On the Finsler metrics obtained as limits of chessboard structures
aa r X i v : . [ m a t h . A P ] A p r ON THE FINSLER METRICS OBTAINED AS LIMITS OFCHESSBOARD STRUCTURES
MICOL AMAR, GRAZIANO CRASTA, AND ANNALISA MALUSA
Abstract.
We study the geodesics in a planar chessboard structure with two values 1and β >
1. The results for a fixed structure allow us to infer the properties of the Finslermetrics obtained, with an homogenization procedure, as limit of oscillating chessboardstructures. Introduction
In this paper we deal with optical paths in a dioptric material with parallel geometryand a chessboard structure on transversal planes. Further to a bidimensional reduction,we fix the optical features of the composite material in terms of its refractive index: given β >
1, let us define on [0 , × [0 ,
2) the function(1) a β ( x, y ) = ( β, if ( x, y ) ∈ [(0 , × (1 , ∪ [(1 , × (0 , , , otherwise , and extend it by periodicity to a function defined on R which we still denote by a β .Hence, normalizing to 1 the speed of light in the vacuum, light travels in the dioptricmaterial with a speed 1 /a β .Since we are dealing with a system employing only refraction, Fermat’s principle dictatesthat the optical paths between two points minimize the optical path length, which coincideswith the time spent. Thus, in an homogeneous material, where the speed of light isconstant, the optical path is a segment. Moreover, Fermat’s principle leads to Snell’s lawof refraction, which completely describes the optical paths in layered materials (see [ ? ] fora comprehensive introduction on the principles of optics). The explicit description of theoptical paths in the chessboard structure becomes harder since, for example, no necessarycondition prescribes their behaviour at corners.From a geometrical viewpoint, we are interested in the description of the geodesics in theRiemannian structure ( R , a β ), that hereafter will be called standard chessboard structure .We shall refer to light squares or to dark squares , when, respectively, a β = 1 or a β = β .In the mathematical model, a virtual path emanated from the origin is a solution u : [0 , T ] → R to the differential inclusion(2) ( u ′ ( t ) ∈ G β ( u ( t )) , ≤ t ≤ T,u (0) = 0 , where the set–valued map (with nonconvex values) G β is defined by G β ( x, y ) = 1 a β ( x, y ) ∂B (0) , ( x, y ) ∈ R Date : April 28, 2008.
Key words and phrases.
Minimum time problems, Fermat’s Principle, Finsler metrics. (see e.g. [ ? ] for an introduction to differential inclusions).Fermat’s Principle states that a ray of light from O to ξ is a solution of the minimumtime problem with target ξ (3) T β ( ξ ) = inf { T ≥ ∃ u ( · ) solution of (2), with u ( T ) = ξ } . In [ ? ] it is proved that the infimum in (3) is reached. Moreover, if u ( · ) is a solution of (2)satisfying u ( T ) = ξ , then the optical length of the curve Γ = { u ( t ); 0 ≤ t ≤ T } is L β (Γ) = Z T a β ( u ( t )) | u ′ ( t ) | dt = T. Then optical paths are geodesic curves in the Riemannian structure ( R , a β ). We underlineagain that the global minimization procedure considers only refracted rays, excluding allreflected rays, because a reflected ray is never a global geodesics. Hence our results havea reasonable physical meaning either if the number of interfaces traversed in the periodicmedia is not very high, or for β near to 1, since in both cases the reflected light can beneglected. Anyhow the results are intended as a depiction of curves of minimal length inthe Riemannian structure ( R , a β ).A starting point is the elementary observation that, for β >
2, any geodesic joining twopoints in the light material (i.e. in the set { a β = 1 } ) is a “light path”, i.e. it never crossesthe dark material. With little more work it is not difficult to prove the same conclusion if β > √
2, provided that the endpoints of the geodesic have integer coordinates. Moreover,the value β = √ β ≤ √ n + j, j ), n , j ∈ Z , that we call light vertices . For β ≥ p / β c ∈ ( p / , √ β such that the optical path joining O with the light vertex (3 ,
1) has the same length (2 + √
2) of the optimal light paths.More precisely, we show that for β > β c the geodesics are optimal light paths, whereas for p / < β < β c the minimal curves are constructed concatenating the maximal numberof translations of the optical path joining O with (3 , < β < p / { ≤ K y ≤ x } , where K = K ( β )is an odd positive integer, whose value diverges to + ∞ as β approaches 1.The knowledge of the minimal length of curves joining the origin to light vertices in thechessboard structure is enough to characterize the so-called homogenized model. Namely,the results described above give information about the optical paths in an inhomoge-neous dioptric material whose observed refractive index, at a mesoscopic level, is givenby the chessboard structure, that is the observed refractive index at a given scale ε > a εβ ( x, y ) = a β ( x/ε, y/ε ). We are interested in the behavior of the optical length ofgeodesics when ε → IMITS OF CHESSBOARD STRUCTURES 3
We already know that, at the scale ε , a virtual path emanated from the origin is asolution u : [0 , T ] → R of the differential inclusion(4) u ′ ( t ) ∈ a εβ ( u ( t )) ∂B (0) , ≤ t ≤ T,u (0) = 0 , and Fermat’s Principle states that a ray of light from O to ξ is a solution of the minimumtime problem with target ξ (5) T εβ ( ξ ) = inf { T ≥ ∃ u ( · ) solution of (4), with u ( T ) = ξ } . We are interested in the characterization of the limit, as ε →
0, of the minimum timeproblems (4)–(5).The minimum time problems can be rephrased in terms of minimum problems of theCalculus of Variations. Let us denote by L εβ the length functional in the chessboardstructure corresponding to a εβ , that is L εβ ( u ) = Z a εβ ( u ( t )) | u ′ ( t ) | dt , u ∈ AC ([0 , , R ) , and by d εβ (0 , ξ ) the distance between O and ξ in such a structure, that is(6) d εβ (0 , ξ ) = inf n L εβ ( u ) : u ∈ AC ([0 , R ) s.t. u (0) = 0 , u (1) = ξ o . If u ( · ) is a solution of (4) satisfying u ( T ) = ξ , then T equals the optical length of thecurve Γ = { u ( t ); 0 ≤ t ≤ T } so that(7) T εβ ( ξ ) = d εβ (0 , ξ ) , ξ ∈ R . The advantage of this formulation is that the asymptotic behavior of d εβ can be discussedin terms of Γ–convergence of the functionals F εβ (see e.g. [ ? ] for an introduction to Γ–convergence). In [ ? ] it was shown that the sequence ( L εβ ) Γ–converges in AC ([0 , , R )(w.r.t. the L topology) to the functional L homβ ( u ) = Z Φ β ( u ′ ( t )) dt , u ∈ AC ([0 , , R ) , where Φ β : R → [0 , + ∞ ) is a convex, positively 1-homogeneous function, such that | ξ | ≤ Φ β ( ξ ) ≤ β | ξ | for every ξ ∈ R . As a consequence, Φ β turns out to be a homogeneousFinsler metric in R (see e.g. [ ? ] for an introduction to Finsler geometry).In [ ? ] it is shown that Φ β is not a Riemannian metric in R for every β >
1. In thispaper we shall refine this result proving that the optical unit ball { Φ β ≤ } is neitherstrictly convex nor differentiable (see Theorem 6.4 and Corollary 6.5 below).Since Φ β is characterized by Φ β ( ξ ) = lim ε → + d εβ (0 , ξ ) , then the limit of the minimum time problems (4)–(5) is given by T β ( ξ ) = inf { T ≥ ∃ u ( · ) such that Φ β ( u ′ ( t )) = 1 , u (0) = 0 , u ( T ) = ξ } = Φ β ( ξ ) . This is what we have called homogenized model related to the chessboard structure.Since Φ β is positively 1-homogeneous, it is completely determined by the geometry of theoptical unit sphere { Φ β = 1 } , which is, in some sense, a generalized geometric wavefrontwith source point located at O . Moreover, due to elementary symmetry properties, it isenough to describe the set { Φ β = 1 } ∩ { ≤ y ≤ x } . M. AMAR, G. CRASTA, AND A. MALUSA
Starting from our results on chessboard structures, we obtain that, if β ≥ β c , then thehomogenized metric is(8) Φ β ( x, y ) = ( √ −
1) min {| x | , | y |} + max {| x | , | y |} , ∀ ( x, y ) ∈ R , and the geometric wavefront { Φ β = 1 } is the regular octagon inscribed in the unit circle,whose vertices lie on the coordinate axes and on the diagonals. This result generalizesthe trivial remark concerning the case β ≥ ? ]). On the other, we obtain thatthe octagonal geometry of the wavefront breaks for β = β c , and the optical unit spherebecomes an irregular polygon with sixteen sides for p / ≤ β < β c (see Figure 11). Theseare two of the main results of our paper; we refer to Theorem 6.3 below for their precisestatement.For 1 < β < p / { K | y | ≤ | x |} and { K | x | ≤ | y |} , where K = K ( β ) is the odd positive integer introducedabove for the chessboard structure. More precisely, we shall show that in these regionsthe boundary { Φ β = 1 } is piecewise flat with corners at the points (1 , , − , , −
1) (see Theorem 6.4). As a consequence, the optical unit ball { Φ β ≤ } is neitherstrictly convex nor differentiable.We conclude this introduction with a warning on the physical interpretation of ourresults concerning the homogenized model. First of all, the geometrical optics approxi-mation is valid if the lengthscale ε is much greater than the wavelength of light. If ε isof the same order of magnitude of the wavelength, then we fall in the domain of photoniccrystals optics, for which the full system of Maxwell equations must be considered. Onthe other hand, even in the range of geometric optics, our global minimization procedureconsiders only refracted rays, excluding all reflected rays. Since, at a macroscopic level,the number of interfaces traversed in the periodic media can be very high, the reflectedlight cannot in general be neglected.In the paper the following notation will be used. • [[ P, Q ]]: closed segment joining
P, Q ∈ R • ]] P, Q ]] = [[
P, Q ]] \ { P } , [[ P, Q [[ = [[
P, Q ]] \ { Q }• [[ P , P , . . . , P n ]]: polygonal line joining the ordered set of points P , P , . . . , P n . • ⌊ t ⌋ = max { k ∈ Z : k ≤ t } . • f s = ∂f∂s : partial derivative of a function f with respect to s • S ( A, B ): Snell path joining A to B (defined in Section 2) • L β (Γ) = L β (Γ): length of Γ in the standard chessboard structure (optical length).2. Snell paths
Let us consider the flat Riemannian structure ( R , a β ), where a β ( x, y ) = ( ⌊ x ⌋ is even ,β if ⌊ x ⌋ is odd , β ∈ R , β > . Note that this structure corresponds to a composite medium whose structure is madeby alternate vertical strips of light and dark material.It is well known that for every pair A = ( x A , y A ), B = ( x B , y B ) there exists a uniquecurve of minimal length (the geodesic curve) joining A to B , which is an affine path in IMITS OF CHESSBOARD STRUCTURES 5 darklight p qB A BA θ θ h Figure 1.
