C Bernardin

P Gonçalves

B Jiménez Oviedo

A MICROSCOPIC MODEL FOR A ONE PARAMETER CLASS OF FRACTIONAL LAPLACIANS WITH DIRICHLET BOUNDARY CONDITIONS

Keywords: Hydrodynamic limit, Heat equation, Boundary conditions, Exclusion with long jumps. 1

We prove the hydrodynamic limit for the symmetric exclusion process with long jumps given by a mean zero probability transition rate with infinite variance and in contact with infinitely many reservoirs with density α at the left of the system and β at the right of the system. The strength of the reservoirs is ruled by κN -θ > 0.

Here N is the size of the system, κ > 0 and θ ∈ . Our results are valid for θ ≤ 0. For θ = 0, we obtain a collection of fractional reaction-diffusion equations indexed by the parameter κ and with Dirichlet boundary conditions. Their solutions also depend on κ. For θ < 0, the hydrodynamic equation corresponds to a reaction equation with Dirichlet boundary conditions. The case θ > 0 is still open. For that reason we also analyze the convergence of the unique weak solution of the equation in the case θ = 0 when we send the parameter κ to zero. Indeed, we conjecture that the limiting profile when κ → 0 is the one that we should obtain when taking small values of θ > 0.

INTRODUCTION

Normal (diffusive) transport phenomena are described by standard random walk models. Anomalous transport, in particular transport phenomena giving rise to superdiffusion, are nowadays encapsulated in the Lévy flights or Lévy walks framework [START_REF] Dubkov | Lévy flight Superdiffusion: An Introduction[END_REF][START_REF] Denisov | [END_REF] and appear in physics, finance, biology ... The term "Lévy flight" was coined by Mandelbrot and is nothing but a random walk in which the step-lengths have a probability distribution that is heavy tailed. A (one-dimensional) Lévy walker moves with a constant velocity v for a heavy-tailed random time τ on a distance x = vτ in either direction with equal probability and then chooses a new direction and moves again. One then easily shows that for Lévy flights or Lévy walks, the space-time scaling limit P(x, t) of the probability distribution of the particle position x(t) is solution of the fractional diffusion equation

∂ t P = -c(-∆) γ/2 P (1.1)
where c is a constant and γ ∈ (1, 2). In physics, the description of anomalous transport phenomena by Lévy walks instead of Lévy flights is sometimes preferred despite the two models having the same scaling limit form provided by (1.1) because the first ones have a finite speed of propagation (see [START_REF] Denisov | [END_REF] for more details). While Lévy walks and Lévy flights are today well known and popular models to describe superdiffusion in infinite systems in various application fields, there has been recently several physical studies pointing out that it would be desirable to have a better understanding of Lévy walks in bounded domains. For bounded domains, boundary conditions and exchange with reservoirs or environment have to be taken into account. A particular interest for this problem is related to the description of anomalous diffusion of energy in low-dimensional lattices [START_REF] Dhar | Heat Transport in low-dimensional systems[END_REF][START_REF] Lepri | Thermal Conduction in classical low-dimensional lattices[END_REF] in contact with reservoirs [START_REF] Dhar | Anomalous transport and current fluctuations in a model of diffusing Levy walkers[END_REF][START_REF] Dhar | Exact solution of a Levy walk model for anomalous heat transport[END_REF][START_REF] Lepri | Density profiles in open superdiffusive systems[END_REF][START_REF] Kundu | Fractional equation description of an open anomalous heat conduction set-up[END_REF]]. It is for example argued in [START_REF] Lepri | Density profiles in open superdiffusive systems[END_REF] that the density profiles of Lévy walkers in a finite box with absorbtion-reflection-creation well reproduces the temperature profile of some chains of harmonic oscillators with conservative momentum-energy noise and thermostat boundaries. It is well established that superdiffusive systems are much more sensitive to the reservoirs and boundaries than diffusive systems but quantitative informations, like the form of the singularities of the profiles at the boundaries, are still missing.

In this work, motivated by these studies, we propose a simple interacting particle system which may be considered as a substitute to Lévy flights in bounded domains with reservoirs when Lévy flights are moreover interacting. Indeed, the previous studies consider only non-interacting cases. The system considered here is composed of interacting Lévy flights on a one-dimensional lattice. More exactly, the system is an exclusion process on a finite lattice of size N with jumps having a distribution in the form p(z) ∼ |z| -(1+γ) , with γ > 0, and in contact with some reservoirs at density α (resp. β) at its left (resp. right boundary). The reservoirs' coupling is modulated by a prefactor κN -θ , κ > 0, θ ∈ . In this work we focus on the case γ ∈ (1, 2) (the case γ > 2 was solved in [START_REF] Bernardin | Slow to Fast infinitely extended reservoirs for the symmetric exclusion process with long jumps[END_REF]) and we also restrict to the case θ ≤ 0. The cases θ > 0, γ ∈ (0, 1] and γ = 2 remain open.

Our main result is the derivation of the hydrodynamic limit for the density of particles for this system. The limiting PDE depends 1 on the value of κ and takes the form of a fractional heat equation with a singular reaction term, see (2.10). The singular reaction term fixes the density on the left to be α and on the right to be β. In our opinion this singular reaction term, which is due to the presence of the reservoirs, should be considered more as a boundary condition than as a reaction term. We obtain in this way a new family of regional fractional Laplacians on [0, 1] with zero Dirichlet boundary conditions indexed by κ and taking the form

κ = -κV 1 , V 1 (u) = r -(u) + r + (u), (1.2)
where r -(u) = c γ γ -1 u -γ and r + (u) = c γ γ -1 (1 -u) -γ and c γ is a constant depending on γ. These operators are symmetric non-positive when restricted to the set of smooth functions compactly supported in (0, 1). For κ = 1, we recover the so-called restricted fractional Laplacian while in the limit κ → 0 we get the so-called regional fractional Laplacian. We recall that since the fractional Laplacian is a non-local operator, the definition of a fractional Laplacian with Dirichlet boundary conditions is not obvious from a modeling point of view. In the PDE's literature several candidates have been proposed, for instance, "restricted fractional Laplacian", "spectral fractional Laplacian", "Neumann Fractional Laplacian " [2,[START_REF] Vázquez | Recent progress in the theory of Nonlinear Diffusion with Fractional Laplacian Operators[END_REF], but often without a clear physical interpretation. A probabilistic interpretation of these operators is sometimes possible and may enlighten their meaning. The restricted fractional Laplacian (κ = 1) corresponds to the generator of a γ-Lévy stable process killed outside of (0, 1), while the regional fractional Laplacian (κ = 0) corresponds to the generator of a censored γ-Lévy stable process on (0, 1) [START_REF] Bogdan | Censored stable processes[END_REF][START_REF] Guan | The reflected α-symmetric stable processes and regional fractional Laplacian[END_REF]. For κ = 0, 1 we could rely on the Feynman-Kac formula but we do not pursue this issue here. As mentioned above our reservoirs are regulated by the parameters κN -θ , κ > 0 and in this work we focus on the case 1 In the diffusive case γ > 2 the limiting PDE is given by the heat equation with Dirichlet boundary conditions [START_REF] Bernardin | Slow to Fast infinitely extended reservoirs for the symmetric exclusion process with long jumps[END_REF]. It does not depend on κ.

θ ≤ 0. The case θ > 0 is quite interesting and we conjecture that for small values of θ > 0 it is given by (2.10) for the choice κ = 0. To support this conjecture, in Theorem 2.13, we analyse the convergence of the profile that we obtained for θ = 0 and which is indexed in κ, when κ → 0 (we also analyse the case κ → ∞ confirming the behaviour obtained from the microscopic system when θ < 0) and indeed, we obtain that the limiting profiles are weak solution of the conjectured equation. We remark that the main problem in analysing the behavior of the microscopic system in this case is at the level of the derivation of the Dirichlet boundary conditions, since the twoblocks estimate does not work. We leave this open problem for a future work. After having obtained the hydrodynamic limits, we have studied their stationary solutions ρκ , which are not explicit apart from the case κ = 1 and the case κ = ∞, i.e. ρ∞ = lim κ→∞ ρκ . These profiles coincide with the profiles of the microscopic system in their non-equilibrium stationary states (see [START_REF] Bernardin | Fractional Fick's law for the boundary driven exclusion process with long jumps[END_REF] for the κ = 1 case). The bounded continuous function ρκ has α and β as boundary conditions and solves in a distributional sense the equation

κ ρκ = -κV 0 , V 0 (u) = αr -(u) + β r + (u).
(

1.3)

There are many recent studies focusing on the regularization properties of fractional operators in bounded domains. Even in this one dimensional setup, the question is in general non trivial. For κ = 1, ρκ can be computed explicitly and it appears that it is smooth in the interior of [0, 1] but has only Hölder regularity equal to γ/2 at the boundaries. For κ = 1, it should be possible to prove the interior regularity of ρκ by some existing methods ( [START_REF] Mou | Interior regularity for regional fractional Laplacian[END_REF]) but the boundary regularity that numerical simulations seem to indicate depends on κ is much more challenging and seems to be open. We prove that as κ → 0, ρκ → ρ0 in a suitable topology and that ρ0 is a weakly harmonic function of the regional fractional Laplacian 0 , i.e. we can take κ = 0 in (1.3). We left these interesting questions for future works. The paper is organized as follows. In Section 2 we introduce the model and we present all the PDE's that will be related to its hydrodynamic limit. We also present the main results of this work, namely the hydrodynamic limit stated in Theorem 2.12, the convergence, when κ → 0 and when κ → ∞, of the hydrodynamical profile in Theorem 2.13 and of the stationary profile in Theorem 2.15. Section 3 is devoted to the proof of Theorem 2.12 while Sections 4 and 5 are dedicated, respectively, to the convergence of the hydrodynamical profile and of the stationary profile. Finally, in Section 6 we prove the uniqueness of all the weak solutions that we consider in this work.

STATEMENT OF RESULTS

2.1. The model. For N ≥ 2 let Λ N = {1, . . . , N -1}, which we refer to as the bulk. The boundary driven exclusion process with long jumps is a Markov process that we denote by {η(t)} t≥0 with state space Ω N := {0, 1} Λ N and it is defined as follows. The configurations of the state space Ω N are denoted by η, so that for x ∈ Λ N , η x = 0 means that the site x is vacant while η x = 1 means that the site x is occupied. Fix γ ∈ (1, 2). Let p : → [0, 1] be a translation invariant transition probability defined by

p(z) = c γ {z =0} |z| γ+1 (2.1)
where c γ > 0 is a normalizing constant. Since γ ∈ (1, 2), we know that p has infinite variance but finite mean.

We consider the process in contact with infinitely many stochastic reservoirs at the left and right of Λ N . We fix the parameters α, β ∈ (0, 1), κ > 0 and θ ≤ 0. Particles can be injected into any site z of the bulk from: the left of 0 at rate ακN -θ p(z) or from the right of N at rate βκN -θ p(z). Particles can be removed from any site of the bulk to: the left of 0 at rate (1 -α)κN -θ p(z) and to the right of N at rate (1 -β)κN -θ p(z). To properly describe the dynamics, at each pair of sites of the bulk {x, y} we associate a Poisson process of intensity one and Poisson processes associated with different bonds are independent. Whenever a clock associated with a bond {x, y} rings, the values of η x and η y are exchanged with rate p( yx)/2. At the boundary the dynamics is described as follows. To each pair of sites {x, y} with x ∈ Λ N and y ≤ 0 (resp. y ≥ N ) we associate a Poisson process of intensity one and of them are independent. If the clock associated with the bond {x, y} rings, the value of

η x changes to 1 -η x with rate κN -θ p(x -y) [(1 -α)η x + α(1 -η x )] (resp. κN -θ p(x -y) [(1 -β)η x + β(1 -η x )]).
The dynamics is illustrated in the figure below.

x y N 0

Left reservoir

Right reservoir

p( y -x) (1 -α) κ N θ p(•) α κ N θ p(•) (1 -β) κ N θ p(•) β κ N θ p(•)
The process is characterized by its infinitesimal generator

L N = L 0 N + κN -θ L N + κN -θ L r N , (2.2)
which acts on functions f : Ω N → as

(L 0 N f )(η) = 1 2 x, y∈Λ N p(x -y)[ f (σ x, y η) -f (η)], (L N f )(η) = x∈Λ N y≤0 p(x -y)c x (η; α)[ f (σ x η) -f (η)], (L r N f )(η) = x∈Λ N y≥N p(x -y)c x (η; β)[ f (σ x η) -f (η)] (2.3)
where

(σ x, y η) z =    η z , if z = x, y, η y , if z = x, η x , if z = y , (σ x η) z = η z , if z = x, 1 -η x , if z = x,
and for a function ϕ : [0, 1] → and for x ∈ Λ N we used the notation

c x (η; ϕ(•)) := η x 1 -ϕ( x N ) + (1 -η x )ϕ( x N ) . (2.4)
We consider the Markov process speeded up in the subdiffusive time scale tΘ(N ) and we use the notation η N t := η(tΘ(N )), so that η N t has infinitesimal generator Θ(N )L N . Although η N t depends on α, β θ and κ, we shall omit these indexes in order to simplify notation.

