Free Boundaries in Rock Mechanics 9783110546163, 9783110544909

Table of contents :
Preface
Contents
1. Introduction
2. The Hele–Shaw problem
3. A joint motion of two immiscible viscous fluids
4. Mathematical models of in-situ leaching
5. Dynamics of cracks in rocks
A. Elements of continuum mechanics
References
Index

Citation preview

Anvarbek Meirmanov, Oleg V. Galtsev, Reshat N. Zimin Free Boundaries in Rock Mechanics

De Gruyter Series in Applied and Numerical Mathematics

| Edited by Rémi Abgrall, Zürich, Switzerland José A. Carrillo de la Plata, London, United Kingdom Jean-Michel Coron, Paris, France Athanassios S. Fokas, Cambridge, United Kingdom

Volume 1

Anvarbek Meirmanov, Oleg V. Galtsev, Reshat N. Zimin

Free Boundaries in Rock Mechanics |

Mathematics Subject Classification 2010 Primary: 35K15, 35D30, 35K55; Secondary: 65M08, 65M06 Authors Prof. Dr. Anvarbek Meirmanov Yachay Tech School of Mathematical Sciences and Information Technology 100119 Urcuqui Ecuador Prof. Oleg Vladimirovich Galtsev Belgorod State National Research University Department of Information Systems Pobedy Street 85 308015 Belgorod Russian Federation Prof. Reshat Narimanovich Zimin Kazakh-British Technical University School of Mathematics and Cybernetics Toli bi 59 050000 Almaty Kazakhstan

ISBN 978-3-11-054490-9 e-ISBN (PDF) 978-3-11-054616-3 e-ISBN (EPUB) 978-3-11-054504-3 Set-ISBN 978-3-11-054617-0 ISSN 2512-1820

Library of Congress Cataloging-in-Publication Data A CIP catalog record for this book has been applied for at the Library of Congress. Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2017 Walter de Gruyter GmbH, Berlin/Boston Typesetting: PTP-Berlin, Protago-TEX-Production GmbH, Berlin Printing and binding: CPI books GmbH, Leck ♾ Printed on acid-free paper Printed in Germany www.degruyter.com

Preface The free-boundary problems are a subset of those partial differential equations devoted to initial boundary-value problems in unknown domains for different types of differential equations. The term unknown domain means that one must define the domain where the solution is to be found, together with a solution of the problem. To define this free boundary we need one more boundary condition for the same differential equations as compared with the regular boundary-value problem. This condition is usually called a free boundary condition. The systematic study of such problems for elliptic differential equations was initiated by V. N. Monakhov in [1], and for the heat equation (the Stefan problem) by L. I. Rubinstein [2]. Later, several books and papers mostly devoted to the Stefan problem [3, 4] have appeared. The free-boundary problems for the Navier–Stokes equations have been intensively studied by V. A. Solonnikov [5, 6]. The most frequently studied free-boundary problems are the Stefan problem, the Hele–Shaw problem, and the Muskat problem. This is because these problems arise from physical processes which are very important from a practical standpoint. For example, the Stefan problem describes phase transitions in pure materials (melting and solidification) and has many applications in metallurgy, and the Hele–Shaw and Muskat problems describe the motion of underground liquids and are very important for hydrology and the oil industry. Of these three problems, the Stefan problem is the most studied. This is confirmed by the fact that of the free-boundary problems the Stefan problem has the largest number of publications. On the other hand, from a practical point of view there are some very important physical processes in rock mechanics which involve free boundaries, and which have been studied only by engineers. For example, in-situ leaching, and the dynamics of cracks in underground rocks. In-situ leaching initially involves drilling holes into an ore deposit, after which explosive or hydraulic fracturing may be used to create open pathways in the deposit for leaching solutions to penetrate. The leaching solution is pumped into the deposit, where it makes contact with the ore and dissolves part of it. The solution, carrying the dissolved ore content, is then pumped to the surface and processed. This method allows the extraction of metals and salts from an ore body without the need for conventional mining involving drill-and-blast, open-cut, or underground mining. But existing mathematical descriptions of this process have been very primitive and involve some postulates about rock dissolution which have no solid basis in classical continuum mechanics. The last problem which we are going to study here arises from the modeling of cracks in underground rocks. Up to now there has been no mathematical model of crack dynamics in underground rocks. For example, for metals this process is well studied [7]. Of course, this is an open question – is there any movement of cracks or not? But if there is, it may explain how may earthquakes occur [8]. In the rest state a DOI 10.1515/9783110546163-001

vi | Preface

single crack may be represented as a connected domain filled by pore fluids. During regular heat impulses, which come from the Earth’s core, the stress on the boundary between fluid and solid skeleton grow up to some limit. After this limit the boundary of the crack begins move (a moving free boundary) and creates strong seismic waves. In the present book we study some theoretical aspects of the Hele–Shaw and Muskat problems, in-situ leaching, and the dynamics of cracks in underground rocks. For the Hele–Shaw problem we prove the existence of the classical solution within some small time interval and the existence and uniqueness of the weak solution. We use standard methods: linearization, exact estimates for the corresponding linear problems, and the Newton–Kantorovich method. The new element here is the linear problem. It consists of a second-order elliptic partial differential equation and of boundary conditions, containing the time derivative of the solution. To get exact estimates of the solutions we use the method suggested by V. A. Solonnikov [6] and the results of K. K. Golovkin [9]. For the Muskat problem we first consider the motion of two different liquids in the pore space of a solid body and prove the existence and uniqueness of the classical solution. Next, under the restriction that the solid body has a periodic structure, we derive the homogenized model, which still remains a free-boundary problem. We apply the same scheme for in-situ leaching and dynamics of cracks. We first derive mathematical models, describing the processes at the pore (microscopic) level and after that we find corresponding homogenized models. Research which forms the basis of the present book has been partially supported by the Science Committee of the Ministry of Education and Science of the Republic of Kazakhstan, Grant 0980/GF4 (Chapter 4), and by the Russian Science Foundation, Grant 14-17-00556 (Chapter 3). Yachay Tech University, San Miguel de Urcuqui, Hacienda San Jose s/n y Proyecto Yachay, Ecuador, February 2017 A. Meirmanov, O. Galtsev, R. Zimin

Contents Preface | v 1

Introduction | 1

2 2.1 2.1.1 2.1.2 2.1.3 2.1.4 2.1.5 2.2 2.2.1 2.2.2

The Hele–Shaw problem | 25 Classical solution to the Hele–Shaw problem | 25 The problem setting | 25 The equivalent problem in a fixed domain | 26 Auxiliary results | 27 Proof of Theorem 2.1 | 28 Proof of Theorem 2.2 | 31 Weak solutions to the Hele–Shaw problem | 42 The problem setting | 42 Proof of Theorem 2.5: existence | 45

3 3.1

A joint motion of two immiscible viscous fluids | 53 A single capillary in an absolutely rigid skeleton: Dirichlet boundary conditions | 53 The problem setting | 53 The main result | 54 Proof of the main result | 54 Smooth initial density: existence and uniqueness | 55 Passage to non-mooth initial data | 59 Existence of a regular free boundary | 60 Uniqueness of the solution | 62 A single capillary of an absolutely rigid skeleton: Neumann boundary conditions | 64 The problem setting | 64 Proof of the main result | 67 Smooth initial density | 68 W 1,1 bounds for density | 74 Passage to nonsmooth initial data, existence of a regular free boundary | 75 Existence of the maximal time interval and uniqueness of the solution | 77 A single capillary of an elastic skeleton: Neumann boundary conditions | 77 The main result | 80 Proof of the main result | 82

3.1.1 3.1.2 3.1.3 3.1.4 3.1.5 3.1.6 3.1.7 3.2 3.2.1 3.2.2 3.2.3 3.2.4 3.2.5 3.2.6 3.3 3.3.1 3.3.2

viii | Contents

3.3.3 3.3.4 3.3.5 3.4 3.4.1 3.4.2 3.4.3 3.5 3.5.1

3.5.2

3.5.3

3.5.4 3.5.5

4 4.1 4.1.1 4.1.2 4.2 4.2.1 4.2.2 4.2.3 4.2.4 4.2.5 5 5.1 5.1.1 5.1.2 5.2 5.2.1

Approximate smooth density | 83 Uniform bounds for the velocity and pressure of fluid | 89 Uniform bounds for density | 96 Generalized Muskat problem | 97 Statement of the problem and main results | 98 Proof of Theorem 3.4 | 101 Proof of Theorem 3.5 | 110 Numerical implementations for the Muskat problem | 110 Numerical implementations for the Muskat problem with Neumann boundary conditions for absolutly rigid skeletons with surface tension | 110 Numerical implementations for the Muskat problem with Neumann boundary conditions for an absolutely rigid skeleton without surface tension | 116 Numerical implementations for the Muskat problem with Neumann boundary conditions for an elastic skeleton without surface tension | 117 Equations for fluid–structure interaction | 124 Numerical implementations for the Muskat problem with Neumann boundary conditions for elastic skeleton with surface tension | 127 Mathematical models of in-situ leaching | 129 Microscopic description | 129 Mathematical model in the form of differential equations | 129 Numerical implementations | 133 Macroscopic description | 139 Mathematical model as a system of integral identities | 139 Homogenization | 142 Mathematical model in a form of differential equations | 143 Initial boundary-value problem, describing in-situ leaching at the macroscopic level | 149 Numerical implementations | 150 Dynamics of cracks in rocks | 155 Accumulation of the energy in a single crack: the microscopic (pore) level | 155 Mathematical model | 155 Basic a-priori estimates | 158 Accumulation of the energy in a single crack: the macroscopic description | 161 Basic axioms | 161

Contents |

5.2.2 5.2.3 5.2.4 5.3 A A.1 A.2 A.3 A.4 A.5 A.6 A.7

Homogenization | 161 A joint motion of the elastic body and the liquid in crack | 163 Accumulation of the energy in a single crack | 164 Macroscopic model of crack propagation | 165 Elements of continuum mechanics | 167 Subject and method of continuum mechanics | 167 Basic definitions and axioms | 170 Continuous motion | 176 Elements of thermodynamics | 180 Some classical models of continuum mechanics | 181 Shock relations | 186 Joint motion of an elastic sold and a viscous liquid | 191

References | 207 Index | 211

ix

1 Introduction In this book we deal with some physical processes in rock mechanics which are described by free-boundary problems. Some of them are well known (the Hele–Shaw and Muskat problems), while some of them are completely new (in-situ leaching and dynamics of cracks in underground rocks). The second chapter is devoted to the Hele–Shaw problem, which is a well-known mathematical model describing the filtration of a pore liquid having a common unknown boundary with a static gas in pores. In its simplest setting the Hele–Shaw problem is formulated as follows: to find a domain Ω(t) = {x ∈ ℝ2 : 0 < x1 < 1, 0 < x2 < u∗ (x1 , t)}, and a liquid pressure p with a liquid velocity υ, which satisfy in Ω(t) for t > 0 the Darcy system of filtration k (1.1) υ = − ∇p, ∇ ⋅ υ = 0. μ On the free boundary Γ(t) = {x ∈ ℝ2 : 0 < x1 < 1, x2 = u∗ (x1 , t)} the normal velocity of the liquid υ n coincides with the normal velocity of the free boundary V n Vn = υn ,

x ∈ Γ(t),

t > 0,

(1.2)

and the pressure p coincides with the constant pressure p0 of the pore gas p = p0 ,

x ∈ Γ(t),

t > 0.

(1.3)

At the bottom S0 = {x ∈ ℝ2 : 0 < x1 < 1, x2 = 0}, p = p∗ > p0 ,

x ∈ S0 ,

t > 0.

(1.4)

We assume that all functions are 1-periodic in the variable x1 , and on the boundaries x1 = 0 and x1 = 1 the periodicity conditions p(0, x2 , t) = p(1, x2 , t),

υ n (0, x2 , t) = υ n (1, x2 , t)

(1.5)

are satisfied. Finally, at the initial time, Γ(0) = Γ0 ,

(u∗ (x1 , 0) = u0 (x1 ), 0 < x1 < 1).

(1.6)

In (1.1)–(1.6) n is the outward to Ω(t) unit normal vector to Γ(t), n=−

∇p , |∇p|

Vn = −

1 ∂p , |∇p| ∂t

υ n = υ ⋅ n,

k is a permeability coefficient and μ is the viscosity of the liquid. DOI 10.1515/9783110546163-002

(1.7)

2 | 1 Introduction

Without loss of generality we may suppose that μ = 1,

k = 1,

p∗ = 1,

p0 = 0.

Then the boundary conditions (1.2), and definition (1.7) of n imply V n = −∇p ⋅ n = |∇p|,

x ∈ Γ(t),

t > 0.

(1.8)

The formulation (1.1)–(1.6) of the problem means that we look for a classical solution. Of course, there is another equivalent formulation of the problem as an integral identity, which does not use the free boundary. Such a formulation is called a weak formulation. To find the weak formulation of the Hele–Shaw problem in the given domain Q = {x : 0 < x1 < 1, 0 < x2 < N}, where N is large enough, we introduce the characteristic function χ(x, t) of the domain Ω(t) ⊂ Q as χ(x, t) = 1 for x ∈ Ω(t),

and

χ(x, t) = 0 for x ∈ Q \ Ω(t),

and the extended pressure u(x, t) u(x, t) = p(x, t) for x ∈ Ω(t),

u(x, t) = 0 for x ∈ Q \ Ω(t).

and

In a sequel we use the following integration by parts: T

∂F ∂F + ∇ ⋅ f ) dx dt ≡ ∫( ∫ ( + ∇ ⋅ f ) dx) dt ∫( ∂t ∂t ΩT

0

Ω(t) T

= − ∫ F(x, 0) dx + ∫( ∫ (−V n F + f ⋅ n) sin α dσ) dt 0

Ω(0)

Γ(t)

≡ − ∫ F(x, 0) dx + ∫ (−V n F + f ⋅ n) sin α dσ dt,

(1.9)

ΓT

Ω(0)

which holds true for any functions f vanishing at S0 , and any functions F, vanishing at t = T . Here Γ T = {(x, t) : x ∈ Γ(t), 0 < t < T}, Ω T = ⋃ Ω(t), 0 0 ∂t ∂t

0 Ω0

0 Ω

as t0 → ∞. Then Π(Ω0 , t0 ) = ∫ α p,f |p f (x, t0 )|2 dx → Π ∗ > 0 = Π(Ω0 , 0), Ω0

as t0 → ∞. It represents exactly the accumulation of the energy in the crack during the heat impact. No material, not even rocks, can sustain a long and growing tension. At some instant of time t∗ the crack starts to break down. That is, it begins moving and changes its configuration. In the last section of the chapter we suggest the macroscopic model of crack propagation involving a hysteresis. We will describe the motion of the fracture after this specific moment t∗ by mean curvature flow [51] D n = σ(P∗ − P)k, (1.69) where D n is the velocity of the moving (free) boundary S = ∂Ω0 toward the outer normal n to S, and k is the mean curvature of S. More precisely, this mechanism is governed by the hysteresis law (Figure 1.11). There are two positions for the state of the fracture. The position M stands for the motion of the fracture and the position R stands for the state of rest. If in the state of rest, the average pressure P achieves the limiting value P∗ , the fracture changes state

M

R

P P* Fig. 1.11: Hysteresis law for the crack propagation.

P*

22 | 1 Introduction

Ground surface displacement (cm)

Ground surface displacement (cm)

40

40

35

35

30

30

25

25

20

20

15

15

10

10

5

5

0

0 10

15

20

25

30

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40

45

50

55

60

10

15

20

25

30

35

40

45

50

Total displacement (km)

Total displacement (km)

t = 15

t = 50

Ground surface displacement (cm)

55

60

Ground surface displacement (cm)

40

40

35

35

30

30

25

25

20

20

15

15

10

10

5

5 0

0 10

15

20

25

30

35

40

45

50

55

60

10

15

20

25

30

35

40

45

50

Total displacement (km)

Total displacement (km)

t = 70

t = 100

55

60

Fig. 1.12: Dynamics of a crack in rock.

from R to M and begins to move. We assume that the product of the average pressure and the volume V of the fracture are the same during the movement: P(t) ⋅ V(t) = const.

