A Survey of Dynamical Billiards

Markus Himmelstrand, mhim@kth.seVictor Wilén, vwilen@kth.se

Department of Mathematics, AnalysisRoyal Institute of Technology (KTH)

Supervisor: Maria Saprykina

Contents

1 The dynamical billiard 11.1 De�nition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 The billiard map 22.1 De�nition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.2 The generating function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

3 Properties of the billiard map 33.1 Periodic orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

4 Billiard in a circular domain 74.1 Properties of billiard in a circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74.2 Billiard in an isosceles triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94.3 Application of the uniform distribution property of the circle billiard . . . . . . . 11

4.3.1 Benfords law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

5 Existence of Birkho� periodic orbits 115.1 Birkho� periodic orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115.2 Proof of Birkho�'s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

6 Caustics 176.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176.2 Homeomorphism of the circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206.3 Poncelet Porism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226.4 String construction of convex caustics . . . . . . . . . . . . . . . . . . . . . . . . 236.5 Proof of the string construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236.6 Non existence of Caustics and global implications . . . . . . . . . . . . . . . . . . 276.7 Nonexistence of caustics at locally �at boundaries . . . . . . . . . . . . . . . . . . 29

6.7.1 Examples and test for tables without caustics . . . . . . . . . . . . . . . . 30

7 Outer billiards 317.1 De�nition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317.2 Properties of the outer billiard map . . . . . . . . . . . . . . . . . . . . . . . . . . 327.3 Bounded orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347.4 Outer billiards and planetary motion . . . . . . . . . . . . . . . . . . . . . . . . . 35

A Calculations 36A.1 Notes for constructing level curves . . . . . . . . . . . . . . . . . . . . . . . . . . 36A.2 Calculation of the Curvature of an Implicit planar curve . . . . . . . . . . . . . . 37

B Basic notions in analysis 37B.1 Metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38B.2 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39B.3 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

1

Abstract

In this report we will present the basic concepts and results of the theory of dynamical billiardswhich idealizes the concept of a volumeless ball re�ecting against the inside of a billiard tablewithout friction.This motion will continue inde�nitely and it is of interest to study its behaviour.

We will show that the study of a billiard system can be reduced to the study of an associatedmap called the billiard map de�ned on a cylindrical phase space. Using this formalism thespeci�c systems where the billiard table is given by a circle, right iscoceles triangle and ellipsewill be studied in some detail along with the existence of peridic points through Birkho�'sfamous theorem and some more novel results such as an instance of Benford's law regardingthe distribution of �rst digits in real-life data. We will also de�ne the concept of a causticand investigate their existence and non-existence which will lead us to the concept of circlehomeomorphisms and will also provide the opportunity to illustrate the systems with somesimulations and yield some more informal and practical insight into the behaviour of thesesystems.

1 The dynamical billiard

We will here introduce the general concept of a dynamical billiard �ow in a bounded region of theplane. It will be seen how the billiard �ow can be described both as a trajectory in continuoustime as well as by considering the points of re�ection at the boundary. The latter option willgive a discrete representation of the �ow that will be a precursor to the billiard map introducedin section 2.

1.1 De�nition

Begin by considering a bounded region Ω in the plane. Denoting its boundary by ∂Ω it will beassumed that there exists a continuously di�erentiable curve

γ : [0, 1] −→ R2

such that γ([0, 1]) = ∂Ω. We also assume that this curve is closed (i.e. that γ(0) = γ(1)) andthat it does not intersect itself anywhere on [0, 1). Introducing coordinate functions x and y byγ(t) = (x(t), y(t)) for t ∈ [0, 1] the arc length parameter s can be de�ned as

s(t) :=

∫ t0

|γ′(t′)| dt′ =∫ t

0

√x′(t′)2 + y′(t′)2 dt′.

De�ning the total arc length as L := s(1) the boundary ∂Ω can now be parametrized with theinterval [0, L] instead of [0, 1]. To avoid introducing confusing notation the boundary curve willstill be denoted by γ. We can now prove the following useful

Lemma 1.1. Denoting by dγds the derivative of γ with respect to arc length it holds that |dγds | = 1

(noting that dγds is tangent to the curve this means thatdγds is a unit tangent to γ).

Proof. ∣∣∣∣dγds∣∣∣∣ =√(

dx

ds

)2+

(dy

ds

)2= {chain rule} =

√(dx

dt

dt

ds

)2+

(dy

dt

dt

ds

)2=

=

√(dx

dt

)2+

(dy

dt

)2dt

ds=

√(dx

dt

)2+

(dy

dt

)21

ds/dt=|γ′(t)||γ′(t)|

= 1

We are now ready to de�ne the concept of a dynamical billiard. Consider a continuous curver(t), t ∈ [0,∞), in Ω with the following properties

1. r(0) ∈ ∂Ω

2. For every t > 0 such that r(t) 6∈ ∂Ω, r is twice di�erentiable at t. More speci�cally, forany such t we assume that r′′(t) = 0 and |r′(t)| = 1. Note that this implies the existenceof a unit vector u such that r′(t′) = u for all t′ in some neighborhood of t. Geometricallythis means that r consists of linear segments where each segment has its endpoints on γ.

3. If t > 0 and r(t) ∈ ∂Ω then there is an � > 0 such that for any t′ ∈ (t − �, t + �) wehave r(t′) 6∈ ∂Ω. Denote, in accordance with the previous point, the unit tangents of r in(t − �, t) and (t, t + �) by u and v respectively and the unit tangent to γ at r(t) by w.Then u will make the same angle with w as w makes with v. This is referred to as thelaw of re�ection.

1

Any r ful�lling the requirements above will be referred to as a billiard trajectory and the set ofall such trajectories constitutes the dynamical billiard system of the region Ω.

The above de�nition of billiards gives a continuous description but it will be more convenientto work with a discrete model. Therefore, let r be a billiard trajectory and de�ne s0 ∈ [0, L]such that r(0) = γ(s0). By the de�nition of r it will follow a linear path until it intersects γ ata new point, say γ(s1) and then continue along a new linear segment. Continuing in this fashionwe get a sequence of points {si}∞i=0 that completely determines the billiard trajectory (the linearsegments of r can be deduced from the law of re�ection). Hence it is shown that the billiardtrajectory is completely determined by its points of re�ection at the boundary.

2 The billiard map

The purpose of this section is to introduce the billiard map as a means of describing the billiardsystem introduced in section 1. We will also prove some important properties about this map,especially for when the billiard is convex. For this purpose the concept of a generating functionfor the billiard system will be introduced. The relationship between the generating function andperiodical orbits will also be investigated.

2.1 De�nition

Assume as in section 1 that Ω is a bounded region and let γ be its continuously di�erentiableboundary curve parametrized by its arc length. We will de�ne a map T on the set M :=[0, L]× [0, π] (L denotes the total arc length of γ as before) by taking (s, θ) ∈M . Then let γ(s′)be the point of intersection with a straight line segment from the point γ(s) making the angle θwith γ at this point (if more than one such intersection exists we just take the one closest to butnot equal to γ(s)). Denote by θ′ the angle that this line segment makes with γ at γ(s′). Thenwe can de�ne T (s, θ) := (s′, θ′). The map T is called the billiard map. Note that the billiardtrajectory starting at γ(s) with initial angle θ can be obtained from the sequence {T k(s, θ)}∞k=0.By what was proved in section 1 any billiard trajectory can be obtained in this manner.

It will be convenient in many situations to visualize the behaviour of the billiard in the socalled phase space. The phase space consists of all possible states of the billiard system, i.e. ofevery point (s, θ) ∈ [0, 1] × [0, π] which represents a point on the boundary and an angle. Atrajectory of a billiard can then be thought of as a point set in the phase space which is invariantunder the billiard map T . Note also that if we consider the interval [0, 1] as the set S1 = R/Z,i.e. the reals modulo the integers, then the phase space can be viewed as a cylinder.

2.2 The generating function

Before we can start proving any properties of the billiard map introduced in section 2.1 we must�rst introduce the important concept of a generating function. This is contained in the following

De�nition 2.1. Let d denote the function d(x,y) :=√

(x1 − y1)2 + (x2 − y2)2 for x,y ∈ R2.Then the generating function H for γ is de�ned for s, s′ ∈ [0, L] as H(s, s′) := −d(γ(s),γ(s′)).

From now on it will be assumed that the region Ω bounded by γ is convex. We can thenprove the following

Theorem 2.1. Let s, s′ ∈ [0, L] and denote by u the unit vector from γ(s) to γ(s′). If θ is theangle that u makes with γ at s and θ′ is the corresponding angle at s′ (see �gure 1) then

2

θ

θ′

s

s′

t(s′)

t(s)

γ

Figure 1: Geometrical situation in theorem 2.1

∂H

∂s= cos θ,

∂H

∂s′= − cos θ′.

Proof.

∂H

∂s=− 1

2√

(x(s)− x(s′))2 + (y(s)− y(s′))2(2(x(s)− x(s′))dx

ds+

+ 2(y(s)− y(s′))dyds

) = − 1|γ(s)− γ(s′)|

(γ(s)− γ(s′)) · dγds

=

=

∣∣∣∣dγds∣∣∣∣ cos θ = cos θ

This proves the �rst half of the theorem. The second half is proven by performing a similarcalculation by di�erentiating with respect to s′ instead.

3 Properties of the billiard map

We start by introducing the change of variables r := − cos θ and from here on consider thebilliard map T as a function on M := [0, L]× [−1, 1] instead (the notation will be unchanged toavoid confusion). The conclusion of theorem 2.1 for convex billiards can then be restated as

∂H

∂s= −r,

∂H

∂s′= r′.

The following de�nition will be crucial to the next theorem.

3

De�nition 3.1. Let f : M → M be a homeomorphism such that f(s, r) = (S(s, r), R(s, r)).Then f is called a twist-map if S(s, r) is strictly increasing in r for any �xed s. If f is adi�eomorphism this requirement can be stated as

∂S

∂r> 0.

