Self-similarity in mathematics andastrophysics

On the Sedov solution for supernova remnants and Hausdorff

dimension

Evi Kruijer

July 17, 2015

Bachelor thesis

Supervisors: dr. Han Peters and dr. Jacco Vink

Korteweg-de Vries Institute for MathematicsAnton Pannekoek Institute for Astronomy

Faculty of science

University of Amsterdam

Abstract

In this thesis, self-similarity is considered in two different settings. The first one is in amathematical context. The definition of Hausdorff measure and dimension are given andsome properties about these will be proved. A theorem that gives a method to easilycalculate the Hausdorff dimension of a self-similar set will be given and proved. In thenext chapter, some known results about dimension in complex dynamics are discussed.In the following chapter, self-similarity is considered in a framework of astrophysics. TheSedov similarity solution is derived and explained, which states that for a certain con-

stant ξ0, rShock = ξ0

(E0t2

ρ

) 15. This solution works under the assumption that the density

of the interstellar medium is homogeneous. Also, the inner structure of a supernovaremnant is plotted for different values of the adiabatic index.

Title: Self-similarity in mathematics and astrophysicsOn the Sedov solution for supernova remnants and Hausdorff dimensionAuthor: Evi Kruijer, evi.kruijer@hotmail.nl, 10358919Supervisors: dr. Han Peters and dr. Jacco VinkSecond graders: dr. Lex Kaper and dr. Jan van MillDate: July 17, 2015

Korteweg-de Vries Institute for MathematicsAnton Pannekoek Institute for AstronomyUniversiteit van AmsterdamScience Park 904, 1098 XH Amsterdamhttp://www.science.uva.nl/math

http://www.astro.uva.nl

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Contents

Introduction 4

1 Hausdorff dimension of self-similar sets 61.1 Hausdorff measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2 Hausdorff dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3 Self-similar sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2 Complex dynamics 212.1 Julia sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.2 The Mandelbrot set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3 The Sedov solution 253.1 Intuıtive approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.2 A mathematical treatment of the Buckingham Pi theorem . . . . . . . . 263.3 The Buckingham Pi theorem applied to supernova remnants . . . . . . . 283.4 The Sedov solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.5 Applying the Sedov solution . . . . . . . . . . . . . . . . . . . . . . . . . 33

4 Discussion 344.1 Mathematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344.2 Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

5 Populaire samenvatting 365.1 Natuurkunde . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.2 Wiskunde . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

Bibliography 38

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Introduction

In this thesis, self-similarity is treated from the viewpoint of (astro)physics and math-ematics. In both settings, self-similarity means something different, but similarity is akeyword in both parts.

The mathematical part will focus on what can be said about the size of (self-similar)sets.

In the first chapter, we will introduce two ways of considering the size of sets. The firstone is a measure, that is comparable with the familiar Lebesgue measure on simple setslike for example open sets. This measure will help to define another way to consider thesize of a self-similar set by giving it a dimension, the Hausdorff dimension. Where theHausdorff measure is comparable to the Lebesgue measure on simply sets, the Hausdorffdimension agrees with the regular topological dimension on such sets. Some propertiesof Hausdorff measure and dimension will also be discussed, which tells us what happensto the Hausdorff measure or dimension of a set if you apply a function to the set. We willconsider how to calculate the Hausdorff measure and dimension for self-similar sets afterhaving defined what being self-similar means in mathematics. The Hausdorff dimensionhas the nice property that it is very easy to calculate for self-similar sets. This is shownby calculating the Hausdorff dimension of the Cantor set, the Sierpinski triangle and theVon Koch snowflake.

In the second chapter, we remain in a mathematical setting but now we consider setsthat are not per se as strictly self-similar as in the first chapter. Those sets can be foundby iterating complex functions and looking at points that will or will not diverge if thefunction is being iterated. The Mandelbrot set and Julia sets are introduced. We willdiscuss results about the Hausdorff dimension of the Mandelbrot set and the Hausdorffmeasure and dimension of Julia sets.

In the next chapter, we move to astrophysics. This chapter is about the shockwaveof a supernova remnant. In a supernova explosion, a lot of energy is released at once,leading to matter being blown away in all directions. This enormous flow of matter willcollide with the interstellar medium and cause a shockwave.

The physisist L. Sedov found a solution for the radius of the shock wave in caseof a homogeneous density of the interstellar medium. Also, when the density of theinterstellar medium is proportinal to a power of r, the distance to the center of thesupernova remnant, the approach he used leads to a correct solution. He used theBuckingham Pi theorem, which is a way to use dimensional analysis. The theorem willbe stated, proven and used in this thesis.

This model for the radius of the shockwave of a supernovaremnant is usefull, sinceit gives a way to calculate the age, radius, energy that was released in the supernovaexplosion or density of the interstellar medium around the supernova, given the other

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three. Instead of the radius of the shockwave of the supernova remnant, it is also possibleto use the speed of the shock wave.

Finally, I would like to thank the people who made it possible for me to write thisthesis. In the first place, I want to thank my supervisors, Han Peters and Jacco Vink,for they have taught me a lot. Not only about mathematics and astrophysics, but alsoabout how to actually do a project and what to do if you don’t find what you hopedto. I have learned a lot and that would not have been possible without them. Also, Iwould like to thank my boyfriend, Luuk Stehouwer, for being so kind when I was stressedand worried, and my family who supported me when I was sure I could never finish mythesis.

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1 Hausdorff dimension ofself-similar sets

For open sets in Rn, the Lebesgue measure is a natural way to consider the ’size’ of aset. The size of the interval (0, 5) ⊂ R is just 5, the size of a sphere with radius 1 in R3 is43π. But not all sets have such a natural size. Take for example the middle third Cantor

set, which can be constructed by taking (0, 1) ⊂ R and then removing the middle 13, and

then removing the middle third of both smaller intervals left and so one.This gives rise to the question whether there is a natural way to define the ’size’ of

such sets, which are not necessarily open. This question motivates the definition ofHausdorff measure and Hausdorff dimension The way that this dimension works, makesit applicable to sets that are self-similar. The idea behind self-similar sets is that theyare made out of copies of themselves scaled by a certain factor. The exact definition ofa self-similar set will be given later.

Most theorems are not only true in Rn but even in all normed spaces, but I willonly consider Rn and subspaces of it. The Hausdorff dimension is different from thetopological dimension and the Hausdorff measure is different from the Lebesgue measure,but we will see that there is still a natural relation between those two.

1.1 Hausdorff measure

In order to define the Hausdorff measure, we need to introduce some notions.

Definition 1. The diameter of a set U ⊂ Rn is defined as

|U | = sup{|x− y| : x, y ∈ U}.

Definition 2. A δ-cover of a set U ⊂ Rn is a countable collection of sets {Ui} such thatthe diameter of all sets Ui is at most δ and U ⊂

⋃i Ui.

Given the definition of a δ-cover and the diameter of a set we can define a functionon the subsets of a space depending on two parameters.

Definition 3. For all U ⊂ Rn and s, δ ∈ R>0 we define

H sδ (U) = inf{

∞∑i=1

|Ui|s : {Ui} is a δ-cover of U}.

6

This is a well defined expression, since the diameter of a set is always non-negative.Using this we can define a measure on U . Since we know that for ε < δ every ε-cover of a set is also an δ-cover of that set, we know that H s

δ (U) as a function of δ isdecreasing.Therefore, it is well defined to take the limit limδ→0 H s

δ (U).

Definition 4. The s-dimensional Haussdorff measure of U is

H s(U) = limδ→0

H sδ (U).

This is indeed a Borel measure:

Lemma 1. The Haussdorff measure is a Borel measure, since it satisfies:

A H s(∅) = 0.

B H s(A) ≤H s(B) when A ⊂ B.

C H s(⋃∞i=1Ai) ≤

∑∞i=1 H s(Ai) for a countable sequence of sets A1, A2, . . . and

H s(⋃∞i=1Ai) =

∑∞i=1 H s(Ai) when the Ai are disjoint Borel sets.

Proof. A This can be seen since any set covers the empty set, and because of coursethe infimum of the diameter of any set is 0, the s-dimensional Hausdorff measureof the empty set is 0.

B Since A ⊂ B any δ-cover of B is also a cover of A, so it is immediately clear thatindeed H s(A) ≤H s(B).

C Let {Bj}j be a δ-cover of⋃∞i=1Ai. Then the inequality follows from the fact that

now for each Ai we have that {Bj ∩ Ai}j is a δ-cover of it.

