U.U.D.M. Project Report 2011:16

Examensarbete i matematik, 15 hpHandledare och examinator: Gunnar BergJuni 2011

Department of MathematicsUppsala University

Ill-posed problems and their applications to climate research

Ina Marcks von Würtemberg

Ill-posed problems and their applications to climate research

Ina Marcks von Würtemberg

June 21, 2011

Abstract

This paper treats ill-posed problems: the historical background of ill-posed prob-

lems and a deeper study of one particular example of an ill-posed problem. The

history of ill-posed problems begins with Jacques Hadamard who conceived the idea

in around 1902, and continues into modern times with various appearances in applied

mathematics. One such ill-posed problem is considered with extra care in an article

by Christer O.Kiselman; the inverse Dirichlet problem of the heat equation on an un-

bounded quadrant, which arises when one tries to calculate past surface temperatures

from a drilling sample in an ice sheet. This paper extends proofs and solving methods

of that article.

Acknowledgements:

I wish to express my gratitude to my supervisor Gunnar Berg for getting me interested inthe subject, and encouraged me throught my work. I would also like to thank Christer O.Kiselman for giving me deeper insight into his article. My appreciation also goes out tomy parents for pushing me onwards, and my friend for his patience with me and my work.

1

Contents

1 Introduction 3

1.1 Jacques Hadamard . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Well-posedness in the sense of Hadamard . . . . . . . . . . . . . . . . . . . . 5

2 The ice problem 8

2.1 Introduction to the ice problem . . . . . . . . . . . . . . . . . . . . . . . . . 82.2 Exponential solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.3 Norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.4 Constructing memory and past functions . . . . . . . . . . . . . . . . . . . . 142.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 APPENDIX 18

3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2

1 Introduction

The french mathematicien Jacques Hadamard raised the question about ill-posed problems.He realized that there was a certain relation between mathematical models and the realitythey depict.

1.1 Jacques Hadamard

Fig 1: Jacques Hadamard

Jacques Hadamard was born 8 december 1865 in Versailles. His father, Amédée Hadamard,was a Latin teacher, and came from a Jewish family with cultural and liberal traditions.His wife, Claire Marie Jeanne Picard, gave piano lessons at home. As a young boy, Jacqueshad terrible tantrums, and his mother was sent to the local police station because of theneighbours complaints about the 'ill-treatment of the child'. When she taught her 4-yearold son to read, Jacques became more calm.

Hadamard and his sisters were brought up in a strict household. The family moved toParis, where the population su�ered from famine during the Franco-Prussian War. In1871, civil war broke out in Paris and during these battles the house where the Hadamardfamily lived was burnt down. During this di�cult period, personal tragedies befell thefamily with the death of two of Jacques' sisters.

To the end of the �fth class, Jacques attended Lycée Charlemagne where his father lectured.He was in the sevent class when he was only eight-years old, two years younger than theother students in his class. From 1875 onwards, he was winning prizes in the nationalcompetition for school pupils, the Concours Généraux, mostly in languages. In that time,he doubted that he had mathematical abilities. Jacques father, Amédée, was in 1875transferred to Lycée Louis-le-Grand after having gained a poor reputation as a teacher.One year later, Jacques began fourth class in Lycée Louis-le-Grand. In 1877, his parentstook the decision to let him repeat the third class. That year, he won �rst prize inLatin translation and second prize in Greek translation and mathematics at the ConcoursGénéral. He regretted all his life that he was placed second in mathematics. The fact isthat the winner of the �rst prize also became a matematician, and Kolmogorov described

3

it in this way: �I came out second, - he said, - but the one who was �rst also become amatematicien, a much weaker one -he was always weaker�.1

In 1884, Jacques had to choose between École Polytechnique and École Normale Supérieureafter been placed �rst in both Écoles. He got 1834 points of 2000 in the examinationfor the École Polytechnique, and he beat all previous records. The École Polytechniqueprepered the students to become engineers, whilst the École Normale had a purely scienti�corientation. He made his decision after an visit to Pasteur, and chose École Normale.

At the École Normale Supériure the students lived in a three-storey building. The studentsdormitores was called piaules, and were small cells equipped with an iron bed, a wardrobe,and a basin for washing. There were no doors between the piaules, only curtains. In thebuilding there were also rooms called thurnes with chairs and tables where the studentsworked. The students spent almost the entire dag together, at lesson and in the thurnesafterwards. Some of the teachers in the École Normale were Jules Tannery, Charles Her-mite, Gaston Darboux, Émile Picard, Paul Appel and Edouard Goursat. It was also inthe École Normale that Hadamard found his lifelong friend Paul Painlevé.

In 1890 he started his career teaching mathematics at the Lycée Bu�on. In the beginning,he did not understand the level of his pupils and had di�culties teaching. One of his �rstpupils was Maurice Fréchet. When Fréchet showed an interest in mathematics, Jacquesencouraged him in giving private lessons and writing mathematical letters to him duringvacations.1892 was a comprehensive year for Hadamard. He obtained his doctorat for the thesis Essaisur l'étude des fonctions données par leur développment de Taylor, a work in the area ofcomplex function theory, in which he developed the �rst general theory on singularities.On June, he married Louise-Anna Trénel who he had known from his childhood, and theylater had �ve children.On December 9, Hadamard obtained the Grand Prix des Sciences Mathématiques for hiswork Determination of the number of prime numbers less than a given quantity, whichpresented important result about entire functions and zeta functions.

