Luisa Arlotti
email: luisa.arlotti@uniud.it

Jacek Banasiak
email: banasiak@ukzn.ac.za

Bertrand Lods
email: bertrand.lods@math.univ-bpclermont.fr

A NEW APPROACH TO TRANSPORT EQUATIONS ASSOCIATED TO A REGULAR FIELD: TRACE RESULTS AND WELL-POSEDNESS

Keywords: Transport equation, Boundary conditions, C0-semigroups, Characteristic curves AMS subject classifications (2000): 47D06, 47D05, 47N55, 35F05, 82C40. 1

come

A new approach to transport equations associated to a regular field: trace results and well-posedness.

INTRODUCTION

In this paper we present new methodological tools to investigate the well-posedness of the general transport equation

∂ t f (x, t) + F (x) • ∇ x f (x, t) = 0 (x ∈ Ω, t > 0), (1.1a)
supplemented by boundary condition

f |Γ -(y, t) = ψ -(y, t), (y ∈ Γ -, t > 0), (1.1b)
and the initial condition f (x, 0) = f 0 (x), (x ∈ Ω).

(1.1c) Here Ω is a sufficiently smooth open subset of R N , Γ ± are suitable boundaries of the phase space and ψ -is a given function of the trace space L 1 (Γ -, dµ -) corresponding to the boundary Γ -(see Section 2 for details).

The present paper is part of a series of papers on transport equations with general vector fields [START_REF] Arlotti | On transport equations driven by a non-divergence-free force field[END_REF][START_REF]On general transport equations with abstract boundary conditions[END_REF] and introduce all the methodological tools that allow us not only to solve the initial-boundary problem (1.1) but also to treat in [START_REF]On general transport equations with abstract boundary conditions[END_REF] the case of abstract boundary conditions relying the incoming and outgoing fluxes, generalizing the results of [START_REF] Beals | Abstract time-dependent transport equations[END_REF].

The main novelty of our approach is that we assume R N to be endowed with a general positive Radon measure µ. Here by a Radon measure we understand a Borel measure (or its completions, see [15, p. 332]) which is finite on compact sets. As we shall see it further on, taking into account such general Radon measure µ leads to a large amount of technical difficulties, in particular in the definition of trace spaces and in the derivation of Green's formula. Moreover, for such a measure µ, it is far from being trivial to identify the vector field F • ∇ x to the time derivative along the characteristic curves (as done in [START_REF] Beals | Abstract time-dependent transport equations[END_REF]Formulae (5.4) & (5.5), p.392]): the main difficulty stemming from the impossibility of applying classical convolution arguments (and the so-called Friedrich's lemma). We overcome this difficulty by introducing new mollification techniques along the characteristic curves. Let us explain in more details our general assumptions:

1.1. General assumption and motivations. The transport coefficient F is a time independent vector field F : R N -→ R N which is (globally) Lipschitz-continuous with Lipschitz constant κ > 0, i.e.

|F (x 1 ) -F (x 2 )| κ|x 1x 2 | for any x 1 , x 2 ∈ R N .

(1.2) Clearly, one can associate a flow (T t ) t∈R to this field F (with the notations of Section 2.1, T t = Θ(•, t, 0)) and we make the following fundamental assumption (known as Liouville's Theorem whenever µ is the Lebesgue measure) on F : Assumption 1. The measure µ is invariant under the flow (T t ) t∈R , i.e. µ(T t A) = µ(A) for any measurable subset A ⊂ R N and any t ∈ R.

Remark 1.1. Notice that, whenever µ is the Lebesgue measure over R N , it is well-known that Assumption 1 is equivalent to div(F (x)) = 0 for any x ∈ R N . More generally, by virtue of [START_REF] Ambrosio | Transport equation and Cauchy problem for BV vector fields[END_REF]Remark 3 & Proposition 4], Assumption 1 holds for a general Borel measure µ provided the field F is locally integrable with respect to µ and divergence-free with respect to µ in the sense that

R N F (T t (x)) • ∇ x f (T t (x))dµ(x) = 0, ∀t ∈ R
for any infinitely differentiable function f with compact support.

A typical example of such a transport equation is the so-called Vlasov equation for which: i) The phase space Ω is given by the cylindrical domain Ω = D × R 3 ⊂ R 6 where D is a sufficiently smooth open subset of R 3 , referred to as the position space, while the so-called velocity space is here given by R 3 . The measure µ is given by dµ(x) = dxdβ(v) where β is a suitable Radon measure on R 3 , e.g. Lebesgue measure over R 3 for continuous models or combination of Lebesgue measures over suitable spheres for the multigroup model. ii) For any x = (x, v) ∈ D × R 3 , F (x) = (v, F(x, v)) ∈ R 6 (1. [START_REF] Ambrosio | Transport equation and Cauchy problem for non-smooth vector fields[END_REF] where F = (F 1 , F 2 , F 3 ) is a time independent force field over D×R 3 satisfying Assumption 1 and (1.2). The free transport case, investigated in [START_REF] Voigt | Functional analytic treatment of the initial boundary value problem for collisionless gases[END_REF][START_REF] Arlotti | Substochastic semigroups for transport equations with conservative boundary conditions[END_REF], corresponds to F = 0. The existence of solution to the transport equation (1.1a) is a classical matter when considering the whole space Ω = R N . In particular, the concept of renormalized solutions allows to consider irregular transport coefficient F (•) (see [START_REF] Diperna | Ordinary differential equations, transport theory and Sobolev spaces[END_REF] and the recent contributions [START_REF] Ambrosio | Transport equation and Cauchy problem for BV vector fields[END_REF][START_REF] Bris | Renormalized solutions of some transport equations with partially W 1,1 velocities and applications[END_REF]) which is of particular relevance in fluid mechanics.

On the other hand, there are few results addressing the initial-boundary value problem (1.1), possibly due to difficulties created by the boundary conditions (1.1b). We mention here the seminal works by C. Bardos [START_REF] Bardos | Problèmes aux limites pour les équations aux dérivées partielles du premier ordre à coefficients réels; théorèmes d'approximation; application à l'équation de transport[END_REF], and by R. Beals and V. Protopopescu [START_REF] Beals | Abstract time-dependent transport equations[END_REF] (see also [START_REF] Greenberg | Boundary Value Problems in Abstract Kinetic theory[END_REF][START_REF] Van Der Mee | Time dependent kinetic equations with collision terms relatively bounded with respect to collision frequency[END_REF]). Let us however mention that the results of [START_REF] Beals | Abstract time-dependent transport equations[END_REF][START_REF] Greenberg | Boundary Value Problems in Abstract Kinetic theory[END_REF] introduce restrictive assumptions on the characteristics of the equation. For instance, fields with 'too many' periodic trajectories create serious difficulties. They are however covered in a natural way by the theory presented here, see Examples 2.5 & 2.6.

Presentation of the results.

In this paper, we revisit and generalize the afore-mentioned results to the general case F = 0 and for a general Radon measure µ. The latter, in particular, leads to numerous technical problems such as e.g. determination of suitable measures µ ± over the 'incoming' and 'outgoing' parts Γ ± of ∂Ω. We provide here a general construction of these 'trace measures' generalizing, and making more precise, the results of [START_REF] Beals | Abstract time-dependent transport equations[END_REF][START_REF] Greenberg | Boundary Value Problems in Abstract Kinetic theory[END_REF]. This construction allows us to establish Proposition 2.12 which allows to compute integrals over Ω via integration along the integral curves of F (•) coming from the boundary ∂Ω, and which is free from some restrictive assumptions of op. cit. In particular, we present a new proof of the Green formula clarifying and removing gaps of the proofs in [START_REF] Beals | Abstract time-dependent transport equations[END_REF][START_REF] Greenberg | Boundary Value Problems in Abstract Kinetic theory[END_REF]. Of course, the boundary condition (1.1b) we treat here is less general than the abstract ones investigated in [START_REF] Beals | Abstract time-dependent transport equations[END_REF][START_REF] Greenberg | Boundary Value Problems in Abstract Kinetic theory[END_REF] but, as we already mentioned it, the tools we introduce here will allow us to generalize, in a subsequent paper [START_REF]On general transport equations with abstract boundary conditions[END_REF], the results of the op. cited by dealing with abstract boundary conditions.

Another major difficulty, when dealing with a general Radon measure µ, is to provide a precise definition of the transport operator T max associated to (1.1). It appears quite natural to define the transport operator T max (with its maximal domain on L 1 (Ω, dµ)) as a weak directional derivative along the characteristic curves in the L 1 -sense. However, it is not clear a priori that any function f for which the weak directional derivative exists in L 1 (Ω, dµ) (with appropriate and minimal class of test-functions) admits a trace over Γ ± . With the aim of proving such a trace result, we provide here a new characterization of the transport operator related to a mild representation of the solution to (1.1). Namely, we prove (Theorem 3.6) that the domain D(T max ) (as defined in Section 3), is precisely the set of functions f ∈ L 1 (Ω, dµ) that admits a representative which is absolutely continuous along almost any characteristic curve.

Note that in the classical case when µ is the Lebesgue measure, such a representation is known to be true [START_REF] Diperna | Ordinary differential equations, transport theory and Sobolev spaces[END_REF]Appendix]. Actually, in this case, one defines the domain D(T max ) as the set of all f ∈ L 1 (Ω, dµ) for which the directional derivative -F • ∇f exists in the distributional sense and belongs to L 1 (Ω, dµ). Then, by convolution arguments, it is well-known that the set

C 1 0 (Ω) ∩ D(T max ) is dense in D(T max ) for the graph norm f = f + F • ∇f .
The question is much more delicate for a general Radon measure µ. Indeed, in such a case, the convolution argument used in the case of the Lebesgue measure does not apply anymore. Our strategy to prove the characterization of T max is also based on a convolution argument but it uses mollification technique along the characteristic curves as developed in Section 3. Such a result shall allow us to obtain a rigorous derivation of Green's formula, clarifying some results of [START_REF] Beals | Abstract time-dependent transport equations[END_REF]. 1.3. Plan of the paper. The organization of the paper is as follows. In Section 2 we introduce main tools used throughout the paper and present the aforementioned new results concerning integration over the characteristic curves of F as well as a new construction of the boundary measures over the 'incoming' and 'outgoing' parts Γ ± of ∂Ω which generalizes and clarifies that of [START_REF] Beals | Abstract time-dependent transport equations[END_REF][START_REF] Greenberg | Boundary Value Problems in Abstract Kinetic theory[END_REF]. In Section 3 we provide a construction of the maximal transport operator T max . It is defined in a weak sense, through its action on suitably defined test functions. The fundamental result of this section shows that any function in the domain D(T max ) admits a representation which is absolutely continuous along almost any characteristic which, in turn, allows for existence of its traces on the outgoing and incoming parts of the boundary. In Section 4 we apply the results of Section 3 to prove well-posedness of the time-dependent transport problem with no reentry boundary conditions associated with T max . Moreover, we consider the corresponding stationary problem and, as a by-product, we recover a new proof of the Green formula.

INTEGRATION ALONG THE CHARACTERISTICS

Characteristic curves.