Hydrodynamic equations.

From now on up to the rest of this article we fix a finite time window [0, T ]. To properly state the hydrodynamic limit, we need to introduce some notations and definitions, which we present as follows: first we abbreviate the Hilbert space L 2 ([0, 1], h(u)du) by L 2 h and we denote its inner product by 〈•, •〉 h and the corresponding norm by • h . When h ≡ 1 we simply write L 2 , 〈•, •〉 and • . For an interval I in and integers m and n, we denote by C m,n ([0, T ] × I) the set of functions defined on [0, T ] × I that are m times differentiable on the first variable and n times differentiable on the second variable, with continuous derivatives. We denote by C ∞ c (I) the set of all smooth real-valued functions defined in I with compact support included in I. The supremum norm is denoted by • ∞ . We also consider the set

C 1,∞ c ([0, T ] × I) of functions G ∈ C 1,∞ ([0, T ] × I) such that G(t, •) ∈ C ∞ c (I) for all t ∈ [0, T ].
An index on a function will always denote a variable, not a derivative. For example, G t (u) means G(t, u). The derivative of G ∈ C m,n ([0, T ] × I) will be denoted by ∂ t G (first variable) and ∂ u G (second variable).

The fractional Laplacian -(-∆) γ/2 of exponent γ/2 is defined on the set of functions

G : → such that ∞ -∞ |G(u)| (1 + |u|) 1+γ du < ∞ (2.5) by -(-∆) γ/2 G (u) = c γ lim →0 ∞ -∞ |u-v|≥ G(v) -G(u) |u -v| 1+γ d v (2.6)
provided the limit exists (which is the case, for example, if G is in the Schwartz space) and where c γ is set in (2.1). Up to a multiplicative constant, -(-∆) γ/2 is the generator of a γ-Lévy stable process. We define the operator by its action on functions G ∈ C ∞ c ((0, 1)), by

∀u ∈ (0, 1), ( G)(u) = c γ lim →0 1 0 |u-v|≥ G(v) -G(u) |u -v| 1+γ d v.
The operator is called the regional fractional Laplacian on (0, 1). The semi inner-

product 〈•, •〉 γ/2 is defined on the set C ∞ c ((0, 1)) by 〈G, H〉 γ/2 = c γ 2 [0,1] 2 (H(u) -H(v))(G(u) -G(v)) |u -v| 1+γ dud v. (2.7)
The corresponding semi-norm is denoted by • γ/2 . Observe that for any G, H ∈ C ∞ c ((0, 1)) we have that 〈G, -H〉 = 〈-G, H〉 = 〈G, H〉 γ/2 .

Recall (1.2). We introduced a family of operators indexed by κ and taking the form κ = -κV 1 , where V 1 was defined in 1.2. Acting on C ∞ c ((0, 1)) these operators are symmetric and non-positive. For κ = 1, we recover the so-called restricted fractional Laplacian (see [START_REF] Vázquez | Recent progress in the theory of Nonlinear Diffusion with Fractional Laplacian Operators[END_REF]):

∀u ∈ (0, 1), -(-∆) γ/2 G (u) = ( G)(u) -V 1 (u)G(u) := ( 1 G)(u),
(2.8) while in the limit κ → 0 we get the regional fractional Laplacian. We rewrite V 1 (u) = r -(u) + r + (u) and V 0 (u) = αr -(u) + β r + (u) where the functions r ± : (0, 1) → (0, ∞) are defined by

r -(u) = c γ γ -1 u -γ , r + (u) = c γ γ -1 (1 -u) -γ . ( 2
f : [0, T ] → γ/2 such that T 0 f t 2 γ/2 d t < ∞.
The spaces L 2 (0, T ; γ/2 0 ) and L 2 (0, T ; L 2 h ) are defined similarly. We now extend the definition of the regional fractional Laplacian on (0, 1), which has been defined on C ∞ ((0, 1)), to the space γ/2 . Definition 2.2. For ρ ∈ γ/2 we define the distribution ρ by 〈 ρ, G〉 = 〈ρ, G〉, G ∈ C ∞ c ((0, 1)). Let us check that ρ is indeed a well defined distribution. Consider a sequence {G n } n≥1 ∈ C ∞ c ((0, 1)) converging to 0 in the usual topology of the test functions. By the integration by parts formula for the regional fractional Laplacian (see Theorem 3.3 in [START_REF] Guan | The reflected α-symmetric stable processes and regional fractional Laplacian[END_REF]) we have for any ρ ∈ γ/2 that 〈 ρ, G n 〉 = 〈ρ, G n 〉 γ/2 . Now using the Cauchy-Schwarz's inequality and the mean value Theorem, we get that 〈 ρ, G n 〉 is bounded from above by a constant times

ρ γ/2 G n γ/2 ρ γ/2 ∂ u G n 2 ∞ [0,1] 2 |u -v| 1-γ dud v
which goes to 0 as n → ∞ since γ ∈ (1, 2). Therefore ρ is a well defined distribution. Above (and hereinafter) we write f (u) g(u) if there exists a constant C independent of u such that f (u) ≤ C g(u) for every u. We will also write

f (u) = O(g(u)) if the condition | f (u)| |g(u)| is satisfied.
Sometimes, in order to stress the dependence of a constant C on some parameter a, we write C(a).

2.3. Hydrodynamic equations. Now, for the following definitions recall the definition of κ given in (1.2) and V 0 from (1.3). Definition 2.3. Let κ ≥ 0 be some parameter and let g : [0, 1] → [0, 1] be a measurable function. We say that ρ κ : [0, T ] × [0, 1] → [0, 1] is a weak solution of the non-homogeneous regional fractional reaction-diffusion equation with Dirichlet boundary conditions given by

   ∂ t ρ κ t (u) = κρ κ t (u) + κV 0 (u), (t, u) ∈ [0, T ] × (0, 1), ρ κ t (0) = α, ρ κ t (1) = β, t ∈ [0, T ], ρ κ 0 (u) = g(u), u ∈ (0, 1), (2.10
)

if : i) ρ κ ∈ L 2 (0, T ; γ/2 ). ii) T 0 1 0 (α-ρ κ t (u)) 2 u γ + (β-ρ κ t (u)) 2 (1-u) γ du d t < ∞ for κ > 0; ρ κ t (0) = α, ρ κ t (1) = β for almost every t ∈ [0, T ], for κ = 0. iii) For all t ∈ [0, T ] and all functions G ∈ C 1,∞ c ([0, T ] × (0, 1)) we have that F Dir (t, ρ κ, G, g) := ρ κ t , G t -〈g, G 0 〉 - t 0 ρ κ s , ∂ s + κ G s ds -κ t 0 〈G s , V 0 〉 ds = 0.
(2.11)

Remark 2.4. Note that item ii) is different for κ > 0 and κ = 0. We can see that the condition for κ = 0 is weaker than the condition for κ > 0. In fact, item i) and item ii) for κ > 0 of the previous definition imply that ρ κ t (0) = α and ρ κ t (1) = β, for almost every t in [0, T ]. Indeed, first note that by item i) we know that ρ t is γ-1 2 -Hölder for almost every t in [0, T ] (see Theorem 8.2 of [START_REF] Di Nezza | Hitchhiker's guide to the fractional Sobolev spaces[END_REF] ). Then, we note that T 0

(ρ κ t (0) -α) 2 γ -1 d t = T 0 lim →0 γ-1 1 (ρ κ t (0) -α) 2 u γ dud t.
By summing and subtracting ρ κ t (u) inside the square in the expression on the right hand side in the previous equality and using the inequality (a + b) 2 ≤ 2a 2 + 2b 2 we get that the right hand side of the previous equality is bounded from above by

2 T 0 lim →0 γ-1 1 (ρ κ t (0) -ρ κ t (u)) 2 u γ dud t + 2 T 0 lim →0 γ-1 1 (ρ κ t (u) -α) 2 u γ dud t. Since ρ t is γ-1
2 -Hölder for almost every t in [0, T ] the first term in the previous expression vanishes. Now, the term on the right hand side in the previous expression is bounded from above by

2 lim →0 γ-1 T 0 1 0 (ρ κ t (u) -α) 2 u γ dud t,
which vanishes as a consequence of item ii). Thus, we have that

T 0 (ρ κ t (0) -α) 2 γ -1 d t = 0,
whence we get that ρ κ t (0) = α for almost every t in [0, T ]. Showing that ρ κ t (1) = β for almost every t in [0, T ] is completely analogous.

Moreover, the existence and uniqueness of a weak solution to the equation above, for κ > 0 does not require the strong form of ii). Nevertheless, in order to prove Theorem 2.13 we need to impose that condition. Remark 2.5. Observe that in the case κ = 1, since 1 = -(-∆) γ/2 we obtain in Definition 2.3 the fractional heat equation with reaction and Dirichlet boundary conditions, i.e.

   ∂ t ρ 1 t (u) = 1 ρ 1 t (u) + V 0 (u), (t, u) ∈ [0, T ] × (0, 1), ρ 1 t (0) = α, ρ 1 t (1) = β, t ∈ [0, T ], ρ 1 0 (u) = g(u)
, u ∈ (0, 1), by (2.8) and (1.2) the notion of item iii) is reduced to

F Dir (t, ρ 1 , G, g) := ρ 1 t , G t -〈g, G 0 〉- t 0 ρ 1 s , ∂ s -(-∆) γ/2 G s ds- t 0 〈G s , V 0 〉 ds = 0, for all t ∈ [0, T ] and all functions G ∈ C 1,∞ c ([0, T ] × (0, 1)).
Definition 2.6. Let κ > 0 be some parameter and let g : [0, 1] → [0, 1] be a measurable function. We say that ρ κ :

[0, T ] × [0, 1] → [0, 1]

is a weak solution of the non-homogeneous reaction equation with Dirichlet boundary conditions given by

   ∂ t ρ κ t (u) = -κρ κ t (u)V 1 (u) + κV 0 (u), (t, u) ∈ [0, T ] × (0, 1), ρ κ t (0) = α, ρ κ t (1) = β, t ∈ [0, T ], ρ κ 0 (u) = g(u), u ∈ (0, 1), (2.12
)

if: i) T 0 1 0 (α-ρ κ t (u)) 2 u γ + (β-ρ κ t (u)) 2 (1-u) γ du d t < ∞. ii) For all t ∈ [0, T ] and all functions G ∈ C 1,∞ c ([0, T ] × (0, 1)) we have F Reac (t, ρ κ, G, g) := ρ κ t , G t -〈g, G 0 〉 - t 0 ρ κ s , ∂ s G s ds + t 0 ρ κ s , G s V 1 ds -κ t 0 〈G s , V 0 〉 ds = 0.
(2.13) Remark 2.7. Note that the explicit solution of (2.12) is given by

ρ∞ (u) + (g(u) -ρ∞ (u))e -t κV 1 (u) , where ρ∞ (u) = V 0 (u) V 1 (u)
. As we will see, the function ρ∞ plays an important role in the proof of our main results, namely, Theorems 2.13 and 2.15.

Lemma 2.8. The weak solutions of (2.10) and (2.12) are unique.

Aiming to concentrate on the main facts, the proof of previous lemma is postponed to Section 6. Definition 2.9. Let κ ≥ 0 be some parameter. We say that ρκ :

[0, 1] → [0, 1]

is a weak solution of the stationary regional fractional reaction-diffusion equation with nonhomogeneous Dirichlet boundary conditions given by

κ ρκ (u) + κV 0 (u) = 0, u ∈ (0, 1), ρκ (0) = α, ρκ (1) = β, (2.14) if: i) ρκ ∈ γ/2 .
ii)

1 0 (α-ρκ (u)) 2 u γ + (β-ρκ (u)) 2 u γ du < ∞ if κ > 0 and ρκ (0) = α, ρκ (1) = β if κ = 0. iii) For any function G ∈ C ∞ c ((0, 1)) we have FDir ( ρκ , G) := ρκ , κ G + κ 〈G, V 0 〉 = 0.
Remark 2.10. We observe that ρ0 is a weak harmonic function for and the interior regularity of this solution is studied in [START_REF] Mou | Interior regularity for regional fractional Laplacian[END_REF], but the regularity at the boundary is unknown.

