Singular limit of a two-phase flow problem in porous medium as the air viscosity tends to zero
aa r X i v : . [ m a t h . NA ] O c t Singular limit of a two-phase flow problem in porousmedium as the air viscosity tends to zero ∗ R. Eymard , M. Henry and D. Hilhorst Abstract
In this paper we consider a two-phase flow problem in porous media and studyits singular limit as the viscosity of the air tends to zero; more precisely, we provethe convergence of subsequences to solutions of a generalized Richards model.
Hydrologists have studied air-water flow in soils, mainly using the so-called Richardsapproximation. At least two hypotheses are physically required for this model to beapplicable: the water pressure in the saturated region must be larger than the atmosphericpressure and all the unsaturated regions must have a boundary connected to the surface.However, in many situations, these hypotheses are not satisfied and a more general two-phase flow model must be considered. This work explores the limit of this general modelas the viscosity of the air tends to zero, which is one of the hypotheses required in theRichards model. To that purpose we prove the existence of a weak solution of the two-phase flow problem and prove estimates which are uniform in the air viscosity. In thispaper, we assume that the air and water phases are incompressible and immiscible. Thegeometric domain is supposed to be horizontal, homogeneous and isotropic. Our startingpoint is the following two-phase flow model, which one can deduce from Darcy’s law(
T P ) u t − div ( k w ( u ) ∇ ( p )) = s w (1 − u ) t − div ( 1 µ k a ( u ) ∇ ( p + p c ( u ))) = s a , ∗ This work was supported by the GNR MoMaS (PACEN/CNRS, ANDRA, BRGM, CEA, EdF, IRSN),France. Universit´e Paris-Est Marne-La-Vall´ee, 5 bd Descartes, Champs-sur-Marne, 77454 Marne-la-Vall´eeCedex 2, France CMI Universit´e de Provence, 39 rue Fr´ed´eric Joliot-Curie 13453 Marseille cedex 13, France CNRS and Laboratoire de Math´ematiques, Universit´e de Paris-Sud 11, F-91405 Orsay Cedex, France u and p are respectively the saturation and the pressure of the water phase, k w and k a are respectively the relative permeabilities of the water and the air phase, µ isthe ratio between the viscosity of the air phase and that of the water phase, p c is thecapillary pressure, s w is an internal source term for the water phase and s a is an internalsource term for the air phase; these source terms are used to represent exchanges with theoutside. We suppose in particular that the physical functions k w , k a and p c only dependon the saturation u of the water phase, and that k w (1) = k a (0) = 1. The aim of thispaper is the study of the limit of the two-phase flow problem as µ ↓ R ) ( u t − div ( k w ( u ) ∇ p ) = s w u = p − c ( p atm − p ) . where the properties of capillary pressure p c = p c ( u ) are describes in hypothesis ( H )below. For the existence and uniqueness of the solution of Richards model together withsuitable initial and boundary conditions as well as qualitative properties of the solutionand methods for numerical approximations we refer to [1], [6], [10], [11]. In this article,we will show that the singular limit as µ ↓ T P ) has theform (
F BP ) ( u t − div ( k w ( u ) ∇ p ) = s w u = 1 or ∇ ( p + p c ( u )) = 0 a.e. in Ω × (0 , T ) . We remark that a solution of ( R ) with u > F BP ).This paper is organized as follows. In Section 2 we present a complete mathematical for-mulation of the problem, and state the main mathematical results, which include a preciseformulation of the singular limit problem. We give a sequence of regularized problemsin Section 3, and prove the existence of a classical solution. In Section 4 we present apriori estimates, which are uniform in an extra regularization parameter δ and in the airviscosity µ . In Section 5, we let δ ↓ µ tendsto zero in Section 6. Finally in Section 7 we propose a finite volume algorithm in a onedimensional context and present a variety of numerical solutions.2 Mathematical formulation and main results
We consider the two-phase flow problem( S µ ) u t = div (cid:18) k w ( u ) ∇ p (cid:19) + f µ ( c ) s − f µ ( u ) s, in Q T , (1 − u ) t = div (cid:18) µ k a ( u ) ∇ ( p + p c ( u )) (cid:19) + (1 − f µ ( c )) s − (1 − f µ ( u )) s, in Q T , Z Ω p ( x, t ) dx = 0 , for t ∈ (0 , T ) , ∇ p.n = 0 , on ∂ Ω × (0 , T ) , ∇ ( p + p c ( u )) .n = 0 , on ∂ Ω × (0 , T ) ,u ( x,
