Explicit speed of convergence of the stochastic billiard in a convex set
aa r X i v : . [ m a t h . P R ] J a n Explicit speed of convergence of the stochastic billiard ina convex set
Ninon Fétique ∗ Abstract
In this paper, we are interested in the speed of convergence of the stochastic billiardevolving in a convex set K . This process can be described as follows: a particle movesat unit speed inside the set K until it hits the boundary, and is randomly reflected,independently of its position and previous velocity. We focus on convex sets in R witha curvature bounded from above and below. We give an explicit coupling for both thecontinuous-time process and the embedded Markov chain of hitting points on the boundary,which leads to an explicit speed of convergence to equilibrium. MSC Classification 2010:
Key words:
Stochastic billiard, invariant measure, coupling, speed of convergence.
Contents Laboratoire de Mathématiques et Physique Théorique (UMR CNRS 7350), Fédération Denis Pois-son (FR CNRS 2964), Université François-Rabelais, Parc de Grandmont, 37200 Tours, France. Email:[email protected] Introduction
In this paper, our goal is to give explicit bounds on the speed of convergence of a process, called"stochastic billiard", towards its invariant measure, under some assumptions that we will detailfurther. This process can be informally described as follows: a particle moves at unit speedinside a domain until it hits the boundary. At this time, the particle is reflected inside thedomain according to a random distribution on the unit sphere, independently on its positionand previous velocity.The stochastic billiard is a generalisation of shake-and-bake algorithm (see [1]), in which thereflection law is the cosine law. In that case, it has been proved that the Markov chain ofhitting points on the boundary has a uniform stationary distribution. In [1], the shake-and-bakealgorithm is introduced for generating uniform points on the boundary of bounded polyhedra.More generally, stochastic billiards can be used for sampling from a bounded set or the boundaryof such a set, through the Markov Chain Monte Carlo algorithms. In that sense, it is thereforeimportant to have an idea of the speed of convergence of the process towards its invariantdistribution.Stochastic billiards have been studied a lot, under different assumptions on the domain in whichit lives and on the reflection law. Let us mention some of these works. In [5], Evans considersthe stochastic billiard with uniform reflection law in a bounded d -dimensional region with C boundary, and also in polygonal regions in the plane. He proves first the exponentially fast totalvariation convergence of the Markov chain, and moreover the uniform total variation Césaroconvergence for the continuous-time process. In [3], the authors only consider the stochasticbilliard Markov chain, in a bounded convex set with curvature bounded from above and witha cosine distribution for the reflection law. They give a bound for the speed of convergenceof this chain towards its invariant measure, that is the uniform distribution on the boundaryof the set, in order to get a bound for the number of steps of the Markov chain required tosample approximatively the uniform distribution. Finally, let us mention the work of Comets,Popov, Schütz and Vachkovskaia [2], in which some ideas have been picked and used in thepresent paper. They study the convergence of the stochastic billiard and its associated Markovchain in a bounded domain in R d with a boundary locally Lipschitz and almost everywhere C . They consider the case of a reflection law which is absolutely continuous with respectto the Haar measure on the unit sphere of R d , and supported on the whole half-sphere thatpoints into the domain. They show the exponential ergodicity of the Markov chain and thecontinuous-time process and also their Gaussian fluctuations. The particular case of the cosinereflection law is discussed. Even if they do not give speeds of convergence, their proofs couldlead to explicit speeds if we write them in particular cases (as for the stochastic billiard in adisc of R for instance). However, as we will mention in Section 2.3, the speed of convergenceobtained in particular cases will not be relevant, since their proof is adapted to their verygeneral framework, and not for more particular and simple domains.The goal of this paper is to give explicit bounds on the speed of convergence of the stochasticbilliard and its embedded Markov chain towards their invariant measures. For that purpose,we are going to give an explicit coupling of which we can estimate the coupling time.In a first part, we study the particular case of the billiard in a disc. In that case, everything isquite simple since all the quantities can be explicitly expressed.Then, in a second part, we extend the results for the case of the stochastic billiard in a compactconvex set of R with curvature bounded from above and below. In that case, we can no more2o explicit computations on the quantities describing the process, since we do not know exactlythe geometry of the convex set. However, thanks to the assumptions on the curvature, we areable to estimate the needed quantities.In both cases, the disc and the convex set, we suppose that the reflection law has a densityfunction which is bounded from below by a strictly positive constant on a part of the sphere.The speed of convergence will obviously depend on it. However, for the convergence of thestochastic billiard process in a convex set, we will need to suppose that the reflection law issupported on the whole half sphere that points inside the domain.At the end of this paper, we briefly discuss the extension of the results to higher dimensions. Notations
We introduce some notations used in the paper: • for A ⊂ R , A denotes the indicator function of the set A , that is A ( x ) is equal to if x ∈ A and otherwise; • for x ∈ R , ⌊ x ⌋ denotes the floor of the real x ; • for x, y ∈ R , we note by k x k the euclidean norm of x and we write h x, y i for the scalarproduct of x and y ; • for A ⊂ R , ∂A denotes the boundary of the set A ; • B r denotes the closed ball of R centred at the origin with radius r , i.e. B r = { x ∈ R : k x k≤ r } ,and S denotes the unit sphere of R , i.e. S = { x ∈ R : k x k = 1 } ; • for I ⊂ R , |I| denotes the Lebesgue measure of the set I ; • for K ⊂ R a compact convex set, we consider the 1-dimensional Hausdorff measure in R restricted to ∂K . Therefore, if A ⊂ ∂K , | A | denotes this Hausdorff measure of A ; • for A ⊂ R , if x ∈ ∂A , we write n x the unitary normal vector of ∂A at x looking into A andwe define S x the set of vectors that point the interior of A : S x = { v ∈ S : h v, n x i ≥ } ; • if two random variables X and Y are equal in law we write X L = Y , and we write X ∼ µ to say that the random variable X has µ for law; • we denote by G ( p ) the geometric law with parameter p . In order to describe the way we will prove the exponential convergences and obtain bounds onthe speeds of convergence, we first need to introduce some notions.Let ν and ∼ ν be two probability measures on a measurable space E . We say that a probabilitymeasure on E × E is a coupling of ν and ∼ ν if its marginals are ν and ∼ ν . Denoting by Γ( ν, ∼ ν ) the set of all the couplings of ν and ∼ ν , we say that two random variables Y and ∼ Y satisfy3 Y, ∼ Y ) ∈ Γ( ν, ∼ ν ) if ν and ∼ ν are the respective laws of Y and ∼ Y . The total variation distancebetween these two probability measures is then defined by k ν − ∼ ν k T V = inf ( Y, ∼ Y ) ∈ Γ( ν, ∼ ν ) P ( Y = ∼ Y ) . For other equivalent definitions of the total variation distance and its properties, see for instance[6].Let ( Y ) t ≥ and ( ˜ Y ) t ≥ be two Markov processes and let define T c = inf n t ≥ Y t = ˜ Y t o thecoupling time of Y and ˜ Y . From the definition of the total variation distance, it immediatelyfollows that kL ( Y t ) − L ( ˜ Y t ) k T V ≤ P ( T c > t ) . Therefore, let T ∗ be a random variable stochastically bigger than T c , T c ≤ st T ∗ , which meansthat P ( T c ≤ t ) ≥ P ( T ∗ ≤ t ) for all t ≥ . If T ∗ has a finite exponential moment, Markov’sinequality gives then, for any λ such that the Laplace transform of T ∗ is well defined: kL ( Y t ) − L ( ˜ Y t ) k T V ≤ P ( T ∗ > t ) ≤ e − λt E (cid:2) e λT ∗ (cid:3) . Thus, if we manage to stochastically bound the coupling time of two stochastic billiards by arandom time whose Laplace transform can be estimated, we get an exponential bound for thespeed of convergence of the stochastic billiard towards its invariant measure.We end this part with a definition that we will use throughout this paper.
Definition 2.1.
Let K ⊂ R be a compact convex set.We say that a pair of random variables ( X, T ) living in ∂K × R + is α -continuous on the set A × B ⊂ ∂K × R + if for any measurable A ⊂ A , B ⊂ B : P ( X ∈ A , T ∈ B ) ≥ α | A || B | . We can also adapt this definition for a single random variable.
Let us now give a precise description of the stochastic billiard ( X t , V t ) t ≥ is a set K .We assume that K ⊂ R is a compact convex set with a boundary at least C .Let e = (1 , be the first coordinate vector of the canonical base of R . We consider a law γ onthe half-sphere S e = { v ∈ S : e · v ≥ } . Let moreover ( U x , x ∈ ∂K ) be a family of rotations of S such that U x e = − n x , where we recall that n x is the normal vector of ∂K at x looking into K .Given x ∈ ∂K , we consider the process ( X t , V t ) t ≥ living in K × S constructed as follows (seeFigure 1): • Let X = x , and V = U X η with η a random vector chosen according to the distribution γ . • Let τ = inf { t > x + tV / ∈ K } and define T = τ . We put X t = x + tV , V t = V for t ∈ [0 , T ) , and X T = x + τ V .Then, let V T = U X T η with η following the law γ .4 x X T v X T V T V T Figure 1: A trajectory of the stochastic billiard in a set K , starting in the interior of K • Let τ = inf { t > X T + tV T / ∈ K } and define T = T + τ . We put X t = X T + tV T , V t = V T for t ∈ [ T , T ) , and X T = X T + τ V T .Then, let V T = U X T η with η following the law γ . • And we start again ...As mentioned in the introduction ( X T n ) n ≥ is a Markov chain living in ∂K and the process ( X t , V t ) t ≥ is a Markov process living in K × S . Remark 2.2.
We can obviously define the continuous-time process starting at any x ∈ K ,what will in fact often do in this paper. If x ∈ K \ ∂K , we have to precise also the initial speed v ∈ S , and we can use the same scheme to construct the process. For x ∈ ∂K , it is equivalent to consider the new speed in S x or to consider the angle in (cid:2) − π , π (cid:3) between this vector speed and the normal vector n x . For n ≥ , we thus denote by Θ n the random variable in (cid:2) − π , π (cid:3) such that r X Tn , Θ n ( n X Tn ) L = V T n , where for x ∈ ∂K and θ ∈ R , r x,θ denotes the rotation with center x and angle θ .We make the following assumption on γ (see Figure 2): Assumption ( H ) : The law γ has a density function ρ with respect to the Haar measure on S e , whichsatisfies: there exist J ⊂ S e symmetric with respect to e and ρ min > such that: ρ ( u ) ≥ ρ min , for all u ∈ J . This assumption is equivalent to the following one on the variables (Θ n ) n ≥ : Assumption ( H ′ ) : θ ∗ U x J ∂Kn x Figure 2: Illustration of Assumptions ( H ) and ( A ) The variables Θ n , n ≥ , have a density function f with respect to the Lebesguemeasure on (cid:2) − π , π (cid:3) satisfying: there exist f min > and θ ∗ ∈ (cid:0) , π (cid:1) such that: f ( θ ) ≥ f min , for all θ ∈ (cid:20) − θ ∗ , θ ∗ (cid:21) . In fact, since these two assumptions are equivalent, we have ρ min = f min and |J | = θ ∗ . In the sequel, we will use both descriptions of the speed vector depending on which is the mostsuitable.
