A PARITY PROBLEM ON THE FREE PATH LENGTH ... - Project Euclid

Functiones et Approximatio XXXV (2006), 19–36

A PARITY PROBLEM ON THE FREE PATH LENGTH OF A BILLIARD IN THE UNIT SQUARE WITH POCKETS Emre Alkan, Andrew H. Ledoan & Alexandru Zaharescu 1 Dedicated to Professor Eduard Wirsing on the occasion of his seventy-fifth birthday Abstract: We present a result on short intervals about the moments of the free path length of the linear trajectory of a billiard in the unit square with small triangular pockets of size ε removed at the corners, in which the trajectory ends in a specified corner pocket. Keywords: Billiards, periodic Lorentz gas, free path length, Farey fractions, visible points, Kloosterman sums.

1. Introduction and statement of results A variety of ergodic and statistical properties of the periodic Lorentz gas and billiards have been intensively studied in the last decades by a number of authors (see for example [8], [9], and [10]). Such systems were introduced by Lorentz [21] in 1905 to investigate the dynamics of electrons in metals. In [4] and [5], a problem on the length of the linear trajectory of a two-dimensional Euclidean billiard generated by the free motion of a single billiard ball subject to elastic reflections on the boundary of the unit square [0, 1]2 ⊂ R2 having vertices (0, 0), (1, 0), (1, 1), and (0, 1), with small pockets of size ε removed at the four corners, is considered. The billiard problem has the point mass moving from the origin along a geodesic line with constant speed and angle θ ∈ [0, π/2] , until it collides with the boundary. At a smooth boundary point, the billiard ball reflects so that the tangential component of its velocity remains the same, while the normal component changes its sign. The reflection is specular, and the trajectory between two such reflections is rectilinear. The motion ends when the billiard ball reaches one of the corner pockets. In [4], the k th moment of the free path length (also called the first exit time by some authors) is estimated for any subinterval of the interval [0, π/2]. The purpose of this paper is to see precisely what can be said about the contribution each corner 2001 Mathematics Subject Classification: Primary 37D50. Secondary 11B57, 11P21, 82C40. 1 Research of the third author is supported by NSF grant number DMS-0456615.

20

Emre Alkan, Andrew H. Ledoan & Alexandru Zaharescu

pocket will make to these moments. Our study is carried out within the framework of number theory provided in [4] and [5]. We consider separately the moments over subsets of the trajectories that end in each corner pocket and determine whether each pocket has asymptotically equivalent contribution to these moments. In order to differentiate between trajectories that end in the corner pockets, we will need to consider questions on the distribution of visible points in the coordinate plane, and hence Farey fractions, with parity constraints. The use of Farey fractions is strongly influenced by geometric ideas and further links the billiard problem to the distribution of inverses in residue classes, in which Kloosterman sums play an important role. For work done in this direction, see [15], [20], [14], [2], [3], [17], and [18]. For surveys of Farey fractions, see [12], [16], [11], and [7]. We begin by introducing some notations. Let lε (θ) denote the length of the trajectory of a particle moving with angle θ ∈ [0, π/2] from the origin. For i, j ∈ {0, 1}, let Ai,j be the set of angles θ for which the trajectory ends at the corner (i, j). For any k > 1, we write the k th moment of lε (θ) over the interval [0, π/2] as a sum of four integrals corresponding to the contribution made by each of the pockets at the corners (i, j) , that is, Z

π 2

0

X

lεk (θ) dθ =

i,j∈{0,1}

Z Ai,j

lεk (θ) dθ.