0) = u ( x ) , for x ∈ Ω , (2.1)(2.2)(2.3)(2.4)(2.5)(2.6)where T is a positive constant, Q T := Ω × (0 , T ) and where we suppose that( H ) Ω is a smooth bounded domain of IR N where the space dimension N is arbitrary , ( H ) u m ∈ (0 , , ( H ) c ∈ L ∞ (Ω × (0 , T )) and u m ≤ c ≤ , ( H ) u ∈ L ∞ (Ω) and u m ≤ u ≤ , ( H ) s ∈ L (Ω) , s ≥ , s ∈ L (Ω) , s ≥ Z Ω ( s ( x ) − s ( x )) dx = 0 , ( H ) k w ∈ C ([0 , , k ′ w ≥ , k w (0) = 0 , k w (1) = 1 and k w ( u m ) > , ( H ) k a ∈ C ([0 , , k ′ a ≤ , k a (1) = 0 , k a (0) = 1 and k a ( s ) > s ∈ [0 , , ( H ) p c ∈ C ([0 , ∪ C ([0 , , p ′ c < s ∈ [0 , ( − k a ( s ) p ′ c ( s )) < + ∞ , ( H ) µ ∈ (0 , . In this model, u and p are respectively the saturation and the pressure of the waterphase, k w and k a are respectively the mobilities of the water phase and the mobility ofthe non-water phase and p c is the capillary pressure. We assume in particular that thepermeability functions k w , k a and the capillary pressure p c only depend on the saturation u of the water phase. Here, we suppose that the flow of the water phase in the reservoiris driven by an injection term f µ ( c ) s and an extraction term f µ ( u ) s where s and s aregiven space dependent functions, c is the saturation of the injected fluid; if c = 1, onlywater will be injected, if c = 0, only air will be injected, whereas a mixture of water andair will be injected if 0 < c <
1. The function f µ is the fractional flow of the water phase,namely f µ ( s ) = k w ( s ) M µ ( s ) , with M µ ( s ) = k w ( s ) + 1 µ k a ( s ) . (2.7)In particular, we remark that f µ ( s ) is non decreasing. (2.8)3ext we introduce a set of notations, which will be useful in the sequel. g ( s ) = − Z s k a ( τ ) p ′ c ( τ ) dτ, (2.9) ζ ( s ) = Z s q k a ( τ ) p ′ c ( τ ) dτ, (2.10) Q µ ( s ) = Z s f µ ( τ ) p ′ c ( τ ) dτ, (2.11)and R µ ( s ) = Z s k a ( τ ) k a ( τ ) + µk w ( τ ) p ′ c ( τ ) dτ, (2.12)for all s ∈ [0 , R µ ( s ) + Q µ ( s ) = p c ( s ) − p c (0) , for all s ∈ [0 , . (2.13) Definition 2.1
The pair ( u µ , p µ ) is a weak solution of Problem ( S µ ) if u µ ∈ L ∞ (Ω × (0 , T )) , with ≤ u µ ≤ in Q T ,p µ ∈ L (0 , T ; H (Ω)) , Z Ω p µ ( x, t ) dx = 0 for almost every t ∈ (0 , T ) ,g ( u µ ) ∈ L (0 , T ; H (Ω)) , with Z T Z Ω u µ ϕ t dxdt = Z T Z Ω k w ( u µ ) ∇ p µ . ∇ ϕdxdt − Z T Z Ω (cid:18) f µ ( c ) s − f µ ( u µ ) s (cid:19) ϕdxdt − Z Ω u ( x ) ϕ ( x, dx, (2.14) and Z T Z Ω (cid:18) − u µ (cid:19) ϕ t dxdt = Z T Z Ω µ k a ( u µ ) (cid:18) ∇ p µ + ∇ p c ( u µ ) (cid:19) . ∇ ϕdxdt − Z T Z Ω (cid:18) (1 − f µ ( c )) s − (1 − f µ ( u µ )) s (cid:19) ϕdxdt − Z Ω (cid:18) − u ( x ) (cid:19) ϕ ( x, dx, (2.15) for all ϕ in C := { w ∈ W , ( Q T ) , w ( ., T ) = 0 in Ω } . Our first result, which we prove in Section 3, is the following
Theorem 2.2
Suppose that the hypotheses ( H ) − ( H ) are satisfied, then there exists aweak solution ( u µ , p µ ) of Problem ( S µ ) . χ by χ ( s ) := ( s ∈ [0 , s = 1 , as well as the graph H ( s ) := ( s ∈ [0 , ,
1] if s = 1 . The main goal of this paper is to prove the following convergence result,
Theorem 2.3
Suppose that the hypotheses ( H ) − ( H ) are satisfied, then there exists asubsequence (( u µ n , p µ n )) n ∈ N of weak solutions of Problem ( S µ n ) and functions u , p , ˆ f suchthat u ∈ L ∞ ( Q T ) , ≤ u ≤ in Q T , ˆ f ∈ L ∞ ( Q T ) , ≤ ˆ f ≤ in Q T ,p ∈ L (0 , T ; H (Ω)) ,k a ( u ) ∇ p c ( u ) ∈ L (Ω × (0 , T )) , and ( u µ n ) n ∈ N tends to u strongly in L ( Q T ) , ( p µ n ) n ∈ N tends to p weakly in L (0 , T ; H (Ω)) , as µ n tends to zero and Z T Z Ω uϕ t dxdt = Z T Z Ω k w ( u ) ∇ p. ∇ ϕdxdt − Z T Z Ω (cid:18) χ ( c ) s − ˆ f s (cid:19) ϕdxdt − Z Ω u ( x ) ϕ ( x, dx, (2.16) for all ϕ ∈ C , where ˆ f ( x, t ) ∈ H ( u ( x, t )) for ( x, t ) ∈ Q T . Moreover we also have that Z T Z Ω (cid:20) k a ( u ) (cid:21) (cid:20) ∇ p + ∇ p c ( u ) (cid:21) dxdt = 0 (2.17) and Z Ω p ( x, t ) dx = 0 , for almost every t ∈ (0 , T ) . (2.18)Formally, u satisfies the following limit problem u t = div (cid:18) k w ( u ) ∇ p (cid:19) + χ ( c )¯ s − ˆ f s, in Q T , ∇ u.n = 0 , on ∂ Ω × (0 , T ) ,u ( x,
0) = u ( x ) , for x ∈ Ω . More precisely the following corollary holds 5 orollary 2.4
Suppose that u < in O = ∪ t ∈ [ τ,T ] Ω t , where τ > and Ω t , for t ∈ [ τ, T ] ,are smooth subdomains of Ω and O is a smooth domain of Ω × [ τ, T ] and that u = 1 in Q T \ O then p ( x, t ) = − p c ( u ( x, t )) + constant ( t ) , for all ( x, t ) ∈ O and u satisfies u t = − div (cid:18) k w ( u ) ∇ p c ( u ) (cid:19) + χ ( c )¯ s, in O ,∂u∂n = 0 , on ∂ O ∩ (cid:18) ∂ Ω × (0 , T ) (cid:19) ,u = 1 , elsewhere on ∂ O ,u ( x,
0) = u ( x ) , for x ∈ Ω . Finally we remark that another form of the limit problem involves a parabolic equation,which is close to the standard Richards equation. Indeed if we set φ ( s ) := p c (0) − p c ( s )and denote by β the inverse function of φ , the function v := φ ( u ) is a weak solution ofthe problem β ( v ) t = div (cid:18) k w ( β ( v )) ∇ v (cid:19) + χ ( c )¯ s − ˆ f s, in Q T , ∇ β ( v ) .n = 0 , on ∂ Ω × (0 , T ) ,β ( v )( x,
