Equations of Motion for Incompressible Viscous Fluids: With Mixed Boundary Conditions [1 ed.] 3030786587, 9783030786588

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Advances in Mathematical Fluid Mechanics

Tujin Kim Daomin Cao

Equations of Motion for Incompressible Viscous Fluids With Mixed Boundary Conditions

Advances in Mathematical Fluid Mechanics Series Editors Giovanni P. Galdi, University of Pittsburgh, Pittsburgh, USA John G. Heywood, University of British Columbia, Vancouver, Canada Rolf Rannacher, Heidelberg University, Heidelberg, Germany

The Advances in Mathematical Fluid Mechanics series is a forum for the publication of high-quality, peer-reviewed research monographs and edited collections on the mathematical theory of fluid mechanics, with special regards to the Navier-Stokes equations and other significant viscous and inviscid fluid models. Titles in this series consider theoretical, numerical, and computational methods, as well as applications to science and engineering. Works in related areas of mathematics that have a direct bearing on fluid mechanics are also welcome. All manuscripts are peer-reviewed to meet the highest standards of scientific literature.

More information about this series at http://www.springer.com/series/5032

Tujin Kim Daomin Cao •

Equations of Motion for Incompressible Viscous Fluids With Mixed Boundary Conditions

Tujin Kim Institute of Mathematics State Academy of Sciences Pyongyang Democratic People’s Republic of Korea

Daomin Cao Guangzhou University and Institute of Applied Mathematics Chinese Academy of Sciences Guanzhou, Beijing, China

ISSN 2297-0320 ISSN 2297-0339 (electronic) Advances in Mathematical Fluid Mechanics ISBN 978-3-030-78658-8 ISBN 978-3-030-78659-5 (eBook) https://doi.org/10.1007/978-3-030-78659-5 Mathematics Subject Classification: 35Q30, 76D03, 76D05, 49J40, 80A20, 47J20, 35A02, 35A15, 76D07, 35R35, 35J87 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This book is published under the imprint Birkhäuser, www.birkhauser-science.com by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

The study of the equations appearing in fluid mechanics is a core topic in the fields of partial differential equations, fluid mechanics and applied sciences. Lots of real phenomena of fluid are described by those equations with different kinds of boundary conditions. We are concerned with various boundary conditions, and nowadays for mathematical analysis of phenomena of fluid, the fluid equations with new boundary conditions are studied. In practice, we are concerned with mixture of various boundary conditions. Now there are many papers dealing with fluid equations with mixed boundary conditions; however, almost all monographs deal with the equations with Dirichlet boundary condition and a few monographs deal with mixture of Dirichlet and stress or Dirichlet and a kind of friction boundary conditions. The purpose of this book is to introduce the recent results and research methods in the field of fluid equations with mixed boundary conditions via one book. For readers broadly ranging from students to engineers and mathematicians involved in the fluid equations, first the fluid equations and the boundary conditions for the Navier-Stokes equations are outlined in the level of senior university students. Then, the recent results for the Navier-Stokes equations and the equations of motion for fluid conducting heat with the mixed boundary conditions originated mainly from authors’ results and the first author’s lectures for postgraduate students in the Institute of Mathematics, State Academy of Sciences, DPR Korea are described in the level of postgraduate students. The style of this book is rather different from other mathematical monographs for the Navier-Stokes equations in the following sense: 1. Many monographs for the fluid equations concentrate on mathematical properties of the equations, whereas this book first intends to show how the fluid equations appearing in the journals are obtained and what kinds of boundary conditions are suitable in various practical situations. 2. In the existing monographs, according to the bilinear forms used for variational formulations of the problems, category of mixture of boundary conditions for fluid is determined, whereas this book almost freely deals with mixture of the

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boundary conditions for fluid relying on some properties on the boundaries of vector fields given on domains, which is the main point to make this book different from others. 3. Many monographs concentrate on mathematical investigation of the formulated variational problems, whereas this book gives attention to equivalence between the original partial differential equations with mixture of various boundary conditions and the corresponding variational problems, especially variational inequalities with one unknown, which is not obvious due to mixture of complicated boundary conditions. The book consists of 10 chapters. Trying to make our book self-contained, in Chap. 1, we give some basic knowledge and known results in analysis used in the book. We do not give the best result for every subject, but sometimes we give results more than just required for subsequent parts of the book to help readers to understand other literature. Readers need not read the whole chapter, and can use the chapter as data to consult when something is quoted in the book. The theory of pseudo-monotone operator in Sect. 1.5 and convex functional in Sect. 1.6 is needed only in Chaps. 7, 9 and 10. In Chap. 2, we first show how the Navier-Stokes equations and the equations of motion for fluid under consideration of heat are derived from the fundamental physical laws. Though the aim of this book is to study the equations of motion for the incompressible fluids, the equations for compressible fluids are discussed together to widen reader’s understanding. Next, we outline some boundary conditions for the Navier-Stokes equations, which would be helpful for readers to understand the meaning of the boundary conditions in this book and to improve the ability to apply these in practice. Then, we consider the widely used existing three kinds of variational formulations for the Navier-Stokes problems with mixed boundary conditions. Also, equivalence between the variational formulations and the original PDE problems is discussed, which is a preparation to understand equivalence between PDE problems with more complicated boundary conditions and their variational formulations in the subsequent chapters. This chapter can serve as an introduction for the students and postgraduate students in the fields of equations describing fluids, mathematical modeling and numerical simulation. In Chap. 3, we first study some properties on the boundary surfaces of vector fields given on domains. To this end, a little knowledge of differential geometry is used, which is contained in the chapter. This is preparation to embed wide various mixed boundary conditions together into variational formulations and the foundation for the whole book, which is the peculiar features of this book different from others. Based on the results, we can consider mixtures of boundary conditions which were not studied in other literature, for example, mixture of the boundary conditions for stress and pressure which is important in practice. Thus, we study the existence and uniqueness of solutions to the steady Navier-Stokes system with mixed boundary conditions which may include seven kinds of boundary conditions together. According to whether the static pressure or total pressure is included in the

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boundary conditions, the research methods and results are different, and so these cases are distinguished in this and next chapters. In Chap. 4, we study the existence and uniqueness of a solution to the non-steadyNavier-Stokes system with mixed boundary conditions corresponding to the steady problem in Chap. 3. In Chap. 5, we study the steady Navier-Stokes system with mixed boundary conditions including friction-type conditions, which may include 11 kinds of boundary conditions together. Unlike Chaps. 3 and 4, due to the friction boundary conditions, the problems are described by the variational inequalities. Sometimes variational inequalities are used as approximate mathematical models for real phenomena; however, in this part equivalence between the variational inequalities with one unknown and the original boundary value problem of PDE with 11 kinds of boundary conditions is studied concretely. Relying on the results for the variational inequalities, we study the existence and uniqueness of solutions to the problems under consideration. In Chap. 6, we study the non-steadyNavier-Stokes system with mixed boundary conditions corresponding to the steady problem of Chap. 5. The problems are reduced to the non-steady variational inequalities. Chapter 7 is devoted to the study of the steady Boussinesq system for heat convection of fluid under mixed boundary conditions including friction conditions. The boundary conditions for fluid are the same as the one in Chap. 5 and the boundary conditions for temperature may include Dirichlet, Neumann and Robin conditions together. The problem is formulated by a simultaneous system of a steady variational inequality for velocity of fluid and a variational equation for temperature. Chapter 8 is devoted to the study of the non-steady problem corresponding to the steady problem in Chap. 7. In Chap. 9, we are concerned with the equation for steady flow of heat conducting incompressible Newtonian fluids with dissipative heating by Joule effect under mixed boundary conditions. The boundary conditions are the same as in Chap. 7. Chapter 10 is devoted to the non-steady problem corresponding to Chap. 9, but the boundary conditions for fluid are restricted to the case of total pressure. At the end of every chapter from Chap. 2, the bibliographical remark for the problem in the chapter is given, which is to provide more information on the research of the problem studied in the chapter. We hope the book is helpful for readers to study the research methods and results for the equations for fluid with mixed boundary conditions and to improve the ability to apply the knowledge. Pyongyang, DPR Korea Guangzhou, P.R. China May 2020

Tujin Kim Daomin Cao

Contents

1

Miscellanea of Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Banach Space, Fixed Point and Basics of Mapping . . . . 1.1.1 Banach Space . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.2 Fixed-Point Theorems . . . . . . . . . . . . . . . . . . . 1.1.3 Basics of Mappings . . . . . . . . . . . . . . . . . . . . 1.2 Lebesgue Space, Convergence . . . . . . . . . . . . . . . . . . . 1.2.1 Lebesgue Space . . . . . . . . . . . . . . . . . . . . . . . 1.2.2 Convergence of Sequences of Functions . . . . . 1.3 Sobolev Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.1 Definition of Sobolev Space . . . . . . . . . . . . . . 1.3.2 Density and Continuation . . . . . . . . . . . . . . . . 1.3.3 Imbedding . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.4 Trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.5 Some Inequalities . . . . . . . . . . . . . . . . . . . . . . 1.4 Space of Abstract Functions . . . . . . . . . . . . . . . . . . . . 1.4.1 Abstract Functions and Its Derivatives . . . . . . . 1.4.2 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Operator Equations and Operator-Differential Equations 1.5.1 Monotone Operator Equation . . . . . . . . . . . . . 1.5.2 Pseudo-Monotone Operator Equation . . . . . . . . 1.5.3 Operator-Differential Equations . . . . . . . . . . . . 1.6 Convex Functional . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Some Elementary Inequalities . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Fluid Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Derivation of Equations for Fluid Motion . . . . . . . . . . . . . . . 2.1.1 Navier-Stokes Equations . . . . . . . . . . . . . . . . . . . . . 2.1.2 Equations of Motion for Fluid Under Consideration of Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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2.2

Boundary Conditions for the Navier-Stokes Equations . . 2.2.1 Boundary Conditions on the Walls . . . . . . . . . . 2.2.2 Boundary Conditions on Symmetric Planes . . . . 2.2.3 Boundary Conditions on Inlets and Outlets . . . . 2.2.4 Outflow Boundary Conditions on Imaginary Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.5 Boundary Conditions on Free Surfaces . . . . . . . 2.3 Bilinear Forms for Hydrodynamics . . . . . . . . . . . . . . . . 2.3.1 Bilinear Forms . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Variational Formulations for Mixed Boundary Value Problems of the Navier-Stokes Equations . 2.4 Bibliographical Remarks . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Fluid Equations . . . . . . . . . . . . . . . . . . . . . . . . 2.4.2 Boundary Conditions of the Navier-Stokes Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.3 Bilinear Forms for Hydrodynamics . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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The Steady Navier-Stokes System . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Properties on the Boundary Surfaces of Vector Fields . . . . . . 3.1.1 The Second Fundamental Form and Shape Operator of Surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.2 Properties on the Boundary Surface of Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Variational Formulations of the Steady Problems . . . . . . . . . 3.3 Existence of Solutions to the Steady Problems . . . . . . . . . . . 3.4 Bibliographical Remark . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Non-steady Navier-Stokes System . . . . . . . . . . . . . . . . . . . 4.1 Existence of a Solution: The Case of Total Pressure . . . . . . 4.1.1 Problem and Variational Formulation . . . . . . . . . . 4.1.2 An Auxiliary Problem by Elliptic Regularization . . 4.1.3 Proof of the Existence of a Solution . . . . . . . . . . . 4.1.4 The Stokes Problem . . . . . . . . . . . . . . . . . . . . . . . 4.2 Existence and Uniqueness of Solutions: The Case of Static Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Existence and Uniqueness of Solutions to Problem I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Existence and Uniqueness of Solutions to Problem II . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.3 Existence and Uniqueness of Solutions for Perturbed Data . . . . . . . . . . . . . . . . . . . . . . . .

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4.3 Bibliographical Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 5

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The Steady Navier-Stokes System with Friction Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Variational Formulations of Problems . . . . . . . . . . . . . 5.1.1 Variational Formulation: The Case of Static Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.2 Variational Formulation: The Case of Total Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.3 Variational Formulation: The Stokes Problem . 5.2 Solutions to Variational Inequalities . . . . . . . . . . . . . . . 5.3 Existence and Uniqueness of Solutions to the Steady Navier-Stokes Problems . . . . . . . . . . . . . . . . . . . . . . . 5.4 Bibliographical Remarks . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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The Non-steady Navier-Stokes System with Friction Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Variational Formulations of Problems . . . . . . . . . . . . . . . . 6.1.1 Variational Formulation: The Case of Total Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.2 Variational Formulation: The Case of Static Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.3 Variational Formulation: The Stokes Problem . . . . 6.2 The Existence and Uniqueness of Solutions to Variational Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Solutions to the Non-steady Navier-Stokes Problems . . . . . 6.3.1 Existence of a Solution: The Case of Total Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.2 Existence of a Unique Solution: The Case of Static Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.3 Existence of a Unique Solution: The Stokes Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Bibliographical Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Steady Boussinesq System . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Problems and Variational Formulations . . . . . . . . . . . . . . . 7.1.1 Variational Formulation: The Case of Static Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.2 Variational Formulation: The Case of Total Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Existence and Uniqueness of Solutions: The Case of Static Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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7.2.1 7.2.2

Existence of a Solution to an Auxiliary Problem Existence and Estimates of Solutions to the Approximate Problem . . . . . . . . . . . . . . . . . . . . 7.2.3 Existence and Uniqueness of a Solution . . . . . . 7.3 Existence of a Solution: The Case of Total Pressure . . . . 7.4 Bibliographical Remarks . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

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The Non-steady Boussinesq System . . . . . . . . . . . . . . . . . . . . . . 8.1 Problems and Assumptions . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Variational Formulations for Problems . . . . . . . . . . . . . . . . 8.2.1 Variational Formulations: The Case of Static Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.2 Variational Formulations: The Case of Total Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Existence and Uniqueness of Solutions: The Case of Static Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.1 Existence and Estimation of Solutions to an Approximate Problem . . . . . . . . . . . . . . . . . 8.3.2 Existence and Uniqueness of a Solution . . . . . . . . 8.4 Existence of a Solution: The Case of Total Pressure . . . . . . 8.4.1 Existence of a Solution to an Approximate Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.2 Existence of a Solution . . . . . . . . . . . . . . . . . . . . . 8.5 Bibliographical Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Steady Equations for Heat-Conducting Fluids . . . . . . . . . . 9.1 Problems and Assumptions . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Variational Formulations for Problems . . . . . . . . . . . . . . . . 9.2.1 Variational Formulation: The Case of Static Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.2 Variational Formulation: The Case of Total Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Existence and Uniqueness of Solutions: The Case of Static Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.1 Existence of a Solution to an Auxiliary Problem . . 9.3.2 A Priori Estimates of Solutions to the Auxiliary Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.3 Passing to Limits . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Existence of a Solution: The Case of Total Pressure . . . . . . 9.5 Bibliographical Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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10 The Non-steady Equations for Heat-Conducting Fluids 10.1 Problem and Variational Formulation . . . . . . . . . . . 10.1.1 Problem and Assumption . . . . . . . . . . . . . 10.1.2 Variational Formulation for Problem . . . . . 10.2 Existence of a Solution . . . . . . . . . . . . . . . . . . . . . 10.2.1 Existence of a Solution to an Approximate Problem . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.2 Estimates of Solutions to the Approximate Problem . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.3 Passing to the Limit . . . . . . . . . . . . . . . . . 10.3 Bibliographical Remarks . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361

Chapter 1

Miscellanea of Analysis

In this chapter, we outline some knowledge of analysis: Banach space, fixed point, Lebesgue and Sobolev spaces, operator and operator-differential equations and convex functional, which will be used in the main part of this book. We do not describe the best results, but to help readers’ understanding sometimes we say more than necessary. The readers who are already acquainted with the elements of functional analysis can skip this chapter and may consult the necessary parts when reading main part of this book.

1.1 Banach Space, Fixed Point and Basics of Mapping 1.1.1 Banach Space Let X be a normed space. A sequence {xn } in X is called a Cauchy sequence iff for every ε > 0 there exists an integer N such that xn − xm  X < ε holds whenever m, n > N . A sequence {xn } in X is convergent to the element x0 if and only if limn→∞ xn − x0  X = 0. A space X is said to be complete if every Cauchy sequence in X converges to an element in X . A complete normed space is called a Banach space. A subset S of a normed space X is said to be dense in M ⊆ X if each x ∈ M is the limit of a sequence of elements of S. The normed space X is called separable if it has a countable dense subset. Denote by X ∗ the space of linear continuous functionals on a normed space X , which is called the dual of X . A norm on the dual X ∗ of a normed space X can be defined by setting x ∗  X ∗ = sup x ∗ , x , x X =1

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 T. Kim and D. Cao, Equations of Motion for Incompressible Viscous Fluids, Advances in Mathematical Fluid Mechanics, https://doi.org/10.1007/978-3-030-78659-5_1

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1 Miscellanea of Analysis

where and in what follows x ∗ , x means the value of a functional x ∗ ∈ X ∗ at an element x ∈ X . Since the fields of real and complex numbers R and C are complete, with the topology induced by this norm X ∗ is a Banach space (no matter whether X is complete). A natural linear injection of a normed space X into its second dual space X ∗∗ = ∗ ∗ (X ) is provided by the mapping J whose value J x at x ∈ X is given by J x, y = y, x ∀y ∈ X ∗ , J is an isometric isomorphism of X into X ∗∗ , and so X ⊂ X ∗∗ . If the range of the isomorphism J is the entire space X ∗∗ , the space X is called reflexive. A reflexive space must be complete and hence a Banach space. For a sequence {xn } of a normed space X if  f, xn − x0 → 0 ∀ f ∈ X ∗ as n → ∞, then {xn } is said to be weakly convergent to x0 . We denote norm convergence of a sequence {xn } to x0 by xn → x0 , and weak convergence by xn  x0 . A sequence {xn } in a normed space X is called a weak Cauchy sequence iff for every ε > 0 and f ∈ X ∗ there exists an integer N such that | f, xn − xm | < ε holds whenever m, n > N . The space X is called weakly complete if every weak Cauchy sequence in X weakly converges to an element in X . For a sequence { f n } of a normed space X ∗ g, f n − f 0 → 0 ∀g ∈ X ∗∗ as n → ∞ means that { f n } weakly converges to f 0 in X ∗ . For a sequence { f n } of a normed space X ∗ if x, f n − f 0 → 0 ∀x ∈ X as n → ∞, ∗

then { f n } is said to be ∗-weakly convergent to f 0 and it is denoted by f n  f . Theorem 1.1 Let M be a convex closed subset of a Banach space X and a sequence {xn } ⊂ M weakly converges to a limit x0 . Then x0 belongs to M, i.e. M is weakly closed. (See Theorem 3.7 of [1].) Theorem 1.2 If a sequence {xn } in a Banach space X weakly converges to an element x, then (1.1) x X ≤ lim inf xn  X . n→∞

(See Theorem 1, Sect. 1, Chap. V of [2].) A weak Cauchy sequence of a Banach space may not have a weak limit in the space, and the Banach space may not be weakly complete. But the following theorem holds.

1.1 Banach Space, Fixed Point and Basics of Mapping

3

Theorem 1.3 Reflexive spaces are weakly complete. In every bounded subset of a reflexive Banach space there exists a weakly convergent sequence. (See Theorem 7, Sect. 1 and Theorem 1, Sect. 2, Chap. V of [2].) Theorem 1.4 If all weakly convergent subsequences of a sequence {xn } of a reflexive space converge to the same element x, then the sequence {xn } weakly converges to x. (See Lemma 5.4, Chap. 1 of [3].) Imbedding of a normed space X into a linear normed space Y means that settheoretic inclusion X ⊂ Y , for which the following inequality is valid x X ≤ CxY ∃C, ∀x ∈ X. We denote by X → Y the imbedding. If X is dense in Y , then f ∈ Y ∗ may be identified with an element in X ∗ , and so we have Y ∗ → X ∗ . If X is reflexive, then this imbedding is dense (see Remark 5.14, Chap.1 of [3]).

1.1.2 Fixed-Point Theorems Theorem 1.5 (Banach fixed-point theorem) Let (X, d) be a complete metric space, M a closed nonempty set of X , and T : M ⊂ X → M be a map such that d(T x, T y) ≤ kd(x, y) ∃k(0 ≤ k < 1), ∀x, y ∈ M. Then, there exists a unique point x ∈ M such that T x = x. Let X be a topological space and M ⊆ X . The set M is called compact iff every open covering of M contains a finite subcover, i.e. finitely many of these open sets already cover M. The set M is called relatively compact iff its closure M is compact. Let X be a complete metric space and M ⊆ X . Then M is relatively compact if and only if for every ε > 0 there is a finite ε-net for M. Let X be a metric space. Then M ⊆ X is relatively compact if and only if for any sequence {xn }∞ n=1 ⊂ M there is a convergent subsequence (see Proposition 1.2.1 of [4]). Definition 1.1 Let X, Y be normed spaces, and T : M ⊆ X → Y an operator. T is called compact if it maps every bounded subset of M into a relatively compact set of Y . If, in addition, T is continuous on M, then it is called completely continuous on this set. When X, Y are normed spaces, for linear operators T the concepts of a compact and a completely continuous operators are the same since every compact linear operator is continuous.

4

1 Miscellanea of Analysis

Remark 1.1 Note that in some literatures the completely continuous operator is called compact. Theorem 1.6 (The Schauder fixed-point theorem) If a completely continuous operator T maps a bounded closed convex set M of a normed linear space X into itself, then there exists at least one point x ∈ M such that T x = x. (See Theorem 5.1.11 of [4].)

1.1.3 Basics of Mappings Let T be a single-valued mapping from a set M to N . Mapping T : M → N is said to be surjective if each element of N is the image of a certain element of M; injective if different elements of M are mapped into different elements of N ; bijective if the mapping is both surjective and injective simultaneously. If T : M → N is bijective, then the mapping is an one-to-one map between M and N . If T : M → N is continuous and bijective, and if its inverse map T −1 : N → M is also continuous, then T is called a homeomorphism and M and N are said to be homeomorphic. Definition 1.2 Let X, Y be Banach spaces and f : U (x) ⊆ X → Y be a map, where U (x) is a neighborhood of x. (1) The map f is said to be Gâteaux differentiable at x iff there exists a linear continuous operator T (x) : X → Y such that f (x + th) − f (x) = t T (x)h + o(t) (t → 0)

(1.2)

for all h with h X = 1 and all real numbers t in some neighborhood of zero. If it exists, this T (x) is called the Gâteaux derivative at x. Put T (x) = f  (x). The Gâteaux differential at x is defined by dG f (x; h) = f  (x)h. (2) The map f is said to be Fréchet differentiable at x iff there exists a linear continuous operator T (x) : X → Y such that f (x + h) − f (x) = T (x)h + o(h X ) (h → 0 X ).

(1.3)

T (x) is called the Fréchet derivative at x and T (x)h is called the Fréchet differential at x. The Gâteaux derivative f  (x) of f at x can be defined equivalently by f  (x)h = lim

t→0

f (x + th) − f (x) . t

Definition 1.3 Let X, Y be Banach spaces, f : U ⊂ X → Y and r ∈ N (the set of natural numbers).

1.1 Banach Space, Fixed Point and Basics of Mapping

5

(1) If U is open, then f is called a C r -map iff f has continuous Fréchet derivatives up to order r on U . (2) If U is arbitrary, then f is called a C r -map iff for each x ∈ U there exists an open neighborhood V (x) such that f can be extended to a C r -map on V (x). Definition 1.4 Let M and N be arbitrary sets in the Banach spaces X and Y , respectively. Let 0 ≤ r ≤ ∞. (1) The mapping f : M → N is called a C r -diffeomorphism iff f is bijective and both f and f −1 are C r -maps. (2) A local C r -diffeomorphism at x0 is a C r -diffeomorphism from some neighborhood U (x0 ) in X onto some neighborhood U ( f (x0 )) in Y . Theorem 1.7 (Local inverse mapping theorem) (Theorem 4.F of [5]) Let f : U (x0 ) ⊆ X → Y is a C 1 -mapping, where both X and Y are Banach spaces. Then f is a local C 1 -diffeomorphism at x0 iff f  (x0 ) : X → Y is bijective. Theorem 1.8 (Theorem 3.4 of [6]) Let X, Y be Banach spaces, T0 be a one-to-one linear continuous operator from X onto Y , Tc be a compact linear operator from X into Y . Then T0 + Tc is a one-to-one operator from X into Y if and only if it is an operator from X onto Y .

1.2 Lebesgue Space, Convergence In this section we consider real-valued functions defined on a (Lebesgue) measurable subset Ω ⊂ Rl , where Rl is l-dimensional Euclidean space.

1.2.1 Lebesgue Space A function u : Ω → R ∪ {±∞} is itself called (Lebesgue) measurable if the set {x : u(x) > a} is measurable for every real a. A function u : Ω → R ∪ {±∞} is said to be countably valued if it assumes at most a countable set of values, assuming each  value is on a measurable subset. A countably valued function is written as u(x) = i ai χ Ai , where χ Ai is the characteristic function of the measurable set Ai . A function u(x) is measurable if and only if it is uniform limit almost everywhere of a sequence of countably valued functions. (See Theorem 2, Sect. 5, Chap. V of [7] or p. 30 of Sect. 2.0 of [8].)   A countably valued function u(x) = i ai χ Ai is integrable if i ai mes (Ai ) < ∞, where mes (Ai ) is (Lebesgue) measure of Ai , and the integral of u(x) is defined by   u(x) d x = ai mes (Ai ). I (u) = Ω

i

6

1 Miscellanea of Analysis

A measurable function u(x) is said to be (Lebesgue) integrable if there exists a sequence {u n (x)} of integrable countably valued functions such that u n (x) → u(x) uniformly outside of a null-measure set and  Ω

|u n (x) − u m (x)| d x → 0 as m, n → ∞.

We define (Lebesgue) integral by 

 Ω

u(x) d x = lim

n→∞ Ω

u n (x) d x.

The definition of integral (or of integrability) does not depend on the particular sequence {u n } chosen. The class of integrable functions on Ω is denoted by L 1 (Ω). Theorem 1.9 (Lebesgue) Let Ω ⊂ Rl be measurable and let {u k } be a sequence of measurable functions converging to u a.e in x ∈ Ω. If there exists a function v(x) ∈ L 1 (Ω) such that |u k (x)| ≤ v(x), k = 1, 2, · · · , then u is integrable and  Ω

 u(x) d x = lim

k→∞ Ω

u k (x) d x.

Definition 1.5 Let Ω be an open subset of Rl and let p be a positive real number. The space L p (Ω) consists of all (Lebesgue) measurable real functions u(x) defined on Ω such that  |u(x)| p d x < ∞. Ω

The space L ∞ (Ω) consists of all measurable functions u(x) defined on Ω such that |u(x)| ≤ M ∃M > 0, for a.e. x ∈ Ω L p (Ω), (1 ≤ p < ∞) and L ∞ (Ω) are Banach spaces, respectively, with norms  u p =

Ω

1/ p |u(x)| p d x

(See Theorem 4.8 of [1].)

and u∞ = ess sup |u(x)|. x∈Ω

1.2 Lebesgue Space, Convergence

7

Let Ω be an open subset of Rl . The support of a continuous function u is defined by supp u = {x; u(x) = 0} ∩ Ω. The space C0∞ (Ω) consists of all infinitely differentiable functions that have compact support in Ω. If 1 ≤ p < ∞, then C0∞ (Ω) is dense in L p (Ω) and L p (Ω) is separable. Theorem 1.10 (The for L p (Ω)) Let 1 ≤ p < ∞. ∗ representation theorem  pRiesz q Then for any f ∈ L (Ω) there exists v ∈ L (Ω), 1/ p + 1/q = 1, such that   f, u =

Ω

v(x)u(x) d x ∀u ∈ L p (Ω)

 ∗ and  f (L p (Ω))∗ = v L q (Ω) . The correspondence between f ∈ L p (Ω) and v ∈ L q (Ω) is linear. L p (Ω) is reflexive if and only if 1 < p < ∞. (See Sect. 4.3 of [1].) Theorem 1.11 (Hölder’s inequality) Let 1 < p < ∞ and let q denote the conjugate exponent defined by 1 1 + = 1. p q If u ∈ L p (Ω) and v ∈ L q (Ω), then uv ∈ L 1 (Ω), and  Ω

|u(x)v(x)| d x ≤ u p vq .

Equality holds if and only if |u(x)| p and |u(x)|q are proportional a.e. in Ω. Theorem 1.12 (cf. Theorem 2.11 of [9]) (complex interpolation of Lebesgue spaces) Let 1 ≤ p < q < r , so that 1 1−θ θ = + q r p for some θ satisfying 0 < θ < 1. If u ∈ L p (Ω) ∩ L r (Ω), then u ∈ L q (Ω) and u(L r ,L q )[θ ] ≡ uq ≤ ur1−θ uθp . Here the functor ( , )[θ] means the complex interpolation (see Definition 2.1.3 of [10] or Sect. 4.1 of [11]).

8

1 Miscellanea of Analysis

1.2.2 Convergence of Sequences of Functions An open and connected subset of Rl is called a domain. Let Ω be a domain. Lemma 1.1 (Lemma 1.19, Chap. 2 of [3]) Let 1 ≤ p < ∞ and a sequence {u n } ⊂ L p (Ω) be such that u n  u in L p (Ω), u n (x) → v(x) a.e. in Ω. Then u = v. Lemma 1.2 Let vn and v be functions of L q (Ω), 1 < q < ∞, such that vn  L q (Ω) ≤ C, vn (x) → v(x) a.e. in Ω. Then, vn  v in L q (Ω). Proof Since L q (Ω), 1 < q < ∞, is reflexive, by Theorem 1.3 there exists a subsequence {vk } ⊂ {vn } such that vk  u in L q (Ω). By Lemma 1.1 the limits of all subsequences of {vn } coincide with v. Thus, by Theorem 1.4 we come to the conclusion.  A set of functions F ⊂ L 1 (Ω) is called uniformly integrable if  lim sup

C→+∞ f ∈F

| f |>C

| f | dμ = 0.

A set consisting of a single integrable function is uniformly integrable since the Lebesgue integral is absolutely continuous. Hence, for any integrable function f 0 , the set of all measurable functions f with | f | ≤ | f 0 | is uniformly integrable. The sequence of functions { f n } is said to converge in measure to a measurable function f if, for every c > 0, one has lim mes{x : | f (x) − f n (x)| ≥ c} = 0.

n→∞

In Lebesgue space Vitali’s convergence theorem (see Theorem 4.5.4 of [12]) is written as follows. Theorem 1.13 Suppose that f is a measurable function and { f n } is a sequence of integrable functions. Then, the following assertions are equivalent: (1) the sequence { f n } converges to f in measure and is uniformly integrable; (2) the function f is integrable and the sequence { f n } converges to f in the space L 1 (Ω).

1.2 Lebesgue Space, Convergence

9

Lemma 1.3 Let Ω be a bounded domain of Rl . If {u n } is such that {u n } is bounded in L ∞ (Ω), u n (x) → u(x) a.e. in Ω, then u n → u in L p (Ω) ∀ p, 1 ≤ p < ∞. Proof Sine Ω is bounded, u n is integrable. By the conditions, {u k } converges to u in measure and is uniformly integrable. By Theorem 1.13, 

 Ω

|u n − u| p d x ≤ M

Ω

|u n − u| d x → 0.



Corollary 1.1 Let Ω be a bounded domain of Rl and μ(ξ ), ξ ∈ R, be a bounded continuous function. If {u n } and {vn } are such that u n (x) → u(x) a.e. in Ω, vn  v in L p (Ω), 1 < p < ∞, then μ(u n )vn  μ(u)v in L p (Ω). Proof By Lemma 1.3, μ(u n ) → μ(u) in L q (Ω),

1 1 + = 1. p q

Thus, μ(u n )vn  μ(u)v in L 1 (Ω), that is,   μ(u n (x))vn (x)φ(x) d x → μ(u(x))v(x)φ(x) d x ∀φ ∈ C0∞ (Ω). Ω

Ω

Since {μ(u n )vn  L p } is bounded and C0∞ (Ω) is dense in L p (Ω), from above we get the asserted conclusion (see Theorem 3, Chap. 5 of [2]).  Corollary 1.2 Let Ω be a bounded domain of Rl and μ(ξ ), ξ ∈ R, be a positive bounded continuous function. If {u n } and {vn } are such that u n (x) → u(x) a.e. in Ω, vn  v in L 2 (Ω), then

    μ(u(x))v(x), v(x) ≤ lim inf μ(u n (x))vn (x), vn (x) . n→∞

Proof By Corollary 1.1

μ(u n )vn 



μ(u)v in L 2 (Ω).

10

1 Miscellanea of Analysis

Then,





μ(u)v

L 2 (Ω)



≤ lim inf μ(u n )vn L 2 (Ω) , n→∞



which implies the asserted conclusion. Lemma 1.4 Let Ω be a bounded domain of Rl . If vε  v in L 2 (Ω) as ε → 0, then

vε 1 + εvε2

 v in L 2 (Ω).

Proof Since       2 1   2 + εvε2 − 2 1 + εvε2 d x ≤ ε − 1 d x ≤ vε2 d x,  1 + εvε2 Ω Ω Ω we have

The sequence √ vε



1+εvε2

1 1 + εvε2

→ 1 in L 2 (Ω).

(1.4)

is bounded and C0∞ (Ω) is dense in L 2 (Ω), and to get the

conclusion, it is enough to prove that   Ω

 vε − v ϕ d x → 0 ∀ϕ ∈ C0∞ (Ω) as ε → 0. 1 + εvε2

Using (1.4), we have   Ω



vε 1 + εvε2



− v ϕ dx =





Ω

vε ϕ

∀ϕ ∈

1 1 + εvε2

C0∞ (Ω)

  − 1 d x + (vε − v)ϕ d x → 0 Ω

as ε → 0,

which implies the asserted conclusion. Lemma 1.5 Let Ω be a domain of Rl . If for a p > 1 vε (x) → v(x) ∈ L p (Ω) a.e. in Ω as ε → 0, then

1  1 in L r (Ω) ∀r (1 < r < ∞). 1 + εvε2



1.2 Lebesgue Space, Convergence

11

1 Proof Since { 1+εv 2 } is bounded, and there exists a subsequence weakly convergent ε to a limit. On the other hand,

1  1 a.e. in Ω. 1 + εvε2 Thus, by Lemma 1.1 and Theorem 1.4 we come to the conclusion.



1.3 Sobolev Space To simplify the discussion, in this section, we also work with real-valued functions. Let Ω be an open subset of Rl with boundary ∂Ω and {ρα } be a family continuous functions defined on Ω. The family of supports {supp ρα } is said to be locally finite on Ω if any compact subset of Ω intersects only finitely many elements of {supp ρα }. Let α = (α1 , · · · , αl ), αi —nonnegative integer, |α| = α1 + · · · + αl , and Dα =

∂ x1α1

∂ |α| . · · · ∂ xlαl

Define the family of semi-norm on C0∞ (Ω) pρ (ϕ) =

 α

sup |ρα (x)D α ϕ(x)|, ϕ ∈ C0∞ (Ω),

(1.5)

x∈Ω

where ρ = {ρα } are taken all families of continuous function supports of which are locally finite. By the family of semi-norms, C0∞ (Ω) becomes a locally convex space, which is denoted by D(Ω). Let D ∗ (Ω) denote the space of linear continuous functional defined on D(Ω), often called the space of distributions on Ω. We denote by ·, · the duality pairing between D ∗ (Ω) and D(Ω) and we remark that when f is a locally integrable function, then f can be identified with a distribution by   f, φ =

Ω

f (x)φ(x) d x ∀φ ∈ D(Ω).

D ∗ (Ω) is given the ∗-weak topology as the dual of D(Ω), i.e. the fact that f n → f in D ∗ (Ω) means  f n − f, ϕ → 0 ∀ϕ ∈ D(Ω). For f ∈ D ∗ (Ω), we define D α f ∈ D ∗ (Ω) by D α f, ϕ = (−1)|α|  f, D α ϕ ∀ϕ ∈ D(Ω).

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1 Miscellanea of Analysis

If f n → f in D ∗ (Ω), then for ϕ ∈ D(Ω) lim D α f n − D α f, ϕ = lim (−1)|α|  f n − f, D α ϕ = 0 ∀ϕ ∈ D(Ω).

n→∞

n→∞

Thus, mapping f → D α f is continuous in D ∗ (Ω).

1.3.1 Definition of Sobolev Space Definition 1.6 Let Ω be an open subset of Rl , s—nonnegative real number and 1 ≤ p ≤ ∞. We define the Sobolev space W s, p (Ω) as follows. (i) s = k (a nonnegative integer), 1 ≤ p ≤ ∞. W k, p (Ω) = {u ∈ L p (Ω) : D α u ∈ L p (Ω) ∀|α| ≤ k}, 1/ p  p D α u L p (1 ≤ p < ∞), uW k, p (Ω) = |α|≤k

uW k,∞ (Ω) =



 ess sup |D α u| .

|α|≤k

x∈Ω

(ii) s = k + σ, 0 < σ < 1, k—nonnegative integer, 1 ≤ p < ∞.   W s, p (Ω) = u ∈ W k, p (Ω) :

 |D α u(x) − D α u(y)| p d xd y < ∞, |α| = k , |x − y|n+σ p Ω×Ω ⎛ ⎞1/ p   |D α u(x) − D α u(y)| p p uW s, p (Ω) = ⎝u k, p + d xd y ⎠ . (Ω) W |x − y|n+σ p Ω×Ω |α|=k

(iii) s = k + σ, 0 < σ < 1, k—nonnegative integer, p = ∞. 

 |D α u(x) − D α u(y)| < ∞ , |x − y|σ |α|=k x,y∈Ω, x = y   |D α u(x) − D α u(y)| p + max ess sup . uW s,∞ (Ω) = u k, p W (Ω) |x − y|σ |α|=k x,y∈Ω, x = y

W s,∞ (Ω) = u ∈ L ∞ : u ∈ W k,∞ (Ω), max

ess sup

The Sobolev spaces are Banach spaces for the norms. s, p For s > 0, we denote by W0 (Ω) the closure of C0∞ (Ω) in W s, p (Ω) and by s, p −s,q (Ω) the dual space of W0 (Ω), where p −1 + q −1 = 1. W If a function f is k times continuously differentiable and its derivatives of order k are Lipschitz continuous, then we say it belongs to the class C k,1 . Definition 1.7 (cf. Definition 1.2.1.1 of [13]) Let Ω be a bounded open subset of Rl . We say that its boundary ∂Ω is continuous (resp. Lipschitz continuous, of class C k , of

1.3 Sobolev Space

13

class C k,1 for some integer k ≥ 0) if for every x ∈ ∂Ω there exists a neighborhood O of x in Rl and new orthogonal coordinates y = (y  , yn ) where y  = (y1 , · · · , yn−1 ), such that (i) O is a hypercube in the new coordinates: O = {y : |yi | < ai , 1 ≤ i ≤ n}; (ii) there exists a continuous (resp. Lipschitz continuous, of class C k , of class C k,1 ) function φ defined in O  = {y  : |yi | < ai , 1 ≤ i ≤ n − 1} that satisfies

an ∀y  ∈ O  , 2 Ω ∩ O = {y : yn < φ(y  )}, |φ(y  )| ≤

∂Ω ∩ O = {y : yn = φ(y  )}. Remark 1.2 Differently from Definition 1.18, Chap. 2 of [3], in Definition 1.2.1.1 of [13] and Definition 1.1 of [14] they did not demand boundedness of Ω. It is not enough for Lipschitz continuity and class C k,1 of ∂Ω. If Ω is bounded, there exist finite family of neighborhood covering ∂Ω, and from local properties we can get Lipschitz continuity and C k,1 of ∂Ω. To examine the boundary values of functions in W s, p (Ω), we consider the spaces of functions defined on open subset Γ of boundary ∂Ω. We assume that Ω is a bounded open subset of Rl with a boundary ∂Ω that is at least Lipschitz continuous. We can view Γ locally as an n − 1-dimensional submanifold of Rl by means of the mapping (1.6) Φ(y  ) = (y  , φ(y  )) from O  onto Γ ∩ O, where φ, O  and O are the ones in Definition 1.7. We assume that ψ is at least a bi-Lipschitz mapping from Ω1 onto Ω2 where Ω1 and Ω2 are bounded open subsets of Rl . This hypothesis ensures that Lebesgue measure is mapped by ψ or ψ −1 to an equivalent image measure. Let u ∈ W s, p (Ω2 ); assume that ψ and ψ −1 are of class C k,1 , where k is an integer ≥ s − 1. Then u ◦ Φ ∈ W s, p (Ω1 ) (Lemma 1.3.3.1 of [13]). It is a justification for the following definition. Definition 1.8 (Definition 1.3.3.2 of [13]) Let Ω be a bounded open subset of Rl with a boundary ∂Ω of class C k,1 , where k is a nonnegative integer. Let Γ be an open subset of ∂Ω. A distribution u on Γ belongs to W s, p (Γ ) with |s| ≤ k + 1 if u ◦ Φ ∈ W s, p O  ∩ Φ −1 (Γ ∩ O) for all possible O and φ fulfilling the assumptions in Definition 1.7.

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1 Miscellanea of Analysis

Let Φ j is defined by (1.6) with φ j instead of φ. Let (O j , Φ j ) Jj=1 be any atlas of Γ such that each couple (O j , Φ j ) satisfies the assumptions of the above definition. Then one possible Banach norm for W s, p (Γ ) is the functional: u → u

W s, p (Γ )

=

J  j=1

u ◦ Φ j W s, p V  ∩Φ −1 (Γ ∩V ) j

j

1/ p

j

.

(1.7)

In the particular case when s ∈ (0, 1), one can easily check that any of the norms defined above is equivalent to 

 |u| dσ + p

Γ

Γ ×Γ

1/ p

|u(x) − u(y)| p dσ (x)dσ (y) |x − y|n−1+sp

.

1.3.2 Density and Continuation Unless otherwise stated we assume that 1 < p < ∞ in the following subsections of this section. Let Ω be open subset of Rl and D(Ω) = {φ|Ω : φ ∈ D(Rl )}.

Theorem 1.14 (Theorem 1.4.2.1 of [13]) If the boundary of Ω is continuous, then D(Ω) is dense in W s, p (Ω) for all s > 0. When s is a nonnegative integer and the boundary is Lipschitz continuous, this theorem is valid for p = 1 (see Theorem 1.2, Chap. 1 of [14]). Theorem 1.15 (Theorem 1.4.2.4 of [13]) Let Ω be a bounded open subset of Rl with a Lipschitz boundary. Then D(Ω) is dense in W s, p (Ω) for 0 < s ≤ 1/ p. For every s > 0, we denote by W s, p (Ω) the space of all distributions in Ω which are restrictions of elements of W s, p (Rl ) (Definition 1.3.2.4 of [13]). In other words, W s, p (Ω) is the space of all u|Ω where u ∈ W s, p (Rl ) and u|Ω is ˜ for all ϕ ∈ D(Ω), where ϕ˜ is the continuation of ϕ by defined by u|Ω , ϕ = u, ϕ zero, outside Ω. A Banach norm on W s, p (Ω) is defined by uW s, p (Ω) =

inf

U ∈W s, p (Rl ) U |Ω =u

U W s, p (Rl ) .

(1.8)

As obvious consequences of the definition, we have the following inclusions: W s, p (Ω) ⊂ W s, p (Ω)

1.3 Sobolev Space

15

((1.3.2.5) of [13]). The following theorem shows a case that W s, p (Ω) = W s, p (Ω). Theorem 1.16 (Theorem 1.4.3.1 of [13]) Let Ω be a bounded open subset of Rl with a Lipschitz boundary. Then for every s > 0 there exists a continuous linear operator Ps from W s, p (Ω) into W s, p (Rl ) such that Ps u|Ω = u.  s, p (Ω) the space of all u ∈ W s, p (Ω) such For every positive s, we denote by W s, p l that u˜ ∈ W (R ), where u˜ is the extension of u by zero outside Ω. Theorem 1.17 (Corollary 1.4.4.5 of [13]) Let Ω be a bounded open subset of Rl with a Lipschitz boundary. Then when s − 1/ p is not integer, we have  s, p (Ω) = W0s, p (Ω), W

(1.9)

and furthermore, when 0 < s < 1/ p we have  s, p (Ω) = W0s, p (Ω) = W s, p (Ω). W

(1.10)

Theorem 1.18 (Corollary 1.4.4.10 of [13]) Let Ω be a bounded open subset of Rl with a Lipschitz boundary. Then for all s > 0, we have    s, p (Ω) = u ∈ W0s, p (Ω) : ρ −σ D α u ∈ L p (Ω), |α| = m W , where ρ(x) is the distance from x to the boundary ∂Ω of Ω and s = m + σ , m integer, σ ∈ [0, 1). 1/2

1/2

Following Sect. 11.5, Chap. 1 of [15], we denote by H00 (Ω), H0 (Ω) and 1/2  1/2,2 (Ω), W01/2,2 (Ω) and W 1/2,2 (Ω). By TheoH (Ω), respectively, the spaces W rem 1.15 1/2 (1.11) H0 (Ω) ≡ H 1/2 (Ω). 1/2

1/2

The space H00 (Ω) is strictly contained in H0 (Ω) with a strictly finer topology 1/2  u H 1/2 (Ω) = u2H 1/2 (Ω) + ρ −1/2 u2L 2 (Ω) . 00

Since C0∞ (Ω) ⊂ H00 (Ω), H00 (Ω) is dense in H 1/2 (Ω). 1/2

1/2

Remark 1.3 Let s ∈ R and 1 < p ≤ ∞. We denote by H ps (Rl ) the space of all distributions in Rl such that   F −1 (1 + |ξ |2 )s/2 F u ∈ L p (Rl ), where F is the Fourier transform operator defined by

16

1 Miscellanea of Analysis

(F u)(ξ ) =

1 (2π )n/2



e−i xξ u(x) d x. Rl

The space H ps is called with different names, i.e. the Bessel potential space [9], the fractional order Sobolev space [9], the generalized Sobolev space [11], the inhomogeneous Sobolev space [16], and in some literature the notation for Sobolev is used for that. It is known that  H ps (Rl ) s : all real number, p = 2, s, p l W (R ) = (1.12) H ps (Rl ) s : integer, 1 < p < ∞. Furthermore, it is proved that for any real s W s, p (Rl ) ⊆ H ps (Rl ) 1 < p ≤ 2, H ps (Rl ) ⊆ W s, p (Rl ) 2 ≤ p < ∞. (See Sect. 1.3.1 of [13].) Define the space H ps (Ω) by H ps (Ω) = {u|Ω : u ∈ H ps (Rl )}.

(1.13)

Then by Theorem 1.16 and (1.12), for a bounded open subset Ω with a Lipschitz boundary, we have  W

s, p

(Ω) =

H ps (Ω) s : all real number, p = 2, H ps (Ω) s : integer, 1 < p < ∞.

(1.14)

Instead of (1.13) in some literatures H ps (Ω) is defined by complex interpolation of L (Ω) and W m, p (Ω). Similarly to Theorem 1.16, under some condition for boundary of Ω, there exists a continuous linear operator Ps from H ps (Ω) into H ps (Rl ). (Noting H ps (Ω) = F s. p,2 (Ω), refer to Sect. 7.69 of [9].) Thus under such definition (1.14) is valid. p

1.3.3 Imbedding If imbedding operator I of a linear normed space into another linear normed space is compact, then the imbedding is called compact. Theorem 1.19 (Theorem 1.4.4.1 of [13]) The following imbeddings hold: W s, p (Rl ) → W t,q (Rl )

(1.15)

1.3 Sobolev Space

17

for t ≤ s and q ≥ p such that s − n/ p = t − n/q; and W s, p (Rl ) → C k,α (Rl )

(1.16)

for k < s − n/ p < k + 1, where α = s − k − n/ p, k a nonnegative integer. Taking into account (1.8), as a consequence we have the conclusions W s, p (Ω) → W (Ω) and W s, p (Ω) → C k,α (Ω) instead of (1.15) and (1.16), where C k,α (Ω) is the space of functions which are k times continuously differentiable and whose derivatives of order k fulfill a uniform Hölder condition of order α throughout Ω. = q1 . On the other hand, we see that In other words, (1.15) is valid when 1p − s−t n for a bounded open subset Ω t,q

W t,q (Ω) → W t,q1 (Ω) for 1 < q1 < q. Then taking into account the fact above, by Theorem 1.16, we have Theorem 1.20 (Sobolev’s imbedding theorem) Let Ω be a bounded open subset of Rl with a Lipschitz boundary and 0 ≤ t < s, 1 < q < ∞. ≤ q1 , then If 1p − s−t n W s, p (Ω) → W t,q (Ω). (1.17) If

1 p

−

s−k n

< 0, where k is a nonnegative integer, then W s, p (Ω) → C k (Ω),

and if, in addition, s −

n p

(1.18)

< k + 1, then

W s, p (Ω) → C k,α (Ω), α = s − k − n/ p.

(1.19)

Remark 1.4 When q = ∞, (1.17) is not valid (cf. Remark 9.2, Chap. 1 of [15]). Theorem 4.47, Corollary 4.53 and Theorem 4.57 of [17] assert that when Ω is Rl or a unbounded open set Ω with a Lipschitz boundary and 1p − ns = 0, W s, p (Ω) → L q (Ω) for every q < ∞. Under the condition p ≤ q < ∞ it is right (see Sect. 4.12 of [9]). Theorem 1.21 (Theorem 1.4.3.2 of [13]) Let s1 > s2 ≥ 0 and assume that Ω is a bounded open subset of Rl with a Lipschitz boundary. Then the imbedding of W s1 , p (Ω) in W s2 , p (Ω) is compact. Theorem 1.22 Let Ω be a bounded open subset of Rl with a Lipschitz boundary. Let 0 < s and 1 < p < ∞. If 1p − ns < q1 , then the imbedding W s, p (Ω) → L q (Ω) is compact. (See Theorem 4.58 of [17].)

18

1 Miscellanea of Analysis

Theorem 1.23 (Theorem 1.4.4.2 of [13]) Let s1 ≥ s and s2 ≥ s be such that either  s1 + s 2 − s ≥ n

1 1 1 + − p1 p2 p



 1 1 − , j = 1, 2 pj p



 1 1 , j = 1, 2, − pj p

 ≥ 0, s j − s > n

or  s1 + s2 − s > n

1 1 1 + − p1 p2 p

 ≥ 0, s j − s ≥ n

then (u, v) → u · v is a continuous bilinear map from W s1 , p1 (Rl ) × W s2 , p2 (Rl ) into W s, p (Rl ). A similar statement holds for Sobolev spaces defined on a bounded open subset of Rl with a Lipschitz boundary (see Sect. 1.4.4 of [13]). Theorem 1.24 (Theorem 1.4.4.6 of [13]) Let Ω be a bounded open subset of Rl with a Lipschitz boundary. Then D α , |α| = 1, is a linear continuous operator from W s, p (Ω) into W s−1, p (Ω) unless s = 1/ p. (s-real number)

1.3.4 Trace When u is a continuous function on Ω, let γ be the operator such that γ u = u|∂Ω . The following trace theorem extends the operator γ to functions in W s, p (Ω) (cf. Definition, Sect. 5.5 of [18]). Theorem 1.25 (Theorem 1.5.1.2 of [13]) Let Ω be a bounded open subset of Rl with a C k,1 boundary. Assume that s − 1/ p is not an integer, s ≤ k + 1, s − 1p = l + σ , 0 < σ < 1, l an integer ≥ 0. Then mapping   ∂ν ∂l u u → γ u, γ , · · · , γ l ∂u ∂ν which is defined for u ∈ C k,1 (Ω) has a unique continuous extension T r : from W

s, p

(Ω) onto

k 

W s− j− p , p (∂Ω). 1

j=0

Theorem 1.26 (Corollary 1.5.1.6 of [13]) Let Ω be a bounded open subset of Rl with a C k,1 boundary. Assume that s ≤ k + 1 and s − 1/ p is not an integer. Let s, p s − 1/ p = l + σ , 0 < σ < 1, l an integer ≥ 0. Then u ∈ W0 (Ω) if and only if u ∈ W s, p (Ω) and ∂u ∂l u γu = γ = · · · = γ l = 0. ∂ν ∂ν

1.3 Sobolev Space

19

By Theorem 1.26, we see that s, p

K er (T r ) = W0 (Ω). Thus, by the open-mapping theorem (see Sect. 5, Chap. 2 of [2]) there exists a continuous linear operator T r −1 : from

k 

 1 s, p W s− j− p , p (∂Ω) onto W s, p (Ω) W0 (Ω) (quotient space).

j=0

(1.20) Theorem 1.27 (Theorem 1.5.1.10 of [13]) Let Ω be a bounded open subset of Rl with a Lipschitz boundary. Then there exists a constant K such that  ∂Ω

   |u| p dσ ≤ K ε1−1/ p |∇u| p d x + ε−1/ p |u| p d x Ω

Ω

∀u ∈ W 1, p (Ω), ∀ε ∈ (0, 1).

1.3.5 Some Inequalities Theorem 1.28 (Friedrichs’ inequality) Let Ω be a bounded domain of Rl with a Lipschitz boundary and Γ ⊂ ∂Ω, mes(Γ ) = 0. Then for u ∈ W 1, p (Ω) ( p ≥ 1) 

 |u| d x ≤ C

n   ∂u 2  p/2

p

Ω

Ω

i=1

∂ xi

  p   d x +  u dσ  . Γ

(See Theorem 1.32 of [19] or Lemma 1.36, Chap. 2 of [3].) Let u = (u 1 , · · · , u n ) ∈ W1, p (Ω) ≡ W 1, p (Ω)n and εi j (u) =

1 2



∂u i ∂x j

+

∂u j ∂ xi

 .

Theorem 1.29 (The first Korn’s inequality) Let Ω be a bounded domain of Rl with a Lipschitz boundary and Γ be a nonempty open subset of ∂Ω. Then for u ∈ 1, p WΓ (Ω) := {v ∈ W1, p (Ω) : v|Γ = 0}, 

p

p

εi j (u) L p (Ω) ≥ CuW1, p (Ω) .

i, j

(See Theorem 6.3.4 of [20] for the case that n = 3, p = 2, and Corollary 4.1 of [21] for the general case.)

20

1 Miscellanea of Analysis

1.4 Space of Abstract Functions 1.4.1 Abstract Functions and Its Derivatives Let X be a Banach space, S be an interval on R and u : S → X . A function u : S → X is said to be finitely valued if it is constant on each of a finite number of disjoint (Lebesgue) measurable sets B j and equal to 0 X on S \ ∪ j B j . A function u : S → X is said to be simple if it is finitely valued and if the set for which u(s) > 0 is of finite measure. u : S → X is said to be countably valued if it assumes at most a countable set of values in X , assuming each value is different from 0 X on a measurable subset. Definition 1.9 A function u(s) is said to be (Bochner) strongly measurable (simply measurable) if there exists a sequence of simple functions strongly convergent to u(s) at a.e. s ∈ S u : S → X is called weakly measurable if, for any f ∈ X ∗ , the numerical function  f, u(s) is (Lebesgue) measurable. v : S → X ∗ is called ∗-weakly measurable if, for any g ∈ X , the numerical function g, v(s) is (Lebesgue) measurable. Remark 1.5 There are some different definitions for strongly measurable function, and some of them are listed below: (i) A function u(s) is said to be strongly measurable if it is the strong limit of a sequence of finitely valued functions almost everywhere. (p. 1681 of [22]) (ii) A function u(s) is said to be strongly measurable if there exists a sequence of countably valued functions converging almost everywhere. (Definition 3.5.4 of [23]) (iii) A function u(s) is called measurable function if there exists a sequence {u n (s)} of countably valued functions such that u n (s) → u(s) uniformly outside a null set (i.e. the set with null measure). (p. 174 of [8]) Definition (i) above is a special case of (ii). u(s) is called almost separably valued if there exists a measurable set B0 of measure zero such that {u(s); s ∈ S − B0 } is separable. By Theorem (Pettis) of Sect. 4, Chap. 5 in [2], u(s) is strongly measurable in sense of Definition 1.9 iff it is weakly measurable and almost separably valued. By Theorem 3.5.3 of [23], for u(s) to be strongly measurable in the sense of Definition (ii), so is it. Thus, Definition 1.9 and Definition (ii) are equivalent. By Corollary 1 of Theorem 3.5.3 of [23], Definitions (ii) and (iii) above are equivalent. Therefore, all definitions in Remark 1.5 are equivalent to Definition 1.9. Theorem 1.30 (Theorem 5.0.11 of [8]) Let X be separable. Then u : S → X is weakly measurable if and only if it is strongly measurable.

1.4 Space of Abstract Functions

21

If X ∗ is separable, then strong measurability of X ∗ -valued function v : S → X ∗ is equivalent to ∗-weak measurability, Remark 1.6 An X ∗ - valued function v(s) is said to be weakly measurable if  f, v(s) is measurable for all f ∈ X ∗∗ . The weak measurability of X ∗ -valued function is in general stronger than ∗-weak measurability. For an example of ∗-weakly measurable but not measurable X ∗ -valued function we refer to Example 5.0.10 of [8]. Theorem 1.30 is valid when X is a normed space. Let u(s) : S → X be a finitely valued function. Let u(s) = xi on Bi , where Bi are disjoint and mes(Bi ) < ∞ for!i = 1, · · · , n and u(s) = 0 on Sn − ∪i Bi . Then we xi mes(Bi ). can define the (Bochner) integral S u(s) ds of u(s) over S by i=1 Define the integral of more general functions. Definition 1.10 A function u(s) : S → X is said to be Bochner integrable, if there exists a sequence of finitely valued functions {u n (s)} which converges to u(s) a.e. in S and  lim u(s) − u n (s) ds = 0. n→∞

S

For any measurable set B ⊂ S, the Bochner integral of u(s) over B is defined by 

 u(s) ds = lim

n→∞

B

χ B (s)u n (s) ds, S

where χ B is the characteristic function of the set B. The definition is justified since the limit on the right above exists and the value of this limit is independent of the approximating sequence of functions {u n } (see Sect. 5, Chap. 5 of [2]). If u(·) is strongly measurable, then u(·) is measurable, and u(·) is Bochner integrable if and only if u(·) is Lebesgue integrable, both integrals standing in the relation







u(s) ds ≤ u(s) ds.



S

S

Theorem 1.31 (Theorem 1.8, Chap. 4 of [3]) Let u : S → X be Bochner integrable and B ⊂ S be a Lebesgue measurable set. Then for any f ∈ X ∗ , 

  f, u(s) ds =  f, B

u(s) ds . B

(See Theorem 8, Appendix E of [18] or Corollary 2, Sect. 5 of Chap. V in [2].) Differently from real functions, not every Banach space valued absolutely continuous functions is the indefinite integral of an integrable function (see Example 5.0.13 of [8]). But the next holds.

22

1 Miscellanea of Analysis

Theorem 1.32 If u(·) : S → X is Bochner integrable in S, then the function v : S → X defined by  t u(s) ds, t0 ∈ S v(t) = t0

is differentiable almost everywhere on S and v  (t) = u(t) for a.e. t ∈ S. (See p. 176 of [8] or Theorem 1.7, Chap. 4 of [3].) Remark 1.7 A Banach space X is called a Gelfand space if every X -valued absolutely continuous function is differentiable almost everywhere. Let X be either (a) reflexive or (b) separable and the dual of another Banach space. Then X is a Gelfand space (Theorem 5.0.14 of [8]). Definition 1.11 The space L p (S; X ), (1 ≤ p < ∞) consists of all X -valued strongly measurable functions u(·) defined in S and such that  u(s) p ds < ∞. S

The space L ∞ (S; X ) consists of all X -valued strongly measurable functions u(·) defined in S and such that u(s) ≤ M ∃M > 0, a.e. s ∈ S. L p (S; X ), (1 ≤ p < ∞) and L ∞ (S; X ) are Banach spaces, respectively, with norms 

1/ p

u L p (S;X ) =

u(s) p ds S

and u L ∞ (0,T ;X ) = esssup u(s). s∈S

Remark 1.8 Let 1 ≤ p < ∞. If Y is a dense subspace of X , then the set {ϕ = φ(t)u : φ ∈ D(0, T ), u ∈ Y } is dense in L p (0, T ; X ) (see Theorem 5.0.27 of [8]). functions Denote by C k ([0, T ]; X ) the space of k-times continuously differentiable  k on [0, T ], 0 < T < ∞. Then, the set of polynomials { p : p(s) = m k=0 ak s , s ∈ [0, T ], ak ∈ X, m = 0, 1, · · · } is dense in the set C([0, T ]; X ) (Weierstrass’ approximation theorem, see Theorem 1.3, Sect. 1, Chap. 4 of [3]). Therefore, the set {ϕ = φ(t)u : φ ∈ C 1 [0, T ], u ∈ X } is dense in L p (0, T ; X ). Also, we can see that the set {ϕ = φ(t)u : φ ∈ C 1 [0, T ], φ(T ) = 0, u ∈ X } is dense in {ϕ ∈ C 1 ([0, T ]; X ) : ϕ(T ) = 0}. Theorem 1.33 Let X 0 , X 1 be Banach spaces such that X 0 → X 1 , 1 ≤ p0 < ∞ and 0 < θ < 1. Then, L p0 (S; X 0 ) ∩ L ∞ (S; X 1 ) → L p (S; (X 0 , X 1 )[θ] )

(1.21)

1.4 Space of Abstract Functions

23

with p = p0 /(1 − θ ), where the functor ( , )[θ] means the complex interpolation (see Definition 2.1.3 of [10]). Proof Since X 0 → X 1 , u ∈ L p (S; X 0 ) ∩ L ∞ (S; X 1 ) is (X 0 , X 1 )[θ] -value strongly measurable and u(t) ∈ (X 0 , X 1 )[θ] at a.e. t ∈ S. Then, θ u(t)(X 0 ,X 1 )[θ ] ≤ u(t)1−θ X 0 u(t) X 1

(see Corollary 2.1.8 of [10]), which yields (1.21).



Complex interpolation between simple Sobolev spaces is as follows. Let Ω be a bounded domain with Lipschitz boundary, s0 = s1 and 0 < θ < 1. Then,   s0 ,2 ∗ (1.22) W (Ω), W s1 ,2 (Ω) [θ] = W s ,2 (Ω), where

s ∗ = (1 − θ )s0 + θ s1 .

(cf. Theorem 6.4.5 of [11]. We used that when p = 2, Sobolev space W s, p equals to Bessel potential space H ps (see (1.12)). All spaces in [11] are defined on R3 ; however, the results are valid for the spaces continuously expandable to the space defined on R3 , for example, the space defined on a bounded domain with Lipschitz boundary, since from the interpolation results for the spaces defined on R3 we can reduce the results for the spaces defined on the restricted domain. We refer to Sect. 1.3.2.) Remark 1.9 In [11, 24] a space L ∞ (S; X ) is defined as follows: If χ B is the characteristic function of the measurable set B ⊂ S, then L ∞ (S; X ) is defined as a completion of v(x) =

N 

a j χ B j (s),

j=1

where B j are disjoint sets, in the norm v∞ = ess supx∈S v(x) X . If above only sets with mes(B j ) < ∞ are admitted, then the completion of those elements is denoted by L 0∞ (S; X ). Then, Remark 3 of Theorem 1.18.4 of [24] and Theorem 5.1.2 of [11] assert without detailed proof the following. Theorem 1.34 Assume that X 0 and X 1 are Banach spaces and that 1 ≤ p0 < ∞, 0 < θ < 1. Then (L p0 (S; X 0 ), L 0∞ (S; X 1 ))[θ] = L p (S; (X 0 , X 1 )[θ] ) with p = p0 /(1 − θ ).

24

1 Miscellanea of Analysis

Let us consider whether the space L ∞ (S; X ) in [11, 24] is the same as the one in Definition 1.11 and L 0∞ (S; X ) coincides with L ∞ (S; X ) when S = [0, T ], T < ∞. The following counterexample shows that the answer is negative. Counterexample (the case that S = [0, T ], T < ∞)  Let l 2 be the space of sequences (a1 , a2 , · · · , ai , · · · ) such that i ai2 < ∞ and i−1 i −1)T (2 −1)T ei = (0, · · · , ai , · · · ) with ai = 1. Let Si = [ (2 2i−1 , 2i ), i = 1, 2, · · · . Define 2 a function u : S → l by u(s) = en ∀s ∈ Sn , u(T ) = e1 . It is easy to verify that when X = l 2 , this function belongs to L ∞ ([0, T ]; X ) in the sense of Definition 1.11. Let us consider a sequence { f n } of finite-value functions f n (s) =

kn 

xi(n) χ B (n) (s) xi(n) ∈ X, Bi(n) − disjoint measurable subsets of [0, T ]. i

i=1

For every n there exists an i n ≤ kn such that Bi(n) intersects with infinite set of Skin n on measurable set with positive measure. Thus, ) > 0. ∀N , ∃m N ≥ N mes(Sm N ∩ Bi(n) n

(1.23)

∈ l 2, On the other hand, since xi(n) n xi(n) = n

∞ 

αl(in ) el ,

l=1

lim αl(in ) = 0.

l→∞

Thus, ∃N , ∀l ≥ N |αl(in ) |
0, and so any sequence { f n } cannot converge to u in the sense of essential supremum norm. Thus, L 0∞ ([0, T ]; X )  L ∞ ([0, T ]; X ). Let us show that C([0, T ]; X 1 ) ⊂ L 0∞ ([0, T ]; X 1 ).

1.4 Space of Abstract Functions

25

 k If T < ∞, then the set of polynomials { p : p(s) = m k=0 ak s , s ∈ [0, T ], ak ∈ ), where X 1 is a Banach space. Thus, for any ε > 0 X 1 } is dense in the set C([0, T ]; X 1 ε k there exists a polynomial p(s) = m k=0 ak s such that sups∈[0,T ] u(s) − p(s) X 1 ≤ ε. We can make a partition Bi = [si , si+1 ] of [0, T ] such that max

k∈{0,1,··· ,m ε }

i

Let us consider a simple function have u(s) −



ε . k=0 ak  X 1

k max |sik − si+1 | ≤ m ε

 i

bi χ Bi (s), where bi =

m ε

k=0

ak sik . Then, we

mε 



 k bi χ Bi (s) = u(s) − ak si χ Bi (s)

i

k=0

i

mε mε  



k

≤ u(s) − p(s) + ak s − ak sik χ Bi (s)

k=0

k=0

i

mε 



≤ε+

ak (s k − sik χ Bi (s)) ≤ 2ε, k=0

i

which shows the asserted conclusion. Let X be a real Banach space. By Theorem 12.9.2 and (12.9.29) of [8] we get Theorem 1.35 Let 1 < p < ∞, X ∗ a separable Banach space. Then to every f ∈ ∗  p L (0, T ; X ) corresponds an unique v f ∈ L q (0, T ; X ∗ ), 1/ p + 1/q = 1, such that  u, f =

u(s), v f (s) ds, ∀u ∈ L p (0, T ; X ) S

and this correspondence is algebraically and metrically isomorphic. By Theorem 12.2.11 and (12.2.13) of [8] we get



Theorem 1.36 Let X ∗ be separable Banach space. Then to every f ∈ L 1 (0, T ; X ) corresponds an unique v f ∈ L ∞ (0, T ; X ∗ ) such that

∗

 u, f =

u(s), v f (s) ds, ∀u ∈ L 1 (0, T ; X ) S

and this correspondence is algebraically and metrically isomorphic. Remark 1.10 The space X = L 1 (0, 1) is separable, but its dual space X ∗ = L ∞ (0, 1) is not separable (Example 5.0.31 of [8]). But if the dual X ∗ of a Banach space X is separable, then X is separable (Lemma of Sect. 2, Chap. 5 of [2]). Thus, if X be a separable reflective Banach space, then from the fact that X ∗∗ = X is separable it follows that X ∗ is separable. Therefore, Theorems 1.35, 1.36 are valid when X is a separable reflective Banach space. When X ∗ is a Gelfand space or a Banach space, for generalizations of Theorems 1.35, 1.36 we refer to Chap. 12 of [8].

26

1 Miscellanea of Analysis

Let S be an open interval, D(S) be the local convex space made from C0∞ (S) by introducing topology with the semi-norms (1.5) and X w denote a Banach space X with weak topology. Definition 1.12 The set L (D(S), X w ) of linear continuous operators from D(S) to X w is denoted by D ∗ (S; X ) and its element is called a distribution on S with value in X . Definition 1.13 The distribution f  ∈ D ∗ (S; X ) defined by f  (ϕ) = − f (ϕ  ) ∀ϕ ∈ D(S) is called derivative of f ∈ D ∗ (S; X ). If f n → f in D ∗ (S; X ), then for y ∈ X ∗ and ϕ ∈ D(S) lim y, f n (ϕ) − f  (ϕ) = lim y, f (ϕ  ) − f n (ϕ  ) = 0.

n→∞

n→∞

Thus, mapping f → f  is continuous in D ∗ (S; X ). Let V be a real separable reflective Banach space and H be a Hilbert space such that V → H and V is dense in it. Identifying H and H ∗ , we have V → H → V ∗ . Let 1 < p ≤ p0 < ∞ and X = L p (S; V ) ∩ L p0 (S; H ), u X = u L p (S;V ) + u L p0 (S;H ) . Then,

(1.25)

1 1 1 1 + = 1, + = 1, p q p0 q0 max{ f 1  L q (S;V ∗ ) ,  f 2  L q0 (S;H ) }.

X ∗ = L q (S; V ) + L q0 (S; H ),  f X ∗ =

inf

f 1 ∈L q (S;V ∗ ) f 2 ∈L q0 (S;H ) f1 + f2 = f

Define a space W by W = {u : u ∈ X, u  ∈ X ∗ }, uW = u X + u   X ∗ .

(1.26)

Theorem 1.37 W ⊂ C(S; H ). If S is a compact interval, then the imbedding of W to C(S; H ) is continuous. The following integration by parts holds. For any u, v ∈ W ,     u(t), v(t)) H − u(s), v(s) H =

 t s

u  (τ ), v(τ ) dτ +

 t s

u(τ ), v  (τ ) dτ s, t ∈ S.

1.4 Space of Abstract Functions

27

From the theorem above we have that when v = u,  1 u(t)2H − u(s)2H = 2



t

u  (τ ), u(τ ) dτ.

(1.27)

s

Remark 1.11 If u  = f 1 + f 2 , f 1 ∈ L q (S; V ), f 2 ∈ L q0 (S; H ), then 

t

u  (τ ), v(τ ) dτ =

s



t



t

 f 1 (τ ), v(τ ) V dτ +

s

( f 2 (τ ), v(τ )) H dτ.

s

(See Lemma 7.3 of [19] for the case of p = p0 , and Theorem 1.17, Chap. 4 of [3] for the general case.)

1.4.2 Compactness Let B be a Banach space. For a positive integer m and 1 ≤ p ≤ ∞ denote  W

m, p

(0, T ; B) =

 ∂m f p f : f, · · · , m ∈ L (0, T ; B) . ∂t

For 0 < σ < 1 and 1 ≤ p ≤ ∞ denote W

σ, p

 (0, T ; B) =



T

f : f ∈ L (0, T ; B),



T

p

0

0

 p  f (t) − f (s) B dtds < ∞ |t − s|σ p+1

with the usual change if p = ∞ (see (iii) of Definition 1.6). For a positive integer m and 0 < σ < 1 denote   ∂ m−1 f ∂m f m+σ, p p σ, p (0, T ; B) = f : f, · · · , m−1 ∈ L (0, T ; B), ∈ W (0, T ; B) . W ∂t ∂t m If f is defined on [0, T ], then the translated function (τh f )(t) = f (t + h) for h > 0 is defined on [−h, T − h], where 0 < T < ∞. Let X, B, Y be Banach spaces such that X → B → Y and the imbedding X → B is compact. Theorem 1.38 (Theorem 5 of [25]) Let 1 ≤ p ≤ ∞ and F be a bounded set of L p (0, T ; X ). If τh f − f  L p (0,T −h;Y ) → 0 as h → 0, uniformly for f ∈ F, then F is relatively compact in L p (0, T ; B) (and in C(0, T ; B) if p = ∞).

28

1 Miscellanea of Analysis

Theorem 1.39 (Corollary 5 of [25]) Let 1 ≤ p ≤ ∞, 1 ≤ r ≤ ∞ and F be a bounded set of L p (0, T ; X ) ∩ W s,r (0, T ; Y ), where s > 0 if r ≥ p and where s > 1/r − 1/ p if r ≤ p. Then F is relatively compact in L p (0, T ; B) (and in C(0, T ; B) if p = ∞).

1.5 Operator Equations and Operator-Differential Equations 1.5.1 Monotone Operator Equation Let X be a separable real reflective Banach space such that X → H , where H is a real separable Hilbert space.. Definition 1.14 The operator A : X → X ∗ is said to be radial continuous if s → A(u + sv), v is continuous on [0, 1] for any fixed u, v ∈ X , hemi-continuous if s → A(u + sv), h is continuous on [0, 1] for any fixed u, v, h ∈ X , demi-continuous if from u n → u in X it follows that Au n  Au in X ∗ , and Lipschitz continuous if Au − Av X ∗ ≤ Mu − v X ∃M > 0, ∀u, v ∈ X. Definition 1.15 Let u, v be any elements of X . The operator A : X → X ∗ is said to be monotone if Au − Av, u − v ≥ 0, strictly monotone if Au − Av, u − v > 0 for u = v, strongly monotone if Au − Av, u − v ≥ αu − v2X α > 0 and coercive if

Au, u → +∞ as u X → ∞. u X

Definition 1.16 The operator A : X → X ∗ is called bounded iff the image of every bounded set of X is bounded in X ∗ . Theorem 1.40 Let A : X → X ∗ be radially continuous, monotone and coercive. Then for any f ∈ X ∗ the set of solutions to Au = f is nonempty, weakly closed and convex.

1.5 Operator Equations and Operator-Differential Equations

29

Theorem 1.41 Let A : X → X ∗ be radially continuous, strictly monotone and coercive. Then there exists A−1 : X ∗ → X which is strictly monotone, bounded and demicontinuous. Theorem 1.42 Let A : X → X ∗ be radially continuous, strongly monotone. Then there exists A−1 : X ∗ → X which is Lipschitz continuous. If, in addition, A is Lipschitz continuous, then A−1 is strongly monotone. (For Theorems 1.40–1.42 we refer to Theorems 2.1, 2.2 and Corollary 2.3 of Chap. 3 of [3] or Theorem 26.A of [26]. If A is continuous and strictly monotone, then for Theorem 1.40 we refer to Theorem 5.16 of [1].) Remark 1.12 If A is linear, strongly monotone and Lipschitz continuous on Hilbert space X , Theorem 1.42 means the Lax-Milgram lemma (Theorem 1, Sect. 6.2 of [18]). Let X be real Hilbert space and a nonlinear functional be defined by a( · ; · , · ) : (w, u, v) ∈ X × X × X → a(w; u, v) ∈ R, where, for w ∈ X , the mapping (u, v) → a(w; u, v) is a bilinear continuous form on X × X . Let us consider the following problem: Find u ∈ X such that a(u; u, v) =  f, v ∀v ∈ X, where f ∈ X ∗ .

(1.28)

Theorems 1.2 and 1.3 of Chap. 4 of [14] are rewritten as follows. Theorem 1.43 Assume that the following hypotheses hold: (i) there exists a constant α > 0 such that a(v; v, v) ≥ αv2X ∀v ∈ X. (ii) the space X is separable and, for all v ∈ X , the mapping u → a(u; u, v) is sequentially weakly continuous on X , i.e. weak − lim u m = u in X implies m→∞

lim a(u m ; u m , v) = a(u; u, v) ∀v ∈ X.

m→∞

Then, problem (1.28) has at least one solution u ∈ X .

30

1 Miscellanea of Analysis

Theorem 1.44 Assume that the following hypotheses hold: (i) there exists a constant α > 0 such that a(w; v, v) ≥ αv2X ∀v, w ∈ X. (ii) there exists a continuous and monotonically increasing function L : R+ → R+ such that for all μ > 0 |a(w1 ; u, v) − a(w2 ; u, v)| ≤ L(μ)u X v X w1 − w2  X ∀u, v ∈ X, ∀w1 , w2 ∈ Oμ (0 X ), where Oμ (0 X ) is μ-neighborhood of 0 X in X . Then, under the condition   f X ∗  L  f  X ∗ /α < 1 2 α problem (1.28) has a unique solution u ∈ X .

1.5.2 Pseudo-Monotone Operator Equation Theorem 1.45 Let A : X → X ∗ be an operator on the real, separable and reflexive Banach space X . Let A be coercive and bounded. If for every sequence {xn } such that xn  x in X, (1.29) lim supAxn , xn − x ≤ 0 n→∞

there exists a subsequence {xk } such that lim inf Axk , xk − v ≥ Ax, x − v ∀v ∈ X, k→∞

(1.30)

then for any f ∈ X ∗ there exists a solution to Au = f. Remark 1.13 If (1.29) implies (not for a subsequence) lim inf Axn , xn − v ≥ Ax, x − v ∀v ∈ X, n→∞

then A is called pseudo-monotone. For the coercive, bounded and pseudo-monotone operators A, the asserted conclusion was proved (see Theorem 27.A of [26] or Theorem 2.7, Chap. 2 of [27]). However, proofs of the facts that A has property (M) (Propo-

1.5 Operator Equations and Operator-Differential Equations

31

sition 2.5, Chap. 2 in [27]) and A is demi-continuous (footnote of (2.27), Chap. 2 of [27]), which guarantee the existence of a solution to Au = f , hold with (1.30) for subsequence, and so we have the conclusion.

1.5.3 Operator-Differential Equations Definition 1.17 Let X 1 , X 2 be spaces and S = [0, T ], T > 0. Mapping D(F) ⊂ (S → X 1 ), F : D(F) → (S → X 2 ), where (S → X i ) is a set of functions from S to X i , is called Volterra operator iff u(s) = v(s) a.e. s ∈ [0, t], t ∈ S implies (Fu)(t) = (Fv)(t) a.e. t ∈ S. Let V be a real separable reflective Banach space and H be a Hilbert space such that V → H and V is dense in it. Now, let the spaces X and W be the ones in (1.25) and (1.26). Theorem 1.46 Let A : X → X ∗ be a radial continuous, monotone, coercive and Volterra operator. Then, for any f ∈ X ∗ and a ∈ H , there exists a unique solution u ∈ W ⊂ C([0, T ]; H ) to u  + A u = f, u(0) = a, and the mapping a → u is continuous from H to C([0, T ]; H ). (See Theorem 1.1, Chap. 6 of [3] or Sect. 8.1 of [19].) ∗

Definition 1.18 Let A be multivalued operator from D(A) ⊂ X to 2 X (the family of subsets of X ∗ ) and its graph G(A) = {(x, y) ∈ X × X ∗ ; y ∈ Ax}. The operator is said to be monotone if y1 − y2 , x1 − x2 ≥ 0 ∀(x1 , y1 ), (x2 , y2 ) ∈ G(A). A monotone operator A is said to be maximal monotone if its graph is not properly ∗ contained in graph of any other monotone operator from X to 2 X . Note that if A is a single-valued operator such that D(A) ⊆ X → X ∗ it is monotone if Au − Av, u − v ≥ 0 ∀u, v ∈ D(A) and it is maximal monotone if, in addition,  f − Av, u − v ≥ 0 ∀v ∈ D(A) implies u ∈ D(A) and Au = f.

32

1 Miscellanea of Analysis

Example 1.1 Let V be a real separable reflective Banach space and H be a Hilbert space such that V → H and V is dense in H . Identifying H and H ∗ , we have V → H → V ∗ . Let 1 < p < ∞ and X = L p (0, T ; V ), u X = u L p (0,T ;V ) . Then, X ∗ = L q (0, T ; V ∗ ), Put

1 1 + = 1. p q

W = {u : u ∈ X, u  ∈ X ∗ }, uW = u X + u   X ∗ .

Then, the operator L : D(L) ⊂ X → X ∗ defined by d , dt

L=

D(L) = {u ∈ W : u(0) = 0}

is a maximal monotone linear operator (see Sect. 2.1 of Chap. 3 in [27]). Theorem 1.47 Let X be a real reflexive Banach space and norms of X and X ∗ be strictly convex. Let a linear operator L, where D(L) is a dense linear subspace of X , be maximal monotone. Let the operator A : D(L) → X ∗ satisfy the following: (i) For every sequence such that xn  x in X , xn , x ∈ D(L), L xn  L x in X ∗ and lim supn→∞ Axn , xn − x ≤ 0, there exists a subsequence xk such that lim inf Axk , xk − v ≥ Ax, x − v ∀v ∈ X ; k→∞

(1.31)

(ii) There exists a function ψ : R+ ≡ [0, ∞) → R+ , which is bounded on any compact interval, and a number θ, 0 ≤ θ < 1, such that

(iii)

A(x) X ∗ ≤ ψ(x X ) + θ L x X ∗ ∀x ∈ D(L);

(1.32)

A(x), x → ∞, as x X → ∞. x X

(1.33)

Then, for any f ∈ X ∗ there exists a solution to

1.5 Operator Equations and Operator-Differential Equations

33

L x + Ax = f. Remark 1.14 When lim inf Axn , xn − v ≥ Ax, x − v ∀v ∈ X, n→∞

instead of (1.31) holds, the asserted conclusion was proved in [28]. In Theorem 1.2, Chap. 3 of [27] it is proved by using Theorem 2.7, Chap. 2 of [27]. However, existence of solutions to the equation holds with (1.31) for subsequence, and so we have the conclusion (cf. Remark 1.13).

1.6 Convex Functional Throughout this section, X is a real Banach space with dual X ∗ . Definition 1.19 A functional f : X → R ≡ R ∪ +∞ is said to be proper if it is not identically equal to ∞. If f (x) ∈ (−∞, +∞) ∀x ∈ X , then it is said to be finite. If a function ϕ : X → R satisfies the inequality   ϕ (1 − λ)x + λy ≤ (1 − λ)ϕ(x) + λϕ(y) ∀x, y ∈ X, ∀λ ∈ [0, 1] then it is said to be convex. The functional ϕ : X → R is said to be lower semi-continuous (l.s.c.) on X if lim inf ϕ(u) ≥ ϕ(v) ∀x ∈ X u→v

or, equivalently, every level subset {x ∈ X : ϕ(x) ≤ λ} is closed. The functional ϕ : X → R is said to be weakly lower semi-continuous if it is lower semi-continuous on the space X endowed with weak topology. Because every level set of a convex functional is convex and every closed convex set is weakly closed (see Theorem 1.1), we may therefore conclude that a proper convex functional is lower semi-continuous if and only if it is weakly lower semicontinuous (see Sect. 1.2 of [29]). As a special case of Definition 1.2 we have Definition 1.20 Given a functional f : X → R, D f (x, h) = lim λ↓0

f (x + λh) − f (x) x, h ∈ X λ

(if it exists) is called the directional derivative of f at x in direction h.

(1.34)

34

1 Miscellanea of Analysis

If D f (x, h) is linear and continuous with respect to h, that is, there exists ∇ f (x) ∈ X ∗ such that D f (x, h) = ∇ f (x), h ∀h ∈ X, the functional f : X → R is said to be Gâteaux differentiable at x ∈ X and ∇ f (x) ∈ X ∗ is called the Gâteaux derivative at x. If D f (x, h) in (1.34) is linear and continuous with respect to h and convergence is uniform in h on bounded subsets, then f is said to be Fréchet differentiable and ∇ f (x) is called the Fréchet derivative of f at x. Remark 1.15 For a finite functional f : X → R, if there exists the Gâteaux derivative at all x ∈ X , then the operator ∇ f : X → X ∗ defined by x → ∇ f (x) is called gradient of f . When a finite functional f has its gradient ∇ f, the functional f is convex if and only if ∇ f : X → X ∗ is monotone (see Theorem 4.4 of [19] or Lemma 4.10, Chap. 3 of [3]). Every monotone gradient operator of a finite functional is demi-continuous (see the footnote9 in p. 113 of [19] or Lemma 4.12, Chap. 3 of [3]). Definition 1.21 Given a l.s.c., convex, proper function ϕ : X → R, the mapping ∗ ∂ϕ : X → 2 X defined by ∂ϕ(x) = {x ∗ ∈ X ∗ : ϕ(y) − ϕ(x) ≥ x ∗ , y − x ∀y ∈ X } is called the subdifferential of ϕ. In general, ∂ϕ is a multivalued operator not everywhere defined. We denote by D(∂ϕ) the set of all x ∈ X for which ∂ϕ(x) = ∅. If a functional ϕ is convex and there exists Gâteaux derivative ∇ϕ(x) at x, then ∂ϕ(x) = ∇ϕ(x) (see Sect. 1.2 of [29]). By the definition of ∂ϕ(x) we see that ϕ(x) = inf{ϕ(y); y ∈ X } if and only if 0 X ∗ ∈ ∂ϕ(x). Example 1.2 Let Ω be a bounded domain, K (Ω) be a nonempty convex closed subset of a Sobolev space H 1 (Ω), Γ be a open subset of ∂Ω and g(x) ∈ L 2 (Γ ). Let a functional Φ(u) : H 1 (Ω) → R be defined by ⎧ ⎨ g(x)u d x Φ(u) = Γ ⎩ +∞

∀u ∈ K (Ω), ∀u ∈ / K (Ω).

Then the functional Φ is proper, convex and weakly lower semi-continuous.

1.6 Convex Functional

35

Theorem 1.48 (Theorem 2.8 of [29]) Let X be a real Banach space and let Φ : X → R be a l.s.c. proper convex function. Then ∂Φ is a maximal monotone operator such ∗ that X → 2 X . In what follows of this section for simplicity let X be a real Hilbert space and X ∗ be identified with X . Let A be a maximal monotone operator in X . For every x ∈ X the equation 0 X ∈ (xε − x) + ε Axε has a unique solution xε . Thus xε = (I + ε A)−1 x, where I is the unit operator on X . The operator Aε : X → X defined by Aε x =

 1 1 (x − xε ) ≡ I − (I + ε A)−1 x ε ε

is called the Yosida approximation of A. Let the functional Φ : X → R be proper, convex and weakly lower semicontinuous (l.s.c.). For every 0 < ε < 1, define a functional Φε by  y − u2X + Φ(u); u ∈ X , Φε (y) = inf 2ε 

y ∈ X,

(1.35)

which is called the Moreau-Yosida approximation (or regularization) of Φ. When ∂Φ : X → 2 X is the subdifferential of Φ in the Hilbert space X , taking into account Theorem 1.48 let Jε = (I + ε∂Φ)−1 . Then the Yosida approximation of ∂Φ is as follows: (∂Φ)ε = ε−1 (I − Jε ) ∀ε > 0. From Theorem 2.9 of [29], we have Theorem 1.49 Let X be a real Hilbert space and let Φ : X → R be a l.s.c. proper convex functional. Then the functional Φε is convex, continuous, Fréchet differentiable, and ∇Φε = (∂Φ)ε . Moreover Φε (y) =

y − Jε y2X + Φ(Jε y) ∀y ∈ X, 2ε

lim Φε (y) = Φ(y), Φ(Jε y) ≤ Φε (y) ≤ Φ(y) ∀y ∈ X.

ε→0

(1.36) (1.37)

36

1 Miscellanea of Analysis

Remark 1.16 The operator ∇Φε is Lipschitz continuous with the constant 1/ε (see Theorem 2.9 and Proposition 2.3 of [29]). The Moreau-Yosida approximation Φε of a l.s.c. proper convex functional Φ is finite, convex, Fréchet differentiable, and so ∇Φε is monotone (see Remark 1.15). If a function u(t) : [0, T ] → X is differentiable at t ∈ [0, T ], then since ∇Φε is Lipschitz continuous and monotone (for notation we refer to Remark 1.17),     ∇Φε u(t + h) − ∇Φε u(t) u(t + h) − u(t) , lim (∇Φε (u(t)) , u(t) = lim h→0 h→0 h h     1 = lim 2 ∇Φε u(t + h) − ∇Φε u(t) , u(t + h) − u(t) h→0 h ≥ 0. (1.38) 



(cf. (5.41), Chap. 1 of [30] or p. 116 of [31]) Remark 1.17 Let V, H be real Hilbert spaces such that V → H and V is dense in H . Then, we have that H ∗ → V ∗ (see Sect. 1.1.1). Denote by  f, v the value of f ∈ V ∗ at v ∈ V . Identifying H and H ∗ by Riesz’s theorem, we have V → H → V ∗ .

(1.39)

In this case, H → V ∗ means that an element u of H is regarded as an element of V ∗ in the sense that (1.40) u, v = (u, v) H ∀v ∈ V. On the other hand, by Riesz’s theorem, there exists an element Ru = gu ∈ V corresponding to u ∈ V ∗ such that u, v = (Ru, v)V = (gu , v)V ∀v ∈ V.

(1.41)

According to inner products of V , the functions gu are different, however always (gu , v)V = (u, v) H ∀v ∈ V. Sometimes, identifying V and V ∗ by Riesz’s theorem, one finds a continuous linear functional f on V as an element of V, and so  f, v = ( f, v)V ∀v ∈ V.

(1.42)

If f ∈ V is not a functional and is an element of V , then it is also an element of H , and is an element of V ∗ by (1.39). In this case, the role of f ∈ V ∗ is represented as  f, v = ( f, v) H ∀v ∈ V,

(1.43)

1.6 Convex Functional

37

which is different from (1.42). To avoid confusion, one can write the left-hand side of (1.42) as R −1 f, v . However, if it is obvious that f represents a functional on V and is not the functional via (1.39), then for simplicity we use (1.42).

1.7 Some Elementary Inequalities Theorem 1.50 (Young’s inequality) Suppose that a function f (x) is continuous, strictly monotone increasing in x ≥ 0, and f (0) = 0. Denote the inverse function of f by f −1 . For a, b > 0, we have 



a

ab ≤

f (x) d x +

0

b

f −1 (x) d x

0

and the equality holds only if b = f (a). (See Sect. 8.3 of [9].) In particular, for f (x) = x p−1 (1 < p < ∞), we have that for s, t ≥ 0 st ≤

tp sq + , p q

1 1 + = 1. p q

By this inequality, we have that for u ∈ L p (Ω) and v ∈ L q (Ω)  Ω

|u(x)v(x)| d x ≤

ε 1 u pp + q−1 vqq p qε

∀ε > 0.

Theorem 1.51 (Bellman-Gronwall’s inequality) Let u(t) and n(t) be real continuous functions on J = [0, T ], f (t) ≥ 0 be integrable on J , and  u(t) ≤ n(t) +

t

f (s)u(s) ds.

α

Then for all t ∈ J  u(t) ≤ n(t) +

t



%

f (s)n(s) exp

0

t

& f (r ) dr ds.

(1.44)

s

In addition, if n(t) is nondecreasing, then  u(t) ≤ n(t) exp

t

f (s) ds,

(1.45)

0

which is called Gronwall’s inequality Remark 1.18 For the proof of (1.45) in the case that f ≡ c ≥ 0 and n(t) is a real and nonnegative function, we refer to Lemma 1.3, Chap. 5 of [3]. For the proof of

38

1 Miscellanea of Analysis

(1.45) in the case that u(t), f (t) are continuous and nonnegative functions and n(t) is a continuous, positive and nondecreasing function, we refer to Theorem 1.1.4 of [32]. Following the proof of Theorem 1.2.1 of [32], we can prove the general case.

References 1. H. Brézis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (Springer, 2011) 2. K. Yosida, Functional Analysis (Springer, 1995) 3. H. Gajewski, K. Gr¨oger, K. Zacharias, Nichtlineare Operatorgleichungen und Operatordifferentialgleichungen (Academic Press, Berlin, 1974) (Russian 1978) 4. P. Drábek, J. Milota, Methods of Nonlinear Analysis (Birkhäser, 2007) 5. E. Zeidler, Nonlinear Functional Analysis and Its Applications I (Springer, 1986) 6. P. Kuˇcera, Basic properties of solution of the non-steady Navier-Stokes equations with mixed boundary conditions in a bounded domain. Ann. Univ. Ferrara 55, 289–308 (2009) 7. A.N. Kolmogorov, S.V. Fomin, Element of Theory of Functions and Functional Analysis (Moscow “Science”, Russian, 1981) 8. H.O. Fattorini, Infinite Dimensional Optimization and Control Theory (Cambridge University Press, Cambridge, 1999) 9. R.A. Adams, J.J.F. Fournier, Sobolev Spaces (Academic Press, 2005) 10. A. Lunardi, Interpolation Theory, 2nd edn. (Edizioni della Normale, Pisa, 2009) 11. J. Bergh, J. Löfström, Interpolation Spaces (Springer, 1976) 12. V.I. Bogachev, Measure Theory I, II (Springer, 2007) 13. P. Grisvard, Elliptic Problems in Nonsmooth Domains (SIAM, Philadelphia, 1985) 14. V. Girault, P.A. Raviart, Finite Element Methods for Navier-Stokes Equations (Springer, 1986) 15. J.L. Lions, E. Magenes, Non-Homogeneous Boundary Value Problems and Applications I (Springer, Berlin, 1972) 16. L. Grafakos, Modern Fourier Analysis (Springer, 2009) 17. F. Demengel, G. Demengel, Functional Spaces for the Theory of Elliptic Partial Differential Equations (Springer, 2012) 18. L.C. Evans, Partial Differential Equations (American Mathematical Society, 2010) 19. T. Roubíˇcek, Nonlinear Partial Differential Equations with Applications (Birkhäuser Verlag, 2005) 20. P.G. Ciarlet, Mathematical Elasticity I: Three-Dimensional Elasticity (North-Holland, 1988) 21. W. Pompe, Korn’s first inequality with variable coefficients and its generalization. Comment. Math. Univ. Carol. 44, 57–70 (2003) 22. The Mathematical Society of Japan, Encyclopedic Dictionary of Mathematics, 2edn. (The MIT Press, Cambridge, 1993) 23. E. Hille, R.S. Phillips, Functional Analysis and Semi-Groups (American Mathematical Society Providence, Rhode Island, 1957) 24. H. Triebel, Interpolation Theory, Function Spaces, Differential Operators (Johann Ambrosius Barth Verlag, 1995) 25. J. Simon, Compact sets in the space L p (0, T ; B). Ann. Math. Pure Appl. 146(4), 65–96 (1987) 26. E. Zeidler, Nonlinear Functional Analysis and Its Applications II/B (Springer, 1990) 27. J.L. Lions, Quelques Méthodes de Résolution des Problèmes aux limites Non linéaires (Dunod, Gauthier-Villars, Paris, 1969) (Russian 1972) 28. H. Brézis, Perturbation non lineaire d’opérateurs maximaux montones. C. R. Acad. Sci. Paris Ser. I(269), 566–569 (1968)

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29. V. Barbu, Nonlinear Differential Equations of Monotone Types in Banach Spaces (Springer, 2010) 30. G. Duvaut, J.L. Lions, Inequalities in Mechanics and Physics (Springer, Berlin, 1976) 31. J. Naumann, On evolution inequalities of Navier-Stokes type in three dimensions. Ann. Math. Pure Appl. 124(4), 107–125 (1980) 32. Y. Qin, Integral and Discrete Inequalities and Their Applications I (Birkhäuser, 2016)

Chapter 2

Fluid Equations

In this chapter, we first show how the Navier-Stokes equations and the equations of motion for fluid under consideration of heat are derived. Next, we outline some boundary conditions for the Navier-Stokes equations, mainly being concerned with the ones dealt with in this book. Last, we consider three kind of bilinear forms for the Stokes and Navier-Stokes equations, variational formulations for the Navier-Stokes problems with mixed boundary conditions and establish the equivalence between the variational formulations and the original PDE problems.

2.1 Derivation of Equations for Fluid Motion In this section, on the basis of conservation laws of mass, momentum and energy in classical mechanics and physics, we derive the equations of motion for fluid. The purpose of this section is for readers to be familiar with different kinds of equations of motion for incompressible fluids as well as for compressible fluids.

2.1.1 Navier-Stokes Equations 2.1.1.1

Acceleration of Fluid Particle

Let (x, t) be coordinates in Rl × (−∞, +∞), l = 2, 3. There are two ways of describing a fluid motion: Lagrangian (material) description and Eulerian (spatial) description.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 T. Kim and D. Cao, Equations of Motion for Incompressible Viscous Fluids, Advances in Mathematical Fluid Mechanics, https://doi.org/10.1007/978-3-030-78659-5_2

41

42

2 Fluid Equations

a kinematic equation of fluid particles x = Φ(x0 , t) =   In Lagrangian description, φ1 (x0 , t), · · · , φl (x0 , t) is given, which represents the location of a particle at t whose position was x0 at the initial instant t = 0. The velocity and acceleration of a fluid particle in the Lagrangian description are given, respectively, by u(x, t) =

∂Φ(x0 , t) du(x, t) ∂ 2 Φ(x0 , t) . and = ∂t dt ∂t 2

In Eulerian description, velocity of fluid particle passing a spatial point x at an instant t, that is, velocity field v(x, t) = (v1 (x, t), · · · , vl (x, t)) is given. In the two descriptions, velocities of fluid particles passing x at instant t are the same, that is, u(x, t) = v(x, t), and acceleration is calculated by dv(Φ(x0 , t), t) du(x, t) ∂ 2 Φ(x0 , t) = = 2 ∂t dt dt ∂v(x, t)  ∂v(x, t) ∂φi (x0 , t) + = · ∂t ∂ xi ∂t i = The total derivative

∂v(x, t) + (v · ∇)v(x, t). ∂t dv ∂v = + (v · ∇)v dt ∂t

(2.1)

is called material derivative of v to emphasize the fact that the derivative is taken following a fluid element. Thus, acceleration of a fluid particle equals to material derivative of velocity field v(x, t). Similarly, we can consider material derivative of other fields, for example, density field ρ(x, t), which is the rate of change of the quantity following a fluid element. The instantaneous curves that are everywhere tangent to the velocity vector are called the streamlines of flow. For unsteady flows the streamline pattern changes with = 0, then the flow is called steady flow. The path line is the trajectory of time. If ∂v ∂t a fluid particle of fixed identity over a period of time. Path lines and streamlines are identical in a steady flow, but not in an unsteady flow. The Eulerian description is used in most problems of fluid flows, but the Lagrangian description is used occasionally, i.e. when one is interested in finding path lines of fixed particles. In this book we use Eulerian description. If Φ(x0 , t) of the kinematic equation is continuous with respect to their independent variables, then particles filling a domain Ωt at instant t fill a new domain Ωt1 at instant t1 . A domain moving with fluid as this is called a material domain and volume of material domain is called material volume. For material domain the following transport theorem is valid. Theorem 2.1 (Sect. 2.1 of [1]) Let Ωt be a material domain moving with fluid of velocity field v(x, t) and f (x, t) be scalar or vector function. Then

2.1 Derivation of Equations for Fluid Motion

d dt





 Ωt

2.1.1.2

f (x, t) d x =

Ωt

43

 ∂ f (x, t) + v · ∇ f (x, t) + f (x, t) div v(x, t) d x. ∂t (2.2)

Conservation of Mass

Let us fix a subdomain W of a domain Ω occupied by a fluid. Let m(W, t) be the mass of fluid in W at instant t and ρ(x, t)—density of fluid at (x, t). In this section, we assume that ρ(x, t) and v(x, t) are differentiable with respect to x and t. Then, the rate of change of fluid mass in the fixed subdomain W is as follows.   d d ∂ρ m(W, t) = (x, t) d x. ρ(x, t) d x = dt dt W W ∂t According to the conservation law of mass, the rate of change of fluid mass in the fixed subdomain W at t equals to the total sum of mass passing through the boundary of W in duration of unit time, and     ∂ρ ρ(x, t) v(x, t) · n ds, (x, t) d x = − W ∂t ∂W where n is outward unit vector normal to the boundary and ds is surface element of the boundary. Thus, we have   W

   ∂ρ(x, t) + div ρ(x, t)v(x, t) d x = 0. ∂t

Since W is arbitrary, from above we get the differential form of law of conservation of mass   ∂ρ(x, t) + div ρ(x, t)v(x, t) = 0, (2.3) ∂t which is called the continuity equation. In view of material derivative, it can be written as dρ(x, t) + ρ(x, t) div v(x, t) = 0. (2.4) dt If volume of any material domain is constant at all time t, the fluid is called incompressible. The rate of change of volume equals to the total sum of volume of fluid passing the boundary of subdomain W in duration of unit time. Since the rate of change of any volume Ωt of incompressible fluid equals to zero, we have d dt



 Ωt

dx =

 ∂Ωt

v(x, t) · n ds =

Thus, incompressibility is equivalent to

Ωt

div v(x, t) d x = 0.

