C 2qR2

r

1XkD1

qk

2k

1 � e�

�2k

R2t

!sin

2kr

R2.

Solution 1.3.20. Define bu .�; t / DZ

Rnu .x; t / e�ix��d�;

the partial Fourier transform of u. Thusbu (formally) satisfies the Cauchy problem´but CD j�j2bu D bf .�; t / � 2 Rn; t > 0bu .�; 0/ Dbg .�/ � 2 Rn

wherebf is the partial Fourier transform of f . Hence

bu .�; t / Dbg .�/ e�Dj�j2t CZ t

0e�Dj�j2.t�s/bf .�; s/ ds.

74 1 Diffusion

We remind that the inverse transform of e�Dj�j2t is19

�D .x; t / D 1

.4�Dt/n=2exp

� jxj24Dt

!and the inverse transform of a product is the convolution of the respective anti-transforms; then weget

u .x; t / DZ

Rn�D .x � y; t / g .y/ dy C

Z t

0

ZRn�D .x � y; t � s/ f .y; s/ ds

� u1 .x; t /C u2 .x; t / .

The analysis of the procedure goes exactly as in Problem 1.2.19 (page 37); the conclusion is that

u .x; t / ! f .x; t /

in L2 .Rn/ as t ! 0.

Solution 1.3.21. We have to solve the following Dirichlet problem (setting D D 1):

B

8<:ut .x; y; z; t/ D �u.x; y; z; t/ .x; y/ 2 R2; 0 < z < 1; t > 0

u.x; y; z; 0/ D g.x; y; z/ .x; y/ 2 R2; 0 < z < 1

u.x; y; 0; t/ D u.x; y; 1; t/ D 0 .x; y/ 2 R2; t > 0

(The Laplacian is intended with respect to the spatial coordinates). In order to solve it we use atwo-dimensional Fourier transform in x and y

bu .�; �; z; t/ DZ

R2u .x; y; z; t/ e�i.x�Cy�/d�d�:

Thenbu .�; �; �; �/ is the (formal) solution to:8<:but �buzz C .�2 C �2/bu D 0 0 < z < 1; t > 0bu .�; �; z; 0/ Dbg .�; �; z/ 0 � z � 1bu .�; �; 0; t/ Dbu .�; �; 1; t/ D 0 t > 0:

If we think of �; � as fixed, we can separate the variables and look for solutions of the formbu.z; t/ D v .z/w .t/ :

In the usual way we find

wn .t/ D ˛n .�; �/ e�.�2C�2/te�n2�2t ; n D 1; 2; : : :

andvn .z/ D ˇn .�; �/ sin .n�z/ ; n D 1; 2; : : :

By superposing the products wn.t/vn.z/ and adding the initial condition, we construct the (formal)solution bu .�; �; z; t/ D

1XnD1

n .�; �/ e�.�2C�2/te�n2�2t sin .n�z/

where

n .�; �/ D 2

Z 1

0bg .�; �; z/ sin .n�z/ dz:

19 Appendix B.


Now we transform back, and call cn .x; y/ the inverse transform of n.�; �/ and also recall that

e�.�2C�2/t is the transform of �1 .x; y; t/. We find:

u .x; y; z; t/ D1XnD1

�ZR�1 .x � x1; y � y1; t / cn .x1; y1/ dx1dy1

e�n2�2t sin .n�z/ .

Solution 1.3.22. a) We will mimic the proof of the case Lu D ut �D�u20.1. Given " > 0, T � " > 0 we set v D u � "t . By contradiction, if

.x0; t0/ 2 QT�" n @pQT�" D � � .0; T � "�were a positive maximum point for v on QT�", we would have:

vt .x0; t0/ � 0; �v .x0; t0/ � 0; ru .x0; t0/ D 0, c .x0; t0/ u .x0; t0/ � 0;

contradicting Lv � �". Therefore

maxQT �"

u � maxQT �"

v C "T � max@pQT �"

v C "T � max@pQT

uC "T:

2. By taking the limit " ! 0 in the previous item we see that if maxQT

u D M > 0, thenmax@pQT

u D M . If now u � 0 on @pQT , then u cannot be positive at some other point.

b) If u; v solve the same problem then w D u� v vanishes on the parabolic boundary, whenceit must be zero everywhere on QT .

c) If jc .x; t /j � K, without assumptions on the sign, we can reduce to the previous situation bysetting

z .x; t / D e�Ktw .x; t / .

In fact,Lz D e�Kt Œwt � a�w C b � rw C .c �K/w� D �Kz

soLz CKz D zt � a�z C b � rz C .c CK/z D 0

and the coefficient of z, i.e. c CK, is � 0. Hence z D 0, implying

w D u � v D 0:

Solution 1.3.23. The idea is to reduce to the diffusion equation Ut D �U by repeated change ofvariables. Let us proceed step by step.

Step 1. We eliminate the reaction term by writing

C .t/ DZ t

0c .s/ ds

and definingw .x; t / D e�C.t/u .x; t / .

The function w solves

wt D a .t/�w C b .t/ � rw; w .x; 0/ D g .x/ :

20 [18, Chap. 2, Sect. 2].

76 1 Diffusion

Step 2. Now we get rid of the transport term, by noting that if

B .t/ DZ t

0b .s/ ds;

it follows@

@tw .z � B .t/ ; t/ D wt .z � B .t/ ; t/ � b .t/ � rw .z � B .t/ ; t/ :

So we setx D z � B .t/

andh .z; t / D w .z � B .t/ ; t/ .

Then h solvesht D a .t/�w, h .z; 0/ D g .z/ .

Step 3. We eliminate the coefficient a .t/ by rescaling in time. Set

A .t/ DZ t

0a .s/ ds:

Asa .s/ � a0 > 0;

A is invertible and we may put

U .z; �/ D h�

z; A�1 .�/�

.

Then

U� D ht1

a;

and therefore U solvesU� D �U , U .z; 0/ D g .z/ .

We can write

U .z; �/ D 1

.4��/3=2

ZR3

exp

´� .z � y/2

4�

μg .y/ dy:

Finally, going back to the original variables, we obtain

u .x; t / D 1

.4�A .t//3=2

ZR3

exp

´C .t/ � .x C B .t/ � y/2

4A .t/

μg .y/ dy:

Solution 1.3.24. a) No, for otherwise .x0; 1/ would be an internal positive maximum and

ut .x0; 1/ D 0; uxx .x0; 1/ � 0;

violating the equationut � uxx D �1:

b) Such a solution cannot exist, because jxj is not regular enough to be taken as initial datumfor the backward equation (see Problem 1.2.7 on page 19).

c) There is no contradiction. The example shows only that, in order to have uniqueness,u .x; t/ ! 0 for any given x is not sufficient. Note, by the way, that u .x; t/ ! 1 as .x; t/ ! .0; 0/

along the parabola x2 D t .


d) If

minu D u .x0; t0/ D m < 0

then t0 > 0 and x0 D 0 or x0 D 1. We may suppose u > m for 0 � t < t0, i.e. that t0 is the firstinstant at which u assumes the value m. By Hopf’s principle (Problem 1.2.11 on page 24)

ux .0; t0/ > 0 or ux .1; t0/ < 0:

Both violate the Robin condition, and thereforem � 0. A similar argument shows that the maximumof u, which is positive, cannot be reached along either of the half-lines x D 0, x D 1. Hence it mustcoincide with the maximum of the initial datum

maxu D max sin�x D 1:

Solution 1.3.25. The concentration c satisfies the equation

ct D Dcxx 0 < x < L, t > 0.

If we denote by i the unit vector along the x-axis, according to Fick’s law the flux entering at x D0 is given byZ

Aq .c .0; t// � i dxdy D

ZA

�Dcx .0; t/ dxdy D �DAcx .0; t/ D C0R0

while the outgoing flow at x D L isZA

q .c .L; t// � i dxdy DZA

�Dcx .L; t/ dxdy D �DAcx .L; t/ D c .L; t/R0:

Therefore we deduce the following Neumann-Robin conditions

cx .0; t/ D �B and cx .L; t/C Ec .L; t/ D 0;

where we denoted

B D C0R0

DAand E D R0

DAI

the problem is also associated to the initial condition c .x; 0/ D c0 .x/.First we determine the stationary solution cSt , which satisfies the conditions´

cStxx D 0 0 < x < L; t > 0

cStx .0; t/ D �B , cStx .L; t/CEcSt .L; t/ D 0 t > 0:

We find

cSt .x/ D B .L � x/C B

E:

Now we analyse the transient function u .x; t/ D c .x; t/ � cSt .x/, which solves the followingproblem 8<

:ut D Duxx 0 < x < L; t > 0

ux .0; t/ D 0, ux .L; t/CEu .L; t/ D 0 t > 0

u .x; 0/ D c0 .x/ � cSt .x/ 0 < x < L:

In this way, we are brought back to homogeneous boundary conditions, and we can use the methodof separation of variables as in Problem 1.2.6 (page 16); we set u.x; t/ D y.x/w.t/ and deduce

w0.t/Dw0.t/ D y00.x/

y.x/D �:

78 1 Diffusion

In particular, w satisfies the equation w0.t/ D �Dw.t/, whose solution is V .t/ D e�Dt , while yis a solution of the eigenvalue problem´

y00 .x/ � �y.x/ D 0

y0 .0; t/ D 0, y0 .L; t/CEy .L; t/ D 0:

If � > 0; the general solution is y .x/ D c1e�p�x C c1e

p�x : The boundary conditions give´

�c1 C c2 D 0

c1.E � p�/e�

p�L C c2.E C p

�/ep�L D 0:

(1.64)

Now we have that

det

�1 1

.E � p�/e�

p�L .E C p

�/ep�L

!D �.E C

p�/e

p�L � .E �

p�/e�

p�L

D .E Cp�/e�

p�L

p� �Ep�CE

� e2p�L

!< 0;

since e2p�L > 1 and .

p��E/=.p�CE/ < 1. System (1.64) has the only solution c1 D c2 D 0.

We can reach the same conclusion even in the case � D 0. If � < 0, we find the conditions´U 0 .0/ D p��c2 D 0

U 0 .L; t/CEU .L; t/ D c1

hcos

�p��L�

� p�� sin�p��L

�iD 0:

So, � satisfies the equationcot

�p��L

�D

p��:

Therefore the solution is

u .x; t/ D1XmD1

ume�Dk2

mt cos.kmx/

where the eigenfunctions cos.kmx/ and the corresponding eigenvalues �m D k2m are related to thepoints km, with 0 < km < m�=L andm > 0, where the two functions f1 .�/ D cotL�; f2 .�/ D �

intersect. Furthermore, each um is the coefficient of the Fourier series21 of u .x; 0/ with respect tothe eigenfunction cos.kmx/, namely

um D 1

˛m

Z L

0u .x; 0/ cos .kmx/ dx ˛m D

Z L

0cos2 .kmx/ dx :

Regarding the concentration c, we finally deduce the formula

c .x; t/ D B

EC B .L � x/C

1XmD1

ume�Dk2

mt cos.kmx/:

Since km > 0 for everym, as t goes to C1 every term of the series converges to zero exponentially,and therefore c settles to the steady solution cSt

c .x; t/ ! C0 C C0R0

DA.L � x/ :

21 Appendix B.


Solution 1.3.26. By the total-probability formula we can write the following difference equationfor the transition probability p D p .x; t/, where x D mh, t D N� W

p .x; t C �/ D 1

2

�1C mC 1

N

p .x C h; t/C 1

2

�1 � m � 1

N

p .x � h; t/

or

p .x; t C �/ D 1

2Œp .x C h; t/C p .x � h; t/�C 1

2

1

N

.x C h/ p .x C h; t/ � .x � h/ p .x � h; t/h

:

Using Taylor formulas we get, after simple calculations,

pt � C o .�/ D 1

2pxxh

2 C o�h2�

C 1

NŒ.xp/x C o .h/�

where p and its derivatives are evaluated at .x; t/ :Dividing by � and letting h; � ! 0, since h2=� D 2D and N� D ; we obtain

pt D Dpxx C 1

.xp/x : (1.65)

To find the required formula forD D D 1, write the equation in the form pt D pxx C xpx Cp.Thus the function u .x; t/ D p .x; t/ e�t satisfies the differential equation in Exercise 1.3.17. Re-calling formula (63) with x0 D 0, we end up with the solution (Fig. 1.17)

p .x; t/ D 1q2��1 � e�2t � exp

´� x2

2�1 � e�2t �

μ: (1.66)

Note that p > 0;R

R p .x; t/ dx D 1 and p .x; t/ ! 0 as t ! 0C; x ¤ 0. Therefore the func-tion in (1.66) is the fundamental solution for the equation (1.65). Also observe the absence of decayin time; indeed, as t ! 1; (1.66) exponentially approaches the standard Gaussian.

�20

2

4

6 12

34

56

7

0

0:5

1

xt

Fig. 1.17 The fundamental solution of eq. (1.65) (compare with Fig. 1.16)

80 1 Diffusion

Solution 1.3.27. The data are radially symmetric, so the solution depends only on the radius r Djxj and on time (prove this fact using uniqueness of the solution). We can separate the variables, asin Problems 1.2.25 and 1.2.5 (pages 46 and 14, respectively). We find:

u.r; t/ D r2

2C3tC 3

8

1XnD1

.�1/nn2�2

²1C 6

n2�2Œ1 � .�1/n�

³e��2

ntcos.�nr/

r; with �n D n� .

Solution 1.3.28. By separation of variables the solution is

u .r; t/ D 2

r

1XnD1

.�1/n�n

sin.�nr/

´q

1 � �2n�e�t � e��2

nt�

� Ue��2nt

μ; with �n D n� .

Solution 1.3.29. a) We are considering a global Cauchy problem:´Pt �D�P D aP .x; y/ 2 R2; t > 0

P.x; y; 0/ D ı2.x; y/M.0/ .x; y/ 2 R2;

where Dirac’s delta is centred at the origin of R2. We argue as in Exercise 1.3.5 and reduce to asimilar problem without reaction term. More precisely, the function u.x; y; t/ D e�atP.x; y; t/solves ´

ut �D�u D 0 .x; y/ 2 R2; t > 0

u.x; y; 0/ D ı2.x; y/M.0/ .x; y/ 2 R2:

Hence u.x; y; t/ D M.0/�D.x; y; t/, and then

P.x; y; t/ D M.0/eat�D.x; y; t/ D M.0/

4�Dtexp

at � jxj2

4Dt

!:

b) We have

M.t/ D M.0/eatZ

R2�D.x; y; t/dx dy D M.0/eat :

c) Passing to polar coordinates yields:ZR2nBR

P .x; y; t/ dxdy DZ C1

R

M.0/

4�Dtexp

at � r2

4Dt

!�2�r dr D M .0/ exp

at � R2

4Dt

!;

for any R. Comparing with M.0/ gives R .t/ D 2tpaD.

d) The metropolitan front advances with constant velocity R0.t/ D 2paD.

2

The Laplace Equation

2.1 Backgrounds

Denote by� a domain (an open connected subset) of Rn and byBr .x/ the open n-dimen-sional ball of radius r and centre x. A C 2 function u is harmonic on � if �u D 0 on �.

• Mean-value property. The function u is harmonic on � if and only if the followingaveraging property holds: for any Br .x/ with Br .x/ � �,

u .x/ D 1

jBr .x/jZBr .x/

u .y/ dy and u .x/ D 1

j@Br .x/[email protected]/

u .� / d�:

• Maximum principle. If� is bounded, u is harmonic on� and continuous on�, theneither u is constant or

min@

u < u.x/ < max@

u for every x 2 �:

A frequently-used consequence: let u; v be harmonic on � (bounded), and continuous on�. If u � v on @� then u � v in �.

• Subharmonic/superharmonic functions. A function u 2 C.�/, � � Rn is calledsubharmonic if for any ball BR .x/ � �

u .x/ � 1

jBR .x/jZBR.x/

u .y/ dy, u .x/ � 1

j@BR .x/[email protected]/

u .y/ dy:

It is called superharmonic if the above inequalities are reversed.Equivalently, u is subharmonic if for any harmonic v onBR .x/ that is greater or equal

to u on the boundary, then u � v in the entire ball. If additionally � is connected, andu 2 C

��

is subharmonic and assumes its maximum at an interior point of �, then umust be constant. Furthermore, the maximum of two subharmonics is subharmonic too.

If u 2 C 2.�/, then u is subharmonic if and only if ��u � 0 in �.


82 2 The Laplace Equation

• Poisson formula. Let u be harmonic on Br .p/ � Rn and continuous on Br .p/ :Then

u .x/ D r2 � jx � pj2!nr

Z@Br .p/

u .� /

j� � xjn d�;

where !n D j@B1 .p/j. Therefore u 2 C1.Br .p//.• Harnack’s inequality. Let u be harmonic and non-negative in Br .p/ � Rn. For any

x 2 Br .p/ we have

rn�2.r � jxj/.r C jxj/n�1 u.p/ � u.x/ � rn�2.r C jxj/

.r � jxj/n�1 u.p/:

In particularmaxBr=2.p/

u � 3n minBr=2.p/

u:

• Liouville’s theorem. A harmonic function u on Rn, with u .x/ � 0 for every x 2 Rn,is constant.Therefore the only harmonic functions on Rn that are bounded either from above or frombelow are the constant functions.

• Fundamental solution and potentials. The function

� .x/ D

8<:

� 1

2�log jxj n D 2;

1

.n � 2/!n1

jxjn�2 n � 3

is the solution of

�� .x/ D ın .x/ in Rn (ın is the n-dimensional Dirac’s delta function at the origin)

and is called the fundamental solution of the Laplace operator.Using � one can reconstruct an arbitrary u 2 C 2 ��,� bounded with regular bound-

ary, as the sum of three terms:

u .x/ DZ@

� .x � � / @�� ud� �Z@

u @�� .x � � / d� �Z

� .x � y/�udy;

called simple-layer potential, double-layer potential, and Newtonian potential respective-ly. Above, @�� D r � �� and �� is the outer unit normal at � 2 @�.

An integral of the form

N .xIf / DZ

� .x � y/ f .y/ dy

is called a Newtonian potential. In three dimensions it represents the electrostatic potentialgenerated by a charge distribution of density f in �. If f 2 C.�/ then N 2 C 1.Rn/. Iff 2 C 1.�/ then N 2 C 2.�/, and ��N D f in �.

2.1 Backgrounds 83

An integral of the form

D .xI/ DZ@

.� / @�� .x � � / d�

is called double-layer potential of . In three dimensions it represents the electrostaticpotential generated by a dipole distribution of momentum on @�. Let � � Rn be abounded, C 2 domain. Then

D .xI 1/ DZ@

@�� .x � � / d� D

8<ˆ:

�1 x 2 ��12

x 2 @�0 x 2 Rn n�:

More generally, let be a continuous function on @�. Then D .xI/ is harmonic onRn n @� and the following jump relations hold for every x 2 @� W

limz!x, z2Rnn

D .zI/ D D .xI/C 1

2 .x/ (2.1)

and

limz!x, z2 D .zI/ D D .xI/ � 1

2 .x/ : (2.2)

On the contrary, normal derivatives across @� are continuous.An integral of the form

S .x; / DZ@

� .x � � / .� / d�

is called single-layer potential of . In three dimensions it represents the electrostatic po-tential generated by a charge distribution of density on @�. If� is a C 2 domain and is continuous on @�, then S is continuous across @� and

�S D 0 in Rn n @�;because one can differentiate inside the integral. On the other hand, the normal deriva-tive of S undergoes a jump when crossing @�. More precisely, for every x 2 @�, settingzt D x C t�x, we have

limt!0C

@�xS .zt ; / DZ@

@�x� .x � � / .� / d� � 1

2 .x/ (2.3)

and

limt!0� @�xS .zt; / D

Z@

@�x� .x � � / .� / d� C 1

2 .x/ : (2.4)

• Green function. � is the fundamental solution for the operator �� on the entire space.We can define a fundamental solution for �� in�, when the latter has, for example, Lip-


schitz boundary; the idea is that the fundamental solution should represent the potentialgenerated by a unit charge at some point y inside a conductor occupying� and connectedto ground at the boundary. We denote by G .x; y/ this function, called the Green functionin �. For any given y it holds

��G .�; y/ D ın.� � y/ in �;

and since the conductor is grounded, we also know that

G .�; y/ D 0; on @�:

Therefore

G .x; y/ D � .x � y/ � g .x; y/where g, as function of x, solves, for any fixed y, the Dirichlet problem²

�g D 0 in �g .�; y/ D � .� � y/ on @�:

2.2 Solved Problems

� 2:2:1 � 2:2:12 W General properties of harmonic functions.� 2:2:13 � 2:2:24 W Boundary-value problems. Solution methods.� 2:2:25 � 2:2:31 W Potentials and Green functions.

2.2.1 General properties of harmonic functions

**Problem 2.2.1 (Analyticity of harmonic functions). Consider a harmonic functionu in a domain � � Rn. Prove the following properties:

a) If BR .p/ � �, for any multi-index ˛ D .˛1; : : : ; ˛n/

jD˛u .p/j � .n j˛j/j˛jRj˛j

[email protected]/

juj ( j˛j D ˛1 C � � � C ˛n) (2.5)

(use induction).

b) Deduce the Taylor expansion

u .x/ D1X

j˛jD0

D˛u .p/˛Š

.x � p/˛ (˛Š D ˛1Š � � �˛nŠ)

for x close enough to p; therefore, u is (real) analytic on �.

c) Deduce that if u has a local maximum or minimum at p 2�, u is constant in �.


Solution. a) Let u be harmonic in � � Rn. Then u belongs to C1.�/ and therefore,for any multi-index ˛ D .˛1; : : : ; ˛n/ the derivative of order ˛ exists:

v D D˛u D @˛1

@x˛1

1

: : :@˛n

@x˛nn

u:

In particular we can compute the Laplacian of v:

�v D �.D˛u/ DnXiD1

@2

@x2i.D˛u/ D

nXiD1

D˛

�@2

@x2iu

D D˛.�u/ D 0

(as u is C1, Schwarz’s theorem allows to swap derivatives).Take w harmonic on �, Br .q/ � �. Since wxj

is harmonic in �, by the mean-valueproperty and the Gauss formula we may write:

ˇwxj

.q/ˇ D n

!nrn

ˇZBr .q/

wxj.y/ dy

ˇD

D n

r� 1

!nrn�1

ˇZ@Br .q/

w .� / jd�

ˇ� n

rmax@Br .q/

jwj : (2.6)

Now we use induction. When j˛j D 1 it suffices to rewrite (2.6) with w D u, r D R,q D p. So suppose (2.5) holds for any j˛j D k (and any p, R such that BR .p/ � �).Using (2.6) with w D D˛u, r D R=.k C 1/, q D p we obtain

ˇ.D˛u/xj

.p/ˇ � n.k C 1/

Rmax

@BR=.kC1/.p/jD˛uj D n.k C 1/

RjD˛u.qmax/j ; (2.7)

where qmax 2 @BR=.kC1/ is the maximum point. Since BkR=.kC1/ .qmax/ � BR .p/ (seeFig. 2.1), we can use equation (2.5), with kR=.k C 1/ replacing R and qmax instead of p

Fig. 2.1 Use of the inductive assumption (Problem 2.2.1)


(by inductive hypothesis). We find

jD˛u .qmax/j � .nk/k

Rk� .k C 1/k

kkmax

@BkR=.kC1/.qmax/juj : (2.8)

From (2.7) and (2.8), together with the fact that

max@BkR=.kC1/.qmax/

juj � [email protected]/

juj

by the maximum principle, we deduce (2.5) for j˛j D k C 1, and hence the claim.

b) Let B2R .p/ � �. Then u is smooth in BR .p/. If 0 < r � R we can use the Taylorexpansion with Lagrange remainder and write, for any x 2 Br .p/,

u .x/ Dk�1Xj˛jD0

D˛u .p/˛Š

.x � p/˛ CRk.x/;

where

Rk.x/ DXj˛jDk

D˛u .p/˛Š

.�x � p/˛

for some �x 2 Br .p/. The known formula1

nk D .1C � � � C 1/k D kŠXj˛jDk

1

˛Š

and part a) give

j�xj � r H) jRk.x/j � .nk/k

Rkmax

@BR.�x/juj � rk � n

k

kŠ

D kk

kŠ

�n2r

R

k„ ƒ‚ …

ak

[email protected]/

juj ;

so there remains to prove that ak ! 0 as k goes to infinity. This actually follows byobserving that

akC1ak

D�1C 1

k

kn2r

R< e

n2r

R� e

3;

and so

akC1 � a1

�e3

�k;

provided we choose r small enough (it suffices to take r D R=.3n2/).

1 This can be proved by induction on n, using Newton’s binomial formula.


c) By definition if p is a local minimum for u, then it is also an absolute minimumfor u restricted on BR.p/, if R is small enough. The maximum principle implies that u isconstant on BR.p/, and by the uniqueness of analytic prolongations u is constant on �.

Problem 2.2.2 (Harmonic functions with polynomial growth). Let u be harmonic inRn, and > 0, C > 0 such that

ju.x/j � C.1C jxj/� for every x 2 Rn: (2.9)

Using Problem 2.2.1 a), prove that u is a polynomial of degree less than or equal to .

Solution. Let k > be an integer. As u is harmonic on Rn, from (2.5) and (2.9) wehave, for any p 2 Rn, R > 0 and j˛j D k,

jD˛u .p/j � .nk/k

[email protected]/

juj � C .nk/k1CR�

Rk:

Letting R go to infinity we see D˛u.p/ D 0, for any p. But this holds for derivatives ofany order greater than , so u is a polynomial of degree less than or equal to .

Problem 2.2.3 (Harmonic polynomials). Find all harmonic polynomials of degree nin two variables.

Solution. Suppose

Pn.x; y/ DnXkD0

ckxn�kyk;

indicates the generic harmonic polynomial of degree n in the variables x; y, with non-zerocoefficients ck . Differentiating

�Pn.x; y/ Dn�2XkD0

ck.n � k/.n � k � 1/xn�k�2yk CnXhD2

chh.h � 1/xn�hyh�2

Dn�2XkD0

ck.n � k/.n � k � 1/xn�k�2yk Cn�2XkD0

ckC2.k C 2/.k C 1/xn�k�2yk

Dn�2XkD0

Œck.n � k/.n � k � 1/C ckC2.k C 2/.k C 1/� xn�k�2yk :

ForPn to be harmonic, each summand must necessarily have zero coefficient, so for any k

ck.n � k/.n � k � 1/C ckC2.k C 2/.k C 1/ D 0;

and then

ckC2 D � .n � k/.n � k � 1/.k C 2/.k C 1/

ck :


Therefore the even coefficients depend on the choice of c0, while the odd ones depend onc1. Let us consider the former ones:

c2 D �n.n � 1/2 � 1 c0 D �n.n � 1/.n � 2/Š

2Š � .n � 2/Š c0 D � n

2

!c0;

c4 D � .n � 2/.n � 3/4 � 3 c2 D n.n � 1/.n � 2/.n � 3/

4 � 3 � 2 � 1 c0 D

D n.n � 1/.n � 2/.n � 3/.n � 4/Š4Š � .n � 4/Š c0 D

n

4

!c0:

By induction

c2h D .�1/h n

2h

!c0; and, analogously, c2hC1 D .�1/h

n

2hC 1

!c1

n:

In conclusion

Pn.x; y/ DnXkD0

Qck n

k

!xn�kyk ; with Qck D

´.�1/hc0 for k D 2h

.�1/h c1

nfor k D 2hC 1

and c0, c1 arbitrary.

Problem 2.2.4 (Subharmonic functions and a variant of Liouville’s theorem). Let ube harmonic on � � Rn. Prove:

a) If F 2 C 2 .R/ is convex then w D F .u/ is subharmonic on �.

b) If � D Rn and ZRn

u2 .x/ dx D M< 1

then u � 0.

Solution. a) Since

wxjD F 0 .u/ uxj

, wxj xjD F 00 .u/ u2xj

C F 0 .u/ uxj xj

it follows

�w DnX

jD1

hF 00 .u/ u2xj

C F 0 .u/ uxj xj

iD F 00 .u/ jruj2 C F 0 .u/�u

D F 00 .u/ jruj2 � 0

as F is convex.

b) Take x 2 Rn and R > 0. Because u is harmonic and F.s/ D s2 is convex, u2 issubharmonic and we may write

u2.x/ � 1

jBR .x/jZBR.x/

u2.y/dy � M

jBR .x/j :


But MjBR.x/j ! 0 for R ! 1, hence u2.x/ D 0 and, since x is arbitrary, u � 0 in Rn.

*Problem 2.2.5 (Liouville theorem for subharmonic functions in the plane). Let u bea (continuous) subharmonic function, bounded from above in R2.

a) Verify that for any " > 0 the function w".x/ D u.x/ � " log jxj satisfies

maxBe

w" D max@Be

w" D max@Be

u;

where Be D ¹x 2 R2 W jxj > 1º.

b) Deduce that u is constant.

Solution. a) Since u.x/ � M for any x, for any given " > 0 there exists R" > 0 suchthat

jxj � R" H) w".x/ � M � " logR" � max@Be

w" (2.10)

(take R" D expŒ.M � max@Bew"/="�). On the other hand v.x/ D log jxj is harmonic on

R2 n ¹0º, sow" is subharmonic on the annulus ¹1 < jxj < R"º, and assumes its maximumon the closure, at points x with jxj D 1 or jxj D R". By (2.10) we deduce

w".x/ � max@Be

w" also when 1 < jxj < R":

b) Letting " ! 0 we have w" ! u uniformly on compact subsets in Be . The previousresult forces

maxBe

u D max@Be

u:

Therefore, for any compact set K � B1

maxKu D max

B1

u D max@B1

u:

So the subharmonic function u has a maximum in the interior, and thus it is constant onK, for any K.

*Problem 2.2.6 (Harnack inequality on compact sets). Consider a domain � in Rn

and a compact subset K � Rn. Prove that there exists a constant > 0, dependingonly on K and �, such that for any harmonic, non-negative function on � one has

maxKu � min

Ku:

Solution. K is compact, so we can find a finite number of balls BRjD BRj

�pj�,

j D 1; : : : ; k inside � such that

i) K is contained in the union of the BRj =2.

ii) ; ¤ BRj =2 \ BRj C1=2 3 zj .


In particular minj Rj decreases (and k increases) as the distance between K and @� getssmaller. From Harnack’s inequality we have

maxBRj =2

u � 3n minBRj =2

u � 3nu.zj / � 3n maxBRj C1=2

u � 32n minBRj C1=2

u:

Take x; y 2 K, and to fix ideas suppose x 2 BRj1=2, y 2 BRj2

=2, with j1 � j2. Iteratingthe previous inequality we get

u.x/ � 3j2�j1C1u.y/ � 3knu.y/:

But x; y were arbitrary, so (with D 3kn) the claim follows.

*Problem 2.2.7 (Series of harmonic functions). Consider harmonic, non-negativefunctions ui , i 2 N, defined on a domain � in Rn. Using Harnack’s inequality showthat if

P1iD0 ui converges at some x0 2 �, then it converges uniformly on any com-

pact set K � �. Deduce that the sum U of the series is non-negative and harmoniceverywhere on �.

Solution. We wish to prove uniform convergence on any given compactK � �. Since� is connected, we may assumeK is connected and that it contains x0 (if not, just chooseanother compact set containing the original one). By Harnack’s inequality (Problem 2.2.6)

maxKui � min

Kui � ui .x0/ :

Then 1XiD0

maxKui .x/ �

1XiD0

ui .x0/ < 1

and by the Weierstrass criterion the seriesP1iD0 ui .x/ converges uniformly on K.

But K � � is arbitrary, so the series converges at any point in �, implying that thesum U is defined on �, and is non-negative as sum of non-negative terms. Also the uiare continuous on�, hence the uniform convergence on compact subsets of� guaranteesthat U is continuous on �.

To show U is harmonic, then, it suffices to show it satisfies the mean-value property2.For any Br .x/ �� ,

1

jBr .x/jZBr .x/

U.y/ dy D 1

jBr .x/jZBr .x/

1XiD0

ui .y/

!dy

D1XiD0

�1

jBr .x/jZBr .x/

ui .y/ dy

D1XiD0

ui .x/ D U.x/;

and the claim follows (we rely on uniform convergence in order to swap sum and integral).

2 With a little extra effort one could prove �U D P�ui D 0.


Problem 2.2.8. Let u be a positive harmonic function in B1 .0/n¹0º � Rn: Show thatthere exists a constant > 0, depending only on n, such that

u .x/ � u .y/

whenever 0 < jxj D jyj � 12:

Solution. Let jxj D jyj D 12

. Since @B1=2 .0/ is a compact subset of B1 .0/ n ¹0º,Harnack’s inequality (Problem 2.2.6 on page 89) gives

u .x/ � u .y/ (2.11)

with > 0, dependent only on n. Now, let 0 < R � 1 and set U .z/ D u .Rz/ : Then Uis harmonic on B1=R .0/ n ¹0º and in particular in B1 .0/ n ¹0º : From (2.11) we infer

u .Rz/ D U .z/ � U .w/ D u .Rw/

whenever jzj D jwj D 12: Letting x D Rz;w D Ry, we get u .x/ � u .y/ whenever

0 < jzj D jwj D R2

� 12:

**Problem 2.2.9 (Hopf principle). Fix � � Rn and u 2 C 2.�/ \ C 1.�/, harmonicand positive on �. Let x0 2 @� be a zero of u. If there exists a ball (Fig. 2.2)

BR .p/ � � such that @� \ BR .p/ D ¹x0º;then @�u.x0/ > 0, where � D .p � x0/=R.

In particular, if @� isC 1 at x0 then � is the inward unit normal to @� at the point x0.

Solution. We set r D jx � pj and consider the annulus

CR D²R

2< r < R

³;

contained in � and touching @� at x0. Let w be the harmonic function on CR that van-ishes on @BR and equals the minimum of u on @BR=2. By the maximum principle u � v

Fig. 2.2 Touching ball at x0


on CR. But since u .x0/ D w .x0/ D 0, we also have

@�u.x0/ D limh!0C

u .x0 C h�/ � u .x0/h

� limh!0C

w .x0 C h�/ � w .x0/h

D @�w.x0/.

Now it suffices to prove @�w.x0/ > 0. The function w is radial, that is

w D w .jx � pj/ D w .r/ ;

and hence it has the form (we consider only n > 2, and leave the case n D 2 to the reader)

w.r/ D C1

rn�2C C2:

Definem D inf

@BR=2

u:

Imposing w .R/ D 0 and w .R=2/ D m, we find

w.r/ D m

2n�2 � 1

"�R

r

n�2� 1

#;

and then

@�w.x0/ D �w0.R/ D m.n � 2/R.2n�2 � 1/ > 0;

whence the claim.

**Problem 2.2.10 (Removable singularities). Let � � R2 be an open bounded do-main, x0 2 � and u 2 C 2.� n ¹x0º/ such that, for some M > 0,

�u.x/ D 0; ju.x/j � M for x 2 � n ¹x0º:Show that u can be extended to a harmonic function at x0, that is, there exists Qu 2C 2.�/ with

Qu.x/ D u.x/ for x ¤ x0; � Qu D 0 in �. (2.12)

(see Exercise 2.3.20 for an improved version of this result).

Solution. Using translations we may suppose x0 D 0. Let R > 0 be so small thatBR D BR.0/ � �. Let v be the solution to´

�v D 0 in BRv D u on @BR:

The maximum principle ensures that the range of v lies between the maximum and mini-mum of u on @BR, so jv.x/j � M .


Now w D u � v is harmonic on BR n ¹0º, it vanishes on @BR and jw.x/j � 2M . Ifwe show that w � 0 on BR n ¹0º, then

Qu.x/ D´u.x/ x 2 � n BRv.x/ x 2 BR

satisfies (2.12). Consider 0 < r < R, Br D Br .0/; the function

h.x/ D 2Mlog .jxj=R/log .r=R/

solves 8<:�h D 0 in BR n Brh D 0 on @BRh D 2M on @Br :

(2.13)

The maximum principle implies that �h � w � h in BR n Br , i.e.

jw.x/j � 2Mlog .jxj=R/log .r=R/

for r � jxj � R:

Fix x ¤ 0. The previous inequality holds for any r � jxj. In particular, then, we let r tendto 0, obtaining jw.x/j D 0. But x was arbitrary, so w is identically zero away from theorigin; as we already remarked, this proves the claim.

Problem 2.2.11 (Radial replacement). Let u be harmonic inB1 .0/n.0/,B1 .0/ � Rn;and suppose u .x/ ! 0 as jxj ! 1: Define the radial replacement of u by the formula

uR .x/ D 1

n!n jxjn�[email protected]/

u .� / d�:

Show that, for some constant C ,

uR .x/ D´C log jxj n D 2

C�jxj2�n � 1

�n � 3

(2.14)

and therefore uR is harmonic on B1 .0/ n .0/a.

a Hint. Set g .jxj/ D uR .x/ and show that rn�1g0 .r/ is constant on .0; 1/.

Solution. Let

g .r/ D 1

n!nrn�1

Z@Br .0/

u .� / d� D 1

n!n

[email protected]/

u .r!/ d!:


Then, uR .x/ D g .jxj/ and g .1/ D 0: We want to show that rn�1g .r/ is constant on.0; 1/ : In fact,

g0 .r/ D 1

n!n

[email protected]/

ru .r!/ � ! d! D 1

n!nrn�1

Z@Br .0/

@�u .� / d�

where � D �=r is the outward unit normal at the point � . Since �u D 0 in B1 .0/ n .0/,by applying Green’s formula to the annulus Ar;s , 0 < r; s < 1 we deduce that

0 DZAr;s

�u [email protected]/

@�u .� / d� �Z@Br .0/

@�u .� / d�:

Therefore

rn�1g0 .r/ D 1

n!n

Z@Br .0/

@�u .� / d� D C .C2 R/

or

g0 .r/ D C

rn�1from which (2.14) follows by integration, taking into account that g .1/ D 0.

*Problem 2.2.12 (Radial replacement of positive harmonic functions). Let u be pos-itive and harmonic on B1 .0/ n .0/ � Rn, and continuously vanishing on @B1 .0/.Denote by uR its radial replacement (see Problem 2.2.11).

a) Show that if u � uR in B1 .0/ n .0/ then u D uR.

b) Show that u coincides with its radial replacement uR:

u .x/ D uR .x/ D´C log jxj n D 2

C�jxj2�n � 1

�n � 3

(2.15)

where C is some constanta.

a Hint. Show that the set of all t 2 Œ0; 1� such that u � tuR in B1 .0/ n .0/ is non-empty, closedand open in Œ0; 1�.

Solution. a) We claim that u � uR implies u D uR: In fact, if u � uR and thereis a point x such that u .x/ > uR .x/, then u .y/ > uR .y/ in a neighbourhood of x on@Bjxj .0/ : Thus, integrating on @Bjxj .0/, we get the contradiction

uR .x/ >�uR�R.x/ D uR .x/

since uR is radial.

b) To show that u � uR in B1 .0/ n .0/ define the set

E D ®t 2 Œ0; 1� W u � tuR in B1 .0/ n .0/¯ :

1. E is not empty because it contains 0.2. E is closed: indeed, if a sequence ¹tmº � E converges to t0, in fact, then

u .x/ � tmuR .x/


for all x 2 B1 .0/ n .0/, and passing to the limit we get u .x/ � t0uR .x/ for all

x 2 B1 .0/ n .0/ : Thus t0 2 E and E is closed.

3. Finally, E is open in Œ0; 1�. Suppose that u � t�uR in B1 .0/ n .0/. Then clearlyu � tuR in B1 .0/ n .0/ for any 0 � t < t�. Set w D u� t�uR � 0. Then w is harmonic,andw D 0 on @B1 .0/ : Ifw � 0 then u D t�uR; as in part a), by integrating and recallingthat .uR/R D uR we obtain that t� D 1, and the proof is completed. Otherwise, w > 0

on B1 .0/ n ¹0º, and by Problem 2.2.8 (page 91),

w .x/ � wR .x/

for all 0 < jxj � 1=2. By the maximum principle w .x/ > wR .x/ on all of B1 .0/ n .0/since w .x/ � wR .x/ D 0 on @B1 .0/ : Thus

u � t�uR � �uR � t�uR� in B1 .0/ n .0/

or, rearranging the terms,

u � �t� C

�1 � t��uR � 0:

Therefore Œ0; t� C .1 � t�// � E and E is open.We deduce thatE D Œ0; 1� : Setting t D 1, we get u � uR, and the conclusion follows

by part a).

2.2.2 Boundary-value problems. Solution methods

Problem 2.2.13 (Mixed problem on a rectangle, separation of variables). On the rect-angle

¹Q D .x; y/ W 0 < x < a; 0 < y < bºsolve the mixed problem:8<

:�u D 0 in Q

u .x; 0/ D 0, u .x; b/ D g .x/ 0 � x � a

u .0; y/ D ux .a; y/ D 0 0 � y � b

where g 2 C 2 .R/, g .0/ D g0 .a/ D 0.

Solution. We seek (nontrivial) harmonic functions of the form u .x; y/ D v .x/w .y/,with v .0/ D v0 .a/ D 0, w .0/ D 0. Substituting in �u D 0 we find v00 .x/w .y/ Cv .x/w00 .y/ D 0, thus

v00 .x/v .x/

D �w00 .y/w .y/

D �

and � constant. The eigenvalue problem for v reads´v00 .x/ � �v .x/ D 0

v .0/ D v0 .a/ D 0;


hence �k D � .2kC1/2�2

4a2 , k � 0, and

vk .x/ D sin.2k C 1/ �x

2a:

With these �k , we get for w the equation

w00 .y/C �kw .y/ D 0;

with w .0/ D 0. Writing the general integral of the ODE as

w .y/ D c1 sinh.2k C 1/ �y

2aC c2 cosh

.2k C 1/ �y

2a;

we see that w .0/ D 0 implies c2 D 0.So we may write the candidate solution in the form

u .x; y/ D1XkD0

ak sin.2k C 1/ �x

2asinh

.2k C 1/ �y

2a: (2.16)

Now we need to find the coefficients ak from the condition u .x; b/ D g .x/. By theassumptions made on g we have

g .x/ D1XkD0

gk sin.2k C 1/ �x

2a

where, after two integrations by parts

gk D 2

a

Z a

0

g .x/ sin.2k C 1/ �x

2adx D � 8a

.2k C 1/2�2

Z a

0

g00 .x/ sin.2k C 1/ �x

2adx;

andP10 jgk j < 1. It will suffice to choose

ak D gk

sinh

.2k C 1/ �b

2a

��1to get u .x; b/ D g .x/.

• Analysis of (2.16). In Q we haveˇak sin

.2k C 1/ �x

2a

ˇsinh

.2k C 1/ �y

2a� jgk j

so the series defining u is uniformly convergent in Q, and hence u 2 C�Q�. Inside the

domain we can differentiate without problem the single terms of the series as many times


as we want, because the coefficients ak go to zero fast enough. This guarantees that u isharmonic on Q. For the same reason, and since g0 .a/ D 0, the Neumann datum alongx D a is also continuously achieved.

Problem 2.2.14 (Poisson-Dirichlet in the disc, separation of variables). Solve the non-homogeneous Dirichlet problem:´

�u .x; y/ D y in B1u D 1 on @B1:

Solution. First, we break up the problem, which is non-homogeneous both in the equa-tion and also in the boundary condition, in two sub-problems each with only one non-homogeneity. Set u D v C w, where´

�v .x; y/ D y in B1v D 0 on @B1;

´�w .x; y/ D 0 in B1w D 1 on @B1:

The latter system immediately tells thatw.x; y/ � 1 is the only solution (by the maximumprinciple). As for the former, we pass to polar coordinates, V.r; �/ D v.r cos �; r sin �/.Then V is 2�-periodic in � , and using the polar expression3 for � we see that V satisfies

Vrr C 1

rVr C 1

r2V�� D r sin �;

with V .1; �/ D 0 and V bounded. The right-hand side y D r sin � suggests to seeksolutions of the type

V.r; �/ D b1.r/ sin �; with b1.1/ D 0 and b1 bounded.

Substituting, we obtain for b1 the equation

b001.r/ sin � C 1

rb01.r/ sin � � 1

r2b1.r/ sin � D r sin �;

i.e. the ODEr2b001 C rb01 � b1 D r3: (2.17)

The associated homogeneous equation is of Euler type, and may be reduced to constantcoefficients by setting s D log r . Alternatively we could seek solutions of the form ra,a 2 R. Substituting, we have

a.a � 1/ra C ara � ra D .aC 1/.a � 1/ra D 0;

whence a D ˙1. The general integral for the homogeneous equation thus reads

C1r C C2r�1 .C1; C2 2 R/:

3 Appendix B.


�0:50

0:5

1�0:5

00:5

0:95

1

1:05

xy

Fig. 2.3 The function u .x; y/ D 1C y�x2 C y2 � 1� =8

Let us now look for a particular non-homogeneous solution to (2.17) of the form Ar3. Wefind

6Ar3 C 3Ar3 � Ar3 D r3;

so A D 1=8. Overall, the general integral of (2.17) is given by

b1.r/ D 1

8r3 C C1r C C2r

�1;

with arbitrary C1, C2. The boundedness of b1 implies C2 D 0, while b1.1/ D 0 forcesC1 D �1=8. Thus

V.r; �/ D 1

8r.r2 � 1/ sin �;

or in Cartesian coordinates (Fig. 2.3)

u.x; y/ D 1C 1

8.x2 C y2 � 1/y:

Problem 2.2.15 (Dirichlet on an annulus, separation of variables). Consider the annu-lus

C1;R D ¹.r; �/ W 1 < r < Rº :a) Given g; h in C 1 .R/ and 2�-periodic, solve the Dirichlet problem8<

:�u D 0 in C1;Ru.1; �/ D g.�/ 0 � � � 2�

u.R; �/ D h.�/ 0 � � � 2�:

b) Study the special case g.�/ D sin � and h.�/ D 1.


Solution. a) As g, h are regular and periodic we can expand them in Fourier series

g.�/ D a0

2CC1XnD1

.an cos n� C bn sin n�/ ;

h.�/ D A0

2CC1XnD1

.An cos n� C Bn sin n�/

with convergent series

C1XnD1

.janj C jbnj/ ;C1XnD1

.jAnj C jBnj/ : (2.18)

The domain symmetry suggests to look for solutions of the type

u.r; �/ D ˛0.r/CC1XnD1

Œ˛n.r/ cos n� C ˇn.r/ sin n�� : (2.19)

The boundary conditions require that at r D 1 and r D R the solution coefficients mustcoincide with the coefficients of g and h respectively. On the other hand, in polar coordi-nates

�u D urr C 1

rur C 1

r2u�� : (2.20)

We have

ur .r; �/ D ˛00.r/CC1XnD1

�˛0n.r/ cos n� C ˇ0n.r/ sin n�

�;

urr .r; �/ D ˛000.r/CC1XnD1

�˛00n.r/ cos n� C ˇ00n.r/ sin n�

�;

u�� .r; �/ D �C1XnD1

n2 Œ˛n.r/ cos n� C ˇn.r/ sin n�� :

Substituting into equation (2.20) we get

˛000.r/C ˛00.r/r

CC1XnD1

²˛00n.r/C 1

r˛0n.r/ � n2

r2˛n.r/

�cos n�C

Cˇ00n.r/C 1

rˇ0n.r/ � n2

r2ˇn.r/

�sin n�

³D 0

from which we obtain the family of problems, for n � 0:8<:˛000.r/C 1

r˛00.r/ D 0

˛0.1/ D a0

2; ˛0.R/ D A0

2;

(2.21)

100 2 The Laplace Equation8<:˛00n.r/C 1

r˛0n.r/ � n2

r2˛n.r/ D 0

˛n.1/ D an; ˛n.R/ D An;

8<:ˇ00n.r/C 1

rˇ0n.r/ � n2

r2ˇn.r/ D 0

ˇn.1/ D bn; ˇn.R/ D Bn:(2.22)

The equation in (2.21) has order one in ˛00, and is solved by C1 C C2 log r . Using theconstraints we find

˛0.r/ D a0

2C A0 � a0

2 logRlog r:

To find the general solution of the (Euler) equations in (2.22) we first seek special solutionsof the form r� , with to be determined. Substituting into (2.22) we obtain

. . � 1/C � n2/r��2 D 0

so D ˙n . The general integral is C1rn C C2r�n. The boundary conditions force4:

˛n.r/ D anKn .r/ r�n C AnHn .r/

� rR

�nand

ˇn.r/ D bnKn .r/ r�n C BnHn .r/

� rR

�n;

where

Hn .r/ D 1 � r�2n1 �R�2n and Kn .r/ D 1 �R�2nr2n

1 �R�2n :

Substituting into (2.19) we finally get the solution. Since

Hn .r/ � 1

1 �R�1 , Kn .r/ � 1

1 �R�1 andr

R< 1,

1

r< 1;

by (2.18), we see that the series in (2.19) is absolutely and uniformly convergent for1 � r � R. Moreover, we can differentiate term by term, and therefore (2.19) is theunique solution. Note that the solution has the form

c0 C c1log r

logRC

1XnD1

.cnrn C dnr

�n/

reminiscent of the Laurent series of an analytic function on the plane.b) The functions g and h are already finite Fourier sums. In particular, using the nota-

tion of part a), the only non-zero coefficients are b1 D 1 and A0 D 2. Hence

˛0.r/ D log r

logR; ˇ1.r/ D 1 �R�2r2

1 �R�2 r�1 D R2 � r2.R2 � 1/r

and ˛n.r/ � ˇm.r/ � 0 for n � 1; m � 2. To sum up, the solution is (Fig. 2.4)

u.r; �/ D log r

logRC R2 � r2.R2 � 1/r sin �;

as a direct computation shows.4 Rearranging terms, with a little patience.


�10

1

2�1

01

�1

0

1

xy

Fig. 2.4 u .r; �/ D log rlog2 C 4�r2

3r sin �

Problem 2.2.16 (Dirichlet problem on the ball). Determine, at least formally, the so-lution of the Dirichlet problem (in spherical coordinates)´

�u D 0 in B1u .1; '; �/ D g .�/ on @B1;

where g is (2�-periodic and) continuous on the ball

B1 D ¹.r; '; �/ D 0 � r < 1, 0 � ' � 2� , 0 < � < �º :

Solution. Since the problem is invariant under rotations about the z-axis, we look forfunctions u D u .r; �/. The Laplace equation for u becomes5

@

@r

�r2@u

@r

C 1

sin �

@

@�.sin �

@u

@�/ D 0

with u.1; �/ D g .�/ and u bounded. Let us separate variables, and seek for solutions ofthe form u .r; �/ D v .�/w .r/. Inserting into the equation and rearranging, we find

1

w

�r2w0

�0 D � 1

v sin �.sin � v0/0.

5 Appendix B.


Setting either side equal a constant �, we find the two eigenvalue problems:

v00 C cos �

sin �v0 C �v D 0, v bounded (2.23)

andr2w00 C 2rw0 � �w D 0, w bounded. (2.24)

Let us consider (2.23). Changing variable from � to x D cos � , we have x in Œ�1; 1� and

v0 D dv

d�D dv

dx

dx

d�D .� sin �/

dv

dx

v00 D d2v

dx2

�dx

d�

2C dv

dx

d2x

d�2D sin2 �

d2v

dx2� cos �

dv

dx:

Substituting into (2.23) we find

sin2 �d2v

dx2� 2 cos �

dv

dxC �v D 0

i.e. �1 � x2� d2v

dx2� 2x dv

dxC �v D 0

which is a Legendre equation6, with

v .�1/ < 1, v .1/ < 1.

The eigenvalues are the integers �n D n .nC 1/, n � 0, and the eigenfunctions, recur-sively defined by the Rodrigues formula

Ln .x/ D 1

2nnŠ

dn

dxn

�x2 � 1�n ,

are called Legendre polynomials. The normalised polynomials

L�n .x/ Dr2nC 1

2Ln .x/

form an orthonormal basis of L2 .�1; 1/. With � D �n equation (2.24) reads

r2w00 C 2rw0 � n .nC 1/w D 0, w bounded,

solved by wn .r/ D crn. Thus we have found the following bounded solutions with sep-arated variables

un .r; �/ D rnLn .cos �/ , n � 0:

6 Appendix A.


Superposing the functions un we can write

u .r; �/ D1XnD0

anrnLn .cos �/ ; (2.25)

so the Dirichlet condition now requires

1XnD0

anLn .cos �/ D g .�/ , 0 � � � �

or, equivalently, 1XnD0

anLn .x/ D g�cos�1 x

�, � 1 � x � 1. (2.26)

As f .x/ D g�cos�1 x

�is continuous on Œ�1; 1�, F 2 L2.�1; 1/ and (2.26) holds (at

least) in L2 .�1; 1/, with

an D 2nC 1

2

Z 1

�1g�cos�1 x

�Ln .x/ dx:

Hence (2.25) is, at least formally, the solution we wanted.

Problem 2.2.17 (Planar reflection). Let

BC1 D ®.x; y/ 2 R2 W x2 C y2 < 1, y > 0

¯denote the upper half-disc and take u 2 C 2.BC1 / \ C.BC1 / harmonic on BC1 withu .x; 0/ D 0. Prove that the (continuous) function

U .x; y/ D´u .x; y/ y � 0

�u .x;�y/ y < 0

obtained through an odd reflection of u with respect to the x-axis, is harmonic on theentire disc B1.

Solution. Call v the solution to´�v .x; y/ D 0 in B1v D U on @B1

which exists, is unique and is defined by Poisson’s formula. The function v .x;�y/ isharmonic in B1, and has on the boundary the same values of v with opposite signs. Setw.x; y/ D v.x; y/C v.x;�y/. We have:´

�w .x; y/ D 0 in B1w D 0 on @B1:


By uniqueness, it must be w � 0. Therefore v.x; y/ D �v.x;�y/, i.e. v is odd withrespect to the horizontal axis, and in particular v.x; 0/ D 0. So, v solves´

�v .x; y/ D 0 in BC1v D u on @BC1 :

As u solves this problem as well, by uniqueness v � u � U on BC1 . But both v and Uare odd in y, so v � U on B1 and then �U D 0 on B1.

Problem 2.2.18 (Poisson kernel on the upper half-plane). Let g 2 L1.R/\ C .R/ bebounded. Show that there is a unique bounded and continuous solution on ¹y � 0º to´

�u.x; y/ D 0 x 2 R; y > 0

u.x; 0/ D g.x/ x 2 R:

Using the partial Fourier transform write a representation formula for u.

Solution. Uniqueness follows from the reflection principle (Problem 2.2.17) and Li-ouville’s theorem. In fact, if u1, u2 are bounded solutions with the same datum g, thedifference w D u1 � u2 is harmonic on y > 0 and has null datum. Extending w to anodd function on y < 0 gives a harmonic, bounded function on the plane, so constant. Theconstant is zero as w vanishes on y D 0; therefore u1 D u2.

For a representation formula, let us set

bu .�; y/ DZ

Re�i�xu .x; y/ dx:

Then cux .�; y/ D i�bu .�; y/ , buxx .�; y/ D ��2bu .�; y/ ;andbu solves ´buyy .�; y/ � �2bu .�; y/ D 0bu.�; 0/ Dbg .�/ :The general integral of the ODE is:

bu .�; y/ D c1 .�/ ej�jy C c2 .�/ e

�j�jy :

The fact that u is bounded says that7 c1 .�/ D 0, while y D 0 forces

c2 .�/ Dbg .�/ :Hence bu .�; y/ Dbg .�/ e�j�jy :7 ej�jy admits no anti-transform.


The anti-transform8 of e�j�jy is the harmonic function (Poisson kernel on the half-plane)

1

�

y

x2 C y2

so we obtain

u .x; y/ D 1

�

ZR

y

.x � s/2 C y2g .s/ ds: (2.27)

Remark 1. Without assuming boundedness there would be infinitely many solutions, forinstance u .x; y/C cy, u.x; y/C cxy, u.x; y/C cex sin y, : : : :

• Analysis of (2.27)��. We will show that (2.27) is effectively the required solution,i.e. that it is harmonic on y > 0, bounded, and that it agrees with g continuously on y D 0.For y > 0 we can differentiate to any order the integrand, and since the Poisson kernel isharmonic, so is u. Now for y > 0,

1

�

ZR

y

.x � s/2 C y2ds D 1

�

ZR

y

x2 C y2dx D 1

�

arctan

�x

y

�C1�1

D 1: (2.28)

Thus

ju .x; y/j � 1

�

ZR

y

.x � s/2 C y2jg .x/j ds � sup

Rjgj

and therefore u is bounded. Let now .x; y/ ! .x0; 0/; given " > 0, we take

jg .s/ � g .x0/j < "for js � x0j < ı". Then

ju .x; y/ � g .x0/j D 1

�

ZR

y

.s � x/2 C y2jg .s/ � g .x0/j ds

D 1

�

Z¹js�x0j<ı"º

� � � ds„ ƒ‚ …I1

C 1

�

Z¹js�x0j�ı"º

� � � ds„ ƒ‚ …I2

:

From (2.28) we infer I1 < " . As for I2, notice that if jx � x0j < ı"=2, the triangleinequality implies

¹s W js � x0j � ı"º �²s W js � xj � ı"

2

³;

so

I2 � y � 2 supR jgj�

Z¹js�xj�ı"=2º

1

.s � x/2 d D 8 supR jgj�ı"

y;

and thence I2 < " for y small enough.

8 Appendix B.


In summary, if .x; y/ is close enough to .x0; 0/,

ju .x; y/ � g .x0/j < 2"or, in other words u .x; y/ ! g .x0/ for .x; y/ ! .x0; 0/.

Remark 2. The above discussion shows that everything carries through ifg is just boundedand continuous, without assuming integrability.

Problem 2.2.19 (Using the reflection principle). Solve in

BC1 D ®x2 C y2 < 1, y > 0

¯the problem ´

�u .x; y/ D 0 in BC1u D g on @BC1 ;

with g continuous on @B1.

Solution. To use reflections we modify u in order to have vanishing datum on thediameter y D 0. Set

h .x/ D

8<:g .1; 0/ for x > 1

g .x; 0/ for � 1 � x � 1

g .�1; 0/ for x < �1:The function thus defined is continuous, bounded on R and coincides with g .x; 0/ alongy D 0, jxj � 1. By Problem 2.2.18 (in particular Remark 2) the function

v .x; y/ D 1

�

ZR

y

.x � s/2 C y2h .s/ ds

is harmonic on y > 0 andv .x; 0/ D g .x; 0/

for �1 � x � 1. Thenw D u � v

is harmonic on BC1 and moreover,

w .x; 0/ D 0 for � 1 � x � 1;

w D g � v �eg on @BC1 \ ¹y > 0º :We extend eg in an odd way on @BC1 \ ¹y < 0º. From the reflection principle and thePoisson formula we deduce

w .x; y/ D 1 � x2 � y22�

Z 2�

0

eg .cos �; sin �/

.x � cos �/2 C .y � sin �/2d�:

The solution is then u D w C v.


Fig. 2.5 Kelvin transform

Problem 2.2.20 (Kelvin transform in R2). Fix a > 0 and x 2 R2 n¹0º. We call Kelvintransform the map sending x to the point

y D Ta.x/ D a2

jxj2 x: (2.29)

a) Verify that the map Ta is regular on R2 n ¹0º, invertible with inverse T �1a D Ta.

b) Prove that if u is harmonic on � � R2 n ¹0º then

v.x/ D u.Ta.x//

is harmonic on T �1a .�/.

Solution. a) In R2n ¹0º, Ta is C1 (as composition of smooth functions) andy DTa .x/ 6D 0 if x ¤ 0 (Fig. 2.5). Let us find the inverse. Computing the modulus of(2.29) gives

jyj D a2

jxj2 jxj D a2

jxj ; i.e. jxj D a2

jyj :

Substituting into (2.29) we obtain y D a2jyj2a4 x D jyj2

a2 x: So whenever y ¤ 0, Ta is invert-ible and

T �1a .y/ D x D a2

jyj2 y D Ta.y/:

b) As x and Ta.x/ are parallel vectors, it is convenient to use polar coordinates: writex D .r; �/, so that

Ta.x/ D .�; �/;

with r� D a2. Setting u D u.�; �/, we have

v.r; �/ D u

�a2

r; �

D u.�; �/; v�� .r; �/ D u

��

�a2

r; �

D u

��.�; �/;

vr .r; �/ D �a2

r2u

�a2

r; �

D ��

2

a2u.�; �/;

vrr .r; �/ D 2a2

r3u

�a2

r; �

C a4

r4u

�a2

r; �

D 2�3

a4u.�; �/C �4

a4u.�; �/;


and so

�v D vrr .r; �/C 1

rvr .r; �/C 1

r2v�� .r; �/

D 2�3

a4u.�; �/C �4

a4u.�; �/ � �

a2�2

a2u.�; �/C �2

a4u�� .r; �/

D �4

a4

�u.�; �/C 1

�u.�; �/C 1

�2u�� .�; �/

D �4

a4�u.�; �/:

Hence, if u is harmonic on its domain then also v is.

Problem 2.2.21 (Using Kelvin transforms). Let Ta denote the Kelvin transform intro-duced in Problem 2.2.20. We write x D .x1; x2/ and y D .y1; y2/.

a) Show that the straight line x1 D a is mapped by Ta onto the circle�y1 � a

2

�2 C y22 D a2

4(2.30)

and the image of the half-plane x1 > a is the corresponding enclosed disc. Similarly,check that the line x2 D a is sent to the circle

y21 C�y2 � a

2

�2 D a2

4: (2.31)

b) On the quadrant Q D ¹x 2 R2 W x1 > 1; x2 > 1º solve the Dirichlet problem8<ˆ:�u.x/ D 0 x 2 Qu.x1; 1/ D 0 x1 > 1

u.1; x2/ D 1 x2 > 1

u bounded in Q:

Check that the solution is uniquea.

c) On� D ¹x 2 R2 W x21 C x22 � x1 < 0; x21 C x22 � x2 < 0º,

intersection of two discs (Fig. 2.7), solve the Dirichlet problem:8<:�v D 0 in �

v D 1 on ¹x W x21 C x22 � x1 D 0; x21 C x22 � x2 < 0ºv D 0 on ¹x W x21 C x22 � x1 < 0; x21 C x22 � x2 D 0º:

a Use the reflection principle of Problem 2.2.17 (page 103) and the result of Problem 2.2.10(page 92).


Solution. a) The Kelvin transform y D Ta.x/ reads, explicitly:

.y1; y2/ D�

a2

x21 C x22x1;

a2

x21 C x22x2

;

with inverse (as seen in the previous problem)

.x1; x2/ D�

a2

y21 C y22y1;

a2

y21 C y22y2

:

Using the latter, the vertical line x1 D a is mapped to:

a2

y21 C y22y1 D a

i.e. (2.30). Analogously, the right half-plane x1 > a is transformed into

a2

y21 C y22y1 > a;

which is the interior of the circle. In general, all vertical lines x1 D b, b ¤ 0, are mappedonto circles centred on the horizontal axis and passing through the origin. The claim abouthorizontal lines x2 D a is identical, so we leave the details to the reader.

b) To find a solution, we start by observing that the domain is a “cone”, i.e. the union ofrays emanating from (1,1). The boundary is made of two rays, on each of which the Dirich-let datum is constant. So it is natural to seek solutions that are constant (and bounded) onhalf-straight lines from .1; 1/; in other words, solutions of the formU.r; �/ D .�/, where.r; �/ are polar coordinates with pole at .1; 1/. If the solution is of this kind, then8<

:�u D Urr .r; �/C 1

rUr .r; �/C 1

r2U�� .r; �/ D 1

r2 00.�/ D 0 0 < � <

�

2 .0/ D 0

.�2/ D 1;

i.e. .�/ D 2�=� . Going back to the Cartesian frame system, we find

u.x1; x2/ D 2

�arctan

�x2 � 1x1 � 1

:

• The same conclusion is reached by noting that the complex function

w D log .z � 1/ D log jz � 1j C i arg .z � 1/is holomorphic on Q. Its imaginary part

arg .z � 1/ D arctan

�x2 � 1x1 � 1


11:5

22:5

33:5

4

1

2

3

4

0

20

40

60

x1x2

Fig. 2.6 u .x; y/ D 2� arctan x2�1

x1�1

is therefore harmonic on Q, and 0 on x2 D 1, x1 > 1, while it equals �=2 on x2 D 1,x1 > 1. The function (Fig. 2.6)

u.x1; x2/ D 2

�arg .z � 1/ D 2

�arctan

�x2 � 1x1 � 1

is therefore the solution we wanted.

• Uniqueness. The difficulties hide in the discontinuity of the datum at .1; 1/, besidesthe fact that the domain is unbounded. Assume that there exist two solutions u and v andset w D u � v. Since the datum is discontinuous, we may assume that u, v, and hencealso w, are continuous only on Q n ¹.1; 1/º. The function w solves8<

:�w.x/ D 0 x 2 Qw.x/ D 0 x 2 @Q n ¹.1; 1/ºw bounded on Q

(as difference of bounded functions, w is bounded). Let us extend w to R2 by reflection,as follows:

w .x; 1 � k/ D �w .x; 1C k/ , x > 1; k > 0 (odd w.r.t. y D 1)

andw .1 � h; y/ D �w .1C h; y/ , h > 0; y 2 R (odd w.r.t. x D 1).

The reflection principle (Problem 2.2.17 on page 103) guarantees that the extended wis harmonic on the whole plane, except at .1; 1/ at most. On the other hand w is bounded,


Fig. 2.7 Domain of Problem 2.2.21 c)

so we may use Problem 2.2.10 (page 92), wherebyw has a removable singularity at .1; 1/:setting w .1; 1/ D 0 gives a harmonic and bounded function on the plane.

By Liouville’s theoremw is constant, and zero on the axes, hence identically zero; thismeans that u coincides with v.

c) By part a) the Kelvin map T1 transforms the half-plane x1 > 1 in the discy21 C y22 � y1 < 0, and the half-plane x2 > 1 into y21 C y22 � y2 < 0, and vice versa(recall that T �11 D T1). Therefore the intersection of the two discs (see Fig. 2.7), i.e. �,is mapped to the quadrant ¹x1 > 1º \ ¹x2 > 1º:

Q D T1.�/:

Set v.x/ D u.T1.x//, with u harmonic as determined above:

v.x1; x2/ D 2

�arctan

�x21 C x22 � x2x21 C x22 � x1

:

The function v is defined on�, and it is not hard to check that it fulfils the boundary con-ditions. By Problem 2.2.20 (page 107), moreover, since u is harmonic also v is harmonic,and must be the required solution.

Problem 2.2.22 (Dirichlet problem on the half-plane, discontinuous data). Solve onthe half-plane: 8<

ˆ:�u.x; y/ D 0 x > 0; y 2 R

u.0; y/ D 1 y > 0

u.0; y/ D �1 y < 0

u bounded on the half-plane.

Verify that the solution is unique, using the reflection principle and Problem 2.2.10(page 92).


Solution. 1st method. As in Problem 2.2.21.b), we try to find a solution that is constanton half-lines emanating from the origin. Set u D u.r; �/ D .�/, so that

�u D urr .r; �/C 1

rur .r; �/C 1

r2u�� .r; �/ D 1

r2 00.�/:

Then solves 8<: 00.�/ D 0 ��

2< � < �

2

.�2/ D 1

.��2/ D �1

so .�/ D 2�� or, in Cartesian coordinates,

u.x; y/ D 2

�arctan

y

x.

2nd method. The complex function

w D log z D log jzj C i arg z

is holomorphic on x > 0. Its imaginary part

arg z D arctany

x

is harmonic on x > 0, and is equal to �=2 and ��=2 along x D 0, y > 0 and x D 0,y < 0, respectively. Hence

u .x; y/ D 2

�arg z D 2

�arctan

y

x:

3rd method. We use the Poisson kernel for the half-plane x > 0 (Problem 2.2.18 onpage 104):

u .x; y/ D 1

�

Z 1

0

x

x2 C .y � s/2 ds � 1

�

Z 0

�1x

x2 C .y � s/2 ds D

D x

�

Z 1

0

�1

x2 C .y � s/2 � 1

x2 C .y C s/2

ds D

D 2

�arctan

y

x:

• Now we check that the problem admits at most one solution. We may argue in anal-ogy to Problem 2.2.21,b), where the details can be found. So let u and v be solutions, andset w D u � v, where 8<

:�w.x; y/ D 0 x > 0; y 2 R

w.0; y/ D 0 y 2 R n ¹0ºw bounded:


Extend w to R2 in an odd way:

Qw.x; y/ D´w.x; y/ x � 0

�w.�x; y/ x � 0:

By the reflection principle (Problem 2.2.17 on page 103) Qw is harmonic on R2 n ¹.0; 0/º,and can be extended harmonically to the origin as well, by Problem 2.2.10 (page 92). AsQw is harmonic and bounded everywhere, by Liouville’s theorem it is constant. Now, since

it vanishes on the x-axis, the constant is necessarily zero: Qw � 0, and u coincides with v.

Problem 2.2.23 (Uniqueness for the exterior Dirichlet problem). Let � � R2 be abounded domain containing the origin and �e D R2 n �. Prove that the followingproblem 8<

:�u D 0 in �eu D g on @�e; with g continuous

u bounded in �e

(2.32)

has at most one solution of class C 2.�e/ \ C.�e/, by answering the following ques-tions:

a) Solve the Dirichlet problem 8<:�v D 0 in Cr;Rv D 0 on @Brv D 1 on @BR;

on the annulus Cr;R D BR n Br , where

Br � � � BR

and Br , BR are centred at the origin.

b) Show that given two solutions u1; u2 of (2.32),

w D´u1 � u2 in �e0 in �

(2.33)

satisfiesjwj � Mv in Cr;R

where v is as in part a) and M is a suitable constant.

c) Let R tend to infinity in the previous inequality, and deduce w � 0. Conclude that(2.32) admits at most one solution.

Solution. a) The Dirichlet problem is solved on the annulus in Problem 2.2.15 (page98). Here the problem is rotationally invariant, thus we seek radially symmetric solutions


(also called radial solutions), which on the plane are of the form

v.�/ D C1 C C2 log �:

The initial conditions forcev.�/ D log .�=r/

log .R=r/:

Note v is non-negative on Cr;R.

b) The function w in (2.33) is continuous on R2, identically zero on � and harmonicand bounded on �e .

Let M be a positive number such that jwj � M in �e . Then

WC D Mv � wsolves 8<

:�WC D 0 in Cr;R \�eWCj@ D Mvj@ � 0

WCj@BRD M � wj@BR

� 0:

By the maximum principle, WC � 0, i.e.

w � Mv

on Cr;R \�e . Given the definition of w, the inequality extends easily to Cr;R. Similarly,let

W� D �Mv � w:The same argument shows W� � 0, so

w � �Mv

on Cr;R. In conclusion:�Mv � w � Mv in Cr;R;

which was our claim.

c) In b) we showed that

jw.�; �/j � Mlog .�=r/

log .R=r/; (2.34)

with r < � < R and r; R such that

Br � � � BR:

In particular, � being bounded, the inequality holds for R arbitrarily large. So fix r , andlet .�; �/ be a given point in �e . If R > �, then .�; �/ 2 Cr;R. Since

log .�=r/

log .R=r/! 0 as R ! C1;


from (2.34) it follows that w.�; �/ D 0, and therefore u1 � u2. Thus, problem (2.32)admits, at most, one solution.

Problem 2.2.24 (Uniqueness for the exterior Neumann problem). Let � � R2 be abounded domain containing the origin and �e D R2 n�. Prove that the problem8<

:�u D 0 in �e@�u D g on @�e; with g continuous

u bounded in �e

(2.35)

has at most one solution of class C 2.�e/\C 1.�e/ up to additive constants, followingthe steps below.

a) Using the Kelvin transform (Problem 2.2.20 on page 107), prove that a solution uof problem (2.35) satisfiesˇ

@u

@xi.x/

ˇ� C

jxj2 .i D 1; 2/

whenever jxj is large enough, for a suitable constant C .

b) Let u1; u2 be solutions of (2.35) and w D u1 � u2 in �e . Suitably integrating byparts and using part a), prove that for large RZ

e\BR

jrw.x/j2 dx � C 0

R:

c) Let R go to infinity and deduce w is constant, whence (2.35) has, up to additiveconstants, at most one solution.

Solution. a) Take a > 0 so large that B D Ba.0/ � �, and hence Be D ¹x 2 R2 Wjxj > aº � �e . Consider the Kelvin transform

y D Ta.x/ D a2

jxj2 x; x D T �1a .y/ D Ta.y/ D a2

jyj2 y:

From Problem 2.2.20 (page 107), Ta.Be/ D B n ¹0º and

v.y/ D u.Ta.y//

is harmonic on that set. As u is bounded, so is v. The result of Problem 2.2.10 (page 92)says that v is extendable to B harmonically. Consequently v 2 C 2.B/, so it has boundedfirst (and second) derivatives on any compact subset of B . In particular there exists a con-stant c0 such that, for j D 1; 2,ˇ

@v

@yj.y/

ˇ� c0 for jyj � a

2:


But

@u

@xi.x/ D

2XjD1

@v

@yj.y/@yj

@xi.x/ D

2XjD1

@v

@yj.y/

�a2

jxj2 ıij � a2xixj

jxj4;

where ıij is the usual Kronecker symbol (ıij D 1 if i D j , 0 otherwise). At this point itis enough to notice that jxixj j � jxj2 (for any i; j ) to have

ˇ@u

@xi.x/

ˇ� 2a2

jxj22X

jD1

ˇ@v

@yj.y/

ˇ� 4a2c0

jxj2 ; for jxj � 2a: (2.36)

It might be useful to notice that the estimates obtained only depend on the fact u is har-monic and bounded (the Neumann condition was not used).

b) Integrating by parts, we haveZ e\BR

'.x/�w.x/ dx D �Z e\BR

rw.x/ � r'.x/ dx CZ@. e\BR/

'.x/@�w.x/ d�;

where ' is an arbitrary smooth function (the formula holds since the domain of integrationis bounded). To obtain the L2 norm of rw we must choose ' D w. Then

Z e\BR

w.x/�w.x/ dx DZ e\BR

jrw.x/j2 dx �Z@. e\BR/

w@�w d�:

Recalling that �w D 0 in � and @�w D 0 on @� we findZ e\BR

jrw.x/j2 dx DZ@BR

w@�w d� � M

Z@BR

jrwj d�;

where M is some constant such that jwj � M on �e (w is bounded as difference ofbounded functions). If R � 2a, from (2.36) we obtain (on @BR we have jxj D R)

Z \BR

jrw.x/j2 dx � 8a2c0M

Z@BR

1

jxj2 d� D 16�a2c0M

R:

c) The inequality holds for any R large enough, so we may take the limit to obtainZ

jrw.x/j2 dx � 0;

i.e. jrw.x/j2 D 0 almost everywhere on �e . By regularity the claim follows.


