COEXISTENCE OF PERIOD 2 AND 3 CAUSTICS FOR DEFORMATIVE NEARLY CIRCULAR BILLIARD MAPS

. For Z 2 − symmetric analytic deformation of the circle (with certain Fourier decaying rate), the necessary condition for the corresponding billiard map to keep the coexistence of period 2 , 3 caustics is that the deformation has to be an isometric transformation.


Introduction
Suppose Ω ⊂ R 2 is a strictly convex domain, with the boundary ∂Ω is C r smooth, r ≥ 2. The billiard problem inside Ω can be described as the following: A massless particle moves with unit speed and no friction following a rectilinear path inside the domain Ω.When the ball hits the boundary, it is reflected elastic-ally according to the law of optical reflection: the angle of reflection equals the angle of incidence.
This problem was first investigated by Birkhoff (see [3]).Later we can see that such trajectories are called broken geodesics, as they correspond to local maximizers of the distance functional.The billiard map can be identified by the correspondence of the positions in one reflection φ : P 0 → P 1 , see Fig. 1.Now let's introduce a coordinate for the billiard map.Suppose that ∂Ω is parametrized by arc length s with the circumference of ∂Ω equal to 1, and v is the angle between P 1 − P 0 and the positive tangent to ∂Ω at P 0 .Then the billiard map can be described as: defined on the closed annulus A = [0, 1] × [0, π].Obviously φ| ∂A = Id, i.e. φ(s, 0) = (s, 0), φ(s, π) = (s, π), ∀s ∈ T.
Let's make a rule for s ∈ T = R/[0, 2π]: we always choose the counter clockwise direction to order the configurations, that means for any q−tuple (s 0 , s 1 , • • • , s q−1 ) with s i ∈ T, q ≥ 2 and 0 ≤ i ≤ q − 1, we can fix a unique configuration (x 0 , x 1 , • • • , x q−1 ) in the universal covering space R, such that x i ≡ s i (mod 2π) for 0 ≤ i ≤ q − 1, 0 ≤ x i+1 − x i < 1 for 0 ≤ i ≤ q − 2, x 0 ∈ [0, 1).This unique configuration implies the following definition: Definition 1.1.For a q−periodic tuple S = (s 0 , s 1 , • • • , s q−1 , s q ) with s 0 = s q , we define the winding number of it by p := x q − x 0 2π ∈ Z + and the rotation number by ρ(S) = p q , where X = (x 0 • • • , x q ) is the configuration corresponding to S defined above.
Then the following action function is well defined F p/q (s) := inf γ∈C p/q (s) and the minimizer γ * obeys the discrete Euler Lagrange equation Moreover, due to (3) the corresponding {(s i , v i )} q i=0 is an orbit of the billiard map φ, see [2].
Remark 1.3.Notice that {(s i , v i )} q i=0 may not be a 'real' periodic orbit, because v 0 = v q could happen.However, if we interpret F p/q (s) as a function defined on T, then the minimizer s * will definitely correspond to a 'real' periodic orbit γ * of the billiard map.Definition 1.4.We call a (possibly not connected) curve Γ ⊂ Ω a caustic if any billiard orbit having one segment tangent to Γ has all its segments tangent to it.Precisely, Γ p/q is an p/q−rational caustic if all the corresponding (noncontractible) tangential orbits are periodic with the rotation number p/q.Remark 1.5.In the remaining part of this paper, we agree that all caustics that we will consider will be smooth and convex; we will refer to such curves simply as caustics.By the Birkhoff 's Theorem of general exact monotone twist maps, such a rational caustic will correspond to a φ−invariant curve in the phase space, i.e. formally we can express by Γ p/q = {(s, g p/q (s)) ∈ A|s ∈ T} which consists of periodic orbits of ratation number p/q.Moreover, Γ p/q is non-contractible and then g p/q (s) is a Lipschitz graph, see [3].
Besides, there is reversibility of the billiard map φ, i.e.
Benefit from this, we can see that Γ p/q = Γ 1−p/q .So in the following we can impose that the rotation number of the caustic belongs to [0, 1/2].
Theorem 1.6.[6] For an exact symplectic twist map, every orbit on a non contractible invariant curve is minimizing in the sense of variation.
By applying this Theorem to the billiard map, we get Corollary 1.7.Suppose the billiard map φ has a p/q−rational caustic, then F p/q (s) equals a constant for all s ∈ T.
Remark 1.8.In [4], the authors defined the width function w(α) of a convex billiard curve by the strip width formed by the tangent rays with the direction α + π/2 and α − π/2, for any unit vector α in R 2 .We called the billiard boundary of constant width, if w(α) is a constant.Observe that billiard boundaries of constant width has 1/2 caustic.
The circle is a trivial example of constant width, of which the 1/2 caustic is the centre point.A popular non trivial example is the Releaux triangle, see [10].
We should remind the readers that elliptic boundary has no 1/2 caustic.Indeed, the elliptic billiard has a first integral, see [8], but the 1/2 periodic orbits just correspond to the rebound along the major (reps.minor) axis, which have separatrix arcs containing homoclinic orbits but not periodic ones.
Organization of this article.In Section 2 we state the rigid constraints on the boundaries to preserve certain rational caustics; we introduce the Projection Theorem and our main conclusion towards it; In Section 3 we get the 1 st and 2 nd order estimate of the action function, which leads to our harmonic equation; In Section 4 we analysis the harmonic equation and give the proof of our main conclusion.In Section 5 we make some heuristic comments and further generalizations of this direction.