44

2 Fluid Equations

div v(x, t) = 0.

(2.5)

The continuity equation (2.3) for incompressible fluid is rewritten dρ(x, t) = 0, dt which means that the density of particles of incompressible fluid is not changed for duration of motion. Thus, if the density of an incompressible fluid at initial instant equals over whole domain, that is, ρ(x, 0) = ρ0 = const (homogeneous), then for any t ≥ 0 the density of fluid ρ(x, t) equals to ρ0 . The problem ⎧ ⎨ dρ(x, t) = 0, dt ⎩ ρ(x, 0) = const is trivial, and for homogeneous incompressible fluids continuity equation (2.4) is equivalent to (2.5). Note that incompressibility of a fluid does not mean the fluid is homogeneous and for inhomogeneous fluid, incompressibility condition (2.5) is not equivalent to continuity equation (2.3).

2.1.1.3

Conservation of Momentum

To derive equations of motion for fluid relying on the conservation of momentum, let us clarify the forces acting on fluid elements. The forces acting on a fluid element can be divided conveniently into two classes, namely, body forces and surface forces. Body forces are those that arise from “action at a distance”, without physical contact. They result from the medium being placed in a certain force field, which can be gravitational, magnetic, electrostatic or electromagnetic in origin. They are distributed throughout the mass of the fluid and are proportional to the mass. Body forces are expressed per unit mass, which is denoted by f . Surface forces are those that are exerted on an area element by the surroundings through direct contact. The surface force acting on a small element of area in a continuous medium depends not only on the magnitude of the area but also on the orientation of the area. The stress, which is force per unit area on a plane, depends not only on the magnitude of the force and orientation of the plane but also on the direction of the force. Thus, specification of stress at a point requires two vectors, one perpendicular to the plane on which the force is acting and the other in the direction of the force. Then state of stress at a point is, in fact, specified by a stress tensor, which has nine components. Let {si j (x)} be stress tensor and n be the outward unit normal vector at point x of the boundary surface of a fluid element. Then the stress on the area element of boundary including x is given by

2.1 Derivation of Equations for Fluid Motion

45

σ (x) = S · n, which can be decomposed into normal and tangent components. Stress depends on deformation of a fluid element, which is described in terms of the strain rate tensor or the deformation rate tensor ⎛

∂v1 ∂ x1

 ⎜   E(v(x)) = εi j (v) = ⎝ 21 ∂∂vx21 +  1 ∂v3 + 2 ∂ x1





∂v1 ∂ x2  ∂v1 ∂ x3

1 ∂v1 + 2 ∂ x2 ∂v2  ∂ x2 1 ∂v3 + 2 ∂ x2

 1  ∂v1 + 2  ∂ x3 1 ∂v2 +  2 ∂ x3∂v3 ∂v2

∂v2 ∂ x1

∂ x3

⎞

∂v3 ∂ x1  ∂v3 ⎟ ∂ x2 ⎠ .

(2.6)

∂ x3

The relation between the stress and deformation in a continuum is called a constitutive equation. When the constitutive equation is the first-order equation, that is, the stress tensor S(v, p) is given by si j = μ2εi j (v) + Cδi j , where δi j is the Kronecker symbol and μ is the dynamic viscosity coefficient, the fluid is called Newtonian fluid. For thermodynamical reasons, C is of the form C = λ div v − p, where p denotes the pressure. Therefore, the stress tensor of Newtonian fluid S(v) is given by si j = 2μεi j (v) + λ div v δi j − pδi j .

(2.7)

(cf. Sect. 5.3 of [2]). By the second law of thermodynamics it is known that μ ≥ 0, 2μ + 3λ ≥ 0 (see Exercise 2.11 of [1]). A fluid is called perfect if μ = λ = 0. (λ + 23 μ is called the coefficient of bulk viscosity. For many applications, the Stokes assumption 2 λ+ μ=0 3 is found to be sufficiently accurate, and can also be supported from the kinetic theory of monatomic gases. In [3], the fluid such that λ = − 23 μ holds is called the Newtonian fluid.) The surface force acting on a fluid element W by the surroundings is  ∂W

S(v, p) · n ds

and the conservation law of momentum on fluid element W is      ∂v(x, t) + (v · ∇)v(x, t) d x = ρ ρ f (x, t) d x + S(v, p) · n ds. ∂t W W ∂W (2.8) From (2.8) we get the differential equation of conservation law of momentum

46

2 Fluid Equations

 ρ

 ∂v(x, t) + (v · ∇)v(x, t) = ρ f (x, t) + div S(v, p), ∂t

(2.9)

 T where div S(v, p) is the vector div S1 (v, p), div S2 (v, p), div S3 (v, p) ,   Si (v, p) = si1 (v, p), si2 (v, p), si3 (v, p) , i = 1, 2, 3. By (2.7) for Newtonian fluid we have 

   ∂v ρ + (v · ∇)v = ρ f (x, t) + div 2μE(v) + λ∇(div v) − ∇ p. ∂t

(2.10)

Since for vectors w, u ⎞ w1 u 1 w1 u 2 w1 u 3 w ⊗ u = ⎝w2 u 1 w2 u 2 w2 u 3 ⎠ , w3 u 1 w3 u 2 w3 u 3 ⎛

we have div (w ⊗ u) = w div u + (u · ∇)w. Then, div (ρv ⊗ v) = div (v ⊗ ρv) = v div (ρv) + ρ(v · ∇)v.

(2.11)

In view of (2.11) and continuity equation (2.3), Eq. (2.10) is rewritten as follows:     ∂(ρv) + div (ρv) ⊗ v = ρ f (x, t) + div 2μE(v) + λ∇(div v) − ∇ p. (2.12) ∂t Since for Ei (v) = (εi1 (v), εi2 (v), εi3 (v)) (i = 1, 2, 3)   ∂ div 2Ei (v) = Δvi + div v ∂ xi

(2.13)

(see (2.77)), in the case that μ = const (2.10) and (2.12) become, respectively,  ρ

 ∂v + (v · ∇)v = ρ f (x, t) + μΔv + (μ + λ)∇(div v) − ∇ p ∂t

(2.14)

and   ∂(ρv) + div (ρv) ⊗ v = ρ f (x, t) + μΔv + (μ + λ)∇(div v) − ∇ p. ∂t

(2.15)

2.1 Derivation of Equations for Fluid Motion

2.1.1.4

47

The Navier-Stokes Equations

Now, from (2.14), (2.15) and (2.3) we get the system of equations of motion for the Newtonian fluid  ⎧    ∂v ⎪ ⎪ + (v · ∇)v = ρ f (x, t) + div 2μE(v) + λ∇(div v) − ∇ p, ⎨ρ ∂t ⎪ ⎪ ⎩ ∂ρ + div (ρv) = 0 ∂t

(2.16)

or ⎧ ∂(ρv)   ⎪ + div (ρv) ⊗ v = ρ f (x, t) + μΔv + (μ + λ)∇(div v) − ∇ p, ⎨ ∂t (2.17) ⎪ ⎩ ∂ρ + div (ρv) = 0. ∂t From (2.16) we have the system of equations of motion for the incompressible Newtonian fluids ⎧     ∂v ⎪ ⎪ + (v · ∇)v = ρ f (x, t) + div 2μE(v) − ∇ p, ⎪ρ ⎪ ⎪ ∂t ⎨ div v = 0, ⎪ ⎪ ⎪ ⎪ ∂ρ(x, t) ⎪ ⎩ + ∇ρ(x, t) · v(x, t) = 0. ∂t

(2.18)

For the incompressible fluid with a constant density, continuity equation (2.3) is equivalent to (2.5), and from (2.18) we have ⎧ ⎨ ∂v − div 2νE(v) + (v · ∇)v + 1 ∇ p = f (x, t), ∂t ρ ⎩ div v = 0,

(2.19)

where ν = μ/ρ is called kinematic viscosity coefficient. In view of (2.13), we have the equations of motion for incompressible Newtonian fluids with constant kinematic viscosity and constant density, the Navier-Stokes equations ⎧ ⎨ ∂v − νΔv + (v · ∇)v + 1 ∇ p = f (x, t), ∂t ρ ⎩ div v = 0.

(2.20)

On the other hand, in the case that l = 2, for convenience, vector v = (v1 (x1 , x2 ), v2 (x1 , x2 )) is identified with v¯ = (v1 (x1 , x2 ), v2 (x1 , x2 ), 0). Then, we have

48

2 Fluid Equations

  ∂v2 ∂v1 , rot v = rot v¯ ≡ 0, 0, − ∂ x1 ∂ x2 and in the case that l = 2, 3 − Δv = rot rot v − ∇ (div v), 1 (v · ∇)v = rot v × v + ∇ |v|2 . 2

(2.21)

By using the relation above, (2.20) is rewritten as ⎧ ⎨ ∂v + ν rot rot v + (v · ∇)v + 1 ∇ p = f (x, t), ∂t ρ ⎩ div v = 0, ⎧ ⎨ ∂v − νΔv + rot v × v + 1 ∇q = f (x, t), ∂t ρ ⎩ div v = 0,

(2.22)

where q = 21 ρ|v|2 + p is called total pressure or Bernoulli’s pressure. To avoid confusion, sometimes p is called the static pressure. Remark 2.1 Due to the conservation law of momentum we get a dynamic equation (2.9). There is another fundamental law of dynamics in classical mechanics, the law of conservation of angular momentum. For an arbitrary continuous medium obeying the continuity equation (2.3) and the dynamical equation (2.9), the symmetry of stress tensor is equivalent to that the equation by conservation of angular momentum holds (see Theorem 2.1 of [1]). Thus, for Newtonian fluid with symmetry stress tensor, a dynamical equation by conservation of angular momentum is equivalent to (2.9). Symmetry of stress tensor is true, however, under a crucial assumption that all internal momenta of the continuum are neglected which might not be sufficient for some materials. Therefore, for modeling of liquid crystals, micropolar fluid and ferrofluid an extra equation for conservation of angular momentum is used (see [4–6] and references therein). Remark 2.2 Any displacement of a rigid body can be expressed by composition of a translation and a rotation. A velocity field v can be regarded as a velocity field of a rigid body motion if and only if E(v(x, t)) = 0 (see Theorem 5.1 of [2]). Remark 2.3 Laws of conservation of momentum and angular momentum are universal, but constitutive laws (equations) depend on physical properties of materials under consideration. Whenever different nonlinear constitutive equations are taken for materials, then, correspondingly, different equations for motion of non-Newtonian fluids are obtained.

2.1 Derivation of Equations for Fluid Motion

49

2.1.2 Equations of Motion for Fluid Under Consideration of Heat 2.1.2.1

Conservation of Energy

Let f (x, t) be a scalar function and ρ(x, t) a density field. Then by the transport theorem we have     ∂ d (ρ f )(x, t) + div (ρ f v)(x, t) d x ρ f dx = dt Ωt ∂t Ω   t ∂ρ ∂f f +ρ + div (ρv) f + ρv · ∇ f d x = ∂t ∂t Ω   t  ∂ρ  ∂ f  + div (ρv) f + ρ + v · ∇ f d x. = ∂t ∂t Ωt Hence, in view of the continuity equation, we get d dt



 Ωt

ρ f dx =

Ωt

   ∂ f  d ρ f d x. ρ + v · ∇ f dx = ∂t dt Ωt

(2.23)

Let e(x, t) be internal energy of unit mass of fluid at (x, t). The total energy of a material Ωt at time t is the sum of its kinetic energy and of its internal  domain  energy: Ωt ρ e + 21 |v|2 d x. By (2.23) the rate of change of the total energy of Ωt is as follows:     1 1 2 d d e + |v|2 d x. ρ e + |v| d x = ρ dt Ωt 2 dt 2 Ωt This equals to sum of the work done by body force and by surface force, heat flux through the boundary and heat radiated by the inside heat source for the duration of unit time. Let g(x, t) be mass density of inside heat source and χ be a surface density of heat. It is known that χ is expressed by χ = −q · n, where n is the unit outward normal on the boundary and q is the heat flux vector at (x, t). Therefore, taking into account symmetry of S, we have

50

2 Fluid Equations

 Ωt

ρ

 1 d e + |v|2 d x dt 2     = ρ f · v dx + (S(v)n) · v ds − q · n ds + ρg d x Ω ∂Ω ∂Ωt Ωt  t  t   = ρ f · v dx + div (S1 · v, · · · , Sl · v) d x − div q d x + Ωt

Ωt

Ωt

Ωt

ρg d x.

(2.24)

Taking into account (2.1) and making the Rl -inner product of (2.9) with v, we have 1 d 2 (2.25) ρ |v| = ρ f (x, t) · v + div S(v) · v. 2 dt Substituting (2.25) into (2.24), we have  Ωt

ρ

d e dx = − dt

 

+

Ωt

Ωt

div S(v) · v d x  div (S1 · v, · · · , Sl · v) d x −

Ωt

 div q d x +

Ωt

ρg d x. (2.26)

By symmetry of (2.6), div (S1 · v, · · · , Sl · v) = div S(v) · v +

 ij

si j

 ∂vi = div S(v) · v + si j εi j (v), ∂x j ij (2.27)

and from (2.26) and (2.27) we get a differential equation for internal energy   ∂e + (v · ∇)e = si j εi j (v) − div q + ρg. ρ ∂t ij 

(2.28)

In view of (2.7), we have from (2.28) a differential equation for internal energy of Newtonian fluid 

 ∂e ρ + (v · ∇)e = 2μE(v) : E(v) + λ(div v)2 − p div v − div q + ρg, (2.29) ∂t  where E(v) : E(v) = i j εi j εi j . The first three terms on the right-hand side express change of internal energy per unit time by deformation of fluid. Relying on the rate of change of total energy on a material domain Ωt , we got (2.24). Now using a fixed element ω, not a material domain, we can get another equation. For this we additionally need to take care of the flow of total energy through the boundary of fluid element ∂ω. Then, we have

2.1 Derivation of Equations for Fluid Motion

51

       d 1 1 ρ e + |v|2 d x = ρ e + |v|2 (v · n) ds + ρ f · v dx dt ω 2 2 ∂ω ω    (S(v)n) · v ds − q · n ds + ρg d x + ∂ω

∂Ωt

ω

    1 div ρv e + |v|2 d x + ρ f · v dx = 2 ω ω    + div (S1 · v, · · · , Sn · v) d x − div q d x + ρg d x, 

ω

ω

ω

which yields the differential equation      1 1 ∂  ρ e + |v|2 + div ρ e + |v|2 v ∂t 2 2 = ρ f · v + div (S1 · v, · · · , Sn · v) − div q + ρg.

(2.30)

In view of (2.27), we have from (2.30) the differential equation for total energy      1 ∂  1 ρ e + |v|2 + div ρ e + |v|2 v ∂t 2 2  (2.31) = ρ f · v + div S(v) · v + si j εi j (v) − div q + ρg. ij

For Newtonian fluid, taking into account (2.7), we have from (2.31)     ∂   1 2  1 ρ e+ |v| + div ρv e + |v|2 v ∂t 2 2 = ρ f · v + div S(v) · v + 2μE(v) : E(v) + λ(div v)2 − p div v − div q + ρg.

(2.32)

Using (2.7), we can rewrite (2.32) as      1 1 ∂  ρ e + |v|2 + div ρv e + |v|2 v ∂t 2 2 = ρ f · v + div (2μE(v)) · v − ∇ p · v + λ∇(div v) · v + 2μE(v) : E(v) + λ(div v)2 − p div v − div q + ρg, which yields the differential equation for total energy of Newtonian fluid     1 ∂   1 2  ρ e+ |v| + div ρv e + |v|2 + pv ∂t 2 2 = ρ f · v + div (2μE(v)) · v + λ∇(div v) · v

(2.33)

+ 2μE(v) : E(v) + λ(div v)2 − div q + ρg. Let us show that (2.33) is equivalent to (2.29). To this end, we rewrite the terms on the left-hand side of (2.33).

52

2 Fluid Equations

 ∂ρ 1 ∂e ∂ρ |v|2 ∂v ∂  ρ e + |v|2 = e+ρ + + ρv , ∂t 2 ∂t ∂t ∂t 2 ∂t    1 2 1 div ρv e + |v| + pv = div (ρv)e + ρ(v · ∇)e + div (ρv) |v|2 2 2 + ρv · (v · ∇)v + ∇ p · v + pdiv v.

(2.34)

Taking into account (2.3) and ∂v dv = + (v · ∇)v, dt ∂t de ∂e = + (v · ∇)e dt ∂t (see (2.2)) and adding two equations of (2.34), we have     ∂   1 2  1 ρ e+ |v| + div ρv e + |v|2 + pv ∂t 2 2 dv ∂e 1 d|v|2 de + ρv = ρ( + (v · ∇)e) + ρ + ∇ p · v + p div v. =ρ dt dt ∂t 2 dt (2.35) By (2.25) and (2.7), we have ρ

1 d 2 |v| = ρ f (x, t) · v + div S(v) · v 2 dt = ρ f · v + div (2μE(v)) · v − ∇ p · v + λ∇(div v) · v.

Thus, we get (2.29) from (2.33) taking into account the above and (2.35). The relation between the heat flux vector and temperature in thermodynamics is also called a constitutive equation. We take Fourier’s law as the constitutive equation q = −k∇θ,

(2.36)

where k is thermal conductivity and θ (x, t) is temperature of fluid. Denoting the specific heat of fluid by γ and assuming the specific internal energy of the fluid is proportional to its temperature, we have e = γ θ.

(2.37)

Specific heat, thermal conductivity and viscosity of fluid, in general, depend on the temperature of fluid. Taking into account (2.37), for the incompressible Newtonian fluid with constant density we have from (2.29)

2.1 Derivation of Equations for Fluid Motion

53

     ∂ γ (θ )θ (x, t) − div κ(θ )∇θ + v · ∇(γ (θ )θ ) − 2ν(θ )E(v) : E(v) = g, ∂t (2.38) where κ = ρk . In the left-hand side of (2.38) the second term expresses heat flow by conducting, the third one expresses heat transport by fluid movement (convection) and the fourth one represents dissipation of kinetic energy by viscous friction of fluid (the Joule effect). As in [3], in this book we shall reserve the term convection to describe heat transport by fluid movements, and so (v · ∇)v describing movement of fluid from place to place is called “advection” (see pp. 55–56 of [3]). Such terminology is suitable with the terms: the convective heat transfer, convection diffusion, heat convection and so on.

2.1.2.2

Equations of Motion for Fluid Under Consideration of Heat

From (2.16), (2.17) and (2.29) we get the system of equations of Newtonian fluid under consideration of internal energy  ⎧    ∂v ⎪ ⎪ ρ + (v · ∇)v = ρ f (x, t) + div 2μE(v) + λ∇(div v) − ∇ p, ⎪ ⎪ ∂t ⎪ ⎪ ⎨ ∂ρ + div (ρv) = 0, ⎪ ∂t ⎪  ⎪ ⎪ ⎪ ∂e ⎪ ⎩ρ + (v · ∇)e = 2μE(v) : E(v) + λ(div v)2 − p div v − div q + ρg ∂t (2.39) or, equivalently, ⎧   ∂(ρv) ⎪ ⎪ + div (ρv) ⊗ v = ρ f (x, t) + μΔv + (μ + λ)∇(div v) − ∇ p, ⎪ ⎪ ∂t ⎪ ⎪ ⎨ ∂ρ + div (ρv) = 0, ∂t ⎪ ⎪  ⎪ ⎪ ∂e ⎪ ⎪ ⎩ρ + (v · ∇)e = 2μE(v) : E(v) + λ(div v)2 − p div v − div q + ρg. ∂t (2.40) Remark 2.4 If we use the equation of total energy instead of the equation of internal energy, we get the system of equations of motion of Newtonian fluids under consideration of total energy

54

2 Fluid Equations

⎧   ∂(ρv) ⎪ ⎪ + div (ρv) ⊗ v = ρ f (x, t) + μΔv + (μ + λ)∇(divv) − ∇ p, ⎪ ⎪ ∂t ⎪ ⎪ ⎪ ⎪ ∂ρ ⎨ + div (ρv) = 0, ∂t ⎪ ⎪ ∂     1 2  1 2 ⎪ ⎪ ⎪ ⎪ ∂t ρ e + 2 |v| + div ρv e + 2 |v| + pv = ρ f · v + div (2μE(v)) · v ⎪ ⎪ ⎩ + λ∇(div v) · v + 2μE(v) : E(v) + λ(div v)2 − div q + ρg. (2.41) The total number of equations constituting (2.39) (or (2.40)), (2.36) and (2.37) is less than the total number of unknown functions, and so one usually needs, in addition, a relation in terms of pressure, density and temperature. Then, the left-hand side of the new system obtained from (2.41) is written as ∂u + div F(u), ∂t which is convenient for study of equations for compressible fluid. When the density of fluid is constant and it is independent of temperature, by (2.19) and (2.38), we have the system of equations of motion of incompressible Newtonian fluids under consideration of heat ⎧ ∂v   1 ⎪ − div 2ν(θ )E(v) + (v · ∇)v + ∇ p = f (x, t), ⎪ ⎪ ρ ⎪ ⎨ ∂t div v = 0, ⎪ ⎪ ⎪  ⎪ ∂  ⎩ γ (θ )θ (x, t) − div (κ(θ )∇θ ) + (v · ∇(γ (θ )θ )) − 2ν(θ )E(v) : E(v) = g(x, t). ∂t

(2.42)

If a fluid is in the gravitational field and density is not constant, then there is buoyancy. For slightly compressible homogeneous fluids in gravitational field, we simplify the equations for the mass and momentum conservation laws as follows: (i) ρ is assumed to be constant (= ρ0 ) everywhere in the equations, except in the gravity term; (ii) In the term of gravity, ρ is replaced by a linear function of θ ρ = ρ0 − α(θ − θ0 ), where ρ0 and θ0 are the average value of the density and temperature. If there is not any body force other than gravity, then f is the gravitational acceleration gg . Then density of buoyancy is −α(θ − θ0 )gg , and we have an approximation of the conservation of momentum 1 1 ∂v + (v · ∇)v − νΔv + ∇ p = [ρ0 − α(θ − θ0 )]gg . ∂t ρ0 ρ0

2.1 Derivation of Equations for Fluid Motion

55

Replacing the first equation of (2.42) with the above and neglecting energy dissipation by viscous friction of fluid, we have ⎧ ∂v 1 ⎪ + (v · ∇)v − νΔv + ∇ p = [1 − α1 (θ − θ0 )]gg , ⎪ ⎪ ⎪ ∂t ρ 0 ⎨ div v = 0, ⎪ ⎪ ⎪ ⎪ ⎩ ∂ γ θ (x, t) − div (κ∇θ ) + (v · ∇(γ θ )) = g, ∂t

(2.43)

which is used for heat convection in gravitational field and is called Boussinesq equations. Putting [1 − α1 (θ − θ0 )]gg = f (1 − α0 θ ) and adding the work α1 θ f · v done by the force α1 θ f due to heat expansion to the heat equation, we have ⎧ ∂v   1 ⎪ − div 2ν(θ )E(v) + (v · ∇)v + ∇ p = f (1 − α0 θ ), ⎪ ⎪ ⎪ ρ0 ⎨ ∂t div v = 0, ⎪ ⎪ ⎪ ⎪ ⎩ ∂ γ θ (x, t) − div (κ∇θ ) + (v · ∇(γ θ )) − 2νE(v) : E(v) = α θ f · v + g, 1 ∂t (2.44) which is a system of equations of motion for incompressible Newtonian fluid under consideration of dissipation of energy in gravitational field [7].

2.2 Boundary Conditions for the Navier-Stokes Equations When solving the Navier-Stokes equations, appropriate initial and boundary conditions need to be applied. The derivation of suitable boundary conditions for flow problems is not obvious and depends on the physics which is to be modeled. In this section, without deriving boundary conditions from point of view of physics, we outline those concerned in this book for the Stokes equations ∂v − μΔv + ∇ p = f, div v = 0 ∂t

in Ω ⊂ Rl , l = 2, 3,

(2.45)

and the Navier-Stokes equations ∂v − μΔv + (v · ∇)v + ∇ p = f, div v = 0 ∂t

in Ω.

(2.46)

56

2 Fluid Equations

Note that Eq. (2.46) is the one with ρ = 1 in (2.20), and p + 21 |v|2 is the total pressure instead of p + ρ 21 |v|2 . Let v = (v1 , · · · , vl ) be a solution to (2.45) or (2.46). The strain tensor E(v) is the matrix with components εi j (v) =

1 (∂x vi + ∂xi v j ) 2 j

and the stress tensor S(v, p) is the one with components si j = − pδi j + 2μεi j (v). In the present book the boundary surfaces concerned by us are pieces of boundary of 3-D or 2-D bounded connected domains, and so we can assume the surfaces are oriented. Thus, let n be an outward unit normal vector on a boundary and τ be unit vector tangent to the boundary. Then the stress vector on the boundary is σ (v, p) = S · n ≡ − pn + 2μE(v) · n,

(2.47)

the value of normal component of the stress vector is σn (v, p) = σ · n and its tangent component is στ = σ − σn · n. Replacing the static pressure p in the stress tensor by the total pressure (Bernoulli’s pressure) p + 21 |v|2 , we have the total stress tensor S t (v, p) with components 1 sit j = −( p + |v|2 )δi j + 2μεi j (v). 2 Then the total stress vector on the boundary is σ t (v, p) = S t · n, the value of normal component of the total stress is σnt (v, p) = σ t · n and its tangent component is στt = σ t − σnt · n.

(2.48)

2.2 Boundary Conditions for the Navier-Stokes Equations

57

Note that the tangent components of stress and total stress are equal, i.e. στ = στt . Denote the value of normal component v · n of velocity v by vn , and the tangent component of v is vτ = v − vn n. In some literatures, the terms “stress” and “total stress”, respectively, are used for (2.6) and (2.47), but in this book we will use the term total stress for (2.48). From mathematical point of view, there are four kinds of boundary conditions: (i) (ii) (iii) (iv)

Dirichlet boundary condition assigning variable value (velocity, pressure). Neumann boundary condition assigning (normal) gradient of variable. Combination of Dirichlet and Neumann boundary condition. Periodic boundary conditions assuming equality between values of a variable on distant parallel two planes.

From physical point of view, there are two kinds of boundary conditions: (i) Natural boundary conditions. These are boundary conditions on the real boundary of a domain occupied by fluid, for example, no-slip condition on the wall, threshold slip condition on the wall and so on. (ii) Artificial boundary conditions. These are boundary conditions on imaginary boundaries of fluid domain, for example, a boundary condition on symmetric planes usually taken to reduce computational burden, Navier slip-with-friction boundary condition on a surface close to a real rough wall and so on. According to relationship between fluid domain under consideration and surroundings, the boundaries of fluid domain may be classified as follows: (i) (ii) (iii) (iv)

Wall, Symmetric plane, Inlet and outlet, Free surface.

Boundary conditions first of all must be able to reflect to a mathematical formulation of problem and must be consistent with real phenomena. Some boundary condition involved to a mathematical formulation has not a correct physical meaning, which is used as an artificial boundary condition for numerical practice. In such a case for mathematical result to approximate the real phenomenon it is important where the artificial boundary is drawn. We outline what kinds of boundary conditions are used on every class of boundaries above when the domain of fluid is not variable along with time t.

58

2 Fluid Equations

2.2.1 Boundary Conditions on the Walls In practice, some domain of fluid has a solid wall boundary and changes of momentum, velocity and scalar quantities near the wall are serious. Thus, boundary condition on the wall must be given in accordance with physical situations.

2.2.1.1

Stick Boundary Condition on the Walls

Most solid surfaces are impermeable to fluid and the fluid sticks to the surfaces. Thus, there is no slip and no penetration, and the fluid particles on the wall move with the velocity of wall w: v = w. If the walls do not move, then stick condition on the wall Γ D ⊂ ∂Ω is the homogeneous Dirichlet condition (2.49) v = 0 on Γ D . Pressure on the wall is seriously variable according to the flow situation, and usually the pressure boundary condition on the wall is not given.

2.2.1.2

Slip Boundary Conditions on the Walls

When a wall is impermeable to fluid and fluid is not stick, a slip boundary condition may be used. Accurate experiments clearly demonstrate that the no-slip boundary condition postulated in most hydrodynamics investigations may be far to be fulfilled (see [8]). If a fluid slips without friction on the wall Γ , then Navier slip boundary condition vn = 0, στ (v) = 0 on Γ

(2.50)

is used. Sometimes free slip on a surface Γ is expressed by a vorticity boundary condition (2.51) vn = 0, rot v × n = 0 on Γ. In some papers the boundary condition (2.51) is called Navier slip-type condition [9–13], slip condition [14] or Hodge boundary condition [15, 16]. On flat portions of the boundary, the vorticity boundary condition (2.51) coincides with the classical Navier slip boundary condition (2.50), but usually these are different as a quantity depending on the shape of boundary surface (cf. Sect. 6 of [10], Remark 1.1 of [14], Remark 3.1 of [17]). The vorticity boundary condition (2.51) is also used on a portion of boundary of a tank containing fluid, which is closed by a membrane (see [18]).

2.2 Boundary Conditions for the Navier-Stokes Equations

59

C.-L. Navier claimed that the tangential component of the stress at the boundary should be proportional to the tangential velocity. This condition is expressed by vn = 0, στ (v) + αv = 0, α : constant or function on Γ,

(2.52)

which is called Navier slip-with-friction boundary condition (see [19–21]). The Navier slip-with-friction boundary condition was rigorously justified as a homogenization of the no-slip condition on a rough boundary and might be more realistic than the classical no-slip boundary condition when the wall is rough. The condition is used for simulations of flows near rough boundaries, such as in aerodynamics, in weather forecast, in haemodynamics, etc. When an artificial boundary condition near the real boundary with periodic micro-roughness is used, usually α is a matrix, which maps each vector tangent to boundary into such a vector. From now on, when we are concern with (2.52) with matrix α, we will always regard the matrix as being such a one. If the shape operator of the boundary Γ (see Sect. 3.1) is used instead α/2 in (2.52), then the condition (2.52) becomes (2.51) (see [13] and Sect. 3.1).

2.2.1.3

Threshold Slip Boundary Conditions on the Walls

According to Tresca friction law between two solids, there is not any slip on the surface when the magnitude of tangent stress is less than a threshold, but when the magnitude of tangent stress reaches the threshold, slipping can occur, the magnitude of tangent stress is not greater than the threshold and the direction of velocity is opposite to the tangent stress. Such phenomena between walls and fluids are expressed by (2.53) vn = 0, |στ (v)| ≤ gτ , στ (v) · vτ + gτ |vτ | = 0 on Γ, where gτ is a threshold of tangent stress for slip. This condition is called Tresca friction boundary condition or threshold slip boundary condition. Using subdifferential operator, we can rewrite (2.53) as vn = 0, −στ (v) ∈ gτ ∂|vτ |, 

where ∂|z| =

(2.54)

z = 0, z ∈ Rl , {w ∈ R , |w| ≤ 1} z = 0, z ∈ Rl . z |z|

l

Physical and experimental backgrounds of such boundary conditions are mentioned in several papers [22–24].

60

2.2.1.4

2 Fluid Equations

Leak Boundary Conditions on the Walls

Similarly to the threshold slip boundary condition, leak boundary conditions of friction type are used when one wants to model a flow problem involving a leak of flow through the boundary. Assume that the tangent component of velocity on a potion of boundary vanishes, there is not any leak through the surface when the magnitude of normal stress on a portion of boundary surface is less than a given threshold, leak through the surface can occur if the magnitude of normal stress reaches the threshold and the magnitude of normal stress is not greater than the threshold. Such phenomena on the boundary surface are expressed by vτ = 0, |σn (v)| ≤ gn , σn (v)vn + gn |vn | = 0 on Γ,

(2.55)

where gn (> 0) is a threshold of normal stress to show leak [24–27]. Using subdifferential operator, we can rewrite (2.55) as vτ = 0, −σn (v) ∈ gn ∂|vn |.

(2.56)

For physical and experimental backgrounds of such boundary conditions, we refer to [24, 27]. The boundary condition (2.55) or (2.56) is called threshold leak boundary condition. Another threshold leak boundary condition, which is based on the normal total stress, (2.57) vτ = 0, |σnt (v)| ≤ gn , σnt (v)vn + gn |vn | = 0 on Γ may be also used. (Note that since στ (v, p) = στt (v, p) = 2μεnτ (v), there is not another slip boundary condition based on στt (v, p).) The boundary conditions (2.55) and (2.57) mean that according to direction of normal stress, fluid may penetrate out or into through boundary. The stress on a boundary surface is density of force applied to the boundary from surroundings. Note n is outward unit vector on a boundary. Therefore, if σn (v) > 0, which means that direction of force acting on the boundary surface of domain of fluid by surrounding is outward, then the value of density of force reacting on the boundary by fluid equals to −σn (v) < 0, and (2.55) implies vn < 0, which means that fluid penetrates into through the boundary. If σn (v) < 0, which means that direction of force acting on the boundary surface of domain of fluid by surrounding is inward, then the value of density of force reacting on the boundary by fluid equals to −σn (v) > 0, which means that the reacting force is outward. On the other hand (2.55) implies vn > 0, which means that fluid penetrates out through the boundary. In practice, we are encountered with one-sided leak of fluid, for example, sand layers, a semipermeable membranes and so on. Assume that the tangent component of velocity on a portion of boundary vanishes, the fluid can only leak out through boundary, there is not any flow through the boundary

2.2 Boundary Conditions for the Navier-Stokes Equations

61

when −σn (v) is less than a threshold g+n (> 0), a leak out can occur if −σn (v) reaches g+n (> 0) and −σn (v) cannot be greater than the threshold. We can describe such a phenomenon by vτ = 0, vn ≥ 0, σn (v) + g+n ≥ 0, (σn (v) + g+n )vn = 0,

(2.58)

which is called threshold leak out boundary condition [28, 29]. In contrast, assume that a fluid can only leak into through boundary, there is not any flow through the boundary when −σn (v) is greater than −g−n (for a threshold g−n > 0), but there can be a leak into if −σn (v) is same as −g−n , and −σn (v) cannot be less than −g−n . Such a phenomenon on the boundary surface is expressed by vτ = 0, vn ≤ 0, σn (v) − g−n ≤ 0, (σn (v) − g−n )vn = 0,

(2.59)

which is called threshold leak into boundary condition. Other threshold one-sided leak boundary conditions vτ = 0, vn ≥ 0, σnt (v) + g+n ≥ 0, (σnt (v) + g+n )vn = 0, vτ = 0, vn ≤ 0, σnt (v) − g+n ≤ 0, (σnt (v) − g+n )vn = 0

(2.60)

based on the normal total stress may be also used. The threshold leak into or out boundary condition is called the one-sided leak boundary conditions. The threshold slip, leak and one-sided leak boundary conditions are called the boundary conditions of friction type.

2.2.2 Boundary Conditions on Symmetric Planes If flow field and geometry of domain of fluid is symmetric with respect to a plane, then to reduce the amount of computation for the problem a half of domain with the symmetry plane as an imaginary boundary may be under consideration. The normal velocity and normal gradients of all variables on a symmetry plane are zero, and tangent stress on the symmetric plane vanishes. Therefore, on the symmetric plane, the boundary conditions (2.61) vn = 0, στ (v) = 0 or vn = 0, rot v × n = 0 can be used.

(2.62)

62

2 Fluid Equations

2.2.3 Boundary Conditions on Inlets and Outlets If there are inlets and outlets on the boundary of fluid domain, then Dirichlet boundary conditions for velocity may be used. But pressure boundary condition is more common because it is difficult to know velocity profile except special cases. In some real-life situations, it is natural to prescribe the value of the pressure on some part of the boundary, as, for instance, in case of pipelines, blood vessels, different hydraulic systems involving pumps, etc. Thus, it seems perfectly reasonable to impose the pressure boundary condition on inlets and outlets for the Navier-Stokes system [30]. In practice according to measurement instruments, we can obtain the static pressure p or total pressure 21 |v|2 + p (Bernoulli’s pressure), and both the static pressure [30, 31] and total pressure [32, 33] may be used for boundary conditions on the real inlets and outlets. When pressure boundary conditions are used, direction of fluid flow (usually flow is orthogonal to the boundary) is given together. But it is known that the static pressure boundary condition and direction of flow are not enough to determine a velocity field of the Navier-Stokes equations. Moreover, when one of static pressure (instead of total pressure), stress (instead of total stress) or the outflow boundary conditions (see the next subsection) is given on a portion of boundary, for the initial boundary value problems of the Navier-Stokes equations only existence of a unique local-intime solution and a unique solution on a given interval for small given data (in what follows we call it a solution for small data) are proved. From the mathematical point of view, the main difficulty of such problems results from the fact that in the process of a priori estimation of velocity, the term   (v · ∇)v, v Ω arising from the nonlinear (advection) term (v · ∇)v is not canceled since the normal component of velocity on the portion of boundary does not vanish. From the mechanical point ofview, it is explained by the fact that the kinetic energy of fluid (with density 1) 21 Ω |v(t, x)|2 d x is not controlled by the data of problem and uncontrolled “backward flow” can take place at the portion of boundary (see Preface in [34]). If the total pressure condition is given, then mathematical treatment is more easy because by 1 (v · ∇)v = rot v × v + ∇ |v|2 2 the Navier-Stokes equations (2.46) are rewritten as 1 ∂v − μΔv + rot v × v + ∇( p + |v|2 ) = f, ∇ · v = 0 in Ω, ∂t 2 and in the process of a priori estimation of solutions the term   rot v × v, v Ω

(2.63)

2.2 Boundary Conditions for the Navier-Stokes Equations

63

arising from the nonlinear term rot v × v vanishes by a property of the mixed product of vectors. As a boundary condition on in/out-stream surfaces, a combination of the tangent component of the velocity and the normal component of stress vτ = 0, σn (v, p) = f

(2.64)

is also used [35]. If the velocity field v of the Navier-Stokes equations on a domain is obtained, then by de Rham Theorem (Proposition 2.3), pressure p is determined with different constants. If, moreover, pressure condition on a portion of boundary is given, then the constant is determined and pressure is uniquely determined when the solution is smooth enough for existence of traces of velocity and pressure on the portion of boundary. Also, if a boundary condition including pressure p as addend (stress, total stress, normal stress) on a portion of boundary is given, then the pressure on the domain is determined uniquely when the solution is smooth enough (see Theorem 4.1 of [36] and Theorems 5.1, 5.3).

2.2.4 Outflow Boundary Conditions on Imaginary Boundary To reduce amount of computation, imaginary boundaries may be drawn inside flow. On the portions of imaginary boundary where flow is out, velocity and pressure boundary conditions are not used since the velocity and pressure on the potion of imaginary boundary are not known prior to prescription of solution. In such portions of imaginary boundary, the free outflow boundary conditions μ

∂v − pn = 0 ∂n

(2.65)

[34, 37–47], which is called “do nothing” boundary condition, and μE(v)n − pn = 0

(2.66)

[47, 48] are used. “Do nothing” boundary condition (2.65) results from variational formulation based on Dirichlet bilinear form (∇v, ∇u) obtained by integrating by parts (Δv, u) (see (2.86)) and does not have a real physical meaning, but is rather used in truncating large physical domains to get smaller computational domains. The condition is appropriate where the exit flow is close to a fully developed one and the normal gradient for velocity is close to zero. Thus, the artificial boundary for the outflow boundary condition must be placed downstream fully and stream line similar to parallel (see [37]).

64

2 Fluid Equations

2.2.5 Boundary Conditions on Free Surfaces Sometimes, we are encountered with a boundary between two fluids, which is called free surface. A common example occurs when a liquid film flows down an inclined plane. The surface of the liquid film in contact with the surrounding gas is a fluidfluid interface. Other examples include the interface between a liquid drop and the surrounding continuous phase or that between two liquid layers. Free surfaces are unknowns to be determined and require two boundary conditions to be applied. (i) A kinematic condition: This condition relates the motion of the free interface to the fluid velocities at the free surface. Let v = (v1 , v2 , v3 ) be the velocity field of one fluid and the position of a free surface be written in an implicit form f (x, t) = 0. Fluid particles on the free surface do not move across the free surface, therefore we can write  ∂f ∂f ∂f + + v · ∇ f = 0, vi = ∂t ∂ xi ∂t i

(2.67)

which is a kinematic condition. When a free surface is written in the form x3 = h(x1 , x2 , t), the kinematic boundary condition is v3 = For steady problems, we have

∂f ∂t

∂f ∂f ∂f + v1 + v2 . ∂t ∂ x1 ∂ x2 = 0, and the kinematic condition is written as v · n = 0,

where n is normal unit vector on the surface, since ∇ f is a normal vector on the surface given by f (x) = 0. This condition means that there is no flow through the free surface. Note that there can be a flow tangent to the free surface. (ii) A dynamic condition: This condition is concerned with force balance on the free surface. When one is concerned with a free surface, if the capillary (surface) force is neglected, then the fluid is called a heavy fluid and if the capillary force is essential, then the fluid is called a capillary fluid. For a heavy fluid, the traction exerted by fluid (1) onto fluid (2) is equal and opposite to the traction exerted by fluid (2) onto fluid (1). Since the outward normal vectors on the boundary between two domains of fluids are opposite, we have the dynamic boundary condition σ (1) = −σ (2) . If n is the normal unit vector oriented from fluid (1) to fluid (2) on the surface, then by (2.47) we have (2.68) σ (1) ≡ S (1) · n = −σ (2) ≡ S (2) · n,

2.2 Boundary Conditions for the Navier-Stokes Equations

65

where S (i) is the stress tensor of fluid (i). If fluid (2) is air, then the dynamic condition on the free surface of fluid (1) is rewritten as (see (1.16), Sect. 3.1 of [49]) σ ≡ − pn + 2μE(v) · n = − pa n,

(2.69)

where pa is the air pressure. For free surface of capillary fluids, surface tensions must be taken into account. The pressure jump induced by surface tension is given by sk, where s is the surface tension and k is twice the mean curvature of the free surface (see Sect. 3.1.1). Surface tension acts like a tensioned membrane at the free surface and tries to minimize the surface area. Hence the pressure inside of fluid including the center of curvature tends to be higher than other. Thus, the dynamic boundary condition for the capillary fluid on free surfaces is S (1) · n + skn = S (2) · n,

(2.70)

where k > 0 if the centers of curvature lie inside fluid (1) when the free surface is a sphere or a circle. Moreover, if a free surface is convex (concave) (see Definition 3.2.) with respect to the normal unit vector from fluid (1) to fluid (2) on the surface, then k ≥ 0 (k ≤ 0) (see Chap. 4 of [3] and Sect. 3.1.1). In what follows let us further consider heavy fluids. If a free surface to be determined is steady and a fluid may move freely along the free surface, then we have a free boundary condition v · n = 0, (2.71) σ (v, p) = 0, which is useful in many physical problems, especially in mathematical modeling of coating flows, flows of melted semiconductors (see [50] and references therein). When one fluid is air, sometimes free surfaces are regarded as fixed surfaces. Then, the kinematic boundary condition vanishes and the boundary condition on the fixed surface is (2.69). Neglecting the press of air, we get σ (v, p) = 0, which in some papers is called Neumann boundary condition (see [51] and references therein, [15]). If there is no flow across a fixed surface and the fluid flows along the surface, then a boundary condition v · n = 0, (2.72) στ (v, p) = f τ

66

2 Fluid Equations

may be used. In practice we deal with mixture of some kinds of boundary conditions. For a channel flow a mixture of Dirichlet condition v = 0 on the wall and “do nothing” condition on the outlet is used. But for a channel flow with a rough boundary surface a mixture of Dirichlet condition, the Navier slip-with-friction boundary condition and “do nothing” condition may be used. For a flow in a vessel with in/outlet a mixture of Dirichlet condition v = 0 on the wall and pressure conditions on the inlet/outlet is used. But for the flow in a vessel with in/outlet and a free surface a mixture of Dirichlet condition, a Neumann condition 2νε(v)n − pn = 0 and pressure conditions may be used. There is a vast body of literature for the Stokes and Navier-Stokes problems with mixed boundary conditions, and several variational formulations have been used for them.

2.3 Bilinear Forms for Hydrodynamics In this section, we consider three types of bilinear form reduced from (Δv, u) by integration by parts and outline the variational formulations of the Navier-Stokes problems with mixed boundary conditions. From now on, in this book we use the following notations. Let Ω be an open subset of Rl , l = 2, 3. When X is a Banach space, X = X l and X∗ is the dual of X. Let W k,α (Ω) be Sobolev spaces, H k (Ω) = W k,2 (Ω), and so Hk (Ω) = {H k (Ω)}l , H0k (Ω) = {W0k,2 (Ω)}l . An inner product and norm in the space L2 (Ω) or L 2 (Ω) are, respectively, denoted by (· , ·) and · . · , · means the duality pairing between a Sobolev space X and its dual one. Also, (· , ·)Γi is an inner product in the L2 (Γi ) or L 2 (Γi ), and · , ·Γi means the 1 1 1 1 duality pairing between H 2 (Γi ) and H− 2 (Γi ) or between H 2 (Γi ) and H − 2 (Γi ) (with 1

an exception in Remark 5.1 concerned with the duality pairing between H002 (Γi ) and 1

(H002 (Γi ))∗ ). The inner product and norms in Rl , respectively, are denoted by (· , ·)Rl and | · |. Sometimes a · b or ab is used for inner product in Rl between a and b. For simplicity, when A is a matrix and n is vector, sometimes we write A · n by An.

2.3 Bilinear Forms for Hydrodynamics

67

2.3.1 Bilinear Forms 2.3.1.1

Dirichlet Bilinear Form

Unless otherwise specified, let Ω be a bounded domain with Lipschitz boundary of Rl , l = 2, 3, and φ ∈ H 1/2 (∂Ω). Then there exists a continuous linear operator H 1/2 (∂Ω) → H 1 (Ω) (lift operator), and for every φ ∈ H 1/2 (∂Ω) there exists a w ∈ H 1 (Ω) with (2.73) w|∂Ω = φ, w H 1 (Ω) ≤ c φ H 1/2 (∂Ω) , where c is independent of φ (Theorem 1.5.1.2 of [52] or (1.3.15)). Let us introduce the space   H (div; Ω) = v ∈ L2 (Ω); div v ∈ L 2 (Ω) , which is a Hilbert space for the inner product (v, u) H (div;Ω) = (v, u) + (div v, div u). If Ω is an open subset with Lipschitz boundary of Rl (not necessarily bounded), then ¯ l is dense in H (div; Ω) (see Theorem 2.4, Chap. 1 of [53]). the space D(Ω) For u ∈ H (div; Ω), let us consider the functional on φ ∈ H 1/2 (∂Ω) defined by Fu (φ) = (div u, w) + (u, ∇w). By (2.73) we have   |Fu (φ)| ≤ c u L2 + div u L 2 φ H 1/2 (∂Ω) , which shows that the functional Fu is linear and continuous on H 1/2 (∂Ω). Since (div u, w) + (u, ∇w) = (u · n, φ)∂Ω for u ∈ C∞ (Ω), where n is the outward unit normal vector on ∂Ω, we can define a trace of u · n γn u ∈ H −1/2 (∂Ω) as a generalization of u · n of smooth functions u. Thus, we have Proposition 2.1 Let Ω be a bounded domain with Lipschitz boundary of Rl . The ¯ l can be extended by continuity to a mapping γn : v → v · n|∂Ω defined on D(Ω) linear and continuous mapping, from H (div; Ω) into H −1/2 (∂Ω). Moreover, the following Green’s formula holds: (div u, w) + (u, ∇w) = γn u, w∂Ω ∀u ∈ H (div; Ω), ∀w ∈ H 1 (Ω).

(2.74)

68

2 Fluid Equations

(cf. Theorem 2.5 and (2.17), Chap. 1 of [53]). By Proposition 2.1, for v ∈ H1 (Ω) with Δv ∈ L2 (Ω) and w ∈ H1 (Ω), we have − (Δv, w) =



(∇vi , ∇wi ) −

i



∂n vi , wi ∂Ω ,

(2.75)

i

where ∂n vi = γn (∇vi ). Having in mind the fact that γn (∇vi ) = functions vi , in what follows we write (2.75) as



∂vi  ∂n ∂Ω

− (Δv, w) = (∇v, ∇w) − ∂n v, w∂Ω .

for smooth

(2.76)

We call (∇v, ∇w) Dirichlet bilinear form.

2.3.1.2

Strain Bilinear Form

When v ∈ H1 (Ω), let us prove   ∂ div v in H −1 (Ω), div 2Ei (v) = Δvi + ∂ xi

(2.77)

where Ei (v) = (εi1 (v), εi2 (v), εi3 (v)), i = 1, 2, 3. If q ∈ C0∞ (Ω), then we have 



2 Ω

(div Ei (v))q dω = −2 

=−

 Ω

∇vi · ∇q dω −

Ω

Ei (v) · ∇q dω = −

   ∂vi

+

∂v j  ∂q dω ∂ xi ∂ x j

∂x j  ∂v ∂v · ∇q dω = Δvi · q dω + div · q dω. ∂ xi ∂ xi Ω Ω Ω



Ω

j

Since C0∞ (Ω) is dense in H01 (Ω), from above we get (2.77). Let v ∈ H2 (Ω) and u ∈ H1 (Ω). By (2.74) and (2.77), we have − (Δvi , u i ) = 2(Ei (v), ∇u i ) − 2(Ei (v) · n, u i )∂Ω + (div

∂v , u i ). ∂ xi

(2.78)

Since the tensor E(v) is symmetric, we have      ∂u i  Ei (v), ∇u i = εi j (v), E(v), ∇u = ∂x j i i j =

  ∂u j   1  ∂u i εi j (v), = E(v), E(u) . + 2 ∂x j ∂ xi i j

Taking into account (2.79), we have from (2.78)

(2.79)

2.3 Bilinear Forms for Hydrodynamics

69

−(Δv, u)Ω = 2(E(v), E(u)) + (∇(div v), u) − 2(E(v)n, u)∂Ω = 2(E(v), E(u)) − (div v, div u) − 2(E(v)n, u)∂Ω + (div v, u · n)∂Ω . (2.80)   If v ∈ H1 (Ω), Δv ∈ L2 (Ω) and div v = 0, then by (2.77) div Ei (v) ∈ L 2 (Ω), and there exists γn (Ei (v)) ∈ H −1/2 (∂Ω). Therefore for v ∈ H1 (Ω) with Δv ∈ L2 (Ω), div v = 0, and u ∈ H1 (Ω), we have − (Δv, u)Ω = 2(E(v), E(u)) − 2(E(v)n, u)∂Ω ,

(2.81)

where E(v)n|∂Ω = (γn (E1 (v)), γn (E2 (v)), γn (E3 (v)) ∈ H−1/2 (∂Ω). We call (E(v), E(u)) the strain bilinear form.

2.3.1.3

Vorticity Bilinear Form

Let us introduce the space   H (rot; Ω) = v ∈ L2 (Ω); rot v ∈ L2 (Ω) . When Ω is a domain with Lipschitz boundary of Rl (not necessarily bounded), ¯ l is dense in H (rot; Ω) (Theorem 2.10, Chap. 1 of [53]). D(Ω) Proposition 2.2 Let Ω be a domain with Lipschitz boundary of R3 (not necessarily ¯ 3 can be extended by bounded). The mapping γτ : u → u × n|∂Ω defined on D(Ω) continuity to a linear and continuous mapping, from H (rot; Ω) into H−1/2 (∂Ω). Moreover, the following Green’s formula holds: (rot u, w)Ω − (u, rot w)Ω = − γτ u, w∂Ω ∀w ∈ H1 (Ω).

(2.82)

(cf. Theorem 2.11, Chap. 1 of [53]). In what follows for convenience we write γτ u by u × n. Whenever we are concerned with vorticity of vector fields with l = 2, for convenience as before, vector u = (u 1 (x1 , x2 ), u 2 (x1 , x2 )) is identified with u¯ = (u 1 (x1 , x2 ), u 2 (x1 , x2 ), 0). Thus, for u = (u 1 (x1 , x2 ), u 2 (x1 , x2 )), v = (v1 (x1 , x2 ), v2 (x1 , x2 )) and n = (n 1 , n 2 ), we think that

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  ∂u 2 ∂u 1 , rot u = rot u¯ ≡ 0, 0, − ∂ x1 ∂ x2 u × n = u¯ × n¯ = (0, 0, u 1 n 2 − u 2 n 1 ),    ∂v2  ∂v2 ∂v1  ∂v1  , n1 ,0 , − − rot v × n = rot v¯ × n¯ = −n 2 ∂ x1 ∂ x2 ∂ x1 ∂ x2  (rot v, rot u)Ω = rot v¯ · rot u¯ d x, Ω  (rot v × n, u)∂Ω = (rot v¯ × n) ¯ · u¯ dσ. ∂Ω

Corollary 2.1 Let Ω be a bounded domain with Lipschitz boundary of Rl , l = 2, 3. If v ∈ H1 (Ω), Δv ∈ L2 (Ω) and div v = 0, then there exists γτ (rot v) ≡ rot v × n ∈ H−1/2 (∂Ω) and the following holds: −(Δv, u)Ω = (rot v, rot u)Ω − rot v × n, u∂Ω .

(2.83)

Proof Since −Δv = rot rot v − grad(div v) (see (2.21)) and div v = 0, we have rot rot v ∈ L2 (Ω). If l = 3, then by Proposition 2.2 we come to the conclusion. If l = 2. Let n be the outward unit normal on ∂Ω and n¯ = (n 1 , n 2 , 0). On the ¯ u. ¯ Then by Proposidomain Ω˜ = Ω × (0, 1) ⊂ R3 let us consider vector fields v, tion 2.2 we have −(Δv, ¯ u) ¯ Ω˜ = (rot v, ¯ rot u) ¯ Ω˜ − (rot v¯ × n, ˜ u) ¯ ∂ Ω˜ , where

(2.84)

    ∂ Ω˜ = ∂Ω × (0, 1) Ω × {0} Ω × {1}

˜ and n˜ is the vectors rot v¯ and n˜ are orthogonal to  outward normal on ∂ Ω. Since both  Ω × {0} Ω × {1}, rot v¯ × n˜ = 0 on Ω × {0} Ω × {1}. Then, noting that n˜ = n¯ on ∂Ω × (0, 1), we have  (rot v¯ × n, ˜ u) ¯ ∂ Ω˜ =



∂Ω×(0,1)

(rot v¯ × n) ¯ · u¯ dσ =

∂Ω

(rot v¯ × n) ¯ · u¯ dσ = (rot v × n, u)∂Ω .

Also, 

 Δv¯ · u¯ d x = Δv¯ · u¯ d x = (Δv, u)Ω , Ω˜ Ω   (rot v, ¯ rot u) ¯ Ω˜ = rot v¯ · rot u¯ d x = rot v¯ · rot u¯ d x = (rot v, rot u)Ω .

(Δv, ¯ u) ¯ Ω˜ =

Ω˜

Ω

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71

Therefore, from (2.84) we get (2.83) for l = 2.



We call (rot v, rot u) the vorticity bilinear form.

2.3.2 Variational Formulations for Mixed Boundary Value Problems of the Navier-Stokes Equations Now, we will briefly outline the main points of variational formulations for the Stokes and Navier-Stokes problems with mixed boundary conditions. In this subsection from now on, let Ω be a bounded domain with Lipschitz boundary of Rl , l = 2, 3. Let us introduce the following spaces of divergence-free functions:     V = φ ∈ D(Ω)l ; div φ = 0 , V = v ∈ H10 (Ω); div v = 0 . The space V is dense in V for the norm of H1 (Ω) (Corollary 2.5, Chap. 1 of [53]). The following proposition which is a simplified version of de Rham’s theorem (see Proposition 1.1, Chap. 1 of [54]) holds. Proposition 2.3 (Theorem 2.3, Chap. 1 of [53]) If f ∈ H−1 (Ω) satisfies

f, φ = 0 ∀φ ∈ V , then there exists p ∈ L 2 (Ω) such that f = ∇ p. If Ω is connected, then p is unique up to an additive constant.

2.3.2.1

Variational Formulation Based on Dirichlet Bilinear Form

Let ∂Ω = Γ D ∪ Γ N , where Γ D , Γ N are open subsets of ∂Ω such that Γ D ∩ Γ N = ∅, and   V D = v ∈ H1 (Ω); div v = 0, v|Γ D = 0 . Let us consider the problem ⎧ − μΔv + (v · ∇)v + ∇ p = f in Ω, ⎪ ⎪ ⎪ ⎪ ⎪ in Ω, ⎨ div v = 0 v|Γ D = 0, ⎪ ⎪ ⎪ ⎪   ⎪ ⎩ μ ∂v − pn |Γ N = g. ∂n

(2.85)

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2 Fluid Equations

Assuming that (v, p) is smooth and making duality product with u ∈ V D in the first equation, by (2.76) we get μ(∇v, ∇u) + (v · ∇)v, u = f, u + g, uΓ N ∀u ∈ V D .

(2.86)

Thus, when f ∈ V∗D , g ∈ H−1/2 (Γ N ) are given, a function v ∈ V D satisfying (2.86) is called a solution to problem (2.85). To show that the concept of solution is well defined, we will prove that if a solution v is smooth (v ∈ H2 (Ω), f ∈ L2 (Ω)), then there exists a p such that (v, p) satisfies (2.85). By (2.76) we get from (2.86) − μ(Δv, u) + (v · ∇)v, u + μ ∂n v, uΓ N = f, u + g, uΓ N ∀u ∈ V D . (2.87) Taking any u ∈ V , we have from (2.87) (−μΔv + (v · ∇)v − f, u) = 0. By Proposition 2.3, there exists a unique function P ∈ L 2 (Ω) such that and − μΔv + (v · ∇)v − f = −∇ P.

 Ω

P dx = 0 (2.88)

Since v ∈ H2 (Ω), f ∈ L2 (Ω) and (v · ∇)v ∈ L2 (Ω), we have that ∇ P ∈ L2 (Ω), which shows that P ∈ H 1 (Ω). Substituting (2.88) into (2.87) and using integration by parts, we have (2.89)

μ∂n v − Pn − g, uΓ N = 0. 1/2 For any tangent vector field φ ∈ H00 (Γ N ) on Γ N there exists its continuation φ˜ ∈ 1/2 H (∂Ω) such that φ˜ = 0 on ∂Ω\Γ N (see Theorem 1.18). There exists a solution u ∈ H1 (Ω) to the Stokes problem

⎧ ⎪ ⎨ − Δu + ∇ p = 0, div u = 0, ⎪ ⎩ u |∂Ω = φ˜ (see Theorem IV.1.1 of [55]), and u ∈ V D . Taking such u in (2.89), we see that the tangent components of μ∂ n v − Pn − g vanish. Similarly, for any normal vector field 1/2 φ ∈ H00 (Γ N ) such that Γ N φ · n ds = 0 there exists a u ∈ V D such that u|Γ N = φ. Using such u in (2.89), we can verify that the values of the normal components of μ∂n v − Pn − g must be the same. Therefore, we have μ

∂v − Pn − g = cn on Γ N , ∂n

(2.90)

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73

where c is some constant. Putting p = P + c, we get the boundary condition on Γ N . Thus, taking into account v ∈ V D , we know that (v, p) satisfies (2.85). If Γ D = ∅, then by Friedrichs’ inequality the norm v V D is equivalent to ∇v L2 and (2.91) (∇ v, ∇ v) ≥ c v 2H1 (Ω) , which is fundamental for study of the problem (2.86).

2.3.2.2

Variational Formulation Based on Strain Bilinear Form   3 Let ∂Ω = Γ D ∪ i=1 Γ i , where Γ D , Γi are open subsets of ∂Ω such that Γ D ∩ Γi = ∅, Γi ∩ Γ j = ∅ for i = j and   V D = v ∈ H1 (Ω); div v = 0, v|Γ D = 0, vτ |Γ1 = 0, vn |Γ2 = 0 . Let us consider the problem ⎧ − μΔv + (v · ∇)v + ∇ p = f in Ω, ⎪ ⎪ ⎪ ⎪ ⎪ div v = 0 in Ω, ⎪ ⎪ ⎪ ⎨ v| = 0, ΓD ⎪ vτ |Γ1 = 0, (− p + 2μεnn (v))|Γ1 = φ1 , ⎪ ⎪ ⎪ ⎪ ⎪ vn |Γ2 = 0, 2(μεnτ (v) + αvτ )|Γ2 = φ2 , ⎪ ⎪ ⎩ (− pn + 2μεn (v))|Γ3 = φ3 ,

(2.92)

where εn (v) = E(v) · n, εnn (v) = (E(v) · n, n)Rl , εnτ (v) = E(v) · n − εnn (v)n. Assuming that (v, p) is smooth and making duality product with u ∈ V D in the first equation, by (2.81) we get 2μ(E(v), E(u)) + (v · ∇), u + 2(αvτ , u)Γ2 = f, u + φ1 , u n Γ1 + φ2 , uΓ2 + φ3 , uΓ3 ∀u ∈ V D .

(2.93)

Thus, when f ∈ V∗D , φ1 ∈ H −1/2 (Γ1 ), φ2 ∈ H−1/2 (Γ2 ), φ3 ∈ H−1/2 (Γ3 ) are given, a function v ∈ V D satisfying (2.93) is called a solution to problem (2.92). If we are concerned with

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2 Fluid Equations

⎧ − μΔv + (v · ∇)v + ∇ p = f in Ω, ⎪ ⎪ ⎪ ⎪ ⎪ in Ω, ⎪ ⎪ div v = 0 ⎪ ⎪ ⎪ v| = 0, ⎪ Γ ⎪ ⎨ D    1 2  |v| v + 2με | = 0, − p + (v)  = φ1 , τ Γ nn 1 ⎪ Γ1 ⎪ 2 ⎪ ⎪ ⎪ ⎪ vn |Γ2 = 0, 2(μεnτ (v) + αvτ )|Γ2 = φ2 , ⎪ ⎪ ⎪   ⎪ ⎪ ⎪ −  p + 1 |v|2 n + 2με (v)  = φ , ⎩  n 3 Γ3 2

(2.94)

which includes total stress boundary conditions, then by 1 (v · ∇)v = rot v × v + ∇ |v|2 2 we can rewrite the Navier-Stokes equations as 1 ∂v − μΔv + rot v × v + ∇( p + |v|2 ) = f, div v = 0 in Ω, ∂t 2

(2.95)

and using (2.81) for a smooth solution we get 2μ(E(v), E(u)) + rot v × v, u + 2(αvτ , u)Γ2 = f, u + φ1 , u n Γ1 + φ2 , uΓ2 + φ3 , uΓ3 ∀u ∈ V D .

(2.96)

Thus, when f ∈ V∗D , φ1 ∈ H −1/2 (Γ1 ), φ2 ∈ H−1/2 (Γ2 ), φ3 ∈ H−1/2 (Γ3 ) are given, a function v ∈ V D satisfying (2.96) is called a solution to problem (2.94). To show that the concepts of solution are well defined, as above we can prove that if a solution v is smooth (v ∈ H2 (Ω), f ∈ L2 (Ω)), then there exists a p such that (v, p) satisfies (2.92) or (2.94). For more general cases, we will prove such a fact in Chap. 5. If Γ D = ∅, then by Korn’s and Friedrichs’ inequalities the norm v V D is equivalent to E(v) L2 and (2.97) (E(v), E(v)) ≥ c v 2H1 (Ω) , which is fundamental for study of the problems (2.93) and (2.96).

2.3.2.3

Variational Formulation Based on Vorticity Bilinear Form

  2 Let ∂Ω = Γ D ∪ i=1 Γ i , where Γ D , Γi , i = 1, 2, are open subsets of ∂Ω such that Γ D ∩ Γi = ∅, Γi ∩ Γ j = ∅ for i = j and   V D = v ∈ H1 (Ω); div v = 0, v|Γ D = 0, vτ |Γ1 = 0, vn |Γ2 = 0 .

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75

Let us consider the problem ⎧ − μΔv + (v · ∇)v + ∇ p = f in Ω, ⎪ ⎪ ⎪ ⎪ ⎪ in Ω, ⎪ ⎨ div v = 0 v|Γ D = 0, ⎪ ⎪ ⎪ vτ |Γ1 = 0, − p|Γ1 = φ1 , ⎪ ⎪ ⎪ ⎩ vn |Γ2 = 0, (rot v × n)|Γ2 = φ2 /μ.

(2.98)

Assuming that (v, p) is smooth and making duality product with u ∈ V D in the first equation, by (2.83) we get μ(rot v, rot u) + (v · ∇)v, u = f, u + φ1 , u n Γ1 + φ2 , uΓ1 ∀u ∈ V D , (2.99) where (rot v × n, u)|Γ1 = 0 was used. Thus, when f ∈ V∗D , φ1 ∈ H −1/2 (Γ1 ) and φ2 ∈ H−1/2 (Γ2 ) are given, a function v ∈ V D satisfying (2.99) is called a solution to problem (2.98). Like (2.96), we can get a variational formulation for the problem with the total pressure boundary condition instead of the static pressure boundary condition. For the discussion of that the concepts of solutions are well defined, we refer to Chap. 5. Let Ω ⊂ R3 , ∂Ω ∈ C 1,1 or Ω be a convex polyhedron, Γ D = ∅ and Γ¯1 ∩ Γ¯2 = ∅. Then, by Lemma 4.1 of [56] (rot v, rot v) ≥ c v 2H1 (Ω) ,

(2.100)

which is fundamental for study of the problem (2.99). When Ω ⊂ R3 , ∂Ω ∈ C 2 and Γ D = ∅, (2.100) is also valid without the condition Γ¯1 ∩ Γ¯2 = ∅ (see Lemma 2 of [57]). As shown above, relying on the Dirichlet bilinear form a(v, u) = (∇v, ∇u),

(2.101)

we can get a variational formulation for the Navier-Stokes problem with Dirichlet and the outlet boundary conditions. Relying on strain bilinear form a(v, u) = 2i, j (εi j (v), εi j (u)),

(2.102)

we can get a variational formulation for the problem with Dirichlet and the stress (total stress) boundary conditions, whereas relying on vorticity bilinear form a(v, u) = (rot v, rot u)

(2.103)

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2 Fluid Equations

we get one for the problem with Dirichlet, static pressure (total pressure) and vorticity boundary conditions.

2.4 Bibliographical Remarks 2.4.1 Fluid Equations For Eulerian and Lagrangian descriptions of movement of particles, one can see Chap. 3 of [3] or Chap. 1 of [2]. For proof of the transport theorem, we refer to Chap. 2 of [1] or Chap. 5 of [58], and for derivation of equations using conservation laws we refer to Chap. 4 of [3], Chap. 1 of [59] or [60]. For papers studying (2.16) and (2.17), respectively, we refer to [61–64]. For papers dealing with (2.39) and (2.40), respectively, see [65–68]. For papers discussing (2.41) we refer to [69–71] and refer to [72] for the incompressible fluid. In [73] Eq. (2.19) is considered and most papers for incompressible Newtonian fluid are concerned with Eq. (2.20). For papers investigating (2.42), we refer to [72, 74–81]. There are many papers dealing with (2.43), for example, [82–84]. For papers dealing with (2.44), we refer to [7, 85].

2.4.2 Boundary Conditions of the Navier-Stokes Equations We refer to Chap. 4 of [3] and Chap. 2 of [59] for explanation of some boundary conditions. The stick boundary condition was formulated by G. Stokes in 1845 and the Navier slip-with-friction boundary condition was suggested by C.-L. Navier in 1823. (cf. Introduction of [86]). In [87] the Navier slip-with-friction boundary condition has been derived rigorously from the boundary condition at the kinetic level (Boltzmann equation) for compressible fluids. For papers dealing with vorticity boundary condition see [9–11, 13, 14, 18, 88, 89]. By introducing the concept of energy-preserving boundary conditions, in [90] a rational derivation of a large class of nonstandard boundary conditions containing several different artificial boundary conditions is given. In 1990s, the leak and slip boundary conditions with threshold have been introduced by H. Fujita (see Introduction of [25]). Physical and experimental backgrounds of the threshold slip boundary conditions are mentioned in several papers (see [8, 22–24]). For physical backgrounds of the threshold leak boundary conditions, we refer to [24, 27, 91]. For non-Newtonian fluid equations with friction slip boundary conditions, we refer to [92–95].

2.4 Bibliographical Remarks

77

In [96], the Navier-Stokes problem with one-sided threshold leak boundary condition based on the total stress was considered as an example of application of a variational inequality. The one-sided threshold leak boundary conditions based on the stress for the Navier-Stokes equations were considered first in [28]. For similar one-sided boundary conditions of elasticity, we refer to [97], Sect. 5.4.1 of Chap. 3 in [98]. “Do nothing” boundary condition was introduced in [99].

2.4.3 Bilinear Forms for Hydrodynamics For papers applying the Dirichlet bilinear form (2.101), we refer to [43, 48, 91, 100, 101]. For papers utilizing the strain bilinear form (2.102), we refer to [24–27, 35, 73, 91, 102–109]. In [73], another equivalent variational formulation, where strain, pressure, velocity and vorticity are unknown functions, also is given. For the papers relying on the vorticity bilinear form (2.103), see [18, 30, 32, 33, 56, 57, 110–116]. In Sect. 1 of [114], the Dirichlet bilinear form (2.101) instead of the vorticity bilinear form (2.103) is used since two bilinear forms (2.101) and (2.103) for polygon or polyhedral domain under some boundary conditions are equal (see [117]). When one deals with the boundary condition for pressure or vorticity on a portion of boundary, there are other variational formulations using three unknown functions v, p and ω, where ω = rot v, (see [31, 118–120]) for the two-dimensional case and v, p and a vector potential for the three-dimensional case (see [121]).

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Chapter 3

The Steady Navier-Stokes System

In this chapter, we are concerned with the steady Navier-Stokes systems with mixed boundary conditions involving Dirichlet, pressure, vorticity, stress and normal derivative of velocity together. As we have seen in Sect. 2.3.2, according to what kinds of bilinear forms for variational formulation are used, types of boundary conditions under consideration together are different. The variational formulations in Sect. 2.3.2 do not reflect, for example, the boundary conditions for stress and pressure together, but this case is important in practice. To include wider boundary conditions together, we first study the relations among strain, vorticity, normal derivative of velocity, and shape of boundary surfaces. Using the relations on the boundary surfaces, we reflect all boundary conditions into variational formulations of the problems. Then we prove the existence and uniqueness of solutions to the problems.

3.1 Properties on the Boundary Surfaces of Vector Fields In this section, we first recall ways describing the shape of surface and study the relations among strain, vorticity, normal derivative of vector fields and shape of a surface when the vector fields near a surface are tangent or normal on the surface.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 T. Kim and D. Cao, Equations of Motion for Incompressible Viscous Fluids, Advances in Mathematical Fluid Mechanics, https://doi.org/10.1007/978-3-030-78659-5_3

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3 The Steady Navier-Stokes System

3.1.1 The Second Fundamental Form and Shape Operator of Surface 3.1.1.1

The Second Fundamental Form of a Surface

When γ : (α, β) → Rl is a parametrized curve, if its derivative γ  (t) at every t ∈ (α, β) is a unit vector, then γ is said to be a unit-speed curve. If the natural parameter (length) of curve is taken as the parameter t, then the curve is a unit-speed curve. A subset M of R3 is called a surface if, for every point p ∈ M, there is an open set U ⊂ R2 and an open set W ⊂ R3 containing p such that M ∩ W is homeomorphic to U . A subset of a surface M of form M ∩ W , where W is an open subset of R3 , is called an open subset of M. A homeomorphism σ : U → M ∩ W as in this definition is called a surface patch for parametrization of the open subset M ∩ W of M. If function σ : U → R3 for a surface M is in class C k (or C k,1 ), then the surface is called class C k −surface (or C k,1 -surface), and a class C 1 -surface is called smooth. If a vector is tangent to a curve in M passing through p, then it is called a tangent vector to the surface M at the point p ∈ M. For a surface patch σ (ξ, ζ ) for a surface M, σξ (ξ0 , ζ0 ) =

∂σ  ∂σ  and σζ (ξ0 , ζ0 ) = ∂ξ (ξ0 ,ζ0 ) ∂ζ (ξ0 ,ζ0 )

are tangent at the point p(ξ0 , ζ0 ) ∈ M to the surface M. If, moreover, ξ and ζ are, respectively, the lengths of curves σ (ξ, ζ0 ) and σ (ξ0 , ζ ) (natural parameters), then those are tangent units at the point p(ξ0 , ζ0 ) ∈ M. The set of all vectors tangent to M at p is called the tangent space T p M of M at p. The tangent space T p M of a surface M at a point p ∈ M is completely determined by giving a unit vector orthogonal to it at p, called a unit normal to M at p. There are, of course, two unit normal vectors, but choosing a surface patch σ : U → R3 containing point p leads to a definite choice, namely, nσ =

σξ × σζ . σξ × σζ 

This is called the standard unit normal of the surface patch σ at point p. To be specific, unless otherwise stated we always take the standard unit normal. If in a neighborhood of (ξ0 , ζ0 ) ∈ U a surface patch σ : U → R3 is in class C 1 , then a unit normal field n of class C always exists in a neighborhood of σ (ξ0 , ζ0 ) ∈ M. But it may not be possible to extend n to all of M, e.g. Möbius band. If a smooth unit normal field n defined on all of M is chosen, then the surface M is said to be oriented.

3.1 Properties on the Boundary Surfaces of Vector Fields

85

Now we are interested in studying the shape of surfaces in R3 . The objects describing the shape of a surface M ⊂ R3 are the second fundamental form and the shape operator (Weingarten map) of the surface. Suppose that σ is a surface patch for M ⊂ R3 with the standard unit normal n. As the parameters (ξ, ζ ) of σ change to (ξ + Δξ, ζ + Δζ ), the surface moves away from the tangent plane through σ (ξ, ζ ). Making inner product in R3 between the vector σ (ξ + Δξ, ζ + Δζ ) − σ (ξ, ζ ) and the normal vector n(ξ, ζ ) at point σ (ξ, ζ ) ∈ M, we get the deviation of σ along n from its tangent plane through σ (ξ, ζ ) ∈ M:   σ (ξ + Δξ, ζ + Δζ ) − σ (ξ, ζ ) · n(ξ, ζ ). By the two variable form of Taylor’s theorem, σ (ξ + Δξ, ζ + Δζ ) − σ (ξ, ζ )    1 = σξ Δξ + σζ Δζ + σξ ξ (Δξ )2 + 2σξ ζ Δξ Δζ + σζ ζ (Δζ )2 + o (Δξ )2 + (Δζ )2 , 2

  where o (Δξ )2 + (Δζ )2 )/((Δξ )2 + (Δζ )2 tends to zero as (Δξ )2 + (Δζ )2 → 0. Note that σξ and σζ are tangent to the surface, hence perpendicular to n, and so the deviation of σ in the direction of the unit vector n from its tangent plane is    1 L(Δξ )2 + 2K Δξ Δζ + N (Δζ )2 + o (Δξ )2 + (Δζ )2 , 2

(3.1)

where L = σξ ξ (ξ, ζ ) · n(ξ, ζ ), K = σξ ζ (ξ, ζ ) · n(ξ, ζ ), N = σζ ζ (ξ, ζ ) · n(ξ, ζ ). (3.2) At point σ (ξ, ζ ) ∈ M σξ (ξ, ζ ) · n(ξ, ζ ) = 0, σζ (ξ, ζ ) · n(ξ, ζ ) = 0, which yields by differentiation that L = σξ ξ (ξ, ζ ) · n(ξ, ζ ) = −σξ (ξ, ζ ) · n ξ (ξ, ζ ), K = σξ ζ (ξ, ζ ) · n(ξ, ζ ) = −σξ (ξ, ζ ) · n ζ (ξ, ζ ) = −σζ (ξ, ζ ) · n ξ (ξ, ζ ), N = σζ ζ (ξ, ζ ) · n(ξ, ζ ) = −σζ (ξ, ζ ) · n ζ (ξ, ζ ).

(3.3)

The expression L(Δξ )2 + 2K Δξ Δζ + N (Δζ )2

(3.4)

is called the second fundamental form of the surface patch σ . Remark 3.1 If in a neighborhood of a point p ∈ M the surface is deviated from the tangent plane to the side opposite to the unit normal, then the quadratic form (3.4) at p ∈ M is negative.

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3 The Steady Navier-Stokes System

Since all vectors w ∈ T p M are expressed as w = λσξ + μσζ , denote by dξ(w) = λ, dζ (w) = μ for w ∈ T p M. Let us define a symmetric bilinear form on T p M by   v, w p,M := Ldξ(v)dξ(w) + K dξ(v)dζ (w) + dξ(w)dζ (v) + N dζ (v)dζ (w) ∀v, w ∈ T p M. (3.5) Let a surface patch σ˜ (ξ˜ , ζ˜ ) be a reparametrization of a surface patch σ (ξ, ζ ) with reparametrization map (ξ, ζ ) = Φ(ξ˜ , ζ˜ ). Denote the Jacobian matrix of the reparametrization map (ξ˜ , ζ˜ ) → (ξ, ζ )

 ∂ξ

∂ξ  ∂ ξ˜ ∂ ζ˜ ∂ζ ∂ζ ∂ ξ˜ ∂ ζ˜

by J (Φ). Then we can prove that

L˜ K˜

K˜ N˜



= ±J (Φ)t

L K K N

J (Φ)

(3.6)

with the plus sign if det(J (Φ)) > 0 and the minus sign if det(J (Φ)) < 0. From (3.6), we can see that the second fundamental form of a surface patch is unchanged by a reparametrization of the patch preserving its orientation (which means det(J (Φ)) > 0) (see Exercise 6.1.4 of [1]).

3.1.1.2

The Shape Operator of Surface

Let M be a smooth surface and f : M → R be a smooth function on M. This means that f = f ◦ σ is smooth for all patches σ : U → M. For p ∈ M and X ∈ T p M, the directional derivative of f at p in the direction X , denoted by ∇ X f , is defined as follows. Let δ : (−ε, ε) → M ⊂ R3 be any smooth curve in M such that δ(0) = p and δ  (0) = X . Then, ∇X f =

d f ◦ δ(t)|t=0 . dt

The directional derivative is well defined, i.e. independent of the particular choice of δ, and linear.

3.1 Properties on the Boundary Surfaces of Vector Fields

87

A vector field along a surface M is a rule which assigns a vector (need not be tangent to M) to each point of M, p ∈ M → Y ( p) ∈ T p R3 , the space of 3-D vectors with the origin at p. Then when X ∈ T p M where T p R3 means  and Y ( p) = Y1 ( p), Y2 ( p), Y3 ( p) ∈ T p M, we define an operator X ∈ T p M → ∇ X Y ( p) ∈ T p R3 by

  X ∈ T p M → ∇ X Y1 ( p), ∇ X Y1 ( p), ∇ X Y1 ( p) ∈ T p R3 .

Let M be a surface, p ∈ M and n be a smooth unit normal vector field defined along a neighborhood W ⊂ M of p. Let us consider the operator X ∈ T p M → ∇ X n( p).   Since n( p), n( p) = 1, we know that ∇ X (n( p), n( p)) = 0, and (∇ X n( p), n( p)) + (n( p), ∇ X n( p)) = 0, which implies 2(∇ X n( p), n( p)) = 0. Therefore, ∇ X n( p) ⊥ n( p) and ∇ X n ∈ T p M.

Definition 3.1 Let M be a surface, p ∈ M and n be a smooth unit normal defined along a neighborhood W ⊂ M of p. The map S : X ∈ T p M → T p M defined by S(X ) = ∇ X n( p) is called the shape operator of M at p (Weingarten map). Remark 3.2 In some books, the operator −S is called the shape operator of M at p. Let us introduce a bilinear form L on T p M     L(v, w) = − S(v), w = − ∇v n( p), w ∀v, w ∈ T p M.

(3.7)

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3 The Steady Navier-Stokes System

Proposition 3.1 (Proposition 7.2.2 of [1]) Let p be a point of a surface M, let σ (ξ, ζ ) be a surface patch of M with p in its image and let v, w p,M be the bilinear form defined by (3.5). Then, for any v, w ∈ T p M, L(v, w) = v, w p,M ,

(3.8)

where L , K , N are as in (3.3). Proof Since both sides of (3.8) define bilinear forms on T p M and v = dξ(v)σξ + dζ (v)σζ , w = dξ(w)σξ + dζ (w)σζ , it suffices to verify that (3.8) is valid when v and w are σξ or σζ . Recalling that dξ(σξ ) = dζ (σζ ) = 1 and dξ(σζ ) = dζ (σξ ) = 0, we only need to show that L(σξ , σξ ) = L , L(σξ , σζ ) = L(σζ , σξ ) = K , L(σζ , σζ ) = N .

(3.9)

Taking into account (3.3), we have L(σξ , σξ ) = −(∇σξ n( p), σξ ) = −n ξ · σξ = L . In the same way as above, the other relations in (3.9) can be proved.



Corollary 3.1 (Corollary 7.2.4 of [1]) The shape operator is self-adjoint. In what follows let ξ, ζ be natural parameters. The curves σ (ξ, ζ0 ) and σ (ξ0 , ζ ) intersect at point p(ξ0 , ζ0 ), and σξ (ξ, ζ0 ) = e1 and σζ (ξ0 , ζ ) = e2 are the tangent unit vectors at point p(ξ0 , ζ0 ). Thus the shape operator is expressed by the following matrix:

−L −K S= , (3.10) −K −N where −L = e1 ·

∂n ∂n ∂n ∂n , −K = e2 · = e1 · , −N = e2 · . ∂e1 ∂e1 ∂e2 ∂e2

Since the shape operator S is self-adjoint, there exists an orthonormal basis {e1 , e2 } of T p M and real numbers λ1 , λ2 such that Se1 = λ1 e1 , Se2 = λ2 e2 , and the matrix for S at p ∈ M is

λ1 0 . S= 0 λ2

3.1 Properties on the Boundary Surfaces of Vector Fields

89

The trace tr (S) of matrix S is invariable with respect to orthogonal transformation. On the other hand, if σ (ξ, ζ ) is a surface patch of M with p = σ (ξ0 , ζ0 ) ∈ M, σξ | p = e1 and σζ | p = e2 , then by (3.3) −σξ ξ · n = σξ · n ξ = λ1 , −σζ ζ · n = σζ · n ζ = λ2 . Since σξ ξ and σζ ζ are, respectively, curvature vectors at the p of the curves σ (ξ, ζ0 ) and σ (ξ0 , ζ ), σξ ξ · n and σζ ζ · n are the normal curvatures. Therefore, for any orthogonal basis 21 tr (S)( p) is called the mean curvature at p ∈ M. Let us consider plane curves and their curvatures. If γ (t) is a unit-speed plane curve with parameter t, then its curvature k(t) at the point γ (t) is defined to be |γ  (t)|. Let n be the normal on γ . Since γ  (t) · n = 0, γ  (t) · n = −γ  (t) · n  (t) ≡ −e ·

dn , de

(3.11)

where e is the tangent unit vector. Taking into account above, we call e · dn the signed de curvature of γ . In this book, the surfaces under consideration are pieces of boundary of 3-D or 2-D bounded connected domains, and so the surfaces are oriented and the outsides of surfaces are fixed. Definition 3.2 If a piece of surface Γ on a neighborhood of x ∈ Γ is on the opposite (same) side of the outward normal vector with respect to tangent plane (line for l = 2) at x or coincides with the tangent plane, then the surface is said to be convex (concave) at x. If the surface is convex (concave) at all x ∈ Γ , then Γ is said to be convex (concave).  Lemma 3.1 If Γ = j Γ j , Γ j are convex (concave), S p is the shape operator of Γ at p ∈ Γ and k( p) is twice the mean curvature at p ∈ Γ , then for v ∈ T p Γ the quadratic forms (S p v, v)Rl−1 and (k( p)v, v)Rl−1 are positive (negative). Proof Let us consider only the case of convexity, since the case of concavity can be proved in the same way. Let l = 3. Since surface Γ j is convex, then at every point of Γ j the quadratic form L(Δξ )2 + 2K Δξ Δζ + N (Δζ )2 is negative (cf. Remark 3.1), and so the matrix S is positive at every point of Γ j . By the argument as above the case that l = 2 is proved. Since k( p) = tr (S), we have the second conclusion. 

90

3 The Steady Navier-Stokes System

3.1.2 Properties on the Boundary Surface of Vector Fields Let Γ be a surface (curve for l = 2) of C 2 and v be a vector field of C 2 on a domain of Rl near Γ . Let n be the outward unit normal on Γ . Theorem 3.1 Suppose that v · n|Γ = 0. Then, at every point of the surface Γ the following hold: 1 (rot v × n, τ )Rl − (S v, ˜ τ˜ )Rl−1 , 2

∂v (rot v × n, τ )Rl = + (S v, ˜ τ˜ )Rl−1 , ,τ ∂n Rl

(E(v)n, τ )Rl =

(E(v)n, τ )Rl

1 = 2



∂v ,τ ∂n

Rl

(3.12)

(3.13)

1 ˜ τ˜ )Rl−1 , − (S v, 2

(3.14) ∂v

where E(v) denotes the matrix with the components εi j (v) = 21 ( ∂∂vx ij + ∂ xij ), τ is tangent vectors on Γ , S is the shape operator of the surface Γ (the matrix (3.10)) for l = 3 and the signed curvature of Γ for l = 2, and v, ˜ τ˜ are expressions of the vectors v, τ in a local orthogonal curvilinear coordinates on Γ . Proof Let l = 3. For vector field v(x) = (v1 (x), v2 (x), v3 (x)) denoted in orthogonal coordinates x, let ⎛ ∂v1 ⎞ · · · ∂∂vx31 ∂ x1 J (v) = ⎝ · · · · · · · · · ⎠ . ∂v1 · · · ∂∂vx33 ∂ x3 Since E(v)n =

  1 1 J (v)T + J (v) n = J (v)T − J (v) n + J (v)n, 2 2 J (v)T − J (v) = rot v × n,

we have E(v)n =

1 rot v × n + J (v)n. 2

(3.15) (3.16)

(3.17)

On the other hand, n(x) ∈ C 1 since Γ ∈ C 2 . Let us make a vector field on a small enough neighborhood of p ∈ Γ in R3 as follows. Consider a family of curves orthogonal to Γ none of which intersect with all the others. Assign tangent unit vector to every point of the lines. Then, the vector coincides with the unit normal n(x) at x ∈ Γ . Denote the vector field again by n(x). Since v(x) · n(x)|Γ = 0, the surface Γ is a contour of the scalar function v · n, and so ∇(v · n) is orthogonal on Γ . Thus,

3.1 Properties on the Boundary Surfaces of Vector Fields

91

τ · ∇(v · n)|Γ = 0.

(3.18)

∇(v · n) = J (v)n + J (n)v, and from (3.18) we get τ · J (v)n = −τ · J (n)v on Γ.

(3.19)

Since τ · J (n)v = J (n)T τ · v = ∇τ n · v = (S τ˜ , v) ˜ Rl−1 = (S v, ˜ τ˜ )Rl−1 on Γ,

(3.20)

where the fact that the shape operator is self-adjoint (Corollary 3.1) was used, from (3.17), (3.19) and (3.20) we have (3.12). Since ∂v , (J (v)T − J (v))n = rot v × n, J (v)T n = ∂n we get

(rot v × n, τ ) =

∂v ,τ ∂n

− (J (v)n, τ ),

(3.21)

which, together with (3.19) and (3.20), implies (3.13). Formulas (3.12) and (3.13) imply (3.14). When l = 2, putting v = (v1 (x1 , x2 ), v2 (x1 , x2 ), 0) and considering a cylindrical surface with the line Γ as a section, we get the conclusion.  Remark 3.3 The bilinear form (S v, ˜ u) ˜ Rl−1 for vectors u, v tangent to the surface is independent of choice of orthogonal curvilinear coordinate system on the surface (see (3.6)). Theorem 3.2 On the surface Γ the following holds: (E(v)n, n)Rl =

∂v ,n ∂n

.

(3.22)

= −(k(x)v, n)Rl + div v,

(3.23)

Rl

If v · τ |Γ = 0, then (E(v)n, n)Rl =

∂v ,n ∂n

Rl

where k(x) = 2 · mean curvature. Proof Let n(x) be the vector field expanded on a domain near Γ as in the proof of Theorem 3.1. Since E(v)n =

1 1 ∂v 1 1 J (v)T n + J (v)n = + J (v)n, 2 2 2 ∂n 2

92

3 The Steady Navier-Stokes System

(J (v)n, n) Rl =

l  l  ∂v j j=1 i=1

∂ xi

n j ni =

∂v ,n ∂n

, Rl

we have (3.22). In a neighborhood W ⊂ R3 of a point x0 ∈ Γ let us introduce local curvilinear coordinates (y1 (x1 , x2 , x3 ), y2 (x1 , x2 , x3 ), y3 (x1 , x2 , x3 )) (if l = 2 then omit y3 ) such that the coordinate lines (y1 ), (y2 ) and (y3 ) are orthogonal at all points to each other, the linear (y3 ) is the outward unit normal n on Γ and the surface y3 = 0 coincides with Γ , where x = (x1 , x2 , x3 ) is the original orthogonal coordinates. Denote the unit vector of ∇x yk by ek , then e1 , e2 are tangent on Γ and e3 = n. At x0 ∈ Γ let us calculate div v(x0 ). Denote vτ = v − (v · n)n. Then ∂v ∂v ∂v (x0 ) · n (x0 ) · e1 + (x0 ) · e2 + ∂e1 ∂e2 ∂n ∂vτ ∂vn n ∂vτ ∂vn n = (x0 ) · e1 + (x0 ) · e1 + (x0 ) · e2 + (x0 ) · e2 (3.24) ∂e1 ∂e1 ∂e2 ∂e2 ∂v + (x0 ) · n. ∂n

div v(x0 ) =

Since v · n = 0,

∂vτ ∂vτ (x0 ) · e1 + (x0 ) · e2 = 0. ∂e1 ∂e2

(3.25)

Also, ∂vn n ∂n ∂vn n ∂vn (x0 ) · e1 + (x0 ) · e2 = (x0 )n · e1 + vn (x0 ) · e1 ∂e1 ∂e2 ∂e1 ∂e1 ∂n ∂vn + (x0 )n · e2 + vn (x0 ) · e2 ∂e2 ∂e2   ∂n ∂n = · e1 + · e2 vn ∂e1 ∂e2 = tr(S)v · n = 2(mean curvature of Γ ) v · n. (3.26) (see Lemma 7 of [2]). Taking into account (3.25) and (3.26), we have from (3.24)

which implies (3.23).

∂v ,n ∂n

Rl

= −(k(x)v, n)Rl + div v on Γ,

(3.27) 

Remark 3.4 If Γ ∈ C 2,1 , then elements of the matrix S belong to C 1 (Γ ) and so does k(x).

3.1 Properties on the Boundary Surfaces of Vector Fields

93

For elements in the divergence-free spaces, Dirichlet, strain and vorticity bilinear forms are not equal in usual. Below we show that for more general bilinear forms in space H1 (Ω) when the bilinear forms are equal to each other.  3 Let Ω be a bounded domain such that ∂Ω ∈ C 0,1 , ∂Ω = i=1 Γ i , Γi = j Γi j , Γi j ∈ C 2 and   HΓ1 1 (Ω) = u ∈ H1 (Ω) : u |Γ1 = 0, u · τ |Γ2 = 0, u · n |Γ3 = 0 . Corollary 3.2 Assume that Γi j for i = 2, 3 are pieces of plane (straight segments for 2-D). Then, (∇v, ∇u) = (rot v, rot u) + (div v, div u) = 2(E(v), E(u)) − (div v, div u) ∀v, u ∈ HΓ1 1 (Ω).

(3.28)

¯ ∩ Proof By density of smooth functions, it suffices to prove for v, u ∈ C 2 (Ω) 1 HΓ1 (Ω). By integration by parts, we get −(Δv, u)Ω = (∇v, ∇u) −

∂v ,u ∂n

Γ2 ∪Γ3

.

(3.29)

On the other hand, using the facts that −Δv = rot rot v − grad(div v), (rot v, u) − (v, rot u) = −(v × n, u)∂Ω (see (2.82)), we get −(Δv, u)Ω = (rot v, rot u)Ω + (div v, div u) − (rot v × n, u)Γ2 ∪Γ3 − (div v, u · n)Γ2 ∪Γ3 = (rot v, rot u)Ω + (div v, div u) − (rot v × n, u)Γ3 − (div v, u · n)Γ2 ,

(3.30)

where u · τ |Γ2 = 0 and u · n |Γ3 = 0 were used. Since S ≡ 0 on Γ3 , by (3.13) we have

∂v ,u . (rot v × n, u)Γ3 = (3.31) ∂n Γ3 Since k(x) = tr (S) ≡ 0 on Γ2 and v · τ |Γ2 = 0, multiplying (3.27) by (u · n), we get (div v, u · n)Γ2 =

∂v ,u ∂n

Γ2

From (3.29)–(3.32), the first equality of (3.28) follows. Also, using

.

(3.32)

94

3 The Steady Navier-Stokes System

Δvi = 2div Ei (v) − div

∂v , where Ei (v) = (εi1 (v), · · · , εil (v)), ∂ xi

(Δvi , u i ) = −2(Ei (v), ∇u i ) + 2(Ei (v) · n, u i )∂Ω − (div

∂v , ui ) ∂ xi

((2.77) and (2.78)), by the fact that the tensor E(v) is symmetric we get −(Δv, u)Ω = 2(E(v), E(u)) + (∇(div v), u) − 2(E(v)n, u)Γ2 ∪Γ3 = 2(E(v), E(u)) − (div v, div u) − 2(E(v)n, u)Γ2 ∪Γ3 + (div v, u · n)Γ2 . (3.33) By (3.14) and (3.22), we have

−2(E(v)n, u)Γ3

∂v ,u =− ∂n

and −2(E(v)n, u)Γ2 = −2

∂v ,u ∂n

(3.34) Γ3

Γ2

.

(3.35)

From (3.32)–(3.35), we obtain −(Δv, u)Ω = 2(E(v), E(u)) − (div v, div u) −

∂v ,u ∂n

Γ2 ∪Γ3

Thus, (3.36) and (3.29) imply the second equality of (3.28).

.

(3.36) 

Remark 3.5 For polygon or polyhedron, the first equality of (3.28) follows from Theorem 4.1 of [3].

3.2 Variational Formulations of the Steady Problems Let l (i) Ω be a bounded domain 7 of R , l = 2, 3; Γ i , Γi ∩ Γ j = ∅ for i = j; (ii) ∂Ω ∈ C 0,1 , ∂Ω = i=1 (iii) Γi = j Γi j , where Γi j are connected open subsets of ∂Ω and Γi j ∈ C 2 , i = 2, 3, 4, 5, 7.

We consider the Stokes equations −μΔv + ∇ p = f, div v = 0 in Ω and the Navier-Stokes equations

(3.37)

3.2 Variational Formulations of the Steady Problems

− μΔv + (v · ∇)v + ∇ p = f, div v = 0 in Ω

95

(3.38)

with the following boundary conditions: (1) v|Γ1 = h 1 , (2) vτ |Γ2 = 0, − p|Γ2 = φ2 , (3) vn |Γ3 = 0, rot v × n|Γ3 = φ3 /μ, (4) vτ |Γ4 = h 4 , (− p + 2μεnn (v))|Γ4 = φ4 , (5) vn |Γ5 = h 5 , 2(μεnτ (v) + αvτ )|Γ5 = φ5 , α : a matrix, (6) (− pn + 2μεn (v))|Γ6 = φ6 ,  ∂v  · n  = φ7 , (7) vτ |Γ7 = 0, − p + μ Γ7 ∂n or

(3.39)

(1) v|Γ1 = h 1 , 1 (2) vτ |Γ2 = 0, −( p + |v|2 )|Γ2 = φ2 , 2 (3) vn |Γ3 = 0, rot v × n|Γ3 = φ3 /μ,   1  (4) vτ |Γ4 = h 4 , − p − |v|2 + 2μεnn (v)  = φ4 , Γ4 2 (5) vn |Γ5 = h 5 , 2(μεnτ (v) + αvτ )|Γ5 = φ5 , α : a matrix,   1  (6) − pn − |v|2 n + 2μεn (v)  = φ6 , Γ6 2  1 2 ∂v  · n  = φ7 , (7) vτ |Γ7 = 0, − p − |v| + μ Γ7 2 ∂n

(3.40)

where (i) vn = v · n, vτ = v − (v · n)n; (ii) εn (v) = E(v) · n, εnn (v) = (E(v) · n, n)Rl , εnτ (v) = E(v) · n − εnn (v)n; (iii) h i , φi , αi j (components of matrix α) are given functions or vectors of functions. The boundary condition (3.40) is obtained from (3.39) by replacing the static pressure p with the total pressure p + 1/2|v|2 . Taking into account (v · ∇)v = rot v × v + ∇ 21 |v|2 and rewriting the Navier-Stokes equation as 1 ∂v − μΔv + rot v × v + ∇( p + |v|2 ) = f, div v = 0 in Ω, ∂t 2 we are concerned with the total pressure and total stress. Then we study the Stokes equation with boundary conditions (3.39), whereas we study the Navier-Stokes equations with boundary condition (3.39) or (3.40).