A crucial role in our study is played by the characteristic curves associated to the field F . Precisely, for any x ∈ R N and t ∈ R, consider the initial-value problem

   d ds X(s) = F (X(s)), (s ∈ R); X(t) = x. (2.1)
Since F is Lipschitz continuous on R N , Eq. (2.1) has a unique global in time solution and this allows to define the flow-mapping Θ :

R N × R × R → R N , such that, for (x, t) ∈ R N × R, the mapping: X(•) : s ∈ R -→ Θ(x, t, s)
is the only solution of Eq. (2.1). Being concerned with solutions to the transport equation (1.1) in the region Ω, we have to introduce the definition of stay times of the characteristic curves in Ω:

Definition 2.1. For any x ∈ Ω, define τ ± (x) = inf{s > 0 ; Θ(x, 0, ±s) / ∈ Ω}, with the convention that inf ∅ = ∞, and set τ

(x) = τ + (x) + τ -(x).
In other words, given x ∈ Ω, I x = (-τ -(x), τ + (x)) is the maximal interval for which Θ(x, 0, s) lies in Ω for any s ∈ I x and τ (x) is the length of the interval I x . Notice that 0 τ ± (x) ∞. Thus, the function Θ restricted to the set

Λ := (x, t, s) ; x ∈ Ω, t ∈ R , s ∈ (t -τ -(x), t + τ + (x))
is such that Θ(Λ) = Ω. Note that here we do not assume that the length of the interval

I x = (-τ -(x), τ + (x)) is finite. In particular, I x = R for any stationary point x of F , i.e. F (x) = 0. If τ (x) is finite, then the function X : s ∈ I x -→ Θ(x, 0, s) is bounded since F is Lipschitz continuous.
Moreover, still by virtue of the Lipschitz continuity of F , the only case when τ ± (x) is finite is when Θ(x, 0, ±s) reaches the boundary ∂Ω so that Θ(x, 0, ±τ ± (x)) ∈ ∂Ω. We note that, since F is Lipschitz around each point of ∂Ω, the points of the set {y ∈ ∂Ω ; F (y) = 0} (introduced in [START_REF] Beals | Abstract time-dependent transport equations[END_REF][START_REF] Greenberg | Boundary Value Problems in Abstract Kinetic theory[END_REF]) are equilibrium points of the F and cannot be reached in finite time.

Remark 2.2. We emphasize that periodic trajectories which do not meet the boundaries have τ ± = ∞ and thus are treated as infinite though geometrically they are bounded.

Finally we mention that it is not difficult to prove that the mappings τ ± : Ω → R + are lower semicontinuous and therefore measurable, see e.g., [7, p. 301] The flow Θ(x, t, s) defines, at each instant t, a mapping of the phase space Ω into R N . Through this mapping, to each point x there corresponds the point x s,t = Θ(x, t, s) reached at time s by the point which was at x at the 'initial' time t. The flow Θ, restricted to Λ, has the properties:

Proposition 2.3. Let x ∈ Ω and t ∈ R be fixed. Then, (i) Θ(x, t, t) = x. (ii) Θ(Θ(x, t, s 1 ), s 1 , s 2 ) = Θ(x, t, s 2 ), ∀s 1 , s 2 ∈ (t -τ -(x), t + τ + (x)). (iii) Θ(x, t, s) = Θ(x, t -s, 0) = Θ(x, 0, s -t), ∀s ∈ (t -τ -(x), t + τ + (x)). (iv) |Θ(x 1 , t, s) -Θ(x 2 , t, s)| exp(κ|t -s|)|x 1 -x 2 | for any x 1 , x 2 ∈ Ω, s -t ∈ I x 1 ∩ I x 2 .
An important consequence of (iii) above is that Θ(x, 0, s) = Θ(x, -s, 0) for any x ∈ Ω, 0 s τ + (x). Therefore, from now on, to shorten notations we shall denote Φ(x, t) = Θ(x, 0, t), ∀t ∈ R, so that Φ(x, -t) = Θ(x, t, 0), t ∈ R. We define the incoming and outgoing part of the boundary ∂Ω through the flow Φ:

Definition 2.4. The incoming Γ -and the outgoing Γ + parts of the boundary ∂Ω are defined by:

Γ ± := {y ∈ ∂Ω ; ∃x ∈ Ω, τ ± (x) < ∞ and y = Φ(x, ±τ ± (x)) } . (2.2)
Properties of Φ and of τ ± imply that Γ ± are Borel sets. It is possible to extend the definition of τ ± to Γ ± as follows. If x ∈ Γ -then we put τ -(x) = 0 and denote τ + (x) the length of the integral curve having x as its left end-point; similarly if x ∈ Γ + then we put τ + (x) = 0 and denote τ -(x) the length of the integral curve having x as its right endpoint. Note that this definition implies that τ ± are measurable over Ω ∪ Γ -∪ Γ + .

Let us illustrate the above definition of Γ ± by two simple 2D examples:

Example 2.5 (Harmonic oscillator in a rectangle). Let Ω = (-a, a) × (-ξ, ξ) with a, ξ > 0 and let us consider the harmonic oscillator force field

F (x) = (v, -ω 2 x), for any x = (x, v) ∈ Ω (2.3)
where ω > 0. We take as µ the Lebesgue measure over R 2 and, since F is divergence-free, Assumption 1 is fulfilled. In this case, for any

x 0 = (x 0 , v 0 ) ∈ Ω, the solution (x(t), v(t)) = Φ(x 0 , t) to the characteristic equation d dt X(t) = F (X(t)) , X(0) = x 0 , given by Φ(x 0 , t) = x 0 cos(ωt) + v 0 ω sin(ωt) ; -x 0 ω sin(ωt) + v 0 cos(ωt) , is such that ω 2 x 2 (t) + v 2 (t) = ω 2 x 2 0 + v 2 0 , t ∈ (-τ -(x 0 ), τ + (x 0
)) which means that the integral curves associated to F are ellipses centered at (0, 0) and oriented in the counterclockwise direction. Now,

∂Ω = {-a} × [-ξ, ξ] {a} × [-ξ, ξ] [-a, a] × {-ξ} [-a, a] × {ξ}
and it is easy to check that

Γ ± = {±a} × (-ξ, 0] {∓a} × [0, ξ) [0, a) × {±ξ} (-a, 0] × {∓ξ} .
Notice that Γ + ∩ Γ -= {(a, 0), (0, ξ), (-a, 0), (0, -ξ)} and ∂Ω \ (Γ + ∪ Γ -) = {(a, ξ), (a, -ξ), (-a, ξ), (-a, ξ)} is a discrete set (of linear Lebesgue measure zero).

Example 2.6 (Hamonic oscillator in a stadium). Consider now the two-dimensional phase space (where R 2 is still endowed with the Lebesgue measure µ):

Ω = {x = (x, v) ∈ R 2 ; x 2 + v 2 < 2 and -1 < v < 1}
and consider the harmonic oscillator force field F given by (2.3) with ω = 1 for simplicity. Then, the integral curves associated to F are circles centered at (0, 0) and oriented in the counterclockwise direction. In this case, one can see that

Γ ± = {(x, -1) ; -1 < ± x 0} ∪ {(x, 1) ; 0 ± x < 1}.
In particular, one sees that ∂Ω\ Γ + ∪Γ -= (x, v) ∈ R 2 ; x 2 + v 2 = 2 ; -1 v 1 is a 'big' part of the boundary ∂Ω (with positive linear Lebesgue measure). Notice also that τ + (x) = +∞ for any x = (x, v) with

x 2 + v 2 < 1.
The main aim of the present discussion is to represent Ω as a collection of characteristics running between points of Γ -and Γ + so that the integral over Ω can be split into integrals over Γ -(or Γ + ) and along the characteristics. However, at present we cannot do this in a precise way since, in general, the sets Γ + and Γ -do not provide a partition of ∂Ω as there may be 'too many' characteristics which extend to infinity on either side. Since we have not assumed Ω to be bounded, Γ -or Γ + may be empty and also we may have characteristics running from -∞ to +∞ such as periodic ones. Thus, in general, characteristics starting from Γ -or ending at Γ + would not fill the whole Ω and, to proceed, we have to construct an auxiliary set by extending Ω into the time domain and use the approach of [START_REF] Beals | Abstract time-dependent transport equations[END_REF] which is explained below.

2.2. Integration along characteristics. For any 0 < T < ∞, we define the domain

Ω T = Ω × (0, T )
and the measure dµ T = dµ ⊗ dt on Ω T . Consider the vector field over Ω T :

Y = ∂ t + F (x) • ∇ x = A (ξ) • ∇ ξ
where A (ξ) = (F (x), 1) for any ξ = (x, t). We can define the characteristic curves of A as the solution ξ(s) = (X(s), θ(s)) to the system d ds ξ(s) = A (ξ(s)), i.e.

d ds X(s) = F (X(s)), d ds θ(s) = 1, (s ∈ R), with X(0) = x, θ(0) = t.
It is clear that the solution ξ(s) to the above system is given by

X(s) = Φ(x, s), θ(s) = s + t,
and we can define the flow of solution Ψ(ξ, s) = (Φ(x, s), s+t) associated to A and the existence times of the characteristic curves of Y are defined, for any ξ = (x, t) ∈ Ω T , as

ℓ ± (ξ) = inf{s > 0, (Φ(x, ±s), ±s + t) / ∈ Ω T }.
The flow Ψ(•, •) enjoys, mutatis mutandis, the properties listed in Proposition 2.3 and µ T is invariant under Ψ. Moreover, since A is clearly Lipschitz continuous on Ω T , no characteristic of Y can escape to infinity in finite time. In other words, all characteristic curves of Y now have finite lengths. Indeed, if Φ(x, ±s) does not reach ∂Ω, then the characteristic curve Ψ(ξ, ±s) enters or leaves Ω T through the bottom Ω × {0}, or through the top Ω × {T } of it. Precisely, it is easy to verify that for ξ = (x, t) ∈ Ω T we have

ℓ + (ξ) = τ + (x) ∧ (T -t) and ℓ -(ξ) = τ -(x) ∧ t,
where ∧ denotes minimum. This clearly implies sup{ℓ ± (ξ) ; ξ ∈ Ω T } T. Define now

Σ ±, T = {ζ ∈ ∂Ω T ; ∃ξ ∈ Ω T such that ζ = Ψ(ξ, ±ℓ ± (ξ))}.
The definition of Σ ±, T is analogous to Γ ± with the understanding that now the charateristic curves correspond to the vector field A . In other words, Σ -, T (resp. Σ +, T ) is the subset of ∂Ω T consisting of all left (resp. right) limits of characteristic curves of A in Ω T whereas Γ -(resp. Γ + ) is the subset of ∂Ω consisting of all left (resp. right) limits of characteristic curves of F in Ω. The main difference (and the interest of such a lifting to Ω T ) is the fact that each characteristic curve of A does reach the boundaries Σ ±, T in finite time. The above formulae allow us to extend functions ℓ ± to Σ ±, T in the same way as we extended the functions τ ± to Γ ± . With these considerations, we can represent, up to a set of zero measure, the phase space Ω T as

Ω T = {Ψ(ξ, s) ; ξ ∈ Σ -, T , 0 < s < ℓ + (ξ)} = {Ψ(ξ, -s) ; ξ ∈ Σ +, T , 0 < s < ℓ -(ξ)}. (2.4)
With this realization we can prove the following:

Proposition 2.7. Let T > 0 be fixed. There are unique positive Borel measures dν ± on Σ ±,T such that dµ T = dν + ⊗ ds = dν -⊗ ds.

