An error estimate for the parabolic approximation of multidimensional scalar conservation laws with boundary conditions

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An error estimate for the parabolic approximation of multidimensional scalar conservation laws with

boundary conditions

Jerome Droniou, Cyril Imbert, Julien Vovelle

To cite this version:

Jerome Droniou, Cyril Imbert, Julien Vovelle. An error estimate for the parabolic approximation of multidimensional scalar conservation laws with boundary conditions. Annales de l’Institut Henri Poincaré (C) Non Linear Analysis, Elsevier, 2004, 21 (5), pp.689-714. �10.1016/j.anihpc.2003.11.001�.

�hal-00018746�

An error estimate for the parabolic approximation of multidimensional scalar conservation laws with boundary

conditions

J. Droniou ^∗ C. Imbert ^† J. Vovelle ^‡ 13/10/2003

Abstract

We study the parabolic approximation of a multidimensional scalar conservation law with initial and boundary conditions. We prove that the rate of convergence of the viscous approximation to the weak entropy solution is of order η

^1/3

, where η is the size of the artificial viscosity. We use a kinetic formulation and kinetic techniques for initial-boundary value problems developed by the last two authors in a previous work.

R´ esum´ e

Nous ´ etudions l’approximation parabolique d’une loi de conservation scalaire multi-dimensionnelle avec conditions initiales et aux limites. Nous prouvons que la vitesse de convergence de l’approxima- tion visqueuse vers la solution entropique est de l’ordre de η

^1/3

, o` u η est la taille de la viscosit´ e arti- ficielle. Nous utilisons une formulation et des techniques cin´ etiques d´ evelopp´ ees pour des probl` emes au bord par les deux derniers auteurs dans un travail pr´ ec´ edent.

keywords: conservation law, initial-boundary value problem, error estimates, parabolic approximation, kinetic techniques.

AMS classification : 35L65, 35D99, 35F25, 35F30, 35A35

1 Introduction

Let Ω be a bounded open subset of R ^d with Lipschitz continuous boundary. Let n(x) be the outward unit normal to Ω at a point x ∈ ∂Ω, Q = (0, + ∞ ) × Ω and Σ = (0, + ∞ ) × ∂Ω. We consider the following multidimensional scalar conservation law

∂ t u + divA(u) = 0 in Q, (1a)

with the initial condition

u(0, x) = u 0 (x), ∀ x ∈ Ω, (1b)

and the boundary condition

u(s, y) = u b (s, y), ∀ (s, y) ∈ Σ. (1c)

∗

D´ epartement de Math´ ematiques, CC 051, Universit´ e Montpellier II, Place Eug` ene Bataillon, 34095 Montpellier cedex 5, France. email: droniou@math.univ-montp2.fr

†

D´ epartement de Math´ ematiques, CC 051, Universit´ e Montpellier II, Place Eug` ene Bataillon, 34095 Montpellier cedex 5, France. email: imbert@math.univ-montp2.fr

‡

CMI, Universit´ e de Provence, 39, rue Joliot-Curie, 13 453 Marseille cedex 13, France. email: vovelle@cmi.univ-mrs.fr

It is known that entropy solutions must be considered if one wants to solve scalar conservation laws (Equation (1a) is replaced by a family of inequalities — see [8] for the Cauchy problem) and that the Dirichlet boundary conditions are to be understood in a generalized sense (see [1] for regular initial and boundary conditions and [11] for merely bounded data).

In this paper, we estimate the difference between the weak entropy solution of (1) and the smooth solution of the regularized parabolic equation

∂ t v + divA(v) = η∆v in Q, (2)

satisfying the same initial and boundary conditions. Throughout the paper, we make the following hypotheses on the data: the flux function A belongs to C

²

(R), the initial condition u 0 is in C

²

(Ω), the boundary ∂Ω of the domain Ω is C

²

, the boundary condition u b belongs to C

²

(Σ). In that case, there exists a unique solution v ^η (regular outside { 0 } × ∂Ω) to the problem (2)-(1b)-(1c).

The aim of this paper is to prove the following error estimate.

Theorem 1 Suppose that Ω is C

²

, A ∈ C

²

(R), u 0 ∈ C

²

(Ω) and u b ∈ C

²

(Σ). Let u be the weak entropy solution of (1) and let v ^η be the solution of the approximate parabolic problem (2)-(1b)-(1c). Let T 0 > 0;

there exists a constant C only depending on (Ω, u b , u 0 , A, T 0 ) such that, for all t ∈ [0, T 0 ],

|| u(t) − v ^η (t) ||

^L1

(Ω) 6 Cη ^1/3 . (3)

We now recall what is known about error estimates for approximations of conservation laws.

In the case where the function u is smooth (a feature which, we recall, requires the data to be smooth, compatible and the time T 0 to be small enough), error estimates of order

η ^1/2 if the boundary is characteristic

η if the boundary is not characteristic (4)

in L ^∞ (0, T ; L

¹

(Ω)) have been given (see Gues [5], Gisclon and Serre [3], Grenier and Gues [4], Joseph and LeFloch [7], Chainais-Hillairet and Grenier [2] and references therein). The technique of boundary layer analysis developed in those articles is devoted to the investigation of the initial-boundary value problem for systems of conservation laws (and not only for a single equation). Roughly speaking, the viscous approximation v ^η is decomposed as v ^η = u + c ^η + (remainder) where c ^η characterizes the boundary layer which appear in the vicinity of ∂Ω. Estimates on v ^η − u are then consequences of estimates on c ^η + (remainder) (see Appendix 8.1).

To our knowledge, there does not exist other techniques of analysis which would confirm the error estimate (4). On the contrary, many techniques have been set and improved to analyse the error of approximation for the Cauchy Problem (Ω = R ^d ) for conservation laws (and results of sharpness of error estimates have also been delivered). The first error estimate for the Cauchy problem is given by Kuznetˇcov in 1976 [9]: an adaptation of the proof of the result of comparison between two weak entropy solutions given by Kruˇzkov [8] yields an error estimate of order 1/2 in the L

¹

-norm. The reader interested in more precise, more general and more recent results is invited to consult the compilation made by Tang [14], the introduction of [13], and references therein.

We establish here Estimate (3) for arbitrary times T 0 ; in particular, the possible occurence of shocks is

taken into account: u is the weak entropy solution to Problem (1) and has no more regularity, in general,

than the ones stated in Proposition 1. As a consequence, u may be irregular in the vicinity of ∂Ω and

this constitutes an obstacle to the analysis of the rate of convergence of v ^η . To circumvent this obstacle,

we use the kinetic formulation of [6] (an adaptation to boundary problems of the kinetic formulation

introduced in [10]) and adapt the technique of error estimate developed by Perthame for the analysis of

the Cauchy Problem [12]. We then obtain a rate of convergence of 1/3. The accuracy or non-sharpeness

of this order (compare to (4)) remains an open problem for us.

The paper is dedicated to the proof of Theorem 1. It is organized as follows. We begin with some prelim- inaries, mainly to state (or recall) the kinetic formulations of both hyperbolic and parabolic equations. In order to enlight the key ideas of this rather technical proof, we present its skeleton in Subsection 2.4. In Section 3, we obtain a first estimate in the interior of the domain; then, in Sections 4 and 5, we transport the equations so that Ω becomes a half space and we regularize them in order to use the solution of one of them as a test function in the other. Eventually, in Section 6, we conclude the proof of Theorem 1 by getting an estimate of the boundary term which appears at the end of Section 5.

2 Preliminaries

In order to clarify computations, we drop the superscript η in v ^η and simply write v for the approximate solution. We prove Theorem 1 in several steps.

2.1 Known estimates on u and v

We gather in the following proposition the estimates we will need to prove Theorem 1. We refer to [1]

for a proof of these results.