A Snell path in the layered material and the equivalent Snellpath with a single interface.every vertical strip {⌊ x ⌋ = k } , k ∈ Z . Moreover, at every interface between two strips, thechange of slope is governed by the Snell’s Law of refraction(9) sin θ = β sin θ , where θ and θ are the angles of incidence with the interface from the light strip and thedark strip, respectively (see, e.g., [ ? , § ? , § Snell path joining A and B and willbe denoted by S ( A, B ). In order to fix the ideas, we shall always assume that x A < x B ,and y A ≤ y B . We shall refer to the positive quantity x B − x A as the thickness of the Snellpath.Let p , q ≥ S ( A, B ), so that x B − x A = p + q , and let h be the vertical height h = y B − y A .Since h = p tan θ + q tan θ (see Figure 1), and (9) holds true, then ˆ σ ( p, q, h ) = sin θ isimplicitly determined in term of p , q and h by the constraint(10) p ˆ σ √ − ˆ σ + q ˆ σ p β − ˆ σ = h . Clearly, ˆ σ ( p, q, h ) is a strictly decreasing function w.r.t. p and q , while it is a strictlyincreasing function w.r.t. h .Finally, the optical length of the Snell path S ( A, B ), given by p/ cos θ + βq/ cos θ , canbe expressed in terms of p , q and h taking again into account (9):(11) L ( p, q, h ) := p q − ˆ σ ( p, q, h ) + β q q β − ˆ σ ( p, q, h ) . Lemma 2.1.
Let ˆ σ and L be defined by (10) and (11) respectively. Then L p ( p, q, h ) = p − ˆ σ , L q ( p, q, h ) = q β − ˆ σ , for every p , q ≥ , and for every h ∈ R .Proof. Differentiating (10) w.r.t. p , we getˆ σ √ − ˆ σ + p ˆ σ p (1 − ˆ σ ) / + β q ˆ σ p ( β − ˆ σ ) / = 0 . M. AMAR, G. CRASTA, AND A. MALUSA q(t) t p(t) A B
Figure 2.
The solid line corresponds to the Snell path S ( A, B ), while thedashed line is a minimal path joining A and B without crossing the darksquares.Hence L p ( p, q, h ) = 1 √ − ˆ σ + p ˆ σ p ˆ σ (1 − ˆ σ ) / + β q ˆ σ p ˆ σ ( β − ˆ σ ) / = 1 √ − ˆ σ − ˆ σ √ − ˆ σ = p − ˆ σ . By an analogous computation one obtains the expression for L q . (cid:3) Remark . By Lemma 2.1, it follows that, given the thickness τ = p + q and the height h ∈ R of a Snell path we have that ddq L ( τ − q, q, h ) = − p − ˆ σ + q β − ˆ σ > . The geometrical meaning of this formula is clear: for Snell paths with fixed thickness, themore is the thickness of the dark material crossed, the more is the optical length of thepath. 3.
The normalized length
Let us consider now R endowed with the standard chessboard structure. Definition 3.1.
The n -th light diagonal , n ∈ Z , is the straight line D n of equation y = x − n . A light vertex is a point having integer coordinates and belonging to a lightdiagonal. Definition 3.2.
Given two points A and B in the same horizontal strip { y ∈ [ r, r + 1] } , r ∈ Z , the Snell path joining A to B (in the chessboard structure) is the geodesic S ( A, B )in the corresponding parallel layer structure a ( x + r, y ).We are interested in the properties of the Snell paths starting from a light vertex A (say A = O , without loss of generality) and ending in a point B on the other side of thehorizontal strip containing A (say B = ( x B , x B >
0) (see Figure 2).If 0 < x B ≤
1, clearly S ( O, B ) = [[
O, B ]] is the unique geodesic joining O to B . Onthe other hand, S ( O, B ) need not to be a geodesic when x B >
1. Namely, we alreadyknow that for β large enough the optical length of S ( O, B ) is strictly greater than theoptical length of the path obtained by a concatenation of horizontal segments on the lines x = 0 or x = 1, with total length x B −
1, and a segment on a light diagonal, with length √
2. In this section we discuss the behavior of the difference L β ( S ( O, B )) − x B + 1 − √ IMITS OF CHESSBOARD STRUCTURES 7 t Σ H t, Β L Figure 3.
Plot of ˆ σ ( t, β ), β = 1 . x B ≥
1. To this aim, for t ≥ β ≥
1, with some abuse of notation we defineˆ σ ( t, β ) = ˆ σ ( p ( t ) , q ( t ) , q ( t ) = ( ⌊ t +1 ⌋ if ⌊ t ⌋ is odd ,t − ⌊ t ⌋ if ⌊ t ⌋ is even , and p ( t ) = t + 1 − q ( t ). Recall that ˆ σ ( t, β ) is determined by the constraint(13) p ( t ) ˆ σ √ − ˆ σ + q ( t ) ˆ σ p β − ˆ σ − . As a consequence we have(14) ˆ σ t ( t, β ) = − ˆ σp t √ − ˆ σ − ˆ σq t p β − ˆ σ p (1 − ˆ σ ) / + β q ( β − ˆ σ ) / < , t > , t N , since p t ( t ), q t ( t ) ≥
0, and p t ( t )+ q t ( t ) = 1 for every t > t N . Thus the map t ˆ σ ( t, β )is strictly decreasing in [0 , + ∞ ), and satisfies ˆ σ (0 , β ) = 1 / √
2, lim t → + ∞ ˆ σ ( t, β ) = 0 (seeFigure 3).Moreover, as a straightforward consequence of the Implicit Function Theorem, we havethat ˆ σ β ( t, β ) > t >
0. In particular we have(15) ˆ σ ( t, β ) > ˆ σ ( t,
1) = 1 p (1 + t ) + 1 , ∀ t > , ∀ β > . Definition 3.3.
We shall call normalized length the function(16) l ( t, β ) := p ( t ) q − ˆ σ ( t, β ) + β q ( t ) q β − ˆ σ ( t, β ) − t − √ . Notice that l ( t, β ) is the length of the Snell path joining the origin (0 ,
0) with the point( t + 1 ,
1) normalized by subtracting the minimal length of the paths joining the same twopoints without crossing the dark squares (see Figure 2).In order to simplify the notation we introduce the sets I L = [ k ∈ N (2 k + 1 , k + 2) , I D = [ k ∈ N (2 k, k + 1) . If t ∈ I L then the last segment of the Snell path is in the interior of a light square, while,if t ∈ I D , it is in the interior of a dark square. M. AMAR, G. CRASTA, AND A. MALUSA
The basic properties of the normalized length are collected in the following lemma.
Lemma 3.4.
The following properties hold. (i) l t ( t, β ) = p − ˆ σ ( t, β ) − for every t ∈ I L ; (ii) l t ( t, β ) = p β − ˆ σ ( t, β ) − for every t ∈ I D ; (iii) l ( · , β ) is strictly convex in any interval of I L ∪ I D ; (iv) l ( · , β ) is strictly monotone decreasing in any interval of I L ; (v) if β ≥ p / , then l ( · , β ) is strictly monotone increasing in any interval of I D ; (vi) if < β < p / , then there exists a unique t > , characterized by ˆ σ ( t , β ) = p β − , and such that l t ( t, β ) < , ∀ t ∈ [0 , t ) ∩ I D ,l t ( t, β ) > , ∀ t ∈ ( t , + ∞ ) ∩ I D . Proof.
The derivatives in (i) and (ii) follow from Lemma 2.1, upon observing that p t = 1and q t = 0 in I L , while p t = 0 and q t = 1 in I D . Clearly (i) implies (iv), while (ii) and thefact that 0 ≤ ˆ σ ( t, β ) ≤ / σ ( · , β ) is a decreasing function. (cid:3) In conclusion, since l (0 , β ) = 0 for every β ≥
1, by Lemma 3.4 we have that, for β ≥ p / l ( t, β ) > t ∈ (0 ,
1) and the local minima of l ( · , β ) are attained at t = 2 k , k ∈ N , corresponding to the Snell paths ending in the light vertices (see Figures 5 and 4).On the contrary, if β < p /
2, a new local minimum for l ( · , β ) may appear (see Figures 6and 7). One may wonder if l ( t , β ) is an absolute minimum for some β . The followingresult shows that this is never the case. (We warn the reader that the proof is rather longand technical, and can be skipped in a first reading.) Theorem 3.5.