Proof. For any δ > 0, define E δ as the set of all bounded Borel subsets E of Σ -,T such that ℓ + (ξ) > δ for any ξ ∈ E. Let us now fix E ∈ E δ . For all 0 < σ δ put

E σ = {Ψ(ξ, s) ; ξ ∈ E, 0 < s σ}. Clearly E σ is a measurable subset of Ω T . Define the mapping h : σ ∈ (0, δ] → h(σ) = µ T (E σ ) with h(0) = 0. If σ 1 and σ 2 are two positive numbers such that σ 1 + σ 2 δ, then E σ 1 +σ 2 \ E σ 1 = {Ψ(ξ, s) ; ξ ∈ E, σ 1 < s σ 1 + σ 2 } = {Ψ(η, σ 1 ) ; η ∈ E σ 2 }.
The properties of the flow Ψ (see Proposition 2.3) ensure that the mapping η → Ψ(η, σ 1 ) is one-to-one and measure preserving, so that

µ T (E σ 1 +σ 2 \ E σ 1 ) = µ T (E σ 2 ) = h(σ 2 ). Since E σ 1 +σ 2 = E σ 1 ∪ (E σ 1 +σ 2 \ E σ 1 ), we immediately obtain h(σ 1 + σ 2 ) = h(σ 1 ) + h(σ 2 ) for any σ 1 , σ 2 > 0 with σ 1 + σ 2 δ. (2.5)
This is the well-known Cauchy equation, though defined only on an interval of the real line. It can be solved in a standard way using non-negativity instead of continuity, yielding:

h(σ) = c E σ for any 0 < σ δ
where c E = h(δ)/δ. We define ν -(E) = c E . It is not difficult to see that, with the above procedure, the mapping ν -(•) defines a positive measure on the ring E = δ>0 E δ of all the Borel subsets of Σ -,T on which the function ℓ + (ξ) is bounded away from 0. Such a measure ν -can be uniquely extended to the σ-algebra of the Borel subsets of Σ -,T (see e.g. [12, Theorem A, p. 54]).

Consider now a Borel subset E of Σ -,T and a Borel subset I of R + , such that for all ξ ∈ E and s ∈ I we have 0 < s < ℓ + (ξ). Then

E × I = {Ψ(ξ, s) ; ξ ∈ E, s ∈ I} ⊂ Ω T .
Thanks to the definition of ν -(•), we can state that µ T (E × I) = ν -(E)meas(I) where meas(I) denotes the linear Lebesgue measure of I ⊂ R. This shows that dµ T = dν -⊗ ds. Similarly we can define a measure ν + on Σ +,T and prove that dµ T = dν + ⊗ ds. The uniqueness of the measures dν ± is then obvious.

Remark 2.8. Note that the above construction of the Borel measures dν

± differs from that of [11, Lemmas XI.3.1 & 3.2], [9, Propositions 7 & 8]
which , moreover, only apply when µ is absolutely continuous with respect to the Lebesgue measure. Our construction is much more general and can also be generalized to the case of a non-divergence force field F , [START_REF] Arlotti | On transport equations driven by a non-divergence-free force field[END_REF].

Next, by the cylindrical structure of Ω T , and the representation of Σ ±,T as

Σ -, T = (Γ -× (0, T )) ∪ Ω × {0} and Σ +, T = (Γ + × (0, T )) ∪ Ω × {T },
the measures dν ± over Γ ± × (0, T ) can be written as dν ± = dµ ± ⊗ dt, where dµ ± are Borel measures on Γ ± . This leads to the following Lemma 2.9. There are unique positive Borel measures dµ ± on Γ ± such that, for any f ∈ L 1 (Ω T , dµ T )

Ω T f (x, t)dµ T (x, t) = T 0 dt Γ + dµ + (y) τ -(y)∧t 0 f (Φ(y, -s), t -s)ds + Ω dµ(x) τ -(x)∧T 0 f (Φ(x, -s), T -s)ds, (2.6)
and

Ω T f (x, t)dµ T (x, t) = T 0 dt Γ - dµ -(y) τ + (y)∧(T -t) 0 f (Φ(y, s), t + s)ds + Ω dµ(x) τ + (x)∧T 0 f (Φ(x, s), s)ds.
(2.7)

The above fundamental result allows to compute integrals over the cylindrical phase-space Ω T through integration along the characteristic curves. Let us now generalize it to the phase space Ω. Here the main difficulty stems from the fact that the characteristic curves of the vector field F are no longer assumed to be of finite length. In order to extend Lemma 2.9 to possibly infinite existence times, first we prove the following: Lemma 2.10. Let T > 0 be fixed. Then, τ + (x) < T for any x ∈ Ω if and only if τ -(x) < T for any x ∈ Ω.

Proof. It is easy to see that τ + (x) < T for any x ∈ Ω is equivalent to τ (x) < T for any x ∈ Ω and this is also equivalent to τ -(x) < T for any x ∈ Ω.

Hereafter, the support of a measurable function f defined on Ω is defined as Suppf = Ω \ ω where ω is the maximal open subset of Ω on which f vanishes dµ-almost everywhere. Proposition 2.11. Let f ∈ L 1 (Ω, dµ). Assume that there exists τ 0 > 0 such that τ ± (x) < τ 0 for any x ∈ Supp(f ). Then,

Ω f (x)dµ(x) = Γ + dµ + (y) τ -(y) 0 f (Φ(y, -s)) ds = Γ - dµ -(y) τ + (y) 0 f (Φ(y, s))ds. (2.8)
Proof. For any T > τ 0 , define the domain

Ω T = Ω × (0, T ). Since T < ∞, it is clear that f ∈ L 1 (Ω T , dµdt) and, by (2.6), we get T Ω f (x)dµ(x) = T 0 dt Γ + dµ + (y) t∧τ -(y) 0 f (Φ(y, -s))ds+ Ω dµ(x) τ -(x) 0 f (Φ(x, -s))ds.
Since the formula is valid for any T > τ 0 , differentiating with respect to T leads to the first assertion. The second assertion is proved in the same way by using formula (2.7).

To drop the finiteness assumption on τ ± (x), first we introduce the sets

Ω ± = {x ∈ Ω ; τ ± (x) < ∞}, Ω ±∞ = {x ∈ Ω ; τ ± (x) = ∞},
and

Γ ±∞ = {y ∈ Γ ± ; τ ∓ (y) = ∞}.
Then Proposition 2.12. Given f ∈ L 1 (Ω, dµ), one has

Ω ± f (x)dµ(x) = Γ ± dµ ± (y) τ ∓ (y) 0 f (Φ(y, ∓s)) ds, (2.9)
and

Ω ± ∩Ω ∓∞ f (x)dµ(x) = Γ ± ∞ dµ ± (y) ∞ 0 f (Φ(y, ∓s)) ds.
(2.10)

Proof. Assume first f 0. Let us fix T > 0. It is clear that x ∈ Ω satisfies τ + (x) < T if and only if x = Φ(y, -s), with y ∈ Γ + and 0 < s < T ∧ τ -(y). Then, by Proposition 2.11, {τ + (x)<T } f (x)dµ(x) = Γ + dµ + (y) T ∧τ -(y) 0 f (Φ(y, -s))ds.
Since f 0, the inner integral is increasing with T and, using the monotone convergence theorem, we let T → ∞ to get

Ω + f (x)dµ(x) = Γ + dµ + (y) τ -(y) 0 f (Φ(y, -s)) ds
which coincides with (2.9). We proceed in the same way with integration on Γ -and get the second part of (2.9). Next we consider the set

∆ = {x ∈ Ω ; x = Φ(y, -s), y ∈ Ω +∞ , 0 < s < T }. Proposition 2.11 asserts that ∆ f (x)dµ(x) = Ω +∞ dµ + (y) T 0 f (Φ(y, -s))ds.
Letting again T → ∞, we get (2.10). We extend the results to arbitrary f by linearity.

Finally, with the following, we show that it is possible to transfer integrals over Γ -to Γ + :

Proposition 2.13. For any ψ ∈ L 1 (Γ -, dµ -), Γ -\Γ -∞ ψ(y)dµ -(y) = Γ + \Γ +∞ ψ(Φ(z, -τ -(z)))dµ + (z).
(2.11)

Proof. For any ǫ > 0, let f ǫ be the function defined on Ω + ∩ Ω -by

ψ ǫ (x) =    ψ(Φ(x, -τ -(x))) τ + (x) + τ -(x) if τ -(x) + τ + (x) > ǫ, 0 else. Since ψ ǫ ∈ L 1 (Ω + ∩ Ω -, dµ), Eqs.
(2.9) and (2.10) give

Ω + ∩Ω - ψ ǫ (x)dµ(x) = {τ + (y)>ǫ}\Γ -∞ dµ -(y) τ + (y) 0 ψ(y) ds τ + (y) = {τ + (y)>ǫ}\Γ -∞ ψ(y)dµ -(y).
In the same way,

Ω + ∩Ω - ψ ǫ (x)dµ(x) = {τ -(y)>ǫ}\Γ +∞ dµ + (y) τ -(y) 0 ψ(Φ(y, -τ -(y))) ds τ -(y) = {τ -(y)>ǫ}\Γ +∞ ψ(Φ(y, -τ -(y)))dµ -(y), which leads to {τ -(y)>ǫ}\Γ +∞ ψ(Φ(y, -τ -(y)))dµ + (y) = {τ + (y)>ǫ}\Γ -∞ ψ(y)dµ -(y)
for any ǫ > 0. Passing to the limit as ǫ → 0 we get the conclusion.

We end this section with a technical result we shall need in the sequel (see Lemma 3.3): Proposition 2.14. Let K be a compact subset of Ω. Denote

K ± := {y ∈ Γ ± ; ∃t 0 ∈ R such that Φ(y, ±t) ∈ K for any t t 0 } . Then µ ± (K ± ) = 0.
Proof. Let K be a fixed compact subset of Ω. Applying Eq. (2.9) or (2.10) to the function

f (x) = χ K (x), one has ∞ > µ(K) K - dµ -(y) ∞ 0 χ K (Φ(y, t))dt.
(2.12)

By definition, if y ∈ K -, then for some t 0 ∈ R, χ K (Φ(y, t)) = 1 for any t t 0 . Therefore,

∞ 0 χ K (Φ(y, t)) = ∞, ∀y ∈ K -.
Inequality (2.12) implies that µ -(K -) = 0. One proves the result for K + in the same way.

THE MAXIMAL TRANSPORT OPERATOR AND TRACE RESULTS

The results of the previous section allow us to define the (maximal) transport operator T max as the weak derivative along the characteristic curves. To be precise, let us define the space of test functions Y as follows: Definition 3.1 (Test-functions). Let Y be the set of all measurable and bounded functions ψ : Ω → R with compact support in Ω and such that, for any x ∈ Ω, the mapping

s ∈ (-τ -(x), τ + (x)) -→ ψ(Φ(x, s)) is continuously differentiable with x ∈ Ω -→ d ds ψ(Φ(x,
lim n→∞ t ± n = τ ± (y)
and ψ(Φ(y, ±t ± n )) = 0 ∀n ∈ N.

