Mathematical Analysis of the Navier-Stokes Equations -- Cetraro, Italy 2017 [1 ed.] 9783030362256

Table of contents :
Preface......Page 6
Contents......Page 7
1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations......Page 8
1.1 Basics on Distributions, Fourier Transforms and Sobolev Spaces......Page 10
1.2 The Vector Valued Setting......Page 18
1.3 Cauchy Problems and Semigroups......Page 24
1.4 Sectorial Operators and Bounded H∞-Functional Calculus......Page 32
1.5 Fractional Powers......Page 40
1.6 Operator-Valued H∞-Calculus, R-Boundedness, Fourier Multipliers and Maximal Lp-Regularity......Page 43
1.7 Quasilinear Evolution Equations......Page 55
1.8 Balance Laws......Page 61
1.9 The Stokes Equation in a Half Space......Page 64
1.10 The Stokes and Hydrostatic Stokes Equations on Domains......Page 78
1.11 Nonlinear Stability of Ekman Boundary Layers......Page 88
1.12 Fluid-Rigid Body Interaction Problems for Compressible Fluids......Page 100
1.13 Two-Phase Free Boundary Value Problems for a Class of Non-Newtonian Fluids......Page 106
1.14 Nematic Liquid Crystals......Page 118
1.15 Global Strong Well-Posedness of the Primitive Equations......Page 127
1.16 Justification of the Hydrostatic Approximation......Page 132
1.17 Notes......Page 139
References......Page 147
2.1 Introduction......Page 154
2.1.1 Notation and Preliminaries......Page 155
2.2.1 Weak Solutions......Page 157
2.2.2 Strong Solutions......Page 159
2.2.3 Weak-Strong Uniqueness......Page 163
2.2.4 Regular and Singular Times......Page 164
2.2.5.1 Dimensions......Page 166
2.2.5.2 Dimension of the Set of Singular Times......Page 168
2.3 Serrin's Local Regularity Result for uL5+(Q*)......Page 170
2.3.1 Step 1: Show that ωL∞(Qs*), s

Citation preview

Lecture Notes in Mathematics 2254 CIME Foundation Subseries

Matthias Hieber James C. Robinson   Yoshihiro Shibata

Mathematical Analysis of the Navier-Stokes Equations Cetraro, Italy 2017 Giovanni P. Galdi · Yoshihiro Shibata Editors

Lecture Notes in Mathematics

2254

Editors-in-Chief: Jean-Michel Morel, Cachan Bernard Teissier, Paris Advisory Editors: Karin Baur, Leeds Michel Brion, Grenoble Camillo De Lellis, Princeton Alessio Figalli, Zurich Annette Huber, Freiburg Davar Khoshnevisan, Salt Lake City Ioannis Kontoyiannis, Cambridge Angela Kunoth, Cologne Ariane Mézard, Paris Mark Podolskij, Aarhus Sylvia Serfaty, New York Gabriele Vezzosi, Florence Anna Wienhard, Heidelberg

More information about this subseries at http://www.springer.com/series/3114

Matthias Hieber • James C. Robinson • Yoshihiro Shibata

Mathematical Analysis of the Navier-Stokes Equations Cetraro, Italy 2017 Giovanni P. Galdi • Yoshihiro Shibata Editors

Authors Matthias Hieber Department (FB) of Mathematics Technische Universit¨at Darmstadt Darmstadt, Germany

James C. Robinson Mathematics Institute University of Warwick Coventry, UK

Yoshihiro Shibata Department of Mathematics Waseda University Tokyo, Japan

Editors Giovanni P. Galdi MEMS Department University of Pittsburgh Pittsburgh, PA, USA

Yoshihiro Shibata Department of Mathematics Waseda University Tokyo, Japan

ISSN 0075-8434 ISSN 1617-9692 (electronic) Lecture Notes in Mathematics C.I.M.E. Foundation Subseries ISBN 978-3-030-36225-6 ISBN 978-3-030-36226-3 (eBook) https://doi.org/10.1007/978-3-030-36226-3 Mathematics Subject Classification (2010): Primary: 35Q30, 76D05, 35Q35; Secondary: 65Mxx, 65Nxx © Springer Nature Switzerland AG 2020 This work is subject to copyright. All rights are reserved 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 Springer imprint is published by the registered company Springer Nature Switzerland AG. The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

As is well known, the Navier–Stokes equations constitute one of the most active and attractive areas of research, from both theoretical and applied viewpoints. In particular, especially over the last two decades, they have been the focus of a number of fundamental mathematical contributions from different perspectives. Probably, this rapid growth may be also due to the circumstance that, since the year 2000, the question of existence of global, regular solutions corresponding to initial data of unrestricted size has been declared as one of the main open Millennium Problems by the Clay Mathematical Institute. Yet, in spite of all efforts, to date, the involved mathematicians unanimously agree that very little is still known about these equations and that their deep secrets are still far from being uncovered. The principal objective of the CIME school on “Mathematical Analysis of the Navier–Stokes Equations: Foundations and Overview of Basic Open Problems,” in Cetraro, September 4–8 2017, was to provide series of lectures devoted to several fundamental and diverse aspects of the Navier–Stokes equations. The present volume collects some of them, including boundary layers, fluid–solid interactions, free surface and complex fluid problems (Professor Matthias Hieber, TU Darmstadt, Germany); questions of existence, uniqueness, and regularity (Professor James C. Robinson, University of Warwick, UK); and local and global well-posedness and asymptotic behavior for free boundary problems (Professor Yoshihiro Shibata, Waseda University, Japan). It is our distinct pleasure to thank all lecturers and participants for their enthusiastic—scientific and social—contribution to the success of the school, the Fondazione CIME and its scientific committee for giving us the opportunity to organize this event, and all the staff for their invaluable help. Pittsburgh, PA, USA Tokyo, Japan

Giovanni P. Galdi Yoshihiro Shibata

v

Contents

1

Analysis of Viscous Fluid Flows: An Approach by Evolution Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . Matthias Hieber

1

2 Partial Regularity for the 3D Navier–Stokes Equations . . . . . . . . . . . . . . . . . 147 James C. Robinson 3 R Boundedness, Maximal Regularity and Free Boundary Problems for the Navier Stokes Equations . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 193 Yoshihiro Shibata

vii

Chapter 1

Analysis of Viscous Fluid Flows: An Approach by Evolution Equations Matthias Hieber

Preface This course of lectures discusses various aspects of viscous fluid flows ranging from boundary layers and fluid structure interaction problems over free boundary value problems and liquid crystal flow to the primitive equations of geophysical flows. We will be mainly interested in strong solutions to the underlying equations and choose as mathematical tool for our investigations the theory of evolution equations. The models considered are mainly represented by semi- or quasilinear parabolic equations and from a modern point of view it is hence natural to investigate the underlying equations by means of the maximal Lp -regularity approach. For this reason, we start these lectures by an introduction to Cauchy problems and sectorial operators. The latter are the starting point for the functional calculus of bounded, holomorphic functions, which will be then extended to the operatorvalued H ∞ -calculus. The extension leads us to the notion of R-bounded families of operators, which will be the key for boundedness results of the H ∞ -calculus in this setting. It implies the Kalton–Weis theorem on the closedness of the sum of two commuting, sectorial operators and via the extension of Mikhlin’s theorem to Banach spaces having the UMD-property, also the characterization theorem of maximal Lp -regularity for parabolic evolution equations in terms of R-boundedness of its resolvent. We then proceed with quasilinear parabolic evolution equations and present three important results for these equations: local well-posedness, the generalized principle of linearized stability implying under suitable assumptions the global existence of strong solutions for data close to an equilibrium point and on the existence of global strong solutions in the presence of compact embeddings and strict Lyapunov functionals.

M. Hieber () Department (FB) of Mathematics, TU Darmstadt, Darmstadt, Germany e-mail: [email protected] © Springer Nature Switzerland AG 2020 G. P. Galdi, Y. Shibata (eds.), Mathematical Analysis of the Navier-Stokes Equations, Lecture Notes in Mathematics 2254, https://doi.org/10.1007/978-3-030-36226-3_1

1

2

M. Hieber

Coming back to the main aims to these notes, well-posedness results for viscous fluid flows, we start the second part with a discussion of balance laws for general heat conducting fluids and deduce from there the fundamental equations of viscous fluid flows, the incompressible and compressible Navier–Stokes equations. We continue with the analysis of the Stokes equation in a half space Rn+ within the Lp -setting. The associated operator, the negative Stokes operator, is shown to be a sectorial operator on Lp (Rn+ ). As a consequence of the results in Part A we obtain the property of maximal Lp –Lq -regularity for the Stokes equation on Rn+ . A localization procedure yields then the corresponding regularity results for the Stokes equations on standard domains. We then consider stability questions for Ekman boundary layers. The latter are explicit stationary solutions of the Navier–Stokes equations in the rotational framework. We then show that the Ekman layer is asymptotic stable provided the Reynolds number involved is small enough. We continue with moving and free boundary value problems. In fact, a fluid-rigid body interaction problem will be discussed for the situation of compressible fluids. Maximal regularity of the linearized equation in Lagragian coordinates allows us to prove a local wellposedness result for strong solutions. Strong solutions for the two-phase problem for generalized Newtonian fluids are again obtained by a fixed point argument in the associated space of maximal regularity. We finally study also the primitive equations, which are a model for oceanic and atmospheric flows and are derived from the Navier–Stokes equations by assuming a hydrostatic balance for the pressure term. We show that these equations are globally strongly well-posedness for arbitrary large initial data lying in critical spaces. Finally, we show that the primitive equations may be obtained as the limit of anisotropically scaled Navier– Stokes equations. The approach to the results concerning the primitive equations are based again on maximal Lp -regularity estimates, this time for the hydrostatic Stokes operator. My sincere thanks go to the Fondazione CIME for their very kind support and hospitality during a Summer Course on ‘Mathematical Analysis of the Navier– Stokes Equations: Foundations and Overview of Basic Open Problems’ held at Cetraro in September 2017, in which these lectures notes were developed.