Let us now informally describe the idea of the couplings used to explicit the speeds of conver-gence of our processes to equilibrium. They will be explain explicitly in Sections 3 and 4.To get a bound on the speed of convergence of the Markov chain recording the location ofhitting points on the boundary of the stochastic billiard, the strategy is the following. Weconsider two stochastic billiard Markov chains with different initial conditions. We estimatethe number of steps that they have to do before they have a strictly positive probability toarrive on the same place at a same step. In particular, it is sufficient to know the number ofsteps needed before the position of each chain charges the half of the boundary of the set onwhich they evolve. Then, their coupling time is stochastically smaller than a geometric timewhose Laplace transform is known.The case of the continuous-time process is a bit more complicated. To couple two stochasticbilliards, it is not sufficient to make them cross in the interior of the set where they live. Indeed,if they cross with a different speed, then they will not be equal after. So the strategy is tomake them arrive at the same place on the boundary of the set at the same time, and then6hey can always keep the same velocity and stay equal. We will do this in two steps. First,we will make the two processes hit the boundary at the same time, but not necessary at thesame point. This will take some random time, that we will be able to quantify. And secondly,with some strictly positive probability, after two bounces, the two processes will have hit theboundary at the same point at the same time. We repeat the scheme until the first success.This leads us to a stochastic upper bound for the coupling time of two stochastic billiards.Obviously, the way that we couple our processes is only one way to do that, and there aremany as we want. Let us for instance describe the coupling constructed in [2]. Consider twostochastic billiard processes evolving in the set K with different initial conditions. Their firststep is to make the processes hit the boundary in the neighbour of a good x ∈ ∂K . Thiscan be done after n bounces, where n is the minimum number of bounces needed to connectany two points of the boundary of K . Once the two processes have succeed, they are in theneighbour of x , but at different times. Then, the strategy used by the authors of [2] is to makethe two processes do goings and comings between the neighbour of x and the neighbour ofanother good y ∈ ∂K . Thereby, if the point y is well chosen, the time difference between thetwo processes decreases gradually, while the positions of the processes stay the same after onegoing and coming. However, the number of goings and comings needed to compensate for thepossibly big difference of times could be very high. This particular coupling is therefore welladapted for sets whose boundary can be quite "chaotic", but not for convex sets with smoothboundary as we consider in this paper. In this section, we consider the particular case where K is a ball: K = B r , for some fixed r > .In that case, for each n ≥ , the couple ( X T n , V T n ) ∈ ∂ B r × S can be represented by a couple (Φ n , Θ n ) ∈ [0 , π ) × (cid:2) − π , π (cid:3) as follows (see Figure 3): • to a position x on ∂ B r corresponds an unique angle φ ∈ [0 , π ) . The variable Φ n nominatesthis unique angle associated to X n , i.e. (1 , Φ n ) are the polar coordinate of X n . • at each speed V T n we associate the variable Θ n introduced in Section 2.2, satisfyingAssumption ( H ′ ) .Remark that for all n ≥ , the random variable Θ n is independent of Φ k for all k ∈ { , n } . Wealso recall that the variables Θ n , n ≥ , are all independent.In the sequel, we do not care about the congruence modulo π : it is implicit that when wewrite Φ , we consider its representative in [0 , π ) .Let us state the following proposition that links the different random variables together. Proposition 3.1.
For all n ≥ we have: τ n = 2 r cos(Θ n − ) and Φ n = π + 2Θ n − + Φ n − (1) Proof.
The relationships are immediate with geometric considerations.7 T n V T n Φ n Θ n Figure 3: Definition of the variables Φ n and Θ n in bijection with the variables X T n and V T n In this section, the goal is to obtain a control of the speed of convergence of the stochasticbilliard Markov chain on the circle. For this purpose, we study the distribution of the positionof the Markov chain at each step.Let Φ = φ ∈ [0 , π ) . Proposition 3.2.
Let (Φ n ) n ≥ be the stochastic billiard Markov chain evolving on ∂B r , satis-fying assumption ( H ′ ) .We have f Φ ( u ) ≥ f min , ∀ u ∈ I = [ π − θ ∗ + φ , π + θ ∗ + φ ] . Moreover, for all n ≥ , for all η , · · · , η n such that η ∈ (0 , θ ∗ ) , and for k ∈ { , · · · , n − } , η k +1 ∈ (cid:0) , nθ ∗ − P n − k =2 η k (cid:1) , we have f Φ n ( u ) ≥ (cid:18) f min (cid:19) n η n · · · η , ∀ u ∈ I n = " ( n − π − nθ ∗ + φ + n X k =2 η k , ( n − π + nθ ∗ + φ − n X k =2 η k . Proof.
Since the Markov chain is totally symmetric, we do the computations with φ = 0 . • Case n = 2 :We have, thanks to (1), Φ = π + 2Θ + φ = π + 2Θ . Thus, for any measurable bounded8unction g , we get: E [ g (Φ )] = E [ g ( π + 2Θ )] = Z π − π g ( π + 2 x ) f ( x )d x ≥ f min Z θ ∗ − θ ∗ g ( π + 2 x ) d x = f min Z π + θ ∗ π − θ ∗ g ( u )d u. We deduce: f Φ ( u ) ≥ f min , ∀ u ∈ [ π − θ ∗ , π + θ ∗ ] . • Induction: let suppose that for some n ≥ , f Φ n ( u ) ≥ c n for all u ∈ [ a n , b n ] . Then,using the relationship (1) and the independence between Θ n and Φ n we have, for anymeasurable bounded function g : E [ g (Φ n +1 )] = E [ g ( π + 2Θ n + Φ n )] ≥ f min c n Z θ ∗ − θ ∗ Z b n a n g ( π + 2 θ + x )d x d θ. Using the substitution u = π + 2 θ + x in the integral with respect to x and Fubini’stheorem, we have: E [ g (Φ n +1 )] ≥ f min c n Z π + θ ∗ + b n π − θ ∗ + a n Z θ ∗ − θ ∗ ( u − π − b n ) ≤ θ ≤ ( u − π − a n ) ! g ( u )d u, and we deduce the following lower bound of the density function f Φ n +1 of Φ n +1 : f Φ n +1 ( u ) ≥ f min c n (cid:12)(cid:12)(cid:12)(cid:12)(cid:20) − θ ∗ , θ ∗ (cid:21) ∩ (cid:20)
12 ( u − π − b n ) ,
12 ( u − π − a n ) (cid:21)(cid:12)(cid:12)(cid:12)(cid:12) , for all u ∈ [ π − θ ∗ + a n , π + θ ∗ + b n ] .When u is equal to one extremal point of this interval, this lower bound is equal to . How-ever, let η n +1 ∈ (cid:0) , ( b n − a n ) (cid:1) , we have, for u ∈ [ π − θ ∗ + a n + η n +1 , π + θ ∗ + b n − η n +1 ] : f Φ n +1 ( u ) ≥ f min c n η n +1 . The results follows immediately.By choosing a constant sequence for the η k , k ≥ in the Proposition 3.2, we immediatelydeduce: Corollaire 3.3.
For all n ≥ , for all ε ∈ (0 , θ ∗ ) , we have f Φ n ( u ) ≥ (cid:18) f min (cid:19) n ε n − , ∀ u ∈ J n = [( n − π − nθ ∗ + φ + ( n − ε, ( n − π + nθ ∗ + φ − ( n − ε ] . Let ( J n ) n ≥ defined as in Corollary 3.3. We put J = I with I defined in Proposition 3.2.9 heorem 3.4. Let (Φ n ) n ≥ be the stochastic billiard Markov chain on the circle ∂ B r , satisfyingassumption ( H ′ ) .There exists a unique invariant probability measure ν on [0 , π ) for the Markov chain (Φ n ) n ≥ ,and we have:1. if θ ∗ > π , for all n ≥ , k P (Φ n ∈ · ) − ν k T V ≤ (1 − f min (2 θ ∗ − π )) n − ,
2. if θ ∗ ≤ π , for all n ≥ and all ε ∈ (0 , θ ∗ ) , k P (Φ n ∈ · ) − ν k T V ≤ (1 − α ) nn − , where n = (cid:22) π − ε θ ∗ − ε ) (cid:23) + 1 and α = (cid:16) ε (cid:17) n − f min n (2 n θ ∗ − n − ε − π ) . Proof.
The existence of the invariant measure is immediate thanks to the compactness of ∂ B r (see [4]). The following proof leads to its uniqueness and the speed of convergence.Let (Φ n , Θ n ) n ≥ and ( ˜Φ n , ˜Θ n ) n ≥ be two versions of the process described above, with initialpositions φ and ˜ φ on ∂ B r .In order to couple Φ n and ˜Φ n at some time n , it is sufficient to show that the intervals J n and ˜ J n corresponding to Corollary 3.3 have a non empty intersection. Since these intervals areincluded in [0 , π ) , a sufficient condition to have J n ∩ ˜ J n = ∅ is that the length of these twointervals is strictly bigger than π .Let ε ∈ (0 , θ ∗ ) . We have |J | = | ˜ J | = 2 θ ∗ , and for n ≥ , |J n | = | ˜ J n | = 2 nθ ∗ − n − ε. Therefore the length of J n is a strictly increasing function of n (which in intuitively clear). • Case 1: θ ∗ > π . In that case we have |J | = | ˜ J | > π . Therefore we can construct acoupling (cid:16) Φ , ˜Φ (cid:17) such that we have, using Proposition 3.2: P (cid:16) Φ = ˜Φ (cid:17) ≥ f min (cid:12)(cid:12)(cid:12) J ∩ ˜ J (cid:12)(cid:12)(cid:12) ≥ f min θ ∗ − π )= f min (2 θ ∗ − π ) . • Case 2: θ ∗ ≤ π . Here we need more jumps before having a positive probability to couple Φ n and ˜Φ n . Let thus define n = min { n ≥ nθ ∗ − n − ε > π } = (cid:22) π − ε θ ∗ − ε ) (cid:23) + 1 . Φ n obtained in Corollary 3.3, we deducethat we can construct a coupling (cid:16) Φ n , ˜Φ n (cid:17) such that: P (cid:16) Φ n = ˜Φ n (cid:17) ≥ (cid:18) f min (cid:19) n ε n − (cid:12)(cid:12)(cid:12) J n ∩ ˜ J n (cid:12)(cid:12)(cid:12) ≥ (cid:18) f min (cid:19) n ε n − n θ ∗ − n − ε − π )= (cid:16) ε (cid:17) n − ( f min ) n (2 n θ ∗ − n − ε − π ) . To treat both cases together, let define m = θ ∗ > π + (cid:18)(cid:22) π − ε θ ∗ − ε ) (cid:23) + 1 (cid:19) θ ∗ ≤ π . and α = f min (2 θ ∗ − π ) θ ∗ > π + (cid:16) ε (cid:17) m − ( f min ) m (2 m θ ∗ − m − ε − π ) θ ∗ ≤ π . We get: k P (Φ n ∈ · ) − ν k T V ≤ P (cid:16) Φ n = ˜Φ n (cid:17) ≤ P (cid:16) Φ ⌊ nm ⌋ m = ˜Φ ⌊ nm ⌋ m (cid:17) ≤ (1 − α ) ⌊ nm ⌋ ≤ (1 − α ) nm − . We assume here that the constant θ ∗ introduced in Assumption ( H ′ ) satisfies θ ∗ ∈ (cid:18) π , π (cid:19) . This condition on θ ∗ is essential in the proof of Theorem 3.7 to couple our processes with "twojumps". However, if θ ∗ ∈ (cid:0) , π (cid:3) we can adapt our method (see Remark 3.9). Notation : Let x ∈ ∂ B r . We write T xn an Φ xn respectively for the hitting time of ∂ B r andthe position of the Markov chain after n steps, and that started at position x .Let us remark that the distribution of T xn does not depend on x since we consider here thestochastic billiard in the disc, which is rotationally symmetric. Therefore, we allow us to omitthis x when it is not necessary for the comprehension. Proposition 3.5.
