mathematical and numerical analysis for non-equilibrium two phase

Mathematical and numerical analysis for non-equilibriumtwo phase flow models in porous mediaCao, X.

Published: 22/03/2016

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Citation for published version (APA):Cao, X. (2016). Mathematical and numerical analysis for non-equilibrium two phase flow models in porous mediaEindhoven: Technische Universiteit Eindhoven

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Mathematical and NumericalAnalysis for Non-EquilibriumTwo Phase Flow Models in

Porous Media

Xiulei Cao

A catalogue record is available from the Eindhoven University of Technology Library

ISBN: 978-90-386-4025-9

Copyright © 2016 by X. Cao, Eindhoven, The Netherlands.All rights are reserved. No part of this publication may be reproduced, stored in a retrievalsystem, or transmitted, in any form or by any means, electronic, mechanical, photocopy-ing, recording or otherwise, without prior permission of the author.

Mathematical and Numerical Analysis forNon-Equilibrium Two Phase Flow Models in

Porous Media

PROEFSCHRIFT

ter verkrijging van de graad van doctor aan deTechnische Universiteit Eindhoven, op gezag van derector magnificus prof.dr.ir. F.P.T. Baaijens, voor een

commissie aangewezen door het College voorPromoties, in het openbaar te verdedigenop dinsdag 22 maart 2016 om 16.00 uur

door

Xiulei Cao

geboren te Jilin, China

Dit proefschrift is goedgekeurd door de promotoren en de samenstelling vande promotiecommissie is als volgt:

voorzitter: prof.dr. J. de Vlieg1e promotor: prof.dr. I.S. Pop (Universiteit Hasselt)2e promotor: prof.dr.ir. C.J. van Duijnleden: prof.dr.-ing. R. Helmig (Universität Stuttgart)

prof.dr. B. Schweizer (Technische Universität Dortmund)prof.dr. B. Wohlmuth (Technische Universität München)prof.dr.ir. E.H. van Brummelenprof.dr.ir. B. Koren

Het onderzoek of ontwerp dat in dit proefschrift wordt beschreven is uitge-voerd in overeenstemming met de TU/e Gedragscode Wetenschapsbeoefe-ning.

Abstract

Flow models in porous media are encountered in many real life applications of highestsocietal relevance. Typically, such models are built on so-called equilibrium assumptions,where besides physical laws like mass balance or the Darcy’s law, algebraic relationshipsare assumed between model unknowns. In this thesis we consider non-equilibrium models,where dynamic effects and hysteresis are taken into account in the capillary pressure -saturation relationship.

Such models provide solutions that are ruled out by equilibrium models: saturationovershoot, or the development of fingers during infiltration in homogeneous media. Sucheffects have been observed experimentally. The focus here is on the mathematical andnumerical analysis for such models.

We start with simplified, scalar model of pseudo-parabolic type which incorporatesdynamic capillary effects. For this, we prove the uniqueness of a weak solution.

Then we consider the full two-phase porous media flow model, again incorporatingdynamic capillary effects. In this case, we prove the existence of a solution for cases thatinclude degeneracy (the vanishing of highest order terms).

Furthermore, for this model a multipoint flux approximation method is proposed.We give a rigorous convergence proof for both saturation and fluxes. The numericalexperiments show a robust convergence of the method even for irregular grids.

Next we consider heterogeneous media consisting of homogeneous blocks. For thiscase, we derive the conditions to be imposed for the coupling of the models defined ineach block at the separating interface.

For the two-phase flow model involving both dynamic capillarity and hysteresis, weprove again the uniqueness of a solution. The final chapter is a numerical approximation:the moisture transport in concrete during wetting-drying cycles.

Keywords: Non-equilibrium, dynamic effects, hysteresis, degeneracy, multipoint flux ap-proximation.MSC 2010: 35K65, 65J05, 65N08, 65N12, 65Z05, 76S05.

i

Contents

1 Introduction 11.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Mathematical models . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Outline of the thesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Uniqueness of a weak solution for the scalar model 92.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 Uniqueness of the weak solution . . . . . . . . . . . . . . . . . . . . . . 122.3 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3 Existence of weak solutions to the two-phase model 173.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.2 Assumptions and known results . . . . . . . . . . . . . . . . . . . . . . . 193.3 Existence, uniqueness of solutions in the regularized case . . . . . . . . . 22

3.3.1 The weak solution concept . . . . . . . . . . . . . . . . . . . . . 223.3.2 The time discretization . . . . . . . . . . . . . . . . . . . . . . . 243.3.3 A priori estimates . . . . . . . . . . . . . . . . . . . . . . . . . . 283.3.4 Existence of weak solutions to the regularized problem . . . . . . 333.3.5 Uniqueness of the weak solution for Problem Pδ . . . . . . . . . . 35

3.4 Existence of weak solutions for Problem P . . . . . . . . . . . . . . . . . 363.5 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4 Finite volume scheme 534.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.2 The weak solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.3 The finite volume scheme . . . . . . . . . . . . . . . . . . . . . . . . . . 56

4.3.1 Meshes and notations . . . . . . . . . . . . . . . . . . . . . . . . 564.3.2 The scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574.3.3 A priori estimates and existence of the fully discrete solution . . . 59

4.4 Convergence of the scheme . . . . . . . . . . . . . . . . . . . . . . . . . 654.4.1 Compactness results . . . . . . . . . . . . . . . . . . . . . . . . . 65

iii

iv CONTENTS

4.4.2 Convergence results . . . . . . . . . . . . . . . . . . . . . . . . . 704.5 Numerical results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

5 Two phase flow in heterogeneous porous media 795.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 795.2 Mathematical model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 805.3 Conditions at the interface . . . . . . . . . . . . . . . . . . . . . . . . . 83

5.3.1 Constant saturation at the coarse side of the interface . . . . . . 855.3.2 Non-constant saturation at the coarse side of the interface . . . . 895.3.3 Comparison of extended pressure conditions with static case . . . 97

5.4 Numerical schemes and examples . . . . . . . . . . . . . . . . . . . . . . 995.4.1 Linear numerical scheme . . . . . . . . . . . . . . . . . . . . . . 995.4.2 Fully implicit scheme . . . . . . . . . . . . . . . . . . . . . . . . 1025.4.3 Numerical results . . . . . . . . . . . . . . . . . . . . . . . . . . 105

5.5 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

6 Uniqueness of a solution for the hysteresis model 1116.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1116.2 Uniqueness of the weak solution . . . . . . . . . . . . . . . . . . . . . . 1136.3 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

7 Moisture transport in concrete 1237.1 Mathematical model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1237.2 Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

7.2.1 Standard model: one diffusion coefficient . . . . . . . . . . . . . 1257.2.2 Hysteretic model . . . . . . . . . . . . . . . . . . . . . . . . . . 1267.2.3 Numerical results . . . . . . . . . . . . . . . . . . . . . . . . . . 128

7.3 Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

Summary 131

Curriculum Vitae 133

List of publications 135

Acknowledgments 137

Bibliography 139

Chapter 1

Introduction

1.1 Motivation

Many processes in the real life are involving flow in porous media: oil recovery, biologicalsystems, technological applications (batteries, catalytic converters), or underground wastedisposal (CO2 sequestration). Understanding such processes require the development andanalysis of appropriate mathematical models, completed by numerical simulations.

Single and multi-phase porous media flow models are commonly written in terms ofquantities like phase saturation, velocity and pressures. Besides balance equations (massconservation), the mathematical models are including functional relationships relatingthe quantities of interest. Traditionally, these constitutive equations are obtained underequilibrium assumptions: for example, for a given medium and at a given phase saturation,the capillary pressure is fixed. This assumes a static distribution of phases in the poresand disregards the history of the system (hysteresis).

However, the relations measured under infiltration and drainage are different, andexperiments have proven limitations of the standard approaches. For instance, the ex-periments reported in the work of DiCarlo [38] show that in homogeneous media, thesaturation profiles may become non-monotonic during infiltration (see Figure 1.1). In thetwo dimensional case, F. Rezanezhad and the coauthors [93] have given an experimentalsetup and results presenting a fingering behavior (see Figure 1.2).

1

2 CHAPTER 1. INTRODUCTION

Figure 1.1 Water saturation overshoot for different fluxes at the inlet taken from [38]. Observethe non-monotonic saturation profiles encountered at large fluxes q.

(a) (b)

Figure 1.2 The setup (a) and flow path (b) in the experiments reported in [93]. Observe thefingers appearing in the right.

Such results are ruled out by classical models, assuming equilibrium relationships.Specifically, for the experiments as reported in [38], standard models would predict mono-tonic saturation profiles but can not explain the fingering effect. This testifies includingnon-equilibrium effects in the mathematical models.

1.2 Mathematical modelsAs discussed, the mathematical models which describe two phase flow in porous me-dia consist of the mass conservation law, the Darcy’s law and constitutive relationshipsbetween the difference of phase pressures and the phase saturation (the proportion ofthe void space that is occupied by that phase in a reference elementary volume). Underequilibrium conditions, denoting the non-wetting and wetting phase pressures by pn and

1.2. MATHEMATICAL MODELS 3

pw, the capillary pressure, defined as the phase pressure difference, is assumed to be amonotone function of the wetting saturation Sw:

pn − pw = pc(Sw). (1.1)

The function pc is obtained experimentally.With Swr and Snr being so-called irreducible wetting phase saturation and residual

non-wetting phase saturation (an amount of wetting/non-wetting phase that can not beremoved from the porous media under standard drainage or imbibition), one defines theeffective water saturation Se as

Se = Sw − Swr1− Swr − Snr

. (1.2)

In [22], Brooks and Corey proposed the relation

pc(Se) = PbS− 1λ

e , (1.3)

here Pb and λ are characteristic constants of the porous medium. λ is a number whichcharacterizes the pore-size distribution. In [58], Van Genuchten proposed

pc(Se) = 1a

(S− 1m

e − 1) 1n

, (1.4)

where a, m and n are given parameters. For both models above, it is clear that thecapillary pressure is a decreasing function of Sw. However, experimental results haveinvalided this assumption. For example, the breakthrough curves of water saturation andpressure differences in Figure 1.3 are reported in [17]. According to these curves, wecan easily conclude that the difference of phase pressures is not a monotone function ofsaturation any more.

Wat

er

satu

rati

on

[-]

time [s] time [s]

Pn-P

w, [

kPa]

Figure 1.3 The water saturation (left) and the phase pressure difference (right) in theexperiment reported in [17]. Observe the non-monotone dependence.

Alternatively to (1.3) and (1.4), in [60] Hassanizadeh and Gray proposed a model that


includes dynamic effects. Play-type hysteresis is included later in [13], leading to

pn − pw ∈ pc(Sw)− γsign(∂tSw)− τ∂tSw. (1.5)

Here γ, τ ≥ 0 are two parameters or even functions depending on Sw. The multi-valuedsign function is defined by

sign(ξ) =

1 ξ > 0,[−1, 1] ξ = 0,−1 ξ < 0.

In [17], the τ(Sw) is proposed by the experimental results (see Figure 1.4), while Camps-Roach and the co-authors introduced it in [26] (see Figure 1.5).

Figure 1.4 The parameter function τ(Sw), obtained experimentally in [17].

Figure 1.5 The function τ(Sw), measured experimentally in [26].

1.2. MATHEMATICAL MODELS 5

By the mass balance equation for each of the fluid phases, one has (see [12,61])

φ∂Sα∂t

+∇ · qα = fα (α = w, n). (1.6)

Here φ, qα (α = w, n) denote the porosity [−], the volumetric velocity of the α phase[ms ], fα [ 1

s ] is the source of the phase α. For simplicity, the source is assumed to be 0here. The volumetric velocity qα is deduced from the Darcy’s law as

qα = − k

µαkrα(Sα)∇Ψα, (1.7)

where k [m2] is the absolute permeability of the porous medium, Ψα [kg ·m−1 · s−2] thephase potential, µα [kg ·m−1 · s−1] the viscosity and krα the relative permeability of theα phase. Normally, krα is assumed to be a function of Sw. In [24], Burdine proposed as

krw = S2+3λλ

e , (1.8)

for wetting phase andkrn = (1− Se)2(1− S

2+λλ

e ), (1.9)

for non-wetting phase. Additionally, Ψw and Ψn are given by

Ψw = pw + ρwgz, Ψn = pn + ρngz, (1.10)

with ρα (α = w, n), g denoting density [kg · m−3] and the gravitational acceleration[m · s−2]. Further, z is the vertical coordinator, being position in the upward direction.

For two phase porous media flows, one has

Sw + Sn = 1. (1.11)

Combining the equations (1.6)-(1.7) and (1.10)-(1.11), together with (1.5), one obtainsthe following system in terms of the unknown triple (Sw, pw, pn):

φ∂tSw −∇ · (k

µwkrw(Sw)∇pw)−∇ · ( k

µwkrwρwgz) = 0, (1.12)

−φ∂tSw −∇ · (k

µnkrn(Sw)∇pn)−∇ · ( k

µnkrnρngz) = 0, (1.13)


Assuming k constant, we derive the system to a dimensionless form by introducing thereference quantities T , L, Pr and defining

t := t

T, x := x

L, pn := pn

Pr, pw := pw

Pr, pc := pc

Pr, (1.15)


andkw := kPrT

L2φ

krwµw

, kn := kPrT

L2φ

krnµn

. (1.16)

With the dimensionless quantities

−→g1 := ρwgzL

Pr, −→g2 := ρngzL

Pr, γ := γ

TPr, τ := τ

TPr, (1.17)

we obtain the following system:

∂tSw −∇ · (kw(Sw)∇pw)−∇ · (kw−→g1) = 0, (1.18)

−∂tSw −∇ · (kn(Sw)∇pn)−∇ · (kn−→g2) = 0, (1.19)


Observe that kw, kn or pc may vanish, or become ∞ whenever Sw = 0 or Sw = 1. Suchcases we call degenerate. If instead, m,M > 0 exist such that m ≤ kw, kn, pc ≤ M , wespeak about non-degenerate cases.

(1.18) - (1.20) can be written in a different form. Summing (1.18) and (1.19) gives

−∇ ·Q = 0, (1.21)

with Q = kw(Sw)∇pw + kw(Sw)−→g1 + kn(Sw)∇pn + kn(Sw)−→g2 denoting the total flow.Then, one can replace (1.18) or (1.19) by (1.21). In particular cases e.g. in one spatialdimension, (1.21) implies Q = q (constant w.r.t spatial parameter). Then, the system(1.18) - (1.20) reduces to

∂Sw∂t

+∇ ·(q

kwkw + kn

)+∇ ·

((−→g2 −−→g1) kwkn

kw + kn

)= −∇ ·

( kwknkw + kn

∇(pn − pw)), (1.22)


Moreover, in the absence of hysteretic effects (γ = 0), one obtains the pseudo-parabolicequation so-called scalar model:

∂Sw∂t

+∇ ·(q

kwkw + kn

)+∇ ·

((−→g2 −−→g1) kwkn

kw + kn

)= −∇ ·

( kwknkw + kn

∇pc(Sw))

+∇ ·( kwknkw + kn

∇(τ ∂Sw∂t

)). (1.24)

The existence of weak solutions for this model has been proved in [77]. The result includesdegenerate cases i.e. when kw or kn may become 0. In [53], one obtains the existenceand uniqueness of a weak solution for a linear highest order term model. The equivalenceof different formulations is showed in [52]. Traveling wave solutions are analyzed in [40].

1.3. OUTLINE OF THE THESIS 7

Numerical schemes are discussed in [62,63].Returning to the original system (1.18) - (1.20), we distinguish types of non-equilibrium

models:A) τ := τ(Sw), γ = 0

A result in this thesis is the existence of weak solutions, including degenerate cases,when kw or kn may vanish.B) τ = 0, γ 6= 0

This is a so called play-type hysteresis model, yielding two different pn − pw curves(see Figure 1.6), an imbibition curve and drainage one: pn−pw = pc(Sw)−γ, pn−pw =pc(Sw) + γ. Here γ may be a function of Sw. Existence and stability analysis for suchmodels are obtained in [10,96].

Figure 1.6 Two capillary pressure - saturation curves including hysteresis.

C) τ 6= 0, γ 6= 0This is the full model including dynamic effects and play-type hysteresis. [68] and [71]

deal with the existence of weak solutions for this model in non-degenerate cases. In [96],Schweizer extended the result to degenerate cases. Numerical schemes are discussedin [68,71,92]. In this thesis, we will show the uniqueness of a weak solution to this model.

1.3 Outline of the thesisThis thesis is structured as follows:

In Chapter 2, without degeneracy, we provide the uniqueness of a weak solution forthe scalar model. We first transform the scalar model into an equivalent form, then provethe uniqueness of a weak solution to the equivalent system which leads to the uniquenessof a weak solution for the original model. To do so, we use a Green function approach.

In Chapter 3, we analyze a degenerate elliptic-parabolic system with the dynamiceffects τ = τ(Sw) in the phase-pressure difference and without hysteresis. We firsttransform the system (1.12) - (1.14) with γ = 0 into a two-equation system as in [35].


Then, by regularizing the coefficients, the existence and uniqueness of weak solutionsin the non-degenerate case is obtained. Further, the coefficients are assumed to havea certain structure which has physical meaning when the system becomes degenerate.Finally, we let the regularization parameter go to zero to show the existence of weaksolutions under degenerate case.

In Chapter 4, still with γ = 0 and τ constant, we present a finite volume method forthe simulation of the solution. The method is based on a multi-point flux approximation.An energy estimate is derived for the numerical solution, and compactness argumentsare used to prove the convergence to the weak solution as the mesh size tends to zero.Finally, we present some numerical results to confirm the convergence proved rigorously.

Chapter 5 discusses the two phase flow in heterogeneous media. The dynamic ef-fects are taken into account in phase pressure difference. We consider a one-dimensionalheterogeneous case, with two adjacent homogeneous blocks separated by an interface.The absolute permeability is assumed constant, but different in each block. This maylead to the entrapment of the non-wetting phase (say oil) when flowing from the coarsematerial into the fine material. We derive the interface conditions coupling the modelsin each homogeneous block. In doing so, the interface is approximated by a thin porouslayer, and its thickness is then passed to zero. Such results have been obtained earlier forstandard models, based on the equilibrium relationship between the capillary pressure andthe saturation. Then, oil is trapped until its saturation on the coarse material side of theinterface exceeds an entry value. In the non-equilibrium case, the situation is different.Due to the dynamic effects, oil may still flow into the fine material even after the sat-uration drops under the entry point, and this flow may continue for a certain amount oftime that is proportional to the non-equilibrium effects. This suggests that operating in adynamic regime reduces the account of oil trapped at interfaces, leading to an enhancedoil recovery. Finally, we present some numerical results supporting the theoretical findings.

In Chapter 6, both the dynamic effects and hysteresis are included in the capillarypressure, which is the full model of (1.12) - (1.14). The existence of weak solutions forthis model has been proved in [68]. In this chapter, we define an auxiliary elliptic systemto show the uniqueness of a weak solution for this problem.

In Chapter 7, we consider a practical situation related to the durability analysis ofconcrete. Specifically, moisture transport in concrete is discussed. Moisture transport inmarine environment, where drying and wetting cycles occur, leads chloride to penetrateinto reinforced concrete structures. When chloride reaches the rebars, corrosion canappear leading to a decrease of the service life time of the structures. To predict thislife time, one needs to understand the moisture transport. This process is modeled invarious ways. For example, in [8, 37, 75, 84, 98] moisture transport is described in termsof a single diffusion coefficient. In [74, 101], two different diffusion coefficients are usedfor drying and wetting. In this chapter we compare the two approaches for alternatingdrying/wetting cycles and present a numerical scheme. In addition to the modeling, anexperimental set-up is proposed to validate the models.

Chapter 2

Uniqueness of a weak solutionfor the scalar model

2.1 Introduction

In this chapter, we consider the scalar model which is proposed in porous media undernon-equilibrium conditions:

∂tu+∇ · F(u) +∇ · (H(u)∇(pc(u)− τ∂tu)) = 0. (2.1)

Here the two-phase flow model includes dynamic effects in the capillary pressure. If underequilibrium conditions, the difference of the pressures in the two phases (wetting andnon-wetting) is a decreasing function in terms of the (say) wetting phase saturation u

pn − pw = pc(u). (2.2)

Existence and uniqueness as such models are analyzed in [35,69]. However, the equilibriumassumption does not hold in several real life applications, like paper drying. Further,experimental evidence for the non-validity of the equilibrium assumption are provided e.g.in [17, 38, 64]). In this case, dynamic effects have to be included. An example in thissense is the model proposed in [60]:

pn − pw = pc(u)− τ∂tu. (2.3)

Here τ > 0 is a parameter accounting for the dynamic effects.Inspired by this, we consider here a simplified model obtained from the mass conserva-

tion and the Darcy’s law for each phase (see [12,61]), assuming that the medium is fully

This chapter is a collaborative work with I.S. Pop and it has been published in Applied MathematicsLetters, 46(2015): 25-30.

9

10CHAPTER 2. UNIQUENESS OF A WEAK SOLUTION FOR THE SCALAR MODEL

saturated by the two phases. Generally, this leads to a system of two equations, a para-bolic one for the wetting phase saturation, and an elliptic one for the total flux. Here weassume the total flux known, and focus on the mass balance for the wetting phase. Withthe phase pressure difference in (2.3), the model reads as (2.1) (see [28,77]). It is definedin a bounded and connected domain Ω in Rd (d ≥ 1) with a given time interval (0, T ].Further, by Ω we mean the closure of Ω, and by ∂Ω its boundary. In the above, F andH denote the water fractional flow function and the capillary induced diffusion function,and pc is the equilibrium capillary pressure (see (2.2) and (2.3)). These are non-linearfunctions defined for the physically relevant interval u ∈ [0, 1]. For the mathematicalanalysis, we extend H, pc and F continuously to the entire R. Throughout this chapter,we assume

A1: H: R −→ R is Lipschitz continuous, and a h0 exists such that

0 < h0 ≤ H(u) if 0 < u < 1, and H(u) = h0 otherwise. (2.4)

A2: pc ∈ C1(R) is a decreasing function, andmp,Mp exist such that 0 < mp ≤ |p′

c(u)| ≤Mp < +∞, for all u ∈ R.

A3: F: R −→ Rd is Lipschitz continuous.

A4: The functions F and H are bounded, |F(u)|+ |H(u)| ≤M < +∞, for all u ∈ R.

To complete the model, the initial and boundary conditions are given

u(·, 0) = u0, and u = 0, at ∂Ω. (2.5)

For the initial data, we assume

A5: u0 ∈ C0,α(Ω) for some α ∈ (0, 1], u0 = 0 at ∂Ω and u0 ∈ [0, 1] in Ω.

Further, the domain Ω has a smooth boundary:

A6: Ω is a C1,α domain.

The motivation for (2.1) is the two phase flow in porous media. Generally, suchmodels are not only non-linear, but may also degenerate. Whenever one of the phase isnot present. More precise, functional relationships between quantities: the permeability,pressure, and saturation lead to cases when, e.g. H(u) = 0 for u = 0 or u = 1 (see[12, 34, 82]). Then the highest order term on (2.1) is vanishing, and the equation is notparabolic anymore. This makes the analysis and numerical simulation of such problemscomplicated. A usual technique to overcome the difficulties related to degeneracy isregularization. For example, one can approximate non-linearities by regular permeabilities,which are bounded away from 0 and ∞. The uniqueness result obtained here works inthe regularized cases, and was still an open issue even for regular cases.

2.1. INTRODUCTION 11

In this chapter we prove the uniqueness of a weak solution to (2.1) with the giveninitial and boundary conditions. This solution solvesProblem P: Find u ∈W 1,2(0, T ;W 1,2

0 (Ω)), such that u(·, 0) = u0, ∇∂tu ∈ L2(0, T ;L2(Ω)d) and∫ T

0

∫Ω∂tuφdxdt−

∫ T

0

∫ΩF(u) · ∇φdxdt

−∫ T

0

∫ΩH(u)∇pc(u) · ∇φdxdt+ τ

∫ T

0

∫ΩH(u)∇∂tu · ∇φdxdt = 0, (2.6)

for any φ ∈ L2(0, T ;W 1,20 (Ω)).

Note the non-linearity appearing in the highest order term, ∇ · (H(u)∇∂tu). For thelinear case, when only ∆∂tu is appearing, existence and uniqueness results are obtainedin [53]. Also we refer to [28, 77, 78] for the existence of weak solutions to the nonlinearProblem P. However, the uniqueness in the case considered here is still an open issue.To overcome the difficulty related to the non-linearity appearing in the highest orderterm, we use an additional unknown p (see [14, 52]). Closest to our work are the recentresults in [14]. There, the uniqueness is obtained for a similar, even degenerate model,but in the absence of convective terms, and more important, when the non-linearitiesappear only under all derivatives e.g. ∂t∆ψ(u). Apart from some particular cases (i.e.H ≡ constant), the model considered here involves the term ∇ · (H(u)∇∂tu), whichcannot be transformed to the form in [14] . Furthermore, the uniqueness in [14] is provedfor an alternative formulation when (2.1) is written as a system. In the non-degeneratecase, the equivalence between (2.1) and its reformulation as a system (like (2.7)-(2.8))is proved in [52], while in the degenerate case, it is still open. Thus, uniqueness resultsin [14], whenever the model considered here matches the frame work there, gives alsouniqueness for (2.1). However, this is only for particular cases, as mentioned. Besides, inthis chapter, we provide an alternative uniqueness proof, based on Green functions.

To avoid the confusion between u given by Problem P and the solution pair of thefollowing system, we denote the saturation latter by s. With this, (2.1) can be rewrittenformally as the system

∂ts+∇ · F(s) +∇ · (H(s)∇p) = 0, (2.7)p = pc(s)− τ∂ts. (2.8)

For the sake of presentation we finally assume

A7: pc(0) = 0.

Remark 2.1.1. A7 can be avoided: if pc(0) 6= 0, one defines p = pc(s) − pc(0) − τ∂tsin (2.8).

Finally, we mention that throughout this chapter, C denotes a generic constant. For thissystem, a weak solution is a pair (s, p) solving


Problem Ps: Find s ∈W 1,2(0, T ;L2(Ω)) and p ∈ L2(0, T ;W 1,20 (Ω)), such that s(·, 0) =

u0 and

(∂ts, φ)− (F(s),∇φ)− (H(s)∇p,∇φ) = 0, (2.9)(p, ψ) = (pc(s), ψ)− τ(∂ts, ψ), (2.10)

for any φ ∈ L2(0, T ;W 1,20 (Ω)), ψ ∈ L2(0, T ;L2(Ω)). Following [52], the Problems P and

Ps are equivalent. Specifically, under the assumptions A1-A4, if (s, p) is a solution toProblem Ps, then u = s solves Problem P. Conversely, if u solves Problem P, then (s, p)with s = u and p defined in (2.8) solves Problem Ps. This equivalence will be exploitedbelow.

2.2 Uniqueness of the weak solution

In this section we prove the uniqueness of weak solutions for Problem Ps. The equivalenceresult also provides uniqueness for Problem P. The first step is to obtain the essentialboundedness of ∇p:

Lemma 2.2.1. The solution component p of Problem Ps satisfies ∇p ∈ L∞((0, T ]×Ω).

Proof. As shown in [28], if u solves Problem P, one has u ∈ L∞(0, T ;W 1,20 (Ω)), ∂tu ∈

L∞(0, T ;W 1,20 (Ω)). In view of the equivalent result, from (2.10) one has p ∈ L∞(0, T ;

W 1,20 (Ω)). Further, for almost every t, one can use (2.8) to eliminate ∂ts in (2.7)

∇ · (H(s)∇p) + 1τp = 1

τpc(s)−∇ · F(s). (2.11)

Note that given s ∈ L∞(0, T ;W 1,20 (Ω)), (2.11) is a linear elliptic equation in p with the

right hand side 1τpc(s)−∇ · F(s) ∈ L2(Ω). Then as in [35,70], one obtains

‖p‖L∞(0,T ;C0,β(Ω)) ≤ C, 0 < β ≤ 1. (2.12)

Then for almost every t ∈ (0, T ] and for almost every x, y ∈ Ω, x 6= y, by using (2.8), weobtain

p(t, x)− p(t, y)|x− y|β

= pc(u)(x)− pc(u)(y)|x− y|β

− τ∂tu(t, x)− u(t, y)|x− y|β

. (2.13)

Since p ∈ C0,β(Ω), this implies supx,y∈Ω,x6=y

|p(t,x)−p(t,y)||x−y|β ≤ C.

Define w : (0, T ]× Ω2 → R:

w = s(t, x)− s(t, y)|x− y|β

. (2.14)

2.2. UNIQUENESS OF THE WEAK SOLUTION 13

Clearly, a ξ exists such that

τ∂tw − p′

c(ξ)w = p(t, x)− p(t, y)|x− y|β

. (2.15)

Multiplying w in the two sides of (2.15), and integrating from 0 to t (t is arbitrary in(0, T ]) give us

τ

2w2 −

∫ t

0p′

c(ξ)w2dt =∫ t

0

p(t, x)− p(t, y)|x− y|β

wdt+ τ

2

(w(0, ·)

)2. (2.16)

According to A2 and A5, by using the Cauchy - Schwarz inequality, one has

τw2 ≤ τ(u0(x)− u0(y)|x− y|β

)2

+∫ t

0

(p(t, x)− p(t, y)|x− y|β

)2+∫ t

0w2dt

≤ C +∫ t

0w2dt. (2.17)

By Gronwall’s inequality we obtainw2 ≤ C, (2.18)

for any time t ∈ (0, T ]. This implies

|s(t, x)− s(t, y)||x− y|β

≤ C, for almost every x, y ∈ Ω, for every t. (2.19)

Let Ωc be the subset of Ω, where (2.19) holds everywhere. Clearly, m(Ω\Ωc) = 0. Forany x ∈ Ω\Ωc, we can find a sequence xnn∈N ∈ Ωc converging to x, and define

s(t, x) = limn→+∞xn∈Ωc

s(t, xn). (2.20)

According to (2.19), s(t, x) does not depend on the choice of xnn∈N. Then, one hass ∈ C0,β(Ω) (see [48]). Finally, by Theorem 8.33 and Corollary 8.35 in Chapter 8 in [59],we get

|p|1,β ≤ C(|p|0 + |pc(s)|0 + |F(s)|0,β), (2.21)

implying ∇p ∈ L∞((0, T ]× Ω).

To show the uniqueness of system (2.7) - (2.8): let g ∈ L2(Ω) and define Gg as thesolution of the elliptic problem (see [53]):

−τ∇(H(u)∇Gg) +Gg = g, in Ω, (2.22)

with Gg = 0 at the boundary ∂Ω. Here u is a solution of Problem P. It is easy to show

Gg ∈W 1,20 (Ω) and ‖Gg‖W 1,2(Ω) ≤ C‖g‖L2(Ω). (2.23)


We have the following result:

Theorem 2.2.1. Under the assumptions A1-A7, Problem P has a unique solution u.

Proof. As discussed above, the uniqueness of a solution to Problem P follows directly fromthe uniqueness to Problem Ps. Assume now Problem Ps has two solutions, (u, pu), (v, pv),then one has

(∂t(u− v), φ)− (H(u)∇pu −H(v)∇pv,∇φ)− (F(u)− F(v),∇φ) = 0, (2.24)

(pu − pv, ψ)− (pc(u)− pc(v), ψ) + τ(∂t(u− v), ψ) = 0, (2.25)

for any φ ∈ L2(0, T ;W 1,20 (Ω)), ψ ∈ L2(0, T ;L2(Ω)).