Part A: Parabolic Evolution Equations

This part of these notes mainly concerns the theory of evolution equations. Aiming for applications to viscous fluid flows we are mainly interested in parabolic evolution equations. We start by introducing the basic concept of distributions, the Schwartz space S and its dual space S  , the Fourier transform on S and S  and discuss also briefly various types of function spaces and their basic properties. Furthermore, we define Fourier multipliers for Lp (Rn ). The celebrated theorem due to Mikhlin on Lp -

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

3

boundedness of translations invariant operators on Lp (Rn ) for 1 < p < ∞ as well as its analogue in the period setting will be used later on in many occasions. This theorem allows us further to introduce the Hilbert transform as well as the Riesz transforms as bounded operators on Lp (R) and Lp (Rn ), respectively. In Sect. 1.2 we generalize this concept to the vector-valued setting, i.e. given a Banach space X, we discuss X-valued distributions, the Bochner integral and basic theorems for singular integral operators with operator-valued kernels. Banach spaces having the UMD-property will be definded as spaces for which the Hilbert transform acts as a bounded operator on Lp (R; X) for some p ∈ (1, ∞). In the sequel, we consider in Sect. 1.3 semigroups of operators and their generators. The classical theorems due to Hille-Yosida and Lumer-Philipps will be proven as well as the characterization theorem and smoothing properties for holomorphic semigroups. In Sect. 1.4 we introduce sectorial operators. They are the starting point for the functional calculus of bounded, holomorphic functions, which, including important examples, will be investigated in this section. An extendend H ∞ -calculus allows us to define fractional powers of sectorial operators and their properties in an elegant way. Section 1.6 deals with the operator-valued H ∞ -calculus. The extension of the scalar-valued H ∞ -calculus to the X-valued functions leads us to the notion of Rbounded families of operators. This notion will be the key for the boundedness result of the H ∞ -calculus in this setting. It implies the Kalton–Weis theorem on the closedness of the sum of two commuting, sectorial operators and via the extension of Mikhlin’s theorem to Banach spaces having the UMD-property, also the characterization theorem of maximal Lp -regularity for parabolic evolution equations in terms of R-boundedness of its resolvent. The final section of this first part discusses quasilinear parabolic evolution equations. Here we present the approach based on the theory of maximal Lp regularity. This property will be the key property, when presenting three important results for these equations: local well-posedness, the generalized principle of linearized stability implying under suitable assuptions the global existence of strong solutions for data close to an equilibrium point and on the existence of global strong solutions in the presence of compact embeddings and strict Lyapunov functionals.

1.1 Basics on Distributions, Fourier Transforms and Sobolev Spaces In this section we collect basic facts on distributions, the Fourier transforms and function spaces and also introduce the notation being used later on. Distributions and Fourier Transforms Let us begin with the notion of a multiindex. A multiindex α = (α1 , . . . , αn ) ∈ Nn0 is an ordered n-tuple of nonnegative integers and we denote by |α| = α1 + . . .+ αn the

4

M. Hieber

order of α. For x = (x1 , . . . , xn ) ∈ Rn and a multiindex α, we set x α = x1α1 · · · xnαn . For a multiindex α, ∂ α f denotes the derivative ∂1α1 · · · ∂nαn f of some function f on Rn . We denote by D(Rn ) (or by Cc∞ (Rn )) the space of all complex-valued ∞ C -functions on Rn with compact support and by S(Rn ) the Schwartz space of all smooth, rapidly decreasing functions on Rn , i.e. S(Rn ) := {ϕ ∈ C ∞ (Rn ) : ϕm,α < ∞ for all m ∈ N0 , α ∈ Nn0 }, where ϕm,α := sup (1 + |x|m )|∂ α ϕ(x)|). x∈Rn

The family of all seminorms  · m,α defines a topology and S(Rn ) equipped with this topology becomes a Fréchet space. The space Cc∞ (Rn ) is a dense subspace of S(Rn ). We call D(Rn ) the space of all distributions, i.e. linear maps f : ϕ →< ϕ, f > of D(Rn ) into C such that for each compact set K ⊂ Rn there exist m ∈ N and a constant C > 0 such that | < ϕ, f > | ≤ C sup sup |∂ α ϕ(x)| |α|≤m x∈Rn

for all ϕ ∈ D(Rn ) with supp ϕ ⊂ K. We denote by S(Rn ) the space of all temperate distributions, i.e. continuous linear maps from S(Rn ) into C. Observe that S(Rn ) is then embedded in a natural way into D(Rn ). We equipp D(Rn ) with the topology arising from the duality with D(Rn ), i.e. a net (fj ) of distributions converges to 0 in D(Rn ) if and only if < ϕ, fj >→ 0 for all ϕ ∈ D(Rn ). It is well known that any locally integrable function f : Rn → C may be identified with a distribution and that any function f belonging to Lp (Rn ) for 1 ≤ p ≤ ∞ or the constant function 1 are temperate distributions. In the following, we describe in which way differentiation, Fourier transform and convolution can be extended from functions to distributions. To this end, let g : Rn → C be a C ∞ -function. Then, obviously, ϕ · g ∈ D(Rn ) for ϕ ∈ D(Rn ). The derivatives ∂j f (j = 1, . . . , n) of a distribution f ∈ D(Rn ) are defined in D(Rn ) by < ϕ, ∂j f >:= − < ∂j ϕ, f >,

ϕ ∈ D(Rn ).

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

5

Note also that ∂j maps S(Rn ) into itself and that this notation is consistent when differentiable functions are identified with distributions. Integration by parts shows for higher order derivatives we have < ϕ, ∂ α f >:= (−1)|α| < ∂ α ϕ, f >,

ϕ ∈ D(Rn ).

Given functions f and g, the convolution f ∗ g of f with g is defined by  (f ∗ g)(x) :=

Rn

f (x − y)g(y)dy

whenever the integral exists. Note that ψ ∗ ϕ ∈ S(Rn ) and that the map ψ → ψ ∗ ϕ is continuous provided ϕ, ψ ∈ S(Rn ). Hence, the convolution ϕ ∗ f of ϕ ∈ S(Rn ) with a tempered distribution f ∈ S(Rn ) can be defined as < ψ, ϕ ∗ f >:=< ψ ∗ ϕ, ˇ f >,

ψ ∈ S(Rn ),

where ϕ(x) ˇ := ϕ(−x). In this case, ϕ ∗ f ∈ S(Rn ) . Note that ϕ ∈ S(Rn ), f ∈ S(Rn )

∂ α (ϕ ∗ f ) = (∂ α ϕ) ∗ f,

for all α. We now consider the Fourier transform and begin with its classical definition in L1 (Rn ). Definition 1.1.1 For f ∈ L1 (Rn ), the Fourier transform F f of f is defined by fˆ(ξ ) := (F f )(ξ ) := where x · ξ :=

 Rn

e−ix·ξ f (x)dx,

ξ ∈ Rn ,

n

j =1 xj ξj .

The Fourier inversion theorem, see e.g. [69], says that the Fourier transform F is a linear and topological isomorphism of S(Rn ) and that F −1 is given by (F −1 ϕ)(ξ ) = (2π)−n (F ϕ)(−ξ ),

ϕ ∈ S(Rn ), ξ ∈ Rn .

The Fourier transform hence induces an isomorphism of S(Rn ) by < ϕ, F f >:=< F ϕ, f >, which is also denoted by F .

ϕ ∈ S(Rn ), f ∈ S(Rn ) ,

(1.1.1)

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M. Hieber

The following properties of the Fourier transform, which are elementary for functions, extend due to the above setting to distributions. Lemma 1.1.2 For ϕ ∈ S(Rn ) and f ∈ S(Rn ) the following assertions hold. ˇ f > F −1 f = (2π)−n (F f )ˇ = (2π)−n F fˇ, where < ϕ, fˇ >:=< ϕ, F ∂ α f = (iξ )α F f, F (ϕ ∗ f ) = (F ϕ) · (F f ). The following Plancherel’s theorem is a very classical result on Fourier transforms. Proposition 1.1.3 (Plancherel) Let ϕ, ψ ∈ S(Rn ). Then < F ϕ, F ψ >= (2π)n < ϕ, ψ > and the Fourier transform F extends to a bounded linear operator on L2 (Rn ) such that (2π)−n/2 F is unitary. Note that ψ denotes the complex conjugate of ψ ∈ S(Rn ). Fourier Multipliers We are now considering the concept of so-called Fourier multipliers. Assume that m ∈ L∞ (Rn ) takes values in C. For ϕ ∈ S(Rn ), define m ϕ ∈ S(Rn ) by < ψ, m ϕ >:=< m,  ϕ · ψ >. We then look for conditions on the function m such that the mapping ϕ → (m ϕ )ˇ,

ϕ ∈ S(Rn )

becomes continuous on Lp (Rn ) for 1 ≤ p ≤ ∞. Definition 1.1.4 Let 1 ≤ p ≤ ∞. A function m ∈ L∞ (Rn ) is called a Fourier multiplier for Lp (Rn ) if F −1 (m ϕ ) ∈ Lp (Rn ) for ϕ ∈ S(Rn ) and there exists a constant C > 0 such that F −1 (m ϕ )p ≤ Cϕp ,

ϕ ∈ S(Rn ).

Given p ∈ [1, ∞) and a Fourier multiplier m for Lp (Rn ), the map ϕ → extends to a bounded, linear operator

F −1 (m ϕ)

Tm : f → F −1 (mf) on Lp (Rn ). If p = ∞, the extension is weak∗ -continuous. The space consisting of all Fourier multipliers for Lp (Rn ) is denoted by Mp (Rn ). Equipped with the norm mMp (Rn ) := Tm L(Lp (Rn )) ,

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

7

Mp (Rn ) is a Banach space. It follows from Plancherel’s theorem that M2 (Rn ) = L∞ (Rn ). A very useful sufficient conditions for a given function m to belong to Mp (Rn ) for 1 < p < ∞ is given by the Mikhlin multiplier theorem. In order to formulate this result, let j = min{k ∈ N, k > n2 } and consider the Banach space MM := {m : Rn → C; m ∈ C j (Rn \ {0}), |m|M < ∞}, where the norm | · |M is defined by |m|M := max

sup

|α|≤j ξ ∈Rn \{0}

|ξ |α |∂ α m(ξ )|.

Then the following holds. Theorem 1.1.5 (Mikhlin) Let 1 < p < ∞. Then MM → Mp (Rn ). For proofs and generalizations of the results stated above and for more information on this topic we refer to [11, 69, 139, 140, 148]. Consider one instance of Mikhlin’s theorem for n = 1. For ε > 0, we define (1/t)ε ∈ L1loc (R) by (1/t)ε (τ ) := τ −1 χ[|τ |≥ε] (τ ),

τ ∈ R,

so that  < (1/t)ε , ϕ >=

|τ |≥ε

ϕ(τ ) dτ τ

for ϕ ∈ S(R).

Then there exists a unique temperate distribution pv(1/t), called the principal value of 1/t, such that  < pv(1/t), ϕ >:= lim

ε→0 |τ |≥ε

ϕ(τ ) dτ, τ

ϕ ∈ S(R).

For u ∈ S(R) we define the Hilbert transform H u of u by H u :=

1 pv(1/t) ∗ u, π

u ∈ S(R).

The symbol m of the translation invariant operator H is given by (pv(1/t))ˆ = −iπsign. Then sign ∈ MM and by Mikhlin’s theorem, sign ∈ Mp (R) for 1 < p < ∞. We thus have the following result.

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M. Hieber

Proposition 1.1.6 Let 1 < p < ∞. Then the Hilbert transform is a bounded linear operator on Lp (R). The Riesz transforms are n-dimensional analogues of the Hilbert transform, with properties analogous to those of the Hilbert transform on R. More precisely, for j = 1, . . . , n consider the functions mj (ξ ) := −i

ξj , |ξ |

ξ ∈ Rn \{0},

and define the j -th Riesz transform Rj by ϕ ), Rj ϕ := F −1 (mj 

ϕ ∈ S(Rn ).