To show that for Borel sets equality holds, I will use a theorem from Mattila [7].Define for A,B ⊂ Rn distance d(A,B) in the usual way, so d(A,B) = inf{|x− y| :x ∈ A, y ∈ B}. Accorcing to [7], an outer measure µ is a Borel measure if

µ(A ∪B) = µ(A) + µ(B)

for all A,B ⊂ X with d(A,B) > 0.

But that is something that we can prove. Let A,B ⊂ Rn such that d(A,B) > 0,

say 0 < δ < d(A,B)2

and let {Ui} be any δ-cover of A ∪ B. It is reasonable to start

with a δ with an upper limit, since our only interest is limδ→0. Since δ < d(A,B)2

none of the sets Ui can cover both a part of A and a part of B, we can split

∞∑i=1

|Ui|s =∑

Ui∩A 6=∅

|Ui|s +∑

Ui∩B 6=∅

|Ui|s,

so indeed taking limδ→0 we get H s(A ∪B) = H s(A) + H s(B).

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Note that the Hausdorff measure does not depend on the ambient space. This meansthat if A ⊂ Rn ⊂ Rn+j with n, j ∈ N≥1, then H s(A) is the same in Rn as in Rn+j.This is clear, since a cover of A in Rn can be extended to a cover of A in Rn+j that hasthe same diameter. To do so, the same sets to cover A can be used where the extra jcoordinates can be chosen zero.

The Hausdorff measure is not linear, but the scaling is straightforward:

Lemma 2. Let U ⊂ Rn, λ ∈ R>0 and a ∈ Rn. Say λU = {λx : x ∈ U} and a + U ={x+ a|x ∈ U}. Then

A H s(λU) = λsH s(U)

B H s(a+ U) = H s(U)

Proof. A Let {Ui} be a δ-cover of U . Then {λUi} is a λδ-cover of λU , since forx ∈ λU we know 1

λx ∈ U , so there is a Ui such that 1

λx ∈ Ui, so x ∈ λUi. Then

H sλδ(λU) ≤

∑i

|λUi|s = λs∑i

|Ui|s ≤ λsH sδ (U)

because this is just one δ-cover of U . Taking the limit of δ to 0 gives thatH s(λU) ≤ λsH s(U). The inequality the other way follows by replacing λ by1λ. This gives H s( 1

λU) ≤ ( 1

λ)sH s(U). Taking U = λU now gives the desired

inequality λsH s(U) ≤H s(λU).

B Let {Ui} be a δ-cover of U . Then {a + Ui} is a δ-cover of a + U , since for x ∈ Uwe have x ∈ Ui for some i, so x+ a ∈ Ui + a for this i. But |Ui| = |Ui + a| since atranslation conserves distances, so

H s(U) = limδ→0

inf{∞∑i=1

|Ui|s : {Ui} is a δ-cover of U}

= limδ→0

inf{∞∑i=1

|Ui|s : {Ui} is a δ-cover of U + a} = H s(U + a)

For Holder continuous functions it is also possible to derive an inequality between themeasure of a set and the measure of its image.

Definition 5. A Holder function of exponent α is a function f : X → Y such that thereis a constant c such that for all x1, x2 ∈ X

|f(x1)− f(x2)| ≤ c|x1 − x2|α.

Here, 0 < α ≤ 1.

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The Holder functions cannot be defined for α < 0, but they could be defined for α > 1.However, the only Holder functions with α > 1 are the constant functions and for thatreason they are not of interest.

Lemma 3. Let U ⊂ Rn and f : U → Rm a Holder function of exponent α. Then for alls ∈ R>0 we have

Hsα (f(U)) ≤ c

sαH s(U).

Proof. Let {Ui} be a δ-cover of U . We know that |f(U∩Ui)| ≤ c|Ui|α, so now {f(U∩Ui)}is a cδα-cover of f(U). Then also∑

i

|f(U ∩ Ui)|sα ≤ c

cα

∑i

|Ui|s,

which meansH

sαcδα(f(U)) ≤ c

sαH s

δ (U).

Taking limδ→0 givesH

sα (f(U)) ≤ c

sαH s(U).

1.2 Hausdorff dimension

Using the s-dimensional Hausdorff measure, it is possible to define a so-called Hausdorffdimension. In order to do so, notice that for a given set U and 0 < δ < 1 H s

δ (U) isnon-increasing with s, hence also H s(U) is non-increasing. Taking t > s and {Ui} aδ-cover of U it is true that ∑

i

|Ui|t

δt≤∑i

|Ui|s

δs,

or equivalently∑

i |Ui|t ≤ δt−s∑

i |Ui|s. If we start with H sδ <∞ and take δ → 0 we get

that H t(U) = 0. This means that H s(U) as a function of s has only two possibilities:it starts at infinity, and at a certain point it ’jumps’ to 0, or it starts at some finite valueand equals zero for all s > 0. The value of s for which H s(U) ’jumps’ is what is calledthe Hausdorff dimension of U .

∞

0 dimH(U)

H s(U)

s

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Definition 6. The Hausdorff dimension of a set U , written as dimH(U), is defined as

dimH(U) = inf{s ∈ R>0|H s(U) = 0} = sup{s ∈ R>0|H s(U) =∞}.

At s = dimH(U), the value of H s(U) can be 0, ∞ or anything in between.A few properties of the Hausdorff measure have an immediate effect on the Hausdorff

dimension.From H s(E) ≤H s(F ) if E ⊂ F it follows that in this case also dimH(E) ≤ dimH(F ),

so the Hausdorff dimension is monotone.Hausdorff dimension has countable stability, which means that for a countable collec-

tion of sets U1, U2, . . . , one has dimH(⋃i Ui) = supi{dimH(Ui)}. Because of monotonic-

ity, it is true that dimH(⋃i Ui) ≥ dimH(Ui) for all i, but when s > dimH(Ui) for all i, it

is clear that also dimH(⋃i Ui) ≤ s.

The Hausdorff dimension of a countable set will always be zero, since for a singlepoint set {x} it holds that H 0({x}) = 1, so dimH({x}) = 0, and since the Hausdorffdimension is countable stable, any countable set has Hausdorff measure zero.

The Hausdorff measure and the Lebesgue measure are for an integer n the same upto multiplying with a constant, i.e. H s(U) = Cµ(U), where µ is the Lebesgue measure.To show that there is ineed such a contant, we need to show first that the hausdorff andLebesgue measure are regular and uniformly distributed.

Definition 7. A Borel measure µ on X is called Borel regular if for all A ⊂ X thereexists a Borel set B ⊂ X such that A ⊂ B and µ(A) = µ(B).

Definition 8. A Borel regular measure µ on X is called uniformly distribured if

0 < µ(B(x, r)) = µ(B(y, r)) <∞

for all x, y ∈ X, 0 < r <∞.

Lemma 4. The n-dimensional Hausdorff measure H n is a Borel regular uniformlydistributed measure on Rn.

Proof. Firstly, it is regular. Let U ⊂ Rn. Now, choose for all i ∈ N a δ-cover {Ui,j}jwith δ = 1

iand

∑∞i=1 |Ui,j|n < H n

1i

+ 12. Now the set ∩i ∪j Ui,j = V ⊂ Rn is a set such

that U ⊂ V and H n(U) = H n(V ).To see that it is also uniformly distributed, apply part B of lemma 2.

The Lebesgue measure is Borel regular and uniformly distributed as wekk. The proofis similar to the proof of lemma 4.

We now have the theorem

Theorem 1. Let µ, ν be Borel regular uniformly distributed measures on a seperablemetric space X. Then there is a constant such that µ = cν.

Proof. See for example Mattila theorem 3.4, [7].

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An equivalent definition of the Hausdorff measure can be obtained by using onlyn-dimensional balls to cover sets.

Lemma 5. For all U ⊂ Rn, we have

H sδ (U) = Bs

δ(U) = inf

{∞∑i=1

|Ui|s : {Ui} is a δ-cover of U and Ui is a

n-dimensional open ball}.

Proof. The inequality H sδ (U) ≤ Bs

δ(U) is easy to see. Any δ-cover using only open ballsover which the infimum is taken for Bs

δ is also a regular δ-cover, so it is in the set overwhich the infimum is taken for H s

δ .For the other inequality: Let {Ui} be a δ-cover of U . But if we now replace each Ui

by a open ball Bi with radius 12|Ui| that contains Ui, we get a δ-cover of U with open

balls and we know that still |Ui| = |Bi| so∑∞

i=1 |Ui|s =∑∞

i=1 |Bi|s.