Further, we rewrite the above system as follows

(∂t(u− v), φ)− (F(u)− F(v),∇φ)−(H(u)(∇pu −∇pv),∇φ)− ((H(u)−H(v))∇pv,∇φ) = 0, (2.26)

(pu − pv, ψ)− (pc(u)− pc(v), ψ) + τ(∂t(u− v), ψ) = 0. (2.27)

Taking g = u− v into (2.22), one gets Gu−v ∈W 1,20 (Ω) and

(Gu−v, λ) + τ(H(u)∇Gu−v,∇λ) = (u− v, λ), (2.28)

for any λ ∈W 1,20 (Ω). Further, we have

‖Gu−v‖W 1,2(Ω) ≤ C‖u− v‖L2(Ω). (2.29)

Setting φ = Gu−v in (2.26) gives

(∂t(u− v), Gu−v)− (H(u)(∇pu −∇pv),∇Gu−v)−((H(u)−H(v))∇pv,∇Gu−v)− (F(u)− F(v),∇Gu−v) = 0, (2.30)

and choose λ = pu − pv in (2.28), we find that

−(H(u)∇(pu − pv),∇Gu−v) = −1τ

(u− v, pu − pv) + 1τ

(Gu−v, pu − pv). (2.31)

Substituting (2.31) into (2.30) leads to

(∂t(u− v), Gu−v)−1τ

(u− v, pu − pv) + 1τ

(Gu−v, pu − pv)

−((H(u)−H(v))∇pv,∇Gu−v)− (F(u)− F(v),∇Gu−v) = 0. (2.32)

Taking the test function ψ = Gu−v in (2.27), we obtain

(∂t(u− v), Gu−v) + 1τ

(Gu−v, pu − pv) = 1τ

(pc(u)− pc(v), Gu−v). (2.33)


Further, substituting (2.33) into (2.32) gives

1τ

(pc(u)− pc(v), Gu−v)−1τ

(u− v, pu − pv)

−((H(u)−H(v))∇pv,∇Gu−v)− (F(u)− F(v),∇Gu−v) = 0. (2.34)

Setting ψ = u− v in (2.27) gives

−1τ

(u− v, pu − pv) = (∂t(u− v), u− v)− 1τ

(pc(u)− pc(v), u− v). (2.35)

Combining (2.34) and (2.35) to eliminate − 1τ (u−v, pu−pv) and integrating the resulting

equation in time over (0, t), with t ∈ (0, T ] arbitrary lead to∫ t

0(∂t(u− v), u− v)dz − 1

τ

∫ t

0(u− v, pc(u)− pc(v))dz −

∫ t

0(F(u)− F(v),∇Gu−v)dz

=− 1τ

∫ t

0(Gu−v, pc(u)− pc(v))dz +

∫ t

0((H(u)−H(v))∇pv,∇Gu−v)dz. (2.36)

We proceed by estimating each term called T1, T2, T3, T4, T5. For T1, since u(0) = v(0) =u0, one has ∫ t

0

∫Ω∂t(u− v)(u− v)dxdz = 1

2‖(u− v)(t)‖2L2(Ω). (2.37)

For T2, by A2, one obtains

−1τ

∫ t

0

∫Ω

(u− v)(pc(u)− pc(v))dxdz ≥ 0. (2.38)

Similarly, since F is Lipschitz continuous, and according to (2.29), we get the estimatesfor T3 and T4:

|1τ

∫ t

0

∫Ω

(F(u)− F(v))∇Gu−vdxdz| ≤C

τ

∫ t

0‖u− v‖2L2(Ω)dz, (2.39)

|1τ

∫ t

0

∫ΩGu−v(pc(u)− pc(v))dxdz| ≤ C

τ

∫ t

0‖u− v‖2L2(Ω)dz. (2.40)

Finally, for the last term T5, by using the Cauchy - Schwarz inequality, A1 and ∇p ∈L∞((0, T ]× Ω), we obtain∫ t

0

∫Ω

(H(u)−H(v))∇pv · ∇Gu−vdxdz ≤ C∫ t

0

∫Ω|(H(u)−H(v)) · ∇Gu−v|dxdz

≤ C∫ t

0‖u− v‖2L2(Ω)dz.


Summarizing the above leads to

‖(u− v)(·, t)‖2L2(Ω) ≤ C∫ t

0‖u− v‖2L2(Ω)dz. (2.41)

Since t is arbitrary, the Gronwall lemma shows that

‖(u− v)(·, t)‖2L2(Ω) = 0, for all t, (2.42)

implying the uniqueness for Problem Ps and therefore for Problem P as well.

2.3 ConclusionsIn this chapter, we have proved the uniqueness of a weak solution to a pseudo-parabolicequation modeling two-phase flow in porous media, and including dynamic effects in thecapillary pressure. The major difficulty is in the non-linearity of the third order derivativeterm. The proof uses the equivalence of the original problem with a mixed form of it. Bythis, the third order derivative term is avoided.

Chapter 3

Existence of weak solutions tothe two-phase model

3.1 Introduction

We analyze the existence and, where appropriate, uniqueness of a weak solution to theelliptic - parabolic system:

∂tu+∇ · (kn(u)∇p)−∆θ(u) = 0, (3.1)

∇ · (k(u)∇p) +∇ · (kw(u)∇(τ(u)∂tu)) = 0, (3.2)

with k = kw + kn. The equations hold in Q := (0, TM ] × Ω. Here Ω is a boundeddomain in Rd (d = 1, 2, 3), having Lipschtiz continuous boundary, and TM > 0 is a givenmaximal time. The unknowns are u and p. The work is motivated by two-phase flow inporous media (e.g. oil and water).

The system (3.1) - (3.2) models two phase flow in porous media, with dynamic effectsin the phase pressure difference. It is obtained by including the Darcy’s law for both phasesin the mass conservation laws. With w, n being indices for the wetting, respectively, non-wetting phase, the mass conservation equations are (see [12,61]):

φ∂sα∂t

+∇ · qα = 0, α = w, n. (3.3)

The coefficient φ represents the porosity of the porous medium, while sα and qα denotethe saturation and the volumetric velocity of the α phase. The volumetric velocity qα is

This chapter is a collaborative work with I.S. Pop and it has been published in Journal of DifferentialEquations, 260 (2016): 2418-2456.

17

18CHAPTER 3. EXISTENCE OF WEAK SOLUTIONS TO THE TWO-PHASE MODEL

deduced from the Darcy’s law as

qα = − k

µαkrα(sα)∇pα, α = w, n, (3.4)

where k is the absolute permeability of the porous medium, pα the pressure, µα theviscosity and krα the relative permeability of the α phase. The specific function of krα isassumed to be known. Substituting (3.4) in (3.3) gives

φ∂sα∂t−∇ · ( kkrα

µα∇pα) = 0, α = w, n. (3.5)

We assume that only two phases are present

sw + sn = 1. (3.6)

To complete the model, one commonly assumes a relationship between the phase pressuredifference and sw. Under equilibrium assumption, this is

pn − pw = pc(sw),

with a given function pc = pc(·). Experimental results [38] have, however, proved thelimitation of this assumption. Alternatively, in [60] the following relation is proposed:

pn − pw = pc(sw)− τ(sw)∂sw∂t

. (3.7)

The damping function τ as well as the function pc, which represents the capillary pressureunder equilibrium condition, are assumed to be known. Summing the two equations from(3.5) and making use of (3.6) gives:

∇ · F = 0, (3.8)

where F = kkrwµw∇pw + kkrn

µn∇pn denotes the total flow.

Introducing the normalized relative permeabilities

kα := krαµα

, α = w, n,

and with k = kw + kn, we also define the fractional flow function

fw(sw) := kwk.

3.2. ASSUMPTIONS AND KNOWN RESULTS 19

Then, we follow [5,33], and define the global pressure p

p = pn −∫ sw

CD

fw(z)p′

c(z)dz, (3.9)

which leads to the following expression of the water pressure

pw = p+∫ sw

CD

fw(z)p′

c(z)dz − pc(sw) + τ(sw)∂tsw. (3.10)

Here CD ∈ (0, 1) is a constant that will be used as the boundary value of water saturation.Furthermore, define the complementary pressure θ as the integral (Kirchhoff) transform-ation

θ(sw) = −∫ sw

CD

kwknk

(z)p′

c(z)dz. (3.11)

Then from (3.5) for the wetting phase, by using (3.6) and (3.9) gives

φ∂tsw +∇ · (kkn∇p)−∇ · (k∇θ(sw)) = 0. (3.12)

Finally, (3.8) becomes

∇ · (kk∇p) +∇ ·(kkw∇(τ∂tsw)

)= 0. (3.13)

The system (3.12)-(3.13) is in dimensional form. Taking Lr, Tr, and Pr as characteristicvalues for the length, time, global pressure, respectively, scaling the space variable x withLr, the time t with Tr and the pressure p with Pr, and assuming

TrL2r

= φ

kPr,

we obtain the following system:

∂tu+∇ · (kn∇p)−∆θ(u) = 0, (3.14)

∇ · (k∇p) +∇ · (kw∇(τ∂tu)) = 0, (3.15)

with u = sw.

3.2 Assumptions and known results

For the non-linearities appearing in the model, we refer to [12, 61, 82]. The special as-sumptions are given below. Here we mention the typical choices of permeability in theliterature ( [22,24])

kw(u) = uα, kn(u) = (1− u)β , with α, β > 1,


and−p′

c(u) = u−λ, λ > 1.

These are defined for u ∈ [0, 1] (the physical range).The model (3.14) - (3.15) is completed by the initial condition

u(0, ·) = u0 in Ω, (3.16)

and the boundary conditions

u = CD, p = 0 at ∂Ω, for all t > 0, (3.17)

where u0 is a given function and the constant CD satisfies 0 < CD < 1.

Remark 3.2.1. To avoid an excess of technicalities, the pressure boundary condition in(3.17) is assumed to be constant. Other types of boundary conditions, as Neumann canbe considered. Similarly, the boundary values of u may be non-constant, but should bebounded away from 0 and 1.

In this chapter we assume

A1: The functions kw, kn [0, 1] −→ [0, 1] are C1, kw is an increasing function, and kn isdecreasing. pc ∈ C1((0, 1],R+) is a decreasing function. τ ∈ C1([0, 1),R+) satisfiesthe following:

i: There exists τ0 > 0 such that τ ≥ τ0 for all u ∈ [0, 1).

ii: There exist u∗ ∈ [0, 1) , ω > 0 and Cτ > 0 such that τ(u) = Cτ (1− u)−ω forall u ∈ [u∗, 1). The restriction on ω > 0 will be given in assumption A5 below.

Furthermore, we assume −p′ckw ≤ C < +∞.

These functions (see also [12]) are defined on (0, 1), which is the physically relevant range.For the analysis, we extend them to R in the following way:

a: kw(u) = kw(0), if u ≤ 0, kw(u) = kw(1), if u ≥ 1.

b: kn(u) = kn(0), if u ≤ 0, kn(u) = kn(1), if u ≥ 1.

c: 1p′c(u) = 1

p′c(0) , if u ≤ 0, 1p′c(u) = 1

p′c(1) , if u ≥ 1.

d: 1/τ(u) = 1/τ(0), if u ≤ 0, 1/τ(u) = 1/τ(1), if u ≥ 1.

Remark 3.2.2. By A1, p′c and τ may become unbounded. Then an extension by con-tinuity is not meaningful. However, this extension makes sense for the reciprocals 1

p′c,

1τ .

3.2. ASSUMPTIONS AND KNOWN RESULTS 21

Remark 3.2.3. The assumptions on kw, kn and p′c are commonly used functions in porousmedia literature. For the behavior of τ , we refer to [26,94]. The experimental results thereshow that τ is bounded away from 0. Moreover, the plots close to full water saturationsuggest a rapid increase of τ close to u = 1. Furthermore, such a profile is obtained in [18]by means of upscaling techniques. This motivated considering the present framework forthe existence proof. At the same time, we mention that other types of τ−curves havealso been proposed in e.g. [56, 57]. The existence proof in such cases remains an openresearch question.

In the proofs below, we will use the functions (see [28,77]) G,Γ : R→ R ∪ ±∞

G(u) =∫ u

CD

τk

kwkn(z)dz, Γ(u) =

∫ u

CD

G(z)dz, (3.18)

and the function T (u) : R→ R ∪ ±∞

T (u) =∫ u

0τ(z)dz + CT , (3.19)

where CT is a constant specified below. Clearly, Γ is a convex function satisfying Γ(0) =G(0) = 0, implying that

Γ(u) ≥ 0, for all u ∈ R. (3.20)

A2: The initial condition u0 satisfies u0 ∈ W 1,20 (Ω) + CD,

∫Ω Γ(u0) < +∞, T (u0) ∈

W 1,2(Ω).

The problem considered here is an extension of scalar, two-phase flow models studiedin [28], while only one pseudo-parabolic equation is considered in the unknown u. Themodel in [28] can be derived from the present one assuming that the total flow is known(see [5, 52]). After non-dimensionlization and some algebraic manipulation one gets

∂tu+∇ · F−∆θ(u)− τ∇ · (kwknk∇∂tu) = 0. (3.21)

The existence and uniqueness of the weak solution have been shown in [52] in the linearcase of the third order derivative term. In the nonlinear and non-degenerate case of thethird order derivative term, we refer to [16] which proved the existence of a solutionand uniqueness in one and two dimensional cases. When the coefficients also depend onthe time derivative of the unknown, the existence of weak solutions has been analyzedin [95]. The present work is closer to [77], where a degenerate pseudo parabolic equationmodeling one-phase flow is considered. The existence is proved based on regularization.For uniqueness but in the non-degenerate case, we refer to [14, 30]. Here we build onsimilar ideas in [77] to prove for the existence for the two-phase model.

Under equilibrium assumptions (τ = 0), such models are analyzed e.g. in [6, 35,69]. For non-equilibrium, two-phase flow models are analyzed in [68], where existence


is obtained for non-degenerate cases, but including hysteresis effects, and in [32], whereuniqueness is proved again in non-degenerate cases.

In what follows, we analyze first the regularized model, and prove the existence anduniqueness of a solution, as well as a priori energy estimates. This is achieved by Rothe’smethod [65]. Finally, we pass the regularization parameter to 0 to obtain the existence ofweak solutions in the degenerate case. Note that uniqueness remains open.

3.3 Existence, uniqueness of solutions in the regularizedcase

3.3.1 The weak solution concept

We use the standard spaces L2(Ω), W 1,2(Ω) and W 1,20 (Ω) in the theory of partial differ-

ential equation. By (·, ·) we mean the scalar product and ‖·‖ stands for the correspondingnorm in L2(Ω), or where needed, in (L2(Ω))d. Furthermore, L2(0, TM ;X) denotes theBochner space of X -valued functions. Let us now define the set

V := W 1,20 (Ω) + CD.

Inspired by [77], a weak solution to the model (3.14)-(3.17) solves the followingProblem P: Find u ∈ L2(0, TM ;V ), such that ∂tu ∈ L2(Q) and p ∈ L2(0, TM ;W 1,r∗

0 (Ω))(for some particular r∗ ∈ (1, 2)), such that u(0, ·) = u0,

√kn∇p ∈ L2(0, TM ;L2(Ω)d),√

kw(∇p+∇∂tT (u)) ∈ L2(0, TM ;L2(Ω)d), and

(∂tu, φ)− (kn∇p,∇φ) + (∇θ(u),∇φ) = 0, (3.22)

(k∇p,∇ψ) + (kw∇∂tT (u),∇ψ) = 0, (3.23)

for any φ, ψ ∈ L2(0, TM ;W 1,20 (Ω)).

Before dealing with the degenerate case, we consider first the regularized case. Tothis aim, we define:

a’: kwδ(u) = kw(u+ δ), if 0 ≤ u ≤ 1− δ, kwδ(u) = kw(δ), if u < 0, kwδ(u) = kw(1),if u > 1− δ.

b’: knδ(u) = kn(u− δ), if δ ≤ u ≤ 1, knδ(u) = kn(0), if u < δ, knδ(u) = kn(1− δ),if u > 1.

c’: −p′cδ(u) = −p′c(u+ δ), if 0 ≤ u ≤ 1− δ, −p′cδ(u) = −pc′(δ), if v < 0, −p′cδ(u) =−p′c(1), if u > 1− δ.

3.3. EXISTENCE, UNIQUENESS OF SOLUTIONS IN THE REGULARIZED CASE 23

d’:

τδ(u) :=

τ(u), if u < u∗,

τ(u∗), if u ∈ [u∗, u∗ + δ],τ(u− δ), if u ∈ [u∗ + δ, 1),τ(1− δ), if u ≥ 1.

Furthermore, we let

Gδ(u) =∫ u

CD

τδkδkwδknδ

(z)dz, Γδ(u) =∫ u

CD

Gδ(z)dz, (3.24)

kδ(u) = kwδ(u) + knδ(u), fwδ = kwδkδ

, θδ(u) = −∫ u

CD

kwδknδkδ

(z)p′

cδ(z)dz, (3.25)

Tδ(u) =∫ u

0τδ(z)dz + CTδ. (3.26)

Here CTδ is a constant which will be specified in Section 3.4.

We now consider the regularized approximation of (3.14) - (3.15)

∂tuδ +∇ · (knδ∇pδ)−∆θδ(uδ) = 0, (3.27)

∇ · (kδ∇pδ) +∇ · (kwδ∇∂tTδ(uδ)) = 0, (3.28)

with the following initial and boundary conditions

uδ(0, ·) = u0 in Ω,

uδ = CD, pδ = 0 at ∂Ω, for all t > 0.

We start by proving the existence of a solution for the followingProblem Pδ: Find uδ ∈ L2(0, TM ;V ) and pδ ∈ L2(0, TM ;W 1,2

0 (Ω)), such that ∂tuδ ∈L2(Q), uδ(0, ·) = u0, ∇∂tTδ(uδ) ∈ L2(0, TM ;L2(Ω)d) and∫ TM

0(∂tuδ, φ)dt−

∫ TM

0(knδ∇pδ,∇φ)dt+

∫ TM

0(∇θδ(uδ),∇φ)dt = 0, (3.29)

∫ TM

0(kδ∇pδ,∇ψ)dt+

∫ TM

0(kwδ∇∂tTδ(uδ),∇ψ)dt = 0, (3.30)

for any φ, ψ ∈ L2(0, TM ;W 1,20 (Ω)).

The proof is based on the method of Rothe [65]. In what follows we use the elementaryinequality

ab ≤ 12σa

2 + σ

2 b2, for any a, b ∈ R, σ > 0. (3.31)


3.3.2 The time discretization

With N ∈ N, let h = TM/N and consider the Euler implicit discretization of (3.29)-(3.30).Problem Pnδ : Given un−1

δ ∈ V (n = 1, 2...N), find unδ ∈ V and pnδ ∈ W1,20 (Ω), such

that(unδ − u

n−1δ

h, φ)− (knδ(unδ )∇pnδ ,∇φ) + (∇θδ(unδ ),∇φ) = 0, (3.32)

(kδ(unδ )∇pnδ ,∇ψ) + (kwδ(unδ )∇Tδ(unδ )− Tδ(un−1

δ )h

,∇ψ) = 0, (3.33)

for any φ, ψ ∈W 1,20 (Ω). Here (·, ·) means L2 inner product.

Lemma 3.3.1. Under the assumptions A1, A2, Problem Pnδ has a solution.

Proof. We start with a finite dimensional approximation (Galerkin), for which we provethe existence of a solution to Problem Pnδ . To this aim, we use Lemma 1.4 (p. 164in [100]). Then we pass to the limit for the discrete solution, and use compactness toshow the existence of a solution for Problem Pnδ .

Let vm∞m=1 be the countable basis of the separable space W 1,20 (Ω). Set Wm =

spanvm(k = 1, ...,m). Then, given α1, ..., αm ∈ R, β1, ..., βm ∈ R, define v =∑mi=1 αivi ∈ Wm, w =

∑mi=1 βivi ∈ Wm, and set ς = v + CD. A finite dimensional

solution (ς, w) ∈ Wm + CD × Wm of (3.32)-(3.33) satisfies

(ς − un−1

δ

h, φ)− (knδ(ς)∇w,∇φ) + (∇θδ(ς),∇φ) = 0, (3.34)

(kδ(ς)∇w,∇ψ) + (kwδ(ς)∇Tδ(ς)− Tδ(un−1

δ )h

,∇ψ) = 0, (3.35)

for any φ, ψ ∈Wm.To avoid an excess of notations in the first part, we do not use any different indices

for the solution pair (ς, w) in the finite dimensional case. For i = 1, ...,m, define

βi = (kδ(ς)∇w,∇vi) + 1h

(kwδ(ς)∇(Tδ(ς)− Tδ(un−1δ )),∇vi), (i = 1, 2, ...,m). (3.36)

Note that if (ς, w) is a solution pair to (3.34)-(3.35), one has βi = 0 for all i. Byw =

∑mi=1 βivi, one clearly has

m∑i=1

βiβi = (kδ(ς)∇w,∇w) + 1h

(kwδ(ς)∇(Tδ(ς)− Tδ(un−1δ )),∇w)

=∥∥∥√kδ(ς)∇w∥∥∥2

+ 1h

(τδ(ς)kwδ(ς)∇ς,∇w)− 1h

(τδ(un−1δ )kwδ(ς)∇un−1

δ ,∇w).

(3.37)


Further, we define g : R→ R by

g(ς) =∫ ς

0

τδkwδknδ

(z)dz.

By A1, there exist 0 < M0 ≤M1 <∞ possibly depending on δ such that

g(ς)ς ≥M0ς2, |g(ς)| ≤M1|ς|. (3.38)

As above, for i = 1, ...,m, we define

li = (g(ς), vi)‖vi‖2

(i = 1, 2, ...,m),

which immediately implies

g(ς) =m∑i=1

livi. (3.39)

Defineli = (

ς − un−1δ

h,vih

)− 1h

(knδ(ς)∇w,∇vi) + 1h

(∇θδ(ς),∇vi), (3.40)

as in above , if (ς, w) is a solution pair, then li = 0 for all i. Similarly, since ∇g(ς) =τδkwδknδ

(ς)∇ς, we get

m∑i=1

li li = 1h2 (ς − un−1

δ , g(ς))− 1h

(knδ(ς)∇w,τδkwδknδ

(ς)∇ς)

+ 1h

(∇θδ(ς),τδkwδknδ

(ς)∇ς). (3.41)

Adding (3.37) and (3.41) yields

m∑i=1

βiβi +m∑i=1

li li =∥∥∥√kδ(ς)∇w∥∥∥2

+ 1h2 (ς − un−1

δ , g(ς))

+ 1h


(ς)∇ς)− 1h


δ ,∇w).

(3.42)

The existence of a solution is provided if the sum on the left in (3.42) becomes positivefor (l1, ..., lm) and (β1, ..., βm) sufficiently large. To prove this, denote the terms on the


right by T1, T2, T3, T4. Note that T1 is positive. For T2, we use (3.38) to obtain

1h2 (ς − un−1

δ , g(ς)) = 1h2 (ς, g(ς))− 1

h2 (un−1δ , g(ς))

≥ M0

h2 · ‖ς‖2 − 1

2h2 ·M2

1M0

∥∥un−1δ

∥∥2 − M0

2h2 · ‖ς‖2

= M0

2h2 · ‖ς‖2 − 1

2h2 ·M2

1M0

∥∥un−1δ

∥∥2. (3.43)

Recalling (3.25), T3 becomes

1h


(ς)∇ς) = 1h

(−kwδ · knδkδ

p′

cδ(ς)∇ς,τδkwδknδ

∇ς)

= 1h

∥∥∥∥ kwδ√kδ√−τδp

′cδ(ς)∇ς

∥∥∥∥2.

For T4, one gets

− 1h


δ ,∇w) ≥ − 12h2

∥∥∥∥∥τδ(un−1δ )kwδ(ς)√kδ(ς)

∇un−1δ

∥∥∥∥∥2

−12

∥∥∥√kδ(ς)∇w∥∥∥2.

(3.44)Using (3.43)-(3.44), the Poincare and Cauchy-Swarchz inequalities lead to

m∑i=1

βiβi +m∑i=1

li li ≥M2

2 ‖∇w‖2 + M0

2h2 ‖ς‖2 + M3

h‖∇ς‖2

− 12h2


∇un−1δ

∥∥∥∥∥2

− 12h2

M21

M0

∥∥un−1δ

∥∥2

≥C(Ω)M2

2 ‖w‖2 + M0

2h2 ‖ς‖2 + M3

h‖∇ς‖2

− 12h2


∇un−1δ

∥∥∥∥∥2

− 12h2

M21

M0

∥∥un−1δ

∥∥2. (3.45)

Note that M2 is independent of δ, and M3 may depend on δ, but they all not dependon the dimension of the space Wm. We know that (3.36) and (3.40) define a mapping

ζm : R2m → R2m by ζm

(l

β

)=(

l

β

), which is continuous by (A1). As ‖l‖ +

‖β‖ (e.g. in the standard Euclidean norm) becomes large enough, ζm is strictly positive,and Lemma 1.4 in [100] (p. 164) guarantees that ζm has a zero, that means there existsa solution to the discrete system. We now pass to the limit as m → ∞. Denoting by(um, pm) ∈ Wm + CD × Wm the finite-dimensional solution (ς, w) obtained above,


from (3.45), we get

C(Ω)M2

2 ‖∇pm‖2 + M0

2h2 ‖um‖2 + M3

h‖∇um‖2

≤ 12h2

∥∥∥∥∥τδ(un−1δ )kwδ(um)√kδ(um)

∇un−1δ

∥∥∥∥∥2

+ 12h2

M21

M0

∥∥un−1δ

∥∥2

≤C,

where C may depend on δ, but not on the dimension m. This means that um is uniformlybounded in V , pm is uniformly bounded in W 1,2

0 (Ω), so we can find u ∈ V, p ∈W 1,20 (Ω),

such that, um u weakly in V and pm p weakly inW 1,20 (Ω). The compact embedding

of W 1,2(Ω) into L2(Ω) gives um → u and pm → p strongly in L2(Ω).

We show that the limit pair (u, p) ∈ (V ×W 1,20 (Ω)) is a solution of (3.32) and (3.33).

By the boundedness of knδ, the sequence knδ(um)∇pm is bounded uniformly in L2(Ω)(w.r.t. m), so it has a weak limit χ. We identify this limit as knδ(u)∇p implying that

(knδ(um)∇pm,∇φ) −→ (knδ(u)∇p,∇φ), for any φ ∈W 1,20 (Ω).

To do so, we choose the test function φ ∈ C∞0 (Ω), clearly, one yields

(knδ(um)∇pm,∇φ) = (∇pm, knδ(u)∇φ) + (∇pm, (knδ(um)− knδ(u)∇φ)),

and since knδ(u) ∈ L∞(Ω), one has

(∇pm, kδ(u)∇φ) −→ (∇p, knδ(u)∇φ).

We now show that the limit of (∇pm, (knδ(um)−knδ(u))∇φ) is 0. Since ∇pm is boundeduniformly in (L2(Ω))d, one has

|(∇pm, (knδ(um)− knδ(u)∇φ))| ≤ ‖∇pm‖ · ‖(knδ(um)− knδ(u))∇φ‖

≤ C(∫

Ω(knδ(um)− knδ(u))2|∇φ|2dx

) 12

.

Because knδ is bounded, we get

|(knδ(um)− knδ(u))∇φ| ≤ C|∇φ| in Ω.

Further, since um −→ u strongly in L2(Ω), in the view of the continuity of knδ we have

knδ(um) −→ knδ(u) a.e.


Then, by the Dominated Convergence Theorem(∫Ω

(knδ(um)− knδ(u))2|∇φ|2dx) 1

2

−→ 0.

Therefore, for any φ ∈ C∞0 (Ω), since knδ(um)∇pm → χ (weakly inW 1,2(Ω)), due to thedensity of C∞0 (Ω) inW 1,2(Ω), this allows identifying χ = knδ(u)∇p for any φ ∈W 1,2

0 (Ω).Similarly, we can also prove that

(kδ(um)∇pm,∇φ) −→ (kδ(u)∇p,∇φ),

(τδ(un−1δ )kδ(um)∇un−1,∇φ) −→ (τδ(un−1

δ )kδ(u)∇un−1,∇φ),

(τδ(um)kwδ(um)∇um,∇φ) −→ (τδ(u)kwδ(u)∇u,∇φ),

(θδ(um)∇um,∇φ) −→ (θδ(u)∇u,∇φ),

for any φ ∈W 1,20 (Ω).

From now on a solution pair to Problem Pnδ is denoted by (unδ , pnδ ).

3.3.3 A priori estimates

Having established the existence for the time discrete problems, we now prove the existenceof a solution to Problem Pδ. To achieve this, we use the elementary result.

Lemma 3.3.2. Let k ∈ 1, ..., N , m ≥ 1. Given two sets of real vectors ak,bk ∈ Rm (k =1, ..., N), one has

N∑k=1

(ak − ak−1, ak) = 12(|aN |2 − |a0|2 +

N∑k=1|ak − ak−1|2).

We have the following:

Lemma 3.3.3. A C > 0, not depending on h exists such that, for any N∗ ∈ 1, 2, ..., N ,one has ∫

ΩΓδ(uN

∗

δ )dx+ 12

∥∥∥∇Tδ(uN∗δ )∥∥∥2≤ C,

N∗∑n=1

∥∥∇(Tδ(unδ )− Tδ(un−1δ ))

∥∥2 + h

N∗∑n=1

∥∥∥∥√−p′cδτδ(unδ )∇unδ∥∥∥∥2≤ C,

∥∥∥uN∗δ ∥∥∥2+ 1

2

∥∥∥∇uN∗δ ∥∥∥2+ h

N∗∑n=1‖∇unδ ‖

2 ≤ C. (3.46)

Proof. Taking φ =∫ unδ

CD

τδkδkwδknδ

(z)dz in (3.32) and ψ =∫ unδ

CD

τδkwδ

(z)dz in (3.33) (both


being in W 1,20 (Ω)) gives

(unδ − u

n−1δ

h,

∫ unδ

CD

τδkδkwδknδ

(z)dz)− (knδ(unδ )∇pnδ ,τδkδkwδknδ

(unδ )∇unδ )

+ (∇θδ(unδ ), τδkδkwδknδ

(unδ )∇unδ ) = 0, (3.47)

and

(kδ(unδ )∇pnδ ,τδkwδ

(unδ )∇unδ ) + (kwδ(unδ )∇Tδ(unδ )− Tδ(un−1

δ )h

,1

kwδ(unδ )∇Tδ(unδ )) = 0.

(3.48)With Gδ defined in (3.24), in view of its monotonicity we have

(unδ − un−1δ ) ·Gδ(unδ ) ≥ Γδ(unδ )− Γδ(un−1

δ ). (3.49)

Adding (3.47) and (3.48) and using (3.49) leads to

1h

∫Ω

(Γδ(unδ )− Γδ(un−1δ ))dx+ (∇θδ(unδ ), τδkδ

kwδknδ(unδ )∇unδ )

+ 1h

(∇(Tδ(unδ )− Tδ(un−1δ )),∇Tδ(unδ )) ≤ 0. (3.50)

Summing up (3.50) for n = 1 to N∗ gives

∫Ω

(Γδ(uN∗

δ )− Γδ(u0))dx+ h

N∗∑n=1

(∇θδ(unδ ), τδkδkwδknδ

(unδ )∇unδ )

+N∗∑n=1

(∇(Tδ(unδ )− Tδ(un−1δ )),∇Tδ(unδ )) ≤ 0. (3.51)

Since θ′δ(unδ ) = −knδkwδkδ

(unδ ) · p′cδ(unδ ), one has

h

N∗∑n=1

(∇θ(unδ ), τδkδkwδknδ

(unδ )∇unδ ) = h

N∗∑n=1

∥∥∥∥√−p′cδτδ(unδ )∇unδ∥∥∥∥2.

According to Lemma 3.3.2, we get

N∗∑n=1

(∇(Tδ(unδ )− Tδ(un−1δ )),∇Tδ(unδ )) =1

2

N∗∑n=1


∥∥2

+ 12

∥∥∥∇Tδ(uN∗δ )∥∥∥2− 1

2∥∥∇Tδ(u0)

∥∥2,


which implies

∫Ω

Γδ(uN∗

δ ) + h

N∗∑n=1

∥∥∥∥√−p′cδτδ(unδ )∇unδ∥∥∥∥2

+12

N∗∑n=1


∥∥2 + 12

∥∥∥∇Tδ(uN∗δ )∥∥∥2≤ Γδ(u0) + 1

2∥∥∇Tδ(u0)

∥∥2.

Recalling (3.20) , (A1) and (A2), uniformly w.r. t. δ, it holds that∫Ω

Γδ(u0)dx =∫

Ω

∫ u0

CD

∫ u

CD

τδkδkwδknδ

dzdvdx

=∫

Ω

∫ u0

CD

∫ v

CD

τδkwδ

(z)dzdvdx+∫

Ω

∫ u0

CD

∫ v

CD

τδknδ

(z)dzdvdx

≤∫

Ω

∫ u0

CD

∫ v

CD

τ

kw(z)dzdvdx+

∫Ω

∫ u0

CD

∫ v

CD

τ

kn(z)dzdvdx

=∫

ΩΓ(u0)dx <∞.

Furthermore, we also have∫Ω|∇Tδ(u0)|2dx =

∫Ωτ2δ |∇u0|2dx ≤

∫Ωτ2|∇u0|2dx =

∫Ω|∇T (u0)|2dx ≤ C.

Therefore, we obtain ∫Ω

Γδ(uN∗

δ )dx+ 12

∥∥∥∇Tδ(uN∗δ )∥∥∥2≤ C,

and12

N∗∑n=1


∥∥2 + h

N∗∑n=1

∥∥∥∥√−p′cδτδ(unδ )∇unδ∥∥∥∥2≤ C.

By the definition of τδ, this immediately gives

12

∥∥∥∇uN∗δ ∥∥∥2+ h

N∗∑n=1‖∇unδ ‖

2 ≤ C.

To complete the proof, we use the Poincare’s inequality∥∥∥uN∗δ ∥∥∥L2(Ω)

≤ ‖uN∗

δ − CD‖L2(Ω) + ‖CD‖L2(Ω) ≤ C(Ω)∥∥∥∇uN∗δ ∥∥∥

L2(Ω)+ C ≤ C.


Lemma 3.3.4. For any N∗ ∈ 1, 2, ..., N, we have

N∗∑n=1


∥∥2 +N∗∑n=1‖∇(unδ − un−1

δ )‖2 ≤ Ch,

N∗∑n=1

∥∥Tδ(unδ )− Tδ(un−1δ )

∥∥2 +N∗∑n=1

∥∥unδ − un−1δ

∥∥2 ≤ Ch,

h

N∗∑n=1‖∇pnδ ‖

2 ≤ C, (3.52)

where C is independent of h.

