Sticky Brownian motions and a probabilistic solution to a two-point boundary value problem
aa r X i v : . [ m a t h . P R ] O c t Sticky Brownian motions and a probabilisticsolution to a two-point boundary value problem
Thu Dang Thien NguyenOctober 16, 2018
Abstract
In this paper, we study a two-point boundary value problem consist-ing of the heat equation on the open interval (0 ,
1) with boundary con-ditions which relate first and second spatial derivatives at the boundarypoints. Moreover, the unique solution to this problem can be representedprobabilistically in terms of a sticky Brownian motion. This probabilisticrepresentation is attained from the stochastic differential equation for asticky Brownian motion on the bounded interval [0 , Keywords: Two-point boundary value problem; Sticky Brownian motions.
Let u : [0 , → [0 , u ∈ C , ([0 , × (0 , ∞ )) solving the following problem u t = 12 u rr , lim t ↓ u ( r, t ) = u ( r ) , for r ∈ (0 , , (1.1) ddt u (0 , t ) = 12 u r (0 , t ) , ddt u (1 , t ) = − u r (1 , t ) . (1.2)Since it is shown in Chapter 6 and Chapter 7 of [7] that there exists a Markovprocess B having a generator A = d /dx on (0 ,
1) with extension by continuityto the points 0 and 1 and restriction to the domain D ( A ) = { f ∈ C ([0 , , f ′′ ( x ) + ( − − x f ′ ( x ) = 0 for x ∈ { , }} , then if u ∈ C ([0 , r, t ) ∈ [0 , × (0 . ∞ ), by u ( r, t ) = E r (cid:2) u ( B ( t )) (cid:3) , where E r stands for the expectation with respect to the process B starting from r . However, when the continuity at the boundary points 0 and 1 of the initialdatum u is relaxed, let us consider the heat equation along with the initialdatum (1.1) and the Dirichlet boundary conditions u (0 , t ) = v , − , u (1 , t ) = v , + for t > , (1.3)1here v , ± ∈ [0 , B absorbed at 0 and 1 asfollows u ( r, t ) = E r (cid:2) u ( B ( t )) τ>t (cid:3) + v , − P r (cid:0) τ = τ ≤ t (cid:1) + v , + P r (cid:0) τ = τ ≤ t (cid:1) , (1.4)where τ a is the first time when the Brownian motion B hits a and τ = τ ∧ τ .Since B is absorbed whenever it reaches 0 and 1, we are just interested in theboundary conditions of the form (1.3).In [11], Pang and Stroock investigate the existence of the solution to the heatequation with the initial datum (1.1) and the boundary conditions (1.2) underthe assumption on the discontinuity of the initial datum u at the boundaries.For the boundary conditions (1.2), we are concerned about the mass flux at eachboundary. The solution in C , ([0 , × (0 , ∞ )) established through a Brownianmotion sticky at 0 and 1 is unique if it satisfies further thatlim t ↓ u (0 , t ) = v , − , lim t ↓ u (1 , t ) = v , + . (1.5)The reason for the requirement (1.5) is that the stiky Brownian motion spendsa positive amount of time at 0 and 1 with positive probability.Moreover, suppose that at each boundary point 0 and 1, there is a reservoirof mass v ± ( t ) ∈ C [0 , ∞ ) changing in time. The boundary conditions (1.2)describe that the mass flux at 0 and 1 equals the mass change in the left andright reservoir respectively. If we identify v − ( t ) = u (0 , t ) and v + ( t ) = u (1 , t )then it can be checked that Z u ( r, t ) dr + v − ( t ) + v + ( t ) = Z u ( r ) dr + v , − + v , + , ∀ t > . This implies the conservation of mass.Now in the current paper, we will examine an analogous boundary valueproblem but depending on a parameter ǫ >
0. The role of ǫ will be explainedlater. For any fixed ǫ >
0, let us consider the heat equation with the initialdatum u ǫ ∈ C (0 ,
1) taking values in [0 ,