This will be referred to as the twist property.

We can now prove the following

Theorem 3.1. The billiard map T is area-preserving and has the twist-property for convexbilliards.

Proof. Introduce the coordinate functions S and R for (s, r) ∈M by T (s, r) := (S(s, r), R(s, r)).To prove that T is area-preserving de�ne the function H̃(s, r) := H(s, S(s, r)). Di�erentiating

H̃ with respect to s then gives

∂H̃

∂s= {chain rule} = ∂H

∂s+∂H

∂s′∂S

∂s= −r +R∂S

∂s.

Di�erentiating with respect to r gives

H̃

∂r=∂H

∂s′∂S

∂r= R

∂S

∂r.

By now calculating ∂2H̃

∂s∂r from the two expressions above we get

− 1 + ∂2H

∂r∂s′∂S

∂s+∂H

∂s′∂2S

∂r∂s=∂R

∂s

∂S

∂r+R

∂2S

∂s∂r⇒

− 1 + ∂R∂r

∂S

∂s+R

∂2S

∂r∂s=∂R

∂s

∂S

∂r+R

∂2S

∂s∂r⇒

|JT | = ∂S∂s

∂R

∂r− ∂S∂r

∂R

∂s= 1.

This proves that T is area-preserving. To prove that T has the twist property we must provethat ∂S∂r > 0. According to the chain rule

∂S

∂r=dθ

dr

∂S

∂θ=

1

sin θ

∂S

∂θ.

Also, by the de�nition of the partial derivative, we have

∂S

∂θ:= lim

∆θ→0

S(s, θ + ∆θ)− S(s, θ)∆θ

= lim∆θ→0

∆s′

∆θ.

Now consider the situation in �gure 2. Applying the sine theorem to the triangle made bythe three marked corners gives

|γ(s′)− γ(s)|sinα

=|γ(s′ + ∆s′)− γ(s′)|

sin ∆θ.

Observe that by applying Taylor's theorem we can write

sin ∆θ = ∆θ +O(∆θ3)

4

s θ

∆θθ′

s′

s′ + ∆s′

α

Figure 2: Geometric proof of the twist property for convex billiards.

and that

|γ(s′ + ∆s′)− γ(s′)| = |∆s′γ′(s′) +O(∆s′2)| = ∆s′ +O(∆s′2).

This �nally gives

|γ(s′)− γ(s)|sinα

=∆s′ +O(∆s′2)∆θ +O(∆θ3)

→ ∂S∂r

as ∆θ → 0.

This proves that ∂S∂r =|γ(s′)−γ(s)|

sinα > 0 and �nishes the proof.

It is also worth noting that the billiard map is a homeomorphism on M (i.e. continuouswith a continuous inverse). That the billiard map is onto M is easily veri�ed by taking a point(s, θ) ∈M and letting (s′, θ′) = T (s, π−θ). By construction of T it follows that T (s′, θ′) = (s, θ)(see left part of �gure 3). That T is one-to-one follows by assuming that T (s, θ) = T (s′, θ′) andobserving that ∆θ := |Θ(s′, θ′) − Θ(s, θ)| is nonzero if s 6= s′ (see the geometric constructionin the right part of �gure 3). Hence s = s′ and by the twist property of the billiard map itnow follows that θ = θ′, proving that T is one-to-one. Since the billiard map is continuous andthe inverse of any continuous surjective function de�ned on a compact set is also continuous itfollows that the billiard map is a homeomorphism on M .

θ

s

θ′ s′

π − θs′′

θ′ s′θ

s

∆θ

Figure 3: Proof that the billiard map is one-to-one and onto.

3.1 Periodic orbits

We will now start to look at another important application of the generating function: periodicorbits for convex billiards. If T is the billiard map, then a point (s, θ) is periodic with period

5

N if TN (s, θ) = (s, θ) and for any 0 < n < N it holds that Tn(s, θ) 6= (s, θ). It will be shownthat for convex billiards the periodic orbits can be described as extreme points of the generatingfunction. But �rst we prove the following

Lemma 3.1. Let s0, s1, s2 ∈ [0, L] be distinct points for a convex billiard. De�ne for s ∈ [0, L]

H̃(s1) := H(s0, s1) +H(s1, s2)

where H denotes the generating function as usual. Then the points s0, s1 and s2 are consecutivepoints of the same billiard trajectory if and only if s1 is an extreme point to H̃.

α0

αα′

α2s2

s1

s0

Figure 4: Three consecutive points on the boundary of the billiard region.

Proof. The situation is illustrated in �gure 4. Di�erentiating H̃ with respect to s and usingtheorem 2.1 gives

dH̃

ds1=∂H

∂s′(s0, s1) +

∂H

∂s(s1, s2) = − cosα+ cosα′.

This together with the law of re�ection proves the lemma.

Now consider a sequence of k distinct points {si}k−1i=0 . De�ne the generating function H̃ onthis sequence as

H̃(s0, . . . , sk−1) :=

k−2∑i=0

H(si, si+1) +H(sk−1, s0).

Then we can prove the following

Theorem 3.2. The sequence {si}k−1i=0 is a k-periodic orbit in a convex billiard if and only if itis an extreme point for H̃.

Proof. As in the proof of lemma 3.1 di�erentiate H̃ with respect to each of the varaibles si. Thisgives for i = 1

∂H̃

∂s1=∂H

∂s′(s0, s1) +

∂H

∂s(s1, s2)

and similarly for any other i. These k equations are all zero if and only if the sequence is anextreme point for H̃ and by lemma 3.1 this happens if and only if all the points lie on the samebilliard trajectory hence proving the theorem.

6

4 Billiard in a circular domain

4.1 Properties of billiard in a circle

Figure 5

One of the simplest examples of a billiard sys-tem is the one that exhibits the greatest degreeof symmetry, that is when the domain of thesystem is a circle Cr = {(x, y) ∈ R2|x2 + y2 =r2}. In this very special case many of the gen-eral properties of convex billiards such as ex-istence of periodic orbits and caustics whichwe will discuss later in the article can be d-educed with relative ease on account of thecorresponding billiard map being quite easyto express and analyse explicitly which oftenis not the case for more exotic tables. The rea-son we will begin with this system is threefold;it serves to illustrate the sort of questions onemay ask about a billiard system, certain orbits in general billiards are in fact isomorphic to thaton a circle, and it also has direct application outside the speci�c subject of billiards.

The �rst step to analysing this system is appropriately to reduce the dimension of the corre-sponding phase space that a given orbit occupies which can be done by observing from Figure 5or explicitly from the billiard map T : (s, θ) 7→ (s+ 2Rθ, θ) for a circle of radius R that the angleof re�ection at the boundary, our coordinate θ, remains constant under the billiard �ow. Thisimplies that for any orbit containing a point (s, θ) the remaining orbit will be constrained to thesub circle S1 × {θ} on the cylindrical phase space, and as such the billiard map T on C can bereplaced with a new map Rα on a circle S

1 which moves points a constant distance α = 2Rθalong the boundary at every iteration. We will without loss of generality consider the case wherethe circle is of circumference 1, R = 1/2π. Or to summarize the dynamics can be studied on thebasis of the function:

Rα : S1 → S1, Rα(x) = x+ α (mod 1).

If α is rational such that p/q is its reduced fraction it is immediately realized that the corre-sponding orbit will be periodic with period q and be carried p revolutions around the circlebefore returning to the starting point. There are thus in�nitely many periodic orbits of this kindbut what will occupy our primary attention is not this case but rather what happens when therotation number is irrational (i.e the most 'realistic' case), seeing as the orbit that such a mapgenerates cannot be periodic and hence is more interesting. The three properties of such orbitsthat will be derived and discussed below are that they will be dense, uniformly distributed,and in fact ergodic.

Theorem 4.1. If the rotation number α /∈ Q then every positive semiorbit is dense.

Remark: Geometrically this simply states that if the billiard ball never returns to the samepoint it will pass arbitrarily close to every other point on the boundary.

Proof. Fix ε > 0, and partition S1 into k = b1/εc + 1 intervals {∆i} of equal length 1/k < εnumbered in sequential order where interval ∆1 has the starting point x and at the border. Nowconsidering the partial orbit of the �rst k + 1 points {Riα(x)}ki=0 at least two points Riα(x) andRi+nα must (by the pigeon hole principle) lie in the same interval and therefore be at a distance

7

less than ε from each other, d(Ri(x), Ri+n(x)) < ε. Due to symmetry Rnα will therefore move anypoint a distance less than ε and a �nite number of iterates starting at any point will thereforedynamically partition the circle into intervals of size less than ε which proves denseness. Thesymmetry argument can also be made explicit by considering

d(Rnα(y), y) = d(RnαRy−x(x), Ry−x(x))

= d(Ry−xRnα(x), Ry−x(x))

= d(Rnα(x), x)

using the fact that rotations preserve distances and commute.

Though this tells us a great deal about the dynamics it does not exclude that the orbit mightaccumulate at certain points or subintervals along the circle, and before anything else we need tomake the concept of frequency more precise, even if it basicly can be understood as the fractionof points that visit a given region. We will choose the presentation of Hasselblatt and Katok [1,p.100�103]

De�nition 4.1. For any arc ∆ ⊆ S1 the relative frequency of visits (to this region) is

f(∆) = limn→∞

F∆(x, n)

n

whereF∆(x, n) = |{k ∈ Z | 0 ≤ k ≤ n, Rkα(x) ∈ ∆}|.

Remark: Note that in the case of rational α and a period of q the relative frequency f(∆)is just the fraction of the �rst q points that lie in ∆. What it now means that the orbit of thismap is uniform is encapsulated by the following theorem

Theorem 4.2. For any point x ∈ S1 with an orbit generated by an irrational rotation α we havethat for any arc ∆ ⊂ S1 that the frequency of visits to that region is equal to the arc length `(∆)

f(∆) = `(∆).

Proof. We begin by establishing that there is an upper limit to the relative frequency of visitsto a given arc.