Proposition 1 Assume that Ω is C

²

, A ∈ C

²

(R), u 0 ∈ C

²

(Ω) and u b ∈ C

²

(Σ). There exists C > 0 only depending on (Ω, u b , u 0 , A, T 0 ) such that

1. the functions u, v : [0, T 0 ] → L

¹

(Ω) are C-Lipschitz continuous 2. for all t ∈ (0, T 0 ), R

Ω | ∂ t v(t, · ) | 6 C

3. for all t ∈ [0, T 0 ], | u(t, · ) |

^BV

(Ω) 6 C and | v(t, · ) |

^BV

(Ω) 6 C.

2.2 Notations

Let us introduce some local charts of Ω. Since Ω is C

²

and bounded, we can find a finite cover { O i } i∈{0,...,n}

of Ω by open sets of R ^d such that O 0 ⊂ Ω and that, for all i ∈ { 1, . . . , n } , there exists a C

²

-diffeomorphism h i : O i → B ^d (the unit ball in R ^d ) satisfying

• ∂Ω ⊂ ∪ ⁿ i=1 O i ;

• h i (O i ∩ ∂Ω) = B ^d−1 := B ^d ∩ (R ^d−1 × { 0 } );

• h i (O i ∩ Ω) = B ^d ₊ := B ^d ∩ (R ^d−1 × (0, + ∞ )).

Let (λ i ) i∈{0,...,n} be a partition of the unity on Ω, subordinate to the cover { O i } i∈{0,...,n} .

In the following, when a quantity appears with a bar above, it denotes something related to the boundary of Ω (possibly transported on B ^d−1 by a chart): either a variable on ∂Ω or the value of a function on this boundary. The values of a function φ at t = 0 are denoted by φ ^(t=0) .

Here are other general notations, related to the regularization of the equations. Let θ ∈ C ^∞

_c

(]1/2, 1[; R ⁺ ) be such that R

R

θ = 1 and define, for τ > 0, θ τ ( · ) = _τ ¹ θ( _τ ^· ) (right-decentred regularizing kernel). When necessary, we define regularizing kernels ρ µ in space (either the whole space or on the (transported) boundary of Ω) or space-time variables; when such a kernel on R ^N (N = d − 1, N = d or N = d + 1) is given and f is a function defined and locally integrable on a set S ⊂ R ^N , we denote, for z ∈ R ^N ,

f ^µ (z) = Z

S

f (r)ρ µ (z − r) dr ,

i.e. f ^µ is the convolution of ρ µ by the extension of f by 0 outside S. We have then, for all φ ∈ L

¹

(R ^N ) with compact support,

Z

S

f (φ ⋆ ρ ˇ µ ) = Z

R^N

f ^µ φ (where ˇ ρ µ (z) = ρ µ ( − z)).

2.3 Kinetic formulations of (1) and (2)

The function sgn ₊ is defined by sgn ₊ (s) = 0 if s 6 0 and sgn ₊ (s) = 1 if s > 0; similarly, sgn ₋ (s) = − 1 if s < 0 and sgn − (s) = 0 if s > 0. Let D = sup( || u b || ^∞ , || u 0 || ^∞ ).

Let us recall the kinetic formulation of (1) obtained in [6]: there exists a bounded nonnegative measure m ∈ M ⁺ (Q × R ξ ), which has a compact support with respect to ξ, and two nonnegative measurable functions m ^b ₊ , m ^b ₋ ∈ L ^∞

loc (Σ × R ξ ) such that the function m ^b ₊ vanishes for ξ ≫ 1 (resp. the function m ^b ₋ vanishes for ξ ≪ − 1) and such that the functions f ± (t, x, ξ) = sgn _± (u(t, x) − ξ) associated with u satisfy, for any φ ∈ C ^∞

_c

(R ^d+2 )

Z

Q×

Rξ

f ± (∂ t + a · ∇ )φ + Z

Ω×

Rξ

f _± ⁰ φ ^(t=0) + Z

Σ×

Rξ

( − a · n)f _± ^τ φ = Z

Q×

Rξ

∂ ξ φdm (5)

where a = A ^′ , f _± ⁰ (x, ξ) = sgn _± (u 0 (x) − ξ) and f _± ^τ (t, x, ξ) = sgn _± (u(t, x) − ξ) satisfies

( − a · n)f _± ^τ = M f _± ^b + ∂ ξ m ^b _± (6)

with f _± ^b (t, x, ξ) = sgn _± (u b (t, x) − ξ) and M the Lipschitz constant of the flux function A on [ − D, D].

This formula is the kinetic formulation of the BLN condition (see [1]).

We next give a kinetic formulation for the approximate solution. Consider two test functions ϕ ∈ C ^∞

_c

(R t ×R ^d ), ψ ∈ C ^∞

_c

(R ξ ) and define E(α) = R

ψ(ξ)sgn _± (α − ξ)dξ and H(α) = R

a(ξ)ψ(ξ)sgn _± (α − ξ)dξ.

Note that E ^′ = ψ and H ^′ = E ^′ a. Now multiply the equation ∂ t v + divA(v) = η △ v by ϕ(t, x)ψ(v(t, x)) = ϕ(t, x)E ^′ (v(t, x)), integrate over Q and integrate by part (using the fact that v is C

²

outside { 0 } × ∂Ω)

Z

Q

E(v)∂ t ϕ + H (v) · ∇ ϕ + Z

Ω

E(u 0 )ϕ ^(t=0) − Z

Σ

H (u b ) · nϕ

= Z

Q

ηE ^′ (v) ∇ v · ∇ ϕ − Z

Σ

ηE ^′ (u b ) ∇ v · nϕ + Z

Q

ηE ^′′ (v) |∇ v | ² ϕ.

Using the definition of E and H, we obtain, denoting g ± (t, x, ξ) = sgn _± (v(t, x) − ξ), Z

Q×

Rξ

g ± (∂ t + a · ∇ )φ − Z

Q×

Rξ

ηδ v ∇ v · ∇ φ + Z

Ω×

Rξ

f _± ⁰ φ ^(t=0) + Z

Σ×

Rξ

G ± φ = Z

Q×

Rξ

∂ ξ φdq (7) where φ(t, x, ξ) = ϕ(t, x)ψ(ξ) and

G ± = ( − a · n)f _± ^b + ηδ u

b

∇ v · n q = ηδ v |∇ v | ² > 0

(notice that the support of q is compact with respect to ξ). Using a classical argument relying on convolution techniques, we claim that (7) holds true for any test function φ ∈ C ^∞

_c

(R ^d+2 ).

Remark 1 i) Since f + , f ₊ ⁰ , f ₊ ^τ and m vanish for ξ ≫ 1, Equation (5) with f + holds true when the

support of the test function φ is merely lower bounded (and not necesseraly compact) with respect to

ξ. Similarly, we can apply (7) with g − to test functions the support of which is only upper bounded

with respect to ξ. Notice also that, in all the following, though we write integrals in ξ on the whole

of R ξ , the integrands we consider are null outside a fixed compact (namely [ − D, D]) of R ξ ; we use

this in some estimates, without recalling it.

ii) Equations (5) and (7) can be applied to certain test functions which are not fully regular but have some monotony properties with respect to ξ, provided we replace the equality by an inequality (the sign of which is given by the monotony of the test function). More precisely, we consider, in the fol- lowing, test functions of the kind φ(t, x, ξ) = R ∞

0

R

Ω ϕ(t, x, s, y)sgn _± (W (s, y) − ξ) dy ds, where W is bounded and ϕ is regular and has a fixed sign; we can approximate sgn _± by some non-decreasing and regular functions sgn _±,δ ; then, applying (5) or (7) to φ δ (t, x, ξ) = R ∞

0

R

Ω ϕ(t, x, s, y)sgn _±,δ (W (s, y) − ξ) dy ds, which is regular and has the same monotony properties as φ (with respect to ξ), we no- tice that the right-hand side has a fixed sign; then, passing to the limit δ → 0, we see that these inequalities are satisfied with φ.