1] as follows u ǫt = 12 u ǫrr , lim t ↓ u ǫ ( r, t ) = u ǫ ( r ) . (1.6)Let us impose a reservoir of mass ρ ǫ ± ( t ) ∈ C [0 , ∞ ) at each boundary. Weobserve that if u ǫ (0 , · ) is larger than ρ ǫ − ( · ), the rate of the mass change in the leftreservoir increases in time. Moreover, the larger this difference is, the faster therate increases. We will use the factor ǫ − to emphasize this property. Although ρ ǫ − ( · ) may be different from u ǫ (0 , · ), the difference between them becomes 0 as ǫ goes to 0. An analogous phenomenon also happens at the boundary point 1.Then for a parameter ǫ >
0, this fact can be described in a rigorous way ddt ρ ǫ − ( t ) = ǫ − (cid:2) u ǫ (0 , t ) − ρ ǫ − ( t ) (cid:3) (1.7) ddt ρ ǫ + ( t ) = ǫ − (cid:2) u ǫ (1 , t ) − ρ ǫ + ( t ) (cid:3) . (1.8)2urthermore, since we will look for a solution in C , ((0 , × (0 , ∞ )), theboundary conditions that implies the conservation of mass are expressed in aweak form ρ ǫ − ( t ) = ρ ǫ − (0) + lim l → lim t ↓ Z tt u ǫr ( l, s ) ds (1.9) ρ ǫ + ( t ) = ρ ǫ + (0) − lim l → lim t ↓ Z tt u ǫr ( l, s ) ds. (1.10)For a fixed ǫ >
0, if we require that u ǫ (0 , · ) , u ǫ (1 , · ) : [0 , ∞ ) → [0 ,
1] and u ǫ (0 , · ) , u ǫ (1 , · ) ∈ C (0 , ∞ ), then there exist unique solutions u ǫ ∈ C , ((0 , × (0 , ∞ )) , ρ ǫ ± ∈ C (0 , ∞ ) ∩ C ([0 , ∞ )) to the boundary value problem (1.6)-(1.10).The first part of the current paper is devoted to arrive at this result. Next, bythe suitably chosen initial data, identifying the unique limits of these solutions as ǫ → u ∈ C , ([0 , × (0 , ∞ )) satisfying (1.1)-(1.5).Moreover, in the second part of this paper, based on the fact that a stickyBrownian motion on the half line [0 , ∞ ) solves a stochastic differential equationas verified in [5], we will give an analogous characterization for a sticky Brownianmotion on the bounded interval [0 , Remark 1.1.
In [10], we consider a system of particles moving according tothe simple symmetric exclusion process in the channel [1 , N ] with reservoirs atthe boundaries. The reservoirs of size N are also particle systems which can beexchanged with the ones in the channel. The hydrodynamic limit equation we ob-tain for this particle system is the two-point boundary value problem mentionedabove. From the physical point of view, the unique solution to this problem isthe limit of a sequence of the one-body correlation functions for an appropriatelyconstructed interacting particle system. Furthermore, by duality technique, onecan also express the correlation function in terms of a sticky random walk. Sincethe convergence of a sequence of rescaled sticky random walks to a sticky Brow-nian motion can be shown based on the arguments presented in [1], it leads usto a probabilistic representation of this unique solution. Let us set U := C , ((0 , × (0 , ∞ )) and H := C (0 , ∞ ) ∩ C ([0 , ∞ )). Theorem 2.1.
For any fixed ǫ > , let u ǫ ∈ C (0 , with values in [0 , and ρ ǫ ± (0) = v ǫ , ± ∈ [0 , . If u ǫ (0 , · ) , u ǫ (1 , · ) : [0 , ∞ ) → [0 , and u ǫ (0 , · ) , u ǫ (1 , · ) ∈ C (0 , ∞ ) , then there exists a unique ( u ǫ , ρ ǫ − , ρ ǫ + ) ∈ U × H × H which satisfies thefollowing problem u ǫt = 12 u ǫrr , lim t ↓ u ǫ ( r, t ) = u ǫ ( r ) (2.1) ddt ρ ǫ − ( t ) = ǫ − (cid:2) u ǫ (0 , t ) − ρ ǫ − ( t ) (cid:3) (2.2) ddt ρ ǫ + ( t ) = ǫ − (cid:2) u ǫ (1 , t ) − ρ ǫ + ( t ) (cid:3) (2.3)3 ǫ − ( t ) = ρ ǫ − (0) + lim l → lim t ↓ Z tt u ǫr ( l, s ) ds (2.4) ρ ǫ + ( t ) = ρ ǫ + (0) − lim l → lim t ↓ Z tt u ǫr ( l, s ) ds. (2.5)The existence and uniqueness of ( u ǫ , ρ ǫ ± ) ∈ U × H × H can be shown byapplying the same technique as presented in Proposition 3.10, [10], where weaim to arrive at an integral equation and then construct its unique solutioninductively by using the contraction mapping theorem. Proof.