If we consider the semiorbit {Riα(x)}ni=0 and the set of points A that lie in an interval ∆then by denseness of the orbit it is always possible to �nd an N such that Rnα(A) lies completelywithin another arc ∆′ so long as `(∆) ≤ `(∆′). That is F∆′(x, n+N) ≥ F∆(x, n) which meansthat if F (∆) := lim supn F∆(x, n)/n then F (∆) ≤ F (∆′) whenever `(∆) ≤ `(∆′).

Assume that a �xed arc ∆ has length `(∆) = 1/k for some integer k and divide S1 into k− 1disjoint arcs {∆i} of equal length 1/(k− 1). Then by the above argument F (∆) ≤ F (∆i) for allthe arcs and

(k − 1)F (∆) ≤k−1∑i=1

F (∆i) = 1

where the �nal equality follows from the fact that⋃

∆i = S1 and so we see that F (∆) ≤ 1/(k−1).

Now take any arc ∆ and cover it by arcs of length 1/k, in total n = b`(∆)kc+ 1 from whichwe have n/k − `(∆) ≤ 1/k. Now

F (∆) ≤n∑i=1

F (∆i) ≤n

k − 1≤ `(∆) k

k − 1+

1

k − 1.

8

So letting k → ∞ yields F (∆) ≤ `(∆). Now it remains only to establish F (∆) ≥ `(∆) whichactually follows from the above argument since from F∆(x, n) = n − F (∆c) we see F (∆) =1− F (∆c) so F (∆) = 1− F (∆c) ≥ `(∆) which completes the proof.

This we refer to as the uniformity property as the orbit is uniformly distributed along theboundary. Though interesting in its own right this can also straight forwardly be extended toshow that the system is ergodic, that is the mean of any Riemann integrable function g : S1 → S1taken over any non periodic orbit is the same as its average over all of S1 whenever the rotationnumber is irrational.

limn→∞

1

n

n∑i=0

g(Riα(x)) =

∫S1g(x) dx (α 6∈ Q,∀x ∈ S1)

which follows from the fact that every such function can be approximated arbitrarily well bystep functions and Theorem 4.2 can be cast in step function form

limn→∞

1

n

n∑i=1

χ∆(Riα(x)) =

∫S1χ∆(x) dx (α 6∈ Q,∀x ∈ S1)

where χ∆ is the characteristic function that is 1 on the set ∆ and 0 elsewhere, whereupon thedetails of proof are relatively straightforward [1, p.103,104].

Remark An alternate approach to this result which is less elementary but which can be usedto extend this equidistribution theorem into higher dimensional circles (tori) Tn = S1×· · ·×S1isthe Kroenecker-Weyl method [1, p.103,104] but it is outside the scope of this text.

4.2 Billiard in an isosceles triangle

Figure 6: Folding of triangle

In this article we review primarily results concerning con-vex billiards with smooth boundaries as re�ections becomesunde�ned at points where the tangent is discontinuous andthough this issue can be corrected by de�ning that the par-ticle terminate or re�ect according to some auxiliary law thebilliard map would not remain continuous under such a cor-rection rendering many of the standard arguments inappli-cable.

Nevertheless polygonal billiards with certain symmetriesare still quite tractable to study directly and as an exam-ple we here review the dynamics of the billiard inside a rightisosceles triangle which illustrates the useful methods of mir-roring and folding. Note that the operation of re�ection ata boundary is geometrically the same as continuing the tra-jectory over the boundary and folding it back over it. Con-sequently the motion can be studied by instead mirroringthe billiard table itself such that the trajectory remains a s-traight line in a plane covered by triangles produced by folding the triangles in all possible ways.In general such a folding procedure would cause folded images to overlap but in a few limitedcases such as this one and as can be seen in Figure 6 the pattern of unfolding repeats in terms ofsquare blocks containing 8 distinct orientations of the original triangle meaning that the billiard

9

in a triangle is isomorphic to straight line �ow on a torus (that is a surface periodic along twoaxis).

In terms of continuous time �ow the orbit on the torus can be written as 1

φt : S1 × S1 → S1 × S1

φt(x, y) = (x+ αt, y + βt) (mod 2)

where (α, β) is the initial velocity if the particle in the billiard.

Lemma 4.1. The return maps to a given boundary segment on the isosceles triangle is iso-morphic to a rotation on a circle and the corresponding rotation is irrational if and only if thefraction of the components of the initial velocity is irrational.

Proof. Case 1. Considering �rst the vertical line segments on the triangle we see from thegeometry that the linear �ow on the torus will lie on the two lines corresponding to this statefor all times t = (β/α)k where k is any integer. Note that the orientation of the two segmentsx = 0 and x = 1 are di�erent but moving from one of these lines to the other is done at timeincrements of t = 1/α where φ1/α acts as translation along the y-axis of the circle {x = 1} × S1with a distance of β/α. Or to summarize φ1/α ' Rβ/α.

Case 2. For the horizontal boundary segment the analysis is exactly the same where thetranslation is instead α/β.

Case 3 Similarly as in the above argument the diagonal segment on the billiard correspondsto the diagonal segments x − y = 0 and y + x = 1 correspond to where we want to �nd thesmallest t that brings the system back to the same line. We only consider return to x − y = 0and realize that this corresponds to solving (y + βt) − (x + αt) = 1 which assuming we starton y − x = 0 results in t = 1/(α − β). Since the diagonal has irrational length 2

√2 we need

to check that the resulting translation (rotation) is irrational with respect to this length. Letγ = α/β then both translations are a distance γ′ = γ (mod 2) along each axis and the totaldistance traversed along the diagonal is

√2γ′ which is irrational with respect to 2

√2 if and only

if γ is irrational. (Note γ ∈ Q⇒ γ′ ∈ Q since modulo leaves the decimal sequence invariant andthe decimal of an irrational is irrational).

This makes it possible to make an immediate categorization

Theorem 4.3. Given an initial velocity (α, β) in an isosceles billiard the trajectory is periodicif γ = α/β is rational and aperiodic, dense, and uniformly distributed on each boundary segmentif γ is irrational.

Proof. Follows immediately from Lemma 4.1 and applying the result to Theorems 4.1 and 4.2respectively.

This argument has of course assumed that the trajectory does not pass though a cornerwhere re�ection might be considered unde�ned. For an introduction to the subject of polygonalbilliards see chapter 7 in [2].

1where φt(x) = φ(x, t) is a common notation in dynamics to distinguish between the dynamical variable t andthe initial condition x

10

4.3 Application of the uniform distribution property of the circle bil-

liard

We will here discuss an interesting property regarding the distribution of the �rst integer ofpowers of two. The result can be stated as

Theorem 4.4. Let n be a nonnegative integer and let k ∈ {1, 2, . . . , 9}. The probability that kis the �rst digit of 2n can then be written as

log10(1 +1

k).

Proof. First we observe that k is the �rst digit of 2n if and only if there exists a nonnegativeinteger q such that k10q ≤ 2n < (k + 1)10q. Taking the logarithm of this expression gives

log10 k + q ≤ n log10 2 < log10 +q.

Let {x} denote the fractional part of any real number x. Then the above inequality implies that{n log10 2} ∈ [log10 k, log10(k+1)) := I. Now note that log10 2 is an irrational number. Then theequidistribution of the circle billiard implies that the sequence {n log10 k} is equidistributed andhas a uniform probability distribution. This implies that the probability for a number in the seriesto be in I is proportional to the length of the interval, i.e. proportional to log10(k+1)− log10 k =log10(1 +

1k ) which proves the theorem.

4.3.1 Benfords law

The distribution that we derived in section 4.3 is called the Benford distribution and occurs inan empirical law called Benfords law regarding the frequency of �rst digits in real-life data. Thename comes from F. Benford who provided many examples of this phenomenon. The law wasnot however discovered by Benford but by S. Newcomb in 1881. It states that the �rst digitsof many real-life data are Benford distributed. An example of this phenomenon can be foundin �gure 7 where the relative frequency of �rst digits in the populations of the world divided bycountries has been plotted against the Benford distribution (the data is taken from [3] for theyear 2011).

5 Existence of Birkho� periodic orbits

We will here prove the existence of at least two periodic orbits of an arbitrary period in convexbilliards. The method of proof relies on the generating function introduced in section 2.2 andthe fact that (in convex billiards) periodic orbits are extreme points of this function. Indeed theexistence of the �rst periodic orbit will be found from a straightforward application of this andthe fact that continuous real-valued functions assume a minimum on compact sets. The otherorbit will be harder to �nd and relies on the so called mini-max principle which loosely statedsays that between two minima there must be a saddle point. This saddle point will be proven tobe the second orbit.

5.1 Birkho� periodic orbits

We will �rst introduce the notion of a Birkho� periodic orbit. Let γ be a convex billiardcurve parametrized by the unit interval [0, 1]. Fix a positive integer n and consider an orderedsequence of distinct vertices n in [0, 1] denoted by (x1, . . . , xn). We will view the numbers xi as

11

Figure 7: Relative frequency of �rst digits in the world's population by countries plotted against thefunction log10(1 +

1x)

x1

x2

x3 x4

x5

Figure 8: The green orbit corresponds to (x1, x2, x3, x4, x5) while the red orbit corresponds to(x1, x3, x5, x2, x4).

real numbers modulo the integers. These correspond to polygonal paths in the billiard with ndistinct vertices. Note that the ordering of the vertices is relevant when determining the orbitcorresponding to it (see �gure 8). We call this an orbit of period n.

We will now distinguish between di�erent orbits consisting of the same set of vertices byintroducing the notion of rotation number for these con�gurations. Fixing the orientation forγ to be counterclockwise we can �nd the smallest numbers ti ∈ (0, 1) such that xi = xi−1 + tifor i 6= 1 and x1 = xn + tn where addition is taken modulo 1 (observe that the numbers ti areuniquely de�ned). Since the con�guration corresponds to a closed polyogonal path we have thatt1 + . . . + tn ∈ Z. This number will be denoted by ρ and is called the rotation number of thecon�guration. Calculating ρ for the con�gurations in �gure 8 we see that the dashed path has

12

rotational number 2 while the dotted path has rotational number 1.We can now make the following

De�nition 5.1. Let p and q be positive integers and let (x1, . . . , xq) be an ordered set of distinctvertices in [0, 1]. If the orbit corresponding to these vertices has rotational number p and is aq-periodic orbit of the billiard γ we call it a Birkho� periodic orbit of class (p, q).