0) = u ( x ) , for x ∈ Ω , with ˆ f ∈ H ( β ( v )). 6 Existence of a solution of an approximate problem ( S µδ ) of Problem ( S µ ) Let δ be an arbitrary positive constant. In order to prove the existence of a solution ofProblem ( S µ ) we introduce a sequence of regularized problems ( S µδ ), namely( S µδ ) u t = div (cid:18) k w ( u ) ∇ p (cid:19) + f µ ( c δ ) s δ − f µ ( u ) (cid:18) s δ + Z − Ω ( s δ − s δ ) dx (cid:19) , in Ω × (0 , T ) , (1 − u ) t = div (cid:18) µ k a ( u ) ∇ ( p + p c ( u )) (cid:19) + (cid:18) − f µ ( c δ ) (cid:19) s δ − (cid:18) − f µ ( u ) (cid:19)(cid:18) s δ + Z − Ω ( s δ − s δ ) dx (cid:19) , in Ω × (0 , T ) , Z Ω p ( x, t ) dx = 0 , for t ∈ (0 , T ) , ∇ p.n = 0 , on ∂ Ω × (0 , T ) , ∇ ( p + p c ( u )) .n = 0 , on ∂ Ω × (0 , T ) ,u ( x,
0) = u δ ( x ) , for x ∈ Ω , (3.1)(3.2)(3.3)(3.4)(3.5)(3.6)where u δ , c δ , s δ and s δ are smooth functions such that u δ tends to u in L (Ω) and c δ , s δ and s δ tend respectively to c , s and s in L ( Q T ), as δ ↓
0. In particular we suppose thatthere exists a positive constant C such that s δ ≥ , s δ ≥ Z Ω s δ + Z Ω s δ ≤ C. (3.7)Moreover we suppose that u δ , c δ satisfy0 < u m ≤ u δ ≤ − δ < < u m ≤ c δ ≤ − δ < Q T . (3.9)Adding up (3.1) and (3.2) we deduce the equation − div (cid:18) M µ ( u ) ∇ p + 1 µ k a ( u ) ∇ ( p c ( u )) (cid:19) = s δ − s δ − Z − Ω ( s δ − s δ ) dx. (3.10)We formulate below an equivalent form of Problem ( S µδ ). To that purpose we define theglobal pressure, P , by P := p + R µ ( u ) = p + Z u k a ( τ ) k a ( τ ) + µk w ( τ ) p ′ c ( τ ) dτ,
7o that (3.10) gives − div (cid:18) M µ ( u ) ∇P (cid:19) = s δ − s δ − Z − Ω ( s δ − s δ ) dx. (3.11)We rewrite the equation (3.1) of Problem ( S µδ ) as u t = ∆ ψ µ ( u ) + div (cid:18) f µ ( u ) M µ ( u ) ∇P (cid:19) + f µ ( c δ ) s δ − f µ ( u ) (cid:18) s δ + Z − Ω ( s δ − s δ ) dx (cid:19) , (3.12)where ψ µ ( s ) = − µ Z s k a ( τ ) k w ( τ ) M µ ( τ ) p ′ c ( τ ) dτ (3.13)is continuous on [0 ,
1] and differentiable on [0 , f µ ( u µδ ) and addingthe result to (3.12) we deduce that u t = ∆ ψ µ ( u ) + M µ ( u ) ∇ f µ ( u ) . ∇P + (cid:18) f µ ( c δ ) − f µ ( u ) (cid:19) s δ . This yields a problem equivalent to ( S µδ ), namely( ˜ S µδ ) u t = ∆ ψ µ ( u ) + M µ ( u ) ∇ f µ ( u ) . ∇P + [ f µ ( c δ ) − f µ ( u )] s δ , in Ω × (0 , T ) , − div (cid:18) M µ ( u ) ∇P (cid:19) = s δ − s δ − Z − Ω ( s δ − s δ ) dx, in Ω × (0 , T ) , Z Ω P ( x, t ) dx = Z Ω R µ ( u ( x, t )) dx, for t ∈ (0 , T ) , ∇P .n = 0 , on ∂ Ω × (0 , T ) , ∇ ψ µ ( u ) .n = 0 , on ∂ Ω × (0 , T ) ,u ( x,
0) = u δ ( x ) , for x ∈ Ω . (3.14)(3.15)(3.16)(3.17)(3.18)(3.19)In order to prove the existence of a smooth solution of ( S µδ ), we introduce the set K := { u ∈ C α, α ( Q T ) , u m ≤ u ≤ − δ } , where α ∈ (0 , Lemma 3.1
Assume ( H ) − ( H ) then there exists ( u µδ , P µδ ) solution of ( ˜ S µδ ) such that u µδ ∈ C α, α ( Q T ) , u m ≤ u µδ ≤ − δ and P µδ , ∇P µδ ∈ C α, α ( Q T ) and ∆ P µδ ∈ C α, α ( Q T ) . T be the map defined for all V ∈ K by T ( V ) = W , where W is the uniquesolution of the elliptic problem,( Q V ) − div (cid:18) M µ ( V ) ∇ W (cid:19) = s δ − s δ − Z − Ω ( s δ − s δ ) dx, in Ω × (0 , T ) , Z Ω W ( x, t ) dx = Z Ω R µ ( V ( x, t )) dx, for t ∈ (0 , T ) , ∇ W.n = 0 , on ∂ Ω × (0 , T ) . By standard theory of elliptic system (see [7] Theorem 3.2 p 137), we have that | W | α, α Q T + |∇ W | α, α Q T ≤ D | V | α, α Q T + D . (3.20)For W solution of ( Q V ) we consider T defined by T ( W ) = ˆ V , where ˆ V is the solutionof the parabolic problem,( Q W ) ˆ V t = ∆ ψ µε ( ˆ V ) + M µ ( ˆ V ) ∇ f µ ( ˆ V ) . ∇ W + [ f µ ( c δ ) − f µ ( ˆ V )] s δ , in Ω × (0 , T ) , ∇ ψ µε ( ˆ V ) .n = 0 , on ∂ Ω × (0 , T ) , ˆ V ( x,
0) = u δ ( x ) , in Ω . ¿From the standard theory of parabolic equations, we have that | ˆ V | α, α Q T ≤ D (cid:18) | W | α, α Q T + |∇ W | α, α Q T (cid:19) + D . (3.21)Moreover defining by L the parabolic operator arising in ( Q W ), namely L ( ˆ V )( x, t ) := ˆ V t − ∆ ψ µε ( ˆ V ) − M µ ( ˆ V ) ∇ f µ ( ˆ V ) . ∇ W − [ f µ ( c δ ) − f µ ( ˆ V )] s δ , we remark that (2.8), the property (3.9) of c δ and the fact that, by (3.7), s δ is positiveimply that L ( u m ) ≤ L (1 − δ ) ≥ . (3.22)Setting T := T ◦ T , the inequalities (3.22) ensure that T maps the convex set K intoitself. Moreover we deduce from (3.21) that T ( K ) is relatively compact in K .Next, we check that T is continuous. Suppose that a sequence ( V m ) m ∈ N converges to alimit V ∈ K in C α, α ( Q T ), as m → ∞ . Since ( V m ) m ∈ N is bounded in C α, α ( Q T ), itfollows from (3.20) that the sequence ( W m := T ( V m )) m ∈ N , where W m is the solution of( Q V m ), is bounded in C α, α ( Q T ), so that as m → ∞ , W m converges to the unique solu-tion W of Problem ( Q V ) in C β, β ( Q T ) for all β ∈ (0 , α ). Moreover W ∈ C α, α ( Q T ).Further it also follows from (3.20) that ( ∇ W m ) m ∈ N is bounded in C α, α ( Q T ), sothat the solution ˆ V m = T ( W m ) of Problem ( Q W m ) is bounded in C α, α ( Q T ). Sinceˆ V m = T ( W m ) = T ( V m ), ( T ( V m )) m ∈ N converges to the unique solution ˆ V of Problem ( Q W )9n C β, β ( Q T ) for all β ∈ (0 , α ), as m → ∞ , so that ˆ V = T ( W ) = T ◦ T ( V ). There-fore we have just proved that ( T ◦ T ( V m )) m ∈ N converges to T ◦ T ( V ) in C β, β ( Q T )for all β ∈ (0 , α ), as m → ∞ , which ensures the continuity of the map T . It follows fromthe Schauder fixed point theorem that there exists a solution ( u µδ , P µδ ) of ( ˜ S µδ ) such that u µδ ∈ K ∩ C α, α ( Q T ) and P µδ , ∇P µδ , ∈ C α, α ( Q T ), ∆ P µδ ∈ C α, α ( Q T ).This concludes the proof of Lemma 3.1. Moreover we deduce from Lemma 3.1 the existenceof a solution of ( S µδ ), namely Corollary 3.2
Assume the hypotheses ( H ) − ( H ) then there exists ( u µδ , p µδ ) solution of ( S µδ ) such that u µδ ∈ C α, α ( Q T ) , u m ≤ u µδ ( x, t ) ≤ − δ (3.23) and p µδ , ∇ p µδ ∈ C α, α ( Q T ) , ∆ p µδ ∈ C α, α ( Q T ) . In view of (2.8) and (3.23) we deduce the following bounds0 = f µ (0) ≤ f µ ( u µδ ( x, t )) ≤ f µ (1) , (4.1)0 = f µ (0) ≤ f µ ( c δ ( x, t )) ≤ f µ (1) , (4.2)0 < k w ( u m ) ≤ k w ( u µδ ( x, t )) ≤ k w (1) = 1 , (4.3)0 = k a (1) ≤ k a ( u µδ ( x, t )) ≤ k a (0) = 1 , (4.4)0 < k w ( u m ) ≤ M µ ( u µδ ) , (4.5) p c (1) ≤ p c ( u µδ ( x, t )) ≤ p c (0) , (4.6) p c (1) − p c (0) ≤ R µ ( u µδ ( x, t )) ≤ , (4.7) p c (1) − p c (0) ≤ Q µ ( u µδ ( x, t )) ≤ , (4.8)for all ( x, t ) ∈ Ω × (0 , T ). Next we state some essential a priori estimates. Lemma 4.1
Let ( u µδ , p µδ ) be a solution of Problem ( S µδ ) . There exists a positive constant C , which only depends on Ω , k w , k a and T such that Z T Z Ω k a ( u µδ ) |∇ p µδ + ∇ p c ( u µδ ) | dxdt ≤ Cµ, (4.9) Z T Z Ω |∇ p µδ | dxdt ≤ C, (4.10)10 nd ≤ − Z T Z Ω ∇ g ( u µδ ) . ∇ p c ( u µδ ) dxdt ≤ C, (4.11) Z T Z Ω |∇ ζ ( u µδ ) | dxdt ≤ C, (4.12) Z T Z Ω |∇ g ( u µδ ) | dxdt ≤ C. (4.13)Proof: We first prove (4.9). Multiplying (3.11) by P = p µδ + R µ ( u µδ ) and integrating theresult on Q T = Ω × (0 , T ) we obtain Z Q T M µ ( u µδ ) |∇ ( p µδ + R µ ( u µδ )) | ≤ h Z Q T ( s δ − s δ ) + h Z Q T ( p µδ + R µ ( u µδ )) , (4.14)for all h >
0. Moreover we have by Poincar´e-Wirtinger inequality that Z Q T ( p µδ + R µ ( u µδ )) ≤ C (cid:20) Z Q T |∇ ( p µδ + R µ ( u µδ )) | + (cid:18) Z Q T p µδ + R µ ( u µδ ) (cid:19) (cid:21) . Using (3.3) and (4.7), it follows that Z Q T ( p µδ + R µ ( u µδ )) ≤ C Z Q T |∇ ( p µδ + R µ ( u µδ )) | + C , which we substitute into (4.14) with h = k w ( u m )2 C to deduce, also in view of (3.7) and(4.5), that Z Q T |∇ ( p µδ + R µ ( u µδ )) | ≤ C and Z Q T | p µδ + R µ ( u µδ ) | ≤ C . (4.15)Furthermore multiplying (3.1) by p µδ and (3.2) by p µδ + p c ( u µδ ), adding up both results andintegrating on Q T we obtain − Z Q T ( u µδ ) t p c ( u µδ ) + Z Q T k w ( u µδ ) |∇ p µδ | + 1 µ k a ( u µδ ) |∇ p µδ + ∇ p c ( u µδ ) | = I, (4.16)where I := Z Q T (cid:20) f µ ( c δ ) s δ − f µ ( u µδ ) (cid:18) s δ + Z − Ω ( s δ − s δ ) (cid:19)(cid:21) p µδ dxdt + Z Q T (cid:20) (1 − f µ ( c δ )) s δ − (cid:18) − f µ ( u µδ ) (cid:19)(cid:18) s δ + Z − Ω ( s δ − s δ ) (cid:19)(cid:21) ( p µδ + p c ( u µδ )) dxdt. We check below that first term on the left-hand-side of (4.16) and I are bounded. Denotingby P c a primitive of p c we have that Z Q T p c ( u µδ )( u µδ ) t = Z Ω Z T ∂∂t [ P c ( u µδ )] . P c is continuous and u µδ is bounded this gives (cid:12)(cid:12)(cid:12)(cid:12) Z Q T p c ( u µδ )( u µδ ) t dxdt (cid:12)(cid:12)(cid:12)(cid:12) ≤ C . (4.17)Moreover we have using (3.3) and (2.13) that I = Z Q T (cid:18) p µδ + R µ ( u µδ ) (cid:19) ( s δ − s δ ) dxdt (4.18) − Z Q T R µ ( u µδ ) (cid:20) f µ ( c δ ) s δ − f µ ( u µδ ) s δ + (cid:18) − f µ ( u µδ ) (cid:19)(cid:18)Z − Ω ( s δ − s δ ) (cid:19)(cid:21) dxdt + Z Q T (cid:20) (1 − f µ ( c δ )) s δ − (cid:18) − f µ ( u µδ ) (cid:19)(cid:18) s δ + Z − Ω ( s δ − s δ ) (cid:19)(cid:21)(cid:20) Q µ ( u µδ ) + p c (0) (cid:21) dxdt. In view of ( H ), (3.7), (4.1), (4.2), (4.7) and (4.8) we obtain I ≤ C Z Q T | p µδ + R µ ( u µδ ) | + C . This together with (4.15) yields I ≤ C C + C . Substituting this into (4.16) and alsousing (4.17) we obtain that Z Q T k w ( u µδ ) |∇ p µδ | + 1 µ k a ( u µδ ) |∇ p µδ + ∇ p c ( u µδ ) | dxdt ≤ C , (4.19)which implies (4.9). In view of (4.3), we also deduce from (4.19) the estimate (4.10).Next we prove (4.11). By the definition (2.9) of g , we obtain from (3.10) that − div (cid:18) M µ ( u µδ ) ∇ p µδ (cid:19) + 1 µ ∆ g ( u µδ ) = s δ − s δ − Z − Ω ( s δ − s δ ) dx. (4.20)Multiplying (4.20) by f µ ( u µδ ) and subtracting the result from (3 .