So since the Lebesgue measure µ and the Hausdorff measure H n are Borel regularand uniformly distributed measures on Rn and Rn is seperable, there is a constant cnsuch that for all U ⊂ Rn it holds that H n(U) = cnµ(U).

We can prove that this constant is cn = πn2

2n(n2)!

.

Theorem 2. For a set U ⊂ Rn the n-dimensional Hausdorff measure and the Lebesgue

measure are related by H n(U) = πn2

2n(n2)!µ(U).

Proof. Let Bn be a n-dimensional ball with radius 1. For all ε > 0, δ > 0 there exists aδ-cover of Bn that consists of open balls {Bi} with radii {ri} and that satisfies∑

i

µ(Bi) ≤ µ(Bn) + ε.

But we know that for any δ-cover of Bn by balls,∑i

(ri)n ≥

(n2)!

πn2

µ(Bn).

The Hausdorff measure, in its definition with covers by balls from lemma 5, tells us that

H n(Bn) = limδ→0

inf{∑i

(2ri)n}.

But we know that

inf

{∑i

(2ri)n

}=

2n(n2)!

πn2

µ(Bn)

so

H n(Bn) =2n(n

2)!

πn2

µ(Bn).

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Since we know how the Hausdorff measure scales and we can cover all sets with balls, itfollows that for all U ⊂ Rn we have

H n(U) =πn2

2n(n2)!µ(U).

The property with the Holder function of exponent α as in lemma 2 can also easilybe transferred to Hausdorf dimension.

Lemma 6. For U ⊂ Rn and f : U → Rm such that for all x1, x2 ∈ U we have|f(x1)− f(x2)| ≤ c|x1 − x2|α it holds that dimH(f(U)) ≤ 1

αdimH(U).

Proof. This proof is just applying the proof of lemma 3.

Holder functions with exponent 1 are called Lipschitz functions and those are veryinteresting in the context of Hausdorff dimension.

Definition 9. A Lipschitz function is a function f : U → Rm such that there exists aconstant c ∈ R>0 such that for all x, y ∈ U |f(x)− f(y)| ≤ c|x− y|.

A bi-Lipschitz function is a function f : U → Rm such that there exists constantsc1, c2 ∈ R>0 such that for all x, y ∈ U it holds that c1|x− y| ≤ |f(x)− f(y)| ≤ c2|x− y|.

Examples of functions that are both Lipschitz and bi-Lipschitz are the constant func-tion and the identity function. A function that is Lipschitz, but not bi-Lipschitz is thecosinus, since | cos(x)− cos(y)| ≤ |x− y| but there cannot be a strict positive constantsuch that c|x− y| ≤ | cos(x)− cos(y)| because for example cos(0) = cos(2π).

Lemma 7. 1. If f is a Lipschitz function then dimH(f(U)) ≤ dimH(U).

2. If f is a bi-Lipschitz function then dimH(f(U)) = dimH(U).

Proof. 1. This follows from lemma 4 with f a Holder function of exponent 1.

2. This follows from lemma 4 by also considering f−1 for the other inequality.

Hence Hausdorff measure is invariant under bi-Lipschitz transformations.Another way to measure the dimension of a set is the Minkowski dimension, also called

the box-counting dimension or the Minkowski-Bouligand dimension.

Definition 10. Let U ⊂ Rn be non-empty and bounded. Let N(U, ε) be the smallestnumber of n-dimensional balls with radius equal to ε that are needed to cover U . Thenthe upper Minkowski dimension of U is defined by

dimM(U) = lim supε↓0

log(N(U, ε))

log(1ε)

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and the lower Minkowski dimension is defined by

dimM(U) = lim infε↓0

log(N(U, ε))

log(1ε)

.

If these are equal, then the Minkowski dimension is defined by the value that these take,so

dimM(U) = limε↓0

log(N(U, ε))

log(1ε)

.

For many sets, including the self-similar sets that we will define in the next section, theMinkowski dimension and the Hausdorff dimension are equal. This will not be provenin this thesis, but for further reading [5] is recommended.

1.3 Self-similar sets

In general, it is difficult to calculate the Hausdorff dimension of a set. For self-similarsets, there is an easy way to find the Hausdorff dimension. In order to do so, we firstneed to have an exact definition of self-similar sets.

Definition 11. A map S : Rn → Rn is called a similarity if there exists a 0 < c < 1such that for all x, y ∈ Rn

|S(x)− S(y)| = c|x− y|.

Here, the constant c is called the ratio of S.

Note that the identity function is not a similarity, since its ratio would be 1. Con-tracting a set with a certain factor is a similarity, where the factor by which you contractthe set is the ratio of the similarity. Of course, this is only true as long as you the ratioof is between 0 and 1.

Using these maps, we can give an exact definition of a self-similar set.

Definition 12. A set U ⊂ Rn is called self-similar if there exists a countable collectionof similarities {Si} with ratios {ci} such that U =

⋃i Si(U).

An example of a set that is self-similar is the middle thirdCantor set, explained in theintroduction.

Example 1. Let U be the Cantor set, S1 : R→ R, x 7→ 13x and S2 : R→ R, x 7→ 1

3x+ 2

3.

Then S1 and S2 are similarities which both have ratio 13, and

⋃i=1,2 Si(U) = U

A weaker condition for a function than being a similarity is being a contraction.

Definition 13. A map S : D → D with D ⊂ Rn is called a contraction if there existsa c > 0 such that for all x, y ∈ D we have the inequality

|S(x)− S(y)| ≤ c|x− y|.

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This means that contractions are always Lipschitz functions. Contractions can be usedto prove a theorem that will later help prove a theorem about the Hausdorff dimensionof a self-similar set. This theorem needs an unusual definition of distance.

Definition 14. Let D be a subset of Rn and let S be the class of all compact andnon-empty subsets of D and A,B ∈ S . Then the Hausdorff distance can be defined by

dH(A,B) = inf{δ > 0 : A ⊂ Bδ and B ⊂ Aδ}

where Aδ = {x ∈ D : ∃a ∈ D such that |x− a| < δ}.Equivalent to this definition is

dH(A,B) = max{supa∈A

infb∈B

d(a, b), supb∈B∈a∈A d(a, b)}.

Lemma 8. The δ-parallel body distance is indeed a distance, since it satisfies

• dH(x, x) ≥ 0

• dH(x, y) = 0 ⇐⇒ x = y

• dH(x, y) = dH(y, x)

• dH(x, z) ≤ dH(z, y) + dH(y, z)

Proof. • This is clear, since we only consider δ > 0.

• First, dH(x, x) = 0 since x ⊂ xδ for all δ ≥ 0.

Also, if dH(x, y) = 0, then inf{δ > 0 : x ⊂ yδ and y ⊂ xδ}, which implies x = y.

• This is true by definition.

• Let x, y, z ∈ S and δ1, δ2 such that x ⊂ yδ1 and y ⊂ xδ1 and also y ⊂ zδ2 andz ⊂ yδ2 . Then it follows that x ⊂ zδ1+δ2 and z ⊂ xδ1+δ2 , from which this inequalityfollows.

Knowing this, we can formulate the following theorem:

Theorem 3. Let D ⊂ Rm and let S1, . . . , Sn : D → D be contractions such that for allx, y ∈ D it holds that |Si(x)− Si(y)| ≤ ci|x− y| with all ci < 1.

Then there exists a unique non-empty bounded compact set F that is invariant underthe contractions, such that F =

⋃ni=1 Si(F ).

If for all E ∈ S we define S(E) =⋃ni=1 Si(E) then F =

⋂∞k=1 S

k(E) for all E suchthat Si(E) ⊂ E for all 1 ≤ i ≤ n. Here, Sk(E) is used to denote the kth iterate of thefunction S.

14

Proof. I will follow the proof from [5]. Notice that since all Si are contractions, they areLipschitz and continuous. Since the image of a compact set under a continuous map iscompact again and a finite union of compact sets is again compact, S transforms setsfrom S into other sets in S . Now let E ∈ S be a set such that Si(E) ⊂ E for all i.There is always such a set, since for example one can take the entire space intersectedwith a ball of a radius large enough. When Si(E) ⊂ E for all i, also S(E) ⊂ E. Fromthat it follows that Sk(E) ⊂ Sk−1(E) for all k > 1. This creates a decreasing sequence ofnon-empty compact sets {Sk(E)}k. Now the intersection ∩∞i=1S

k(E) has to be compactand non-empty too. This intersection will be the invariant set F .