(1.70)

Thus, when the crack propagates and the volume of the crack decreases, the pressure P(t) inside the crack increases up to value P∗ , and after that the crack returns to the position R. Obviously the movement of the fracture creates seismic waves, which may reach the Earth’s surface.

1 Introduction

|

23

We may describe this stage of the process by Lamé’s system of elasticity ατ ϱs

∂2 w = ∇ ⋅ (α λ 𝔻(x, w) − p𝕀) ∂t2

(1.71)

for the displacements w and pressure p of the rock material in the domain Ω, coupled with the mean curvature flow (1.69) for the free boundary S of the domain Ω0 . The problem is completed with the postulate (1.70), which gives boundary conditions (α λ 𝔻(x, w) − p𝕀) ⋅ n = −P(t)n

(1.72)

for the Lamé equations at the boundary S. Corresponding numerical implementations confirm this suggested model (Figure 1.12). In Appendix A we concisely list the main notions of continuum mechanics following [52].

2 The Hele–Shaw problem In this chapter we prove the existence of weak and classical solutions to the Hele– Shaw problem under general conditions of the problem’s setting. To prove the existence of the classical solution we use von Mises variables, which transform the problem into an equivalent problem for a completely nonlinear parabolic equation in a fixed domain with a nonstandard boundary condition on the image of the free boundary, which involves a time derivative of the solution. The solvability of this nonlinear problem is based on the linearization and exact estimates for the solutions to the elliptic equations with nonstandard boundary condition on the given boundary with the time derivative of the solution. In turn, these estimates are based upon the method suggested by V. A. Solonnikov and use results of K. K. Golovkin [9]. The weak solution to the Hele–Shaw problem is obtained as a limit of smooth solutions to nondegenerate parabolic equations for some artificial enthalpy depending on a small parameter ε. To pass to the limit as ε → 0, we derive uniform in ε BV estimates for this enthalpy and use standard compactness results.

2.1 Classical solution to the Hele–Shaw problem 2.1.1 The problem setting In this section we consider the Hele–Shaw problem, which is a free boundary problem when a moving liquid in pores has a joint unknown boundary with a static pore gas. In its simplest setting, the Hele–Shaw problem is formulated as follows: to define a domain Ω(t) = {x ∈ ℝ2 : 0 < x1 < 1, 0 < x2 < u∗ (x1 , t)}, and a liquid pressure p with a liquid velocity υ, which satisfy in Ω(t) for t > 0 the Darcy system of filtration k (2.1) υ = − ∇p, ∇ ⋅ υ = 0. μ On the free boundary Γ(t) = {x ∈ ℝ2 : 0 < x1 < 1, x2 = u∗ (x1 , t)} the normal velocity of the liquid υ n coincides with the normal velocity of the free boundary V n , Vn = υn ,

x ∈ Γ(t),

t > 0,

(2.2)

and the pressure p coincides with the constant pressure p0 of the pore gas, p = p0 ,

x ∈ Γ(t),

t > 0.

(2.3)

At the bottom, S0 = {x ∈ ℝ2 : 0 < x1 < 1, x2 = 0} p = p∗ > p0 , DOI 10.1515/9783110546163-003

x ∈ S0 ,

t > 0.

(2.4)

26 | 2 The Hele–Shaw problem

We assume that all functions are 1-periodic in the variable x1 , and on the boundaries x1 = 0 and x1 = 1 periodicity conditions p(0, x2 , t) = p(1, x2 , t),

υ n (0, x2 , t) = υ n (1, x2 , t)

(2.5)

are satisfied. Finally, at the initial time, Γ(0) = Γ0 ,

(u∗ (x1 , 0) = u0 (x1 ), 0 < x1 < 1).

(2.6)

In (2.1)–(2.6) n is an outward to Ω(t) unit normal vector to Γ(t), n=−

∇p , |∇p|

Vn = −

1 ∂p , |∇p| ∂t

υ n = υ ⋅ n,

k is a permeability coefficient, and μ is the viscosity of the liquid. Without loss of generality we may suppose that μ = 1,

k = 1,

p∗ = 1,

p0 = 0.

2.1.2 The equivalent problem in a fixed domain Let us introduce the following new independent variables: t = t,

y1 = x1 ,

y2 = p(x, t).

(2.7)

This change of variables transforms Ω(t) to the unit square Q = {y ∈ ℝ2 : 0 < y1 < 1, 0 < y2 < 1}. A new unknown function u(y, t) = x2

(2.8)

satisfies in Q T = {y ∈ Q, 0 < t < T} the following initially-boundary value problem: Ψ(u) ≡

1 + p21 ∂ ∂2 u − ( ) = 0, p2 ∂2 y1 ∂y2

(y, t) ∈ Q T ,

∂u 1 + p21 = 0, (y, t) ∈ S0T , + ∂t p2 ∂u ∂u u(0, y2 , t) = u(1, y2 , t), (0, y2 , t) = (1, y2 , t), ∂y1 ∂y1

(2.9)

Ψ0 (u) ≡

u(y1 , 1, t) = 1, u(y1 , 0, 0) = u0 (y1 ),

(y, t) ∈ S1T , 0 < y1 < 1.

(2.10) t > 0,

(2.11) (2.12) (2.13)

2.1 Classical solution to the Hele–Shaw problem

Here p1 =

∂u , ∂y1

|

27

∂u . ∂y2

p2 =

Equations (2.9) and (2.10) are derived with the help of relations p1 ∂p 1 ∂p 1 ∂u ∂p =− , = , =− , ∂x1 p2 ∂x2 p2 ∂t p2 ∂t ∂ 1 ∂ ∂ ∂ p1 ∂ ∂ 1 ∂ = , = − , = . ∂x2 p2 ∂y2 ∂x1 ∂y1 p2 ∂y2 ∂t p2 ∂t S iT = S i × (0, T),

S i = {y : 0 < y1 < 1, y2 = i},

i = 0, 1.

Theorem 2.1. Let u0 ∈ ℂ3+α (S0 ) and be 1-periodic in the variable x1 . Then the problem (2.9)–(2.13) has a unique solution u ∈ ℂ2+α,β (Q T∗ ) for a sufficiently small time interval (0, T∗ ).

2.1.3 Auxiliary results First, we consider an initially linear boundary-value problem for elliptic equations 2

Lu ≡ ∑ a ij (y, t) i,j=1

L0 u ≡

2 ∂2 u ∂u + ∑ a i (y, t) = f(y, t), ∂y i ∂y j i=1 ∂y i

2 ∂u ∂u = f0 (y, t), + ∑ b i (y, t) ∂t i=1 ∂y i

u(0, y2 , t) = u(1, y2 , t),

x ∈ QT ,

(y, t) ∈ S0T ,

∂u ∂u (0, y2 , t) = (1, y2 , t), ∂y1 ∂y1

u(y, t) = u1 (y, t),

(2.15)

t > 0,

(y, t) ∈ S1T ,

u(y1 , 0, 0) = u0 (y1 ),

(2.14)

(2.16) (2.17)

0 < y1 < 1.

(2.18)

In what follows we suppose that all functions are 1-periodic in the variable x1 . Theorem 2.2. Let a ij , a i , f ∈ ℂk−2+α,β (Q T ), b i , f0 ∈ ℂk−1+α,β (S0T ), u1 ∈ ℂk+α,β (S1T ), u0 ∈ ℂk+α (S0 ) for k ⩾ 2, 0 < β < 2α , and n

n

0 < α|ξ|2 = α ∑ ξ i2 ⩽ ∑ a ij (y, t)ξ i ξ j , i=1

a ij (y, t) = a ji (y, t),

i,j=1

(y, t) ∈ Q T ,

b2 (y, t) ⩽ 0,

(y, t) ∈ S0T .

Then the problem (2.14)–(2.18) has a unique solution u ∈ ℂk+α,β (Q T ), and (k+α,0)

|u|Q T

(k+α,β)

|u|Q T

(k−2+α,0)

⩽ C(|f|Q T

(k−2+α,β)

⩽ C(|f|Q T

(k−1+α,0)

+ |f0 |S0

T

(k−1+α,β)

+ |f0 |S0

T

(k+α,0)

+ |u1 |S1

T

(k+α,β)

+ |u1 |S1

T

(k+α)

),

(2.19)

(k+α)

).

(2.20)

+ |u0 |S0 + |u0 |S0

In (2.19) and (2.20) constants C depend only on the corresponding norms a ij , a i , a, b i .

28 | 2 The Hele–Shaw problem Theorem 2.3 (Newton–Kantorovich method, [53]). Let 𝕏, 𝕐 be two Banach spaces, operator Φ : 𝕏 → 𝕐 be differentiable in some sphere B r (u0 ) = {u ∈ 𝔹 : ‖u − u0 ‖ < r}, and its first derivative be Lipschitz continuous: ‖Φ󸀠 (u1 ) − Φ󸀠 (u2 )‖ < L‖u1 − u2 ‖. Let the inverse operator (Φ󸀠 (u0 ))−1 to the operator Φ󸀠 (u0 ) also exist, and ‖(Φ󸀠 (u0 ))−1 ‖ = M, Then for h
0.

Moreover, due to the maximum principle ∂p0 (x) < −C0 < 0, ∂x2 Therefore (3+α)

|u0 |Q

< C1 ,

x ∈ ΓT .

∂u0 (y) < −C1 < 0, ∂y2

y ∈ Q,

(2.21)

where C1 depends only on C0 . To prove the theorem we use the Newton–Kantorovich method (Theorem 2.3). Namely, let 𝕏 = ℂ2+α,0 (Q T ),

𝕐 = ℂ2+α,0 (S1T ) × ℂα,0 (Q T ) × ℂ1+α,0 (S0T ) × ℂ2+α (S0 ), Φ(u) = {Φ1 (u), Ψ(u), Ψ0 (u), Φ0 (u)},

where Φ1 (u) = u(y, t),

(y, t) ∈ S1T ,

Φ0 (u) = u(y, 0),

y ∈ S0 .

2.1 Classical solution to the Hele–Shaw problem

Then

|

29

Φ : 𝕏 → 𝕐, 1 + (p01 )2

Φ(u0 ) = {0, 0,

(p02 )2

, u0 } ,

and the problem (2.9)–(2.13) is reduced to the equation Φ(u) = υ0 ,

(2.22)

where υ0 = (0, 0, 0, u0 ). So, we only need to check all conditions of Theorem 2.3. The first condition requires the smoothness of the operator Φ, which is ensured by the construction. The second condition implies the unique solvability of the corresponding linear initial boundary-value problem. Linearization of (2.9)–(2.13) at u = u0 gives us Ψ 󸀠 (u0 )⟨υ⟩ ≡

1 + (p01 )2 ∂υ p01 ∂υ ∂2 υ ∂ + + (−2 ) = f(y, t) ∂2 y1 ∂y2 p02 ∂y1 (p02 )2 ∂y2

Ψ 󸀠 (u0 )⟨υ⟩ ≡

1 + (p01 )2 ∂υ p0 ∂υ ∂υ − = f0 (y, t), + 2 10 ∂t p2 ∂y1 (p02 )2 ∂y2 Φ󸀠1 (u0 )⟨υ⟩ ≡ υ = u1 (y, t), Φ󸀠0 (u0 )⟨υ⟩ ≡ υ = u0 (y),

υ(0, y2 , t) = υ(1, y2 , t),

(y, t) ∈ Q T ,

(y, t) ∈ S0T , (2.23)

(y, t) ∈ S1T , y ∈ S0 ,

t = 0,

∂υ ∂υ (0, y2 , t) = (1, y2 , t), ∂y1 ∂y1

t > 0.

Due to (2.21) the problem (2.23) satisfies all conditions of Theorem 2.2. Therefore, operator Φ󸀠 (u0 ) has a bounded inverse operator, and (3+α,0)

|υ|Q T

(1+α,0)

⩽ C(|f|Q T

(2+α,0)

+ |f0 |S0

T

(3+α,0)

+ |u1 |S1

T

(3+α)

+ |u0 |S0

).

The third condition ‖(Φ󸀠 (u0 ))−1 ⟨Φ(u0 ) − υ0 ⟩‖ = k,

Φ(u0 ) = (0, 0,

means that solution w to the problem Φ󸀠 (u0 )⟨w⟩ = Φ(u0 ) − υ0 ,

1 + (p01 )2 (p02 )

, u0 )

(2.24)

30 | 2 The Hele–Shaw problem

or

p01 ∂w 1 + (p01 )2 ∂w ∂ ∂2 w + + (−2 ) = 0 (y, t) ∈ Q T , ∂2 y1 ∂y2 p02 ∂y1 (p02 )2 ∂y2 1 + (p01 )2 p0 ∂w 1 + (p01 )2 ∂w ∂w − = , + 2 10 0 ∂t p2 ∂y1 (p2 )2 ∂y2 (p02 )2 w(y, t) = 0,

(y, t) ∈ S1T ,

w(y, 0) = 0,

y ∈ S0 ,

(y, t) ∈ S0T , (2.25)

∂w ∂w (0, y2 , t) = (1, y2 , t), ∂y1 ∂y1

w(0, y2 , t) = w(1, y2 , t),

t > 0,

(2+α,0)

has a small norm |w|Q T = k. We prove it for the short time interval (0, T∗ ) using the following interpolation inequalities [54]: (k,0)

⩽ C(|u|Q T

(α,0)

⩽ C(|u|Q T

|u|Q T ⟨Du k ⟩Q T

(k+α,0)

1

(k−1,0)

) 1+α (|u|Q T

(k+1,0) α

(k,0) 1−α

) (|u|Q T )

α

) 1+α , k ⩾ 1,

,

β ⩾ 0,

or, in an equivalent form, (k,0)

⩽ ε|u|Q T

(α,0)

⩽ ε|u|Q T

|u|Q T ⟨Du k ⟩Q T

(k+α,0)

+ C(ε)|u|Q T ,

(0)

(k+1,0)

+ C(ε)|u|Q T ,

(0)

k ⩾ 1,

(2.26)

β ⩾ 0,

which obviously imply the desired relation (2+α,0)

|u|Q T

(2+α,0)

⩽ C ((|u|Q T

1+2α

(3,0) α

(2+α,0)

+ (|u|Q T ) (|u|Q T According to (2.24),

α2

(0)

) (1+α)2 (|u|Q T ) (1+α)2

(3+α,0)

|w|Q T

)

(1−α)(1+2α) (1+α)2

(0)

(1−α)α2

(0)

α2

(|u|Q T ) (1+α)2 ) ⩽ C(|u|Q T ) (1+α)2 .

(3+α)

⩽ C|u0 |S0

= C2 .

(2.27)

(2.28)

Therefore, (2+α,0)

|w|Q T

(0)

α2

⩽ C3 (C2 )(|w|Q T ) (1+α)2 .

(2.29)

w(y, 0) = 0

(2.30)

Note also that is a solution of the linear elliptic equation, satisfying the homogeneous boundary conditions.

2.1 Classical solution to the Hele–Shaw problem

|

31

No coefficients of the system (2.25) depend on the time t. Therefore, we may differentiate all equations and boundary conditions with respect to time and get for ω = ∂w/∂t the following problem: p01 ∂ω 1 + (p01 )2 ∂ω ∂2 ω ∂ + + (−2 ) = 0, ∂2 y1 ∂y2 p02 ∂y1 (p02 )2 ∂y2

(y, t) ∈ Q T ,

p0 ∂ω 1 + (p01 )2 ∂ω ∂ω − = 0, + 2 10 ∂t p2 ∂y1 (p02 )2 ∂y2

(y, t) ∈ S0T ,

ω(y, t) = 0,

(y, t) ∈ S1T ,

ω(y, 0) = ω0 (y), ω(0, y2 , t) = ω(1, y2 , t), where ω0 = −2

y ∈ S0 ,

∂ω ∂ω (0, y2 , t) = (1, y2 , t), ∂y1 ∂y1

t > 0,

p01 ∂w 1 + (p01 )2 ∂w + ∈ ℂ2+α (S0 ). p02 ∂y1 (p02 )2 ∂y2

Thus, due to (2.24),

(2+α,0)

|ω|Q T

(3+α)

⩽ C|u0 |S0

= C2 .