In 1893 Hadamard moved to Bourdeaux with his wife. Until 1897 he was lecturer atFaculté des Sciences of Bourdeaux, during this time, he published 29 papers in di�erentmathematical topics, one of them was his famous proof of the Prime Number Theorem.While in Bordeaux, he got politically involved in the Dreyfus a�air, in which the frenchjewish o�cer Alfred Dreyfus was in 1894 falsely accused to sending military documents tothe Germans. He was found guilty and was sentenced to life imprisonment on the Devil'sIsland o� the Guyana. The Dreyfus a�air split France into two camps: Dreyfusards andanti-Dreyfusards. Jacques was emotional involved as he was a second cousin of Dreyfuswife Lucie. He was certain that Dreyfus was innoncent, and so was the writer Émile Zola.In 1898, Zola wrote a open letter J'accuse to the President of the French Republic, whichwas published in the newspaper L'Aurore. In that letter he accuses o�cers and generals ofhaving forged documents to get Dreyfus convicted. The family returned to Paris in 1897,where he become lecturer at the Collège de France. One year later he awarded the PrixPoncelet of the Académie des Sciences.

1Quoted in [3], p.18

4

From 1913 and onwards Hadamard held his Séminaire Hadamard at the Collège de France,which rapidly gained a prestigous reputation.In the 1920's and the 1930's Hadamard wrote several papers about mathematical education,including articles and letters about teaching metods.In November 1923 he become Doctor ès Sciences with mention très honorable.During the First World War he lost his two oldest sons. In 1941, when France was invadedby Germans, the Hadamard family escaped to the United States,and Jacques Hadamardobtained a non-permanent post at Columbia University.In 1944, Hadamards third son Mathieu died at the front of Tripolitaine. After his sonsdeath, he wanted to live closer to France so the family moved to London.In 1945 the Hadamard family returned to Paris. His book The psychology of invention inthe mathamatical �eld was published by Princeton University and collected his thoughtson the mechanisms of scienti�c reasoning and the psychology of creativity.Hadamard's wife Louise died in 1960. Two years later, his grandson Étienne was killed ina climbing accident. After that, Hadamard did not leave the house anymore.

On the occasion of the 50th anniversairy of his election to the Académie in 1962, Hadamardwas awarded the Gold medal of the Académie des Sciences. It was his last award.He died peacefully in his home on October 17, 1963 in Paris.

Among his closeds friends, Hadamard was descibed as a homourous and witty man. Everytime Einstein was in Paris, he visited the Hadamard family to play violin in an amateurorchestra in their home. During these visits, the two men spoke more about music thanabout relativity.2

1.2 Well-posedness in the sense of Hadamard

A boundary value problem in mathematical physics is said to be well posed in the sense ofHadamard if it satis�es all of the following criteria:3

• there exists a solution

• the solution is unique

• the solution depends continuously on the data4

These three conditions are all determined by the mechanical or physical origin of theproblem. The �rst condition describes the consistency of the mathematical model, thesecond re�ects the de�niteness of the real situation. The third condition expresses thestability of the equation, a small change in the equation or in the side conditions give riseto a small change in the solution. This stability is desired in mathematical physics wherethe models and data are approximations of reality, if the error in the approximation issmall then the error in the result will also be small.

2Fig1 is taken from the public domain, and this section comes from [3], p.3-p.297.3This de�nition is found in [3].4In [2] Hadamard wrote: �... problèmes se présentait comme parfaitement bien posé, je veux dire comme

possible et déterminé.

5

5 At �rst, Hadamard insisted that the third condition is only important for the Cauchyproblem. In his book La théorie des équations aux dérivées partielles he explained that thethird condition was added by Hilbert and Courant, and that he adopted their view.In his article Sur les problemès aux dérivées partielles et leur signi�cation physique (1902),Hadamard introduce the concept well posedness of boundary value problems. In thisarticle he wrote that to the varied problems of mathematical physics there correspondtwo general types of boundary conditions for partial di�erential equations, the Dirichletcondition and the Cauchy problem. Both of these problems can be parfaitement bien posé,it means possible et determinée6. The physical origin of the problem is accordeing toHadamard related to the two last properties. He thought that one should take note thatthese circumstances are intimately connected. For two problems that seems to be similarone can be possible while the other is impossible depending on the correspondance to aphysical problem. For example, he considers the Laplace equation

∂2u

∂x2+∂2u

∂y2+∂2u

∂z2= 0 (1)

which is the Dirichlet problem presented under physical applications and the Dirichletproblem is well-posed. Consider instead the Cauchy problem of (1), i.e. determine forwhich x ≥ 0 we have a solution to the equation such that for x = 0

u = f(y, z),∂u

∂x= g(y, z)

where f and g are two given functions. Physically, this problem is always possible sincef and g are analytic functions. In the general case, when f and g are not analytic, it isdi�erent. Letting g be determined by an almost analytic function f , Hadamard shows thatthe problem is not always solvable.He also considered the wave equation

∂2u

∂x2+∂2u

∂y2+∂2u

∂z2− ∂2u

∂t2= 0 (2)

for which the Cauchy problem with initial data at t = 0 is uniquely solvable. Hadamardpointed to the importance in not concluding that the Cauchy problem for equation (2) iswell-posed. Take the case when f and g are independent of t. Then, if the solution isunique, it is also independent of t. But, problem (2) is then reduced to (1), which meansthat the Cauchy problem is in general impossible.

When Hadamard wrote his de�nition of well-posed problems, the notation well-posednessof boundary value problems was not considered as natural as today. Many mathematiciansthought that the Cauchy problem was already attended to by the Cauchy-Kovalevskayatheorem:The hypothesis, for the case of one equation (with similar results for systems of equations),is

∂ku

∂tk= f(t, y1, · · · , yn,

∂u

∂t,∂u

∂y1, · · · , ∂

ku

∂ykn).