96

3 The Steady Navier-Stokes System

For convenience in what follows, the Navier-Stokes problems with boundary conditions (3.39) and (3.40) are, respectively, called the case of static pressure and the case of total pressure. Let V = {u ∈ H1 (Ω) : div u = 0, u |Γ1 = 0, u τ |(Γ2 ∪Γ4 ∪Γ7 ) = 0, u n |(Γ3 ∪Γ5 ) = 0}, VΓ 237 = {u ∈ H1 (Ω) : div u = 0, u τ |(Γ2 ∪Γ7 ) = 0, u n |Γ3 = 0}. Let us consider variational formulations based on the bilinear form (E(v), E(u)). Taking into account (2.81) and applying Theorems 3.1 and 3.2 on Γi (i = 2, 3, 7), for v ∈ H2 (Ω) ∩ VΓ 237 and u ∈ V, we have 7 −(Δv, u) = 2(E(v), E(u)) − 2(E(v) · n, u)∪i=2 Γi

= 2(E(v), E(u)) + 2(k(x)v, u)Γ2 − (rot v × n, u)Γ3 7 + 2(S v, ˜ u) ˜ Γ3 − 2(εn (v), u)∪i=4 Γi

= 2(E(v), E(u)) + 2(k(x)v, u)Γ2 − (rot v × n, u)Γ3 + 2(S v, ˜ u) ˜ Γ3 − 2(εnn (v), u · n)Γ4 − 2(εnτ (v), u)Γ5 − 2(εn (v), u)Γ6

∂v ,u − + (k(x)v, u)Γ7 . ∂n Γ7

(3.41)

Also, for p ∈ H 1 (Ω) and u ∈ V we have 7 (∇ p, u) = ( p, u · n)∪i=2 Γi

= ( p, u n )Γ2 + ( p, u n )Γ4 + ( pn, u)Γ6 + ( p, u n )Γ7 ,

(3.42)

where u · n |Γ3 ∪Γ5 = 0 was used. By (3.41) and (3.42), we have −μ(Δv, u) + (∇ p, u) ˜ u) ˜ Γ3 + μ(k(x)v, u)Γ7 = 2μ(E(v), E(u)) + 2μ(k(x)v, u)Γ2 + 2μ(S v,   + ( p, u · n)Γ2 − μ(rot v × u, u)Γ3 − (− p + 2μεnn (v)), u · n Γ 4    ∂v  · n , un − 2μ(εnτ (v), u)Γ5 − (− pn + 2μ(εn (v)), u)Γ6 − − p + μ . Γ7 ∂n

(3.43)

Assume that the following holds. Assumption 3.1 There exists a function U ∈ H1 (Ω) such that div U = 0, U |Γ1 = h 1 , Uτ |(Γ2 ∪Γ7 ) = 0, Un |Γ3 = 0, Uτ |Γ4 = h 4 , Un |Γ5 = h 5 . 1

1

Also, f ∈ V∗ , φi ∈ H − 2 (Γi ), i = 2, 4, 7, φi ∈ H− 2 (Γi ), i = 3, 5, 6, αi j ∈ L ∞ (Γ5 ), and Γ1  = ∅.

3.2 Variational Formulations of the Steady Problems

97

Then, in view of (3.43), we get a variational formulation for the Stokes problem (3.37), (3.39): Problem 3.1 Find v such that v − U ∈ V, ˜ u) ˜ Γ3 + 2(α(x)vτ , u)Γ5 + μ(k(x)v, u)Γ7 2μ(E(v), E(u)) + 2μ(k(x)v, u)Γ2 + 2μ(S v,   =  f, u + φi , u n Γi + φi , u Γi ∀u ∈ V. i=2,4,7

i=3,5,6

(3.44) Also, we get a variational formulation for the Navier-Stokes problem of the case of static pressure: Problem 3.2 Find v such that v − U ∈ V, 2μ(E(v), E(u)) + (v · ∇)v, u + 2μ(k(x)v, u)Γ2 + 2μ(S v, ˜ u) ˜ Γ3 + 2(α(x)vτ , u)Γ5 + μ(k(x)v, u)Γ7   =  f, u + φi , u n Γi + φi , u Γi ∀u ∈ V. i=2,4,7

(3.45)

i=3,5,6

On the other hand, taking (v · ∇)v = rot v × v + 21 grad|v|2 into account, we get a variational formulation for the Navier-Stokes problem of the case of total pressure: Problem 3.3 Find v such that v − U ∈ V, 2μ(E(v), E(u)) + rot v × v, u + 2μ(k(x)v, u)Γ2 + 2μ(S v, ˜ u) ˜ Γ3 + 2(α(x)vτ , u)Γ5 + μ(k(x)v, u)Γ7   =  f, u + φi , u n Γi + φi , u Γi ∀u ∈ V. i=2,4,7

(3.46)

i=3,5,6

To show that the formulations above are reasonable, we can prove that if solutions v are smooth (v ∈ H2 (Ω), f ∈ L2 (Ω)), then there exists p such that (v, p) satisfies the original problems. We will do it for more general cases including the boundary conditions of friction type in Chap. 5. Next, let us consider variational formulations using the bilinear form (∇v, ∇u). Assume that Γ6 = ∅ and the boundary conditions are as follows:

98

3 The Steady Navier-Stokes System

(1) v|Γ1 = h 1 , (2) vτ |Γ2 = 0, − p|Γ2 = φ2 , (3) vn |Γ3 = 0, rot v × n|Γ3 = φ3 /μ, (4 ) vτ |Γ4 = 0, (− p + μεnn (v))|Γ4 = φ4 ,

(3.47)

(5) vn |Γ5 = 0, 2(μεnτ (v) + αvτ )|Γ5 = φ5 , α : a matrix, ∂v (7 ) (− pn + μ )|Γ7 = φ7 , φ7 : vector. ∂n Let VΓ 1−5 = {u ∈ H1 (Ω) : div u = 0, u |Γ1 = 0, u τ |(Γ2 ∪Γ4 ) = 0, u · n |(Γ3 ∪Γ5 ) = 0}, VΓ 2−5 = {u ∈ H1 (Ω) : div u = 0, u τ |(Γ2 ∪Γ4 ) = 0, u · n |(Γ3 ∪Γ5 ) = 0}. Applying Theorems 3.1 and 3.2 on Γi (i = 2, 3, 4, 5), for v ∈ H2 (Ω) ∩ VΓ 2−5 and u ∈ VΓ 1−5 , we have −(Δv, u) = (∇v, ∇u) −

∂v ,u ∂n

∂Ω

= (∇v, ∇u) + (k(x)v, u)Γ2 − (rot v × n, u)Γ3 + (S v, ˜ u) ˜ Γ3

∂v ,u − (εn (v), u)Γ4 − (εn (v), u)Γ5 + (S v, ˜ u) ˜ Γ5 − ∂n Γ7 = (∇v, ∇u) + (k(x)v, u)Γ2 − (rot v × n, u)Γ3 + (S v, ˜ u) ˜ Γ3

∂v ,u − (εnn (v), u · n)Γ4 − 2(εnτ (v), u)Γ5 − (S v, ˜ u)| ˜ Γ5 − . ∂n Γ7 (3.48) Using (3.48) and (3.42) yields −μ(Δv, u) + (∇ p, u) = μ(∇v, ∇u) + μ(k(x)v, u)Γ2 + μ(S v, ˜ u) ˜ Γ3 + μ(k(x)v, u)Γ7   + ( p, u · n)Γ2 − (rot v × u, u)Γ3 − (− p + μεnn (v)), u · n Γ4  ∂v   , un . − 2μ(εnτ (v), u)Γ5 − − pn + μ Γ7 ∂n (3.49) In view of (3.49), we get a variational formulation for the Stokes problem (3.37), (3.47): Problem 3.4 Find v ∈ VΓ 2−5 such that

3.2 Variational Formulations of the Steady Problems

99

v|Γ1 = h 1 , μ(∇v, ∇u) + μ(k(x)vτ , u)Γ2 + μ(S v, ˜ u) ˜ Γ3 + 2(α(x)vτ , u)Γ5 − μ(S v, ˜ u) ˜ Γ5   =  f, u + φi , u n Γi + φi , u Γi ∀u ∈ VΓ 1−5 . i=2,4,7

i=3,5

(3.50) For existence of solutions to the problems of (3.44), (3.45) and (3.46), coercivity of the quadratic form corresponding to the bilinear form 2μ(E(v), E(u)) + 2μ(k(x)v, u)Γ2 + 2μ(S v, ˜ u) ˜ Γ3 + μ(k(x)v, u)Γ7 is important (see the next section), and so is coercivity of the quadratic form corresponding to ˜ u) ˜ Γ3 − μ(S v, ˜ u) ˜ Γ5 μ(∇v, ∇u) + μ(k(x)v, u)Γ2 + μ(S v, for problem (3.50). Thus, from the point of view of coercivity, two bilinear forms 2(E(v), E(u)) and (∇v, ∇u) seem to be similar. However, relying on the bilinear form (∇v, ∇u), we cannot reflect the boundary conditions (6) in (3.39), (3.40) into variational formulations, because we know the relation between strain and normal derivative of vector fields on the boundary only when the vector fields are tangent or orthogonal to the boundary. The conditions (4 ), (7 ) of (3.47) are slightly different from (4), (7) of (3.39). Remark 3.6 Since S ≡ 0 on the flat surface Γ , the vorticity boundary condition vn |Γ = 0, rot v × n|Γ = φ/μ and the Navier slip condition vn |Γ = 0, μεnτ (v)|Γ = φ are equivalent. Remark 3.7 Condition (7’) of (3.47) (with φ7 = 0) is “do nothing” condition, but (7) of (3.39) (with φ7 = 0) is rather different from “do nothing” condition and we cannot replace (7) with (7’). Let us consider why (7) of (3.39) cannot be replaced by (7’) of (3.47). Above relying on the bilinear form 2(E(v), E(u)) and integrating by parts (−μΔv + ∇ p, u), we get boundary integral (−2μ(E(v)n, u)∂Ω + ( p, u · n)∂Ω . Then, in order to reflect the boundary conditions into Problems 3.1 and 3.2, using vτ = 0 or vn = 0 and applying Theorems 3.1 or 3.2, we transform the boundary integrals on Γi , i = 2, 3, 7. (cf. (3.43)). Concretely, under condition vτ |Γ7 = 0 we have 

− pn + μ

Usually, vτ = 0 does not imply and (7) of (3.39) we have

∂v  ∀u with u τ = 0. ,u ∂n Γ7 ∂v ∂n

(3.51)

· τ = 0, but by virtue of the conditions u τ = 0

100

3 The Steady Navier-Stokes System

  ∂v  ∂v , u Γ7 = − p + μ n, u n Γ7 = φ7 , u n Γ7 ∀u with u τ = 0. ∂n ∂n (3.52)   ∂v Thus, replacing − p + μ ∂n · n, u n Γ7 by φ7 , u n Γ7 , we reflect the boundary condition (7) of (3.39) into Problems 3.1 and 3.2. ∂v )|Γ7 = φ7 with a vector φ7 . TransLet us replace (7) of (3.39) by (− pn + μ ∂n   ∂v forming as above and replacing − pn + μ ∂n , u Γ7 by φ7 , u Γ7 , we can come to a formal variational formulation. But when a solution v is smooth enough, going back from the formal variational formulation to the original problem, we come to 

− pn + μ

(− pn + μ

∂v , u)Γ7 = φ7 , u Γ7 ∀u with u τ = 0. ∂n

(3.53)

If we have (3.53) without u τ = 0, then from (3.53) we can get − pn + μ

∂v = φ7 on Γ7 . ∂n

But due to u τ = 0, we get only (− pn + μ

∂v , n)Γ7 =< φ7 , n >Γ7 . ∂n

This shows that the formal variational formulation is not equivalent to the original ∂v condition on Γ7 and equivalent to (− p + μ ∂n n)|Γ7 = φ7 · n. Remark 3.8 “Do nothing” boundary condition (7 ) with φ7 = 0 is used in truncating large physical domains to smaller computational domains by assuming parallel flow. If the flow is parallel near the boundary, then (7) of (3.39) is the same as (7’). In the outlet the boundary condition μE(v)n − pn = 0 is also used. But, to our knowledge it seems not known whether this condition and “do nothing” condition are equivalent or not. As “do nothing” boundary condition, the boundary condition (4 ) vτ = 0, μE(v)n − pn = φ4 also results from the variational formulation based on (∇v, ∇u). By Theorem 3.2, we know that for divergence-free flows orthogonal to the boundary, two conditions (4 ) and (7 ) are equivalent in variational formulations above.

3.3 Existence of Solutions to the Steady Problems When V is a Banach space, the zero element of V is denoted by 0V and O M (0V ) means M-neighborhood of 0V . For the Stokes problem we have

3.3 Existence of Solutions to the Steady Problems

101

Theorem 3.3 Assume that the surfaces Γ2 j , Γ3 j , Γ7 j are convex and α is a positive matrix. Then, under Assumption 3.1 there exists a unique solution to Problem 3.1 for the steady Stokes system with mixed boundary condition (3.39) for any f and φi , i = 2, · · · , 7. Proof Having in mind Assumption 3.1 and putting v = w + U , we get a new problem equivalent to Problem 3.1: Find w ∈ V such that 2μ(E(w), E(u)) + 2μ(k(x)w, u)Γ2 + 2μ(S w, ˜ u) ˜ Γ3 + 2(α(x)w, u)Γ5 + μ(k(x)w, u)Γ7 = −2μ(E(U ), E(u)) − 2μ(k(x)Uτ , u)Γ2 − 2μ(SU˜ , u) ˜ Γ3 − 2(α(x)Uτ , u)Γ5   − μ(k(x)U, u)Γ7 +  f, u + φi , u · n Γi + φi , u Γi ∀u ∈ V. i=2,4,7

i=3,5,6

(3.54)

Define a linear operator A : V → V∗ by Aw, u =2μ(E(w), E(u)) + 2μ(k(x)w, u)Γ2 + 2μ(S w, ˜ u) ˜ Γ3 + 2(α(x)w, u)Γ5 + μ(k(x)w, u)Γ7 ∀w, u ∈ V.

(3.55)

Then, by Remark 3.4 and Assumption 3.1, we have |Aw, u | ≤ K wV · uV ∀w, u ∈ V.

(3.56)

By Korn’s inequality 2μ(E(w), E(w)) ≥ δw2V ∃δ > 0 (see Theorem 1.29), Assumption 3.1 and Lemma 3.1, we have Aw, w ≥ β1 w2V ∃β1 > 0.

(3.57)

Define an element F ∈ V∗ by ˜ Γ3 − 2(α(x)Uτ , u)Γ5 F, u = − 2μ(E(U ), E(u)) − 2μ(k(x)U, u)Γ2 − 2μ(SU˜ , u)   − μ(k(x)U, u)Γ7 +  f, u + φi , u · n Γi + φi , u Γi i=2,4,7

i=3,5,6

∀u ∈ V. (3.58) Thus, in view of (3.56)–(3.58), by the Lax-Milgram lemma we come to our assertion.  Remark 3.9 In (3.54), we used (α(x)w, u)Γ5 instead of (α(x)wτ , u)Γ5 since w = wτ on Γ5 , and for simplicity from now on we will use such expressions.

102

3 The Steady Navier-Stokes System 1

Remark 3.10 Since n(x) ∈ C1 (Γ i ), u · n ∈ H 2 (Γi ) (see Theorem 1.23). Also, 1 2

H (Γi ) = H0 (Γi ) (see (1.11)), and so for φi ∈ H−1/2 (Γi ), i = 3, 5, 6, and φi ∈ H −1/2 (Γi ), i = 2, 4, 7, respectively, φi , u Γi and φi , u n Γi are meaningful. When the surfaces Γ2 j , Γ3 j and Γ7 j are not convex, if the absolute value of twice the mean curvature k(x) and the norm of the shape operator S are small enough, then by Korn’s inequality estimate (3.57) is valid, and so is the assertion. 1 2

For the Navier-Stokes problem of the case of total pressure, we have Theorem 3.4 Suppose that Assumption 3.1 holds and that the surfaces Γ2 j , Γ3 j , Γ7 j are convex, α is a positive matrix and U L3 (Ω) is small enough. Then, there exists a solution to Problem 3.3 for the steady Navier-Stokes system (the case of total pressure) for any f and φi , i = 2 − 7. If U = 0,  f V∗ , φi  H − 21 (Γ ) , i = 2, 4, 7, i and φi H− 21 (Γ ) , i = 3, 5, 6, are small enough, then the solution is unique. i

Proof In the same way as in the proof of Theorem 3.3, we get a new problem equivalent to Problem 3.3: Find w ∈ V such that 2μ(E(w), E(u)) + rot w × w, u + rot U × w, u + rot w × U, u ˜ u) ˜ Γ3 + 2(α(x)w, u)Γ5 + μ(k(x)w, u)Γ7 + 2μ(k(x)w, u)Γ2 + 2μ(S w, = −2μ(E(U ), E(u)) − rot U × U, u − 2μ(k(x)U, u)Γ2 − 2μ(SU˜ , u) ˜ Γ3   − 2(α(x)Uτ , u)Γ5 − μ(k(x)U, u)Γ7 +  f, u + φi , u n Γi + φi , u Γi i=2,4,7

i=3,5,6

∀u ∈ V.

(3.59) Define a a(w; v, u) : V × V × V → R by a(w; v, u) =2μ(E(v), E(u)) + rot w × v, u + rot U × v, u + rot w × U, u + 2μ(k(x)v, u)Γ2 + 2μ(S v, ˜ u) ˜ Γ3 + 2(α(x)v, u)Γ5 + μ(k(x)v, u)Γ7 ∀w, v, u ∈ V. (3.60) On the other hand, for any w ∈ V we have rot w × w, w = 0, rot U × w, w = 0,  |rot w × U, w | ≤ |(rot w × U ) · w| d x ≤ γ w2V · U L3 .

(3.61)

Ω

Also, by Korn’s inequality, 2μ(E(v), E(v)) ≥ δv2V .

(3.62)

Therefore, if δ − γ U L3 (Ω) = β2 > 0, then by Assumption 3.1, Lemma 3.1, (3.61) and (3.62), we have

3.3 Existence of Solutions to the Steady Problems

a(v; v, v) ≥ β2 v2V β2 > 0.

103

(3.63)

Define an element F ∈ V∗ by F, u = − 2μ(E(U ), E(u)) − rot U × U, u − 2μ(k(x)U, u)Γ2 − 2μ(SU˜ , u) ˜ Γ3 − 2(α(x)Uτ , u)Γ5 − μ(k(x)U, u)Γ7 +  f, u   φi , u n Γi + φi , u Γi ∀u ∈ V. + i=2,4,7

(3.64)

i=3.5,6

Now, let us prove that if w k  w in V as k → ∞, then a(wk ; w k , u) → a(w; w, u) ∀u ∈ V. (3.65) First, let us prove that rot wk × w k , u → rot w × w, u ∀u ∈ V as k → ∞.

(3.66)

Indeed, rot wk × w k , u − rot w × w, u = rot w k × (w k − w), u + rot (w k − w) × w, u .

(3.67)

Let us estimate the first term on the right-hand side of (3.67).   rot wk × (w k − w), u  ≤ γ1

 Ω

|rot w k | · |(w k − w)| · |u| d x

(3.68)

≤ γ2 rot w k L2 · (w k − w)L3 · uL6 . Since wk → w in L3 (Ω) and rot w k is bounded in L2 (Ω), by virtue of (3.68) the first term on the right-hand side of (3.67) converges to zero as k → ∞. By the imbedding of the space V into L6 (Ω), we have that wi u j ∈ L 2 (Ω), i, j = 1, · · · , l, for any w, v ∈ V. Also, since wk  w in V as k → ∞, it follows that rot wk  rot w in L2 (Ω). Then, the second term on the right-hand side of (3.67) converges to zero as k → ∞. Thus, we get (3.66). All terms except rot w × w, u in a(w; w, u) are linear with respect to w, and so it is easy to check their convergence as k → ∞. Therefore, we have proved (3.65). By (3.63) and (3.65), there exists a solution to (3.59) (see Theorem 1.43), and so (3.46) has a solution. Let us now prove uniqueness. Taking into account U = 0 and rot w × v, v = 0, we get from (3.60) a(w; v, v) ≥ βv2V β > 0 ∀w, v ∈ V.

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3 The Steady Navier-Stokes System

We can get the following estimates: |rot w1 × u, v − rot w2 × u, v | ≤ γ3 rot w1 − rot w2 L2 · uL3 · vL6 ≤ γ4 w1 − w2 V · uV · vV ∀u, v, w1 , w2 ∈ V,     FV∗ ≤ M1  f V∗ + φi  H − 21 (Γ ) + φi H− 21 (Γ ) , i

i=2,4,7

i

i=3,5,6

where M1 depends on mean curvature, shape operator, μ and α. V∗ γ4 < 1, then by virtue of Theorem 1.44 the solution is unique.  Thus, if F β2 For the Navier-Stokes problem of the case of static pressure, we have Theorem 3.5 Suppose that Assumption 3.1 holds and α is a positive matrix. Let Γ2 = Γ4 = Γ6 = Γ7 = ∅, the surfaces Γ3 j are convex and U L3 (Ω) is small enough. Then, there exists a solution to Problem 3.2 for the steady Navier-Stokes system for any f and φi , i = 3, 5. If, moreover, U H1 ,  f V∗ , φi H− 21 (Γ ) , i = 3, 5, are small i enough, then the solution is unique. Proof In the same way as in the proof of Theorem 3.3, we get a new problem equivalent to Problem 3.2: Find w ∈ V such that 2μ(E(w), E(u)) + (w · ∇)w, u + (U · ∇)w, u + (w · ∇)U, u + 2μ(S w, ˜ u) ˜ Γ3 + 2(α(x)w, u)Γ5 = −2μ(E(U ), E(u)) − (U · ∇)U, u − 2μ(SU˜ , u) ˜ Γ3  − 2(α(x)Uτ , u)Γ5 +  f, u + φi , u Γi ∀u ∈ V.

(3.69)

i=3,5

Define a nonlinear operator A : V → V∗ by Aw, u =2μ(E(w), E(u)) + (w · ∇)w, u + (U · ∇)w, u + (w · ∇)U, u + 2μ(S w, ˜ u) ˜ Γ3 + 2(α(x)w, u)Γ5 ∀w, u ∈ V. (3.70) Using the fact that wn = 0, vn = 0 on any parts of the boundary, we can prove that (w · ∇)u, v = −(w · ∇)v, u ∀w, v ∈ V, u ∈ H1 (Ω), and so we have (w · ∇)w, w = 0, (w · ∇)U, w = −(w · ∇)w, U . Then, |(U · ∇)w, w + (w · ∇)U, w | ≤ γ1 w2V · U L3 .

(3.71)

3.3 Existence of Solutions to the Steady Problems

105

If δ − γ1 U L3 = β3 > 0, then by Korn’s inequality (3.62), Assumption 3.1, Lemma 3.1, (3.70) and (3.71) we have Aw, w ≥ β3 w2V , β3 > 0.

(3.72)

Define F ∈ V∗ by F, u = − 2μ(E(U ), E(u)) − (U · ∇)U, u − 2μ(SU˜ , u) ˜ Γ3  − 2(α(x)Uτ , u)Γ5 +  f, u + φi , u Γi ∀u ∈ V.

(3.73)

i=3,5

Now, by virtue of (3.72), (3.73) in the same way as in the proof of Theorem 3.4 we have the existence of a solution. Uniqueness can also be proved as in the proof of Theorem 3.4.  For the Navier-Stokes problem of the case of static pressure, we have Theorem 3.6 Suppose that Assumption 3.1 holds, α is a positive matrix, the surfaces Γ2 j , Γ3 j , Γ7 j are convex and U H1 (Ω) is small enough. Then, when f and φi , i = 2, · · · , 7, are small enough, there exists a unique solution to Problem 3.2 for the steady Navier-Stokes system in a neighborhood of U in H1 (Ω). Proof In the same way as in the proof of Theorem 3.5, we get the following new problem equivalent to Problem 3.2: Find w ∈ V such that 2μ(E(w), E(u)) + (w · ∇)w, u + (U · ∇)w, u + (w · ∇)U, u ˜ u) ˜ Γ3 + 2(α(x)w, u)Γ5 + μ(k(x)w, u)Γ7 + 2μ(k(x)w, u)Γ2 + 2μ(S w, = −2μ(E(U ), E(u)) − (U · ∇)U, u − 2μ(k(x)U, u)Γ2 − 2μ(SU˜ , u) ˜ Γ3  − 2(α(x)Uτ , u)Γ5 − μ(k(x)U, u)Γ7 +  f, u + φi , u n Γi +



i=2,4,7

φi , u Γi ∀u ∈ V.

i=3,5,6

Define operators A, B : V → V∗ and an element F ∈ V∗ , respectively, by

(3.74)

Aw, u = 2μ(E(w), E(u)) + (U · ∇)w, u + (w · ∇)U, u + 2μ(k(x)w, u)Γ2 + 2μ(S w, ˜ u) ˜ Γ3 + 2(α(x)w, u)Γ5 + μ(k(x)w, u)Γ7 ∀w, u ∈ V, Bw, u = (w · ∇)w, u ∀w, u ∈ V, F, u = −2μ(E(U ), E(u)) − (U · ∇)U, u − 2μ(k(x)U, u)Γ2 − 2μ(SU˜ , u) ˜ Γ3  − 2(α(x)Uτ , u)Γ5 − μ(k(x)U, u)Γ7 +  f, u + φi , u n Γi +

 i=3,5,6

i=2,4,7

φi , u Γi ∀u ∈ V.

106

3 The Steady Navier-Stokes System

Then, we can rewrite (3.74) as follows: Aw, u = F, u − Bw, u ∀u ∈ V.

(3.75)

We can see that |Aw, u | ≤ mwV · uV ,

(3.76)

Aw, w ≥ β4 w2V , ∃β4 > 0, ∀w, u ∈ V. Also |Bw1 − Bw2 ,u | ≤ δ Mw1 − w2 V · uV

(3.77)

∀wi ∈ O M (0V ) ≡ {v ∈ V : vV ≤ M}, ∀u ∈ V,

    FV∗ ≤ M1 U 2H1 +  f V∗ + φi  H − 21 (Γ ) + φi H− 21 (Γ ) , i

i=2,4,7

i

i=3,5,6

(3.78) where M1 depends on the mean curvature, shape operator, μ and α. By the Lax-Milgram lemma and (3.76), for any fixed z ∈ O M (0V ) there exists a unique solution to the problem Aw = F − Bz,

(3.79)

and by (3.76) and (3.77) the solution w to (3.79) is estimated as follows: wV ≤

1 1 (FV∗ + BzV∗ ) ≤ (FV∗ + δ M 2 ). β4 β4

Thus, if FV∗ and M are small enough, then the map z → w is strict contract in O M (0V ), and so there exists a unique solution to (3.75) in O M (0V ), which shows our assertion.  In the same way as in Theorem 3.3, for the the Stokes system with boundary condition (3.47), we have Theorem 3.7 Assume that Γ6 = ∅, the surfaces Γ2 j , Γ3 j are convex, Γ5 j is concave and α is a positive matrix. Then, under Assumption 3.1 without the condition Uτ |Γ7 = 0 there exists a unique solution to Problem 3.4 for the stationary Stokes system with mixed boundary condition (3.47) for any f and φi , i = 2, · · · , 5, 7. Remark 3.11 Assuming that the surfaces Γ2 j , Γ3 j are convex and Γ5 j are concave, we can obtain existence and uniqueness of solution to the Navier-Stokes problem with boundary condition (3.47) formulated similarly to Problem 3.4.

3.4 Bibliographical Remark

107

3.4 Bibliographical Remark Let us point out that explanation for the second fundamental form of surface and shape operator comes mainly from [1], but the shape operator in this book has the sign opposite to the one in [1] and so are the signed curvature and mean curvature. The remaining content of Sect. 3.1 is taken from Sect. 2 of [4]. Formula (3.12) is a generalization of Lemma 2.1 for 2-D in [5] and for three-dimensional case it is known in the form 2 (E(v)n, τ )R3 = (rot v × n, τ )R3 − 2v ·

∂n ∂τ

in [6, 7]. For a direct elementary proof of (3.12) not using the knowledge of differential geometry in Sect. 3.1, we refer to Theorem 2.1 and Lemma A.1 of [4]. The proof of (3.23) follows from Lemma 7 of [2]. The content of Sect. 3.2 is from Sect. 3 of [4] and the content of Sect. 3.3 is taken from Sect. 4 of [4]. Some mistakes in [4] are corrected here. Based on the strain bilinear form, the Stokes and Navier-Stokes problems with the mixture of Dirichlet and stress boundary conditions were studied. For such a monograph, we refer to [8]. In a polyhedral domain with Γi = ∅, i = 2, 3, 7,, the Stokes problem in [9, 10] and the Navier-Stokes problem in [10, 11] were studied. In [12], the Navier-Stokes problem with Γi = ∅, i = 2, 3, 4, 5, 7, was studied. In [13], by the potential theory, a mixed boundary value problem for the NavierStokes equations in a bounded Lipschitz two-dimensional domain with Γi = ∅, i = 2, 3, 4, 6, 7, was studied. Reducing the original problem to a boundary integral equation, [14] studied the Stokes problem in a class of Lipschitz domains with Γi = ∅, i = 2, 3, 4, 7. In [15], smoothness of a solution to the Navier-Stokes problems with Γi = ∅, i = 2, 4, 5, 6, 7, or Γi = ∅, i = 2, 3, 4, 6, 7, was studied. Applying the vorticity bilinear form, someone studied the Stokes and NavierStokes problems with the mixture of Dirichlet, pressure and vorticity boundary conditions. The Stokes and Navier-Stokes problems with Γi = ∅, i = 4, 5, 6, 7 in [16–19] and [20], where for the Navier-Stokes problem the total pressure condition is used on Γ2 . In [21, 22], the Navier-Stokes problems with Γi = ∅, i = 2, 4, 5, 6, 7, were studied. Relying on the Dirichlet bilinear form, someone studied the Navier-Stokes problems with the mixture of Dirichlet, “do nothing” condition (see [23]). Before [4] no one has considered mixed boundary problems for the Stokes and Navier-Stokes equations with both stress and pressure boundary conditions.

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3 The Steady Navier-Stokes System

References 1. A. Pressley, Elementary Differential Geometry (Springer, 2010) 2. U. Kangro, R. Nicoaides, Divergence boundary conditions for vector Helmholtz equations with divergence constraints. Math. Model. Numer. Anal. 33(3), 479–492 (1999) 3. M. Costabel, A coecive bilinear form for Maxwell’s equations. J. Math. Anal. Appl. 157, 527–541 (1991) 4. T. Kim, D. Cao, Some properties on the surfaces of vector fields and its application to the Stokes and Navier-Stokes problems with mixed boundary conditions. Nonlinear Anal. 113, 94–114 (2015). Erratum, ibid 135, 249–250 (2016) 5. T. Clopeau, A. Mikeli´c, R. Robert, On the vanishing viscosity limit for the 2D incompressible Navier-Stokes equations with the friction type boundary conditions. Nonlinearity 11, 1625– 1636 (1998) 6. H. Beirão da Veiga, F. Crispo, Sharp inviscid limit results under Navier type boundary conditions, An Lp theory. J. Math. Fluid Mech. 12(3), 397–411 (2010) 7. H. Beirão da Veiga, F. Crispo, A missed persistence property for the Euler equations, and its effect on inviscid limits. Nonlinearity 25(6), 1661–1669 (2012) 8. N.D. Kopachevsky, S.G. Krein, Operator Approach to Linear Problems of Hydrodynamics 1, 2 (Springer Basel AG 2001, 2003) 9. V. Maz’ya, J. Rossmann, Lp estimates of solutions to mixed boundary value problems for the Stokes system in polyhedral domains. Math. Nachr. 280(7), 751–793 (2007) 10. J. Rossmann, Mixed boundary value problems for Stokes and Navier-Stokes systems in polyhedral domains. Oper. Theory Adv. Appl. 193, 269–280 (2009) 11. V. Maz’ya, J. Rossmann, Mixed boundary value problems for the Navier-Stokes system in polyhedral domains. Arch. Rational Mech. Anal. 194, 669–712 (2009) 12. M. Orlt, A.-M. Sändrig, Regularity of viscous Navier-Stokes flows in nonsmooth domains, Boundary value problems and integral equations in nonsmooth domains. Lect. Note Pure Appl. Math. 167, 185–201 (1995) 13. A. Russo, G. Starita, A mixed problem for the steady Navier-Stokes equations. Math. Comput. Model. 49, 681–688 (2009) 14. R. Brown, I. Mitrea, M. Mitrea, M. Wright, Mixed boundary value problems for the Stokes system. Trans. Am. Math. Soc. 362(3), 1211–1230 (2010) 15. C. Ebmeyer, J. Frehse, Steady Navier-Stokes equations with mixed boundary value conditions in three-dimensional Lipschitzian domains. Mathematische Annalen 319, 349–381 (2001) 16. C. Begue, C. Conca, F. Murat, O. Pironneau, A nouveau sur les equations de Stokes et de Navier-Stokes avec des conditions aux limites sur la pression. C. R. Acad. Sci. Paris Ser. I(304), 23–28 (1987) 17. C. Begue, C. Conca, F. Murat, O. Pironneau, Les Equations de Stokes et de Navier-Stokes Avec Des Condition Sur La Pression. Nonlinear Partial Diff. Equat. and Their Appl. College de France, Seminar IX, 179–384 (1988) 18. C. Conca, F. Murat, O. Pironneau, The Stokes and Navier-Stokes equations with boundary conditions involving the pressure. Japan J. Math. 20(2), 279–318 (1994) 19. C. Conca, C. Pares, O. Pironneau, M. Thiriet, Navier-Stokes equations with imposed pressure and velocity fluxes. Inter. J. Numer. Meth. Fluids 20, 267–287 (1995) 20. A.A. Illarionov, A.Yu. Chebotarev, Solvability of a mixed boundary value problem for the stationary Navier-Stokes equations. Diff. Equat. 37(5), 724–731 (2001) 21. C. Bernardi, F. Hecht, R. Verfürth, A finite element discretization of the three-dimensional Navier-Stokes equations with mixed boundary conditions. ESAIM: Math. Model. Numer. Anal. 43, 1185–1201 (2009) 22. V. Girault, Incompressible finite element methods for Navier-Stokes equations with nonstandard boundary conditions in R 3 . Math. Comput. 51(183), 55–74 (1988) 23. M. Beneš, Solutions to the mixed problem of viscous incompressible flows in a channel. Arch. Math. 93, 287–297 (2009)

Chapter 4

The Non-steady Navier-Stokes System

In this chapter we are concerned with the non-steady Navier-Stokes equations and Stokes equations with mixed boundary conditions including conditions for velocity, pressure, stress, vorticity and Navier slip condition together. As in Sect. 3.2, relying on the result in Sect. 3.1, we embed all these boundary conditions into variational formulations of problems. For the problem with boundary conditions involving the total pressure and total stress, by a transformation of unknown functions and a penalty method we can turn the problem into an elliptic operator equation for functions defined in the time-spatial domain. In this case we need not assume that the given data are small enough. For the problem with boundary conditions involving the static pressure and the stress, we prove the existence of a unique solution for small data under a compatibility condition at initial time. We can also prove that if a smooth solution is given, then under the compatibility condition for the small perturbed data there exists a unique solution. Section 4.1 is devoted to the problems with boundary conditions involving the total pressure and the total stress, and Sect. 4.2 is concerned with the problems with boundary conditions involving the static pressure and the stress.

4.1 Existence of a Solution: The Case of Total Pressure Let Ω satisfy the following conditions: l (i) Ω be a bounded domain 7 of R , l = 2, 3, 0,1 (ii) ∂Ω ∈ C , ∂Ω = i=1 Γ i , Γi ∩ Γ j = ∅ for i = j,

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 T. Kim and D. Cao, Equations of Motion for Incompressible Viscous Fluids, Advances in Mathematical Fluid Mechanics, https://doi.org/10.1007/978-3-030-78659-5_4

109

110

4 The Non-steady Navier-Stokes System

 (iii) Γi = j Γi j , where Γi j are connected open subsets of ∂Ω and Γi j ∈ C 2,1 , i = 2 − 5, 7, and Q = Ω × (0, T ), Σi = Γi × (0, T ), 0 < T < ∞.

4.1.1 Problem and Variational Formulation In this section we study the Navier-Stokes problem ⎧ ∂v ⎪ ⎪ ⎨ ∂t − μΔv + (v · ∇)v + ∇ p = f, div v = 0, ⎪ ⎪ ⎩ v(0) = v0

(4.1)

with boundary conditions including the total pressure (1) v|Γ1 = h 1 , 1 (2) vτ |Γ2 = 0, −( p + |v|2 )|Γ2 = φ2 , 2 (3) vn |Γ3 = 0, rot v × n|Γ3 = φ3 /μ,   1 (4) vτ |Γ4 = h 4 , − p − |v|2 + 2μεnn (v) = φ4 , Γ4 2 (5) vn |Γ5 = h 5 , 2(μεnτ (v) + αvτ )|Γ5 = φ5 , α : a matrix,   1 (6) − pn − |v|2 n + 2μεn (v) = φ6 , Γ6 2  1 2 ∂v  · n = φ7 , (7) vτ |Γ7 = 0, − p − |v| + μ Γ7 2 ∂n

(4.2)

where vn = v · n, vτ = v − (v · n)n, εn (v) = E(v)n, εnn (v) = (E(v)n, n)Rl , εnτ (v) = ε(v)n − εnn (v)n and h i , φi , α jk (components of matrix α) are given functions or vectors of functions of x, t on Σi . Define V = {u ∈ H1 (Ω) : div u = 0, u |Γ1 = 0, u τ |(Γ2 ∪Γ4 ∪Γ7 ) = 0, u n |(Γ3 ∪Γ5 ) = 0}, ¯ : div u = 0, u |Σ1 = 0, u τ |(Σ2 ∪Σ4 ∪Σ7 ) = 0, u n |(Σ3 ∪Σ5 ) = 0}, (Q) = {u ∈ C2 ( Q) 2 V(Q) = L (0, T ; V), W(Q) = {the completion of (Q) in the space H1 (Q)}. Assume that the following holds:

4.1 Existence of a Solution: The Case of Total Pressure

111

Assumption 4.1 There exists a function U ∈ H1 (Q) ∩ L∞ (Q) such that div U = 0, U |Σ1 = h 1 , Uτ |(Σ2 ∪Σ7 ) = 0, Un |Σ3 = 0, Uτ |Σ4 = h 4 , Un |Σ5 = h 5 . 1

1

Also, f ∈ L 2 (0, T ; V∗ ), φi ∈ L 2 (0, T ; H − 2 (Γi )), i = 2, 4, 7 φi ∈ L 2 (0, T ; H− 2 (Γi )), i = 3, 5, 6, αi j ∈ L ∞ (0, T ; L ∞ (Γ5 )), v0 − U (x, 0) ∈ H , where H is closure of V in L2 (Ω), and Γ1  = ∅.

Applying Theorems 3.1, 3.2 on Γi j (i = 2, 3, 7) and taking (v · ∇)v = rot v × v + 21 grad|v|2 into account (cf. (3.46)), we get a variational formulation for problem (4.1), (4.2): Problem 4.1 Find v such that v − U ∈ V(Q),

T



T ∂u  dt + v, − 2μ(E(v), E(u)) dt +

rot v × v, u dt ∂t Q 0 0

T

T + 2μ(k(x)v, u)Γ2 dt + 2μ(S v, ˜ u) ˜ Γ3 dt 0 0

T

T 2(α(t, x)vτ , u)Γ5 dt + μ(k(x)v, u)Γ7 dt + 0 0

T

T 

T 

f, u dt +

φi , u n Γi dt +

φi , uΓi dt = (v0 , u(0)) + 0

0

i=2,4,7

0

i=3,5,6

∀u ∈ (Q) with u(x, T ) = 0. (4.3) Theorem 4.1 Under Assumption 4.1 for any U , f and φi , i = 2, · · · , 7, there exists a solution to Problem 4.1 such that ess sup v ≤ c. t∈(0,T )

For the proof of Theorem 4.1, we transform problem (4.3) into an equivalent one. Taking into account Assumption 4.1 and putting v = z + U , from Problem 4.1 we get an equivalent new problem: Find z ∈ V(Q) such that

112

4 The Non-steady Navier-Stokes System

−

 Q



∂u  dt + z, ∂t



T

2μ(E(z), E(u)) dt +

0

T

rot z × z, u dt

0

T 2μ(k(x)z, u)Γ2 dt (rot z × U + rot U × z, u) dt + 0 0

T

T

T 2μ(S z˜ , u) ˜ Γ3 dt + 2(α(t, x)z, u)Γ5 dt + μ(k(x)z, u)Γ7 dt + 0 0 0

T

T

T ∂u  dt − U, 2μ(E(U ), E(u)) dt − (rot U × U, u) dt = ∂t 0 0 0 (4.4)

T

T 2μ(k(x)U, u)Γ2 dt − 2μ(SU˜ , u) ˜ Γ3 dt − 0 0

T

T

T 2(α(x)Uτ , u)Γ5 dt − μ(k(x)U, u)Γ7 +

f, u dt − 0 0 0

T 

T 

φi , u n Γi dt +

φi , uΓi dt + (v0 , u(0)) + +

0

T

i=2,4,7

0

i=3,5,6

∀u ∈ (Q) with u(x, T ) = 0. In (4.4) let us make again a change of the unknown function by w = ek1 t z where k1 is a constant to be determined in Lemma 4.1 later. Then we have



∂u ∂ uˆ −k1 t ∂u d xdt = − d xdt = − d xdt − k1 − z e w w w uˆ d xdt, ∂t ∂t Q ∂t Q Q Q



 ∂ uˆ  ∂u ∂u d xdt = d xdt = + k1 uˆ d xdt, U U ek1 t e−k1 t U e k1 t ∂t ∂t ∂t Q Q Q

where uˆ = e−k1 t u. Substituting these into (4.4), we see that the problem to find a solution to (4.3) is equivalent to the following problem. Find w ∈ V(Q) such that

4.1 Existence of a Solution: The Case of Total Pressure

113

T

T ∂ uˆ  dt + 2μ (E(w), E(u)) ˆ dt + e−k1 t (rot w × w, u) ˆ dt ∂t 0 0 0

T

T

T

˜ˆ Γ dt + rot w × U + rot U × w, uˆ dt + 2μ(k(x)w, u)Γ2 dt + 2μ(S w, ˜ u) 3

−

T

w,

0

+

T



0

2(α(t, x)w, u) ˆ Γ5 dt +

T 0

0

μ(k(x)w, u) ˆ Γ7 dt − k1

T

T

T

0

0

(w, u) ˆ dt

  ∂ uˆ (E(U¯ ), E(u)) ˆ dt − (rot U¯ × U, u) ˆ dt U¯ + k1 uˆ d xdt − 2μ ∂t Q 0 0

T

T ˜¯ u) ˜ˆ Γ dt 2μ(k(x)U¯ , u) ˆ Γ2 dt − 2μ(SU, − 3

=

0

− +

T 0

0

2(α(t, x)U¯ τ , u) ˆ Γ5 dt −

T



0

i=2,4,7

φ¯ i , uˆ n Γi dt +

T

0

T 0

μ(k(x)U, u) ˆ Γ7 + 

T 0

f¯, u ˆ dt

φ¯ i , u ˆ Γi dt + (v0 , u(x, 0))

i=3,5,6

∀uˆ ∈ (Q), with u(x, ˆ T ) = 0,

(4.5) where γ¯ = ek1 t γ for any γ . Therefore, for the proof of Theorem 4.1 it suffices to prove the existence of a solution w to problem (4.5) and ess sup w ≤ c. t∈(0,T )

(4.6)

4.1.2 An Auxiliary Problem by Elliptic Regularization First we will consider the following auxiliary problem by elliptic regularization. For any positive integer m, we find the functions w m ∈ W(Q) satisfying the following:

114

4 The Non-steady Navier-Stokes System





T 1 ∂w m ∂u m · −w d xdt + 2μ (E(w m ), E(u)) dt m ∂t ∂t 0 Q

T

T

+ rot w m × U + rot U × w m , u dt e−k1 t (rot w m × w m , u) dt + 0 0

T

T

T 2μ(k(x)w m , u)Γ2 dt + 2μ(S w˜m , u) ˜ Γ3 dt + 2(α(t, x)w m , u)Γ5 dt + 0 0 0

T

T μ(k(x)w m , u)Γ7 dt − k1 (w m , u) dt + (w m (T ), u(T )) + 0 0

T

T

  ∂u + k1 u d xdt − 2μ (E(U¯ ), E(u)) dt − (rot U¯ × U, u)Ω(t) dt = U¯ ∂t Q 0 0

T

T

T ˜ ¯ ¯ − 2μ(k(x)U , u)Γ2 dt − 2μ(SU, u) ˜ Γ3 dt − 2(α(t, x)U¯ τ , u)Γ5 dt 0 0 0

T

T 

T ¯ ¯ μ(k(x)U , u)Γ7 dt +

f , u dt +

φ¯ i , u n Γi dt − 0

+ 0

0 T



0

i=2,4,7

φ¯ i , uΓi dt + (v0 , u(x, 0))

i=3,5,6

∀u ∈ W(Q). (4.7) For (4.7) we have the following result on the existence and uniqueness of the solution: Lemma 4.1 Under Assumptions 4.1, there exists a constant k1 independent of m such that there exists a unique solution to problem (4.7). Proof Define an operator Am from W(Q) into its dual space by

T  1 ∂z ∂u  ∂u  , dt − dt z, m ∂t ∂t ∂t 0 0

T

T + 2μ (E(z), E(u)) dt + e−k1 t (rot z × z, u) dt 0 0

T

T 2μ(k(x)z, u)Γ2 dt + (rot z × U + rot U × z, u) dt + (4.8) 0 0

T

T 2μ(S z˜ , u) ˜ Γ3 dt + 2(α(t, x)z, u)Γ5 dt + 0 0

T

T μ(k(x)z, u)Γ7 dt − k1 (z, u) dt + (z(T ), u(T )) +



Am z, u =

T

0

0

∀z, u ∈ W(Q). And also define an element F ∈ W(Q)∗ by

4.1 Existence of a Solution: The Case of Total Pressure

115



T   ∂u ¯

F, u = U + k1 u d xdt − 2μ (E(U¯ ), E(u)) dt ∂t 0 Q

T

T

T ˜¯ u) − (rot U¯ × U, u) dt − 2μ(k(x)U¯ , u)Γ2 dt − 2μ(SU, ˜ Γ3 dt 0 0 0

T

T

T 2(α(t, x)U¯ τ , u)Γ5 dt − μ(k(x)U¯ , u)Γ7 dt +

f¯, u dt − 0 0 0

T 

T 

φ¯ i , u n Γi dt +

φ¯ i , uΓi dt + (v0 , u(x, 0)) + 0

0

i=2,4,7

i=3,5,6

∀u ∈ W(Q). (4.9) Now, let us consider the existence of a solution to the following problem: Am w m = F,

(4.10)

which is equivalent to the existence of a solution to the auxiliary problem (4.7). For all z ∈ W(Q) we have

Am z, z =

T

T

T 1 ∂z ∂z  ∂z  dt − dt + 2μ z, (E(z), E(z)) dt , m ∂t ∂t ∂t 0 0 0

T

T

T 2μ(k(x)z, z)Γ2 dt + 2μ(S z˜ , z˜ )Γ3 dt + (rot z × U, z) dt + 0 0 0

T

T

T 2(α(t, x)z, z)Γ5 dt + μ(k(x)z, z)Γ7 dt − k1

z(t) 2 dt + z(T ) 2 , + 0

0

0

(4.11)

where (rot z × z, z) = 0 and (rot U × z, z) = 0 were used. Integrating by parts yields

−

z Q

1 ∂z d xdt =

z(0) 2 − z(T ) 2 for z ∈ W(Q). ∂t 2

(4.12)

By (4.11) and (4.12), for all z ∈ W(Q) we have

Am z, z =

T

T

T  2 1  ∂z  2 Σi j εi j (z) dt + (rot z × U, z) dt   dt + 2μ 0 m ∂t 0 0

T

T

T (4.13) 2μ(k(x)z, z)Γ2 dt + 2μ(S z˜ , z˜ )Γ3 dt + 2(α(t, x)z, z)Γ5 dt + 0 0 0

T

T

1

z(0) 2 + z(T ) 2 . μ(k(x)z, z)Γ7 dt − k1

z(t) 2 dt + + 2 0 0

116

4 The Non-steady Navier-Stokes System

By Korn’s inequality,

T

Σi j εi j (z) dt ≥ c1

T

2

2 0

0

z 2H1 (Ω) dt,

c1 > 0.

(4.14)

By Young’s inequality and Assumption 4.1,



0

T



T

μc1 T 2

z H1 (Ω) dt + c2

z 2 dt. (rot z × U, z) dt ≤ 2 0 0

(4.15)

On the other hand, by virtue of Remark 3.4 and Assumption 4.1 there exists a constant M such that

S(x) ∞ , k(x) ∞ , α L ∞ (0,T ;L ∞ (Γ5 )) ≤ M. Therefore, there exists a constant c3 such that



T 2μ(k(x)z, z)Γ2 dt + 2μ(S z˜ , z˜ )Γ3 dt 0 0 0 T

T  μc1

z 2H1 (Ω) + c3 z 2 dt ∀z ∈ W(Q) μ(k(x)z, z)Γ7 dt ≤ + 4 0 0 (4.16) (see Theorem 1.27). Taking −k1 large enough in (4.13) independently of m, from (4.13)–(4.16) we have, therefore T



2(α(t, x)z, z)Γ5 dt +

T



Am z, z ≥ c4 z 2W(Q) , ∃c4 > 0, ∀z ∈ W(Q),

(4.17)

where c4 depends on m. Now, let us prove that if z k z in W(Q) as k → ∞, then

Am z k , u → Am z, u ∀u ∈ W(Q).

(4.18)

First, let us prove that

T 0

e−k1 t (rot z k × z k , u) dt →

T 0

e−k1 t (rot z × z, u) dt ∀u ∈ W(Q) as k → ∞.

Indeed, straightforward calculation gives

(4.19)

4.1 Existence of a Solution: The Case of Total Pressure



T

e



−k1 t

0

T

e−k1 t (rot z × z, u) dt =

T −k1 t e e−k1 t (rot (z k − z) × z), u) dt. (rot z k × (z k − z), u) dt +

(rot z k × z k , u) dt −



117

0

T

0

0

(4.20) By the embedding of H 1 (Q) in L 4 (Q) we have e−k1 t zu ∈ L2 (Q) and rot (z k − z) 0 in L2 (Q) as z k z in W(Q). Thus, the second integral on the right-hand side of (4.20) converges to zero when k → ∞. Let us consider the first integral on the right-hand side of (4.20). For any ε ≥ 0 we can choose u ε ∈ (Q) such that

u − u ε W(Q) ≤ ε. Then,

T

0

e−k1 t (rot z k × (z k − z), u) dt



=

T

e

−k1 t

(rot z k × (z k − z), u ε ) dt +

0

T

e−k1 t (rot z k × (z k − z), u − u ε ) dt.

0

(4.21)

Since z k → z in L2 (Q) as k → ∞,



T 0

e−k1 t (rot z k × (z k − z), u ε ) dt

(4.22)

≤ C ∇z k L2 (Q) z k − z L2 (Q) u ε L∞ (Q) → 0 as k → ∞. Also, since {z k − z} is bounded in W(Q), we have



0

T

e−k1 t (rot z k × (z k − z), u − u ε ) dt ≤ C ∇z k L2 (Q) z k − z L4 (Q) u − u ε L4 (Q) ≤ C z k W(Q) z k − z W(Q) u − u ε W(Q) ≤ Cε.

(4.23)

From (4.21)–(4.23), we see that the first integral on the right-hand side of (4.20) goes to zero when k → ∞, and so we get (4.19). It is easy to check that other terms in Am z k , u converge when k → ∞. This fact together with (4.19) implies (4.18). By (4.17) and (4.18), there exists a solution to (4.10) (see Theorem 1.43), and therefore the assertion is proved. 

118

4 The Non-steady Navier-Stokes System

4.1.3 Proof of the Existence of a Solution Now, we prove Theorem 4.1. To do this we need to establish two lemmas first. Lemma 4.2 If w m ∈ W(Q) are solutions to problem (4.7) with the k1 in Lemma 4.1, then

1 ∂w m ∂u d xdt → 0 ∀u ∈ W(Q) as m → ∞. (4.24) Q m ∂t ∂t Proof By (4.10), (4.13)–(4.16), we have



1 ∂w m 2 μc1 T 1 m

w (x, 0) 2 + w m (x, T ) 2 d xdt +

w m 2H1 (Ω) dt + ∂t 4 0 2 Q m   ≤ F, w m . (4.25) On the other hand,

∂w d xdt = (U¯ , w m (x, T )) − (U¯ , w m (x, 0)) − U¯ ∂t Q m

Q

∂ U¯ m w d xdt ∂t

(4.26)

∀w ∈ W(Q). m

Taking (4.26) into account in (4.9) and applying Young’s inequality to the righthand side of (4.25), we have  m 2

T

T   1  1 m  ∂w  dt + μc1

w (x, 0) 2 + wm (x, T ) 2 ≤ c,

wm 2H1 (Ω) dt +   ∂t 8 0 8 0 m

(4.27) where c is independent of m and depends on U¯ , φ¯ i , v0 , f¯, S, k and k1 . m 2  d xdt ≤ c, we can get (4.24). Indeed, by Hölder’s inequality Using Q m1 ∂w ∂t  21 

2 m 1 ∂w m 2 ∂u 1 ∂w ∂u 1 √ d xdt , m ∂t ∂t d xdt ≤ √m m ∂t d xdt · Q Q Q ∂t

which shows (4.24).



Lemma 4.3 If wm ∈ W(Q) are solutions to problem (4.7) guaranteed by Lemma 4.1, then {wm } are relatively compact in L2 (Q). Proof Putting w¯ m (x, t) = 0 on (Ω × (−T, 2T )) \ Q, let us make w¯ m (x, t) an extension of wm . Thus, by virtue of (4.27) we get

4.1 Existence of a Solution: The Case of Total Pressure

119

1 ∂ w¯ m 2 ∂t d xdt ≤ c, Ω×(0,T ) m

T  m 2 w¯  1 dt ≤ c,

H ()

0

(4.28)

w¯ m (x, 0) ≤ c,

w¯ m (x, T ) ≤ c. For 0 < h < T let whm (x, t) =

1 h



t+h

w¯ m (x, s) ds.

t

Then, whm | Q ∈ W (Q) and (whm )t :=

∂whm (x, t) 1 m w¯ (x, t + h) − w¯ m (x, t) . = ∂t h

Replacing u by whm in (4.7), we have

T

T

T  m 

m

1 ∂w (x, t) , (whm )t dt − w , (whm )t dt + 2μ E(wm ), E(whm ) dt ∂t 0 m 0 0

T  e−k1 t rot wm × wm + rot wm × U + rot U × w m , whm dt + 0

+

T 0



2μ k(x)wm , whm Γ dt + 2

T 0



2μ S w˜m , w˜hm Γ dt 3

T

T

T

+ 2 α(t, x)wm , whm Γ dt + μ k(x)wm , whm Γ dt − k1 (wm , whm ) dt 7 5 0 0 0

T   F(t), whm (t) d xdt, = 0

(4.29)

where whm (T ) = 0 was used and F is the one in (4.9). Assuming w(x, ¯ t) ∈ C1 (Ω¯ × [0, T ]), we estimate 1 m

Ω×(0,T )

2 1 [w(x, d xdt. ¯ t + h) − w(x, ¯ t)] h

Applying Hölder’s inequality, we have

120

4 The Non-steady Navier-Stokes System

1 m

2 1 t+h ∂ w(x, ¯ s) ds d xdt 2 h ∂s t Ω×(0,T −h)

t

2 1 1 ∂ w(x, ¯ s) ds + w(x, ¯ T ) + d xdt m Ω×(T −h,T ) h 2 T ∂s



¯ s) 2 1 1 t+h ∂ w(x, 1 2

w(x, ¯ T ) 2 ≤ ds d xdt + m Ω×(0,T −h) h t ∂s mh

T 2 1 2 ∂ w(x, ¯ s) + ds · h d xdt m h 2 Ω×(T −h,T ) t ∂s

∂ 2 1 1 (T − h) ≤ ¯ T ) 2 ¯ t) d xdt + 2 w(x, w(x, hm ∂t Ω×(0,T )

2 ∂  ¯ t) d xdt + 2h w(x, Ω×(0,T ) ∂t

2 ∂   1 1 (T + h) ¯ t) d xdt + 2 w(x, ≤ ¯ T ) 2 . w(x, hm Ω×(0,T ) ∂t



(4.30)

Since C1 (Ω¯ × [0, T ]) is dense in H1 (Ω × (0, T )), by (4.28) and (4.30) for any w¯ m ∈ H1 (Ω × (0, T )) we have 1

1 1  m  w¯ (x, t) − w¯ m (x, t − h) 2 d xdt 2 ≤ √c . m Ω×(0,T ) h h

(4.31)

By Hölder’s inequality, we get from (4.27) and (4.31)

 1 ∂w m (x, t) 1  m m · w¯ (x, t) − w¯ (x, t − h) d xdt m ∂t h Q  1 ∂w m 2  21  1 1   21  2 ≤ w¯ m (x, t) − w¯ m (x, t − h) d xdt d xdt ∂t m Q h Q m c ≤√ . h (4.32) Using (4.28), we have

2μ

(E(w m ), E(whm )) dt ≤ c

 1 t+h   

w m H1 ()  w¯ m (x, s) ds  1 dt H () h t 0 0

t+h

T 1 √ 1 2 ≤c

w m H1 () √

w¯ m (x, s) 2H1 () ds dt ≤ c/ h. h 0 t (4.33)  T −k t

Now, let us estimate 0 e 1 rot w m × w m , whm dt. Since T

T

4.1 Existence of a Solution: The Case of Total Pressure



t+h

√ 

w L (Ω) ds ≤ h m

t

3

t

t+h

121

w m 2L3 () ds

 21

√ ≤ c h w¯ m 2L 2 (0,T ;H1 (Ω)) ,

by Hölder’s inequality and (4.28),

t+h T   c T  e−k1 t rot wm × wm , whm dt ≤

rot wm · wm L6

wm L3 ds dt h 0 0 t

T c c

wm 2H1 () dt ≤ √ . ≤ √ h 0 h

(4.34)

In the same way, we get



√

rot w m × U + rot U × w m , whm dt ≤ c/ h , 0

T √ (w m , whm ) dt ≤ c/ h. − k1 T

(4.35)

0

Taking into account







∂whm ∂ U¯ m d xdt = U¯ (T ), whm (T ) − U¯ (0), whm (0) − wh d xdt U¯ ∂t Q Q ∂t

T ¯

∂U m  m ¯ , wh dt, = − U (0), wh (0) − ∂t 0

we have

T

T ¯  ∂U , whm dt + (U¯ k1 whm ) dt ∂t 0 0

T

T

T

E(U¯ ), E(whm ) dt − (rot U¯ × U, whm ) dt − 2μ(k(x)U¯ , whm )Γ2 dt − 2μ



 F, whm = − U¯ (0), whm (0) −

− +

T 0

T 0

0

˜¯ w˜ m ) dt − 2μ(SU, h Γ3

f¯, whm  dt +

T 0

T



0

i=2,4,7

Similarly, we have

0

0

2(α(t, x)U¯ , whm )Γ5 dt −

φ¯ i , whm · nΓi dt +

T 0

μ(k(x)U¯ , whm )Γ7 dt

T



0

i=3,5,6

φ¯ i , whm Γi dt + (v0 , whm (0)).

F, w m  ≤ √c . h h

Let us estimate

−

wm · Q

 1 m w¯ (x, t + h) − w¯ m (x, t) d xdt. h

(4.36)

122

4 The Non-steady Navier-Stokes System

Using −ab = − 21 [(a + b)2 − a 2 − b2 ], we have 1 h

−



T 0



w¯ m (t), w¯ m (x, t + h) − w¯ m (x, t) Ω dt

T

T    m  m 1 w (x, t)2 dt − 1 w¯ (x, t + h)2 dt = 2h 0 2h 0

T   1 w¯ m (x, t + h) − w¯ m (x, t)2 dt + 2h 0

T

T 2   m  m 1 1   w¯ (x, t)2 dt w (x, t) dt − = 2h 0 2h h

T   m 1 w¯ (x, t + h) − w¯ m (x, t)2 dt + 2h 0

T

T 2   m  m 1 1   w (x, t)2 dt w (x, t) dt − = 2h 0 2h h

T   m 1 w¯ (x, t + h) − w¯ m (x, t)2 dt + 2h 0

T   m 1 w¯ (x, t + h) − w¯ m (x, t)2 dt. ≥ 2h 0

(4.37)

Formulas (4.29) and (4.32)–(4.37) imply

T

√   m w¯ (x, t + h) − w¯ m (x, t)2 dt ≤ c[ h + ω(h) 23 ] for T > h > 0.

0



Therefore, by Theorem 1.38 we have the conclusion.

Proof of Theorem 4.1. Let {wm (Q)} be the sequence of solutions to (4.7) guaranteed by Lemma 4.1. By (4.27) {w m (Q)} is bounded in V(Q). Then by Lemma 4.3, we can choose its subsequence {w k (Q)} such that w k (Q) → w ∈ V(Q) in L2 (Q). First, let us prove that for u ∈ (Q)

T

e 0

−k1 t



rot w k × w k , u dt →



T 0

e−k1 t (rot w × w, u) dt as k → ∞. (4.38)

4.1 Existence of a Solution: The Case of Total Pressure

123

We can write

T

T

e−k1 t rot w k × w k , u dt − e−k1 t (rot w × w, u) dt 0 0

T

T

−k1 t k k rot w × (w − w), u dt + e e−k1 t rot (w k − w) × w, u dt. = 0

0

(4.39) Since {rot wk } is bounded in L2 (Q), w k → w in L2 (Q) and u ∈ L∞ (Q), the first integral on the right-hand side of (4.39) converges to zero as k → ∞. Meanwhile, since e−k1 t wu ∈ L2 (Q) and wk w in V(Q), the second integral on the right-hand side of (4.39) converges to zero. Thus, (4.38) holds. It is easy to verify the convergence of other terms in (4.7) when k → ∞. Thus, passing to limit as k → ∞ in (4.7) with k instead of m, by Lemma 4.2 we have (4.5). The estimate (4.6) can be obtained in the same way as in the proof of (4.2) in [1]. 

4.1.4 The Stokes Problem Let us consider the Stokes problem ⎧ ⎨ ∂v − μΔv + ∇ p = f, div v = 0, ∂t ⎩ v(0) = v0

(4.40)

with boundary condition (1) v|Γ1 = h 1 , (2) vτ |Γ2 = 0, − p|Γ2 = φ2 , (3) vn |Γ3 = 0, rot v × n|Γ3 = φ3 /μ, (4) vτ |Γ4 = h 4 , (− p + 2μεnn (v))|Γ4 = φ4 , (5) vn |Γ5 = h 5 , 2(μεnτ (v) + αvτ )|Γ5 = φ5 ,

(4.41)

(6) (− pn + 2μεn (v))|Γ6 = φ6 , ∂v · n)|Γ7 = φ7 . (7) vτ |Γ7 = 0, (− p + μ ∂n Applying Theorems 3.1 and 3.2 on Γi j (i = 2, 3, 7) (and so using (3.41) and (3.42)), we get a variational formulation for problem (4.40), (4.41): Problem 4.2 Find v such that

124

4 The Non-steady Navier-Stokes System

v − U ∈ V(Q),

T

T

T ∂u  dt + 2μ v, (E(v), E(u)) dt + 2μ (k(x)v, u)Γ2 dt − ∂t 0 0 0

T

T

T (S v, ˜ u) ˜ Γ3 dt + 2 (α(t, x)vτ , u)Γ5 dt + μ(k(x)v, u)Γ7 + 2μ 0 0 0

T

T 

T  = (v0 , u(0)) +

f, u dt +

φi , u n Γi dt +

φi , uΓi dt 0

0

0

i=2,4,7

i=3,5,6

∀u ∈ (Q) with u(x, T ) = 0. (4.42) Theorem 4.2 Under Assumption 4.1 there exists a unique solution to Problem 4.2 for the non-stationary Stokes system with mixed boundary condition (4.41) and the solution belongs to C([0, T ]; L2 (Ω)). Proof Put v = z + U , w = ek1 t z, where k1 is taken as in Lemma 4.1 under consideration of the fact that there is not any nonlinear term. Then, we get the new problem equivalent to Problem 4.2: Find w ∈ L 2 (0, T ; V) such that −

T

T ∂ uˆ  dt + 2μ (E(w), E(u)) ˆ dt + 2μ(k(x)w, u) ˆ Γ2 dt ∂t 0 0 0

T

T ˜ˆ Γ dt + 2μ(S w, ˜ u) 2(α(t, x)w, u) ˆ Γ5 dt + 3

T

w,

0

+

0

0

μ(k(x)w, u) ˆ Γ7 dt − k1

T 0

(w, u) ˆ dt

T ∂ U¯ uˆ d xdt − 2μ (E(U¯ ), E(u)) ˆ dt Q Q ∂t 0

T

T

T ˜¯ u) ˜ˆ Γ dt − 2μ(k(x)U¯ , u) ˆ Γ2 dt − 2μ(SU, 2(α(t, x)U¯ τ , u) ˆ Γ5 dt − 3

=

T

k1 U¯ uˆ d xdt −



0

− +

T 0

μ(k(x)U¯ , u)Γ7 +

T



0

i=3,5,6

T 0

0

f¯, u ˆ dt +

T



0

i=2,4,7



φ¯ i , u ˆ Γi dt + v0 − U (x, 0), u(x, 0)

0

φ¯ i , uˆ n Γi dt ∀uˆ ∈ (Q), u(x, ˆ T ) = 0,

(4.43) where γ¯ = ek1 t γ for γ . Existence of a solution to (4.43) is proved in much the same way as Theorem 4.1 without applying the condition U ∈ L∞ (Q) and Lemma 4.3, which are needed for the nonlinear term. To complete the proof, by Assumption 4.1 it is enough to prove that the solutions to (4.43) belong to C([0, T ]; H ) and are unique. When w ∈ L 2 (0, T ; V) is a solution to (4.43), estimating

4.1 Existence of a Solution: The Case of Total Pressure



T

I ≡



125

˜ˆ Γ − 2μ(E(w), E(u)) ˆ − 2μ(k(x)w, u)Γ2 − 2μ(S w, ˜ u) 3

0

− 2(α(t, x)w, u) ˆ Γ5 − μ(k(x)w, u) ˆ Γ7

− k1 (w, u) ˆ + k1 (U¯ , u) ˆ −

 ¯  ∂U , uˆ ∂t

˜ˆ Γ − 2(α(t, x)U¯ τ , u) ¯˜ u) − 2μ(E(U¯ ), E(u)) ˆ − 2μ(k(x)U¯ , u) ˆ Γ2 − 2μ(SU, ˆ Γ5 3    − μ(k(x)U, u)Γ7 + f¯, u ˆ +

φ¯ i , uˆ n Γi +

φ¯ i , u ˆ Γi dt i=2,4,7

i=3,5,6

∀uˆ ∈ L 2 (0, T ; V), we get |I | ≤ K u ˆ L 2 (0,T ;V) ∀uˆ ∈ L 2 (0, T ; V). This means that I is a continuous linear functional on L 2 (0, T ; V). Thus, there exists a F ∈ L 2 (0, T ; V∗ ) such that

T

I =

F, u ˆ dt.

(4.44)

0

Taking any φ ∈ D(0, T ) and u ∈ V ∩ C 2 (Ω) and putting uˆ = φu, by definition of w  ∈ D ∗ (0, T ; V ∗ ) (see Definition 1.13), (4.43), (4.44) and Theorem 1.31 we have 

w (φ), u = −

T









T

wφ dt, u = −

0





w, φ u dt =

0

T

F, φu dt

0

=



T

 Fφ dt, u ,

0

which means w  = F ∈ L 2 (0, T ; V∗ ). Therefore, w ∈ C([0, T ]; H ) and

0

T

w, w   dt =

 1

w(T ) 2 − w(0) 2 . 2

(4.45)

Let w1 , w2 be solutions to (4.43) corresponding to the same given data and w = w1 − w2 . Then, by (4.43) we have

T 0

T

T

w  , u ˆ dt + 2μ (E(w), E(u)) ˆ dt + 2μ(k(x)w, u)Γ2 dt dt 0 0

T

T ˜ˆ Γ + (4.46) 2μ(S w, ˜ u) 2(α(t, x)w, u) ˆ Γ5 dt + 3 0 0

T

T μ(k(x)w, u) ˆ Γ7 dt − k1 (w, u) ˆ dt = 0 for uˆ = φ · u. + 0

0

126

4 The Non-steady Navier-Stokes System

Since the set {uˆ = φu : φ ∈ D(0, T ), u ∈ V ∩ C 2 (Ω)} is dense in L 2 (0, T ; V) (see Remark 1.8), (4.46) is valid for uˆ = w. Thus, from (4.45) and (4.46) it follows that

T

w 2V dt ≤ 0, β > 0

w(T ) 2 + β 0

(cf. (4.14), (4.16)), which shows that w ≡ 0, that is, the solution to (4.43) is unique. 

4.2 Existence and Uniqueness of Solutions: The Case of Static Pressure In this section we study the Navier-Stokes problem ⎧ ∂v ⎪ ⎪ ⎨ ∂t − μΔv + (v · ∇)v + ∇ p = f, div v = 0, ⎪ ⎪ ⎩ v(0) = v0

(4.47)

with boundary conditions including the static pressure. We are concerned with the Problems I and II, which are distinguished according to boundary conditions. Problem I is the one with the boundary conditions (1) v|Γ1 = h 1 , (2) vτ |Γ2 = 0, − p|Γ2 = φ2 , (3) vn |Γ3 = 0, rot v × n|Γ3 = φ3 /μ, (4) vτ |Γ4 = h 4 , (− p + 2μεnn (v))|Γ4 = φ4 , (5) vn |Γ5 = h 5 , 2(μεnτ (v) + αvτ )|Γ5 = φ5 , (6) (− pn + 2μεn (v))|Γ6 = φ6 ,  ∂v  · n = φ7 , (7) vτ |Γ7 = 0, − p + μ Γ7 ∂n and Problem II is the one with the conditions

(4.48)

4.2 Existence and Uniqueness of Solutions: The Case of Static Pressure

127

(1) v|Γ1 = h 1 , (2) vτ |Γ2 = 0, − p|Γ2 = φ2 , (3) vn |Γ3 = 0, rot v × n|Γ3 = φ3 /μ, (4) vτ |Γ4 = h 4 , (− p + μεnn (v))|Γ4 = φ4 ,

(4.49)

(5) vn |Γ5 = h 5 , 2(μεnτ (v) + αvτ )|Γ5 = φ5 ,  ∂v  (6) − pn + μ = φ7 ∂n Γ7 together Γ6 = ∅. Condition (6) of (4.49) is “do nothing” condition, but (7) of (4.48) is rather different from “do nothing” condition (see Remark 3.7). For Problems I and II stated above (for the corresponding perturbation problems in Sect. 4.2.3), respectively, conditions Γi j ∈ C 2,1 , i = 2, 3, 7, and Γi j ∈ C 2,1 , i = 2, · · · , 5, are used since Theorems 3.1, 3.2 are applied for variational formulations for Problem I and Problem II, respectively, on Γi j ∈ C 2,1 (i = 2, 3, 7) and Γi j ∈ C 2,1 (i = 2, · · · , 5). For the proof of the existence of a unique solution to Problems I and II we use the Local inverse mapping theorem (see Theorem 1.7).

4.2.1 Existence and Uniqueness of Solutions to Problem I We use the following notation:   V = u ∈ H1 (Ω) : div u = 0, u|Γ1 = 0, u τ |Γ2 ∪Γ4 ∪Γ7 = 0, u n |Γ3 ∪Γ5 = 0 and

  VΓ 237 (Ω) = u ∈ H1 (Ω) : div u = 0, u τ |Γ2 ∪Γ7 = 0, u n |Γ3 = 0 .

Denote by H the completion of V in the space L2 (Ω). Throughout this section V = {u ∈ H1 (Ω) : div u = 0}. Let X = {w ∈ L 2 (0, T ; V); w  ∈ L 2 (0, T ; V), w  ∈ L 2 (0, T ; V∗ )},

w X = w L 2 (0,T ;V) + w  L 2 (0,T ;V) + w  L 2 (0,T ;V∗ ) , Y = {w ∈ L 2 (0, T ; V∗ ); w  ∈ L 2 (0, T ; V∗ )},

w Y = w L 2 (0,T ;V∗ ) + w  L 2 (0,T ;V∗ ) , ∗

W = {w ∈ L 2 (0, T ; V); w  ∈ L 2 (0, T ; V), w  ∈ L 2 (0, T ; V )},

w W = w L 2 (0,T ;V) + w  L 2 (0,T ;V) + w  L 2 (0,T ;V∗ ) . Here and in what follows w  means the derivative with respect to t of w(t).

128

4 The Non-steady Navier-Stokes System

For Problem I, we will use the following assumptions: Assumption 4.2 f, f  ∈ L 2 (0, T ; V∗ ), φi , φi ∈ L 2 (0, T ; H − 2 (Γi )), i = 2, 4, 7, 1 φi , φi ∈ L 2 (0, T ; H− 2 (Γi )), i = 3, 5, 6, αi j ∈ L ∞ (Γ5 ), where αi j are components of the matrix α, and Γ1 = ∅. 1

Assumption 4.3 There exists a function U ∈ W such that div U = 0, U |Γ1 = h 1 , Uτ |Γ2 ∪Γ7 = 0, Un |Γ3 = 0, Uτ |Γ4 = h 4 , Un |Γ5 = h 5 . Also, U (0, x) − v0 ∈ V. 1

Remark 4.1 For the particular situation such that h 4 , h 5 = 0, h 1 (t, x) ∈ H002 (Γ1 ),  1 2 2 Γ1 h 1 (t, x) · n d x = 0 for every fixed t, and h 1 (t, x) ∈ C (0, T ; H (Γ1 )), applying the existence and estimate of a solution to the steady Stokes problem with nonhomogeneous Dirichlet boundary condition (see Theorem IV.1.1 [2]), we can prove the existence of a function U above. Taking (3.43) into account, we get the following variational formulation for Problem I (with boundary conditions (4.48)): Problem 4.3 Find v such that v − U ∈ L 2 (0, T ; V), v(0) = v0 ,

v  , u + 2μ(E(v), E(u)) + (v · ∇)v, u + 2μ(k(x)v, u)Γ2 + 2μ(S v, ˜ u) ˜ Γ3 + 2(α(x)vτ , u)Γ5 + μ(k(x)v, u)Γ7   = f, u +

φi , u n Γi +

φi , uΓi for all u ∈ V. i=2,4,7

(4.50)

i=3,5,6

Using Assumption 4.3 and taking v = z + U , we deduce the following problem equivalent to Problem 4.3: Find z such that z ∈ L 2 (0, T ; V), z(0) = z 0 ≡ v0 − U (0) ∈ V,

z  , u + 2μ(E(z), E(u)) + (z · ∇)z, u + (U · ∇)z, u + (z · ∇)U, u ˜ u) ˜ Γ3 + 2(α(x)z, u)Γ5 + μ(k(x)z, u)Γ7 + 2μ(k(x)z, u)Γ2 + 2μ(S z, = −(U  , u) − 2μ(E(U ), E(u)) − (U · ∇)U, u − 2μ(k(x)U, u)Γ2 − 2μ(SU˜ , u) ˜ Γ3 − 2(α(x)Uτ , u)Γ5 − μ(k(x)U, u)Γ7 + f, u  

φi , u n Γi +

φi , uΓi for all u ∈ V. + i=2,4,7

i=3,5,6

(4.51)

4.2 Existence and Uniqueness of Solutions: The Case of Static Pressure

129

Now, define an operator A0 : V → V∗ by

A0 y, u =2μ(E(y), E(u)) + 2μ(k(x)y, u)Γ2 + 2μ(S y˜ , u) ˜ Γ3 + 2(α(x)y, u)Γ5 + μ(k(x)y, u)Γ7 for all y, u ∈ V.

(4.52)

Lemma 4.4 ∃δ > 0, ∃k0 ≥ 0; A0 u, u ≥ δ u 2V − k0 u 2 for all u ∈ V. Proof By Korn’s inequality we have 2μ(E(u), E(u)) ≥ β u 2V ∃β > 0, for all u ∈ V.

(4.53)

By Remark 3.4 and Assumption 4.2, there exists a constant M such that

S(x) ∞ , k(x) ∞ , α(x) ∞ ≤ M, and so there exists a constant c0 (depending on β) such that 2μ(k(x)z, z)Γ + 2μ(S z˜ , z˜ )Γ + μ(k(x)z, z)Γ + 2(α(x)y, u)Γ 2 3 7 5 β ≤ z 2H1 (Ω) + c0 z 2 dt for all z ∈ V 2 (see Theorem 1.27). Set δ = the asserted conclusion.

β 2

(4.54)

and k0 = c0 . Then, by (4.53) and (4.54) we come to 

Remark 4.2 In the process of proving Lemma 4.4, we see that if Γi = ∅, i = 2, 3, 7, or these are unions of pieces of planes (segments in case of 2-D) and Γ5 = ∅ or α(x) = 0, then we can take k0 = 0. When k0 > 0, if k0 is not small enough, then the operator defined by (4.52) is not positive, and so let us transform the unknown function to get a positive operator A in (4.56) below. Now, let k0 be the constant in Lemma 4.4 and put z = e−k0 t z. Then, since e−k0 t z  = z  + k0 z, we get the following problem equivalent to problem (4.51): Find z such that

130

4 The Non-steady Navier-Stokes System

z ∈ L 2 (0, T ; V), z(0) = v0 − U (0) ∈ V,

z  (t), u + 2μ(E(z(t)), E(u)) + ek0 t (z(t) · ∇)z(t), u + (U (t) · ∇)z(t), u + (z(t) · ∇)U (t), u + k0 (z(t), u) + 2μ(k(x)z(t), u)Γ2 + 2μ(S z˜ (t), u) ˜ Γ3 + 2(α(x)z(t), u)Γ5 + μ(k(x)z(t), u)Γ7  = e−k0 t − (U  (t), u) − 2μ(E(U (t)), E(u)) − (U (t) · ∇)U (t), u − 2μ(k(x)U (t), u)Γ2 − 2μ(SU˜ (t), u) ˜ Γ3 − 2(α(x)U (t)τ , u)Γ5  − μ(k(x)U (t), u)Γ7 + f (t), u +

φi (t), u n Γi +



φi (t), uΓi



i=2,4,7

for all u ∈ V.

i=3,5,6

(4.55)

Define operators A, AU (t) : V → V∗ , respectively, by

Av, u = A0 v, u + (k0 v, u) for all v, u ∈ V,

(4.56)

and

AU (t)v, u = (U (t, x) · ∇)v, u + (v · ∇)U (t, x), u for all v, u ∈ V, (4.57) where A0 is the operator by (4.52) and k0 is the one in Lemma 4.4. Since U ∈ W , we have U ∈ C([0, T ]; H1 (Ω)) and so such a definition makes sense. Then, the operator A is positive definite, and this fact is used in (4.67). Define an operator B(t) : V → V∗ and an element F(t) ∈ V ∗ , respectively, by

B(t)v, u = ek0 t (v · ∇)v, u for all v, u ∈ V,

(4.58)

and 

F(t), u =e−k0 t − (U  (t), u) − 2μ(E(U (t)), E(u)) − (U (t) · ∇)U (t), u ˜ Γ3 − 2(α(x)U (t)τ , u)Γ5 − 2μ(k(x)U (t), u)Γ2 − 2μ(SU˜ (t), u)    − μ(k(x)U (t), u)Γ7 + f (t), u +

φi (t), u n Γi +

φi (t), uΓi i=2,4,7

i=3,5,6

for all u ∈ V.

(4.59) Then, (4.55) is written as

4.2 Existence and Uniqueness of Solutions: The Case of Static Pressure

131

z ∈ L 2 (0, T ; V), z(0) = v0 − U (0) ∈ V, z  (t) + (A + AU (t) + B(t)) z(t) = F(t).

(4.60)

Now, define operators L , AU , L U , B : X → Y , C : X × X → Y and an element F ∈ Y by

(Lz)(t), u = z  (t), u + Az(t), u for all z ∈ X , for all u ∈ V,

( AU z)(t), u = AU (t)z(t), u for all z ∈ X , for all u ∈ V,

(L U z)(t), u = z  (t), u + (A + AU (t))z(t), u for all z ∈ X , for all u ∈ V,

( Bz)(t), u = B(t)z(t), u for all z ∈ X , for all u ∈ V,

C(w, z)(t), u = ek0 t (w(t) · ∇)z(t), u + ek0 t (z(t) · ∇)w(t), u for all w, z ∈ X , for all u ∈ V, (F)(t) = F(t). (4.61) For AU and F, we have Lemma 4.5 C is a bilinear continuous operator such that X × X → Y . Under Assumptions 4.2 and 4.3, AU is a linear continuous operator such that X → Y and F ∈ Y . Proof Obviously, C is bilinear. When w ∈ X ,   w ∈ L ∞ (0, T ; V), w L ∞ (0,T ;V) ≤ c w L 2 (0,T ;V) + w  L 2 (0,T ;V) and by virtue of Hölder’s inequality and Sobolev’s imbedding theorem (Theorem 1.20) kt e 0 (w · ∇)z, u + ek0 t (z · ∇)w, u ≤ c( w L3 ∇z L2 u L6 + z L3 ∇w L2 u L6 ) ≤ c w V z V u V for all w, z, u ∈ V. Thus,

C(w, z) L 2 (0,T ;V∗ ) ≤ c w L ∞ (0,T ;V) z L 2 (0,T ;V) ≤ c w X · z X .

(4.62)

Also, since | C(w, z) (t), u| =ek0 t k0 (w · ∇)z, u + k0 (z · ∇)w, u + (w  · ∇)z, u + (w · ∇)z  , u + (z  · ∇)w, u + (z · ∇)w  , u , taking (4.62) into account, we have

132

4 The Non-steady Navier-Stokes System

C(w, z) L 2 (0,T ;V∗ ) ≤ c C(w, z) L 2 (0,T ;V∗ )

  + c w  L 2 (0,T ;V) + w L ∞ (0,T ;V) )( z  L 2 (0,T ;V) + z L ∞ (0,T ;V) ≤ c w X · z X . (4.63) (4.62) and (4.63) imply

C(w, z) Y ≤ c w X · z X .

(4.64)

By the same argument as above, we have

AU z Y ≤ c U W · z X .

(4.65)

By Assumption 4.2, Remark 3.4 and the trace theorem, we can see that F ∈ Y .  Lemma 4.6 The operator L defined by Lz = (z  (0), Lz) for z ∈ X is a linear continuous and one-to-one operator from X onto H × Y . Proof The linearity of L is obvious. The fact z ∈ X implies that z  ∈ C([0, T ]; H ),

z  C([0,T ];H ) ≤ c z X , and so we see that a map z ∈ X → z  (0) ∈ H is continuous. Clearly, z  Y ≤ c z X . Also, by Assumption 4.2, Remark 3.4 and the trace theorem, (4.66) | Av, u| ≤ c v V · u V for all v, u ∈ V. Equation (4.66) implies that the mapping z ∈ X → Az ∈ Y is continuous. Therefore, L is continuous. Next, we show that L is a one-to-one and surjective operator from X onto H × Y . First, let us prove that this operator is injective. For this, it is enough to prove that the inverse image of (0 H , 0Y ) ∈ H × Y by the operator L is 0X . By Lemma 4.4 and (4.56), we can find δ > 0 such that

Av, v ≥ δ v 2V for all v ∈ V.

(4.67)

By (4.66) and (4.67), for any q ∈ V∗ there exists a unique solution y ∈ V to the following problem: Ay = q. (4.68) Let z ∈ X be the inverse image of (0 H , 0Y ) ∈ H × Y by L. Then since z  (0) = 0 H , putting t = 0 in the first equation of (4.61) we get

Az(0), u = 0 for all u ∈ V, where z(0) = z(0, x). This means that z(0) is a unique solution to (4.68) for q = 0V∗ , i.e. z(0) = 0V . Putting w = z  , we get w(0) = z  (0) = 0 H . Taking Lz = 0 into account and differentiating the first equation of (4.61), we have

4.2 Existence and Uniqueness of Solutions: The Case of Static Pressure

w  (t), u + Aw(t), u = 0 for all u ∈ V.

133

(4.69)

The operator A defined by (Aw)(t) = Aw(t) ∀w ∈ L 2 (0, T ; V) satisfies all conditions of Theorem 1.46. Thus, for problem (4.69) with an initial condition w(0) ∈ H there exists a unique solution w such that w ∈ L 2 (0, T ; V), w  ∈ L 2 (0, T ; V∗ ). Since w(0) = 0 H , we have w = 0, which means z = 0X since z(0) = 0V . Second, let us prove that L is surjective. Let (w0 , g) ∈ H × Y . Since g ∈ Y , we have g(0) ∈ V∗ . Then, by (4.66) and (4.67), there exists a unique solution z 0 ∈ V to problem (4.70) Az 0 = g(0) − w0 . Let us consider problem

!

w  + Aw = g  , w(0) = w0 .

(4.71)

By Theorem 1.46, there exists a unique solution w such that w ∈ L 2 (0, T ; V), w  ∈ L 2 (0, T ; V∗ ) to problem (4.71). Now, put

z = z0 +

t

w(s) ds,

(4.72)

0

where z 0 is the solution to (4.70). Then, z  = w and z ∈ X . Integrating both sides of the first one in (4.71) from 0 to t and using (4.72), we have

w(t), u + Az(t), u − [ w0 , u + Az 0 , u] = g(t), u − g(0), u for all u ∈ V.

(4.73)

Taking (4.70), (4.72) into account, from (4.73) we get

z  (t), u + Az(t), u = g(t), u for all u ∈ V.

(4.74)

This means that z ∈ X defined by (4.72) is the inverse image of (w0 , g) ∈ H × Y  by the operator L, i.e. L is surjective. Lemma 4.7 Under Assumption 4.3, let U (0, x) V be small enough. The operator L U defined by L U z = (z  (0), L U z) for z ∈ X is a linear continuous one-to-one operator from X onto H × Y . Proof If z ∈ X , then z ∈ C([0, T ]; V) and  

z C([0,T ];V) ≤ c z L 2 (0,T ;V) + z  L 2 (0,T ;V) .

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4 The Non-steady Navier-Stokes System

By virtue of this fact and Lemma 4.5, the operator AU ∈ (X → H × Y ) defined by AU z = (0 H , AU z) is continuous. Thus, the operator L U defined on X is linear continuous. As in Lemma 3.5 of [3] it is proved that the operator AU ∈ (X → Y ) is compact. Thus, AU ∈ (X → H × Y ) is also compact. Since L U = L + AU , by virtue of Theorem 1.8 and Lemma 4.6 we see that in order to prove that the operator L U is one-to-one from X onto H × Y , it is enough to prove that L U is one-to-one from X into H × Y . To prove the last fact it is enough to show that the inverse image of (0 H , 0Y ) by L U is 0X . By Hölder’s inequality and Sobolev’s imbedding theorem     (U (t, x) · ∇)v, v + (v · ∇)U (t, x), v ≤ K 0 v V U (t, x) H1 v V . Thus, if U (0, x) V is so small that U (0, x) H1 ≤ (4.75) imply

δ , 2K 0

(4.75)

then (4.66), (4.67) and



 A + AU (0) v, u ≤ c v V · u V for all v, u ∈ V, 

 δ A + AU (0) v, v ≥ v 2V for all v ∈ V. 2

(4.76)

By (4.76), for any q ∈ V∗ there exists a unique solution y ∈ V to (A + AU (0))y = q.

(4.77)

Let z ∈ X be the inverse image of (0 H , 0Y ) by L. Then z  (0) = 0 H , and putting t = 0 from the third one in (4.61) we get 

 A + AU (0) z(0), u = 0 for all u ∈ V,

where z(0) = z(0, x). This means that z(0) is the unique solution to (4.77) with q = 0V∗ , i.e. z(0) = 0V . Therefore, z ∈ X satisfies !



z  (t) + A + AU (t) z(t) = 0, z(0) = 0V .

(4.78)

Now, making duality pairing with z(t) on both sides of z  (t) + Az(t) = −AU (t)z(t) and taking (4.67) into account and using Gronwall’s inequality, we can prove z = 0X as in Lemma 3.8 of [3]. The proof is thus completed.  Lemma 4.8 Under Assumption 4.3 the operator T defined by T z = z  (0), (L U

+ B)z for z ∈ X is continuously differentiable, T (0X ) = (0 H , 0Y ) and the Fréchet derivative of T at 0X is L U .

4.2 Existence and Uniqueness of Solutions: The Case of Static Pressure

135

Proof It is easy to verify that T (0X ) = (0 H , 0Y ). Since the operator L U is linear, its Fréchet derivative is the same as itself. Therefore, if B is continuously differentiable, then so is T . For any w, z ∈ X ,





B(w + z) − Bw (t) = ek0 t w(t) · ∇ z(t) + ek0 t z(t) · ∇ w(t) + ( Bz)(t).

By (4.64), we get c z 2X

Bz Y ≤ lim = 0. →0 z X

z X →0 z X

lim

z X

Then, put



C(w, z)(t) ≡ ek0 t w(t) · ∇ z(t) + ek0 t z(t) · ∇ w(t) = ( Bw z)(t). By Lemma 4.5 Bw : X → Y is continuous, and it is the Fréchet derivative of B at w and also continuous with respect to w. Thus, T is continuously differentiable. Also from the formula above we can see that the Fréchet derivative of B at 0X is  zero. Therefore, the Fréchet derivative of T at 0X is L U . Let us consider problem

A + AU (0) + B(0) u = q.

(4.79)

Lemma 4.9 Assume that U (0, x) V is small enough. If the norm of q ∈ V ∗ is small enough, then there exists a unique solution to (4.79) in some O M (0V ). Proof Since U (0, x) V is small enough, by (4.76), for any fixed z ∈ V there exists a unique solution to problem

A + AU (0) w = q − B(0)z.

(4.80)

On the other hand, | B(0)w1 − B(0)w2 , u| ≤K M w1 − w2 V · u V for all wi ∈ O M (0V ), for all u ∈ V.

(4.81) Owing to (4.76) the solution w to (4.80) is estimated as follows:

w V ≤

2

2

q V∗ + B(0)z V∗ ≤ q V∗ + K M 2 . δ δ

Thus, if q V∗ and M are small enough, then the operator (z → w) maps O M (0V ) into itself and by (4.81) this operator is strictly contract. Therefore, in O M (0V ) there exists a unique solution to (4.80). Thus, we come to the asserted conclusion. 

136

4 The Non-steady Navier-Stokes System

One of main results of this section is the following: Theorem 4.3 Suppose that Assumptions 4.2 and 4.3 hold. Assume that U W and the norms of f, f  , φi , φi in the spaces where they belong to are small enough. If w0 ≡ F(0) − (A + AU (0) + B(0))z 0 ∈ H (4.82) (compatibility condition at the initial time), where z 0 = v0 − U (0, ·), and w0 H is small enough, then there exists a unique solution to Problem 4.3 in the space W . Proof First, let us prove the existence of a solution. If U W and the norms of f, f  , φi , φi in the spaces that they belong to are small enough, then F Y is also small enough. By virtue of Lemmas 4.7, 4.8 and the Local inverse mapping theorem (Theorem 1.7), for any R1 > 0 small enough if F Y , R are small enough and w1 ∈ O R (0 H ), then there exists a unique z ∈ O R1 (0X ) such that

z  (t) + A + AU (t) + B(t) z(t) = F(t), (4.83) z  (0) = w1 ∈ O R (0 H ). Putting t = 0 in (4.83), we get

F(0) − A + AU (0) + B(0) z(0) = w1 ∈ O R (0 H ). On the other hand, if U W is small enough, then so is U (0, x) V . Thus, when

F(0) − w1 V ∗ is small enough, by Lemma 4.9 there exists a unique solution z 0 ∈ O R2 (0V ) for some R2 > 0 to

A + AU (0) + B(0) z 0 = F(0) − w1 .

(4.84)

Since z(0) V ≤ c z X , we can choose R1 such that z(0) ∈ O R2 (0V ), and we have

z(0) = z 0 . Therefore, if F Y is small enough, F(0) − A + AU (0) + B(0) z 0 belongs to H and its norm is small enough, then z ∈ X , the solution to (4.83), is a solution to problem !



z  (t) + A + AU (t) + B(t) z(t) = F(t), z(0) = z 0 .

(4.85)

By definitions of A, AU (t), B(t), F, the solution z of (4.85) is also a solution to (4.55) which is equivalent to (4.51). Thus, ek0 t z + U ∈ W is a solution to (4.50). Second, let us prove uniqueness. We argue by contradiction. Let v1 , v2 be two solutions to (4.49) corresponding to the same data. Putting w = v1 − v2 , we have

4.2 Existence and Uniqueness of Solutions: The Case of Static Pressure

137

w ∈ L 2 (0, T ; V), w(0) = 0,

w  , u + 2μ(E(w), E(u)) + (v1 · ∇)w, u + (w · ∇)v2 , u + 2μ(k(x)w, u)Γ2 ˜ u) + 2μ(S w, ˜ Γ3 + 2(α(x)w, u)Γ5 + μ(k(x)w, u)Γ7 = 0 for all u ∈ V. (4.86) Putting w = e−k0 t w, where k0 is the constant in Lemma 4.4, we get e−k0 t w = w  + k0 w. Then, we have w ∈ L 2 (0, T ; V), w(0) = 0,

w  , u + 2μ(E(w), E(u)) + (v1 · ∇)w, u + (w · ∇)v2 , u + k0 (w, u) + 2μ(k(x)w, u)Γ2 + 2μ(S w, ˜ u) ˜ Γ3

(4.87)

+ 2(α(x)w, u)Γ5 + μ(k(x)w, u)Γ7 = 0 ∀u ∈ V, which is equivalent to (4.86). By Lemma 4.4, 2μ(E(w), E(w)) + k0 (w, w) + 2μ(k(x)w, w)Γ2 + 2μ(S w, ˜ w) ˜ Γ3 + 2(α(x)w, w)Γ5 + μ(k(x)w, w)Γ7 ≥ δ w 2V . Taking this into account, we can prove w = 0X as in Lemma 3.8 of [3]. Thus, uniqueness of solutions is proved, and so the proof is completed.  Remark

4.3 Let us consider more precisely the condition that F(0) − A + AU (0) + B(0) z 0 belongs to H and its norm is small enough. By (4.56)–(4.61), we have 



 F(0) − A + AU (0) + B(0) z 0 , u =  − (U  (0, x), u) − 2μ(E(U (0, x)), E(u)) − (U (0, x) · ∇)U (0, x), u − 2μ(k(x)U (0, x), u)Γ2 − 2μ(SU˜ (0), u) ˜ Γ3 − 2(α(x)U (0, x)τ , u)Γ5    − μ(k(x)U (0, x), u)Γ7 + f (0), u +

φi (t), u n Γi +

φi (t), uΓi i=2,4,7

 − 2μ(E(z 0 ), E(u)) + 2μ(k(x)z 0 , u)Γ2 + 2μ(S z˜0 , u) ˜ Γ3

i=3,5,6

 + 2(α(x)z 0 , u)Γ5 + μ(k(x)z 0 , u)Γ7   − (U (0, x) · ∇)z 0 , u + (z 0 · ∇)U (0, x), u − (z 0 , ∇)z 0 , u for all u ∈ V. (4.88) Taking into account the fact that U (0, x) + z 0 = v0 , U  (0, x) ∈ L2 (Ω) and its norm is small enough, from (4.88) we can see that the condition mentioned above is equivalent to the condition w0 ∈ O R (0 H ) for R > 0 small enough, where w 0 is defined by

138

4 The Non-steady Navier-Stokes System

w 0 , u ≡ f (0), u +

 i=2,4,7

φi (0, x), u n Γi +



φi (0, x), uΓi

i=3,5,6

 − 2μ(E(v0 ), E(u)) + 2μ(k(x)v0 , u)Γ2 + 2μ(S v˜0 , u) ˜ Γ3 + 2(α(x)v0 , u)Γ5  + μ(k(x)v0 , u)Γ7 + (v0 · ∇)v0 , u + k0 (v0 , u) for all u ∈ V. (4.89) Remark 4.4 If Γi = ∅, i = 2, · · · , 5, 7, then the problem is reduced to the one in [4] where a local-in-time solution was studied. In this case k0 = 0 (see Remark 4.1), and condition (4.82) is the same as (25) in [4]. And our condition for U is also the same as the one in [4].