Proof. Let ψ ∈ Y be given and let K = Supp(ψ). For any y ∈ Γ -with τ + (y) < ∞ one has Φ(y, τ + (y)) ∈ Γ + and, since K is compact in Ω, ψ(Φ(y, τ + (y)) = 0 and the existence of a sequence (t + n ) n converging to τ + (y) with the above property is clear. Now, Proposition 2.14 applied to K shows that there exists a set Γ ′ -⊂ Γ -with µ -(Γ \ Γ ′ -) = 0 and such that, for any y ∈ Γ ′ -, there is a sequence (t + n ) n converging to ∞ such that Φ(y, t n ) / ∈ K for any n ∈ N. This proves the result. The statement for Γ + is proved in the same way.

In the next step we define the transport operator (T max , D(T max )).

Definition 3.4 (Transport operator T max ). The domain of the maximal transport operator

T max is the set D(T max ) of all f ∈ L 1 (Ω, dµ) for which there exists g ∈ L 1 (Ω, dµ) such that Ω g(x)ψ(x)dµ(x) = Ω f (x) d ds ψ(Φ(x, s)) s=0 dµ(x), ∀ψ ∈ Y.
In this case, g =: T max f. Remark 3.5. Of course, in some weak sense, T max f = -F • ∇f . Precisely, for any ϕ ∈ C 1 0 (Ω), the following formula holds:

Ω (F (x) • ∇ϕ(x)) f (x)dµ(x) = Ω T max f (x)ϕ(x)dµ(x).
3.1. Fundamental representation formula: mild formulation. Recall that, if f 1 and f 2 are two functions defined over Ω, we say that f 2 is a representative of f 1 if µ{x ∈ Ω ; f 1 (x) = f 2 (x)} = 0, i.e. when f 1 (x) = f 2 (x) for µ-almost every x ∈ Ω. The following fundamental result provides a characterization of the domain of D(T max ):

Theorem 3.6. Let f ∈ L 1 (Ω,

µ). The following are equivalent:

(1) There exists g ∈ L 1 (Ω, µ) and a representative f ♯ of f such that, for µ-almost every x ∈ Ω and any -τ -(x) < t 1 t 2 < τ + (x):

f ♯ (Φ(x, t 1 )) -f ♯ (Φ(x, t 2 )) = t 2 t 1 g(Φ(x, s))ds. (3.2) (2) f ∈ D(T max ). In this case, g = T max f .
The proof of the theorem is made of several steps. The difficult part of the proof is the implication (2) =⇒ [START_REF] Amann | Ordinary Differential Equations. An introduction to nonlinear analysis[END_REF]. It is carried out through several technical lemmas based upon mollification along the characteristic curves (recall that, whenever µ is not absolutely continuous with respect to the Lebesgue measure, no global convolution argument is available). Let us make precise what this is all about. Consider a sequence (̺ n ) n of one dimensional mollifiers supported in [0, 1], i.e. for any n

∈ N, ̺ n ∈ C ∞ 0 (R), ̺ n (s) = 0 if s / ∈ [0, 1/n], ̺ n (s) 0 and 1/n 0 ̺ n (s)ds = 1.
Then, for any f ∈ L 1 (Ω, dµ), define the (extended) mollification:

̺ n ⋄ f (x) = τ -(x) 0 ̺ n (s)f (Φ(x, -s))ds.
As we shall see later, such a definition corresponds precisely to a time convolution over any characteristic curves (see e.g. (3.4)). Note that, with such a definition, it is not clear a priori that ̺ n ⋄ f defines a measurable function, finite almost everywhere. It is proved in the following that actually such a function is integrable.

Lemma 3.7. Given f ∈ L 1 (Ω, dµ), ̺ n ⋄ f ∈ L 1 (Ω, dµ) for any n ∈ N. Moreover, ̺ n ⋄ f f , ∀f ∈ L 1 (Ω, dµ), n ∈ N. (3.3)
Proof. One considers, for a given f ∈ L 1 (Ω, dµ), the extension of f by zero outside Ω:

f (x) = f (x), ∀x ∈ Ω, f (x) = 0 ∀x ∈ R N \ Ω. Then f ∈ L 1 (R N , dµ).
Let us consider the transformation:

Υ : (x, s) ∈ R N × R → Υ(x, s) = (Φ(x, -s), -s) ∈ R N × R.
As a homeomorphism, Υ is measure preserving for pure Borel measures. It is also measure preserving for completions of Borel measures (such as a Lebesgue measure) since it is measurepreserving on Borel sets and the completion of a measure is obtained by adding to the Borel σ-algebra all sets contained in a measure-zero Borel sets, see [

(x, s) ∈ R N × R → f (Φ(x, -s))
is measurable as the composition of Υ with the measurable function (x, s) → f (x). Define now

Λ = {(x, s) ; x ∈ Ω, 0 < s < τ -(x)}, Λ is a measurable subset of R N × R.
Therefore, the mapping

(x, s) ∈ R N × R -→ f (Φ(x, -s))χ Λ (x, s)̺ n (s)
is measurable. Since ̺ n is compactly supported, it is also integrable over R N × R and, according to Fubini's Theorem

[̺ n ⋄ f ](x) := R f (Φ(x, -s))χ Λ (x, s)̺ n (s)ds = τ -(x) 0 ̺ n (s)f (Φ(x, -s))ds
is finite for almost every x ∈ Ω the and the associated application

̺ n ⋄ f is integrable. Let us prove now (3.3). Since |̺ n ⋄ f | ̺ n ⋄ |f |, to show that ̺ n ⋄ f ∈ L 1 (Ω, dµ)
, it suffices to deal with a nonnegative function f ∈ L 1 (Ω, dµ). One sees easily that, for any y ∈ Γ -and any 0 < t < τ + (y),

(̺ n ⋄ f )(Φ(y, t)) = t 0 ̺ n (s)f (Φ(y, t -s))ds = t 0 ̺ n (t -s)f (Φ(y, s))ds.
(3.4) Thus,

τ + (y) 0 [̺ n ⋄ f ](Φ(y, t))dt = τ + (y) 0 ds τ + (y) s ̺ n (s)f (Φ(y, t -s))dt = τ + (y)∧1/n 0 ̺ n (s)ds τ + (y)-s 0 f (Φ(y, r))dr.
Therefore,

0 τ + (y) 0 [̺ n ⋄ f ](Φ(y, t))dt 1/n 0 ̺ n (s)ds τ + (y) 0 f (Φ(y, r))dr = τ + (y) 0 f (Φ(y, r))dr, ∀y ∈ Γ -, n ∈ N so that Γ - dµ -(y) τ + (y) 0 [̺ n ⋄ f ](Φ(y, t))dt Γ - dµ -(y) τ + (y) 0 f (Φ(y, r))dr.
This proves, thanks to Proposition 2.12, that

Ω - [̺ n ⋄ f ]dµ Ω - f dµ. (3.5)
Now, in the same way:

Ω + ∩Ω -∞ [̺ n ⋄ f ](x)dµ(x) = Γ +∞ dµ + (y) ∞ 0 [̺ n ⋄ f ](Φ(y, -t))dt = Γ +∞ dµ + (y) ∞ 0 dt ∞ 0 ̺ n (s)f (Φ(y, -s -t))ds = Γ +∞ dµ + (y) ∞ 0 dt ∞ t ̺ n (r -t)f (Φ(y, -r))dr.
so that

Ω + ∩Ω -∞ [̺ n ⋄ f ](x)dµ(x) = Γ +∞ dµ + (y) ∞ 0 f (Φ(y, -r))dr r 0 ̺ n (r -t)dt Γ +∞ dµ + (y) ∞ 0 f (Φ(y, -r))dr i.e. Ω + ∩Ω -∞ ̺ n ⋄ f (x)dµ(x) Ω + ∩Ω -∞ f (x)dµ(x). (3.6)
Finally

Ω +∞ ∩Ω -∞ [̺ n ⋄ f ](x)dµ(x) = Ω +∞ ∩Ω -∞ dµ(x) ∞ 0 ̺ n (s)f (Φ(x, -s))ds = ∞ 0 ̺ n (s)ds Ω +∞ ∩Ω -∞ f (Φ(x, -s))dµ(x).
Now, from Assumption 1, for any s 0, As it is the case for classical convolution, the family (̺ n ⋄ f ) n approximates f in L 1 -norm:

Ω +∞ ∩Ω -∞ f (Φ(x, -s))dµ(x) = Ω +∞ ∩Ω -∞ f (x)dµ(x), so that Ω +∞ ∩Ω -∞ [̺ n ⋄ f ](x)dµ(x) = Ω +∞ ∩Ω -∞ f (x)dµ(x). ( 3
Proposition 3.8. Given f ∈ L 1 (Ω, dµ), lim n→∞ Ω (̺ n ⋄ f )(x) -f (x) dµ(x) = 0. (3.8)
Proof. According to (3.3) and from the density of C 0 (Ω) in L 1 (Ω, dµ), it suffices to prove the result for any f continuous over Ω and compactly supported. Splitting f into positive and negative parts, f = f + -f -, one can also assume f to be nonnegative. From the continuity of both f and Φ(•, •), one has

K n := Supp(̺ n ⋄ f ) = x ∈ Ω , ∃s 0 ∈ Supp(̺ n ) such that Φ(x, -s 0 ) ∈ Supp(f ) .
Moreover, it is easily seen that K n+1 ⊂ K n for any n 1. Finally, it is clear that

K 1 ⊂ {x ∈ Ω ; ∃y ∈ Supp(f ) with |x -y| d} where d = sup{|Φ(x, s) -x| ; 0 s 1 ; x ∈ Supp(f )} < ∞. Therefore, K 1 is compact. Set now O n := K n ∪ Supp(f ) and O - n = {x ∈ O n ; τ -(x) < 1/n}. Noticing that µ(O 1 )
O - n |̺ n ⋄ f (x)|dµ(x) ε ∀n n 0 . Now, noticing that Supp(̺ n ⋄ f -f ) ⊂ O n , one has for any n n 0 , Ω |̺ n ⋄ f -f |dµ = On |̺ n ⋄ f -f | 2ε + On\O - n |̺ n ⋄ f -f |dµ. For any x ∈ O n \ O - n , since ̺ is supported in [0, 1/n], one has [̺ n ⋄ f ](x) -f (x) = 1/n 0 ̺ n (s)f (Φ(x, -s))ds -f (x) = 1/n 0 ̺ n (s) (f (Φ(x, -s)) -f (x)) ds.
Note that, thanks to Gronwall's lemma,

|Φ(x, -s) -x| L κ (exp(ks) -1) L κ (exp(κ/n) -1), ∀x ∈ O 1 , s ∈ (0, 1/n) where L = sup{|F (x)|, x ∈ O 1 }. Since f is uniformly continuous on O 1 , it follows that lim n→∞ sup |f (Φ(x, -s) -f (x)| ; x ∈ O 1 , s ∈ (0, 1/n) = 0
from which we deduce that there exists some n 1 0, such that

|̺ n ⋄ f (x) -f (x)| ε for any x ∈ O n \ O -
n and any n n 1 . One obtains then, for any n n 1 ,

Ω |̺ n ⋄ f -f |dµ 2ε + εµ(O n \ O - n ) 2ε + εµ(O 1 )
which proves the result.