5The rest of this pharagraph comes mostly from Hadamards original article Sur les problemès aux

dérivées partielles et leur signi�cation physique from 1902.6admissible, the problems are solvable and admits a unique solution

6

where f is an analytic function of its arguments, which vary in some neighboorhood ofthe origin. Analytic functions ϕi(y1, · · · , yn), 0 ≤ i ≤ k − 1 are given near the point(y1, · · · , yn, t) = 0, for t = 0.For the Cauchy problem, the initial conditions are

u(0, y1, · · · , yn) = ϕ0(y1, .., yn),

· · ·∂k−1u

∂tk−1(0, y1, · · · , yn) = ϕk−1(y1, · · · , yn).

The theorem says that it exists a unique solution which is analytic at the origin.Hadamard raised the question of proper posedness in understanding the limitations of thistheorem. 7 In his book based on lectures he gave at Yale University in 1920, he wrote thatthe result is correct, but not for the entire general case. That is because the hypothesis,the Cauchy data and the coe�cients in the equation is described by analytic functions, andthe theorem is often false when the hypothesis is not satis�ed. Most mathematicians ofthat day considered the Cauchy-Kovlevskaya theorem to hold for both analytic and smoothfunctions. Hadamard questioned this, thinking about the fact that only the problems thatcorrespond to a physical phenomena is well-posed. He returned to the Dirichlet problemfor a hyperbolic function several times in his work, and during his lifetime he had donemany variations of this problem.

After Hadamard, other mathematicians continued the work of ill-posed problems. Thequestion of uniquess is easier than the question of existence, and more work is done in thisarea. In the 1950's and the beginning of 1960's many proofs of uniqueness of improperlyposed Cauchy problems appeared in the literature, as a consequence of unique continu-ation theorems. Far more important than the question of uniqueness is the question ofcontinous dependence (implying uniqueness) and approximations of solutions. Early on,mathematicians admitted that many problems with physical interest turned out impossiblewhen tackled directly. To �nd appropriate approximative solutions, they adopted a inverseor semi-inverse method. In the late 1930's matematiciens began to focus on the variety ofquestions that occured while treating the inverse problem. Generally, these problems fallsinto the category of ill-posed problems.

One example: Suppose that we have an initial boundary value problem for the equation:

a∂u

∂t= ∆u,

where a is an arbitrary constant, and ∆ is the Laplace operator. The question is then:given the value of the solution at other suitable points in space-time (in additon to initialboundary value and data), is it possible to determine a? And if this turns out to be thecase, is it possible to determine necessary and su�cient conditions for determine a?

Other cases of improperly posed problems can occur: Often, the region where the equationof the problem is de�ned is unknown. It is also frequent that in some point of the boundaryin the region it is impossible to measure desired data.

7This paragraph comes from [3], p.435

7

Returning to improperly posed Cauchy problems (or initial boundary problems) for whichsolutions do not exist globally unless strong compatibility relations hold among the data. Itmight be the case that a solution exists for particular data, but does not satisfy Hadamard's3:rd criterion.But in practise, the necessary information to determine continuability is not always avail-able, since the data is obtained by measurement and are therefore not precise ( i.e. thegeometry, the coe�cients or boundary values of the solution etc.). The result must allowthis possible error in data.Typical for improperly posed problems is that there exists at most one solution. Generally,it is impossible to verify if the necessairy conditions for ill posed problems are satis�ed ornot.8

2 The ice problem

2.1 Introduction to the ice problem

In the ice caps of Greenland and Antarctica, the temperature of the past is preserved deepdown in the ice. In 1995, D.Dahl-Jensen and team of researchers drilled a hole with depth3028,6 (13 cm radius) in the ice caps of Greenland, as a part of the Greenland Ice CoreProject (GRID). In 1998, they developed a Monte Carlo inverse metod to �t the data andinfer past climate. The authors made 3,300,000 forward calculations and selected 2000that gave the best �t to the temperatures recorded in the hole in 1995. The mesurementswere used to reconstruct past temperatures on the ice surface for the last 50,000 years.9

Consider a model for the ice sheets G = {(t, x, y, z) ∈ R4; t ≤ 0, z ≤ ρt}. Here t is thetime, x and y are the horizontal coordinates, and z is the ice depth, counted negativelyunder the surface. ρ is the nonnegative accumulation constant for the snow. It is oforder 10−9 so it can be taken to be zero for shorter periods. In this model we neglectthe terrestrial heat �ow from the underlying bedrock, and we also simplify and consider−∞ < x, y <∞ and −∞ < z ≤ ρt. The temperature functions u are continuous complexvalued functions on G which are of class C2 in the interior of G.

The heat equation is∂u

∂t= κ

(∂2u

∂x2+∂2u

∂y2+∂2u

∂z2

)where κ is a positive constant called thermal di�usivity. For ice at −4◦C, the thermaldi�usivity is 1.04 · 10−6m2/s. It is natural to assume that the temperature functions areindependent of x and y. We then get a new domain G1 which is only depending of thetime and the depth

G1 = {(t, z) ∈ R2; t ≤ 0, z ≤ ρt},

and the heat equation reduces to∂u

∂t= κ

∂2u

∂z2. (3)

8This chapter comes mostly from [3], and the last two paragraphs comes from [5].9This paragraph comes from [4].

