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

Examensarbete i matematik, 15 hp

Handledare och examinator: Gunnar Berg

Juni 2011

Department of Mathematics

Ill-posed problems and their applications

to climate research

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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:

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Contents

1 Introduction 3

1.1 Jacques Hadamard . . . 3

1.2 Well-posedness in the sense of Hadamard. . . 5

2 The ice problem 8 2.1 Introduction to the ice problem . . . 8

2.2 Exponential solutions. . . 10

2.3 Norms . . . 12

2.4 Constructing memory and past functions. . . 14

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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 reality they 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, Jacques had terrible tantrums, and his mother was sent to the local police station because of the neighbours complaints about the 'ill-treatment of the child'. When she taught her 4-year old son to read, Jacques became more calm.

Hadamard and his sisters were brought up in a strict household. The family moved to Paris, where the population suered from famine during the Franco-Prussian War. In 1871, civil war broke out in Paris and during these battles the house where the Hadamard family lived was burnt down. During this dicult period, personal tragedies befell the family with the death of two of Jacques' sisters.

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

In 1884, Jacques had to choose between École Polytechnique and École Normale Supérieure after been placed rst in both Écoles. He got 1834 points of 2000 in the examination for the École Polytechnique, and he beat all previous records. The École Polytechnique prepered the students to become engineers, whilst the École Normale had a purely scientic orientation. 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 students dormitores 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 the building there were also rooms called thurnes with chairs and tables where the students worked. The students spent almost the entire dag together, at lesson and in the thurnes afterwards. 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 in the École Normale that Hadamard found his lifelong friend Paul Painlevé.

In 1890 he started his career teaching mathematics at the Lycée Buon. In the beginning, he did not understand the level of his pupils and had diculties teaching. One of his rst pupils was Maurice Fréchet. When Fréchet showed an interest in mathematics, Jacques encouraged him in giving private lessons and writing mathematical letters to him during vacations.

1892 was a comprehensive year for Hadamard. He obtained his doctorat for the thesis Essai sur l'étude des fonctions données par leur développment de Taylor, a work in the area of complex 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 they later had ve children.

On December 9, Hadamard obtained the Grand Prix des Sciences Mathématiques for his work Determination of the number of prime numbers less than a given quantity, which presented important result about entire functions and zeta functions.

In 1893 Hadamard moved to Bourdeaux with his wife. Until 1897 he was lecturer at Faculté des Sciences of Bourdeaux, during this time, he published 29 papers in dierent mathematical topics, one of them was his famous proof of the Prime Number Theorem. While in Bordeaux, he got politically involved in the Dreyfus aair, in which the french jewish ocer Alfred Dreyfus was in 1894 falsely accused to sending military documents to the Germans. He was found guilty and was sentenced to life imprisonment on the Devil's Island o the Guyana. The Dreyfus aair split France into two camps: Dreyfusards and anti-Dreyfusards. Jacques was emotional involved as he was a second cousin of Dreyfus wife 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, which was published in the newspaper L'Aurore. In that letter he accuses ocers and generals of having 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 Prix Poncelet of the Académie des Sciences.

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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 invaded by Germans, the Hadamard family escaped to the United States,and Jacques Hadamard obtained a non-permanent post at Columbia University.

In 1944, Hadamards third son Mathieu died at the front of Tripolitaine. After his sons death, 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 in the mathamatical eld was published by Princeton University and collected his thoughts on the mechanisms of scientic reasoning and the psychology of creativity.

Hadamard's wife Louise died in 1960. Two years later, his grandson Étienne was killed in a 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, Hadamard was 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. Every time Einstein was in Paris, he visited the Hadamard family to play violin in an amateur orchestra in their home. During these visits, the two men spoke more about music than about 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 of Hadamard if it satises 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 the problem. The rst condition describes the consistency of the mathematical model, the second reects the deniteness of the real situation. The third condition expresses the stability of the equation, a small change in the equation or in the side conditions give rise to a small change in the solution. This stability is desired in mathematical physics where the models and data are approximations of reality, if the error in the approximation is small 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 denition is found in [3].

