New York Journal of Mathematics
New York J. Math. 9 (2003) 303–330.
The average length of a trajectory in a certain
billiard in a flat two-torus
F. P. Boca, R. N. Gologan, and A. Zaharescu
Abstract. We remove a small disc of radius ε > 0 from the flat torus T2 and
consider a point-like particle that starts moving from the center of the disk
with linear trajectory under angle ω. Let τε (ω) denote the first exit time of the
particle. For any interval I ⊆ [0, 2π), any r > 0, and any δ > 0, we estimate
the moments of τε on I and prove the asymptotic formula
1
τεr (ω) dω = cr |I|ε−r + Oδ (ε−r+ 8 −δ )
as ε → 0+ ,
I
where cr is the constant
12
π2
1/2
x(xr−1 + (1 − x)r−1 ) +
0
1 − (1 − x)r
1 − (1 − x)r+1
−
rx(1 − x)
(r + 1)x(1 − x)
dx.
A similar estimate is obtained for the moments of the number of reflections in
the side cushions when T2 is identified with [0, 1)2 .
Contents
1. Introduction and main results
2. Farey fractions and Kloosterman sums
3. The second moment of the first exit time for cross-like scatterers
4. The rth moment of the first exit time for cross-like scatterers
5. The moments of the number of reflections
6. The case of circular scatterers
References
303
307
310
314
321
324
329
1. Introduction and main results
For each 0 < ε <
1
2
we consider the region
Zε = {z ∈ R2 ; dist(z, Z2 ) > ε}
Received April 29, 2003.
Mathematics Subject Classification. 11B57; 11P21; 37D50; 58F25; 82C40.
Key words and phrases. Periodic Lorentz gas; average first exit time.
All three authors were partially supported by ANSTI grant C6189/2000.
ISSN 1076-9803/03
303
304
F. P. Boca, R. N. Gologan, and A. Zaharescu
and the first exit time (also called free path length by some authors)
τε (z, ω) = inf{τ > 0 ; z + τ ω ∈ ∂Zε },
z ∈ Zε , ω ∈ T,
of a point-like particle which starts moving from the point z with linear trajectory,
velocity ω, and constant speed equal to 1. This is the model of the periodic twodimensional Lorentz gas, intensively studied during the last decades (see [2], [7],
[8], [9], [10], [11], [12], [13], [14], [15], [16], [17], [18], [20], [21], [29], [31], [32] for a
non-exhaustive list of references). The phase space of the system consists in the
range of the initial position and velocity and is one of the spaces Yε × T with the
normalized Lebesgue measure, or Σ+
ε = {(x, y) ∈ ∂Yε × T ; ω · nx > 0} with the
normalized Liouville measure.
Equivalently, one can consider the billiard table Yε = Zε /Z2 obtained by removing pockets of the form of quarters of a circle of radius ε from the corners. The
reflections in the side cushions are specular and the motion ends when the point-like
particle reaches one of the pockets at the corners. In this setting τε (z, ω) coincides
with the exit time from the table (see Figure 1).
This paper considers the situation where the trajectory starts at the origin O =
(0, 0). In this case the phase space only consists in the range of the initial velocity
of the particle. It is given by the one-dimensional
torus T and can be reduced,
for obvious symmetry reasons, to the interval 0, π4 . From the point of view of
Diophantine approximation this corresponds to a homogeneous problem. We shall
be concerned with estimating the moments of the first exit time τε (ω) = τε (O, ω)
as ε → 0+ when the phase space is the range [0, π4 ] of the velocity ω. This question
was raised by Ya. G. Sinai in a seminar at the Moskow University in 1981. We
answer the question by supplying asymptotic formulas with explicit main term and
error for all the moments of τε in short intervals as follows:
Theorem 1.1. For any interval I ⊆ [0, π4 ] and any r, δ > 0, one has
1
2
Or,δ (ε 8 −δ ) if r =
r
r
τε (ω) dω = cr |I| +
as ε → 0+ ,
ε
1
−δ
4
Or,δ (ε
) if r = 2
I
where
12
cr = 2
π
1/2
r−1
x(x
0
+ (1 − x)
r−1
1 − (1 − x)r+1
1 − (1 − x)r
−
)+
rx(1 − x)
(r + 1)x(1 − x)
The mean free path length is in this case
4
π
π/4
c1
12 ln 2
0.421383
τε (ω) dω ∼
= 2·
≈
.
ε
π
2ε
ε
0
Note also that
12
lim+ cr = − 2
π
r→0
1/2
0
ln(1 − x)
dx = 1.
x(1 − x)
dx.
Length of a trajectory in a certain billiard
305
To prove Theorem 1.1 we first replace the circular scatterers by cross-like scatterers [m − ε, m + ε] × {n} ∪ {m} × [n − ε, n + ε], m, n ∈ Z2 \ {(0, 0)}.1 We denote
by lε (ω) the free path length in this situation, and first prove:
Theorem 1.2. For any interval I ⊆ [0, π4 ] and any r, α, δ > 0, one has
1
Or,δ (ε 2 −2α−δ + |I|εα ) if r = 2
dx
r
r
+
ε
as ε → 0+ .
lε (ω) dω = cr
1
cosr x
if r = 2
Oδ (ε 2 −δ )
I
I
We consider the probability measures μ
Iε and μIε on [0, ∞), defined by
1
1
f ε
τε (ω) dω, μIε (f ) =
f εlε (ω) dω,
f ∈ Cc ([0, ∞)).
μ
Iε (f ) =
|I|
|I|
I
I
√
Their supports are all contained in [0, 2] as a result of Lemma 3.1. Moreover, we
infer from Theorems 1.1 and 1.2 that their moments of order n ∈ N∗ are of the
form2
1
εn
1
On,δ (ε 8 −δ );
μ
Iε (X n ) =
τεn (ω) dω = cn +
|I|
|I|
I
n 1
1
dx
ε
cn
On,δ (ε 6 −δ ).
+
lεn (ω) dω =
μIε (X n ) =
n
|I|
|I|
cos x |I|
I
I
These asymptotic formulas show in particular that μ
Iε (X n ) and μIε (X n ) converge to
the main terms as ε → 0+ . The Banach-Alaoglu and Stone-Weierstrass theorems
now lead to:
√
Corollary 1.3. There exist probability measures μ
and μ
I on [0, 2] such that
μ
Iε → μ
and
μIε → μI
weakly as ε → 0+ .
Moreover, the moments of μ
and μI are
∞
tn d
μ(t) = cn
0
and respectively
∞
tn dμI (t) =
0
cn
|I|
dx
.
cosn x
I
ε (ω) in the side
Besides, we estimate the average of the number of reflections R
cushions of the billiard table in the case of circular scatterers and prove:
Theorem 1.4. For any interval I ⊆ [0, π4 ] and any r, δ > 0, one has
εr (ω) dω = cr (sin x + cos x)r dx + Or,δ (ε 18 −δ ) as ε → 0+ .
εr R
I
I
1 Actually it is not hard to see that for ω ∈ [0, π ] the result for cross-like scatterers is asymp4
totically the same as when using vertical scatterers {m} × [n − ε, n + ε].
2 We denote N = {0, 1, 2, . . . } and N∗ = {1, 2, 3, . . . }.
306
F. P. Boca, R. N. Gologan, and A. Zaharescu
ε
ε
ε
ε
ω
ε
ε
Oε
ε
Figure 1. The trajectory of the billiard
Again, we first consider the case of cross-like (or vertical) scatterers, let Rε (ω)
denote the number of reflections in the side cushions of the billiard table in this
case, and prove:
Theorem 1.5. For any interval I ⊆ [0, π4 ] and any r, α, δ > 0, one has
1
εr Rεr (ω) dω = cr (1 + tan x)r dx + Or,δ (ε 2 −2α−δ + |I|εα )
as ε → 0+ .