Since mj ∈ MM for all j = 1, . . . , n, we have the following result. Proposition 1.1.7 Let 1 < p < ∞ and j = 1, . . . , n. Then the Riesz transforms Rj are bounded, linear operators on Lp (Rn ). It is useful to note that the Riesz transforms satisfy n 

Rj2 = −I d.

j =1

We will also consider Fourier multipliers for the space Lp ((−π, π)n ) for 1 < p < ∞. We say that a sequence (ak )k∈Zn ⊂ C is said to be a Fourier multiplier on Lp ((−π, π)n ) if there exists a constant C > 0 such that        i  ak ck ei  p ≤ C c e (1.1.2)  p   k n n k∈Zn

L ((−π,π) )

k∈Zn

L ((−π,π) )

for any sequence (ck )k∈Zn with ck = 0 for only finitely many k ∈ Zn . Given a Fourier multiplier (ak )k∈Zn for Lp ((−π, π)n ), the mapping  k∈Zn

ck ei →



ak ck ei

k∈Zn

extends uniquely to a bounded operator Ta on Lp ((−π, π)n ) with norm Ta  being the greatest lower bound of the set of all constants such that (1.1.2) holds for any finite sequence (ck )k∈Zn ⊂ C. A classical result due to Marcinkiewicz gives a sufficient criteria for a sequence to be a Fourier multiplier for Lp ((−π, π)n ) for 1 < p < ∞. In order to formulate this result, let Dν = Iν1 × . . . × Iν n for ν ∈ Nn0 and denote by (Dν )ν∈Nn0 be a dyadic decomposition of Zn , where I0 = {0} and Ij = {m ∈ Z : 2j −1 ≤ |m| ≤ 2j } for j ∈ N.

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

9

Theorem 1.1.8 (Marcinkiewicz) Let 1 < p < ∞ and (ak )k∈Zn ⊂ C be a sequence satisfying sup varDν a < ∞.

ν∈Nn0

Then the sequence (ak )k∈Zn is a Fourier multiplier for Lp ((−π, π)n ) and there exists a constant C = C(n, p) > 0 such that Ta L(Lp ((−π,π)n )) ≤ C sup varDν a. ν∈Nn0

The following Corollary of Theorem 1.1.8 will be used in Sect. 1.16. Corollary 1.1.9 Let 1 < p < ∞, n ∈ N and (ak )k∈Zn ⊂ C be a sequence such that ak = m(k) for all k ∈ Zn \ {0} and some function m ∈ C n (Rn \ {0}). Suppose that [m] :=

sup sup |ξ γ D γ m(ξ )| < ∞.

γ ∈{0,1}n ξ =0

Then the sequence (ak )k∈Zn is a Fourier multiplier for Lp ((−π, π)n ) and there exists a constant C = C(n, p) > 0 such that Ta L(Lp ((−π,π)n )) ≤ C max{[m], |a0|}. Function Spaces on Domains The concept of distributions described above can be extended to the case of distributions on an open subset  of Rn . We denote by D() := Cc∞ () the space of test functions on , i.e. C ∞ -functions of compact support in . The space D() of distributions on  is defined to be the space of all linear functionals f on D() such that for each compact set K ⊂  there exists m ∈ N and a constant C > 0 such that | < ϕ, f > | ≤ C sup sup |∂ α ϕ(x)| |α|≤m x∈

for all ϕ ∈ D() with suppϕ ⊂ K. Again, locally integrable functions on  can be identified with distributions and the derivatives ∂j of a distribution f are defined by < ϕ, ∂j f >:= − < ∂j ϕ, f >,

ϕ ∈ D().

The Sobolev spaces W m,p () are defined for m ∈ N and 1 ≤ p ≤ ∞ by W m,p () := {f ∈ Lp () : ∂ α f ∈ Lp () for all α ∈ Nn0 with |α| ≤ m},

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M. Hieber

where ∂ α f is understood in the sense of distributions. Equipped with the norm f W m,p () := (



p

∂ α f p )1/p ,

|α|≤m

the space W m,p () becomes a Banach space. Moreover, the closure of D() in m,p W m,p () is denoted by W0 (), i.e. m,p

W0

() := Cc∞ ()

·W m,p

.

For p = 2, one often uses the notation H m () := W m,2 () and H0m () := W0m,2 (). Then, equipped with the equivalent norm f H m () := (



∂ α f 22 )1/2 ,

|α|≤m

H m () is a Hilbert space with the inner product (f |g)H m () =

 

∂ α f ∂ α gdx.

|α|≤m 

By Plancherel’s Theorem 1.1.3 and Lemma 1.1.2, H m (Rn ) = {f ∈ L2 (Rn ) : ξ α F f ∈ L2 (Rn ) for all multiindices α with |α| ≤ m}, and H m () coincides also with the space consisting of all functions in L2 (Rn ) such that ξ → (1 + |ξ |2 )m/2 F f (ξ ) ∈ L2 (Rn ). The analysis of viscous fluid flow described in Part B relies on various types of function spaces. In particular, we mention here (a) (b) (c) (d) (e)

 m,p () for m ∈ N0 and 1 ≤ p ≤ ∞, the homogenous Sobolev spaces W s,p the Bessel-Potential spaces H () for s ∈ R and 1 < p < ∞, s () for s ∈ R and 1 ≤ p, q ≤ ∞, the Besov spaces Bp,q s () for s ∈ R and 1 ≤ p, q ≤ ∞, the Triebel-Lizorkin spaces Fp,q the real interpolation spaces   Wps () = W m,p (), W m+1,p () θ,p

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

11

for m ∈ N0 , s ∈ (m, m + 1), θ = s − [s] and 1 < p < ∞, often called Sobolev-Slobodeckij spaces, (f) and the complex interpolation spaces

H s,p () = W m,p (), W m+1,p () θ for 1 ≤ p ≤ ∞, 0 < θ < 1, s = (1 − θ )m + θ (m + 1) = m + θ , defined on open sets  ⊂ Rn . We do not give precise definitions of these spaces here and refer the reader e.g. to the a classical monograph by Triebel [149] or the recent one by Amann [12].

1.2 The Vector Valued Setting As before, we start this section by introducing basic facts on distributions and Fourier transforms, now however in the vector-valued setting. Banach Space Valued Distributions We note that the proofs of the assertions listed below on distributions are only straightforward modifications of the ones for the scalar case described in Sect. 1.1. Let X be a Banach space. We denote by S(Rn ; X) the Schwartz space of smooth rapidly decreasing X-valued functions on Rn . Then d

d

D(Rn ; X) → S(Rn ; X) → E(Rn ; X), where D(Rn ; X) is the space of all X–valued C ∞ -functions on Rn with compact supports, as usual equipped with the inductive limit topology, and E(Rn ; X) := C ∞ (Rn ; X) equipped with the bounded convergence topology. The space S  (Rn ; X) of X–valued temperate distributions is defined by S  (X) := L(S(Rn ); X); the spaces E  (Rn ; X) and D (Rn ; X) are defined analogously. Then E  (Rn ; X) → S  (Rn ; X) → D (Rn ; X). Finally, the space OM (Rn ; X) of all X–valued slowly increasing smooth functions on Rn consists of all u ∈ E(Rn ; X) such that, given α ∈ Nn0 , there exists mα ∈ N and Cα > 0 such that ∂ α u(x) ≤ Cα (1 + |x|2)mα ,

x ∈ Rn .

Observe that S(Rn ; X) → OM (Rn ; X) → S  (Rn ; X). We also write S(Rn ) := S(Rn ; K), S  (Rn ) := S  (Rn ; K) and OM (Rn ) := OM (Rn ; K), where K ∈ {C, R}. We next turn our attention to the Bochner integral. To this end, given an interval 2 I in R, bounded or unbounded, or a rectangle In in R , we call a function g : I → X a simple function if it is of the form g(t) = j =1 xj χSj (t) for some n ∈ N, xj ∈ X

12

M. Hieber

and Lebesgue measurable sets Sj ⊂ I with finite Lebesgue measure μ(Sj ); g is called a step function if Sj can be chosen to be an interval. A function f : I → X is called measurable if there exists a sequence of simple functions gn such that f (t) = lim n→∞ gn (t) for almost all t ∈ I . For a simple function g : I → X of the form g = nj=1 xj χSj we define  g(t)dt := I

n 

xj μ(Sj ).

j =1

A function f : I → X is then called Bochner integrable if there exist simple functions gn such that gn → f pointwise almost everywhere and  limn→∞ I f (t) − gn (t)dt = 0. In this case, the Bochner integral of f on I is given by 

 f (t)dt := lim

n→∞ I

I

gn (t)dt.

The class of Bochner integrable functions can be characterized as follows. Proposition 1.2.1 (Bochner) A function f : I → X is Bochner integrable if and only if f is measurable and f  is integrable. If f is Bochner integrable, then       f (t)dt  ≤ f (t)dt. I

I

For 1 ≤ p < ∞, we denote by Lp (I ; X) the space of all measurable functions f : I → X such that f p :=



f (t)p dt

1/p

< ∞.

I

Furthermore, we set L∞ (I ; X) to be the space of all measurable functions f : I → X such that f ∞ := ess sup f (t) < ∞. t ∈I

With the usual identifications, Lp (I ; X) becomes a Banach space for all 1 ≤ p ≤ ∞. We note that the above theory of integration works in the same way, when the interval I is replaced by a measurable set in Rn . It follows from Fubini’s theorem that for 1 ≤ p < ∞ there is an isometric isomorphism between Lp (I × ; X) and Lp (I ; Lp (; X)) for any measurable set  ⊂ Rn given by f → g, where (g(s))(t) := f (s, t).

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

13

We further note that for 1 ≤ p < ∞ the norm of a function f ∈ Lp (Rn ; X) can be calculated by f Lp (Rn ;X)   = sup 

Rn

   < g(x), f (x) > dx  : g ∈ Lp (Rn ; X ), gLp (Rn ;X ) ≤ 1 . (1.2.1)

Given u ∈ L1 (Rn ; X) we define the Fourier transform of u by  F u(ξ ) := u(ξ ˆ ) :=

Rn

e−i u(x)dx,

ξ ∈ Rn ,

 where as above < x, ξ >= nj=1 xj ξj . The Fourier inversion theorem guarantees that the Fourier transform F is an isomorphism on S(Rn ; X) and that ˆˇ F −1 u = (2π)−n uˇˆ = (2π)−n u,

u ∈ S(Rn ; X)

(1.2.2)

where u(x) ˇ := u(−x) for x ∈ Rn denotes the reflection of u. The Fourier transform uˆ := F u of u ∈ S  (Rn ; X) is defined by u(ϕ) ˆ := u(ϕ), ˆ

ϕ ∈ S(Rn ).

Defining u(ϕ) ˇ := u(ϕ) ˇ for u ∈ D (Rn ; X) and ϕ ∈ D(Rn ) it follows that F is an  isomorphism on S (Rn ; X) and that (1.2.2) holds for u ∈ S  (Rn ; X). Let ϕ ∈ S(Rn ) and u ∈ S  (Rn ; X). Then the convolution u ∗ ϕ of u and ϕ is defined by (u ∗ ϕ)(x) := u(τx ϕ), ˘

x ∈ Rn ,

where τa ϕ(x) := ϕ(x − a) for x ∈ Rn and a ∈ Rn . Moreover, the convolution theorem states that (u ∗ ϕ)ˆ = uˆ ϕ. ˆ Finally, let H be a Hilbert space. Then L2 (Rn ; H ) is a Hilbert space with respect to the inner product  (u|v)2 := (u|v)L2 (Rn ;H ) :=

Rn

u · v dx.