2.4 Main ideas of the proof

We present here formal manipulations which enable to understand the key steps of the proof. Let (t, x) 7→ ϕ(t, x) be a non-negative regular function. Plugging φ = ϕg − in (5) and φ = ϕf + in (7), we

obtain Z

Q×

Rξ

f + (∂ t + a · ∇ )(ϕg − ) + Z

Σ×

Rξ

( − a · n)f ₊ ^τ f ₋ ^b ϕ 6 0 and

Z

Q×

Rξ

g − (∂ t + a · ∇ )(ϕf + ) + Z

Σ×

Rξ

( − a · n)f ₋ ^b f ₊ ^τ ϕ − η Z

Q×

Rξ

δ v ∇ v · ∇ (f + ϕ) + η Z

Σ×

Rξ

δ u

_b

∇ v · nf + ϕ 6 0 (since f ₊ ⁰ f ₋ ⁰ = 0). Summing these inequalities and integrating by parts, it comes

Z

Q×R

ξ

f + g − (∂ t + a · ∇ )ϕ 6 − Z

Σ×R

ξ

( − a · n)f ₊ ^τ f ₋ ^b ϕ − η Z

Σ×R

ξ

δ u

_b

∇ v · nf + ϕ + η Z

Q×R

ξ

δ v ∇ v · ∇ (f + ϕ).

Taking ϕ(t, x) = ω ζ (t) with (ω ζ ) ζ>0 which converges to the characteristic function of [0, T ] and ω _ζ ^′ → − δ T , this gives

Z

Ω

(u − v) ⁺ (T, x) dx = Z

Ω×

Rξ

( − f + g − ) ^{(t=T )} 6 −

Z

[0,T]×∂Ω×

Rξ

( − a · n)f ₊ ^τ f ₋ ^b − η Z

[0,T]×∂Ω×

Rξ

δ u

b

∇ v · nf + + η Z

[0,T]×Ω×

Rξ

δ v ∇ v · ∇ f + . (8) The functions f + and g − are not regular enough to justify such manipulations, which are therefore performed with f ₊ ^ε and g ₋ ^ν , regularized versions of these applications. The smoothing of g − is purely technical and we immediately let ν → 0; at the contrary, the way we define f ₊ ^ε is crucial for the proof.

A decentralizing regularization allows to get rid of the second term of the right-hand side of (8); the size of the regularization being ε, ||∇ f ₊ ^ε || ^∞ is bounded by C/ε and the last term of (8) is of order η/ε.

There remains to estimate the first term of the right-hand side of (8), which is the aim of a whole section (Section 6); the idea is to re-use the kinetic equation satisfied by v.

3 Estimate in the interior of the domain

In this section, we let λ = λ 0 (we drop the subscript 0) and K := supp(λ 0 ). In order to obtain an estimate on the interior of the domain, we need to localize using λ, regularize both kinetic equations in order to combine them, proceeding as we did when proving the Comparison Theorem in [6]. This step is more or less classical.

Let α > 0 and 0 < ε < dist(K, ∂Ω); denote γ ε (x) = Q d

i=1 θ ε (x i ). Taking φ ∈ C ^∞

_c

(R ^d+2 ) with support in R × K × R ξ and using φ ⋆ ( ˇ γ ε ⊗ θ ˇ α ) ( ¹ ) — notice that this function is null on the boundary of Ω — as a

1

Here and after, the tensorial product is used to recall that ˇ γ

_ε

and ˇ θ

_α

use different variables (for example, ˇ γ

_ε

⊗ θ ˇ

_α

(t, x) = ˇ

γ

ε

(x) ˇ θ

α

(t)) and the convolution product ⋆ never involves the kinetic variable ξ.

test function in (5) with f + , we find Z

R^d+2

f ₊ ^α,ε (∂ t + a · ∇ )φ + Z

R^d+2

f ₊ ^{0 ε} ⊗ θ α φ = Z

R^d+2

∂ ξ φ dm ^α,ε (9)

(where m ^α,ε is the convolution in (t, x) of γ ε ⊗ θ α by the extention of m by 0 outside Q × R ξ ). We next regularize the equation satisfied by g − , using the same method but different parameters β > 0 and 0 < ν < dist(K, ∂Ω): we obtain for the same φ’s

Z

R^d+2

g ₋ ^β,ν (∂ t + a · ∇ )φ + Z

R^d+2

f ₋ ^{0 ν} ⊗ θ β φ − η Z

Q×

Rξ

δ v ∇ v · ( ∇ φ) ⋆ ( ˇ γ ν ⊗ θ ˇ β ) = Z

R^d+2

∂ ξ φ dq ^β,ν . (10) Suppose that φ ∈ C ^∞

_c

(R ^d+1 ) is non-negative with support in R × K and apply (9) to the test function

− g ^β,ν ₋ (t, x, ξ)φ(t, x) and (10) to − f ₊ ^α,ε (t, x, ξ)φ(t, x), and sum the two equations; using the fact that − f ₊ ^α,ε and − g ^β,ν ₋ are non-decreasing with respect to ξ, we find, after some integrate by parts,

− Z

R^d+2

f ₊ ^α,ε g ₋ ^β,ν (∂ t + a · ∇ )φ + η Z

Q×

Rξ

δ v ∇ v · ( ∇ (f ₊ ^α,ε φ)) ⋆ ( ˇ γ ν ⊗ θ ˇ β )

− Z

R^d+2

h f ₊ ^{0 ε} ⊗ θ α g ^β,ν ₋ + f ₋ ^{0 ν} ⊗ θ β f ₊ ^α,ε i

φ > 0. (11) Thanks to the decentred regularization, f ₊ ^α,ε (t, x, ξ) is null if t 6 α/2; hence, for β 6 α/2, f ₋ ^{0 ν} ⊗ θ β f ₊ ^α,ε ≡ 0.

Moreover, the function which associates t with Z

Ω×

Rξ

f ₊ ^{0 ε} (x, ξ)g − (t, x, ξ)φ(t, x) dx dξ = Z

Ω

Z

Ω

Z

Rξ

f ₊ ⁰ (y, ξ)g − (t, x, ξ)γ ε (x − y)φ(t, x) dξ dy dx

= −

Z

Ω

Z

Ω

(u 0 (y) − v(t, x)) ⁺ γ ε (x − y)φ(t, x) dξ dy dx (12) is continuous (because v ∈ C ([0, T 0 ]; L ¹ (Ω)) and we have g − (0, · , · ) = f ₋ ⁰ . Therefore, letting β, ν and α successively tend to zero in (11), we have

Z

Q×

Rξ

( − f ₊ ^ε g − )(∂ t + a · ∇ )φ + η Z

Q×

Rξ

δ v ∇ v · ∇ (f ₊ ^ε φ) − Z

Ω×

Rξ

f ₊ ^{0 ε} f ₋ ⁰ φ ^(t=0) > 0.

Choose T ∈ [0, T 0 ] and let φ(t, x) = λ(x)w β (t) where w β (t) = R +∞

t−T θ β (r)dr; we obtain Z

Q×R

ξ

( − f ₊ ^ε g − )[ − θ β (t − T )λ + w β a · ∇ λ] + η Z

Q×R

ξ

δ v ∇ v · ∇ (f ₊ ^ε λ)w β − Z

Ω×R

ξ

f ₊ ^{0 ε} f ₋ ⁰ λ > 0.

The function t → R

Ω×

Rξ

( − f ₊ ^ε g − )(t, x, ξ)λ(x) dx is continuous (it is similar to (12)); thus, letting β → 0,

− Z

Ω×

Rξ

( − f ₊ ^ε g − ) ^(t=T) λ + Z

Q

^T

×

Rξ

( − f ₊ ^ε g − )a · ∇ λ + Z

Q

^T

×

Rξ

ηδ v ∇ v · ∇ (f ₊ ^ε λ) − Z

Ω×

Rξ

f ₊ ^{0 ε} f ₋ ⁰ λ > 0 where Q ^T = (0, T ) × Ω. We therefore obtain

T 1 6 T 2 + T IC + T D (13)

where

T 1 = Z

Ω×

Rξ

( − f ₊ ^ε g − ) ^(t=T) λ , T 2 =

Z

Q

^T

×R

ξ

( − f ₊ ^ε g − )a · ∇ λ , T D =

Z

Q

^T

×R

ξ

ηδ v ∇ v · ∇ (f ₊ ^ε λ) , T IC = −

Z

Ω×R

ξ

f ₊ ^{0 ε} f ₋ ⁰ λ.