1) we deduce that( u µδ ) t = 1 µ f µ ( u µδ )∆ g ( u µδ )+ div (cid:18) k w ( u µδ ) ∇ p µδ (cid:19) − f µ ( u µδ ) div (cid:18) M µ ( u µδ ) ∇ ( p µ ) (cid:19) + s δ [ f µ ( c δ ) − f µ ( u µδ )] . (4.21)Moreover using the definition (2.7) of f µ and M µ we note that div (cid:18) M µ ( u µδ ) f µ ( u µδ ) ∇ p µδ (cid:19) = div (cid:18) k w ( u µδ ) ∇ p µδ (cid:19) = M µ ( u µδ ) ∇ ( f µ ( u µδ )) . ∇ p µδ + f µ ( u µδ ) div (cid:18) M µ ( u µδ ) ∇ ( p µ ) (cid:19) , which we substitute into (4.21) to obtain( u µδ ) t − µ f µ ( u µδ )∆ g ( u µδ ) − M µ ( u µδ ) ∇ ( f µ ( u µδ )) . ∇ p µδ = s δ [ f µ ( c δ ) − f µ ( u µδ )] . (4.22)12e set D µ ( a ) := p c ( a ) f µ ( a ) − Q µ ( a ) , (4.23)for all a ∈ [0 , Q µ we have ∇ D µ ( u µδ ) = p c ( u µδ ) ∇ ( f µ ( u µδ )).Substituting this into (4.22), which we have multiplied by p c ( u µδ ), we deduce that p c ( u µδ )( u µδ ) t − µ f µ ( u µδ ) p c ( u µδ )∆ g ( u µδ ) − M µ ( u µδ ) ∇ D µ ( u µδ ) . ∇ p µδ = p c ( u µδ ) s δ [ f µ ( c δ ) − f µ ( u µδ )] . (4.24)Multiplying (4.20) by D µ ( u µδ ), adding the result to (4.24) and also using the fact that div (cid:18) M µ ( u µδ ) D µ ( u µδ ) ∇ p µδ (cid:19) = M µ ( u µδ ) ∇ D µ ( u µδ ) . ∇ p µδ + D µ ( u µδ ) div (cid:18) M µ ( u µδ ) ∇ p µδ (cid:19) , we deduce that p c ( u µδ )( u µδ ) t − µ (cid:18) f µ ( u µδ ) p c ( u µδ ) − D µ ( u µδ ) (cid:19) ∆ g ( u µδ ) − div (cid:18) M µ ( u µδ ) D µ ( u µδ ) ∇ p µδ (cid:19) = p c ( u µδ ) s δ (cid:20) f µ ( c δ ) − f µ ( u µδ ) (cid:21) + D µ ( u µδ ) (cid:18) s δ − s δ − Z − Ω ( s δ − s δ ) (cid:19) . (4.25)Integrating (4.25) on Q T and using the fact that the definition (4.23) of D µ implies p c ( u µδ ) f µ ( u µδ ) − D µ ( u µδ ) = Q µ ( u µδ ) , we obtain Z Q T p c ( u µδ )( u µδ ) t dxdt − µ Z Q T Q µ ( u µδ )∆ g ( u µδ ) dxdt = J, (4.26)where J := Z Q T p c ( u µδ ) s δ [ f µ ( c δ ) − f µ ( u µδ )] dxdt + Z Q T (cid:18) p c ( u µδ ) f µ ( u µδ ) − Q µ ( u µδ ) (cid:19)(cid:18) s δ − s δ − Z − Ω ( s δ − s δ ) (cid:19) dxdt. It follows from (4.1), (4.2), (4.6), (4.8) and (3.7) that | J | ≤ C . Substituting this into(4.26) and also using (4.17) we obtain that0 ≤ − µ Z Q T ∇Q µ ( u µδ ) . ∇ ( g ( u µδ )) dxdt ≤ C . (4.27)Furthermore we remark that 1 µ f µ ( u µδ ) ≥ k w ( u m )2 , ∇Q µ ( u µδ ) = f µ ( u µδ ) ∇ p c ( u µδ ) yields0 ≤ − Z Q T ∇ p c ( u µδ ) ∇ ( g ( u µδ )) dxdt ≤ C . (4.28)By the definition (2.10) of ζ , we have −∇ p c ( u µδ ) ∇ g ( u µδ ) = |∇ ζ ( u µδ ) | . This together with(4.28) implies (4.11) and (4.12), which in view of (4.4) gives (4.13). This completes theproof of Lemma 4.1.In what follows we give estimates of differences of space translates of p µδ and g ( u µδ ). Weset for r ∈ IR + sufficiently small:Ω r = { x ∈ Ω , B ( x, r ) ⊂ Ω } . Lemma 4.2
Let ( u µδ , p µδ ) be a solution of Problem ( S µδ ) ; there exists a positive constant C such that Z T Z Ω r (cid:12)(cid:12)(cid:12)(cid:12) p µδ ( x + ξ, t ) − p µδ ( x, t ) (cid:12)(cid:12)(cid:12)(cid:12) ( x, t ) dxdt ≤ Cξ (4.29) and Z T Z Ω r (cid:12)(cid:12)(cid:12)(cid:12) g ( u µδ )( x + ξ, t ) − g ( u µδ )( x, t ) (cid:12)(cid:12)(cid:12)(cid:12) dxdt ≤ Cξ , (4.30) where ξ ∈ IR N and | ξ | ≤ r. Proof: The inequalities (4.29) and (4.30) follow from (4.10) and (4.13) respectively.Next we estimate differences of time translates of g ( u µδ ). Lemma 4.3
Let ( u µδ , p µδ ) be a solution of Problem ( S µδ ) then there exists a positive con-stant C such that Z T − τ Z Ω [ g ( u µδ )( x, t + τ ) − g ( u µδ )( x, t )] dxdt ≤ Cτ, (4.31) for all τ ∈ (0 , T ) . Proof: We set A ( t ) := Z Ω [ g ( u µδ )( x, t + τ ) − g ( u µδ )( x, t )] dx. Since g is a non decreasing Lipschitz continuous function with the Lipschitz constant C g we have that A ( t ) ≤ C g Z Ω [ g ( u µδ ( x, t + τ )) − g ( u µδ ( x, t ))][ u µδ ( x, t + τ ) − u µδ ( x, t )] dx ≤ C g Z Ω [ g ( u µδ ( x, t + τ )) − g ( u µδ ( x, t ))] (cid:20)Z t + τt ( u µδ ) t ( x, θ ) dθ (cid:21) dx ≤ C g Z Ω Z t + τt (cid:20) g ( u µδ ( x, t + τ )) − g ( u µδ ( x, t )) (cid:21)(cid:20) div ( k w ( u µδ ) ∇ p µδ ) + f µ ( c δ ) s δ − f µ ( u µδ ) (cid:18) s δ + Z − Ω ( s δ ( y ) − s δ ( y )) dy (cid:19)(cid:21) ( x, θ ) dθdx, A ( t ) ≤ C g (cid:26) Z t + τt Z Ω (cid:12)(cid:12)(cid:12)(cid:12) k w ( u µδ )( x, θ ) ∇ p µδ ( x, θ ) ∇ g ( u µδ )( x, t + τ ) (cid:12)(cid:12)(cid:12)(cid:12) dxdθ + Z t + τt Z Ω (cid:12)(cid:12)(cid:12)(cid:12) k w ( u µδ )( x, θ ) ∇ p µδ ( x, θ ) ∇ g ( u µδ )( x, t ) (cid:12)(cid:12)(cid:12)(cid:12) dxdθ + (cid:12)(cid:12)(cid:12)(cid:12) Z Ω (cid:20) g ( u µδ )( x, t + τ ) − g ( u µδ )( x, t ) (cid:21) K ( x, t, τ ) dx (cid:12)(cid:12)(cid:12)(cid:12)(cid:27) , (4.32)where K ( x, t, τ ) := Z t + τt (cid:18) f µ ( c δ ( x, θ )) s δ ( x ) − f µ ( u δ ( x, θ )) (cid:20) s δ ( x ) + Z − Ω ( s δ ( y ) − s δ ( y )) dy (cid:21)(cid:19) dθ. (4.33)Next we estimate the right hand side of (4.32). Using (4.3) we have that Z t + τt Z Ω (cid:12)(cid:12)(cid:12)(cid:12) k w ( u µδ )( x, θ ) ∇ p µδ ( x, θ ) ∇ g ( u µδ )( x, t + τ ) (cid:12)(cid:12)(cid:12)(cid:12) dxdθ ≤ (cid:18) Z t + τt Z Ω |∇ p µδ ( x, θ ) | dxdθ + Z t + τt Z Ω |∇ g ( u µδ )( x, t + τ ) | dxdθ (cid:19) ≤ (cid:18) Z t + τt Z Ω |∇ p µδ ( x, θ ) | dxdθ + τ Z Ω |∇ g ( u µδ )( x, t + τ ) | dx (cid:19) . (4.34)Similarly we have that Z t + τt Z Ω (cid:12)(cid:12)(cid:12)(cid:12) k w ( u µδ )( x, θ ) ∇ p µδ ( x, θ ) ∇ g ( u µδ )( x, t ) (cid:12)(cid:12)(cid:12)(cid:12) dxdθ ≤ (cid:18) Z t + τt Z Ω |∇ p µδ ( x, θ ) | dxdθ + τ Z Ω |∇ g ( u µδ )( x, t ) | dx (cid:19) . (4.35)Moreover using (4.1) and (4.2) we obtain from the definition (4.33) of K that | K ( x, t, τ ) | ≤ Z t + τt (cid:20) | s δ | + | s δ | + Z − Ω | s δ − s δ | dx (cid:21) dθ ≤ (cid:20) | s δ | + | s δ | + Z − Ω | s δ − s δ | dx (cid:21) τ. This together with (3 .
7) and the fact that the function g ( u µδ ) is bounded uniformly on µ and δ yields (cid:12)(cid:12)(cid:12)(cid:12) Z Ω (cid:20) g ( u µδ )( x, t + τ ) − g ( u µδ )( x, t ) (cid:21) K ( x, t, τ ) (cid:12)(cid:12)(cid:12)(cid:12) dx ≤ ˜ Cτ. (4.36)Substituting (4.34), (4.35) and (4.36) into (4.32) we deduce that A ( t ) ≤ C g (cid:18) Z t + τt Z Ω |∇ p µδ ( x, θ ) | dxdθ + τ Z Ω |∇ g ( u µδ )( x, t + τ ) | dx + τ Z Ω |∇ g ( u µδ )( x, t ) | dx + ˜ Cτ (cid:19) , , T − τ ] to obtain Z T − τ A ( t ) dt ≤ C g (cid:18) Z T − τ Z t + τt Z Ω |∇ p µδ ( x, θ ) | dxdθdt + τ Z T Z Ω |∇ g ( u µδ ) | dxdt + ˜ Cτ T (cid:19) ≤ C g (cid:18) τ Z T Z Ω |∇ p µδ ( x, θ ) | dxdθ + τ Z T Z Ω |∇ g ( u µδ ) | dxdt + ˜ Cτ T (cid:19) . In view of (4.10) and (4.13) we deduce (4.31), which completes the proof of Lemma 4.3. δ ↓ . Letting δ tend to 0, we deduce from the estimates given in Lemmas 4.1 and 4.2 theexistence of a weak solution of Problem ( S µ ). More precisely, we have the followingresult, Lemma 5.1
There exists a weak solution ( u µ , p µ ) of Problem ( S µ ) , which satisfies Z T Z Ω (cid:20) k a ( u µ ) (cid:21) (cid:20) ∇ p µ + ∇ p c ( u µ ) (cid:21) dxdt ≤ Cµ, (5.1) Z T Z Ω |∇ p µ | dxdt ≤ C, (5.2) Z T Z Ω |∇ g ( u µ ) | dxdt ≤ C, (5.3) Z T Z Ω r [ g ( u µ )( x + ξ, t ) − g ( u µ )( x, t )] dxdt ≤ Cξ , (5.4) where ξ ∈ IR N and | ξ | ≤ r . Moreover the following estimate of differences of timetranslates holds Z T − τ Z Ω [ g ( u µ )( x, t + τ ) − g ( u µ )( x, t )] dxdt ≤ Cτ, (5.5) for all τ ∈ (0 , T ) . Proof: We deduce from (4.10), (4.30) and (4.31) that there exist functions ˆ g µ and p µ anda subsequence (( u µδ n , p µδ n )) n ∈ N of weak solutions of Problem ( S µδ n ) such that( g ( u µδ n )) n ∈ N tends to ˆ g µ strongly in L ( Q T ) , (5.6)( p µδ n ) n ∈ N tends to p µ weakly in L (0 , T ; H (Ω)) , as δ n tends to zero. Thus for a subsequence, which we denote again by δ n , we have that( g ( u µδ n )) n ∈ N tends to ˆ g µ for almost ( x, t ) ∈ Q T . (5.7)16sing the fact that g is bijective we deduce that( u µδ n ) n ∈ N tends to u µ := g − (ˆ g µ ) strongly in L ( Q T ) and almost everywhere in Q T , (5.8)as δ n tends to zero. Moreover we have in view of (4.13) and (5.6) that ∇ g ( u µδ n ) tends to ∇ g ( u µ ) weakly in L ( Q T ) as δ n ↓
0, so that by the definition (2.9) of gk a ( u µδ n ) ∇ p c ( u µδ n ) tends to k a ( u µ ) ∇ p c ( u µ ) weakly in L ( Q T ) as δ n ↓
0. (5.9)Letting δ n tend to 0 in (3.23) we deduce that u m ≤ u µ ( x, t ) ≤ . (5.10)Moreover we deduce from (3.3) that Z Ω p µ ( x, t ) dx = 0 , for almost every t ∈ (0 , T ) . (5.11)Multiplying (3.1) by ϕ ∈ C , integrating by parts and letting δ n tend to 0 we obtain Z T Z Ω u µ ϕ t dxdt = Z T Z Ω k w ( u µ ) ∇ p µ . ∇ ϕdxdt − Z T Z Ω (cid:18) f µ ( c ) s − f µ ( u µ ) s (cid:19) ϕdxdt − Z Ω u ( x ) ϕ ( x, dx, (5.12)where we have used that u δ tends to u in L (Ω) and that c δ , s δ and s δ tend respectivelyto c , s and s in L ( Q T ) as δ ↓