8

Fig 2: Diagram illustrating the domain and boundary for u(t, z)

In our domain G1 we have two temperature functions. h(t) = u(t, ρt), t ≤ 0 describes thetemperature on the ice surface, and the current temperature in the drilled hole is describedby v(z) = u(0, z), z ≤ 0. These two functions construct two rays

S1 = {(t, ρt) ∈ R2; t ≤ 0}; S2 = {(0, z) ∈ R2; z ≤ 0}

which are exactly the boundary of G1.

From this we can construct two problems:

De�nition 2.1: The direct problem

Given a function on the surface of the ice for all past moments in time, �nd the presenttemperature at all depths z ≤ 0. Thus, given h(t) = u(t, ρt) for t ≤ 0, �nd v(z) = u(0, z)for z ≤ 0 for a suitable temperature function u.10

Nonuniqueness in the direct problem:There are many temperatures such that h(t) = u(t, ρt) = 0 for all t ≤ 0. A general solutionto the heat equation is

u(t, z) = C(eκβ2t+βz − eκγ2t+γz),

where β is arbitrary number and γ = −β − ρ/κ, and there exists a bounded solution onlywhen u(t, z) = 0.

Uniqueness for the direct problem:Given a function h ∈ C(R−) there is at most one temperature function u in G1 such thatu(t, ρt) = h(t), t ∈ R−, in the class of temperature functions satisfying

(−t1)−1/2 supz≤ρt1

|u(t1, z)| → 0 as t1 → −∞

here R− = {t ∈ R; t ≤ 0}. 11

10comes from de�nition in the article [1] page 311from the article [1] page 4

9

The quantity supz≤ρt1 |u(t1, z)| has to be �nite for all t1, which means that the solutionis bounded on every ray {(t1, z) ∈ R2; z ≤ ρt1}. But this presumes that the solutionu(t, z) is de�ned in the entire plane. If the solution is de�ned only for z ≤ ρt and vanisheswhen z = ρt, we can extend u(t, z) by mirroring, to show that the same conditions is usedon u(t, z) and that the solution is unique. Let us de�ne U(t, z) as

U(t, z) ={u(t, z), z ≤ ρt−u(t, 2ρt− z)V (t, z), z > ρt

Here, V (t, z) = eAt+Bz and choose A = ρ2/κ,B = −ρ/κ. Let H denote the heat operator∂/∂t = κ∂2/∂z2. Using elementary di�erential calculus, we then get

(HU)(t, z) = −(Hu)(t, 2ρt− z)V (t, z)− 2uz(t, 2ρt− z)(ρV (t, z) + κVz(t, z))−u(t, 2ρt− z)(HV )(z, t), z > ρt

When (Hu)(t, 2ρt− z) = (HV )(z, t) = 0, ρV (t, z) +κVz = 0 with A and B as before. U iscontinuously di�erentiable and satis�es the heat equation, and the solution is zero on theline z = ρt. This is proved in Appendix 3.1.

De�nition 2.2: The inverse problem

Given a function in the drilled hole in the ice at the present time, �nd the temperatureat the surface for all moments in the past. Thus, given v(z) = u(0, z) for z ≤ 0, �ndh(t) = u(t, ρt) for t ≤ 0 for a suitable temperature function u.12

It has been shown by A.N. Tikhonov that there is no uniqueness in the inverse problemand it is also known that two di�erent solutions must di�er by an unbounded function.When we later introduce the class U(G1) there is uniqueness for the inverse problem in thecase of an analytic memory function.

2.2 Exponential solutions

An exponential solution to the reduced heat equation is

u(t, z) = eAt+Bz, (4)

i� A = κB2, where A and B are complex constants.So, the horizontal function can be expressed as h(t) = eA+ρBt and the vertical function isv(t) = eBz. Consider

u(t, z) = e(iα−ρ(β+iγ))t+(β+iγ)z,

where β, γ are real numbers and α is a complex number. From (4) we have that A =iα− ρ(β + iγ) and B = β + iγ. The horizontal and vertical functions then becomes

h(t) = u(t, ρt) = eiαt; v(z) = u(0, z) = e(β+iγ)z

which is a solution to (3) if and only if

α = (2κβ + ρ)γ; γ2 = β2 +ρβ

κ. (5)

12The de�nition comes from article [1] p.3

10

For simplicity, assume that it is not snowing, i.e ρ = 0. The equations above is thensimpli�ed to

α = ±2κβ2, γ = ±β. (6)

where β ≥ 0. Working with temperatures, it is reasonable to assume that h(t) = eiαt isbounded, and α should be real (comes from the fact that β, γ, κ, ρ are real, and equation(5)).

For β ≥ 0, there are two exponential solutions

u1(t, z) = eiαt+(β+iγ)z, α ≥ 0; u2(t, z) = eiαt+(β−iγ)z, α ≤ 0

which have damping like eβz as z → −∞ (when β = 0 coinciding and constant) α is givenby (5). When assuming that ρ = 0, the equation above is simpli�ed to

u1(t, z) = eiαt+β(1+i)z, α ≥ 0; u2(t, z) = eiαt+β(1−i)z, α ≤ 0

α is given by (6)

De�nition 2.3: The two general functions u1(t, z) and u2(t, z) have a temporal periodp= 2π/|α|, an attenuation parameter= β ≥ 0 and a spatial period q= 2π/|γ|.

The attenuation parameter describes how the temperature changes when going downwardsin the ice.