We now estimate these terms. We have T 1 =

Z

K

Z

Ω

"

Z

Rξ

( − f + (T, y, ξ)g − (T, x, ξ)) dξ

#

λ(x)γ ε (x − y) dy dx

= Z

K

Z

Ω

(u(T, y) − v(T, x)) ⁺ λ(x)γ ε (x − y) dy dx

>

Z

K

Z

Ω

(u(T, x) − v(T, x)) ⁺ λ(x)γ ε (x − y) dy dx − Z

K

Z

Ω

(u(T, x) − u(T, y)) ⁺ λ(x)γ ε (x − y) dy dx.

But, if x ∈ K, we have, by choice of ε, x − Ω ⊃ supp(γ ε ), hence R

Ω γ ε (x − y) dy = 1. Moreover, since u(T, · ) ∈ BV (Ω), by Lemma 2 (see the appendix),

Z

K

Z

Ω

(u(T, x) − u(T, y)) ⁺ λ(x)γ ε (x − y) dy dx 6 Z

Ω

Z

Ω | u(T, x) − u(T, y) | γ ε (x − y) dy dx 6 Cε.

Hence,

T 1 >

Z

Ω

(u(T, x) − v(T, x)) ⁺ λ(x) dx − Cε.

Next, reasoning as for T 1 , T 2 =

Z T

0

Z

K

Z

Ω

Z

Rξ

( − f + (t, y, ξ)g − (t, x, ξ))γ ε (x − y)a(ξ) · ∇ λ(x) dy dx dt 6 C

Z T

0

Z

Ω

Z

Ω

(u(t, y) − v(t, x)) ⁺ γ ε (x − y) dy dx dt 6 Cε + C

Z T

0

Z

Ω

(u(t, x) − v(t, x)) ⁺ dx dt.

Let us estimate the diffusion term T D . First, we write: T D = T _D ¹ + T _D ² with T _D ¹ =

Z

Q

^T

×R

ξ

ηδ v ∇ v · f ₊ ^ε ∇ λ = Z

Q

^T

×R

ξ

η ∇ v · Z

Rξ

δ v f ₊ ^ε

!

∇ λ 6 η Z

Q

^T

|∇ v ||∇ λ | 6 Cη and

T _D ² = Z

Q

^T

×

Rξ

ηδ v ∇ v · λ ∇ f ₊ ^ε 6 Cη Z

Q

^T

|∇ v | (t, x) sup

ξ |∇ f ₊ ^ε | (t, x, ξ) dtdx.

But ∇ f ₊ ^ε (t, x, ξ) = R

Ω f + (t, y, ξ) ∇ γ ε (x − y) dy, so that |∇ f ₊ ^ε | (t, x, ξ) 6 ||∇ γ ε || L

¹

(R

^d

) 6 C/ε. Hence, T _D ² 6 Cη

ε Z

Q

^T

|∇ v | 6 Cη ε .

Using Lemma 2, a straightforward computation gives T IC 6 Cε. We finally gather the different estimates in (13) and get, for all ε,

Z

Ω

(u(T, x) − v(T, x)) ⁺ λ(x)dx 6 C ε + η

ε + C

Z T

0

Z

Ω

(u(t, x) − v(t, x)) ⁺ dx dt.

Minimizing on ε, we obtain (recall that λ = λ 0 here) Z

Ω

(u(T, x) − v(T, x)) ⁺ λ 0 (x)dx 6 C √ η + C Z T

0

Z

Ω

(u(t, x) − v(t, x)) ⁺ dx dt. (14)

4 Transport and regularization of the kinetic equations

In order to estimate (u(T, · ) − v(T, · )) ⁺ near the boundary of Ω, we choose a chart (O i , h i , λ i ) and we transport the equations to B ₊ ^d . In the following, we drop the subscript i.

4.1 Transport of the kinetic equations

We now write the kinetic equations satisfied by u and v once they have been transported on B ₊ ^d . Consider a test function Ψ ∈ C ^∞

_c

(R × B ^d × R ξ ) and set φ(t, x, ξ) = Ψ(t, h(x), ξ) ∈ C

²_c

(R t × O × R ξ ). Next, extend φ by 0 to get a function φ ∈ C

²_c

(R ^d+2 ) and plug it into (5) ₊ (φ is not C ^∞ but is regular enough to be taken as a test function in this equation). This gives

Z

R

×O×

Rξ

f +

h (∂ t Ψ) ◦ h + a · h ^{′ T} ( ∇ Ψ) ◦ h i +

Z

O×

Rξ

f ₊ ⁰ Ψ ^(t=0) ◦ h + Z

R

×(∂Ω∩O)×

Rξ

( − a · n)f ₊ ^τ Ψ ◦ h

= Z

R

×O×

Rξ

(∂ ξ Ψ) ◦ h dm.

Through the change of variables y = h(x), and by definition of the measure on Σ, we obtain Z ∞

0

Z

B

^d₊

Z

Rξ

| Jh ⁻¹ | f + ◦ h ⁻¹ (∂ t Ψ + h ^′ ◦ h ⁻¹ a · ∇ Ψ) + Z

B

^d₊

Z

Rξ

| Jh ⁻¹ | f ₊ ⁰ ◦ h ⁻¹ Ψ ^(t=0) +

Z ∞

0

Z

B

^d−1

Z

Rξ

( − a · nf ₊ ^τ ) ◦ h ⁻¹ Ψ

∂h ⁻¹

∂x 1 ∧ · · · ∧ ∂h ⁻¹

∂x d−1

dx 1 . . . dx d−1

= Z ∞

0

Z

B

^d₊

Z

Rξ

(∂ ξ Ψ) d(h ∗ m).

In the following, we adopt the notations

j(x) = | Jh ⁻¹ (x) | and H (x) = h ^′ ◦ h ⁻¹ (x) and l(x) =

∂h ⁻¹

∂x 1 ∧ · · · ∧ ∂h ⁻¹

∂x d−1

(x).

Moreover, for any function r(t, x, ξ), we write ˜ r(t, x, ξ) for r(t, h ⁻¹ (x), ξ). Therefore, the previous equality reads

Z ∞

0

Z

B

^d₊

Z

Rξ

j f ˜ + (∂ t Ψ + Ha · ∇ Ψ) + Z

B

₊^d

Z

Rξ

j f ˜ ₊ ⁰ Ψ ^(t=0) + Z ∞

0

Z

B

^d−1

Z

Rξ

l( − a · n) ˜ ˜ f ₊ ^τ Ψ

= Z ∞

0

Z

B

₊^d

Z

Rξ

∂ ξ Ψ d(h ∗ m). (15) Similar computations are achieved on the kinetic equation satisfied by v. We obtain

Z ∞

0

Z

B

^d₊

Z

Rξ

j g ˜ − (∂ t Ψ + Ha · ∇ Ψ) + Z

B

₊^d

Z

Rξ

j f ˜ ₋ ⁰ Ψ ^(t=0) +

Z ∞

0

Z

B

^d−1

Z

Rξ

l( − a · n) ˜ ˜ f ₋ ^b Ψ + Z ∞

0

Z

B

^d−1

Z

Rξ

l Dδ ˜ u ˜

_b

Ψ + η

Z ∞

0

Z

B

^d₊

Z

Rξ

Zδ ˜ v ˜ · ∇ Ψ = Z ∞

0

Z

B

^d₊

Z

Rξ

∂ ξ Ψ d(h ∗ q) (16) where D(t, x) = η ∇ v(t, x) · n(x) and Z (t, x) = − h ^′ (x) ∇ v(t, x). Notice that

Z is bounded in L ¹ ((0, T ) × Ω) for all T > 0. (17)

This property, as well as the Lipschitz continuity [0, ∞ ) → L

¹

of u and v (with a Lipschitz constant for

v independent of η) and the bounds on | u(t, · ) |

^BV

and | v(t, · ) |

^BV

, are conserved by the transport by h.