In particular, (0)

|w(y, t + τ) − w(y, t)| |τ|β |w(y, t + τ) − w(y, t)| 1−β (0) ⩽ ( max ) ⋅ T∗ ⩽ |ω|Q T ⋅ T∗ ⩽ C2 T∗ . |τ| (y,t)∈Q T∗ ,τ>0

|w|Q T =

max

(y,t)∈Q T∗ ,τ>0

Gathering all these together we arrive at α2 (1+α)2

k ⩽ C4 (C2 )T∗

,

which finally proves the statement of the theorem for a sufficiently small time interval (0, T∗ ).

2.1.5 Proof of Theorem 2.2 2.1.5.1 Model problem in a half-space To prove the theorem we first derive a-priori estimates for the problem (2.14)–(2.18), then prove the solvability of this problem for the Laplace equation and, finally, using a continuation along the parameter, the solvability for the general problem. Let u(x, t) be a solution to the Poisson equation ∆u = f(x, t),

x ∈ ℝ2+ = {x ∈ ℝ2 : x2 > 0},

t > 0,

(2.31)

32 | 2 The Hele–Shaw problem

satisfying the following boundary condition and initial conditions ∂u ∂u = f0 (x, t), − ∂t ∂x2

x ∈ ℝ10 = {x ∈ ℝ2 : x2 = 0},

u(x, 0) = u0 (x),

t > 0,

x ∈ ℝ10 .

(2.32) (2.33)

As a first step we consider the Dirichlet problem for the Laplace equation: ∆u0 = f(x, t),

x ∈ ℝ2+ ,

(k+α,0) ⟨u0 ⟩(ℝ2 ) + T

⩽

u0 (x) = u0 (x),

(k−2+α,0) C(⟨f⟩(ℝ2 ) + T

x ∈ ℝ10 ,

(2.34)

(k+α) ⟨u0 ⟩ℝ1 ). 0

+

(2.35)

Then the difference υ = u − u0 satisfies the following initial boundary-value problem: ∆υ = 0,

x ∈ ℝ2+ ,

∂υ ∂υ = F0 (x, t), − ∂t ∂x2

x ∈ ℝ10 ,

υ(x, 0) = 0,

x ∈ ℝ10 ,

F0 (x1 , t) = f0 (x1 , t) − and

(k−1+α,0)

⟨F0 ⟩(ℝ1 )

t > 0,

0 T

(k−2+α,0)

⩽ C(⟨f⟩(ℝ2 )

+ T

∂u0 ∂x2

(2.36) t > 0,

(2.37) (2.38)

(x1 , 0),

(k−1+α,0)

+ ⟨f0 ⟩(ℝ1 )

0 T

(2.39)

(k+α)

+ ⟨u0 ⟩ℝ1 0

),

(2.40)

for all T > 0 and all k ⩾ 2. Here, C = C(k), (ℝ2+ )T = ℝ2+ × (0, T), (ℝ10 )T = ℝ10 × (0, T). Using the Fourier transform ∞

̂ 1 , x2 , t) = υ(y

1 ∫ υ(x1 , x2 , t) e−ix1 y1 dx1 , √2π

(2.41)

−∞

we reduce the problem to the initial boundary-value problem for ordinary differential equation in the interval x2 > 0: ∂ 2 υ̂ − |y1 |2 υ̂ = 0, ∂2 x2 ∂ υ̂ ∂ υ̂ = F̂ 0 (y1 , t), − ∂t ∂x2 ̂ 0) = 0, υ(x,

x2 > 0,

t > 0,

x2 = 0,

t > 0,

x2 = 0.

It is easy to verify that the solution of problem (2.42) has the form t

̂ 1 , x2 , t) = ∫ F̂ 0 (y1 , τ) e−|y1 |(t−τ) dτ e−|y1 |x2 υ(y 0

(2.42)

2.1 Classical solution to the Hele–Shaw problem |

and

∂ 2 υ̂ = |y1 |2 û2 , ∂2 x2 t

υ̂ = ∫ F̂ 0 (y1 , τ) e−|y1 |(t−τ) dτ, 0

33

x2 > 0;

∂ υ̂ ∂ υ̂ = F̂ 0 , − ∂t ∂x2

x2 = 0.

These formulas imply the following representation of u(x, t) in terms of convolution integrals: t

υ(x, t) = ∫ F0 (x1 , τ) ∗ 0

∂G (x1 , x2 + t − τ) dτ. ∂x2

(2.43)

Here, ∗ denotes the convolution with respect the variable x1 : ∞

(w ∗ υ)(x1 ) = ∫ w(x1 − y1 )υ(y1 ) dy1 , −∞

and

1 ln|x|. π To estimate the solution u to the problem (2.31)–(2.33), we use the method introduced by V. Solonnikov in [55] and the following result due to K. Golovkin [9]: G(x) =

Theorem 2.4. Let n

mi

∆ m (h)u(x, t) = ∑ ∑ (−1)m i −k i C mi i u(x + hk i e i ) k

(2.44)

i=1 k i =0

be a finite difference, where m = (m1 , . . . , m n ), m i ⩾ 0, |m| = m1 + ⋅ ⋅ ⋅ + m n , and (e1 , . . . , e n ) be an orthonormal base in ℝn . (k+α) Then the seminorm ⟨u⟩ℝn , 0 < α < 1, is equivalent to T

∑

sup {sup|h|−(k+α) |∆ m (h)u(x, t)|}

|m|=r>k+1 (x,t)∈ℝT n

(2.45)

h>0

for any r > k. Lemma 2.1. The solution (2.43) to the problem (2.31)–(2.33) satisfies the following estimates: (k+α,0)

⟨u⟩(ℝ2 )

+ T

(k+α,β)

⟨u⟩(ℝ2 )

+ T

(k−2+α,0)

⩽ C(⟨f⟩(ℝ2 )

+ T

(k−2+α,0)

⩽ C(⟨f⟩(ℝ2 )

+ T

(k−1+α,0)

+ ⟨f0 ⟩(ℝ1 )

0 T

(k−1+α,0)

+ ⟨f0 ⟩(ℝ1 )

0 T

(k+α)

+ ⟨u0 ⟩ℝ1 0

(k+α)

+ ⟨u0 ⟩ℝ1 0

),

(2.46)

)

(2.47)

for all t > 0, k ⩾ 2, and any β < 2α . Here (ℝ2+ )T = ℝ2+ × (0, T), (ℝ10 )T = ℝ10 × (0, T), and the constant C depend on k, T , and N , which is the maximal diameter of support of the problem’s data with respect to spacial variables.

34 | 2 The Hele–Shaw problem

Proof. Let ∂ k−1 υ

w(x1 , t) =

(x , 0, t), Φ0 (x1 , t) = k−1 1

∂x1

(α,0)

⟨Φ0 ⟩(ℝ1 ) ⩽ 0 T

Then, t

∞

0

−∞

∂ k−1 F0

(x , t), G0 (ξ1 , τ) =

1 ∂x1k−1 (k−1+α,0) C⟨F0 ⟩(ℝ1 ) . 0 T

τ2

τ , + ξ12

1 ∫ dτ ∫ Φ0 (x1 − ξ1 , t − τ)G0 (ξ1 , τ) dξ1 , 2π

w(x1 , t) =

where υ is given by (2.36)–(2.39) and (2.43). Due to the properties of the Laplace equation it suffices to evaluate w only at S0 . It is easy to check, that G0 (y1 h, τh) =

1 G0 (y1 , τ), h

(2)

∆ y1 (h)(G0 (y1 h, τh)) = G0 (y1 h + 2h, τh) − 2G0 (y1 h + h, τh) + G0 (y1 h, τh) 1 = (G0 (y1 + 2, τ) − 2G0 (y1 + 1, τ) + G0 (y1 , τ)) h 1 (2) = ∆ y1 (1)(G0 (y1 , τ)), h (2)

1

(2)

3

∆ y1 (1)(G0 (y1 , τ)) ∼ (y21 + τ2 )− 2 , ∆ y1 (1)(G0 (y1 , τ)) ∼ (y21 + τ2 )− 2 ,

as y21 + τ2 → 0, as y21 + τ2 → ∞,

∞∞

󵄨 (2) 󵄨 ∫ ∫ 󵄨󵄨󵄨∆ x1 (h)(G0 (x1 , t))󵄨󵄨󵄨 dx1 dt = (x1 = hy1 , t = hτ) ∞∞

0 0

󵄨 (2) 󵄨 = ∫ ∫ 󵄨󵄨󵄨∆ y1 (h)(G0 (hy1 , hτ))󵄨󵄨󵄨h2 dy1 dτ 0 0 ∞∞

󵄨 (2) 󵄨 = |h| ∫ ∫ 󵄨󵄨󵄨∆ y1 (1)(G0 (y1 , τ))󵄨󵄨󵄨 dy1 dτ ⩽ C|h|. 0 0

Therefore, (k+α,0)

⟨υ⟩(ℝ1 )

0 T

= ∑

sup

m=3 (x,t)∈(ℝ0 )T 1

{sup|h|−(1+α) |∆ m (h)w(x1 , t)|} h>0

∞∞

(α,0) (k−1+α,0) 󵄨 (2) 󵄨 ⩽ C{sup|h|−1 ∫ ∫ 󵄨󵄨󵄨∆ x1 (h)(G0 (x1 , t))󵄨󵄨󵄨 dx1 dt}⟨Φ0 ⟩(ℝ1 ) ⩽ C⟨F0 ⟩(ℝ1 ) , h>0

0 T

0 0

which proves (2.46). To prove (2.47) we use the following lemma [28, p. 78].

0 T

2.1 Classical solution to the Hele–Shaw problem |

Lemma 2.2. Let υ ∈ ℂ0,β (Q T ) and Then

∂υ ∂x i

∂υ α,0 (Q ), T ∂x i ∈ ℂ α , and 2

∈ ℂα,γ (Q T ) with γ < ⟨

35

0 < α, β < 1.

∂υ (α,γ) ∂υ (α,0) (0,β) ⩽ C (⟨υ⟩Q T + ⟨ ⟩ ⟩ ). ∂x i Q T ∂x i Q T

In fact, υ= and

∂ k−1 u ∂x1k−1

∈ ℂ1+α,0 ((ℝ10 )T ),

∂k u ∂υ ∈ ℂ0 ((ℝ10 )T ). = ∂t ∂t∂x1k−1 Therefore, υ ∈ ℂ0,γ ((ℝ10 )T ) for any γ < 1, and due to the previous lemma, ∂k u ∂υ = ∈ ℂ(α,β) ((ℝ10 )T ) ∂x1 ∂x1k

for β= or for any β
k + 1, m2 = 2s + 1 > k + 1, n = m1 + 2s. Then, ∞

∆ m (h)u(x1 , x2 , t) =

1 m m ∫ (∆1 1 (h)u0 )(∆2 2 (h)G0 ) dξ1 2π −∞ ∞

=

1 m +2s ∫ (∆1 1 (h)u0 )(∆12 (h)G0 ) dξ1 , 2π −∞

m

where ∆ i i (h) is a finite difference with respect to the variable x i , i = 1, 2.

(2.48)

36 | 2 The Hele–Shaw problem

Therefore, ∞

(k+α,0)

|∆ m (h)u(x1 , x2 , t)| ⩽ C⟨u0 ⟩(ℝ1 )

0 T

󵄨 󵄨 |h|k+α ∫ 󵄨󵄨󵄨(∆12 (h)G0 )󵄨󵄨󵄨 dx1 −∞ ∞

(k+α,0)

= C⟨u0 ⟩(ℝ1 )

0 T

|h|k+α ∫ −∞ ∞

(k+α,0)

⩽ C⟨u0 ⟩(ℝ1 )

0 T

|h|k+α ∫ −∞

for |m| = m1 + m2 = k + 2. Thus,

|hx21 − hx22 − hx2 | ((x2 + h)2 + x21 )(x21 + x22 )

dx1

|h| (k+α,0) dx1 ⩽ C⟨u0 ⟩(ℝ1 ) |h|k+α , 0 T (h2 + x21 )

∞

1 |m| D u(x1 , x2 , t) = ∫ D1 u0 (x1 − ξ1 , t)G0 (ξ1 , x2 ) dξ1 2π m

−∞

and

(k+α,0)

⟨Du(x1 , x2 , t)⟩(ℝ2 )

+ T

(k+α,0)

⩽ C⟨u0 ⟩(ℝ1 )

0 T

|h|k+α .

Here, |m| = m1 + m2 ⩽ k, m

m

D m u(x1 , x2 , t) = D1 1 D2 2 u(x1 , x2 , t) =

∂|m| m m u(x 1 , x 2 , t). ∂x1 1 ∂x2 2

The last representation permits us to consider finite differences with respect to time, and to evaluate Holder norms with respect to time of all derivatives of u via the corresponding norms of u0 and prove (2.47).

2.1.5.2 General linear problem Lemma 2.3. Under the conditions of Theorem 2.3, any solution u(x, t) of the problem (2.14)–(2.18) satisfies the following a-priori estimates: (k+α,0)

|u|Q T

(k−2+α,0)

⩽ C(|f|Q T

(k−1+α,0)

+ |f0 |S0

(k+α,0)

+ |u1 |S1

T

T

(k+α)

+ |u0 |S0

(0)

+ |u|Q T ),

(2.49)

where the constant C depends only on the corresponding norms a ij , a i , a, b i . Proof. First, note that due to the supposed periodicity of all the problem’s functions with respect to the variable y1 , any solution of the problem will be periodic with respect to y1 . Therefore, we may extend the solution over the boundaries {y1 = i}, i = 0, 1 and consider it in the domain Q δ = {y ∈ ℝ2+ : −

1 2

where solution u(y, t) will be smooth.

< y1 < 32 , δ < y2 < 1} ⊃ Q,

2.1 Classical solution to the Hele–Shaw problem

|

37

Now, let φ1 (y, t), . . . , φ m (y, t), φ n ⩾ 0, φ n ∈ ℂ∞ (ℝ2T ), be a partition of unity of the domain Q T : m

∑ φ n (y, t) = 1,

y ∈ QT .

n=1

We can always assume that all diameters of supports Ω n ⊂ (ℝ2+ )T = ℝ2+ × (0, T) of functions φ n are less than some small number δ, which we can choose later. By means of this partition of unity we represent the solution u(y, t) as m

u(y, t) = ∑ u n (y, t),

u n (y, t) = u(y, t)φ n (y, t).

n=1

Multiplying equality Lu = f by φ n we put 2

Mu n ≡ ∑ a ij (y, t) i,j=1

∂2 u (y, t) = F n (y, t), ∂y i ∂y j

2

F n = fφ n + ∑ a ij ( i,j=1

(2.50)

2 ∂2 φ n ∂u ∂φ n ∂u +u φn . )−∑ ∂y i ∂y j ∂y i ∂y j ∂y i i=1

Using interpolation inequalities (2.25) we may estimate F n as (k−2+α,0)

|F n |ℝ2

T

(k−2+α,0)

⩽ C|f|Q T

(k+α,0)

+ δ|u|Q T

(0)

+ C1 (δ)|u|Q T .

(2.51)

If Ω n ⊂ Q󸀠T , then we extend u n (y, t) as zero outside of Ω n , and represent equation (2.50) as 2

M 0n u n ≡ ∑ a ij (y n , t n ) i,j=1

∂2 u (y, t) = Ψ n (y, t) ∂y i ∂y j

2

≡ F n (y, t) + ∑ (a ij (y n , t n ) − a ij (y, t)) i,j=1

∂2 u (y, t), ∂y i ∂y j

(2.52)

with some (y n , t n ) ∈ Ω n . The orthogonal transformation y = T(n) ⟨z⟩, and the following stretching of variables x i = a i y i , a i > 0, i = 1, 2, u n (y, t) = w n (z, t) = υ n (x, t) reduces (2.52) to the Poisson equation ∆υ n = Φ n (x, t),

z ∈ ℝ2 ,

t > 0,

Φ n (x, t) = Ψ n (y, t).

The properties of υ n are well known, and (k+α,0)

|υ n |ℝ2

T

(k−2+α,0)

⩽ C|Φ n |ℝ2

T

.