Consider the quotients

|α|2κβ2

=√

1 +ρ

κβ

(1 +

ρ

2κβ

);

γ2

β2= 1 +

ρ

κβ(7)

which are obtained from (5). These two quotients increase with ρ when β is �xed but ifwe instead consider

|α|2κβ2

/γ2

β2=

1 + ρ2κβ√

1 + ρκβ

=

√1 +

ρ2

4κβ(κβ + ρ)

this does not vary so much with ρ for �xed β. This means that the relation between p andq is not so sensitive for changes of values of ρ. Given p, when ρ increase the attenuationparameter becomes smaller.When ρ = 0 we have that β = γ = 2π/q. So, for one spatial period the attenuation isequal to e−2π meaning that the variation is in phase with that on the surface, and theamplitude is reduced by a factor of e2π ≈ 535.

The table below lists some spatial periods for di�erent periods of time. In the table, onlythe case ρ = 0 is considered, when ρ > 0 the numerical values are di�erent. The periodof the wave is h(t) = eiαt, with frequency α = 2π/p measured in Hertz (inverse seconds),the attenuation parameter is equal to

√α/(2κ) expressed in inverse meters. κ is the

thermal di�usivity for ice at temperature −4◦C. For the spatial period, q, consider againthe quotient (|α|/2κβ2)/(γ2/β2) and let ρ = 0

|α|2κβ2

/γ2

β2=|α|

2κγ2=

q2

4πκp= 1

11

These equations can easily be obtained by (7) and De�nition 4.1. So, the spatial period qin the table is equal to

√4πκp.

Temporal period, p Frequency, α Attenuation parameter, β Spatial period, qs Hz= s−1 m−1 m

24 h = 8.64 · 104 7.27 · 10−5 5.91 1.06326.2422 days = 3.1557 · 107 1.99 · 10−7 0.309 20.350, 000 years = 1.5778 · 1012 3.98 · 10−12 1.38 · 10−3 4541100, 000 years = 3.1557 · 1012 1.99 · 10−12 9.78 · 10−4 6425

13

For example, let the temporal period be 24h and consider the diurnal variations at the sur-face. The frequency α is then 2π/86, 400s−1, and we have that the attenuation parameterβ =

√α/(2κ) is approximatively 5.91m−1. So, the spatial period is 1, 06m which means

that the diurnal variation is reduced by a factor of 535. If we instead consider one halfspatial period, meaning that q = 0.53m, the amplitude on the surface is only reduced bya factor of eπ ≈ 23. In conclusion, when the amplitude at the depth of one spatial periodis barely measurable the variation of one half spatial period is still quite large and can notbe neglected. At the depth of 20.3 meters both the diurnal and the annual variations canbe negligible.

2.3 Norms

A bounded solution of the heat equation is

u(t, z) =∑

aαe(iα−ρ(β+iγ))t+(β+iγ)z, (t, z) ∈ R2.

Here, α, β and γ are related by (5), and only �nitely many of the coe�cients aα are notzero. Expressed as a sum, the history and the memory functions are

h(t) =∑α∈R

aαeiαt, t ≤ 0; v(z) =

∑aαe

(β+iγ)z, z ≤ 0.

De�nition 2.4: Let U(G1) denote the space of all generalized trigonometric polynomialsrestricted to G1,

u(t, z) =∑α∈R

aαeiαt+(β+iγ)z =

∑β,γ∈R

bβ+iγeiαt+(β+iγ)z, (t, z) ∈ G1,

where α, β, γ are related by (3), and the coe�cients are related by aα = bβ+iγ for thesetriples (α, β, γ). In the case when ρ = 0 we can have bβ+iγ 6= 0 only if γ = ±β. 14

De�nition 2.5: De�ne the U-norm of u ∈ U(G1) as

‖u‖U =∑α

w(α)|aα| =∑β,γ

w̃(β + iγ)|bβ+iγ |,

13For more detailed table see [1] p.7.14This de�nition is De�nition 8.1 in [1].

12

where w : R →]0,∞[ is a weight function with property w(α) = w̃(β + iγ). Here, α, β, γare related by (5).

De�nition 2.6: The H-norm of h is de�ned as

‖h‖H =∑α∈R

w(α)|aα|

where H ={h(t) =

∑α∈R aαe

iαt| t ∈ R−}. Only �nitely many of aα are not zero.

The coe�cients aα are given by

aα = lim|I|→+∞

1|I|

∫Ih(t)e−iαtdt, α ∈ R. (8)

Here I = [r, s] is a subinterval of R− with length s − r, which is assumed to be positiveand is denoted by |I|.15 The coe�cients are uniquely de�ned by the function values.A proof of equation 8 can be found in Appendix 3.2.

De�nition 2.7: The V-norm of v is de�ned as

‖v‖V =∑β,γ∈R

w̃(β + iγ)|bβ+iγ |.

where V ={v(z) =

∑β,γ∈R bβ+iγe

β+iγ | z ∈ R−}, and all but �nitely many of the coe�-

cients bβ+iγ are zero.

For �nding the coe�cients bβ+iγ = bβ±iγ , v is extended to the whole complex plane as anentire function w, thus w(z) is given by the same formula for all z ∈ C, and w(z) = v(z)when z ≤ 0. For simplicity, only the case ρ = 0 is considered. De�ne

v1(z) = w(12

(1 + i)z) =∑β≥0

bβ+iβeiβz +

∑β>0

bβ−iβeβz, z ∈ R−

and

v2(z) = w(12

(1− i)z) =∑β≥0

bβ+iβeβz +

∑β>0

bβ−iβe−iβz, z ∈ R−

The coe�cients are given by the formulas

bθ+iθ = lim|I|→+∞

1|I|∑I

v1(z)e−iθzdz, θ ≥ 0

bθ−iθ = lim|I|→+∞

1|I|∑I

v2(z)eiθzdz, θ > 0

I = [r, s] is an interval with r < s ≤ 0. The proof is analogous with the proof of (8).

So, it is possible to �nd the coe�cients in the expansion of v, but the way we did it is notsuitable for calculations. Since in practice we will only have measurements from an intervalof �nite lenght, it would be desirable to approximate the coe�cients from this informationalone.

15This is Proposition 9.1 in [1].

13

2.4 Constructing memory and past functions

Constructing the memory function M:Given h ∈ H at the ice surface for all past moments in time, there is a unique memoryv = M(h) ∈ V which gives all present temperatures in the hole. 16

The memory function M(h) can be obtained in this way:Let (aα)α of h be de�ned as (aα)α := {aαn |n = 1, · · · , N}. These coe�cients are deter-mined by (8). From (5) we can also obtain (bβ+iγ)β,γ . Then v can be constructed by theformula v(z) =

∑β,γ∈R bβ+iγe

(β+iγ)z, z ∈ R−.

Construction the past function P:Given v ∈ V of the temperature v(z) at all depths in the hole at the present time, thereis a unique past h = P(v) that gives h(t) at the surface of the ice for all past moments intime. 17

To obtain the past function P(v):Using analytic continuation, determine v1 and v2. The indexed families of coe�cients(bβ+iγ)β , (bβ−iγ)β and (aα)α are then determined. Finally, h(t) is constructed ash(t) =

∑α∈(R) aαe

iαt, t ∈ R.

The model that we have constructed have three di�cultes:To determine the past function P(v) we have to use analytic continuation from v to obtainv1 and v2. When numerical data is approximated by an analytic function, there can be caseswhen it does not exist a solution. So, by Hadamard's �rst criteria, analytic continuationis an ill-posed problem. All the other steps in the construction of the past function is wellde�ned.It is also important to �nd a good approximation of v ∈ V to given �nitely many temper-ature measurements Vz1 , ..., Vzp .The constructed norms H and V requires that the functions is de�ned on R−, and it isdesirable to change them to norms de�ned on a bounded interval I. The norm H is quitenaturally de�ned, but the norm V can be seen as unusual since the exponential part isdismissed in the de�nition. It is desirable to express the H- and V-norm in terms of thesupremum norm over a bounded interval. If we consider the trigonometric polynomial

hα,ε(t) =1εei(α+ε)t − 1

εeiαt, t ∈ R−, α ∈ R, ε 6= 0

we see that this polynomial converges uniformly to an exponential polynomial, hα,ε → iteiαt

as ε→ 0 when t is bounded. So, for any bounded interval I it is not possible to make theestimate ‖h‖H ≤ C‖h|I‖∞. But if the frequencies are kept apart, the estimation holds fora su�cently long interval I ⊂ R−.

Theorem 2.8: For all functions h ∈ H, h(t) =∑aαe

iαt, t ∈ R−, we have an estimate

‖h‖∞ ≤ A‖h‖H, where A =(

infαw(α)

)−1

here the in�mum is taken over �nitely many α such that aα 6= 0.16This is Theorem 12.1 in [1].17This is Theorem 13.1 in [1].

14

Conversely, if the interval I ⊂ R− is long enought, and the frequencies are kept apart,then

‖h‖H ≤ C‖h|I‖∞for some constant C, more precisely

(1− c)‖h‖H ≤∑θ

w(θ)‖h|I‖∞, c =1|I|∑θ

w(θ) supα 6=θ

2w(α)|α− θ|

where the frequencies α, θ are taken so that aα, aθ 6= 0. An estimate is obtained when I isso large that c < 1. 18

Proof of this Theorem can be found in Appendix 3.3.

So, let n the number of frequencies α such that the coe�cient aα is not zero. If we takethe weight w(α) = 1, we see from Fig4. in Appendix 3.5 that

c =2n

|I| infα 6=θ |α− θ|< 1

so we have

|I| > 2ninfα 6=θ |α− θ|

and putting γ = infα 6=θ |α− θ|, the interval has to be larger than 2n/γ for obtain anestimate.

An example: Let us choose three frequencies, with temporal periods pα =10 000,20 000 and 40 000 years. The corresponding frequencies aα are equal to 2π/pα (rememberDe�nition 2.3). So,

1γ

= supα6=θ

1|aα − aθ|

= supα 6=θ

pαpθpαpθ

· 12π/pα − 2π/pθ

=1

2πsupα 6=θ

pαpθpθ − pα

=40, 000

2πyears

We then have that γ = 2π/40, 000. It means that the interval |I| > 6/γ ≈ 38, 198 years.Hence, the length of the interval is depending on the number of frequencies, few frequenciesgives a short interval.

The following theorem shows that the V-norm also can be expressed by the supremumnorm over a bounded interval. Because of the damping factors eβz, we need a strongseparation of the attenuation parameters.

Theorem 2.10: For any two functions v ∈ V of the form

v(z) = b0 +n∑1

(bje

(βj+iβj)z + cje(βj−iβj)z

), z ≤ 0,

with attenuation parameters β0 = 0 < β1 < · · · < βn and coe�cients b0, · · · , bn, c1, · · · , cn,we have an estimate

‖v‖∞ ≤ ‖v‖V18This is Theorem 14.1 in [1]

15

if we use weights wj = 1 for all j. Conversily, assume that the function is real valued,so that b0 ∈ R and cj = b̄j , and that the parameters are well separated in the sense thatβj ≤ σβj+1 for some number σ < 1

3 , or, a little weaker, σ < 13(1 + 2σn). Then

‖v − b0‖V ≤1

cosθ‖v|I − b0‖∞,

where θ = σπ(1− σn−1)/(1− σ) < π/2 and I is the interval I = [sn, 0] with

sn = −2πβ1

1− σn

1− σ≥ −2π

β1

11− σ

> −3πβ1,

and the weights are de�ned by

wj = eβjsn , j = 1, · · · , n.

It is possible to pass to other weights from this weight.19

So, this theorem give us the desired estimation of the V-norm by the supremum norm. Iis here the bounded interval, with length sn, and σ is a separation parameter. A completeproof of this theorem using induction can be found in Kiselman's article [1].

Fig 3: Figure illustrating the inequality Reζ ≥ cos θ|ζ|.

We know that Reζ = cos(arg ζ)·|ζ|, and if we suppose that 0 < θ < π/2 and θ ≥ arg ζ ≥ −θwhere arg : C→]− π, π[ this gives that Reζ ≥ cos θ|ζ|,as illustrated in Fig 3.When σ < 1

3 we can choose θ = πσ/(1− σ).

An example: If the longest temporal period that we are interested in is p1 = 50, 000 years,we have from the table in chapter 2.2 that the longest spatial period is q1 = 4541 meters.From the fact that α = 2π/p and β =

√α/2κ we get q1 =

√4π2/β2

1 . Since all thecoe�cients are positive, we get q1 = 2π/β1. Choosing the separation parameter σ = 0.3, itfollows from Theorem 2.10 that ‖v − b0‖V ≤ 4.5‖v|I − b0‖∞, with θ ≈ 1.35 and I = [sn, 0]where sn ≥ −q1 · (1/0.7) ≈ −6487 meters.20

19This is Theorem 15.1 in article [1].20This chapter comes mostly from [1].

16

2.5 Summary

The ice problem shows the great di�culties that can arise when dealing with ill-posedproblems. In chapter 2.1 we saw that for the inverse problem there are too many solutions.We then introduced the classes H and V and later saw in chapter 2.3 that the problembecame well-posed for analytic functions v(t) but this becomes problematic if we want toconsider measured data since analytic continuation is an ill-posed problem with a wholeclass of solutions (all analytic and matching the measured data).

In practise, the V-norm can not be used because it presupposes that v is a trigonometricpolynomial. If the frequencies are well separeted and the interval is large enought we canapproximate using the supremum norm. Then we can �t the data into a trigonometricpolynomial and from the polynomial obtain coe�cients that can be used to obtain anapproximate memory function v(t). We see that this approximation can be made arbitrarilygood, since if the di�erence measured in the supremum norm is made smaller, then thedistance in the V-norm becomes smaller and since that norm is of the same magnitude as theH-norm it also becomes smaller. This is if the trigonometric polynomials are restrictionsof the same function in U(G1).

We saw in the last example that one has to drill a hole of depth at most 6487 meters to beable to estimate the V-norm for temporal periods up to 50,000 years. This is unproblematicsince the greatest ice depth on Greenland is about 3,000 meters, and even on Antarcticathe greatest ice depth is only about 4,500 meters. What might be problematic thoughis that a temporal period of 50,000 years corresponds to a spatial period of 4541 meters,so the question might then be if these depths are enough to measure such larger periods,especially if one takes in to account the parameters that where neglected in the beginning,such as geothermic heat.

17

3 APPENDIX

3.1

Proves that the z-derivative of U from the right is equal to that of u from the left on theline z = ρt.

Let U(t, z) be de�ned as

U(t, z) ={u(t, z), z ≤ ρt−u(t, 2ρt− z)V (t, z), z > ρt

where u(t, z) solves ∂u∂t = κ∂

2u∂z2

u(t, ρt) = 0u(0, z) = v(z)

and V (t, z) = eρ(ρt−z)

κ .

Let U+z(t, z) = limh→0+U(t,z+h)−U(t,z)

h and U−z(t, z) = limh→0−U(t,z+h)−U(t,z)

h . Then thederivative Uz exists and is continuous on the line z = ρt if U+z(t, ρt) = U−z(t, ρt).

⇒ U−z(t, ρt) = uz(t, ρt)

U+z(t, ρt) =∂

∂z(−u(t, 2ρt− z)V (t, z))

∣∣∣∣z=ρt

= (uz(t, 2ρt− z)V (t, z) +ρ

κu(t, 2ρt− z)V (t, z))

∣∣∣∣z=ρt

= uz(t, 2ρt− ρt)V (t, ρt)︸ ︷︷ ︸1

+ρ

κu(t, 2ρt− ρt)︸ ︷︷ ︸

u(t,ρt)=0

V (t, ρt)︸ ︷︷ ︸1

= uz(t, ρt)

Hence U−z(t, ρt) = U+z(t, ρt) �

3.2

Proves equation 8:

h(t) =∑

α∈R aαeiαt, where aα are zero for all but �nitely many α, implies lim|I|→+∞

1|I|∫I h(t)e−iθtdt

18

= lim|I|→+∞

1|I|

∫I

(∑α∈R

aαeiαt)e−iθtdt

= lim|I|→+∞

1|I|

∫I

(aθeiθt +∑α 6=θ

aαeiαt)e−iθtdt

= lim|I|→+∞

1|I|

∫I

(aθ +∑α 6=θ

aαei(α−θ)t)dt

= lim|I|→+∞

1|I|

(aθ

∫Idt+

∑α 6=θ

aα

∫Iei(α−θ)tdt

)21

= lim|I|→+∞

1|I|

(aθ|I|+

∑α 6=θ

aα

[ei(α−θ)t

i(α− θ)

]I︸ ︷︷ ︸

<∞

)

= lim|I|→+∞

aθ

= aθ

�

3.3

Proves Theorem 2.8:

The �rst part of the Theorem follows from the de�nition of h(t) and the de�nition of theH-norm

|h(t)| ≤∑α

|aα| =∑α

w(α)|aα|w(α)−1 ≤ ‖h‖H supαw(α)−1.

To proove the second part of Theorem 2.8 we need to de�ne a new function ϕ(ζ)

De�nition 2.9: Let ϕ(ζ) be an entire function de�ned as

ϕ(ζ) =

{1−e−ζζ , ζ ∈ C \ {0}

1, ζ = 0

then

|ϕ(ζ)| ≤

{1Reζ , ζ ∈ C,Re > 01+e−Reζ

|ζ| , ζ ∈ C \ {0}

The proof of this inequality can be found in Appendix 3.4.

We want to �nd a suitable estimation of |aθ| in terms of ‖h‖H and ‖h|I‖∞. For this, weneed to know the mean value MI,θ of h(t)e−θt over an interval I.

19

The mean value MI,θ of h(t)e−θt over an interval I = [r, s] ⊂ R− is obtained

MI,θ =1|I|

∫Ih(t)e−iθtdt

= aθ +∑α 6=θ

aα|I|

∫Iei(α−θ)tdt

= aθ +∑α 6=θ

aα|I|

[ei(α−θ)t

i(α− θ)

]sr

= aθ +∑α 6=θ

aαei(α−θ)s − ei(α−θ)r

i|I|(α− θ)

= aθ +∑α 6=θ

aαei(α−θ)s 1− e−i(α−θ)|I|

i|I|(α− θ)︸ ︷︷ ︸ϕ(i(|I|(α−θ)))

where ϕ : C→ C is the function in De�nition 2.9.So now we can estimate |aθ| in terms of |MI,θ|.

|aθ| = |MI,θ − (MI,θ − aθ)| ≤ |MI,θ|+ |MI,θ − aθ|

where

|MI,θ| =1|I|∣∣ ∫

Ih(t)e−iθtdt

∣∣≤ 1|I|∑I

|h(t)| |e−iθt|︸ ︷︷ ︸=1

dt

=1|I|

∫I|h(t)|dt

≤ 1|I||I| · ‖h(t)‖∞

= ‖h(t)‖∞

and

|MI,θ − aθ| ≤∑α 6=θ|aαϕ(i|I|(α− θ))|

=∑α 6=θ

w(α)|aα||ϕ(i|I|(α− θ))|

w(α)

≤ supα 6=θ

|ϕ(i|I|(α− θ))|w(α)

‖h‖H

De�ne

cθ = |I| supα 6=θ

|ϕ(i|I|(α− θ))|w(α)

20

By de�nition of ϕ we have that|ϕ(iη)| ≤ 2/|η|

this gives

cθ ≤ supα 6=θ

2w(α)|α− θ|

.

So we have

|aθ| ≤ |MI,θ|+ |MI,θ − aθ| ≤ |MI,θ|+ cθ|I|−1‖h‖H ≤ ‖h|I‖∞ + cθ|I|−1‖h‖H.

Multiplication by w(θ) gives

w(α)|aθ| ≤ w(θ)‖h|I‖∞ + cθ|I|−1w(θ)‖h‖H.

and summing over all θ

‖h‖H(

1− 1|I|∑θ

cθw(θ))≤(∑

θ

w(θ)‖h|I‖∞)

we get the desired estimation. �

3.4

Proves De�nition 2.9:

ϕ(ζ) is de�ned as

ϕ(ζ) =

{1−e−ζζ , ζ ∈ C r {0}

1, ζ = 0.

We have that

|1− e−ζ | ≤ |1|+ |e−ζ |22

= 1 + |e−Reζ−iImζ |= 1 + |e−Reζ | · |e−iImζ |︸ ︷︷ ︸

1

= 1 + e−Reζ

And

|ϕ(ζ)| =∣∣∣∣ ∫ 0

−1eζtdt

∣∣∣∣≤

∫ 0

−1|eζt|dt23 =

∫ 0

−1eReζtdt

=

[eReζt

]0−1

Reζ=

(1− e−Reζ)Reζ

≤ 1Reζ

, if Reζ > 0

Hence,

|ϕ(ζ)| ≤

{1+e−Reζ

|ζ| , ζ ∈ C r {0}1Reζ , ζ ∈ C, Reζ > 0

�

21

3.5

Fig 4: Figure illustrating the relation between sup, inf and their inverses.

22

References

[1] Kiselman,C.O. 2009. Frozen History: Reconstructing the Climate of the Past, inCialdea,A., Lanzara,F.,Ricci,P.E. eds. Analysis, Partial Di�erential Equations and Ap-plications. Basel, Birkhäuser p. 97-114.

[2] Jacques Hadamard. 1902. Sur les problèmes aux dérivées partielles et leur signi�cationphysique. Princeton University Bulletin. Vol. 13, p.49-52.

[3] Vladimir Maz'ya, Tatyana Shaposhnikova. 1998. Jacques Hadamard, A Universal Math-ematician. American math.Soc.

[4] D.Dahl-Jensen, K.Mosegaard et al. 1998. Past Temperatures Directly from the Green-land Ice Sheets. Science 282, p.268-271.

[5] L.E. Payne. 1973. Some general remarks on improperly posed problems for partial dif-ferential equations. Lecture Notes in Mathematics 316, p.1-30.

23