In Section 6 we will prove the following lemma.

Lemma 2.11.

There exists a unique weak solution of (2.14).

Statement of results.

First we want to state the hydrodynamic limit of the process {η N t } t≥0 with state space Ω N and with infinitesimal generator Θ(N )L N defined in (2.2). Let + be the space of positive measures on [0, 1] with total mass bounded by 1 equipped with the weak topology. For any configuration η ∈ Ω N we define the empirical measure π N (η, du)

:= π N ,κ (η, du) in Ω N by π N (η, du) = 1 N -1 x∈Λ N η x δ x N (du) , (2.15)
where δ a is a Dirac mass at a ∈ [0, 1] and

π N t (η, du) := π N (η N t , du). Let g : [0, 1] → [0, 1] be a measurable function. We say that a sequence of probabil- ity measures {µ N } N ≥1 in Ω N is associated to the profile g if for any continuous function G : [0, 1] → and every δ > 0 lim N →∞ µ N η ∈ Ω N : 1 N x∈Λ N G x N η x - 1 0 G(u)g(u)du > δ = 0.
We denote by µ N the probability measure in the Skorohod space ([0, T ], Ω N ) induced by the Markov process η N t and the initial measure µ N in Ω N and we denote by µ N the expectation with respect to µ N . Let { N } N ≥1 be the sequence of probability measures on the Skorohod space ([0, T ], + ) induced by the Markov process {π N t } t≥0 and by µ N .

At this point we are ready to state the hydrodynamic limit of the process η N t . Theorem 2.12. (Hydrodynamic limit) Let g : [0, 1] → [0, 1] be a measurable function and let {µ N } N ≥1 be a sequence of probability measures in Ω N associated to g. Then, for any 0 ≤ t ≤ T ,

lim N →∞ µ N η N • ∈ ([0, T ], Ω N ) : 1 N x∈Λ N G x N η x (tΘ(N )) - 1 0 G(u)ρ κ t (u)du > δ = 0,
where the time scale is given by Θ(N ) = N γ+θ and ρ κ t is the unique weak solution of:

• (2.12) with κ = κ, if θ < 0; • (2.10) with κ = κ, if θ = 0.
Once the hydrodynamic limit is obtained, we would like to know how the weak solution ρ κ t and the stationary solution ρκ behave as κ goes to 0 or ∞ and this is the purpose of Theorems 2.13 and 2.15 stated below. This limiting profile will give us an idea of what to expect at the hydrodynamics level when we consider our microscopic dynamics in contact with reservoirs whose strength is regulated by κ/N θ and when θ > 0 as in [START_REF] Bernardin | Slow to Fast infinitely extended reservoirs for the symmetric exclusion process with long jumps[END_REF]. As mentioned in the introduction we do not analyze the system in this regime but we conjecture that for small positive values of θ > 0 (that corresponds to slow reservoirs) the hydrodynamic limit should be given by the weak solution of (2.10) with κ = 0. Theorem 2.13. Let ρ 0 : [0, 1] → [0, 1] be a measurable function. Further, let ρ κ be the weak solution of (2.10) with κ = κ and with initial condition ρ 0 which is independent of κ and let ρκ t := ρ κ t/κ , for all t ∈ [0, T ]. Then i) ρ κ converges strongly to ρ 0 in L 2 (0, T ; γ/2 ) as κ goes to 0, where ρ 0 is the weak solution of (2.10) with κ = 0 and initial condition

ρ 0 . ii) If ρ 0 -ρ∞ ∈ γ/2 then ρκ converges strongly to ρ ∞ in L 2 (0, T ; L 2 V 1
) as κ goes to ∞, where ρ ∞ is the weak solution of (2.12) with κ = 1 and initial condition ρ 0 .

Remark 2.14. The convergence in Theorem 2.13 is also true in L 2 (0, T ; L 2 ). In fact, we will see that a crucial step in the proof of the theorem is to show that ρ κ converges strongly in L 2 (0, T ;

L 2 ). Convergence in i) is also true in L 2 (0, T ; L 2 V 1
) and it is a consequence of the fractional Hardy's inequality (see e.g. [START_REF] Dyda | A fractional order Hardy inequality[END_REF]).

Theorem 2.15. Let ρκ be the weak solution of (2.14). Then, i) ρκ converges strongly to ρ0 in γ/2 as κ goes to 0, where ρ0 is the weak solution of (2.14)

with κ = 0. ii) ρκ converges strongly to ρ∞ in L 2 V 1
as κ goes to ∞, where ρ∞ is given in Remark 2.7.

PROOF OF THEOREM 2.12: HYDRODYNAMIC LIMIT

The proof of this theorem follows the usual approach of convergence in distribution of stochastic processes: we prove tightness of the sequence { N } N ≥1 and then we prove uniqueness of the limiting point, which we denote by . These two results combined give the convergence of { N } N ≥1 to , as N → ∞. In order to characterize the limiting point , we prove that all limiting points of the sequence { N } N ≥1 are concentrated on trajectories of measures that are absolutely continuous with respect to the Lebesgue measure and whose density ρ κ t is a weak solution of the hydrodynamic equation as given in Definition 2.3. From the uniqueness of the weak solutions of this equation, namely Lemma 2.11, we conclude that { N } N ≥1 has a unique limit point .

First, in the following subsection we explain how the item iii) in Definition 2.3 appears. In Subsection 3.2 we prove that { N } N ≥1 is tight, then in Subsection 3.3 we obtain energy estimates which are crucial to ensure the uniqueness of the limiting point. We conclude this section with the characterization of the limiting point (in Subsection 3.4).

Heuristics for the hydrodynamic equations.

In order to make the presentation simple, let us fix a function G : [0, 1] → which does not depend on time and has compact support included in (0, 1). By Dynkin's formula (see Lemma A.5.1 in [START_REF] Kipnis | Scaling Limits of Interacting Particle Systems[END_REF]) we have that

M N t (G) = 〈π N t , G〉 -〈π N 0 , G〉 - t 0 Θ(N )L N 〈π N s , G〉ds, (3.1)
is a martingale with respect to the natural filtration { t } t≥0 where t := σ({η(s)} s≤t ) for all t ∈ [0, T ].

Above, for an integrable function G : [0, 1] → , we used the notation 〈π N t , G〉 to represent the integral of G with respect the measure π N t :

π N t , G = 1 N -1 x∈Λ N G x N η x (tΘ(N )).
In the previous expression, we are using a measure π N t and a function G, therefore, this notation should not be mistaken with the one used for the inner product in L 2 . Note that L N η x is equal to

y∈Λ N p(x -y)[η y -η x ] + κ N θ y≤0 p(x -y)[α -η x ] + κ N θ y≥N p(x -y)[β -η x ].
Therefore, a simple computation shows that

Θ(N )L N (〈π N , G〉) = Θ(N ) N -1 x∈Λ N ( N G)( x N )η x + κΘ(N ) (N -1)N θ x∈Λ N G( x N ) r - N ( x N )(α -η x ) + r + N ( x N )(β -η x ) , (3.2)
where, we denote by N G the continuous function on [0, 1] which is defined as the linear interpolation of the function

( N G)( x N ) = y∈Λ N p( y -x) G( y N ) -G( x N ) , (3.3) for all x ∈ Λ N with ( N G)(0) = ( N G)(1) = 0.
We also define the functions r ± N : [0, 1] → as the linear interpolation of the function

r - N ( x N ) = y≥x p( y), r + N ( x N ) = y≤x-N p( y), (3.4)
for all x ∈ Λ N with r ± N (0) = r ± N ( 1 N ) and r ± N (1) = r ± N ( N -1 N )
. By Lemma 3.3 in [START_REF] Bernardin | Fractional Fick's law for the boundary driven exclusion process with long jumps[END_REF] we have that lim

N →∞ N γ (r - N )(u) = r -(u), lim N →∞ N γ (r + N )(u) = r + (u) (3.5)
uniformly in [a, 1 -a] for a ∈ (0, 1) and we also can deduce from that lemma that lim

N →∞ N γ ( N G)(u) = ( G)(u) (3.6)
uniformly in [a, 1 -a], for all functions G with compact support included in [a, 1 -a]. Now, we are going to analyse all the terms in (3.2) for θ ≤ 0. Thus, we will be able to see how the different boundary conditions appear on the hydrodynamic equations given in Subsection 2.3 from the underlying particle system.

3.1.1. The case θ < 0. In this regime we take Θ(N ) = N γ+θ and a function G ∈ C ∞ c (0, 1). By using (3.6) we have that the first term on the right hand side of (3.2) vanishes since θ < 0. Now, the second term on the right hand side in (3.2) is equal to κ〈α -

π N t , N γ G r - N 〉 + κ〈β -π N t , N γ G r + N 〉. By (3.5) the previous expression converges, as N goes to ∞, to κ 1 0 (α -ρ κ t (u))G(u)r -(u)du + κ 1 0 (β -ρ κ t (u))G(u)r + (u)du = -κ 1 0 ρ κ t (u)G(u)V 1 (u)du + κ 1 0 G(u)V 0 (u)du.
3.1.2. The case θ = 0. In this regime we take N γ and a function G ∈ C ∞ c (0, 1). The first term on the right hand side in (3.2) can be replaced, thanks to (3.6) by

〈π N t , G〉 → 1 0 ( G)(u)ρ κ t (u)du,
as N goes to ∞. Similarly, the second term on the right hand side of (3.2) is equal to

κ〈α -π N t , N γ G r - N 〉 + κ〈β -π N t , N γ Gr + N 〉 which converges, as N goes to ∞, to κ 1 0 (α -ρ κ t (u))G(u)r -(u)du + κ 1 0 (β -ρ κ t (u))G(u)r + (u)du = -κ 1 0 ρ κ t (u)G(u)V 1 (u)du + κ 1 0 G(u)V 0 (u)du.
This intuitive argument is rigorously proved in Subsection 3.4.

Tightness.

In this subsection we prove that the sequence { N } N ≥1 is tight. We use the usual approach (see, for example, Proposition 4.1.6 and 4.1.7 in [START_REF] Kipnis | Scaling Limits of Interacting Particle Systems[END_REF]), which says that is enough to show that, for all > 0 lim δ→0 lim sup

N →∞ sup τ∈ T ,τ≤δ µ N η N • ∈ ([0, T ], Ω N ) : 〈π N τ+τ , G〉 -〈π N τ , G〉 > = 0, (3.7)
for any function G belonging to C([0, 1]) . Above T is the set of stopping times bounded by T and we implicitly assume that all the stopping times are bounded by T , thus, τ + τ should be read as (τ + τ) ∧ T . Indeed, we prove below that (3.7) is true for any function G in C 2 c ((0, 1)), by using an L 1 approximation procedure(a similar argument as done in [START_REF] Bernardin | Slow to Fast infinitely extended reservoirs for the symmetric exclusion process with long jumps[END_REF]), we can extend this class of functions to functions G ∈ C([0, 1]).

Proposition 3.1. The sequence of measures { N } N ≥1 is tight with respect to the Skorohod topology of ([0, T ], + ).

Proof. Note that, we are going to prove (3.7) for functions G in C 2 c ((0, 1)). Recall from (3.1) that M N t (G) is a martingale with respect to the natural filtration { t } t≥0 . In order to prove (3.7) it is enough to show that lim δ→0 lim sup

N →∞ sup τ∈ T ,τ≤δ µ N τ+τ τ Θ(N )L N 〈π N s , G〉ds = 0 (3.8)
and lim δ→0 lim sup

N →∞ sup τ∈ T ,τ≤δ µ N M N τ (G) -M N τ+τ (G) 2 = 0. ( 3
r - N ( x N ) + r + N ( x N ) < ∞ (3.10) (since γ > 1)
, we can bound from above the second term at the right hand side in (3.2) by a constant times Θ(N )N -1-θ . Considering the different values of θ we see that such term is bounded from above by a constant. Then we have that

|Θ(N )L N (〈π N s , G〉)| 1 (3.11)
for any s ≤ T , which trivially implies (3.8).