Given < β < p / , let t > be as in Lemma 3.4(vi). Then l (2 k + 2 , β ) ≤ l ( t , β ) , where k = min { k ∈ N : t ≤ k + 2 } . Moreover, the strict inequality holds if t = 2 k + 2 .Proof. If t ∈ [2 k + 1 , k + 2], then by Lemma 3.4(iv),(vi), l (2 k + 2 , β ) < l ( t, β ) for every t ∈ [0 , k + 2), and the result is straightforward.Let us now consider the case t ∈ (0 , k = 0.Recalling that ˆ σ ( t , β ) = p β −
1, we obtain that l ( t , β ) = p − β + p β − − √ l (2 , β ) = 2 q − σ + β q β − σ − − √ − σ σ q − σ + σ q β − σ − = 2 q − σ + q β − σ − − √ − σ , where σ = ˆ σ (2 , β ), and we have used the constraint (13) satisfied by σ . Hence, denotingby(17) ϕ ( σ, β ) = 2 p − σ + q β − σ − − σ − q − β − q β − , we have to show that l (2 , β ) − l ( t , β ) = ϕ ( σ , β ) < ϕ ( σ, β ) is strictly monotone decreasing w.r.t. σ in [0 , ϕ ( σ , β ) < ϕ (1 / √ , β ) for every β ≥
1. In addition, ϕ (1 / √ , β ) IMITS OF CHESSBOARD STRUCTURES 9 is a strictly monotone increasing function w.r.t. β , so that we get ϕ ( σ , β ) < ϕ (1 / √ , β ) < ϕ (1 / √ , q /
2) = 5 √ − r − √ < , for every β ∈ (1 , p / t ∈ (0 , t ∈ (2 k , k + 1) with k ≥ l (2 k + 2 , β ) = l ( t , β ) + Z k +1 t l t ( t, β ) dt + Z k +22 k +1 l t ( t, β ) dt = Z k +1 t (cid:18)q β − ˆ σ ( t, β ) − (cid:19) dt + Z k +22 k +1 (cid:18)q − ˆ σ ( t, β ) − (cid:19) dt , our aim is to prove that(18) Z k +1 t q β − ˆ σ ( t, β ) dt + Z k +22 k +1 q − ˆ σ ( t, β ) dt < k + 2 − t . We split the proof of (18) into three steps.
Step 1.
Setting f ( k, t, γ ) = 2 k + 1 + ( β − t + β − β q (2 k + 2) + ( t + 1) ( β − − ( k + 1) log q γ − e − / ( k +1) √ γ + 1 (19)and c = 1 / (1 − ˆ σ (2 k + 1 , β ) ), we show that if f ( k , t , c ) ≥ Step 2.
Setting(20) g ( b ) = 1 + 3 b − b √ c + (1 − e − ) √ c − √ c , c = c ( b ) = 1 + 916 ( b − , then f ( k , t , c ) ≥ g ( β ). Step 3. g ( b ) > b ∈ (1 , p / Proof of Step 1.
Let us consider the functions ψ ( t ) = q β − ˆ σ ( t, β ) , χ ( t ) = q − ˆ σ ( t, β ) . Recalling (14), and taking into account that p t ( t ) = 0 and q t ( t ) = 1 for t ∈ [ t , k + 1),we obtain ψ ′ = − ˆ σ t ˆ σ p β − ˆ σ = ˆ σ β − ˆ σ · p (1 − ˆ σ ) / + β q ( β − ˆ σ ) / < ˆ σ p β − ˆ σ β ( t + 1) , where in the last inequality we have used the fact that the function b b / ( b − ˆ σ ) / isstrictly monotone decreasing, and p + q = t + 1. In conclusion we obtain that ψ satisfiesthe differential inequality(21) ψ ′ < t + 1 ψ − β ( t + 1) ψ , t ∈ [ t , k + 1) ,ψ ( t ) = 1 . Similarly, recalling that p t ( t ) = 1 and q t ( t ) = 0 for t ∈ (2 k + 1 , k + 2), we obtain χ ′ = − ˆ σ t ˆ σ √ − ˆ σ = ˆ σ − ˆ σ · p (1 − ˆ σ ) / + β q ( β − ˆ σ ) / < ˆ σ p p − ˆ σ , and, since p ≥ k + 1, we conclude that χ satisfies the differential inequality(22) χ ′ < χ − χ k + 1 , t ∈ (2 k + 1 , k + 2) ,χ (2 k + 1) = p − β + ψ (2 k + 1) . Solving the Cauchy problems associated to the differential inequalities (21), (22), we get ψ ( t ) ≤ s β + (cid:18) − β (cid:19) ( t + 1) ( t + 1) , t ∈ [ t , k + 1] , (23) χ ( t ) ≤ s (cid:18) − β + ψ − (cid:19) e − t − k − / ( k +1) , t ∈ [2 k + 1 , k + 2] . (24)As a consequence of these estimates we obtain Z k +1 t q β − σ ( t, β ) dt ≤ β (cid:18)q (2 k + 2) + ( t + 1) ( β − − β ( t + 1) (cid:19)Z k +22 k +1 q − σ ( t, β ) dt ≤ k + 1) log q c − e − / ( k +1) √ c + 1 , where c = 1 / (1 − β + ψ (2 k + 1) ), which concludes the proof of the Step 1. Proof of Step 2.
From (23) and the fact that t > k , we have that ψ (2 k + 1) ≤ β (2 k + 2) p (2 k + 2) + ( β − t + 1) < β (2 k + 2) p (2 k + 2) + ( β − k + 1) , so that c = 11 − β + ψ (2 k + 1) ≥ β − (cid:18) k + 12 k + 2 (cid:19) − ( β − (cid:18) k + 12 k + 2 (cid:19) > β − (cid:18) k + 12 k + 2 (cid:19) ≥ β −
1) = c ( β ) . Moreover, it can be easily checked that the function γ log q γ − e − / ( k +1) √ γ + 1 , γ > t ( β − t − β q (2 k + 2) + ( t + 1) ( β − , t ∈ (2 k , k + 1) IMITS OF CHESSBOARD STRUCTURES 11 is strictly monotone increasing. Hence we have that f ( k , t , c ) > f ( k , k , c ( β )) = 2 k β − β q (2 k + 2) + (2 k + 1) ( β − β − ( k + 1) log q c − e − / ( k +1) √ c + 1 , where c = c ( β ) is defined as in (20).In addition, the functions k kβ − β q (2 k + 2) + (2 k + 1) ( β − ,k ( k + 1) log q c − e − / ( k +1) √ c + 1 , are strictly monotone increasing for k ≥
1. Hence we get f ( k , t , c ) > β − β q
16 + 9( β − − p c − e − √ c + 1 ! . Finally log p c − e − √ c + 1 ! ≤ p c − e − − √ c √ c + 1= √ c q − e − ) − c c − √ c + 1 ≤
12 (1 − e − ) 1 − √ c √ c , so that the proof of Step 2 is complete. Proof of Step 3.
We have that g ′ ( b ) = 6 b − b √ b + 36(1 − e − ) b (7 + 9 b ) / , and, for 1 < b ≤ p / ddb b − b √ b ! = 6 − b b (7 + 9 b ) / = 6 − b √ b b b ≥ − √ √
41 1316 > , and ddb (cid:18) b (7 + 9 b ) / (cid:19) = −
16 18 b − b ) / < , b > . Then we get g ′ ( b ) > −
254 + 72 √ / (1 − e − ) > , ∀ < b ≤ q / . Hence g ( b ) > g (1) = 0 for all 1 < b ≤ p /
2, and Step 3 is proved. (cid:3)
Now we focus our attention to the study of the sequence l (2 k, β ), k ∈ N . Given k ∈ N ,we set δ ( k, β ) = l (2 k + 2 , β ) − l (2 k, β ), that is(25) δ ( k, β ) = k + 2 √ − τ + β ( k + 1) p β − τ − k + 1 √ − σ − β k p β − σ − , where τ = τ ( k, β ) := ˆ σ (2 k + 2 , β ) and σ = σ ( k, β ) := ˆ σ (2 k, β ) are implicitly defined by( k + 2) τ √ − τ + ( k + 1) τ p β − τ = 1 , (26) ( k + 1) σ √ − σ + k σ p β − σ = 1 . (27)By the monotonicity of the function ˆ σ ( · , β ) (see inequality (14)) it follows that τ < σ . Remark . While the sign of l ( t, β ) gives a comparison between the optical lengths ofthe Snell path S ( O, ( t + 1 , O, (1 , , (1 + t, δ ( k, β )gives a comparison between the optical lengths of S ( O, (2 k +3 , S ( O, (2 k +1 , ∪ [[(2 k + 1 , , (2 k + 3 , Remark . Given β >
1, consider the function ˜ l : [0 , + ∞ ) → R , affine on each interval[2 k, k + 2] and such that ˜ l (2 k ) = l (2 k, β ), k ∈ N . Then the derivative of ˜ l ( t ), for t ∈ (2 k, k + 2), is given by δ ( k, β ) / δ ( k, β ) = L β ( S ( O, (2 k + 3 , − L β ( S ( O, (2 k + 1 , −
2, and it is clear that L β ( S ( O, (2 k + 3 , − L β ( S ( O, (2 k + 1 , ∼ β + 1 for k → + ∞ , one expects that δ ( k, β ) ∼ β −
1. A more precise result is the following.
Theorem 3.8.
Let β > be fixed. Then ( δ ( k, β )) k is a strictly monotone increasingsequence and (28) δ ( k, β ) = ( β − − β β + 1) 1 k + O (cid:18) k (cid:19) , k → + ∞ . Proof.
We can define τ , σ , and δ respectively through (26), (27) and (25) as smoothfunctions of k ∈ R , k ≥
0. Differentiating δ w.r.t. k , we get δ k ( k, β ) = 1 √ − τ + ( k + 2) τ k τ (1 − τ ) / + β p β − τ + ( k + 1) β τ k τ ( β − τ ) / − √ − σ − ( k + 1) σ k σ (1 − σ ) / − β p β − σ − kβ σ k σ ( β − σ ) / . On the other hand, differentiating the constraints (26) and (27) we obtain the identities( k + 2) τ k (1 − τ ) / + ( k + 1) β τ k ( β − τ ) / = − τ √ − τ − τ p β − τ , (29) ( k + 1) σ k (1 − σ ) / + kβ σ k ( β − σ ) / = − σ √ − σ − σ p β − σ . (30)Hence, being τ < σ , we get δ k ( k, β ) = p − τ − p − σ + q β − τ − q β − σ > . In order to determine the behavior of δ ( k, β ) for k large, notice that, setting ε = 1 /k and e σ ( ε ) = σ (1 /ε, β ), (27) becomes(1 + ε ) e σ √ − e σ + e σ p β − e σ = ε , IMITS OF CHESSBOARD STRUCTURES 13 that is e σ is implicitly defined by f ( e σ ) − ε = 0 , f ( t ) = t √ − t + t p β − t − t √ − t . One has f (0) = 0, f ′ (0) = β + 1 β , f ′′ (0) = 2 β + 1 β , so that e σ (0) = 0 , e σ ′ (0) = ββ + 1 , e σ ′′ (0) = − β ( β + 1) , and hence(31) e σ ( ε ) = ββ + 1 ε − β ( β + 1) ε + O ( ε ) ε → + , that is(32) σ ( k, β ) = ββ + 1 1 k − β ( β + 1) k + O (cid:18) k (cid:19) , and(33) τ ( k, β ) = σ ( k + 1 , β ) = e σ (cid:18) k + 1 (cid:19) = e σ (cid:18) k − k + O (cid:16) k (cid:17)(cid:19) = ββ + 1 1 k − ββ + 1 + β ( β + 1) ! k + O (cid:18) k (cid:19) . Finally we have δ ( k, β ) = ( k + 2) (cid:18) τ (cid:19) + β ( k + 1) τ β ! − ( k + 1) (cid:18) σ (cid:19) − βk σ β ! − O (cid:18) k (cid:19) = ( β − − β β + 1) 1 k + O (cid:18) k (cid:19) , completing the proof. (cid:3) Definition 3.9.