Let ( X t , V t ) t ≥ be the stochastic billiard process in the ball B r satisfyingAssumption ( H ′ ) with θ ∗ ∈ (cid:0) π , π (cid:1) .We denote by f T the density function of T . Let η ∈ (cid:0) , r (cid:0) − cos (cid:0) θ ∗ (cid:1)(cid:1)(cid:1) . We have f T ( x ) ≥ δ for all x ∈ [4 r cos (cid:18) θ ∗ (cid:19) + η, r − η ] , here δ = 2 f r sin (cid:0) θ ∗ (cid:1) min (cid:26) θ ∗ − arccos (cid:18) cos (cid:18) θ ∗ (cid:19) + η r (cid:19) ; arccos (cid:16) − η r (cid:17)(cid:27) . (2) Proof.
If the density function f is supported on (cid:2) − θ ∗ , θ ∗ (cid:3) , it is immediate to observe that r cos (cid:0) θ ∗ (cid:1) ≤ T ≤ r . But let be more precise.Let g : R → R be a bounded measurable function. Let us recall that, thanks to (1), T =2 r (cos(Θ ) + cos(Θ )) with Θ , Θ two independent random variables with density function f .We have, using Assumption ( H ′ ) : E [ g ( T )] = E [ g (2 r (cos(Θ ) + cos(Θ )))] ≥ f Z θ ∗ − θ ∗ Z θ ∗ − θ ∗ g (2 r (cos( u ) + cos( v ))) d u d v = 4 f Z θ ∗ Z θ ∗ g (2 r (cos( u ) + cos( v ))) d u d v. The substitution x = 2 r (cos( u ) + cos( v )) in the integral with respect to u gives then: E [ g ( T )] ≥ f Z θ ∗ Z r (1+cos( v ))2 r ( cos ( θ ∗ ) +cos( v ) ) g ( x ) 12 r sin (cid:0) arccos (cid:0) x r − cos( v ) (cid:1)(cid:1) d x d v. Fubini’s theorem leads to E [ g ( T )] ≥ f r Z r r cos ( θ ∗ ) Z θ ∗ q − (cid:0) x r − cos( v ) (cid:1) x r − < cos( v ) < x r − cos ( θ ∗ )d v g ( x )d x. We then deduce a lower-bound for the density function of T : f T ( x ) ≥ f r Z θ ∗ q − (cid:0) x r − cos( v ) (cid:1) x r − < cos( v ) < x r − cos ( θ ∗ )d v x ∈ ( r cos ( θ ∗ ) , r ) . Let x ∈ (cid:0) r cos (cid:0) θ ∗ (cid:1) , r (cid:1) . Cutting the interval (cid:0) r cos (cid:0) θ ∗ (cid:1) , r (cid:1) at point r (cid:0) (cid:0) θ ∗ (cid:1)(cid:1) , weget: f T ( x ) ≥ f r Z θ ∗ q − (cid:0) x r − cos( v ) (cid:1) x r − < cos( v ) < x r − cos ( θ ∗ )d v x ∈ ( r cos ( θ ∗ ) , r ( ( θ ∗ ))]+ 2 f r Z θ ∗ q − (cid:0) x r − cos( v ) (cid:1) x r − < cos( v ) < x r − cos ( θ ∗ )d v x ∈ [ r ( ( θ ∗ )) , r )= 2 f r Z θ ∗ arccos ( x r − cos ( θ ∗ )) 1 q − (cid:0) x r − cos( v ) (cid:1) d v x ∈ ( r cos ( θ ∗ ) , r ( ( θ ∗ ))]+ 2 f r Z arccos ( x r − ) q − (cid:0) x r − cos( v ) (cid:1) d v x ∈ [ r ( ( θ ∗ )) , r ) . v ∈ (cid:0) arccos (cid:0) x r − cos (cid:0) θ ∗ (cid:1)(cid:1) , θ ∗ (cid:1) we have cos( v ) ≤ x r − cos (cid:0) θ ∗ (cid:1) , and for v ∈ (cid:0) , arccos (cid:0) x r − (cid:1)(cid:1) we have cos( v ) ≤ . We thus have: f T ( x ) ≥ f r sin (cid:0) θ ∗ (cid:1) (cid:18) θ ∗ − arccos (cid:18) x r − cos (cid:18) θ ∗ (cid:19)(cid:19)(cid:19) x ∈ ( r cos ( θ ∗ ) , r ( ( θ ∗ ))]+ 2 f r arccos (cid:0) x r − (cid:1)q xr (cid:0) − x r (cid:1) x ∈ [ r ( ( θ ∗ )) , r ) . We can observe than the lower bound of f T is strictly positive for x ∈ (cid:0) r cos (cid:0) θ ∗ (cid:1) , r (cid:1) , butis equal to when x is one of the extremal points of this interval. Let therefore introduce η ∈ (cid:0) , r (cid:0) − cos (cid:0) θ ∗ (cid:1)(cid:1)(cid:1) . We have: • for x ∈ [4 r cos (cid:0) θ ∗ (cid:1) + η, r (cid:0) (cid:0) θ ∗ (cid:1)(cid:1) ] we have f r sin (cid:0) θ ∗ (cid:1) (cid:18) θ ∗ − arccos (cid:18) x r − cos (cid:18) θ ∗ (cid:19)(cid:19)(cid:19) ≥ f r sin (cid:0) θ ∗ (cid:1) θ ∗ − arccos r cos (cid:0) θ ∗ (cid:1) + η r − cos (cid:18) θ ∗ (cid:19)!! = 2 f r sin (cid:0) θ ∗ (cid:1) (cid:18) θ ∗ − arccos (cid:18) cos (cid:18) θ ∗ (cid:19) + η r (cid:19)(cid:19) • for x ∈ (cid:2) r (cid:0) (cid:0) θ ∗ (cid:1)(cid:1) , r − η (cid:3) we have f r arccos (cid:0) x r − (cid:1)q xr (cid:0) − x r (cid:1) ≥ f r arccos (cid:0) r − η r − (cid:1)s r ( ( θ ∗ )) r (cid:18) − r ( ( θ ∗ )) r (cid:19) = 2 f r arccos (cid:0) − η r (cid:1)q(cid:0) (cid:0) θ ∗ (cid:1)(cid:1) (cid:0) − cos (cid:0) θ ∗ (cid:1)(cid:1) = 2 f r sin (cid:0) θ ∗ (cid:1) arccos (cid:16) − η r (cid:17) . The result follows immediately.
Notation : For x ∈ ∂ B r , we denote by ϕ x the unique angle in [0 , π ) describing the positionof x on ∂ B r . Proposition 3.6.
Let ( X t , V t ) t ≥ be the stochastic billiard process in B r satisfying Assumption ( H ′ ) with θ ∗ ∈ (cid:0) π , π (cid:1) .For all ε ∈ (cid:0) , θ ∗ (cid:1) , the pair (Φ x , T x ) is f r sin ( θ ∗ ) -continuous on ( ϕ x − θ ∗ + 4 ε, ϕ x + θ ∗ − ε ) × (cid:0) r cos (cid:0) θ ∗ (cid:1) , r cos (cid:0) θ ∗ − ε (cid:1)(cid:1) for all x ∈ ∂ B (0 , r ) . roof. By symmetry of the process, it is sufficient to prove the lemma for x ∈ ∂ B r such that ϕ x = 0 , what we do.Let ε ∈ (cid:0) , π (cid:1) , A ⊂ ( − θ ∗ + 4 ε, θ ∗ − ε ) and ( r , r ) ⊂ (cid:0) r cos (cid:0) θ ∗ (cid:1) , r cos (cid:0) θ ∗ − ε (cid:1)(cid:1) .Let us recall that Φ = 2Θ +2Θ and T = 2 r (cos(Θ )+cos(Θ )) , where Θ , Θ are independentvariables with density function f . We thus have: P (cid:0) Φ ∈ A, T ∈ ( r , r ) (cid:1) = P (2Θ + 2Θ ∈ A, r (cos(Θ ) + cos(Θ )) ∈ ( r , r ))= Z π − π Z π − π u +2 v ∈ A cos( u )+cos( v ) ∈ ( r r , r r ) f ( u ) f ( v )d u d v ≥ f Z θ ∗ − θ ∗ Z θ ∗ − θ ∗ u + v ∈ A cos ( u + v ) cos ( u − v ) ∈ ( r r , r r )d u d v. Let us consider g : ( u, v ) ∈ (cid:20) − θ ∗ , θ ∗ (cid:21) (cid:18) u + v , u − v (cid:19) . We have (cid:20) − θ ∗ , θ ∗ (cid:21) ⊂ g (cid:20) − θ ∗ , θ ∗ (cid:21) ! , and | det Jac g | = 12 . With this substitution, and using Fubini’s theorem, we get: P (cid:0) Φ ∈ A, T ∈ ( r , r ) (cid:1) ≥ f Z θ ∗ − θ ∗ Z θ ∗ − θ ∗ x ∈ A cos( x ) cos( y ) ∈ ( r r , r r )2d x d y = 4 f Z θ ∗ − θ ∗ Z θ ∗ cos( x ) cos( y ) ∈ ( r r , r r )d y x ∈ A d x. We now do the substitution z = cos( x ) cos( y ) in the integral with respect to d y : P (cid:0) Φ ∈ A, T ∈ ( r , r ) (cid:1) ≥ f Z θ ∗ − θ ∗ Z cos( x )cos ( θ ∗ ) cos( x ) z ∈ ( r r , r r ) 1 p cos ( x ) − z d z x ∈ A d x ≥ f Z θ ∗ − θ ∗ Z cos( x )cos ( θ ∗ ) cos( x ) z ∈ ( r r , r r ) 1sin (cid:0) θ ∗ (cid:1) d z x ∈ A d x ≥ f sin (cid:0) θ ∗ (cid:1) Z θ ∗ − ε − θ ∗ + ε Z cos ( θ ∗ − ε ) cos ( θ ∗ ) z ∈ ( r r , r r )d z x ∈ A d x = f r sin (cid:0) θ ∗ (cid:1) ( r − r ) | A | , A ⊂ [ − θ ∗ + 4 ε, θ ∗ − ε ) and ( r , r ) ⊂ (cid:0) r cos (cid:0) θ ∗ (cid:1) , r cos (cid:0) θ ∗ − ε (cid:1)(cid:1) .This ends the proof.Let fix η ∈ (0 , r (cid:0) − (cid:0) θ ∗ (cid:1)(cid:1) ) and ε ∈ (cid:0) , θ ∗ − π (cid:1) (the condition θ ∗ > π ensures that wecan take such η and ε ).Let define h = 4 r (cid:18) − cos (cid:18) θ ∗ (cid:19)(cid:19) − η − r = 2 r (cid:18) − (cid:18) θ ∗ (cid:19)(cid:19) − η > (3)and α = f r sin (cid:0) θ ∗ (cid:1) (4 θ ∗ − π − ε )2 r (cid:18) cos (cid:18) θ ∗ − ε (cid:19) − cos (cid:18) θ ∗ (cid:19)(cid:19) = f sin (cid:0) θ ∗ (cid:1) (4 θ ∗ − π − ε ) (cid:18) cos (cid:18) θ ∗ − ε (cid:19) − cos (cid:18) θ ∗ (cid:19)(cid:19) (4) Theorem 3.7.