In order to prove (3.9), by Dynkin's formula (see Appendix 1 in [START_REF] Kipnis | Scaling Limits of Interacting Particle Systems[END_REF]) we know that

M N t (G) 2 - t 0 Θ(N ) L N 〈π N s , G〉 2 -2〈π N s , G〉L N 〈π N s , G〉 ds,
is a martingale with respect to the natural filtration { t } t≥0 . By Lemma A.1 we get that the term inside the time integral in the previous expression is equal to

Θ(N ) (N -1) 2 x< y∈Λ N G x N -G y N 2 p(x -y)(η y (sΘ(N )) -η x (sΘ(N ))) 2 + κΘ(N ) (N -1) 2 N θ x∈Λ N G x N 2 (1 -2η x (sΘ(N )))r - N ( x N )(α -η x (sΘ(N ))) + κΘ(N ) (N -1) 2 N θ x∈Λ N G x N 2 (1 -2η x (sΘ(N )))r + N ( x N )(β -η x (sΘ(N ))).
(3.12)

Since the first derivative of G is bounded it is easy to see that the absolute value of (3.12) is bounded from above by a constant times

Θ(N ) (N -1) 4 x, y∈Λ N (x -y) 2 p(x -y) + κΘ(N ) (N -1) 2 N θ x∈Λ N G x N 2 r - N ( x N ) + r + N ( x N ) . (3.13) Note that (x -y) 2 p(x -y) 1 because γ > 1, so that Θ(N ) (N -1) 4 x, y∈Λ N (x -y) 2 p(x -y) Θ(N )N -2 = (N γ-2 ).
By (3.10), the remaining terms in (3.13) are (Θ(N )N -θ -2 ) so that (3.13) is (N γ-2 ).

Thus, since τ is a stopping time and γ < 2 we have that lim δ→0 lim sup

N →∞ sup τ∈ T ,τ≤δ µ N M N ,G τ -M N ,G τ+τ 2 = lim δ→0 lim sup N →∞ sup τ∈ T ,τ≤δ µ N τ+τ τ Θ(N ) L N 〈π N s , G〉 2 -2〈π N s , G〉L N 〈π N s , G〉 ds = 0.
Therefore, we have proved (3.7) for functions G in C 2 c ((0, 1)) and as we have said in the beginning of the subsection this is enough to conclude tightness.

Energy Estimate.

We prove in this subsection that any limit point of the sequence { N } N ≥1 is concentrated on trajectories π κ t (u)du with finite energy, i.e., π κ belongs to L 2 (0, T ; γ/2 ). Moreover, we prove that π κ t satisfies item ii) in Definition 2.3. The latter is the content of Theorem 3.2 stated below. Fix a limit point of the sequence { N } N ≥1 and assume, without of loss of generality, that the sequence N converges to as N goes to ∞. Theorem 3.2. The probability measure is concentrated on trajectories of measures of the form π κ t (u)du, such that for any interval I ⊂ [0, T ] the density 2 -Hölder for all t ∈ I. By taking I = [0, T ] in item i) of Theorem 3.2 we can see that π κ ∈ L 2 (0, T ; γ/2 ). Moreover, from item ii) of Theorem 3.2, we claim that

π κ satisfies i) I π κ t 2 γ/2 d t |I|(κ + 1), if θ = 0. ii) I 1 0 (α -π κ t (u)) 2 u γ + (β -π κ t (u)) 2 (1 -u) γ du d t |I| κ + 1 κ , if θ ≤ 0.
I π κ t -ρ∞ 2 V 1 d t |I| κ + 1 κ (3. 14
)
where ρ∞ is given in Remark 2.7. Note that

I π κ t -ρ∞ 2 V 1 d t = c γ γ -1 I 1 0 (π κ t (u) -ρ∞ (u)) 2 u γ + (π κ t (u) -ρ∞ (u)) 2 (1 -u) γ dud t.
(3.15)

By summing and subtracting α inside the first square in the expression on the right hand side in (3.15), β in the second one and using the fact that (a + b) 2 ≤ 2(a 2 + b 2 ) we get that (3.15) is bounded from above by

2c γ γ -1 I 1 0 (π κ t (u) -α) 2 u γ + (π κ t (u) -β) 2 (1 -u) γ dud t + 2c γ γ -1 I 1 0 (α -ρ∞ (u)) 2 u γ + (β -ρ∞ (u)) 2 (1 -u) γ dud t. ( 3
2c γ γ -1 (β -α) 2 |I| 1 0 (u γ + (1 -u) γ ) -1 du 1.
Before we prove Theorem 3.2, we establish some estimates on the Dirichlet form which are needed in due course.

Estimates on the Dirichlet form. Let

h : [0, 1] → [0, 1] be a function such that α ≤ h(u) ≤ β, for all u ∈ [0, 1]
, and assume that h(0) = α and h(1) = β. Let ν N h be the inhomogeneous Bernoulli product measure on Ω N with marginals given by ν N h {η : η x = 1} = h x N . We denote by H N (µ|ν N h ) the relative entropy of a probability measure µ on Ω N with respect to the probability measure ν N h . It is easy to prove the existence of a constant C 0 , such that

H N (µ N |ν N h ) ≤ C 0 N . ( 3
.17) (see for example [START_REF] Bernardin | Slow to Fast infinitely extended reservoirs for the symmetric exclusion process with long jumps[END_REF]). We remark here that the restriction α = 0 and β = 1 comes from last estimate since the constant C 0 given above is given by C 0 = -log(α ∧ (1 -β)). On the other hand, for a probability measure µ on Ω N and a density function f :

Ω N → [0, ∞) with respect to µ we introduce D 0 N ( f , µ) := 1 2 x, y∈Λ N p( y -x) I x, y ( f , µ), (3.18) D N ( f , µ) := x∈Λ N y≤0 p( y -x) I α x ( f , µ) = x∈Λ N r - N ( x N )I α x ( f , µ) (3.19)
and D r N ( f , µ) is the same as D N ( f , µ) but in I α x ( f , µ) the parameter α is replaced by β and r - N is replaced by r + N . Above, we used the following notation

I x, y ( f , µ) := f (σ x, y η) -f (η) 2 dµ, I α x ( f , µ) := c x (η; α) f (σ x η) -f (η) 2 dµ
where c x (η, α) is given in (2.4) with ϕ(•) ≡ α; and I β x is the same as I α x when the parameter α is replaced by β.

Our goal is to express, for the measure ν N h , a relation between the Dirichlet form defined by 〈L N f , f 〉 ν N h and the quantity

D N ( f , ν N h ) := (D 0 N + κN -θ D N + κN -θ D r N )( f , ν N h ).
More precisely, we have the following result.

Lemma 3.4. For any positive constant B and any density function f with respect to ν N

h , there exists a constant C > 0 (independent of f and N ) such that

Θ(N ) N B 〈L N f , f 〉 ν N h ≤ - Θ(N ) 4N B D N ( f , ν N h ) + CΘ(N ) N B x, y∈Λ N p( y -x) h( x N ) -h( y N ) 2 + CκΘ(N ) N θ +1 B x∈Λ N h( x N ) -α 2 r - N ( x N ) + h( x N ) -β 2 r + N ( x N ) . (3.20)
The proof of this statement is similar to the one in Section 5 of [START_REF] Bernardin | Slow to Fast infinitely extended reservoirs for the symmetric exclusion process with long jumps[END_REF] and thus it is omitted. Moreover, note that as a consequence of the previous lemma, for a function h such that α ≤ h(u) ≤ β and h Lipschitz we have that

Θ(N ) N B 〈L N f , f 〉 ν N h ≤ - Θ(N ) 4N B D N ( f , ν N h ) + Θ(N )N -γ C(κN -θ + 1) B . ( 3

.21)

Lemma 3.5. For any density f with respect to ν N h , any x ∈ Λ N and any positive constant A x , we have that

〈η x -α, f 〉 ν N h 1 4A x I α x ( f , ν N h ) + A x + h( x N ) -α .
The same result holds if α is replaced by β.

The proof of Lemma 3.5 is omitted since is similar to the one of Lemma 5.5 in [START_REF] Bernardin | Slow to Fast infinitely extended reservoirs for the symmetric exclusion process with long jumps[END_REF]. Note that in the case α ≤ h ≤ β and Lipschitz we get

〈η x -α, f 〉 ν N h 1 4A x I α x ( f , ν N h ) + A x + x N .
3.3.2. Proof of Theorem 3.2. First item: π κ ∈ L 2 (0, T ; γ/2 ) -almost surely. Recall that in this case θ = 0 and the system is speeded up in the sub-diffusive time scale

Θ(N ) = N γ . Let > 0 be a small real number. Let F ∈ C 0,∞ c (I × [0, 1] 2 )
, where the I is a subinterval of [0, T ]. Observe that by the entropy inequality

µ N    I N γ-1 x, y∈Λ N |x-y|≥ N F t ( x N , y N )p( y -x)(η y (t N γ ) -η x (t N γ ))d t   
is bounded from above by

C 0 + 1 N log e η [| I N G N (t,η t N γ ) d t|] ν N h (dη) (3.22)
where

G N (t, η) = N γ-1 x, y∈Λ N |x-y|≥ N F t ( x N , y N )p( y -x)(η y -η x )
and by Jensen's inequality we can bound last expression from above by

C 0 + 1 N log ν N h e | I N G N (t,η t N γ ) d t| . (3.23)
Since e |x| ≤ e x + e -x and lim sup

N →∞ 1 N log(a N + b N ) ≤ max lim sup N →∞ 1 N log(a N ), lim sup N →∞ 1 N log(b N ) ,
we can remove the absolute value from expression (3.23). By Feynman-Kac's formula (see Lemma 7.3 in [START_REF] Baldasso | Exclusion process with slow boundary[END_REF]), we finally have that

µ N    I N γ-1 x, y∈Λ N |x-y|≥ N F t ( x N , y N )p( y -x)(η y (t N γ ) -η x (t N γ ))d t    ≤ C 0 + I sup f N γ-1 x, y∈Λ N |x-y|≥ N F t ( x N , y N )p( y -x) (η y -η x ) f (η)dν N h + N γ-1 L N f , f ν N h d t (3.24)
where the supremum is taken over all densities f on Ω N with respect to ν N h . Note that, by a change of variables, we have that

N γ-1 x, y∈Λ N |x-y|≥ N F t ( x N , y N )p( y -x) (η y -η x ) f (η)dν N h =N γ-1 x, y∈Λ N |x-y|≥ N F a t ( x N , y N )p( y -x) (η y -η x ) f (η)dν N h =N γ-1 x, y∈Λ N |x-y|≥ N F a t ( x N , y N )p( y -x) η y ( f (η) -f (σ x, y η)) dν N h +N γ-1 x, y∈Λ N |x-y|≥ N F a t ( x N , y N )p( y -x) η x f (η) (θ x, y (η) -1) dν N h (3.25)
where θ x, y (η) =

dν N h (σ x, y η) dν N h (η)
and F a is the antisymmetric part of F , i.e. for all t ∈ I and

(u, v) ∈ [0, 1] 2 F a t (u, v) = 1 2 F t (u, v) -F t (v, u) .
Observe that F a t (u, u) = 0. By Young's inequality, the fact that f is a density and |η y | ≤ 1, we have that, for any A > 0, the third term in (3.25) is bounded from above by a constant times

N γ-1 A x, y∈Λ N |x-y|≥ N F a t x N , y N 2 p( y -x) + N γ-1 A x, y∈Λ N |x-y|≥ N p( y -x)I x, y ( f , ν N h ) ≤ c γ A N 2 x, y∈Λ N |x-y|≥ N F a t x N , y N 2 | x N - y N | 1+γ + 2N γ-1 A D 0 N ( f , ν N h ).
Since h is Lipschitz we have that sup

η∈Ω N |θ x, y (η) -1| = |x-y| N
. By Young's inequality and the fact that f is a density, for any A > 0, the last term in (3.25) is bounded from above by