4In [2] Hadamard wrote: ... problèmes se présentait comme parfaitement bien posé, je veux dire comme

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5 At rst, Hadamard insisted that the third condition is only important for the Cauchy

problem. In his book La théorie des équations aux dérivées partielles he explained that the third condition was added by Hilbert and Courant, and that he adopted their view. In his article Sur les problemès aux dérivées partielles et leur signication physique (1902), Hadamard introduce the concept well posedness of boundary value problems. In this article he wrote that to the varied problems of mathematical physics there correspond two general types of boundary conditions for partial dierential equations, the Dirichlet condition 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 to

Hadamard related to the two last properties. He thought that one should take note that these circumstances are intimately connected. For two problems that seems to be similar one can be possible while the other is impossible depending on the correspondance to a physical 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 Dirichlet problem is well-posed. Consider instead the Cauchy problem of (1), i.e. determine for which 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 since f and g are analytic functions. In the general case, when f and g are not analytic, it is dierent. Letting g be determined by an almost analytic function f, Hadamard shows that the 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. Hadamard pointed to the importance in not concluding that the Cauchy problem for equation (2) is well-posed. Take the case when f and g are independent of t. Then, if the solution is unique, it is also independent of t. But, problem (2) is then reduced to (1), which means that the Cauchy problem is in general impossible.

When Hadamard wrote his denition of well-posed problems, the notation well-posedness of boundary value problems was not considered as natural as today. Many mathematicians thought that the Cauchy problem was already attended to by the Cauchy-Kovalevskaya theorem:

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 ∂yk n ).

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

dérivées partielles et leur signication physique from 1902.

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where f is an analytic function of its arguments, which vary in some neighboorhood of the 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 this theorem. 7 In his book based on lectures he gave at Yale University in 1920, he wrote that

the result is correct, but not for the entire general case. That is because the hypothesis, the Cauchy data and the coecients in the equation is described by analytic functions, and the theorem is often false when the hypothesis is not satised. Most mathematicians of that day considered the Cauchy-Kovlevskaya theorem to hold for both analytic and smooth functions. Hadamard questioned this, thinking about the fact that only the problems that correspond to a physical phenomena is well-posed. He returned to the Dirichlet problem for a hyperbolic function several times in his work, and during his lifetime he had done many variations of this problem.

After Hadamard, other mathematicians continued the work of ill-posed problems. The question of uniquess is easier than the question of existence, and more work is done in this area. In the 1950's and the beginning of 1960's many proofs of uniqueness of improperly posed 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 of continous dependence (implying uniqueness) and approximations of solutions. Early on, mathematicians admitted that many problems with physical interest turned out impossible when tackled directly. To nd appropriate approximative solutions, they adopted a inverse or semi-inverse method. In the late 1930's matematiciens began to focus on the variety of questions that occured while treating the inverse problem. Generally, these problems falls into 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 initial boundary value and data), is it possible to determine a? And if this turns out to be the case, is it possible to determine necessary and sucient conditions for determine a? Other cases of improperly posed problems can occur: Often, the region where the equation of the problem is dened is unknown. It is also frequent that in some point of the boundary in the region it is impossible to measure desired data.

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Returning to improperly posed Cauchy problems (or initial boundary problems) for which solutions do not exist globally unless strong compatibility relations hold among the data. It might be the case that a solution exists for particular data, but does not satisfy Hadamard's 3: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. the geometry, the coecients or boundary values of the solution etc.). The result must allow this 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 satised or not.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 deep down in the ice. In 1995, D.Dahl-Jensen and team of researchers drilled a hole with depth 3028,6 (13 cm radius) in the ice caps of Greenland, as a part of the Greenland Ice Core Project (GRID). In 1998, they developed a Monte Carlo inverse metod to t the data and infer past climate. The authors made 3,300,000 forward calculations and selected 2000 that gave the best t to the temperatures recorded in the hole in 1995. The mesurements were 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 the