(1.1)

According to Theorem 1.2 in [4] (p. 305), the k th moment of lε (θ) over any fixed subinterval [α, β] ⊆ [0, π/4] is asymptotic to a constant depending on k , α , and β only, times ε−k . Thus, for any k , α , β , δ > 0, Z

β

α

Z lεk (θ) dθ = ck ε−k

β α

1 dx + Ok,δ (ε−k+ 6 +δ ), cosk x

(1.2)

where the constant ck is proved in Theorem 1.1 from [4] (p. 304) to be ck =

12 π2

Z

1 2

0

  1 − (1 − x)k 1 − (1 − x)k+1 x(xk−1 + (1 − x)k−1 ) + − dx. kx(1 − x) (k + 1)x(1 − x)

We will see that the k th moment of lε (θ) splits up asymptotically into three equivalent parts for each pocket at the corners (1, 0), (1, 1), and (0, 1) , while the pocket at the corner (0, 0) makes no contribution at all to these moments. More precisely, we will prove the following result. Theorem 1.1. For any subinterval [α, β] ⊆ [0, π/4], any k > 1 , any δ > 0 , and any ε > 0 , we have 

Z Ai,j ∩[α,β]

lεk (θ) dθ

=

0 ck ε−k 3

Rβ

dx α cosk x

if (i, j) = (0, 0), 1 + Ok,δ (ε−k+ 14 −δ ) otherwise.

On the free path length of a billiard in the unit square

21

2. Counting inverses with parity constraints In this section, we prove some lemmas that will be used in the proof of Theorem 1.1 in Section 3. An important tool employed in [1], [4], [5], and [6], to estimate sums over primitive lattice points is the Weil-Sali´e type [22] inequality 1

1

|K(m, n; q)|  σ0 (q) gcd(m, n, q) 2 q 2 ,

(2.1)

proved in [19] and [13], for complete Kloosterman sums   X mx + n¯ x K(m, n; q) = e q x (mod q) gcd(x,q)=1

in the presence of an integer albeit not necessarily prime modulus q . Here σ0 is the “number of divisors” function and x ¯ denotes the multiplicative inverse of x (mod q). The bound (2.1) is used to prove, for a fixed integer q > 2 and any subintervals I, J ⊂ [0, q), the estimate (see Lemma 1.7 in [1], p. 445) Nq (I, J) := |{(x, y) ∈ I × J: xy ≡ 1 =

(mod q)}| (2.2)

1 ϕ(q) |I||J| + Oδ (q 2 +δ ), q2

where ϕ stands for Euler’s totient function. Thus, the arithmetic problem concerning the number of solutions of the congruence xy ≡ 1 (mod q) , where (x, y) ∈ I × J , is reduced to the estimate of exponential sums. We remark that the ordinary incomplete Kloosterman sums   X mx + n¯ x KI (m, n; q) = e q x∈I gcd(x,q)=1

may be written in terms of the complete Kloosterman sums, so that the inequality (2.1) gives (see Lemma A2 in [6], p. 1823) 1

1

|KI (0, n; q)| δ gcd(n, q) 2 q 2 +δ . This bound is used to prove, for any integer j , the estimate (see Proposition A3 in [6], p. 1823) Nq,j (I, J) := |{(x, y) ∈ I × J: gcd(x, q) = 1, xy = j 1 1 ϕ(q) = 2 |I||J| + Oδ (q 2 +δ gcd(j, q) 2 ). q

(mod q)}| (2.3)

Estimates (2.2) and (2.3) allow us to immediately deduce the following key technical tool.

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Emre Alkan, Andrew H. Ledoan & Alexandru Zaharescu

Lemma 2.1. Let i, j ∈ {0, 1} and δ > 0. Assume that q > 1 is an integer such that q ≡ i (mod 2) and I, J ⊂ [0, q) . Denote by Vq,j (I, J) the number of pairs of integers (a, b) ∈ I × J for which ab ≡ 1 (mod q) and a ≡ j (mod 2). Then

where

Vq,j (I, J) := |{(a, b) ∈ I × J: ab ≡ 1 (mod q), a ≡ j 1 ηi,j ϕ(q) = |I||J| + Oδ (q 2 +δ ), 2q 2  if (i, j) = (0, 0), 0 ηi,j = 1 if (i, j) = (0, 1),  1/2 if (i, j) = (1, 0) or (1, 1).