N γ-1 A x, y∈Λ N |x-y|≥ N F a t x N , y N 2 p( y -x) + A N γ-1 x, y∈Λ N |x-y|≥ N p( y -x) |x-y| N 2 = c γ A N 2 x, y∈Λ N |x-y|≥ N F a t x N , y N 2 | x N - y N | 1+γ + A c γ N 2 x, y∈Λ N |x-y|≥ N 1 | x N - y N | γ-1 .
Recall (3.21), so that by choosing A = 8 and B = 1 and using the two results above we have just proved that (3.24) is bounded from above by C 0 plus

c γ (8 + 1 A ) N 2 x = y∈Λ N F a t ( x N , y N ) 2 | x N - y N | 1+γ + C(κ + 1) + c γ A A ,
where

A := sup >0 sup N ≥1 1 N 2 x, y∈Λ N |x-y|≥ N 1 | x N - y N | γ-1 < ∞
since γ < 2. Therefore, we have proved that there exist constants A and B (independent of > 0, N ≥ 1, and

F ∈ C ∞ c (I × [0, 1] 2 )) such that µ N I N γ-1 x, y∈Λ N |x-y|≥ N F t ( x N , y N )p( y -x)(η N y (t) -η N x (t)) d t = µ N I -2c γ 〈π N t , g N t 〉 d t ≤ I A N 2 x, y∈Λ N |x-y|≥ N c γ F a t ( x N , y N ) 2 | x N - y N | 1+γ d t + B |I|(κ + 1). (3.26)
Above the function g N is defined on I × [0, 1] by

g N t (u) = 1 N y∈Λ N 1 y N -u ≥ F a t u, y N |u - y N | 1+γ
and it is a discretization of the smooth function g defined on (t, u) ∈ I × [0, 1] by

g t (u) = 1 0 1 {|v-u|≥ } F a t (u, v) |u -v| 1+γ d v. Let Q = {(u, v) ∈ [0, 1] 2 ; |u -v| ≥ }.
Observe first that for symmetry reasons we have that for any integrable function π,

1 0 π(u)g t (u)du = Q (π(v) -π(u))F a t (u, v) |u -v| 1+γ dud v.
By taking the limit as N → ∞ in (3.26), we conclude that there exist constants

C > 0 independent of F ∈ C 0,∞ c (I × [0, 1] 2 ) and > 0 such that   I Q (π κ t (v) -π κ t (u))F a t (u, v) |u -v| 1+γ -C F a t (u, v) 2 |u -v| 1+γ dud vd t   |I|(κ + 1).
Using similar arguments as the ones in the proof of Lemma 6.1 of [START_REF] Bernardin | Slow to Fast infinitely extended reservoirs for the symmetric exclusion process with long jumps[END_REF] we can insert the supremum over F inside the expectation above, so that

  sup F I Q (π κ t (v) -π κ t (u))F a t (u, v) |u -v| 1+γ -C F a t (u, v) 2 |u -v| 1+γ dud vd t   |I|(κ + 1). Since the function (u, v) ∈ [0, 1] 2 → π(v) -π(u)
is antisymmetric we may replace F a by F in the previous variational formula, i.e. (3.27)

  sup

F I Q (π κ t (v) -π κ t (u))F t (u, v) |u -v| 1+γ -C F t (u, v)
Consider the Hilbert space 2 ([0, 1] 2 , dµ ) where µ is the measure whose density with respect to Lebesgue measure is given by (u, v) ∈ [0, 1] 2 → |u-v|≥ |u -v| -(1+γ) . By taking

Π κ : (t; u, v) ∈ I × [0, 1] 2 → π κ t (v) -π κ t (u), the previous formula implies that I [0,1] 2 Π κ t (u, v) 2 dµ (u, v)d t |I|(κ + 1). (3.28)
Letting → 0, by the monotone convergence theorem, we conclude that

I [0,1] 2 (π κ t (v) -π κ t (u)) 2 |u -v| 1+γ dud vd t < ∞ almost surely.
Second item:

I 1 0 (α -π κ t (u)) 2 u γ + (β -π κ t (u)) 2 (1 -u) γ du d t < ∞ almost surely.
Now we have to prove that the function (t, u) → π κ t (u) -α is in the space L 2 (I × (0, 1), d t ⊗ dµ), where µ is the measure whose density with respect to the Lebesgue measure is given by u ∈ (0, 1) → 1 u γ . A similar argument to the one used in the first item shows that the function (t, u) → π κ t (u)-β belongs to L 2 ([0, T ]×(0, 1), d t ⊗dµ ), where µ is the measure whose density with respect to the Lebesgue measure is given by

u ∈ [0, 1] → 1 (1 -u) γ .
Let ν N h be the Bernoulli product measure corresponding to a profile h which is Lipschitz such that h(0 ]). As in the beginning of the proof of Theorem 3.2, using the entropy and Jensen's inequalities we get that

) = α ≤ h(u) ≤ β = h(1) for all u ∈ [0, 1]. Let G ∈ C ∞ c (I ×[0, 1
µ N   I N γ-1 x∈Λ N G t r - N x N (η x (tΘ(N )) -α)d t   ≤ H N (µ N |ν N h ) N + 1 N log µ N e N | I N γ-1 x∈Λ N G t r - N x N (η x (tΘ(N ))-α)d t|
. Now, using (3.17) and Feynman-Kac's formula (see Lemma 7.3 of [START_REF] Baldasso | Exclusion process with slow boundary[END_REF]) we obtain that

µ N   I N γ-1 x∈Λ N G t r - N x N (η x (tΘ(N )) -α)d t   ≤ C 0 + I sup f N γ-1 x∈Λ N (G t r - N ) x N 〈η x -α, f 〉 ν N h + Θ(N )N -1 L N f , f ν N h d t, (3.29)
where the supremun is taken over all the densities f on Ω N with respect to ν N h . Using (3.21) with B = 1 we can bound from above the second term on the right hand side of (3.29) by

- Θ(N ) 4N D N ( f , ν N h ) + CΘ(N )N -γ (κN -θ + 1),
and from Lemma 3.5 with

A x = 1 κ G t
x N the first term on the right side of (3.29) is bounded from above by

C N γ-1 κ x∈Λ N r - N x N G t x N 2 + C(κ + 1).
Taking N → ∞ we can conclude that there exists a constant C > 0 independent of G and of t such that

I 1 0 (π κ t (u) -α)G t (u) |u| γ - C κ G 2 t (u) |u| γ dud t |I|(κ + 1).
From Lemma 6.1 in [START_REF] Bernardin | Slow to Fast infinitely extended reservoirs for the symmetric exclusion process with long jumps[END_REF] we can insert the supremum over G inside the expectation above, and we get sup

G I 1 0 (π κ t (u) -α)G t (u) |u| γ - C κ G 2 t (u) |u| γ dud t |I|(κ + 1). (3.30)
The previous formula implies that

I 1 0 (π κ t (u) -α) 2 |u| γ dud t < ∞
almost surely. Similarly, we get

I 1 0 (π κ t (u) -β) 2 |u| γ dud t < ∞ almost surely.

Conclusion

. By Definition 2.3, the two steps above allow us to show that is concentrated on trajectories of measures whose density is a weak solution of the corresponding hydrodynamic equation (see Proposition 3.6). By uniqueness of the weak solution (see Lemma 2.8) we get that is unique. Indeed, we have that = δ {ρ κ t (u)du} (Dirac mass). Then, by using the latter, we compute the expectation in (3.28) and (3.30) and we are done.

Characterization of limit points.

In the present subsection we characterize all limit points of the sequence { N } N ≥1 , which we know that exist from the results of Subsection 3.2. Let us assume without loss of generality, that { N } N ≥1 converges to . Since there is at most one particle per site, it is easy to show that is concentrated on trajectories of measures absolutely continuous with respect to the Lebesgue measure, i.e. π κ t (du) = ρ κ t (u)du. Indeed, for any t ∈ [0, T ] and for any function G : [0, 1] → we have that

|〈π N t , G〉| ≤ 1 N -1 x∈Λ N |G( x n )|.
Also, we know that for any continuous function G the functional π ∈ + → 〈π, G〉 is continuous. Then, taking N → ∞, we obtain that

|〈π t , G〉| ≤ 1 0 |G(u)|du.
which implies that π t is absolutely continuous with respect to the Lebesgue measure.

In Proposition 3.6 below we prove, for each range of θ , that is concentrated on trajectories of measures whose density satisfies a weak form of the corresponding hydrodynamic equation. Moreover, we have seen in Theorem 3.2 that is concentrated on trajectories of measures whose density satisfies the energy estimate, i.e. ρ κ ∈ L 2 (0, T ; γ/2 ) when θ = 0 and

T 0 1 0 (α -ρ κ t (u)) 2 u γ + (β -ρ κ t (u)) 2 (1 -u) γ dud t < ∞,
for any θ ≤ 0. Since a weak solution of the hydrodynamic equation (2.10) is unique we have that is unique and takes the form of a Dirac mass.

Proposition 3.6. If is a limit point of { N } N ≥1 then 1. if θ < 0:

π • : F Reac (t, ρ κ , G, g) = 0, ∀t ∈ [0, T ], ∀G ∈ C 1,2 c ([0, T ] × [0, 1]) = 1. 2. if θ = 0: π • : F Dir (t, ρ κ , G, g) = 0, ∀t ∈ [0, T ], ∀G ∈ C 1,2 c ([0, T ] × [0, 1]) = 1.
Proof. Note that in order to prove the proposition, it is enough to verify, for δ > 0 and G in the corresponding space of test functions, that

π • ∈ T + : sup 0≤t≤T |F θ (t, ρ κ , G, g)| > δ = 0,
for each θ , where F θ stands for F Reac if θ < 0 and F Dir if θ = 0 . Indeed, we have that

F θ (t, ρ κ , G, g) = ρ κ t , G t -〈g, G 0 〉 - t 0 ρ κ s , ∂ s + {θ =0} G s ds + {θ ≤0} κ t 0 ρ κ s , G s V 1 ds -{θ ≤0} κ t 0 〈G s , V 0 〉 ds = 0. (3.31)
From here on, in order to simplify notation, we will erase π • from the sets that we have to look at. By definition of F θ above we can bound from above the previous probability by the sum of

sup 0≤t≤T |F θ (t, ρ κ , G, ρ 0 )| > δ 2 (3.32)
and

|〈ρ 0 -g, G 0 〉| > δ 2 .
We note that last probability is equal to zero since is a limit point of { N } N ≥1 and N is induced by µ N which is associated to g. Now we deal with (3.32). Since for θ ≤ 0 the function G s has compact support included in (0, 1) the singularities of V 0 and V 1 are not present, thus from Proposition A.3 of [START_REF] Franco | Hydrodynamical behavior of symmetric exclusion with slow bonds[END_REF], the set inside the probability in (3.32) is an open set in the Skorohod topology. Therefore, from Portmanteau's Theorem we bound (3.32) from above by lim inf

N →∞ N sup 0≤t≤T |F θ (t, ρ κ , G, ρ 0 )| > δ 2 .
Summing and subtracting t 0 Θ(N )L N 〈π N s , G s 〉ds to the term inside the previous absolute value, recalling (3.1) and the definition of N , we can bound the previous probability from above by the sum of the next two terms

µ N sup 0≤t≤T M N t (G) > δ 4 and µ N sup 0≤t≤T t 0 Θ(N )L N 〈π N s , G s 〉ds - t 0 π N s , {θ =0} G s ds + {θ ≤0} κ t 0 〈ρ s , G s 〉 V 1 ds -{θ ≤0} κ t 0 〈G s , V 0 〉 ds > δ 4 .
(3.33) By Doob's inequality we have that

µ N sup 0≤t≤T M N t (G) > δ 4 1 δ 2 µ N T 0 Θ(N ) L N 〈π N s , G〉 2 -2〈π N s , G〉L N 〈π N s , G〉 ds .
In the proof of Proposition 3.1 we have proved that the term inside the time integral in the previous expression is (N γ-2 ). Then, using the fact that γ < 2 we have that last probability vanishes as N → ∞. It remains to prove that (3.33) vanishes as N → ∞.

For that purpose, we recall (3.2) and we bound (3.33) from above by the sum of the following terms

µ N sup 0≤t≤T t 0 Θ(N ) N -1 x∈Λ N N G s ( x N )η N x (s)ds - t 0 π N s , {θ =0} G s ds > δ 2 4 , (3.34) µ N sup 0≤t≤T t 0 κΘ(N ) N θ (N -1) x∈Λ N (G s r - N )( x N )(α -η N x (s)) -{θ ≤0} κ 1 0 (G s r -)(u)(α -ρ κ s (u))du ds > δ 2 4 (3.35) and µ N sup 0≤t≤T t 0 κΘ(N ) N θ (N -1) x∈Λ N (G s r + N )( x N )(β -η N x (s)) -{θ ≤0} κ 1 0 (G s r + )(u)(β -ρ κ s (u))du ds > δ 2 4 . (3.36)
For θ = 0 from (3.6) we have that (3.34) goes to 0 as N → ∞. For θ ≤ 0 we have that from (3.6) and 3.5 the boundary terms (3.35) and (3.36) go to 0 as N → ∞. This finishes the proof Proposition 3.6.