(2.53)

38 | 2 The Hele–Shaw problem

Coming back to original variables, one has (k+α,0)

|u n |ℝ2

T

(k−2+α,0)

⩽ C|Ψ n |ℝ2

T

(k+α,0)

⩽ C(sup|a ij (y n , t) − a ij (y, t)||u n |ℝ2 Ωn

T

(k+α,0)

⩽ C(δ β |u n |ℝ2

T

(0)

(k,0)

+ |u n |ℝ2

T

(k−2+α,0)

+ C2 (δ)|u|Q T + |f|Q T

(k−2+α,0)

+ |F n |ℝ2

T

)

).

Therefore, for sufficiently small δ (k+α,0)

|u n |ℝ2

T

(k−2+α,0)

⩽ C(|f|Q T

(0)

+ |u|Q T ).

(2.54)

Now let S(n,0) = Ω n ⋂ S0 ≠ 0, and y n ∈ S(n,0) . Then M 0n u n = Ψ n ,

(y, t) ∈ (ℝ2+ )T ,

(2.55)

and L00 u n ≡

2 ∂u n ∂u n = ψ n (y, t), + ∑ b i (y n , t n ) ∂t ∂y i i=1

(y, t) ∈ S0T = S0 ,

u n (y1 , 0, 0) = u n,0 (y1 ) = u0 (y1 )φ n (y1 , 0, 0),

(2.56) (2.57)

where 2

ψ n = f0 φ n + ∑ (b i (y n , t n ) − b i (y, t)) i=1 (k−2+α0)

|F n |ℝ2

T

(k−1+α,0)

|ψ n |(ℝ1 )

0 T

(k−2+α,0)

⩽ C|f|Q T

T

(k+α,0)

+ δ|u|Q T

(k−1+α,0)

⩽ C|f0 |S0

2 ∂u n ∂φ n ∂φ n + un ( + ∑ b i (y, t) ). ∂y i ∂t ∂y i i=1

(k+α,0)

+ δ|u n |Q T

(0)

+ C1 (δ)|u|Q T , (0)

+ C1 (δ)|u|Q T .

(2.58)

As before, the orthogonal transformation and the stretching of variables with another orthogonal transformation (which does not change the Poisson equation and returns the obtained half-space in the first two transformations to the original position x2 = 0) reduce (2.55) to the Poisson equation for the new function υ n (x, t) = u n (y, t) in the half-space x2 > 0 for t > 0, and boundary and initial conditions (2.56)–(2.57) at x2 = 0 to ∆υ n = Φ n (x, t), x2 = 0, t > 0, 2 ∂υ n ∂υ n = ϕ n (x, t), + ∑ ci ∂t ∂x m i=1

x2 = 0,

t > 0,

υ n (x1 , 0, 0) = υ n,0 (x1 ). Here Φ n (x, t) = Ψ n (y, t), ϕ n (x, t) = ψ n (y, t), and υ n,0 (x1 ) = u n,0 (y1 ).

(2.59)

2.1 Classical solution to the Hele–Shaw problem

|

39

Finally, the change of variables t = τ, y1 = x1 − c1 t,

y2 = x2 − (c2 + 1)t,

υ n (x, t) = u n (y, t)

reduces the problem (2.55), (2.56), with a corresponding initial condition, to the standard form (2.32)–(2.34). Therefore, (k+α,0)

|u n |(ℝ2 )

+ T

(k−2+α,0)

⩽ C(|f|Q T

(k−1+α,0)

+ |f0 |S0

(k+α,0)

+ |u1 |S1

T

T

(k+α)

+ |u|Q T ),

(k+α)

+ |u|Q T ),

+ |u0 |S0

(0)

(2.60)

and due to the equality u = ∑m n=1 u n , (k+α,0)

|u|Q T

(k−2+α,0)

⩽ C(|f|Q T

(k−1+α,0)

+ |f0 |S0

(k+α,0)

+ |u1 |S1

T

T

+ |u0 |S0

(0)

(2.61)

which is exactly (2.49). Lemma 2.4 (The maximum principle). Under the conditions of Theorem 2.2, any solution u(x, t) of the problem (2.14)–(2.18) satisfies the maximum principle in the following form: (0) (α,0) (α,0) (α,0) (0) |u|Q T ⩽ C(|f|Q T + |f0 |S0 + |u1 |S1 + |u0 |S0 ), (2.62) T

T

where the constant C depends only on the corresponding norms a ij , a i , a, b i . Proof. Let

υ(0, y2 , t) = υ(1, y2 , t), Theorem 3.1 [54] implies

Lυ = f,

(y, t) ∈ Q T ,

υ = 0,

(y, t) ∈ S0T ,

υ = 0,

(y, t) ∈ S1T ,

(2.63)

∂υ ∂υ (0, y2 , t) = (1, y2 , t), ∂y1 ∂y1 (2+α,0)

|υ|Q T

0 < t < T.

(α,0)

⩽ C|f|Q T .

(2.64)

The function w = (u − υ)e−t solves the following initial boundary-value problem Lw = 0, L0 w + w =

(y, t) ∈ Q T ,

∂w ∂w ∂w + b2 + w = F0 , + b1 ∂t ∂y1 ∂y2

(y, t) ∈ S0T ,

(y, t) ∈ S1T ,

w = u1 ,

(2.65)

w(y1 , 0, 0) = 0, w(0, y2 , t) = w(1, y2 , t),

∂w ∂w (0, y2 , t) = (1, y2 , t), ∂y1 ∂y1

0 < y2 < 1, where F0 = e−t (f0 − b1

0 < t < T,

∂υ ∂υ − b2 ), ∂y1 ∂y2

(y, t) ∈ S0T .

40 | 2 The Hele–Shaw problem

According to the maximum principle, w has no extremum in Q T . If w achieves its extremum on S1T , then (0)

(0)

|w|Q T ⩽ |u1 |S1 . T

If w achieves its positive maximum at (y0 , t0 ) ∈ S0T , t0 > 0, then at this point ∂w ⩾ 0, ∂t

∂w = 0, ∂y1

∂w ⩽ 0, ∂y2

0 ⩽ L0 w,

Therefore,

w ⩽ L0 w + w = F0 .

(0)

w(y0 , t0 ) ⩽ |F0 |S0 ; T

the same situation with a minimum of w(y, t): (0)

w(y0 , t0 ) ⩾ −|F0 |S0 . T

Thus,

(0)

(0)

|w|Q T ⩽ |F0 |S0 . T

If an extremum of w achieves at y0 ∈

S0 ,

t0 = 0, then

(0)

(0)

|w|Q T ⩽ |u0 |S0 . Gathering all these together we arrive at the statement of the lemma. Estimates (2.49) and (2.62) prove (2.19). Lemma 2.5. Under the conditions of Theorem 2.2 there exists a unique solution of the problem (2.14)–(2.18) satisfying (2.19). Proof. To prove this lemma we use the method of continuation along the parameter. Let ∂υ ∂υ − L λ υ = λLυ + (1 − λ)∆υ, L0λ υ = λL0 + (1 − λ) ( ). ∂t ∂y2 For λ = 0, the problem L0 υ ≡ ∆υ = f,

(y, t) ∈ Q T ,

υ(0, y2 , t) = υ(1, y2 , t), υ(y, t) = u1 (y, t),

∂υ ∂υ = f0 , (y, t) ∈ Q T , − ∂t ∂y2 ∂υ ∂υ (0, y2 , t) = (1, y2 , t), t > 0, ∂y1 ∂y1 L00 υ ≡

(y, t) ∈ S1T ,

υ(y1 , 0, 0) = u0 (y1 ),

0 < y1 < 1

has a unique solution for any f, f0 , u1 , u0 under the conditions of Theorem 2.2. This is proved similarly as for the problem (2.36)–(2.38).

2.1 Classical solution to the Hele–Shaw problem |

41

Let Λ be the set of all λ for which the problem L λ υ = f,

(y, t) ∈ Q T ,

(y, t) ∈ Q T ,

∂υ ∂υ (0, y2 , t) = (1, y2 , t), ∂y1 ∂y1

υ(0, y2 , t) = υ(1, y2 , t), υ(y, t) = u1 (y, t),

L0λ υ = f0 ,

(y, t) ∈ S1T ,

υ(y1 , 0, 0) = u0 (y1 ),

t > 0,

(2.66)

0 < y1 < 1

has a unique solution. This set is not empty, because it contains point λ = 0. Let λ0 ∈ Λ. Using the contraction mapping theorem we prove that the problem (2.66) has a unique solution υ λ for |λ − λ0 | < ε with sufficiently small ε. So, the set Λ is open. Estimates (2.19) for υ λ permit us to prove that the set Λ is closed. Therefore, Λ = [0, 1], which proves our statement. Lemma 2.6. Under the conditions of Theorem 2.2 any solution u(y, t) of the problem (2.14)–(2.18) satisfies (2.20). Proof. In the same way as in Lemma 2.1, we prove that u0 = u(y1 , 0, t) ∈ ℂk+α,β (S0T ). Now let ∆ t (h)υ =

υ(y, t + τ) − υ(y, t) . |τ|β

To prove this lemma we apply estimates (2.19) for the difference w(y, t) = ∆ t (h)u(y, t), which satisfies the following Dirichlet problem: Lw = Φ,

(y, t) ∈ Q T ,

2

Φ = ∆ t (h)f − ∑ (∆ t (h)a i,j ) i,j=1

2 ∂2 u ∂u − ∑ (∆ t (h)a i ) , ∂y i ∂y j i=1 ∂y i

w(y, t) = ∆ t (h)u (y1 , t),

(y, t) ∈

w(y, t) = ∆ t (h)u1 (y1 , t),

(y, t) ∈

0

(k−2+α,0)

|Φ|Q T

(k−2+α,β)

⩽ C(|f|Q T

(k−1+α,β)

+ |f0 |S0

(k+α,β)

+ |u1 |S1

T

T

(2.67)

S0T , S1T , (k+α)

+ |u0 |S0

).

(2.68)

Due to the conditions of Theorem 2.2 and properties of u0 , ∆ t (h)a ij ,

∆ t (h)a i ∈ ℂk−2+α,0 (Q T ),

∆ t (h)u0 ∈ ℂk+α,0 (S0T ),

∆ t (h)υ1 ∈ ℂk+α,0 (S1T ).

Applying estimates (2.19) one has (k+α,0)

|w|Q T

(k−2+α,β)

⩽ C(|f|Q T

(k−1+α,β)

+ |f0 |S0

T

(k+α,β)

+ |u1 |S1

T

(k+α)

+ |u0 |S0

).

(2.69)

42 | 2 The Hele–Shaw problem

Therefore (k+α,β)

|u|Q T

(k+α,0)

= sup (|w|Q T |τ|>0

(k−2+α,β)

⩽ C(|f|Q T

) (k−1+α,β)

+ |f0 |S0

(k+α,β)

+ |u1 |S1

T

T

(k+α)

+ |u0 |S0

).

(2.70)

2.2 Weak solutions to the Hele–Shaw problem 2.2.1 The problem setting In this section we consider weak solutions to the Hele–Shaw problem. First, we recall the classical formulation of the problem in its simplest setting. Let Q = {x ∈ ℝ2 : 0 < x1 < 1, 0 < 0 < x2 < N}, where N is large enough. One must define a domain Ω(t) = {x ∈ Q : 0 < x1 < 1, 0 < x2 < u∗ (x1 , t)} and a liquid pressure p with a liquid velocity υ which satisfy in Ω(t) for t > 0 the Darcy system of filtration k (2.71) υ = − ∇p, ∇ ⋅ υ = 0. μ On the free boundary Γ(t) = {x ∈ ℝ2 : 0 < x1 < 1, x2 = u∗ (x1 , t)} the normal velocity of the liquid υ n coincides with the normal velocity of the free boundary V n : Vn = υn ,

x ∈ Γ(t),

t > 0,

(2.72)

and the pressure p coincides with the constant pressure p0 of the pore gas: p = p0 ,

x ∈ Γ(t),

t > 0.

(2.73)

At the bottom S0 = {x ∈ ℝ2 : 0 < x1 < 1, x2 = 0}, p = p∗ > p0 ,

x ∈ S0 ,

t > 0.

(2.74)

We assume that all functions are 1-periodic in the variable x1 , and that on the boundaries x1 = 0 and x1 = 1 periodicity conditions p(0, x2 , t) = p(1, x2 , t),

υ n (0, x2 , t) = υ n (1, x2 , t)

(2.75)

are satisfied. Finally, at the initial time moment, Γ(0) = Γ0 ,

(u∗ (x1 , 0) = u0 (x1 ), 0 < x1 < 1).

(2.76)

2.2 Weak solutions to the Hele–Shaw problem |

43

In (2.71)–(2.76) n is an outward to Ω unit normal vector to Γ(t): n=−

∇p , |∇p|

Vn = −

1 ∂p , |∇p| ∂t

υ n = υ ⋅ n,

(2.77)

k is a permeability coefficient, and μ is the viscosity of the liquid. Without loss of generality we may suppose that μ = 1,

k = 1,

p∗ = 1,

p0 = 0.

Then boundary conditions (2.72), (2.77) imply V n = −∇p ⋅ n = |∇p|,

x ∈ Γ(t),

t > 0.

(2.78)

Our aim is to find some equivalent formulation of the problem as an integral identity which does not use the free boundary. To do this we introduce the characteristic function χ(x, t) of the domain Ω(t) as χ(x, t) = 1 for x ∈ Ω(t),

and

χ(x, t) = 0 for x ∈ Q \ Ω(t)

and the extended pressure u(x, t) as u(x, t) = p(x, t) for x ∈ Ω(t),

and

u(x, t) = 0 for x ∈ Q \ Ω(t).

In a sequel we use the following formula for integration by parts: ∫(

∂F + ∇ ⋅ f ) dx dt = − ∫ F(x, 0) dx + ∫ (−V n F + f ⋅ n) sin α dσ dt, ∂t

ΩT

(2.79)

ΓT

Ω(0)

which holds true for any functions f vanishing at S0 , and any functions F, vanishing at t = T . Here, Γ T = {(x, t) : x ∈ Γ(t), 0 < t < T}, and α is an angle between the time axis and the normal ν to Γ T . Next we consider the equality 0 = ∫ φ ∆u dx dt, ΩT

which holds true for any smooth function φ, vanishing at S0 . We also suppose that φ vanishes at t = T . Then integration by parts and relations (2.77), (2.78) give us 0 = ∫ φ∇u ⋅ n sin α dσ dt − ∫ ∇φ ⋅ ∇u dx dt ΓT

ΩT

= − ∫ φ|∇u| sin α dσ dt + ∫ u∆φ dx dt ΓT

ΩT

= − ∫ φV n sin α dσ dt + ∫ u∆φ dx dt. ΓT

QT

44 | 2 The Hele–Shaw problem

Thus ∫ φV n sin α dσ dt = ∫ u∆φ dx dt. ΓT

(2.80)

QT

Formula (2.79) with F = φ and f = 0 implies ∂φ dx dt = − ∫ φ(x, 0) dx − ∫ V n φ sin α dσ dt, ∂t

∫ ΩT

ΓT

Ω(0)

or ∫χ

∂φ dx dt = − ∫ χ(x, 0)φ(x, 0) dx − ∫ V n φ sin α dσ dt. ∂t

QT

Q

ΓT

Therefore, ∫ (χ

∂φ + u∆φ) dx dt = − ∫ χ(x, 0)φ(x, 0) dx. ∂t

QT

(2.81)

Q

The last identity holds true for any smooth functions φ, vanishing at S0 , and at t = T , and does not contain free boundary Γ T . It is exactly this identity which we use as a definition of the weak solution to the Hele–Shaw problem. Definition 2.1. We say that a pair of measurable bounded functions {u, χ} is a weak solution to the problem (2.71)–(2.76), if it satisfies the integral identity (2.81) and boundary conditions (2.75) in the form u(x 1 , 0, t) = u(x1 , 1, t),

∂u ∂u (0, x2 , t) = (1, x2 , t), ∂x2 ∂x2

(2.82)

with boundary conditions u(x, t) = 1,

(x, t) ∈ S0T ,

u(x, t) = 0,

(x, t) ∈ S1T ,

(2.83)

and relations χ = Φ(u),

Φ(u) = 0 for u ⩽ 0,

Φ(u) = 1 for u > 0,

and 0 ⩽ Φ(u) ⩽ 1. (2.84)

In (2.83) S 1T = {x ∈ ℝ2 : 0 < x1 < 1, x2 = 1}. As 𝔹𝕍(Q) we denote the functional space of all measurable functions χ(x), such that ‖χ‖𝔹𝕍(Q) ≡ ‖χ‖1,Q + sup ∫ |h|>0

Q

|χ(x + h) − χ(x)| dx < ∞, |h|

(2.85)

and as u0 (x) we denote a function, which is identically equal to zero in Q \ Ω(0) and which solves the Laplace equation in Ω(0) with boundary conditions (2.82), boundary condition (2.83) at S0 , and equal to zero at Γ0 = Γ(0). Theorem 2.5. Let χ ̊ ∈ 𝔹𝕍(Q), u0 ∈ 𝕃2 (Q), and (2.86) hold true. Then the problem (2.71)–(2.76) has a unique weak solution: χ ∈ 𝕃∞ ((0, T); 𝔹𝕍(Q)), u ∈ 𝕃2 ((0, T); 𝕎12 (Q)).