(mod 2)}|

Proof. It is easily seen that Vq,0 (I, J) = 0, if (i, j) = (0, 0). Since Vq,0 (I, J) + Vq,1 (I, J) = |{(a, b) ∈ I × J: ab ≡ 1 (mod q)}|, if (i, j) = (0, 1) , then by (2.2) Vq,1 (I, J) =

1 ϕ(q) |I||J| + Oδ (q 2 +δ ). 2 q

If (i, j) = (1, 0) , then by (2.3) Vq,0 (I, J) = |{(x, y) ∈ I/2 × J: gcd(x, q) = 1, xy ≡ ¯2 (mod q)}| 1 ϕ(q) = Nq,¯2 (I/2, J) = |I||J| + Oδ (q 2 +δ ). 2q 2 Lastly, if (i, j) = (1, 1), then by (2.2) 1 ϕ(q) Vq,1 (I, J) = Nq (I, J) − Vq,0 (I, J) = |I||J| + Oδ (q 2 +δ ). 2q 2 This completes the proof of the lemma. Next, we note some important corollaries of Lemma 2.1. We give a detailed proof for Lemma 2.2 below, and indicate the modifications needed for Lemma 2.3. For each subintervals I, J ⊂ R and each C 1 function f : I × J → R , we denote kf k∞ = sup |f (x, y)|, (x,y)∈I×J

kDf k∞ =

sup (x,y)∈I×J

! ∂f ∂f (x, y) + (x, y) . ∂x ∂y

Lemma 2.2. Let i, j ∈ {0, 1} and δ > 0. Let ηi,j be as in Lemma 2.1. Assume that q, T > 1 are integers, q ≡ i (mod 2), and f : I × J → R is a C 1 function with I, J ⊂ [0, q) . Then ZZ X ηi,j ϕ(q) f (a, b) = f + Eq,I,J,f,T , 2q 2 (a,b)∈I×J ab≡1 (mod q) a≡j (mod 2)

where

I×J

  1 3 |I||J| Eq,I,J,f,T δ T 2 q 2 +δ kf k∞ + T kDf k∞ q 2 +δ + . T2

On the free path length of a billiard in the unit square

23

Proof. With the proof of Lemma 2.2 from [4] (pp. 309–310) as a guide, we observe on the one hand that if T > q , the error is larger than the sum to estimate. Hence, there is nothing to prove. On the other hand, if T < q , we approximate the function f (x, y) by a constant whenever (x, y) ∈ Ir × Js by partitioning the intervals I and J , respectively, into T intervals, I1 , . . . , IT and J1 , . . . , JT , of equal size |Ir | = |I|/T and |Js | = |J|/T . For each pair of indices (r, s), we choose a point (xrs , yrs ) ∈ Ir × Js for which ZZ f = |Ir ||Js |f (xrs , yrs ). (2.4) Ir ×Js

Now for (x, y) ∈ Ir × Js , the mean-value theorem gives  f (x, y) = f (xrs , yrs ) + O((|Ir | + |Js |)kDf k∞ ) = f (xrs , yrs ) + O

 q kDf k∞ , T

from which follows X

f (a, b) =

(a,b)∈I×J ab≡1 (mod q) a≡j (mod 2)

T X

X

r,s=1

(x,y)∈Ir ×Js xy≡1 (mod q) x≡j (mod 2)

T X

f (x, y)

"



qkDf k∞ = Vq,j (Ir , Js ) f (xrs , yrs ) + O T r,s=1

#

Since Ir , Js ⊂ [0, q) , Lemma 2.1 applies “infinitesimally” and produces Vq,j (Ir , Js ) =

1 ηi,j ϕ(q) |Ir ||Js | + Oδ (q 2 +δ ). 2q 2

In view of (2.4), the main term on the right side of (2.5) becomes T 1 ηi,j ϕ(q) X |Ir ||Js |f (xrs , yrs ) + Oδ (T 2 q 2 +δ kf k∞ ) 2q 2 r,s=1 ZZ 1 ηi,j ϕ(q) = f + Oδ (T 2 q 2 +δ kf k∞ ), 2q 2 I×J

whereas the error term there will be     1 3 qkDf k∞ ϕ(q) |I||J|  T 2 q 2 +δ + 2 |I||J|  T kDf k∞ q 2 +δ + . T q T2 Piecing this together gives the required result.