Given β >
1, we shall denote by k c ( β ) the integer number defined by(34) k c ( β ) = min { k ∈ N : δ ( k, β ) > } . By the very definition, we have that l (2 k c ( β ) , β ) ≤ l (2 k, β ) for every k ∈ N (see alsoRemark 3.7). Moreover, by Lemma 3.4 and Theorem 3.5, the absolute minimum of l ( · , β )is attained at a point t = 2 k , k ∈ N . Hence l (2 k c ( β ) , β ) (i.e., the normalized length of theSnell path joining the origin with the right-top vertex of the (2 k c +1)–th square) minimizesthe normalized length l ( · , β ) among all the paths remaining in a single horizontal strip.Now we want to study k c ( β ), β >
1. As a preliminary step we investigate the behaviorof δ ( k, · ) for a given k . Lemma 3.10.
Let k ∈ N be fixed. Then the function β δ ( k, β ) is strictly increasing in [1 , + ∞ ) . Moreover δ ( k, < and δ ( k, √ > . Proof.
Differentiating δ w.r.t. β , we get δ β ( k, β ) = ( k + 2) τ τ β (1 − τ ) / + ( k + 1) β β − τ ( β − τ ) / + ( k + 1) β τ τ β ( β − τ ) / − ( k + 1) σσ β (1 − σ ) / − kβ β − σ ( β − σ ) / − k β σσ β ( β − σ ) / . (35)Differentiating (26) and (27) w.r.t. β , we obtain( k + 2) τ τ β (1 − τ ) / + ( k + 1) β τ τ β ( β − τ ) / = ( k + 1) τ β ( β − τ ) / (36) ( k + 1) σσ β (1 − σ ) / + k β σσ β ( β − σ ) / = k τ β ( β − σ ) / . (37)Substituting (36) and (37) into (35), we conclude that δ β ( k, β ) = ( k + 1) β p β − τ − kβ p β − σ . We have to show that(38) ( k + 1) β p β − τ − kβ p β − σ > , ∀ k ∈ N . For every κ ∈ R , let us denote by s ( κ ) the unique function implicitly defined by(39) ( κ + 1) s √ − s + κ s p β − s = 1so that s ( k ) = σ , s ( k + 1) = τ . Since inequality (38) clearly holds true for k = 0, it isenough to show that ddκ κ q β − s ( κ ) = β − s + κss κ ( β − s ) / > ∀ κ ∈ R , κ ≥ . Differentiating (39) w.r.t. κ (see also (29)), we get s κ ( κ ) = − s √ − s + 1 p β − s κ + 1(1 − s ) / + κβ ( β − s ) / . Moreover, using again (39), we have1 √ − s + 1 p β − s = 1 κs − κ √ − s < κs , so that β − s + κss κ = β − s − κs √ − s + 1 p β − s κ + 1(1 − s ) / + κβ ( β − s ) / > β − s κβ (cid:18) κβ − s q β − s (cid:19) > , IMITS OF CHESSBOARD STRUCTURES 15 where the last inequality can be easily checked recalling that 0 < s <
1, while β , κ ≥ δ ( k, β )is strictly increasing w.r.t. β .Taking into account that, by (27), σ (0 , √
2) = √ /
2, we easily get that δ (0 , √ >
0, sothat, by Theorem 3.8, δ ( k, √ > δ ( k, <
0, we note that, from (26) and (27), we get τ ( k,
1) = 1 p k + 3) , σ ( k,
1) = 1 p k + 1) , so that δ ( k,
1) = q k + 3) − q k + 1) − < , concluding the proof. (cid:3) Remark . As an easy consequence of Lemma 3.10 we obtain that k c ( β ) is a nonin-creasing function of β .Thanks to Lemma 3.10, the following definition makes sense. Definition 3.12.
For every k ∈ N we shall denote by β ck the unique number in (1 , √ δ ( k, β ck ) = 0.By Theorem 3.8 we have that δ ( j, β ck ) < , ∀ j < k, δ ( j, β ck ) > , ∀ j > k, hence(40) l (2 k, β ck ) = l (2 k + 2 , β ck ) < l (2 j, β ck ) ∀ j ∈ N \ { k, k + 1 } . In particular we have that(41) l (2 k + 2 , β ck ) = l (2 k, β ck ) ≤ l (0 , β ck ) = 0 , where we have taken into account that l (0 , β ) = 0, for every β ≥ Lemma 3.13.
The sequence ( β ck ) k ∈ N is strictly decreasing and lim k → + ∞ β ck = 1 .Proof. Given k ∈ N , by the monotonicity of δ w.r.t. k stated in Theorem 3.8 we have δ ( k, β ck +1 ) < δ ( k + 1 , β ck +1 ) = 0 , where the last equality follows from the very definition of β ck +1 . Again, the definition of β ck and the monotonicity of δ w.r.t. β stated in Lemma 3.10 imply that β ck +1 < β ck .In order to prove the last part of the thesis, given k ∈ N , let us define the functions f k ( s, β ) := ( k + 1) √ − s + β k p β − s − k − √ ,g k ( s, β ) := ( k + 1) s √ − s + k s p β − s − ,h k ( s, β ) := f k ( s, β ) − s g k ( s, β ) = ( k + 1) p − s + k q β − s − k − √ s , (42)where s ∈ [0 ,
1) and β >
1. Since ˆ σ (2 k, β ) is the unique solution of g k ( s, β ) = 0, wehave that h k (ˆ σ (2 k, β ) , β ) = f k (ˆ σ (2 k, β ) , β ) = l (2 k, β ). Moreover h ks ( s, β ) = − g k ( s, β ) and g ks ( s, β ) >
0, so that h k ( · , β ) is a strictly concave function in [0,1] which attains its absolutemaximum at s = ˆ σ (2 k, β ). Hence l (2 k, β ) = h k (ˆ σ (2 k, β ) , β ) = max s ∈ [0 , h k ( s, β ) > h k (0 , β ) = ( β − k + 1 − √ , ∀ β > . Then, from (41), we have0 ≥ l (2 k + 2 , β ck ) > ( β ck − k + 1) + 1 − √ < β ck < √ − k + 1and the conclusion follows. (cid:3) The first values ( β ck ) (up to the fifth digit) are listed in the following table. k β ck β ≥ p / S ( O, ( t, t ∈ (0 ,
1) is never ageodesic. This property will be crucial for the results in Section 5. The following resultgives the position of p / ≃ . β ck . Lemma 3.14. β c < p / < β c .Proof. From (43) we have β c < √ − < r , β c < √ . In order to prove the inequality p / < β c , let f , g , h be the functions defined in (42)for k = 1, and let σ ∈ (0 ,
1) be the unique solution of g ( s, β c ) = 0. Since0 = δ (0 , β c ) = l (2 , β c ) = f ( σ , β c ) , we have that h ( σ , β c ) = 0. Moreover h β ( s, β ) = β p β − s > . Hence the inequality p / < β c can be obtained showing that(44) h s, r ! = 2 p − s + r − s − − √ s < , ∀ s ∈ (0 , . The inequality (44) easily follows observing that2 p − s + 23 s ≤ √ , r − s + 13 s ≤ r , ∀ s ∈ (0 , , so that h s, r ! ≤ √
10 + r − − √ < , ∀ s ∈ (0 , , which completes the proof. (cid:3) We summarize the previous analysis in the following result, which is depicted in Fig-ures 4–7.
Corollary 3.15.
For every β > s ≥ l ( s, β ) = min k ∈ N l (2 k, β ) < l ( t, β ) ∀ t ∈ [ k ∈ N (2 k, k + 2) . Moreover the following hold.
IMITS OF CHESSBOARD STRUCTURES 17 t H t, Β L t H t, Β L Figure 4.
Plot of l ( t, β ), β = 1 .
26 ( β > β c ) t H t, Β L t H t, Β L Figure 5.
Plot of l ( t, β ), β = 1 .
23 ( p / < β < β c ) t H t, Β L t H t, Β L Figure 6.
Plot of l ( t, β ), β = 1 . β c < β < p / k c ( β ) = 0 for every β > β c , and k c ( β ) = k + 1 for every β ∈ ( β ck +1 , β ck ] , k ∈ N ; (ii) if β > β c , then l (0 , β ) < l ( t, β ) for every t > ; (iii) if β ∈ ( β ck +1 , β ck ) , k ∈ N , then l (2 k + 2 , β ) < l ( t, β ) for all t ∈ [0 , + ∞ ) \ { k + 2 } ; (iv) l (2 k, β ck ) = l (2 k + 2 , β ck ) < l ( t, β ck ) for every k ∈ N , and t ∈ [0 , + ∞ ) \ { k, k + 2 } ; (v) if β ≥ p / , then l ( t, β ) > for t ∈ (0 , . (vi) for every β > , l (2 k c ( β ) , β ) < l ( t, β ) for every t > k c ( β ) .Proof. Formula (45) summarizes the results in Lemma 3.4 and Theorem 3.5.Let us now prove (i). By Lemma 3.10 and Theorem 3.8 we have δ ( j, β ) ≤ δ ( k, β ) ≤ δ ( k, β ck ) = 0 , ∀ j ∈ N , j ≤ k β ≤ β ck , (46) δ ( j, β ) ≥ δ ( k + 1 , β ) > δ ( k + 1 , β ck +1 ) = 0 , ∀ j ∈ N , j ≥ k + 1 , β > β ck +1 , (47)
20 40 60 80 100 t - - - - - - H t, Β L t - - - - H t, Β L Figure 7.