Main conclusion and Scheme of the proof
Based on previous section's setting, a natural question can be asked: How far can we constraint the boundary to preserve certain rational caustics?Recently, Avila, de Simoi and Kaloshin proved the following result: Theorem 2.1.[1] There exist e 0 > 0 and ε 0 > 0 such that for any 0 ≤ e ≤ e 0 , 0 ≤ ε < ε 0 , any rationally integrable C 39 −smooth domain Ω so that ∂Ω is C 39 ε−close to the ellipse E e with the eccentricity e is also an ellipse.
This implies that locally ellipse with small eccentricity is the only possible integrable convex billiard boundaries.The proof of this Theorem relies on an improved 1 st order estimate of the action function.To help the readers to get a clearer understanding of that, let us start by exploring the integrable infinitesimal deformations of a circle.Suppose Ω 0 be the unit disk, and the polar coordinates on the plane be (r, θ).Let Ω ε be a one-parameter family of deformations given by ∂Ω ε = {r = 1 + εn(θ) + O(ε 2 )}, where the Fourier expansion of n: The following Theorem is available: Theorem 2.2 (Ramirez-Ros [9]).If Ω ε has an integrable rational caustic Γ 1/q of rotation number 1/q for all sufficiently small ε, then n kq = n kq = 0 for all k ∈ N.
If we impose that Ω ε is rationally integrable for all 1/q with q > 2 and sufficiently small ε, then the above theorem implies that n k = n k = 0 for all k > 2, i.e. we establish n(θ) by: Remark 2.3.As in [1], we can find that in (6) n 0 corresponds to a homothety, n * 1 corresponds to a shift in the direction θ 1 and n * 2 corresponds to a deformation into an ellipse of small eccentricity with the major axis coincides with the direction θ 2 .
It's remarkable that in the above theorem, one have to face ε → 0 as q → ∞.So they need to replace the infinitesimal deformation by a fixed lower bound such that ε > ε 0 ; You can see [1] for more technical details.
A natural generalization of previous Theorem is reducing to 'finitely many' rational caustics, based on which we can still get the integrability of the billiard maps: The Projected Conjecture.In a C r (r = 2, • • • , ∞, w) neighborhood of the circle there is no other billiard domain of constant width and preserving 1/3 caustics.Definition 2.4.Denote by BZ q the manifold of codimension infinity containing all the strictly convex billiard boundaries preserving the 1/q caustic Γ q , q ≥ 2.
Here we propose a possible approach to prove it, still we start from the infinitesimal deformation of the circle: Step 1. Find the tangent bundle of BZ q at the circle.Let n(θ) and m(θ) be functions with θ ∈ T 1 = R/[0, 2π] given by the Fourier series In the polar angle coordinate, for sufficiently small ε the perturbed domain can be denoted by Denote by T q the tangent space of BZ q at the circle, and T ⊥ q be the orthogonal complementary of T q .Due to Theorem 2.2, T q consists of functions given by Fourier coefficients whose indices are not divisible by q.
Remark 2.5.Notice that ε 0 (n) may be not uniform for different n ∈ C r (T, R), even though n C r = 1 is imposed.So we need a similar approach as [1] to avoid the the collapse of the infinitesimal ε 0 (•).To keep the consistency and readability, we don't consider this part in this paper.
Following aforementioned strategy, we claim our main conclusion.Before we doing that, let's formalize the symbol system first: • γ 0 ∈ R 2 be the unit circle and γ ε ∈ R 2 be a deformation, and in the polar coordinate r 0 , r ε represent the corresponding axis length.
• s ∈ [0, 2π] be the arc length variable and θ ∈ [0, 2π] be the rotational angle; • Suppose γ t (θ) is a curve of strictly convex boundaries in R 2 with the parameter t ∈ [0, 1] and starting from the circle, i.e. γ 0 (θ) = 1.If γ t (θ) is C 3 smooth of t, then we can expand the curve in the polar coordinate by We can assume n, m ∈ C w (T, R, ρ), ρ > 0.Moreover, by rescaling t, we can make • Recall that two different deformation γ t and γ t may be homogenous by a rigid transformation on R 2 space (parallel shift, rotation), so we just need to choose a representation by imposing • Besides, we can fix the perimeter by 2π for all the deformations, i.e.
with the modular function w : and there exists a subsequence {k i } ∞ i=1 , such that then there exists ε 0 (n) such that for all 0 < ε ≤ ε 0 , the deformed boundary couldn't persist the coexistence of 1/2, 1/3-caustics.
Proof.This conclusion is proved in Section 4, Theorem 4.2.
Proof.This conclusion is proved in Section 4, Theorem 4.12.
Let's define the action function of q−periodic configuration by P q (θ, Ω ε ) = maximal perimeter of a q − gon inscribed into the domain Ω ε starting and ending at θ, if we consider the arc length variable s as a function of the angle variable θ.Recall that s ∈ [0, 2π] → θ ∈ [0, 2π] is a diffeomorphism via the following: for some c, where is the averaging of 1/q−frequency of m(θ).
Proof.See section 3 for details.