For ( r, t ) ∈ [0 , × (0 , ∞ ), we denoteΘ( r, t ) = + ∞ X n = −∞ √ πt e − ( r +2 n )22 t . Then as a result of Theorem 6.3.1, [4], for any fixed ǫ >
0, ( r, t ) ∈ (0 , × (0 , ∞ ), the function u ǫ ∈ C , ((0 , × (0 , ∞ )) defined by the following expression u ǫ ( r, t ) = Z u ǫ ( r ′ )[Θ( r − r ′ , t ) − Θ( r + r ′ , t )] dr ′ − Z t ∂ Θ ∂r ( r, t − s ) u ǫ (0 , s ) ds + Z t ∂ Θ ∂r ( r − , t − s ) u ǫ (1 , s ) ds (2.6)satisfies the linear heat equation (2.1) with boundary values u ǫ (0 , · ) , u ǫ (1 , · ) andthe initial datum u ǫ .Thus (2.4) and (2.5) give us ρ ǫ − ( t ) = v ǫ , − + lim l → lim t ↓ Z tt Z u ǫ ( r ′ ) " ∂ Θ ∂r ( l − r ′ , s ) − ∂ Θ ∂r ( l + r ′ , s ) dr ′ ds − Z t Θ(0 , t − s ) u ǫ (0 , s ) ds + Z t Θ(1 , t − s ) u ǫ (1 , s ) ds, (2.7)and similarly, ρ ǫ + ( t ) = v ǫ , + − lim l → lim t ↓ Z tt Z u ǫ ( r ′ ) " ∂ Θ ∂r ( l − r ′ , s ) − ∂ Θ ∂r ( l + r ′ , s ) dr ′ ds + Z t Θ(1 , t − s ) u ǫ (0 , s ) ds − Z t Θ(0 , t − s ) u ǫ (1 , s ) ds. (2.8)On the other hand, since it follows from (2.2) and (2.3) that ρ ǫ − ( t ) = e − ǫ − ( t − t ) ρ ǫ − ( t ) + Z tt ǫ − e − ǫ − ( t − s ) u ǫ (0 , s ) ds (2.9) ρ ǫ + ( t ) = e − ǫ − ( t − t ) ρ ǫ + ( t ) + Z tt ǫ − e − ǫ − ( t − s ) u ǫ (1 , s ) ds (2.10)4hen we obtain the following system Z t h Θ(0 , t − s ) + ǫ − e − ǫ − ( t − s ) i u ǫ (0 , s ) ds + Z t − Θ(1 , t − s ) u ǫ (1 , s ) ds = f ǫ − ( t ) + v , − (1 − e − ǫ − t ) Z t − Θ(1 , t − s ) u ǫ (0 , s ) ds + Z t h Θ(0 , t − s ) + ǫ − e − ǫ − ( t − s ) i u ǫ (1 , s ) ds = f ǫ + ( t ) + v , + (1 − e − ǫ − t ) , (2.11)where f ǫ − ( t ) = lim l → lim t ↓ Z tt Z u ǫ ( r ′ ) (cid:20) ∂ Θ ∂r ( l − r ′ , s ) − ∂ Θ ∂r ( l + r ′ , s ) (cid:21) dr ′ ds,f ǫ + ( t ) = − lim l → lim t ↓ Z tt Z u ǫ ( r ′ ) (cid:20) ∂ Θ ∂r ( l − r ′ , s ) − ∂ Θ ∂r ( l + r ′ , s ) (cid:21) dr ′ ds. Now multiplying both sides of the first equation of the above system by( x − t ) − / and integrating with respect to t from 0 to x yield Z x r π u ǫ (0 , s ) ds + Z x (cid:20) Z X n ≥ √ π p y (1 − y ) e − (2 n )22 y ( x − s ) dy + Z xs ǫ − √ x − t e − ǫ − ( t − s ) dt (cid:21) u ǫ (0 , s ) ds + Z x (cid:20) Z − X n ≥ √ π p y (1 − y ) e − (2 n − y ( x − s ) dy (cid:21) u ǫ (1 , s ) ds = Z x √ x − t (cid:16) f ǫ − ( t ) + v , − (1 − e − ǫ − t ) (cid:17) dt. (2.12)If u ǫ (0 , · ) , u ǫ (1 , · ) ∈ C (0 , ∞ ) and take values in [0 ,
1] on [0 , ∞ ) then thisenables us to define for x > ψ ǫ − ( x ) = Z x u ǫ (0 , s ) ds, ψ ǫ + ( x ) = Z x u ǫ (1 , s ) ds and thus ddx ψ ǫ − ( x ) = u ǫ (0 , x ) , ddx ψ ǫ + ( x ) = u ǫ (1 , x ) . For α := p /π , applying integration by parts in (2.12) gives us ψ ǫ − ( x ) = Z x α (cid:20) Z X n ≥ − √ π p y (1 − y ) e − (2 n )22 y ( x − s ) (2 n ) y ( x − s ) dy − ǫ − √ x − s + 2 ǫ − √ x − s − Z xs ǫ − √ x − t e − ǫ − ( t − s ) dt (cid:21) ψ ǫ − ( s ) ds + Z x (cid:20) Z X n ≥ α √ π p y (1 − y ) e − (2 n − y ( x − s ) (2 n − y ( x − s ) dy (cid:21) ψ ǫ + ( s ) ds Z x α √ x − t (cid:16) f ǫ − ( t ) + v , − (1 − e − ǫ − t ) (cid:17) dt. Making a similar argument as above for the second equation of the system(2.11) leads us to consider the equation (cid:20) ψ ǫ − ( x ) ψ ǫ + ( x ) (cid:21) = (cid:20) F ǫ − ( x ) F ǫ + ( x ) (cid:21) + Z x (cid:20) K ǫ − ( x − s ) K ǫ + ( x − s ) K ǫ + ( x − s ) K ǫ − ( x − s ) (cid:21) (cid:20) ψ ǫ − ( s ) ψ ǫ + ( s ) (cid:21) ds, (2.13)where K ǫ − ( t ) = α (cid:20) Z X n ≥ − √ π p y (1 − y ) e − (2 n )22 yt (2 n ) yt dy − ǫ − √ t + 2 ǫ − √ t − Z t ǫ − √ t − σ e − ǫ − σ dσ (cid:21) ,K ǫ + ( t ) = Z X n ≥ − α √ π p y (1 − y ) e − (2 n − yt (2 n − yt dy,F ǫ ± ( x ) = Z x α √ x − t (cid:16) f ǫ ± ( t ) + v , ± (1 − e − ǫ − t ) (cid:17) dt. Applying the same technique as introduced in Proposition 3.10, [10], one canverify the following result.