4.2 Transport of the BLN condition

We state here the only consequence of (6) that we use in the following.

Let Ψ ∈ C ^∞

_c

(R × B ^d−1 × R ξ ) be non-negative and non-decreasing with respect to ξ. The function φ(t, x, ξ) = Ψ(t, h(x), ξ)(1 − f ₊ ^b (t, x, ξ)) is non-decreasing with respect to ξ (since Ψ and 1 − f ₊ ^b are non-negative and non-decreasing with respect to ξ). Hence, (6) implies

Z ∞

0

Z

∂Ω∩O

Z

Rξ

( − a · n)f ₊ ^τ Ψ ◦ h(1 − f ₊ ^b ) 6 Z ∞

0

Z

∂Ω∩O

Z

Rξ

M f ₊ ^b (1 − f ₊ ^b )Ψ ◦ h.

But f ₊ ^b (1 − f ₊ ^b ) = 0 so that, transporting this equation with h ⁻¹ on B ^d−1 , we deduce that, for all Ψ ∈ C ^∞

_c

(R × B ^d−1 × R ξ ) which is non-negative and non-decreasing with respect to ξ,

Z ∞

0

Z

B

^d−1

Z

Rξ

l( − a · n) ˜ ˜ f ₊ ^τ Ψ 6 Z ∞

0

Z

B

^d−1

Z

Rξ

l( − a · n) ˜ ˜ f ₊ ^τ f ˜ ₊ ^b Ψ. (18) We also need to understand how the unit normal is transported by the chart (O, h).

Lemma 1 For all y ∈ B ^d−1 and all X ∈ R ^d , we have l(y)X · n(y) = ˜ − j(y)(H (y)X ) d , where (H(y)X ) d

is the d-th coordinate of H (y)X . Proof of Lemma 1

Let ψ ∈ C ^∞

_c

(B ^d ) and φ = ψ ◦ h ∈ C

²_c

(O) (extented by 0 outside O). Integrating by parts, we have Z

Ω

X · ∇ φ(x) dx = Z

∂Ω

φ(x)X · n(x) dσ(x).

Since ∇ φ(x) = h ^′ (x) ^T ∇ ψ(h(x)), transporting these integrals by h (all the integrands are null outside O), we find

Z

B

₊^d

j(x)H (x)X ·∇ ψ(x) dx = Z

B

₊^d

X · (h ^′ (h ⁻¹ (x))) ^T ∇ ψ(x) | Jh ⁻¹ (x) | dx = Z

B

^d−1

ψ(x)X · n(h ⁻¹ (x))l(x) dx.

Another integrate by parts then yields Z

B

^d−1

ψ(x)X · n(h ⁻¹ (x))l(x) dx = Z

B

^d−1

( − j(x)(H (x)X ) d )ψ(x) dx − Z

B

₊^d

div(jHX)(x)ψ(x) dx.

(the unit normal to B ₊ ^d on B ^d−1 is (0, . . . , 0, − 1)). Taking first ψ ∈ C ^∞

_c

(B ₊ ^d ), we see that div(jHX) = 0 on B ₊ ^d ; thus, for all ψ ∈ C ^∞

_c

(B ^d ), R

B

^d−1

ψ(x)X · n(h ⁻¹ (x))l(x) dx = R

B

^d−1

( − j(x)(H (x)X) d )ψ(x) dx, which concludes the proof.

4.3 Regularization of the transported equations

From now on, we work on B ^d and we thus simply write r for ˜ r. Let K := supp(λ) (compact subset of B ^d ). We now regularize equations (15) and (16).

For ε > 0, we denote γ _ε (x) = Q d−1

i=1 θ ε (x i ); we take ε d > 0 and we denote ε = (ε, ε d ), γ ε (x) = γ _ε (x)θ ε

d

(x d ).

We choose ε+ε d < dist(K, ∂B ^d ). Let Ψ ∈ C

²_c

(R× B ^d ×R ^ξ ) with support in R× K ×R ^ξ ; then, Ψ⋆ ( ˇ γ ε ⊗ θ ˇ α ) is compactly supported in R × B ^d × R ξ . Using Ψ ⋆ ( ˇ γ ε ⊗ θ ˇ α ) in (15), we get

Z

R^d+2

(jf + ) ^α,ε ∂ t Ψ + (jf + H) ^α,ε a · ∇ Ψ + Z

R^d+2

(jf ₊ ⁰ ) ^ε ⊗ θ α Ψ +

Z

R^d+2

(l( − a · n)f ₊ ^τ ) ^α,ε ⊗ θ ε

d

Ψ = Z

R^d+2

∂ ξ Ψ d(h ∗ m) ^α,ε . (19)

The same test function with parameters β and ν in (16) gives Z

R^d+2

(jg − ) ^β,ν ∂ t Ψ + (jg − H ) ^β,ν a · ∇ Ψ + Z

R^d+2

(jf ₋ ⁰ ) ^ν ⊗ θ β Ψ +

Z

R^d+2

(l( − a · n)f ₋ ^b ) ^β,ν ⊗ θ ν

d

Ψ + Z ∞

0

Z

B

^d−1

Z

Rξ

lDδ u

b

Ψ ⋆ ( ˇ γ ν ⊗ θ ˇ β ) + η

Z ∞

0

Z

B

^d₊

Z

Rξ

Zδ v · ∇ Ψ ⋆ ( ˇ γ ν ⊗ θ ˇ β ) = Z

R^d+2

∂ ξ Ψ d(h ∗ q) ^β,ν . (20)

5 Combination of the equations and new estimates

The next step consists in combining the two preceding kinetic equations. Choose a non-negative regular function φ(t, x), with support in ] − ∞ , T 0 ] × K, and apply ( − jf + ) ^α,ε (t, x, ξ)φ(t, x) as a test function in (20) and ( − jg − ) ^β,ν (t, x, ξ)φ(t, x) as a test function in (19). These two test functions are non-decreasing with respect to ξ so that, summing the results, we get U ₁ ^β,ν + U ₂ ^β,ν + U ₃ ^β,ν + U ₄ ^β,ν + U ₅ ^β,ν + U ₆ ^β,ν > 0, where

U ₁ ^β,ν = Z

R^d+2

(jf + ) ^α,ε (∂ t ( − jg − ) ^β,ν φ + ( − jg − ) ^β,ν ∂ t φ) + ( − jg − ) ^β,ν (∂ t (jf + ) ^α,ε φ + (jf + ) ^α,ε ∂ t φ) U ₂ ^β,ν =

Z

R^d+2

(jf + H) ^α,ε a · ∇ ( − jg − ) ^β,ν φ + ( − jg − ) ^β,ν ∇ φ +( − jg − H ) ^β,ν a · ( ∇ (jf + ) ^α,ε φ + (jf + ) ^α,ε ∇ φ) U ₃ ^β,ν =

Z

R^d+2

(jf ₊ ⁰ ) ^ε ⊗ θ α ( − jg − ) ^β,ν φ + ( − jf ₋ ⁰ ) ^ν ⊗ θ β (jf + ) ^α,ε φ U ₄ ^β,ν = −

Z ∞

0

Z

B

^d−1

Z

Rξ

lDδ u

_b

((jf + ) ^α,ε φ) ⋆ (ˇ γ ν ⊗ θ ˇ β ) U ₅ ^β,ν =

Z

R^d+2

(l( − a · n)f ₊ ^τ ) ^α,ε ⊗ θ ε

_d

( − jg − ) ^β,ν φ + ( − l( − a · n)f ₋ ^b ) ^β,ν ⊗ θ ν

_d

(jf + ) ^α,ε φ U ₆ ^β,ν = η

Z ∞

0

Z

B

^d₊

Z

Rξ

δ v Z · ∇ ((jf + ) ^α,ε φ) ⋆ (ˇ γ ν ⊗ θ ˇ β ).

5.1 Passing to the limit in β and ν

We study the limits of U ₁ ^β,ν , . . . , U ₆ ^β,ν as β and ν tend to 0.

