Distribution of the linear flow length in a honeycomb in the small-scatterer limit
aa r X i v : . [ m a t h . D S ] J un DISTRIBUTION OF THE LINEAR FLOW LENGTH IN A HONEYCOMBIN THE SMALL-SCATTERER LIMIT
FLORIN P. BOCAA bstract . We study the statistics of the linear flow in a punctured honeycomb lattice, or equivalentlythe free motion of a particle on a regular hexagonal billiard table with holes of equal size at thecorners and obeying the customary reflection rules. In the small-scatterer limit we prove the existenceof the limiting distribution of the free path length with randomly chosen origin of the trajectory andexplicitly compute it.
1. I ntroduction
From the regular hexagon of unit size remove circular holes of small radius ε > H ε of area | H ε | = √ − πε . For each pair ( x , ω ) ∈ H ε × [0 , π ]consider a point particle moving at unit speed on a linear trajectory, with specular reflections whenreaching the boundary. The time τ hex ε ( x , ω ) it takes the particle to reach one of the holes is called the free path length (or first exit time ). Equivalently, one can consider the unit honeycomb tessellationof the euclidean plane, with “fat points” (obstacles or scatterers) of radius ε centered at the vertices m e + n j , m . n (mod 3), of the lattice Λ = Z e + Z j = Z (cid:16) / √ / (cid:17) , j = (cid:0) , √ (cid:1) , and a particlemoving at unit speed and velocity ω on a linear trajectory until it hits one of the obstacles (seeFigure 1). If the initial position x is always chosen in a fundamental domain, the first hitting timecoincides with τ hex ε ( x , ω ). In this paper we are interested in estimating the probability P hex ε ( ξ ) = π | H ε | (cid:12)(cid:12)(cid:12)(cid:8) ( x , ω ) ∈ H ε × [0 , π ] : ετ hex ε ( x , ω ) > ξ (cid:9)(cid:12)(cid:12)(cid:12) , ξ ∈ [0 , ∞ ) , (1.1)that ετ hex ε ( x , ω ) > ξ as ε → + . We will prove that Φ hex ( ξ ) = lim ε → + P hex ε ( ξ ) exists for all ξ > Φ hex ( ξ ).The version of this problem where the initial point is chosen to be the center of the hexagon hasbeen solved in [3]. The square lattice analog of estimating (1.1) has a longer history originatingin the work of H. A. Lorentz [14] and G. P´olya [19]. A complete solution was given in [7]. Adetailed history and presentation of various ideas and tools involved in this and related problems,as well as a description of recent developments in the study of the periodic Lorentz gas, including[10, 11, 12, 16, 17, 18], is provided in [13, 15].One additional di ffi culty encountered here is the absence of a theory of continued fractions in thecase of the hexagonal tessellation. To bypass this obstacle we shall deform this tessellation, as in[3], into Z = { ( m , n ) ∈ Z , m . n (mod 3) } . The three-strip partition of the unit square employed[9, 7] in the situation of the square lattice, or equivalently the corresponding tiling of R shown inFigure 6, will be useful here. However, the presence of certain (mod 3) constraints translates here Date : May 31, 2010. • • C • C • C ′ • C • C ′ • C • C ′ • C • C ′ • C • C ′ • C • C ′ ◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦ ω x F igure
1. The free path in a hexagonal billiard and respectively in a hexagonal latticein the existence of a positive proportion of angles with very long trajectories. This leads to a largenumber of (non-redundant) cases that have to be analyzed individually. The main result is
Theorem 1.
There exists a decreasing continuous function Φ hex : [0 , ∞ ) → (0 , ∞ ) , Φ hex (0) = , Φ hex ( ∞ ) = , such that for any δ > , as ε → + , P hex ε ( ξ ) = Φ hex ( ξ ) + O δ (cid:0) ε − δ (cid:1) , ∀ ξ > , uniformly for ξ in compact subsets of [0 , ∞ ) . Moreover, there exist constants C , C > such thatC ξ Φ hex ( ξ ) C ξ , ∀ ξ ∈ [1 , ∞ ) . (1.2)Estimate (1.2) is discussed in Remark 2 of Section 5. The repartition function Φ hex can be explic-itly computed as Φ hex ( ξ ) = π G ξ √ ! , (1.3)with G ( ξ ) obtained by adding all terms ζ (2) c I G ( ∗ ) I , Q from formulas (5.22), (6.6), (6.7), (6.8), (6.9),(6.10), (6.11), (7.1), (7.2), (7.5), (7.6), (7.8), (7.11), (7.12), (7.13), (8.1), (8.2), (8.5), (8.6), (8.8),(8.9), (8.10), (8.11), (8.12), (8.13), (8.14), (9.1), (9.2), (9.4), (9.5), (9.6), (9.7), (9.9), (9.10), (9.11),(9.12), (9.13), (9.14), (9.15) and (9.16).In the case of the square lattice only the term from (5.22) arises, with a di ff erent constant and no √ scaling for ξ . The limiting distribution Φ (cid:3) also satisfies (1.2).In the case of a lattice it was actually proved in [16] that for every x ∈ R \ Q , lim ε → + π (cid:12)(cid:12)(cid:12)(cid:8) ω ∈ [0 , π ) : ετ (cid:3) ε ( x , ω ) > ξ (cid:9)(cid:12)(cid:12)(cid:12) = Φ (cid:3) ( ξ ). It would be interesting to know whether a similar result holds truefor a generic choice of x in the case of the honeycomb.The analog problem about the free path length in a regular polygon with n sides ( n , , , HE LINEAR FLOW IN A HONEYCOMB 3 F hex F ƒ F igure
2. The limiting repartition functions Φ hex and Φ (cid:3)
2. T ranslating the problem to the square lattice with mod 3 constraints
For manifest symmetry reasons it su ffi ces to consider x ∈ H ε and ω ∈ (cid:2) , π (cid:3) , or equivalently t = tan ω ∈ (cid:2) , √ (cid:3) . We will simply write τ hex ε ( x , ω ) = τ hex ε ( x , t ). As in [3] consider the lattice Z M , M = (cid:16) / √ / (cid:17) , and the linear transformation T x = x M − on R : T ( x , y ) = x − y √ , y √ ! = ( x ′ , y ′ ) . This maps the vertices (cid:0) q + a , a √ (cid:1) of the grid of equilateral triangles of unit side onto the vertices( q , a ) of the square lattice Z . The vertices of the honeycomb are mapped exactly into Z , thesubset of elements of Z with q . a (mod 3) (see Figure 3). The points of the x -axis are fixed by T . The circular scatterers S q , a ,ε = ( x , y ) + ε (cos θ, sin θ ) with ( x , y ) = (cid:0) q + a , a √ (cid:1) are mappedonto ellipsoidal scatterers ( x ′ , y ′ ) + ε (cid:0) cos θ − sin θ √ , θ √ (cid:1) centered at ( x ′ , y ′ ) = ( q , a ) = T ( x , y ).The channel of width w = ε , bounded by the two lines of slope t = tan ω and tangent to the circle S q , a ,ε , is mapped (see Figure 4) onto the channel of width w ′ = ε ′ cos ω ′ , bounded by the two linestangent to the ellipse T ( S q , a ,ε ), of slope t ′ = tan ω ′ = Ψ ( t ), where Ψ : " , √ → [0 , , t ′ = Ψ ( t ) = t √ − t , t = Ψ − ( t ′ ) = t ′ √ t ′ + . The intersection of these two channels and the x -axis is the horizontal segment centered at the origin,of length 2 ε sin ω = w sin ω = w ′ sin ω ′ = ε ′ tan ω ′ . In particular ε ′ = ε ′ ( ω, ε ) = ε tan ω ′ sin ω = ε cos( π/ + ω ) . (2.1)We can first replace each circular scatterer S q , a ,ε by the segment ˜ S q , a ,ε centered at ( x , y ), of slope π and length ε cos( π/ + ω ) = ε ′ (see Figures 3 and 4). Indeed, this change will result in altering, for FLORIN P. BOCA H L H L F igure
3. The free path length in the honeycomb and in the deformed honeycombF igure
4. Change of scatterers under the linear transformation T each ω , the free path length τ hex ε ( x , t ) to the free path length ˜ τ hex ε ( x , t ) corresponding to the latermodel by a quantity lesser than 4 √ ε , which is insignificant for the final result.Next we apply T to transport the problem from the honeycomb to the square lattice with con-gruence (mod 3) constraints (or in the opposite direction through T − ). The unit regular hexagon H centered at the origin is mapped to the hexagon T ( H ) which contains (0 ,
0) in Figure 4. Actually itwill be more convenient to replace T ( H ) by the fundamental domain F consisting of the union of HE LINEAR FLOW IN A HONEYCOMB 5 the square [0 , and of its translates [ − , × [0 ,
1) and [0 , × [ − , ε ′ = ε ′ ( ω, ε ) be as in(2.1), t ′ = tan ω ′ as above, and consider the vertical segment V ε ′ = { } × [ − ε ′ , ε ′ ]. Consider˜ q (cid:3) ε ′ ( x ′ , t ′ ) = inf { n ∈ N : x ′ + ( n , nt ′ ) ∈ Z + V ε ′ } , the horizontal free path length in the square lattice with vertical scatterers of (nonconstant) length2 ε ′ centered at points ( x ′ , y ′ ) = ( q , a ) ∈ Z . Consider also q (cid:3) ε ( x ′ , t ′ ), the horizontal free path in thesquare lattice with vertical scatterers of constant length 2 ε centered at points ( x ′ , y ′ ) = ( q , a ) ∈ Z .Clearly q (cid:3) ε + ( x ′ , t ′ ) ˜ q (cid:3) ε ′ ( x ′ , t ′ ) q (cid:3) ε − ( x ′ , t ′ ) when t ′ belongs to an interval I ′ and ε − ε ′ = ε ′ ( t ′ ) ε + , ∀ t ′ ∈ I ′ .For each angle ω ′ the transformation T maps H onto F and T − preserves the structure of channelsin the corresponding three-strip partition from [1, 7, 9] (see also the expository paper [13]). Removalof vertical scatterers V q , a ,ε ′ = T ( ˜ S q , a ,ε ) with q ≡ a (mod 3) in the Z picture results in dividing thecorresponding channel of the three-strip partition from the square lattice model into several sub-channels, and in the occurrence of longer trajectories associated with them. This is transported by T − back to the honeycomb model. The key observation here is that, by the Rule of Sines,˜ τ hex ε ( x , t )sin(2 π/ = ˜ q (cid:3) ε ′ ( T x , t ′ )sin( π/ − ω ) = ˜ q (cid:3) ε ′ ( T x , t ′ )cos( π/ + ω ) . This shows that˜ τ hex ε ( x , t ) > ξε ⇐⇒ ˜ q (cid:3) ε ′ ( T x , t ′ ) > ξ cos( π/ + ω ) ε √ = ξ ′ ε ′ , ξ ′ : = ξ √ , (2.2)leading to χ (cid:0) ξε , ∞ (cid:1)(cid:16) ˜ τ hex ε ( x , t ) (cid:17) = χ (cid:0) ξ ′ ε ′ , ∞ (cid:1)(cid:16) ˜ q (cid:3) ε ′ ( T x , t ′ ) (cid:17) , ∀ x ∈ H , ∀ t ∈ " , √ . For each interval I = [tan ω , tan ω ] ⊆ (cid:2) , √ (cid:3) one has ε − I : = ε cos( π/ + ω ) ε ′ = ε cos( π/ + ω ) ε + I : = ε cos( π/ + ω ) ,ε ± I = (cid:0) + O | I | ) (cid:1) ε. Employing now (2.2) and the fact that ε ˜ q (cid:3) ε ( T x , t ′ ) is non-decreasing, and taking ξ − I : = ξ ′ ε − I ε + I ξ ′ ξ + I : = ξ ′ ε + I ε − I , ξ ± I = (cid:0) + O ( | I | ) (cid:1) ξ ′ , we infer χ (cid:0) ξ + I ε + I , ∞ (cid:1)(cid:16) q (cid:3) ε + I ( T x , t ′ ) (cid:17) = χ (cid:0) ξ ′ ε − I , ∞ (cid:1)(cid:16) q (cid:3) ε + I ( T x , t ′ ) (cid:17) χ (cid:0) ξ ′ ε ′ , ∞ (cid:1)(cid:16) q (cid:3) ε + I ( T x , t ′ ) (cid:17) χ (cid:0) ξ ′ ε ′ , ∞ (cid:1)(cid:16) ˜ q (cid:3) ε ′ ( T x , t ′ ) (cid:17) = χ (cid:0) ξε , ∞ (cid:1)(cid:16) ˜ τ hex ε ( x , t ) (cid:17) χ (cid:0) ξ ′ ε ′ , ∞ (cid:1)(cid:16) q (cid:3) ε − I ( T x , t ′ ) (cid:17) χ (cid:0) ξ ′ ε + I , ∞ (cid:1)(cid:16) q (cid:3) ε − I ( T x , t ′ ) (cid:17) = χ (cid:0) ξ − I ε − I , ∞ (cid:1)(cid:16) q (cid:3) ε − I ( T x , t ′ ) (cid:17) . Consider G I ,ε ( ξ ) : = Z Ψ ( I ) dt ′ t ′ + t ′ + Z F d x ′ χ (cid:0) ξε , ∞ (cid:1)(cid:16) q (cid:3) ε ( x ′ , t ′ ) (cid:17) . (2.3) FLORIN P. BOCA
Applying the change of variable ( x ′ , t ′ ) = (cid:0) T x , Ψ ( t ) (cid:1) and employing (2.3), d x = √ d x ′ and dtt + = √ · dt ′ t ′ + t ′ + we infer34 G I ,ε + I ( ξ + I ) Z Ψ ( I ) dt ′ t ′ + t ′ + Z F d x ′ χ (cid:0) ξ + I ε + I , ∞ (cid:1)(cid:16) q (cid:3) ε + I ( x ′ , t ′ ) (cid:17) e P I ,ε ( ξ ) : = " H × I χ (cid:0) ξε , ∞ (cid:1)(cid:16) ˜ τ hex ε ( x , ω ) (cid:17) d x d ω = Z I dtt + Z H d x χ (cid:0) ξε , ∞ (cid:1)(cid:16) ˜ τ hex ε ( x , t ) (cid:17) Z Ψ ( I ) dt ′ t ′ + t ′ + Z F d x ′ χ (cid:0) ξ − I ε − I , ∞ (cid:1)(cid:16) q (cid:3) ε − I ( x ′ , t ′ ) (cid:17) = G I ,ε − I ( ξ − I ) . To simplify notation we simply denote G I , / (2 Q ) by G I , Q throughout. We shall employ the follow-ing result, whose proof occupies the remaining part of the paper. Theorem 2.
Let c , c ′ > such that c + c ′ < . For every interval I ⊆ [0 , of length | I | ≍ Q c , every ξ > and δ > , uniformly for ξ in compact subsets K of [0 , ∞ ) , G I , Q ( ξ ) = c I ζ (2) G ( ξ ) + O δ, K (cid:0) E c , c ′ ,δ ( Q ) (cid:1) , (2.4) where G ( ξ ) is the 40 term sum described below (1.3) andc I = Z I dtt + t + , c [0 , = π √ , E c , c ′ ,δ ( Q ) = Q max (cid:8) c ′ − , − c − c ′ (cid:9) + δ . (2.5) Proof.
Proof of Theorem 1. Let Q − = Q − I : = j ε − I k + ε − I + ε − I ε − : = Q − ε − I . Then q (cid:3) ε − I ( x ′ , t ′ ) q (cid:3) ε − ( x ′ , t ′ ), so χ ( ξ − I /ε − I , ∞ ) (cid:0) q (cid:3) ε − I ( x ′ , t ′ ) (cid:1) χ ( ξ − I /ε − I , ∞ ) (cid:0) q (cid:3) ε − ( x ′ , t ′ ) (cid:1) . Similarly, taking Q + = Q + I : = j ε + I k , ε + I ε + : = Q + ε + I − ε + I , we have χ ( ξ + I /ε + I , ∞ ) (cid:0) q (cid:3) ε + I ( x ′ , t ′ ) (cid:1) > χ ( ξ + I /ε + I , ∞ ) (cid:0) q (cid:3) ε + ( x ′ , t ′ ) (cid:1) . On the otherhand ε ± ε ± I = + O ( ε ), hence ξ ± I ε ± I = ξ ± I ε ± · ε ± ε ± I = (cid:0) + O ( | I | ) (cid:1) ξ ′ ε ± . We now infer G I , Q + (cid:16)(cid:0) + O ( | I | ) (cid:1) ξ ′ (cid:17) G I ,ε + I ( ξ + I ) e P I ,ε ( ξ ) G I ,ε − I ( ξ − I ) G I , Q − (cid:16)(cid:0) + O ( | I | ) (cid:1) ξ ′ (cid:17) . (2.6)Partition now the interval (cid:2) , √ (cid:3) as a union of N = (cid:2) ε − c (cid:3) intervals I j = [tan ω j , tan ω j + ] of equallength | I j | = N ≍ ε c , with 0 < c < Q ± j = Q ± I j . The intervals Ψ ( I j ) partition [0 ,
1] and | Ψ ( I j ) | ≍ ε c . Applying (2.6), Theorem 2 and the property of G of being HE LINEAR FLOW IN A HONEYCOMB 7
Lipschitz on the compact K we infer P hex ε ( ξ ) = π | H | N X j = e P I j ,ε ( ξ ) = √ π N X j = G I j , Q ± j ( ξ ) = √ πζ (2) N X j = c I j G (cid:16) + O ( | I | ) (cid:17) ξ √ ! + O δ, K (cid:0) N ε − max { c − , − c − c } (cid:1) = c [0 , √ πζ (2) G ξ √ ! + O δ, K (cid:0) ε − c + ε − max { c + c − , − c }− δ (cid:1) , uniformly for ξ in compacts of [0 , ∞ ). Taking c = c = we find P hex ε ( ξ ) = π G ξ √ ! + O δ, K (cid:0) ε − δ (cid:1) , as stated in Theorem 1 and in (1.3). (cid:3)
3. S ome number theoretical estimates
In this section we review and prove some number theoretical estimates that will be further used toestimate certain sums over integer lattice points with congruence constraints. The principal Dirichletcharacter (mod ℓ ) will be denoted by χ . The number of divisors of N is denoted by σ ( N ). Lemma 1 (Lemma 2.2 of [2]) . For each function f ∈ C [0 , N ] with total variation T N f , X q N ( q ,ℓ ) = f ( q ) = ϕ ( ℓ ) ℓ Z N f ( x ) dx + O (cid:16) ( k f k ∞ + T N f ) σ ( N ) (cid:17) . Lemma 2 (Lemma 2.1 of [3]) . For each function V ∈ C [0 , N ] , X q N ( q ,ℓ ) = ϕ ( q ) q V ( q ) = C ( ℓ ) Z N V ( x ) dx + O ℓ (cid:16) ( k V k ∞ + T N V ) log N (cid:17) , where C ( ℓ ) = ϕ ( ℓ ) ζ (2) ℓ Y p ∈P p | ℓ − p ! − = ζ (2) Y p ∈P p | ℓ + p ! − = ϕ ( ℓ ) ℓ L (2 , χ ) . We need a more precise form of Lemma 1, as follows:
Lemma 3.
Suppose that ( r , ℓ ) = . For each function V ∈ C [0 , N ] , X q Nq ≡ r (mod ℓ ) ϕ ( q ) q V ( q ) = C ( ℓ ) ϕ ( ℓ ) Z N V ( x ) dx + O ℓ (cid:16) ( k V k ∞ + T N V ) log N (cid:17) . In particular X q Nq ≡± ϕ ( q ) q V ( q ) ∼ ζ (2) Z N V ( x ) dx , X q Nq ≡ ϕ ( q ) q V ( q ) ∼ ζ (2) Z N V ( x ) dx . FLORIN P. BOCA
Proof.