0. Similarly, multiplying (3.2) by ϕ ∈ C , integrating byparts and letting δ n tend to 0 we deduce that Z T Z Ω (cid:18) − u µ (cid:19) ϕ t dxdt = 1 µ Z T Z Ω (cid:18) k a ( u µ ) ∇ p µ + ∇ g ( u µ ) (cid:19) . ∇ ϕdxdt − Z T Z Ω (cid:18) (1 − f µ ( c )) s − (1 − f µ ( u µ )) s (cid:19) ϕdxdt − Z Ω (cid:18) − u ( x ) (cid:19) ϕ ( x, dx, (5.13)which since ∇ g ( u µ ) = k a ( u µ ) ∇ p c ( u µ ) coincides with (2.15). Next we prove (5.1). We firstcheck that k a ( u µδ n ) ∇ p µδ n tends to k a ( u µ ) ∇ p µ weakly in L ( Q T ) , (5.14)as δ n tends to 0. Let ϕ ∈ L ( Q T ), we have that (cid:12)(cid:12)(cid:12)(cid:12) Z Q T (cid:18) k a ( u µδ n ) ∇ p µδ n − k a ( u µ ) ∇ p µ (cid:19) ϕ dxdt (cid:12)(cid:12)(cid:12)(cid:12) ≤ | I δ n | + | I δ n | , (5.15)where I δ n := Z Q T (cid:18) k a ( u µδ n ) − k a ( u µ ) (cid:19) ∇ p µδ n ϕ dxdt I δ n = Z Q T k a ( u µ ) ϕ (cid:18) ∇ p µδ n − ∇ p µ (cid:19) dxdt. Using the fact that ∇ p µδ n converges to ∇ p µ weakly in L ( Q T ) as δ n ↓
0, we deduce, since k a ( u µ ) ϕ ∈ L ( Q T ), that | I δ n | tends to 0 as δ n ↓
0. (5.16)Moreover we have by (4.10) that | I δ n | ≤ (cid:18) Z Q T (cid:12)(cid:12)(cid:12)(cid:12) k a ( u µδ n ) − k a ( u µ ) (cid:12)(cid:12)(cid:12)(cid:12) ϕ dxdt (cid:19) / (cid:18) Z Q T |∇ p µδ n | dxdt (cid:19) / ≤ C (cid:18) Z Q T (cid:12)(cid:12)(cid:12)(cid:12) k a ( u µδ n ) − k a ( u µ ) (cid:12)(cid:12)(cid:12)(cid:12) ϕ dxdt (cid:19) / . Since (cid:12)(cid:12)(cid:12)(cid:12) k a ( u µδ n ) − k a ( u µ ) (cid:12)(cid:12)(cid:12)(cid:12) ϕ ≤ ϕ and since k a ( u µδ n ) tends to k a ( u µ ) almost everywhere,we deduce from the Dominated Convergence Theorem that I δ n tends to 0 as δ n ↓
0. Thiswith (5.16) implies (5.14), which with (5.9) gives that k a ( u µδ n ) (cid:20) ∇ p µδ n + ∇ p c ( u µδ n ) (cid:21) tends to k a ( u µ ) (cid:20) ∇ p µ + ∇ p c ( u µ ) (cid:21) weakly in L ( Q T ) . (5.17)The functional v Z Q T v dxdt is convex and lower semi continuous from L ( Q T ) to R therefore it is also weakly l.s.c. (see [2] Corollary III.8) and thus we deduce from (4.4),(4.9) and (5.17) that Z Q T (cid:20) k a ( u µ ) (cid:21) (cid:20) ∇ p µ + ∇ p c ( u µ ) (cid:21) dxdt ≤ lim inf δ n ↓ Z Q T (cid:20) k a ( u µδ n ) (cid:21) (cid:20) ∇ p µδ n + ∇ p c ( u µδ n ) (cid:21) dxdt ≤ lim inf δ n ↓ Z Q T k a ( u µδ n ) (cid:20) ∇ p µδ n + ∇ p c ( u µδ n ) (cid:21) dxdt ≤ Cµ, which coincides with (5.1). Finally, we deduce respectively from (4.10), (4.13), (4.30) and(4.31) the estimates (5.2), (5.3), (5.4) and (5.5). This concludes the proof of Lemma 5.1. µ ↓ . The goal of this section is to prove Theorem 2.3. We first deduce from the estimates(5.2), (5.4) and (5.5) that there exists a couple of functions ( u, p ) and a subsequence(( u µ n , p µ n )) n ∈ N such that( u µ n ) n ∈ N tends to u strongly in L ( Q T ) and almost everywhere in Q T , ( p µ n ) n ∈ N tends to p weakly in L (0 , T ; H (Ω)) , µ n tends to zero. Moreover since0 ≤ f µ n ( u µ n ) ≤ , there exists a function ˆ f ∈ L ( Q T ) with 0 ≤ ˆ f ≤ f µ nm ( u µ nm )) n m ∈ N of ( f µ n ( u µ n )) n ∈ N such that ( f µ nm ( u µ nm )) n m ∈ N tends to ˆ f weakly in L ( Q T ) as µ n m tendsto zero. Moreover we deduce respectively from (5.10) and (5.11) that 0 ≤ u ≤ Z Ω p ( x, t ) dx = 0 , for almost every t ∈ (0 , T ) , which gives (2.18). As it is done in Section 6 in the proof of (5.1), one can first check that k a ( u µ nm )( ∇ p µ nm + ∇ p c ( u µ nm )) tends to k a ( u )( ∇ p + ∇ p c ( u )) weakly in L ( Q T ) , as µ n m ↓ µ n m tendsto zero into (5.12) we obtain, since lim µ nm ↓ f µ nm ( s ) = χ ( s ) for all s ∈ [0 , Z T Z Ω uϕ t dxdt = Z T Z Ω k w ( u ) ∇ p. ∇ ϕdxdt − Z T Z Ω (cid:18) χ ( c ) s − ˆ f s (cid:19) ϕdxdt − Z Ω u ( x ) ϕ ( x, dx, which coincides with (2.16) and concludes the proof of Theorem 2.3. In this section we present numerical simulations in one space dimension. To that purposewe apply the finite volume method, which we present below. To begin with, we rewrite theequations (2.1) and (2.2) in the case that Ω = (0 , x, t ) ∈ (0 , × (0 , T ) u µt = ∂ x (cid:18) k w ( u µ ) ∂ x p µ (cid:19) + f µ ( c ) s − f µ ( u µ ) s, (7.1)(1 − u µ ) t = ∂ x (cid:18) µ k a ( u µ ) ∂ x ( p µ + p c ( u µ )) (cid:19) + (1 − f µ ( c )) s − (1 − f µ ( u µ )) s. (7.2)Adding up both equations and using the boundary conditions (2.4) and (2.5) we obtain ∂ x p µ = − k a ( u µ ) k a ( u µ ) + µk w ( u µ ) ∂ x ( p c ( u µ )) . (7.3)Substituting (7.3) into (7.1) yields u µt = − ∂ x (cid:20) f µ ( u µ ) k a ( u µ ) µ ∂ x ( p c ( u µ )) (cid:21) + f µ ( c ) s − f µ ( u µ ) s. (7.4)19oreover we deduce from (7.3) and the definition (2.12) of R µ that ∂ x p µ = − ∂ x ( R µ ( u µ )),so that in view of (2.3) we have p µ ( x, t ) = −R µ ( u µ )( x, t ) + Z R µ ( u µ )( y, t ) dy. (7.5)In the sequel, we compare numerically the solution u µ of (7.4) with the solution u of thelimit equation in the case that u <