The first term U ₁ ^β,ν Integrating by parts, we have

U ₁ ^β,ν = Z

R^d+2

(jf + ) ^α,ε ( − jg − ) ^β,ν ∂ t φ and thus, as β → 0 and ν → 0,

U ₁ ^β,ν → Z ∞

0

Z

B

^d₊

Z

Rξ

(jf + ) ^α,ε ( − jg − )∂ t φ. (21)

The second term U ₂ ^β,ν

The first step, here, is to get H out of the regularizations (jf + H) ^α,ε and (jg − H) ^β,ν . To do this, we

notice that, for all (t, x, ξ) ∈ R × B ^d × R, since H is C

¹

,

| (jf + H ) ^α,ε (t, x, ξ) − H (x)(jf + ) ^α,ε (t, x, ξ) |

=

Z ∞

0

Z

B

^d₊

j(y)f + (s, y, ξ)H (y)θ α (t − s)γ ε (x − y) ds dy

− H(x) Z ∞

0

Z

B

₊^d

j(y)f + (s, y, ξ)θ α (t − s)γ ε (x − y) ds dy

6 Z ∞

0

Z

B

^d₊

j(y)f + (s, y, ξ) | H (y) − H(x) | θ α (t − s)γ ε (x − y) ds dy 6 C(ε + ε d )

Z ∞

0

Z

B

^d₊

θ α (t − s)γ ε (x − y) ds dy 6 C(ε + ε d ) (here, “ | · | ” is a matrice norm). Hence,

Z

R

×B

^d

×

Rξ

(jf + H ) ^α,ε a · ∇ ( − jg − ) ^β,ν φ + ( − jg − ) ^β,ν ∇ φ

− Z

R

×B

^d

×

Rξ

(jf + ) ^α,ε Ha · ∇ ( − jg − ) ^β,ν φ + ( − jg − ) ^β,ν ∇ φ

6 C(ε + ε d ) ||∇ ( − jg − ) ^β,ν ||

^L1

(]−∞,T

0

]×K×[−D,D]) || φ || ∞ + || ( − jg − ) ^β,ν ||

^L∞

(R

^d+2

) ||∇ φ || ∞ (22) (recall that supp(φ) ⊂ ] − ∞ , T 0 ] × K). But, by Lemma 3 (see the appendix), for all s ∈ R ⁺ ,

Z

K

Z D

−D |∇ (jsgn ₋ (v(s, · ) − ξ)) ^ν | 6 C(1 + | v(s, · ) |

^BV

(B

^d₊

) ) 6 C.

Therefore,

||∇ ( − jg − ) ^β,ν ||

^L1

(]−∞,T

0

]×K×[−D,D]) 6 Z T

0

−∞

Z ∞

0

Z

K

Z D

−D |∇ (jsgn ₋ (v(s, · ) − ξ)) ^ν | θ β (t − s) ds dt 6 C

Z T

0

0

Z ∞

0

θ β (t − s) ds dt 6 C. (23)

Noticing that, thanks to φ, the integrals in U ₂ ^β,ν are in fact on R × B ^d × R ξ , we deduce from (22), (23) and similar estimates for the second part of U ₂ ^β,ν that U ₂ ^β,ν is equal to

Z

R

×B

^d

×

Rξ

(jf + ) ^α,ε Ha · ( ∇ ( − jg − ) ^β,ν φ + ( − jg − ) ^β,ν ∇ φ) + ( − jg − ) ^β,ν Ha · ( ∇ (jf + ) ^α,ε φ + (jf + ) ^α,ε ∇ φ) + O ((ε + ε d + ν + ν d )( || φ || ∞ + ||∇ φ || ∞ ))

= Z

R×B^d

×R

ξ

φHa · ∇ ((jf + ) ^α,ε ( − jg − ) ^β,ν ) + 2(jf + ) ^α,ε ( − jg − ) ^β,ν Ha · ∇ φ + O ((ε + ε d + ν + ν d )( || φ || ^∞ + ||∇ φ || ^∞ ))

= Z

R×B^d

×R

ξ

(jf + ) ^α,ε ( − jg − ) ^β,ν (2Ha · ∇ φ − div(φHa)) + O ((ε + ε d + ν + ν d )( || φ || ^∞ + ||∇ φ || ^∞ )) (we used the fact that φ has a compact support in R × B ^d ). Letting β and ν tend to 0, this gives

lim sup

β,ν→0

U ₂ ^β,ν 6 O ((ε + ε d )( || φ || ∞ + ||∇ φ || ∞ )) + Z ∞

0

Z

B

^d₊

Z

Rξ

(jf + ) ^α,ε ( − jg − )(2Ha · ∇ φ − div(φHa)). (24)

The third, fourth and fifth terms

By the choice of a decentred convolution kernel, we have for β and ν d small enough

θ β ( · )(jf + ) ^α,ε ( · , x, ξ) ≡ 0, ((jf + ) ^α,ε φ) ⋆ (ˇ γ ν ⊗ θ ˇ β ) ≡ 0, θ ν

d

( · )(jf + ) ^α,ε (t, x, · , ξ) ≡ 0.

Therefore, as β and ν go to 0, U ₃ ^β,ν →

Z ∞

0

Z

B

^d₊

Z

Rξ

(jf ₊ ⁰ ) ^ε ⊗ θ α ( − jg − )φ (25)

U ₄ ^β,ν → 0 (26)

U ₅ ^β,ν → Z ∞

0

Z

B

^d₊

Z

Rξ

(l( − a · n)f ₊ ^τ ) ^α,ε ⊗ θ ε

d

( − jg − )φ. (27)

The sixth term U ₆ ^β,ν

Since (jf + ) ^α,ε φ is regular, we have U ₆ ^β,ν → η

Z ∞

0

Z

B

₊^d

Z

Rξ

δ v Z · ∇ ((jf + ) ^α,ε φ). (28) as β and ν tend to 0.

Using (21), (24), (25), (26), (27) and (28) in U ₁ ^β,ν + U ₂ ^β,ν + U ₃ ^β,ν + U ₄ ^β,ν + U ₅ ^β,ν + U ₆ ^β,ν > 0, we obtain as β and ν go to 0

− Z ∞

0

Z

B

₊^d

Z

Rξ

(jf + ) ^α,ε ( − jg − )∂ t φ 6 C((ε + ε d )( || φ || ∞ + ||∇ φ || ∞ )) +

Z ∞

0

Z

B

^d₊

Z

Rξ

(jf + ) ^α,ε ( − jg − )(2Ha · ∇ φ − div(φHa)) +

Z ∞

0

Z

B

^d₊

Z

Rξ

(jf ₊ ⁰ ) ^ε ⊗ θ α ( − jg − )φ + Z ∞

0

Z

B

₊^d

Z

Rξ

(l( − a · n)f ₊ ^τ ) ^α,ε ⊗ θ ε

d

( − jg − )φ +η

Z ∞

0

Z

B

^d₊

Z

Rξ

δ v Z · ∇ ((jf + ) ^α,ε φ). (29)

5.2 Choice of φ and continuation of the estimates

We now take T ∈ [0, T 0 ] and φ(t, x) = λ(x)w β (t), where w β (t) = R ∞

t−T θ ˇ β (s) ds (notice that w β has its support in ] − ∞ , T 0 ]). The function w β converges, as β → 0, to the characteristic function of ] − ∞ , T ] and w ^′ _β (t) = − θ ˇ β (t − T ) converges to − δ T . Since t → R

B

^d₊

R

Rξ

(jf + ) ^α,ε (t, x, ξ)( − jg − )(t, x, ξ)λ(x) dx dξ is continuous (it is similar to (12)), we deduce from (29) that

T ₁ ^α,ε ≤ T ₂ ^α,ε + T ₃ ^α,ε + T ₄ ^α,ε + T ₅ ^α,ε + C(ε + ε d ) (30)

where

T ₁ ^α,ε = Z

B

₊^d

Z

Rξ

((jf + ) ^α,ε ( − jg − )) ^{(t=T )} λ T ₂ ^α,ε =

Z T

0

Z

B

₊^d

Z

Rξ

(jf + ) ^α,ε ( − jg − )Y T ₃ ^α,ε =

Z T

0

Z

B

₊^d

Z

Rξ

(jf ₊ ⁰ ) ^ε ⊗ θ α ( − jg − )λ T ₄ ^α,ε = η

Z T

0

Z

B

₊^d

Z

Rξ

δ v Z · ∇ ((jf + ) ^α,ε λ) T ₅ ^α,ε =

Z T

0

Z

B

₊^d

Z

Rξ

(l( − a · n)f ₊ ^τ ) ^α,ε ⊗ θ ε

d

( − jg − )λ

and Y = 2Ha · ∇ λ − div(λHa) ∈ L ^∞ ((0, T 0 ) × B ^d ₊ × R ξ ). Our aim is to obtain an inequality of the kind of (14); we now estimate each term T _i ^α,ε .