When ( k , ℓ ) = k the multiplicative inverse of k (mod ℓ ). Let G = U ( Z /ℓ Z ) denotethe multiplicative group of units of Z /ℓ Z and b G be the group of characters χ : G → T , extended asmultiplicative functions on N . Set V d ( x ) = V ( xd ). By Schur’s orthogonality relations for characters,for every x , s ∈ N with ( s , ℓ ) = ϕ ( ℓ ) X χ ∈ b G χ ( x ) χ ( ¯ s ) = ϕ ( ℓ ) X χ ∈ b G χ ( x ) χ ( s ) = x ≡ s (mod ℓ ) , x . s (mod ℓ ) . (3.1)Taking s = r and summing over x = q N , we infer by M ¨obius summation, with q = md , X q Nq ≡ r (mod ℓ ) ϕ ( q ) q V ( q ) = ϕ ( ℓ ) X q N ϕ ( q ) q V ( q ) X χ ∈ b G χ ( q ) χ (¯ r ) = ϕ ( ℓ ) X q N X d | q µ ( d ) d V ( q ) X χ ∈ b G χ ( q ) χ (¯ r ) = ϕ ( ℓ ) X d N ⌊ N / d ⌋ X m = X χ ∈ b G µ ( d ) d V d ( m ) χ ( md ) χ (¯ r ) = ϕ ( ℓ ) X χ ∈ b G χ (¯ r ) X d N µ ( d ) χ ( d ) d ⌊ N / d ⌋ X m = V d ( m ) χ ( m ) . (3.2)We split the inner sum above according to whether χ = χ or χ , χ . Employing Lemma 1 for thefunction V d we find that the contribution of the former is1 ϕ ( ℓ ) X d N µ ( d ) χ ( d ) d ϕ ( ℓ ) ℓ Z ⌊ N / d ⌋ V d + O (cid:16) ( k V d k ∞ + T N / d V d ) log N (cid:17)! = ℓ X d N µ ( d ) χ ( d ) d Z N V + O ℓ (cid:16) ( k V k ∞ + T N V ) log N (cid:17) = ℓ L (2 , χ ) + O ℓ (cid:16) N (cid:17)! Z N V + O ℓ (cid:16) ( k V k ∞ + T N V ) log N (cid:17) = C ( ℓ ) ϕ ( ℓ ) Z N V + O ℓ (cid:16) ( k V k ∞ + T N V ) log N (cid:17) . When χ , χ we find by partial summation and P´olya-Vinogradov (or a weaker inequality) that theinnermost sum in (3.2) is ≪ ℓ T N V + k V k ∞ , so the total contribution of non-principal characters in(3.2) is ≪ ℓ k V k ∞ log N . The proof is complete. (cid:3) Lemma 4 (Proposition A4 of [6]) . Assume that q > and h are integers, I and J are intervals oflength less than q, and f ∈ C ( I × J ) . For any integer T > and any δ > X a ∈I , b ∈J ( a , q ) = ab ≡ h (mod q ) f ( a , b ) = ϕ ( q ) q " I×J f ( x , y ) dx dy + E , HE LINEAR FLOW IN A HONEYCOMB 9 with
E ≪ δ T k f k ∞ q + δ ( h , q ) + T k∇ f k ∞ q + δ ( h , q ) + k∇ f k ∞ |I| |J | T , where k f k ∞ and k∇ f k ∞ denote the sup-norm of f and respectively (cid:12)(cid:12)(cid:12) ∂ f ∂ x (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12) ∂ f ∂ y (cid:12)(cid:12)(cid:12) on I × J . For q , ℓ positive integers denote A ( q , ℓ ) = q , ℓ ) = , Q p ∈P p | ( q ,ℓ ) (cid:16) − p (cid:17) − if ( q , ℓ ) > . Lemma 5.
Suppose that ( r , ℓ ) = . For any interval I , uniformly in |I| , X x ∈I , ( x , q ) = x ≡ r (mod ℓ ) = A ( q , ℓ ) ℓ · ϕ ( q ) q |I| + O ( σ ( q )) . Proof.
Without loss of generality assume I = [1 , N ]. As in the proof of Lemma 3 take s = r andsum in (3.1) over x ∈ I with ( x , q ) =
1. By M ¨obius inversion with x = md we infer X x ∈I , ( x , q ) = x ≡ r (mod ℓ ) = ϕ ( ℓ ) X x ∈I ( x , q ) = X χ ∈ b G χ ( x ) χ (¯ r ) = ϕ ( ℓ ) X χ ∈ b G X x ∈I χ ( x ) χ (¯ r ) X d | qd | x µ ( d ) = ϕ ( ℓ ) X d | q µ ( d ) X x ∈I d | x X χ ∈ b G χ ( x ) χ (¯ r ) = ϕ ( ℓ ) X d | q µ ( d ) ⌊ N / d ⌋ X m = X χ ∈ b G χ ( md ) χ (¯ r ) . (3.3)The contribution of χ = χ to (3.3) is1 ϕ ( ℓ ) X d | q µ ( d ) χ ( d ) X m ⌊ N / d ⌋ ( m ,ℓ ) = = ϕ ( ℓ ) X d | q µ ( d ) χ ( d ) ϕ ( ℓ ) ℓ (cid:18) Nd + O (1) (cid:19)! = |I| ℓ X d | q µ ( d ) χ ( d ) d + O ( σ ( q )) . (3.4)In the contribution of non-principal characters to (3.3),1 ϕ ( ℓ ) X d | q µ ( d ) X χ ∈ b G χ , χ χ ( d ) χ (¯ r ) ⌊ N / d ⌋ X m = χ ( m ) , (3.5)the innermost sum is ≪ ℓ ≪ ℓ σ ( q ). The statement follows now because the sum in (3.4) is equal to P d | q µ ( d ) d = ϕ ( q ) q when ( q , ℓ ) =
1, while when ( q , ℓ ) >
1, writing q = p α · · · p α r r ˜ q with p , . . . , p r prime divisors of ℓ and ( ˜ q , ℓ ) =
1, this sum is equal to X d | ˜ q µ ( d ) d = ϕ ( ˜ q )˜ q = ϕ ( q ) q r Y i = − p i ! − = A ( q , ℓ ) ϕ ( q ) q , as desired. (cid:3) We also need a slight extension of Lemma 4. Suppose that ( r , ℓ ) = x the multi-plicative inverse of x (mod ℓ q ) when ( x , ℓ q ) =
1. The Kloosterman type sums K ( m , n ; ℓ q ) : = X x (mod ℓ q )( x ,ℓ q ) = e mx + n ¯ x ℓ q ! , e K r ( m , n ; ℓ q ) : = X x (mod ℓ q )( x , q ) = , x ≡ r (mod ℓ ) e mx + n ¯ x ℓ q ! , K I , r ( m , n ; ℓ q ) : = X x ∈I , ( x , q ) = x ≡ r (mod ℓ q ) e mx + n ¯ x ℓ q ! , will be used to estimate e N q ,ℓ, r , h ( I , I ) : = (cid:8) ( x , y ) ∈ I × I : ( x , q ) = , x ≡ r (mod ℓ ) , xy ≡ h (mod ℓ q ) (cid:9) Lemma 6.
When ( r , ℓ ) = , for any interval I of length less than q, | K I , r (0 , n ; ℓ q ) | ≪ ℓ,δ ( n , q ) q + δ . Proof.
We write K I , r (0 , n ; ℓ q ) = X x ∈I , ( x , q ) = x ≡ r (mod ℓ ) e n ¯ x ℓ q ! = X x (mod ℓ q )( x , q ) = , x ≡ r (mod ℓ ) e n ¯ x ℓ q ! X y ∈I ℓ q X k (mod ℓ q ) e k ( y − x ) ℓ q ! = ℓ q X k (mod ℓ q ) e K r ( − k , n ; ℓ q ) X y ∈I e ky ℓ q ! , and | e K r ( m , n ; ℓ q ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X x (mod ℓ q )( x ,ℓ q ) = e mx + n ¯ x ℓ q ! ℓ X j (mod ℓ ) e j ( x − r ) ℓ ! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ℓ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j (mod ℓ ) e (cid:18) − jr ℓ (cid:19) K ( m + jq , n ; ℓ q ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) max j | K ( m + jq , n ; ℓ q ) | . Employing (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X y ∈I e ky ℓ q ! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) min ( |I| + , (cid:13)(cid:13)(cid:13) k ℓ q (cid:13)(cid:13)(cid:13) ) , (3.6) | K (0 , n ; ℓ q ) | = | c ℓ q ( n ) | ( n , ℓ q ) ( n , q ) ℓ ≪ ℓ ( n , q ) q , and the Weil estimate | K ( m + jq , n ; ℓ q ) | σ ( ℓ q )( m + jq , n , ℓ q ) ( ℓ q ) ≪ ℓ,δ ( n , q ) q + δ , we infer | K I , r (0 , n ; ℓ q ) | |I| + ℓ q | e K r (0 , n ; ℓ q ) | + ℓ q X k (mod ℓ q ) k , | e K r ( − k , n ; ℓ q ) | (cid:13)(cid:13)(cid:13) k ℓ q (cid:13)(cid:13)(cid:13) ≪ ℓ,δ ( n , q ) q + δ + ℓ q · ( n , q ) q + δ ℓ q log( ℓ q ) ≪ ℓ ( n , q ) q + δ , Here k x k = dist( x , Z ), x ∈ R , and c q ( n ) = K ( n , q ) = K (0 , n ; q ) is the Ramanujan sum. HE LINEAR FLOW IN A HONEYCOMB 11 as desired. (cid:3)
Lemma 7.
Suppose that ( r , ℓ ) = . For any intervals I and I of length less than q, any integer hand any δ > , e N q ,ℓ, r , h ( I , I ) = A ( q , ℓ ) ℓ · ϕ ( q ) q |I | |I | + O ℓ,δ (cid:16) ( h , q ) q + δ (cid:17) . Proof.
We write e N q ,ℓ, r , h ( I , I ) = X x ∈I , y ∈I ( x , q ) = , x ≡ r (mod ℓ ) ℓ q X k (mod ℓ q ) e k ( y − h ¯ x ) ℓ q ! = M + E , where ¯ x denotes the multiplicative inverse of x (mod ℓ q ) and E = ℓ q X k (mod ℓ q ) k , X y ∈I e ky ℓ q ! K I , r (0 , − hk ; ℓ q ) , M = ℓ q X x ∈I , y ∈I ( x , q ) = , x ≡ r (mod ℓ ) = ℓ q A ( q , ℓ ) ℓ · ϕ ( q ) q |I | + O ( σ ( q )) ! (cid:16) |I | + O (1) (cid:17) = A ( q , ℓ ) ℓ · ϕ ( q ) q |I | |I | + O ℓ ( σ ( q )) . (3.7)From (3.6), Lemma 6 and ( hk , ℓ q ) ( h , q )( k , q ) ℓ , we infer as in the proof of [6, Proposition A.3] | E | ≪ ℓ,δ q − + δ X k (mod ℓ q ) k , ( hk , ℓ q ) / (cid:13)(cid:13)(cid:13) k ℓ q (cid:13)(cid:13)(cid:13) ≪ ℓ,δ q − + δ ( h , q ) X k (mod ℓ q ) k , ( k , q ) / (cid:13)(cid:13)(cid:13) k ℓ q (cid:13)(cid:13)(cid:13) ≪ q + δ ( h , q ) X k ( q − / ( k , q ) / k ≪ δ ( h , q ) q + δ . (3.8)The statement follows now from (3.7) and (3.8). (cid:3) Lemma 8.
Suppose that ( r , ℓ ) = . Let I , I be intervals of length less than q, f ∈ C ( I × I ) ,and h ∈ Z . For any integer T > and any δ > , X x ∈I , y ∈I ( x , q ) = , x ≡ r (mod ℓ ) xy ≡ h (mod ℓ q ) f ( x , y ) = A ( q , ℓ ) ℓ · ϕ ( q ) q " I ×I f ( u , v ) du dv + E , with E ≪ ℓ,δ T k f k ∞ q + δ ( h , q ) + T k∇ f k ∞ q + δ ( h , q ) + k∇ f k ∞ |I ||I | T . Proof.
This plainly follows from Lemma 7 as in the proof of [4, Lemma 2.2]. (cid:3)
Only the case ℓ = x ≡ r . X x ∈I , y ∈I ( x , q ) = , x ≡ r (mod 3) xy ≡ h (mod 3 q ) f ( x , y ) ∼ ϕ ( q ) q " I ×I f ( x , y ) dx dy ! · if 3 ∤ q , if 3 | q .
4. C oding the linear flow in Z and the three - strip partition To keep notation short denote x ∨ y = max { x , y } , x ∧ y = min { x , y } , x + = max { x , } . Consider c , c ′ , δ, ξ and I as in the statement of Theorem 2 and E c , c ′ ,δ ( Q ) as in (2.5). For ( A Q ), ( B Q )sequences of real numbers we write A Q ≅ B Q when A Q = B Q + O δ,ξ (cid:0) E c , c ′ ,δ ( Q ) (cid:1) , uniformly for ξ incompact subsets of [0 , ∞ ). Our primary aim is to estimate the quantity G I , Q ( ξ ) from Theorem 2,associated with lattice points from Z with corresponding vertical scatterers of width 2 ε = Q , as Q → ∞ .It is useful to recall first the approach and notation from [7]. F ( Q ) denotes the set of Fareyfractions of order Q , consisting of rational numbers γ = aq , 0 < a q Q , with ( a , q ) = I will be first partitioned into intervals I γ = ( γ, γ ′ ) with γ, γ ′ consecutive in F I ( Q ) : = F ( Q ) ∩ I . Each interval I γ is further partitioned into subintervals I γ, k , k ∈ Z , defined as I γ, k = ( t k , t k − ] , I γ, = ( t , u ] , I γ, − k = ( u k − , u k ] , k ∈ N , where t k = a k − ε q k , u k = a ′ k + ε q ′ k , k ∈ N , and q k = q ′ + kq , a k = a ′ + ka , q ′ k = q + kq ′ , a ′ k = a + ka ′ , k ∈ Z , satisfy the fundamental relations a k − q k − a k q k − = = a k − q − aq k − , a ′ k q ′ k − − a ′ k − q ′ k = = a ′ q ′ k − − a ′ k − q ′ , k ∈ Z , ε q k ∧ ε q ′ k > ε ( q + q ′ ) > , k > . Consider also t : = tan ω and γ k = a k q k , k ∈ N . As it will be seen shortly, the coding of the linear flow is considerably more involved than in thecase of the square lattice. As a result our attempt of providing asymptotic results for the repartitionof the free path length will require additional partitioning for each of the interval I γ, k . For symmetryreasons the mediant intervals ( γ, γ ] and ( γ , γ ′ ] will contribute by the same amount to the mainterm, so we shall only consider t ∈ ( γ, γ ] and redefine I γ, : = ( t , t − ] , t − : = γ = a ′ + aq ′ + q . This explains the appearance of the factor 2 in formula (2.4).As in [7, Section 3] we shall consider , when t = tan ω ∈ I γ, k , k ∈ N , w A ( t ) = a + ε − qt = q ( u − t ) , w C k ( t ) = a k − − ε − q k − t = q k − ( t k − − t ) , w B k ( t ) = q k t − a k + ε = q k ( t − t k ) ∈ [0 , ε ] , Which are not geometrically obvious but become apparent after translating G I , Q ( ξ ) into sums involving(sub)intervals (of) I γ, k and Farey fractions from F I ( Q ). Here we use t = tan ω as variable and use dtt + t + instead of d ω . HE LINEAR FLOW IN A HONEYCOMB 13 representing the widths of the bottom, center, and respectively top channels A , C k , B k , of thethree-strip partition of [0 , (see Figures 5 and 6). Clearly ε = w A ( t ) + w B k ( t ) + w C k ( t ) , = qw A ( t ) + q k + w C k ( t ) + q k w B k ( t ) , ∀ t ∈ I γ, k . Recall [7, 9] that in the case of the square lattice the three weights corresponding to ω are given by W A ( t ) = ( q − ξ Q ) + w A ( t ) , W B k ( t ) = ( q k − ξ Q ) + w B k ( t ) , W C k ( t ) = ( q k + − ξ Q ) + w C k ( t ) . (4.1)They reflect the area of the parallelogram of height given respectively by w A , w C k or w B k , and lengthgiven by the distance from ξ Q to the bottom of the corresponding sub-channel (if ξ Q is lesser thanthe total length of the sub-channel). (0 , − ε )(0 , ε ) A C k B k ◦◦ ◦◦ ◦◦◦◦◦◦ ( q k − , a k − − ε )( q k − , a k − + ε )( q , a + ε )( q , a − ε ) ( q k + , a k + − ε )( q k + , a k + + ε )( q k , a k + ε )( q k , a k − ε ) ω F igure
5. The three-strip partition of R / Z when t ∈ I γ, k F igure
6. The tiling S ω of the plane (shaded region represents R / Z )The range for q k = q ′ + kq , respectively q ′ k , will be q k ∈ I q , k : = (cid:0) Q + ( k − q , Q + kq (cid:3) , respectively q ′ k ∈ I q ′ , k . Denote by r , r k , r ′ , r ′ k ∈ Z / Z the remainders (mod 3) of q − a , q k − a k , q ′ − a ′ , q ′ k − a ′ k respectively.The equality a ′ q − aq ′ = r , r k , r k + ) ∈ ( Z / Z ) can beequal to zero. Similarly, at most one element of the triple ( r ′ , r ′ k , r ′ k + ) can be zero.To ascertain the contribution of the slope t = tan ω ∈ I γ, k to G I , Q ( ξ ), we should look at the tiling S ω defined by the three-strip partition of R (shown in Figure 6), but also at S ← ω , its left-horizontaltranslate by (1 , S ↓ ω , its down-vertical translate by (0 , m , n ) ∈ Z with ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦◦ ◦◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦◦ ◦ ◦◦ ◦ ω O ◦◦ F igure
7. The channels C , C ← and C ↓ m ≡ n (mod 3) are being removed, sinks are going to arise in the channels. This phenomenon willlead to frequent occurrence of trajectories much longer then in the case of the square lattice. Acareful analysis of the bottom of the channels A , C k and B k is required when the correspondingslit where trajectory ends in the case of the square lattice has been removed. Besides, there is amanifest di ff erence between the three situations where the channel originates at O = (0 ,
0) (thiswill be referred to as C O contribution), at ( − ,
0) ( C ← contribution), or at ( − ,
0) ( C ↓ contribution),resulting from the di ff erent congruence conditions (mod 3) satisfied by the centers of removedslits. This is shown in Figure 6 where the small circles centered at lattice points ( m , n ) with m . n (mod 3) represent the vertical slits of width 2 ε = Q . We were not able to spot any symmetry thatwould reduce the analysis to only one of these three types of channels. The contributions to Φ hex ofthe five types of situations that we analyze seem to be quite di ff erent (see Figure 19).To attain a better visualization of the structure of channels we shall represent the slope tan ω horizontally. The possible situations are shown in Figure 8, where dotted lines indicate that thecorresponding slit has been removed from S ω in the case of C O , from S ← ω in the case of C ← andrespectively from S ↓ ω in the case of C ↓ .To clarify the terminology, by “slit q k ” etc. we will mean the slit which is centered at some latticepoint ( q k , m ) and which intersects the channel that is analyzed (there is at most one such point forgiven q k ). HE LINEAR FLOW IN A HONEYCOMB 15 ( − , , , − , − , , − , , , − , − , , − , − , , , − r , r k , r k + ) C O C ← C ↓ F igure
8. The removed slits for C O , C ← and C ↓
5. T he contribution of channels whose slits are not removed
This resembles the situation of the square lattice and will be discussed in this section. The moreintricate situation of the channels where bottom slits are removed will be analyzed in Sections 6-9.When the “first” slit (i.e. the one corresponding to q for A , q k + for C k , respectively q k for B k ) isnot removed, the table from Figure 8 shows that the corresponding weight is described in the tablefrom Figure 9.The weights W A , W B k and W C k are given by (4.1) as in the case of the square lattice. Thecumulative contribution is G (0) I , Q ( ξ ) = X ( α,β ) ∈ ( Z / Z ) \{ (0 , } G (0) I , Q ,α,β ( ξ ) , with G (0) I , Q ,α,β ( ξ ) given by ∞ X k = X γ ∈F I ( Q ) q − a ≡ α (mod 3) q k − a k ≡ β (mod 3) Z t k − t k (cid:0) ( q − ξ Q ) + w A ( t ) + ( q k + − ξ Q ) + w C k ( t ) + ( q k − ξ Q ) + w B k ( t ) (cid:1) dtt + t + . (5.1) Remark 1..