Proof. Testing in both (3.32) and (3.33) with h(Tδ(unδ )−Tδ(un−1δ )), adding the resulting

gives

(unδ − un−1δ , Tδ(unδ )− Tδ(un−1

δ )) + h(kwδ(unδ )∇pnδ ,∇(Tδ(unδ )− Tδ(un−1δ )))

+∥∥∥∥√kwδ(unδ )∇(Tδ(unδ )− Tδ(un−1

δ ))∥∥∥∥2

+ h(∇θδ(unδ ),∇(Tδ(unδ )− Tδ(un−1δ ))) = 0.

(3.53)

With ψ = hpnδ in (3.33), we have

h

∥∥∥∥√kδ(unδ )∇pnδ∥∥∥∥2

= −(kwδ(unδ )∇(Tδ(unδ )− Tδ(un−1δ )),∇pnδ )

≤

∥∥∥∥∥ kwδ(unδ )√kδ(unδ )

∇(Tδ(unδ )− Tδ(un−1δ ))

∥∥∥∥∥ ·∥∥∥∥√kδ(unδ )∇pnδ

∥∥∥∥ , (3.54)

giving

−h(kwδ(unδ )∇pnδ ,∇(Tδ(unδ )− Tδ(un−1δ ))) ≤



∥∥∥∥∥2

.

Then, (3.53) becomes


δ )) +∥∥∥∥√kwδ(unδ )∇(Tδ(unδ )− Tδ(un−1

δ ))∥∥∥∥2

+h(∇θδ(unδ ),∇(Tδ(unδ )− Tδ(un−1δ ))) ≤



∥∥∥∥∥2

. (3.55)


In view of the definition kδ, one has∥∥∥∥√kwδ(unδ )∇(Tδ(unδ )− Tδ(un−1δ ))

∥∥∥∥2−



∥∥∥∥∥2

=

∥∥∥∥∥√kwδknδkδ

(unδ )∇(Tδ(unδ )− Tδ(un−1δ ))

∥∥∥∥∥2

.

Further, since

h∣∣(∇θ(unδ ),∇(Tδ(unδ )− Tδ(un−1

δ )))∣∣ ≤1

2



∥∥∥∥∥2

+ h2

2


(unδ )p′

cδ(unδ )∇unδ

∥∥∥∥∥2

,

(3.55) yields


δ )) + 12



∥∥∥∥∥2

≤ h2

2


(unδ )p′

cδ(unδ )∇unδ

∥∥∥∥∥2

≤ h2

2 |kwδp′

cδ|∞

∥∥∥∥∥√

knδkδτδ

(unδ )√−p′cδτδ(unδ )∇unδ

∥∥∥∥∥2

.

(3.56)

By (3.46) and (A1) and (A2), we have


δ )) +∥∥∇(Tδ(unδ )− Tδ(un−1

δ ))∥∥2 ≤ Ch2. (3.57)

This leads to

N∗∑n=1


δ )) +N∗∑n=1


∥∥2 ≤ Ch.

By the definition of T , a ξ exists, such that


δ )) =∥∥∥√τδ(ξ)(unδ − un−1

δ )∥∥∥2≤ Ch2.

Similarly, a ξ exists, such that


δ )) =

∥∥∥∥∥∥ 1√τδ(ξ)

(Tδ(unδ )− Tδ(un−1δ ))

∥∥∥∥∥∥2

≤ Ch2.


Then we obtain

N∗∑n=1


∥∥2 +N∗∑n=1


∥∥2 ≤ Ch.

Furthermore, according to (A1), (3.46) and (3.57), one also has

N∗∑n=1‖∇(unδ − un−1

δ )‖2 ≤ Ch.

Finally, (3.52) follows from (3.54) and (3.57).

3.3.4 Existence of weak solutions to the regularized problem

To show the existence of a solution to Problem Pδ, we consider linear interpolation intime:

TδN (t) = Tδ(un−1δ ) + t− tn−1

h(Tδ(unδ )− Tδ(un−1

δ )),

UδN (t) = un−1δ + t− tn−1

h(unδ − un−1

δ ),

and piecewise constant functions in time

TδN (t) = Tδ(unδ ), UδN (t) = unδ , PδN (t) = pnδ ,

for t ∈ (tn−1, tn], n = 1, 2, ..., N. Clearly, TδN ∈ L2(Q), UδN ∈ L2(0, TM ;V ), PδN (t) ∈L2(0, TM ;W 1,2

0 (Ω)).We have the following result:

Theorem 3.3.1. Under the assumptions A1 and A2, Problem Pδ has a solution (uδ, pδ).

Proof. According to the priori estimates in Lemma 3.3.3 and Lemma 3.3.4, we have∫ TM

0‖TδN (t)‖2L2(Ω) dt =

N∑n=1

∫ tn

tn−1

∥∥∥∥Tδ(un−1δ ) + t− tn−1

h(Tδ(unδ )− Tδ(un−1

δ ))∥∥∥∥2

L2(Ω)dt

≤ 2N∑n=1

∫ tn

tn−1

(∥∥Tδ(un−1δ )

∥∥2L2(Ω) +


∥∥2L2(Ω)

)dt

≤ C, ∫ TM

0‖∇TδN (t)‖2L2(Ω) dt ≤ C,∫ TM

0‖∂tUδN (t)‖2L2(Ω) dt = 1

h

N∑n=1


∥∥2L2(Ω) ≤ C,


∫ TM

0‖∂tTδN (t)‖2L2(Ω) dt = 1

h

N∑n=1


∥∥2L2(Ω) ≤ C,

and ∫ TM

0‖∂t∇TδN (t)‖2L2(Ω) dt =

N∑n=1

∫ tn

tn−1

∥∥∥∥ 1h∇(Tδ(unδ )− Tδ(un−1

δ ))∥∥∥∥2

L2(Ω)dt

= 1h

N∑n=1


∥∥2L2(Ω)

≤ C.

In the same way, one obtains similar estimates∫ TM

0‖UδN (t)‖2L2(Ω) dt+

∫ TM

0‖∇UδN (t)‖2L2(Ω) dt

+∫ TM

0‖∂tUδN (t)‖2L2(Ω) dt+

∫ TM

0‖∂t∇UδN (t)‖2L2(Ω) dt ≤ C.

Therefore, the sequences TδNN∈N and UδNN∈N are uniformly bounded inW 1,2(0, TM ;W 1,2(Ω)), so there exist two sub-sequences (still denoted by TδN and UδN ) whichconverge weakly to some T ∗δ ∈W 1,2(0, TM ;W 1,2(Ω)) and uδ ∈W 1,2(0, TM ;V ).For any φ, ψ ∈ L2(0, TM ;W 1,2

0 (Ω)), (3.32)-(3.33) give∫ TM

0

∫Ω∂tUδNφdxdt−

∫ TM

0

∫Ωknδ(UδN )∇PδN∇φdxdt+

∫ TM

0

∫Ω∇θδ(UδN )∇φdxdt = 0,

(3.58)∫ TM

0

∫Ωkδ(UδN )∇PδN∇ψdxdt+

∫ TM

0

∫Ωkwδ(UδN )∇∂tTδN∇ψdxdt = 0. (3.59)

Clearly, TδN (t)→ T ∗δ strongly in L2(Q). By Lemma 3.2 in [73], it follows that TδN (t)→T ∗δ strongly in L2(Q) as well and a similar conclusion can be drawn for UδN and uδ. By thecontinuity of kδ, kwδ, knδ, we also have knδ(UδN )→ knδ(uδ), kwδ(UδN )→ kwδ(uδ), andkδ(UδN )→ kδ(uδ). Now we show that ∂t∇T ∗δ = ∂t∇Tδ(uδ). Since TδN (t) = Tδ(unδ ) =Tδ(UδN ) converges to T ∗δ , by the definition of Tδ, we also find that Tδ(UδN )→ Tδ(uδ),then we have T ∗δ = Tδ(uδ).

Similarly, for PδN (t), we have a pδ such that

∇PδN ∇pδ weakly in L2(0, T ;L2(Ω)d).

As in the proof of Lemma 3.3.1, we get

knδ(UδN )∇PδN knδ(uδ)∇pδ weakly in L2(0, TM ;L2(Ω)d),kδ(UδN )∇PδN kδ(uδ)∇pδ weakly in L2(0, TM ;L2(Ω)d),


kwδ(UδN )∇∂tTδN kwδ(uδ)∇∂tTδ(uδ) weakly in L2(0, TM ;L2(Ω)d),

θ′

δ(UδN )∇UN θ′

δ(uδ)∇uδ weakly in L2(0, TM ;L2(Ω)d).

Combining the results, we obtain that (uδ, pδ) is the solution pair of Problem Pδ.

3.3.5 Uniqueness of the weak solution for Problem Pδ

After having obtained the existence of a weak solution, we show its uniqueness. To doso, we consider the following system:

∂ts−∇ · (kwδ(s)∇pw) = 0, (3.60)

−∂ts−∇ · (knδ(s)∇pn) = 0, (3.61)

pn − pw = pcδ(s)− pcδ(CD)− ∂tTδ(s). (3.62)

Formally, this is equivalent to (3.27) - (3.28), in the sense that uδ = s and the globalpressure p is given by (3.9) or (3.10), often finding pw, pn in (3.60) - (3.62). Clearly, theinitial and boundary conditions should be compatible with the original ones:

s(0, ·) = u0 in Ω,

pw = pn = 0, at ∂Ω, for all t > 0.

Furthermore, to prove the uniqueness, we also assume

A3: Ω is a C0,ε domain for some 0 < ε ≤ 1.

A4: u0 ∈ C0,ε(Ω).

A weak solution of (3.60)-(3.62) solvesProblem Pe: Given s(0, ·) = u0, find pw ∈ L2(0, TM ;W 1,2

0 (Ω)), pn ∈ L2(0, TM ;W 1,20 (Ω))

and s ∈ L2(0, TM ;L2(Ω)), such that

(∂ts, φ) + (kwδ(s)∇pw,∇φ) = 0, (3.63)

−(∂ts, ψ)− (knδ(s)∇pn,∇ψ) = 0, (3.64)

(pn − pw, ρ) = (pcδ(s)− pcδ(CD), ρ)− (∂tTδ(s), ρ), (3.65)

for all φ, ψ ∈ L2(0, T ;W 1,20 (Ω)) and ρ ∈ L2(0, T ;L2(Ω)).

The equivalence above can be made more precisely.

Lemma 3.3.5. Problem Pe and Problem Pδ are equivalent. Specifically, (uδ, pδ) is asolution to Problem Pδ if and only if (s, pw, pn) solves Problem Pe, with s = uδ, pw =pδ+

∫ uδCD

fwδ(z)p′

cδ(z)dz−pc(uδ)+pcδ(CD)+τ(uδ)∂tuδ, pn = pδ+∫ uδCD

fwδ(z)p′

cδ(z)dz.

Proof. The proof follows the ideas in [52], see [88] for the underlying ideas.


Theorem 3.3.2. If assumptions A1-A4 are satisfied, then there exists at most one solu-tion (uδ, pδ) for Problem Pδ.

Proof. The uniqueness of a solution for Problem Pe is proved in [32]. Since Problems Peand Pδ are equivalent, this immediately concludes the proof.

3.4 Existence of weak solutions for Problem P

Below we obtain the existence result for the degenerate case. More precise, the nonlinearfunctions kw, kn, pc should behave as stated below.

A5: There exist α > 0, β > 0, λ > 0, ω > 0, and for different dimensions, we assumethe followingif d = 3: α ≥ λ > α/3 + 10/3, ω > 5/2, e := ω + β > 5,if d = 2 or d = 1: α ≥ λ > 4, ω > 2, e > 4.

With this, for some constants Cw, Cp, Cn, Cτ > 0 there hold:

limu0

kw(u) · u−α = Cw, (3.66)

limu0

p′

c(u) · uλ = −Cp, (3.67)

limu1

kn(u) · (1− u)−β = Cn, (3.68)

limu1

τ(u) · (1− u)ω = Cτ . (3.69)

To avoid unnecessary technical complications, we restrict the proof to the cases kw =uα, kn = (1 − u)β , p′c(u) = −uλ, and τ(u) = (1 − u)−ω. We note that this type ofbehavior is commonly encountered in the porous media literature [12, 22, 61, 82]. Lessstandard is the function τ . The structure considered here agrees with the one proposedin [17,18,26,94]. Then the regularized τ becomes

τδ(u) =

1, if u < δ,

(1 + δ − u)−ω, if δ ≤ u < 1,δ−ω, if u ≥ 1.

More general cases, in particular for τ , but still respecting assumption A1, A2 can beconsidered as well. Then all integrals appearing below should be decomposed into twoparts, one integral up to u∗, and the second from u∗ to u. To avoid an excess of technicalcalculation we restrict the proof to the case given above. However, in the general casethe existence proof would rely on the same ideas.

Since in (A1) we assume that the product kw · p′

c is uniformly bounded in R, thisimplies α ≥ λ. According to the definition of τ and its regularization, one has T (u) and

3.4. EXISTENCE OF WEAK SOLUTIONS FOR PROBLEM P 37

Tδ(u) as follows:

T (u) =

u+ 1

ω−1 , if u < 0,1

ω−1 (1− u)1−ω, if 0 ≤ u < 1,+∞, if u ≥ 1,

Tδ(u) =

u+ 1

ω−1 − δ, if u < δ,1

ω−1 (1 + δ − u)1−ω, if δ ≤ u < 1,u−1δω + 1

ω−11

δω−1 , if u ≥ 1.

Remark 3.4.1. Choosing CT = 1/(ω − 1) in (3.19) and CTδ = 1/(ω − 1)− δ in (3.26)gives simple expressions for T (u) = 1

ω−1 (1 − u)1−ω and Tδ(u) = 1ω−1 (1 + δ − u)1−ω

(u ∈ [δ, 1)). Nevertheless, the choice of CT and CTδ has no importance for the proof.

Here we define the characteristic function

χ(a,b) =

1, if v ∈ (a, b),0, if v /∈ (a, b).

Then we have the following results

Lemma 3.4.1. Let suppose the hypotheses (A1), (A2) and (A5). Then there exists aconstant C > 0 independent of δ, such that for the first component of the weak solutionpair (uδ, pδ) to Problem Pδ, one has

‖(|uδ|+ δ)2−α‖L∞(0,TM ;L1(Ω)) ≤ C, (3.70)

‖(|1− uδ|+ δ)2−e‖L∞(0,TM ;L1(Ω)) ≤ C, (3.71)

‖∇Tδ(uδ)‖L∞(0,TM ;L2(Ω)) + ‖∇uδ‖L∞(0,TM ;L2(Ω)) ≤ C, (3.72)∥∥∥∥√−p′cδτδ∇uδ∥∥∥∥L2(0,TM ;L2(Ω))

≤ C. (3.73)

Proof. With t ∈ (0, TM ], taking the test function φ = χ(0,t)

∫ uδ

CD

τδkδkwδknδ

(z)dz in (3.29)

and ψ = χ(0,t)

∫ uδ

CD

τδkwδ

(z)dz in (3.30) gives

∫ t

0(∂tuδ,

∫ uδ

CD

τδkδkwδknδ

(z)dz)dt−∫ t

0(∇pδ,

τδkδkwδ∇uδ)dt+

∫ t

0

∥∥∥∥√−p′cδτδ∇uδ∥∥∥∥2dt = 0,

(3.74)∫ t

0(∇pδ,

τδkδkwδ∇uδ)dt+

∫ t

0(∇∂tTδ(uδ),∇Tδ(uδ))dt = 0. (3.75)


Adding (3.74) and (3.75) gives∫ t

0

∫Ω∂tuδ

∫ uδ

CD

τδkδkwδknδ

(z)dzdxdt+∫ t

0

∥∥∥∥√−p′cδτδ∇uδ∥∥∥∥2dt+1

2

∫ t

0

d

dt‖∇Tδ(uδ)‖2dt = 0.

(3.76)By the definition of kδ, one has∫

Ω∂tuδ

∫ uδ

CD

τδkδkwδknδ

(z)dzdx =∫

Ω∂tuδ

∫ uδ

CD

τδkwδ

(z)dzdx+∫

Ω∂tuδ

∫ uδ

CD

τδknδ

(z)dzdx.

The following identities hold a.e.

∂tuδ

∫ uδ

CD

τδkwδ

(z)dz = ∂t

(uδ

∫ uδ

CD

τδkwδ

(z)dz −∫ uδ

CD

zτδkwδ

(z)dz),

∂tuδ

∫ uδ

CD

τδknδ

(z)dz = ∂t

((uδ − 1)

∫ uδ

CD

τδknδ

(z)dz −∫ uδ

1(z − 1) τδ

knδ(z)dz

).

As in [77], we define the functions Ewδ, Enδ : R→ R

Ewδ(y) =∫ y

CD

∫ v

CD

τδkwδ

dzdv + (CD + δ)2−α

(1− α)(2− α) + (CD + δ)1−α

α− 1 CD

=y∫ y

CD

τδkwδ

dz −∫ y

CD

τδz

kwδdz + (CD + δ)2−α

(1− α)(2− α) + (CD + δ)1−α

α− 1 CD,

Enδ(y) =∫ y

1

∫ v

CD

τδknδ

dzdv + δ2−e

(1− e)(2− e)

=(y − 1)∫ y

CD

τδknδ

dz −∫ y

1

(z − 1)τδknδ

dz + δ2−e

(1− e)(2− e) ,

and

Ewδ(y) =∫ y

CD

∫ v

CD

1kwδ

(z)dzdv + (CD + δ)2−α

(1− α)(2− α) + (CD + δ)1−α

α− 1 CD. (3.77)

The choice of these terms is justified by the following calculations. Recalling (A5), for0 ≤ uδ ≤ 1− δ we have∫ uδ

CD

∫ v

CD

1kwδ

(z)dzdv = (uδ + δ)2−α

(1− α)(2− α) + (CD + δ)1−α

α− 1 uδ

− (CD + δ)2−α

(1− α)(2− α) −(CD + δ)1−α

α− 1 CD.


Similarly, for uδ < 0, one has∫ uδ

CD

∫ v

CD

1kwδ

(z)dzdv =δ−α

2 u2δ + (CD + δ)1−α − δ1−α

α− 1 uδ + δ2−α

(1− α)(2− α)

− (CD + δ)2−α

(1− α)(2− α) −(CD + δ)1−α

α− 1 CD,

and for uδ > 1− δ, we get∫ uδ

CD

∫ v

CD

1kwδ

(z)dzdv =(uδ − (1− δ))2

2 + (CD + δ)1−α − 1α− 1 uδ + 1

(1− α)(2− α)

+ 1− δα− 1 −

(CD + δ)2−α

(1− α)(2− α) −(CD + δ)1−α

α− 1 CD.

Note that the calculations above hold for the choice kw(u) = u−α. If, instead, kw behaveslike in (3.66), there the expressions on the right in the above are dominating terms, thereminders being regular w.r.t. δ.In this way Ewδ(uδ) becomes

Ewδ(uδ) =

δ−α

2 u2δ + (CD+δ)1−α−δ1−α

α−1 uδ + δ2−α

(1−α)(2−α) , for uδ < 0,(uδ+δ)2−α

(1−α)(2−α) + (CD+δ)1−α

α−1 uδ, for 0 ≤ uδ ≤ 1− δ,(uδ−(1−δ))2

2 + (CD+δ)1−α−1α−1 uδ + 1

(1−α)(2−α) + 1−δα−1 , for uδ > 1− δ.

We note that

Ewδ(uδ) ≥ E0wδ(uδ) :=

δ−α

2 u2δ + δ2−α

(1−α)(2−α) , for uδ < 0,(uδ+δ)2−α

(1−α)(2−α) , for 0 ≤ uδ ≤ 1− δ,(uδ−(1−δ))2

2 + 1(1−α)(2−α) , for uδ > 1− δ,

whereas E0wδ(uδ) ≥ C(|uδ|+ δ)2−α (C > 0 independent of δ).

In the same way, for Enδ, since 1τδ· knδ = (1 − z + δ)e (see (A5)), for δ ≤ uδ ≤ 1,

one has ∫ uδ

1

∫ v

CD

τδknδ

dzdv =(1− uδ + δ)2−e

(1− e)(2− e) −δ2−e

(1− e)(2− e)

+ (1− CD + δ)1−e

e− 1 (1− uδ).

Further, for uδ < δ, one gets∫ uδ

1

∫ v

CD

τδknδ

dzdv =(uδ − δ)2

2 + 1(1− e)(2− e) + 1− δ

e− 1

+ (1 + δ − CD)1−e − 1e− 1 (1− uδ)−

δ2−e

(1− e)(2− e) ,


and for uδ > 1, we have∫ uδ

1

∫ v

CD

τδknδ

dzdv =δ−e

2 (uδ − 1)2 + ( δ1−e

e− 1 −(1 + δ − CD)1−e

e− 1 )(uδ − 1).

Then Enδ(v) rewrites

Enδ(uδ) =

(uδ−δ)2

2 + 1(1−e)(2−e) + 1−δ

e−1 + (1+δ−CD)1−e−1e−1 (1− uδ), for uδ < δ,

(1−uδ+δ)2−e

(1−e)(2−e) + (1−CD+δ)1−e

e−1 (1− uδ), for δ ≤ uδ ≤ 1,δ−e(uδ−1)2

2 + ( δ1−e−(1+δ−CD)1−e

e−1 )(uδ − 1) + δ2−e

(1−e)(2−e) , for uδ > 1,

Enδ(uδ) ≥ E0nδ(uδ) :=

(uδ−δ)2

2 + 1(1−e)(2−e) + 1−δ

e−1 , for uδ < δ,(1−uδ+δ)2−e

(1−e)(2−e) , for δ ≤ uδ ≤ 1,δ−e

2 (uδ − 1)2 + δ2−e

(1−e)(2−e) , for uδ > 1,(3.78)

andE0nδ(uδ) ≥ C(|1− uδ|+ δ)2−e, (3.79)

with C > 0 independent of δ.

Substitute Ewδ+Enδ into (3.76) instead of∫

Ω∂tuδ

∫ uδ

CD

τδkwδknδ

(z)dzdx, then we have

∫ΩEwδ(uδ(t))dx+

∫ΩEnδ(uδ(t))dx+

∫ t

0‖√−p′cδτδ∇uδ‖

2dt+ 12‖∇Tδ(uδ(t))‖

2

=∫

ΩEwδ(u0)dx+

∫ΩEnδ(u0)dx+ 1

2‖∇Tδ(u0)‖2. (3.80)

As the proof of Lemma 3.3.3, one has

Γδ(u0) =∫ u0

CD

∫ v

CD

τδkwδ

(z)dzdv +∫ u0

CD

∫ v

CD

τδknδ

(z)dzdv ≤ C,

and ∫Ω|∇Tδ(u0)|2dx =

∫Ωτ2δ |∇u0|2dx ≤

∫Ωτ2|∇u0|2dx =

∫Ω|∇T (u0)|2dx ≤ C.

These lead to∫ΩEwδ(u0)dx =

∫Ω

∫ u0

CD

∫ v

CD

τδkwδ

dzdvdx+∫

Ω

(CD + δ)2−α

(1− α)(2− α) + (CD + δ)1−αCD(α− 1) dx

≤ C.


Similarly, we also have∫ΩEnδ(u0)dx =

∫Ω

∫ u0

CD

∫ v

CD

τδknδ

dx−∫

Ω

∫ 1

CD

∫ v

CD

τδknδ

dx+∫

Ω

δ2−e

(1− e)(2− e)dx

=∫

Ω

∫ u0

CD

∫ v

CD

τδknδ

dx+∫

Ω

(1− CD + δ)2−e

(1− e)(2− e) dx

+∫

Ω

(1− CD + δ)1−e(1− CD)(e− 1) dx

≤C.

Observe that Ewδ is convex, allowing a minimum at uδ = CD, then we have Ewδ ≥(CD+δ)2−α

(1−α)(2−α) + (CD+δ)1−α

α−1 CD > 0. And Enδ is also positive obtained from (3.78) and(3.79). These give the estimates from (3.80):∫

ΩEwδ(uδ) +

∫ΩEnδ(uδ) ≤ C.

Since τδ is far away from 0, then one has∫ΩEwδ(uδ) ≤ C.

These lead to ∫ TM

0‖√−p′cδτδ∇uδ‖

2dt ≤ C, (3.81)∫Ω

(|uδ|+ δ)2−αdx ≤ C, (3.82)∫Ω

(|1− uδ|+ δ)2−edx ≤ C, (3.83)

‖∇Tδ(uδ)‖2 + ‖∇uδ‖2 ≤ C.

Lemma 3.4.2. Under the assumptions (A1), (A2) and (A5), there exists a constantC > 0 independent of δ, such that the weak solution pair (uδ, pδ) of Problem Pδ satisfies:∫ TM

0‖∂tuδ‖2dt+

∫ TM

0‖ 1√

τδ(uδ)∂tTδ(uδ)‖2dt

+∫ TM

0‖√kwδknδkδ

∇∂tTδ(uδ)‖2dt ≤ C, (3.84)

∫ TM

0‖√knδ∇pδ‖2 +

∫ TM

0‖√kwδ∇(pδ + ∂tTδ(uδ))‖2 ≤ C. (3.85)


Proof. Testing by ∂tTδ(uδ) both in (3.29) and (3.30), adding the resulting gives∫ TM

0(∂tuδ, ∂tTδ(uδ))dt+

∫ TM

0(kwδ∇pδ,∇∂tTδ(uδ))dt

+∫ TM

0(∇θδ(uδ),∇∂tTδ(uδ))dt+

∫ TM

0‖√kwδ∇∂tTδ(uδ)‖2dt = 0. (3.86)

Further, taking ψ = pδ in (3.30) gives∥∥∥√kδ∇pδ∥∥∥2= −(kwδ∇∂tTδ(uδ),∇pδ) ≤

∥∥∥∥ kwδ√kδ∇∂tTδ(uδ)∥∥∥∥ · ∥∥∥√kδ∇pδ∥∥∥ ,

implying

−(kwδ∇pδ,∇∂tTδ(uδ)) ≤∥∥∥∥ kwδ√kδ∇∂tTδ(uδ)

∥∥∥∥2.

Then (3.86) becomes∫ TM

0(∂tuδ, ∂tTδ(uδ))dt+

∫ TM

0

∥∥∥√kwδ∇∂tTδ(uδ)∥∥∥2dt

+∫ TM

0(∇θδ(uδ),∇∂tTδ(uδ))dt ≤

∫ TM

0

∥∥∥∥ kwδ√kδ∇∂tTδ(uδ)∥∥∥∥2dt. (3.87)

Further, one has

∥∥∥√kwδ∇∂tTδ(uδ)∥∥∥2−∥∥∥∥ kwδ√kδ∇∂tTδ(uδ)

∥∥∥∥2=


∇∂tTδ(uδ)

∥∥∥∥∥2

,

and

|(∇θ(uδ),∇∂tTδ(uδ))| ≤12


∇∂tTδ(uδ)

∥∥∥∥∥2

+ 12


(−p′

cδ)∇uδ

∥∥∥∥∥2

. (3.88)


Then (3.87) leads to

∫ TM

0(∂tuδ, ∂tTδ(uδ))dt+ 1

2

∫ TM

0


∇∂tTδ(uδ)

∥∥∥∥∥2

dt

≤12

∫ TM

0

∥∥∥∥∥−√kwδknδkδ

p′

cδ∇uδ

∥∥∥∥∥2

dt

≤12

∫ TM

0|kwδp

′

cδ|∞

∥∥∥∥∥√

knδkδτδ

√−p′cδτδ∇uδ

∥∥∥∥∥2

dt. (3.89)

By using∥∥∥√p′cδτδ∇uδ∥∥∥2

≤ C, |kwδp′

cδ|∞ ≤ C, and since by (A1), 1τδ

is bounded, wehave ∫ TM

0(∂tuδ, ∂tTδ(uδ))dt+ 1

2

∫ TM

0


∇∂tTδ(uδ)

∥∥∥∥∥2

dt ≤ C. (3.90)

Then, by (3.88), this particularly implies∫ TM

0|(∇θ(uδ),∇∂tTδ(uδ))| ≤ C. (3.91)

Clearly,∫ TM

0‖√τδ∂tuδ‖2 dt =

∫ TM

0

∥∥∥∥ 1√τδ∂tTδ(uδ)

∥∥∥∥2dt =

∫ TM

0(∂tuδ, ∂tTδ(uδ))dt ≤ C.

Testing again (3.29) with φ = ∂tTδ(uδ), we have∫ TM

0(∂tuδ, ∂tTδ(uδ))dt−

∫ TM

0(knδ∇pδ,∇∂tTδ(uδ)) dt

+∫ TM

0(∇θδ(uδ),∇∂tTδ(uδ)) dt = 0. (3.92)

Choosing now ψ = pδ + ∂tTδ(uδ) in (3.30) gives∫ TM

0(knδ∇pδ,∇pδ)dt+

∫ TM

0(knδ∇pδ,∇∂tTδ(uδ))dt

+∫ TM

0‖√kwδ∇(pδ + ∂tTδ(uδ))‖2dt = 0. (3.93)


Adding the equations (3.92) and (3.93), and using (3.91), we find∫ TM

0‖√τδ∂tuδ‖2dt+

∫ TM

0‖√knδ∇pδ‖2dt+

∫ TM

0‖√kwδ∇(pδ + ∂tTδ(uδ))‖2dt

≤∫ TM

0|(∇θδ(uδ),∇∂tTδ(uδ))|dt ≤ C, (3.94)

which concludes the proof.

Furthermore, by (A1), from (3.73), one gets∫ TM

0‖√−p′cδ(uδ)∇uδ‖

2dxdt+∫ TM

0‖√τδ(uδ)∇uδ‖2dxdt ≤ C. (3.95)

With the notation:[uδ]1−δ0 = max0,min1− δ, uδ,

since ∇∫ uδ

0

√−pcδ ′(z)dz =

√−pcδ ′(uδ)∇uδ ∈ L2(Q), recalling that (A5) implies λ > 2

for all d, and∫ [uδ]1−δ0

0(z + δ)−λ/2dz = 1

1− λ/2(([uδ]1−δ0 + δ)1−λ/2 − δ1−λ/2),

we have∇([uδ]1−δ0 + δ)1−λ/2 ∈ L2(Q),

and by (3.95) it is bounded uniformly w.r.t. δ. Further, since the trace of [uδ]1−δ0 + δ on∂Ω is CD+δ, applying the Poincare’s inequality for

([uδ]1−δ0 +δ

)1−λ/2−(CD+δ

)1−λ/2,

one immediately obtains that

‖([uδ]1−δ0 + δ)1−λ/2‖2L2(0,TM ;W 1,2(Ω)) ≤ C,

for some δ-independent C. By Sobolev Embedding Theorem, one obtains

([uδ]1−δ0 + δ)1−λ/2 ∈ L2(0, TM ;C(Ω)), if d = 1,

([uδ]1−δ0 + δ)1−λ/2 ∈ L2(0, TM ;Lr(Ω)), for any r ∈ (1,+∞), if d = 2,

([uδ]1−δ0 + δ)1−λ/2 ∈ L2(0, TM ;L2dd−2 (Ω)), if d > 2,

and the respective norms are bounded uniformly w.r.t. δ.Similarly, for

[uδ]1δ := maxδ,min1, uδ,

one has(1− [uδ]1δ + δ)1−ω/2 ∈ L2(0, TM ;C(Ω)), if d = 1, (3.96)


(1− [uδ]1δ + δ)1−ω/2 ∈ L2(0, TM ;Lr(Ω)), for any r ∈ (1,+∞), if d = 2,

(1− [uδ]1δ + δ)1−ω/2 ∈ L2(0, TM ;L2dd−2 (Ω)), if d > 2. (3.97)

Lemma 3.4.3. For γw, γn, γτ chosen appropriately, the functions(kwδ([uδ]1−δ0 )

)−γw,(

knδ([uδ]1δ))−γn

, and(τδ([uδ]1δ)

)γτare in L1(Q) and have uniformly bounded norms

w.r.t. δ.

Proof. We detail the proof for knδ and τδ, the arguments for kwδ being identical to thosein [77]. To do so, we consider the cases d = 1, 2, and 3 separately (d > 3 being similarto d = 3). We start with the case d = 3, and choose γn = 1

β ( 2β3 + 5ω

3 −103 ). By (A5),

one gets γn > 1. Applying Hölder inequality, for p = 3, q = 32 , one gets for a.e. t:∫

Ω(1− [uδ]1δ + δ)−γnβdxdt =

(∫Ω

(1− [uδ]1δ + δ)(1−ω/2)2dx

)dt

·(∫

Ω(1− [uδ]1δ + δ)(2−e)·2/3dx

)≤(∫

Ω(1− [uδ]1δ + δ)(1−ω/2)6dx

)1/3dt

·(∫

Ω(1− [uδ]1δ + δ)2−edx

)2/3.