We note that Plancherel’s theorem carries over to the situation of Hilbert space valued functions. More precisely, the following holds.

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M. Hieber

Proposition 1.2.2 (Plancherel’s Theorem in Hilbert Spaces) Let H be a Hilbert space and u, v ∈ S(Rn ; H ). Then (u| ˆ v) ˆ 2 = (2π)n (u|v)2 Moreover, (2π)−n/2 F is a unitary operator on L2 (Rn ; H ). Singular Integrals: Classical Theory Singular integral operators will be an important tool in the analysis of the Stokes equation in the following sections. We now state two results on the Lp -boundedness of singular integral operators in the vector valued setting. Consider a kernel operator of the form  (Tf )(x) =

Rn

K(x, y)f (y)dμ(y),

(1.2.3)

where the kernel K is singular near x = y, and (1.2.3) is meaningful only in some limiting sense. More precisely, we assume that for f ∈ L2 (Rn ; X) with compact support, the integral in (1.2.3) converges absolutely for a. a. x ∈ (suppf )C and that (1.2.3) holds for these x, where the kernel K is assumed to belong to L1loc (Rn × Rn , L(X, Y )). Theorem 1.2.3 Let X and Y be Banach spaces and let T be defined as in (1.2.3). Assume that there exists a constant C > 0 such that (i) Tf L2 (Rn ;Y ) ≤ Cf L2 (Rn ;X) for all f ∈ L2 (Rn ; X) (ii) There exists a constant r > 1 such that for all y1 ∈ Rn , δ > 0  B(y1

,rδ)C

  K(x, y1 ) − K(x, y2) L(X,Y ) dμ(x) ≤ C if y2 ∈ B(y1 , δ)

Then, for all 1 < p < 2, there exists a constant Cp (depending only on C and p) such that Tf Lp (Rn ;Y ) ≤ Cp f Lp (Rn ;X) , for f ∈ Lp (Rn ; X). We note that the proof of the scalar-valued version of Theorem 1.2.3 given e.g. in [140] extends without any difficulties to the situation considered above. Remark 1.2.4 It is worthwhile to point out the following variant of Theorem 1.2.3. Assume that T defined as in (1.2.3) satisfies the condition Tf Lr (Rn ;Y ) ≤ Cf Lr (Rn ;X) ,

f ∈ Lr (Rn ; X)

for some r ∈ (1, ∞) and condition (ii) of Theorem 1.2.3. Then T admits a bounded extensions to Lp for 1 < p < r. The proof follows the lines of the proof given in [140] and is left to the reader.

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15

Consider now the special case of convolution operators. More specifically, let X and Y be Banach spaces and suppose that K ∈ L1loc (Rn \ {0}, L(X, Y )). Then the integral  Tf (x) :=

Rn

K(x − y)f (y)dy

(1.2.4)

is well defined for f ∈ L∞ (Rn ; X) with compact support and x ∈ (suppf )C . For operators of the form (1.2.4), the condition (ii) of Theorem 1.2.3 is equivalent to the Hörmanders’s condition    K(x − y) − K(x)dx ≤ C < ∞, y ∈ Rn \{0}. (1.2.5) |x|>2|y|

The following result due to Benedek, Calderón and Panzone says that operators of the form (1.2.4) are bounded operators on Lp (Rn ; X), for 1 < p < ∞, provided this holds for some r ∈ (1, ∞) and condition (1.2.5) is satisfied. Theorem 1.2.5 (Benedek, Calderón, Panzone) [16] Suppose that T ∈ L(Lr (Rn ; X) Lr (Rn ; Y )) for some r ∈ (1, ∞). Assume that T may be represented by (1.2.4) for f ∈ L∞ (Rn ; X) with compact support and x ∈ ( supp f )C , and that (1.2.5) is satisfied. Then T admits a bounded extension to Lp (Rn ; X) for p ∈ (1, ∞). Moreover, there exists a constant C such that Tf Lp (Rn ;Y ) ≤ Cf Lp (Rn ;X) ,

1 < p < ∞.

Proof We note that for p ∈ (1, r) the assertion follows from Remark 1.2.4 since for translation invariant operators of the form (1.2.4), the condition ii) of Theorem 1.2.3 is equivalent to (1.2.5). The remaining case where p ∈ (r, ∞) follows by a duality argument. 

Hilbert Transform and UMD-Spaces An interesting application of Theorem 1.2.5 concerns the Hilbert transform on Lp (R; X). We already saw in the previous Sect. 1.1 that the Hilbert transform acts boundedly on Lp (R). As in the scalar case, we define for u ∈ S(Rn ; X) the Hilbert transform H u of u by H u :=

1 pv(1/t) ∗ u, π

u ∈ S(Rn ; X).

For the time being, assume that for some p ∈ (1, ∞) we have H uLp (R;X) ≤ CuLp (R;X) for u ∈ S(R; X). Then there exists a unique extension of H to a bounded operator on Lp (R; X). This extension, also denoted by H , is called

16

M. Hieber

 the Hilbert transform on Lp (R; X). Since H u = −isign(·)u, ˆ it follows from Plancherel’s Theorem 1.2.2 that (H u|H v)2 =

1  1 (H u|H v)2 = (u| ˆ v) ˆ 2 = (u|v)2 2π 2π

for u, v ∈ S(R; X). By density, H is an isomorphism on L2 (R; X) satisfying H −1 v = −H v for v ∈ L2 (R; X). We summarize these observations in part (a) of the following proposition. Proposition 1.2.6 Let X be a Banach space. (a) The Hilbert transform is a unitary operator on L2 (R; X) provided X is a Hilbert space. (b) Suppose that the Hilbert transform is bounded on Lp (R; X) for some p ∈ (1, ∞). Then it is bounded on Lq (R; X) for all q ∈ (1, ∞). Proof Note that assertion (b) is an immediate consequence of Theorem 1.2.5 since  |x|>2|y|

  K(x − y) − K(x)dx =

 |x|>2|y|

  

1 1  − dx ≤ C. x−y x 

Let us remark that the Hilbert transform is not bounded on Lp (R; X), in general. Banach spaces for which the Hilbert transform is, however, bounded for some (and then all) p ∈ (1, ∞) play an important role in the regularity theory of parabolic evolution equations. We finish this section with the definition of such spaces. Definition 1.2.7 A Banach space X is called a UMD-space if the Hilbert transform is bounded on Lp (R; X) for some (and then all) p ∈ (1, ∞). The word U MD stands for the property of unconditional martingale differences and goes back to the work of Burkholder [24] and Bourgain [20]. For a comprehensive treatment of these spaces we refer to the monographs [87] and [8]. The following remark collects some basic facts and examples of U MD-spaces. Remarks 1.2.8 (a) Let us remark here that every finite dimensional Banach space or any Hilbert space is a U MD-space. (b) U MD spaces are reflexive. The converse is not true. (c) If X is a U MD space and (, μ) a σ -finite measure space, then Lp (, μ; X) has the U MD-property provided 1 < p < ∞. (d) All closed linear subspaces of U MD-spaces are U MD spaces. (e) L1 (, μ) or spaces of continuous functions C(K) do not have the U MD property.

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

17

1.3 Cauchy Problems and Semigroups We start this section by investigating the well-posedness of the abstract Cauchy problem  u (t)

(ACP )

u(0)

= Au(t),

t ≥ 0,

= x,

(1.3.1)

where A is a densely defined, closed operator on a Banach space X. Note that A is called closed whenever for a sequence (xj ) ∈ D(A) with xj → x and Axj → y it follows that x ∈ D(A) and Ax = y. Cauchy Problems and Semigroups The following definition of a C0 -semigroup and its generator is fundamental for our approach. Definition 1.3.1 A C0 -semigroup on a Banach space X is a strongly continuous function T : R+ → L(X) satisfying T (0) = I and T (t + s) = T (t)T (s),

s, t ≥ 0.

Its generator A is defined by T (t)x − x t →0 t

Ax := lim with domain D(A) := {x ∈ X : limt →0

T (t )x−x t

exists }.

Then D(A) is dense in X and A is a closed, linear operator. The operator A is called the infinitesimal generator of T since A coincides with the strong derivative of T in 0. Furthermore, we call a C0 -semigroup a bounded semigroup, if there exists a constant M > 0 such that T (t) ≤ M for all t > 0; it is called a contraction semigroup provided T (t) ≤ 1 for all t > 0. In the following lemma we collect basic properties of semigroups and their generators. Lemma 1.3.2 Let T be a C0 -semigroup on X with generator A and x ∈ D(A). Then (a) T (t)x ∈ D(A) for all t ≥ 0, (b) AT (t)x = T (t)Ax for all t ≥ 0,

18

M. Hieber

(c) the mapping t → T (t)x is differentiable for all t > 0 and d T (t)x = AT (t)x, dt

t > 0.

For the proof of these basic properties, we refer e.g. to [13], Chapter 3.1. Before proceeding further, some comments about resolvents and the resolvent set of A are in order. Given λ ∈ C and an operator A on X, λ is said to belong to the resolvent set (A) of A if λ − A is invertible. In this case we write R(λ, A) = (λ − A)−1 . Note that if (A) is non-empty, then A is closed. The function R(·, A) : (A) → L(X) is the resolvent of A. The spectrum of A is defined as σ (A) := C\(A); the spectral bound is given by s(A) := sup{Reλ : λ ∈ σ (A)}. The point spectrum σp (A) of A consists of all the eigenvalues of A, i.e., σp (A) := {λ ∈ C : ker (λ − A) = {0}}. We now collect basic properties of resolvents in the following lemma. Lemma 1.3.3 Let A be an operator on X. Then (a) (A) is open and σ (A) is closed in C. (b) If μ ∈ (A) and λ ∈ C with |λ − μ| < ||R(μ, A)||−1 , then λ ∈ (A) and R(λ, A) =

∞ 

(μ − λ)n R(μ, A)n+1 .

n=0

(c) R(·, A) is holomorphic on (A) and R(μ, A)(n) = (−1)n n!R(μ, A)n+1 ,

n ∈ N.

(d) It λ, μ ∈ (A), then R(λ, A) − R(μ, A) = (μ − λ)R(λ, A)R(μ, A).