Putting x = q − a, y = q k , and employing q k − a k = q k − + aq k q = xy − q , we see thatthe summation conditions in (5.1) are equivalent to ( x , q ) = , x ≡ α (mod 3) , and xy ≡ β q + q ) . Note that ( β q + , q ) = and sum as follows: ( − , , W A + W B k W A + W C k W C k + W B k (1 , − , W A + W B k W C k + W B k W A + W C k ( − , , − W A + W C k W A + W C k + W B k W B k (1 , , W A + W C k W B k W A + W C k + W B k (0 , − , − W C k + W B k W A + W C k + W B k W A (0 , , W C k + W B k W A W A + W C k + W B k ( − , − , W A + W C k + W B k W A + W B k W C k (1 , , − W A + W C k + W B k W C k W A + W B k ( r , r k , r k + ) C O C ← C ↓ F igure
9. The contribution of channels whose slits are not removed • When α , Lemma 8 may be applied ( because ( x , q ) = , followed by Lemma 3. • When α = we have β , and x = x, ˜ x ∈ q (1 − I ) . Furthermore, β q + ≡ , soq ≡ − β (mod 3) and one may first sum, as in Lemma 4, over ( ˜ x , y ) and the conditions q − a = x = x , ˜ x ∈ q (1 − I ) , ( ˜ x , q ) = , q ≡ − β (mod 3) , ˜ xy ≡ β q + (mod q ) , y ∈ I q , k , (5.2) and then sum over q ∈ [1 , Q ] with q ≡ − β (mod 3) employing Lemma 3.In all situations where sums over γ ∈ F I ( Q ) with q − a ≡ α (mod 3) and q k − a k ≡ β (mod 3) , ( α, β ) , (0 , , have to be evaluated, the resulting constant will be ζ (2) . The following elementary estimate will be used throughout.
Lemma 9.
For b , c , h , λ, µ ∈ R such that b , c , c + h and λ c + µ = , there exist θ ′ , θ = θ c , h ∈ [ − , such that Z c + hc λ t + µ t + t + du = h λ b + b + + h θ ( | λ | + | µ | ) + h θ ′ | b − c || λ | . Proof.
This follows immediately applying Taylor’s formula twice: Z c + hc dtt + t + = hc + c + − h (2 c + c + c + + ξ ′ h , | ξ ′ | , Z c + hc t dtt + t + = hc + c + − h (2 c + c + c + ! c + h c + c + + ξ ′′ h , | ξ ′′ | , and employing (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) c + c + − b + b + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | b − c | . (cid:3) Together with 0 < q − Q q ∧ q ′ Lemma 9 yields ( t − = γ ) Z t − t w B ( t ) dtt + t + = Z γ t q ′ t − a ′ + ε t + t + dt = ( q − Q ) Q q ′ q ( γ + γ + + O Q ! . (5.3) HE LINEAR FLOW IN A HONEYCOMB 17 Z t − t w C ( t ) dtt + t + = Z γ γ qt − at + t + dt − Z t γ qt − at + t + dt − Z γ t q ′ t − a ′ + ε t + t + dt = q ( γ − γ ) − ( t − γ ) γ + γ + − q ′ ( γ − t ) γ + γ + + O (cid:16) q ( γ − γ ) + q ′ ( γ − t ) + q ′ ( γ − t ) ( t − γ ) (cid:17) = q − Q Q q ′ q ( γ + γ + Q − q q + Q − qq ′ ! + O Q q ! . (5.4) Z t − t w A ( t ) dtt + t + = Z u t a + ε − qtt + t + dt − Z u γ a + ε − qtt + t + dt = q ( u − t ) − ( u − γ ) γ + γ + + O (cid:16) q ( u − t ) + q ( u − t ) ( u − γ ) (cid:17) = ( q − Q ) ( q + q ′ )2 Q q ′ q ( γ + γ + + O Q qq ′ ! . (5.5)For every k > Z t k − t k w B k ( t ) dtt + t + = Z t k − t k q k t − a k + ε t + t + dt = ( Q − q ) Q q k − q k ( γ + γ + + O qq k − q k ! . (5.6) Z t k − t k w C k ( t ) dtt + t + = Z t k − t k a k − − ε − q k − tt + t + dt = ( Q − q ) Q q k − q k ( γ + γ + + O qq k − q k ! . (5.7) Z t k − t k ε dtt + t + = t k − − t k Q ( γ + γ + + O (cid:16) ε ( t k − − t k )( t k − γ ) + ε ( t k − − t k ) (cid:17) = Q − qQ q k − q k ( γ + γ + + O Qqq k − q k ! . (5.8) Z t k − t k w A ( t ) dtt + t + = Z t k − t k ε − w B k ( t ) − w C k ( t ) t + t + dt = Q − q Q q k − q k ( γ + γ + q k + − Qq k + q k − Qq k − ! + O qq ′ q k − q k ! . (5.9)The total contribution of the error terms from (5.3)-(5.7) and (5.9) to G (0) I , Q ( ξ ) is ≪ X γ ∈F I ( Q ) qq ′ ) + ∞ X k = X γ ∈F I ( Q ) q ′ q k − q k + qq k − q k ! X γ ∈F I ( Q ) qq ′ ) + X γ ∈F I ( Q ) qq ′ ∞ X k = q k − q k = X γ ∈F I ( Q ) qq ′ ) Q X γ ∈F I ( Q ) qq ′ Q (cid:18) | I | + Q (cid:19) | I | Q . When summing over a family of intervals I that partition [0 ,
1] this adds up to O ( Q − ), thus all errorterms above can be discarded below. We emphasize that, since the contribution of each interval I γ, k is | I γ, k | , we can remove one element of F I ( Q ) for every I . Applying Remark 1 to the inner sum in (5.1) with x = q − a ∈ q (1 − I ), y = q k ∈ I q , k , andemploying formulas (5.3)-(5.9) we find G (0) I , Q ,α,β ( ξ ) = Q X q = ∞ X k = X x ∈ q (1 − I ) , y ∈I q , k x ≡ α (mod 3) xy ≡ β q + q ) f q ( x , y ) + Q X q = X x ∈ q (1 − I ) , y ∈ ( Q − q , Q ] x ≡ α (mod 3) xy ≡ β q + q ) g q ( x , y ) , (5.10)with C functions f q and g q on R × ( R + \ { q , ξ Q − q , ξ Q } ) given by f q ( x , y ) = q q + ( q − x ) F q ( y ) , g q ( x , y ) = q q + ( q − a ) G q ( y ) , F q ( y ) = Q − qQ ( y − q ) y y + q − Qy + y − Qy − q ! ( q − ξ Q ) + + ( Q − q ) Q ( y − q ) y ( y − ξ Q ) + y − q + ( y + q − ξ Q ) + y ! , G q ( y ) = y + q − QQ y ( y + q ) Q − q − yy + q + Q − qy ! ( y + q − ξ Q ) + + ( y + q − Q ) Q y ( y + q ) y + qy ( q − ξ Q ) + + ( y − ξ Q ) + ! . The innermost sums in (5.10) will be estimated employing Lemmas 4 or 8. We first need to bound k f q k ∞ , k∇ f q k ∞ on q (1 − I ) × I q , k , k >
1, and respectively k g q k ∞ , k∇ g q k ∞ on q (1 − I ) × ( Q − q , Q ].From y > Q in the first case and y + q − Q Q Q y + q Q in the second one we find for all y ∈ I q , k , k > f q ( x , y ) F q ( y ) Q − qQ ( y − q ) y q + ( Q − q ) (cid:18) + qy − q + qy (cid:19)! Q − q ) Q ( y − q ) y , (5.11) | F ′ q ( y ) | ξ + Q − q ) Q ( y − q ) y , (5.12)and for all y ∈ ( Q − q , Q ]:0 g q ( x , y ) G q ( y ) Q ( y + q ) · y + q ) + Q ( y + q ) · (cid:0) (2 y + q ) q + y (cid:1) = Q , (5.13) | G ′ q ( y ) | Qy . (5.14)From (5.11) we infer ∞ X k = k F q k I q , k Q − q ) Q ∞ X k = (cid:0) Q + ( k − q (cid:1)(cid:0) Q + ( k − q (cid:1) = Qq , which leads in turn to Q X q = q + δ ∞ X k = k f q k q (1 − I ) ×I q , k Q Q X q = q − + δ Q + δ . HE LINEAR FLOW IN A HONEYCOMB 19
From (5.12) we infer ∞ X k = k F ′ q k I q , k ξ + Q − q ) Q ∞ X k = (cid:0) Q + ( k − q + (cid:1) (cid:0) Q + ( k − q + (cid:1) ξ + Q ∞ X k = (cid:0) Q + ( k − q + (cid:1)(cid:0) Q + ( k − q + (cid:1) ξ + Qq ( Q − q + , which leads to ∞ X k = k∇ f q k q (1 − I ) ×I q , k ∞ X k = q k F q k I q , k + k F ′ q k I q , k ! ξ + q ( Q − q + , and finally gives Q X q = q + δ ∞ X k = k∇ f q k q (1 − I ) ×I q , k ≪ ξ X q Q q − / + δ Q − q + + X Q q Q q − / + δ Q − q + Q Q X q = q − + δ + (cid:18) Q (cid:19) − + δ Q X n = n ≪ Q − + δ log Q , (5.15) Q X q = q ∞ X k = k∇ f q k q (1 − I ) ×I q , k ≪ ξ Q X q = Q − q + ≪ log Q . (5.16)Taking T = [ Q c ′ ] we infer upon (5.15) and (5.16), with E c , c ′ ,δ ( Q ) as in Theorem 2, Q X q = ∞ X k = T q + δ (cid:16) T k f q k q (1 − I ) ×I q , k + q k∇ f q k q (1 − I ) ×I q , k (cid:17) + q | I ||I q , k | T k∇ f q k q (1 − I ) ×I q , k ! ≪ ξ,δ T Q − + δ + T Q − + δ log Q + | I | T − log Q ≪ E c , c ′ ,δ ( Q ) , (5.17)uniformly for ξ in compact subsets of [0 , ∞ ). From (5.13) and (5.14) we also get Q X q = T q + δ (cid:16) T k g q k q (1 − I ) ×I q , + q k∇ g q k q (1 − I ) ×I q , (cid:17) + q | I | T k∇ g q k q (1 − I ) ×I q , ! ≪ δ E c , c ′ ,δ ( Q ) . (5.18)When α ,
0, Lemma 8 applies to the innermost sums in (5.10). Upon (5.17) and (5.18) we find G (0) I , Q ,α,β ( ξ ) ≅ X q Q ∤ q ϕ ( q )9 q · qc I ∞ X k = Z I q , k F q ( y ) dy + Z QQ − q G q ( y ) dy ≅ X q Q | q ϕ ( q )6 q · qc I ∞ X k = Z I q , k F q ( y ) dy + Z QQ − q G q ( y ) dy . (5.19)Applying Lemma 3 with ℓ = q in (5.19) and making the substitution ( q , y ) = ( Qu , Qw ) we gather G (0) I , Q ,α,β ( ξ ) ≅ · + · ! c I ζ (2) · Z du Z − u dw F (0 . ( ξ ; u , w ) + Z ∞ dw F (0 . ( ξ ; u , w ) ! , with F (0 . and F (0 . as in (5.22). The total contribution of cases α , G (0) I , Q ( ξ ) is obtained bymultiplying the quantity above by 6.When α = β = ± G (0 I , Q , , ( ξ ) + G (0) I , Q , , − ( ξ ) ≅ X q Q ( q , = ϕ ( q ) q · qc I Z ∞ Q F q ( y ) dy + Z QQ − q G q ( y ) dy ! ≅ c I ζ (2) Z du Z − u dw F (0 . ( ξ ; u , w ) + Z ∞ dw F (0 . ( ξ ; u , w ) ! . (5.20)Applying Lemma 2 to (5.20) we finally find G (0) I , Q ( ξ ) ≅ c I ζ (2) Z du Z − u dw F (0 . ( ξ ; u , w ) + Z ∞ dw F (0 . ( ξ ; u , w ) ! , (5.21)where F (0 . ( ξ ; u , w ) = ( w + u − (2 w + u ) w ( w + u ) ( u − ξ ) + + ( w + u − w ( w + u ) ( w − ξ ) + + w + u − w ( w + u ) − u − ww + u + − uw ! ( w + u − ξ ) + , F (0 . ( ξ ; u , w ) = − u ( w − u ) w w + u − w + w − w − u ! ( u − ξ ) + + (1 − u ) ( w − u ) w ( w − ξ ) + + (1 − u ) ( w − u ) w ( w + u − ξ ) + . (5.22) Remark 2..
For each angle ω the weight W γ, k ( t ) = W hex γ, k ( t ) is clearly no larger than the weightW (cid:3) γ, k ( t ) from the situation where no slit is being removed (and which corresponds up to some scalingto the case of the the unit square). Since the corresponding limiting distribution Φ (cid:3) satisfies (1.2)[8, 7, 11] , it follows that Φ hex satisfies the second inequality in (1.2) . The first inequality in (1.2) follows for instance fromG (0) ( ξ ) : = Z du Z ∞ dw (1 − u ) ( w − u ) w ( w − ξ ) + > Z du (1 − u ) Z ∞ ξ v dv ( v + ξ ) = ξ .
6. C hannels with removed slits . T he case r = r k = ± r = r k =
1. Then r k + = r + r k = − C ← and the C ↓ contributions. We shall sum as in Remark 1 above with x = q − a ∈ q (1 − I ), α = y = q k ∈ I q , k , β =
1, considering X ∗ = X ∗ γ ∈F I ( Q ) q − a ≡ q k − a k ≡ . The table in Figure 8 shows that the weights W γ, k ( t ) for ( C ← , r = r k =
1) and for ( C ↓ , r = r k = − C ↓ , r = r k =
1) and ( C ← , r = r k = − Note that because of the scaling of ξ this does not imply Φ hex ( ξ ) Φ (cid:3) ( ξ ) (see also Figure 2). HE LINEAR FLOW IN A HONEYCOMB 21 the corresponding contribution for r = r k = − r = r k = β by − β and α by − α , which will produce the same mainterm and error size). As a result we shall only take r = r k = The C ← contribution. The slits q and q k are removed, while 2 q , q k + , q k + , 2 q k and q k + q k + are not because 2 r = r k + = r k = r k + = r k + r k + = . T ( • ), respectively B ( • ), the height of the top, respectively bottom, of the slit • with respect to the top of the strip B k ,with positive downwards direction. Since B ( q k + ) = w B k + w C k < T (2 q ) = w B k + w C k ) < B ( q k + ) = w B k + w C k and B (2 q k ) = w B k − w A − w C k < T ( q k + + q k ) = w B k − w A , the slits 2 q , q k + , q k + , q k + + q k and 2 q k lock all channels B k , C k and A . Two cases arise: ( − , − ε )( − , ε ) ( − , − ε )( − , ε ) q q ξ Q q k q k + q k + q q ξ Q q k q k + q k + A C k B k A C k B k w A w B k w C k w B k w C k w A w B k w B k w C k w C k w A F igure
10. The case t ∈ I γ, k , ( C ← , r = r k = w B k + w C k < w A respectively w B k + w C k > w A w B k < w A ( ⇐⇒ t = tan ω < γ k + ) . In this case B k is locked by the slit q k + while A islocked by the slits q k + , q k + and 2 q . Two subcases arise:(I) w B k + w C k < w A ( ⇐⇒ t < a + ε q ). Then B ( q k + ) < T (2 q ) < ε , so A is locked by the slits q k + , q k + and 2 q (see left-hand side of Figure 10). The widths of the relevant three sub-channels of A are (from bottom to top) 2 ε − T (2 q ) = a + ε − qt ), w C k and w B k , so W γ, k ( t ) = W (1) γ, k ( t ) = a + ε − qt ) · (2 q − ξ Q ) + ∧ q + w C k ( t ) · ( q k + − ξ Q ) + ∧ q + w B k ( t ) · ( q k + − ξ Q ) + ∧ q + w B k ( t ) · ( q k + − ξ Q ) + ∧ q k . (6.1)(II) w B k < w A < w B k + w C k ( ⇐⇒ t > a + ε q ). Then B ( q k + ) = w B k + w C k < ε < T (2 q ) = w B k + w C k , so A is locked by the slits q k + and q k + (see right-hand side of Figure 10). Thewidths of the relevant two sub-channels of A are (from bottom to top) 2 ε − B ( q k + ) = a k + − q k + t and w B k , hence in this case W γ, k ( t ) = W (2) γ, k ( t ) = ( a k + − q k + t ) · ( q k + − ξ Q ) + ∧ q + w B k ( t ) · ( q k + − ξ Q ) + ∧ q + w B k ( t ) · ( q k + − ξ Q ) + ∧ q k . (6.2) Only A and B k have to taken into account here because C k has been already considered in the previous section. w A < w B k ( ⇐⇒ t > γ k + ) . In this case A is locked by the slits q k + and B k by q k + , q k + + q k and 2 q k . Two subcases arise:(III) w A < w B k < w A + w C k ( ⇐⇒ γ k + < t < a k − ε q k ). Then B (2 q k ) < < T ( q k + ), so B k islocked by the slits q k + and q k + + q k . The widths of the relevant two sub-channels of B k are w A and T ( q k + ) = q k + t − a k + , hence in this case W γ, k ( t ) = W (3) γ, k ( t ) = w A ( t ) · ( q k + − ξ Q ) + ∧ q + w A ( t ) · ( q k + − ξ Q ) + ∧ q k + ( q k + t − a k + ) · ( q k + + q k − ξ Q ) + ∧ q k . (6.3)(IV) w A + w C k < w B k ( ⇐⇒ a k − ε q k < t ). Then 0 < B (2 q k ) < T ( q k + ), so B k is locked by the slits q k + , q k + + q k and 2 q k . The widths of the relevant three sub-channels of B k are w A , T ( q k + ) − B (2 q k ) = w C k and B (2 q k ) = q k t − a k + ε ), showing that in this case W γ, k ( t ) = W (4) γ, k ( t ) = w A ( t ) · ( q k + − ξ Q ) + ∧ q + w A ( t ) · ( q k + − ξ Q ) + ∧ q k + w C k ( t ) · ( q k + + q k − ξ Q ) + ∧ q k + q k t − a k + ε ) · (2 q k − ξ Q ) + ∧ q k . (6.4)6.2. The C ↓ contribution. Since r = , r k = , r k + r k + = r + r k + = . r k + = . q k + is notremoved either and the slits q k + = q + q k + , q k + q k + and 2 q k + lock the central channel (see Figure11). Furthermore, ordering w B k , B ( q k + q k + ) = w B k − w A , w B k + w C k and T ( q k + ) = w B k + w C k − w A we find as in Subsection 6.1 that the corresponding weight W γ, k ( t ) is given by ( w B k − w A )( q k + q k + − ξ Q ) + ∧ q k + + ( w A + w C k − w B k )(2 q k + − ξ Q ) + ∧ q k + if w A < w B k < w A + w C k , w C k ( q k + q k + − ξ Q ) + ∧ q k + if w A + w C k < w B k , ( w A − w B k )( q k + − ξ Q ) + ∧ q k + + ( w B k + w C k − w A )(2 q k + − ξ Q ) + ∧ q k + if w B k < w A < w B k + w C k , w C k ( q k + − ξ Q ) + ∧ q k + if w B k + w C k < w A , = W (5) γ, k ( t ) : = w C k ( t ) · ( q k + − ξ Q ) + ∧ q k + if t < a + ε q ∧ γ k + , W (6) γ, k ( t ) : = ( a k + − q k + t ) · ( q k + − ξ Q ) + ∧ q k + + qt − a − ε ) · (2 q k + − ξ Q ) + ∧ q k + if a + ε q < t < γ k + , W (7) γ, k ( t ) : = ( q k + t − a k + ) · ( q k + q k + − ξ Q ) + ∧ q k + + a k − ε − q k t ) · (2 q k + − ξ Q ) + ∧ q k + if γ k + < t < a k − ε q k , W (8) γ, k ( t ) : = w C k ( t ) · ( q k + q k + − ξ Q ) + ∧ q k + if a k − ε q k ∨ γ k + < t . (6.5)6.3. Estimating the total contribution. Only the contribution of C k needs to be taken into account here. HE LINEAR FLOW IN A HONEYCOMB 23 (0 , − − ε )(0 , − + ε ) q q k ξ Qq k + q k + q k + q k + q k + A C k B k w A + w C k − w B k w B k − w A w B k w A F igure