time, x and y are the horizontal coordinates, and z is the ice depth, counted negatively under the surface. ρ is the nonnegative accumulation constant for the snow. It is of order 10−9 so it can be taken to be zero for shorter periods. In this model we neglect

the terrestrial heat ow from the underlying bedrock, and we also simplify and consider −∞ < x, y < ∞and −∞ < z ≤ ρt. The temperature functions u are continuous complex valued 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 diusivity. For ice at −4◦C, the thermal

diusivity is 1.04 · 10−6m2/s. It is natural to assume that the temperature functions are

independent of x and y. We then get a new domain G1 which is only depending of the

time and the depth

G1 = {(t, z) ∈ R2; t ≤ 0, z ≤ ρt},

and the heat equation reduces to

∂u ∂t = κ

∂2u

∂z2. (3)

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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 the

temperature on the ice surface, and the current temperature in the drilled hole is described by 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: Denition 2.1: The direct problem

Given a function on the surface of the ice for all past moments in time, nd the present temperature 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 solution to 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 only when 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 that

u(t, ρt) = h(t), t ∈ R−, in the class of temperature functions satisfying

(−t1)−1/2 sup z≤ρt1

|u(t1, z)| → 0 as t1→ −∞

here R−= {t ∈ R; t ≤ 0}. 11

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The quantity supz≤ρt1|u(t1, z)| has to be nite for all t1, which means that the solution

is bounded on every ray {(t1, z) ∈ R2; z ≤ ρt1}. But this presumes that the solution

u(t, z)is dened in the entire plane. If the solution is dened only for z ≤ ρt and vanishes when z = ρt, we can extend u(t, z) by mirroring, to show that the same conditions is used on u(t, z) and that the solution is unique. Let us dene 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 dierential 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 is

continuously dierentiable and satises the heat equation, and the solution is zero on the line z = ρt. This is proved in Appendix 3.1.

Denition 2.2: The inverse problem

Given a function in the drilled hole in the ice at the present time, nd the temperature at the surface for all moments in the past. Thus, given v(z) = u(0, z) for z ≤ 0, nd h(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 problem and it is also known that two dierent solutions must dier by an unbounded function. When we later introduce the class U(G1)there is uniqueness for the inverse problem in the

case 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+ρBtand the vertical function is

v(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)

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For simplicity, assume that it is not snowing, i.e ρ = 0. The equations above is then simplied to

α = ±2κβ2, γ = ±β. (6)

where β ≥ 0. Working with temperatures, it is reasonable to assume that h(t) = eiαt is

bounded, 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 given

by (5). When assuming that ρ = 0, the equation above is simplied 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)

Denition 2.3: The two general functions u1(t, z) and u2(t, z) have a temporal period

p= 2π/|α|, an attenuation parameter= β ≥ 0 and a spatial period q= 2π/|γ|.

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

Consider the quotients |α| 2κβ2 = r 1 + ρ κβ  1 + ρ 2κβ  ; γ 2 β2 = 1 + ρ κβ (7)

which are obtained from (5). These two quotients increase with ρ when β is xed but if we instead consider |α| 2κβ2  γ2 β2 = 1 +2κβρ q 1 +κβρ = s 1 + ρ 2 4κβ(κβ + ρ)

this does not vary so much with ρ for xed β. This means that the relation between p and q is not so sensitive for changes of values of ρ. Given p, when ρ increase the attenuation parameter becomes smaller.

When ρ = 0 we have that β = γ = 2π/q. So, for one spatial period the attenuation is equal to e−2π meaning that the variation is in phase with that on the surface, and the

amplitude is reduced by a factor of e2π ≈ 535.

The table below lists some spatial periods for dierent periods of time. In the table, only the case ρ = 0 is considered, when ρ > 0 the numerical values are dierent. The period of the wave is h(t) = eiαt, with frequency α = 2π/p measured in Hertz (inverse seconds),

the attenuation parameter is equal to pα/(2κ) expressed in inverse meters. κ is the thermal diusivity for ice at temperature −4◦C. For the spatial period, q, consider again

the quotient (|α|/2κβ2)/(γ22)and let ρ = 0

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These equations can easily be obtained by (7) and Denition 4.1. So, the spatial period q in the table is equal to√4πκp.