The first term T ₁ ^α,ε We have

T ₁ ^α,ε = Z

B

^d₊

Z ∞

0

Z

B

^d₊

j(y)j(x)(u(s, y) − v(T, x)) ⁺ θ α (T − s)γ ε (x − y)λ(x) dy ds dx

>

Z

B

^d₊

Z ∞

0

Z

B

^d₊

j(y)j(x)(u(T, x) − v(T, x)) ⁺ θ α (T − s)γ ε (x − y)λ(x) dy ds dx

− Z

B

^d₊

Z ∞

0

Z

B

^d₊

j(y)j(x)(u(T, x) − u(s, y)) ⁺ θ α (T − s)γ ε (x − y)λ(x) dy ds dx. (31) Lemma 2 and Proposition 1 give

Z

B

^d₊

Z ∞

0

Z

B

^d₊

| u(T, x) − u(s, y) | θ α (T − s)γ ε (x − y) dy ds dx 6 C(ε + ε d + α). (32) Since j is bounded from below by j > 0, we have

Z

B

₊^d

Z ∞

0

Z

B

₊^d

j(y)j(x)(u(T, x) − v(T, x)) ⁺ θ α (T − s)γ ε (x − y)λ(x) dy ds dx

> j ² Z

B

^d₊

λ(x)(u(T, x) − v(T, x)) ⁺ Z ∞

0

θ α (T − s) ds Z

B

^d₊

γ ε (x − y) dy

! dx

> j ² Z

K∩B

^d₊

∩{x

d

>ε

d

}

λ(x)(u(T, x) − v(T, x)) ⁺ Z ∞

0

θ α (T − s) ds Z

B

₊^d

γ ε (x − y) dy

! dx (recall that K is the support of λ).

If T > α, then R ∞

0 θ α (T − s) ds = R T

−∞ θ α = 1. Moreover, if x ∈ K and x d > ε d , we have ]0, ε[ ^d−1 × ]0, ε d [ ⊂ x − B ₊ ^d (indeed, if z ∈ ]0, ε[ ^d−1 × ]0, ε d [ then, since x ∈ K, we have x − z ∈ B ^d and, since x d > ε d > z d , x − z ∈ B ₊ ^d ); hence, for those x’s, R

B

₊^d

γ ε (x − y) dy = 1 since the support of γ ε is contained in ]0, ε[ ^d−1 × ]0, ε d [.

Thus, for T > α, j ²

Z

B

₊^d

∩{x

d

>ε

d

}

λ(x)(u(T, x) − v(T, x)) ⁺ dx 6

Z

B

₊^d

Z ∞

0

Z

B

₊^d

j(y)j(x)(u(T, x) − v(T, x)) ⁺ θ α (T − s)γ ε (x − y)λ(x) dy ds dx.

Since u and v are bounded, Z

B

₊^d

∩{x

d6εd

}

λ(x)(u(T, x) − v(T, x)) ⁺ dx 6 Z

B

₊^d

∩{x

d6εd

}

C dx 6 Cε d . Hence, if T > α,

j ² Z

B

₊^d

λ(x)(u(T, x) − v(T, x)) ⁺ dx 6

Z

B

^d₊

Z ∞

0

Z

B

^d₊

j(y)j(x)(u(T, x) − v(T, x)) ⁺ θ α (T − s)γ ε (x − y)λ(x) dy ds dx + Cε d . (33)

Equations (31), (32) and (33) give, if T > α, T ₁ ^α,ε > − C(ε + ε d + α) + j ²

Z

B

^d₊

λ(x)(u(T, x) − v(T, x)) ⁺ dx.

But u and v are Lipschitz continuous [0, T 0 ] → L

¹

(B ₊ ^d ) (with a Lipschitz constant not depending on η) and equal to u 0 at t = 0; hence, for T 6 α,

Z

B

₊^d

λ(x)(u(T, x) − v(T, x)) ⁺ dx 6 C Z

B

₊^d

| u(T, x) − u 0 (x) | dx + C Z

B

^d₊

| v(T, x) − u 0 (x) | dx 6 Cα.

Therefore, T ₁ ^α,ε being non-negative, we have, for all T ∈ [0, T 0 ], T ₁ ^α,ε > − C(ε + ε d + α) + j ²

Z

B

^d₊

λ(x)(u(T, x) − v(T, x)) ⁺ dx. (34)

The second term T ₂ ^α,ε We have

T ₂ ^α,ε = Z T

0

Z

B

₊^d

Z ∞

0

Z

B

₊^d

Z

Rξ

j(y)j(x)f + (s, y, ξ)( − g − (t, x, ξ))Y (t, x)θ α (t − s)γ ε (x − y) dξ dy ds dx dt

6 C Z T

0

Z

B

₊^d

Z ∞

0

Z

B

^d₊

(u(s, y) − v(t, x)) ⁺ θ α (t − s)γ ε (x − y) dy ds dx dt

6 C Z T

0

Z

B

₊^d

Z ∞

0

Z

B

^d₊

(u(s, y) − u(t, x)) ⁺ θ α (t − s)γ ε (x − y) dy ds dx dt +C

Z T

0

Z

B

₊^d

Z ∞

0

Z

B

₊^d

(u(t, x) − v(t, x)) ⁺ θ α (t − s)γ ε (x − y) dy ds dx dt 6 C(ε + ε d + α) + C

Z T

0

Z

B

₊^d

(u(t, x) − v(t, x)) ⁺ dx dt

(we used (32) with T = t).

Therefore,

T ₂ ^α,ε 6 C(ε + ε d + α) + C Z T

0

Z

B

^d₊

(u(t, x) − v(t, x)) ⁺ dx dt. (35) The third term T ₃ ^α,ε

We write

T ₃ ^α,ε = Z T

0

Z

B

₊^d

Z

B

^d₊

Z

Rξ

j(y)j(x)f ₊ ⁰ (y, ξ)( − g − (t, x, ξ))θ α (t)γ ε (x − y)λ(x) dξ dy dx dt

6 C Z T

0

Z

B

₊^d

Z

B

₊^d

(u 0 (y) − v(t, x)) ⁺ θ α (t)γ ε (x − y) dy dx dt.