2.2.3 Potentials and Green functions

Problem 2.2.25 (Newtonian potential). Compute the Newtonian potential of a homo-geneous mass distribution (of density D 1) on the disc B1 centred at the origin.

Solution. The required potential (defined up to additive constants) is

u .x; y/ D � 1

4�

ZB1

logh.x � �/2 C .y � �/2

id�d�

and it is C 1 on the entire R2. For an explicit expression, first we observe that the prob-lem is invariant under rotations and therefore u has radial symmetry. Thus u D u .r/,r2 D x2 C y2, and

�u D´

�1 0 � r < 1

0 r > 1

together withu .1�/ D u .1C/ , ur .1�/ D ur .1C/ . (2.37)

For r > 1; u is a radial harmonic functions, thus

u.r/ D a log r C b; r > 1:

On the other hand, from

urr C 1

rur D �1;

we easily find

u .r/ D c C d log r � 1

4r2; r < 1:

Since u is bounded for r < 1, it follows d D 0. By (2.37) we obtain

c � 1

4D b and a D �1

2:

Choosing c D 1=4, then, we find

u .r/ D´14

�1 � r2� r � 1

�12

log r r > 1:

Problem 2.2.26. Let u D u .x/ be the double-layer potential with density on thecircle C D ¹x D .x1; x2/ W � D 1º ; �2 D x21 C x22 : Outside the circle u is given bythe harmonic function

u .x/ D �x1x2�4

:

Compute :


Solution. Note that since the double-layer potential of a constant is zero outside thedisk bounded by C , is determined up to an additive constant.

The relation between u and is given, at any point z D .z1; z2/ on the circle, by theformula

uE .z/ � uI .z/ D .z/

whereuE .z/ D lim

x!z;>1u .x/ ; uI .z/ D lim

x!z;<1u .x/ :

We know that uE .z/ D �z1z2: To compute uI we use the continuity of the normalderivative @u

@across C . We find

@uE

@�D @uI

@�.z/ D 2z1z2: (2.38)

We deduce that u is harmonic inside C and satisfies the Neumann condition (2.38). Notethat

RC z1z2 D 0, so that the Neumann problem has solutions, differing by additive con-

stants. Therefore inside C we find u .x/ D x1x2 C k, k 2 R, and

.z/ D uE .z/ � uI .z/ D �2z1z2 C k k 2 R:

Problem 2.2.27. Let f 2 C.R2/ be a function with compact support K and define

u .x/ D � 1

2�

ZR2

log jx � yj f .y/ dy:

Show that

u .x/ D �M2�

log jxj CO.jxj�1/; as jxj ! C1where

M DZK

f .y/ dy:

Consequence: u .x/ ! 0 at infinity if and only if M D 0:

Solution. We haveˇu .x/C M

2�log jxj

ˇD 1

2�

ˇZR2

Œlog jx � yj � log jxj� f .y/ dy

ˇ� 1

2�

ZK

ˇlog

� jx � yjjxj

ˇjf .y/j dy:

Since K is compact we may assume that K � BR .0/, R � 1. If y 2 K then jyj � R. Letjxj � 2R. Then

jx � yj � jxj � jyj � jxj �R � jxj2

� 1;

so the last integral is finite.


Assume jx � yj � jxj. Since log.1C t / � t , t > �1,

0 < log

� jx � yjjxj

� log

� jxj C jyjjxj

� log

�1C R

jxj

� R

jxj :

Now let jx � yj � jxj. We have, since jyj � jxj=2,

jxjjx � yj � 1C jyj

jxj � jyj � 1C 2jyjjxj ;

so that ˇlog

� jx � yjjxj

ˇD log

� jxjjx � yj

� log

�1C 2jyj

jxj

� 2R

jxj :Thus ˇ

u .x/C M

2�log jxj

ˇ� 2R

� jxjZK

jf .y/j dy;

which yields the desired result.

Problem 2.2.28 (Computing Green functions). Determine the Green functions of theLaplace operator for the following sets.

a) The half-plane PC D ¹x D .x1; x2/ W x2 > 0º.

b) The disc B1 D ®x D .x1; x2/ W x2 C y2 < r

¯.

c) The half-disc BC1 D ®x D .x1; x2/ W x2 C y2 < r , x2 > 0

¯.

Solution. a) For any given y 2 PC, the Green function G D G .x; y/ is harmonic onthe half-plane, G .x; y/ D 0 on x2 D 0, and

�xG .x; y/ D �ı2 .x � y/ ; (2.39)

where ı2 .x � y/ is the Dirac distribution at y. Let us use the method of images. We knowthat the fundamental solution

� .x � y/ D � 1

2�log jx � yj

satisfies (2.39). If y D .y1; y2/, we define ey D .y1;�y2/, the mirror image of y withrespect to the y1-axis. The function � .x �ey/ is harmonic on PC and coincides with� .x � y/ on x2 D 0. Hence

G .x; y/ D � .x � y/ � � .x �ey/ .

b) Using again the method of images, define

y� D T1 .y/ D y

jyj2


to be the Kelvin image of y ¤ 0. For jxj D 1,

jx � y�j2 D 1 � 2x � y

jyj2 C 1

jyj2

D 1

jyj2�1 � 2x � y C jyj2

�D 1

jyj2 jx � yj2 .

If y ¤ 0, set

G .x; y/ D � 1

2�¹log jx � yj � log.jyj jx � y�j/º :

This givesG .x; y/ D 0

for jxj D 1, y ¤ 0 and�xG .x; y/ D �ı .x � y/ in B1:

When y D 0, we simply define

G .x; 0/ D � 1

2�log jxj .

When x ¤ 0 and y ! 0, note that

G .x; y/ ! G .x; 0/ :

c) Call GB1the Green function for the disc B1 of the previous part, and defineey D .y1;�y2/. Then

GCB1.x; y/ D GB1

.x; y/ �GB1.x;ey/ :

*Problem 2.2.29 (Symmetry of Green functions). Let G.x; y/ be the Green functionassociated to the Laplace operator on a bounded, smooth domain � � R3. Prove that

G.x1; x2/ D G.x2; x1/

for any pair x1; x2 2 �.

Solution. First recall that (in dimension three) the Green function G can be written as

G.x; y/ D � .x � y/ � g .x; y/ D 1

4�jx � yj � g.x; y/;

where g .x; �/ is harmonic on � for fixed x, continuous on � and satisfies

g .x; �/ D � .x � �/ on @�:

In particular, G.x; �/ is non-negative on �, null on @� and

G.x; y/ � 1

4� jx � yj in � ��. (2.40)


If x1 D x2 the claim is trivial, so we take x1 6D x2. We carve out of � two ballsBr .x1/, Br .x2/ with radius r small enough so to be disjoint. On the resulting domain

�r D � n .Br .x1/ [ Br .x2//;the functions u.y/ D G.x1; y/, v.y/ D G.x2; y/ are harmonic and vanish on @�. Hencewe can invoke Green’s identity:Z

r

.v�u � u�v/ dx DZ@ r

.v@�u � u@�v/ d�:

Because of our choice for u and v the latter reduces toZ@Br .x1/[@Br .x2/

.v@�u � u@�v/ d� D 0

i.e. Z@Br .x1/

.v@�u � u@�v/ d� DZ@Br .x2/

.u@�v � v@�u/ d�: (2.41)

Let us compute the limit of the left-hand side when r ! 0. Since v is smooth near x1, wehave jrvj � M on @Br .x1/ provided r is small enough. From (2.40) we also have

0 � u � 1

4�ron @Br .x1/ .

ThenˇZ@Br .x1/

u@�v d�

ˇ�Z@Br .x1/

u j@�vj d� � M

4�r4�r2 D Mr ! 0 for r ! 0:

On the other handZ@Br .x1/

v@�ud� D 1

4�

Z@Br .x1/

v@�

�1

j� � x1jd� C

Z@Br .x1/

v@�g.x1; � / d�:

The last integrand is a smooth function in a neighbourhood of x1, so the integral tends to0 as r ! 0. Moreover

@�

�1

j� � x1j

D r�

1

j� � x1j

� � D � � x1j� � x1j3 � � � x1

j� � x1j D 1

j� � x1j2 ;

and then

1

4�

Z@Br .x1/

v@�

�1

j� � x1jd� D 1

4�

Z@Br .x1/

1

j� � x1j2 v d� D

D 1

j@Br .x1/jZ@Br .x1/

v d� ! v.x1/ for r ! 0:

Overall, we have foundZ@Br .x1/

.v@�u � u@�v/ d� ! v.x1/ for r ! 0: (2.42)


Similar computations showZ@Br .x2/

.u@�v � v@�u/ d� ! u.x2/ for r ! 0: (2.43)

Taking the limit in (2.41), and with the aid of (2.42) and (2.43), we finally get

v.x1/ D u.x2/;

that is, the desired symmetry for G:

G .x2; x1/ D G .x1; x2/ :

Remark. The Green function for the Laplace operator is always symmetric, in any dimen-sion. The proof is identical to the above one.

*Problem 2.2.30 (Poisson formula, double-layer potential). Recover the Poisson for-mula in the plane by representing the solution of´

�u D 0 in BRu D g on @BR

as double-layer potential.

Solution. We have to find a function W @BR ! R such that the solution u of thegiven Dirichlet problem reads

u .x/ DZ@BR

@

@�

�� 1

2�log jx � � j

.� / d� D 1

2�

Z@BR

.x � � / � � .� /

jx � � j2 .� / d�:

Let us briefly recall the properties of double-layer potentials. First of all, if is continu-ous, the above u is harmonic on BR. In fact, if x 62 @BR the denominator of the integrandnever vanishes, so we can differentiate inside the integral, obtaining that u is harmonic.The unknown density is determined so to fulfill the boundary condition. We remind thatif x 2 BR, z 2 @BR and x ! z, then

u .x/ ! 1

2�

Z@BR

.z � � / � � .� /

jz � � j2 .� / d� � 1

2.z/:

From u .x/ ! g .z/, we obtain the integral equation

1

2�

Z@BR

.z � � / � � .� /

jz � � j2 .� / d� � 1

2.z/ D g .z/ .


Now observe �.� / D �=R (note that � 2 @BR, so j� j D R). Substituting we get

g .z/C 1

2.z/ D 1

2�R

Z@BR

z � � � j� j2jzj2 � 2z � � C j� j2 .� / d� D

D 1

2�R

Z@BR

z � � �R22.R2 � z � � /

.� / d� D � 1

4�R

Z@BR

.� / d�:

(2.44)

We have to compute the integral in the last term to finally get . To this end we integrate(2.44) on @BR:Z

@BR

g .� / d� D 2�R ��

� 1

4�R

Z@BR

.� / d�

� 1

2

Z@BR

.� /d�;

whence Z@BR

.� / d� D �Z@BR

g .� / d�:

Substituting in (2.44) gives

.z/ D �2g.z/C 1

2�R

Z@BR

g .� / d�:

Now that we have , we revert to the initial definition of u, and recall that if x 2 BR,

1

2�

Z@BR

.x � � / � � .� /

jx � � j2 d� D 1:

Then

u .x/ D 1

2�

Z@BR

.x � � / � � .� /

jx � � j2�2g.� /C 1

2�R

Z@BR

g .� / d�

�d� D

D 1

2�

Z@BR

�2.x � � / � � .� /

jx � � j2 g.� /d� � 1

2�R

Z@BR

g .� / d�:

Substituting �.� / D �=R, we obtain

D 1

2�R

Z@BR

�2x � � C 2R2

jx � � j2 g.� /d� � 1

2�R

Z@BR

g .� / d� D

D 1

2�R

Z@BR

R2 � jxj2 C jx � � j2jx � � j2 g.� /d� � 1

2�R

Z@BR

g .� / d� D

D R2 � jxj22�R

Z@BR

g.� /

jx � � j2 d�;

which is precisely the Poisson formula on the plane.


*Problem 2.2.31 (Non-homogeneous Poisson-Dirichlet problem). Prove the represen-tation formula

u .x/ D �Z@

h .� / @�G .x; � / d� �Z

f .y/G .x; y/ dy;

for the solution of the problem ²�u D f in �u D h on @�

where G is the Green function of � � R2.

Solution. Recall that in dimension two

G.x; y/ D � .x � y/ � g.x; y/D � 1

2�log jx � yj � g.x; y/;

where g.x; �/ solves ´�yg .x; y/ D 0 y 2 �g .x; y/ D � 1

2�log jx � yj y 2 @�:

The function u can be written as sum of three potentials (Newtonian, double- and singlelayer)

u .x/ DZ@

@�u .� / �jx � � jd� �Z@

h .� / @��jx � � jd� �Z

f .y/ �jx � yjdy:

At the same time, applyingZ

. �' � '� /dx DZ@

. @�' � '@� /d� (2.45)

to' D u and D g.x; �/;

gives

0 D �Z@-g .x; � / @�u .� / d� C

Z@

h .� / @�g .� / d� CZ

g .y/ f .y/ dy: (2.46)

Adding up (2.46) and (2.45) furnishes straightaway the claim.


2.3.1. (Mixed problem on the square, separation of variables) Solve, on the square

¹Q D .x; y/ W 0 < x < 1; 0 < y < 1º ;


the problem 8<:�u D 0 in Q

uy .x; 0/ D sin �x2 , u .x; 1/ D 0 0 � x � 1

u .0; y/ D ux .1; y/ D 0 0 � y � 1:

2.3.2. Determine all harmonic functions on the annulus

a < r < b (r2 D x2 C y2)

that satisfy the following boundary conditions:

a) u .a; �/ D 0, u .b; �/ D cos � .b) u .a; �/ D cos � , u .b; �/ D U sin 2� .

2.3.3. Let B1;2 D ¹.r; �/ 2 R2 W 1 < r < 2º. Discuss whether the Neumann problem8<:�u D �1 in B1;2u� D cos � on r D 1

u� D �.cos �/2 on r D 2;

has solutions, depending on the parameter � 2 R.

2.3.4. (Neumann-Robin problem on a half-strip) Solve, on the half-strip

S D ¹.x; y/ W 0 < x < 1; 0 < y < 1º

the problem 8<:�u D 0 in Q

uy .x; 0/ D uy .x; 1/C hu .x; 1/ D 0 0 � x � 1u .0; y/ D g .y/ , u .1; y/ D 0 0 � y � 1

with h > 0.

2.3.5. Show that for any integer n � 1

u.x; y/ D 1

n2sinh ny cosnx

solves the Cauchy problem 8<:�u D 0 in x 2 R; y > 0

u.x; 0/ D 0 for x 2 R

uy.x; 0/ D 1n cosnx for x 2 R:

Deduce that the Cauchy problem for the Laplace operator is not well posed, because the solutiondoes not depend continuously on the data.

2.3.6. Let u D u .x/ be harmonic on Rn. Using the mean-value property show that the function

v .x/ D u .Mx/

is harmonic on Rn, where M is a rotation Rn (represented by an orthogonal matrix).


2.3.7. Let Q denote the square

¹.x; y/ W �1 � x � 1, � 1 � y � 1ºand Li , i D 0; : : : 3; its sides, labelled counter-clockwise from the bottom one

L0 D ¹.x;�1/ W �1 < x < 1º:The solution u of 8<

:�u D 0 in Q

u D 1 on L0u D 0 on Li ; i D 1; 2; 3;

(2.47)

is continuous and bounded onQ, except at the corners p D .�1; 0/ and q D .1; 0/. Compute u.0; 0/.

2.3.8. Verify that

u.x; y/ D 1 � x2 � y21 � 2x C x2 C y2

is harmonic on B1.0; 0/. Since the numerator vanishes on @B1.0; 0/, should we not have u � 0, byuniqueness? Explain this (apparent) contradiction.

2.3.9. Let � be a bounded domain in Rn and suppose u 2 C 2 .�/ \ C��

is a solution of theequation �u D u3 � u with u D 0 on @�. Show that juj � 1:

2.3.10. Let u � 0 be harmonic on B4.0; 0/ � R2 with u.1; 0/ D 1. Using Harnack’s inequalityfind upper and lower bounds for u.�1; 0/.

2.3.11. Determine for which ˛ 2 R the function

u.x/ D jxj˛

is subharmonic on Rn n ¹0º.

2.3.12. Let u be a harmonic function on the open set � � R2, continuous on �. Call .x0; y0/ apoint in � where u.x0; y0/ D 2 and let E1 be the set

E1 D ¹.x; y/ 2 � W u � 1º:Show that @E1 cannot consist only of a (smooth) closed curve contained in �.

2.3.13. Solve, on

BC1 D B1 \ ¹y > 0º D ¹.x; y/ 2 R2 W x2 C y2 < 1; y > 0ºthe Dirichlet problem: 8<

:�u.x; y/ D 0 in �

u.x; 0/ D 0 �1 � x � 1

u.x; y/ D y3 x2 C y2 D 1; y � 0:

2.3.14. (Neumann problem and reflection principle) Consider

BC1 D°.x; y/ 2 R2 W x2 C y2 < 1, y > 0

±


and u 2 C 2.BC1 / \ C.BC1 /; harmonic on BC1 , with uy .x; 0/ D 0:

a) Prove that

U .x; y/ D´u .x; y/ y � 0

u .x;�y/ y < 0;

obtained by evenly reflecting u about the x-axis, is harmonic on the whole B1.

b) Let u be the solution to the mixed problem8<:�u .x; y/ D 0 in BC1u .x; y/ D x2 on @BC1 , y > 0

uy .x; 0/ D 0 �1 � x � 1

on BC1 D ®x2 C y2 < 1, y > 0

¯. Compute u .0; 0/.

** 2.3.15. Consider the Neumann problem´�u .x; y/ D 0 in PCuy .x; 0/ D g .x/ x 2 R

(2.48)

on the half-plane PC D ¹.x; y/ W y > 0º ; where g is smooth, zero outside an interval Œa; b�, andR ba g D 0. Prove that the problem possesses a bounded solution u 2 C 1.P

C/, unique up to an

additive constant, and find it explicitly.

2.3.16. (Dirichlet problem on a circular sector. Counterexample to Hopf’s principle) a) In thesector

S˛ D ¹.r; �/ W r < 1; 0 < � < ˛ < 2�º ;solve the Dirichlet problem8<

:�u D 0 r < 1; 0 < � < ˛

u.r; 0/ D u.r; ˛/ D 0 r � 1

u.1; �/ D g.�/ 0 � � � ˛

where g is regular and g.0/ D g.˛/ D 0.

b) If ˛ < � , show that there exist positive harmonic functions on S˛ that vanish, together with theirgradients, at the origin.

2.3.17. Using the known Poisson formula for the disc and the Kelvin transform, find the Poissonformula for the problem (written in polar coordinates):8<

:�u D 0 in Beu.1; �/ D g.�/ 0 � � � 2�; with g 2 C 1 and 2�-periodic

juj � M in Be

on the exterior of the circle Be D ¹x 2 R2 W jxj > 1º.


2.3.18. Let B1 be the unit ball centred at the origin in R3, and u be the solution to Dirichletproblem ´

�u.x; y; z/ D 0 in B1u.x; y; z/ D x4 C y4 C z4 on @B1:

Compute the maximum and minimum values of u on B1.

2.3.19. In relation to Problem 2.2.17 (page 103), state and prove the reflection principle in dimen-sion three. Deduce that the Dirichlet problem on the half-space has at most one bounded solution.

2.3.20. (Removable singularities) Given the ballB D B1.0/ � Rn, let u be harmonic onB n¹0ºand such that, as jxj ! 0 8<

:u .x/

log jxj ! 0 n D 2

jxjn�2u .x/ ! 0 n � 3:

Prove that the possible discontinuity at 0 is removable and, therefore, that u can be defined at 0 soto become harmonic on the whole B (see Problem 2.2.10 on page 92).

2.3.21. Referring to Problem 2.2.5 (page 89), give examples of:

a) Non-constant subharmonic functions on R2 that are bounded from below.

b) Non-constant subharmonic functions on R3 that are bounded.

2.3.22. Suppose that the functions ui , i 2 N, are harmonic on the domain � � Rn, and thatthey form a pointwise-monotone increasing sequence. Show that if the sequence converges at somepoint x0 2 �, then it converges uniformly on every compact setK � �. Deduce that the limit U isharmonic on �.

2.3.23. (Kelvin transform in R3) Take x 2 R3 n ¹0º, a > 0, and set

Ta.x/ D a2

jxj2 x:

Verify that Ta is a smooth map of R3 n ¹0º to itself, with smooth inverse, and that if u is harmonicon � � R3 then

v.x/ D a

jxju.Ta.x//

is harmonic on T �1a .�/.

2.3.24. (Exterior Dirichlet problem) Define Be D ¹x 2 R3 W jxj > 1º. Using the Kelvin trans-form in three dimensions, solve the problem8<

:�u D 0 in Beu D g on @Beu .x/ ! A for jxj ! 1;

where g is continuous. Examine in detail the case g .x/ D x1.


2.3.25. (Uniqueness for the exterior Robin problem) Let� � R2 be a smooth, bounded domain,and �e D R2 n�. Prove that the Robin problem8<

:�u D 0 in �e@�uC ˛u D g on @�e ; with g continuous and ˛ � 0

u bounded in �e

has, at most, one solution in C 2.�e/ \ C 1.�e/.2.3.26. Consider the Neumann problem on the exterior domain �e � R38<

:�u D 0 in �e@�u D g on @�eu ! 0 for jxj ! 1:

(2.49)

Show that if ju .x/j � M jxj�1�", with " > 0, as jxj ! 1, thenZ@ gd� D 0

is a necessary condition for solving (2.49).

2.3.27. (A variant of Liouville’s theorem) Prove that a harmonic function u on Rn such thatZRn

jru.x/j2 dx < C1 (2.50)

is constant.

2.3.28. (Sublinearity of stress functions) Denote by � the cross-section of a cylinder parallel tothe z-axis. An external force produces a shear stress on each section. If �1 and �2 denote the scalarcomponents of the stress on the planes .x; z/ and .y; z/, there is a function v D v.x; yI z/ (calledstress function) such that

vx D �1; vy D �2:

In suitable units v solves the problem´vxx C vyy D �2 in �

v D 0 on @�:

Assuming v 2 C 2.�/ \ C 1.�/, prove that jrvj2 assumes its maximum on @�.

2.3.29. (Bôcher’s theorem). Let u be harmonic and positive onB1 .0/n.0/ � Rn, and continuousup to the boundary @B1 .0/. Show that:

u .x/ D´a log jxj C v .x/ n D 2

a jxj2�n C v .x/ n � 3(2.51)

for some constant a � 0; where v is harmonic and continuous on B1 .0/.

2.3.30. The potential u D u .x; y/ of a distribution of mass inside the disc r2 D x2 C y2 < 1 isgiven by

u .r/ D �

8

�1 � r4

�:


Determine the density and the Newtonian potential for r � 1.

2.3.31. Compute the Newtonian potential of a homogeneous distribution of mass (density D 1)in the ball B1 � R3, at the origin.

2.3.32. Let u D u .x/ be the single-layer potential with density on the unit circle C D¹x W jxj D 1º. Outside the circle u is given by the harmonic function

u .x/ D x1

�2

�1C x2

�2

;

where x D .x1; x2/ and � D jxj. Compute .

2.3.33. Consider the harmonic function

u .x1; x2/ D x1

x21 C x22

outside a bounded smooth domain � in R2. Can u be represented as a double-layer potential withbounded density on @�?

2.3.34. Determine Green functions for these sets:

a) The half-space PC D ¹x D .x1; x2; x3/ W x3 > 0º.b) The ball B1 D ®

x D .x1; x2; x3/ W x2 C y2 C z2 < r¯

.

c) The half-ball BC1 D ®x D .x1; x2; x3/ W x2 C y2 C z2 < r , x3 > 0

¯:

2.3.1 Solutions


u .x; y/ D 2

�

°sinh

�y

2� tanh

�

2cosh

�y

2

±sin

�x

2:

Solution 2.3.2. The solutions are (Fig. 2.8):

a) u .r; �/ D b

b2 � a2 r � a2

r

!cos � .

b) u .r; �/ D a2

b2 C a2

r � b2

r

!cos � C b2U

b2 C a2

r2 C a4

r2

!sin 2�:

Solution 2.3.3. We have to check the compatibility conditionZB1;2

�u.x/ dx DZ@B1;2

@�u.s/ ds;

that is

�jB1;2j D �Z@B1

cos � ds CZ@B2

� cos2 � ds;

or�3� D 2��:


�1:5 �1 �0:5 0 0:5 1 1:5 2 �2�1

0

1

2

�1

0

1

x

y

Fig. 2.8 u .r; �/ D 23

�r � 1

r

�cos �

Thus the problem can not be solved unless � D �3=2. Separating the variables we find that for� D �3=2 the problem is solved by

u.r; �/ D aC 1

2log r � 1

4r2 C 1

3

�r C 4

r

cos � � 1

5

�r2 C 1

r2

cos 2�;

with a 2 R arbitrary.


u .x; y/ D1XnD1

8<: 2�h2 C �2n

�hC

�h2 C �2n

� Z 1

0g .z/ cos�nz dz

9=; e��nx cos�ny

where �n are the positive solutions of � tan� D h.

Solution 2.3.5. As ˇ1

ncosnx

ˇ� 1

n;

the datum uy.x; 0/ D 1n cosnx tends to 0 uniformly on R. On the other hand for large n the solution

may become arbitrarily large, even at points .x; y/ with jyj extremely small. In fact, given a smallı > 0 we have

limn!1u .0; ı/ D lim

n!11

n2sinh nı D C1

and thus the solution does not depend continuously on the data.

Solution 2.3.6. As composition of continuous functions, v is continuous on Rn. Hence it sufficesto prove that v fulfils one of the averaging formulas on BR.x/, for any x 2 Rn and any R > 0; for


instance:

u.x/ D 1

jBR .x/jZBR.x/

v .y/ dy D 1

jBR .x/jZBR.x/

u .My/ dy:

As M is orthogonal (MT D M�1), we have jdetM j D 1. Setting z D My then,

dz D j detM jdy D dy;

and we can rewrite

1

jBR .x/jZBR.x/

v .y/ dy D 1

jBR .Mx/jZBR.Mx/

u .z/ dz D u.Mx/ D v.x/;

where the mean-value property of u was employed (as jBR .x/j D jBR .Mx/j). So, v satisfies themean-value property and is therefore harmonic on Rn.

Solution 2.3.7. Suppose for the moment that the solution of problem (2.47) is unique. The valueu .0; 0/ could be computed directly from the analytic expression of u obtained from variable sepa-ration. A better way to proceed exploits the domain’s symmetry, and avoids explicit computations,as follows. Let M W R2 ! R2 be the clockwise �=2-rotation. By Exercise 2.3.6 (page 125) thefunction

u1 .x/ D u.Mx/

is harmonic on the square, it equals 1 on L1 and 0 on the other sides. Analogously,

u2.x/ D u.M 2x/;

is 1 along L2 and 0 on the rest, while

u3.x/ D u.M 3x/

is 1 on L3 and 0 elsewhere. But then v D uC u1 C u2 C u3 is a solution to´�v D 0 in Q

v D 1 on @Q;

and additionally bounded and continuous onQ without the vertices. Since we are assuming unique-ness, we immediately get v.x/ � 1. At the same time M0 D 0, so v.0/ D 4u.0/, and then

u.0; 0/ D 1

4:

We are left to prove uniqueness. The problem is that the Dirichlet datum is discontinuous at thecorners p D .�1; 0/ and q D .0; 1/, so, a priori, we cannot use the maximum principle. Yet we caninvoke the reflection principle as follows. Let U1; U2 be harmonic, continuous on Q except at thecorners p; q. Then w D U2 � U1 is harmonic on Q, continuous on Q n ¹p; qº and equal 0 on theboundary minus the two vertices. Moreover, applying Problem 2.2.10 (page 92) to the symmetricextension ofw, depicted in Fig. 2.9, we knoww is extendable with continuity to the corners as well.To sum up, w 2 C.Q/ is harmonic and null on the boundary, hence null overall, and thereforeU1 D U2 on Q.


Fig. 2.9 Symmetric extension of w (Exercise 2.3.7)

Solution 2.3.8. The function u is indeed harmonic on the disc, as a computation shows, but it isunbounded, hence not continuous, at .1; 0/, where the denominator vanishes. The maximum princi-ple cannot be used.

Solution 2.3.9. If the maximum of u is attained at a point x0 and u .x0/ > 1 we have

u3 .x0/ � u .x0/ > 0 and �u .x0/ � 0;

a contradiction. Therefore u � 1; analogously we could prove that u � �1.

Solution 2.3.10. Let x D .1; 0/, y D .�1; 0/. As u is harmonic and non-negative, it is possibileto use Harnack’s inequality. In this case R D 4 and jxj D jyj D 1. Then

3

5u.0; 0/ � u.˙1; 0/ � 5

3u.0; 0/;

from which

u.�1; 0/ � 5

3u.0; 0/ � 25

9u.1; 0/ D 25

9

and

u.�1; 0/ � 3

5u.0; 0/ � 9

25u.1; 0/ D 9

25:

It would have been possible to apply Harnack’s inequality directly, by considering a disc centred at.1; 0/. If so, the larger disc where u satisfies the assumptions is B3.1; 0/, jx � yj D 2 and Harnacktells, since u.1; 0/ D 1,

1

5� u.�1; 0/ � 5:

The estimate thus obtained, however, is worse than the previous one (no surprise, since we used theharmonicity of u only on a subset of B4.0; 0/).


Solution 2.3.11. Note u is C 2 on Rn n ¹0º, for any ˛. Since u is radial, u.r/ D r˛ , we cancompute the Laplacian as9:

�u D urr C n � 1r

ur D ˛.˛ � 1/r˛�2 C .n � 1/˛r˛�2 D .˛2 C .n � 2/˛/r˛�2:

Therefore �u � 0 whenever ˛2 C .n � 2/˛ � 0; i.e. for ˛ � 0 or ˛ � �nC 2.

Solution 2.3.12. This is an application of the maximum principle. Let us suppose, by contradic-tion, that @E1 is a closed curve contained in�. Since u is continuous and equals 1 on @E1, it solves´

�u D 0 in E1u D 1 on @E1:

By the maximum principle, u � 1 onE1. But by definition x0 2 E1 and u.x0/ D 2 by assumption,a contradiction.

Solution 2.3.13. The solution is unique, because BC1 is a Lipschitz domain and the boundary da-tum is continuous and so the maximum principle applies. To find it we use the reflection principle(Problem 2.2.17 on page 103) and solve´

�u.x; y/ D 0 in B1u.x; y/ D y3 on @B1:

We have chosen the datum on @B1 \ ¹y < 0º in order to have an odd function in y. In polarcoordinates, we have8<

:urr C 1

rur C 1

r2u�� D 0 r < 1; 0 � � � 2�

u.1; �/ D sin3 � D 3

4sin � � 1

4sin 3� 0 � � � 2�:

The boundary datum is a finite trigonometric sum, thus we seek solutions of the type

u.r; �/ D b1.r/ sin � C b3.r/ sin 3�:

Substituting we find b1.r/ D ˇ1r and b3.r/ D ˇ3r3, and the boundary condition produces ˇ1 D

3=4, ˇ3 D �1=4, so

v.r; �/ D 3

4r sin � � 1

4r3 sin 3�:

But v.r; 0/ D v.r; �/ D 0, hence the restriction of v to B1 \ ¹y � 0º is the solution to the originalproblem. Going back to Cartesian coordinates, the solution reads

u.x; y/ D 1

4y.3 � 3x2 C y2/:

9 Appendix B.


• As an alternative approach notice that the datum is a degree-three polynomial. From Problem 2.2.3(page 87) we know that a homogeneous harmonic polynomial of degree three looks like

P .x; y/ D ax3 C 3bx2y � 3axy2 � by3:On @B1 D ®

x2 C y2 D 1¯

P .x; y/ D ax�1 � y2

�C 3by

�1 � y2

�� by3 D ax

�1 � y2

�C 3by � 4by3.

The solution is found by fixing a D 0, b D �1=4, and setting

u .x; y/ D P .x; y/C 3

4y D �3

4x2y C 1

4y3 C 3

4y

is precisely the solution found earlier.

Solution 2.3.14. a) Let v denote the solution to´�v .x; y/ D 0 in B1v D U on @B1

which we know exists, is unique and is given by the Poisson formula. The function v .x;�y/ isharmonic on B1 and agrees with v on the boundary, so

v .x; y/ D v .x;�y/ ;implying

vy .x; y/ D �vy .x;�y/and in particular

vy .x; 0/ D �vy .x; 0/ :Therefore vy .x; 0/ D 0, making v the solution on BC1 to the same mixed problem that u solves.

Then u D v on BC1 , and because it is even in y, it coincides with U on B1. In conclusion, U isharmonic on B1.

b) Define

U .x; y/ D´u .x; y/ y � 0

u .x;�y/ y < 0:

Then U is harmonic on B1 and coincides with u on BC1 . In polar coordinates:

u.r; �/ D 1 � r22�

Z �

��cos2 '

r2 C 1 � 2r cos .'��/ d'

and

u .0; �/ D U .0; �/ D 1

2�

Z �

��cos2 ' d' D 1

2.

Solution 2.3.15. Let u and v be bounded solutions of (2.48) in C 1.PC/, and set w D u � v.

Extend w evenly on y < 0. As w D 0 on y D 0, the new function is bounded and harmonic onR2 by the reflection principle. Liouville’s theorem ensures w is constant, and so u, v differ by aconstant.


To find a solution explicitly we may proceed in two ways.

1st method. We use the Fourier transform. Set

bu .�; y/ DZ

Re�i�xu .x; y/ dx:

Then bux .�; y/ D i�bu .�; y/ , buxx .�; y/ D ��2bu .�; y/andbu solves ´buyy .�; y/ � �2bu .�; y/ D 0; y > 0buy.�; 0/ Dbg .�/ :The general integral of the ODE is

bu .�; y/ D c1 .�/ ej�jy C c2 .�/ e

�j�jy :

The boundedness10 of u forces c1 .�/ D 0, while the constraint on y D 0 requires

�c2 .�/ Dbg .�/whence bu .�; y/ D �bg .�/ e�j�jyj�j : (2.52)

Now,ˇbgˇ is integrable on R and has null average on .a; b/, so

bg .0/ DZ b

ag .x/ dx D 0:

Moreover g is smooth, so when � ! 0

bg .�/ bg0 .0/ � .

Formula (2.52) defines

u .x; y/ D � 1

2�

ZRbg .�/ e�j�jyj�j ei�x d�;

as a bounded function for y � 0:

ju .x; y/j � 1

2�

ZR

ˇbg .�/ˇj�j d� < 1:

We will show below that the anti-transform ofe�j�jy

j�j is

� 1

2�log

�x2 C y2

�:

10 ej�jy has no inverse Fourier transform.


Given this, the Neumann problem is solved by

u .x; y/ D 1

2�

ZR

logŒ.x � z/2 C y2� g .z/ dz.

Remark. To compute the anti-transform of e�j�jyj�j we observe the following:

1. e�j�jy is the transform (with respect to x) of11

1

�

y

x2 C y2:

2. xu .x/ transforms to

id

d�bu .�/

and the transform (in x) of1

�

2xy

x2 C y2

is

2id

d�e�j�jy D �2iy sign .�/ e�j�jy :

3. If h .x; y/ D log�x2 C y2

�,

y

�hx .x; y/ D 1

�

2xy

x2 C y2

and bhx D i�bh:Putting all this together yields

i�y

�bh .�/ D �2iy sign .�/ e�j�jy

i.e.e�j�jy

j�j D � 1

2�bh .�; y/ :

We deduce that the anti-transform ofe�j�jy

j�j must be

� 1

2�log

�x2 C y2

�:

2nd method. Let v be the harmonic conjugate to u, so that

f .z/ D u.x; y/C iv.x; y/; (z D x C iy)

is an analytic function on C. The Cauchy-Riemann equations ux D vy , uy D �vx imply that vx isharmonic and vx .x; 0/ D �g .x/. Now, the formula for the solution of the Dirichlet problem (see

11 Appendix B.


Problem 2.2.18 on page 104) gives

vx .x; y/ D 1

�

ZR

y

x2 C y2g .x � s/ ds D �uy .x; y/ :

A primitive (in y) of y

x2Cy2 is given by 12 log

�x2 C y2

�, and therefore

u .x; y/ D � 1

2�

ZR

log�x2 C y2

�g .x � s/ ds;

which is the unique solution up to additive constant.

Solution 2.3.16. We seek solutions with separated variables:

un D w .r/ v .�/ :

Then v solves the eigenvalue problem

v00 .�/C �v .�/ D 0, v .0/ D 0, v .˛/ D 0.

We get � D n2�2=˛2, n > 0, with eigenfunctions

vn .�/ D sin�n�˛��

.

For w, instead, we have

r2w00.r/C rw0.r/ � n2�2

˛2w.r/ D 0

with w bounded, and thenwn.r/ D crn�=˛ :

To comply with the Dirichlet datum on r D 1, 0 < � < ˛, we take linear combinations

u.r; �/ DC1XnD1

cnrn�=˛ sin

�n�˛��:

Necessarily

u.1; �/ DC1XnD1

cn sin�n�˛��

D g .�/

and therefore cn D gn, where

gn D 2

˛

Z ˛

0g.�/ sin

�n�˛��d�

are the coefficients of the sine expansion of g:

b) We choose the special datum

g .�/ D sin��˛��:

The corresponding solutionu.r; �/ D r�=˛ sin

��˛��


is positive on S˛ . Note

ur .r; �/ D �

˛r.��˛/=˛ sin

��˛��;

u� .r; �/ D �

˛r�=˛ cos

��˛��;

and so

jruj2 D .ur /2 C 1

r2.u� /

2 D �2

˛2r2.��˛/=˛ :

Hence we see that

limr!0 jruj2 D

8<:0 for ˛ < �

1 for ˛ D �

C1 for ˛ > �:

As a consequence, for ˛ < � the gradient of u at the origin is zero, and hence no version of Hopf’sprinciple holds (irrespective of the orientation given to the outward normal at that point; as a matterof fact, this domain contains no ball in the sense of Problem 2.2.9 on page 91). Note, furthermore,that for ˛ > � , u is an example of a harmonic function, defined on an open set, that is continuousbut not C 2 on the closure (it is not even differentiable at (0,0)!).

Solution 2.3.17. The Kelvin transform (Problem 2.2.20 on page 107)

y D T1.x/ D x=jxj2

gives T1.�/ D B1 n ¹0º. Set v.y/ D u.y=jyj2/: Then v is harmonic and bounded on B1 n ¹0º (u isbounded). By Problem 2.2.10 (page 92) v admits a harmonic extension to the entire ball B1, whichmeans that ´

�v D 0 in B1v D g on B1:

By Poisson’s formula we have

v.y/ D 1 � jyj22�

Z@B1

g.� /

jy � � j2 d�:

Anti-transforming,

y D x

jxj2 , u .x/ D 1

jxjv

x

jxj2!:

Since

jxjˇˇ x

jxj2 � �

ˇˇ D jx � � j ;

we finally get

u.x/ D jxj2 � 12�

Z@B1

g.� /

jx � � j2 d� .

Solution 2.3.18. The function u belongs to C 2.B1/\C.B1/ and is harmonic, thus by the maxi-mum principle it attains extreme values on @B1. As uj@B1

is known, what we have is a constrainedoptimisation problem. The constraint

g.x; y; z/ D x2 C y2 C z2 D 1


is regular (the gradient rg has no zeroes), so we can use Lagrange multipliers and find the stationarypoints of

ˆ.x; y; z; �/ D x4 C y4 C z4 � 2�.x2 C y2 C z2 � 1/:The system to be solved is 8<

ˆ:x.x2 � �/ D 0

y.y2 � �/ D 0

z.z2 � �/ D 0

x2 C y2 C z2 D 1;

and it is algebraic. For symmetry reasons we can consider only the solutions p2

2;

p2

2; 0

!;

p3

3;

p3

3;

p3

3

!; and .1; 0; 0/;

corresponding to the critical values 1/2, 1/3, 1. The maximum of u is therefore 1, the minimum 1/3.

Solution 2.3.19. The reflection principle in R3 can be formulated in the following way:Define

BC1 D°.x; y; z/ 2 R3 W x2 C y2 C z2 < 1, z > 0

±and let u 2 C 2.BC1 /\C.BC1 / be harmonic onBC1 and such that u .x; y; 0/ D 0. Then the function

U .x; y; z/ D´u .x; y; z/ z � 0

�u .x; y;�z/ z < 0;

generated by an odd reflection in the variable z, is harmonic on the whole ball B1.The proof is exactly the same as in Problem 2.2.17 (page 103), so we will recall only the major

steps. Call v the solution to ´�v .x; y; z/ D 0 in B1v D U on @B1

and setw.x; y; z/ D v.x; y; z/C v.x; y;�z/:

Then w is harmonic on B1 and vanishes on the boundary, and by uniqueness w � 0. Hencev.x; y; z/ D �v.x; y;�z/, and v.x; y; 0/ D 0. Consequently v and u solve the same Dirichletproblem on BC1 , so again v � u � U on BC1 . Now both v and U are odd with respect to z, whencev � U on B1. In particular U is harmonic on B1.

If the Dirichlet problem on z > 0 had two bounded solutions u and v, the difference w D u� vwould vanish on the plane z D 0. Extending w to z > 0 in an odd way would produce a boundedharmonic function in R3. By Liouville’s theorem w would be constant, hence zero (for it wouldvanish on z D 0).

Solution 2.3.20. We shall follow the ideas used in Problem 2.2.10 (page 92), and provide theproof details of the proof for the case n � 3, leaving it to the reader to address dimension two. Callv the solution (smooth and bounded) of´

�v D 0 in B

v D u on @B


and set w D u � v. Then w is harmonic on B n ¹0º, it vanishes on @B and

jxjn�2w.x/ ! 0 for jxj ! 0: (2.53)

If we show that w.x/ D 0 for any x ¤ 0 the claim is immediate. So let 0 < r < 1 and

Mr D maxjxjDr jw.x/j:

Then (2.53) reads

limr!0C

rn�2Mr D 0: (2.54)

Call h the solution to 8<:�h D 0 in B n Brh D 0 on @B

h D Mr on @Br :

It is not hard to check that

h.x/ D Mr�jxj3 C 2jxj2 � 2

r2�n � 1 D jxj2�n � 11 � rn�2 r

n�2Mr :

As w D 0 on @B , the maximum principle implies

�h.x/ � w.x/ � h.x/

on B n Br . Fix x ¤ 0. The previous inequality holds for any r � jxj. Let r tend to 0 and exploit(2.54), so that

jw.x/j � jxj2�n � 11 � rn�2 .r

n�2Mr / ! 0 for r ! 0;

i.e. w.x/ D 0.

Solution 2.3.21. a) For example, u.x/ D jxj2 satisfies the requirements.

b) We saw in Problem 2.2.5 (page 89) that such a function cannot exist in dimension two, becausethe fundamental solution in R2 is unbounded at infinity. This fact suggests using the fundamentalsolution outside a bounded set to construct an example. An example is the C 2 function

u.x/ D´

�jxj3 C 2jxj2 � 2 jxj � 1

�jxj�1 jxj � 1:

Another example is given by

u.x/ D²�1 jxj � 1

�jxj�1 jxj � 1;

which coincides with max¹�jxj�1;�1º.

Solution 2.3.22. We can use Problem 2.2.7 (page 90): in fact, if we set vi D uiC1 � ui , thenvi is harmonic and non-negative on �, and the convergence of the sequence reduces to that of the(telescopic) series

P1iD0 vi .


Solution 2.3.23. Checking the smoothness of Ta and T�1a goes exactly as in the two-dimensionalcase (Problem 2.2.20 on page 107), and so we leave it to the reader. To verify that

v .x/ D a

jxju .Ta.x// D a

jxju a2

jxj2 x

!is harmonic, it is convenient to pass to spherical coordinates. Write x D .r; �; /, 0 � � � 2� ,0 � � � , so Ta.x/ D .�; �; / with r� D a2. Recall that

�v D vrr C 2

rvr C 1

r2

1

sin2 v�� C v C cos

sin v

�� vrr C 2

rvr C 1

r2�Sv:

Since x and Ta.x/ are collinear vectors, the Kelvin transform leaves the ‘spherical part’ �S of theoperator � invariant (�S is called Laplace-Beltrami operator). For u D u.�; �; / we have

v.r; �; / D a

ru

a2

r; �;

!

vr .r; �; / D � a

r2u

a2

r; �;

!� a3

r3u

a2

r; �;

!

vrr .r; �; / D 2a

r3u

a2

r; �;

!C 4a3

r4u

a2

r; �;

!C a5

r5u

a2

r; �;

!

�Sv.r; �; / D a

r�Su

a2

r; �;

!:

Substituting

�v D a5

r5u

a2

r; �;

!C 2a3

r4u

a2

r; �;

!C a

r3�Su

a2

r; �;

!

D �5

a5

�u.�; �; /C 2

�u.�; �; /C 1

�2�Su .�; �; /

D

D �5

a5�u:

In conclusion, if u is harmonic on its domain, so is v.

Solution 2.3.24. Since u ! A at infinity, the problem has a unique solution. First note that

eu .jxj/ D a

�1 � 1

jxj

is harmonic on Be , becomes 0 when jxj D 1 andeu .jxj/ ! A for jxj ! 1. Set w D u �eu; sow ! 0 at infinity. The Kelvin transform T1 maps Be bijectively onto the unit ball minus the origin.Moreover, every point of @Be is fixed. Setting

v.y/ D 1

jyjw

y

jyj2!;

we obtain that v is harmonic on the punctured ball. Since w .jxj/ ! 0 at infinity, we have

jyj v.y/ ! 0 as jyj ! 0;


and then Exercise 2.3.18 implies that v extends harmonically to the whole ball B1, with datum g on@B1 (we still call v the extension). By Poisson’s formula

v.y/ D 1 � jyj24�

Z@B1

g.� /

jy � � j3 d�:

Transforming back gives

y D x

jxj2 , w .x/ D 1

jxjv

x

jxj2!:

Since

jxjˇˇ x

jxj2 � �

ˇˇ D jx � � j ;

we then obtain

w.x/ D jxj2 � 14�

Z@B1

g.� /

jx � � j3 d� .

In conclusion

u .jxj/ D a

�1 � 1

jxj

C jxj2 � 14�

Z@B1

g.� /

jx � � j3 d� .

In case g .x/ D x1, v .y/ D y1 is the harmonic extension of v to ¹0 � jyj < 1º. The anti-transformation produces

u.x/ D a

�1 � 1

jxj

C x1

jxj3 :

Solution 2.3.25. One can proceed exactly as in Problem 2.2.24 (page 115). Let u1; u2 be twosolutions. Thenw D u1�u2 is harmonic on�e and has zero Robin datum on @�e . As in 2.2.24.a),ˇ

@w

@xi.x/

ˇ� C

jxj2

for jxj large enough; in fact, the estimate does not depend on the condition on @�e . We multiply�w D 0 by w, integrate on �e \ BR by parts to obtain

0 DZ e\BR

w.x/�w.x/ dx DZ e\BR

jrw.x/j2 dx �Z@. e\BR/

w.x/@�w.x/ d�

just like in 2.2.24.b). Since w is bounded, i.e. jwj � M , we inferZ e\BR

jrw.x/j2 dx DZ@ e

w.x/@�w.x/ d� CZ@BR

w.x/@�w.x/ d� �

� �Z@ e

˛w2.x/ d� CM

Z@BR

ˇrw.x/ � x

jxjˇd� �

has ˛ � 0

i� M

�2�R

Z@BR

jrw.x/j2 d�1=2

�

� M

�2�R

Z@BR

C

jxj4 d�1=2

D C0R�3=2:


Taking the limit for R ! C1, we find:Z e

jrw.x/j2 dx D 0:

As w was assumed C 1 up to the boundary, rw.x/ D 0 on �e ; i.e. w is constant. Since@�w C ˛w D 0 on @�e , it follows that the constant is zero.

Solution 2.3.26. TakeR > 0, so that� � BR.0/. Integrate the equation over�e \BR.0/ to getZ@ g d� D

Z¹jxjDRº

@�ud�:

Suppose R is large, so that BR=2.x/ � �e for any jxj D R. Problem 2.2.1 a) (page 84) gives us theestimate

jru.x/j � p32n

Rmax

@BR=2.x/juj � p

32n

Rmax

R=2�jyj�3R=2 juj :

Using the decay estimate for u, we can writeˇZ@ g d�

ˇ�Z¹jxjDRº

jruj d� � 4�R2 � p32n

R

M � 21C"R1C" ! 0 for R ! C1:

Solution 2.3.27. Observe that

v.x/ D jru.x/j2is subharmonic, as sum of squares of harmonic functions (Problem 2.2.4 on page 88). Then, for anyx 2 Rn,

v.x/ � 1

jBR .x/jZBR.x/

v .y/ dy;

i.e., using (2.50),

jru.x/j2 � 1

!nRn

ZBR.x/

jru.y/j2dy ! 0 for R ! C1:

Therefore ru.x/ � 0 in Rn, and the claim follows.

Solution 2.3.28. Observe first that

u.x; y/ D v.x; y/C .x2 C y2/=2

solves �u D 0. Hence u is in C1.�/, whence also

v D u � .x2 C y2/=2

is in C1.�/. The derivatives vx ; vy are continuous on �, and harmonic, so

jrvj2 D v2x C v2y

is subharmonic, being a sum of squares of harmonic functions. It follows that jrvj2 is maximisedon @�.


Solution 2.3.29. We present the solution for n D 2; the proof for n > 2 is analogous. Let v bethe harmonic extension of u to B1 .0/, that is �v D 0 in B1, v D u on @B1. Since u is continuouson @B1, then v is bounded. Then w .x/ D u .x/� v .x/� log jxj is harmonic on B1 .0/ n .0/, w D 0

on @B1 .0/, and it is positive by the maximum principle, since w ! C1 as jxj ! 0:

Therefore w satisfies the hypotheses in Problem 2.2.12 (page 94), and

w .x/ D u .x/ � v .x/ � log jxj D C log jxj ;or

u .x/ D .C C 1/ log jxj C v .x/ :

Solution 2.3.30. The relationship between the Newtonian potential in the plane and the generatingdensity is

�u D urr C 1

rur D �:

Since �u D �2�r2, we deduce .r/ D 2�r2:

The potential (defined up to additive constants) is C 1 in R2, harmonic for r > 1 and radially sym-metric, so it must be of the form

a log r C b:

Furthermore,u .1�/ D u .1C/ , ur .1�/ D ur .1C/ . (2.55)

Thenb D 0 and a D ��

2

and the potential for r > 1 is u .r/ D ��2 log r .

Solution 2.3.31. Up to additive constants the potential is

u .x; y; z/ D 1

4�

ZB1

1h.x � �/2 C .y � �/2 C .z � �/2

i1=2 d�d�d�:This is C 1 on R3. For an explicit expression note u is radially symmetric, u D u .r/, r2 Dx2 C y2 C z2, (the problem is invariant under rotations) and solves

�u D´

�1 0 � r < 1

0 r > 1

withu .1�/ D u .1C/ , ur .1�/ D ur .1C/ . (2.56)

Radial harmonic functions in R3 have the form

a

rC b;

while radial solutions of �u D �1 satisfy

urr C 2

rur D �4�


which have general integral

v .r/ D c1 C c2

r� 1

6r2:

But u is bounded for r < 1, so c2 D 0. From (2.56)

c1 � 1

6D aC b and � a D �1

3:

Choosing b D 0 (zero potential at infinity), we have

u .r/ D

8<:1

2

1 � r2

3

!r � 1

1

3rr > 1:

Solution 2.3.32. The relationship between u and is given, at any point z D .z1; z2/ on thecircle, by the formula

@uI

@�.z/ � @uE

@�.z/ D .z/

where (the limits are taken radially)

@uE

@�.z/ D lim

x!z;>1

@uE

@�.x/ ;

@uI

@�.z/ D lim

x!z;<1

@uI

@�.x/ :

We know that @uE

@.z/ D �z1 � 2z1z2: To compute @u

I

@we first use the continuity of u across C .

We finduE .z/ D uI .z/ D z1 C z1z2: (2.57)

We deduce that u is harmonic inside C and satisfies the Dirichlet condition (2.57). Therefore insideC we find u .x/ D x1 C x1x2, and so

@uI

@�.z/ D z1 C 2z1z2:

Finally,

.z/ D @uI

@�.z/ � @uE

@�.z/ D 2z1 C 4z1z2:

Solution 2.3.33. The answer is no. In fact, let

u .x/ DZ@

hx � � ; �� ijx � � j3 .� / d� x D .x1; x2/

with j .� /j � M: Assume � � BR .0/. Then j� j � R for � 2 @� and, if jxj � 2R, we have

jx � � j � jxj � j� j � jxj2:

Using jhx � � ; �� ij � jx � � j, we infer

ju .x/j �Z@

ˇˇ hx � � ; �� i

jx � � j3 .� /

ˇˇ d� �

Z@

.� /

jx � � j2 d� � 4M j@�jjxj2 :


In particular, we should have

ju .x1; 0/j D 1

jx1j � 4M j@�jjx1j2

which is a contradiction for x1 large.

Solution 2.3.34. a) The Green function G D G .x; y/ is harmonic on the half-space, G .x; y/ D0 on x3 D 0 for any given y 2PC, and

�xG .x; y/ D �ı3 .x � y/ (2.58)

where ı3 .x � y/ is the Dirac distribution at y. Let us use the methods of images. The fundamentalsolution

� .x � y/ D � 1

4� jx � yjsatisfies (2.39). Setey D .y1; y2;�y3/ to be the mirror image of y D .y1; y2; y3/with respect to theplane y3 D 0. The function � .x �ey/ is harmonic on PC and coincides with � .x � y/ on x3 D 0.Hence

G .x; y/ D � .x � y/ � � .x �ey/ .

b) We invoke the method of images once more. This time let, for y ¤ 0,

y� D T1 .y/ D y

jyj2

be the mirror of y under the Kelvin transform. As in the case n D 2, for jxj D 1 we haveˇx � y�

ˇ2 D 1 � 2x � y

jyj2 C 1

jyj2 D 1

jyj2�1 � 2x � y C jyj2

�D 1

jyj2 jx � yj2 .

If y ¤ 0 define

G .x; y/ D 1

4�

²1

jx � yj � 1

jyj jx � y�j³:

Then G .x; y/ D 0 for jxj D 1, y ¤ 0, and

�xG .x; y/ D �ı3 .x � y/ in B1:

When y D 0, put

G .x; 0/ D 1

4�

²1

jxj � 1³

.

Note how, for x ¤ y and y ! 0,G .x; y/ ! G .x; 0/ :

c) Denote by GB1the Green function for the ball B1 constructed in part b). Set ey D

.y1; y2;�y3/. ThenGCB1.x; y/ D GB1

.x; y/ �GB1.x;ey/ :

3

First Order Equations

3.1 Backgrounds

The first part of the present chapter is devoted to first order scalar conservation laws, ofthe type

ut C q.u/x D 0; (3.1)

where q is a smooth funciton. One seeks solutions u D u.x; t/ defined on the half-plane¹t � 0º, and typically subject to a (Cauchy or) initial condition

u.x; 0/ D g.x/; x 2 R:

The equation represents a convection or trasport model. The velocity v is related to theflux function q by the relationship

q.u/ D v.u/u:

• Characteristics and local solution. The straight lines

x.t/ D q0.g.�//t C �; (3.2)

along which u is constant, are the characteristic lines for the equation (3.1). If only onecharacteristic passes through the point .x; t/, and this “starts ” from .�; 0/, then

u.x; t/ D g.�/:

One says that “the characteristic carries the datum g .�/”.Since u.x.t/; t/ D g.�/ along the characteristic (3.2), for any t small enough the

solution u is implicitly defined by the equation

u D g�x � q0.u/t� . (3.3)


150 3 First Order Equations

• Rankine-Hugoniot conditions. When two characteristics, carrying distinct data, in-tersect, they cause a jump discontinuity in the solution, and (3.3) is no longer valid. Thesolution should be interpreted suitably, in weak sense. The discontinuity curve � is calleda shock wave. If � is regular and has equation x D s.t/, the Rankine-Hugoniot conditionshold: if we call uC, u� the limit values of u as it approaches � from the right and leftrespectively, one has

s0.t/ D q.uC.s.t/; t// � q.u�.s.t/; t//uC.s.t/; t/ � u�.s.t/; t/ :

Let q be strictly concave or convex.

• Rarefaction waves. In the areas of the upper half-plane ¹t > 0º not reached by thecharacteristics carrying the initial datum it is usually possible to construct the solution (de-fined coherently with all other areas where it is defined) as a rarefaction wave. The wavecentred at .x0; t0/, given by

u.x; t/ D R

�x � x0t � t0

; where R D .q0/�1 (the inverse of q0/;

is a rarefaction wave centred at .x0; t0/.

• An entropy condition along the shock wave is:

q.uC.s; t// < s0 .t/ < q.u�.s; t//:

Its purpose is to select among the weak solutions the physically significant one.

The second part of the chapter is devoted to first order quasilinear equations of thetype

a.x; y; u/ux C b.x; y; u/uy D c.x; y; u/; (3.4)

with a, b, c being C 1 functions on their domain.

• If u D u .x; y/ is a C 1 solution and P0 D .x0; y0; z0/ is a point on the graph(z0 D u .x0; y0/), then the vector .a .P0/ ; b .P0/ ; c .P0// is tangent to the graph of u atP0. The solutions graphs are therefore integral surfaces for the vector field .a; b; c/, i.e.unions of integral curves (characteristic curves) solving the characteristic system

dx

dtD a .x; y; z/ ,

dy

dtD b .x; y; z/ ;

dz

dtD c .x; y; z/ ; (3.5)

where z .t/ D u .x .t/ ; y .t//.If c � 0 then z must be constant on the plane curves solutions of the reduced charac-

teristic systemdx

dtD a .x; y; z/ ,

dy

dtD b .x; y; z/ I

if, moreover, a D a.u/ and b � 1 the curves are straight lines, in agreement with conser-vation laws (which correspond to this situation, with t replacing y).

3.1 Backgrounds 151

• A Cauchy problem for (3.5) is given by assigning the value of u along a curve ofthe xy-plane or, equivalently, by requiring to find an integral surface for the vector field.a; b; c/ that contains a given space curve �0. If �0 has parametric equations

x .s/ D f .s/ ; y .s/ D g .s/ ; z .s/ D u.x.s/; y.s// D h .s/ ; s 2 I; (3.6)

one has to solve the characteristic system (3.5) with the family of initial conditions (3.6),depending on the parameter s 2 I . Under our assumptions, for any given s, the Cauchyproblem (3.5), (3.6) has exactly one solution

x D X .s; t/ ; y D Y .s; t/ ; z D Z .s; t/ ; (3.7)

for t in a neighbourhood of 0. If the first two equations can be solved for suitable variabless D S.x; y/, t D T .x; y/, the solution of equation (3.4) reads

z D u.x; y/ D Z.S.x; y/; T .x; y//:

Let

P0 D .x0; y0; z0/ D .X.s0; 0/; Y.s0; 0/; Z.s0; 0// D .f .s0/; g.s0/; h.s0//;

and consider the Jacobian

J .s0; 0/ D det

�@sX .s0; 0/ @sY .s0; 0/

@tX .s0; 0/ @tY .s0; 0/

D det

�f 0 .s0/ g0 .s0/a .P0/ b .P0/

:

Based on the inverse function theorem, the following possibilities can occur:

a) J .s; 0/ is different from zero in I . Then in a neighbourhood of �0 there is a uniquesolution u D u .x; y/ for the Cauchy problem, defined parametrically by (3.7).

b) J .s0; 0/ D 0 for some s0 2 I . A C 1 solution in a neighbourhood of P0 may existonly if �0 is characteristic at P0, i.e.

rank

�f 0 .0/ g0 .0/ h0 .0/a .P0/ b .P0/ c .P0/

D 1:

If this is not the case, then, there are no C 1 solutions in a neighbourhood of P0 (theremight exist less regular solutions).

In particular, if �0 is characteristic for any s 2 I there are infinitely manyC 1 solutionsin a neighbourhood of �0.

• A function ' D '.x; y; z/ of class C 1 is a first integral for the characteristic system(3.5) if it is constant along characteristic curves, in other words if

r' � .a; b; c/ � 0:

If ' and are two independent first integrals for (3.5) (the vectors r' and r are every-where linearly independent), the general solution z D u .x; y/ of the quasilinear equation


is defined implicitly byF .' .x; y; z/ ; .x; y; z// D 0;

where F D F .h; k/ is an arbitrary C 1 function such that Fh'z C Fk z ¤ 0:

3.2 Solved Problems

� 3:2:1 � 3:2:11 W Conservation laws and applications.� 3:2:12 � 3:2:21 W Characteristics for linear and quasilinear equations.

3.2.1 Conservation laws and applications

Problem 3.2.1 (Burgers equation, shock waves). Study the global Cauchy problem forBurgers equation ´

ut C uux D 0 x 2 R; t > 0

u.x; 0/ D g.x/ x 2 R

where

a) g.x/ D

8<:1 x < �1�1=2 �1 < x < 1�1 x > 1;

b) g.x/ D´0 x � 0; x > 1

2x 0 � x < 1:

Solution. a) The Burgers equation is a conservation law of the type

ut C q .u/x D 0

with q.u/ D u2=2 and q0.u/ D u.The characteristic emanating from the point .�; 0/ on the xt -plane, along which the

solution is constant and equals g.�/, has equation

x D q0.g.�//t C � D g.�/t C � D

8<:t C � � < �1�12t C � �1 < � < 1

�t C � � > 1:

As q0 is increasing (q is convex) and g has decreasing discontinuities, the characteris-tic slopes decrease when crossing the datum discontinuities. Then the characteristics thenintersect, for small times, near x D �1 and also x D 1 (Fig. 3.1).

Therefore from both points we have shock waves x D s.t/, which can be determinedusing the Rankine-Hugoniot condition

s0.t/ D q.uC.s.t/; t// � q.u�.s.t/; t//uC.s.t/; t/ � u�.s.t/; t/ D 1

2ŒuC.s.t/; t/C u�.s.t/; t/�:


Fig. 3.1 Characteristics for Problem 3.2.1 a) (small times)

Near .x; t/ D .�1; 0/ we have u� � 1, uC � �1=2, so´s01.t/ D 1

4

s1.0/ D �1 whence x D s1.t/ D 1

4t � 1:

Similarly, near .x; t/ D .1; 0/, u� � �1=2, uC � �1 and´s02.t/ D �3

4

s2.0/ D 1whence x D s2.t/ D �3

4t C 1:

Consequently, for small times, the solution u.x; t/ equals �1=2 for

1

4t � 1 < x < �3

4t C 1:

As t increases, this interval gets smaller, until it disappears for t D 2 (and x D �1=2). Atthis point the two shock waves collide, and the surviving characteristics carry the datumu� � 1 (left) and uC � �1 (right); this generates a third shock curve x D s3.t/, where´

s03.t/ D 0

s3.2/ D �12

thus x D s3.t/ D �12:

Overall, the only entropic solution is (Fig. 3.2)

u.x; t/ D

8<:1 x < min

�14t � 1;�1

2

��12

14t � 1 < x < �3

4t C 1

�1 x < max��3

4t C 1;�1

2

�:

b) In this case the characteristics are

x D´� � � 0; � > 1

2�t C � 0 � � < 1:

In particular, u.x; t/ � 0 as x � 0, t � 0. When 0 < � < 1, if t is small, the implicitsolution is given in implicit form by

u D g�x � q0.u/t� D 2.x � ut/;


�1 1

u D 12

u D �1u D 1

xD

1 4t

�1

x D �34 t C

1x

D�1 2

x

t

Fig. 3.2 Characteristic lines for Problem 3.2.1 a)

whence

u.x; t/ D 2x

2t C 1:

Alternatively, from the characteristic x D 2�t C � we find

� D x

2t C 1; and hence u.x; t/ D g.�/ D 2x

2t C 1:

As before, the decreasing discontinuity of g at x D 1, plus the convexity of q, cause theformation of a shock wave x D s.t/ satisfying the Rankine-Hugoniot condition. Sincehere u�.x; t/ D 2x=.2t C 1/, uC � 0, we have

8<:s01.t/ D s.t/

2t C 1s1.0/ D 1:

The (ordinary) equation is linear and with separate variables. Integrating and imposing theinitial condition gives s.t/ D p

2t C 1. The required solution is thus (Fig. 3.3)

u.x; t/ D8<:0 x � 0; x >

p2t C 1

2x

2t C 10 � x <

p2t C 1:


Fig. 3.3 Characteristic lines for Problem 3.2.1 b)

Problem 3.2.2 (Burgers equation, rarefaction vs. shock). Solve the problem´ut C uux D 0 x 2 R; t > 0

u.x; 0/ D g.x/ x 2 R

where

a) g.x/ D

8<:1 x � �1�x �1 � x < 0

1 x > 1;

b) g.x/ D

8<:0 x < 0

1 0 < x < 1

0 x > 1:

Solution. a) As in the previous problem the characteristics are

x D q0.g.�//t C � D g.�/t C � D´t C � � � �1 or � > 0

��t C � �1 � � < 0:

This time, though, g has an increasing discontinuity at x D 0; since q.u/ D u2=2 isconvex (and hence q0 is increasing), the slope of the characteristic has an increasing jumpwhen � crosses 0 from left to right. Hence we expect that a region of the xt -plane willnot be met by any characteristic. In this case the only entropic solution in this region isa rarefaction wave. On the other hand the characteristics corresponding to �1 � � < 0

form a family of straight lines through the point .x; t/ D .0; 1/. Consequently, for t < 1,the solution is constructed by taking

� D x

1 � t and consequently u.x; t/ D g.�/ D � x

1 � t :

So for t < 1, no other characteristic enters the sector between the characteristics x D 0

and x D t , and the solution is given by a rarefaction wave. In general, a rarefaction wavecentred at .x0; t0/ has equation

u .x; t/ D R

�x � x0t � t0

where R .y/ D �

q0��1

.y/ :


�1

xD t

� 1

xDt�

p t

xD t

u D xt�1

u D 1

u D xt

u D 1

x

t

Fig. 3.4 Characteristic lines for Problem 3.2.2 a)

Since here R .y/ D y, we find

u.x; t/ D x

t; 0 � x � t; t < 1.

Alternatively we may put � D 0 and g.�/ D u.x; t/ in the characteristics equation to get

x D u.x; t/t

and hence u D x=t . Note that a rarefaction wave is constant along the straight linesthrough the origin, also called characteristics.

When t > 1 the characteristics carrying u� � 1 hit the rarefaction characteristics,along which uC.x; t/ D x=t , and generate a shock curve � satisfying8<:s01.t/ D s.t/

2t

s1.1/ D 0:

This gives s.t/ D t � pt . Note that � does not meet the characteristic x D t . Finally, we

have (Figs. 3.4 and 3.5)

u.x; t/ D

8ˆ<ˆ:

1 x � t � 1 for t < 1

1 x < t � pt for t � 1

x=.t � 1/ t � 1 � x � 0 for t < 1

x=t max.0; t � pt / < x � t

1 x � t:

b) The function g has an increasing jump at x D 0 and a decreasing one at x D 1.Since q is convex, we expect a rarefaction wave at .0; 0/; after some time, the latter inter-


�1 0

t D 0

1 1

x a � 1 0 a

t D a < 1

1 1

x

0 1

t D 1

1 1

x a � pa a

t D a > 1

1

a�paa

1

x

Fig. 3.5 Solution to Problem 3.2.2 a), at various times

acts with a shock wave emanating from .1; 0/. The characteristic line from the point .�; 0/is

x D � C q0.g.�//t D � C g.�/t D´� � < 0 or � > 1

t C � 0 < � < 1:

By varying � we deduce immediately the following properties for the solution (Fig. 3.6):

• u.x; t/ equals 0 when x < 0 (vertical characteristics).

• The characteristics x D 0 and x D t bound the region occupied by a rarefaction wavecentred at the origin, at least until some instant time t0 to be determined.

• From .1; 0/ starts a shock wave; on the right u.x; t/ is 0, while on the left, at least untilt0, u.x; t/ equals 1.

• For times larger than t0 the shock interacts on the left with the rarefaction wave.

If we argue as in the previous situation, the rarefaction wave is

u.x; t/ D x

t; 0 � x � t .

Concerning the shock wave, for small t we have uC D 0 and u� D 1, so´s0.t/ D 1

2

s.0/ D 1;and thus x D s.t/ D 1

2t C 1:

What we have said holds until the characteristic x D t intersects the shock curve, that is


up to t0 D 2. For later times we still have a shock wave with uC D 0, but now

u�.s; t/ D s

t;

corresponding to the value of u carried by the rarefaction wave. Therefore8<:s0.t/ D s.t/

2ts.2/ D 2:

The ODE is linear, and with separated variable, and has one solution

s.t/ D p2t :

Summarising,

u.x; t/ D

8ˆ<ˆ:

0 x � 0

xt

0 � x < min�t;

p2t�

1 t � x < 12t C 1; with t < 2

0 x > max�12t C 1;

p2t�:

The shock speed is 1=2 until t D 2 and then becomes negative, �1=2t3=2. The strengthequals the jump value of u across the shock, i.e. 1 until t D 2, and then fades to zero ast ! 1 (Fig. 3.6).

1 2

2

xD t

xD

p 2t

xD

t2

C1

u D 0

u D 1

u D xt

u D 0

x

t

0 1

t D 01

x

0 a a2 C 1

t D a < 21

x

0p2a

t D a � 2 p2=a

x

Fig. 3.6 Problem 3.2.2 b): characteristics and shock wave (left); solution at various times (right)


Problem 3.2.3 (Non-extendability). Consider the Cauchy problem:´ut C u2ux D 0 x 2 R; t > 0

u.x; 0/ D x x 2 R:

a) Check whether the family of characteristics admits an envelope.

b) Find an explicit formula for the solution and discuss whether it may be extended tothe whole half plane ¹t > 0º.

Solution. a) The PDE is written as conservation law with q.u/ D u3=3, q0.u/ D u2.Note how the initial datum g.x/ D x is unbounded when x ! ˙1. The characteristicfrom .�; 0/ has equation:

x D � C q0.g.�//t D � C �2t:

To establish whether this family, depending on �, admits an envelope, we must solve forx and t the system ´

x D � C �2t

0 D 1C 2�t:

The second equation is just the first one differentiated with respect to �. The parameter� can be eliminated and we find that the envelope lies in the quadrant x < 0, t > 0 andcoincides with the hyperbola 4xt D �1 (Fig. 3.7).

b) The solution u D u .x; t/ is defined implicitly by

u D g�x � q0 .u/ t�

at least for small times. In our case, since g .x/ D x, we find

u D x � u2t .

4xt D �1x

t

Fig. 3.7 Envelope of characteristics for Problem 3.2.3


Solving for u, we get

u˙.x; t/ D �1˙ p1C 4xt

2t; x � � 1

4t:

Let us determine limt!0C u˙ .x; t/. For given x:

limt!0C

u� .x; t/ D limt!0C

�1 � p1C 4xt

2tD �1;

while

limt!0C

uC .x; t/ D limt!0C

�1C p1C 4xt

2tD lim

t!0C4xt

2t�1C p

1C 4xt� D x.

Only uC satisfies the initial condition, and is therefore the unique solution, defined in theregion ¹x � �1=4tº and regular inside. This region is bounded above by the envelope ofthe characteristics, which becomes a barrier beyond which the characteristics do not carryinitial data. Moreover, since the initial datum tends to �1 as � ! �1, and the charac-teristics tend to flatten horizontally, there is no coherent way to extend the definition of ubeyond the envelope, in the quadrant x < 0, t > 0.

On the contrary, the formula

u .x; t/ D �1C p1C 4xt

2t

defines the solution on x � 0, t � 0 as well.

Problem 3.2.4 (A traffic model, vehicle path). The following problem models whathappens at a traffic light:8<

:�t C vm

�1 � 2

m

��x D 0 x 2 R; t > 0

�.x; 0/ D´�m x < 0

0 x > 0;

where � is the density of cars, �m the maximum density, vm the maximum speed al-lowed. Determine the solution and calculate:

a) The density of cars at the light for any t > 0.

b) The time taken by a car placed at x0 < 0 at time t D 0 to get past the light.

Solution. a) The equation is written as conservation law with

q .�/ D �v .�/ D vm�

�1 � �

�m


where v .�/ is the speed when the cars are in an area of density �. The characteristicthrough .�; 0/ is

x D vm

�1 � 2� .�; 0/

�m

t C �:

When � < 0 we findx D �vmt C �:

Thus in the region x < �vmt we have � .x; t/ D �m. When � > 0

x D vmt C �

and if x > vmt we have � .x; t/ D 0. In the sector �vmt � x � vmt we can join thevalues �m and 0 with a rarefaction wave centred at the origin. Setting

q0 .�/ D vm

�1 � 2�

�m

D y

we can find the inverse function

R .y/ D .q0/�1 .y/ D �m2

�1 � y

vm

;

and the rarefaction wave is

� .x; t/ D R�xt

�D �m

2

�1 � x

vmt

:

To sum up, the solution is

�.x; t/ D

8<:�m x < �vmtm

2

�1 � x

vmt

��vmt � x � vmt

0 x > vmt:

Therefore the vehicle density at the traffic light is

�.0; t/ D �m2;

constant in time.

b) In the present model the speed of a vehicle at x at time t depends only on the density:

v.�/ D vm

�1 � �

�m

:

Denote by x D x.t/ the law of motion of the car, with x.0/ D x0 < 0. Initially the cardoes not move, until time t0, with x0 D �vmt0; after t0 the car moves within the regionof the rarefaction wave as long as x.t/ < vmt , in particular before it reaches the trafficlight; after that it moves with constant speed vm. Therefore, after t0 and before reaching


jx0jvm

4jx0jvm

x0

t

x

Fig. 3.8 Problem 3.2.4: path of the car starting from x D x0 < 0 at time t D 0. The traffic light isreached at time t D 4jx0j=vm

the light, x solves the Cauchy problem´x0.t/ D v.�.x.t/; t// D vm

2

�1C x.t/

vmt

�x.t0/ D �vmt0:

Integrating the (linear) equation gives

x.t/ D vm.t � 2pt0t /;

and hence x.t/ D 0 for t D 4t0 D 4jx0j=vm (Fig. 3.8).