11. The case t ∈ I γ, k , ( C ↓ , r = r k = w A < w B k < w A + w C k q k + > Q. In this case k > t k − < a k − ε q k < γ k + < a + ε q . The cumulative contribution of C ← and C ↓ when r = r k = r = r k = − G (1 . I , Q ( ξ ) = ∞ X k = X ∗ q k ∈I q , k q k + > Q Z t k − t k W (1) γ, k ( t ) + + W (5) γ, k ( t ) t + t + dt = G (1 . . I , Q ( ξ ) + G (1 . . I , Q ( ξ ) + G (1 . . I , Q ( ξ ) , with G (1 . . I , Q ( ξ ) = ∞ X k = X ∗ q k ∈I q , k q k + > Q (cid:16) ( q k + − ξ Q ) + ∧ q + ( q k + − ξ Q ) + ∧ q k (cid:17) Z t k − t k w B k ( t ) dtt + t + , G (1 . . I , Q ( ξ ) = ∞ X k = X ∗ q k ∈I q , k q k + > Q (2 q − ξ Q ) + ∧ q Z t k − t k a + ε − qt ) dtt + t + , G (1 . . I , Q ( ξ ) = ∞ X k = X ∗ q k ∈I q , k q k + > Q (cid:16) ( q k + − ξ Q ) + ∧ q + ( q k + − ξ Q ) + ∧ q k + (cid:17) Z t k − t k w C k ( t ) dtt + t + . This quantity is estimated employing (5.6), (10.1) and (5.7). The cumulative contribution of theerror terms from those formulas to P I G (1 . I , Q ( ξ ) is ≪ X I X γ ∈F I ( Q ) ∞ X k = Qqq k − X I X γ ∈F I ( Q ) Qq q ′ + qq ′ ! Q X I X γ ∈F I ( Q ) qq ′ | I | Q , so they can be discarded. Following the outline from the end of Section 5 we find G (1 . I , Q ( ξ ) ≅ c I ζ (2) Z du Z ∞ − u dw F (1 . ( ξ ; u , w ) , (6.6) where F (1 . ( ξ ; u , w ) = (1 − u ) ( w − u ) w · (cid:16) ( w + u − ξ ) + ∧ u + ( w + u − ξ ) + ∧ w (cid:17) + − u ( w − u ) w w + u − w − u + w + u − w ! · (2 u − ξ ) + ∧ u + (1 − u ) ( w − u ) w · (cid:16) ( w + u − ξ ) + ∧ u + ( w + u − ξ ) + ∧ ( w + u ) (cid:17) . q k + Q. In this case a + ε q γ k + a k − ε q k . Furthermore, when k > a + ε q t k γ k + a k − ε q k t k − if q k + Q − q , and t k a + ε q γ k + a k − ε q k t k − if 2 Q − q q k + Q . When k = ff erence is that γ = t − a − ε q = a ′ − ε q ′ . In this case the cumulative contribution of C ← and C ↓ arising from (6.1)-(6.5) when r = r k = r = r k = − G (1 . I , Q ( ξ ) = G (1 . . I , Q ( ξ ) + · · · + G (1 . . I , Q ( ξ ) , with G (1 . . I , Q ( ξ ) = ∞ X k = X ∗ q k ∈I q , k q k + Q Z t k − ak − ε qk W (4) γ, k ( t ) + W (8) γ, k ( t ) t + t + dt , G (1 . . I , Q ( ξ ) = ∞ X k = X ∗ q k ∈I q , k q k + Q Z ak − ε qk γ k + W (3) γ, k ( t ) + W (7) γ, k ( t ) t + t + dt , G (1 . . I , Q ( ξ ) = ∞ X k = X ∗ q k ∈I q , k q k + Q − q Z γ k + t k W (2) γ, k ( t ) + W (6) γ, k ( t ) t + t + dt , G (1 . . I , Q ( ξ ) = ∞ X k = X ∗ q k ∈I q , k Q − q < q k + Q Z a + ε q t k W (1) γ, k ( t ) + W (5) γ, k ( t ) t + t + dt , G (1 . . I , Q ( ξ ) = ∞ X k = X ∗ q k ∈I q , k Q − q < q k + Q Z γ k + a + ε q W (2) γ, k ( t ) + W (6) γ, k ( t ) t + t + dt . The total contribution of the error terms from (10.2), (10.3) and (10.4) to P I G (1 . . I , Q ( ξ ) is ≪ X I X γ ∈F I ( Q ) ∞ X k = qq k − + q q k − q k ! X I X γ ∈F I ( Q ) qq ′ + Qq q ′ + qq ′ ) + q q ′ ! Q X γ ∈F ( Q ) qq ′ + Q Q X q = ϕ ( q ) q + X I | I | Q ≪ Q c − , showing that they can be discarded in the sequel. The same holds for G (1 . . I , Q ( ξ ) , . . . , G (1 . . I , Q ( ξ ),where for G (1 . . I , Q ( ξ ) and G (1 . . I , Q ( ξ ) one uses the fact that k takes exactly one value as a result of q k + ∈ (2 Q − q , Q ]. HE LINEAR FLOW IN A HONEYCOMB 25
Employing (10.2), (10.3), (10.4), and proceeding as in Section 5 we find G (1 . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z − u dw F (1 . . ( ξ ; u , w ) , (6.7)with F (1 . . ( ξ ; u , w ) = (2 − u − w ) w − u ) w · (cid:16) (2 w + u − ξ ) + ∧ w + (2 w + u − ξ ) + ∧ ( w + u ) (cid:17) + − u − w w − u ) w w + u − w + w − w − u ! · (cid:16) ( w + u − ξ ) + ∧ u + ( w + u − ξ ) + ∧ w (cid:17) + (2 − u − w ) w − u ) w · (2 w − ξ ) + ∧ w . Employing (10.5), (10.6) and (10.7) we find G (1 . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z − u dw F (1 . . ( ξ ; u , w ) , (6.8)with F (1 . . ( ξ ; u , w ) = (2 − u − w ) w ( w + u ) · (cid:16) (2 w + u − ξ ) + ∧ w + (2 w + u − ξ ) + ∧ ( w + u ) (cid:17) + − u − w w ( w + u ) w + u − w + u + w + u − w ! · (cid:16) ( w + u − ξ ) + ∧ u + ( w + u − ξ ) + ∧ w (cid:17) + (2 − u − w ) w ( w + u ) · (2 w + u − ξ ) + ∧ ( w + u ) . Employing (10.8), (10.9) and (10.10) we find G (1 . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z − u − u dw F (1 . . ( ξ ; u , w ) , (6.9)with F (1 . . ( ξ ; u , w ) = ( w + u − w ( w + u ) · (cid:16) ( w + u − ξ ) + ∧ u + ( w + u − ξ ) + ∧ w (cid:17) + w + u − w ( w + u ) − u − ww + u + − u − ww ! · (2 w + u − ξ ) + ∧ ( w + u ) + ( w + u − w ( w + u ) · (cid:16) ( w + u − ξ ) + ∧ u + ( w + u − ξ ) + ∧ ( w + u ) (cid:17) . Employing (10.13)-(10.17) we find G (1 . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z − u − u dw F (1 . . ( ξ ; u , w ) , (6.10) with F (1 . . ( ξ ; u , w ) = ( w + u − u w · (cid:16) ( w + u − ξ ) + ∧ u + ( w + u − ξ ) + ∧ w (cid:17) + w + u − uw − u − w u + − uw ! · (cid:16) ( w + u − ξ ) + ∧ u + ( w + u − ξ ) + ∧ ( w + u ) (cid:17) + ( w + u − uw · (2 u − ξ ) + ∧ u . Employing (10.17), (10.18) and (10.19) we find G (1 . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z − u − u dw F (1 . . ( ξ ; u , w ) , (6.11)with F (1 . . ( ξ ; u , w ) = (2 − u − w ) u ( w + u ) · (cid:16) ( w + u − ξ ) + ∧ u + ( w + u − ξ ) + ∧ ( w + u ) (cid:17) + − u − w u ( w + u ) w + u − w + u + w + u − u ! · (cid:16) ( w + u − ξ ) + ∧ u + ( w + u − ξ ) + ∧ w (cid:17) + (2 − u − w ) u ( w + u ) · (2 w + u − ξ ) + ∧ ( w + u ) .
7. C hannels with removed slits . T he case r = r k = ±
1. The C O contributions for ( r , r k ) = (0 ,
1) and respectively ( r , r k ) = (0 , − C ← contribution for ( r , r k ) = (0 ,
1) coincides with the C ↓ contribution for ( r , r k ) = (0 , − X ∗ = X ∗ γ ∈F I ( Q ) q ≡ a (mod 3) q k . a k (mod 3) . It su ffi ces to only analyze the C O and C ← contributions of ( r , r k ) = (0 , β to take both values − C O contribution when ( r , r k ) = (0 , ± C ← contribution for ( r , r k ) = (0 ,
1) and of the C ↓ contribution for( r , r k ) = (0 , − The C O contribution. The situation is described in Figure 12. Two cases arise:7.1.1. w A w B k ( ⇐⇒ t > γ k + ) . In this case k >
1. Since γ k + t k − , we must also have q k + Q . The channel A is locked by the slit q k + and W γ, k ( t ) = w A ( t ) · ( q k + − ξ Q ) + ∧ q , withcontribution G (2 . I , Q ( ξ ) = ∞ X k = X ∗ q k ∈I q , k q k + Q ( q k + − ξ Q ) + ∧ q Z t k − γ k + w A ( t ) dtt + t + , estimated in Subsection 11.1.1 as ≅ c I ζ (2) Z du Z − u dw − u − w ( w − u )( w + u ) w + u − w + u + w − w − u ! · ( w + u − ξ ) + ∧ u . (7.1) HE LINEAR FLOW IN A HONEYCOMB 27 w A > w B k ( ⇐⇒ t < γ k + ) and ξ Q < q k + . In this situation we have W γ, k ( t ) = w B k ( t )( q k + − ξ Q ) + (cid:0) w A ( t ) − w B k ( t ) (cid:1) q , with contribution (according to whether γ k + t k − or t k − < γ k + ) G (2 . I , Q ( ξ ) = ∞ X k = X ∗ q k ∈I q , k ξ Q < q k + Q Z γ k + t k W γ, k ( t ) dtt + t + + ∞ X k = X ∗ q k ∈I q , k q k + > ( ξ ∨ Q Z t k − t k W γ, k ( t ) dtt + t + . Employing (10.8), (10.9), respectively (5.6), (11.2) and the procedure described at the beginningof Appendix 2 we find G (2 . I , Q ( ξ ) ≅ c I ζ (2) Z du Z − u ( ξ ∨ ∧ − u dw F (2 . . ( ξ ; u , w ) + Z ∞ ξ ∨ − u dw F (2 . . ( ξ ; u , w ) ! , (7.2)with F (2 . . ( ξ ; u , w ) = ( w + u − w ( w + u ) (cid:18) w + uw − ξ w + u (cid:19) , F (2 . . ( ξ ; u , w ) = (1 − u ) ( w + u − ξ )( w − u ) w + (1 − u ) u ( w − u ) w w + u − w + w + u − w − u ! . (0 , − ε )(0 , ε ) q q q k ξ Qq k + q k + q k + A C k B k w C k w B k w C k w B k w A + w C k F igure
12. The case t ∈ I γ, k , r = r k = ± C O , w B k < w A , n = N = w A > w B k ( ⇐⇒ t < γ k + ) and ξ Q > q k + . Consider the integer N for which q k + N ξ Q < q k + N + , that is 1 N : = j ξ Q − q k q k . We will keep N > q Q fixed and sum over y = q k ∈ J q , N : = (cid:0) ξ Q − ( N + q , ξ Q − Nq (cid:3) . Let n : = j w A + w C k w B k + w C k k = j a k − q k tqt − a k >
1, so γ k + n + < t γ k + n and B ( q k + n ) = w B k + n ( w B k + w C k ) ε < B ( q k + n + ). This shows that the channel A is locked by the slits q k + , . . . , q k + n , q k + n + . Supposethat N > ξ . Then q k ξ Q − Nq N ( Q − q ), showing that q k + N NQ and γ k + N < t k . So n < N and ξ Q > q k + N > q k + n + , showing that in this case W γ, k , N ( t ) = ∀ t ∈ I γ, k . It remains that N < ξ . Thefollowing cases arise:(1) B ( q k + N ) ε < B ( q k + N + ) ( ⇐⇒ γ k + N + < t γ k + N ). In this case W γ, k , N ( t ) = W (1) γ, k , N ( t ) : = (cid:0) ε − B ( q k + N ) (cid:1) ( q k + N + − ξ Q ) = ( a k + N − q k + N t )( q k + N + − ξ Q ) . (7.3)(2) B ( q k + N + ) ε ( ⇐⇒ t γ k + N + ). In this case W γ, k , N ( t ) = W (2) γ, k ( t ) : = (cid:0) w B k ( t ) + w C k ( t ) (cid:1) ( q k + N + − ξ Q ) + (cid:0) ε − B ( q k + N + ) (cid:1) q = − ξ Q (cid:0) w B k ( t ) + w C k ( t ) (cid:1) . (7.4)We find G (2 . I , Q ( ξ ) = X N <ξ ∞ X k = X ∗ q k ∈I q , k ∩J q , N Z I γ, k ∩ ( γ k + N + ,γ k + N ] W (1) γ, k , N ( t ) dtt + t + + Z I γ, k ∩ ( γ,γ k + N + ] W (2) γ, k , N ( t ) dtt + t + = G (2 . . I , Q ( ξ ) + G (2 . . I , Q ( ξ ) . Note that I γ, k ∩ ( γ k + N + , γ k + N ] = ∅ if q k < N ( Q − q ) or Q + ( N + Q − q ) q k , ( t k , γ k + N ] if N ( Q − q ) q k < ( N + Q − q ) , ( γ k + N + , γ k + N ] if ( N + Q − q ) q k < Q + N ( Q − q ) , ( γ k + N + , t k − ] if Q + N ( Q − q ) q k < Q + ( N + Q − q ) , and respectively I γ, k ∩ ( γ, γ k + N + ] = ∅ if q k < ( N + Q − q ) , ( t k , γ k + N + ] if ( N + Q − q ) q k < Q + ( N + Q − q ) , I γ, k if Q + ( N + Q − q ) q k . Summing as in (5.2) and employing (11.3), (11.4), (11.5) and Lemmas 4 and 2, then changing y + Nq to y and making the substitution ( q , y ) = ( Qu , Qw ), we find G (2 . . I , Q ( ξ ) ≅ c I ζ (2) X N <ξ Z du (Z [ ξ − u ,ξ ] ∩ [ N , N + − u ] dw ( w + u − ξ )( w − N ) ( w − Nu ) w + Z [ ξ − u ,ξ ] ∩ [ N + − u , N + dw w + u − ξ ( w + u ) w + Z [ ξ − u ,ξ ] ∩ [ N + , N + − u ] dw ( w + u − ξ )( N + − u − w ) (cid:0) w − ( N + u (cid:1) ( w + u ) w + u + w − ( N + w − ( N + u !) . (7.5) HE LINEAR FLOW IN A HONEYCOMB 29
In a similar way (but replacing y + ( N + q by y ) we infer from (11.6) and (11.7) that G (2 . . I , Q ( ξ ) ≅ c I ζ (2) X N <ξ Z du Z [ ξ,ξ + u ] ∩ [ N + , N + dw w − ( N + (cid:0) w − ( N + u (cid:1) w · − ξ w − ξ (1 − u ) w − ( N + u ! + c I ζ (2) X N <ξ Z du Z [ ξ,ξ + u ] ∩ [ N + , ∞ ) dw − u (cid:0) w − ( N + u (cid:1)(cid:0) w − ( N + u (cid:1) · − ξ (1 − u ) w − ( N + u − ξ (1 − u ) w − ( N + u ! . (7.6)The inner integrals in (7.5) and the first one in (7.6) can be nonzero only when N > ξ − ξ − < N < ξ or ξ − < N < ξ − The C ← contribution. In this case we shall analyze the contribution of the channels B k and C k . All slits q k + n , n ∈ Z , are removed, while neither q nor any of 2 q k + nq = q k + q k + n , n ∈ Z , is beingremoved. Since T (2 q k ) = − w A − w C k < B (2 q k + nq ) = w B k − w A − w C k + n ( w B k + w C k ), itfollows that B k ∪ C k is locked exactly by two of the slits 2 q k + nq , n >
0. To make this precise let n : = j w A + w C k w B k + w C k k = j a k + a k − − ε − ( q k + q k − ) tqt − a k >
0. We have 0 < B (2 q k + n q ) = w B k − w A − w C k + n ( w B k + w C k ) w B k + w C k . The situation is described in Figure 13. Consider also λ k , n : = a k + a k + n − − ε q k + q k + n − ց γ as n → ∞ , (7.7)so λ k , n + < t λ k , n . Note that λ , > t − = γ > λ , and λ k , > t k − when k >
1, showing that forevery k > λ k , n + , λ k , n ] cover I γ, k . The following cases arise:7.2.1. 0 < B (2 q k + n q ) w B k ( ⇐⇒ λ k , n + < t γ k + n ) . In this case C k is locked by the slit2 q k + ( n + q and B k by the slits 2 q k + n q = q k + q k + n and 2 q k + ( n + q = q k + q k + n + . Thewidths of the three relevant sub-channels of B k ∪C k are (from bottom to top) w C k , w B k − B (2 q k + n q ) = a k + n − q k + n t , B (2 q k + n q ) = w B k − ( a k + n − q k + n t ), and so W γ, k ( t ) = W (1) γ, k , n ( t ) = w C k ( t ) · ( q k + + q k + n − ξ Q ) + ∧ q k + + w B k ( t ) · ( q k + q k + n − ξ Q ) + ∧ q k + ( a k + n − q k + n t ) · (cid:16) ( q k + + q k + n − ξ Q ) + ∧ q k − ( q k + q k + n − ξ Q ) + ∧ q k (cid:17) . w B k < B (2 q k + n q ) w B k + w C k ( ⇐⇒ γ k + n < t λ k , n ) . In this case B k is locked by the slit q k + q k + n and C k by the slits q k + q k + n and q k + q k + n + . The widths of the three relevant sub-channelsof B k ∪ C k are w B k + w C k − B (2 q k + n q ) = w C k − ( q k + n t − a k + n ), B (2 q k + n q ) − w B k = q k + n t − a k + n , w B k , and so W γ, k ( t ) = W (2) γ, k , n ( t ) = w C k ( t ) · ( q k + + q k + n − ξ Q ) + ∧ q k + + w B k ( t ) · ( q k + q k + n − ξ Q ) + ∧ q k − ( q k + n t − a k + n ) · (cid:16) ( q k + + q k + n − ξ Q ) + ∧ q k + − ( q k + q k + n − ξ Q ) + ∧ q k + (cid:17) . The following five cases arise:(I) q k < ( n − Q − q ). Then λ k , n < t k , thus I γ, k ∩ ( λ k , n + , λ k , n ] = ∅ . (0 , − ε )(0 , ε ) q q k q k + q q k + q q k + q ξ Qq k + q k + q k + A C k B k w B k + w C k w B k + w C k F igure
13. The case t ∈ I γ, k , r = r k = C ← , n = w B k < B (2 q k + n q ) w B k + w C k (II) ( n − Q − q ) q k < n ( Q − q ). Then n > γ k + n < t k λ k , n < t k − in both cases k > k =
0. The corresponding contribution is G (2 . . I , Q ( ξ ) = ∞ X n = ∞ X k = X ∗ q k ∈I q , k ( n − Q − q ) q k < n ( Q − q ) Z λ k , n t k W (2) γ, k , n ( t ) dtt + t + . Employing (11.8)-(11.11), Lemmas 8 and 3, and the change of variable ( q , y ) = ( Qu , Qw ) we find(according to whether k = k > G (2 . . I , Q ( ξ ) ≅ c I ζ (2) ∞ X n = Z − n − du Z n − − u ) dw F (2 . . n ( ξ ; u , w ) + c I ζ (2) ∞ X n = Z − n − du Z n (1 − u )( n − − u ) dw F (2 . . n ( ξ ; u , w ) + c I ζ (2) ∞ X n = Z − n − n − du Z n (1 − u )1 dw F (2 . . n ( ξ ; u , w ) , (7.8)where F (2 . . n ( ξ ; u , w ) = (cid:0) w − ( n − − u ) (cid:1) w (cid:0) w + ( n − u (cid:1) · (2 w + nu − ξ ) + ∧ w + w − ( n − − u ) w (cid:0) w + ( n − u (cid:1) − uw − n (1 − u ) − ww ! · (cid:0) w + ( n + u − ξ (cid:1) + ∧ ( w + u ) + w − ( n − − u ) w (cid:0) w + ( n − u (cid:1) + n (1 − u ) − w w + ( n − u + n (1 − u ) − ww ! · (2 w + nu − ξ ) + ∧ ( w + u ) . HE LINEAR FLOW IN A HONEYCOMB 31 (III) n ( Q − q ) q k < Q + n ( Q − q ). Upon λ , < γ = t − λ , = a ′ − ε q ′ we see that I γ, k ∩ ( λ k , n + , λ k , n ] = ( λ k , n + , γ k + n ] ∪ ( γ k + n , λ k , n ] when k > k = n >