4.2.2 Existence and Uniqueness of Solutions to Problem II Let   V1 = u ∈ H1 (Ω) : div u = 0, u|Γ1 = 0, u τ |(Γ2 ∪Γ4 ) = 0, u n |(Γ3 ∪Γ5 ) = 0 ,   VΓ 2−5 = u ∈ H1 (Ω) : div u = 0, u τ |(Γ2 ∪Γ4 ) = 0, u n |(Γ3 ∪Γ5 ) = 0 . Denote by H1 the completion of V1 in the space L2 (Ω). By Theorems 3.1 and 3.2, for v ∈ H2 (Ω) ∩ VΓ 2−5 (Ω), p ∈ H 1 (Ω) and u ∈ V1 we have −μ(Δv, u) + (∇ p, u) = μ(∇v, ∇u) + μ(k(x)v, u)Γ2 + μ(S v, ˜ u) ˜ Γ3 + μ(k(x)v, u)Γ7

+ ( p, u · n)Γ2 − (rot v × u, u)Γ3 − (− p + μεnn (v)), u · n Γ4  ∂v  , un . − 2μ(εnτ (v), u)Γ5 − − pn + Γ7 ∂n (4.90) Let X1 = {w ∈ L 2 (0, T ; V1 ); w  ∈ L 2 (0, T ; V1 ), w  ∈ L 2 (0, T ; V∗1 )},

w X 1 = w L 2 (0,T ;V1 ) + w  L 2 (0,T ;V1 ) + w  L 2 (0,T ;V∗1 ) , Y1 = {w ∈ L 2 (0, T ; V∗1 ); w  ∈ L 2 (0, T ; V∗1 )},

w Y 1 = w L 2 (0,T ;V∗1 ) + w  L 2 (0,T ;V∗1 ) . Unlike Problem I, for Problem II we do not require the condition vτ |Γ7 = 0, and so instead of Assumptions 4.2 and 4.3, we use the following assumptions: Assumption 4.4 Assumption 4.2 holds with φ7 , φ7 ∈ L 2 (0, T ; H− 2 (Γ7 )) instead 1 of φ7 , φ7 ∈ L 2 (0, T ; H − 2 (Γ7 )). 1

Assumption 4.5 There exists a function U ∈ W such that

4.2 Existence and Uniqueness of Solutions: The Case of Static Pressure

139

div U = 0, U |Γ1 = h 1 , Uτ |Γ2 = 0, Un |Γ3 = 0, Uτ |Γ4 = h 4 , Un |Γ5 = h 5 , where W is the same as in the previous section. Also, U (0, x) − v0 ∈ V1 . Applying (4.90), we get the following variational formulation for Problem II (with boundary conditions (4.49)): Problem 4.4 Find v such that v − U ∈ L 2 (0, T ; V1 ), v(0) = v0 ,    v , u + μ(∇v, ∇u) + (v · ∇)v, u + μ(k(x)v, u)Γ2 + μ(S v, ˜ u) ˜ Γ3 + 2(α(x)vτ , u)Γ5 − μ(S v, ˜ u) ˜ Γ5  

φi , u n Γi +

φi , uΓi for all u ∈ V1 . = f, u + i=2,4

(4.91)

i=3,5,7

Taking into account Assumption 4.4 and putting v = z + U , we get the following problem equivalent to Problem 4.4: Find z such that z ∈ L 2 (0, T ; V1 ), z(0) ≡ v0 − U (0) ∈ V1 ,

z  , u + μ(∇z, ∇u) + (z · ∇)z, u + (U · ∇)z, u + (z · ∇)U, u ˜ u) ˜ u) ˜ Γ3 + 2(α(x)z, u)Γ5 − μ(S z, ˜ Γ5 + μ(k(x)z, u)Γ2 + μ(S z, = − U  , u − μ(∇U, ∇u) − (U · ∇)U, u − μ(k(x)U, u)Γ2 − μ(SU˜ , u) ˜ Γ3   ˜ − 2(α(x)Uτ , u)Γ5 + μ(SU , u) ˜ Γ5 + f, u +

φi , u n Γi +

φi , uΓi i=2,4

i=3,5,7

for all u ∈ V1 .

(4.92) Define an operator A01 : V1 → V∗1 by

A01 y, u = μ(∇ y, ∇u) + μ(k(x)y, u)Γ2 + μ(S y˜ , u) ˜ Γ3 + 2(α(x)y, u)Γ5 − μ(S y˜ , u) ˜ Γ5 for all y, u ∈ V1 .

(4.93) By virtue of the same argument as in the proof of Lemma 4.4 we get Lemma 4.10 ∃δ > 0, ∃k1 ≥ 0:

A01 u, u ≥ δ u 2V1 − k1 u 2 for all u ∈ V1 . Putting z = e−k1 t z, where k1 is the constant in Lemma 4.10, and using the fact that e−k1 t z  = z  + k1 z, we get the following problem equivalent to (4.92):

140

4 The Non-steady Navier-Stokes System

Find z such that z ∈ L 2 (0, T ; V1 ), z(0) = z 0 ≡ v0 − U (0) ∈ V1 ,

z  (t), u + μ(∇z, ∇u) + ek1 t (z(t) · ∇)z(t), u + (U (t) · ∇)z(t), u + (z(t) · ∇)U (t), u + k1 (z(t), u) + μ(k(x)z(t), u)Γ2 + μ(S z˜ (t), u) ˜ Γ3 + 2(α(x)z(t), u)Γ5 − μ(S z˜ (t), u) ˜ Γ5  = e−k1 t − (U  (t), u) − μ(∇U, ∇u) − (U (t) · ∇)U (t), u ˜ Γ3 − 2(α(x)U (t), u)Γ5 − μ(SU˜ (t)τ , u) ˜ Γ5 − μ(k(x)U (t), u)Γ2 − μ(SU˜ (t), u)    + f (t), u +

φi (t), u n Γi +

φi (t), uΓi for all u ∈ V1 . i=2,4

i=3,5,7

(4.94) Define operators A1 , A1U (t) by

A1 v, u = A01 v, u + (k1 v, u) for all v, u ∈ V1 ,

(4.95)

A1U (t)v, u = (U (t, x) · ∇)v, u + (v · ∇)U (t, x), u for all v, u ∈ V1 , (4.96) where A01 is the one defined in (4.93). U ∈ W implies U ∈ C([0, T ]; H1 (Ω)), and such definitions make sense. Also, define an operator B1 (t) : V1 → V∗1 by

B1 (t)v, u = ek1 t (v · ∇)v, u for all v, u ∈ V1 .

(4.97)

Define an element F1 ∈ Y1 by 

F1 (t), u = e−k1 t − U  (t), u − μ(∇U (t), ∇u) − (U (t) · ∇)U (t), u − μ(k(x)U (t), u)Γ2 − μ(SU˜ (t), u) ˜ Γ3 − 2(α(x)U (t)τ , u)Γ5 + μ(SU˜ (t), u) ˜ Γ5    + f, u +

φi , u n Γi +

φi , uΓi for all u ∈ V1 . i=2,4

i=3,5,7

(4.98) Now, in the same way as in the proof of Theorem 4.3 we can prove the following theorem which is one of the main results of this section. Theorem 4.4 Suppose that Assumptions 4.4 and 4.5 hold. Assume that U W and the norms of f, f  , φi , φi in the spaces where they belong to are small enough. If (compatibility condition at the initial instance) w1 ≡ F1 (0) − (A1 + A1U (0) + B1 (0))z 0 ∈ H1 ,

(4.99)

4.2 Existence and Uniqueness of Solutions: The Case of Static Pressure

141

where z 0 = v0 − U (0, ·), and w1 H1 is small enough, then in the space W there exists a unique solution to Problem 4.4. Remark 4.5 By the same argument as in Remark 4.3, we can see that the condition (4.99) is equivalent to the condition w 1 ∈ H1 , where w 1 ∈ V1∗ is defined by

w 1 , u = f (0), u +

 i=2,4

φi (0, x), u n Γi +



φi (0, x), uΓi

i=3,5,7

 − μ(∇v0 , ∇u) + μ(k(x)v0 , u)Γ2 + μ(S v˜0 , u) ˜ Γ3 + 2(α(x)v0 , u)Γ5  − μ(S v˜0 , u) ˜ Γ5 + (v0 · ∇)v0 , u + k1 (v0 , u) for all u ∈ V1 , (4.100) with k1 in Lemma 4.10. Remark 4.6 If U ≡ 0 and Γi = ∅, i = 2, · · · , 5, then problem (4.91) is reduced to one in [3]. In this case k1 = 0 (see Remark 4.2). If v0 ∈ Hl/2 (Ω), then (v0 · ∇)v0 ∈ L2 (Ω). Thus, the condition above for w 1 being in H1 is the same as one of conditions of Theorems 3.5–3.8 of [3], but we do not demand v0 ∈ Hr0 (Ω), r0 > 2l .

4.2.3 Existence and Uniqueness of Solutions for Perturbed Data In [3] it is proved that if a solution satisfying smoothness and a compatibility condition is given, then there exists a unique solution for small perturbed data satisfying the compatibility condition. In this section we get such results for the Problems I and II. In our results the conditions for a given solution is essentially the same as the one in [3], but the smoothness condition for the initial functions in the compatibility condition for small perturbed data is weaker than the one in [3] (see Remark 4.8). r Let V 0 = {u ∈ Hr0 (Ω) : div u = 0}, r0 > l/2, and   ∗ r W = w ∈ L 2 (0, T ; V); w  ∈ L 2 (0, T ; V), w  ∈ L 2 (0, T ; V ), w(0) ∈ V 0 ,

w W = w L 2 (0,T ;V) + w  L 2 (0,T ;V) + w  L 2 (0,T ;V∗ ) + w(0) Vr0 . Let us first consider Problem I. Let W (x, t) ∈ W be a given solution to Problem I. Let v be the solution for the data perturbed except h i , and put v = z + W . Then, we get a problem for z: Find z such that

142

4 The Non-steady Navier-Stokes System

z ∈ L 2 (0, T ; V), z(0) = z 0 ≡ v0 − W (0, x) ∈ V,

z  , u + 2μ(E(z), E(u)) + (z · ∇)z, u + (W · ∇)z, u + (z · ∇)W, u ˜ u) ˜ Γ3 + 2(α(x)z, u)Γ5 + μ(k(x)z, u)Γ7 + 2μ(k(x)z, u)Γ2 + 2μ(S z,   = f, u +

φi , u n Γi +

φi , uΓi for all u ∈ V, i=2,4,7

i=3,5,6

(4.101) where z 0 , f, φi are perturbations of corresponding data. Remark 4.7 The proofs of this section are similar to the ones in the preceding subsections. The main difference is that unlike U (0, x) in the preceding subsections we do not assume smallness of W (0, x). Define an operator A02 : V → V∗ by

A02 y, u =2μ(E(y), E(u)) + 2μ(k(x)y, u)Γ2 + 2μ(S y˜ , u) ˜ Γ3 + 2(α(x)y, u)Γ5 + μ(k(x)y, u)Γ7 + (W (0, x) · ∇)y, u + (y · ∇)W (0, x), u for all y, u ∈ V.

(4.102)

For A02 we have Lemma 4.11 There exists δ > 0 and k2 ≥ 0 such that

A02 u, u ≥ δ u 2V − k2 u 2 for all u ∈ V. Proof By Korn’s inequality, there exists β > 0 such that 2μ(E(u), E(u)) ≥ β u 2V for all u ∈ V.

(4.103)

By Remark 3.4, there exists a constant M such that

S(x) ∞ , k(x) ∞ , α(x) ∞ ≤ M. Then, there exists a constant c0 (depending on β) such that 2μ(k(x)u, u)Γ + 2μ(S u, ˜ u) ˜ Γ3 + μ(k(x)u, u)Γ7 + 2(α(x)u, u)Γ5 2 β ≤ u 2H1 (Ω) + c0 u 2 for all u ∈ V 4

(4.104)

(see Theorem 1.27). Let us estimate (W (0, x) · ∇)u, u + (u · ∇)W (0, x), u. Since W (0, x) ∈ C(Ω),   (W (0, x) · ∇)u, u ≤ β u 2 1 + c1 u 2 . H (Ω) 8

(4.105)

4.2 Existence and Uniqueness of Solutions: The Case of Static Pressure

143

Taking div u = 0 into account, we get   (u · ∇)W (0, x), u =



=

j

∂Ω

Ω

uj

∂ W (0, x) u dx ∂x j

(u · n)(W (0, x) · u) dΓ −



j

Ω

uj

∂u W (0, x) d x. ∂x j

Estimating the first term in the right-hand side of the equality above as in (4.104) and applying Hölder’s inequality in the second term, we have   (u · ∇)W (0, x), u ≤ β u 2 1 + c2 u 2 . H (Ω) 8

(4.106)

Taking δ = β2 , k2 = c0 + c1 + c2 , from (4.103)–(4.106) we get the asserted conclusion.  Put z = e−k2 t z, where k2 is a constant in Lemma 4.11. Then e−k2 t z  = z  + k2 z, and we have the following problem equivalent to (4.101): Find z such that z ∈ L 2 (0, T ; V), z(0) = z 0 = v0 − W (0) ∈ V,

z  (t), u + 2μ(E(z(t)), E(u)) + ek2 t (z(t) · ∇)z(t), u + (W (t) · ∇)z(t), u + (z(t) · ∇)W (t), u + k2 (z(t), u) + 2μ(k(x)z(t), u)Γ2 + 2μ(S z˜ (t), u) ˜ Γ3 + 2(α(x)z(t), u)Γ5 + μ(k(x)z(t), u)Γ7     −k2 t

f (t), u + =e

φi (t), u n Γi +

φi (t), uΓi for all u ∈ V. i=2,4,7

i=3,5,6

Define operators A2 , A W (t) : V → V∗ , respectively, by

(4.107)

A2 y, u =2μ(E(y), E(u)) + 2μ(k(x)y, u)Γ2 + 2μ(S y˜ , u) ˜ Γ3 + 2(α(x)y, u)Γ5 + μ(k(x)y, u)Γ7 + k2 (y, u) for all y, u ∈ V, (4.108) and

A W (t)v, u = (W (t, x) · ∇)v, u + (v · ∇)W (t, x), u for all v, u ∈ V, (4.109) where k2 is the constant in Lemma 4.11. W ∈ W implies W ∈ C([0, T ]; H1 (Ω)), and such definitions are well. In the proof of Lemma 4.11 it is clear that

A2 u, u ≥

3β

u 2V . 4

(4.110)

144

4 The Non-steady Navier-Stokes System

Also, by Lemma 4.11 we have 

 β A2 + A W (0) u, u ≥ u 2V . 4

(4.111)

Define an operator B2 (t) : V → V∗ by

B2 (t)v, u = ek2 t (v · ∇)v, u for all v, u ∈ V.

(4.112)

Define operators L 2 , A W , L 2W , B2 : X → Y , C2 : X × X → Y and an element F2 ∈ Y by

(L 2 z)(t), u = z  (t), u + A2 z(t), u ∀z ∈ X , ∀u ∈ V,

( A W z)(t), u = A W (t)z(t), u ∀z ∈ X , ∀u ∈ V, 



(L 2W z)(t), u = z  (t), u + A2 + A W (t) z(t), u ∀z ∈ X , ∀u ∈ V,

( B2 z)(t), u = B2 (t)z(t), u ∀z ∈ X , ∀u ∈ V,

C2 (w, z)(t), u = ek2 t (w · ∇)z, u + ek2 t (z · ∇)w, u ∀z ∈ X , ∀u ∈ V,    

(F2 )(t), u = e−k2 t f (t), u +

φi (t), u n Γi +

φi (t), uΓi ∀u ∈ V. i=2,4,7

i=3,5,6

(4.113) By the same argument as in Lemma 4.5 we get Lemma 4.12 C2 is a bilinear continuous operator such that X × X → Y . Under Assumption 4.2 A W is a linear continuous operator such that X → Y and F2 ∈ Y . Using (4.110) instead of (4.67), as Lemma 4.6 we get Lemma 4.13 The operator L 2 defined by L 2 z = (z  (0), L 2 z) for z ∈ X is a linear continuous one-to-one operator from X onto H × Y . Now, using (4.111) without assuming the fact that W (0, x) V is small enough, as in Lemma 4.7 we prove the following: Lemma 4.14 The operator L 2W defined by L 2W z = (z  (0), L 2W z) for z ∈ X is a linear continuous one-to-one operator from X onto H × Y . Proof As in Lemma 3.5 of [3] the operator A W : X → Y is compact. Thus, A W : X → H × Y defined by A W z = {0 H , A W z} is also compact. Since L 2W = L 2 + A W , in order to get the asserted conclusion by virtue of Theorem 1.8 it suffices to prove that L 2W is one-to-one from X into H × Y . To prove the last statement it is enough to show that the inverse image of (0 H , 0Y ) by L 2W is 0X . It is easy to verify that 

 A2 + A W (0) v, u ≤ c v V · u V for all v, u ∈ V.

(4.114)

By (4.111) and (4.114), for any q ∈ V∗ there exists a unique solution y ∈ V to

4.2 Existence and Uniqueness of Solutions: The Case of Static Pressure

(A2 + A W (0))y = q.

145

(4.115)

Let z ∈ X be the inverse image of (0 H , 0Y ) by L. Then, z  (0) = 0 H , and putting t = 0 from the third one of (4.113) we get 

 A2 + A W (0) z(0), u = 0 for all u ∈ V,

where z(0) = z(0, x). This means that z(0) is the unique solution to (4.115) with q = 0V∗ , i.e. z(0) = 0V . Therefore, z ∈ X satisfies !



z  (t) + A2 + A W (t) z(t) = 0, z(0) = 0V .

(4.116)

Now, using (4.116) and Gronwall’s inequality, as in Lemma 3.8 of [3] we can prove z = 0X . The proof is thus completed.  By the same argument as in Lemma 4.8 we get

Lemma 4.15 The operator T2 defined by T2 z = z  (0), (L 2W + B2 )z for z ∈ X is continuously differentiable, T2 0X = (0 H , 0Y ) and the Fréchet derivative of T2 at 0X is L 2W . Let us now turn to the following problem:

A2 + A W (0) + B2 (0) u = q.

(4.117)

Using (4.111) without assuming the fact that W (0, x) V is small enough, as in Lemma 4.9 we can prove Lemma 4.16 If the norm of q ∈ V∗ is small enough, then there exists a unique solution to (4.117) in some O M (0V ). Using Lemmas 4.12–4.15 and Theorem 1.7 (Local inverse mapping theorem), in the same way as in Theorem 4.3 we get Theorem 4.5 Suppose that Assumption 4.2 holds and the norms of f, f  , φi , φi in the spaces that they belong to are small enough. If w2 ≡ F2 (0) − (A2 + A2W (0) + B2 (0))z 0 ∈ H, (4.118) where z 0 = v0 − U (0, ·), and w2 H is small enough, then there exists a unique solution to (4.101) in the space W . Remark 4.8 By the same argument as in Remark 4.3, we can see that the condition (4.118) is equivalent to the condition w 2 ∈ H1 , where w 2 ∈ V1∗ is defined by

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4 The Non-steady Navier-Stokes System



w 2 , u = f (0), u +



φi (0, x), u n Γi +

i=2,4

φi (0, x), uΓi

i=3,5,7

 − 2μ(E(z 0 ), E(u)) + 2μ(k(x)z 0 , u)Γ2 + 2μ(S z˜ 0 , u) ˜ Γ3 + 2(α(x)z 0 , u)Γ5 + μ(k(x)z 0 , u)Γ7 + (W (0, x) · ∇)z 0 , u  + (z 0 · ∇)W (0, x), u + k2 (z 0 , u) + (z 0 · ∇)z 0 , u for all u ∈ V (4.119) with k2 in Lemma 4.11. Now let us consider Problem II. Let W (x, t) ∈ W be a given solution to Problem II. Let v be the solution for the data perturbed except h i and put v = z + W . Then, we get a problem for z as follows: Find z such that z ∈ L 2 (0, T ; V1 ), z(0) = z 0 ≡ v0 − W (0, x) ∈ V1 ,

z  , u + μ(∇z, ∇u) + (z · ∇)z, u + (W · ∇)z, u + (z · ∇)W, u ˜ u) ˜ u) ˜ Γ3 + 2(α(x)z, u)Γ5 − μ(S z, ˜ Γ5 + μ(k(x)z, u)Γ2 + μ(S z,   = f, u +

φi , u n Γi +

φi , uΓi for all u ∈ V1 , i=2,4

(4.120)

i=3,5,7

where z 0 , f, φi are perturbations of corresponding data. By the same argument as in Theorem 4.5 we have Theorem 4.6 Suppose that Assumption 4.2 holds and the norms of f, f  , φi , φi in the spaces that they belong to are small enough. Define an element w3 ∈ V1∗ by

w3 , u = f (0), u +

 i=2,4

φi (0, x), u n Γi +



φi (0, x), uΓi

i=3,5,7

 − μ(∇z 0 , ∇u) + μ(k(x)z 0 , u)Γ2 + μ(S z˜ 0 , u) ˜ Γ3 + 2(α(x)z 0 , u)Γ5 − μ(S z˜ 0 , u) ˜ Γ5 + (W (0, x) · ∇)z 0 , u + (z 0 · ∇)W (0, x), u  + k3 (z 0 , u) + (z 0 · ∇)z 0 , u for all u ∈ V1 , (4.121) where k3 is a constant determined as in Lemma 4.11. If w3 ∈ O R (0 H1 ) for R > 0 small enough, then there exists a unique solution to (4.120) in the space W . Remark 4.9 If Γi = ∅, i = 2, · · · , 5, then problem (4.120) is reduced to the one in [3]. If z 0 ∈ Hl/2 (Ω), then (z 0 · ∇)z 0 , (W (0, x) · ∇)z 0 , (z 0 · ∇)W (0, x) ∈ L2 (Ω) and k3 z 0 ∈ L2 (Ω). Thus, the last four terms on the right-hand side of (4.121) do not give any effect to the condition for w3 being in H1 , and so the conditions in Theorem 4.6 are the same as one of conditions of Theorems 3.5–3.8 of [3]. Thus, Theorem 4.6 guarantees existence of a unique solution under a condition weaker than the one in [3]. Note that taking W (t, x) ≡ 0 in Theorems 4.5 and 4.6, we can not get Theorems 4.3 and 4.4, since there h i = 0.

4.2 Existence and Uniqueness of Solutions: The Case of Static Pressure

147

Remark 4.10 Let Hk (Ω) = (W k,2 (Ω))l be a Sobolev spaces on Ω with dimension l, V be a divergence-free subspace of H1 satisfying appropriate boundary conditions, H —the closure of V in (L 2 (Ω))l , V r0 (Ω) = V ∩ Hr0 (Ω), where r0 > l/2, X = {w ∈ L 2 (0, T ; V ); w  ∈ L 2 (0, T ; V ), w  ∈ L 2 (0, T ; V ∗ )}, Y = {w ∈ L 2 (0, T ; V ∗ ); w  ∈ L 2 (0, T ; V ∗ )} and A : V → V ∗ —the Stokes operator. Concerned with the linearized problem of !

u  (t) + (A + B)u(t) = f (t), u(0) = ϕ,

in [3] it is proved that a map u → {u(0), Lu ≡ u  + Au} is linear continuous oneto-one from X = {u ∈ X : u(0) ∈ V r0 (Ω)} onto Y = {[ϕ, h] : ϕ ∈ V r0 (Ω), h ∈ Y , h(0) − Aϕ ∈ H } and the compatibility condition f (0) − Aϕ ∈ H, ϕ ∈ V r0 (Ω) for the existence of a unique solution to the nonlinear problem is obtained. However, in Sect. 4.2 it is proved that for a modified operator A a map u → {u  (0), Lu ≡ u  + Au} is linear continuous one-to-one from X onto H × Y (Lemma 4.6). Then, it is obtained the compatibility condition f (0) − Aϕ − B(0)ϕ ∈ H without ϕ ∈ V r0 (Ω) for the existence of unique solution to a nonlinear problems. Since B(0)ϕ = (ϕ · ∇)ϕ, for ϕ ∈ V l/2 (Ω) we get B(0)ϕ ∈ H , and so from point of view of smoothness of the initial functions, the compatibility conditions in Sect. 4.2 are weaker than the one in [3].

4.3 Bibliographical Remarks The content of Sect. 4.1 is the one of Sect. 5 of [5] and the content of Sect. 4.2 is taken from [6]. The Navier-Stokes equations on a domain decreasing in time with mixture of Dirichlet condition, the total pressure and vorticity conditions were studied in [1]. For a more general problem we refer to [7]. For the 2-D Navier-Stokes equations with mixture of the Dirichlet and pressure boundary conditions, existence of a unique solution for small data was proved in [8]. The Navier-Stokes equations with mixture of Dirichlet condition and stress condition were studied. In [4] under smoothness condition and a compatibility condition

148

4 The Non-steady Navier-Stokes System

of data at the initial time the existence of a unique local-in-time solution to the 3-D Navier-Stokes equations was discussed. In [9] for the Navier-Stokes equations on the polyhedral domain with mixture of Dirichlet condition, Navier slip condition and stress condition a local-in-time solution was considered. Here smoothness of solutions to the corresponding steady Stokes problem was used essentially. The Navier-Stokes equations with mixture of Dirichlet condition and the outflow boundary condition were studied. For 2-D Navier-Stokes equations a local-in-time solution in [10] and a solution for small data were obtained in [11]. Here also smoothness of solutions to the corresponding 2-D steady linear problem is important. In [3] it was proved that if under a compatibility condition at initial time there exists a unique solution, then so does for small perturbed data. This result shows that under the compatibility condition there exists a unique solution for small data. Smoothness of initial function in the compatibility condition of [3] is stronger than the one in [4, 12]. The Navier-Stokes equations with mixture of Dirichlet condition and the static pressure boundary condition were studied. In [8] for the 2-D Navier-Stokes problem existence of a unique solution for small data was established.

References 1. G. Łukaszewicz, On the Navier-Stokes equations in time dependent domains and with boundary conditions involving the pressure. J. Math. Sci. Univ. Tokyo 4(3), 529–550 (1997) 2. G.P. Galdi, An Introduction to the Mathematical Theory of the Navier-Stokes Equations (Springer, Berlin, 2011) 3. P. Kuˇcera, Basic properties of solution of the non-steady Navier-Stokes equations with mixed boundary conditions in a bounded domain. Ann. Univ. Ferrara 55, 289–308 (2009) 4. P. Kuˇcera, Z. Skalák, Local solutions to the Navier-Stokes equations with mixed boundary conditions. Acta Appl. Math. 54, 275–288 (1998) 5. T. Kim, D. Cao, Some properties on the surfaces of vector fields and its application to the Stokes and Navier-Stokes problems with mixed boundary conditions. Nonlinear Anal. 113, 94–114 (2015). Erratum, ibid 135, 249–250 (2016) 6. T. Kim, D. Cao, Existence of solutions to the heat convection equations in a time-dependent domain with mixed boundary conditions. J. Math. Sci. Univ. Tokyo 22, 531–568 (2015) 7. T. Kim, D. Cao, A non-steady system with friction boundary conditions for flow of heatconducting incompressible viscous fluids. J. Math. Anal. Appl. 484, 123676 (2020) 8. S. Maruši´c, On the Navier-Stokes system with pressure boundary condition. Ann. Univ. Ferrara 53, 319–331 (2007) 9. M. Beneš, Mixed initial-boundary value problem for the three-dimensional Navier-Stokes equations in polyhedral domains. Discret. Contin. Dyn. Syst. Suppl. 02, 135–144 (2011) 10. M. Beneš, P. Kuˇcera, Non-steady Navier-Stokes equations with homogeneous mixed boundary conditions and arbitrarily large initial condition. Carpathian J. Math. 23(1–2), 32–40 (2007) 11. M. Beneš, P. Kuˇcera, Solutions to the Navier-Stokes equations with mixed boundary conditions in two-dimensional bounded domains. Math. Nachr. 289(2–3), 194–212 (2016) 12. Z. Skalák, P. Kuˇcera, An existence theorem for the Boussinesq equations with non-Dirichlet boundary conditions. Appl. Math. 45(2), 81–98 (2000)

Chapter 5

The Steady Navier-Stokes System with Friction Boundary Conditions

In this chapter we are concerned with the steady Navier-Stokes systems with mixed boundary conditions which may include Tresca slip condition, leak boundary condition, one-sided leak boundary conditions, velocity, pressure, vorticity, stress and normal derivative of velocity together. Relying on the results in Sect. 3.1 and using the strain bilinear form, we embed all these boundary conditions into variational formulations of corresponding problems. In Sect. 5.1 for every problem, which is distinguished according to boundary conditions, we first get the variational formulation which consists of five formulas with five unknown functions, that is, using velocity, tangent stress on slip surface, normal stress on leak surface, normal stresses on one-sided leak surfaces together as unknown functions. To show that such formulations are well defined, we note that when the solution is smooth enough, these variational formulations are equivalent to the original PDE problems. It is shown that if the boundary condition for pressure or stress on a portion of boundary where there is flux is given, then the pressure is determined uniquely. Then, we get variational inequalities with one unknown function (velocity) equivalent to the variational formulations with five unknown functions, by which the Navier-Stokes problem with 11 kinds of boundary conditions is reduced to one variational inequality. In Sect. 5.2 we study three kinds of variational inequality obtained in Sect. 5.1. In Sect. 5.3, using the results in Sect. 5.2, we study the existence, uniqueness and estimates of solutions to the Navier-Stokes problems with 11 kinds of boundary conditions. For the problem with boundary conditions involving the static pressure and stress, the existence of a unique solution is proved when the data are small enough. For the problem with boundary conditions involving the total pressure and total stress, the existence and estimate of solutions are proved without smallness of the data of problem.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 T. Kim and D. Cao, Equations of Motion for Incompressible Viscous Fluids, Advances in Mathematical Fluid Mechanics, https://doi.org/10.1007/978-3-030-78659-5_5

149

150

5 The Steady Navier-Stokes System with Friction Boundary Conditions

From now on throughout this book the following notations are used. When f ∈ H −1/2 (Γi ), if  f, wΓi ≥ 0 (≤ 0) ∀w ∈ C0∞ (Γi ) with w ≥ 0, then we denoted by f ≥ 0 (≤ 0).

5.1 Variational Formulations of Problems 11 l 0,1 Let Ω be a bounded  domain of R , l = 2, 3. ∂Ω ∈ C , ∂Ω = ∪i=1 Γ i , Γi ∩ Γ j = ∅2 for i = j, Γi = j Γi j , where Γi j are connected open subsets of ∂Ω and Γi j ∈ C for i = 2, 3, 7 and Γi j ∈ C 1 for others (see Remark 5.2).

We are concerned with the Problems I and II for the Navier-Stokes equations − μΔv + (v · ∇)v + ∇ p = f, div v = 0,

(5.1)

which are distinguished according to boundary conditions. Problem I is one with the following boundary conditions: (1) v|Γ1 = h 1 , (2) vτ |Γ2 = 0, − p|Γ2 = φ2 , (3) vn |Γ3 = 0, rot v × n|Γ3 = φ3 /μ, (4) vτ |Γ4 = h 4 , (− p + 2μεnn (v))|Γ4 = φ4 , (5) vn |Γ5 = h 5 , 2(μεnτ (v) + αvτ )|Γ5 = φ5 , α : a matrix, (6) (− pn + 2μεn (v))|Γ6 = φ6 , ∂v · n)|Γ7 = φ7 , (7) vτ |Γ7 = 0, (− p + μ ∂n (8) vn = h 8 , |στ (v)| ≤ gτ , στ (v) · vτ + gτ |vτ | = 0 on Γ8 , (9) vτ = h 9 , |σn (v)| ≤ gn , σn (v)vn + gn |vn | = 0 on Γ9 , (10) vτ = 0, vn ≥ 0, σn (v) + g+n ≥ 0, (σn (v) + g+n )vn = 0 on Γ10 , (11) vτ = 0, vn ≤ 0, σn (v) − g−n ≤ 0, (σn (v) − g−n )vn = 0 on Γ11 , and Problem II is one with the following boundary conditions:

(5.2)

5.1 Variational Formulations of Problems

151

(1) v|Γ1 = h 1 , (2) vτ |Γ2 = 0, −( p + 1/2|v|2 )|Γ2 = φ2 , (3) vn |Γ3 = 0, rot v × n|Γ3 = φ3 /μ, (4) vτ |Γ4 = h 4 , (− p − 1/2|v|2 + 2μεnn (v))|Γ4 = φ4 , (5) vn |Γ5 = h 5 , 2(μεnτ (v) + αvτ )|Γ5 = φ5 , α : a matrix, (6) (− pn − 1/2|v|2 n + 2μεn (v))|Γ6 = φ6 , ∂v · n)|Γ7 = φ7 , (7) vτ |Γ7 = 0, (− p − 1/2|v|2 + μ ∂n (8) vn = h 8 , |στt (v)| ≤ gτ , στt (v) · vτ + gτ |vτ | = 0 on Γ8 ,

(5.3)

(9) vτ = h 9 , |σnt (v)| ≤ gn , σnt (v)vn + gn |vn | = 0 on Γ9 , (10) vτ = 0, vn ≥ 0, σnt (v) + g+n ≥ 0, (σnt (v) + g+n )vn = 0 on Γ10 , (11) vτ = 0, vn ≤ 0, σnt (v) − g−n ≤ 0, (σnt (v) − g−n )vn = 0 on Γ11 , where and in what follows εn (v) = E(v)n, εnn (v) = (E(v)n, n)Rl , εnτ (v) = E(v)n − εnn (v)n and h i , φi , αkl (components of matrix α) are given functions or vectors of functions. σn and σnt are, respectively, the normal components of stress and total stress on a boundary surface, that is, σn = σ · n and σnt = σ t · n. Also, στ (v, p) = σ (v, p) − σn (v, p)n, στt (v, p) = σ t (v, p) − σnt (v, p)n. Throughout this book we always assume that gτ ∈ L 2 (Γ8 ), gn ∈ L 2 (Γ9 ), g+n ∈ L (Γ10 ), g−n ∈ L 2 (Γ11 ) and that gτ > 0, gn > 0, g+n > 0, g−n > 0, for a.e. x of the portions of boundary. 2

For Problem II the static pressure p and the stress in the boundary conditions for Problem I are, respectively, changed with the total pressure and the total stress. Thus, as in Chaps. 3, 4 for convenience in what follows, the Navier-Stokes problems with boundary conditions (5.2) and (5.3) are, respectively, called the case of static pressure and the case of total pressure. We also consider the Stokes equations −μΔv + ∇ p = f, div v = 0

(5.4)

with the boundary conditions (5.2), which is Problem III. Assuming that f ∈ L2 (Ω), φi ∈ L 2 (Γi ), i = 2, 4, 7, and φi ∈ L2 (Γi ), i = 3, 5, 6, formally we introduce the following: Definition 5.1 A function (v, p) ∈ H2 (Ω) × H 1 (Ω) is called a solution to Problem I (or Problem II ) if (5.1) holds in L2 (Ω) and each of (5.2) (or (5.3)) holds in L 2 (Γi ) or L2 (Γi ). A function (v, p) ∈ H2 (Ω) × H 1 (Ω) is called a solution to Problem III if (5.4) holds in L2 (Ω) and each of (5.2) holds in L 2 (Γi ) or L2 (Γi ).

152

5 The Steady Navier-Stokes System with Friction Boundary Conditions

Usually, it is difficult (or impossible) to find solutions in the sense of the definition above, and we will consider variational formulations of the problems.

5.1.1 Variational Formulation: The Case of Static Pressure Let   V = u ∈ H1 (Ω) : div u = 0, u|Γ1 = 0, u τ |(∪i=2,4,7,9,10,11 Γi ) = 0, u n |(∪i=3,5,8 Γi ) = 0 ,   VΓ 237 (Ω) = u ∈ H1 (Ω) : div u = 0, u τ |(Γ2 ∪Γ7 ) = 0, u n |Γ3 = 0 ,   K (Ω) = u ∈ V : u n |Γ10 ≥ 0, u n |Γ11 ≤ 0 . (5.5) Taking into account (2.81) and applying Theorems 3.1 and 3.2 on Γi j (i = 2, 3, 7), for v ∈ H2 (Ω) ∩ VΓ 237 (Ω) and u ∈ V we have 11 −(Δv, u) = 2(E(v), E(u)) − 2(E(v)n, u)∪i=2 Γi

= 2(E(v), E(u)) + 2(k(x)v, u)Γ2 − (rot v × n, u)Γ3 + 2(S v, ˜ u) ˜ Γ3 − 2(εnn (v), u n )Γ4 − 2(εnτ (v), u)Γ5 − 2(εn (v), u)Γ6   ∂v − + (k(x)v, u)Γ7 − 2(εnτ (v), u)Γ8 ,u ∂n Γ7 − 2(εnn (v), u n )Γ9 − 2(εnn (v), u n )Γ10 − 2(εnn (v), u n )Γ11 . (5.6) Also, for p ∈ H 1 (Ω) and u ∈ V we have 11 (∇ p, u) = ( p, u n )∪i=2 Γi = ( p, u n )Γ2 + ( p, u n )Γ4 ∪Γ7 ∪Γ9 ∪Γ10 ∪Γ11 + ( pn, u)Γ6 ,

(5.7) where u n |Γ3 ∪Γ5 ∪Γ8 = 0 was used. We assume that the following holds: Assumption 5.1 (1) There exists a function U ∈ H1 (Ω) such that div U = 0, U |Γ1 = h 1 , Uτ |(Γ2 ∪Γ7 ) = 0, Un |Γ3 = 0, Uτ |Γ4 = h 4 , Un |Γ5 = h 5 , U |Γ8 = h 8 n, U |Γ9 = h 9 , U |Γ10 ∪Γ11 = 0. (2) f ∈ V∗ , φi ∈ H − 2 (Γi ), i = 2, 4, 7, φi ∈ H− 2 (Γi ), i = 3, 5, 6, αi j ∈ ∞ L (Γ5 ), and Γ1 = ∅. (3) If Γi , where i is 10 or 11, is nonempty, then at least one of {Γ j : j ∈ {2, 4, 7, 9 − 11}\{i}} is nonempty and there exists a diffeomorphism in C 1 between Γi and Γ j . 1

1

5.1 Variational Formulations of Problems

153

Having in mind Assumption 5.1 and putting v = w + U , by (5.6) and (5.7) we can see that solutions (v, p) of the problem (5.1), (5.2) in the sense of Definition 5.1 satisfy the following: ⎧ v − U = w ∈ K (Ω), ⎪ ⎪ ⎪ ⎪ ⎪ 2μ(E(w), E(u)) + (w · ∇)w, u + (U · ∇)w, u + (w · ∇)U, u ⎪ ⎪ ⎪ ⎪ ⎪ + 2μ(k(x)w, u)Γ2 + 2μ(S w, ˜ u) ˜ Γ3 + 2(α(x)w, u)Γ5 + μ(k(x)w, u)Γ7 ⎪ ⎪ ⎪ ⎪ ⎪ − 2μ(ε (w + U ), u) + ( p − 2μεnn (w + U ), u n )Γ9 ∪Γ10 ∪Γ11 nτ Γ8 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ = −2μ(E(U ), E(u)) − (U · ∇)U, u − 2μ(k(x)U, u)Γ2 ⎪ ⎪ ⎨ − 2μ(SU˜ , u) ˜ Γ3 − 2(α(x)Uτ , u)Γ5 − μ(k(x)U, u)Γ7 +  f, u   ⎪ ⎪ φ , u  + φi , uΓi ∀u ∈ V, + ⎪ i n Γ i ⎪ ⎪ ⎪ i=2,4,7 i=3,5,6 ⎪ ⎪ ⎪ ⎪ ⎪ |στ (v)| ≤ gτ , στ (v) · vτ + gτ |vτ | = 0 on Γ8 , ⎪ ⎪ ⎪ ⎪ ⎪ |σn (v)| ≤ gn , σn (v)vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪ ⎪ ⎪ σn (v) + g+n ≥ 0, (σn (v) + g+n )vn = 0 on Γ10 , ⎪ ⎪ ⎩ σn (v) − g−n ≤ 0, (σn (v) − g−n )vn = 0 on Γ11 . (5.8) Define a01 (·, ·), a11 (·, ·, ·) and F1 ∈ V ∗ by a01 (w, u) = 2μ(E(w), E(u)) + (U · ∇)w, u + (w · ∇)U, u + 2μ(k(x)w, u)Γ2 + 2μ(S w, ˜ u) ˜ Γ3 + 2(α(x)w, u)Γ5 + μ(k(x)w, u)Γ7 ∀w, u ∈ V, a11 (w, u, v) = (w · ∇)u, v ∀w, u, v ∈ V, F1 , u = −2μ(E(U ), E(u)) − (U · ∇)U, u − 2μ(k(x)U, u)Γ2 − 2μ(SU˜ , u) ˜ Γ3 − 2(α(x)Uτ , u)Γ5 − μ(k(x)U, u)Γ7 +  f, u   φi , u n Γi + φi , uΓi ∀u ∈ V. + i=2,4,7

(5.9)

i=3,5,6

Then, taking into account στ (v) = 2μεnτ (v), σn (v) = − p + 2μεnn (v) and (5.8), we introduce the following variational formulation for problem (5.1), (5.2).

Problem I-VE. Find (v, στ , σn , σ+n , σ−n ) ∈ U + K (Ω) × L2τ (Γ8 ) × L 2 (Γ9 ) × H −1/2 (Γ10 ) × H −1/2 (Γ11 ) such that

154

5 The Steady Navier-Stokes System with Friction Boundary Conditions

⎧ ⎪ ⎪ v − U = w ∈ K (Ω), ⎪ ⎪ ⎪ a01 (w, u) + a11 (w, w, u) − (στ , u τ )Γ8 − (σn , u n )Γ9 ⎪ ⎪ ⎪ ⎪ ⎪ − σ+n , u n Γ10 − σ−n , u n Γ11 = F1 , u ∀u ∈ V, ⎪ ⎨ |στ | ≤ gτ , στ · vτ + gτ |vτ | = 0 on Γ8 , ⎪ ⎪ ⎪ |σn | ≤ gn , σn vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪ ⎪ ⎪ σ ⎪ +n + g+n ≥ 0, σ+n + g+n , vn Γ10 = 0 on Γ10 , ⎪ ⎪ ⎩ σ−n − g−n ≤ 0, σ−n − g−n , vn Γ11 = 0 on Γ11 ,

(5.10)

where L2τ (Γ8 ) is the subspace of L2 (Γ8 ) consisting of functions such that[ (u, n)L2 (Γ8 ) = 0. Remark 5.1 If u ∈ H 1 (Ω) and u|∂Ω = 0 on O(Γi )\Γ i , where O(Γi ) is an open 1

subset of ∂Ω such that Γ i ⊂ O(Γi ), then u|Γi ∈ H002 (Γi ) (see Sect. 1.3.2). Since 1

1

1

1

H002 (Γi ) → H02 (Γi ) and H 2 (Γi ) = H02 (Γi ) (see (1.11)), 1

1

H002 (Γi ) → H 2 (Γi ) → H − 2 (Γi ) → (H002 (Γi ))∗ . 1

1

1

Thus, under condition u|∂Ω = 0 on O(Γi )\Γ i , for φi ∈ (H002 (Γi )) a dual prod1 uct φi , uΓi is well defined. However, if φi ∈ H − 2 (Γi ), then without knowing that u|∂Ω = 0 on O(Γi )\Γ i the dual product φi , uΓi is meaningful. Therefore, under (2) of Assumption 5.1 the dual products on Γi in (5.8) is meaningful. The following theorem shows that the variational formulation above is well defined. Theorem 5.1 Let Assumption 5.1 hold. If (v, p) is a solution in the sense of Definition 5.1 of the problem (5.1), (5.2), then (v, στ |Γ8 , σn |Γ9 , σn |Γ10 , σn |Γ11 ) is a solution to Problem I-VE. Conversely, if f ∈ L2 (Ω) and Problem I-VE has a smooth solution (v, στ , σn , σ+ , σ− ) such that v ∈ H2 (Ω), σ+ ∈ L 2 (Γ10 ) and σ− ∈ L 2 (Γ11 ), then there exists p ∈ H 1 (Ω) such that (v, p) is a solution to the problem (5.1), (5.2). Moreover, if at least one of the sets Γi , i = 2, 4, 6, 7, is nonempty, then p is unique. Proof It suffices to prove the conversion from Problem I-VE to the problem (5.1)– (5.2). Let v be a solution smooth enough as above to Problem I-VE. From (5.9) and (5.10) we have 2μ(E(v), E(u)) + (v · ∇)v, u + 2μ(k(x)v, u)Γ2 + 2μ(S v, ˜ u) ˜ Γ3 + 2(α(x)vτ , u)Γ5 + μ(k(x)v, u)Γ7 − (στ , u τ )Γ8 − (σn , u n )Γ9   − σ+n , u n Γ10 − σ−n , u n Γ11 − φi , u n Γi − φi , uΓi i=2,4,7

=  f, u ∀u ∈ V.

i=3,5,6

(5.11)

5.1 Variational Formulations of Problems

155

From (5.6) we get 2μ(E(v), E(u)) = − μ(Δv, u) − 2μ(k(x)v, u)Γ2 + μ(rot v × n, u)Γ3 − 2μ(S v, ˜ u) ˜ Γ3 + 2μ(εnn (v), u · n)Γ4 + 2μ(εnτ (v), u)Γ5 + 2μ(εn (v), u)Γ6   ∂v ,u +μ − μ(k(x)v, u)Γ7 + 2μ(εnτ (v), u)Γ8 + 2μ(εnn (v), u)Γ9 ∂n Γ7

(5.12)

+ 2μ(εnn (v), u)Γ10 + 2μ(εnn (v), u)Γ11 .

Combining (5.11) with (5.12) yields (−μΔv + (v · ∇)v − f, u) + μ(rot v × n, u)Γ3 + 2μ(εnn (v), u · n)Γ4   ∂v ,u + 2μ(εnτ (v), u)Γ5 + 2(α(x)vτ , u)Γ5 + 2μ(εn (v), u)Γ6 + μ ∂n Γ7 + 2μ(εnτ (v), u)Γ8 + 2μ(εnn (v), u n )Γ9 + 2μ(εnn (v), u n )Γ10 + 2μ(εnn (v), u n )Γ11 − (στ , u τ )Γ8 − (σn , u n )Γ9 − σ+n , u n Γ10 − σ−n , u n Γ11   − φi , u n Γi − φi , uΓi = 0. i=2,4,7

i=3,5,6

(5.13)

Taking any u ∈ C0∞ (Ω) with div u = 0, we have (−μΔv + (v · ∇)v − f, u) = 0, and by Proposition 2.3 there exists a unique P ∈ L 2 (Ω) such that

 Ω

− μΔv + (v · ∇)v − f = −∇ P.

P d x = 0 and (5.14)

Since v ∈ H2 (Ω), f ∈ L2 (Ω) and (v · ∇)v ∈ L2 (Ω), we have that ∇ P ∈ L2 (Ω), which shows that P ∈ H 1 (Ω). Substituting (5.14) into (5.13), integrating by parts and taking into account (5.7), we have (see Remark 1.17) (−P − φ2 , u n )Γ2 + μ(rot v × n − φ3 /μ, u)Γ3 + (−P + 2μεnn (v) − φ4 , u n )Γ4 + (2μεnτ (v) + α(x)vτ − φ5 , u)Γ5 + (−Pn + 2μεn (v) − φ6 , u)Γ6

∂v · n − φ7 , u n Γ7 + (2μεnτ (v) − στ , u)Γ8 + − P +μ ∂n + (−P + 2μεnn (v) − σn , u n )Γ9 + (−P + 2μεnn (v) − σ+n , u n )Γ10 + (−P + 2μεnn (v) − σ−n , u n )Γ11 = 0, (5.15)

∂v ∂v , u Γ7 = (μ ∂n · n, u n Γ7 were used. where (v, u)Γ5 = (vτ , u)Γ5 and (μ ∂n For every i = 3, 5, 8, respectively, let us take any u ∈ V such that u n |Γi = 0 and u|(∂Ω\Γi ) = 0. The set of traces of such functions on Γi includes any tangent vector 1/2 fields ϕ ∈ H00 (Γi ) (see Sect. 2.3.2.1), and so from (5.15) we get

156

5 The Steady Navier-Stokes System with Friction Boundary Conditions

rot v × n = φ3 /μ on Γ3 , 2μεnτ (v) + α(x)vτ − φ5 = 0 on Γ5 ,

(5.16)

2μεnτ (v) − στ = 0 on Γ8 . If for all i = 2, 4, 6, 7, 9, 10, 11, Γi = ∅, then putting p = P + c, where c is any constant, we get a solution (v, p) to problem (5.1), (5.2). Assume that among Γi , i = 2, 4, 6, 7, 9, 10, 11, at least one is nonempty. For every i = 2, 4, 7, 9, 10, 11, respectively, let us take any u ∈ V such that u τ |Γi = 0 and u|(∂Ω\Γi ) = 0. The set of traces of such functions on Γi include any normal vector  1/2 fields ϕ ∈ H00 (Γi ) such that Γi ϕ d x = 0 (see Theorem IV.1.1 of [1]), and so from (5.15) we have that for some constants ci , i = 2, 4, 7, 9, 10, 11, respectively, − P − φ2 = c2 on Γ2 , − P + 2μεnn (v) − φ4 = c4 on Γ4 , ∂v · n − φ7 = c7 on Γ7 , − P +μ ∂n − P + 2μεnn (v) − σn = c9 on Γ9 ,

(5.17)

− P + 2μεnn (v) − σ+n = c10 on Γ10 , − P + 2μεnn (v) − σ−n = c11 on Γ11 . (See the proof of (2.90).) Taking any u ∈ V such that u|(∂Ω\Γ6 ) = 0 and arguing as in the proof of (2.90), we have from (5.15) that for a constant c6 − Pn + 2μεn (v) − φ6 = c6 n on Γ6 .

(5.18)

Let us prove that indeed, all ci are equal to one constant c. For example, assume that Γ2 and Γ4 are nonempty. Taking any u ∈ V such that u|∂Ω = 0 on ∂Ω \ (Γ2 ∪ Γ4 ), we get from (5.15)   c2

Γ2

u n d x + c4

Γ4

u n d x = 0,

  which implies c2 = c4 = c since Γ2 u n d x = − Γ4 u n d x. Thus, from (5.14), (5.17) and (5.18) we see that p = P + c satisfies −μΔv + (v · ∇) + ∇ p = f and all the following boundary conditions:

5.1 Variational Formulations of Problems

157

− p = φ2 on Γ2 , − p + 2μεnn (v) = φ4 on Γ4 , − pn + 2μεn (v) = φ6 on Γ6 , ∂v − p+μ · n = φ7 on Γ7 , ∂n − p + 2μεnn (v) = σn on Γ9 , − p + 2μεnn (v) = σ+n on Γ10 ,

(5.19)

− p + 2μεnn (v) = σ−n on Γ11 . By virtue of (5.10), (5.16) and (5.19), all conditions in (5.2) are satisfied. Therefore, (v, p) is a solution to problem (5.1), (5.2). If at least one of the sets Γi , i = 2, 4, 6, 7, is nonempty, then the equality with the given φi above on Γi holds, and p is unique.  We will find a variational inequality equivalent to Problem I-VE. Let (v, στ , σn , σ+n , σ−n ) be a solution of Problem I-VE. Subtracting the second formula of (5.10) with u = w from the second one of (5.10), we get a01 (w, u − w) + a11 (w, w, u − w) − (στ , u τ − wτ )Γ8 − (σn , u n − wn )Γ9 − σ+n , u n − wn Γ10 − σ−n , u n − wn Γ11 = F1 , u − w ∀u ∈ V. (5.20) Define the functionals φτ , φn , φ+ , φ− , respectively, by  φτ (η) =  φn (η) =

Γ8

Γ9

gτ |η| d x ∀η ∈ L2τ (Γ8 ), gn |η| d x ∀η ∈ L 2 (Γ9 ),

 φ+ (η) =

Γ10

(5.21) g+n η d x ∀η ∈ L 2 (Γ10 ),



φ− (η) = −

Γ11

g−n η d x ∀η ∈ L 2 (Γ11 ).

Since if u ∈ K (Ω), then u|Γ8 ∈ L2τ (Γ8 ), u n |Γ9 ∈ L 2 (Γ9 ), u n |Γ10 ∈ L 2 (Γ10 ) and u n |Γ11 ∈ L 2 (Γ11 ), in what follows for convenience we use the notation φτ (u) = φτ (u|Γ8 ), φn (u) = φn (u n |Γ9 ), φ+ (u) = φ+ (u n |Γ10 ), φ− (u) = φ− (u n |Γ11 ) ∀u ∈ K (Ω). Define a functional Φ : V → R by

158

5 The Steady Navier-Stokes System with Friction Boundary Conditions

 Φ(u) =

φτ (u) + φn (u) + φ+ (u) + φ− (u)

∀u ∈ K (Ω),

+∞

∀u ∈ / K (Ω).

(5.22)

Then, Φ is proper, convex and weakly lower semi-continuous (cf. Example 1.2). Note Φ ≥ 0 since u n |Γ10 ≥ 0 and u n |Γ11 ≤ 0 for u ∈ K (Ω). 11 Γi . Taking into account By Assumption 5.1, wτ = vτ on Γ8 and wn = vn on ∪i=9 the fact that gτ |vτ | + στ · vτ = 0 and |στ | ≤ gτ on Γ8 , we have that φτ (u) − φτ (w) + (στ , u τ )Γ8 − (στ , wτ )Γ8   = (gτ |u τ | + στ · u τ ) d x − (gτ |wτ | + στ · wτ ) d x Γ8 Γ8   = (gτ |u τ | + στ · u τ ) d x − (gτ |vτ | + στ · vτ ) d x ≥ 0 ∀u ∈ K (Ω). Γ8

Γ8

Taking into account the fact that gn |vn | + σn · vn = 0 and |σn | ≤ gn on Γ9 , in the same way we have

Also,

φn (u) − φn (w) + (σn , u n )Γ9 − (σn , wn )Γ9 ≥ 0.

(5.23)

φ+ (u) − φ+ (w) + σ+n , u n Γ10 − σ+n , wn Γ10 = g+n + σ+n , u n Γ10 − g+n + σ+n , wn Γ10 ≥ 0,

(5.24)

where the facts that u n ≥ 0, σ+n + g+n ≥ 0 and σ+n + g+n , vn Γ10 = 0, wn = vn on Γ10 were used. In the same way, we have φ− (u) − φ− (w) + σ−n , u n Γ11 − σ−n , wn Γ11 ≥ 0.

(5.25)

By virtue of (5.22)–(5.25), we have Φ(u) − Φ(w) ≥ − (στ , u τ − wτ )Γ8 − (σn , u n − wn )Γ9 − σ+n , u n − wn Γ10 − σ−n , u n − wn Γ11

∀u ∈ V.

(5.26)

Therefore, we get from (5.20) and (5.26) a01 (w, u − w) + a11 (w, w, u − w) + Φ(u) − Φ(w) ≥ F1 , u − w ∀u ∈ V. (5.27) Thus, we come to the following formulation associated with Problem I (the case of static pressure) by a variational inequality. Problem I-VI Find v = w + U such that a01 (w, u − w) + a11 (w, w, u − w) + Φ(u) − Φ(w) ≥ F1 , u − w ∀u ∈ V, (5.28)

5.1 Variational Formulations of Problems

159

where a01 , a11 , F1 are the same as in (5.9), U is the same as in Assumption 5.1 and Φ is as in (5.22). To prove equivalence of Problem I-VI and Problem I-VE we need first to establish the following result: Lemma 5.1 For ψ ∈ C0∞ (Γi ), i = 10, 11, there exists a function u ∈ V such that u n |Γi = ψ, uV ≤ Ci ψ H 1/2 (Γi ) , where Ci are independent of ψ. Proof By (3) of Assumption 5.1 if Γ10 ∪ Γ11 = ∅, then, for example, Γ2 = ∅ and there exists a diffeomorphism y = f i (x) ∈ C 1 from Γi onto Γ2 . Define ϕ(y) at point y ∈ Γ2 corresponding to point x ∈ Γi by ϕ(y) = D f1(x) ψ( f i−1 (y)), where D f i (x) is Jacobian of the transformation f i . Then, 



1 ψ( f i−1 (y))D f i (x) d x = D f i (x)

 ψ(x) d x,

(5.29)

   1   ψ(x) H 21 (Γ ) ≤ ci ψ(x) H 21 (Γ ) . ϕ(y) H 21 (Γ ) ≤   D f (x)  2 i i i C(Γi )

(5.30)

Γ2

ϕ(y) dy =

Γi

Γi

and

When ψ ∈ C0∞ (Γi ), define a function φ ∈ H1/2 (∂Ω) on ∂Ω as follows: φ × n|Γ2 ∪Γi = 0, φ n |Γ2 = −ϕ, φ n |Γ10 = ψ, φ|(∪i=1,3−9,11 Γi ) = 0. Thus, by (5.29) Stokes problem

 ∂Ω

φ n ds = 0. Then, there exists a solution u ∈ W1,2 (Ω) to the ⎧ ⎪ ⎨ − Δu + ∇ p = 0, div u = 0, ⎪ ⎩ u|∂Ω = φ

and uV ≤ cφH1/2 (∂Ω) . (cf. Theorem IV.1.1 of [1]). Taking into account (5.30), we come to the asserted  estimation with Ci = 1 + ci . Thus u is the asserted function. Problem I-VE and Problem I-VI are equivalent in the following sense: Theorem 5.2 If (v, στ , σn , σ+n , σ−n ) is a solution to Problem I-VE, then v is a solution to Problem I-VI.

160

5 The Steady Navier-Stokes System with Friction Boundary Conditions

Conversely, if v is a solution to Problem I-VI, then there exist στ , σn , σ+n , σ−n such that (v, στ , σn , σ+n , σ−n ) is a solution to Problem I-VE. Proof We already showed that if (v, στ , σn , σ+n , σ−n ) is a solution to Problem I-VE, then v is a solution to Problem I-VI. Thus, it remains to prove the second assertion of theorem. Since the functional Φ is proper, we have from (5.28) v − U = w ∈ K (Ω)

(5.31)

because if w ∈ / K (Ω), then the left hand side of (5.28) is −∞, which is a contradiction to the fact that the right-hand side is finite. Let ψ ∈ V8−11 ≡ {u ∈ V : u|Γ8 ∪Γ9 ∪Γ10 ∪Γ11 = 0} (⊂ K (Ω)). Putting u = w + ψ, u = w − ψ and taking into account φτ (w) = φτ (w + ψ), φn (w) = φn (w + ψ), φ+ (w) = φ+ (w + ψ), φ− (w) = φ− (w + ψ), we get from (5.22) and (5.28) a01 (w, ψ) + a11 (w, w, ψ) ≥ F1 , ψ, a01 (w, −ψ) + a11 (w, w, −ψ) ≥ F1 , −ψ ∀ψ ∈ V8−11 , which imply a01 (w, ψ) + a11 (w, w, ψ) = F1 , ψ ∀ψ ∈ V8−11 .

(5.32)

When u ∈ V10−11 ≡ {u ∈ V : u|Γ10 ∪Γ11 = 0} (⊂ K (Ω)), the set {(u|Γ8 , u n |Γ9 )} is a subspace of L2τ (Γ8 ) × L 2 (Γ9 ), where u n |Γ9 is u|Γ9 · n. Define a functional σ ∗ on the set by 

 σ ∗ , (u|Γ8 , u n |Γ9 ) = a01 (w, u) + a11 (w, w, u) − F1 , u ∀u ∈ V10−11 .

(5.33)

This functional is well defined. Because if u, u 1 ∈ V10−11 are such that (u|Γ8 , u|Γ9 ) = (u 1 |Γ8 , u 1 |Γ9 ), then since u − u 1 ∈ V8−11 , by (5.32) a01 (w, u − u 1 ) + a11 (w, w, u − u 1 ) − F1 , u − u 1  = 0, that is, a01 (w, u) + a11 (w, w, u) − F1 , u = a01 (w, u 1 ) + a11 (w, w, u 1 ) − F1 , u 1 ,

5.1 Variational Formulations of Problems

and so by (5.33)

161

    ∗ σ , (u|Γ8 , u n |Γ9 ) = σ ∗ , (u 1 |Γ8 , u 1n |Γ9 ) .

This function is linear. Putting u = w + ψ, where ψ ∈ V10−11 , and taking into account φ+ (w + ψ) = φ+ (w), φ− (w + ψ) = φ− (w), we have from (5.33) and (5.28)   − σ ∗ , (ψ|Γ8 , ψn |Γ9 ) = − [a01 (w, ψ) + a11 (w, w, ψ) − F1 , ψ] ≤ Φ(w + ψ) − Φ(w) (5.34) = φτ (w + ψ) − φτ (w) + φn (w + ψ) − φn (w)   ≤ gτ |ψ|Γ8 d x + gn |ψ|Γ9 d x ∀ψ ∈ V10−11 . Γ8

Γ9

Putting u = w − ψ, in the same way we have  ∗  σ , (ψ|Γ8 , ψn |Γ9 ) = [a01 (w, ψ) + a11 (w, w, ψ) − F1 , ψ] ≤ φτ (w − ψ) − φτ (w) + φn (w − ψ) − φn (w)   ≤ gτ |ψ|Γ8 d x + gn |ψ|Γ9 d x ∀ψ ∈ V10−11 . Γ8

(5.35)

Γ9

By (5.34) and (5.35), we can see that σ ∗ is a bounded linear functional with a norm not greater than 1 on a subspace of L1gτ (Γ8 ) × L 1gn (Γ9 ), where L1gτ (Γ8 ), L 1gn (Γ9 ) are, respectively, the spaces of functions integrable with weights gτ , gn on Γ8 and Γ9 . By the Hahn-Banach theorem the functional can be extended as a functional on L1gτ (Γ8 ) × L 1gn (Γ9 ) norms of which is not greater than 1. Therefore, there exist elements στ ∈ L∞1 (Γ8 ) with στ L∞1 (Γ8 ) ≤ 1 and σn ∈ L ∞1 (Γ9 ) with σn  L ∞1 (Γ9 ) ≤ gτ

1, which imply

gτ

gn

gn

|στ | ≤ gτ , |σn | ≤ gn

(5.36)

and 

 σ ∗ , (u|Γ8 , u n |Γ9 ) = στ , u|Γ8 Γ8 + σn , u n |Γ9 Γ9

∀u ∈ V10−11 . 1

(5.37) 1

When u ∈ V, the set {(u n |Γ10 , u n |Γ11 )} is a subspace of H 2 (Γ10 ) × H 2 (Γ11 ). Define a functional σ1∗ on the set V by

162

5 The Steady Navier-Stokes System with Friction Boundary Conditions



 σ1∗ , (u n |Γ10 , u n |Γ11 ) =

a01 (w, u) + a11 (w, w, u) − στ , u|Γ8 Γ8 − σn , u|Γ9 Γ9 − F1 , u ∀u ∈ V. (5.38) This functional is also well defined. Because if u, u 1 ∈ V are such that (u Γ10 , u|Γ11 ) = (u 1 |Γ10 , u 1 |Γ11 ), then since u − u 1 ∈ V10−11 (Ω), by (5.33) and (5.37)   a01 (w, u − u 1 ) + a11 (w, w, u − u 1 ) − στ , (u − u 1 )|Γ8 Γ8   − σn , (u − u 1 )|Γ9 − F1 , u − u 1  Γ9       ∗ 1 − σn , (u − u 1 )|Γ9 = σ , ((u − u )|Γ8 , (u − u 1 )|Γ9 ) − στ , (u − u 1 )|Γ8 Γ8

Γ9

= 0,

and so by (5.38)     ∗ σ1 , (u n |Γ10 , u n |Γ11 ) = σ1∗ , (u 1n |Γ10 , u 1n |Γ11 ) . The functional σ1∗ is linear. Let us prove next its continuity. Let u be the function corresponding to ψ ∈ C0∞ (Γ10 ) by Lemma 5.1. Then, by Lemma 5.1 we have from (5.38)   | σ1∗ , (ψ, 0) |   ≤ C wV uV + w2V uV + (στ  L 2 (Γ8 ) + στ  L 2 (Γ9 ) )u + F1 V∗ uV   ≤ C wV  + w2V + (στ  L 2 (Γ8 ) + στ  L 2 (Γ9 ) ) + F1 V∗ · ψ H 21 (Γ ) . 10 (5.39) Also assuming that u is the function corresponding to ψ ∈ C0∞ (Γ11 ) by Lemma 5.1, we have   | σ1∗ , (0, ψ) |   ≤ C wV uV + w2V uV + (στ  L 2 (Γ8 ) + στ  L 2 (Γ9 ) )u + F1 V∗ uV   ≤ C wV + w2V + (στ  L 2 (Γ8 ) + στ  L 2 (Γ9 ) ) + F1 V∗ · ψ H 21 (Γ ) . 11 (5.40) 1 1/2 Since H0 (Γi ) = H 2 (Γi ), i = 10, 11, (cf. (1.11)), (5.39) and (5.40) show that the 1 1 functional σ1∗ is continuous on the subspace of H 2 (Γ10 ) × H 2 (Γ11 ) mentioned above. Thus, by the Hahn-Banach theorem the functional is extended as a functional 1 1 on H 2 (Γ10 ) × H 2 (Γ11 ). Therefore, there exists an element (σ+n , σ−n ) ∈ H −1/2 (Γ10 ) × H −1/2 (Γ11 ) such that       ∗ σ1 , (u|Γ10 , u|Γ11 ) = σ+n , u|Γ10 Γ10 + σ−n , u|Γ11 Γ11

∀u ∈ V.

(5.41)

When ψ ≥ 0 is such that ψ ∈ C0∞ (Γ10 ), let u ∈ K (Ω) be the function asserted in Lemma 5.1. Putting u = w + u, by (5.28) we have

5.1 Variational Formulations of Problems

163

a01 (w, u) + a11 (w, w, u) + Φ(w + u) − Φ(w) − F1 , u ≥ 0.

(5.42)

On the other hand, by (5.38), (5.41) and property of u, a01 (w, u) + a11 (w, w, u) − F1 , u = σ+n , ψΓ10 and so from (5.42) we have that σ+n , ψΓ10 + Φ(w + u) − Φ(w) ≥ 0.

(5.43)

By (5.21), (5.22) and the property of u, we get Φ(w + u) − Φ(w) = g+n , ψΓ10 , which combined with (5.43) gives σ+n , ψΓ10 + (g+n , ψ)Γ10 ≥ 0, that is, σ+n + g+n ≥ 0.

(5.44)

When ψ ≤ 0 is such that ψ ∈ C0∞ (Γ11 ), let u ∈ K (Ω) be the function asserted in Lemma 5.1. Then, in the same way we have that σ−n , −ψΓ11 − (g−n , −ψ)Γ11 ≥ 0, that is, σ−n − g−n ≤ 0.

(5.45)

From (5.38) and (5.41), we have a01 (w, u) + a11 (w, w, u) − (στ , u τ )Γ8 − (σn , u)Γ9 − σ+n , uΓ10 − σ−n , uΓ11 = F1 , u ∀u ∈ V. (5.46) Putting u = 0 in (5.28) and taking into account (5.46) with u = w, we have (στ , w)Γ8 + (σn , w)Γ9 + σ+n , wn Γ10 + σ−n , wn Γ11 + φτ (w) + φn (w) + φ+ (w) + φ− (w) ≤ 0, that is, 

 Γ8

(στ wτ + gτ |wτ |) ds +

Γ9

(σn wn + gn |wn |) ds

+ σ+n + g+n , wn Γ10 + σ−n − g−n , wn Γ11 ≤ 0.

(5.47)

164

5 The Steady Navier-Stokes System with Friction Boundary Conditions

Since on Γ8 , Γ9 , Γ10 and Γ11 , respectively, wτ = vτ , wn = vn , wn = vn ≥ 0 and wn = vn ≤ 0, taking into account (5.36), (5.44), (5.45), by (5.47) we have στ vτ + gτ |vτ | = 0, σn vn + gn |vn | = 0, σ+n + g+n , vn  = 0, σ−n − g−n , vn  = 0.

(5.48)

Therefore, by virtue of (5.31), (5.36), (5.44)–(5.46) and (5.48), we come to the conclusion. 

5.1.2 Variational Formulation: The Case of Total Pressure Taking (v · ∇)v = rot v × v + 21 grad|v|2 into account and putting v = w + U , by (5.6), (5.7) and Assumption 5.1 we can see that solutions (v, p) of the problem (5.1), (5.3) in the sense of Definition 5.1 satisfy the following: ⎧ v − U = w ∈ K (Ω), ⎪ ⎪ ⎪ ⎪ ⎪ 2μ(E(w), E(u)) + rot w × w, u + rot U × w, u ⎪ ⎪ ⎪ ⎪ ⎪ + rot w × U, u + 2μ(k(x)w, u)Γ2 + 2μ(S w, ˜ u) ˜ Γ3 ⎪ ⎪ ⎪ ⎪ ⎪ + 2(α(x)w, u)Γ5 + μ(k(x)w, u)Γ7 − 2μ(εnτ (w + U ), u)Γ8 ⎪ ⎪ ⎪ ⎪

1 ⎪ ⎪ ⎪ + p + |v|2 − 2μεnn (w + U ), u n Γ9 ∪Γ10 ∪Γ11 ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ = −2μ(E(U ), E(u)) − rot U × U, u − 2μ(k(x)U, u)Γ2 ⎪ ⎪ ⎨ − 2μ(SU˜ , u) ˜ Γ3 − 2(α(x)Uτ , u)Γ5 − μ(k(x)U, u)Γ7 ⎪   ⎪ ⎪ ⎪ +  f, u + φi , u n Γi + φi , uΓi ⎪ ⎪ ⎪ ⎪ i=2,4,7 i=3,5,6 ⎪ ⎪ ⎪ ⎪ ⎪ ∀u ∈ V, ⎪ ⎪ ⎪ t t ⎪ ⎪ |σ (v)| ≤ g , σ · v + g |v | = 0 on Γ , τ τ τ τ 8 ⎪ τ ⎪ τ ⎪ t t ⎪ ⎪ |σ (v)| ≤ g , σ (v)v + g |v | = 0 on Γ n n n n 9, ⎪ n n ⎪ ⎪ t t ⎪ ⎪ σn (v) + g+n ≥ 0, (σn (v) + g+n )vn = 0 on Γ10 , ⎪ ⎪ ⎩ t σn (v) − g−n ≤ 0, (σnt (v) − g−n )vn = 0 on Γ11 . Define a02 (·, ·), a12 (·, ·, ·) and F2 ∈ V∗ , respectively, by

(5.49)

5.1 Variational Formulations of Problems

165

a02 (w, u) = 2μ(E(w), E(u)) + rot U × w, u + rot w × U, u + 2μ(k(x)w, u)Γ2 + 2μ(S w, ˜ u) ˜ Γ3 + 2(α(x)w, u)Γ5 + μ(k(x)w, u)Γ7 a12 (w, u, v) = rot w × u, v

∀w, u ∈ V, ∀w, u, v ∈ V,

F2 , u = −2μ(E(U ), E(u)) − rot U × U, u − 2μ(k(x)U, u)Γ2 − 2μ(SU˜ , u) ˜ Γ3 − 2(α(x)Uτ , u)Γ5 − μ(k(x)U, u)Γ7   +  f, u + φi , u n Γi + φi , uΓi ∀u ∈ V. i=2,4,7

(5.50)

i=3,5,6

Then, taking into account στt (v) = 2μεnτ (v), 1 σnt (v) = −( p + |v|2 ) + 2μεnn (v) 2 and (5.49), we introduce the following variational formulation for the problem (5.1), (5.3).

t t , σ−n ) ∈ U + K (Ω) × L2τ (Γ8 ) × L 2 (Γ9 ) × Problem II-VE. Find (v, στt , σnt , σ+n 1 1 H − 2 (Γ10 ) × H − 2 (Γ11 ) such that ⎧ v − U = w ∈ K (Ω), ⎪ ⎪ ⎪ ⎪ ⎪ a02 (w, u) + a12 (w, w, u) − (στt , u τ )Γ8 − (σnt , u n )Γ9 ⎪ ⎪ ⎪    t  t ⎪ ⎪ − σ+n , u n Γ10 − σ−n , u n Γ11 = F2 , u ∀u ∈ V, ⎪ ⎪ ⎨ |στt | ≤ gτ , στt · vτ + gτ |vτ | = 0 on Γ8 , ⎪ ⎪ ⎪ |σnt | ≤ gn , σnt vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪   t ⎪ t ⎪ σ+n + g+n ≥ 0, σ+n + g+n , vn Γ10 = 0 on Γ10 , ⎪ ⎪ ⎪ ⎪   t ⎩ t σ−n − g−n ≤ 0, σ−n − g−n , vn Γ11 = 0 on Γ11 .

(5.51)

In the same way as in Theorem 5.1 we have Theorem 5.3 Let Assumption 5.1 hold. If (v, p) is a solution in the sense of Definition 5.1 of the problem (5.1), (5.3), then (v, στt |Γ8 , σnt |Γ9 , σnt |Γ10 , σnt |Γ11 ) is a solution to Problem II-VE. Conversely, if f ∈ L2 (Ω) and Problem II-VE has a smooth solution (v, στt , σnt , t σ+ , σ−t ) such that v ∈ H2 (Ω), σ+t ∈ L 2 (Γ10 ) and σ−t ∈ L 2 (Γ11 ), then there exists p ∈ H 1 (Ω) such that (v, p) is a solution to the problem (5.1), (5.3). Moreover, if at least one of the sets Γi , i = 2, 4, 6, 7, is nonempty, then p is unique. Then, in the same way as in Problem I we get Problem II-VI (the case of total pressure) formulated by a variational inequality and can prove that the problem is equivalent to Problem II-VE.

166

5 The Steady Navier-Stokes System with Friction Boundary Conditions

Problem II-VI. Find v = w + U such that a02 (w, u − w) + a12 (w, w, u − w) + Φ(u) − Φ(w) ≥ F2 , u − w ∀u ∈ V, (5.52) where a02 , a12 , F2 are in (5.50) and Φ is defined by (5.22). t t Theorem 5.4 If (v, στt , σnt , σ+n , σ−n ) is a solution to Problem II-VE, then v is a solution to Problem II-VI. t t , σ−n Conversely, if v is a solution to Problem II-VI, then there exist στt , σnt , σ+n t t t t such that (v, στ , σn , σ+n , σ−n ) is a solution to Problem II-VE.

5.1.3 Variational Formulation: The Stokes Problem In the same way as in Problem I we get the following equivalent formulations of Problem III for the Stokes equations with boundary condition (5.2).

Problem III-VE. Find (v, στ , σn , σ+n , σ−n ) ∈ U + K (Ω) × L2τ (Γ8 ) × L 2 (Γ9 ) × 1 1 H − 2 (Γ10 ) × H − 2 (Γ11 ) such that ⎧ v − U = w ∈ K (Ω), ⎪ ⎪ ⎪ ⎪ ⎪ a03 (w, u) − (στ , u τ )Γ8 − (σn , u n )Γ9 − σ+n , u n Γ10 − σ−n , u n Γ11 = F3 , u ⎪ ⎪ ⎪ ⎪ ⎪ ∀u ∈ V, ⎪ ⎨ |στ | ≤ gτ , στ · vτ + gτ |vτ | = 0 on Γ8 , ⎪ ⎪ ⎪ |σn | ≤ gn , σn vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪ ⎪ ⎪ σ+n + g+n ≥ 0, σ+n + g+n , vn  = 0 on Γ10 , ⎪ ⎪ ⎪ ⎩ σ−n − g−n ≤ 0, σ−n − g−n , vn  = 0 on Γ11 , (5.53) where a03 (w, u) = 2μ(E(w), E(u)) + 2μ(k(x)w, u)Γ2 + 2μ(S w, ˜ u) ˜ Γ3 + 2(α(x)w, u)Γ5 + μ(k(x)w, u)Γ7 ∀w, u ∈ V, F3 , u = −2μ(E(U ), E(u)) − 2μ(k(x)U, u)Γ2 − 2μ(SU˜ , u) ˜ Γ3 − 2(α(x)Uτ , u)Γ5   − μ(k(x)U, u)Γ7 +  f, u + φi , u n Γi + φi , uΓi ∀u ∈ V. i=2,4,7

i=3,5,6

(5.54) Problem III-VI Find v such that v − U = w ∈ K (Ω), a03 (w, u − w) + Φ(u) − Φ(w) ≥ F3 , u − w ∀u ∈ V, where the functionals Φ is defined by (5.22).

(5.55)

5.1 Variational Formulations of Problems

167

Remark 5.2 Using the strain bilinear form and applying Theorems 3.1 and 3.2 on Γi j , i = 2, 3, 7, we embedded all the boundary conditions into variational formulations of problems. For this, we used condition Γi j ∈ C 2 , i = 2, 3, 7. The condition Γi j ∈ C 1 is for the existence of a diffeomorphism in C 1 between Γi and Γ j in (3) of Assumption 5.1.

5.2 Solutions to Variational Inequalities In this section we study some variational inequalities obtained in Sect. 5.1. We denote by X →→ Y compact imbedding of a space X into Y . Theorem 5.5 Let X, X 1 be real separable Hilbert spaces such that X →→ X 1 , and X ∗ be dual space of X . Assume the followings: (1) Φ : X → [0, +∞] is a proper lower semi-continuous convex functional such that Φ(0 X ) = 0. (2) a0 (·, ·) ∈ (X × X → R) is a bilinear form such that |a0 (u, v)| ≤ K u X v X ∀u, v ∈ X, a0 (u, u) ≥ αu2X ∃α > 0, ∀u ∈ X. (3) a1 (·, ·, ·) ∈ (X 1 × X × X → R) is a triple linear functional such that a1 (w, u, u) = 0 ∀w ∈ X 1 , ∀u ∈ X, |a1 (w, u, v)| ≤ K w X 1 u X v X , ∀w ∈ X 1 , ∀u, v ∈ X. Then for each f ∈ X ∗ there exists a solution to the variational inequality a0 (v, u − v) + a1 (v, v, u − v) + Φ(u) − Φ(v) ≥  f, u − v ∀u ∈ X

(5.56)

and all solutions v satisfy the estimate v X ≤

1  f X ∗ . α

(5.57)

If moreover Kc  f  X ∗ < 1, α2

(5.58)

then solution is unique, where c is the constant in  ·  X 1 ≤ c ·  X . Proof Fix w ∈ X 1 . Let us consider the following variational inequality: a0 (v, u − v) + a1 (w, v, u − v) + Φ(u) − Φ(v) ≥  f, u − v ∀u ∈ X,

(5.59)

168

5 The Steady Navier-Stokes System with Friction Boundary Conditions

where f ∈ X ∗ . From existing results there exists a unique solution to (5.59) (see Theorem 10.5 of [2]). Let v1 , v2 be the solutions corresponding to f 1 , f 2 instead of f . Then, under consideration of condition (2) it is easy to verify that v1 − v2  X ≤

1  f1 − f2  X ∗ . α

(5.60)

Now, let us consider the operator which maps w to the solution v of (5.59) T ∈ (X 1 → X ) : w → T (w) = v. Taking into account condition (1), we can easily verify that the solution corresponding to f = 0 X ∗ is 0 X . Thus, we have from (5.60) v X ≤

1  f  X ∗ ∀w ∈ X 1 . α

(5.61)

Note that this estimate is independent of w. Denote by v1 and v2 , respectively, the solutions to (5.59) corresponding to w1 and w2 . Then a0 (v1 , u − v1 ) + a1 (w1 , v1 , u − v1 ) + Φ(u) − Φ(v1 ) ≥  f, u − v1  ∀u ∈ X, a0 (v2 , u − v2 ) + a1 (w2 , v2 , u − v2 ) + Φ(u) − Φ(v2 ) ≥  f, u − v2  ∀u ∈ X. (5.62) Putting u = v2 and u = v1 , respectively, in the first formula and the second one of (5.62), and adding two formulas, we get a0 (v1 − v2 , v2 − v1 ) + a1 (w1 , v1 , v2 − v1 ) + a1 (w2 , v2 , v1 − v2 ) ≥ 0.

(5.63)

From (5.63), conditions (2), (3) of Theorem and (5.61), we get 1 |a1 (w1 , v1 , v2 − v1 ) − a1 (w2 , v1 , v2 − v1 ) α + a1 (w2 , v1 , v2 − v1 ) − a1 (w2 , v2 , v2 − v1 )| 1 1 ≤ |a1 (w1 − w2 , v1 , v2 − v1 )| + |a1 (w2 , v2 − v1 , v2 − v1 )| α α K ≤ w1 − w2  X 1 v1  X v2 − v1  X α K  f X ∗ ≤ w1 − w2  X 1 v2 − v1  X ∀w1 , w2 ∈ X 1 , α2

v2 − v1 2X ≤

which implies v2 − v1  X ≤

K  f X ∗ w1 − w2  X 1 ∀w1 , w2 ∈ X 1 . α2

(5.64)

5.2 Solutions to Variational Inequalities

169

By (5.61), (5.64) and Schauder fixed-point theorem (Theorem 1.6), there exists a solution to (5.56). And any solution is a fixed point of operator T, and by (5.61) all solutions satisfy the estimate (5.57). If (5.58) holds, then the operator T : w ∈ X → v ∈ X is contract, and so we come to the last conclusion.  Let us next study variational inequalities when condition (3) of the above theorem is weakened. Theorem 5.6 Let X be a real separable Hilbert space. Assume the followings: (1) Condition (1) of Theorem 5.5 holds. (2) Condition (2) of Theorem 5.5 holds. (3) a1 (·, ·, ·) ∈ (X × X × X → R) is a triple linear functional such that |a1 (w, u, v)| ≤ K w X u X v X , ∀w, u, v ∈ X. If f is small enough, then in O M (0 X ), where M is determined in (5.73), there exists a unique solution to the variational inequality a0 (v, u − v) + a1 (v, v, u − v) + Φ(u) − Φ(v) ≥  f, u − v ∀u ∈ X.

(5.65)

Proof Fix w ∈ X . Let us consider a variational inequality a0 (v, u − v) + a1 (w, w, u − v) + Φ(u) − Φ(v) ≥  f, u − v ∀u ∈ X, (5.66) where f ∈ X ∗ . Defining an element a1 (w) ∈ X ∗ by a1 (w), u = a1 (w, w, u) ∀u ∈ X, by condition (3) we have a1 (w) X ∗ ≤ K w2X ∀w ∈ X.

(5.67)

Then, (5.66) can be rewritten as follows: a0 (v, u − v) + Φ(u) − Φ(v) ≥  f − a1 (w), u − v ∀u ∈ X.

(5.68)

By the same argument as in the proof of Theorem 5.5, there exists a unique solution vw to (5.68) and vw  ≤

1 1 ( f  X ∗ + a1 (w) X ∗ ) ≤ ( f  X ∗ + K w2X ), α α

where (5.67) was used and α is the one of Theorem 5.5. Now, let us consider the operator which maps w to the solution of (5.68)

(5.69)

170

5 The Steady Navier-Stokes System with Friction Boundary Conditions

T ∈ (X → X ) : w → T (w) = v. Denote by v1 and v2 , respectively, the solutions to (5.66) corresponding to w1 , w2 ∈ O M (0 X ), where M is to be determined below. Then v1 − v2  X ≤

1 a1 (w1 ) − a1 (w2 ) X ∗ . α

(5.70)

By condition (3) a1 (w1 ) − a1 (w2 ) X ∗ ≤ K (w2 − w1  X w2  X + w1  X w1 − w2  X ) . (5.71) Thus, by (5.70) and (5.71), K (w2 − w1  X w2  X + w1  X w1 − w2  X ) α 2K M w2 − w1  X ∀w1 , w2 ∈ O M (0 X ). ≤ α

v1 − v2  X ≤

Therefore, if M is taken such that ⎧ 1 ⎪ ⎨ M = ( f  X ∗ + K M 2 ), α ⎪ ⎩ 2K M < 1 α

(5.72)

(5.73)

(If α is large and  f  X ∗ is small enough, then such a choice is possible.), then by (5.69) and (5.72) the operator T on O M (0 X ) is contract, and so there exists a unique solution to (5.65).  Theorem 5.7 Let X be a real separable Hilbert space and X ∗ be its dual space. Assume that (1) Φ : X → R is a finite weak continuous convex functional, Φε : X → R is convex such that Φε (v) → Φ(v) uniformly on X as ε → 0, Gâteaux derivative DΦε ≡ Aε ∈ (X → X ∗ ) is weak continuousand Aε (0 X ) = 0 X [∗ . (2) a(·, ·, ·) ∈ (X × X × X → R) is a form such that when w ∈ X, (u, v) → a(w; u, v) is bilinear on X × X, a(v, v, v) ≥ αv2X ∃α > 0, ∀v ∈ X and when vm  v in X, a(vm , vm , u) → a(v, v, u) ∀u ∈ X and lim inf a(vm , vm , vm ) ≥ a(v, v, v). m→∞

5.2 Solutions to Variational Inequalities

171

Then, for f ∈ X ∗ there exists a solution to a variational inequality a(v, v, u − v) + Φ(u) − Φ(v) ≥  f, u − v ∀u ∈ X

(5.74)

satisfying the estimate v X ≤

1  f X ∗ . α

(5.75)

Proof First let us prove the existence of a solution to a variational equation a(v, v, u) + Aε (v), u =  f, u ∀u ∈ X.

(5.76)

We will rely on Theorem 1.43. Let {wn } be a basis of X and denote by X m the subspace of X spannedby w1 , · · · , wm . m μi wi ∈ X m satisfying We will find vm = i=1 a(vm , vm , u) + Aε (vm ), u =  f, u ∀u ∈ X m .

(5.77)

Define Fm ∈ (X m → X m ) by (Fm (v), wi ) = a(v, v, wi ) + Aε (v), wi  −  f, wi , 1 ≤ i ≤ m.

(5.78)

Since Gâteaux derivative of finite convex functional is monotone (see Remark 1.16) and Aε (0 X ) = 0 X ∗ , we get Aε (u) − Aε (0 X ), u − 0 X  = Aε (u), u ≥ 0 ∀u ∈ X. Thus, a(u, u, u) + Aε (u), u ≥ αu2X ∀u ∈ X.

(5.79)

Combining (5.78) with (5.79) yields (Fm (v), v) ≥ (αv X −  f  X ∗ )v X ∀v ∈ X m . Therefore, (Fm (v), v) ≥ 0 ∀v ∈ X with v X =

(5.80)

 f X ∗ . α

And by virtue of assumption (2), Fm is continuous in X m . Thus, there exists a solution vεm to problem (5.77). By (5.80), we have that 0 = (Fm (vεm ), vεm ) ≥ (αvεm  X −  f  X ∗ )vεm  X for all solution vεm to (5.77), which implies

172

5 The Steady Navier-Stokes System with Friction Boundary Conditions

vεm  X ≤

1  f X ∗ . α

(5.81)

Note this estimation is independent of ε, m. Thus, from {vεm } we can extract a subsequence {vεm p } such that vεm p  vε in X as p → +∞. By the assumptions of theorem, a(vεm p , vεm p , u) + Aε (vεm p ), u → a(vε , vε , u) + Aε (vε ), u ∀u ∈ X. (5.82) By virtue of (5.77), (5.82) and (5.81), we see that vε is a solution to (5.76) and satisfies 1 (5.83) vε  X ≤  f  X ∗ . α Subtracting the following two formulas which are got from (5.76) a(vε , vε , u) + Aε (vε ), u =  f, u ∀u ∈ X, a(vε , vε , vε ) + Aε (vε ), vε  =  f, vε  and taking into account that Φε (u) − Φε (vε ) ≥ Aε (vε ), u − vε  which is due to convexity of Φε , we come to the following inequality: a(vε , vε , u − vε ) + Φε (u) − Φε (vε ) ≥  f, u − vε  ∀u ∈ X.

(5.84)

By (5.83) we can choose {vεk } such that vεk  v ∗ in X as εk → 0.

(5.85)

By virtue of assumption (1), |Φεk (vεk ) − Φ(v ∗ )| ≤ |Φεk (vεk ) − Φ(vεk )| + |Φ(vεk ) − Φ(v ∗ )| → 0 as εk → 0, and so

Φεk (vεk ) → Φ(v ∗ ) as εk → 0.

(5.86)

Φεk (u) → Φ(u) ∀u ∈ X as εk → 0.

(5.87)

Also

By virtue of assumption (2),

5.2 Solutions to Variational Inequalities

173

a(vεk , vεk , u) → a(v ∗ , v ∗ , u) ∀u ∈ X as εk → 0, lim inf a(vεk , vεk , vεk ) ≥ a(v ∗ , v ∗ , v ∗ ).

(5.88)

k→∞

Taking into account (5.86)–(5.88), we get from (5.84) a(v ∗ , v ∗ , u − v ∗ ) + Φ(u) − Φ(v ∗ ) ≥  f, u − v ∗  ∀u ∈ X. By (5.83) we have v ∗  X ≤

1  f X ∗ . α

Therefore, we finish our proof.

(5.89) 

Remark 5.3 The estimate of solutions in Theorem 5.5 is for all solutions of the problem, but the one in Theorem 5.7 is for the solution whose existence is guaranteed by the theorem.

5.3 Existence and Uniqueness of Solutions to the Steady Navier-Stokes Problems In this section, on the basis of the results in Sect. 5.2, we study the problems formulated in Sect. 5.1. Our first result is the following: Theorem 5.8 Let Assumption 5.1 hold, the surfaces Γ2 j , Γ3 j , Γ7 j be convex (see Definition 3.2), α positive and U H1 (Ω) small enough. Then, when f and φi , i = 2, · · · , 7, are small enough, in a neighborhood of U in H1 (Ω) there exists a unique solution to Problem I-VI (for the steady Navier-Stokes problem of the case of static pressure). Proof Define a functional Φ by (5.22). Trace operator is continuous and sum of convex functions is also convex. Thus, the functional satisfies condition (1) of Theorem 5.6. Set w = v − U , where U is the function in Assumption 5.1. Let a01 (·, ·), a11 (·, ·, ·) and F1 ∈ V∗ be the same as in (5.9). By Korn’s inequality 2μ(E(w), E(w)) ≥ δw2V , δ > 0.

(5.90)

On the other hand, applying Hölder’s inequality, for w ∈ V we have |(U · ∇)w, w + (w · ∇)U, w| ≤ γ w2V · U H1 (Ω) .

(5.91)

Therefore, if δ − γ U H1 (Ω) = β1 > 0, then by (5.90), (5.91), Assumption 5.1 and Lemma 3.1 we have

174

5 The Steady Navier-Stokes System with Friction Boundary Conditions

a01 (u, u) ≥ β1 u2V ∀u ∈ V.

(5.92)

|a01 (u, v)| ≤ cuV vV ∀u, v ∈ V.

(5.93)

It is easy to verify that

By (5.92) and (5.93), a0 (u, v) satisfies condition (2) of Theorem 5.6. By Hölder’s inequality we can see |a11 (w, u, v)| ≤ cwV uV vV ∀w, u, v ∈ V.

(5.94)

which means that a11 (w, u, v) satisfies condition (3) of Theorem 5.6. Also   F1 V∗ ≤ M1 U H1 + U 2H1 +  f V∗ + φi  i=2,4,7

H

− 21

(Γi )



+

i=3,5,6

φi 

H

− 21

(Γi )

 ,

(5.95) where M1 depends on mean curvature of Γ7 , shape operator of Γ3 , μ and α. By Theorem 5.6, if U H1 ,  f V∗ , φi  H − 21 (Γ ) , i = 2, 4, 7, and φi H− 21 (Γ ) , i = i i 3, 5, 6, are small enough, then there exists a unique solution w ∈ K (Ω) to a01 (w, u − w) + a11 (w, w, u − w) + Φ(u) − Φ(w) ≥ F1 , u − w ∀u ∈ K (Ω).

Since v = w + U is solution, we come to the asserted conclusion.

(5.96) 

Theorem 5.9 Let Assumption 5.1 hold, the surfaces Γ2 j , Γ3 j , Γ7 j be convex, α positive and U H1 (Ω) small enough. Then, for any f φi , i = 2 ∼ 7, there exists a solution v to Problem II-VI (the steady Navier-Stokes problem for the case of total pressure) in a neighborhood of U in H1 (Ω) and all solutions satisfy v − U H1 ≤

 M1 U H1 + U 2H1 +  f V∗ δ − γ U H1  (5.97)   + φi  H − 21 (Γ ) + φi H− 21 (Γ ) , i

i=2,4,7

i

i=3,5,6

where δ, γ , M1 are as in (5.98), (5.99), (5.107). If U H1 ,  f V∗ , φi  H − 21 (Γ ) , i = 2, 4, 7, and φi H− 21 (Γ ) , i = 3, 5, 6, are small i i enough, then the solution is unique. Proof Define a functional Φ(u) by (5.22). Then, Φ satisfies condition (1) of Theorem 5.5. Let a02 (·, ·), a12 (·, ·, ·) and F2 ∈ V∗ be as in (5.50). By Korn’s inequality

5.3 Existence and Uniqueness of Solutions to the Steady Navier-Stokes Problems

2μ(E(w), E(w)) ≥ δw2V , δ > 0.

175

(5.98)

On the other hand, for any w ∈ V we have rot U × w, w = 0, |rot w × U, w| ≤ γ w2V · U H1 (Ω) .

(5.99)

Therefore, if δ − γ U H1 (Ω) = β1 > 0, then by (5.98), (5.99), Assumption 5.1 and Lemma 3.1 we have (5.100) a02 (u, u) ≥ β1 u2V ∀u ∈ V. It is easy to verify |a02 (u, v)| ≤ cuV vV ∀u, v ∈ V.

(5.101)

Then, (5.100) and (5.101) imply that a02 (u, v) satisfies condition (2) of Theorem 5.5. By a property of mixed product of vectors, 2

a12 (w, u, u) = rot w × u, u = 0 ∀w ∈ V 3 (Ω), ∀u ∈ V,

(5.102)

where   2 2 V 3 (Ω) = u ∈ H 3 (Ω) : div u = 0, u|Γ1 = 0, u τ |(∪i=2,4,7,9 Γi ) = 0, u · n|(∪i=3,5,8 Γi ) = 0 .

On the other hand, by density argument we get a12 (w, u, v) = rot w × u, v = −rot w, v × u.

(5.103)

1

When u, v ∈ V, v × u ∈ H 2 (Ω) and v × uH 13 (Ω) ≤ c1 v × uH 21 (Ω) ≤ cvV uV

(5.104)

(cf. Theorem 1.23). Also, if w ∈ V 3 (Ω), then rot w ∈ H− 3 (Ω) and 2

1

rot wH− 13 (Ω) ≤ cwH 23 (Ω) 1

(5.105)

1

(see Theorem 1.24). Since H03 (Ω) = H 3 (Ω) (see Theorem 1.15), by (5.103)– (5.105) we get 2

|a12 (w, u, v)| ≤ K wV 23 (Ω) uV vV ∀w ∈ V 3 (Ω) ∀u, v ∈ V.

(5.106)

176

5 The Steady Navier-Stokes System with Friction Boundary Conditions 2

2

Since V →→ V 3 (Ω), setting X = V, X 1 = V 3 (Ω), we can see that by (5.102) and (5.106) a11 (w, u, v) satisfies condition (3) of Theorem 5.5. Also, we have   F2 V∗ ≤ M1 U H1 + U 2H1 +  f V∗ + φi  i=2,4,7

H

− 21

(Γi )

+



φi 

i=3,5,6

 H

− 21

(Γi )

,

(5.107) where M1 depends on the mean curvature, shape operator, μ and α. Therefore, by Theorem 5.5, we have the existence and estimate of solutions to a02 (w, u − w) + a12 (w, w, u − w) + Φ(u) − Φ(w) ≥ F2 , u − w ∀u ∈ V. Since v = w + U is a solution to the given problem, we have the existence of solutions and the estimate (5.97). If U H1 ,  f V∗ , φi  H − 21 (Γ ) , i = 2, 4, 7, and φi H− 21 (Γ ) , i = 3, 5, 6, are small i i enough, then the solution is unique.  Let us consider now a special case of the Navier-Stokes problem with boundary condition (5.2) in which there is no flux across boundary except Γ1 , Γ8 . Theorem 5.10 Let Assumption 5.1 hold, Γi = ∅ (i = 2, 4, 6, 7, 9, 10, 11), the surfaces Γ3 j be convex, α positive and U H1 (Ω) small enough. Then, for any f and φi , i = 3, 5, there exists a solution v to Problem I-VI (the steady Navier-Stokes problem for the case of static pressure) and all solutions satisfy v − U H1 ≤

   M1 U 2H1 +  f V∗ + φi H− 21 (Γ ) , i δ − γ U H1 i=3,5

(5.108)

where δ, γ and M1 are as in (5.90), (5.91) and (5.95), respectively. In addition, if  f V∗ , φi H− 21 (Γ ) , i = 3, 5, are small enough, then the solution i is unique. Proof Define a functional Φ(u) = φτ (u) by (5.21) and (5.22). Then, the functional satisfies condition (1) of Theorem 5.6. Let w = v − U , U be a function in Assumption 5.1 and a01 (·, ·), a11 (·, ·, ·) and F1 ∈ V∗ be as in (5.9). We can see that condition (2) in Theorem 5.5 is satisfied (see proof of Theorem 5.8). By the condition of theorem, 2

a11 (w, u, u) = (w · ∇)u, u = 0 ∀w ∈ V 3 (Ω), ∀u ∈ V.

(5.109)

By Hölder’s inequality we can see 2

|a11 (w, u, v)| ≤ K wV 23 (Ω) uV vV ∀w ∈ V 3 (Ω), ∀u, v ∈ V.

(5.110)

5.3 Existence and Uniqueness of Solutions to the Steady Navier-Stokes Problems

177

By (5.109) and (5.110), a11 (w, u, v) satisfies condition (3) of Theorem 5.5. Applying Theorem 5.5 to a01 (w, u − w) + a11 (w, w, u − w) + Φ(u) − Φ(w) ≥ F1 , u − w ∀u ∈ K (Ω),



we come to the asserted conclusion.

Remark 5.4 Assumption Γi = ∅, i = 2, 4, 6, 7, 9 − 11, is only used to get (5.109). Applying Theorem 5.7, next we revisit the problem concerned within Theorem 5.10. We will get a generalization of methods used in papers based on smooth approximation of functional in variational inequalities (see [3]). Lemma 5.2 Let X, Y be reflexive Banach spaces, an operator i : X → Y be completely linear continuous, j : Y → R be convex and Gâteaux derivative D j (y) = a(y) for y ∈ Y . Then, Φ(v) ≡ j (iv) : X → R is convex, DΦ(v) ≡ A(v) = i ∗ a(iv), where i ∗ is the operator adjoint to i, and A : X → X ∗ is weak continuous. Proof It is easy to verify the convexity of Φ. Φ(v + tu) − Φ(u) j (i(v + tu)) − j (iu) = lim t→0 t t = a(iv), iuY = i ∗ a(iv), u X ∀v, u ∈ X,

A(v), u X = lim

t→0

which means A(v) = i ∗ a(iv). Let vn  v in X . Since gradient of a finite convex functional is monotone and demi-continuous (see Remark 1.15) and ivn → iv in Y , A(vn ), u X = i ∗ a(ivn ), u X = a(ivn ), iuY → a(iv), iuY = i ∗ a(iv), u X

that is, DΦ = A : X → X ∗ is weak continuous.