Temporal period, p Frequency, α Attenuation parameter, β Spatial period, q

s Hz= s−1 m−1 m 24 h = 8.64 · 104 7.27 · 10−5 5.91 1.06 326.2422days = 3.1557 · 107 1.99 · 10−7 0.309 20.3 50, 000years = 1.5778 · 1012 3.98 · 10−12 1.38 · 10−3 4541 100, 000years = 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 parameter

β =pα/(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 half spatial period, meaning that q = 0.53m, the amplitude on the surface is only reduced by a factor of eπ ≈ 23. In conclusion, when the amplitude at the depth of one spatial period

is barely measurable the variation of one half spatial period is still quite large and can not be neglected. At the depth of 20.3 meters both the diurnal and the annual variations can be negligible.

2.3 Norms

A bounded solution of the heat equation is

u(t, z) =Xaαe(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 coecients aα are not

zero. Expressed as a sum, the history and the memory functions are

h(t) = X

α∈R

aαeiαt, t ≤ 0; v(z) =

X

aαe(β+iγ)z, z ≤ 0.

Denition 2.4: Let U(G1) denote the space of all generalized trigonometric polynomials

restricted to G1, u(t, z) = X α∈R aαeiαt+(β+iγ)z = X β,γ∈R bβ+iγeiαt+(β+iγ)z, (t, z) ∈ G1,

where α, β, γ are related by (3), and the coecients are related by aα = bβ+iγ for these

triples (α, β, γ). In the case when ρ = 0 we can have bβ+iγ 6= 0 only if γ = ±β. 14

Denition 2.5: Dene the U-norm of u ∈ U(G1) as

kukU = X α w(α)|aα| = X β,γ ˜ w(β + iγ)|bβ+iγ|,

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

Denition 2.6: The H-norm of h is dened as khkH=

X

α∈R

w(α)|aα|

where H = h(t) = Pα∈Raαeiαt| t ∈ R− . Only nitely many of aα are not zero.

The coecients aα are given by

aα= lim |I|→+∞ 1 |I| Z I h(t)e−iαtdt, α ∈ R. (8)

Here I = [r, s] is a subinterval of R− with length s − r, which is assumed to be positive

and is denoted by |I|.15 The coecients are uniquely dened by the function values.

A proof of equation 8 can be found in Appendix 3.2. Denition 2.7: The V-norm of v is dened as

kvkV =

X

β,γ∈R

˜

w(β + iγ)|bβ+iγ|.

where V = v(z) = Pβ,γ∈Rbβ+iγeβ+iγ| z ∈ R−

, and all but nitely many of the coe-cients bβ+iγ are zero.

For nding the coecients bβ+iγ = bβ±iγ, v is extended to the whole complex plane as an

entire 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. Dene

v1(z) = w( 1 2(1 + i)z) = X β≥0 bβ+iβeiβz+ X β>0 bβ−iβeβz, z ∈ R− and v2(z) = w( 1 2(1 − i)z) = X β≥0 bβ+iβeβz+ X β>0 bβ−iβe−iβz, z ∈ R−

The coecients are given by the formulas bθ+iθ = lim |I|→+∞ 1 |I| X I v1(z)e−iθzdz, θ ≥ 0 bθ−iθ = lim |I|→+∞ 1 |I| X 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 coecients in the expansion of v, but the way we did it is not suitable for calculations. Since in practice we will only have measurements from an interval of nite lenght, it would be desirable to approximate the coecients from this information alone.

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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 memory v =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 dened as (aα)α := {aαn|n = 1, · · · , N }. These coecients are

deter-mined by (8). From (5) we can also obtain (bβ+iγ)β,γ. Then v can be constructed by the

formula v(z) = Pβ,γ∈Rbβ+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, there is a unique past h = P(v) that gives h(t) at the surface of the ice for all past moments in time. 17

To obtain the past function P(v):

Using analytic continuation, determine v1 and v2. The indexed families of coecients

(bβ+iγ)β, (bβ−iγ)β and (aα)α are then determined. Finally, h(t) is constructed as

h(t) =P

α∈(R)aαeiαt, t ∈ R.