I
I
We may also consider the probability
ε and εRε ,
with the random variables εR
1
ε (ω) dω, ν I (f ) =
νεI (f ) =
f εR
ε
|I|
I
measures νεI and νεI on [0, ∞) associated
and defined by
1
f εRε (ω) dω,
f ∈ Cc ([0, ∞)).
|I|
I
From Theorems 1.4 and 1.5 we derive:
√
Corollary 1.6. There exist probability measures νI and νI on [0, 2] such that
νεI → νI
and
νεI → ν I
as ε → 0+ .
Moreover, the moments of νI and ν I are
∞
cn
n
I
t d
ν (t) =
(sin x + cos x)n dx,
|I|
0
I
and respectively
∞
tn dν I (t) =
0
cn
|I|
(1 + tan x)n dx.
I
one gets formulas similar to the ones in Theorems 1.1, 1.2,
In the case I ⊆
1.4 and 1.5 after performing a symmetry with respect to a diagonal of the square,
i.e., replacing (α, β) by ( π2 − β, π2 − α).
The proofs make use of techniques employed in the study of the spacings between
Farey fractions, pioneered in [23], [24], [25], and furthered recently in [3], [4], [1],
[ π4 , π2 ]
307
Length of a trajectory in a certain billiard
[28] where estimates for Kloosterman sums are being used. The first step consists
in proving Theorems 1.2 and 1.5, which refer to the case of cross-like or vertical
scatterers. In this case one can directly take advantage of the fact that the intervals
a+ε
a
1
Ia/q = [ a−ε
q , q ], with q Farey fraction of order Q = [ ε ], provide a covering of
[0, 1] such that two intervals Ia/q and Ia /q overlap if and only if aq and aq are
consecutive Farey fractions of order Q.
Finally, the case of circular scatterers is settled by partitioning the range I into
[ε−θ ] intervals of equal size for a convenient value of the exponent θ, and replacing
the small circles of radius ε first by vertical scatterers of type {m}×[n−ε− (m, n), n+
ε, n+
ε] for appropriate choices
ε+ (m, n)], and finally by scatterers of type {m}×[n−
of ε± (m, n) and ε.
It should be possible in theory to compute the densities of the limit measures from
their moments using either the Cauchy transform or the inverse Mellin transform.
An attempt of this kind does not seem to easily lead however to a tractable formula
for these densities. The convergence of the measures μ
Iε and νεI was proved in a
different way and the limit measures were explicitly computed in [5].
Techniques using Farey fractions and Kloosterman sums were recently used in
[6] to establish the existence, and compute the distribution, of the free path length
for the periodic two-dimensional Lorentz gas in the small-scatterer limit in the case
where the trajectory does not necessary start from the origin, and one averages
over both initial position and initial velocity.
This is the final version of the paper with the same title, circulated as preprint
arXiv math.NT/0110208.
2. Farey fractions and Kloosterman sums
For each integer Q ≥ 1, let FQ denote the set of Farey fractions of order Q, i.e.,
irreducible fractions in the interval (0, 1] with denominator ≤ Q. The number of
Farey fractions of order Q in an interval J ⊆ [0, 1] can be expressed as
#(J ∩ FQ ) =
Recall that if
a
q
<
a
q
Q2 |J|
+ O(Q ln Q).
2ζ(2)
are two consecutive elements in FQ , then
a q − aq = 1
and
q + q > Q.
Conversely, if q, q ∈ {1, . . . , Q} and q + q > Q, then there are a ∈ {1, . . . , q − 1}
and a ∈ {1, . . . , q − 1} such that aq < aq are consecutive elements in FQ . Proofs
of these well-known properties of Farey fractions can be found for instance in [26],
[23], [30].
<
>
, and respectively by FQ
, the set
Throughout the paper we shall denote by FQ
of pairs ( aq , aq ) of consecutive elements in FQ with q < q , and respectively with
308
F. P. Boca, R. N. Gologan, and A. Zaharescu
q > q . We also set
Z2pr = {(a, b) ∈ Z2 ; gcd(a, b) = 1};
J
J
=
and
=
a/q
<
(a/q,a /q )∈FQ
a/q∈J
a/q
;
>
(a/q,a /q )∈FQ
a/q∈J
ΔQ = {(x, y) ∈ Z2pr ; 0 < x, y ≤ Q, x + y > Q};
Rm,n = [m, m + 1] × [n, n + 1],
m, n ∈ R.
For each region R in R and each C function f : R → C, we denote
∂f
∂f
f ∞,R = sup |f (x, y)|, Df ∞,R = sup
∂x (x, y) + ∂y (x, y) .
2
1
(x,y)∈R
(x,y)∈R
The notation f g means the same thing as f = O(g); that is, there exists
an absolute constant c > 0 such that |f | ≤ cg for all values of the variable under
consideration. When the constant depends on a parameter δ, this dependence will
be indicated by writing f δ g. The notation f g will mean that f g and
g f simultaneously.
We shall be mainly interested in consecutive Farey fractions aq < aq in FQ with
the property that, say, aq belongs to a prescribed interval J ⊆ [0, 1]. The equality
a q −aq = 1 yields a = q − q¯ , where x̄ denotes the unique integer in {1, 2, . . . , q −1}
(1)
for which xx̄ = 1 (mod q). Thus aq ∈ J = [t1 , t2 ] is equivalent to q¯ ∈ Jq :=
[(1 − t2 )q, (1 − t1 )q]. Moreover, aq ∈ J is equivalent to q̄ ∈ Jq := [t1 q , t2 q ], where
this time q̄ denotes the multiplicative inverse of q (mod q ).
An important device employed in [3], [4], [1] to estimate sums over primitive
lattice points is the Weil type [33] estimate
(2.1)
(2)
1
1
|S(m, n; q)| τ (q) gcd(m, n, q) 2 q 2
on complete Kloosterman sums
S(m, n; q) =
x∈[1,q]
gcd(x,q)=1
mx + nx̄
,
e
q
in the presence of an integer albeit not necessarily prime modulus q, proved in [27]
(see also [19]). The bound from (2.1) can be used (see [4, Lemma 1.7]) to prove the
estimate
1
ϕ(q)
Nq (I, J ) = 2 |I| |J | + Oδ (q 2 +δ )
(2.2)
q
for the number Nq (I, J ) of pairs of integers (x, y) ∈ I × J for which xy = 1
(mod q), whenever I and J are intervals which contain at most q integers.
We shall use the following slight improvement of Corollary 1 and Lemma 8 in
[4]. The proof follows literally the reasoning from Lemmas 2, 3 and 8 in [4].
Length of a trajectory in a certain billiard
309
Lemma 2.1. Let Ω ⊆ [1, R] × [1, R] be a convex region and let f be a C 1 function
on Ω. Then:
1
(i)
f (a, b) =
f (x, y) dx dy + ER,Ω,f ,
ζ(2)
2
(a,b)∈Ω∩Zpr
Ω
where
ER,Ω,f f ∞,Ω R ln R +
Df ∞,Rm,n ln R.