It is non-empty, so left to prove is its uniqueness and that it is invariant. To see thatit is indeed invariant, consider

S(F ) = S(∩∞i=1Sk(E)) = ∩∞i=1S(Sk(E)) = ∩∞i=1S

k(E).

This leaves only the uniqueness. First, let A,B ∈ S . Notice that if δ is such thatSi(B) ⊂ (Si(A))δ for all i, also ∪ni=1Si(B) ⊂ (∪∞i=1Si(A))δ. Using this, we get theinequality

dH(S(A), S(B)) = dH(∪ni=1(A),∪ni=1(B))

≤ maxidH(Si(A), Si(B))

≤ (maxici)dH(A,B).

Now let A,B ∈ S both be invariant. So we have S(A) = A, S(B) = B, the inequalitydH(S(A), S(B)) ≤ (maxi ci)dH(A,B) and we know that ci < 1 for all i. That is onlytrue when dH(A,B) = 0, and since dH is a distance, it is only true when A = B, whichcompletes the proof.

Definition 15. A mass distribution is a measure ν on a bounded subset A of Rn suchthat 0 < ν(A) <∞.

Lemma 9. Let ν be a mass distribution on a non-empty set F ⊂ Rn and let there be ans such that there are c, δ > 0 such that ν(U) ≤ c|U |s for all U such that |U | < δ. ThenH s(F ) ≥ ν(F )1

cand s ≤ dimH(F ).

Proof. Let {Ui} be a cover of F . Then we know that

0 < ν(F ) = ν(∪iUi) ≤∑i

ν(Ui) ≤ c∑i

|Ui|s.

Now taking the infimum and the limit limδ→0 we get

ν(F ) ≤ c∑i

|Ui|s = cH s(U)

so indeed H s(F ) ≥ ν(F )1c.

A lemma that is a bit more geometrical is the following:

15

Lemma 10. Let {Vi} be a set of disjoint open subsets of Rn. Assume that each Vicontains a ball with radius a1r and that it is contained in a ball of radius a2r, so a1r ≤|Vi| ≤ a2r for all i. Then any ball of radius r intersects no more than (1+2a2

a1)n sets Vi.

Proof. Let B be a ball of radius r and center b. Let Vi be such that B ∩ Vi 6= ∅. ThenVi is contained in a ball with center b and radius r + 2a2r = r(1 + 2a2). The number of

balls of radius a1r, that fit in a ball of radius r(1 + 2a2) is rn(1+2a2)n

an1 rn = (1+2a2

a1)n. Since

any Vi with B ∩Vi 6= ∅ contains a ball of radius a1r and should be contained in a ball ofradius (1 + 2a2)r cocentric with B, this is the maximum amount of closures of Vi’s thatintersect B.

So now we can tell something about the dimension of a self-similar set. Since anysimilarity is, of course, a contraction and it has a ratio smaller than 1, we can also applytheorem 3 here.

Theorem 4. Suppose S1, . . . , Sm : Rn → Rn are similarities with ratios c1, . . . , cmand that there exists a non-empty bounded open set V such that ∪mi=1Si(V ) ⊂ V andSi(V ) ∩ Sj(V ) = ∅ whenever i 6= j. Let F be self-similar, i.e. F =

⋃mi=1 Si(F ). Let s be

such that∑m

i=1 csi = 1.

Then dimH(F ) = s and 0 < H s(F ) <∞.

Proof. I will follow the proof from [5]. The proof will consist of two parts. In the firstpart, an upper limit for the Hausdorff dimension will be derived, and in the second parta lower limit will be derived.

Notice that those Si are also contractions as in theorem 3, which implies that the setF is the only invariant set under these similarities. Let s be such that

∑mi=1 c

si = 1. For

any set A ⊂ Rn, write Ai1,...,ik = Si1 ◦ · · · ◦ Sik(A). Then, if we take k ∈ N≥1 and Jk allsets {i1, . . . , ik} with 1 ≤ ij ≤ m, we can see that

F =⋃Jk

Fi1,...,ik

since we assumed that F =⋃mi=1 Si(F ). Now consider

∑Jk|Fi1,...,ik |s. Since we know

that the ratios of the similarities S1, . . . , Sm are c1, . . . , cm we can calculate:∑Jk

|Fi1,...,ik |s =∑Jk

(ci1 · · · · · cik)s|F |s = (∑i1

ci1)s · · ·

∑ik

(cik)s|F |s = |F |s.

Now let δ > 0. Since ci < 1 for all i, we know that there must be a k ∈ N≥0 such that|Fi1,...,ik | ≤ maxi{cki } ≤ δ. Since

⋃JkFi1,...,ik covers F , this gives an upper bound for the

Hausdorff measure: It implies H sδ (F ) ≤ |F |s for all s, which implies that H s(F ) ≤ |F |s,

because F is bounded.This gives therefore an upper bound for the Hausdorff measure. Now we will derive a

lower bound.Let I be the set of all infinite but countable sequences I = {(i1, i2, . . . )|1 ≤ ij ≤ m}.

Call the collection of sequences starting with i1, . . . , ik with 1 ≤ ij ≤ m and the otherterms free Ii1,...,ik , so Ii1,...,ik = {(i1, . . . , ik, qk+1, . . . )|1 ≤ qj ≤ m}.

16

Let ν be a mass distribution on I such that ν(Ii1...,ik) = (ci1 · · · · · cik)s. Because∑mi=1 c

si = 1, it follows that ν(Ii1,...,ik) =

∑mi=1 ν(Ii1,...,ik,i), so indeed this is a mass

distribution on I with ν(I) = 1.Now we will transfer this mass distribution ν to a mass distribution ν ′ on F . This can

be done because there is a natural way to relate F and I. One can define a map from Ito subsets of F ,

f : (i1, i2, . . . ) 7→∞⋂k=1

Fi1,...,ik

and this map will be the idea behind the transfer of the mass dirstibution on I to a massdistribution on F .

We define for any A ⊂ F

ν ′(A) = ν({(i1, i2, . . . ) :∞⋂k=1

Fi1,...,ik ∈ A})

for which also ν ′(F ) = 1.Next, we want to apply lemma 9, so we need to check that ν ′ satisfies the conditions

stated.Let V be the open, non-empty and bounded set such that ∪mi=1Si(V ) ⊂ V . Then also

V ⊃ ∪mi=1Si(V ) = S(V ). Since V is non-empty, bounded, closed and a subset of Rn, itis compact. Theorem 3 implies that iterating S on V will converge to F . From that, itbecomes clear that F ⊂ V and also we must have Fi1,...,ik ⊂ V i1,...,ik .

Now let B be a ball of radius r, where 0 < r < 1. We will make an estimation for themeasure of this ball, ν ′(B). This is possible by looking at Vi1,...,ik such that |Vi1,...,ik | iscomparable to |B| = 2r and Vi1,...,ik ∪ F 6= ∅ and Vi1,...,ik ∪B 6= ∅.

Since 0 < r < 1 and 0 < ci < r for all i, we can do the following: For all (i1, i2, . . . ) ∈ I,let ik be the first k for which

(minici)r ≤ ci1 · ci2 . . . cik ≤ r.

For any element of I we get exactly one finite sequence (i1, . . . , ik) in this way. The setof all these sequences will be called Q.

We know that all Vi’s are disjoint for i = 1, . . . ,m. Then also Vi1,...,ik,j are disjoint forj = 1, . . . ,m. As a consequence, even {Vi1,...,ik : (i1, . . . , ik) ∈ Q} is a set of disjoint sets.In the same way, we know that because F =

⋃mi=1 Si(F ), also F ⊂

⋃mj=1 Fi1,...,ik,j. The

consequence here is that F ⊂⋃Q Fi1,...,ik . For this last set, it holds that

⋃Q Fi1,...,ik ⊂⋃

Q V i1,...,ik .Next, we will need the geometrical result as in lemma 10. Choose a1, a2 such that there

exists an r > 0 for which V contains a ball of radius a1r and is itself contained in a ballof radius a2r. Let (i1, . . . , ik) ∈ Q and notice that Vi1,...,ik now contains a ball of radiusci1ci2 . . . cika1r, so it certainly contains a ball of radius (mini c

ki )a1r. It is contained in a

ball of radius ci1ci2 . . . cika2r, but since ai < 1 it is also just contained in the ball of radiusa2r. Write Q1 for all sequences (i1, . . . , ik) ∈ Q wuch that B ∩ V i1,...,ik 6= ∅. Lemma 10tells us that the amount of sequences in Q1 is at most ( 1+2a2

a1 mini ci)n. Say ( 1+2a2

a1 mini ci)n = q.