. (2.5)

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Emre Alkan, Andrew H. Ledoan & Alexandru Zaharescu

Lemma 2.3. Let i, j ∈ {0, 1} and δ > 0. Let ηi,j be as in Lemma 2.1. Assume that q, T1 , T2 > 1 are integers, q ≡ i (mod 2), and f : I ×J → R is a C 1 function with I, J ⊂ [0, q) . Then ZZ X ηi,j ϕ(q) f (a, b) = f + Eq,I,J,f,T , 2q 2 (a,b)∈I×J ab≡1 (mod q) a≡j (mod 2)

where

Eq,I,J,f,T δ T1 T2 q

1 2 +δ

I×J

  3 |I||J| +δ 2 kf k∞ + (T2 kDx f k∞ + T1 kDy f k∞ ) q + . T1 T2

Proof. Only the case T < q is interesting. We approximate the function f (x, y) by a constant whenever (x, y) ∈ Ir × Js by partitioning the intervals I and J , respectively, into T1 intervals I1 , . . . , IT1 and T2 intervals J1 , . . . , JT2 of equal size |Ir | = |I|/T1 and |Js | = |J|/T2 . For (x, y) ∈ Ir × Js , by the mean-value theorem X

f (a, b) =

(a,b)∈I×J ab≡1 (mod q) a≡j (mod 2)

=

T1 X T2 X

X

r=1 s=1

(x,y)∈Ir ×Js xy≡1 (mod q) x≡j (mod 2)

T1 X T2 X

f (x, y)

"



Vq,j (Ir , Js ) f (xrs , yrs ) + O

r=1 s=1

!#  1 1 + qkDf k∞ . T1 T2

By virtue of Lemma 2.1, the main term is T

T

1 X 2 1 ηi,j ϕ(q) X |Ir ||Js |f (xrs , yrs ) + Oδ (T1 T2 q 2 +δ kf k∞ ) 2q 2 r=1 s=1 ZZ 1 ηi,j ϕ(q) = f + Oδ (T1 T2 q 2 +δ kf k∞ ), 2q 2

I×J

while the error term is



1 1 + T1 T2



ϕ(q)  qkDf k∞ T1 T2 q + 2 |I||J| q   3 |I||J| +δ 2  (T2 kDx f k∞ + T1 kDy f k∞ ) q + . T1 T2



1 2 +δ

Hence, the required result follows obviously. We will need one other key summation formula. Lemma 2.4. Let i ∈ {0, 1} . Assume that 0 < a < b are real numbers and f is a continuous piecewise C 1 function on the interval [a, b]. Then  ! Z b Z b X ϕ(k) %i 0 |f | , f (k) = f + O log(2 + b) kf k∞ + k ζ(2) a a a max(q , q ) ⇐⇒ 6 1 and =0 q q ⇐⇒ ⇐⇒

Let

 Jq,2 =

and observe that

Q+a ¯ < 2q and Q − a ¯ 2Q/3.

(Q − q, 2q − Q) if q > 2Q/3, ∅ if q 6 2Q/3,



 Q−a ¯ (i) q < q ⇐⇒ = 0 , so that q 00 = a ¯;  q  Q+a ¯ (ii) q 0 < q ⇐⇒ = 1 , so that q 0 = q − a ¯. q Thus, we have   X X k 1 1 ε ε Si,j,2 = q2 + − − (q 2 + a2 ) 2 −1 q(q − a ¯) q¯ a q−a ¯ a ¯ 00

2Q/3