(m,n)∈Z2
Rm,n ⊂Ω
(ii) For any interval J ⊆ [0, 1] one has
|J|
f (a, b) =
f (x, y) dx dy + FR,Ω,f,J ,
ζ(2)
2
(a,b)∈Ω∩Zpr
b̄∈Ja
Ω
where
3
FR,Ω,f,J δ f ∞,Ω mf R 2 +δ + f ∞,Ω length(∂Ω) ln R +
Df ∞,Rm,n ln R
2
(m,n)∈Z
Rm,n ∈Ω
for any δ > 0, where b̄ denotes3 the multiplicative inverse of b (mod a), Ja
(1)
(2)
is either Ja or Ja , and mf is an upper bound for the number of intervals
of monotonicity of each of the functions y → f (x, y).
The proof of (ii) relies on (2.2). We also note the following important corollary
of (2.2), which will be often employed in this paper and in the subsequent work
from [5] and [6].
Lemma 2.2. Assume that q ≥ 1 is an integer, I and J are intervals which contain
at most q integers, and f : I ×J → R is a C 1 function. Then for any integer T > 1
one has
ϕ(q)
f (a, b) = 2
f (x, y)dxdy + Eq,I,J ,f,T ,
q
a∈I, b∈J
ab=1(mod q)
I×J
where
|I| |J | Df ∞
T
denotes the L∞ -norm on I × J .
1
3
Eq,I,J ,f,T δ T 2 q 2 +δ f ∞ + T q 2 +δ Df ∞ +
for all δ > 0. Here · ∞
Proof. If T ≥ q, then the error is larger than the sum to estimate and there is
nothing to prove.
If T < q, we partition the intervals I and J respectively into T intervals
|J |
I1 , . . . , IT and J1 , . . . , JT of equal size |Ii | = |I|
T and |Jj | = T . The idea is
to approximate f (x, y) by a constant whenever (x, y) ∈ Ii × Jj . For, we choose for
each pair of indices (i, j) a point (xij , yij ) ∈ Ii × Jj for which
(2.3)
f = |Ii | |Jj |f (xij , yij ).
Ii ×Jj
3 When
writing b̄ ∈ Ja we implicitly assume that gcd(a, b) = 1.
310
F. P. Boca, R. N. Gologan, and A. Zaharescu
For (x, y) ∈ Ii × Jj the mean value theorem gives
f (x, y) = f (xij , yij ) + O (|Ii | + |Jj |)Df ∞
(2.4)
q
Df ∞ .
= f (xij , yij ) + O
T
This gives in turn
(2.5)
f (a, b) =
T
f (x, y)
i,j=1 (x,y)∈Ii ×Jj
xy=1(mod q)
a∈I,b∈J
ab=1(mod q)
=
qDf ∞
.
Nq (Ii , Jj ) f (xij , yij ) + O
T
i,j=1
T
Since each interval Ii and Jj contains at most q integers, estimate (2.2) applies
to them and gives
(2.6)
Nq (Ii , Jj ) =
1
ϕ(q)
|Ii | |Jj | + Oδ (q 2 +δ ).
2
q
As a result of (2.6) and (2.3), the main term in (2.5) becomes
T
1
ϕ(q) |Ii | |Jj |f (xij , yij ) + Oδ (T 2 q 2 +δ f ∞ )
q 2 i,j=1
1
ϕ(q)
f + Oδ (T 2 q 2 +δ f ∞ ),
= 2
q
I×J
while the error term in (2.5) will be
3
|I| |J |
qDf ∞ ϕ(q)
2 12 +δ
+δ
+ Tq2
≤ Df ∞
.
|I| |J | + T q
T
q2
T
3. The second moment of the first exit time for cross-like
scatterers
Throughout this section we keep 0 < ε < 12 fixed, and take
1
1
= the integer part of .
Q = Qε =
ε
ε
We also denote
Z2∗ = Z2 \ {(0, 0)},
Cε = {0} × [−ε, ε] ∪ [−ε, ε] × {0},
Vε = {0} × [−ε, ε],
lε (ω) = inf{τ > 0 ; (τ cos ω, τ sin ω) ∈ Cε + Z2∗ }
tP = the slope of the line OP ,
(x, y) = x2 + y 2 ,
x, y ∈ R.
Length of a trajectory in a certain billiard
311
N’
W’
A’
q’
N’(q’,
o (a+ ε) q )
S’
111111
000000
N
000000
111111
aq
000000
111111
W Wo (q, q- ε )
000000
111111
A
000000
111111
000000
111111
000000
111111
S
000000
111111
1
O
Figure 2. The case q < q Let
Cε = Cε + {(q, a) ; a/q ∈ FQ }
denote the translates of Cε at all integer points with slope in FQ .
For each point A(q, a) with aq ∈ FQ we construct a vertical segment N S of length
2ε and a horizontal segment W E of length 2ε, both centered at A.
Performing symmetries with respect to the integer vertical and horizontal lines,
the problem translates into a covering version in R2 . It is clear that one can discard
the points (q , a ) with gcd(q , a ) = d > 1, which are already hidden by qd , ad .
The trajectory will now originate at O = (0, 0) and end when it reaches one of
the components (q, a) + Cε of Cε , aq ∈ FQ , as seen in the next elementary but useful
lemma.
Lemma 3.1. Any ray of direction ω ∈ [0, π4 ] which originates at O inevitably in
tersects Cε . Moreover, if γ = aq < γ = aq are two consecutive Farey fractions in
FQ and tan ω ∈ [γ, γ ], then the ray of direction ω intersects either (q, a) + Cε or
(q , a ) + Cε and does not intersect any other component of Cε .4
Proof. We shall utilize the inequalities q + q ≥ Q + 1 >
getting
tA =
1
ε
≥ Q ≥ max{q, q },
a
a − ε
a+ε
a
≤ tS =
≤ tA = .
< tN =
q
q
q
q
In the case q < q , we set {W0 } = OW ∩ N S and {N0 } = ON ∩ N S (see
Figure 2 and note that a < a ), inferring that
4 Equivalently,
the intervals Ia/q = [ a−ε
,
q
Ia/q and Ia /q overlap if and only if
a
q
and
a+ε
], aq
q
a
q
∈ FQ , cover [0, 1] and two such intervals
are consecutive elements in FQ .
312
F. P. Boca, R. N. Gologan, and A. Zaharescu
N
11111111111
00000000000
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
0000
1111
0000
1111
00000000000
11111111111
0000
1111
0000
1111
00000000000
11111111111
0000
1111
0000
1111
00000000000
11111111111
0000
1111
0000
1111
00000000000
11111111111
0000
1111
0000
1111
O 11111111111
00000000000
So
W
N
ε
(q, (a’- )q )
q’
(a’- ε)q
111111111
000000000
S’o(q,
000000000
111111111
q’
W
000000000
111111111
A
000000000
111111111
000000000
111111111
000000000
111111111
N’
000000000
111111111
S
000000000
111111111
000000000
111111111
A’
W’
000000000
111111111
000000000
111111111
S’
000000000
111111111
000000000
111111111
000000000
111111111
000000000
111111111
000000000
111111111
000000000
111111111
S
N’
W’
S’o
A
A’
S’
)
O
Figure 3. The case q < q and tS ≤ tW , respectively q < q and
tS > tW .
arctan
γ
(3.1)
lε2 (ω) dω = 2 area(OAN ) + 2 area(ON0 A ) − 2 area(AW0 W )
arctan γ
(a + ε)q = εq + q a −
+ O(ε2 )
q
q − ε(q 2 − q 2 )
+ O(ε2 ).