(1.3.2)

The above Eq. (1.3.2) is called the resolvent equation. Consider now an arbitrary C0 -semigroup T with generator A. Then it is not difficult to prove that there exist constants M, ω > 0 such that T (t) ≤ Meωt for all t > 0 and that  ∞ R(λ, A)x = e−λt T (t)xdt, x ∈ X, Re λ > ω, 0

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19

where R(λ, A) = (λ − A)−1 . This means in particular that λ ∈ (A) for all λ ∈ (ω, ∞). One can prove the stronger assertion that generators of C0 -semigroups are precisely those operators whose resolvent is a Laplace transform. More precisely, we have the following result. Proposition 1.3.4 ([13, Thm. 3.1.7]) Let T : [0, ∞) → L(X) be a strongly continuous function, ω ∈ R and A be an operator on X such that (ω, ∞) ⊂ (A) and  ∞ R(λ, A)x = e−λt T (t)xdt, x ∈ X, Re λ > ω. 0

Then T is a C0 -semigroup on X with A its generator. After the more algebraic definition of a C0 -semigroup in Definition 1.3.1 and the approach based on the Laplace transform in Proposition 1.3.4 we now show that the well-posedness of (ACP0 ) is very closely related to the fact that A is the generator of a C0 -semigroup. By a classical solution of (1.3.1) we understand a function u ∈ C 1 (R+ ; X) such that u(t) ∈ D(A) for all t ≥ 0 and (1.3.1) holds for all t ≥ 0. If a classical solution exists, then u(0) = x ∈ D(A). Proposition 1.3.5 Let A be a closed operator on X. Then A generates a C0 semigroup if and only if ρ(A) = ∅ and for all x ∈ D(A) there exists unique, classical solution of (ACP ). It is thus interesting to characterize generators of C0 -semigroups by properties of the operators A or their resolvents. In the following, we first prove the Hille-Yosida theorem, which characterizes generators of C0 -semigroups in terms of a resolvent estimate for real λ. Theorem 1.3.6 (Hille-Yosida) Let A be a densely defined operator on X. Then A generates a C0 -semigroup T on X satisfying T (t) ≤ 1 for all t ≥ 0 if and only if (0, ∞) ⊂ ρ(A) and λR(λ, A) ≤ 1,

λ > 0.

(1.3.3)

Proof Observe first that if A is the generator of a contraction semigroup T , then the assertion follows immediately from Proposition 1.3.4. Conversely, assume that A satisfies (1.3.3). For λ > 0, we define the Yosida approximation of A by Aλ := λ2 R(λ, A) − λI d = λAR(λ, A). Then, for x ∈ D(A), Aλ x → Ax as λ → ∞.

20

M. Hieber

Since Aλ is a bounded operator on X, we may define the semigroup Tλ generated by Aλ by Tλ (t) := et Aλ = e−λt eλ

2 t R(λ,A)

= e−λt

∞  (λ2 t)j j =0

j!

R(λ, A)j .

The assumption ||λR(λ, A)|| ≤ 1 for all λ > 0 implies that ||Tλ (t)|| ≤ 1. Thus Tλ is a contraction semigroup on X having Aλ as its generator. Next, due to the resolvent Eq. (1.3.2), Aμ Tλ (t) = Tλ (t)Aμ ,

λ, μ > 0, t > 0.

Hence, for x ∈ D(A) we thus obtain by Lemma 1.3.2(c),  Tλ (t)x−Tμ (t)x = 0

t

d [Tμ (t −s)Tλ (s)x]ds = ds



t

Tμ (t −s)Tλ (s)(Aλ x−Aμ x)ds.

0

Thus ||Tλ (t)x − Tμ (t)x|| ≤ t||Aλ x − Aμ x|| → 0 as λ, μ → ∞. Therefore, T (t)x := lim Tλ (t)x λ→∞

exists for all t ≥ 0 and all x ∈ D(A). Since ||Tλ (t)|| ≤ 1 for all λ > 0 and all t > 0, the above limit exists for all x ∈ X and thus T is a contraction semigroup on X. The proof of the fact that A is the generator of T is left to the reader as an exercise.  There is a second characterization of contraction semigroups which turns out to be quite useful when dealing with differential operators. To this end, we call an operator A in X dissipative if (λ − A)x ≥ λx for all x ∈ D(A), λ > 0.

(1.3.4)

We then have the following so called Lumer–Phillips theorem. Theorem 1.3.7 (Lumer–Phillips) Let A be a densely defined operator on X. Then A generates a C0 -semigroup of contractions on X if and only if A is dissipative and (λ − A)D(A) = X for some (or all) λ > 0. Proof Assume that A is the generator of a contraction semigroup. Then the assertion follows from the Hille-Yosida Theorem 1.3.6. In order to prove the converse implication, note that by assumption λ0 − A is invertible and that ||R(λ0 , A)|| ≤ λ−1 0 for some λ0 > 0. Hence,  := (A)∩(0, ∞)

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

21

is non empty, and thus A is closed. Proving furthermore that  is open and closed in (0, ∞) it follows that  = (0, ∞) and thus (0, ∞) ⊂ (A). The assertion follows then from (1.3.4) and the Hille-Yosida Theorem. 

Remark 1.3.8 Dissipative operators acting on Hilbert spaces H may be characterized as follows. Denote by (·|·) the inner product in H and let A be an operator on H . Then A is dissipative if and only if Re(Ax|x) ≤ 0 for all x ∈ D(A). Holomorphic Semigroups We now turn our attention to bounded holomorphic semigroups. To this end, for φ ∈ (0, π), we define the sector θ in the complex plane by θ := {λ ∈ C\{0} : | arg λ| < θ }. The definition of such semigroups reads as follows. Definition 1.3.9 Let θ ∈ (0, π2 ]. A C0 -semigroup T is called a bounded holomorphic semigroup of angle θ if T has a bounded holomorphic extension to θ  for each θ  ∈ (0, θ ). If we do not want to specify the angle, we call T a bounded holomorphic semigroup if T is a bounded holomorphic semigroup of angle θ for some θ ∈ (0, π2 ]. The following characterization theorem for holomorphic semigroups is of fundamental importance for many parabolic problems. Generators of bounded analytic semigroups are closely related to so-called sectorial operators, which we will study in more detail in the following section. For this reason, we will also postpone the proof of the following Theorem 1.3.10 to this section; see Remark 1.4.8(a). Theorem 1.3.10 Let A be an operator in X and θ ∈ (0, π2 ]. Then the following assertions are equivalent. (i) A generates a bounded holomorphic semigroup of angle θ . (ii) θ+ π2 ⊂ ρ(A) and sup

λR(λ, A) < ∞ for all ε > 0.

λ∈θ + π −ε 2

We remark that if one is not interested in the angle of holomorphy in the above theorem, it suffices to verify condition (ii) above in a right half plane, only. Corollary 1.3.11 An operator A on X generates a bounded holomorphic semigroup on X if and only if {z ∈ C : Rez > 0} ⊂ ρ(A) and M := sup λR(λ, A) < ∞.

(1.3.5)

Reλ>0

In order to prove the assertion of this corollary it suffices by the above Theorem 1.3.10 to show that condition (1.3.5) implies assertion (ii) of Theorem 1.3.10. To this end, set c := 1/2M and for s ∈ R\{0} and −c|s| < r ≤ 0 let λ := c|s| + r + is.

22

M. Hieber

Then |λ − (r + is)| = c|s| ≤ 1/2R(λ, A)−1 . It follows that r + is ∈ ρ(A) and (r + is)R(r + is, A) ≤ 2M(c + 1). Thus (ii) is satisfied with θ = arctan c. Next, we define holomorphic semigroups as follows.



Definition 1.3.12 An operator A is said to generate a holomorphic semigroup on X if there exists ω ≥ 0 such that A − ω generates a bounded holomorphic semigroup. The following examples of holomorphic semigroups are of special interest in the following sections. Examples 1.3.13 (a) (Selfadjoint Operators). Every selfadjoint operator in a Hilbert space H which is bounded above by ω is the generator of a bounded holomorphic semigroup of angle π/2 on H satisfying T (z) ≤ eωRez ,

Rez > 0.

In fact, by the spectral theorem we may assume that H = L2 (, μ) and that A is given by Af = m · f for f ∈ D(A) = {f ∈ H : mf ∈ H } and a measurable function m :  → (−∞, ω]. Thus T (z)f (x) := ezm(x) f (x) for Rez > 0 and x ∈  defines a bounded, holomorphic semigroup on H with generator A. (b) (The Laplacian and the Gaussian semigroup). The solution of the heat equation ut − u = 0, t > 0 on Rn with initial data u(0) = f ∈ X is governed by the Gaussian semigroup T , which for X being one of the spaces Lp (Rn ) for 1 ≤ p < ∞, C0 (Rn ) or BU C(Rn ) can be represented as  2 −n/2 T (t)f (x) := (4πt) f (x − y)e−|y| /4t dy, t > 0, f ∈ X, x ∈ Rn . Rn

This semigroup is a bounded holomorphic semigroup on X of angle π/2. Its generator is the Laplacian X on X equipped with maximal domain, i.e. D(X ) = {f ∈ X : f ∈ X} and X f = f. If X = Lp (Rn ) and 1 < p < ∞, then by Mikhlin’s Theorem 1.1.5, D(X ) coincides with W 2,p (Rn ). Mikhlin’s theorem implies also that in this case σ () = (−∞, 0]. (c) (The Dirichlet Laplacian D in Lp (Rn+ ) for 1 < p < ∞). Consider the Dirichlet Laplacian D in Lp (Rn+ ) for 1 < p < ∞ defined by Au = u,

1,p

D(A) = W 2,p (Rn+ ) ∩ W0 (Rn+ ).

(1.3.6)

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23

Then A is densely defined, injective and has dense range. Let E : Lp (Rn+ ) → Lp (Rn ) be the extension operator by 0 and R : Lp (Rn ) → Lp (Rn+ ) be the restriction of a function u ∈ Lp (Rn ) to Lp (Rn+ ). Then, for λ ∈ C with | arg(λ)| < π, R(λ, A) = RR(λ, )E − RSR(λ, )E,

(1.3.7)

where S denotes the reflection of a function in the normal coordinate, i.e., S is given by Su(x1 , . . . , xn−1 , xn ) = u(x1 , . . . , xn−1 , −xn ). Hence, D generates a bounded holomorphic semigroup on Lp (Rn+ ). d) (The Dirichlet Laplacian D in spaces of bounded functions). Using (1.3.7) and Example (b), it follows that the Dirichlet Laplacian generates a bounded holomorphic semigroup also on BU C(Rn+ ), C0 (Rn+ ) and L∞ (Rn+ ), which in the latter case is, however, not strongly continuous. Note that its domain does not coincide with the spaces given in (1.3.6). When compared with arbitrary C0 -semigroups, holomorphic semigroups show many particular properties. The so-called smoothing effect of holomorphic semigroups can be shown elegantly by means of the functional calculus developed in the following Sect. 1.4. We hence postpone the proof of the following Theorem 1.3.14 again to Sect. 1.4; see Remark 1.4.8(b). Theorem 1.3.14 Let A be the generator of a bounded holomorphic semigroup on X, x ∈ X and n ∈ N. Then sup t n An et A  < ∞, t >0

and there exists a unique function u ∈ C ∞ ((0, ∞); X) ∩ C([0, ∞); X) ∩ C 1 ((0, ∞), D(A)) satisfying the Cauchy problem u (t) = Au(t) for t > 0 and u(0) = x. The following perturbation result for generators of holomorphic semigroups is in particular useful when treating lower order perturbations of differential operators. Theorem 1.3.15 (Relatively Bounded Perturbations) Let A be the generator of a holomorphic semigroup on X. Assume that B : D(A) → X is an operator such that for every ε > 0 there exists a constant b ≥ 0 such that Bx ≤ εAx + bx,

x ∈ D(A).