We saw that, for a given f ∈ L 1 (Ω, dµ), ̺ n ⋄ f is also integrable (n ∈ N). Actually, we shall see that ̺ n ⋄ f is even more regular than f :

Lemma 3.9. Given f ∈ L 1 (Ω, dµ), set f n = ̺ n ⋄ f , n ∈ N. Then, f n ∈ D(T max ) with [T max f n ](x) = - τ -(x) 0 ̺ ′ n (s)f (Φ(x, -s))ds, x ∈ Ω. Proof. Set g n (x) = - τ -(x) 0 ̺ ′ n (s)f (Φ(x, -s))ds, x ∈ Ω.
It is easy to see that g n ∈ L 1 (Ω, dµ). Now, given ψ ∈ Y, let us consider the quantity

I = Ω f n (x) d ds ψ(Φ(x, s)) s=0 dµ(x).
One has to prove that I = Ω g n (x)ψ(x)dµ(x). We split the above integral over Ω into three integrals I -, I + and I ∞ over Ω -, Ω -∞ ∩ Ω + and Ω +∞ ∩ Ω -∞ respectively. Recall that, for any x ∈ Ω -, there is some y ∈ Γ -and some t ∈ (0, τ + (y)) such that x = Φ(y, t). In such a case

d ds ψ(Φ(x, s)) s=0 = d dt ψ(Φ(y, t)). (3.9)
Then, according to Prop. 2.12 and Eq. (3.4):

I -= Γ - dµ -(y) τ + (y) 0 f n (Φ(y, t)) d dt ψ(Φ(y, t))dt = Γ - dµ -(y) τ + (y) 0 d dt ψ(Φ(y, t))dt t 0 ̺ n (t -s)f (Φ(y, s))ds = Γ - dµ -(y) τ + (y) 0 f (Φ(y, s))ds τ + (y) s d dt ψ(Φ(y, t))̺ n (t -s)dt. (3.10)
Let us now investigate more carefully this last integral. Let y ∈ Γ -be fixed. If τ + (y) < ∞ then, since ψ is compactly supported, we have ψ(Φ(y, τ + (y))) = 0 and integration by part (together with ̺ n (0) = 0) leads to

τ + (y) s d dt ψ(Φ(y, t))̺ n (t -s)dt = - τ + (y) s ̺ ′ n (t -s)ψ(Φ(y, t))dt. If now τ + (y) > ∞, then, since ̺ n is supported in [0, 1/n], one has τ + (y) s d dt ψ(Φ(y, t))̺ n (t -s)dt = s+1/n s d dt ψ(Φ(y, t))̺ n (t -s)dt = - τ + (y) s ̺ ′ n (t -s)ψ(Φ(y, t))dt
Finally, we obtain,

I -= - Γ - dµ -(y) τ + (y) 0 f (Φ(y, s))ds τ + (y) s ψ(Φ(y, t))̺ ′ n (t -s)dt = - Γ - dµ -(y) τ + (y) 0 ψ(Φ(y, t))dt t 0 ̺ ′ n (s)f (Φ(y, t -s))ds.
Using again Prop. 2.12, we finally get

I -= Ω - g n (x)ψ(x)dµ(x).
One proves in the same way that

I + = Ω + ∩Ω -∞ f n (x) d ds ψ(Φ(x, s)) s=0 dµ(x) = Ω + ∩Ω -∞ g n (x)ψ(x)dµ(x).
It remains to consider

I ∞ = Ω +∞ ∩Ω -∞ f n (x) d ds ψ(Φ(x, s)) s=0 dµ(x).
One has

I ∞ = Ω +∞ ∩Ω -∞ d ds ψ(Φ(x, s)) s=0 dµ(x) ∞ 0 ̺ n (t)f (Φ(x, -t))dt = ∞ 0 ̺ n (t)dt Ω +∞ ∩Ω -∞ d ds ψ(Φ(x, s)) s=0 f (Φ(x, -t))dµ(x).
For any x ∈ Ω +∞ ∩ Ω -∞ and any t 0, setting y = Φ(x, -t), one has y ∈ Ω -∞ ∩ Ω +∞ and d ds ψ(Φ(x, s)) s=0 = d dt ψ(Φ(y, t)) from which Liouville's Theorem (Assumption 1) yields

Ω +∞ ∩Ω -∞ d ds ψ(Φ(x, s)) s=0 f (Φ(x, -t))dµ(x) = Ω +∞ ∩Ω -∞ d dt ψ(Φ(y, t))f (y)dµ(y).
Therefore,

I ∞ = Ω +∞ ∩Ω -∞ f (y)dµ(y) ∞ 0 ̺ n (t) d dt ψ(Φ(y, t))dt = - Ω +∞ ∩Ω -∞ f (y)dµ(y) ∞ 0 ̺ ′ n (t)ψ(Φ(y, t))dt = - ∞ 0 ̺ ′ n (t)dt Ω +∞ ∩Ω -∞ f (y)ψ(Φ(y, t))dµ(y).
Arguing as above, one can "turn back" to the x variable to get

Ω +∞ ∩Ω -∞ f (y)ψ(Φ(y, t))dµ(y) = Ω +∞ ∩Ω -∞ f (Φ(x, -t))ψ(x)dµ(x),
i.e.

I ∞ = - Ω +∞ ∩Ω -∞ ψ(x)dµ(x) ∞ 0 ̺ ′ n (t)f (Φ(x, -t))dt = Ω +∞ ∩Ω -∞ ψ(x)g n (x)dµ(x)
and the Lemma is proven.

Remark 3.10. Notice that Proposition 3.8 together with Lemma 3.9 prove that D(T max ) is a dense subset of L 1 (Ω, dµ). Now, whenever f ∈ D(T max ), one has the following more precise result:

Proposition 3.11. If f ∈ D(T max ), then [T max (̺ n ⋄ f )](x) = [̺ n ⋄ T max f ](x), (x ∈ Ω , n ∈ N). (3.11)
Before proving this result, we need the following very simple lemma: Lemma 3.12. For any ψ ∈ Y and any n ∈ N, define

χ n (x) = τ + (x) 0 ̺ n (s)ψ(Φ(x, s))ds, x ∈ Ω.
Then, χ n belongs to Y.

Proof. Since τ + is measurable and ̺ n is compactly supported, it is easy to see that χ n is measurable and bounded over Ω. Now, for any x ∈ Ω, and any t ∈ (τ -(x), τ + (x)), one has

χ n (Φ(x, t)) = τ + (x) t ̺ n (s -t)ψ(Φ(x, s))ds.
It is clear then from the properties of ̺ n that the mapping t ∈ (-τ

-(x), τ + (x)) → χ n (Φ(x, t)) is continuously differentiable with d dt χ n (Φ(x, t)) = - τ + (x) t ̺ ′ n (s -t)ψ(Φ(x, s))ds = τ + (x) t ̺ n (s -t) d ds [ψ(Φ(x, s))] ds.
(3.12) In particular, for t = 0, one gets

d dt χ n (Φ(x, t)) t=0 = - τ + (x) 0 ̺ ′ n (s)ψ(Φ(x, s))ds.
Since ̺ ′ n is compactly supported and ψ ∈ Y, the application x ∈ Ω -→ d dt χ n (Φ(x, t)) t=0 is measurable and bounded.

PROOF OF PROPOSITION 3.11. We use the notations of Lemma 3.9. Since ̺ n ⋄ T max f ∈ L 1 (Ω, dµ), it suffices to show that

Ω f n (x) d ds ψ(Φ(x, s)) s=0 dµ(x) = Ω ψ(x)[̺ n ⋄ T max f ](x)dµ(x), ∀ψ ∈ Y.
Here again, we shall deal separately with the integrals over

Ω -, Ω + ∩ Ω -∞ and Ω +∞ ∩ Ω -∞ .
Let χ n be defined as in Lemma 3.12, as we already saw it (see (3.12)), for any y ∈ Γ -, and any

0 < s < τ + (y), d ds χ n (Φ(y, s)) = τ + (y) s ̺ n (t -s) d dt [ψ(Φ(y, t))]dt. Consequently, according to (3.10), Ω - f n (x) d ds ψ(Φ(x, s)) s=0 dµ(x) = Γ - dµ(y) τ + (y) 0 f (Φ(y, r)) d dr χ n (Φ(y, r))dr = Ω - f (x) d ds χ n (Φ(x, s)) s=0 dµ(x) = Ω - χ n (x)[T max f ](x)dµ(x)
where, for the two last identities, we used (3.9) and the fact that χ n ∈ Y. Now, using Prop. 2.12

Ω - χ n (x)[T max f ](x)dµ(x) = Ω - [T max f ](x)dµ(x) τ + (x) 0 ̺ n (r)ψ(Φ(x, r))dr = Γ - dµ -(y) τ + (y) 0 ψ(Φ(y, s))ds s 0 ̺ n (s -t)[T max f ](Φ(y, t))dt.
Therefore, Eq. (3.4) leads to

Ω - χ n (x)[T max f ](x)dµ(x) = Γ - dµ -(y) τ + (y) 0 ψ(Φ(y, s))[̺ n ⋄ T max f ](Φ(y, s))ds = Ω - ψ(x) [̺ n ⋄ T max f ] (x)dµ(x).
The integrals over Ω + ∩ Ω -∞ and Ω -∞ ∩ Ω +∞ are evaluated in the same way.

We are in position to prove the following Proposition 3.13. Let f ∈ L 1 (Ω, dµ) and

f n = ̺ n ⋄ f , n ∈ N. Then, for µ --a. e. y ∈ Γ -, f n (Φ(y, s)) -f n (Φ(y, t)) = t s [T max f n ](Φ(y, r))dr ∀0 < s < t < τ + (y). (3.13)
In the same way, for almost every z ∈ Γ + ,

f n (Φ(z, -s)) -f n (Φ(z, -t)) = t s T max f n (Φ(z, -r))dr, ∀0 < s < t < τ -(z).
Proof. We focus only on (3.13), the second assertion following the same lines. Since f ∈ L 1 (Ω -, dµ), Proposition 2.12 implies that the integral τ + (y) 0

|f (Φ(y, r))|dr exists and is finite for µ --almost every y ∈ Γ -. Therefore, for µ --almost every y ∈ Γ -and any 0 < t < τ + (y), the quantities t 0 ̺ n (t -s)f (Φ(y, s))ds and t 0 ̺ ′ n (t -s)f (Φ(y, s))ds are well-defined and finite. Moreover, thanks to Eq. (3.4) Lemma 3.9, they are respectively equal to f n (Φ(y, t)) and [T max f n ](Φ(y, t)). In particular, the mapping t ∈ (0, τ + (y)) → [T max f n ](Φ(y, t)) ∈ R is continuous. It is then easy to see that, for any 0 < s < t < τ + (y)

t s [T max f n ](Φ(y, r))dr = - t s dr r 0 ̺ ′ n (r -u)f (Φ(y, u))du = - s 0 f (Φ(y, u))du t s ̺ ′ n (r -u)dr - t s f (Φ(y, u))du t u ̺ ′ n (r -u)dr = - t 0 f (Φ(y, u))̺ n (t -u)du + s 0 f (Φ(y, u))̺ n (s -u)du,
which is nothing but (3.13).