=
q
In the case q > q one has a < a. Moreover,
a − ε
a
,
≤ max{tW , tS } ≤ γ = tA .
tA = γ ≤ min(tW , tS ) = min
q−ε
q
This shows that any ray of slope tan ω ∈ [γ, γ ] intersects either (q, a) + Cε or
(q , a ) + Cε and no other component of Cε (see Figures 2 and 3).
Besides, we estimate the average of the second moment of the length lε (ω) of the
trajectory when tan ω ∈ [γ, γ ].
When tS ≤ tW (i.e., a +q > ε+ 1ε ), we set {S0 } = OS ∩AW , {S0 } = OS ∩N S,
and note (see the first picture in Figure 3) that
arctan
γ
(3.2)
lε2 (ω) dω = 2 area(OA S ) + 2 area(OAS0 ) − 2 area(AS0 S0 )
arctan γ
(a − ε)q
= εq + q
− a + O(ε2 )
q
q − ε(q 2 − q 2 )
=
+ O(ε2 ).
q
Length of a trajectory in a certain billiard
0110
0
1
10 10
1010 10
0
1
10
1010
10
(0,Q) (1,Q) (x,Q)
(x,Q-x)
313
(Q,Q)
(Q/2,Q/2)
(0,0)
(x,0)
(Q,0)
Figure 4. The region ΩQ
When tS > tW , we set {W0 } = OW ∩ N S, {S0 } = OS ∩ N S, and get (see the
second picture in Figure 3)
arctan
γ
lε2 (ω) dω = 2 area(OA S ) + 2 area(OAS0 ) − 2 area(AW W0 )
(3.3)
arctan γ
=
q − ε(q 2 − q 2 )
+ O(ε2 ).
q
We consider the region
ΩQ = {(x, y) ∈ R2 ; 1 ≤ x ≤ y ≤ Q, x + y > Q}
and the function
y + ε(x2 − y 2 )
y(1 − εy)
+ εx,
(x, y) ∈ ΩQ .
=
x
x
Consider also I = [α, β] ⊆ [0, π4 ], take t1 = tan α, t2 = tan β, and let J =
[t1 , t2 ] ⊆ [0, 1]. For (x, y) ∈ ΩQ one has x > Q − y > 1ε − 1 − y, which gives
y
1 − εy < ε(x + 1) ≤ 2εx. It is also seen that 1 − εy ≥ 1 − Q
≥ 0. As a result we
find that f ∞,ΩQ ≤ 3. Since ε2 #FQ < 1, formulas (3.1), (3.2), (3.3) provide
f (x, y) =
lε2 (ω) dω = 2
(3.4)
J
arctan
a/q arctan
I
a
q
lε2 (ω) dω + O(1) = 2
J
f (q, q ) + O(1).
a/q
a
q
To master the latest sum, we aim to apply Lemma 2.1 to ΩQ . With the notation
from Section 2, relation (3.4) yields
(3.5)
lε2 (ω) dω = 2
f (a, b) + O(1).
I
(a,b)∈ΩQ
b̄∈Ja(1)
314
F. P. Boca, R. N. Gologan, and A. Zaharescu
We also see that for (x, y) ∈ ΩQ one has ∂f
∂x = ε −
∂f |1−2εy|
1
=
≤ x ; hence
∂y
x
(3.6)
Df ∞,Ra,b (a,b)∈Z2
Ra,b ⊂Ω̄Q
Q
Q
x=1 y=max{Q−x,x}
y(1−εy) x2
≤ ε+
2εy
x
≤
3
x
and
1
1 = Q.
x
x=1
Q
Now we can apply Lemma 2.1 (ii) to the sum from (3.5), and employ (3.6) and
mf ≤ 2, to infer that
3
2(t2 − t1 )
2
(3.7)
lε (ω) dω =
f (x, y) dx dy + Oδ (Q 2 +δ ).
ζ(2)
I
ΩQ
When α = 0 and β =
π
4,
Lemma 2.1 (i) improves upon the error in (3.7) to
π/4
lε2 (ω) dω =
(3.8)
2
ζ(2)
0
f (x, y) dx dy + O(Q ln Q).
ΩQ
In summary, (3.7), (3.8) and the equality
1 + 2 ln 2 2
Q
f (x, y) dx dy =
12
ΩQ
lead to:
Theorem 3.2. (i)
π/4
1 + 2 ln 2
ε
lε2 (ω) dω =
+ O(ε| ln ε|)
π2
2
0
(ii) For any 0 ≤ α < β ≤
β
ε
2
lε2 (ω) dω =
π
4
as ε → 0+ .
and δ > 0, one has
1
(1 + 2 ln 2)(tan β − tan α)
+ Oδ (ε 2 −δ )
π2
as ε → 0+ .
α
Part (i) of this result was already proved in [22].
4. The r th moment of the first exit time for cross-like
scatterers
In this section we estimate the average of the first exit time for cross-like scatterers,
1.2. The first step towards estimating the integral
r thus proving
r−2 Theorem
2
l
(ω)
dω
=
l
(ω)l
(ω)
dω
consists in approximating lεr−2 by a step function.
ε
I ε
I ε
π
We take I = [α, β] ⊆ [0, 4 ], t1 = tan α, t2 = tan β, J = [t1 , t2 ] ⊆ [0, 1]. For
consecutive Farey fractions aq < aq from J ∩ FQ , where Q = 1ε , we denote
ω1 = arctan
a
,
q
ω2 = arctan
a+ε
,
q
ω2 = arctan
a − ε
,
q
ω3 = arctan
a
.
q
The function lεr−2 will be approximated by the constants lεr−2 (ω1 ) = (q, a)r−2
on [ω1 , ω2 ] and by lεr−2 (ω3 ) = (q , a )r−2 on [ω2 , ω3 ] when q < q , and respectively
Length of a trajectory in a certain billiard
315
by lεr−2 (ω1 ) on [ω1 , ω2 ] and by lεr−2 (ω3 ) on [ω2 , ω3 ] when q > q . To be precise, we
set
ω2
ω3
J
J
r−2
2
r−2
(q, a)
lε (ω) dω +
(q , a )
lε2 (ω) dω,
Ar,J,ε=
a/q
Br,J,ε=
J
a/q
ω1
ω2
(q, a)
r−2
a/q
lε2 (ω) dω
+
J
ω2
ω3
(q , a )
r−2
a/q
ω1
lε2 (ω) dω,
ω2
Sr,J,ε = Ar,J,ε + Br,J,ε .
Next we estimate the quantities
J ω3
(1)
r
Er,J,ε = lε (ω) dω − Ar,J,ε (1)
≤ Er,[0,1],ε ,
(2)
≤ Er,[0,1],ε .
a/q ω1
and respectively
(2)
Er,J,ε
J ω3
r
=
lε (ω) dω − Br,J,ε a /q
ω1
An inspection of the case q < q in the proof of Lemma 3.1 leads to
sup lεr−2 (ω) − lεr−2 (ω1 ) ≤ (q, a + ε)r−2 − (q − ε, a)r−2 r εQr−3 ,
ω∈[ω1 ,ω2 ]
and to
r−2
(a + ε)q r−2
r−2
r−2
sup
lε (ω) − lε (ω3 ) ≤ (q , a )
−
q
,
q
ω∈[ω2 ,ω3 ]
(a + ε)q Qr−3
Qr−3 r a −
.
q
q
Therefore
(4.1)
ω3
ω2
ω3
r
r−2
2
r−2
2
lε (ω) dω − lε (ω1 ) lε (ω) dω − lε (ω3 ) lε (ω) dω ω1
ω1
ω2
r q 2 (ω2 − ω1 )εQr−3 + q 2 (ω3 − ω2 )
a
But ω2 −ω1 ≤ a+ε
q −q =
side in (4.1) is
ε
q
and ω3 −ω2 ≤
r qε2 Qr−3 +
As a result we infer that
(4.2)
(1)
Er,[0,1],ε
r−2
= Or Q
a
q
− a+ε
q =
<
1
qq ,
so the right-hand
Qr−2
Qr−2
.