17

This gives a possible estimate for ν ′(B), namely:

ν ′(B) = ν ′(B ∩ F ) ≤ ν({(i1, i2, . . . ) :m⋂i=1

fi1,...,ik ⊂ F ∩B}) ≤ ν({⋃Q1

Ii1,...,ik})

But because we chose (i1, . . . , ik) such that c1c2 . . . ck ≤ r we can calculate that

ν({⋃Q1

Ii1,...,ik}) ≤∑Q1

ν(Ii1,...,ik) =∑Q1

(ci1 . . . cik)s ≤

∑)Q1r

s ≤ rsq.

So the estimate we have now is that ν ′(B) ≤ rsq. Because q is independent of the setswe chose, we can apply lemma 9 again with δ any number and c = q to obtain thatH s(F ) > 0 and s ≤ dimH(F ).

Combining this with the upper bound for the Hausdorff dimension derived earlier, weget dimH(F ) = s and even 0 < H s(F ) <∞.

This theorem is important, since it can be used to easily derive the Hausdorff dimen-sion for self-similar sets.

Example 2. For example, take again the middle third Cantor set. It can be seen thatthe Cantor set is self-similar by considering the similarities

S1(x) =1

3x S2(x) =

1

3x+

2

3.

Both these similarities have ratio 13, so the Hausdorff dimension of the Cantor set is the

s for which

(1

3)s + (

1

3)s = 1,

so s = log(2)log(3)

.

Was it reasonable to assume that the Hausdorff dimension of the Cantor set wouldbe about 0.6? Is it what you could expect, since the Hausdorff dimension agrees withthe intuitive topological dimension? The Cantor set is obviously not a plane, so itsdimension has to be smaller than 2. One could day that it only contains points and istotally disconnected, so it has to have dimension 0, like a point. But on the other hand,it has uncountably many elements, which might lead you to think that it has dimension1, like a part of the real line. The value log(2)

log(3)that was found is in between these two

values, which makes some sense if tought about it like this.A very similar example is the Sierpinski triangle.

Example 3. The Sierpinsky triangle looks like this:One starts with a filled equilateral trangle. The middles of the sides are connected and

the middle triangle is excluded. In every next step, the middle triangle of all trianglesthat are left are excluded.

Intuitively, the Hausdorff dimension will be greater than one, since there is uncount-ably many line, but smaller than two, since it does not contain any surfaces.

18

Here, there are three similarities needed, taking (0, 0) to be the left bottom side ofthe triangle and the length of each side to be 1:

S1((x, y)) =

(1

2x,

1

2y

)S2((x, y)) =

(1

2x+

1

2,1

2y

)S3((x, y)) =

(1

2x+

1

4,1

2y +

1

2

)These similarities each have ratio 1

2so the Hausdorff dimension is the s such that

3(1

2)s = 1

which leads to the Hausdorff dimension being log(3)log(2)

. This value is between one and two,as was expected.

In these examples, all the similarities have the same ratio. But that is not always thecase. As long as a set is self-similar, calculating the Hausdorff dimension is still easy ascan be seen in the next example.

Example 4. This example is a slight variation on the Von Koch snowflake.Start with the following shape:

which is a line of length two, the a line going one up and one to the right, a line goingone back down and one to the right with a line of length two attached to it.

In the following step, all lines will be replaced by this figure itself:

19

and in every next step, all lines will again be replaced by the figure we started with.Now, there are four similarities in this situation. Since there is also some rotation in

this situation, the coordinates used are complex.

S1(x+ iy) =1

3(x+ iy) S2(x+ iy) = 2 +

√2

6eπ4i(x+ iy)

S3(x+ iy) = 3 +

√2

6e−

π4i(x+ iy) S4(x+ iy) =

1

3(x+ iy) + 4

In this case, two of the similarities have ratio 13

and the other two have ratio√26

. Thismeans that for the Hausdorff dimension s, we have

2(1

3)s + 2(

√2

6)s = 1

which can also be solved without any real effort, and gives that the dimension of thisset is − log(2)

log( 13+

√2

6). As could be expected, this set has its dimension between 1 and 2.

20

2 Complex dynamics

The self-similar sets in the previous chapter were all linearly self-similar. But one canalso consider sets that are not that strict self-similar. For example sets that are madeup out of copies of themself that are not only smaller, but also slightly bent. Or setsmade out of copies of themselves where not all parts of the set are scaled by the samefactor. Considering more sets might give us more insight in the theory of self-similarsets. The sets that we will consider are not rare, since they occur often in the field ofcomplex dynamical analysis. An interesting collection of sets are the Julia sets. Theyare encountered when looking at iterating complex functions.

2.1 Julia sets

Let f : C → C be a polynomial with complex coefficients, so f(z) = a0 + a1z + a2z2 +

· · ·+ anzn. Write fk = f ◦ · · · ◦ f for k times applying f . Let w be a periodic point such

that fp(w) = w for some p > 0. Let λ be the value of the pth complex derivative of fin w, so (fp)′(w) = λ. We call w superattractive if λ = 0. If 0 < |λ| < 1, w is calledattractive. If |λ| = 1 then w is called indifferent and if |λ| > 1 then w is repelling.

Definition 16. The Julia set of a complex polynomial f is the closure of the set ofrepelling periodic points. It is denoted as J(f), so it can be written as

J(f) = {x ∈ C : ∃p > 0 such that fp(x) = x and |(fp)′(x)| > 1}.

A lot of equivalent definitions are used, because calculating the Julia set can be easierwhen another definition is used. An example of this is the following:

Theorem 5. Let f be a complex polynomial and let K be the set

K = {z ∈ C : {fn(z)} is bounded}.

Then the Julia set of f will beJ(f) = ∂K.

Proof. For the proof, see for example Milnor [8].

For some easy sets, the Julia set is easy to calculate. In the following example we willuse the equivalent definition of a Julia set from theorem 5. Let us consider the complexpolynomial f(z) = z2.

21

Figure 2.1: Julia set of the polynomial f(z) = z2 − 0.41 + 0.3i.

Example 5. Let f be the complex polynomial f(z) = z2. For all x ∈ C with |x| < 1,we will certainly have that |f(x)| < |x| and so for those points, {fn(x) : n ∈ N} willbe bounded. For all x ∈ C with norm |x| > 1, the norm will increase: |f(x)| > |x|.This means that for those points, {fn(x) : x ∈ N} is not bounded. Finally, the pointswith norm 1 will be mapped to other points with norm 1, so these points will have{fn(x) : n ∈ N} bounded. Now the set K from theorem 5 in this case will be D1 = {x ∈C : |x| ≤ 1}. The boundary of this set is now the Julia set, so

J(f) = S1 = {x ∈ C : |x| = 1}.

2.2 The Mandelbrot set

A specific kind of polynomials are the polynomials of the form f(z) = z2 + c for someconstant c ∈ C. For the Julia set coresponding with this f , we write Jc. Using thesepolynomials we can define a new set, which is called the Mandelbrot set.

Definition 17. The Mandelbrot set M is defined as

M = {c ∈ C : Jc is connected}= {c ∈ C : {fn(0) : n ∈ N} is bounded}.

These definitions are equivalent, as proven in [5].Julia sets are often called fractals, as is the Mandelbrot set. Note that every point of

the Mandelbrot set corresponds with a polynomial f(z) = z2 + c, so it also correspondswith a Julia set Jc.

The self-similarity is not strict, as can be seen by comparing figure 2.2 to figure 2.3.In figure 2.3 we see a copy of the Mandelbrot set that can be found when enlarging the

22

Figure 2.2: The Mandelbrot set

Figure 2.3: A baby Mandelbrot set

Mandelbrot set. The baby Mandelbrot set looks like the Mandelbrot set, but it is notonly smaller but also slightly deformed.