∀u ∈ X,



Theorem 5.11 Let Assumption 5.1 hold, Γi = ∅ (i = 2, 4, 6, 7, 9, 10, 11), the surfaces Γ3 j be convex, α positive and U H1 (Ω) small enough. Then, for any f and φi , i = 3, 5, there exists a solution v to Problem I-VI (the steady Navier-Stokes problem for the case of static pressure) and the solution satisfies the estimate (5.108). Proof Define an operator i : V → L2τ (Γ8 ) by iu = u|Γ8 and a functional Φ : V → R 1 by Φ(v) ≡ φτ (iv), where φτ is as in (5.21). Since the trace operator V → H 2 (∂Ω) 1 2 is continuous and H 2 (∂Ω) →→ L (∂Ω), the operator i is compact, and by Lemma 5.2 Φ : V → R is weak continuous and convex. Define a functional Φε : V → R by

178

5 The Steady Navier-Stokes System with Friction Boundary Conditions

Φε (v) = φτ ε (iv),  φτ ε (η) = gτ ρε (η) ds, Γ8  |η| − ε/2 |η| > ε, ρε (η) = |η|2 /2ε |η| ≤ ε . Since |φτ ε (η) − φτ (η)| ≤

(5.111)

ε |gτ | ∀η ∈ L2τ (Γ8 ) 2

(cf. Lemma 2.1 of [3]), we have |Φε (v) − Φ(v)| ≤

ε |gτ | ∀v ∈ V. 2

(5.112)

Also, φτ ε is convex, and so its Gâteaux derivative is demi-continuous. Thus, by Lemma 5.2 DΦε ≡ Aε ∈ (V → V∗ ) is weak continuous. By this fact together (5.112), condition (1) of Theorem 5.7 is satisfied. Under the assumption of theorem a01 (·, ·), a11 (·, ·, ·) and F1 ∈ V∗ of (5.9) are as follows: a01 (u, v) = 2μ(E(u), E(v)) + (U · ∇)u, v + (u · ∇)U, v + 2μ(S u, ˜ v) ˜ Γ3 + 2(α(x)u, v)Γ5 ∀u, v ∈ V, a11 (w, u, v) = (w · ∇)u, v ∀w, u, v ∈ V, F1 , u = −2μ(E(U ), E(u)) − (U · ∇)U, u − 2μ(SU˜ , u) ˜ Γ3  − 2(α(x)Uτ , u)Γ5 +  f, u + φi , uΓi ∀u ∈ V.

(5.113)

i=3,5

By Korn’s inequality, 2μ(E(u), E(u)) ≥ δu2V , δ > 0.

(5.114)

On the other hand, for any u ∈ V we have |(U · ∇)u, u + (u · ∇)U, u| ≤ γ u2V · U H1 (Ω) .

(5.115)

Therefore, if δ − γ U H1 (Ω) = β1 > 0, then by (5.114), (5.115), Assumption 5.1 and Lemma 3.1 we have a01 (u, u) ≥ β1 u2V ∀u ∈ V.

(5.116)

Under condition Γi = ∅, i = 2, 4, 6, 7, 9, 10, 11, it is easy to verify that a11 (v, v, v) = 0 ∀v ∈ V.

(5.117)

5.3 Existence and Uniqueness of Solutions to the Steady Navier-Stokes Problems

179

Let a(w, u, v) = a01 (u, v) + a11 (w, u, v). Then, by (5.116) and (5.117) we have a(v, v, v) ≥ β1 u2V ∀v ∈ V.

(5.118)

Let us prove that when vm  v in V, there exists a subsequence {vm p } such that a(vm p , vm p , u) → a(v; v, u) ∀u ∈ V as p → ∞.

(5.119)

To this end, first let us prove that when vm  v in V, there exists a subsequence {vm p } such that a01 (vm p , u) → a01 (v, u) ∀u ∈ V as p → ∞.

(5.120)

Since Ui u j ∈ L 2 (Ω), i, j = 1, 2, 3, and ∂i vm  ∂i v in L 2 (Ω), we have (U · ∇)vm , u → (U · ∇)v, u as m → ∞.

(5.121)

By Hölder’s inequality, |((vm − v) · ∇)U, u| ≤ cvm − vL3 (Ω) ∇U L2 (Ω) uL6 (Ω) . Since H 1 (Ω) →→ L 3 (Ω), we can choose a subsequence {vm p } such that vmp → v in L3 (Ω). Then, we have (vm p · ∇)U, u → (v · ∇)U, u

as m p → ∞.

(5.122)

It is easy to verify the convergence of other terms. Thus, using (5.121) and (5.122), we have (5.120). Using Hölder’s inequality and a11 (v, u, w) = −a11 (v, w, u), we have |a11 (vm , vm , u) − a11 (v, v, u)| ≤ |a11 (vm , vm , u) − a11 (v, vm , u)| + |a11 (v, vm , u) − a11 (v, v, u)|

≤ c vm − vL3 (Ω) ∇vm L2 (Ω) uL6 (Ω) + vL6 (Ω) ∇uL2 (Ω) vm − vL3 (Ω) ∀u ∈ V.

Thus, we have a11 (vm p , vm p , u) → a11 (v, v, u) ∀u ∈ V as m p → ∞. Combining (5.120) and (5.123) yields (5.119). Let us prove that

(5.123)

180

5 The Steady Navier-Stokes System with Friction Boundary Conditions

lim inf a(vm p , vm p , vm p ) ≥ a(v, v, v). m→∞

(5.124)

By lower semi-continuity of norm lim inf 2μ(E(vm ), E(vm )) ≥ 2μ(E(v), E(v)) as vm  v in V. m→∞

(5.125)

It is easy to prove that 2μ(S v˜m , u) ˜ Γ3 + 2(α(x)vm , u)Γ5 → 2μ(S v, ˜ u) ˜ Γ3 + 2(α(x)v, u)Γ5 ∀u ∈ V. (5.126) Using Hölder’s inequality and a11 (v, vm , u) = −a11 (v, u, vm ), we have |a11 (vm , vm , vm ) − a11 (v, v, v)| ≤ |a11 (vm , vm , vm ) − a11 (v, vm , vm )| + |a11 (v, vm , vm ) − a11 (v, vm , v)| + |a11 (v, vm , v) − a11 (v, v, v)| ≤ c vm − vL3 (Ω) ∇vm L2 (Ω) vm L6 (Ω) + vL6 (Ω) ∇vm L2 (Ω) vm − vL3 (Ω)

+ vL6 (Ω) ∇vL2 (Ω) vm − vL3 (Ω) ,

which implies a11 (vm p , vm p , vm p ) → a11 (v, v, v) as m p → ∞.

(5.127)

From (5.124)–(5.127), we have (5.124). By virtue of (5.118), (5.119) and (5.124), condition (2) of Theorem 5.7 is satisfied. Therefore, by Theorem 5.7 we have the existence and estimate of a solution w ∈ V to a01 (w, u − w) + a11 (w, w, u − w) + φτ (u) − φτ (w) ≥ F1 , u − w ∀u ∈ V. (5.128) Since v = w + U is a solution, we come to the asserted conclusion.  Remark 5.5 The estimate of solution of Theorem 5.11 may not be true for all solutions, and so Theorem 5.11 is weaker than Theorem 5.10. Let us consider Problem III for the Stokes system. Theorem 5.12 Let Assumption 5.1 hold, the surfaces Γ2 j , Γ3 j , Γ7 j be convex and α positive. Then, for any f φi , i = 2, · · · , 7, there exists a unique solution v to Problem III-VI for the steady Stokes problem with mixed boundary condition (5.2) and the following estimate holds: v − U H1 ≤

   M1  U H1 +  f V∗ + φi  H − 21 (Γ ) + φi H− 21 (Γ ) , i i δ i=2,4,7 i=3,5,6 (5.129)

5.3 Existence and Uniqueness of Solutions to the Steady Navier-Stokes Problems

181

where δ and M1 are, respectively, as in (5.98) and (5.107) (for F3 instead of F2 ). If v1 , v2 are solutions, respectively, to Problem-III-VI with gτ 1 , gn1 , g+n1 , g−n1 , f 1 , h i1 , φi1 and gτ 2 , gn2 , g+n2 , g−n2 , f 2 , h i1 , φi2 , then v1 − v2 H1 ≤

M1  U1 − U2 H1 +  f 1 − f 2 V∗ + gτ 1 − gτ 2 Lτ2 (Γ8 ) δ + gn1 − gn2  L 2 (Γ9 ) + g+n1 − g+n2  L 2 (Γ10 ) + g−n1 − g−n2  L 2 (Γ11 )    + U1 − U2 H1 , + φi1 − φi2  − 1 + φi1 − φi2  − 1 i=2,4,7

H

2

(Γi )

i=3,5,6

H

2

(Γi )

j

(5.130)

where U j , j = 1, 2, are the functions in Assumption 5.1 with h i instead h i . Proof By an argument similar to the proof of Theorem 5.6 we can apply the well known result for variational inequality (see Theorem 10.5 of [2]) a03 (w, u − w) + Φ(u) − Φ(w) ≥ F3 , u − w ∀u ∈ X,

(5.131)

where Φ(u) is defined by (5.22) and a03 (v, u), F3 are as in (5.54). Thus, we have the existence of a unique solution and estimate (5.129). If v1 = w1 + U1 , v2 = w2 + U2 are solutions corresponding to the given data, we get a03 (w1 , u − w1 ) + Φ1 (u) − Φ1 (w1 ) ≥ F31 , u − w1 , a03 (w2 , u − w2 ) + Φ2 (u) − Φ2 (w2 ) ≥ F32 , u − w2  ∀u ∈ V,

(5.132)

j

where Φ j (u), F3 , j = 1, 2, are the one corresponding to U j , gτ j , gn j , g+n j , g−n j , j j f j , h i , φi . Putting u = w2 , u = w1 , respectively, in the first and second inequality in (5.132) and adding those obtained, we have a03 (w1 − w2 , w2 − w1 ) + Φ1 (w2 )−Φ1 (w1 ) + Φ2 (w1 ) − Φ2 (w2 ) ≥ F31 − F32 , w2 − w1 .

(5.133)

By Korn’s inequality and Lemma 3.1 we have a03 (w1 − w2 , w1 − w2 ) ≥ δw1 − w2 2V .

(5.134)

From (5.133) and (5.134) we have w1 − w2 2V ≤

 1 1 |F3 − F32 , w2 − w1 | + |Φ1 (w2 ) − Φ1 (w1 ) + Φ2 (w1 ) − Φ2 (w2 )| . δ

(5.135)

Since w1 , w2 ∈ K (Ω),

182

5 The Steady Navier-Stokes System with Friction Boundary Conditions

  Φ1 (w2 ) − Φ1 (w1 ) = gτ 1 (|w2τ | − |w1τ |) ds + gn1 (|w2n | − |w1n |) ds Γ8 Γ9   + g+n1 (w2n − w1n ) ds − g−n1 (w2n − w1n ) ds, Γ Γ11  10  Φ2 (w2 ) − Φ2 (w1 ) = gτ 2 (|w2τ | − |w1τ |) ds + gn2 (|w2n | − |w1n |) ds Γ Γ  8  9 + g+n2 (w2n − w1n ) ds − g−n2 (w2n − w1n ) ds. Γ10

Γ11

(5.136)

Subtracting two formulas of (5.136), we have |Φ1 (w2 ) − Φ1 (w1 ) + Φ2 (w1 ) − Φ2 (w2 )| ≤ gτ 1 − gτ 2 L2 (Γ8 ) w2τ − w1τ L2 (Γ8 ) + gn1 − gn2  L 2 (Γ9 ) w2n − w1n  L 2 (Γ9 ) τ

τ

+ g+n1 − g+n2  L 2 (Γ10 ) wn2 − wn1  L 2 (Γ10 ) + g−n1 − g−n2  L 2 (Γ11 ) w2n − w1n  L 2 (Γ11 ) ≤ M gτ 1 − gτ 2 L2 (Γ8 ) + gn1 − gn2  L 2 (Γ9 ) + g+n1 − g+n2  L 2 (Γ10 ) τ

+ g−n1 − g−n2  L 2 (Γ11 ) w2 − w1 V .

(5.137)

By (5.135) and (5.137), we have w1 − w2 V ≤

M 1 F3 − F32 V∗ + gτ 1 − gτ 2 L2 (Γ8 ) + gn1 − gn2  L 2 (Γ9 ) τ δ

 + g+n1 − g+n2  L 2 (Γ10 ) + g−n1 − g−n2  L 2 (Γ11 ) ,

from which we get (5.130).



Remark 5.6 The estimates of solutions to the problems (5.97), (5.108) and (5.129) are independent of the thresholds gτ , gn , g+n , g−n . (See (8) of [4] and (25) of [3].)

5.4 Bibliographical Remarks The content of Chap. 5 is taken from [5]. Till now, for the Stokes and Navier-Stokes problems with friction type boundary conditions rather simple cases were studied. More clearly, problems with the Dirichlet boundary condition on a portion of boundary and either Tresca slip condition or the leak condition on the other portion have been dealt with. In [6] the existence of solutions to the steady Stokes and Navier-stokes equations with the homogeneous Dirichlet boundary condition on a portion of boundary and leak or threshold slip boundary condition on the other portion was studied. Also, [7– 9] concerned with the steady or non-steady Stokes equations with the homogeneous Dirichlet boundary condition and leak boundary condition.

5.4 Bibliographical Remarks

183

When a portion of boundary with Dirichlet boundary condition and another moving portion where nonlinear slip occurs are separated, the existence, uniqueness and continuous dependence on the data were studied for the steady Stokes equations in [9]. In [10] when a portion of boundary with Dirichlet boundary condition and another portion with slip condition are separated, the existence of strong solution to the steady Stokes equations was established. In [11] when a portion with homogeneous Dirichlet boundary condition and another portion with nonlinear boundary condition are separated, for the steady Stokes equations a relation between a regularized problem and the original problem, the regularity of solution were discussed. In [12] for the steady Navier-Stokes equations, the existence, uniqueness and continuous dependence on the data were considered when a portion of boundary with Dirichlet boundary condition and another moving portion where nonlinear slip occurs are separated. In [13] the existence and uniqueness of a local solution to the steady Navier-Stokes problem with homogeneous Dirichlet boundary condition and one of friction boundary conditions was investigated. In [4] the existence and uniqueness of solution to the steady rotating Navier-Stokes equations are studied when boundary consists of a portion with homogeneous Dirichlet boundary condition and other portions with a threshold slip. In [3] under similar boundary condition the steady Navier-Stokes problem is studied. With exception of [13] in all above-mentioned papers dealing the Navier-Stokes problem with friction boundary conditions, ones approximated the functionals in the considering variational inequalities with smooth one resulting to study of operator equation and it’s convergence. Numerical solution methods were studied for the Stokes and Navier-Stokes problems with friction boundary conditions. For the 2-D steady Stokes problems we refer to [14–18] and for the 3-D steady Stokes problems see [19]. For the 2-D steady Navier-Stokes problem we refer to [20–23].

References 1. G.P. Galdi, An Introduction to the Mathematical Theory of the Navier-Stokes Equations (Springer, Berlin, 2011) 2. C. Baiocchi, A. Capelo, Variational and Quasivariational Inequalities: Applications to FreeBoundary Problems (Wiley, Chichester, 1984) (Russian, 1988) 3. Y. Li, K. Li, Existence of the solution to stationary Navier-Stokes equations with nonlinear slip boundary conditions. J. Math. Anal. Appl. 381, 1–9 (2011) 4. R. An, K. Li, Variational inequality for the rotating Navier-Stokes equations with subdifferential boundary conditions. Comput. Math. Appl. 55, 581–587 (2008) 5. T. Kim, D. Cao, The steady Navier-Stokes and Stokes systems with mixed boundary conditions including one-sided leaks and pressure. Methods Appl. Anal. 23, 329–364 (2016) 6. H. Fujita, A mathematical analysis of motions of viscous incompressible fluid under leak or slip boundary conditions. RIMS Kokyuroku 888, 199–216 (1994) 7. H. Fujita, Non-stationary Stokes flows under leak boundary conditions of friction type. J. Comput. Math. 19, 1–8 (2001) 8. H. Fujita, A coherent analysis of Stokes flows under boundary conditions of friction type. J. Comput. Appl. Math. 149, 57–69 (2002)

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9. C.L. Roux, Steady Stokes flows with threshold slip boundary conditions. Math. Model. Methods Appl. Sci. 15, 1141–1168 (2005) 10. N. Saito, On the Stokes equations with the leak and slip boundary conditions of friction type: regularity of solutions. Publ. RIMS (Kyoto University) 40, 345–383 (2004) 11. N. Saito, H. Fujita, Regularity of solutions to the Stokes equation under a certain nonlinear boundary condition. Lect. Notes Pure Appl. Math. 223, 73–86 (2001) 12. C.L. Roux, A. Tani, Steady solutions of the Navier-Stokes equations with threshold slip boundary conditions. Math. Methods Appl. Sci. 30, 595–624 (2007) 13. F. Saidi, On the Navier-Stokes equations with the slip boundary conditions of friction type: regularity of solution. Math. Model. Anal. 12, 389–398 (2007) 14. M. Ayadi, M.K. Gdoura, T. Sassi, Mixed formulation for Stokes problem with Tresca friction. C. R. Acad. Sci. Paris Ser. I 348, 1069–1072 (2010) 15. T. Kashiwabara, On a finite element approximation of the Stokes problem under a leak boundary condition of friction type. Jpn. J. Ind. Appl. Math. 30, 227–261 (2013) 16. Y. Li, K. Li, Penalty finite element method for Stokes problem with nonlinear slip boundary conditions. Appl. Math. Comput. 204, 216–226 (2008) 17. Y. Li, K. Li, Locally stabilized finite element method for Stokes problem with nonlinear slip boundary conditions. J. Comput. Math. 28, 826–836 (2010) 18. Y. Li, K. Li, Uzawa iteration method for Stokes type variational inequality of the second kind. Acta Math. Appl. Sin. Engl. Ser. 27, 303–316 (2011) 19. T. Kashiwabara, Finite element method for Stokes equations under leak boundary condition of friction type. SIAM J. Numer. Anal. 52, 2448–2469 (2013) 20. R. An, Comparisons of Stokes/Oseen/Newton iteration methods for Navier-Stokes equations with friction boundary conditions. Appl. Math. Model. 38, 5535–5544 (2014) 21. Y. Li, R. An, Penalty finite element method for Navier-Stokes equations with nonlinear slip boundary conditions. Int. J. Numer. Methods Fluids 69, 550–566 (2012) 22. Y. Li, R. An, Two-level pressure projection finite element methods for Navier-Stokes equations with nonlinear slip boundary conditions. Appl. Numer. Math. 61, 285–297 (2011) 23. Y. Li, R. An, Two-level variational multiscale finite element methods for Navier-Stokes type variational inequality problem. J. Comput. Appl. Math. 290, 656–669 (2015)

Chapter 6

The Non-steady Navier-Stokes System with Friction Boundary Conditions

In this chapter we are concerned with the non-steady Navier-Stokes and Stokes problems corresponding to the steady problems in Chap. 5. In Sect. 6.1 relying on the results of Sect. 3.1, we embed all boundary conditions to variational formulations. We get variational inequalities with one unknown which are equivalent to the original PDE problems for the smooth solutions. In Sect. 6.2 we study the existence and uniqueness of solutions to the variational inequalities obtained in Sect. 6.1. In Sect. 6.3 using the results of Sect. 6.2, we get the existence, uniqueness and estimates of solutions to the Navier-Stokes and Stokes problems with the boundary conditions. For the problem with boundary conditions involving the total pressure and total stress, the existence of a solution without restriction of data of the problem is proved. For the problem with boundary conditions involving the static pressure and stress, under a compatibility condition at the initial time for the small data it is proved that there exists a unique solution on the given interval of time.

6.1 Variational Formulations of Problems 11 l 0,1 Let Ω be a bounded ∅  domain of R , l = 2, 3. ∂Ω ∈ C , ∂Ω = ∪i=1 Γ i , Γi ∩ Γ j =2,1 for i = j, Γi = j Γi j , where Γi j are connected open subsets of ∂Ω and Γi j ∈ C for i = 2, 3, 7 and Γi j ∈ C 1 for others. Also, let Q = Ω × (0, T ), Σi = Γi × (0, T ), 0 < T < ∞.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 T. Kim and D. Cao, Equations of Motion for Incompressible Viscous Fluids, Advances in Mathematical Fluid Mechanics, https://doi.org/10.1007/978-3-030-78659-5_6

185

186

6 The Non-steady Navier-Stokes System with Friction Boundary Conditions

Remark 6.1 In Chap. 5 for the boundary of Ω it is assumed that Γi j ∈ C 2 , i = 2, 3, 7, and those are convex in Theorems 5.8–5.12, whereas in this chapter it is assumed that Γi j ∈ C 2,1 , i = 2, 3, 7, and convexity of Γi j , i = 2, 3, 7, is not assumed in the following theorems. In the following chapters for the steady problems the assumptions as in Chap. 5 are used, whereas for the non-steady problems the assumptions as in this chapter are used (cf. Remark 5.2). We are concerned with Problems I and II for the non-steady Navier-Stokes problem ⎧ ∂v ⎪ ⎪ ⎨ ∂t − μΔv + (v · ∇)v + ∇ p = f, (6.1) div v = 0, ⎪ ⎪ ⎩ v(0) = v0 , which are distinguished according to boundary conditions. Problem I is the one with the boundary conditions (5.3) (the case of total pressure) and Problem II is the one with the boundary conditions (5.2) (the case of static pressure). We also consider the Stokes problem ⎧ ∂v ⎪ ⎪ ⎨ ∂t − μΔv + ∇ p = f, div v = 0, ⎪ ⎪ ⎩ v(0) = v0

(6.2)

with the boundary conditions (5.2), which is Problem III. Assuming that f ∈ L 2 (0, T ; L2 (Ω)), φi ∈ L 2 (0, T ; L 2 (Γi )), i = 2, 4, 7, and φi ∈ L 2 (0, T ; L2 (Γi )), i = 3, 5, 6, formally we introduce the following: Definition 6.1 A function (v, p) ∈ L 2 (0, T ; H2 (Ω)) × L 1 (0, T ; H 1 (Ω)) is called a solution to Problem I (or Problem II ) if the first two equations of (6.1) hold in L 1 (0, T ; L2 (Ω)), v(0) = v0 in L2 (Ω) and each of (5.3) (or (5.2)) holds in L 1 (0, T ; L 2 (Γi )) or L 1 (0, T ; L2 (Γi )). A function (v, p) ∈ L 2 (0, T ; H2 (Ω)) × L 2 (0, T ; H 1 (Ω)) is called a solution to Problem III if the first two equations of (6.2) hold in L 2 (0, T ; L2 (Ω)), v(0) = v0 in L2 (Ω) and each of (5.2) holds in L 2 (0, T ; L 2 (Γi )) or L 2 (0, T ; L2 (Γi )). Remark 6.2 Under the conditions for (v, p) and f , we have that v ∈ C([0, T ]; L2 (Ω)) (see Lemma 1.1 of Chap. III of [1]), and so the condition that v(0) = v0 in L2 (Ω) is meaningful.

6.1 Variational Formulations of Problems

187

6.1.1 Variational Formulation: The Case of Total Pressure Let V, K (Ω) be the same as in (5.5) and H : the closer of V in L2 (Ω), V(Q) = L 2 (0, T ; V) VΓ 237 (Ω) = {u ∈ H1 (Ω) : div u = 0, u τ |(Γ2 ∪Γ7 ) = 0, u n |Γ3 = 0}, K (Q) = {u ∈ L 2 (0, T ; V) : u ∈ L 1 (0, T ; V∗ ); u n |Γ10 ≥ 0, u n |Γ11 ≤ 0}, ¯ : div u = 0, u |Σ1 = 0, u τ |(Σ2 ∪Σ4 ∪Σ7 ) = 0, u n |(Σ3 ∪Σ5 ) = 0}. (Q) = {u ∈ C2 ( Q) We will use the following assumption: Assumption 6.1 The followings hold: (1) There exists a function U ∈ W 1,2 (0, T ; H1 ) ∩ L∞ (Q) such that div U = 0, U |Σ1 = h 1 , Uτ |(Σ2 ∪Σ7 ) = 0, Un |Σ3 = 0, Uτ |Σ4 = h 4 , Un |Σ5 = h 5 , U |Σ8 = h 8 n, U |Σ9 = h 9 , U |Σ10 ∪Σ11 = 0, where Σi = Γi × (0, T ). 1 (2) f ∈ L 2 (0, T ; V∗ ), φi ∈ L 2 (0, T ; H − 2 (Γi )), i = 2, 4, 7, φi ∈ L 2 (0, T ; 1 H− 2 (Γi )), i = 3, 5, 6, αi j ∈ L ∞ (Γ5 ), v0 − U (x, 0) ∈ H , and Γ1 = ∅. (3) of Assumption 5.1 holds. Remark 6.3 On the Γ10 (Γ11 ) outflow (inflow) only is possible. Thus, to guarantee div u = 0 we assume that if Γ10 = ∅ (Γ11 = ∅), then for example Γ2 = ∅. In Sect. 5.1 for proof of equivalence of variational formulations between variational inequalities, (3) of Assumption 5.1 was used via Lemma 5.1 to get existence of σ+n , σ−n such that σ+n + g+n ≥ 0, σ−n − g−n ≤ 0 from variational inequalities, which is also necessary for Theorems 6.2 and 6.4. Taking (v · ∇)v = rot v × v + 21 grad|v|2 into account and putting v = w + U , by (5.6), (5.7) and Assumption 6.1 we can see that solutions (v, p) of the problem (6.1), (5.3) in the sense of Definition 6.1 satisfy the following:

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6 The Non-steady Navier-Stokes System with Friction Boundary Conditions

⎧ v − U = w ∈ K (Q), v0 − U (x, 0) = w(0), ⎪ ⎪ ⎪ ⎪  ∂w  ⎪ ⎪ ⎪ , u + 2μ(E(w), E(u)) + rot w × w, u + rot U × w, u ⎪ ⎪ ⎪ ⎪ ∂t ⎪ ⎪ ⎪ + rot w × U, u + 2μ(k(x)w, u)Γ2 + 2μ(S w, ˜ u) ˜ Γ3 ⎪ ⎪ ⎪ ⎪ ⎪ + 2(α(x)w, u)Γ5 + μ(k(x)w, u)Γ7 − 2μ(εnτ (w + U ), u)Γ8 ⎪ ⎪ ⎪ ⎪   1 ⎪ ⎪ ⎪ + p + |v|2 − 2μεnn (w + U ), u n Γ9 ∪Γ10 ∪Γ11 ⎪ ⎪ 2 ⎪ ⎪ ∂U

⎪ ⎨ , u − 2μ(E(U ), E(u)) − rot U × U, u − 2μ(k(x)U, u)Γ2 (6.3) =− ∂t ⎪ ⎪ ⎪ ⎪ − 2μ(SU˜ , u) ˜ Γ3 − 2(α(x)Uτ , u)Γ5 − μ(k(x)U, u)Γ7 ⎪ ⎪ ⎪   ⎪ ⎪ ⎪ + f, u +

φi , u n Γi +

φi , uΓi ∀u ∈ V, ⎪ ⎪ ⎪ ⎪ i=2,4,7 i=3,5,6 ⎪ ⎪ ⎪ ⎪ ⎪ |στt (v)| ≤ gτ , στt · vτ + gτ |vτ | = 0 on Γ8 , ⎪ ⎪ ⎪ ⎪ ⎪ |σnt (v)| ≤ gn , σnt (v)vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪ t t ⎪ ⎪ ⎪ σn (v) + g+n ≥ 0, (σn (v) + g+n )vn = 0 on Γ10 , ⎪ ⎩ t σn (v) − g−n ≤ 0, (σnt (v) − g−n )vn = 0 on Γ11 . Define a01 (t, ·, ·), a11 (·, ·, ·) and F1 (t) ∈ V∗ by a01 (t, w, u) = 2μ(E(w), E(u)) + rot U (t) × w, u + rot w × U (t), u + 2μ(k(x)w, u)Γ2 + 2μ(S w, ˜ u) ˜ Γ3 + 2(α(x)w, u)Γ5 + μ(k(x)w, u)Γ7 ∀w, u ∈ V, a11 (w, u, v) = rot w × u, v ∀w, u, v ∈ V, ∂U (t) , u − 2μ(E(U (t)), E(u)) − rot U (t) × U (t), u

F1 (t), u = − ∂t − 2μ(k(x)U (t), u)Γ2 − 2μ(SU ˜(t), u) ˜ Γ3

(6.4)

− 2(α(x)U (t)τ , u)Γ5 − μ(k(x)U (t), u)Γ7 + f (t), u  

φi (t), u n Γi +

φi (t), uΓi ∀u ∈ V. + i=2,4,7

i=3,5,6

Then, taking into account 1 στt (v) = 2μεnτ (v), σnt (v) = −( p + |v|2 ) + 2μεnn (v) 2 and (6.3), we introduce the following variational formulation for problem (6.1), (5.3). t t , σ−n ) ∈ L2τ (Γ8 ) × Problem I-VE. Find v=U +w ∈ (U + K (Q)) and (στt , σnt , σ+n 2 − 21 − 21 L (Γ9 ) × H (Γ10 ) × H (Γ11 ) in a.e. t ∈ (0, T ) such that w(0) = v0 − U (0) and

6.1 Variational Formulations of Problems

189

⎧ ∂w  ⎪ ⎪ , u + a01 (t, w, u) + a11 (w, w, u) − (στt , u τ )Γ8 − (σnt , u n )Γ9 ⎪ ⎪ ⎪ ∂t ⎪

t t ⎪ ⎪ ⎪ − σ+n , u n Γ10 − σ−n , u n Γ11 = F1 , u ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ∀u ∈ L 2 (0, T ; V), |στt | ≤ gτ , ⎪ ⎪ ⎪ t ⎪ ⎪ ⎪ |σn | ≤ gn , ⎪ ⎪ ⎪ t ⎪ σ+n + g+n ⎪ ⎪ ⎪ ⎪ ⎩σt − g −n

−n

(6.5)

στt · vτ + gτ |vτ | = 0 on Γ8 , σnt vn + gn |vn | = 0 on Γ9 ,

t ≥ 0, σ+n + g+n , vn Γ10 = 0 on Γ10 ,

t ≤ 0, σ−n − g−n , vn Γ11 = 0 on Γ11 ,

where and in what follows L2τ (Γ8 ) is the subspace of L2 (Γ8 ) consisting of functions such that (u, n)L2 (Γ8 ) = 0. Remark 6.4 Under (2) of Assumption 6.1 the dual products on Γi in (6.4) and (6.5) is meaningful. For more detail we refer to Remark 5.1. In the same way as Theorem 5.1, we have Theorem 6.1 Let Assumption 6.1 hold. If (v, p) is a solution in the sense of Definition 6.1 of the problem (6.1), (5.3), then (v, στt |Γ8 , σnt |Γ9 , σnt |Γ10 , σnt |Γ11 ) is a solution to Problem I-VE. Conversely, if f ∈ L 2 (0, T ; L2 (Ω)) and Problem I-VE has a smooth solution (v, στt , σnt , σ+t , σ−t ) such that v ∈ L 2 (0, T ; H2 (Ω)), στt ∈ L 2 (0, T ; L2 (Γ8 )), σnt ∈ L 2 (0, T ; L 2 (Γ9 )), σ+t ∈ L 2 (0, T ; L 2 (Γ10 )) and σ−t ∈ L 2 (0, T ; L 2 (Γ11 )), then there exists p ∈ L 1 (0, T ; H 1 (Ω)) such that (v, p) is a solution to the problem (6.1), (5.3). Moreover, if at least one of the sets Γi , i = 2, 4, 6, 7, is nonempty, then p is unique. We will find a variational inequality corresponding to Problem I-VE. Let (v, στ , σn , σ+n , σ−n ) be a solution of Problem I-VE. Subtracting the first formula of (6.5) with u = w from the first formula of (6.5), we get  , u − w + a01 (t, w, u − w) + a11 (w, w, u − w) − (στt , u τ − wτ )Γ8 ∂t t t

− (σnt , u n − wn )Γ9 − σ+n , u n − wn Γ10 − σ−n , u n − wn Γ11

 ∂w

(6.6)

= F1 , u − w. Let Φ : V → R be the functional defined by (5.22). Define a functional Ψ (u) by Ψ (u) =

⎧ ⎨ ⎩

T

Φ(u(t)) dt

i f Φ(u(t)) ∈ L 1 (0, T ),

0

+∞

other wise.

In the same way as in Problem I of Chap. 5, we get from (6.6)

(6.7)

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6 The Non-steady Navier-Stokes System with Friction Boundary Conditions

 ∂w ∂t

 , u − w + a01 (t, w, u − w) + a11 (w, w, u − w) + Φ(u) − Φ(w)

(6.8)

≥ F1 , u − w.

If u − w ∈ (Q), then 

T

 ∂w ∂t

0

  , u − w dt = 

T

 ∂(w − u)

T

 ∂u

0

= 0

∂t

  , w − u dt + 

T 0

 ∂u ∂t

 , u − w dt

1 , u − w dt + (w0 − u(0)2 − w(T ) − u(T )2 ). ∂t 2

(6.9)

Define operators A1 (t) : V → V∗ and B1 : V × V → V∗ by

A1 (t)w, u = a01 (t, w, u) ∀w, u ∈ V,

B1 (w, u), v = a11 (w, u, v) ∀w, u, v ∈ V.

(6.10)

If w ∈ L 2 (0, T, V) ∩ L ∞ (0, T ; L2 (Ω)), then w ∈ L 4 (0, T, L3 (Ω)) (see (1.21)) 4 and B1 (w, w) ∈ L 3 (0, T, V∗ ). Thus, when u ∈ L 4 (0, T ; V), the integral T

B1 (w, w), w = 0. 0 B1 (w, w), u − w dt is well defined since Thus, neglecting 21 (w(T ) − u(T )2 ), from (6.8) and (6.9) we come to a variational inequality corresponding to Problem I (the case of total pressure). Problem I-VI. Find v such that v − U ≡ w ∈ L 2 (0, T ; V) ∩ L ∞ (0, T ; H ) and  T

u + A1 (t)w(t) + B1 (w(t), w(t)) − F1 (t), u(t) − w(t) dt + Ψ (u) − Ψ (w) 0

1 ≥ − w0 − u(0)2 ∀u ∈ L 4 (0, T ; V) with u ∈ L 2 (0, T ; V∗ ). 2 (6.11) If the solutions to Problem I-VI is smooth as much as v − U ≡ w ∈ L 2 (0, T ; V), w ∈ L 2 (0, T ; V∗ ), then we can see that the solutions satisfy 

T



w (t) + A1 (t)w(t) + B1 (w(t), w(t)) − F1 (t), u(t) − w(t) dt + Ψ (u) − Ψ (w) ≥ 0

0

∀u ∈ L 4 (0, T ; V).

(6.12) Especially, for the cases of 2-D and small data we will study such solutions to Problem I. Thus, we come to another formulation associated with Problem I by a variational inequality. Problem I-VI’. Find v such that v − U ≡ w ∈ L 2 (0, T ; V), w ∈ L 2 (0, T ; V∗ ), w(0, x) = v0 − U (0, x) and (6.12) is satisfied.

6.1 Variational Formulations of Problems

191

Remark 6.5 (Remark, p. 114 of [2]) The inequality (6.12) is equivalent to

w (t) + A1 (t)w(t) + B1 (w(t), w(t)) − F1 (t), u − w(t) + Φ(u) − Φ(w(t)) ≥ 0 for a.e. t ∈ [0, T ], ∀u ∈ K (Ω). (6.13) Further, the function w in (6.12) also satisfies the inequality (6.11). + F1 , u − w, by the argument in the proof In (6.13) putting F, u − w = − ∂w ∂t t t of Theorem 5.2 we can obtain existence of (στt , σnt , σ+n , σ−n ) ∈ L2τ (Γ8 ) × L 2 (Γ9 ) × −1/2 −1/2 t t t (Γ10 ) × H (Γ11 ) at a.e. t ∈ (0, T ) such that (v, στ , σnt , σ+n , σ−n ) is a soluH tion to Problem I-VE. Thus, we have t t , σ−n ) is a solution to Problem I-VE such that v − Theorem 6.2 If (v, στt , σnt , σ+n ∗ 2 U ∈ L (0, T ; V ), then v is a solution to Problem I-VI’. t t , σ−n Conversely, if v is a solution to Problem I-VI’, then there exist στt , σnt , σ+n t t t t for a.e. t ∈ (0, T ) such that (v, στ , σn , σ+n , σ−n ) is a solution to Problem I-VE.

6.1.2 Variational Formulation: The Case of Static Pressure In the same way as in Problem I, we will get the variational formulations of Problem II for the Navier-Stokes equations with boundary condition (5.2). Having in mind Assumption 6.1 and putting v = w + U , by (5.6) and (5.7) we can see that solutions (v, p) of problem (6.1), (5.2) in the sense of Definition 6.1 satisfy the following: ⎧ v − U = w ∈ K (Q), ⎪ ⎪ ⎪ ⎪ ∂w

⎪ ⎪ ⎪ , u + 2μ(E(w), E(u)) + (w · ∇)w, u + (U · ∇)w, u ⎪ ⎪ ∂t ⎪ ⎪ ⎪ ⎪ + (w · ∇)U, u + 2μ(k(x)w, u)Γ2 + 2μ(S w, ˜ u) ˜ Γ3 ⎪ ⎪ ⎪ ⎪ ⎪ + 2(α(x)w, u) + μ(k(x)w, u) − 2μ(ε (w + U ), u)Γ8 ⎪ Γ5 Γ7 nτ ⎪ ⎪ ⎪ ⎪ + ( p − 2μεnn (w + U ), u n )Γ9 ∪Γ10 ∪Γ11 ⎪ ⎪ ⎪ ⎪ ∂U

⎪ ⎪ ⎨ , u − 2μ(E(U ), E(u)) − (U · ∇)U, u − 2μ(k(x)U, u)Γ2 =− ∂t ⎪ ⎪ − 2μ(SU˜ , u) ˜ Γ3 − 2(α(x)Uτ , u)Γ5 − μ(k(x)U, u)Γ7 ⎪ ⎪ ⎪   ⎪ ⎪ ⎪ + f, u +

φi , u n Γi +

φi , uΓi ∀u ∈ V, ⎪ ⎪ ⎪ ⎪ i=2,4,7 i=3,5,6 ⎪ ⎪ ⎪ ⎪ ⎪ |στ (v)| ≤ gτ , στ (v) · vτ + gτ |vτ | = 0 on Γ8 , ⎪ ⎪ ⎪ ⎪ ⎪ |σn (v)| ≤ gn , σn (v)vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪ ⎪ ⎪ σn (v) + g+n ≥ 0, (σn (v) + g+n )vn = 0 on Γ10 , ⎪ ⎪ ⎩ σn (v) − g−n ≤ 0, (σn (v) − g−n )vn = 0 on Γ11 . (6.14)

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6 The Non-steady Navier-Stokes System with Friction Boundary Conditions

Define a02 (t, ·, ·), a12 (·, ·, ·) and F2 (t) ∈ V∗ by a02 (t, w, u) = 2μ(E(w), E(u)) + (U (t) · ∇)w, u + (w · ∇)U (t), u + 2μ(k(x)w, u)Γ2 + 2μ(S w, ˜ u) ˜ Γ3 + 2(α(x)w, u)Γ5 + μ(k(x)w, u)Γ7 ∀w, u ∈ V, a12 (w, u, v) = (w · ∇)u, v ∀w, u, v ∈ V,  ∂U (t) 

F2 (t), u = − , u − 2μ(E(U (t)), E(u)) − (U (t) · ∇)U (t), u ∂t − 2μ(k(x)U (t), u)Γ2 − 2μ(SU ˜(t), u) ˜ Γ3

(6.15)

− 2(α(x)U (t)τ , u)Γ5 − μ(k(x)U (t), u)Γ7 + f (t), u  

φi (t), u n Γi +

φi (t), uΓi ∀u ∈ V. + i=2,4,7

i=3,5,6

Then, taking into account στ (v) = 2μεnτ (v), σn (v) = − p + 2μεnn (v) and (6.14), we introduce the following variational formulation for problem (6.1), (5.2).   Problem II-VE. Find v = U +w ∈ U + K (Q) and (στ , σn , σ+n , σ−n ) ∈ L2τ (Γ8 ) × L 2 (Γ9 ) × H −1/2 (Γ10 ) × H −1/2 (Γ11 ) in a.e. t ∈ (0, T ) such that w(0) = v0 − U (0) and ⎧ ∂w  ⎪ ⎪ , u + a02 (t, w, u) + a12 (w, w, u) − (στ , u τ )Γ8 − (σn , u n )Γ9 ⎪ ⎪ ∂t ⎪ ⎪ ⎪ ⎪ ⎪ − σ+n , u n Γ10 − σ−n , u n Γ11 = F2 , u ∀u ∈ L 2 (0, T ; V), ⎪ ⎨ |στ | ≤ gτ , στ · vτ + gτ |vτ | = 0 on Γ8 , ⎪ ⎪ ⎪ |σn | ≤ gn , σn vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ σ+n + g+n ≥ 0, σ+n + g+n , vn Γ10 = 0 on Γ10 , ⎪ ⎩ σ−n − g−n ≤ 0, σ−n − g−n , vn Γ11 = 0 on Γ11 .

(6.16)

In the same way as in Theorem 5.1 we have Theorem 6.3 Let Assumption 6.1 hold. If (v, p) is a solution in the sense of Definition 6.1 of the problem (6.1), (5.2), then (v, στ |Γ8 , σn |Γ9 , σn |Γ10 , σn |Γ11 ) is a solution to Problem II-VE.

6.1 Variational Formulations of Problems

193

Conversely, if f ∈ L 2 (0, T ; L2 (Ω)) and Problem II-VE has a smooth solution (v, στ , σn , σ+ , σ− ) such that v ∈ L 2 (0, T ; H2 (Ω)), στ ∈ L 2 (0, T ; L2 (Γ8 )), σn ∈ L 2 (0, T ; L 2 (Γ9 )), σ+ ∈ L 2 (0, T ; L 2 (Γ10 )) and σ− ∈ L 2 (0, T ; L 2 (Γ11 )), then there exists p ∈ L 1 (0, T ; H 1 (Ω)) such that (v, p) is a solution to the problem (6.1), (5.2). Moreover, if at least one of the sets Γi , i = 2, 4, 6, 7, is nonempty, then p is unique. We will find a variational inequality corresponding to Problem II-VE. Define operators A2 (t) : V → V∗ and B2 : V × V → V∗ by

A2 (t)w, u = a02 (t, w, u) ∀w, u ∈ V,

B2 (w, u), v = a12 (w, u, v) ∀w, u, v ∈ V.

(6.17)

Unlike Problem I, in this problem the property B2 (w, w), w = 0 fails and for T w ∈ L 2 (0, T ; V) integral 0 B2 (w, w), w ds is not meaningful. Thus, we will find more smooth solution w ∈ L 4 (0, T ; V), such that w ∈ L 2 (0, T ; V∗ ). Thus, as Problem I-VI’ we come to the following variational inequality. Problem II-VI. (the case of static pressure) Find v such that w ≡ v − U ∈ L 4 (0, T ; V), ∂w ∈ L 2 (0, T ; V∗ ), w(0) = v0 − U (0, x) and ∂t 

T 0

∂w + A2 (t)w(t) + B2 (w(t), w(t)) − F2 (t), u(t) − w(t) dt + Ψ (u) − Ψ (w) ≥ 0 ∂t ∀u ∈ L 4 (0, T ; V),

(6.18) where Ψ is defined by (6.7). As Theorem 6.2 we have the following theorem: Theorem 6.4 If (v, στ , σn , σ+n , σ−n ) is a solution to Problem II-VE such that v − U ∈ L 2 (0, T ; V∗ ), then v is a solution to Problem II-VI. Conversely, if v is a solution to Problem II-VI, then there exist στ , σn , σ+n , σ−n for a.e. t ∈ (0, T ) such that (v, στ , σn , σ+n , σ−n ) is a solution to Problem II-VE.

6.1.3 Variational Formulation: The Stokes Problem In the same way as Problem I we get the equivalent formulations of Problem III for the Stokes equation with boundary condition (5.2).   Problem III-VE. Find v = U + w ∈ U + K (Q) and (στ , σn , σ+n , σ−n ) ∈ 1 1 L2τ (Γ8 ) × L 2 (Γ9 ) × H − 2 (Γ10 ) × H − 2 (Γ11 ) in a.e. t such that w(0) = v0 − U (0) and

194

6 The Non-steady Navier-Stokes System with Friction Boundary Conditions

⎧ ∂w  ⎪ ⎪ , u + a03 (w, u) − (στ , u τ )Γ8 − (σn , u n )Γ9 − σ+n , u n Γ10 − σ−n , u n Γ11 ⎪ ⎪ ∂t ⎪ ⎪ ⎪ ⎪ ⎪ = F3 , u ∀u ∈ L 2 (0, T ; V), ⎪ ⎨ |στ | ≤ gτ , στ · vτ + gτ |vτ | = 0 on Γ8 , ⎪ ⎪ ⎪ |σn | ≤ gn , σn vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪ ⎪ ⎪ σ+n + g+n ≥ 0, σ+n + g+n , vn  = 0 on Γ10 , ⎪ ⎪ ⎩ σ−n − g−n ≤ 0, σ−n − g−n , vn  = 0 on Γ11 , (6.19) where ˜ u) ˜ Γ3 a03 (w, u) = 2μ(E(w), E(u)) + 2μ(k(x)w, u)Γ2 + 2μ(S w, + 2(α(x)w, u)Γ5 + μ(k(x)w, u)Γ7 ∀w, u ∈ V,  ∂U  , u − 2μ(E(U ), E(u)) − 2μ(k(x)U, u)Γ2

F3 , u = − ∂t − 2μ(SU˜ , u) ˜ Γ3 − 2(α(x)Uτ , u)Γ5 − μ(k(x)U, u)Γ7 + f, u  

φi , u n Γi +

φi , uΓi ∀u ∈ V. + i=2,4,7

i=3,5,6

Problem III-VI. Find v such that w ≡ v − U ∈ L 2 (0, T ; V), w(0) = v0 − U (0) and  0

T

∂w ∂t

(6.20)

∂w ∂t

∈ L 2 (0, T ; V∗ ),

+ A3 w(t) − F3 (t), u(t) − w(t) dt + Ψ (u) − Ψ (w) ≥ 0

(6.21)

∀u ∈ L (0, T ; V), 2

where A3 : V → V∗ is defined by

A3 w, u = a03 (w, u) ∀w, u ∈ V.

(6.22)

As Theorem 6.2 we have Theorem 6.5 If (v, στ , σn , σ+n , σ−n ) is a solution to Problem III-VE such that v − U ∈ L 2 (0, T ; V∗ ), then v is a solution to Problem III-VI. Conversely, if v is a solution to Problem III-VI, then there exist στ , σn , σ+n , σ−n at a.e. t ∈ (0, T ) such that (v, στ , σn , σ+n , σ−n ) is a solution to Problem III-VE.

6.2 The Existence and Uniqueness of Solutions to Variational Inequalities In this section we study the variational inequalities obtained in Sect. 6.1.

6.2 The Existence and Uniqueness of Solutions to Variational Inequalities

195

Let us start with the existence of a solution w ∈ L 2 (0, T ; V) to the following problem: 

T

u (t) + A(t)w(t) + B(w(t), w(t)) − F(t), u(t) − w(t) dt + Ψ (u) − Ψ (w)

0

1 ≥ − w0 − u(0)2 ∀u ∈ L 4 (0, T ; V) with u ∈ L 2 (0, T ; V∗ ). 2 (6.23) Theorem 6.6 Assume that (1) For any u, v ∈ V, A(t)u, v is continuous with respect to t and

A(t)u, u ≥ c1 u2V − c2 u2 ∃c1 > 0, c2 ≥ 0, ∀u ∈ V, | A(t)u, v | ≤ c3 uV vV ∃c3 > 0, ∀u, v ∈ V, lim inf A(t)u m , u m  ≥ A(t)u, u as u m  u in V; m→∞

(2) B(u, v) is bilinear continuous from V × V into V∗ and 1

1

| B(w, v), z | ≤ K wH1 v 2 vH2 1 zH1 ,

B(w, v), v = 0; (3) F ∈ L 2 (0, T ; V∗ ); (4) The functional Φ : V → R is proper, convex, weakly lower semi-continuous and Φ(u) ≥ 0 ∀u ∈ V, Φ(0V ) = 0. Then, for any initial function w0 ∈ H there exists a solution to (6.23) satisfying w(t)2 ≤ K 1 (w0 2 + F2L 2 (0,T ;V∗ ) ), w2L 2 (0,T ;V) ≤ K 1 (w0 2 + F2L 2 (0,T ;V∗ ) ),

(6.24)

where K 1 is independent of Φ. Proof For every ε > 0, let Φε be the Moreau-Yosida approximation of Φ and ∇Φε be Fréchet derivative of Φε (see Sect. 1.6). mLet {v j , j = 1, 2, · · · } be a basis of the space V. We find a solution wm = j=1 g jm (t)v j to problem







⎧ wm (t)), v j + ∇Φε (wm (t)), v j ⎪ ⎨ wm (t), v j + A(t)wm (t), v j + B(wm (t),

= F(t), v j , ⎪ ⎩ wm (0) = w0m , (6.25)

196

6 The Non-steady Navier-Stokes System with Friction Boundary Conditions

m where wm (t) = dtd wm (t) and w0m is such that w0m = i=1 k j v j → w0 in H as m → ∞. The solutions to (6.25) depend on ε, but for convenience of notation here and in what follows we use subindex m instead of subindex m, ε. For a tm there exists an absolute continuous function g jm (t) on [0, tm ) and by the first a priori estimate of (6.30) continuation of the function beyond tm is possible. Thus, we can see that tm = T. Multiplying (6.25) by g jm and summing over j = 1, · · · , m, yield 1 dwm (t)2 + A(t)wm (t) + B(wm (t), wm (t)) + ∇Φε (wm (t)), wm (t) = F(t), wm (t) . 2 dt

(6.26) By (1.37) and condition (4) we know that Φε (0V ) = 0. Also, since Φε is convex, continuous and Fréchet differentiable, we have Φε (v) − Φε (wm (t)) ≥ ∇Φε (wm (t)), v − wm (t) ∀v ∈ V,

(6.27)

and so by Φε (0V ) = 0, 0 ≤ Φε (wm (t)) ≤ ∇Φε (wm (t)), wm (t) .

(6.28)

By virtue of conditions (1), (2) and (6.28), we have from (6.26) 1 dwm (t)2 + c1 wm (t)2V + Φε (wm (t)) ≤ F(t), wm (t) + c2 wm (t)2 , 2 dt d 1 wm (t)2 + 2c1 wm (t)2V + 2Φε (wm (t)) ≤ F2V∗ + c1 wm (t)2V + 2c2 wm (t)2 , dt c1

and 1 d wm (t)2 + c1 wm (t)2V + 2Φε (wm (t)) ≤ F2V∗ + 2c2 wm (t)2 . (6.29) dt c1 We have from (6.29) 1 F2L 2 (0,T ;V∗ ) )e2c2 T ≤ C1 (w0 2 + F2L 2 (0,T ;V∗ ) ), c1 2c2 1 1 ≤ (w0 2 + F2L 2 (0,T ;V∗ ) )e2c2 T + 2 F2L 2 (0,T ;V∗ ) c1 c1 c1

wm (t)2 ≤ (w0 2 + wm (t)2L 2 (0,T ;V)

≤ C2 (w0 2 + F2L 2 (0,T ;V∗ ) ) and



T 0

Φε (wm (t)) dt ≤ C3 ,

(6.30) (6.31)

6.2 The Existence and Uniqueness of Solutions to Variational Inequalities

197

where Ci , i = 1, 2, 3, are independent of m, ε and Φ. From (1.36), (1.37) and (6.31) we have  T

0

wm (t) − Jε (wm (t))2V dt ≤ c7 ε

(6.32)

with c7 independent of m and  ε.  Multiplying (6.25) by g jm (t) − g jm (s) , s ∈ (0, T ), and summing over j = 1, · · · , m, we have 1 dwm (t) − wm (s)2 + A(t)wm (t) + B(wm (t), wm (t)) − F(t), wm (t) − wm (s) 2 dt = ∇Φε (wm (t)), wm (s) − wm (t) ≤ Φε (wm (s)) − Φε (wm (t)) ≤ Φε (wm (s)). By conditions (1) and (2), from the above it follows that dwm (t) − wm (s)2 ≤Φε (wm (s)) + A(t)wm (t), wm (s) + B(wm (t), wm (t)), wm (s) dt + F(t), wm (t) − wm (s) + c2 wm (t)2 .

(6.33) Let us integrate every terms in (6.33) first with respect to t from s to s + h and then with respect to s from 0 to T , where wm (t) = 0 when t ∈ (T, T + h). We get 



T 0

dwm (t) − wm (s)2 dtds = dt

s+h s



T

wm (s + h) − wm (s)2 ds.

(6.34)

0

By (6.31), 

T

0



s+h

 Φε (wm (s)) dtds ≤ h

s

T

Φε (wm (s)) ds ≤ c6 h.

(6.35)

0

By conditions (1) and (6.30) we have 

T

0



s+h

√ 

A(t)wm (t), wm (s) dtds ≤ c8 h

s

0

T

wm (s)2V ds.

(6.36)

By conditions (2) and (6.30) we have  0

T



s+h

B(wm (t), wm (t))wm (s) dtds

s



T

≤K 0



s+h s

(6.37) 3

1

1

wm (t)V2 wm (t) 2 wm (s)V dtds ≤ c9 h 4 .

198

6 The Non-steady Navier-Stokes System with Friction Boundary Conditions

Using (6.30) we have 

T

0





s+h



s

T



0

T

F(t), wm (t) dtds ≤ 0

s+h

| F(t), wm (t)| 

T

F(t), −wm (s) dtds ≤ K

s





t−h s+h

 wm (s)V

0

t

 ds dt ≤ K h, √

(6.38)

F(t)V∗ dtds ≤ c9 h.

s

By virtue of (6.34)–(6.38), uniformly with respect to m and ε 

T

1

wm (s + h) − wm (s)2 ds ≤ O(h 4 ),

(6.39)

0

which implies that the set {wm } is relatively compact in L 2 (0, T ; H ) (see Theorem 1.38). Therefore, by (6.30) and (6.39), there exists w and a subsequence {wm k } such that ∗ wm k  w in L ∞ (0, T ; H ), wm k → w in L 2 (0, T ; H ),

(6.40)

wm k  w in L (0, T ; V) 2

when m k → ∞ and ε → 0 (note that for convenience we used subindex m instead of m, ε).  1 On the other hand, putting v = M j=1 k j (t)v j , where k j (t) ∈ C [0, T ] and Mpositive integer, multiply (6.25) by k j (t) and sum for j = 1, · · · , M. Then, multiply (6.25) by g jm (t) and sum for j = 1, · · · , m. Substituting the resulting equations, we have

wm (t) + A(t)wm (t) + B(wm (t), wm (t)) + ∇Φε (wm ), v(t) − wm (t) (6.41) = F(t), v(t) − wm (t) . Since  T 0

wm (t), v(t) − wm (t) dt =  T

1 1 v (t), v(t) − wm (t) dt − wm (T ) − v(T )2 + wm (0) − v(0)2 , 2 2 0



we have 

T 0



v (t) + A(t)wm (t) + B(wm (t), wm (t)) + ∇Φε (wm (t)) − F(t), v(t) − wm (t) dt 1 1 = − wm (0) − v(0)2 + wm (T ) − v(T )2 . 2 2

(6.42)

6.2 The Existence and Uniqueness of Solutions to Variational Inequalities

199

Taking into account (6.27), we have from (6.42) 



T

0

v (t) + A(t)wm (t) + B(wm (t), wm (t)) − F(t), v(t) − wm (t) dt  T   1 Φε (v(t)) − Φε (wm (t)) dt ≥ − wm (0) − v(0)2 . + 2 0

(6.43)

Since Φ(v) ≥ Φε (v) and Φ(Jε wm (t)) ≤ Φε (wm (t)) (see (1.37)), we have from (6.43) 

T

0



v (t) + A(t)wm (t) + B(wm (t), wm (t)) − F(t), v(t) − wm (t) dt  T 1 Φ(Jε wm (t)) dt ≥ − wm (0) − v(0)2 . + Ψ (v) − 2 0

(6.44)

By (6.32) it follows that Jε wm k  w in L 2 (0, T ; V) as m k → ∞ and ε → 0, and  lim inf

m k →∞,ε→0 0

T



T

Φ(Jε wm k (t)) dt ≥

Φ(w(t)) dt = Ψ (w).

(6.45)

B(w(t), w(t)), v(t) dt

(6.46)

0

It is not difficult to prove that 

T



B(wm k (t), wm k (t)), v(t) dt →

0



T 0

as m k → ∞ and ε → 0. We give a proof below for convenience of the readers.  0

T



( B(wm k (t), wm k (t)), v(t) − B(w(t), w(t)), v(t)) dt 

T

= 0



B(wm k (t), wm k (t) − w(t)), v(t) dt +



T 0



B(wm k (t) − w(t), w(t)), v(t) dt

≡ I1 + I2 .

(6.47) By condition (2), Hölder’s inequality with exponents 2, 4, 4 and (6.40) we have  |I1 | ≤ K sup v(t)H1 t∈[0,T ]

T 0

1

1

wm k (t)H1 wm k (t) − w(t)H2 1 wm k (t) − w(t) 2 dt 1

1

≤ K sup v(t)H1 wm k  L 2 (0,T ;H1 ) wm k − w L2 2 (0,T ;H1 ) wm k − w L2 2 (0,T ;L2 ) → 0. t∈[0,T ]

(6.48) By condition (2),

200  

6 The Non-steady Navier-Stokes System with Friction Boundary Conditions



T 0

1 1 

B(z(t), w(t)), v(t) dt  ≤ K sup v(t)H1 z(t) L 2 (0,T ;H1 ) w 2 2 w L2 2 (0,T ;L2 ) L (0,T ;H1 )

t∈[0,T ]

≤ K 1 z(t) L 2 (0,T ;H1 ) ,

which means that 

T

z(t) ∈ L 2 (0, T ; V) →

B(z(t), w(t)), v(t) dt

0

is a continuous linear on L 2 (0, T ; V), that is, there exists a f ∈ L 2 (0, T ; V)∗ such that  T

B(z(t), w(t)), v(t) dt = z(t), f (t)

0

and so by (6.40) it follows that |I2 | → 0 as m k → ∞ and ε → 0.

(6.49)

By (6.47)–(6.49) we get (6.46).

Since lim inf m k →∞,ε→0 A(t)wm k (t), wm k (t) ≥ A(t)w(t), w(t) , by (6.45) and (6.46) we have from (6.44) 

T



v (t) + A(t)w(t) + B(w(t), w(t)) − F(t), v(t) − w(t) dt + Ψ (v) − Ψ (w)

0

1 ≥ − w(0) − v(0)2 . 2 3 2

(6.50)

1 2

On the other hand, B(w(t), w(t)) ≤ K w(t)V w(t) (cf. condition(2)) 4 ∗ 3 and w(t) ∈ L ∞ (0, T ; L2 (Ω)), and we can get that B(w(t), w(t))  M ∈ L (0, T ; V ). Therefore, taking into account the fact that the set {v = i=1 k j (t)v j ; k j (t) ∈ C 1 [0, T ]} is dense in {L 4 (0, T ; V) : u ∈ L 2 (0, T ; V∗ )} (see Remark 1.8), we see that (6.50) is valid for all v ∈ {L 4 (0, T ; V) : u ∈ L 2 (0, T ; V∗ )}. From (6.30) we get estimates (6.24).  V∗

Remark 6.6 The operator A1 (t) in Problem I-VI is not symmetric and depends on time t. Thus, unlike Theorem 1 of [2] we can not use eigenfunctions of a symmetric operator as the basis for Galërkin method. Next, let us study the following problem: ⎧ ⎪ ⎪ ⎪ ⎨

T



w (t) + A(t)w(t) + B(w(t), w(t)) − F(t), u(t) − w(t) dt + Ψ (u) − Ψ (w) ≥ 0

0

⎪ ⎪ ⎪ ⎩

∀u ∈ L 4 (0, T ; V), w(0) = w0 .

(6.51)

6.2 The Existence and Uniqueness of Solutions to Variational Inequalities

201

Condition (2) for B(w, w) in the following theorem is for 2-D problem corresponding to Problem I (the case of total pressure). Theorem 6.7 Assume that

(1) Let l = 2. For any u, v ∈ V, A(t)u, v and A (t)u, v are continuous with respect to t and

A(t)u, u ≥ c1 u2V − c2 u2 ∃c1 > 0, c2 ≥ 0, ∀u ∈ V, | A(t)u, v | ≤ c3 uV vV ∃c3 > 0, ∀u, v ∈ V,

| A (t)u, v | ≤ c4 uV vV ∃c4 ≥ 0, ∀u, v ∈ V;

(6.52)

(2) B(u, v) is bilinear continuous from V × V into V∗ and 1

2

| B(w, v), z | ≤ K wH1 vL4 z 3 zH3 1 ,

B(w, v), v = 0; (3) F, F ∈ L 2 (0, T ; V∗ ); (4) The functional Φ : V → R is proper, weakly lower semi-continuous and Φ(u) ≥ 0 ∀u ∈ V, Φ(0V ) = 0 ; (5) w0 ∈ V, Φ(w0 ) = 0 and (A(0)w0 + B(w0 , w0 ) − F(0)) ∈ H (a compatibility condition at initial time). Then, there exists a unique solution w to (6.51) such that w ∈ C([0, T ]; V), w ∈ L 2 (0, T ; V) ∩ L ∞ (0, T ; H ), w(0) = w0 ,

(6.53)

and the estimates (6.24) and the following hold: w (t)2 ≤ K 1 , w 2L 2 (0,T ;V) ≤ K 1 ,

(6.54)

where K 1 depends on A(0)w0 + B(w0 , w0 ) − F(0), w0 , F1 2L 2 (0,T ;V∗ ) , F1 2L 2 (0,T ;V∗ ) and is independent of Φ. Without loss of generality, Proof Let {v j , j = 1, 2, · · · } be a basis of the space V. m gim (t)vi to problem we assume that v1 = w0 . We find a solution wm = i=1 









wm (t), v j + A(t)wm (t), v j + B(wm (t), wm (t)), v j + ∇Φε (wm (t)), v j = F, v j , wm (0) = w0 .

(6.55)

202

6 The Non-steady Navier-Stokes System with Friction Boundary Conditions

By the same argument as in the proof of Theorem 6.6, there exists continuously differentiable function gm on the interval [0, T ]. Also, as (6.30) in the proof of Theorem 6.6 we have wm (t)2 ≤ C1 (w0 2 + F2L 2 (0,T ;V∗ ) ), wm 2L 2 (0,T ;V) ≤ C2 (w0 2 + F2L 2 (0,T ;V∗ ) ),

(6.56)

where C1 , C2 are independent of m, ε. We are going to derive a priori estimates for functions wm . In the same way as in the proof of Theorem 6.6 we see that there exists gim (t) on the interval [0, T ].

a solution Owing to F ∈ L 2 (0, T ; V∗ ), the function F, v j is absolute continuous with respect

to t. Therefore, by the Lipschitz continuity of ∇Φε and continuity of A (t)u, v with (t) is in fact absolute continuous. respect to t, gim Setting t = 0 in (6.55), multiplying by gim (0) and summing over i = 1, · · · , m, we get







wm (0)2 + A(0)wm (0), wm (0) + B(wm (0), wm (0)), wm (0) + ∇Φε (w0 ), wm (0) (6.57)

= F(0), wm (0) .

By conditions (4) and (5), for any u ∈ V we have Φ(u) ≥ Φ(w0 ), and so ∇Φε (w0 ) = 0. Thus, we have from (6.57) wm (0) ≤ A(0)w0 + B(w0 , w0 ) − F(0).

(6.58)

Differentiating (6.55) yields







wm (t), v j + A (t)wm (t), v j + A(t)wm (t), v j + B(wm (t), wm (t)), v j





(6.59) + B(wm (t), wm (t)), v j + (∇Φε (wm )) , v j = F , v j , where (∇Φε (wm )) = for i, we have

d (∇Φε (wm (t))). dt

Multiplying (6.59) by gim (t) and summing





(t), wm (t) + A (t)wm (t), wm (t) + A(t)wm (t), wm (t) wm





+ B(wm (t), wm (t)), wm (t) + (∇Φε (wm )) , wm (t) = F , wm (t) ,

(6.60)



where the fact m ), wm = 0 was used. Using monotonicity of ∇Φε , we

that B(w m , w know that (∇Φε (wm )) , wm ≥ 0 (see (1.38)). By condition (2), (6.56) and Young’s inequality with exponents 6/5, 6, we get

6.2 The Existence and Uniqueness of Solutions to Variational Inequalities

203

2

1 2| B(wm , wm ), wm | ≤ 2K wm V wm L4 wm  3 wm V3

≤ 2K wm V wm L4 wm  3 c1 ≤ wm 2V + K 1 wm 6L4 wm 2 . 4 1

5/3

Thus, we have from (6.60) d c4 wm (t)2 + c1 wm 2V + c1 wm 2V − c1 wm 2V − wm 2V dt c1 c1 2 4 c1 2 2 2 ≤ wm V + F1 V∗ + 2c2 wm  + wm V + K 2 wm 2 wm 6L4 , 4 c1 4 and so d c1 c4 4 wm (t)2 + wm 2V ≤ wm 2V + F 2V∗ + (2c2 + K 2 wm 6L4 )wm 2 . dt 2 c1 c1 (6.61) Integrating (6.61) yields wm (t)2 +



t 0

c1 c4 4 (0)2 + wm 2L 2 (0,T ;V) + F1 2L 2 (0,T ;V∗ ) wm (s)2V ds ≤wm 2 c1 c1 (6.62)  t   2c2 + K 2 wm (s)6L4 wm + (s)2 ds. 0

By (6.56),

wm ∈ L 2 (0, T ; V) ∩ L ∞ (0, T ; L2 (Ω))

and {wm } is bounded in the space above. Since V → Lq (Ω) for any 1 < q < ∞ when l = 2, by virtue of Theorems 1.12 and 1.33 we have that wm ∈ L 6 (0, T ; L4 (Ω)) and {wm } is bounded in L 6 (0, T ; L4 (Ω)). Taking into account (6.56), (6.58) and Gronwall’s inequality (see Theorem 1.51), we have from (6.62) wm (t)2 ≤ C4 e

t 0





2c2 +K 2 wm (s)6L4 ds

,

(6.63)

where C4 depends on Aw0 + B(w0 , w0 ) − F(0), w0 , F1 2L 2 (0,T ;V∗ ) , F1 2L 2 (0,T ;V∗ ) and is independent of ε, m, Φ. From (6.56), (6.62) and (6.63) we have  0

T

wm (s)2V ds ≤ C5 ,

(6.64)

204

6 The Non-steady Navier-Stokes System with Friction Boundary Conditions

where C5 is similar to C4 . Therefore, by (6.56), (6.63) and (6.64) wm (t) ≤ const ∀t ∈ [0, T ], ∀m, ∀ε > 0; wm  L 2 (0,T ;V) ≤ const ∀m, ∀ε > 0;

(6.65)

wm (t) ≤ const ∀t ∈ [0, T ], ∀m, ∀ε > 0; wm  L 2 (0,T ;V) ≤ const ∀m, ∀ε > 0.

Extracting a subsequence and passing to limit in (6.55) as m → ∞, and then letting ε → 0 and using (6.27), (6.45), similarly to the proof of Theorem 6.6 (from (6.41)) we can prove the existence of a solution w ∈ L 2 (0, T ; V) to (6.51) satisfying that w ∈ L 2 (0, T ; V) ∩ L ∞ (0, T ; H ). From (6.56), (6.63) and (6.64), we get the estimates (6.24) and (6.54). Let us prove uniqueness next. Let w1 , w2 be two solutions asserted in Theorem 6.7. Then, taking into account (6.13) and Remark 6.5, we have from (6.51)

w1 (t) + A(t)w1 (t) + B(w1 (t), w1 (t)) − F(t), w2 (t) − w1 (t) + Φ(w2 (t)) − Φ(w1 (t)) ≥ 0,

w2 (t) + A(t)w2 (t) + B(w2 (t), w2 (t)) − F(t), w1 (t) − w2 (t) + Φ(w1 (t)) − Φ(w2 (t)) ≥ 0,

which imply

w1 (t) − w2 (t), w1 (t) − w2 (t) + A(t)(w1 (t) − w2 (t)), w1 (t) − w2 (t) ≤ | B(w1 (t), w1 (t)) − B(w2 (t), w2 (t)), w1 (t) − w2 (t) |. (6.66) By virtue of conditions (1) and (2) we have 1 ∂(w1 (t) − w2 (t) + A(t)(w1 (t) − w2 (t)), w1 (t) − w2 (t) 2 ∂t ≤ |B(w1 (t), w1 (t)) − B(w2 (t), w2 (t)), w1 (t) − w2 (t)|, 1 ∂w1 (t) − w2 (t)2 + c1 w1 (t) − w2 (t)2V 2 ∂t 5

1

≤ c2 w1 (t) − w2 (t)2 + w2 (t)L4 w1 (t) − w2 V3 w1 (t) − w2  3 ≤ c2 (w1 (t) − w2 (t)2 + c1 w1 (t) − w2 2V + K w2 (t)6L4 w1 (t) − w2 2 . (6.67) We have from (6.67)  (w1 (t) − w2 (t) ≤ const 2

0

t

(c2 + K w2 (s)6L4 )w1 (s) − w2 (s)2 ds. (6.68)

6.2 The Existence and Uniqueness of Solutions to Variational Inequalities

205

Since w2 ∈ C([0, T ]; V), by Gronwall’s inequality we have that w1 (t) = w2 (t) for all t ∈ [0, T ].  The following theorem is for small data in (6.51) corresponding to Problem I (the case of total pressure). Theorem 6.8 Assume that

(1) For any u, v ∈ V, A(t)u, v and A (t)u, v are continuous with respect to t and

A(t)u, u ≥ c1 u2V − c2 u2 , ∃c1 > 0, c2 ≥ 0, ∀u ∈ V, | A(t)u, v | ≤ c3 uV vV ∃c3 > 0, ∀u, v ∈ V,

| A (t)u, v | ≤ c4 uV vV ∃c4 > 0, ∀u, v ∈ V; (2) B(u, v) is bilinear continuous from V × V into V∗ and 1

1

B(w, v), u ≤ c5 wV vV u 2 uV2 ,

B(w, v), v = 0; (3) F, F ∈ L 2 (0, T ; V∗ ); (4) The functional Φ : V → R is proper, weakly lower semi-continuous and Φ(u) ≥ 0 ∀u ∈ V, Φ(0V ) = 0 ; (5) w0 ∈ V, Φ(w0 ) = 0 and (A(0)w0 + B(w0 , w0 ) − F(0)) ∈ H (a compatibility condition at the initial time). If w0 V ,  ( A(0)w0 + B(w0 , w0 ) − F(0))  H and norms of F, F in the spaces where they belong to are small enough, then there exists a unique solution w to (6.51) such that (6.69) w ∈ C([0, T ]; V), w ∈ L 2 (0, T ; V) ∩ L ∞ (0, T ; H ), and (6.24), (6.54) hold. Proof First, let us prove the existence of the asserted solution. Let {v j , j = 1, 2, · · · } be a basis of the space Without loss of generality, we V. m gim (t)vi to problem assume that v1 = w0 . We find a solution wm = i=1 









wm (t), v j + A(t)wm (t), v j + B(wm , wm ), v j + ∇Φε (wm ), v j = F, v j , wm (0) = w0 . (6.70) As (6.30) in the proof of Theorem 6.6 we get wm (t)2 ≤ C1 (w0 2 + F2L 2 (0,T ;V∗ ) ), wm 2L 2 (0,T ;V) ≤ C2 (w0 2 + F2L 2 (0,T ;V∗ ) ),

(6.71)

206

6 The Non-steady Navier-Stokes System with Friction Boundary Conditions

where C1 , C2 are independent of m and ε. (t) is For a t˜m there exists a differentiable function gim (t) on [0, t˜m ] such that gim absolute continuous (cf. proof of Theorem 6.7). Multiplying (6.70) by gim (t) and summing over i = 1, · · · , m, yield ⎧

⎪ ⎨ wm (t), wm (t) + A(t)wm (t), wm (t) + B(wm (t), wm (t)), wm (t) + ∇Φε (wm ), wm (t) = F, wm (t) , ⎪ ⎩ wm (0) = w0 .

(6.72)

We are going to derive a priori estimates for wm . (0) and adding on Setting t = 0 in (6.70), multiplying the resulting equation by gim i = 1, · · · , m, we get



wm (0)2 + A(0)wm (0), wm (0) + B(wm (0), wm (0)), wm (0)



(6.73) + ∇Φε (w0 ), wm (0) = F(0), wm (0) . By conditions (4), (5) for any u ∈ V, Φ(u) ≥ Φ(w0 ), and so ∇Φε (w0 ) = 0. Thus, we have from (6.73) wm (0) ≤ A(0)w0 + B(w0 , w0 ) − F(0).

(6.74)

On the other hand, taking into account (6.28), we have from (6.72) c1 wm (t)2V ≤FV∗ wm (t)V +c2 wm (t)2 − (wm (t), wm (t)), and so c1 wm (t)V ≤ F L ∞ (0,T ;V∗ ) + α1 wm (t) + c2 α1 wm (t),

(6.75)

where and in what follows of this section α1 is the one in  ·  ≤ α1  · V . Differentiating (6.70) yields







wm (t), v j + A(t) wm (t), v j + A(t)wm (t), v j + B(wm (t), wm (t)), v j





(6.76) + B(wm (t), wm (t)), v j + (∇Φε (wm )) , v j = F , v j .

(t), summing for i and observing the fact that (∇Φε (wm )) , Multiplying (6.76) by gim wm  ≥ 0 due to monotonicity of ∇Φε (see (1.38)), by conditions (2), (3) we have

2c2 √ c1 2 d 2 2 (t)2 + (c1 − 2c5 α1 wm (t)V )wm V + c1 wm V − wm V − 4 wm 2V wm dt 2 c1 c1 2 2 2 2 ≤ wm V + F V∗ + 2c2 wm  , 2 c1

6.2 The Existence and Uniqueness of Solutions to Variational Inequalities

207

√ where | B(wm , wm ), wm | ≤ c5 α1 wm V wm 2V , B(wm , wm ), wm = 0 and

c1 2c2 2| A(t) wm (t), wm (t) | ≤ wm 2V + 4 wm 2V 2 c1 were used. Thus, 2c2 √ 2 d 2 2 wm (t)2 + (c1 − 2c5 α1 wm (t)V )wm V ≤ F 2V∗ + 2c2 wm  + 4 wm 2V . dt c1 c1

(6.77)

If wm (0)V = w0 V
0 for t ∈ [0, tm ]. Therefore, we have from (6.74) and (6.77) that wm (t)2 ≤Aw0 + B(w0 , w0 ) − F(0)2    t 2 T 2c2 T + F 2V∗ dt + 4 wm (s)2V ds + 2c2 wm (s)2 ds c1 0 c1 0 0 for t ∈ [0, tm ], from which by Gronwall’s inequality and (6.71) it follows that

2c2 2 wm (t)2 ≤ Aw0 + B(w0 , w0 ) − F(0)2 + 4 wm 2L 2 (0,T ;V) + F 2L 2 (0,T ;V∗ ) e2c2 T c1 c1 2C   2c 2 w0 2 + F2L 2 (0,T ;V∗ ) ≤ Aw0 + B(w0 , w0 ) − F(0)2 + 4 c1

2 + F 2L 2 (0,T ;V∗ ) e2c2 T . c1

(6.79)

Using (6.71) and (6.79), we have from (6.75)  1 F L ∞ (0,T ;V∗ ) + (α1 wm (t) + c2 α1 wm (t)) c1    2 1 F L ∞ (0,T ;V∗ ) + α1 Aw0 + B(w0 , w0 ) − F(0) + α1 ≤ F  L 2 (0,T ;V∗ ) ec2 T c1 c1 

  2c42 C2 c2 T  (w0  + F L 2 (0,T ;V∗ ) ) . + c2 α1 C1 + α1 e c1

wm (t)V ≤

(6.80) Putting

208

6 The Non-steady Navier-Stokes System with Friction Boundary Conditions    1 2 F  L 2 (0,T ;V∗ ) ec2 T β≡ F L ∞ (0,T ;V∗ ) + α1 Aw0 + B(w0 , w0 ) − F(0) + α1 c1 c1 

  2c42 C2 c2 T  + c2 α1 C1 + α1 e (w0  + F L 2 (0,T ;V∗ ) ) , c1

(6.81) we have from (6.80) wm (t)V ≤ β ∀t ∈ [0, tm ).

(6.82)

By the assumption of theorem, we can assume that β is small so that √ c1 c1 − 2c5 α1 β ≥ . 2 Now, let us prove that if w(0)V ≤

c√ 1 4c5 α1

(6.83)

and β satisfies (6.83), then for any m

√ c1 − 2c5 α1 wm (t)V > 0 ∀t ∈ [0, T ].

(6.84)

To this end, let us assume that there exists an m and t m (≤ T ) such that √ c1 − 2c5 α1 wm (t)V > 0 ∀t ∈ [0, t m ), √ c1 − 2c5 α1 wm (t m )V = 0.

(6.85)

Then, by (6.82) and (6.83), we have √ c1 c1 − 2c5 α1 wm (t)V > ∀t ∈ [0, t m ), 2

(6.86)

which is a contradiction to the second formula of (6.85), and so we get (6.84). From (6.83) we have (6.82) with tm = T . Therefore, by (6.79), (6.83), (6.77) and (6.86) we have that wm (t)V ≤ const ∀t ∈ [0, T ], ∀m, ∀ε > 0, wm (t) ≤ const ∀t ∈ [0, T ], ∀m, ∀ε > 0, wm  L 2 (0,T ;V) ≤ const ∀m, ∀ε > 0.

(6.87)

Extracting a subsequence and passing to limit in (6.70) as m → ∞, and then ε → 0 and using (6.27), (6.45), in a similar way as in the proof of Theorem 6.6 (from (6.41)) we can prove the existence of a solution w to (6.51) such that w ∈ L 2 (0, T ; V) ∩ L ∞ (0, T ; H ), and so w ∈ C([0, T ]; V). By the same argument as in Theorem 6.7 we get the estimates for solutions. It only remains to prove the uniqueness of solution to the problem. Let w1 , w2 be two solutions asserted in Theorem 6.7. Then, as (6.67) we have from (6.51)

6.2 The Existence and Uniqueness of Solutions to Variational Inequalities

209

1 ∂w1 (t) − w2 (t)2 + c1 w1 (t) − w2 (t)2V 2 ∂t 3

1

≤ c2 w1 (t) − w2 (t)2 + w2 (t)V w1 (t) − w2 V2 w1 (t) − w2  2 ≤ c2 w1 (t) − w2 (t)2 + c1 w1 (t) − w2 2V + K w2 (t)4V w1 (t) − w2 2 . (6.88) Equation (6.88) implies  w1 (t) − w2 (t) ≤ const 2

0

t

(c2 + K w2 (s)4V )w1 (s) − w2 (s)2 ds,

which by Gronwall’s inequality deduces w1 (t) = w2 (t) for all t ∈ [0, T ].



Remark 6.7 Following Theorem 3.7 of [1], to study existence of a unique solution to a variational inequality under smallness condition of data in Theorem 2 of [2] the author assumed that F ∈ L ∞ (0, T ; V∗ ) ∩ L 1 (0, T ; H ), F ∈ L 1 (0, T ; H ). The right-hand side of (6.15) for the definition of F2 includes some boundary integrals: ˜ Γ3 − 2(α(x)Uτ , u)Γ5 − 2μ(k(x)U, u)Γ2 − 2μ(SU˜ , u)   − μ(k(x)U, u)Γ7 +

φi , u n Γi +

φi , uΓi ∀u ∈ V. i=2,4,7

i=3,5,6

1

For fixed t let us assume that φi (t) ∈ H 2 (Γi ), i = 2, 4, 7. Due to u τ |Γi = 0, u · n ∈ H − 2 (Γi ), u · n H − 21 (Γ ) ≤ K u H 1

i

(see Proposition 2.1), and so | φi , u n Γi | ≤ φi  H 21 (Γ ) u n  H − 21 (Γ ) ≤ K φi  H 21 (Γ ) u H ≤ K 1 u H . i

i

i

Then, there exist elements f i ∈ H such that φi , u n Γi = ( f i , u). Thus, for φi ∈ 1 L 2 (0, T ; H 2 (Γi )), i = 2, 4, 7, there exist f i ∈ L 2 (0, T ; H ) such that  i=2,4,7

φi , u n Γi =



f i , u H ∀u ∈ V.

i=2,4,7

But, for other boundary integrals we can not find such functions belonging to L 2 (0, T ; H ).

210

6 The Non-steady Navier-Stokes System with Friction Boundary Conditions

Therefore, the assumption such that F ∈ L 1 (0, T ; H ), F ∈ L 1 (0, T ; H ) is not suitable for our problems with mixed boundary conditions, and in Theorems 6.8 and 6.9 we assume that F ∈ L 2 (0, T ; V∗ ), F ∈ L 2 (0, T ; V∗ ). The following theorem is for small data in (6.51) corresponding to Problem II (the case of static pressure). Theorem 6.9 Assume that

(1) For any u, v ∈ V, A(t)u, v, A (t)u, v are continuous with respect to t and

A(t)u, u ≥ c1 u2V − c2 u2 ∃c1 > 0, c2 ≥ 0, ∀u ∈ V, | A(t)u, v | ≤ c3 uV vV ∃c3 > 0, ∀u, v ∈ V,

| A (t)u, v | ≤ c1 α2 uV vV ∃α2 (0 ≤ α2 < 1), ∀u, v ∈ V; (2) B(u, v) is bilinear continuous from V × V into V∗ and 1 c5 1

B(w, v), u ≤ √ wV vV u 2 uV2 ; α1

(3) F, F ∈ L 2 (0, T ; V∗ ); (4) The functional Φ : V → R is proper lower weak semi-continuous and Φ(u) ≥ 0 ∀u ∈ V, Φ(0V ) = 0 ; (5) w0 ∈ V, Φ(w0 ) = 0 and (A(0)w0 + B(w0 , w0 ) − F(0)) ∈ H (a compatibility condition at the initial time). If w0 V ,  ( A(0)w0 + B(w0 , w0 ) − F(0))  H and norms of F, F in the spaces that they belong to are small enough, then there exists a unique solution v to problem (6.51) such that w ∈ C([0, T ]; V), w ∈ L 2 (0, T ; V) ∩ L ∞ (0, T ; H ).

(6.89)

Proof First, let us prove the existence of the asserted solution. Let {v j , j = 1, 2, · · · } be a basis of V. Without loss of generality, we assume that m gim (t)vi to problem v1 = w0 . We find a solution wm = i=1 









wm (t), v j + A(t)wm (t), v j + B(wm , wm ), v j + ∇Φε (wm ), v j = F, v j , wm (0) = w0 , (6.90) which gives us an ordinary differential system for gim (t), i = 1, · · · , m. In the same way as in Theorem 6.7 we see that for a t˜m > 0 there exists a solution gim (t) on the (t) is absolute continuous. If wm (t) is bounded, then gim (t) interval [0, t˜m ] and gim

6.2 The Existence and Uniqueness of Solutions to Variational Inequalities

211

can be extended over t˜m . Under smallness of data we will find estimate for wm (t) below, by which we can say that t˜m = T. Multiplying (6.90) by gim (t) and adding for i = 1, · · · , m, yield

⎧ ⎪ ⎨ wm (t), wm (t) + A(t)wm (t), wm (t) + B(wm , wm ), wm (t) + ∇Φε (wm ), wm (t) = F, wm (t) , ⎪ ⎩ wm (0) = w0 .

(6.91) We will find a priori estimates for wm (t)2 + wm (t)2 . By virtue of (6.28), we have from (6.91) d wm (t)2 + 2c1 wm (t)2V − 2c5 wm (t)3V dt 1 F2V∗ + c1 (1 − α2 )wm (t)2V + 2c2 wm (t)2 , ≤ c1 (1 − α2 ) and so   d wm (t)2 + c1 − 2c5 wm (t)V wm (t)2V + c1 α2 wm (t)2V dt 1 F2 2V∗ + 2c2 wm (t)2 . ≤ c1 (1 − α2 ) (6.92) (0) and adding for Setting t = 0 in (6.90), multiplying the resulting equation by gim i = 1, · · · , m, we get





wm (0)2 + A(0)wm (0), wm (0) + B(wm (0), wm (0)), wm (0) + ∇Φε (w0 ), wm (0) (6.93)

(0) . = F(0), wm

By conditions (4), (5) we have that for any u ∈ V, Φ(u) ≥ Φ(w0 ), and so ∇Φε (w0 ) = 0. Thus, we have from (6.91) wm (0) ≤ A(0)w0 + B(w0 , w0 ) − F(0).

(6.94)

On the other hand, taking into account (6.28), we have from (6.91) c1 wm (t)2V ≤ F(t)V∗ wm (t)V + c5 wm (t)3V + c2 wm (t)2 − (wm (t), wm (t)), and so c1 wm (t)V ≤ F(t)V∗ + c5 wm (t)2V + (α1 wm (t) + c2 α1 wm (t)), (6.95) where α1 is in  ·  ≤ α1  · V .

212

6 The Non-steady Navier-Stokes System with Friction Boundary Conditions

Differentiating (6.90) yields







wm (t), v j + A (t)wm (t), v j + A(t)wm (t), v j + B(wm (t), wm (t)), v j





(6.96) + B(wm (t), wm (t)), v j + (∇Φε (wm )) , v j = F , v j .

(t), summing for i and observing the fact that (∇Φε (wm )) , Multiplying (6.96) by gim wm  ≥ 0 (see (1.38)), by conditions (2), (3) we have

d w (t)2 + (c1 − 4c5 wm (t)V )wm (t)2V + c1 wm (t)2V dt m − c1 α2 wm (t)2V − c1 α2 wm (t)2V 1 F (t)2V∗ + 2c2 wm (t)2 , ≤ c1 (1 − α2 )wm (t)2V + c1 (1 − α2 )



where | B(wm , wm ), wm + B(wm , wm ), wm | ≤ 2c5 wm V wm 2V was used. Thus, d w (t)2 + (c1 −4c5 wm (t)V )wm (t)2V − c1 α2 wm (t)2V dt m 1 F (t)2V∗ + 2c2 wm (t)2 . ≤ c1 (1 − α2 )

(6.97)

Adding (6.92) and (6.97) yields    d  wm (t)2 + wm (t)2 + (c1 − 4c5 wm (t)V ) wm (t)2V + wm (t)2V dt     1 F(t)2V∗ + F (t)2V∗ + 2c2 wm (t)2 + wm (t)2 . ≤ c1 (1 − α2 ) (6.98) If wm (0)V = w0 V < 4cc15 , then there exists a tm such that c1 − 4c5 wm (t)V ≥ 0 for t ∈ [0, tm ]. Therefore, we from (6.94) and (6.98) have 

   wm (t)2 + wm (t)2 ≤ α1 w0 2 + Aw0 + B(w0 , w0 ) − F(0)2  T   1 F2V∗ + F 2V∗ dt + c1 (1 − α2 ) 0  t   wm (s)2 + wm (s)2 ds + 2c2 0

for t ∈ [0, tm ], from which by Gronwall’s inequality we get   wm (t)2 + wm (t)2 ≤ α1 w0 2V + Aw0 + B(w0 , w0 ) − F(0)2 +

 

1 F2L 2 (0,T ;V∗ ) + F 2L 2 (0,T ;V∗ ) e2c2 T . c1 (1 − α2 ) (6.99)

6.2 The Existence and Uniqueness of Solutions to Variational Inequalities

213

Using the estimate, we will obtain a quadratic inequality satisfied by wm (t)V . Put √ β ≡F L ∞ (0,T ;V∗ ) + max{α1 , c2 α1 } 2 α1 w0 2V + Aw0 + B(w0 , w0 ) − F(0)2   21 c2 T 1 F2L 2 (0,T ;V∗ ) + F 2L 2 (0,T ;V∗ ) + e . c1 (1 − α2 ) (6.100) Taking into account  √  1  wm (t) + wm (t) ≤ 2 wm (t)2 + wm (t)2 2 , we have from (6.99)   √ wm (t) + wm (t) ≤ 2 α1 w0 2V + Aw0 + B(w0 , w0 ) − F(0)2   1/2 c2 T 1 F2L 2 (0,T ;V∗ ) + F 2L 2 (0,T ;V∗ ) + e . c1 (1 − α2 ) (6.101) Taking into account (6.101) and (6.100), we have from (6.95) a quadratic inequality for wm (t)V 0 ≤ β − c1 wm (t)V + c5 wm (t)2V ∀t ∈ [0, tm ],

(6.102)

which is the one we want. By the assumption of theorem, we can assume that β is small so that c12 − 4c5 β > Now, let us prove that if w(0)V ≤

4c12 . 9

(6.103)

√

c1 −

c1 − 4c5 wm (t)V >

c12 −4c5 β (< 6cc15 ), 2c5

then for any m, ε

c1 ∀t ∈ [0, T ]. 4

Taking into account c1 − 4c5 wm (0)V >

(6.104)

c1 , 3

let us assume that there exists an m and t m (≤ T ) such that c1 − 4c5 wm (t)V >

c1 c1 ∀t ∈ [0, t m ) and c1 − 4c5 wm (t m )V = . (6.105) 4 4

Setting y = wm (t)V in (6.102), we get

214

6 The Non-steady Navier-Stokes System with Friction Boundary Conditions

0 ≤ β − c1 y + c5 y 2 ∀t ∈ [0, t m ].

(6.106)

By virtue of (6.103), there exist two real roots of quadratic polynomial β − c1 y + c5 y 2   y1 =

c1 −

c12 − 4c5 β

2c5

and y2 =

c1 +

c12 − 4c5 β

2c5

,

and on the intervals [0, y1 ] and [y2 , +∞) (6.106) holds. Thus, by the continuity of wm (t)V with respect to t we have from w(0)V ∈ [0, y1 ] that wm (t)V ∈ [0, y1 ] ∀t ∈ [0, t m ), that is, wm (t)V ≤

c1 −

 c12 − 4c5 β 2c5

Thus, c1 − 4c5 wm (t)V >


0; wm  L 2 (0,T ;V) ≤ const ∀m, ∀ε > 0; wm (t) ≤ const ∀t ∈ [0, T ], ∀m, ∀ε > 0;

(6.108)

wm  L 2 (0,T ;V) ≤ const ∀m, ∀ε > 0. Extracting a subsequence, passing to limit in (6.41) as m → ∞ and then letting ε → 0 and using (6.27) and (6.45), by a way similar to the proof of Theorem 6.6 we can prove the existence of a solution w to (6.51) such that w ∈ L 2 (0, T ; V) ∩ L ∞ (0, T ; H ), and so w ∈ C([0, T ]; V). As in Theorem 6.7 we can prove uniqueness.  Remark 6.8 In Sect. 4.2 the non-steady Navier-Stokes equations with boundary condition (5.2) with Γi = ∅, i = 8, 9, 10, 11, is considered. For the proof of existence of a unique solution, the following property of the linear problem obtained by neglecting the nonlinear term B(w, w) from 

w (t) + Aw + B(w, w) = f (t), u(0) = ϕ,

was essential. Property: (Lemma 4.6) When X = {w ∈ L 2 (0, T ; V); w ∈ L 2 (0, T ; V), w ∈ L 2 (0, T ; V∗ )} and Y = {w ∈ L 2 (0, T ; V∗ ); w ∈ L 2 (0, T ; V∗ )}, a map u → {u (0), Lu ≡ u + Au} is linear continuous one-to-one from X onto H × Y .

6.2 The Existence and Uniqueness of Solutions to Variational Inequalities

215

But, in the case of Problem II in this chapter, owing to the boundary conditions on Γi (= ∅), i = 8, 9, 10, 11, neglecting the nonlinear term B(w, w), we only get a nonlinear problem ⎧ ∂w

⎪ ⎪ ⎨ ∂t , u − w + Aw, u − w + Φ(u) − Φ(w) ≥ f, u − w for a.e. t ∈ (0, T ) ∀u ∈ V, ⎪ ⎪ ⎩ w(0) = w0 ∈ V, which seems not to have the property above (cf. Proposition 6.1), and so we can not use the method in Sect. 4.2.

6.3 Solutions to the Non-steady Navier-Stokes Problems 6.3.1 Existence of a Solution: The Case of Total Pressure Let us study Problem I-VI. Theorem 6.10 Under Assumption 6.1 there exists a solution to Problem I-VI (the non-steady Navier-Stokes problem for the case of total pressure) for any f, φi , i = 2 ∼ 7 and initial function v0 such that v0 − U (0) ∈ H . Moreover the following estimate for the solution holds: v(t) − U (t) ≤ K ∀t ∈ [0, T ], v − U  L 2 (0,T ;V) ≤ K ,

(6.109)

where  ∂U   2 K = v0 − U (0) +  ∗ + U  L 2 (0,T ;H1 ) + U  L 2 (0,T ;H1 ) U  L ∞ (0,T ;H1 ) ∂t L (0,T ;V )

  . +  f  L 2 (0,T ;V∗ ) + φi  2 + φi  2 −1 −1 i=2,4,7

L (0,T ;H

2

(Γi ))

i=3,5,6

L (0,T ;H

2

(Γi ))

Proof By Assumption 6.1 we can prove that F1 ∈ L 2 (0, T ; V∗ ) and F1 ∈ L 2 (0, T ; V∗ ). Define an operator A1 (t) : V → V∗ by

A1 (t)v, u = a01 (t, v, u) ∀v, u ∈ V,

(6.110)

where a01 (t, v, u) is the one in (6.4). Then, A1 (t)v, u is continuous with respect to t. By Korn’s inequality,

216

6 The Non-steady Navier-Stokes System with Friction Boundary Conditions

2Σi j εi j (z)2 ≥ C1 z2H1 (Ω) ,

C1 > 0.

(6.111)

Since rot U (t) × w, w = 0, by Young’s inequality and Assumption 6.1, we have | rot U (t) × w, w + rot w × U (t), w| ≤

μC1 w2H1 (Ω) + C2 w2 , 4

C2 > 0 (6.112) for any t ∈ [0, T ]. On the other hand, by virtue of Remark 3.4 and Assumption 6.1 there exists a constant M such that S(x)∞ , k(x)∞ , α L ∞ (Γ5 ) ≤ M. Therefore, there exists C3 such that   μ(k(x)z, z)Γ + 2μ(S z˜ , z˜ )Γ + (α(x)z, z)Γ + μ(k(x)z, z)Γ  ≤ μC1 z2 1 + C3 z2 2 3 7 5 H (Ω) 4 ∀z ∈ V

(6.113) (see Theorem 1.27). Setting c2 = C2 + C3 , we have therefore from (6.110)–(6.113)

A1 (t)z, z ≥ c1 z2V − c2 z2 , c1 =

μC1 > 0, ∀z ∈ V. 2

(6.114)

By Hölder’s inequality and Assumption 6.1, we have | rot U (t) × u, v + rot u × U (t), v| ≤ K uV vV ∀u, v ∈ V

(6.115)

for any t ∈ [0, T ]. Also, we can prove that | A1 (t)u, v | ≤ K uV vV ∀u, v ∈ V.

(6.116)

Let us prove that lim inf A1 (t)u m , u m  ≥ A1 (t)u, u as u m  u in V. m→∞

(6.117)

By (6.4) we have

A1 (t)u m , u m  ≡ a01 (t, u m , u m ) = 2μ(E (u m ), E (u m )) + rot U (t) × u m , u m  + rot u m × U (t), u m  + 2μ(k(x)u m , u m )Γ2 + 2μ(S u˜ m , u˜ m )Γ3 + 2(α(x)u m , u m )Γ5 + μ(k(x)u m , u m )Γ7 .

(6.118) Taking into account lim inf (E(u m ), E(u m )) ≥ (E(u), E(u)) as u m  u in V m→∞

6.3 Solutions to the Non-steady Navier-Stokes Problems

217

(see (1.1)), from (6.118) we get (6.117). Define a bilinear operator B1 : V × V → V∗ by

B1 (z, v), u = a11 (z, v, u) ∀z, v, u ∈ V,

(6.119)

where a11 (t, v, u) is the one in (6.4). Then, by a property of mixed production of vectors,

B1 (w, v), v = rot w × v, v = 0. (6.120) Using Hölder’s inequality, imbedding theorems and interpolation inequality vL3 ≤ 1

1

K vL2 2 vL2 6 , we have | B1 (w, v), z | = | rot w × v, z| ≤ K rot wL2 vL3 zL6 1

1

≤ K wV v 2 vV2 zV .

(6.121)

By the definition of the functional Φ, we know Φ(u) ≥ 0 and Φ(0V ) = 0. Let F1 be the one in (6.4). Then, by Assumption 6.1 we can prove  ∂U   2 F1  L 2 (0,T ;V∗ ) ≤M1  ∗ + U  L 2 (0,T ;H1 ) + U  L 2 (0,T ;H1 ) |U  L ∞ (0,T ;H1 ) ∂t L (0,T ;V )   +  f  L 2 (0,T ;V∗ ) + φi  2 + φi  2 −1 −1 i=2,4,7

L (0,T ;H

2

(Γi ))

i=3,5,6

L (0,T ;H

2

(Γi ))

.

(6.122) By virtue of (6.114), (6.116) and (6.120)–(6.122), we can verify that all conditions of Theorem 6.6 are satisfied, and we come to the conclusion.  Theorem 6.11 Assume that (1) In addition to Assumption 6.1, ∂t∂ U ∈ W 1,2 (0, T ; H1 (Ω)); 1 (2) f, f ∈ L 2 (0, T ; V∗ ), φi ∈W 1,2 (0, T ; H − 2 (Γi )), i = 2, 4, 7, 1 φi ∈ W 1,2 (0, T ; H− 2 (Γi )), i = 3, 5, 6; (3) w0 ≡ v0 − U ∈ V, w0 |∪i=8−11 Γi = 0 and (A1 (0)w0 + B1 (w0 , w0 ) − F1 (0)) ∈ H,

and

(6.123)

where F1 is the one in (6.4) and A1 , B1 are in (6.110), (6.119), respectively. If l = 2, then there exists a unique solution to Problem I-VI’ satisfying (6.109) and v − U ≡ w ∈ C([0, T ]; V), w ∈ L 2 (0, T ; V) ∩ L ∞ (0, T ; H ).