PROOF OF THEOREM 2.13

For easy understanding of the proof of items i) and ii) of Theorem 2.13, we first establish some notation and prove some lemmata.

Recall the function ρ∞ introduced in Remark 2.7 which can be rewritten as

ρ∞ (u) = βu γ + α(1 -u) γ u γ + (1 -u) γ .
It is easy to see that ρ∞ (0) = α and ρ∞ (1) = β. Moreover, it is not difficult to see that ρ∞ ∈ C 1 ([0, 1]) and that lim u→0

( ρ∞ (u)) u 2-γ = lim u→1 ( ρ∞ (u)) (1 -u) 2-γ = 0,
and from Lemma 7.2 of [START_REF] Guan | The reflected α-symmetric stable processes and regional fractional Laplacian[END_REF] we conclude that

ρ∞ γ/2 < ∞. (4.1)
By the fractional Hardy's inequality (see e.g. [START_REF] Dyda | A fractional order Hardy inequality[END_REF]) and the fact that

V 1 ( 1 2 ) ≤ V 1 (u) for all u ∈ (0, 1) we know that g g V 1 g γ/2 (4.2)
for any g ∈ γ/2 0 , and where g V 1 is defined in the beginning of Section 2.3. In order to prove items i) and ii) of Theorem 2.13 we first guarantee the existence of weak solutions of equation (2.10) with κ = 0 and (2.12), (see Lemma 4.1 and 4.3 below), then we establish the convergence in L 2 (0, T ; L 2 ) (see Lemma 4.2 and 4.4) which will allow us to conclude. Lemma 4.1. Let ρ 0 : [0, 1] → [0, 1] be a measurable function. Then, there exists a weak solution of (2.10) with κ = 0 and initial condition ρ 0 .

Proof. The strategy of the proof is to construct the solution as the limit of ρ κ , as κ → 0, where ρ k is the weak solution of (2.10) with initial condition ρ 0 and κ = κ.

By item i) in Theorem 3.2, for any κ > 0 we know that

I ρ κ t 2 γ/2 d t |I|(κ + 1) (4.3) for any interval I ⊂ [0, T ]. We define ∀t ∈ [0, T ], ∀u ∈ [0, 1], ϕ κ t (u) := ρ κ t (u) -ρ∞ (u). (4.4)
Since we are interested in small values of κ, say κ ≤ 1, from (4.3), (4.1) and the fact (a + b) 2 ≤ 2a 2 + 2b 2 , it is not difficult to see that

I ϕ κ t 2 γ/2 d t |I|, (4.5)
thus we have that ϕ κ ∈ L 2 (0, T ; γ/2 0 ). It is also easy to see that ϕ κ satisfies

〈ϕ κ t , G t 〉-〈ϕ 0 , G 0 〉- t 0 ϕ κ s , ( + ∂ s ) G s ds+κ t 0 〈ϕ κ s , G s 〉 V 1 ds- t 0 〈 ρ∞ , G s 〉ds = 0 (4.6) for all t ∈ [0, T ], for any function G ∈ C 1,∞ c ([0, T ] × (0, 1
)) and where ϕ 0 (u) = ρ 0 (u) -ρ∞ (u). From (4.5) we conclude that there exists a subsequence of (ϕ κ ) κ∈(0,1) converging weakly to some element ϕ 0 ∈ L 2 (0, T ; γ/2 0 ) as κ → 0. We claim that ρ 0 := ρ∞ + ϕ 0 is the desired solution. Indeed, first note that since the norm • γ/2 is weakly lower-semicontinuous we have that

I ϕ 0 t 2 γ/2 d t |I|. (4.7)
By using (a + b) 2 ≤ 2a 2 + 2b 2 we have that

I ρ 0 t 2 γ/2 d t ≤ 2 I ρ∞ 2 γ/2 d t + 2 I ϕ 0 t 2 γ/2 d t |I|.
Taking I = [0, T ], we have that ρ 0 satisfies item i) of Definition 2.3. Since ϕ 0 ∈ L 2 (0, T ;

γ/2 0 ), it is easy to see that ρ 0 t (0) = ρ∞ (0) = α and ρ 0 t (1) = ρ∞ (1) = β for almost every t ∈ [0, T ]. Then, item ii) for κ = 0 in Definition 2.3 is satisfied. In order to verify that ρ 0 satisfies item iii) in Definition 2.3 we first integrate (4.6) over [0, t]. Thus we have that )). Taking κ → 0, by weak convergence and Lebesgue's dominated convergence theorem we get from the previous equality that

t 0 〈ϕ κ s , G s 〉ds -t〈ϕ 0 , G 0 〉 - t 0 s 0 ϕ κ r , ( + ∂ r ) G r d r ds +κ t 0 s 0 〈ϕ κ r , G r 〉 V 1 d r ds - t 0 s 0 〈 ρ∞ , G r 〉d r ds = 0 for any function G ∈ C 1,∞ c ([0, T ] × (0, 1
t 0 〈ϕ 0 s , G s 〉ds -t〈ϕ 0 , G 0 〉 - t 0 s 0 ϕ 0 r , ( + ∂ r ) G r -〈 ρ∞ , G r 〉d r ds = 0.
Now, taking the derivative with respect to t in the previous equality we get that ϕ 0 satisfies

〈ϕ 0 t , G t 〉 -〈ϕ 0 , G 0 〉 - t 0 〈ϕ 0 s , + ∂ s G s 〉 ds - t 0 〈 ρ∞ , G s 〉ds = 0, (4.8)
for all t ∈ [0, T ]. Then, item iii) with κ = 0 in Definition 2.3 follows from (4.8), the definition of ρ 0 and ρ∞ Lemma 4.2. Let ρ 0 : [0, 1] → [0, 1] be a measurable function. Let ρ κ be the weak solution of (2.10) with initial condition ρ 0 and κ = κ. Then, ρ κ converges strongly to ρ 0 in L 2 (0, T ; L 2 ) as κ goes to 0, where ρ 0 is the weak solution of (2.10) with κ = 0 and initial condition ρ 0 .

Proof. Note that is enough to show that

t 0 ρ κ s -ρ 0 s 2 ds t 2 κ,
for all t ∈ [0, T ]. By Lemma 4.1 we know that ρ 0 = ρ∞ + ϕ 0 . Then, last inequality is equivalent to

t 0 ϕ κ s -ϕ 0 s 2 ds t 2 κ. ( 4.9)
By subtracting (4.8) from (4.6) and calling δ k t := ϕ κ t -ϕ 0 t we obtain that

〈δ κ t , G t 〉 - t 0 δ κ s , ( + ∂ s ) G s ds = -κ t 0 〈ϕ κ s , G s 〉 V 1 ds (4.10) for any function G ∈ C 1,∞ c ([0, T ] × (0, 1)). Let {H κ n } n≥1 be a sequence of functions in C 1,∞ c ([0, T ] × (0, 1)) converging to δ κ as n → ∞ with respect to the norm of L 2 (0, T ; γ/2 0 ) and for n ≥ 1, let G κ n (s, u) = t s H κ n (r, u)d r.
We claim that by plugging G n into (4.10) and taking n → ∞ we get that

t 0 δ κ s 2 ds + 1 2 t 0 δ κ s ds 2 γ/2 = -κ t 0 ϕ κ s , t s δ κ r d r V 1 ds. (4.11)
We leave the justification of the equality above to the end of the proof. Now, by using successively the Cauchy-Schwarz's inequality we have that

t 0 δ κ s 2 ds + 1 2 t 0 δ κ s ds 2 γ/2 ≤ κ t 0 ϕ κ s V 1 t s δ κ r d r V 1 ds κ t 0 ϕ κ s 2 γ/2 ds t 0 t s δ κ r d r 2 γ/2
ds.

(4.12)

In the last inequality of the previous expression we used (4.2). By the triangular inequality we have that

t 0 t s δ κ r d r 2 γ/2
ds is bounded from above by

t 0 t s δ κ r γ/2 d r 2 ds ≤ t t 0 t 0 δ κ r 2 γ/2 d r ds t 2 t 0 ϕ κ r 2 γ/2 + ϕ 0 r 2 γ/2 d r. ( 4.13)
In the first inequality in the previous display we used the Cauchy-Schwarz's inequality and in the second inequality we used the Minkowski's inequality and the inequality (a + b) 2 ≤ 2(a 2 + b 2 ). Using (4.5) and (4.7), we get from (4.12) and (4.13) the result. We conclude this proof justifying (4.11). Note that it is enough to show i) lim

n→∞ t 0 〈δ κ s , (∂ s G κ n )(s, •)〉ds = - t 0 δ κ s 2 ds. ii) lim n→∞ t 0 〈δ κ s , G κ n (s, •)〉ds = - 1 2 t 0 δ κ s ds 2 γ/2 . iii) lim n→∞ t 0 ϕ κ s , G κ n (s, •) V 1 ds = t 0 ϕ κ s , t s δ κ r d r V 1
ds.

For i) we rewrite

t 0 〈δ κ s , (∂ s G κ n )(s, •)〉ds as - t 0 〈δ κ s , H κ n (s, •)〉 ds = - t 0 δ κ s , H κ n (s, •) -δ κ s ds - t 0 δ κ s 2 ds.
Observe then that by the Cauchy-Schwarz's inequality we have

T 0 δ κ s , H κ n (s, •) -δ κ s ds ≤ T 0 δ κ s H κ n (s, •) -δ κ s ds ≤ T 0 δ κ s 2 ds T 0 H κ n (s, •) -δ κ s 2 ds which goes to 0 as n → ∞ since H κ n → δ κ s in L 2 (0, T ; γ/2 0 ).
For ii), since G n has compact support included in (0, 1), we can use the integration by parts formula for the regional fractional Laplacian (see Theorem 3.3 in [START_REF] Guan | The reflected α-symmetric stable processes and regional fractional Laplacian[END_REF]) which permits to write

t 0 〈δ κ s , G κ n (s, •)〉ds = - t 0 δ κ s , G κ n (s, •) γ/2
ds.

Then we have

t 0 δ κ s , G κ n (s, •) γ/2 ds = t 0 δ κ s , t s δ κ r d r γ/2 ds + t 0 δ κ s , G κ n (s, •) - t s δ κ r d r γ/2 ds = 0≤s<r≤t 〈δ κ s , δ κ r 〉 γ/2 dsd r + t 0 δ κ s , t s H κ n (r, •) -δ κ r d r γ/2 ds = 1 2 [0,t] 2 〈δ κ s , δ κ r 〉 γ/2 dsd r + t 0 δ κ s , t s H κ n (r, •) -δ κ r d r γ/2 ds = 1 2 t 0 δ κ s ds 2 γ/2 + t 0 δ κ s , t s H κ n (r, •) -δ κ r d r γ/2
ds.

To conclude the proof of ii) it is sufficient to show that the term at the right hand side of last expression vanishes as n goes to ∞. This is a consequence of a successive use of Cauchy-Schwarz's inequalities:

t 0 δ κ s , t s H κ n (r, •) -δ κ r d r γ/2 ds ≤ t 0 δ κ s γ/2 t s H κ n (r, •) -δ κ r d r γ/2 ds ≤ t 0 δ κ s γ/2 t s H κ n (r, •) -δ κ r γ/2 d r ds ≤ t 0 δ κ s γ/2 t 0 H κ n (r, •) -δ κ r γ/2 d r ds = t 0 δ κ s γ/2 ds t 0 H κ n (r, •) -δ κ r γ/2 d r ≤ t t 0 δ κ s 2 γ/2 ds t 0 H κ n (r, •) -δ κ r 2 γ/2 d r ---→ n→∞ 0. (4.14)
To prove iii) we rewrite

t 0 〈ϕ κ s , G κ n (s, •)〉 V 1 ds as t 0 ϕ κ s , t s H κ n (r, •) -δ κ r d r V 1 ds + t 0 ϕ κ s , t s δ κ r d r V 1
ds and, to conclude the proof it is sufficient to show that the term at the left hand side of last expression vanishes as n → ∞. This is a consequence of a successive use of the Cauchy-Schwarz's inequality as in (4.14), with • γ/2 replaced by • V 1 and Hardy's inequality:

t 0 ϕ κ s , t s {H κ n (r, •) -δ κ r }d r V 1 ds ≤ t 0 ϕ κ s V 1 t s H κ n (r•) -δ κ r d r V 1 ds ≤ t 0 ϕ κ s V 1 t s H κ n (r, •) -δ κ r V 1 d r ds ≤ t 0 ϕ κ s V 1 t 0 H κ n (r, •) -δ κ r V 1 d r ds = t 0 ϕ κ s V 1 ds t 0 H κ n (r, •) -δ κ r V 1 d r ≤ t t 0 ϕ κ s 2 V 1 ds t 0 H κ n (r, •) -δ κ r 2 V 1 d r ≤ C t t 0 ϕ κ s 2 γ/2 ds t 0 H κ n (r, •) -δ κ r 2 γ/2 d r ---→ n→∞ 0
where in the last inequality we used the fractional Hardy's inequality (see (4.2)).