Let ( X t , V t ) t ≥ be the stochastic billiard process in B r satisfying Assumption ( H ′ ) with θ ∗ ∈ (cid:0) π , π (cid:1) .There exists a unique invariant probability measure on B r × S for the process ( X t , V t ) t ≥ .Moreover let η ∈ (0 , r (cid:0) − (cid:0) θ ∗ (cid:1)(cid:1) ) and ε ∈ (cid:0) , θ ∗ − π (cid:1) . For all t ≥ and all λ < λ M wehave k P ( X t ∈ · , V t ∈ · ) − χ k T V ≤ C λ e − λt , where λ M = min ( r log (cid:18) − δh (cid:19) ; 14 r log − (1 − δh ) + p (1 − δh ) + 4 δh (1 − α )2 δh (1 − α ) !) . (5) and C λ = αδh e λr − e λr (1 − δh ) − e λr δh (1 − α ) , with δ , h and α respectively given by (2) , (3) and (4) . Remark 3.8.
The following proof of this theorem is largely inspired by the proof of Theorem2.2 in [2].Proof.
The existence of the invariant probability measure comes from the compactness of thespace B r × S . The following proof show its uniqueness and gives the speed of convergence ofthe stochastic billiard to equilibrium.Let ( X t , V t ) t ≥ and ( ˜ X t , ˜ V t ) t ≥ be two versions of the stochastic billiard with ( X , V ) = ( x , v ) ∈B r × S and ( ˜ X , ˜ V ) = (˜ x , ˜ v ) ∈ B r × S .We recall the definition of T and ˜ T and define w, ˜ w as follows: T = inf { t ≥ x + tv / ∈ K } , w = x + T v ∈ ∂ B r , and ˜ T = inf { t ≥ x + t ˜ v / ∈ K } , ˜ w = ˜ x + ˜ T ˜ v ∈ ∂Br.
15e are going to couple ( X t , V t ) and ( ˜ X t , ˜ V t ) in two steps: we first couple the times, so that thetwo processes hit ∂ B r at a same time, and then we couple both position and time.In the sequel, we write X aT n or ˜ X aT n for the position of the Markov chain at time T n when itstarts at position a ∈ ∂ B r . Similarly, we write T an and ˜ T an for the successive hitting times of ∂ B r of the processes. Step 1 . Proposition 3.5 ensures that T w and ˜ T ˜ w are both δ -continuous on [4 r cos (cid:0) θ ∗ (cid:1) + η, r − η ] .Therefore, the variables T + T w and ˜ T + ˜ T ˜ w are δ -continuous on [ T + 4 r cos (cid:0) θ ∗ (cid:1) + η, T + 4 r − η ] ∩ [ ˜ T + 4 r cos (cid:0) θ ∗ (cid:1) + η, ˜ T + 4 r − η ] , with (cid:12)(cid:12)(cid:12) [ T + 2 r cos (cid:0) θ ∗ (cid:1) + η, T + 4 r − η ] ∩ [ ˜ T + 2 r cos (cid:0) θ ∗ (cid:1) + η, ˜ T + 4 r − η ] (cid:12)(cid:12)(cid:12) ≥ h since | T − ˜ T |≤ r .Note that the condition θ ∗ > π has been introduced to ensure that this intersection is non-empty.Thus, there exists a coupling of T + T w and ˜ T + ˜ T ˜ w such that P ( E ) ≥ δh, where E = n T + T w = ˜ T + ˜ T ˜ w o . On the event E we define T c = T + T w .On the event E c , we can suppose, by symmetry that T + T w ≤ ˜ T + ˜ T ˜ w . In order to try againto couple the hitting times, we need to begin at times whose difference is smaller than r . Letthus define m = min n n > T + T w + T X T Tw n > ˜ T + ˜ T ˜ w o and ˜ m = 0 . We then have (cid:12)(cid:12)(cid:12)(cid:12)(cid:16) T + T w + T X T Tw m (cid:17) − (cid:18) ˜ T + ˜ T ˜ w + ˜ T ˜ X ˜ T
0+ ˜ T ˜ w ˜ m (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ r .Defining Z = X T + T w , Z = X T + T w + T Z m , ˜ Z = ˜ X ˜ T + ˜ T ˜ w , ˜ Z = ˜ X ˜ T + ˜ T ˜ w + ˜ T ˜ Z m , we obtain as previously: P ( E | E c ) ≥ δh, where E = n T + T w + T Z m + T Z = ˜ T + ˜ T ˜ w + ˜ T ˜ Z ˜ m + ˜ T ˜ Z o . On the event E c ∩ E we define T c = T + T w + T Z m + T Z . We thus have T c L = T + R + R ,with R , R independent variables with distribution f T .We then repeat the same procedure. We thus construct two sequences of stopping times ( m k ) k ≥ , ( ˜ m k ) k ≥ and a sequence of events ( E k ) k ≥ satisfying P (cid:0) E k | E c ∩ · · · ∩ E ck − (cid:1) ≥ δh. On the event E c ∩· · ·∩ E ck − ∩ E k we define T c as previously, and we have T c L = T + R + · · · + R k with R , · · · , R k independent variables with distribution f T . By construction, T c is the cou-pling time of the hitting times of the boundary.16 tep 2 . Let us now work conditionally on T c .Let define y = X wT c and ˜ y = ˜ X ˜ wT c in order to simplify the notations. By construction of T c , y and ˜ y are on ∂ B r . We define N c = min (cid:8) n > X wT n = y (cid:9) , i.e. T c is the time at which thechain starting at w hit the boundary for the N c -th time.Proposition 3.6 ensures that the couples (cid:0) X yT , T y (cid:1) and (cid:16) ˜ X ˜ yT , ˜ T ˜ Xy (cid:17) are both f r sin ( θ ∗ ) -continuouson the set (( ϕ y − θ ∗ + 4 ε, ϕ y + θ ∗ − ε ) ∩ ( ϕ ˜ y − θ ∗ + 4 ε, ϕ ˜ y + θ ∗ − ε )) × (cid:0) √ r, (cid:0) π − ε (cid:1) r (cid:1) ,with | ( ϕ y − θ ∗ + 4 ε, ϕ y + θ ∗ − ε ) ∩ ( ϕ ˜ y − θ ∗ + 4 ε, ϕ ˜ y + θ ∗ − ε ) | ≥ θ ∗ − π − ε . Note thatthe condition θ ∗ > π implies in particular that the previous intersection in non-empty.Therefore we can construct a coupling such that P (cid:16) F | E c ∩ · · · ∩ E cN c − ∩ E N c (cid:17) ≥ α, where F = n X yT = ˜ X ˜ yT and T y = ˜ T ˜ y o . On the event F we define T c = T c + T y .If F does not occur, we can not directly try to couple both position and time since the twoprocesses have not necessarily hit ∂ B r at the same time. We thus have to couple first the hittingtimes, as we have done in step 1.Let suppose that on (cid:16) E c ∩ · · · ∩ E cN c − ∩ E N c (cid:17) ∩ F c , we have T y ≤ ˜ T ˜ y (the other case can betreated in the same way thanks to the symmetry of the problem). Let define ℓ = min (cid:26) n > T y + T X Ty n > ˜ T ˜ y (cid:27) and ˜ ℓ = 0 We clearly have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) T y + T X Ty ℓ − ˜ T ˜ y + ˜ T ˜ X ˜ T ˜ y ˜ ℓ !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ r . Therefore, we can start again: we try tocouple the times at which the two processes hit the boundary, and then to couple the positionsand times together.Finally, the probability that we succeed to couple the positions and times in "one step" is: P (cid:18)(cid:18) ∪ k ≥ (cid:0) E c ∩ · · · ∩ E ck − ∩ E k (cid:1)(cid:19) ∩ F (cid:19) = P (cid:18) F (cid:12)(cid:12)(cid:12)(cid:12) ∪ k ≥ (cid:0) E c ∩ · · · ∩ E ck − ∩ E k (cid:1) (cid:19) P (cid:18) ∪ k ≥ (cid:0) E c ∩ · · · ∩ E ck − ∩ E k (cid:1)(cid:19) = P (cid:18) F (cid:12)(cid:12)(cid:12)(cid:12) ∪ k ≥ (cid:0) E c ∩ · · · ∩ E ck − ∩ E k (cid:1) (cid:19) ≥ α. Thus, the coupling time ˆ T of the couples position-time satisfies: ˆ T ≤ st T + G X k =1 G k X l =1 T k,l + T k G ∼ G ( α ) , G , G , · · · ∼ G ( δh ) are independent geometric variables, and (cid:0) T k,l (cid:1) k,l ≥ , (cid:0) T k (cid:1) k ≥ are independent random variables, independent from the geometric variables, withdistribution f T .Let λ ∈ (0 , λ M ) , with λ M defined in equation (5). Since all the random variables T k,l and T k , k, l ≥ , are almost surely smaller than two times the diameter of the ball B r , and since T isalmost surely smaller than this diameter, we have: P (cid:16) ˆ T > t (cid:17) ≤ e − λt E h e λ ˆ T i ≤ e λ ( T − t ) E exp λ G X k =1 G k X l =1 T k,l + T k ≤ e λ (2 r − t ) E G Y k =1 G k Y l =1 exp ( λ r ) exp ( λ r ) = e λ (2 r − t ) E " G Y k =1 E h e λr ( G k +1) i . Now, using the expression of generating function of a geometric random variable we get: P (cid:16) ˆ T > t (cid:17) ≤ e λ (2 r − t ) E " G Y k =1 ∞ X l =1 e λr ( l +1) δh (1 − δh ) l − ! = e λ (2 r − t ) E "(cid:18) e λr δh − e λr (1 − δh ) (cid:19) G = e λ (2 r − t ) α e λr δh − e λr (1 − δh ) 11 − e λr δh (1 − α )1 − e λr (1 − δh ) = e − λt α e λr δh − e λr (1 − δh ) − e λr δh (1 − α ) . This calculations are valid for λ > such that the generating functions are well defined,that is for λ > satisfying e λr (1 − δh ) < and e λr δh (1 − α )1 − e λr (1 − δh ) < . The first condition is equivalent to λ < r log (cid:0) − δh (cid:1) .The second condition is equivalent to δh (1 − α ) s + (1 − δh ) s − < with s = e λr . It gives s where ∆ = (1 − δh ) +4 δh (1 − α ) > .And finally we get λ < r log ( s ) .Therefore, the estimation for P (cid:16) ˆ T > t (cid:17) is indeed valid for all λ ∈ (0 , λ M ) . The conclusion ofthe theorem follows immediately. Remark 3.9. If θ ∗ ∈ (cid:0) , π (cid:3) , Step of the proof of Theorem 3.7 fails: the intervals on whichthe random variables T + T w and ˜ T + ˜ T ˜ w are continuous can have an empty intersection. imilarly, in Step , the intersection of the intervals on which the couples (cid:18) X X wT c T , T X wT c (cid:19) and (cid:18) ˜ X ˜ X ˜ wT c T , ˜ T ˜ X ˜ wT c (cid:19) are continuous can be empty if θ ∗ ≤ π .However, instead of trying to couple the times or both positions and times in two jumps, wejust need more jumps to do that. Therefore, the method and the results are similar in the case θ ∗ ≤ π , the only difference is that the computations and notations will be much more awful. We make the following assumption on the set K in which the stochastic billiard evolves: Assumption ( K ) : K is a compact convex set with curvature bounded from above by C < ∞ andbounded from below by c > .This means that for each x ∈ ∂K , there is a ball B with radius C included in K and a ball B containing K , so that the tangent planes of K , B and B at x coincide (see Figure 4). In fact,for x ∈ ∂K , the ball B is the ball with radius C and with center the unique point at distance C from x in the direction of n x . And B is the one with the center at distance c from x in thedirection of n x .In this section, we consider the stochastic billiard in such a convex K .Let us observe that the case of the disc is a particular case. Moreover, Assumption ( K ) excludesin particular the case of the polygons: because of the upper bound C on the curvature, theboundary of K can not have "corners", and because of the lower bound c , the boundary cannot have straight lines.In the following, D will denote the diameter of K , that is D = max {k x − y k : x, y ∈ ∂K } . Notation : We define l x,y = y − x k x − y k = − l y,x and we denote by ϕ x,y the angle between l x,y andthe normal n x to ∂K at the point x (see Figure 5).The following property, proved by Comets and al. in [2], gives the dynamics of the Markovchain ( X T n ) n ≥ defined in Section 2.2 Proposition 4.1.
The transition kernel of the chain ( X T n ) n ≥ is given by: P (cid:0) X T n +1 ∈ A | X T n = x (cid:1) = Z A Q ( x, y )d y where Q ( x, y ) = ρ ( U − x l x,y ) cos( ϕ y,x ) k x − y k . Ke C c B B Figure 4: Illustration of Assumption ( K ) xy ϕ y,x ϕ x,y l y,x K Figure 5: Definition of the quantities ϕ x,y and l y,x for x, y ∈ ∂K Theorem 4.2.
Let K ∈ R satisfying Assumption ( K ) with diameter D . Let ( X T n ) n ≥ be thestochastic billiard Markov chain on ∂K verifying Assumption ( H ) .There exists a unique invariant measure ν on ∂K for ( X T n ) n ≥ .Moreover, recalling that θ ∗ = |J | in Assumption ( H ) , we have:1. if θ ∗ > C | ∂K | , for all n ≥ , k P ( X T n ∈ · ) − ν k T V ≤ (cid:18) − q min (cid:18) θ ∗ C − | ∂K | (cid:19)(cid:19) n − ;
2. if θ ∗ ≤ C | ∂K | , for all n ≥ and all ε ∈ (cid:0) , θ ∗ C (cid:1) , k P ( X T n ∈ · ) − ν k T V ≤ (1 − α ) nn − where n = $ | ∂K | − ε θ ∗ C − ε % + 1 and α = ( 4 θ ∗ C ) n − q min n (cid:18) (cid:18) n θ ∗ C − ( n − ε (cid:19) − | ∂K | (cid:19) with q min = cρ min cos (cid:0) θ ∗ (cid:1) CD .
Proof.
Once more, the existence of the invariant measure is immediate since the state space ∂K of the Markov chain is compact. The following shows its uniqueness and gives the speedof convergence of ( X T n ) n ≥ towards ν .Let ( X T n ) n ≥ and ( ˜ X T n ) n ≥ be two versions of the Markov chain with initial conditions x and ˜ x on ∂K . In order to have a strictly positive probability to couple X T n and ˜ X T n at time n , it issufficient that their density functions are bounded from below on an interval of length strictlybigger than | ∂K | . Let us therefore study the length of set on which f X Tn is bounded from belowby a strictly positive constant.Let x ∈ ∂K . For v ∈ S x , we denote by h x ( v ) the unique point on ∂K seen from x in thedirection of v . We firstly get a lower bound on | h x ( U x J ) | , the length of the subset of ∂K seenfrom x with a strictly positive density.It is easy to observe, with a drawing for instance, the following facts: • | h x ( U x J ) | increases when k x − h x ( n x ) k increases, • | h x ( U x J ) | decreases when the curvature at h x ( n x ) increases, • | h x ( U x J ) | decreases when | ϕ h x ( n x ) ,x | increases.Therefore, | h x ( U x J ) | is minimal when k x − h x ( n x ) k is minimal, when the curvature at h x ( n x ) ismaximal, and then equal to C , and finally when ϕ h x ( n x ) ,x = 0 . Moreover, the minimal value of k x − h x ( n x ) k is C since C is the upper bound for the curvature of ∂K . The configuration that21 = h x ( n x ) xθ ∗ h x ( U x J ) C Figure 6: Worst scenario for the length of h x ( U x J ) makes the quantity | h x ( U x J ) | minimal is thus the case where x and h x ( n x ) define a diameteron a circle of diameter C (see Figure 6). We immediately deduce a lower bound for | h x ( U x J ) | : | h x ( U x J ) |≥ θ ∗ × C = 4 θ ∗ C .
This means that the density function f X T of X T is strictly positive on a subset of ∂K oflength at least θ ∗ C .Let now ε ∈ (cid:0) , θ ∗ C (cid:1) . As it has been done in Section 3 for the disc, we can deduce that for all n ≥ , the density function f X Tn is strictly positive on a set of length at least nθ ∗ C − n − ε = nθ ∗ C − n − ε .Let define, for x ∈ ∂K and n ≥ , J nx the set of points of ∂K that can be reached from x in n bounces by picking for each bounce a velocity in J .We now separate the cases where we can couple in one jump, and where we need more jumps. • Case 1: θ ∗ > C | ∂K | . In that case we have, for all x ∈ ∂K , |J x |≥ θ ∗ C > | ∂K | , and we canthus construct a coupling ( X T , ˜ X T ) such that: P (cid:16) X T = ˜ X T (cid:17) ≥ q min (cid:12)(cid:12)(cid:12) J x ∩ ˜ J x (cid:12)(cid:12)(cid:12) ≥ q min × (cid:18) θ ∗ C − | ∂K | (cid:19) = q min (cid:18) θ ∗ C − | ∂K | (cid:19) , where q min is a uniform lower bound of Q ( a, b ) with a ∈ ∂K and b ∈ h a ( U a J ) , i.e. q min ≤ min a ∈ ∂K,b ∈ h a ( U a J ) Q ( a, b ) . Let thus give an explicit expression for q min . Let a ∈ ∂K and b ∈ h a ( U a J ) . We have Q ( a, b ) ≥ ρ min cos ( ϕ b,a ) D .
We could have cos ( ϕ b,a ) = 0 if a and b were on a straight part of ∂K , which is not possiblesince the curvature of K is bounded from below by c . Thus, the quantity cos ( ϕ b,a ) is22 a c n b δϕ b,a K ab θ ∗ C δ min Figure 7: Illustration for the calculation of a lower bound for cos ( ϕ b,a ) with a ∈ ∂K and b ∈ h a ( U a J ) minimal when a and b are on a part of a disc with curvature c . In that case, cos ( ϕ b,a ) = δc , where δ is the distance between a and b (see the first picture of Figure 7). Since b ∈ h a ( U a J ) , we have δ ≥ δ min := ( θ ∗ ) C (see the second picture of Figure 7). Finallywe get Q ( a, b ) ≥ cρ min cos (cid:0) θ ∗ (cid:1) CD =: q min . • Case 2: θ ∗ ≤ C | ∂K | . In that case, we need more than one jump to couple the two Markovchains. Therefore, defining n = min (cid:26) n ≥ nθ ∗ C − n − ε > ∂K (cid:27) = $ | ∂K | − ε θ ∗ C − ε % + 1 , we get that the intersection J n x ∩ ˜ J n ˜ x is non-empty, and then we can construct X T n and ˜ X T n such that the probability P (cid:16) X T n = ˜ X T n (cid:17) is strictly positive. It remains toestimate a lower bound of this probability.First, we have (cid:12)(cid:12)(cid:12) J n x ∩ ˜ J n ˜ x (cid:12)(cid:12)(cid:12) ≥ (cid:18) n θ ∗ C − n − ε − | ∂K | (cid:19) = 4 (cid:18) n θ ∗ C − ( n − ε (cid:19) − | ∂K | . x ∈ { x , ˜ x } and y ∈ J n x ∩ ˜ J n ˜ x . We have: Q n ( x, y ) ≥ Z h x ( U x J ) Z h z ( U z J ) · · · Z h zn − ( U zn − J ) Q ( x, z ) Q ( z , z ) · · · Q ( z n − , y )d z d z · · · d z n − ≥ ( 4 θ ∗ C ) n − q min n . We thus deduce: P (cid:16) X T n = ˜ X T n (cid:17) ≥ ( 4 θ ∗ C ) n − q min n (cid:12)(cid:12)(cid:12) J n x ∩ ˜ J n ˜ x (cid:12)(cid:12)(cid:12) ≥ ( 4 θ ∗ C ) n − q min n (cid:18) (cid:18) n θ ∗ C − ( n − ε (cid:19) − | ∂K | (cid:19) . We can now conclude, including the two cases: let define m = θ ∗ > C | ∂K | + $ | ∂K | − ε θ ∗ C − ε % + 1 ! θ ∗ ≤ C | ∂K | and α = q min (cid:18) θ ∗ C − | ∂K | (cid:19) θ ∗ > C | ∂K | +( 4 θ ∗ C ) m − q min m (cid:18) (cid:18) m θ ∗ C − ( m − ε (cid:19) − | ∂K | (cid:19) θ ∗ ≤ C | ∂K | . We have proved that we can construct a coupling (cid:16) X T m , ˜ X T m (cid:17) such that P (cid:16) X T m = ˜ X T m (cid:17) ≥ α , and then we get k P ( X T n ∈ · ) − ν k T V ≤ (1 − α ) nm − . In this section, we suppose |J | = θ ∗ = π . Proposition 4.3.