Still, it is possible to prove things about its Hausdorff dimension. There is a strongrelation between the Hausdorff dimension of the boundary of the Mandelbrot set andthe Hausdorff dimension of certain Julia sets. Using this relation, the Japanese mathe-matician M. Shishikura proved by the end of the twentieth century the following aboutthe Mandelbrot set:

23

Theorem 6. Let ∂M denote the boundary of the Mandelbrot set. Then

dimH(∂M) = 2.

Also, there exists a c ∈ C such that

dimH(Jc) = 2.

Proof. See [10].

The second result, that there exists a x ∈ C such that dimH(Jc) = 2 follows quicklyafter it has been proven that dimH(∂M) = 2.

It remained a question what the 2-dimensional Hausdorff measure of those sets was.In 2008, X. Buff and A. Cheritat proved the following:

Theorem 7. There exists a c ∈ C such that

H 2(Jc) > 0.

Proof. See [3].

This is a remarkable result, since at that time it was only known that H 2(Jc) ≥ 0for specific values of c. However, the relation between Julia sets and the boundary ofthe Mandelbrot set turned out to be not strong enough to conclude from this that alsoH 2(∂M) > 0. Proving that H 2(∂M) > 0 remains an open question today.

24

3 The Sedov solution

L. Sedov was a physisist from the Sovjet Union. He solved a problem about the shockwave when an atomic bomb explodes. The conditions also applied to a supernova ex-plosion and the solution is generally referred to as the Sedov solution, or even the Sedovstage of supernova remnant evolution. This solution was discovered simultaniously byother physisists independently. Because of that, it is sometimes called the Sedov-Taylorsolution. The problem that is considered is a supernova remnant. In the supernovaexplosion, a certain amount of energy is released all at once and this energy is assumedto remain constant, so no energy gets lost or added. Given this energy and the densityof the interstellar medium, the Sedov solution gives the radius of the supernova as afunction of the time.

It is assumed here that all the energy is available at once. From the moment itis released, it causes particles to move. This is not true all the time in a supernovaremnant. Usually, the evolution of supernova remnants is divided into four phases,the ejecta dominated phase, the Sedov-Taylor phase, the snowplough phase and themerging phase.The Sedov sulution is valid in the Sedov-Taylor phase [11]. There existother solutions which do not assume all energy is released at once, but the energy isreleased to the outer shell of the supernova remnants over time.

The situation is assumed to be spherical symmetric. This is a reasonable assumptiongiven the scale of this problem versus the distribution of matter in space. Anotherassuption is that the gas is perfect, having no viscosity and being non-heat conducting.

3.1 Intuıtive approach

An intuıtive, but not completely correct approach to the Sedov solution can be found ifone assumes that E = 1

2mv2.

Assume the density profile around the supernova explosion is ρ(r) = β · rα whereβ, α ∈ R are constants. In this case, the mass within a radius R equals

M(R) =

∫ R

0

β · rα · 4π · r2dr

=

[4πβ · (3 + α)

1

2 + α

]R0

=4πβ

2 + αR3+α

25

If we now use the simplified formula E = 12M(R)v2 with E = E0 which is constant,

and we know that v is the speed of the shock wave, so v = dRdt

, we get

M(R)(dR

dt)2 = 2E0

or, equivalently, √4πβ

2 + αR3+α

dR

dt=√

2E0

which can be rewritten as

R3+α2dR

dt=

√2E0

2 + α

4πβ

This equation can be solved since

R3+α2dR

dt=

d

dtR

5+α2 and

√2E0

2 + α

4πβ=

d

dt

√2E0

2 + α

4πβt.

That gives us that

R5+α2 =

√2E0

2 + α

4πβt+ C

where C is some constant. If one sets t such that R(0) = 0 this leads to

R(t) =

(2E0

2 + α

4πβ

) 15+α

t2

5+α .

When we take α = 0 and β = ρ0, we get the Sedov solution back. The reason thatthis derivation is simple but also incorrect, is because it assumes that all of the mass ismoving with the shock wave velocity. In fact, this is not true and the correct solution canbe found when instead of E = 1

2mv2 we use the correct formula E = 0.5

∫ρ(r)v(r)2dr.

Now, the v(r) needs to have properties that make the solution self-similar. The problemhere is that the constant that is found is not correct, but the proportionality is correct:

when ρ ∝ rα, then also RShock ∝ t2

5+α . In this case, it is possible to determine the density

profile, since now v ∝ 25+α

t2

5+α−1. Indeed, such velocities are sometimes measured, for

example in Jacco, 2008.To obtain a correct derivation, dimensional analysis is needed. In the next section,

the Buckingham Pi theorem that will be used is explained.

3.2 A mathematical treatment of the Buckingham

Pi theorem

In physics, dimensional analysis can be usefull to solve certain problems. A way to usedimensional analysis is the Buckingham Pi theorem. The dimensions mentioned in thistheorem are physical dimenions.

26

Theorem 8 (Buckingham Pi theorem). Let f : Rn → R be a function that satisfiesf(q1, q2, . . . , qn) = 0 where q1, . . . , qn are physical variables which can be expressed interms of k ≤ n independent physical units. Then the equation can is equivalent to anequation g(π1, π2, . . . , πn−k) = 0. The new variables πi are of the form qi11 · qi22 · · · · ·qinn and these are dimensionless. This also makes the equation g(π1, π2, . . . , πn−k) = 0dimensionless [2].

Before we prove this theorem, we will discuss in more detail what it means. Takefor example a function depending on distance x, speed v and time t. Distance has canbe expressed in meters, speed can be expressed in meters per second and time can beexpressed in seconds. Hence the physical variables here can be expressed in just twoindependent physical units, seconds and meters. Therefore the theorem tells us that ifwe have an equation f(x, v, t) = 0, it can be rewritten as a dimensionless equation in3− 2 = 1 variable.

To prove this, it is necessary to consider physical dimensions, but in a mathematicalway.

Proof. Let f(q1, q2, . . . , qn) = 0 be an equation depending on the n (physical) variablesq1, q2, . . . , qn and let those variables depend on the k physical units y1, y2, . . . , yk. Thisequation must, of course, have correct dimensions to begin with.

Notice that a physical unit that a variable can have is always of the form ya11 ·ya22 ·· · ··yakk

and will never be a sum of terms. Adding two quantities in different units is not allowedin physics, but multiplying is.

Now it is possible to construct a function that matches variables with their units. Thiswill be a (n×k)−matrix which is called the dimensional matrix D. Its (i, j)th entry willbe the power of the ith unit in the jth variable. This means that if qj = yj11 · y

j22 · · · · · y

jkk ,

then the (i, j)th entry will be ji.This leads to a function d : Rn → Rk v 7→ Dv.If a variable is dimensionless, it has to be of the form y01 ·y02 · · · · ·y0k, which means that

it has to be in the kernel of the dimensional matrix. Because dim(ker) + dim(Im) = n itfollows that dim(ker) = n− dim(Im) = n− k. This shows that there exist indeed n− kindependent dimensionless variables, which we call π1, π2, . . . , πn−k. These independentdimensionless variables will have the property that they can be written as πi = yi11 · yi22 ·· · · · yikk with the vector

y =

i1i2...ik

belonging to the kernel of D.

What the theorem now states is that we can create a function f such that the followingdiagram commutes:

27

Rn

Rn−k

RRk

g f

fd

Here, the function g : Rn → Rn−k sends q1, q2, . . . , qn to π1, π2, . . . , πn−k by multiplyingthem and taking powers. This is a well defined function, since we know that πi =qi11 · qi22 · · · · · qinn .

Now given a function f(q1, . . . , qn) = 0 and such a g, we know that this f indeedexists, since the image of g is in the kernel of d.

This prove is giving a way to construct the dimensionless variables and is thereforconstructive. Thus in the case of a function depending on distance, speed and time wecan apply this strategy. The dimensional matrix will look like this:

D =

(x v t

m 1 1 0s 0 −1 1

)We expect to find one dimensionless variable as stated above. In other words, we nowwant to find the kernel of the matrix D and see that it is one dimensional.

Calculating the kernel gives that

ker(D) = R

−111

So the dimensionless variable we get here is x−1vt. Note that here we made a choice

for one vector from the kernel, equivalently we could have chosen x−2v2t2 or xv−1t−1 or,in general, (x−1vt)c with c ∈ R.

3.3 The Buckingham Pi theorem applied to

supernova remnants

The Sedov solution is based on dimensional analysis and in particular the BuckinghamPi theorem.