2.2 Weak solutions to the Hele–Shaw problem | 45

2.2.2 Proof of Theorem 2.5: existence For any ε > 0 we put Φ ε ∈ ℂ∞ (−∞, ∞), Φ ε (s) = −εs,

s < 0;

Φ󸀠ε (s) > 0;

0 < Φ ε (s) < 1,

Φ ε (s) = 1 + ε(s − ε), Ψ ε (s) =

and

Φ−1 ε (s),

0 ⩽ s ⩽ ε;

s > ε;

Ψ ε (Φ ε (s)) ≡ s.

As approximate solutions to the problem (2.71)–(2.76) we consider functions χ ε = Φ ε (u ε ), satisfying the following problem: ∂u ε = ∆u ε , (x, 0) ∈ Q T , ∂t ∂u ε ∂u ε u ε (x1 , 0, t) = u ε (x1 , 1, t), (0, x2 , t) = (1, x2 , t), ∂x2 ∂x2 Φ󸀠ε (u ε )

u ε (x, t) = 1,

(x, t) ∈ S0T ,

u ε (x, t) = 0,

u ε (x, 0) = ů ε (x),

(x, t) ∈ S1T ,

x ∈ Q;

(2.86)

∂χ ε = ∇ ⋅ (Ψ ε󸀠 (χ ε )∇χ ε ), (x, 0) ∈ Q T , ∂t ∂χ ε ∂χ ε χ ε (x1 , 0, t) = χ ε (x1 , 1, t), (0, x2 , t) = (1, x2 , t), ∂x2 ∂x2 χ ε (x, t) = 1 + ε(1 − ε),

(x, t) ∈ S0T ,

χ ε (x, 0) = χ ̊ε (x),

χ ε (x, t) = 0,

(x, t) ∈ S1T ,

x ∈ Q,

̊ ̊ ̊ where ů ε (x) = (M ε u)(x) = u0 , x ∈ Ω(0), is a mollifier of u(x), χ ̊ε (x) = Φ ε (ů ε ), u(x) ̊ u(x) = 0, x ∈ Ω \ Ω(0), and u0 solves the problem ∆u0 = 0,

x ∈ Ω(0),

u0 (x) = 1,

x ∈ S0 ,

u0 (x) = 0,

x ∈ Γ(0).

The simple energy estimate for u0 and definition of ů result in 2 ̊ dx = ∫ |∇u0 (x)|2 dx ⩽ β, ∫|∇u(x)| Q

Ω(0)

(2.87)

2 ̊ dx dt ⩽ C(Ω(0))β, ∫|∇ů ε (x)|2 dx ⩽ C(Ω(0)) ∫|∇u(x)| Q

Q

where β depends only on Ω(0) (or ‖χ0 ‖𝔹𝕍(Q) ). Next we multiply the differential equation in (2.86) for u ε by parts over domain Q × (0, t0 ):

∂u ε ∂t

and integrate by

t0 󵄨󵄨 ∂u 󵄨󵄨󵄨2 1 1 󵄨 ε (x, t)󵄨󵄨󵄨 dx dt + ∫|∇u ε (x, t0 )|2 dx = ∫|∇ů ε (x)|2 dx ⩽ C(Ω(0))β. ∫ ∫ Φ󸀠ε (u ε ) 󵄨󵄨󵄨 󵄨󵄨 ∂t 󵄨󵄨 2 2 0 Q

Q

Q

46 | 2 The Hele–Shaw problem The last estimate, and the inequality Φ󸀠ε (u ε ) ⩾ 0 imply max ∫|∇u ε (x, t)|2 dx ⩽ C(Ω(0))β.

(2.88)

0 n. According to estimates (3.11), this implies the strong convergence of the corresponding subsequence of velocities {u(m) } in the space L∞ (0, T; W 2,q (Ω)). Estimates (3.12) guarantee strong convergence of {u(m) } in the space L∞ (0, T; C1,λ (Ω)) with λ = 1 − nq as well. Passing to the limit in the corresponding equations (3.1)–(3.3) for the functions u(m) , ρ(m) , and p(m) as m → ∞, one sees that the limit functions satisfy the original problem. Instead of the original transport equation (3.3), the density ρ satisfies the corresponding integral identity ∫ ΩT

∂φ dφ ρ dx dt ≡ ∫ ( + u ⋅ ∇φ) ρ dx dt dt ∂t ΩT

= − ∫ ρ0 (x)φ(x, 0) dx

(3.31)

Ω

for any smooth test function φ such that φ(x, T) = 0 and φ|S = 0.

3.1.6 Existence of a regular free boundary Let us now prove that there exists a regular free boundary which divides the domain Ω into two subdomains Ω+ (t) and Ω− (t) such that { ρ+ , x ∈ Ω+ (t), ρ(x, t) = { ρ− , x ∈ Ω− (t). { The reasoning is based on the explicit representation of the solution to the transport equation (3.3) using translation along its characteristics dX (3.32) = u(X(ξ, t), t), X(ξ, 0) = ξ, ξ ∈ Γ0 , dt with the velocity field u from the solution of the original problem (cf. [31]). Note that the characteristics X (m) (ξ, t) of the transport equation (3.3) for the (m) smooth density ρ(m) which corresponds to ρ0 of (3.30) converge uniformly to those defined by (3.32). For this reason, whenever a point x is not in Γ(t) = {X(ξ, t): ξ ∈ Γ0 }, it has a neighborhood U δ (x) ⊆ {X(ξ, t) : ρ0 (ξ) = const, dist(ξ, G0 ) ⩾ δ},

δ > 0,

3.1 A single capillary in an absolutely rigid skeleton: Dirichlet boundary conditions |

61

separated from Γ(t) by a positive distance. For m large enough, each characteristic X (m) that arrives to U δ (x) at time t starts from a point ξ with the same initial value (m) of density ρ0 (ξ ) = ρ0 (ξ), so ρ(m) is constant on U(x). By strong L q convergence of densities, the limit ρ(x, t) is also constant on U δ (x) and assumes one of the values ρ± . Thus, the complement in Ω to Γ(t) ∪ S consists of two open sets Ω± (t) = {ρ(x, t) = ρ± }. To show that the moving boundary Γ(t) is a smooth surface, we consider in more detail characteristics (3.32) that start from the initial surface Γ(0) = Γ0 . By estimate (3.12) the velocity field is C1,λ smooth, and it remains to show that the corresponding surface Γ(t) = {x : x = X(ξ, t)} belongs to the same class C1,λ . Differentiating (3.32) in parameter ξ , we get a Cauchy problem for the derivatives (∂/∂ξ j )X k : 󵄨 n ∂X i 󵄨󵄨󵄨 ∂X k d ∂X i 󵄨󵄨 = δ ij , = 0, ( ) − ∑ ∇k u i dt ∂ξ j ∂ξ j ∂ξ j 󵄨󵄨󵄨t=0 k=1 j

where δ k is the Kronecker symbol. This is another transport equation, so the derivatives considered are bounded for t > 0. We arrive at a similar conclusion considering the quotients of finite differences M ijs =

1 ∂X i ∂X i (t; ξ + he s ) − (t; ξ)) , ( ∂ξ j h λ ∂ξ j

where (e1 , . . . , e n ) is the standard basis in ℝn . The pertinent Cauchy problem is n d s s + B sij , M ij = ∑ A sik M kj dt k=1

M ijs (ξ, 0) = 0,

and the coefficients A sik = ∇k u i (X(t; ξ + he s ), t) B sij =

1 n ∂X k (t; ξ) ∑ (∇k u i (X(t; ξ + he s ), t) − ∇k u i (X(t; ξ), t)) ∂ξ j h λ k=1

s | are bounded, and ∂X k /∂ξ j are Hölder continuous in paare bounded. Therefore |M kj rameter ξ . Consequently, the map induced by (3.32) is smooth, for each t > 0:

x = X(ξ, t) ∈ C1+λ .

(3.33)

Note that for incompressible fluids the Jacobian J(t) = det( ∂X ∂ξ ) of the map (3.32) preserves its value: J(t) = J(0) = 1, t ∈ [0, T], by (3.2) and the Euler formula d J(t) = ρ∇ ⋅ u. dt Thus, we can use (3.33) to obtain for all t ∈ [0, T] and (ξ1 , ξ2 ) ∈ Ω0 the estimates 1 |ξ1 − ξ2 | ⩽ |X(ξ1 , t) − X(ξ2 , t)| ⩽ C|ξ1 − ξ2 | C

62 | 3 A joint motion of two immiscible viscous fluids with a positive constant C. This means that at any time t > 0 the free boundary has no common points with the given boundary if the two boundaries had no common points at the initial moment t = 0.

3.1.7 Uniqueness of the solution The last step in the proof of Theorem 3.1 is to show that the solution of the boundary and initial-value problem (3.1)–(3.5) is unique. Suppose that the problem has two solutions (u(i) , p(i) , ρ(i) ), i = 1, 2. Let us consider the boundary-value problem for their difference u = u(1) − u(2) ,

p = p(1) − p(2) ,

ρ = ρ(1) − ρ(2) .

The difference satisfies the equations ∆u = ∇p + αρe, ∇ ⋅ u = 0, u|S = 0, ∂ρ (1) + u ⋅ ∇ρ = −∇ ⋅ (ρ(2) u), ρ(x, 0) = 0, ∂t

(3.34)

where the last equation is fulfilled as an integral identity: for any smooth function φ such that φ(x, T) = 0 T

T

I0 ≡ − ∫ ∫ ρ

dφ dx dt = ∫ ∫ ρ(2) (u ⋅ ∇φ) dx dt ≡ I1 , dt

0 Ω

0 Ω

and the full derivative (d/dt)φ = (∂/∂t)φ + u(1) ⋅ ∇φ is calculated along the velocity field u(1) . Using the fact that ρ(2) is piecewise constant, one can rewrite the term I1 as I1 =

∫

ρ+ (u ⋅ ∇φ) dx dt +

ρ(2) =ρ+

∫

ρ− (u ⋅ ∇φ) dx dt

ρ(2) =ρ−

T

= β ∫ dt ∫ φ(u ⋅ n) ds, 0

Γ (2) (t)

where β = ρ+ − ρ− and n is the normal unit vector to the surface Γ (2) (t) which divides the domain Ω into two subdomains Ω+ (t).and Ω− (t), ρ(2) (x, t) = ρ± in Ω± (t). On the other hand, one can use the Lagrangian coordinates (ξ, t) → (x, t) defined by characteristics of the transport equation in (3.34), dx = u(1) (x, t), dt

x(ξ, 0) = ξ.

3.1 A single capillary in an absolutely rigid skeleton: Dirichlet boundary conditions |

63

In Lagrangian coordinates, the integral I0 becomes T

̂ t) I0 = − ∫ ∫ ρ(ξ,

∂ φ̂ (ξ, t) dξ dt ∂t

0 Ω

̂ t) = ρ(x(ξ, t), t) and φ(ξ, ̂ t) = φ(x(ξ, t), t)). (here ρ(ξ, As in the proof of Lemma 4.1 in [28, Chapter 3], choose the “Lagrangian” test function as the time average t

̂ t) = η̂ h (ξ, t) = φ(ξ,

1 ̂ τ) dτ ∫ η(ξ, h t−h

and put t+h

ρ̂ h (ξ, t) =

1 ̂ τ) dτ. ∫ ρ(ξ, h t

It is easily seen that I0 = − ∫ ρ̂ h (ξ, t)

∂ η̂ ∂ ρ̂ ̂ t) h (ξ, t) dξ dt. (ξ, t) dξ dt = ∫ η(ξ, ∂t ∂t

ΩT

(3.35)

ΩT

Arguing along the customary lines, we can choose for the test function η̂ in (3.35) any bounded function ψ̂ which vanishes for t > t0 , t0 < T − h. For ψ̂ = sign ρ̂ h (ξ, t), identity (3.35) becomes I0 = ∫|ρ̂ h (ξ, t0 )| dξ = ∫|ρ h (x, t0 )| dx. Ω

Ω

Therefore t0 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ∫|ρ h (x, t0 )| dx ⩽ I1 ⩽ α 󵄨󵄨󵄨 ∫ φ(u ⋅ n) ds dt󵄨󵄨󵄨 ⩽ K1 ∫‖u(t)‖W 1,1 (Ω) dt. 󵄨󵄨 󵄨󵄨 (2) 0 Ω Γ t0

Next, we pass to the limit as h → 0. We finally get the inequality t0

‖ρ(t0 )‖L1 (Ω) = ∫|ρ(x, t0 )| dx ⩽ K1 ∫‖u(t)‖W 1,1 (Ω) dt.

(3.36)

0

Ω

We consider once again the Stokes equations for the function u. One can see that the force term g = αρe defines a continuous linear functional in the space W 1,r (Ω) for r > n. Indeed, 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 |⟨g ⋅ u⟩| = 󵄨󵄨󵄨󵄨∫ g ⋅ u dx󵄨󵄨󵄨󵄨 = 󵄨󵄨󵄨󵄨α ∫ ρ(e ⋅ u) dx󵄨󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 Ω Ω ⩽ K2 ‖ρ(t)‖L1 (Ω) max|u(x, t)| ⩽ K3 ‖ρ(t)‖L1 (Ω) ‖u(t)‖W 1,r (Ω) . x∈Ω

64 | 3 A joint motion of two immiscible viscous fluids ∗

Therefore (see [63, p. 226]) g ∈ W0−1,r (Ω), and consequently u(t) ∈ W 1,r (Ω), where 1 < r∗ = r−1 r < n. Moreover, u(t) satisfies the inequality ∗

‖u(t)‖W 1,r∗ (Ω) ⩽ K4 ‖ρ(t)‖L1 (Ω) . Combined with (3.36), this estimate yields the inequality d ‖ρ(t)‖L1 (Ω) ⩽ K5 ‖ρ(t)‖L1 (Ω) , dt which shows that ρ = 0. Thus, the solution of the original problem (3.1)–(3.5) is unique.

3.2 A single capillary of an absolutely rigid skeleton: Neumann boundary conditions 3.2.1 The problem setting As in the previous section, we consider the flow of two immiscible viscous fluids with different constant densities in a single capillary Ω = {x ∈ ℝ2 : − 1 < x1 < 1, −h < x2 < h}. The evolution is driven by input pressure and the force of gravity. More precisely, in this problem one must find velocity u(x, t) = (u1 (x, t), u2 (x, t)) ∈ ℝ2 , pressure p(x, t), and density ρ(x, t) from the system of equations for velocity and pressure μ∆u − ∇p + gρe = 0,

(3.37)

∇ ⋅ u = 0,

(3.38)

where μ = const is a viscosity of liquids, e is a given unit vector and g is acceleration due to gravity, and the transport equation for density dρ ∂ρ ∂ρ ≡ + ∇ ⋅ (ρu) = + u ⋅ ∇ρ = 0. dt ∂t ∂t

(3.39)

At the initial moment t = 0, the density is piecewise constant and assumes two positive values characterizing the distinct phases of the flow: { ρ+ , ρ(x, 0) = ρ0 (x) = { ρ− , {

x ∈ Ω+ (0), x ∈ Ω− (0),

ρ± = const,

ρ− > ρ+ > 0.