As a consequence, one gets the following result : Proposition 3.14. For any f ∈ D(T max ), there exists some functions f ± ∈ L 1 (Ω ± , dµ) such that f ± (x) = f (x) for µalmost every x ∈ Ω ± and, for µ --almost every y ∈ Γ -:

f -(Φ(y, s)) -f -(Φ(y, t)) = t s [T max f ](Φ(y, r))dr ∀0 < s < t < τ + (y), (3.14)
while, for µ + -almost every z ∈ Γ + :

f + (Φ(z, -s)) -f + (Φ(z, -t)) = t s [T max f ](Φ(z, -r))dr ∀0 < s < t < τ -(z).
Proof. Define, for any n 1, f n = ̺ n ⋄f , so that, from Propositions 3.11 and 3.8, lim n→∞ f nf + T max f n -T max f = 0. In particular,

lim n→∞ Ω - |f n (x) -f (x)| + | [T max f n ](x) -[T max f ](x)| dµ(x) = 0.
Then Eq. (2.9) yields

Γ - dµ -(y) τ + (y) 0 |f n (Φ(y, s)) -f (Φ(y, s))| ds + Γ - dµ -(y) τ + (y) 0 |[T max f n ](Φ(y, s)) -[T max f ](Φ(y, s))| ds -→ n→∞ 0
since T max f and T max f n both belong to L 1 (Ω, dµ). Consequently, for almost every y ∈ Γ -(up to a subsequence, still denoted by f n ) we get

f n (Φ(y, •)) -→ f (Φ(y, •)) T max f n (Φ(y, •)) -→ [T max f ](Φ(y, •)) in L 1 ((0, τ + (y)) , ds)
as n → ∞. Let us fix y ∈ Γ -for which this holds. Passing again to a subsequence, we may assume that f n (Φ(y, s)) converges (pointwise) to f (Φ(y, s)) for almost every s ∈ (0, τ + (y)).

Let us fix such a s 0 . Then,

f n (Φ(y, s 0 )) -f n (Φ(y, s)) = s s 0 [T max f n ](Φ(y, r))dr ∀s ∈ (0, τ + (y)).
Now, the right-hand-side has a limit as n → ∞ so that the first term on the left-hand side also must converge as n → ∞. Thus, for any s ∈ (0, τ + (y)), the limit

lim n→∞ f n (Φ(y, s)) = f -(Φ(y, s))
exists and, for any 0 < s < τ + (y)

f -(Φ(y, s)) = f -(Φ(y, s 0 )) - s s 0 [T max f ](Φ(y, r))dr.
It is easy to check then that f -(x) = f (x) for almost every x ∈ Ω -. The same arguments lead to the existence of f + .

The above result shows that the mild formulation of Theorem 3.6 is fulfilled for any

x ∈ Ω -∪ Ω + . It remains to deal with Ω ∞ := Ω -∞ ∩ Ω +∞ . Proposition 3.15. Let f ∈ D(T max ). Then, there exists a set O ⊂ Ω ∞ with µ(O) = 0 and a function f defined on {z = Φ(x, t), x ∈ Ω ∞ \ O, t ∈ R} such that f (x) = f (x) µ-almost every x ∈ Ω ∞ and f (Φ(x, s)) -f (Φ(x, t)) = t s [T max f ](Φ(x, r))dr, ∀ x ∈ Ω ∞ \ O, s < t.
Proof. Since (x, t) → (z, t) = (Φ(x, t), t) is a measurable and measure preserving mapping from Ω ∞ × R onto itself, Propositions 3.8 and 3.11 give

lim n→∞ Ω∞ dµ(x) I k |f n (Φ(x, t)) -f (Φ(x, t))| dt = 0 (3.15) lim n→∞ Ω∞ dµ(x) I k |T max f n (Φ(x, t)) -T max f (Φ(x, t))| dt = 0, (3.16)
for any

I k = [-k, k], k ∈ N.
This shows, in particular, that there is (a maximal) E ⊂ Ω ∞ with µ(E) = 0 such that, for almost every x ∈ Ω ∞ \ E and any bounded interval I ⊂ R:

I |f (Φ(x, t))|dt + I |[T max f ](Φ(x, t))|dt < ∞
and we can argue as in Proposition 3.13 that

f n (Φ(x, s)) -f n (x) = - s 0 T max f n (Φ(x, r))dr, ∀s ∈ R.
Proposition 3.8 yields the existence of a subsequence (f np ) p and a µ-null set

A 0 with E ⊂ A 0 ⊂ Ω ∞ such that lim p→∞ f np (x) = f (x), ∀x ∈ Ω ∞ \ A 0 . Now, for any k ∈ N, lim p→∞ Ω∞ dµ(x) I k T max f np (Φ(x, t)) -T max f (Φ(x, t)) dt = 0
so that, there is a subsequence (depending on k) and a µ-null set

A k with A 0 ⊂ A k ⊂ Ω ∞ such that lim p (k) →∞ I k T max f np (k) (Φ(x, t)) -T max f (Φ(x, t)) dt = 0, ∀x ∈ Ω ∞ \ A k . Let x ∈ Ω ∞ \ A k and |s| < k be fixed. From f np (k) (Φ(x, s)) -f np (k) (x) = - s 0 T max f np (k) (Φ(x, r))dr
we deduce that the limit lim p (k) →∞ f np (k) (Φ(x, s)) exists and is equal to

lim p (k) →∞ f np (k) (Φ(x, s)) = f (x) - s 0 T max f (Φ(x, r))dr.
We define then f by

f (Φ(x, s)) = lim p (k) →∞ f np (k) (Φ(x, s)), x ∈ Ω ∞ \ A k , |s| < k and defining O = k 1 A k , we get the result.
Before the proof of Theorem 3.6, we have to establish existence of the trace on Γ -.

Proposition 3.16. Let f satisfies condition (1) of Theorem 3.6. Then

lim t→0+ f ♯ (Φ(y, t))
exists for almost every y ∈ Γ -. Similarly, lim t→0+ f ♯ (Φ(y, -t)) exists for almost every y ∈ Γ + .

Proof. First we note that there is

Ω -⊂ Ω -with µ(Ω -\ Ω -) = 0 such that (3.2) is valid any x ∈ Ω -. Let Γ -= {y ∈ Γ -; y = Φ(x, τ -(x)), x ∈ Ω -}. It is easy to see that µ -(Γ -\ Γ -) = 0.
Indeed, otherwise, by (2.9), there would be a subset of Ω -of positive µ-measure, not intersecting Ω -, which would contradict (3.2). Consequently, any x ∈ Ω -can be written as x = Φ(y, τ -(y)), y ∈ Γ -and (3.2) can be recast as

f ♯ (Φ(y, t)) -f ♯ (Φ(y, t 0 )) = t 0 t g(Φ(y, s)ds. (3.17)
for almost any y ∈ Γ -, where 0 < t t 0 < τ + (y). Using again (2.9), s → g(Φ(y, s) is integrable on (0, τ + (y) for almost any y ∈ Γ -. Consequently, for almost every y ∈ Γ -we can pass to the limit in (3.17) with t → 0; it is easy to check that this limit does not depend on t 0 . The existence of lim t→0+ f ♯ (Φ(y, -t)) for a. e. y ∈ Γ + follows by the same argument.

The above proposition allows to define the trace operators. Definition 3.17. For any f ∈ D(T max ), define the traces B ± f by

B + f (y) := lim t→0+ f ♯ (Φ(y, -t)) and B -f (y) := lim t→0+ f ♯ (Φ(y, t))
for any y ∈ Γ ± for which the limits exist, where f ♯ is a suitable representative of f . PROOF OF THEOREM 3.6. To prove that (2) =⇒ (1), given f ∈ D(T max ), set

f ♮ (x) =      f -(x) if x ∈ Ω -, f + (x) if x ∈ Ω + ∩ Ω -∞ , f (x) if x ∈ Ω -∞ ∩ Ω +∞ ,
where f ± are given by Proposition 3.14 while f is provided by Prop. 3.15. Then, it is clear that for any x ∈ Ω and any -τ -(x) < t 1 t 2 < τ + (x)

f ♯ (Φ(x, t 1 )) -f ♯ (Φ(x, t 2 )) = t 2 t 1 [T max f ](Φ(x, s))ds and (3.2) is proven.
Let us now prove that (1) =⇒ [START_REF] Ambrosio | Transport equation and Cauchy problem for BV vector fields[END_REF]. Let us fix ψ ∈ Y, one has

Ω - f (x) d ds ψ(Φ(x, s)) s=0 dµ(x) = Γ - dµ -(y) τ + (y) 0 f (Φ(y, t)) d dt ψ(Φ(y, t))dt = Γ - dµ -(y) τ + (y) 0 f ♯ (Φ(y, t)) d dt ψ(Φ(y, t))dt.
Notice that since both Ω -f (x) d ds ψ(Φ(x, s)) s=0 dµ(x) and Ω -ψ(x)g(x)dµ(x) exist, Proposition 2.12 and Fubini's Theorem, the integrals Further, for almost every y ∈ Γ -, according to (3.2),

τ + (y) 0 f ♯ (Φ(
f ♯ (Φ(y, t)) = B -f (y) - t 0 g(Φ(y, r))dr, ∀t ∈ (0, τ + (y)).
Integration by parts, using the fact that ψ(Φ(y, 0)) = ψ(Φ(y, t n )) = 0 for any n, leads to Finally, we get

Ω - f (x) d ds ψ(Φ(x, s)) s=0 dµ(x) = Γ - dµ -(y) τ + (y) 0 ψ(Φ(y, t))g(Φ(y, t))dt = Ω - g(x)ψ(x)dµ(x). (3.19)
Using now parametrization over Γ + , we prove in the same way that

Ω + ∩Ω -∞ f (x) d ds ψ(Φ(x, s)) s=0 dµ(x) = Ω + ∩Ω -∞ g(x)ψ(x)dµ(x). (3.20)
It remains now to evaluate A := Ω +∞ ∩Ω -∞ f (x) d ds ψ(Φ(x, s)) s=0 dµ(x). According to Assumption 1

A = Ω +∞ ∩Ω -∞ f ♯ (Φ(x, t)) d dt ψ(Φ(x, t))dµ(x), ∀t ∈ R.
Let us integrate the above identity over (0, 1), so that

A = Ω -∞ ∩Ω +∞ dµ(x) 1 0 f ♯ (Φ(x, t)) d dt ψ(Φ(x, t))dt.
Let us fix x ∈ Ω -∞ ∩ Ω +∞ . For any t ∈ (0, 1), one has f ♯ (Φ(x, t)) = f ♯ (x) -t 0 g(Φ(x, s))ds and integration by parts yields

1 0 f ♯ (Φ(x, t)) d dt ψ(Φ(x, t))dt = 1 0 ψ(Φ(x, t))g(Φ(x, t))dt -ψ(x)f ♯ (x) + ψ(Φ(x, 1)) f ♯ (x) - 1 0 g(Φ(x, s))ds = 1 0 ψ(Φ(x, t))g(Φ(x, t))dt + ψ(Φ(x, 1))f ♯ (Φ(x, 1)) -ψ(x)f ♯ (x)
where we used again (3.2). Integrating over Ω -∞ ∩ Ω +∞ we see from Liouville's Theorem (Assumption 1) that

Ω -∞ ∩Ω +∞ ψ(Φ(x, 1))f ♯ (Φ(x, 1))dµ(x) = Ω -∞ ∩Ω +∞ ψ(x)f ♯ (x)dµ(x), i.e. A = Ω -∞ ∩Ω +∞ dµ(x) 1 0 ψ(Φ(x, t))g(Φ(x, t))dt
which, thanks to Liouville's Theorem, is nothing but

Ω +∞ ∩Ω -∞ f (x) d ds ψ(Φ(x, s)) s=0 dµ(x) = Ω -∞ ∩Ω +∞ g(x) ψ(x)dµ(x). ( 3
B -f (y) = f ♯ (Φ(y, t)) + t 0 [T max f ](Φ(y, s))ds, ∀t ∈ (0, τ + (y)),
where f ♯ is a suitable representative of f . An analogous formula holds for B + f . Lemma 2.9 provides the existence of Borel measures dµ ± on Γ ± , which allow us to define the natural trace spaces associated to Problem (1.1), namely, L 1 ± := L 1 (Γ ± , dµ ± ). However, the traces B ± f , f ∈ D(T max ), not necessarily belong to L 1 ± .