Plot of l ( t, β ), β = 1 .
009 ( β c < β < β c )and we have the strict inequality in (46) if β = β ck . Hence, recalling the definition of k c ( β )given in (34), we conclude that (i) holds.Moreover, since for every n , m ∈ N , n < m , we have l (2 m, β ) = l (2 n, β ) + m − X j = n δ ( j, β ) , as a consequence of (46) and (47), for every k ∈ N we get0 = l (0 , β ) < l (2 j, β ) , ∀ j ∈ N , j = 0 , β > β c (48) l (2 k + 2 , β ) < l (2 j, β ) , ∀ j ∈ N , j = k + 1 , β ∈ ( β ck +1 , β ck )(49) l (2 k, β ck ) = l (2 k + 2 , β ck ) < l (2 j, β ck ) , ∀ j ∈ N , j = k, k + 1 , . (50)Hence (ii), (iii), and (iv) follow from (45). Finally (v) follows from Lemma 3.4(v), whereas(vi) is a direct consequence of (i)–(iv). (cid:3) Up to now we have described the behavior of the normalized length of Snell pathsstarting from a light vertex and remaining in a horizontal strip. The following result dealswith the normalized length of any Snell path remaining in a horizontal strip.
Lemma 3.16.
Given x ∈ R , τ ≥ , r ∈ Z , let S ( A, B ) be the Snell path joining A = ( x, r ) to B = ( x + τ, r + 1) . Then (51) L β ( S ( A, B )) − τ + 1 − √ ≥ l (2 k c ( β ) , β ) . Moreover i) if β = β ck for every k ∈ N , then the equality in (51) holds if and only if τ =2 k c ( β ) + 1 , and x = 2 n , n ∈ Z ; ii) if β = β ck for some k ∈ N , then the equality in (51) holds if and only if τ ∈{ k c ( β ) − , k c ( β ) + 1 } , and x = 2 n , n ∈ Z .Proof. By Remark 2.2 we have that(52) L β ( S ( A, B )) ≥ L β ( S (0 , C )) , C = ( τ, , since S (0 , C ) crosses a quantity of dark material not greater then the one crossed by anyother Snell path with thickness τ . On the other hand, if τ ∈ [0 , L β ( S ( O, C )) = p τ > τ − √ , so that, by (52), L β ( S ( A, B )) − τ + 1 − √ > ≥ l (2 k c ( β ) + 2 , β ) . IMITS OF CHESSBOARD STRUCTURES 19
Moreover, we stress that the equality in (51) never occurs when τ ∈ [0 , τ ≥
1, then L β ( S (0 , C )) = l ( τ, β ) + τ + √ , so that(53) L β ( S ( A, B )) − τ + 1 − √ ≥ l ( τ, β ) ≥ l (2 k c ( β ) , β ) . It remains to discuss the occurrence of the equality in (53) for τ ≥ β = β ck , k ∈ N , then l ( · , β ) has its strict absolute minimum point at t = 2 k c ( β ), sothat l ( τ − , β ) = l (2 k c ( β ) , β ) if and only if τ = 2 k c ( β ) + 1.Moreover, for τ = 2 k c ( β ) + 1, the equality L β ( S ( A, B )) = L β ( S (0 , C )) holds if andonly if x = 2 n , n ∈ Z . Namely, if p x denotes the thickness of light material crossed by S ( A, B ), then p x ≤ k c ( β ) + 1 = p (2 k c ( β ) + 1) (since the total thickness is 2 k c ( β ) + 1), and p x = k c ( β ) + 1 if and only if A is a light vertex (i.e. x = 2 n , n ∈ Z ).If β = β ck for some k ∈ N , then the conclusion in (ii) follows from (53) and the fact thatthe absolute minimum of l ( · , β ck ) is attained both for t = 2 k c ( β ) −
1, and t = 2 k c ( β )+1. (cid:3) We conclude this section by stating some properties, which will be useful in the secondpart of Section 4, of the following generalization of the normalized length introduced in(16). Let q ( t ), t ≥
0, be the function defined in (12). For 0 < h ≤
1, let p ( t, h ) = t + h − q ( t ).Given β >
1, let ˜ σ ( t, β, h ) be the unique solution of the implicit equation(54) p ( t, h ) ˜ σ √ − ˜ σ + q ( t ) ˜ σ p β − ˜ σ − h = 0 , and let us define the function(55) ˜ l ( t, β, h ) = p ( t, h ) q − ˜ σ ( t, β, h ) + β q ( t ) q β − ˜ σ ( t, β, h ) − t − h √ . The function ˜ l is the normalized length of a Snell path starting from the point ( − h, − h )and ending in ( t, σ ( t, β,
1) and ˜ l ( t, β,
1) coincide with thefunctions ˆ σ ( t, β ) and l ( t, β ) defined in (13) and (16) respectively. The function t ˜ l ( t, β, h )has the same qualitative properties of ˜ l ( · , β,
1) studied at the beginning of this Section.More precisely the derivative(56) ˜ l t ( t, β, h ) = (p − ˜ σ ( t, β, h ) − , if t ∈ I L , p β − ˜ σ ( t, β, h ) − , if t ∈ I D , is a monotone increasing function both in the set I L and in the set I D (see Lemma 2.1).Concerning the derivative ˜ l h w.r.t. h , we have the following result. Lemma 3.17.
The function h ˜ l ( t, β, h ) is strictly convex and monotone decreasing in (0 , for every t > , β > , and (57) ˜ l h ( t, β, h ) = q − ˜ σ ( t, β, h ) + ˜ σ ( t, β, h ) − √ . Proof.
Since p h ( t, h ) = 1 we have that(58) ˜ l h ( t, β, h ) = 1 √ − ˜ σ + p ˜ σ ˜ σ h (1 − ˜ σ ) / + qβ ˜ σ ˜ σ h ( β − ˜ σ ) / − √ . Differentiating (54) w.r.t. h , we get(59) p ˜ σ h (1 − ˜ σ ) / + β q ˜ σ h ( β − ˜ σ ) / = 1 − ˜ σ √ − ˜ σ . A A
Figure 8.
Substituting (59) in (58), we obtain (57). It is straightforward to check that the function˜ σ
7→ √ − ˜ σ + ˜ σ is strictly increasing in (0 , / √ t > σ < / √
2, it follows that ˜ l h ( t, β, h ) <
0, for h ∈ (0 , σ h >
0, so that the function h ˜ σ ( t, β, h ) is strictly increasing in (0 ,
1) for every t > β >
1. Hence, ˜ l h is strictly increasing for h ∈ (0 , l ( t, β, h ) isstrictly convex w.r.t. h . (cid:3) General properties of the geodesics in the chessboard structure
Up to now we have investigated the properties of a geodesic joining two points on thesides of one horizontal strip in the chessboard structure. In this section we will study theproperties of a geodesic starting from the origin O , crossing an arbitrary large number ofhorizontal strips, and ending in a light vertex (2 n + j, j ), n, j ∈ N . If n = 0 or j = 0 , thenthe unique geodesic from the origin to the point (2 n + j, j ) is the segment joining the twopoints. Hence we shall further assume that j, n ≥ n + j, j ), n, j ∈ N , n, j ≥ . The basic properties of Γ are listed in the following two propositions.
Proposition 4.1.
Let Γ be as in (60). Then the following properties hold. (i) Let H be a closed half plane such that ∂H is either the line x = k , or y = k , ora light diagonal D k , for some k ∈ Z . Let e Γ ⊆ Γ be a path, with endpoints e A , e B ,such that e Γ ⊆ H and e A , e B ∈ ∂H . Then e Γ = [[ e A, e B ]] . (ii) Let A = ( x A , y A ) ∈ Γ and let Γ − , Γ + ⊆ Γ be the two paths joining O to A and A to (2 n + j, j ) respectively. Then the following bounds hold: – if A ∈ Z × Z then Γ − ⊆ [0 , x A ] × [0 , y A ] and Γ + ⊆ [ x A , n + j ] × [ y A , j ] ; – if in addition A is a light vertex, then Γ − ⊆ { ( x, y ) ∈ R : 0 ≤ y ≤ y A , y ≤ x ≤ y − y A + x A } , Γ + ⊆ { ( x, y ) ∈ R : y A ≤ y ≤ j, y − y A + x A ≤ x ≤ y + 2 n } (see Figure 8). (iii) Let Q be the interior of a light or a dark square, and assume that Γ ∩ Q = ∅ . Then Γ ∩ Q is a segment. As a consequence, Γ = ∪ Ni =1 [[ P i − , P i ]] , where P = (0 , , P N = (2 n + j, j ) , P i = P k for i = k , and P i belongs to the boundary of a squarefor every i = 0 , . . . , N . IMITS OF CHESSBOARD STRUCTURES 21 (iv)
Let θ i denote the oriented angle between the horizontal axis and [[ P i − , P i ]] , i =1 , . . . , N . Then θ i ∈ [0 , π/ . Moreover, if either P i − or P i is a light vertex, then θ i ∈ [0 , π/ .Proof. Property (i) is a straightforward consequence of the local minimality of Γ and thefact that a segment [[ e A, e B ]] contained in the lines x = k , or y = k , or D k , for some k ∈ Z is the unique geodesic from e A to e B .In order to prove (ii), we first observe thatΓ ⊆ W . = { ( x, y ) ∈ R : 0 ≤ y ≤ j, y ≤ x ≤ n + y } . Namely, if this is not the case, there exists an open half plane H such that H ∩ W = ∅ , H ∩ Γ = ∅ , and ∂H is one of the lines y = 0, y = j , D , D n , a contradiction with (i).Then, if A ∈ Γ ⊆ W has the stated requirements, then the bounds in (ii) can be obtainedreasoning as above with half planes with boundary given by a line of the type x = x A , or y = y A , or D ( x A − y A ) / .Property (iii) is a necessary condition for minimality, see, e.g., [ ? , Section IV].In order to prove (iv), we notice that, by (ii), θ i ∈ [0 , π/
2] whenever either P i − or P i is in Z × Z , and θ i ∈ [0 , π/
4] if either P i − or P i is a light vertex. In particular θ , θ N ∈ [0 , π/ θ i − ∈ [0 , π/
2] and P i − Z × Z , then θ i ∈ [0 , π/ P i − is in the interior of a side of a square,so that the Snell’s law (9) holds. (cid:3) Remark . Since a β = 1 on the boundary of the squares, then Proposition 4.1(iii) canbe improved observing that the intersection of Γ with the closure of a light square is asegment.In what follows we will be interested in the intersections Γ ∩ D k , k = 1 , . . . , n . Proposition 4.3.
