On a pseudoparabolic problem with constraint

On a pseudoparabolic problem with constraint∗ Stanislav N. ANTONTSEV†, Departamento de Matem´atica, Universidade da Beira Interior Covilh˜a 6201-001 (Portugal) G´erard GAGNEUX‡, Robert LUCE‡ and Guy VALLET‡ Laboratoire de Math´ematiques Appliqu´ees UMR 5142 IPRA BP 1155 64 013 Pau cedex (France)

Abstract This work deals with the study to a nonlinear degenerated pseudoparabolic problem. Arising from the modelling of sedimentary basins formation, the equation degenerates in order to take implicitly into account a constraint on the time-derivative of the unknown. An existence result of a solution with an adapted compactness result and qualitative properties of the solutions are proposed (localization, finite speed of propagation, etc.). Keywords: Pseudoparabolic, compactness, finite speed propagation.

1

Introduction

The original motivation of this study is the mathematical analysis of general models arising from geological basin formation. The initial model has been developed by the Institut Fran¸cais du P´etrole and it takes into account sedimentation, transport and accumulation, erosion phenomena at large scales in time and space. The main feature of these models is characterized by a constraint on the time-derivative of the solution. This constraint leads us to consider an original class of conservation laws, a priori parabolic or pseudoparabolic, revealing some hyperbolic comportment because of a diffusive coefficient that depends on the time-derivative of the solution. A more precise description of these models have been exposed, on the one hand, for the multilithological case, by R. Eymard et al. [?] and V. Gervais et al. [?]; on the second hand, by S. N. Antontsev et al. [?, ?], G. Gagneux et al. [?, ?] and G. Vallet [?] for the mathematical aspect ∗ AMS

subject classifications: 35K70, 35R35, 35B99 address: [email protected]. ‡ Email address: [email protected], [email protected], [email protected]. † Email

1

of the monolithological case and R. Eymard and al. [?] for a theoretical and numerical approach of an inverse problem. For a brief presentation of the model, let us consider a sedimentary basin whose base, denoted by Ω, is a fixed domain of Rd (d = 1, 2) with a Lipschitzian boundary Γ. The sediment height, denoted by u, naturally satisfies the mass balance equation ∂t u + div {~q} = 0 in ]0, T [×Ω, where ~q is proportional to ∇[u + τ ∂t u] according to a Darcy-Barenblatt’s law (see C. Cuesta et al. [?]). Following D. Granjeon [?], in the framework of a perfect physical equilibrium, ~ ∝ the flux is considered to be proportional to the slope by writing that ~q = S −∇u, i.e. τ = 0. Taking into account a balancing of the slope leads us to ~ + τ ∂t S, ~ where consider ~q = S~e with the first order kinetic equation S~e = S τ > 0 is time-scaled. Moreover, in a sedimentary basin formation process, sediments must first be produced in situ by weathering effects prior to be transported by surfacing erosion. Thus, R. Eymard et al. [?] introduce a maximum erosion rate E such that −∂t u ≤ E in ]0, T [×Ω, where E takes into account the composition, the structure and the age of the sediments. Then, the authors propose a sediment → flux given by − q = −λ ∇[u + τ ∂t u] where λ, 0 ≤ λ ≤ 1, is playing the role of a flux limiter, with the relevant law of state in the form: 1 − λ ≥ 0, ∂t u + E ≥ 0, (1 − λ) (∂t u + E) = 0 a.e. in ]0, T [×Ω. Therefore, S. N. Antontsev et al. [?], G. Gagneux et al. [?] and G. Vallet [?] propose a conservative formulation that contains implicitly (see Remark ??) the above mathematical modelling of λ:

i.e.

0 ∈ ∂t u − div(H(∂t u + E)∇[u + τ ∂t u]) in ]0, T [×Ω, 0 = ∂t u − div(λ∇[u + τ ∂t u]) where λ ∈ H(∂t u + E) in ]0, T [×Ω,

where H denotes the maximal monotone graph of the function of Heaviside. Existence and uniqueness results for a solution to the above differential inclusion are still open problems. Our purpose is to analyse a modified model where H is replaced by a continuous function a, an approximation Yosida of H for example. An existence result of a solution to a nonlinear degenerate pseudoparabolic equation by the way of a compactness argument is presented, generalizing the first result of G. Gagneux and G. Vallet in [?]. Then, a qualitative study of the local hyperbolic behaviour, via finite speed propagation properties, of the problem is presented. This study points out the strong influence of the regularity of the function a, in the neighbourhood of 0+ , on these descriptive properties.

1.1

Presentation of the model

Let us consider a Lipschitzian bounded domain Ω, with boundary Γ separated in two nontrivial distinct parts Γs and Γe . For any positive T , let us denote Q :=]0, T [×Ω. 2

Therefore, we are interested in the following pseudoparabolic problem: ∂t u − div(a(∂t u + E)∇[u + τ ∂t u]) = 0 in

Q,

(1)

with the initial height given by: u(0, x) = u0 (x), x ∈ Ω, and the following boundary conditions: −a(∂t u + E)∇[u + τ ∂t u].~n = 0 −a(∂t u + E)∇[u + τ ∂t u].~n ∈ β(∂t u)

on on

Γe , Γs .

β is assumed to be the real graph of a maximal monotone operator (cf. J.-L. Lions [?] p.422 and references therein) such that: i) 0 ∈ β(0), ∀x, y ∈ R, ∀(ξ, θ) ∈ β(x) × β(y), (ξ − θ)(x − y) ≥ 0 (2) (in particular, ∀x > 0, β(x) ⊂ [0, +∞[ and β(−x) ⊂] − ∞, 0]). ii) There exists C1 , C2 in R+∗ satisfying the growth control condition: C1 θ2 ≤ xθ + C2 .

∀x ∈ R, ∀θ ∈ β(x),

(3)

1

iii) There exists a sequence (fk )k∈N ⊂ C (R), compatible with the growth control, such that: fk (0) = 0, fk0 > 0 and fk (±∞) = ±∞ and

fk−1 → β −1 uniformly on any compact set.

(4)

In the sequel τ > 0, E : [0, T ] → R is a non negative continuous function and a is a continuous function defined on R that satisfies 0 ≤ a ≤ M,

a(0) = 0,

∀x > 0, a(x) > 0 and a(−x) = 0 by extention

in order to have implicitly ∂t u + R xE ≥ 0 a.e. in Q (Cf. Rmq ??-ii). For any real x, denote A(x) = 0 a(s) ds , and for technical reasons (the motivation is to derive various ways of approximating the Heaviside graph by Yosida regularized functions), one assumes that a◦A−1 has got on R+ a continuity modulus which square is an Osgood function; for simplicity, we consider the case where a ◦ A−1 is H¨ older continuous with exponent 21 . Note that if a(x) ∼+ cxα 0

then the hypothesis on a ◦ A−1 imply that α ≥ 1.