Proposition 2.2.
The equation (2.13) has a unique solution ( ψ ǫ − , ψ ǫ + ) ∈ C (0 , T ] × C (0 , T ] for any T > . As a consequence of the above result, there exists a unique solution ( u ǫ (0 , · ) , u ǫ (1 , · ))to the system (2.11) for any ǫ > u ǫ , ρ ǫ − , ρ ǫ + ) ∈ U × H × H to our main problem.Let us now identify the limits of sequences of functions u ǫ , ρ ǫ ± as ǫ goes to 0up to subsequences. Since the uniqueness of the limits can be verified, we obtainthe identification of the limits u, v ± for the original sequences. Moreover, theboundary conditions can be attained in a strong form in view of the continuousdifferentiability of v ± . More precisely, the limit u ∈ C , ([0 , × (0 , ∞ )) is theunique solution to a two-point boundary value problem. Theorem 2.3.
Let u ∈ C (0 , with values in [0 , and v , ± ∈ [0 , . Thereexists a unique u ∈ C , ([0 , × (0 , ∞ )) which solves the following problem u t = 12 u rr , u (1 , t ) = v + ( t ) , u (0 , t ) = v − ( t ) , lim t ↓ u ( r, t ) = u ( r ) (2.14) with v ± ( t ) such that for any t > , ddt v − ( t ) = 12 u r (0 , t ) , ddt v + ( t ) = − u r (1 , t ) , lim t ↓ v ± ( t ) = v , ± . (2.15) Proof.
For any fixed ǫ >
0, let ( u ǫ , ρ ǫ ± ) be the unique solution obtained inTheorem 2.1. At the initial time, the sequence u ǫ is chosen such that it converges6niformly to u on any compact set of (0 ,
1) as ǫ →
0. Moreover, let us selectthe sequences v ǫ , ± which converge to v , ± , respectively, as ǫ → δ, T >
0, there exists a constant C such that for any s, t ∈ [ δ, T ], | ρ ǫ ± ( t ) − ρ ǫ ± ( s ) | ≤ C | t − s | , thus the sequences ( ρ ǫ ± ) ǫ are uniformly equicontinuous. Moreover, (2.9) and(2.10) imply the uniform boundedness of these sequences on [ δ, T ]. Therefore,there exist subsequences ρ ǫ k ± converge uniformly on [ δ, T ] to v ± , respectively.Notice that v ± ∈ C (0 , ∞ ).On the other hand, by (2.9), we claim that for any t > ǫ → (cid:12)(cid:12) ρ ǫ − ( t ) − u ǫ (0 , t ) (cid:12)(cid:12) = 0 . Indeed, since u ǫ (0 , · ) is continuous at t > δ ′ > ǫ and t such that (cid:12)(cid:12) u ǫ (0 , s ) − u ǫ (0 , t ) (cid:12)(cid:12) < ǫ for any s ∈ [ t − δ ′ , t ]. Therefore, ourclaim follows from the following estimate (cid:12)(cid:12) ρ ǫ − ( t ) − u ǫ (0 , t ) (cid:12)(cid:12) ≤ e − ǫ − t (cid:12)(cid:12) v , ± − u ǫ (0 , t ) (cid:12)(cid:12) + Z t ǫ − e − ǫ − ( t − s ) (cid:12)(cid:12) u ǫ (0 , s ) − u ǫ (0 , t ) (cid:12)(cid:12) ds ≤ e − ǫ − t + Z t − δ ′ ǫ − e − ǫ − ( t − s ) ds + ǫ Z tt − δ ′ ǫ − e − ǫ − ( t − s ) ds. Analogously, it can be obtained from (2.10) that for any t > ǫ → (cid:12)(cid:12) ρ ǫ + ( t ) − u ǫ (1 , t ) (cid:12)(cid:12) = 0 , then the subsequences u ǫ k (0 , · ) , u ǫ k (1 , · ) converge pointwise on (0 , ∞ ) to v ∓ ,respectively.Hence, for any ( r, t ) ∈ (0 , × (0 , ∞ ), the corresponding subsequence u ǫ k converges to u given by u ( r, t ) = Z u ( r ′ )[Θ( r − r ′ , t ) − Θ( r + r ′ , t )] dr ′ − Z t ∂ Θ ∂r ( r, t − s ) v − ( s ) ds + Z t ∂ Θ ∂r ( r − , t − s ) v + ( s ) ds. (2.16)One can easily checked that u ∈ C , ((0 , × (0 , ∞ )). As already shownin Chapter 6 of [4], the limit u solves the linear heat equation