Then A + B generates a holomorphic semigroup. Proof Assume first that A generates a bounded analytic semigroup. By Theorem 1.3.10 there exists θ ∈ (0, π/2] such that θ+ π2 ⊂ ρ(A) and

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M. Hieber

supλ∈θ + π λR(λ, A) =: M < ∞. It follows from the assumption that, given 2

ε > 0, there exists b ≥ 0 such that for x ∈ X

||BR(λ, A)x|| ≤ ε(M + 1)||x|| +

bM ||x||, λ

λ ∈ θ+ π2 .

Choosing ε small enough, it follows that ||BR(λ, A)|| ≤ q < 1 for |λ| sufficiently large. Hence I d − BR(λ, A) is invertible and since λ − (A + B) = (I d − BR(λ, A))(λ − A),

λ ∈ θ+ π2 ,

it follows that λ − (A + B) is invertible for all λ ∈ θ+ π2 with λ sufficiently large C and that ||R(λ, A + B)|| ≤ |λ| for all λ ∈ θ+ π2 with λ sufficiently large. The general case follows by applying the above proof to A − ω. 

Real Interpolation Spaces We conclude this section with some remarks concerning the real interpolation space (X, D(A))θ,p which is defined for generators A of analytic semigroups T as follows. Let us first recall from Theorem 1.3.14 that there exists a constant C > 0 such that et A x + tAet A x ≤ Cx for x ∈ X and t ∈ [0, 1]. Further, for θ ∈ (0, 1) and p ∈ (1, ∞) we set φθ,p (x, t) := t 1−θ−1/p Aet A xX . Definition 1.3.16 For θ ∈ (0, 1) and p ∈ (1, ∞) the real interpolation space (X, D(A))θ,p is defined as   X, D(A) θ,p := {x ∈ X : φθ,p (x, ·) ∈ Lp (0, 1)} equipped with the norm xθ,p := x + φθ,p (x, ·)Lp (0,1). Remarks 1.3.17 space (a) We remark first that the space (X, D(A))θ,p is obviously  an intermediate  between X and D(A) in the sense that D(A) ⊂ X, D(A) θ,p ⊂ X with continuous embeddings. (b) The real interpolation space (X, D(A))θ,p equipped with the above norm is a Banach space. This can be seen by standard methods.

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25

(c) Replacing the interval (0, 1) in the above definition by the interval (0, T ) for   some T > 0 leads to the same space X, D(A) θ,p as above, however with an equivalent norm. (d) It is sometimes more convenient to denote the above real interpolation space by DA (θ, p), i.e. we set DA (θ, p) := X, D(A) θ,p with θ and p as above.

1.4 Sectorial Operators and Bounded H ∞ -Functional Calculus In this section we introduce and investigate the class of sectorial operators, a class of operators which play an important role in the analysis of the Stokes equation. We also introduce the concept of a bounded H ∞ -functional calculus for sectorial operators. Sectorial Operators and Bounded H ∞ -Calculus Throughout this section, X always denotes a complex Banach space. For θ ∈ (0, π) we recall that the sector θ in the complex plane is given by θ := {λ ∈ C\{0} : | arg λ| < θ }. Definition 1.4.1 (Sectorial Operators) Let X be a Banach space and A be a closed, densely defined linear operator on X. Then A is called sectorial of angle φ ∈ (0, π) if (S1) A is injective and has dense range, (S2) σ (A) ⊂ φ and sup{λR(λ, A)L(X) : λ ∈ C\φ  } < ∞ for all φ  ∈ (φ, π). The class of sectorial operators on X of angle φ will be denoted by Sφ (X), and S(X) denotes the class of sectorial operators of some angle φ. Moreover, we call the infimum of all φ for which (S2) is satisfied, the spectral angle φA of A. The following elementary properties of sectorial operators will be useful later on. Lemma 1.4.2 Let A be a sectorial operator on X. Then, for x ∈ X limt →∞ t (t + A)−1 x = x, limt →0 t (t + A)−1 x = 0, limt →∞ A(t + A)−1 x = 0, limt →0 A(t + A)−1 x = x. In particular, D(Ak ) ∩ R(Ak ) is dense in X for all k ∈ N. The proof of the above assertions is based on the identity I d − t (t + A)−1 = A(t + A)−1

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M. Hieber

and on the fact that the sets {t (t + A)−1 : t > 0} and {A(t + A)−1 : t > 0} are bounded in L(X). Given x ∈ X and n, k ∈ N, define xn,k := (1 + n−1 A)−k Ak (n−1 + A)−k x. Then xn,k ∈ D(Ak ) ∩ R(Ak ) and xn,k → x, as n → ∞, by the above assertion. We now describe the construction of a functional calculus for sectorial operators which is being inspired by the classical Dunford calculus. To this end, for θ ∈ (0, π], we introduce the space H ∞ (θ ) of holomorphic functions as H ∞ (θ ) = {f : θ → C, f holomorphic and bounded}.

(1.4.1)

Equipped with the norm |f |θ∞ = sup{|f (λ)| : | arg λ| < θ },

(1.4.2)

H ∞ (θ ) is a Banach algebra. For the time being, assume that A ∈ S(X) is bounded and invertible. Fix θ > φA . Then the classical Dunford calculus for bounded linear operators applies. In this situation the spectrum σ (A) is a compact subset of θ and choosing a closed path A in θ surrounding σ (A) counterclockwise, we define f (A) :=

1 2πi

 f (λ)R(λ, A)dλ,

f ∈ H ∞ (θ ).

(1.4.3)

A

Noting that A is compact, the above integral is well defined. It is well known that this formula defines an algebra homomorphism from H ∞ (θ ) to L(X). Relation (1.4.3) will be the starting point of our definition of a boundeded holomorphic functional calculus for sectorial operators A on X. First we consider functions f ∈ H ∞ (θ ) having some decay at 0 and infinity. More precisely, for λ λ ∈ θ define (λ) := (1+λ) 2 and assume that |f (λ)| ≤ C|(λ)|ε ,

λ ∈ θ

(1.4.4)

for some C > 0 and ε > 0. For θ ∈ (0, π) we then set H0∞ (θ ) := {f ∈ H ∞ (θ ) : there exists C, ε > 0 such that (1.4.4) holds}. (1.4.5) Given a sectorial operator A ∈ S(X) and f ∈ H0∞ (θ ) where θ > φA , we define  1 f (A) := f (λ)R(λ, A)dλ, 2πi ∂θ 

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

27

where φA < θ  < θ . It then follows that this integral is absolutely convergent in L(X). This definition of f (A) allows further to define a functional calculus for A on H0∞ (θ ) having the following properties. Proposition 1.4.3 Let X be a Banach space and A ∈ Sφ (X). For φ < θ  < θ and f ∈ H0∞ (θ ) set A (f ) := f (A) :=

1 2πi

 f (λ)R(λ, A)dλ.

(1.4.6)

∂θ 

Then A : H0∞ (θ ) → L(X) is a linear mapping with the following properties: (a) the above integral is independent of the particular choice of θ  ∈ (φ, θ ). (b) A (f · g) = A (f )A (g), f, g ∈ H0∞ (θ ). (c) Let (fn ), f ∈ H0∞ (θ ) be uniformly bounded and assume that fn (λ) → f (λ) for λ ∈ θ . Then lim A (fn · g) = A (f · g) in L(X)

n→∞

for all g ∈ H0∞ (θ ). / θ , then (d) If f (λ) = λ(μ1 − λ)−1 (μ2 − λ)−1 with μ1 , μ2 ∈ f (A) = AR(μ1 , A)R(μ2 , A). Note that the functional calculus developed in Proposition 1.4.3 does not yield a satisfying calculus so far, since it does not apply to standard functions such as λ → (μ − λ)−1 . However, we will improve the above construction by the following approximation procedure. To this end, we introduce the following approximate identity. Example 1.4.4 (Approximate Identity) In the situation of Proposition 1.4.3, define for n ≥ 2 the functions n : θ → C by n (λ) := n(n + λ)−1 − (1 + nλ)−1 . Then n (A) = n(n + A)−1 − n−1 (n−1 + A)−1 . In fact, n ∈ H0∞ (θ ) for all n ≥ 2 and the above equality follows from Proposition 1.4.3(d). Moreover, the range of n (A)k equals D(Ak ) ∩ Rg(Ak ) for all k ∈ N and limn→∞ n (A)k x = x for all x ∈ X. The latter assertion follows from Lemma 1.4.2 since limn→∞ n (A)x = x and hence limn→∞ n (A)k x = x. We will now construct a closed extension of A defined in Proposition 1.4.3 whose domain HA∞ (θ ) will consist of all f ∈ H ∞ (θ ) for which f (A) can be defined as a bounded operator on X. More precisely, for A ∈ S(X), θ > φA and f ∈ H0∞ (θ ) set |||f |||A := f H ∞ (θ ) + f (A)L(X)

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M. Hieber

and define HA∞ (θ ) to be the class of functions f ∈ H ∞ (θ ) for which there exists a sequence (fn ) ⊂ H0∞ (θ ) with fn (λ) → f (λ) for all λ ∈ θ and supn∈N |||fn (A)|||A < ∞. Then HA∞ (θ ) is a subalgebra of H ∞ (θ ). We will now show that for f ∈ ∞ HA (θ ) the limit A (f )(x) := lim A (fn )(x), n→∞

x∈X

exists and defines a functional calculus on HA∞ (θ ). Proposition 1.4.5 (McIntosh’s Convergence Lemma) Let A ∈ S(X) and θ > φA . Then there exists an extension A : HA∞ (θ ) → L(X) of A defined in Proposition 1.4.3 satisfying the following properties: (a) A is linear and multiplicative. (b) For μ ∈ / θ , rμ (λ) = (μ − λ)−1 ∈ HA∞ (θ ) and A (rμ ) = R(μ, A). (c) Let f ∈ H ∞ (θ ) and fn ∈ HA∞ (θ ) such that fn (λ) → f (λ) for all λ ∈ θ and |||fn |||A < C for all n ∈ N and some C > 0. Then f ∈ HA∞ (θ ), limn→∞ A (fn )(x) = A (f )(x) for all x ∈ X and A  ≤ C. For a proof of this lemma we refer to [113] or [95]. We are now in the position to introduce the concept of operators having a bounded H ∞ -calculus on X. Definition 1.4.6 Let A ∈ S(X) and θ ∈ (φA , π). Then A is said to admit a bounded H ∞ (θ )-calculus if there exists a constant C > 0 such that for all f ∈ H0∞ (θ ).

f (A)L(X) ≤ Cf H ∞ (θ )

(1.4.7)

∞ . The The infimum of such θ is called the H ∞ -angle of A and is denoted by φA ∞ class of sectorial operators on X admitting a bounded H -calculus for some angle θ ∈ (0, π) is denoted by H∞ (X).

Remarks 1.4.7 (a) Note that due to Proposition 1.4.5, for θ > φA , A ∈ S(X) admits a bounded H ∞ (θ )-calculus if and only if there exists a bounded algebra homomorphism H ∞ (θ ) → L(X), f → f (A), satisfying the convergence properties (c) of Proposition 1.4.5 and such that f (A) =

1 2πi

 f (λ)R(λ, A)dλ ∂θ 

for f ∈ H0∞ (θ ) and θ > θ  > φA . (b) We also remark that A ∈ S(X) admits a bounded H ∞ -calculus of angle θ > φA if and only if H ∞ (θ ) = HA∞ (θ ).