Let K ⊂ R satisfying Assumption ( K ) . Let ( X t , V t ) t ≥ the stochastic billiardprocess evolving in K and verifying Assumption ( H ) with |J | = π .For all x ∈ ∂K , the first hitting-time T x of ∂K starting at point x is cρ min -continuous on (cid:2) , C (cid:3) .Proof. Let x ∈ ∂K . Let us recall that the curvature of K is bounded from above by C , whichmeans that for each x ∈ ∂K , there is a ball B with radius C included in K so that the tangentplanes of K and B at x coincide. Therefore, starting from x , the maximal time to go onanother point of ∂K is bigger than C (the diameter of the ball B ).That is why we are going to prove the continuity of T x on the interval (cid:2) , C (cid:3) . Let thus ≤ r ≤ R ≤ C .Let Θ be a random variable living in (cid:2) − π , π (cid:3) such that the velocity vector (cos(Θ) , sin(Θ)) follows the law γ .The time T x being completely determined by the velocity V and thus by its angle with respectto n x , it is clear that there exist − π ≤ θ ≤ θ ≤ θ ≤ θ ≤ π such that we have: P ( T x ∈ [ r, R ]) = P (Θ ∈ [ θ , θ ] ∪ [ θ , θ ]) . ( H ) on the law γ , and since we assume here that |J | = π , thedensity function of Θ is bounded from below by ρ min on (cid:2) − π , π (cid:3) . It gives: P ( T x ∈ [ r, R ]) ≥ ρ min ( θ − θ + θ − θ ) . Moreover, since the curvature is bounded from below by c , there exists a ball B with radius c containing K so that the tangent planes of K and B at x coincide. And it is easy to see thatthe differences θ − θ and θ − θ are larger than the difference α − α where α and α arethe angles corresponding to the distances r and R starting from x and to arrive on the ball B .The time of hitting the boundary of B is equal to d ∈ (cid:2) , C (cid:3) if the angle between n x and thevelocity is equal to arccos (cid:0) cd (cid:1) . We thus deduce: P ( T x ∈ [ r, R ]) ≥ ρ min (cid:18) arccos (cid:16) cr (cid:17) − arccos (cid:18) cR (cid:19)(cid:19) ≥ ρ min (cid:12)(cid:12)(cid:12)(cid:12) cr − cR (cid:12)(cid:12)(cid:12)(cid:12) = ρ min c ( R − r ) , where we have used the mean value theorem for the second inequality.Let us introduce some constants that will appear in the following results.Let β > and δ > such that | ∂K | − max { δ ; β + δ } > .Let ε ∈ (cid:0) , min { β ; C } (cid:1) such that h > where h = δD (cid:18) βc (cid:19) − εM, (6)with M = 2 (cid:18) C − ε + 1 β − ε + C (cid:19) . (7)Let us remark that M is non decreasing with ε , so that it is possible to take ε small enoughto have h > . Proposition 4.4.
Let K ⊂ R satisfying Assumption ( K ) with diameter D . Let ( X t , V t ) t ≥ thestochastic billiard process evolving in K and verifying Assumption ( H ) with |J | = π .Let x, ˜ x ∈ ∂K with x = ˜ x .There exist R > , R > and J ∗ ⊂ ∂K , with | J ∗ | < hε , such that the couples ( X xT , T x ) and ( ˜ X ˜ x ˜ T , ˜ T ˜ x ) are both η -continuous on J ∗ × ( R , R ) , with η = 12 (cid:16) cρ min D (cid:17) (cid:18) C − ε (cid:19) ( β − ε ) . Moreover we have R − R ≥ hε − | J ∗ | ) . Remark 4.5.
The following proof is largely inspired by the proof of Lemma 5.1 in [2]. roof. Let x, ˜ x ∈ ∂K , x = ˜ x . Let us denotes by ∆ x ˜ x the bisector of the segment defined by thetwo points x and ˜ x . The intersection ∆ x ˜ x ∩ ∂K contains two points, let thus define ¯ y the onewhich achieves the larger distance towards x and ˜ x (we consider this point of intersection sincewe need in the sequel to have a lower bound on k x − ¯ y k and k ˜ x − ¯ y k ).Let t ∈ I g ( t ) be a parametrization of ∂K with g (0) = ¯ y , such that k g ′ ( t ) k = 1 for all t ∈ I .Consequently, the length of an arc satisfies length ( g | [ s,t ] ) = || g ( t ) − g ( s ) || = | t − s | . We canthus write I = [0 , | ∂K | ] , and g (0) = g ( | ∂K | ) . Note that the parametrization g is C thanks toAssumption ( K ) .In the sequel, for z ∈ ∂K , we write s z (or t z ) for the unique s ∈ I such that g ( s ) = z . And for A ⊂ ∂K , we define I A = { t ∈ I : g ( t ) ∈ A } .Let define, for s, t ∈ I and w ∈ { x, ˜ x } : ϕ w ( s, t ) = k w − g ( s ) k + k g ( s ) − g ( t ) k . Lemma 4.6.
There exists an interval I ∗ β,δ ⊂ I , satisfying | I ∗ β,δ | < hε , such that for w ∈ { x, ˜ x } : | ∂ s ϕ w ( s, t ) |≥ h, for s ∈ B ε ¯ y and t ∈ I ∗ β,δ , where B ε ¯ y = { s ∈ I ; | s − s ¯ y |≤ ε } . We admit this lemma for the moment and prove it after the end of the current proof.Let suppose for instance that ∂ s ϕ w ( s, t ) is positive for s ∈ B ε ¯ y and t ∈ I ∗ β,δ , for w = x and w = ˜ x . If one or both of ∂ s ϕ x ( s, t ) and ∂ s ϕ ˜ x ( s, t ) are negative, we just need to consider | ϕ x | or | ϕ ˜ x | , and everything works similarly.We thus have, by the lemma: ∂ s ϕ w ( s, t ) ≥ h, for s ∈ B ε ¯ y and t ∈ I ∗ β,δ . Let us now define: r = sup t ∈ I ∗ β,δ inf s ∈ B ε ¯ y ϕ x ( s, t ) and r = inf t ∈ I ∗ β,δ sup s ∈ B ε ¯ y ϕ x ( s, t ) and ˜ r = sup t ∈ I ∗ β,δ inf s ∈ B ε ¯ y ϕ ˜ x ( s, t ) and ˜ r = inf t ∈ I ∗ β,δ sup s ∈ B ε ¯ y ϕ ˜ x ( s, t ) . Since s ϕ x ( s, t ) and s ϕ ˜ x ( s, t ) are strictly increasing on B ε ¯ y for all t ∈ I ∗ β,δ , we deduce that,considering B ε ¯ y as the interval ( s , s ) , r = sup t ∈ I ∗ β,δ ϕ x ( s , t ) , r = inf t ∈ I ∗ β,δ ϕ x ( s , t ) , ˜ r = sup t ∈ I ∗ β,δ ϕ ˜ x ( s , t ) , ˜ r = inf t ∈ I ∗ β,δ ϕ ˜ x ( s , t ) . Lemma 4.7.