(6.124)

In the case of l = 3, if w0 V , A1 w0 + B1 (w0 , w0 ) − F1 (0), U W 1,2 (0,T ;H1 ) , 2  ∂t∂ 2 U  L 2 (0,T ;(H1 )∗ ) and norms of f, f , φi in the spaces of condition (2) are small

218

6 The Non-steady Navier-Stokes System with Friction Boundary Conditions

enough, then there exists a unique solution v to Problem I-VI’ satisfying (6.124) and (6.109). Proof We know that the first two of condition (6.52) of Theorem 6.7 are satisfied (see (6.114), (6.116)). By (6.110), we know that  

∂a01 (t, v, u)  ∂U (t) ∂U (t)  A 1 (t)v, u = = rot × v, u + rot v × ,u , ∂t ∂t ∂t

which show that A 1 (t)v, u is Lipschitz continuous with respect to t. Also,

 ∂U (t)   

 2 vV uV + rot vL2  ∂U (t)  1 uV | A 1 (t)v, u | ≤ rot L H ∂t ∂t and the third one of condition (6.52) of Theorem 6.7 is satisfied since   ∂ ∂  U (t) 1,2  U (t) . 1 ≤ K C([0,T ];H ) W (0,T ;H1 ) ∂t ∂t By Hölder’s inequality and interpolation inequality, we have | B1 (w, v), z | = | rot w × v, z| ≤ K rot wL2 vL4 zL4 2

1

≤ K wV vL4 z 3 zV3

(6.125)

for l = 2 and | B1 (w, v), z | = | rot w × v, z| ≤ K rot wL2 vL6 zL3 1

1

≤ K wV vV z 2 zV2

(6.126)

for l = 3. Since

F1 , u = −

∂U

, u − 2μ(E(U ), E(u)) − rot U × U, u

∂t − 2μ(k(x)U, u)Γ2 − 2μ(SU˜ , u) ˜ Γ3 − 2(α(x)Uτ , u)Γ5   − μ(k(x)U, u)Γ7 + f, u +

φi , u n Γi +

φi , uΓi , i=2,4,7

by the conditions we have

i=3,5,6

F1 , F1 ∈ L 2 (0, T ; V∗ ).

It is easy to verify that condition (4) of Theorem 6.7 is satisfied. By the condition w0 |∪i=8−11 Γi = 0 and (5.22), we know Φ(w0 ) = 0, which shows that condition (5) of Theorem 6.7 is satisfied. Thus, in the case of 2-D by Theorem 6.7 we come to the conclusion.

6.3 Solutions to the Non-steady Navier-Stokes Problems

219

1 If  ∂U  , U  L 2 (0,T ;H1 ) are small and U ∈ C(0, T ; H1 ), then under ∂t L 2 (0,T ;(H )∗ ) smallness conditions of the theorem for f, φi we can know that F1  L 2 (0,T ;V∗ ) is small. 2 Also if  ∂∂tU2  L 2 (0,T ;(H1 )∗ ) ,  ∂t∂ U  L 2 (0,T ;H1 ) are small and U ∈ C(0, T ; H1 ), then under smallness conditions of the theorem for f, φi we can know that F1  L 2 (0,T ;V∗ ) 2 is small. Thus, if U W 1,2 (0,T ;H1 ) ,  ∂t∂ 2 U  L 2 (0,T ;(H1 )∗ ) and norms of f, f , φi in the spaces of condition (2) are small enough, then norms of F1 , F1 in the space L 2 (0, T ; V∗ ) are small. Therefore, in case of 3-D by Theorem 6.8 we come to the conclusion. 

Remark 6.9 From (6.109) we know that the estimates for solutions are independent of the thresholds of friction conditions gτ , gn , g+n , g−n .

6.3.2 Existence of a Unique Solution: The Case of Static Pressure Let us study now Problem II-VI. Theorem 6.12 Assume that ∈ W 1,2 (0, T ; H1 (Ω)) hold; (1) Assumption 6.1 and dU dt 1 (2) f, f ∈ L 2 (0, T ; V∗ ), φi ∈ W 1,2 (0, T ; H − 2 (Γi )), i = 2, 4, 7, and φi ∈ 1 W 1,2 (0, T ; H− 2 (Γi )), i = 3, 5, 6; (3) w0 ≡ v0 − U ∈ V and w0 |∪i=8−11 Γi = 0 and (A2 (0)w0 + B2 (w0 , w0 ) − F2 (0)) ∈ H,

(6.127)

where A2 , B2 and F2 are the one, respectively, (6.17) and (6.15). If w0 V ,  (A2 (0)w0 + B2 (w0 , w0 ) − F2 (0)) , U W 1,2 (0,T ;H1 ) , U W 1,2 (0,T ;H1 ) and norms of f, f , φi in the spaces that they belong to are small enough, then there exists a unique solution v to Problem II-VI such that v − U ≡ w ∈ C([0, T ]; V), w ∈ L 2 (0, T ; V) ∩ L ∞ (0, T ; H ).

(6.128)

Proof First, let us prove the existence of the desired solution. The operator A2 (t) : V → V∗ was defined in (6.17) by

A2 (t)v, u = a02 (t, v, u) ∀v, u ∈ V.

(6.129)

Then, A2 (t)v, u is continuous with respect to t. Note the operator A2 (t) is not symmetric. By Korn’s inequality 2Σi j εi j (z)2 ≥ C1 z2H1 (Ω) , If U C([0,T ;H1 ) is small enough, then

C1 > 0.

(6.130)

220

6 The Non-steady Navier-Stokes System with Friction Boundary Conditions

μC1 wH1 (Ω) uH1 (Ω) . 4

(6.131)

μC1 > 0, c2 > 0 ∀z ∈ V. 2

(6.132)

| (U (t) · ∇)w, u + (w · ∇)U (t), u| ≤ By (6.130), (6.131) and (6.113), we have

A2 (t)z, z ≥ c1 z2V − c2 z2 c1 = Using (6.130), we can prove that

| A2 (t)u, v | ≤ c3 uV vV ∀u, v ∈ V.

(6.133)

By (6.129), we know

 

 ∂U (t) ∂U (t)  · ∇ w, u + (w · ∇) ,u , A 2 (t)w, u = ∂t ∂t

which shows that A 2 (t)w, u is continuous with respect to t. Also, U (t)C([0,T ;H1 ) is small enough, then  

   ∂U (t)

∂U (t)  | A 2 (t)w, u | =  · ∇ w, u + (w · ∇) ,u  ∂t ∂t ≤ c1 αwV uV , α < 1.

(6.134)

The bilinear operator B2 : V × V → V∗ was defined in (6.17) by

B2 (w, v), u = a12 (z, v, u) ∀w, v, u ∈ V.

(6.135)

Then, by virtue of Hölder’s inequality, imbedding theorems and interpolation inequal1

1

ity uL3 ≤ K u 2 uL2 6 we have | B2 (w, v), u | = | (w · ∇v), u| ≤ wL6 ∇vL2 uL3 1

1

≤ K wV vV u 2 uV2 .

(6.136)

Since

F2 , u = −

 ∂U

 , u − 2μ(E(U ), E(u)) − rot U × U, u

∂t − 2μ(k(x)U, u)Γ2 − 2μ(SU˜ , u) ˜ Γ3 − 2(α(x)U, u)Γ5   − μ(k(x)U, u)Γ7 + f, u +

φi , u n Γi +

φi , uΓi , i=2,4,7

by Conditions (1), (2) we have

i=3,5,6

6.3 Solutions to the Non-steady Navier-Stokes Problems

F2 , F2 ∈ L 2 (0, T ; V∗ ).

221

(6.137)

As in the proof of Theorem 6.11 we know that under conditions of the theorem F2  L 2 (0,T ;V∗ ) and F2  L 2 (0,T ;V∗ ) are also small enough. As in Theorem 6.11, we can verify that other conditions of Theorem 6.9 are satisfied as well. Therefore, by Theorem 6.9 we come to the asserted conclusion. 

6.3.3 Existence of a Unique Solution: The Stokes Problem To study Problem III-VI, we use the following known result. To state it we need to introduce a problem first. Let A be a linear continuous and symmetric operator from V to V∗ satisfying the coercivity condition

Au, u + αu ≥ ωu2V ω > 0, α ∈ R. We find w ∈ L 2 (0, T ; V) ∩ C([0, T ]; H ) ∩ W 1,2 (0, T ; V∗ ) such that ⎧  ∂w  ⎪ , u − w + Aw, u − w + Φ(u) − Φ(w) ≥ f, u − w ⎪ ⎨ ∂t for a.e. t ∈ (0, T ) ∀u ∈ V, ⎪ ⎪ ⎩ w(0) = w0 ∈ V,

(6.138)

where f ∈ L 2 (0, T ; V∗ ). Proposition 6.1 (Theorem 5.1 of [3]) Let f ∈ L 2 (0, T ; V∗ ) and w0 ∈ V be such that (6.139) {Aw0 + ∂Φ(w0 ) − f (0)} ∩ H = ∅. Then, problem (6.138) has a unique solution w ∈ W 1,2 ([0, T ]; V)∩W 1,∞ ([0, T ]; H ) and the map (w0 , f ) → w is Lipschitz from H × L 2 (0, T ; V∗ ) to C([0, T ]; H ) ∩ L 2 (0, T ; V). If f ∈ W 1,2 ([0, T ]; V∗ ) and Φ(w0 ) < ∞, then problem (6.138) has a unique solution w ∈ W 1,2 ([0, T ]; H ) ∩ Cw ([0, T ]; V). Theorem 6.13 Assume that (1) U ∈ L 2 (0, T ; H1 ), ∂U ∈ L 2 (0, T ; (H1 )∗ ); ∂t 1 2 ∗ (2) f ∈ L (0, T ; V ), φi ∈ L 2 (0, T ; H 2 (Γi )), i = 2, 4, 7, and φi ∈ L 2 (0, T ; − 21 H (Γi )), i = 3, 5, 6; (3) w0 ≡ v0 − U ∈ V and w0 |∪i=8−11 Γi = 0 and (A3 w0 − F3 (0)) ∈ H, where A3 and F3 are, respectively, as in (6.22) and (6.20).

(6.140)

222

6 The Non-steady Navier-Stokes System with Friction Boundary Conditions

Then, there exists a unique solution v to Problem III-VI for the non-steady Stokes problem with mixed boundary condition (5.2) and v − U C([0,T ];L2 )∩L 2 (0,T ;H1 )  ∂U   2 ≤ K w0  + U  L 2 (0,T ;H1 ) +  1 ∗ +  f  L 2 (0,T ;V∗ ) ∂t L (0,T ;(H ) )

  + φi  L 2 (0,T ;H − 21 (Γ ) + φi  L 2 (0,T ;H− 21 (Γ ) . i

i

i=2,4,7

i=3,5,6

(6.141) If v1 , v2 are solutions, respectively, to Problem-III-VI corresponding to data v01 , gτ 1 , gn1 , g+n1 , g−n1 , f 1 , h i1 , φi1 and v02 , gτ 2 , gn2 , g+n2 , g−n2 , f 2 , h i2 , φi2 , then v1 −v2 C([0,T ];L2 )∩L 2 (0,T ;H1 ) ≤ K v01 − v02 V + U1 − U2 C([0,T ];L2 ) + U1 − U2  L 2 (0,T ;H1 ) ∂U1 − U2  L 2 (0,T ;(H1 )∗ ) +  f 1 − f 2  L 2 (0,T ;V∗ ) + gτ 1 − gτ 2 Lτ2 (Γ8 ) ∂t + gn1 − gn2  L 2 (Γ9 ) + g+n1 − g+n2  L 2 (Γ10 ) + g−n1 − g−n2  L 2 (Γ10 )

  + φi1 − φi2  L 2 (0,T ;H − 21 (Γ )) + φi1 − φi2  L 2 (0,T ;H− 21 (Γ )) , +

i

i=2,4,7

i

i=3,5,6

(6.142) j where U j , j = 1, 2, are the functions in Assumption 6.1 with h i instead of h i . Proof By (6.111) and (6.113), we have

A3 w, w ≥ c1 w2V − c2 w2 , c1 > 0, c2 > 0.

(6.143)

Also, we can easily prove | A3 w, u| ≤ c3 wV uV .

(6.144)

By definition of F3 in (6.20), we know  ∂U   2 F3  L 2 (0,T ;V∗ ) ≤c  1 ∗ + U  L 2 (0,T ;(H1 )∗ ) +  f  L 2 (0,T ;V∗ ) ∂t L (0,T ;(H ) )

  + φi  L 2 (0,T ;H − 21 (Γ )) + φi  L 2 (0,T ;H− 21 (Γ )) . i

i=2,4,7

i

i=3,5,6

(6.145) By the condition w0 |∪i=8−11 Γi = 0, we know Φ(w0 ) = 0. Since Φ(u) ≥ 0, we have that ∂Φ(w0 ) = 0. Thus, taking into account Remark 6.5, by Proposition 6.1 we have existence of a unique solution and estimate (6.141). If v1 = w1 + U1 , v2 = w2 + U2 are solutions corresponding to the given data, we get

6.3 Solutions to the Non-steady Navier-Stokes Problems

 ∂w

223

 + A3 w1 , u − w1 + Φ1 (u) − Φ1 (w1 ) ≥ F31 , u − w1 , ∂t  ∂w  2 + A3 w2 , u − w2 + Φ2 (u) − Φ2 (w2 ) ≥ F32 , u − w2 , ∂t 1

(6.146)

where Φ j (u), F3 j , j = 1, 2, are one corresponding to U j , gτ j , gn j , g+n j , g−n j , f j , j j h i , φi . Putting u = w2 and u = w1 , respectively, in the first and second one of (6.146), and adding those, we have  ∂(w − w )  1 2 + A3 (w1 − w2 ), w2 − w1 + Φ1 (w2 ) − Φ1 (w1 ) + Φ2 (w1 ) − Φ2 (w2 ) ∂t (6.147) ≥ F31 − F32 , w2 − w1 .

From (6.143) and (6.147) we have ∂w1 − w2 2 + 2c1 w1 − w2 2V ≤ F31 − F32 2V∗ + w2 − w1 2V ∂t

+ 2|Φ1 (w2 ) − Φ1 (w1 ) + Φ2 (w1 ) − Φ2 (w2 )| + 2c2 w1 − w2 2 . (6.148) Since w1 , w2 ∈ K (Ω),   Φ1 (w2 ) − Φ1 (w1 ) = gτ 1 (|w2τ | − |w1τ |) ds + gn1 (|w2n | − |w1n |) ds Γ8 Γ9   + g+n1 (w2n − w1n ) ds − g−n1 (w2n − w1n ) ds, Γ10 Γ11   Φ2 (w2 ) − Φ2 (w1 ) = gτ 2 (|w2τ | − |w1τ |) ds + gn2 (|w2n | − |w1n |) ds Γ8 Γ9   + g+n2 (w2n − w1n ) ds − g−n2 (w2n − w1n ) ds. Γ10

Γ11

(6.149)

Subtracting two formulas in (6.149) yields |Φ1 (w2 ) − Φ1 (w1 ) + Φ2 (w1 ) − Φ2 (w2 )| ≤ G 12 +

c1 w2 − w1 2V , 2

(6.150)

where G 12 =

2 gτ 1 − gτ 2 2L2 (Γ8 ) + gn1 − gn2 2L 2 (Γ9 ) τ c1

+ g+n1 − g+n2 2L 2 (Γ10 ) + g−n1 − g−n2 2L 2 (Γ11 ) .

By (6.148) and (6.150) we have

224

6 The Non-steady Navier-Stokes System with Friction Boundary Conditions  c1 t w1 − w2 2V ≤ 2 0  t  t F31 (s) − F32 (s)2V∗ ds + G 12 t + 2c2 w1 (s) − w2 (s)2 ds. w10 − w20 2 +

w1 (t) − w2 (t)2 +

0

0

(6.151)

By Gronwall’s inequality, we have from (6.151)  w1 (t) − w2 (t)2 ≤ w10 − w20 2 + 0

w1 (t) − w2 (t)2L 2 (0,T ;V)

T

F31 (s) − F32 (s)2V∗ ds + G 12 T e2c2 T ,

 ≤ K w10 − w20 2 +

T 0

F31 (s) − F32 (s)2V∗ ds + G 12 .

(6.152) Also F31 − F32 V∗ is estimated as (6.145), and so from (6.152) we get estimate (6.142). 

6.4 Bibliographical Remarks The content of Chap. 6 is taken from [4]. One mistake in [4] is corrected here. In [5–7] the non-steady Stokes equations with the homogeneous Dirichlet boundary condition and leak boundary condition were studied. In [8] the existence of a weak solution and a local-in-time strong solution to the non-steady Navier-Stokes problem was studied when boundary consists of a portion with homogeneous Dirichlet boundary condition and another portion with the leak condition based on the total stress. In [2] a variational inequality for the NavierStokes problem with homogeneous Dirichlet boundary condition and one-sided leak condition based on the total stress was considered. Also, in [9] a variational inequality for the Navier-Stokes problems with homogeneous Dirichlet boundary condition and one-sided boundary conditions based on total pressure was studied. When boundary consists of separated portions with homogeneous Dirichlet boundary condition and with nonlinear slip condition or leak condition, in [10] existence of a unique strong solution to the non-steady Navier-Stokes problem was studied. In the case of nonlinear slip condition, existence of a unique strong solution to 2-D problem and a unique local-in-time strong solution to 3-D problem was proved. In the case of leak condition, existence of a unique local-in-time strong solution to 3-D problem was proved. The semi-discrete finite element approximation to the time-dependent 2-D NavierStokes equations with mixture of homogeneous Dirichlet boundary condition and Tresca slip boundary conditions was discussed in [11]. For other kinds of non-steady fluid equations with mixture of friction slip boundary conditions and Dirichlet condition, we refer to [12–15].

References

225

References 1. R. Temam, Navier-Stokes Equations (North-Holland, Amsterdam, 1985) 2. J. Naumann, On evolution inequalities of Navier-Stokes type in three dimensions. Ann. Math. Pure Appl. 124(4), 107–125 (1980) 3. V. Barbu, Nonlinear Differential Equations of Monotone Types in Banach Spaces (Springer, Berlin, 2010) 4. T. Kim, F. Huang, The non-steady Navier-Stokes systems with mixed boundary conditions including friction conditions. Methods Appl. Anal. 25, 13–50 (2018) 5. H. Fujita, Non-stationary Stokes flows under leak boundary conditions of friction type. J. Comput. Math. 19, 1–8 (2001) 6. H. Fujita, Variational inequalities and nonlinear semi-groups applied to certain nonlinear problems for the Stokes equation. RIMS Kokyuroku 1234, 70–85 (2001) 7. H. Fujita, A coherent analysis of Stokes flows under boundary conditions of friction type. J. Comput. Appl. Math. 149, 57–69 (2002) 8. R. An, Y. Li, K. Li, Solvability of Navier-Stokes equations with leak boundary conditions. Acta Math. Appl. Sin. Engl. Ser. 25, 225–234 (2009) 9. D.S. Konovalova, Subdifferential boundary value problems for the nonstationary NavierStokes and Stokes equations. Differ. Equ. 36, 875–885 (2000) 10. T. Kashiwabara, On a strong solution of the non-stationary Navier-Stokes equations under slip or leak boundary conditions of friction. J. Differ. Equ. 254, 756–778 (2013) 11. Y. Li, R. An, Semi-discrete stabilized finite element methods for Navier-Stokes equations with nonlinear slip boundary conditions based on regularization procedure. Numer. Math. 117, 1–36 (2011) 12. M. Boukrouche, I. Boussetouan, L. Paoli, Non-isothermal Navier-Stokes system with mixed boundary conditions and friction law: uniqueness and regularity properties. Nonlinear Anal. 102, 168–185 (2014) 13. M. Boukrouche, I. Boussentoua, L. Paoli, Existence for non-isothermal fluid flows with Tresca’s friction and Cattaneo’s heat law. J. Math. Anal. Appl. 427, 499–514 (2015) 14. M. Boukrouche, G. Łukaszewicz, On global in time dynamics of a planar Bingham flow subject to a subdifferential boundary conditions. Discret. Contin. Dyn. Syst. 34, 3969–3983 (2014) 15. J.K. Djoko, P.A. Razafimandimby, Analysis of the Brinkman-Forchheimer equations with slip boundary conditions. Appl. Anal. 93, 1477–1494 (2014)

Chapter 7

The Steady Boussinesq System

In this chapter we are concerned with the steady Boussinesq system with mixed boundary conditions. The boundary conditions for fluid may include Tresca slip, leak condition, one-sided leak conditions, velocity, pressure, vorticity, stress together and the conditions for temperature may include Dirichlet, Neumann and Robin conditions together. We will get variational formulations consisting of a variational inequality for velocity and a variational equation for temperature, which are equivalent to the original PDE problems for smooth solutions. Then, we will study the existence of solutions to the variational problems. For the problem with boundary conditions involving the static pressure and stress, it is proved that if the data of problem are small enough, then there exists a unique solution. For the problem with boundary conditions involving the total pressure and total stress, the existence of a solution is proved as well without smallness of the data.

7.1 Problems and Variational Formulations Let us consider the steady Boussinesq system   ⎧ ⎪ ⎨ − 2∇ · μ(θ )E(v) + (v · ∇)v + ∇ p = (1 − α0 θ ) f, div v = 0, ⎪   ⎩ − ∇ · κ(θ )∇θ + v · ∇(γ (θ )θ ) = g

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 T. Kim and D. Cao, Equations of Motion for Incompressible Viscous Fluids, Advances in Mathematical Fluid Mechanics, https://doi.org/10.1007/978-3-030-78659-5_7

(7.1)

227

228

7 The Steady Boussinesq System

under mixed boundary conditions. Here v, p and θ are, respectively, velocity, pressure and temperature, and α0 —parameter for buoyancy effect, f —body force, g—heat source, κ(θ )—thermal conductivity, γ (θ )—specific heat of the fluid. The viscosity, thermal conductivity and specific heat of fluid depend on the temperature. Let Ω be a bounded domain of Rl , l = 2, 3. ∂Ω ∈ C 0,1 , 11 Γi = ΓD ∪Γ R ∂Ω = ∪i=1

 and Γ D ∩ Γ R = ∅, Γi ∩ Γ j = ∅ for i = j, Γi = j Γi j , where Γi j are connected open subsets of ∂Ω, Γi j ∈ C 2 for i = 2, 3, 7 and Γi j ∈ C 1 for others. (See Remark 6.1.) For temperature we are concerned with the boundary conditions (1) θ |Γ D = 0,   ∂θ + β(x)θ Γ R = g R (x), β(x), g R (x) − given functions on Γ R . (2) κ(θ ) ∂n (7.2) According to boundary conditions for fluid, the Problems I and II are distinguished. Problem I is the one with the following boundary conditions (the case of static pressure) (1) v|Γ1 = 0, (2) vτ |Γ2 = 0, − p|Γ2 = φ2 , (3) vn |Γ3 = 0, rot v × n|Γ3 = φ3 /μ(θ ), (4) vτ |Γ4 = 0, (− p + 2μ(θ )εnn (v))|Γ4 = φ4 , (5) vn |Γ5 = 0, 2(μ(θ)εnτ (v) + αvτ )|Γ5 = φ5 , α : a matrix, (6) (− pn + 2μ(θ )εn (v))|Γ6 = φ6 , ∂v · n)|Γ7 = φ7 , (7) vτ |Γ7 = 0, (− p + μ(θ ) ∂n (8) vn = 0, |στ (θ, v)| ≤ gτ , στ (θ, v) · vτ + gτ |vτ | = 0 on Γ8 , (9) vτ = 0, |σn (θ, v, p)| ≤ gn , σn (θ, v, p)vn + gn |vn | = 0 on Γ9 , (10) vτ = 0, vn ≥ 0, σn (θ, v, p) + g+n ≥ 0, (σn (θ, v, p) + g+n )vn = 0 on Γ10 , (11) vτ = 0, vn ≤ 0, σn (θ, v, p) − g−n ≤ 0, (σn (θ, v, p) − g−n )vn = 0 on Γ11 ,

(7.3)

and Problem II is the one with the following boundary conditions (the case of total pressure)

7.1 Problems and Variational Formulations

229

(1) v|Γ1 = 0, (2) vτ |Γ2 = 0, −( p + 1/2|v|2 )|Γ2 = φ2 , (3) vn |Γ3 = 0, rot v × n|Γ3 = φ3 /μ(θ), (4) vτ |Γ4 = 0, (− p − 1/2|v|2 + 2μ(θ)εnn (v))|Γ4 = φ4 , (5) vn |Γ5 = 0, 2(μ(θ)εnτ (v) + αvτ )|Γ5 = φ5 , α : a matrix, (6) (− pn − 1/2|v|2 n + 2μ(θ)εn (v))|Γ6 = φ6 ,

(7.4)

∂v · n)|Γ7 = φ7 , (7) vτ |Γ7 = 0, (− p − 1/2|v| + μ(θ) ∂n (8) vn = 0, |στt (θ, v)| ≤ gτ , στt (θ, v) · vτ + gτ |vτ | = 0 on Γ8 , 2

(9) vτ = 0, |σnt (θ, v, p)| ≤ gn , σnt (θ, v, p)vn + gn |vn | = 0 on Γ9 , (10) vτ = 0, vn ≥ 0, σnt (θ, v, p) + g+n ≥ 0, (σnt (θ, v, p) + g+n )vn = 0 on Γ10 , (11) vτ = 0, vn ≤ 0, σnt (θ, v, p) − g−n ≤ 0, (σnt (θ, v, p) − g−n )vn = 0 on Γ11 .

Difference between (7.3) and (5.2), or (7.4) and (5.3), is that μ in (7.3) and (7.4) depends on θ . We use the following assumption. Assumption 7.1 Assume the followings. (1) Γ2 j , Γ3 j and Γ7 j are convex. Γ1 = ∅, Γ D = ∅ and   Γ R ⊂ ∪i=1,3,5,8 Γi .

(7.5)

(2) (3) of Assumption 5.1 holds. (3) For the functions of (7.1) f ∈ L3 (Ω), g ∈ L 6/5 (Ω) and μ ∈ C(R), 0 < μ0 ≤ μ(ξ ) ≤ μ1 < ∞ ∀ξ ∈ R; κ ∈ C(R), 0 < κ0 ≤ κ(ξ ) ≤ κ1 < ∞ ∀ξ ∈ R; γ ∈ C(R), |γ (ξ )| ≤ γ0 ∀ξ ∈ R.

(7.6)

(4) For the functions of (7.2), (7.3) and (7.4), g R ∈ L 4/3 (Γ R ); β0 ≥ β(x) ≥ 0, β0 − a constant, β(x) − measurable; φi ∈ H − 2 (Γi ), i = 2, 4, 7, φi ∈ H− 2 (Γi ), i = 3, 5, 6; 1

1

(7.7)

the matrix α is positive, αi j ∈ L ∞ (Γ5 ). Remark 7.1 For the first part of (2) in Assumption 7.1 we refer to Remark 6.3. In this chapter this assumption is also necessary to guarantee equivalence between Problem I-VE and Problem I-VI, and between Problem II-VE and Problem II-VI. Assuming that f ∈ L2 (Ω), g ∈ L 2 (Ω), g R ∈ L 2 (Γ R ), φi ∈ L 2 (Γi ), i = 2, 4, 7, and φi ∈ L2 (Γi ), i = 3, 5, 6, formally we introduce the following

230

7 The Steady Boussinesq System

Definition 7.1 A function (v, p, θ ) ∈ H2 (Ω) × H 1 (Ω) × H 2 (Ω) is called a solution to Problem I (or Problem II) if the first two equations and the last equation of (7.1), respectively, hold in L2 (Ω) and L 2 (Ω), and each of (7.2) and (7.3) (or (7.4)), respectively, holds in L 2 (Γ D ), L 2 (Γ R ), L 2 (Γi ) and L2 (Γi ). Let V, K (Ω) be the same as in (5.5) and

 (Ω) = y ∈ W 1,2 (Ω) : y|Γ D = 0 . WΓ1,2 D Since Γ1 = ∅ and Γ D = ∅, by Korn’s and Poincaré’s inequalities we use (v, u)V = (E(v), E(u)), (y, z)WΓ1,2 (Ω) = (∇ y, ∇ z).

(7.8)

D

As in Chap. 5, we will get variational formulations for Problems I, II. Integrating by parts, taking into account (2.79), and applying Theorems 3.1 and 3.2 on Γi j (i = 2, 3, 7), for v ∈ H2 (Ω) ∩ V, θ ∈ H 2 (Ω) and u ∈ V we have

−2 ∇ · (μ(θ)E (v)), u = 2(μ(θ)E (v), E (u)) − 2(μ(θ)E (v)n, u)∪11

i=2 Γi

= 2(μ(θ)E (v), E (u)) + 2(μ(θ)k(x)v, u)Γ2 − (μ(θ)rot v × n, u)Γ3 + 2(μ(θ)S v, ˜ u) ˜ Γ3 − 2(μ(θ)εnn (v), u n )Γ4 − 2(μ(θ)εnτ (v), u)Γ5  ∂v  − 2(μ(θ)εn (v), u)Γ6 − μ(θ) , u + (μ(θ)k(x)v, u)Γ7 − 2(μ(θ)εnτ (v), u)Γ8 Γ7 ∂n − 2(μ(θ)εnn (v), u n )Γ9 − 2(μ(θ)εnn (v), u n )Γ10 − 2(μ(θ)εnn (v), u n )Γ11 .

(7.9) For p ∈ H 1 (Ω) and u ∈ V we have 11 (∇ p, u) = ( p, u n )∪i=2 Γi = ( p, u n )Γ2 + ( p, u n )Γ4 ∪Γ7 ∪Γ9 ∪Γ10 ∪Γ11 + ( pn, u)Γ6 , (7.10) (Ω), by (7.2) we where u n |Γ3 ∪Γ5 ∪Γ8 = 0 was used. For θ ∈ H 2 (Ω) and ϕ ∈ WΓ1,2 D have

 ∂θ 

−∇ · (κ(θ)∇θ), ϕ = (κ(θ)∇θ, ∇ϕ) − κ(θ) , ϕ Γ = (κ(θ)∇θ, ∇ϕ) + (βθ − g R , ϕ)Γ R . R ∂n

(7.11) (Ω) we have By (7.5), vn = 0 on Γ R , and so for v ∈ V, θ ∈ H 1 (Ω) and ϕ ∈ WΓ1,2 D

v · ∇(γ (θ )θ ), ϕ = (vn γ (θ )θ, ϕ)Γ R − (γ (θ )θ v, ∇ϕ) = −(γ (θ )θ v, ∇ϕ). (7.12)

7.1 Problems and Variational Formulations

231

7.1.1 Variational Formulation: The Case of Static Pressure By (7.9)–(7.12), we can see that solutions in the sense of Definition 7.1 (v, p, θ ) of the problem (7.1), (7.2), (7.3) satisfy the following (cf. Remark 3.9). ⎧ 2(μ(θ)E(v), E(u)) + (v · ∇)v, u + 2(μ(θ)k(x)v, u)Γ2 ⎪ ⎪ ⎪ ⎪ ⎪ + 2(μ(θ)S v, ˜ u) ˜ Γ3 + 2(α(x)v, u)Γ5 + (μ(θ)k(x)v, u)Γ7 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ − 2(μ(θ)ε (v), u)Γ8 + ( p − 2μ(θ )εnn (v), u n )Γ9 ∪Γ10 ∪Γ11 nτ ⎪ ⎪ ⎪   ⎪ ⎪ ⎪ = (1 − α0 θ ) f, u +

φi , u n Γi +

φi , uΓi ∀u ∈ V, ⎪ ⎪ ⎪ ⎨ i=2,4,7 i=3,5,6 ⎪ (κ(θ)∇θ, ∇ϕ) − (γ (θ)θv, ∇ϕ) + (βθ, ϕ)Γ R = g R , ϕΓ R + g, ϕ ∀ϕ ∈ WΓ1,2 (Ω), ⎪ ⎪ D ⎪ ⎪ ⎪ ⎪ |στ (θ, v)| ≤ gτ , στ (θ, v) · vτ + gτ |vτ | = 0 on Γ8 , ⎪ ⎪ ⎪ ⎪ ⎪ |σn (θ, v, p)| ≤ gn , σn (θ, v, p)vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪ ⎪ ⎪ σ ⎪ n (θ, v, p) + g+n ≥ 0, (σn (θ, v, p) + g+n )vn = 0 on Γ10 , ⎪ ⎪ ⎩ σn (θ, v, p) − g−n ≤ 0, (σn (θ, v, p) − g−n )vn = 0 on Γ11 .

(7.13)

Define a0 (θ ; ·, ·), a1 (·, ·, ·) and f 1 ∈ V∗ by a0 (θ ; w, u) = 2(μ(θ )E(w), E(u)) + 2(μ(θ )k(x)w, u)Γ2 + 2(μ(θ )S w, ˜ u) ˜ Γ3 + 2(α(x)w, u)Γ5 + (μ(θ )k(x)w, u)Γ7 ∀w, u ∈ V, θ ∈ W 1,2 (Ω), a1 (v, w, u) = (v · ∇)w, u ∀v, w, u ∈ V,  

φi , u n Γi +

φi , uΓi ∀u ∈ V.

f 1 , u = i=2,4,7

i=3,5,6

˜ ·, ·) and g1 ∈ (WΓ1,2 (Ω))∗ by Define b0 (θ; D

(7.14)

˜ θ, ϕ) = (κ(θ)∇θ, ˜ ∇ϕ) + (β(x)θ, ϕ)Γ R ∀θ˜ , θ ∈ W 1,2 (Ω), ∀ϕ ∈ WΓ1,2 (Ω), b0 (θ; D

g1 , ϕ = g R , ϕΓ R + g, ϕ ∀ϕ ∈ WΓ1,2 (Ω). D

(7.15) Then, taking into account στ (θ, v) = 2μ(θ )εnτ (v), σn (θ, v, p) = − p + 2μ(θ ) εnn (v) and (7.13), we introduce the following variational formulation for the problem (7.1)–(7.3). Problem I-VE. Find (v, θ, στ , σn , σ+n , σ−n ) ∈ K (Ω) × WΓ1,2 (Ω) × L2τ (Γ8 )× D 2 −1/2 −1/2 L (Γ9 ) × H (Γ10 ) × H (Γ11 ) such that

232

7 The Steady Boussinesq System

⎧ a0 (θ ; v, u) + a1 (v, v, u) − (στ , u τ )Γ8 − (σn , u n )Γ9 ⎪ ⎪ ⎪ ⎪ ⎪ − σ+n , u n Γ10 − σ−n , u n Γ11 − f − α0 θ f, u = f 1 , u ∀u ∈ V, ⎪ ⎪ ⎪ ⎪ 1,2 ⎪ ⎪ ⎨ b0 (θ ; θ, ϕ) − γ (θ )θ v, ∇ϕ = g1 , ϕ ∀ϕ ∈ WΓ D (Ω), (7.16) |στ | ≤ gτ , στ · vτ + gτ |vτ | = 0 on Γ8 , ⎪ ⎪ ⎪ ⎪ |σn | ≤ gn , σn vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪ ⎪ ⎪ σ +n + g+n ≥ 0, σ+n + g+n , vn Γ10 = 0 on Γ10 , ⎪ ⎪ ⎩ σ−n − g−n ≤ 0, σ−n − g−n , vn Γ11 = 0 on Γ11 , where L2τ (Γ8 ) is the subspace of L2 (Γ8 ) consisting of functions such that (u, n)L2 (Γ8 ) = 0. Remark 7.2 Under (4) of Assumption 7.1 the duality products f 1 , u of (7.14) is meaningful (see Remark 5.1). Theorem 7.1 Let Assumption 7.1 hold. If (v, p, θ ) is a solution in the sense of Definition 7.1 of the problem (7.1)–(7.3), then (v, θ, στ |Γ8 , σn |Γ9 , σn |Γ10 , σn |Γ11 ) is a solution to Problem I-VE. Conversely, if f ∈ L2 (Ω) and Problem I-VE has a smooth solution (v, θ, στ , σn , σ+ , σ− ) such that v ∈ H2 (Ω), θ ∈ H 2 (Ω), σ+ ∈ L 2 (Γ10 ) and σ− ∈ L 2 (Γ11 ), then there exists p ∈ H 1 (Ω) such that (v, p, θ ) is a solution to the problem (7.1)–(7.3). Moreover, if at least one of the sets Γi , i = 2, 4, 6, 7, is nonempty, then p is unique. Proof From the problem (7.1)–(7.3) we deduced Problem I-VE, and thus it is suffices to prove conversion from Problem I-VE to the problem (7.1)–(7.3). By Theorem 5.1 there exists a p such that (v, p) satisfies (7.1) and the boundary condition (7.3), and p is unique under the additional condition above. In a routine way (see Sect. 1, Ch. 2 of [1]) we can prove that θ satisfies (7.1) and the boundary condition (7.2).  We will find another variational formulation consisting of a variational inequality and a variational equation, which is equivalent to Problem I-VE. Let (v, θ, στ , σn , σ+n , σ−n ) be a solution of Problem I-VE. Subtracting the first formula of (7.16) with u = v from the first formula of (7.16), we get a0 (θ; v, u − v) + a1 (v, v, u − v) − (στ , u τ − vτ )Γ8 − (σn , u n − vn )Γ9 − σ+n , u n − vn Γ10 − σ−n , u n − vn Γ11 − f − α0 θ f, u − v = f 1 , u − v ∀u ∈ V.

(7.17) Let Φ : V → R be the functional defined by (5.22).Then, Φ is proper, convex, weakly lower semi-continuous and nonnegative. By Theorem 5.2, under Assumption 7.1 for fixed θ the problem

7.1 Problems and Variational Formulations

⎧ a0 (θ ; v, u) + a1 (v, v, u) − (στ , u τ )Γ8 − (σn , u n )Γ9 ⎪ ⎪ ⎪ ⎪ ⎪ − σ+n , u n Γ10 − σ−n , u n Γ11 − f − α0 θ f, u = f 1 , u, ⎪ ⎪ ⎨ |στ | ≤ gτ , στ · vτ + gτ |vτ | = 0 on Γ8 , ⎪ |σn | ≤ gn , σn vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪ ⎪ σ ⎪ +n + g+n ≥ 0, σ+n + g+n , vn Γ10 = 0 on Γ10 , ⎪ ⎩ σ−n − g−n ≤ 0, σ−n − g−n , vn Γ11 = 0 on Γ11

233

(7.18)

is equivalent to the following variational inequality. Find v ∈ V such that a0 (θ ; v, u − v) + a1 (v, v, u − v)+Φ(u) − Φ(v) − f − α0 θ f, u − v ≥ f 1 , u − v ∀u ∈ V.

(7.19)

Therefore, we have the following variational formulation equivalent to Problem I-VE which consists of a variational inequality for velocity and a variational equation for temperature. (Ω) such that Problem I-VI. Find (v, θ ) ∈ V × WΓ1,2 D ⎧ ⎪ ⎨ a0 (θ ; v, u − v) + a1 (v, v, u − v) + Φ(u) − Φ(v) − f − α0 θ f, u − v (7.20) ≥ f 1 , u − v ∀u ∈ V, ⎪ ⎩ b (θ ; θ, ϕ) − γ (θ )θ v, ∇ϕ = g , ϕ ∀ϕ ∈ W 1,2 (Ω). 0 1 ΓD

7.1.2 Variational Formulation: The Case of Total Pressure Taking (v · ∇)v = rot v × v + 21 grad|v|2 into account, by (7.9)–(7.12) we can see that solutions in the sense of Definition 7.1 (v, p, θ ) of problem (7.1), (7.2), (7.4) satisfy the following. ⎧ 2(μ(θ )E(v), E(u)) + rotv × v, u + 2(μ(θ )k(x)v, u)Γ2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ + 2(μ(θ )S v, ˜ u) ˜ Γ3 + 2(α(x)v, u)Γ5 + (μ(θ )k(x)v, u)Γ7 ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ − 2(μ(θ )εnτ (v), u)Γ8 + ( p + |v|2 − 2μ(θ )εnn (v), u n )Γ9 ∪Γ10 ∪Γ11 ⎪ ⎪ 2 ⎪ ⎪   ⎪ ⎪ ⎪ ⎪ = (1 − α0 θ ) f, u +

φi , u n Γi +

φi , uΓi ∀u ∈ V, ⎪ ⎪ ⎨ i=2,4,7 i=3,5,6 ⎪ ⎪ (κ(θ )∇θ, ∇ϕ) − (γ (θ )θ v, ∇ϕ) + (βθ, ϕ)Γ R = g R , ϕΓ R + g, ϕ ∀ϕ ∈ WΓ1,2 (Ω), ⎪ ⎪ D ⎪ ⎪ ⎪ t ⎪ |στ (θ, v)| ≤ gτ , στt (θ, v) · vτ + gτ |vτ | = 0 on Γ8 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ |σnt (θ, v, p)| ≤ gn , σnt (θ, v, p)vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪ ⎪ ⎪ σ t (θ, v, p) + g ≥ 0, (σ t (θ, v, p) + g )v = 0 on Γ , ⎪ ⎪ +n +n n 10 n ⎪ n ⎪ ⎩ t σn (θ, v, p) − g−n ≤ 0, (σnt (θ, v, p) − g−n )vn = 0 on Γ11 .

(7.21)

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7 The Steady Boussinesq System

Define a2 (·, ·, ·) by a2 (v, u, w) = rot v × u, w ∀v, u, w ∈ V.

(7.22)

Then, taking into account 1 στt (θ, v) = 2μ(θ )εnτ (v), σnt (θ, v, p) = −( p + |v|2 ) + 2μ(θ )εnn (v) 2 and (7.21), we introduce the following variational formulation for problem (7.1), (7.2) (7.4). t t Problem II-VE. Find (v, θ, στt , σnt , σ+n , σ−n ) ∈ K (Ω) × WΓ1,2 (Ω) × L2τ (Γ8 )× D L 2 (Γ9 ) × H −1/2 (Γ10 ) × H −1/2 (Γ11 ) such that

⎧ a0 (θ ; v, u) + a2 (v, v, u) − (στt , u τ )Γ8 − (σnt , u n )Γ9 ⎪ ⎪

t t ⎪ ⎪ ⎪ − σ , u − σ , u − f − α0 θ f, u = f 1 , u ∀u ∈ V, n n ⎪ +n −n Γ Γ11 ⎪ 10 ⎪ ⎪ ⎪ ⎪ b0 (θ ; θ, ϕ) − γ (θ )θ v, ∇ϕ = g1 , ϕ ∀ϕ ∈ WΓ1,2 (Ω), ⎪ D ⎨ |στt | ≤ gτ , ⎪ ⎪ t ⎪ ⎪ ⎪ |σn | ≤ gn , ⎪ ⎪ t ⎪ ⎪ σ+n + g+n ⎪ ⎪ ⎪ ⎩ t σ−n − g−n

στt · vτ + gτ |vτ | = 0 on Γ8 , σnt vn + gn |vn | = 0 on Γ9 ,

t ≥ 0, σ+n + g+n , vn Γ10 = 0 on Γ10 ,

t ≤ 0, σ−n − g−n , vn Γ11 = 0 on Γ11 .

(7.23)

Relying on Theorem 5.3, in the same way as in Theorem 7.1 we have Theorem 7.2 Let Assumption 7.1 hold. If (v, p, θ ) is a solution in the sense of Definition 7.1 of the problem (7.1), (7.2), (7.4), then (v, θ, στt |Γ8 , σnt |Γ9 , σnt |Γ10 , σnt |Γ11 ) is a solution to Problem I-VE. Conversely, if f ∈ L2 (Ω) and Problem I-VE has a smooth solution (v, θ, στt , σnt , t σ+ , σ−t ) such that v ∈ H2 (Ω), θ ∈ H 2 (Ω), σ+t ∈ L 2 (Γ10 ) and σ−t ∈ L 2 (Γ11 ), then there exists p ∈ H 1 (Ω) such that (v, p, θ ) is a solution to the problem (7.1), (7.1), (7.4). Moreover, if at least one of the sets Γi , i = 2, 4, 6, 7, is nonempty, then p is unique. Then, using Theorem 5.4, in the same way as used for Problem I-VI we get Problem II-VI equivalent to Problem II-VE which consists of a variational inequality for velocity and a variational equation for temperature. (Ω) such that Problem II-VI. Find (v, θ ) ∈ V × WΓ1,2 D ⎧ ⎪ ⎨ a0 (θ ; v, u − v) + a2 (v, v, u − v) + Φ(u) − Φ(v) − f − α0 θ f, u − v ≥ f 1 , u − v ∀u ∈ V, (7.24) ⎪ ⎩ 1,2 b0 (θ ; θ, ϕ) − γ (θ )θ v, ∇ϕ = g1 , ϕ ∀ϕ ∈ WΓ D (Ω).

7.1 Problems and Variational Formulations

235

where a2 (·, ·, ·) is the one in (7.22). In view of the results above, we will study Problem I-VI and Problem II-VI in the following sections.

7.2 Existence and Uniqueness of Solutions: The Case of Static Pressure The aim of this section is to prove the following Theorem 7.3 Under Assumption 7.1 assume that f, φi , i = 2, · · · , 7, g, g R are small enough (depending on α0 ) in the spaces in (3), (4) of Assumption 7.1 (see (7.51)). Then, there exists a solution (v, θ ) to Problem I-VI such that μ0 , K   θ W 1,2 (Ω) ≤ c g R  L 4/3 (Γ R ) + g L 6/5 (Ω) ,

vV ≤

(7.25)

where K is the one in (7.26) below. If μ(θ ), κ(θ ) and γ (θ ) are independent of θ and  f L3 is small enough, then the solution satisfying vV ≤ c, θ WΓ1,2 (Ω) ≤ c for a constant c small enough is D unique.

7.2.1 Existence of a Solution to an Auxiliary Problem Since

  |a1 (v, v, u)| = | (v · ∇)v, u | ≤ K v2V uV ∀v, u ∈ V,

(7.26)

define a 1 (v) ∈ V∗ by

a 1 (v), u = a1 (v, v, u) ∀v, u ∈ V. Define γε (t) by γε (t) :=

γ (t)t t ∈ R, ε > 0. (1 + ε|γ (t)|)(1 + ε|t|)

Then, |γε (t)| ≤

1 , |γε (t)| ≤ |γ (t)||t| ≤ γ0 |t|, γε (t) → γ (t)t ε2

as ε → 0. (7.27)

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7 The Steady Boussinesq System

For every ε > 0, let Φε be the Moreau-Yosida approximation of Φ and ∇Φε be Fréchet derivative of Φε . We first consider an auxiliary problem involving two parameters δ, ζ concerned with the norm of velocity (which is useful when there is fluid flux across a portion of boundary), one parameter λ concerned with the norm of temperature (which is useful to deal with buoyancy effect) and a parameter ε for approximation. (Ω) Problem I-VIA. Let δ > 0, ζ > 0, λ > 0 and ε > 0. Find (v, θ ) ∈ V × WΓ1,2 D such that ⎧

δ ⎪ ⎪ ⎪ a0 (θ ; v, u) + max{δ, a (v) ∗ } a1 (v, v, u) + ∇Φε (v), u ⎪ 1 V ⎪ ⎪ ⎨    λ − 1− α0 θ f, u = f 1 , u ∀u ∈ V, (7.28) ⎪ max{λ, θ } ⎪ ⎪ ⎪ ⎪ ζ ⎪ ⎩ b0 (θ ; θ, ϕ) −

γε (θ )v, ∇ϕ = g1 , ϕ ∀ϕ ∈ WΓ1,2 (Ω). D max{ζ, vV } Theorem 7.4 There exists a solution (vε , θε ) ∈ V × WΓ1,2 (Ω) to Problem I-VIA. D Proof Let H = V × WΓ1,2 (Ω). Define an operator A : H → H ∗ by D

δ a1 (v, v, u) + ∇Φε (v), u max{δ, a 1 (v)V∗ }    α0 λ − 1− θ f, u + b0 (θ ; θ, φ) max{λ, θ } ζ −

γε (θ )v, ∇φ ∀(v, θ ), (u, φ) ∈ H . max{ζ, vV } (7.29) Let us first check that operator A is well-defined. By the definition of Φε , Φε (0V ) = 0 and ∇Φε (0V ) = 0. Since ∇Φε is Lipschitz continuous with the constant ε−1 (see Remark 1.16),

A (v, θ ), (u, φ) =a0 (θ ; v, u) +





∇Φε (v), u = ∇Φε (v) − ∇Φε (0V ), u ≤ ε−1 vV uV .

(7.30)

By (7.26) we have

δ a1 (v, v, u) ≤ δuV . ∗ max{δ, a 1 (v)V }

(7.31)

Also, by Hölder’s inequality we have  By (7.27) we have

 α0 λ θ f, u ≤ cα0 λ f L3 uV . max{λ, θ }

(7.32)

7.2 Existence and Uniqueness of Solutions: The Case of Static Pressure



cζ ζ

γε (θ )v, ∇φ ≤ 2 φW 1,2 . max{ζ, vV } ε

237

(7.33)

Estimation of other terms of (7.29) is easy, and so operator A is well defined. Then, existence of a solution to Problem I-VIA is equivalent to existence of a solution to   f A (v, θ ) = F , F = 1 . g1 We will use Theorem 1.45 to prove the existence of a solution to the equation above. To this end, we need to show that A satisfies the requirement in Theorem 1.45. (i) Let us prove that A is coercive, i.e.

1 A (v, θ ), (v, θ ) → ∞ as (v, θ )H → ∞. (v, θ )H Since Γ2 j , Γ3 j , Γ7 j are convex (see Lemma 3.1) and the matrix α is positive, we have from (7.14) (7.34) a0 (θ ; v, v) ≥ 2μ0 v2V . Since the operator ∇Φε is monotone and ∇Φε (0V ) = 0,



∇Φε (v), v = ∇Φε (v) − ∇Φε (0V ), v − 0V ≥ 0.

(7.35)

Taking into account (7.31)–(7.35), we have from (7.29)



δ a1 (v, v, v) + ∇Φε (v), v max{δ, a 1 (v)V∗ }    ζ α0 λ − 1− θ f, v + b0 (θ; θ, θ) −

γε (θ)v, ∇θ max{λ, θ} max{ζ, vV }   cζ ≥ min{2μ0 , κ0 } v2V + θ2 1,2 − δv −  f V∗ vV − cλ f L3 vV − 2 θW 1,2 WΓ ΓD ε D

A (v, θ), (v, θ) = a0 (θ; v, v) +

∀(v, θ) ∈ H ,

(7.36) which implies coercive property of A . (ii) Taking into account (7.30)–(7.33), we have from (7.29) A (v, θ )H ∗ =

sup

(u,φ)H =1



A (v, θ ), (u, φ)

 1 ζ  ≤ c vV + δ + vV +  f L3 + λ f L3 + θ WΓ1,2 + 2 D ε ε ∀(v, θ ) ∈ H , (7.37) which shows that A maps bounded sets of H into bounded sets of H ∗ . (iii) Let {(vk , ηk )} be a sequence such that

238

7 The Steady Boussinesq System

(vk , ηk )  (v, η) in H , lim sup A (vk , ηk ), (vk , ηk ) − (v, η) ≤ 0. k→∞

By taking a subsequence and denoting with the same subindex if necessary, we may assume vk → v in Ls (Ω)(1 ≤ s < 6) and a.e. in Ω, ηk → η in L s (Ω)(1 ≤ s < 6) and a.e. in Ω

as k → ∞.

(7.38)

Since a0 (ηk ; vk − v, vk − v) = a0 (ηk ; vk , vk − v) − a0 (ηk ; v, vk − v), b0 (ηk ; ηk − η, ηk − η) = b0 (ηk ; ηk , ηk − η) − b0 (ηk ; η, ηk − η), by (7.29) we have   min{μ0 ,κ0 } vk − v2V + ηk − ηW 1,2 ≤ A (vk , ηk ), (vk , ηk ) − (v, η) ΓD

− a0 (ηk ; v, vk − v) − b0 (ηk ; η, ηk − η)

δ a1 (vk , vk , vk − v) − ∇Φε (vk ), vk − v max{δ, a 1 (vk )V∗ }    ζ α0 λ ηk f, vk − v +

γε (ηk )vk , ∇(ηk − η). + 1− max{λ, ηk } max{ζ, vk V }

−

(7.39)

By Corollary 1.1 we have that a0 (ηk ; v, vk − v) = 2(μ(ηk )E (v), E (vk − v)) + 2(μ(ηk )k(x)v, vk − v)Γ2 + 2(μ(ηk )S v, ˜ v˜k − v) ˜ Γ3 + 2(α(x)v, vk − v)Γ5 + (μ(ηk )k(x)v, vk − v)Γ7 → 0, (7.40)     b0 (ηk ; η, ηk − η) = κ(ηk )∇(η), ∇(ηk − η) + β(x)η, (ηk − η) Γ → 0 R

as k → ∞. Also,

δ a1 (vk , vk , vk − v) ≤ vk L4 ∇vk L2 vk − vL4 → 0 as k → ∞. max{δ, a 1 (vk )V∗ }

(7.41)

Since ∇Φε is monotone,





− ∇Φε (vk ), vk − v = − ∇Φε (vk ) − ∇Φε (v), vk − v − ∇Φε (v), vk − v

≤ − ∇Φε (v), vk − v → 0 as k → ∞. (7.42) By (7.38), the followings hold.

7.2 Existence and Uniqueness of Solutions: The Case of Static Pressure

239



 α0 λ ηk f, vk − v ≤ cηk L3  f L3 vk − vL3 → 0, max{λ, ηk }   ζ |

γε (ηk )vk , ∇(ηk − η)| ≤ | γε (ηk ) − γε (η) vk , ∇(ηk − η)| max{ζ, vk V } + | γε (η)(vk − v), ∇(ηk − η)| + | γε (η)v, ∇(ηk − η)| ≤ cγε (ηk ) − γε (η) L 3 vk V ηk − ηW 1,2 ΓD

+ cγε (η) L 6 (vk − v)L3 ηk − ηW 1,2 ΓD

+ γε (η)v, ∇(ηk − η) → 0

(7.43) as k → ∞, where the fact that by Lemma 1.3 γε (ηk ) → γε (η) in L 3 (Ω) as k → ∞ was used. It is easy to prove convergence of other terms on the right hand side of (7.39). Thus, by (7.39)–(7.43) we have   lim sup min{μ0 , κ0 }| vk − v2V + ηk − ηWΓ1,2 D

k→∞

≤ lim sup A (vk , ηk ), (vk , ηk ) − (v, η) ≤ 0, k→∞

which implies

(vk , ηk ) → (v, η) in H as k → ∞, vk → v, ηk → η a.e. in Ω as k → ∞.

(7.44)

By definition of A ,

A (vk , ηk ),(vk , ηk ) − (u, φ) = a0 (ηk ; vk , vk − u)

δ + a1 (vk , vk , vk − u) + ∇Φε (vk ), vk − u ∗ max{δ, a 1 (vk )V }    (7.45) α0 λ ηk f, vk − u + b0 (ηk ; ηk , ηk − φ) − 1− max{λ, ηk } ζ γ (ηk )vk , ∇(ηk − φ) ∀(u, φ) ∈ H . − max{ζ, vk V } Taking into account (7.44), by Corollary 1.2 we have lim inf a0 (ηk ; vk , vk ) ≥ a0 (η; v, v), k→∞

lim inf b0 (ηk ; ηk , ηk ) ≥ b0 (η; η, η).

(7.46)

k→∞

 αλ  α0 λ α0 λ 0   2≤ The sequence {max{λ,η η } converges a.e. in Ω to η and η k k } max{λ,η} max{λ,η } L k k α0 λ 2 α0 λ, and so this sequence weakly converges to max{λ,η} η in L (see Lemma 1.1). Since f ∈ L3 and (v − u) ∈ L6 , we have f · (v − u) ∈ L 2 . Thus, taking into account the first formula of (7.43), we have

240

7 The Steady Boussinesq System

 lim

k→∞

 α0 λ ηk f, vk − u max{λ, ηk }     α0 λ α0 λ = lim ηk f, vk − v + lim ηk f, v − u k→∞ max{λ, ηk } k→∞ max{λ, ηk }   α0 λ = η f, v − u . (7.47) max{λ, η}

Using the second formula in (7.43), in the same way we get 

   ζ ζ γε (ηk )vk , ∇(ηk − φ) = γε (η)v, ∇(η − φ) . k→∞ max{ζ, vk V } max{ζ, vV } (7.48) Similarly, we have lim

δ δ a1 (vk , vk , vk − u)= a1 (v, v, v − u). ∗ k→∞ max{δ, a 1 (vk )V } max{δ, a 1 (v)V∗ } (7.49) Since ∇Φε is monotone and lim







∇Φε (vk ),vk − u = ∇Φε (vk ), vk − v + ∇Φε (vk ), v − u





≥ ∇Φε (vk ) − ∇Φε (v), vk − v + ∇Φε (v), vk − v + ∇Φε (vk ), v − u ,





lim ∇Φε (vk ), vk − u ≥ ∇Φε (v), v − u .

we have

k→∞

(7.50)

Thus, by (7.46)–(7.50) we have existence of a subsequence {(vk , ηk )} such that lim inf A (vk , ηk ), (vk , ηk ) − (u, φ) ≥ A (v, η), (v, η) − (u, φ). k→∞

Therefore, by virtue of Theorem 1.45 we come to the conclusion.



7.2.2 Existence and Estimates of Solutions to the Approximate Problem In this section the following theorem is proved. Theorem 7.5 If g1 (WΓ1,2 )∗  f L3 +  f + f 1 V∗ ≤ D

μ20 , K cα0

(7.51)

where K is the one in (7.26) and cα0 is the one in (7.63) below, then there exists a solution (vε , θε ) ∈ V × WΓ1,2 (Ω) to the following problem D

7.2 Existence and Uniqueness of Solutions: The Case of Static Pressure



241





a0 (θε ; vε , u) + a1 (vε , vε , u) + ∇Φε (vε ), u − (1 − α0 θε ) f, u = f 1 , u ∀u ∈ V, b0 (θε ; θε , ϕ) − γε (θε )vε , ∇ϕ = g1 , ϕ

∀ϕ ∈ WΓ1,2 (Ω), D

(7.52)

and the solution satisfies: μ0 , K θε WΓ1,2 (Ω) ≤ cg1 (WΓ1,2 )∗ . vε V ≤ D

(7.53)

D

Proof Let (vε , θε ) be solutions to (7.28). Putting ϕ = θε in the second formula of (7.28), we have ζ

γε (θε )vε , ∇θε  = g1 , θε . max{ζ, vV } (7.54)

(κ(θε )∇θε , ∇θε ) + (β(x)θε , θε )Γ R − Let us first prove

γε (θε )vε , ∇θε  = 0. To this end, define

 (t) :=

t

(7.55)

γε (s) ds, t ∈ R.

0

Then,  ∈ C 1 (R) and ∇(θ ) = γε (θ )∇θ, (θ ) ∈ W 1,2 (Ω) ∀θ ∈ W 1,2 (Ω), (θ )|Γ D = 0 ∀θ ∈ WΓ1,2 (Ω). D

(7.56)

Taking into account vε · n|Γ R = 0, by (7.56) we have 

γε (θ )vε , ∇θε  =

 Ω

γε (θε )vε · ∇θε d x =

Ω

vε · ∇(θε ) = 0,

which means (7.55). Next, we estimate other terms in (7.54). g1 , θε  ≤ κ0 θε 2 1 + cg1 2 1,2 ∗ , H (WΓ ) 4 D (β(x)θε , θε )Γ R ≥ 0.

(7.57)

By (7.54), (7.54) and (7.57), we have θε 2W 1,2 ≤ ΓD

2c g1 2(W 1,2 )∗ , ΓD κ0

(7.58)

242

7 The Steady Boussinesq System

which implies θε  ≤ c1 g1 (WΓ1,2 )∗ .

(7.59)

D

Putting λ = c1 g1 (WΓ1,2 )∗ D

and taking into account (7.59), we have λ = 1, max{λ, θε } 

(7.60)

 α0 λ θε f, u ≤ c1 α0 g1 (WΓ1,2 )∗  f L3 uV . D max{λ, θε }

(7.61)

Putting u = vε in the first equation of (7.28), we have

δ a1 (vε , vε , vε ) + ∇Φε (vε ), vε max{δ, a 1 (vε )V∗ }    α0 λ − 1− θε f, vε = f 1 , vε . max{λ, θε }

a0 (θε ; vε , vε ) +

(7.62)

Thus taking into account (7.26), (7.34), (7.35) and (7.61), we have from (7.62) 2μ0 vε 2V ≤ a0 (θε ; vε , vε ) δ |a1 (vε , vε , vε )| + |α0 θε f, vε | + | f + f 1 , vε | ≤ max{δ, a 1 (vε )V∗ }   ≤ K vε 3V + cα0 g1 (WΓ1,2 )∗  f L3 +  f + f 1 V∗ vε V . D (7.63) Note that the estimate above is independent of δ, but cα0 depends on α0 . This implies   0 ≤ K vε 2V − 2μ0 vε V + cα0 g1 (WΓ1,2 )∗  f L3 +  f + f 1 V∗ . D

Let us consider a quadratic polynomial for x > 0 related to the inequality above K x 2 − 2μ0 x + a. If 0 ≤ K a ≤ μ20 , then there exists a nonnegative minimum root x1 (≤ maximum root x2 . Thus, we can see that if g1 (WΓ1,2 )∗  f L3 +  f + f 1 V∗ ≤ D

then

μ20 , K cα0

μ0 ) K

and a

(7.64)

7.2 Existence and Uniqueness of Solutions: The Case of Static Pressure

vε V ≤

μ0 or vε V ≥ x2 . K

243

(7.65)

On the other hand, we have from (7.63) another estimation under consideration of δ   2μ0 vε 2V ≤ a0 (θε ; vε , vε ) ≤ δvε V +cα0 g1 (WΓ1,2 )∗  f L3 +  f + f 1 V∗ vε V , D

which implies vε V ≤

  1  δ + cα0 g1 (WΓ1,2 )∗  f L3 +  f + f 1 V∗ . D 2μ0

(7.66)

 2 μ2 In view of (7.65), let us take δ = K μK0 = K0 . Thus, in view of (7.51), we have from (7.66) vε V ≤

μ0 δ 1 μ20 = . + 2μ0 2μ0 K K

(7.67) μ2

By (7.67) under the condition (7.51) we have that a 1 (vε )V∗ ≤ K vε 2V ≤ K0 (see (7.26)), and so we get δ = 1. (7.68) max{δ, a 1 (vε )V∗ } Taking ζ =

μ0 , K

by (7.67) we get ζ = 1. max{ζ, vε V }

(7.69)

By (7.60), (7.68) and (7.69), we see that under condition (7.64) (vε , θε ) satisfies (7.52). By virtue of (7.67) and (7.58) we get (7.53). 

7.2.3 Existence and Uniqueness of a Solution First, by passing to the limit of solutions in Theorem 7.5, we will prove existence of a solution to Problem I-VI. Owing to (7.53) we can extract subsequences, which are denoted as before, such that vε  v in V, vε → v in Lq , 1 ≤ q < 6, θε  θ in H 1 (Ω), θε → θ in L q (Ω), 1 ≤ q < 6,

(7.70)

244

7 The Steady Boussinesq System

as ε → 0. Subtracting the first formula of (7.52) with u = vε from the first formula of (7.52), we have

a0 (θε ; vε , u − vε ) + a1 (vε , vε , u − vε ) + ∇Φε (vε ), u − vε

(7.71) − (1 − α0 θε ) f, u − vε = f 1 , u − vε  ∀u ∈ V. By Corollaries 1.1 and 1.2, we have a0 (θε ; vε , u) → a0 (θ ; v, u) as ε → 0, lim inf a0 (θε ; vε , vε ) ≥ a0 (θ ; v, v), ε→0

which imply that lim sup a0 (θε ; vε , u − vε ) ≤ a0 (θ ; v, u − v).

(7.72)

a1 (vε , vε , u − vε ) → a1 (v, v, u − v) as ε → 0.

(7.73)

ε→0

It is easy to prove

Since Φε is convex, continuous and Fréchet differentiable, we have Φε (u) − Φε (vε ) ≥ ∇Φε (vε ), u − vε  ∀u ∈ V,

(7.74)

which, together with (1.37), implies Φε (u) − Φ(Jε vε ) ≥ ∇Φε (vε ), u − vε  ∀u ∈ V.

(7.75)

Since Φ(0V ) = 0, by (1.37) we get Φε (0V ) = 0, and so we have from (7.74) Φε (vε ) ≤ ∇Φε (vε ), vε .

(7.76)

On the other hand, putting u = vε in the first formula of (7.52), we have



a0 (θε ; vε , vε ) + a1 (vε , vε , vε ) + ∇Φε (vε ), vε = (1 − α0 θε ) f, vε + f 1 , vε . (7.77) From (7.76) and (7.77) we have

a0 (θε ; vε , vε ) + a1 (vε , vε , vε ) + Φε (vε ) ≤ (1 − α0 θε ) f, vε + f 1 , vε , from which we get

7.2 Existence and Uniqueness of Solutions: The Case of Static Pressure

245

  |Φε (vε )| ≤ c g1 (WΓ1,2 )∗  f L3 +  f L3 +  f 1 V∗ vε V + |a1 (vε , vε , vε )|. D (7.78) By virtue of (1.36), (7.26), (7.66) and (7.78), we have     μ0 μ3  + 02 2ε, vε − Jε vε 2V ≤ c g1 (WΓ1,2 )∗  f L3 +  f L3 +  f 1 V∗ vε V D K K which shows that since vε  v in V, Jε vε  v in V as ε → 0. Then, by virtue of lower weak semi-continuity of Φ(v), lim inf Φ(Jε vε ) ≥ Φ(v).

(7.79)

Φε (u) → Φ(u) as ε → 0.

(7.80)

ε→0

By (1.37) we have

Taking into account (7.79) and (7.80), we have from (7.75) Φ(u) − Φ(v) ≥ lim sup ∇Φε (vε ), u − vε  ∀u ∈ V. ε→0

(7.81)

Using











| θε f, vε − θ f, v | ≤ | θε f, vε − θ f, vε | + | θ f, vε − θ f, v | ≤ θε − θ  L 3  f L2 vε L6 + θ  L 6  f L2 vε − vL3 ,

(7.82)

we can prove



(1 − α0 θε ) f, u − vε → (1 − α0 θ ) f, u − v as ε → 0.

(7.83)

It is easy to prove

f 1 , u − vε  → f 1 , u − v as ε → 0.

(7.84)

By virtue of (7.72), (7.73), (7.81), (7.83) and (7.84), from (7.71) we get

a0 (θ; v, u − v) + a1 (v, v, u − v) + Φ(u) − Φ(v) − (1 − α0 θ) f, u − v ≥ f 1 , u − v ∀u ∈ V,

which is the first formula in (7.20). We will get the second equation in (7.20). By Corollary 1.1, we have (Ω) as ε → 0. b0 (θε ; θε , ϕ) → b0 (θ ; θ, ϕ) ∀ϕ ∈ WΓ1,2 D

(7.85)

246

7 The Steady Boussinesq System

Let us prove (Ω) as ε → 0.

γε (θε )vε , ∇ϕ → γ (θ )θ v, ∇ϕ ∀ϕ ∈ WΓ1,2 D

(7.86)

By Hölder’s inequality we have | γε (θε )vε , ∇ϕ − γ (θ )θ v, ∇ϕ| ≤ | γε (θε )vε , ∇ϕ − γ (θ )θ vε , ∇ϕ| + | γ (θ )θ vε , ∇ϕ − γ (θ )θ v, ∇ϕ| ≤ γε (θε ) − γ (θ )θ  L 3 vε L6 ∇ϕL2 + γ (θ )θ  L 4 vε − vL4 ∇ϕL2 . (7.87) By the definition of γε (t) we have   γε (θε ) − γ (θ)θ  L 3 ≤  

  γ (θε )θε − γ (θ)θ   3 (1 + ε|γ (θε )|)(1 + ε|θε |) L    ≤ γ (θε )θε − γ (θ)θ  L 3 + ε γ (θ)θ |γ (θε )| + |θε | + ε|γ (θε )||θε | 

L3

.

(7.88) By virtue of Lemma 1.3 we know that γ (θε ) converges to γ (θ ) in space L p (Ω) (∀ p, 1 < p < ∞, ) as ε goes to zero. Thus, from (7.87) and (7.88) we get (7.86). By virtue of (7.85) and (7.86), from the second formula in (7.52) we get the second formula in (7.20). Estimates (7.25) follow from (7.53). Next, let us prove uniqueness of solutions. Suppose that there are two solutions (v1 , θ1 ) and (v2 , θ2 ). Since μ is independent of θ by the condition, denoting a0 (·; v, u) by a0 (v, u) yields a0 (v1 , v2 − v1 ) + a1 (v1 , v1 , v2 − v1 ) + Φ(v2 ) − Φ(v1 ) − f − α0 θ1 f, v2 − v1  ≥ f 1 , v2 − v1 , a0 (v2 , v1 − v2 ) + a1 (v2 , v2 , v1 − v2 ) + Φ(v1 ) − Φ(v2 ) − f − α0 θ2 f, v1 − v2  ≥ f 1 , v1 − v2 , which imply   a0 v1 − v2 , v1 − v2 ≤ |α0 ||(θ1 − θ2 ) f, v1 − v2 )| + |a1 (v1 , v1 , v1 − v2 ) − a1 (v2 , v2 , v1 − v2 )|.

(7.89)

By virtue of (7.89), we have 2μv1 − v2 2V μ ≤ v1 − v2 2V + c f 2L3 θ1 − θ2 2 2 + |a1 (v1 − v2 , v1 , v1 − v2 ) + a1 (v2 , v1 − v2 , v1 − v2 )| μ ≤ v1 − v2 2V + c f 2L3 θ1 − θ2 2 + c1 (v1 V + v2 V )v1 − v2 2V 2

7.2 Existence and Uniqueness of Solutions: The Case of Static Pressure

247

and so 3μ v1 − v2 2V ≤ c f 2L3 θ1 − θ2 2 + c1 (v1 V + v2 V )v1 − v2 2V . 2

(7.90)

Since κ(θ ), γ (θ ) are independent of θ , put κ(θ ) = κ, γ (θ ) = cv . Then, from (κ∇θ1 , ∇ϕ) + (β(x)θ1 , ϕ)Γ R − cv v1 θ1 , ∇ϕ1  = g1 , ϕ, (κ∇θ2 , ∇ϕ) + (β(x)θ2 , ϕ)Γ R − cv v2 θ2 , ∇ϕ = g1 , ϕ we have κ(∇θ1 − ∇θ2 , ∇θ1 − ∇θ2 ) + (β(x)(θ1 − θ2 ), θ1 − θ2 )Γ R − cv v1 (θ1 − θ2 ), ∇(θ1 − θ2 ) − cv (v1 − v2 )θ2 , ∇(θ1 − θ2 ) = 0.

(7.91)

Using cv v1 (θ1 − θ2 ), ∇(θ1 − θ2 ) = 0 (see (7.12) with γ (θ )=const), we obtain from (7.91) κ∇θ1 − ∇θ2 2 ≤ and so

cv c κ v1 − v2 2V θ2 2W 1,2 + ∇θ1 − ∇θ2 2 , ΓD κ 2

κ cv c θ1 − θ2 2W 1,2 ≤ v1 − v2 2V θ2 2W 1,2 . ΓD 2 κ

(7.92)

Therefore adding (7.90) and (7.92), we get min{

 3μ κ  , } v1 − v2 2V + θ1 − θ2 2W 1,2 2 2

≤ c f 2L3 θ1 − θ2 2 + c1 (v1 V + v2 V )v1 − v2 2V +

cv c v1 − v2 2V θ2 2W 1,2 . ΓD κ (7.93)

Thus, if vi V , θ2 WΓ1,2 and  f L3 are small, then we have from the above that D v1 = v2 and θ1 = θ2 . 

7.3 Existence of a Solution: The Case of Total Pressure Theorem 7.6 Under Assumption 7.1 there exists a solution (v, θ ) to Problem II-VI such that    φi Γi + g R  L 4/3 (Γ R ) + g L 6/5 (Ω) , vV ≤ c  f L3 + θ W 1,2 (Ω)

i=2−7

 ≤ c g R  L 4/3 (Γ R ) + g L 6/5 (Ω) . 

(7.94)

248

7 The Steady Boussinesq System

First, we look for solutions to an auxiliary problem: Problem II-VIA. Let ζ > 0, λ > 0 and ε > 0. Find (v, θ ) ∈ V × W 1,2 (Ω) such that (Ω) and η = θ − θ D ∈ WΓ1,2 D ⎧  

 α0 λ ⎪ ⎪ a θ f, u (θ ; v, u) + a (v, v, u) + ∇Φ (v), u − 1 − ⎪ 0 2 ε ⎪ ⎪ max{λ, θ } ⎨ = f 1 , u ∀u ∈ V, (7.95) ⎪ ⎪ ⎪ ζ ⎪ ⎪

γε (θ )v, ∇ϕ = g1 , ϕ ∀ϕ ∈ WΓ1,2 (Ω). ⎩ b0 (θ ; θ, ϕ) − D max{ζ, vε V } Theorem 7.7 There exists a solution (vε , θε ) ∈ V × W 1,2 (Ω) to Problem II-VIA. Proof Let H = V × WΓ1,2 (Ω). Define an operator A : H → H ∗ by D



A (v, η), (u, φ) =a0 (η; v, u) + a2 (v, v, u) + ∇Φε (v), u    α0 λ − 1− θ f, u + b0 (θ ; θ, ϕ) max{λ, θ } ζ −

γε (η + θ D )v, ∇ϕ ∀(v, η), (u, φ) ∈ H . max{ζ, vV } Using

a2 (v, v, v) = 0, |a2 (v, v, u)| ≤ K v2V uV , |a2 (vε , vε , vε − u)| ≤ c∇vε L2 vε L4 vε − uL4 ,

respectively, in (7.36), (7.37) and (7.41), we can verify that the proof of Theorem 7.4 for Problem I-VIA is valid for Problem II-VIA. Thus, we come to the asserted conclusion.  Theorem 7.8 There exists a solution (vε , θε ) ∈ V × W 1,2 (Ω) to the following problem 





a0 (θε ; vε , u) + a2 (vε , vε , u) + ∇Φε (vε ), u − (1 − α0 θε ) f, u = f 1 , u ∀u ∈ V, b0 (θε ; θε , ϕ) − γε (θε )vε , ∇ϕ = g1 , ϕ ∀ϕ ∈ WΓ1,2 (Ω), D

(7.96)

and the solution satisfies:  c  g1 (WΓ1,2 )∗  f L3 +  f V∗ +  f 1 V∗ , D 2μ0 1,2 1,2 θε WΓ (Ω) ≤ cg1 (WΓ )∗ . vε V ≤ D

D

(7.97)

7.3 Existence of a Solution: The Case of Total Pressure

249

Proof Let (vε , θε ) be solutions to (7.95). In the same way as in (7.54)–(7.58) we have 2c (7.98) θε 2W 1,2 ≤ g1 2(W 1,2 )∗ , ΓD ΓD κ0 which implies θε (x) ≤ c1 g1 (WΓ1,2 )∗ .

(7.99)

D

Putting λ = c1 g1 (WΓ1,2 )∗ , we have from the first equation of (7.95) D

 

 a0 (θε ; vε , u) + a2 (vε , vε , u) + ∇Φε (vε ), u − 1 − α0 θε f, vε = f 1 , u ∀u ∈ V,

(7.100) which is the first formula of (7.96). Putting u = vε in (7.100), we have

 

a0 (θε ; vε , vε ) + a2 (vε , vε , vε ) + ∇Φε (vε ), vε − 1 − α0 θε f, vε = f 1 , vε . (7.101) Taking into account a2 (vε , vε , vε ) = 0, (7.34), (7.35) and (7.99), we have from (7.101)   2μ0 vε 2V ≤ a0 (θ ; v, v) ≤ c α0 c1 g1 (WΓ1,2 )∗  f L3 +  f V∗ +  f 1 V∗ vε V , D

which implies vε V ≤

 c  α0 c1 g1 (WΓ1,2 )∗  f L3 +  f V∗ +  f 1 V∗ . D 2μ0

(7.102)

Taking the right hand side of (7.102) as ζ in (7.95), we get the second equation of (7.96). By (7.98) and (7.102), we get (7.97).  Now repeating the arguments in Sect. 7.2.3 with the solutions of Theorem 7.8, we complete proof of Theorem 7.6.  Remark 7.3 The Eq. (9.1) of Chap. 9 is more general than (7.1), and from the results of Chap. 9 some results for (7.1) with the boundary conditions (7.2), (7.3) or (7.2), (7.4) can be obtained. However the result for the case of total pressure demands that the parameter for buoyancy effect α0 is small enough in accordance with the data of problem (see (9.138), and the result for the case of static pressure demands that the data of problem satisfy two smallness conditions together (see (9.50) and (9.97)).

7.4 Bibliographical Remarks The content of Chap. 7 is taken from [2].

250

7 The Steady Boussinesq System

Several papers are concerned with (7.1). In [3, 4] under homogeneous Dirichlet boundary condition for velocity and mixture of non-homogeneous Dirichlet and Neumann conditions for temperature existence of a solution to (7.1) was studied. In [5] under non-homogeneous Dirichlet boundary condition for velocity and mixture of non-homogeneous Dirichlet and Neumann conditions for temperature, where smoothness of boundary data is weaker than [3, 4], existence of a solution to (7.1) was obtained. In [6] under mixture of non-homogeneous Dirichlet, total pressure and vorticity boundary conditions for fluid and mixture of non-homogeneous Dirichlet, Neumann and Robin conditions for temperature the existence of a solution was established. In [7] variational inequalities for Navier-Stokes type operators were studied, which can describe (7.1) with one-sided flow boundary conditions for fluid and heat on a portion of boundary. In [8] under homogeneous Dirichlet boundary condition for velocity and mixture of non-homogeneous Dirichlet and homogeneous Neumann conditions for temperature the existence and uniqueness and smoothness of a weak solution were discussed. In [9] when the boundary consists of several connected components, boundary value problem of (7.1) with non-homogeneous Dirichlet boundary condition was studied. In [10] when the boundary consists of several connected components, Dirichlet problem of (7.1) under a weaker condition than [9] was investigated. In [11] Dirichlet problem of (7.1) for arbitrarily large and very weak boundary data was studied.

References 1. H. Gajewski, K. Gr¨oger, K. Zacharias, Nichtlineare Operatorgleichungen und Operatordifferentialgleichungen (Academic-Verlag Berlin, 1974) (Russian 1978) 2. T. Kim, The steady Boussinesq system with mixed boundary conditions including conditions of friction type. Applications of Mathematics, to appear 3. H. Morimoto, On the existence of weak solutions of equations of natural convection. J. Fac. Sci. Univ. Tokyo, Sect. IA 36, 87–102 (1989) 4. H. Morimoto, On the existence and uniqueness of the stationary solution to the equation of natural convection. Tokyo J. Math. 14, 217–226 (1991) 5. E.J. Villamizar-Roa, M.A. Rodríguez-Bellido, M.A. Rojas-Medar, The Boussinesq system with mixed nonsmooth boundary data. C. R. Acad. Sci. Paris, Ser. I 343, 191–196 (2006) 6. G.V. Alekseev, A.B. Smishliaev, Solvability of the boundary value problems for Boussinesq equations with inhomogeneous boundary conditions. J. Math. Fluid Mech. 3(1), 18–39 (2001) 7. A.Yu. Chebotarev, Variational inequalities for Navier-Stokes type operators and one-sided problems for equations of viscous heat-conducting fluids. Math. Notes 70, 264–274 (2001) 8. D.A. Kovtunov, Solvability of the stationary heat convection problem for a high-viscosity fluid. Diff. Equat. 45(1), 73–85 (2009) 9. H. Morimoto, Heat convection equation with nonhomogeneous boundary conditions. Funkcialai Ekvacioj 53, 213–229 (2010) 10. P. Acevedo, C. Amrouche, C. Conca, Boussinesq system with non-homogeneous boundary conditions. Appl. Math. Lett. 53, 39–44 (2016) 11. H. Kim, The existence and uniqueness of very weak solutions of the stationary Boussinesq system. Nonlinear Anal. 75, 317–330 (2012)

Chapter 8

The Non-steady Boussinesq System

In this chapter we are concerned with the non-steady Boussinesq problem corresponding to the steady problem in Chap. 7. The formulations consist of a non-steady variational inequality for velocity and a non-steady variational equation for temperature. For the problem with boundary conditions involving the static pressure and stress, it is proved that if the data of problem are small enough and compatibility conditions at the initial time for velocity and temperature are satisfied, then there exists a unique solution on the given interval. For the problem with boundary conditions involving the total pressure and total stress, the existence of a solution is proved without restriction on the data of problem.

8.1 Problems and Assumptions The non-steady Boussinesq system with initial condition for heat convection is as follows. ⎧   ∂v ⎪ ⎪ − 2∇ · μ(θ )E(v) + (v · ∇)v + ∇ p = (1 − α0 θ ) f, ⎪ ⎪ ∂t ⎪ ⎪ ⎪ ⎨ div v = 0, (8.1)   ∂θ ⎪ ⎪ ⎪ − ∇ · κ(θ )∇θ + v · ∇θ = g, ⎪ ⎪ ∂t ⎪ ⎪ ⎩ v(0) = v0 , θ (0) = θ0 . Let Ω be as in Chap. 7 with Γi j ∈ C 2,1 for i = 2, 3, 7 instead of Γi j ∈ C 2 . (See Remark 6.1) Let Q = Ω × (0, T ), 0 < T < ∞. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 T. Kim and D. Cao, Equations of Motion for Incompressible Viscous Fluids, Advances in Mathematical Fluid Mechanics, https://doi.org/10.1007/978-3-030-78659-5_8

251

252

8 The Non-steady Boussinesq System

The boundary conditions for temperature are as follows: (1) θ |Γ D = 0,   ∂θ + β(x)θ Γ R = g R (t, x), β(x), g R (t, x) − given functions. (2) κ(θ ) ∂n (8.2) Problems I and II are distinguished according to boundary conditions for fluids. Problem I is the one with boundary condition (7.3) assumed that μ and κ are independent of θ (the case of static pressure) and Problem II is the one with boundary condition (7.4) (the case of total pressure). Let V, K (Ω) be the same as in (5.5) and H : completion in L2 (Ω) of V, HK : closure in L2 (Ω) of K (Ω), K (Q) = {u ∈ L 2 (0, T ; V) : u  ∈ L 1 (0, T ; V∗ ); u n |Γ10 ≥ 0, u n |Γ11 ≤ 0}, WΓ1,2 (Ω) = {y ∈ W 1,2 (Ω) : y|Γ D = 0}. D (Ω). We use the inner products in (7.8) for V and WΓ1,2 D We will use the following two kinds of assumptions according to different boundary conditions. The assumption for Problem I is stronger than the one for Problem II, and consequently smoothness of solution with respect to t for Problem I is stronger than the one in Problem II. Assumption 8.1 (for the case of static pressure) We assume the followings. (1) Γ1 = ∅, Γ D = ∅ and   Γ R ⊂ ∪i=1,3,5,8 Γi .

(8.3)

(2) (3) of Assumption 5.1 holds. (3) For the functions of (8.1) f ∈ W 1,∞ (0, T ; L3 (Ω)),   g ∈ W 1,2 0, T ; (WΓ1,2 )∗ , D

(8.4)

μ and κ are independent of θ. (4) For the functions of (8.2) and (7.3), g R ∈ W 1,2 (0, T ; L 4/3 (Γ R )), β1 ≥ β(x) ≥ 0, β1 − a constant, β(x) − measurable, φi ∈ W 1,∞ (0, T ; H −1/2 (Γi )), i = 2, 4, 7, φi ∈ W (0, T ; H αi j ∈ L ∞ (Γ5 ). 1,∞

−1/2

(Γi )), i = 3, 5, 6,

(8.5)

8.1 Problems and Assumptions

253

Assumption 8.2 (for the case of total pressure) Let (1) and (2) of Assumption 8.1 hold and suppose the followings. (3’) For the functions of (8.1), f ∈ L ∞ (0, T ; L3 (Ω)),  g ∈ L 2 0, T ; (WΓ1,2 )∗ ), D μ ∈ C(R), 0 < μ0 ≤ μ(ξ ) ≤ μ1 < ∞ ∀ξ ∈ R,

(8.6)

κ ∈ C(R), 0 < κ0 ≤ κ(ξ ) ≤ κ1 < ∞ ∀ξ ∈ R. (4’) For the functions of (8.2) and (7.4), g R ∈ L 2 (0, T ; L 4/3 (Γ R )); β1 ≥ β(x) ≥ 0, β1 − a constant, β(x) − measurable; φi ∈ L 2 (0, T ; H − 2 (Γi )), i = 2, 4, 7, 1

φi ∈ L 2 (0, T ; H αi j ∈ L ∞ (Γ5 ).

− 21

(8.7)

(Γi )), i = 3, 5, 6;

Remark 8.1 For (2) of Assumption 8.1 we refer to Remark 6.3. In this chapter this assumption is also necessary to guarantee equivalence between Problem I-VE and Problem I-VI (in Remark 8.4), and between Problem II-VE and Problem II-VI.

8.2 Variational Formulations for Problems In this section we first give variational formulations for the problems above. Taking (Ω), into account (8.3) and vn |Γ3 ∪Γ5 ∪Γ8 = 0, for v ∈ V, θ ∈ W 1,2 (Ω) and ϕ ∈ WΓ1,2 D we have (v · ∇θ, ϕ) = (vn θ, ϕ)Γ R − (θ v, ∇ϕ) = −(θ v, ∇ϕ). (8.8)

8.2.1 Variational Formulations: The Case of Static Pressure By (7.9)–(7.11) and (8.8), as in Sect. 7.1 we can see that smooth solutions (v, p, θ ) of problem (8.1), (8.2), (7.3) satisfy the following system.

254

8 The Non-steady Boussinesq System

⎧ ∂v ⎪ ⎪ ˜ u) ˜ Γ3 ( , u) + 2(μE (v), E (u)) + (v · ∇)v, u + 2(μk(x)v, u)Γ2 + 2(μS v, ⎪ ⎪ ∂t ⎪ ⎪ ⎪ ⎪ ⎪ + 2(α(x)v, u)Γ5 + (μk(x)v, u)Γ7 − 2(μεnτ (v), u)Γ8 + ( p − 2μεnn (v), u n )Γ9 ∪Γ10 ∪Γ11 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ = (1 − α0 θ) f, u + φi , u n Γi + φi , uΓi ∀u ∈ V, ⎪ ⎪ ⎪ ⎪ i=2,4,7 i=3,5,6 ⎪ ⎨ ∂θ ( , ϕ) + (κ∇θ, ∇ϕ) − (θv, ∇ϕ) + (βθ, ϕ)Γ R = g R , ϕΓ R + g, ϕ ∀ϕ ∈ WΓ1,2 (Ω), ⎪ D ⎪ ⎪ ∂t ⎪ ⎪ ⎪ ⎪ |στ (θ, v)| ≤ gτ , στ (θ, v) · vτ + gτ |vτ | = 0 on Γ8 , ⎪ ⎪ ⎪ ⎪ ⎪ |σn (v, p)| ≤ gn , σn (v, p)vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪ ⎪ ⎪ σ ⎪ n (v, p) + g+n ≥ 0, (σn (v, p) + g+n )vn = 0 on Γ10 , ⎪ ⎪ ⎩ σn (v, p) − g−n ≤ 0, (σn (v, p) − g−n )vn = 0 on Γ11 .

(8.9) Define a01 (·, ·), a11 (·, ·, ·) and f 1 (t) ∈ V∗ by a01 (w, u) = 2(μE(w), E(u)) + 2(μk(x)w, u)Γ2 + 2(μS w, ˜ u) ˜ Γ3 + 2(α(x)w, u)Γ5 + (μk(x)w, u)Γ7 ∀w, u ∈ V, a11 (v, u, w) = (v · ∇)u, w ∀v, u, w ∈ V, φi (t), u n Γi + φi (t), uΓi ∀u ∈ V.  f 1 (t), u = i=2,4,7

(8.10)

i=3,5,6

Define b1 (·, ·) and g1 (t) ∈ (WΓ1,2 (Ω))∗ by D (Ω), b1 (θ, ϕ) = (κ∇θ, ∇ϕ) + (β(x)θ, ϕ)Γ R ∀θ ∈ W 1,2 (Ω), ϕ ∈ WΓ1,2 D (Ω). g1 (t), ϕ = g R (t), ϕΓ R + g(t), ϕ ∀ϕ ∈ WΓ1,2 D

(8.11)

Remark 8.2 Under (4) of Assumption 8.1 the duality product  f 1 , u of (8.10) is meaningful (see Remark 5.1). By (8.4) and (8.5), )∗ ). f 1 ∈ W 1,∞ (0, T ; V∗ ), g1 ∈ W 1,2 (0, T ; (WΓ1,2 D

(8.12)

Then, taking into account στ (θ, v) = 2μεnτ (v), σn (θ, v, p) = − p + 2μεnn (v) and (8.9), we introduce the following variational formulation for problem (8.1), (8.2), (7.3). (Ω)) and (στ , σn , σ+n , σ−n ) ∈ Problem I-VE. Find v ∈ K (Q), θ ∈ L 2 (0, T ; WΓ1,2 D 1 1 L2τ (Γ8 ) × L 2 (Γ9 ) × H − 2 (Γ10 ) × H − 2 (Γ11 ) for a.e. t ∈ (0, T ) such that v(0) = v0 , θ (0) = θ0 and

8.2 Variational Formulations for Problems

255

⎧ ∂v ⎪ ⎪  , u + a01 (v, u) + a11 (v, v, u) − (στ , u τ )Γ8 − (σn , u n )Γ9 ⎪ ⎪ ∂t ⎪ ⎪ ⎪ ⎪ ⎪ − σ+n , u n Γ10 − σ−n , u n Γ11 −  f − α0 θ f, u =  f 1 , u ∀u ∈ L 2 (0, T ; V), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪  ∂θ , ϕ + b (θ, ϕ) − θv, ∇ϕ = g , ϕ ∀ϕ ∈ L 2 (0, T ; W 1,2 (Ω)), ⎨ 1 1 ΓD ∂t (8.13) ⎪ ⎪ |στ | ≤ gτ , στ · vτ + gτ |vτ | = 0 on Γ8 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ |σn | ≤ gn , σn vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪ ⎪ σ+n + g+n ≥ 0, σ+n + g+n , vn Γ10 = 0 on Γ10 , ⎪ ⎪ ⎪ ⎩ σ−n − g−n ≤ 0, σ−n − g−n , vn Γ11 = 0 on Γ11 ,

where L2τ (Γ8 ) is the subspace of L2 (Γ8 ) consisting of functions such that (u, n)L2 (Γ8 ) = 0. Remark 8.3 We showed that smooth solutions to the problem (8.1), (8.2), (7.3) are solutions to Problem I-VE. Assume that a solution to Problem I-VE is smooth  enough v ∈ L 2 (0, T ; H2 (Ω)), v  ∈ L 1 (0, T ; L2 (Ω)), θ ∈ L 2 (0, T ; H 2 (Ω)), θ  ∈ L 1 (0, T ; L 2 (Ω)), στ ∈ L 2 (0, T ; L2 (Γ 8 )), σn ∈ L 2 (0, T ; L 2 (Γ9 )), σ+ ∈ L 2 (0, T ; L 2 (Γ10 )) and σ− ∈ L 2 (0, T ; L 2 (Γ11 )) and f ∈ L 2 (0, T ; L2 (Ω)). Then, for a.e. t ∈ (0, T ) there exists p(t) ∈ H 1 (Ω) (see Theorem 5.3) such that p ∈ L 1 (0, T ; H 1 (Ω)) and (v, p, θ ) satisfies the first two equations of (8.1) and the boundary condition (7.3) in suitable spaces. By a standard way, it is proved that (v, θ ) satisfies the third equation of (8.1), the boundary condition (8.2) and the initial conditions. In this sense, Problem I-VE is equivalent to the problem (8.1), (8.2), (7.3). We will find another variational formulation consisting of a variational inequality and a variational equation, which is equivalent to Problem I-VE if the solution is smooth enough (see Remark 8.4). For fixed θ , let us consider the problem ⎧ ∂v ⎪ ⎪  , u + a01 (v, u) + a11 (v, v, u) − (στ , u τ )Γ8 − (σn , u n )Γ9 − σ+n , u n Γ10 ⎪ ⎪ ∂t ⎪ ⎪ ⎪ ⎪ ⎪ − σ−n , u n Γ11 − ( f − α0 θ f, u =  f 1 , u, ∀u ∈ L 2 (0, T ; V), ⎪ ⎨ |στ | ≤ gτ , στ · vτ + gτ |vτ | = 0 on Γ8 , ⎪ ⎪ ⎪ |σn | ≤ gn , σn vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪ ⎪ ⎪ σ+n + g+n ≥ 0, σ+n + g+n , vn Γ10 = 0 on Γ10 , ⎪ ⎪ ⎩ σ−n − g−n ≤ 0, σ−n − g−n , vn Γ11 = 0 on Γ11 . (8.14) Subtracting the first formula of (8.14) with u = v from the first one of (8.14), we get 

∂v , u − v + a01 (v, u − v) + a11 (v, v, u − v) − (στ , u τ − vτ )Γ8 − (σn , u n − vn )Γ9 ∂t − σ+n , u n − vn Γ10 − σ−n , u n − vn Γ11 −  f − α0 θ f, u − v =  f 1 , u − v ∀u ∈ V.

(8.15)

256

8 The Non-steady Boussinesq System

Let Φ : V → R be the functional defined by (5.22). Then, the functional Φ is proper, convex, lower weak semi-continuous and nonnegative. Define a functional Ψ (u) by Ψ (u) =

⎧

⎨ ⎩

T

Φ(u(t)) dt

if Φ(u(t)) ∈ L 1 (0, T ),

0

+∞

(8.16)

otherwise.

In the same way as Problem I of Sect. 5.1, from (8.15) we get  ∂v ∂t

, u − v + a01 (v, u − v) + a11 (v, v, u − v) + Φ(u) − Φ(v)

(8.17)

≥ (1 − α0 θ ) f, u − v +  f 1 , u − v.

(Ω) → Define operators A1 : V → V∗ , B1 : V × V → V∗ and C1 : WΓ1,2 D 1,2 ∗ (WΓ D (Ω)) , respectively, by A1 v, u = a01 (v, u) ∀v, u ∈ V, B1 (v, u), w = a11 (v, u, w) ∀v, u, w ∈ V, C1 θ, ϕ = b1 (θ, ϕ) ∀θ, ϕ ∈

(8.18)

WΓ1,2 (Ω). D

If v is a solutions to (8.14), then we can see that the solution satisfies (see (6.18))

T

v  (t) + A1 v(t) + B1 (v(t), v(t)) − (1 − α0 )θ (t) f (t) − f 1 (t), u(t) − v(t) dt

0

+ Ψ (u) − Ψ (v) ≥ 0

∀u ∈ L 4 (0, T ; V). (8.19)

Therefore, we have the following variational formulation of the problem (8.1), (8.2), (7.3) which consists of a variational inequality for velocity and a variational equation for temperature.    Problem I-VI. Find (v, θ ) ∈ L ∞ (0, T ; H ) ∩ L 2 (0, T ; V) × L ∞ (0, T ; L 2 (Ω)) ∩  (Ω)) such that L 2 (0, T ; WΓ1,2 D ⎧ T ⎪ ⎪ v  + A1 v(t) + B1 (v(t), v(t)) − (1 − α0 θ) f − f 1 , u(t) − v(t) dt ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎨ + Ψ (u) − Ψ (v) ≥ 0 ∀u ∈ L 4 (0, T ; V), (8.20)

 T  ∂θ ⎪ ⎪ 1,2 2 ⎪ ⎪ θ(t), ϕ − θv, ∇ϕ − g , ϕ dt = 0 ∀ϕ ∈ L (0, T ; W ), , ϕ + C 1 1 ⎪ ΓD ⎪ ∂t ⎪ 0 ⎪ ⎩ v(0) = v0 , θ(0) = θ0 .

8.2 Variational Formulations for Problems

257

Remark 8.4 If the solution to Problem I-VI is smooth as much as v ∈ L 2 (0, T ; V), v  ∈ L 2 (0, T ; V∗ ), then the first one of (8.20) is equivalent to v  (t) + A1 v(t) + B1 (v(t), v(t)) − (1 − α0 θ ) f − f 1 , u − v(t) + Φ(u) − Φ(v(t)) ≥ 0 for a.e. t ∈ [0, T ], ∀u ∈ K (Ω)

(8.21)

− (1 − α0 θ ) f − (Remark, p. 114 of [1]). In (8.17) putting F1 , u − v = − ∂v ∂t f 1 , u − v, by Theorem 5.4 we can see the existence of (στ , σn , σ+n , σ−n ) ∈ L2τ (Γ8 ) × L 2 (Γ9 ) × H −1/2 (Γ10 ) × H −1/2 (Γ11 ) for a.e. t ∈ (0, T ) such that (v, θ, στ , σn , σ+n , σ−n ) is a solution to Problem I-VE.

8.2.2 Variational Formulations: The Case of Total Pressure Since (v · ∇)v = rot v × v + 21 grad|v|2 , by (7.9)–(7.11) and (8.8), we can see that smooth solutions (v, p, θ ) of problem (8.1), (8.2), (7.4) satisfy the following. ⎧  ∂v  ⎪ , u + 2(μ(θ)E (v), E (u)) + rotv × v, u + 2(μ(θ)k(x)v, u)Γ2 ⎪ ⎪ ⎪ ∂t ⎪ ⎪ ⎪ ⎪ + 2(μ(θ)S v, ˜ u) ˜ Γ3 + 2(α(x)v, u)Γ5 + (μ(θ)k(x)v, u)Γ7 ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ − 2(μ(θ)εnτ (v), u)Γ8 + ( p + |v|2 − 2μ(θ)εnn (v), u n )Γ9 ∪Γ10 ∪Γ11 ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ ⎪ = (1 − α0 θ) f, u + φi , u n Γi + φi , uΓi ∀u ∈ V, ⎪ ⎨ i=2,4,7

i=3,5,6

 ∂θ  ⎪ ⎪ ⎪ ⎪ (Ω), , ϕ + (κ(θ)∇θ, ∇ϕ) − (θv, ∇ϕ) + (βθ, ϕ)Γ R = g R , ϕΓ R + g, ϕ ∀ϕ ∈ WΓ1,2 ⎪ ⎪ D ∂t ⎪ ⎪ ⎪ t ⎪ t ⎪ |στ (θ, v)| ≤ gτ , στ (θ, v) · vτ + gτ |vτ | = 0 on Γ8 , ⎪ ⎪ ⎪ ⎪ t t ⎪ |σ ⎪ n (θ, v, p)| ≤ gn , σn (θ, v, p)vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪ t ⎪ ⎪ σn (θ, v, p) + g+n ≥ 0, (σnt (θ, v, p) + g+n )vn = 0 on Γ10 , ⎪ ⎪ ⎩ t σn (θ, v, p) − g−n ≤ 0, (σnt (θ, v, p) − g−n )vn = 0 on Γ11 .