Due to (3.71) and (3.97), we have (1 − [uδ]1δ + δ)2−e ∈ L∞(0, TM ;L1(Ω)) and (1 −[uδ]1δ + δ)1−ω/2 ∈ L2(0, TM ;L6(Ω)), and the norms are bounded uniformly w.r.t. δ.This implies:∫ TM

0

∫Ω

(1− [uδ]1δ + δ)−γnβdxdt ≤∫ TM

0

(∫Ω

(1− [uδ]1δ + δ)(1−ω/2)6dx

)1/3dt

·(

max0≤t≤TM

∫Ω

(1− [uδ]1δ + δ)2−edx

)2/3

≤C.

With γτ = 1ω ( 2β

3 + 5ω3 −

103 ), the estimate for

(τδ([uδ]1δ)

)γτfollows similarly.

For d = 2, we choose any r > max(2α − 4)/λ, (e− 2)/(ω − 2), (2e− 4)/(e− 4) anddefine

γn = − 1β

(4− 4

r− ω − e(1− 2

r)),

respectively,γτ = − 1

ω

(4− 4

r− ω − e(1− 2

r)),


and we apply Hölder inequality for p = r2 and q = r

r−2 to obtain for a.e. t

∫Ω

(1− [uδ]1δ + δ)−γnβdxdt ≤(∫

Ω(1− [uδ]1δ + δ)(1−ω/2)rdx

)2/rdt

·(∫

Ω(1− [uδ]1δ + δ)2−edx

)r−2/r,

and ∫Ω

(1− [uδ]1δ + δ)−γτωdxdt ≤(∫

Ω(1− [uδ]1δ + δ)(1−ω/2)rdx

)2/rdt

·(∫

Ω(1− [uδ]1δ + δ)2−edx

)r−2/r.

Then, the proof continues as before.Finally, for d = 1, we take γn = (ω + e − 4)/β and γτ = (ω + e − 4)/ω. Similarly, byassumption (A5), we have γn, γτ > 1. Then, use (3.71), (3.96) to estimate for a.e. t∫

Ω(1− [uδ]1δ + δ)−γnβdx =

∫Ω

(1− [uδ]1δ + δ)(1−ω/2)2(1− [uδ]1δ + δ)2−edx

≤(max

Ω(1− [uδ]1δ + δ)1−ω/2

)2

·∫

Ω(1− [uδ]1δ + δ)2−edx ∈ L1(0, TM ),

and ∫Ω

(1− [uδ]1δ + δ)−γτωdx =∫

Ω(1− [uδ]1δ + δ)(1−ω/2)2(1− [uδ]1δ + δ)2−edx

≤(max

Ω(1− [uδ]1δ + δ)1−ω/2

)2

·∫

Ω(1− [uδ]1δ + δ)2−edx ∈ L1(0, TM ),

and the proof follows again as before.

Now we obtain further estimates for (uδ, pδ).

Lemma 3.4.4. Let d = 1, 2, or 3 and assume (A1), (A2) and (A5). There existr1, r2, r3 ∈ (1, 2) and C > 0 independent of δ, such that the weak solution pair (uδ, pδ)satisfies for all δ > 0

‖∂tTδ(uδ)‖Lr1 (Q) + ‖∇pδ‖Lr2 (Q) + ‖∇(pδ + ∂tTδ(uδ))‖Lr3 (Q) ≤ C.

Proof. The proof uses the estimates in Lemma 3.4.3 and distinguishes as before three


cases, d = 1, 2, and 3. We start with the latter.

Case 1: d =3

By (3.84), one has 1√τδ∂tTδ(uδ) ∈ L2(Q). So here we show for any r1 ∈ (1, 2), one

has ∂tTδ(uδ) ∈ Lr1(Q). Moreover, for appropriately chosen r1, the corresponding normis bounded uniformly w.r.t. δ. To see this, we apply Hölder inequality to get∫ TM

0

∫Ω|∂tTδ(uδ)|r1dxdt =

∫ TM

0

∫Ω| 1√τδ∂tTδ(uδ)|r1τ

r1/2δ dxdt

≤

(∫ TM

0

∫Ω

1τδ|∂tTδ(uδ)|2dxdt

)r1/2

·

(∫ TM

0

∫Ωτr1/(2−r1)δ dxdt

)1−r1/2

.

The first integral on the right hand side is bounded by (3.84), the second we recallLemma 3.4.3 and choose r1 such that r1

2−r1= γτ = 1

ω ( 2β3 + 5ω

3 −103 ), which, by (A5),

r1 = 2β+5ω−10)3β+4ω−5 , satisfies r1 ∈ (1, 2).

Similarly, one also has the estimate∫ TM

0

∫Ω|∇pδ|r2dxdt =

∫ TM

0

∫Ω|√knδ∇pδ|r2k

−r2/2nδ dxdt

≤

(∫ TM

0

∫Ωknδ|∇pδ|2dxdt

)r2/2

·

(∫ TM

0

∫Ωk−r2/(2−r2)nδ dxdt

)1−r2/2

.

We obtainr2 = 2(3ω + 2e− 10)

(5e− 10) ,

and following (A5), one has r2 ∈ (1, 2).Similarly, one has∫ TM

0

∫Ω|∇(pδ + ∂tTδ(uδ))|r3dxdt =

∫ TM

0

∫Ω|√kwδ∇(pδ + ∂tTδ(uδ))|r3(

√kwδ)−r3/2dxdt

≤

(∫ TM

0

∫Ωkwδ|∇∂t(pδ + Tδ(uδ))|2dxdt

)r3/2

·

(∫ TM

0

∫Ω

(kwδ)−r3/(2−r3)dxdt

)1−r3/2

≤C,


then for this case, we have

r3 = 2(3λ+ 2α− 10)3λ+ 5α− 10 ∈ (1, 2),

when α > 5 and λ > 10/3 + α/3, this implies ∇(pδ + ∂tTδ(uδ)) ∈ Lr3(Q).

Case 2: d =2Similarly, for two dimensional case, we show for any r1 ∈ (1, 2), one has ∂tTδ(uδ) ∈Lr1(Q) and the corresponding norm is bounded uniformly w.r.t. δ. Since one has

∫ TM

0

∫Ω|∂tTδ(uδ)|r1dxdt ≤

(∫ TM

0

∫Ω

1τδ|∂tTδ(uδ)|2dxdt

)r1/2

·

(∫ TM

0

∫Ωτr1/(2−r1)δ dxdt

)1−r1/2

.

Then, for any r > 2(e− 2)/(2ω + e− 4), we solve

r1 = 2(ω + e− 2(e− 2)/r − 4)(2ω + e− 2(e− 2)/r − 4) ∈ (1, 2),

which implies ∂tTδ(uδ) ∈ Lr1(Q).Using the same way, we also get ∇pδ ∈ Lr2(Q) for

r2 = 2(ω + e− 2(e− 2)/r − 4)(ω + e+ β − 2(e− 2)/r − 4) ,

and ∇(pδ + ∂tTδ(uδ)) ∈ Lr3(Q) for

r3 = 2(4− λ− α+ 2(α− 2)/r)(4− λ− 2α+ 2(α− 2)/r) .

Case 3: d =1The proof follows as before, we have ∂tTδ(uδ) ∈ Lr1(Q) for

r1 = 2(ω + e− 4)(2ω + e− 4) ∈ (1, 2),

∇pδ ∈ Lr2(Q), forr2 = 2(ω + e− 4)

(ω + e+ β − 4) ∈ (1, 2),

and furthermore, ∇(pδ + ∂tTδ(uδ)) ∈ Lr3(Q) for

r3 = 2(4− λ− α)(4− λ− 2α) ∈ (1, 2).


Then we have the estimates

‖∂tTδ(uδ)‖Lr1 (Q) ≤ C, with r1 ∈ (1, 2), (3.98)

‖∇pδ‖Lr2 (Q) ≤ C, with r2 ∈ (1, 2), (3.99)

‖∇(pδ + ∂tTδ(uδ))‖Lr3 (Q) ≤ C, with r3 ∈ (1, 2). (3.100)

With r∗ = minr1, r2, r3, by Lemmas 3.4.1, 3.4.2 and 3.4.4, one obtains the ex-istence of a subsequence δ 0 (still denoted by δ) and of u ∈ W 1,2(Q), T ∗ ∈W 1,r∗(0, TM ;W 1,r∗(Ω)) and p ∈ L2(0, TM ;W 1,r∗(Ω)), such that

uδ −→ u strongly in L2(Q), (3.101)∂tuδ ∂tu weakly in L2(Q), (3.102)∇uδ ∇u weakly in L2(Q), (3.103)Tδ T ∗ weakly in W 1,r∗(Q), (3.104)Tδ −→ T ∗ strongly in Lq(Q), (3.105)

∇∂tTδ ∇∂tT ∗ weakly in Lr∗(Q), (3.106)pδ p weakly in W 1,r∗(Q), (3.107)

where q = +∞, if d = 1, q = dr∗

d−r∗ , if d = 2 or d = 3 (see [48]).In the remaining, we prove that T ∗ = T (u) a.e., and that (u, p) is a solution pair

to Problem P. But before doing so, we also prove that the limit u above is essentiallybounded by 0 and 1.

Theorem 3.4.1. The limit u ∈W 1,2(Q) satisfies 0 ≤ u ≤ 1 a.e. in Q.

Proof. Given t ∈ (0, TM ], let Ω−δ,ε(t) be the support of [uδ(t, ·) + ε]− (the negative cutof uδ(t, ·) + ε). As follows from Lemma 3.4.1 , a C > 0 exists such that, for all δ > 0,one has ∫

ΩEwδ(uδ)dx =

∫Ω

∫ uδ

CD

∫ v

CD

1kwδ

(z)dzdvdx

+∫

Ω

(CD + δ)2−α

(1− α)(2− α) + (CD + δ)1−α

(α− 1) CDdx

≤C,∫ΩEnδ(uδ)dx =

∫Ω

∫ uδ

1

∫ v

CD

τδknδ

(z)dzdvdx+∫

Ω

δ2−e

(1− e)(2− e)dx ≤ C.


Since the constant arguments in the last two integrals are positive, this gives∫Ω

∫ uδ

CD

∫ v

CD

1kwδ

(z)dzdvdx+∫

Ω

∫ y

1

∫ v

CD

τδknδ

dzdvdx ≤ C.

Then we get

C ≥∫

Ω

∫ uδ

0

∫ v

CD

1kwδ

(z)dzdvdx

=∫

Ω

δ−α

2 u2δ + (CD + δ)1−α − δ1−α

(α− 1) uδ + δ2−α

(1− α)(2− α)dx

≥∫

Ω−δ,ε(t)

δ−α

2 ε2 + δ1−α − (CD + δ)1−α

(α− 1) ε+ δ2−α

(1− α)(2− α)dx.

Let now δ 0, this immediately implies that

meas(Ω−0,ε(t)) = 0,

with Ω−0,ε(t) having the same definition as Ω−δ,ε(t), but now for the function u. Sinceuδ → u in C((0, TM );L2(Ω)) (by (3.102), (3.103) and the compact embedding see [2],Theorem 4.12), thus uδ → u a.e. in Ω, for all t. This holds for every ε > 0, hence u ≥ 0.Similarly, if uδ > 1 + ε, use the bounds on

∫Ω

∫ y

1

∫ v

CD

τδknδ

dzdv dx, we obtain u ≤ 1.

Remark 3.4.2. For two phase flow model, 0 ≤ u ≤ 1 means that the saturation remainsin the physical range. Note that this only holds due to the degeneracy encountered foru = 0 or u = 1.

Finally, we obtain the existence of a solution for Problem P.

Theorem 3.4.2. Let assumptions (A1), (A2) and (A5) be satisfied, then there exists asolution pair (u, p) for Problem P.

Proof. We start by identifying T ∗ as T (u). To do so, define [v]1− = min1, v and letT (f) = T ∗ a.e. We proceed as in [81] and consider first the inverse function of T−1.According to the definition of T and Tδ, one has

f = T−1(T ∗) =

T ∗ − 1ω−1 , if T ∗ < 1

ω−1 ,

1−(

1(ω−1)T∗

) 1ω−1

, if T ∗ ≥ 1ω−1 .

(3.108)

Clearly,

[uδ]1− = T−1δ (Tδ) =

Tδ − 1ω−1 + δ, if Tδ < 1

ω−1 ,

1−(

1(ω−1)Tδ

) 1ω−1 + δ, if Tδ ≥ 1

ω−1 .(3.109)


Now we prove that [uδ]1− → f strongly in L2(Q), and hence a.e. in Q. Since T−1(·) isLipschitz continuous, by (3.109) one has∫

Q

|f − [uδ]1−|dxdt =∫Q

|T−1(T ∗)− T−1(Tδ) + T−1(Tδ)− T−1δ (Tδ)|dxdt

≤ C∫Q

|T ∗ − Tδ|dxdt+∫Q

|T−1(Tδ)− T−1δ (Tδ)|dxdt.

Clearly, the first integral above approaches 0 as δ 0. For the second we note that, by(3.108) and (3.109), one has T−1

δ (Tδ) − T−1(Tδ) = δ, for any argument Tδ. With this,the second integral also approaches 0 as δ 0. This means that

[uδ]1− → f a.e. in Q.

However uδ → u a.e. in Q (by the strong convergence in L2(Q)). This immediately gives[uδ]1− → [u]1− a.e. In the view of Theorem 3.4.1, we also have [u]1− = u a.e. Therefore,u = f a.e., and consequently, we have

T (u) = T ∗.

Having identified T ∗ by T (u), (3.106) gives ∇∂tTδ(uδ) ∇∂tT (u).Then, according to knδ(·) Lipschitz continuous, one has

knδ(uδ)→ knδ(u) a.e. in Q.

Further, since knδ(u) converges pointwise to kn(u), and

|knδ(u)| ≤ C,

uniformly with w.r.t. δ. Then by Dominated Convergence Theorem, we obtain

knδ(u)→ kn(u) a.e. in Q.

Therefore, we have

knδ(uδ)→ kn(u) strongly in L2(Q).

Similarly, we also have√knδ(uδ) converges to

√kn(u) strongly in L2(Q). In the same

fashion, since kw, k and θ are Lipschitz continuous, on also obtains

kwδ(uδ)→ kw(u),kδ(uδ)→ k(u),θδ(uδ)→ θ(u),


strongly in L2(Q).By the (3.85), a g ∈ L2(Q) exists, such that√

kwδ(uδ)∇(pδ + ∂tTδ(uδ)) g weakly in L2(Q), for δ 0.

To identify g, we consider φ ∈ C∞0 (Ω) arbitrarily, and note that φ√kwδ(uδ)→ φ

√kw(u)

strongly in Lq(Q), for any q ∈ [1,∞). This follows as above, using the uniform bounded-ness of kwδ, kw and the pointwise convergence of kwδ(uδ) to kw(u). Further, by (3.106)and (3.107), one gets ∇(pδ + ∂tTδ(uδ)) → ∇(p + ∂tT (u)) weakly in Lr∗(Q). Taking qsuch that 1/q + 1/r∗ = 1 gives by weak-strong convergence argument∫ TM

0(kwδ∇(pδ + ∂tTδ(uδ)),∇φ)dt→

∫ TM

0(kw∇(p+ ∂tT (u)),∇φ)dt. (3.110)

This is sufficient to identify g =√kw(u)∇(p+ ∂tT (u)).

We have now all the ingredients to pass to the limit (δ 0) in the integrals appearingin Problem Pδ. Using the convergence results above, it is straightforward to show that(u, p) solves Problem P.

3.5 ConclusionsIn this chapter, we have proved the existence of a weak solution to degenerate ellipticparabolic system modeling two-phase flow in porous media, and including dynamic effectsin the capillary pressure. The major difficulty is the degeneracy of the non-linear thirdorder derivative term. We get the estimate for the third order derivative term by apply-ing the structures of relative permeabilities, capillary pressure and the dynamic dampingfactor. By compactness arguments, we show the existence of a solution for the originalproblem.

Chapter 4

Finite volume scheme

4.1 Introduction

In this chapter, we define and analyze a finite volume method for a two phase flow modelin a porous medium:

∂tu−∇ · (ko(u)∇p) = 0, (4.1)∂t(1− u)−∇ · (kw(u)∇p) = 0, (4.2)

p− p = pc(u) + τ∂tu, (4.3)

which are defined in Q := Ω × (0, T ], where Ω is a bounded subset of R2, T is a givenmaximal time. The unknowns u, p and p are the non-wetting phase saturation, the non-wetting phase and wetting phase pressures. The equations (4.1), (4.2) are obtained bycombing the mass balance and the Darcy’s law (see [12, 61, 82]). The permeabilitiesko(·), kw(·) for non-wetting phase and wetting phase are given monotone functions. Thegravity is neglected in the model. In order to close the above system, we prescribe theinitial and boundary conditions

u(0, ·) = u0, in Ω, (4.4)p = p = 0, at ∂Ω for t > 0, (4.5)

where u0 is a given function, which will be specified later.Equation (4.3) expresses the phase pressure difference p − p, as a function of u and

∂tu. In classical models (see [12,61,69]), one assumes

p− p = pc(u),

This chapter is a collaborative work with S.F. Nemadjieu, I.S. Pop and has been submitted to SIAMJournal of Numerical Analysis.

53

54 CHAPTER 4. FINITE VOLUME SCHEME

where pc, the capillary pressure is a monotone function of saturation u. This however,holds only if measurements are obtained under equilibrium condition. Experiments (see[17, 38]) have invalidated this assumption, whenever flow is more rapid. One possibleextension is (4.3) as proposed in [60], where τ is a positive damping factor (τ > 0).

Standard models, obtained for τ = 0 have been intensively investigated in the reservoirsimulation. In this sense, existence and uniqueness of the weak solutions are provedin [69] , but assuming that initial and boundary conditions bounded away from 0. Thishas been extended to the case of arbitrarily chosen saturation for initial and boundaryconditions. The existence can be found in [6, 35] and the uniqueness of weak solutionwas proved in [35]. For numerical schemes, we refer to ( [7, 21, 23, 39, 47, 49, 83, 88,89]), where finite element method, mixed finite element method, discontinuous Galerkinmethod are analyzed, or linear iterative schemes are investigated. In particular, for finitevolume schemes, we refer to [29, 51, 76]. Whenever τ > 0, (4.1) - (4.3) becomes aso-called non-equilibrium model. In this case, the existence and uniqueness of a weaksolution is obtained in [28, 30, 53, 77], but in a simplified context when the total flowis assumed to be known. This allows reducing one equation in (4.1) - (4.3). In thiscase, but in the heterogeneous case, if no entry pressure presents, numerical schemes arediscussed in [63]. Also, variational inequality approaches have been considered in [62]for situations including an entry pressure. Further, we refer to [41] which gives thecoupling conditions analysis. In [85], they consider numerical algorithms for unsaturatedflow in highly heterogeneous media for this model. For the full model, the existence anduniqueness of the weak solutions are proved in [32, 68], but assuming that the equationsare non-degenerate (i.e. all non-linearities are bounded away from 0 or +∞). The authorsin [66] present discontinuous Galerkin scheme for this case. The authors have given thenumerical investigations in [57] in heterogeneous case. In the degenerate case, we referto [31], which gives the existence of weak solutions for the model in a equivalent form.

In this chapter, we show that the approximate solution of the system (4.1) - (4.3)obtained by a multi-point flux approximation finite volume scheme converges to its weaksolution. The rest of the chapter is organized as follows. In Section 4.2, we give theassumptions on the data and the define the weak solution. We introduce the finite volumescheme in Section 4.3, and show the existence of the numerical solution. In Section 4.4,we prove the convergence of the scheme by compactness arguments. In the last section,we present some numerical results that confirm the theoretically obtained convergence.

4.2 The weak solutionTo investigate the system (4.1) - (4.5), we make the following assumptions

• (A1) Ω is an open, bounded and connected polygonal domain in R2 with Lipschitzcontinuous boundary ∂Ω. Ω denotes the closure of Ω.

• (A2) The functions ko and kw: R → R are C1, and there exists δ > 0 such that

4.2. THE WEAK SOLUTION 55

δ ≤ ko(u), kw(u) ≤ 1 for all u ∈ R. We assume ko to be an increasing functionwith ko(u) = δ for u ≤ 0 and ko(u) = 1 for u ≥ 1. Also kw will be considered tobe a decreasing function with kw(u) = 1 for u ≤ 0 and kw(u) = δ for u ≥ 1.

• (A3) pc : R→ R is an increasing function of u, pc ∈ C1, pc(0) = 0 and there existmp,Mp > 0 such that mp ≤ p

′

c(u) ≤Mp <∞.

• (A4) τ > 0 is a positive constant.

• (A5) The initial condition u0 is in C1(Ω) ∩W 1,20 (Ω).

Remark 4.2.1. It is not necessary to take pc(0) = 0. We just expect to obtain a consistentboundary condition for u|∂Ω = 0. If pc(0) 6= 0, one can impose p|∂Ω = pc(0) or define a’new non-wetting phase pressure’ p := p− pc(0) + pc(u) + ∂tu to make sure that u = 0at the boundary (see [52]). Furthermore, the proofs here can be extended easily to othertypes of boundary conditions like non-homogeneous Dirichlet or Neumann.Remark 4.2.2. The choice of u0 ∈ C(Ω) is for the ease of presentation, since theproposed discretization of the gradients involves continuous functions. While these areavailable pointwise approximations due to the spaces where these are sought, takingu0 ∈ W 1,2(Ω)\C0(Ω) would not be sufficient to define its discrete gradient, This ishowever, needed in the proof, but not for the scheme itself. If u0 is not continuous, thenone may take into convolution with a grid-size dependent mollifier.

Furthermore, we define Pc as

Pc(u) =∫ u

0pc(s)ds. (4.6)

Clearly, by (A3), Pc is convex and for all u ∈ R

Pc(u) ≥ 0, with Pc(0) = 0. (4.7)

Also one haspc(a)(a− b) ≤ Pc(a)− Pc(b) for all a, b ∈ R. (4.8)

In the following, we define the solution for the system (4.1) - (4.5):Definition 4.2.1. (u, p, p) is a weak solution of the model (4.1) - (4.5) if u ∈W 1,2(0, T ;L2(Ω)), p, p ∈ L2(0, T ;W 1,2

0 (Ω)), and for any φ, ψ ∈ L2(0, T ;W 1,20 (Ω)), λ ∈ L2(0, T ;

L2(Ω)) there hold

(∂tu, φ) + (ko∇p,∇φ) = 0, (4.9)−(∂tu, ψ) + (kw∇p,∇ψ) = 0, (4.10)

(p− p, λ) = (pc(u), λ) + τ(∂tu, λ). (4.11)

As mentioned before, existence and uniqueness results can be found in [31, 32, 68].Note that by (A3) we obtain u ∈W 1,2(0, T ;W 1,2

0 (Ω)) (see [52]).


4.3 The finite volume schemeIn this section, we introduce a finite volume scheme for the system (4.1) - (4.5), thengive the a priori estimates.

4.3.1 Meshes and notationsTo introduce the finite volume scheme to the system, we consider an admissible mesh(see [50] pp. 38).

Definition 4.3.1. Let Ω be an open bounded polygonal subset of R2. An admissiblefinite volume mesh of Ω, denoted by T is a family of triangular disjoint subsets of Ω suchthat two triangles may either be disjoint, or share a node, or a full edge. The set of alledges including the boundary ones is denoted by E . The geometric centers of the trianglesform the set P. In other words:

• The closure of the union of all the triangles is Ω;

• For any K ∈ T , there exists a subset EK of E such that ∂K = K \K = ∪σ∈EK σ.Furthermore, E = ∪K∈T EK .

• For any (K,L) ∈ T 2 with K 6= L, either the 1-dimension Lebesgue measure ofK ∩ L is 0 or K ∩ L =

∑σ for some σ ∈ E . There exists a subset EK of E such

that ∂K = K \K = ∪σ∈EK σ.

• The family P = xKK∈T is such that xK ∈ K (for all K ∈ T ) and it is thegeometric center of the volume K.

Further, we assume:

• (A6) mp,Mp satisfy 4√mpMp ≥ mp +Mp. The angles θ of any triangle K ∈ T

satisfy arccos( 2√mpMp

mp+Mp≤ θ ≤ π − arccos( 2

√mpMp

mp+Mp).

Remark 4.3.1. If pc(·) is a linear function with respect to u, the assumption (A6) canbe relaxed to 0 < θ < π, which is practically fulfilled by any triangular mesh.

Throughout this chapter, the following notations are used: the mesh size is defined bysize(T ) = supdiam(K), K ∈ T . The sets of interior and boundary edges are denotedby Eint, resp. Eext: Eint = σ ∈ E ;σ 6⊂ ∂Ω, and Eext = σ ∈ E ;σ ⊂ ∂Ω. For a triangle,we denote by m(K) its measure. We introduce some notations for the triangle K ∈ T(see Figure 4.1). Pi, Pj , Pk denote the vertices of the triangle K, xK is the geometriccenter of K, Pi,j , Pj,k, Pk,i are the midpoints of the segments PiPj , PjPk, PkPi. Pi/2,jis the point on PiPj which satisfies m(PiPi/2,j)/m(PiPj) = 1/3, similar to Pi/2,k. Welet Pv stand for the set of all vertices Pi, PM for all edges midpoint, PT for all pointsPi/2,j introduced above. We use Kr (r = i, j, k) to denote the quadrilateral determinedby Pr, xK and the midpoints Pr,·, P·,r of the edges. Let σ1

Kidenote the segment PiPi,j ,

4.3. THE FINITE VOLUME SCHEME 57

σ2Ki

denote the segment PiPk,i. Then we denote eσ1Ki

= −−−−−→xKPi/2,j , eσ2Ki

= −−−−−−→xKPi/2,k asthe vectors. Let nσ1

Kiand nσ2

Kibe the normal vectors to PiPj and PkPi outward to Ki.

Finally, we define two vectors µσ1Ki, µσ2

Kias following (see [72, 80]). Observe that µσ1

Ki

and nσ1Ki, respectively µσ2

Kiand nσ2

Kiare parallel.

µσ1Ki· eσ1

Ki= 1,

µσ1Ki· eσ2

Ki= 0,

µσ2Ki· eσ1

Ki= 0,

µσ2Ki· eσ2

Ki= 1.

(4.12)

Figure 4.1 A triangular finite volume and the associated nodes, edges and vectors.

4.3.2 The scheme

To define the scheme, some notations are needed.

Definition 4.3.2. Let Ω be an open bounded polygonal subset of R2, T be an admissiblemesh in Section 4.3.1. h = T

N denotes the time step for any N ∈ N and tn denotes thetime at t = nh for n ∈ 0, ..., N. Let X(T , h) be the set of functions that are piecewiseconstant in both time and space, i.e. v from Ω × (0, Nh) to R such that there exists afamily of real values vnK ,K ∈ T , n ∈ 0, ..., N, with v(x, t) = vnK for a.e. x ∈ K,K ∈ T and for a.e. t ∈ (nh, (n+ 1)h], n ∈ 0, ..., N − 1.

Further, for considering discrete gradients, additional values at edges σ will be needed.To give the full discretization for the system (4.1) - (4.5), we use unK ,K ∈ T , n ∈0, ..., N to denote the discrete approximation of u, the value unK is the approximationof u(xK , nh), and the same for pnK , pnK . For given K ∈ T and with Pr being oneof its nodes, r ∈ i, j, k a counterclockwise ordering, un

σ1Kr

, pnσ1Kr

, pnσ1Kr

denote the


approximations of u(xPr/2,r+1 , tn), p(xPr/2,r+1 , t

n), p(xPr/2,r+1 , tn) and it’s similar for

unσ2Kr

, pnσ2Kr

, pnσ2Kr

.

Observe that, due to (4.12), given a vector v ∈ R2 one has

v = (v · eσ1Kr

) µσ1Kr

+ (v · eσ2Kr

) µσ2Kr. (4.13)

This inspires the definition of discrete gradient: for K ∈ T and r ∈ i, j, k, let the valuesvK , vσ1

Kr, vσ2

Krbe given, the discrete gradient in the quadrilateral Kr is

∇KrvK := (vσ1Kr− vK) · µσ1

Kr+ (vσ2

Kr− vK) · µσ2

Kr. (4.14)

Then for any n = 0, 1, ..., N − 1, we define the scheme as follows

m(K)un+1K − unK

h(4.15)

=ko(un+1K )

∑r=i,j,k

(m(σ1

Kr )(

(pn+1σ1Kr

− pn+1K ) µσ1

Kr+ (pn+1

σ2Kr

− pn+1K ) µσ2

Kr

)· nσ1

Kr

+ m(σ2Kr )(

(pn+1σ1Kr

− pn+1K ) µσ1

Kr+ (pn+1

σ2Kr

− pn+1K ) µσ2

Kr

)· nσ2

Kr

),

−m(K)un+1K − unK

h(4.16)

=kw(un+1K )

∑r=i,j,k

(m(σ1

Kr )(

(pn+1σ1Kr

− pn+1K ) µσ1

Kr+ (pn+1

σ2Kr

− pn+1K ) µσ2

Kr

)· nσ1

Kr

+ m(σ2Kr )(

(pn+1σ1Kr

− pn+1K ) µσ1

Kr+ (pn+1

σ2Kr

− pn+1K ) µσ2

Kr

)· nσ2

Kr

),

pn+1K − pn+1

K = pc(un+1K ) + τ

un+1K − unK

h, (4.17)

for all K ∈ T . Similarly, at each edge σ ∈ Eint, we impose

pn+1σIdKr

− pn+1σIdKr

= pc(un+1σIdKr

) + τ

un+1σIdKr

− unσIdKr

h(Id = 1, 2), (4.18)

and the flux continuity of each phase

ko(un+1K )

((pn+1σ1Kr

− pn+1K ) µσ1

Kr+ (pn+1

σ2Kr

− pn+1K ) µσ2

Kr

)· nK|L

+ko(un+1L )

((pn+1σ1Lr

− pn+1L ) µσ1

Lr+ (pn+1

σ2Lr

− pn+1L ) µσ2

Lr

)· nL|K = 0, (4.19)


kw(un+1K )

((pn+1σ1Kr

− pn+1K ) µσ1

Kr+ (pn+1

σ2Kr

− pn+1K ) µσ2

Kr

)· nK|L

+kw(un+1L )

((pn+1σ1Lr

− pn+1L ) µσ1

Lr+ (pn+1

σ2Lr

− pn+1L ) µσ2

Lr

)· nL|K = 0. (4.20)

Here L is the neighboring element of K sharing the edge σ, and nL|K is the unit normalvector from L into K. Whenever σ ∈ Eext, the values pσ1

Kr, pσ2

Kr, pσ1

Kr, pσ2

Krare set to

0. One takes un+1σIdKr

= 0 (Id = 1, 2) for any K ∈ T and r = i, j, k such that σIdKr ∈ Eext.Also, flux continuity holds each half edge σ and for the discrete gradient is used.

Initially, we take u0K = u0(xK) for any K ∈ T . This makes sense since u0 ∈ C(Ω). If

u0 /∈ C(Ω), then one takes as explanation in Remark 4.2.2, u0T = ηT ∗ u0, where η is the

standard mollifier ( [48]). Clearly, since u0 ∈ W 1,20 (Ω), one has ‖u0

T − u0‖W 1,2(Ω) → 0as size(T ) → 0.

4.3.3 A priori estimates and existence of the fully discrete solutionIn this section, we discuss the fully discrete solution to (4.15) - (4.20). We first providesome elementary results that will be used later.

Lemma 4.3.1. Let m ≥ 1 and aj ,bj ∈ Rm be an m-dimensional real vectors, j ∈0, ..., N. We have the following identities:

N∑j=1〈aj − aj−1,

N∑n=j

bn〉 =N∑j=1〈aj ,bj〉 − 〈a0,

N∑j=1

bj〉, (4.21)

N∑j=1〈aj − aj−1, aj〉 = 1

2(|aN |2 − |a0|2 +N∑j=1|aj − aj−1|2), (4.22)

N∑j=1〈N∑j=n

aj , an〉 = 12 |

N∑j=1

aj |2 + 12

N∑j=1|aj |2, (4.23)

where < ·, · > is the inner product in Rm.

Lemma 4.3.2. Discrete Gronwall inequality: If yn, fn and gn are nonnegativesequences and

yn ≤ fn +∑

0≤k<ngkyk, for all n ≥ 0,

then

yn ≤ fn +∑

0≤k<nfkgk exp(

∑k<j<n

gj), for all n ≥ 0.

The existence of a solution to the discrete system (4.15) - (4.20) can be obtained bya Leray - Schauder argument, as done in [76]. One has


Lemma 4.3.3. Let n ∈ 0, 1, ..., N −1, and assume un given. Then there exists a solu-tion (un+1

K , un+1σ1Kr

, un+1σ2Kr

, pn+1K , pn+1

σ1Kr

, pn+1σ2Kr

, pn+1K , pn+1

σ1Kr

, pn+1σ2Kr

)K∈T to the discrete system(4.15) - (4.20).

Without entering into details, the proof requires a priori estimates, which are obtainedbelow.

Lemma 4.3.4. A C > 0 not depending on h or size(T ) exists such that, for any N∗ ∈0, ..., N − 1 we have the following:

N∗∑n=0

h∑K∈T

ko(un+1K )

∑r=i,j,k

m(Kr)∇Kr pn+1K · ∇Kr pn+1

K (4.24)

+N∗∑n=0

h∑K∈T

kw(un+1K )

∑r=i,j,k

m(Kr)∇Krpn+1K · ∇Krpn+1

K

+τN∗∑n=0

h∑K∈T

m(K)(un+1

K − unKh

)2+∑K∈T

m(K)Pc(uN∗+1

K ) ≤ C.

Proof. We start by proving the following:

µσ1Kr

=m(σ1

Kr) · nσ1

Kr

m(Kr), µσ2

Kr=

m(σ2Kr

) · nσ2Kr

m(Kr). (4.25)

To see this, we refer to Figure 4.1 and take without losing of generality r = i. Note thatm(PixKPk,i) = m(PixKPi,j) = 1

6m(PiPjPk) since xK is the geometric center and Pk,i,Pi,j are midpoints. This gives m(Ki) = 1

3m(PiPjPk). With θi being the angle spannedby e

σIdKi

and µσIdKi

, the matching height of xK to σIdKi is |eσIdKi

|cosθi = 1|µσIdKi

| , due to

(4.12). Therefore, one has

m(Ki) = 2m(PixKPi,j) = 1|µσIdKi|m(σIdKi) (Id = 1, 2),

so|µσIdKi

| =m(σIdKi)m(Ki)

.

This immediately implies (4.25) since µσIdKi

and nσIdKi

are parallel and have the same sense.

Then multiplying (4.15) by pn+1K , (4.19) by m(σ1

Kr)pn+1σ1Kr

and m(σ2Kr

)pn+1σ2Kr

respect-ively, adding the three equalities and summing the resulting over K ∈ T , we find that

−∑K∈T

m(K)(un+1K − unK)pn+1

K = h∑K∈T

ko(un+1K )

∑r=i,j,k


K .

(4.26)


Similarly, we also obtain∑K∈T

m(K)(un+1K − unK)pn+1

K = h∑K∈T

kw(un+1K )

∑r=i,j,k


K .

(4.27)

Adding (4.26) and (4.27) gives

h∑K∈T

ko(un+1K )

∑r=i,j,k


K (4.28)

+h∑K∈T

kw(un+1K )

∑r=i,j,k


K

+∑K∈T

m(K)(un+1K − unK)(pn+1

K − pn+1K ) = 0.

Further, multiplying (4.17) by m(K)(un+1K −unK) and summing the resulting over K ∈ T

leads to ∑K∈T

m(K)(pn+1K − pn+1

K )(un+1K − unK) (4.29)

=∑K∈T

m(K)pc(un+1K )(un+1

K − unK) + τ∑K∈T

m(K)un+1K − unK

h(un+1K − unK).

Using this into (4.28) gives

h2∑K∈T

ko(un+1K )

∑r=i,j,k


K (4.30)

+h2∑K∈T

kw(un+1K )

∑r=i,j,k


K

+τ∑K∈T

m(K)(un+1K − unK)2 + h

∑K∈T

m(K)pc(un+1K )(un+1

K − unK) = 0.

Recalling (4.8), one gets

h2∑K∈T

ko(un+1K )

∑r=i,j,k


K (4.31)

+h2∑K∈T

kw(un+1K )

∑r=i,j,k


K

+τ∑K∈T

m(K)(un+1K − unK)2 + h

∑K∈T

m(K)Pc(un+1K ) ≤ h

∑K∈T

m(K)Pc(unK).


Summing the above equation from 0 to N∗ for any N∗ ∈ 0, ..., N − 1 gives

N∗∑n=0

h∑K∈T

kw(un+1K )

∑r=i,j,k


K (4.32)

+N∗∑n=0

h∑K∈T

ko(un+1K )

∑r=i,j,k


K

+τN∗∑n=0

h∑K∈T

m(K)(un+1

K − unKh

)2+∑K∈T

m(K)Pc(uN∗+1

K ) ≤∑K∈T

m(K)Pc(u0K).

This proof is then concluded by using the continuity of pc(·) and (A5), yielding∑K∈T

m(K)Pc(u0K) ≤ C.

To obtain estimates in terms of the discrete gradients of the saturation, we first provethe result below:

Lemma 4.3.5. Given α, β ∈ [mp,Mp] and two vectors a,b such that the angle in betweenis γ ∈ [arccos( 2

√mpMp

mp+Mp), π − arccos( 2

√mpMp

mp+Mp)], one has

α|a|2 + β|b|2 + (α+ β)|a| |b|cosγ ≥ 0. (4.33)

Proof. The case b = 0 is trivial. If b 6= 0, let x = |a||b| . Then, the proof reduces to

showing thatαx2 + (α+ β)cosγ x+ β ≥ 0,

for all x ∈ R. Since |cosγ| ≤ 2√mpMp

mp+Mp, one has

∆ = (α+ β)2(cosγ)2 − 4αβ

≤ (α+ β)2 4mpMp

(mp +Mp)2 − 4αβ

= 4α2( mpMp

(mp +Mp)2 (1 + α

β)2 − β

α

).

Observing that mpMp≤ α

β ≤Mp

mp, one immediately sees that ∆ ≤ 0, which concludes the

proof.

Now we can provide the a-priori estimates


Lemma 4.3.6. If h < τ , for any N∗ ∈ 0, ..., N − 1 it holds∑K∈T

∑r=i,j,k

m(Kr)|∇KruN∗+1

K |2 ≤ C, (4.34)

where C is independent of h, size(T ), or N∗.

Proof. Subtracting (4.17) from (4.18) gives

h(pn+1σIdKr

− pn+1K )− h(pn+1

σIdKr

− pn+1K )

=h(pc(un+1σIdKr

)− pc(un+1K )) + τ(un+1

σIdKr

− un+1K )− τ(un+1

σIdKr

− unK) (Id = 1, 2). (4.35)

Multiplying (4.35) by m(Kr)µσIdKr∇Krun+1

K , adding the resulting for Id = 1 and 2 andsumming over r ∈ i, j, k, K ∈ T and n ∈ 0, ..., N∗ for any fixed N∗ < N gives

N∗∑n=0

h∑K∈T

∑r=i,j,k

m(Kr)(∇Kr pn+1K −∇Krpn+1

K ) · ∇Krun+1K

=N∗∑n=0

h∑K∈T

∑r=i,j,k

m(Kr)∇Krpc(un+1K ) · ∇Krun+1

K

+N∗∑n=0

τ∑K∈T

∑r=i,j,k

m(Kr)(∇Krun+1K −∇KrunK) · ∇Krun+1

K .

Applying on the left the Cauchy-Schwarz inequality and using Lemma 4.3.1 for the lastterm on the right gives

τ

2∑K∈T

∑r=i,j,k

m(Kr)|∇KruN∗+1

K |2 + τ

2

N∗∑n=0

∑K∈T

∑r=i,j,k

m(Kr)|∇Krun+1K −∇KrunK |2

+N∗∑n=0

h∑K∈T

∑r=i,j,k


K

≤ τ

2∑K∈T

∑r=i,j,k

m(Kr)|∇Kru0K |2 + 1

2

N∗∑n=0

h∑K∈T

∑r=i,j,k

m(Kr)|∇Kr pn+1K −∇Krpn+1

K |2

+ 12

N∗∑n=0

h∑K∈T

∑r=i,j,k

m(Kr)|∇Krun+1K |2.

By Lemma 4.3.4, the second term on the right is bounded uniformly in h, size(T ) and


N∗. This gives:

τ

2∑K∈T

∑r=i,j,k

m(Kr)|∇KruN∗+1

K |2 + τ

2

N∗∑n=0

∑K∈T

∑r=i,j,k

m(Kr)|∇Krun+1K −∇KrunK |2

+N∗∑n=0

h∑K∈T

∑r=i,j,k


K

≤C + 12

N∗∑n=0

h∑K∈T

∑r=i,j,k

m(Kr)|∇Krun+1K |2.

Furthermore, the third term on the left is positive. To see this, observe that for anyn ∈ 0, ..., N∗, K ∈ T , r ∈ i, j, k there exist ξ1, ξ2 ∈ R such that

N∗∑n=0

h∑K∈T

∑r=i,j,k


K

=N∗∑n=0

h∑K∈T

∑r=i,j,k

m(Kr)((pc(un+1

σ1Kr

)− pc(un+1K )

)µσ1

Kr+(pc(un+1

σ2Kr

)− pc(un+1K )

)µσ2

Kr

)·((un+1σ1Kr

− un+1K

)µσ1

Kr+(un+1σ2Kr

− un+1K

)µσ2

Kr

)=

N∗∑n=0

h∑K∈T

∑r=i,j,k

m(Kr)(p′

c(ξ1)((un+1σ1Kr

− un+1K

)|µσ1

Kr|)2

+ p′

c(ξ2)((un+1σ2Kr

− un+1K

)|µσ2

Kr|)2

+(p′

c(ξ1) + p′

c(ξ2))(un+1σ1Kr

− un+1K

)|µσ1

Kr| ·(un+1σ2Kr

− un+1K

)|µσ2

Kr| · cos(π − θ)

).

Note that to avoid an excess of notions, we omitted any additional indices for ξ1, ξ2,which actually depend on the particular n, K or r. Observing that γ the angle betweenµσ1

Krand µσ2

Kr, satisfies γ = π − θ and by (A6), |cosγ| ≤ 2

√mpMp

mp+Mp, using (A3) and

Lemma 4.3.5, one immediately gets that the above is positive. This gives

τ − h2

∑K∈T

∑r=i,j,k

m(Kr)|∇KruN∗+1

K |2 ≤ C + 12

N∗−1∑n=0

h∑K∈T

∑r=i,j,k

m(Kr)|∇Krun+1K |2,

and the conclusion is a direct consequence of Lemma 4.3.2.

4.4. CONVERGENCE OF THE SCHEME 65

4.4 Convergence of the scheme

4.4.1 Compactness results

To prove the convergence we recall Definition 3.2 and use the time-space discrete valuesto construct a sequence of triples defined in Ω× (0, T ]:

vT ,h(x, t) = vnK for all x ∈ K and t ∈ (nh, (n+ 1)h], n = 0, ..., N − 1. (4.36)

Further, we will use the discrete version of the seminorm in the space L2(0, T ;W 1,2(Ω)).

Definition 4.4.1. (Discrete seminorms) For v ∈ X(T , h) enriched with values (vnσ1Kr

,

vn+1σ2Kr

)|K ∈ T , r = i, j, k, define

|v(·, t)|1,T =(∑

K

∑r=i,j,k

m(Kr)(|vnσ1

Kr

− vnK |2|µσ1Kr|2 + |vnσ2

Kr

− vnK |2|µσ2Kr|2))1/2

,

for all t ∈ (nh, (n+ 1)h], n = 0, ..., N − 1, and

|v|1,T ,h =(

N∑n=0

h∑K

∑r=i,j,k

m(Kr)(|vnσ1

Kr

− vnK |2|µσ1Kr|2 + |vnσ2

Kr

− vnK |2|µσ2Kr|2))1/2

.

Note that | · |1,T and | · |1,T ,h are the discrete counterparts of the gradient norms forfunctions inW 1,2(Ω), respectively L2(0, T ;W 1,2(Ω)). Clearly, these are seminorms in thecorresponding spaces. Following the Lemma 4.3.4 and Lemma 4.3.6, we have

Lemma 4.4.1. Under assumption (A3), if (uT ,h, pT ,h, pT ,h) ∈ (X(T , h)))3 solves thesystem (4.15) - (4.20), one has

|pT ,h|21,T ,h + |pT ,h|21,T ,h + |uT ,h|21,T ,h ≤ C, and |uT ,h|21,T ≤ C for all t ∈ (0, T ],

where C does not depend on size(T ) or h.

The following is a discrete counterpart of the Poincare inequality. Before stating it,let (vK , vσ1

Ki, vσ2

Ki, vσ1

Kj, vσ2

Kj, vσ1

Kk, vσ2

Kk) be given 7-tuples for any K ∈ T satisfying

vσ1Kr

= vσ2Kr

= 0 whenever σ1Kr, σ2Kr∈ Eext. Let X0

T be the space of piecewise functionv : Ω → R, v|K = vK , endowed with the discrete gradient ∇Kv by using the additionalvalues, and ‖ · ‖L2(Ω) we mean the L2-norm of the piecewise constant v. Then we have

Lemma 4.4.2. (Discrete Poincare inequality) A constant C > 0 depending on Ω,but not on size(T ) exists such that

‖v(·)‖2L2(Ω) ≤ C|v|21,T .


Proof. We essentially apply the technique in [50]. For σ ∈ E , define χσ from R2×R2 to0, 1 as

χσ(x, y) :=

1, [x, y] ∩ σ 6= ∅,0, [x, y] ∩ σ = ∅.

(4.37)

Let e be a given vector and x ∈ Ω. Let Dx be the semi-line having the origin x andcontinuing in the direction of e. Let y(x) be such that y(x) ∈ Dx∩∂Ω and [x, y(x)] ⊂ Ω,where [x, y(x)] = βx + (1 − β)y(x), β ∈ [0, 1] (i.e. y(x) is the first point where Dxmeets ∂Ω). For y(x) such that χσ(x, y(x)) = 1 and if σ ∈ Eint, let K|L be the trianglesadjacent to σ, where K is the one closest to x. Let vK , vL be the corresponding values.Also, let vσ be the value from v

σIdKr

, I = 1, 2, r = i, j, k corresponding to the past ofσ in K intersected by Dx. For σ ∈ Eext, one takes vL = vσ = 0. Also if the intersectionpoint happens to be vertex in P or edge midpoint in PM , then one takes arbitrary vσ,vK , vL. The choice is not important as finally, we integrate for x ∈ Ω.

Along [x, y(x)], we define

Dσv :=

|vσ − vK |+ |vσ − vL|, if σ ∈ Eint and χσ(x, y) = 1,|vσ − vK |, if σ ∈ Eext and χσ(x, y) = 1,0, if χσ(x, y) = 0.

(4.38)

Let now K ∈ T arbitrary. For a.e. x ∈ K, one has

|vK | ≤∑σ∈E

χσ(x, y(x))Dσv. (4.39)

Using the Cauchy - Schwarz inequality, this gives

|vK |2 ≤∑σ∈E

χσ(x, y(x)) (Dσv)2

dσcσ

∑σ∈E

χσ(x, x+ η)dσcσ, (4.40)

for a.e. x ∈ R2, where cσ = |nσ · e|, nσ denotes a unit normal vector to σ, and

dσ :=

1

|µσK |+ 1|µσL |

, if σ ∈ Eint,1

|µσK |, if σ ∈ Eext.

(4.41)

As in [50], we show that for a.e. x ∈ Ω∑σ∈E

χσ(x, y(x))dσcσ ≤ diam(Ω). (4.42)

Given now e and x and assuming that Dx does not go through any vertex. AssumingσL ∈ L ∩ ∂Ω, L ∈ T , let xσx ∈ Ω be the perpendicular foot of XLxσx to σL. Since the


control volumes are convex, there exists xC ∈ K such that∑σ∈E

χσ(x, y(x))dσcσ ≤ |(xC − xσx) · e|. (4.43)

Further, using xC , xσx ∈ Ω gives (4.42). Integrating (4.40) over Ω and using to (4.42)this gives ∑

K∈T

∫K

|vK |2dx

≤diam(Ω)( ∑σ∈Eint

∫Ωχσ(x, y(x))dx (|vσ − vK |+ |vσ − vL|)2

dσcσ

+∑σ∈Eext

∫Ωχσ(x, y(x))dx |vσ − vK |

2

dσcσ

).

Since∫

Ω χσ(x, y(x))dx ≤ diam(Ω)m(σ)cσ, one has

∑K∈T

∫K

|vK |2dx

≤2(diam(Ω))2( ∑σ∈Eint

m(σ)(|µσK ||vσ − vK |2 + |µσL ||vσ − vL|2)

+∑σ∈Eext

m(σ)(|µσK ||vσ − vK |2).

Recalling (4.25), Definition 4.4.1 and Lemma 4.4.1, we have

‖v‖2L2(Ω) ≤ C|v|21,T .

With this lemma, one has uniformly with respective to T and h

‖uT ,h‖2L2(0,T ;L2(Ω)) + ‖pT ,h‖2L2(0,T ;L2(Ω)) + ‖pT ,h‖2L2(0,T ;L2(Ω)) ≤ C.

Now we show the following lemma about space translations.

Lemma 4.4.3. Given the trianglarization T and v ∈ X0T , let v be the extension of v by

0 to the entire R. Then for any η ∈ R2, one has

‖v(·+ η)− v(·)‖2L2(R2) ≤ 2|v|21,T |η|(|η|+ Csize(T )), (4.44)

with C > 0 only depending on Ω and not on v, η or T .


Proof. For σ ∈ E , using χσ as defined in (4.37), and for η ∈ R2, one has

|v(x+ η)− v(x)| ≤∑σ∈E

χσ(x, x+ η)Dσv, for a.e. x ∈ Ω,

where K, L are the volumes adjacent to σ. Following again [50], but defining dσ as in(4.41), one obtains

|v(x+ η, t)− v(x, t)|2 (4.45)

≤( ∑σ∈Eint

χσ(x, x+ η) (|vσ − vK |+ |vσ − vL|)2

dσcσ

+∑σ∈Eext

χσ(x, x+ η) |vσ − vK |2

dσcσ

)·∑σ∈E

χσ(x, x+ η)dσcσ,

for a. e. x ∈ R2. Here cσ = |nσ · η|η| |, and nσ denotes a unit normal vector to σ. First,by [50] there exists C > 0, only depending on Ω such that∑

σ∈Eχσ(x, x+ η)dσcσ ≤ |η|+ Csize(T ), for a.e. x ∈ R2. (4.46)

Further, observe that for all σ ∈ E ,∫R2χσ(x, x+ η)dx ≤ m(σ)cσ|η|.

Therefore, integrating (4.45) over R2 and using (4.25) one gets

‖v(·+ η, ·)− v(·, ·)‖2L2(R2)

≤∫R2

∑σ∈Eint

χσ(x, x+ η) (|vσ − vK |+ |vσ − vL|)2

dσcσ

∑σ∈E

χσ(x, x+ η)dσcσdx

+∫R2

∑σ∈Eext

χσ(x, x+ η) |vσ − vK |2

dσcσ

∑σ∈E

χσ(x, x+ η)dσcσdx

≤( ∑σ∈Eint

m(σ)dσ

(|vσ − vK |+ |vσ − vL|)2 +∑σ∈Eext

m(σ)dσ|vσ − vK |2

)|η|(|η|+ Csize(T ))

≤2( ∑σ∈Eint

m(σ)(|µσK ||vσ − vK |2 + |µσL ||vσ − vL|2)

+∑σ∈Eext

m(σ)|µσK ||vσ − vK |2)|η|(|η|+ Csize(T ))

=2|v|21,T |η|(|η|+ Csize(T )),

which completes the proof.

The result in Lemma 4.4.3 extends straightforwardly to the case where v is time


dependent as well, namely if v is piecewise constant in the space-time volumes as in thecase of X(T , h) elements. Clearly, when estimating the L2(0, T ;L2(R2)) norm, in thiscase the norm |v|21,T ,h will appear on the right. We continue with the estimates for thetime translations:

Lemma 4.4.4. Let un+1K , n = 0, ..., N − 1 be the u components of the solution of

(4.15) - (4.20) and uT ,h the extension to Ω × (0, T ] defined in (4.36). A C > 0 existssuch that for any ξ ∈ (0, T )

‖uT ,h(·, ·+ ξ)− uT ,h(·, ·)‖2L2(Ω×(0,T−ξ)) ≤ C.

Proof. Letting

B(t) =∫

Ω

(uT ,h(x, t+ ξ)− uT ,h(x, t)

)2dx,

for t ∈ (0, T − ξ), one has obviously∫Ω×(0,T−ξ)

(uT ,h(x, t+ ξ)− uT ,h(x, t)

)2dxdt =

∫ T−ξ

0B(t)dt.

With n0(t), n1(t) ∈ 0, ..., N − 1 such that n0(t)h ≤ t ≤ (n0(t) + 1)h and n1(t)h ≤t+ ξ ≤ (n1(t) + 1)h, B rewrites

B(t) =∑K∈T

m(K)(un1(t)K − un0(t)

K

)2=∑K∈T

m(K)( n1(t)−1∑n=n0(t)

un+1K − unK

)2. (4.47)

Using the Cauchy-Schwarz Inequality gives

B(t) ≤ N∑K∈T

m(K)n1(t)−1∑n=n0(t)

(un+1K − unK

)2. (4.48)

Defining χ(n; t, t+ ξ) as

χ(n; t, t+ ξ) =

1, if nk ∈ (t, t+ ξ],0, if nk /∈ (t, t+ ξ],

(4.48) becomes

B(t) ≤ NN−1∑n=0

χn(t, t+ ξ)∑K∈T

m(K)(un+1K − unK

)2.


Since 0 ≤∫ T−ξ

0 χn(t, t+ ξ)dt ≤ ξ, we have

∫ T−ξ

0B(t)dt ≤ Nξ

N−1∑n=0

∑K∈T

m(K)(un+1K − unK

)2.

Following (4.31) and according to (A5), one has

τ

N−1∑n=0

∑K∈T

m(K)(un+1K − unK

)2≤ h

∑K∈T

m(K)Pc(u0K),

which gives ∫ T−ξ

0B(t)dt ≤ CTξ,

and the proof is concluded.

4.4.2 Convergence results

In this section, we show the convergence of the finite volume scheme. Following the apriori estimates obtained above, one has

Theorem 4.4.1. There exists a sequence (Tm, hm) such that size(Tm) → 0, hm → 0as m → ∞, and the triple (uTm,hm , pTm,h, pTm,hm) converges weakly in L2(Q) to thesolution (u, p, p) in the sense of Definition 4.2.1. Moreover, uTm,hm converges stronglyto u in L2(0, T ;L2(Ω)).

Proof. Lemma 4.4.1 gives that (uT ,h, pT ,h, pT ,h) is bounded uniformly in L2(0, T ;L2(Ω)).This gives immediately the existence of a sequence (Tm, hm) and of a triple (uTm,hm , pTm,h,pTm,hm) such that it converges weakly to a triplet (u, p, p) in L2(Q). Then, by Lemma4.4.3, Lemma 4.4.4 and Theorem 3.11 in [50], we obtain u ∈W 1,2(0, T ;W 1,2

0 (Ω)), p, p ∈L2(0, T ;W 1,2

0 (Ω)). Furthermore, Lemma 4.4.3 and Lemma 4.4.4 and the Kolmogorov-M. Riesz-Frécht Theorem (Theorem 4.26 in [20]) also give the strong convergence:uTm,hm → u as m → ∞. In the following, we show (u, p, p) is the weak solution ofProblem P. To do so, we let φ, ψ ∈ C2(Ω× [0, T ]) such that φ = ψ = 0 on ∂Ω× [0, T ],φ(·, T ) = ψ(·, T ) = 0. For λ, we make the assumption as λ ∈ C1(Ω × [0, T ]),λ(·, T ) = 0, which means that pointwise values make sense. Multiplying (4.17) byhmλTm,hm(xK , (n + 1)hm)m(K), summing the resulting for n ∈ 0, ..., N − 1 and


K ∈ T gives

N−1∑n=0

hm∑K∈Tm

m(K)(pn+1K − pn+1

K )λ(xK , (n+ 1)hm) (4.49)

=N−1∑n=0

hm∑K∈Tm

m(K)pc(un+1K )λ(xK , (n+ 1)hm)

+ τ

N−1∑n=0

∑K∈T

m(K)(un+1K − unK)λ(xK , (n+ 1)hm).

We denote the last term of (4.49) by T1 and rewrite it as

T1 =τN−1∑n=0

∑K∈Tm

m(K)(un+1K − unK)λ(xK , (n+ 1)hm)

=τN−1∑n=1

∑K∈Tm

m(K)unK(λ(xK , nhm)− λ(xK , (n+ 1)hm))

+τ∑K∈Tm

m(K)(uNKλ(xK , T )− u0

Kλ(xK , hm)).

First We have λ(xK , T ) = 0. Then according to the property of the initial condition, onehas ∑

K∈Tm

m(K)u0Kλ(xK , hm) −→

∫Ωu0(x)λ(x, 0)dx as m −→∞.

Further, since λ ∈ C1(Ω× [0, T ]), λ(·, T ) = 0, one has

N−1∑n=0

∑K∈Tm

m(K)un+1K (λ(xK , (n− 1)hm)− λ(xK , nhm))→

∫ T

0

∫Ωu(x, t)∂tλ as m→∞.

Similarly, since pTm,hm − pTm,hm converges weakly to p− p, one has

N−1∑n=0

∑K∈Tm

(pn+1K − pn+1

K )λ(xK , nh) −→∫ T

0

∫Ω

(p− p)λdtdx as m −→∞.

From the above, one gets that (u, p, p) satisfies (4.11).Furthermore, given ϕ ∈ (C∞0 (Ω× [0, T )))2, for any K ∈ T and n ∈ 0, ..., N set

ϕnK = ϕ(xK , tn), divKϕnK = 1m(K)

∑L∈NK

2∑Id=1

m(σIdK|L) ϕnK · nK|L. (4.50)

Letting χK×[tn,tn+1) be characteristic function of K × [tn, tn+1), we use (4.50) to define


Now we concentrate on A =∫ T

0∫

Ω ko(u)∇p · ∇φdxdt. To do so, we define thediscretization and approximation of φ denoted by Φ and φT ,h:

Φn+1K = φ(xK , (n+ 1)h), K ∈ T , n ∈ 0, ...N − 1,

Φn+1σ = φ(xσ, (n+ 1)h), σ ∈ E , n ∈ 0, ...N − 1,

φT ,h = Φn+1K , x ∈ K, t ∈ (nh, (n+ 1)h) for all n = 1, ..., N − 1.

(4.53)

Then multiplying (4.15) by hmφn+1K := hmφ(xK , (n + 1)h) and (4.19) by hmφn+1

σIdKr

:=

hmφ(xσIdKr

, (n+1)hm) and summing overK ∈ Tm and n ∈ 0, N−1, with r = Pi, Pj , Pk

and Id = 1, 2, one has

ATm,hm

=N−1∑n=0

∑K∈Tm

∑r=i,j,k

∫ tn+1

tn

∫Kr

ko(un+1K )

((pn+1σ1Kr

− pn+1K ) · µσ1

Kr+ (pn+1

σ2Kr

− pn+1K ) · µσ2

Kr

)·

((φn+1σ1Kr

− φn+1K ) · µσ1

Kr+ (φn+1

σ2Kr

− φn+1K ) · µσ2

Kr

)dxdt.

=∫ T

0

∫Ωko(uTm,hm)∇Tm pTm,hm∇TmφTm,hmdxdt.

Then we have

T2 =ATm,hm −A

=∫ T

0

∫Ωko(uTm,hm)∇Tm pTm,hm · ∇TmφTm,hm −

∫ T

0

∫Ωko(u)∇p · ∇φ

=∫ T

0

∫Ωko(uTm,hm)∇Tm pTm,hm∇TmφTm,h −

∫ T

0

∫Ωko(u)∇T pTm,hm · ∇TmφTm,hm︸︷︷︸

T21

+∫ T

0

∫Ωko(u)∇Tm pTm,hm · ∇TmφTm,hm −

∫ T

0

∫Ωko(u)∇p · ∇φ︸︷︷︸

T22

.

By the assumption (A2), the compactness of∇T pT ,h, the regularity of φ and uTm,hm → u

as m→∞, we easily obtain

T21 → 0 as m→∞. (4.54)

Furthermore, for T22 since we have ∇Tm pTm,hm → ∇p, it is also easily obtained that

T22 → 0 as m→∞, (4.55)

which together with (4.54) implies ATm,hm converges weakly to A as m → ∞. In the


same way, one gets the convergence for∫ T

0∫

Ω kw(u)∇p · ∇φdxdt, which concludes theproof.

4.5 Numerical results

We start with a simple test problem defined in Ω = (0, 1)× (0, 1) and for t > 0. Further,we take ko(u) = kw(u) = 1 and pc(u) = u, leading to

∂tu−∆p = 0, (4.56)∂t(1− u)−∆p = 0, (4.57)p− p = u+ τ∂tu. (4.58)

To close the system, we prescribe the boundary conditions:

p = p = 0 at ∂Ω,

the initial condition

u(x, y, 0) = sin(2πx) · sin(3πy).

In this case, an explicit solution can be found:

u(x, y, t) = exp( −13π2t

2 + 13τπ2 ) · sin(2πx) · sin(3πy),

p(x, y, t) = exp( −13π2t

2 + 13τπ2 ) · sin(2πx) · sin(3πy) · (12 −

13τπ2

2(2 + 13τπ2) ), and

p(x, y, t) = exp( −13π2t

2 + 13τπ2 ) · sin(2πx) · sin(3πy) · (−12 + 13τπ2

2(2 + 13τπ2) ).

This is used to test the convergence of the method. Recall that the convergence is provedbased on compactness arguments, without having rigorous error estimates. Nevertheless,for this specific example, since an explicit solution is known, we can estimate the order ofthe scheme as follows.

After constructing a mesh and taking uniform time step, we refine it uniformly threetimes by halving the mesh size and time step. First we consider a uniform mesh as shownin Figure 4.2 (left). For each of the discretization parameters, we compute the L2 andW 1,2 errors at t = 1/16:

EuT ,h =(∫

Ω(u(x, t)−uT ,h(x, t))2dx

)1/2, EpT ,h =

(∫Ω

(∇p(x, t)−∇T pT ,h(x, t))2dx)1/2

.

(4.59)The results in Table 4.1 refer to EuT ,h , EpT ,h, which are representative. All other errors

4.5. NUMERICAL RESULTS 75

have similar behavior. We estimate the order by computing

α = log2(EuT ,h

EuT /2,h/2), β = log2(

EpT ,hEpT /2,h/2

). (4.60)

Based on this the scheme is first order convergent in both L2 and W 1,2 norm. Observethat the order in the approximation of the gradient is the same as the L2-order, this beinga consequence of the multipoint flux approximation.

One of the advantage of the proposed scheme is that, theoretically, it is robust withrespect to the meshing. Since pc is linear, (A6) brings no restriction for the meshing. Toevaluate the behavior of the scheme for non-uniform meshes and anisotropic cases, weuse as starting point the non-uniform mesh in Figure 4.2 (right). Note that this mesh isbuilt without any connection with the solution, such as rapid changes in the magnitude ofthe gradient. The results presented in Table 4.2 show practically no change in the orderof the scheme.

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Figure 4.2 The uniform mesh (left) and nonuniform mesh (right).

No. of cells EuT ,h α EpT ,h β

43 × 8 8.6546× 10−4 – 5.7211× 10−3 –44 × 8 4.3880× 10−4 0.9799 2.8433× 10−3 1.008745 × 8 2.2097× 10−4 0.9897 1.4191× 10−3 1.002646 × 8 1.1088× 10−4 0.9949 7.0912× 10−4 1.0009

Table 4.1 Convergence results for uniform mesh, τ = 1.



43 × 8 8.7661× 10−4 – 7.8924× 10−3 –44 × 8 4.4138× 10−4 0.9899 3.8945× 10−3 1.019045 × 8 2.2160× 10−4 0.9941 1.9390× 10−3 1.006146 × 8 1.1104× 10−4 0.9969 9.6800× 10−4 1.0022

Table 4.2 Convergence results for nonuniform mesh, τ = 1.

The first test involved a scalar permeability. However, the multipoint flux approx-imation considered here applies to anisotropic cases too. To see this, we consider thefollowing problem:

∂tu−∇ · (K∇p) = 0, (4.61)∂t(1− u)−∇ · (K∇p) = 0, (4.62)

p− p = u+ τ∂tu. (4.63)

With k1 = 1, k2 = 5, K is defined as

K =(k1 00 k2

). (4.64)

The boundary conditions remain unchanged:

p = p = 0 at ∂Ω,

as the initial condition

u(x, y, 0) = sin(2πx) · sin(3πy).

Again, an explicit solution can be found:

u(x, y, t) = exp( −49π2t

2 + 49τπ2 ) · sin(2πx) · sin(3πy),

p(x, y, t) = exp( −49π2t

2 + 49τπ2 ) · sin(2πx) · sin(3πy) · (12 −

49τπ2

2(2 + 49τπ2) ), and

p(x, y, t) = exp( −49π2t

2 + 49τπ2 ) · sin(2πx) · sin(3πy) · (−12 + 49τπ2

2(2 + 49τπ2) ).

For the numerical tests we carried out the same steps as before: two meshes (uniformand non-uniform) are refined successively three times. We compute the same errors, andobserve that even in the anisotropic case, the scheme still remains first order convergencefor both meshes and practically does not loose accuracy. The results are given in Tables4.3 and 4.4.

4.5. NUMERICAL RESULTS 77


43 × 8 8.7790× 10−4 – 1.8338× 10−3 –44 × 8 4.4681× 10−4 0.9777 9.0883× 10−4 1.012745 × 8 2.2544× 10−4 0.9869 4.5329× 10−4 1.003546 × 8 1.1324× 10−4 0.9934 2.2647× 10−4 1.0011

Table 4.3 Convergence results for uniform mesh, τ = 1 and in the anisotropic case.


43 × 8 8.8311× 10−4 – 3.1188× 10−3 –44 × 8 4.4791× 10−4 0.9794 1.4995× 10−3 1.056545 × 8 2.2569× 10−4 0.9889 7.3982× 10−4 1.019246 × 8 1.1330× 10−4 0.9942 3.6810× 10−4 1.0071

Table 4.4 Convergence results for nonuniform mesh, τ = 1 and in the anisotropic case.

For the final test, we mention that one of the known features of the model (4.1)- (4.3) is that their solution does not satisfy a maximum principle. Instead, effects likesaturation overshoot can be observed both experimentally [17,38] and analytically [40,99].To investigate this aspect, we present some numerical experiments carried out with therelative permeability functions

kw(u) = u1.5, ko = (1− u)1.5.

These are commonly encountered in modeling two-phase flows in porous media, when udenotes the water saturation. For the equilibrium capillary pressure, we still take a linearfunction

pc(u) = 1− u.

We consider the domain Ω = (−5, 10)× (0, 10). The initial condition is shown in Figure4.3,

u0 = (ur − ul)/(1 + exp(−4x) + ul), for all (x, y) ∈ Ω.

Here ul, ur ∈ [0, 1] are two constant values, ul = 0.9, ur = 0.1. Observe that u0 doesnot depend on y.


−5 0 5 1005

100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Figure 4.3 The initial saturation.

At the lateral boundaries, we assume that the normal flux is 0,

−ko(s)py = −kw(s)py = 0, along (−5, 10)× 0 and (−5, 10)× 10.

At the inflow and outflow boundary, we assume a given, constant total flux in the xdirection:

−ko(s)px − kw(s)px = 1, along −5 × (0, 10) and 10 × (0, 10).

Further, we assume that the saturation is given there,

u(−5, y) = ul, u(10, y) = ur for any y ∈ (0, 10),

and compute the pressures accordingly.The numerical approximation of the saturation is displayed in Figure 4.4 for two times.

Observe that the saturation exceeds its maximal initial and boundary values, which is theso-called overshoot effect. The results are in good agreement with the profiles in theone-spatial dimensional case obtained e.g. in [40].

−5 0 5 100

5

10

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

−5 0 5 100

5

10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Figure 4.4 The saturation at t = 1 (left) and t = 3 (right) with τ = 1.

Chapter 5

Two phase flow inheterogeneous porous media

5.1 Introduction

Two-phase (wetting/non-wetting) flows are widely encountered in various real-life pro-cesses. A few prominent examples are water driven oil recovery, or geological sequestrationof CO2. These processes involve large spatial scales, and (rock) heterogeneities appearnaturally. In this chapter we consider a simplified situation, where the medium consists oftwo different homogeneous blocks with different permeabilities (coarse and fine), whichare separated by an interface. This makes the transition from one material to anothernot smooth, and appropriate conditions have to be imposed at the interface for couplingthe models written in each of the two blocks. In particular, when the underlying modelsare involving entry pressures to describe the dependency of the capillary pressure on thephase saturations, the non-wetting phase may remain trapped in the coarse block at theinterface.

This situation has been analyzed in [42,46], but for the case when the phase pressuredifference depends on the, say, wetting phase phase saturation and the medium itself.These are standard models, for which the dependency between various quantities are de-termined under equilibrium conditions. Therefore, such models are also called equilibriummodels. In the paper mentioned above, regularization arguments (i.e. approximating theinterface by a thin porous layer ensuring a smooth transition between the two homogen-eous blocks) are employed to derive appropriate coupling conditions between the modelsin the two sub-domains. The resulting conditions are the flux continuity and an extendedpressure condition. We refer to [14,25,27] for the mathematical analysis of such models,where the existence of weak solutions has been analyzed. Further, the case of many layers

This chapter is a collaborative work with C.J. van Duijn, I.S. Pop and it has been published inTransport in Porous Media, 110(2015): 1-26.

79

80 CHAPTER 5. TWO PHASE FLOW IN HETEROGENEOUS POROUS MEDIA

is studied in [43,45], where homogenization techniques are applied for deriving an effectivemodel. Such kind of models are also studied in [3, 19, 54, 55, 88, 90], where appropriatenumerical schemes are studied.

Various experiments [17,38] have invalidated the equilibrium assumptions, and motiv-ated considering non-equilibrium approaches. Here we focus on models involving dynamiceffects in the phase pressure difference, as proposed in [60]. In this case, we follow theideas in [42,46] and derive the coupling conditions at the interface separating the two ho-mogeneous blocks. When compared to the equilibrium case, a striking difference appears.In the former the non-wetting phase can only flow into the fine block if its saturation atthe coarse block side of the interface exceeds an entry value, in the latter situation thisflow can appear for lower saturation values. This is due to the dynamic effects in thephase pressure difference and can reduce the amount of non-wetting phase that remainstrapped at the interface.

The models including dynamic effects in the phase pressure difference lead to so-called pseudo-parabolic problems. For such models, but posed in homogeneous domains,existence and uniqueness of weak solutions are obtained in [30, 52, 77]. The case ofvanishing capillary effects and the connection to hyperbolic conservation laws is studiedin [40, 44]. For dynamic capillarity models in the heterogeneous case, but in the absenceof an entry pressure, numerical schemes are discussed in [61]. This situation is similarto the case analyzed in [36], where the interface is replaced by a discontinuity in theinitial conditions. Also, variational inequality approaches have been considered in [62] forsituations including an entry pressure. However, the conditions are simply postulated andno derivation is presented.

In Section 5.2 we present the mathematical model. For simplicity we consider the casewhen only the absolute permeability is different in the two blocks, all other parametersbeing the same. In Section 5.3, we derive the coupling conditions at the interface. Theseare the flux continuity and an extended pressure continuity. In Section 5.4, we discuss dif-ferent numerical approaches, and present numerical experiments that confirm the analysisin Section 5.3. We give the conclusion in the last section.

5.2 Mathematical model

We consider the flow of two immiscible and incompressible phases in a one-dimensionalheterogeneous porous medium. Letting sw denote the saturation of the wetting fluid, andsn the saturation of the non-wetting fluid, one has 0 ≤ sw, sn ≤ 1. The porous mediumis assumed to be saturated by the two fluids,

sw + sn = 1. (5.1)

5.2. MATHEMATICAL MODEL 81

Mass balance holds for each phase (see [61,82]),

φ∂sα∂t

+ ∂qα∂x

= 0, α = w, n, (5.2)

where φ is the porosity assumed constant, and qα denotes the volumetric velocity of thephase α. These velocities satisfy the Darcy’s law

qα = −k(x)krα(sα)µα

∂pα∂x

, α = w, n, (5.3)

where k(x) is the absolute permeability of the porous medium, pα the pressure, µαthe viscosity, and krα the relative permeability of the α phase. The functions krα areassumed known. Gravity effects are disregarded, as they have no influence on the interfaceconditions derived here. Substituting (5.3) into (5.2) gives

φ∂sα∂t− ∂

∂x

(k(x)krα

µα

∂pα∂x

)= 0, α = w, n. (5.4)

In standard models, the phase pressure difference depends on the saturation, which isdetermined experimentally. An example in this sense is the Leverett relationship

pn − pw = pc(x, sw) = σ

√φ

k(x)J(sw), (5.5)

where σ is the interfacial tension, and J a decreasing function.

The relationship in (5.5) is determined by measurements carried out under equilibriumcondition. In other words, before measuring the pressure and saturation in a representativeelementary volume, fluids have reached equilibrium and are at rest. However, processesof interest may not satisfy this condition, and dynamic effects have to be included. Al-ternatively to (5.5), in [60] the following model is proposed

pn − pw = pc(x, sw)− τ ∂sw∂t

. (5.6)

The damping factor τ is assumed to be known and constant. Summing the two equationsin (5.4) and using (5.1), one gets

∂q

∂x= 0, (5.7)

where q = − k(x)krwµw

∂pw∂x −

k(x)krnµn

∂pn∂x denotes the total flow. In the one-dimension case,

this means that q is constant in space. Here we assume it is constant in time as well, andis positive. This allows reducing the two-phase flow model to a scalar equation in termsof, say s = sw. After rescaling the space x with L, the time t with T , and using the


reference value K for the absolute permeability and σ√

φK for the pressure, one obtains

∂s

∂t+ ∂F

∂x= 0, (5.8)

where the F denotes the dimensionless flux of the wetting phase

F = qfw(s) +Nck(x)λ(s) ∂∂x

(J(s)√k(x)

−Ncτ∂s

∂t

). (5.9)

Here q = qTφL > 0, and k(x) = k(x)

K . Further, fw is the fractional flow function of thewetting phase,

fw = krw(s)krw(s) + krn(s)/M ,

with the mobility ratioM = µn/µw, the capillary number Nc = σ√φK

µnqL, the dimensionless

damping factor τ = τµn

(qσφ

)2and λ(s) = krn(s)fw(s). For simplicity, we assume here

Nc = 1, as different values of Nc would not have any influence on the conditions derivedbelow.

Throughout this work, we make the following assumptions

(A1) krw, krn: [0, 1]→ [0, 1] are continuous differentiable functions satisfying

a) krw is strictly increasing such that krw(0) = 0 and krw(1) = 1;

b) krn is strictly decreasing such that krn(0) = 1 and krn(1) = 0;

(A2) J is (0, 1] → R is continuous differentiable, decreasing which satisfies J ′ < 0 on(0, 1], J(1) ≥ 0 and lim

s0J(s) = +∞.

We consider a simple heterogeneous situation, where two adjacent homogeneous blocksare separated by an interface located at x = 0. For the ease of presentation, we assumethat all parameters and functional dependencies except the absolute permeability are thesame. For the latter, we have

k(x) =

k−, if x < 0 (the coarse medium),k+, if x > 0 (the fine medium).

Here k− > k+ > 0 are given.Throughout this chapter, the non-wetting phase may be oil, or CO2 or any other

phase with non-wetting phase properties. For simplicity, below, we use as oil non-wettingphase. Also, by pressure we actually mean the phase pressure difference.

5.3. CONDITIONS AT THE INTERFACE 83

5.3 Conditions at the interface

The model (5.8)-(5.9) is a parabolic equation, where the factor 1√kappears under a second

order spatial derivative. Since k has a jump discontinuity at the interface, the model isonly valid in each of the two blocks, and coupling conditions at x = 0 have to be derived.Commonly, the pressure continuity is taken as a second condition. However, this is shownto be inappropriate for entry-pressure models in the absence of dynamic effects (τ = 0).This statement is made rigorous in [42] by regularizing k. Specifically, the interface isreplaced by a thin layer in which k decays continuously from k− to k+. Next to k, wewill use the quantity

h(x) =√k(x).

Clearly,

h(x) =

h− =√k−, if x < 0,

h+ =√k+, if x > 0,

and h− > h+ > 0.For given ε > 0, we approximate the interface x = 0 by the interval [−ε, ε] (the thin

layer) and h by a smooth function hε, such that hε is monotonic in the small interval[−ε, ε]. Specifically, the discontinuous function h is now approximated by the smoothfunction hε, such that

hε(x) =

h−, for x < −ε,h(xε ), for− ε < x < ε,

h+, for x > ε.

The function h is smooth and monotonic on [−ε, ε] (see Figure 5.1).

Coarse block Fine block

Figure 5.1 The function hε.


With the given ε the solution corresponding to the regularized problem is denoted bysε, the corresponding flux by Fε. In the expression for the flux, we replace h by hε. Bytaking y = x

ε we rescale [−ε, ε] to [−1, 1]. We define vε(y, t) = vε(xε , t) = sε(x, t), andinvestigate its behavior when ε 0. First, for x ∈ [−ε, ε] (and thus y ∈ [−1, 1]), from(5.8) one gets

∂

∂tvε(y, t) + 1

ε

∂

∂yFε(y, t) = 0. (5.10)

Assuming ∂vε∂t bounded uniformly with respect to ε, passing ε to 0 gives lim

ε0∂Fε∂y (y, t) = 0,

implyinglimε0

Fε(−1, t) = limε0

Fε(1, t).

In fact, this is nothing but the flux continuity at the interface, which is as expected.For the second condition we consider again y ∈ [−1, 1]. From (5.9), one has

Fε − qfw(vε) = h2ε λ(vε)

1ε

∂

∂y

(J(vε)hε− τ ∂vε

∂t

).

As before, let now ε 0, assume that vε(y, t)→ v(y, t) and that during the limit process,the flux Fε is bounded uniformly in ε. Then, this gives

λ(v) ∂∂y

(J(v)h(y) − τ

∂v

∂t

)= 0, in Ω, (5.11)

where Ω denotes the strip Ω = (y, t) : −1 < y < 1, t > 0 (see Figure 5.2).

Coarse block

Fine block

Figure 5.2 The strip Ω.

Along the boundary of Ω, we define

s−(t) := s(0−, t) = v(−1, t),s+(t) := s(0+, t) = v(1, t),

for t > 0.


The goal is to understand how s−(t) and s+(t) are related, as well as p−(t) = J(s−(t))h− −

τ ∂s−(t)∂t , p+(t) = J(s+(t))

h+ − τ ∂s+(t)∂t . These are nothing but the left and right saturation

and pressure at the interface. Note that (5.11) is an ordinary differential equation int, where y can be seen as a parameter. To define an initial condition, we let s0(y) bea smooth function s0 : [−1, 1] → R satisfying s0(−1) = s−(0) and s0(+1) = s+(0).Clearly, the choice of s0 is not unique. Below we investigate the relation between s−(t)and s+(t), and its dependence of the regularization of h. Since λ(v) > 0, for 0 < v < 1,and λ(0) = λ(1) = 0, from (5.11), one has

v = 0, or v = 1,

or if 0 < v < 1,∂

∂y

(J(v)h(y) − τ

∂v

∂t

)= 0. (5.12)

In other words

• J(v)h(y) − τ

∂v

∂tis constant in y,whenever 0 < v < 1;

• v(·, t) is continuous with respect to y, for y[−1, 1];

• v(y, ·) is C1 with respect to t, for t ≥ 0.

For the sake of understanding, we consider some particular cases.

5.3.1 Constant saturation at the coarse side of the interface

We let s− ∈ (0, 1) and assume s−(t) = s− for all t. Let s0 : [−1, 1] −→ (0, 1) bethe given initial value, not necessarily compatible to s−: s0(−1) 6= s− in general. Weconstruct a solution for which the set Ωc = (y, t) ∈ Ω, 0 < v(y, t) < 1 is connected.From (5.12), if (y, t) ∈ Ω, v solves the autonomous initial value problem:

(Py)τ

∂v

∂t= 1h(y)

(J(v)− h(y)

h−J(s−)

), for t > 0,

v(y, 0) = s0(y).(5.13)

By (A2) and the assumption on h,(Py)has a unique solution locally. Let s∗ be defined

byJ(s∗)h−

= J(1)h+ . (5.14)

We consider the cases s− > s∗ and s− ≤ s∗, separately.

• Case 1: s− > s∗


Note thatJ(s−)

h− <J(s∗)h− =

J(1)h+ .

Since h is decreasing function in y, there exists a unique y∗ ∈ (−1, 1) such that

J(s−)h−

= J(1)h(y∗) <

J(1)h+ . (5.15)

We study now the long time behavior of v(y, t). We distinguish the following sub-cases.

a) For y ∈ (−1, y∗], define s∞(y) as the unique solution of

J(s∞(y))h(y) = J(s−)

h−. (5.16)

Clearly, s∞(y) is the equilibrium point for(Py)and satisfies s∞ > s− (see Figure 5.3).

Further, standard phase plane arguments show that, regardless of s0(y) ∈ (0, 1),limt→∞

v(y, t) = s∞(y). If s0(y) > s∞(y), the solution v(y, t) decreases continuously fromv(y, 0) = s0(y) to v(y,∞) = s∞(y). If s0(y) < s∞(y), the solution v(y, t) increasescontinuously from v(y, 0) = s0(y) to v(y,∞) = s∞(y). Note that s∞(−1) = s−, andthat s∞(y) is strictly increasing in y up to s∞(y∗) = 1. For y > y∗, we have s∞(y) = 1.

b) For y > y∗, one has

τ∂v

∂t= J(v)h(y) −

J(s−)h−

≥ J(1)h(y) −

J(s−)h−

> 0. (5.17)

This implies the solution v(y, t) increases in t and reaches v = 1 in finite positive time(see Figure 5.4)

0 < t∗(y) < τ(1− s0(y))J(1)h(y) −

J(s−)h−

.

Note: if s0(y) is non-decreasing, then t∗(y) is decreasing to t∗(1) > 0.


Figure 5.3 The functions J(v)h(y) for various y ∈ [−1, 1] for s− > s∗.

Figure 5.4 The solution for s− > s∗.

The long time behavior of v is summarized inProposition 5.3.1. Assume s0(y) ∈ (0, 1) for all y ∈ [−1, 1] and let s− > s∗, where s∗is defined in (5.14). With y∗ in (5.15), one has:a) if y ∈ [−1, y∗) then lim

t→∞v(y, t) = s∞(y), where s∞(y) ∈ (0, 1] is given by (5.16).

Further, s∞(·) is strictly increasing from s− = s∞(−1) to 1 = s∞(y∗);

b) if y ∈ (y∗, 1], then there exists t∗(y) > 0 such that v(y, ·) is increasing in t for allt < t∗ <∞, and v(y, t) = 1 for all t ≥ t∗(y). Moreover, if s0(·) is non-decreasing,then t∗(·) is decreasing to t∗(1) > 0.

Corollary 5.3.1. In particular, at y = 1, we have for s+(t) = v(1, t). s+(t) increasesto 1 for t ∈ (0, t∗(1)), where 0 < t∗(1) < τ(1 − s0(y))/(J(1)

h+ − J(s−)h− ), s+(t) = 1 for

t ≥ t∗(1), as presented in Figure 5.5.


Figure 5.5 Behavior of s+(t) for t > 0: s+ is increasing to 1 for t < t∗(1), and s+(t) = 1 fort ≥ t∗(1).

This allows constructing an extended pressure condition, similar to [42]. Consider thepressure

p(y, t) := J(v(y, t))h(y) − τ ∂v

∂t, (5.18)

observe that

p(−1, t) = p−(t) = J(s−)h−

, and p(+1, t) = p+(t) = J(s+)h+ − τ ∂s

+

∂t.

With the entry pressurep+e := J(1)

h+ , (5.19)

since s− > s∗, one hasJ(s−)h−

< p+e .

By Corollary 5.3.1, we obtain the condition:p−(t) = p+(t), for 0 < t < t∗(1),s+(t) = 1, for t ≥ t∗(1).

(5.20)

In other words, the pressure remains continuous for t < t∗(1) and oil keeps flowing into thefine material although p− is below the entry pressure. This effect is due to incorporatingdynamic effects in the phase pressure difference, and would not be possible in their absence(τ = 0).

• Case 2: s− < s∗


For any y ∈ [−1, 1], with s∞(y) defined by (5.16), one has

J(s∞(y))h(y) = J(s−)

h−>J(s∗)h−

= J(1)h+ . (5.21)

Therefore, there exists an s ∈ (s−, 1), such that

J(s∞(y))h(y) = J(s−)

h−= J(s)

h+ ,

which, in view of the monotonicity of h and J , gives s∞ < s < 1. Consequently, thiscase is similar to the sub-case y < y∗ before.

Proposition 5.3.2. Assume s0(y) ∈ (0, 1) for all y ∈ [−1, 1] and let s− < s∗ (see(5.14)). Then,

limt→∞

v(y, t) = s∞(y) for all y ∈ (−1, 1),

with s∞(y) given by (5.16). In particular, for y = 1, one can define s∞+ = s∞(1) as theunique solution of

J(s∞+ )h+ = J(s−)

h−.

Corollary 5.3.2. At y = 1, we have for s+(t) = v(1, t):

limt→∞

s+(t) = s∞+ ∈ (0, 1).

In other words, the pressure remains continuous at the interface for all t > 0:

p+(t) = J(s+(t))h+ − τ ∂s

+

∂t= J(s−)

h−= p−. (5.22)

Note that unlike in (5.20), the pressure remains continuous for all t > 0.All results refer to the case J(1) > 0, hence to the entry pressure model. If instead,

J(1) = 0 (no entry pressure), s∗ = 1 and the analysis before leads to s− < 1, andp−(t) = p+(t) for all t > 0 (see [61]).

Another special case is when τ 0. Then, the time t∗(y) in Proposition 5.3.1 andCorollary 5.3.1 approaches to 0, and if s− > s∗ the pressure becomes discontinuousinstantaneously. This is, in fact, exactly the behavior in [42] for equilibrium models.

5.3.2 Non-constant saturation at the coarse side of the interface

The results before are obtained for a constant saturation at the coarse side of the interface.Here we generalize these results. We let p±(t) = J(s±(t))

h± −τ ∂s±(t)∂t be the phase pressure

difference at the two sides of the interface and derive an extended pressure conditionsimilar to (5.20) and (5.22). With the entry pressure p+

e defined in (5.19), we assumep−(t) given, and distinguish the following cases.


• Case 1: p−(t) > p+e for all t > 0

Again, for y ∈ (−1, 1), v(y, t) solves

(Py)

J(v(y, t))h(y) − τ ∂v

∂t= p−(t), for t > 0,

v(y, 0) = s0(y).(5.23)

Assume s0(y) ∈ (0, 1), then the equation holds in the neighborhood of t = 0. We showthat in this case, v(y, t) < 1 for any t, and hence p− = p(y, t), for any y ∈ (−1, 1).Assume v(y, t∗) = 1 for some t∗ <∞. For t < t∗ one has

τ∂v

∂t= 1h(y)

(J(v(y, t))− h(y)p−(t)

).

By the monotonicity of h,

h(y)p−(t) > h+p−(t) > h+p+e = J(1),

implying that, at t = t∗, one has

τ∂v

∂t(t∗) = J(1)

h(y) − p−(t) < 0.

This shows that v(y, ·) cannot grow to 1 for t t∗. Therefore no finite t∗ exists, suchthat v(y, t∗) = 1, implying v(y, t) ∈ (0, 1) for all t. We have proved

Proposition 5.3.3. Assume s0(y) ∈ (0, 1) for all y ∈ [−1, 1] and let p−(t) > p+e for all

t > 0. Then for all y ∈ [−1, 1] and t > 0, one has v(y, t) ∈ (0, 1).

Corollary 5.3.3. At y = 1, we get s+(t) = v(1, t) ∈ (0, 1) for all t > 0, and thereforethe pressure remains continuous at both sides of the interface for all t > 0

p−(t) = p+(t). (5.24)

As in Subsection 5.3.1, if the model involves no entry pressure (J(1) = 0), one hasp+e = 0. Then p−(t) ≥ p+

e = 0, and the pressure is continuous for all t > 0.

• Case 2: 0 < p−(t) < p+e for all t > 0

We assume first that an ε exists such that 0 < p−(t) < p+e − ε for all t > 0. Further,

we assume that the initial condition s0 : (−1, 1)→ R is non-decreasing and smooth, andsatisfies s0(y) ∈ (0, 1) for any y. We have

Proposition 5.3.4. Assume there exists ε > 0 such that 0 < p−(t) < p+e −ε for all t > 0,


and that s0 is non-decreasing and smooth. Then there exists a t∗ > 0 such thatp−(t) = p+(t), for 0 < t < t∗,

s+(t) = 1, for t ≥ t∗.(5.25)

Proof. Clearly, if v(y, t) < 1, then v(y, ·) solves

(Py)

J(v(y, t))h(y) − τ ∂v

∂t= p−(t), for t > 0,

v(y, 0) = s0(y).(5.26)

First we prove the monotonicity of v(·, t). Specifically, for all t such that v(·, t) < 1uniformly in y, one also has v(·, t) is non-decreasing. To see this, we differentiate (5.26)with respect to y to obtain

τ∂2v

∂t∂y− J

′(v)h(y)

∂v

∂y= − h

′

h2 J(v).

For a fixed y, this has the general form:

τ u = fu+ b,

with u = ∂v∂y , f = J

′(v)

h(y) < 0, b = − h′

h2 J(v). Clearly, u(0) = s′

0 ≥ 0. Assuming that at > 0 exists such that u(t) = 0 and u(t) > 0 for any t ∈ (0, t], one gets:

u(t) = f(t)u(t) + b(t) = b(t) > 0.

On the other hand, one has

u(t) = lim∆t0

u(t)− u(t−∆t)∆t ≤ 0,

which contradicts the above.

Now we proceed by proving the conclusion of the proposition. Since p−(t) < p+e − ε,

by the continuity of h, there exists δ > 0 such that

h(y)h+ p−(t) < p+

e − ε/2, for 1− δ < y < 1.

Hence, for y ∈ (1− δ, 1) and t > 0, we have

J(v)− h(y)p−(t) > J(v)− h+p+e + εh+/2 = J(v)− J(1) + εh+/2.

This impliesτh(y)∂v

∂t> J(v)− J(1) + εh+/2 > εh+/2.


Clearly, a finite t = t(y) > 0 exists such that v(y, t(y)) = 1 and v(y, t) for t < t(y). Bythe monotonicity of v and s0, taking t∗ = t(1), one has v(y, t) < 1 for any y ∈ [−1, 1]and t < t∗. Therefore, v(y, t) solves (Py) for all t < t∗ and all y, hence p(y, t) remainscontinuous, in particular, p−(t) = p+(t).

Observe that, for fixed y ∈ (−1, 1), v(y, t) solving(Py)and w(y, t) solving

(P′

y

)τh(y)∂w∂t

= J(w)− h(y)(p+e − ε), t > 0,

w(y, 0) = s0(y),

are ordered. Specifically, as long as both v and w remain less 1, one has w(y, t) ≤ v(y, t).To see this, we let u = v − w, and define u− = min0, u. Then, subtracting theequations in

(Py)and

(P′

y

), and multiplying the result by u− gives

h(y)2

∂

∂t(u−)2 = (J(v)− J(w))u− − h(y)(p−(t)− (p+

e − ε))u− ≤ 0.

Here we have used the monotonicity of J . Since at t = 0 one has u−(y, 0) = 0, integratingin time gives u−(y, t) = 0, implying the ordering. Hence w is a lower bound for v, andtherefore analyzing w makes sense. Let y∗ ∈ (−1, 1) be defined by

h(y∗) = J(1)p+e − ε/2

= h+ J(1)J(1)− ε/2h+ .

Then for y > y∗, a δ > 0 exists such that h(y) = h(y∗) − δ < h(y∗), as long asw(y, t) < 1, one has

τ∂w

∂t(y, t) = J(w)

h(y) −(p+e −ε) ≥

J(1)h(y∗)−(p+−ε)+

( 1h(y)−

1h(y∗)

)J(1) = δ

h(y)h(y∗)J(1) > 0.

This shows that w reaches the cut-off value w = 1 at finite time t = t(y). If the functions0(y) is constant, it is straightforward to show that t(y) is strictly decreasing to t(1) > 0(see Figure 5.6). At y = y∗, w = 1 satisfies the equation. Therefore, w ≡ 1 is anequilibrium solution, implying that w(y∗, t) solving

(P′

y

)satisfies w(y∗, t) < 1 for all t,

and thuslimyy∗

t(y) =∞.

Remark 5.3.1. In fact, the argument above is an alternative proof for Proposition 5.3.4.Since s+(t) = v(1, t) > w(1, t), and w(1, t(1)) = 1 for t(1) <∞, (5.25) follows immedi-ately.


Figure 5.6 The function t(y).

Based on the analysis before, we discuss particular examples where oil trapping mayoccur. Defining p(y, t) := J(v)

h(y) − τ∂v∂t , in the transition region (the blown up interface)

Ω = (y, t) : −1 < y < 1, t > 0, we haveλ(v) ∂∂y

(J(v)h(y) − τ

∂v∂t

)= 0,

p(t) = p−(t), t > 0,v(y, 0) = s0(y), −1 < y < 1,

which implies that either v = 0 or v = 1, or

h(y)τ ∂v∂t

= J(v)− h(y)p−(t). (5.27)

To understand the trapping, we now take s0(y) = 1, for −1 < y < 1, and give thepressure at the coarse side of the interface for t > 0. We assume the following behavior(see Figure 5.7):

Figure 5.7 The function p−(t).


a) There exists T1 > 0 such that 0 < p−(t) < p−e for t ∈ (0, T1), where p−e = J(1)h− .

In this case we have

J(1)− h(y)p−(t) = h−p−e − h(y)p−(t) > 0.

Let y ∈ [−1, 1] be fixed. Assume that a finite t > 0 and δ > 0 exist such that v(y, t) = 1for all t < t and v(y, t) < 1 for t ∈ (t, t + δ), then one has ∂v

∂t (y, t) ≤ 0. Whereas byassumption for t ∈ (t, t+ δ), one has

∂v

∂t(y, t) = 1

τ

(J(1)h(y) − p

−(t))> 0.

This gives a contradiction.

b) There exists a T2 > T1 such that p−e < p−(t) < p+e , for t ∈ (T1, T2).

Since p−e < p−(t) < p+e , one has

J(1)h−

< p−(t) < J(1)h+ .

For T1 < t < T2, we define y(t) by

h(y(t)) := J(1)p−(t) .

Note that the definition makes sense, as p−e < p−(t) < p+e implies 1

h− < p−(t)J(1) < 1

h+ , andh(·) is a monotone, continuous interpolation between h+ and h−. Since p− is increasingin [T1, T2], this implies that y(·) is increasing and y(T1) = −1, y(T2) = 1. Then, fory > y(t), we have

J(1)− h(y)p−(t) > J(1)− h(y(t))p−(t) = 0. (5.28)

Similarly, for y < y(t), we have

J(1)− h(y)p−(t) < 0. (5.29)

Furthermore, since v(y, T1) = 1 for all y ∈ (−1, 1), from (5.28), (5.29), one get for allt ∈ (T1, T2)

v(y, t) = 1, for y > y(t), and v(y, t) < 1, for y < y(t).

Thus y = y(t) defines a free boundary in the transition region, separating regions wherev = 1 from regions where v < 1.

Remark 5.3.2. In this context, for t < T2 one has s+(t) = v(1, t) = 1, and oil remainstrapped in the coarse medium at the interface.


c) There exists T3 > T2 such that p− > p+e for t ∈ (T2, T3).

Based on the above analysis, we have v(y, T2) < 1 for all y ≤ 1, since y(T2) = 1. Further,one also has

J(1)− h(y)p−(t) < J(1)− h(y)p+e < J(1)− h+p+

e = 0.

As before, one cannot obtain v(y, t) = 1 for some t > T2, since at t, it holds

∂v

∂t(y, t) = 1

τ

(J(1)h(y) − p

−(t))< 0.

Therefore, v(y, t) < 1 for− 1 < y < 1 and T2 < t < T3. Consequently, we have

p−(t) = p+(t) for T2 < t < T3.

Remark 5.3.3. Next to the pressure continuity, this shows that oil starts flowing into thefine region for t > T2.

d) For all t > T3, p− < p+e .

We assume p−(·) decreasing and limt→∞

p−(t) = p∞ ∈ (p−e , p+e ). Given y ∈ [−1, 1], we

compare the solution v(y, t) and w(y, t) of

h(y)τ ∂v∂t

= J(v)− h(y)p−(t),

h(y)τ ∂w∂t

= J(w)− h(y)p∞,

for all t > T3, with v(y, T3) = w(y, T3) < 1.Since J is decreasing and p−(t) > p∞, one gets v(y, t) ≤ w(y, t) for −1 < y < 1, t >

T3. Further, there exists y∗ ∈ (−1, 1) such that

h(y∗) = J(1)p∞

∈ (h+, h−).

This gives for y > y∗

J(1)− h(y)p∞ > 0.

As before, there exists t(y) <∞, such that w(y, t) = 1 for t > t(y).Similarly, for y < y∗, one has

J(1)− h(y)p∞ < 0,

implying w(y, t) < 1 for all t > T3. In this case,

limt→∞

w(y, t) = s∞(y),


where s∞(y) is defined byJ(s∞(y)) = h(y)p∞.

Observe that, since h(y) > h(y∗), J(s∞(y)) = h(y)h(y∗)J(1) > J(1), so s∞(y) ∈ (0, 1). For

y = −1 one hasJ(s∞(−1)) = h−

h(y∗)J(1) < h−

h+ J(1) = J(s∗),

giving s∞(−1) > s∗. Further, if p∞ = p+e = J(1)

h+ , then s∞(−1) = s∗.

This analysis shows that, if p−(t) behaves as in Figure 5.7, and p∞ ∈ (p−e , p+e ), a

T ∗ < ∞ exists such that s+(t) = 1, for t > T ∗, and p−(t) = p+(t) for T3 < t < T ∗.This means up to T ∗, oil flows into the fine material, while trapping occurs for t > T ∗.This behavior is sketched in Figure 5.8.

Remark 5.3.4. Compared to the equilibrium case (τ = 0), a striking difference appears.If τ = 0 for a pressure p−(t) behaving as in Figure 5.7, no oil flows into the fine layer forany t > T3. If τ > 0, oil continues flowing for t > T3 and up to a time T ∗ < ∞, thedelay time. This delay appears, as discussed, if lim

t→∞p−(t) = p∞ ∈ (p−e , p+

e ).

The following result extends the statement in the remark to a more general situation.

Proposition 5.3.5. Let p−(t) ≤ p+e be such that

∫∞0 (p+

e −p−(t))dt > τ and let s0(y) ∈(0, 1). Further, let s+ be the solution ofh+τ

ds+

dt= J(s+)− h+p−(t), for t > 0,

s+(0) = s0.

Then there exists T ∗ <∞ such that s0 < s+(t) < 1 for all 0 < t < T ∗ and s+(T ∗) = 1.

Proof. Since p−(t) ≤ p+e = J(1)

h+ , and J(s+) is strictly decreasing, we have

J(s+)− h+p−(t) > 0, for s+ ∈ (0, 1).

Therefore s+(t) is strictly increasing whenever s+ < 1. Furthermore, we have

h+τds+

dt> h+(p+

e − p−(t)),

giving for all t such that s+(t) < 1 and

s+(t) > s0 + 1τ

∫ t

0(p+e − p−(ζ))dζ.

For convenience, define f(t) = 1τ

∫ t0 (p+

e − p−(ζ))dζ. Clearly, f(0) = 0, ∂tf ≥ 0 andf(∞) > 1. Hence, there exists t∗ < ∞ for which f(t∗) = 1 − s0. Consequently, there


exists T ∗ < t∗ for which s+(T ∗) = 1, and since s+ is increasing, one has s0 < s+(t) < 1for all t < T ∗.

Remark 5.3.5. A lower bound for s+ is w = w(t), the solution ofh+τdw

dt= J(w)− J(1), for t > 0,

w(0) = s0.

Since J : (0, 1] → R+ is locally Lipschitz, then for all t > 0, w(·) is strictly increasing,w(t) < 1, and lim

t→∞w(t) = 1. By a comparison argument, s(t) > w(t) for all t > 0.

5.3.3 Comparison of extended pressure conditions with static case

In this section, we show the difference in the pressure conditions appearing in the equi-librium and non-equilibrium models between static case and dynamic case. As provedin [42], if s− ≥ s∗, then one has

s+ = 1,

and no oil flows into the fine medium. Further, if s− < s∗, then s+ < 1, but the pressurecontinuous:

[p] = p−(t)− p+(t) = 0.

Then oil flows into the fine medium.Conversely, given p−(t) = J(s−(t))

h− the pressure at the interface on the coarse side andassuming that

p−(t) ≤ p+e = J(1)

h+ ,

it impliesJ(s−) ≤ h−

h+ J(1) = J(s∗),

and therefore s− ≥ s∗. In this case, one also has s+ = 1.Similarly, p−(t) > p+

e implies s− < s∗, and the pressure is continuous:

p−(t) = p+(t).

Referring to Figure 5.7, if p−(t) ≥ p+e for t > T2, the matching conditions in static

and dynamic case are the same. Specifically, s+ = 1 for t < T2, and p−(t) = p+(t)for t ∈ (T2, T3). Assuming now p−(t) is decreasing for t > T3, with p−(T3) = p+

e and


limt→∞

p−(t) = p∞ ∈ (p−e , p+e ), we have (also see Figure 5.8):

s+(t) = 1, for t > T3 in the equilibrium case τ = 0,s+(t) < 1, for t ∈ (T3, T

∗), and s+(t) = 1 for t ≥ T ∗ in the non-equilibrium case.

In other words, a delay (T ∗ − T3) appears in the non-equilibrium case before trappingoccurs.

In fact, this delay can be infinite. An extreme situation, when s(t) < 1 for all t > 0,can be constructed. To see this, we assume J : (0, 1] → R locally Lipschitz, and studythe behavior of s+ solvingh+τ

ds+

dt= J(s)− h+p−(t), for t > 0,

s(0) = s0 ∈ (0, 1),

with appropriately chosen p− satisfying p−(t) < p+e for all t > 0.

First, note that an L > 0 exists such that for all s ∈ [s0, 1]

0 ≤ J(s)− J(1) ≤ L(1− s).

Hence, an upper bound to s+ is the solution u ofτdu

dt= L

h+ (1− u) + (p+e − p−(t)), for t > 0,

u(0) = s0.

Let now v = 1− u. Then

τdv

dt+ L

h+ v = −(p+e − p−(t)), t > 0.

This gives

v(t) = (1− s0)e− Lτh+ t − 1

τ

∫ t

0e Lτh+ (z−t)(p+

e − p−(z))dz,

and the upper bound for s+ reads

u(t) = 1− (1− s0)e− Lτh+ t + 1

τ

∫ t

0e Lτh+ (z−t)(p+

e − p−(z))dz.

Thus, if p−(t) < p+e is such that∫ ∞

0e Lτh+ z(p+

e − p−(z))dz ≤ (1− s0)τ,

we obtain s+(t) < 1 for all t > 0, and consequently, the pressure remains continuous forall t > 0, whereas oil flows into the fine layer.

5.4. NUMERICAL SCHEMES AND EXAMPLES 99

Figure 5.8 Saturation inside the thin layer appximating the interface.

5.4 Numerical schemes and examples

In this section, we provide some numerical examples to illustrate how the dynamic effectsinfluence the flow of the oil across the interface between two homogeneous blocks. Forsimplicity, in (5.8)-(5.9), we take φ = 1. This gives

∂s

∂t+ ∂F

∂x= 0, (5.30)

F = qfw(s) + h2λ(s) ∂∂x

(J(s)h(x) − τ

∂s

∂t

), (5.31)

for t > 0, and x ∈ (−l, l). The boundary and initial conditions are given below.

5.4.1 Linear numerical scheme

For the discretization of (5.30) and (5.31) we decompose the interval (−l, l) into 2N + 1cells : −l = x−N−1/2 < x−N+1/2 < ... < x−1/2 < x1/2 < ... < xN−1/2 < xN+1/2 = l,where the grid is uniform with ∆x = 2l

2N+1 , we let xj = j∆x for j ∈ −N −1/2, ..., N +1/2. The discontinuity of h(x) is at x = 0. With ∆t > 0 a given time step, the fullydiscrete scheme is:

sni−1/2 − sn−1i−1/2

∆t = −Fni − Fni−1

∆x ,


where sn−1i−1/2 is the approximation of s(x, t) at x = xi−1/2 and at t = tn = n∆t. Since

q > 0, if i 6= 0 the upwind flux Fni at x = i∆x and t = tn is defined as

Fni = qfw(sn−1i−1/2)+h+/−λ(

sn−1i−1/2 + sn−1

i+1/2

2 ) ·(J′(sn−1i−1/2 + sn−1

i+1/2

2 )(sni+1/2 − s

ni−1/2

∆x )

−h+/−τ

∆x (sni+1/2 − s

n−1i+1/2

∆t −sni−1/2 − s

n−1i−1/2

∆t )).

Here h+/− means h− if i < 0, or h+ if i > 0.

At i = 0, we introduce two saturation unknowns s−,n, s+,n and define the F−,n, andF+,n as

F−,n = qfw(sn−1−1/2) + h−λ(s−,n−1) ·

[J′(s−,n−1)

(s−,n − sn−1/2

∆x/2

)

− h−τ

∆x/2

(s−,n − s−,n−1

∆t −sn−1/2 − s

n−1−1/2

∆t

)],

F+,n = qfw(s+,n−1) + h+λ(s+,n−1) ·[J′(s+,n−1)

(sn+1/2 − s

+,n

∆x/2

)

− h+τ

∆x/2

(sn+1/2 − s

n−1+1/2

∆t − s+,n − s+,n−1

∆t

)].

At the interface we also define the left and right discretized pressures

p+,n = J(s+,n)h+ − τ s

+,n − s+,n−1

∆t and p−,n = J(s−,n)h−

− τ s−,n − s−,n−1

∆t .

By using the extended pressure condition discussed before one has

(p−,n − p+,n)(1− s+,n) = 0. (5.32)

Defining

g(s−,n, s+,n) = J(s−,n)h−

− J(s+,n)h+ − τ s

−,n − s+,n

∆t ,

and

Cn−1 := C(s−,n−1, s+,n−1) = τ

∆t (s+,n−1 − s−,n−1), (5.33)

(5.32) implies either s+,n = 1, or the pressure continuity

g(s−,n, s+,n)− C(s−,n−1, s+,n−1) = 0.


Obviously, ∂1g > 0 and ∂2g < 0. Further, given s− ∈ (0, 1), one has

lims+0

g(s−, s+) = −∞,

andg(s−, 1) = J(s−)

h−− J(1)

h+ + τ

∆t (1− s−) ≤ J(1)

h−− J(1)

h+ ,

since g(s−, ·) is strictly increasing and continuous. For any Cn−1 ∈ (−∞, g(s−, 1)], thereexists a unique s+ = s+(s−) such that

g(s−, s+(s−)) = Cn−1.

Also, note that g(s−, 1) is decreasing in s−,

lims−0

g(s−, 1) = +∞, g(1, 1) = J(1)h−− J(1)

h+ ,

and therefore g(·, 1) : (0, 1] → [g(1, 1),+∞) is one to one. Observe that if cn−1 >

g(s−, 1), discretized pressure becomes discontinuous at the interface x = 0. By (5.32),one obtains s+n = 1. In this way, we have actually constructed the curves in the(0, 1]× (0, 1] square:

if Cn−1 > g(1, 1), then Γ(Cn−1) =

(s−, s+)|s− ∈ (0, D(Cn−1)], g(s−, s+) = Cn−1

∪

(s−, 1)|s− ∈ (D(Cn−1), 1],

if Cn−1 ≤ g(1, 1), then Γ(Cn−1) =

(s−, s+)|s− ∈ (0, 1], g(s−, s+) = Cn−1,

(5.34)where D(·) is the inverse of g(·, 1).

Below we give a property of the discretized extended pressure condition:

Proposition 5.4.1. If τ∆t (s

+,n−1 − s−,n−1) > J(1)h− −

J(1)h+ , then τ

∆t (s+,n − s−,n) >

J(1)h− −

J(1)h+ .

Proof. If p−,n 6= p+,n, one has s+,n = 1, and obviously,

τ1− s−,n

∆t ≥ 0 > J(1)h−− J(1)

h+ .

If p−,n = p+,n, then we have

J(s−,n)h−

− J(s+,n)h+ + τ(s+,n − s−,n)

∆t = Cn−1 ≥ J(1)h−− J(1)

h+ . (5.35)

In this case, if s+,n ≥ s−,n, one has

τ(s+,n − s−,n)∆t ≥ 0 > J(1)

h−− J(1)

h+ ,


otherwise, s+,n < s−,n implies

J(s−,n)h−

− J(s+,n)h+ < 0.

Together with (5.35), we yield

τ(s+,n − s−,n)∆t >

J(1)h−− J(1)

h+ ,

which concludes the proof.

5.4.2 Fully implicit scheme

Here a nonlinear, implicit scheme is discussed as an alternative to the linear one. Nextto improved stability properties, for this scheme we can prove that s±,n, the saturationat the interface, remain between 0 and 1, a property that is not guaranteed for the linearscheme. To construct the scheme we first define the decreasing function β : R→ R by

β(s) =∫ s

0λ(z)J

′(z)dz,

and rewrite the flux in (5.31) as

F = qfw(s) + h∂β(s)∂x

− τh2λ(s) ∂∂x

(∂s∂t

).

As before, for i 6= 0 we write

sni−1/2 − sn−1i−1/2

∆t = −Fni − Fni−1

∆x , (5.36)

but now the upwind flux Fni becomes

Fni = qfw(sni−1/2)+ h±

∆x

(β(sni+1/2)− β(sni−1/2)

)−τ(h±)2

∆x∆t λ((sn−1i+1/2 + sn−1

i−1/2)/2)(

(sni+1/2 − sn−1i+1/2)− (sni−1/2 − s

n−1i−1/2)

).

As before, by h± we mean h− if i < 0, or h+ if i > 0. Further, if i = 0, the flux isdefined on each side of the interface as

F−,n = qfw(sn− 12) + 2h−

∆x

(β(s−,n)− β(sn−1/2)

)− 2τ(h−)2

∆x∆t λ(s−,n−1)(

(s−,n − s−,n−1)− (sn−1/2 − sn−1−1/2)

),


and

F+,n = qfw(s+,n) + 2h+

∆x

(β(sn+1/2)− β(s+,n)

)− 2τ(h+)2

∆x∆t λ(s+,n−1)(

(sn+1/2 − sn−1+1/2)− (s+,n − s+,n−1)

).

For having a conservative scheme, the two expressions should be equal. Combined withthe pressure condition (5.32), and viewing sn±1/2 as well as the saturation values sn−1

i+1/2and s±,n−1 as known, this provides a nonlinear system with s±,n as unknowns. Below weshow that this system has a unique solution pair in the square [0, 1]2.

The condition F−,n = F+,n can be written as

R(s−,n, s+,n) = B(sn−1/2, sn+1/2), (5.37)

where

R(s−,n, s+,n) = qfw(s+,n)− 2∆x

(h+β(s+,n) + h−β(s−,n)

)+ 2τ

∆x∆t

((h+)2λ(s+,n−1)s+,n + (h−)2λ(s−,n−1)s−,n

),

(5.38)

B = qfw(sn−1/2)− 2∆x

(h−β(sn−1/2) + h+β(sn+1/2)

)+ 2τ

∆x∆t

((h−)2λ(s−,n−1)(s−,n−1 + (sn−1/2 − s

n−1−1/2))

+(h+)2λ(s+,n−1)(s+,n−1 + (sn+1/2 − sn−1+1/2))

).

Using (5.36), B becomes

B = qfw(sn−1/2)− 2∆x

(h−β(sn−1/2) + h+β(sn+1/2)

)+ 2τ

∆x∆t

((h−)2λ(s−,n−1)

(s−,n−1 − ∆t

∆x (F−,n − Fn−1))

+(h+)2λ(s+,n−1)(s+,n−1 − ∆t

∆x (Fn1 − F+,n)) ).

(5.39)

Obviously, R is increasing in both arguments and one has

0 = R(0, 0)≤ R(s−, s+)

≤ R(1, 1) = q − 2∆x (h+ + h−)β(1) + 2τ

∆x∆t [(h+)2λ(s+,n−1) + (h−)2λ(s−,n−1)].

Note that, if both s−,n−1, s+,n−1 take the values 0 or 1, the last terms in (5.39) vanish,giving

B =qfw(sn−1/2)− 2∆x

(h−β(sn−1/2) + h+β(sn+1/2)

).

Since β is decreasing, in this case one has 0 ≤ B ≤ R(1, 1). Further, if s−,n−1 is not


equal to 0 or to 1, with ∆t small enough one gets

0 ≤ s−,n−1 − ∆t∆x (F−,n − Fn−1) ≤ 1,

and analogously for s+,n−1. From (5.39), this shows again that 0 ≤ B ≤ R(1, 1). Thisgives the following:

Lemma 5.4.1. For a sufficiently small time step ∆t, with Cn−1 = C(s−,n−1, s+,n−1)defined in (5.33) and for R and B in (5.37) and (5.39), the system

R(s−,n, s+,n) = B,

(g(s−,n, s+,n)− Cn−1)(1− s+,n) = 0,(5.40)

has a unique solution pair (s−,n, s+,n) ∈ [0, 1]2.

Proof. The set Γ(Cn−1) introduced in (5.34) contains pairs satisfying the pressure con-dition (5.32). In this set, we seek a pair (s−,n0 , s+,n

0 ) such that

R(s−,n0 , s+,n0 ) = B(s−,n0 , s+,n

0 ).

If such a pair exists, it solves the system (5.40). Since g is increasing in the first argumentand decreasing in the second one, long as both s−,n and s+,n are below 1, the curveΓ(Cn−1) is a graph of a general non-decreasing function (see Figure 5.9). Similarly, forB ∈ [0, R(1, 1)], the set R(·, ·) = B is the graph of a decreasing function in the s−,ns+,n-plane. Also, observe that both curves are continuous. Therefore, these two curves haveat most one intersection point inside the square [0, 1]2, implying that (5.40) has a uniquesolution pair.

Note that if the solution pair provided above lies inside [0, 1)2, then one has pressurecontinuity across the interface, p−,n = p+,n. Otherwise, the solution pair lies on theboundary of the square [0, 1]2. Moreover, assuming that initially one has s−,0 ≤ s+,0,implying that at the interface separating the two blocks more oil is present at the coarsematerial side then at the fine material side, repeating the proof of Proposition 5.4.1 oneobtains that s−,n ≤ s+,n for all n. This means that pressure discontinuity at the interfacecan only occur if no oil is present at the fine material side, s+,n = 1, which is preciselythe discrete pressure condition in (5.32).


Figure 5.9 The curves g(·, ·) = C and R(·, ·) = B.

Remark 5.4.1. The construction above assumes that sn−1/2, sn1/2 are known. In fact,these are part of the solution computed implicitly, at time step tn. This means that,actually, sn−1/2, sn1/2 and consequently B depend on s−,n, s+,n. However, decoupling thecalculation of the solution pair s±,n from the effective time stepping suggests an iterativeprocedure for the implicit scheme: using, say, the values s±,n−1 as starting point, computesni+1/2 by solving (5.36) away from the interface, and use sn−1/2, sn1/2 to update s±,n.

5.4.3 Numerical results

In this section, we give some numerical results obtained with the semi-implicit methodin Subsection 5.4.1, which is much easier to implement. Here we used the followingfunctions and parameters:

krw(s) = s2, krn(s) = (1− s)2, J(s) = s−1, Nc = 1, M = 1, h− = 1, h+ = 0.5.

−1 −0.5 0 0.5 10

0.2

0.4

0.6

0.8

1

x

Oil

sat

urat

ion

so−

so+

soentry

Figure 5.10 Initial oil saturation s0o.


The tests are done in the interval (−1, 1). Further, we present the results in termsof the oil/nonwettig phase saturation so = 1− s, as this is the phase for which trappingmay occur. The initial oil saturation is hat shaped (see Figure 5.10)

so(x, 0) = s0o :=

0, −1 < x < −0.34,0.9, −0.34 ≤ x ≤ −0.12,0, −0.12 < x < 1.

At x = ±1 we take homogeneous boundary conditions, ∂xs(±1, t) = 0. This mimics thesituations when an oil blob in the coarse layer is displaced by water. After a certain time,the oil reaches the interface. Note that initially no oil is present in the fine medium.

−1 −0.5 0 0.5 10

0.2

0.4

0.6

0.8

1

x

Oil

sat

urat

ion

so−

so+

soentry

−1 −0.5 0 0.5 1

0

0.05

0.1

0.15

x

Oil

flux

Fo −

Fo +

−1 −0.5 0 0.5 1

1.2

1.4

1.6

1.8

2

x

Pre

ssur

e

p −

p +

Figure 5.11 τ = 0, t = 0.7: oil saturation (left), oil flux (middle) and pressure difference (right).

Before discussing the results we recall that the saturation s∗ defined in (5.14) is thelimit saturation allowing for pressure continuity in the equilibrium models (τ = 0). Thisalso defines an entry saturation for the oil, sentry = 1−s∗. For the equilibrium model, oilflows into the fine material only if so > sentry at the coarse material side of the interface,and it remains trapped if so ≤ sentry. Figure 5.11, displaying the results obtained forτ = 0 at t = 0.7, confirm this statement. At the coarse side of the interface, one hass−o < sentry (picture on the left) and the oil flux is 0 there (picture in the middle). Thismeans that no oil enters into the fine medium. Further, the pressure is discontinuous atthe interface (picture on the right).

−1 −0.5 0 0.5 10

0.2

0.4

0.6

0.8

1

x

Oil

satu

ratio

n

so−

so+

soentry

−1 −0.5 0 0.5 10

0.05

0.1

0.15

0.2

0.25

0.3

x

Oil

flux

Fo −

Fo +

−1 −0.5 0 0.5 1

1.5

2

2.5

3

3.5

x

Pre

ssur

e

p − p +

Figure 5.12 τ = 1, t = 0.7: oil saturation (left), oil flux (middle) and pressure difference (right).

The case τ = 1, presented in Figure 5.12, shows a different situation. In the leftpicture, although s−o < sentry, one still has s+

o > 0, meaning that oil has already entered


in the fine medium. This is also confirmed by the middle picture, displaying a non-zerothe oil flux at the interface. Finally, the picture on the right shows that the pressure iscontinuous.

−1 −0.5 0 0.5 10

0.2

0.4

0.6

0.8

1

x

Oil

sat

urat

ion

so−

so+

soentry

−1 −0.5 0 0.5 1−0.01

−0.005

0

0.005

0.01

x

Oil

flux

Fo − F

o +

−1 −0.5 0 0.5 1

1.2

1.4

1.6

1.8

2

x

Pre

ssur

e

p −p +

Figure 5.13 τ = 0, t = 4: oil saturation (left), oil flux (middle) and pressure difference (right).

−1 −0.5 0 0.5 10

0.2

0.4

0.6

0.8

1

x

Oil

satu

ratio

n

so−

so+

soentry

−1 −0.5 0 0.5 1−0.01

0

0.01

0.02

0.03

0.04

0.05

0.06

x

Oil

flux

Fo −

Fo +

−1 −0.5 0 0.5 1

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

x

Pre

ssur

e

p − p +

Figure 5.14 τ = 1, t = 4: oil saturation (left), oil flux (middle) and pressure difference (right).

Figure 5.13, presents the results for τ = 0 and at t = 4. Then oil has flown into thefine medium (picture on the left). The flux and the pressure are both continuous (middleand right pictures). At the same time, but with τ = 1, we observe that more oil has flowninto the fine media (left picture of Figure 5.14). As expected, the flux and pressure arecontinuous as well.

−1 −0.5 0 0.5 10

0.2

0.4

0.6

0.8

1

x

Oil

sat

urat

ion

so−

so+

soentry

−1 −0.5 0 0.5 10

0.2

0.4

0.6

0.8

1

x

Oil

satu

ratio

n

so−

so+

soentry

−1 −0.5 0 0.5 10

0.2

0.4

0.6

0.8

1

x

Oil

satu

ratio

n

so−

so+

soentry

Figure 5.15 t = 4, oil saturation: τ = 0 (left), τ = 1 (middle) and τ = 10 (right).

The results above suggest that the amount of oil flowing into the fine material increaseswith τ . To understand this behaviour we compare the oil saturation obtained for τ = 0,τ = 1 and τ = 10, all at the same time t = 4. The profiles in Figure 5.15 show that, forτ = 0, little oil has flown into the fine media, and this amount is higher for τ = 1. In


both cases, s−o , the oil saturation at the coarse side of the interface, already exceeds theentry saturation sentry. However, for τ = 10, s−o < sentry, but oil has still flown into thefine material. On expects that the oil flow into the fine material will take longer for thelargest value of τ , in agreement with Corollary 5.3.1.

Finally, we observe that in the equilibrium case τ = 0 one can determine the maximalamount of oil that can be trapped at the interface, see [42]. Having this in mind, wechoose again a hat-shaped initial saturation

so(x, 0) = s0o :=

0, −1 < x < −0.34,0.4, −0.34 ≤ x ≤ −0.12,0, −0.12 < x < 1,

where the total amount of oil equals the maximal amount that can be trapped for equilib-rium models. With this initial data, we compute the numerical solutions for three valuesof τ , namely 0, 10 and 30. Figure 5.16 shows the results at t = 400, when practically allsolutions have reached a steady state and no oil flow is encountered anymore.

−1 −0.5 0 0.5 10

0.2

0.4

0.6

0.8

1

x

Oil

sat

urat

ion

so−

so+

soentry

−1 −0.5 0 0.5 10

0.2

0.4

0.6

0.8

1

x

Oil

satu

ratio

n

so−

so+

soentry

−1 −0.5 0 0.5 10

0.2

0.4

0.6

0.8

1

x

Oil

satu

ratio

n

so−

so+

soentry

Figure 5.16 t = 400, oil saturation: τ = 0 (left), τ = 10 (middle) and τ = 30 (right).

The left picture shows the result for τ = 0. In this case, the entire amount of oil istrapped in the coarse medium, and the oil saturation s−o is matches the entry saturationsentry. No oil has flown at all into the fine material. As following from the middle picture,for τ = 10, one has s−o < sentry, and the oil remaining trapped in the coarse material isless than in the equilibrium case. The situation becomes more obvious for the solutioncorresponding to τ = 30. The oil saturation s−o has decayed further, and the trapped oilis less than in the previous cases. This is again in agreement with the analysis in Section5.3.2.

5.5 ConclusionsWe have considered a non-equilibrium model for two-phase flow in heterogeneous porousmedia, where dynamic effects are included in the phase pressure difference. A simplesituation is considered, where the medium consists of two adjacent homogeneous blocks.We obtain the conditions coupling the models in each of the two sub-domains. The first

5.5. CONCLUSIONS 109

condition is, as expected, flux continuity, whereas the second is an extended pressurecondition extending the results in [46] for the standard two-phase flow model.

In the equilibrium case, if an entry pressure model is considered, oil can flow into thefine material only if its saturation exceeds an entry point. In the non-equilibrium caseinstead, the non-wetting phase may flow even if the oil saturation at the coarse side of theinterface is below the entry point, and amount of oil remaining trapped at the interfaceis less than in the case of equilibrium models.

Finally, two different numerical schemes are discussed, and different numerical exper-iments are presented to sustain the theoretical findings.

Chapter 6

Uniqueness of a solution for thehysteresis model

6.1 IntroductionIn this chapter, we consider the full non-equilibrium mathematical model for two-phaseflow in a porous medium. Two immiscible fluid phases (for example, water – the wettingphase, and oil – the non-wetting one) are flowing through a porous medium occupying abounded, connected domain Ω ⊂ Rd (d ≥ 1). Ω and ∂Ω denote the closure, respectivelyboundary of Ω. We let t ∈ (0, T ] be the time variable, where T > 0 is a given maximaltime. The phase pressures are denoted by pw, pn. The non-wetting phase saturation is s.We assume the porous medium is saturated by the two phases, so no other flowing phaseis present. This means, the wetting phase saturation is 1− s. Then from the Darcy’s lawand mass conservation for each fluid, one obtains (see [12,61])

−∂ts−∇ · (kw(s)∇pw)−∇ · (kw(s)−→g1) = 0, (6.1)

∂ts−∇ · (kn(s)∇pn)−∇ · (kn(s)−→g2) = 0. (6.2)

Here−→g1 ,−→g2 ∈ Rd are the gravity vectors in direction−−→ed = (0, ..., 0,−1) ∈ Rd. kw(s), kn(s)

are the relative permeabilities, two nonlinear functions depending on s. The system isclosed by the relation between the phase pressures and saturation. Standardly, one as-sumes that

pn − pw = pc(s),

pc being a given increasing function in s (see [69]). These are so-called equilibrium models.However, experiments [17,38,64] have invalidated such models. In particular, one shoulddistinguish between infiltration, when the wetting phase is displacing the non-wetting

This chapter is a collaborative work with I.S. Pop and it has been published in Computers & Math-ematics with Applications, 69 (2015): 688-695.

111

112 CHAPTER 6. UNIQUENESS OF A SOLUTION FOR THE HYSTERESIS MODEL

phase, and the opposite process, drainage. The switch between the two situations aboveis achieved in so-called hysteresis models. Further, experiments for which classical, equi-librium models would predict monotone infiltration saturation profiles, actually revealednon-monotone profiles (saturation overshoot) (see [40]). Such profiles are instead allowedin dynamic capillarity models. Both effects mentioned above are included in:

pn − pw ∈ pc(s) + γ(x)sign(∂ts) + τ∂ts. (6.3)

The second term on the right hand side models play-type hysteresis (see [13,103]), whilethe last one accounts for dynamic capillarity (see [60]). In (6.3), γ ≥ 0, τ > 0 are given,sign is the multi-valued graph:

sign(ξ) =

1 if ξ > 0,[−1, 1] if ξ = 0,−1 if ξ < 0.

(6.4)

Following [68], we also consider the Lipschitz continuous function Ψ : R× Ω→ R

Ψ(ξ, x) =

ξ−γ(x)τ for ξ > γ(x),0 for ξ ∈ [−γ(x), γ(x)],

ξ+γ(x)τ for ξ < −γ(x).

(6.5)

Clearly, for a. e. ξ ∈ R, one has

0 ≤ ∂ξΨ(ξ, x) ≤ 1/τ. (6.6)

With this, (6.3) rewrites∂ts = Ψ(pn − pw − pc(s), x). (6.7)

The model (6.1), (6.2), (6.7) is complemented by the initial and boundary conditions:

s(0, ·) = s0, (6.8)

pn = pw = 0 at ∂Ω, for all t ≥ 0. (6.9)

Remark 6.1.1. Other boundary conditions are possible (see Remark 6.2.1), but for clarity,we restrict the presentation to (6.9).

We make the following assumptions:

A1: The functions kw, kn : R → R are Lipschitz continuous. Further, δ,Mk > 0 exist,such that δ ≤ kn(s), kw(s) ≤Mk < +∞, for all s ∈ R.

A2: pc(·) ∈ C1(R) is increasing and Lipschitz continuous, there exist mp,Mp > 0, suchthat mp ≤ ‖pc‖Lip ≤Mp < +∞, for all s ∈ R.


A3: Ω is a C1,β domain with 0 < β ≤ 1.

A4: γ(x) ∈ C0,1(Ω).

A5: s0 ∈ C0,β(Ω).

Remark 6.1.2. Commonly, the permeabilities and capillary pressure model encounteredin the literature ( [22,24]) are

kn(s) = sp, kw(s) = (1− s)q, with p, q > 1,

andpc(s) = (1− s)−

1µ , µ > 1, for s ∈ [0, 1].

Then (A1) and (A2) are not satisfied when s approaches 0 or 1. We consider here aregularized approximation of these functions.

Below we use standard notation in the theory of partial differential equations. For anyh ∈ C0,β : Ω→ R, we define the norm

‖h‖C(Ω) := supx∈Ω|h(x)|,

the βth - Hölder semi-norm of h

[h]C0,β(Ω) := supx,y∈Ω,x 6=y

|h(x)− h(y)||x− y|β

,

and the βth - Hölder norm of h

‖h‖C0,β(Ω) := ‖h‖C(Ω) + [h]C0,β(Ω).

The existence of a solution for such models is studied in [68, 71, 96], for both one-phase and two-phase models. The analysis covers degenerate cases, i.e. when k(s) = 0,or pc(s) = ∞ for some s. For γ = 0, existence for such models is obtained in [52, 77].Uniqueness results are much fewer. In [52], this is obtained, but only for a scalar modelwith linear higher order term. Also for a scalar nonlinear model, but in a form that doesnot match such porous media flow models, uniqueness is obtained in [14]. Closest to thepresent work is [30], where uniqueness is obtained for a scalar, non-degenerate model.

Here we prove the uniqueness of a (weak) solution to the model (6.1), (6.2) and(6.7), describing two-phase porous media flow. The proof uses essential bounds for thepressures, which are obtained in the first part.

6.2 Uniqueness of the weak solutionIn this section, we provide a rigorous proof of the uniqueness of weak solutions to (6.1),(6.2), (6.7). We use common notations for function spaces, namely, L2, W 1,2, W 1,2

0 ,


and Bochner space L2(0, T ;X). We follow [68] and consider weak solutions solvingProblem P: Given s0 satisfying (A5), find pn ∈ L2(0, T ;W 1,2

0 (Ω)), pw ∈ L2(0, T ;W 1,20 (Ω))

and s ∈W 1,2(0, T ;L2(Ω)), such that s(·, 0) = s0 in Ω, and

(∂ts, φ) + (kn(s)∇pn,∇φ) + (kn(s)−→g2 ,∇φ) = 0, (6.10)

(−∂ts, ψ) + (kw(s)∇pw,∇ψ) + (kw(s)−→g1 ,∇ψ) = 0, (6.11)

(∂ts, ρ) = (Ψ(pn − pw − pc(s), x), ρ), (6.12)

for any φ, ψ ∈ L2(0, T ;W 1,20 (Ω)) and ρ ∈ L2(0, T ;L2(Ω)).

In [68], the hysteresis is modeled by considering (6.3) valid a.e. This immediatelyimplies that (6.7) holds a.e. and further (6.12). The weak solution introduced by ProblemP is, in fact, equivalent to the weak solution concept in [68]. Here to prove the uniquenessof the weak solution, some intermediate results are needed. We start with essential boundsfor the gradients of pn and pw.

Lemma 6.2.1. Let (pn, pw, s) be a weak solution to Problem Pe, then one has∇pn,∇pw ∈L∞((0, T ]× Ω).

Proof. First we show that ‖∇pn‖L2(Ω) ∈ L∞(0, T ) and ‖∇pw‖L2(Ω) ∈ L∞(0, T ). Takingφ = pn in (6.10), ψ = pw in (6.11) and adding the resulting equations give

(∂ts, pn − pw) + ‖√kn(s)∇pn‖2L2(Ω) + ‖

√kw(s)∇pw‖2L2(Ω)

+ (kn(s)−→g2 ,∇pn) + (kw(s)−→g1 ,∇pw) = 0. (6.13)

For the first term of (6.13), we note that (6.3) holds almost everywhere. Then sincesign(ξ)ξ ≥ 0 for any ξ ∈ R, one has∫

Ω∂ts(pn − pw) ≥

∫Ωτ |∂ts|2dx+

∫Ωpc(s)∂tsdx ≥

τ

2‖∂ts‖2L2(Ω) −

12τ

∫Ω|pc(s)|2dx.

(6.14)Further, since s ∈ L∞(0, T ;L2(Ω)) (see [68,71]), by using the Cauchy-Schwarz inequality,(A1) and (A2), (6.13) gives

‖∂ts‖2L2(Ω) + ‖∇pn‖2L2(Ω) + ‖∇pw‖2L2(Ω) ≤ C, for a.e. t ∈ (0, T ]. (6.15)

Then for almost every time t, we have the following elliptic equations with respect topressure pn, pw:

−∇ · (kn(s)∇pn) = −∂ts+∇ · (kn(s)−→g2), (6.16)

−∇ · (kw(s)∇pw) = ∂ts+∇ · (kw(s)−→g1). (6.17)

Then, since for almost every time t, we have pn, pw ∈ W 1,20 (Ω), kn(s)−→g2 , kw(s)−→g1 ∈


L∞(Ω), ∂ts ∈ L2(Ω), and (A3), by using Theorem 14.1 in [70] gives for almost every t,

‖pn‖C0,α(Ω) + ‖pw‖C0,α(Ω) ≤ C, (6.18)

here α is independent on t.

Furthermore, we use pn, pw, s to define

P = pn − pw − pc(s). (6.19)

Based on this, we also define χ:

χ(P1,P2,x) =Ψ(P1,x)−Ψ(P2,x)

P1−P2if P1 6= P2,

0, if P1 = P2.(6.20)

Obviously, one has 0 ≤ χ(P1,P2,x) ≤ 1τ .

Since pc ∈ C1, from (6.7), for almost every x, y ∈ Ω (x 6= y) and t > 0, a ζ dependingon x, y, t exists, such that

∂ts(t, x)− s(t, y)|x− y|α

=Ψ(P (x, t), x

)−Ψ

(P (t, y), x

)|x− y|α

+Ψ(P (t, y), x

)−Ψ

(P (t, y), y

)|x− y|α

=χ(P (x,t),P (y,t),x)

(P (t, x)− P (t, y)|x− y|α

)+ Ψ(P (t, y), x)−Ψ(P (t, y), y)

|x− y|α

=χ(P (x,t),P (y,t),x)

(pn(t, x)− pn(t, y)|x− y|α

− pw(t, x)− pw(t, y)|x− y|α

− p′

c(ζ) · s(t, x)− s(t, y)|x− y|α

)+ Γ(t, x, y), (6.21)

whereΓ(t, x, y) = Ψ(P (t, y), x)−Ψ(P (t, y), y)

|x− y|α. (6.22)

By (A2) - (A4), and since pn, pw ∈ C0,α(Ω) for almost every t, one gets

|Γ(t, x, y)|+ supx,y∈Ω,x 6=y

|pn(t, x)− pn(t, y)||x− y|α

+ supx,y∈Ω,x 6=y

|pw(t, x)− pw(t, y)||x− y|α

≤ C, (6.23)

for some C > 0 not depending on t, x, y.With w : (0, T ]× Ω2 → R as

w = s(t, x)− s(t, y)|x− y|α

, (6.24)

one has∂tw = fw + g, (6.25)


wheref(t, x) = −χ(P (x,t),P (y,t),x) · p

′

c(ζ),

and

g(t, x) = χ(P (x,t),P (y,t),x)(pn(t, x)− pn(t, y)

|x− y|α− pw(t, x)− pw(t, y)

|x− y|α) + Γ(t, x, y).

Note that (6.20), (6.23) and (A2) give f, g ∈ L∞((0, T ] × Ω). Multiplying (6.25) by wand integrating from 0 to t lead to

12w

2(t) =∫ t

0fw2(z)dz +

∫ t

0gw(z)dz + 1

2

(s0(x)− s0(y)|x− y|α

)2. (6.26)

Since f, g ∈ L∞((0, T ]× Ω) and s0 ∈ C0,α(Ω) from (A5), we have

w2(t) ≤ C(1 +∫ t

0w2dz), for every t. (6.27)

Using Gronwall’s inequality yields w ≤ C, implying that

|s(t, x)− s(t, y)||x− y|α

≤ C, for almost every x, y ∈ Ω, for every t. (6.28)

Let Ωc be the subset of Ω, where (6.28) holds everywhere. Clearly, Ω\Ωc is zero measured.For any x ∈ Ω\Ωc, we consider a sequence xnn∈N ∈ Ωc converging to x, and define

s(t, x) = limn→+∞

s(t, xn). (6.29)

In the view of (6.28), s(t, x) does not depend on the choice of xnn∈N. With this choice,s ∈ C0,α(Ω) (see [48]).

Finally, by Theorem 8.33 and Corollary 8.35 in [59] (see also [35]), and recalling theuniform time estimates in (6.18), we get for almost every t:

‖pn‖C1,α(Ω) ≤ C(‖pn‖C(Ω) + ‖Ψ‖C(Ω) + ‖kn(s)‖C0,α(Ω)), (6.30)

‖pw‖C1,α(Ω) ≤ C(‖pw‖C(Ω) + ‖Ψ‖C(Ω) + ‖kw(s)‖C0,α(Ω)), (6.31)

here C is independent on t, then, the above estimates imply ∇pn,∇pw ∈ L∞((0, T ] ×Ω).

Remark 6.2.1. Actually, the uniqueness proof only requires that pn, pw have L∞(Ω ×(0, T ]) gradients. Such results can be obtained with other boundary conditions thanhomogeneous Dirichlet. We refer to Theorem 16.1 and Theorem 16.2 in [70] for similarresults, in the case of homogeneous Neumann boundary conditions.

Here we show the solution to Problem P is unique. To do so, we use the weak solution


pair (Ga−b, Ga−b) to the elliptic system (see [28,30,52]):

−∇ · (kn(b)∇Ga−b) + χ(a,b,x)(Ga−b +Ga−b) = χ(a,b,x)(a− b), (6.32)

−∇ · (kw(b)∇Ga−b) + χ(a,b,x)(Ga−b + Ga−b) = χ(a,b,x)(a− b), (6.33)

Ga−b = Ga−b = 0, at ∂Ω, (6.34)

with a, b ∈ L2(Ω).The existence and uniqueness follow the Lax-Milgram Theorem. Using the properties

of χ, kw, kn, one immediately gets

‖Ga−b‖2W 1,2(Ω) ≤ C‖a− b‖2L2(Ω), and ‖Ga−b‖2W 1,2(Ω) ≤ C‖a− b‖

2L2(Ω). (6.35)

These functions are introduced because the uniqueness proof below is based on contradic-tion. Specially, assuming the existence of two solutions, we estimate their difference. Indoing so, we need to deal with difference as in the derivatives of nonlinearities applied tothese solutions, or when these appear as multiplications. The functions introduced abovewill be used in these estimates.

Theorem 6.2.1. Problem P has at most one solution.

Proof. Let (u, pun, puw) and (v, pvn, pvw) be the two solutions of Problem P, and since Ψ isLipschitz, for almost every (x, t) ∈ ΩT , then one has

(∂t(u− v), φ) + (kn(v)∇(pun − pvn),∇φ)+((kn(u)− kn(v))∇pun,∇φ) + ((kn(u)− kn(v))−→g2 ,∇φ) = 0, (6.36)

− (∂t(u− v), ψ) + (kw(v)∇(puw − pvw),∇ψ)+ ((kw(u)− kw(v))∇puw,∇ψ) + ((kw(u)− kw(v))−→g1 ,∇ψ) = 0, (6.37)

and

(∂t(u− v), ρ) =(χ(u,v,x)

((pun − pvn)− (puw − pvw)− (pc(u)− pc(v))

), ρ), (6.38)

for any φ, ψ ∈ L2(0, T ;W 1,20 (Ω)), ρ ∈ L2(0, T ;L2(Ω)).

According to the functions defined in (6.32) - (6.33), and the estimates in (6.35), weknow that Gu−v, Gu−v ∈W 1,2

0 (Ω) and

(χ(u,v,x)Gu−v, λ) + (χ(u,v,x)Gu−v, λ) + (kn(v)∇Gu−v,∇λ) = (χ(u,v,x)(u− v), λ),(6.39)

(χ(u,v,x)Gu−v, λ) + (χ(u,v,x)Gu−v, λ) + (kw(v)∇Gu−v,∇λ) = (χ(u,v,x)(u− v), λ),(6.40)

for any λ, λ ∈W 1,20 (Ω).


Testing with φ = Gu−v in (6.36), and ψ = Gu−v in (6.37), one has

(∂t(u− v), Gu−v) + (kn(v)∇(pun − pvn),∇Gu−v)+((kn(u)− kn(v))∇pun,∇Gu−v) + ((kn(u)− kn(v))−→g2 ,∇Gu−v) = 0, (6.41)

(∂t(u− v), Gu−v)− (kw(v)∇(puw − pvw),∇Gu−v)−((kw(u)− kw(v))∇puw,∇Gu−v)− ((kw(u)− kw(v))−→g1 ,∇Gu−v) = 0. (6.42)

Choosing λ = pun − pvn in (6.39) and λ = puw − pvw in (6.40) gives

(kn(v)∇Gu−v,∇(pun − pvn))=(χ(u,v,x)(u− v), pun − pvn)− (χ(u,v,x)Gu−v, p

un − pvn)− (χ(u,v,x)Gu−v, p

un − pvn),

(6.43)

(kw(v)∇Gu−v,∇(puw − pvw))=(χ(u,v,x)(u− v), puw − pvw)− (χ(u,v,x)Gu−v, p

uw − pvw)− (χ(u,v,x)Gu−v, p

uw − pvw).

(6.44)

Substitute (6.43) into (6.41) and (6.44) into (6.42) to replace the terms (kn(v)∇Gu−v,∇(pun − pvn)) and (kw(v)∇Gu−v,∇(puw − pvw)), we find that

(∂t(u− v), Gu−v)− (χ(u,v,x)Gu−v, pun − pvn)

−(χ(u,v,x)Gu−v, pun − pvn) + (χ(u,v,x)(u− v), pun − pvn)

+((kn(u)− kn(v))∇pun,∇Gu−v) + ((kn(u)− kn(v))−→g2 ,∇Gu−v) = 0, (6.45)

(∂t(u− v), Gu−v) + (χ(u,v,x)Gu−v, puw − pvw)

+(χ(u,v,x)Gu−v, puw − pvw)− (χ(u,v,x)(u− v), puw − pvw)

−((kw(u)− kw(v))∇puw,∇Gu−v)− ((kw(u)− kw(v))−→g1 ,∇Gu−v) = 0. (6.46)

Taking ρ = u− v into (6.38) yields

− (χ(u,v,x)(pun − pvn), u− v) + (χ(u,v,x)(puw − pvw), u− v)=− (∂t(u− v), u− v)− (χ(u,v,x)(pc(u)− pc(v)), u− v). (6.47)

Adding (6.45), (6.46), and (6.47), we eliminate (χ(u,v,x)(pun−pvn), u−v)−(χ(u,v,x)(puw−


pvw), u− v) and have the following result:

(∂t(u− v), Gu−v) + (χ(u,v,x)Gu−v, puw − pvw)− (χ(u,v,x)Gu−v, p

un − pvn)

+(∂t(u− v), Gu−v) + (χ(u,v,x)Gu−v, puw − pvw)− (χ(u,v,x)Gu−v, p

un − pvn)

+(∂t(u− v), u− v) + (χ(u,v,x)(pc(u)− pc(v)), u− v) + ((kn(u)− kn(v))∇puw,∇Gu−v)−((kw(u)− kw(v))∇pun,∇Gu−v) + ((kn(u)− kn(v))−→g2 ,∇Gu−v)−((kw(u)− kw(v))−→g1 ,∇Gu−v) = 0. (6.48)

Further, taking ρ = Gu−v and ρ = Gu−v in (6.38), respectively, one has


un − pvn)

=− (χ(u,v,x)Gu−v, pc(u)− pc(v)), (6.49)

and


un − pvn)

=− (χ(u,v,x)Gu−v, pc(u)− pc(v)). (6.50)

Substituting (6.49) and (6.50) into (6.48) leads to

(∂t(u− v), u− v) + (χ(u,v,x)(u− v), pc(u)− pc(v))−(χ(u,v,x)Gu−v, pc(u)− pc(v))− (χ(u,v,x)Gu−v, pc(u)− pc(v))+((kn(u)− kn(v))∇pun,∇Gu−v)− ((kw(u)− kw(v))∇puw,∇Gu−v)+((kn(u)− kn(v))−→g2 ,∇Gu−v)− ((kw(u)− kw(v))−→g1 ,∇Gu−v) = 0.

For any t ∈ (0, T ], integrating (6.51) from 0 to t gives us∫ t

0(∂t(u− v), u− v) +

∫ t

0(χ(u,v,x)(u− v), pc(u)− pc(v))

−∫ t

0(χ(u,v,x)Gu−v, pc(u)− pc(v))−

∫ t

0(χ(u,v,x)Gu−v, pc(u)− pc(v))

+∫ t

0((kn(u)− kn(v))∇pun,∇Gu−v)−

∫ t

0((kw(u)− kw(v))∇puw,∇Gu−v)

+∫ t

0((kn(u)− kn(v))−→g2 ,∇Gu−v)−

∫ t

0((kw(u)− kw(v))−→g1 ,∇Gu−v) = 0. (6.51)

We proceed by estimating each term called T1, T2, T3, T4, T5, T6, T7, T8. For T1, sinceu(0) = v(0) = s0, one has∫ t

0

∫Ω∂t(u− v)(u− v)dxdz = 1

2‖(u− v)(t)‖2L2(Ω). (6.52)


For the terms T2, T3, T4, according to (6.6), (A2) and (6.35), we obtain∫ t

0(χ(u,v,x)(u− v), pc(u)− pc(v)) ≥ 0, (6.53)

|∫ t

0(χ(u,v,x)Gu−v, pc(u)− pc(v))| ≤ C

∫ t

0‖(u− v)(·, z)‖2L2(Ω)dz, (6.54)

|∫ t

0(χ(u,v,x)Gu−v, pc(u)− pc(v))| ≤ C

∫ t

0‖(u− v)(·, z)‖2L2(Ω)dz. (6.55)

Furthermore, for T5 and T6, since ∇pn,∇pw ∈ L∞((0, T ]× Ω), then by using Cauchy -Schwarz inequality, one has the following estimates:

|∫ t

0((kn(u)− kn(v))∇pun,∇Gu−v)| ≤ C

∫ t

0‖(u− v)(·, z)‖2L2(Ω)dz, (6.56)

|∫ t

0((kw(u)− kw(v))∇puw,∇Gu−v)| ≤ C

∫ t

0‖(u− v)(·, z)‖2L2(Ω)dz. (6.57)

Similarly, one also has the estimates for T7 and T8:

|∫ t

0((kn(u)− kn(v))−→g2 ,∇Gu−v)| ≤ C

∫ t

0‖(u− v)(·, z)‖2L2(Ω)dz, (6.58)

|∫ t

0((kw(u)− kw(v))−→g1 ,∇Gu−v)| ≤ C

∫ t

0‖(u− v)(·, z)‖2L2(Ω)dz. (6.59)

Finally, summarizing the above leads to

‖(u− v)(·, t)‖2L2(Ω) ≤ C∫ t

0‖(u− v)(·, z)‖2L2(Ω)dz. (6.60)

By Gronwall’s inequality, we have ‖(u− v)(·, t)‖2L2(Ω) = 0. Since t is arbitrary, this givesu = v a.e. in Ω and for all t ∈ (0, T ].

To show that pun = pvn, puw = pvw, we use (6.36) and (6.37). Since u = v, one has

(kn(u)∇(pun − pvn),∇φ) = 0, (6.61)

(kw(u)∇(puw − pvw),∇ψ) = 0, (6.62)

for any φ, ψ ∈W 1,20 (Ω), for almost every t. The rest of the proof follows straightforwardly

by taking φ = pun − pvn, ψ = puw − pvw, and recalling that pun, pvn, puw, pvw have equal traceson ∂Ω.

6.3. CONCLUSIONS 121

6.3 ConclusionsIn this chapter, we have proved the uniqueness of weak solutions to a non-degeneratesystem which models two-phase flow in porous media including hysteresis and dynamiceffects in the capillary pressure. In doing so, we use arguments based on Green’s function.

Chapter 7

Moisture transport in concrete

7.1 Mathematical model

In this chapter, the moisture transport (one phase flow) in concrete is described based onthe Darcy’s law and mass balance

v = − kµkr∇pc (Darcy), (7.1)

where v is the volumetric velocity (m/s), k the intrinsic permeability of the concrete(m2), kr the relative permeability (−), µ the viscosity (kg ·m−1 · s−2).

∂

∂t(φs) +∇ · v = 0 (mass balance), (7.2)

where φ is the porosity of the concrete (−), s the water saturation (−), and t the time(s). Commonly, kr and pc are monotone functions of the saturation s, see [86].

By substituting (7.1) into (7.2), we have

∂

∂t(φs)−∇ · ( k

µkr∇pc) = 0. (7.3)

Furthermore, the following is assumed in this chapter:

• The sample is homogeneous.

• There is no flow at the boundaries except the boundary at the right.

• The initial saturation is constant.

This chapter is a collaborative work with A. Taher, I. Pop, A.J.J van der Zanden, H.J.H Brouwersand it has been published in Chemistry and Materials Research, 5(2013): 86-90.

123

124 CHAPTER 7. MOISTURE TRANSPORT IN CONCRETE

For the boundary at the right, a periodic repetition of wetting and drying cycles is assumed.In view of these assumptions, the lateral flow is 0. This allows reducing the model to onedimension. The sample occupies then the interval.

The following conditions are imposed

s = si, at t = 0, in (0, L), (7.4)

k

µkr∂pc∂x

= 0, x = 0, t > 0, (7.5)

s = sb(t), x = L, t > 0, (7.6)

where si is the initial saturation, sb(t) the saturation at the right boundary. In the contextof a one-dimensional model, (7.3) becomes

∂

∂t(φs)− ∂

∂x

(k

µkr∂pc∂x

)= 0. (7.7)

To make the equation dimensionless, the following notations are introduced

x := x

L, t := t

Tr, pc := pc

P, (7.8)

where L, Tr, P are characteristic values for the length, time, and capillary pressure.Substitute (7.8) into (7.7) and set

TrL2 = φµ

kP, (7.9)

then (7.7) becomes∂s

∂t= ∂

∂x

(D(s) ∂s

∂x

). (7.10)

HereD(s) = kr(s) ·

∂pc(s)∂s

. (7.11)

By using the Kirchhoff transformation

β(s) =∫ s

0D(z)dz, (7.12)

(7.10) becomes∂s

∂t= ∂xxβ(s), (7.13)

while the initial and boundary conditions are

s(x, 0) = si,∂s

∂x(0, t) = 0, s(1, t) = sb(t), (7.14)

7.2. APPROACH 125

where si is a constant number between 0 (in case of dried concrete) and 1 (in case of fullsaturated concrete). sb(t) is a periodic function, which simulates wetting/drying cycles

sb(t) =

1 t ∈ (0, Tw] + k · Tp,0 t ∈ (Tw, Td] + k · Tp,

(7.15)

where Td is the dimensionless drying time, Tw is the dimensionless wetting time, Tp =Td + Tw, is the period of one cycle, and k is any natural numbers.

7.2 Approach

Two modeling approaches for the moisture transport in concrete are compared in thischapter. The first model uses the same diffusion coefficient for both wetting and dryingphase. The second model uses two diffusion coefficients, one for the drying D−(s) andone for the wetting D+(s).

In the literature [74], these following coefficients are determined experimentally

D+(s) = 10−10 · e6s, (7.16)

D−(s) = 10−10 ·(

0.025 + 0.9751 + ( 1−s

0.208 )6

), (7.17)

which are plotted in Figure 7.1.

7.2.1 Standard model: one diffusion coefficient

A standard model considers the same diffusion coefficient for both wetting and drying. Inview of (7.16) and (7.17), the following averages are considered

Da(s) = D+(s) +D−(s)2 , (7.18)

Dg(s) =√D+(s) ·D−(s). (7.19)

Here Da(s) is the arithmetic average of the diffusion coefficients, Dg(s) the geometricaverage of the diffusion coefficients.


Figure 7.1 Various diffusion coefficients as a function of saturation.

The coefficients introduced in (7.18) and (7.19) are then used for the Kirchhoff trans-formation in (7.10). For the resulting model, (7.13), an implicit numerical scheme [88] iscombined with a linear iterative procedure [90].

7.2.2 Hysteretic model

Here the hysteretic model in [13] is adapted

∂s

∂t= ∂xxp. (7.20)

In the above, p has to include the switch between two diffusion coefficients, in this case,p reads

p ∈ pc(s) + γ(s)sign(∂ts), (7.21)

where pc(s), γ(s) and sign(∂ts) are defined as

pc(s) = β+(s) + β−(s)2 , (7.22)

γ(s) = β+(s)− β−(s)2 , (7.23)

and

sign(∂ts) =

1 ∂ts > 0,−1 ∂ts < 0.

(7.24)

β+/−(s) are the Kirchhoff transformations for the wetting/drying diffusion coefficients

β+/−(s) =∫ s

0D+/−(z)dz. (7.25)

7.2. APPROACH 127

When solving the system, sign(∂ts) is replaced by the regularization signδ(∂ts) (see Figure7.2)

signδ(∂ts)

δ∂ts+ δ2 − 1 if ∂ts < −δ,

∂tsδ if − δ < ∂ts < δ,

δ∂ts+ δ2 + 1 if ∂ts > δ,

(7.26)

where 0 < δ << 1 is the regularization parameter.

−2 −1 0 1 2−1

−0.5

0

0.5

1

z

sign

(a)

−2 −1 0 1 2−1.5

−1

−0.5

0

0.5

1

1.5

z

sign

δ

(b)

Figure 7.2 The graphs of (a.) sign(z) and (b.) signδ(z).

Solving the above equations directly using finite difference method leads to incorrectresults. Alternatively, the inverse φδ(r) of the regularized signδ(∂ts) is used as consideredin [97]

φδ(r) :=

−δ + r+1

δ if r < −δ,δr if − δ < r < δ,

δ + r−1δ if r > δ.

(7.27)

For the time discretization, let ∆t be the time step, and pm ≈ p(m · ∆t), sm ≈s(m ·∆t). The implicit discretization of (7.20) and (7.21) are:

sm+1 − sm

∆t = ∂xxpm+1, (7.28)

sm+1 − sm

∆t = φδ

(pm+1 − pc(sm+1)γ(sm)

), (7.29)

where m = 0, 1, ..., s0 = si. Further, at x = 0, ∂xpm+1 = 0 and at x = L, sm+1 =sb((m+ 1)∆t).

In the equations above, the right side of (7.29) is implicit for sm+1, due to thenonlinear function of pc(s). Therefore, pc(sm+1) is simplified for a linear function asinvestigated in [92] by a Taylor expansion given by

pc(sm+1) ≈ pc(sm) + p′

c(sm) · (sm+1 − sm). (7.30)


Then the discretized form becomes

sm+1 − sm

∆t = ∂xxpm+1, (7.31)

sm+1 − sm

∆t = φδ

(pm+1 − pc(sm)− p′c(sm) · (sm+1 − sm)γ(sm)

). (7.32)

For discretizing the spatial derivatives, finite difference method is applied.

7.2.3 Numerical results

The model is implemented in Matlab. Here, only one cycle is computed, which is oneday for wetting and six days for drying. The initial condition (si) in this case is assumedto be 0.5 saturation. Further, the following is used in the numerical scheme: δ = 10−5,∆x = 10−3 and ∆t = 10−6. The comparison between the results of the standard modelwith one diffusion coefficient and the results of the hysteretic model with two diffusioncoefficients are shown in Figure 7.3.

Figure 7.3 Comparison of the models.

The results show that there is a large difference between the three models. To validatethe models, experimental work is needed, which will be explained in the next section.

7.3 ExperimentMortar specimens with a water cement ratio of 0.5 and cement type of CEM I 42.5 N areused in this chapter to validate the models. Mortar specimens are prepared by castingthem in PVC tubes with a diameter of 100 mm. After one day, the mortar is demoulded

7.3. EXPERIMENT 129

and cured for 28 days. The side of the specimens is sealed with epoxy to ensure a onedimensional flux. At one of the two open surfaces of the specimen, the condition ischanged to simulate cycles. Wetting is simulated by contacting the surface with waterduring one day and drying by blowing dry air with a certain flow at the surface during6 days. After one drying/wetting cycle (one week), the moisture content is determinedby weighing the specimen. This mass is compared with the calculated mass from themoisture profiles of the models for validation. Figure 7.4 shows the experimental setup.

Figure 7.4 Sealed mortar specimens encountered with wetting cycle for one day and dryingcycle for six days.

Summary

Mathematical and Numerical Analysis for Non-EquilibriumTwo Phase Flow Models in Porous MediaThis thesis deals with models for the two phase flow in porous media under non-equilibriumconditions. Porous media models are derived by the mass balance and Darcy’s law, andcompleted by the relationship between phase saturation and pressures. For the standardcase, these constitutive equations are obtained under equilibrium assumptions: e.g. fora given medium and at a given phase saturation, the capillary pressure is fixed and theso-called capillary pressure is a monotone function of saturation.

This obtained equilibrium approach is invalidated by many experimental results. Forexample, non-monotonic saturation profiles are observed for infiltration processes in ahomogeneous medium whereas standard models rule out such profiles. In such a case,non-equilibrium models which include hysteretic effects are proposed. There are at leasttwo kinds of hysteresis: the play-type hysteresis and the one including dynamic effects.

Without play-type hysteresis and under the assumption of a known total flow, the originalmodel has been transformed into a scalar model — a pseudo parabolic equation. In sucha case, the existence of weak solutions for this model is a known result. However, theuniqueness was still an open question. This is answered positively in Chapter 2 of this work.

Next, for the original two phase flow model including dynamic effects, we investigate anelliptic-parabolic system. We do this even in the degenerate case, when nonlinearitiesmultiplying the higher order term in the model may vanish, depending on the unknownsof the model. We first deal with the regularized model when all nonlinearities are assumedto be bounded away from 0 by a factor δ. Finally, the existence of weak solutions in de-generate case is obtained when letting the regularization parameter go to zero.

In Chapter 4, we propose a multipoint flux approximation finite volume scheme to approx-imate the solution of the models. By giving energy estimates and compactness arguments,the convergence of the numerical approximation to the weak solution has been proved, as

131

132 SUMMARY

the mesh size and time step tend to zero. Finally, we present some numerical results toconfirm the convergence proved rigorously. Moreover, we calculate the numerical errorsfor saturation and gradient of pressures, which show that they all have a first order con-vergence, even in the anisotropic cases and for irregular meshes. At the end, we presentsome numerical experiments to simulate the saturation overshoot, which has been ob-served both experimentally and analytically.

In Chapter 5, we investigate the two phase flow in heterogeneous media, where dynamiceffects are taken into account. The two phase flow model reduces to one equation in thecase of one spatial dimension. We consider a domain which combines two homogeneoussubdomains. After doing so, the behavior of the solution compared to the one for equi-librium models is discussed and illustrated numerically.

For the two-phase flow model including dynamic effects and play-type hysteresis, the ex-istence of weak solutions for such non-equilibrium models under non-degenerate cases isalso a known result, but the uniqueness was again an open question. In Chapter 6, wegive the proof for the uniqueness. The difficulty is the strong coupling of the two phaseflow differential equations and the play-type hysteresis term. We prove the uniqueness ofa weak solution by means of an auxiliary dual system.

In the last chapter of this thesis, we consider a practical situation related to the durabilityanalysis of concrete. Here we investigate two different diffusion coefficients which shouldbe considered while drainage, respectively, imbibition. Finally, we propose a model includ-ing play-type hysteresis in diffusion coefficients and give some numerical calculations forthis model.

Curriculum Vitae

Xiulei Cao was born on April 02, 1985 in Jilin, China. After finishing his schooling fromQianan 7th School, he started his Bachelor studies in Computational Mathematics inJilin University, Jilin, China. After his bachelor studies, he pursued his Master degreeof Computational Mathematics in Jilin University. During his Master studies, he wasawarded first prize for two times.

From October 2011, he started a PhD project at the Center for Analysis, Scientific comput-ing and Applications (CASA) of the Department of Mathematics and Computer Science,Eindhoven University of Technology, the Netherlands under the supervision of Prof. I. S.Pop and Prof. C. J. van Duijn. The results of this PhD research are presented in thisthesis.

133

134 CURRICULUM VITAE

List of Publications

Journal articles:1. A. Taher, X. Cao, I.S. Pop, A.J.J. van der Zanden and H.J.H. Brouwers. Mois-

ture transport in concrete during wetting/drying cycles. Chemistry and MaterialsResearch. 5:86-90, 2013.

2. X. Cao and I.S. Pop. Uniqueness of weak solutions for a pseudo-parabolic equationmodeling two phase flow in porous media. Appl. Math. Lett., 46:25-30, 2015.

3. X. Cao and I.S. Pop. Two-phase porous media flows with dynamic capillary effectsand hysteresis: uniqueness of weak solutions. Comput. Math. Appl., 69(7):688-695, 2015.

4. C.J. van Duijn, X. Cao and I.S. Pop. Two-phase flow in porous media: dynamiccapillarity and heterogeneous media. Transp. Porous Med., 110(3):1-26, 2015.

5. X. Cao and I.S. Pop. Degenerate two-phase porous media flow model with dynamiccapillarity. J. Differ. Equ., 260(3):2418-2456, 2016.

Preprints/Work in process:1. X. Cao, S.F. Nemadjieu and I.S. Pop. A multipoint flux approximation finite volume

scheme for two phase porous media flow with dynamic capillarity. CASA Report15-33, Eindhoven University of Technology, preprint, 2015.

2. X. Cao, M. Kimura and I.S. Pop. Finite element approximation of non-equilibriumflow in porous media. In preparation.

135

136 LIST OF PUBLICATIONS

Acknowledgments

First of all, I would like to thank my supervisor Prof. Iuliu Sorin Pop who is also myfirst promoter for his constant patience, advice and trust for the last four years. His en-couragement has given me self-confidence. Without him, this thesis would not have beenpossible. I would like to express my gratitude to him for inspiring me and guiding metirelessly for the entire period of my PhD. I would also like to thank my second promoterProf. Hans van Duijn for sharing his invaluable knowledge when we were working on apaper.

I would like to take this opportunity to thank Prof. Harald van Brummelen, Prof. RainerHelmig, Prof. Barry Koren, Prof. Ben Schweizer and Prof. Barbara Wohlmuth for beingin the committee of my defense ceremony. I feel honored that you have agreed to refereethis thesis.

I am thankful to Prof. Majid Hassanizadeh and Prof. Rainer Helmig for many nice sug-gestions and for giving me scientific motivation. The NUPUS environment was inspiring.I would also like to thank Prof. Mark Peletier, who has taught me how to improve mypresentation skills. I want to thank Prof. Masato Kimura for giving me the opportunityto visit Kanazawa University and his supervision there. Thanks are also to Prof. MichalBeneš who read the thesis and gave helpful comments during my stay in Japan. I wouldlike to thank Prof. Florin Adrian Radu and Prof. Kundan Kumar for many discussionsand useful suggestions and for their help in broadening my network. Thanks are to mycollaborators: Azee, Ton van der Zanden, Prof. Jos Brouwers, Prof. Hans van Duijn andSimplice.

I want to thank all the CASA members, who provided a nice working atmosphere. Inparticular, many thanks to the secretaries, Enna and Marèse, who helped in all kinds ofadministrative work. I am thankful to my office mates Sangye, Ian, Sarah, René and myformer office mate Bogdan for keeping up the good and positive atmosphere in our office.Thanks are to Upanshu, Thomas and Sangye whom I have talked a lot with and learneda lot from. Koondanibha, it was a pleasure to travel with you to Spain for the conferenceand I had a wonderful time. Thanks are to Sarah, Deepesh, Nitin, Nikhil and Pranav for

137

138 ACKNOWLEDGMENTS

the nice party time. Behnaz and Prof. Mikko Karttunen, thank you for the help whenlooking for a postdoctoral position.

Thanks are to Liu Lei, Lu Shengnan, Yang Yusen and Wang Jiquan for being wonderfulroommates, for the fun time together. Hou Qingzhi, Li Guangliang, Ma Ming, Sangay,Sangye and Thomas, thank you for those lovely dinners.

I would like to thank my Chinese friends in the Netherlands: Wang Wenhan, Sui Huapeng,Zhang Shiqiang and Ran Shenghai. Fan Yabin, Zhuang Luwen, Yin Xiaoguang and QinChaozhong, I thank all of you that we share a lot of happy time when we have attendedthe NUPUS conferences. I would also like to thank all my friends in China: Gu Yuanye,Liu Jiaji, Cao Xianhui, Guo Hexin, Li Yonggang, Niu Chunhu, Sun Duo, Wu Yongdi andXu Jian for enriching my personal life.

Last but not the least, I express my gratitude to my parents, my brother and cousins, whohave always helped and encouraged me a lot. I would like to thank my wife Zhao Xiukunfor her support during all these years.

Xiulei Cao.Eindhoven, February 2016.

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