Note that in the case that p and q are not relatively prime we can �nd an n-periodic orbitthat is a multiple of an orbit with period less than n. Hence we will always assume that p andq are relatively prime.

We can now state the main result of this section concerning the existence of Birkho� periodicorbits.

Theorem 5.1. Let γ be a strictly convex billiard that is C2 and �x p and q as positive integersthat are relatively prime. Then there exists at least two distinct Birkho� periodic orbits of class(p, q).

We will refer to this as Birkho�'s theorem.

5.2 Proof of Birkho�'s theorem

We begin by denoting the space of all q-periodic polygonal paths with rotation number p inscribedin γ by Mp,q, i.e. (s0, . . . , sq−1) ∈Mp,q if and only if the path given by connecting si with si+1for 0 ≤ i ≤ q− 2 and sq−1 with s0 is rotation number p. We will consider its closure M := Mp,q.Note that M also contains degenerate paths for which two vertices are equal and hence have aperiod less than q. An important part of our proof will be to assure that the orbits we �nd areindeed not degenerate.

Now de�ne the generating function for these con�gurations as (for notation see section 2.2)

H̃(s0, . . . , sq−1) :=

q−2∑i=0

H(si, si+1) +H(sq−1, s0).

x∗1 = x∗2

x∗3x∗q

x′

Figure 9: Creating a longer q-periodic polygon from a (q − 1)-periodic polygon.

We recall that q-periodic orbits correspond to extreme points of this functional and we will�nd our �rst orbit by �nding a minimum of H̃. To do this we observe that M is closed and

13

bonded in Rq and hence M is a compact set by the Heine-Borel theorem. This implies, since H̃is a continuous functional and continuous functions assume extreme values on compact sets, thatH̃ assumes a minimum on M . Denote the vertices of this orbit by (x∗1, . . . , x

∗q). If we can prove

that this orbit is not degenerate we have found our �rst orbit of class (p, q). Observe that thegenerating function is, by de�nition, the negative of the length of the orbit. Now assume that theorbit is degenerate and let x∗1 = x

∗2 and that the other vertices are distinct. Then we can choose

a new vertex x′ between x2 and x∗q and create a new polygonal path (x

′, x∗2, . . . , x∗q) ∈ M (see

�gure 9). Then, by the triangle inequality, this new polygonal path is strictly longer than theoriginal orbit and hence this orbit is not a minimum for the generating fucntion. This contradictsthe orbit was chosen and hence proves that it is nondegenerate. This is our �rst orbit and wewill continue to refer to it as x∗ = (x∗1, . . . , x

∗q).

To �nd the second orbit we will look for another extreme point of the generating functionin M . This point will be a saddle point between two minimum. To get a second minimumof the generating function we make a cyclic permutation of the indices of the �rst orbit, sayx∗∗ = (x∗2, . . . , x

∗q , x∗1). This corresponds geometrically to the same orbit but is a distinct point

in M .Now de�ne the set N := {(x1, . . . , xq)|x∗i ≤ xi ≤ x∗i+1, i 6= q;x∗q ≤ xq ≤ x∗1 (mod 1)} (note

that N ⊂M). Consider the space C of all continuous functions

f : [0, 1] −→ N

such that f(0) = x∗ and f(1) = x∗∗. Since H̃ ◦ f is continuous for any f ∈ C it assumesa maximum on the closed interval [0, 1]. We can assume that this maximum is greater than

H̃(x∗) = H̃(x∗∗) for any f ∈ C since otherwise H̃ is constant on f and any point along f wouldcorrespond to a new orbit of class (p, q). Having assumed this de�ne

Lminmax = inff∈C

maxt∈[0,1]

H̃(f(t)).

We will now show that H̃ assumes this value at some point in N . To prove this choose a sequence{fn} in C such that

Lminmax ≤ maxt∈[0,1]

H̃(fn(t)) < Lminmax +1

n

(this can be done by de�nition of the in�mum) and choose points tn ∈ [0, 1] such that maxt∈[0,1] H̃(fn(t)) =H(fn(tn)). Now consider the sequence {fn(tn)} in N and observe that it is bounded since N isbounded. Hence there exists a convergent subsequence {fn′(tn′)} that converges to some u ∈ Nsince N is closed. Then the following inequality holds

|H̃(u)− Lminmax| ≤ |H̃(u)− H̃(fn′(tn′))|+ |H̃(fn′(tn′))− Lminmax|.

Take some � > 0 and choose a positive integer K such that

|H̃(u)− H̃(fK′(tK′))| < �.

This can be done since H̃ is continuous and the sequence converges to u. Also observe that thesecond term can be approximated by

|H̃(fn′(tn′))− Lminmax| <1

n′

which follows by construction of the sequence {fn(tn)}. Hence if K ′ is chosen such that 1K′ < �then for K ′′ = max(K,K ′) it follows that

|H̃(u)− Lminmax| ≤ |H̃(u)− H̃(fK′′(tK′′))|+ |H̃(fK′′(tK′′))− Lminmax| < 2�.

14

Figure 10: Illustration of the gradient �ow of H̃ on the boundary of N .

Since � > 0 was arbitrary it follows that H̃(u) = Lminmax. Hence the set H̃−1(Lminmax) is

nonempty and it will be shown that this set must contain another (p, q) orbit.

To prove this we will �rst show that the gradient of H̃ is nonzero on the boundary of Nexcept at the points x∗ and x∗∗; indeed, we will show that it points in the outward direction(see �gure 10). This follows from the following simple argument. Remember that the �rst orbitis x∗ = (x∗1, . . . , x

∗q) and let (s1, . . . , sq) be an element of N . Now supppose that sk = x

∗k for

some k (this corresponds to being on the boundary of N). We will calculate the directional

derivative ∂H̃∂sk . Using theorem 2.1 this can be calculated as (here θxy denotes the angle that the

line segment from x to y makes with the billiard table at x, see �gure 11)

∂H̃

∂sk=∂H(sk−1, sk)

∂sk+∂H(sk, sk+1)

∂sk= cos θsksk+1 − cos θsksk+1 .

Now observe that by the strict convexity of the billiard table it follows that

cos θx∗k+1x∗k > cos θx∗ksk−1 = cos θsksk−1 .

From the twist-property of the billiard map T = (S,R) it follows that

S(sk,− cos θsksk+1) = sk+1 < x∗k+2 = S(x∗k+1,− cos θx∗k+1x∗k+2) =

= S(sk,− cos θx∗k+1x∗k).

Again by the twist-property this implies that

− cos θsksk+1 < − cos θx∗k+1x∗k < − cos θsksk−1which implies the conclusion. This analysis can be performed on the rest of the boundaries in asimilar manner and proves that the gradient of H̃ is outward pointing.

Now we are ready to show that H̃−1(Lminmax) contains an extreme point of H̃. Assuming

that this is false it follows that there exists an � > 0 such that H̃−1(Lminmax − �, Lminmax + �)contains no extreme points of H̃. Observe that if this did not hold then there would exist asequence {un} of extreme points of H̃ such that

H̃(un) ∈ (Lminmax − �, Lminmax + �).

15

sk = x∗k+1

x∗k+2

sk+1

x∗k

sk−1

θsksk−1

θsk−1sk

Figure 11: Proof that the gradient of H̃ is pointing outward on the boundary.

Since the sequence is bounded there would exist a convergent subsequence {un′} that convergesto some u. The continuity of H̃ and its derivative would then imply that H̃(u) = Lminmax and

∇H̃(u) = 0 which would contradict the assumption that H̃−1(Lminmax) contains no extremepoints.

Choosing � > 0 such that H̃−1(Lminmax−�, Lminmax+�) contains no extreme points it followsthat there exists δ > 0 such that ‖∇H̃‖ > δ on H̃−1(Lminmax − �, Lminmax + �). Now takeu ∈ H̃−1(−∞, Lminmax + �) and �x some t > 0. Then from the de�nition of gradient of H̃ itfollows that

H̃

(u− t ∇H̃

‖∇H̃‖

)= H̃(u)−∇H̃ · t ∇H̃

‖∇H̃‖+O(t2) =

= H̃(u)− t‖∇H̃‖+O(t2).

Now assume that u ∈ H̃−1(Lminmax − �, Lminmax + �). It follows that

H̃

(u− t ∇H̃

‖∇H̃‖

)= H̃(u)− t‖∇H̃‖+O(t2) < H̃(u)− tδ +O(t2).

By the de�nition of the big-O notation it follows that

limt→0

|O(t2)|t

= 0

hence t can be chosen such that O(t2) < tδ2 which implies that

H̃

(u− t ∇H̃

‖∇H̃‖

)≤ H̃(u)− tδ

2< Lminmax + �−

tδ

2.

This proves that if we take � = tδ2 then H̃(u) < Lminmax for u ∈ H̃−1(Lminmax − �, Lminmax + �).

Now taking u ∈ H̃−1(−∞, Lminmax − �) it follows that

H̃

(u− t ∇H̃

‖∇H̃‖

)≤ H̃(u) +O(t2) ≤ Lminmax − �+O(t2) < Lminmax − �+

tδ

2= Lminmax.

Now take f ∈ C such that maxt∈[0,1] H̃(f(t)) < Lminmax + � and de�ne f̃(s) = f(s)− t ∇H̃(f(s))‖∇H̃(f(s))‖ .

Then f̃ ∈ C and by the previous calculation it follows that maxt∈[0,1]H̃(f̃(t)) < Lminmax contraryto the de�nition of Lminmax. Hence the set H̃

−1(Lminmax) contains an extreme point u of H̃.