WELL-POSEDNESS FOR INITIAL AND BOUNDARY-VALUE PROBLEMS

4.1. Absorption semigroup. From now on, we will denote X = L 1 (Ω, dµ) endowed with its natural norm • X . Let T 0 be the free streaming operator with no re-entry boundary conditions:

T 0 ψ = T max ψ, for any ψ ∈ D(T 0 ),
where the domain D(T 0 ) is defined by

D(T 0 ) = {ψ ∈ D(T max ) ; B -ψ = 0}.
We state the following generation result:

Theorem 4.1. The operator (T 0 , D(T 0 )) is the generator of a nonnegative C 0 -semigroup of contractions (U 0 (t)) t 0 in L 1 (Ω, dµ) given by

U 0 (t)f (x) = f (Φ(x, -t))χ {t<τ -(x)} (x), (x ∈ Ω, f ∈ X),
where χ A denotes the characteristic function of a set A.

Proof. The proof is divided into three steps:

• Step 1. Let us first check that the family of operators (U 0 (t)) t 0 is a nonnegative contractive C 0 -semigroup in X. Thanks to Proposition 2.3, we can prove that, for any f ∈ X and any t 0, the mapping U 0 (t)f : Ω → R is measurable and the semigroup properties U 0 (0)f = f and U 0 (t)U 0 (s)f = U 0 (t + s)f (t, s 0) hold. Let us now show that U 0 (t)f X f X . We have

U 0 (t)f X = Ω + |U 0 (t)f |dµ + Ω -∩Ω +∞ |U 0 (t)f |dµ + Ω -∞ ∩Ω +∞ |U 0 (t)f |dµ.
Propositions 2.12 and 2.3 yield

Ω + |U 0 (t)f |dµ = Γ + dµ + (y) τ -(y) 0 |U 0 (t)f (Φ(y, -s))|ds = Γ + dµ + (y) max(0,τ -(y)-t) 0 |f (Φ(y, -s -t))|ds Γ + dµ + (y) max(t,τ -(y)) t |f (Φ(y, -r))|dr Ω + |f |dµ.
In the same way we obtain

Ω -∩Ω +∞ |U 0 (t)f |dµ = Γ -∞ dµ -(y) ∞ 0 |U 0 (t)f (Φ(y, s))|ds = Ω -∩Ω +∞ |f |dµ,
and

Ω -∞ ∩Ω +∞ |U 0 (t)f |dµ = Ω -∞ ∩Ω +∞ |f |dµ.
This proves contractivity of U 0 (t). Let us now show that U 0 (t)f is continuous, i.e.

lim t→0 U 0 (t)f -f X = 0.
It is enough to show that this property holds for any f ∈ C 0 (Ω). In this case, lim t→0 U 0 (t)f (x) = f (x) for any x ∈ Ω. Moreover, sup x∈Ω |U 0 (t)f (x)| sup x∈Ω |f (x)| and the support of U 0 (t)f is bounded, so that the Lebesgue dominated convergence theorem leads to the result. This proves that (U 0 (t)) t 0 is a C 0 -semigroup of contractions in X. Let A 0 denote its generator.

• Step 2. To show that D(A 0 ) ⊂ D(T 0 ), fix f ∈ D(A 0 ), λ > 0 and g = (λ -A 0 )f. Then,

f (x) = τ -(x) 0 exp(-λt) g(Φ(x, -t))dt, (x ∈ Ω). (4.1)
To prove that f ∈ D(T max ) with T max f = A 0 f , it suffices to prove that

Ω (λf (x) -g(x))ψ(x)dµ(x) = Ω f (x) d ds ψ(Φ(x, s)) s=0 dµ(x), ∀ψ ∈ Y.
Let us fix ψ ∈ Y, set ϕ(x) := d ds ψ(Φ(x, s)) s=0 and write

Ω f (x)ϕ(x)dµ(x) = Ω + f (x)ϕ(x)dµ(x) + Ω +∞ ∩Ω - f (x)ϕ(x)dµ(x) + Ω +∞ ∩Ω -∞ f (x)ϕ(x)dµ(x) = I 1 + I 2 + I 3 .
We first deal with I 1 . For any y ∈ Γ + and t ∈ (0, τ -(y)) we have ϕ(Φ(y, -t)) = -d dt ψ(Φ(y, -t)) and f (Φ(y, -t)) = τ -(y) t exp(-λ(s -t))g(Φ(y, -s))ds. Then, by Proposition 2.12,

I 1 = - Γ + dµ + (y) τ -(y) 0 d dt ψ(Φ(y, -t))dt τ -(y) t exp(-λ(s -t))g(Φ(y, -s))ds = - Γ + dµ + (y) τ -(y) 0 g(Φ(y, -s))ds s 0 exp(-λ(s -t)) d dt (ψ(Φ(y, -t))) dt = Γ + dµ + (y) τ -(y) 0 g(Φ(y, -s))× × λ s 0 exp(-λ(s -t))ψ(Φ(y, -t))dt -ψ(Φ(y, -s)) ds
where we used the fact that ψ(Φ(y, 0)) = 0 for any y ∈ Γ + since ψ is compactly supported. Thus

I 1 = λ Γ + dµ + (y) τ -(y) 0 ψ(Φ(y, -t))dt τ -(y) t exp(-λ(s -t))g(Φ(y, -s))ds - Γ - dµ + (y) τ -(y) 0 g(Φ(y, -s))ψ(Φ(y, -s))ds = Γ + dµ + (y) τ -(y) 0 ψ(Φ(y, -t)) λf (Φ(y, -t)) -g(Φ(y, -t)) dt.
Using again Proposition 2.12, we obtain

I 1 = Ω + (λf (x) -g(x)) ψ(x)dµ(x). (4.2)
Arguing in a similar way, we prove that

I 2 = - Ω -∩Ω +∞ (λf (x) -g(x)) ψ(x)dµ(x). (4.3)
Finally, since

f (x) = ∞ 0 exp(-λt)g (Φ(x, -t)) dt for any x ∈ Ω -∞ ∩ Ω +∞ ,
one has

I 3 = Ω -∞ ∩Ω +∞ ϕ(x)dµ(x) ∞ 0 exp(-λt)g(Φ(x, -t))dt = ∞ 0 exp(-λt)dt Ω -∞ ∩Ω +∞ ϕ(x)g(Φ(x, -t))dµ(x). Now, Assumption 1 asserts that Ω -∞ ∩Ω +∞ ϕ(x)g(Φ(x, -t))dµ(x) = Ω -∞ ∩Ω +∞ g(x)ϕ(Φ(x, t))dµ(x), ∀t 0,
and, since ϕ(Φ(x, t)) = d dt ψ(Φ(x, t)), finally

I 3 = Ω -∞ ∩Ω +∞ g(x)dµ(x) ∞ 0 exp(-λt) d dt (ψ(Φ(x, t))) dt = - Ω -∞ ∩Ω +∞ g(x)ψ(x)dµ(x) + λ Ω -∞ ∩Ω +∞ g(x)dµ(x) ∞ 0 exp(-λt)ψ(Φ(x, t))dt.
Using again Assumption 1, this finally gives

I 3 = - Ω -∞ ∩Ω +∞ (g(x) -λf (x)) ψ(x)dµ(x). (4.4) Combining (4.2)-(4.4) leads to Ω f (x) d ds ψ(Φ(x, s)) s=0 dµ(x) = - Ω (g(x) -λf (x)) ψ(x)dµ(x)
which proves that f ∈ D(T max ) and (λ -T max )f = g. Next, for y ∈ Γ -and 0 < t < τ + (y) we write t = τ -(Φ(y, t)) and, by Proposition 2.3 and (4.1), we obtain

f (Φ(y, t)) = t 0 exp(-λ(t -s)) g(Φ(y, s))ds. (4.5)
Consequently, lim t→0 + f (Φ(y, t)) = 0 a.e. y ∈ Γ -, i.e. B -f = 0 so that f ∈ D(T 0 ) and

A 0 f = T 0 f = λf -g. • Step 3. Now let us show the converse inclusion D(T 0 ) ⊂ D(A 0 ). Let f ∈ D(T 0 ).
Changing possibly f on a set of zero measure, we may write f = f ♯ , where f ♯ is the representative of f given by Theorem 3.6. Then, for any x ∈ Ω and any 0 t < τ -(x)

f (Φ(x, -t)) -f (x) = t 0 [T max f ](Φ(x, -r))dr
which, according to the explicit expression of U 0 (t), means that

U 0 (t)f (x) -f (x) = t 0 U 0 (r)T max f (x)dr (4.6)
for any x ∈ Ω and t < τ -(x). Letting t converge towards τ -(x) we obtain

f (x) = - τ -(x) 0 [T max f ](Φ(x, -r))dr.
In particular, Eq. (4.6) holds true for any x ∈ Ω and any t τ -(x). Arguing exactly as in [16, p. 38], the pointwise identity (4.6) represents the X-integral, i.e, U 0 (t -and g ∈ X be given. Then the function

)f -f = t 0 U 0 (r)T max f dr in L 1 (Ω, dµ). Consequently, f ∈ D(A 0 ) with A 0 f = T max f.
f (x) = τ -(x) 0 exp(-λt) g(Φ(x, -t))dt + χ {τ -(x)<∞} exp(-λτ -(x))u(Φ(x, -τ -(x)))
is a unique solution f ∈ D(T max ) of the boundary value problem:

(λ -T max )f = g, B -f = u, (4.7)
where λ > 0. Moreover, B + f ∈ L 1 + and

B + f L 1 + + λ f X u L 1 -+ g X , (4.8)
with equality sign if g 0 and u 0.