with boundaryvalues v ± and the initial datum u . Moreover, we observe thatlim t ↓ v ± ( t ) = lim t ↓ lim k →∞ ρ ǫ k ± ( t ) = lim k →∞ lim t ↓ ρ ǫ k ± ( t )= lim k →∞ ρ ǫ k ± (0) = lim k →∞ v ǫ k , ± = v , ± . Taking the limit of both sides of (2.7) and (2.8) along the subsequences ρ ǫ k ± with our choice of the sequences u ǫ , v ǫ , ± yields v − ( t ) = v , − + lim l → lim t ↓ Z tt u r ( l, s ) dsv + ( t ) = v , + − lim l → lim t ↓ Z tt u r ( l, s ) ds. (2.17)7ore precisely, u satisfies (2.14) with v ± ( t ) such that lim t ↓ v ± ( t ) = v , ± andthe boundary conditions (2.17).So far we have just identified the limit of the sequence u ǫ up to a subsequence.Let us now consider other subsequences ρ ǫ m ± such that they converge uniformlyto ˆ v ± , respectively. Then the corresponding limit ˆ u of the subsequence u ǫ m can be given by the same expression as in (2.16), where v ± are replaced by ˆ v ± .Applying the same argument as before, we deduce that ˆ u also solves the problem(2.14) with the boundary conditions (2.17), where we replace v ± by ˆ v ± .We denote ¯ v ± = v ± − ˆ v ± . The boundary conditions (2.17) imply that v − ( t ) = v , − + lim l → lim t ↓ Z tt Z u ( r ′ ) (cid:20) ∂ Θ ∂r ( l − r ′ , s ) − ∂ Θ ∂r ( l + r ′ , s ) (cid:21) dr ′ ds − Z t Θ(0 , t − s ) v − ( s ) ds + Z t Θ(1 , t − s ) v + ( s ) ds, (2.18)and similarly, v + ( t ) = v , + − lim l → lim t ↓ Z tt Z u ( r ′ ) (cid:20) ∂ Θ ∂r ( l − r ′ , s ) − ∂ Θ ∂r ( l + r ′ , s ) (cid:21) dr ′ ds + Z t Θ(1 , t − s ) v − ( s ) ds − Z t Θ(0 , t − s ) v + ( s ) ds. (2.19)This implies¯ v − ( t ) = − Z t Θ(0 , t − s )¯ v − ( s ) ds + Z t Θ(1 , t − s )¯ v + ( s ) ds, and similarly,¯ v + ( t ) = Z t Θ(1 , t − s )¯ v − ( s ) ds − Z t Θ(0 , t − s )¯ v + ( s ) ds. By setting V ( t ) = ¯ v − ( t )+ ¯ v + ( t ), the two above expressions allow us to attainthat V ( t ) = Z t (cid:0) Θ(1 , t − s ) − Θ(0 , t − s ) (cid:1) V ( s ) ds. This gives us V ( t ) = 0 , ∀ t >
0, by applying Gronwall’s inequality. Sincefor any t >
0, ¯ v ± ( t ) ∈ [0 , v ± ( t ) = ˆ v ± ( t ) and u ( r, t ) = ˆ u ( r, t ) for all( r, t ) ∈ [0 , × (0 , ∞ ). This leads us to the uniqueness of the solution to theproblelm (2.14) with the boundary conditions (2.17).Hence, we have verified that u is the pointwise limit of the sequence u ǫ for( r, t ) ∈ (0 , × (0 , ∞ ) as ǫ goes to 0.Moreover, in Section 3.6, [10], it can be shown that v ± ∈ C (0 , ∞ ). Thus u ∈ C , ([0 , × (0 , ∞ )) and now one can rewrite the boundary conditions (2.17)in the strong form (2.15). It completes the proof of Theorem 2.3.8 Sticky Brownian motion
Sticky random walk ( X ( t )) t ≥ moving on [0 , N + 1] ∩ N is a continuous timerandom walk with jump rates c ( x, x ±
1) = 12 , ∀ x ∈ [1 , N ] ∩ N and c (0 ,