Let Γ be as in (60), and let P i , i = 0 , . . . , N be as in Proposition4.1(iii). Then the following properties hold. (i) For every k = 0 , . . . , n there exist ≤ η k ≤ ζ k ≤ j such that Γ ∩ D k = [[ A k , B k ]] , A k = (2 k + η k , η k ) , B k = (2 k + ζ k , ζ k ) . Moreover A = (0 ,
0) = P , B n =(2 n + j, j ) = P N , and ζ k ≤ η k +1 for every k = 0 , . . . , n − . (ii) If A k = B k then η k and ζ k are integers (that is A k and B k are light vertices). (iii) Given k = 1 , . . . , n , let i = 1 , . . . N be such that A k ∈ ]] P i − , P i ]] . Then ≤ θ i <π/ . If in addition A k is not a light vertex, then θ i = 0 . The same properties holdif B k ∈ [[ P i − , P i [[ .Proof. It is clear that Γ ∩ D k = ∅ for every k = 1 , . . . , n . Moreover, by Proposition 4.1(i)and Remark 4.2, Γ ∩ D k is either a single point, or a segment joining two light vertices.The inequality ζ k ≤ η k +1 , k = 0 , . . . , n − θ i ∈ [0 , π/ A k is a light vertex, then by Proposition 4.1(ii) with A = A k we get θ i ∈ [0 , π/ θ i = π/ P i − , A k ]] ⊆ Γ ∩ D k , in contradictionwith the definition of A k .Assume now that A k belongs to the interior of a light square. Then, by (ii), Γ ∩ D k = { A k } , so that θ i < π/
4. Finally θ i = 0, otherwise Γ has to be an horizontal segment, dueto Snell’s Law (9). (cid:3) Definition 4.4.
Given k = 0 , . . . , n , we say that Γ cuts the light diagonal D k if the points A k and B k , defined in Proposition 4.3, coincide and belong to the interior of a light square. A A = A B B B B
Figure 9.
Construction of e Γ, k c ( β ) = 1.For every r = 1 , . . . , j , we shall denote by Γ r the curve(61) Γ r = Γ ∩ { r − < y < r } . By Proposition 4.1(i), the intersection A := Γ r ∩ { y = r − } is a single point, as well asfor B := Γ r ∩ { y = r } . The curve Γ r is a Snell path joining the two points A and B , andlying in a single horizontal strip. Remark . In what follows we shall assume, without loss of generality, that Γ is aSnell path starting from the origin. Namely, if this is not the case, Γ = S ( C , C )where C = ( x ,
0) and C = ( x , < x < x . Let us denote by p , q respectivelythe thickness of the light zone and of the dark zone crossed by Γ , and by p , q theanalogous quantities for the Snell path S ( O, C ), C = ( x − x , p + q = p + q , and p ≤ p , so that, by Remark 2.2, L β (Γ ) ≥ L β ( S ( O, C )). Hence the curve S ( O, C ) ∪ [ C , C ] ∪ (Γ ∩ { y ≥ } ) is a geodesic.The following result is another fairly general property of the geodesics based on thebehavior of the function l ( t, β ) studied in the previous section. Proposition 4.6.
Let β > be given, and let k c ( β ) be as in Definition 3.9 (see alsoCorollary 3.15(i)). For every r = 1 , . . . , j the curve Γ r defined in (61) intersects at most k c ( β ) light diagonals.Proof. Set, as above, A = Γ r ∩ { y = r − } = ( x A , r −
1) and B = Γ r ∩ { y = r } = ( x B , r )respectively the starting and the ending point of the Snell path Γ r .Assume by contradiction that Γ r intersects more than k c ( β ) light diagonals, so that t := x B − x A − > k c ( β ).Let k ∈ N be the smallest integer such that 2 k + r − ≥ x A , and let m ∈ N be thelargest integer such that 2 m + r ≤ x B . Set A := (2 n + r − , r − B := (2 m + r, r ). Itis clear that A and B lie respectively on the first and the last light diagonal intersectedby Γ r (see Figure 9), so that, by assumption, m − n ≥ k c ( β ).Let us denote by ∆ A = 2 k + r − − x A and ∆ B = x B − m − r . It is not restrictive toassume that ∆ A ≤ ∆ B . The points A = ( α, r −
1) and B = ( α ′ , r ) defined by(T1) A = A , B = (2 m + r + ∆ A + ∆ B , r ), if either 0 < ∆ A ≤ ∆ B ≤
1, or 0 < ∆ A ≤ , < ∆ B ≤ , ∆ A + ∆ B < A = (2 k + r − − ∆ A − ∆ B , r − , B = B , if either 1 ≤ ∆ A ≤ ∆ B ≤
2, or0 < ∆ A ≤ , < ∆ B ≤ , ∆ A + ∆ B ≥ α ′ − α − t > m − k , and either A or B is a light vertex. Moreover,from a direct inspection we can check that the thickness of the dark zone from A to B is not greater that the one from A to B , so that by Remark 2.2 we conclude that(62) L β ( S ( A , B )) ≤ L β ( S ( A, B )) . Let us consider the case (T1), so that A = A , and let B ′ = (2 k + 2 k c ( β ) + r, r ).The assumption m − k ≥ k c ( β ) implies that B ′ lies on the left of B (possibly the two IMITS OF CHESSBOARD STRUCTURES 23 points coincide). Let us consider the following new paths: Γ ′ = S ( A , B ′ ) ∪ [[ B ′ , B ]], andΓ = S ( A , B ). Since 2 k c ( β ) < t , from Corollary 3.15(vi) we deduce that L β (Γ ′ ) = l (2 k c ( β ) , β ) + t + √ < l ( t , β ) + t + √ L β (Γ ) . Finally, setting e Γ = [[
A, A ]] ∪ S ( A , B ′ ) ∪ [[ B ′ , B ]] and noticing that the segments [[ A, A ]]and [[ B, B ]] have the same length, we have that L β ( e Γ) = L β (Γ ′ ) < L β (Γ ) ≤ L β (Γ r ) , where the last inequality follows from (62).The analysis of the case (T2) can be carried out in a similar way, with obvious modifi-cations. (cid:3) Geodesics of the chessboard structure ( β ≥ p / ) In this section we shall restrict our analysis to the case β ≥ p /
2, and we shall provide acomplete description of the geodesics joining two light vertices in the chessboard structure.The case β < p / β > β c , any geodesic Γ joining the origin with thepoint (2 n + j, j ) is a finite union of segments, connecting light vertices, and lying on lightdiagonals or on horizontal lines. Theorem 5.1.
Let Γ be a geodesic as in (60). If β > β c , then the points A k , B k arelight vertices for every k = 0 , . . . , n , and [[ B k − , A k ]] is an horizontal segment for every k = 1 , . . . , n .Proof. By Corollary 3.15(i) we have that k c ( β ) = 0, so that, by Proposition 4.6, Γ nevercuts a light diagonal. Hence the points A k , B k defined in Proposition 4.3(i) are verticesof light squares, for every k = 0 , . . . , n . Given r = 1 , . . . , n , let B r − be the exit pointfrom D r − and A r be the access point to D r , and let i , i ′ = 1 , . . . N be such that B r − ∈ [[ P i − , P i [[, and A r ∈ ]] P i ′ − , P i ′ ]]. We have that θ i = θ i ′ = 0, that is both [[ P i − , P i ]]and [[ P i ′ − , P i ′ ]] lie on the horizontal sides of the squares. Indeed, by Proposition 4.3(iii), θ i ′ , θ i ∈ [0 , π/ θ i or θ i ′ belong to (0 , π/
4) , then Γ should contain a Snellpath starting from a vertex of a light square and lying in a horizontal strip, a contradictionwith Corollary (3.15)(ii) and the local optimality of Γ.Then θ i = θ i ′ = 0, and, by Remark 4.2, there exists two points A , B such that thesegments [[ B r − , B ]] and [[ A, A r ]] are horizontal, with length greater than or equal to 1, andthey are contained in Γ. In order to complete the proof we have to show that [[ B, A ]] isa horizontal segment. Assume by contradiction that this is not the case, that is B = A ′ ,where A ′ is the intersection of the line containing [[ A, A r ]] with D r − . In this case the lengthof the polygonal line [[ B r − , A ′ , A r ]] is less than the length of any curve joining B r − with A r and containing [[ B r − , B ]] and [[ A, A r ]], in contradiction with the local minimality ofΓ. (cid:3) Definition 5.2. An S –path is a Snell path joining the point (2 m + k, k ) to the point(2 m + k + 3 , k + 1) for some m , k ∈ N . We shall denote its normalized length by λ = l (2 , β ) = 2 q − σ + β q β − σ − − √ , where σ = ˆ σ (2 , β ) is implicitly defined by2 σ q − σ + σ q β − σ − . The optical length of an S –path will be denoted by Λ = λ + 2 + √ Remark . It is straightforward to check that √ < Λ < β √
10, and, by the verydefinition of β c , Λ = 2 + √ β = β c . Moreover, 1 / √ < σ < β/ √ S –paths will play a fundamental rˆole in the analysis of the geodesics for p / ≤ β < β c (see Theorem 5.4 below). Namely for β in this range the S –paths have the minimalnormalized length among all the Snell paths starting from a light vertex (2 m + k, k ) andreaching a point on the line y = k + 1 (see Corollary 3.15). Theorem 5.4.