(3.40)

Time t enters the equations for velocity as a parameter, so it needs no initial condition. The boundary condition on the lateral part S0 = {x ∈ ℝ2 : − 1 < x1 < 1, x2 = ±h} of the boundary S = ∂Ω is u(x, t) = 0. (3.41) The boundary condition on the “entrance” S− = {x ∈ ℝ2 : x1 = −1, −h < x2 < h} ⊂ S and “exit” S+ = {x ∈ ℝ2 : x1 = 1, −h < x2 < h} ⊂ S are ℙ(u, p) ⋅ n = −p0 n,

ρ = ρ± ,

x ∈ S± .

(3.42)

3.2 A single capillary of an absolutely rigid skeleton: Neumann boundary conditions |

65

Here, 𝔻(u) = 12 (∇u + (∇u)∗ ),

ℙ(u, p) = 2μ𝔻(u) − p𝕀,

where 𝕀 is a unit tensor, p0 (x) is a given function, and n = (1, 0) is a unit normal vector to S± . Note, that one needs the boundary condition for the density only at points x ∈ S± , where ±u ⋅ n < 0. The initial and boundary conditions for density are equivalent to specifying the surface Γ0 that separates the two subdomains Ω± (0) initially occupied by different fluids. For the sake of simplicity we suppose that Γ (0) = {x ∈ Ω : x1 = 0, −h < x2 < h}, Ω± = {x ∈ Ω : 0 < ±x1 < 1}. If the velocity u(x, t) is sufficiently smooth, then the Cauchy problem dx = u(x, t), dt

x|t=t0 = ξ

t > t0 ,

(3.43)

determines a mapping x = γ(ξ , t; u; t0 ),

γ : Ω → Ω.

(3.44)

In particular, the free boundary Γ(t) is determined as a set Γ(t) = {x ∈ Ω : x = γ(ξ , t; u; 0), ξ ∈ Γ (0) }, and subdomains Ω± (t) = {x ∈ Ω : ρ(x, t) = ρ± } as sets Ω± (t) = {x ∈ Ω : x = γ(ξ , t; u; 0), ξ ∈ Ω± } ⋂ {x ∈ Ω : x = γ(ξ , t; u; t0 ), ξ ∈ S± (0), t0 > 0}. The problem treated here is that of finding the velocity u(x, t), pressure p(x, t), and density ρ(x, t) from the above equations and initial and boundary data. Note that it is non-linear because of the coupling term u ⋅ ∇ρ in (3.39). To simplify our considerations we pass to the homogeneous boundary conditions ℙ(u, p) ⋅ n = 0,

x ∈ S±

(3.45)

by introducing a new pressure p → p − p0 (x): μ∆u − ∇p = f ≡ ∇p0 − gρe,

(3.46)

where ∇p0 is a bounded function: |∇p0 (x, t)| < P0 = const.

(3.47)

It is shown below that the evolution described by the above equations preserves the existence of two subdomains Ω± (t), each occupied by one of the fluids, that are separated at time t > 0 by a regular free boundary Γ(t). Thus, the problem studied is equivalent to finding u, p, and the moving boundary Γ(t). Let 1 1 Ω (m) = {x ∈ Ω : − 1 + m < x1 < 1 − m }, where m > 1 (i.e. any real positive number). Our principal result is the following theorem.

66 | 3 A joint motion of two immiscible viscous fluids

Theorem 3.2. Under condition (3.47) the problem (3.38)–(3.41), (3.45)–(3.46) has a unique solution in the interval [0, T) for some T > 0. The elements of this solution have the following properties. (i) For arbitrary positive m ∈ ℕ, q > 2, and λ = 1 − 2q , the velocity u and pressure p satisfy the regularity conditions u ∈ L∞ (0, T; W 2,q (Ω(m) )) ∩ L∞ (0, T; C1,λ (Ω(m) )),

p ∈ L∞ (0, T; W 1,q (Ω(m) )),

equations (3.38), (3.46) almost everywhere in Ω T = Ω × (0, T), boundary condition (3.41) in the usual sense, and boundary condition (3.44) in the sense of distributions as an integral identity ∫(ℙ(u(t), p(t)) : 𝔻(φ) + f ⋅ φ) dx = 0

(3.48)

Ω

for almost all 0 < t < T and for any smooth solenoidal functions φ vanishing at x ∈ S0 . (ii) The free boundary Γ(t) is a surface of class C1,λ at each time t ∈ [0, T), and the normal velocity V n (x, t) of the free boundary in the direction of its normal n at position x is uniformly bounded: sup |V n (x, t)| < ∞. t∈(0,T) x∈Γ(t)

(iii) The density ρ has bounded variation: (m)

ρ ∈ L∞ (0, T; BV(Ω(m) )) ∩ BV(Ω T ), and satisfies the transport equation (3.38) in the sense of distributions ∫ ρ( ΩT

∂ψ + u ⋅ ∇ψ) dx dt = − ∫ ρ0 (x)ψ(x, 0) dx ∂t

(3.49)

Ω

for any smooth functions ψ, vanishing at t = T and x ∈ S± . The time T of the existence of the classical solution depends on the behavior of the free boundary Γ(t). Namely, let δ± (t) be the distance between Γ(t) and the boundary S± and δ(t) = min(δ− (t), δ+ (t)). Then δ(t) > 0 for all 0 < t < T and δ(t) → 0 as t → T .

3.2 A single capillary of an absolutely rigid skeleton: Neumann boundary conditions |

67

Throughout this section, we use the customary notation of function spaces and norms (see e.g. [63]). Thus, for 1 < q < ∞ u ∈ L q (Ω) ⇒ (‖u‖L q (Ω) )q = ∫|u|q dx < ∞, Ω 2 󵄨󵄨 󵄨q 󵄨 ∂u 󵄨󵄨󵄨 u ∈ W 1,q (Ω) ⇒ (‖u‖W 1,q (Ω) )q = ∫(|u|q + ∑ 󵄨󵄨󵄨 󵄨 ) dx < ∞, 󵄨 ∂x i 󵄨󵄨󵄨 i=1 󵄨 Ω

2 󵄨󵄨 2 󵄨󵄨q 󵄨 ∂ u 󵄨󵄨 󵄨󵄨 ) dx < ∞, u ∈ W 2,q (Ω) ⇒ (‖u‖W 2,q (Ω) )q = ∫(|u|q + ∑ 󵄨󵄨󵄨󵄨 󵄨 ∂x i x j 󵄨󵄨󵄨 i,j=1 󵄨 Ω T q

u ∈ L q ((0, T); B) ⇒ ∫‖u(t)‖B dt < ∞. 0

For q = ∞ u ∈ L∞ ((0, T); B) ⇒ sup ‖u(t)‖B < ∞. 0 0. More precisely, we put { ρ+ , (ε) ρ 0 (x) = { ρ− , {

−1 ⩽ x1 ⩽ −ε, ε ⩽ x1 ⩽ 1,

(ε)

and ρ− < ρ0 (x) < ρ+ for −ε < x1 < ε. The function class M consists of all continuous functions ρ̃ ∈ C(Ω T ) such that ̃ t) ⩽ ρ+ . ρ− ⩽ ρ(x,

(3.50)

Now we define the following two linear operators. The first of them transforms a “frozen” density into the corresponding field of velocities: M ∋ ρ̃ 󳨃→ υ = U[ρ]̃ ∈ L∞ (0, T; W 2,q (Ω(m) )).

The second operator describes the evolution of density driven by a “frozen” velocity field (and starts from the initial smooth density specified in the beginning of the subsection): (ε) L∞ (0, T; W 2,q (Ω(m) )) ∋ U 󳨃→ ϱ = R[ρ0 , υ] ∈ L∞ (Ω T ). Namely, the operator U transforms ρ̃ into the solution of ̃ μ∆υ − ∇p = f ̃ ≡ ∇p0 − g ρe,

x ∈ Ω, 0 < t < T,

(3.51)

∇ ⋅ υ = 0,

x ∈ Ω, 0 < t < T,

(3.52)

υ(x, t) = 0,

x ∈ S , 0 < t < T,

(3.53)

0

±

ℙ(υ, p) ⋅ n = 0,

x ∈ S , 0 < t < T.

(ε)

(3.54) (ε)

The operator ϱ = R[ρ0 , υ], which depends on the given initial density ρ0 , transforms υ into the solution of the initial boundary-value problem ∂ϱ + υ ⋅ ∇ϱ = 0, ∂t (ε) ϱ(x, 0) = ρ0 (x), ±

ϱ(x, t0 ) = ρ ,

x ∈ Ω(m) , 0 < t < T, x∈Ω

at x ∈

(m)

S±m ,

(3.55)

,

(3.56)

0 < t0 < T,

(3.57)

where ± υ1 (x, t0 ) < 0, υ = (υ1 , υ2 ). Here S±m = {x ∈ Ω : x1 = ± (1 −

1 m )} .

3.2 A single capillary of an absolutely rigid skeleton: Neumann boundary conditions

| 69

Outside of Ω(m) we put { ρ+ , ϱ(x, t) = { ρ− , {

−1 ⩽ x1 ⩽ −1 + 1−

1 m

1 m,

0 < t < T,

⩽ x1 ⩽ 1, 0 < t < T,

(3.58)

Let Γ (±ε) (υ; t) = {x ∈ Ω : x = γ(ξ , t; υ; 0), ξ ∈ Γ (±ε) },

Γ (±ε) = {ξ ∈ Ω : ξ1 = ±ε},

where γ(ξ , t; υ; t0 ) has been defined in (3.43)–(3.44). In this problem we restrict ourselves to the time range T m satisfying the following condition: for all ρ̃ ∈ M and ε > 0 and there exist some ρ̃ ∈ M and ε > 0 such that Γ (ε) (υ; T m ) touches S+m , or Γ (−ε) (υ; T m ) touches S−m . It is clear that T m ⩽ T m+1 ∀m > 0. (ε)

For a smooth initial density ρ0 , the original problem of finding u and ρ from (3.38)– (3.41), (3.45)–(3.46) reduces to finding a fixed point of the superposition of these two linear operators, the operator F = R ∘ U defined as def

(ε)

̃ ∈ M. M ∋ ρ̃ 󳨃→ F[ρ]̃ = (R ∘ U)[ρ]̃ = R[ρ 0 , U[ρ]] We will show that the conditions of the Schauder fixed point theorem are satisfied for the operator F in the time interval (0, T m ) where T m > T0 and T0 > 0 do not depend on m, ρ̃ ∈ M , and ε.

3.2.3.1 Continuity of F Below we use notation C for positive constants whose values does not depend on m and ε, and notation K for positive constants whose values does not depend on ε. (a) For each function ρ̃ ∈ M , the linear problem (3.51)–(3.54) has a unique solution υ ∈ L∞ (0, T; W 1,2 (Ω)) ∩ L∞ (0, T; W 2,q (Ω(m) )), p ∈ L∞ (0, T; L2 (Ω)) ∩ L∞ (0, T; W 1,q (Ω(m) )), and for each q ∈ (1, ∞) and each value of the parameter t ∈ [0, T] the solution admits the estimates ̃ L (Ω) , ‖p(t)‖L2 (Ω) + ‖υ(t)‖W 1,2 (Ω) ⩽ C‖f (t)‖ 2 ̃ ‖p(t)‖W 1,q (Ω(m) ) + ‖υ(t)‖W 2,q (Ω(m) ) ⩽ K‖f (t)‖L (Ω) . q

(3.59) (3.60)

These results originate in [31, 61, 62, 60, 63] and [27, Chapter 3, § 5]. In fact, the first estimate is the well-known result for the unique weak solution of this problem in the form of the integral identity ∫(ℙ(υ, p) : 𝔻(φ) + f ̃ ⋅ φ) dx = 0 Ω

for fixed t ∈ (0, T) and for any smooth solenoidal functions φ, vanishing at x ∈ S0 .

70 | 3 A joint motion of two immiscible viscous fluids

This identity contains equation (3.51) and boundary condition (3.54). To prove (3.60) we consider an infinitely smooth function χ(4m) (x) such that x ∈ Ω(4m)

{ 1, χ(4m) (x) = { 0, {

x ∈ Ω \ Ω(8m) .

It is easy to see that ̃ , μ∆(χ(4m) υ) − ∇(χ(4m) p) = f (4m) ∇ ⋅ (χ (χ

(4m)

(4m)

υ) = φ

(4m)

,

υ)(x, t) = 0,

x ∈ G(4m) , 0 < t < T, x∈G

(4m)

x ∈ ∂G

(3.61)

, 0 < t < T,

(4m)

(3.62)

, 0 < t < T,

(3.63)

where Ω(8m) ⊂ G(4m) ⊂ Ω(16m) , ∂G(4m) ∈ C∞ , and ̃ f (4m) = 2μ∇χ(4m) ⋅ ∇υ + μυ∆χ(4m) − p∇χ(4m) + f ̃χ(4m) ,

φ(4m) = ∇χ(4m) ⋅ υ.

In fact, one just differentiates all terms in the left-hand sides of equations (3.61) and (3.62) and uses corresponding equations (3.51) and (3.52) to get the right-hand sides of (3.61) and (3.62). Estimates (3.59) provide ̃ f (4m) ∈ L∞ (0, T; L2 (G(4m) )),

φ(4m) ∈ L∞ (0, T; W 1,2 (G(4m) )) :

̃ L (Ω) . ̃ (t)‖L2 (G(4m) ) + ‖φ(4m) (t)‖W 1,2 (G(4m) ) ⩽ K‖f (t)‖ ‖f (4m) 2 Therefore, according to the above-mentioned works, the pair {(χ(4m) υ), (χ(4m) p)} as a solution of the problem (3.61)–(3.63) satisfies estimates ‖(χ(4m) p)(t)‖W 1,2 (G(4m) ) + ‖(χ(4m) υ)(t)‖W 2,2 (G(4m) ) ̃ ̃ L (Ω) . ⩽ ‖f (4m) (t)‖L2 (G(4m) ) + ‖φ(4m) (t)‖W 1,2 (G(4m) ) ⩽ K‖f (t)‖ 2

(3.64)

These results are typical for elliptic systems and may be explained by the Dirichlet problem for the Poisson equation ∆u = f,

x ∈ G,

u(x) = 0,

x ∈ ∂G.

Namely, if f ∈ L q (G), then u ∈ W 2,q (G) and ‖u‖W 2,q (G) ⩽ ‖f‖L q (G)

for q > 1.