Lemma 4.3. Let ρ 0 : [0, 1] → [0, 1] be a measurable function. Consider the function ρ ∞ t = ρ∞ + (ρ 0 -ρ∞ )e -t V 1 . If g ∞ := ρ 0 -ρ∞ ∈ γ/2 , then i) ρ ∞ ∈ L 2 (0, T ; γ/2 ) .
ii) ρ ∞ is a weak solution of (2.12) with initial condition ρ 0 .

Proof. For i) note that by using the inequality (a + b) 2 ≤ 2a 2 + 2b 2 we get that

T 0 ρ ∞ t 2 γ/2 d t ≤ 2T ρ∞ 2 γ/2 + 2 T 0 g ∞ e -t V 1 2 γ/2 d t.
Since ρ∞ γ/2 < ∞ (see (4.1)) it is enough to prove that the term on the right hand side of last expression is finite. Note that g ∞ e -t V 1 2 γ/2 is equal to

c γ 2 [0,1] 2 g ∞ (u)e -t V 1 (u) -g ∞ (v)e -t V 1 (v) 2 |u -v| γ+1 dud v = c γ 2 [0,1] 2 g ∞ (u) e -t V 1 (u) -e -t V 1 (v) + (g ∞ (u) -g ∞ (v)) e -t V 1 (v) 2 |u -v| γ+1 dud v.
Using the fact that (a + b) 2 ≤ 2a 2 + 2b 2 and that |g ∞ (u)| ≤ 2 for any u ∈ [0, 1] we get that last expression is less than 8

e -t V 1 2 γ/2 + 2 g ∞ 2 γ/2 . Note that the term 8 e -t V 1 2 γ/2
can be written as

4c γ [0,1] 2 u v -t V 1 (w)e -t V 1 (w) dw 2 |u -v| γ+1 dud v =4c γ [0,1] 2 u v t γ w r -(w) - γ 1-w r + (w) e -t V 1 (w) dw 2 |u -v| γ+1 dud v.
Using again (a + b) 2 ≤ 2a 2 + 2b 2 and the fact that e -t V 1 (w) ≤ e -t r ± (w) for any w ∈ [0, 1], we get that last expression is bounded from above by

8c γ [0,1] 2 u v γ w t r -(w)e -t r -(w) dw 2 |u -v| γ+1 + u v γ 1-w t r + (w)e -t r + (w) dw 2 |u -v| γ+1 dud v =16c γ [0,1] 2 u v γ w t r -(w)e -t r -(w) dw 2 |u -v| γ+1 dud v.
In the last equality we used a symmetry argument. We can write last expression as

C γ t 2-2γ γ [0,1] 2 u v w γ-2 (t r -(w)) 2γ-1 γ e -t r -(w) dw 2 |u -v| γ+1 dud v,
where

C γ = 16c 2-γ γ γ γ 4γ-2 γ . Since the function E γ : [0, ∞) → [0, ∞) defined as E γ (z) = z 2γ-1 γ e -z is bounded from above by E γ 2γ-1 γ
we can bound last expression from above by

C γ t 2-2γ γ E 2 γ ( 2γ-1 γ ) [0,1] 2 u v w γ-2 dw 2 |u -v| γ+1 dud v = C γ t 2-2γ γ E 2 γ ( 2γ-1 γ )(γ -2) -2 [0,1] 2 u γ-1 -v γ-1 2 |u -v| γ+1 dud v,
which is finite from (7.2) in the proof of Lemma 7.2 of [START_REF] Guan | The reflected α-symmetric stable processes and regional fractional Laplacian[END_REF]. Thus, we have that

8 e -t V 1 2 γ/2 t 2-2γ γ . (4.15)
Therefore, if g ∞ ∈ γ/2 then we conclude that

T 0 ρ ∞ t 2 γ/2 d t T ρ∞ 2 γ/2 + T g ∞ 2 γ/2 + T 0 t 2-2γ γ d t T ρ∞ 2 γ/2 + T g ∞ 2 γ/2 + T 2-γ γ ,
which is finite since γ < 2.

For ii), since ρ ∞ is the solution of (2.12) then it satisfies item ii) of Definition 2.6. In order to see that ρ ∞ satisfies item i) of Definition 2.6, note that using (a + b) 2 ≤ 2a 2 + 2b 2 we have that

T 0 1 0 α -ρ ∞ t (u) 2 u γ + β -ρ ∞ t (u) 2 (1 -u) γ dud t ≤ 2T 1 0 (α -ρ∞ (u)) 2 u γ + (β -ρ∞ (u)) 2 (1 -u) γ du + 8γ c γ T 0 e -t V 1 2 V 1 d t = 2T (β -α) 2 1 0 (u γ + (1 -u) γ ) du + 8γ c γ T 0 e -t V 1 2 V 1 d t ≤ 2 γ (β -α) 2 T + 8γ c γ T 0 e -t V 1 2 V 1 d t.
For the term on the right hand side of last expression we first see that we can extend continuously the function e -t V 1 in such a way that it vanishes at 0 and at 1. There exists a constant C 2 (see 4.2) such that the previous expression is bounded from above by

2 γ (β -α) 2 T + 8γC 2 2 c γ T 0 e -t V
∞ t = (ρ 0 -ρ∞ )e -t V 1 . It is not difficult to see that φκ t satisfies 〈 φκ t , G t 〉 -〈ϕ 0 , G 0 〉 - t 0 〈 φκ s , ∂ s G s 〉 ds + t 0 〈 φκ s , G s 〉 V 1 ds - 1 κ t 0 〈 ρκ s , G s 〉ds = 0 (4.18)
for all functions G ∈ C 1,∞ c ([0, T ] × (0, 1)). Then, calling δk := φκϕ ∞ we have that

〈 δκ t , G t 〉 - t 0 δκ s , 1 κ + ∂ s G s ds + t 0 δκ s , G s V 1 = 1 κ t 0 〈ρ ∞ s , G s 〉 γ/
+ 1 2 t 0 δκ s ds 2 V 1 = 1 κ t 0 ρ ∞ s , t s δκ r d r γ/2
ds.

By neglecting terms we get that

t 0 ρκ s -ρ ∞ s 2 ds = t 0 δκ s 2 ds ≤ 1 κ t 0 ρ ∞ s , t s δκ r d r γ/2
ds.

Then it is suffices to show that

1 κ t 0 ρ ∞ s , t s δκ r d r γ/2 ds 1 κ
To do so, we start by using twice the Cauchy-Schwarz's inequality we have that the term at the left hand side of the previous expression is bounded from above by

1 κ t 0 ρ ∞ s γ/2 t s δκ r d r γ/2 ds ≤ 1 κ t 0 ρ ∞ s 2 γ/2 ds t 0 t s δκ r d r 2 γ/2
ds.

Since by hypothesis ρ 0 -ρ∞ ∈ γ/2 we know from item i) of Lemma 4.3 that ρ ∞ ∈ L 2 (0, T ; γ/2 ). Thus, from the latter and by the triangular inequality, the right hand side in the previous expression can be bounded from above by a constant times

1 κ t 0 t s δκ r γ/2 d r 2 ds 1 κ t t 0 δκ r γ/2 d r 2 .
By using again the Cauchy-Schwarz's inequality, the term on the right hand side in the last expression is bounded from above by

1 κ t 2 t 0 δκ r 2 γ/2 d r = 1 κ t 2 t 0 ρκ r -ρ ∞ r 2 γ/2 d r 1 κ 2t 2 t 0 ρκ r 2 γ/2 + ρ ∞ r 2 γ/2 d r.
In the last inequality we used the Minkowski's inequality and the fact that (a + b) (4.4). Note that it is enough to show (4.9) with • replaced with • γ/2 . From (4.10) we obtain, for > 0, that Taking → 0, using Lebesgue's differentiation theorem (see Theorem 1.35 in [START_REF] Roubíček | Nonlinear partial differential equations with applications[END_REF]) and the fact that δ κ 0 = 0 (since the initial condition for ρ κ and ρ 0 is the same) we get that In the second inequality above we used the fact that ρ ∞ ∈ L 2 (0, T ; γ/2 ) (see item i) of Lemma 4.3) and (4.29), while in the third inequality of we used Minkoski's inequality and the fact that (a + b) 2 ≤ 2a 2 + 2b 2 . And finally, the last inequality of (4.36) is true since ρ ∞ ∈ L 2 (0, T ; γ/2 ) and item i) of Theorem 3.2. Then, by a simple computation we have that

〈δ κ t+ , G t+ 〉 -〈δ κ t , G t 〉 - t+ t 〈δ κ s , ( + ∂ s ) G s 〉 ds = -κ t+ t 〈ϕ κ s , G s 〉 V 1 ds (4.20) for any function G ∈ C 1,∞ c ([0, T ] × [0, 1]). Let {H κ n } n≥1 be a sequence of functions in C 1,∞ c ([0, T ], (0, 1)) converging to δ κ with respect to the norm of L 2 (0, T ; γ/2 0 ) as n → ∞. Now, for n ≥ 1, we define the test function G κ n (u) =
V 1 : T 0 φκ t -ϕ ∞ t 2 V 1 d t 1 κ . ( 4
〈 δκ t+ , G t+ 〉 -〈 δκ t , G t 〉 - t+ t 〈 δκ s , 1 κ + ∂ s G s 〉 ds + t+ t 〈 δκ s , G s 〉 V 1 ds = 1 κ t+ t 〈ρ ∞ s , G s 〉 γ/2 ds (4.30) for any function G ∈ C 1,∞ c ([0, T ] × [0, 1]). Let { Ĥκ n } n≥1 be a sequence of functions in C 1,∞ c ([0, T ], (0, 1
T 0 t 0 δκ t 2 V 1 d t dt 1 κ . ( 4
Recall that 〈•, •〉 V 1 (resp. • V 1 )
is the scalar product (resp. the norm) corresponding to the Hilbert space L 2 V 1 . Then, it follows that for almost every time s ∈ [0, T ] the continuous function ρκ s is equal to 0 and we conclude the uniqueness of the weak solutions to (2.10). .

iii)

lim n→∞ T 0 ρκ s , G κ n (s, •) V 1 ds = 1 2 T 0 ρκ s ds 2 V 1 < ∞.
Proof. The proof of this lemma is quite similar to the proof of items i), ii) and iii) in the proof of Lemma 4.2. For that reason we just sketch the main steps of the proof and we leave the details to the reader. For i) we have that which goes to 0 as n → ∞.

For ii), we first use the integration by parts formula for the regional fractional Laplacian (see Theorem 3.3 in [START_REF] Guan | The reflected α-symmetric stable processes and regional fractional Laplacian[END_REF]) to get Now, note that the term on the right hand side of last expression vanishes as n → ∞ as a consequence of a successive use of Cauchy-Schwarz's inequalities. The proof of iii) is similar to the proof of ii) by using the fractional Hardy's inequality (see (4.2)) and since C ∞ c ((0, 1)) is dense in H γ/2 0 we have that any g ∈ H γ/2 0 is also in the space L 2 V 1 and that (4.2) remains valid for g. In particular, we have that the right hand side of iii) is finite. We have

T 0 ρκ s , G κ n (s, •) V 1 ds = 1 2 T 0 ρκ s ds 2 V 1 + T 0 ρκ s , T s H κ n (t, •) -ρκ t d t V 1
ds.