Let Γ be a geodesic as in (60). If p / ≤ β < β c , then the points A k , B k are light vertices for every k = 0 , . . . , n . Moreover B k − is connected to A k either by anhorizontal segment or by an S –path for every k = 1 , . . . , n .Proof. By Remark 4.5 we can assume, without loss of generality, that A = B = (0 , i = min { r ∈ { , . . . , n } ; A r is a light vertex } . Notice that the index i in (63) is well defined, since as a consequence of Proposition 4.3(i)and (ii) at least A n is a light vertex. Moreover, if i ≥
2, then Γ cuts every light diagonal D k , k = 1 , . . . , i −
1, that is, the points A k and B k coincide and they are not light vertices.Let us denote by Γ ′ the portion of Γ joining A = B = (0 ,
0) to A i . We are going toprove that i = 1 , andΓ ′ is either an horizontal segment or an S –path . (64)Once these properties are proved, then B is a light vertex and, repeating the procedure n times, we reach the conclusion.Let m ∈ N be such that A i = (2 i + m, m ). If m = 0, then B is connected to A i by anhorizontal segment, so that i = 1 and (64) holds. If m = 1, then by Corollary 3.15(ii) andthe local minimality of Γ, we conclude that i = 1 and Γ ′ is and S –path. It remains toprove that the case m ≥ k = 0 , . . . , i let r k = 1 , . . . , m be such that A k = (2 k + η k , η k ) ∈ Γ r k , whereΓ r is the Snell path defined in (61). Let us denote by C k = ( x k , r k −
1) and C ′ k = ( x ′ k , r k )respectively the starting and the ending point of Γ r k , let δ k = x ′ k − x k be the thickness ofΓ r k , and define t − k = 2 k + r k − − x k , t + k = x ′ k − k − r k , h k = η k − ⌊ η k ⌋ , so that δ k = t − k + t + k + 1 (see Figure 10).For k = 0, we have that r = 1, C = B = (0 ,
0) and C ′ = ( δ , δ >
1. We claim that 2 < δ <
3. Namely, by Corollary 3.15(v) and (iii), we have ( L β (Γ ) = l ( δ , β ) + √ δ − > √ δ − L β ([[ O, (1 , , ( δ , , if δ ∈ (1 , , L β (Γ ) > l (2 , β ) + √ δ − ≥ L β ( S (0 , (3 , ∪ [[(3 , , ( δ , , if δ > , hence, by the local minimality of Γ , we cannot have neither 1 < δ ≤ δ >
3. For k = i the same arguments show that 2 < δ i <
3. Hence, from Proposition 4.6, we havethat 1 = r < r < · · · < r i = m . IMITS OF CHESSBOARD STRUCTURES 25 C k A C k k t k+k t − Figure 10.
Finally, by Lemma 3.17,˜ l ( t − k , β, h k ) > ˜ l ( t − k , β, , ˜ l ( t + k , β, − h k ) > ˜ l ( t + k , β, , k = 1 , . . . , i − , ( i ≥ l is the function defined in (55). Moreover, being Γ a geodesic, we have ˜ l ( t − k , β, h k ) < l ( t + k , β, − h k ) <
0, so that by Corollary 3.15(v), we have t − k , t + k >
1. On theother hand, from Corollary 3.15 and Proposition 4.6, we cannot have t − k ≥ t + k ≥ r k would intersect more than one light diagonal. In conclusion, we have that t − k , t + k ∈ (1 , δ k ∈ (3 , k = 1 , . . . , i − i ≥ i X k =0 δ k . By construction and from the estimates above we have∆ ≤ i + m, m ≥ i + 1 , i + 1 < ∆ < i + 1 . It is straightforward to show that L β Γ ′ \ i [ k =0 Γ r k ! ≥ L β ([[(∆ , i + 1) , (2 i + m, m )]]) , so that L β (Γ ′ ) ≥ i X k =0 L β (Γ r k ) + q (2 i + m − ∆) + ( m − i − . L β (Γ r k ) can be estimated using the function ˜ l . For k = 0 and k = i we have that L β (Γ r ) = ˜ l ( δ − , β,
1) + δ − √ , L β (Γ r i ) = ˜ l ( δ i − , β,
1) + δ i − √ , (65)while, for k = 1 , . . . , i −
1, ( i ≥ L β (Γ r k ) = ˜ l ( t − k , β, h k ) + ˜ l ( t + k , β, − h k ) + δ k − √ . From Lemma 3.4(iii) and Lemma 3.17 we have that˜ l ( t, β, h ) > ˜ l h ( t, β, h −
1) + ˜ l t (2 , β, t −
2) + λ for every ( t, h ) ∈ (1 , × [0 , λ = ˜ l (2 , β, l t (2 , β,
1) = q − σ − , ˜ l h ( t, β,
1) = q − ˜ σ ( t, β,
1) + ˜ σ ( t, β, − √ . where σ = ˜ σ (2 , β, t ˜ σ ( t, β,
1) is a decreasing function for t ≥
0, and the map s
7→ √ − s + s − √ ≤ s ≤ / √
2, we finally obtain(67) ˜ l ( t, β, h ) > λ + (cid:18)q − σ − (cid:19) ( t −
2) + (cid:18)q − σ + σ − √ (cid:19) ( h − for every ( t, h ) ∈ (1 , × [0 , L β (Γ r ) > λ + (cid:18)q − σ − (cid:19) ( δ −
3) + δ − √ , L β (Γ r i ) > λ + (cid:18)q − σ − (cid:19) ( δ i −
3) + δ i − √ , L β (Γ r k ) > λ + (cid:18)q − σ − (cid:19) ( δ k − − q − σ − σ + δ k − √ , (68)( k = 1 , . . . , i − i ≥ i X k =0 L β (Γ r k ) > ∆ q − σ + σ + (cid:18) √ − q − σ − σ + 2 λ (cid:19) i and L β (Γ ′ ) > ∆ q − σ + σ + (cid:18) √ − q − σ − σ + 2 λ (cid:19) i + q (2 i + m − ∆) + ( m − i − . (69)Let us now consider the path Γ ′′ starting from the origin, obtained by the concatenationof i S -paths and the segment connecting the point (3 i, i ) to A i = (2 i + m, m ). Since thissegment connects two points on the light diagonal D i , its length is ( m − i ) √
2, so that L β (Γ ′′ ) = i Λ + ( m − i ) √ i ( λ + 2 + √
2) + ( m − i ) √ . We are going to show that L β (Γ ′′ ) < L β (Γ ′ ), in contradiction with the local minimality ofΓ ′ . We have that L β (Γ ′ ) − L β (Γ ′′ ) > (cid:18) √ − q − σ − σ + λ (cid:19) i − (cid:18) √ − q − σ − σ (cid:19) + q µ + µ − µ q − σ − µ (cid:18) √ − q − σ (cid:19) (70)where µ = 2 i + m − ∆ ≥ , µ = m − i − ≥ . Since 3 i + 1 < ∆ and m ≥ i + 1 we have that(71) 0 ≤ µ < µ . Moreover(72) q µ + µ − µ q − σ − µ (cid:18) √ − q − σ (cid:19) = µ ϕ (cid:18) µ µ (cid:19) , where ϕ ( s ) = p s − s q − σ − (cid:18) √ − q − σ (cid:19) . Since 0 < σ < / √
2, we have that 1 / √ < q − σ <
1. It can be easily checked that ϕ ′ ( s ) < s ∈ [0 , ϕ (1) < ϕ ( s ) , ∀ s ∈ [0 , . From (70), (71), (72) and (73) we thus get L β (Γ ′ ) − L β (Γ ′′ ) > (cid:18) √ − q − σ − σ + λ (cid:19) i − (cid:18) √ − q − σ − σ (cid:19) . (74) IMITS OF CHESSBOARD STRUCTURES 27
We claim that 2 + √ − q − σ − σ + λ > √ − q − σ − σ > . (75)The second inequality in (75) easily follows from the fact that 0 < σ < / √
2. Concerningthe first one, by the very definition of λ , and since b b / q b − σ is an increasingfunction in [1 , + ∞ ), we have that (cid:18) √ − q − σ − σ + λ (cid:19) − (cid:18) √ − q − σ − σ (cid:19) = − q − σ + 2 q − σ + β q β − σ − √ ≥ − q − σ + 2 q − σ + 3 / q / − σ − √ ψ ( σ ) . It can be checked that ψ is strictly increasing in (0 , σ > / √ ψ ( σ ) > ψ (cid:18) √ (cid:19) = √ √ √ − √ > , and (75) is proved.Since i ≥
1, it is straightforward to check that (74) and (75) imply that L β (Γ ′ ) −L β (Γ ′′ ) >
0, in contradiction with the local minimality of Γ ′ . (cid:3) Theorem 5.5.