1, namely u t = − ∂ x (cid:18) k w ( u ) ∂ x p c ( u ) (cid:19) + χ ( c )¯ s. (7.6)We discretize the time evolution equation (7.4) together with the initial condition andthe homogeneous Neumann boundary condition. The time explicit finite volume schemeis defined by the following equations in which K > J > i ∈ { , ..., [1 / J ] } by[ U µ ] i = u µ ( i J , . (7.7)(ii) For i ∈ { , ..., [1 / J ] } and for n ∈ { , ..., [ T / K ] } the discrete equation is given by1 K (cid:18) [ U µ ] n +1 i − [ U µ ] ni (cid:19) = [ F µ ] ni +1 − [ F µ ] ni + f µ ( C ni ) S ni − f µ ([ U µ ] ni ) S ni , (7.8)where [ F µ ] ni = − J (cid:18) p c ([ U µ ] ni +1 ) − p c ([ U µ ] ni (cid:19) k w ([ U µ ] ni +1 ) k a ([ U µ ] ni ) µk w ([ U µ ] ni +1 ) + k a ([ U µ ] ni ) . (iii) For n ∈ { , ..., [ T / K ] } the discrete Neumann condition is defined by[ F µ ] n = 0 and [ F µ ] n [1 / J ] = 0 . (7.9)The numerical scheme (7.7)-(7.9) allows to build an approximate solution, u J , K : [0 , × [0 , T ] → IR for all i ∈ { , ..., [1 / J ] } and all n ∈ { , ..., [ T / K ] } , which is given by u J , K ( x, t ) = u ni , for all x ∈ ( i J , ( i + 1) J ] and for all t ∈ ( n K , ( n + 1) K ] . (7.10)In order to also compute the pressures, we propose the following discrete equation corre-sponding to (7.5) [ P µ ] ni = −R ([ U µ ] ni ) + J Σ [1 / J ] j =1 R ([ U µ ] nj ) . (7.11)Finally, setting p µg ( x, t ) = p µ ( x, t ) + p c ( u µ )( x, t ) we deduce that([ P µg ]) ni = −R ([ U µ ] ni + p c ([ U µ ] ni ) + J Σ [1 / J ] j =1 R ([ U µ ] nj ) , (7.12)20or all i ∈ { , ..., [1 / J ] } and all n ∈ { , ..., [ T / K ] } . Similarly we propose a finite volumescheme corresponding to the equation (7.6), namely1 K (cid:18) U n +1 i − U ni (cid:19) = F ni +1 − F ni + χ ( C ni ) S ni , (7.13)where F ni = − J (cid:18) p c ( U ni +1 ) − p c ( U ni (cid:19) k w ( U ni +1 ) , for all ( i, n ) ∈ { , ..., [1 / J ] } × { , ..., [ T / K ] } . For the numerical computation we take µ = 10 − , p c ( z ) = 0 , √ − z , k a ( z ) = (1 − z ) , k w ( z ) = √ z and s ( z ) = δ ( z ), s ( z ) = δ ( z ), where δ a is the Dirac function at the point a .Furthermore u µ is given by the line with crosses, p µg is given by the lines with diam andthe limit function u corresponds to the continuous line. First test case:
The case that c = 0 , u = 1 on [0 , t = 0 ,
01 thefollowing pictures
Figure 1 : t=0,01
We note that, for µ small, the functions u and u µ are very close. Here we only startwith water and inject a mixture of water and air. The air immediately invades the wholedomain. Figure 1 illustrates the result which we proved in this paper, namely that u µ tends to the solution u of the limit equation (7.6) as µ tends to 0 and moreover that thepressure p µa is constant. This is indeed the case since u < Second test case:
The case that c = 0 , u ( x ) = ( , , / , / ,
1] . We obtainthe following pictures for t = 0 ,
01 and for t = 0 , .9 0.71.0 0.50.40.3 0.8 0.90.10.0 0.6 1.00.80.70.60.50.40.30.20.10.0 0.2 Figure 2 : t=0,01
Figure 3 : t=0,1
The injection of a mixture of water and air ( c = 0 .
7) takes place in a region of low watersaturation. We first remark that both functions u µ and u evolve very slowly. Here againwe have that u ( x, t ) < x, t ) ∈ (0 , × (0 , T ) and we remark that the graphs ofthe two functions u µ and u nearly coincide. Third test case:
The case that c = 1 and u ( x ) = ( , , / , / ,
1] . We obtain thefollowing pictures for t = 0 ,
01 and for t = 0 , .9 0.71.0 0.50.40.3 0.8 0.90.10.0 0.6 1.00.80.70.60.50.40.30.20.10.0 0.2 Figure 4 : t=0,01
Figure 5 : t=0,1
Here only water is injected; note that the saturation u µ evolves rather fast.23 cknowledgement We would like to thank Professor Rapha`ele Herbin from the University of Provence forher interest in this work.
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