(6.5)

To conclude the proof of iii) it is sufficient to prove that the term on the right hand side of last expression vanishes as n → ∞. But this is a consequence of a successive use of the Cauchy-Schwarz inequalities and Hardy's inequality, from which we get The proof of the uniqueness of the weak solutions of (2.10) for κ = 0 is analogous, the difference is that only the first two items in Lemma 6.1 above are required. The uniqueness of the weak solutions of (2.12) is analogous as well, in this case only items i) and iii) in Lemma 6.1 above are required.

6.2. Proof of Lemma 2.11. Recall (5.1). As we will see below, by Lax-Milgram's Theorem (see [START_REF] Brezis | Functional analysis, Sobolev spaces and partial differential equations[END_REF]), there exists a unique function φκ ∈ γ/2 0 which is solution of (5.1). Then, it is not difficult to see that ρκ := φκ + ρ∞ is the desired weak solution of (2.14). For that purpose, let a κ : From Lax-Milgram Theorem, in order to conclude the existence and uniqueness it is enough to prove that a κ is coercive and continuous. For κ > 0, we can easily see that

a κ(G, G) ≥ min{1, κV 1 ( 1 2 )} G 2 γ/2 + G 2 = min{1, κV 1 ( 1 2 )} G 2 γ/2 0 .
For κ = 0, since on γ 0 the norms • γ/2 and • γ/2 are equivalent we have that

a 0 (G, G) = G 2 γ/2 G 2 γ/2 0 .
Therefore a κ is coercive for κ ≥ 0. Moreover, by using the Cauchy-Schwarz's inequality we obtain that

|a κ(F, G)| ≤ F γ/2 G γ/2 + κ( F V 1 G V 1 ).
From the fractional Hardy's inequality (see (4.2)) we have that

|a κ(F, G)| (κ + 1)( F γ/2 G γ/2 )
and since on γ/2 0 the norms • γ/2 and • γ/2 are equivalent, we conclude that the bilinear form a κ is continuous for κ ≥ 0. This end the proof.

. 9 )

 9 By using (3.5),(3.6) and the fact that G ∈ C 2 c ((0, 1)) we can bound the expression in (3.2) by a constant. By using the fact that |η N x (s)| ≤ 1 and

  x≥1

2 |u-

 2 v| 1+γ dud vd t   |I|(κ + 1).

1 )/ 2 0

 12 Note that, it is easy to showthat C 1,∞ c ([0, T ] × (0, 1)) is dense in L 2 (0, T ; γ/2 0 ). Let {H κ n } n≥1 be a sequence of functions in C 1,∞ c ([0, T ] × (0,1)) converging to ρκ with respect to the norm of L 2 (0, T ;1) as n → ∞. For n ≥ 1, we define the test functions ∀t ∈ [0, T ], ∀u ∈ [0, 1], G κ n (t, u) = T t H κ n (s, u) ds.Plugging G κ n into (6.1) and letting n → ∞ we conclude by Lemma 6.1 below that

Lemma 6 . 1 .

 61 Let {G κ n } n≥1 be defined as above. We have i)lim n→∞ T 0 ρκ s , (∂ s G κ n )(s, •) ds = -

  ∂ s G κ n )(s, •) ds = T 0 ρκ s , H κ n (s, •) -ρκ s ds +

  a κ(F, G) = 〈F, G〉 γ/2 + κ〈F, G〉 V 1 . (6.6)

  Remark 3.3. It follows from item i) of the previous and from Theorem 8.2 of [14] that

	π κ t is, almost surely,	γ-1

  Let ρ 0 : [0, 1] → [0, 1] be a measurable function, such that ρ 0 -ρ∞ ∈ γ/2 . Furthermore, let ρ κ and ρ ∞ be the weak solutions of (2.10) and (2.12), respectively, and with the same initial condition ρ 0 . Let ρκ t := ρ κ t/κ , for all t ∈ [0, T ]. Then ρκ converges strongly to ρ ∞ in L 2 (0, T ; L 2 ), as κ goes to ∞.

							1 2 γ/2 d t.			(4.16)
	Thus, we obtain the desired result by using (4.15).
	Lemma 4.4. Proof. It is enough to show that
	0	t	ρκ s -ρ ∞ s	2 ds =	0	t	φκ s -ϕ ∞ s	2 ds	1 κ	,	(4.17)
	for all t ∈ [0, T ] where φκ

t = ρκ t -ρ∞ and ϕ

Proof of item i) of Theorem 2.13. Recall

												2 ≤
	2a 2 + 2b 2 . Now, since	t 0	ρκ r	2 γ/2 d r κ (this is due to item i) of Theorem 3.2 and a
	change of variables) and ρ ∞ ∈ L 2 (0, T ; γ/2 ) we can see that
	1 κ	2t 2	0	t	ρκ r	2 γ/2 + ρ ∞ r	2 γ/2 d r	1 κ	κ + 1	1 κ	,
	and we are done.
	4.1.

ϕ κ t defined in

  )) converging to δκ with respect to the norm of L 2 (0, T ; ∈ [0, T ]. Integrating the previous expression over [0, T ] and using the Cauchy-Schwarz's inequality we get that

	Therefore, by using (4.33) and (4.34) in (4.32) we get that
	t 0	1	t	t+	δκ r d r		2 V 1	d t ≤	1 κ			0 t	1	t	t+		ρ ∞ s	2 γ/2 dsd t	0 t	1	t	t+	δκ r d r	γ/2 2	d t
										+	1		t+	δκ t	2 d t +	1	δκ t	2 d t.
	t n → ∞. Now, for n ≥ 1 we define the test functions Ĝκ 0 n (u) = 1 t+ t Taking → 0, using Lebesgue's differentiation theorem (see Theorem 1.35 in [21]) and (4.35) γ/2 0 ) as Ĥκ n (r, u)d r. Plugging Ĝκ n into (4.30) and taking n → ∞, a similar argument to the one of the proof the fact that δκ 0 = 0 we get that
	of Lemma 4.2 allows to get
	1 δκ t+ -δκ t , t 0 By neglecting the term κ t+ t δκ δκ t 2 V 1 d t ≤ r d r + κ 1 κ + 1 t+ t δκ r d r 1 t 0 1 t+ t δκ r d r ρ ∞ t+ t t 2 V 1 = 2 for all t T 0 t 0 δκ t 2 V 1 d t dt ≤ T t 1 κ 0 0 ρ ∞ t 2 γ/2 d t dt δκ r d r 2 γ/2 d t 1 κ t+ t in (4.31) and then integrating over [0, t] 2 t δκ t 2 γ/2 d t + δκ t 2 , 0 γ/2 ρ ∞ s , 1 t+ t δκ r d r γ/2 ds. (4.31) T t T 0 0 δκ t 2 δκ t 2 dt γ/2 d t dt + 0 γ/2 we get that t 0 1 t+ t δκ r d r 2 V 1 d t ≤ 1 κ t 0 t+ t ρ ∞ s , 1 κ T 0 T 0 δκ t 2 γ/2 d t dt + 1 , κ t+ 1 t δκ r d r γ/2 ds d t -t 1 κ 2T T 0 ρκ t 2 γ/2 + ρ ∞ t 2 γ/2 d t + 1 , κ t+ 1 0 δκ t+ -δκ t , t δκ r d r d t. (4.32) 1 κ (κ + 2) + 1 κ .
	Now we use twice the Cauchy-Schwarz's inequality in order to get that the first term (4.36)
	on the right hand side in the previous expression is bounded from above by
		1 κ t 0	t	t+	ρ ∞ s	γ/2	1	t	t+	δκ r d r	γ/2	ds d t
		≤	1 κ		0 t	t	t+	ρ ∞ s	2 γ/2 dsd t		0 t		t	t+	1	t	t+	δκ r d r	γ/2 2	ds d t	(4.33)
		≤ κ		0 t		t	t+	ρ ∞ s	2 γ/2 dsd t			0 t			1	t	t+	δκ r d r	γ/2 2	d t.
	By a similar argument as the one in the proof of item i) of Theorem 2.13 we have that
	the second term on the right hand side in (4.32) is bounded from above by
										1			t+	δκ t	2 d t +	1	δκ t	2 d t.	(4.34)
												t									0

ACKNOWLEDGEMENTS

This work has been supported by the projects EDNHS ANR-14-CE25-0011, LSD ANR-15-CE40-0020-01 of the French National Research Agency (ANR) and of the PHC Pessoa Project 37854WM. B.J.O. thanks Universidad Nacional de Costa Rica and L'institut Français d'Amérique centrale-IFAC for financial support through his Ph.D grant. P.G. thanks FCT/Portugal for support through the project UID/MAT/04459/2013. This project has received funding from the European Research Council (ERC) under the European Union's Horizon 2020 research and innovative programme (grant agreement No 715734). This work was finished during the stay of P.G. at Institut Henri Poincaré -Centre Emile Borel during the trimester "Stochastic Dynamics Out of Equilibrium". P.G. thanks this institution for hospitality and support. The authors thank the Program Pessoa of Cooperation between Portugal and France with reference 406/4/4/2017/S.

PROOF OF THEOREM 2.15

In this section we prove items i) and ii) of Theorem 2.15. Now we are interested in analyzing the convergence of the stationary solution ρκ as κ → 0 and κ → ∞. From Definition 2.9, for κ ≥ 0, and for φκ = ρκ -ρ∞ we have that φκ ∈ γ/2 0 and 〈 φκ , -G〉 + κ〈 φκ , G〉 V 1 = I ρ∞ (G), (5.1) for any test function G of compact support included in (0, 1). Above I ρ∞ :

→ is a linear form defined by I ρ∞ (G) = 〈 ρ∞ , G〉. Moreover, this linear form is continuous. Indeed, using integration by parts given in Proposition 3.3 in [START_REF] Guan | The reflected α-symmetric stable processes and regional fractional Laplacian[END_REF] we have that

(5.2)

Above we used the Cauchy-Schwarz's inequality and the fact that ρ∞ γ/2 is finite (see

Then it is enough to analyze the behavior of φκ . We claim that we can take G = φκ in (5.1). The justification is postponed to the end of the proof. Whence, from (5.2) we have that φκ 2 γ/2 + κ φκ 2

from where we conclude that φκ γ/2 < ∞. Plugging this back into (5.3) we get that )). We claim that we can take G = φκ -φ0 in the previous equality. The proof is analogous to the one done at the end of this section. Thus, we get that φκ -

From (5.4) and fractional Hardy's inequality given in (4.2) we have that

from where we conclude that φκ -φ0 γ/2 κ. Then φκ converges to φ0 , as κ → 0 in the • γ/2 norm. So far we proved item i).

Remark 5.1. From fractional Hardy's inequality (see 4.2) the convergence is also true in

) φκ -φ0 we conclude that the convergence also holds in L 2 .

For item ii), by (5.4) we get that φκ V 1 → 0 and so φκ → 0 as k → ∞. We conclude this proof by showing that we can take G = φκ in (5.1). Indeed, since

0 , there exists a sequence { Hκ n } n≥1 in C ∞ c ((0, 1)) converging to φκ , i.e, Hκ n -φκ γ/2 → 0 as n → ∞. Observe that as a result of the latter and (4.2) we also have Hκ

Using the Cauchy-Schwarz's inequality we have that

Hκ n -φκ γ/2 , all going to 0 as n → ∞. Thus, we can rewrite (5.1) as

). Now it is enough to take n → ∞.

UNIQUENESS OF WEAK SOLUTIONS

In this section we prove Lemmas 2.8 and 2.11. For Lemma 2.8, we only focus in the proof of the uniqueness for the weak solutions of (2.10) for κ = κ > 0. The proof of the uniqueness of the weak solutions of (2.10) for κ = 0 and (2.12) is analogous, the difference is that only the first two items in Lemma 6.1 below are required. Finally, in Subsection 6.2 we prove Lemma 2.11.

6.1. Proof of Lemma 2.15. Let ρ κ,1 and ρ κ,2 two weak solutions of (2.10) with the same initial condition and let us denote ρκ = ρ κ,1 -ρ κ,2 . For almost every t ∈ [0, T ], we identify ρκ t with its continuous representation on [0, 1]. Therefore, by Remark 2.4 we have ρκ

Proof. For i) we have, by definition of L 0 N , that

In order to prove ii), note that (σ x η) x -η x (σ x η) y -η y is equal to zero, for all x ∈ . Thus, by definition of L r N , we have that