2
q
q2
Q
ϕ(q)
q=1
1−εq qq Qr−3
.
q
q2
= Or (Qr−2 ln Q).
316
F. P. Boca, R. N. Gologan, and A. Zaharescu
In the case q < q we get (in both subcases tS ≤ tW and tS > tW )
r−2
r−2
(a − ε)q r−2
lε (ω) − lε (ω1 ) ≤ q,
− (q, a − ε)r−2
sup
q
ω∈[ω1 ,ω2 ]
(a − ε)q
Qr−3
r−3
r
−
a
+
ε
Q
≤
q
q
and
sup
ω∈[ω2 ,ω3 ]
r−2
lε (ω) − lεr−2 (ω3 ) ≤ (q , a )r−2 − (q , a − ε)r−2 r εQr−3 .
1
a
− aq = 1−εq
Employing also ω2 − ω1 = a q−ε
qq ≤ qq and ω3 − ω2 = q −
we get in the case q < q the estimate
ω3
ω2
ω3
r
r−2
2
r−2
2
lε (ω) dω − lε (ω1 ) lε (ω) dω − lε (ω3 ) lε (ω) dω ω1
(4.3)
(2)
Er,[0,1],ε
= Or
(q,q )∈ΔQ
Qr−2
q 2
=
ε
q ,
ω2
ω1
r
Hence
a −ε
q
Qr−1
Qr−3
Qr−2
+
.
2
qq
q
q
r−2
= Or Q
Q
ϕ(q )
q 2
= Or (Qr−2 ln Q).
q =1
ω3
Since the contribution of a single term ω1 lεr (ω) dω is infer from (4.2) and (4.3) that
(4.4)
lεr (ω) dω = Sr,J,ε + Or (ε1−r ).
max{q,q }r
qq ≤ Qr−1 , we
I
Next, we adjust the second integral in the expression of Ar,J,ε , writing
2
2
a
a
1
2
2
2
2
+1 =q
a +q =q
+ +1
q
q
qq
2 2
q
1
=
a + + q2
q
q
2 q
a
2
2
=
a +q +O q
q
2
q
a
2
2
=
(a + q ) 1 + O 2
q
q (a + q 2 )
2
q
1
2
2
=
(a + q ) 1 + O
q
qq 2
q
1
2
2
=
.
(a + q ) 1 + O
q
Q
Length of a trajectory in a certain billiard
This gives in turn
(q , a )
(4.5)
r−2
=
In a similar way
(q, a)r−2 =
(4.6)
q
q
q
q
r−2
r−2
(q, a)
r−2
1 + Or
1 Q
317
.
1 .
(q , a )r−2 1 + Or
Q
For further use, it is also worth to note
1
1 + aq
1 + aq 1 + O( Q
)
1 + aq
1
=
(4.7)
.
1+O
=
1
1 + ( aq )2
Q
1 + ( aq )2 1 + O( Q
)
1 + ( aq )2
Making use of (see the proof of Lemma 3.1)
ω2
lε2 (ω) dω = 2 area(OAN ) + O(ε2 ) = εq + O(ε2 ),
ω1
ω3
lε2 (ω) dω = 2 area(ON0 A ) =
q (1 − εq )
,
q
ω2
and (4.5), we see that Ar,J,ε can be expressed as
r−2 1 J
q (1 − εq )
r−2 q
1 + Or
(4.8)
(q, a)
q
q
Q
a/q
+ε
J
(q, a)r−2 q + Or (1).
a/q
If x̄ denotes the inverse of the integer x (mod q) in [1, q], then a q − aq = 1 gives
< ε, the error in the first sum in (4.8) is r Q2 Qr−1 εQ−1 =
a = q − q¯ . Since 1−εq
q
Qr−1 , and so Ar,J,ε is equal up to an error term of order Or (Qr−1 ) to
J
q r−1
εq r
(4.9)
(q, a)r−2 εq + r−1 − r−1
q
q
a/q
=
Q
(q − x̄) + q
2
2
r−2
2
q=1 max{q,Q−q}<x≤Q
x̄∈Jq(1)
xr−1
εxr
εq + r−1 − r−1 ,
q
q
(1)
with Jq as defined in Section 2.
When q > q , we employ (see the proof of Lemma 3.1)
ω2
ω1
ω3
ω2
lε2 (ω) dω = 2 area(OAS0 ) + O(ε2 ) =
lε2 (ω) dω = 2 area(OA S ) = εq ,
(1 − εq)q
+ O(ε2 ),
q
318
F. P. Boca, R. N. Gologan, and A. Zaharescu
together with Qr−2 ε2 #FQ ≤ Qr−2 , relation (4.6), and the fact that a q − aq = 1
implies a = q̄ (mod q ), to infer that Br,J,ε is expressible as
J
J q r−2
r−2 (1 − εq)q
(q, a)
+
(q, a)r−2 εq + Or (Qr−1 )
q
q
a /q
a /q
r−1
J
q
εq r
=
(q , a )r−2 εq + r−1 − r−1 + Or (Qr−1 )
q
q
a /q
=
Q
2
x̄ + q
2
r−2
2
q =1 max{q ,Q−q }<x≤Q
(2)
x̄∈Jq
xr−1
εxr
εq + r−1 − r−1
q
q
+ Or (Qr−1 ),
where x̄ denotes the inverse of an integer x (mod q ) in [1, q ]. Changing notation,
Br,J,ε can be rewritten as
(4.10)
Q
x̄2 + q 2
r−2
2
εq +
q=1 max{q,Q−q}<x≤Q
x̄∈Jq(2)
xr−1
εxr
− r−1
r−1
q
q
+ Or (Qr−1 ).
By (4.4), (4.9) and (4.10), we infer that
(4.11)
lεr (ω) dω = Tr,J,ε + Or (Qr−1 ),
I
where Tr,J (ε) = S1 (Q) + S2 (Q), with
(4.12)
Sk (Q) =
Q
fk (x, x̄, q),
k = 1, 2,
q=1 max{Q−q,q}<x≤Q
x̄∈Jq(k)
and
r−2
xr−1
εxr
f2 (x, y, z) = y 2 + z 2 2 εz + r−1 − r−1 ,
z
z
f1 (x, y, z) = f2 (x, z − y, z).
For each q ∈ [1, Q], the functions fk (·, ·, q), defined on [1, Q] × [1, q], manifestly
satisfy the estimates
(4.13)
fk (·, ·, q)∞ r
Qr−1
q
and
(4.14)
Dfk (·, ·, q)∞ r
Qr−1
Qr−2
≤
.
q
q2
Thus we may consider in Lemma 2.2 for each q ∈ [1, Q] the function fk (·, ·, q), the
(k)
intervals I = (max{Q − q, q}, Q] and J = Jq with |I| ≤ q and |J | = q|I|, and
Length of a trajectory in a certain billiard
319
take T = [Qα ], to infer that the inner sum in (4.12) can be expressed as
Q
ϕ(q)
q2
dx
dy fk (x, y, q)
max{Q−q,q} Jq(k)
1
1
+ Oδ,r (q − 2 +δ Qr−1+2α + q − 2 +δ Qr−1+α + |I|Qr−1−α ).