Since the gradient of H̃ is nonzero this extreme point cannot lie on the boundary of N henceproving that u is a proper (p, q) orbit which �nishes the proof.

16

6 Caustics

6.1 Introduction

One property of a class of billiard which can be used to analyse the dynamics of a billiard mapis if there exists caustics. A caustic, geometrically, is a curve such that if the trajectory of thebilliard tangents the curve at a singe point then all subsequent trajectories are also be tangentsthe same curve. Note that caustic need not be con�ned inside the billiard table and the pointof tangency too may lie outside it were the trajectory extended to a complete line stretching toin�nity. The existence of caustics is not a general property, it is easy to show that x4 + y4 = 1has no convex caustics but a class of tables which do have an in�nite number of caustics is thebilliard inside an ellipse x2/a2 + y2/b2 = 1 and will therefore demand some special attention.

As is hinted by Figure 12 every orbit in an ellipse either has a confocal ellipse or a hyperbolaas its caustic and the �gure also displays how the orbits in phase space correspondingly occupyinvariant sets of curves instead of spreading out more uniformly. It is common that caustics alsodirectly refers to these curves themselves. Before continuing with a more general discussion we

0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

2.5

Figure 12: Simulation of the �rst 100 iteration of a billiard in an ellipse with the corresponding pointsin phase space as angle relative to tangent over parametrizing angle where the trajectory remains tangentto a (convex) ellipse or a (non convex) hyperbola.

show formally the existence of caustics in the case of an ellipse, which is special in the sense thatevery single orbit has a corresponding caustics and fall into two primary characteristic cases:ellipse and hyperbola. There does exist a third degenerate case where the trajectory passesthough the foci but it is not strictly speaking tangency in that case. For this discussion we willrequire a simple lemma.

Lemma 6.1. Any trajectory that passes though one foci is re�ected though the second foci. Thatis the lines corresponding to the string construction make equal angles with the tangent of the

17

ellipse

Proof. The proof is straightforward. Let f1 and f2 be the foci of the ellipse. Then given f(x) =|x− f1|+ |x− f2| we have that ∇f is normal to the ellipse along it and evaluating the gradientwe see that it is the sum of the unit vectors from the foci to x meaning their sum is normal theellipse which corresponds to them making equal angles with the tangent.

p1

f1

f ′1

p0

f ′2

f2

p2

ab

p1

f1

f ′1

p0

f ′2f2

p2a

b

Figure 13: Geomtric construction of tangential a confocal ellipse (left) and hyperbola (right)

We will use this to show the three cases where an ellipse

Theorem 6.1. If a trajectory passes outside the two foci of the ellipse in one iteration it willremain tangent to a confocal ellipse.

Proof. Let p1p0p2 be a billiard trajectory not passing between the two foci f1 and f2 of theellipse (see the left �gure in Figure 13. We are going to construct two ellipses such that the linesegments p1p0 and p0p2 are corresponding tangents and �nally that the two ellipsis are equal.For the �rst ellipse mirror f1 in p1p0 to form a new point f

′1, and then denote by a the point

at the intersection of the lines p1p0 and f′1f2. Now a lies on the ellipse formed by a string of

length `(f ′1f2) attached to the two original foci and due to the fact that f′1 was constructed by

mirroring f in a will also be a point of tangency.Now if the exact same construction of a point f ′2 is performed by mirroring f2 in p2p0 and

taking b to be the similar point of intersection it can too is seen that b lies at a point of tangencyof an ellipse formed by a string of length `(f ′2f1)

`(f ′1f2) = `(f′2f1) by the following argument; since f1p0f2 also make equal angles with the

boundary of the ellipse (Lemma 6.1) the angles ∠p1p0f1 and ∠p2p0f2 are equal. From this wesee that the triangles 4f ′1f2p0 and 4f ′2f1p0 have have at least two sides of equal length and oneangle in common and hence they are equivalent and the ellipses are the same. By induction theorbit thus remains tangent to the same confocal ellipse.

Theorem 6.2. If the trajectory passes between the two foci then the trajectory is and will remaintangent to a confocal hyperbola.

Proof. The construction is nearly exactly the same as in the case of the ellipse, see the right�gure in Figure 13 . The points of intersection, a of p1p0 and f

′1f2 and b of p2p0 and f

′2f1 now

lie on a hyperbola de�ned such that the di�erence between the distances between a point on thehyperbola to the foci is `(f ′1f2) = `(f

′2f1)

18

αiαi+1

γi

βi+1γi+1

βi

pi

pi+1

pi+2

Figure 14: Convergence to greater axis

This leaves a third possibility not illustrated in 12and that is if the trajectory passes though one of thefoci. The basic behaviour is already present in Lemma6.1, the trajectory is bound to always pass though thefoci. The �rst such orbit is the two periodic orbit alongthe major axis of the ellipse passing though both focibut as we show below the other orbits that do not lieon the axis still converge to it.

Theorem 6.3. If the trajectory passes though one of thefoci while not lying on the major axis it will converge tothe major axis.

Proof. The proof utilizes the geometric construction inFigure 14 where we can associate with each point piand triangle with corners at the foci and our result willfollow from the fact that the triangle becomes �attenedat every iteration. Consider the related sequences {αi}∞i=0, {βi}∞i=0, and {γi}∞i=0. By basicgeometry we have αi+1 + βi+1 < π and since βi < π − αi = βi+1 we see that {βi}∞i=0 is amonotonically increasing sequence and since it is geometrically bounded by π it must converge,and it remains to show that the limit is π.

Now since convergent sequences are Cauchy there exist for every ε > 0 some N such that suchthat βN+1−βN < ε and reversing the argument from before βN+1−βN = π−αN−βN = γN < εand we have that γi → 0 which means βi → π and αi → 0

In the phase space these results correspond to every orbit being constrained on invariantsets of curves. In the case of the ellipse it is possible to recover these curves explicitly as levelcurves a subset of which can be seen in Figure 15. This image is commonly used to illustratehow perturbing a circle into an ellipse preserves the presence of invariant curves but yields asigni�cantly more complicated appearance. One method of explicitely deriving the function inquestion is provided in the appendix, A.1

Figure 15

We now return to one of the implicaitons of Figure 12 and Figure 15 where one may notethat the caustic corresponding to an ellipse yields a closed non self intersecting curve in the phase

19

space. It is relatively straightforward to deduce that this is always the case when the caustic isconvex which we summarize as a lemma

Lemma 6.2. Every convex caustic corresponds to an invariant circle in the phase space.

proof (sketch). Assuming there exists a convex caustic then for every points s along the tableboundary there are only two re�ection angles θ1 and θ2 that result in tangency with the caustic,and choosing a speci�c orientation every s has a unique θ(s). Upon varying s, θ(s) variescontinuously if the caustic is smooth forming a closed curve (s, θ(s)).

If we start with an invariant circle (s, θ) we may recreate the geometric caustic from theenvelope of a family of rays starting from the boundary in the direction of the re�ection anglesθ.

A famous result by V.F Lazutkin, later improved by R. Douady proves that a large range oftables have convex caustics [4] [5]

Theorem 6.4. If the border of the table is C6 and strictly convex then there exists a collectionof smooth convex caustics close to the border of the table of the table whose union has positivearea.

These speci�c caustics are also sometimes referred to as whisper galleries in references to theacoustic phenomenon where two people standing close to a wall in an oval room may communicatewithout shouting as the soundwave becomes concentrated close to the wall much like the billiardball according to this theorem.

Futhermore, this fact that convex caustics are circles and invariant under the billiard maphas important local implication as the fact that the billiard map is homeomorphic on all of thephase plane means that orbits corresponding to caustics can be studied using the machinery ofcircle homeomorphisms2.

The basic concept is similar to the methods employed to the circle billiard in the previoussection as one isolates the �elds of study to a sub circle (Figure 5 which indeed correspondedto a geometric caustic.

The existence of invariant closed curves in general also has important global implication forthe whole billiard system as they partition the phase space into invariant disjoint subsets whichprevents any orbit, the implications of which are treated in more closely in section 6.6 while webegin to review the local implications now.

6.2 Homeomorphism of the circle

Much of the theory we shall review brie�y in this section is due to the work of mathematicianHenri Poincaré who studied circle homeomorphisms in the later half of the 19th century througha powerful characterizing number, which while it might not be trivial to compute, can be usedto classify di�erent behaviours and show what dynamics may coexist and what may not. [1,p.125-133].

In order to formulate the results of interest with some precision we review some classicalde�nitions useful for describing functions on circles in terms of functions on the line. Similarlyto how a periodic function F : R → R can be projected onto the circle forming an functionf : S1 → S1, by for example f(x) = F (bxc), it is possible to lift certain functions de�ned on acircle to functions de�ned on R.

2A homeomorphism is a continuous function that has a continuous inverse.

20

Proposition 6.1. If f : S1 → S1 is continuous then there exists a function F : R→ R, uniqueup to the addition of an integer, called a lift such that

f ◦ π = π ◦ F

where π : R → S1 projects the real line onto the circle. deg(f) = F (x + 1) − F (x) is called thedegree of f and is independent of the lift and x. If f is a homeomorphism then |deg(f)| = 1.

De�nition 6.1. If deg(f) = 1 then we say that the map is orientation preserving, and if deg(f) =−1 we say that it is orientation reversing.

The notion of degree corresponds to how the map stretches and orients the circle.

Theorem 6.5. Let f : S1 → S1 be a circle homeomorphism and F : R → R a lift. The thenumber

ρ(F ) = limn→∞

Fn(x)− xn

exists and is independent of x. The number depends on the lift only to the extent ρ(F )− ρ(F̃ =F − F̃ ∈ Z.

Since this number is well de�ned up to an integer one usually takes the fractional partρ(f) = {ρ(F )}, a number which is called the rotation number a name which hints to the factthat in the case of a circle rotation the rotation number reduces to the rotation number α. Asone notes immediately the limit could equally well be taken without −x but is taken for historicalreasons of how proofs are best constructed, and to better illustrate ρ as the mean distance a mapf translates any given point.

Three things the rotation number directly implies are

1. If ρ(f) = 0 then f has a �xed point

2. If ρ(f) is rational if and only if f has periodic points.

3. If ρ(f) = p/q ∈ Q then every periodic orbit has the same period q and F q(x) = x+ p, thatis it moves x p revolutions around the circle.

Note that these three properties already hold for rotations on a circle, the third of which impliesfor billiards that all periodic orbits that tangent the same caustic must have the same periodand similar geometric structure. Unlike for circle rotations rational rotation number does notgenerally exclude the possibility of aperiodic orbits but it does allow the following classi�cationtheorem

Theorem 6.6. If f : S1 → S1 is a circle homeomorphism such that ρ(f) = p/q ∈ Q then thereare two possible orbits

1. If f has exactly one periodic point then every other point on the circle is heteroclinic totwo points under fq.

2. If f has more than one periodic orbit then each non periodic orbit is heteroclinic to twopoints on two di�erent periodic orbits.

It is perhaps the possibility of the type 2 behaviour that holds the highest interest as anorbit converging a periodic point might be considered unexpected. This convergent behaviourhas already been encountered in the ellipse with the orbit converging to the major axis and thetwo 2-periodic points along it but this did not strictly a consequence of the above theorem as

21

the corresponding curve is not a circle or caustic in the strictest sense. However one may stillwish to investigate if there are other orbits along caustics with periodic points that show thisconvergent behaviour. For the elliptical caustics however the answer is in the negative as if onetrajectory is tangent to a given ellipse then all the orbits tangent to it are periodic with the sameperiod, a consequence of the geometrical Poncelet closure theorem reviewed in the next section.

6.3 Poncelet Porism

The Poncelet Porism is a purely geometrical result with greater generality than the fact that allorbits in an elliptic billiard table that stay tangent to an elliptic caustic have the same period.However it has this fact as an immediate consequence when considered together with theorem6.1 [6, p.397-399]

Theorem 6.7 (Poncelet porism). Let Γ and γ be nested ellipses. If γ can be circumscribed byan n-gon with corners in Γ, then every point on Γ is a vertex of such an n-gon.

Proof. Let Γ be the outer ellipse, and let γ the inner ellipse. First we de�ne two functions Rγ,Γ(x)and Lγ,Γ(x) which are the distances to tangency, ergo the length of the line segments from x ∈ Γto γ along the two left and right lines that tangent γ seen in Figure 16 (a)

Now we let Γ(s) be parametrized with curve length and let T (s) be coordinate correspondingto the point Γ(T (s)) de�ned so that the line Γ(s)Γ(T ) tangents γ in the right hand orientation.If s ∈ R corresponds to a point that generates a n-sided polygon this corresponds to Tn(s) = s(mod L) where L is the total length of Γ.

Our goal will be to �nd a monotone function g which rescales the parametrization s so that(g◦T )(s) = g(s)+C. If such a function exists then we may de�ne a new function G = (g◦T ◦g−1)which acts as translation G(x) = x + C. Now from Gn(x) = (g ◦ Tn ◦ g−1)(x) = x + nC wesee that if g−1(x) is a n-periodic point of T then x is an n-periodic point of G but since Gis just translation then all coordinates x ∈ R are n-periodic points of G, and inverting to getT = g−1 ◦G ◦ g we have that every point s ∈ R is a periodic point of T .

Now we simply show the existence of such function g but before continuing note that wemay always assume that the outer ellipse is a circle, for if it was not we could apply an a�netransformation rescaling and rotating the con�guration as to convert the outer ellipse to a circle.Such a transformation maps parallel lines into parallel lines so if we could inscribe a polygonin the original con�guration the image of the polygon would still be properly inscribed in theimages of the ellipses.

Working with a circle has the consequence that any line connecting to points along Γ willmake equal angles with the boundary. Now, if we start with s and choose a coordinate s′ closeto s, and let T and T ′ be the image coordinates construct two triangles such as in Figure 16where in the limit where s′ is in�nitesimally close to s we have from the law of sines

|Γ(T ′)− Γ(T )|sin ε

=Lγ,Γ(Γ(T ))

sin(π − α),|Γ(s′)− Γ(s′)|

sin ε=Rγ,Γ(Γ(s))

sinα.

which in di�erential notation can be cast as

dT

Lγ,Γ(Γ(T ))=

ds

Rγ,Γ(Γ(s))or

dT

ds=Lγ,Γ(Γ(T ))

Rγ,Γ(Γ(s)).

Now we once again remind ourselves that a�ne transforms preserve the ratio of two line segmentsalong the same line (as they are stretched equally). This means that the right hand fraction of

22

Γ(s)

Γ(s′)

Γ(T )

Γ(T ′)α

α

ε

ε

Lγ,Γ(Γ(T ))

Rγ,Γ(Γ(s))

Lγ,Γ(x)

Rγ,Γ(x)

(a) (b)

x

Figure 16

the rightmost expression is invariant under an a�ne transform Λ which transforms the innerellipse γ into a circle. If it already is a circle we may just apply the identity, and so we have

dT

ds=LΛγ,ΛΓ(ΛΓ(T ))

RΛγ,ΛΓ(ΛΓ(s))(1)

but by symmetry and the de�nition of the distances to tangency L, and R we have that LΛγ,ΛΓ ≡RΛγ,ΛΓ when Λγ is a circle and we may denote this common function DΛγ,ΛΓ, and we canconstruct the monotone function

g(s) =

∫ s0

dy

DΛγ,ΛΓ(ΛΓ(y))

where we see that (1) is equivalent to

d[g(T (s))]

d[g(s)]= 1

and the proof is complete.

6.4 String construction of convex caustics

In this section we will prove that to any strictly convex curve γ in the plane (i.e. the curve isconvex and |γ′′(t)| > 0 everywhere) we can �nd another curve Γ such that it has γ as caustic.This will be done with the so called string construction. Intuitively, the method of proof can bedescribed as taking a string of some length L′, where L′ is greater than the length L of γ, and apen. Then we put the string around γ and pull it tight at one point with the pen. Now, keepingthe string tight, we move the pen around γ, thus tracing out a new curve Γ. This new curve willhave γ as caustic.

6.5 Proof of the string construction

Let γ be parametrized by the interval [0, 1] and choose L to be greater than the arc length ofγ. Fix some point t ∈ [0, 1] and let ut be the unit tangent to γ at γ(t). Now consider the line

23

γ(t)

γ(t′1)

γ(t′2)

l(r1)

l(r2)

Figure 17: String construction of curve with caustic γ.

l(r) := γ(t) + rut for r ≥ 0. For each r ≥ 0 we can choose a point γ(t′) such that γ(t′) − l(r)is tangent to γ at γ(t′) (see �gure 17). The uniqueness of this t′ is guaranteed by the strictconvexity of γ.

Now de�ne a function L(r) as

L(r) := r + |l(r)− γ(t′)|+ L(t, t′).

s1

s2

s3

p

Figure 18: Approximation of γ with polygonal curves to prove that L(r) is strictly increasing.

Here L(t, t′) denotes the arc length from γ(t) to γ(t′) going counterclockwise in �gure 17. Wewill prove that L(r) is strictly increasing and continuous. To prove that it is strictly increasingwe must show that the section r1t

′1t′2 is shorter than r1r2t

′2. Denoting the intersection of the two

dashed lines in �gure 17 by p we can note that it su�ces to prove that the section t′1t′2 is shorter

than t′1pt′2. Now consider the geometrical construction in �gure 18. The dashed curve s1s3s2 is

constructed by intersecting the segment s1s2 of γ at the middlepoint, denoted by s3. This curveis clearly shorter than the segment s1ps2. In the same manner we construct the dotted curve bybisecting each of the two new segments. This dotted curve is also shorter than s1ps2. Continuingin this fashion we construct a sequence of polygonal curves denoted by {qn} that are all strictlyshorter than the segment s1ps2.

We now want to show that this sequence of polygonal curves has the section s1s2 of γ aslimit. To do this we note that γ is uniformly continuous. This implies that for any � > 0 we can

24

�

Figure 19: Illustration of uniform convergence of the polygonal curves to a segment of γ

choose a δ > 0 such that for any |t − t′| < δ we have |γ(t) − γ(t′)| < �. Hence if we choose apositive integer N big enough the polygonal curve qn for n ≥ N will ful�ll (see �gure 19)

maxs1≤t≤s2

|γ(t)− qn(t)| < �.

This proves that the sequence {qn} is uniformly convergent to γ and hence if we denote by Lnthe length of the polygonal curves and by Ls1s2 the length of the segment of γ between s1 ands2 then we have

limn→∞

Ln = Ls1s2 .

But this implies that the segment s1s2 of γ is strictly less than the segment s1ps2 which provesthat L(r) is strictly increasing.

Now to show that L(r) is continuous we look at its value for r + ∆r. Let t′ + ∆t′ be thecorresponding point on γ. Then we can write

L(r + ∆r) = r + ∆r + |l(r + ∆r)− γ(t′ + ∆t′)|+ L(t, t′ + ∆t′).

To prove that L(r) is continuous we must show that

lim∆r→0

L(r + ∆r) = L(r).

Since L(r) is a sum of continuous functions it su�ces to prove that ∆t′ → 0 as ∆r → 0. Toprove this we assume that ∆s′ > � always for some � > 0 to prove a contradiction. This implies,since γ is strictly convex, that |ut′ − ut′+∆t′ | > δ for some δ > 0. But this implies (see theconstruction in �gure 17) that ∆r cannot go to zero, hence we have a contradiction. This impliesthat L(r) is continuous.

Now �x L′ > L where L is the length of γ. Having proved that L(r) is a continuous andstrictly increasing function such that L(0) = L we can choose rt > 0 such that L(rt) = L

′. Nowde�ne Γ(t) := l(rt) = γ(t) + rtut. De�ning Γ(t) for each t ∈ [0, 1] we get a parametrized set inthe plane. We will show that this set is a continuously di�erentiable curve such that it has γ asits caustic.

It is obvious by construction that Γ(t) 6= Γ(t′) for any t 6= t′ except for Γ(0) = Γ(1), henceΓ is closed and does not intersect itself. To prove that Γ is di�erentiable we must evaluate theexpression Γ(t+ ∆t)− Γ(t). By the de�nition of Γ we have

Γ(t+ ∆t)− Γ(t) = (γ(t+ ∆t)− γ(t)) + (rt+∆tut+∆t − rtut).

25

Since γ is assumed to be continuously di�erentiable we can write

γ(t+ ∆t)− γ(t) = ∆tγ′(t) +O(∆t2).

We also have

rt+∆tut+∆t − rtut = rt+∆t(ut + ∆tu′t +O(∆t2))− rtut == (rt+∆t − rt)ut + rt+∆t∆tu′t +O(∆t2).

Γ(t)Γ(t+ ∆t)

t

t+ ∆t

t′

t′ + ∆t′

Figure 20: Geometrical proof that Γ is di�erentiable.

This shows that it su�ces to prove that rt+∆t → rt if ∆t→ 0. But this follows from the factthat if ∆t → 0 then ut+∆t → ut which in turn implies that ut′+∆t′ → ut′ (see �gure 20). Thisgives the desired conclusion by the construction of the curve. Hence Γ is a di�erentiable curve.

Γx

y

γ

a b

Figure 21: Proof that γ is a caustic of Γ.

To prove that Γ has γ as caustic we will follow the procedure in [2, p.73,74]. Take a pointx ∈ Γ and let y be a reference point on γ. Now denote by f(x) and g(x) respectively thedistances from x to y by going around γ in a counterclockwise and clockwise manner respectively

26

(see �gure 21). Then Γ is by construction a level curve of f + g and hence ∇(f + g) = 0. Wenow want to prove that ∇f is the unit vector along ax. To do this we consider a level curve off , say f = c for some constant c, through the point x. Then if a = γ(t) for some t we can writex = γ(t) + (c − t)γ′(t). This implies that x′ = (c − t)γ′′(t). But then, since t is an arc lengthparameter, we have that γ′ and γ′′ are orthogonal (this follows by di�erentiating the expression1 = |γ′(t)| = γ′(t) ·γ′(t)) which in turn implies that x′ is orthogonal to ax since γ′ is parallell toax. Hence we have proven that the level curve of f is perpendicular to ax at x which must implythat ∇f is parallell to ax since we are working in the plane. That the directional derivative of fis one follows from that ∆f = ∆t, the change in the arc length parameter as we move x alongΓ. Hence it follows that ∇f is a unit tangent along ax. Similarly one proves that ∇g is a unittangent along bx. But this together with ∇f +∇g = 0 implies that ∇f and ∇g must make thesame angle with the tangent at x (see �gure 22). By de�nition this proves that γ is a caustic ofΓ.

∇f ∇g

Figure 22: ∇f and ∇g make equal angles with Γ at x.

6.6 Non existence of Caustics and global implications

The existence of convex caustics is not generally guaranteed. For most tables a given orbit willnot remain tangent to a single curve and in fact a long standing conjecture by Birkho� has beenthat the billiard in an ellipse is the only table where almost every orbit belongs to an invariantcircle supporting which at least partial results have been made [2, p.95].

In this section we illustrate how caustics not only determine the dynamics of the trajectoriestangent to them but also have consequences for orbits far away from the corresponding curves inphase space. The implication itself comes from the fact that invariant curves partition the phasespace and prevents orbits from exploring more than a limit subset which immediately excludesergodicity. We will investigate this by considering sets with invariant curves as bounderies andtherefore will have need of reminding ourselves of the following result.

Lemma 6.3. Homeorphisms preserve boundaries

Proof. Use the de�nition of the boundary ∂A = A\A◦ and the fact that the order of applying ahomeomorphism and operations of intersection, closure, and interior are interchangeable.

f(∂A) = f(A)\f(A◦) = f(A)\f(A)◦ = ∂f(A).

27

And so if a set has invariant curves as it's boundary we may immediately derive our parti-tioning result

Theorem 6.8. Let A be region of phase between two invariant circles Γ1 and Γ2. Then givenany point in A its entire orbit is con�ned in A. That is T (A) = A.

Proof. Make the identi�cation of phase space with the annulus where the region in question isa sub-annulus, the intersection of the set bounded by the outer boundary and the unboundedpart of the inner boundary. We have from the Jordan Curve theorem that circles divide theplane into two connected sets and so the set bounded by two circles i connected. Since T iscontinuous the image T (A) has to be connected but there is only one connected set with Γ1 andΓ2 as boundaries so the set must be invariant under T

Remark In the case of convex caustics this means that a trajectory which at one instant iscontained between the boundary and the interior caustic will remain in this region forever, andany trajectory passing though the caustic will continue to pass though it. Of course there

What we have considered so far have been the fact that caustics tend to make the dynamicmore well ordered and possible to characterize, but we will now take a step in the oppositedirection showing how regions of the phase space without caustics can exhibit a great deal ofcomplexity. Let a billiard map have two invariant circles with di�erent rotational numbers ρsuch that the sub cylinder bounded by these curves is free of invariant circles. We call such aregion a region of instability and it can be shown that it contains at least one orbit that movesarbitrarily close to both the upper and lower circle. For the discussion we require a relativelystrong theorem by Birkho� [7, p.430] concerning functions de�ned on semi-unbounded cylinderA = S1 × [0,∞) but which is also applicable to our bounded phase space

Theorem 6.9. Let f : A → A be an orientation preserving twist homeomorphism and U ⊂NW (f) an f -invariant open relatively compact set containing S1×{0} with connected boundary.Then ∂U is the graph of a continuous function ψ : S1 → (0,∞)

Where NW (f) is the set of so called nonwandering points, points such that every neigh-bourhood of a point U of it eventually self intersects after a �nite number of iterations φn(U)∩U 6=∅. For the billiard map T , this is always satis�ed for any subset since NW (T ) = A as area p-reservation guarantees this by the Poincaré recurrence theorem, and also compactness is simpleto assume since the closed cylinder is compact.

And by this one can state the following theorem[7, p.433]

Theorem 6.10. Suppose C1 and C2 bound a region of instability of an area preserving twist mapT and let ε > 0. Then there exists a point p in the ε-neighborhood Wε,1 of C1 with an iterate inthe ε-neighborhood of W2,ε of C2.

Proof. If not then there would exist some ε such that V0 =⋃n∈N f

n(W1,ε) is disjoint from W2,ε.Our aim will be to show that if this is the case then the upper boundary of V0, Γ, is the graph ofa continuous function of S1 and hence isomorphic to a circle. Since V0 is dynamically invariantunder T by de�nition, T (V0) = V0 and homeomorphism preserve boundaries we would haveT (∂V ) = ∂T (V ) = ∂V , and the boundary segment would hence be an invariant circle whichwould be a contradiction since the region of instability was supposed to be free of invariantcircles.

To show that upper boundary of V0 is a graph we will apply theorem 6.9 but it is not certainthat V0 would satisfy all the necessary properties. For example it is quite possible that V0 mightcontain some holes such that its whole boundary would not be connected. For this reason we

28

construct another set with Γ as its boundary which is certain to have a connected boundary (ina modi�ed sense).

Let W be the connected component between C2 and Γ which exists since V0 is disjoint fromW2,ε. We are certain that W does not contain any holes since V0 must be connected on accountof the fact that every sub image fn(W1,ε) contains C1. W is therefore an annulus, and to removethe second part of the boundary C2 from the discussion we take the area between C1 and C2consider it to be an annulus in the R2-plane with C2 as its inner circle whereupon we deform C2to a single point and complete W with a point such that the resulting set is a simply connectedset in the plane surrounded by an an annulus V0.

This con�guration where ∂W is embedded in ∂V0 is special in the sense that it immediatelyimplies that ∂W = Γ is connected by the following Lemma A.7.8 in [7, p.737].

Lemma 6.4. Suppose O1, O2 ⊂ R2 are disjoint open sets such that ∂O1 ⊂ ∂O2. Then ∂O1 isconnected.

Now we take the part of V = (W )c contained below C2 on the ellipse and and see that it is aninvariant set with a connected boundary, also containing S1 × {0} and so by Birkho�'s theoremits (upper) boundary is an invariant circle which completes the necessary contradiction.

This means that there exists at least one orbit which has a relatively complicated behaviourmoving from the lower circle to the upper circle meaning one can expect some complexity in thephase space, and has an immediate corollary

Corollary 6.1. If a table has no convex caustic then there exists an orbit such that the angularcoordinate θ takes values in the whole range (0, π)

Proof. If there are no caustics then there are no invariant circles except for the boundary circlesS1 × {0} and S1 × {1} and these have di�erent rotation numbers 0 and 1 so all of phase spaceis a region of instability. Then one may take ε to be arbitrarily small whereupon the statementfollows from Theorem 6.10.

Figure 23: Simulation of 50 itera-tions inside the billiard x4 + y2 = 1

which has no convex caustics

In Figure 23 we see an example of an orbit in a tablethat has no convex caustics where the freedom in angle ofre�ection is immediately evident. This is one motivation forrelaying another simple result due to Mather which providesa very simple condition under which a billiard table has noconvex caustics and thus 6.10 is applicable.

With this as a partial motivation we now review a resultwhich establishes that the curvature of the billiard table de-termines whether the existence of convex caustics can exist.

6.7 Nonexistence of caustics at locally �at

boundaries

We will here prove another important result about the exis-tence of caustics in convex billiards. In section 6.4 we provedthat we could always create a billiard table that had a givenstrictly convex curve as caustic. Here we will prove that ifwe �atten the boundary of the billiard table, i.e. we set thecurvature κ := |Γ′′(t)| = 0 at some point t, then the table has no convex caustics. This result isa direct consequence of an important equation from geometrical optics which can be shown tohold in billiards as well. We state it here without proof.