Problem 3.2.5 (Traffic model; normalised density). Let � be the vehicle density in themodel of Problem 3.2.4. Normalise the density by setting u.x; t/ D �.x; t/=�m, so that0 � u � 1. Check that u solves

ut C vm.1 � 2u/ux D 0; x 2 R; t > 0: (3.8)

Determine the solution to (3.8) with initial condition

u.x; 0/ D g.x/ D

8<:1=3 x � 0

1=3C 5x=12 0 � x � 1

3=4 x � 1:

Solution. The initial datum g D g .x/ is shown in Fig. 3.9, left.Elementary computations show that u satisfies eq. (3.8). We have q0.u/ D vm.1�2u/

and hence q.u/ D vm.u � u2/. The characteristic issued from .�; 0/ is

x D � C vm.1 � 2g.�//t; (3.9)


Fig. 3.9 Initial datum and characteristics for Problem 3.2.5

i.e.

x D � C 1

3vmt for � � 0

x D � C�1

3� 5

6�

vmt for 0 � � � 1

x D � � 1

2vmt for � � 1:

We can see that the characteristics meet, creating a shock wave. The starting pointof it is the point with smallest time coordinate, at which the characteristics intersect for0 � � � 1. In this case the characteristics form a pencil depending on the parameter �,and the pencil base point, where all characteristics with 0 � � � 1 meet, is

.x0; t0/ D�2

5;6

5vm

:

This is shown in Fig. 3.9, right.The shock curve � , of equation, say, x D s.t/, is thus emanating from .2=5; 6=.5vm//.

On the right of� uC D 3=4, while on the left u� D 1=3. The Rankine-Hugoniot conditiongives

s0.t/ D q.uC/ � q.u�/uC � u� D � 1

12vm:

Since s .6=.5vm// D 2=5 we get the straight line

s.t/ D 1

2� 1

12vmt:

Thus we have found the (entropic) solution, for t > t0 D 6=.5vm/.Suppose now t < t0. To compute the solution in the region

S D².x; t/ W 0 � t <

6

5vm;1

3vmt � x � 1 � 1

2vmt

³;


bounded by the characteristics from � D 0 and � D 1, we solve for � the characteristicsequation. We get

� D 6x � 2vmt6 � 5vmt ; 0 � � � 1;

from which, u being constant along characteristics,

u.x; t/ D g.�/ D 1

3C 5

12

6x � 2vmt6 � 5vmt D 4C 5x � 5vmt

2.6 � 5vmt / in S:

Another way to proceed would be to use the formula

u D g .x � vm.1 � 2u/t/

which gives u in implicit form. Substituting the expression of g in the internal 0 < x < 1we find

u D 1

3C 5

12.x � .1 � 2u/ vmt / :

Solving for u, we obtain the previous formula. In summary:

u.x; t/ D

8ˆ<ˆ:

1

3x < min

²1

3vmt;

1

2� 1

12vmt

³4C 5x � 5vmt2.6 � 5vmt /

1

3vmt � x � 1 � 1

2vmt

3

4x > max

²1 � 1

2vmt;

1

2� 1

12vmt

³:

* Problem 3.2.6 (Traffic in a tunnel). A realistic model for the velocity inside a verylong tunnel is

v.�/ D´vm 0 � � � �c

� log .�m=�/ �c � � � �m

where � is the vehicles density and � D vm

log.m=c/. Note v is continuous also at the

point �c D �me�vm=�, which represents a critical density, below which drivers are

free to cruise at the maximum speed allowed. Practical values are �c D 7 cars/Km,vm D 90 Km/h, �m D 110 cars/Km, vm=� D 2:75.

Suppose the tunnel entrance is placed at x D 0, and that prior to the tunnel opening(at time t D 0) a queue has formed. The initial datum is

�.x; 0/ D g.x/ D´�m x < 0

0 x > 0:

a) Determine the traffic density and velocity, and sketch the graphs of these functions.

b) Determine and sketch on the xt -plane the path of a car initially at x D x0 < 0, thencompute how long it takes it to enter the tunnel.


�c �m

vm

v D v.�/

� �c �m

q D q.�/

�

Fig. 3.10 Velocity and flux function for the traffic in a tunnel

Solution. a) By using the usual convective model the problem to solve reads8<:�t C q0.�/�x D 0 x 2 R; t > 0

�.x; 0/ D g .x/ D´�m x < 0

0 x > 0;

whereq.�/ D �v.�/

and hence (e�vm=� D �c=�m)

q0.�/ D´vm 0 � � < �c

� Œlog .�m=�/ � 1� �c < � � �m:

The graphs of v and q in terms of the density � are shown in Fig. 3.10. Notice how q0jumps at � D �c :

q0.��c / D vm and q0.�Cc / D vm � �:The characteristic from .�; 0/, i.e. the line x D � C q0.g.�//t , is

x D � � �t for � < 0; and x D � C vmt for � > 0:

Therefore we obtain immediately the solution is certain regions:

�.x; t/ D �m for x < ��t:

It remains to find � in the sector

S D ¹.x; t/ W ��t � x � vmtº:

For this we recall that q0 is discontinuous at � D �c :

q0.��c / D vm and q0.�Cc / D vm � �:


This suggests writing S D S1 [ S2, with

S1 D ¹.x; t/ W ��t � x � .vm � �/tº;

where �c < � � �m, and

S2 D ¹.x; t/ W .vm � �/t � x � vmtº;

where 0 < � � �c .In S1 we proceed as follows. When �c < � � �m we have

q00.�/ D ��=� < 0;

so that q is strictly concave. Since the initial datum is decreasing we seek a solution in theform of a rarefaction wave, centred at the origin, that attains continuously the value �m onthe line x D ��t . The wave is given by � .x; t/ D R .x=t/ where R D .q0/�1. To find Rwe solve for � the equation

q0 .�/ D �

log

��m�

� 1

�D y:

This givesR .y/ D �m exp

��1 � y

�

�and hence we find

� .x; t/ D �m exp��1 � x

�t

�in the region

�� x

t� vm � �:

Notice that � D �c on the straight line x D .vm � �/t . In S2, where � � �c , we haveq0.�/ D vm. Thus, q is not strictly convex of concave, and there is no possibility to con-struct a solution via a rarefaction wave. Changing perspective, we construct the entropicsolution by solving the equation in the “quadrant” ¹x > .vm � �/t; t > 0º, prescribingthe values � D �c on x D .vm ��/t and 0 on t D 0. We have already found � D 0 whenx > vmt (Fig. 3.11). In the sector S2 � is constant along the characteristics

x D vmt C k;

that carry the value � D �c D c�vm=�.To sum up,

�.x; t/ D

8<ˆ:�m x � ��t�me

�.1Cx=.�t// ��t � x � .vm � �/t�me

�vm=� .vm � �/t � x < vmt

0 x > vmt:


x D ��t xD.vm

� �/t

x D vmt

� D �m

� D �me�.1C x

�t /

� D �c

� D 0

x

t

��t .vm � �/t vmt

�m�me

�.1C x�t /

�c

x

Fig. 3.11 Problem 3.2.6 b): characteristics (left); solution at time t (right)

In Fig. 3.11 (on the right) we see the density behaviour at a given time: it decreases fromits maximum value (at zero speed) to reach the critical density (maximum speed). Notethat the solution is discontinuous only along x D vmt . This type of discontinuity is calledcontact discontinuity

b) Consider the vehicle initially placed at x0 < 0. We want to describe its trajectoryon the xt -plane. Observe first that the car will not move until time t0 D jx0j=� (Fig. 3.12).At that moment it enters the region S where the velocity is

v.�.x; t// D � log .e1Cx=�t / D �C x

t:

If x D x.t/ denotes the vehicle path, we have8<:x0.t/ D �C x.t/

tx.t0/ D x0:

The equation is linear, and integrating gives

x.t/ D �t

�log

�t

jx0j � 1:

jx0j�

ejx0j�

x0x D ��t

t

x

Fig. 3.12 Trajectory of a car in Problem 3.2.6


The car enters the tunnel at the time T such that x.T / D 0. The required lapse is then

T D ejx0j�

:

Problem 3.2.7 (Shock formation in a traffic model). Let u, 0 � u � 1 be the nor-malised density that solves the following traffic problem:´

ut C vm.1 � 2u/ux D 0 x 2 R; t > 0

u.x; 0/ D g.x/ x 2 R:

Assume g 2 C 1.R/, that g0 has a unique maximum point x1 and that

g0.x1/ D maxRg0.x/ > 0:

a) Study the qualitative behaviour of the characteristics and deduce that the solutiondevelops a shock.

b) Verify that for small times u is defined implicitly by

u D g.x � vmt .1 � 2u//:Deduce that the first instant ts at which the shock forms (critical time) is the firsttime for which

1 � 2vmtg0.x � vmt .1 � 2u// D 0:

c) Show that the initial point .xs; ts/ of the shock belongs to the characteristic �x1

emanating from .x1; 0/, and

ts D 1

2vmg0.x1/:

In case vm D 1, g.x/ D 34

�2�

arctan x C 1�; analyse numerically the graph of u at

various times and interpret the results.

Solution. a) The characteristic �� from the point .�; 0/ has equation

x D � C .1 � 2g.�//vmt: (3.10)

Under the given hypotheses g is strictly increasing in a neighbourhood of x1, thus thecharacteristics starting in the neighbourhood meet, generating a shock.

b) On �� we know that u.x; t/ D g.�/, and from (3.10) we find

� D x � .1 � 2g.�//vmt:


Hence

u.x; t/ D g.x � .1 � 2u.x; t//vmt /:Now we verify when the equation

h.x; t; u/ D u � g.x � .1 � 2u/vmt / D 0; (3.11)

really defines an implicit function u of x and t . The sufficient conditions provided by theimplicit function theorem are the following:

1. h is C 1, true because g is C 1.

2. (3.11) can be solved at some point, in fact

h.x; 0; g .x// D g .x/ � g.x/ D 0

at all points on the x-axis.

3. Finally,

hu.x; t; u/ D 1 � 2vmtg0.x � .1 � 2u/vmt / ¤ 0: (3.12)

As g0 is either negative, or bounded when positive, equation (3.12) is always true forsmall times.

As long as (3.12) holds, by the implicit function theorem, equation (3.11) defines a uniquefunction u D u .x; t/ in C 1 .R/. This solution cannot develop (shock) discontinuities. Onthe other hand the same inverse function theorem gives a formula for ux :

ux .x; t/ D �hx.x; t; u/hu.x; t; u/

D g0.x � .1 � 2u/vmt /1 � 2vmtg0.x � .1 � 2u/vmt / : (3.13)

So if ts > 0 is the first instant for which hu is zero (for some x D xs), necessarily

ux .x; t/ ! 1 as .x; t/ ! .xs; ts/

since the numerator of (3.13) does not vanish at .xs; ts/ (it goes to .2vmts/�1). Thereforets must be the critical time, i.e. when the shock starts.

c) Let us find ts . Consider the characteristic �� . For any .x; t/ 2 ��

x � .1 � 2u.x; t/vmt / D �;

so that (3.12) fails when

hu.x; t; u.x; t// D 1 � 2vmtg0.�/ D 0; i.e. t D 1

2vmg0.�/:

From part a) we know that ts is the smallest (positive) t for which the previous equation


�1 1

3=4

t D 0

x

u

�1 1

t D :5

x

u

�1 1

3=4

t D 1

x

u

�1 xs 1

3=4

ts D �=3

x

u

s.1:5/ 1

3=4

t D 1:5

x

u

s.2/ 1

3=4

t D 2

x

u

Fig. 3.13 Curve implicitly defined by the equation u� 34

�2� arctan.x � .1� 2u/vmt /C 1

� D 0 atvarious times. The abscissas s .1:5/ and s .2/ denote the shock positions obtained by the equal-arearule [18, Chap. 4, Sect. 4]

holds. By assumption g0.x1/ � g0.�/ for any �, therefore .xs; ts/ belongs to �x1, and

moreover

ts D 1

2vmg0.x1/;

xs D x1 C 1

2g0 .x1/.1 � 2g .x1// .

In case

g .x/ D 3

4

2

�arctan x C 1

�;

the curve defined implicitly by (3.11) evolves as in Fig. 3.13.


Problem 3.2.8 (Envelope of characteristics and shock formation). Consider theCauchy problem: ´

ut C q.u/x D 0 x 2 R; t > 0

u.x; 0/ D g.x/ x 2 R:

Suppose q 2 C 2.R/, q00 < 0 and g 2 C 1.R/, with

g.x/ D

8<:g.x/ D 0 x � 0

g0.x/ > 0 0 < x < 1

g.x/ D 1 x � 1:

a) Show that the family of characteristics

x D q0.u/t C � D q0.g.�//t C �; � 2 .0; 1/admit an envelope.

b) Determine the point .xs; ts/ of the envelope with smallest time coordinate, and showthat this is the point where the shock originates from. Recover the result of Prob-lem 3.2.7.

c) Show that .xs; ts/ is a singular point for the envelope, meaning that the tangentvector at .xs; ts/ is zero (assume q and g are regular enough.)

Solution. a) Figure 3.14 shows the envelope of the characteristics

x D q0.g.�//t C �;

� 2 .0; 1/, in two particular cases.

To check the existence of an envelope, we consider the system´x D q0.g.�//t C �

0 D q00.g.�//g0.�/t C 1 D 0

where the second equation is the derivative of the first with respect to �. As q00 < 0 andg0 > 0 for � 2 .0; 1/, we have q00.g.�//g0.�/ < 0 and the envelope is given by theparametric equations

xinv .�/ D � � q0.g.�//q00.g.�//g0.�/

, tinv .�/ D � 1

q00.g.�//g0.�/;

obtained by solving for � the system in the variables x and t .

b) The shock forms in correspondence to the point .xs; ts/ of the envelope with small-est time coordinate, because that is the first point where two characteristics meet. As


1 x

t

1 x

t

Fig. 3.14 Problem 3.2.8, envelope of characteristics with a cusp, in the case q .u/ D u � u2 and:g .�/ D .1 � cos.��//=2 (left); g .�/ D 5�2e�2� (right)

g0.0/ D g0.1/ D 0 and q00.g.�//g0.�/ < 0 for 0 < � < 1, the function

z .�/ D �q00.g.�//g0.�/

has a positive maximum at some �M 2 .0; 1/. From the second equation

ts D min�2.0;1/

1

z .�/D 1

z .�M /.

For Problem 3.2.7 (page 168) we have

q.u/ D vm.u � u2/q0 .u/ D vm .1 � 2u/q00 D �2vm < 0:

In a neighbourhood of x1, the positive maximum of g0, we have g0 > 0, so that the char-acteristics starting there have an envelope. From

z .�/ D �q00.g.�//g0.�/ D 2vmg0 .�/

we deduce �M D x1, and the solution has a shock starting at time

ts D 1

2vmg0 .x1/;

confirming the result in Problem 3.2.7.


c) To check that .xs; ts/, origin of the shock and “origin ” of the envelope, is singular,we need to show that

dx

d�and

dt

d�

vanish at � D �M . Assume q has three derivatives and g two. Then

dx

d�D �q0.g.�//

z2 .�/z0 .�/ D � dt

d�:

Since z has a (positive) maximum at � D �M , we have

z0 .�M / D 0

and the derivatives vanish. The shock starts at the singular points of the envelope (cusps).

Problem 3.2.9 (Non-homogeneous conservation laws). Consider the problem´ut C q.u/x D f .u; x; t/ x 2 R; t > 0

u.x; 0/ D g.x/ x 2 R:

a) Let x D x.t/ be a characteristic for the homogeneous equation (f D 0) and set

z.t/ D u.x.t/; t/:

Which Cauchy problems do x.t/ and z.t/ solve?

b) Supposing f and g bounded, define weak solutions for the problem.

c) Deduce the Rankine-Hugoniot conditions for a shock curve x D s.t/.

Solution. a) Set z D u .x .t/ ; t/. We have

z0 .t/ D ut .x .t/ ; t/C ux .x .t/ ; t/ x0 .t/ ;

and by the conservation law

ut .x .t/ ; t/C ux .x .t/ ; t/ q0 .z .t// D f .z .t/ ; x .t/ ; t/ :

The characteristic from .�; 0/ solves

x0 .t/ D q0 .z .t// , x .0/ D �

while z satisfies the Cauchy problem

z0 .t/ D f .z .t/ ; x .t/ ; t/ , z .0/ D g .�/ ;

which uniquely determines u along the characteristic, under the usual smoothness assump-tion on f .


b) We mimic the procedure for homogeneous equations. Let us multiply the equationby a test function ' 2 C 10 .D/, which is C 1 with compact support K contained in

D D ¹.x; t/ 2 R2 W t � 0º;

and integrate over D, obtainingZD

.q.u/x C ut /' dxdt DZD

f .u; x; t/' dxdt:

The integrals are finite because the support of ' is bounded. The notion of weak solutionis found essentially by integrating by parts. Interpreting q.u/x C ut as the divergence ofthe vector field .q.u/; u/ we may apply Green’s theorem:Z

K

.q.u/x C ut /' dxdt D �ZK

Œq.u/'x C u' t � dxdt �Z

Ru.x; 0/'.x; 0/ dx

CZ@K\¹t>0º

Œq.u/'n1 C u'n2� ds

where .n1; n2/ is the outer unit normal to @K and ds the infinitesimal length element. Thelast integral is zero (' is continuous, hence null on @K \ ¹t > 0º). So we define weaksolution a locally bounded function u such thatZD

Œq.u/'x C u' t � dxdt CZ

Ru' dx C

ZD

f .u; x; t/' dxdt for any ' 2 C 10 .D/:

As for the homogeneous situation, a weak solution which isC 1 in R�¹t � 0º is a classicalsolution as well.

c) Suppose a curve � , x D s.t/, splits an open set V � ¹t > 0º into two disjointsubdomains

V � D ¹.x; t/ W x < s.t/º and V C D ¹.x; t/ 2 V W x > s.t/º:

Assume that u is a weak solution, which isC 1 in the closures V � and V C separately, witha jump discontinuity along � . In particular, this implies

ut C q.u/x D f .u; x; t/

in V � and V C. If .x; t/ 2 � , write uC.x; t/ for the limit of u when approaching � onthe right, u�.x; t/ for the limit from the left. Pick a test function ', with support in V thatintersects � . From part b)

�ZV�

Œq.u/'x C u' t � dxdt �ZVC

Œq.u/'x C u' t � dxdt DZV

f .u; x; t/' dxdt:


Since u is regular on V �, V C, we can invoke Green’s theorem on the integrals on the left.Recalling that ' D 0 on @V ˙ n �:

�ZV˙

Œq.u/'x C u' t � dxdt DZV˙.q.u/x C ut /' dxdt

Z�

�q.u˙/n1 C u˙n2

�' ds

DZV

f .u; x; t/' dxdt Z�

�q.u˙/n1 C u˙n2

�' ds

where .n1; n2/ is the outward unit normal to � with respect to V C (we used the fact thatut C q.u/x D f .u; x; t/ on V ˙). Substituting into the definition of weak solution, wefind Z

�

�.q.uC/ � q.u�//n1 C .uC � u�/n2

�' ds D 0:

Since ' is arbitrary, and the jumps q.uC/� q.u�/, uC � u� are continuous along � , wededuce

.q.uC/ � q.u�//n1 C .uC � u�/n2 D 0 along �:

On the other hand, if s 2 C 1 we have

.n1; n2/ D 1p1C s0.t/2

.�1; s0.t//;

so

s0 D q.uC/ � q.u�/uC � u� :

The Rankine-Hugoniot condition coincides with the one for the non-homogeneous case.

Problem 3.2.10 (Fluid in a porous tube). Consider a cylindrical tube, infinitely long,placed along the x-axis, containing a fluid moving to the right. Denote by � D �.x; t/

the fluid density, and suppose that the speed at each point depends on the density byv D 1

2�. Assume, further, that the tube wall is made of a porous material that leaks at

the rate H D k�2 (mass per unit length, per unit time).

a) Deduce that if � is smooth, it satisfies

�t C�1

2�2x

D �k�2:

b) Compute the solution with �.x; 0/ D 1 and the corresponding characteristics.

Solution. a) We are dealing with a transport model. The leaking rate H leads to writethe conservation law

�t C q.�/x D �H D �k�2:Due to the convective nature of motion the flow is described by

q.�/ D v.�/� D 1

2�2;

yielding the required equation.


b) From Problem 3.2.9 a), indicating by x D x .t/ the characteristic from .0; �/ andsetting z D �.x.t/; t/, we have´

x0.t/ D z.t/ x.0/ D �

z0.t/ D �kz2.t/ z.0/ D 1:

From the second equation we get

z.t/ D 1=.kt C 1/Ias the latter does not depend on �; we may write

� .x; t/ D 1

1C kt:

The characteristics are parallel logarithms:

x .t/ D 1

kln .1C kt/C �:

** Problem 3.2.11 (A saturation problem). Suppose a certain substance is poured intoa semi-infinite container (aligned along the axis x � 0) with a solvent; the substanceconcentration u D u.x; t/ is governed by the equation

ux C .1C f 0.u//ut D 0 with u.x; 0/ D 0; x > 0; t > 0:

At the entrance (x D 0) the substance is maintained at the concentration

g.t/ D´c0˛t 0 � t � ˛

c0 t � ˛:.c0; ˛ > 0/

Study the evolution of u if one takes

f .u/ D u

1C u(Langmuir isothermal . > 0/)

and discuss the case where ˛ tends to zeroa.

a See [28, Vol. 1, Chap. 6.4], also for the physical-chemical interpretation of the model.

Solution. First of all let us remark that, compared to the conservation laws seen so far,the roles of x and t are exchanged. We have

q0.u/ D 1C f 0.u/ D 1C

.1C u/2;

and since q is concave and g increasing, we expect a shock. The characteristics are thelines

t D .1C f 0.u//x C k D�1C

.1C u/2

x C k k 2 R:


In particular the characteristics from the point .�; 0/ on the x-axis are the parallel lines

t D .1C /.x � �/:

Those from .0; �/, on the t -axis, 0 � � � ˛, are the lines

t D�1C

.1C c0�=˛/2

x C �; if 0 � � � ˛;

which we rewrite as �1C c0

˛��2.t � � � x/ D x: (3.14)

Since1C

.1C c0�=˛/2� .1C /

these and the lines from the x-axis will end up meeting along a shock curve. Equations(3.14), moreover, have decreasing slope as � increases, thus they will have an envelope.The first part of the shock wave is contained in the cusp region bounded by the branches ofthe envelope of (3.14) (Fig. 3.15), and starts from the point C D .xs; ts/ of the envelopewith minimum time coordinate, i.e. the cusp itself.

The characteristics from .0; �/, � � ˛, are the parallel lines:

t D .1C f 0.c0//x C �:

As1C f 0 .c0/ D 1C

.1C c0/2< .1C /;

also these interact, along the shock wave, with the characteristics issued from the horizon-tal axis.

1

C.xs ; ts/

x

t

Fig. 3.15 Envelope in Problem 3.2.11 (˛ D D 1; c0 D 7)


To find the envelope let us differentiate (3.14) in � ; we find

��1C c0

˛��2 C 2c0

˛

�1C c0

˛��.t � � � x/ D 0:

The envelope is found by solving the system8<:t � � � x D x

�1C c0

˛��2

2c0

˛.t � � � x/ D

�1C c0

˛��:

The parametric equations t D t .�/, x D x .�/ can be determined by dividing the equa-tions by one another (note that the second one implies t � � � x ¤ 0); we find:

x D�1C c0

˛��3 ˛

2c0;

hence

t D�1C

�1C c0

˛��2

x C �

D ˛

2c0

�1C c0

˛��

1C c0

˛��2 C

C � � h .�/ :

The shock wave starts at the time coordinate of the cusp,

ts D min0��˛ h .�/ :

Since h .�/ is the product of increasing terms in � , the minimum of h is reached for � D0, whence

ts D ˛.1C /

2c0; xs D ˛

2c0:

The shock curve t D t .x/ is found by means of the Rankine-Hugoniot conditions:

dt

dxD ŒuC f .u/�C�

Œu�C�D 1C

1C u�(3.15)

(recall uC D 0). It remains to compute u D u� along the shock. To this purpose, observethat a shock point .x; t/ comes, for times close to ts , from the characteristic issued fromthe t axis, of equation

t D � C�1C

.1C u/2

x (3.16)

where� D g�1.u/ D ˛u=c0:

Then we substitute u, obtained from (3.16), into (3.15). To simplify the computations wethink of (3.15), (3.16) as equations in the parameter u along the shock line. Differentiating


(3.16) in u we get

dt

duD d�

du� 2

.1C u/3x C

�1C

.1C u/2

dx

duD

D ˛

c0� 2

.1C u/3x C

�1C

.1C u/2

dx

du:

From (3.15), on the other hand,

dt

duD dt

dx

dx

duD�1C

1C u

dx

du;

so �1C

1C u

dx

duD ˛

c0� 2

.1C u/3x C

�1C

.1C u/2

dx

du:

Simplifying,u

.1C u/2dx

duC 2

.1C u/3x D ˛

c0; (3.17)

which is a linear equation in x D x.u/, that we can easily integrate. An alternative way isto observe that

d

du

�u

1C u

2D d

du

�1 � 1

1C u

2D 2u

.1C u/3:

Multiply (3.17) by u. We find

d

du

"x

�u

1C u

2#D ˛u

c0

and integrate using the initial condition x.0/ D xs D ˛=.2c0/, to get

x.u/ D ˛

2c0.1C u/2

and then

u D u� D �1C�2c0x

˛

1=2; (3.18)

along the shock curve. The formula indicates that the jump in u along the shock (the shockstrength) grows like

px. Finally, substituting into (3.16), a little manipulations give

t D ˛u

c0C�1C

.1C u/2

x D x C

�2˛

c0x

1=2� ˛

2c0(3.19)

showing that the shock curve is a parabola.From (3.15), the shock starts off with speed

dx

dtD .1C /�1


and accelerates until u reaches its maximum c0. By (3.18), (3.19) this corresponds to thepoint P of coordinates

xP D ˛

2c0.1C c0/

2 , tP D ˛

2c0

h.1C c0/

2 C .1C 2c0/i

.

From there on, the shock wave travels with constant speed

dx

dtD�1C

1C c0

�1along a straight line. Since it passes through P , it is easy to see that the equation of theshock past P is

t D�1C

1C c0

x C ˛

2.

Note that the first characteristic to carry u� D c0 on the shock curve corresponds to � D˛ (Fig. 3.16):

t D ˛ C�1C

.1C c0/2

x:

If ˛ ! 0, the shock starts in the origin and travels along

t D�1C

1C c0

x:

The values of u are shown in Fig. 3.16. u� D u�.x; t/ is the unique positive solution ofequation (3.15) with � D ˛u=c0.

c0

g

t

˛

C

P

u D 0

u D c0

u D u�

x

t

Fig. 3.16 Initial profile, shock curve and characteristics for Problem 3.2.11


3.2.2 Characteristics for linear and quasilinear equations

Problem 3.2.12 (Two-dimensional Cauchy problem). Solve´ux C xuy D y .x; y/ 2 R2

u.0; y/ D cos y y 2 R:

Solution. We parametrise the curve carrying (in R3) the data by

f .s/ D 0; g.s/ D s; h .s/ D cos s:

Set x D x.t/, y D y.t/ and z D u.x.t/; y.t//, and write the associated characteristicsystem 8<

:x0.t/ D 1

y0.t/ D x.t/

z0.t/ D y.t/

with initial conditionx.0/ D 0; y.0/ D s; z.0/ D cos s:

The first equation gives x.t/ D tC c, i.e. x.t/ D t using the initial condition. The secondequation becomes

y0.t/ D t;

hence y.t/ D t2=2C c, and then again y.t/ D t2=2C s. The third equation reads

z0.t/ D t2

2C s

thus z.t/ D t3=6C st C cos s. In terms of the variable s we have, overall:8<ˆ:

x D X.s; t/ D t

y D Y.s; t/ D t2

2C s

z D Z.s; t/ D t3

6C st C cos s:

At this point we can solve for s D S.x; y/, t D T .x; y/ in the first two equations, andsubstitute into the third one to obtain the solution:

u .x; y/ D z .T .x; y/; S.x; y// :

The assumption in the implicit function theorem holds for the initial datum: in fact,

J.s; t/ D det

�Xs YsXt Yt

D det

�0 1

1 t

D �1 ¤ 0;

for any t , hence also for t D 0. Here the computation is direct, since easily

t D x, and s D y � t2=2 D y � x2=2;


giving

u.x; y/ D x3

6C x

�y � x2

2

C cos

�y � x2

2

D xy � 1

3x3 C cos

�y � x2

2

:

The solution is defined in the whole R2.

Problem 3.2.13 (Two-dimensional Cauchy problem). Compute, where defined, thesolution u D u.x; y/ of ´

uux C yuy D x

u.x; 1/ D 2x:

Solution. The curve carrying the data is

f .s/ D s, g .s/ D 1, h .s/ D 2s.

Write x D x.t/, y D y.t/, z D u.x.t/; y.t//. The characteristic system reads8<:x0.t/ D z.t/

y0.t/ D y.t/

z0.t/ D x.t/

with initial conditionx.0/ D s; y.0/ D 1; z.0/ D 2s:

The second equation and y.0/ D 1 immediately give y.t/ D et . Differentiating the first,and using the third equation we write

x00.t/ D x.t/; x.0/ D s; x0.0/ D 2s:

The general integral is c1etCc2e�t , hence x.t/ D 32set� 1

2se�t and z.t/ D 3

2setC 1

2se�t .

Altogether 8<:x D X.s; t/ D 3

2set � 1

2se�t

y D Y.s; t/ D et

z D Z.s; t/ D 32set C 1

2se�t :

(3.20)

Let us see if we can make t D T .x; y/, s D S.x; y/ explicit from the first two. We have

J.s; t/ D det

�Xs YsXt Yt

D det

�32et � 1

2e�t 0

32set C 1

2se�t et

D 3

2e2t � 1

2:

On the initial curve J.s; 0/ D �1, always non-zero. Yet J D 0 for t D � logp3, any s.

Hence we expect to encounter some problems near�X�s;� log

p3�; Y

�s;� log

p3��

D .0;p3=3/:

At any rate, as et D y, the first equation in (3.20) implies

s

�3y

2� 1

2y

D x;


i.e.

s D 2xy

3y2 � 1 , t D log y:

Substituting into the third equation of (3.20) we find

u.x; y/ D 2xy

3y2 � 1�3y

2C 1

2y

D x

3y2 C 1

3y2 � 1 :As expected, the solution is defined only when

y >

p3

3

(recall that the initial datum is given on y D 1).

Problem 3.2.14 (Compatibility conditions on the data). Consider the Cauchy problem

a.x; y/ux � uy D �uon

D D ¹.x; y/ W y > x2ºsubject to

u.x; x2/ D g.x/:

a) Discuss whether the problem has solutions.

b) Study the case a.x; y/ D y=2, g.x/ D exp .�cx2/ as c 2 R varies.

Solution. a) We begin by examining the characteristic points on the parabola .s; s2/.Consider the determinant

J.s; 0/ D det

�Xs.s; 0/ Ys.s; 0/

Xt .s; 0/ Yt .s; 0/

D det

�1 2s

a�s; s2

� �1

D �2a �s; s2� s � 1;

where X.s; t/, Y.s; t/, Z.s; t/ is the solution of the characteristic system8<:x0 D a.x; y/

y0 D �1z0 D z

with x.0/ D s, y.0/ D s2, z.0/ D g.s/.If J .s; 0/ ¤ 0, the characteristic directions are never tangent and the problem admits

a local C 1 solution in a neighbourhood of the parabola. Suppose now that for some s0

J.s0; 0/ D 2a�s0; s

20

�s0 C 1 D 0:

Necessarily s0 ¤ 0. If a local C 1 solution around�s0; s

20

�exists, then the space curve�

s; s2; g .s/�

must be characteristic at s D s0. This means that the rank of�Xs.s; 0/ Ys.s; 0/ Zs.s; 0/

Xt .s; 0/ Yt .s; 0/ Zt .s; 0/

D�

1 2s0 g0 .s0/�1=2s0 �1 �g .s0/


is 1, i.e.g0 .s0/ D 2s0g .s0/ : (3.21)

If (3.21) does not hold there are no C 1 solutions in any neighbourhood of�s0; s

20

�: There

might be less regular solutions of course.

b) In the case a .x; y/ D y=2 and g .s/ D exp��cs2�, we have

2a�s; s2

�s C 1 D s3 C 1

and there is only one characteristic point: .�1; 1/. Since g0 .s/ D �2csg .s/, (3.21) isequivalent to

2c D �2and the problem can admit regular solutions around .�1; 1/ only if c D �1. If so,g .s/ D exp

�s2�

and the solution is u .x; y/ D ey , as is easy to verify.

Problem 3.2.15 (First integrals). Use the method of first integrals to find a solutionu D u.x; y/ for

uuy D �yin terms of an arbitrary function.

Solution. Set x D x.t/, y D y.t/, z D u.x.t/; y.t// and write (formally) the charac-teristic system as follows:

dx

0D dy

zD �dz

y. (3.22)

We have to find two independent first integrals ' .x; y; z/ and .x; y; z/. From (3.22) weinfer

dx D 0; y dy D �z dz:The first one gives

x D c1;

so'.x; y; z/ D x

is a first integral (x remains constant on the trajectories of the characteristic system). Fromthe other one

y2 C z2 D c2;

showing that .x; y; z/ D y2 C z2

is another first integral (for any y and z, also zero). To check the independence of ' and we must verify that their gradients are linearly independent vector fields. In fact

r'.x; y; z/ D .1; 0; 0/; r .x; y; z/ D .0; 2y; 2z/

are independent except at the origin. Overall, the general integral z D u.x; y/ is definedimplicitly by

F.'.x; y; z/; .x; y; z// D F.x; y2 C z2/ D 0;


with F smooth and arbitrary. Supposing Fx ¤ 0 we may write

x D f�y2 C z2

�for any smooth f . Hence the solution are surfaces of revolution about the x-axis.

Problem 3.2.16 (Euler equation). Determine the solutions of the Euler equation

xux C yuy D nu: (3.23)

where n > 0.

Solution. Let us define x D x.t/, y D y.t/, z D u.x.t/; y.t// and write the charac-teristic system in the form

dx

xD dy

yD dz

nz:

We seek two (independent) first integrals, and observe that each of the equations

dx

xD dy

yand

dx

xD dz

nz

contain two of the three unknowns only. The first one gives log jxj D log jyj C c, soyx

D c1 and

'.x; y; z/ D y

x

is then a first integral when x ¤ 0 (clearly x=y is a first integral when y ¤ 0). The secondequation can be treated similarly (still for x ¤ 0)

z

xnD c2;

thus another first integral is .x; y; z/ D z

xn:

For x ¤ 0 the first integrals are independent, for their gradients

r' D .�x�2y; x�1; 0/ and r D .�nx�n�1z; 0; x�n/

are linearly independent. The general solution z D u.x; y/ is given, implicitly, by theequation

F .'; / D F�yx;z

xn

�D 0;

with F arbitrary and smooth. If F ¤ 0 we can write

z D u.x; y/ D xnf�yx

�;


with f smooth and arbitrary. Thus the solutions are positively homogeneous functions ofdegree n. Indeed, for any l > 0,

u.lx; ly/ D .lx/n f

�lx

ly

D lnu.x; y/

As a matter of fact it can be proved in an elementary way that a function is homogeneousof degree n if and only if it satisfies the Euler equation (3.23).

Problem 3.2.17 (Cauchy problem). Solve the problem

ux C uy D u u.x; 0/ D cos x

in two different ways: a) using the characteristic system, and b) via first integrals.

Solution. a) The parametric equations of the initial curve are

f .s/ D s; g .s/ D 0; h .s/ D cos s.

Write x D x.t/, y D y.t/, z D z.x.t/; y.t//, and the characteristic system is8<:x0.t/ D 1 x.0/ D s

y0.t/ D 1 y.0/ D 0

z0.t/ D z.t/ z.0/ D cos s:

The solutions 8<:x.t; s/ D t C s

y.t; s/ D t

z.t; s/ D et cos s;

immediately give t D y, s D x � y. Hence the solution is

z D u.x; y/ D ey cos.x � y/:b) Using first integrals, let us rewrite the characteristic system as

dx D dy D dz

z:

Regrouping as dx D dy, dx D dz=z, we obtain two first integrals

'.x; y; z/ D x � y, and .x; y; z/ D ze�x :

Asr' D .1;�1; 0/ and r D .�ze�x ; 0; e�x/

the gradients are linearly independent, making

z D u.x; y/ D exf .x � y/


for any smooth f , the general solution. Imposing the initial condition

exf .x/ D cos x;

i.e.f .x/ D e�x cos x;

we recover the solution of part a).

Problem 3.2.18 (Characteristic variables). Consider the first-order linear equation

a.x; y/ux C b.x; y/uy C c.x; y/u D f .x; y/ (3.24)

where a, b, c, f are C 1 functions on the open set D � R2, and a ¤ 0 on D. Let'.x; y/ D k, k 2 R, be the family of characteristic curves for the reduced equation

aux C buy D 0:

Let D .x; y/ be a smooth function independent from ', that is:

det

�'x 'y x y

¤ 0 in D: (3.25)

Show that the change of variables´� D '.x; y/

� D .x; y/(3.26)

transforms (3.24) in an ordinary equation of the form

w� C P.�; �/w D Q.�; �/:

Deduce a formula for the general solution of (3.24).

Solution. By (3.25) the transformation (3.26) is locally invertible, thus we can expressx; y as smooth functions of � and �:

x D ˆ.�; �/; y D ‰.�; �/:

Setw.�; �/ D u.ˆ.�; �/;‰.�; �//

oru.x; y/ D w.'.x; y/; .x; y//:

We haveux D w�'x C w� x; uy D w�'y C w� y ;

and substituting into (3.24) we obtain

.a'x C b'y/w� C .a x C b y/w� C C.�; �/w D F.�; �/; (3.27)

where, for simplicity, we have set

C.�; �/ D c.ˆ.�; �/;‰.�; �//; F.�; �/ D f .ˆ.�; �/;‰.�; �//:


Now observe the following:1. a'x C b'y D 0 on D. In fact, along the characteristics '.x; y/ D k,

'x dx C 'y dy D 0:

On the other hand, along the characteristics ' D k we have the system

dx D a dt; dy D b dt;

so that �'x aC 'y b

�dt D 0;

and thus the claim.2. a x C b y ¤ 0 in D. This is due to the independence of and ', i.e. equation

(3.25), implying

�ba

D 'x'y

¤ x y

:

Setting

D.�; �/ D a.x; y/ x.x; y/C b.x; y/ y.x; y/ˇxDˆ.�;�/; yD‰.�;�/ ;

from 1. and 2. we obtain that (3.27) now reads

w� C C.�; �/

D.�; �/w D F.�; �/

D.�; �/;

which is an (ordinary) linear equation of order one in w.�; �/, for any given � . The generalsolution is

w.�; �/ D exp

��ZC.�; �/

D.�; �/d�

�Z

exp

�ZC.�; �/

D.�; �/d�

F.�; �/

D.�; �/d�CG.�/

�;

with G smooth. Now it suffices to go back to the original variables to obtain the solution

u.x; y/ D w.'.x; y/; .x; y//:

Problem 3.2.19 (Using characteristic variables). Using the method seen in the previ-ous problem determine the solution of

xux � yuy D u � y in D D ¹.x; y/ 2 R2 W y > 0ºsuch that u.x; y/ D y on the parabola x D y2 (y > 0).

Solution. We adopt the strategy (and notation) of the previous exercise. The charac-teristics of xux � yuy D 0 solve the differential equation

dx

xD �dy

y;


and are then given by '.x; y/ D xy D k: Let us choose (for example) .x; y/ D y. Thischoice is admissible, for

det

�'x 'y x y

D det

�y x

0 1

D y > 0 on D;

and therefore ' and are independent inD. We now make the change of variables (withy > 0)

� D xy, � D y

with inverse (� > 0) x D �=�, y D �. Then

C.�; �/ D �1; F.�; �/ D ��; D.�; �/ D ��and w.�; �/ D u.�=�; �/ solves the equation

w� C 1

�w D 1:

Hence

w.�; �/ D �

2C G.�/

�;

i.e.

u.x; y/ D y

2C G.xy/

y;

for any smooth G. To find G we impose the boundary condition: along x D y2 we musthave

y D y

2C G.y3/

y;

whence G.y3/ D y2=2, i.e. G.z/ D z2=3=2. The solution is then

u.x; y/ D 1

2.y C x2=3y�1=3/:

Remark. No neighbourhood of the origin contains a solution to the Cauchy problem. Infact, since x D 0 is a characteristic tangent to x D y2 at .0; 0/, the origin is a characteristicpoint. Differentiating u

�y2; y

� D y we obtain

2yux.y2; y/C uy.y

2; y/ D 1: (3.28)

Suppose y ¤ 0 for the moment. Restricting the PDE on the parabola and simplifyinggives

yux.y2; y/ � uy.y2; y/ D 0: (3.29)

The last two conditions uniquely determine ux; uy on the parabola provided the matrices�2y 1

y �1

and

�2y 1 1

y �1 0


have the same rank. By continuity this condition must hold at y D 0 as well. But there theranks are different, and the solution does not exist.

Problem 3.2.20 (Heat exchanger). The simplest heat exchanger is modelled by acylindrical tube at temperature Ta D Ta.t/ (in Celsius degrees) immersed in a fluidof temperature T .x; t/. The fluid flows in the tube at constant speed v m/s (metersper seconds). Denote by U the heat transfer coefficient through the walls (unit: cals�1m�2(ıC/�1), by Cp the heat capacity per unit of volume (in cal m�3(ıC/�1), by A(m2) the area and by p (m) the perimeter of the cross section. We assume that U andCp are constant.

a) Write down the differential equation governing the evolution of the temperature T .

b) Solve the problem knowing that´T .x; 0/ D T0.x/ 0 � x � L

T .0; t/ D g.t/ t > 0:

Solution. a) We start by writing the equation of thermal equilibrium between the crosssections at x and x C�x, relative to the time interval .t; t C�t/. We know that:

• The heat exchanged (better, absorbed) through the walls of the tube is linear (Newton’slaw of cooling)

Up�x.Ta � T /�t (calories).

• The heat flowing in, due to the temperature difference at x, x C �x, is given by(Fourier’s law)

vACpŒT .x; t/ � T .x C�x; t/��t (calories).

• The variation of heat in this portion, between times t and t C�t , is

A�xCpŒT .x; t/ � T .x; t C�t/� (calories).

All this together gives

A�xCpŒT .x; t/ � T .x; t C�t/� DvACpŒT .x; t/ � T .x C�x; t/��t C Up�x.Ta.t/ � T /�t:

Divide by �x�t and take the limit as �x;�t ! 0. We obtain:

@T

@tC v

@T

@xD h.Ta.t/ � T /; where h D Up

ACp:

b) The characteristics (actually, their projections on the xt -plane) are the straight lines

x � vt D k:


It is natural to divide the region Œ0; L�� Œ0;C1/ in two parts: the first part between the x-axis and the line x D vt , where T is determined by T0, and corresponding to vt < x � L;the second one above the line x D vt , corresponding to 0 � x � min.vt; L/, where Tdepends on g.

To determine the solution in vt < x � L we consider the characteristic x D vt C �,0 � � � L, and set z.t/ D T .vt C �; t/. Using the differential equation, we see that zsolves

z0.t/ D vTx.vt C �; t/C Tt .vt C �; t/ D h.Ta.t/ � z.t//with

z.0/ D T .�; 0/ D T0.�/:

Therefore

z.t/ D T0.�/e�ht C h

Z t

0

Ta.s/e�h.t�s/ ds

and since � D x � vt , we find

T .x; t/ D T0.x � vt/e�ht C h

Z t

0

Ta.s/e�h.t�s/ ds; vt < x � vL:

Now let us examine the points .x; t/ for which vt � x. Solve the equation of the charac-teristic for t

t D x

vC �

and setu.x/ D T

�x;x

vC �

�:

Thendu

dxD Tx

�x;x

vC �

�C 1

vTt

�x;x

vC �

�D h

v

�Ta

�xv

C ��

� u�

withu.0/ D T .0; �/ D g.�/:

Therefore

u.x/ D g.�/e�hx=v C h

v

Z x

0

Ta

� sv

C ��e�h.x�s/=v ds;

and, since � D t � xv

, we have

T .x; t/ D g�t � x

v

�e�hx=v C h

v

Z x

0

Ta

�t C s � x

v

�e�h.x�s/=v ds.

Finally, setting

� D t C s � xv

; d� D 1

vds

we obtain

T .x; t/ D g�t � x

v

�e�hx=v C h

Z t

t�xv

Ta .�/ e�h.t��/ d�; 0 � x � min.vt; L/:


Remark. In caseg.t/ D g0; Ta.t/ D Ta

are constant, for x > vt we find

T .x; t/ D g0e�hx=v C Ta

�1 � e�hx=v

�;

the stationary state, which is reached everywhere along the tube as soon as t � L=v.

Problem 3.2.21 (Three-dimensional Cauchy problem). Compute, where defined, thesolution u D u.x; y; z/ to ´

ux C xuy � uz D u

u.x; y; 1/ D x C y:

Solution. We parametrise the surface carrying the data by

f .r; s/ D s; g .r; s/ D r; h .r; s/ D 1; k .r; s/ D s C r

then write x D x.t/, y D y.t/, z D z.t/, v D u.x.t/; y.t/; z.t//, and the characteristicsystem reads 8<

ˆ:x0.t/ D 1 x.0/ D s

y0.t/ D x.t/ y.0/ D r

z0.t/ D �1 z.0/ D 1

v0.t/ D v.t/ v.0/ D s C r:

The first and third equations give x.t/ D t C s and z.t/ D �t C 1. The last one isdecoupled, and implies v.t/ D .s C r/et . The second one now reads

y0.t/ D t C s;

and by the initial condition y.t/ D t2=2C st C r . Summarising,8<ˆ:x D X.r; s; t/ D t C s

y D Y.r; s; t/ D t2=2C st C r

z D Z.r; s; t/ D �t C 1

v D V.r; s; t/ D .s C r/et :

(3.30)

To have .t; r; s/ in terms of .x; y; z/, we get t D 1 � z from the third, s D x � t D x Cz � 1 from the first and

r D y � t2=2 � st D y � .1 � z/2=2 � .1 � z/.x C z � 1/


from the second. Substituting into the fourth equation we find

u.x; y; z/ Dxz C y C 1

2.z2 � 1/

�e1�z :

The solution is defined and smooth in the whole R3. In this case the general sufficientcondition for solving the first three equations of system (3.30) would have been

J.r; s; t/ D det

0@Xr Yr ZrXs Ys ZsXt Yt Zt

1A D det

0@0 1 0

1 t 0

1 t C s �1

1A D 1 ¤ 0;

In fact, in our case J.r; s; t/ D 1.

Problem 3.2.22 (Clairaut’s equation). Consider the Clairaut equation

xux C yuy C f .ux; uy/ D u:

a) Verify that u.x; yI a; b/ D ax C by C f .a; b/ is a solution for any .a; b/ 2 R2.

b) In the case

f .a; b/ D a2 C b2

2

use the envelope of the solutions of part a) to generate a nonlinear solution in x andy and a solution depending on an arbitrary function.

Solution. a) Immediate: ux D a; uy D b and hence

xux C yuy C f .ux; uy/ D ax C by C f .a; b/ D u:

b) With the given f the equation becomes

xux C yuy C u2x C u2y

2D u;

while the family of solutions reads

G.x; y; uI a; b/ D u �ax C by C a2 C b2

2

�D 0:

Let us find its envelope: we have to eliminate a; b from the system

G D 0; Ga D 0; Gb D 0

i.e. 8<:u D ax C by C a2Cb2

2

x C a D 0

y C b D 0;


showing that the envelope is

u.x; y/ D �12.x2 C y2/:

It is easy to see, by direct computation, that u is indeed a solution.

Remark. A two-parameter family of solutions z D '.x; yI a; b/ is called complete inte-gral if the vector fields �

'a; 'xa; 'ya�

and .'b; 'xb; 'yb/

are linearly independent. Given a complete integral, we can recover a solution depend-ing on a function by setting, for instance, b D w .a/ and writing the envelope ofz D '.x; yI a;w .a//. For instance, choosing w.a/ D a gives, for the Clairaut equationwith f .a; b/ D a2Cb2

2,´

u D a.x C y/C a2

0 D x C y C 2awhence u D � .x C y/2

4:


3.3.1. Solve the global Cauchy problem for the Burgers equation ut C uux D 0 with initialdatum g.x/ given by:

a)

8<:1 x � 0

1 � x 0 < x < 1

0 x � 1:

b)

8<:1 x < 0

2 0 < x < 1

0 x < 1:

c)

´jxj jxj � 1

1 jxj � 1:

3.3.2. Consider the traffic equation (Problem 3.2.4 on page 160), with initial density

g.x/ D´a�m x < 0

�m x > 0:

For 0 < a < 1, determine the characteristics, the possible shock waves, and find the solutions onthe half-plane .x; t/, t > 0.

3.3.3. Solve the global Cauchy problem for the traffic equation (Problem 3.2.4 on page 160) withinitial density

g.x/ D

8<ˆ:�m=4 x < �1�m=3 �1 < x � 0

.2C x/�m=6 0 � x � 1

�m=2 x � 1:

3.3.4. Solve the following problem, variant of the tunnel Problem 3.2.6 (page 164)´�t C q.�/x D 0 x 2 R; t > 0

�.x; 0/ D g.x/;


where

q.�/ D � log

�1

�

and g.x/ D

´1 x < 0

e�4 x > 0:

Determine the path of the vehicle initially at x0 D �1.

3.3.5. Generalise Problem 3.2.7 (page 168) to´ut C q.u/x D 0 x 2 R; t > 0

u.x; 0/ D g.x/ x 2 R;

with q 2 C 2.R/ concave.

3.3.6. Study 8<:cx C q .c/t D �c x > 0; t > 0

c.x; 0/ D 0 x > 0

c.0; t/ D c0 t > 0

where q .c/ D �c C c

cC1�

and c0 > 0 is constant.

3.3.7. Compute, where defined, the solutions to the following (two-dimensional) Cauchy prob-lems:

a) xux C uy D y, u .x; 0/ D x2.b) ux � 2uy D u, u .0; y/ D y.c) ux C 3u2uy D 1, u .x; 0/ D 0.

3.3.8. Write the general solution of (depending on an arbitrary function):

a) ux � puuy D 0. b)

1

yux C uy D u2.

3.3.9. Find a solution z D u.x; y/ of

yux � xuy D 2xyu; u D s2 along � D ¹.s; s/ W s 2 Rº:3.3.10. Compute the general solution of

xux � yuy C u D y on D D ¹.x; y/ 2 R2 W x; y > 0ºin two ways: a) by using characteristic variables (Problem 3.2.18 on page 187) and b) by using firstintegrals.

3.3.11. (Picone) LetD D ®.x; y/ W x2 C y2 � 1

¯be the closed unit disc in R2 and u 2 C 1 .D/

be a solution toa .x; y/ ux C b .x; y/ uy D �u on D.

Suppose the coefficients a and b are continuous on D and satisfy

a .x; y/ x C b .x; y/ y > 0 on @D:

If > 0, show that u � 0 on D.

3.3.12. (Buckley-Leverett equation) Consider the equation

�St D H 0.S/q � rS;


where S D S.x; t /, x 2 R3 and

H.S/ D �k1.S/1

�k1.S/

1C k2.S/

2

�1:

Solve the equation when

k1.S/

1D 1 � S; k2.S/

2D S; � D 1

2; q D

�1

2e�x1 ;

1

2e�x2 ;

1

2e�x3

;

with initial condition S.x; 0/ D g.x/.

3.3.13. Solve

.cy � bz/ux C .az � cx/uy C .bx � ay/uz D 0

and interpret geometrically the solution (seek a suitable combination of the characteristic equationsthat gives integrable equations).

3.3.14. a) Let F be the family of surfaces in R3 given in implicit form by w .x; y; z/ D c

depending on the parameter c. Find the family of orthogonal surfaces, those that intersect eachmember of F perpendicularly.

b) Determine the family of orthogonal surfaces to the family of paraboloids z � x2 � y2 D c:

3.3.15. The concentration u of a substance contained in a semi-infinite tube (along the semiaxisx � 0) solves

ut C vux D �ku;where v and k are positive constants. Compute u knowing that

u.x; 0/ D f .x/ for x > 0; u.0; t/ D g.t/ for t > 0:

3.3.16. The concentration u of a substance contained in a semi-infinite tube (along the semiaxisx � 0) solves

ut C vux D �ku˛ ;with v, k, ˛ ¤ 1 positive constants. Compute u knowing that

u.x; 0/ D f .x/ for x > 0; u.0; t/ D g.t/ for t > 0:

3.3.17. (Separation of variables) Consider the PDE

a .x/ u2x C b .y/ u2y D f .x/C g .y/ ;

with a, b continuous on some interval I .

a) Determine, at least formally, a complete integral by seeking solutions of the form u .x; y/ Dv .x/C w .y/.

b) Exploit this result for solving u2x C u2y D 1.


3.3.18. (Burgers equation, decreasing datum) Study the problem

´ut C uux D 0 x 2 R; t > 0

u.x; 0/ D g.x/ x 2 R;where g.x/ D

8<:1 x < 0

1 � x2 0 � x � 1

0 x > 1:

3.3.19. Consider the model of Problem 3.2.11 on page 176. Set D 1 and find the solution onthe quadrant x > 0, t > 0, with u.x; 0/ D 0 and

u.0; t/ D g.t/ D´c0 0 � t � 1

0 t > 1:

3.3.1 Solutions

Solution 3.3.1. a) With the given datum we deduce that the characteristics have equation8<:x D � C t if � � 0

x D � C .1 � �/t if 0 < � < 1

x D � if � � 1:

All characteristics based on the segment 0 < � < 1 cross the point .1; 1/; therefore, for 0 < t < 1

they do not collide. This fact means that in this time interval the solution has no discontinuity. A dis-continuity originates at the point .1; 1/ because of the collision between the vertical characteristics(carrying the datum u� D 0) and the characteristics x D � C t (carrying uC D 1). Observing thatq� D 0 and qC D 1=2 are the fluxes from the left and the right of the discontinuity, respectively,according to the Rankine-Hugoniot condition, the shock curve solves the problem´

Ps D 1=2

s.1/ D 1;

and therefore s.t/ D .t C 1/=2.In the region S D ¹0 � x < 1; 0 � t < xº the solution is implicitly defined by the equation

u D g.x � q0.u/t/: In this case, given that g.x/ D 1 � x, we deduce

u.x; t/ D 1 � x1 � t :

The solution of the problem is

u.x; t/ D

8<:1 if x � t < 1 or x < .t C 1/=2 with t � 1

.1 � x/=.1 � t / if 0 � t � x < 1

0 if t < 2x � 1 with x > 1:

b) For the assigned initial condition, the characteristics have equation8<:x D � C t if � < 0

x D � C 2t if 0 < � < 1

x D � if � > 1:

The domain S D ¹t < x < 2tº is reached by no characteristics; we can define the solution u asa rarefaction wave that connects the values u D 1 and u D 2, transported by the characteristics


x D t and x D 2t . In this case, the rarefaction wave based at the point .0; 0/ is u D x=t . Alongthe straight lines x D ht , with 1 < h < 2, u is constant. The collision of the characteristics createsa discontinuity x D s.t/ that propagates following the Rankine-Hugoniot condition. At least forsmall times t , the shock curve is due to a discontinuity created by the impact between the valueuC D 2 (transported along the characteristics x D �C2t ) and the value u� D 0 (along the verticalcharacteristics). Taking into account that q.u/ D u2=2, those data are respectively associated to thefluxes qC D q.uC/ D 2 and q� D q.u�/ D 0. Therefore, the shock curve is given by the solutionof the problem ´

Ps D 1

s.0/ D 1i.e. s.t/ D t C 1:

For times t > 1, namely, from the point .2; 1/ the vertical characteristics x D � , with � > 2, in-teract with the rarefaction wave u D x=t . In this situation, the Rankine-Hugoniot condition has tobe solved using uC D s=t and qC D s2=2t2; hence, the shock curve is determined by the Cauchyproblem ´

Ps D s=2t

s.1/ D 2:

The solution of the Cauchy problem is s.t/ D 2pt . For times t > 4, i.e. from the point .4; 4/, the

shock curve interacts with the characteristics x D �C t , � < 0, that carry the data u D 1. Thereforeit is deviated again; in this case we have uC D 1, qC D q.uC/ D 1=2, and´

Ps D 1=2

s.4/ D 4i.e. s.t/ D t

2C 2:

Then the shock curve is made of three arcs connected at the points .2; 1/ and .4; 4/. The solution is

u.x; t/ D

8ˆ<ˆ:

1 if x < t < 4 or x < t=2C 2 with t > 4

x=t if t � x � 2t; with x � 2pt

2 if 2t � x < t C 1

0 if t C 1 < x < 2 or t < x2=4 with 2 � x < 4

or t < 2x � 4 with x � 4:

c) Reasoning as above we find

u.x; t/ D

8ˆ<ˆ:

1 x � t � 1 for t < 1

1 x < 1C t � p2C 2t for t � 1

xt�1 t � 1 � x � 0 for t < 1xtC1 max.0; 1C t � p

2C 2t/ < x � t C 1

1 x � t C 1:

Solution 3.3.2. The characteristic from .�; 0/ has equation

x D � C q0.g.�//t D � C vm

�1 � 2�

�m

t:

Substituting the initial datum we find

x D � C vm.1 � 2a/t; for � < 0


andx D � � vmt; for � > 0:

As 0 < a < 1, vm.1�2a/ > �vm, and thus a shock wave originates from (0,0). From the Rankine-Hugoniot condition we find

s0 D q.�C/ � q.��/�C � �� D vm

�m

�C.�m � �C/ � ��.�m � ��/�C � �� D vm

�m

��m � �C � ��

�:

Substituting�C D �m; �

� D a�m

we obtain ´s0 D �avms.0/ D 0;

and thens.t/ D �avmt:

Therefore, for any 0 < a < 1, we have

u.x; t/ D´a�m x < �avmt�m x > �avmt:

Solution 3.3.3. Since q.�/ D vm�.1 � �=�m/ is concave and g increasing we expect multipleshocks. The characteristics have equation

x D q0.g.�//t C � D vm

�1 � 2g.�/

�m

t C � D

8<ˆ:vmt=2C � � < �1vmt=3C � �1 < � � 0

vmt .1 � �/=3C � 0 � � � 1

� � � 1:

The Rankine-Hugoniot condition for the shock wave x D s.t/ may be written

s0.t/ D q.�C.s.t/; t// � q.��.s.t/; t//�C.s.t/; t/ � ��.s.t/; t/ D vm

"1 � �C.s.t/; t/C ��.s.t/; t/

�m

#:

The first shock is determined by the collision of characteristics coming from points � < �1 on theleft and points �1 < � < 0 on the right. The starting point is S1 D .�1; 0/. Substituting �� m=4

and �C � �m=3, we obtain´s01 D 5vm=12

s1.0/ D �1; and hence x D s1.t/ D 5

12vmt � 1:

Solving for � the equation of the characteristics corresponding to 0 � � � 1, we find

� D vmt � 3xvmt � 3 ; and hence �.x; t/ D g.�/ D vmt � x � 2

2.vmt � 3/ �m:

This expression holds when t < 3=vm, and all these characteristics meet at

S2 D�1;

3

vm

:

From S2 another shock starts, for which �C � �m=2 and �� m=3. The latter follows froms1.3=vm/ D 1=4 < 1, whence the wave with density �m=4 reaches x D 1 in a subsequent time.


We find: ´s02 D vm=6

s2.3=vm/ D 1;hence x D s2.t/ D 1

6vmt C 1

2:

The two shock curves meet at a point .x; t/ solution of the system´x D 5

12vmt � 1x D 1

6vmt C 12 ;

i.e. at S3 D�3

2;6

vm

:

The third and final shock is born at S3, where �C � �m=2 and �� m=4:´s03 D vm=4

s3.6=vm/ D 3=2;yielding x D s3.t/ D 1

4vmt:

Summing up, the solution is:

• For 0 � t <3

vm, �.x; t/ D

8<ˆ:�m=4 x < s1.t/

�m=3 s1.t/ < x � vmt=3

�m.vmt � x � 2/=.2vmt � 6/ vmt=3 � x � 1

�m=2 x � 1:

• For3

vm� t <

6

vm, �.x; t/ D

8<:�m=4 x < s1.t/

�m=3 s1.t/ < x < s2.t/

�m=2 x > s2.t/:

• For t � 6

vm, �.x; t/ D

´�m=4 x < s3.t/

�m=2 x > s3.t/:

Solution 3.3.4. Since

q0.�/ D �1 � log �;

the problem to solve is 8<:�t � .1C log �/�x D 0 x 2 R; t > 0

�.x; 0/ D´1 x < 0

e�4 x > 0:

The characteristic from .�; 0/, the line x D � C q0.g.�//t , coincides with

x D � � t for � < 0

and withx D � C 3t for � > 0:

Therefore, using the initial data, we get

�.x; t/ D´1 x < �te�4 x > 3t:

There remains to find � in the region

¹.x; t/ W �t � x � 3tº:We have

q00.�/ D �1=� < 0;


and since the initial datum is decreasing, we look for an entropic solution in the form of a rarefactionwave centred at the origin. If we set R .y/ D .q0/�1 .y/, the wave is defined by

�.x; t/ D R�xt

�.

Solving for � the equationq0.�/ D �1 � log � D y

we findR .y/ D e�.1Cy/

and hence�.x; t/ D e�.1Cx=t/ for � 1 � x

t� 3:

Summing up,

�.x; t/ D

8<:1 x � �te�.1Cx=t/ �t � x � 3t

e�4 x > 3t:

To study the path of a vehicle, recall that the speed at a given density � is given by

v.�/ D q.�/=� D � log �:

Suppose now that a car is initially at x0 D �1. It will not move (maximum density, v.1/ D 0) untilx0 D �1 < �t , i.e. until time t0 D 1. At that instant the car path enters the region of the rarefactionwave, where the speed is

v.�.x; t// D � log.e�.1Cx=t// D 1C x

t:

If x D x.t/ denotes the car path, we have8<:x0 D 1C x

tx.1/ D �1:

The equation is linear, and integrating we have

x.t/ D t .log t � 1/:This holds until

x.t/ D t .log t � 1/ < 3tthat is, up to t D t1 D e4. At this moment the car enters (and remains) in the region where it travelsat constant speed v.e�4/ D 4, therefore the law of motion is

x.t/ D 4t � e4:Therefore

x.t/ D

8<:

�1 0 � t � 1

t.log t � 1/ 1 � t � e4

4t � e4 t � e4:

Note that the trajectory is not only continuous, but also C 1.

Solution 3.3.5. The characteristic �� , from the point .�; 0/, is

x D � C q0.g.�//t;and on �� we have u.x; t/ D g.�/. Hence

� D x � q0.g.�//t D x � q0.u.x; t//t


can be substituted to giveu.x; t/ D g.x � q0.u.x; t//t/:

But g is C 1, and u.x; 0/ D g.x/, hence we can use the implicit function theorem to solve for u. Set

h.x; t; u/ D u � g.x � q0.u/t/ D 0:

Then the theorem applies provided

hu.x; t; u/ D 1C g0.x � q0.u/t/q00.u/t ¤ 0:

Consider the characteristic �� . For any .x; t/ 2 ��x � q0.u/t D �;

hencehu.x; t; u.x; t// D 1C g0.�/q00.u/t D 1C g0.�/q00.g.�//t:

We deduce that hu D 0 if and only if

t D � 1

g0.�/q00.g.�// :

The starting time ts for the shock is the smallest t for which the above equation holds. We find tson the characteristic corresponding to �m, where �m realises

max�g0.�/q00.g.�//:

Therefore the sufficient conditions for the shock to begin are

q00 < 0; g0 > 0(q is concave by assumption).

Solution 3.3.6. Let us begin finding the characteristics. Since

q.c/ D c C c

c C 1;

we have

q0.c/ D 1C 1

.c C 1/2:

The characteristic system, with datum on x > 0, is8<:x0.s/ D 1 x.0/ D �

t 0.s/ D 1C 1

.c C 1/2t .0/ D 0

c0.s/ D �c c.0/ D 0:

Thenx D s C � , t D 2s and c D 0:

The characteristics from the x-axis are the lines

t D 2.x � �/; with � > 0;

and carrycC D 0:


Consider now the datum on t > 0. The characteristic system reads8<:x0.s/ D 1 x.0/ D 0

t 0.s/ D 1C 1

.c C 1/2t .0/ D �

c0.s/ D �c c.0/ D c0:

Immediately x D s and c D c0e�s , whence

c�.x; t/ D c0e�x :

The corresponding characteristics (aside, the precise expression of the datum is unnecessary to de-termine the solution) can be found dividing the second and third equations and integrating. We find

�dt D�1

cC 1

c.c C 1/2

dc

and thenZdv

v.v C 1/2DZ �

1

v� 1

v C 1� 1

.v C 1/2

dv D ln

v

v C 1C 1

v C 1C constant;

so

t D � � lnc2

c C 1� 1

c C 1C k D � C 2x � ln

1

c0e�x C 1� 1

c0e�x C 1C k0

(where k0 is chosen so that t .0/ D �). In any case, as already noticed, the computation was su-perfluous. Indeed, from the characteristic system we see directly that along the characteristics fromt > 0

dt

dx

ˇxD0

D 1C 1

.c0 C 1/2< 2

for any c0 > 0. As a consequence the characteristics from t > 0 meet those coming from x > 0, atleast for � and � small. Since the two characteristics families carry pointwise-distinct data, there isa shock curve t D s.x/. From the Rankine-Hugoniot conditions we get (see Problem 3.2.9)

s0.x/ D q.cC/ � q.c�/cC � c� D 1C 1

c0e�x C 1:

Thus we have Zdx

c0e�x C 1DZ

exdx

c0 C exD log

�c0 C ex

�C k;

and since s.0/ D 0, the shock wave has equation

t D s.x/ D x C logc0 C ex

c0 C 1:

In conclusion, the solution is

c.x; t/ D

8<ˆ:0 0 � t < x C log

c0 C ex

c0 C 1

c0e�x t > x C log

c0 C ex

c0 C 1:


Solution 3.3.7. a) Set x D x.t/, y D y.t/, z D u.x.t/; y.t//. The characteristic system is8<:x0.t/ D x.t/ x.0/ D s

y0.t/ D 1 y.0/ D 0

z0.t/ D y.t/ z.0/ D s2from which

8<:x D X.s; t/ D set

y D Y.s; t/ D t

z D Z.s; t/ D t2

2 C s2:

We find t D y, s D xe�y , and then u.x; t/ D y2

2 C x2e�2y .

b) The characteristic system is8<:x0.t/ D 1 x.0/ D 0

y0.t/ D �2 y.0/ D s

z0.t/ D z.t/ z.0/ D s

from which

8<:x D X.s; t/ D t

y D Y.s; t/ D �2t C s

z D Z.s; t/ D set :

This gives t D x, s D 2x C y and u.x; t/ D .2x C y/ex .

c) The characteristic system is8<:x0.t/ D 1 x.0/ D s

y0.t/ D 3z2.t/ y.0/ D 0

z0.t/ D 1 z.0/ D 0

from which

8<:x D X.s; t/ D t C s

y D Y.s; t/ D t3

z D Z.s; t/ D t:

We obtain, t D y1=3, and then u.x; t/ D y1=3.

Solution 3.3.8. a) Set x D x.t/, y D y.t/, z D u.x.t/; y.t//, and write the characteristic systemin the form

dx

1D dyp

zD dz

1; so that dy D p

zdz; dx D dz:

We find the first integrals

'.x; y; z/ D 2

3z3=2 � y; .x; y; z/ D x � z:

The general solution z D u.x; y/ is given implicitly by the equation

F

�2

3z3=2 � y; x � z

D 0;

for some smooth F (and z > 0).

b) The characteristic system

dx

1=yD dy

1D dz

z2; gives dy D dz

z2; dx D dy

y:

We find the first integrals

'.x; y; z/ D y C 1

z; .x; y; z/ D x � log jyj;

and the general solution z D u.x; y/ is given implicitly by the equation

F

�y C 1

z; x � log jyj

D 0;

for arbitrary F , smooth (and yz ¤ 0).


Solution 3.3.9. The characteristic system8<:x0.t/ D y.t/ x.0/ D s

y0.t/ D �x.t/ y.0/ D s

z0.t/ D 2x.t/y.t/z.t/ z.0/ D s2;

is solved by

8<:x D X.s; t/ D s.cos t C sin t /

y D Y.s; t/ D s.cos t � sin t /

z D Z.s; t/ D s2es2 sin2t :

One sees that x2 C y2 D 2s2; while x2 � y2 D 2s2 sin 2t: Therefore

u.x; t/ D x2 C y2

2e.x

2�y2/=2:

Solution 3.3.10. a) The characteristics for

xux � yuy D 0

are (see Problem 3.2.19 on page 188)

'.x; y/ D xy D k:

Let us choose, again, .x; y/ D y. We know that ' and are independent. Choose new variables(for x; y > 0) ´

� D xy

� D y

with inverse (�; � > 0) x D �=�, y D �. According to the notation of Problem 3.2.18 (page 187)we have

C.�; �/ D 1; F.�; �/ D �; D.�; �/ D ��;and then

w.�; �/ D u.�=�; �/

solves

w� � 1

�w D �1:

In this wayw.�; �/ D �.G.�/ � log �/;

i.e.u.x; y/ D y.G.xy/ � log y/

for an arbitrary function G (smooth).

b) We write the characteristic system in the form

dx

xD �dy

yD du

y � u :The first equality immediately gives the first integral

'.x; y; z/ D xy;

while the second, written asdu

dyD �1C u

y,

gives the first integral, independent from the previous one,

.x; y; z/ D u

yC log y:


As ' does not depend on u we can write the solution in the form D G .'/ directly,

u

yC log y D G.xy/:

This completely agrees with part a).

Solution 3.3.11. Let .x0; y0/ be a global maximum point for u onD. If .x0; y0/ is insideD thenru.x0; y0/ D 0, and from the differential equation u .x0; y0/ D 0. If, instead, .x0; y0/ 2 @D, thenru.x0; y0/ may not vanish, but it certainly points outwards. On the other hand, the unit outwardnormal to @D is the vector .x0; y0/ itself, and � � .a; b/ D x0a.x0; y0/Cy0b.x0; y0/ > 0 says that.a; b/, too, points outwards. Therefore

0 � ru.x0; y0/ � .a.x0; y0/; b.x0; y0// D �u.x0; y0/;and since > 0, in either case

maxD

u � 0:

In the same way (or writing �u instead of u and repeating the argument) we see

minDu � 0;

making u identically zero.

Solution 3.3.12. We obtain H.S/ D S � 1, thus the problem to solve reads8<:

3XiD1

e�xiSxi� St D 0 x 2 R3; t > 0

S.x; 0/ D f .x/:

The associated characteristic system (for z.v/ D S.x.v/; t.v//) is8<:x0i .v/ D e�xi .v/ xi .0/ D si ; i D 1; 2; 3

t 0.v/ D �1 t.0/ D 0

z0.v/ D 0 z.0/ D g.s1; s2; s3/;

from which 8<:xi .v/ D log.v C esi /; i D 1; 2; 3

t.v/ D �vz.v/ D g.s1; s2; s3/:

For any isi D log .tCexi /;

and finallyS.x; t / D g.log.t C ex1/; log.t C ex2 /; log.t C ex3//:

Solution 3.3.13. As the equation does not contain zero-order terms, umust be constant along thesolutions of the reduced characteristic system, which can be written as8<

:dx D .cy � bz/ dtdy D .az � cx/ dtdz D .bx � ay/ dt:


In order to decouple the equations and find two remaining first integrals, let us multiply the first oneby a, the second by b and the third one by c, then add. We find

a dx C b dy C c dz D 0;

which gives the first integral'.x; y; z/ D ax C by C cz:

Multiplying the equations by x, y, z, respectively, and adding up, yields

x dx C y dy C z dz D 0;

and thus a first integral, independent from the previous one (for .x; y; z/ ¤ 0), is given by

.x; y; z/ D x2 C y2 C z2:

The general solution is then

u.x; y; z/ D F .'.x; y; z/; .x; y; z// D F.ax C by C cz; x2 C y2 C z2/;

for arbitrary smooth function F .

Geometrical interpretation. The level sets of the first integral '

ax C by C cz D c1

are parallel planes, orthogonal to the unit vector

.a0; b0; c0/ D .a; b; c/pa2 C b2 C c2

:

The level sets of x2 C y2 C z2 D c2

are spheres centred at the origin. The solution u is constant on the intersection curves (characteris-tics) of the two families, which are circles lying on the given planes and centred along the line r

x

a0 D y

b0 D z

c0 ;

orthogonal to every plane. Therefore the level surfaces of u are, in implicit form,

a0x C b0y C c0z D G.x2 C y2 C z2/:

This equation defines, for any given G, a surface of revolution S about the line r . Given P D.x; y; z/ on S , in fact, the modulus of the left-hand side ja0xC b0y C c0zj equals the distance of Pfrom r , and the equation shows that this distance depends on the norm of P .

Solution 3.3.14. a) The simplest way to express the orthogonality of two surfaces is to imposethat the normal vector fields be orthogonal. For level surfaces defined by a function w a normal issimply given by the gradient of w. We indicate with '.x; y; z/ D c the equation of the orthogonalfamily. We must have

0 D rw � r' D wx'x C wy'y C wz'z :

The reduced characteristic system is

dx

wxD dy

wyD dz

wz:


If '1, '2 are first integrals for this system, then the family we are looking for is

F .'1 .x; y; z/ ; '2 .x; y; z// D 0;

with F arbitrary and smooth.

b) This is a family of paraboloids of revolution about the z-axis. The equation of the orthogonalsurfaces

' .x; y; z/ D c

is�2x'x � 2y'y C 'z D 0

whose reduced characteristic system is

�dx2x

D �dy2y

D dz:

The first integrals are

'1.x; y; z/ D y

x; '2.x; y; z/ D 2z C log y;

and since '1 does not depend on z, the solution can be written in the form

z D �12

log y CG�yx

�for arbitrary smooth G.