(8.22) Define a02 (θ ; ·, ·), a12 (·, ·, ·) and f 2 ∈ V ∗ by ˜ a02 (θ˜ ; w, u) = 2(μ(θ˜ )E(w), E(u)) + 2(μ(θ)k(x)w, u)Γ2 + 2(μ(θ˜ )S w, ˜ u) ˜ Γ3 1,2 ˜ + 2(α(x)w, u)Γ5 + (μ(θ)k(x)w, u)Γ7 ∀w, u ∈ V, ∀θ˜ ∈ WΓ (Ω), D

a12 (v, u, w) = rot v × u, w ∀v, u, w ∈ V, φi , u n Γi + φi , uΓi ∀u ∈ V.  f 2 , u = i=2,4,7

i=3,5,6

Define b2 (θ ; ·, ·) and g2 ∈ (WΓ1,2 (Ω))∗ by D

(8.23)

258

8 The Non-steady Boussinesq System

˜ θ, ϕ) = (κ(θ)∇θ, ˜ b2 (θ; ∇ϕ) + (β(x)θ, ϕ)Γ R ∀ θ˜ , θ, ϕ ∈ WΓ1,2 (Ω), D (Ω). g2 , ϕ = g R , ϕΓ R + g, ϕ ∀ϕ ∈ WΓ1,2 D

(8.24)

By (8.6) and (8.7), )∗ ). f 2 ∈ L 2 (0, T ; V∗ ), g2 ∈ L 2 (0, T ; (WΓ1,2 D

(8.25)

Then, taking into account 1 στt (θ, v) = 2μ(θ )εnτ (v), σnt (θ, v, p) = −( p + |v|2 ) + 2μ(θ )εnn (v) 2 and (8.22), we are led to the following variational formulation for problem (8.1), (8.2) (7.4). (Ω)) and Problem II-VE. Find v ∈ K (Q), θ ∈ L ∞ (0, T ; L 2 (Ω)) ∩ L 2 (0, T ; WΓ1,2 D 2 − 21 − 21 t t t t 2 (στ , σn , σ+n , σ−n ) ∈ Lτ (Γ8 ) × L (Γ9 ) × H (Γ10 ) × H (Γ11 ) for a.e. t ∈ (0, T ) such that v(0) = v0 , θ (0) = θ0 and ⎧ ∂v ⎪ ⎪ , u + a02 (θ ; v, u) + a12 (v, v, u) − (στt , u τ )Γ8 − (σnt , u n )Γ9 ⎪ ⎪ ∂t ⎪    t  t ⎪ ⎪ ⎪ − σ+n , u n Γ10 − σ−n , u n Γ11 −  f − α0 θ f, u =  f 2 , u ∀u ∈ V, ⎪ ⎪ ⎪ ⎪  ∂θ ⎪ ⎪ ⎪ ⎨ , ϕ + b2 (θ ; θ, ϕ) − θ v, ∇ϕ = g2 , ϕ ∀ϕ ∈ WΓ1,2 (Ω), D ∂t (8.26) ⎪ |στt | ≤ gτ , στt · vτ + gτ |vτ | = 0 on Γ8 , ⎪ ⎪ ⎪ ⎪ ⎪ |σnt | ≤ gn , σnt vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪  t  ⎪ t ⎪ ⎪ σ+n + g+n ≥ 0, σ+n + g+n , vn Γ10 = 0 on Γ10 , ⎪ ⎪ ⎪   t ⎩ t σ−n − g−n ≤ 0, σ−n − g−n , vn Γ11 = 0 on Γ11 , where L2τ (Γ8 ) is the subspace of L2 (Γ8 ) consisting of functions such that (u, n)L2 (Γ8 ) = 0. Remark 8.5 Similarly to Remark 8.3, Problem II-VE is equivalent to the problem (8.1), (8.2), (7.4). Define operators A2 (θ˜ ) : V → V∗ and B2 : V × V → V∗ , respectively, by ˜ u = a02 (θ; ˜ v, u) ∀v, u ∈ V, θ˜ ∈ WΓ1,2 (Ω), A2 (θ)v, D B2 (v, u), w = a12 (v, u, v) ∀v, u, w ∈ V.

(8.27)

Let functional Ψ be defined by (8.16). Then, in the same way as in Problem I-VI of Sect. 6.1 we find a variational inequality for velocity. Then we have another variational formulation consisting of a variational inequality for velocity and a variational equation for temperature, which is equivalent to Problem II-VE if the solution is

8.2 Variational Formulations for Problems

259

smooth enough.    Problem II-VI. Find (v, θ ) ∈ L ∞ (0, T ; H ) ∩ L 2 (0, T ; V) × L ∞ (0, T ; L 2  (Ω)) ∩ L 2 (0, T ; WΓ1,2 (Ω)) such that D ⎧ T ⎪ ⎪ u  + A2 (θ)v(t) + B2 (v(t), v(t)) − (1 − α0 θ) f − f 2 , u(t) − v(t) dt ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ + Ψ (u) − Ψ (v) ≥ − 1 v0 − u(0)2 ∀u ∈ L 4 (0, T ; V) with u  ∈ L 2 (0, T ; V∗ ), 2 (8.28)

T  ⎪  ⎪ ∂ϕ ⎪ ⎪ ⎪ − θ, + b2 (θ; θ, ϕ) − θv, ∇ϕ − g2 , ϕ dt ⎪ ⎪ ∂t ⎪ 0 ⎪ ⎪ ⎩ = θ0 (x), ϕ(x, 0) ∀ϕ ∈ C 1 ([0, T ]; WΓ1,2 (Ω)) with ϕ(·, T ) = 0. D

8.3 Existence and Uniqueness of Solutions: The Case of Static Pressure The aim of this section is to prove the following Theorem 8.1 (the case of static pressure) Let Assumption 8.1 be satisfied. Suppose that (1) The norms of f, φi , i = 2, · · · , 6, g, g R in the spaces that they belong to are small enough (depending on α0 ) ; (2) v0 ∈ V, Φ(v0 ) = 0; (Ω); (3) θ0 ∈ WΓ1,2 D   (4) A1 v0 + B1 (v0 , v0 ) − f 1 (0) ∈ H (compatibility condition at the initial time for velocity); (5) v  0 V , A1 v0 + B1 (v0 , v0 ) − (1 − α0 θ0 ) f (0) − f 1 (0) are small enough; (6) C1 θ0 + v0 · ∇θ0 − g1 (0) ∈ L 2 (Ω) (compatibility condition at the initial time for temperature); (7) θ0 WΓ1,2 (Ω) and C1 θ0 + v0 · ∇θ0 − g1 (0) L 2 (Ω) are small enough. D Then there exists a solution (v, θ ) to (8.20) such that v ∈ C([0, T ]; V), v  ∈ L 2 (0, T ; V) ∩ L ∞ (0, T ; H ), θ ∈ C([0, T ]; WΓ1,2 ), θ  ∈ L 2 (0, T ; WΓ1,2 (Ω)) ∩ L ∞ (0, T ; L 2 (Ω)). D D

(8.29)

Furthermore, the solution satisfying vV ≤ c, θ WΓ1,2 ≤ c for a constant c > 0 D small enough is unique. Theorem 8.1 will be proved in the next two subsections.

260

8 The Non-steady Boussinesq System

8.3.1 Existence and Estimation of Solutions to an Approximate Problem We first consider an approximate problem for (8.20). For every ε > 0, let Φε be the Moreau-Yosida approximation of Φ and ∇Φε be Fréchet derivative of Φε (see Sect. 1.6). By the fact that Γ2 j , Γ3 j , Γ7 j are in C 2.1 (Γi j ) and 4) of Assumption 8.1, there exists a constant M such that S(x)∞ , k(x)∞ , α L ∞ (Γ5 ) ≤ M. Thus, there exists c∗ such that  μ  2 μ(k(x)z, z)Γ2 + 2μ(S z˜ , z˜ )Γ3 + (α(x)z, z)Γ5 + μ(k(x)z, z)Γ7  ≤ z2V + c∗ z2 4 ∀z ∈ V

(8.30)

(see Theorem 1.27). Thus, we have 7μ u2V − c∗ u2 ∀u ∈ V, 4 |A1 u, v| ≤ c1 uV vV ∃c1 > 0, ∀u, v ∈ V

(8.31)

|B1 (v, u), w| ≤ c2 vV uV wV ,

(8.32)

A1 u, u ≥

and where the operators A1 , B1 are the ones in (8.18). Let {u j , j = 1, 2, · · · } and {ϕ j , j = 1, 2, · · · } be, respectively, bases of the space V and WΓ1,2 (Ω). Without loss of generality, we assume that D   u 1 = v0 , ϕ1 = θ0 as in [1]. We are to find a solution vm = mj=1 g jm (t)u j , θm = mj=1 r jm (t)ϕ j to problem ⎧   ∂vm ⎪ ⎪ + 2 μE (vm ), E (u j ) + (vm · ∇)vm , u j  + 2(μk(x)vm , u j )Γ2 , u j ⎪ ⎪ ∂t ⎪ ⎪   ⎪ ⎪ ⎪ + 2(μS v˜m , u˜ j )Γ3 + 2(α(x)vm , u j )Γ5 + (μk(x)vm , u j )Γ7 + ∇Φε (vm (t)), u j ⎪ ⎨ = (1 − α0 θm ) f, u j  +  f 1 , u j , ⎪ ⎪  ∂θ ⎪ ⎪ m ⎪ ⎪ ⎪ ∂t , ϕ j + (κ∇θm , ∇ϕ j ) + (β(x)θm , ϕ j )Γ R − vm θm , ∇ϕ j  = g1 , ϕ j , ⎪ ⎪ ⎪ ⎩ vm (0) = v0 , θm (0) = θ0 .

(8.33) which gives us an ordinary differential system for g jm (t), r jm (t), j = 1, · · · , m. The solutions to (8.33) depend on ε, but for convenience of notation hereafter we use subindex m instead of subindex m, ε. For a tm there exist absolute continuous functions g jm (t) and r jm (t) on [0, tm ). Since f ∈ W 1,∞ (0, T ; L3 (Ω)),

8.3 Existence and Uniqueness of Solutions: The Case of Static Pressure

261

f 1 ∈ W 1,∞ (0, T ; V∗ ), g1 ∈ W 1,2 (0, T ; (WΓ1,2 )∗ ) and ∇Φε is Lipschitz continuous, D   g jm (t) and r jm (t) are in fact absolute continuous. If vm (t), θm (t) are bounded and vm (t), θm (t) are integrable, then g jm (t), r jm (t) are prolonged over tm . Under smallness of the data of problem and compatibility condition of the data at the initial instant we will find estimates for vm (t) and θm (t) below, by which we obtain (8.82) and see that tm = T. Multiplying the first and second equation of (8.33), respectively, by g jm (t), ϕ jm (t) and adding for i = 1, · · · , m, we get ⎧   ∂vm ⎪ ⎪ , vm + 2 μE (vm ), E (vm ) + (vm · ∇)vm , vm  + 2(μk(x)vm , vm )Γ2 ⎪ ⎪ ∂t ⎪ ⎪ ⎪ ⎪ ⎪ + 2(μS v˜m , v˜m )Γ3 + 2(α(x)vm , vm )Γ5 + (μk(x)vm , vm )Γ7 + ∇Φε (vm (t)), vm  ⎪ ⎨ = (1 − α0 θm ) f, vm  +  f 1 , vm , ⎪ ⎪  ∂θ ⎪ ⎪ m ⎪ ⎪ , θm + (κ∇θm , ∇θm ) + (β(x)θm , θm )Γ R − vm θm , ∇θm  = g1 , θm , ⎪ ⎪ ∂t ⎪ ⎪ ⎩ vm (0) = v0 , θm (0) = θ0 .

(8.34) We will find a priori estimate for I (t) := vm (t)2 + vm (t)2 + θm (t)2 + θm (t)2 . Since Φε is convex, continuous and Fréchet differentiable, we have Φε (u) − Φε (vm (t)) ≥ ∇Φε (vm (t)), u − vm (t) ∀u ∈ V.

(8.35)

On the other hand, by condition (2) of the theorem and (1.37), we have that Φε (0V ) = 0. Thus 0 ≤ Φε (vm (t)) ≤ ∇Φε (vm (t)), vm (t). (8.36) Also, 2|−α0 θm (t) f, vm (t)| ≤ c|α0 |θm (t)2W 1,2  f 2L3 + ΓD

μ vm (t)2V . 4

(8.37)

By virtue of (8.31), (8.32), (8.36) and (8.37), we have from the first equation of (8.34) d 7μ vm (t)2 + vm (t)2V − 2c2 vm (t)3V + 2Φε (vm (t)) dt 2 ≤ c|α0 |θm (t)2

WΓ1,2 D

 f 2L3 + c f 2L3 + c f 1 2V∗ +

μ vm (t)2V + 2c∗ vm (t)2 , 2

where c∗ and c2 are, respectively, the one in (8.30) and (8.32), and so

262

8 The Non-steady Boussinesq System

  d vm (t)2 + 3μ − 2c2 vm (t)V vm (t)2V + 2Φε (vm (t)) dt ≤ c|α0 |θm (t)2W 1,2  f 2L3 + c f 2L3 + c f 1 2V∗ + 2c∗ vm (t)2 . ΓD

(8.38) Here and the rest of this section constants independent of the data of problem are written by c with the exceptions of c∗ , c2 . Setting t = 0 in the first equation of (8.33), multiplying the resulting equation by g jm (0) and adding for j = 1, · · · , m, we get vm (0)2 + A1 vm (0), vm (0) + B1 (vm (0), vm (0)), vm (0) + ∇Φε (v0 ), vm (0) = (1 − α0 θ0 ) f (0), vm (0) +  f 1 (0), vm (0). (8.39) By condition (2) of theorem, for any u ∈ V we have Φ(u) ≥ Φ(v0 ) = 0, which by (1.37) implies Φε (v0 ) = 0 and ∇Φε (v0 ) = 0. Then, we have from (8.39) vm (0) ≤ A1 v0 + B1 (v0 , v0 ) − (1 − α0 θ0 ) f (0) − f 1 (0),

(8.40)

which is valid by the compatibility condition at the initial time for velocity (condition (4)) and the conditions for θ0 , f . On the other hand, taking into account (8.31), (8.32) and (8.36), we have from the first equation of (8.34) 7μ vm (t)2V ≤2c2 vm 3V + 2(1 − α0 θm (t)) f (t), vm (t) 2 + 2 f 1 (t), vm (t) + 2c∗ vm 2 − 2(vm (t), vm (t)), and so 3μvm (t)V ≤2c2 vm (t)2V + c f (t)L3 + c|α0 |θm (t) f (t)L3 + c f 1 (t)V∗ + (2δvm (t) + 2c∗ δvm (t)),

(8.41)

where δ is such that  ·  ≤ δ · V . Since vm θm , ∇θm  = 0 by (8.8), we get from the second equation of (8.34)   d θm (t)2 + 2κθm (t)2W 1,2 + 2(β(x)θm , θm )Γ R = 2 g1 , θm (t) . Γ dt D

(8.42)

We have from (8.42) d 1 θm (t)2 + κθm (t)2W 1,2 + (β(x)θm , θm )Γ R ≤ g1 2(W 1,2 )∗ Γ ΓD dt κ D

(8.43)

and θm (t) +

t

2

1 κ|∇θm (s)| d xds ≤ θ0  + κ Ω 2

0



2

0

t

g1 (s)2(W 1,2 )∗ ds. ΓD

(8.44)

8.3 Existence and Uniqueness of Solutions: The Case of Static Pressure

263

Setting t = 0 in the second equation of (8.33), multiplying the resulting equation by r jm (0) and adding for j = 1, · · · , m, we get θm (0)2 + b1 (θ0 , θm (0)) + (v0 · ∇θ0 , θm (0)) = g1 (0), θm (0),

(8.45)

where −(v0 θ0 , ∇θm (0)) = (v0 · ∇θ0 , θm (0)) was used. We have from (8.45) θm (0) ≤ C1 θ0 + v0 · ∇θ0 − g1 (0),

(8.46)

which is valid by the compatibility condition at the initial time for temperature (condition (6)). On the other hand, taking into account vm θm , ∇θm  = 0, we have from the second equation of (8.34) κθm (t)2W 1,2 ≤ g1 (WΓ1,2 )∗ θm (t)WΓ1,2 + δ1 θm (t)WΓ1,2 θm (t), ΓD

D

D

D

where δ1 is such that  ·  ≤ δ1  · WΓ1,2 , and so D

θm (t)WΓ1,2 ≤ D

 1 g1 (WΓ1,2 )∗ + δ1 θm (t) . D κ

(8.47)

substituting (8.44) into (8.41), we have 3μvm (t)V ≤c2 vm (t)2V + c f  L ∞ (0,T ;L3 )  1/2

1 T + c|α0 | θ0 2 + g1 2(W 1,2 )∗ ds  f  L ∞ (0,T ;L3 ) (8.48) ΓD κ 0 +  f 1  L ∞ (0,T ;V∗ ) + max{2δ, 2c∗ δ}(vm (t) + vm (t)). Differentiating the first equality of (8.33) with respect to t yields 

       vm (t), v j + A1 vm (t), v j + (B1 (vm (t), vm (t)) , v j + (∇Φε (vm )) , v j       (8.49) = − α0 θm (t) f, v j − (1 − α0 θm (t)) f  , v j + f 1 , v j .

Multiplying (8.49) by g jm (t) and summing for j yield       vm (t), vm (t) + A1 vm (t), vm (t) + (B1 vm (t), vm (t)) , vm (t) + (∇Φε (vm )) , vm (t)    = −α0 θm (t) f, vm (t) − (1 − α0 θm (t)) f  , vm (t) +  f 1 , vm (t).

Calculating (B((vm (t), vm (t))) , we have |(B1 vm (t), vm (t)) , vm (t)| = |B1 (vm , vm ), vm  + B1 (vm , vm ), vm | ≤ 2c2 vm V vm 2V ,

(8.50)

(8.51)

264

8 The Non-steady Boussinesq System

where c2 is the one in (8.32). Also, by Hölder’s and Young’s inequalities we have μ  v (t)2V , 8 m μ + vm (t)2V , 8

2|α0 θm (t) f, vm (t)| ≤ c|α0 |θm (t)2W 1,2  f 2L3 + ΓD

2|α0 θm (t) f  , vm (t)| ≤ c|α0 |θm (t)2W 1,2  f  2L3 ΓD

μ ≤ c f + vm (t)2V , 2| f 8 μ     2 2| f 1 (t), vm (t)| ≤ c f 1 (t)V∗ + vm (t)2V . 8 

(t), vm (t)|



(t)2L3

(8.52)

Taking into account (8.31), (8.51), (8.52) and the fact that (∇Φε (vm )) , vm  ≥ 0 (see (1.38)), we have from (8.50) d  μ vm (t)2 + (3μ − 4c2 vm (t)V )vm (t)2V + vm (t)2V dt 2 ≤ cα0 θm (t)2W 1,2  f 2L3 + c|α0 |θm (t)2W 1,2  f  2L3 ΓD

ΓD

+ c f  2L3 + c f 1 2V∗ + 2c∗ vm (t)2 +

μ  v (t)2V , 2 m

that is, d  v (t)2 + (3μ − 4c2 vm (t)V )vm (t)2V dt m ≤ c|α0 |θm (t)2W 1,2  f 2L3 + c|α0 |θm (t)2W 1,2  f  2L3 ΓD

(8.53)

ΓD

+ c f  2L3 + c f 1 2V∗ + 2c∗ vm (t)2 . Differentiating the second equality of (8.33) with respect to t, we have 

         θm (t), ϕ j + C1 θm (t), ϕ j − vm θ, ∇ϕ j − vm θ  , ∇ϕ j = g1 , ϕ j .

(8.54)

Multiplying (8.54) by r jm (t) and adding for j, we have  (t), θ  (t) + C θ  (t), θ  (t) − v  θ , ∇θ  (t) − v θ  , ∇θ  (t) = g  , θ  (t). θm m 1 m m m m m m m 1 m

(8.55)

On the other hand, we get 2|vm θ, ∇θm (t)| ≤

c  2 v  θ 2W 1,2 + κθm (t)2W 1,2 . ΓD ΓD κ m V

(8.56)

Taking into account vm θ  , ∇θm (t) = 0 (see (8.8)) and (8.56), we have from (8.55)

8.3 Existence and Uniqueness of Solutions: The Case of Static Pressure

265

d  θ (t)2 + 2κθm (t)2W 1,2 ΓD dt m c  2 κ ≤ vm V θ 2W 1,2 + κθm (t)2W 1,2 + cg1 2(W 1,2 )∗ + θm (t)2W 1,2 . ΓD ΓD ΓD ΓD κ 4 (8.57) Rewriting (8.57) yields d  3κ  c θm (t)2 + θm (t)2W 1,2 ≤ vm 2V θ 2W 1,2 + cg1 2(W 1,2 )∗ . ΓD ΓD ΓD dt 4 κ

(8.58)

Adding (8.38), (8.53), (8.43) and (8.58), we have     d c   (t)2V + μ − θm 2 1,2 vm (t)2V I (t) + (2μ − 4c2 vm (t)V ) vm (t)2V + vm WΓ dt κ D   3κ + (κ − c|α0 | f 2L3 − c|α0 | f  2L3 )θm 2 1,2 + − c|α0 | f 2L3 θm 2 1,2 WΓ WΓ 4 D D ≤ c( f 2L3 +  f  2L3 ) + c( f 1 2V∗ +  f 1 2V∗ ) + c(g1 2

(WΓ1,2 )∗

+ g1 2

(WΓ1,2 )∗

D

)

D

 + 2c∗ (vm (t)2 + vm (t)2 ).

(8.59) Integrating (8.59) yields

t     c   (2μ − 4c2 vm (s)V ) vm (t)2V + vm (s)2V + μ − θm 2 1,2 vm (s)2V W κ ΓD 0    3κ + (κ − c|α0 | f 2L3 − c|α0 | f  2L3 )θm 2 1,2 + − c|α0 | f 2L3 θm 2 1,2 ds WΓ W 4 ΓD D

t  ≤ I (0) + F(t) + 2c∗ (vm (s)2 + vm (s)2 ) ds,

I (t) +

0

(8.60)

where F(t) :=ct ( f 2W 1,∞ (0,T ;L3 ) +  f 1 2W 1,∞ (0,T ;V∗ ) ) + cg1 2W 1,2 (0,t;(W 1,2 )∗ ) . ΓD

(8.61)

By (8.40) and (8.46) we have I (0) ≤v0 2 + A1 v0 + B1 (v0 , v0 ) − (1 − α0 θ0 ) f (0) − f 1 (0)2 + θ0 2 + C1 θ0 + v0 · ∇θ0 − g1 (0)2 .

(8.62)

By the condition of theorem, we can assume  f W 1,∞ (0,T ;L3 ) to be so small that κ − c|α0 | f (t)2L3 − c|α0 | f  (t)2L3 ≥ 0 at a.e. t ∈ [0, T ], 3κ − c|α0 | f (t)2L3 ≥ 0 at a.e. t ∈ [0, T ]. 4

(8.63)

266

8 The Non-steady Boussinesq System

If vm (0)V = v0 V < and

μ 2c2

θm (0)WΓ1,2 = θ0 WΓ1,2 < D

D

μκ c

(8.64)

(8.65)

are valid, then there exists a tm such that on [0, tm ] 2μ − 4c2 vm (t)V ≥ 0, c μ − θm (t)WΓ1,2 ≥ 0. D κ

(8.66)

Therefore, taking into account (8.62), by Gronwall’s inequality we have  I (t) ≤ v0 2 + A1 v0 + B1 (v0 , v0 ) − (1 − α0 θ0 ) f (0) − f 1 (0)2  + θ0 2 + C1 θ0 + v0 · ∇θ0 − g1 (0)2 + F(T ) e2c∗ t

(8.67)

on the all intervals of t satisfying (8.66). Using this estimate, we will obtain a quadratic inequality satisfied by vm (t)V . Put  β := v0 2 + A1 v0 + B1 (v0 , v0 ) − (1 − α0 θ0 ) f (0) − f 1 (0)2  (8.68) + θ0 2 + C1 θ0 + v0 · ∇θ0 − g1 (0)2 + F(T ) e2c∗ T . Obviously, β depends only on the data of problem. Then, when f satisfies (8.63), we can get from (8.67)  √  1   vm (t) + vm (t) ≤ 2 vm (t)2 + vm (t)2 2 ≤ 2β,  θm (t) ≤ β

(8.69)

on [0, tm ] where (8.66) holds. Let the data of problem be so small that  μ c (g1 W 1,2 (0,T ;(WΓ1,2 )∗ ) + δ1 β) ≤ . 2 D κ 2 By (8.47) and (8.69), for the small data of problem we have on [0, tm ]  c μ c θm (t)WΓ1,2 ≤ 2 (g1 W 1,2 (0,T ;(WΓ1,2 ))∗ + δ1 β) ≤ , D D κ κ 2 which implies

(8.70)

8.3 Existence and Uniqueness of Solutions: The Case of Static Pressure

μ−

c μ θm (t)WΓ1,2 ≥ ∀t ∈ [0, tm ]. D κ 2

267

(8.71)

Therefore, for such small data of problem that (8.70) is valid, if 2μ − 4c2 vm (t)V ≥ 0 ∀t ∈ [0, tm + γ ], γ > 0, tm + γ ≤ T, then owing to (8.71), step by step, we have μ−

μ c θm (t)WΓ1,2 ≥ ∀t ∈ [0, tm + γ ]. D κ 2

(8.72)

From above we see that for the small data of problem satisfying (8.63)–(8.65) and (8.70), μ c (8.73) μ − θm (t)WΓ1,2 ≥ D κ 2 is valid on the interval where the first inequality of (8.66) holds. Put

 1/2 1 T γ := f (t) L ∞ (0,T ;L3 ) + c|α0 | θ0 2 + g1 2(W 1,2 )∗ ds  f (t) L ∞ (0,T ;L3 ) ΓD κ 0  +  f 1 (t) L ∞ (0,T ;V ∗ ) + max{2δ, 2c∗ δ} 2β. (8.74) By (8.44), (8.69) and (8.41), for the small data satisfying (8.63)–(8.65) and (8.70) we have a quadratic inequality for vm (t)V , which is the one we want, 0 ≤ γ − 3μvm (t)V + 2c2 vm (t)2V

(8.75)

on the intervals where the first inequality of (8.66) is satisfied. By the conditions of theorem, we can assume that the data of problem are so small that (8.63)–(8.65) and (8.70) are valid and γ satisfies the following inequality 9μ2 − 8c2 γ > 4μ2 .

(8.76)

Now, let us prove that if v0 V ≤

3μ −

 9μ2 − 8c2 γ  μ  < , 4c2 4c2

(8.77)

then for any m 2μ − 4c2 vm (t)V ≥ μ ∀t ∈ [0, T ]. Since 2μ − 4c2 v0 V > μ, on a interval [0, tm ] 2μ − 4c2 vm (t)V ≥ μ.

(8.78)

268

8 The Non-steady Boussinesq System

Let us prove that if the first inequality of (8.66) is valid on an interval [0, t m ], then more strong estimate 2μ − 4c2 vm (t)V ≥ μ ∀t ∈ [0, t m ]

(8.79)

is valid. To this end, putting y = vm (t)V in (8.75) (which is valid on the interval where the first inequality of (8.66) holds when (8.63)–(8.65) and(8.70) are valid), we get (8.80) 0 ≤ γ − 3μy + 2c2 y 2 on [0, t m ]. By virtue of (8.76), there exist two real roots of polynomial γ − 3μy + c2 y 2 y1 =

3μ −



9μ2 − 8c2 γ 4c2

and y2 =

3μ +



9μ2 − 8c2 γ , 4c2

and on the intervals [0, y1 ] and [y2 , +∞) (8.80) holds. Thus, by continuity of vm (t)V with respect to t we have from v(0)V ∈ [0, y1 ] that vm (t)V ∈ [0, y1 ] ∀t ∈ [0, t m ], that is, vm (t)V ≤

3μ −

 9μ2 − 8c2 γ μ < ∀t ∈ [0, t m ]. 4c2 4c2

Thus, 2μ − 4c2 vm (t)V > μ ∀t ∈ [0, t m ],

(8.81)

which shows (8.79). Thus, step by step we see that the first inequality of (8.66) is valid on [0, T ] and so is (8.78). If (8.77) is valid, then so is (8.64). Therefore, for the small data satisfying (8.63), (8.65), (8.70), (8.76) and (8.77), we have also (8.73) on [0, T ]. By (8.78) and (8.73), we have μ ∀t ∈ [0, T ], ∀m, ∀ε > 0, vm (t)V ≤ 4c2 (8.82) c μ θm (t)WΓ1,2 ≤ ∀t ∈ [0, T ], ∀m, ∀ε > 0. D κ 2 Note that β, γ depend on α0 , and the restriction of data depends on α0 . Then, by (8.67) and (8.60), we have vm (t) ≤ const ∀t ∈ [0, T ], ∀m, ∀ε > 0, vm  L 2 (0,T ;V) ≤ const ∀m, ∀ε > 0, θm (t) ≤ const ∀t ∈ [0, T ], ∀m, ∀ε > 0, θm  L 2 (0,T ;WΓ1,2 ) ≤ const ∀m, ∀ε > 0. D

By (8.82),

(8.83)

8.3 Existence and Uniqueness of Solutions: The Case of Static Pressure



T

Φε (vm (t)) dt ≤ const ∀m, ∀ε > 0,

269

(8.84)

0

and so by (1.36) and (1.37),

T 0

vm (t) − Jε (vm (t))2V dt ≤ cε ∀m, ∀ε > 0.

(8.85)

8.3.2 Existence and Uniqueness of a Solution In this subsection we complete proof of Theorem 8.1. Let us first prove existence of a solution. Due to (8.82) and (8.83), we can extract a subsequence from {(vm , θm )} obtained in preceding subsection, which is still denoted with the subindex as before, such that vm → v in C([0, T ]; V), vm  v  in L 2 (0, T ; V), ∗

vm  v  in L ∞ (0, T ; H ), θm → θ

in C([0, T ]; WΓ1,2 ), D

(8.86)

θm  θ  in L 2 (0, T ; WΓ1,2 ), D ∗

θm  θ  in L ∞ (0, T ; L 2 (Ω)) when m → ∞ and ε → 0 (note that for convenience in the preceding subsection we used subindex  m instead of m, ε). 1 Put u = M j=1 k j (t)u j , where k j (t) ∈ C [0, T ], M is a positive integer. Multiply the first equation of (8.33) by k j (t) and sum for j = 1, · · · , M. Then, multiply the first equation of (8.33) by g jm (t) and sum for j = 1, · · · , m. Subtracting the resulting equations yields vm (t) + A1 vm (t) + B1 (vm (t), vm (t)) + ∇Φε (vm ), u(t) − vm (t) = (1 − α0 θ ) f + f 1 , u(t) − vm (t).

(8.87)

Since Φε is convex, continuous and Fréchet differentiable, we have Φε (u(t)) − Φε (vm (t)) ≥ ∇Φε (vm (t)), u − vm (t). Taking into account (8.88), we have from (8.87)

(8.88)

270



8 The Non-steady Boussinesq System T

0

vm (t) + A1 vm (t) + B1 (vm (t), vm (t)) − (1 − α0 θm ) f − f 1 , u(t) − vm (t) dt

T   Φε (u(t)) − Φε (vm (t)) dt ≥ 0. + 0

(8.89) Since Φ(u) ≥ Φε (u) and Φ(Jε wm (t)) ≤ Φε (wm (t)) (see (1.37)), we have from (8.89)

T

0

vm (t) + A1 vm (t) + B1 (vm (t), vm (t)) − (1 − α0 θm ) f − f 1 , u(t) − vm (t) dt

T Φ(Jε vm (t)) dt ≥ 0. (8.90) + Ψ (u(t)) − 0

By (8.85), we have that Jε vm → v in L 2 (0, T ; V) as m → ∞ and ε → 0, and so by lower semi-continuity of Φ we have

lim

m→∞,ε→0 0

T

Φ(Jε vm (t)) dt ≥

T

Φ(v(t)) dt.

(8.91)

0

By standard way we can prove that

T



T

B1 (vm (t), vm (t)), v(t) dt →

0

B1 (v(t), v(t)), v(t) dt

(8.92)

0

as m → ∞ and ε → 0. Since limm k →∞,ε→0 A1 vm (t), vm (t) = A1 v(t), v(t), by (8.91) and (8.92), we have from (8.90)

T

v  (t) + A1 v(t) + B1 (v(t), v(t)) − (1 − α0 θ ) f − f 1 , u(t) − v(t) dt

0

+ Ψ (u) − Ψ (v) ≥ 0.

(8.93)

Since vm (0) = v0 , by (8.86) it is obvious that v(0) = v0 .  and the set {u = M B1 (v(t), v(t)) ∈ L ∞ (0, T ; V∗ ) j=1 k j (t)u j ; 1 4 k j (t) ∈ C [0, T ], M : positive integer} is dense in L (0, T ; V), and so (8.93) is valid for all u ∈ L 4 (0, T ; V). By (8.86), from the second equation of (8.33) we can get

0

T

 ∂θm ∂t

 , ϕ + (κ∇θm , ∇ϕ) + (β(x)θm , ϕ)Γ R −vm θm , ∇ϕ dt =



T

g1 , ϕ dt,

0

ϕ ∈ L 2 (0, T ; WΓ1,2 ). D It is easy to see that

(8.94)

8.3 Existence and Uniqueness of Solutions: The Case of Static Pressure

271

(κ∇θm , ∇ϕ) → (κ∇θ, ∇ϕ) for a.e. t ∈ [0, T ]. Also,

   vm θm , ∇ϕ − vθ, ∇ϕ dt

T 0

≤

T



T

vm − vL6 θm  L 3 ∇ϕL2 dt +

v(θm − θ ), ∇ϕ dt

T |v(θm − θ ), ∇ϕ| dt. ≤ vm − v L ∞ (0,T ;V) θm  L 2 (0;T ;L 3 ) ϕ L 2 (0,T ;WΓ1,2 ) + 0

0

D

0

(8.95) Since v∇ϕ ∈ L 2 (0, T ; L 6/5 (Ω)), by (8.86) we have that as m → ∞ and ε → 0, T 0 |v(θm − θ ), ∇ϕ| dt → 0. Thus,

T





T

vm θm , ∇ϕ dt →

0

vθ, ∇ϕ dt.

0

It is easy to prove that

T



T

(β(x)θm , ϕ)Γ R dt →

0

(β(x)θ, ϕ)Γ R dt.

0

Therefore, we have from (8.94)

T 0

 ∂θ ∂t



, ϕ + (κ∇θ, ∇ϕ) + (β(x)θ, ϕ)Γ R

 − vθ, ∇ϕ dt =

T

g1 , ϕ dt.

0

(8.96) Since θm (0) = θ0 , by (8.86) it is obvious that θ (0) = θ0 . Therefore, we have proved the existence of a solution. Next, let us prove uniqueness of a solution. Let (v1 , θ1 ), (v2 , θ2 ) be two solutions to Problem I-VI satisfying inequality (8.82) instead of approximate solutions. Then, taking (8.29) into account (Remark 8.4), we have from (8.21) v1 (t) + A1 v1 (t) + B1 (v1 (t), v1 (t)) − (1 − α0 θ1 ) f − f 1 , v2 (t) − v1 (t) + Φ(v2 (t)) − Φ(v1 (t)) ≥ 0, v2 (t) + A1 v2 (t) + B1 (v2 (t), v2 (t)) − (1 − α0 θ2 ) f − f 1 , v1 (t) − v2 (t) + Φ(v1 (t)) − Φ(v2 (t)) ≥ 0, which imply

272

8 The Non-steady Boussinesq System

v1 (t) − v2 (t), v1 (t) − v2 (t) + A1 (v1 (t) − v2 (t)), v1 (t) − v2 (t) ≤ |α0 ||(θ1 − θ2 ) f, v1 (t) − v2 (t))| + |B1 (v1 (t), v1 (t)) − B1 (v2 (t), v2 (t)), v1 (t) − v2 (t)|.

(8.97) By virtue of (8.31) and (8.32), we have 7μ d ((v1 (t) − v2 (t)2 ) + v1 (t) − v2 (t)2V dt 2 μ ≤ 2c∗ v1 (t) − v2 (t)2 + v1 (t) − v2 (t)2V + c f 2L3 θ1 − θ2 2 2 + 2|B(v1 (t) − v2 (t), v1 (t)), v1 (t) − v2 (t)| + 2|B(v2 (t), v1 (t) − v2 (t)), v1 (t) − v2 (t)| μ ≤ 2c∗ v1 (t) − v2 (t)2 + v1 (t) − v2 (t)2V + c f 2L3 θ1 − θ2 2 2 + 2c2 (v1 (t)V + v2 (t)V )v1 (t) − v2 (t)2V , where c2 is the one in (8.32). By (8.82), 2c2 (v1 (t)V + v2 (t)V ) ≤ μ, and so we have dv1 (t) − v2 (t)2 + 2μv1 (t) − v2 (t)2V ≤ 2c∗ v1 (t) − v2 (t)2 + c f 2L3 θ1 − θ2 2 . dt

(8.98)

Also, from , ϕ + (κ∇θ1 , ∇ϕ) + (β(x)θ1 , ϕ)Γ R − v1 θ1 , ∇ϕ1  = g1 , ϕ, ∂t  ∂θ 2 , ϕ + (κ∇θ2 , ∇ϕ) + (β(x)θ2 , ϕ)Γ R − v2 θ2 , ∇ϕ = g1 , ϕ ∂t

 ∂θ

1

we have  ∂θ − θ 1 2 , θ1 − θ2 + κ(∇θ1 − ∇θ2 , ∇θ1 − ∇θ2 ) + (β(x)(θ1 − θ2 ), θ1 − θ2 )Γ R ∂t − v1 (θ1 − θ2 ), ∇(θ1 − θ2 ) − (v1 − v2 )θ2 , ∇(θ1 − θ2 ) = 0. (8.99) Taking into account v1 (θ1 − θ2 ), ∇(θ1 − θ2 ) = 0 (see (8.8)), by (8.56) we have c d θ1 − θ2 2 + 2κ∇θ1 − ∇θ2 2 ≤ v1 (t) − v2 (t)2V θ2 (t)2 1,2 + κ∇θ1 − ∇θ2 2 , WΓ dt κ D

and so

8.3 Existence and Uniqueness of Solutions: The Case of Static Pressure

d c (θ1 (t) − θ2 (t)2 ) ≤ v1 (t) − v2 (t)2V θ2 (t)2W 1,2 . ΓD dt κ By (8.82),

273

(8.100)

c μ θ2 (t)2W 1,2 ≤ . ΓD κ 2

Therefore adding (8.98) and (8.100), we get d (v1 (t) − v2 (t)2 + θ1 (t) − θ2 (t)2 ) dt    ≤ 2c∗ + c f 2L3 v1 (t) − v2 (t)2 + θ1 (t) − θ2 (t)2 . (8.101) We have from (8.101) v1 (t) − v2 (t)2 + θ1 (t) − θ2 (t)2

t    ≤ 2c∗ + 2c f 2L3 v1 (s) − v2 (s)2 + θ1 (s) − θ2 (s)2 ds,

(8.102)

0

which implies v1 (t) = v2 (t), θ1 (t) = θ2 (t) for all t ∈ [0, T ], and the proof is complete. 

8.4 Existence of a Solution: The Case of Total Pressure Theorem 8.2 (the case of total pressure) Let Assumption 8.2 be satisfied, v0 ∈ HK  and θ0 ∈ L 2 (Ω). Then there exists a solution (v, θ )∈ L ∞ (0, T ; H ) ∩L 2 (0, T ; V) ×   ∞ (Ω)) to (8.28). L (0, T ; L 2 (Ω)) ∩ L 2 (0, T ; WΓ1,2 D As in the preceding section, we will show Theorem 8.2 in the next two subsections.

8.4.1 Existence of a Solution to an Approximate Problem We first consider an approximate problem for (8.28). For every 0 < ε < 1, let a functional Φε be Moreau-Yosida approximation of Φ. Let {u j , j = 1, 2, · · · } and {ϕ j , j = 1, 2, · · · } be, respectively, bases of the space V and WΓ1,2 (Ω). Without D  loss of generality,we assume that u 1 = v0 , ϕ1 = θ0 . We find a solution vm = mj=1 g jm (t)u j , θm = mj=1 r jm (t)ϕ j to the following problem

274

8 The Non-steady Boussinesq System

⎧   ∂vm ⎪ ⎪ + 2 μ(θm )E (vm ), E (u j ) + rotvm × vm , u j  + 2(μ(θm )k(x)vm , u j )Γ2 , u j ⎪ ⎪ ∂t ⎪ ⎪ ⎪ ⎪ ⎪ + 2(μ(θm )S v˜m , u˜ j )Γ3 + 2(α(x)vm , u j )Γ5 + (μ(θm )k(x)vm , u j )Γ7 + ∇Φε (vm (t)), u j  ⎪ ⎨ = (1 − α0 θm ) f, u j  +  f 2 , u j , ⎪ ⎪  ∂θ ⎪ ⎪ m ⎪ ⎪ ⎪ ∂t , ϕ j + (κ(θm )∇θm , ∇ϕ j ) + (β(x)θm , ϕ j )Γ R − vm θm , ∇ϕ j  = g2 , ϕ j , ⎪ ⎪ ⎪ ⎩ vm (0) = v0 , θm (0) = θ0 .

(8.103) which gives us a system for g jm (t), r jm (t), j = 1, · · · , m. The solutions to (8.103) depend on ε, but for convenience of notation here and in what follows of this section we use subindex m. For a tm there exist absolute continuous functions g jm (t) and r jm (t) on [0, tm ). If vm (t), θm (t) are bounded and vm (t), θm (t) are integrable, then g jm (t), r jm (t) are prolonged over tm . We will find estimates (8.118) below, by which we see that tm = T. Multiplying the first and second equations of (8.103), respectively, by g jm (t) and ϕ jm (t) and adding for i = 1, · · · , m, we get ⎧   ∂vm ⎪ ⎪ + 2 μ(θm )E (vm ), E (vm ) + rot vm × vm , vm  + 2(μ(θm )k(x)vm , vm )Γ2 , v m ⎪ ⎪ ∂t ⎪ ⎪ ⎪ ⎪ ⎪ + 2(μ(θm )S v˜m , v˜m )Γ3 + 2(α(x, t)vm , vm )Γ5 + (μ(θm )k(x)vm , vm )Γ7 ⎪ ⎨ + ∇Φε (vm (t)), vm  + α0 θm f, vm  =  f, vm  +  f 2 , vm , (8.104) ⎪ ⎪  ∂θ ⎪ ⎪ m ⎪ ⎪ ⎪ ∂t , θm + (κ(θm )∇θm , ∇θm ) + (β(x)θm , θm )Γ R − vm θm , ∇θm  = g2 , θm , ⎪ ⎪ ⎪ ⎩ vm (0) = v0 , θm (0) = θ0 .

Let us estimate terms on the left hand side above. It is easy to see  2 μ(θm )E(vm ), E(vm ) dt ≥ 2μ0 vm 2V .

(8.105)

By the fact that Γ2 j , Γ3 j , Γ7 j are in C 2.1 (Γi j ) and Assumption 8.2, there exists a constant M such that S(x)∞ , k(x)∞ , αL∞ (Γ5 ) ≤ M.

(8.106)

Thus,     2(μ(θm )k(x)v, v)Γ2 + 2(μ(θm )S v˜m , v˜m )Γ3 + 2(α(x)vm , vm )Γ5 + (μ(θm )k(x)vm , vm )Γ7  ≤

μ0 vm 2V + k11 vm 2 2

(8.107)

(see Theorem 1.27). Obviously, rot vm × vm , vm  = 0.

(8.108)

8.4 Existence of a Solution: The Case of Total Pressure

275

Since Φε is convex, continuous and Fréchet differentiable, we have Φε (y) − Φε (x) ≥ ∇Φε (x), y − x ∀x, y ∈ V.

(8.109)

Thus, ∇Φε (vm (t)), 0V − vm (t) ≤ Φε (0V ) − Φε (vm (t)) ≤ −Φε (vm (t)), and ∇Φε (vm (t)), vm (t) ≥ Φε (vm (t)).

(8.110)

Also, by the Hölder inequality we have   κ0   α0 θm f, vm  dt  ≤ k12 vm (t)2 + θm (t)2W 1,2 , ΓD 4

(8.111)

where k12 = c|α0 | f 2L ∞ (0,T ;L3 ) . | f, wm  +  f 2 , wm | ≤ Also, we have

μ0 wm 2V + c( f 2L3 +  f 2 2V∗ ). 2

 κ(θm )∇θm , ∇θm ) ≥ κ0 θm 2W 1,2 ,

(8.112)



ΓD

(β(x)θm , θm )Γ R ≥ 0.

(8.113)

By (8.8) we have wm θm , ∇θm  = 0. Also, |g2 , θm | ≤

(8.114)

κ0 θm (t)2W 1,2 + cg2 2(W 1,2 )∗ . ΓD ΓD 4

(8.115)

Taking k1 = k11 + k12

(8.116)

and using (8.105)–(8.115), we have from (8.104) d d vm (t)2 + θm (t)2 + 2μ0 vm 2V + κ0 θm 2W 1,2 + Φε (vm (t)) ΓD dt dt   2 2 ≤ c  f (t)L3 +  f 2 (t)V∗ + g2 (t)2(W 1,2 )∗ + 2k1 vm (t)2 . ΓD

Applying Gronwall’s inequality, we have from (8.117)

(8.117)

276

8 The Non-steady Boussinesq System

vm (t)2 + θm (t)2

t

 f (s)2L3 +  f 2 (s)2V∗ + g2 (s)2 ≤ v0 2 + θ0 2 + vm 2L 2 (0,T ;V) 

0 2 + θm  2 L (0,t;WΓ1,2 ) D



T

≤ c v0  + θ0  +

0

2

2

0

T



(WΓ1,2 )∗ D



 f (t)2L3 +  f 2 (t)2V∗ + g2 (t)2

 Φε (vm (t)) dt ≤ c v0 2 + θ0 2 +

T 0



 ds e2k1 t ,



(WΓ1,2 )∗

 f (t)2L3

+  f 2 (t)2V∗

 dt ,

D



+ g2 (t)2

(WΓ1,2 )∗

 dt .

D

(8.118) Note that c in (8.118) depends on T and f (via k12 ), but independent of m and ε. Since Φ is nonnegative, by (1.36) and the third inequality of (8.118), we have

0

T

vm (t) − Jε (vm (t))2V dt ≤ cε

(8.119)

with c independent of ε. Multiplying the first equation of (8.103) by g jm (t) − g jm (s), summing for j and taking into account (8.109), we have 1 dvm (t) − vm (s)2 2 dt + A2 (θm )vm (t) + B2 (vm (t), vm (t)) + α0 θm (t) f (t) − f (t) − f 2 (t), vm (t) − vm (s) = ∇Φε (vm (t)), vm (s) − vm (t) ≤ Φε (vm (s)) − Φε (vm (t)) ≤ Φε (vm (s)),

(8.120) where the operators A2 (θm ), B2 are the ones in (8.27). By (8.107) and (8.106), we have 3μ0 vm (t)2V − k11 vm (t)2 , 2 |A2 (θm )vm (t), vm (s)| ≤ cvm (t)V vm (s)V ,

A2 (θm )vm (t), vm (t) ≥

(8.121)

where k11 is the one in (8.107). Taking into account (8.121) and the fact that B2 (vm (t), vm (t)), vm (t) = 0, we have from (8.120) 1 dvm (t) − vm (s)2 ≤Φε (vm (s)) + A2 (θm )vm (t), vm (s) + B2 (vm (t), vm (t)), vm (s) 2 dt + −α0 θm (t) f (t) + f (t) + f 2 (t), vm (t) − vm (s) + k11 vm (t)2 .

(8.122) Let us integrate every term of (8.122) first with respect to t from s to s + h and then with respect to s from 0 to T , where vm (t) = 0 when t ∈ (T, T + h). It is clear that

8.4 Existence of a Solution: The Case of Total Pressure



T



0

dvm (t) − vm (s)2 dtds = dt

s+h

s



T

277

vm (s + h) − vm (s)2 ds.

(8.123)

0

By the third inequality of (8.118),

T 0



s+h



T

Φε (vm (s)) dtds ≤ h

s

Φε (vm (s)) ds ≤ c1 h.

(8.124)

0

By (8.118) and (8.122), we have 

 

T

0



s+h

  A2 (θm )vm (t), vm (s) dtds  ≤ c

s

T



0



vm (t)V dt ds

s

√ √ vm (s)V ( hvm  L 2 (0,T ;V) ) ds ≤ c2 h.

T

≤c

s+h

vm (s)V

0

(8.125) 1

1

Since wL3 ≤ K wL2 2 wL2 6 (see Theorem 1.12), we have |B2 (v, w), z| = |rot v × w, z| ≤ K rot vL2 wL3 zL6 1

1

≤ K vV w 2 wV2 zV ,

(8.126)

and so by (8.118) we have 

 

T

0



s+h

  B2 (vm (t), vm (t))vm (s) dtds 

s



T

≤K



0

s T

≤K

s+h

vm (s)V

0

3

1

vm (t)V2 vm (t) 2 vm (s)V dtds 

s+h

s

vm (t)2V dt

 43 

s+h

vm (t)2 dt

 14

ds

s

1

≤ c3 h 4 . (8.127) Also, by (8.118) we have 

  

 

T



T



0

0

s+h



 ( f + f 2 )(t), vm (t) dtds  ≤

s+h

  ( f + f 2 )(t), −vm (s) dtds  ≤ K

s

s

In the same way, we get

T

 |( f + f 2 )(t), vm (t)|

0

T

vm (s)V √ ≤ c5 h. 0





t

 ds dt ≤ c4 h,

t−h s+h

( f + f 2 )(t)V∗ dtds

s

(8.128)

278

8 The Non-steady Boussinesq System

  T s+h √   α0 θm f, vm (t) − vm (s) dtds  ≤ c6 h + c7 h,  0 s  T s+h    k11 vm (t)2 dtds  ≤ c8 h.  0

(8.129)

s

Note that constants ci , i = 1, · · · , 8, are independent of m and ε. By virtue of (8.123)–(8.129), uniformly with respect to m, ε

T

1

vm (s + h) − vm (s)2 ds ≤ O(h 4 ),

(8.130)

0

and the set {vm } is relatively compact in L 2 (0, T ; W 10 ,2 (Ω)) (see Theorem 1.38). Also, we have 9

|(κ(θm )∇θm , ∇ϕ)| ≤ κ1 ∇θm L2 ϕWΓ1,2 (Ω) , D

|(β(x)θm , ϕ)Γ R | ≤ cθm WΓ1,2 (Ω) ϕWΓ1,2 (Ω) , D

D

(8.131)

|vm θm , ∇ϕ| ≤ cvm V θm  L 3 ϕWΓ1,2 (Ω) . D

By (8.131), we have from the second equation of (8.103)  ∂θ     m  , ϕ  ≤ c θm WΓ1,2 (Ω) + vm 2V + θm 2W 1,2 (Ω) + g2 (WΓ1,2 )∗ ϕWΓ1,2  ΓD D D D ∂t 1,2 ∀ϕ ∈ WΓ D (Ω). Hence, by (8.118) we see that   θm ∈ L 1 0, T ; (WΓ1,τ )∗ , D  ≤ c, θ    1,τ

(8.132)

m L 1 0,T ;(W )∗ Γ D

where c is independent of m and ε. Thus, the set {θε } is relatively compact in 9 L 2 (0, T ; W 10 ,2 (Ω)) (see Theorem 1.39).

8.4.2 Existence of a Solution Having obtained solutions to (8.103), which are actually a sequence of approximating solutions to (8.28), we can extract subsequences, which are still denoted as before, such that as m → ∞, ε → 0,

8.4 Existence of a Solution: The Case of Total Pressure

279

vm  v in L 2 (0, T ; V), ∗

vm  v in L ∞ (0, T ; H ), vm → v in L 2 (0, T ; W 10 ,2 (Ω)), 9

θm  θ in L 2 (0, T ; W D1,2 (Ω)),

(8.133)

∗

θm  θ in L ∞ (0, T ; L 2 (Ω)), θm → θ in L 2 (0, T ; W 10 ,2 (Ω)). 9

 1 On the other hand, putting u = M j=1 k j (t)u j , where k j (t) ∈ C [0, T ] and M is a positive integer, let us multiply the first equation of (8.103) by k j (t) and sum for j = 1, · · · , M. Then, let us multiply the first equation of (8.103) by g jm (t) and sum for j = 1, · · · , m. Substituting the resulting equations yields

T

 ∂vm ∂t

0

 + A2 (θm )vm (t) + B2 (vm (t), vm (t)) + ∇Φε (vm (t)), u(t) − vm (t) dt

T −α0 θm f + f + f 2 , u(t) − vm (t) dt. = 0

(8.134) Since

T vm (t), u(t) − vm (t) dt 0

T 1 1 u  (t), u(t) − vm (t) dt − vm (T ) − u(T )2 + vm (0) − u(0)2 , = 2 2 0 taking into account (8.109), we have from (8.134)

T

u  (t) + A2 (θm )vm (t) + B2 (vm (t), vm (t)), u(t) − vm (t) dt

T

T   −α0 θm f + f + f 2 , u(t) − vm (t) dt + Φε (u(t)) − Φε (vm (t)) dt −

0

0

1 ≥ − vm (0) − u(0)2 . 2

0

(8.135) Since Φε (u) ≤ Φ(u) and Φ(Jε vm (t)) ≤ Φε (vm (t)) (see (1.37)), we have from (8.135)

280



8 The Non-steady Boussinesq System T

0

u  (t) + A2 (θm )vm (t) + B2 (vm (t), vm (t)), u(t) − vm (t) dt

T −α0 θm f + f + f 2 , u(t) − vm (t) dt + Ψ (u) − − 0

T

 Φ(Jε vm (t)) dt

0

1 ≥ − vm (0) − u(0)2 . 2

(8.136)

By (8.133) and Corollary 1.1, we get

T



T

A2 (θm )vm (t), u(t) dt ≡

0



T

a02 (θm (t); vm (t), u(t)) dt →

0

A2 (θ)v(t), u(t) dt.

0

(8.137)

Owing to (8.133), vm → v in L 2 (0, T ; L2 (∂Ω)). Thus taking a subsequence if necessary, by Lemma 1.3 we deduce 2(μ(θm )k(x)vm , vm )Γ2 + 2(μ(θm )S v˜m , v˜m )Γ3 + 2(α(x)vm , vm )Γ5 + (μ(θm )k(x)vm , vm )Γ7 → 2(μ(θ)k(x)v, v)Γ2 + 2(μ(θ)S v, ˜ v) ˜ Γ3 + 2(α(x)v, v)Γ5 + (μ(θ)k(x)v, v)Γ7 .

Therefore, taking into account lim inf 2(μ(θm )E(vm ), E(vm )) ≥ 2(μ(θ )E(v), E(v)) (see Corollary 1.2), we have lim infA2 (θm )vm (t), vm (t) ≥ A2 (θ )v(t), v(t).

(8.138)

From (8.137) and (8.138), it follows

T

lim inf m→∞ ε→0



T

A2 (θm )vm (t), u(t) − vm (t) dt ≤

0

A2 (θ )v(t), u(t) − v(t) dt.

0

(8.139) By (8.119) and (8.133), we have that Jε (vm )  v in L 2 (0, T ; V) as m → ∞, ε → 0. Since the functional Φ : V → R is weakly lower semi-continuous, we have

T

lim inf m→∞ ε→0



T

Φ(Jε vm (t)) dt ≥

0

Φ(v(t)) dt ≡ Ψ (v).

(8.140)

0

It is not difficult to prove that

0

T



T

B2 (vm (t), vm (t)), u(t) dt → 0

B2 (v(t), v(t)), u(t) dt

(8.141)

8.4 Existence of a Solution: The Case of Total Pressure

281

as m → ∞, ε → 0 (see (6.46)–(6.49)). Let us prove that

T



T

α0 θm f, u(t) − vm (t) dt →

0

α0 θ f, u(t) − v(t) dt → 0 as m → ∞, ε → 0.

0

By (8.133), we have



T

(8.142)

(α0 θm − α0 θ ) f, u(t) dt → 0

(8.143)

0

and

  α0 θm f, vm (t) − α0 θ f, v(t) dt 0

T

T α0 (θm − θ ) f, vm (t) dt + α0 θ f, vm (t) − v(t) dt → 0. = T

0

(8.144)

0

By (8.143) and (8.144) we get (8.142). It is obvious that B2 (v(t), v(t)), v(t) = 0. Therefore, by (8.139)–(8.141) and (8.142), we have from (8.136)

T

u  (t) + A2 (θ )v(t) + B2 (v(t), v(t)) − (1 − α0 θ ) f − f 2 , u(t) − v(t) dt

0

1 + Ψ (u) − Ψ (v) ≥ − v0 − u(0)2 . 2 3

(8.145)

Since B2 (v(t), v(t))V∗ ≤ K v(t)V2 v(t) 2 (see (8.126)) and v ∈ L ∞ (0, T ; 4 2 L we see that B2 (v, v) ∈ L 3 (0, T ; V∗ ). Therefore, by density of the set {u = (Ω)), M 1 4  2 ∗ j=1 k j (t)u j , k j (t) ∈ C [0, T ], M=1, 2, · · · } in {L (0, T ; V) : u ∈L (0, T ; V )} (see Remark 1.8), the inequality (8.145) is valid for all u ∈ {L 4 (0, T ; V) : u  ∈ L 2 (0, T ; V∗ )}. Thus, the first formula of (8.28) holds.  1 k Putting ϕ(t) = M j=1 j (t)ϕ j , where k j (t) ∈ C [0, T ], k j (T ) = 0, multiplying the second equation of (8.103) by k j (t) and summing up for j = 1, · · · , M, we obtain

t

t

t    ∂ϕ  ds + θm , (κ(θm ∇θm , ∇ϕ) ds + (β(x)θm , ϕ)Γ R ds θm (t), ϕ(t) − ∂t 0 0 0

t

t     g2 , ϕ ds ∀t ∈ [0, T ]. vm θm , ∇ϕ ds = θm (0), ϕ(0) + − 0

By Corollary 1.1 we have

1

0

(8.146)

282

8 The Non-steady Boussinesq System





t

t

(κ(θm )∇θm , ∇ϕ) ds →

0

(κ(θ )∇θ, ∇ϕ) ds as m → ∞, ε → 0. (8.147)

0

Since W 10 ,2 (Ω) ⊂ L 4 (Ω) (see Theorem 1.20), we have 9

vm → v in L 2 (0, T ; L4 (Ω)),

θm → θ in L 2 (0, T ; L 4 (Ω)).

(8.148)

By (8.148), we have

t 0

≤

|vm θm , ∇ϕ − vθ, ∇ϕ| ds

t 0

|(vm − v)θm , ∇ϕ| ds +

t 0

|v(θm − θ), ∇ϕ| ds

  ≤ vm − v L 2 (0,T ;L4 ) θm  L 2 (0,T ;L 4 ) + v L 2 (0,T ;L4 ) θm − θ L 2 (0,T ;L 4 )  ∇ϕ L ∞ (0,T ;L2 ) → 0,

which implies

t

t vm θm , ∇ϕ ds → vθ, ∇ϕ ds as m → ∞, ε → 0. 0

(8.149)

0

Therefore taking into account (8.147) and (8.149), we have from (8.146)

T

T

T  ∂ϕ  dt + − θ, (κ(θ )∇θ, ∇ϕ) dt + (β(x)θ, ϕ)Γ R dt ∂t 0 0 0 (8.150)

t

T     g2 , ϕ ds. vθ, ∇ϕ dt = θ (0), ϕ(0) + − 0

0

M

Since the set {ϕ(t) = j=1 k j (t)ϕ j : k j (t) ∈ C 1 [0, T ], k j (T ) = 0, M = 1, 2, · · · } is dense in {ϕ ∈ C 1 ([0, T ]; WΓ1,2 ) : ϕ(T ) = 0} (see Remark 1.8), we deduce from D (8.150) the second equation of (8.28).  Remark 8.6 The Eq. (10.1) of Chap. 10 is more general than (8.1), and from the result of Chap. 10 a result for (8.1) with the boundary conditions (8.2), (7.4) can be got, however the result demands that the parameter for buoyancy effect α0 is small enough in accordance with the data of problem and the solution includes “defect measure” as in [2].

8.5 Bibliographical Remarks The content of Chap. 8 is taken from [3]. In [4, 5] existence and uniqueness (for 2-D) of a solution to the problem were shown under homogeneous Dirichlet boundary condition for velocity and mixture of non-homogeneous Dirichlet and Neumann boundary conditions for temperature.

8.5 Bibliographical Remarks

283

In [6] for the problem with non-homogeneous Dirichlet boundary conditions for velocity and temperature the existence of time periodic solution was shown (see [7]). In [8–11] problem (8.1) on the time dependent domain was studied under nonhomogeneous Dirichlet boundary condition for velocity and temperature. In [12, 13] the problem on exterior domains with homogeneous Dirichlet boundary condition for velocity and non-homogeneous Dirichlet boundary condition for temperature was discussed. In [14] problem (8.1) was studied under mixture of non-homogeneous Dirichlet and stress boundary conditions for fluid and mixture of non-homogeneous Dirichlet, Neumann and Robin boundary conditions for temperature. They proved existence of a unique local-in-time solution under a compatibility condition at initial time (see (27) and (31) of [14]). In [15] problem (8.1) in cylindrical pipe with inflow and outflow was studied under slip boundary conditions for velocity and the Neumann conditions for temperature. There it was proved the existence of a solution on the given interval when norms of derivatives in the direction along the cylinder of the initial velocity, initial temperature and the external force are small enough. In [16] the existence of a solution to problem (8.1) on the time dependent domain was obtained under mixture of Dirichlet condition of velocity, total pressure and vorticity boundary conditions for fluid and mixture of Dirichlet, Neumann and Robin boundary conditions for temperature. In [17] on 3-D channel under mixture of homogeneous Dirichlet boundary condition and “do nothing” condition for fluid and mixture of homogeneous Dirichlet and Neumann boundary conditions for temperature, local existence and global uniqueness of strong solutions were studied.

References 1. J. Naumann, On evolution inequalities of Navier-Stokes type in three dimensions. Ann. Math. Pure Appl. 124(4), 107–125 (1980) 2. J. Naumann, J. Wolf, Existence of weak solutions to the equations of natural convection with dissipative heating. Advances in Mathematical Fluid Mechanics, pp. 367–384 (Springer, 2010) 3. T. Kim, The non-steady Boussinesq system with mixed boundary conditions including conditions of friction type. Inter. J. Diff. Equat. 2020 Article ID 6096531 (2020) 4. H. Morimoto, On non-stationary Boussinesq equations. Proc. Japan Acad. 67 Ser. A, 159–161 (1991) 5. H. Morimoto, Non-stationary Boussinesq equations. J. Fac. Sci. Univ. Tokyo Sect. IA Math. 39, 61–75 (1992) 6. H. Morimoto, Heat convection equation with nonhomogeneous boundary conditions. Funkcialai Ekvacioj 53, 213–229 (2010) 7. H. Morimoto, Survey on time periodic problem for fluid flow under inhomogeneous boundary condition. Discrete Cont. Dyn. Syst. Ser. S 5, 631–639 (2012) 8. H. Inoue, M. Ôtani, Strong solutions of initial boundary value problems for heat convection equations in noncylindrical domains. Nonlinear Anal. 24, 1061–1090 (1995) 9. H. Inoue, M. Ôtani, Periodic problems for heat convection equations in noncylindrical domains. Funkcialaj Ekvacioj 40, 19–39 (1997) 10. K. Oeda, On the initial value problem for the heat convection equation of Boussinesq approximation in a time-dependent domain. Proc. Japan Acad. 64 Ser. A, 143–146 (1988)

284

8 The Non-steady Boussinesq System

11. K. Oeda, Remarks on the periodic solution of the heat convection equation in a perturbed annulus domain. Proc. Japan Acad. 73 Set. A, 21–25 (1997) 12. K. Oeda, Periodic solutions of the heat convection equations in exterior domains. Proc. Japan Acad. 73 Set. A, 49–54 (1997) 13. K. Oeda, N. Matsuda, Initial value problems for the heat convection equations in exterior domains. Tokyo J. Math. 21, 359–375 (1998) 14. Z. Skalák, P. Kuˇcera, An existence theorem for the Boussinesq equations with non-Dirichlet boundary conditions. Appl. Math. 45(2), 81–98 (2000) 15. P. Kacprzyk, Long-time existence of solutions to the Navier-Stokes equations with inflowoutflow and heat convection. Math. Meth. Appl. Sci. 35, 1000–1013 (2012) 16. T. Kim, D. Cao, Existence of solutions to the heat convection equations in a time-dependent domain with mixed boundary conditions. J. Math. Sci. Univ. Tokyo 22, 531–568 (2015) 17. M. Beneš, The “do nothing” problem for Boussinesq fluids. Appl. Math. Lett. 31, 25–28 (2014)

Chapter 9

The Steady Equations for Heat-Conducting Fluids

In this chapter we are concerned with the equation for steady flow of heat-conducting incompressible Newtonian fluids with dissipative heating under mixed boundary conditions. The boundary conditions for fluid may include Tresca slip, leak condition, one-sided leak conditions, velocity, pressure, vorticity, stress together and the conditions for temperature may include Dirichlet, Neumann and Robin conditions together. On the basis of results of Sect. 3.1, we get variational formulations consisting of a variational inequality for velocity and a variational equation for temperature, which are equivalent to the original PDE problems for smooth solutions. Then, we study the existence of solutions to the variational problems. To this end, we first study the existence of solutions to auxiliary problems including one parameter for approximation and two or three parameters concerned with the norms of velocity and temperature. Then we determine the parameters concerned with the norms of velocity and temperature in accordance with the data of problems, and we get the existence of solutions by passing to limits as the parameter for approximation goes to zero. For the problem of the case of static pressure it is proved that if the body force and boundary data for fluid are small enough and buoyancy effect and energy dissipation effect due to expansion are small enough, then there exists a solution. However, for the problem of the case of total pressure it is proved that if buoyancy effect and energy dissipation effect due to expansion are small enough, then there exists a solution.

9.1 Problems and Assumptions The system studied in this chapter is as follows:   ⎧ ⎪ ⎨ − 2∇ · μ(θ)E (v) + (v · ∇)v + ∇ p = (1 − α0 θ) f, div v = 0, ⎪ ⎩ − ∇ · (κ(θ)∇θ) + v · ∇(γ (θ)θ) − α2 μ(θ)E (v) : E (v) = α1 θ f · v + g. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 T. Kim and D. Cao, Equations of Motion for Incompressible Viscous Fluids, Advances in Mathematical Fluid Mechanics, https://doi.org/10.1007/978-3-030-78659-5_9

(9.1)

285

286

9 The Steady Equations for Heat-Conducting Fluids

Here α1 is parameter for dissipation of energy due to expansion, α2 is a nonnegative real number and others are the same as in Chap. 7. The term μ(θ )E(v) : E(v) represents the dissipation of energy due to viscosity (the Joule effect). Let Ω be as in Chap. 7. (See Remark 6.1.) For temperature we consider the following boundary conditions (1) θ |Γ D = θ D |Γ D , θ D − a given function on Ω,   ∂θ + β(x)θ Γ R = g R (x), β(x), g R (x) − given functions on Γ R . (2) κ(θ ) ∂n (9.2) As before, Problems I and II are distinguished according to boundary conditions for fluid. Problem I is the one with the boundary conditions (7.3) (the case of static pressure) and Problem II is the one with boundary conditions (7.4) (the case of total pressure). We use the following assumption. Assumption 9.1 We assume the followings. (1) Γ2 j , Γ3 j and Γ7 j are convex. Γ1 = ∅, Γ D = ∅ and   Γ R ⊂ ∪i=1,3,5,8 Γi .

(9.3)

(2) (3) of Assumption 5.1 holds. (3) For the functions of (9.1) f ∈ Lt (Ω), t > 3, g ∈ L 6/5 (Ω) and μ ∈ C(R), 0 < μ0 ≤ μ(ξ ) ≤ μ1 < ∞ ∀ξ ∈ R; κ ∈ C(R), 0 < κ0 ≤ κ(ξ ) ≤ κ1 < ∞ ∀ξ ∈ R;

(9.4)

γ ∈ C(R), |γ (ξ )| ≤ γ0 ∀ξ ∈ R. (4) For the functions of (9.2), (7.3) and (7.4), θ D ∈ W 1,2 (Ω), ∇θ D ∈ L ∞ (Ω), θ D ≥ 0,

g R ∈ L 4/3 (Γ R );

β0 ≥ β(x) ≥ 0, β0 − a constant, β(x) − measurable; φi ∈ H − 2 (Γi ), i = 2, 4, 7, φi ∈ H− 2 (Γi ), i = 3, 5, 6; the matrix α is positive, α jk ∈ L ∞ (Γ5 ). 1

1

(9.5)

Remark 9.1 For (2) of Assumption 9.1 we refer to Remark 7.1.

9.2 Variational Formulations for Problems In this section for every problem above we first obtain a variational formulation which consists of six formulas with six unknown functions. Then, for every problem we get another variational formulation equivalent to the variational formulation obtained,

9.2 Variational Formulations for Problems

287

which consists of one variational inequality for velocity and a variational equation for temperature. Let V, K (Ω) be the same as in (5.5) and 1, p

WΓ D (Ω) = {y ∈ W 1, p (Ω) : y|Γ D = 0}. (Ω). We will use inner products in (7.8) for V and WΓ1,2 D

9.2.1 Variational Formulation: The Case of Static Pressure By (7.9)–(7.12), we can see that smooth solutions (v, p, θ ) of problem (9.1), (9.2), (7.3) satisfy the following (cf. Remark 3.9). ⎧ 2(μ(θ)E(v), E(u)) + (v · ∇)v, u + 2(μ(θ)k(x)v, u)Γ2 ⎪ ⎪ ⎪ ⎪ ⎪ + 2(μ(θ)S v, ˜ u) ˜ Γ3 + 2(α(x)v, u)Γ5 + (μ(θ)k(x)v, u)Γ7 ⎪ ⎪ ⎪ ⎪ ⎪ − 2(μ(θ)εnτ (v), u)Γ8 + ( p − 2μ(θ )εnn (v), u n )Γ9 ∪Γ10 ∪Γ11 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ = (1 − α0 θ ) f, u + φi , u n Γi + φi , u Γi ∀u ∈ V, ⎪ ⎪ ⎪ ⎪ ⎪ i=2,4,7 i=3,5,6 ⎪ ⎪ ⎪ ⎪ (κ(θ)∇θ, ∇ϕ) − (γ (θ)θv, ∇ϕ) − (α μ(θ )|E(v)|2 , ϕ) + (βθ, ϕ) − (α θ f · v, ϕ) ⎨ ΓR 2 1 ⎪ = g R , ϕ Γ R + g, ϕ ∀ϕ ∈ WΓ1,∞ (Ω), ⎪ ⎪ D ⎪ ⎪ ⎪ ⎪ (θ, v)| ≤ g , σ (θ, v) · v + g |v | = 0 on Γ8 , |σ ⎪ τ τ τ τ τ τ ⎪ ⎪ ⎪ ⎪ |σ (θ, v, p)| ≤ g , σ (θ, v, p)v + g |v ⎪ n n n n n n | = 0 on Γ9 , ⎪ ⎪ ⎪ ⎪ ⎪ σn (θ, v, p) + g+n ≥ 0, (σn (θ, v, p) + g+n )vn = 0 on Γ10 , ⎪ ⎪ ⎪ ⎪ ⎪ σ ⎪ n (θ, v, p) − g−n ≤ 0, (σn (θ, v, p) − g−n )vn = 0 on Γ11 , ⎪ ⎩ θ |Γ D = θ D |Γ D on Γ D .

(9.6)

Define a0 (θ ; ·, ·), a1 (·, ·, ·) and f 1 ∈ V∗ by a0 (θ ; w, u) = 2(μ(θ )E(w), E(u)) + 2(μ(θ )k(x)w, u)Γ2 + 2(μ(θ )S w, ˜ u) ˜ Γ3 + 2(α(x)w, u)Γ5 + (μ(θ )k(x)w, u)Γ7 ∀w, u ∈ V, ∀θ ∈ W 1,2 (Ω), a1 (v, w, u) = (v · ∇)w, u ∀v, w, u ∈ V, φi , u n Γi + φi , u Γi ∀u ∈ V. f 1 , u = i=2,4,7

i=3,5,6

˜ ·, ·) and f 2 ∈ (WΓ1,2 (Ω))∗ by Define b0 (θ; D

(9.7)

288

9 The Steady Equations for Heat-Conducting Fluids

˜ θ, ϕ) = (κ(θ)∇θ, ˜ b0 (θ; ∇ϕ) + (β(x)θ, ϕ)Γ R ∀θ˜ , θ ∈ W 1,2 (Ω), ϕ ∈ WΓ1,2 (Ω), D (Ω). f 2 , ϕ = g R , ϕ Γ R + g, ϕ ∀ϕ ∈ WΓ1,2 D

(9.8)

Then, taking into account στ (θ, v) = 2μ(θ )εnτ (v), σn (θ, v, p) = − p + 2μ(θ )εnn (v) and (9.6), we introduce the following variational formulation for problem (9.1), (9.2), (7.3). Problem I-VE. Find (v, θ, στ , σn , σ+n , σ−n ) ∈ K (Ω) ×





1,r (Ω) × L2 (Γ ) τ 8 1≤r < 32 W

× L 2 (Γ9 ) × H −1/2 (Γ10 ) × H −1/2 (Γ11 ) such that θ |Γ D = θ D |Γ D and

⎧ a0 (θ ; v, u) + a1 (v, v, u) − (στ , u τ )Γ8 − (σn , u n )Γ9 ⎪ ⎪ ⎪ ⎪ ⎪ − σ+n , u n Γ10 − σ−n , u n Γ11 − f − α0 θ f, u = f 1 , u ∀u ∈ V, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ b0 (θ ; θ, ϕ) − γ (θ )θ v, ∇ϕ − α2 μ(θ )|E(v)|2 , ϕ − α1 θ f · v, ϕ = f 2 , ϕ ⎪ ⎪ ⎪ ⎨ ∀ϕ ∈ W 1,∞ (Ω), ⎪ ⎪ |στ | ≤ gτ , ⎪ ⎪ ⎪ ⎪ ⎪ |σn | ≤ gn , ⎪ ⎪ ⎪ ⎪ ⎪ σ+n + g+n ⎪ ⎪ ⎩ σ−n − g−n

ΓD

στ · vτ + gτ |vτ | = 0 on Γ8 , σn vn + gn |vn | = 0 on Γ9 , ≥ 0, σ+n + g+n , vn Γ10 = 0 on Γ10 , ≤ 0, σ−n − g−n , vn Γ11 = 0 on Γ11 ,

(9.9) where L2τ (Γ8 ) is the subspace of L2 (Γ8 ) consisting of functions such that (u, n)L2 (Γ8 ) = 0.

Remark 9.2 As in Definition 7.1, except that the third equality of (9.1) holds in L 1 (Ω), define a solution to the problem (9.1), (9.2), (7.3). Then, as in Theorem 7.1, Problem I-VE is equivalent to the problem (9.1), (9.2), (7.3), which shows that Problem I-VE is well-defined. We will find another variational formulation consisting of a variational inequality and a variational equation, which is equivalent to Problem I-VE. Let (v, θ, στ , σn , σ+n , σ−n ) be a solution of Problem I-VE. Subtracting the first formula of (9.9) with u = v from the first formula of (9.9), we get a0 (θ ; v, u − v) + a1 (v, v, u − v) − (στ , u τ − vτ )Γ8 − (σn , u n − vn )Γ9 − σ+n , u n − vn Γ10 − σ−n , u n − vn Γ11 − f − α0 θ f, u − v = f 1 , u − v ∀u ∈ V.

(9.10)

9.2 Variational Formulations for Problems

289

Let Φ : V → R be the functional defined by (5.22). Then, Φ is proper, convex lower weak semi-continuous and nonnegative. By Theorem 5.2, under Assumption 9.1 for a fixed θ , the problem ⎧ a0 (θ ; v, u) + a1 (v, v, u) − (στ , u τ )Γ8 − (σn , u n )Γ9 ⎪ ⎪ ⎪ ⎪ ⎪ − σ+n , u n Γ10 − σ−n , u n Γ11 − f − α0 θ f, u = f 1 , u , ⎪ ⎪ ⎪ ⎨ |σ | ≤ g , σ · v + g |v | = 0 on Γ , τ τ τ τ τ τ 8 ⎪ |σ | ≤ g , σ v + g |v | = 0 on Γ , n n n n n n 9 ⎪ ⎪ ⎪ ⎪ ⎪ σ + g ≥ 0, σ + g , v

= 0 on Γ10 , +n +n +n +n n Γ ⎪ 10 ⎪ ⎩ σ−n − g−n ≤ 0, σ−n − g−n , vn Γ11 = 0 on Γ11

(9.11)

is equivalent to the following variational inequality. Find v ∈ V such that a0 (θ ; v, u − v) + a1 (v, v, u − v) + Φ(u) − Φ(v) − f − α0 θ f, u − v ≥ f 1 , u − v ∀u ∈ V.

(9.12)

Therefore, we have the following variational formulation equivalent to Problem IVE, which consists of a variational inequality for velocity and a variational equation for temperature.

 1,r Problem I-VI. Find (v, θ ) ∈ V × (Ω) such that θ |Γ D = θ D |Γ D and 3 W 1≤r < 2

⎧ ⎪ ⎪ a0 (θ ; v, u − v) + a1 (v, v, u − v) + Φ(u) − Φ(v) − f − α0 θ f, u − v ≥ f 1 , u − v ⎪ ⎪ ⎪ ∀u ∈ V, ⎨ ⎪ b0 (θ ; θ, ϕ) − γ (θ )θ v, ∇ϕ − α2 μ(θ )|E(u)|2 , ϕ − α1 θ f · v, ϕ = f 2 , ϕ ⎪ ⎪ ⎪ ⎪ ⎩ ∀ϕ ∈ WΓ1,∞ (Ω).

(9.13)

D

9.2.2 Variational Formulation: The Case of Total Pressure Since (v · ∇)v = rot v × v + 21 grad|v|2 , by (7.9)–(7.12) we can verify that smooth solutions (v, p, θ ) of problem (9.1), (9.2), (7.4) satisfy the following.

290

9 The Steady Equations for Heat-Conducting Fluids

⎧ 2(μ(θ)E(v), E(u)) + rotv × v, u + 2(μ(θ)k(x)v, u)Γ2 ⎪ ⎪ ⎪ ⎪ ⎪ + 2(μ(θ)S v, ˜ u) ˜ Γ3 + 2(α(x)v, u)Γ5 + (μ(θ)k(x)v, u)Γ7 ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ − 2(μ(θ)εnτ (v), u)Γ8 + ( p + |v|2 − 2μ(θ )εnn (v), u n )Γ9 ∪Γ10 ∪Γ11 ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ ⎪ = (1 − α θ ) f, u + φ , u

Γi + φi , u Γi ∀u ∈ V, ⎪ n 0 i ⎪ ⎪ ⎪ i=2,4,7 i=3,5,6 ⎪ ⎪ ⎪ ⎪ ⎨ (κ(θ)∇θ, ∇ϕ) − (γ (θ)θv, ∇ϕ) − (α μ(θ )|E(v)|2 , ϕ) + (βθ, ϕ) − (α θ f · v, ϕ) 2

⎪ ⎪ = g R , ϕ Γ R + g, ϕ ∀ϕ ∈ WΓ1,∞ (Ω), ⎪ ⎪ D ⎪ ⎪ ⎪ t (θ, v)| ≤ g , σ t (θ, v) · v + g |v | = 0 on Γ , ⎪ |σ ⎪ τ τ τ τ 8 τ τ ⎪ ⎪ ⎪ t t ⎪ ⎪ |σn (θ, v, p)| ≤ gn , σn (θ, v, p)vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪ ⎪ ⎪ σ t (θ, v, p) + g+n ≥ 0, (σ t (θ, v, p) + g+n )vn = 0 on Γ10 , ⎪ n n ⎪ ⎪ ⎪ ⎪ ⎪ σnt (θ, v, p) − g−n ≤ 0, (σnt (θ, v, p) − g−n )vn = 0 on Γ11 , ⎪ ⎪ ⎪ ⎩ θ | Γ D = θ D |Γ D .

ΓR

1

(9.14)

Let us define a2 (·, ·, ·) by a2 (v, u, w) = rot v × u, w ∀v, u, w ∈ V.

(9.15)

Then, taking into account (9.14) and 1 στt (θ, v) = 2μ(θ )εnτ (v), σnt (θ, v, p) = −( p + |v|2 ) + 2μ(θ )εnn (v), 2 we get the following variational formulation for problem (9.1), (9.2), (7.4).

 t t 1,r , σ−n ) ∈ K (Ω) × (Ω) Problem II-VE. Find (v, θ, στt , σnt , σ+n 3 W 1≤r < 2

× L2τ (Γ8 ) × L 2 (Γ9 ) × H −1/2 (Γ10 ) × H −1/2 (Γ11 ) such that θ |Γ D = θ D |Γ D and 

t ⎧ a0 (θ ; v, u) + a2 (v, v, u) − (στt , u τ )Γ8 − (σnt , u n )Γ9 − σ+n , u n Γ10 ⎪ ⎪ ⎪ 

t ⎪ ⎪ − σ−n , u n Γ11 − f − α0 θ f, u = f 1 , u ∀u ∈ V, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ b0 (θ ; θ, ϕ) − γ (θ )θ v, ∇ϕ − α2 μ(θ )|E(v)|2 , ϕ − α1 θ f · v, ϕ ⎪ ⎪ ⎪ ⎪ ⎨ = f 2 , ϕ ∀ϕ ∈ WΓ1,∞ (Ω), D t t ⎪ |στ | ≤ gτ , στ · vτ + gτ |vτ | = 0 on Γ8 , ⎪ ⎪ ⎪ ⎪ ⎪ |σnt | ≤ gn , σnt vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪ 

t ⎪ t ⎪ ⎪ σ+n + g+n ≥ 0, σ+n + g+n , vn Γ10 = 0 on Γ10 , ⎪ ⎪ ⎪ 

t ⎩ t σ−n − g−n ≤ 0, σ−n − g−n , vn Γ11 = 0 on Γ11 .

(9.16)

Remark 9.3 Similarly to Remark 9.2, Problem II-VE is equivalent to problem (9.1), (9.2), (7.4).

9.2 Variational Formulations for Problems

291

Then, using Theorem 5.4, in the same way as Problem I-VI we get Problem II-VI equivalent to Problem II-VE which consists of a variational inequality for velocity and a variational equation for temperature.

 1,r (Ω) such that θ |Γ D = θ D |Γ D and Problem II-VI. Find (v, θ ) ∈ V × 3 W 1≤r < 2

⎧ ⎪ ⎪ a0 (θ ; v, u − v) + a2 (v, v, u − v) + Φ(u) − Φ(v) − f − α0 θ f, u − v ⎪ ⎪ ⎨ ≥ f 1 , u − v ∀u ∈ V, ⎪ b0 (θ ; θ, ϕ) − γ (θ )θ v, ∇ϕ − α2 μ(θ )|E(v)|2 , ϕ − α1 θ f · v, ϕ ⎪ ⎪ ⎪ ⎩ = f 2 , ϕ ∀ϕ ∈ WΓ1,∞ (Ω), D

(9.17)

where a2 (·, ·, ·) is the one defined in (9.15) and Φ is defined by (5.22).

9.3 Existence and Uniqueness of Solutions: The Case of Static Pressure In this section we prove the main result for the case of static pressure. Theorem 9.1 Let Assumption 9.1 hold and assume that (1) f, φi , i = 2, · · · , 7, are small enough in the spaces in (3), (4) of Assumption 9.1 (see (9.50)), (2) max{|α0 |, |α1 |} is small enough in accordance with f, φi (i = 2, · · · , 7), g, g R , θ D (see (9.97)). Then, there exists a solution (v, θ ) to Problem I-VI such that vV ≤

μ0 , K



μ 0  f Lt + g R  L 4/3 (Γ R ) + g L 6/5 (Ω) , θ − W 1,2 (Ω) ≤ c K    |∇θ |r d x ≤ cK σr/(1−σ ) 1 + θ D 2W 1,2 ∀r (1 < r < 3/2), Ω

where θ − (x) = min{θ (x), 0}, K is the constant in (9.18) below, K σ is the one in . (9.90) and σ = 3−2r 3−r

9.3.1 Existence of a Solution to an Auxiliary Problem Since

  |a1 (v, v, u)| = | (v · ∇)v, u | ≤ K v2V uV ∀v, u ∈ V,

(9.18)

292

9 The Steady Equations for Heat-Conducting Fluids

define a 1 (v) ∈ V∗ by a 1 (v), u = a1 (v, v, u) ∀v, u ∈ V. Define γε (t) by γε (t) :=

γ (t)t t ∈ R, ε > 0. (1 + ε|γ (t)|)(1 + ε|t|)

Then, |γε (t)| ≤

1 , |γε (t)| ≤ |γ (t)||t| ≤ γ0 |t|, γε (t) → γ (t)t as ε → 0. ε2

(9.19)

For every ε > 0, let Φε be the Moreau-Yosida approximation of Φ and ∇Φε be Fréchet derivative of Φε . Similarly to Sect. 7.2.1, we first consider an auxiliary problem involving two parameters δ, ζ concerned with the norm of velocity (which is useful when there is fluid flux across a portion of boundary), one parameter λ concerned with the norm of temperature (which is useful to deal with buoyancy effect and energy dissipation effect due to expansion) and a parameter ε for approximation. , 6). Find (v, θ ) ∈ V × Problem I-VIA. Let δ > 0, ζ > 0, λ > 0, ε > 0 and q ∈ ( 12 5 1,2 1,2 W (Ω) such that η = θ − θ D ∈ WΓ D (Ω) and ⎧

 δ ⎪ a0 (θ ; v, u) + a1 (v, v, u) + ∇Φε (v), u ⎪ ⎪ ⎪ max{δ, a 1 (v)V∗ } ⎪ ⎪ 

 ⎪ ⎪ α0 λ ⎪ ⎪ − 1 − θ f , u = f 1 , u ∀u ∈ V, ⎪ ε ⎨ max{λ, θ 2L q }   (9.20) ⎪ ζ |E(v)|2 ⎪ ⎪ b γ (θ ; θ, ϕ) − (θ )v, ∇ϕ − α μ(θ ) , ϕ ε 2 ⎪ 0 ⎪ max{ζ, vV } 1 + ε|E(v)|2 ⎪ ⎪ ⎪   ⎪ ⎪ α1 λ ⎪ ⎩ θ f ε · v, ϕ = f 2 , ϕ ∀ϕ ∈ WΓ1,2 (Ω), − D max{λ, θ 2L q } where f ε ∈ L∞ (Ω) is such that  f − f ε Lt ≤ ε. Theorem 9.2 There exists a solution (vε , θε ) ∈ V × W 1,2 (Ω) to Problem I-VIA. Proof Let r be such that 2 1 + = 1. q r Since q ∈ ( 12 , 6), we see that 5 ator A : H → H ∗ by

3 2

(9.21)

< r < 6. Let H = V × WΓ1,2 (Ω). Define an operD

9.3 Existence and Uniqueness of Solutions: The Case of Static Pressure

 δ a1 (v, v, u) + ∇Φε (v), u max{δ, a 1 (v)V∗ }  

α0 λ (η + θ D ) f ε , u + b0 (η + θ D ; η + θ D , φ) − 1− 2 max{λ, η + θ D  L q }   ζ |E(v)|2 − ,φ γε (η + θ D )v, ∇φ − α2 μ(η + θ D ) max{ζ, vV } 1 + ε|E(v)|2   α1 λ (η + θ D ) f ε · v, φ ∀(v, η), (u, φ) ∈ H . − 2 max{λ, η + θ D  L q }

293

 A (v, η), (u, φ) = a0 (η + θ D ; v, u) +

(9.22)

Let us show that the operator A is well-defined. By definition of Φε it follows that Φε (0V ) = 0 and ∇Φε (0V ) = 0. Since ∇Φε is Lipschitz continuous with the constant ε−1 (see Remark 1.16),      ∇Φε (v), u  =  ∇Φε (v) − ∇Φε (0V ), u  ≤ ε−1 vV uV .

(9.23)

By (9.18) we have   

 δ  a1 (v, v, u) ≤ δuV . max{δ, a 1 (v)V∗ }

(9.24)

On the other hand, it is easily checked that by (9.19)   

 cζ ζ  γε (η + θ D )v, ∇φ  ≤ 2 φW 1,2 . max{ζ, vV } ε

(9.25)

The trivial inequality ab ≤ max{a 2 , b2 } implies that √

λ η + θ D  L q ≤ 1. max{λ, η + θ D 2L q }

(9.26)

Taking into account (9.21) and (9.26), we have  

  

α0 λ max{λ, η + θ D 2L q } α1 λ max{λ, η + θ D 2L q } ≤

 (η + θ D ) f ε , u  ≤   (η + θ D ) f ε · v, φ 

α1 λ max{λ, η + θ D 2L q }

α0 λ η + θ D  L q  f ε Lq uLr max{λ, η + θ D 2L q } √ ≤ cα0 λ f ε Lq uLr ,

√ η + θ D  L q  f ε L∞ vLr φ L q ≤ cα1 λ f ε L∞ vLr φ L q .

(9.27)

Estimation of other terms is easy, and so operator A is well defined. Then, the existence of a solution to Problem I-VIA is equivalent to the one of a solution to

294

9 The Steady Equations for Heat-Conducting Fluids

A (v, η) = F , F =

  f1 . f2

By applying Theorem 1.45, we will prove the existence of a solution to the equation above. To this end, we verify that A satisfies assumptions of Theorem 1.45. (i) Let us first prove that A is coercive, i.e.,

 1 A (v, η), (v, η) → ∞ as (v, η)H → ∞. (v, η)H Since Γ2 j , Γ3 j , Γ7 j are convex and the matrix α is positive, it follows from (9.7) (see Lemma 3.1) that (9.28) a0 (η + θ D ; v, v) ≥ 2μ0 v2V . Taking into account (9.24), (9.25), (9.28) and the first formula of (9.27), we have

 A (v, η), (v, η) = a0 (η + θ D ; v, v) +

δ a1 (v, v, v) max{δ, a 1 (v)V∗ } 



 α0 λ + ∇Φε (v), v − 1 − (η + θ D ) f ε , v 2 max{λ, η + θ D  L q }

ζ γε (η + θ D )v, ∇η max{ζ, vV }     α1 λ |E(v)|2 − α2 μ(η + θ D ) (η + θ D ) f ε · v, η ,η − 1 + ε|E(v)|2 max{λ, η + θ D 2L q }     3 1 ≥ min{2μ0 , κ0 } v2V + η2 1,2 − δv − c  f ε 2Lq + θ D 2 2 + 2 + ∇Φε (v), v L WΓ 4 ε D   α1 λ cζ (η + θ D ) f ε · v, η ∀(v, η) ∈ H . − 2 η 1,2 − 2 W ε max{λ, η + θ D  L q } ΓD + b0 (η + θ D ; η, η) + b0 (η + θ D ; θ D , η) −

(9.29)

Since the operator ∇Φε is monotone and ∇Φε (0V ) = 0, we have

  ∇Φε (v), v = ∇Φε (v) − ∇Φε (0V ), v − 0V ≥ 0.

(9.30)

Since (η + θ D ) f · v, η = (η + θ D ) f · v, (η + θ D ) − (η + θ D ) f · v, θ D , by (9.21) and (9.26) we have   

α1 λ

 

(η + θ D ) f ε · v, η  max{λ, η + θ D 2L q } λη + θ D 2L q λη + θ D  L q ≤c  f ε L∞ vLr + c  f ε L∞ vLr θ D  L q max{λ, η + θ D 2L q } max{λ, η + θ D 2L q } √ ≤ cλ f ε L∞ vLr + c λ f ε L∞ vLr θ D  L q .

(9.31)

9.3 Existence and Uniqueness of Solutions: The Case of Static Pressure

295

By virtue of (9.29)–(9.31), we conclude that

 1 cζ A (v, η), (v, η) ≥ min{2μ0 , κ0 } v2V + η2 1,2 − δv − 2 η 1,2 WΓ WΓ 4 ε D D

√ 1 2 − cλ f ε L∞ vV − c λ f ε L∞ vV θ D  L q − c  f ε Lt + θ D 2 2 + 2 L ε ∀(v, η) ∈ H ,

which implies coercivity of A . (ii) Taking into account (9.23)–(9.27), we have from (9.22) A (v, η)H

∗

=

sup

(u,φ)H =1

 A (v, η), (u, φ)

√ cζ 1 ≤ c vV + δ + 2 + vV +  f ε L2 + λ f ε Lq + η + θ D  1,2 WΓ ε ε D √ 1 + vL2 + + λ f ε L∞ vV ∀(v, η), (u, φ) ∈ H , ε

(9.32)

which shows that A maps bounded sets of H into bounded sets of H ∗ . (iii) Let {(vk , ηk )} be a sequence such that (vk , ηk )  (v, η) in H , lim sup A (vk , ηk ), (vk , ηk ) − (v, η) ≤ 0. k→∞

By taking a subsequence and denoting with same subindex if necessary, we may assume vk → v in Ls (Ω) (1 ≤ s < 6) and a.e. in Ω, ηk → η in L s (Ω) (1 ≤ s < 6) and a.e. in Ω

as k → ∞.

(9.33)

Since a0 (ηk + θ D ; vk − v, vk − v) = a0 (ηk + θ D ; vk , vk − v) − a0 (ηk + θ D ; v, vk − v), b0 (ηk + θ D ; ηk − η, ηk − η) = b0 (ηk + θ D ; ηk + θ D , ηk − η) − b0 (ηk + θ D ; η + θ D , ηk − η), by (9.22) we have

296

9 The Steady Equations for Heat-Conducting Fluids

  min{μ0 , κ0 } vk − v2V + ηk − ηWΓ1,2 ≤ A (vk , ηk ), (vk , ηk ) − (v, η) D

− a0 (ηk + θ D ; v, vk − v) − b0 (ηk + θ D ; η + θ D , ηk − η)

 δ − a1 (vk , vk , vk − v) − ∇Φε (vk ), vk − v ∗ max{δ, a 1 (vk )V } 

 α0 λ + 1− (ηk + θ D ) f ε , vk − v 2 max{λ, ηk + θ D  L q } (9.34) ζ + γε (ηk + θ D )vk , ∇(ηk − η) max{ζ, vk V }   |E(vk )|2 , η − η + α2 μ(θ ) k 1 + ε|E(vk )|2   α1 λ (η + θ ) f · v , η − η . + k D ε k k max{λ, ηk + θ D 2L q } Now let us estimate all terms except the first one on the right hand side of (9.34). By Corollary 1.1 it follows that as k → ∞ a0 (ηk + θ D ; v, vk − v) = 2(μ(ηk + θ D )E(v), E(vk − v)) ˜ v˜k − v) ˜ Γ3 + 2(μ(ηk + θ D )k(x)v, vk − v)Γ2 + 2(μ(ηk + θ D )S v, + 2(α(x)v, vk − v)Γ5 + (μ(ηk + θ D )k(x)v, vk − v)Γ7 → 0,   b0 (ηk + θ D ; η + θ D , ηk − η) = κ(ηk + θ D )∇(η + θ D ), ∇(ηk − η)   + β(x)(η + θ D ), (ηk − η) Γ → 0. R

(9.35)

Also, as k → ∞   

 δ  a1 (vk , vk , vk − v) max{δ, a 1 (vk )V∗ } δ ≤ vk L4 ∇vk L2 vk − vL4 → 0. max{δ, a 1 (vk )V∗ }

(9.36)

Since ∇Φε is monotone, as k → ∞



  − ∇Φε (vk ), vk − v = − ∇Φε (vk ) − ∇Φε (v), vk − v − ∇Φε (v), vk − v

 ≤ − ∇Φε (v), vk − v → 0. (9.37) By (9.26) and (9.33), the followings hold.

9.3 Existence and Uniqueness of Solutions: The Case of Static Pressure        

α0 λ max{λ, ηk + θ D 2L q } α1 λ

297

 √  (ηk + θ D ) f ε , vk − v  ≤ c λ f ε L∞ vk − vL3 → 0,  √  (ηk + θ D ) f ε · vk , ηk − η  ≤ c λ f ε L∞ vk L6 ηk − η L r → 0,

max{λ, ηk + θ D 2L q }  ζ γε (ηk + θ D )vk , ∇(ηk − η)  max{ζ, vk V }   ≤ | γε (ηk + θ D ) − γε (η + θ D ) vk , ∇(ηk − η) |

+ | γε (η + θ D )(vk − v), ∇(ηk − η) | + | γε (η + θ D )v, ∇(ηk − η) | ≤ cγε (ηk + θ D ) − γε (η + θ D ) L 3 vk V ηk − η

WΓ1,2 D

+ cγε (η + θ D ) L 6 (vk − v)L3 ηk − η

WΓ1,2 D

+ γε (η + θ D )v, ∇(ηk − η) → 0

(9.38) as k → ∞, where the fact that by Lemma 1.3 γε (ηk + θ D ) → γε (η + θ D ) in L 3 (Ω) as k → ∞ was used. It is easy to prove convergence to zero of other terms in the right hand side of (9.34). Thus, by (9.33)–(9.38) we conclude that   lim sup min{μ0 , κ0 }| vk − v2V + ηk − ηWΓ1,2 D

k→∞

≤ lim sup A (vk , ηk ), (vk , ηk ) − (v, η) ≤ 0, k→∞

which implies

(vk , ηk ) → (v, η) in H as k → ∞, vk → v, ηk → η a.e. in Ω as k → ∞.