2. When k = n = λ , , γ ]. The cumulative contribution is thus G (2 . . I , Q ( ξ ) = ∞ X n = ∞ X k = X ∗ q k ∈I q , k n ( Q − q ) q k < Q + n ( Q − q ) Z γ k + n λ k , n + W (1) γ, k , n ( t ) dtt + t + + Z λ k , n γ k + n W (2) γ, k , n ( t ) dtt + t + + ∞ X n = X ∗ q ′ ∈I q , q ′ > n ( Q − q ) Z γ n λ , n + W (1) γ, , n ( t ) dtt + t + + Z λ , n γ n W (2) γ, , n ( t ) dtt + t + + X ∗ q ′ ∈I q , Z γ λ , W (1) γ, , ( t ) dtt + t + = G (2 . . . I , Q ( ξ ) + G (2 . . . I , Q ( ξ ) + G (2 . . . I , Q ( ξ ) . For fixed k we have q k − QQ − q < n q k Q − q , so n can take at most 1 + j QQ − q k QQ − q values. Employing(11.12)-(11.15) we find G (2 . . . I , Q ( ξ ) ≅ c I ζ (2) ∞ X n = Z du Z n (1 − u ) + n (1 − u ) ∨ dw F (2 . . n ( ξ ; u , w ) , (7.9)where F (2 . . n ( ξ ; u , w ) = − u (cid:0) w + ( n − u (cid:1) (2 w + nu ) · + ( n + − u ) − w w + nu + + n (1 − u ) − w w + ( n − u ! · (cid:0) w + ( n + u − ξ (cid:1) + ∧ ( w + u ) + − u (cid:0) w + ( n − u (cid:1) (2 w + nu ) w − ( n − − u )2 w + ( n − u + w − n (1 − u )2 w + nu ! · (2 w + nu − ξ ) + ∧ w + (cid:0) w − n (1 − u ) (cid:1) ( w + nu )(2 w + nu ) · (cid:16)(cid:0) w + ( n + u − ξ (cid:1) + ∧ w − (2 w + nu − ξ ) + ∧ w (cid:17) − (cid:0) + n (1 − u ) − w (cid:1) ( w + nu ) (cid:0) w + ( n − u (cid:1) · (cid:16) (2 w + ( n + u − ξ ) + ∧ ( w + u ) − (2 w + nu − ξ ) + ∧ ( w + u ) (cid:17) . Employing (11.13)-(11.16) we find G (2 . . . I , Q ( ξ ) ≅ c I ζ (2) ∞ X n = Z du Z n (1 − u ) ∧ dw F (2 . . n ( ξ ; u , w ) . (7.10)From (7.9) and (7.10) we gather G (2 . . . I , Q ( ξ ) + G (2 . . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z − u dw F (2 . . ( ξ ; u , w ) + c I ζ (2) ∞ X n = Z n (1 − u ) + n (1 − u ) dw F (2 . . n ( ξ ; u , w ) . (7.11)Employing (11.17), (11.18), (11.19) we find G (2 . . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z − u dw F (2 . . . ( ξ ; u , w ) , (7.12) where F (2 . . . ( ξ ; u , w ) = w + u − w + u )(2 w + u ) − u − w w + u + − u − ww + u ! · (2 w + u − ξ ) + ∧ ( w + u ) + ( w + u − ( w + u ) (2 w + u ) · (2 w + u − ξ ) + ∧ w + ( w + u − ( w + u )(2 w + u ) · (2 w + u − ξ ) + ∧ w . (IV) Q + n ( Q − q ) q k < Q + ( n + Q − q ). Then k > n + Q − q k + n + Q − q . In thiscase I γ, k ∩ ( λ k , n + , λ k , n ] = ( λ k , n + , t k − ] and t k − < γ k + n , so the corresponding contribution is G (2 . . I , Q ( ξ ) = ∞ X n = ∞ X k = X ∗ q k ∈I q , k Q + n ( Q − q ) q k < Q + ( n + Q − q ) Z t k − λ k , n + W (1) γ, k , n ( t ) dtt + t + . Note also that n can only take the value n = j q k − QQ − q k for each k . Employing (11.20), (11.21), (11.22)we find G (2 . . I , Q ( ξ ) ≅ c I ζ (2) ∞ X n = Z du Z ( n + − u ) + n (1 − u ) + dw F (2 . . n ( ξ ; u , w ) , (7.13)where F (2 . . n ( ξ ; u , w ) = (cid:0) + ( n + − u ) − w (cid:1) ( w − u )(2 w + nu ) · (cid:0) w + ( n + u − ξ (cid:1) + ∧ ( w + u ) + (cid:0) + ( n + − u ) − w (cid:1) ( w − u ) (2 w + nu ) · (2 w + nu − ξ ) + ∧ w + + ( n + − u ) − w ( w − u )(2 w + nu ) w − n (1 − u )2 w + nu + w − n (1 − u ) − w − u ! · (cid:0) w + ( n + u − ξ (cid:1) + ∧ w . (V) q k > Q + ( n + Q − q ). Then k > λ k , n + > t k − and I γ, k ∩ ( λ k , n + , λ k , n ] = ∅ .8. C hannels with removed slits . T he case r k = C O contribution of ( r , r k ) = (1 ,
0) and of ( r , r k ) = ( − , q − a ≡ q − a ≡ − C ← contribution of ( r , r k ) = (1 ,
0) and the C ↓ contribution of ( r , r k ) = ( − ,
0) (see Figure 8).So it su ffi ces to take ( r , r k ) = (0 ,
1) in the sequel, doubling the C O and the C ← contributions. Thistime we consider X ∗ = X ∗ γ ∈F I ( Q ) q − a ≡ q k ≡ a k (mod 3) . The channel C O . The slit q k is removed, while q + nq k , n >
0, are not, as shown in Figure 14.The following three cases arise:
HE LINEAR FLOW IN A HONEYCOMB 33 (0 , − ε )(0 , ε ) q q k q k q k ξ Qq k + q + q k q + q k A C k B k w C k w A w B k + w C k w A + w C k w A + w C k F igure
14. The case t ∈ I γ, k , r k = C O , w A < w B k , n = N = w A > w B k ( ⇐⇒ t γ k + ) . The channel B k is locked by the slit q k + and W γ, k ( t ) = w B k ( t ) · ( q k + − ξ Q ) + ∧ q k , with contribution G (3 . I , Q ( ξ ) = ∞ X k = X ∗ q k ∈I q , k q k + Q Z γ k + t k W γ, k ( t ) dtt + t + + ∞ X k = X ∗ q k ∈I q , k q k + > Q Z t k − t k W γ, k ( t ) dtt + t + . Employing (10.8) (which also holds for k =
0) and (5.6) we find G (3 . I , Q ( ξ ) ≅ c I ζ (2) Z du Z − u − u dw F (3 . . ( ξ ; u , w ) + Z ∞ − u dw F (3 . . ( ξ ; u , w ) ! , (8.1)where F (3 . . ( ξ ; u , w ) = ( w + u − ( w + u ) w · ( w + u − ξ ) + ∧ w , F (3 . . ( ξ ; u , v ) = (1 − u ) ( w − u ) w · ( w + u − ξ ) + ∧ w . w A < w B k ( ⇐⇒ t > γ k + ) and ξ Q < q k + . In this case k > q k + Q , and W γ, k ( t ) = w A ( t ) · ( q k + − ξ Q ) + ∧ q k + (cid:0) w B k ( t ) − w A ( t ) (cid:1) q k , with contribution G (3 . I , Q ( ξ ) = ∞ X k = X ∗ q k ∈I q , k ξ Q < q k + Q Z t k − γ k + W γ, k ( t ) dtt + t + . Employing (11.1) and (12.2) we find G (3 . I , Q ( ξ ) ≅ c I ζ (2) Z du Z − u (1 ∨ ( ξ − u )) ∧ (2 − u ) dw F (3 . ( ξ ; u , w ) , (8.2) where F (3 . ( ξ ; u , w ) = − u − w ( w − u )( w + u ) w + u − w + u + w − w − u ! · ( w + u − ξ ) + ∧ w + (2 − u − w ) w ( w − u ) ( w + u ) . w A < w B k ( ⇐⇒ t > q k + ) and q k + ξ Q. Again k > q k + Q . Consider theinteger N for which q + Nq k ξ Q < q + ( N + q k , that is 1 N = j ξ Q − qq k k ξ . Consider also n : = j w B k + w C k w A + w C k k = j qt − aa k − q k t k >
1, and let λ k , n : = a + na k q + nq k ր γ k as n → ∞ , (8.3)hence λ k , n t < λ k , n + and T ( q + ( n + q k ) < T ( q + n q k ). The channel B k is locked exactlyby the slits q k + = q + q k , . . . , q + n q k , q + ( n + q k (see Figure 14) and W γ, k ( t ) is given by N > n , (cid:0) w B k + w C k − n ( w A + w C k ) (cid:1)(cid:0) q + ( n + q k − ξ Q (cid:1) if n = N , ( w A + w C k ) (cid:0) q + ( N + q k − ξ Q (cid:1) + (cid:0) w B k + w C k − ( N + w A + w C k ) (cid:1) q k if N < n , or equivalently by t ∈ I γ, k ∩ ( t k , λ k , N ] , W (1) γ, k , N ( t ) : = (cid:0) q k + + Nq k − ξ Q (cid:1)(cid:0) ( q + Nq k ) t − a − Na k (cid:1) if t ∈ I γ, k ∩ ( λ k , N , λ k , N + ] , W (2) γ, k ( t ) : = − ξ Q (cid:0) w A ( t ) + w C k ( t ) (cid:1) if t ∈ I γ, k ∩ ( λ k , N + , t k − ] . (8.4)We shall start by keeping N > q Q fixed, and summing over y = q k ∈ J q , N : = (cid:18) ξ Q − qN + , ξ Q − qN (cid:21) . Since λ k , = γ k + > t k and γ k > t k − for all k , nonzero contribution only arises from one of thefollowing two subcases:(I) N ( q k − Q ) Q − q < ( N + q k − Q ) . Then t k < γ k + λ k , N t k − < λ k , N + and the contributionis G (3 . . I , Q ( ξ ) = X N ξ ∞ X k = X ∗ q k ∈I q , k ∩J q , N Q + Q − qN + < q k Q + Q − qN Z t k − λ k , N W (1) γ, k , N ( t ) dtt + t + . In this case N can only take the value N = j Q − qq k − Q k for fixed k . Employing (12.3) we find G (3 . . I , Q ( ξ ) ≅ c I ζ (2) X N ξ Z du Z [ + − uN + , + − uN ] ∩ h ξ − uN + , ξ − uN i dw F (3 . . N ( ξ ; u , w ) , (8.5)where F (3 . . N ( ξ ; u , w ) = (cid:0) − u − N ( w − (cid:1) (cid:0) ( N + w + u − ξ (cid:1) ( w − u ) ( Nw + u ) . (II) ( N + q k − Q ) Q − q . Then t k < γ k + λ k , N < λ k , N + t k − . The contribution G (3 . . I , Q ( ξ ) = X N ξ ∞ X k = X ∗ q k ∈I q , k ∩J q , N q k Q + Q − qN + Z λ k , N + λ k , N W (1) γ, k , N ( t ) dtt + t + + Z t k − λ k , N + W (2) γ, k ( t ) dtt + t + HE LINEAR FLOW IN A HONEYCOMB 35 is estimated upon (12.4) and (12.7) as G (3 . . I , Q ( ξ ) ≅ c I ζ (2) X N ξ Z du Z [ , + − uN + ] ∩ h ξ − uN + , ξ − uN i dw F (3 . . N ( ξ ; u , w ) , (8.6)where F (3 . . N ( ξ ; u , w ) = ( N + w + u − ξ ( Nw + u ) (cid:0) ( N + w + u (cid:1) + − u − ( N + w − w − u ) (cid:0) ( N + w + u (cid:1) − ξ ( N + w + u − ξ ( w − w − u ! . In this case N ∈ {⌊ ξ ⌋ − , ⌊ ξ ⌋} su ffi ces as a result of the inequality ξ − u < N + − u .8.2. The channel C ← . In this case all slits q + nq k are removed, while slits 2 q + nq k , n >
0, are not.Exactly two of the later ones lock A ∪ C k . Letting n : = j w B k + w C k w A + w C k k = j a k − − ε − q k − ta k − q k t k >
0, we have w B k T (2 q + n q k ) = w B k + w C k ) − n ( w A + w C k ) < ε , so A ∪ C k is locked exactly by the slits2 q + n q k and 2 q + ( n + q k . The situation is described in Figure 15. Consider also µ k , n : = a + na k + ε q + nq k and ν k , n : = a + na k q + nq k . (8.7)When q k Q , µ k , n ր γ k > t k − as n → ∞ and µ k , = a + ε q t k − , ∀ k >
0. We also have µ k , n − t < µ k , n and the following two situations can arise:(I) w B k T (2 q + n q k ) < w B k + w C k ( ⇐⇒ µ k , n − t < ν k , n ) . Then n > A ∪ B k are (from bottom to top) w A , w B k + w C k − T (2 q + n q k ) = a + n q k − ( q + n q k ) t , T (2 q + n q k ) − w B k = w C k − (cid:0) a + n a k − ( q + n q k ) t (cid:1) , yielding W γ, k ( t ) = W (1) γ, k , n ( t ) = w A ( t ) · (2 q + n q k − ξ Q ) + ∧ q + w C k ( t ) · ( q k + + n q k − ξ Q ) + ∧ q k + − (cid:0) a + n a k − ( q + n q k ) t ) (cid:1) · (cid:16) ( q k + + n q k − ξ Q ) + ∧ q k + − (2 q + n q k − ξ Q ) + ∧ q k + (cid:17) . ( − , − ε )( − , ε ) q q q k q k + q + q k ξ Q q + q k q + q k q k q + q k q + q k A C k B k w A + w C k w A + w C k w A + w C k F igure
15. The case t ∈ I γ, k , r = r k = r k + = − C ← , n = (II) w B k + w C k T (2 q + n q k ) < ε ( ⇐⇒ ν k , n t < µ k , n ) . Then n > A ∪ C k are: 2 ε − T (2 q + n q k ) = w A − (cid:0) ( q + n q k ) t − a − n q k (cid:1) , T (2 q + n q k ) − ( w B k + w C k ) = ( q + n q k ) t − a − n a k , w C k , yielding W γ, k ( t ) = W (2) γ, k , n ( t ) = w A ( t ) · (2 q + n q k − ξ Q ) + ∧ q + w C k ( t ) · ( q k + + n q k − ξ Q (cid:1) + ∧ q k + + (cid:0) ( q + n q k ) t − a − n a k (cid:1) · (cid:16)(cid:0) q k + + n q k − ξ Q (cid:1) + ∧ q − (2 q + n q k − ξ Q ) + ∧ q (cid:17) . The following three cases arise:8.2.1. n = ⇐⇒ t < a + ε q ) . In this case A is locked by the slits 2 q and q k + . One sees that W γ, k ( t ) = W (2) γ, k , ( t ). When q k + Q we have a + ε q < t k with zero contribution, so we must take q k + > Q − q . When q k + > Q we have k > a + ε q > t k − , with contribution G (3 . . I , Q ( ξ ) = ∞ X k = X ∗ q k ∈I q , k q k + > Q Z t k − t k W (2) γ, k , ( t ) dtt + t + . Employing (5.6), (5.7), (5.9) and qt − a = w B k ( t ) + w C k ( t ) we find G (3 . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z ∞ − u dw F (3 . . ( ξ ; u , w ) , (8.8)where F (3 . . ( ξ ; u , w ) = (1 − u ) (2 w − u )( w − u ) w · (cid:16) ( w + u − ξ ) + ∧ u − (2 u − ξ ) + ∧ u (cid:17) + − u ( w − u ) w w + u − w + w − w − u ! · (2 u − ξ ) + ∧ u + (1 − u ) ( w − u ) w · ( w + u − ξ ) + ∧ ( w + u ) . When 2 Q − q < q k + Q we have t k < a + ε q t k − and q k takes exactly one value. The correspondingcontribution G (3 . . I , Q ( ξ ) = ∞ X k = X ∗ q k ∈I q , k Q − q < q k + Q Z a + ε q t k W (2) γ, k , ( t ) dtt + t + G (3 . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z − u − u dw F (3 . . ( ξ ; u , w ) , (8.9)where F (3 . . ( ξ ; u , w ) = w + u − uw − uw + − u − w u ! · ( w + u − ξ ) + ∧ ( w + u ) + ( w + u − w − u + uw · ( w + u − ξ ) + ∧ u + ( w + u − uw · (2 u − ξ ) + ∧ u . HE LINEAR FLOW IN A HONEYCOMB 37 n > and k = . Then µ , = a ′ + a + ε q + q ′ > γ = t − = ν , > a + ε q , with contribution G (3 . . I , Q ( ξ ) = X ∗ q > Q / q ′ > Q − q ) Z γ a + ε q W (1) γ, , ( t ) dtt + t + . Employing (12.8), (12.9), (12.10) we find G (3 . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z − u dw F (3 . . ( ξ ; u , w ) , (8.10)where F (3 . . ( ξ ; u , w ) = (2 − u − w )(3 w + u − u ( w + u ) · ( w + u − ξ ) + ∧ u + (2 − u − w ) u ( w + u ) · (2 w + u − ξ ) + ∧ ( w + u ) + (2 − u − w ) u ( w + u ) · ( w + u − ξ ) + ∧ ( w + u ) . n > and k > . Note first that µ k , n t k when q k Q − q ) n + and t k − < µ k , n − when q k > Q + Q − qn . In both cases I γ, k ∩ [ µ k , n − , µ k , n ] has measure zero, so we shall only consider q k ∈ h Q − q ) n + , Q + Q − qn i . The following four subcases arise:(I) Q − q ) n + < q k Q − q ) n ∧ (cid:16) Q + Q − qn + (cid:17) . Since q k > Q we have Q Q − q ) n , so n = q Q .Furthermore µ k , < t k < ν k , < µ k , < t k − and the contribution is G (3 . . I , Q ( ξ ) = ∞ X k = X ∗ q k ∈I q , k , q Q / q k Q − q ) ∧ Q − q Z ν k , t k W (1) γ, k , ( t ) dtt + t + + Z µ k , ν k , W (2) γ, k , ( t ) dtt + t + . Employing (12.11)-(12.14) we find G (3 . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z − u dw F (3 . . ( ξ ; u , w ) + c I ζ (2) Z du Z − u dw F (3 . . ( ξ ; u , w ) , (8.11)where F (3 . . ( ξ ; u , w ) = ( w + u − (3 w + u ) w ( w + u )( w + u ) · ( w + u − ξ ) + ∧ u + ( w + u − w ( w + u ) − uw + − u − ww + u ! − ( w + u − w ( w + u ) ) · (2 w + u − ξ ) + ∧ ( w + u ) + ( w + u − w ( w + u ) · ( w + u − ξ ) + ∧ ( w + u ) + ( w + u − ( w + u )( w + u ) · (2 w + u − ξ ) + ∧ u . (II) Q − q ) n < q k Q + Q − qn + . Then t k < µ k , n − < ν k , n < µ k , n t k − , with contribution G (3 . . I , Q ( ξ ) = ∞ X k = ∞ X n = X ∗ q k ∈I q , k Q − q ) n < q k Q + Q − qn + Z ν k , n µ k , n − W (1) γ, k , n ( t ) dtt + t + + Z µ k , n ν k , n W (2) γ, k , n ( t ) dtt + t + . Employing (12.15)-(12.19) we find (according to whether n = n > G (3 . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z − u − u dw F (3 . . ( ξ ; u , w ) + c I ζ (2) Z du Z − u u dw F (3 . . ( ξ ; u , w ) + c I ζ (2) ∞ X n = Z du Z + − un + dw F (3 . . n ( ξ ; u , w ) , (8.12)where F (3 . . n ( ξ ; u , w ) = (cid:0) n ( w − + u (cid:1) ( nw + u )( nw + u ) · (cid:0) ( n + w + u − ξ (cid:1) + ∧ u + (cid:0) − u − n ( w − (cid:1) ( nw + u ) (cid:0) ( n − w + u (cid:1) · ( nw + u − ξ ) + ∧ ( w + u ) + − w ( nw + u ) (cid:0) ( n − w + u (cid:1) ( n − w − + u ( n − w + u + n ( w − + unw + u ! − (cid:0) n ( w − + u (cid:1) ( nw + u )( nw + u ) · ( nw + u − ξ ) + ∧ u + ( − w ( nw + u ) (cid:0) ( n − w + u (cid:1) − u − n ( w − n − w + u + − u − ( n + w − nw + u ! − (cid:0) − u − n ( w − (cid:1) ( nw + u ) (cid:0) ( n − w + u (cid:1) · (cid:0) ( n + w + u − ξ (cid:1) + ∧ ( w + u ) . (III) Q + Q − qn + < q k Q − q ) n . This gives n = µ k , t k < ν k , = γ k + < t k − < µ k , , and G (3 . . I , Q ( ξ ) = ∞ X k = X ∗ q k ∈I q , k Q − q < q k Q − q ) Z ν k , t k W (1) γ, k , n ( t ) dtt + t + + Z t k − ν k , W (2) γ, k , n ( t ) dtt + t + . Employing (5.7), (5.9), (10.9) and (12.20) we find G (3 . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z − u − u dw F (3 . . ( ξ ; u , w ) , (8.13) Note that 2( Q − q ) < q k Q + Q − q implies q > Q . HE LINEAR FLOW IN A HONEYCOMB 39 where F (3 . . ( ξ ; u , w ) = ( − u ( w − u ) w w + u − w + w − w − u ! − (2 − w − u ) ( w − u ) ( w + u ) ) · ( w + u − ξ ) + ∧ u + ( (1 − u ) ( w − u ) w − ( w + u − ( w + u ) w ) · (2 w + u − ξ ) + ∧ ( w + u ) + ( w + u − ( w + u ) w · ( w + u − ξ ) + ∧ ( w + u ) + (2 − w − u ) ( w − u ) ( w + u ) · (2 w + u − ξ ) + ∧ u . (IV) Q − q ) n ∨ (cid:16) Q + Q − qn + (cid:17) < q k Q + Q − qn . For fixed k there is only one value n can take, namely j Q − qq k − Q k . We have t k < µ k , n − ν k , n t k − < µ k , n and G (3 . . I , Q ( ξ ) = ∞ X k = ∞ X n = X ∗ q k ∈I q , k Q − q ) n ∨ (cid:16) Q + Q − qn + (cid:17) < q k Q + Q − qn Z ν k , n µ k , n − W (1) γ, k , n ( t ) dtt + t + + Z t k − ν k , n W (2) γ, k , n ( t ) dtt + t + . Employing (12.21)-(12.25) we find G (3 . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z − u − u dw F (3 . . ( ξ ; u , w ) + c I ζ (2) Z du Z − u − u dw F (3 . . ( ξ ; u , w ) + c I ζ (2) ∞ X n = Z du Z + − un + − un + dw F (3 . . n ( ξ ; u , w ) , (8.14)where F (3 . . n ( ξ ; u , w ) = (cid:0) − u − n ( w − (cid:1) ( w − u )( nw + u ) (cid:0) ( n − w + u (cid:1) · (cid:0) ( n + w + u − ξ (cid:1) + ∧ ( w + u ) + − u − n ( w − w − u ) (cid:0) ( n − w + u (cid:1) ( n − w − + u ( n − w + u + w − w − u ! − (cid:0) − u − n ( w − (cid:1) ( w − u ) ( nw + u ) · ( nw + u − ξ ) + ∧ u + (cid:0) − u − n ( w − (cid:1) ( nw + u ) (cid:0) ( n − w + u (cid:1) · ( nw + u − ξ ) + ∧ ( w + u ) + (cid:0) − u − n ( w − (cid:1) ( w − u ) ( nw + u ) · (cid:0) ( n + w + u − ξ (cid:1) + ∧ u .