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(c) Let A ∈ S(X) and θ > φA . Assume that f ∈ H0∞ (θ ) is in addition analytic in a neighborhood of 0. Then f ∈ HA∞ (θ ) and A (f ) =

1 2πi

 f (λ)R(λ, A)dλ, ∂(δ)

where (δ) := θ  ∪ B(0, δ) for θ  ∈ (φA , θ ) and δ being small enough so that f is analytic on (δ). Note that in this situation A (f ) ≤

1 2π



|f (λ)||λ|−1 dλ.

sup λR(λ, A) ∂(δ)

λ∈∂(δ)

(d) Since n (λ) → 1 and n (A)x → x for all x ∈ X by Example 1.4.4 we have 1 ∈ HA∞ (θ ) and A (1) = I d. (e) Let A be a sectorial operator in a Banach space X such that D(A ) is dense in ∞ = φ∞. X . Then A ∈ H∞ (X) if and only if A ∈ H∞ (X ) and φA A Note that for a given sectorial operator A the class HA∞ (θ ) contains many interesting functions; in particular it contains the functions λ → λn e−λz . They allow us to derive elegant proofs of the basic theorems for bounded holomorphic semigroups by the functional calculus described above as follows. Remarks 1.4.8 Let A ∈ S(X) with φA < π/2. (a) For z ∈ C with | arg(z)| < π/2 − φA set ez (λ) := e−λz . Then ez ∈ HA∞ (θ ) for θ ∈ (φA , π2 − | arg(z)|) by Remark 1.4.7(c). Setting T (z) := ez (A), we obtain the assertion of Theorem 1.3.10 thus by our functional calculus, this remark and by noting that the semigroup property T (z1 )T (z2 ) = T (z1 + z2 ) follows from ez1 ez2 = ez1 +z2 . (b) Similarly, we obtain the assertion of Theorem 1.3.14 again from Remark 1.4.7 (c) bynoting that {t n An et A : t > 0} = {ϕ(tA) : t > 0} with ϕ(λ) = λn e−λ and since |ϕ(tλ)| dλ| |λ| < ∞ and independent of t > 0. Examples d and the We now present two basic examples of operators, the derivative operator dt ∞ p Laplacian , respectively, both admitting a bounded H -calculus on L -spaces. Example 1.4.9 (The Time Derivative on R, R+ and (0, T )) (a) The derivative operator on Lp (R). For 1 < p < ∞ define AR on Lp (R) by AR u =

d u, dt

D(AR ) = W 1,p (R).

It is easy to see that AR is densely defined, injective and has dense range. In order to show that AR is sectorial, consider for λ ∈ C with | arg λ| > θ > π2 and

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g ∈ S(R) the equation λu − AR u = g. Applying the Fourier transform yields λ u(τ ) − iτ u(τ ) =  g (τ ) for all τ ∈ R. This equation admits the unique solution  u = mλ g with mλ (τ ) = (λ − iτ )−1 . By Mikhlin’s Theorem 1.1.5, the operator Tmλ given by Tmλ f = F −1 mλ F f for f ∈ S(R) has a continuous extension to C Lp (R) satisfying Tmλ  ≤ |λ| for some C > 0 independent of λ. Moreover, we have R(λ, AR ) = Tmλ for all such λ. Hence AR is a sectorial operator in Lp (R) of angle π2 . Furthermore, let f ∈ H0∞ (θ ) for θ > π2 and u ∈ S(R). Then 1 f (AR ) = 2πi

 ∂θ 

f (λ)R(λ, AR )dλ

for θ  ∈ ( π2 , θ ) and applying the Fourier transform yields F (f (AR )u)(τ ) =

1 2πi



f (λ)(λ − iτ )−1 u(τ )dλ = f (iτ ) u(τ ),

τ ∈ R.

∂θ 

Cauchy’s theorem and Mikhlin’s multiplier theorem imply that τ → f (iτ ) defines a Fourier multiplier for Lp (R) and we verify that f (AR )L(Lp (R)) ≤ Cf H ∞ (θ ) ,

f ∈ H0∞ (θ ),

d for some C > 0. We hence proved that the derivative operator AR = dt in Lp (R) equipped with the domain W 1,p (R) admits a bounded H ∞ -calculus on Lp (R) with ∞ = π. φA 2 R

(b) The time derivative on Lp (R+ ) Consider next the derivative operator AR+ defined on Lp (R+ ) with domain 1,p W0 (R+ ). As before, we see that AR+ is densely defined, injective and has dense range. Given u ∈ Lp (R+ ), we call Eu ∈ Lp (R) the extension of u to R by 0. Obviously, E : Lp (R+ ) → Lp (R) is a continuous operator. The restriction of 1,p 1,p E to W0 is obviously again continuous from W0 (R+ ) to W 1,p (R). Further, the restriction operator R : Lp (R) → Lp (R+ ) defined by Ru := u|R+ is also continuous. Hence, for λ ∈ C with | arg λ| > π2 , the resolvent R(λ, AR+ ) of AR+ is given by R(λ, AR+ ) = RR(λ, AR )E. Thus AR+ is sectorial of angle obtain

π 2.

Furthermore, for f ∈ H0∞ (θ ) with θ >

f (AR+ )v = Rf (AR )Ev,

v ∈ Lp (R+ ).

π 2

we

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31

It thus follows that the derivative operator AR+ in Lp (R+ ) equipped with the 1,p ∞ = π. domain W0 (R+ ) admits a bounded H ∞ -calculus on Lp (R+ ) with φA 2 R +

(c) The Time Derivative on

Lp (0, T ).

For T > 0 the derivative operator A(0,T ) on Lp (0, T ) is defined by Au := u ,

D(A(0,T ) ) = {u ∈ W 1,p (0, T ) : u(0) = 0}

In this case A(0,T ) is even surjective. Denoting by E(0,T ) and R(0,T ) the extension and restriction operators, from Lp (0, T ) → Lp (R+ ) and Lp (R+ ) → Lp (0, T ), respectively, we see that for λ ∈ C with | arg λ| > π2 , the resolvent R(λ, A(0,T ) ) of A(0,T ) is given by R(λ, A(0,T ) ) = R(0,T ) R(λ, AR+ )E(0,T ) . As above f (A(0,T ) )v = R(0,T ) f (AR+ )E(0,T ) v,

v ∈ Lp (0, T ), f ∈ H0∞ (θ ),

and thus the derivative operator A(0,T ) in Lp (0, T ) is invertible and admits a ∞ bounded H ∞ -calculus on Lp (0, T ) with φA = π2 . (0,T ) Example 1.4.10 (The Negative Laplacian on Rn and Rn+ ) (a) The negative Laplacian − in Lp (Rn ). As a second example, we consider the negative Laplacian − in Lp (Rn ) for 1 < p < ∞ defined as Au = −u,

D(A) = W 2,p (Rn ).

Then − is densely defined and injective. Weyl’s lemma implies that A has dense range in Lp (Rn ). As above we see that for λ ∈ C with | arg λ| > θ > 0 and g ∈ S(Rn ) the equation λu − Au = g has a unique solution u given by Tmλ g = F −1 mλ F g, where mλ is now given by mλ (ξ ) = (λ + |ξ |2 )−1 for ξ ∈ Rn . Mikhlin’s theorem implies that mλ ∈ Mp (Rn ) and that λmλ Mp ≤ C for some C independent of θ . Hence − is sectorial with angle φA = 0. Similarly as above we see that for f ∈ H0∞ (θ ) with θ > 0 and u ∈ S(Rn ) F (f (A)u)(ξ ) =

1 2πi



f (λ)(λ+|ξ |2 )−1 u(ξ )dλ = f (−|ξ |2 ) u(ξ ), ξ ∈ Rn . ∂θ 

Again, Cauchy’s theorem and Mikhlin’s multiplier theorem imply that ξ → f (−|ξ |2 ) defines a Fourier multiplier for Lp (Rn ) and we verify that f (A)L(Rn ) ≤ Cf H ∞ (θ ) ,

f ∈ H0∞ (θ ),

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M. Hieber

for some C > 0. Hence, the negative Laplacian A = − in Lp (Rn ) with ∞ = 0. domain W 2,p (Rn ) admits a bounded H ∞ -calculus on Lp (Rn ) with φA n p (b) The negative Dirichlet Laplacian −D in L (R+ ). Consider the negative Dirichlet Laplacian −D in Lp (Rn+ ) for 1 < p < ∞ defined as in Example 1.3.13(c) by Au = −u,

1,p

D(A) = W 2,p (Rn+ ) ∩ W0 (Rn+ ).

Formula (1.3.7) implies that f (A) = Rf (−)E − RSf (−)E, where − is defined as in (a). Hence, −D admits a bounded H ∞ -calculus on ∞ = 0. Lp (Rn+ ) with φA Further Results We conclude this section with stating several classes of operators allowing for a bounded H ∞ -calculus. For a proof of these assertions, we refer e.g. to [75, 95] or [88]. Proposition 1.4.11 (a) Let H be a Hilbert space, A ∈ S(H ), and assume that −A generates a contraction semigroup on H . Then A has a bounded H ∞ -calculus on H and ∞ =φ . φA A (b) Let −A be the generator of a positive contraction semigroup or a bounded group on Lp (), where 1 < p < ∞ and  ⊂ Rn is open. Then A has a bounded H ∞ (θ )-calculus on Lp () for all θ ∈ (π/2, π]. Furthermore, let −A be an elliptic differential operators of order 2m on Lp () subject to general boundary conditions, where  ⊂ Rn is a domain of regularityclass C 2m . Then, roughly speaking, A admits a bounded H ∞ -calculus on Lp () provided the top-order coefficients of A are Hölder continuous. For details, see [37]. We mention here explicitly only the examples of the Dirichlet and Neumann Laplacian. To this end, let  ⊂ Rn , n ∈ N, be a bounded domain with boundary of class C 2 , 1 < p < ∞ and let −D be the negative Dirichlet Laplacian on Lp () defined by 1,p

D u = u for u ∈ D(D ) = W 2,p () ∩ W0 (). We also define the negative Laplacian subject to Neumann boundary conditions −N as N u = u

for u ∈ D(N ) = {u ∈ W 2,p () : ∂ν u = 0 on ∂}.

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33

Proposition 1.4.12 The negative Dirichlet Laplacian −D as well as the negative Neumann Laplacian −N admit a bounded H ∞ -calculus on Lp () for 1 < p < ∞ ∞ ∞ with φ− = φ− = 0. D N Remark 1.4.13 In Sect. 1.14 on nematic liquid crystal flow we consider the Neumann Laplacian also in the space H 1,p (), where  ⊂ Rn is above. We then set 1N u = u

for u ∈ D(N ) = {u ∈ W 3,p () : ∂ν u = 0 on ∂}.