(9.39)

By the definition of A , for (u, φ) ∈ H A (vk , ηk ), (vk , ηk ) − (u, φ) δ a1 (vk , vk , vk − u) max{δ, a 1 (vk )V∗ } 

 

α0 λ (ηk + θ D ) f, vk − u + ∇Φε (vk ), vk − u − 1 − 2 max{λ, ηk + θ D  L q }   ζ + b0 (ηk + θ D ; ηk + θ D , ηk − φ) − γ (ηk + θ D )vk , ∇(ηk − φ) max{ζ, vk V }     α1 λ |E(vk )|2 (η , η − φ − + θ ) f · v , η − φ . − α2 μ(θk ) ε k k D k k 1 + ε|E(vk )|2 max{λ, ηk + θ D 2L q } = a0 (ηk + θ D ; vk , vk − u) +

(9.40)

Taking into account (9.39), by Corollary 1.2 we have lim inf a0 (ηk + θ D ; vk , vk ) ≥ a0 (η + θ D ; v, v), k→∞

lim inf b0 (ηk + θ D ; ηk , ηk ) ≥ b0 (η + θ D ; η, η). k→∞

(9.41)

298

9 The Steady Equations for Heat-Conducting Fluids

α0 λ The sequence { max{λ,ηα0 λ+θ 2 } (ηk + θ D )} converges a.e. in Ω to max{λ,η+θ (η + 2 k D Lq D Lq }   √ α λ 0 θ D ) and  (ηk + θ D ) q ≤ α0 λ, and so this sequence weakly con2 max{λ,ηk +θ D  L q }

L

verges in L q (see Lemma 1.2). Then by virtue of this fact and the first formula of (9.38), we have  lim

k→∞

 α0 λ (η + θ ) f , v − u k D ε k max{λ, ηk + θ D 2L q }   α0 λ = lim (η + θ ) f , v − v k D ε k k→∞ max{λ, ηk + θ D 2 q } L   α0 λ + lim (η + θ ) f , v − u k D ε k→∞ max{λ, ηk + θ D 2 q } L   α0 λ = (η + θ ) f , v − u , D ε max{λ, η + θ D 2L q } 

where the fact that owing to (9.21) f ε · (v − u) ∈ L 3 ⊂ L q , q1 + Using the second formula in (9.38), in the same way we get  lim

k→∞

1 q

(9.42)

= 1, was used.

 α1 λ (η + θ ) f · v , η − φ k D ε k k max{λ, ηk + θ D 2L q }   α1 λ = (η + θ ) f · v, η − φ . D ε max{λ, η + θ D 2L q }

(9.43)

It is easy to prove convergence of other terms on the right hand side of (9.40). Thus, by (9.41)–(9.43) we have existence of a subsequence {(vk , ηk )} such that lim inf A (vk , ηk ), (vk , ηk ) − (u, φ) ≥ A (v, η), (v, η) − (u, φ) . k→∞

Therefore, we can apply Theorem 1.45 to obtain the conclusion.



9.3.2 A Priori Estimates of Solutions to the Auxiliary Problem Let us choose q0 such that 12 1 1 1 < q0 < 3. + + ≤ 1, q0 t 3 5

(9.44)

Since t > 3 (see (3) of Assumption 9.1), such a choice is possible. Then, q0 satisfies the condition for q in Problem I-VIA.

9.3 Existence and Uniqueness of Solutions: The Case of Static Pressure

299

√ Lemma 9.1 If v ∈ V and |α1 | λ ≤ 1, then for all θε satisfying the second formula (Ω) the following estimate holds. of (9.20) and the condition θε − θ D ∈ WΓ1,2 D

θε− WΓ1,2 (Ω) ≤ c  f ε Lt vV + g R  L 4/3 (Γ R ) + g L 6/5 (Ω) ,

(9.45)

D

where constant c is independent of ε. Proof Since θε |Γ D = θ D |Γ D ≥ 0, a function ϕ(x) = θε− (x) ≡ min{θε (x), 0} is admissible in the second formula of (9.20), and we have ζ γε (θε )v, ∇θε− max{ζ, vV }     α1 λ |E(v)|2 − α2 μ(θ ) θε f ε · v, θε− = f 2 , θε− . , θε− − 2 2 1 + ε|E(v)| max{λ, θε  L q0 }

(κ(θε )∇θε ,∇θε− ) + (β(x)θε , θε− )Γ R −

Let us prove

γε (θε )v, ∇θε− = 0.

To this end, define

 (t) :=

t

(9.46)

(9.47)

γε (s) ds, t ∈ R.

0

Then,  ∈ C 1 (R) and ∇(θ ) = γε (θ )∇θ, (θ ) ∈ W 1,2 (Ω) ∀θ ∈ W 1,2 (Ω), (θ )|Γ D = 0 ∀θ ∈ WΓ1,2 (Ω). D

(9.48)

Since ∇θε− = 0 on Ω + = {x : θ (x) ≥ 0} and v · n|Γ R = 0, by (9.48) we have γε (θ )v, ∇θε−

 =

Ω

γε (θε− )v

·

∇θε−

 dx =

Ω

v · ∇(θε− ) = 0,

which means (9.47). Also,  |E(v)|2 − ≥ 0, , θ 1 + ε|E(v)|2 ε   α1 λ  −  θ f · v, θ   ≤ c f ε Lt vV θε− WΓ1,2 ε ε ε D max{λ, θε 2L q0 } κ0 − 2 ≤ θε W 1,2 + c f ε 2Lt v2V , ΓD 4    f 2 , θ −  ≤ κ0 θ − 2 1 + c f 2 2 1,2 ∗ , ε (WΓ ) 4 ε H D − (β(x)θε , θε )Γ R ≥ 0,  − α2 μ(θ )

(9.49)

300

9 The Steady Equations for Heat-Conducting Fluids

where to get the second inequality (9.26) and (9.44) were used. By (9.46)–(9.49) we have 2c

 f ε 2Lt v2V +  f 2 2(W 1,2 )∗ , θε− 2W 1,2 ≤ ΓD ΓD κ0 

which implies (9.45). √ Lemma 9.2 If |α0 | λ ≤ 1 and ( f Lt +  f 1 V∗ )
0 concerned with the inequality above K x 2 − 2μ0 x + a. If 0 ≤ K a ≤ μ20 , then there exists a nonnegative minimum root x1 (≤ maximum root x2 . Thus, we can know that if ( f ε Lt +  f 1 V∗ ) ≤ then vε V ≤

μ0 ) K

and a

μ20 , K c0

μ0 or vε V ≥ x2 . K

(9.56)

On the other hand, we have from (9.54) another estimation under consideration of δ 2μ0 vε 2V ≤ a0 (θε ; vε , vε ) ≤ δvε V + c0 ( f ε Lt +  f 1 V∗ )vε V , which implies vε V ≤

 1  δ + c0 ( f ε Lt +  f 1 V∗ ) . 2μ0

In view of (9.56), we can take δ = K If ( f Lt +  f 1 V∗ )
0 (1 + (θ − d0 )+ )σ

(9.62)

belongs to WΓ1,2 (Ω) and 0 ≤ ϕ(x) ≤ 1 a.e. in Ω. D Taking ϕ of (9.62) and ζ satisfying (9.51), we have from the second equation of (9.20) 

 |∇(θ − d0 )+ |2 ∇(θ − d0 )+ d x + (βθ, ϕ) − σ γε (θ )v · dx Γ R (1 + (θ − d0 )+ )1+σ (1 + (θ − d0 )+ )1+σ Ω Ω  |E(v)|2

1 = α2 μ(θ ) 1− + )σ d x 2 (1 + (θ − d ) 1 + ε|E(v)| Ω 0   



α1 λ 1 1 d x + f2 , 1 − + θ f · v 1 − ε + σ + σ 2 (1 + (θ − d0 ) ) (1 + (θ − d0 ) ) max{λ, θ  q } Ω

σ

κ(θ )

L 0

≡ I1 + I2 + I3 .

(9.63) Note that ϕ(x) = 0 at x such that θ (x) ≤ d0 , and (β(x)θ, ϕ)Γ R ≥ 0. Let us show the third term on the left hand side of (9.63) vanishes. To this end, define

9.3 Existence and Uniqueness of Solutions: The Case of Static Pressure



t

σ (t) := 0

303

γε (s + d0 ) ds, t ≥ 0. (1 + s)1+σ

Then, σ ∈ C 1 (R) and σ ((θ − d0 )+ )|Γ D = 0, ∇σ ((θ − d0 )+ ) = γε (θ )

∇(θ − d0 )+ , (1 + (θ − d0 )+ )1+σ

(9.64)

where the fact that if θ (x) − d0 ≥ 0, then (θ (x) − d0 )+ + d0 = θ (x) was used. Taking into account the fact that vn = 0 on Γ R (see (9.3)) and the first equality of (9.64), we have   ∇(θ − d0 )+ γε (θ )v · d x = v · ∇σ ((θ − d0 )+ ) = 0. (9.65) + )1+σ (1 + (θ − d ) 0 Ω Ω It is easily seen that |I1 | ≤ cv2V .

(9.66)

√ Since max{|α0 |, |α1 |} λ ≤ 1, by (9.26) we have |I2 | ≤

√ λ θ q0  f ε Lt vL3 ≤ c f ε Lt vV . max{λ, θ 2L q0 }

(9.67)

Also, √   |I3 | ≤ c mesΩ f 2 (WΓ1,2 )∗ ≤ c g R  L 4/3 (Γ R ) + g L 6/5 (Ω) .

(9.68)

D

By (9.65)–(9.68), we have from (9.63)  σ

|∇θ |2 d x ≤ c v2V +  f ε Lt vV + g R  L 4/3 (Γ ) + g L 6/5 (Ω) . + 1+σ R {x;θ≥d0 } (1 + (θ − d0 ) )

(9.69)

Next, taking ϕ = min{θ − θ D , d0 } admissible in the second formula of (9.20), we have

304

9 The Steady Equations for Heat-Conducting Fluids 

κ0

{x;θ−θ D 0, λ > 0, ε > 0 and q ∈ ( 12 5 1,2 1,2 W (Ω) such that θ − θ D ∈ WΓ D (Ω) and ⎧ 

 

α0 λ ⎪ ⎪ a θ f (θ ; v, u) + a (v, v, u) + ∇Φ (v), u − 1 − , u ⎪ 0 2 ε ε ⎪ ⎪ max{λ, θ 2L q } ⎪ ⎪ ⎪ ⎪ ⎪ = f 1 , u ∀u ∈ V, ⎨   ζ |E(v)|2 ⎪ b0 (θ ; θ, ϕ) − γε (θ )v, ∇ϕ − α2 μ(θ ) , ϕ ⎪ ⎪ ⎪ max{ζ, vε V } 1 + ε|E(v)|2 ⎪ ⎪   ⎪ ⎪ λ α 1 ⎪ ⎪ θ f ε · v, ϕ = f 2 , ϕ ∀ϕ ∈ WΓ1,2 (Ω), − ⎩ D max{λ, θ 2L q } (9.126) where f ε ∈ L∞ (Ω) is such that  f − f ε Lt ≤ ε. Theorem 9.5 There exists a solution (vε , θε ) ∈ V × W 1,2 (Ω) to Problem II-VIA. (Ω). Define an operator A : H → H ∗ by Proof Let H = V × WΓ1,2 D



 A (v, η), (u, φ) = a0 (η + θ D ; v, u) + a2 (v, v, u) + ∇Φε (v), u  

α0 λ (η + θ D ) f ε , u + b0 (η + θ D ; η + θ D , ϕ) − 1− 2 max{λ, η + θ D  L q }   ζ |E(v)|2 − γε (η + θ D )v, ∇ϕ − α2 μ(η + θ D ) ,ϕ max{ζ, vε V } 1 + ε|E(v)|2   α1 λ (η + θ D ) f ε · v, ϕ − 2 max{λ, η + θ D  L q } ∀(v, η), (u, φ) ∈ H .

Note that instead of max{δ,aδ1 (v)V∗ } a1 (v, v, u) in Problem I-VIA, for Problem II-VIA the term a2 (v, v, u) is used.

9.4 Existence of a Solution: The Case of Total Pressure

Using

315

a2 (v, v, v) = 0, |a2 (v, v, u)| ≤ K v2V uV , |a2 (vε , vε , vε − u)| ≤ c∇vε L2 vε L4 vε − uL4 ,

respectively, in (9.29), (9.32) and (9.36), we can see that the proof of Theorem 9.2 for Problem I-VIA is valid for Problem II-VIA. Thus, we come to the asserted conclusion.  As (9.44) let us choose q0 such that 12 1 1 1 < q0 < 3. + + ≤ 1, q0 t 3 5 √ Lemma 9.4 If |α0 | λ ≤ 1, then there exists a parameter ζ such that ζ =1 max{ζ, vε V }

(9.127)

for all small ε and solutions (vε , θε ) of (9.126). √ Proof Since |α0 | λ ≤ 1, we have   

α0 λ max{λ, θε 2L q0 }

  θε f ε , u  ≤

√ c λ max{λ, θε 2L q0 }

θε  L q0  f ε Lt uL6 ≤ c f ε Lt uV∗ .

(9.128)

Putting u = vε in the first equation of (9.126), we have  

a0 (θε ; vε , vε ) + a2 (vε , vε , vε ) + ∇Φε (vε ), vε − 1 −

 α0 λ θ f , v ε ε ε max{λ, θε 2L q0 }

= f 1 , vε . (9.129) Taking into account a2 (vε , vε , vε ) = 0, (9.30) and (9.128), we have from (9.129) 2μ0 vε 2V ≤ a0 (θε ; vε , vε ) ≤



1−

 α0 λ θε f ε , v + f 1 , vε 2 max{λ, θε  L q0 }

≤ c( f ε L 65 +  f ε Lt +  f 1 V∗ )vε V , which implies vε V ≤

c ( f ε Lt +  f 1 V∗ ). 2μ0

Since f ε → f in Lt (Ω), again we may assume that  f ε Lt ≤ 1 +  f Lt for all ε. Therefore, c vε V ≤ (1 +  f Lt +  f 1 V∗ ). (9.130) 2μ0

316

9 The Steady Equations for Heat-Conducting Fluids

Putting ζ =

c (1 2μ0

+  f Lt +  f 1 V∗ ), we come to the asserted conclusion.



√ Lemma 9.5 If max{|α0 |, |α1 |} λ ≤ 1, then under the parameter ζ by Lemma 9.4 there exists a λ2 independent of ε such that  (9.131) θε  L q0 ≤ λ2 . If moreover 1 < r < 23 , then 

  3 − 2r ) , 1 + θ D 2W 1,2 , σ = |∇θε |r d x ≤ cL r/(1−σ σ 3−r Ω

(9.132)

where L σ is the one in (9.135) below. Proof By Lemma 9.1 (which is valid for the second formula of (9.126)), we have   θε− 2W 1,2 ≤ c  f ε Lt vε V + g R  L 4/3 (Γ R ) + g L 6/5 (Ω) .

(9.133)

Using (9.133) and arguing as in (9.61)–(9.89) yield 

|∇θε |2 dx + 1+σ (1 + (θ Ω ε − d0 ) ) 

2  1 ≤c 1+ 1 + vε 2V +  f ε Lt vε V + g R  L 4/3 (Γ ) + g L 6/5 (Ω) R σ 

2 +θ D 2 1,2 + vε 2V +  f ε Lt vε V + g R  L 4/3 (Γ ) + g L 6/5 (Ω) vε 2V . W

R

(9.134) Combining (9.134) with (9.130), we have 

|∇θε |2

+ 1+σ Ω (1 + (θε − d0 ) )



≤c



1+

1 

σ

dx

1 + (1 +  f Lt +  f 1 V∗ )2 +  f Lt (1 +  f Lt +  f 1 V∗ )

2 +g R  L 4/3 (Γ ) + g L 6/5 (Ω) + θ D 2 1,2 + (1 +  f Lt +  f 1 V∗ )2 W R  2 

2 1 +  f Lt +  f 1 V∗ +  f Lt (1 +  f Lt +  f 1 V∗ ) + g R  L 4/3 (Γ ) + g L 6/5 (Ω) R

≡ Lσ .

(9.135) Using (9.135), in the same way as in (9.94) we have   0) (1 + θ D W 1,2 ) + θ D W 1,2 ≡ L σ0 , θε  L q0 ≤ c L 1/(1−σ σ0 where L σ0 is the one with σ0 instead of σ in L σ of (9.135). Putting

(9.136)

9.4 Existence of a Solution: The Case of Total Pressure

 we get (9.131). Now, for 1 < r < 23 putting σ = (9.93), we have (9.132).

λ 2 = L σ0 , 3−2r 3−r

317

(9.137)

and repeating the arguments of (9.91)– 

Fixing λ = λ2 , under the condition  max{|α0 |, |α1 |} λ2 ≤ 1,

(9.138)

by (9.136) and (9.137) we have λ2 = 1 ∀ε > 0. max{λ2 , θε 2L q0 } Therefore, by virtue of Lemmas 9.4 and 9.5 we have Theorem 9.6 If

 max{|α0 |, |α1 |} λ2 ≤ 1,

then there exists a solution (vε , θε ) ∈ V × W 1,2 (Ω) to the following problem ⎧

  a0 (θε ; vε , u) + a2 (vε , vε , u) + ∇Φε (vε ), u − (1 − α0 θε ) f ε , u = f 1 , u ⎪ ⎪ ⎪ ⎪ ⎪ ∀u ∈ V, ⎪ ⎨   |E(vε )|2 ⎪ , ϕ − α1 θε f ε · vε , ϕ ⎪ b0 (θε ; θε , ϕ) − γε (θε )vε , ∇ϕ − α2 μ(θε ) 2 ⎪ 1 + ε|E(vε )| ⎪ ⎪ ⎪ ⎩ = f 2 , ϕ ∀ϕ ∈ WΓ1,2 (Ω), D (Ω), θε − θ D ∈ WΓ1,2 D and the solution satisfies: c (1 +  f Lt +  f 1 V∗ ), 2μ0

θε− WΓ1,2 (Ω) ≤ c  f Lt vε V + g R  L 4/3 (Γ R ) + g L 6/5 (Ω) , D    3 ) |∇θ |r d x ≤ cL r/(1−σ 1 + θ D 2W 1,2 ∀r, 1 < r < , σ 2 Ω

vε V ≤

where σ =

3−2r . 3−r

Now repeating the arguments as in Sect. 9.3.3 with the solutions of Theorem 9.6, we complete the proof of Theorem 9.4. 

318

9 The Steady Equations for Heat-Conducting Fluids

9.5 Bibliographical Remarks The content of Chap. 9 is taken from [1]. Several papers concerning (9.1) have appeared. In [2] the problem with α0 = α1 = 0 was studied under homogeneous Dirichlet boundary conditions for velocity and mixture of non-homogeneous Dirichlet condition and homogeneous Neumann condition for temperature. In [3] when |α0 |, |α1 | are small enough or α0 = 0, αα01 > 1, the existence of a solution to the problems was shown under homogeneous Dirichlet boundary conditions for velocity and mixture of non-homogeneous Dirichlet and homogeneous Neumann conditions for temperature. In [4] for problem on an open channel domain when α0 = α1 = 0, local solvability was studied under mixture of Dirichlet boundary condition of velocity and the free outflow condition for fluid and mixture of non-homogeneous Dirichlet and Neumann boundary conditions for temperature. In [5] a modified problem on an open channel domain, where the buoyancy term (1 − α0 θ ) f is changed by ρ(θ ) f, 0 < ρ(θ ) < ρ1 (const), α1 = 0 and the viscosity, specific heat and thermal conductivity are independent of temperature, was considered. For the problem under mixture of Dirichlet boundary condition of velocity and the free outflow condition for fluid and mixture of non-homogeneous Dirichlet and homogeneous Neumann boundary conditions for temperature it was proved that if the data of problem are small enough, then there exists a unique strong solution. In [6] when the viscosity, specific heat and thermal conductivity are independent of temperature, the steady problem on 2-D bounded domain was considered. When the body force is small enough, the existence of a strong solution was proved under mixture of Dirichlet boundary condition of velocity, tangent stress and stress condition for fluid and mixture of non-homogeneous Dirichlet and homogeneous Neumann boundary conditions for temperature. In several papers the non-Newtonian fluid were studied. In [7] for non-Newtonian fluid with α0 = α1 = 0, under homogeneous Dirichlet boundary condition for velocity and mixture of homogeneous Dirichlet and Robin conditions for temperature existence of a solution was studied. In [8] for the problems of a generalized Newtonian fluid with α0 = α1 = 0 regularity of weak solutions was proved under homogeneous Dirichlet boundary conditions of velocity and temperature. In [9] for the problems of a non-Newtonian fluid with α0 = α1 = 0 existence of a solutions was proved under homogeneous Dirichlet boundary conditions for velocity and temperature on a portion of boundary and a generalized Navier slip and Robin conditions for velocity and temperature on another portion of boundary. In [10] for the problem of a non-Newtonian fluid with heat sources allowed in L 1 and even as measures, under homogeneous Dirichlet boundary conditions for velocity and Robin condition for temperature the existence of a distributional solution was shown for sufficiently small data.

9.5 Bibliographical Remarks

319

But all results above exclude Newtonian fluid owing to conditions for nonlinear terms for strain.

References 1. T. Kim, D. Cao, Mixed boundary value problems of the system for steady flow of heatconducting incompressible viscous fluids with dissipative heating. Meth. Appl. Anal. 27, 87– 124 (2020) 2. J. Naumann, Existence of weak solutions to the equations of stationary motion of heatconducting Incompressible viscous fluids. Progress in Nonlinear Differential Equations and Their Applications, vol. 66 (Birkhäuser, 2005), pp. 373–390 3. J. Naumann, M. Pokorny, J. Wolf, On the existence of weak solutions to the equations of steady flow of heat-conducting fluids with dissipative heating. Nonlinear Anal. RWA 13, 1600–1620 (2012) 4. M. Beneš, P. Kuˇcera, On the Navier-Stokes flows for heat-conducting fluids with mixed boundary conditions. J. Math. Anal. Appl. 389, 769–780 (2012) 5. M. Beneš, A note on regularity and uniqueness of natural convection with effects of viscous dissipation in 3D open channels. Z. Angew. Math. Phys. 65, 961–975 (2014) 6. M. Beneš, A note on the regularity of thermally coupled viscous flows with critical growth in nonsmooth domains. Math. Meth. Appl. Sci. 36, 1290–1300 (2013) 7. L. Consiglieri, Stationary weak solutions for a class of non-Newtonian fluids with energy transfer. Int. J. Non-Linear Mech. 32, 961–972 (1997) 8. L. Consiglieri, T. Shilkin, Regularity of stationary weak solutions in the theory of generalized Newtonian fluid with energy transfer. J. Math. Sci. 115, 2771–2788 (2003) 9. L. Consiglieri, A ( p − q) coupled system in elliptic nonlinear problems with nonstandard boundary conditions. J. Math. Anal. Appl. 340, 183–196 (2008) 10. T. Roubíˇcek, Steady-state buoyancy-driven viscous flow with measure data. Mathematica Bohemica 126, 493–504 (2001)

Chapter 10

The Non-steady Equations for Heat-Conducting Fluids

In this chapter we are concerned with a non-steady system for motion of incompressible Newtonian heat-conducting fluids with mixed boundary conditions. The boundary condition for fluid is the case of total pressure and the boundary conditions for temperature may include Dirichlet, Neumann and Robin conditions together. On the basis of results of Sect. 3.1, we get a variational formulation for the problem. The variational formulation consists of a time-dependent variational inequality for velocity due to the boundary conditions of friction type and a variational equation for temperature. Then we prove the existence of a solution to the problem. It is proved that if the buoyancy effect and energy dissipation effect due to expansion are small enough in accordance with data of problem, then there exists a solution with “defect measure”.

10.1 Problem and Variational Formulation 10.1.1 Problem and Assumption We are concerned with the following non-steady problem ⎧   ∂v ⎪ ⎪ − 2∇ · μ(θ )E(v) + (v · ∇)v + ∇ p = (1 − α0 θ) f, ⎪ ⎪ ∂t ⎪ ⎪ ⎪ ⎨ div v = 0, ∂θ ⎪ ⎪ ⎪ − ∇ · (κ(θ)∇θ ) + v · ∇θ − α2 μ(θ )E(v) : E(v) = α1 θ f · v + g, ⎪ ⎪ ∂t ⎪ ⎪ ⎩ v(0) = v0 , θ(0) = θ0 .

(10.1)

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 T. Kim and D. Cao, Equations of Motion for Incompressible Viscous Fluids, Advances in Mathematical Fluid Mechanics, https://doi.org/10.1007/978-3-030-78659-5_10

321

322

10 The Non-steady Equations for Heat-Conducting Fluids

Due to the dissipation of energy μ(θ )E(v) : E(v), it is more difficult to study the system (10.1) than the Boussinesq system. Let Ω be as in Chap. 8. (See Remark 6.1.) For temperature we are concerned with the boundary conditions (1) θ |Γ D = 0,   ∂θ + β(x)θ Γ R = g R (t, x), (2) κ(θ ) ∂n β(x), g R (t, x) − given functions on Γ R , (0, T ) × Γ R .

(10.2)

The boundary condition for fluid is as in (7.4). Let V, K (Ω) be the same as in (5.5) and H : completion in L2 (Ω) of V, HK : closure in L2 (Ω) of K (Ω), K (Q) = {u ∈ L 2 (0, T ; V) : u  ∈ L 2 (0, T ; V∗ ); u n |Γ10 ≥ 0, u n |Γ11 ≤ 0}, ¯ : div u = 0, u| = 0, u τ |( ∪ 4 ∪ 7 ) = 0, u · n|( ∪ ) = 0}, (Q) = {u ∈ C2 ( Q) 1 2 3 5 1, p

WΓ (Ω) = {y ∈ W 1, p (Ω) : y|Γ D = 0}, D ¯ : y|Γ = 0}. CΓ1 D (Ω) = {y ∈ C 1 (Ω) D

We use inner products in (7.8) for V and WΓ1,2 (Ω). D We use the following assumption. Assumption 10.1 Assume the followings. (1) Γ1 = ∅, Γ D = ∅ and   Γ R ⊂ ∪i=1,3,5,8 Γi .

(10.3)

(2) (3) of Assumption 5.1 holds.   )∗ and (3) For the functions of (10.1) f ∈ L ∞ (Q), g ∈ L 1 0, T ; (WΓ1,2 D μ ∈ C(R), 0 < μ0 ≤ μ(ξ ) ≤ μ1 < ∞ ∀ξ ∈ R, κ ∈ C(R), 0 < κ0 ≤ κ(ξ ) ≤ κ1 < ∞ ∀ξ ∈ R, v0 ∈ HK , θ0 ∈ L (Ω). 1

(4) For the functions of (10.2), (7.4)

(10.4)

10.1 Problem and Variational Formulation

323

g R ∈ L 1 (0, T ; L 4/3 (Γ R )), β1 ≥ β(x) ≥ 0, β1 − a constant, β(x) − measurable, φi ∈ L 2 (0, T ; H − 2 (Γi )), i = 2, 4, 7, φi ∈ L 2 (0, T ; H− 2 (Γi )), i = 3, 5, 6, αi j ∈ L ∞ (Γ5 ). (10.5) 1

1

10.1.2 Variational Formulation for Problem In this section we first give variational formulations for the problem above. Taking (v · ∇)v = rot v × v + 21 grad|v|2 into account, by (7.9)–(7.11), (8.8), we can see that smooth solutions (v, p, θ ) of problem (10.1), (10.2), (7.4) satisfy the following. ⎧ ∂v

⎪ ⎪ , u + 2(μ(θ )E(v), E(u)) + rotv × v, u + 2(μ(θ )k(x)v, u)Γ2 ⎪ ⎪ ∂t ⎪ ⎪ ⎪ ⎪ + 2(μ(θ )S v, ˜ u) ˜ Γ3 + 2(α(x)v, u)Γ5 + (μ(θ )k(x)v, u)Γ7 ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ − 2(μ(θ )εnτ (v), u)Γ8 + ( p + |v|2 − 2μ(θ )εnn (v), u n )Γ9 ∪Γ10 ∪Γ11 ⎪ ⎪ ⎪ 2 ⎪   ⎪ ⎪ ⎪ ⎪ = (1 − α θ ) f, u + φ , u φi , uΓi ∀u ∈ V, 0 i n Γi + ⎪ ⎪ ⎪ ⎨ i=2,4,7 i=3,5,6 ∂θ

⎪ , ϕ + (κ(θ )∇θ, ∇ϕ) − (θ v, ∇ϕ) − (α2 μ(θ )|E(v)|2 , ϕ) + (βθ, ϕ)Γ R ⎪ ⎪ ⎪ ∂t ⎪ ⎪ ⎪ ⎪ − (α1 θ f · v, ϕ) = g R , ϕΓ R + g, ϕ ∀ϕ ∈ WΓ1,∞ (Ω), ⎪ D ⎪ ⎪ ⎪ t t ⎪ ⎪ |στ (θ, v)| ≤ gτ , στ (θ, v) · vτ + gτ |vτ | = 0 on Γ8 , ⎪ ⎪ ⎪ ⎪ ⎪ |σnt (θ, v, p)| ≤ gn , σnt (θ, v, p)vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪ ⎪ ⎪ σnt (θ, v, p) + g+n ≥ 0, (σnt (θ, v, p) + g+n )vn = 0 on Γ10 , ⎪ ⎪ ⎩ t σn (θ, v, p) − g−n ≤ 0, (σnt (θ, v, p) − g−n )vn = 0 on Γ11 . (10.6) Define a0 (θ ; ·, ·), a1 (·, ·, ·) and f 1 ∈ V∗ by a0 (θ ; w, u) = 2(μ(θ )E(w), E(u)) + 2(μ(θ )k(x)w, u)Γ2 + 2(μ(θ )S w, ˜ u) ˜ Γ3 + 2(α(x)w, u)Γ5 + (μ(θ )k(x)w, u)Γ7 ∀w, u ∈ V, θ ∈ W 1,2 (Ω), a1 (v, u, w) = rot v × u, w ∀v, u, w ∈ V,   φi , u n Γi + φi , uΓi ∀u ∈ V.  f 1 , u = i=2,4,7

i=3,5,6

˜ ·, ·) and f 2 ∈ (WΓ1,2 (Ω))∗ by Define b0 (θ; D

(10.7)

324

10 The Non-steady Equations for Heat-Conducting Fluids

˜ θ, ϕ) = (κ(θ)∇θ, ˜ b0 (θ; ∇ϕ) + (β(x)θ, ϕ)Γ R ∀θ, θ˜ ∈ W 1,2 (Ω), ϕ ∈ WΓ1,2 (Ω), D (Ω).  f 2 , ϕ = g R , ϕΓ R + g, ϕ ∀ϕ ∈ WΓ1,2 D

(10.8)

By (10.4) and (10.5), f 1 ∈ L 2 (0, T ; V∗ ),

f 2 ∈ L 1 (0, T ; (WΓ1,2 )∗ ). D

(10.9)

Then, taking into account 1 στt (θ, v) = 2μ(θ )εnτ (v), σnt (θ, v, p) = −( p + |v|2 ) + 2μ(θ )εnn (v) 2 and (10.6), we introduce the following variational formulation for problem (10.1), (10.2), (7.4). Problem VE. Find v ∈ K (Q), θ ∈ L ∞ (0, T ; L 1 (Ω)) ∩ L r (0, T ; WΓ1,rD (Ω)) for all 1 1 t t , σ−n ) ∈ L2τ (Γ8 ) × L 2 (Γ9 ) × H − 2 (Γ10 ) × H − 2 (Γ11 ) r ∈ [1, 5/4) and (στt , σnt , σ+n for a.e. t ∈ (0, T ) such that v(0) = v0 , θ (0) = θ0 and ⎧ ∂v ⎪ ⎪ , u + a0 (θ ; v, u) + a1 (v, v, u) − (στt , u τ )Γ8 − (σnt , u n )Γ9 ⎪ ⎪ ∂t ⎪ ⎪ ⎪ t t ⎪ − σ+n , u n Γ10 − σ−n , u n Γ11 −  f − α0 θ f, u =  f 1 , u ∀u ∈ V, ⎪ ⎪ ⎪ ⎪

⎪ ∂θ ⎪ ⎪ , ϕ + b0 (θ ; θ, ϕ) − θ v, ∇ϕ − α2 μ(θ )|E(v)|2 , ϕ − α1 θ f · v, ϕ ⎪ ⎪ ⎨ ∂t ⎪ ⎪ ⎪ t ⎪ ⎪ ⎪ |στ | ≤ gτ , ⎪ ⎪ ⎪ ⎪ |σnt | ≤ gn , ⎪ ⎪ ⎪ t ⎪ ⎪ σ+n + g+n ⎪ ⎪ ⎩ t σ−n − g−n

(Ω), =  f 2 , ϕ ∀ϕ ∈ WΓ1,∞ D στt · vτ + gτ |vτ | = 0 on Γ8 , σnt vn + gn |vn | = 0 on Γ9 , t ≥ 0, σ+n + g+n , vn Γ10 = 0 on Γ10 , t ≤ 0, σ−n − g−n , vn Γ11 = 0 on Γ11 ,

(10.10) where L2τ (Γ8 ) is the subspace of L2 (Γ8 ) consisting of functions such that (u, n)L2 (Γ8 ) = 0. Remark 10.1 As in Remark 8.3, for solutions smooth enough Problem VE is equivalent to the problem (10.1), (10.2), (7.4). We will find another variational formulation consisting of a variational inequality and a variational equation, which is equivalent to Problem VE if the solution is smooth enough (see Remark 10.2). For fixed θ , let us consider the problem

10.1 Problem and Variational Formulation

325

⎧ ∂v ⎪ ⎪ , u + a0 (θ ; v, u) + a1 (v, v, u) − (στ , u τ )Γ8 − (σn , u n )Γ9 ⎪ ⎪ ∂t ⎪ ⎪ ⎪ ⎪ − σ+n , u n Γ10 − σ−n , u n Γ11 − ( f − α0 θ f, u =  f 1 , u, ⎪ ⎪ ⎨ t |στ | ≤ gτ , στt · vτ + gτ |vτ | = 0 on Γ8 , ⎪ ⎪ ⎪ |σnt | ≤ gn , σnt vn + gn |vn | = 0 on Γ9 , ⎪ ⎪ ⎪ t ⎪ σ t + g+n ≥ 0, σ+n + g+n , vn Γ10 = 0 on Γ10 , ⎪ ⎪ ⎪ +n ⎩ t t σ−n − g−n ≤ 0, σ−n − g−n , vn Γ11 = 0 on Γ11 .

(10.11)

Subtracting the first formula of (10.11) with u = v from the first one of (10.11), we get ∂v

, u − v + a0 (θ ; v, u − v) + a1 (v, v, u − v) − (στt , u τ − vτ )Γ8 − (σnt , u n − vn )Γ9 ∂t t ,u − v  t − σ+n n n Γ10 − σ−n , u n − vn Γ11 −  f − α0 θ f, u − v =  f 1 , u − v ∀u ∈ V.

(10.12)

Let Φ : V → R be the functional defined by (5.22). Then, the functional Φ is proper, convex, lower weak semi-continuous and nonnegative. Define a functional Ψ (u) by Ψ (u) =

⎧ ⎨ ⎩

T

Φ(u(t)) dt

i f Φ(u(t)) ∈ L 1 (0, T ),

0

+∞

(10.13)

other wise.

In the same way as in Problem I of Chap. 5, we get from (10.7) and (10.12) ∂v ∂t

, u − v + a0 (θ ; v, u − v)+a1 (v, v, u − v) + Φ(u) − Φ(v)

(10.14)

≥ (1 − α0 θ ) f, u − v +  f 1 , u − v.

If u − v ∈ (Q), then 

T

 T

, u − v dt = ∂t 0  T =

∂v

∂(v − u)

T

∂u

, u − v dt

∂t 0 1 , u − v dt + (v0 − u(0)2 − v(T ) − u(T )2 ). ∂t 2 0 (10.15) Define operators A(θ ) : V → V∗ and B : V × V → V∗ , respectively, by 0

∂u

∂t



, v − u dt +

A(θ )v, u = a0 (θ ; v, u) ∀w, u ∈ V, B(v, u), w = a1 (v, u, w) ∀v, u, w ∈ V.

(10.16)

326

10 The Non-steady Equations for Heat-Conducting Fluids

If v ∈ L 2 (0, T ; V) ∩ L ∞ (0, T ; L2 ), then v ∈ L 4 (0, T ; L3 ) (see  T (1.21)) and B(v, v) ∈ 4 L 3 (0, T ; V∗ ). Thus, when u ∈ L 4 (0, T ; V), the integral 0 B(v, v), u − v dt is meaningful since B(v, v), v = 0. Combining (10.14), (10.15), and neglecting 21 (w(T ) − u(T )2 ), we come to the following variational inequality corresponding to problem (10.11) (see (6.11), or (1.7) of [1], (2.1) of [2]). Find v ∈ L 2 (0, T ; V) ∩ L ∞ (0, T ; H ) such that  T 0

u  + A(θ )v(t) + B(v(t), v(t)) − (1 − α0 θ ) f − f 1 , u(t) − v(t) dt + Ψ (u) − Ψ (v) 1 ≥ − v0 − u(0)2 ∀u ∈ L 4 (0, T ; V) with u  ∈ L 2 (0, T ; V∗ ). 2

(10.17)

If (v, θ, στ , σn , σ+n , σ−n ) is a solution of Problem VE, then (v, θ ) satisfies the problem above. Therefore, we have the following variational formulation of the problem which consists of a variational inequality for velocity and a variational equation for temperature.    Problem VI. Find (v, θ ) ∈ L ∞ (0, T ; H ) ∩ L 2 (0, T ; V) × L ∞ (0, T ; L 1 (Ω)) ∩  L r (0, T ; WΓ1,rD ) ∀r ∈ [1, 5/4) such that ⎧ T ⎪ ⎪ ⎪ u  + A(θ )v(t) + B(v(t), v(t)) − (1 − α0 θ ) f − f 1 , u(t) − v(t) dt + Ψ (u) − Ψ (v) ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎨ ≥ − v0 − u(0)2 ∀u ∈ L 4 (0, T ; V) with u  ∈ L 2 (0, T ; V∗ ), 2  ⎪ ⎪ T  ∂ϕ ⎪ ⎪ − θ, + b0 (θ ; θ, ϕ) − θ v, ∇ϕ − α2 μ(θ )|E(u)|2 , ϕ − α1 θ f · v, ϕ ⎪ ⎪ ∂t ⎪ 0 ⎪ ⎪  ⎪ ⎪ ⎩ −  f 2 , ϕ dt = θ0 (x), ϕ(x, 0) ∀ϕ ∈ C 1 (0, T ; CΓ1 (Ω)) with ϕ(·, T ) = 0. D

(10.18) Remark 10.2 If the solutions to Problem VI is smooth such that v ∈ L 2 (0, T ; V), v  ∈ L 2 (0, T ; V∗ ), then v satisfies (10.14), which is equivalent to v  (t) + A(θ )v(t) + B(v(t), v(t)) − (1 − α0 θ ) f − f 1 , u − v(t) + Φ(u) − Φ(v(t)) ≥ 0 for a.e. t ∈ [0, T ], ∀u ∈ K (Ω)

(10.19)

 − (1 − α0 θ ) f − (Remark, p. 114 of [1]). In (10.14) putting F1 , u − v = − ∂v ∂t f 1 , u − v, by Theorem 5.4 we can get the existence of (στ , σn , σ+n , σ−n ) ∈ L2τ (Γ8 ) × L 2 (Γ9 ) × H −1/2 (Γ10 ) × H −1/2 (Γ11 ) for a.e. t ∈ (0, T ) such that (v, θ, στ , σn , σ+n , σ−n ) satisfies (10.10).

10.1 Problem and Variational Formulation

327

It is desirable to get a solution satisfying (10.18), but we can only prove the existence of (v, θ ) with a “defect measure” in the second equation of (10.18) The main result of this chapter is the following Theorem 10.1 Let Assumption 10.1 be satisfied. If |α0 | + |α1 | is small  enough in accordance with the date of the problem, then there exists (v, θ ) ∈ L ∞ (0, T ; H )    ∩L 2 (0, T ; V) × L ∞ (0, T ; L 1 (Ω)) ∩ L r (0, T ; WΓ1,rD ) ∀r ∈ [1, 5/4) and a Radon measure μ such that ⎧ T ⎪ ⎪ u  + A(θ )v(t) + B(v(t), v(t)) − (1 − α0 θ ) f − f 1 , u(t) − v(t) dt + Ψ (u) − Ψ (w) ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ≥ − v0 − u(0)2 ∀u ∈ L 4 (0, T ; V) with u  ∈ L 2 (0, T ; V∗ ), ⎪ ⎪ ⎪ 2 ⎪ ⎨   T ∂ϕ − θ, + b0 (θ ; θ, ϕ) − θ v, ∇ϕ − α2 μ(θ )|E(u)|2 , ϕ − α1 θ f · v, ϕ ⎪ ⎪ ∂t ⎪ 0 ⎪ ⎪  T ⎪  ⎪ ⎪ ⎪ ⎪ ϕ dμ −  f 2 , ϕ dt = θ0 (x), ϕ(x, 0) + ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎩ ∀ϕ ∈ C 1 (0, T ; CΓ1 (Ω)) with ϕ(·, T ) = 0. D

(10.20)

Moreover, (v, θ ) satisfies the estimate  sup v(t)2 + v2 2

t∈[0,T ]

L (0,T ;V)

+ ess sup θ (t) L 1 (Ω) + t∈[0,T ]

 φi  ≤ C1 v0 ,  f ∞ , i=2,4,7

 Q

1

|∇θ |r d xdt + ,



− L 2 (0,T ;H 2 (Γi )) i=3,5,6

|∇θ |2 d xdt (1 + |θ |)1+δ Q

φi 

1

− L 2 (0,T ;H 2 (Γi ))

δ, |α|, θ0  L 1 (Ω) , g R  L 1 (0,T ;L 4/3 (Γ )) , g 1 L (0,T ;(W 1,2 )∗ ) R

,



ΓD

5 − 4r 5 . ∀1 ≤ r < , ∀0 < δ < 4 3

(10.21)

10.2 Existence of a Solution 10.2.1 Existence of a Solution to an Approximate Problem We first consider a problem approximating (10.18). For every 0 < ε < 1, let Φε be the Moreau-Yosida approximation of Φ and ∇Φε be Fréchet derivative of Φε .

328

10 The Non-steady Equations for Heat-Conducting Fluids

Let v0 ∈ W1,6 (Ω) ∩ K (Ω), θ0 ∈ WΓ1,6 (Ω) and 0 < ε ≤ 1. Also assume that D 2 4/3 2 )∗ ), and so f 2 ∈ L 2 (0, T ; (WΓ1,2 )∗ ). g R ∈ L (0, T ; L (Γ R )), g ∈ L (0, T ; (WΓ1,2 D D Problem VEA. Find v ∈ L 6 (0, T ; V) ∩ C([0, T ]; H ), θ ∈ L 2 (0, T ; WΓ1,2 ) ∩ C([0, T ]; D 2 L (Ω)) such that ⎧

∂v 4 ⎪ ⎪ ⎪ ∂t , u + 2 [μ(θ ) + εvV ]E(v), E(u) + rotv × v, u + 2(μ(θ )k(x)v, u)Γ2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ + 2(μ(θ )S v, ˜ u) ˜ Γ3 + 2(α(x)v, u)Γ5 + (μ(θ )k(x)v, u)Γ7 + ∇Φε (v(t)), u ⎪ ⎪ ⎪

⎪ ⎪ θ α 0 ⎪ ⎪ ∀u ∈ L 6 (0, T ; V), ⎪ ⎨ = (1 − 1 + εθ 2 ) f, u +  f 1 , u

∂θ ⎪ |E(v)|2 ⎪ ⎪ ⎪ ⎪ ∂t , ϕ + (κ(θ )∇θ, ∇ϕ) + (β(x)θ, ϕ)Γ R − vθ, ∇ϕ − α2 μ(θ ) 1 + ε|E(v)|2 , ϕ ⎪ ⎪ ⎪ ⎪ α θ

⎪ ⎪ 1 ⎪ = f · v, ϕ +  f 2 , ϕ ∀ϕ ∈ L 2 (0, T ; WΓ1,2 (Ω)), ⎪ ⎪ 2 D ⎪ 1 + εθ ⎪ ⎪ ⎩ v(0) = v0 , θ (0) = θ0 .

(10.22) Let

), V := L 6 (0, T ; V) × L 2 (0, T ; WΓ1,2 D

1/2 (·, ·)V =  · 2L 6 (0,T ;V) +  · 2L 2 (0,T ;W 1,2 ) . ΓD

Then

  V ∗ = L 6/5 (0, T ; V∗ ) × L 2 0, T ; (WΓ1,2 )∗ , D (·, ·)V ∗ =  · 2L 6/5 (0,T ;V∗ ) +  · 2 2  L

0,T ;(WΓ1,2 )∗



1/2

.

D

Theorem 10.2 There exists a solution (vε , θε ) ∈ V to Problem VEA. ˆ Proof In (10.22) let us make changes of the unknown functions by v = ek1 t w, ˆ θ = θ, where k1 is a constant to be determined later (in (10.37)). Owing to the changes, differently from Ch. 9 we need not assume that Γ2 j , Γ3 j , Γ7 j are convex and the matrix α is positive. Then we have the following problem equivalent to Problem VEA. )∩ C([0, T ]; L 2 (Ω)) Problem VEA’. Find wˆ ∈ L 6 (0, T ; V) ∩ C([0, T ]; H ), θˆ ∈ L 2 (0, T ; WΓ1,2 D such that

10.2 Existence of a Solution

329

⎧

∂ wˆ ⎪ ˆ + e4k1 t εw ⎪ , u + 2 [μ(θ) ˆ 4V ]E(w), ˆ E(u) + ek1 t rotwˆ × w, ˆ u ⎪ ⎪ ⎪ ∂t ⎪ ⎪ ⎪ ˜ˆ u) ⎪ + k1 (w, ˆ u) + 2(μ(θˆ )k(x)w, ˆ u)Γ2 + 2(μ(θˆ )S w, ˜ Γ3 ⎪ ⎪ ⎪ ⎪ ⎪ −k t 1 ˆ ⎪ + 2(α(x)w, ˆ u)Γ5 + (μ(θ )k(x)w, ˆ u)Γ7 + e ∇Φε (ek1 t w(t)), ˆ u ⎪ ⎪ ⎪ ⎪

⎪ ⎪ α0 θˆ ⎪ ⎪ = e−k1 t (1 − f ), u + e−k1 t  f 1 , u ⎪ ⎪ ˆ2 ⎪ 1 + ε θ ⎪ ⎪ ⎪ ⎪ ∀u ∈ L 6 (0, T ; V), ⎪ ⎨ ˆ ⎪ ∂ θ , ϕ + (κ(θ)∇ ˆ θˆ , ∇ϕ) + (β(x)θˆ , ϕ)Γ R − ek1 t wˆ θˆ , ∇ϕ ⎪ ⎪ ⎪ ∂t ⎪ ⎪ ⎪

⎪ |E(w)| ˆ 2 ⎪ ⎪ ˆ − α μ( θ) , ϕ ⎪ 2 ⎪ ⎪ e−2k1 t + ε|E(w)| ˆ 2 ⎪ ⎪ ⎪ α θˆ

⎪ ⎪ 1 ⎪ ⎪ f · w, ˆ ϕ +  f 2 , ϕ = e2k1 t ⎪ ⎪ ⎪ 1 + εθˆ 2 ⎪ ⎪ ⎪ ⎪ ⎪ ∀ϕ ∈ L 2 (0, T ; WΓ1,2 (Ω)), ⎪ D ⎪ ⎪ ⎩ ˆ = θ0 . w(0) ˆ = v0 , θ(0) (10.23) Define an operator L : D(L ) → V ∗ by   D(L ) := (v, θ ) ∈ V , (v  , θ  ) ∈ V ∗ , v(0) = 0, θ (0) = 0 , L (v, θ ) = (v  , θ  ), (·, ·) D(L ) := (·, ·V + ·, ·V ∗ ). Then, L is a linear, maximal monotone operator (see Example 1.1). Putting wˆ = w + v0 , θˆ = θ + θ0 , we will consider an evolution problem with zero initial conditions. Define an operator Aε : D(L ) → V ∗ by 

 Aε (w, θ ), (u, φ) =

 T

ˆ 4 ]E(w), ˆ E(u) 2 [μ(θˆ ) + e4k1 t εw 0

V

+ ek1 t rotwˆ × w, ˆ u + k1 (w, ˆ u) + 2(μ(θˆ )k(x)w, ˆ u)Γ2 ˜ˆ u) + 2(μ(θˆ )S w, ˜ Γ3 + 2(α(x)w, ˆ u)Γ5 + (μ(θˆ )k(x)w, ˆ u)Γ7

α θˆ 0 + e−k1 t ∇Φε (ek1 t w(t)), ˆ u + e−k1 t f, u 1 + ε θˆ 2 ˆ ∇φ) + (β(x)θˆ , φ)Γ − ek1 t wˆ θ, ˆ ∇φ + (κ(θˆ )∇ θ, R

 2 α1 θˆ |E(w)| ˆ − α2 μ(θˆ ) −2k t , φ − e2k1 t f · w, ˆ φ dt 2 1 + ε|E(w)| e ˆ 1 + ε θˆ 2 ∀(w, θ ) ∈ D(L ), (u, φ) ∈ V .

(10.24)

330

10 The Non-steady Equations for Heat-Conducting Fluids

Estimate (10.47) to be given later shows that this operator is well-defined. Then, the existence of a solution to Problem VEA’ is equivalent to one of a solution to   −k t e 1 ( f + f1 ) . L (w, θ ) + Aε (w, θ ) = F , F = f2 Applying Theorem 1.47, we will prove existence of a solution to the equation above. To this end, we need to check that Aε satisfies assumptions of Theorem 1.47. (i) Let us prove property (iii) of Theorem 1.47 for Aε (w, θ ):   1 Aε (w, θ ), (w, θ ) → ∞ as (w, θ )V → ∞. (w, θ )V

(10.25)

We have that  Aε (w, θ ), (w, θ ) =



 T

2 [μ(θˆ ) + e4k1 t εw ˆ 4V ]E(w), ˆ E(w) 0

ˆ w + k1 (w, ˆ w) + 2(μ(θˆ )k(x)w, ˆ w)Γ2 + ek1 t rotwˆ × w, ˜ˆ w) + 2(μ(θˆ )S w, ˜ Γ3 + 2(α(x)w, ˆ w)Γ5 + (μ(θˆ )k(x)w, ˆ w)Γ7 α θˆ

0 + e−k1 t ∇Φε (ek1 t w(t)), ˆ w + e−k1 t f, w 1 + ε θˆ 2 ˆ θ, ˆ ∇θ ) + (β(x)θˆ , θ )Γ − ek1 t wˆ θˆ , ∇θ  + (κ(θ)∇

(10.26)

R



 α θˆ |E(w)| ˆ 2 1 − α2 μ(θˆ ) −2k t , θ − e2k1 t f · w, ˆ θ dt. 2 2 ˆ 1 e + ε|E(w)| ˆ 1 + εθ

Let us now estimate each term on the right hand side above. By Hölder’s inequality we have  T

2 [μ(θˆ ) + e4k1 t εw ˆ 4V ]E(w), ˆ E(w) dt 0

≥ 2μ0 w2 2

L (0,T ;V)

+

 T

2 μ(θˆ )E(vˆ0 ), E(w) dt 0

 T  T

2 e4k1 t w ˆ 4V E(w), ˆ E(w) ˆ dt − ε 2 e4k1 t w ˆ 4V E(w), ˆ E(v0 ) dt +ε 0

≥ μ0 w2 2

0

+ 2εw + v0 6 6

L (0,T ;V) L (0,T ;V) 2 6 ≥ μ0 w 2 + εw 6 − K1, L (0,T ;V) L (0,T ;V)

− 2cεw + v0 5 6

L (0,T ;V)

v0  L 6 (0,T ;V) − kT v0 2

(10.27)

where K 1 depends on v0 V , ε, T . By the fact that Γ2 j , Γ3 j , Γ7 j are in C 2.1 and Assumption 10.1, there exists a constant M such that S(x)∞ , k(x)∞ , α L ∞ (Γ5 ) ≤ M.

(10.28)

10.2 Existence of a Solution

331

Thus,  T          2(μ(θˆ )k(x)w, w Γ + 2 μ(θˆ )S w, ˜ w˜ Γ + 2 α(x)w, w Γ + μ(θˆ )k(x)w, w Γ dt   2

0

3

5

1 ≤ μ0 w2 2 + k11 w2 2 L (0,T ;V) L (0,T ;H ) 8

7

(10.29)

(see Theorem 1.27) and  T

   ˜ Γ3 + 2(α(x)v0 , w)Γ5 + (μ(θˆ )k(x)v0 , w)Γ7 dt  2(μ(θˆ )k(x)v0 , w)Γ2 + 2(μ(θˆ )S v˜0 , w)  0

≤

1 μ0 w2 2 + K2, L (0,T ;V) 8

(10.30)

where K 2 depends on v0 V , μ0 , μ1 , T . Since rot w × v, v = 0,  T   T       ek1 t rotwˆ × w, ˆ w dt  =  ek1 t rot w × v0 , w + rot v0 × v0 , w dt   0

0

1 + K 3 + k12 w2 2 , ≤ μ0 w2 2 L (0,T ;V) L (0,T ;H ) 8

(10.31)

where K 3 depends on v0 W1,6 (Ω) , μ0 , μ1 , T . Let us estimate  T  T e−k1 t ∇Φε (ek1 t w(t)), ˆ w dt = e−k1 t ∇Φε (ek1 t w(t)), ˆ ek1 t wˆ − ek1 t v0  dt. 0

0

Since Φε is convex, continuous and Fréchet differentiable, by (1.37) we have ∇Φε (ek1 t w(t)), ˆ ek1 t v0 − ek1 t w ˆ ≤ Φε (ek1 t v0 ) − Φε (ek1 t w(t)) ˆ ≤ Φ(ek1 t v0 ). Thus, 

T

e−k1 t ∇Φε (ek1 t w(t)), ˆ w dt ≥ −



0

T

e−k1 t Φ(ek1 t v0 ) dt = −K 4 ,

0

K 4 ≥ 0. (10.32)

Also, we have  T  0

 

 (κ(θ )∇θ, ∇θ ) dt ≥ κ0 θ 2 2 , L (0,T ;W 1,2 )

 T  0

 T 0

ΓD

  κ0 θ 2 2 (κ(θ )∇θ0 , ∇θ ) dt  ≤ + K5, L (0,T ;WΓ1,2 ) 6 D

(β(x)θ, θ )Γ R dt ≥ 0,

 κ  T   0 θ 2 2 (β(x)θ0 , θ )Γ R dt  ≤ + K6,  L (0,T ;WΓ1,2 ) 6 0 D

(10.33)

332

10 The Non-steady Equations for Heat-Conducting Fluids

where K 5 , K 6 depend on θ0 W 1,6 (Ω) and T . Since wθ, ∇θ  = 0 (see (8.8)),   

  T     −ek1 t wˆ θˆ , ∇θ  dt  =  −ek1 t wθ ˆ 0 , ∇θ  dt  0  T     = −ek1 t wθ0 , ∇θ  + v0 θ0 , ∇θ  dt 

T

0

(10.34)

0

≤

κ0 θ 2L 2 (0,T ;W 1,2 ) + K 7 + k13 w2L 2 (0,T ;H ) . ΓD 6

It is easy to get 

T



ˆ α2 μ(θ)

0

Since

|E(w)| ˆ 2 c , θ dt ≤ θ  L 2 (0,T ;L 2 (Ω)) . e−2k1 t + ε|E(w)| ˆ 2 ε

(10.35)

ξ 1 ≤ √ ∀ξ ∈ [0, ∞), 1 + εξ 2 2 ε

we have   

 c  f, w dt  ≤ √ w L 2 (0,T ;H ) , ε 1 + εθˆ 2 0  T α θˆ

 K k14 κ0 1 8  + w2L 2 (0,T ;H ) + θ 2L 2 (0,T ;W 1,2 ) . e2k1 t f · w, ˆ θ dt  ≤ 2 ΓD ˆ ε ε 6 1 + εθ 0 (10.36) Taking k14 , (10.37) k1 = k11 + k12 + k13 + ε T

e−k1 t

α θˆ 0

by (10.27)–(10.36) we have from (10.26) 

 Aε (w, θ ), (w, θ ) μ

κ0  0 −k1 T e w2L 2 (0,T ;V) + w6L 6 (0,T ;V) + θ 2L 2 (0,T ;W 1,2 ) ≥ min , ε, ΓD 2 2 7  c K8 c − √ w L 2 (0,T ;H ) − θ  L 2 (0,T ;L 2 (Ω)) ∀(w, θ ) ∈ D(L ), − Ki − ε ε ε i=1 (10.38) which implies (10.25). (ii) Let us prove property (ii) of Theorem 1.47 for Aε (w, θ ). Taking into account (10.28) and applying an inequality |a + b| p ≤ 2 p (|a| p + |b| p ), p ∈ (1, ∞), we have

10.2 Existence of a Solution 1 u L 6 (0,T ;V)

  

T 0

333



ˆ + e4k1 t εw ˆ 2 [μ(θ) ˆ 4V ]E (w), ˆ E (u) + k1 (w, ˆ u) + 2(μ(θ)k(x) w, ˆ u)Γ2

   ˜ˆ u) ˆ w, ˆ + 2(μ(θ)S ˜ Γ3 + 2(α(x)w, ˆ u)Γ5 + (μ(θ)k(x) w, ˆ u)Γ7 dt      ≤ c w ˆ L 2 (0,T ;V) + w ˆ 6L 6 (0,T ;V) ≤ c K + w L 2 (0,T ;V) + w6L 6 (0,T ;V) ,

(10.39)

where K depends on v0 . By (1.37) and Φ(0V ) = 0, we know that Φε (0V ) = 0. Also, since Φε is convex, continuous and Fréchet differentiable, we have ∇Φε (0V ) = 0V∗ . Then, since the operator ∇Φε is Lipschitz continuous with the constant ε−1 (see Remark 1.16), we have 1 u L 6 (0,T ;V)

  T   e−k1 t ∇Φε (ek1 t w(t)), ˆ u dt   0

=

1 u L 6 (0,T ;V)

  T   e−k1 t ∇Φε (ek1 t w(t)) ˆ − ∇Φε (0V ), u dt   0

 T 1 ≤ w(t) ˆ V uV dt u L 6 (0,T ;V) 0 ε

c c ˆ L 2 (0,T ;V) ≤ K + w L 2 (0,T ;V) . ≤ w ε ε

(10.40)

1

Similarly, we have  T 

  |E(w)| ˆ 2   ˆ θˆ , ∇φ) + (β(x)θˆ , φ)Γ − α2 μ(θˆ ) (κ(θ)∇ , φ dt   R −2k t 2 1 + ε|E(w)| φ 2 e ˆ 1,2 0 L (0,T ;W ) 1

ΓD

ˆ ≤ c θ

L 2 (0,T ;WΓ1,2 ) D



≤ c K + θ 

+

1

ε

L 2 (0,T ;WΓ1,2 )

+

D

1

, ε

(10.41)

where K depends on θ0 . On the other hand, taking into account w(0) = 0, θ (0) = 0 and applying (1.27), for (w, θ ) ∈ D(L ) we have (w, θ ) ∈ C([0, T ]; H ) × C([0, T ]; L 2 (Ω)), wC([0,T ];H ) ≤ cw   L 6/5 (0,T ;V∗ ) w L 6 (0,T ;V) , 1/2

1/2

 1/2

θ C([0,T ];H ) ≤ cθ 

L 2 0,T ;(WΓ1,2 )∗ D

(10.42)

 θ 1/2 2

L (0,T ;WΓ1,2 )

.

D

Also, by Theorem 1.12 1/2

1/2

wL3 (Ω) ≤ cwL2 (Ω) ∇wL2 (Ω) 1/4

3/4

θ  L 4 (Ω) ≤ cθ  L 2 (Ω) ∇θ L2 (Ω)

∀w ∈ V, 1, p

∀θ ∈ WΓ D (Ω).

(10.43)

334

10 The Non-steady Equations for Heat-Conducting Fluids

Applying (10.42) and (10.43), by Hölder’s inequality with exponents 4, 4/3 and Young’s inequality we have  T  T      ek1 t rotwˆ × w, ˆ u dt  ≤ c rotw ˆ L2 w ˆ L3 uL6 dt  0

0

1/2

≤ cw ˆ C([0,T ];H )

 T 0

∇ w ˆ

3/2 uV dt L2

≤ cw  

1/4 1/4 3/2 w ˆ 6 w ˆ 6 u L 4/3 (0,T ;V) L (0,T ;V) L (0,T ;V) L 6/5 (0,T ;V∗ ) 1/4 7/4  ≤ cw  6/5 w ˆ 6 u L 6 (0,T ;V) L (0,T ;V) L (0,T ;V∗ )

≤

1 4



w   L 6/5 (0,T ;V∗ ) + K w ˆ

7/3 u L 6 (0,T ;V) . L 6 (0,T ;V)

(10.44)

Applying Hölder’s inequality with exponents 8, 8/3, 2 and Young’s inequality with exponents 4, 4/3, in the same way we have   

T

   ek1 t wˆ θˆ , ∇φ dt  ≤ c

T

w ˆ L4 θˆ L4 ∇φL2 dt  T 1/4 1/4 3/4 3/4 ∇ w ˆ L2 ∇ θˆ L2 φWΓ1,2 dt ≤ cw ˆ C([0,T ];H ) θˆ C([0,T ];L 2 ) 0

≤

0

D 0 1/8 1/8 1/8 3/4  1/8  ˆ L 6 (0,T ;V) θˆ  L 2 (0,T ;(W 1,2 )∗ ) θˆ  L 2 (0,T ;W 1,2 ) w ˆ L 6 (0,T ;V) cwˆ  L 6/5 (0,T ;V∗ ) w Γ Γ D

D

3/4 × θˆ  L 2 (0,T ;W 1,2 ) φ L 2 (0,T ;WΓ1,2 ) ΓD

≤ ≤

D

1/8 7/8 1/8 7/8 cwˆ   L 6/5 (0,T ;V∗ ) w ˆ L 6 (0,T ;V) θˆ   L 2 (0,T ;(W 1,2 )∗ ) θˆ  L 2 (0,T ;W 1,2 ) φ L 2 (0,T ;WΓ1,2 ) D Γ Γ



1/2 δw   L 6/5 (0,T ;V∗ )

+

7/6 K δ w ˆ L 6 (0,T ;V)

D

D

7/6  1/2 δθ  L 2 (0,T ;(W 1,2 )∗ ) + K δ θˆ  L 2 (0,T ;W 1,2 ) ΓD

ΓD

× φ L 2 (0,T ;WΓ1,2 ) D

2

1  1/2 7/6 δw   L 6/5 (0,T ;V∗ ) + K δ w ˆ L 6 (0,T ;V) 2

2  1/2 7/6 · φ L 2 (0,T ;WΓ1,2 ) + δθ   L 2 (0,T ;(W 1,2 )∗ ) + K δ θˆ  L 2 (0,T ;W 1,2 )

≤

ΓD

ΓD

D

  7/3 7/3 ≤ δ 2 w   L 6/5 (0,T ;V∗ ) + K δ2 w ˆ L 6 (0,T ;V) + δ 2 θ   L 2 (0,T ;(WΓ1,2 )∗ ) + K δ2 θˆ  L 2 (0,T ;W 1,2 ) D

× φ

L 2 (0,T ;WΓ1,2 ) D

ΓD

. (10.45)

10.2 Existence of a Solution

335

Also,   

 c  f, u dt  ≤ √ u L 2 (0,T ;H ) , 2 ε 1 + εθˆ 0  T α θˆ

 c 1  e2k1 t f · w, ˆ φ dt  ≤ √ (K + w L 2 (0,T ;H ) )φ L 2 (0,T ;WΓ1,2 ) . 2 D ˆ ε 1 + εθ 0 (10.46) Owing to (10.39)–(10.41), (10.44), (10.35), (10.46) and (10.45) with δ = 21 , we have from (10.24) T

e−k1 t

Aε (w, θ )V

∗

α θˆ 0

=

sup

(u,φ)V =1

 Aε (w, θ ), (u, φ)



K 1 ≤c K+ + w L 2 (0,T ;V) + w L 2 (0,T ;V) + w6 6 + θ  2 1, p L (0,T ;V) L (0,T ;WΓ ) ε ε D

1 7/3 7/3 7/3 + K w ˆ 6 + K w ˆ 6 + θ  L 2 (0,T ;L 2 (Ω)) + K θˆ  2 1,2 L (0,T ;V) L (0,T ;V) ε L (0,T ;WΓ ) D

 1  w  L 6/5 (0,T ;V∗ ) + θ   2 L (0,T ;(WΓ1,2 )∗ ) 2 D K 1 1 ≤c K+ + θ  2 + w L 2 (0,T ;V) + θ  L 2 (0,T ;L 2 (Ω)) + w6 6 1, p L (0,T ;V) L (0,T ;WΓ ) ε ε ε D

1 7/3 7/3 ∗ L (w, θ ) + w 6 + θ  2 , + V L (0,T ;V) 2 L (0,T ;WΓ1,2 ) +

D

(10.47)

which shows the property (ii) for Aε (w, θ ). (iii) Let us prove property (i) of Theorem 1.47 for Aε (w, θ ). Let {(wk , θk )} ⊂ D(L ) be a sequence such that (wk , θk )  (w, θ ) ∈ D(L ) in V , L (wk , θk )  L (w, θ ) in V ∗ , lim supAε (wk , θk ), (wk , θk ) − (w, θ ) ≤ 0.

(10.48)

k→∞

By taking a subsequence if necessary, we may assume that wk → w in L 6 (0, T ; Ls (Ω)) (1 ≤ s < 6) and a.e. in Q θk → θ in L 2 (0, T ; L s (Ω)) (1 ≤ s < 6) and a.e. in Q when k → ∞. Then, by (10.24) we have

(10.49)

336

10 The Non-steady Equations for Heat-Conducting Fluids  c min{μ0 , κ0 }

 Q



= c min{μ0 , κ0 } ≤

wk − w2V + θk − θ   Q

 WΓ1,2

dt

D

ˆ ˆ 2V + θˆk − θ wˆ k − w



WΓ1,2

dt

D

 T

2 [μ(θˆk ) + e4k1 t εwˆ k 4V ]E(wˆ k ), E(wk − w) dt 0

− +

 T  T

  2 μ(θˆk )E(w), ˆ E(wk − w) dt − 2e4k1 t εwˆ k 4V E(wˆ k ), E(wk − w) dt 0

 T  0

 κ(θˆk )∇ θˆk , ∇θk − ∇θ dt −

 ≤ Aε (wˆ k , θˆk ), (wk − w, θk − θ ) − 

+ − −

 T 0

 T 0



 T  0

0

 ˆ ∇θk − ∇θ dt κ(θˆk )∇ θ,

 T

2 μ(θˆk )E(w), ˆ E(wk − w) dt 0

2e4k1 t ε wˆ k 4V E(w), ˆ E(w − wk ) dt − e−k1 t ∇Φε (ek1 t wˆ k (t)), wk − w dt −

 T  0



κ(θˆk )∇ θˆ , ∇θk − ∇θ dt −

 T 0

 T

0  T

ek1 t rotwˆ k × wˆ k , wk − w dt

e−k1 t

0

α θˆ

0 k f, wk − w dt 2 ˆ 1 + ε θk

(k1 (wˆ k , wk − w) + 2(μ(θˆk )k(x)wˆ k , wk − w)Γ2

+ 2(μ(θˆk )S w˜ˆ k , w˜ k − w) ˜ Γ3 + 2(α(x)wˆ k , wk − w)Γ5 + (μ(θˆk )k(x)wˆ k , wk − w)Γ7 + (β(x)θˆk , θk − θ )Γ R − ek1 t wˆ k θˆk , ∇(θk − θ )

 α θˆ |E(wˆ k )|2 1 k − α2 μ(θˆk ) −2k t , θk − θ − e2k1 t f · wˆ k , θk − θ dt, 2 2 ˆ 1 e + ε|E(wˆ k )| 1 + ε θk

(10.50)

where k1 is the one in (10.37) and c is a positive constant. Now let us estimate all terms except the first one on the right hand side of (10.50). 



2 e4k1 t εwˆ k 4V E(wˆ k ), E(w − wk ) dt

T

lim sup k→∞

0



= lim sup  ≤

k→∞ T



k→∞

T

≤ 0

T

k→∞

2e4k1 t ε lim sup wˆ k 4V · E(w) ˆ 2 − lim inf E(wˆ k )2 dt k→∞

0





2 e4k1 t εwˆ k 4V E(wˆ k ), E(wˆ − wˆ k ) dt



2e4k1 t ε lim sup wˆ k 4V · lim sup E(wˆ k ), E(wˆ − wˆ k ) dt

0

≤

0

T

k→∞

2e4k1 t ε lim sup wˆ k 4V · E(w) ˆ 2 − E(w) ˆ 2 dt = 0. k→∞

Taking into account Corollary 1.1, by (10.48) we know

(10.51)

10.2 Existence of a Solution

 

T

337



2 μ(θˆk )E(w), ˆ E(wk − w) dt → 0,

0 T

  κ(θˆk )∇ θˆ , ∇θk − ∇θ dt → 0.

(10.52)

0

Also, by (10.48) and (10.49) we have that  0

T

ek1 t rotwˆ k × wˆ k , wk − w dt ≤ cwˆ k L4 (Q) ∇ wˆ k L2 (Q) wk − wL4 (Q) → 0 (10.53)

as k → ∞. Since ∇Φε is monotone, we have 

T

− 0

e−k1 t ∇Φε (ek1 t wˆ k (t)), wk − w dt  T e−2k1 t ∇Φε (ek1 t wˆ k (t)), ek1 t wˆ k − ek1 t w ˆ dt =− 0  T e−2k1 t ∇Φε (ek1 t wˆ k (t)) − ∇Φε (ek1 t w), ˆ ek1 t wˆ k − ek1 t w ˆ dt =− 0  T e−2k1 t ∇Φε (ek1 t w), ˆ ek1 t wˆ k − ek1 t w ˆ dt − 0  T e−2k1 t ∇Φε (ek1 t w), ˆ ek1 t wk − ek1 t w dt → 0 as k → ∞. ≤− 0

(10.54)

By (10.49) we have that  

T 0 T 0

α θˆ

0 k f, w − w dt → 0, k 1 + εθˆk2 α θˆ

1 k e2k1 t f · w ˆ , θ − θ dt → 0 k k 1 + εθˆk2 e−k1 t

as k → ∞. By taking a sequence if necessary, we get from (10.48) wk → w in L 6 (0, T ; Wα,2 (Ω)) (2/3 < α < 1), θk → θ

in L 2 (0, T ; W α,2 (Ω)) (2/3 < α < 1).

Then, by the trace theorem and Remark 3.4 we have

(10.55)

338

10 The Non-steady Equations for Heat-Conducting Fluids



T

(k1 (wˆ k , wk − w) + 2(μ(θˆk )k(x)wˆ k , wk − w)Γ2 + 2(μ(θˆk )S w˜ˆ k , w˜ k − w) ˜ Γ3

− 0

+ 2(α(x)wˆ k , wk − w)Γ5 + (μ(θˆk )k(x)wˆ k , wk − w)Γ7 + (β(x)θˆk , θk − θ )Γ R

 |E(wˆ k )|2 − α2 μ(θˆk ) −2k t , θ − θ dt → 0 as k → ∞. k e 1 + ε|E(wˆ k )|2 (10.56) Now let us prove that by taking subsequences there holds 

 T ˆ e wˆ k θk , ∇(θk − θ ) dt = ek1 t (wˆ k − w) ˆ θˆk , ∇(θk − θ ) dt 0 0  T  T k1 t ˆ ˆ ˆ ∇(θk − θ ) dt → 0 e w( ˆ θk − θ ), ∇(θk − θ ) dt + ek1 t wˆ θ, + T

k1 t

0

0

as k → ∞. (10.57) Since {wˆ k } is bounded in L 6 (0, T ; V), taking into account (10.48) and applying Theorem 1.39, we have that the set is relatively compact in the space L 5 (0, T H9/10 (Ω)) ⊂ L5 (Q), where imbedding H9/10 (Ω) → L5 (Ω) was used. Since {θk } is bounded in L 2 (0, T ; W 1,2 (Ω)) ∩ L ∞ (0, T ; L 2 (Ω)), by Theorem 1.33 and (1.22) it is bounded in L 10/3 (0, T ; W 3/5,2 (Ω)) ⊂ L 10/3 (Q) as well and thus by Theorem 1.39 the set {θk } is also relatively compact in L 3 (Q). Consequently, there exist the subsequences such that wˆ k → wˆ in L5 (Q), {θk } is bounded in L 10/3 (Q), θk → θ in L 3 (Q). Applying Hölder’s inequalities with exponents 5, 10/3, 2 for the first term and with exponents 6, 3, 2 for the second term of the right hand side of (10.57), we get convergence to zero of the two terms. The convergence to zero of the last term of the right hand side of (10.57) is obvious. Taking into account (10.51)–(10.57) and Assumption (10.48), we have from (10.50)    wk − w2V + θk − θ WΓ1,2 dt c min{μ0 , κ0 } lim sup k→∞

Q

D

  ≤ lim sup Aε (wˆ k , θˆk ), (wk − w, θk − θ ) ≤ 0, k→∞

which implies that

10.2 Existence of a Solution

339

wk → w in L 2 (0, T ; V), θk → θ in L 2 (0, T ; WΓ1,2 ), D ∇wk → ∇w a.e. in Q, ∇θk → ∇θ a.e. in Q.

(10.58)

When (u, φ) ∈ V , by definition of A 

 Aε (wk , θk ), (wk − u, θk − φ)  T

2 [μ(θˆk ) + e4k1 t εw ˆ 4V E(wˆ k ), E(wk − u) + ek1 t rotwˆ k × wˆ k , wk − u = 0

+ k1 (wˆ k , wk − u) + 2(μ(θˆk )k(x)wˆ k , wk − u)Γ2 + 2(μ(θˆk )S w˜ˆ k , w˜ k − u) ˜ Γ3 + 2(α(x)wˆ k , wk − u)Γ5 + (μ(θˆk )k(x)wˆ k , wk − u)Γ7 α θˆ

0 k + e−k1 t ∇Φε (ek1 t wˆ k (t)), wk − u − e−k1 t f, w − u k 1 + εθˆk2 + (κ(θˆk )∇ θˆk , ∇(θk − φ)) + (β(x)θˆk , θk − φ)Γ R − ek1 t wˆ k θˆk , ∇(θk − φ) α θˆ

 |E(wˆ k )|2 1 k 2k1 t − α2 μ(θˆk ) −2k t θ − φ − e f · w ˆ , θ − φ dt. k k k e 1 + ε|E(wˆ k )|2 1 + εθˆk2 (10.59) By Corollary 1.1 and (10.58), we have 

T

lim

k→∞ 0

  2 μ(θˆk )E(wˆ k ), E(wk − u) dt =



T

  2 μ(θˆ )E(w), ˆ E(w − u) dt.

0

(10.60) The sequence {e4k1 t wˆ k 4V εi j (wˆ k )} is bounded in L 6/5 (0, T ; L 2 (Ω)) and εi j (u) ∈ L 6 (0, T ; L 2 (Ω)), and by taking a subsequence and denoting with the same subindex, we see that  T  T

ˆ 4V E(w), 2e4k1 t wˆ k 4V E(wˆ k ), E(u) ˆ dt → 2e4k1 t w ˆ E(u) ˆ dt. lim k→∞ 0

0

(10.61)     ˆ E(w) ˆ (see Exercises 2, ˆ 4V E(w), Since lim inf k→∞ wˆ k 4V E(wˆ k ), E(wˆ k ) ≥ w (c), p. 173 of [3]), by (10.60) and (10.61) we have 

T

lim inf k→∞

0



2 [μ(θˆk )+e4k1 t εwˆ k 4V E(wˆ k ), E(wk − u) dt  T

ˆ + e4k1 t εw 2 [μ(θ) ˆ 4V ]E(w), ˆ E(w − u) dt. ≥ 0

(10.62) Since ek1 t rotwˆ k × wˆ k , wk − u = ek1 t rotwˆ k × wˆ k , −v0 − u, by (10.58) we can see that

340

10 The Non-steady Equations for Heat-Conducting Fluids



T



T

e rotwˆ k × wˆ k , wk − u dt → k1 t

0

ek1 t rotwˆ × w, ˆ w − u dt.

(10.63)

0

Also, by (10.58) we can see that  T

−

0

e−k1 t ∇Φε (ek1 t wˆ k (t)), wk − u dt → −

 T 0

e−k1 t ∇Φε (ek1 t w(t)), ˆ w − u dt.

(10.64)

Since α0 θˆk α0 θˆ α0 θˆk → a.e. in Q and is bounded in L 2 (0, T ; L 2 ), 2 2 ˆ ˆ 1 + εθk 1 + εθ 1 + εθˆk2 we have

α0 θˆk α0 θˆ  in L 2 (0, T ; L 2 ) 2 1 + εθˆk 1 + εθˆ 2

(10.65)

(see Lemma 1.2). Owing to (10.58) and (10.65), we have 

α θˆ

0 k f, w − u dt k 1 + εθˆk2 0  T  T α θˆ α θˆ

0 k 0 k −k1 t −k1 t e f, w − w dt + e f, w − u dt = k 1 + εθˆk2 1 + εθˆk2 0 0  T α θˆ

0 e−k1 t f, w − u dt. → 1 + εθˆ 2 0 (10.66) Using (10.58) and (10.65), we have T

 T 0

e−k1 t

e2k1 t

 T

α θˆ α θˆ 1 k 1 k f · wˆ k , θk − φ dt = e2k1 t f · wˆ k , θk − θ dt 2 2 1 + ε θˆk 1 + ε θˆk 0  T  T α θˆ α θˆ

1 k 1 k e2k1 t f · wˆ k − w, θ dt + e2k1 t f · w, ˆ θ − φ dt + 2 2 1 + ε θˆk 1 + ε θˆk 0 0  T

α θˆ 1 → e2k1 t f · w, ˆ θ − φ dt. 1 + ε θˆ 2 0

(10.67)

By Corollary 1.1 and (10.58) ˆ θˆ , ∇(θ − φ)). (κ(θˆk )∇ θˆk , ∇(θk − φ)) → (κ(θ)∇

(10.68)

It is easy to prove convergence of other terms in the right hand side of (10.59). Thus, by (10.60)–(10.64) and (10.66)–(10.68), we have the existence of a subsequence {(vk , θk )} such that

10.2 Existence of a Solution

341

    lim inf Aε (wk , θk ), (wk − u, θk − φ) ≥ Aε (w, θ ), (w − u, θ − φ) , k→∞

(10.69)

by which property (i) of Theorem 1.47 for Aε (w, θ ) is proved. Therefore, by Theorem 1.47 there exists a solution to Problem VEA’, which gives us the conclusion of Theorem 10.2. 

10.2.2 Estimates of Solutions to the Approximate Problem Let 0 < ε < 1. For v0 ∈ HK , θ0 ∈ L 1 (Ω) and f 2 ∈ L 1 (0, T ; (WΓ1,2 )∗ ), there exists D  1,6 1,2 ∗  1,6 2 v0ε ∈ W (Ω) ∩ K (Ω), θ0ε ∈ WΓ D (Ω) and f 2ε ∈ L 0, T ; (WΓ D ) such that v0 − v0ε  H ≤ ε, θ0 − θ0ε  L 1 (Ω) ≤ ε,  f 2 − f 2ε  L 1 (0,T ;(WΓ1,2 )∗ ) ≤ ε. D (10.70) Let (vε (t), θε (t)) be a solution to Problem VEA with the initial function (v0ε , θ0ε ) and f 2ε ∈ L 2 (0, T ; (WΓ1,2 )∗ ) instead of f 2 . D Define Aε (θ ) : V → V∗ by   Aε (θ )v, w = A(θ )v, w + 2εv4V E(v), E(w) ∀v, w ∈ V, where A(θ ) is the one in (10.16). Then, we get from (10.22) ∂v (t)

ε , u(t) + Aε (θε )vε (t) + B(vε (t), vε (t)), u(t) + ∇Φε (vε (t)), u(t) ∂t

α θ 0 f, u(t) +  f + f 1 , u(t) ∀u ∈ L 6 (0, T ; V). =− 1 + εθ 2 (10.71) By (10.28) there exists c∗ such that   ν(k(x)z, z)Γ + 2ν(S z˜ , z˜ )Γ + (α(x)z, z)Γ + ν(k(x)z, z)Γ  2 3 5 7 ≤ μ0 z2V + c∗ z2 ∀z ∈ V

(10.72)

(see Theorem 1.27). Remark 10.3 Note that c∗ depends on the shape of Γi , i = 2, 3, 7, and the norm of matrix α on Γ5 , and so for the fixed domain Ω it depends only on the norm of α. By (10.5), (10.7), (10.15) and (10.72), we have Aε (θε )z, z ≥ μ0 z2V + 2εz6V − c∗ z2 ∀z ∈ V,   |Aε (θε )z, w| ≤ c2 zV + εz5V wV ∀z, w ∈ V.

(10.73)

Putting u = vε in (10.71) and taking into account (10.73), we have from (10.71)

342

10 The Non-steady Equations for Heat-Conducting Fluids

 t   μ0 vε 2V + 2εvε 6V ds + ∇Φε (vε (s)), vε (s) ds 0 0  t  t 1 (10.74) vε 2 ds + c ( f 2∞ +  f 1 2V∗ ) ds ≤ v0ε 2 + c∗ 2 0 0  t |θε ||vε | d xds. + |α0 | f ∞

1 vε (t)2 + 2



t

Ω

0

Since Φε is convex, continuous and Fréchet differentiable, we have Φε (u) − Φε (vε (t)) ≥ ∇Φε (vε (t)), u − vε (t) ∀u ∈ V,

(10.75)

and so by Φε (0V ) = 0 0 ≤ Φε (vε (t)) ≤ ∇Φε (vε (t)), vε (t).

(10.76)

From (10.74) and (10.76) we have 

t

vε (t) + 2 2

  μ0 vε 2V + 2εvε 6V ds + 2



t

Φε (vε (t)) ds  t 2 2 2 vε (s)2 ds ≤ v0ε  + 2c f ∞ T + 2c f 1  L 2 (0,T ;V∗ ) + 2c∗ 0  t |θε ||vε | d xds ∀t ∈ [0, T ]. + |α0 | f ∞ 0

0

0

Ω

(10.77)

For fixed 0 < δ < 1, define

1 sign ξ, ξ ∈ R, Φ(ξ ) := 1 − (1 + |ξ |)δ  ξ Φ(τ ) dτ for ξ ≥ 0, Ψ (ξ ) := 0  0 − ξ Φ(τ ) dτ for ξ < 0. Then

δ ∀ξ ∈ R, (1 + |ξ |)1+δ |Φ(ξ )| ≤ 1, 0 ≤ Φ  (ξ ) ≤ δ ∀ξ ∈ R, Ψ  (ξ ) = Φ(ξ ),  1  1 − (1 + |ξ |)1−δ ∀ξ ∈ R, Ψ (ξ ) = |ξ | + 1−δ |ξ | 2(1−δ)/δ − ≤ Ψ (ξ ) ≤ |ξ | ∀ξ ∈ R. 2 1−δ Φ  (ξ ) =

(10.78)

10.2 Existence of a Solution

343

From (10.22) we get ∂θ

ε

∂t

, ϕ +(κ(θε )∇θε , ∇ϕ) + (β(x)θε , ϕ)Γ R − vε θε , ∇ϕ − α2 μ(θε ) =

α θ

  1 ε f · vε , ϕ + f 2ε , ϕ 1 + εθε2

|E(vε )|2 ,ϕ 2 1 + ε|E(vε )|

(10.79)

∀ϕ ∈ L 2 (0, T ; WΓ1,2 (Ω)). D

Since θε ∈ L 2 (0, T ; WΓ1,2 (Ω)) ∩ C([0, T ]; L 2 (Ω)), we have D (Ω), Ψ (θε (t)) ∈ WΓ1,2 (Ω) for a.e. t ∈ [0, T ], Φ(θε (t)) ∈ WΓ1,2 D D vε θε , ∇Φ(θε (t)) = −(vε · ∇θε ), Φ(θε (t))

(10.80)

= −vε , ∇Ψ (θε (t)) = div vε , Ψ (θε (t)) = 0, where v · n|Γ R = 0 (see (10.3)) and Ψ (θε ) = 0 on Γ D were used. Also    t

∂θε , Φ(θε ) ds = Ψ (θε (t)) d x − Ψ (θ0ε ) d x ∀t ∈ [0, T ], ∂t Ω 0 Ω (κ(θε )∇θε , ∇Φ(θε )) = κ(θε )|∇θε |2 Φ  (θε ) d x,

(10.81)

Ω

(β(x)θε , Φ(θε ))Γ R ≥ 0. Putting ϕ(t) = Φ(θε (t)) in (10.79) and using (10.80) and (10.81), we have 

 t κ(θε )|∇θε |2 Φ  (θε ) d x + (β(x)θε , Φ(θε ))Γ R ds 0 0 Ω Ω  t 

|E(vε )|2 α2 μ(θε ) Ψ (θ0ε ) d x + , Φ(θ ) = ε ds 1 + ε|E(vε )|2 Ω 0  t  t   α1 θε f 2ε , Φ(θε ) ds.  f · v , Φ(θ ) ds + + ε ε 2 0 1 + εθε 0 (10.82) By virtue of (10.78) and (10.81), we have from (10.82) Ψ (θε (t)) d x +

 t

 t 1 |∇θε |2 2(1−δ)/δ θε (t) L 1 (Ω) − meas Ω + δ κ0 d xds 2 1−δ (1 + |θε |)1+δ 0 Ω  t vε 2V ds ≤ θ0ε  L 1 (Ω) + α2 μ1 0  t  t |θε ||vε | d xds + c  f 2ε (WΓ1,2 )∗ ds. + |α1 | f ∞ 0

Multiplying (10.83) by

μ0 α2 μ1

Ω

0

D

(10.83) and adding the resulting formula to (10.77), we obtain

344

10 The Non-steady Equations for Heat-Conducting Fluids

 t   μ0 vε 2V + 2εvε 6V ds vε (t)2 + θε (t) L 1 (Ω) + 0  t  t |∇θε |2 + Φε (vε (t)) ds + δ d xdt 1+δ Ω (1 + |θε |) 0 0  t  t ≤ cΛδ + c∗ vε (s)2 ds + c1 (|α0 | + |α1 |) f ∞ |θε ||vε | d xds), 0

0

Ω

0 < δ < 1, (10.84) where Λδ = cδ + v0 2 + θ0  L 1 (Ω) +  f 2∞ T +  f 1 2L 2 (0,T ;V∗ ) +  f 2  L 1 (0,T ;(WΓ1,2 )∗ ) . D (10.85) By Hölder’s inequality and Young’s inequality, c1 (|α0 | + |α1 |) f ∞

 t

|θε ||vε | d xds  t

1/2  t

1/2 θε 2L 6/5 ds vε 2V ds ≤ c1 (|α0 | + |α1 |) f ∞ 0 0  t  μ0 t 2 ≤ c1 (μ0 )(|α0 | + |α1 |) f ∞ θε  L 6/5 ds + vε 2V ds. 2 0 0 (10.86) , we have By Theorem 1.12 with θ = 13 18 0

Ω

13/18

θε  L 6/5 ≤ θε  L 1

5/18

θε  L 5/2 .

(10.87)

Taking into account (10.86) and (10.87), we deduce from (10.84)  t  μ0  vε 2V + 2εvε 6V ds 2 0  t  t |∇θε |2 Φε (vε (t)) ds + δ d xdt + 1+δ 0 0 Ω (1 + |θε |)  t  t 13/9 5/9 vε (s)2 ds + c1 (μ0 )(|α0 | + |α1 |) f ∞ θε (s) 1 θε (s) 5/2 ds. ≤ cΛδ + c∗ L L

vε (t)2 + θε (t) L 1 (Ω) +

0

(10.88)

0

For 0 < δ < 1 define ηε :=

|θε | a.e. in Q. (1 + |θε |)(1+δ)/2

Then, (1 + |θε |)(1−δ)/2 ≤ 1 + ηε , |∇ηε | ≤

|∇θε | a.e. in Q. (1 + |θε |)(1+δ)/2

(10.89)

10.2 Existence of a Solution

345

Taking δ = 16 , we have from the first inequality of (10.89) 5/6

θε (s) L 5/2 ≤ c6 (1 + ηε 2L 6 ). Thus, we have from (10.88)  t  μ0  vε 2V + 2εvε 6V ds vε (t)2 + θε (t) L 1 (Ω) + 2 0  t   |∇θε |2 1 t + Φε (vε (t)) ds + d xdt 6 0 Ω (1 + |θε |)7/6 0  t vε (s)2 ds ≤ cΛ1/6 + c∗ 0  t 13/9 θε (s) L 1 (1 + ηε 2L 6 )2/3 ds. + c1 (μ0 )(|α0 | + |α1 |) f ∞

(10.90)

0

By Gronwoll’s inequality, we have from (10.90)  t

13/9 vε (t)2 ≤ cΛ1/6 + c1 (|α0 | + |α1 |) f ∞ θε (s) 1 (1 + ηε 2 6 )2/3 ds ec∗ t . L

L

0

(10.91)

It follows from (10.90) and (10.91) that  t  μ0  vε 2V + 2εvε 6V ds vε (t)2 + θε (t) L 1 (Ω) + 2 0  t   |∇θε |2 1 t + Φε (vε (t)) ds + d xdt 6 0 Ω (1 + |θε |)7/6 0 ≤ cΛ1/6 (1 + c∗ ec∗ T T )  t 13/9 θε (s) L 1 (1 + ηε (s)2L 6 )2/3 ds. + (1 + c∗ ec∗ T T )c1 (|α0 | + |α1 |) f ∞ 0

(10.92)

Put φ(t) := θε C([0,t];L 1 ) + ηε 2L 2 (0,t;L 6 ) . Taking into account (10.89) and using Hölder’s inequality with exponents 3/2, 3, we have from (10.92) φ(t) ≤ θε C([0,t];L 1 ) + c

 t 0

≤ cΛ1/6 (1 + c∗ ec∗ T T )

|∇θε |2 d xds 7/6 Ω (1 + |θε |)  2/3 13/9 t + ηε 2 2 , L (0,t;L 6 ) C[0,t];L 1 )

+ (1 + c∗ ec∗ T T )c1 (|α0 | + |α1 |) f ∞ T 1/3 θε (s)

and so

346

10 The Non-steady Equations for Heat-Conducting Fluids

 19/9 t + φ(t) ≤ K 0 + (1 + c∗ ec∗ T T )c5 (|α0 | + |α1 |) f ∞ T 1/3 t + φ(t) , (10.93) where K 0 is cΛ1/6 (1 + c∗ ec∗ T T ) + T . By (10.85), without loss of generality, we may assume that φ(0) ≤ K 0 . If (1 + c∗ ec∗ T T )c(|α0 | + |α1 |) f ∞ T 1/3 ≤

1 (2K 0 )(1−19/9) , 3

(10.94)

then by Lemma 1 of Appendix of [4] we have from (10.93) φ(t) ≤ 2cΛ1/6 (1 + c∗ ec∗ T T ) + T ∀t ∈ [0, T ].

(10.95)

By virtue of (10.92) and (10.95), using (10.94) we have vε (t)2 + θε (t) L 1 (Ω) +

 t  μ0 2

0

and



T

 vε 2V + 2εvε 6V ds+

 t

|∇θε |2 d xdt 1+δ 0 Ω (1 + |θε |) ≤ C(Λ, c∗ )

Φε (vε (t)) dt ≤ C(Λ, c∗ ),

(10.96)

(10.97)

0

where C(Λ, c∗ ) is independent of ε, Φ. Since Φ is nonnegative, by (1.36) and (10.97) we have  T

0

vε (t) − Jε (vε (t))2V dt ≤ c7 ε

(10.98)

with c7 independent   of ε. Taking u(t) = vε (t) − vε (s) , s ∈ (0, T ), in (10.71), we then have 1 dvε (t) − vε (s)2 2 dt + Aε (θε )vε (t) + B(vε (t), vε (t)) +

α0 θε f − f − f , v (t) − v (s) 1 ε ε 1 + εθε2

= ∇Φε (vε (t)), vε (s) − vε (t) ≤ Φε (vε (s)) − Φε (vε (t)) ≤ Φε (vε (s)). (10.99) Then by virtue of (10.73), we have from (10.99) 1 dvε (t) − vε (s)2 ≤Φε (vε (s)) + Aε vε (t), vε (s) + B(vε (t), vε (t)), vε (s) 2 dt

α0 θε f + f + f , v (t) − v (s) + c∗ vε (t)2 , + − 1 ε ε 1 + εθε2 (10.100) where B(vε (t), vε (t)), vε (t) = 0 was used.

10.2 Existence of a Solution

347

Let us integrate each term of (10.100) respectively first with respect to t from s to s + h and then with respect to s from 0 to T , where vε (t) = 0 when t ∈ (T, T + h). 

T



0

dvε (t) − vε (s)2 dtds = dt

s+h s



T

vε (s + h) − vε (s)2 ds.

(10.101)

0

By (10.97) 



T 0

s+h



T

Φε (vε (s)) dtds ≤ h

s

Φε (vε (s)) ds ≤ c6 h.

(10.102)

0

By (10.96) and (10.73), we have  T  s+h 0

s

≤c ≤c

Aε (θ )vε (t), vε (s) dtds

 T 0

 T

vε (s)V

 s+h  s



vε (t)V + εvε (t)5V dt ds

 T √ vε (s)V ( hvε  L 2 (0,T ;V) ) ds + c vε (s)V εh 1/6 vε 5 6

0 ≤ c8 h 1/6 ,



L (0,T ;V)

0

ds

(10.103) 1

1

where c8 is independent of ε. Since wL3 ≤ K wL2 2 wL2 6 , |B(v, w), z| = |rot v × w, z| ≤ K rot vL2 wL3 zL6 1

1

≤ K vV w 2 wV2 zV ,

(10.104)

and so by (10.96) we have 

T 0



s+h

B(vε (t), vε (t))vε (s) dtds

s



T

≤K 

0

s+h s

T

≤K



0

3

1

vε (t)V2 vε (t) 2 vε (s)V dtds

vε (s)V



s+h s

vε (t)2V dt

43 

s+h

vε (t)2 dt

41

ds

s

1

≤ c9 h 4 , (10.105) where vε (t)2 ≤ C(Λ, c∗ ) (see (10.96)) was used. Also, by (10.96) we have

348

10 The Non-steady Equations for Heat-Conducting Fluids

 T  s+h 0

s

0

s

 T  s+h

( f + f 1 )(t), vε (t) dtds ≤

 T 0

( f + f 1 )(t), −vε (s) dtds ≤ K

 |( f + f 1 )(t), vε (t)|  T 0

vε (s)V

 s+h s

√ ≤ c9 h.

 t t−h

 ds dt ≤ K h,

( f + f 1 )(t)V∗ dtds

(10.106)

In the same way we get  0

T



s+h

s

α θ

√ 0 ε f, v (t) − v (s) dtds ≤ K h + c9 h, ε ε 2 1 + εθε 

T 0



s+h

(10.107)

c∗ vε (t)2 dtds ≤ K h.

(10.108)

s

By virtue of (10.100)–(10.108), uniformly with respect to ε 

T

1

vε (s + h) − vε (s)2 ds ≤ O(h 6 ),

(10.109)

0

which implies that the set {vε } is relatively compact in L 2 (0, T ; W 10 ,2 (Ω)) (see Theorem 1.38). Therefore, by (10.96), there exists v and a subsequence {vεk } such that as εk → 0, 9

∗

vεk  v in L ∞ (0, T ; H ), vεk → v in L 2 (0, T ; W 10 ,2 (Ω)), 9

(10.110)

vεk  v in L 2 (0, T ; V). We will show that θε is relative compact in L r (0, T ; W α,r ) ∀r (1 < r < 5/4), ∀α (0 < α < 1). To this end, we need to get an estimate on ∇θε . Let 1 ≤ r < 3/2. By 1 1 (10.96), inequalities |a + b| p ≤ 2 p (|a| p + |b| p ), |a| + |b| ≤ (|a| p + |b| p ) p , p ∈ 2 2 (1, ∞), and Hölder’s inequality with exponents r , 2−r , we have 

 Q

r/2  (2−r )/2  |∇θε |2 r (1+δ)/(2−r ) d xdt 1 + |θ d xdt |) ε 1+δ Q (1 + |θε |) Q  

(2−r )/2 , ≤ 2r (1+δ)/2 C(Λ, c∗ )r/2 (mes Q)(2−r )/2 + |θε |r (1+δ)/(2−r ) d xdt

|∇θε |r d xdt ≤

Q

(10.111) where C(Λ, c∗ ) is the one in (10.96). To use the property W 1,r (Ω) ⊂ L q (Ω), let us take q such that 3r . Let us take δ0 > 0 such that q = 3−r 1 < r (1 + δ0 )/(2 − r ) < q,

1 r

−

1 3

= q1 , i.e.

10.2 Existence of a Solution

349

which holds if 0 < δ0 < (3 − 2r )/(3 − r ). Set s = r (1 + δ0 )/(2 − r ) and take λ such that 1 λ 1−λ = + . s 1 q Then, λ =

q−s , s(q−1)

0 < λ < 1 and by Theorem 1.12

1−λ λ η L s ≤ cηλL 1 η1−λ ∀η ∈ WΓ1.rD . L q ≤ cη L 1 ∇ηLr

(10.112)

Therefore, by (10.96) and (10.112) we have 

|θε |r (1+δ0 )/(2−r ) d xdt ≤ cC(Λ, c∗ )λs



T

 Ω

0

Q

|∇θε |r d x

(1−λ)s/r dt.

We will fix r and δ0 so that 1≤r