Solution 3.3.15. The characteristics of the reduced equation are the straight lines

x � vt D �; � 2 R:

If we setz.t/ D u.vt C �; t/

then z solves the ODE

z0.t/ D vux.vt C �; t/C ut .vt C �; t/ D �ku.vt C �; t/ D �kz.t/;from which

z.t/ D Ce�kt :We have to recover C , as � varies, in terms of the initial datum. If � > 0 (i.e. in the region x > vt )the corresponding characteristic is issued from .�; 0/, so

f .�/ D z.0/ D C:

In this caseu.x; t/ D f .�/e�kt D f .x � vt/e�kt :

Conversely, if � < 0, it is more convenient to write

� D � �v:

The characteristic issued from .0; �/ is given by

t � x

vD �; with � > 0:


Sinceg.�/ D z.�/ D Ce�k� ;

we see thatu.x; t/ D g .�/ ek�e�kt D g

�t � x

v

�e�kx=v :

Thus, the solution is

u.x; t/ D8<:f .x � vt/ exp.�kt/ x > vt

g�t � x

v

�exp

��kxv

�x < vt:

Solution 3.3.16. Exactly as in the previous exercise (which we follow), the characteristics of thereduced equation are the lines

x � vt D �; � 2 R:

Setz.t/ D u.vt C �; t/

so thatz0.t/ D vux.vt C �; t/C ut .vt C �; t/ D �ku˛.vt C �; t/ D �kz˛.t/:

Integrating (the equation is an ODE with separable variables) gives (˛ ¤ 1)

z1�˛.t/1 � ˛ D �kt C C;

whence (for some other C )

z.t/ D u.vt C �; t/ D ..˛ � 1/kt C C/1=.1�˛/:If � > 0 the characteristic is issued from .�; 0/, so

f .�/ D z.0/ D C 1=.1�˛/:If � < 0, the characteristic from .0; �/ is given by

t � x

vD �; with � > 0:

We findg.�/ D z.�/ D .k.˛ � 1/� C C/1=.1�˛/ :

Solving for C in the two cases and substituting, we finally obtain

u.x; t/ D

8<:®.˛ � 1/kt C Œf .x � vt/�1�˛¯1=.1�˛/ 0 � vt < x°.˛ � 1/kxv C �

g�t � x

v

��1�˛±1=.1�˛/0 � x < vt:

Solution 3.3.17. a) Substituteu .x; y/ D v .x/C w .y/

into the equation to find

a .x/ .v0/2 C b .y/ .w0/2 D f .x/C g .y/

i.e.a .x/ .v0/2 � f .x/ D �b .y/ .w0/2 C g .y/ :

The two sides of the equation depend on different variables, and hence they are constant:

a .x/ .v0/2 � f .x/ D �b .y/ .w0/2 C g .y/ D ˛:


We then solve separatelya .x/ .v0/2 � f .x/ D ˛

and�b .y/ .w0/2 C g .y/ D ˛;

to obtain four solutions

u .x; y/ D ˙"Z x

x0

sf .s/C ˛

a .s/ds ˙

Z x

x0

sg .s/ � ˛b .s/

ds

#C ˇ (3.31)

with ˛; ˇ constant.

b) In caseu2x C u2y D 1;

then a D b D 1 and we can take f D 0, g D 1. Then

.v0.x//2 D 1 � .w0.y//2 D ˛

so necessarily ˛ � 0. Setting ˛ D sin2 from (3.31) we find the complete integral

u .x; y/ D x sin C y cos C ˇ;

with ; ˇ arbitrary constants.

Solution 3.3.18. The equation is of the type

ut C q .u/x D 0

withq.u/ D u2=2 and q0.u/ D u:

The function q is convex, and g is decreasing, thus we expect that the solution will present a shockwave originating from the point with smallest time coordinate t0, on the envelope of characteristics.The characteristic line from the point .�; 0/ is

x D � C q0.g.�//t D � C g.�/t:

For small times, then, the characteristics carry the datum u D 1 for � < 0, and u D 0 for � > 1.When 0 � � � 1, instead, we have

x D � C .1 � �2/t: (3.32)

To find the envelope we differentiate the previous equation in � , obtaining

1 � 2�t D 0; i.e. � D 1

2t.

As 0 � � � 1, we have t � 1=2. Substituting into (3.32) we obtain

x D t C 1

4t:

The point with minimum time coordinate (t0 D 1=2) is x0 D 1, the starting point of the shockcurve. If x D s .t/ denotes the equation of the shock curve, the Rankine-Hugoniot conditions give

s0.t/ D q.uC.s.t/; t// � q.u�.s.t/; t//uC.s.t/; t/ � u�.s.t/; t/ D 1

2ŒuC.s.t/; t/C u�.s.t/; t/�:


1

enve

lopesh

ock

.1; 1=2/

u D 1 u D 0

x

t

Fig. 3.17 Characteristics for Exercise 3.3.18.

Near the point .1; 1=2/ we have uC D 0. To find u� note that .x; t/, lying on the shock curve,belongs to the characteristic

x D � C .1 � �2/t; with � D 1 � p1C 4t2 � 4xt2t

:

(And not � D .1C p1C 4t2 � 4tx/=.2t/. Why?) On this characteristic,

u�.x; t/ D 1 � �2 D 2tx � 1C p1C 4t2 � 4xt2t2

:

The shock wave, until it meets the characteristic x D t , solves the Cauchy problem

s0.t/ D 2ts � 1C p1C 4t2 � 4st4t2

, s .1=2/ D 1

which needs to be solved numerically1.When the shock wave from .1; 1=2/ intersects the characteristic x D t , u� equals 1 and the

shock wave travels with velocitys0 .t/ D 1=2

and is a half-line (Fig. 3.17).

Solution 3.3.19. Comparing with Problem 3.2.11 on page 176 we have

q.u/ D uC f .u/ D uC u

1C u

and the problem to solve is8<:ux C

�1C 1

.1C u/2

ut D 0 x > 0; t > 0

u.x; 0/ D 0 x > 0

u.0; t/ D g.t/ t > 0;

1 Despite the function is not Lipschitz, it can be proved that the Cauchy problem with initial condi-tion .1; 1=2/ has exactly one solution.


where

g.t/ D´c0 0 � t < 1

0 t > 1:

The characteristics are the lines

t D�1C 1

.1C u/2

x C �:

In particular, the equation of the characteristics are:

• t D 2.x � �/, if based on .�; 0/.• t D 2x C � , if based on .0; �/, � > 1.

• t D�1C 1

.1Cc0/2

�x C � , if based on .0; �/, 0 < � < 1.

Notice that the slope of the characteristics of the last family is smaller than 2. This reveals thepresence of a rarefaction wave centred at .0; 1/ and of a shock curve from the origin. The rarefactionwave is defined by

u.x; t/ D R� x

t � 1�; with R .s/ D �

q0��1

.s/

in the region on the left of the shock curve:

2 <t � 1x

< 1C 1

.1C c0/2:

As �q0��1

.s/ D .s � 1/�1=2 � 1;we have

u.x; t/ D R� x

t � 1�

D� x

t � 1 � 1��1=2 � 1 D

rx

t � x � 1 � 1:Concerning the shock, near the origin

uC D 0 and u� D c0:

The Rankine-Hugoniot condition then gives

dt

dxD q.uC/ � q.u�/

uC � u� D 1C 1

1C c0:

Therefore the equation of the shock curve is

t D�1C 1

1C c0

x:

The shock travels along a straight line until it hits the characteristic from .0; 1/, given by

t D�1C 1

.1C c0/2

x C 1:

The intersection point is

.x0; t0/ D .1C c0/

2

c0;2C 3c0 C c20

c0

!:

After that point, in particular when t > xC1, we always have uC D 0; this time, though, the shockinteracts on the left with the rarefaction wave

u� Dr

x

t � x � 1 � 1:


c0

g

t

1

u D 0

u D c0

u D R�t�1x

�

u D 0

x

t

Fig. 3.18 Initial profile, shock and characteristics for Exercise 3.3.19

The ODE of the shock is then

dt

dxD 1C 1

1C u� D 1Crt � x � 1

x:

The right-hand side suggests the substitution

z.x/ D t .x/ � x � 1z0.x/ D t 0.x/ � 1:

Remembering

t .x0/ D t0;

we obtain that z solves 8<ˆ:dz

dxDrz

x

z

.1C c0/

2

c0

!D 1

c0:

Separating variables and integrating between x0 and x, we find:

pz �

s1

c0D p

x �s.1C c0/

2

c0;

that is to say pt � x � 1 D p

x � pc0

and then

t D 2x � 2pc0x C 1C c0:


To sum up, the shock curve � is

t D8<:�1C 1

1Cc0

�x 0 � x � .1Cc0/

2

c0

2x � 2pc0x C 1C c0 x � .1Cc0/2

c0:

It is not hard to see � belongs to C 1�RC

�.

The solution is zero on the right of � . On the left we have (see Fig. 3.18):

u.x; t/ D

8<ˆ:c0

�1C 1

1Cc0

�x < t �

�1C 1

.1Cc0/2

�x C 1;r

x

t � x � 1 � 1�1C 1

.1Cc0/2

�x C 1 � t � 2x C 1;

0 t � 2x C 1:

4

Waves

4.1 Backgrounds

We shall recall some basic and frequent results in the general theory of second-order waveequations in two variables.

• A global Cauchy problem (n D 1) asks to determine u D u .x; t/ such that´ut t � c2uxx D 0 x 2 R, t > 0

u .x; 0/ D g .x/ ; ut .x; 0/ D h .x/ x 2 R:

The solution is given by d’Alembert’s formula

u .x; t/ D 1

2Œg.x C ct/C g .x � ct/�C 1

2c

Z xCct

x�cth .y/ dy: (4.1)

If g 2 C 2 .R/ and h 2 C 1 .R/, u is C 2 on the half-space R � Œ0;1/: there is no regular-ising effect. The solution has the form

F .x � ct/CG .x C ct/ ;

that is, it is the superposition of two waves moving with speed c in opposite directions.The information provided by the initial data propagates along the characteristics

x ˙ ct D constant.

In particular, the solution at the point .x; t/ depends only on the value of h on the entireinterval Œx � ct; x C ct � and those of g at the endpoints.

• Classification of linear equations of order two. Given an equation

auxx C 2buxy C cuyy C dux C euy C hu D f


216 4 Waves

one calls principal part of the differential operator on the left the bulk of second-orderterms

a .x; y/ @xx C 2b .x; y/ @xy C c .x; y/ @yy :

Given a domain � in the xt -plane, the equation is said to be:

a) Hyperbolic if b2 � ac > 0:b) Parabolic if b2 � ac D 0:

c) Elliptic if b2 � ac < 0.

A hyperbolic equation has two families of real characteristics � .x; y/ D constant, where� solve

a�2x C 2b�x�y C c�2y D 0:

A parabolic equation admits one family of real characteristics, solutions to the same equa-tion. If the equation is elliptic there are no real characteristics.

• Global Cauchy problem (n � 2). In dimension n � 2 the global Cauchy problemreads ´

ut t � c2�u D 0 x 2 Rn; t > 0

u .x; 0/ D g .x/ ; ut .x; 0/ D h .x/ x 2 Rn:

If n D 3, g 2 C 3 �R3� and h 2 C 2 �R3�, then the only C 2 solution on R3 � Œ0;C1/ isprovided by Kirchhoff’s formula

u .x; t / D @

@t

264 1

4�c2t

Z¹jx�� jDctº

g .� / d�

375C 1

4�c2t

Z¹jx�� jDctº

h .� / d�:

In case n D 2, g 2 C 3 �R2� and h 2 C 2 �R2�, the only C 2 solution on R2 � Œ0;C1/ isdetermined by Poisson’s formula

u .x; t / D 1

2�c

8<: @

@t

Z¹jx�yj�ctº

g .y/ dyqc2t2 � jx � yj2

CZ

¹jx�yj�ctº

h .y/ dyqc2t2 � jx � yj2

9>=>; :• Domain dependence. For n D 3, Kirchhoff’s formula shows that u .x; t / depends

only on the values of the data assumed on the sphere®� 2 R3 W jx � � j Dct¯ :

In dimension n D 2, the solution at .x; t / depends on the values of the data assumed onthe disc ®

y 2 R2 W jx � yj � ct¯:


4.2 Solved Problems

� 4:2:1 � 4:2:10 W One-dimensional waves and vibrations.� 4:2:11 � 4:2:16 W Canonical forms. Cauchy and Goursat problems.� 4:2:17 � 4:2:22 W Higher-dimensional problems.

4.2.1 One-dimensional waves and vibrations

Problem 4.2.1 (Pinched string). A guitar string (initially at rest) is pinched at the mid-point and released. Denoting the string density by � and the tension by �; formulate themathematical model and write the solution as superposition of standing waves.

Solution. Let L be the string length and suppose that the string at rest lies along thex-axis between 0 and L. Denote by u.x; t/ the displacement from the rest position of thepoint x at time t , and let a be the initial displacement of x D L=2. The initial configurationof the string, once it is pinched in the middle, is described by the function

g.x/ D a � 2a

L

ˇx � L

2

ˇD´2ax=L 0 � x � L=2

2a.L � x/=L L=2 � x � L:

If a is small with respect to the length and we ignore the string weight, u solves

ut t � c2uxx D 0

where c D p�=� is the travelling speed of waves along the string. The fixed endpoints

impose homogeneous Dirichlet conditions at the boundary of the interval, while the initialrest status means that the initial velocity is zero. All this gives the following model:8<

:ut t � c2uxx D 0 0 < x < L; t > 0

u.0; t/ D u.L; t/ D 0 t � 0

u.x; 0/ D g.x/; ut .x; 0/ D 0 0 � x � L:

In order to write the solution as linear combination of standing waves we separate vari-ables, and seek (non-zero) solutions u.x; t/ D v.x/w.t/. Substituting into the equationand separating the variables gives

1

c2w00.t/w.t/

D v00.x/v.x/

:

The two sides must be equal to a constant � 2 R. In particular, keeping in mind theDirichlet conditions, v will solve the eigenvalue problem´

v00.x/ � �v.x/ D 0 0 < x < L

v.0/ D v.L/ D 0:

218 4 Waves

It is easy to check that this has only the zero solutions if � � 0. For � < 0 the generalintegral is

v.x/ D C1 cos�x

p��

C C2 sin�x

p��

.

Imposing the boundary conditions gives C1 D 0 and

sin�L

p��

D 0:

Thus there are non-trivial solutions if and only if Lp�� D k� , with k D 1; 2 : : : : To

summarise, we have infinitely many solutions

vk.x/ D sin

�k�

Lx

; with �k D �k

2�2

L2; k D 1; 2; : : : :

The corresponding wk then solve

w00k.t/C c2k2�2

L2wk.t/ D 0;

whence

wk.t/ D ak cos

�ck�

Lt

C bk sin

�ck�

Lt

:

The standing wavesuk .x; t/ D wk.t/vk.x/

are the normal modes of vibration of the string, each with frequency k D ck=2L. Wecan then represent the solution as superposition of normal modes:

u.x; t/ DC1XkD1

ak cos

�ck�

Lt

C bk sin

�ck�

Lt

�sin

�k�

Lx

:

Supposing we can differentiate the series, we find

ut .x; t/ DC1XkD1

ck�

L

�ak sin

�ck�

Lt

C bk cos

�ck�

Lt

�sin

�k�

Lx

:

The initial conditions force, for 0 � x � L,

C1XkD1

ak sin

�k�

Lx

D g.x/;

C1XkD1

ck�

Lbk sin

�k�

Lx

D 0:

Consequently bk D 0 for any k, while the ak are found by expanding in Fourier series thefunction g on Œ0; L�. A formal solution is

u.x; t/ DC1XkD1

ak cos

�ck�

Lt

sin

�k�

Lx

(4.2)


with

ak D 2

L

Z L

0

g.x/ sin

�k�

Lx

dx:

As the initial datum is symmetric with respect to x D L=2, and using the elementaryidentity

sin.k� � ˛/ D .�1/kC1 sin˛;

we find

ak D

8<:4

L

Z L=2

0

2a

Lx sin

�k�

Lx

dx k odd

0 k even ;

thus, with k D 2hC 1, we have

a2hC1 D

D 8a

L2

"�Lx

.2hC 1/�cos

�.2hC 1/�

Lx

ˇL=20

C L

.2hC 1/�

Z L=2

0

cos

�.2hC 1/�

Lx

dx

#

D 8a

L2

� L2

2.2hC 1/�cos

�.2hC 1/

�

2

�C L2

.2hC 1/2�2sin�.2hC 1/

�

2

��D 8a

.2hC 1/2�2.�1/h:

Altogether

u.x; t/ D 8a

�2

C1XhD0

.�1/h.2hC 1/2

cos

�c�.2hC 1/

Lt

sin

�.2hC 1/�

Lx

:

This expression says that only modes that are symmetric with respect to x D L=2 can beactivated by the initial profile. Note how the series converges uniformly in Œ0; L� (Weier-strass criterion), but the second derivatives in x and t cannot be computed by swappingderivation and series, because g is not regular, having a non-smooth point at x D L=2.The solution thus found is classical only formally; the correct way to interpret it is in aproper (distributional) weak sense (see Chap. 6).

Problem 4.2.2 (Reflection of waves). Consider the problem8<:ut t � c2uxx D 0 0 < x < L; t > 0

u.x; 0/ D g.x/, ut .x; 0/ D 0 0 � x � L

u .0; t/ D u .L; t/ D 0 t � 0.

a) Define suitably the datum g outside the interval Œ0; L�, and use d’Alembert’s for-mula to represent the solution as superposition of traveling waves.

b) Examine the physical meaning of the result and the relationship with the method ofseparation of variables.

220 4 Waves

Solution. a) The idea is to extend the Cauchy data to the whole R, so that the corre-sponding global Cauchy problem has a solution vanishing on the lines x D 0, x D L. Ifwe restrict this solution to Œ0; L� � ¹t > 0º we will find what we want. We indicate by Qgand Qh the extended data and set out to solve8<

:ut t � c2uxx D 0 x 2 R; t > 0

u.x; 0/ D Qg.x/ x 2 R

ut .x; 0/ D Qh.x/ x 2 R

via d’Alembert’s formula

u.x; t/ D 1

2Œ Qg.x � ct/C Qg.x C ct/�C 1

2c

Z xCct

x�ctQh.s/ ds:

As u automatically satisfies the vibrating string equation (at least formally), we shouldchoose Qg and Qh so to satisfy the initial/boundary conditions. In our case the simplest ex-tension of h is Qh D 0. As for g, we must have8

<ˆ:

u.x; 0/ D Qg.x/ D g.x/ 0 � x � L

u.0; t/ D 1

2Œ Qg.�ct/C Qg.ct/� D 0 t > 0

u.L; t/ D 1

2Œ Qg.L � ct/C Qg.LC ct/� D 0 t > 0:

Therefore, for any s,

Qg.s/ D � Qg.�s/; Qg.LC s/ D � Qg.L � s/: (4.3)

The first condition implies that Qg must be an odd function. Moreover

Qg.s C 2L/ D Qg.LC .LC s// D � Qg.L � .LC s// D � Qg.�s/ D Qg.s/;so Qg is 2L-periodic. Then we may define Qg to be the 2L-periodic function whose restric-tion to Œ�L;L� is given by

Qg.s/ D´g.s/ 0 < s < L

�g.�s/ �L < s < 0:The solution of the initial problem is then

u.x; t/ D 1

2Œ Qg.x � ct/C Qg.x C ct/� for 0 � x � L; t � 0: (4.4)

• Physical meaning of (4.4). Let us divide the strip Œ0; L� � .0;C1/ in regions sepa-rated by characteristic segments as in Fig. 4.1.

We analyse (4.4) starting from points P D .x0; t0/ in region 1. The direct and inversecharacteristic emanating from P meet the x-axis at a D x0 � ct0 and b D x0 C ct0


Fig. 4.1 Regions in the xt -plane for Problem 4.2.2

respectively. Since a; b lie in Œ0; L�,

u.x0; t0/ D 1

2Œ Qg.x0 C ct0/C Qg.x0 � ct0/� D 1

2Œg.x0 C ct0/C g.x0 � ct0/�

and the perturbation at P is the average of a direct and an inverse wave determined by thedatum g (at a and b).

Take now P D .x0; t0/ in region 2 (Fig. 4.2). The point b D x0 C ct0, foot of theinverse characteristic emanating from P , belongs to Œ0; L�, so Qg.x0Cct0/ D g.x0Cct0/.The direct characteristic through P meets the x-axis at a D x0 � ct0 < 0. Buteg is odd,so Qg.a/ D �eg.�a/ D �g .�a/ i.e.

Qg.x0 � ct0/ D �g.�x0 C ct0/:

Fig. 4.2 Reflection of an inverse wave

222 4 Waves

Fig. 4.3 Reflection of a direct wave

This means that the value Qg.x0 � ct0/ comes from an inverse wave determined by thedatum at �a 2 Œ0; L� that reaches the left end at t D �a=c, is reflected, changes sign, andthen maintains its value until t0. Thus u .x0; t0/ is the superposition of an inverse wavefrom b and an inverse wave from �a, reflected at .0;�a=c/ and with opposite sign.

The argument is similar for any P D .x0; t0/ in region 3. The point a D x0�ct0, footof the direct characteristic through P , belongs to Œ0; L�, so Qg.x0 � ct0/ D g.x0 � ct0/.The inverse characteristic from P intersects the x-axis at b D x0 C ct0 > L. The secondrelation in (4.3) implies

Qg.x0 C ct0/ D �eg.2L � x0 � ct0/ D �g.2L � x0 � ct0/:This time Qg.x0 � ct0/ arises from a direct wave determined by the datum at 2L� b, thatreaches the right endpoint at time t D .b � L/ =c, gets reflected, changes sign and staysconstant until t0. Hence u .x0; t0/ is the superposition of a direct wave from a and a directwave from 2L � b, reflected at .0; .b � L/ =c/ and with opposite sign (Fig. 4.3).

Figure 4.4 should clarify the meaning of u in other regions. For instance, at a point inregion 5 the wave is the superposition of a direct wave that is reflected and changes signat the right end, and a direct wave that is reflected twice, changing sign first at the rightend then at the left end.

Fig. 4.4 More reflections


• Relation with the method of separation of variables. Formula (4.4) can be obtainedalso by separating variables. In fact, if we proceed as in Problem 4.2.1, and assuming thatg is expanded in series of sines on Œ0; L�, the solution is

u.x; t/ DC1XkD1

ak cos

�ck�

Lt

sin

�k�

Lx

; with ak D 2

L

Z L

0

g.x/ sin

�k�

Lx

dx:

(4.5)The ak are the Fourier coefficients of the 2L-periodic odd function Qg, which coincideswith g on Œ0; L�. With the notation introduced, this means

C1XkD1

ak sin

�k�

Ls

D Qg.s/; for any s 2 R: (4.6)

On the other hand, known trigonometric addition formulas transform (4.5) into

u.x; t/ DC1XkD1

ak1

2

sin

�k�

Lx C ck�

Lt

C sin

�k�

Lx � ck�

Lt

�D

D 1

2

"C1XkD1

ak sin

�k�

L.x C ct/

CC1XkD1

ak sin

�k�

L.x � ct/

#;

and from (4.6) we deduce (4.4).

Problem 4.2.3 (Equipartition of energy). Let u denote the solution to the followingglobal Cauchy problem for the vibrating string:8<

:�ut t � �uxx D 0 x 2 R; t > 0

u.x; 0/ D g.x/ x 2 R

ut .x; 0/ D h.x/ x 2 R:

Assume g and h are regular functions that vanish outside a compact interval Œa; b�.Prove that after a sufficiently long time T

Ecin.t/ D Epot .t/ for any t � T:

Solution. Let us recall the expressions for the kinetic and potential energy for smalltransverse vibrations of an infinite elastic string:

Ecin D 1

2

ZR�u2t dx; Epot D 1

2

ZR�u2x dx;

where � is the linear density of mass, � the tension (constant along the string), and

224 4 Waves

c D p�=� is the wave speed along the string. D’Alembert’s formula reads

u.x; t/ D F.x C ct/CG.x � ct/;

where

F.s/ D 1

2

g.s/C 1

c

Z s

0

h.v/ dv

�; G.s/ D 1

2

g.s/ � 1

c

Z s

0

h.v/ dv

�:

So we have

ux.x; t/ D F 0.x C ct/CG0.x � ct/; ut .x; t/ D c�F 0.x C ct/ �G0.x � ct/� ;

and then

Epot D 1

2

ZR��F 0.x C ct/CG0.x � ct/�2 dx D

D 1

2

ZR�h�F 0.x C ct/

�2 C �G0.x � ct/�2 C 2F 0.x C ct/G0.x � ct/

idx

Ecin D 1

2

ZRc2�0

�F 0.x C ct/�G0.x � ct/�2 dx D

D 1

2

ZR�h�F 0.x C ct/

�2 C �G0.x � ct/�2 � 2F 0.x C ct/G0.x � ct/

idx:

To prove the claim it suffices, for t large enough, that the product

F 0.x C ct/G0.x � ct/ D 1

4

g0.x C ct/C 1

ch.x C ct/

��g0.x � ct/ � 1

ch.x � ct/

�vanishes identically. We exploit the fact that the data are zero outside Œa; b�: if F 0.x Cct/ ¤ 0 then

a < x C ct < bI (4.7)

in the same way G0.x � ct/ ¤ 0 forces

a < x � ct < b: (4.8)

Therefore F 0.x C ct/G0.x � ct/ ¤ 0 implies, by subtracting (4.7) and (4.8), a � b <

2ct < b � a: Consequently, if

t > T D b � a2c

;

the product is zero and the kinetic energy equals the potential energy.


Problem 4.2.4 (Global Cauchy problem – impulses). Find the formal solution to theproblem 8<

:ut t � c2uxx D 0 x 2 R; t > 0

u.x; 0/ D g.x/ x 2 R

ut .x; 0/ D h.x/ x 2 R

with the following initial data:

a) g.x/ D 1 if jxj < a, g.x/ D 0 if jxj > a; h.x/ D 0.

b) g.x/ D 0; h.x/ D 1 if jxj < a, h.x/ D 0 if jxj > a.

Solution. a) As h is identically zero, d’Alembert’s formula reads

u.x; t/ D 1

2Œg.x C ct/C g.x � ct/� :

We then need to distinguish the regions in the plane where jx˙ct j ? a. The possible casesare described below (see the corresponding regions in Fig. 4.5 starting from the right):

• x > aC ct . A fortiori, then, x > a � ct , and u.x; t/ D 0.• max¹a�ct;�aCctº < x < aCct . Here g.x�ct/ D 1 and g.xCct/ D 0. Thereforeu.x; t/ D 1=2.

• min¹a � ct;�a C ctº < x < max¹a � ct;�a C ctº. Both contributions are positiveand u.x; t/ D 1.

• �aC ct < x < a � ct (so t < a=c). Both contributions are positive and u.x; t/ D 1.• a � ct < x < �aC ct (so t > a=c). Both contributions vanish and u.x; t/ D 0.• �a � ct < x < min¹a � ct;�a C ctº. Now g.x � ct/ D 0 and g.x C ct/ D 1, sou.x; t/ D 1=2.

• x < �a � ct . This implies x < �aC ct and u.x; t/ D 0.

Fig. 4.5 Solution of Problem 4.2.4 a)

226 4 Waves

Fig. 4.6 Solution of Problem 4.2.4 b)

b) This time

u.x; t/ D 1

2c

Z xCct

x�cth.s/ ds;

and arguing case by case as before, we obtain (Fig. 4.6):

• x > aC ct . Then u.x; t/ D 0.

• max¹a � ct;�aC ctº < x < aC ct . We have

u.x; t/ D 1

2c

Z a

x�ctds D a � x C ct

2c:

• �aC ct < x < a � ct (so t < a=c). We have

u.x; t/ D 1

2c

Z xCct

x�ctds D t:

• a � ct < x < �aC ct (hence t > a=c). Here

u.x; t/ D 1

2c

Z a

�ads D a

c:

• �a � ct < x < min¹a � ct;�aC ctº. Then

u.x; t/ D 1

2c

Z xCct

�ads D aC x C ct

2c:

• x < �a � ct . It follows u.x; t/ D 0.


Problem 4.2.5 (Forced vibrations). Consider the problem8<:ut t � uxx D f .x; t/ 0 < x < L; t > 0

u.x; 0/ D ut .x; 0/ D 0 0 � x � L

u.0; t/ D u.L; t/ D 0 t � 0;

(4.9)

with f 2 C 3.Œ0; L� � Œ0;C1// and f .0; t/ D f .L; t/ D fxx.0; t/ D fxx.L; t/ forany t � 0.

a) Solve the problem by the separation of variables. Show that the expression found isthe only classical solution.

b) Study in detail the case

f .x; t/ D g.t/ sin��xL

�:

Solution. a) The Dirichlet conditions are homogeneous so, recalling the computationsof Problem 4.2.1 on page 217, we look for solutions of the form

u.x; t/ DC1XkD1

wk.t/ sin .klx/ ; (4.10)

where l D �=L for short. Note that u satisfies automatically the boundary conditions.Assuming we can differentiate term by term, twice, we find

ut t .x; t/ DC1XkD1

w00k.t/ sin .klx/ ;

uxx.x; t/ D �C1XkD1

k2l2wk.t/ sin .klx/ :

By a formal substitution, then,

C1XkD1

�w00k.t/C k2l2wk.t/

�sin .klx/ D f .x; t/: (4.11)

So now we express f in series of sines in x. For that, we extend f to the strip Œ�L;L� �¹t > 0º as an odd function of x. We find

f .x; t/ DC1XkD1

fk.t/ sin .klx/ where fk.t/ D 2

L

Z L

0

f .�; t/ sin .kl�/ d�: (4.12)

By the Cauchy conditions (applied to each summand of the Fourier series), equation (4.11)is equivalent to the system of infinitely many ODEs´

w00k.t/C k2l2wk.t/ D fk.t/

wk.0/ D 0; w0k.0/ D 0

.k � 1/ : (4.13)

228 4 Waves

The general integral of the homogeneous equation is

wk.t/ D C1 cos .klt/C C2 sin .klt/ :

To find a particular solution of the complete equation we use the method of variation ofconstants, by seeking solution of the type

wk.t/ D C1.t/ cos .klt/C C2.t/ sin .klt/

together withC 01.t/ cos .klt/C C 02.t/ sin .klt/ D 0:

Substituting into equation (4.13) we find that C1, C2 solve´C 01.t/ cos .klt/C C 02.t/ sin .klt/ D 0

�klC 01.t/ sin .klt/C klC 02.t/ cos .klt/ D fk.t/:

With a little patience we obtain the kth solution of system (4.13), given by

wk.t/ D 1

kl

Z t

0

sin.kl�/fk.t � �/ d�:

Substituing back in (4.10) provides

u.x; t/ D 2

�

C1XkD1

1

ksin .klx/

Z t

0

Z L

0

f .�; t � �/ sin .kl�/ sin.kl�/ d�d�: (4.14)

• Analysis of the solution. With the assumptions made on f , after three integrationsby parts of (4.12) we obtain

fk.t/ D 2L2

�3k3

Z L

0

fxxx.�; t/ cos.kl�/ d�

so that for any t 2 Œ0; T �,

jfk.t/j � 2L3

�3k3max

Œ0;L��Œ0;T �jfxxxj; jwk.t/j � 2L4T

�4k4max

Œ0;L��Œ0;T �jfxxxj:

Therefore the Fourier series (4.12) converges uniformly, and (4.14) can be differentiatedtwice term by term. The expression (4.14) is then the unique solution of problem (4.9).

b) The forcing term corresponds to the first fundamental oscillation mode, and the de-pendence of the amplitude upon time is given by g. The regularity assumptions hold. Wehave

fk.t/ D 2

Lg .t/

Z L

0

sin .lx/ sin .klx/ dx D 0 if k � 2

while

f1.t/ D 2

Lg .t/

Z L

0

sin2 .lx/ dx D g .t/ :


Therefore wk.t/ D 0, k � 2, while

w1.t/ D L

�

Z t

0

sin .l�/ g.t � �/ d�:

From (4.10) we find

u.x; t/ D L

�sin��Lx��Z t

0

sin��L��g.t � �/ d�

:

The string reacts to the forcing term by vibrating with the first fundamental mode, whoseamplitude depends upon the convolution integral.

Problem 4.2.6 (Semi-infinite string with fixed end). Consider the problem8<:ut t � c2uxx D 0 x > 0; t > 0

u.x; 0/ D g.x/; ut .x; 0/ D h.x/ x � 0

u.0; t/ D 0 t � 0;

with g; h regular, g.0/ D 0.

a) Extend suitably the initial data to R and use d’Alembert’s formula to write a repre-sentation formula for the solution.

b) Interpret the solution in the case h.x/ D 0 and

g.x/ D´

cos .x � 4/ jx � 4j � �

20 otherwise.

Solution. a) Let us look for Qg and Qh, defined on R and extending g and h to x < 0.The d’Alembert solution

u.x; t/ D 1

2Œ Qg.x C ct/C Qg.x � ct/�C 1

2c

Z xCct

x�ctQh.s/ ds (4.15)

must satisfyu.0; t/ D 0

for any t > 0, and therefore we have the necessary condition

1

2Œ Qg.ct/C Qg.�ct/�C 1

2c

Z ct

�ctQh.s/ ds D 0:

The easiest way to satisfy this condition is to require the Qg- and Qh-summands to vanishseparately, which happens if we extend g and h in an odd way.

Therefore the solution is given by (4.15) with

Qg.s/ D´g.s/ s � 0

�g.�s/ s < 0

230 4 Waves

and

Qh.s/ D´h.s/ s � 0

�h.�s/ s < 0:

b) As h D 0, the solution reduces to

u.x; t/ D 1

2Œ Qg.x C ct/C Qg.x � ct/� :

The initial datum can be understood as the superposition of two sinusoidal waves (withcompact support and amplitude 1=2) that at t D 0 start to travel in opposite directionswith speed c. Since x C ct is always positive, we have

Qg.x C ct/ D g.x C ct/

for any .x; t/, while

x � ct � 0 for any x 2h4 � �

2; 4C �

2

iif t � 8 � �

2c;

x � ct � 0 for any x 2h4 � �

2; 4C �

2

iif t � 8C �

2c:

Therefore we distinguish several intervals of time:

• 0 < t < �2c

. The impulses start as opposite, but continue to interact in a neighbourhoodof the point x D 4.

• �2c< t < 8��

2c. Same as above, but the impulses do not interfere with each other.

• 8��2c

< t < 8C�2c

. The impulse heading left reaches the fixed end and is reflected,turning upside down1 (and interfering with itself).

• t > 8C�2c

. The impulse moving leftwards has turned completely upside down. Thestring profile is given by two impulses of same shape, one positive and one negative,at a distance of 8, travelling towards the right at speed c.

These phases are shown in Fig. 4.7.

Problem 4.2.7 (Forced vibrations of a semi-infinite string). A semi-infinite string isinitially at rest along the axis x � 0, and fixed at x D 0. An external force f D f .t/

sets it in motion.

a) Write the mathematical model governing the vibrations.

b) Solve the problem using the Laplace transform in t , assuming that the transform ofu is bounded as s tends to C1.

1 Special case of Problem 4.2.2 (page 219).


Fig. 4.7 The solution to Problem 4.2.6 .c D 1/ at different values of t

Solution. a) Indicating with u.x; t/ the string profile at time t , the model reads8<:ut t � c2uxx D f .t/ x > 0; t > 0

u.x; 0/ D ut .x; 0/ D 0 x � 0

u.0; t/ D 0 t � 0:

b) Set

U.x; s/ D L.u.x; �//.s/ DZ C1

0

u.x; t/e�stdt , s � 0;

the t -Laplace transform of u. Transforming the equation and using the initial conditionswe find2

�c2Uxx.x; s/C s2U.x; s/ D F.s/; (4.16)

(where F D L.f /), a second order ODE for the function x 7! U.x; s/, with constantcoefficients. The general integral of the homogeneous equation is

A.s/e�sx=c C B.s/esx=c .s � 0/ :

As F.s/ doe not depend on x it is easy to check that

F.s/=s2

2 Recall that L.ut / D sU.x; s/ � u.x; 0/; L.ut t / D s2U.x; s/ � su.x; 0/ � ut .x; 0/:

232 4 Waves

is a particular solution of (4.16). Hence

U.x; s/ D F.s/

s2C A.s/e�sx=c C B.s/esx=c :

But since we need U.x; s/ bounded as s ! C1, necessarily B D 0. At the same timethe homogeneous condition at the boundary point x D 0 implies

0 D U.0; s/ D A.s/C F.s/

s2

whence

A.s/ D �F.s/s2

:

Therefore

U.x; s/ D F.s/ � 1 � e�sx=cs2

:

To transform back we recall that

LŒt �.s/ D 1=s2;

so by the ‘time-shift’ formula (Appendix B) we obtain

LŒ.t � a/H .t � a/�.s/ D e�as

s2

(H is Heaviside’s step function). Setting a D x=c, we finally obtain

u.x; t/ DZ t

0

f .t � �/h� �

�� x

c

�H�� x

c

�id�: (4.17)

Problem 4.2.8 (Vibrations of a hanging chain). In this problem we shall find theequation governing the small (plane) vibrations of a hanging chain of length L. Callu D u.x; t/ the displacement from the horizontal position and � the linear density ofmass (a constant). Let us assume that the chain is completely flexible (that is, no resis-tance to deformations) and that the oscillations are only transverse (the chain moveson a vertical plane).

a) Denote by �.x C �x/ and �.x/ the tensions at points x C �x and x relatively tosome small interval .x; xC�x/ on the chain; these tensions are the forces acting onthat portion of chain from below and above respectively. Argue as for the vibratingstring and show that, up to first order approximation,

j�.x/j D �.x/ D �gx

(where g is the acceleration of gravity).

b) Show that small vibrations are governed by the equation

ut t D g.xuxx C ux/:


τ

τ

Fig. 4.8 Tension vectors for the hanging chain

Solution. a) Indicate by ˛.x/, ˛.x C �x/ the angles between the vertical directionand the tension vector at x; x C�x, respectively (Fig. 4.8). There is no vertical motion,so the resulting force cannot have a vertical component:

�.x C�x/ cos .˛.x C�x// � �g.x C�x/ D �.x/ cos .˛.x// � �gx

i.e.,�.x C�x/ cos .˛.x C�x// � �.x/ cos .˛.x// D �g�x:

Divide by �x and let it tend to 0, so that

d

dxŒ�.x/ cos .˛.x//� D �g

and then�.x/ cos .˛.x// D �gx:

Having assumed the vibrations are small, we may write ˛.x/ 0, and at first-order ap-proximation we may write

�.x/ D �gx:

b) Use Newton’s law for the horizontal component of the force; noting thatsin .˛.x// tan .˛.x// ux , we infer

��xut t D �.x C�x/ sin .˛.x C�x// � �.x/ sin .˛.x// DD �.x C�x/ux.x C�x/� �.x/ux.x/:

Dividing by �x and taking the limit �x �! 0, we obtain

�ut t D d

dxŒ�.x/ux� D � 0.x/ux C �.x/uxxI

234 4 Waves

as �.x/ D �gx (by part a)), u satisfies

ut t D gux C gxuxx D g .ux C xuxx/ :

Problem 4.2.9 (Hanging chain – separation of variables). In relation to the previousproblem solve (by separation of variables)8<

:ut t D g.xuxx C ux/ 0 < x < L; t > 0

u.x; 0/ D f .x/; ut .x; 0/ D h.x/ 0 � x � L

u.L; t/ D 0; ju.0; t/j bounded t � 0:

Solution. Set u.x; t/ D v.x/w.t/ and substitute into the differential equation to find

v.x/w00.t/ D g.xv00.x/C v0.x//w.t/;

whence1

g

w00.t/w.t/

D xv00.x/C v0.x/v.x/

D �:

If � � 0 we get only the zero solution (check this), so we might as well set � D �2.This gives the two problems

w00.t/C 2gw.t/ D 0; w.t/ bounded (4.18)

andxv00.x/C v0.x/C 2v.x/ D 0; v.L/ D 0; v.0/ bounded: (4.19)

The general integral of (4.18) is

w.t/ D A cos .pgt/C B sin .

pgt/:

The equation for v is �xv0.x/

�0 C 2v.x/ D 0; 0 < x < L:

As p.x/ D x vanishes at the origin, we are looking at a singular Sturm-Liouville prob-lem3. A change of variables reduces this equation to a Bessel equation. Define s D 2

px

and V.s/ D v.s2=4/. Then

v0.x/ D 1pxV 0.s/; v00.x/ D 1

xV 00.s/ � 1

2xpxV 0.s/:

3 Appendix A.


By substituting into (4.19) we see that V solves

s2V 00 C sV 0 C 2s2V D 0; V .2pL/ D 0; V .0/ bounded;

which in turn (once simplified) is a parametric Bessel equation4 of order 0. The eigenfunc-tions are

Vj .s/ D J0

�˛j

2pLs

with corresponding eigenvalues j D ˛j =2

pL;

where the ˛j are the infinitely many zeroes of J0, ordered increasingly. The eigenfunctionsform a Hilbert basis for the space L2w.0; 2

pL/, with weight w .x/ D x, and moreover

Z 2pL

0

J0

�˛j

2pLs

J0

�˛k

2pLs

s ds D 0; for j ¤ k;

and Z 2pL

0

J 20

�˛j

2pLs

s ds D 2LJ 21 .˛j /:

Substituting back we obtain

vj .x/ D J0

�˛j

rx

L

; j D 1; 2; : : :

and1

LJ 21 .˛j /

Z L

0

J0

�˛j

rx

L

J0

�˛k

rx

L

dx D ıjk;

so the vj form a Hilbert basis in L2 .0; L/ . The functions

uj .x; t/ D J0

�˛j

rx

L

Aj cos

�rg

L

˛j

2t

C Bj sin

�rg

L

˛j

2t

�are called the normal modes of the vibrating chain. The value j D 1 gives the fundamentalmode, with fundamental frequency

1 D ˛1

4�

rg

L:

The general solution is then

u.x; t/ DC1XjD1

uj .x; t/:

4 Appendix A.

236 4 Waves

The coefficients Aj and Bj appear in the power series expansion of f; h in the basis®vj¯.

From u.x; 0/ D f .x/ we have

Aj D 1

LJ 21 .˛j /

Z L

0

f .x/J0

�˛j

rx

L

dx;

while ut .x; 0/ D h.x/ gives

Bj D 2

˛jJ21 .˛j /

1pgL

Z L

0

h.x/J0

�˛j

rx

L

dx:

Problem 4.2.10 (Sound waves in a pipe). LetP1; P2 be two identical, cylindrical organpipes of length L. Assume that its axis is the segment Œ0; L� along the z-direction. PipeP1 is stopped (closed) at z D 0 and open at z D L, whereas P2 is open at both ends.

Pressing a key makes pressurised air move through the pipes. Which pipe producesthe note of higher pitch (i.e., higher frequency)?

Solution. We choose to describe the movement of air inside the pipe, assumed one-dimensional, by means of the condensation5 s. Recall that if �0 denotes the air density atrest, we define the condensation to be the ratio

s .z; t/ D � .z; t/ � �0�0

which is a solution tost t � c2szz D 0;

where c D pdp=d� and p is the pressure. If u D u.z; t/ is the air velocity, we also have

the relation6

ut D �c2sz : (4.20)

Let us consider P1: at the stopped end s .0; t/ D 0, while at z D L the velocity u is zero,so (4.20) implies sz .L; t/ D 0. To sum up, the condensation for P1 solves:´

st t � c2szz D 0 0 < z < L

s .0; t/ D 0, sz .0; t/ D 0 t > 0

under suitable initial conditions. Let us separate variables and set s .z; t/ D v .z/w .t/.Then

v00 .z/v .z/

D w00 .t/c2w .t/

D ��2

5 Alternatively one could use the velocity potential.6 [18, Chap. 5, Sect. 7].


with � a positive constant. Keeping into account the boundary constraints, v solves theeigenvalue problem

v00 C �2v D 0, v0 .0/ D v .L/ D 0:

There are infinitely many independent eigenfunctions

vk .z/ D cos

�k C 1

2

�z

L

�; k D 0; 1; 2; : : :

For each k we have a wk .t/, solution of w00k.t/C �2kc

2wk.t/ D 0, which we write as:

wk .t/ D cos

�k C 1

2

�ct

LC ˇk

�; k D 0; 1; 2; : : :

The general solution is found by superposing the sk .z; t/ D vk .z/wk .t/:

s .z; t/ D1XkD0

Ak cos

�k C 1

2

�z

L

�cos

�k C 1

2

�ct

LC ˇk

�

in which the coefficients are determined by the initial excitation mode of the air. At anyrate the amplitudes Ak , i.e. the Fourier coefficients of the initial profile, tend to zero ask ! 1, so the pitch of P1 is determined by the fundamental harmonic, corresponding to

s0 .z; t/ D v0 .z/w0 .t/ D A0 cos��z2L

�cos

��ct

2LC ˇ0

with frequency

f0 D c

4L.

Concerning P2, s solves´st t � c2szz D 0 0 < z < L

s .0; t/ D 0, s .0; t/ D 0 t > 0

again with suitable initial conditions. We proceed in the same way and find the solution

s .z; t/ D1XkD1

Bk cos

�k�z

L

cos

�k�ct

LC k

;

whose fundamental harmonic is

s1 .z; t/ D B1 cos��zL

�cos

��ct

LC 1

238 4 Waves

with fundamental frequencyg0 D c

2L.

In conclusion, the open pipe produces a sound of double frequency, essentially becausethe closed end allows for twice as many wavelengths inside the pipe, and therefore halvesthe frequency.

4.2.2 Canonical forms. Cauchy and Goursat problems

Problem 4.2.11 (Characteristics and general solution). Determine the type of the fol-lowing linear equation of order two (in two variables)

2uxx C 6uxy C 4uyy C ux C uy D 0

and compute its characteristics. Reduce it to canonical form and find the general solu-tion.

Solution. As32 � 2 � 4 D 1 > 0

the equation is hyperbolic. The principal part factorises as

2uxx C 6uxy C 4uyy D 2.@x � 2@y/.@x � @y/u:

Using the techniques of Chap. 3 we solve

�x � 2�y D 0

to get the family of (real) characteristics �.x; y/ D 2x � y D constant. Moreover,

x � y D 0

gives the characteristic family .x; y/ D x � y D constant.Another way to proceed would be to look for y D y.x/ and to solve the characteristics

ODE

2

�dy

dx

2� 6dy

dxC 4 D 0;

givingdy

dxD 1 or

dy

dxD 2

so y � x D c1 or y � 2x D c2.

To write the equation in normal form we change coordinates by setting´� D 2x � y� D x � y i.e.

´x D � � �y D � � 2�:


Set now U.�; �/ D u.� � �; � � 2�/, so that u.x; y/ D U.2x � y; x � y/, and then

ux D 2U� C U�; uy D �U� � U�;uxx D 4U�� C4U��CU��; uxy D �2U�� 2U��U��; uyy D U�� C2U��CU��:The equation for U is 4U�� C U� D 0, or equivalently�

U��

D 1

4U� :

Integrating in � first, we obtain U�.�; �/ D e�=4f .�/ with f arbitrary (and regular), andthen, integrating in � , we find

U.�; �/ D e�=4F.�/CG.�/;

for some regular functions F;G. Returning to the original variables, the general solutionreads

u.x; y/ D e.x�y/=4F.2x � y/CG.x � y/:

Problem 4.2.12 (Euler-Tricomi equation). Determine the characteristics of the Tri-comi equation

ut t � tuxx D 0:

Solution. First of all we note that the equation is hyperbolic, so the characteristics arereal, only if t > 0 (it is parabolic for t D 0, a set with empty interior and not characteristicat any point). In the hyperbolic situation, the Tricomi operator factorises as

ut t � tuxx D .@t � pt@x/.@t C p

t@x/u;

and the characteristics are �.x; t/ D constant, .x; t/ D constant, with

� t � pt�x D 0 t C p

t x D 0:

The methods of the previous chapter lead to the general solution of these first-order equa-tions:

�.x; t/ D F�3x C 2t3=2

� .x; t/ D G

�3x � 2t3=2

�with F;G arbitrary. Therefore the characteristic curves have equation

3x ˙ 2t3=2 D constant for t � 0:

Problem 4.2.13 (Cauchy problem). Solve, if possibile, the Cauchy problem8<:uyy � 2uxy C 4ex D 0 .x; y/ 2 R2

u.x; 0/ D '.x/ x 2 R

uy.x; 0/ D .x/ x 2 R:

240 4 Waves

Solution. First let us try to put the equation in normal form in order to write the generalintegral, and then we shall discuss the initial condition. It is easy to see that the equationis hyperbolic and its principal part decomposes

uyy � 2uxy D @y.@y � 2@x/u:

Thus we can compute immediately the two characteristic families

x D constant and x C 2y D constant:

To reduce to normal form we change the coordinates by putting´� D x

� D x C 2yi.e.

8<:x D �

y D �� C �

2:

Set u.x; y/ D U.x; x C 2y/, so

uy D 2U�; uxy D 2U�� C 2U��; uyy D 4U��:

The equation for U is

U�� D e� ;

hence

U� D e� C f .�/ and U.�; �/ D �e� C F.�/CG.�/

with F and G arbitrary. Back in the original variables, the general integral is

u.x; y/ D 2yex C F.x C 2y/CG.x/

where the summand xex has been absorbed into the function G.Now we come to the Cauchy conditions. To begin with, these are given on the straight

line y D 0, which is not characteristic at any of its points, so the problem is well posed(at least locally). Substituting, we have´

'.x/ D u.x; 0/ D F.x/CG.x/

.x/ D uy.x; 0/ D 2ex C 2F 0.x/:

The second relation gives

F.x/ D �ex C 1

2

Z x

c

.s/ ds (c arbitrary),

whence

G.x/ D '.x/C ex � 1

2

Z x

c

.s/ ds


and finally

u.x; y/ D 2yex � exC2y C 1

2

Z xC2y

c

.s/ ds C '.x/C ex � 1

2

Z x

c

.s/ ds D

D �1C 2y � e2y� ex C '.x/C 1

2

Z xC2y

x

.s/ ds:

Problem 4.2.14 (Characteristics and general integral). Determine the characteristicsof

t2ut t C 2tuxt C uxx � ux D 0:

Reduce the equation to canonical form and find the general solution.

Solution. The equation is parabolic, since the principal part decomposes as

t2ut t C 2tuxt C uxx D .t@t C @x/2u:

Thus the unique characteristic family is �.x; t/ D constant, where � solves the first-orderequation

t� t C �x D 0: (4.21)

Using the methods of the previous chapter, we compute

�.x; t/ D g .te�x/ ;

g arbitrary, so �.x; t/ D constant reads

te�x D constant:

Let D .x/ be a smooth function, to be chosen later on, such that 0 > 0. Thus, itsinverse �1 is well defined, and we can set´

� D te�x

� D .x/i.e.

´x D �1.�/t D � expŒ �1.�/�:

Define U.�; �/ D u. �1.�/; � expŒ �1.�/�/, or in other words

u.x; y/ D U.te�x ; .x//:

Thenux D �te�xU� C 0U�; ut D e�xU� ;

uxx D te�xU� C t2e�2xU�� 2 0te�xU�� C . 0/2U�� C 00U�;

uxt D �e�xU� � te�2xU�� C 0e�xU��; ut t D e�2xU�� :

242 4 Waves

Substituting into the original equation, we get

. 0/2U�� C . 00 � 0/U� D 0:

Now, if we pick � D .x/ D ex , the second summand vanishes, and 0 > 0. ThereforeU solves U�� D 0, and then

U.�; �/ D F.�/CG.�/�;

with F and G arbitrary. Returning to the original variables, we finally find

u.x; t/ D F.te�x/CG.te�x/ex :

Problem 4.2.15 (A maximum principle). Let u D u .x; t/ be a function such that

Lu D ut t � uxx � 0 (4.22)

on the characteristic (triangular) domain

T D ¹.x; t/ : x > t , x C t < 1, t > 0º :

Assume that u 2 C 2 �T � and prove that:

a) If ut .x; 0/ � 0, for 0 � x � 1, then

maxT

u D max0�x�1u .x; 0/ :

b) If ut .x; 0/ < 0, for 0 � x � 1, or Lu < 0 on T , then

u .x; t/ < max0�x�1u .x; 0/ , for any .x; t/ 2 T:

Solution. a) Fix a point C in T and consider the characteristic triangle TC of verticesA;B;C , as in Fig. 4.9. We integrate on TC the inequality Lu � 0:Z

TC

.ut t � uxx/ dxdt � 0:

From Green’s formulaZTC

.ut t � uxx/ dxdt DZ@CTC

.�ut dx � ux dt/


where @CTC is the boundary of TC , oriented counter-clockwise. Notice that dt D 0 alongAB , dx D �dt on BC and dx D dt on CA. Therefore:Z

@CTC

.�ut dx � ux dt/ D

D �ZAB

ut dx CZBC

.ut dt C ux dx/ �ZCA

.ut dt C ux dx/

D �ZAB

ut dx CZBC

du �ZCA

du

D �Z xB

xA

ut .x; 0/ dx C .u .C / � u .B// � .u .A/ � u .C//

D �Z xB

xA

ut .x; 0/ dx C 2u .C / � u .B/ � u .A/ :

So we may write7

u .C/ D u .B/C u .A/

2CZ xB

xA

ut .x; 0/ dx CZTC

.ut t � uxx/ dxdt: (4.23)

As ut .x; 0/ � 0 and ut t � uxx � 0, we deduce

u .C/ � u .B/C u .A/

2� max0�x�1u.

But C was chosen arbitrarily in T , so the claim follows.

Fig. 4.9 Characteristic triangles

7 By this method one also recovers d’Alembert’s formula.

244 4 Waves

b) If ut .x; 0/ < 0 for 0 � x � 1 or Lu < 0 on T , then (4.23) implies

u .C/ <u .B/C u .A/

2� max0�x�1u

for C 2 T .

Problem 4.2.16 (Goursat problem). Consider the problem´uxy C �u D 0 x > 0; y > 0

u.x; 0/ D u.0; y/ D 1 x � 0; y � 0(4.24)

where � > 0, on the quadrant R2C D ¹.x; y/ W x � 0; y � 0º. Observe that the dataare defined on the characteristics x D 0 and y D 0, and for this reason the problem iscalled characteristic or Goursat problem.

a) Assume that u 2 C 2.R2C/ and reduce the problem to the integral equation

u.x; y/ D 1 � �Z x

0

Z y

0

u.�; �/ d�d�: (4.25)

b) Show that the recursive sequence´u0.x; y/ D 1

unC1.x; y/ D 1 � � R x0

R y0un.�; �/ d�d�

(4.26)

is uniformly convergent on compact sets in R2C.

c) Setu1 D lim

n!C1un:

Show that u1 is a solution (C 2 on the quadrant) of (4.25), and also that

u1.x; y/ D J0

�2p�xy

�;

where J0 is Bessel’s function of order 0.

d) Prove that the solution is unique.

Solution. a) We will show that the Goursat problem (4.24) and the integral equation(4.25) are equivalent. Let u be a C 2 solution to the problem (4.24). Integrating in x be-tween 0 and x gives

uy.x; y/ � uy.0; y/ D ��Z x

0

u.�; y/ d�:


Integrating once more, this time from 0 to y in the other variable, gives

u.x; y/ � u.0; y/ � u.x; 0/C u.0; 0/ D ��Z x

0

Z y

0

u.�; �/ d�d�:

As u.x; 0/ D u.0; y/ D 1, we obtain (4.25). Vice versa, if u 2 C�R2C

�satisfies (4.25),

a direct computation shows that u.x; 0/ D u.0; y/ D 1, and by the fundamental theoremof calculus u 2 C 2 �R2C�. By differentiating (4.25) twice, we prove the claim.

b) Let us compute the first few terms of the sequence explicitly:

u1.x; y/ D 1 � �Z x

0

Z y

0

d�d� D 1 � �xy

u2.x; y/ D 1 � �Z x

0

Z y

0

.1 � ��/ d�d� D 1 � �xy C �2x2y2

4D u1.x; y/C �2

x2y2

.2Š/2

u3.x; y/ D 1 � �Z x

0

Z y

0

�1 � �xy C �2

x2y2

4

d�d� D u2.x; y/ � �3 x

3y3

.3Š/2:

Using an induction argument, we have

unC1.x; y/ D un.x; y/C .�1/nC1�nC1 xnC1ynC1

..nC 1/Š/2

so

u1.x; y/ DC1XnD0

.��/n.nŠ/2

xnyn:

This expression is a power seriesPanx

nyn with convergence radius

limn!C1

ˇan

anC1

ˇD limn!C1

1

�.nC 1/2 D C1 .� > 0/ :

The uniform convergence theorem for power series guarantees uniform convergence onany set of the type ¹.x; y/ W 0 � xy � M/º, and therefore on any compact subset of R2C.In particular, u1 2 C �R2C�.

c) Exploiting the uniform convergence on compact subsets in R2C once more, we may,for any given .x; y/, pass to the limit in the recursive relation (4.26), obtaining

u1.x; y/ D 1 � �Z x

0

Z y

0

u1.�; �/ d�d�.

By a), u1 2 C 2�R2C

�and it solves the Goursat problem. To finish, since by definition

(see Appendix A)

J0.z/ DC1XnD0

.�1/n..n/Š/2

z2n

246 4 Waves

we may write

u1.x; y/ DC1XnD0

.�1/n..n/Š/2

.p�xy/2n D J0.

p�xy/:

d) Suppose u1 and u2 are Goursat solutions (for the same �). Thenw D u1�u2 solvesthe integral equation

w.x; y/ D ��Z x

0

Z y

0

w.�; �/ d�d�:

Therefore we can write

w.x; y/ D ��Z x

0

Z y

0

��

Z s

0

Z t

0

w.�; �/ d�d�

�dsdt: (4.27)

Integrating by parts we haveZ y

0

Z t

0

w.�; �/ d�

�dt D

t

Z t

0

w.�; �/ d�

�y0

�Z y

0

tw.�; t/ dt

DZ y

0

.y � t /w.�; t/ dt

and, similarly, Z x

0

Z s

0

w.�; t/ d�

�dt D

Z x

0

.x � s/w.s; t/ ds:Substituting into (4.27), we get

w.x; y/ D �2Z x

0

Z y

0

.x � s/.y � t /w.s; t/ dsdt:

By iterating the argument we obtain

w.x; y/ D ��3Z x

0

Z y

0

.x � s/2.y � t /22 � 2 w.s; t/ dsdt

and, by induction,

w.x; y/ D .��/nZ x

0

Z y

0

.x � s/n.y � t /n.nŠ/2

w.s; t/ dsdt

for any n � 1. Now let us fix a compact set

K D ¹.x; y/ 2 R2C W 0 � x � k; 0 � y � kºand set M D maxK jwj. As .x � s/n.y � t /n � k2n, we have

jw.x; y/j � j�jnk2nC2M.nŠ/2

! 0 for n ! C1:

Therefore w � 0 on any compact set in R2C, and therefore everywhere in R2C.


4.2.3 Higher-dimensional problems

Problem 4.2.17 (Circular membrane). Consider a membrane

B1 D ¹.x; y/ 2 R2 W x2 C y2 � 1ºat rest with fixed boundary. If the weight is negligible and there are no external forcesacting on it, the membrane vibrates according to the following problem (written in polarcoordinates, given the round symmetry):8<:ut t � c2

�urr C 1

rur C 1

r2u��

D 0 0 < r < 1; 0 � � � 2�; t > 0

u.r; �; 0/ D g.r; �/; ut .r; � ; 0/ D h.r; �/ 0 < r < 1; 0 � � � 2�

u.1; �; t/ D 0 0 � � � 2�; t > 0:

Using separation of variables, find a representation formula for the solution u in thecase h D 0; g D g.r/.

Solution. To start with, we remark that the solution (unique under reasonable assump-tions on the data) is radially symmetric: indeed, if not, we might construct different solu-tions to the same problem simply by rotating a given solution (all data have a sphericalsymmetry!), contradicting uniqueness. Therefore u D u.r; t/ solves8<

:ut t � c2

�urr C 1

rur

D 0 0 < r < 1; t > 0

u.r; 0/ D g.r/; ut .r; 0/ D 0 0 < r < 1

u.1; t/ D 0 t > 0:

We seek solution of the type u.r; t/ D v.r/w.t/ and, for the time being, such thatw0 .0/ D 0 and

v.0/ finite, v.1/ D 0:

Substituting in the equation, we get

v.r/w00.t/ � c2�v00.r/C 1

rv0.r/

w.t/ D 0.

Now we separate the variables and deduce

1

c2w00.t/w.t/

D rv00.r/C v0.r/rv.r/

D ;

with constant. In particular v solves´.rv0/0 � rv D 0

v.0/ bounded v.1/ D 0:

248 4 Waves

If � 0 the only solution is v � 0. In fact, by multiplying the ODE by v and integrat-ing by parts on the interval .0; 1/, then taking into account the boundary conditions, weeventually find

0 DZ 1

0

.rv0/0v dr � Z 1

0

rv2 dr D �Z 1

0

Œ.v0/2 C v2�r dr:

If � 0, the last integrand is non-negative and therefore v � 0 on .0; 1/.If, conversely, D ��2, the ODE is

v00 C v0

rC �2v D 0;

that is a parametric Bessel equation of order 0 (Appendix A). The eigenfunctions, forminga basis of the Hilbert space L2w.0; 1/ with weight w .r/ D r , are

un.r/ D J0 .�nr/ ;

where J0 is the zero-order Bessel function and �1; �2; : : : are its zeroes. On the other handthe equation for w is

w00.t/C c2�2nw.t/ D 0Ithis equation, together with the initial conditionw0 .0/ D 0 is solved by the one-parameterfamily

wn.t/ D an cos .c�nt /:

Thus we have found the fundamental modes of vibration of the membrane:

un .r; t/ D an cos .c�nt /J0 .�nr/ .

The solution of the original problem is then obtained by superposing those modes, that is

u.r; t/ DC1XnD1

an cos.c�nt /J0 .�nr/ :

To satisfy the initial condition u .r; 0/ D g .r/, as well, the coefficients an must satisfythe equation

C1XnD1

anJ0 .�nr/ D g.r/

whence

an D 2

J 21 .�n/

Z 1

0

sg.s/J0.�ns/ ds

in terms of the Bessel function of order 1, J1.


Problem 4.2.18 (Dissipative term). Consider the problem´ut t .x; t /C kut .x; t / D c2�u.x; t / x 2 R2; t > 0

u.x; 0/ D 0; ut .x; 0/ D g.x/ x 2 R2:

a) Determine ˛ 2 R so thatv.x; t / D e˛tu.x; t /

solves an equation without first-order term (but with zero-order one) on R2�¹t > 0º.

b) Find ˇ 2 R so that

w.x1; x2; x3; t / D w.x; x3; t / D eˇx3v.x; t /

solves an equation with second-order terms only, on R3 � ¹t > 0º.

c) Determine the solution u of the original problem.

Solution. a) Let us writeu.x; t / D e�˛tv.x; t /;

so that

ut D �˛e�˛tv C e�˛tvt ; ut t D ˛2e�˛tv � 2˛e�˛tvt C e�˛tvt t ; �u D e�˛t�v:

Substituting into the equation, we find

e�˛t�vt t C .k � 2˛/vt C .˛2 � k˛/v� D c2e�˛t�v:

Consequently, if we put ˛ D k=2, then

v.x; t / D ekt=2u.x; t / solves vt t � k2

4v D c2�v:

b) We writew.x; t3; t / D e�ˇx3w.x; t /

and then

wt t D eˇx3vt t ; wx1x1C wx2x2

/ D eˇx3�vx1x1

C vx2x2

�:

Substituting gives

e�ˇx3wt t D c2e�ˇx3

�wx1x1

C wx2x2C k2

4c2w

:

Now notice that wx3x3D ˇ2w. If we choose ˇ D k=.2c/, we obtain that

w.x; x3; t / D ekx3=.2c/v.x; t / solves wt t D c2�3w;

where the operator �3 is the Laplacian in three dimensions.

250 4 Waves

c) Sincew.x; x3; t / D ek.x3Ctc/=2cu.x; t /;

we can find the initial conditions for w. Since

w.x; x3; 0/ D ekx3=2cu.x; 0/ D 0

wt .x; x3; 0/ D k

2ekx3=2cu.x; 0/C ekx3=2cu.x; 0/ D ekx3=2cg.x/;

from Kirchhoff’s formula we obtain

v.x; t / D w.x; 0; t/ D 1

4�c2t

Z@Bct .x;0/

ek�3=2cg.�1; �2/ d�;

where d� is the surface element of @Bct .x; 0/. We can attain a more significant formulaby reducing the integral on the sphere @Bct .x; 0/ to a double integral on the disc

Kct .x/ D°

y D .y1; y2/ W jy � xj2 < c2t2±:

The equations on the upper and lower hemispheres are

y3 D ˙pc2t2 � r2

�r2 D jy � xj2

�where

d� Ds1C r2

c2t2 � r2 dy1dy2 D ctpc2t2 � r2 dy1dy2:

Hence

v.x; t / D 1

4�c

ZKct .x/

hekpc2t2�r2=2c C e�k

pc2t2�r2=2c

i g.y1; y2/pc2t2 � r2 dy1dy2

D 1

2�c

ZKct .x/

cosh

k

pc2t2 � r22c

!g.y1; y2/pc2t2 � r2 dy1dy2:

and finallyu.x; t / D e�kt=2v.x; t /:

Problem 4.2.19 (Fundamental solution in dimension 2). Write the fundamental solu-tion to the two-dimensional wave equation, with and without dissipation.

Solution. In order to determine the fundamental solution we remind that the Cauchyproblem ´

ut t .x; t / � c2�u.x; t / D 0 x 2 R2; t > 0

u.x; 0/ D 0; ut .x; 0/ D h.x/ x 2 R2


is solved by the Poisson integral

u.x; t / D 1

2�c

ZBct .x/

h.y/pc2t2 � jx � yj2 dy D

D 1

2�c

ZR2

H�c2t2 � jx � yj2�pc2t2 � jx � yj2 h.y/dy

where H is Heaviside’s function. The expression for the solution K.x; z; t / is found bychoosing h to be the Dirac distribution in the variable z. We find

K.x; z; t / D 1

2�c

H�c2t2 � jx � zj2�pc2t2 � jx � zj2

D8<:

1

2�c

1pc2t2 � jx � zj2 for jx � zj2 < c2t2

0 for jx � zj2 > c2t2:This solution describes the perturbation created by an initial unit impulse, correspondingto the data

u .x; 0/ D 0 and ut .x; 0/ D ı .x � z/ :

Note that at a given time t every point inside the disc®.x; t / W jx � zj2 < c2t2¯

is affected by the oscillation. In the case the equation has a dissipative term kut as well,from the previous problem we recover

Kdiss.x; z; t / D e�kt=2

2�ccosh

kpc2t2 � jx � zj2

2c

!H�c2t2 � jx � zj2�pc2t2 � jx � zj2

as the fundamental solution.

Problem 4.2.20 (An application of Kirchhoff’s formula). Let us consider´ut t � c2�u D 0 x 2 R3; t > 0

u.x; 0/ D g.x/; ut .x; 0/ D h.x/ x 2 R3;

where g and h are respectively C 3 and C 2 in R3 � Œ0;C1/. Suppose that g, h havesupport contained in the ball B.0/. Describe the support of the solution u at each timeinstant t .

Solution. By Kirchhoff’s formula the problem is solved by

u.x; t / D @

@t

1

4�c2t

Z@Bct .x/

g.� / d�

�C 1

4�c2t

Z@Bct .x/

h.� / d�

252 4 Waves

The support condition on the initial data may be written as

jyj � � H) g.y/ D 0; h.y/ D 0:

From the solution expression we infer that u vanishes at the point .x; t / if

@Bct .x/ \ B .0/ D ;:This happens when either

jxj C � < ct or jxj � � > ct:The solution support at time t is therefore contained in the ball jxj � ct C � for t � �=c,and in the spherical shell

¹x 2 R3 W ct � � � jxj � ct C �º for t > �=c;

which grows in any radial direction at speed c.

Problem 4.2.21 (Focalised discontinuity). Determine the solution to´ut t � c2�u D 0 x 2 R3; t > 0

u.x; 0/ D 0; ut .x; 0/ D h.jxj/ x 2 R3;

where r D jxj and

h.r/ D´1 0 � r � 1

0 r > 1:

Solution. Note that initially the solution is continuous at x D 0. If we imagine h ex-tended to the entire real axis in an even way, w, being radial, has the form

w .r; t/ D F .r C ct/

rC G .r � ct/

r:

Now, w .r; 0/ D 0 implies F .r/ D �G .r/. The second initial condition then givesG 0 .r/ D �rh .r/ =2c, whence

G .r/ D � 1

2c

Z r

0

sh .s/ ds CG .0/

and then

w .r; t/ D F .r C ct/

rC G .r � ct/

rD 1

2cr

Z rCct

r�ctsh .s/ ds:

To computew we must consider various regions in the rt -plane (Fig. 4.10). On the triangle

¹.r; t/ W �1 < r � ct < r C ct < 1º


Fig. 4.10 Solution to Problem 4.2.21

(where, in particular, t < 1=c) we have h D 1, soZ rCct

r�ctsh .s/ ds D

s2

2

�rCctr�ct

D 2crt:

If r ¤ 0 then w .r; t/ D t , which holds at r D 0, too, by continuity. In particular

w .0; t/ ! 1

cif t !

�1

c

�:

If t > 1=c, on the region

¹.r; t/ W r � ct < �1, r C ct > 1ºwe have Z rCct

r�ctsh .s/ ds D

Z 1

�1s ds D 0:

Therefore w .r; t/ D 0 and w is discontinuous at .0; 1=c/: this fact is a consequence ofinitial discontinuity of wt on the sphere r D 1 focalized at r D 0 at time t D 1=c.

On the sector ¹.r; t/ W r � ct > 1º and ¹.r; t/ W r C ct < �1º h is null, sow .r; t/ D 0.Finally, on the strip

¹.r; t/ W �1 < r � ct < 1, r C ct > 1ºwe have

w .r; t/ D 1

2cr

Z 1

r�ctsh .s/ ds D 1

2cr

Z 1

r�cts ds D 1 � .r � ct/2 ;

4cr

while on the strip¹.r; t/ W r � ct < �1, � 1 < r C ct < 1º

254 4 Waves

we have (Fig. 4.10)

w .r; t/ D 1

2cr

Z rCct

�1sh .s/ ds D 1

2cr

Z rCct

�1s ds D .r C ct/2 � 1

4cr:

Problem 4.2.22 (Reflection of acoustic waves). A plane harmonic of amplitude A andfrequency !I , moving at speed c on the plane xz, hits the plane z D 0 and is reflected,see Fig. 4.11. Determine the (velocity) potentials of the incoming and the reflectedwaves (use complex variables).

Solution. In general, a plane harmonic has a potential of the form

� .x; z; t/ D A exp ¹i .k1x C k2y C k3z � !t/º .! > 0/ :

The wave moves in the direction of the vector k D .k1; k2; k3/, called wavenumber vec-tor, with velocity != jkj. In the present case kI D .k1; 0; k3/, and the potential of thewave is

�I .x; z; t/ D A exp ¹i .k1x C k3z � !I t /º :Let c denote the wave’s speed. Then

jkI j D !I

c;

so, looking at Fig. 4.11, we obtain

k1 D �!Ic

sin �I , k2 D !I

ccos �I

and then�I .x; z; t/ D A exp

°i!I

��zc

cos �I C x

csin �I � t

�±.

To determine the reflected potential �R .x; z; t/ we look for a solution to

� t t � c2 .�xx C �zz/ D 0

Fig. 4.11 Reflection of acoustic waves


of the form� .x; z; t/ D �I .x; z; t/C �R .x; z; t/

together with the condition

@�

@zD 0; i.e.

@�R@z

D �@�I@z

; on the plane z D 0;

due to the reflection. Then

�R .x; z; t/ D B exp°i!R

�zc

cos �R C x

csin �R � t

�±works. The reflection condition imposes

�A cos �I exp°i!I

�zc

cos �I � t�±

C B cos �R exp°i!R

�zc

cos �R � t�±

D 0:

This equation must hold for z 2 R and t � 0, so

A D B , �I D �R, !I D !R.

The reflected wave has the same amplitude or frequency, and the angle of incidence equalsthe angle of reflection.


4.3.1. Consider the problem8<:ut t � c2uxx D 0 0 < x < L; t > 0

u.x; 0/ D g.x/; ut .x; 0/ D h.x/ 0 � x � L

u.0; t/ D 0; u.L; t/ D B t > 0:

a) Determine the stationary solution u0.b) Find a formal expression for u.c) Establish whether u tends to u0 as t ! 1.

4.3.2. Consider the problem8<:ut t � c2uxx D 0 0 < x < L; t > 0

u.x; 0/ D g.x/, ut .x; 0/ D 0 0 � x � L

ux .0; t/ D ux .L; t/ D 0 t � 0.

a) Define suitably g outside the interval Œ0; L�, and use d’Alembert’s formula to represent the solu-tion as superposition of travelling waves.

b) Explain the physical meaning of the result.

256 4 Waves

4.3.3. Solve the (Cauchy-Dirichlet) problem8<:ut t � c2uxx D f .x; t/ 0 < x < L; t > 0

u.x; 0/ D ut .x; 0/ D 0 0 � x � L

u.0; t/ D u.L; t/ D 0 t � 0;

formally, in the following cases:

a) f .x; t/ D e�t sin��xL

�.

b) f .x; t/ D xe�t .

4.3.4. Find the formal solution to the (Cauchy-Neumann) problem8<:ut t � c2uxx D f .x; t/ 0 < x < L; t > 0

u.x; 0/ D ut .x; 0/ D 0 0 � x � L

ux.0; t/ D ux.L; t/ D 0 t � 0

wheref .x; t/ D e�t cos

��x2L

�:

4.3.5. Let u D u .x; t/ solve

Lu D ut t � uxx � h2u � 0, h > 0 (4.28)

on the characteristic triangle

T D ¹.x; t/ : x > t , x C t < 1, t > 0º :

Prove that if u 2 C 2 �T �, and

u .x; 0/ � M < 0, ut .x; 0/ � 0

for 0 � x � 1, thenu < 0 on T .

4.3.6. Consider the free transverse vibrations of a semi-infinite string. Assume that the initialposition and velocity are known, and that the end slides along a transverse straight guide perpendic-ular to the strip. Find a mathematical model describing the motion and write the expression of thesolution (see also Problem 4.2.6 on page 229).