1) = c ( N + 1 , N ) = 12 N .Let Y be a simple symmetric random walk on Z starting from x . Recall thatthe sequence of rescaled random walks N − Y ( N t ) converges uniformly almostsurely on compact intervals of [0 , ∞ ) to a Brownian motion B, B = r ∈ [0 , , F , P ), see [8].We denote by Y rf the simple random walk Y reflected at 0 and N + 1. Letus call T (0 , N + 1; t ; Y rf ) = Z t (cid:0) Y rf ( s )=0 + Y rf ( s )= N +1 (cid:1) ds the local time spent by Y rf at 0 and N + 1. Then it is shown in [10] that thesticky random walk X can be realized by setting X (cid:16) t + (2 N − T (0 , N + 1; t ; Y rf ) (cid:17) = Y rf ( t ) . (3.1) Theorem 3.1. (2 N − N − T (0 , N + 1; N t ; Y rf ) converges uniformly almostsurely on compact intervals of [0 , ∞ ) to L t which is the local time at and ofthe reflecting Brownian motion B rf on [0 , . Moreover, the rescaled sticky ran-dom walk N − X ( N t ) converges uniformly almost surely on compact intervalsof [0 , ∞ ) to the sticky Brownian motion B st on [0 , defined as B st ( t + L t ) = B rf ( t ) . (3.2) Proof.
We know that the continuous time random walk Y can be defined by Y ( t ) = S N ( t ) , where S is a simple symmetric discrete time random walk and N is a Poisson process of parameter 1. Let us denote by µ ( m ) k the number of visits to m ∈ Z in the first k steps of the random walk S . Then for µ k := X m ∈ Z µ ( m ( N +1)) k ,we can write2 N − N T (0 , N + 1; N t ; Y rf ) = 2 N − N Z N t (cid:0) Y rf ( s )=0 + Y rf ( s )= N +1 (cid:1) ds = 2 N − N Z N t ( N +1) − Y ( s ) ∈ Z ds = 2 N − N µ N ( N t ) X k =1 G k , where G k are independent exponential random variables with parameter 1. Wenext verify the following result. Proposition 3.2.
For any
T > and m ∈ Z , sup t ∈ [0 ,T ] (cid:12)(cid:12)(cid:12)(cid:12) N − N µ N ( N t ) X k =1 G k − N µ N ( N t ) (cid:12)(cid:12)(cid:12)(cid:12) a.s. −→ . roof. As a consequence of the Borel-Cantelli lemma, it is enough to verify thatfor any
T > ε > X N →∞ P (cid:18) sup t ∈ [0 ,T ] (cid:12)(cid:12)(cid:12)(cid:12) N − N µ ( m ( N +1)) N ( N t ) X k =1 G k − N µ ( m ( N +1)) N ( N t ) (cid:12)(cid:12)(cid:12)(cid:12) > ε (cid:19) < ∞ . (3.3)This follows from applying Doob’s martingale inequality and Markov’s in-equality.On the other hand, making use of the same arguments as in [1] leads us tothe fact that for any T > m ∈ Z ,sup t ∈ [0 ,T ] (cid:12)(cid:12)(cid:12)(cid:12) N µ ( m ( N +1))[ N t ] − L mt ( B ) (cid:12)(cid:12)(cid:12)(cid:12) a.s. −→ , where L mt ( B ) stands for the local time at m of the Brownian motion B . Itfollows that sup t ∈ [0 ,T ] (cid:12)(cid:12)(cid:12)(cid:12) N µ N ( N t ) − X m ∈ Z L mt ( B ) (cid:12)(cid:12)(cid:12)(cid:12) a.s. −→ . Combining with the above proposition, we can conclude thatsup t ∈ [0 ,T ] (cid:12)(cid:12)(cid:12)(cid:12) N − N µ N ( N t ) X k =1 G k − X m ∈ Z L mt ( B ) (cid:12)(cid:12)(cid:12)(cid:12) a.s. −→ . Since it can be checked that2 X m ∈ Z L mt ( B ) = 2 X m ∈ Z L mt ( B ) + 2 X m ∈ Z L m +1 t ( B ) = L t ( B rf ) + L t ( B rf ) = L t , then for any T >
0, this implies almost surely thatlim N →∞ sup t ∈ [0 ,T ] (cid:12)(cid:12)(cid:12)(cid:12) N − N T (0 , N + 1; N t ; Y rf ) − L t (cid:12)(cid:12)(cid:12)(cid:12) = 0 . For any
T >
0, applying again the same arguments as in [1] yields P ( lim N →∞ sup t ∈ [0 ,T ] (cid:12)(cid:12) N − X ( N t ) − B st ( t ) | = 0) = 1 . The sticky Brownian motion on the bounded interval [0 ,
1] also solves a stochas-tic differential equation similar to the one on the half line [0 , ∞ ) as consideredin [5]. Proposition 3.3.