Consider the Fourier expansion of
then the following property holds: Lemma 2.9.In the above notations, if Recall that m (q) (θ) = l m ql exp(iql • θ) contains only harmonics divisible by q, that means D q (θ, ε) has to contain only harmonics divisible by q as well, i.e.
2.1.Obstruction to coexistence of two rational caustics.In particular, for q = 2, 3 we want to show that for all n ∈ (T 2 ∩ T 3 ) \ 0 the functions 3 √ 3 4 D 2 (θ, ε) and D 3 (θ, ε) have at least one Fourier harmonic divisible by 6 whose coefficients are different.Once we did this, the contradiction with the action functions will lead to our main conclusion.Proof.Indeed, If we take q = 2 and p = 3, then the necessary condition to preserve 1/2, 1/3 caustics is that P 2 , P 3 should be both constant, which leads to Once again we can take the averaging and get 3 (θ).This is a necessary condition for boundary Ω ε to preserve both 1/2 and 1/3 caustics.
The Fourier coefficients of D q (θ) are obtained through a convolution of Fourier coefficients of n.We will compute the corresponding formula in the next section.

Evaluation of P q action function
Let's start with the circle Ω 0 and the q-gon for some q ≥ 2. Suppose the deformation in the polar coordinate has the form Consider a q-perimeter function P ε (θ, Ω ε ) as defined above.For ε = 0 we have Let (z 0 0 , . . ., z 0 q−1 ) be the right q-gon, i.e. θ 0 k = θ 0 0 + 2kπ/q.For small ε we compute We postpone the computation of η k and ξ k .Consider the k-th edge between z k and z k+1 .
Taking a dot product of z ε k − z ε k+1 with itself we have Summing over k we get , where z ε 0 /|z ε 0 | = e iθ = z 0 0 /|z 0 0 | (for convenience we can interpret the unit vector in R 2 as a complex number in C, this greatly simplifies our formulas without confusion).Also we can denote the unit vector | by e k for short.
Recall that z ε q = z ε 0 and z 0 q = z 0 0 , which leads to η 0 = η q and ξ 0 = ξ q .Combining e k • η k and −e k−1 • η k and observing that (e k − e k−1 ) is the outer normal vector to the boundary ∂Ω 0 at z 0 k , then we have ) or, equivalently, the component of η k (resp.ξ k ) normal to the boundary at z 0 k .
3.1.The leading term in the case of the circle.Consider the leading term of the expansion for q ≥ 2.Here the bold n(•) is the normal vector, not the 1 st jet of the deformation !Then we get This is because and ), which leads to the 1 st order equation: Lemma 3.1.The 1 st order estimate of P q (θ, Ω ε ) obeys P q (θ, Ω ε ) = P q (θ, Ω 0 ) + 2εq sin π q n (q) (θ) + O(ε 2 ), q ≥ 2.

3.2.
The second order in the case of the circle.In order to get a 2 nd order estimate for P q (θ, Ω ε ), we need to get the exact expression of η k and ξ k .Due to (14), we get and Recall that θ 1 0 = θ 1 q = 0, then On the other side, where we used the estimate Then we turn back to (15) and get Simplify it: we finally get a triple-diagonal linear equation group: with θ 1 0 = θ 1 q = 0.This is a general formula holding for all q ≥ 3.
Remark 3.2.For q = 2, θ 1 1 = n (θ + π) − n (θ), θ 1 0 = 0. Mention that in this case (19) is invalid, but we can use that n ε (θ ε 0 ) n ε (θ ε 1 ).For q = 3, we can solve (19) by Recall that which is solved from (19).Notice also that and Substitute them in P q and we get the ε 2 -term by Substituting for some computable matrix A q×q , B q×q , C q×q we have Recall that m (q) (s) = k∈Z m kq exp(iqks) and n(s) = k∈Z n k exp iks, we have In the Appendix you can find a detailed calculation of D (2) and D (3) .
From previous corollary we also get the following property: Lemma 3.6.Let's simplify the notation by Proof.This is a direct arithmetic observation.