4.3.7. Solve Problem 4.2.7 (page 230) with

f .t/ D �g

(that is, when the oscillations are forced by gravity).

4.3.8. Solve the previous problem in case the string is not initially at rest, but rather

ut .x; 0/ D 1:


4.3.9. Analyse numerically Problems 4.2.8, 4.2.9 in case L D 1 m, g D 9:8 m/s2, h D 0 and

f .x/ D

8<:x

1000 � x � 1

2.1 � x/100

1

2� x � 1.

In particular:

a) Determine the first three frequencies of the vibration.

b) Compute the amplitudes of the first three modes.

c) Plot, for some values of t , the approximate sum of the first three modes.

4.3.10. Determine the type of the second-order equation

uyy � 2uxy C 2ux � uy D 4ex

and compute the characteristics. Determine, if possible, the general solution.

4.3.11. Determine the type of the second-order equation

uxy C yuy � u D 0;

and compute the characteristics. Determine, if possible, the general integral.

4.3.12. Let � denote the curve x D ' .y/ with ' 2 C 2 .0;1/, continuous at y D 0, ' .0/ D 0,and such that '0 .x/ � 0 for x > 0. Consider the problem:8<

:uxy D F .x; y/ .x; y/ 2 Q D ¹x > ' .y/ ; y > 0ºu .x; 0/ D g .x/ x > 0

u .' .y/ ; y/ D h .y/ y > 0;

where F 2 C.Q/, h; g 2 C 2 .0;1/ continuous at y D 0.

a) Integrate on the rectangle R of Fig. 4.12 and find a representation for u in terms of F , g, h.

b) Determine conditions on g, h that make u a solution belonging to C 2.Q/ \ C.Q/.c) Discuss uniqueness.

4.3.13. Consider the domain

S D ¹.x; t/ W �t < x < t; t > 0º :

Solve 8<:ut t � uxx D 0 on S

u .x; x/ D g .x/ , u .x;�x/ D h .x/ x � 0

g .0/ D h .0/ :

258 4 Waves

Fig. 4.12 Fig. for Exercise 4.3.12

4.3.14. Establish whether the following problems are well or ill posed.

a) Characteristic Cauchy problem8:8<:ut t � uxx D 0 on the half-space x > t

u .x; x/ D f .x/ x 2 R

@�u .x; x/ D g .x/ x 2 R;

where

D 1p2.1;�1/

(unit normal to the characteristic x D t).b) Boundary values on the quadrant:8<

:ut t � uxx D 0 x > 0, t > 0

u .x; 0/ D f .x/ x � 0

u .0; t/ D g .t/ t � 0:

4.3.15. Consider the following model for a string with internal friction:8<:�ut t � �uxx D uxxt 0 < x < L, t > 0

u .x; 0/ D f .x/ , ut .x; 0/ D g .x/ 0 < x < L

u .0; t/ D u .L; t/ D 0 t � 0,

(4.29)

where �; �; are positive constants.

a) Define

E .t/ D 1

2

Z L

0

h�u2t C �u2x

idx:

8 The Cauchy data are assigned on a characteristic curve.


Assuming u sufficiently regular, prove that

E0 .t/ � 0

and interpret the result.

b) Use part a) to prove a uniqueness theorem for problem (4.29).

c) Using separation of variables establish whether u .x; t/ ! 0 as t ! C1.

4.3.16. Given � > 0, consider the problem

´ut t .x; y; t/ D uxx.x; y; t/C uyy.x; y; t/C �u.x; y; t/ .x; y/ 2 R2; t > 0

u.x; y; 0/ D f .x; y/; ut .x; y; 0/ D g.x; y/ .x; y/ 2 R2:

a) Determine k so that the function

v.x; y; z; t/ D ekzu.x; y; t/

solves

vt t D vxx C vyy C vzz :

b) Use part a) to represent the solution.

4.3.17. (Plane acoustic waveguide) Consider the region in space bounded by the planes x D 0,x D d (acoustic waveguide). Which are the plane harmonics, polarised on the plane xz, that canpropagate in the region? Analyse their properties.

4.3.18. (Sound waves in a pipe) Study acoustic waves of speed c and angular frequency ! thattravel in a semi-infinite cylindrical pipe of radius a. Decide, in particular, which modes can propa-gate without damping effects as a varies.

4.3.1 Solutions

Solution 4.3.1. a) The stationary solution is the time-independent function that solves the prob-lem with the Dirichlet conditions only. Thus u0 D u0.x/ solves´

u000 D 0 0 < x < L

u0.0/ D 0; u0.L/ D B;

i.e. u0.x/ D BLx:

b) To apply variable separation we need homogeneous Dirichlet conditions. To this purpose setU.x; t/ D u.x; t/ � u0.x/: Then U solves8<

:Ut t � c2Uxx D 0 0 < x < L; t > 0

U.x; 0/ D g.x/ � B

Lx; Ut .x; 0/ D h.x/ 0 � x � L

U.0; t/ D U.L; t/ D 0 t > 0:

260 4 Waves

We want to find solutions U.x; t/ D v.x/w.t/. Following Problem 4.2.1 (page 217) we can write

U.x; t/ DC1XkD1

ak cos

�ck�

Lt

C bk sin

�ck�

Lt

�sin

�k�

Lx

:

For t D 0 and 0 � x � L we have

C1XkD1

ak sin

�k�

Lx

D g.x/ � B

Lx;

C1XkD1

ck�

Lbk sin

�k�

Lx

D h.x/;

so the Fourier coefficients are determined by

ak D 2

L

Z L

0

�g.x/ � B

Lx

sin

�k�

Lx

dx; bk D 2

ck�

Z L

0h.x/ sin

�k�

Lx

dx:

c) The function u found in part b) is .2L=c/-periodic in time, so it admits limit for t ! 1 ifand only if it is constant in time, i.e. if and only if g.x/ D u0.x/ and h.x/ D 0. In all other casesthe solution oscillates indefinitely and does not converge to the stationary solution.

Solution 4.3.2. Let us proceed as in Problem 4.2.2 on page 219, by extending the Cauchy data toR, so that the corresponding global Cauchy problem has a solution with vanishing spatial derivativealong x D 0, x D L. By restricting this solution to the strip Œ0; L� � ¹t > 0º we get the requiredsolution. Indicate by Qg and Qh the extended data. The global solution is given by d’Alembert’s formula

u.x; t/ D 1

2Œ Qg.x � ct/C Qg.x C ct/�C 1

2c

Z xCct

x�ctQh.s/ ds:

We must choose Qg; Qh to satisfy both the initial and the boundary Neumann conditions. The simplestpossibility for h is to set Qh D 0. As for g we must have8

<ˆ:u.x; 0/ D Qg.x/ D g.x/ 0 � x � L

ux.0; t/ D 1

2

� Qg0.�ct/C Qg0.ct/� D 0 t > 0

u.L; t/ D 1

2

� Qg0.L � ct/C Qg0.LC ct/� D 0 t > 0:

Therefore, for any s,

Qg0.s/ D � Qg0.�s/; Qg0.LC s/ D � Qg0.L � s/: (4.30)

The first condition says that Qg0 must be odd, henceeg is even. By the second condition

Qg0.s C 2L/ D Qg0.LC .LC s// D � Qg0.L � .LC s// D � Qg0.�s/ D Qg0.s/;

so Qg0 is 2L-periodic. But Qg0 is odd, so its mean value on the period interval is zero: consequentlyalsoeg is 2L-periodic, and

Qg.s/ D´g.s/ 0 < s < L;

�g.�s/ �L < s < 0:


The solution to the original problem is then

u.x; t/ D 1

2Œ Qg.x � ct/C Qg.x C ct/� for 0 � x � L; t � 0: (4.31)

The interpretation of (4.31) follows from arguments similar to those of Problem 4.2.2 (page219). The only difference is that, here, the reflection at the ends does not change the overall sign,because the prolongation of g is even.

Solution 4.3.3. Formula (4.14) gives

a) u.x; t/ D L2

L2 C c2�2

e�t � cos

�c�Lt�

C L

c�sin�c�Lt��

sin��Lx�:

b) u.x; t/ D 2L

�

C1XkD1

.�1/kC1L2k.L2 C c2�2k2/

e�t � cos

�ck�

Lt

C L

ck�sin

�ck�

Lt

�sin

�k�

Lx

:

Solution 4.3.4. Following the solution of Problem 4.2.5 on page 227:

u.x; t/ D 4L2

4L2 C c2�2

e�t � cos

� c�2Lt�

C 2L

c�sin� c�2Lt��

cos� �2Lx�:

Solution 4.3.5. Let C be a point in T and consider the characteristic triangle TC determined byA;B;C , see Fig. 4.9 (page 243). We proceed as in Problem 4.2.15 (page 242) and integrate on TCthe inequality Lu � 0. We find

u .C / � u .B/C u .A/

2CZ 1

0ut .x; 0/ dx C 1

2

ZTC

h2u dxdt: (4.32)

As u .x; 0/ � M < 0 for 0 � x � 1, by continuity u .x; t/ stays negative, at least for t > 0 smallenough. If, by contradiction, u did not remain negative everywhere on T , there would exist a pointC D .x0; t0/ at which it vanished for the first time, i.e.

u .x0; t0/ D 0 and u .x; t/ < 0 for t < t0.

By (4.32), since u .A/ � M , u .B/ � M and u < 0 on TC , we would have

0 D u .C / � M < 0;

a contradiction. Therefore u < 0 on the whole T .

Solution 4.3.6. Call u.x; t/ the string shape at time t . The model is then

8<:ut t � c2uxx D 0 x > 0; t > 0

u.x; 0/ D g.x/; ut .x; 0/ D h.x/ x � 0

ux.0; t/ D 0 t � 0:

262 4 Waves

We shall mimic Problem 4.2.6 (page 229) and seek functions Qg and Qh defined on R, coinciding withg and h on x > 0, and also such that the function

u.x; t/ D 1

2Œ Qg.x C ct/C Qg.x � ct/�C 1

2c

Z xCct

x�ctQh.s/ ds

(from d’Alembert’s formula) satisfies ux.0; t/ D 0 for any t > 0. We compute

ux.x; t/ D 1

2

� Qg0.x C ct/C Qg0.x � ct/�C 1

2c

h Qh.x C ct/ � Qh.x � ct/i;

and the boundary condition reads

1

2

� Qg0.ct/C Qg0.�ct/�C 1

2c

h Qh.ct/ � Qh.�ct/i

D 0:

It suffices to extend h as an even function and g0 as an odd one (i.e g to become even).

Solution 4.3.7. From (4.17) we get

u.x; t/ D �gZ t

0

h� �

�� x

c

�H�� x

c

�id� D �1

2gt2 C g

Z t

0

�� x

c

�H�� x

c

�d�:

The last integral is zero if t � x=c, and equals

1

2

�t � x

c

�2; if t >

x

c:

Hence

u.x; t/ D8<:�g

2

ht2 � �

t � xc

�2i0 � x � ct

�g2 t2 x � ct:

The interpretation is rather easy. The portion of string placed on x � ct is only subject to gravity.The fixed end affects only the portion between x D 0 and x D ct (and the wave propagates withspeed c).

Solution 4.3.8. We resort to what we did for Problem 4.2.7 (page 230), and arrive at the ODE

�c2Uxx.x; s/C s2u.x; s/ D F.s/C 1

which, once integrated, gives

U.x; s/ D .1C F.s//1 � e�xs=c

s2:

Now we can anti-transform and obtain

u.x; t/ D t ��t � x

c

�H�t � x

c

�� g

2

t2 �

�t � x

c

�2H�t � x

c

��

D

8<:x

c� g

2

t2 �

�t � x

c

�2�0 � x � ct

t � g

2t2 x � ct:


Solution 4.3.9. a) From Problem 4.2.9 on page 234 the fundamental frequencies are given by

j D ˛j

4�

rg

L

where the ˛j are the zeroes of the Bessel function J0. These are, approximately,

˛1 D 2:40483 : : : , ˛2 D 5:52008 : : : , ˛3 D 8:65373 : : :

so that

1 D 0:5990 : : : ; 2 D 1:3750 : : : ; 3 D 2:1558 : : : :

b) Again by Problem 4.2.9, the first fundamental modes are

uj .x; t/ D Aj J0�˛j

px�

cos

p9:8˛j

2t

!j D 1; 2; 3

where

Aj D 1

J 21�˛j� ´Z 1=2

0

x

100J0�˛j

px�dx C

Z 1

1=2

.1 � x/100

J0�˛j

px�dx

μ:

Inserting the values obtained in part a) we get

A1 D 0:003874 : : : ; A2 D �0:005787 : : : ; A3 D 0:002371 : : : :

The amplitudes are much smaller than 1 (the rest length).

c) See Fig. 4.13.

Solution 4.3.10. The equation is hyperbolic, with characteristics x D constant and x C 2y Dconstant. The general solution is

u.x; y/ D 2ex C e.xC2y/=2 ŒF .x/CG.x C 2y/�

where F and G are two arbitrary functions.

Solution 4.3.11. The equation is hyperbolic, with characteristics defined by x D constant andy D constant (the equation is already in normal form). Set v D uy and differentiate in y:

vxy C yvy C v � v D 0;

i.e. .vy/x C yvy D 0. Integrating in x gives vy D f .y/e�xy , and then integrating in y twice,

u.x; y/ D yG.x/CG0.x/CZ y

0.y � �/e�x�f .�/ d�;

with f;G arbitrary.

264 4 Waves

Fig. 4.13 Positions of the hanging chain (Exercise 4.3.9) at various instants

Solution 4.3.12. a) Integrating on R we have:

ZRuxy .�; �/ d�d� D

Z y

0d�

Z x

'.y/uxy .�; �/ d� D

Z y

0

�uy .x; �/ � uy .' .y/ ; �/

�d�

D u .x; y/ � u .x; 0/ � u .' .y/ ; y/C u .' .y/ ; 0/ .

Using the data and the equation we can write the following formula for u:

u .x; y/ D h .y/C g .x/ � g .' .y//CZ y

0

Z x

'.y/F .�; �/ d�d�: (4.33)

b) It is easy to verify that (4.33) defines a solution u 2 C 2 .Q/ \ C�Q�

precisely whenh .0/ D g .0/.

c) Suppose u; v are solutions in C 2 .Q/ \ C�Q�

of the same problem. Their differencew D u � v solves wxy D 0 with vanishing data. If we integrate on the rectangle defined bypoints .a; b/ ; .x; b/ ; .a; y/ ; .x; y/ ; with ' .y/ < a < x and 0 < b < y, we find, with similar


computations,

0 DZRwxy .�; �/ d�d� D w .x; y/ � w .a; y/ � w .x; b/C w .a; b/ :

Passing to the limit a ! ' .y/ and b ! 0, we see that w .a; y/, w .x; b/ and w .a; b/ converge tothe origin, and therefore w .x; y/ D 0, proving uniqueness.

Solution 4.3.13. If g; h 2 C 2 .0;1/ are continuous at zero, there exists a unique solution con-

tinuous on S with second derivatives continuous on S , given by

u .x; t/ D g

�x C t

2

C h

�x � t2

� g .0/ :

Solution 4.3.14. a) The solution has the form u .x; t/ D F .x C t /CG .x � t /. We impose theCauchy data on the straight line x D t :

u .x; x/ D F .2x/CG .0/ D f .x/ ; (4.34)

which impliesF .z/ D f

�z2

��G .0/ ;

and1p2Œux .x; x/ � ut .x; x/� D p

2G0 .0/ D g .x/ :

The problem can be solved only if the Neumann datum g is constant: g .x/ D k. In this case thereare infinitely many solutions

u .x; t/ D f

�x C t

2

C kp

2.x � t /CG .x � t /

where G is any function such that G .0/ D G0 .0/ D 0. The problem is, therefore, ill posed.

b) This problem, too, is ill posed. Every function

u .x; t/ D F .x C t / � F .x � t /with F even (F .�z/ D F .z/) will solve the problem with zero data.

Solution 4.3.15. a) Let us differentiate the integrand function9:

E 0 .t/ DZ L

0Œ�utut t C �uxuxt � dx:

The boundary conditions force ut .0; t/ D ut .0; L/ D 0, whenceZ L

0uxuxtdx D Œuxut �

L0 �

Z L

0uxxutdx D �

Z L

0uxxutdx;

9 This is allowed if, for instance, u has continuous second derivatives on 0 � x � L, t � 0.

266 4 Waves

and from the ODE we compute

E 0 .t/ DZ L

0ut Œ�ut t � �uxx � dx D

Z L

0utuxxtdx

D Œutuxx �L0 �

Z L

0u2xtdx D �

Z L

0u2xtdx � 0:

Interpretation: E .t/ is the total mechanical energy at time t . The formula

E 0 .t/ D �Z L

0u2xt .x; t/dx

shows that the string dissipates energy at rate � R L0 u2xtdx. In the model of a string small vibra-tions, ux represents the relative displacement of the string particles, and uxt controls the variationof ux . The term � R L0 u2xtdx is therefore the kinetic energy per unit time that is dissipated becauseof the internal friction of the string particles.

b) If u, v solve (4.29), the quantity w D u� v also solves (4.29), with data f .x/ D g .x/ D 0.Call E .t/ the energy associated to w. From part a) we have E 0 .t/ � 0 and E .0/ D 0, sincewx .x; 0/ D f 0 .x/ D 0 and wt .x; 0/ D g .x/ D 0. Consequently E .t/ D 0 for any t > 0, whichimplies

wx D wt D 0

for any 0 < x < L and every t > 0. But thenw is constant. Being initially zero, it must be identicallyzero, so u D v.

c) We seek solutions of the type u .x; t/ D v .x/w .t/. Substituting into the equation, we obtain

�v .x/w00 .t/ � �v00 .x/w .t/ D v00 .x/w0 .t/ :

Now we separate variables and set c2 D �=� and "2 D =�:

v00 .x/v .x/

D w00 .t/c2w .t/C "2w0 .t/ D ��2:

The eigenvalue problem for v is v00 .x/C �2v .x/ D 0, v .0/ D v .L/ D 0, solved by

vn .x/ D sin�nx, �n D n�

L; n D 1; 2; : : : :

For w we therefore have

w00 .t/C "2�2nw0 .t/C c2�2nw .t/ D 0:

The general integral depends on the sign of

ın D "4�2n � 4c2.

When ın < 0, i.e. 1 � n < 2cL=.�"2/, we have

wn .t/ D exp

�"

2�2n2

t

!"an sin

�npjınj2

t

!C bn cos

�npjınj2

t

!#:


If there is an n such that ın D 0, then

wn .t/ D .an C bnt / exp

�"

2�2n2

t

!:

When ın > 0

wn .t/ D an exp

�"2�2n C �n

pın

2t

!C bn exp

�"2�2n � �n

pın

2t

!:

The solution to (4.29) will be

u .x; t/ D1XnD1

wn .t/ sin�nx:

Now assume that f and g can be expanded in sine series on Œ0; L�; the Fourier coefficients an andbn will be determined by the data. To see if u .x; t/ ! 0 as t ! 1 there is no need to compute thecoefficients explicitly. Suppose, in fact, that f; g are C 1 .R/, so that

1XnD1

.janj C jbnj/ < 1:

Then

ın < 0 H) jwn .t/j � .janj C jbnj/ exp

�"

2�2n2

t

!� .janj C jbnj/ exp

�"

2�2

2L2t

!;

ın D 0 H) jwn .t/j � .janj C t jbnj/ exp

�"

2�2n2

t

!� .janj C t jbnj/ exp

�"

2�2

2L2t

!;

ın > 0 H) �"2�2n C �npın D "2�2n

�1C

s1 � 4c2

"4�2n

!� �2c

2

"2< 0

so

jwn .t/j � janj exp

�c

2

"2t

!C jbnj exp

�"

2�2

2L2t

!:

It is easy to conclude u .x; t/ ! 0 as t ! 1.

Solution 4.3.16. For better clarity we denote by �2; �3 the Laplacians in two, three dimensionsrespectively.

a) We have

vt t ��3v Dhut t ��2u � k2u

iekz D 0 if k D

p�:

b) The function v introduced in a) solves the Cauchy problem´vt t ��3v D 0 .x; y; z/ 2 R3; t > 0

v.x; y; z; 0/ D f .x; y/ekz ; vt .x; y; z; 0/ D g.x; y/ekz .x; y; z/ 2 R3

268 4 Waves

(k D p�). Kirchhoff’s formula gives

v.x; y; z; t/ D @

@t

1

4�t

Z@Bt .x;y;z/

f .�; �/ek� d�

�C 1

4�t

Z@Bt .x;y;z/

g.�; �/ek� d�

and then

u .x; y; t/ D v .x; y; 0; t/ DD @

@t

1

4�t

Z@Bt .x;y;0/

f .�; �/ek� d�

�C 1

4�t

Z@Bt .x;y;0/

g.�; �/ek� d�:

The integrals on the surface @Bt .x; y; 0/ can be reduced to the disc Kt .x; y/, as in Problem 4.2.18on page 249.

Solution 4.3.17. It is clear that there can be waves with wave vector parallel to the (horizontal)z-axis; their potential is

� .z; t/ D A exp ¹i .k3z � !t/º :But we may also imagine waves moving in different directions and being reflected on the walls ofthe waveguide. They are polarised on the xz-plane, so we write them as sums of an incoming waveand a reflected one: equivalently, we represent them by means of a potential of the form (Fig. 4.14)

� .x; z; t/ D A exp°i!�xc

cos �I C z

csin �I � t

�±C A exp

°i!��xc

cos �I C z

csin �I � t

�±where we used the result of Problem 4.2.22 on page 254.

The wave will get reflected if the homogeneous Neumann conditions on the walls

�x .0; z; t/ D �x .d; z; t/ D 0

are fulfilled. While the first one clearly holds, the other one gives

exp

²i!d

ccos �I

³D exp

²�i !d

ccos �I

³

Fig. 4.14 Reflections along the waveguide


which implies!d

ccos �I D �!d

ccos �I C 2n� , n D 0; 1; 2; : : :

and socos �I D n�c

!d. (4.35)

As 0 < cos �I � 1, there are real solutions only for integer numbers n � N , where N is themaximum integer such that

n � !d

�c.

Consequently there is only a finite number N of incidence angles �I at which a plane harmonic canpropagate along the waveguide. Each angle corresponds actually to one mode of the waveguide.Given n, we have

sin �I Ds1 � n2�2c2

!2d2and

!

csin �I D

s!2

c2� n2�2

d2

and so

k3 Ds!2

c2� n2�2

d2(4.36)

is the wavenumber along z. By Euler’s relation 2 cos˛ D ei˛ C e�i˛ , ˛ D n�=d , we can writethe potential for the nth mode in a more significant way:

�n .x; z; t/ D 2A cosn�x

dexp ¹i .k3z � !t/º : (4.37)

When n D 0 we recover the wave moving along z at speed c. When 0 < n � N the wave behavesas a standing wave along x and a travelling one along z.

Complex solutions to (4.35) correspond to purely imaginary wavenumbers k3 D i�, � > 0. Thepotential (4.37) refers to waves that decay exponentially as z ! C1 (called evanescent waves).

Remark. An alternative way to look at this is the following. Fix the wavenumber k3 and try to findthe possible travelling frequencies. By (4.36) there exists a cut-off frequency

!c D �c

d

and a sequence of resonant frequencies

!n D c

sk23 C n2�2

d2

corresponding to the harmonics that can propagate.

Solution 4.3.18. We use cylindrical coordinates r D px2 C y2; �; z with 0 � � � 2� , z 2 R.

Waves of angular frequency ! moving along the pipe can be described by the potential

ˆ.r; �; z; t/ D ' .r; �; z/ e�i!t ;

solution to

ˆt t � c2²ˆrr C 1

rˆr C 1

r2ˆ�� Cˆzz

³D 0 (4.38)

270 4 Waves

inside the cylinder. Moreover, the zero normal velocity on the walls enforces the homogeneousNeumann condition

'r .a; �; z/ D 0:

Substituting ' .r; �; z/ e�i!t in (4.38) we find for ' the equation

'rr C 1

r'r C 1

r2'�� C 'zz C !2

c2' D 0: (4.39)

Ler us determine the modes by analysing solutions with separate variables:

' .r; �; z/ D u .r/ v .�/w .z/

with u bounded, u0 .a/ D 0 and v .�/ 2��periodic. Substituting in (4.39) and dividing by ', we get

u00 .r/C 1

ru0 .r/

u .r/C 1

r2v00 .�/v .�/

C w00 .z/w .z/

C !2

c2D 0:

We set

u00 .r/C 1

ru0 .r/

u .r/C 1

r2v00 .�/v .�/

D �w00 .z/w .z/

� !2

c2D ˛

and define

D !2

c2� ˛;

so thatw00 .z/C w .z/ D 0: (4.40)

From

u00 .r/C 1

ru0 .r/

u .r/C 1

r2v00 .�/v .�/

D ˛

we deducer2u00 .r/C ru0 .r/ � r2˛u .r/

u .r/D �v

00 .�/v .�/

D ˇ:

As v is 2�-periodic, byv00 .�/C ˇv .�/ D 0

we haveˇ D m2 with m � 0 integer

andv .�/ D v0 cos Œm .� � �0/� .

Then the equation for u is

u00 .r/C 1

ru0 .r/C

�˛ � m2

r2

!u .r/ D 0: (4.41)

If ˛ D ��2 (we shall consider only this case, for simplicity), equation (4.41) is a parametric Besselequation10 of order m. The only bounded solutions are

um .r/ D u0Jm .�r/

10 Appendix A.


where Jm is the Bessel function of the first kind of order m. The constraint u0 .a/ D 0 reads

J 0m .�a/ D 0.

There are infinitely many zeroes �m1 < �m2 < : : : < �mp < : : : of J 0m, all found in standardtextbooks. The first one is zero. Here are a few other positive roots, approximated to two digits:

�01 D 3:83 �02 D 7:02 �03 D 10:17

�11 D 1:84 �12 D 5:33 �13 D 8:54

�21 D 3:05 �22 D 6:71 �23 D 9:97:

So we have infinitely many solutions to (4.41) of the form

ump .r/ D u0Jm��mpr=a

�.

If the angular frequency is given, the values of are determined by

mp D !2

c2� �2mp

a2: (4.42)

When the right-hand side in (4.42) is positive,

!2

c2� �2mk

a2D k2; k > 0;

the wave

ˆmk .r; �; z; t/ D AmkJm

��mk

ar

cos Œm .� � �0/� exp ¹i.kz � !t/º

travels undamped along the pipe. If, instead,

!2

c2� �2mk

a2D �k2;

the corresponding mode (bounded for z > 0)

ˆmk .r; �; z; t/ D AmkJm

��mk

ar

cos Œm .� � �0/� exp ¹�kz � i!t/º

becomes weaker as it propagates in the pipe.Equation (4.42) shows that as the pipe radius decreases, more and more modes are damped, to

the point that if the radius is small enough only one mode survives. In case m D 0, in fact, �00 D0 and the corresponding mode is the plane harmonic

ˆ00 .r; �; z; t/ D A exp°i!.

z

c� t /

±which propagates irrespective of the pipe’ size. Therefore, as soon as

! < �11c

aD 1:84

c

a;

only the plane wave will travel, while all others are “absorbed” by the pipe.

5

Functional Analysis

5.1 Backgrounds

We shall recall some basic facts on Hilbert spaces, distributions and Sobolev spaces.

Banach and Hilbert spaces

• A Banach space is a complete normed vector space1 X . ‘Normed’ means that there is apositive function defined on X , called the norm,

k�k D k�kX W X ! RC

with the following three properties: positivity (kxk D 0 () x D 0), homogeneity(k�xk D j�j kxk) and the triangle inequality

kx C yk � kxk C kyk :‘Complete’ means that every Cauchy sequence2 converges in X . Given a norm, the dis-tance between two vectorx x and y is given by kx � yk.

Typical examples of Banach spaces are the spaces Lp .�/, 1 � p � 1, of Lebesgue-measurable functions f defined on3 � � Rn that are Lebesgue p-integrable, i.e.

kf kLp. / D�Z

jf jp1=p

< 1 for 1 � p < 1;

orkf kL1. / D ess sup jf j < 1 if p D 1:

1 We shall only deal with vector spaces over the reals.2 ¹xnº � X is called a Cauchy sequence if and only if kxn � xmk ! 0 as m; n ! 1.3 Most of times � will be a domain, that is a connected open subset of Rn.


274 5 Functional Analysis

• Hölder’s inequality. Let f 2 Lp.�/, g 2 Lq.�/, with 1 � p; q � C1, andp�1 C q�1 D 1 (p and q are called conjugate). Then the following inequality holdsZ

juvj dx ��Z

jujp dx1=p �Z

jvjq dx1=q

:

In case p D q D 2 this goes under the name of the Cauchy-Schwarz inequality.

• A pre-Hilbert space is a vector space H equipped with an inner product (also dotproduct, especially in the case dimH < 1)

.�; �/ D .�; �/H W H �H ! R

with the properties of positivity (.x; x/ � 0, and .x; x/ D 0 if and only if x D 0), bilin-earity4, symmetry (.x; y/ D .y; x/). The inner product induces a norm

kxk D .x; x/1=2:

If H is complete with respect to the induced norm it is called a Hilbert space.Most prominently, L2.�/ is a Hilbert space with inner product

.u; v/L2. / DZ

uv dx:

In particular, by positivity, ifR uv dx D 0 for any v 2 L2.�/ then u � 0 almost every-

where in � (to prove it just take v D u). By density, actually, it suffices that the integralvanishes for any v 2 C10 .�/.

• Parallelogram rule. In a Hilbert space H one always has

2kuk2 C 2kvk2 D kuC vk2 C ku � vk2:This relation characterises those norms that come from an inner product.

• Orthogonal decomposition theorem. Let H be a Hilbert space and V a closed sub-space inH . For every x 2 H there exists a unique PV x in V (the projection of x onto V )such that

kPV x � xk D infv2V kv � xk:

Moreover, setting QV x D x � PV x, we have .v;QV x/ D 0 for all v 2 V and

kxk2 D kPV xk2 C kQV xk2 (Pythagoras’ theorem).

• Linear maps. If F is a linear map between two Hilbert spacesH1,H2, the continuityof F is equivalent to its boundedness: there exists a constant C such that

kFukH2� CkukH1

for any u 2 H1. The smallest of such constants defines the norm of F .

4 That is, linearity in both arguments.

5.1 Backgrounds 275

• Dual space and Riesz’s theorem. The set of real linear, continuous functionals on aHilbert space H is called the dual space of H , and is denoted by H�.

Riesz’s representation theorem states that, for any F 2 H� there exists a unique ele-ment u 2 H such that

.u; v/ D Fv for any v 2 H:Moreover, kukH D kF kH� .

• Weak convergence. Let H be a Hilbert space. A sequence ¹xnº � H is said to con-verge weakly, or in the weak sense, to Nx (written xn * Nx), if .xn; v/ ! . Nx; v/ for anyv 2 H as n ! 1. If a sequence converges strongly it converges weakly, but the converseis not true.

If xn * Nx then ¹xnº is bounded and ¹kxnkº is lower semi-continuous; that is:

k Nxk � lim inf kxnk :• Compactness and weak compactness. A subset K in a normed space X is said to

be compact if any open covering of K admits a finite sub-covering. In a normed space,K is compact if and only if is sequentially compact, i.e. if every sequence in K admits asubsequence converging in K. The set K is called relatively compact (or precompact)if its closure is compact. This is equivalent to the following: K is relatively compactif and only if, for any " > 0, K is contained in the union of finitely many balls of ra-dius ".

An operator T W H1 ! H2 between two Hilbert spaces is called compact if it mapsbounded sets to relatively compact subsets, i.e. T .B/ � H2 is compact for any B � H1bounded. In particular, if T is linear and compact then it is bounded and therefore con-tinuous; if T is not linear it may be compact but not continuous. We shall call completelycontinuous a compact and continuous operator.

Every bounded set is sequentially weakly relatively compact: any bounded sequenceadmits a weakly convergent subsequence.

Distributions

Let� � Rn be a domain. One denotes by D.�/ the vector space of test functions. Theseare smooth (infinitely differentiable) functions that have compact support5. Test functionscome equipped with the following notion of convergence: vk ! v in D.�/ if the sup-port of any vk is contained in one single compact set, and all derivatives D˛vk convergeuniformly in � to D˛v.

The space of distributions on �, denoted by D 0 .�/, is the set of continuous linearfunctionals on D.�/. If F 2 D 0 .�/ we shall write

hF; vi

5 The support of a continuous map is the closure of the set of its non-zero values.


to indicate how F acts on a test function v. In particular, F is continuous for

hF; vki ! hF; vi

if vk ! v in D .�/. By using strings of m test functions one defines, in a completelysimilar manner, vector-valued distributions D 0 .�I Rm/.

A function u in L1loc.�/ may be seen as a distribution, under the canonical identifica-tion

hu; vi DZ

uv dx:

• Derivative of a distribution. Take F 2 D 0 .�/. The derivative @xiF is the distribu-

tion defined by the formula

h@xiF; vi D �hF; @xi

vi; 8v 2 D .�/ :

A remarkable result is: suppose F 2 D 0.�/ and rF is the zero distribution; then F is aconstant function (� is a domain, hence connected).

• Rapidly-decreasing functions and tempered distributions. Denote by S .Rn/ thespace of functions v 2 C1 .Rn/ that decrease ‘very quickly’ when jxj ! 1, i.e. suchthat, for any m 2 N and every multi-index ˛ D .˛1; : : : ; ˛n/,

D˛v .x/ D o�jxj�m� ; jxj ! 1:

(We use the customary shortcut D˛ D @˛1x1: : : @

˛nxn

.)If ¹vkº � S .Rn/ and v 2 S .Rn/ we write vk ! v in S .Rn/ if, for every multi-

indices ˛;ˇ;

xˇD˛vk ! xˇD˛v uniformly in Rn:

A distribution T 2 D 0 .Rn/ is called tempered if

hT; vki ! 0

for any ¹vkº � D .Rn/ such that vk ! 0 in S .Rn/. The set of tempered distributions isdenoted by S 0 .Rn/ :

• Fourier transforms of tempered distributions. Let T 2 S 0.Rn/. The Fourier transformbT D F ŒT � is the tempered distribution defined by

hbT ; vi D hT;bvi; 8v 2 S.Rn/:

Sobolev spaces

Let � � Rn be a domain. The Sobolev spaces we will use most are: the Hilbert space

H 1.�/ D ®u 2 L2.�/ W ru 2 L2.�I Rn/

¯

5.1 Backgrounds 277

(all derivatives are meant in distributional sense) with the inner product

.u; v/H1. / DZ

Œru � rv C uv� dx;

and its closed subspace H 10 .�/ D D.�/

H1. /. Elements of H 1

0 .�/ have zero traceon @�. In a similar way one may define the Hilbert spaces Hm.�/ of functions whosederivatives up to order m, included, belong to L2.�/.

• Sobolev spaces in R. We have the inclusion

H 1.a; b/ � C.Œa; b�/

(this is false in dimension n � 2). Furthermore, for any u 2 H 1 .a; b/ the FundamentalTheorem of Calculus holds:

u .y/ � u .x/ DZ y

x

u0 .s/ ds a � x � y � b:

• Poincaré inequality and equivalent norms. If � is bounded there exists a constantCP , only depending on � and n, such that

kukL2. / � CP krukL2. /

for any u 2 H 10 .�/. This fact allows to use in H 1

0 .�/ the equivalent norm krukL2. /.This will be always our choice, unless otherwise stated.

Poincaré’s inequality holds on other subspaces of H 1.�/ (such as the space of func-tions with zero average).

• The dual of H 10 . The dual of H 1

0 .�/, denoted by H�1.�/, consists of elements thatcan be written (albeit not uniquely) as f Cdiv f, where f 2 L2.�I R/ and f 2 L2.�I Rn/.

• Traces. Let � be a bounded and Lipschitz domain (or half-space) and set � D @�.Then there is a well-defined trace operator

0 W H 1.�/ ! L2 .�/ ;

which is linear, continuous, and such that

1. 0u D uj� , if u 2 C ��.2. k0ukL2.�/ � C� kukH1. /, with C� independent of u:

(Abusing notations we shall write uj� instead of 0u even when u 62 C.�/.) The spaceof traces of functions in H 1.�/, i.e. the image of 0, is denoted by

H 1=2 .@�/ D ®uj� W u 2 H 1.�/

¯:


It is a Hilbert space with norm

kwkH1=2.�/ D inf®kukH1. / W u 2 H 1.�/; uj� D w

¯:

For any u 2 H 1.�/, we have

kuj�kL2.�/ � C�kuj�kH1=2.�/ � C�kukH1.�/:

• Rellich’s theorem. Let � be a bounded Lipschitz domain. Then the embeddingH 1.�/ ,! L2.�/ is compact.

5.2 Solved Problems

� 5:2:1 � 5:2:11 W Hilbert spaces.� 5:2:12 � 5:2:20 W Distributions.� 5:2:21 � 5:2:30 W Sobolev spaces.

5.2.1 Hilbert spaces

Problem 5.2.1 (The space of continuous functions). a) Let C.Œ�1; 1�/ denote thespace of real, continuous functions on the interval Œ�1; 1�, with norm

kf k D kf kC.Œ�1;1�/ D maxt2Œ�1;1�

jf .t/j:

Check that C.Œ�1; 1�/ is a Banach space, but that its norm cannot be induced byan inner product (thus C.Œ�1; 1�/ cannot be a Hilbert space). Hint. If the norm isinduced by an inner product the parallelogram rule must hold.

b) Let C ?.Œ�1; 1�/ be the space of real, continuous functions on Œ�1; 1� with norm

kf k? D kf kL2.�1;1/ D�Z 1

�1jf .t/j2 dt

1=2:

Check that although this norm comes from an inner product, it does not make thespace complete (so that neither C ?.Œ�1; 1�/ is a Hilbert space). Hint. Show that

fn.t/ D

8<:0 �1 � t � 0

nt 0 � t � 1n

1 1n

� t � 1;

is a Cauchy sequence in C ?.Œ�1; 1�/ but it converges to the Heaviside function H

(H .x/ D 0 for x < 0, H .x/ D 1 for x � 0).


Solution. a) It is enough to show that C .Œ�1; 1�/ is complete; in fact, let¹fnº � C .Œ�1; 1�/ be a Cauchy sequence for the sup norm. Given " > 0, then, forn > m � N ."/ we have

jfn .x/ � fm .x/j < ", for any x 2 Œ�1; 1� : (5.1)

Equation (5.1) indicates that for any x 2 Œ�1; 1� the sequence of real numbers ¹fn .x/º isa Cauchy sequence, so

limn!1 fn .x/ D f .x/

exists and is finite. Passing to the limit in (5.1) as n ! 1, we obtain

jf .x/ � fm .x/j < ", for any x 2 Œ�1; 1� ; (5.2)

i.e. fn ! f uniformly in Œ�1; 1�. As uniform convergence of a sequence of continuousfunctions implies the continuity of the limit, we deduce f 2 C .Œ�1; 1�/. Every Cauchysequence in C .Œ�1; 1�/ therefore converges to some element of C .Œ�1; 1�/, which is thuscomplete.

In order to show that the norm can not be induced by an inner product, we show thatthe parallelogram rule fails. For this we need to find two functions f; g for which

kf C gk2 C kf � gk2 ¤ 2kf k2 C 2kgk2:For example, let

f D 1C x and g D 1 � x:Then

k1C xk D k1� xk D 2 and k.1C x/C .1� x/k D k.1C x/� .1� x/k D 2;

thus the parallelogram rule does not hold and the space is not Hilbert.

b) The given norm is the norm of L2.�1; 1/, induced by the inner product

.f; g/L2.�1;1/ DZ 1

�1f .t/g.t/ dt;

as is easy to see. Let us consider the given sequence. To fix ideas suppose m > n. Then

kfn � fmk2? DZ 1=m

0

..m � n/t/2 dt CZ 1=n

1=m

.1 � nt/2 dt D

D .m � n/23m3

C 1

3n

�1 � n

m

�3 � 1

3mC 1

3n:

Consequently, for n;m ! C1, kfn�fmk? ! 0 and the sequence is a Cauchy sequencefor this norm. Analogously, if H is Heaviside’s step function,

kH � fnk2? DZ 1=n

0

.1 � nt/2 dt D 1

3n;


so fn converges to H in L2. As H … C ?.Œ�1; 1�/ (it is discontinuous at x D 0), fn isa Cauchy sequence in C ?.Œ�1; 1�/ that does not converge in C ?.Œ�1; 1�/, so the space isnot complete.

Problem 5.2.2 (Projections in a Hilbert space). Set H D L2.�1; 1/ and consider thesubspace V of odd functions:

V D ¹u 2 H W u.�t / D �u.t/ for (almost) every t 2 .�1; 1/º:a) Verify that the orthogonal decomposition theorem holds.

b) Determine V ? and then write the expressions for PV f , QV f for an arbitrary f 2H .

Solution. a) AsH is a Hilbert space, we only need to check that V is a closed subspacein H . Since a linear combination of odd functions is odd, V is a linear space. To prove itis closed we must check that the limit (inL2 .�1; 1/-norm) of a sequence of odd functionsis still odd:

¹unº � V; un ! Nu implies Nu 2 V:For that we can, for instance, use the fact that if a sequence converges in L2 it admits asubsequence that converges almost everywhere. So there exists

®unk

¯such that

unk.t/ ! Nu.t/

for almost every t 2 .�1; 1/; then

0 D unk.t/C unk

.�t / ! Nu.t/C Nu.�t / for almost every t 2 .�1; 1/;whence Nu 2 V .

b) To identify V ? we seek all u 2 L2.�1; 1/ such thatR 1�1 uv dt D 0 for any v 2 V .

Take u 2 V ?; then:

0 DZ 1

�1u.t/v.t/ dt D

Z 0

�1u.t/v.t/ dt C

Z 1

0

u.t/v.t/ dt D

D �Z 1

0

u.�t /v.t/ dt CZ 1

0

u.t/v.t/ dt DZ 1

0

Œu.t/ � u.�t /�v.t/ dt:

Again, by the freedom in choosing v, we deduce

u.t/ � u.�t / D 0 a.e. in .0; 1/

and so V ? is the subspace of even functions. To find the projections onto V and V ? of afunction f it is enough to decompose f as the sum of its even and odd parts

f .t/ D f .t/C f .�t /2

C f .t/ � f .�t /2

:


Since an even function and an odd one are always orthogonal with respect to theL2.�1; 1/scalar product, we obtain that

PV f .t/ D f .t/ � f .�t /2

; QV f .t/ D f .t/C f .�t /2

:

Problem 5.2.3 (Orthogonal splitting). Set Q D .0; 1/ � .0; 1/, H D L2.Q/, andconsider the subspace

V D ¹u 2 H W u.x; y/ D v.x/ a.e., with v 2 L2.0; 1/º:Determine the projection operators PV , PV? . Decompose the function f .x; y/ D xy.

Solution. Note that, given u 2 H , PV u is the best approximation (in L2 sense) of uby means of a function depending only on x.

By the orthogonal decomposition theorem, the decomposition exists if V is closed. It iseasy to see that V is a linear space, so we just check its closure. Setting un.x; y/ D vn.x/,we suppose un converges to Nu in L2.Q/. The claim is that Nu depends only on x and be-longs to L2 .0; 1/. SinceZ 1

0

Z 1

0

Œvn .x/ � vm .x/�2dxdy DZ 1

0

Œvn .x/ � vm .x/�2dx,

the sequence ¹vnº is Cauchy inL2 .0; 1/ and hence converges to v D v .x/, v 2 L2 .0; 1/.But Z 1

0

Œvn .x/ � v .x/�2dx DZ 1

0

Z 1

0

Œvn .x/ � v .x/�2dxdy;

and the uniqueness of limits implies v D Nu, making V closed.Let us find V ?. This subspace consists of functions u 2 H such that for any v in

L2.0; 1/

0 DZQ

u.x; y/v.x/ dxdy DZ 1

0

�Z 1

0

u.x; y/ dy

v.x/ dx:

We deduce

V ? D²u 2 H W

Z 1

0

u.x; y/ dy D 0, for almost every x 2 .0; 1/³:

Now it is immediate to see that for any f 2 H

PV f .�/ DZ 1

0

f .�; y/ dy;

QV f .�/ D f .�; y/ �Z 1

0

f .�; y/ dy:


In particular, taking f .x; y/ D xy,

xy D 1

2x C

�xy � 1

2x

;

where the first summand belongs to V and the one in brackets lives in V ?.

Problem 5.2.4 (Norms of functionals and Riesz’s theorem). GivenH D L2.0; 1/ andu 2 H , consider the functional

Fu DZ 1=2

0

u.t/ dt:

a) Check that F is well defined and belongs to H�.

b) Compute kF kH� in two ways, first using the definition and then Riesz’s theorem.

Solution. a) Clearly, if u 2 L2.0; 1/ then u 2 L1.0; 1=2/. Moreover, by Schwarz’sinequalityˇ

ˇZ 1=2

0

u.t/ dt

ˇˇ �

Z 1=2

0

ju.t/j dt � Z 1=2

0

1 dt

!1=2� Z 1=2

0

ju.t/j2 dt!1=2

�p2

2

�Z 1

0

ju.t/j2 dt1=2

Dp2

2kukH :

(5.3)

Therefore F is a well-defined functional on H and is bounded. Being linear it is alsocontinuous, and belongs to H�.

b) We recall that by definition

kF kH� D supkukH�1

jFuj:

From (5.3) we immediately deduce kF kH� � p2=2. Let us show kF kH� D p

2=2, byexhibiting a function with unitary norm such that Fu D p

2=2. It is reasonable to chooseu so to concentrate its norm on the interval Œ0; 1=2�. In particular, if we pick

u.t/ D

8<:

p2 0 < t <

1

2

01

2< t < 1;

we have kukH D 1 and Fu D p2=2, whence the claim.


An alternative method is to invoke Riesz’s theorem. SinceF 2 H�, the theorem yieldsthe existence (and uniqueness) of an f 2 H such that

Fu DZ 1

0

f .t/u.t/ dt and kF kH� D kf kH :

Therefore

f .t/ D

8<:1 0 < t <

1

2

01

2< t < 1;

and then

kF kH� Dp2

2:

Problem 5.2.5 (Operators: norm, invertibility, eigenvalues). Let H D L2.0;C1/

and consider the linear operator

.Lu/ .t/ D u.2t/

from H to itself.

a) Prove that L is continuous and compute its norm.

b) Compute the (real) eigenvalues of L, if any.

Solution. a) L is linear, so we have to prove its boundedness. As

kLukH D�Z C1

0

u2.2t/ dt

1=2D�1

2

Z C1

0

u2.s/ ds

1=2D

p2

2kukH ; (5.4)

L is bounded (hence continuous), and has normp2=2.

b) We must find non-zero solutions to

Lu D �u; i.e. u.2t/ D �u.t/; a.e. t > 0: (5.5)

We will show that (5.5) implies that u � 0, thus proving that L has no real eigenvalue.Computing the norms of both sides and exploiting (5.4) we find

p2

2kuk D j�jkuk:

Consequently if u is not identically zero, � D ˙p2=2. Observe that applying L to both

sides of (5.5) givesu .4t/ D L.Lu/ D L.�u/ D �2u:

Applying L n times, we find that the eigenfunction u satisfies

u.2nt / D �nu.t/ (5.6)


for almost every t > 0 and any integer n � 0. But

u .t/ D u

�2n

t

2n

D �nu .2�nt / ;

sou .2�nt / D ��nu .t/

and (5.6) holds for any n 2 Z. Therefore to find u we only need to know its value on, say,Œ1; 2/. In particularZ 2nC1

2n

u2.s/ ds D 2nZ 2

1

u2.2nt / dt D 2nZ 2

1

�2nu2.t/ dt DZ 2

1

u2.t/ dt:

By writing.0;C1/ D

[n2Z

Œ2n; 2nC1/;

we have

kuk2H DZ C1

0

u2.t/ dt DXn2Z

Z 2nC1

2n

u2.t/ dt DXn2Z

Z 2

1

u2.t/ dt:

If u is an eigenfunction, then0 < kukH < C1:

On the other hand the above series is the sum of infinitely many identical terms, and con-verges if and only if its (only) term vanishes identically, i.e. u D 0 almost everywhere on.1; 2/. By property (5.6), we have u D 0 almost everywhere on .0;C1/. To sum up, noteven for � D ˙p

2=2 the equation Lu D �u has non-trivial solutions. Consequently Ldoes not have real eigenvalues.

**Problem 5.2.6 (Compactness). a) Write H D l2, i.e. the Hilbert space

l2 D8<:x D ¹xnºn�1 W

Xn�1

x2n < 19=;

with inner product .x; y/l2 D Pn�1 xnyn. Given M > 0, define the set

K D8<:x 2 l2 W

Xn�1

x2n < M ,Xn�1

n2x2n < M

9=; :Prove that for any given " > 0 there exists a finite-dimensional subspace V" � H

such thatdist .x; V"/ D kx � PV"

xk < " for any x 2 K:b) Deduce K is relatively compact in H (use the definition of compactness involving

coverings by "-spheres).


Solution. a) We can interpret H as an infinite-dimensional generalisation of Rn: se-quences, in fact, can be seen as vectors with an infinite number of coordinates, and thedefinition of norm is nothing else but the infinite-dimensional version of Pythagoras’ the-orem! It can be proved that a Hilbert basis (a complete orthonormal system) for H isgiven by the infinitely many vectors ei D .xn/, with xi D 1, xn D 0 8n ¤ i . Thefinite-dimensional subspaces ofH easiest to describe are those whose elements have onlya finite number of non-zero coordinates. Said better, set

Vk D ¹x 2 H W xn D 0 for any n � kº :

We invite the reader to check Vk is a vector subspace of dimension k � 1 (hence closed),and that, for any x 2 H , its distance from Vk is

kx � PVkxk D

C1XnDk

x2n

!1=2: (5.7)

Making the distance between an element of H and Vk arbitrarily small is the same asmaking small the kth remainder of the corresponding series. By definition of K, for anyx 2 K and any natural number k, we have

M >

C1XnD1

n2x2n �C1XnDk

n2x2n � k2C1XnDk

x2n:

Recalling (5.7) we have

x 2 K ) kx � PVkxk �

pM

k:

Now it suffices to choose k >pM=" to conclude.

b) We recall that a set is called relatively compact if its closure is compact. In partic-ular a set A � H is relatively compact when for any " > 0 there is a finite number ofpoints y1; : : : ; yN such that

A �N[iD1

B".yi /

where B" .yi / is the ball of radius " centred at yi . Fix " > 0 and

k > 2

pM

":

By part a) every point in K has distance less than "=2 apart from Vk . Consider

F D K \ Vk;


a bounded subset (since x 2 F implies kxk2Vk� M ) of the finite-dimensional space Vk ,

and hence relatively compact. By the characterisation mentioned earlier there areN pointsy1; : : : ; yN such that

F �N[iD1

B"=2.yi /:

Take x 2 K. Then PVkx 2 F and there is an index i for which PVk

x 2 B"=2.yi /: Thus

kx � yik � kx � PVkxk C kPVk

x � yik � "

2C "

2D "

i.e. x 2 B".yi /. Since the argument holds for any x 2 K, we have proved

K �N[iD1

B".yi /; i.e. K is relatively compact.

Problem 5.2.7 (Weak convergence). Let H be a Hilbert space and ¹unº � H a se-quence such that

un * Nuand

either kunk ! k Nuk or kunk � k Nuk:Show that un converges (strongly) to Nu.

Solution. We recall that un converges weakly to Nu if .un; v/ ! . Nu; v/ for any v 2 H .In particular, choosing v D Nu gives .un; Nu/ ! . Nu; Nu/ D k Nuk2. We obtain

kun � Nuk2 D .un; un/ � 2.un; Nu/C . Nu; Nu/ D kunk2 � k Nuk2 C 2�k Nuk2 � .un; Nu/�„ ƒ‚ …

!0:

In both cases we deduce kun � Nuk ! 0, i.e. un ! Nu strongly.

Problem 5.2.8 (Nonlinear compact operator). Write H D L2.0; 1/ and consider thenonlinear operator

T Œf �.t/ DZ 1

0

.t C f .s//2 ds:

a) Show T is well defined and continuous from H to H .

b) Using the criterium for subsets of L2.0; 1/ (see [18, Chap. 6]), prove that T is com-pact.

Solution. a) If f 2 H then

jT Œf �.t/j DZ 1

0

.t C f .s//2 ds � 2

Z 1

0

�t2 C f 2.s/

�ds � 2C 2kf k22; (5.8)


and T is well defined on H . Let us check that the range of T is contained in H . By (5.8)

kT Œf �k22 DZ 1

0

Z 1

0

.t C f .s//2 ds

�2dt � �

2C 2kf k22�2;

so T W H ! H . To prove continuity (T is not linear, so we really need a proof) let f; gbe in H . Then

kT Œf � � T Œg�k22 DZ 1

0

Z 1

0

.t C f .s//2 � .t C g.s//2 ds

�2dt D

DZ 1

0

Z 1

0

.2t C f .s/C g.s// .f .s/ � g.s// ds�2dt �

�Z 1

0

Z 1

0

.2t C f .s/C g.s//2 ds

��Z 1

0

.f .s/ � g.s//2 ds�dt �

� �4C 2kf C gk22

�2 � kf � gk22

(we have used Schwarz’s inequality and, in the last step, (5.8)). Hence, if f ! g in Hwe have T Œf � ! T Œg� in H , so T is continuous.

b) Write

F D ¹f 2 H W kf k2 � M º :

In order to prove that T ŒF � is relatively compact in H we shall use the criterion for L2-compactness, and check whether T ŒF � is bounded and there exist C , ˛ such that

kT Œf �.� C h/ � T Œf �.�/k2 � C jhj˛ (5.9)

for any f 2 F vanishing outside .0; 1/. As we saw in part a), from (5.8) we have, for anyf 2 F ,

kT Œf �k22 � �2C 2kf k22

�2 � �2C 2M 2

�2;

and so T ŒF � is bounded. But we also have

kT Œf �.� C h/ � T Œf �.�/k22 �Z 1

0

Z 1

0

.t C hC f .s//2 � .t C f .s//2 ds

�2dt D

DZ 1

0

Z 1

0

�2h .t C f .s//C h2

�ds

�2dt � h2

Z 1

0

Z 1

0

.hC 2t C 2f .s// ds

�2dt

and Schwarz’s inequality plus (5.8) give (5.9) with ˛ D 2.


Problem 5.2.9 (Iterated projectionsa). Let V and W be closed subspaces in a Hilbertspace H . Define the sequence ¹xnº of projections as follows: x0 2 H is given, and

x2nC1 D PW x2n, x2nC2 D PV x2nC1 when n � 0.

Prove the following assertions:

a) V \W D ¹0º implies xn ! 0.

b) V \W ¤ ¹0º implies xn ! PV\W x0.

Hint. a) Show, in this order: kxnk decreases, xn * 0 and kxnk2 D .x2n�1; x0/; b)Reduce to to previous case by subtracting PV?\W x0.

a This problem is related to Schwarz’s alternating method, see Problem 6.2.15, Chap. 6 (page361).

Solution. a) The idea – as we are looking at projections – is that the sequence of normsshould decrease. Therefore ¹xnº will be bounded, with finite limit, in particular equal tozero. Let us check that kxnk decreases. Since

.xnC1; xn/ D .xnC1; xnC1/ D kxnC1k2 ;for every n � 0, we have

kxnC1 � xnk2 D kxnC1k2 � 2.xnC1; xn/C kxnk2D kxnC1k2 � 2 kxnC1k2 C kxnk2D � kxnC1k2 C kxnk2

so kxnk � kxnC1k and kxnk # l � 0. Moreover

kxnC1 � xnk ! 0:

We will show that xn * 0 weakly and then l D 0 (see Problem 5.2.7 on page 286).Let

®x2nk

¯be any weakly convergent subsequence of ¹x2nº: x2nk

* x, with x inV . As also x2nkC1 * x (in fact the distance between x2nk

and x2nkC1 tends to 0) onehas x 2 W , and so x D 0. But the subsequence is arbitrary, so x2n * 0. Similarly for¹x2nC1º, and altogether xn * 0.

Now, to fix ideas let us assume xn 2 V , and therefore xn�1 2 W . As orthogonalprojections are symmetric operators,

kxnk2 D .xn; xn�1/ D .xn; PW xn�2/ D .PW xn; xn�2/ D .xnC1; xn�2/.Iterating the argument gives

kxnk2 D .xnC1; xn�2/ D .xnC2; xn�3/ D � � � D .x2n�1; x0/and since x2n�1 * 0 we see that kxnk ! 0.

b) If V \W ¤ ¹0º we set

z0 D x0 � PV\W x0.

The sequence starting with z0 and generated by projecting on V andW as before, is givenby

zn D xn � PV\W x0:


It is easy to see that PV\W zn D 0, and then zn belongs to V n �V ? \W �or to

W n �V \W ?�, whose intersection reduces to ¹0º. In this way one falls back to the pre-vious situation, whence zn ! 0, that is xn ! PV\W x0.

Problem 5.2.10 (Projection onto a convex set). LetH be a Hilbert space and K � H

a (non-empty) closed, convex subset. Prove, along the lines of the Projection Theorem[18, Chap. 6, Sect. 4], that for any x 2 H there is a unique PKx 2 K such that

kPKx � xk D infv2K kv � xk:

Verify that PKx is uniquely determined by the variational inequality

.x � PKx; v � PKx/ � 0 for any v 2 K: (5.10)

Solution. Defined D inf

v2K kv � xkand let ¹vnº � K be a minimising sequence, i.e. for any n

vn 2 K; d2 � kvn � xk2 � d2 C 1

n: (5.11)

We shall prove that vn is a Cauchy sequence. The parallelogram rule for x � vn, x � vmgives

2kx � vnk2 C 2kx � vmk2 D kvn � vmk2 C 4

��x � vn C vm

2

��2 :Since K is convex and vn and vm belong to K, also .vn C vm/=2 belongs to K and itsdistance from x is greater than or equal to d . So we can write

kvn � vmk2 D 2kx � vnk2 C 2kx � vmk2 � 4��x � vn C vm

2

��2 �

� 2d2 C 2

nC 2d2 C 2

m� 4d2 D 2

nC 2

m:

Then kvn � vmk ! 0 asm; n tend to infinity, and the sequence ¹vnº is Cauchy sequence.But H is complete, so there exists w 2 H with vn ! w in H . Keeping into account theclosure of K (and the norm continuity) we have, from (5.11),

w 2 K; kx � wk D d:

To show uniqueness for w consider w0 2 K, with kx �w0k D d . The parallelogram rule(for x � w, x � w0) gives (again: K convex implies .w C w0/=2 2 K)

kw � w0k2 D 2kx � wk2 C 2kx � w0k2 � 4��x � w C w0

2

��2 � 2d2 C 2d2 � 4d2 D 0

i.e. w D w0. Hence, to any x 2 H we may associate a unique element w D PKx 2 K,the projection of x to K, that achieves the shortest distance.

Let us now prove that PKx satisfies (5.10). Fix v 2 K and set, for 0 � t � 1,

u D tPKx C .1 � t /v:


The convexity of K forces u 2 K and also

kx � PKxk2 � kx � uk2 D kx � PKx � .1 � t /.v � PKx/k2 DD kx � PKxk2 C .1 � t /2kv � PKxk2 � 2.1 � t /.x � PKx; v � PKx/:

That is, for t < 1,

0 � .1 � t /kv � PKxk2 � 2.x � PKx; v � PKx/:Passing to the limit as t ! 1� we find (5.10).

Vice versa, let y 2 K be such that

.x � y; v � y/ � 0; for any v 2 K:In particular, with v D PKx, we have

0 � .x � y; PKx � y/ D .x � PKx C PKx � y; PKx � y/ D .x � PKx; PKx � y/C kPKx � yk2:We have just shown .x � PKx; PKx � y/ � 0, so now kPKx � yk2 � 0 and then y DPKx. Therefore PKx is characterised by (5.10).

Problem 5.2.11 (Projection onto a convex set in L2). Let� be a domain in Rn,H DL2 .�/ and

K D ®v 2 L2 .�/ I a � v .x/ � b a.e. in �

¯with a; b given constants.

a) Verify thatK is closed and convex. Using Problem 5.2.10 deduce that the projectionOu D PK .u/ of any u 2 L2 .�/ is characterised by the variational inequalityZ

Œ Ou .x/ � u .x/� Œv .x/ � Ou .x/� dx � 0; 8v 2 K: (5.12)

b) Define � D E� [E0 [EC, where

E� D ¹x W Ou .x/ < u .x/º ; E0 D ¹x W Ou .x/ D u .x/º ; EC D ¹x W Ou .x/ > u .x/º :Prove that, a.e. in �,

Ou .x/ D

8<:a in EC

u .x/ in E0b in E�:

(5.13)

c) Deduce that (5.12) is equivalent to the pointwise inequality:

Œ Ou .x/ � u .x/� Œv .x/ � Ou .x/� � 0 a.e. in �, 8v 2 K (5.14)

and thus find the formula, valid a.e. in �,

Ou .x/ D PŒa;b� .u .x// D min ¹b;max ¹a; u .x/ºº D max ¹a;min ¹b; u .x/ºº :(5.15)

Solution. a) To prove that K is closed we have to show that if ¹unº is a sequence inL2 such that a � un � b and un ! Nu in L2, then a � Nu � b. For that it is enough to


notice that L2 convergence implies pointwise convergence almost everywhere of at leastone subsequence: hence a � Nu.x/ � b almost everywhere, and then the claim follows.

As for convexity, fix u1, u2 in K. For 0 � � � 1

a D �aC .1 � �/a � �u1.x/C .1 � �/u2.x/ � �b C .1 � �/b D b:

Therefore any convex combination of elements ofK belongs toK, and we can apply Prob-lem 5.2.10. In particular (5.12) is another way of writing (5.10) in the present context.

b) First, observe that all three sets are measurable and defined up to zero-measure sets.By contradiction, suppose there exists an A � EC; of positive measure, with Ou .x/ > a.Define

v .x/ D´a in A

Ou .x/ in �nA:Then v 2 K andZ

Œ Ou .x/ � u .x/� Œv .x/ � Ou .x/� dx DZA

Œ Ou .x/ � u .x/� Œa � Ou .x/� dx < 0;

which contradicts (5.12). The same can be done for E�.

c) By (5.13) we have, for any v 2 K:

Œa � u .x/� Œv .x/ � a� � 0 a.e. in EC,

Œb � u .x/� Œv .x/ � b� � 0 a.e. in E�

and then (5.14) follows. The latter, in turn, indicates that

Ou .x/ D PŒa;b�u .x/

a.e. in�, and therefore (5.15) is a consequence of the following one-dimensional formula,which is not hard to prove:

8z 2 R; PŒa;b� .z/ D min ¹b;max ¹a; zºº D max ¹b;min ¹a; zºº :

5.2.2 Distributions

Problem 5.2.12 (Distributions and Fourier series). Prove that

C1XkD1

ck sin kx

converges in D 0.R/ if the numerical sequence ¹ckº grows ‘slowly’, meaning that thereexists p 2 R, C > 0 such that jckj � Ckp for every k � 1.


Solution. We need to prove that for any v 2 D .R/, the series

1XkD1

ck

ZRv .x/ sin kx dx (5.16)

converges. Let N be an integer greater than pC 2; we can assume that N is even and setN D 2n. Take any test function v, whose support is contained in some interval, say Œa; b�.Notice that v and all its derivatives vanish at a and b. If we integrate N times by parts wefindZ

Rv .x/ sin kx dx D

Z b

a

v .x/ sin kx dx D 1

k

Z b

a

v0 .x/ cos kx dx

D � 1

k2

Z b

a

v00 .x/ sin kx dx D .�1/nkN

Z b

a

v.N/ .x/ sin kx dx:

Hence ˇZRv .x/ sin kx dx

ˇ� .b � a/

kNmax

ˇv.N/

ˇandˇ

ck

ZRv .x/ sin kx dx

ˇ� C .b � a/max

ˇv.N/

ˇ 1

kN�p� C .b � a/max

ˇv.N/

ˇ 1k2

for N > p C 2. The series (5.16) is therefore convergent.

Problem 5.2.13 (Support of a distribution). a) Prove that if F 2 D 0.Rn/, v 2D.Rn/ and v vanishes on an open set containing supp.F /, then hF; vi D 0.

b) Is it true that hF; vi D 0 if v vanishes only on the support of F ?

Solution. a) Call K the support of F ; K is the complement of the largest open set �such that, if v is a test function with support in �, then hF; vi D 0. But if v is zero on anopen set containing K then supp.v/ must be contained in � and therefore hF; vi D 0.

b) This is false. It suffices to take F D ı0 and v a test function such that

v .0/ D 0; v0 .0/ ¤ 0:

The support of F is K D ¹0º and v vanishes on K, but

hF; vi D hı0; vi D �hı; v0i D �v0.0/ ¤ 0:

Problem 5.2.14 (Differential equation in D 0). Given a 2 C1.R/, solve the equation

G0 C aG D ı0 in D 0.R/:


Solution. First observe that, if v 2 D.R/ and g 2 C1.R/, then gv 2 D.R/. Hencefor any F 2 D 0.R/ we can define the distribution gF 2 D 0.R/ by

hgF; vi D hF; gvi; 8v 2 D.R/:

As a 2 C1 .R/, also h .x/ D expR x0a .s/ ds belongs to C1 .R/, and we can multiply

both sides of the equation by h, obtaining

d

dxŒG .x/ h.x/� D hı0:

But

hhı0; vi D hı0; hvi D �hı; h0v C hv0i D �h0 .0/ v .0/ � h .0/ v0 .0/D �a .0/ v .0/ � v0 .0/ D h�a .0/ ı C ı0; vi;

so thatd

dxŒG .x/ h.x/� D �a .0/ ı C ı0:

The primitives of �a .0/ ı C ı0 are given by

�a .0/H C ı C c;

where H is the Heaviside step function and c 2 R, so we have

G .x/ expZ x

0

a .s/ ds D �a .0/H C ı C c:

The general solution of the differential equation is then

G .x/ D .�a .0/H C ı C c/ e�R x

0 a.s/ds D ı C .�a .0/H C c/ e�R x

0 a.s/ds :

Problem 5.2.15 (Rapidly decreasing functions). Verify that:

a) The function v .x/ D e�jxj2 belongs to S.Rn/.

b) The function v .x/ D e�jxj2 sin�ejxj2

�does not belong to S.Rn/.

Solution. a) Take v .x/ D e�jxj2 . Each derivative D˛v of order k has the form

.polynomial of degree k in x1; : : : ; xn/ e�jxj2 :

Therefore whichever m 2 N we consider, we have

limjxj!1

jxjmD˛v .x/ D 0

and D˛v .x/ D o�jxj�m�, so that v decreases rapidly at infinity.


b) Let v .x/ D e�jxj2 sin�ejxj2

�. Then

Dxjv .x/ D 2xj

h�e�jxj2 sin

�ejxj

2�C cos�ejxj

2�iand so

limjxj!1

Dxjv .x/ does not exist.

We conclude that v … S .Rn/.

Problem 5.2.16 (Distributions and tempered distributions). Verify that ifF 2 D 0.Rn/has compact support then F 2 S 0.Rn/.

Solution. Let F have compact support K � Rn and take ¹vkº � D .Rn/ such thatvk ! 0 in S .Rn/. We have to show that hF; vki ! 0. Choose open sets A and B so thatK � A � B: If w is a test function with support in B and w � 1 on A, then for any othertest function v

hF; vi D hF; vwi.In fact, the test function z D v � vw vanishes on A and hence has support in the comple-ment of K; consequently hF; zi D hF; vi � hF; vwi D 0.

Since (check this fact) vkw ! 0 in D .Rn/, we conclude

hF; vki D hF; vkwi ! 0

and therefore F is tempered.

Problem 5.2.17 (Lp functions and tempered distributions). a) Verify that, for any1 � p � C1, Lp.Rn/ � S 0.Rn/.

b) Show that if uk ! u in Lp.Rn/ then uk ! u in S 0.Rn/.

Solution. Take f 2 Lp .Rn/, 1 � p � 1 and ¹vkº � D .Rn/ such that vk ! 0 inS .Rn/. If q D p= .p � 1/ is the conjugate exponent to p, we have

hf; vki DZ

Rn

f .x/.1C jxj/.nC1/=q .1C jxj/.nC1/=qvk .x/ dx;

and by Hölder’s inequality

jhf; vkij � kf kLp supRn

h.1C jxj/.nC1/=q jvk .x/j

i�ZRn

1

.1C jxj/.nC1/ dx1=q

:

But ZRn

1

.1C jxj/.nC1/ dx D !n

Z C1

0

�n�1

.1C �/.nC1/d� < 1;


and moreover, vk ! 0 in S .Rn/ gives

supRn

h.1C jxj/.nC1/=q jvk .x/j

i! 0:

Thus, overall hf; vki ! 0. This says that f is tempered.

b) Take uk ! 0 in Lp .Rn/, v 2 S .Rn/, 1 � p � 1 and q D p= .p � 1/. Arguingas above we obtain

jhuk; vij � kukkLp„ ƒ‚ …!0

supRn

h.1C jxj/.nC1/=q jv .x/j

i„ ƒ‚ …

<1

�ZRn

1

.1C jxj/.nC1/ dx1=q

„ ƒ‚ …<1

;

so huk ; vi ! 0, and then uk ! 0 in S 0 .Rn/.

Problem 5.2.18 (Fourier transform of a periodic distribution). Let u 2 D 0 .R/ be peri-odic of period T . Compute the Fourier transform of u, then anti-transform the formulaobtained and discuss.

(We remind that F is said periodic of period T if

hF .x C T / ; vi WD hF; v .x � T /i D hF; vifor any test function v, and that every periodic distribution is tempered.)

Solution. As u is periodic,

u .x C T / � u .x/ D 0:

Taking the Fourier transform we find�eiT � � 1�bu .�/ D 0. (5.17)

The zeroes of eiT � �1 are �n D 2�n=T , n 2 Z, all simple (multiplicity one). The generalsolution of (5.17) is given by the following formula6:

bu .�/ DXn2Z

cnı

�� 2n�

T

(5.18)

where cn 2 C are arbitrary constants, and the series converges7 in D 0 .R/. As the inverseFourier transform of ı .� � a/ is e�iax=2� , when we transform back, we find the Fourierseries

u .x/ D 1

2�

Xn2Z

cn exp

��i 2n�x

T

;

which holds in D 0 .R/ (and also in S 0 .R/). The numbers cn=2� are therefore the Fouriercoefficients of the expansion of u.

6 [18, Chap. 7, Sect. 5].7 It can be proved that it also converges in S 0 .R/ as well.


Remark. Computing the Fourier coefficientsbun D cn=2� is a rather delicate matter. Arelatively easy case is the following. Suppose we can decompose

u D u1 C u2

using T -periodic distributions u1; u2 such that u1 2 L1loc .R/. If x0 does not belong to thesupport of u2, and we writeeu2 for the restriction of u2 to .x0; x0 C T /, we have

bun D 1

T

Z x0CT

x0

u1 .x/ e�i2n�x=T dx C heu2; e�i2n�x=T i: (5.19)

Note that the choice u D u1 gives the usual formula. An explicit example can be foundin Exercise 5.3.12 on page 311.

Problem 5.2.19 (Rapidly decreasing functions and tempered distributions). Supposeu 2 L1loc.R

n/ and let P D P.x/ be a polynomial withu

P2 L1.Rn/:

Prove that u 2 S 0.Rn/.

Solution. Take u 2 L1loc .Rn/ and P a polynomial such that

u

P2 L1 .Rn/. Let

¹vkº � D .Rn/ be such that vk ! 0 in S .Rn/. In particular Pvk ! 0 uniformly inRn. Then

jhu; vkij DˇZ

Rn

u .x/ vk .x/ dx

ˇDˇZ

Rn

u .x/P .x/

P .x/ vk .x/ dx

ˇ� sup

RnjPvk j

ZRn

ˇu .x/P .x/

ˇdx ! 0;

because u=P 2 L1 .Rn/. Consequently u is tempered.

Problem 5.2.20 (Global Cauchy problem for the wave equation). Let g; h 2 D .Rn/.

a) Show that u 2 C 2 .Rn � Œ0;C1/ is a solution of the global Cauchy problem (n D1; 2; 3) ´

ut t � c2�u D 0 x 2 Rn, t > 0

u .x; 0/ D g .x/ ; ut .x; 0/ D h .x/ x 2 Rn(5.20)

if and only if u0 .x; t / D H .t/ u .x; t / (the extension of u to zero for t < 0) is asolution in D 0.Rn � R/ of the equation

u0t t � c2�u0 D g .x/˝ ı0 .t/C h .x/˝ ı .t/ . (5.21)

b) Let K D K .x; t / be the fundamental solution of the wave equation in dimensionn D 1; 2; 3: Deduce that K0 .x; t / D H .t/K .x; t / satisfies

K0t t � c2�K0 D ı .x/˝ ı .t/ in D 0.Rn � R/.