Proof. Let us write f = f 1 + f 2 with f 1 (x) = τ -(x) 0
exp(-λt) g(Φ(x, -t))dt, and

f 2 (x) = χ {τ -(x)<∞} exp(-λτ -(x))u Φ (x, -τ -(x)) , x ∈ Ω.
According to Theorem 4.1,

f 1 = (λ -T 0 ) -1 g, i.e. f 1 ∈ D(T max ) with (λ -T max )f 1 = g and B -f 1 = 0. Therefore, to prove that f is a solution of (4.7) it suffices to check that f 2 ∈ D(T max ), (λ -T max )f 2 = 0 and B -f 2 = u. It is easy to see that f 2 ∈ L 1 (Ω,
Ω f 2 (x) d ds ψ(Φ(x, s)) s=0 dµ(x) = Γ - dµ -(y) τ + (y) 0 f 2 (Φ(y, t)) d dt ψ(Φ(y, t))dt = Γ - u(y)dµ -(y) τ + (y) 0 exp(-λt) d dt ψ(Φ(y, t))dt.
For almost every y ∈ Γ -, we compute the integral over (0, τ + (y)) by parts, which yields

f 2 ∈ D(T max ) with T max f 2 = λf 2 . Also, f 2 (Φ(y, t)) = exp(-λt)u(y), y ∈ Γ -, 0 < t < τ + (y) (4.9)
from which we see that B -f 2 = u. Consequently, f is a solution to (4.7). To prove that the solution is unique, it is sufficient to prove that the only solution h ∈ D(T max ) to (λ -T max )h = 0, B -h = 0, is h = 0. This follows from the fact that such a solution h actually belongs to D(T 0 ) if λ ∈ ̺(T 0 ). Finally, it remains to prove (4.8). For simplicity, we denote the representative of f i , i = 1, 2, defined in Proposition 3.16, with the same letter. Using (4.9) and the fact that f 2 vanishes on Ω -∞ , from (2.9) we get Combining this with (4.10) leads to

λ f 2 X + B + f 2 L 1 + = u L 1 -. (4.11)
Now, let us show that B + f 1 ∈ L 1 + and B + f 1 L 1 + + λ f 1 X g X . For any y ∈ Γ + and 0 < t < τ -(y), we see, as above, that f 1 (Φ(y, -t)) =

λ f 1 = λ Ω f 1 dµ = λ Ω + f 1 dµ + λ Ω -∩Ω +∞ f 1 dµ + λ Ω -∞ ∩Ω +∞ f 1 dµ.
Using similar arguments to those used in the study of f 2 , we have

λ Ω + f 1 dµ = Γ + dµ + (y)
τ -(y) 0 g(Φ(y, -t)) (1exp(-λt)) dt, which, by Proposition 2.12, implies λ Ω + f 1 dµ = Ω + g dµ -Γ + B + f 1 dµ + . Similar argument shows that λ Ω -∩Ω +∞ f 1 dµ = Ω -∩Ω +∞ g dµ, while the equality λ

Ω -∞ ∩Ω +∞ f 1 dµ = Ω -∞ ∩Ω +∞ g dµ,
is a direct consequence of the invariance of µ with respect to Φ(•, t). This shows that λ f X = g X -B + f L 1 + for g 0. In general, defining Remark 4.3. Notice that, in order to get the existence and uniqueness of the solution f to (4.7), it is not necessary for u to belong to L 1 (Γ -, dµ -). Indeed, we only have to make sure that f 2 ∈ L 1 (Ω, dµ), i.e., from (4.10), Γ -|u(y)| 1 -e -λτ + (y) dµ -(y) < ∞. Of course, to get (4.8), the assumption u ∈ L 1 (Γ -, dµ -) is necessary.

Let us note that, with the notation of Theorem 4.2, we have

Γ + B + f dµ + + λ Ω f dµ = Γ - u dµ -+ Ω g dµ.
(4.12)

Indeed, for nonnegative u and g, (4.8) turns out to be precisely (4.12). Then, for arbitrary u ∈ L 1 and g ∈ X, we get (4.12) by splitting functions into positive and negative parts. This leads to the following generalization of Green's formula: Proof. For given f ∈ D(T max ), we obtain the result by setting u = B -f ∈ L 1 -and g = (λ -T max )f ∈ X in Eq. (4.12). Remark 4.5. If dµ is the Lebesgue measure on R N , the above formula leads to a better understanding of the measures dµ ± . Indeed, comparing it to the classical Green's formula (see e.g.

APPENDIX: ABOUT THE CLASS OF TEST-FUNCTIONS

We answer in this Appendix a natural question concerning the definition of the class of testfunctions Y. Precisely, we prove that two test-functions equal µ-almost everywhere are such that their derivatives (in the sense of (3.1)) also coincide µ-almost everywhere. To prove our claim, it clearly suffices to prove that, given ψ ∈ Y such that ψ(x) = 0 for µ-a. e. x ∈ Ω, one has ϕ(x) = 0 for µ-a. e. x ∈ Ω where ϕ(x) = d ds ψ(Φ(x, s)) s=0 . Let E := x ∈ Ω ; ψ(x) = 0 and ϕ(x) = 0 .

It is clear that E is measurable and that one has to prove that µ(E) = 0. It is no loss of generality to assume that E is bounded. We observe that for any x ∈ E, there exists δ x > 0 such that ψ(Φ(x, t)) = 0, ∀ 0 < |t| < δ x .

(A.1)

Let us split E as follows

E = E ∩ Ω -∪ E ∩ Ω + ∩ Ω -∞ ∪ E ∩ Ω +∞ ∩ Ω -∞ := E -∪ E + ∪ E ∞
and prove that µ(E -) = µ(E + ) = µ(E ∞ ) = 0.

(1) First consider E -. Since ψ(x) = 0 for µ-a. e. x ∈ Ω -and using the fact that any x ∈ Ω - can be written as x = Φ(y, t) for some y ∈ Γ -and 0 < t < τ + (y), we observe that, for µ -a. e. y ∈ Γ -, ψ(Φ(y, t)) = 0 for almost every (in the sense of the Lebesgue measure in R) 0 < t < τ + (y). For such a y ∈ Γ -, continuous differentiability of t → ψ(Φ(y, t)) implies ψ(Φ(y, t)) = 0 for any 0 < t < τ + (y). This means, according to (A.1) that, for µ --a. e. y ∈ Γ -, Φ(y, t) / ∈ E for any 0 < t < τ + (y). Since

µ(E ∩ Ω -) = Γ - dµ -(y) τ + (y) 0 χ E (Φ(y, t))dt
we see that µ(E -) = 0. (2) In the same way, using Γ + instead of Γ -, we show that µ(E ∩ Ω + ∩ Ω -∞ ) = 0.

(3) It remains to prove that µ(E ∞ ) = 0. In accordance with (A.1), we define for, any n ∈ N,

E n := x ∈ E ∞ ; δ x 1/n = x ∈ E ∞ ; ψ(Φ(x, t)) = 0, ∀ 0 < |t| < 1/n .
According to Assumption 1, it is easy to see that µ(E n ) = 0 for any n ∈ N since ψ(x) = 0 for µ-a.e. x ∈ Ω. Moreover, E 1 ⊂ E 2 ⊂ . . . ⊂ E n ⊂ E n+1 ⊂ . . ., and

∞ n=1 E ∞ \ E n = ∅.

. 7 )

 7 Combining (3.5),(3.6) and (3.7), one finally gets ̺ n ⋄ f f .

  is finite, one can see easily that lim n µ(O - n ) = 0. Since sup x∈Ω |̺ n ⋄f (x)| sup x∈Ω |f (x)|, for any ε > 0, there exists n 0 1 such that O - n |f (x)|dµ(x) ε, and

tn 0 f

 0 ♯ (Φ(y, t)) d dt ψ(Φ(y, t))dt = tn 0 g(Φ(y, t))ψ(Φ(y, t))dt. Consequently, for µ -almost every y ∈ Γ -: τ + (y) 0 f ♯ (Φ(y, t)) d dt ψ(Φ(y, t))dt = τ + (y) 0 ψ(Φ(y, s))g(Φ(y, t))dt. (3.18)

  .21) Combining(3.19),(3.20) and (3.21), we obtainΩ f (x) d ds ψ(Φ(x, s)) s=0 dµ(x) = Ω g(x)ψ(x)dµ(x), ∀ψ ∈ Ywhich exactly means that f ∈ D(T max ) with g = T max and the proof is complete.Corollary 3.18. Traces B ± f on Γ ± can be defined for any f ∈ D(T max ). For µ --almost any y ∈ Γ -we have

4. 2 .Theorem 4 . 2 .

 242 Green's formula. The above result allows us to treat more general boundary-value problem: Let u ∈ L 1

λ Ω |f 2

 2 |dµ = λ Ω - |f 2 |dµ = λ Γ - dµ -(y) τ + (y) 0 e -λt |u(y)|dt = Γ - |u(y)| 1 -e -λτ + (y) dµ -(y).

( 4 . 10 )

 410 Define h : y ∈ Γ --→ h(y) = |u(y)|e -λτ + (y) . It is clear that h vanishes on Γ -∞ and h(y) |u(y)| for a.e. y ∈ Γ -. In particular, h ∈ L 1 -and, according to (2.11),Γ - h(y)dµ -(y) = Γ -\Γ -∞ h(y)dµ -(y) = Γ + \Γ +∞ h(Φ(z, -τ -(z)))dµ + (z) = Γ + \Γ +∞ e -λτ -(z) |u(Φ(z, -τ -(z)))|dµ + (z) = Γ + |B + f 2 (z)|dµ + (z) = B + f 2 L 1 + .

τ

  -(y) t exp(-λ(s -t))g(Φ(y, -s))ds.This shows thatB + f 1 (y) = lim t→0 + f 1 (Φ(y, -t)) = τ -(y) 0 exp(-λs))g(Φ(y, -s))ds.According to Proposition 2.12, exp(-λ(s-t))|g(Φ(y, -s))| |g(Φ(y, -s))|, implies B + f 1 ∈ L 1 + .Let us now assume g 0. Then f 1 0 and hence

F 1 (

 1 λs) |g(Φ(x, -s)| ds, x ∈ Ω,we obtain B + f 1 L 1 + + λ f 1 X B + F 1 L 1 + + λ F 1 X = g X which,combined with (4.11), gives (4.8).

Proposition 4 . 4 (

 44 Green's formula). Let f ∈ D(T max ) satisfies B -f ∈ L 1 -. Then B + f ∈ L 1 + and Ω T max f dµ = Γ - B -f dµ --Γ + B + f dµ +

[ 8 ]

 8 ), we see that the restriction of dµ ± on the setΣ ± = {y ∈ ∂Ω ; ±F (y) • n(y) > 0} equals |F (y) • n(y)| dγ(y),where dγ(•) is the surface Lebesgue measure on ∂Ω.

  Notice that the class of test-functions Y is not defined as a subset of L ∞ (Ω, dµ); that is, we do not identify functions equal µ-almost everywhere. It is however a natural question to know whether two test-functions coinciding µ-almost everywhere are such that there derivatives (defined by (3.1)) do coincide µ-almost everywhere. We provide a positive answer to this question at the end of the paper (see Appendix). Let ψ ∈ Y be given. For µ ∓ -almost any y ∈ Γ ∓ there exists a sequence (t ± n ) n (depending on y) such that

	Remark 3.2. An important property of test-functions is the following consequence of Proposition 2.14:
	Lemma 3.3.
	s))	measurable and bounded.	(3.1)
	s=0

  --almost every y ∈ Γ -. Let us prove that these two integrals coincide for almost-every y ∈ Γ -. According to Lemma 3.3, for almost every y ∈ Γ -, there is a sequence (t n ) n (depending on y) such that ψ(Φ(y, t n )) = 0 and t n → τ + (y). Thus,

	y, t)) f ♯ (Φ(y, t)) dt d d dt τ + (y) are well-defined for µ τ + (y) 0 and 0 g(Φ(y, t))ψ(Φ(y, t))dt = lim ψ(Φ(y, t))dt and ψ(Φ(y, t))dt = lim n→∞ n→∞	τ + (y) tn 0 f ♯ (Φ(y, t)) g(Φ(y, t))ψ(Φ(y, t))dt d dt ψ(Φ(y, t))dt tn 0 0 ψ(Φ(y, t))g(Φ(y, t))dt.

  dµ) (see also(4.10)). To prove that f 2 ∈ D(T max ) one argues as in the proof of Theorem 4.1. Precisely, let ψ ∈ Y, noticing that f 2 vanishes outside Ω -, one has thanks to (4.9)

Since we assumed