Coming back to the definition of χ(4m) we obtain ̃ L (Ω) . ‖p(t)‖W 1,2 (Ω(4m) ) + ‖υ(t)‖W 2,2 (Ω(4m) ) ⩽ K‖f (t)‖ 2

(3.65)

Combined with the embedding theorem for the pair of spaces W 1,2 (Ω(4m) ) and L q (Ω(4m) ) for any q > 2 (see [28, Chapter II, § 3]) we arrive at ̃ L (Ω) . ‖p(t)‖L q (Ω(4m) ) + ‖υ(t)‖W 1,q (Ω(4m) ) ⩽ K‖f (t)‖ 2

(3.66)

3.2 A single capillary of an absolutely rigid skeleton: Neumann boundary conditions |

71

Now we use estimates in W 2,q and repeat for the function χ(m) and domains Ω(2m) ⊂ G(m) ⊂ Ω(4m) , ∂G(m) ∈ C∞ and get ̃ (t)‖L (G(m) ) + ‖φ(m) (t)‖W 1,q (G(m) ) ⩽ ‖f (m) ̃ (t)‖L (Ω(4m) ) + ‖φ(m) (t)‖W 1,q (Ω(4m) ) ‖f (m) q q ̃ L (Ω) + ‖f (t)‖ ̃ L (Ω) ) ⩽ K(‖f (t)‖ 2

q

̃ L (Ω) , ⩽ K‖f (t)‖ q

‖(χ(m) p)(t)‖W 1,q (G(m) )

̃ (t)‖L (Ω(4m) ) + ‖φ(m) (t)‖W 1,q (Ω(4m) ) + ‖(χ(m) υ)(t)‖W 2,q (G(m) ) ⩽ ‖f (m) q ̃ ⩽ K‖f (t)‖L (Ω) , (3.67) q

̃ L (Ω) . ‖p(t)‖W 1,q (Ω(m) ) + ‖υ(t)‖W 2,q (Ω(m) ) ⩽ K‖f (t)‖ q

(3.68)

Combined again with the embedding theorem for the pair of spaces W 2,q (Ω(m) ) 󳨅→ C1,λ (Ω(m) ) (see [28, Chapter II, § 3]), this estimate shows that for q > 2 and λ = 1 − 2/q ̃ L (Ω) . ‖υ(t)‖C1,λ (Ω(m) ) ⩽ C‖υ(t)‖W 2,q (Ω(m) ) ⩽ K‖f (t)‖ q

(3.69)

Whenever ρ̃ is continuous with respect to t, the above estimates imply that υ = U[ρ]̃ is continuous as a function of t with values in W 2,q (Ω(m) ) (or C1,λ (Ω(m) )), or as a realvalued function of t and x: υ ∈ C(0, T; W 2,q (Ω(m) )) ⊂ C(0, T; C1,λ (Ω(m) )). These last estimates and the arbitrary choice of m show that υ ∈ C(0, T; W 2,q (Ω)) ⊂ C(0, T; C1,λ (Ω)).

(3.70)

(b) We now establish the existence of a regular solution ϱ to the transport equation (ε) (ε) (3.55) for a smooth initial density ρ0 ∈ C∞ (Ω) such that ρ− ⩽ ϱ0 (x) ⩽ ρ+ . ̃ we find the starting point (ξ , t0 ) Given the velocity field υ = U[ρ], ξ = γ−1 (x, t; υ; t0 ) ∈ Ω( 1−ε ) 1

of the characteristic of (3.55) which hits x = γ(ξ , t; υ; t0 ) ∈ Ω at time t: ∂γ = υ(γ, t), ∂t

t ≠ t 0 ,

γ(ξ , t0 ; υ; t0 ) = ξ ,

(3.71)

where ξ = γ−1 (x, t; υ; t0 ). By construction, t0 = 0 for ξ ∈ Ω(m) and t0 ⩾ 0 for ξ ∈ S±m . For given υ = (υ1 , υ2 ) let Ω̃ (m) (t) = Ω(m) ⋃ {(ξ , t0 ) : ξ ∈ S±m , 0 < t0 < t such that ± υ1 (ξ , t0 ) < 0}. Then for all 0 < t < T m transformations (m) γ̃ : Ω̃ (m) (t) → Ω ,

̃ , t) = γ(ξ , t; υ; 0) x = γ(ξ

for ξ ∈ Ω

̃ , t) = γ(ξ , t; υ; t0 ) for ξ ∈ x = γ(ξ

(m)

S±m ,

, 0 < t0 < t

72 | 3 A joint motion of two immiscible viscous fluids

and γ̃−1 : Ω

(m)

→ Ω̃ (m) (t),

ξ = γ̃−1 (x, t)

are continuously differentiable if υ posessess the regularity properties (3.69), typical of solutions to (3.51)–(3.54): 󵄨󵄨 󵄨󵄨 ∂γ ̃ t)‖C1,λ (Ω̃ (m) (t)) + ‖γ̃−1 (⋅, t)‖C1,λ (Ω(m) ) ⩽ K. sup 󵄨󵄨󵄨󵄨 (ξ , t)󵄨󵄨󵄨󵄨 + ‖γ(⋅, (3.72) 󵄨 Ω(m) 󵄨 ∂t To see this, it suffices to consider the Cauchy problem the linear system 󵄨 ∂γ i 󵄨󵄨󵄨 󵄨󵄨 = δ ij , ∂ξ j 󵄨󵄨󵄨t=t

2 ∂ ∂γ i ∂υ i ∂γ k , ( )= ∑ ∂t ∂ξ j ∂x k ∂x j k=1

γ = (γ1 , γ2 )

0

̃ and use estimates (3.69) for υ = U[ρ]. Estimates for the inverse to the matrix (∂γ i /∂ξ j ) follow from the estimates for the proper matrix and the fact that the Jacobian |∂γ/∂ξ | preserves its value due to incompressibility. The above provides for the existence of a unique solution of (3.71) and an explicit representation for solution of (3.55)–(3.57) for 0 < t < T m using (3.71) is [23, Chapter II, § 2] (ε) ϱ̃(ε) (x, t) = ρ0 (γ̃−1 (x, t)). (3.73) This shows that ϱ̃(ε) (x, t) is uniformly bounded: (ε)

ρ− ⩽ ϱ̃(ε) (x, t) = ρ0 (γ̃−1 (x, t)) ⩽ ρ+ .

(3.74)

Now we find the time T m using transformation γ̃ of the domain Ω( 1−ε ) onto the domain 1 Q(ε) (υ; t) = {x ∈ Ω : x = γ(ξ , t; υ; 0), ξ ∈ Ω( 1−ε ) }. 1

By construction, ρ− < ϱ̃(ε) (x, t) < ρ+

for x ∈ Q(ε) (υ; t)

and ϱ̃(ε) (x, t) = ρ± In fact, if x ∈

Ω(m)

for x ∈ Ω \ Q(ε) (υ; t).

\ (Q(ε) (υ; t)), then x = γ(ξ , t; υ; 0)

where ξ ∈ Ω(m) \ (Ω( 1−ε ) )

x = γ(ξ , t; υ; t0 )

where ξ ∈ S±m ,

1

or 0 ⩽ t0 < t.

In both cases ϱ̃(ε) (x, t) = ρ± . Outside of Ω(m) , ϱ̃(ε) (x, t) has been already defined by (3.58). Thus, one defines T m from Q(ε) (υ; t) ⊂ Ω(m)

for 0 < t < T m ,

Q(ε) (υ; T m ) ⋂ S±m ≠ 0.

(3.75)

3.2 A single capillary of an absolutely rigid skeleton: Neumann boundary conditions

| 73

(ε)

The density inherits from ρ0 the existence of bounded derivatives: 󵄨󵄨 󵄨󵄨 ∂ ϱ̃(ε) (ε) (x, t)󵄨󵄨󵄨󵄨 + |∇ϱ̃(ε) (x, t)|) ⩽ K sup|∇ρ0 (x)|. sup (󵄨󵄨󵄨󵄨 ∂t 󵄨 󵄨 x (x,t)

(3.76)

This follows from representation (3.73), equation (3.71), and estimates (3.72). Note, that that due to (3.69) the time range T m is bounded from below independently on ε and ρ̃ ∈ M : for all ρ̃ ∈ M and ε > 0.

T m ⩾ T0 > 0

(3.77)

To prove it we just note that |υ(x, t)| ⩽ K , and due to this restriction cross-sections of Γ (ε) (t) by the planes {x2 = const} are bounded from below by the line x1 = −ε − Kt. This fact results in 1 − 2ε 1 1 T0 ⩾ ⩾ for ε ⩽ . K 2K 4 The same restriction is valid for Γ (−ε) (t). (c) The operator U is evidently continuous and linear. To see that the operator F = ̃ R ∘ U is continuous we must prove the continuity of R[υ] for υ = U[ρ]. (ε) (ε) To do that we consider two different solutions ϱ̃1 = R[υ1 ] and ϱ̃2 = R[υ2 ] to (3.55), (3.56) and put (ε) (ε) ϱ̂ = ϱ̃1 − ϱ̃2 , υ̂ = υ1 − υ2 . (3.78) The function ϱ̂ solves the inhomogeneous transport equation ∂ ϱ̂ (ε) + υ1 ⋅ ∇ϱ̂ = −υ̂ ⋅ ∇ϱ̃2 ∂t

(3.79)

with the homogeneous initial condition. Integration of the last equation along characteristics results in (ε)

̂ ̂ t)| ⩽ T m sup|∇ϱ̃2 (x, t)| ⋅ sup|υ(x, sup|ϱ(x, t)|, (x,t)

(x,t)

(x,t)

(3.80)

which proves the continuity of F. Moreover, estimate (3.76) shows that F is completely continuous, and estimate (3.74) shows that F transforms M into itself. Applying the Schauder fixed point theorem we find at least one fixed point ρ ε of F, which defines a solution {ρ ε , p ε u ε }, u ε = U[ρ ε ] to the problem (3.38)–(3.41), (ε) (3.45)–(3.46) for smooth initial density ρ0 in the time interval (0, T m ) such that for all positive integer m and for all t ∈ (0, T m ) ρ− ⩽ ρ(ε) (x, t) ⩽ ρ+ , ‖p (t)‖L2 (Ω) + ‖u (t)‖ ε

ε

W 1,2 (Ω)

(3.81)

⩽ C‖f (t)‖L2 (Ω) , ε

‖p (t)‖W 1,q (Ω(m) ) + ‖u (t)‖W 2,q (Ω(m) ) + ‖u (t)‖C1,λ (Ω(m) ) ⩽ K‖f (t)‖L q (Ω) , ε

ε

ε

where f ε = ∇p0 − gρ ε e.

ε

(3.82) (3.83)

74 | 3 A joint motion of two immiscible viscous fluids

This solution satisfies equations (3.38) and (3.46) in the usual sense almost everywhere in Ω T = Ω × (0, T), boundary condition (3.41) in the usual sense, and boundary condition (3.45) in the sense of the integral identity ∫(ℙ(u ε (t), p ε (t)) : 𝔻(φ) + f ε ⋅ φ) dx = 0

(3.84)

Ω

for almost all 0 < t < T and for any smooth solenoidal functions φ, vanishing at x ∈ S0 . The density ρ ε is defined as { ρ(ε) (ξ ), ρ(ε) (x, t) = { 0 ρ± , {

x = γ( ξ , t; u ε ; 0),

ξ ∈ Ω( 1−ε ) , 1

x ∈ Ω \ Q(ε) (u ε ; t).

(3.85)

3.2.4 W 1,1 bounds for density For a smooth initial density ρ0 , both velocity and density are smooth in the time interval (0, T m ), and the derivatives ρ i = ∂ρ/∂x i (we omit for simplicity index ε) satisfy the equations 2 ∂ρ0 ∂ρ i (x), + u ⋅ ∇ρ i = ∑ a ij ρ j , ρ i (x, 0) = ∂t ∂x i j=1 where a ij = −∂u j /∂x i , i, j = 1, 2, u = (u1 , u2 ). By the previous estimates (3.84) |a ij (x, t)| ⩽ K,

for i, j = 1, 2,

x ∈ Ω(m) , 0 < t < T m .

(3.86)

Note also that ρ i ≡ 0 in Ω \ Ω(m) . Therefore, ρ i u vanishes on the boundary ∂Ω(m) . We multiply the i-th equation by ρi √(ρ i )2 + δ2

with δ > 0,

and integrate over Ω(m) . The result is 2 d ρi ρj dx, ∫ √(ρ i )2 + δ2 dx + ∫ u ⋅ ∇Ψ(ρ i ) dx = ∫ ∑ a ij dt √(ρ i )2 + δ2 j=1 Ω(m)

Ω(m)

Ω(m)

ρi

where Ψ(ρ i ) = ∫0 (s2 + δ2 )−1/2 ds. Since u is solenoidal and ρ i u = 0 on ∂Ω(m) , ∫ u ⋅ ∇Ψ(ρ i ) dx = − ∫ Ψ(ρ i )∇ ⋅ u dx + ∫ Ψ(ρ i )u ⋅ n dS = 0 Ω(m)

Ω(m)

∂Ω(m)

(here n is the unit normal to ∂Ω(m) ), and we arrive at 2 ρi ρj d dx. ∫ √(ρ i )2 + δ2 dx = ∫ ∑ a ij dt 2 + δ2 √ j=1 (ρ ) i (m) (m) Ω Ω

3.2 A single capillary of an absolutely rigid skeleton: Neumann boundary conditions

| 75

One can sum these identities in i and get the equation 2 2 d ρi ρj dx. ∫ ∑ √(ρ i )2 + δ2 dx = ∫ ∑ a ij dt √(ρ i )2 + δ2 i=1 i,j=1 Ω(m)

Ω(m)

Let 2

z(t) = ∫ ∑ √(ρ i )2 + δ2 dx. Ω(m)

i=1

Then the last relation by means of estimates (3.86) transforms to 2 dz ⩽ ∫ ∑ |a ij | |ρ j | dx ⩽ Kz, dt i,j=1

z(0) = z0 ,

Ω(m)

where 2

z0 = ∫ ∑ √ ( Ω(m)

i=1

∂ρ0 2 ) + δ2 dx ⩽ C ∫(|∇ρ0ε | + 1) dx ⩽ C. ∂x i

(3.87)

Ω

Applying the Gronwall inequality and keeping in mind equalities ρ i (x, t) ≡ 0 in Ω \ Ω(m) , we establish that for each t ∈ (0, T m ) ∫|∇ρ ε (x, t)| dx ⩽ z(t) ⩽ K ∫(|∇ρ0ε | + 1) dx ⩽ K. Ω

(3.88)

Ω

By the transport equation (3.39) 󵄩󵄩 ∂ρ ε 󵄩󵄩 󵄩 󵄩󵄩 ⩽ max |u ε (x, t)| ‖∇ρ ε (t)‖L1 (Ω) . (t)󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩L1 (Ω) x∈Ω(m) 󵄩󵄩 ∂t

(3.89)

Combining this estimate and (3.88), we conclude finally that max‖ρ ε (t)‖L∞ (Ω) ⩽ C, t

󵄩󵄩 ∂ρ ε 󵄩󵄩 󵄩 󵄩 + ‖∇ρ ε (t)‖L1 (Ω) ) ⩽ K‖∇ρ0ε ‖L1 (Ω) ⩽ K. (t)󵄩󵄩󵄩 max (󵄩󵄩󵄩 󵄩󵄩L1 (Ω) 󵄩󵄩 ∂t t

(3.90) (3.91)

3.2.5 Passage to nonsmooth initial data, existence of a regular free boundary Let {u ε , p ε , ρ ε } be the solution of problem (3.38)–(3.41), (3.45)–(3.46) that corresponds to the initial density ρ0ε . The estimates (3.90), (3.91) allow us to extract a subsequence {ρ ε } (we preserve the same notation for this subsequence for the sake of simplicity) which converges strongly in L r (Ω T m ) to ρ(x, t) for some r > 1. This can be done by the embedding theorem for the pair of spaces W 1,1 (Ω T m ) 󳨅→ L r (Ω T m ). Since the densities ρ ε (x, t) satisfy (3.90), the same subsequence {ρ ε } converges strongly in L∞ ((0, T m ); L q (Ω)) for each q > 2. According to estimates (3.82)–(3.84),

76 | 3 A joint motion of two immiscible viscous fluids

this implies for every positive integer m the strong convergence of the corresponding subsequence of velocities {u ε } to some function u in the space L∞ (0, T; W 2,q (Ω(m) )) ∩ L∞ (0, T; C1,λ (Ω(m) )) ∩ L∞ (0, T; W 1,2 (Ω)) with λ = 1 − 2/q and the strong convergence of the corresponding subsequence of pressures {p ε } to some function p in the space L∞ (0, T; L2 (Ω)) ∩ L∞ (0, T; W 1,q (Ω(m) )). Indeed, for ρ = ρ ε1 − ρ ε2 the difference u = u ε1 − u ε2 , p = p ε1 − p ε2 satisfies the problem μ∆u − ∇p + gρe = 0, u = 0,

x∈S , 0

∇ ⋅ u = 0,

ℙ(u, p) ⋅ n = 0,

x ∈ Ω, x ∈ S± ,

and estimates (3.82)–(3.84) for the solution of this system imply the above mentioned convergence. The direct limiting procedure as ε → 0 in equations (3.46), (3.38) and boundary condition (3.41) for functions {u ε , p ε } shows that the pair {u, p} satisfies these equations almost everywhere in Ω T and boundary condition (3.41). Passing to the limit as ε → 0 in the integral identity (3.84) in the form T

∫ ξ(t)(∫(ℙ(u ε (t), p ε (t)) : 𝔻(φ) + f ε ⋅ φ) dx) dt = 0 0

Ω

for any smooth function ξ(t), one sees that the limit functions satisfy the original problem and the boundary condition (3.45) in a weak sense (3.37). It is clear that for all ε > 0 functions u ε and ρ ε satisfy the integral identity (3.38) with initial function ρ0ε (x). The limit there as ε → 0 results in the integral identity (3.38) for functions u and ρ with initial function ρ0 (x). On the other hand, the density ρ ε is defined by (3.85) and ρ ε (x, t) = ρ± for fixed 0 < t < T m outside of Q ε (u ε ; t). By construction the curves Γ (±ε) (u ε ; t) are the part of the boundary ∂Q ε (u ε ; t) and due to (3.72) Γ (±ε) (u ε ; t) → Γ(t) as ε → 0. Let Γ(u ε ; t) = {x ∈ Ω : x = γ(ξ , t; u ε ; 0), ξ ∈ Γ (0) }. Then Γ(u ε ; t) ⊂ Q ε (u ε ; t), and by the same reason Γ(u ε ; t) → Γ(t) = {x ∈ Ω : x = γ(ξ , t; u; 0), ξ ∈ Γ (0) } as

ε → 0.