1.2

Definition of a solution

By considering weak solutions, one is looking for u a priori in H 1 (Q), with ∂t u in L2 (0, T ; H 1 (Ω)), such that for any v in H 1 (Ω) and for a.e. t in ]0, T [, Z Z 0 ∈ {∂t uv + a(∂t u + E)∇[u + τ ∂t u]∇v} dx + β(∂t u)v dσ. (5) Ω

Γs

i.e. there exists a measurable function θ in β(∂t u) such that Z Z {∂t uv + a(∂t u + E)∇[u + τ ∂t u]∇v} dx + θv dσ = 0 Ω

Γs

together with the initial condition u|t=0 = u0 a.e. in Ω. 3

Remark 1 1. The problem can be written in the non-classical form:  ∂t u − τ ∆A(∂t u + E) − div(a(∂t u + E)∇u) = 0 in Q,     ∂  t u + E ≥ 0 in Q and ]0, T [×Γ, −τ ∂n A(∂t u + E) − a(∂t u + E)∂n u = 0 on ]0, T [×Γe ,   −τ ∂n A(∂t u + E) − a(∂t u + E)∂n u ∈ β(∂t u) on ]0, T [×Γs ,    u|t=0 = u0 a.e. in Ω. 2. Given by extension that a(x) = 0 for any negative x, the suitable test function v = −(∂t u + E)− leads, for a.e. t in ]0, T [, to: Z Z Z [(∂t u + E)− ]2 dx − θ(∂t u + E)− dσ = − (∂t u + E)− E dx ≤ 0 Ω

Γs

Ω

where θ ∈ β(∂t u). Since −θ(∂t u + E)− ≥ 0, one gets ∂t u + E ≥ 0 a.e. in Q. 3. If u0 ∈ H 1 (Ω), then u belongs to L∞ (0, T, H 1 (Ω)). Let us consider ε > 0 R∂ u ds . and vε = 0 t a(s+E)+ε Then, the variational formulation gives Z Z ∂t u ds a(∂t u + E) 0 = {∂t u + ∇[u + τ ∂t u]∇∂t u} dx a(s + E) + ε a(∂t u + E) + ε Ω 0 Z Z ∂t u ds + θ dσ, a(s + E) + ε Γs 0 where θ ∈ β(∂t u). R∂ u R∂ u ds ds ≥ M1+ε |∂t u|2 and θ 0 t a(s+E)+ε ≥ 0, one gets Since ∂t u 0 t a(s+E)+ε that Z a(∂t u + E) |∂t u|2 + ∇[u + τ ∂t u]∇∂t u} dx. 0≥ { M + ε a(∂ t u + E) + ε Ω Note that, when  → 0+ , a(∂t u + E) ∇[u + σ∂t u]∇∂t u → ∇[u + τ ∂t u]∇∂t u a(∂t u + E) + ε

a.e. in Ω,

since ∇[∂t u + E] = ∇∂t u = 0 a.e. in the set {∂t u + E} = 0 and since ∂t u + E ≥ 0. Moreover, given that a(∂t u + E) ∇[u + σ∂ u]∇∂ u a.e. in Ω, t t ≤ |∇[u + τ ∂t u]∇∂t u| a(∂t u + E) + ε the theorem of dominated convergence leads to: Z 1 2 ||∂t u||L2 (Ω) + ∇[u + τ ∂t u]∇∂t u dx ≤ 0. M Ω 4

(6)

Thus, 1 1 d ||∂t u||2L2 (Ω) + ||∇u||2L2 (Ω)d + τ ||∇∂t u||2L2 (Ω)d ≤ 0 M 2 dt and for a.e. t in ]0, T [,

≤

1 1 ||∂t u||2L2 (0,t,L2 (Ω)) + ||∇u(t)||2L2 (Ω)d + τ ||∇∂t u||2L2 (0,t,L2 (Ω)d ) M 2 1 2 ||∇u0 ||L2 (Ω)d . 2

Since for any t, ||u(t)||L2 (Ω) ≤ ||u0 ||L2 (Ω) +

√

t||∂t u||L2 (0,t,L2 (Ω))

the result holds. Note that this is also true for any t since u ∈ Cs ([0, T ], H 1 (Ω)), i.e. u(t) exists in H 1 (Ω) for any t. Moreover, one may remark that t 7→ ||∇u(t)||L2 (Ω)d is a non increasing function. 4. If u0 ∈ H 1 (Ω), then ∂t u belongs to L∞ (0, T, H 1 (Ω)) . Let us consider again inequality (??) to obtain with the above estimations 1 1 τ ||∂t u||2L2 (Ω) + ||∇∂t u||2L2 (Ω)d ≤ ||∇u0 ||2L2 (Ω)d M 2 2τ and conclude.

1.3 1.3.1

Existence of a strong solution A uniqueness lemma

The key to understanding compactness properties in the regularizing procedure is the following assertion: Lemma 1 Let us consider κ in H 1 (Ω), E a real number, β a monotone graph (not necessary maximal) and b an essentially bounded nonnegative continuous function such that Z x 1 2 ∀x, y ∈ R, |b(x) − b(y)| ≤ c|B(x) − B(y)| where B(x) = b(s) ds. 0

Then, there exists at most one solution w in H 1 (Ω) such that for any v in H 1 (Ω), Z Z 0 ∈ {wv + b(w + E)∇[κ + w]∇v} dx + β(w)v dσ. (7) Ω

Γs

5

Proof. If one denotes by w1 and w2 two admissible solutions. Let us set ξ =   1 if t ≥ µ, µ ln( te B(w1 +E)−B(w2 +E) and, for a given µ > 0, pµ (t) = µ ) if t ∈ [ e , µ],  µ 0 if t ≤ e . pµ is a non-increasing lipschitzian function with a lipschitz-constant equal to µe . Therefore, v = pµ [ξ] is a suitable test function and it comes Z 1 b(w1 + E) − b(w2 + E) 0 = { |∇ξ|2 + ∇κ∇ξ} dx µ ξ ξ Ω∩{ e ≤ξ≤µ} Z Z + (w1 − w2 )pµ (ξ) dx + [θ1 − θ2 ]pµ (ξ) dσ, Ω

Γs

where θi ∈ β(wi ). Since (θ1 − θ2 )pµ (ξ) = (θ1 − θ2 )(w1 − w2 )

pµ (ξ) ≥ 0, (w1 + E) − (w2 + E)

so, Z

Z (w1 − w2 )pµ (ξ) dx +

Ω

Z ≤ Ω∩{ µ e ≤ξ≤µ}

Z ≤ C

Ω∩{ µ e ≤ξ≤µ}

1 |∇ξ|2 dx ξ

√ c ξ |∇κ∇ξ| dx ξ Z 2 |∇κ| dx +

Ω∩{ µ e ≤ξ≤µ}

Ω∩{ µ e ≤ξ≤µ}

1 |∇ξ|2 dx 2ξ

and the solution is unique by considering the limits when µ goes to 0+ .