29

γ

Γ

αα

ff

Figure 24: Disappearence of caustics at zero curvature.

Theorem 6.11. Suppose that Γ has a caustic γ completely contained within Γ. Take a pointγ(t) on γ and let ut be the unit tangent to γ at this point. Re�ect this tangent in Γ, say Γ(s)according to the law of re�ection and let γ(t′) be the point of tangency with γ of this re�ectedtangent. Denote by α the re�ection angle and let f := |γ(t)−Γ(s)| and f := |Γ(s)−γ(t′)|. Thenthe following relation holds beteween α, f and f :

1

f+

1

f=

2κ

sinα.

Here κ denotes the curvature of Γ at Γ(s).

This equation is often called the mirror equation and we will use it to prove

Theorem 6.12. Let Γ be a convex billiard and suppose κ = 0 at some point Γ(s). Then Γ hasno convex caustics γ.

Proof. Suppose that a convex caustic γ exists and let Γ(s) be the point for which κ = 0. Then,since both curves are convex I can �nd a point γ(t) such that the straight line from γ(t) to Γ(s)is tangent to γ (see �gure 24). Denoting the length of this line segment by f we then re�ectthis tangent at Γ(s) according to the law of re�ection to get another tangent to γ at some pointγ(t′). Denoting the respective length by f the mirror equation gives at Γ(s) that

κ =sinα

2

(1

f+

1

f

)> 0.

This contradicts our assumption on κ and hence Γ can have no convex caustics.

6.7.1 Examples and test for tables without caustics

We may apply this to prove a straight forward result for a class of tables given by algebraiccurves

Proposition 6.2. Of all the tables given by the implicit curves

x2n + y2n = 1

where n ≥ 1, only n = 1 has convex caustics

30

Proof. The curvature of an implicit curve f(x, y) = C is given by

κ = −fxxf

2y − 2fxyfxfy + f2xfyy

(f2x + f2y )

3/2

which is derived in appendix A.2. Evaluating the numerator for f(x, y) = x2n + y2n we see thatit is

−22n2(2n− 1)(x2(n−1)y2(2n−1) + x2(2n−1)y2(n−1))

which is 0 at the points of intersection with the coordinate axis x = 0 and y = 0 if n ≥ 1.

This argument can of course be extended to all polynomial algebraic curves where there areno mixed terms xiyj to complicate it. The nonexistence of caustics of course does not excludethe existence of non convex caustics and the associated invariant curves.

We now take the opportunity to in a novel way illustrate how the phase space may look inthe absence of convex caustics, though it should be stressed that the phase space of any giventable has no reason to resemble another one.

Take the orbit starting at the rightmost point of x4 + y4 = 1 such that the initial ray makesthe initial angle π/6 with the boundary then the coordinates in phase space of the �rst 10 000iteration is displayed in Figure 25 which wanders close to both the upper and lower circlesbut stays out of a number of circular regions which other simulations indicate correspond tonon-convex caustics encircling periodic points.

7 Outer billiards

In this section we will make a short introduction to the subject of outer billiards. This systemis closely related to the ordinary billiards but instead of letting the ball re�ect at the inside ofthe table we will now consider the case where the billiard ball re�ects on the outside of the tablein a manner which will be speci�ed. A new question that arises in context of these billiards iswheter the orbit will be bounded or not. We will make no attempt to any rigorous treatmentof these systems but will instead discuss some interesting results regarding them and sketch theproofs of some of them. This presentation will follow the one given in [2, p.147�160].

7.1 De�nition

Consider a continuous closed simple convex curve in the plane and denote it by γ. Now takea point u in the plane outside of γ. The outer billiard re�ection of this point is de�ned byconsidering the two tangent lines from u to γ (assuming they exist). By convention we choosethe tangent line to the right of γ as seen from u (see �gure 26). Assuming that this tangent lineis tangent to γ at exactly one point on γ denoted by v consider the length from L := |u − v|.By extending the tangent line from v a distance L a new point w outside of γ is reached. De�neT (u) = w where T is called the outer billiard map. The map T is well-de�ned everywhere thatthere exists a tangent line from u to γ to the right of γ that intersects the curve at a uniquepoint. It is easily seen that if for example the billiard table contains straight line segments thenthe outer billiard map will not be de�ned at points that lie at extensions of these segments. Onecan compare this to the situatiuon in ordinary billiards when the ball hits a corner of the curveand the motion becomes unde�ned.

31

-4 -3 -2 -1 0 1 2 3 40

0.5

1

1.5

2

2.5

3

3.5

Figure 25: 10000 iterations inside the table x4 + y4 = 1 in the coordinates re�ection angle over a curveparametrization

u

T (u)

γ

Figure 26: Construction of the outer billiard map

7.2 Properties of the outer billiard map

In this section we will present some properties of the outer billiard map along with some sketchesof the proofs. First we note that like the ordinary billiard map the outer billiard map is area-preserving for smooth boundaries.

Theorem 7.1. Assume that γ is smooth (continuously di�erentiable) and let T be the outer

32

billiard map. Then T is area-preserving.

Proof.

A

A′

C

C ′

B

B′

D

D′X X

′Y

Figure 27: Proof sketch of area preservation.

Let X and Y be two �close� points on γ and extend the tangents of these points by somepositive real number r (see �gure 27). Denote by A, A′, B and B′ the endpoints of these tangents.Note that the billiard map T maps the segment AA′ to BB′. Now take � > 0 and repeat thisconstruction for r − � and denote the new endpoints of the resulting tangents by C, C ′, D andD′ respectively. The billiard map T now maps AA′C ′C to BDD′B′ and we want to show thatthis is done in an area-preserving fashion in the limit X → X ′. Denoting by δ the angle betweenAB and A′B′ we can calculate the area of A′AY and C ′CY which gives

|A′AY | ≈ r2δ

2,

and

|C ′CY | ≈ (r − �)2δ

2=r2δ

2+�2δ

2− r�δ.

Using this the area of AA′C ′C is given by

|AA′C ′C| ≈ r�δ − �2δ

2≈ r�δ.

A similar calculation gives the same for BDD′B′ and hence the area-preservation is proven.

Remark: By using an argument similar to the one in theorem 7.1 a result similar to thestring construction for inner billiards can be proven. In the case of outer billiards the theoremstates that given an outer billiard curve γ and a convex curve Γ that is invariant under the outerbilliard map T one can recover the curve γ from Γ. The idea is to �x a number c > 0 and thenconsider all line segments that intersects Γ by cutting out sections of area c from Γ. The curveγ is then taken to be the envelope of all these segments (see �gure 28).

Next we turn to periodic orbits for strictly convex and smooth outer billiards. A periodicorbit for an outer billiard table is a polygon circumscribed around the billiard curve γ such thateach side is bisected by its point of tangency at γ (see �gure 29). Like for inner billiards, periodicorbits for convex outer billiards can be found to be equivalent to extreme points of a functionalde�ned on the space of all circumscribing polygons. In this case the functional is the area of thepolygon. Note that for inner billiards it was the perimeter length of the inscribed polygon thatwas used.

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γ

Γ

Figure 28: String construction for outer billiards.

Figure 29: Example of a periodic orbit for outer billiards.

7.3 Bounded orbits

In this section we will brie�y account for an interesting question that arises in the context of outerbilliards; for which tables are the orbits bounded? This distinguishes the outer billiard systemsfrom the inner billiards systems since in the latter case the orbits are trivially bounded by thebilliard curve itself. This question has still not been resolved in the general case but solutionshave been given for two large classes of billiard curves: C5 curves and rational polygons. Thelatter is de�ned as all polygons that can be described as a�ne images of polygons whose verticeshave integer coordinates (remember that an a�ne transformation is a transformation of the formx 7→ ax+b where a and b are real numbers). We state the result for these classes of tables withoutproof.

Theorem 7.2. If the billiard curve γ is C5 and strictly convex then every orbit of the outerbilliard is bounded. This conclusion also holds if γ is a rational polygon.

Remark: The result for rational polygons can actually be made stronger by de�ning anotherclass of polygons called quasi-rational polygons. We refer to [2, p.158�160] for a further discussionof these polygons. As an example of a table for which associated outer billiard has unboundedorbits one can take the semi-circle. In [8] it is proved that for this particular system there existsorbits escaping to in�nity.

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7.4 Outer billiards and planetary motion

One of the reasons that outer billiards became popular was due to an interpretation by J. Moserof these systems as a model for planetary motion. To motivate this we consider a billiard curveγ and choose a starting point u �far away� from γ. Now taking the second iterate of the billiardmap T , T 2, of this point we can notice that the point seems to move in a centrally symmetricway with γ as center of motion (see �gure 30). More precisely one can prove that this motionsatis�es Kepler's second law; that the area per time unit that the positional vector of the movingpoint sweeps over is constant. For a further discussion of this topic see [2, p.155�157].

γ

u

T 2(u)

T 4(u)T 6(u)

Figure 30: Illustration of outer billiard movement as planetary motion.

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A Calculations

A.1 Notes for constructing level curves

g(ϕ)h(ϕ)

α

γ(ϕ)

Figure 31

There exists a well known integral of the billiard inside an elliptic table

x2

a2+y2

b2= 1

which is essentially a function of the cartesian position (x, y) of the billiard ball, and its cartesianvelocity (u, v) that is constant over every orbit. This integral is

b2xu+ a2yv

and is relatively straightforward to derive, see [2, p. 54].The basic idea is to transform the billiard integral xu/a2 + yv/b2 into the normal variables

of the billiard map of arclength s, and angle relative to tangent θ. Instead of using s directly weuse the traditional parametrizing variable ϕ such that the curve of the ellipse is spanned by

γ(ϕ) = (x, y) = (a cosϕ, b sinϕ)

where both variables x and y in the integral are reduced to sines and cosines. The relativedi�culty lies in expressing the velocity (u, v) in terms of ϕ and θ. To do this we propose thefollowing constructions expressing (u, v) in the form (