We have r < r and ˜ r < ˜ r .Moreover, there exist R , R with ≤ R < R satisfying R − R ≥ hε − | I ∗ β,δ | ) , such that ( r , r ) ∩ ( ˜ r , ˜ r ) = ( R , R ) . We admit this result to continue the proof, and will give a demonstration later.We can now prove that the pairs (cid:0) X xT , T x (cid:1) and (cid:16) ˜ X ˜ x ˜ T , ˜ T ˜ x (cid:17) are both η -continuous on I ∗ β,δ × ( R , R ) with some η > that we are going to define after the computations.We first prove that (cid:0) X xT , T x (cid:1) is η -continuous on I ∗ β,δ × ( r , r ) . By the same way we can26 x ˜ x ∆ x, ˜ x ¯ y ¯ y u Figure 8: Upper bound for the distance k w − ¯ y k , w ∈ { x, ˜ x } prove that (cid:16) ˜ X ˜ x ˜ T , ˜ T ˜ x (cid:17) is η -continuous on I ∗ β,δ × (˜ r , ˜ r ) . These two facts imply immediately thecontinuity with ( R , R ) since the interval ( R , R ) is included in ( r , r ) and (˜ r , ˜ r ) .Let ( u , u ) ⊂ ( r , r ) and A ⊂ I ∗ β,δ . We have: P (cid:0) X xT ∈ A, T x ∈ ( u , u ) (cid:1) ≥ Z I A Z B ε ¯ y Q ( x, g ( s )) Q ( g ( s ) , g ( t )) ϕ x ( s,t ) ∈ ( u ,u ) d s d t. Let s ∈ B ε ¯ y and t ∈ I ∗ β,δ . We now give a lower bound of Q ( x, g ( s )) and Q ( g ( s ) , g ( t )) .Proposition 4.1 gives: Q ( x, g ( s )) = ρ ( U − x l x,g ( s ) ) cos (cid:0) ϕ g ( s ) ,x (cid:1) k x − g ( s ) k≥ cρ min D (cid:18) C − ε (cid:19) , where we have used the same method as in he proof of Theorem 4.2 (with Figure 7) to getthat cos (cid:0) ϕ g ( s ) ,x (cid:1) ≥ k x − g ( s ) k c , and then the fact that k x − g ( s ) k≥ C − ε . Let us prove thislatter. With the notations of Figure 8, by Pythagore’s theorem we have, for ¯ y ∈ { ¯ y , ¯ y } , k x − ¯ y k = (cid:16) k x − ˜ x k (cid:17) + k u − ¯ y k . Moreover, since the curvature of K is bounded by C , it followsthat k ¯ y − ¯ y k≥ C , and then max {k u − ¯ y k ; k u − ¯ y k} ≥ C . We deduce: max {k x − ¯ y k ; k x − ¯ y k} ≥ C . Therefore, by the definition of ¯ y , we have k x − ¯ y k≥ C . Thus, the reverse triangle inequalitygives, for s ∈ B ε ¯ y , k x − g ( s ) k≥ C − ε .By the same way, since k g ( t ) − g ( s ) k≥ β − ε , we have: Q ( g ( s ) , g ( t )) ≥ cρ min D ( β − ε ) . Therefore we get: P (cid:0) X xT ∈ A, T x ∈ ( u , u ) (cid:1) ≥ a Z I A Z B ε ¯ y ϕ x ( s,t ) ∈ ( u ,u ) d s d t, a = (cid:16) cρ min D (cid:17) (cid:18) C − ε (cid:19) ( β − ε ) . (8)Let define, for t ∈ I ∗ β,δ : M x,t ( u , u ) := (cid:8) s ∈ B ε ¯ y : ϕ x ( s, t ) ∈ ( u , u ) (cid:9) . Using the fact that s ϕ x ( s, t ) is strictly increasing on B ε ¯ y for t ∈ I ∗ β,δ we get ( ϕ − w ( s, t ) standsfor the inverse function of s ϕ x ( s, t ) ): | M x,t ( u , u ) | = (cid:12)(cid:12)(cid:8) s ∈ B ε ¯ y : s ∈ (cid:0) ϕ − x ( u , t ) , ϕ − x ( u , t ) (cid:1)(cid:9)(cid:12)(cid:12) = (cid:12)(cid:12) ( s , s ) ∩ (cid:0) ϕ − x ( u , t ) , ϕ − x ( u , t ) (cid:1)(cid:12)(cid:12) . By definition of r and r , and since ( u , u ) ⊂ ( r , r ) we have: ϕ x ( s , t ) ≤ r ≤ u and ϕ x ( s , t ) ≥ r ≥ u , and since s ϕ x ( s, t ) is strictly increasing: s ≤ ϕ − x ( u , t ) and s ≥ ϕ − x ( u , t ) . Therefore we deduce: | M x,t ( u , u ) | = (cid:12)(cid:12)(cid:0) ϕ − x ( u , t ) , ϕ − x ( u , t ) (cid:1)(cid:12)(cid:12) = (cid:12)(cid:12) ϕ − x (( u , u ) , t ) (cid:12)(cid:12) ≥
12 ( u − u ) . For the last inequality we have used the following property. Let ψ : R R a function. Iffor all x ∈ [ a , a ] we have c < ψ ′ ( x ) < c with < c < c < ∞ , then for any interval I ⊂ [ ψ ( a ) , ψ ( a )] , we have c − | I |≤ | ψ − ( I ) |≤ c − | I | . In our case, the Cauchy-Schwarz inequalitygives ∂ s ϕ x ( s, t ) ≤ (see Equation (9) for the expression of ∂ s ϕ x ( s, t ) ).Finally we get, with a given by (8): P (cid:0) X xT ∈ A, T x ∈ ( u , u ) (cid:1) ≥ a Z A
12 ( u − u )d z = a u − u ) | A | , which proves that (cid:0) X xT , T x (cid:1) is a -continuous on I ∗ β,δ × (˜ r , ˜ r ) .Thanks to the remarks before, the proof is completed with η = a and J = I ∗ β,δ .Let us now give the proofs of Lemma 4.6 and 4.7 that we have admitted so far. Proof of Lemma 4.6.
We use the notations introduced at the end of the proof of Proposition4.4.We have, for s, t ∈ I : ∂ s ϕ w ( s, t ) = (cid:28) g ( s ) − w k g ( s ) − w k + g ( s ) − g ( t ) k g ( s ) − g ( t ) k , g ′ ( s ) (cid:29) . (9)28y the definition of g , we note that g ′ ( s ) is a director vector of the tangent line of ∂K at point g ( s ) .It is easy to verify that for w ∈ { x, ˜ x } , there exists a unique t ∈ I \ { s ¯ y } such that ∂ s ϕ w ( s ¯ y , t ) = 0 . (10)For w = x (resp. w = ˜ x ), we denote by t z x (resp. t z ˜ x ) this unique element of I . With ournotations we thus have g ( t z x ) = z x and g ( t z ˜ x ) = z ˜ x .Let w ∈ { x, ˜ x } . We have: ∂ t ∂ s ϕ w ( s, t ) = ∂ t (cid:18)(cid:28) g ( s ) − g ( t ) k g ( s ) − g ( t ) k , g ′ ( s ) (cid:29)(cid:19) = 1 k g ( t ) − g ( s ) k (cid:18) − h g ′ ( t ) , g ′ ( s ) i + (cid:28) g ( t ) − g ( s ) k g ( t ) − g ( s ) k , g ′ ( t ) (cid:29) (cid:28) g ( t ) − g ( s ) k g ( t ) − g ( s ) k , g ′ ( s ) (cid:29)(cid:19) . Let us look at the term in parenthesis. Let us denote by [ u, v ] the oriented angle between thevectors u, v ∈ R . We have: − h g ′ ( t ) , g ′ ( s ) i + (cid:28) g ( t ) − g ( s ) k g ( t ) − g ( s ) k , g ′ ( t ) (cid:29) (cid:28) g ( t ) − g ( s ) k g ( t ) − g ( s ) k , g ′ ( s ) (cid:29) = − cos ( [ g ′ ( t ) , g ′ ( s ) ] ) + cos ( [ g ( t ) − g ( s ) , g ′ ( t ) ] ) cos ( [ g ( t ) − g ( s ) , g ′ ( s ) ] )= − cos ( [ g ′ ( t ) , g ′ ( s ) ] ) + 12 cos ( [ g ( t ) − g ( s ) , g ′ ( t ) ] − [ g ( t ) − g ( s ) , g ′ ( s ) ] )+ 12 cos ( [ g ( t ) − g ( s ) , g ′ ( t ) ] + [ g ( t ) − g ( s ) , g ′ ( s ) ] )= − cos ( [ g ′ ( t ) , g ′ ( s ) ] ) + 12 cos ( [ g ′ ( s ) , g ′ ( t ) ] )+ 12 cos ( [ g ( t ) − g ( s ) , g ′ ( t ) ] + [ g ( t ) − g ( s ) , g ′ ( s ) ] )= −
12 cos ( [ g ′ ( t ) , g ′ ( s ) ] ) + 12 cos ( [ g ( t ) − g ( s ) , g ′ ( t ) ] + [ g ( t ) − g ( s ) , g ′ ( s ) ] )= − sin (cid:18)
12 ( [ g ( t ) − g ( s ) , g ′ ( t ) ] + [ g ( t ) − g ( s ) , g ′ ( s ) ] + [ g ′ ( t ) , g ′ ( s ) ] ) (cid:19) × sin (cid:18)
12 ( [ g ( t ) − g ( s ) , g ′ ( t ) ] + [ g ( t ) − g ( s ) , g ′ ( s ) ] − [ g ′ ( t ) , g ′ ( s ) ] ) (cid:19) = − sin ( [ g ( t ) − g ( s ) , g ′ ( s ) ] ) sin ( [ g ( t ) − g ( s ) , g ′ ( t ) ] ) . Therefore we get ∂ t ∂ s ϕ w ( s, t ) = − k g ( t ) − g ( s ) k sin ( [ g ( t ) − g ( s ) , g ′ ( s ) ] ) sin ( [ g ( t ) − g ( s ) , g ′ ( t ) ] ) , and then | ∂ t ∂ s ϕ w ( s, t ) | = 1 k g ( t ) − g ( s ) k | sin ( [ g ( t ) − g ( s ) , g ′ ( s ) ] ) sin ( [ g ( t ) − g ( s ) , g ′ ( t ) ] ) | = 1 k g ( t ) − g ( s ) k (cid:12)(cid:12) cos (cid:0) ϕ g ( s ) ,g ( t ) (cid:1) cos (cid:0) ϕ g ( t ) ,g ( s ) (cid:1)(cid:12)(cid:12) t ∈ I such that | t − s ¯ y |≥ β (we recall that β is introduced at the beginning of the section).Using once more Figure 7, we get, as we have done in the proof of Theorem 4.2: | ∂ t ∂ s ϕ w ( s, t ) | ≥ k g ( t ) − g ( s ) k (cid:18) βc (cid:19) ≥ D (cid:18) βc (cid:19) . (11)Using Equations (10) and Equation (11), the mean value theorem gives: for t ∈ I such that | t − s ¯ y |≥ β and | t − t z w |≥ δ ( δ is introduced at the beginning of the section), | ∂ s ϕ w ( s ¯ y , t ) | = | ∂ s ϕ w ( s ¯ y , t ) − ∂ s ϕ w ( s ¯ y , t z w ) | ≥ D (cid:18) βc (cid:19) | t − t z w |≥ δD (cid:18) βc (cid:19) . (12)We want now such an inequality for s ∈ I near from s ¯ y . We thus compute: ∂ s ϕ w ( s, t ) = 1 k w − g ( s ) k + 1 k g ( s ) − g ( t ) k + (cid:28) g ( s ) − w k g ( s ) − w k + g ( s ) − g ( t ) k g ( s ) − g ( t ) k , g ′′ ( s ) (cid:29) − k w − g ( s ) k (cid:28) w − g ( s ) k w − g ( s ) k , g ′ ( s ) (cid:29) − k g ( s ) − g ( t ) k (cid:28) g ( s ) − g ( t ) k g ( s ) − g ( t ) k , g ′ ( s ) (cid:29) . We immediately deduce, using the Cauchy-Schwarz inequality, and the fact that k g ′ ( s ) k = 1 forall s ∈ I : | ∂ s ϕ w ( s, t ) | ≤ k w − g ( s ) k + 1 k g ( s ) − g ( t ) k + 2 k g ′′ ( s ) k + 1 k w − g ( s ) k + 1 k g ( s ) − g ( t ) k≤ (cid:18) k w − g ( s ) k + 1 k g ( s ) − g ( t ) k + C (cid:19) , where we recall that C is the upper bound on the curvature of K .Let now t ∈ I such that | t − s ¯ y |≥ β and | t − t z w |≥ δ , and let s ∈ I such that | s − s ¯ y |≤ ε . Withsuch s and t we have | t − s |≥ β − ε . Moreover, we have already seen in proof of Proposition4.4 (with Figure 