Solution. a) Recall that for or every ' 2 D .Rn � R/ we have, by definition of tensorproduct,˝

g .x/˝ ı0 .t/ ; ' .x; t/˛ D ˝

g .x/ ;˝ı0 .t/ ; ' .x; t /

˛˛ D � hg .x/ ; hı .t/ ; ' t .x; t /iiD � hg .x/ ; ' t .x; 0/i D �

ZRg .x/ ' t .x; 0/ dx;

hh .x/˝ ı .t/ ; ' .x; t/i D hh .x/ ; hı .t/ ; ' .x; t /iiD hh .x/ ; ' .x; 0/i D

ZRh .x/ ' .x; 0/ dx:

Let now u0 satisfy (5.21). This means that,ZR

ZRn

u0�' t t � c2�'� dtdx D �

ZRn

g .x/ ' t .x; 0/ dxCZ

Rn

h .x/ ' .x; 0/ dx:

(5.22)Integrating twice by parts the first integral, we get, taking into account that u0 D u0 .x; t /D H .t/ u .x; t / ;

Z C1

0

ZRn

u�' t t � c2�'� dxdt D

Z C1

0

ZRn

�ut t � c2�u�' dxdt

�Z

Rn

u .x; 0/ ' t .x; 0/ dxCZ

Rn

ut .x; 0/ ' .x; 0/ dx: (5.23)

Comparing (5.22) and (5.23) we infer thatZ C1

0

ZRn

�ut t � c2�u�' dxdt �

ZRn

u .x; 0/ � g .x/�' t .x; 0/ dx

CZ

Rn

Œut .x; 0/ � h .x/� ' .x; 0/ dx D 0 (5.24)

for every ' 2 D .Rn � R/. Take ' 2 D .Rn � .0;C1//. Then ' t .x; 0/ D ' .x; 0/ D 0

and (5.24) gives Z C1

0

ZRn

�ut t � c2�u�' dxdt D 0:

The arbitrariness of ' implies that ut t � c2�u D 0 in Rn � .0;C1/. Now we can goback to ' 2 D .Rn � R/. Since ut t � c2�u D 0 we get from (5.24)

�Z

Rn

u .x; 0/ � g .x/�' t .x; 0/ dx C

ZRn

Œut .x; 0/ � h .x/� ' .x; 0/ dx D 0: (5.25)

Choose now 0; 1 2 D .Rn/ and b .t/ 2 D .R/ such that b .0/ D 1 and b0 .0/ D 0:

The function

.x; t / D . 0 .x/C t 1 .x// b .t/


belongs to D .Rn � R/ and .x; 0/ D 0 .x/ , t .x; 0/ D 1 .x/ . Inserting into(5.25) we obtain

�Z

Rn

Œu .x; 0/ � g .x/� 1 .x/ dx CZ

Rn

Œut .x; 0/ � h .x/� 0 .x/ dx D 0:

The arbitrariness of 0; 1 implies u .x;0/ D g .x/ and ut .x; 0/ D h .x/.Conversely, let u satisfy (5.20): A double integration by parts, using the initial condi-

tions, gives

0 DZ C1

0

ZRn

u�' t t � c2�'� dxdt C

ZRn

g .x/ ' t .x; 0/ dx �Z

Rn

h .x/ ' .x; 0/ dx

which is equivalent to (5.22) since u0 D 0 for t < 0.b) It is enough to remember thatK is a solution of the wave equations with initial data

K .x; 0/ D 0, Kt .x; 0/ D ı .x/.

5.2.3 Sobolev spaces

Problem 5.2.21 (Singularities in Sobolev spaces). Consider the function

u.x; y/ D�p

x2 C y2�˛

and let D D ¹.x; y/ 2 R2 W x2 C y2 < 1º be the unit disc. Determine the values ˛ 2R for which:

a) u 2 L2.D/. b) u 2 H 1.D/. c) u 2 H�1.D/.

Solution. a) It is enough to compute the integral of u2 and check whether it is finite.In polar coordinates

kuk22 DZD

.x2 C y2/˛ dxdy DZ 2�

0

d�

Z 1

0

r2˛C1 dr D 2�

Z 1

0

r2˛C1 dr:

The integral is finite if and only if 2˛ C 1 > �1. Hence

u 2 L2.D/ () ˛ > �1:

b) First of all umust be in L2.D/, i.e. ˛ > �1. Next, the gradient of u (in the sense ofdistributions) must belong to L2.DI R2/. When ˛ � 0 we have u … C 1 �D�, so, a priori,distribution derivatives may not coincide with classical ones. However, for ˛ > �1, thiscan be proved (see Exercise 5.3.15 on page 312). Thus, in such case,

ux.x; y/ D ˛x.x2 C y2/.˛�2/=2; uy.x; y/ D ˛y.x2 C y2/.˛�2/=2:


Hence

kruk22 DZD

˛2.x2 C y2/˛�1 dxdy D 2�˛2Z 1

0

r2˛�1 dr:

Therefore 2˛ � 1 > �1 or ˛ D 0, and so

u 2 H 1.D/ if and only if ˛ � 0:

c) For u to belong to the dual of H 10 .D/ we must find a constant C satisfyingˇZ

D

uv dxdy

ˇ� C

�ZD

jrvj2 dxdy1=2

for any v 2 H 10 .D/:

As we are in dimension 2, for any 1 � p < C1 we have8 H 1.D/ � Lp.D/. FromPoincaré’s inequality we infer the existence of a constant CP (depending on D and p)such that any v 2 H 1

0 .D/ fulfils

kvkLP .D/ � CP krvkL2.DIR2/:

If q > 1 is the conjugate to p, the above inequality combines with Hölder’s inequality togive ˇZ

D

uv dxdy

ˇ��Z

D

uq dxdy

1=q �ZD

vp dxdy

1=p�

� CP

�ZD

uq dxdy

1=q �ZD

jrvj2 dxdy1=2

:

This inequality shows that if u 2 Lq.D/, with q > 1, then u 2 H�1.D/. So let us see forwhich numbers ˛, u belongs to Lq.D/ (for some q > 1). AsZ

D

.x2 C y2/q˛=2 dxdy D 2�

Z 1

0

rq˛C1 dr

we must have q˛ C 1 > �1. This is true (for some q > 1) as soon as ˛ > �2. Theconverse statement can be proved, too: when ˛ � �2 the function u does not belong tothe dual of H 1

0 .D/. It suffices to pick v 2 D.D/ � H 10 .D/ satisfying

v � 0 in D; v � 1 in D0 D².x; y/ 2 R2 W x2 C y2 <

1

2

³;

to obtain (note that u � 0 in D)ZD

uv dxdy �ZD0udxdy D 2�

Z 1=2

0

r˛ � r dr D C1 if ˛ � �2:

8 By Sobolev’s embedding theorem, see [18, Chap. 7, Sect. 10].


To sum up,u 2 H�1.D/ if and only if ˛ > �2:

Problem 5.2.22 (Singularities in trace spaces). Consider the function

u.x; y/ D�p

x2 C y2�˛

and letDC D ¹.x; y/ 2 R2 W x > 0; y > 0; x2 C y2 < 1º. Find all ˛ 2 R for which:

a) u 2 L2.@DC/. b) u 2 H 1.@DC/. c) u 2 H 1=2.@DC/.

Solution. a) The function u has a singularity only at the origin and when ˛ < 0; inparticular, it is smooth along the arc

¹.x; y/ W x > 0; y > 0; x2 C y2 D 1º � @DC:

On the segmentS D ¹.x; 0/ W 0 < x < 1º � @DC

we have

kuk2L2.S/

DZ 1

0

u2.x; 0/ dx DZ 1

0

x2˛ dx;

and the integral converges if 2˛ > �1. By symmetry this holds also on ¹.0; y/ W 0 <y < 1º, therefore

u 2 L2.@DC/ if and only if ˛ > �12:

b) As in the previous exercise, for ˛ > �1=2, the distributional derivatives of u areclassical. Hence

kuxk2L2.S/

DZ 1

0

˛2x2˛�2 dx:

The integral is finite if ˛ D 0 or 2˛ � 2 > �1. Then

u 2 H 1.@DC/ if and only if ˛ D 0 or ˛ >1

2:

c) The domain is Lipschitz so we can resort to the theory of traces. We saw in the pre-vious problem that the extension of u toDC, defined by the same algebraic expression, isin H 1.D/ (whence in H 1.DC/) for ˛ � 0. Immediately, then,

u 2 H 1=2.@DC/ provided ˛ � 0:

Problem 5.2.23 (Norm estimates in L1.R/). Prove that u 2 H 1.R/ implies u 2L1.R/, and

kukL1.R/ � kukH1.R/:


Solution. Let us prove the inequality for C1 functions with compact support. By den-sity the result will hold in general (recall thatH 1.R/ D H 1

0 .R/). In fact, take u 2 D.R/,and observe that

u2.t/ DZ t

�12u.s/u0.s/ ds � 2

�Z t

�1u2.s/ ds

1=2 �Z t

�1u02.s/ ds

1=2�Z t

�1u2.s/ ds C

Z t

�1u02.s/ ds

(in the last line we used the elementary inequality 2ab � a2 C b2). We can now take thesupremum over t and write

u 2 D.R/ ) kukL1.R/ � kukH1.R/: (5.26)

Let now ¹unº � D.R/ and un ! u in H 1.R/. Equation (5.26) implies

kun � umkL1.R/ � kun � umkH1.R/;

so ¹unº is a Cauchy sequence in L1.R/. Therefore un converges to u also in L1.R/,and (5.26) holds for u 2 H 1 .R/ as well.

Problem 5.2.24 (Norm estimates in L1.a; b/). Prove that u 2 H 1.a; b/ implies u 2L1.a; b/, and

kukL1.a;b/ � CkukH1.a;b/

with C D max¹.b � a/�1=2; .b � a/1=2º.

Solution. Since u is continuous on the compact set Œa; b�, it is bounded. We must showthe existence of a constant C such that

ju.t/j � C.kuk2 C ku0k2/

for any t 2 Œa; b� and any u 2 H 1.a; b/. To this end, let us remark that by the mean-valuetheorem and the continuity of u we may fix a point � 2 Œa; b� with

u2.�/ D 1

b � aZ b

a

u2.s/ ds D 1

b � akuk22:

Hence

ju.t/j Dˇu.�/C

Z t

�

u0.s/ dsˇ

� 1pb � akuk2 C

Z b

a

ju0.s/j ds �

� 1pb � akuk2 C

pb � aku0k2 � C.kuk2 C ku0k2/;


where

C D max

²1pb � a ;

pb � a

³:

Problem 5.2.25 (Riesz’s theorem and the Dirac ı measure). Let ı denote the one-dimensional Dirac delta measure at 0. Prove that ı belongs to the dual of H 1.�1; 1/.Determine the element ofH 1.�1; 1/ (which exists by Riesz’s theorem) representing ıwith respect to the inner product of H 1.�1; 1/.

Solution. Clearly ı is a linear functional that is well defined on H 1.�1; 1/, since thefunctions in that space are continuous and it makes sense to take their value at any point.The continuity of ı follows from Problem 5.2.24, where it is proved that

jv.0/j � p2.kvk2 C kv0k2/ for any v 2 H 1.�1; 1/:

Thus ı belongs to dual of H 1. By Riesz’s representation theorem there is a function u 2H 1.�1; 1/ such thatZ 1

�1�u0.t/v0.t/C u.t/v.t/

�dt D v.0/; for any v 2 H 1.�1; 1/:

To identify this function we suppose u is regular on Œ�1; 0� and Œ0; 1�, separately. Integrat-ing by parts, we findZ 0

�1�u0.t/v0.t/C u.t/v.t/

�dt D u0.0�/v.0/�u0.�1/v.�1/�

Z 0

�1�u00.t/ � u.t/� v.t/ dt

andZ 1

0

�u0.t/v0.t/C u.t/v.t/

�dt D u0.1/v.1/ � u0.0C/v.0/ �

Z 1

0

�u00.t/ � u.t/� v.t/ dt:

If we add the two relations we find, after rearranging the terms,

�Z 1

�1�u00.t/ � u.t/� v.t/ dt C v .0/

�u0 .0�/ � u0 �0C�� u0.�1/v.�1/C u0.1/v.1/ D v .0/:

By choosing v with support in .�1; 0/, we get:Z 0

�1�u00.t/ � u.t/� v.t/ dt D 0;

forcing u00.t/ � u.t/ D 0 in .�1; 0/, as v is arbitrary. Analogously, u00.t/ � u.t/ D 0 in.0; 1/. If we now take a general v, we are left with

v .0/�u0 .0�/ � u0 �0C�� u0.�1/v.�1/C u0.1/v.1/ D v .0/ :


Since the values v .0/ ; v .1/ and v .�1/ are arbitrary, we deduce the “transmission” con-dition

u0 .0�/ � u0 �0C� D 1;

and the Neumann conditionsu0.�1/ D u0.1/ D 0:

Summarising, we have reduced to the following problem:´u00.t/ � u0.t/ D 0 when � 1 < t < 0; 0 < t < 1u0.�1/ D u0.1/ D 0; u0.0�/ � u0.0C/ D 1:

The general integral of the ODE is a linear combination of exponentials, so

u.t/ D´A1e

t C A2e�t �1 � t � 0

B1et C B2e

�t 0 � t � 1:

The boundary conditions and the continuity at 0 give8<ˆ:A1e

�1 � A2e D 0

B1e � B2e�1 D 0

A1 � A2 � B1 C B2 D 1

A1 C A2 D B1 C B2

whence

A2 D B1 D 1

2.e2 � 1/ ; A1 D B2 D e2

2.e2 � 1/and finally

u.t/ D

8<:

1

2.e2 � 1/�e2Ct C e�t

� �1 � t � 0

1

2.e2 � 1/�et C e2�t

�0 � t � 1:

It is not hard to see that u 2 H 1.�1; 1/ (it is piecewise C 1), and integrating by parts oneproves (do it) that this is indeed the required function. (Uniqueness is a consequence ofRiesz’s theorem.)

Problem 5.2.26 (Riesz’s theorem and integral average). Prove that the functional

Lu DZ 1

0

u.t/ dt

belongs to H�1.0; 1/. Determine the element that represents it in H 10 .0; 1/.


Solution. L is a linear and continuous functional on L2.0; 1/ and therefore, a fortiori,onH 1

0 .0; 1/. In particular, continuity follows from the Schwarz and Poincaré inequalities:ˇZ 1

0

u.t/ dt

ˇ��Z 1

0

dt

1=2 �Z 1

0

ju.t/j2 dt1=2

� kukL2.0;1/ � CP ku0kL2.0;1/:

By Riesz’s theorem there is a unique u 2 H 10 .0; 1/ such thatZ 1

0

u0.t/v0.t/ dt DZ 1

0

v.t/ dt; for all v 2 H 10 .0; 1/:

To find u explicitly we proceed formally, supposing u regular (with at least two continuousderivatives in Œ0; 1�), so to be able to integrate by parts. We obtain

u0.1/v.1/ � u0 .0/ v .0/ �Z 1

0

u00.t/v.t/ dt DZ 1

0

v.t/ dt; for any v 2 H 10 .0; 1/

whence Z 1

0

Œu00.t/C 1�v.t/ dt D 0 for any v 2 H 10 .0; 1/:

As v was arbitrary, we infer

u00.t/ D �1 in .0; 1/

with u .1/ D u .0/ D 0. Then

u.t/ D 1

2.t � t2/:

By reading the argument backwards we see easily that u representsL for the inner productof H 1

0 .0; 1/.

Problem 5.2.27 (Riesz’s theorem). Let f and g be regular functions and consider thefunctional

Lv DZ 1

0

�f .t/v0.t/C g.t/v.t/

�dt:

Prove that L belongs to the dual of H 1.0; 1/.Calling u the element of H 1.0; 1/ representing L, write a boundary-value problem

that determines u uniquely.Solve the problem explicitly in case f .t/ D t .t � 1/ and g.t/ D 2t .

Solution. The functional L is linear, so we just have to prove its boundedness. BySchwarz’s inequality

jLvj �Z 1

0

�ˇf .t/v0.t/

ˇC jg.t/v.t/j� dt � kf kL2.0;1/kv0kL2.0;1/ C kgkL2.0;1/kvkL2.0;1/;


and therefore, as long as f and g are square-integrable, L is continuous. By Riesz’s the-orem there exists a unique u 2 H 1.0; 1/ such thatZ 1

0


�dt D

Z 1

0


�dt (5.27)

for any v 2 H 1.0; 1/. Mimicking the previous exercises we assume u is sufficiently reg-ular so to allow for integration by parts. ThenZ 1

0

u0.t/v0.t/ dt D u0.1/v.1/ � u0.0/v.0/ �Z 1

0

u00.t/v.t/ dt

and Z 1

0

f .t/v0.t/ dt D f .1/v.1/ � f .0/v.0/ �Z 1

0

f 0.t/v.t/ dt:

Substituting into (5.27) givesZ 1

0

.�u00 C uC f 0 � g/v dt C Œu0.1/ � f .1/�v.1/ � Œu0.0/ � f .0/�v.0/ D 0

for any v 2 H 1.0; 1/. Choosing v vanishing at the endpoints, we obtainZ 1

0

.�u00 C uC f 0 � g/v dt D 0;

forcing

u00 � u D f 0 � g in .0; 1/ :

Then, for general v 2 H 1.0; 1/, we are left with

Œu0.1/ � f .1/�v.1/ � Œu0.0/ � f .0/�v.0/ D 0

which gives

u0.1/ D f .1/ and u0.0/ D f .0/.

In conclusion, u is solves ´u00.t/ � u.t/ D f 0.t/ � g.t/u0.0/ D f .0/; u0.1/ D f .1/:

In case we take f .t/ D t2 � t and g .t/ D 2t , we find

u00 � u D �1, u0.0/ D u0.1/ D 0:

The solution is u .t/ � 1.


Problem 5.2.28 (Projections). Define H D H 1.�1; 1/ and

V D ¹u 2 H W u.0/ D 0º:After proving V is a closed subspace inH , compute the projection onto V of f , wheref .t/ D 1 on Œ�1; 1�.

Solution. That V is a subspace is straightforward (recall that functions in H are con-tinuous, so pointwise values make sense). We have to prove that V is closed, i.e. that ifun.0/ D 0 and un ! Nu in H , then Nu.0/ D 0. For this we observe that convergence in Himplies uniform convergence. As seen in Problem 5.2.24 (page 301), in fact,

kukL1.�1;1/ � p2kukH .

In turn, uniform convergence implies pointwise convergence everywhere on Œ�1; 1�. Inparticular if un ! Nu in H then un.0/ ! Nu.0/ and V is closed. The projection theo-rem guarantees the existence (and uniqueness) of the projection PV f . To find the latterexplicitly, we may minimise over V the quadratic functional

E.u/ D kf � uk2H DZ 1

�1�.u0.t//2 C .1 � u.t//2� dt:

We know that the minimum u D PV f satisfies the necessary condition

.u � f; v/H DZ 1

�1�u0.t/v0.t/ � .1 � u.t//v.t/� dt D 0; for any v 2 H: (5.28)

Assuming u regular and integrating by parts, we findZ 1

�1u0.t/v0.t/ dt D u0.t/v.t/

ˇ1�1 �

Z 1

�1u00.t/v.t/ dt;

and so

u0.1/v.1/ � u0.�1/v.�1/ �Z 1

�1�u00 C 1 � u� v dt D 0; (5.29)

for any v 2 H 1.�1; 1/. In particular, the equality must hold for every v 2 H 10 .�1; 1/, so

that Z 1

�1�u00 C 1 � u� v dt D 0; for any v 2 H 1

0 .�1; 1/;which implies

u00 C 1 � u D 0; a.e. in .�1; 1/ .

But then the integral in (5.29) is always zero, and we are left with

u0.1/v.1/ � u0.�1/v.�1/ D 0; for any v 2 H 1.�1; 1/;


whenceu0.�1/ D u0.1/ D 0:

As u.0/ D 0 (in fact u 2 V ) we find that u solves´u00.t/ � u.t/ D �1 �1 < t < 0; 0 < t < 1u0.�1/ D u0.1/ D 0; u.0/ D 0:

Then

u.t/ D´1C A1e

t C A2e�t �1 � t � 0

1C B1et C B2e

�t 0 � t � 1;

with 8<:A1e

�1 � A2e D 0

B1e � B2e�1 D 0

1C A1 C A2 D 1C B1 C B2 D 0:

We find

A2 D B1 D � 1

e2 C 1; A1 D B2 D � e2

e2 C 1

and finally

u.t/ D

8<:1 � 1

e2 C 1

�e2Ct C e�t

� �1 � t � 0

1 � 1

e2 C 1

�et C e2�t

�0 � t � 1:

Since there is a unique element inH 1.�1; 1/ satisfying (5.28), it follows that u coincideswith PV f .

Problem 5.2.29 (Piecewise defined function). Define

QC D ®.x; y/ 2 R2I 0 < x < 1; 0 < y < 1¯

Q� D ®.x; y/ 2 R2I �1 < x < 0; 0 < y < 1¯

and Q D QC [Q� [ ¹x D 0, 0 < y < 1º, then consider

u .x; y/ D´xy2 in QC

1 � x2y in Q�:

Compute the gradient of u in the sense of distributions. Decide whether u 2 H 1 .Q/.

Solution. On QC, and separately on Q�, u is a polynomial, and thus it defines a dis-tribution; note that u jumps by �1 horizontally (for example) across the vertical segment¹x D 0, 0 < y < 1º. Let v be a test function and let us compute the distributional gradient


of u. By definition

hux ; vi D �hu; vxi D �ZQ

u .x; y/ vx .x; y/ dxdy D

D �Z 1

0

dy

Z 1

0

xy2vx .x; y/ dx �Z 1

0

dy

Z 0

�1.1 � x2y/ vx .x; y/ dx:

Integrating by parts in x, we find, as v .1; y/ D 0:

hux ; vi DZ 1

0

dy

Z 1

0

y2v .x; y/ dx �Z 1

0

v .0; y/ dy �Z 1

0

dy

Z 0

�12xy v .x; y/ dx:

Set

w1 .x; y/ D´y2 in QC

�2xy in Q�

and let F D ı ˝ 1 be the distribution9

hF; vi DZ 1

0

v .0; y/ dy:

Thenux D w1 � ı ˝ 1.

The distribution �ı˝ 1 has support on the discontinuity curve of u, and signals the pres-ence of a �1 jump across the y-axis in the direction x > 0. Let us compute uy . Bydefinition

huy ; vi D �hu; vyi D �ZQ

u .x; y/ vy .x; y/ dxdy D

D �Z 1

0

dx

Z 1

0

xy2vy .x; y/ dy �Z 0

�1dx

Z 1

0

.1 � x2y/ vy .x; y/ dy:

Integrating by parts in y, since v .x; 0/ D v .x; 1/ D 0, we have

hux ; vi DZ 1

0

dx

Z 1

0

2xy v .x; y/ dy �Z 0

�1dx

Z 1

0

x2 v .x; y/ dy:

Defining

w2 .x; y/ D´2xy in QC

�x2 in Q�;we then obtain

uy D w2.

There are no distributions concentrated on the discontinuity, because along the verticaldirection u is not discontinuous. Whereas uy 2 L2 .Q/, the distribution representing uxcannot be identified with a function of L2 .Q/. Therefore u … H 1 .Q/.

9 This is a Dirac distribution along the segment x D 0, 0 < y < 1.


Problem 5.2.30 (Gluing functions inH 1). Let� � Rn be a regular bounded domain,divided into two regular domains �1, �2 with a regular interface � (hypersurface):

� D �1 [�2 [ �; �1 \�2 D �1 \ � D �2 \ � D ;:Take u1 2 H 1.�/; u2 2 H 1.�/. Under which conditions does the function

u.x/ D´u1.x/ x 2 �1u2.x/ x 2 �2

belong to H 1.�/ ?

Solution. As ui 2 L2.�i /, i D 1; 2 we immediately have

u 2 L2.�/

and so u 2 D 0.�/. If we want u 2 H 1 .�/, ru must by represented (as a distribution)by some vector w 2 L2 .�I Rn/, i.e.

hru;Fi DZ

w � F dx

for any vector field F 2 D.�I Rn/. Since ru1 and ru2 must be elements ofL2 .�1I Rn/and L2 .�2I Rn/ respectively, the candidate w should be defined by

w D´

ru1 in �1ru2 in �2

with the proviso that u should not “jump” when crossing � (see Problem 5.2.29). A nat-ural condition is to demand that the traces of u1 and u2 on � coincide. Let us check thisfact. By definition of distribution gradient

hru;Fi D �hu; div Fi D �Z

u div F dx D �Z 1

u1 div F dx �Z 2

u2 div F dx

(5.30)(� has measure zero, being regular). Let � denote the unit normal to � , pointing outwardwith respect to �1 (so that �� points outward with respect to �2). As F D 0 on @�, thedivergence theorem in H 1 tells thatZ

1

u1 div F dx DZ�

u1F � � d� �Z 1

ru1 � F dx (5.31)

and Z 2

u2 div F dx D �Z�

u2F � � d� �Z 2

ru2 � F dx: (5.32)


If u1 D u2 on � , we find, from (5.30), (5.31) and (5.32),

hru;Fi DZ

w � F dx:

The condition u1 D u2 is therefore sufficient (and also necessary10) for u 2 H 1 .�/.


5.3.1. Extend the functions of H D L2.0; 2�/ to the entire R in a periodic way. Consider thesubspace of 2�=3-periodic functions

V D²u 2 H W u

�t C 2

3�

D u.t/ for almost every t 2 R

³:

Check that V is closed in H , then determine V ? and the projections PV , PV? .

5.3.2. In H D L2.0; 1/ consider the closed subspace V of quadratic polynomials. ComputePV f when f .t/ D t3.

5.3.3. Set Q D .0; 1/ � .0; 1/;H D L2.Q/, and consider subspaces

K D ¹u 2 H W u.x; y/ D constantº;

V D´u 2 H W u.x; y/ D v.x/, v 2 L2.0; 1/;

Z 1

0v.x/dx D 0

μ;

W D´u 2 H W u.x; y/ D w.y/, w 2 L2.0; 1/;

Z 1

0w.y/dy D 0

μ:

Verify thatS D K ˚ V ˚W

is closed and determine the projections PS , PS? . Decompose the function f .x; y/ D xy accord-ingly.

5.3.4. Take H D L2.B1/ (B1 being the unit disc in R2) and consider the linear operator on H :

.Lu/ .x; y/ D u.y;�x/:a) Verify L is continuous and compute its norm.b) Find the possible (real) eigenvalues.

5.3.5. Let ¹xkº � R be a sequence going to C1. Prove that

C1XkD1

ckı.x � xk/

converges in D 0.R/ irrespective of the behaviour of the real sequence ¹ckº.

10 We leave this for the reader to check.


5.3.6. Let f : R ! R denote the 2�-periodic extension of

f0 .x/ D � � x2

0 < x < 2�:

a) Write the Fourier series of f and check that it converges in D 0 .R/ to f .b) Deduce the (remarkable) formula

1XnD1

cosnx D �12

C �

1XnD�1

ı .x � 2�n/ ; valid in D 0 .R/ : (5.33)

5.3.7. Give an example of sequence of test functions ¹vkº � D.R/ that converges to 0 in S.R/,but not in D.R/.

5.3.8. Prove that any polynomial is a tempered distribution, whilst ex 62 S 0.R/.

5.3.9. Show that the following distributions belong to S 0.R/:a) p: v:.1=x/.b) comb.x/.

5.3.10. a) Let Br � R3 be the ball with radius r and centre at the origin, and T 2 D 0 �R3� bedefined by

hT; vi DZ@Br

v d�

(a sort of ‘delta’ function spread over the spherical surface @Br ; it coincides with ı .jxj � r/).Prove that T is a tempered distribution and compute its Fourier transform.

b) Use part a) to find the fundamental solution to the equation

ut t ��u D 0 in R3:

5.3.11. Prove the following result: if there exist m and C such that

jcnj < C �1C jnjm�for any n 2 Z, then the distribution

F DC1XnD�1

cnı .x � n/

belongs to S 0.R/ 11.

5.3.12. Let combT be the distribution

combT .x/ DXn2Z

ı .x � nT / ;

known as the Dirac comb.

11 The condition is also necessary: if F is tempered, there exist m and C such that jcnj �C�1C jnjm�, for any integer n.


a) Prove that combT is T -periodic; using Problem 5.2.18 on page 295 (and the remark therein!)compute its Fourier series.

b) Deduce the formulas:

1combT D 2�

Tcomb2�=T ;

and (Poisson’s formula) Xn2Z

bv .nT / D 2�

T

Xn2Z

v

�2�n

T

for any v 2 S 0 .R/.

5.3.13. a) Take f 2 H1 .0; �/. Discuss the possibility of expanding f in a series of cosines, orsines, only.

b) Compute the best Poincaré constant CP , for which

kukL2.0;L/ � CP ku0kL2.0;L/ for any u 2 V;in the cases V D H1

0 .0; L/ and V D ¹u 2 H1.0; L/ W u.0/ D 0º.

5.3.14. Relying on Problem 5.2.6 (page 284) and the previous exercise, prove that the embeddingH1.0; �/ ,! L2.0; �/ is compact.

5.3.15. Let ˛ > �1 and consider the map

u.x; y/ D .x2 C y2/˛=2;

defined (almost everywhere) on the unit disc in R2. Prove that even if u is not C 1, its distributiongradient coincides with the usual gradient.

5.3.16. Let B1 be the unit disc centred at the origin in R2. Prove that

v.x; y/ D log

log

1C 1p

x2 C y2

!!� log .log 2/

belongs to H10 .B1/.

5.3.17. (Heisenberg’s Uncertainty Principle) Let 2 H1.R/ be such that xŒ .x/�2 ! 0 as

jxj ! 1, andR

R Œ .x/�2 dx D 1. Prove that

1 � 4

ZRx2 j .x/j2 dx

ZR

ˇ 0 .x/

ˇ2dx:

(Interpretation: if is Schrödinger’s wavefunction, the first integral above measures the total densityof a particle, the second one is the total momentum).

5.3.18. True of false?

a) If u 2 H1 .R/, then u .x/ ! 0 as x ! ˙1.b) If u 2 H1 .Rn/, n � 2, then u .x/ ! 0 as jxj ! C1.c) If � � Rn is an unbounded domain, Poincaré’s inequalityZ

v2 � CP

Z

jrvj2 8v 2 H10 .�/

does not hold.


5.3.19. Let � be bounded and Lipschitz. Prove that if a 2 L1 .�/, a � 0 a.e. andR a.x/ dx > 0, then

kuk? D�Z .jruj2 C a .x/ u2/ dx

1=2is equivalent to kukH1. / :

5.3.20. Suppose� is bounded and Lipschitz, and f W @� ! R is Lipschitz on @�with Lipschitzconstant L.

a) Prove thatQf .x/ D min

@ ¹f .y/C L jx � yjº

is Lipschitz in Rn and coincides with f on @�:

b) Deduce f 2 H1=2 .@�/ :

5.3.21. Solve Problem 5.2.25 on page 302 with H1.R/ replacing H1.�1; 1/.5.3.22. Let ı be the Dirac delta function at 0. Prove that ı belongs to H�1.�1; 1/. Determine

the representative element in H10 .�1; 1/ according to Riesz’s theorem.

5.3.23. Prove that the functional

Lu DZ 1

0u0.t/ dt

belongs to the dual of H1.0; 1/. What is the Riesz’s element representing it in H1.0; 1/?

5.3.24. a) Given f; g regular, consider the functional

Lv DZ 1

0


�dt:

Prove that L 2 H�1.0; 1/. Letting u 2 H10 .0; 1/ be the element representing it, write a

boundary-value problem that determines u uniquely.

b) Compute u explicitly if f D g D 1.

5.3.25. Set H D H10 .�1; 1/ and

V D ¹u 2 H W u.0/ D 0º:

Prove that V is closed in H , and compute the projection onto V of f , where f .t/ D 1 � t2 onŒ�1; 1�.

5.3.26. (Subspaces of H10 .�/) Let Rn � � D �1 [�2 be a domain, and take V D H1

0 .�/

with inner product .u; v/H1

0. /

D R ru � rv. Define

V1 D°u 2 H1

0 .�1/ , u D 0 in �n�1±

and V2 D°u 2 H1

0 .�2/ , u D 0 in �n�2±

.


Prove that V1 and V2 are closed subspaces in V , and

V ?1 \ V ?2 D ¹0º:

5.3.1 Solutions

Solution 5.3.1. V is a subspace and is closed, since L2 convergence implies convergence almost

everywhere of one subsequence. Take u 2 V ?. Then for any v 2 V , we can write

0 DZ 2�

0u.t/v.t/ dt D

Z 2�=3

0u.t/v.t/ dt C

Z 4�=3

2�=3u.t/v.t/ dt C

Z 2�

4�=3u.t/v.t/ dt:

Changing variable in the last two integrals and using the periodicity of v, we find

0 DZ 2�=3

0

u.t/C u

�t C 2

3�

C u

�t C 4

3�

�v.t/ dt:

This equation holds for any v 2 L2.0; 2�=3/, so we deduce that

u.t/C u

�t C 2

3�

C u

�t C 4

3�

D 0 a.e. in .0; 2�=3/ :

Integrating on .2�=3; 4�=3/ or .4�=3; 2�/, rather than on .0; 2�=3/, and repeating the above ar-gument, we finally deduce that

V ? D²u 2 L2.0; 2�/ W u.t/C u

�t C 2

3�

C u

�t C 4

3�

D 0 a.e. in .0; 2�/

³:

Hence

PV f .t/ D 1

3

f .t/C f

�t C 2

3�

C f

�t C 4

3�

�and

PV?f .t/ D 1

3

2f .t/ � f

�t C 2

3�

� f

�t C 4

3�

�:

Solution 5.3.2. We have to minimise the distance in L2 .0; 1/ between the cubic functionp .t/ D t3 and some polynomial of order two to be determined. The space

V D°at2 C bt C c W a; b; c 2 R

±is closed as finite-dimensional. So we have to minimiseZ 1

0

�t3 � at2 � bt � c

�2dt

as a; b; c vary12. By the orthogonal decomposition theorem, setting PV t3 D At2 C Bt C C and

g.t/ D t3 � PV t3, we haveZ 1

0g.t/.at2 C bt C c/ dt D 0 for any a; b; c:

12 Clearly, one could compute the integral directly and then minimise the resulting function of threevariables.


In particular, the above equation holds when two of the coefficients are zero and the third one is 1.This gives three equations

0 DZ 1

0g.t/ dt D

Z 1

0.t3 � At2 � Bt � C/ dt D 1

4� A

3� B

2� C

0 DZ 1

0g.t/t dt D

Z 1

0.t4 � At3 � Bt2 � Ct/ dt D 1

5� A

4� B

3� C

2

0 DZ 1

0g.t/t2 dt D

Z 1

0.t5 � At4 � Bt3 � Ct2/ dt D 1

6� A

5� B

4� C

3;

i.e. the system 8<:4AC 6B C 12C D 3

15AC 20B C 30C D 12

12AC 15B C 20C D 10:

With a little work we find A D 3=2, B D �3=5 and C D 1=20, whence

PV f D 3

2t2 � 3

5t C 1

20:

Solution 5.3.3. To prove S is closed we may argue as in Problem 5.2.3 (page 281) for inspiration.

We wish to characterise S?. If h 2 S? thenZQh.x; y/.v.x/C w.y/C k/ dxdy D 0

for any v, w and k. In particular, choosing w � 0, we deduce thatZ 1

0.v.x/C k/

"Z 1

0h.x; y/dy

#dx D 0:

Since every function u 2 L2.0; 1/ can be written in the form u.x/ D v.x/ C k with v 2 V andk D R 1

0 u.x/ dx, we infer that Z 1

0h.x; y/ dy D 0 a.e. in .0; 1/:

Analgously, we deduce that Z 1

0h.x; y/ dx D 0 a.e. in .0; 1/:

We conclude that

S? D´h 2 L2.Q/ W

Z 1

0h.x; y/ dx D

Z 1

0h.x; y/ dy D 0 a.e. in .0; 1/

μ:

Set

PSf .x; y/ D vf .x/C wf .y/C kf :


Let us determine vf . For any g 2 L2.0; 1/ we have, since f � PSf 2 S?,

0 DZQ

�f .x; y/ � .vf .x/C wf .y/C kf /

�g.x/ dxdy D

DZ 1

0

"Z 1

0f .x; y/ dy � vf .x/ � kf

#g.x/ dx

(recallR 10 wf .y/ dy D 0), so thatZ 1

0f .x; y/ dy � vf .x/ � kf D 0

and

vf .x/ DZ 1

0f .x; y/ dy � kf

almost everywhere. In the same way, for any g 2 L2.0; 1/ we have

0 DZQ

�f .x; y/ � .vf .x/C wf .y/C kf /

�g.y/ dxdy D

DZ 1

0

"Z 1

0f .x; y/ dx � wf .y/ � kf

#g.y/ dy;

and therefore

wf .y/ DZ 1

0f .x; y/ dx � kf :

Finally,

0 DZQ

�f .x; y/ � .vf .x/C wf .y/C kf /

�dxdy D

ZQf .x; y/ dxdy � kf ;

and

kf DZQf .x; y/ dxdy:

To sum up,

PSf .x; y/ DZ 1

0f .x; y/ dy C

Z 1

0f .x; y/ dx �

ZQf .x; y/ dxdy;

and PS?f D f � PSf . In particular,

xy D 2x C 2y � 14„ ƒ‚ …2S

C 4xy � 2x � 2y C 1

4„ ƒ‚ …2S?

:

Solution 5.3.4. a) L is linear, so we have to prove it is bounded. As

kLuk2H DZB1

.Lu/2.x; y/ dxdy DZB1

u2.y;�x/ dxdy D kuk2H ;

L is continuous and kLk D 1.


b) Let � 2 R, v 6� 0 be such thatLv D �v:

Taking norms and using part a) we obtain j�j D 1, so the only possible eigenvalues are � D ˙1.Now we have to find non-zero functions vC and v� satisfying

vC.y;�x/ D vC.x; y/ or v�.y;�x/ D �v�.x; y/:For instance: vC.x; y/ D x2 C y2 and v�.x; y/ D xy show that ˙1 do occur as eigenvalues.

Solution 5.3.5. Take ¹ckº � R. We have to show that for any v 2 D .R/, the series

1XkD1

ckhı .x � xk/ ; vi (5.34)

converges. Notice that the support of v is compact (bounded, in particular) and xk ! C1. Hence,if we take a large enough k0, for any k > k0 the points xk will not belong to the support of v,and therefore v .xk/ D 0 for k > k0. But then the series in practice reduces to a finite number ofsummands, and its sum is obviously finite.

Solution 5.3.6. a) As f is odd we can expand in sine series

f 1XnD1

sinnx

n:

The equality is understood in L2loc .R/, and therefore in D 0 .R/.b) Let us denote by �A the characteristic function of the set A:

�A.x/ D´1 if x 2 A0 if x 62 A:

Due to the periodicity, f can be written as

f .x/ D1X

nD�1f .x/�Œ2n�;2.nC1/�/.x/ D

1XnD�1

f .x � 2n�/�Œ2n�;2.nC1/�/.x/

D1X

nD�1� � .x � 2n�/

2�Œ2n�;2.nC1/�/.x/

and so 1XnD1

sinnx

nD

1XnD�1

� � .x � 2n�/2

�Œ2n�;2.nC1/�/.x/; (5.35)

a relation valid in D 0 .R/. Now, series that converge in D 0 .R/ can be differentiated term by term, togive another convergent series in D 0 .R/. Then (5.33) follows from (5.35), by differentiating eachterm and making use of the relation

d

dxg.x/�Œa;b/.x/ D �g.b�/ı.x � b/C g.aC/ı.x � a/C g0.x/�Œa;b/.x/;

valid for any g 2 C 1.a; b/ with finite limits for x ! aC and x ! b� (prove this last fact).


Solution 5.3.7. By comparing the two definitions of convergence one sees that convergence in S

does not require any assumption on supp.vk/. One example is the following: fix a test function vequal zero outside Œ0; 1�, and set

vk .x/ D v .x � k/ e�x :For any k � 1, vk is a test function that vanishes outside Œk; k C 1�. In particular, the support of vkis not contained in any given compact set, and vk does not converge in D.R/. On the other handfor any m;p � 0

supR

ˇxmDpvk .x/

ˇ ! 0 as k ! C1(prove it), so vk ! 0 in S .R/.

Solution 5.3.8. It suffices to show that any monomial that has the form P .x/ D x˛1

1 x˛2 � � � x˛nn

is a tempered distribution. Letm � 0 denote its degree and ¹vkº � D .Rn/ be such that vk ! 0 inS .Rn/. Notice that

jP .x/j D ˇx˛1

1 x˛2 � � � x˛nn

ˇ � jxjmand that the function

h .x/ D jxjm .1C jxj/�.mCnC1/is integrable in Rn. In fact if !n denotes the measure of the surface of the unit sphere in Rn,Z

Rn

jxjm.1C jxj/mCnC1 dx D !n

Z C1

0

�mCn�1.1C �/mCnC1

d� D M < 1:

By definition of convergence in S .Rn/,

supRn

h.1C jxj/mCnC1 jvk .x/j

i! 0 as k ! C1:

Therefore

jhP; vkij DˇZ

RnP .x/ vk .x/ dx

ˇ�Z

Rn

jxjm.1C jxj/mCnC1 .1C jxj/mCnC1 jvk .x/j dx

� M supRn

h.1C jxj/mCnC1 jvk .x/j

i! 0;

so P defines a tempered distribution.Now set u .x/ D ex and take v 2 D.R/ non-negative, equal to e�x on Œ�1; 1� and zero outside

Œ�2; 2�. Considervk .x/ D v .x � k/ e�x :

We argue as in Exercise 5.3.7 to find that vk ! 0 in S .R/. Yet

hu; vki DZ kC2

k�2v .x � k/ dx �

Z kC1

k�1dx D 2 6! 0:

In conclusion, u is not a tempered distribution.

Solution 5.3.9. a) Let us show that F 2 S 0 .Rn/ implies F 0 2 S 0 .Rn/. If ¹vkº � D .Rn/ andvk ! 0 in S .Rn/, then also v0

k! 0 in S .Rn/. Therefore

hF 0; vki D �hF; v0ki ! 0


because F is tempered. By virtue of Problem 5.2.19 on page 296, u .x/ D log jxj is tempered, andhence its derivative is given by p: v:.1=x/.

b) The Dirac comb is a tempered distribution. To see it, let ¹vkº � D .R/ be such that vk ! 0

in S .R/. Then

hcomb; vki DC1XnD�1

vk .n/ :

Convergence in S .R/ guarantees that n2vk .n/ ! 0 uniformly in n when k ! 1. Given " > 0

we can then writen2 jvk .n/j < "

for k � k0 large enough. Therefore when k � k0

jhcomb; vkij � "

C1XnD�1

1

n2� C"

and so hcomb; vki ! 0.

Solution 5.3.10. a) T is tempered because its support @Br is compact. By definition, given anyv 2 S

�R3�,

hbT ; vi D hT;bvi DZ@Br

bv d� DZ@Br

d�

ZR3e�ix�� v .x/ dx D

DZ

R3v .x/

�Z@Br

e�ix�� d�dx:

Hence bT .�/ DZ@Br

e�i�� d�:

To deal with the integral we pass to spherical coordinates .r; �; /with vertical axis coinciding with�, 0 < � < 2� , 0 < < � . Set j�j D � and observe that on @Br

� � � D r� cos

and d� D r2 sin d , soZ@Br

e�i��d� D 2�r2Z �

0e�ir cos sin d D 4�r

r

�sin r�:

In conclusion bT .�/ D 4�rsin r j�j

j�j :

b) The fundamental solution of the wave equation in dimension n D 3 is the solution of theproblem ´

ut t ��u D 0 in R3 � .0;C1/

u.x; 0/ D 0; ut .x; 0/ D ı.x � y/ in R3:

Since the equation is translation-invariant we can assume y D 0. We set

Ou.�; t / DZ

R3e�i��xu.x; t / dx;


the Fourier transform of u in the spatial variables. We find that Ou solves´Out t C j�j2 Ou D 0 t > 0

Ou.�; 0/ D 0; Out .�; 0/ D 1;

for any � 2 R3. The general solution is Ou.�; t / D C1ei j�jt C C2e

�i j�jt . The initial conditionimpose

C1 C C2 D 0 and C1i j�j � C2i j�j D 1;

so

Ou.�; t / D 1

2i j�j�ei j�jt � e�i j�jt

�D sin .j�jt /

j�j :

At this stage we can use part a) (with r D t ) and get

u.x; t / D ı.jxj � t /4�t

so, in the end,

K.x; y; t / D ı.jx � yj � t /4�t

:

Solution 5.3.11. Consider

F DX

cnı .x � n/ ; where cn <�1C jnjm� ;

and a sequence ¹vkº � D .R/ such that vk ! 0 in S .R/. Then

hF; vki DC1XnD�1

cnvk .n/ :

By definition of convergence in S .R/ we have

nmC2vk .n/ ! 0

uniformly in n, as k ! 1. Given " > 0, we write

nmC2 jvk .n/j < "

for some large k � k0. When k � k0, then,

jhF; vkij � "

C1XnD�1

.1C nm/

nmC2 � C"

and hF; vki ! 0.

Solution 5.3.12. a) Let us check combT is T -periodic. Take a test function v and compute:

hcombT .x C T / ; vi D hcombT ; v .x � T /i DXn2Z

v .nT � T / DXn2Z

v .nT / D hcombT ; vi:


By Problem 5.2.18 on page 295, the Fourier series of u D combT is given in D 0 .R/ by the formula:

combT .x/ DXn2Z

bun exp

��i 2n�x

T

:

To computebun we use the remark on page 296: the point x0 D �T=2 does not belong to the supportof combT , and the restriction of combT to

.x0; x0 C T / D .�T=2; T=2/coincides with ı .x/. By (5.19), with u1 D 0,

bun D 1

Thı; e�i2n�x=T i D 1

T

whence the Fourier series reads

combT .x/ D 1

T

Xn2Z

exp

��i 2n�x

T

:

b) From (5.18), sincebun D cn=2� , we have

1combT .�/ D 2�

T

Xn2Z

ı

�� 2n�

T

D 2�

Tcomb2�=T .�/ .

If now v 2 S 0 .R/:h 1combT ; vi D hcombT ;bvi D

Xn2Z

bv .nT /while

2�

Thcomb2�=T ; vi D 2�

T

Xn2Z

v

�2�n

T

;

giving the Poisson formula.

Solution 5.3.13. As f 0 2 L2 .0; �/, by considering the even and odd extensions to .��; �/ wecan write, in the two cases

f 0 .x/ D1XnD0

an cosnx and f 0 .x/ D1XnD1

bn sin nx

respectively, where convergence is in L2 .0; �/. Note13

a0 D 1

�

Z �

0f 0 D f .�/ � f .0/

�:

Since Fourier series can be integrated term by term, in the first case we have:

f .x/ D f .0/C1XnD0

Z x

0an cosnx dx D f .0/C f .�/ � f .0/

�x C

1XnD1

An sin nx

13 The functions of H1 .a; b/ are absolutely continuous, so the Fundamental Theorem of Calculusholds.


with An D an=n. Assumingf .�/ D f .0/ D 0;

it follows

f .x/ D1XnD1

An sin nx

with the series converging uniformly on the entire R.In the other case, term-by-term integration shows that

f .x/ D f .0/C ˇ C1XnD1

Bn cosnx

where

Bn D �bnn

and ˇ D1XnD1

bn

n:

Each function inH1 .0; �/ can therefore be expanded in cosines, and the series converges uniformlyon the whole R.

b) Take u 2 H10 .0; L/. Repeating the steps of part a) for 2L-periodic functions we obtain

u D1XnD1

An sin�n�Lx�; u0 D

1XnD1

an cos�n�Lx�;

with an D n�

LAn. Parseval’s identity gives

kuk2L2.0;L/

D1XnD1

A2n and ku0k2L2.0;L/

D1XnD1

a2n D1XnD1

�n�L

�2A2n �

��L

�2 1XnD1

A2n:

Hence

kukL2.0;L/ � L

�ku0kL2.0;L/ for any u 2 H1

0 .0; L/:

Note that if A1 D 1 and An D 0 when n � 2, that is u D sin �Lx, the above inequalities become

equalities, so the constant CP D L=� is optimal.Let now

V D ¹u 2 H1.0; L/ W u.0/ D 0º:It is easy to see14 that

Qu.x/ D´u.x/ 0 � x � L

u.L � x/ L � x � 2L

belongs to H10 .0; 2L/, and that

k QukL2.0;L/ D p2kukL2.0;L/, k Qu0kL2.0;2L/ D p

2ku0kL2.0;2L/:

Previous arguments show

k QukL2.0;2L/ � 2L

�k Qu0kL2.0;2L/;

14 This fact is left to the reader.


so that

kukL2.0;L/ � 2L

�ku0kL2.0;L/ for any u 2 V:

Solution 5.3.14. We have to prove that the unit ball in H1.0; �/;

B D°u 2 H1.0; �/ W ku0k22 C kuk22 � 1

±;

is relatively compact in L2.0; �/. To do that we recall, from Exercise 5.3.13, that we can expand uin cosine Fourier series:

u DC1XnD0

B0 cosnt

where

B0 D 1

�

Z �

0u.t/ cos nt dt and Bn D 2

�

Z �

0u.t/ cosnt dt;

and the series converges uniformly in R. Parseval’s identity gives

kuk22 D 2B20 CC1XnD0

B2n :

The map that associates to a function u the sequence of its Fourier coefficients Bn is an isomor-phism between the vector spaces L2.0; 2�/ and l2; it is also an isometry of the underlying metricstructures. Therefore the claim follows provided we prove that B can be identified with a subspaceof l2, in analogy to the space K of Problem 5.2.6 (page 284).

From Exercise 5.3.13 we know that Bn D �bn=n, where the bn, n � 1, are the Fourier coeffi-cients of the sine series of u0. What is more, Parseval’s identity gives

ku0k22 DC1XnD1

n2B2n :

In terms of Fourier series, then,

ku0k22 C kuk22 � 1 () 2B20 CC1XnD1

B2n CC1XnD1

n2B2n � 1:

The ball B � H1 .0; �/ can be identified with the subset of l2

K D´.Bn/ :

C1XnD0

B2n CC1XnD1

n2B2n � 1

μ:

(the factor 2 in front of B0 is irrelevant). By Problem 5.2.6 the proof is complete.

Solution 5.3.15. As˛ > �1we have already seen (Problem 5.2.21 on page 298) thatu 2 L2.B1/,so u 2 D 0.B1/. Let us call v the partial derivative of u with respect to x in distributional sense, andkeep the notation ux for the ordinary derivative (defined outside the origin). We compute v. Taking


' 2 D.B1/, we have

hv; 'i D �hu; 'xi D �ZB1

u.x; y/'x.x; y/ dxdy D �ZB"

u'x dxdy �ZB1nB"

u'x dxdy D

D �ZB"

u'x dxdy �Z@B"

u' x ds CZB1nB"

ux' dxdy;

where x denotes the h-component of the unit normal to @B", pointing away from B1 n B" (recall' D 0 on @B1). Now,ˇZ

B"

u'x dxdy

ˇ� C

ZB"

juj dxdy D C 0Z "

0r˛C1 dr D C 00"˛C2 ! 0;

and ˇZ@B"

u' x ds

ˇ� C

Z@B"

juj dxdy D C 0Z 2�

0"˛ d� D C 00"˛C1 ! 0

as " ! 0, since ˛ > �1. Overall,

hv; 'i DZB1

ux' dxdy:

The same argument for uy will then lead to the claim.

Solution 5.3.16. a) First remember thatZ 1

2ta logb t dt

is finite if and only if either a < �1, or a D 1 and b < �1.Exactly as in the previous exercise one sees that the distributional derivatives of v coincide with

the classical ones, so that

rv.x; y/ D

log

1C 1p

x2 C y2

!!�1� 1C 1p

x2 C y2

!�1� .2x; 2y/

.x2 C y2/3=2:

In polar coordinatesZB1

v2.x; y/ dxdy D 2�

Z 1

0

hlog log .1C r�1/

i2rdr < C1

(the integrand is finite on a neighbourhood of the origin) andZB1

jrv.x; y/j2 dxdy D 8�

Z 1

0

1

log2 .1C r�1/ � 1

r.1C r/2dr:

Setting t D 1C r�1, we find:

1

8�

ZB1

jrv.x; y/j2 dxdy DZ 1

2

1

log2 t� t � 1t2

dt < 1:

But v.x; y/ D 0 on @B1, so v 2 H10 .B1/.


Solution 5.3.17. We can write

0 D x 2.x/ˇC1�1 D

ZR

d

dxŒx 2.x/� dx D

ZR

j .x/j2 dx C 2

ZRx .x/ 0.x/ dx;

from which

1 D �2Z

Rx .x/ 0.x/ dx � 2

�ZRx2j .x/j2 dx

1=2:

�ZR

j 0.x/j2 dx1=2

:

Squaring the latter gives the required inequality.

Solution 5.3.18. a) True. First, we prove that the limit exists. In fact

u2 .x/ � u2 .0/ D 2

Z x

0u .s/ u0 .s/ ds;

and the limit exists since u � u0 2 L1.R/. Then

limx!C1u2 .x/ D l � 0:

We claim that l D 0. If, on the contrary, l were positive, given " > 0 such that l � " > 0 we wouldhave

u2 .x/ > l � "for x > N D N ."/, and thenZ

Ru2 .x/ dx �

Z C1

Nu2 .x/ dx >

Z C1

N.l � "/dx D C1;

contradicting the fact that u 2 L2 .R/.b) False, in general, for any n > 1. For instance, let n D 3. In the ball BR .x0/ consider the

radial function

uR .x/ D

8<:1 if jx � x0j < R

2R

jx � x0j � 1 ifR

2� jx � x0j � R:

Notice that uR .x/ D 0 if jx � x0j D R, while uR .x/ D 1 for jx � x0j D R=2. ThereforeuR is continuous in BR .x0/ and vanishes on the boundary. Moreover 0 � uR � 1, and forR=2 � jx � x0j � R,

jruR .x/j D R

jx � x0j2 � 2

R.

Now we will construct a function u 2 H1�R3�

without limit for jxj ! C1. Let us choose, forany integer k � 1,

xk D .k; 0; 0/ and Rk D 1

k2

and define

u .x/ D´uRk

.x/ in BRk.xk/ , k � 1

0 if x … [k�1BRk.xk/


where

uRk.x/ D

8<:1 if jx � xk j < Rk

2Rk

jx � xk j � 1 ifRk

2� jx � xk j � Rk :

As 0 � uRk� 1,Z

R3u2 D

Xk�1

ZBRk

.xk/u2Rk

<4�

3

Xk�1

R3k D 4�

3

Xk�1

1

k6< 1.

Additionally,ˇruRk

ˇ � 2

Rk;

ZR3

jruj2 DXk�1

ZBRk

.xk/

ˇruRk

ˇ2<8�

3

Xk�1

Rk D 2Xk�1

1

k2< 1.

Hence u lives in H1�R3�

and is bounded; yet jxk j D k ! C1 and u .xk/ D 1 ¹ 0, while ifyk D .0; 0; k/ then jyk j D k ! C1 and u .yk/ D 0. The limit of u .x/ when jxj ! 1 does notexist.

c) False. To have an inequality of Poincaré type it is enough that the domain is bounded in onedirection only. When n > 1, for example, we may consider the strip

� D°�

x0; xn� 2 Rn: x0 2 Rn�1; 0 < xn < d

±.

Take v 2 D .�/. As v�x0; 0

� D 0,

v2�x0; xn

� D�Z xn

0vxn

�x0; s

�ds

2� xn

Z xn

0v2xn

�x0; s

�ds � xn

Z d

0v2xn

�x0; s

�ds

(using the Cauchy-Schwarz inequality). Integrating in xn over .0; d/ and in x0 over Rn�1 givesZ v2 � d2

2

Z v2xn

� d2

2

Z

jrvj2 :

A density argument allows to extend the inequality to v 2 H10 .�/.

Solution 5.3.19. We have to prove there are positive constants C1 < C2 such that

C1kukH1. / � kuk? � C2kukH1. /: (5.36)

On the one handZ .jruj2 C a .x/ u2/ dx �

Z

jruj2 dx C kakL1. /

Z u2 dx;

and the second inequality in (5.36) holds with C2 Dq

max�1; kakL1. /

�.

To prove the other inequality, it is enough to find a constant C such thatZ u2 dx � C

Z .jruj2 C a .x/ u2/ dx:


Suppose not. Then for every n 2 N there exists un 2 H1.�/, un 6� 0, such that

1

n

Z u2n dx �

Z .jrunj2 C a .x/ u2n/ dx:

We may suppose kunkL2. / D 1. Therefore, since a � 0,Z

jrunj2 ! 0;

Z a .x/ u2n dx ! 0;

Z u2n dx ! 1: (5.37)

On the other hand, un is bounded in H1.�/, thus un * Nu in H1.�/ and, by Rellich’s theorem,un ! Nu in L2.�/ strongly (up to extracting a subsequence). Using the weak semicontinuity of thenorm, we infer Z

jr Nuj2 � lim inf

Z

jrunj2 D 0;

Z

Nu2 D limZ u2n D 1;

and Nu is a non-zero constant. We claim thatZ a .x/ u2n dx !

Z a .x/ Nu2 dx D Nu2

Z a .x/ dx > 0:

This would contradict (5.37), hence concluding the proof. The claim follows from the Cauchy-Schwarz inequality, which impliesˇZ

a .x/ u2n dx �

Z a .x/ Nu2 dx

ˇ�Z a .x/

ˇu2n � Nu2

ˇdx

� kakL1. /kun C NukL2. /kun � NukL2. /

� 2kakL1. /kun � NukL2. / ! 0;

since un ! Nu in L2.�/.

Solution 5.3.20. a) Let x1; x2 2 Rn, and y2 2 @� be such that

Qf .x2/ D f .y2/C Ljx2 � y2j:By definition, Qf .x1/ � f .y2/C Ljx1 � y2j, therefore

Qf .x1/ � Qf .x2/ � L .jx1 � y2j � jx2 � y2j/ � Ljx1 � x2jIchanging the role of x1 and x2 one obtains that Qf is Lipschitz on Rn.

Now take x 2 @�. As f is Lipschitz, for y 2 @�f .x/ � f .y/C Ljx � yj;

and so f .x/ � Qf .x/. At the same time, choosing y D x in Qf gives Qf .x/ � f .x/ right away. Hencef � Qf j@ .

b) Qf is Lipschitz, so Qf j 2 H1.�/. But then its trace on @�, i.e. f , belongs to H1=2.�/.

Solution 5.3.21. If we follow Problem 5.2.25 (page 302) we must seek u 2 H1.R/ such thatZR


�dt D v.0/ for any v 2 H1.R/:


Supposing u is regular enough to integrate by parts on .�1; 0/ and .0;1/, we may exploit Exer-cise 5.3.18, a), to rewrite the above expression as

�Z

R

�u00.t/ � u.t/� v.t/ dt C v .0/

hu0 .0�/ � u0

�0C�i

D v .0/ :

Let us choose first v zero at 0, and then an arbitrary v; we find that u must solveú00.t/ � u0.t/ D 0 t ¤ 0

limjxj!1 u.x/ D 0; u0.0�/ � u0.0C/ D 1

(we have used Exercise 5.3.18, a) again). The solution has the form

u.t/ DÁ1e

t C A2e�t t � 0

B1et C B2e

�t t � 0;

with 8<:A2 D B1 D 0

A1 � A2 � B1 C B2 D 1

A1 C A2 D B1 C B2:

Hence the candidate for representing ı in H1.R/ is

u.t/ D

8<:1

2e�t t � 0

1

2et t � 0:

The reader should check that the above function is indeed the function required.

Solution 5.3.22. The fact that ı is a linear, continuous functional on H10 .�1; 1/ follows from

arguments akin to those of Problem 5.2.25 on page 302. By Riesz’s theorem there exists u 2H10 .�1; 1/ such that Z 1

�1u0.t/v0.t/ dt D v.0/ for any v 2 H1

0 .�1; 1/:

To be able to integrate by parts we suppose u is regular on Œ�1; 0� and Œ0; 1�. ThenZ 1

�1u0.t/v0.t/ dt D

Z 0

�1u0.t/v0.t/ dt C

Z 1

0u0.t/v0.t/ dt D

D .u0.0�/ � u0.0C//v.0/ �Z 0

�1u00.t/v.t/ dt �

Z 1

0u00.t/v.t/ dt:

As v is arbitrary, we have for u the following:ú00.t/ D 0 when � 1 < t < 0 and 0 < t < 1;

u.�1/ D u.1/ D 0; u0.0�/ � u0.0C/ D 1:

Then

u.t/ DÁ1 C A2t �1 � t � 0

B1 C B2t 0 � t � 1:


The boundary conditions together with the continuity at 0 force8<ˆ:A1 � A2 D 0

B1 C B2 D 0

A2 � B2 D 1

A1 D B1;

so A1 D A2 D B1 D 1=2, B2 D �1=2 and ultimately

u.t/ D 1

2.1 � jt j/ :

Solution 5.3.23. We are in the same situation of Problem 5.2.27 (page 304), with f � 1 andg � 0. The function u that should represent L will solve´

u00.t/ � u.t/ D 0 0 < x < 1

u0.0/ D u0.1/ D 1;that is, u.t/ D 1

1C e

�et C e1�t

�:

Solution 5.3.24. a) As in Problem 5.2.27 on page 304, it is easy to prove L is linear and contin-uous. We seek u 2 H1

0 .0; 1/ so thatZ 1

0u0.t/v0.t/ dt D

Z 1

0


�dt; for any v 2 H1

0 .0; 1/:

Assume u regular enough. ThenZ 1

0u0.t/v0.t/ dt D u0.1/v.1/ � u0.0/v.0/ �

Z 1

0u00.t/v.t/ dt D �

Z 1

0u00.t/v.t/ dt

and Z 1

0f .t/v0.t/ dt D f .1/v.1/ � f .0/v.0/ �

Z 1

0f 0.t/v.t/ dt D �

Z 1

0f 0.t/v.t/ dt:

Therefore Z 1

0

��u00.t/C f 0.t/ � g.t/� v.t/ dt D 0 for any v 2 H10 .0; 1/;

which implies that the Riesz’s element u is the solution of the problem´u00.t/ D f 0.t/ � g.t/u.0/ D u.1/ D 0:

b) The problem becomes´u00.t/ D �1u.0/ D u.1/ D 0;

with solution u.t/ D 1

2.t � t2/:

Solution 5.3.25. According to Problem 5.2.28 on page 306 the space V is closed in H . To findPV f we must minimise over V the quadratic functional

E.u/ D kf � uk2H DZ 1

�1.2t � u0.t//2 dt:


The minimum u D PV f must satisfy the necessary condition

.f � u; v/H1

0.�1;1/ D

Z 1

�1.2t � u0.t//v0.t/ dt D 0 for any v 2 H1

0 .�1; 1/:

Assume we can integrate by parts:Z 1

�1.2t � u0.t//v0.t/ dt D �

Z 1

�1.2 � u00.t//v.t/ dt D 0

for any v 2 H . The candidate projection u solves´u00.t/ D 2 for � 1 < t < 0; 0 < t < 1u.�1/ D u.1/ D 0; u.0/ D 0:

Therefore

u.t/ D´t2 C A1t C A2 �1 � t � 0

t2 C B1t C B2 0 � t � 1;

where 8<:1 � A1 C A2 D 0

1C B1 C B2 D 0

A2 D B2 D 0:

A direct computation yieldsu.t/ D t2 � jt j:

Solution 5.3.26. If u 2 H10 .�1/, the zero extension of u on �n�1 belongs to H1

0 .�/ (Prob-lem 5.2.30 on page 309), making V1 a subspace of V . To check its closure take vn 2 V1 convergingto v0 2 V . In particular, ¹vnº is a Cauchy sequence in H1

0 .�1/ and hence vn ! v0 in H10 .�1/.

Let v0 be the zero extension of v0 on �n�1. Then v0 2 V1 and

vn ! v0

in V . Consequently v0 D v0 2 V1, so that V1 is closed. The same reasoning goes through for V2.Proving V ?1 \ V ?2 D ¹0º amounts to showing15 that V1 C V2 is dense in V , which in turn is a

consequence ofD .�1/C D .�2/ D D .�/ :

So we shall prove the latter. Take any v 2 D .�/; we aim to construct v1 2 D .�1/ andv2 2 D .�2/ such that v D v1 C v2. Call K the support of v. Then

dist .K; @�/ D d > 0:

For j D 1; 2 define

Aı=2j D

²x 2 �j W dist

�x; @�j

�>ı

2

³;

in such a way thatK � A1 [ A2:

The functionsv1 D v

w1

w1 C w2and v2 D v

w2

w1 C w2

15 Prove it.


will do the job, provided we find functions w1 2 D .�1/ and w2 2 D .�2/ such that w1 D w2and w1 C w2 > 0 in K. Let us concentrate on w1; the argument is completely similar for w2. Callz D � .A1/ the characteristic function of Ad=21 and take the mollifier16

� .x/ D

8<:c exp

1

jxj2 � 1

!0 � jxj < 1

0 jxj � 1

where c D �RRn �

��1. Set

�" .x/ D "�n��x"

�and then take

w1 D �" � z:For " D d=2 the function w1 has the desired properties.

16 [18, Chap. 7, Sect. 2].

6

Variational Formulations

6.1 Backgrounds

We start by recalling the basic facts about variational problems.

• Bilinear forms and coercivity. Let V be a Hilbert space. A map

B W V � V ! R

is called a bilinear form if it is linear in either argument. The form

B�.u; v/ D B.v; u/

is called adjoint form of B . B is:

! Self-adjoint (or symmetric) if

B.v; u/ D B.u; v/:

! Continuous if and only if it is bounded, i.e. there is a constant M > 0 such that

jB.u; v/j � MkukV kvkV :

! Coercive if there is a constant a > 0 such that

B.u; u/ � akuk2V ;

for every u, v 2 V .Let

V ,! H D H� ,! V �

be a Hilbert triplet: V is continuously embedded in H , dense in H; H is identified withits dual H�, and it is dense and continuously embedded in V �,


334 6 Variational Formulations

The form B is weakly coercive with respect to the Hilbert triplet ¹V;H; V �º if thereexist �0 2 R and a > 0 such that

B.u; u/C �0kuk2H � akuk2Vfor any u 2 V .

• Abstract variational problem (stationary case): determine u 2 V such that

B .u; v/ D Fv; 8v 2 V (6.1)

where B is a bilinear form on V and F 2 V �:

Theorem 6.1 (Lax-Milgram). If B is continuous and coercive with coercivity constanta, there exists a unique solution Nu to problem (6.1), and the following stability estimateholds:

k NukV � 1

akF kV � :

Moreover, if B is self-adjoint, then Nu is the unique minimiser of the ‘energy’ functional

E.v/ D 1

2B.v; v/ � Fv; for v 2 V:

Theorem 6.2 (Fredholm alternative). Let B be a weakly coercive bilinear form on Vw.r.t. the Hilbert triplet ¹V;H; V �º, with V , H separable and V compactly embedded inH , and let F 2 V �.

Suppose N .B/, N .B�/ denote the solution subspaces (eigenspaces) of the homoge-neous problems

B .u; v/ D 0; 8v 2 V and B� .w; v/ D 0; 8v 2 V:Then

a) dim N .B/ D dim N .B�/ D d < 1.b) Problem (6.1) has a solution if and only if Fw D 0 for any w 2 N .B�/.

• Dirichlet eigenvalues of the operator ��. Let � � Rn be a bounded Lipschitzdomain and consider the eigenvalue problem

��u D �u; u 2 H 10 .�/:

There exists a sequence of numbers

0 < �1 < �2 � �3 � : : : ;

�k ! C1, for which the problem has non-trivial solutions if and only if � D �k . The cor-responding eigenfunctions (suitably normalized) form an orthonormal system in L2.�/and an orthogonal system in H 1

0 .�/. In particular, the first eigenvalue �1 is simple, i.e.the first eigenspace has the form

¹t'1 W t 2 Rºand '1 can be chosen strictly positive on �.

Similar results hold for the other boundary conditions.

6.1 Backgrounds 335

• Abstract variational problem (first-order evolution case): let ¹V;H; V �º be a Hilberttriplet with V , H separable and B .�; �; t / a bilinear form on V (no assumption on the t -dependence is needed, except for measurability). Find a function t 7! u.t/ such that

u 2 L2 .0; T IV / ; Pu 2 L2 �0; T IV ��and 8<:

d

dt.u .t/ ; v/H C B .u .t/ ; v; t/ D .f .t/ ; v/H 8v 2 V;

u .0/ D g

for almost every t in .0; T /, in distributional sense on .0; T / :

Theorem 6.3. If f 2 L2 .0; T IH/, g 2 H and B is continuous and weakly coercive,uniformly1 in t , the variational problem has a unique solution.

• Abstract variational problem (second-order evolution case): given a Hilbert triplet¹V;H; V �º with V , H separable and B .�; �/ bilinear, symmetric on V , find a functiont 7! u.t/ such that

u 2 L2 .0; T IV / ; Pu 2 L2 .0; T IH/ ; Ru 2 L2 �0; T IV ��and 8<

ˆ:d2

dt2.u .t/ ; v/H C B .u .t/ ; v/ D .f .t/ ; v/H 8v 2 V;

u .0/ D g

Pu .0/ D h

for almost every t 2 .0; T /, in distributional sense on .0; T /.

Theorem 6.4. If f 2 L2 .0; T IH/, g 2 V , h 2 H and B is continuous, self-adjoint andweakly coercive, the variational problem has a unique solution.

A typical Hilbert triplet is given by ¹V;H; V �º where

H 10 .�/ � V � H 1 .�/ ;

and H D L2 .�/.

1 The continuity constant M and the coercivity constant a do not depend on t .


6.2 Solved Problems

� 6:2:1 � 6:2:6 W One-dimensional problems.� 6:2:7 � 6:2:17 W Elliptic problems.� 6:2:18 � 6:2:24 W Evolution problems.

6.2.1 One-dimensional problems

Problem 6.2.1 (Dirichlet conditions). Write the variational formulation of the prob-lem: ´

.x2 C 1/u00 � xu0 D sin 2�x 0 < x < 1

u.0/ D u.1/ D 0:

Show that it has a unique solution u 2 H 10 .0; 1/ and find a constant C for which

ku0kL2.0;1/ � C:

Solution. As the Dirichlet conditions are homogeneous, we choose test functions inH 10 .0; 1/ (which is the closure, for the usual norm, of C 10 .0; 1/). Multiply the equation by

v 2 H 10 .0; 1/ and integrate. We find:Z 1

0

�.x2 C 1/u00.x/ � xu0.x/� v.x/ dx D

Z 1

0

sin .2�x/v.x/ dx: (6.2)

Integrating by parts we may writeZ 1

0

.x2 C 1/u00.x/v.x/ dx

D �.x2 C 1/u0.x/v.x/

�10

�Z 1

0

u0.x/d

dx

�.x2 C 1/v.x/

�dx D

D �Z 1

0

�.x2 C 1/u0.x/v0.x/C 2xu0.x/v.x/

�dx

and then substituting into (6.2),Z 1

0

�.x2 C 1/u0.x/v0.x/C 3xu0.x/v.x/

�dx D �

Z 1

0

sin .2�x/v.x/ dx:

Setting

B.u; v/ DZ 1

0

�.x2 C 1/u0.x/v0.x/C 3xu0.x/v.x/

�dx;

F v D �Z 1

0

sin .2�x/v.x/ dx;


we obtain the following variational formulation: find u 2 H 10 .0; 1/ such that

B.u; v/ D Fv for any v 2 H 10 .0; 1/: (6.3)

Note that if u 2 C 2, v 2 C 1 satisfy (6.3), we can repeat the integration the other wayaround to obtainZ 1

0

�.x2 C 1/u00.x/ � xu0.x/ � sin .2�x/

�v.x/ dx for all v 2 C 10 .0; 1/:

As v is arbitrary, we deduce

.x2 C 1/u00.x/ � xu0.x/ � sin .2�x/ D 0 in .0; 1/

so that u is the classical solution to the initial problem. This indicates that the variationalformulation is coherent with the classical one for regular solutions. For the analysis weneed the Lax-Milgram theorem. Clearly, B is bilinear and F is linear. As for the continu-ity, by the Schwarz and Poincaré inequalities (the latter holds here with CP D 1=� , seeExercise 5.3.13, Chap. 5, page 312), we get

jB.u; v/j � kx2 C 1kL1ku0kL2kv0kL2 C k3xkL1ku0kL2kvkL2 ��2C 3

�

ku0kL2kv0kL2

(where Lp D Lp.0; 1/) and

jFvj � k sin 2�xkL2kvkL2 �p2

2�kv0kL2 :

Concerning the coercivity of B we have to estimate (from below)

B.u; u/ DZ 1

0

�.x2 C 1/.u0/2 C 3xu0u/

�dx:

Noting thatuu0 D .u2/0=2;

and integrating by parts, using Poincaré’s inequality, we can writeZ 1

0

3xuu0 dx DZ 1

0

3

2x.u2/0 dx D �

Z 1

0

3

2u2 dx

D �32

kuk2L2 � � 3

2�2ku0k2

L2 :

Since x2 C 1 � 1,B.u; u/ � aku0k2

L2 ;


where a D 1 � 32�2 > 0. Hence B is coercive and we can apply Lax–Milgram to obtain

the existence of a unique solution u to (6.3). Moreover,

ku0kL2 � 1

akF kH�1 �

�1 � 3

2�2

p2

2�;

which is the required inequality.

Problem 6.2.2 (Neumann conditions). Write the variational formulation of the prob-lem: ´

exu00 C exu0 � cu D 1 � 2x 0 < x < 1

u0.0/ D u0.1/ D 0:

Discuss the existence of solutions as the real parameter c varies.

Solution. The equation can be written as

�.exu0/0 C cu D 2x � 1:

Given the Neumann boundary conditions, we choose V D H 1.0; 1/ as test functions.Multiply the equation by a generic test function v and integrate over .0; 1/ WZ 1

0

��.exu0/0 C cu�v dx D

Z 1

0

.2x � 1/v dx:

Integrating by parts now givesZ 1

0

��.exu0/0 C cu�v dx D ��exu0v�1

0CZ 1

0

�exu0v0 C cuv

�dx:

Setting

B.u; v/ DZ 1

0

�exu0v0 C cuv

�dx; F v D

Z 1

0

.2x � 1/v dx;

gives the following variational formulation: determine u 2 V such that

B.u; v/ D Fv for any v 2 V: (6.4)

To check the coherence of this formulation with the classical one, observe that if u 2 C 2is a weak solution, integrating by parts givesZ 1

0

�ex.u00 C u0/ � cuC 2x � 1� v dx��exu0v�1

0D 0 for every v 2 C 1.0; 1/: (6.5)

In particular, the relation (6.5) must hold for every v that vanishes at the boundary: that isZ 1

0

�exu00 C exu0 � cuC 2x � 1� v dx D 0 for every v 2 C 10 .0; 1/;


which forcesexu00 C exu0 � cuC 2x � 1 D 0 on .0; 1/ ;

i.e. u solves the starting equation. Substituting in (6.5) produces

�eu0.1/v.1/C u0.0/v.0/ D 0 for every v 2 C 1.0; 1/;hence u0.0/ D u0.1/ D 0. In other words a regular weak solution to (6.4) is indeed aclassical solution to the initial problem, so the variational formulation is correct.