1.4



The univoque case

Let us assume in this section that β ∈ C 1 (R) with β(0) = 0 and β 0 > 0. 1.4.1

Semi-discretized non degenerated processes

Let us consider h > 0, u0 in H 1 (Ω), E ≥ 0 (let us assume E constant to simplify slightly the notations, E = 5m/M yr for example in R. Eymard et al. [?]) and for a given positive α, aα = max(α, a). Proposition 1 There exists a unique uα in H 1 (Ω) such that for any v in H 1 (Ω), Z uα − u0 )v dσ 0 = β( h Γs Z uα − u0 uα − u0 uα − u0 + { v + aα ( + E)∇[uα + τ ]∇v} dx. h h h Ω

6

0 Proof. The demonstration is classical since if one denotes by w = u−u h , the 1 1 problem is lead to: find w in H (Ω) such that, for any v in H (Ω), Z Z {wv + (h + τ )∇Aα (w + E)∇v + aα (w + E)∇u0 ∇v} dx + β(w)v dσ = 0

Ω

Γs

where Aα a bi-lipschitzian continuous function. On the one hand, Schauder-Tikhonov fixed point theorem and the compactness1 of the trace operator from H 1 (Ω) into L2 (Γ) are used for the existence of a solution in the hilbertian separable framework. Firstly, the assertion is proved with βn = max(−n, min(β, n)) where n ∈ N. Then, thanks to a priori estimates, one is able to pass to the limits with n towards +∞ since Zw

Z βn (w) Γs

ds dσ ≥ aα (s + E)

Z

0

βn (w) Γs

C1 w dσ ≥ M M

Z

βn2 (w) dσ −

Γs

C2 mes(Γs ). M

On the other hand, the uniqueness of the solution comes from Lemma ?? with u0 b = (h + τ ) aα , κ = h+τ and the graph x 7→ {β(x)}.  1.4.2

Semi-discretized degenerated problem

Let us consider h > 0, u0 in H 1 (Ω), E ≥ 0 and in the sequel, a(x) = 0 for any nonnegative real x. Proposition 2 There exists a unique u in H 1 (Ω) such that, Z u − u0 u − u0 u − u0 1 v + a( + E)∇[u + τ ]∇v} dx ∀v ∈ H (Ω), 0 = { h h h Ω Z u − u0 + β( )v dσ. (8) h Γs Moreover,

u−u0 h

+ E ≥ 0 a.e. in Ω.

0 + E)− . Proof. Note that the last inequality is obvious by using v = −( u−u h

Let us note wα =

uα −u0 h

uα −u0 h

and consider v =

R 0

ds aα (s+E)

in the equation of

proposition (??) to obtain uα − u0 { h Ω

Z 0 ≥

Z +

β( Γs

uα −u0 h

Z

ds uα − u0 uα − u0 + ∇[uα + τ ]∇ } dx aα (s + E) h h

0

uα − u0 ) h

uα −u0 h

Z

ds dσ. aα (s + E)

0

1 Remind

that the embedding of H 1 (Ω) into H s (Ω) is compact for any trace operator of H s (Ω) into L2 (Γ) is continuous.

7

1 2

< s < 1 and the

Since

uα −u0 h

uα −u0 h

R 0

uα − u0 ) β( h Γs

Z

ds aα (s+E)

≥

1 uα −u0 2 | M| h

and

uα −u0 h

Z

ds dσ aα (s + E)

Z ≥ Γs

0

≥

β(

C1 M

Z Γs

uα − u0 uα − u0 ) dσ h hM

β2(

uα − u0 C2 ) dσ − mes(Γs ), h M

one gets C2 1 mes(Γs ) ≥ [||uα ||2H 1 (Ω) + ||uα − u0 ||2H 1 (Ω) − ||u0 ||2H 1 (Ω) ] (9) M 2h uα − u0 2 C1 uα − u0 2 1 uα − u0 2 + || ||L2 (Ω) + τ || ||H 1 (Ω) + ||β( )||L2 (Γs ) . M h h M h 0 Therefore, ( uα −u )α and (uα )α are bounded sequences in H 1 (Ω) and passing to h 0 limits is possible since up to a sub-sequence uα −u converges a.e. in Ω and Γ. h At last, the uniqueness can be proved by using Lemma ?? with b = (h + τ ) a, u0 and the graph x 7→ {β(x)}. κ = h+τ 

1.4.3

Semi-discretized degenerated differential inclusion

Proposition 3 There exists a pair (u, λ) in H 1 (Ω) × L∞ (Ω) such that λ ∈ 0 H( u−u h + E) where H is the maximal monotone graph of the function of Heaviside, satisfying for any v in H 1 (Ω), Z Z u − u0 u − u0 u − u0 β( { v + λ∇[u + τ ]∇v} dx + )v dσ = 0. (10) h h h Ω Γs Proof. Let us assume in this section that for any given positive ε, a = aε = max(0, min(1, 1ε Id)). The corresponding solutions are denoted by uε and thanks to the above inequality, a sub-sequence still indexed by ε can be extracted, such that uε converges weakly in H 1 (Ω) towards u, strongly in L2 (Ω) and a.e. in Ω and Γ . Moreover, u−u0 + E ≥ 0 a.e. in Ω. h 0 Furthermore, Aε converges uniformly towards (.)+ and then Aε ( uε −u +E) conh u−u0 1 verges weakly in H (Ω) towards h + E. Up to a new sub-sequence, let us denote by λ the weak−∗ limit in L∞ (Ω) of 0 0 aε ( uε −u + E) and remark that inevitably λ = 1 a.e. in { u−u + E > 0} i.e. h h u−u0 λ ∈ H( h + E). Since for any v in H 1 (Ω), Z uε − u0 uε − u0 uε − u0 { v + (h + τ )∇Aε ( + E)∇v + aε ( + E)∇u0 ∇v} dx h h h Ω Z uε − u0 + β( )v dσ = 0, h Γs 8

passing to the limits leads to Z Z u − u0 u − u0 1 ∀v ∈ H (Ω), 0 = v dx + (h + τ )∇ ∇v + λ∇u0 ∇v} dx h h Ω Ω Z u − u0 )v dσ, + β( h Γs and to u − u0 u − u0 v + λ∇[u + τ ]∇v} dx + { h h Ω

Z

1

∀v ∈ H (Ω),

Z β( Γs

u − u0 )v dσ = 0, h

0 = 0 a.e. in {λ 6= 1}. since ∇ u−u h

1.4.4



Existence of a solution

Inductively, the following result can be proved: T let us consider N ∈ N∗ with h = N , u0 in H 1 (Ω) and E k ≥ 0 for any integer k. Proposition 4 For u0 = u0 , there exists a unique sequence (uk )k in H 1 (Ω) such that for any v in H 1 (Ω), uk+1 − uk uk+1 − uk uk+1 − uk v + a( + E k )∇[uk+1 + τ ]∇v} dx { h h h Ω Z uk+1 − uk + β( )v dσ. h Γs

Z 0

=

Moreover,

uk+1 −uk h

+ E k ≥ 0 a.e. in Ω, and

1 uk+1 − uk 2 uk+1 − uk 2 || ||L2 (Ω) + τ || ||H 1 (Ω) M h h 1 + [||uk+1 ||2H 1 (Ω) + ||uk+1 − uk ||2H 1 (Ω) − ||uk ||2H 1 (Ω) ] ≤ 0. 2h

(11)

A priori estimations leading to the main result In order to prove this result, for any sequence (vk )k ⊂ L2 (Ω), let us note in the sequel vh =

N −1 X k=0

v k+1 1[kh,(k+1)h[

and veh =

N −1 X k=0

[

v k+1 − v k (t − kh) + v k ]1[kh,(k+1)h[ . h

Lemma 2 (uh )h and (e uh )h are bounded sequences in L∞ (0, T, H 1 (Ω)) and ∂t u eh is a bounded sequence in L∞ (0, T, H 1 (Ω)) and then in L∞ (0, T, L2 (Γs )).

9

Proof. This result comes from (??) for the first part and since 2∇(uk+1 + τ

uk+1 − uk uk+1 − uk uk+1 − uk 2 1 )∇ ≥ τ |∇ | − |∇uk+1 |2 h h h τ

implies that 1 uk+1 − uk 2 τ uk+1 − uk 2 1 || ||L2 (Ω) + || ||H 1 (Ω) ≤ ||uk+1 ||2H 1 (Ω) . M h 2 h 2τ 

Existence of a solution Our aim is to prove the following result, under the hypothesis of lemme ??: Proposition 5 There exists u in H 1 (Q) with ∂t u in L2 (0, T, H 1 (Ω)) such that for any v in H 1 (Ω) and for a.e. t in ]0, T [, Z Z {∂t uv + a(∂t u + E)∇[u + τ ∂t u]∇v} dx + β(∂t u)v dσ = 0. Ω

Γs

Moreover, u belongs to W 1,∞ (0, T, H 1 (Ω)). Proof. Since (e uh )h is bounded in H 1 (0, T, H 1 (Ω)), there exists a sub-sequence, still indexed by h such that for any t, u eh (t) * u(t) in H 1 (Ω). h Moreover, if t ∈ [kh, (k + 1)h[, ||u (t) − u eh (t)||H 1 (Ω) = ||e uh (kh) − u eh (t)||H 1 (Ω) ≤ R (k+1)h ||∂t u eh (s)||H 1 (Ω) ds ≤ Ch. Then, for any t, uh (t) * u(t) in H 1 (Ω). kh h Since ∂t u e is bounded in L∞ (0, T, H 1 (Ω)), it follows that there exists a measurable set Z ⊂ [0, T ] with L([0, T ]\Z) = 0, such that for any t in Z, ∂t u eh (t) is 1 a bounded sequence in H (Ω). Therefore, up to a sub-sequence indexed by ht , ∂t u eht * ξ(t) in H 1 (Ω), strongly 2 in L (Ω) and a.e. in Ω with ξ(t) + E(t) ≥ 0 a.e. in Ω, strongly in L2 (Γ) and a.e. in Γ by using the compactness of the trace operator from H 1 (Ω) into L2 (Γ). Let us note that for any t in [kh, (k + 1)h[, ∀v ∈ H 1 (Ω), Z 0 = {∂t u eh v + a(∂t u eh + E h )∇[uh + τ ∂t u eh ]∇v} dx (12) Ω Z + β(∂t u eh )v dσ. Γs

Given that, a(∂t u eht + E)∇v converges towards a(ξ(t) + E)∇v in L2 (Ω)d and ht that ∇[u + τ ∂t u eht ] converges weakly towards ∇[u(t) + τ ξ(t)] in L2 (Ω)d , and thanks to the hypothesis on β, ξ(t) is a solution to the problem: at time t, find w in H 1 (Ω) with w + E(t) ≥ 0 a.e. in Ω , such that for any v in H 1 (Ω), Z Z {wv + a(w + E(t))∇[u(t) + τ w]∇v} dx + β(w)v dσ = 0. (13) Ω

Γs

10

Thanks to Lemma ?? with b = τ a in R+ , κ = τ1 u(t) and the monotone graph x 7→ {β(x)}, the solution ξ(t) is unique and all the sequence ∂t u eh (t) converges 1 towards ξ(t) weakly in H (Ω). The function ξ :]0, T [→ H 1 (Ω) is weakly measurable (indeed, for any f in (H 1 )0 (Ω), t 7→< f, ξ(t) > is the limit of the sequence of measurable functions t 7→< f, ∂t u eh (t) >) and consequently ξ is a measurable function thanks to Pettis theorem (K. Yosida [?] p. 131), since H 1 (Ω) is a separable set. For any v in L2 (0, T, H 1 (Ω)), (∂t u eh (t), v(t)) converges a.e. in ]0, T [ towards h (ξ(t), v(t)). Moreover, |(∂t u e (t), v(t))| ≤ C||v(t)||H 1 (Ω) a.e. since the sequence (∂t u eh )h is bounded in L∞ (0, T, H 1 (Ω)). Thus, the weak convergence in L2 (0, T, H 1 (Ω)) of ∂t u eh towards ξ can be proved. And one gets that ξ = ∂t u. At last, for t a.e. in ]0, T [, passing to limits in (??) leads to: for any v in L2 (0, T, H 1 (Ω)) Z Z {∂t uv + a(∂t u + E)∇[u + ε∂t u]∇v} dxdt + β(∂t u)v dσdt = 0 (14) Q

]0,T [×Γs

and to the existence of a solution.

1.5



The multivoque case

Let us prove in this section that if β is the maximal monotone graph presented in the introduction, then Proposition 6 There exists u in W 1,∞ (0, T, H 1 (Ω)) such that for any v in H 1 (Ω) and for a.e. t in ]0, T [, Z Z 0 ∈ {∂t uv + a(∂t u + E)∇[u + τ ∂t u]∇v} dx + β(∂t u)v dσ. (15) Ω

Γs

Proof. Following J.-L. Lions idea in [?] p.422 et passim, one denotes by βn = max(−n, min(fn , n)). Associated with this sequence, one has a sequence (un ) of solutions to (??). Of course, this sequence is bounded in H 1 (0, T ; H 1 (Ω)) and the same kind of demonstration can be done in order to prove the existence of a solution to (??). Let us considering the notations of the previous section. Then, for any t in Z and for the sub-sequence indexed by nt such that ∂t unt * ξ(t) in H 1 (Ω), strongly in L2 (Γ) and a.e. in Γ, one has for any v in H 1 (Ω), Z Z {∂t unt v + a(∂t unt + E)∇[unt + τ ∂t unt ]∇v} dx + βnt (∂t unt )v dσ = 0. Ω

Γs

Therefore, one gets that βnt (∂t unt ) converges weakly in L2 (Γs ) towards an element denoted by φ. The same kind of demonstration than the one given p. 424 by J.-L. Lions [?] leads to φ = β −1 (ξ(t)) and proves that ξ(t) is a 11

solution to the problem: at time t, find w in H 1 (Ω) with w + E(t) ≥ 0 a.e. in Ω, such that for any v in H 1 (Ω), Z Z {wv + a(w + E(t))∇[u(t) + τ w]∇v} dx + β(w)v dσ 3 0. (16) Ω

Γs

Then, one gets a result of existence of a solution in the same way, according to the proposition ??.  Now, we examine more carefully the sets of points at which equation (??) degenerates, i.e. the Ld -measurable sets Ns = {x ∈ Ω, ∂t u(s) + E = 0}, s ∈ ]0, T [, when the function a is smooth enough in the neighbourhood of 0+ . We will see that for a.e. s in ]0, T [, Ns , the sets where the weathering rate E is maximum, are Ld - negligible if and only if E > 0. In this sense, the case E = 0 (sedimentation process) is singular. Proposition 7 Assume2 that lim

x→0+

a(x) x

< +∞ then:

if E = 0, u = u0 is the unique solution, else, in the relevant case, E > 0, for s a.e. in ]0, T [, Ln {∂t u(s) + E = 0} = 0, i.e. ∂t u + E > 0 a.e. in Ω. Moreover, if β(−E)∩]−∞, 0[6= ∅, for s a.e. in ]0, T [, Hn−1 {∂t u(s)+E = 0} = 0, i.e. ∂t u + E > 0 a.e. in Γ. Proof. Let us consider the test function v = p (∂t u+E) in equation (??) where for any x, p (x) = max(0, min( x , 1)). Since β is monotone, for any θ ∈ β(−E), one gets, for a.e. t in ]0, T [, Z Z E p (∂t u + E) dx − θp (∂t u + E) dσ Ω Γs Z ≥ {(∂t u + E)p (∂t u + E) + a(∂t u + E)∇[u + τ ∂t u]∇p (∂t u + E)} dx. Ω

Thanks to the hypothesis on a and the Saks lemma, it comes Z Z Z + + E sgn0 (∂t u + E) dx − θsgn0 (∂t u + E) dσ ≥ |∂t u + E| dx. Ω

Γs

Ω

On the one hand, if E = 0, since 0 ∈ β(0), one gets |∂t u| = 0 a.e. and the only solution is the trivial one: u(t, x) = u0 (x). On the other hand, if E > 0, test function v = 1 − p (∂t u + E) in equation (??) leads to Z Z + E {1 − sgn0 (∂t u + E)} dx + θ(sgn+ 0 (∂t u + E) − 1) dσ Ω Γs Z ≤ ∂t u + E − |∂t u + E| dx. Ω 2 Note

that if a(x) ∼ cxα then the hypothesis on a ◦ A−1 imply that α ≥ 1. 0+

12

Since θ is nonpositive and ∂t u+E nonnegative, it comes E.Ln {∂t u+E = 0} ≤ 0 and the result holds. Moreover, if θ can be chosen negative, the last part of the property holds too. 

1.6

Localization properties of solutions. Finite speed of propagation

In this section we analyze localization properties of weak solutions of considered problem following to ideas of the book S. N. Antonsev et al. [?]. We start consider local properties of solutions to generalized equation ∂t u − ∂x (a(∂t u, t)) (∂x u + τ ∂xt u)) = 0 in Q =]0, T [×Ω, Ω =]0, 1[.

(17)

We assume that there exists a local weak solution of equation (??) satisfying to the boundary and initial conditions a(∂t u, t)) (∂x u + τ ∂xt u)|x=0 = 0, u(x, 0) = u0 (x), x ∈ [0, 1],

(18)

u0 (x) = 0, x ∈ [0, ρ0 ], ρ0 < 1.

(19)

where Assume that the function a(r, t)) satisfies to conditions: ∀(r, t) ∈ R × R+ Z r Z r ds ds 1 2−α , a(r, t) (20) |r| ≤r 0 ≤ a, ≤ Ca |r|, Ca 0 a(s, t) 0 a(s, t) with some constants Ca and α ∈]0, 1[. Remark 2 It is easy to verify that the function a(s) = |s|α , 0 < α < 1, satisfies (??).

1.6.1

Energy relations

Let us introduce the function  1  k(x − ρ) ψk (ρ, x) =  0

for x > ρ + 1/k, for x ∈ [ρ, ρ + 1/k], for x < ρ, k = 1, 2..

R  ∂ u ds Substituting v = ψk 0 t a(s,t) into integral identity (Remark ??.3) as the test -function, we have ! # Z 1" Z ∂t u Z ρ+1/k ds ∂t u + (∂x u + τ ∂xt u) ∂xt u ψk dx + k I(s, t)ds = 0, a(s, t) 0 0 ρ (21)

13

where ∂t u(x,t)

Z I(x, t) = a(∂t u(x, t), t) 0

ds a(s, t)

! (∂x u(x, t) + τ ∂xt u(x, t)) .

It follows from the Lebesgue theorem that for a.e. ρ ∈ (0, ρ0 ) Z

ρ+1/k

lim k

k→∞

I(s, t)ds = I(ρ, t). ρ

Passing into (??) to the limit on k → +∞, we come to the energy relation ! # Z ∂t u Z ρ" ds ∂t u + (∂x u + τ ∂xt u) ∂xt u dx = −I(ρ, t). a(s, t) 0 0 Integrating last relation with respect to t, and repeating of arguments of Remark ??, we obtain the inequalities ! # Z ρ" Z ∂t u Z ρ ds τ 1 2 ∂t u + |∂xt u| dx = |I| + |∂x u|2 dx,(22) a(s, t) 2 2τ 0 0 0 ! # Z tZ ρ" Z ∂t u ds 2 + τ (∂xt u) dxdt (23) ∂t u a(s, t) 0 0 0 Z Z t 1 ρ (∂x u)2 dx ≤ |I(ρ, s)|ds. + 2 0 0 Multiplying (??) by τ2 , adding to (??) and using (??) and (??), we come to the energy inequality Z tZ ρ Z ρ   2−α 2−α |∂t u| + (∂xt u)2 dxdt + |∂t u| + |∂x u(x, t)|2 + (∂xt u)2 dx 0

0

0

(24) 

Z

≤ C(τ ) J(ρ, t) +

t

 J(ρ, s)ds ,

0

where |I(ρ, t)| ≤ CJ(ρ, t) = C|∂t u(ρ, t)| (|∂x u(ρ, t)| + |∂xt u(ρ, t)|) = C (J1 + J2 ) . (25) We introduce the energy functions Z tZ ρ  2−α E(ρ, t) ≡ |∂t u| + (∂xt u)2 dxdt (26) 0

Z b(ρ, t) ≡

ρ



0 2−α

|∂t u|

 + |∂x u(x, t)|2 + (∂xt u)2 dx

0

14

(27)

satisfying Z t

∂E(ρ, t) Eρ = = ∂ρ

2−α

|∂t u|

 + (∂xt u)2 dt > 0,

0

∂( sup E(ρ, t))  0≤t≤t0 + (∂xt u)2 dt = , ∂ρ 0≤t≤t0 0  ∂b(ρ, t)  2−α bρ = = |∂t u| + |∂x u(x, t)|2 + (∂xt u)2 > 0. ∂ρ t0

Z



sup Eρ =

1.6.2

2−α

|∂t u|

Global estimate for energy functions

For ρ = 1 the inequalities (??), (??) take the forms ! # Z ∂t u Z 1" Z 1 ds τ 1 2 ∂t u |∂x u|2 dx, + |∂xt u| dx ≤ a(s, t) 2 2τ 0 0 0 Z tZ

"

1

ds a(s, t)

!

1 (∂x u) dx ≤ 2

Z

Z

∂t u

∂t u 0

0

0

1 + 2

1

Z

2

0

# 2

+ τ (∂xt u)

dxdt

1

(∂x u0 )2 dx,

0

and give to us the estimate Z E(ρ, t) + b(ρ, t) ≤ C(Ca , τ )

1

(∂x u0 )2 = K.

(28)

0

1.6.3

Ordinary differential inequality for energy functions

In notations (??) and (??) the inequality (??) takes the form   Z t E(ρ, t) + b(ρ, t) ≤ C J1 + J2 + (J1 + J2 )ds .

(29)

0

Next we would use the following inequalities γ

Z

|∂t u(x, t)| ≤ γ

x γ−1

|∂t u|

ρ

Z

2−α

|∂xt u|dx ≤ γ

|∂t u|

0

 21 Z dx

0

ρ 2

|∂xt u| dx

 12 ,

0

(30) ρ

Z ≤C

 γ1
1 2−α 2 = C (b) γ , |∂t u| + |∂xt u| dx

0

with γ = (4 − α)/2 and Z |∂x u(x, t)| ≤

t

|∂tx u|dt ≤ t

1 2

Z

0

2

|∂xt u| dt 0

15

t

 12

1

1

≤ t 2 (Eρ ) 2 .

(31)

Remind that ∂x u(ρ, 0) = 0, if ρ ≤ ρ0 . Applying (??) and (??), we can evaluate J1 and J2 in the following way 1

1

1

J1 = |∂t u||∂x u| ≤ Ct 2 (b) γ (Eρ ) 2 , 1

(32)

1

J2 = |∂t u||∂xt u| ≤ C (b) γ (bρ ) 2 .

(33)

 1  1 1 1 1 J = J1 + J2 ≤ C t 2 (b) γ (Eρ ) 2 + (b) γ (bρ ) 2 .

(34)

Finally

Next using (??) and (??), we obtain the estimates Z

t

t

Z J1 ds

t

≤ C J2 ds

0

|∂t u|2 dt

≤

0

|∂x u| ds

(35)

0

 12  12 Z t  21 Z t  12 Z t 2 1 γ 2 b ds t (Eρ ) ds b ds Eρ ds ≤ Ct ,

t

Z

2

2 γ

0 t

 12

t

|∂t u| ds 0

0

Z

Z

2

|∂t u||∂x u|ds ≤

=

0

 21 Z

t

Z

0

 21 Z

t

|∂xt u|2 dt

0

 21 (36)

0

0

Z t Z

2−α

≤

|∂t u| 0

2

+ |∂xt u|




! 21

 γ2

ρ

dx

Z

1

t

(Eρ ) 2 ≤

dt

2

 21

(b) γ dt

0

0

Joining (??), (??)-(??), we come to E(ρ, t) + b(ρ, t) " 1 2

1 γ

1 2

1 γ

1 2

≤ C t (b) (Eρ ) + (b) (bρ ) +

t

1 2

(37)  21

t

Z

Eρ ds

+ (Eρ )

1 2

! Z

t

2 γ

 12 #

b ds

0

.

0

Next we will be considering last inequality for ∈ [0, t0 ] with some t0 ∈ (0, 1) which will be chosen later. Applying the Young inequality with γ > 1, in the following way  1  γ 1 1 1 1 t 2 (b) γ (Eρ ) 2 + (b) γ (bρ ) 2 ≤ εb + C(ε)(Eρ + bρ ) 2(γ−1) , ε ∈ (0, 1), Z

t

2 γ

 12

(b) dt

1 2

(Eρ ) ≤ C

(Eρ )

γ 2(γ−1)

Z

t

+

 γ2 !

2 γ

(b) dt

.