8) that k w − g ( s ) k≥ C − ε for s ∈ B ε ¯ y . Therefore, for such s and t : | ∂ s ϕ w ( s, t ) |≤ (cid:18) C − ε + 1 β − ε + C (cid:19) = M > . (13)Using once again the mean value theorem with Equations (12) and (13), we deduce that for all t ∈ I such that | t − s ¯ y |≥ β and | t − t z w |≥ δ , and for all s ∈ I such that | s − s ¯ y |≤ ε : | ∂ s ϕ w ( s, t ) | ≥ δD (cid:18) βc (cid:19) − εM = h > . Let now take I ∗ β,δ ⊂ I \ { s ¯ y , t z x , t z ˜ x } an interval of length strictly smaller than hε (this conditionon the length of I ∗ β,δ is not necessary for the lemma, but for the continuation of the proof ofthe proposition), and such that for all t ∈ I ∗ β,δ , | t − t z x |≥ δ , | t − t z ˜ x |≥ δ and | t − s ¯ y |≥ β . Inorder to ensure that I ∗ β,δ is not empty, we take β and δ such that | ∂K | − max { δ ; β + δ } > .Indeed, it is necessary that one of the intervals ”( t z x , t z ˜ x )” , ”( t z x , s ¯ y )” and ”( s ¯ y , t z ˜ x )” at whichwe removes a length β or δ on the good extremity, is not empty. And since the larger of theseintervals has a length at least ∂K , we obtain the good condition on β and δ .We thus just proved that | ∂ s ϕ w ( s, t ) | ≥ h for s ∈ B ε ¯ y and t ∈ I ∗ β,δ , which is the result of thelemma. 30 roof of Lemma 4.7. Let first prove that r < r . We do it only for r and r since it is thesame for ˜ r and ˜ r . We have: r − r = inf t ∈ I ∗ β,δ ϕ x ( s , t ) − sup t ∈ I ∗ β,δ ϕ x ( s , t )= inf t ∈ I ∗ β,δ ϕ x ( s , t ) − inf t ∈ I ∗ β,δ ϕ x ( s , t ) − sup t ∈ I ∗ β,δ ϕ x ( s , t ) − inf t ∈ I ∗ β,δ ϕ x ( s , t ) ! ≥ h ( s − s ) − sup t ∈ I ∗ β,δ | ∂ t ϕ x ( s , t ) | ! (cid:12)(cid:12) I ∗ β,δ (cid:12)(cid:12) ≥ hε − (cid:12)(cid:12) I ∗ β,δ (cid:12)(cid:12) , and this quantity is strictly positive since | I ∗ β,δ | < hε by construction.For the first inequality, we have used the mean value theorem twice, and for the last inequality,we have used the fact that sup t ∈ I ∗ β,δ | ∂ t ϕ x ( s , t ) | = sup t ∈ I ∗ β,δ (cid:12)(cid:12)(cid:12)D g ( t ) − g ( s ) k g ( t ) − g ( s ) k , g ′ ( t ) E(cid:12)(cid:12)(cid:12) ≤ thanks tothe Cauchy-Schwarz inequality.Let us now prove that the intersection ( r , r ) ∩ (˜ r , ˜ r ) is not empty.Let t ∈ I ∗ β,δ , we have: r − ϕ x ( s ¯ y , t ) = inf t ∈ I ∗ β,δ ϕ x ( s , t ) − ϕ x ( s ¯ y , t )= inf t ∈ I ∗ β,δ ϕ x ( s , t ) − inf t ∈ I ∗ β,δ ϕ x ( s ¯ y , t ) − (cid:18) ϕ x ( s ¯ y , t ) − inf t ∈ I ∗ β,δ ϕ x ( s ¯ y , t ) (cid:19) ≥ h ( s − s ¯ y ) − | I ∗ β,δ | = hε − | I ∗ β,δ | > , once again thanks to the mean value theorem. Similarly we have ϕ x ( s ¯ y , t ) − r = ϕ x ( s ¯ y , t ) − sup t ∈ I ∗ β,δ ϕ x ( s , t )= ϕ x ( s ¯ y , t ) − sup t ∈ I ∗ β,δ ϕ x ( s ¯ y , t ) − sup t ∈ I ∗ β,δ ϕ x ( s , t ) − sup t ∈ I ∗ β,δ ϕ x ( s ¯ y , t ) ! ≥ −| I ∗ β,δ | + h ( s ¯ y − s )= hε − | I ∗ β,δ | > . Moreover, since ¯ y ∈ ∆ x, ˜ x , we have ϕ x ( s ¯ y , t ) = ϕ ˜ x ( s ¯ y , t ) , and we thus can prove the sameinequalities with ˜ r and ˜ r instead of r and r .Finally we thus get that the interval ( R , R ) = ( r , r ) ∩ ( ˜ r , ˜ r ) is well defined and R − R ≥ (cid:0) hε − | I ∗ β,δ | (cid:1) . Remark 4.8.
The fact that |J | = π is here to ensure that the process can go from x and ˜ x to ¯ y in the proof of Proposition 4.4. If |J | < π , since x and ˜ x are unspecified and ¯ y can thereforebe everywhere on ∂K , nothing ensures that this path is available.
31e can now state the following theorem on the speed of convergence of the stochastic billiardin the convex set K . Theorem 4.9.
Let K ⊂ R satisfying Assumption ( K ) with diameter D . Let ( X t , V t ) t ≥ thestochastic billiard process evolving in K and verifying Assumption ( H ) with |J | = π .There exists a unique invariant probability measure χ on K × S for the process ( X t , V t ) t ≥ .Moreover, let define n and p by (14) and (15) with ζ ∈ (cid:0) , C (cid:1) . Let consider η , I ∗ β,δ , R , R asin Proposition 4.4 and Lemma 4.6, and let define κ by (16) .For all t ≥ and all λ < λ M : k P ( X t ∈ · , V t ∈ · ) − χ k T V ≤ C λ e − λt , where λ M = min ( D log (cid:18) − p (cid:19) ; 12 D log − (1 − p ) + p (1 − p ) + 4 p (1 − κ )2 p (1 − κ ) !) and C λ = pκ e λD − e λD (1 − p ) − e λD p (1 − κ ) . Proof.
As previously, the existence of an invariant probability measure for the stochastic billiardprocess comes from the compactness of K × S . The following proof ensures its uniqueness andgives an explicit speed of convergence.Let ( X t , V t ) t ≥ and ( ˜ X t , ˜ V t ) t ≥ be two versions of the stochastic billiard with ( X , V ) = ( x , v ) ∈ K × S and ( ˜ X , ˜ V ) = (˜ x , ˜ v ) ∈ K × S .We define (or recall the definition for T and ˜ T ): T = inf { t ≥ , x + tv / ∈ K } , w = x + T v ∈ ∂K, and ˜ T = inf { t ≥ , ˜ x + t ˜ v / ∈ K } , ˜ w = ˜ x + ˜ T ˜ v ∈ ∂K. Step 1 . From Proposition 4.3, we deduce that for all x ∈ ∂K and all ζ ∈ (cid:0) , C (cid:1) , T xn is ( cρ min ) n ζ n − -continuous on the interval Γ n = (cid:2) ( n − ζ , nC − ( n − ζ (cid:3) .Let thus ζ ∈ (cid:0) , C (cid:1) and let define n = min { n ≥ | Γ n | > D } = $ D − ζ (cid:0) C − (cid:1) % + 1 . (14)The variables T + T wn and ˜ T + ˜ T ˜ wn are both ( cρ min ) n ζ n − -continuous on (cid:2) T + ( n − ζ , T + nC − ( n − ζ (cid:3) ∩ h ˜ T + ( n − ζ , ˜ T + nC − ( n − ζ i . Since (cid:12)(cid:12)(cid:12) T − ˜ T (cid:12)(cid:12)(cid:12) ≤ D , this intersection is non-empty and its length is larger that n C − n − ζ − D .Let define p = ( cρ min ) n ζ n − (cid:18) n C − n − ζ − D (cid:19) . (15)Using the fact that the for all w ∈ ∂K , T wn ≤ n D almost surely, we deduce that we canconstruct a coupling such that the coupling-time T c of T + T wn and ˜ T + ˜ T ˜ wn satisfies: T c ≤ st T + n DG G ∼ G ( p ) . Step 2 . Once the coupling of these times has succeed, we try to couple the couples (cid:18) X X wT c T , T X wTc (cid:19) and (cid:18) ˜ X ˜ X ˜ wT c ˜ T , ˜ T ˜ X ˜ wT c (cid:19) . By the Proposition 4.4, we can construct a coupling suchthat P (cid:18) X X wT c T = ˜ X ˜ X ˜ wT c ˜ T and T X wT c = ˜ T ˜ X ˜ wT c (cid:19) ≥ η | I ∗ β,δ | ( R − R ) . Defining κ = η | I ∗ β,δ | ( R − R ) , (16)we get that the entire coupling-time of the two processes satisfies: ˆ T ≤ st T + G X l =1 (cid:0) n DG l + n D (cid:1) = T + G X l =1 (cid:0) n D ( G l + 1) (cid:1) where G as a geometric distribution with parameter κ and the ( G l ) l ≥ are independent geometricrandom variables with parameter p , and independent of G .Finally, we get P (cid:16) ˆ T > t (cid:17) ≤ e − λt pκ e λD − e λD (1 − p ) − e λD p (1 − κ ) , for all λ ∈ (0 , λ M ) . All the results presented in this paper are in dimension . However, the ideas developed herecan be adapted to higher dimensions. Let us briefly explain it. Stochastic billiard in a ball of R d Let us first look at the stochastic billiard ( X, V ) in a ball B ⊂ R d with d ≥ .As we have done in Section 3, we can represent the Markov chain ( X T n , V T n ) n ≥ by anotherMarkov chain. Indeed, for n ≥ , the position X T n ∈ ∂ B can be uniquely represented byits hyperspherical coordinates: a ( d − -tuple (Φ n , · · · , Φ d − n ) with Φ n , · · · , Φ d − n ∈ [0 , π ) and Φ d − n ∈ [0 , π ) .Similarly, for n ≥ , the vector speed V T n ∈ (cid:8) v ∈ S d − : v · n X Tn ≥ (cid:9) can be represented by itshyperspherical coordinates.Thereby, we can give relations between the different random variables as in Proposition 3.1, andin theory, we can do explicit computations to get lower bounds on the needed density function.Then the same coupling method in two steps can be applied. Nevertheless, it could be difficultto manage the computations in practice when the dimension increases.33 tochastic billiard in a convex set K ⊂ R d To get bounds on the speed of convergence of the stochastic billiard ( X, V ) in a convex set K ⊂ R d , d ≥ , satisfying Assumption ( K ) , we can apply exactly the same method as inSection 4. The main difficulty could be the proof of the equivalent of Proposition 4.4. But itcan easily be adapted, and we refer to the proof of Lemma 5.1 in [2], where the authors leadthe proof in dimension d ≥ . Acknowledgements.
The author thanks Hélène Guérin and Florent Malrieu for their helpin this work.This work was supported by the Agence Nationale de la Recherche project PIECE 12-JS01-0006-01.
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