9. C hannels with removed slits . T he case r k + = X ∗ = X γ ∈F I ( Q ) q . a (mod 3) q k + ≡ a k + (mod 3) . We shall actually sum as in Remark 1 of Section 5 over x = q − a ∈ q (1 − I ), x ≡ ± y = q k + ∈ I q , k + , β = xy ≡ q ). The contributions arising from x ≡ x ≡ − The channel C O . In the formulas for C O and ( r , r k ) = (1 , − r , r k ) = ( − , r , r k , r k + ) = (1 , − ,
0) and doubleits C O contribution. The slits q + ℓ q k + , l >
1, are not removed and lock the channel C k (see Figure16). The following two situations arise: (0 , − ε )(0 , ε ) q q k ξ Qq k + q + q k + q k + q k + q k + q + q k + q k + q + q k + q k + q k + A C k B k w A − w B k w C k w A − w B k w A − w B k F igure
16. The case t ∈ I γ, k , r = ± r k = ∓ r k + = C O , w A > w B k , N = n = w A > w B k ( ⇐⇒ t < γ k + ) . We split the analysis in the following two cases:(I) ξ Q < q k + . Then (since w A − w B k > w C k ⇐⇒ t < a + ε q ) we find W γ, k ( t ) = W (1) γ, k ( t ) : = w C k ( t ) · ( q k + − ξ Q ) ∧ q k + if t < a + ε q , W (2) γ, k ( t ) : = (cid:0) ( w A ( t ) − w B k ( t ) (cid:1) · ( q k + − ξ Q ) ∧ q k + + (cid:0) w C k ( t ) − w A ( t ) + w B k ( t ) (cid:1) q k + if t > a + ε q . • When q k + > Q we have k > t k − < γ k + < a + ε q . with contribution G (4 . . . I , Q ( ξ ) = ∞ X k = X ∗ q k + ∈I q , k + q k + > ( ξ Q − q ) ∨ Q Z t k − t k W (1) γ, k ( t ) dt estimated upon (5.7) as G (4 . . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z ∞ ∨ ( ξ − u ) ∨ (1 + u ) dw (1 − u ) ( w − u )( w − u ) · ( w + u − ξ ) + ∧ w . (9.1) HE LINEAR FLOW IN A HONEYCOMB 41 • When q k + Q we have a + ε q γ k + t k − and t k < a + ε q ⇐⇒ q k + > Q for all k >
0. In thiscase the contribution G (4 . . . I , Q ( ξ ) = ∞ X k = X ∗ q k + ∈I q , k + Q > q k + > ( ξ ∨ Q − q Z a + ε q t k W (1) γ, k ( t ) dt + Z γ k + a + ε q W (2) γ, k ( t ) dt + ∞ X k = X ∗ q k + ∈I q , k + Q − q > q k + >ξ Q − q Z γ k + t k W (2) γ, k ( t ) dt is estimated upon (10.15), (10.16), (10.18) (which also holds for k = G (4 . . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z ξ ∨ − u ) ∧ dw F (4 . . . . ( ξ ; u , w ) + c I ζ (2) Z du Z − u (( ξ − u ) ∨ ∧ (2 − u ) dw F (4 . . . . ( ξ ; u , w ) , (9.2)where F (4 . . . . ( ξ ; u , w ) = w + u − u ( w − u ) (cid:18) − uw − u + − w u (cid:19) + (2 − w ) u w ! · ( w + u − ξ ) ∧ w + (2 − w ) uw , F (4 . . . . ( ξ ; u , w ) = w − w − u − u − ww − u + − ww ! + ( w − ( w − u ) w · ( w + u − ξ ) ∧ w . (II) ξ Q > q k + . Let N be the unique integer for which q + Nq k + ξ Q < q + ( N + q k + = q k + + Nq k + , that is 1 N : = j ξ Q − qq k + k ξ . The corresponding range of q k + is y = q k + ∈ J q , N : = (cid:18) ξ Q − qN + , ξ Q − qN (cid:21) . We take n : = j w C k w A − w B k k = j a k − − ε − q k − ta k + − q k + t k >
1. Then t > a + ε q . We only need a + ε q γ k + , that is q k + Q . In this case we take λ k , n : = na k + − a k − + ε nq k + − q k − ր γ k + t k − as n → ∞ , (9.3)hence λ k , n t < λ k , n + and T ( q + n q k + ) = w B k + w C k − n ( w A − w B k ) > w B k > T (cid:0) q + ( n + q k + (cid:1) .This shows that C k is locked by the slits q + q k + , . . . , q + ( n + q k + and W γ, k ( t ) = W γ, k , N ( t ) is givenby n < N , (cid:0) w C k − N ( w A − w B k ) (cid:1)(cid:0) q + ( N + q k + − ξ Q (cid:1) if n = N , ( w A − w B k ) (cid:0) q + ( N + q k + − ξ Q (cid:1) + (cid:0) w C k − ( N + w A − w B k ) (cid:1) q k + if n > N , = t ∈ I γ, k ∩ ( γ, λ k , N ] , W (1) γ, k , N ( t ) : = (cid:0) q + ( N + q k + − ξ Q (cid:1) · (cid:0) ( Nq k + − q k − ) t − Na k + + a k − − ε (cid:1) if t ∈ I γ, k ∩ ( λ k , N , λ k , N + ] , W (2) γ, k ( t ) : = w C k ( t ) q k + − (cid:0) w A ( t ) − w B k ( t ) (cid:1) ( ξ Q − q ) if t ∈ I γ, k ∩ ( λ k , N + , γ k + ] . The following three subcases arise:(II ) q k + Q + Q − qN + . Then λ k , N + t k , W γ, k ( t ) = W (2) γ, k ( t ), ∀ t ∈ ( t k , γ k + ], with contribution G (4 . . I , Q ( ξ ) = X N ξ ∞ X k = X ∗ q ξ Qq k + ∈I q , k + ∩J q , N q k + Q + Q − qN + Z γ k + t k W (2) γ, k ( t ) dtt + t + . Employing (5.4), (13.1), (10.9) (which also holds for k =
0) we find G (4 . . I , Q ( ξ ) ≅ c I ζ (2) X N ξ Z ξ ∧ du Z [ , + − uN + ] ∩ h ξ − uN + , ξ − uN i dw F (4 . . ( ξ ; u , w ) , (9.4)where F (4 . . ( ξ ; u , w ) = w − w − u − uw − u + − ww ! − ( w − ( ξ − u )( w − u ) w . (II ) Q + Q − qN + < q k + Q + Q − qN . In this case λ k , N t k < λ k , N + . The contribution G (4 . . I , Q ( ξ ) = X N ξ ∞ X k = X ∗ q ξ Qq k + ∈I q , k + ∩J q , N Q + Q − qN + < q k + Q + Q − qN Z λ k , N + t k W (1) γ, k , N ( t ) dtt + t + + Z γ k + λ k , N + W (2) γ, k ( t ) dtt + t + is estimated upon (13.4), (13.5), (13.6), (13.9) as G (4 . . I , Q ( ξ ) ≅ c I ζ (2) X N ξ Z ξ ∧ du Z [ + − uN + , + − uN ] ∩ h ξ − uN + , ξ − uN i dw F (4 . . N ( ξ ; u , w ) , (9.5)where F (4 . . N ( ξ ; u , w ) = (2 − w ) (cid:0) (2 N + w + u − ξ (cid:1) ( Nw + u ) w + (cid:0) ( N + w − + u − (cid:1)(cid:0) ( N + w + u − ξ (cid:1) ( w − u )( Nw + u ) − wNw + u + − u − N ( w − w − u ! . (II ) Q + Q − qN < q k + Q . In this case t k < λ k , N . The contribution G (4 . . I , Q ( ξ ) = X N ξ ∞ X k = X ∗ q ξ Qq k + ∈I q , k + ∩J q , N Q + Q − qN < q k + Q Z λ k , N + λ k , N W (1) γ, k , N ( t ) dtt + t + + Z γ k + λ k , N + W (2) γ, k ( t ) dtt + t + is estimated upon (13.4)-(13.7) as G (4 . . ( ξ ) ≅ c I ζ (2) X N ξ Z ξ ∧ du Z [ + − uN , ] ∩ h ξ − uN + , ξ − uN i dw F (4 . . N ( ξ ; u , w ) , (9.6)where F (4 . . N ( ξ ; u , w ) = (2 − w ) ( Nw + u ) ( N + w + u − ξ ( N − w + u + (2 N + w + u − ξ w ! . HE LINEAR FLOW IN A HONEYCOMB 43 w A < w B k ( ⇐⇒ t > γ k + ) . In this situation k > γ k + t k − , so q k + Q . Two casesarise:(I) ξ Q < q k . Then W γ, k ( t ) = w C k ( t ) q k + , ∀ t ∈ ( γ k + , t k − ], with contribution G (4 . . I , Q ( ξ ) = ∞ X k = X ∗ q + ξ Q < q k + Q Z t k − γ k + W γ, k ( t ) dtt + t + . Employing (12.1) we find G (4 . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z ∧ ( u + ξ ∨ dw (2 − w ) ( w − u ) w . (9.7)(II) ξ Q > q k . Consider the unique integer 0 N = j ξ Q − q k q k + k ξ for which q k + Nq k + < ξ Q q k + ( N + q k + = q k + + Nq k + , so the range of q k + is y = q k + ∈ J q , N : = (cid:18) ξ Q + qN + , ξ Q + qN + (cid:21) . This time we take n : = (cid:22) w C k w B k − w A (cid:23) = j a k − − ε − q k − tq k + t − a k + k > λ k , n : = na k + + a k − − ε nq k + + q k − ց γ k + > t k as n → ∞ , (9.8)hence λ k , n + < t λ k , n and B ( q k + n q k + ) = w B k + n ( w B k − w A ) w B k + w C k < B (cid:0) q k + ( n + q k + (cid:1) .This shows that C k is locked by the slits q k + q k + , . . . , q k + ( n + q k + and W γ, k ( t ) = W γ, k , N ( t ) isgiven by n < N , (cid:0) w C k − N ( w B k − w A ) (cid:1)(cid:0) q k + ( N + q k + − ξ Q (cid:1) if n = N , ( w B k − w A ) (cid:0) q k + ( N + q k + − ξ Q (cid:1) + (cid:0) w C k − ( N + w B k − w A ) (cid:1) q k + if n > N , = t ∈ ( λ k , N , t k − ] , W (3) γ, k , N ( t ) : = (cid:0) q k + ( N + q k + − ξ Q (cid:1) · (cid:0) Na k + + a k − − ε − ( Nq k + + q k − ) t (cid:1) if t ∈ ( λ k , N + , λ k , N ] , W (4) γ, k ( t ) : = w C k ( t ) q k + − (cid:0) w B k ( t ) − w A ( t ) (cid:1) ( ξ Q − q k ) if t ∈ ( γ k + , λ k , N + ] . Since λ k , = t k − , the corresponding contribution is given by G (4 . . I , Q ( ξ ) = X N ξ ∞ X k = X ∗ q k + Qq k + ∈I q , k + ∩J q , N Z λ k , N λ k , N + W (3) γ, k , N ( t ) dtt + t + + Z λ k , N + γ k + W (4) γ, k ( t ) dtt + t + . Employing (13.12), (13.13), (13.14) we find G (4 . . I , Q ( ξ ) ≅ c I ζ (2) X N ξ Z du Z [1 + u , ∩ h ξ + uN + , ξ + uN + i dw F (4 . . N ( ξ ; u , w ) , (9.9)where F (4 . . N ( ξ ; u , w ) = (2 − w ) (cid:0) ( N + w − u (cid:1) ( N + w − u − ξ ( N + w − u + (2 N + w − u − ξ w ! . The channel C ← when r = and r k = − . The contributions of ( C ← , r = , r k = −
1) andof ( C ↓ , r = − , r k =
1) have the same main and error terms, so we shall consider below the formersituation and double the result. The slits q + nq k + are removed, while 2 q + nq k + are not, n > B (2 q ) = w B k + w C k ) > B ( q k + ) = w C k + w B k . Again, two cases arise:9.2.1. w A w B k ( ⇐⇒ t > γ k + ) . The slit q k + locks the channel A and W γ, k ( t ) = w A ( t ) · ( q k + − ξ Q ) + ∧ q . We must have γ k + < t k − , so k > q k + Q . The corresponding contribution is G (4 . . I , Q ( ξ ) = ∞ X k = X ∗ q k + ∈I q , k + q k + Q ( q k + − ξ Q ) + ∧ q Z t k − γ k + w A ( t ) dtt + t + . Employing (13.2) we find G (4 . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z + u dw − w ( w − u ) w w − w + w − u − w − u ! · ( w − ξ ) + ∧ u . (9.10)9.2.2. w A > w B k ( ⇐⇒ t < γ k + ) . Consider first the sub-channel of A of width w B k locked by theslit q k + . Its contribution is G (4 . . I , Q ( ξ ) = ∞ X k = X ∗ q k + ∈I q , k + q k + Q ( q k + − ξ Q ) + ∧ q Z γ k + t k w B k ( t ) dtt + t + + ∞ X k = X ∗ q k + ∈I q , k + q k + > Q ( q k + − ξ Q ) + ∧ q Z t k − t k w B k ( t ) dtt + t + . Employing (10.8) and (5.6) we find G (4 . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z dw F (4 . . . ( ξ ; u , w ) + Z ∞ dw F (4 . . . ( ξ ; u , w ) ! , (9.11)where F (4 . . . ( ξ ; u , w ) = ( w − ( w − u ) w · ( w − ξ ) + ∧ u , F (4 . . . ( ξ ; u , w ) = (1 − u ) ( w − u ) ( w − u ) · ( w − ξ ) + ∧ u . The remaining part ˜ A of A (of total width w A − w B k ) is locked by the slits 2 q + n q k + and2 q + ( n + q k + , with n uniquely determined by 2 ε > w B k + w C k ) − n ( w A − w B k ) > w B k + w C k ,or equivalently n = j w C k w A − w B k k = j a k − − ε − q k − ta k + − q k + t k >
0. The widths of the relevant sub-channels (frombottom to top) are 2 ε − w B k + w C k ) + n ( w A − w B k ) = ( n + a k + − a k − + ε − (cid:0) ( n + q k + − q k − (cid:1) t and w C k − n ( w A − w B k ) = ( n q k + − q k − ) t − n a k + + a k − − ε . The following two subcases arise:(I) n = ⇐⇒ t < a + ε q ). From t k < a + ε q we infer q k + > Q . In this situation we have W γ, k , ( t ) = a + ε − qt ) · (2 q − ξ Q ) + ∧ q + w C k ( t ) · (2 q + q k + − ξ Q ) + ∧ q , HE LINEAR FLOW IN A HONEYCOMB 45 ( − , − ε )( − , ε ) q qq k q k + q + q k + q + q k + q + q k + q k + = q + q k + q k + q k + q k + q k + q + q k + q + q k + q + q k + ξ Q A C k B k w A − w B k w A − w B k w B k w A − w B k F igure
17. The case t ∈ I γ, k , r = r k = − r k + = C ← , w A > w B k , n = q k + > Q ⇐⇒ t k − < γ k + ) G (4 . . I , Q ( ξ ) = ∞ X k = X ∗ q k + ∈I q , k + Q − q < q k + Q Z a + ε q t k W γ, k , ( t ) dtt + t + + ∞ X k = X ∗ q k + ∈I q , k + q k + > Q Z t k − t k W γ, k , ( t ) dtt + t + . Employing (10.14), (10.15), (10.16), (5.7), (10.1) we find G (4 . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z − u dw F (4 . . . ( ξ ; u , w ) + Z ∞ dw F (4 . . . ( ξ ; u , w ) ! , (9.12)where F (4 . . . ( ξ ; u , w ) = w + u − u ( w − u ) − uw − u + − w u ! · ( w + u − ξ ) + ∧ u + ( w + u − u ( w − u ) · (2 u − ξ ) + ∧ u , F (4 . . . ( ξ ; u , w ) = (1 − u ) ( w − u )( w − u ) · ( w + u − ξ ) + ∧ u + − u ( w − u )( w − u ) w − w − u + w + u − w − u ! · (2 u − ξ ) + ∧ u . (II) n > ⇐⇒ t > a + ε q ). From a + ε q γ k + we infer q k + Q . Taking λ k , n as in (9.3) we obtain,when t ∈ [ λ k , n , λ k , n + ), W γ, k , n ( t ) = (cid:0) ( n + a k + − a k − + ε − (cid:0) ( n + q k + − q k − ) t (cid:1) · (2 q + n q k + − ξ Q ) + ∧ q + (cid:0) ( n q k + − q k − ) t − n a k + + a k − − ε (cid:1) · (cid:0) q + ( n + q k + − ξ Q (cid:1) + ∧ q . Ordering λ k , n , λ k , n + and t k , the corresponding contribution takes the form G (4 . . I , Q ( ξ ) = ∞ X n = ∞ X k = X ∗ q k + ∈I q , k + Q + Q − qn + < q k + Q + Q − qn Z λ k , n + t k W γ, k , n ( t ) dtt + t + + X ∗ q k + ∈I q , k + Q + Q − qn < q k + Q Z λ k , n + λ k , n W γ, k , n ( t ) dtt + t + . Employing (13.8)-(13.11) we find G (4 . . I , Q ( ξ ) ≅ c I ζ (2) ∞ X n = Z du Z + − un + − un + dw F (4 . . . n ( ξ ; u , w ) + c I ζ (2) ∞ X n = Z du Z + − un dw F (4 . . . n ( ξ ; u , w ) , (9.13)where F (4 . . . n ( ξ ; u , w ) = (cid:0) ( n + w − + u − (cid:1) ( w − u ) ( nw + u ) · ( nw + u − ξ ) + ∧ u + ( n + w − + u − w − u )( nw + u ) − wnw + u + − u − n ( w − w − u ! · (cid:0) ( n + w + u − ξ (cid:1) + ∧ u , F (4 . . . n ( ξ ; u , w ) = (2 − w ) ( nw + u ) (cid:0) ( n − w + u (cid:1) · ( nw + u − ξ ) + ∧ u + (2 − w ) ( nw + u ) (cid:0) ( n − w + u (cid:1) · (cid:0) ( n + w + u − ξ (cid:1) + ∧ u . The channel C ← when r = − and r k = . The ( C ← , r = − , r k =