The desired smoothness of Γ(t) follows from (3.72). Thus, Γ(t) divides Ω into two subdomains Ω± (t) and, due to properties of the obtained limits, ρ ε (x, t) = ρ± outside of Γ(t).

3.3 A single capillary of an elastic skeleton: Neumann boundary conditions

| 77

3.2.6 Existence of the maximal time interval and uniqueness of the solution By definition δ(T m ) ⩾

1 m.

Therefore we may extend characteristics x = γ(ξ , t; u; 0),

ξ ∈ Γ (0) ,

which form Γ(t), up to time T , where δ(t) ↘ 0,

as

t ↗ T.

The uniqueness of the solution is proved in the same way as in [32].

3.3 A single capillary of an elastic skeleton: Neumann boundary conditions We consider a flow of two immiscible viscous fluids with different constant densities in a single capillary Q f ⊂ Q ⊂ ℝ2 , where Q is a unit cube. Suppose for simplicity that Q = {x : − 1 < x i < 1, i = 1, 2},

Q f = {x : − 1 < x1 < 1, − 12 < x2 < 12 } .

In dimensionless variables he evolution of flow is driven by the input pressure and the force of gravity . More precisely, in this problem one has to find the velocity u f (x, t), pressure p f (x, t), and density ρ f (x, t) of the nonhomogeneous liquid in Q f , and displacements u s (x, t) and pressure p s (x, t) of an elastic skeleton in Q s = Q \ Q f from the following system of differential equations ∇ ⋅ ℙf + ρ f e = 0,

∇ ⋅ u f = 0,

x ∈ Q f , 0 < t < T,

∇ ⋅ ℙs + ρ s e = 0,

∇ ⋅ u = 0,

x ∈ Q s , 0 < t < T,

s

dρ f ∂ρ f ∂ρ f ≡ + ∇ ⋅ (ρ f u f ) = + u f ⋅ ∇ρ f = 0, dt ∂t ∂t

x ∈ Q f , 0 < t < T,

(3.92) (3.93)

where ℙf = 2μ𝔻(u f ) − p f 𝕀,

𝔻(u f ) =

1 (∇u f + (∇u f )∗ ), 2

ℙs = 2λ𝔻(u s ) − p s 𝕀, μ = const is a viscosity of liquids, λ = const is a Lamé’s coefficient, e is a given vector, ρ s is a density of the solid body, and 𝕀 is a unit tensor. The mass and momentum conservation laws dictate the coincidence of velocities and normal tensions in the liquid and solid components uf =

∂u s , ∂t

ℙf ⋅ n = ℙs ⋅ n

on the common boundary S = ∂Q f ∩ ∂Q s with unit normal vector n.

(3.94)

78 | 3 A joint motion of two immiscible viscous fluids The boundary condition on the lateral part S0 = {x2 = ±1} of the boundary ∂Q for 0 < t < T has the form u s (x, t) = 0. (3.95) At the “entrance” and “exit” boundaries S± = {x ∈ ∂Q : x1 = ∓1} ℙs ⋅ e1 = −p+ (x)e1 , x ∈ S+s , x ∈ S−s ,

ℙs ⋅ e1 = 0,

ℙf ⋅ e1 = −p+ (x)e1 , x ∈ S+f , 0 < t < T, x ∈ S−f , 0 < t < T,

ℙf ⋅ e1 = 0,

(3.96)

where p+ (x) is a given function, S±f = S± ∩ ∂Q f , S±s = S± ∩ ∂Q s , and e i is the unit vector of the x i axis for i = 1, 2. To simplify our considerations we pass to the homogeneous boundary conditions at S± ℙi ⋅ e1 = 0, x ∈ S±i , i = f, s, 0 < t < T, (3.97) by introducing a new pressure p f → p f − p0 (x), p0 (x) = 12 p+ (x)(1 − x1 ).

(3.98)

With this new pressure the dynamics equations take the form ∇ ⋅ ℙf + f + ρ f e = 0,

∇ ⋅ u f = 0,

x ∈ Q f , 0 < t < T;

∇ ⋅ ℙs + f = 0,

∇ ⋅ u = 0,

x ∈ Q s , 0 < t < T,

s

(3.99)

where f (x) = (1 − χ(x))ρ s e + ∇p0 (x)

(3.100)

and χ(x) = 1, for x ∈ Q f ,

and

χ(x) = 0, for x ∈ Q s .

Finally u s (x, 0) = 0,

x ∈ S0 .

(3.101)

The initial and boundary conditions for density are equivalent to specifying the surface Γ0 that separates two subdomains Q±f (0) initially occupied by different fluids. For the sake of simplicity we suppose that Γ (0) = {x ∈ Q f : x1 = h(x2 ), − 12 < x2 < 12 }

(3.102)

and −

1 2

+ δ < h(x2 )
t0 ,

x|t=t0 = ξ

(3.106)

determines a mapping x = γ(ξ , t; u f ; t0 ),

γ : Qf → Qf .

(3.107)

In particular, the free boundary Γ(t) is determined as a set Γ(t) = {x ∈ Q f : x = γ(ξ , t; u f ; 0),

ξ ∈ Γ(0)},

and subdomains Q±f (t) = {x ∈ Q f : ρ f (x, t) = ρ± } as sets Q±f (t) = {x ∈ Q f : x = γ(ξ , t; u f ; 0), ξ ∈ Q±f (0)} ∩ {x ∈ Q f : x = γ(ξ, t; u f ; t0 ), ξ ∈ S±f , t0 > 0}. The problem treated here is that of finding the velocity u f (x, t) and pressure p f (x, t) of the liquid in pores, the displacement u s (x, t) and pressure p s (x, t) of the solid skeleton, and the density ρ f (x, t) of the liquid from the above equations and initial and boundary data. Note that it is nonlinear because of the coupling term u f ⋅ ∇ρ f in (3.93). It is shown below that the evolution described by the above equations preserves the existence of two subdomains Q±f (t), each occupied by one of the fluids, that are separated at time t > 0 by a regular free boundary Γ(t). Thus, the problem studied is equivalent to finding {u f , p f , u s , p s }, and the moving boundary Γ(t).

80 | 3 A joint motion of two immiscible viscous fluids

3.3.1 The main result Throughout this section, we use the customary notation of function spaces and norms (see e.g. [28]). Thus, for 1 < q < ∞ 1 q

u ∈ L q (Ω) ⇒ ‖u‖q,Ω = (∫|u|q dx) < ∞, Ω

u ∈ L∞ (Ω) ⇒ ‖u‖∞,Ω = lim ‖u‖q,Ω < ∞, q→∞

u∈

W q1 (Ω)

⇒

(1) ‖u‖q,Ω

󵄨󵄨 ∂u 󵄨󵄨q q 󵄨 󵄨󵄨 = (∫|u| dx) + ∑ (∫ 󵄨󵄨󵄨 󵄨󵄨 dx) < ∞, 󵄨󵄨 ∂x i 󵄨󵄨 i=1 1 q

q

1

2

Ω

Ω

u ∈ W̊ 1q (Ω) ⇒ u ∈ W q1 (Ω), u∈

W ql (Ω)

⇒

(l) ‖u‖q,Ω

and

u(x) = 0,

Dm u =

∂ m1 x 1 ⋅ ⋅ ⋅ ∂ m n x n

,

1 q

= (∫|u| dx) + ∑ (∫|D u| dx) < ∞, q

m

q

|m|=l Ω

Ω

∂|m| u

x ∈ ∂Ω,

1 q

m = (m1 , . . . , m n ),

m i ⩾ 0,

|m| = m1 + ⋅ ⋅ ⋅ + m n .

Next we introduce the space of functions with noninteger derivatives. To make it easy we consider half-spaces ℝ2f = {x = (x1 , x2 ) ∈ ℝ2 : |x1 | < ∞, x2 > 12 } , ℝ2s = {x = (x1 , x2 ) ∈ ℝ2 : |x1 | < ∞, x2 < 12 } , with the boundary ℝ = {x ∈ ℝ2 : |x1 | < ∞, x2 = 12 } . l− 1

The space W2 2 (ℝ) is a space of all functions υ(x1 ) with a finite norm (l− 1 ) ‖υ‖2,ℝ2

∞

= ( ∫ |ξ|

2l−1

1 2

̂ |υ(ξ)| dξ ) , 2

−∞

where υ̂ is a Fourier transformation of υ: ∞

1 ̂ = υ(ξ) ∫ υ(x1 )e−iξx1 dx1 . √2π −∞

According to [28, Chapter 2, Theorem 2.3] (l− 1 )

(l− 1 )

(l)

‖υ‖2,ℝ2 ⩽ C1 ‖υ‖2,ℝ2 ⩽ C2 ‖υ‖2,ℝ2 , j

j = f, s.

For smooth functions we define the following norms: (0)

(α)

|u|Ω = sup|u(x)|, ⟨u⟩Ω = sup x∈Ω

x∈Ω

|u(x) − u(y)| . |x − y|α

(3.108)

3.3 A single capillary of an elastic skeleton: Neumann boundary conditions |

81

We say that the function u(x) belongs to the space C α (Ω), if (α)

(0)

(α)

|u|Ω = |u|Ω + ⟨u⟩Ω < ∞, the function u(x) belongs to the space C k (Ω), if k

(k)

(0)

|u|Ω = ∑ |D m u|Ω < ∞, |m|=0

and the function u belongs to the space C k+α (Ω), if (k+α)

|u|Ω

k

(k)

(α)

= |u|Ω + ∑ |D m u|Ω < ∞. |m|=0

We say that the surface Γ ∈ Ω is C k+α regular, if in local coordinates it presents by C k+α regular functions. If u = u(x, t) and u(x, t) ∈ 𝔹 for all 0 < t < T , then T q

u ∈ L q ((0, T); 𝔹) ⇐⇒ ∫‖u(⋅, t)‖B dt < ∞, 0

and for q = ∞ u ∈ L∞ ((0, T); 𝔹) ⇐⇒ sup ‖u(⋅, t)‖B < ∞. 0 1,

J(s) = J(−s),

J ∈ C∞ (−∞, +∞),

∫ J(|x|) dx = 1, ℝ2

and consider the following approximate problem: ∇ ⋅ ℙεf + ρ εf e + f = 0, ∇ ⋅ ℙεs + f = 0,

∇ ⋅ u s,ε = 0,

u f,ε = ℙεi ⋅ e1 = 0,

∂u s,ε

∂t x ∈ S±i ,

∇ ⋅ u f,ε = 0,

x ∈ Q s , 0 < t < T,

ℙεf ⋅ n = ℙεs ⋅ n,

,

x ∈ Q f , 0 < t < T,

i = f, s,

x ∈ S,

x ∈ S, 0 < t < T,

u s,ε (x, t) = 0,

ℙf (u f,ε , p εf ) = 2μ𝔻(u f,ε ) − p εf 𝕀,

u s,ε (x, 0) = 0,

(3.113)

x ∈ S0 , 0 < t < T,

ℙs (u s,ε , p εs ) = 2λ𝔻(u s,ε ) − p εs 𝕀,

∂ρ ε + υ ε ⋅ ∇ρ ε = 0, x ∈ Q, 0 < t < T, ∂t ρ ε (x, t) = ρ±0 , x ∈ S±f , 0 < t < T, ρ ε (x, 0) = ρ0ε (x), (1)

(3.114)

x ∈ Q,

(2)

υ ε = M ε (M ε (u f,ε )). Here,

∞

(1)

M ε (υ) =

1 |t − τ| ) υ(x, τ) dτ. ∫J( ε ε 0

Definition 3.1. We say that a set of functions {u f,ε , p εf , u s,ε , p εs , ρ ε } u i,ε ∈ L∞ ((0, T); W21 (Q i )), ρ ε ∈ C1 (Q T ),

p εi ∈ L∞ ((0, T); L2 (Q i )),

i = f, s,

Q T = Q × (0, T),

is a weak solution of the problem (3.113), (3.114), if it satisfies the integral identity ∫ ℙf (u f,ε , p εf ) : 𝔻(φ) dx + ∫ ℙs (u s,ε , p εs ) : 𝔻(φ) dx Qf

Qs

= ∫ ρ ε (e ⋅ φ) dx + ∫ f ⋅ φ dx, Qf

Q

(3.115)

84 | 3 A joint motion of two immiscible viscous fluids for almost all t ∈ (0, T), and for arbitrary smooth functions φ(x), vanishing at S0 , and the problem (3.114) in the usual sense. To solve the problem (3.113), (3.114) we use the Schauder fixed point theorem [64]. Let M be the set of all continuous functions ρ̃ ∈ C(G),

G = Q f × (0, T),

such that ̃ t) ⩽ ρ+ . ρ− ⩽ ρ(x,

(3.116)

For fixed ε > 0 we define the following nonlinear operator: ̃ ρ = Φ(ρ),

Φ : M → M,

(3.117)

where {u f , p f , u s , p s , ρ} is a weak solution of the nonlinear boundary-value problem ∇ ⋅ ℙf + ρ̃ ε e + f = 0, ∇ ⋅ ℙs + f = 0,

∇ ⋅ u = 0,

uf = ℙi ⋅ e1 = 0,

s

∂u s

,

∇ ⋅ u f = 0,

x ∈ Q f , 0 < t < T,

x ∈ Q s , 0 < t < T,

ℙf ⋅ n = ℙs ⋅ n,

∂t x ∈ S±i , i = f, s,

u s (x, 0) = 0,

x ∈ S,

x ∈ S, 0 < t < T,

u s (x, t) = 0,

x ∈ S0 , 0 < t < T,

∂ρ + υ ε ⋅ ∇ρ = 0, x ∈ Q, 0 < t < T, ∂t ρ(x, t) = ρ±0 , x ∈ S±f , 0 < t < T, ρ(x, 0) = ρ0ε (x),

(3.118)

(3.119)

x ∈ Q,

where (1)

̃ ρ̃ ε (x, t) = M ε (ρ), ℙf = 2μ𝔻(u f ) − p f 𝕀,

(1)

(2)

υ ε = M ε (M ε (u f )), ℙs = 2λ𝔻(u s ) − p s 𝕀.

(2)

The properties of the mollifier M ε and continuity equation for u imply the continuity equation for υ ε : ∇ ⋅ υ ε = 0, x ∈ Q f , 0 < t < T. Lemma 3.1. Under conditions of Theorem 3.3 the problem (3.118) has a unique weak solution u i ∈ L∞ ((0, T); W21 (Q i )),

i = f, s,

∂u s ∈ L2 ((0, T); W21 (Q s )), ∂t

3.3 A single capillary of an elastic skeleton: Neumann boundary conditions |

85

satisfying the following estimates: T

∫ ∫ 𝔻(u f ) : 𝔻(u f ) dx dt + max ∫ 𝔻(u s (x, t)) : 𝔻(u s (x, t)) dx ⩽ C, 0