The model that we have constructed have three dicultes:

To determine the past function P(v) we have to use analytic continuation from v to obtain v1and v2. When numerical data is approximated by an analytic function, there can be cases

when it does not exist a solution. So, by Hadamard's rst criteria, analytic continuation is an ill-posed problem. All the other steps in the construction of the past function is well dened.

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 dened on R−, and it is

desirable to change them to norms dened on a bounded interval I. The norm H is quite naturally dened, but the norm V can be seen as unusual since the exponential part is dismissed in the denition. It is desirable to express the H- and V-norm in terms of the supremum norm over a bounded interval. If we consider the trigonometric polynomial

hα,(t) = 1 e i(α+)t 1 e iαt, t ∈ R−, α ∈ R,  6= 0

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

as  → 0 when t is bounded. So, for any bounded interval I it is not possible to make the estimate khkH ≤ Ckh|Ik∞. But if the frequencies are kept apart, the estimation holds for

a sucently long interval I ⊂ R−.

Theorem 2.8: For all functions h ∈ H, h(t) = P aαeiαt, t ∈ R−, we have an estimate

khk∞≤AkhkH, where A = inf

α w(α)

−1

here the inmum is taken over nitely many α such that aα 6= 0.

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

then

khkH≤ Ckh|Ik∞

for some constant C, more precisely (1 − c)khkH≤ X θ w(θ)kh|Ik∞, c = 1 |I| X θ w(θ) sup α6=θ 2 w(α)|α − θ|

where the frequencies α, θ are taken so that aα, aθ 6= 0. An estimate is obtained when I is

so 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 coecient aα is not zero. If we take

the weight w(α) = 1, we see from Fig4. in Appendix 3.5 that

c = 2n

|I| infα6=θ|α − θ| < 1 so we have

|I| > 2n

infα6=θ|α − θ|

and putting γ = infα6=θ|α − θ|, the interval has to be larger than 2n/γ for obtain an

estimate.

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 (remember

Denition 2.3). So, 1 γ = supα6=θ 1 |aα− aθ| = supα6=θ pαpθ pαpθ · 1 2π/pα− 2π/pθ = 1 2πsupα6=θ pαpθ pθ− pα = 40, 000 2π years 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 frequencies gives a short interval.

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

separation of the attenuation parameters.

Theorem 2.10: For any two functions v ∈ V of the form v(z) = b0+

n

X

1

bje(βj+iβj)z+ cje(βj−iβj)z, z ≤ 0,

with attenuation parameters β0= 0 < β1 < · · · < βn and coecients b0, · · · , bn, c1, · · · , cn,

we have an estimate

kvk∞≤ kvkV

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if we use weights wj = 1 for all j. Conversily, assume that the function is real valued,

so that b0 ∈ R and cj = ¯bj, and that the parameters are well separated in the sense that

βj ≤ σβj+1 for some number σ < 13, or, a little weaker, σ < 13(1 + 2σn). Then

kv − b0kV ≤

1

cosθkv|I− b0k∞, where θ = σπ(1 − σn−1)/(1 − σ) < π/2 and I is the interval I = [s

n, 0] with sn= − 2π β1 1 − σn 1 − σ ≥ − 2π β1 1 1 − σ > − 3π β1 , and the weights are dened 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. I is here the bounded interval, with length sn, and σ is a separation parameter. A complete

proof 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 ≥ −θ where 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 = pα/2κ we get q1 = p4π2/β12. Since all the

coecients are positive, we get q1= 2π/β1. Choosing the separation parameter σ = 0.3, it

follows from Theorem 2.10 that kv − b0kV ≤ 4.5kv|I− b0k∞, with θ ≈ 1.35 and I = [sn, 0]