1) and ( C ↓ , r = , r k = − q k + nq k + areremoved, while 2 q + nq k + are not. Two cases arise:9.3.1. w B k w A ( ⇐⇒ t γ k + ) . The slit q k + locks the channel B k and W γ, k ( t ) = w B k ( t ) · ( q k + − ξ Q ) + ∧ q k . The contribution G (4 . . I , Q ( ξ ) = ∞ X k = X ∗ q k + ∈I q , k + q k + Q ( q k + − ξ Q ) + ∧ q k Z γ k + t k w B k ( t ) dtt + t + + ∞ X k = X ∗ q k + ∈I q , k + q k + > Q ( q k + − ξ Q ) + ∧ q k Z t k − t k w B k ( t ) dtt + t + G (4 . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z dw F (4 . . . ( ξ ; u , w ) + Z ∞ dw F (4 . . . ( ξ ; u , w ) ! , (9.14)where F (4 . . . ( ξ ; u , w ) = ( w − ( w − u ) w · ( w − ξ ) + ∧ ( w − u ) , F (4 . . . ( ξ ; u , w ) = (1 − u ) ( w − u ) ( w − u ) · ( w − ξ ) + ∧ ( w − u ) . HE LINEAR FLOW IN A HONEYCOMB 47 ( − , − ε )( − , ε ) q q k q k + q k q k + q k + q k + q k + q k + q k + q k + ξ Q A C k B k w B k − w A w B k − w A w A F igure
18. The case t ∈ I γ, k , r = − r k = r k + = C ← , w A < w B k , n = w A < w B k ( ⇐⇒ t > γ k + ) . Then k > q k + Q . Consider first the sub-channel of B k of width w A locked by q k + , with contribution G (4 . . I , Q ( ξ ) = ∞ X k = X ∗ q k + ∈I q , k + q k + Q ( q k + − ξ Q ) + ∧ q k Z t k − γ k + w A ( t ) dtt + t + ≅ c I ζ (2) Z du Z + u dw − w ( w − u ) w w − w + w − u − w − u ! · ( w − ξ ) + ∧ ( w − u ) . (9.15)The remaining sub-channel ˜ B k of B k (of width w B k − w A ) is locked by the slits 2 q k + n q k + and2 q k + ( n + q k + , with n uniquely determined by 0 B (2 q k ) + n ( w B k − w A ) < w B k − w A = T ( q k + ),or equivalently n : = j w C k w B k − w A k = j a k − − ε − q k − tq k + t − a k + k >
0. The situation is described in Figure 18.The widths of the relevant two sub-channels of ˜ B k are (from top to bottom) B (2 q k + n q k + ) = (cid:0) ( n + q k + + q k − (cid:1) t − ( n + a k + − a k − + ε and w B k − w A − B (2 q k + n q k + ) = n a k + + a k − − ε − ( n q k + + q k − ) t . Taking this time λ k , n as in (9.8) we find, for t ∈ ( λ k , n + , λ k , n ], W γ, k , n ( t ) = (cid:16)(cid:0) ( n + q k + + q k − (cid:1) t − (cid:0) ( n + a k + + a k − − ε (cid:1)(cid:17) · (2 q k + n q k + − ξ Q ) + ∧ q k + (cid:16) n a k + + a k − − ε − ( n q k + + q k − ) t (cid:17) · (cid:0) q k + ( n + q k + − ξ Q (cid:1) + ∧ q k . Since λ k , = t k − the corresponding contribution is given by G (4 . . I , Q ( ξ ) = ∞ X n = ∞ X k = X ∗ q k + ∈I q , k + q k + Q Z λ k , n λ k , n + W γ, k , n ( t ) dtt + t + . Employing (13.14) and (13.15) we find G (4 . . I , Q ( ξ ) ≅ c I ζ (2) Z du Z + u dw ∞ X n = F (4 . . n ( ξ ; u , w ) , (9.16)where F (4 . . n ( ξ ; u , w ) = (2 − w ) (cid:0) ( n + w − u (cid:1) (cid:0) ( n + w − u (cid:1) · (cid:0) ( n + w − u − ξ (cid:1) + ∧ ( w − u ) + (2 − w ) (cid:0) ( n + w − u (cid:1)(cid:0) ( n + w − u (cid:1) · (cid:0) ( n + w − u − ξ (cid:1) + ∧ ( w − u ) . F igure
19. The individual contributions of G (0) , . . . , G (4) to Φ hex HE LINEAR FLOW IN A HONEYCOMB 49
10. A ppendix For k > Z t k − t k a + ε − qt ) dtt + t + = − Z t k − γ qt − at + t + dt − Z t k γ qt − at + t + dt ! + Z t k − t k ε t + t + dt = Q − q Q q k − q k ( γ + γ + q k + − Qq k − + q k + − Qq k ! + O Qq q k − ! . (10.1) For k > q k + Q : Z t k − ak − ε qk w C k ( t ) dtt + t + = (2 Q − q k + ) Q q k − q k ( γ + γ + + O Qqq k ! . (10.2) Z t k − ak − ε qk q k t − a k + ε ) dtt + t + = (2 Q − q k + ) Q q k − q k ( γ + γ + + O Q qq k ! . (10.3) Z t k − ak − ε qk w A ( t ) dtt + t + = Z u ak − ε qk a + ε − qtt + t + dt − Z u t k − a + ε − qtt + t + dt = Q − q k + Q q k − q k ( γ + γ + q k + q k + − Q q k + q k − Qq k − ! + O q q k − q k ! . (10.4) Z ak − ε qk γ k + w A ( t ) dtt + t + = Z u γ k + a + ε − qtt + t + dt − Z u ak − ε qk a + ε − qtt + t + dt = Q − q k + Q q k q k + ( γ + γ + q k + − Qq k + + q k + q k + − Q q k ! + O q q k q k + ! . (10.5) Z ak − ε qk γ k + q k + t − a k + t + t + dt = (2 Q − q k + ) Q q k q k + ( γ + γ + + O qq k q k + ! . (10.6) Z ak − ε qk γ k + a k − ε − q k t ) dtt + t + = (2 Q − q k + ) Q q k q k + ( γ + γ + + O qq k q k + ! . (10.7) Z γ k + t k w B k ( t ) dtt + t + = ( q k + − Q ) Q q k q k + ( γ + γ + + O qq k q k + ! . (10.8) Z γ k + t k w A ( t ) − w B k ( t ) t + t + dt = Z γ k + t k a k + − q k + tt + t + dt = ( q k + − Q ) Q q k q k + ( γ + γ + + O qq k q k + ! . (10.9) For k > q k + Q − q : Z γ k + t k qt − a − ε ) t + t + dt = Z γ k + a + ε q qt − a − ε ) t + t + dt − Z t ka + ε q qt − a − ε ) t + t + dt = q k + − Q Q q k q k + ( γ + γ + Q − q k + q k + + Q − q k + q k ! + O q q k + ! . (10.10) For k > Q − q < q k + Q it follows that 0 q k + − Q q , k = j Q − q − q ′ q k , and Z a + ε q t k w A ( t ) dtt + t + = Z u t k a + ε − qtt + t + dt − Z u a + ε q a + ε − qtt + t + dt = ( q k + − Q )( q k + q k + − Q )8 Q qq k ( γ + γ + + O q q ′ q k ! . (10.11) Z a + ε q t k qt − at + t + dt = Z a + ε q γ qt − at + t + dt − Z t k γ qt − at + t + dt = ( q k + − Q )( q k − + Q )8 Q qq k ( γ + γ + + O Q q ! . (10.12) Z a + ε q t k w B k ( t ) dtt + t + = ( q k + − Q ) Q q q k ( γ + γ + + O Q qq k ! . (10.13) Z a + ε q t k a + ε − qt ) dtt + t + = ( q k + − Q ) Q qq k ( γ + γ + + O Qqq ′ q k ! . (10.14) Z a + ε q t k w C k ( t ) dtt + t + = Z t k − t k a k − − ε − q k − tt + t + dt − Z t k − a + ε q a k − − ε − q k − tt + t + dt = q k + − Q Q qq k ( γ + γ + Q − qq k + Q − q k + q ! + O Qqq k ! if k > . (10.15) Z a + ε q t w C ( t ) t + t + dt = Z γ ′ t a ′ − q ′ tt + t + dt − Z γ ′ a + ε q a ′ − q ′ tt + t + dt − Z u t a + ε − qtt + t + dt − Z u a + ε q a + ε − qtt + t + dt = q − Q Q qq ′ ( γ + γ + Q − qq ′ + Q − q q ! + O Q qq ′ ! . (10.16) Z γ k + a + ε q w B k ( t ) dtt + t + = Z γ k + t k q k t − a k + ε t + t + dt − Z a + ε q t k q k t − a k + ε t + t + dt = Q − q k + Q qq k + ( γ + γ + q k + − Qq k + + q k + − Q q ! + O Q qq k q k + ! . (10.17) Z γ k + a + ε q a k + − q k + tt + t + dt = (2 Q − q k + ) Q q q k + ( γ + γ + + O q q k + ! . (10.18) HE LINEAR FLOW IN A HONEYCOMB 51 Z γ k + a + ε q qt − a − ε ) dtt + t + = (2 Q − q k + ) Q qq k + ( γ + γ + + O Qq q k + ! . (10.19) Z γ k + a + ε q w C k ( t ) dtt + t + = (2 Q − q k + ) q k + Q q q k + ( γ + γ + + O (cid:16) q q k − (cid:17) if k > , O (cid:16) q q ′ (cid:17) if k = . (10.20) A ppendix C O contribution when ( r , r k ) = (0 , . Here W (1) γ, k , n and W (2) γ, k are as in (7.3)and (7.4). For k > q k + Q : Z t k − γ k + w A ( t ) dtt + t + = Z u γ k + a + ε − qtt + t + dt − Z u t k − a + ε − qtt + t + dt = Q − q k + Q q k − q k + ( γ + γ + q k + − Qq k + + q k − Qq k − ! + O Qq q k + ! . (11.1)Summing as in (5.2) and employing Lemma 4 we infer G (2 . I , Q ( ξ ) = X β ∈{− , } X q Qq ≡− β (mod 3) ∞ X k = X x = x ∈ q (1 − I ) , ( ˜ x , q ) = y = q k ∈I q , k , y Q − q ˜ xy ≡ β q + (mod q ) ( y + q − ξ Q ) + ∧ q Z t k − γ k + w A ( t ) t + t + dt ≅ c I C (3)3 X β ∈{− , } ϕ ( q ) q Z Q − qQ Q − q − y Q ( y − q )( y + q ) y + q − Qy + q + y − Qy − q ! · ( y + q − ξ Q ) + ∧ q dy . Estimate (7.1) follows applying Lemma 2 and making the change of variable ( q , y ) = ( Qu , Qw ). For k > q k + > Q : Z t k − t k w A ( t ) − w B k ( t ) t + t + dt = Z γ k + t k a k + − q k + tt + t + dt − Z γ k + t k − a k + − q k + tt + t + dt = Q − q Q q k − q k ( γ + γ + q k + − Qq k + q k + − Qq k − ! + O Q qq k q k + ! . (11.2) When n > n ( Q − q ) q k ( n + Q − q ): Z γ k + n t k W (1) γ, k , n ( t ) dtt + t + = ( q k + n + − ξ Q ) Z γ k + n t k a k + n − q k + n tt + t + dt = ( q k + n − nQ ) ( q k + n + − ξ Q )2 Q q k q k + n ( γ + γ + + O ξ qq k q k + n ! . (11.3)Here and below ≪ ξ means uniformly for ξ in compact subsets of [0 , ∞ ). When n > n + Q − q ) q k Q + n ( Q − q ): Z γ k + n γ k + n + W (1) γ, k , n ( t ) dtt + t + = ( q k + n + − ξ Q ) Z γ k + n γ k + n + a k + n − q k + n tt + t + dt = q k + n + − ξ Q q k + n q k + n + ( γ + γ + + O ξ qq k + n q k + n + ! . (11.4) When k > Q + n ( Q − q ) q k Q + ( n + Q − q ): Z t k − γ k + n + W (1) γ, k , n ( t ) dtt + t + = ( q k + n + − ξ Q ) Z γ k + n γ k + n + a k + n − q k + n tt + t + dt − Z γ k + n t k − a k + n − q k + n tt + t + dt ! = (cid:0) ( n + Q − q k + n + (cid:1) ( q k + n + − ξ Q )2 Q q k − q k + n + ( γ + γ + Qq k + n + + q k + n − ( n + Qq k − ! + O ξ qq k − q k + n q k + n + ! . (11.5) When n > n + Q − q ) q k Q + ( n + Q − q ): Z γ k + n + t k dtt + t + = q k − ( n + Q − q ) Qq k q k + n + ( γ + γ + + O Qqq k + n + ! , Z γ n + k + t k w B k ( t ) + w C k ( t ) t + t + dt = Z γ k + n + γ qt − at + t + dt − Z t k γ qt − at + t + dt = q k − ( n + Q − q )2 Q q k q k + n + ( γ + γ + Qq n + k + + Q − qq k ! + O Q q n + k + ! , yielding Z γ k + n + t k W (2) γ, k ( t ) dtt + t + = q k + n + − ( n + Q Qq k q k + n + ( γ + γ + q k + n + − ξ Qq k + n + + q k − ξ ( Q − q ) q k ! + O ξ Qqq k + n + . (11.6) When q k > Q + ( n + Q − q ) formulas (5.8) and (5.9) yield (here k > Z t k − t k W (2) γ, k ( t ) dtt + t + = Q − q Qq k − q k ( γ + γ + − ξ (cid:18) Q − qq k − + Q − qq k (cid:19)! + O ξ qq k − q k ! . (11.7) C ← contribution when ( r , r k ) = (0 , . Here λ k , n is as in (7.7). ( n − Q − q ) q k n ( Q − q ). In this case n >
2. When k > Z λ k , n t k w C k ( t ) dtt + t + = Z t k − t k a k − − ε − q k − tt + t + dt − Z t k − λ k , n a k − − ε − q k − tt + t + dt = q k − ( n − Q − q )2 Q q k ( q k + q k + n − )( γ + γ + Q − qq k + ( n + Q − q k + n q k + q k + n − ! + O qq k − q k ! . (11.8) Z λ k , n t k w B k ( t ) dtt + t + = Z λ k , n t k q k t − a k + ε t + t + dt = (cid:0) q k − ( n − Q − q ) (cid:1) Q q k ( q k + q k + n − ) ( γ + γ + + O qq k q k + n ! . (11.9) HE LINEAR FLOW IN A HONEYCOMB 53 Z λ k , n t k q k + n t − a k + n t + t + dt = Z λ k , n γ k + n q k + n t − a k + n t + t + dt − Z t k γ k + n q k + n t − a k + n t + t + dt = q k − ( n − Q − q )2 Q q k ( q k + q k + n − )( γ + γ + ( n + Q − q k + n q k + q k + n − + n ( Q − q ) − q k q k ! + O qq k q k + n ! . (11.10)When k = Z λ , n t w C ( t ) dtt + t + = Z λ , n γ qt − at + t + dt − Z t γ qt − at + t + dt − Z λ , n t q ′ t − a ′ + ε t + t + dt = q n − − ( n − Q Q q ′ ( q ′ + q n − )( γ + γ + ( n + Q − q n q ′ + q n − + Q − qq ′ ! + O Qq q n − ! . (11.11)Since n only takes one value n = j Q + q k − Q − q k for fixed k , the error terms in (11.8)-(11.11) do not play arole in the final asymptotic formula. When k > n ( Q − q ) q k Q + n ( Q − q ): Z λ k , n λ k , n + w C k ( t ) dtt + t + = Z t k − λ k , n + a k − − ε − q k − tt + t + dt − Z t k − λ k , n a k − − ε − q k − tt + t + dt = Q − q Q ( q k + q k + n − )( q k + q k + n )( γ + γ + ( n + Q − q k + n + q k + q k + n + ( n + Q − q k + n q k + q k + n − ! + O n − q q k ! . (11.12) Z λ k , n λ k , n + w B k ( t ) dtt + t + = Z λ k , n t k q k t − a k + ε t + t + dt − Z λ k , n + t k q k t − a k + ε t + t + dt = Q − q Q ( q k + q k + n − )( q k + q k + n )( γ + γ + q k + n − − ( n − Qq k + q k + n − + q k + n − nQq k + q k + n ! + O qq k q k + n ! . (11.13) Z γ k + n λ k , n + a k + n − q k + n tt + t + dt = ( q k + n − nQ ) Q q k + n ( q k + q k + n ) ( γ + γ + + O qq k + n ! . (11.14) Z λ k , n γ k + n q k + n t − a k + n t + t + dt = (cid:0) ( n + Q − q k + n (cid:1) Q q k + n ( q k − + q k + n ) ( γ + γ + + O qq k + n ! . (11.15) When k = n > t < λ , n + < γ n < λ , n t k − . Analog formulas as(11.12)-(11.15) hold, with same main terms and Z λ , n λ , n + w C ( t ) dtt + t + = Z λ , n γ qt − at + t + dt − Z λ , n + γ qt − at + t + dt − Z λ , n t q ′ t − a ′ + ε t + t + dt − Z λ , n + t q ′ t − a ′ + ε t + t + dt ! = Q − q Q ( q ′ + q n − )( q ′ + q n )( γ + γ + ( n + Q − q n + q ′ + q n + ( n + Q − q n q ′ + q n − ! + O qq ′ q n ! . (11.16) When k = n = t λ , < γ λ , . We have Z γ λ , w B ( t ) dtt + t + = Z γ t q ′ t − a ′ + ε t + t + dt − Z λ , t q ′ t − a ′ + ε t + t + dt = ( q − Q ) ( q ′ + q )2 Q q ( q ′ + q ) ( γ + γ + + O Q qq ′ ! . (11.17) Z γ λ , a − q tt + t + dt = ( q − Q ) Q q ( q ′ + q ) ( γ + γ + + O Q q ! . (11.18) Z γ λ , w C ( t ) dtt + t + = Z γ ′ λ , a ′ − q ′ tt + t + dt − Z γ ′ γ a ′ − q ′ tt + t + dt − Z a + ε q λ , a + ε − qtt + t + dt + Z a + ε q γ a + ε − qtt + t + dt = q − Q Q q ( q ′ + q )( γ + γ + Q − q q ′ + q + Q − q q ! + O Q qq ′ ! . (11.19) When Q + n ( Q − q ) q k Q + ( n + Q − q ): Z t k − λ k , n + w C k ( t ) dtt + t + = (cid:0) ( n + Q − q k + n + (cid:1) Q q k − ( q k + q k + n ) ( γ + γ + + O Qqq k − q k + n ! . (11.20) Z t k − λ k , n + w B k ( t ) dtt + t + = Z t k − t k q k t − a k + ε t + t + dt − Z λ k , n + t k q k t − a k + ε t + t + dt = ( n + Q − q k + n + Q q k − ( q k + q k + n )( γ + γ + Q − qq k − + q k + n − nQq k + q k + n ! + O Q qq k ! . (11.21) Z t k − λ k , n + a k + n − q k + n tt + t + dt = Z γ k + n λ k , n + a k + n − q k + n tt + t + dt − Z γ k + n t k − a k + n − q k + n tt + t + dt = ( n + Q − q k + n + Q q k − ( q k + q k + n )( γ + γ + q k + n − nQq k + q k + n + q k + n − ( n + Qq k − ! + O Q qq k + n ! . (11.22) A ppendix C O contribution when ( r , r k ) = (1 , . Here λ k , n is as in (8.3) and W (1) γ, k , n and W (2) γ, k as in (8.4). HE LINEAR FLOW IN A HONEYCOMB 55