Summing up over q ∈ [1, Q], we arrive at
Sk (Q) =
(4.15)
Q
ϕ(q)
1
g(q) + Oδ,r (Qr− 2 +2α+δ + |I|Qr−α ),
q
q=1
where
1
g(z) =
z
Q
(1−t
1 )z
dx
max{Q−z,z}
1
=
z
Q
dy f1 (x, y, z)
(1−t2 )z
t2 z
dx
dy f2 (x, y, z),
z ∈ [1, Q].
t1 z
max{Q−z,z}
The formulas
d
dz
Q bz
Q bz
1
1
dx dy h(x, y, z) = − 2
dx dy h(x, y, z)
z
z
z
az
z
+
1
z
Q
dx
z
+
d
dz
b
z
az
bz
dy
1
∂h
(x, y, z) −
∂z
z
az
Q
h(x, bz, z) dx −
dx
Q−z
bz
dy h(x, y, z)
a
z
1
=− 2
z
Q
Q
bz
dx
Q−z
b
z
h(x, az, z) dx,
bz
dx
Q−z
+
Q
z
az
1
+
z
h(z, y, z) dy
az
Q
z
1
z
bz
dy h(x, y, z)
az
1
∂h
(x, y, z) +
dy
∂z
z
az
Q
h(x, bz, z) dx −
Q−z
bz
h(Q − z, y, z) dy
az
a
z
Q
h(x, az, z) dx,
Q−z
r−1
and the estimates (4.13) and (4.14) show that |g (z)| r Q z . As a result we get
Q |g (z)| dz r Qr−1 ln Q. It is also clear that g∞ r Qr−1 , so we are in the
1
320
F. P. Boca, R. N. Gologan, and A. Zaharescu
position of being able to apply Lemma 2.3 in [3] to g, collecting
Q
ϕ(q)
q
q=1
1
g(q) =
ζ(2)
Q
Q
g∞ +
g(z) dz + O
1
=
1
ζ(2)
|g (z)| dz ln Q
1
Q
g(z) dz + Or (Qr−1 ln2 Q).
1
Comparing the previous relation with (4.15), we infer that both S1 (Q) and S2 (Q)
can now be expressed as
1
ζ(2)
(4.16)
Q
1
g(z) dz + Oδ,r (Qr− 2 +2α+δ + |I|Qr−α ).
1
Taking into account (4.11) and (4.12), we gather
lεr (ω) dω
(4.17)
Q
2
=
ζ(2)
1
g(z) dz + Oδ,r (Qr− 2 +2α+δ + |I|Qr−α ).
1
I
Integrating with respect to y in the formula that gives g and changing then z
into Qz, we may express the main term in (4.17) as
Kr,I
ζ(2)
(4.18)
Q
Q
dz
1
dx
xr−1
εxr r−1
ε+ r − r z
z
z
max{Q−z,z}
Kr,I Qr+1
=
ζ(2)
1
1
dz
1/Q
dx
εz
r−1
εxr
xr−1
−
,
+
Qz
z
max{1−z,z}
where
Kr,I
t2
r−2
= 2 (1 + t2 ) 2 dt = 2
t1
dx
.
cosr x
I
Up to an error term of order Or (Q−2 ), the double integral in the right-hand side
of (4.18) is given by
1/2
1/2
1/2
ε
1 − (1 − z)r
1 − (1 − z)r+1
1
r−1
r−1
dz −
dz.
+ (1 − z) ) dz +
ε z(z
rQ
z(1 − z)
r+1
z(1 − z)
0
0
0
1
Since Qr+1 = εr+1
+ Or ( ε1r ), we now infer from (4.17) and (4.18) the equality
1
dx
cr
+ Oδ,r (ε−r+ 2 −2α−δ + |I|ε−r+α ),
(4.19)
lεr (ω) dω = r
r
ε
cos x
I
I
with cr as in Theorem 1.1. The proof of Theorem 1.2 is now complete.
321
Length of a trajectory in a certain billiard
5. The moments of the number of reflections
We take as before Q = [ 1ε ] and keep up with the notation from the beginning
of Section 4. An inspection of the proof of Lemma 3.1 shows that if aq < aq are
consecutive in FQ , then
⎧⎧
a
⎪
if tan ω ∈ [ aq , q−ε
)
⎪
⎪
⎨q + a
⎪
⎪
⎪
a
a+ε
⎪
if q < q ;
q
+
a
+
1
if
tan
ω
∈
[
,
⎪
q−ε
q )
⎪⎪
⎪
⎩ ⎪
a
⎪
q + a
if tan ω ∈ [ a+ε
⎪
q , q ]
⎪
⎪
⎨ q+a
if tan ω ∈ [ aq , a q−ε
)
Rε (ω) =
if q > q and tS ≤ tW ;
a −ε a
⎪
q
+
a
if
tan
ω
∈
[
,
]
⎪
q
q
⎪
⎧
⎪
⎪
a
⎪
if tan ω ∈ [ aq , q−ε
)
⎪⎪
⎨q + a
⎪
⎪
⎪
a
a
⎪ q + a + 1 if tan ω ∈ [ q−ε , q−ε
if q > q and tS > tW .
)
⎪
⎪
⎪
⎪
⎩
⎩ q + a
if tan ω ∈ [ a −ε , a ]
q
q
A first immediate remark is that we may replace q + a + 1 by q + a in the above
formulas, since the contribution of the corresponding arcs is small, as we see from
| arctan x − arctan y| ≤ |x − y|, and from
a + ε
ε(q − a − ε)
1
a
−
=
ε
≤1
q
q−ε
q(q − ε)
q
a/q∈FQ
and
a/q∈FQ
a − ε
a
=
−
q
q−ε
a/q∈FQ
a/q∈FQ
a/q∈FQ
1 − ε(q + a ) + ε2
q (q − ε)
a/q∈FQ
1
= 1.
qq As a result we may write
(1)
(2)
Rεr (ω) dω = Tr,J,ε + Tr,J,ε + Or (Qr−1 ),
I
where we set
(1)
Tr,J,ε
=
a
a+ε
− arctan
(q + a) arctan
q
q
a/q
J
a
a+ε
+
(q + a )r arctan − arctan
q
q
J
r
a/q
and
(2)
Tr,J,ε
=
J
a/q
a − ε
a
(q + a) arctan
− arctan
q
q
J
a
a − ε
r
+
.
(q + a ) arctan − arctan
q
q
r
a/q
Employing
arctan(x + h) − arctan x =
h
h
+ O(h2 ) =
+ O(h2 )
1 + x2
1 + (x + h)2
322
F. P. Boca, R. N. Gologan, and A. Zaharescu
we now arrive at
Rεr (ω) dω = Sr,J,ε + Tr,J,ε + Or (Qr−1 ),
(5.1)
I
where
(5.2)
Sr,J,ε =
J
(5.3)
Tr,J,ε =
J
(q + a) +
1
a/q
1−εq
qq + ( aq )2
1
ε
q
r
(q + a) +
1−εq qq + ( aq )2
r
1 + ( aq )2
a/q
and
ε
q
r
r
(q + a )
1 + ( aq )2
(q + a )
.