For the analysis of the problem let us try to apply the Lax-Milgram theorem, or, ifnot possible, Fredholm’s alternative. That F is linear and B bilinear is clear. As for thecontinuity, write L2 D L2.0; 1/, so

jFvj � k2x � 1kL2kvkL2 �p3

3

�kv0kL2 C kvkL2

�and

jB.u; v/j � kexkL1ku0kL2kv0kL2 C jcjkukL2kvkL2 � max¹e; jcjºkukV kvkV ;hence F belongs to V � and B is continuous. Coercivity: as ex � 1 on Œ0; 1�, we have

B.u; u/ DZ 1

0

�ex.u0/2 C cu2

�dx � ku0k2

L2 C ckuk2L2 :

If c > 0, B is coercive, with constant min¹1; cº. By the Lax-Milgram theorem we deducethere is a unique solution u to (6.3). Vice versa, if c � 0,

B.u; u/C .1 � c/kuk2L2 � ku0k2

L2 C kuk2L2

and B is only weakly coercive for the triplet ¹V;H D L2.0; 1/; V �º. In fact the embed-ding of V in H is compact, and we can use Fredholm’s alternative, by which (6.4) hassolutions if and only if

Fw DZ 1

0

.1 � 2x/w dx D 0

for every solution w of the adjoint homogeneous problem:

B.v;w/ D 0; for any v 2 V:Note that B.u; v/ D B.v; u/, making the problem self-adjoint. In formulas, if w solvesthe adjoint homogeneous problem, for every v 2 V we haveZ 1

0

�exv0w0 C cvw

�dx D 0;

which is the weak formulation (prove this fact) of the Sturm-Liouville problem´�.exu0/0 D �cu 0 < x < 1

u0.0/ D u0.1/ D 0:


We know2 that such problem admits a sequence of simple eigenvalues ¹�ckº D ¹�kº,with �1 D 0 and �k > 0 for k � 1. In particular, each eigenspace has dimension 1 andis generated by k 2 V , k kk D 1. When �c ¤ �k , the problem has only the zerosolution. In summary,

if c ¤ ��k for every k there exists a unique solution u 2 V to (6.4).

Conversely,

if c D ��k then (6.4) can be solved if and only ifZ 1

0

x k.x/ dx D 0

(in factR 10 k.x/ dx D 0 for every k � 1; why is this?), and then there are infinitely

many solutions u D NuC C k , where Nu is a particular solution.In general the explicit determination of k , and so the verification of the compatibil-

ity conditions, is not elementary. The case c D ��1 D 0 is easy and one can see thatthe solutions of the adjoint homogeneous problem are only the constants (i.e. 1 � 1).Therefore the problem has solutions, forZ 1

0

.1 � 2x/ � 1 dx D 0:

Integrating we get the infinitely many solutions

u.t/ D .x2 C x C 1/e�x C C;

with C an arbitrary constant.

Problem 6.2.3 (Robin-Dirichlet conditions). Write the variational formulation of theproblem ´

cos x u00 � sin x u0 � xu D 1 0 < x < �=6

u0.0/ D �u.0/; u.�=6/ D 0

and discuss existence and uniqueness. Determine a stability estimate for the solution.

Solution. The equation can be written in divergence form

�.cos x u0/0 C xu D �1:Given the homogeneous Dirichlet condition at x D �=6, we choose as functional space

V D ®v 2 H 1.0; �=6/ W v.�=6/ D 0

¯:

Multiplying by v 2 V and integrating we haveZ �=6

0

��.cos xu0/0 C xu/�v dx D

Z �=6

0

�v dx; for every v 2 V:

2 Appendix A.


An integration by parts givesZ �=6

0

�.cos x u0/0v dx D �� cos x u0v��=60

CZ �=6

0

cos x u0v0 dx D

D u0.0/v.0/CZ �=6

0

cos x u0v0 dx D

D �u.0/v.0/CZ �=6

0

cos x u0v0 dx;

where, in the last equality, we used Robin’s condition at x D 0. The weak formulationthus reads: determine u 2 V such that

B.u; v/ D Fv for every v 2 V; (6.6)

where

B.u; v/ DZ �=6

0

�cos x u0v0 C xuv

�dx � u.0/v.0/; F v D

Z �=6

0

�v dx:

As in previous problems, it is possible to check (and we invite the reader to do so) thatif the variational solution u is regular, the classical formulation follows from the weakformulation after integration.

The functional F is linear and B is a bilinear form. Let us check continuity. For everyv 2 V we may write

v .�=6/ � v.0/ DZ �=6

0

v0.t/ dt;

whence (Lp D Lp.0; �=6/)

jv.0/j �Z �=6

0

jv0.x/j dx �r�

6kv0kL2

(in the last step we used Schwarz’s inequality on the product jv0j � 1). The functions in Vsatisfy Poincaré’s inequality

kvkL2 � CP kv0kL2 ;

hence we can employ the equivalent norm kvkV D kv0kL2 . Keeping previous inequalitiesinto account, we deduce

jB.u; v/j � ku0kL2kv0kL2 C kxkL1kukL2kvkL2 C ju.0/jjv.0/j ��1C �

6C 2P C �

6

�ku0kL2kv0kL2 ;

and

jFvj � k1kL2kvkL2 � CP

r�

6kv0kL2 :


Both B and F are continuous. Let us examine the coercivity of B . Recalling that cos x �p3=2 and x � 0 on Œ0; �=6� we can write

B.u; u/ DZ �=6

0

�cos x .u0/2 C xu2

�dx � u2.0/ �

�p3

2ku0k2

L2 � �

6ku0k2

L2 D aku0k2L2

where a Dp32

� �6> 0. Then we can apply the Lax-Milgram theorem, and obtain the

existence of a unique solution u to (6.6). By the same theorem, moreover,

ku0kL2 � 1

akF kV � � CP

p3

2� �

6

!�1r�

6:

Thanks to Exercise 5.3.13, Chap. 5 (on page 312), we know that the best Poincaré constantin V is CP D 1=3. In conclusion,

ku0kL2 �p2�

9 � �p3

0:7043 : : : :

Problem 6.2.4 (Problems on unbounded domains). Write the variational formulationof 8<

:�u00 C 2C x2

1C x2u D 1

1C x2x 2 R

u.x/ ! 0 x ! ˙1and prove existence and uniqueness. Determine an L1.R/-estimate for the solution.

Solution. Problems on unbounded domains are usually more delicate to treat. At pres-ent, however, the space V D H 1 .R/, that coincides with the closure of C10 .R/ in theusual norm, is particularly suited as it incorporates the condition of vanishing at infinity(Exercise 5.3.18 a), on page 312 in Chap. 5). What is more, a function v 2 C10 .R/ hascompact support and so we can multiply by v and integrate on R, without worrying aboutthe integrals convergence. ThenZ C1

�1

�u00 C 2C x2

1C x2u

�v dx D

Z C1

�11

1C x2v dx:

Integrating by parts we getZ C1

�1�u00v dx D � �u0v�C1�1 C

Z C1

�1u0v0 dx;

giving the variational formulation: determine u 2 V such that:

B.u; v/ D Fv for every v 2 V; (6.7)


where

B.u; v/ DZ C1

�1

u0v0 C 2C x2

1C x2uv

�dx; F v D

Z C1

�11

1C x2v dx:

The reader can check that if u is regular and solves (6.7), then it is a classical solution.To apply Lax-Milgram we need to prove the continuity of F , B and the coercivity of

B . As

jFvj �� 1

1C x2

��L2.R/

kvkL2.R/ D p�kvkL2.R/; (6.8)

and

jB.u; v/j � ku0kL2.R/kv0kL2.R/ C��2C x2

1C x2

��L1.R/

kukL2.R/kvkL2.R/ � 2kukV kvkV ;

F and B are continuous. Since infR

�2Cx2

1Cx2

�D 1, we have

B.u; u/ DZ C1

�1

.u0/2 C 2C x2

1C x2u2�dx � kuk2V ;

showing B is coercive, with coercivity constant 1. Consequently the Lax-Milgram theo-rem guarantees existence and uniqueness of the solution.

Concerning the estimate, from3 kukL1.R/ � kukV and (6.8) we obtain kukL1.R/ �kF kV � � � .

Problem 6.2.5 (Legendre equation). Let

V D°v 2 L2 .�1; 1/ W �1 � x2�1=2 v0 2 L2 .�1; 1/

±:

a) Verify that the formula

.u; v/V DZ 1

�1Œuv C �

1 � x2�u0v0� dx (6.9)

defines an inner product making V into a Hilbert space.

b) Study the variational problem

.u; v/V D Fv �Z 1

�1f v dx; for every v 2 V; (6.10)

where f 2 L2 .�1; 1/, and interpret it in the strong sense.

c) Study the eigenvalue problem

.u; v/V D �

Z 1

�1uv dx; for every v 2 V:

3 Problem 5.2.23, Chap. 5 (page 300).


Solution. a) Let us verify the properties of the inner product. That .u; u/V is non-negative is clear, while

.u; u/V DZ 1

�1Œu2 C �

1 � x2� .u0/2� dx D 0

implies u D 0 a.e.4 on .�1; 1/. Hence .u; v/V is an inner product, so we just have to proveV is complete with respect to it. Take ¹unº a Cauchy sequence in V . Then ¹unº is a Cauchy

sequence in L2 .�1; 1/, thus un ! u0 2 L2 .�1; 1/. Analogously, ¹�1 � x2�1=2 u0nº is aCauchy sequence in L2 .�1; 1/ and therefore�

1 � x2�1=2 u0n ! w 2 L2 .�1; 1/ :

On the other hand, if ' 2 D.�1; 1/, thenh�1 � x2�1=2 'i0 2 D.�1; 1/, and

h�1 � x2�1=2 u0n; 'i D �Z 1

�1un

h�1 � x2�1=2 'i0 dx

! �Z 1

�1u0

h�1 � x2�1=2 'i0 dx D h�1 � x2�1=2 u00; 'i

Therefore�1 � x2�1=2 u00 D w 2 L2.�1; 1/, which means V is complete.

b) Denote by k�k and k�kV the L2 .�1; 1/-norm and the norm induced by (6.9). As

jFvj DˇZ 1

�1f v dx

ˇ� kf k kvk � kf k kvkV ;

the functional F defines an element in V �. Riesz’s theorem implies there exists a uniqueu solving (6.10) and

kukV � kf k :For the strong interpretation let us proceed formally and integrate by parts the second termof the inner product; (6.10) reads��1 � x2�u0v�1�1 C

Z 1

�1®u � Œ�1 � x2�u0�0¯ v dx D

Z 1

�1f v dx for every v 2 V;

and as v is arbitrary we have

�Œ�1 � x2�u0�0 C u D f in .�1; 1/ (6.11)

and alsolim

x!˙1�

�1 � x2�u0 .x/ D 0:

The boundary conditions are thus the ‘natural’ homogeneous Neumann conditions. Notethat the limit is necessary because functions in V may be unbounded in a neighbourhoodof ˙1.

4 In fact, u 2 V implies u 2 H1loc.�1; 1/, hence u 2 C.�1; 1/.


c) From part b) the eigenvalue problem

.u; v/V D �

Z 1

�1uv dx for every v 2 V

is equivalent to the (Legendre) equation

�Œ�1 � x2�u0�0 D .� � 1/ u on .�1; 1/ ;with boundary conditions

limx!˙1

�1 � x2�u0 .x/ D 0:

The only solutions5 are Legendre’s polynomialsPn, defined recursively by the (Rodrigues)formula

Pn .x/ D 1

2nnŠ

dn

dxn

�x2 � 1�n ; n 2 N;

with eigenvalues �n D 1C n .nC 1/. For example

P0 .x/ D 1, P1 .x/ D x, P2 .x/ D 1

3

�3x2 � 1� , P4 .x/ D 1

2

�5x3 � 3x� :

Problem 6.2.6 (Variational problem in H 1). Consider the following weak problem:determine u 2 H 1.0; 1/ such thatZ 1

0

u0v0 dt D v.1/ � v.0/ �Z 1

0

.log t /v0 dt; 8v 2 H 1.0; 1/:

Study when it can be solved. Show that there exist infinitely many solutions and deter-mine them explicitely.

Solution. Let

B.u; v/ DZ 1

0

u0v0 dt; F v D v.1/ � v.0/ �Z 1

0

.log t /v0 dt:

First of all, F is a continuous linear functional on V D H 1.0; 1/: indeed, observing thatZ 1

0

log2 t dt DZ 0

�1s2es ds D 2

we have6

jFvj � 2kvkL1 C k log tkL2kv0kL2 ��2C p

2�

kvkH1 :

At the same time, it is easy to check that B is continuous but only weakly coercive on V(see for instance Exercise 6.3.1 on page 381.c)). The homogeneous problem is self-adjoint

5 Appendix A.6 For pointwise estimates of functions in H1.a; b/ we always use Problem 5.2.24 (page 301),Chap. 5.


and has only the constants as solutions. We can apply Fredholm’s alternative, and obtainthat the problem can be solved if and only if Fv D 0 for every v constant. But this com-patibility condition is automatic, so the problem has infinitely many solutions that differby an additive constant.

In order to write the associated boundary-value problem let us rearrange terms:Z 1

0

�u0 � log t

�v0 dt D v.1/ � v.0/; 8v 2 V:

By direct inspection we inferu0 � log t D 1;

from which the solutions areu.t/ D t log t C C:

Notice that u 62 C 1.Œ0; 1�/ (u0.0/ does not exist), so that this problem does not have acorresponding classical version.

6.2.2 Elliptic problems

Problem 6.2.7 (Dirichlet conditions). Write the variational formulation of the problem´��uC c.x/u D f .x/ x 2 �u D 0 x 2 @�;

where � is a regular bounded domain in Rn, f 2 L2.�/. What are the constraintson c that ensure the Lax-Milgram theorem can be applied? Write the correspondingstability estimate.

Solution. The homogeneous Dirichlet conditions suggest to place the weak formula-tion in H 1

0 .�/. So let us multiply by some v 2 H 10 .�/ and integrate7. Since (by Gauss’s

formula)Z

��uv dx D �Z@

@�u v d� CZ

ru � rv dx DZ

ru � rv dx;

we obtain Z

Œru � rv C c.x/uv� dx DZ

f v dx for every v 2 H 10 .�/:

But u, v are in L2, so to make sense of the integral let us assume8 c 2 L1.�/. So set

B.u; v/ DZ

Œru � rv C c.x/uv� dx; F v DZ

f v dx;

7 More precisely, we use v 2 D.�/ and extend the resulting equation to H10 .�/ by density.

8 Or, more sophisticatedly, note that by Sobolev’s embedding (see [18, Chap. 7, Sect. 10]) u and vbelong to Lp

�.�/, p� D 2n=.n � 2/, so it suffices to demand c 2 Lq.�/, with q D n=2 for n �

3 and any q for n D 2.


and then we have the variational formulation: determine u 2 H 10 .�/ such that

B.u; v/ D Fv for every v 2 H 10 .�/: (6.12)

Vice versa, if u 2 C 20 .�/ solves (6.12), we can use Gauss’s formula in the opposite wayZ

Œ��uC c.x/u � f � v dx for every v 2 C 10 .�/;

so that ��u C c.x/u � f D 0 a.e. on �. In particular, even if the coefficients c and fare just continuous, the equality holds everywhere and u is the classical solution.

Now, to use Lax-Milgram theorem we need to understand the continuity of F and Band the coercivity of B (that F and B are linear is straightforward). To this end let usrecall that for every u 2 H 1

0 .�/ we have Poincaré’s inequality

kukL2. / � CP krukL2. /;

which entitles us to use in H 10 .�/ the equivalent norm kukH1

0. / D krukL2. /. As for

the continuity, from Schwarz’s and Poincaré’s inequalities (Lp D Lp�),

jB.u; v/j � krukL2krvkL2 CkckL1kukL2kvkL2 � �1C kckL1C 2P

� krukL2krvkL2

and

jFvj � kf kL2kvkL2 � CP kf kL2krvkL2 :

Coercivity: we have to bound (from below)

B.u; u/ DZ

�jruj2 C c.x/u2�dx:

One possibility is imposing c.x/ � 0 a.e. ona�. ThusB is coercive, with constant a D 1.There are less restrictive conditions on c, though, provided we take a smaller coercivityconstant. In fact, suppose that

c.x/ � �; for almost every x 2 �:

By Poincaré’s inequalityZ

c.x/u2 dx � �kuk2L2 � �C 2P kruk2

L2

and therefore

B.u; u/ � akruk2L2 ;

where

a D 1 � C 2P :


Thus to have a > 0 (coercivity) it is enough to take c not ‘too’ negative9, i.e.

<1

C 2P.

Using Lax-Milgram theorem implies, with the previous hypotheses, that there is a uniquesolution u to (6.12), which satisfies the estimate

krukL2 � 1

akF kH�1 � CP

1 � C 2Pkf kL2 :

Problem 6.2.8 (A minimum problem). Let Q D .0; 1/ � .0; 1/ � R2. Minimise, asv 2 H 1

0 .Q/ varies, the functional

E.v/ DZQ

²1

2jrvj2 � xv

³dxdy:

Write the Euler equation and prove there exists a unique minimiser u 2 H 10 .Q/. Find

an explicit formula for u.

Solution. With v;w 2 H 10 .Q/ we set

B.v;w/ DZQ

rv � rw dxdy and Fv DZQ

xv dxdy;

and write E.v/ D 1

2B.v; v/ � Fv. As B is self-adjoint, u minimises E if and only if it

solves the Euler equationZQ

ru � rv dxdy DZQ

xv dxdy; i.e. B.u; v/ D Fv; 8v 2 H 10 .Q/: (6.13)

We have

jFvj � kxkL2kvkL2 � CPp2

krvkL2

where CP is the Poincaré constant for H 10 .Q/. Hence F is linear, continuous, while B

is bilinear, continuous and coercive (it is the scalar product of H 10 .Q/). Using the Lax-

Milgram theorem (or better, Riesz’s theorem), we obtain existence and uniqueness of thesolution u minimising E.

For an explicit expression, note that (6.13) is the variational formulation of the follow-ing Dirichlet problem: ´

��u D x in Q

u D 0 on @Q:

9 Using Sobolev’s embedding theorem, in line with the previous footnote argument, the role of istaken by kc�kLq , where q is as above and c� .x/ D inf¹0;�c.x/º is the negative part of c.


Thus, we can use the methods of Chap. 2, in particular separation of variables, to computeit. Let us expand the source term in a sines-Fourier series in y, that is

x DC1XnD1

xbn sin .n�y/

where

bn D 2

Z 1

0

sin .n�y/ dy D 2

n�Œ� cos n� C 1� D

´0 for n even

4=n� for n odd,

and let us seek solutions of the form

u.x; y/ DC1XnD1

un.x/ sin .n�y/:

Notice that u satisfies u.x; 0/ D u.x; 1/ D 0 automatically. Differentiating (formally),we obtain

uxx.x; y/ DC1XnD1

u00n sin .n�y/; uyy.x; y/ DC1XnD1

�n2�2un.x/ sin .n�y/

and therefore

�u.x; y/ DC1XnD1

�u00n.x/ � n2�2un.x/

�sin .n�y/ D

C1XnD1

xbn sin .n�y/: (6.14)

Overall, the un solve the boundary-value problems´u00n.x/ � n2�2un.x/ D bnx

un.0/ D un.1/ D 0:

When n is even bn D 0, and un � 0. When n is odd the equation reads

u00n.x/ � n2�2un.x/ D 4

n�x;

whose general integral is

un.x/ D C1en�x C C2e

�n�x � 4

n3�3x:

The boundary conditions force

un.x/ D 4

n3�3

en�x � e�n�xen� � e�n� � x

�:


Finally,

u.x; y/ D 4

�3

C1XkD0

1

.2k C 1/3

sinh ..2k C 1/�x/

sinh ..2k C 1/�/� x

�sin ..2k C 1/�y/: (6.15)

By Weierstrass’s criterion, since

sinh ..2k C 1/�x/

sinh ..2k C 1/�/� 1 for 0 � x � 1; k � 0;

both (6.15) and the series of the partial derivatives (whose terms have order 1=.2kC 1/2)converge uniformly on Q, so u 2 C 10 .Q/ � H 1

0 .Q/. Moreover, also the second-derivatives series converges uniformly on every compact set contained in Q, so we candifferentiate term by term and check that u is the (at this point classical) solution.

The series of second derivatives does not converge uniformly on Q: from (6.14), �uis discontinuous on Q, because it equals x on Q and 0 on y D 1; despite the coefficientsbeing regular, u is not C 2.Q/. This should come as no surprise, because the domain isonly Lipschitz.

Problem 6.2.9 (Riesz’s theorem and Laplacian). Given a bounded regular domain� � Rn, consider H 1

0 .�/ with product

.u; v/ DZ

ru � rv dx;

and let F 2 H�1.�/.

a) Apply Riesz’s theorem to F and deduce a suitable variational problem.

b) Write the corresponding classical problem and interpret it in view of Riesz’s theo-rem.

Solution. a) Using .ru;rv/ as scalar product in H 10 .�/, Riesz’s theorem says that

there is a unique u 2 H 10 .�/ such thatZ

ru � rv dx D Fv for every v 2 H 10 .�/; (6.16)

andkukH1

0. / D kF kH�1. /:

The variational problem (6.16) is thus well posed on H 10 .�/.

b) The space of test functions D .�/ is dense inH 10 .�/, so the variational formulation

of a) is equivalent toZ

ru � r' dx D hF; 'i for every v 2 D .�/ ;


where hF; 'i is the duality pairing between D .�/ and D 0 .�/. Using the language ofdistributions we may write, for any ' 2 D .�/,

0 D hru;r'i � hF; 'i D h��u � F; 'i:Hence

��u D F

in distributional sense. Therefore for every F 2 H�1.�/, u D .��/�1 F is the elementof H 1

0 .�/ for which

Fv D ..��/�1 F; v/H10. / for every v 2 H 1

0 .�/;

and moreoverkF kH�1. / D k .��/�1 F kH1

0. /:

This amounts to say that the canonical isomorphism in Riesz’s theorem is actually theoperator

.��/�1 W H�1.�/ �! H 10 .�/:

Problem 6.2.10 (Bilinear forms on subspaces). Given a bounded and regular domain� � Rn, consider the subspace

V D²u 2 H 1.�/ W

Z

udx D 0

³in H 1.�/ (with usual scalar product).

a) Say to which problem the following variational formulation is associated: find u 2V such that Z

ru � rv dx D 0; for every v 2 V:

b) Use the Lax-Milgram theorem and deduce that the problem has the unique solutionu D 0.

c) Deduce that two solutions of ´��u D f x 2 �@�u D g x 2 @�;

with f 2 L2.�/ and g 2 L2.@�/, differ by a constant.

Solution. a) First observe that V is closed in H 1.�/, so we have the decomposition

H 1.�/ D V ˚ V ?


where V ? is the space of constants10. Consequently, for every v 2 H 1.�/, we can write

v D Qv C Cv where Cv DZ

v dx and Qv 2 V:

Let u 2 V be a solution to the given problem and v 2 H 1.�/. ThenZ

ru � rv dx DZ

ru � r . Qv C Cv/ dx DZ

ru � r Qv dx D 0;

so u solves the equivalent problem: determine u 2 V such thatZ

ru � rv dx D 0; for every v 2 H 1.�/:

Assuming u regular and integrating by parts with Gauss’s formula gives

0 DZ

ru � rv dx DZ@

@�u v d� �Z

��uv dx (6.17)

for every v 2 H 1.�/. In particular,Z

��uv dx D 0 for every v 2 H 10 .�/;

which implies �u D 0 in � . Substituting in (6.17) we findZ@

@�u v d� D 0 for every v 2 H 1.�/;

hence @�u D 0 on @�. To sum up, the weak formulation corresponds to the boundary-value problem 8<

:��u D 0 x 2 �@�u D 0 x 2 @�Z

udx D 0:

b) Recall that Poincaré’s inequality

kukL2. / � CP krukL2. /

holds on the subspace V of H 1 .�/ of functions with zero average on a regular boundeddomain, thus we have the equivalence of the norms. We want to prove that the bilinearform

B.u; v/ DZ

ru � rv dx

10 The proof is left to the reader.


is continuous and coercive on V . From Lax-Milgram, we will deduce that the variationalproblem has only the zero solution. Now,

jB.u; v/j � krukL2. /krvkL2. / � kukH1. /kvkH1. /

and by Poincaré’s inequality, for every �,

B.u; u/ D kruk2L2. /

D �kruk2L2. /

C .1 � �/kruk2L2. /

� �kruk2L2. /

C 1 � �C 2P

kuk2L2. /

D 1

C 2P C 1kuk2

H1. /

(in the end we chose � D .1� �/=C 2P D 1=.C 2P C 1/). From these inequalities the claimfollows.

c) Let u1, u2 be solutions of the given problem. Then w D u1 � u2 solvesZ

rw � rv dx D 0 for every v 2 H 1.�/: (6.18)

Using the orthogonal decomposition of H 1 .�/ seen in a), we may write w D Qw C Cwand v D Qv C Cv . Then (6.18) readsZ

r Qw � r Qv dx D 0 for every Qv 2 V:

But by b) we obtain Qw D 0, i.e. w D Cw , proving the claim.

Problem 6.2.11 (Robin conditions). Let � be a bounded regular domain in Rn.

a) Show that the formula

kuk2� DZ

jruj2 dx CZ@

u2 d�

defines on H 1.�/ an equivalent norm to the standard one.

b) Write the variational formulation of the Robin problem´��u D f x 2 �@�uC u D 0 x 2 @�;

with > 0, and discuss whether it is well posed.

Solution. a) Let us prove the equivalence between kukH1. / and kuk� for functionsin C1.�/. As � bounded and regular (Lipschitz would be enough) we get the equiva-lence in H 1 .�/ by density. So take u 2 C1.�/. One way to proceed is the following.Consider the vector field (on Rn)

F D xn:


Clearly div F D 1, and since � is bounded, jFj � M on �. We can use Gauss’s formulaon u2 and F and writeZ

u2 div F dx DZ@

u2F � � d� �Z

r.u2/ � F dx: (6.19)

Therefore Z

u2 div F dx DZ

u2 dx;ˇZ@

u2F � � d�

ˇ�Z@

u2jFj d� � M

Z@

u2 d�

and ˇZ

r.u2/ � F dx

ˇ�Z

2jujjrujjFj dx � 1

4

Z

u2 dx C 4M 2

Z

jruj2 dx

(we used the elementary fact 2ab � a2 C b2, with a D juj=2 and b D 2M jruj). Substi-tuting in (6.19) gives

3

4

Z

u2 dx � M

Z@

u2 d� C 4M 2

Z

jruj2 dx

i.e.

kuk2H1 D

Z

jruj2 dx CZ

u2 dx

� 19

3M 2

Z

jruj2 dx C 4

3M

Z@

u2 d� � Ckuk2�;

whereC D M max ¹19M; 4º =3. As already mentioned, this holds onH 1.�/ by a densityargument.

Conversely, the theory of traces implies uj@ 2 H 1=2.@�/, and the latter is embeddedin L2.�/. Put in formulas, there exists a constant C� such that, for every u 2 H 1.�/,

kuj@ kL2.@ / � C� kukH1. /

so

kuk� �q1C C 2� kukH1. /:

Overall, C1k � kH1. / � k � k� � C2k � kH1. /, making the norms equivalent. Even-tually, the norm k � k� is induced by the scalar product

.u; v/� DZ

ru � rv dx CZ@

uv d�:

b) By the previous result, let us introduce the Hilbert space H 1� .�/ of H 1 functionsequipped with norm k � k� (and the relative scalar product). Multiplying by v 2 H 1� .�/and integrating: Z

��uv dx DZ

f v dx:


Gauss’s formula and the Robin condition giveZ

��uv dx DZ

ru � rv dx �Z@

@�u v d�

DZ

ru � rv dx C

Z@

u v d�:

So let us set

B.u; v/ DZ

ru � rv dx C

Z@

u v dx; F v DZ

f v dx;

and write a variational formulation: determine u 2 H 1� .�/ such that

B.u; v/ D Fv for every v 2 H 1� .�/:

The bilinear form B satisfies

jB.u; v/j � max¹1; ºj.u; v/�j � max¹1; ºkuk�kvk�;

B.u; u/ DZ

jruj2 dx C

Z@

u2 d� � min¹1; ºkuk2�(in the former we used Schwarz’s inequality for the scalar product in H 1� .�/). Hence Bis continuous and coercive (as > 0) and to use Lax-Milgram we just need to checkthe continuity on H 1� .�/ of the linear functional. As norms are equivalent, this holds forf 2 L2.�/, for example (it suffices that it belong to the dual of H 1.�/). In that case theproblem is always well posed.

Problem 6.2.12 (Neumann conditions). Let � � Rn be a bounded and regular do-main, b 2 L1.�I Rn/ and f 2 L2.�/. Write the variational formulation of´

��uC b.x/ � ru D f x 2 �@�u D 0 x 2 @�;

and discuss when it can be solved.

Solution. Due to the Neumann boundary conditions we choose V D H 1.�/ and for-mulate the problem in weak form as follows: determine u 2 V such that

B.u; v/ DZ

Œru � rv C vb.x/ � ru� dx DZ

f v dx D Fv for every v 2 V:

As f 2 L2.�/ the linear functional F is bounded. Similarly, setting kbkL1. / D b, wehave

jB.u; v/j � krukL2. /krvkL2. / C bkrukL2. /kvkL2. / � .1C b/kukV kvkVand also the bilinear form B is continuous.


The form B cannot be coercive: if u D k ¤ 0; constant, kukL2. / D kpj�j > 0

whileB.u; u/ D 0. YetB is weakly coercive for the Hilbert triple ¹V;H D L2 .�/ ; V �º.In fact by Schwarz’s inequality (and 2ab � a2 C b2):Z

ub � rudx � �bkrukL2. /kukL2. / � �14

kruk2L2. /

� b2kuk2L2. /

(6.20)

so

B.u; u/ DZ

jruj2 dx CZ

ub � rudx � 3

4kruk2

L2. /� b2kuk2

L2. /

i.e.

B.u; u/C�3

4C b2

kukL2. / � 3

4kuk2V

which shows the weak coercivity of B . As V is compactly embedded in H we can useFredholm’s alternative. Therefore the problem has solutions if and only ifZ

f w dx D 0 (6.21)

for every w that solves the homogeneous problemZ

Œrw � rv C wb � rv� dx D 0 for every v 2 V:

This condition cannot be expressed in an elementary manner, except in special cases likeb D 0. In that case, in fact, the only homogeneous solutions are the constants, and (6.21)reduces to Z

f dx D 0.

Remark. Find the mistake in the following argument. If div b D 0, instead of (6.20) wecan write Z

ub � rudx D1

2

Z

b � r.u2/ dx DZ@

u2b � � � d�

whence, if b � � � b0 > 0 (the flow of b across @� is outgoing),

B.u; u/ DZ

jruj2 dx CZ

ub � rudx � kruk2L2. /

C b0kuk2L2.@ /

:

Recalling Problem 6.2.11, the form B is coercive!!

Problem 6.2.13 (Fredholm alternative). LetQ D .0; �/�.0; �/. Discuss the Dirichletproblem ´

�uC 2u D f in Q

u D 0 on @Q:

Examine, in particular, the cases

f .x; y/ D 1 and f .x; y/ D x � �

2:


Solution. Set V D H 10 .Q/with the gradient norm inL2 .Q/, so the weak formulation

is

B.u; v/ DZQ

Œru � rv � 2uv� dxdy DZQ

�f v dxdy � Fv; for every v 2 V:

By Schwarz’s and Poincaré’s inequalities B is bilinear and continuous on V ; in fact

jB.u; v/j � .1C 2C 2P /kukV kvkV :

If f 2 L2.Q/ (as in the given cases) also F is continuous on V . The form B cannot becoercive, since the homogeneous problem

B.u; v/ DZQ

.ru � rv � 2uv/ dxdy D 0; for every v 2 V , (6.22)

which is the variational form of´�uC 2u D 0 in Q

u D 0 on @Q,(6.23)

has nonzero solutions on V (found by separating variables as explained in Chap. 2). Easilywe find

Nu.x; y/ D c sin x sin y c 2 R (6.24)

In other words, � D 2 is a Dirichlet eigenvalue of �� and (6.24) are the eigenfunctions.Inserting u D v D Nu in (6.22) produces

B. Nu; Nu/ D 0

so the bilinear form is not coercive. Accidentally, the eigenfunction sin x siny is positiveonQ. By general principles we deduce that 2 is the first Dirichlet eigenvalue of the opera-tor �� (the smallest one). The other eigenvalues can be computed by variable separation.Let us write u .x; y/ D p .x/ q .y/ and substitute into ��u D �u. A few computationsgive

p00 .x/p .x/

D �q00 .y/q .y/

� �

which breaks into two boundary-value problems

p00 C 2p D 0, p .0/ D p .�/ D 0

andq00 C .� � 2/q D 0; q .0/ D q .�/ D 0.

This gives the eigenvalue sequence

�nm D n2 Cm2; 'nm .x; y/ D sinnx sinmy; n;m � 1, integer.


Back to the variational problem, we cannot use the Lax-Milgram theorem. Let us trywith Fredholm’s alternative. The bilinear form B is weakly coercive; in fact,

B.u; u/C 2

ZQ

u2dxdy D kuk2V

and the Hilbert triple ¹V;L2 .Q/ ; V �º satisfies the assumptions. Hence the problem hassolutions if and only if F vanishes on every solution of the adjoint homogeneous prob-lem. As B is symmetric, the adjoint homogeneous problem coincides with (6.22), so theproblem has a solution if and only ifZ

Q

f .x; y/ sin x sin y dxdy D 0:

When f .x; y/ D 1,RQ sin x sin y dxdy D 4, so no solution exists. Vice versa, when

f .x; y/ D x � �=2 we haveZQ

�x � �

2

�sin x sin y dxdy D 0

and there are infinitely many solutions of the form

u.x; y/ D U.x; y/C c sin x sin y:

In order to find U we write it as superposition of eigenfunctions 'nm, that is11

U .x; y/ DXn;m�1

unm sin nx sinmy

and we require that it satisfies the equation

�U C 2U DXn;m�1

��n2 �m2 C 2�unm sin nx sinmy D x � �

2:

Expand f in sines-Fourier series. The coefficients fnm are

fnm D 4

�2

Z �

0

Z �

0

�x � �

2

�sin nx sinmy dxdy; n;m � 1:

Then

fnm D8<:0 n odd and any m; or m even and any n;

� 8�

1

2h .2k C 1/n D 2h and m D 2k C 1, h � 1, k � 0

11 Chap. 2.


so

unm D8<:0 n odd, or m even;

�4�h .2k C 1/ Œ2 � 4h2 � .2k C 1/2�

n D 2h and m D 2k C 1, h � 1, k � 0:

Finally,

U .x; y/ DX

h�1;k�0

�4�h .2k C 1/ Œ2 � 4h2 � .2k C 1/2�

sin 2hx sin .2k C 1/ y:

Problem 6.2.14 (Fredholm alternative). SetQ D .0; �/� .0; �/. Discuss the problem8<:�uC �u D 1 in Q

u D 0 on @Q n ¹y D 0º�uy.x; 0/ D x for 0 � x � �

where � is a real parameter.

Solution. For the variational formulation it is natural to choose the subspace inH 1.Q/

adapted to the Dirichlet conditions, that is:

V D ®v 2 H 1.Q/ W u D 0 on @Q n ¹y D 0º¯ :

Since the Poincaré inequality kvkL2.Q/ � CP krvkL2.Q/ holds, we can choose an equiv-alent norm kvkV D krvkL2.Q/. Multiply by v 2 V and integrate. The left-hand sidebecomesZ

Q

Œ�uC �u� v dxdy DZ@Q

@�u v ds �ZQ

Œru � rv � �uv� dxdy

D �Z �

0

uy.x; 0/v.x; 0/ dx �ZQ


DZ �

0

xv.x; 0/ dx �ZQ

Œru � rv � �uv� dxdy:

The weak formulation reads: determine u 2 V such that

B.u; v/ D Fv for every v 2 V;where

B.u; v/ DZQ


and

Fu D �ZQ

v dxdy CZ �

0

xv.x; 0/ dx:


From the trace inequality we have kvkL2.@Q/ � C�kvkV . Hence

jFvj � �kvkL2.Q/ C �

r�

3kv .�; 0/ kL2.0;�/ � �

�CP C

r�

3C�

kvkV ;

so the linear functional F is continuous. Similarly, since

jB.u; v/j � krukL2.Q/krvkL2.Q/ C j�jkukL2.Q/kvkL2.Q/ � .1C j�jC 2P /kukV kvkV ;

B is continuous.Take � � 0. Then

B.u; u/ � kuk2Vso B is coercive, too, and the Lax-Milgram theorem guarantees existence and uniquenessof the weak solution.

Conversely, take � > 0. Now we can only guarantee weak coercivity: in fact

B.u; u/C �kuk2L2.Q/

� kuk2V :

Hence we can use Fredholm’s alternative with respect to the Hilbert triple ¹V;L2.Q/; V �º.By Rellich’s theorem, in fact, the embedding of V in L2 .Q/ is compact. As B is sym-metric, the problem admits solutions if and only ifZ

Q

w.x; y/ dxdy �Z �

0

xw.x; 0/ dx D 0 (6.25)

for every w that solves the adjoint homogeneous problem B .w; v/ D 0 for every v 2 V ,which is the weak formulation of8<

:��w D �w in Q

w D 0 on @Q n ¹y D 0º�wy .x; 0/ D 0 0 � x � �:

(6.26)

From the general theory we know that (6.26) has a sequence of eigenvalues 0 < �1 <

�2 � �3 � : : : , while it only has the zero solution for � ¤ �k . Therefore when � ¤ �kthe solution exists and is unique. Vice versa, if � D �k , the problem has solutions if andonly if the compatibility condition (6.25) holds, and then there are infinitely many solu-tions. In this case the eigenvalues can be found by separating the variables, as was donein the previous problem; we find

� D �nm D n2 C .2mC 1/2

4.n � 1, m � 0/

with eigenfunctions

'nm .x; y/ D sin nx cos

�2mC 1

2y

:


Fig. 6.1 Domain for Problem 6.2.15

If � ¤ �nm the problem has one solution only. If � is one of the �nm the problem can besolved if and only ifZ

Q

sin nx cos

�2mC 1

2y

dxdy �

Z �

0

x sin nx dx D 0:

A simple computation shows that the sum on the left is zero for no value of n � 1 andm � 0, thus the compatibility conditions do not hold and so if � is one of the �nm, thereis no solution.

Problem 6.2.15 (Schwarz’s alternating methoda). Referring to Fig. 6.1, let� � R2 bea domain such that� D �1[�2 where�1 and�2 have regular boundary (Lipschitz isenough), with�1\�2 ¤ ;. Set V D H 1

0 .�/with inner product .u; v/ D R ru �rv

and define (see Exercise 5.3.26 on page 313, Chap. 5)

V1 D ®u 2H 1

0 .�1/, u D 0 in � n�1¯

and V2 D ®u 2 H 1

0 .�2/, u D 0 in �n�2¯:

Given f 2 L2 .�/ and an arbitrary element u0 2 H 10 .�/, define u2nC1 (n � 0) and

u2n (n � 1) by the following recursive procedure:´��u2nC1 D f in �1u2nC1 D u2n in .� n�1/ [ @�1 and

´��u2n D f in �2u2n D u2n�1 in .� n�2/ [ @�2:

Prove that un converges in H 10 .�/ to the solution u 2 H 1

0 .�/ of

��u D f in � ;

by showing that

u2nC1 � u D PV?1.u2n � u/ and u � u2n D PV?

2.u � u2n�1/

and using Problem 5.2.9, Chap. 5 (page 288).

a Useful in numerical approximation methods for boundary-value problems.


Solution. The sequence un is built starting from u0, solving the Dirichlet problemone time on �1 and the next on �2 and updating the boundary data with un�1. Note thatu2nC1 � u is harmonic in �1 and every v1 2 V1 vanishes in �1. So we can write

.u2nC1 � u; v1/ D 0 8v1 2 V1; i.e. u2nC1 � u 2 V ?1 .

Adding and subtracting u2n gives, for every n � 0,

.u2nC1 � u2n; v1/ D .u � u2n; v1/; 8v1 2 V1.

As u2nC1 � u2n 2 V1, the last equation amounts to say that

u2nC1 � u2n D PV1.u � u2n/

or (as u � u2n D .u � u2nC1/C .u2nC1 � u2n/)u � u2nC1 D PV?

1.u � u2n/ :

Arguing in a similar manner we find

u � u2n D PV?2.u � u2n�1/ :

Set x0 D u0, x2nC1 D u � u2nC1 and xn D u � u2n. The sequence .xn/ fits intothe iterative pattern of Problem 5.2.9, Chap. 5. Now, un ! u in H 1

0 .�/ is the same asxn ! 0 in V . To use Problem 5.2.9, Chap. 5, we need V1 and V2 to be closed in V andV ?1 \ V ?2 D ¹0º. But this follows from Exercise 5.3.26, Chap. 5.

Problem 6.2.16 (Oblique derivative). Let � � R2 be a bounded domain with C 1

boundary. For b 2 C 1.�/, consider the bilinear form on H 1 .�/:

a .u; v/ DZ

�uxvx C uyvy C buxvy � buyvx C byuxv � bxuyv

�dxdy:

a) Given f 2 L2 .�/, f D .f 1; f2/ 2 L2��I R2

�and g 2 L2 .@�/, interpret in

classical sense the variational problem

a .u; v/ D Fv; for any v 2 H 1 .�/ (6.27)

where Fv D R .f v C f � rv/C R

@ gv ds.

b) Tell under which conditions the problem has solutions. Discuss the case b �constant.

Solution. a) First of all let us write the bilinear form in a clearer way. Setting

A .x; y/ D�

1 b .x; y/

�b .x; y/ 1

and b .x; y/ D �

by ;�bx�


we have

a .u; v/ DZ

ŒA .x; y/ru � rv C .b .x; y/ � ru/ v� dxdy:

Thus (6.27) is the variational formulation of´Lu D � div .A .x; y/ru/C b .x; y/ � ru D f � div f in �;

@A� u � A .x; y/ru � � D g on @�;(6.28)

where � D . 1; 2/ is the outward normal to @�. Let us make (6.28) more explicit. As

A .x; y/ru D �ux C b .x; y/ uy ;�b .x; y/ ux C uy

�we have

div .A .x; y/ru/ D uxx C buyx C bxuy � byux � buxy C uyy

so

� div .A .x; y/ru/C b .x; y/ � ru D ��u.

Introducing the unit vector � D .� 2; 1/, tangent to @�, we obtain:

A .x; y/ru � � D .ux C buy/ 1 C .�bux C uy/ 2 DD ru � � C bru � � D ru � .� C b�/

D @�u

where � D � C b�. Eventually, we have´�u D div f � f in �;

@�uDg on @�:(6.29)

Note that

� � �

j� j D � � � C b�p1C b2

D 1p1C b2

; (6.30)

so the vector � is never tangent to @�; for this reason (6.29) is called an oblique-derivativeproblem.

b) If u is solution to problem (6.29) and c 2 R, u C c solves the same problem; theLax-Milgram lemma does not apply, for the bilinear form, albeit continuous on H 1 .�/,is not coercive. Yet it is weakly coercive. In fact, set M D max jbj, so thatˇZ

.b .x; y/ � rv/ v dxdyˇ

� M

Z

jrvj jvj dxdy � M krvkL2. / kvkL2. /

� 1

2krvkL2. / C M 2

2kvk2L2. /


whence

a .v; v/ DZ

ŒA .x; y/rv � rv C .b .x; y/ � rv/ v� dxdy

� 1

2krvkL2. / � M 2

2kvk2L2. /

and then

a .v; v/CM 2 kvk2L2. / � min

²1

2;M 2

2

³kvk2H1. / .

By Rellich’s theorem the immersion of V D H 1 .�/ in H D L2 .�/ is compact.By Fredholm’s alternative, problem (6.27) can be solved if and only ifF is orthogonal,

in H 1 .�/, to the solution of the adjoint homogeneous problem

a� .w; v/ D 0 for every v 2 H 1 .�/ (6.31)

where

a� .w; v/ D a .v;w/ DZ

ŒA .x; y/rv � rw C .b .x; y/ � rv/w� dxdy D

DZ

�A> .x; y/rw � rv C .b .x; y/ � rv/w� dxdy:

The adjoint equation (6.31) is the variational formulation of´L�w D � div

�A> .x; y/rw� � div.b .x; y/w/ D 0 in �;

@A>

� w � A> .x; y/rw � � D 0 on @�(6.32)

which reduces, with computations of the previous kind, to the following (Robin-like) prob-lem: ´

�w D 0 in �;

@� �w C .@�b/w D 0 on @�;(6.33)

where� � D � � b�:

Problem (6.27) can be solved if and only ifZ

.f w C f � rw/CZ@

gw ds D 0;

for every solution w of (6.31).If b is constant, the solutions to the adjoint homogeneous problem are only the con-

stants. In fact, if w 2 H 1 .�/ solves (6.31), choosing v D w gives

0 D a� .w;w/ DZ

A> .x; y/rw � rw dxdy DZ

jrwj2 dxdy


whence w D is constant. Problem (6.27) can be solved, therefore, if and only ifZ

f CZ@

g ds D 0.

Problem 6.2.17 (Transmission conditions). Let �1 and � be bounded Lipschitz do-mains in Rn such that �1 � � and set �2 D � n �1. On �1 and �2 consider thebilinear forms

ak .u; v/ DZ k

Ak .x/ru � rv dx .k D 1; 2/

where Ak are uniformly elliptic matrices. Suppose the entries of Ak are continuous on�k , while we allow

A .x/ D´A1 .x/ in �1A2 .x/ in �2

to be discontinuous across � D @�1. Let u 2 H 10 .�/ be the variational solution of

a .u; v/ D a1 .u; v/C a2 .u; v/ DZ

f v dx for every v 2 H 10 .�/ (6.34)

with f 2 L2 .�/.Calling uk the restriction of u to�k , determine (formally) what type of problem the

pair u1, u2 satisfies, and in particular which (two) conditions on � express the couplingof u1 and u2.

Solution. If we restrict to v 2 C10 .�k/, (6.34) reduces to

ak .u; v/ DZ

fkv dx;

where fk D fj k, and then

Lku D � div�Ak .x/ruk

�D fk in �k .k D 1; 2/ . (6.35)

On the other hand, as u 2 H 10 .�/, it follows u2 D 0 on @�, and on � the traces of u1

and u2 must coincide (Problem 5.2.30 on page 309, Chap. 5):

u1 D u2 on �

which gives us the first constraint.To find the second condition we proceed formally, multiplying (6.35) by v 2 C10 .�/,

integrating on�k and using Gauss’s formula; if �k is the unit normal of � , pointing away


from �k , and

��k D .Ak/T .x/ �k is the outward unit conormal;

then

�Z@ 1

@uk

@��k

vd� C ak .uk ; v/ DZ k

fkv dxIadding these gives Z

�

v

�@u1

@��1C @u2

@��2

d� D 0:

As v was arbitrary we obtain the (transmission condition)

@u1

@��1C @u2

@��2D 0. (6.36)

Remark. As u 2 H 10 .�/, its trace on � belongs to H 1=2 .�/ but the trace of the normal

and conormal derivatives is, a priori, not well defined. Notwithstanding, for functions suchthat Lku 2 L2 .�k/, k D 1; 2, as in the present case, it is possible to define @��

1u1C@��

2u2

as an element of the dual of H 1=2 .�/ and to give a meaining to (6.36).

6.2.3 Evolution problems

Problem 6.2.18 (Drift-diffusion-reaction). Consider the problem8<:ut � .a .x/ ux/x C b .x/ ux C c .x/ u D f .x; t/ 0 < x < 1; 0 < t < T;

u .x; 0/ D u0 .x/ 0 � x � 1;

u .0; t/ D 0; u .1; t/ D k .t/ 0 � t � T:

a) Modify the unknown function so to reduce to homogeneous Dirichlet conditions.

b) Suitably choosing the functional space, write a weak formulation.

c) Discuss whether the problem is well posed, clarifying the conditions on the coeffi-cients and data, and giving an energy estimate of the solution.

Solution. a) Let us proceed formally. We reduce first to vanishing boundary condi-tions. We interpolate the Dirichlet data on the interval Œ0; 1� using

g .x; t/ D xk .t/

and set w D u � g. Note

g .0; t/ D 0 and g .1; t/ D k .t/ :

Then

ut .x; t/ D wt .x; t/C x Pk .t/ and ux .x; t/ D wx .x; t/C k .t/ :


Substituting into the differential equation gives the following problem for w:8<:wt � .a .x/wx/x C b .x/wx C c .x/w D F .x; t/C k .t/ a0.x/ 0 < x < 1; 0 < t < T;

w .x; 0/ D w0 .x/ 0 � x � 1;

w .0; t/ D 0, w .1; t/ D 0 0 � t � T;

whereF .x; t/ D f .x; t/ � Œb .x/C xc .x/�k .t/ � x Pk .t/

andw0 .x/ D u0 .x/ � k .0/ x:

b) Let .�; �/ be the inner product on L2 .0; 1/ and set

B.u; v/ DZ 1

0

�a .x/ u0 .x/ v0 .x/C b .x/ u0 .x/ v .x/C c .x/ u.x/v .x/

�dx

for every pair u; v 2 V D H 10 .0; 1/. On V we choose ku0kL2.0;1/ as norm.

Now set w .t/ D w .�; t /, interpreting w .x; t/ as function of t with values in V . Aweak formulation is: find w 2 L2 .0; T IV / such that Pw 2 L2 .0; T IV �/ and:

i)d

dt.w .t/ ; v/C B .w .t/ ; v/ D .F .t/ ; v/ � k.t/.a; v0/

for every v 2 V , in distributional sense in .0; T /.

ii) kw .t/ � w0kL2.0;1/ ! 0 as t ! 0C.

c) For the problem to be well posed we require the coefficients a, b and c be bounded,and the diffusion coefficient a positive12; precisely:

ja .x/j ; jb .x/j ; jc .x/j � M; a .x/ � a0 > 0 a.e. on .0; 1/ .

We also need the functional

v 7! .F .t/ ; v/ � k.t/.a; v0/to define, for almost every t , an element in V �. As a, b and c are bounded, it suffices toask

f 2 L2 �0; T IL2 .0; 1/� and k 2 H 1 .0; T / :

Finally, we assume w0, or equivalently u0, belongs to L2 .0; 1/.There remains to examine the bilinear form B . To use the Faedo-Galerkin theory we

have to check if B is continuous and weakly coercive on V .

12 Then the equation is uniformly parabolic.


Using the boundedness of the coefficients and Schwarz’s and Poincaré’s inequalities(L2 D L2.0; 1/), we have:

jB.u; v/j � M

Z 1

0

ˇu0v0 C u0v C uv

ˇdx

� M��u0��

L2

��v0��L2 C ��u0��

L2 kvkL2 C kukL2 kukL2

�� M

�1C CP C C 2P

� ��u0��L2

��v0��L2

so that B is continuous on V .Let us check the weak coercivity:

B .u; u/ DZ 1

0

�a .x/ .u0/2 C b .x/ u0uC c .x/ u2

�dx:

Concerning the middle integral term, we may write, for every " > 0,ˇZ 1

0

b .x/ u0udxˇ

� M��u0��

L2 kukL2 � M"

2

��u0��2L2 C M

2"kuk2L2

so that

B .u; u/ ��a0 � M"

2

��u0��2L2 �

�M

2"CM

kuk2L2 :

Choose " D a0

M. Then, setting �0 D M2

2a0CM we have

B .u; u/ � a0

2

��u0��2L2 � �0 kuk2L2

which entails that the bilinear form

B0 .u; v/ D B .u; v/C �0.u; v/

is coercive on V or, equivalently, B is weakly coercive (with coercivity constant a0=2).Setting z .t/De��0tw .t/, the function z is a solution to:

i 0)d

dt.z .t/ ; v/C B0 .z .t/ ; v/ D .e��0tF .t/ ; v/

for every v 2 V , in distributional sense in .0; T /.

ii0) kz .t/ � w0kL2 ! 0 as t ! 0C.


Under the mentioned hypotheses, the problem has a unique weak solution z. From thetheory we then obtain an energy estimate, written in terms of z .x; t/:

maxt2Œ0;T �

Z 1

0

z2 .x; t/2 dx C a0

Z T

0

Z 1

0

z2x .x; t/ dxdt

� e.�0C1/T´Z T

0

Z 1

0

e��0tF 2 .x; t/ dxdt CZ 1

0

z20 .x/ dx

μ:

At this point it is not hard to transfer the conclusions back to the original problem.

Problem 6.2.19 (Asymptotic stability). Consider8<:ut ��u D 0 x 2 �; t 2 .0; T /u.x; 0/ D g.x/ x 2 �u.� ; t / D 0 � 2 @�; t 2 .0; T /;

where � is bounded and regular, and g 2 L2.�/.a) Write a weak formulation. Using the Faedo-Galerkin method with the subspaces

spanned by the Dirichlet eigenfunctions of the operator �� ([18, Chap. 8, Sect. 3]),deduce the existence of a solution for every t > 0, and find its expression.

b) If u denotes the solution, prove that

krukL2. / � 1p2et

kgkL2. / for every t > 0:

c) If g 2 H 10 .�/, prove that

krukL2. / � e��1tkrgkL2. / for every t > 0;

where �1 is the first eigenvalue of ��.

Solution. a) Let us set u .t/ D u .�; t /, by interpreting u .x; t/ as a function in t withvalues in V D H 1

0 .�/ (this choice is dictated by the boundary conditions). Call .�; �/ theusual scalar product on H D L2.�/ and define

B.u; v/ DZ

ru � rv dx:

A weak formulation is: seek u 2 L2 .0; T IV / such that Pu 2 L2 .0; T IV �/ and

i)d

dt.u .t/ ; v/C B .u .t/ ; v/ D 0

for every v 2 V , in distributional sense on .0; T /.

ii) ku .t/ � gkH ! 0 for t ! 0C.


The conditions for being well posed are fulfilled (the bilinear form is the scalar producton V , and hence continuous and uniformly coercive in t ).

Let 0 < �1 < �2 � �3 � : : : be the eigenvalues of

u 2 V W ��u D �u

and '1; '2; : : : the corresponding normalised eigenfunctions in L2 .�/. These form acomplete orthonormal system in L2.�/ and a complete orthogonal system in V . Recallthat for every v

B.'k ; v/ D �k.'k; v/ (6.37)

whencekr'kk2

L2 D �kk'kk2L2 D �k :

Let Vm be the space spanned by the first m eigenfunctions. Set

um DmXkD1

cm;k.t/'k and g D1XkD1

gk'k :

By (6.37) the Faedo-Galerkin approximating problem is

0 DmXkD1

. Pcm;k .t/ 'k; v/C B�cm;k.t/'k; v

� DmXkD1

�� Pcm;k.t/C �kcm;k.t/�'k; v

�and the coefficient cm;k solves Pcm;k.t/C �kcm;k.t/ D 0 with initial condition cm;k.0/ Dgk . Therefore

cm;k.t/ D gke��k t :

The approximate solutions are then

um .t/ DmXkD1

gke��k t'k

which converge in L2 .0; T ;V / to the solution

u.t/ D1XkD1

gke��k t'k.x/

as m ! 1. In practice we have used variable separation.

b) Since 'k form an orthogonal basis in V , we can write, back in the original notation,

ru.x; t / D1XkD1

gke��k tr'k.x/

and from (6.37) we have

kru.x; t /k2L2. /

D1XkD1

jgkj2�ke�2�k t :


The function f .z/ D ze�az has a maximum at z D 1=a, with f .1=a/ D 1=ae, so forevery k � 1

�ke�2�k t � 1

2et:

Hence

kru.x; t /kL2. / � 1p2et

kgkL2. /, t > 0:

c) If g is in H 10 .�/ we can get a better estimate, since

rg D1XkD1

gkr'k

and so

krgk2L2. /

D1XkD1

gk2kr'kk2

L2. /D

1XkD1

jgkj2�k :

Then

kru.x; t /k2L2. /

D1XkD1

jgk j2�ke�2�k t � e�2�1t

1XkD1

jgkj2�k D e��1tkrgk2L2. /

as required.

Remark. From b) and c) we also see that if f D f .x/ belongs, for instance, to L2 .�/,the solution to the evolution problem8<

:ut ��u D f in � � .0; T /u .x; 0/ D g .x/ in �

u .� ; t / D 0 on @� � .0; T / ;converges in H 1

0 .�/-norm to the solution u1 of the stationary problem´��u1 D f in �

u1 .� / D 0 on @�

at a rate depending on the regularity of the initial datum. In such a case we say that u1 isasymptotically stable.

Problem 6.2.20 (Wave equation, Neumann conditions). Consider8<:ut t � uxx D f .x; t/ in QT D .0; 1/ � .0; T / ;u .x; 0/ D u0 .x/ , ut .x; 0/ D u1 .x/ in Œ0; 1�

ux .0; t/ D ux .1; t/ D 0 for 0 � t � T .

After finding a weak formulation, discuss whether the problem is well posed and givean energy estimate of the solution.


Solution. Set H D L2 .0; 1/ with inner product .v; w/ and V D H 1 .0; 1/. Supposef 2 L2 .QT /, u0 2 H 1 .0; 1/, u1 2 L2 .0; 1/. Finally, let

B.u; v/ DZ 1

0

uxvx dx:

A possible weak formulation is: find a function u .t/ D u .�; t / such that

u 2 L2 .0; T IV / ; Pu 2 L2 .0; T IH/ ; Ru 2 L2 �0; T IV �� ;together with:

i) For every v 2 V ,

d2

dt2.u; v/C B.u; v/ D .f .t/ ; v/

in D 0 .0; 1/ and for almost every t 2 .0; T /.ii) ku .t/ � u0kV ! 0 and kPu .t/ � u1kH ! 0 as t ! 0C.

To check that the problem is well posed we can use the Faedo-Galerkin theory. In factthe bilinear form B.u; v/ is continuous and weakly coercive on V , whilst the functional

Fv D .f .t/ ; v/

is continuous on V . So there exists a unique weak solution u, and

maxŒ0;T �

ku .t/k2V � eT

´ku0kV C ku1kH C

Z T

0

kf .s/k2H ds

μ:

Problem 6.2.21 (Wave equation with concentrated reaction). Consider the problem8<:ut t � uxx C u .x; t/ ı .x/ D f .x; t/ in QT D .a; b/ � .0; T / ;u .x; 0/ D u0 .x/ , ut .x; 0/ D u1 .x/ in Œa; b�

u .a; t/ D u .b; t/ D 0 for 0 � t � T

where ı .x/ is the Dirac distribution centred at the origin.

a) Write a weak formulation.

b) Say if the problem is well posed, and in that case give an energy estimate for thesolution.

Solution. a) If 0 62 .a; b/ the problem reduces to the standard wave equation, sowe assume 0 2 .a; b/. Suppose f 2 L2 .QT /, u0 2 H 1

0 .a; b/, u1 2 L2 .a; b/.Let H D L2 .a; b/, with scalar product .�; �/, and V D H 1

0 .a; b/ with inner product.v; w/V D .ux; vx/ . Recall that V � C .Œa; b�/, so the differential equation can be writ-ten as

ut t � uxx C u .0; t/ D f .x; t/ .


A possible weak formulation is: find u .t/ D u .�; t / such that

u 2 L2 .0; T IV / ; Pu 2 L2 .0; T IH/ ; Ru 2 L2 �0; T IV �� ;and satisfying:

i) For every v 2 V ,

d2

dt2.u; v/C .ux; vx/C u .0; t/ v .0/ D .f; v/

in D 0 .a; b/ for almost every t 2 .a; b/.ii) ku.t/ � u0kV ! 0 and kPu.t/ � u1kH ! 0 as t ! 0C.

b) The bilinear form a .w; v/ D .wx ; vx/ C u .0; t/ v .0/ is continuous and coerciveon V . In fact, by Problem 5.2.24 on page 301 (Chap. 5), we have

ja .w; v/j � .1C .b � a// kwxkH kvxkHand

a .w;w/ D .wx ; wx/C w2 .0/ � kwxk2H .

Hence there exists a unique weak solution u, and moreover

maxŒ0;T �

kux .�; t /k2H � eT

´ku0kV C ku1kH C

Z T

0

kf .�; s/k2H ds

μ: (6.38)

This estimate can be deduced from the general theory, but it can also be found directly:multiply the PDE by ut and integrate on (a; b/:Z b

a

.ut tut � uxxut C u2 .0; t// dx DZ b

a

f .x; t/ ut .x; t/ dx.

Integrating the second term by parts, since ut .a; t/ D ut .b; t/ D 0, we haveZ b

a

uxxut dx DZ b

a

uxuxt dx D 1

2

d

dt

Z b

a

u2x .x; t/ dx:

Furthermore,ˇˇZ b

a

f v dx

ˇˇ � kf .�; t /kH kut .�; t /kH � 1

2kf .�; t /k2H C 1

2kut .�; t /k2H :

Hence

d

dt

Z b

a

Œu2t .x; t/C u2x .x; t/� dx C 2

Z b

a

u2.0; t/ dx � kf .�; t /k2H C kut .�; t /k2H :


Integrating on .0; t/, t � T , we find

kut .�; t /k2H C kux .�; t /k2H� ku1k2H C ku0k2V C

Z t

0

kf .�; s/k2H ds CZ t

0

kut .�; s/k2H ds (6.39)

and in particular

kut .�; t /k2H � ku1k2H C ku0k2V CZ t

0

kf .�; s/k2H ds CZ t

0

kut .�; s/k2H ds:

Gronwall’s lemma13 now gives

kut .�; t /k2H � et²

ku1k2H C ku0k2V CZ t

0

kf .�; s/k2H ds³

and thenZ T

0

kut .�; t /k2H � .eT � 1/´

ku1k2H C ku0k2V CZ T

0

kf .�; s/k2H dsμ:

Substituting in the right-hand side of (6.39) gives back (6.38).To be more rigorous, (6.38) can be proved for the Galerkin approximations, and ex-

tended to the solution by a limiting procedure.

Problem 6.2.22 (Oblique-derivative conditions for the heat equation). Consider theproblem 8<

:ut ��u D f .x; t / in QT D � � .0; T /u .x; 0/ D g .x/ in �

@�uC h .t/ ut D 0 on ST D @� � .0; T /where � is a bounded Lipschitz domain in Rn and � the outward unit normal to @�.Take f 2 L2 .QT /, g 2 H 1 .�/ and h 2 C.Œ0; T �/ such that 0 < h0 � h .t/ � h1.

a) Let H D L2 .�/ and V D H 1 .�/; write a weak formulation.

b) If Vm and um denote the Galerkin approximations of V and u respectively, findenergy estimates for um, @tum and @t .eum/, whereeum is the tracea of um on @�.

c) Deduce existence, uniqueness and an energy estimate for the solution to the originalproblem.

a To avoid confusion, here we will use a specific notation for the trace of a function on theboundary.

13 [18, Chap. 10, Sect. 3].


Solution. This problem does not belong to the so-called standard theory, yet we canuse the latter with a few modifications.

a) As usual we use the notation u .t/ D u .x; t /, so that Pu D @tu, and similarly forany function depending on t . Let us first proceed formally, multiplying by v 2 V andintegrating by parts; keeping the boundary condition into account,Z

Puv dx CZ@

h .t/ Peu .t/ev d�CZ

ru .t/ � rv dx DZ

f .t/ v dx:

As f 2 L2 .QT / and g 2 H 1 .�/, we seek solutions

u 2 L2 .0; T IV / with Pu 2 L2 .0; T IH/ :In particular, u 2 C .Œ0; T �IH/. Due to the presence of an integral involving the trace ofPu .t/ on @� we also require that

Peu 2 L2 �0; T IL2 .@�/� .

Moreover:i) For every v 2 V and a.e. on .0; T / ;

. Pu .t/ ; v/C h .t/ . Peu .t/ ;ev/L2.@ / C B.u .t/ ; v/ D .f .t/ ; v/;

where, as usual, .�; �/ is the inner product on L2 .�/ and B .w; v/ D .rw;rv/. To com-plete the weak formulation we require that

ii) u .t/ ! g in H as t ! 0C.

b) The Galerkin approximate problem for um is then:

. Pum .t/ ; v/C h .t/ . Peum .t/ ;ev/L2.@ / C B.um .t/ ; v/ D .f .t/ ; v/ (6.40)

for every v 2 Vm.If ¹w1; : : : ; wmº is an orthonormal basis of Vm in V and orthogonal in H , set

um .t/ DmXkD1

cmk .t/ wk and Gm DmXkD1

gkwk ; with gk D .g;wk/V :

As g 2 V , then Gm ! g in V . Substituting these expressions into (6.40) and choosingv D ws , s D 1; : : : ; m, we find the following ODE system for the unknown coefficientscmk .t/:

mXkD1

Mks .t/ Pcmk .t/C cms .t/ D fs .t/ , s D 1; : : : ; m (6.41)

where fs .t/ D .f .t/ ; ws/ and

Mks .t/ D kwkk2H ıks C h .t/ .wk ; ws/L2.@ /

with initial conditioncmk .0/ D gk , k D 1; : : : ; m.


The matrix .Mks .t// is positive definite: indeed

Xk;s

Mks�k�s DXk

kwkk2H j�kj2 C h

��Xk

�kwk

��L2.@ /

:

As fs 2 L2 .0; T / and g 2 V , there is a unique solution

.cm1 .t/ ; : : : ; cmm .t// 2 H 1 .0; T I Rm/

of (6.41), corresponding to um 2 H 1 .0; T IV /. Note that Pum .t/ 2 V , whose trace Peum .t/is in L2 .@�/ for almost every t . Setting v D Pum .t/ in (6.40) we may write

kPum .t/k2H C h .t/�� Peum .t/��2L2.@ /

C 1

2

d

dtkrum .t/k2H D .f .t/ ; Pum .t//I

recalling that h .t/ � h0 > 0 and

j.f .t/ ; Pum .t//j � 1

2kf .t/k2H C 1

2kPum .t/k2H ;

it follows

kPum .t/k2H C 2h0�� Peum .t/��2L2.@ /

C d

dtkrum .t/k2H D kf .t/k2H .

Integrating from 0 to t � T , we obtain

kPumk2L2.0;t IH/ C 2h0�� Peum��2L2.0;t IL2.@ //

C krum .t/k2H � krgk2H C kfk2L2.0;t IH/ .(6.42)

This means, in particular, that

Pum is equi-bounded on L2.0; T IH/Peum is equi-bounded on L2.0; T IL2 .@�//rum is equi-bounded on L1.0; T IH/:

Only the norm of um in L2.0; T IH/ is missing. Set v D um .t/ in (6.40), so that

1

2

d

dtkum .t/k2H C h .t/ . Peum .t/ ;eum .t//L2.@ / C krum .t/k2H D .f .t/ ;um .t//: (6.43)

As

keum .t/k2L2.@ / � C��kum .t/k2H C krum .t/k2H

�;

we haveˇh.t/. Peum .t/ ;eum .t//L2.@ /

ˇ � h1�� Peum .t/��L2.@ /

keum .t/kL2.@ /

� C1�� Peum .t/��2L2.@ /

C 1

2

�kum .t/k2H C krum .t/k2H

�;


where C1 D h21C2�

2. Then (6.43) becomes

1

2

d

dtkum .t/k2H C krum .t/k2H � 1

2

�kf .t/k2H C kum .t/k2H

�C C1

�� Peum .t/��2L2.@ /C 1

2

�kum .t/k2H C krum .t/k2H

�;

which implies

d

dtkum .t/k2H � kf .t/k2H C 2 kum .t/k2H C 2C1

�� Peum .t/��2L2.@ /:

Integrating between 0 and t � T , and exploiting (6.42),

kum .t/k2H � kgk2H C kfk2L2.0;t IH/ C 2

Z t

0

kum .s/k2H ds C 2C1�� Peum��2L2.0;t IL2.@ //

� C2

hkgk2V C kfk2L2.0;t IH/

iC 2

Z t

0

kum .s/k2H ds;

where C2 D 1C C1

h0. By Gronwall’s lemma

kum .t/k2H � e2tC2

hkgk2V C kfk2L2.0;T IH/

i;

and therefore

um is bounded in L1.0; T IH/:

c) By the results of b), there is a subsequence, still denoted by um, such that

um * u in L2.0; T IV /Pum * Pu in L2.0; T IH/Peum * Peu in L2.0; T IL2 .@�//.

This is enough to pass to the limit in (6.40) and then conclude that u is the unique weaksolution of the original equation.


Problem 6.2.23 (Wave equation, equipartition of energy). Let � � R3 be a boundeddomain, QT D � � .0; T /, V D H 1

0 .�/, H D L2 .�/. Given u0 2 V and u1 2 H ,call u 2 L2 .0; T IV / the (unique) weak solution to´

Ru .t/ ��u D 0 in QT ;

u .0/ D u0, Pu .0/ D u1

such that Pu 2 L2 .0; T IH/ and Ru 2 L2 .0; T IV �/.a) Let Vm D span¹w1; : : : ; wmº, where the

®wj¯

are the Dirichlet eigenfunctions of��, and ¹umº the Galerkin sequence ([18, Chap. 10, Sect. 7]) approximating u.Prove that

Em .t/ D 1

2kPum .t/k2H C 1

2krum .t/k2H D Em .0/ ; for every t � 0:

b) Setting

Km .t/ D 1

2t

Z t

0

kPum .s/k2H ds, Pm .t/ D 1

2t

Z t

0

krum .s/k2H ds;

deduce from a) that, for every m � 1,

Km .t/ ! Em .0/

2, Pm .t/ ! Em .0/

2as t ! C1: (6.44)

Solution. a) Call .�; �/ the inner product of L2 .�/. The general theory, in the presentcontext, says that um 2 H 2 .0; T ;V / satisfies

. Rum .t/ ; v/C .rum .t/ ;rv/ D 0 (6.45)

for every v 2 V and almost all t 2 .0; T /. As Pum 2 L2 .0; T IV / for almost every givent , we can insert v D Pum .t/ into (6.45); then

. Rum .t/ ; Pum .t//C .rum .t/ ;r Pum .t// D 1

2

d

dtkPum .t/k2H C 1

2

d

dtkrum .t/k2H D 0

whence

Em .t/ � 1

2kPum .t/k2H C 1

2krum .t/k2H D Em .0/ ; t � 0: (6.46)

b) From (6.46),

Km .t/C Pm .t/ D 1

2t

Z t

0

°kPum .s/k2H C krum .s/k2H

±ds D Em .0/ . (6.47)

As um 2 L2 .0; T IV /, for almost every given t , we set v D um .t/ in (6.45); then

. Rum .t/ ;um .t//C krum .s/k2H D 0:


Integrating by parts,Z t

0

. Rum .s/ ;um .s//ds D �Z t

0

. Pum .s/ ; Pum .s//ds C . Pum .t/ ;um .t// � .u1; u0/:

Hence

Pm .t/ �Km .t/ D �.u1; u0/C . Pum .t/ ;um .t//2t

. (6.48)

On the other hand, from (6.46) and Poincaré’s inequality

j. Pum .t/ ;um .t//j � 1

2kPum .t/k2H C 1

2kum .t/k2H � 1

2kPum .t/k2H C C 2P

2krum .t/k2H

� max®1; C 2P

¯Em .0/ .

Therefore Pm .t/ �Km .t/ ! 0 as t ! C1, which together with (6.47) implies

Km .t/ ! Em .0/

2, Pm .t/ ! Em .0/

2as t ! C1.

Remark. Condition (6.44) is also satisfied by the solution u. The proof requires a lit-tle more work, especially for proving (6.46). The difficulty lies in the fact that Pu 2L2 .0; T IH/ and Ru 2 L2 .0; T IV �/, so we cannot insert Pu directly into the weak equation

h Ru .t/ ; vi� C .ru .t/ ;rv/ D 0

where h Ru .t/ ; vi� is the pairing of V and V �. To do that we would need a better regularityfor Pu, for instance Pu 2 L2 .0; T IV /.

Problem 6.2.24 (Mixed boundary conditions). Let

� D ®x 2 R2 W x21 C 4x22 < 4

¯; �D D @� \ ¹x1 � 0º ; �N D @� \ ¹x1 < 0º :

Consider the problem8<ˆ:ut � div .A˛ru/C b � ru � ˛u D x2 in � � .0; T /u.x; t / D 0 on �D � .0; T /�A˛ru � n D cos x1 on �N � .0; T /u.x; 0/ D H .x1/ on �

(6.49)

where

A˛ D1 0

0 ˛ ex21Cx2

2

�; b D

sin.x1 C x2/

x21 C x22

�and H is the Heaviside function. For which values of the parameter ˛ 2 R is the prob-lem parabolic? For such values, give a weak formulation of the problem and deduceexistence and uniqueness of the solution.


Solution. For every � 2 R2, we have

A˛.x1; x2/� � � � �21 C ˛�22 � min.1; ˛/.�21 C �22/: (6.50)

Therefore, the problem is parabolic for every ˛ > 0.We have a mixed problem with homogeneous Dirichlet conditions on the boundary

�D and, then, we are allowed to use the Poincaré inequality; we consider

V D H 10;�D

.�/; with kvk2V D krvk2L2. /

:

Let us proceed formally multiplying the first equation of the problem (6.49) by v 2 V andintegrating

h Pu; vi� CZ

A˛ru � rv dx �Z�N

A˛ru � n v d� CZ

b � ru v dx

� ˛Z

u v dx DZ

x2v dx:

On account of the third equation of problem (6.49), we find

h Pu; vi� CZ

A˛ru � rv dx CZ

b � ru v dx � ˛Z

u v dx

DZ

x2v dx �Z�N

cos x1 v d�:

Define:

B.u; v/ DZ

A˛ru � rv dx CZ

b � ru v dx � ˛Z

u v dx

Fv DZ

x2v dx �Z�N

cos x1 v d�:

The weak formulation of problem (6.49) consists in finding u 2 L2.0; T IV / such thatPu 2 L2.0; T IV 0/, u.0/ D u0 in �, and for every v 2 V , a.e. t in Œ0; T �,

h Pu.t/; vi� C B.u.t/; v/ D Fv:

We prove the continuity of B and F and the weak coercivity of B .We recall that for the functions in V a trace inequality kvk�N

� CT kvkV holds. Then,using also Schwarz’s inequality, we can write:

jFvj ��Z

x22dx1=2

kvkL2. / C�Z

�N

cos2 x1d�

1=2kvk�N

��j�j1=2 C j�N j1=2CT

�kvkV D .

p� C p

3CT /kvkV

where j�j and j�N j denote the measures of the sets � and �N .