For k > q k + Q : Z t k − γ k + w C k ( t ) dtt + t + = (2 Q − q k + ) Q q k − q k + ( γ + γ + + O qq k − q k ! . (12.1) Z t k − γ k + w B k ( t ) − w A ( t ) t + t + dt = Z t k − γ k + q k + t − a k + t + t + dt = (2 Q − q k + ) Q q k − q k + ( γ + γ + + O qq k − q k + ! . (12.2) For k > n ( q k − Q ) Q − q ( N + q k − Q ): Z t k − λ k , n ( q + nq k ) t − ( a + na k ) t + t + dt = (cid:0) ( n + Q − ( q + nq k ) (cid:1) Q q k − ( q + nq k )( γ + γ + + O qq k − ( q + nq k ) ! . (12.3) For k > n + q k − Q ) Q − q : Z λ k , n + λ k , n W (1) γ, k , n ( t ) dtt + t + = q k + + nq k − ξ Q q + nq k )( q k + + nq k ) ( γ + γ + + O ξ qq k ( q + nq k )( q k + + nq k ) ! . (12.4) Z t k − λ k , n + dtt + t + = ( n + Q − q k + − nq k Qq k − ( q k + + nq k )( γ + γ + + O qq k − q k ! . (12.5) Z t k − λ k , n + w A ( t ) + w C k ( t ) t + t + dt = Z γ k λ k , n + a k − q k tt + t + dt − Z γ k t k − a k − q k tt + t + dt = ( n + Q − q k + − nq k Q q k − ( q k + + nq k )( γ + γ + Qq k + + nq k + q k − Qq k − ! + O q k − q k + ! . (12.6) Z t k − λ k , n + W (2) γ, k ( t ) t + t + dt = ( n + Q − q k + − nq k Qq k − ( q k + + nq k )( γ + γ + · q k + + nq k − ξ Qq k + + nq k + q k − − ξ ( q k − Q ) q k − ! + O ξ qq k − ! . (12.7) C ← contribution when ( r , r k ) = (1 , . Here µ k , n and ν k , n are as in (8.7). For q > Q : Z γ a + ε q w C ( t ) dtt + t + = − Z γ t q ′ t − a ′ + ε t + t + dt − Z a + ε q t q ′ t − a ′ + ε t + t + dt + Z γ γ qt − at + t + dt − Z a + ε q γ qt − at + t + dt = (2 Q − q ) q Q q q ( γ + γ + + O Q q q ′ ! . (12.8) Z γ a + ε q w A ( t ) dtt + t + = Z u a + ε q w A ( t ) dtt + t + dt − Z u γ w A ( t ) dtt + t + = (2 Q − q )(3 q − Q )8 Q qq ( γ + γ + + O Q q ! . (12.9) Z γ a + ε q a − q tt + t + dt = (2 Q − q ) Q q q ( γ + γ + + O q Q ! . (12.10) When n = Q − q ) n + q k Q − q ) n ∧ (cid:16) Q + Q − qn + (cid:17) : Z µ k , n t k w A ( t ) dtt + t + = Z u t k a + ε − qtt + t + dt − Z u µ k , n a + ε − qtt + t + dt = ( n + q k − Q − q )2 Q q k (2 q + nq k )( γ + γ + q k + − Qq k + q + nq k − nQ q + nq k ! + O Qq q k ! . (12.11) Z µ k , n t k w C k ( t ) dtt + t + = Z t k − t k a k − − ε − q k − tt + t + dt − Z t k − µ k , n a k − − ε − q k − tt + t + dt = ( n + q k − Q − q )2 Q q k (2 q + nq k )( γ + γ + Q − qq k + ( n + Q − q − ( n + q k q + nq k ! + O Qqq k − q k ! . (12.12) Z ν k , n t k a + na k − ( q + nq k ) tt + t + dt = ( q + nq k − Q ) Q q k ( q + nq k )( γ + γ + + O qq k ! . (12.13) Z µ k , n ν k , n ( q + nq k ) t − ( a + na k ) t + t + dt = ( q + nq k − Q ) Q ( q + nq k )(2 q + nq k ) ( γ + γ + + O qq k ! . (12.14) When Q − q ) n q k Q + Q − qn + , n , k > n > Z µ k , n µ k , n − w A ( t ) dtt + t + = Z u µ k , n − a + ε − qtt + t + dt − Z u µ k , n a + ε − qtt + t + dt = Q − q k Q (2 q + nq k ) (cid:0) q + ( n − q k (cid:1) ( γ + γ + q + ( n − q k − Q )2 q + ( n − q k + q + n ( q k − Q )2 q + nq k ! + O n − q q k ! . (12.15)For n = Q − q k q and µ k , − γ qq k ) as follows: Z µ k , µ k , w A ( t ) dtt + t + = − Z µ k , µ k , qt − a − ε t + t + dt + Z µ k , µ k , ε dtt + t + = Q − q k Q (2 q + q k )(2 q )( γ + γ + q q + q + q k − Q q + q k ! + O Q qq k ! . (12.16) HE LINEAR FLOW IN A HONEYCOMB 57
Since 0 ( n + Q − q − nq k Q − q < q ′ and 0 nq k − ( n + Q Q we find: Z µ k , n µ k , n − w C k ( t ) dtt + t + = Z t k − µ k , n − a k − − ε − q k − tt + t + dt − Z t k − µ k , n a k − − ε − q k − tt + t + dt = Q − q k Q (2 q + nq k ) (cid:0) q + ( n − q k (cid:1) ( γ + γ + · Q − q − n ( q k − Q )2 q + ( n − q k + Q − q − ( n + q k − Q )2 q + nq k ! + O qq k − (cid:0) q + ( n − q k (cid:1) . (12.17) Z ν k , n µ k , n − a + na k − ( q + nq k ) tt + t + dt = (cid:0) Q − q − n ( q k − Q ) (cid:1) Q ( q + nq k ) (cid:0) q + ( n − q k (cid:1) ( γ + γ + + O nqq k (cid:0) q + ( n − q k (cid:1) . (12.18) Z µ k , n ν k , n ( q + nq k ) t − ( a + na k ) t + t + dt = (cid:0) q + n ( q k − Q ) (cid:1) Q ( q + nq k )(2 q + nq k ) ( γ + γ + + O n qq k ! . (12.19)We check that the contribution of error terms is negligible. Note that when n = q > Q − q k > Q − Q − q = Q + q , so q > Q . The errors in (12.15)-(12.19) add up to ≪ ∞ X n = ∞ X k = X γ ∈F ( Q ) q ( n − q q k + ∞ X k = X γ ∈F ( Q ) Q q k + ∞ X n = ∞ X k = X γ ∈F ( Q ) q k + qq k − q k + ∞ X k = X γ ∈F ( Q ) q k + q q k − + ∞ X n = ∞ X k = X γ ∈F ( Q ) q k + n ( n − qq k + ∞ X k = X γ ∈F ( Q ) q k + q q k + ∞ X n = ∞ X k = n q k + | I | · Q ≪ X γ ∈F ( Q ) Qq q ′ + X Q q Q ϕ ( q ) q + Q c − ≪ Q c − . For k > q k + Q : Z t k − γ k + q k + t − a k + t + t + dt = (2 Q − q k + ) Q q k − q k + ( γ + γ + + O qq k − q k + ! . (12.20) For n , k > Q − q ) n ∨ (cid:16) Q + Q − qn + (cid:17) q k Q + Q − qn : Z t k − µ k , n − w A ( t ) dtt + t + = Z u µ k , n − a + ε − qtt + t + dt − Z u t k − a + ε − qtt + t + dt = ( n + Q − q − nq k Q q k − (cid:0) q + ( n − q k (cid:1) ( γ + γ + q + ( n − q k − Q )2 q + ( n − q k + q k − Qq k − ! + O q (cid:0) q + ( n − q k (cid:1) (cid:18) Q − qQ q k − + q (cid:0) q + ( n − q k (cid:1) (cid:19)! . (12.21) When n > q q k − q k . When n = Z t k − µ k , a + ε − qtt + t + dt = − Z t k − a + ε q a + ε − qtt + t + dt + Z t k − a + ε q ε dtt + t + = Q − q k + Q qq k − ( γ + γ + − Q − q k + q k − ! + O Qq q k − + q q k − + Q q q k − ! . (12.22) Z t k − µ k , n − w C k ( t ) dtt + t + = (cid:0) ( n + Q − q − nq k (cid:1) Q q k − (cid:0) q + ( n − q k (cid:1) ( γ + γ + + O qq k − (cid:0) q + ( n − q k (cid:1) . (12.23) Z ν k , n µ k , n − a + na k − ( q + nq k ) tt + t + dt = (cid:0) ( n + Q − q − nq k (cid:1) Q ( q + nq k ) (cid:0) q + ( n − q k (cid:1) ( γ + γ + + O q q k ! . (12.24) Z t k − ν k , n ( q + nq k ) t − ( a + na k ) t + t + dt = (cid:0) ( n + Q − q − nq k (cid:1) Q q k − ( q + nq k )( γ + γ + + O Q qq k ! . (12.25) A ppendix k > q k + Q : Z γ k + t k w C k ( t ) dtt + t + = Z t k − t k a k − − ε − q k − tt + t + dt − Z t k − γ k + a k − − ε − q k − tt + t + dt = q k + − Q Q q k q k + ( γ + γ + Q − qq k + Q − q k + q k + ! + O qq k − q k ! . (13.1) Z t k − γ k + w A ( t ) dtt + t + = Z u γ k + a + ε − qtt + t + dt − Z u t k − a + ε − qtt + t + dt = Q − q k + Q q k − q k + q k + − Qq k + + q k − Qq k − ! + O q q k + ! . (13.2)When q k + > Q : Z γ k + t k a + ε − qt ) dtt + t + = − Z γ k + γ qt − at + t + dt − Z t k γ qt − at + t + dt ! + Z γ k + t k ε dtt + t + = q k + − Q Q q k q k + ( γ + γ + q k + − Qq k + q k + − Qq k + ! + O q q k ! . (13.3) C O and the C ← contributions when ( r , r k ) = (1 , − and w A > w B k . Here λ k , n is as in (9.3). If Q + Q − qn + q k + Q and k >
1, then Z γ k + λ k , n + w C k ( t ) dtt + t + = Z t k − λ k , n + a k − − ε − q k − tt + t + dt − Z t k − γ k + a k − − ε − q k − tt + t + dt = Q − q k + Q q k + ( nq k + + q )( γ + γ + ( n + Q − q k + ) nq k + + q + Q − q k + q k + ! + O qq k − q k + ! . (13.4) HE LINEAR FLOW IN A HONEYCOMB 59
When k = Z γ λ , n + w C ( t ) dtt + t + = − Z γ t q ′ t − a ′ + ε t + t + dt − Z λ , n + t q ′ t − a ′ + ε t + t + dt ! + Z γ γ qt − at + t + dt − Z λ , n + γ qt − at + t + dt ! = Q − q Q q ( nq + q )( γ + γ + ( n + Q − q ) nq + q + Q − q q ! + O Q q q ′ ! . (13.5) When n > Q + Q − qn + q k + Q : Z γ k + λ k , n + w A ( t ) − w B k ( t ) t + t + dt = (2 Q − q k + ) Q q k + ( nq k + + q ) ( γ + γ + + O qq k + ! . (13.6) Z λ k , n + λ k , n ( nq k + − q k − ) t − na k + + a k − − ε t + t + dt = (2 Q − q k + ) Q ( nq k + + q ) (cid:0) ( n − q k + + q (cid:1) ( γ + γ + + O Qq q k + ! . (13.7) When n > Q + Q − qn + q k + Q + Q − qn we have 0 ( n + q k + − Q ) + q − Q Q − qn and Z λ k , n + t k ( n + a k + − a k − + ε − (cid:0) ( n + q k + − q k − (cid:1) tt + t + dt = (cid:0) ( n + q k + − Q ) + q − Q (cid:1) Q q k ( nq k + + q )( γ + γ + + O n qq k q k + ! . (13.8) Z λ k , n + t k ( nq k + − q k − ) t − na k + + a k − − ε t + t + dt = Z λ k , n + λ k , n · · · − Z t k λ k , n · · · = ( n + q k + − Q ) + q − Q Q q k ( nq k + + q )( γ + γ + Q − q k + nq k + + q + Q − q − n ( q k + − Q ) q k ! + O n Qqq k q k + ! . (13.9) When n > Q + Q − qn q k + Q : Z λ k , n + λ k , n ( n + a k + − a k − + ε − (cid:0) ( n + q k + − q k − (cid:1) tt + t + dt = (2 Q − q k + ) Q ( nq k + + q ) (cid:0) ( n − q k + + q (cid:1) ( γ + γ + + O n − n + q q k + ! . (13.10) Z λ k , n + λ k , n ( nq k + − q k − ) t − na k + + a k − − ε t + t + dt = (2 Q − q k + ) Q ( nq k + + q ) (cid:0) ( n − q k + + q (cid:1) ( γ + γ + + O n Qq q k + ! . (13.11) C O and the C ↓ contributions when ( r , r k ) = (1 , − and w A < w B k . Herewe take λ k , n as in (9.8), k > q k + Q . Z λ k , n + γ k + w C k ( t ) dtt + t + = Z t k − γ k + a k − − ε − q k − tt + t + dt − Z t k − λ k , n + a k − − ε − q k − tt + t + dt = (2 Q − q k + ) (cid:0) (2 n + q k + + q k − (cid:1) Q q k + (cid:0) ( n + q k + + q k − (cid:1) ( γ + γ + + O qq k − q k + ! . (13.12) Z λ k , n + γ k + w B k ( t ) − w A ( t ) t + t + dt = Z λ k , n + γ k + q k + t − a k + t + t + dt = (2 Q − q k + ) Q q k + (cid:0) ( n + q k + + q k − (cid:1) ( γ + γ + + O qq k + ! . (13.13) Z λ k , n λ k , n + na k + + a k − − ε − ( nq k + + q k − ) tt + t + dt = (2 Q − q k + ) Q ( nq k + + q k − ) (cid:0) ( n + q k + + q k − (cid:1) ( γ + γ + + O n + qq k − q k + ! . (13.14) Z λ k , n λ k , n + (cid:0) ( n + q k + + q k − (cid:1) t − ( n + a k + − a k − + ε t + t + dt = (2 Q − q k + ) Q ( nq k + + q k − ) (cid:0) ( n + q k + + q k − (cid:1) ( γ + γ + + O n + qq k − q k + ! . (13.15)R eferences [1] S. Blank and N. Krikorian, Thom’s problem on irrational flows , Internat. J. Math. (1993), 721–726.[2] F. P. Boca, C. Cobeli and A. Zaharescu, Distribution of lattice points visible from the origin , Comm. Math. Phys. (2000), 433–470.[3] F. P. Boca and R. N. Gologan,
On the distribution of the free path length of the linear flow in a honeycomb , Ann.Inst. Fourier (2009), 1043–1075.[4] F. P. Boca, R. N. Gologan and A. Zaharescu, The average length of a trajectory in a certain billiard in a flattwo-torus , New York J. Math. (2003), 303–330.[5] F. P. Boca, R. N. Gologan and A. Zaharescu, The statistics of the trajectory of a billiard in a flat two-torus , Comm.Math. Phys. (2003), 53–73.[6] F. P. Boca and A. Zaharescu,
On the correlations of directions in the Euclidean plane , Trans. Amer. Math. Soc. (2006), 1797–1825.[7] F. P. Boca and A. Zaharescu,
The distribution of the free path lengths in the periodic two-dimensional Lorentz gasin the small-scatterer limit , Comm. Math. Phys. (2007), 425–471.[8] J. Bourgain, F. Golse and B. Wennberg,
On the distribution of free path lengths for the periodic Lorentz gas ,Comm. Math. Phys. (1998), 491–508.[9] E. Caglioti and F. Golse,
On the distribution of free path lengths for the periodic Lorentz gas. III , Comm. Math.Phys. (2003), 199–221.[10] E. Caglioti and F. Golse,
The Boltzmann-Grad limit of the periodic Lorentz gas in two space dimensions , C. R.Math. Acad. Sci. Paris (2008), 477–482.[11] P. Dahlqvist,
The Lyapunov exponent in the Sinai billiard in the small scatterer limit , Nonlinearity (1997),159–173.[12] F. Golse, The periodic Lorentz gas in the Boltzmann-Grad limit , Proc. ICM (Madrid, 2006), vol. 3, EMS, Z¨urich2006, pp. 183–201.
HE LINEAR FLOW IN A HONEYCOMB 61 [13] F. Golse,
Recent results on the periodic Lorentz gas , preprint arXiv:0906.0191.[14] H. A. Lorentz,
Le mouvement des ´electrons dans les m´etaux . Arch. N´eerl. (1905), p. 336. Reprinted in Collectedpapers , Vol. , The Hague: Martinus Nijho ff , 1936.[15] J. Marklof, Kinetic transport in crystals , preprint math-ph / The distribution of free path lengths in the periodic Lorentz gas and relatedlattice point problems , preprint arXiv:0706.4395, to appear in Ann. Math.[17] J. Marklof and A. Str¨ombergsson,
The Boltzmann-Grad limit of the periodic Lorentz gas , preprintarXiv:0801.0612, to appear in Ann. Math.[18] J. Marklof and A. Str¨ombergsson,
Kinetic transport in the two-dimensional periodic Lorentz gas , Nonlinearity (2008), 1413–1422.[19] G. P´olya, Zahlentheoretisches und wahrscheinlichkeitstheoretisches ¨uber die sichtweite im walde , Arch. Math.Phys. (1918), 135–142.[20] J. Smillie, C. Ulcigrai, Symbolic coding for linear trajectories in the regular octogon , preprint math.DS / epartment of M athematics , U niversity of I llinois , 1409 W. G reen S treet , U rbana , IL 61801, USAI nstitute of M athematics of the R omanian A cademy , P.O. B ox ucharest , R omania E-mail address ::