The sticky Brownian motion B st , B st (0) = r , defined in (3.2) and the unique solution B to the following stochastic differential equation d B ( t ) = < B ( t ) < dW ( t ) + 12 B ( t )=0 dt − B ( t )=1 dt, B (0) = r, (3.4) for some standard Brownian motion W and r ∈ [0 , , have the same law. Theorem 3.4.
The unique solution u to the boundary value problelm (2.14) , (2.15) can be represented probabilistically, for r ∈ [0 , , t > , as u ( r, t ) = E r [ u ( B ( t )) < B ( t ) < + v , − B ( t )=0 + v , + B ( t )=1 ] , (3.5) where B solves the stochastic differential equation (3.4) . Proposition 3.3 and Theorem 3.4 are verified in the next sections.
The existence and uniqueness in law of the solution to the stochastic differentialequation (3.4) are proved in [13].Therefore, it suffices to verify that the sticky Brownian motion B st and theunique solution B have the same law. The idea to show this is looking fora Skorokhod problem that a suitable time change of the process B and thereflecting Brownian motion satisfy. More precisely, the proof consists of thefollowing steps. Step 1.
First, we show that the stochastic differential equation (3.4) is equiv-alent to the following system d B ( t ) = < B ( t ) < dW ( t ) + 12 dL t ( B ) − dL − t ( B ) ,dL t ( B ) = B ( t )=0 dtdL − t ( B ) = B ( t )=1 dt B (0) = r, (3.6)where L at ( B ) stands for the local time of B at a .It is obvious that the system (3.6) implies (3.4). For the converse, weremark that 0 ≤ B ( t ) ≤ t ≥ B is a solution of(3.4). This follows from using the Ito - Tanaka formula (see Theorem 1.2,[12]), namely B ( t ) − = − Z t B ( s ) < d B ( s ) + 12 L − t ( B ) = 0 , ( B ( t ) − + = Z t B ( s ) > d B ( s ) + 12 L t ( B ) = 0 , where we have used that L − t ( B ) = lim ε ↓ ε Z t − ε ≤ B ( s ) < d [ B ] s = lim ε ↓ ε Z t − ε ≤ B ( s ) < < B ( s ) < ds = 0 ,L t ( B ) = lim ε ↓ ε Z t ≤ B ( s ) < ε d [ B ] s
11 lim ε ↓ ε Z t ≤ B ( s ) < ε < B ( s ) < ds = 0 . In view of this remark and using again the Ito - Tanaka formula, we get B ( t ) = B ( t ) + = r + Z t B ( s ) > d B ( s ) + 12 L t ( B )= r + Z t < B ( s ) < dW ( s ) − Z t B ( s )=1 ds + 12 L t ( B )and this implies dL t ( B ) = B ( t )=0 dt .Similarly, we have − ( B ( t ) −
1) = ( B ( t ) − − = − ( r − − Z t B ( s ) < d B ( s ) + 12 L − t ( B )= − ( r − − Z t < B ( s ) < dW ( s ) − Z t B ( s )=0 ds + 12 L − t ( B )and this implies dL − t ( B ) = B ( t )=1 dt . Hence, we obtain the equivalenceof (3.4) and (3.6). Step 2.