In this problem, the relevant variables are the time t since the supernova explosion,the radius r of the shock wave, the density ρ0 of the interstellar medium and the energyE0 that was released during the explosion.

These are four variables. Time can be expressed in units of seconds, s. Radius canbe expressed in units of meters, m. Density can be expressed in units of kilograms per

28

meter cubed, so kg m−3 and energy can be expressed in units of joule, kg m s−2. Thosefour variables can thus be expressed in terms of three independent variables, meters,seconds and kilograms.

So with the Buckingham Pi theorem, we can find one variable in which we can re-express the relevant equations.

To find the one variable, first we consider the dimensional matrix:

D =

R t E0 ρ0

m 1 0 2 −3s 0 1 −2 0kg 0 0 1 1

with kernel

ker(D) = R

−52−11

So here we find the dimensionles variable to be R−5tE−10 ρ0, or equivalently

(ρ0E0t2

) 15R.

This last variable is the one that is usually used in literature, so it is also the one that

I will use. It will be called ξ =(

ρ0E0t2

) 15R.

3.4 The Sedov solution

As we saw in the previous section, there is a similarity variable ξ such that

ξ =

(ρ0E0t2

) 15

R.

We can express R in terms of ξ, t, ρ0 and E0 to obtain

R = ξ

(E0t

2

ρ0

) 15

.

Now let RShock be the radius of the shock wave. Then there is a certain α such that

RShock =

(αE0t

2

ρ0

) 15

which shows that the radius of the shock wave of the supernova indeed only dependson time. The energy and the (constant) density of the interstellar medium are onlyparameters. To find the velocity with which the shock wave is moving, we take thederivative with respect to t and obtain

vShock =2

5

(αE0

ρ0

) 15

t−35

29

which now, like the radius, only depends on time.It is also possible to determine the value of α. In order to do so, we will have to

consider the Euler equations. The Euler equations in spherical symmetric form are thefollowing [9]:

∂ρ

∂t+

1

r2∂

∂r(r2ρv) = 0 (3.1)

∂v

∂t+ v

∂v

∂r+

1

ρ

∂p

∂r= 0 (3.2)

∂

∂t(pρ−γ) + v

∂

∂r(pρ−γ) = 0. (3.3)

Here, ρ = ρ(r, t) is the density within the shockwave as a function of time and distanceto the center of the shock wave. Also, v = v(r, t) is the speed withing the radiusof the shockwave and p = p(r, t) the pressure. The adiabatic index is denoted byγ. Equation 3.1 is the equation of mass conservation, equation 3.2 is the equation ofconservation of momentum and equation 3.3 is the equation of energy conservation.Since this problem now only depends on time and one geometrical coordinate r, theproblem is one-dimensional.

In the ideal case, these equations can be solved and give ρ(r, t), p(r, t) and v(r, t).We now want to rewrite the equations 3.1, 3.2 and 3.3 in terms of ξ. Therefore

calculate∂ξ

∂r=

∂

∂rr

(ρ0E0t2

) 15

=

(ρ0E0t2

) 15

=ξ

r(3.4)

and also∂ξ

∂t=

∂

∂tr

(ρ0E0t2

) 15

= −2

5r

(ρ0E0

) 15

t−75 = −2

5

ξ

t(3.5)

What we also need is to express ρ(r, t), p(r, t) and v(r, t) in terms of a dimensionlessfunction of ξ. This leads to the following definitions:

ρ(r, t) = Aρ0G(ξ) (3.6)

v(r, t) = Br

tV (ξ) (3.7)

p(r, t) = CP (ξ) (3.8)

Here, A,B and C are constants which will satisfy conditions on the boundary of thesupernova remnant. If we choose A = γ+1

γ−1 , B = 45(γ+1)

and C = 825(γ+1)2

, the boundary

conditions become [1].

G(1) = 1 V (1) = 1 P (1) = 1

We can use these in order to rewrite equation 3.1, 3.2 and 3.3 in terms of ξ.Using the chain rule and equations 3.4 and 3.5, we can derive the following expressions

for the derivatives of ρ, v and p with respect to r and t.

30

∂ρ

∂t=∂ξ

∂t

∂ρ

∂ξ= −2

5

ξ

t

∂ρ

∂ξ∂ρ

∂r=∂ξ

∂r

∂ρ

∂ξ=ξ

r

∂ρ

∂ξ∂v

∂t=∂ξ

∂t

∂v

∂ξ= −2

5

ξ

t

∂v

∂ξ∂v

∂r=∂ξ

∂r

∂v

∂ξ=ξ

r

∂v

∂ξ∂p

∂t=∂ξ

∂t

∂p

∂ξ= −2

5

ξ

t

∂p

∂ξ∂p

∂r=∂ξ

∂r

∂p

∂ξ=ξ

r

∂p

∂ξ

Now first rewrite equation 3.1, 3.2 and 3.3 toFor the equation of mass conservation, we find

∂ρ

∂t+ v

∂ρ

∂r+ ρ

∂v

∂r+

2ρv

r= 0,

for the equation of momentum conservation we find

∂v

∂t+ v

∂v

∂r+

1

ρ

∂ρ

∂r= 0

and finally for the equation of energy conservation we get

−γpρ−γ−1∂ρ∂t

+ ρ−γ∂p

∂t− γvpρ−γ−1∂ρ

∂r+ vρ−γ

∂p

∂t= 0.

We will fill in the equations 3.6, 3.7 and 3.8 where we use the derivatives to obtainFirst equation:

−ξ ∂G(ξ)

∂ξ+

2

γ + 1

(2G(ξ)V (ξ) + ξ

∂G(ξ)V (ξ)

∂ξ

)= 0.

Seccond equation: (2

γ + 1− 1

)∂V (ξ)

∂ξ+

1

G(x)

γ − 1

γ + 1

∂P (ξ)

∂ξ= 0.

And the third equation:

G(ξ)−γ∂P (ξ)

∂ξ− γP (ξ)G(ξ)−(γ+1)∂G(ξ)

∂ξ= 0.

So indeed the equations can be re-written in terms of only ξ and all the dependenceon r, t, ρ and E is gone. For any given value of the adiabatic index γ, this leaves us withthree ordinary differential equations to be solved for three functions of ξ.

31

These expressions will be solved in Mathematica nummerically. The fact that wedon’t find an analytical solution does not mean that having used dimensional analysisis useless, since the form our problem now has makes it much easier to solve. This givesus a profile of the density, pressure and velocity inside the supernova remnant.

This gives the following plot for γ = 53:

Figure 3.1: Plot of the solution of the Euler equations in spherical symmetry rewrittenin terms of one dimensionless variable using γ = 5

3.

and after rescaling and selecting the positive part, we get

Figure 3.2: Normalized plot of the relevant part of the solution of the Euler equationsin spherical symmetry rewritten in terms of one dimensionless variable usingγ = 5

3.

32

Taking γ = 1.3, rescaling and taking the relevant part gives the following plot:

Figure 3.3: Normalized plot of the relevant part of the solution of the Euler equationsin spherical symmetry rewritten in terms of one dimensionless variable usingγ = 1.1.

3.5 Applying the Sedov solution

The Sedov solution can be used to determine several variables. If it is known how muchenergy is released in a supernova explosion, how old and how large a supernova remnantis, the density of the interstellar medium can be calculated. In a similar way, if thedensity is known but the energy is not, the energy can be calculated from the Sedovsolution. When all parameters are known, it is possible to determine wether the densityof the interstellar medium is indeed homogeneous by checking if the Sedov solutionworks.

For example, [6] uses the Sedov model to calculate properties of several supernovaremnants.

Typical values for the energy of a supernova explosion are about 1051 erg or 1044 J.For the interstellar medium, a value of about 10−24g cm−3 is reasonable.

If we assume that the Kepler supernova remnant has this typical energy of 1051 ergs,the interstellar medium in the region of the supernova remnant has indeed a density of10−24g cm−3 and that it is about 420 years old, we find that its radius right now wouldbe approximately

RShock =

(αE0t

2

ρ0

) 15

= 3.7pc.

This estimate has a correct order of magnitude, compared to [12].

33

4 Discussion

In this thesis, self-similarity is discussed in both in a physical as in a mathematicalway. They turned out to be not the same. Physical self-similarity is continuous, as theproblem scales with the time and what you see does not change its shape over time. Themathematical self-similarity now seems to be a discrete version. If you picture a set andzoom in on a right spot, you see the same figure as the big picture at discrete degreesof zooming in.