But v(0, x) = u 0 (x) so that, v being Lipschitz continuous [0, T 0 ] → L

¹

(B ^d ₊ ) (with a Lipschitz constant not depending on η) and u 0 being in BV (B ₊ ^d ), by Lemma 2,

T ₃ ^α,ε 6 C Z T

0

Z

B

₊^d

Z

B

₊^d

| u 0 (y) − u 0 (x) | θ α (t)γ ε (x − y) dy dx dt +C

Z T

0

Z

B

^d₊

Z

B

^d₊

| v(0, x) − v(t, x) | θ α (t)γ ε (x − y) dy dx dt

6 C(ε + ε d + α). (36)

The fourth term T ₄ ^α,ε We have, for all (t, x, ξ),

|∇ ((jf + ) ^α,ε λ)(t, x, ξ) | =

Z ∞

0

Z

B

₊^d

j(y)f + (s, y, ξ)θ α (t − s)( ∇ γ ε (x − y)λ(x) + γ ε (x − y) ∇ λ(x)) dy ds

6 C ||∇ γ ε || L

¹

(

R^d

) + C ||∇ λ || L

^∞

(

R^d

) 6 C ε + C

ε d + C 6 C ε + C

ε d

(recall that ε d 6 1). Hence, by (17), T ₄ ^α,ε 6 η

Z T

0

Z

B

₊^d

| Z | (t, x) sup

ξ |∇ ((jf + ) ^α,ε λ) | (t, x, ξ)

!

dtdx 6 Cη ε + Cη

ε d

. (37)

To sum up, gathering (30), (34), (35), (36) and (37), we have proved so far that Z

B

^d₊

λ(x)(u(T, x) − v(T, x)) ⁺ dx

6 C

ε + ε d + α + η ε + η

ε d

+ C

Z T

0

Z

B

^d₊

(u(t, x) − v(t, x)) ⁺ dx dt + T ₅ ^α,ε . (38) The aim of the following section is to estimate T ₅ ^α,ε . Using boundary layers arguments (see the introduc- tion), we give in subsection 8.1 of the appendix an insight of the reason why this term can be bounded.

However, this is only an insight: since we also want to consider irregular solutions to (1), we cannot in

general estimate T ₅ ^α,ε using boundary layers analysis.

6 Estimate for the boundary term

This estimate is made in several steps. First, using the BLN condition, we introduce f ₊ ^b and give an upper bound T 5

α,ε to T ₅ ^α,ε . Then, we want to see (Ha) d in T 5

α,ε , in order to express T 5

α,ε as a part of the interior term in (16); to this end, we use Lemma 1. Finally, we must regularize the function f ₊ ^b introduced above in order that T 5

α,ε appears in (16) for some regular Ψ. The resulting term S ^α,ε is then estimated.

6.1 Introduction of f ₊ ^b

We have

T ₅ ^α,ε = Z ∞

0

Z

B

^d−1

Z

Rξ

l(y)( − a(ξ) · n(y))f ₊ ^τ (s, y, ξ)Ψ(s, y, ξ) dξ dy ds where

Ψ(s, y, ξ) = Z T

0

Z

B

₊^d

θ α (t − s)γ _ε (x − y)θ ε

_d

(x d )( − j(x)g − (t, x, ξ))λ(x) dx dt.

As ( − g − ), Ψ is non-negative and non-decreasing with respect to ξ. Thus, (18) implies T ₅ ^α,ε 6 T ^α,ε ₅ :=

Z ∞

0

Z

B

^d−1

Z

Rξ

l(y)( − a(ξ) · n(y))f ₊ ^τ (s, y, ξ)f ₊ ^b (s, y, ξ)Ψ(s, y, ξ) dξ dy ds (39)

= Z T

0

Z

B

₊^d

Z

Rξ

j(x)g − (t, x, ξ)Φ 0 (t, x, ξ) dξ dt dx with

Φ 0 (t, x, ξ) = θ ε

d

(x d ) Z ∞

0

Z

B

^d−1

λ(x)l(y)(a(ξ) · n(y))f ₊ ^τ (s, y, ξ)f ₊ ^b (s, y, ξ)θ α (t − s)γ _ε (x − y) dy ds.

6.2 Apparition of Ha

By Lemma 1, we have l(y)(a(ξ) · n(y)) = − j(y)(H(y)a(ξ)) d . Thus, if we define Φ(t, x, ξ) = θ ε

d

(x d )( − (H (x)a(ξ)) d )

Z ∞

0

Z

B

^d−1

λ(y)j(y)f ₊ ^τ (s, y, ξ)f ₊ ^b (s, y, ξ)θ α (t − s)γ _ε (x − y) dy ds, we have

| Φ 0 (t, x, ξ) − Φ(t, x, ξ) | 6

Z ∞

0

Z

B

^d−1

j(y) | (H(y)a(ξ)) d λ(x) − (H(x)a(ξ)) d λ(y) | θ α (t − s)γ _ε (x − y) dy ds θ ε

d

(x d ) 6 C(ε + ε d )θ ε

d

(x d )

(we used the fact that H and λ are Lipschitz continuous) and T 5

α,ε 6 Z T

0

Z

B

₊^d

Z

Rξ

j(x)g − (t, x, ξ)Φ(t, x, ξ) dξ dx dt + C(ε + ε d ). (40)

6.3 Regularization of f ₊ ^b

We now want to replace f ₊ ^b (s, y, ξ) in Φ by a regular approximation. Let sgn _+,δ be a regular non- decreasing function, equal to 0 on R ⁻ , to 1 on [δ, ∞ [, such that | (sgn _+,δ ) ^′ | 6 C/δ and sgn _+,δ → sgn ₊ everywhere as δ → 0. We have, for all (a, b, ξ) ∈ R ³ ,

Z

Rξ

| sgn _+,δ (a − ξ) − sgn ₊ (b − ξ) | dξ 6 | a − b | + δ. (41)

Thus, defining

Φ δ (t, x, ξ) = θ ε

d

(x d )( − (H (x)a(ξ)) d )

× Z ∞

0

Z

B

^d−1

λ(y)j(y)f ₊ ^τ (s, y, ξ)sgn _+,δ (u b (t, x) − ξ)θ α (t − s)γ _ε (x − y) dy ds, we have

| Φ(t, x, ξ) − Φ δ (t, x, ξ) | 6 Cθ ε

d

(x d )

Z ∞

0

Z

B

^d−1

| sgn ₊ (u b (s, y) − ξ) − sgn _+,δ (u b (t, x) − ξ) | θ α (t − s)γ _ε (x − y) dy ds and therefore, by (41),

Z T

0

Z

B

₊^d

Z

Rξ

j(x)g − (t, x, ξ)Φ(t, x, ξ) dξ dx dt − Z T

0

Z

B

^d₊

Z

Rξ

j(x)g − (t, x, ξ)Φ δ (t, x, ξ) dξ dx dt

6 C Z T

0

Z

B

^d₊

Z

Rξ

Z ∞

0

Z

B

^d−1

| sgn ₊ (u b (s, y) − ξ) − sgn _+,δ (u b (t, x) − ξ) |

× θ α (t − s)γ _ε (x − y)θ ε

d

(x d ) dy ds dξ dx dt 6 C

Z T

0

Z

B

^d₊

Z ∞

0

Z

B

^d−1

( | u b (s, y) − u b (t, x) | + δ)θ α (t − s)γ _ε (x − y) dy ds θ ε

d

(x d ) dx dt

6 C(ε + α + δ) Z T

0

Z

B

^d₊

Z ∞

0

Z

B

^d−1

θ α (t − s)γ _ε (x − y)θ ε

d

(x d ) dy ds dx dt 6 C(ε + α + δ)

(we used the fact that u b is Lipschitz continuous). We deduce from this last inequality and (40) that T 5

α,ε

6 Z T

0

Z

B

₊^d

Z

Rξ

j(x)g − (t, x, ξ)Φ δ (t, x, ξ) dξ dx dt + C(ε + ε d + α + δ) 6

Z T

0

Z

B

₊^d

Z

Rξ

j(x)g − (t, x, ξ)(H(x)a(ξ)) d

× Z ∞

0

Z

B

^d−1

λ(y)j(y)f + ^τ (s, y, ξ)θ α (t − s)γ ε (x − y) dy ds

sgn +,δ (u b (t, x) − ξ)( − θ ε

d

(x d )) dξ dx dt +C(ε + ε d + α + δ)

Let Θ ε

_d

(x d ) = R x

d

0 θ ε

_d

(r) dr (we have 0 6 Θ ε

_d

6 1 and Θ ε

_d

= 1 on [ε d , + ∞ [), Γ(t, x, ξ) =

Z ∞

0

Z

B

^d−1

λ(y)j(y)f ₊ ^τ (s, y, ξ)θ α (t − s)γ _ε (x − y) dy ds

sgn _+,δ (u b (t, x) − ξ)(1 − Θ ε

d

(x d ))