Let Γ be a geodesic from the origin to a light vertex ξ = ( x, y ) , with ≤ y ≤ x . (i) If β ≥ β c , then L β (Γ) = x + ( √ − y . (ii) If p / ≤ β ≤ β c , then L β (Γ) = x + (Λ − y, if ≤ y ≤ x/ , Λ − √ x + 3 √ − Λ y, if x/ ≤ y ≤ x , where Λ = Λ ( β ) is the length of an S –path, introduced in Definition 5.2.Proof. (i) For β > β c it is a straightforward consequence of Theorem 5.1.(ii) Let us consider the case p / ≤ β < β c . From Theorem 5.4 we know that Γ is theconcatenation of S –paths and segments joining light vertices, lying on light diagonals oron horizontal lines. Hence we have that L β (Γ) = t Λ + r + d √ , where t is the number of S –paths, r is the number of unit horizontal segments, and d is the number of diagonals of light squares. It is clear that the three numbers t, r, d ∈ N must satisfy the constraints 3 t + r + d = x, t + d = y , so that L β (Γ) = (Λ − − √ t + x + ( √ − y . Since Λ < √ L is minimized by choosing the largest admissible value of t , which is y if y ≤ x/
3, and ( x − y ) / y ≥ x/
3. (We remark that ( x − y ) / P = ( x, y ) is a light vertex.) In conclusion, if y ≤ x/ t = y , d = 0 and r = x − y , whereas if y ≥ x/ t = ( x − y ) / d = (3 y − x ) / r = 0,obtaining (ii).Finally, if β = β c , it is straightforward to check that the lengths of geodesics can becomputed indifferently as in (i) or in (ii). We remark that, in this case, these two formulasgive the same result, since Λ = 2 + √ (cid:3) One may wonder if the previous characterization of the geodesics for p / ≤ β < β c remains valid for β c < β < β c . The following example shows that this is not the case. Example . Let Γ = [[(0 , , (1 , ∪ S ((1 , , (4 , p / ≤ β < β c ,then Γ is a geodesic joining the origin to the point ξ = (4 , t ∈ [0 , t ) = S ((0 , , (1 + t, ∪ S ((1 + t, , (4 , . We have that L ( t ) := L β (Γ( t )) = l ( t, β ) + l (2 − t, β ) . Moreover, for t = 0 we have Γ(0) = Γ, L (0) = Λ + √
2, and L ′ (0) = lim t → + L ( t ) − L (0) t = l + t (0 , β ) − l − t (2 , β )= r β − − q − σ . (76)We recall that, for a given β > σ is the unique zero in (0 ,
1) of the function g ( σ ) = 2 σ √ − σ + σ p β − σ − . Moreover, g is a strictly monotone increasing function in (0 , g (0) = − g ( s ) → + ∞ as s → − . For β ≤ p /
2, let us compute ψ ( β ) = g r − β ! = 2 s − β β − s − β β − − . Since ψ ′ ( β ) = − β p − β (cid:18) β − / + 4(2 β − / (cid:19) < , the map ψ is strictly monotone decreasing in [1 , p / ψ (1) = 1 and ψ (cid:16)p / (cid:17) = − ψ in (1 , p /
2) is ˜ β ≃ . β > β c ≃ . < β < ˜ β , we have that σ < p / − β and, by (76), L ′ (0) <
0. In conclusion, if1 < β < ˜ β , for t > L ( t ) < L (0), and Γ is not a geodesic.6. The homogenized metric
As a direct consequence of Theorem 5.5, we obtain the complete description of thehomogenized metric Φ β for β ≥ p /
2. In the general case we discuss the regularity of thehomogenized metric.In order to make some usefull reductions, we need two remarks on the distance d εβ (0 , ξ )defined in (6). Remark . Since | ξ − η | ≤ d εβ ( η, ξ ) ≤ β | ξ − η | , it can be easily seen that(77) Φ β ( ξ ) = lim ε → + d εβ ( η ε , ξ ε ) , ∀ ξ ∈ R , ∀ ξ ε → ξ , ∀ η ε → . IMITS OF CHESSBOARD STRUCTURES 29
Figure 11.
The homogenized unit ball for β ≥ β c (left) and p / ≤ β <β c (right) Remark . For every ε > η ε = ( ε , ε ). From (77) we have that Φ β ( ξ ) = lim ε → + d εβ ( η ε , ξ + η ε ) , ∀ ξ ∈ R . Since the map ξ d εβ ( η ε , ξ + η ε ) is symmetric w.r.t. the coordinated axis and the diagonalspassing through the origin, it is clear that Φ β has the same symmetries.In what follows, given ( x, y ) ∈ R we denote M = max {| x | , | y |} , m = min {| x | , | y |} . Theorem 6.3.
For every ( x, y ) ∈ R the following hold. (i) If β ≥ β c , then Φ β ( x, y ) = M + ( √ − m . (ii) If p / ≤ β ≤ β c , then Φ β ( x, y ) = M + (Λ − m, if m ≤ M , Λ − √ M + 3 √ − Λ m, if m ≥ M , where Λ = Λ ( β ) is the length of an S –path, introduced in Definition 5.2.Proof. By Remark 6.2 it is enough to consider the case 0 ≤ y ≤ x , so that M = x and m = y . Let ξ = ( x, y ) and, for ε >
0, let us define j = (cid:22) yε (cid:23) , n = (cid:22) x − jε ε (cid:23) , ξ ε = ( x ε , y ε ) = ((2 n + j ) ε, jε ) . Then | y − y ε | < ε, | x − x ε | < ε, so that | ξ − ξ ε | < ε √
5. Moreover d εβ (0 , ξ ε ) is explicitly computed in Theorem 5.5. Since ξ ε → ξ , by (77), the conclusion follows. (cid:3) In the general case we have the following result.
Theorem 6.4.
Let β > be given, and let k c ( β ) be the number defined in Definition 3.9.Then Φ β ( x, y ) = M + ( l (2 k c ( β ) , β ) + √ − m for every ( x, y ) ∈ R belonging to one ofthe cones { (2 k c ( β ) + 1) | y | ≤ | x |} or { (2 k c ( β ) + 1) | x | ≤ | y |} .Proof. From Remark 6.2 it is enough to consider the case 0 ≤ (2 k c ( β )+1) y ≤ x . Moreover,we can assume that y >
0, the case y = 0 being trivial. Since the homogenized metricdepends continuously on β , it is not restrictive to assume β = β ck , in such a way that l (2 k c ( β ) , β ) < l ( t, β ) , ∀ t ≥ , t = 2 k c ( β ) (see Corollary 3.15(iii)). For every 0 < ε < y , let ξ ε = ( x ε , y ε ) be the nearest light vertexto ξ = ( x, y ) below the line x = (2 k c ( β ) + 1) y . Then ξ ε = ((2 n + j ) ε, jε ) for some j ≥ n ≥ k c ( β ) j , and ξ ε → ξ as ε tends to 0.We claim that(78) d εβ (0 , ξ ε ) = x ε + ( l (2 k c ( β ) , β ) + √ − y ε , ∀ < ε < y , so that the result will follows from (77). In order to prove the claim, after a scaling, wehave to depict a geodesic Γ joining the origin to the point (2 n + j, j ) in the standardchessboard structure. Let us define the class S of all Snell paths joining the light vertices(2 m + r − , r −
1) and (2 m + 2 k c ( β ) + r, r ), m, r ∈ Z . We are going to show that Γ has tobe the concatenation of j Snell paths in S and of horizontal segments. As a consequence,since the length of any path in S equals to l (2 k c ( β ) , β ) + 2 k c ( β ) + √
2, whereas the totallength of the horizontal segments is 2( n − k c ( β ) j ), we obtain that (78) holds.Let us define the paths Γ , . . . , Γ j as in (61). For every r = 1 , . . . , j let us denote by( x r , r −
1) and ( x ′ r , r ) the endpoints of Γ r , and let τ r = x ′ r − x r . Thanks to Lemma 3.16,we have that L β (Γ r ) ≥ l (2 k c ( β ) , β ) + τ r − √ , for every r = 1 , . . . , j . Then we get L β (Γ) = j X r =1 L β (Γ r ) + 2 n + j − j X r =1 τ r ≥ n + j + ( l (2 k c ( β ) , β ) + √ − j and again by Lemma 3.16 the equality holds if and only if τ r = 2 k c ( β ) and Γ r ∈ S forevery r = 1 , . . . , j . (cid:3) Corollary 6.5.
For every β > the unit ball of the homogenized metric is not strictlyconvex, and its boundary is not differentiable.Remark . The presence of faces in the optical ball corresponds to nonuniqueness of thegeodesics. More precisely, if F is a face of positive length, and C = { λη ; η ∈ F, λ ≥ } isthe corresponding cone, then for every ξ ∈ C , a function u ∈ AC ([0 , , R ) with u (0) = 0, u (1) = ξ , parameterizes a geodesic if and only if u ′ ( t ) ∈ C for a.e. t ∈ [0 , p ∈ R such that Φ β ( η ) = h p, η i for every η ∈ C , and Φ β ( η ) > h p, η i for every η ∈ R \ C . Hence, if u ′ ( t ) ∈ C for a.e. t ∈ [0 , L homβ ( u ) = Z Φ β ( u ′ ( t )) dt = Z h p, u ′ ( t ) i dt = Φ β ( ξ ) , whereas L homβ ( u ) > Φ β ( ξ ) whenever { t ∈ [0 , u ′ ( t ) C } has positive measure.As a final remark, let us consider the chessboard structure corresponding to the uppersemicontinuous function(79) ˜ a β ( ξ ) = ( β if ξ ∈ ([0 , × [1 , ∪ ([1 , × [0 , a = β instead of 1 on the sides of the squares.In this way we obtain a new family of length functionals e L εβ . In this case the existenceof a geodesic joining the origin with a point ξ ∈ R is not guaranteed. For example, if ξ = ( ε, e L εβ ( u ) > ε for every u ∈ AC ([0 , , R ) such that u (0) = 0 and u (1) = ξ . IMITS OF CHESSBOARD STRUCTURES 31
On the other hand, we can construct a minimizing sequence ( u n ) n such that e L εβ ( u n ) → ε for n → + ∞ , defining u n ( t ) = ( ( εt, t/n ) , if t ∈ [0 , / , ( εt, − t ) /n ) , if t ∈ [1 / , . Nevertheless, the Γ-limit with respect to the L -topology of the functionals ( e L εβ ) coincideswith the Γ-limit L homβ of the functionals ( L εβ ). Namely, the liminf inequality is certainlysatisfied since e L εβ ≥ L εβ . On the other hand, given u ∈ AC ([0 , , R ) and a recoveringsequence ( u ε ) for ( L εβ ), we can construct a recovering sequence (˜ u ε ) for ( e L εβ ) in the followingway: for a given ε , we obtain ˜ u ε modifying u ε in the region where u ε ( t ) belongs to theset S of the sides of squares, in such a way that L εβ (˜ u ε ) < L εβ ( u ε ) + ε , and the set { t ∈ [0 , u ε ∈ S } has vanishing Lebesgue measure. Dipartimento di Metodi e Modelli Matematici, Univ. di Roma I, Via Scarpa, 16 – 00161Roma (Italy)
E-mail address , Micol Amar: [email protected]
Dipartimento di Matematica “G. Castelnuovo”, Univ. di Roma I, P.le A. Moro 2 – 00185Roma (Italy)
E-mail address , Graziano Crasta: [email protected]
E-mail address , Annalisa Malusa:, Annalisa Malusa: