The free boundary represents the region separating the wet and the dry part of the porous medium

THE EVOLUTION DAM PROBLEM FOR NONLINEAR DARCY’S LAW AND DIRICHLET BOUNDARY CONDITIONS *

A. Lyaghfouri

Abstract:In this paper, we study a time dependent dam problem modeling a non- linear fluid flow through a homogeneous or nonhomogeneous porous medium governed by a nonlinear Darcy’s law. We prove existence and uniqueness of a weak solution.

Introduction

The dam problem consists of finding the flow region and the pressure of fluid flow through a porous medium Ω under gravity. The free boundary represents the region separating the wet and the dry part of the porous medium. Assuming the flow governed by a linear Darcy’s law and taking Dirichlet boundary conditions on some part of the boundary, this problem has been widely studied from several points of view both for the stationary and the evolutionary case.

For the stationary case, the first results are due to C. Baiocchi ([5], [6]) who solved the case of rectangular dams by introducing the so called Baiocchi’s trans- formation, which leads him to consider problems of variational inequalities. Then he established existence and uniqueness results. Although this method is not adaptable for the general case, many authors ([7], [8], [16], [23]), used same techniques to treat questions related to heterogeneous or three dimensional rect- angular dams.

Few years after, the steady problem has been studied in the general case by H.W. Alt ([2], [3], [4]), H. Br´ezis, D. Kinderlehrer and G. Stampacchia [9], J. Carrillo and M. Chipot [15]. An existence theorem has been proved and the uniqueness of the solution established up to a certain class of disturbing functions.

The regularity of the free boundary was also investigated.

Received: May 21, 1997; Revised: June 11, 1997.

AMS Mathematical Subject Classification: 35R35, 76S05.

* Supported by the European Science Foundation Scientific Programme on the Mathematical Treatment of Free Boundary Problems.

The evolution dam problem has been solved first by A. Torelli ([24], [25]) in the case of a rectangular domain. He used a similar transformation to the Baiocchi’s one, which allowed him to reduce the problem to a quasi-variational inequality problem. He obtained, in this way, results of existence, uniqueness and regularity of his solution. Unfortunately, this method is not adapted for the general case.

Later, it was G. Gilardi [19] who proved an existence theorem for a weak formulation of the evolution dam problem when the porous medium is assumed to be a general Lipschitz bounded domain of Rⁿ. In [18] E. Dibenedetto and A. Friedman proved an existence theorem in a general way by an other method, both for compressible or incompressible flow. Moreover they proved uniqueness for rectangular dams. The question of uniqueness of the solution, in its generality, remains open until solved by J. Carrillo [14].

In this study, we consider an incompressible fluid flow governed by a gener- alized nonlinear Darcy’s law relating the velocity v of the fluid to its pressure p by:

v=−A(x,∇(p+x_n))

where A is a function defined in Ω×Rⁿ and x = (x₁, ..., x_n) denotes points in Rⁿ.

The prime example of nonlinear Darcy’s laws for a homogeneous porous medium (see [17]) corresponds to theq-Laplacian: A(x, ξ) =|ξ|^q−2ξ. For hetero- geneous media we have the following measurable perturbations of theq-Laplacian:

A(x, ξ) =|a(x)·ξ|^q−2a(x)·ξ, wherea(x) is a measurable positive definite matrix representing the permeability of the medium at x. Note that when q = 2, we rediscover the well known linear Darcy Law.

In addition, we would like to consider a model of Dirichlet boundary condi- tions. The paper is organized as follows: In section 1, we begin by transforming the problem usually stated in terms of the pressure function into a problem for the hydrostatic headu=p+xn. The dry part is described by a bounded func- tion g. Then we give a weaker formulation to our problem. In section 2 and 3 we prove an existence theorem by means of regularization and by using the Tychonoff fixed point theorem. In section 4, we prove some properties of the solutions. In particular for any solution (u, g),u is bounded and A-subharmonic and g is continuous in time variable. In section 5, we assume that q ≤ n+ 1, A(x, ξ) =A(ξ) andA(e)·ν ≤0 on the bottom of the dam. Then from section 4, we derive a monotonicity property for g. Making use of this result, we prove a comparison theorem and with the help of the continuity of g we deduce the uniqueness of the solution.

1 – Statement of the problem

The dam is a bounded locally Lipschitz domain Ω inRⁿ(n≥2) (see Figure 1).

The boundary Γ of Ω is divided in two parts: an impervious part Γ₁and a pervious one Γ₂ which is assumed to be nonempty and relatively open in Γ. Let now T be a positive number, Q = Ω×(0, T) and ϕ a nonnegative Lipschitz function defined inQ. We define:

Σ₁ = Γ₁×(0, T), Σ₂ = Γ₂×(0, T), Σ₃ = (Γ₂×(0, T))∩ {ϕ >0}

and Σ₄= (Γ₂×(0, T))∩ {ϕ= 0}.

Fig. 1

We shall be interested with the problem of finding the pressure p and the saturationχ of the fluid. For convenience, we set: ψ=ϕ+x_n,u =p+x_n and g = 1−χ. Starting from the nonlinear Darcy’s law, the mass conservation law and taking a Dirichlet boundary condition on Σ₂, the flow is governed by the following equations:

(1.1)



































i) u≥x_n, 0≤g≤1, g(u−x_n) = 0 in Q ii) div^³A(x,∇u)−gA(x, e)^´+gt= 0 in Q iii) u=ψ on Σ₂

iv) g(·,0) =g₀ in Ω

v) ^³A(x,∇u)−gA(x, e)^´·ν= 0 on Σ₁ vi) ^³A(x,∇u)−gA(x, e)^´·ν≤0 on Σ₄

whereeis the vertical unit vector ofRⁿ, i.e.e= (0,1) with 0∈Rⁿ⁻¹,ν denoting the outward unit normal to∂Ω, g0 is a given function satisfying 0≤g0 ≤1 and A: Ω×Rⁿ −→ Rⁿ is a mapping that satisfies the following assumptions with some constantsq >1 and 0< α≤β <∞:

(1.2)







the function x 7−→ A(x, ξ) is measurable∀ξ ∈Rⁿ, and the function ξ 7−→ A(x, ξ) is continuous for a.e x∈Ω, for allξ ∈Rⁿ and a.e. x∈Ω

A(x, ξ)·ξ ≥α|ξ|^q , (1.3)

|A(x, ξ)| ≤β|ξ|^q−1 , (1.4)

for allξ, ζ ∈Rⁿ such thatξ6=ζ and a.e. x∈Ω

³A(x, ξ)− A(x, ζ)^´·(ξ−ζ)>0 , (1.5)

∃r >1 : div(A(x, e))∈L^r(Ω). (1.6)

The condition (1.1) i) means that we look for a nonnegative pressure p and g(·, t) characterises the wet region Ω(t) at time t. (1.1) iii) means that the trace pressure at the bottoms of fluid reservoirs is equal to the one of the fluid and equal to the atmospheric one when the boundary of Ω is in contact with the air.

(1.1) iv) is an initial data. (1.1) v) and (1.1) vi) are due to the fact that the flux of fluid vanishes on Σ₁ (since Γ₁ is impervious) and is nonnegative on Σ₄ where the fluid is free to exit from our porous medium.

From the strong formulation (1.1), we are led to consider the following weak formulation:

(P)































Find (u, g)∈L^q(0, T, W^1,q(Ω))×L^∞(Q) such that : i) u≥x_n, 0≤g≤1, g(u−x_n) = 0 a.e. in Q; ii) u=ψ on Σ₂ ;

iii) Z

Q

³A(x,∇u)−gA(x, e)^´· ∇ξ+g ξ_tdx dt+ Z

Ω

g₀(x)ξ(x,0)dx≤0

∀ξ∈W^1,q(Q), ξ= 0 on Σ3, ξ≥0 on Σ4, ξ(x, T) = 0 a.e. in Ω .

2 – A regularized problem

We first introduce the following approximated problem:

(Pε)



























Find uε∈W^1,q(Q) such that : uε=ψ on Σ2 , Z

QA(x,∇uε)·∇ξ+ε|uεt|^q−2u_εt·ξt+G_ε(u_ε)^³ξ_t− A(x, e)·∇ξ^´dx dt=

= Z

ΩGε(uε(x, T))ξ(x, T)dx− Z

Ωg0(x)ξ(x,0)dx

∀ξ∈W^1,q(Q), ξ= 0 on Σ₂ , whereG_ε: L^q(Q) (resp.L^q(Ω)) −→ L^∞(Q) (resp.L^∞(Ω)) is defined by

(2.1) G_ε(v) =











0 if v−x_n≥ε

1−(v−x_n)/ε if 0≤v−x_n≤ε

1 if v−x_n≤0.

Then we have:

Theorem 2.1. Assume thatϕis a nonnegative Lipschitz continuous function and thatAsatisfies (1.2)–(1.5). Then, there exists a solutionu_ε of (P_ε).

Proof: It will be done in three steps:

Step 1: We define

V=ⁿv∈W^1,q(Q) / v= 0 on Σ₂^o and K=ⁿv∈W^1,q(Q) / v=ψon Σ₂^o. For u∈K, we consider the map:

A(u) : W^1,q(Q)−→R, ξ7−→ hA(u), ξi= Z

QA(x,∇u)·∇ξ+ε|ut|^q−2utξtdx dt . Then the operatorA defined byA: u∈K 7−→A(u), satisfies:

Lemma 2.2. If we denote by(W^1,q(Q))⁰the dual space ofW^1,q(Q), we have i) For everyu∈K,A(u)∈(W^1,q(Q))⁰;

ii) A is continuous from K into(W^1,q(Q))⁰; iii) A is monotone and coercive.

Proof: (see [20] for example).

Step 2: Forv∈W^1,q(Q), we consider the map: f_v: W^1,q(Q)−→R, ξ 7−→

Z

Q

G_ε(v)^³A(x, e)·∇ξ−ξt

´dx dt+ Z

Ω

G_ε(v(x, T))ξ(x, T)−g₀(x)ξ(x,0)dx . It is clear thatf_v ∈(W^1,q(Q))⁰. Using Lemma 2.2, we deduce (see [20]) that for everyv∈W^1,q(Q) there exists a unique uε solution of the variational problem (2.2) u_ε ∈K , hAuε, ξi=hfv, ξi ∀ξ∈V .

Step 3: Now, let us consider the map F_ε defined by: F_ε: W^1,q(Q) −→ K, v7−→u_ε. Let us denote by B(0, R(ε)) the closed ball inW^1,q(Q) of center 0 and radiusR(ε). Then we have:

Lemma 2.3.

i) ∃R(ε)>0/ F_ε(B(0, R(ε))) ⊂ B(0, R(ε));

ii) F_ε: B(0, R(ε))−→B(0, R(ε))is weakly continuous.

Proof: i) Note that u_ε−ψ is a suitable test function to (2.2), so:

(2.3)

Z

QA(x,∇uε)· ∇uε+ε|uεt|^qdx dt=

= Z

QA(x,∇uε)· ∇ψ+ε|uεt|^q−2u_εt·ψ_tdx dt +

Z

Q

G_ε(v)A(x, e)· ∇(uε−ψ)dx dt

− Z

QG_ε(v) (u_ε−ψ)_tdx dt− Z

Ωg₀(x) (u_ε−ψ)(x,0)dx +

Z

ΩGε(v(x, T)) (uε−ψ)(x, T)dx .

Using (1.3), (1.4), (2.1), (2.3) and H¨older’s inequality, we get for some con- stantsc_i

min(α, ε) Z

Q|∇uε|^q+|uεt|^qdx dt≤c1|uε|^q−1_1,q +c2|uε|1,q+c3 .

By Poincar´e’s Inequality, this leads for some constantsc⁰_ito|uε|^q_1,q≤c⁰₁|uε|^q−1_1,q + c⁰₂|uε|1,q+c⁰₃ from which we deduce that: |uε|1,q ≤R(ε) where R(ε) is some con- stant depending onε. So we have: F_ε(B(0, R(ε)))⊂B(0, R(ε)). More precisely, we have proved that: F_ε(W^1,q(Q))⊂B(0, R(ε)).

ii) Let (v_i)_i∈I be a generalized sequence in C =B(0, R(ε)) which converges tov in C weakly.

Set uⁱ_ε = F_ε(v_i) and u_ε = F_ε(v). We want to prove that (uⁱ_ε)_i∈I converges to uε weakly in C. Since C is compact with respect to the weak topology, it is enough to show that (uⁱ_ε)_i∈I has u_ε as the unique limit point for the weak topology inC. So letube a weak limit point for (uⁱ_ε)i∈I inC. Using the compact imbedding: W^1,q(Q),→L^q(Q), one can construct a sequence (uⁱ_ε^k)_k∈N such that {ik/ k∈N} ⊂ I, uⁱ_ε^k * u weakly in W^1,q(Q) and uⁱ_ε^k → u strongly inL^q(Q).

Choose uⁱ_ε^k −u_ε as a suitable test function for (2.2) written for uⁱ_ε^k and u_ε. Subtract the equations, so that

(2.4) Z

Q

³A(x,∇uⁱ_ε^k)− A(x,∇uε)^´· ∇(uⁱ_ε^k −u_ε) +

+ε^³|uⁱ_εt^k|^q−2uⁱ_εt^k − |uεt|^q−2u_εt^´·(uⁱ_ε^k−u_ε)_tdx dt=

= Z

Q

³Gε(vik)−Gε(v)^{´ ³}A(x, e)· ∇(uⁱ_ε^k−uε)−(uⁱ_ε^k−uε)t

´dx dt

+ Z

Ω

³G_ε(v_ik(x, T))−G_ε(v(x, T))^´(uⁱ_ε^k−u_ε)(x, T)dx .

Now we have by (1.4), (2.1), H¨older’s inequality and the fact that|uⁱ_ε^k−uε|1,q ≤ 2R(ε):

(2.5)

¯

¯

¯

¯ Z

Q

³G_ε(v_ik)−G_ε(v)^{´ ³}A(x, e)· ∇(uⁱ_ε^k−u_ε)−(uⁱ_ε^k−u_ε)_t^´dx dt

¯

¯

¯

¯≤

≤c1(ε)· |vik−v|q . (2.6)

¯

¯

¯

¯ Z

Ω

³G_ε(v_ik(x, T))−G_ε(v(x, T))^´(uⁱ_ε^k−u_ε)(x, T)dx

¯

¯

¯

¯≤

≤c2(ε)·^¯¯_¯(vik−v) (·, T)^¯¯_¯

q . Now due to (1.5), (2.4)–(2.6) and the compact imbeddings: W^1,q(Q),→L^q(Q) andW^1−1q,q(Ω× {T}),→L^q(Ω× {T}), we get:

(2.7)

k→+∞lim Z

Q

³A(x,∇uⁱ_ε^k)− A(x,∇uε)^´· ∇(uⁱ_ε^k −uε)dx dt= 0,

k→+∞lim Z

Q

³|uⁱ_εt^k|^q−2uⁱ_εt^k − |uεt|^q−2u_εt^´·(uⁱ_ε^k−u_ε)_tdx dt= 0 .

Using (2.7), we deduce that (see [11]) there exists a subsequence of (uⁱ_ε^k) also denoted by (uⁱ_ε^k) such that ∇uⁱ_ε^k → ∇uε and uⁱ_εt^k → u_εt a.e. in Q. Taking into

account the boundedness of (uⁱ_ε^k) in W^1,q(Q), we get: ∇uⁱ_ε^k * ∇uε weakly in L^q(Q) anduⁱ_εt^k * uεt weakly inL^q(Q). So we have uε =u and uε is the unique weak limit point of (uⁱ_ε) in C. Thus uⁱ_ε = F_ε(vⁱ) * u_ε = F_ε(v) weakly in C.

Hence the continuity ofFε holds.

At this step, applying the Tychonoff fixed point theorem on B(0, R(ε)) (see [22]), we derive thatFε has a fixed point. Thus (Pε) has at least one solution.

Let us now show that our sequence (u_ε) is uniformly bounded inL^∞(Q). More precisely we have:

Proposition 2.4. Let u_ε be a solution of (P_ε) and let ε₀ > 0. Then we have for anyε ∈ (0, ε₀) and H such that H ≥max(ε₀+ max{xn,(x⁰, x_n)∈Ω}, max{ψ(x, t),(x, t)∈Σ₂})

(2.9) x_n≤u_ε≤H a.e. in Q .

Proof: i) Since (u_ε−H)⁺ is a suitable test function for (P_ε), we have by (2.1) and the choice ofH

(2.10)

Z

QA(x,∇uε)· ∇(uε−H)⁺+ε|uεt|^q−2u_εt(u_ε−H)⁺_t dx dt=

= Z

QGε(uε)^³A(x, e)· ∇(uε−H)⁺−(uε−H)⁺_t ^´dx dt

− Z

Ω

g₀(x) (u_ε−H)⁺(x,0)dx+ Z

Ω

G_ε(u_ε(x, T)) (u_ε−H)⁺(x, T)dx

=− Z

Ωg₀(x) (u_ε−H)⁺(x,0)dx≤0 . Then we deduce from (1.3) and (2.10)

Z

Qα|∇(uε−H)⁺|^q+ε|(uε−H)⁺_t |^qdx dt≤0, which leads to|∇(uε−H)⁺|=|(uε−H)⁺_t|= 0 a.e. in Q. Thus (u_ε−H)⁺= 0 andu_ε≤H a.e. in Q.

ii) We denote by (·)⁻ the negative part of a function. Thenξ= (uε−xn)⁻ is a test function for (P_ε) and one has by taking into account (2.1):

(2.11) Z

QA(x,∇uε)· ∇(uε−x_n)⁻+ε|uεt|^q−2u_εt·(u_ε−x_n)⁻_t dx dt=

= Z

Ω(u_ε−x_n)⁻(x, T)dx+ Z

QA(x, e)· ∇(u_ε−x_n)⁻dx dt

− Z

Ω

g₀(x) (u_ε−x_n)⁻(x,0)dx− Z

Q

(u_ε−x_n)⁻_t dx dt .

Integrating by part the last term of (2.11), we obtain (2.12)

Z

[uε≤xn]

³A(x,∇uε)−A(x,∇xn)^´·(∇uε−∇xn)+ε|(uε−xn)_t|^qdx dt ≤ 0.

Using (1.5) and (2.12) we conclude that uε≥xna.e. in Q.

Now we give an a priori estimate for ∇uε and u_εt.

Proposition 2.5. Under assumptions of Proposition 2.4, we have for any ε∈(0, ε₀):

(2.13)

Z

Q

³α|∇uε|^q+ε|uεt|^q´dx dt≤C ,

whereC is a constant independent ofε.

Proof: Using the fact that u_ε−ψ is a suitable test function, we get (2.14)

Z

QA(x,∇uε)· ∇uε+ε|uεt|^qdx dt=

= Z

QA(x,∇uε)· ∇ψ dx dt+ Z

Q

ε|uεt|^q−2u_εt·ψ_tdx dt +

Z

QGε(uε)A(x, e)· ∇(uε−ψ)dx dt− Z

QGε(uε) (uε−ψ)tdx dt +

Z

Ω

G_ε(u_ε(x, T)) (u_ε−ψ)(x, T)dx− Z

Ω

g₀(x) (u_ε−ψ)(x,0)dx .

First let us set: E_ε(y) = Z _y

0

(1−H_ε(s))ds and H_ε(s) = 1 ∧ s⁺

ε . We have 0≤(1−H_ε(y))y≤E_ε(y)≤y ∀y≥0 and using (2.9), we get

(2.15)

Z

Q−Gε(u_ε) (u_ε−ψ)_tdx dt=

= Z

Q−G_ε(u_ε) (u_ε−x_n)_tdx dt+ Z

QG_ε(u_ε)ϕ_tdx dt

=− Z

Q

∂

∂tE_ε(u_ε−x_n)dx dt+ Z

Q

G_ε(u_ε)ϕ_tdx dt

= Z

Ω

·

E_ε^³u_ε(x,0)−x_n^´−E_ε^³u_ε(x, T)−x_n^´

¸ dx+

Z

Q

G_ε(u_ε)ϕ_tdx dt≤C .

Next, by (2.9) the last two terms of (2.14) are bounded. So using (1.3), (2.14), (2.15) and H¨older’s inequality, we derive for some constant C > 0:

0 ≤ U_ε ≤ C(1 +U_ε^1/q + U_ε^1/q0) where U_ε = Z

Q

(α|∇uε|^q +ε|uεt|^q)dx dt.

Hence we get (2.13) sinceq, q⁰ >1.

In the following proposition, we show that u_ε satisfies an inequality similar to (P) iii).

Proposition 2.6. Letu_ε be a solution of(P_ε). Then we have:

Z

QA(x,∇uε)·∇ξ+ε|uεt|^q−2u_εt·ξ_t+G_ε(u_ε)^³ξ_t− A(x, e)·∇ξ^´dx dt+

(2.16) +

Z

Ω

g₀(x)ξ(x,0)dx≤0

∀ξ∈W^1,q(Q), ξ= 0 on Σ₃, ξ≥0 on Σ₄, ξ(x, T) = 0 a.e. x∈Ω. Proof: Letξ as in (2.16). For anyδ >0, (^p_δ^ε ∧ξ), where p_ε =u_ε−x_n, is a test function for (P_ε). So we can write

Z

QA(x,∇uε)·∇

µpε

δ ∧ξ

¶

− A(x, e)·∇

µpε

δ ∧ξ

¶

+ε|uεt|^q−2uεt· µpε

δ ∧ξ

¶

t

dx dt+ +

Z

Q

H_ε(u_ε−x_n)A(x, e)·∇

µp_ε δ ∧ξ

¶

dx dt − Z

Q

H_ε(u_ε−x_n) µp_ε

δ ∧ξ

¶

t

dx dt=

= Z

Ω

³1−g₀(x)^´· µp_ε

δ ∧ξ

¶

(x,0)dx .

The first integral in the left side of this equality can be written as Z

[uε−xn<δξ]

³A(x,∇uε)− A(x,∇xn)^´· ∇u_ε−x_n

δ +ε

δ|uεt|^qdx dt+ +

Z

[uε−xn≥δξ]

³A(x,∇uε)− A(x, e)^´· ∇ξ+ε|uεt|^q−2u_εtξ_tdx dt . Using (1.5) and the fact that 1−g0(x)≥0 a.e. x∈Ω, we obtain

(2.17) Z

[uε−xn≥δξ]

³A(x,∇uε)− A(x, e)^´· ∇ξ+ε|uεt|^q−2u_εtξ_tdx dt+ +

Z

QH_ε(u_ε−x_n)A(x, e)· ∇ µp_ε

δ ∧ξ

¶

dx dt− Z

QH_ε(u_ε−x_n) µp_ε

δ ∧ξ

¶

t

dx dt≤

≤ Z

Ω(1−g0(x))ξ(x,0)dx .

Let us show that:

δ→0lim Z

QHε(uε−xn)A(x, e)· ∇

µuε−xn

δ ∧ξ

¶

dx dt= (2.18)

= Z

QHε(uε−xn)A(x, e)· ∇ξ dx dt ,

δ→0lim Z

QH_ε(u_ε−x_n)

µuε−xn

δ ∧ξ

¶

t

dx dt= Z

QH_ε(u_ε−x_n)ξ_tdx dt . (2.19)

Indeed, taking in account (1.6) we can use the divergence formula to get:

Z

QHε(uε−xn)A(x, e)· ∇

µuε−xn

δ ∧ξ

¶

dx dt=

=− Z

Q

div^³H_ε(u_ε−x_n)A(x, e)^´·

µu_ε−x_n δ ∧ξ

¶ dx dt

+ Z

∂Q

H_ε(u_ε−x_n)A(x, e)·ν·

µu_ε−x_n δ ∧ξ

¶

dσ(x, t) .

Since ξ∧ u_ε−xn

δ −→ ξ a.e. on [u_ε > x_n] when δ goes to 0, we obtain by the Lebesgue theorem,

δ→0lim Z

QH_ε(u_ε−x_n)A(x, e)· ∇

µu_ε−x_n δ ∧ξ

¶

dx dt =

= − Z

Q

div^³H_ε(u_ε−xn)A(x, e)^´·ξ dx dt+ Z

∂Q

H_ε(u_ε−xn)A(x, e)·ν·ξ dσ(x, t) =

= Z

Q

H_ε(u_ε−x_n)A(x, e)· ∇ξ dx dt , which proves (2.18). Similarly we establish (2.19).

So combining (2.18)–(2.19) and lettingδ →0 in (2.17) we obtain (2.16) since Z

Ω

ξ(x,0)dx=− Z

Q

ξ_tdx dt.

3 – Existence of a solution

Theorem 3.1. Assume thatϕis a nonnegative Lipschitz continuous function and thatAsatisfies (1.2)–(1.6). Then there exists a solution (u, g) of (P).

The proof will consist in passing to the limit, when ε goes to 0, in (P_ε).

To do this we shall need some lemmas.

First from definition (2.1) of G_ε and estimates (2.9), (2.13) and (1.4), we de- duce the existence of a subsequenceε_k ofεand functions: u∈L^q(0, T, W^1,q(Ω)), g∈L^q0(Q), A₀∈L^q0(Q) such that:

uεk * u weakly in L^q(0, T, W^1,q(Ω)), (3.1)

A(x,∇uεk)* A₀ weakly in L^q0(Q) , (3.2)

G_εk(u_εk)* g weakly in L^q0(Q) . (3.3)

Then we can prove:

Lemma 3.2. Let u, g be defined by (3.1) and (3.3) respectively. Then we have:

i) u=ψ on Σ₂, u≥x_n a.e. inQ;

ii) 0≤g≤1 a.e. inQ.

Proof: We consider the setK₁={v∈L^q(0, T, W^1,q(Ω))/ v≥x_na.e. in Q, v = ψ on Σ2}. K1 is closed and convex in L^q(0, T, W^1,q(Ω)), then it is weakly closed. Since u_εk ∈ K₁, u ∈ K₁ and i) holds. In the same way, we prove that g∈K2 = {v∈L^q0(Q) / 0≤v≤1 a.e. in Q} and ii) holds.

Lemma 3.3. Let u, g be defined by (3.1) and (3.3) respectively. Then we have:

(3.4) g(u−x_n) = 0 a.e. in Q .

Proof: First, note that (3.4) is not an obvious result as it is in the stationary case (see [17]), since we do not know, a priori, whetheru_ε converges strongly to u in L^q(Q) because the imbedding L^q(0, T, W^1,q(Ω)) ,→ L^q(Q) is not compact.

To overcome this difficulty, we are going to prove a strong convergence of the sequence (G_εk(u_εk))_k in a suitable space.

Next, we have for θ∈D(Q),θ≥0 andp_εk =u_εk −x_n 0 ≤

Z

QG_εk(u_εk) (u_εk −x_n)θ dx dt =

= Z

Q∩[0≤p_εk≤εk]

(1−H_εk(p_εk))p_εkθ dx dt ≤ ε_k· |θ|∞· |Q|. So

(3.5) lim

k→+∞

Z

Q

G_εk(u_εk) (u_εk−x_n)θ dx dt= 0 .

To get (3.4), it suffices to prove that:

(3.6) lim

k→+∞

Z

Q

G_εk(u_εk) (u_εk−x_n)θ dx dt= Z

Q

g(u−x_n)θ dx dt .

For this purpose we introduce the function: w_εk = ε_k|uεkt|^q−2u_εkt+G_εk(u_εk).

From (2.13), sinceq⁰ >1, we have:

(3.7) ε_k|uεkt|^q−2u_εkt−→0 strongly in L^q0(Q). Then from (3.3) and (3.7) we deduce that

(3.8) w_εk * g weakly in L^q0(Q) . We are going to prove that:

(3.9) w_εkt* g_t weakly in L^q0(0, T, W^−1,q0(Ω)).

So, let us show thatg_t∈L^q0(0, T, W^−1,q0(Ω)). Let ξ∈D(0, T, W₀^1,q(Ω)). Since ξ is a test function for (Pεk), we obtain, after lettingk→+∞in (Pεk), and taking into account (3.2), (3.3) and (3.7)

Z

Qg ξtdx dt=− Z

Q

³A0−gA(x, e)^´· ∇ξ dx dt

from which we deduce (see [10]) thatg_t=−div(A₀−gA(x, e))∈L^q0(0, T,W^−1,q0(Ω)).

Now we have in the distributional sense:

w_εkt=−div^³A(x,∇u_εk)−G_εk(u_εk)A(x, e)^´ and w_εkt*div^³gA(x, e)−A₀^´=g_t weakly in L^q0(0, T, W^−1,q0(Ω)). So (3.9) holds.

At this stage let us introduce the spaceW defined by:W={v∈L^q0(0, T, L^q0(Ω))/

v_t∈L^q0(0, T, W^−1,q0(Ω))}which is a Banach space for the norme: kvk_Lq0

(0,T,L^q0(Ω))+ kv_tk_Lq0

(0,T,W^−1,q0(Ω)). Since L^q0(Ω) and W^−1,q0(Ω) are reflexifs, the imbedding L^q0(Ω) ,→ W^−1,q0(Ω) being continuous and compact (see [1]), we deduce that (see [20]), the imbedding

(3.10) W ,→L^q0(0, T, W^−1,q0(Ω)) is compact.

Now, the sequencew_εk ∈W and it is bounded in W by (3.8) and (3.9). So up to a subsequence still denoted byεk, we have wεk * wweakly inW. But it is easy to see thatw=g. We then deduce from (3.7) and (3.10)

(3.11) G_εk(u_εk)→g strongly in L^q0(0, T, W^−1,q0(Ω)).

Finally, to conclude, it suffices to remark that: (u_εk−xn)θ *(u−xn)θ weakly in L^q(0,T,W₀^1,q(Ω)). Then by (3.11), we get (3.6). Consequently

Z

Qg(u−xn)θ dxdt= 0

∀θ∈D(Q), θ≥0 and sinceg(u−x_n)≥0 a.e. in Q, we get (3.4).

Remark 3.4. We have (see [20]) the imbedding W ,→C⁰([0, T], W^−1,q0(Ω)) theng∈C⁰([0, T], W^−1,q0(Ω)). In section 4, we shall improve this regularity and prove thatg∈C⁰([0, T], L^p(Ω)), ∀p∈[1,+∞[.

Lemma 3.5. Letu,A₀andgbe defined by (3.1), (3.2) and (3.3) respectively.

Then we have:

(3.12) Z

Q

³A₀−gA(x, e)^´· ∇(u−ψ)ξ dx dt= Z

Q

g(ϕ ξ)_tdx dt ∀ξ∈D(0, T) .

Proof: Let ζ be a smooth function such that d(suppζ,Σ2) > 0 and suppζ ⊂Rⁿ×(τ₀⁰, T −τ₀⁰) for T > τ₀⁰ >0. Then there exists τ₀ > 0 such that:

∀τ ∈ (−τ0, τ0), (x, t) 7−→ ζ(x, t−τ) is a test function for (Pεk). So we get, for allτ ∈(−τ0, τ₀), after letting kgo to +∞

(3.13)

Z

Q

³A₀(x, t)−g(x, t)A(x, e)^´· ∇ζ(x, t−τ)dx dt−

− ∂

∂τ

³Z

Q

g(x, t+τ)ζ(x, t)dx dt^´= 0 since

− Z

Qg(x, t)ζ_t(x, t−τ)dx dt= ∂

∂τ

³Z

Qg(x, t)ζ(x, t−τ)dx dt^´

= ∂

∂τ

³Z

Qg(x, t+τ)ζ(x, t)dx dt^´.

Now it is easy to see that (3.13) still holds for functions ζ inL^q(0, T, W^1,q(Ω)) such thatζ= 0 on Σ₂ and ζ = 0 on Ω×((0, τ₀)∪(T −τ₀, T)). So if we consider ξ∈D(τ₀, T−τ₀),ξ ≥0 and set: ζ = (u−ψ)ξ, we have ∀τ ∈(−τ₀, τ₀)

(3.14) Z

Q

µ³

A₀(x, t)−g(x, t)A(x, e)^´·∇^³(u−ψ)ξ^´−g(ϕ ξ)_t

¶

(x, t−τ)dx dt=

= ∂G

∂τ (τ) withG(τ) =

Z

Q

g(x, t+τ) ((u−x_n)ξ)(x, t)dx dt.

From (3.14) we know that G ∈ C¹(−τ0, τ₀). Moreover by Lemmas 3.2 and 3.3, we getG(τ)≥0 =G(0) ∀τ ∈(−τ0, τ0). So 0 is an absolute minimum forG in (−τ0, τ₀) and

(3.15) ∂G

∂τ(0) = 0.

Combining (3.14) and (3.15) we get (3.12) for all ξ∈D(τ0, T−τ0),ξ ≥0.

Thanks to Lemma 3.5, we are going to prove a result which allows us to pass to the limit in (P_εk).

Lemma 3.6. The sequence(u_εk) (resp.A(x,∇uεk)) converges strongly tou (resp.A(x,∇u)) in L^q(0, T, W^1,q(Ω))(resp. L^q0(Q)).

To prove Lemma 3.6, we need a lemma:

Lemma 3.7. Letu andA₀ defined by (3.1) and (3.2) respectively. Then we have

(3.16) Z

QA(x,∇u)·∇ξ dx dt= Z

QA0(x, t)·∇ξ dx dt ∀ξ ∈L^q(0, T, W^1,q(Ω)). Proof: Letθ∈D(0, T),θ≥0. Choose ξ= (u_εk−ψ)θas a test function for (P_εk) and write (3.12) for ξ =θ. Subtract the equations, we obtain:

(3.17)

Z

QθA(x,∇uεk)· ∇uεkdx dt=

= Z

Q

θ A₀· ∇u dx dt+ Z

Q

³A(x,∇uεk)−A₀^´· ∇ψ θ dx dt

− Z

Qε_k|u_εkt|^q−2u_εkt^³(u_εk−ψ)θ^´

tdx dt +

Z

QA(x, e)·^³Gεk(uεk)∇(uεk−ψ)−g∇(u−ψ)^´θ dx dt

− Z

Q

G_εk(u_εk)^³(u_εk −ψ)θ^´

tdx dt− Z

Q

g(ϕ θ)_tdx dt .

By (3.2) we have:

(3.18) lim

k→+∞

Z

Q

³A(x,∇uεk)−A₀^´· ∇ψ θ dx dt= 0 .

Remark that Z

Q

ε_k|uεkt|^q−2u_εkt·^³(u_εk −ψ)θ^´

tdx dt=

= Z

Qε_k|uεkt|^qθ dx dt+

Z

Qε_k|uεkt|^q−2uεktuεkθtdx dt−

Z

Qε_k|uεkt|^q−2uεkt·(ψ θ)tdx dt.

The first term of the above equality is nonnegative. Moreover using (3.7) and the fact thatu_εk is uniformly bounded, we get:

(3.19) lim−

Z

Qε_k|uεkt|^q−2uεkt·^³(uεk−ψ)θ^´

tdx dt≤0. From (3.4), we deduce:

(3.20) Z

QA(x, e)·^³G_εk(u_εk)∇(u_εk −ψ)−g∇(u−ψ)^´θ dx dt=

= Z

Q

θ G_εk(u_εk)A(x, e)· ∇(uεk−x_n)dx dt

− Z

Q

³Gεk(uεk)−g^´A(x, e)· ∇(ϕ θ)dx dt . By (3.3) the last term of (3.20) goes to 0. Applying the divergence formula, we get:

(3.21)

¯

¯

¯

¯ Z

Qθ Gεk(uεk)A(x, e)· ∇(uεk −xn)dx dt

¯

¯

¯

¯=

=

¯

¯

¯

¯ Z

Q

θA(x, e)· ∇

µZ _{min(uεk−xn,εk)}

0

³1−H_εk(s)^´ds

¶ dx dt

¯

¯

¯

¯

≤εk

µZ

Qθ|div(A(x, e))|dx dt+ Z

∂Ω×(0,T)θ|A(x, e)·ν|dσ(x, t)

¶ .

We obtain from (3.20)–(3.21):

(3.22) lim

k→+∞

Z

QA(x, e)·^³G_εk(u_εk)∇(uεk−ψ)−g∇(u−ψ)^´θ dx dt= 0 . The last two terms of the right hand side of (3.17) can be written:

(3.23) − Z

Q

G_εk(u_εk)^³(u_εk −ψ)θ^´

tdx dt− Z

Q

g(ϕ θ)_tdx dt=

=− Z

QG_εk(u_εk) (u_εk−x_n)θ_tdx dt− Z

QG_εk(u_εk) (u_εk−x_n)_tθ dx dt +

Z

Q

³G_εk(u_εk)−g^´(ϕ θ)_tdx dt .

Arguing as in (3.5), the first term of the right side of (3.23) converges to 0.

Integrating by parts, we can see as in (3.21) that the second term goes also to 0.

Using (3.3), we get (3.24) lim

k→+∞− Z

Q

G_εk(u_εk)^³(u_εk −ψ)θ^´

tdx dt− Z

Q

g(ϕ θ)_tdx dt= 0 . Combining (3.17), (3.18), (3.19), (3.22) and (3.24) we conclude that:

(3.25) lim Z

Q

θA(x,∇uεk)· ∇uεkdx dt ≤ Z

Q

θ A₀(x, t)· ∇u dx dt

∀θ∈D(0, T), θ≥0 . Let now v∈L^q(0, T, W^1,q(Ω)) and θ∈D(0, T) such that θ≥0. Using (1.5) we have:

Z

Qθ^³A(x,∇u_εk)− A(x,∇v)^´·(∇u_εk− ∇v)dx dt≥0 ∀k∈N which can be written for allk∈N:

(3.26) Z

Q

θA(x,∇uεk)· ∇uεkdx dt − Z

Q

θA(x,∇uεk)· ∇v dx dt −

− Z

Q

θA(x,∇v)· ∇(uεk −v)dx dt ≥ 0 . Passing to the limit sup in (3.26) and taking into account (3.1)–(3.2) and (3.25), we get:

(3.27)

Z

Qθ^³A₀(x, t)− A(x,∇v)^´· ∇(u−v)dx dt≥0 .

If we choosev=u±λ ξ withξ ∈L^q(0, T, W^1,q(Ω)) andλ∈[0,1] in (3.27) we obtain, after lettingλgo to 0 and taking into account (1.2) and (1.4)

Z

Qθ^³A₀(x, t)− A(x,∇u)^´· ∇ξ dx dt= 0 ∀θ∈D(0, T), θ≥0,

∀ξ ∈L^q(0, T, W^1,q(Ω)) and by density we get (3.16).

Proof of Lemma 3.6: Taking ξ=θ uin (3.16) with θ∈D(0, T), we get (3.28)

Z

Q

θA(x,∇u)· ∇u dx dt= Z

Q

θ A₀(x, t)· ∇u dx dt .

Using (3.25) and (3.28) we obtain:

(3.29) lim Z

Q

θA(x,∇uεk)· ∇uεkdx dt ≤ Z

Q

θA(x,∇u)· ∇u dx dt . Combining (3.1)–(3.2) and (3.28)–(3.29), one can prove easily:

(3.30) lim

Z

Q

θ^³A(x,∇uεk)− A(x,∇u)^´· ∇(uεk−u)dx dt≤0 .

Now, since ∇uεk * ∇u weakly in L^q(Q), we conclude by (3.30) because A satisfies the Browder’s property (S₊) (see [12]), that ∇uεk → ∇u strongly in L^q(Q). Moreover the mapping L^q(Q) → L^q0(Q), v 7→ A(·, v) being continuous, we deduce thatA(x,∇u_εk) converges strongly toA(x,∇u) inL^q0(Q). Now by the Poincar´e Inequality one can see thatu_εkconverges strongly inL^q(0, T, W^1,q(Ω)).

Proof of Theorem 3.1: It is clear that (P) i) and (P) ii) follow from Lemma 3.2 and Lemma 3.3. Let ξ ∈ W^1,q(Q), ξ = 0 on Σ₃, ξ ≥ 0 on Σ₄ and ξ(x, T) = 0 a.e. in Ω. Lettingkgo to +∞in (2.16) written forξ and using (3.3), (3.7) and Lemma 3.6, we get (P) iii). This achieves the proof of Theorem 3.1.

Remark 3.8. Note that the Lemma 3.7 is sufficient for the proof of Theorem 3.1, however the result of Lemma 3.6 is more precise.

4 – Some properties

Let us first prove a technical lemma which generalizes Lemma 3.5.

Lemma 4.1.Let(u, g)be a solution of (P), letv∈W^1,q(Q)andF∈W_loc^1,∞(R²), such that:

i) F(u−xn, v)∈L^q(0, T, W^1,q(Ω));

ii) F(ψ−xn, v)∈W^1,q(Q);

iii) F(z₁, z₂)≥0 for a.e. (z₁, z₂)∈R²; iv) either ∂F

∂z₁(z₁, z₂)≥0 a.e.(z₁, z₂)∈R², or ∂F

∂z₁(z₁, z₂)≤0 a.e.(z₁, z₂)∈R². Then we have ∀ξ∈D(Ω×(0, T)):

(4.1) Z

Q

³A(x,∇u)−gA(x, e)^´· ∇^³F(u−x_n, v)ξ^´+g^³F(0, v)ξ^´

tdx dt=

= Z

Q

³A(x,∇u)−gA(x, e)^´· ∇^³F(ψ−x_n, v)ξ^´+g^³F(ψ−x_n, v)ξ^´

tdx dt .

Particularly, ifF(ψ−x_n, v)ξ = 0on Σ₂, then (4.2)

Z

Q

³A(x,∇u)−gA(x, e)^´· ∇^³F(u−x_n, v)ξ^´+g^³F(0, v)ξ^´

tdx dt= 0 . Proof:Arguing as in the proof of Lemma 3.5, we get forξ∈D(Rⁿ×(τ₀, T−τ0)), ξ≥0,τ0>0 and ζ = (F(u−xn, v)−F(ψ−xn, v))ξ

(4.3) Z

Q

³A(x,∇u(x, t))−g(x, t)A(x, e)^´· ∇^³F(u−x_n, v)ξ^´(x, t−τ) + +g(x, t)^³F(0, v)ξ^´

t(x, t−τ)dx dt −

− Z

Q

³A(x,∇u(x, t))−g(x, t)A(x, e)^´· ∇^³F(ψ−xn, v)ξ^´(x, t−τ)−

−g(x, t)^³F(ψ−xn, v)ξ)t(x, t−τ)dx dt =

= ∂

∂τG(τ) withG(τ) =

Z

Q

g(x, t+τ) ((F(u−xn, v)−F(0, v))ξ)(x, t)dx dt. Since the integrals on the left hand side of (4.3) are continuous functions on τ, we deduce that G∈C¹(−τ0, τ₀). Using the monotonicity of F and (3.4), we can see that 0 is an extremum forGin (−τ0, τ₀) and

(4.4) ∂G

∂τ(0) = 0. From (4.3) and (4.4) we deduce the Lemma.

From Lemma 4.1, we have:

Corollary 4.2. Let(u, g) be a solution of (P). Then:

Z

QA(x,∇u)· ∇ µ

min

µ(u−x_n−k)⁺

ε ,1

¶ ξ

¶

dx dt= 0

∀ε >0, ∀k≥0, ∀ξ∈D(Rⁿ×(0, T)) such that ξ≥0, ξ= 0onΣ3 .

Proof: It suffices to chooseF(z1, z2) = min(^(z1−k)_ε ⁺,1) in Lemma 4.1 and to take in account (3.4).

Corollary 4.3. Let(u, g) be a solution of (P). Then we haveu∈L^∞(Q).

Proof: Since u≥x_n a.e. inQ, it suffices to prove thatu is bounded above.

So letH be a constant such that:

H≥max µ

maxⁿxn, (x⁰, xn)∈Ω^o, maxⁿψ(x, t), (x, t)∈Σ2

o¶ .

Letξ be a nonnegative function in D(0, T). Then if one apply Lemma 4.1 with F(z₁, z₂) = (z₁−z₂)⁺ andv=H−x_n, we get by taking in account (3.4) and the choice ofH:

(4.5)

Z

Q

ξ(t)A^³x,∇(u−H)^´· ∇(u−H)⁺dx dt= 0 .

Since (u−H)⁺= 0 on Σ2, we deduce from (1.3) and (4.5) that u≤H a.e. in Q.

Theorem 4.4. Let (u, g) be a solution of (P). Then we have in the distri- butional sense:

(4.6) div^³A(x,∇u)−gA(x, e)^´+g_t= 0 . Moreover, ifdiv(A(x, e))≥0in D⁰(Ω), we have:

(4.7) div^³gA(x, e)^´−gt= div^³A(x,∇u)^´≥0 .

Proof: i) Taking ±ξ ∈D(Q) as a test function for (P), we get (4.6).

ii) Letξ ∈D(Q),ξ≥0, then from Corollary 4.2, we have forε >0 andk= 0:

(4.8)

Z

QA(x,∇u)· ∇ µ

min

µu−xn

ε ,1

¶ ξ

¶

dx dt= 0 . Note thatξ = 0 on∂Q, so

(4.9)

Z

QA(x, e)· ∇ õ

1−min

µu−x_n ε ,1

¶¶

ξ

!

dx dt≤0 . Adding (4.8) and (4.9), we get:

1 ε Z

Q∩[u−xn<ε]ξ^³A(x,∇u)− A(x,∇xn)^´(∇u− ∇xn)dx dt+ +

Z

Qmin

µu−xn

ε ,1

¶³

A(x,∇u)− A(x,∇x_n)^´· ∇ξ dx dt ≤

≤ − Z

QA(x, e)· ∇ξ dx dt .