To further simplify the expressions in (5.2) and (5.3) we employ
1 a
1
1
a
a
a
1+ =1+ + =1+ +O
= 1+
1+O
,
q
q
qq
q
Q
q
Q
r r 1 a
a
1+ ,
1 + Or
= 1+
q
q
Q
and
1
Q
a/q∈FQ
1 r
q ≤ Qr−2
qq
a/q∈FQ
to infer that
(5.4)
Sr,J,ε =
and also that
Tr,J,ε =
J
(1 + aq )r
a/q
1 + ( aq )2
εq
J (1 + aq )r a/q
1 + ( aq )2
r−1
1
≤ Qr−1 ,
q
1 − εq r−1
q
+
q
εq r−1 +
1 − εq r−1
q
q
+ Or (Qr−1 ),
+ Or (Qr−1 ).
The main term in (5.4) can now be conveniently expressed as
(5.5)
Ar,J,ε =
Q
f (x, x̄, q),
q=1 max{Q−q,q}<x≤Q
x̄∈Jq(1)
where x̄ is the multiplicative inverse of x in {1, . . . , q − 1}, and this time we set
I = (max{Q − q, q}, Q],
f (x, y) = f (x, y, q) =
(1 +
q2 +
q−y r
q )
q−y 2
( q )
r
J = Jq(1) ,
q(2q − y)
= 2
q + (q − y)2
εq
1 − εx r−1
x
+
q
1 − εx x r−1
r−1
ε+
q
q
.
Length of a trajectory in a certain billiard
323
Since 0 ≤ 1 − εx < 1, it is easy to check that
(2q − y)r
f ∞,I×J (1) 2
r q r−2 ,
q
q + (q − y)2
∂f q(2q − y)r
1
2
·
r q r−3 ,
∂x 2 q2
(1)
q
+
(q
−
y)
∞,I×Jq
∂f ∂
(2q − y)r
r q r−3 .
∂y 2
2
(1)
∂y q + (q − y) ∞
∞,I×Jq
Taking these estimates into account and applying Lemma 2.2 for each q ∈ [1, Q]
(1)
to f = f (·, ·, q), I = (max{Q − q, q}, Q], J = Jq and T = [Qα ], we infer that the
inner sum in (5.5) can be expressed as
Q
ϕ(q)
q2
1 − εx x r−1
q(2q − y)r
ε+
dx
dy
q
q
q 2 + (q − y)2
(1)
max{Q−q,q}
Jq
1
3
+ Or,δ (Q2α q 2 +δ q r−2 + Qα q 2 +δ q r−3 + |I|Q−α q 2 q r−3 )
=
3
Cr,I ϕ(q) r−1
q
gr (q) + Or,δ (Qr− 2 +2α+δ + |I|Qr−1−α ),
q
where
Cr,I =
(1−t
1 )q
1
q r−1
(2q − y)r
1
dy = r−1
2
q + (q − y)2
q
(1−t2 )q
t2
=
(1 + t)r
dt =
1 + t2
t1
t2 q
(q + y)r
dy
q2 + y2
t1 q
(1 + tan x)r dx
I
and
Q
gr (q) =
1 − εx x r−1
ε+
dx.
q
q
max{Q−q,q}
We now arrive at
(5.6)
f (x, x̄, q) =
max{Q−q,q}<x≤Q
x̄∈Jq(1)
3
Cr,I
hr (q) + Or,δ (Qr− 2 +2α+δ + |I|Qr−α−1 ),
q
where
Q
εq r−1 +
hr (q) =
max{Q−q,q}
1 − εx r−1
x
dx.
q
324
F. P. Boca, R. N. Gologan, and A. Zaharescu
Since h∞ Qr−1 and
(5.5) and (5.6) show that
Q
1
|h (q)| dq Qr−1 , Lemma 2.3 in [3] together with
Q
ϕ(q)
Sr,J,ε = Cr,I
q=1
Cr,I
=
ζ(2)
q
1
hr (q) + Or,δ (Qr− 2 +2α+δ + |I|Qr−α )
Q
1
hr (q) dq + Or,δ (Qr− 2 +2α+δ + |I|Qr−α ).
1
Making use of εQ = 1 + O(ε) we arrive by a straightforward computation to
1 Q
r
hr (q) dq = Q
1
0
1 − max{1 − x, x}r
xr−1 1 − max{1 − x, x} +
rx
1 − max{1 − x, x}r+1
dx + Or (Qr−1 ).
−
(r + 1)x
The integral above is seen to coincide with
1/2
0
1 − (1 − x)r
1 − (1 − x)r+1
x xr−1 + (1 − x)r−1 +
−
rx(1 − x)
(r + 1)x(1 − x)
dx =
π 2 cr
,
12
hence
Q
hr (q) dq =
π 2 cr Qr
π 2 cr ε−r
+ Or (Qr−1 ) =
+ Or (ε−r+1 )
12
12
1
and as a result
Sr,J,ε
1
6 π 2 cr ε−r
= 2·
(1 + tan x)r dx + Or,δ (ε−r+ 2 −2α−δ + |I|ε−r+α )
π
12
I
−r 1
cr ε
(1 + tan x)r dx + Or,δ (ε−r+ 2 −2α−δ + |I|ε−r+α ).
=
2
I
By reversing the roles of q and q it is seen in a similar way that
1
cr ε−r
(1 + tan x)r dx + Or (ε−r+ 2 −2α−δ + |I|ε−r+α ).
Tr,J,ε =
2
I
This concludes the estimates of Sr,J,ε and Tr,J,ε . Theorem 1.5 now follows from
(5.1).
6. The case of circular scatterers
Note first that the statements of Theorems 1.2 and 1.5 hold true if we replace
the scatterers Cε + Z2∗ by Vε + Z2∗ .
In this section we consider the circular scatterers Dε + Z2∗ , where
Dε = {(x, y) ∈ R2 ; x2 + y 2 = ε2 }.
Length of a trajectory in a certain billiard
325
(q,a+ ε + )
(q,a)
(q,a- ε- )
Ο
Figure 5. A circular scatterer
For each integer lattice point (q, a), let (q, a ± ε± ) denote the intersections of the
line x = q with the tangents from O to the circle
Dε,q,a = (q, a) + Dε = {(x, y) ∈ R2 ; (x − q)2 + (y − a)2 = ε2 },
where ε± = ε± (q, a) are computed from the equality
a − a±ε± q ε± q
q
ε= a±ε± 2 = q 2 + (a ± ε )2 ,
±
1+
q
which gives in turn
ε2± q 2 = ε2 q 2 + ε2 (a ± ε± )2 ,
or
(q 2 − ε2 )ε2± ∓ 2aε2 ε± − ε2 (q 2 + a2 ) = 0.
The latter provides
ε± = ±
(6.1)
Employing also
aε2
ε 4
+
q + a2 q 2 − a2 ε2 .
q 2 − ε2
q 2 − ε2
q 4 + a2 q 2 − q 4 + a2 q 2 − a2 ε2 ε2 ,
ε 2 ε
ε
= 2 1+O 2
q 2 − ε2
q
q
and
2
aε2
ε
,
=O
2
2
q −ε
q
we arrive at
(6.2)
ε± (q, a) = ε
1+
2
2
ε
a2
ε
ε
=
.
+
O
+
O
a
2
q
q
cos arctan q
q
326
F. P. Boca, R. N. Gologan, and A. Zaharescu
r
Proof of Theorem 1.1. We wish to compare I τεr (ω) dω with I τε (ω) dω where
τε (ω), the smallest τ > 0 for which
(τ cos ω, τ sin ω) ∈
{q} × [a − ε− (q, a), a + ε+ (q, a)],
(q,a)∈Z2∗
denotes the first exit time in the case where the scatterers are the vertical segments
{q} × [a − ε− (q, a), a + ε+ (q, a)]. From Figure 5 it is apparent that
τε (ω) − τε (ω)| ≤ 2ε,
sup |
ω
and so, since supω τε (ω) ≤ supω lε (ω) ≤
√
2
ε ,
we get
√ r−1
r
2
τεr (ω) − τε (ω)| r ε
r ε2−r ,
sup |
ε
ω
which gives
τεr (ω) dω
(6.3)
=
I
r
τε (ω) dω + Or (ε2−r ).