where sn≥ −q1· (1/0.7) ≈ −6487 meters.20

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2.5 Summary

The ice problem shows the great diculties that can arise when dealing with ill-posed problems. 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 problem became well-posed for analytic functions v(t) but this becomes problematic if we want to consider measured data since analytic continuation is an ill-posed problem with a whole class 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 trigonometric polynomial. If the frequencies are well separeted and the interval is large enought we can approximate using the supremum norm. Then we can t the data into a trigonometric polynomial and from the polynomial obtain coecients that can be used to obtain an approximate memory function v(t). We see that this approximation can be made arbitrarily good, since if the dierence measured in the supremum norm is made smaller, then the distance in the V-norm becomes smaller and since that norm is of the same magnitude as the H-norm it also becomes smaller. This is if the trigonometric polynomials are restrictions of the same function in U(G1).

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3 APPENDIX

3.1

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

Let U(t, z) be dened 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) = 0 u(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 the

derivative 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) | {z } 1 +ρ κu(t, 2ρt − ρt)| {z } u(t,ρt)=0 V (t, ρt) | {z } 1 = uz(t, ρt) Hence U−z(t, ρt) = U+z(t, ρt)  3.2 Proves equation 8: h(t) =P

α∈Raαeiαt, where aαare zero for all but nitely many α, implies lim|I|→+∞|I|1

R

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= lim |I|→+∞ 1 |I| Z I (X α∈R aαeiαt)e−iθtdt = lim |I|→+∞ 1 |I| Z I (aθeiθt+ X α6=θ aαeiαt)e−iθtdt = lim |I|→+∞ 1 |I| Z I (aθ+ X α6=θ aαei(α−θ)t)dt = lim |I|→+∞ 1 |I|  aθ Z I dt +X α6=θ aα Z I ei(α−θ)tdt  21 = lim |I|→+∞ 1 |I|  aθ|I| + X α6=θ aα  ei(α−θ)t i(α − θ)  I | {z } <∞  = lim |I|→+∞aθ = aθ  3.3 Proves Theorem 2.8:

The rst part of the Theorem follows from the denition of h(t) and the denition of the H-norm |h(t)| ≤X α |aα| =X α w(α)|aα|w(α)−1≤ khkHsup α w(α)−1. To proove the second part of Theorem 2.8 we need to dene a new function ϕ(ζ) Denition 2.9: Let ϕ(ζ) be an entire function dened as

ϕ(ζ) = ( 1−e−ζ ζ , ζ ∈ C \ {0} 1, ζ = 0 then |ϕ(ζ)| ≤ ( 1 Reζ, ζ ∈ C, Re > 0 1+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 khkH and kh|Ik∞. For this, we

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The mean value MI,θ of h(t)e−θt over an interval I = [r, s] ⊂ R− is obtained MI,θ = 1 |I| Z I h(t)e−iθtdt = aθ+ X α6=θ aα |I| Z I ei(α−θ)tdt = aθ+ X α6=θ aα |I| " ei(α−θ)t i(α − θ) #s r = aθ+ X α6=θ aα ei(α−θ)s− ei(α−θ)r i|I|(α − θ) = aθ+ X α6=θ aαei(α−θ)s 1 − e−i(α−θ)|I| i|I|(α − θ) | {z } ϕ(i(|I|(α−θ)))

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

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

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By denition of ϕ we have that |ϕ(iη)| ≤ 2/|η| this gives cθ ≤ sup α6=θ 2 w(α)|α − θ|. So we have

|aθ| ≤ |MI,θ| + |MI,θ− aθ| ≤ |MI,θ| + cθ|I|−1khkH≤ kh|Ik∞+ cθ|I|−1khkH.

Multiplication by w(θ) gives

w(α)|aθ| ≤ w(θ)kh|Ik∞+ cθ|I|−1w(θ)khkH.

and summing over all θ khkH  1 − 1 |I| X θ cθw(θ)  ≤  X θ w(θ)kh|Ik∞ 

we get the desired estimation. 

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3.5

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References

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

[2] Jacques Hadamard. 1902. Sur les problèmes aux dérivées partielles et leur signication physique. 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.

References

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