the harmonic analysis for n ∈ T 2 ∩ T 3
In this section we analyze the harmonic behaviour of (24).Since n ∈ T 2 ∩ T 3 , then the Fourier coefficients satisfy n 2l = n 3l = 0, ∀l ∈ Z due to Lemma 3.1.Moreover, suppose the Fourier expansion of D q (θ) satisfies then from the second order estimate we get the following harmonic equalities: for all l ∈ Z, As long as one of previous equalities is failed for non trivial n(θ), we would get γ ε / ∈ BZ 2 ∩ BZ 3 and prove the Projected Theorem.Proof.This is a direct conclusion from Corollary 3.5.4.0.1.the polynomial case: pyramid type harmonic equations.Suppose n(θ) is a trigonometric polynomial with the degree be N , i.e.
Proof.We can prove this by contradiction.Suppose 1/2, 1/3 caustics coexist, then (30) should hold.Without loss of generality, we can assume 6P + 1 is the largest integer in Z 2,3 which doesn't exceed N , due to Lemma 3.6, n 6p+1 • n 6q−1 = 0, ∀q, p ∈ Z, −P ≤ q, p ≤ P. Recall that n(θ) is even, so n k = n −k for all k ∈ Z. Due to the claim, there will be only one couple (−a, a), such that n −a • n a = 0.But from (9) we know that n a + n −a = 0 should hold simultaneously.This implies that n a = n −a = 0 and n(θ) = 0. 4.0.2. the general analytic case: avalanche caused by a quantitative control of error terms.In this section we generalize the idea in Theorem 4.2, and prove a similar result for general analytic n(θ).Denote by C w (T, R, ρ) the set of all analytic functions with radius ρ, then it's a Banach space under the analytic norm • ρ .The following estimate of the Fourier coefficients holds: Notice that previous Lemma is not always the optimal estimate of the Fourier coefficients for all functions in C w (T, R, ρ), so we can use the following procedure to find the slowest decaying coefficient sequence, and define the corresponding modular function.
For n(θ) = n k e ikθ consisting of infinitely many terms, since Lemma 4.3 is available, then we can pick k 1 ∈ Z be the index satisfying If there are several candidate index, we can choose the one with the greatest absolute value.Based on the same principle, we can choose k 2 ∈ Z, such that Repeat this process we can get a sequence {k i } ∞ i=1 , and the corresponding coefficient sequence {n ki } ∞ i=1 is just the slowest decaying coefficient sequence of n(θ).Definition 4.4.We call a smooth decreasing, positive function w : [0, +∞) → R + the modular function of n(θ), if w(|k i |) = |n ki | for all i = 1, 2, • • • .Remark 4.5.Notice that the modular function w(x) is not uniquely defined, but any two modular functions w 1 , w 2 corresponding to the same n(θ) should satisfy: Moreover, for n(θ) ∈ C w (T, R, ρ), For any L ∈ {k i } ∞ i=1 , which is a positive subsequence of {k i } ∞ i=1 , there exists a maximal P ∈ Z + , such that L = 6P + 1 or 6P − 1.
Without loss of generality, we just need to consider the first case.The following approximated equations can be derived from (♦): Here we use E ∼ O(e −w(L) L 2 ) is a Fourier reminder term.We can use the notation c (2)−(3) (k, l) = c (2) (k, l) − c (3) (k, l) for short.