4.1 Mathematics

In the first part of the thesis, the focus was mathematics. Hausdorff dimension wasdefined and a theorem to easily calculate it for sets that are self-similar. If the ratios ofthe similarities are ci where the i labels the similarities, then the Hausdorff dimensionof the self-similar set is the value of s for which

n∑i=1

csi .

This theorem made calculating the Hausdorff dimension of the Cantor set an easy task.Also we looked into complex dynamics and some familiar results about Hausdorff

dimension of Julia sets and the Mandelbrot set were given.Some questions in this area are still open tough. It is still unknown what the 2-

dimensional Hausdorff measure of the Mandelbrot set is, since there is no universal wayof calculating the Hausdorff dimension of any set.

4.2 Physics

In the second part of the thesis, the focus was astrophysics. Here we saw a way touse dimensional analysis, namely the Buckingham Pi theorem. Using this theorem, wederived the Sedov similarity solution for the shock wave of a supernova remnant. Thiswell-known result was also derived in a intuıtive way, which gave the right proportion-alities but not the correct constants.

This method is very useful, and the Sedov solution is applied in several ways andpapers.

But in the astrophysics part too, there are still questions to be answered. The mostimportant improvement that could be made right now is to generalize this solutionwhere a homogeneous density of the interstellar medium around the supernova remnant

34

is assumed, to a situation where the density of the interstellar medium around thesupernova remnant can be arbitrary. If it would be possible to find a way to do this forpower laws, other density profiles could be written as their Taylor series. This could beuseful if the supernova remnant is situated in a cavity or nearby molecular clouds. Thismight also work the other way, so the density profile of the interstellar medium arounda supernova remnant could be determined using properties of the supernova remnant.The method discussed in this thesis is not suited for this purpose.

35

5 Populaire samenvatting

In dit project heb ik gekeken naar self-similarity, wat ongeveer vertaald kan worden alszelfgelijkendheid. Dit onderwerp is zowel binnen wiskunde als binnen natuurkunde heelerg interessant. In beide gebieden betekent het wel iets verschillends.

5.1 Natuurkunde

De context waarin ik zelfgelijkendheid bekijk, is de schokgolf van een supernovarest. Eensupernova is een zeer oude ster die aan het einde van zijn leven ontploft. Daarbij komtveel energie in een keer vrij en deze energie zorgt dat er veel matterie naar buiten wordtgeslingerd. Waar astronomen geınteresseerd in kunnen zijn, is wat de straal is van deschokgolf die dit veroorzaakt. Het blijkt dat de straal van de schokgolf en het profielvan de supernova alleen maar afhangen van de tijd. Als je dus een foto zou maken en ergeen schaal bij zou vermelden, is het niet te zeggen op welk tijdstip de foto is gemaakt.

De oplossing van het supernovaprobleem kwam van L. Sedov, een natuurkundige uitde Sovjet-Unie. Het probleem dat hij oploste, was dat van de schokgolf na de ontploffingvan een atoombom. Dit bleek vergelijkbaar te zijn met het plotselinge vrijkomen van zeerveel energie in een supernova. De methode die hij gebruikte was dimensieanalyse, maarop een andere manier dan je misschien gewend bent. Hij keek welke variabelen relevantwaren, en van welke eenheden deze afhingen. De relevante variabelen zijn E0, ρ, r en twaar E0 de energie is die vrijkomt, ρ de dichtheid in de supernova, r de afstand vanafhet middelpunt van de explosie en t de tijd. De eenheden zijn voor energie kg m2 s−2,voor dichtheid kg m−3, voor afstand m en voor tijd s. Dit zijn vier eenheden, maareigenlijk zijn ze slechts afhankelijk van drie onafhankelijke eenheden, namelijk kg,m ens. De zogenaamde Buckingham Pi stelling vertelt dan dat alle vergelijkingen die vandeze variabelen afhangen kunnen worden omgeschreven naar vergelijkingen in 4− 3 = 1varriabele. Die ene variabele is dan zelfs dimensieloos. In dit geval is de dimensielozevariabele, die even ξ genoemd wordt,

ξ = r

(ρ

E0t2

) 15

.

Voor r = rshock is er een specifieke constante waarde van ξ, die we even ξ0 zullen noemen.Dit zorgt ervoor dat we de Sedov similarity solution kunnen opschrijven,

rShock = ξ0

(E0t

2

ρ

) 15

.

36

5.2 Wiskunde

In de natuurkunde betekent zelfgelijkendheid dat je op elk moment naar de steeds groterwordende supernova kan kijken en dat je op elk moment hetzelfde plaatje ziet. In dewiskunde zit het net wat anders. Bekijk bijvoorbeeld eens de Sierpinskidriehoek in devolgende afbeelding:

Deze driehoek is gemaakt door ten eerste een gevulde gelijkzijdige driehoek te tekenen.Vervolgens haal je hier de middelste driehoek uit. In alle volgende stappen haal je demiddelste driehoek uit de driehoeken die nog over zijn. Na dit oneindig lang gedaan tehebben, heb je een sierpinskydriehoek verkregen.

Stel dat je nu inzoomt op de driehoek linksonder. Dan zie je na een tijdje precieshetzelfde plaatje als je aan het begin had. Als je een goede plek kiest, kun je veel langerblijven inzoomen. Ook al zie je dan niet constant hetzelfde plaatje, op bepaalde tijdenzie je wel weer hetzelfde plaatje.

Dat komt omdat de verzameling gemaakt is uit kopieen van zichzelf die iets kleinerzijn. Je kan bijvoorbeeld de driehoek half zo groot maken, drie keer kopieren, een beetjenaar rechts, links en boven schuiven en je hebt je orginele driehoek weer terug. Of jekan de driehoek vier keer zo klein maken, negen keer kopieren en als je ze op de goedeplekken legt, heb je weer je orginele driehoek terug.

Een vraag die wiskundigen zichzelf dan stellen, is welke dimensie zo’n soort figuurlogischerwijs moet hebben. Dimensies zijn hier dus wiskundig bedoeld: Zoals een kubusdimensie drie heeft, een vierkant dimensie twee, een lijn dimensie een en een punt dimen-sie nul. Dat de dimensie van de Sierpinskydriehoek groter is dan nul, is wel duidelijk.Maar moet de dimensie een zijn, omdat wat er uiteindelijk overblijft alleen nog maar uitlijnen bestaat, of twee, omdat er wel oneindig veel lijn is?

Hiervoor is de Hausdorff dimensie bedacht. In het geval van zelfgelijkende verzamelin-gen blijkt dat deze dimensie precies gelijk is aan de waarde van s waarvoor geldt dat∑

i csi = 1 waar de ci’s gelijk zijn aan de mate van verkleining van het origineel. Voor

de sierpinskydriehoek geldt dus

3 ∗ (1

2)s = 1,

oftewel s = log(3)log(2)

.

37

Bibliography

[1] D.L. Book, THe Sedov self-similar point blast solutions on nonuniform media, 1994.

[2] E. Buckingham, On physical similar systems; Illustrations of the use of dimensionalequations, aps journals 4, 1914.

[3] X. Buff and A. Cheritat, Quadratic Julia sets with positive area, 2008.

[4] R. A. Chevalier, Self-similar solutions for the interaction of stellar ejecta with anexternal medium, The Astrophysical Journey 258, 790-797, 1982

[5] K.J. Falconer, Fractal Geometry, Mathematical Foundations and Applications, 1990.

[6] J. P. Hughes, I. Hayashi and K. Koyama, ASCS X-ray spectroscopy of large mag-ellanic cloud supernova remnants and the metal abundances of the large magellaniccloud, The Astrophysical Journey 505, 732-748, 1988.

[7] P. Mattila, Geometry of Sets and Measures in Euclidean Spaces, 1995.

[8] J.W. Milnor, Dynamics in One Complex Variable, 1990.

[9] L. I. Sedov, Similarity and Dimensional Methods in Mechanics, translation from the4th Russian Edition, 1959.

[10] M. Shishikura, The Hausdorff dimension of the boundary of the Mandelbrot set andJulia sets, 1991.

[11] E. van der Swaluw, Supernova Remnants, Pulsar Wind Nebulae and Their Interac-tion, 2001, PhD thesis. University of Utrecht.

[12] J. Vink, The kinematics of Kepler’s supernova remnant as revealed by Chandra,The Astrophysical Journal 689, 231-241, 2008.

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