Next, let us denote K ( t ) = Z t < B ( s ) < ds, κ ( t ) = inf { u ≥ K ( u ) > t } . Applying the time change κ to the first equation of (3.6) yields V ( t ) := B ( κ ( t )) = r + Z κ ( t )0 < B ( s ) < dW ( s ) + 12 L κ ( t ) ( B ) − L − κ ( t ) ( B )= r + Z κ ( t )0 < B ( s ) < dW ( s ) + 12 L t ( V ) − L − t ( V ) . Since Q ( t ) := r + Z κ ( t )0 < B ( s ) < dW ( s ) = r + Z t < B ( κ ( u )) < dW ( κ ( u ))is a continuous local martingale and note that[ Q ] t = Z t < B ( κ ( u )) < dκ ( u ) = Z κ ( t )0 < B ( s ) < ds = t, then Q is a Brownian motion starting from r by P. Levy’s characterizationtheorem. Moreover, due to the explicit expression of the solution to theSkorokhod problem V ( t ) = Q ( t ) + 12 L t ( V ) − L − t ( V ) , (see, e.g., [9], [3], [2]), we obtain V ( t ) = ˜ R ( Q ( t )) , (3.7)where˜ R ( Q ( t )) := Q ( t ) − (cid:2) ( r − + ∧ inf u ∈ [0 ,t ] Q ( u ) (cid:3) ∨ sup s ∈ [0 ,t ] (cid:2) ( Q ( s ) − ∧ inf u ∈ [ s,t ] Q ( u ) (cid:3) . tep 3. Let us denote the Brownian motion Q reflected at 0 and 1 by R ( Q ( t )) := X m ∈ Z | Q ( t ) − m | | Q ( t ) − m |≤ . (3.8)From [6], there is an explicit representation of Z := R ( Q ), the Brownianmotion Q starting from r reflected at two barriers 0 and 1, as follows Z ( t ) = r + Z t (cid:0) Q ( s ) ∈ S m ∈ Z (2 m, m +1) − Q ( s ) ∈ S m ∈ Z (2 m +1 , m +2) (cid:1) dQ ( s )+ X m ∈ Z L mt ( Q ) − X m ∈ Z L m +1 t ( Q ) . Set ˆ Q ( t ) := r + Z t (cid:0) { Q ( s ) ∈ S m ∈ Z (2 m, m +1) } − { Q ( s ) ∈ S m ∈ Z (2 m +1 , m +2) } (cid:1) dQ ( s ).We notice that ˆ Q is a Brownian motion starting from r by P. Levy’s char-acterization theorem. Moreover, X m ∈ Z L mt ( Q ) = 12 L t ( Z ) and X m ∈ Z L m +1 t ( Q ) = 12 L − t ( Z ) . (3.9)Hence, we can write Z ( t ) = ˆ Q ( t ) + 12 L t ( Z ) − L − t ( Z ) . Using again the explicit representation of the solution to the above Sko-rokhod problem gives us Z ( t ) = ˜ R ( ˆ Q ( t )) d = ˜ R ( Q ( t )) . (3.10)It follows from (3.7) and (3.10) that V ( t ) d = Z ( t ) = R ( Q ( t )) . (3.11) Step 4.
Moreover, using the second and the third equation of the system (3.6)gives us κ ( t ) = Z κ ( t )0 < B ( s ) < ds + Z κ ( t )0 B ( s )=0 ds + Z κ ( t )0 B ( s )=1 ds = t + L κ ( t ) ( B ) + L − κ ( t ) ( B )= t + L t ( V ) + L − t ( V ) d = t + L t ( Z ) + L − t ( Z ) . Then in view of (3.11), we deduce that Z ( t ) d = B ( t + L t ( Z ) + L − t ( Z )) . Since Z ( t ) d = B rf ( t ), the proof is complete.13 .4 Proof of Theorem 3.4 For any δ >
0, we fix t ≥ δ . Since the unique solution u ∈ C , ([0 , × (0 , ∞ )),we apply Ito’s formula to the function u ( B ( t ) , t − t ) for t ∈ [0 , t − δ ] and obtainthat u ( B ( t ) , t − t )= u ( r, t ) + Z t u r ( B ( s ) , t − s ) d B ( s ) + Z t u rr ( B ( s ) , t − s ) < B ( s ) < ds + Z t u s ( B ( s ) , t − s ) ds = u ( r, t ) + Z t u r ( B ( s ) , t − s ) < B ( s ) < dW ( s )+ Z t h u r ( B ( s ) , t − s ) + dds v − ( t − s ) i B ( s )=0 ds + Z t h − u r ( B ( s ) , t − s ) + dds v + ( t − s ) i B ( s )=1 ds = u ( r, t ) + Z t u r ( B ( s ) , t − s ) < B ( s ) < dW ( s ) . Let us call M ( t ) := Z t u r ( B ( s ) , t − s ) < B ( s ) < dW ( s ). Then M is a mar-tingale since u ∈ C , ([0 , × (0 , ∞ )). So E r [ M ( t − δ )] = E r [ M (0)] = 0.Hence, u ( r, t ) = E r [ u ( B ( t − δ ) , δ )] . Taking the limit δ ↓ u ( r, t ) = lim σ ↓ E r [ u ( B ( t ) , σ )]= lim σ ↓ E r [ u ( B ( t ) , σ ) < B ( t ) < + u (0 , σ ) B ( t )=0 + u (1 , σ ) B ( t )=1 ]= E r [ u ( B ( t )) < B ( t ) < + v , − B ( t )=0 + v , + B ( t )=1 ] . Since δ > t > Acknowledgements.
I would like to express my sincere gratitude to Prof.Errico Presutti for his great ideas which help me a lot to complete this paper.
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