Since in� we have x21 Cx22 � 4, using once more Schwarz’s and Poincaré’s inequal-ities we have

jB.u; v/j �Z

jux1vx1

C ˛ex21Cx2

2ux2vx2

jdx C ˛

Z

juvj dx

CZ

j sin.x1 C x2/ux1C .x21 C x22/ux2

jjvjdx

� max¹1; ˛e4ºZ

jru � rvjdx C ˛kukL2. /kvkL2. / C 5krukL2. /kvkL2. /

� �max¹1; ˛e4º C ˛C 2P C 5CP

� kukV kvkV :

To estimate the drift term, we use Young’s inequality:ˇZ

ˇ � ru v dx

ˇ�Z

.j sin.x21 C x22/jjux1jjvj C .x21 C x22/juy jjvj/dx �

Z

.jux1jjvj C 4juy jjvj/dx

� "kux1k2L2. /

C 1

4"kvk2

L2. /C"kux2

k2L2. /

C 4

"kvk2

L2. /� "kuk2V C 17

4"kvk2

L2. /:

Thanks to the uniform ellipticity of A˛ proved in (6.50), we deduce that

B.u; u/C �kukL2. � .min.1; ˛/ � �/ kuk2V C�� 17

4"� ˛

kuk2

L2. /:

Now we choose " and, consequently, � in order that the two coefficients appearing in theright-hand side are positive. Hence, the weak coercivity of the bilinear form B is proved.

We conclude that problem (6.49) has a unique weak solution, satisfying the estimate

kukC.Œ0;T �IH/; kukL2.0;T IV /; kPukL2.0;T IV �/ � C.�/:


6.3.1. Establish whether the bilinear forms below are continuous on the given Hilbert spaces, andunder which conditions they are coercive (or weakly coercive).

a) H D Rn, a.x; y/ DnX

i;jD1aij xiyj , where .aij / is a symmetric n � n-matrix.

b) H D H1.0; 1/,

a.u; v/ DZ 1

0A.x/u0v0dx C

Z 1

0B.x/u0v dx C

Z 1

0C.x/uv dx;

A;B; C in L1.a; b/.


c) H D H10 .�/, � domain in Rn,

a.u; v/ DZ ˛.x/ru � rv dx;

with ˛ 2 L1.�/.6.3.2. Given the following boundary-value problems on .0; 1/, write the weak formulations and

discuss the applicability of the Lax-Milgram theorem.

a) �u00 C .e�tu/0 D 4, u.0/ D 1, u.1/ D 2.b) �.4C t2/u00 C 3u D sin t , u0.0/ D 1, u0.1/C u.1/ D 0.c) �u00 C u0 C u D 0, u.0/ D u.1/, u0.0/ D u0.1/.

6.3.3. Discuss, in the two cases below, whether

B.u; v/ D Fv; for every v 2 Vis a well-posed problem. Then write the corresponding boundary-value problem in classical form.

a) V D H1.0; 1/, Fv D v.1/, B.u; v/ DZ 1

0.x C 1/u0v0 dx C u.0/v.0/:

b) V D ¹v2H1 .0; 2/ W v.2/ D 0º,

Fv DZ 2

0f v dx, B.u; v/ D

Z 2

0

hu0v0 � 2x2u0v � 4xuv

idx:

6.3.4. Referring to Problem 6.2.3 (page 340), write a variational formulation of the Robin-Dirich-let problem: ´

Lu � cos x u00 � sin x u0 � xu D f 0 < x < L

u0.0/ D �u.0/; u.L/ D 0;

where 0 < L < �=2 and f 2 L2 .0; L/. Discuss the solvability of the problem.

6.3.5. (Hermite equation) Let

V D°v W R ! R W e�x2=2v 2 L2 .R/ , e�x2=2v0 2 L2 .R/

±:

a) Verify that the formula

.u; v/V DZ

RŒuv C u0v0�e�x2

dx (6.51)

defines an inner product which makes V into a Hilbert space.b) Study the variational problem

.u; v/V D Fv �Z

Rf v e�x2

dx for every v 2 V , (6.52)

with f 2 V , and interpret it in the classical sense.

6.3.6. a) Given� � Rn, a bounded regular domain, f 2 L2.�/ and g 2 L2.@�/, say to whichproblem the following variational formulation is associatedZ

ru � rv dx C

Z uv dx D

Z f v dx C

Z@ gv d�; for every v 2 H1.�/: (6.53)

b) Is this well posed?


6.3.7. Let B1 denote the unit ball in Rn. Consider the subspace of H1.B1/ given by

V D²u 2 H1.B1/ W

Z@B1

ud� D 0

³and the following variational problem: determine u 2 V such thatZ

B1

.ru � rv C uv/ dx DZB1

f v dx; for every v 2 V:

Can one apply the Lax-Milgram theorem? To which problem does the variational formulation cor-respond?

6.3.8. Given a bounded regular domain � � Rn, consider the system

8<:

��u1 C u1 � u2 D f1 in �

��u2 C u1 C u2 D f2 in �

u1 D u2 D 0 on @�:

Write a weak formulation and apply the Lax-Milgram theorem to prove the existence and uniquenessof the solution, for every f D .f1; f2/ 2 L2.�/ � L2.�/.

6.3.9. Let � D .0; 1/ � .0; 1/, and set

�N D ¹.0; y/ W 0 < y < 1º

and �D D @� n �N . Study8<:

� div .A .x; y/ruC bu/ D f in �

u D 0 on �D3ux .0; y/ � 4uy .0; y/C hu D 0 0 < y < 1

where f 2 L2 .�/, h > 0 is constant and

A .x; y/ D�2C sin xy 3

�3 4 � sin xy

, b D .�4x � y;�2x C y/ :

6.3.10. In relationship to Problem 6.2.12 on page 355, consider

´��uC b � ru D f x 2 �@�u D 0 x 2 @�;

with b 2 Rn constant.

a) Multiply the equation by e�b�x, and write it in divergence form.

b) Provide a necessary and sufficient condition for the problem to admit solutions.


6.3.11. Let Q D .0; �/ � .0; �/. In relationship to Problem 6.2.13, study the Dirichlet problem´�uC 2u D div f in Q

u D 0 on @Q;

whenf D .log .sin x/; 0/:

6.3.12. (Alternating method, Neumann conditions) In relationship to Problem 6.2.15 let � D�1 [ �2 � R2 be a domain, where �1, �2 have regular boundary and �1 \ �2 ¤ ;. AdaptSchwarz’s alternating method to build a sequence ¹unº converging inH1 .�/ to the solution of theNeumann problem

��uC u D f in � , @�u D 0 on @�, (6.54)

where f 2 L2 .�/ and � is the outward normal to @�.

6.3.13. Let B1 D ®.x; y/ : x2 C y2 < 1

¯. Write the variational formulations of the following

problems and discuss solvability:

a)

´�u D 0 in B1.x C y/ux C .y � x/ uy D 1C ˛x2 on @B1

˛ 2 R:

b)

´��u D �

x2 � y2�n in B1.x C y/ux C .y � x/ uy D 0 on @B1

n 2 N:

6.3.14. (Obstacle problem) Let � be a bounded, convex domain in Rn and a strictly concavefunction14 on � such that max > 0 and < 0 on @�. Define

K D°v 2 H1

0 .�/ W v � a.e. in �±:

a) Check that K is convex and closed in H10 .�/.

b) Show there is a unique function u 2 K that minimises the functional

J .v/ D 1

2

Z

jruj2 dx

on K, characterised by the following variational inequality:Z .rv � ru/ � rudx �0; for every v 2 K. (6.55)

c) Deduce that, if u 2 H2.�/, then u solves the obstacle problem if and only if

��u � 0, u � � 0 and �u .u � / D 0 a.e. on �:

In particular, u is a harmonic function on every open set where u > .

14 A concave function is always Lipschitz.


6.3.15. Let B1 � R2 be the unit disc centred at the origin,QT D B1� .0; T /, �D and �N opensubsets in @� with �D [ �N D @�. Consider the mixed problem:8<

ˆ:ut � div.xeyru/C .sign .xy/u/y D f in QT ;

u .x; y; 0/ D u0 .x; y/ in B1;

u D g on �D ;

@�uC u D 0 on �N ;

where f D f .x; y; t/, u0 D u0 .x; y/ and g D g .t/ are given functions.

a) Using appropriate function spaces, write a weak formulation.b) Study existence and uniqueness and give a stability estimate.

6.3.16. Let � be a bounded domain in Rn and QT D � � .0; T /. Let �D and �N be open in@� with �D [ �N D @�. Consider the mixed problem:8<

ˆ:ut � div.a .x/ru � bu/ � c .x/ u D 0 in QT ;

u .x; 0/ D g .x/ in �;

u D 0 on �D ;

a .x/ @�uC hu D 0 on �N ;

where g 2 L2 .�/ and h 2 R is constant.

a) Using suitable function spaces, write the weak formulation.b) Under which conditions on the coefficients a, b, c and h is the problem well posed?

6.3.17. The potassium concentration c.x; y; z; t/ in a cell of spherical shape � and radius R,with boundary � , satisfies the evolution problem8<

:ct � div.rc/ � �c D 0 in � � .0; T /rc � � C �c D �cext on � � .0; T /c.x; y; z; 0/ D c0.x; y; z/ on �

(6.56)

where cext is the given external concentration which is constant, � and � are positive numbers and is a strictly positive function. Write the weak formulation and analyse the well-posedness, providingsuitable assumptions on the coefficients and on the data.

6.3.1 Solutions

Solution 6.3.1. a) The form is always continuous (the setting is finite-dimensional), and is coer-cive if and only if the matrix A is positive definite (with coercivity constant equal to the smallesteigenvalue of A). Let us remind that in finite dimensions weak coercivity coincides with coercivity,for there are no dense subspaces in Rn other than Rn itself.

b) The form is continuous (Lp D Lp.0; 1/):

ja.u; v/j � kAkL1ku0kL2kv0kL2 C kBkL1ku0kL2kvkL2 C kCkL1kukL2kvkL2 �� .kAkL1 C kBkL1 C kCkL1/ kukH1kvkH1 :


To prove the coercivity, we have to estimate from below

a.u; u/ DZ 1

0A.x/.u0/2dx C

Z 1

0B.x/u0udx C

Z 1

0C.x/u2 dx:

Here we can provide different conditions. A necessary one is A � A0 > 0. In fact, if A is nega-tive on a subinterval we can construct functions u with kukL2 arbitrarily small and u0 concentratedwhere A is negative, making A.u; u/ negative.

There remains to estimate the mixed term. Let jB .x/j � B0, B 0 .x/ � B1; then by Prob-lem 5.2.24, Chap. 5 (page 301):Z 1

0B.x/u0udx �

B .x/

2u2.x/

�10

�Z 1

0B 0.x/u2 dx

� �B0kuk2L1 � B1 kukL2 � �B0kuk2H1 � B1 kukL2 :

If C .x/ � C0, we can write

a.u; u/ � .A0 � B0/Z 1

0.u0/2dx C .C0 � B1/

Z 1

0u2 dx:

Finally, if A0 � B0 > 0 and C0 � B1 > 0 the form is coercive. Assuming A0 � B0 > 0 we canonly guarantee weak coercivity (for the Hilbert triple ¹H1 .0; 1/ ; L2 .0; 1/ ; .H1 .0; 1//0º).

c) The form is continuous, for

ja.u; v/j � k˛kL1krukL2kkrvkL2 � k˛kL1krukH1kkrvkH1

by Hölder’s inequality. The form is coercive if and only if ˛ � ˛0 > 0 a.e. on �.

Solution 6.3.2. a) The boundary conditions are inhomogeneous, so it is convenient to definew.t/ D u.t/ � t � 1. As w0.t/ D u0.t/ � 1, w00.t/ D u00.t/, the function w solves´

�w00 � e�tw0 D 4C e�t 0 < t < 1

w.0/ D w.1/ D 0:

The weak formulation is:Z 1

0

�w0v0 � e�tw0v� dt D

Z 1

0

�4C e�t

�v dt for every v 2 H1

0 .0; 1/:

Arguing as in Problem 6.2.1 (page 336) we can apply Lax-Milgram, and obtain that the problem iswell posed.

b) Choose V D H1.0; 1/. For v 2 VZ 1

0�.4C t2/u00v dt D �.4C t2/u0vj10 C

Z 1

0

h.4C t2/u0v0 C 2tu0v

idt D

D �5u0.1/v.1/CZ 1

0


idt D

D 5u.1/v.1/C 4v.0/CZ 1

0


idt:


Hence the weak formulationZ 1

0

h.4C t2/u0v0 C 2tu0v C 3uv

idt C 5u.1/v.1/ D

Z 1

0sin t v dt � 4v.0/; 8v 2 H1.0; 1/:

Arguing as in Problems 5.2.25 (page 302) and 5.2.26 (page 303), one can prove that the right-handside is the sum of two continuous linear functionals on H1.0; 1/. Combining the arguments of Ex-ercise 6.3.1.b) with Problem 5.2.24 (page 301), the bilinear form B.u; v/ is continuous. Finally

B.u; u/ DZ 1

0

h.4C t2/.u0/2 C 2tu0uC 3u2

idt C 5u2.1/

DZ 1

0

h.4C t2/.u0/2 C 2u2

idt C 5u2.1/ �

Z 1

0

h4.u0/2 C 2u2

idt;

hence B is coercive and the Lax-Milgram theorem can be applied.c) V will be a suitable subspace of H1.0; 1/, adapted to the boundary (periodicity) conditions.

To find V let us multiply the equation by v 2 H1.0; 1/ and integrate by parts. Formally,Z 1

0�u00v dt D �u0.1/v.1/C u0.0/v.0/C

Z 1

0.u0/2 dt:

Now if u is in H1.0; 1/, we cannot, in general, make pointwise sense of u0. Hence a correct choicefor test functions must annihilate the first two terms on the right. Since we want u0.0/ D u0.1/, itsuffices to take v.0/ D v.1/. So set

V D H1per � ¹v 2 H1.0; 1/ W v.0/ D v.1/º:

Then one can see15 that V is closed in H1.0; 1/, hence a Hilbert space. The weak formulation is

B .u; v/ �Z 1

0

�u0v0 C u0v C uv

�dt D 0; for every v 2 V:

Supposing u regular and a weak solution, it is enough to integrate by parts the other way aroundatodeduce that u is a strong solution, too. The bilinear form B.u; v/ is continuous on V , and asZ 1

0uu0 D 1

2

hu2 .1/ � u2 .0/

iD 0;

it is also coercive. Hence the problem has only the trivial solution.

Solution 6.3.3. a) The problem is well posed thanks to the Lax-Milgram theorem (proceed asin Exercise 6.3.2.b). Let us find the classical formulation, supposing u 2 C 2.0; 1/ \ C 1.Œ0; 1�/.Integrating by parts,

.x C 1/u0vj10 �Z 1

0

�.x C 1/u00v C u0v

�dx C u.0/v.0/ D v.1/; 8v 2 H1.0; 1/;

15 To prove it use the argument of Problem 5.2.28 on page 306.


i.e.Z 1

0

�.x C 1/u00 C u0

�v dx D .2u0.1/ � 1/v.1/C .�u0.0/C u.0//v.0/; 8v 2 H1.0; 1/:

In particular, the previous identity must hold for every v 2 H10 .0; 1/, which forces

.x C 1/u00 C u0 D 0; in .0; 1/ :

Then.2u0.1/ � 1/v.1/C .�u0.0/C u.0//v.0/ D 0 for every v.0/; v.1/ 2 R:

By choosing v zero at one endpoint and non-zero at the other, and then swapping, we see that usolves the mixed (Robin-Neumann) problem8<

:.x C 1/u00 C u0 D 0 on .0; 1/

u0.0/ D u.0/

u0.1/ D 1=2:

We can explicitly solve the problem: integrating once we get

.x C 1/u0 D C1

and the Neumann condition gives C1 D 1; integrating again, we find

u D log .x C 1/C C2

and now the Robin condition gives C2 D 1. Overall,

u.x/ D log .x C 1/C 1:

b) On V we use the standard norm of V D H1 .0; 2/. It is not hard to see (see Exercise 6.3.1.b))thatB is continuous. AlsoF is continuous as soon as (for instance) f 2 L2.0; 2/. For the coercivity,we preliminarily observe that, as u.2/ D 0Z 2

0x2u0udx D 1

2

Z 2

0x2.u2/0 dx D

1

2x2u2

�20

�Z 2

0xu2 dx D �

Z 2

0xu2 dx:

Then

B.u; u/ DZ 2

0Œ.u0/2 � 2xu2� dx:

It is easy to see B is not coercive: for example, by choosing u D 2 � x (which belongs to V ), weget B.u; u/ D �2=3. At the same time

B.u; u/C 5kuk2L2 � kuk2V ;

so the form is weakly coercive. By Fredholm’s alternative, whether the problem has solutions de-pends on the solutions of the homogeneous problem:

B .v;w/ DZ 2

0

hw0v0 � 2x2v0w � 4xwv

idx D 0; for every v 2 V:


To solve this let us write the classical formulation. Assuming w regular,Z 2

0

hw0 � 2x2w

iv0 dx D

hw0v � 2x2wv

i20

�Z 2

0

hw00 � 2x2w0 � 4xw

iv dx

so the adjoint problem (formally) reads

w0.0/v.0/CZ 2

0

hw00 � 2x2w0

idx D 0 for any v 2 V:

As is customary, choosing v 2 H10 .0; 2/ first, and then v 2 V , we obtain

w00 � 2x2w0 D 0 in .0; 2/

and the condition w0 .0/ D 0. Altogether the classical form is8<:w00 � 2x2w0 D 0 in .0; 2/ ;

w0.0/ D 0;

w.2/ D 0:

Integrating once gives e�x2w0 D C1; the Neumann boundary condition implies C1 D 0, whence

w D C2 and then w D 0. As the adjoint problem has only the zero solution, by Fredholm’s alterna-tive the original problem is well posed for every f .

There remains to compute the classical form. Proceeding along the previous lines, we find8<:u00 C 2x2u0 C 4xu D �f in .0; 2/

u0.0/ D 0

u.2/ D 0:

Solution 6.3.4. Note first that the differential equation can be written as

� �cos x u0�0 C xu D �f:

To write a variational formulation, we follow Problem 6.2.3 on page 340, which actually correspondsto the choice L D �=6, f � 1. More precisely, we choose the Hilbert space

V D°v 2 H1 .0; L/ W v .L/ D 0

±with norm kvkV D ��v0��

L2.0;1/(recall that in V a Poincaré inequality holds), and proceeding as

usual we obtain the following variational formulation: to find u 2 V such that

a .u; v/ DZ L

0Œcos x u0v0 C xuv�dx � u .0/ v .0/ D �

Z L

0f vdx 8v 2 V:

To study well-posedness, we first check the hypotheses of the Lax-Milgram theorem. Observe that,since u .0/ D � R L0 u0dx, we can write, using Schwarz’s inequality,

ju .0/j �Z L

0

ˇu0ˇdx � p

L kukV (6.57)


so that

ja .u; v/j � .1C L/ kukV kvkV C L kukL2.0;L/ kvkL2.0;L/ � .1C LC C 2PL/ kukV kvkV :Therefore a is continuous. Now, using (6.57), we have

a .u; u/ DZ L

0Œcosx

�u0�2 C xu2�dx � u2 .0/ � .cosL � L/ kuk2V

and therefore a is coercive if L < L�, where L� ' 0:739 : : : is the solution of the equation

˛ � cosL � L D 0:

Case L < L�. Since clearly f 2 V �, the Lax-Milgram theorem applies and we get existence,uniqueness and the stability estimate

kukV � 1

˛kf kL2.0;L/ (6.58)

(in particular, for L D �6 and f � 1, we recover Problem 6.2.3).

Case L � L�. In this case a is only weakly coercive; to see it, we estimate u .0/ in a differentway, writing

u2 .0/ D �Z L

0

d

dxu2dx D �2

Z L

0uu0dx

and using Young’s inequality to get, for any " > 0 W

u2 .0/ � " kuk2V C 1

"kuk2

L2.0;L/:

Then, choosing (for instance) " D 12 cosL we can write

a .u; u/ � 1

2cosL kuk2V � 2 kuk2

L2.0;L/:

ThereforeQa .u; v/ � a .u; v/C 2 .u; v/L2.0;L/

is coercive and a is weakly coercive.We can apply the Fredholm alternative. We have to examine the homogeneous adjoint problem,

which is, since a is symmetric:´Lw D cos x w00 � sin x w0 � xw D 0 0 < x < L

w0 .0/C w .0/ D 0; w .L/ D 0:(6.59)

We want to show that there exists only one value L D L0 such that problem (6.59) has a nontrivialsolution and this vanishes at x D L0. It is convenient to consider the following related Cauchyproblem: 8<

:Ly D cos x y00 � sin x y0 � xy D 0 ��

2 < x <�2

y .0/ D r

y0 .0/ D �r:(6.60)

From general ODE theory we know that, for any r 2 R, there exists a unique solution yr ofproblem (6.60), analytic in

��2 ;

�2

�. Let us examine the properties of yr .


1. We have

yr D ry1:

In fact both yr and ry1 satisfy (6.60). By uniqueness they coincide. Therefore it is enough to con-sider y1.

2. Let z .x/ D 1 � x � y1. Then:

z .x/ � 0 for every x 2�0;�

2

�:

To prove it, note first that, since sinx � x C x2 � 0 for every x 2 R,

Lz D sin x � x C x2 � Ly1 � 0 in�0;�

2

�: (6.61)

As a consequence, z cannot have a positive maximum at a point in .0; �2 /. Indeed, if x0 2 �0; �2

�and z .x0/ D maxŒ0;1� z, then z .x0/ > 0, z0 .x0/ D 0; z00 .x0/ � 0. Thus,

Lz .x0/ D cos x0 z00 .x0/ � x0z .x0/ < 0

in contradiction with (6.61).Moreover, z .0/ D z0 .0/ D 0 and z > 0 near x D 0: Indeed, differentiating twice the differen-

tial equation for y1 and using y1 .0/ D 1, y01 .0/ D �1, we have

y1 .x/ D 1 � x � 1

4Šx4 C 1

5Šx5 C : : : :

so that

z .x/ D 1

4Šx4 � 1

5Šx5 C : : : > 0

in a positive neighbourhood of x D 0.As a consequence, z cannot become negative in

�0; �2

�; otherwise there would be a point in�

0; �2�

at which z would attain a positive maximum.

3. From 2. we deduce that y1 < 1 � x in�0; �2

�. Therefore y1 must have a zero at some point

L0 2 .0; 1�.Moreover y1 is strictly decreasing in

�0; �2

�since it is easy to check that y1 cannot have a max-

imum or a minimum at a point inside�0; �2

�. Therefore y1 > 0 in Œ0; L0/ and y1 < 0 in

�L0;

�2

�.

Numerical calculations give L0 D 0:915 : : : .

4. From what we have proved so far, we deduce that there exists only one value L D L0 2�0; �2

�such that problem (6.59) has a non-trivial solution. Actually, in this case, the family of solu-

tions to (6.59) is a 1�dimensional vector space generated by y1.

Going back to the original problem, we can draw the following conclusions:

Case 0 < L < L�: For any f 2 L2 .0; 1/, there exists a unique solution u 2 V and

kukV � 1

˛kf kL2.0;L/ (˛ D cosL � L). (6.62)


Case L� � L < �2

, L ¤ L0. For any f 2 L2 .0; 1/, there exist a unique solution u 2 V anda constant C independent of u and f (but depending on L) such that

kukV � C kf kL2.0;L/ . (6.63)

Case L D L0. The problem has a solution only if .f; y1/L2.0;L/ D 0. In this case the familyof solutions is given by

u D NuC cy1 c 2 R

where Nu is a particular solution of (6.3.4).Note that if f is constant, .f; y1/0 ¤ 0, since y1 > 0 in .0; L0/, so that the problem has no

solution.

Solution 6.3.5. a) We can mimic Problem 6.2.5 (page 343), with minor changes.

b) Set

kvk D�Z

Rv2e�x2

dx

1=2and denote by kvkV the norm induced by (6.51). As

jFvj DˇZ

Rf v e�x2

dx

ˇ� kf k kvk � kf k kvkV ;

the functional F defines an element of V �. Riesz’s theorem implies there is a unique solution u to(6.52) and

kukV � kf k :For the classical interpretation let us integrate formally, by parts, the second term of the inner prod-uct; then (6.52) reads

Œe�x2

u0v�C1�1 CZ

RŒue�x2 � .e�x2

u0/0�v dx DZ

Rf v e�x2

dx for every v 2 V

and as v was arbitrary, if we choose v with compact support we find

�.e�x2

u0/0 C ue�x2 D fe�x2

on R;

equivalent to (Hermite’s equation)

u00 � 2xu0 � u D f on R;

and thenlim

x!˙1 e�x2

u0 .x/ D 0

by picking first v equal to zero at ˙1, and then equal to 1 at 1. The boundary conditions arethus homogeneous Neumann conditions, weighted by the Gaussian function.

Solution 6.3.6. a) Assuming u and v regular we can writeZ

ru � rv dx DZ@ @u� v d� �

Z �uv dx:


In particular, since H10 .�/ � H1.�/, the weak formulation impliesZ

Œ��uC u � f � v dx D

Z@ Œg � @�u� v dx D 0; for every v 2 H1

0 .�/;

from which we deduce ��uCu�f D 0 a.e. on�. Substituting back in the variational formulation,we find Z

@ Œg � @�u� v dx D 0; for every v 2 H1.�/;

from which @�u D g a.e. on @�. In conclusion (6.53) is the weak formulation of the problem´��uC u D f in �

@�u D g on @�:

b) Observe that the left-hand side in the weak formulation is the standard scalar product inH1 .�/. At the same time if u 2 H1 .�/, by the trace inequality

kukL2.@ / � CkukH1. /:

Using Schwarz’s inequality we deduce that the linear functional on the right of (6.53) is continuouson H1 .�/, so Riesz’s theorem implies the problem is well posed.

Solution 6.3.7. It is immediate to see that the problem is of the type

B.u; v/ D Fv;

where B is the scalar product in V and F (for f 2 L2.B1/) is continuous onH1 .B1/, so a fortiorion V . In this case we can apply the Lax-Milgram theorem or, directly, Riesz’s theorem.

If f is regular, i.e. f 2 C1�B1�, we know16 that u 2 C1

�B1�, so we may integrate by

parts, Z@B1

@�u v d� CZB1

Œ��uC u � f � v dx D 0 for any v 2 V:

Since H10 .B1/ � V , we deduce, in particular,Z

B1

Œ��uC u � f � v dx D 0 for every v 2 H10 .B1/ :

Consequently ��uC u � f D 0. Substituting in the weak formulation, we getZ@B1

@�u v d� D 0 for any v 2 V: (6.64)

Let us show that this relation holds if and only if @�u is constant on @B1. On one hand, if @�u isconstant, the integral of the product with a function of zero average is clearly null. On the other hand,suppose g D @�u is not constant. Then there exist points � 1 and � 2 on @B1 where g .� 1/ D a andg .� 2/ D b with (say) a < b. By continuity we can find arcs �1, �2 in @� so that

gj�1� a; gj�2

� b:

16 [18, Chap. 8, Sect. 6].


Let w be a C 1.B1/ function, positive on �1, negative on �2, null on @� n .�1 [ �2/ and such thatZ�1

w d� D �Z�2

w d�:

Then w belongs to V and can be used as test function in (6.64):Z@ g w d� D

Z�1

g w d� CZ�2

g w d� � a

Z�1

w d� C b

Z�2

w d�

D .a � b/Z�1

w d� < 0:

This contradicts (6.64). In conclusion, the problem is8<ˆ:

��uC u D f in B

@�u D constant on B1Z@B1

u D 0:

Solution 6.3.8. Multiplying the first equation by v1 2 H10 .�/ and integrating by parts, we obtainZ

Œru1 � rv1 C u1v1 � u2v1� dx D

Z f1v1 dx; for any v1 2 H1

0 .�/:

Similarly, Z Œru2 � rv2 C u1v2 C u2v2� dx D

Z f2v2 dx; for any v2 2 H1

0 .�/:

The previous two equations can be combined into the following oneZ Œru1 � rv1 C ru2 � rv2 C u1v1 � u2v1 C u1v2 C u2v2� dx D

DZ Œf1v1 C f2v2� dx; for every .v1; v2/ 2 H1

0 .�/ �H10 .�/

(just choose vi D 0 to get either). Define the Hilbert space

V D H10 .�/ �H1

0 .�/;

with product.u; v/V D .ru1;rv1/L2. / C .ru1;rv1/L2. /

(where u D .u1; u2/, v D .v1; v2/). Set f D .f1; f2/,

B.u; v/ DZ Œru1 � rv1 C ru2 � rv2 C u1v1 � u2v1 C u1v2 C u2v2� dx

and

F v DZ

f � v dx:


This gives the variational formulation

B.u; v/ D F v for every v 2 V:

Using Schwarz’s and Poincaré’s inequalities we can write

jF vj � kf1kL2kv1kL2 C kf2kL2kv2kL2 � C 2P kukV kvkVand

jB.u; v/j � kru1kL2krv1kL2 C kru2kL2krv2kL2

C .ku1kL2 C ku2kL2/.kv1kL2 C kv2kL2/

� .1C C 2P /kukV kvkV :

Moreover,

B.u; u/ DZ

hjru1j2 C jru2j2 C u21 C u22

idx � kuk2

L2 ;

and therefore B is coercive. It is then possible to use Lax-Milgram and infer existence, uniquenessand a stability estimate for the solution.

Solution 6.3.9. Let us first check that the operator is uniformly elliptic, i.e. that the coefficientsare bounded (true) and there exists a0 > 0 such that

2Xj;kD1

ajk .x; y/ zj zk � a0

�z21 C z22

�; for every .z1; z2/ 2 R2. (6.65)

In fact,

2Xj;kD1

ajk .x; y/ zj zk D .2C sin.xy// z21 C 3z1z2 � 3z1z2 C .4 � sin.xy// z22

� z21 C 3z22 � z21 C z22 ;

thus (6.65) holds with a0 D 1. Let V D H10;�D

.�/ be the space of functions in H1 .�/ vanishingon �D . Poincaré’s inequality holds, so we choose krukL2. / as norm on V . For the variationalformulation, multiply the equation by v 2 V and integrate by parts:Z ŒA .x; y/ru �rvCub �rv� dxdy�

Z�N

ŒA .x; y/ru ��Cub � ��v d� DZ f v dxdy: (6.66)

On �N we have � D .0;�1/ and

A .0; y/ru � � D 3ux .0; y/ � 4uy .0; y/ D �hu .0; y/ub � � D �yu .0; y/ :

Substituting in (6.66) producesZ ŒA .x; y/ru � rv C ub � rv�dxdy C

Z 1

0ŒhC y�u .0; y/ v .0; y/ dy D

Z f v dxdy:


If we denote by B .u; v/ the bilinear form on the left, the variational formulation is: determine afunction u 2 V such that

B .u; v/ D .f; v/; for every v 2 V .

As f 2 L2 .�/, the functional on the left is in V �. Let us verify the continuity of B . By the traceand Poincaré inequalities:

jB .u; v/j � 4C 2� .hC 2/ krukL2 krvkL2

so B is continuous. As for coercivity, recalling that(6.65) holds with a0 D 1,Z A .x; y/ru � rudxdy � kruk2

L2

while, as div b D �3,Z ub � ru D 1

2

Z

b � r.u2/ D �12

Z 1

0yu2 .0; y/ dy C 3

2

Z u2

hence .h > 0/

B .u; u/ � kruk2L2 C 3

2

Z u2 C

Z 1

0ŒhC y

2�u2 .0; y/ dy � kruk2

L2

and coercivity of B follows, with constant 1. The assumptions of the Lax-Milgram theorem hold.Therefore the given problem has a unique solution; what is more,

krukL2. / � CP kf kL2. / ;

where CP is a Poincaré constant.

Solution 6.3.10. a) Recall that if ' is a scalar field and F a vector field, both regular, we have

div.'F/ D ' div F C r' � F:

Since

re�b�x D �e�b�xb

we obtain

�e�b�x�uC e�b�xb � ru D � div�e�b�xru

�:

Hence u solves the divergence problem´� div�e�b�xru� D e�b�xf x 2 �

@�u D 0 x 2 @�:

b) There is a standard way for proving (e.g. following Problem 6.2.12 on page 355) that the bilin-ear form appearing in the weak formulation is (beside continuous) weakly coercive, so Fredholm’s


theory applies. The operator is self-adjoint, and the adjoint homogeneous problem is´� div�e�b�xru� D 0 x 2 �

@�u D 0 x 2 @�:

It is easy to show17 that the solutions are only constants. The starting problem has solutions if andonly if Z

e�b�xf .x/ dx D 0

in which case the solution is defined up to an additive constant.

Solution 6.3.11. The weak formulation is

B.u; v/ DZQŒru � rv � 2uv� D

ZQ

f � rv D Fv; for every v 2 H10 .Q/:

The functional F belongs to H�1 .Q/ if and only if f 2 L2.QI R2/. We have that log t is inL2.0; 1/. Now as

sin x x as x ! 0; sin x � � x for x ! �;

we obtain the required integrability for f.As seen in Problem 6.2.13 (page 356), B is continuous and weakly coercive in H1

0 .Q/ and theadjoint homogeneous problem has only the solution Nu.x; y/ D c sin x sin y, with c 2 R. Fredholm’stheory implies that the problem can be solved if and only if

F Nu D 0:

We have r Nu D .c cos x sin y; c sin x cosy/ and

F Nu D�Z �

0c sin y dy

�Z �

0log .sin x/ cos x dx

D 0:

The problem has infinitely many solutions u.x; y/C c sin x sin y.

Solution 6.3.12. Define V D H1 .�/ and

V1 D°u 2 H1 .�1/ , u D 0 in � n�1

±and V2 D

°u 2 H1 .�2/ , u D 0 in � n�2

±.

Given u0 2 H1 .�/, let u2nC1 (n � 0) and u2n (n � 1) be the solutions of the mixed problems

��u2nC1 C u D f in �1, u2nC1 D u2n on @�1 \�, @�u D 0 on @�1 \ @�and

��u2n C u D f in �2, u2n D u2n�1 on @�2 \�, @�u D 0 on @�2 \ @�respectively. Let us follow the solution of Problem 6.2.15 on page 361 to obtain

u � u2nC1 D PV?

1.u � u2n/ and u � u2n D P

V?1.u � u2n�1/ ,

17 For example as in Problem 6.2.10 (page 351).


where the projections are intended with respect to the inner product .u; v/ D R Œuv C ru � rv�

on V . The function un converges in H1 .�/ to the solution u 2 H1 .�/ of problem (6.54) if (seeProblem 5.2.9, Chap. 5, page 288) V1 and V2 are closed in V and

V ?1 \ V ?2 D ¹0º

i.e. V1 C V2 is dense in V . Let v 2 V . To verify the closure of V1 and V2, we may resort to thetechnique of Exercise 5.3.26 on page 313, Chap. 5. Details are left to the reader. To check the densityof V1CV2 in V , letw1 andw2 be regular withw1 D 0 on�n�1 and�n�2, positive elsewhere.Setting

v1 D w1

w1 C w2v and v1 D w2

w1 C w2v;

we have v1 2 V1, v 2 V2 and v1 C v2 D v. Then V1 C V1 is not only dense, it coincides with V .

Solution 6.3.13. a) First, on @B1 the outward unit normal is � D xi C yj, while � D yi � xj istangential. The unit vector

� D .x C y/ i C .y � x/ jp2

D � C �p2

is never tangent, for � � � D 1=p2. Hence we are dealing with an oblique-derivative problem. In

relationship to Problem 6.2.16, with b D 1,

A .x; y/ D�1 1

�1 1

and b .x; y/ D �by ;�bx

� D .0; 0/ :

The variational formulation reads

a .u; v/ DZB1

A .x; y/ru � rv dxdy DZB1

f v dxdy CZ@B1

gv ds:

Explicitly, we haveZB1

.uxvx C uyvx � uxvy C uyvy/ dxdy DZ@B1

�1C ˛x2

�v ds; for every v 2 H1 .B1/

which can be solved if and only ifZ@B1

�1C ˛x2

�ds D 2� C ˛� D 0

i.e. for ˛ D �2.b) We haveZ

B1

.uxvx C uyvx � uxvy C uyvy/ dxdy DZB1

�x2 � y2

�nv dxdy; for every v 2 H1 .B1/

which can be solved if and only if ZB1

�x2 � y2

�ndxdy D 0

i.e. if and only if n is odd.


Solution 6.3.14. a) If u and v 2 K and s 2 Œ0; 1�, then

.1 � s/ uC sv 2 K;so K is convex. To check that K is closed in H1

0 .�/ consider a sequence ¹vnº � K converging inH10 .�/ to v. In particular, there exists a subsequence

®vnk

¯convergent to v a.e. on �. Therefore

vnk� a.e. on �, which implies v � a.e. on � and hence K is closed in H1

0 .�/.Let us view this in terms of projections: J .v/ represents, up to the factor 1=2, the square of

the distance in H10 .�/ between v and w D 0, and to minimise J is the same as minimising the

distance of K from the origin. The existence of a minimiser for J amounts thus to the existence ofthe projection of the origin on the closed, convex set K. So we can use Problem 5.2.10 on page 289(Chap. 5), and conclude that there is a unique u 2 K such that

J .u/ D minv2K J .v/ .

This element is characterised by

.u; u � v/H1

0. /

� 0; for every v 2 Kcorresponding precisely to (6.55).

c) Set u � v D w in (6.55). Then w � 0 on �. If u 2 H2 .�/ we can integrate by parts so tohave the derivatives on u:Z

w�u � 0; for every w 2 H1

0 .�/ , w � 0 a.e. on �;

which forces (prove it)�u � 0 a.e. on�. Additionally, ifD � � is an open set over which u > ,take � 2 C 10 .D/. If h 2 R, positive or negative, is small enough,

v D uC hv >

on D and v � on �. So we can take it as test function in (6.55) and deduce

h

ZD

ru � r� D 0 for every � 2 C 10 .D/ :

But this means �u D 0 on D.

Remark. When n D 2 we can view the graph of u as an elastic membrane fixed to the boundary of�. J .u/ is proportional to the potential energy of the membrane’s deformation. The problem asksto find the configuration of least energy (equilibrium) under the constraint that the membrane cannotmove below , which acts as an obstacle.

Solution 6.3.15. a) Set

w .x; y; t/ D u .x; y; t/ � g .t/so to have zero Dirichlet datum on �D . The problem for w is8<

ˆ:wt � div.xeyrw/C jxj .sign .y/w/y D F .x; y; t/ in QT ;

w .x; y; 0/ D u0 .x; y/ � g .0/ in B1;

w D 0 on �D ;

@�w C w C g D 0 on �N ;

(6.67)


whereF .x; y; t/ D f .x; y; t/ � Pg .t/ ; U0 .x; y/ D u0 .x; y/ � g .0/ :

For the weak formulation we use V D H10;�D

.B1/, the space of functions in H1 .B1/ withzero trace on �D . The Poincaré inequality holds

kvkL2.B1/� CP krvkL2.B1/

so we choosekvkV D krvkL2.B1/

.

Multiply the equation by v 2 V , integrate on B1 and use Gauss’s formula, keeping in mind themixed boundary conditions; as � D .x; y/ is the unit normal to @B1,Z

B1

Œwtv C xeyrw � rv � jxj sign .y/wvy � dxdy CZ�N

jxyjwv d� DZB1

Fv dxdy:

Set

B .w; v/ DZB1

Œxeyrw � rv � jxj sign .y/wvy � dxdy CZ�N

jxyjwv d�

and let us pass from the notation w .x; y; t/ to

w .t/ W t ! w .�; t / .

As usual .�:�/ is the inner product on L2 .B1/. The weak formulation of (6.67) is: determinew 2 L2 .0; T IV / \ C �Œ0; T � IL2 .B1/� such that Pw 2 L2 .0; T IV �/ and:

i) For every v 2 V ,d

dt.w .t/ ; v/C B .w .t/ ; v/ D .F; v/

in D 0 .B1/ for almost every t 2 .0; T /.ii) w .t/ ! U0 in L2 .B1/ as t ! 0C.

To see if the problem is well posed we check the continuity and weak coercivity of B on V .On B1

e�1 � xey � e, jxyj � 1:

So for every v; z 2 V :

jB .v; z/j � e kvkV kzkV C kvkL2.B1/

��zy��L2.B1/C kvkL2.�N /

kzkL2.�N /:

By the Poincaré and trace inequalities18

jB .v; z/j � .e C CP C C 2� / kvkV kzkV ;and B is continuous on V . Moreover,

B .v; v/ DZB1

Œxey jrvj2 � jxj sign .y/vvy � dxdy CZ�N

jxyj v2 d�

� e�1 kvk2V � ��vy��L2.B1/kvkL2.B1/

:

18 kvkL2.�N /� C� kvkV , see [18, Chap. 7].


By the Poincaré and trace inequalities, plus the elementary fact

ab � 1

2ea2 C e

2b2

we can write

B .v; v/ � 1

ekvk2V � 1

2ekvk2V � e

2kvkL2.B1/

D 1

2ekvk2V � e

2kvkL2.B1/

:

Hence the bilinear form eB .v; z/ D B .v; z/C e

2kvk2

L2.B1/

is coercive with constant 1=2e, so B is weakly coercive.By assuming

f .x; y; t/ 2 L2 .QT / , g 2 H1 .t/ and U0 .x; y/ D u0 .x; y/ ;

the problem has a unique weak solution, and

maxŒ0;T �

ku .�; t /k2H CZ T

0kru .�; t /k2H dt

� ceT

´ku0 � g .0/k2

H10.B1/

CZ T

0

°kf .�; �; s/k2

L2.B1/C . Pg .s//2

±ds

μ:

Solution 6.3.16. a) Set

B .u; v/ DZ Œa .x/ru � rv � ub � rv C c .x/ uv� dxdy C

Z�N

ŒhC b � ��uv d�

and let us pass from using u .x/ to

u .t/ W t ! u .�; t / .

The symbol .�:�/ is the inner product onL2 .�/. The variational formulation goes as usual. Take V DH10;�D

.�/, space of H1 .�/ functions with zero trace on �D , normed by kvkV D krvkL2.B1/.

We want to find u 2 L2 .0; T IV / \ C �Œ0; T � IL2 .�/� such that Pu 2 L2 .0; T IV �/ and:

i) For every v 2 V ,d

dt.u .t/ ; v/C B .u .t/ ; v/ D 0

in D 0 .B1/ and for almost every t 2 .0; T /.ii) u .t/ ! U0 in L2 .�/ as t ! 0C.

b) Let us assume:

ja .x/j ; jb .x/j ; jc .x/j � M , a .x/ � a0 > 0, jdiv b .x/j � M 0, a.e. on �,

hC b � � � 0 a.e. on �N (if h D 0, the flow of b through �N is outgoing).


By the Poincaré and trace inequalities the bilinear form B is continuous (the reader should fill in thedetails). As for weak coercivity, using repeatedly the assumptions on the coefficients:

B .u; u/ �Z Œa .x/ jruj2 � ub � ruC c .x/ u2� C

Z�N

ŒhC b � ��u2 d�

� a0

Z Œjruj2 � 1

2b � r.u2/C c .x/ u2� C

Z�N

ŒhC b � ��u2 d�

D a0

Z

jruj2 CZ Œ1

2div b C c .x/�u2 C

Z�N

ŒhC 1

2b � ��u2 d�

� a0 kruk2V ��1

2M 0 CM

kuk2

L2 � a0 kruk2V � �0 kuk2L2 :

The formeB .u; v/ D B .u; v/C�0.u; v/ is thus coercive on V , which meansB weakly coercive onV . Under the given hypotheses, the general theory applies and the problem has one weak solution.

Solution 6.3.17. A Robin problem is assigned. Since is a positive function, we write

.x/ � 0 > 0; a.e. .x; t / in QT D � � Œ0; T �and this inequality ensures the uniform ellipticity of the equation.

We consider V D H1.�/ and v 2 V , with the usual norm. We recall that in H1.�/ thefollowing trace inequality holds

kvkL2.�/ � CT kvkV :Writing c.t/ D c.�; t / and Pc for ct , let us proceed formally and multiply the first equation of

system (6.56) by v; integrating on �, we find

hPc; vi� CZ rc � rv dx �

Z�rc � � v d� �

Z �cv dx D 0

namely

hPc; vi� CZ rc � rv dx C

Z��cv d� �

Z �cv dx D

Z��cextv d�:

Let us introduce the bilinear form and the functional

B.w; vI t / DZ rw � rv dx C

Z��wv d� �

Z �wv dx; F v D

Z��cextv d�:

The problem can be written in weak form as follows. Find c 2 L2.0; T IV / such that c0 2L2.0; T IV 0/ and:

1. For every v 2 V and for a.e. t 2 Œ0; T �hPc.t/; vi� C B.c.t/; v/ D Fv:

2. c.0/ D c0 in �.

In order to use the theory for the well-posedness of the problem we have to show that B is con-tinuous and weakly coercive and that F is continuous. Using the trace and the Schwarz inequalitieswe find that the functional F is continuous:

jFvj � �cextj�j1=2kvkL2.�/ � 2Rp��cextCT kvkV :


On the other hand, since 2 L1.�/, the bilinear form B is V -continuous, in fact (Lp D Lp.�/)

jB.c; v/j � kkL1krckL2krvkL2 C �kckL2kvkL2 C �C 2T kckV kvkV��kkL1 C � C �C 2T

�kckV kvkV :

Moreover

B.v; v/C �kvkL2 DZ jrcj2dx C

Z�v2 d� C

Z .� � �/v2dx � min.0; � � �/kvk2V ;

hence B is weakly coercive if we choose, say, � D � C 0=2. The constant of weak coercivity isthen 0=2.

Appendix A

Sturm-Liouville, Legendre and Bessel Equations

A.1 Sturm-Liouville Equations

A.1.1 Regular equations

The eigenfunctions of a large class of ODEs form complete orthonormal systems in suit-able Hilbert spaces. To this class belong equations of the form

� �p .x/ u0�0 C q .x/ u D �w .x/ u .x/ ; a < x < b (A.1)

under the assumption that p; p0; q; and w are continuous and positive on Œa; b�. In sucha case (A.1) is called a regular Sturm-Liouville equation. We associate to (A.1) theboundary conditions:

˛u .a/ � ˇp .a/ u0 .a/ D 0

u .b/ � ıp .b/ u0 .b/ D 0;(A.2)

where the coefficients ˛; ˇ; ; ı are real numbers. To avoid trivial situations we shall as-sume the following normalization condition:

˛2 C ˇ2 D 2 C ı2 D 1:

In general, problem (A.1), (A.2) has nontrivial solutions only for special values of the pa-rameter �, called eigenvalues. The corresponding solutions are called eigenfunctions, andthey form the eigenspace associated to �. Let us introduce the (Hilbert) space L2w .a; b/of weighted square-integrable functions u on .a; b/with respect to the weight functionw:

L2w .a; b/ D´u W

Z b

a

u2 .x/w .x/ dx < 1μ

.

Then the following theorem holds:

© Springer International Publishing Switzerland 2015S. Salsa, G. Verzini, Partial Differential Equations in Action. Complements and Exercises,UNITEXT – La Matematica per il 3+2 87, DOI 10.1007/978-3-319-15416-9_A

406 Appendix A Sturm-Liouville, Legendre and Bessel Equations

Theorem A.1. There exists an increasing sequence of positive numbers ¹�j ºj�1 such that�j ! C1 and:

a) Problem (A.1), (A.2) admits a non-trivial solution if and only if � equals one of the �j .

b) For every j , �j is simple, that is the associated eigenspace has dimension 1.

c) The eigenfunction system ¹'j ºj�1 (suitably normalised) is an orthonormal basis inL2w .a; b/.

A.1.2 Legendre’s equation

When the coefficient p is zero, e.g. at a or b, the equation is irregular and the studybecomes more complicated. A classical case is that of Legendre’s equation��

1 � x2�u0�0 C �u D 0 � 1 < x < 1: (A.3)

In the applications this is coupled with boundary conditions of the type

u .�1/ finite, u .1/ finite. (A.4)

Particular solutions to (A.3), (A.4) are Legendre’s polynomials, defined by the Rodriguesformula:

Ln .x/ D 1

2nnŠ

dn

dxn

�x2 � 1�n .n � 0/:

Each polynomial Ln corresponds to the eigenvalue �n

D n .nC 1/. The first four Legen-dre polynomials are

L0 .x/ D 1, L1 .x/ D x, L2 .x/ D 1

2

�3x2 � 1� , L4 .x/ D 1

2

�5x3 � 3x� :

The following theorem holds:

Theorem A.2. In relationship to problem (A.3), (A.4):

a) There exists a non-zero solution if and only if

�D�n D n .nC 1/ ; n D 0; 1; 2; : : : :

b) For every n � 0 the solution corrisponding to �n is unique up to a constant factor, andcoincides with Legendre’s polynomial Ln.

c) The normalised polynomials ´r2nC 1

2Ln

μn�0

form an orthonormal system in L2 .�1; 1/.

A.2 Bessel’s Equation and Functions 407

Theorem A.2 allows to expand any f 2 L2 .�1; 1/ in Fourier-Legendre series:

f .x/ D1XnD0

fnLn .x/ ; where fn D 2nC 1

2

Z 1

�1f .x/Ln .x/ ; dx

with L2 .�1; 1/-convergence. We also have a result about pointwise convergence, in per-fect analogy with Fourier series.

Theorem A.3. If f and f 0 have at most a finite number of jump points in .�1; 1/, then

1XnD0

fnLn .x/ D f .xC/C f .x�/2

for every x 2 .�1; 1/.

A.2 Bessel’s Equation and Functions

A.2.1 Bessel functions

Here is a short summary of the main properties of Bessel functions. First, though, we needto introduce a function interpolating the values of the factorial nŠ: The gamma function� D � .z/ is

� .z/ DZ 1

0

e�t tz�1dt (A.5)

for z complex with Re z > 0. The function � is analytic for Re z > 0 and satisfies thefollowing relationships:

� .z C 1/ D z� .z/

� .z/ � .1 � z/ D �

sin�z.z ¤ 0; 1; 2; : : :/:

In particular,� .nC 1/ D nŠ .n D 0; 1; 2; : : :/

and

�

�nC 1

2

D 1 � 3 � 5 � � � � � 2n � 1

2n

p� .n D 1; 2; : : :/ :

One can define � .z/ for z real, negative and not integer, using

� .z/ D � .z C 1/

z:

In fact, we know how to compute � on .0; 1/, and the formula allows to find � on .�1; 0/.In general, once we know� on .�n;�nC 1/, we can compute it on .�n � 1;�n/. Finally,


�4 �3 �2 �1 1 2 3 4

�5

5

Fig. A.1 Graph of the gamma function on the real axis

coherently with (A.5) we define

� .�2n/ D �1 and � .�2n � 1/ D C1:

In this way � is defined on the entire real axis (Fig. A.1).

Bessel’s function of the first kind and order p, p real, is

Jp .z/ D1XkD0

.�1/k� .k C 1/ � .k C p C 1/

�z2

�pC2k.

In particular, if p D n � 0 is an integer (Fig. A.2):

Jn .z/ D1XkD0

.�1/kkŠ .k C n/Š

�z2

�nC2k:

When p D �n is a negative integer, the first n terms of the series vanish and

J�n .z/ D .�1/n Jn .z/ :

Hence Jn .z/ and J�n .z/ are linearly dependent.If p is not integer, for z ! 0 we have asymptotic behaviours:

Jp .z/ D 1

� .1C p/

�z2

�pCO �zpC2� , J�p .z/ D 1

� .1 � p/�z2

��pCO �z�pC2�so Jp .z/ and J�p .z/ are linearly independent.


5 10 15 20

�0:5

0:5

1

Fig. A.2 Graphs of J0 (solid), J1 (dashed) and J2 (dotted)

Functions of the first kind satisfy a number of identities:

d

dz

�zpJp .z/

� D zpJp�1 .z/ ,d

dz

�z�pJp .z/

� D �z�pJpC1 .z/ : (A.6)

In particularJ 00 .z/ D �J1 .z/ :

From these we also infer that for p D n C 12

(and only in that case), the correspondingBessel functions are elementary. For instance,

J 12.z/ D

r2

�zsin z, J� 1

2.z/ D

r2

�zcos z.

Particularly important are the zeroes of Jp . For any p, there is an infinite increasing se-quence

®˛pj j�1 of positive numbers such that

Jp�˛pj

� D 0 .j D 1; 2; : : :/:

When p is not an integer, every linear combination

c2Jp .z/C c2J�p .z/

is a Bessel function of the second kind. The (standard) function of the second kind is

Yp .z/ D cosp�Jp .z/ � J�p .z/sinp�

.

When p D n is integer, one defines1 (Fig. A.3)

Yn .z/ WD limp!nYp .z/

Note that Yp .z/ ! �1 when z ! 0C.

1 One can prove that the limit exists.


5 10 15 20

�1

�0:5

0:5

Fig. A.3 Graphs of Y0 (solid), Y1 (dashed) and Y2 (dotted)

A.2.2 Bessel’s equation

The Bessel functions Jp ,Yp are solutions of the so-called Bessel equation of order p � 0:

z2y00 C zy0 C �z2 � p2�y D 0:

The general integral is given, for any p � 0, by

y .z/ D c1Jp .z/C c2Yp .z/ .

In the most important applications, one is typically led to solve the (parametric) equation(with parameter �)

z2y00 C zy0 C��2z2 � p2

�y D 0 (A.7)

on a bounded interval .0; a/, with boundary conditions of the sort

y .0/ finite, y .a/ D 0. (A.8)

For these, the following theorem holds.

Theorem A.4. Problem (A.7), (A.8) has nontrivial solutions if and only if

� D �pj D�˛pja

�2.

In that case the solutions are

ypj .z/ D Jp

�˛pjaz�

up to multiplicative constants. Moreover, the normalised functionsp2

aJpC1�˛pj

�ypj


form an orthonormal basis in (w.z/ D z)

L2w .0; a/ D²u W kuk22;w D

Z a

0

u2 .z/ zdz < 1³

by virtue of the orthogonality relations:

2

a2J 2pC1�˛pj

� Z a

0

zJp��pj z

�Jp��pkz

�dz D

´0 j ¤ k

1 j D k:

With Theorem A.4 we can expand any f 2 L2w .0; a/ in Fourier-Bessel series:

f .z/ D1XjD1

fjJp��pj z

�; where fj D 2

a2J 2pC1�˛pj

� Z a

0

zf .z/ Jp��pj z

�dz;

with L2w .0; a/-convergence.Let us compute, for example, the expansion of f .x/ D 1 on the interval .0; 1/, with

p D 0:

fj D 2

J 21�˛pj

� Z 1

0

zJ0�˛0j z

�dz:

Using (A.6),d

dzŒzJ1 .z/� D zJ0 .z/

so we may write Z 1

0

zJ0�˛0j z

�dz D

1

�0jzJ1

�˛0j z

��10

D J1�˛0j

��0j

:

Finally

1 D1XjD1

2

�0jJ1�˛0j

�J0 �˛0j z�with convergence in norm L2w .0; 1/.

Also in this case one can insure pointwise convergence.

Theorem A.5. If f and f 0 have at most finitely many jump discontinuities on .0; a/, then

1XjD1

fjJp��pj z

� D f .zC/C f .z�/2

at every point z 2 .0; a/.

Appendix B

Identities

Here is a compilation of significant formulas and identities of common use.

B.1 Gradient, Divergence, Curl, Laplacian

Let F;u; v be vector fields and f; ' scalar fields, all assumed regular on R3.

Orthogonal Cartesian coordinates

1. gradient:

rf D @f

@xi C @f

@yj C @f

@zk

2. divergence:

div F D @

@xFx C @

@yFy C @

@zFz

3. Laplacian:

�f D @2f

@x2C @2f

@y2C @2f

@z2

4. curl:

curl F Dˇˇ i j k@x @y @zFx Fy Fz

ˇˇ

Cylindrical coordinates

x D r cos �; y D r sin �; z D z .r > 0; 0 � � � 2�/

er D cos � iC sin � j, e� D � sin � iC cos � j; ez D k:

© Springer International Publishing Switzerland 2015S. Salsa, G. Verzini, Partial Differential Equations in Action. Complements and Exercises,UNITEXT – La Matematica per il 3+2 87, DOI 10.1007/978-3-319-15416-9_B

414 Appendix B Identities

1. gradient:

rf D @f

@rer C 1

r

@f

@�e� C @f

@zez

2. divergence:

div F D1

r

@

@r.rFr /C 1

r

@

@�F� C @

@zFz

3. Laplacian:

�f D @2f

@r2C 1

r

@f

@rC 1

r2@2f

@�2C @2f

@z2D 1

r

@

@r

�r@f

@r

C 1

r2@2f

@�2C @2f

@z2

4. curl:

curl F D1

r

ˇˇ er re� ez@r @� @zFr rF� Fz

ˇˇ

Spherical coordinates

x D r cos � sin ; y D r sin � sin ; z D r cos .r > 0, 0 � � � 2� , 0 � � �/

er D cos � sin iC sin � sin jC cos k

e� D � sin � iC cos � j

ez D cos � cos iC sin � cos j� sin k:

1. gradient:

rf D @f

@rer C 1

r sin

@f

@�e� C 1

r

@f

@ e

2. divergence:

div F D @

@rFr C 2

rFr„ ƒ‚ …

radial part

C 1

r

1

sin

@

@�F� C @

@ F C cot F

�„ ƒ‚ …

spherical part

3. Laplacian:

�f D @2f

@r2C 2

r

@f

@r„ ƒ‚ …radial part

C 1

r2

²1

.sin /2@2f

@�2C @2f

@ 2C cot

@f

@

³„ ƒ‚ …

spherical part (Laplace-Beltrami operator)

4. curl:

curl F D 1

r2 sin

ˇˇ er re r sin e�@r @ @�Fr rF r sin Fz

ˇˇ :

B.2 Formulas 415

B.2 Formulas

Gauss’s formulas

The following formulas hold on Rn, n � 2, and we denote by:

• � a bounded domain with regular boundary @� and outward normal �.

• u; v vector fields that are regular1 up to the boundary of �.

• '; regular scalar fields up to the boundary of �.

• d� the infinitesimal surface element of @�:

We have the following formulas:

1.R div u dx D R

@ u � � d� (divergence formula)

2.R r' dx D R

@ '� d�

3.R �' dx D R

@ r' � � d� D R@ @�' d�

4.R div F dx D R

@ F � � d� � R

r � F dx

5.R �' dx D R

@ @�' d� � R r' � r dx (integration by parts)

6.R . �' � '� / dx D R

@ . @�' �'@� / d�

7.R curl u dx D � R@ u � � d�

8.R u � curl v dx D R

- v� curl u dxC R@ .v � u/ � � d�

Identities

div curl u D 0

curl grad' D 0

div .'u/ D ' div uCr' � u

curl .'u/ D ' curl uCr' � u

curl.u � v/ D .v � r/u� .u � r/ vC .div v/ u� .div u/ v

div.u � v/ D curl u � v � curl v � u

r .u � v/ D u � curl v C v � curl u C .u � r/ vC .v � r/u

.u � r/u D curl u � uC 12r juj2

curl curl u D r.div u/ ��u .curl curl D grad div � Laplacian/:

1 C 1��

is enough.

416 Appendix B Identities

B.3 Fourier Transforms

bu .�/ DZ

Ru .x/ e�i�x dx

General formulas

u buu .x � a/ e�ia�bu .�/eiaxu .x/ bu .� � a/

u .ax/ , a > 01

abu� �

a

u0 .x/ i�bu .�/xu .x/ ibu0 .�/.u � v/ .x/ bu .�/bv .�/u .x/ v .x/ .bu �bv/ .�/

Special transforms

u bue�ajxj, a > 0

2a

a2 C �2

1

a2 C x2�

ae�aj�j

e�ax2, a > 0

r�

ae� �2

4a

sin x

xe�jxj arctan

2

�2

�Œ�a;a� .x/ 2sin a�

�ı .x/ 1

1 2�ı .�/

B.4 Laplace Transforms 417

B.4 Laplace Transforms

eu .s/ DZ C1

0

u .t/ e�st dt

General formulas (u.t/ D 0 for t < 0)

u euu .t � a/ , a > 0 e�aseu .s/eatu .t/ , a 2 C eu .s � a/u .at/ , a > 0

1

aeu � s

a

�u0 .t/ seu .s/ � u.0C/u00 .t/ s2eu .s/ � u0.0C/ � su.0C/tu .t/ �eu0 .s/u .t/

t

R C1s

eu .�/ d�R t0u .�/ d�

eu .s/s

.u � v/ .t/ eu .s/ev .s/Special transforms

u euH .t/eat , a 2 C

1

s � aH .t/ sin at , a 2 R

a

s2 C a2

H .t/ cos at , a 2 Rs

s2 C a2

H .t/ sinh at , a 2 Ra

s2 � a2H .t/ cosh at , a 2 R

s

s2 � a2H .t/tn, n 2 N

nŠ

snC1

H .t/t˛, Re˛ > �1 �.˛ C 1/

s˛C1H .t/e�t2 es

2=4R C1s=2

e��2d�

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[13] R. McOwen, Partial Differential Equations: Methods and Applications, Prentice-Hall, New Jersey, 1964.

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[29] O. Scherzer, M. Grasmair, H. Grossauer, M. Haltmeier, and F. Lenzen,Variational methods in imaging, vol. 167 of Applied Mathematical Sciences,Springer, New York, 2009.

[30] L. A. Segel, Mathematics applied to continuum mechanics, vol. 52 of Classicsin Applied Mathematics, Society for Industrial and Applied Mathematics (SIAM),Philadelphia, PA, 2007. Reprint of the 1977 original.

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Collana Unitext – La Matematica per il 3+2

Series Editors:A. Quarteroni (Editor-in-Chief)L. AmbrosioP. BiscariC. CilibertoM. LedouxW.J. Runggaldier

Editor at Springer:F. [email protected]

As of 2004, the books published in the series have been given a volume number.Titles in grey indicate editions out of print.As of 2011, the series also publishes books in English.

A. Bernasconi, B. CodenottiIntroduzione alla complessità computazionale1998, X+260 pp, ISBN 88-470-0020-3

A. Bernasconi, B. Codenotti, G. RestaMetodi matematici in complessità computazionale1999, X+364 pp, ISBN 88-470-0060-2

E. Salinelli, F. TomarelliModelli dinamici discreti2002, XII+354 pp, ISBN 88-470-0187-0

S. BoschAlgebra2003, VIII+380 pp, ISBN 88-470-0221-4

S. Graffi, M. Degli EspostiFisica matematica discreta2003, X+248 pp, ISBN 88-470-0212-5

S. Margarita, E. SalinelliMultiMath – Matematica Multimediale per l’Università2004, XX+270 pp, ISBN 88-470-0228-1

A. Quarteroni, R. Sacco, F.SaleriMatematica numerica (2a Ed.)2000, XIV+448 pp, ISBN 88-470-0077-72002, 2004 ristampa riveduta e corretta(1a edizione 1998, ISBN 88-470-0010-6)

13. A. Quarteroni, F. SaleriIntroduzione al Calcolo Scientifico (2a Ed.)2004, X+262 pp, ISBN 88-470-0256-7(1a edizione 2002, ISBN 88-470-0149-8)

14. S. SalsaEquazioni a derivate parziali - Metodi, modelli e applicazioni2004, XII+426 pp, ISBN 88-470-0259-1

15. G. RiccardiCalcolo differenziale ed integrale2004, XII+314 pp, ISBN 88-470-0285-0

16. M. ImpedovoMatematica generale con il calcolatore2005, X+526 pp, ISBN 88-470-0258-3

17. L. Formaggia, F. Saleri, A. VenezianiApplicazioni ed esercizi di modellistica numericaper problemi differenziali2005, VIII+396 pp, ISBN 88-470-0257-5

18. S. Salsa, G. VerziniEquazioni a derivate parziali – Complementi ed esercizi2005, VIII+406 pp, ISBN 88-470-0260-52007, ristampa con modifiche

19. C. Canuto, A. TabaccoAnalisi Matematica I (2a Ed.)2005, XII+448 pp, ISBN 88-470-0337-7(1a edizione, 2003, XII+376 pp, ISBN 88-470-0220-6)

20. F. Biagini, M. CampaninoElementi di Probabilità e Statistica2006, XII+236 pp, ISBN 88-470-0330-X

21. S. Leonesi, C. ToffaloriNumeri e Crittografia2006, VIII+178 pp, ISBN 88-470-0331-8

22. A. Quarteroni, F. SaleriIntroduzione al Calcolo Scientifico (3a Ed.)2006, X+306 pp, ISBN 88-470-0480-2

23. S. Leonesi, C. ToffaloriUn invito all’Algebra2006, XVII+432 pp, ISBN 88-470-0313-X

24. W.M. Baldoni, C. Ciliberto, G.M. Piacentini CattaneoAritmetica, Crittografia e Codici2006, XVI+518 pp, ISBN 88-470-0455-1

25. A. QuarteroniModellistica numerica per problemi differenziali (3a Ed.)2006, XIV+452 pp, ISBN 88-470-0493-4(1a edizione 2000, ISBN 88-470-0108-0)(2a edizione 2003, ISBN 88-470-0203-6)

26. M. Abate, F. TovenaCurve e superfici2006, XIV+394 pp, ISBN 88-470-0535-3

27. L. GiuzziCodici correttori2006, XVI+402 pp, ISBN 88-470-0539-6

28. L. RobbianoAlgebra lineare2007, XVI+210 pp, ISBN 88-470-0446-2

29. E. Rosazza Gianin, C. SgarraEsercizi di finanza matematica2007, X+184 pp, ISBN 978-88-470-0610-2

30. A. MachìGruppi – Una introduzione a idee e metodi della Teoria dei Gruppi2007, XII+350 pp, ISBN 978-88-470-0622-52010, ristampa con modifiche

31 Y. Biollay, A. Chaabouni, J. StubbeMatematica si parte!A cura di A. Quarteroni2007, XII+196 pp, ISBN 978-88-470-0675-1

32. M. ManettiTopologia2008, XII+298 pp, ISBN 978-88-470-0756-7

33. A. PascucciCalcolo stocastico per la finanza2008, XVI+518 pp, ISBN 978-88-470-0600-3

34. A. Quarteroni, R. Sacco, F. SaleriMatematica numerica (3a Ed.)2008, XVI+510 pp, ISBN 978-88-470-0782-6

35. P. Cannarsa, T. D’AprileIntroduzione alla teoria della misura e all’analisi funzionale2008, XII+268 pp, ISBN 978-88-470-0701-7

36. A. Quarteroni, F. SaleriCalcolo scientifico (4a Ed.)2008, XIV+358 pp, ISBN 978-88-470-0837-3

37. C. Canuto, A. TabaccoAnalisi Matematica I (3a Ed.)2008, XIV+452 pp, ISBN 978-88-470-0871-3

38. S. GabelliTeoria delle Equazioni e Teoria di Galois2008, XVI+410 pp, ISBN 978-88-470-0618-8

39. A. QuarteroniModellistica numerica per problemi differenziali (4a Ed.)2008, XVI+560 pp, ISBN 978-88-470-0841-0

40. C. Canuto, A. TabaccoAnalisi Matematica II2008, XVI+536 pp, ISBN 978-88-470-0873-12010, ristampa con modifiche

41. E. Salinelli, F. TomarelliModelli Dinamici Discreti (2a Ed.)2009, XIV+382 pp, ISBN 978-88-470-1075-8

42. S. Salsa, F.M.G. Vegni, A. Zaretti, P. ZuninoInvito alle equazioni a derivate parziali2009, XIV+440 pp, ISBN 978-88-470-1179-3

43. S. Dulli, S. Furini, E. PeronData mining2009, XIV+178 pp, ISBN 978-88-470-1162-5

44. A. Pascucci, W.J. RunggaldierFinanza Matematica2009, X+264 pp, ISBN 978-88-470-1441-1

45. S. SalsaEquazioni a derivate parziali – Metodi, modelli e applicazioni (2a Ed.)2010, XVI+614 pp, ISBN 978-88-470-1645-3

46. C. D’Angelo, A. QuarteroniMatematica Numerica – Esercizi, Laboratori e Progetti2010, VIII+374 pp, ISBN 978-88-470-1639-2

47. V. MorettiTeoria Spettrale e Meccanica Quantistica – Operatori in spazi di Hilbert2010, XVI+704 pp, ISBN 978-88-470-1610-1

48. C. Parenti, A. ParmeggianiAlgebra lineare ed equazioni differenziali ordinarie2010, VIII+208 pp, ISBN 978-88-470-1787-0

49. B. Korte, J. VygenOttimizzazione Combinatoria. Teoria e Algoritmi2010, XVI+662 pp, ISBN 978-88-470-1522-7

50. D. MundiciLogica: Metodo Breve2011, XII+126 pp, ISBN 978-88-470-1883-9

51. E. Fortuna, R. Frigerio, R. PardiniGeometria proiettiva. Problemi risolti e richiami di teoria2011, VIII+274 pp, ISBN 978-88-470-1746-7

52. C. PresillaElementi di Analisi Complessa. Funzioni di una variabile2011, XII+324 pp, ISBN 978-88-470-1829-7

53. L. Grippo, M. SciandroneMetodi di ottimizzazione non vincolata2011, XIV+614 pp, ISBN 978-88-470-1793-1

54. M. Abate, F. TovenaGeometria Differenziale2011, XIV+466 pp, ISBN 978-88-470-1919-5

55. M. Abate, F. TovenaCurves and Surfaces2011, XIV+390 pp, ISBN 978-88-470-1940-9

56. A. AmbrosettiAppunti sulle equazioni differenziali ordinarie2011, X+114 pp, ISBN 978-88-470-2393-2

57. L. Formaggia, F. Saleri, A. VenezianiSolving Numerical PDEs: Problems, Applications, Exercises2011, X+434 pp, ISBN 978-88-470-2411-3

58. A. MachìGroups. An Introduction to Ideas and Methods of the Theory of Groups2011, XIV+372 pp, ISBN 978-88-470-2420-5

59. A. Pascucci, W.J. RunggaldierFinancial Mathematics. Theory and Problems for Multi-period Models2011, X+288 pp, ISBN 978-88-470-2537-0

60. D. MundiciLogic: a Brief Course2012, XII+124 pp, ISBN 978-88-470-2360-4

61. A. MachìAlgebra for Symbolic Computation2012, VIII+174 pp, ISBN 978-88-470-2396-3

62. A. Quarteroni, F. Saleri, P. GervasioCalcolo Scientifico (5a ed.)2012, XVIII+450 pp, ISBN 978-88-470-2744-2

63. A. QuarteroniModellistica Numerica per Problemi Differenziali (5a ed.)2012, XVIII+628 pp, ISBN 978-88-470-2747-3

64. V. MorettiSpectral Theory and QuantumMechanicsWith an Introduction to the Algebraic Formulation2013, XVI+728 pp, ISBN 978-88-470-2834-0

65. S. Salsa, F.M.G. Vegni, A. Zaretti, P. ZuninoA Primer on PDEs. Models, Methods, Simulations2013, XIV+482 pp, ISBN 978-88-470-2861-6

66. V.I. ArnoldReal Algebraic Geometry2013, X+110 pp, ISBN 978-3-642–36242-2

67. F. Caravenna, P. Dai PraProbabilità. Un’introduzione attraverso modelli e applicazioni2013, X+396 pp, ISBN 978-88-470-2594-3

68. A. de Luca, F. D’AlessandroTeoria degli Automi Finiti2013, XII+316 pp, ISBN 978-88-470-5473-8

69. P. Biscari, T. Ruggeri, G. Saccomandi, M. VianelloMeccanica Razionale2013, XII+352 pp, ISBN 978-88-470-5696-3

70. E. Rosazza Gianin, C. SgarraMathematical Finance: Theory Review and Exercises. From BinomialModel to Risk Measures2013, X+278pp, ISBN 978-3-319-01356-5

71. E. Salinelli, F. TomarelliModelli Dinamici Discreti (3a Ed.)2014, XVI+394pp, ISBN 978-88-470-5503-2

72. C. PresillaElementi di Analisi Complessa. Funzioni di una variabile (2a Ed.)2014, XII+360pp, ISBN 978-88-470-5500-1

73. S. Ahmad, A. AmbrosettiA Textbook on Ordinary Differential Equations2014, XIV+324pp, ISBN 978-3-319-02128-7

74. A. Bermúdez, D. Gómez, P. SalgadoMathematical Models and Numerical Simulation in Electromagnetism2014, XVIII+430pp, ISBN 978-3-319-02948-1

75. A. QuarteroniMatematica Numerica. Esercizi, Laboratori e Progetti (2a Ed.)2013, XVIII+406pp, ISBN 978-88-470-5540-7

76. E. Salinelli, F. TomarelliDiscrete Dynamical Models2014, XVI+386pp, ISBN 978-3-319-02290-1

77. A. Quarteroni, R. Sacco, F. Saleri, P. GervasioMatematica Numerica (4a Ed.)2014, XVIII+532pp, ISBN 978-88-470-5643-5

78. M. ManettiTopologia (2a Ed.)2014, XII+334pp, ISBN 978-88-470-5661-9

79. M. Iannelli, A. PuglieseAn Introduction to Mathematical Population Dynamics.Along the trail of Volterra and Lotka2014, XIV+338pp, ISBN 978-3-319-03025-8

80. V.M. Abrusci, L. Tortora de FalcoLogica. Volume 12014, X+180pp, ISBN 978-88-470-5537-7

81. P. Biscari, T. Ruggeri, G. Saccomandi, M. VianelloMeccanica Razionale (2a Ed.)2014, XII+390pp, ISBN 978-88-470-5725-8

82. C. Canuto, A. TabaccoAnalisi Matematica I (4a Ed.)2014, XIV+508pp, ISBN 978-88-470-5722-7

83. C. Canuto, A. TabaccoAnalisi Matematica II (2a Ed.)2014, XII+576pp, ISBN 978-88-470-5728-9

84. C. Canuto, A. TabaccoMathematical Analysis I (2nd Ed.)2015, XIV+484pp, ISBN 978-3-319-12771-2

85. C. Canuto, A. TabaccoMathematical Analysis II (2nd Ed.)2015, XII+550pp, ISBN 978-3-319-12756-9

86. S. SalsaPartial Differential Equations in Action. FromModelling to Theory (2nd Ed.)2015, XVIII+688, ISBN 978-3-319-15092-5

87. S. Salsa, G. VerziniPartial Differential Equations in Action. Complements and Exercises2015, VIII+422, ISBN 978-3-319-15415-2

The online version of the books published in this series is available atSpringerLink.For further information, please visit the following link:http://www.springer.com/series/5418

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