I
To estimate I τε (ω) dω, we divide the interval I into N = [ε−θ ] intervals of
1
θ
equal size Ij = [ωj , ωj+1 ] with |Ij | = |I|
N ε for some 0 < θ < 2 . Then one has
for all j that
cos ωj
cos ωj+1 θ−1
ε − ε3/2 − ε + ε3/2 ε ,
cos ωj cos ωj+1 −
thus the integers Q+
satisfy
j = ε−ε3/2 + 1 and Qj = ε+ε3/2
−
θ−1
0 < Q+
j − Qj ε
(6.4)
and
1
ε − ε3/2
ε + ε3/2
1
≤
≤
≤ −.
+
cos
ω
cos
ω
Qj
Qj
j
j+1
(6.5)
Furthermore, it follows from (6.2) that there exists ε0 = ε0 (θ) > 0 such that for all
ε < ε0 , all j, and all aq ∈ [tan ωj , tan ωj+1 ], one has
ε − ε3/2
ε + ε3/2
≤ ε± (q, a) ≤
.
cos ωj
cos ωj+1
This implies in conjunction with (6.5), for all
(6.7)
∈ [tan ωj , tan ωj+1 ], the inequalities
1
1
.
+ ≤ ε± (q, a) ≤
Qj
Q−
j
(6.6)
Since Q±
j =
a
q
cos ωj
ε
+ O(εθ−1 ), one has
r
(Q±
j ) =
cosr ωj
+ Or (ε−r+θ ).
εr
327
Length of a trajectory in a certain billiard
The first exit time increases when all the sizes of scatterers decrease. Thus we
infer from (6.6) the inequalities
r
r
(6.8)
l 1 (ω) dω ≤ τε (ω) dω ≤ lr 1 (ω) dω.
Q
Ij
−
j
Q
Ij
Ij
+
j
But by Theorem 1.2 and by (6.7) we may write
1
dx
r
+ Or,δ (ε−r+ 2 −2α−δ + ε−r+θ+α )
(6.9)
lr 1 (ω) dω = cr (Q±
)
j
r
±
cos x
Q
j
Ij
Ij
− 32 −δ
with the better error term ε
for r = 2. Also using
infer that the first integral in (6.9) is expressible as
(6.10)
cr cosr ωj
εr
ω
j+1
dx
Ij cosr x
r |Ij | εθ we
1
dx
+ Or,δ (ε−r+2θ + ε−r+ 2 −2α−δ + ε−r+θ+α ),
cosr x
ωj
3
with the better error term ε−2+θ + ε− 2 −δ for r = 2.
Summing up over j we infer from (6.8), (6.10) and (6.3) that
τεr (ω) dω
(6.11)
ω
j+1
N
dx
cr r
= r
cos ωj
ε j=1
cosr x
ωj
I
+ Or,δ (ε
−r+ 12 −2α−θ−δ
+ ε−r+α + ε−r+θ )
3
with the better error term ε− 2 −θ−δ + ε−2+θ for r = 2.
Finally, we apply the mean value theorem and chose some ξj ∈ [ωj , ωj+1 ] to
evaluate the sum
N
ω
j+1
r
cos ωj
j=1
dx
cosr x
ωj
as
N
(ωj+1 − ωj ) cosr ωj
j=1
cosr
ξj
=
N
(ωj+1 − ωj ) 1 + Or (ωj+1 − ωj )
j=1
=
N
(ωj+1 − ωj ) 1 + Or (εθ )
j=1
= |I| + Or (εθ ).
This implies Theorem 1.1 in conjunction with (6.11) by taking θ = α =
and θ = 41 for r = 2.
1
8
for r = 2
Proof of Theorem 1.4. We proceed along the same line to estimate the moments
Here we denote by R(ω)
of R.
the number of reflections in the side cushions in
the case of vertical scatterers (of variable size) {q} × [a − ε− (q, a), a + ε+ (q, a)],
328
F. P. Boca, R. N. Gologan, and A. Zaharescu
r
(ω) dω differs from
(q, a) ∈ Z2∗ . It is seen as in the proof of Theorem 1.1 that I R
ε
r
2−r
R (ω) dω by an error term of order Or (ε ). One can also show that
I ε
r
r
(6.12)
R 1 (ω) dω ≤ Rε (ω) dω ≤ Rr 1 (ω) dω.
Q
Ij
−
j
Q
Ij
Ij
+
j
Applying now Theorem 1.5 to the vertical scatterers V1/Q± on the intervals Ij =
j
1
8
[ωj , ωj+1 ] of equal size |Ij | = |I|
N ε with θ = α = 8 , and also using (6.7), we find
that
1
cr cosr ωj
Rr 1 (ω) dω =
(1 + tan x)r dx + Or,δ (ε−r+ 4 −δ ),
r
±
ε
Q
j
1
Ij
Ij
and thus
εr (ω) dω
R
(6.13)
ω
j+1
N
1
cr r
= r
cos ωj
(1 + tan x)r dx + Or,δ (ε−r+ 8 −δ ).
ε j=1
ωj
I
By the mean value theorem we find ξj , ηj ∈ Ij such that
N
(6.14)
ω
j+1
r
(1 + tan x)r dx =
cos ωj
j=1
N
(ωj+1 − ωj ) cosr ωj (1 + tan ξj )r ,
j=1
ωj
and respectively
ω
N j+1
r
(sin x + cos x) dx =
(sin x + cos x)r dx
j=1 ω
j
I
=
N
(ωj+1 − ωj )(sin ηj + cos ηj )r =
j=1
N
(ωj+1 − ωj ) cosr ηj (1 + tan ηj )r .
j=1
From
1
cosr ωj = cosr ηj + Or (ωj+1 − ωj ) = cosr ηj + Or (ε 8 )
and
1
(1 + tan ξj )r = (1 + tan ηj )r + Or (| tan ξj − tan ηj |) = (1 + tan ηj )r + Or (ε 8 )
we infer that the sum in (6.14) is equal to
N
1
1
(ωj+1 − ωj ) cosr ηj + Or (ε 8 ) (1 + tan ηj )r + Or (ε 8 )
j=1
1
(sin x + cos x)r dx + Or (ε 8 ).
=
I
This can be combined with (6.13) to collect
εr (ω) dω = cr (sin x + cos x)r dx + Or,δ (ε−r+ 18 −δ ),
R
εr
I
I
which concludes the proof of Theorem 1.4.
Length of a trajectory in a certain billiard
329
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Department of Mathematics, University of Illinois, 1409 W. Green St., Urbana, IL
61801, USA
Institute of Mathematics of the Romanian Academy, P.O.Box 1-764, RO-014700, Bucharest, Romania
fboca@math.uiuc.edu
Institute of Mathematics of the Romanian Academy, P.O.Box 1-764, RO-014700 Bucharest, Romania
Radu.Gologan@imar.ro
Department of Mathematics, University of Illinois, 1409 W. Green St., Urbana, IL
61801, USA
Institute of Mathematics of the Romanian Academy, P.O.Box 1-764, RO-014700, Bucharest, Romania
zaharesc@math.uiuc.edu
This paper is available via http://nyjm.albany.edu:8000/j/2003/9-16.html.