0

0

and choosing ε in a reasonable way, we reduce (??) to next one E(ρ, t) + b(ρ, t) ≤C

(Eρ + bρ )

γ 2(γ−1)

+t

1 2

Z

t

2 γ

 12 Z

b ds 0

 21 Eρ ds

0

16

(38)

t

Z + 0

t

 γ2 ! (b) dt . 2 γ

1

(Eρ ) 2

Assume that sup0≤t≤t0 b(ρ, t) = b(ρ, δ), δ > 0, as for as b(ρ, 0) = 0. We introduce the function δ

Z

Z

ρ

E(ρ) = sup0≤t≤δ E(ρ, t) =



2−α

|∂t u|

 + (∂xt u)2 dxdt,

0

0

which satisfies sup0≤t≤δ

∂(E(ρ, t)) = ∂ρ

δ

Z



2−α

|∂t u|

0

 ∂(E(ρ)) + (∂xt u)2 dt = . ∂ρ

Putting into (??) t = δ, we come to E(ρ) + b(ρ, δ) γ



3

1

≤ C (E ρ + bρ (ρ, δ)) 2(γ−1) + δ 2 b γ (ρ, δ) E ρ


12

 γ + δ 2 b(ρ, δ) .

Applying the Young inequality 3

1

δ 2 b γ (ρ, δ) E ρ


12

≤

γ 1 b(ρ, δ) + C(E ρ )) 2(γ−1) , 2

and choosing δ > 0 in a reasonable way, we come finally to ordinary differential inequality Λν ≤ CΛρ , with Λ = E(ρ) + b(ρ, δ), ν(α) =

2(γ − 1) 2(2 − α) = < 1 ⇔ γ < 2 ⇔ 0 < α. γ 4−α

Integrating last inequality over (ρ, ρ0 ), we come to the desired estimate Λ1−ν (ρ, δ) =

1.6.4



1−ν 1 − ν(α) . sup E(ρ, s) + b(ρ, δ) ≤ Λ1−ν (ρ0 , δ) − (ρ0 − ρ) C(α, τ ) 0≤s≤δ (39)

Finite speed of propagation

Proposition 8 Let u be a weak solution of problem (??), (??) and (?? ) and condition (??) be fulfilled with 0 < α < 1. Then there exist positive numbers δ > 0 and ρ ∈ (0, ρ0 ) such that u(x, t) = 0, 0 ≤ x ≤ ρ, 0 ≤ t ≤ δ, if the global energy E(1, T ) + b(1, T ), satisfying to (??), is small.

17

Proof. According to (??), we have the inequality 1−ν

Λ1−ν (ρ, δ) ≤ (2K)

− (ρ0 − ρ)

1−ν . C

It follows that there exists ρ > 0 such that Λ1−ν (ρ, δ) ≤ 0,

i.e. u(x, t) = 0, 0 ≤ x ≤ ρ, 0 ≤ t ≤ δ

if 1−ν

(2K)

< ρ0

1−ν . C 

1.6.5

Elliptic equation in semi-discretized degenerated processes

In this section we consider a local weak solution of the problem (see Proposition ??) Z Z u − u0 u − u0 u − u0 u − u0 v + a( + E)∇[u + τ ]∇v} dx + β( )v dσ = 0. { h h h h Γs Ω (40) 1 0 Introducing w = uα −u , we reduce the problem to: find w in H (Ω) such that, h ∀v ∈ H 1 (Ω), Z Z {wv + a(w + E)∇[(h + τ )w + u0 ]∇v}} dx + β(w)v dσ = 0. (41) Ω

Γs

For local solutions of equation (??), we can prove the following results: Assume that 1 α |s| ≤ a(s) ≤ C|s|α , 0 < α < 1, C = const > 0, C

(42)

E = 0.

(43)

Let us introduce the notation

Z b(ρ, w) ≡

Bρ (x0 ) = {x ∈ Ω : |x − x0 | < ρ} , Sρ = ∂Bρ (x0 ) Z (w2 + (h + τ )a(w)|∇w|2 )dx, bρ = w2 + (h + τ )a(w)|∇w|2 )dx.

Bρ

Sρ

(44) Let us consider a local weak solution of equation (??) into the ball Bρ∗ (x0 ) ⊂ Ω ⊂ Rn such that Z ∗ b(ρ , w) ≡ (w2 + (h + τ )a(w)|∇w|2 )dx ≤ K. (45) Bρ∗

18

Proposition 9 Let u be a weak solution to (??) and the conditions (??), (??) are fulfilled. Then: a) Assume that u0 ≡ 0 in Bρ0 (x0 ), ρ0 < ρ∗ . Given an arbitrary point x0 ∈ Ω, u(x) ≡ 0 a.e. in Bρ (x0 )

(46)

with ρ given by the expression ρ ≡ (ρ0 − CK)+ , C = C(α, n). b) Assume that u0 ≡ 0 in Bρ0 (x0 ) satisfying Z µ |∇u0 |γ dx ≤ ε (ρ − ρ0 )+ , ρ ∈ [ρ0 , ρ∗ ],

(47)

(48)

Bρ

with some positive constants γ, µ, ε. Then there exist positive constants K∗ , ε∗ such that once K ≤ K ∗ , ε ≤ ε∗ , (49) any weak solution of equation (??) possesses the property u(x) ≡ 0 a.e. in Bρ0 (x0 ).

(50)

Remark 3 If in (??) ρ = 0, then the above statement offers no information on the vanishing set of w. Nonetheless, if the total energy K is small enough, then always ρ > 0 and the vanishing set of w is not empty.

1.6.6

Finite speed of propagation of solution to equation with τ = 0

We consider special (traveling waves) solutions of equation ∂t u = ∂x [a(∂t u) ∂x u] , x ∈ Ω =] − 1, 1[

(51)

in the form: for a given positive λ, u(x, t) = u(λx + t) = u(ξ),

ξ ≥ −1.

(52)

Substituting (??) into (??), we obtain the ordinary differential equation 0

u0 = λ2 (a(u0 ) u0 ) ,

(53)

or first order differential equation with separated variables u = λ2 a(u0 ) u0 + C.

(54)

If a(s) = 0 in ] − ∞, 0[ and a(s) > 0 else with a(s) = 1 if s ≥ s0 > 0. Then, if one notes f (ξ) = ξa(ξ), u = λ2 f (u0 ) + C. 19

f −1 is a graph, f −1 is a function in ]0, ∞[ and f −1 (0) =] − ∞, 0]. Let us consider β = f −1 in ]0, +∞[ and β(s) = 0 if s ≤ 0, let us assume that R R y−C 2 dx dx exists and denote by B(y) = 0 λ2 λβ(x) . 0+ β(x) Since β(x) > 0 for any x > 0 and has linear growth at infinity, B is an increasing continuous function on [C, +∞[, derivable in ]C, +∞[, with lim B(y) = +∞. Thus, B −1 has got the same properties on (0, +∞). Moreover, for any ξ > ξ0 , y 0 (ξ) := (B −1 )0 (ξ−ξ0 ) =

1 B 0 (B −1 (ξ

− ξ0 ))