. Here π i is the coordinate projection to the corresponding component, i = 1, 2.
We can easily prove Lemma 4.6 and Lemma 4.9 from observation.To prove the following Lemma, let's assume lim i→∞ w(i) Proof.Because n(θ) is even, so the modular function w(i) is even for i ∈ Z and we just need to prove this Lemma for 0 < K ≤ 2P .Let's start from the top level, i.e.K = 2P , if we denote by to the pyramid structure of (32).Iterate this inequality we get (33) for all 0 < K ≤ 2P .Due to the symmetry of n(θ) we can generalize to −2P ≤ K < 0. Recall that n(θ) is even and obeys (9), and let L → 0, |n(θ)| C 0 → 0, which contradicts with the assumption |n| C 0 = 1.

Further comments and heuristic improvement
Here we list some facts that guides our further exploration towards the Projected Conjecture.Recall that ( 19) and ( 23) supply as a universal formula for all q ≥ 2, so does Lemma 2.8 and Lemma 2.9.That gives us chance to propose a similar Conjecture: the Elliptic Projected Conjecture: In a C r (r = 2, • • • , ∞, w) neighborhood of the circle there is no other billiard domain preserving both 1/3 and 1/5 caustics.
The strategy to prove this elliptic conjecture is more or less the same with the case of 1/2 and 1/3 caustics.But instead, some arithmetic properties related with the harmonics of n(θ) will be changed, including the exact form of (32).
Another aspect we could do is to generalize our main conclusion to general analytic function space C w (T, R, ρ), or even finite smooth space C r (T, R).The crucial lies in (32), which is a homogeneous quadratic equation group of pyramid type.If we can make a better error control as solving it, we can reduce the decaying speed of the modular function w(i).Some evidence indeed impies so: Here we just give the essence for the proof: the condition we impose is actually for reducing the dimension of (32).The similar idea holds, if we impose that more caustics preserve: Que: In a C r (r = 2, • • • , ∞, w) neighborhood of the circle there is no other billiard domain of constant width, and preserving 1/3, 1/5 caustics.
Que: For any decreasing rational sequence {1/q i } ∞ i=1 with q 1 = 2, there exists a neighborhood of the circle, such that there is no other billiard domain preserving {1/q i } ∞ i=1 caustics.
The former question can be generalized to the case with finitely many caustics preserved; As for the latter one, it can be considered as a generalization of the Theorem in [1].

Figure 1 .
Figure 1.The reflective angle keeps equal to the incident angle for every rebound.

Remark 4 . 7 .
From previous analysis, we can extra get N L ♦ (K) = N L ♦ (−K).Now let's explore the mechanism how the variables n k , n l relate with each other for N L ♦ (K): Definition 4.8.We define the generation of Z 2,3 ∩ [−L, L] by G(k) := P + 1 − max |k| 6 , |k|