= β(

∞

y(ξ) − C B −1 (ξ − ξ0 ) − C ) = β( ). 2 λ λ2

B −1 is a continuous function with B −1 (0) = C, thus (B −1 )0 is also a continuous function and y(ξ0 ) = C

and

y 0 (ξ0 ) = β(

y(ξ0 ) − C ) = 0. λ2

y is a global classical solution (see A. F. Filippov [?]) to y 0 = β(

y−C ) in R λ2

with

y(ξ0 ) = C

(55)

different from the trivial solution y(ξ) = C and satisfying y(ξ) = C for any ξ ≤ ξ0 . Traveling-waves solutions. Consider in the sequel C = ξ0 = 0 and y a non-trivial solution to y y 0 = β( 2 ) in R with y(0) = 0. (56) λ Note that y 0 ≥ 0 and y ≥ 0 then, λ2 f (y 0 ) = y. Let us consider u(t, x) = y(λx + t).

and

∂t u(t, x) = y 0 (λx + t), ∂x u = λy 0 (λx + t) [a(∂t u)∂x u](t, x) = λ[a(y 0 )y 0 ](t + λx) = λf (y 0 )(t + λx),

and then, ∂x [a(∂t u)∂x u](t, x) = λ2 [f (y 0 )]0 (t + λx) = ∂t u(t, x). The initial condition is given by u0 (x) = y(λx) and the boundary conditions u(t, −1) = y(t − λ) and

u(t, 1) = y(t + λ).

At last, note that u0 (x) = 0 in ] − 1, 0] and that u possesses the property of finite speed of propagation (from nonzero disturbances) in the following sense : u(x, t) = 0,

t ≤ −x < 1. λ

20

(57)

1.7

Conclusion and open problems

A solution to the problem related to a strictly positive maximum erosion rate E (really the relevant case)  ∂t u − τ ∆A(∂t u + E) − div(a(∂t u + E)∇u) = 0 in Q,      ∂t u + E ≥ 0 in Q, −τ ∂n A(∂t u + E) − a(∂t u + E)∂n u = 0 in ]0, T [×Γe ,    −τ ∂n A(∂t u + E) − a(∂t u + E)∂n u ∈ β(∂t u) in ]0, T [×Γs ,   u|t=0 = u0 in Ω. has been found in Lip([0, T ], H 1 (Ω)). In order to conclude that this problem is well-posed in the sense of Hadamard, one still has to prove that such a solution is unique. This is still an open problem, mainly due to a behaviour of hysteresis type of the equation. Let us cite a recent paper of Z. Wang and al. [?] where the uniqueness of the solution to a similar equation has been proved. The equation is posed in the one-dimension space case, and the method is based on a Holmgren approach. The above solution is a solution to a perturbation of the real problem since a has to be the graph H of the Heaviside function in order to satisfy the initial law of state. This problem is open too, and to our knowledge , the realistic model appears as a non standard free boundary problem for flux limiter λ by imposing the original unilateral condition 1 − λ ≥ 0, ∂t u + E ≥ 0, (1 − λ) (∂t u + E) = 0 a.e. in ]0, T [×Ω, via the inclusion λ ∈ H(∂t u + E).

References [1] S.N. Antontsev, J.I. D´ıaz and S. Shmarev, ”Energy methods for free boundary problems. Applications to nonlinear PDEs and fluid mechanics.” Progress in Nonlinear Differential Equations and their Applications 48, Basel, Birkh¨ auser, 2002. [2] S. N. Antontsev, G. Gagneux, R. Luce and G. Vallet, A Non Standard Free Boundary Problem Arising From Stratigraphy, Analysis and Applications (to appear). [3] S.N. Antontsev, G. Gagneux and G. Vallet, On some stratigraphic control problems, Prikladnaya Mekhanika Tekhnicheskaja Fisika (Novosibirsk). Vol. 44, No. 6 (2003) 85-94 (in russian). and Journal of Applied Mechanics and Technical Physics (New York). Vol. 44, No. 6 (2003) 821-828. [4] C. Cuesta, C. J. van Duijn and J. Hulshof, Infiltration in porous media with dynamic capillary pressure: Travelling waves. Eur. J. Appl. Math. 11, No. 4 (2000) 381-397. 21

[5] R. Eymard, T. Gallou¨et, Analytical and numerical study of a model of erosion and sedimentation. Siam J. Numer. Anal., 43(6) (2006) 2344-2370. [6] R. Eymard, T. Gallou¨et, D. Granjeon, R. Masson and Q.H. Tran, Multilithology stratigraphic model under maximum erosion rate constraint. Internat. J. Numer. Methods Eng., 60, No. 2 (2004) 527-548. [7] A. F. Filippov, ”Differential equations with discontinuous righthand sides, Mathematics and its Applications” (Soviet series) vol. 18, Kluwer, Dordrecht, 1988. [8] G. Gagneux, R. Luce and G.Vallet, ”A non standard free-boundary problem arising from stratigraphy.” Publication interne du Laboratoire de Math´ematiques Appliqu´ees. UMR-CNRS 5142, Universit´e de Pau. No. 04/33, Pau, 2004. [9] G. Gagneux and G. Vallet, Sur des probl`emes d’asservissements stratigraphiques. ESAIM : Control, optimisation and calculus of variations ” A tribute to Jacques-Louis Lions”, 8 (2002) 715-739. [10] G. Gagneux and G. Vallet, A result of existence for an original convectiondiffusion equation, Revista de la Real Academia de Ciencias, Serie A: Matem´ aticas (RACSAM) 99(1) (2005) 125-131. [11] T. Gallou¨et, Equations satisfaites par des limites de solutions approch´ees, conf´erence pl´eni`ere, 34`eme congr`es d’analyse num´erique, Anglet (Pyr´en´ees Atlantiques), in Canum (2002) 87-96. [12] V. Gervais and R. Masson, Mathematical and numerical analysis of a stratigraphic model. M2AN, Vol. 38, N◦ 4 (2004) 585-611. [13] D. Granjeon, (Institut Fran¸cais du P´etrole) IFP, personal communication (2006). [14] J.-L. Lions, ”Quelques m´ethodes de r´esolution des probl`emes aux limites non lin´eaires.” Dunod, Paris, 1969. [15] G. Vallet, Sur une loi de conservation issue de la g´eologie. C. R. Acad. Sci. Paris, S´er. I, 337 (2003) 559-564. [16] Z. Wang and J. Yin, Uniqueness of solutions to a viscous diffusion equation, Applied Math. Letters, 17(12) (2004) 1317-1322. [17] K. Yosida, ”Functional analysis.” Springer-Verlag. XI, Berlin, 1965.

22