It then can be shown that −1N admits also a bounded H ∞ -calculus on H 1,p () for 1 < p < ∞ with φ ∞ 1 = 0. −N

1.5 Fractional Powers In this section we introduce fractional powers of sectorial operators. The approach we are using is based on an extended H ∞ -calculus. Note that so far we are unable to define operators An , A1/2 or more generally Aα for α ∈ C by the methods developed in Sect. 1.4, since these operators as well as the functions λ → λα are unbounded on suitable sectors θ . We hence extend the functional calculus described in Sect. 1.4 to polynomially bounded functions. Before we start with this extended functional calculus, let us remark that in the sequel we do not assume that 0 ∈ ρ(A). In the following, let θ ∈ (0, π) and let α ≥ 0. Then Hα (θ ) is defined to be the space of all holomorphic functions f : θ → C for which f Hα (θ ) := sup{|(λ)|α |f (λ)| : λ ∈ θ } < ∞, where  is defined as in Sect. 1.4 as (λ) = λ(1 + λ)−2 . For f ∈ Hα (θ ) and k ∈ N with k > α we obtain f (λ) = (λ)−k (k f )(λ). Since the second factor above belongs to H0∞ (θ ), it follows from Proposition 1.4.3 that (k f )(A) is a bounded operator on X. The following definition is thus natural. Definition 1.5.1 Let A ∈ S(X), θ ∈ (φA , π) and α ≥ 0. For f ∈ Hα∞ (θ ) and k ∈ N with k > α set f (A) := (A)−k (k f )(A) D(f (A)) := {x ∈ X : (k f )(A)x ∈ D((A)−k )}. The multiplicativity of the functional calculus implies that f (A) is well defined.

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It is now natural to define the fractional powers Az for z ∈ C by applying the extended functional calculus to the function fz : θ → C, λ → λz = ez log λ , where we use the branch of the logarithm which is holomorphic in C\{R− }. In fact, the estimate |fz (λ)| = |r z ||eiϕz | ≤ |λ|Rez eθ|I mz| ,

λ = reiϕ ∈ θ ,

implies that fz ∈ HRez (θ ) for all θ < π. Definition 1.5.2 For A ∈ S(X), θ ∈ (φA , π) and z ∈ C, the operator Az is defined by Az := fz (A),

z ∈ C.

It is then clear that Az is a well defined, closed and densely defined operator on X. In the following we state the basic properties of the fractional power operators Az . Theorem 1.5.3 For A ∈ S(X), the fractional powers Az , z ∈ C, are closed and injective operators satisfying D(A) = R(A) = X. Moreover, (a) if Rez1 < Rez2 < 0 < Rez3 < Rez4 , then D(Az1 ) ⊂ D(Az2 ), D(Az3 ) ⊃ D(Az4 ), R(Az1 ) ⊂ R(Az2 ), R(Az3 ) ⊃ R(Az4 ). (b) Az1 +z2 = Az1 Az2 for all z1 , z2 ∈ C with Rez1 · Rez2 > 0. (c) (Az )−1 = A−z , z ∈ C. The proof of the above properties of Az is quite technical and we refrain from giving it here. For a detailed proofs of the above result, we refer to [95], Theorem 15.15. For exponents α ∈ R we have the following multiplicative rule. Proposition 1.5.4 Let A ∈ S(X) and α ∈ R with φA |α| < π. Then Aα ∈ S(X) and φAα = |α|φA . Furthermore, (Aα )z = Aαz ,

z ∈ C.

Moreover, for any α, β, γ ∈ R satisfying α < β < γ with γ − α < π/φA , the momentum inequality γ −β

β−α

Aα u ≤ CAα u γ −α Aγ u γ −α ,

u ∈ D(Aα ) ∩ D(Aγ )

is valid for some constant M independent of u.

(1.5.1)

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35

We remark that in the above situation D(A + η)α = D(Aα ) for any η > 0 and that there exists a constant C > 0 such that (A + η)α − Aα u ≤ Cηα u,

u ∈ D(Aα ).

(1.5.2)

If in the situation of Proposition 1.5.4 the operator A admits a bounded H ∞ -calculus on X and α > 0, then Aα also admits a bounded H ∞ -calculus ∞ < π. More precisely, the following result holds. provided αφA ∞ < π. Then Corollary 1.5.5 Let A ∈ H∞ (X), α > 0 such that αφA

Aα ∈ H∞ (X)

∞ ∞ and φA α ≤ αφA .

∞ , π) and  as in In order to prove this assertion, let f ∈ H0∞ (θ ) for θ ∈ (αφA n Example 1.4.4. Then,

R(μ, Aα )n (A)x =

1 2πi



(μ − λα )−1 n (λ)R(λ, A)xdλ ∂θ/α

for x ∈ X, μ ∈ ∂θ and Fubini’s theorem implies that  1 f (λ)n (λ)R(λ, Aα )xdλ f (Aα )n (A)x = 2πi ∂θ   1 2 ) =( f (μ)(μ − λα )−1 n (λ)R(μ, A)xdλdμ 2πi ∂θ/α ∂θ  1 f (λα )n (λ)R(λ, A)xdλ = n (A)f (·α )(A)x. = 2πi ∂θ/α Letting n → ∞, we see that f (Aα )x and f (·α )(A)x coincide and that f (Aα ) = f (·α )(A) ≤ C(·α )H ∞ (θ/α ) ≤ f H ∞ (θ ) ,

f ∈ H0∞ (θ ).

The assertion thus follows. We finish this section by considering the fractional power spaces Xα := (D(Aα ),  · α ),

xα := x + Aα x,

0 < α < 1,

for sectorial operators A on X. If, in addition A ∈ H∞ (X), then it is possible to identify Xα in many cases as a concrete function space by the following theorem. Theorem 1.5.6 Let A ∈ H ∞ (X) for some Banach space X. Then Xα  [X, D(A)]α ,

0 < α < 1,

where [X, D(A)]α denotes the complex interpolation space of order α. For a proof of this fact, we refer e.g. to [148].

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M. Hieber

1.6 Operator-Valued H ∞ -Calculus, R-Boundedness, Fourier Multipliers and Maximal Lp -Regularity In this section we extend the bounded H ∞ -calculus developed in the previous section to the operator-valued setting. To this end, denote by A the subalgebra of L(X) of all bounded operators that commute with resolvents of a given sectorial operator A on a Banach space X. We then consider bounded holomorphic functions F : θ → A. Roughly speaking, we show that the functional calculus for scalar valued, bounded holomorphic functions described in Sect. 1.4 can be extended to operator-valued functions F : θ → A provided the range {F (λ) : λ ∈ θ } of F is not only uniformly bounded, but R-bounded. This notion will be of fundamental importance in the operator-valued setting. Starting from this extension we may then deduce fairly easily a result on the closedness of the sum of two sectorial operators which allows further to characterize the property of maximal Lp -regularity for parabolic evolution equations. We begin with the notion of R-bounded families of operators. R-Bounded Families of Operators Definition 1.6.1 Let X and Y be Banach spaces. A family of operators T ⊂ L(X, Y ) is called R-bounded, if there is a constant C > 0 and p ∈ [1, ∞) such that for each N ∈ N, Tj ∈ T , xj ∈ X and for all independent, symmetric, {−1, 1}valued random variables εj on a probability space (, M, μ), the inequality 

N  j =1

εj Tj xj Lp (;Y ) ≤ C

N 

εj xj Lp (;X)

(1.6.1)

j =1

is valid. The smallest such C is called the R-bound of T , we denote it by R(T ). Some remarks about properties of R-bounded families of operators are in order. Remarks 1.6.2 (a) If T ⊂ L(X, Y ) is R-bounded then it is also uniformly bounded, with sup{T  : T ∈ T } ≤ R(T ). This follows from the definition of R-boundedness with N = 1, since ε1 Lp () = 1. (b) The definition of R-boundedness is independent of p ∈ [1, ∞). This observation follows directly from Kahane’s inequality: For any Banach space X and 1 ≤ p, q < ∞ there is a constant C(p, q, X) such that 

N  j =1

εj xj Lp (;X) ≤ C(p, q, X)

N  j =1

εj xj Lq (;X) ,

(1.6.2)

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

37

for each N ∈ N, xj ∈ X, and for all independent, symmetric, {−1, 1}-valued random variables εj on a probability space (, M, μ). (c) Assume that X and Y are Hilbert spaces. Then T ⊂ L(X, Y ) is R-bounded if and only if T is uniformly bounded. In fact, let T be uniformly bounded by C > 0. Then, choosing p = 2 it follows that 

N  j =1

εj Tj xj 2L2 (;Y ) ≤ C 2 

N  j =1

εj xj 2L2 (;X)

since the εj are independent, hence orthogonal in L2 (). (d) Let X = Y = Lp (G) for some open G ⊂ Rn . Then T ⊂ L(X, Y ) is R-bounded if and only if there is a constant M > 0 such that the following square function estimate holds: (

N 

|Tj fj |2 )1/2 Lp (G)

j =1

≤ M(

N 

fj |2 )1/2 Lp (G) ,

N ∈ N, fj ∈ Lp (G), Tj ∈ T .

(1.6.3)

j =1

This is a consequence of Khintchine’s inequality: For each p ∈ [1, ∞) there is a constant Kp > 0 such that Kp−1 

N 

εj aj Lp () ≤ (

j =1

N 

|aj |2 )1/2 ≤ Kp 

j =1

N 

εj aj Lp () ,

(1.6.4)

j =1

for all N ∈ N, aj ∈ C, and for all independent, symmetric, {−1, 1}-valued random variables εj on a probability space (, M, μ). (e) The following assertion is known as Kahane’s contraction principle. Let N ∈ N, xj ∈ X, εj independent, symmetric, {−1, 1}-valued random variables on a probability space (, M, μ), and αj , βj ∈ C such that |αj | ≤ |βj |, for each j = 1, . . . , N. Then 

N 

αj εj xj 

Lp (;X)

j =1

≤ 2

N 

βj εj xj Lp (;X) .

j =1

The above Remark 1.6.2(d) gives a very useful sufficient condition for the R-boundedness of kernel operators in Lp (G). More precisely, we have the following lemma.

38

M. Hieber

Lemma 1.6.3 Let G ⊂ Rn be open and T = {Tμ : μ ∈ M} ⊂ L(Lp (G; Cm )) a family of kernel operators of the form  [Tμ f ](x) =

kμ (x, y)f (y)dy,

x ∈ G, f ∈ Lp (G; Cm ),

G

which are dominated by a kernel k0 , i.e. |kμ (x, y)| ≤ k0 (x, y),

for a.a. x, y ∈ G, and all μ ∈ M.

Then T ⊂ L(Lp (G; Rm )) is R-bounded, provided T0 is bounded in Lp (G). Proof By Remark 1.6.2(d) we only have to verify the square function estimate (1.6.3). Due to Lp -boundedness of the dominating operator T0 , we have (

N 

|Tj fj |2 )1/2 Lp (G) ≤ (

j =1

N  (T0 |fj |)2 )1/2 Lp (G) j =1

≤ T0 (

N 

|fj |2 )1/2 Lp (G) ≤ T0 L(Lp (G)) (

j =1

N 

|fj |2 )1/2 Lp (G) .

j =1



We remark further that R-bounds behave like norms. More precisely, let X, Y, Z be Banach spaces and T , S ⊂ L(X, Y ) be R-bounded. Then T + S = {T + S : T ∈ T , S ∈ S} is R-bounded as well, and R(T + S) ≤ R(T ) + R(S). Moreover, let T ⊂ L(X, Y ) and S ⊂ L(Y, Z) be R-bounded. Then ST = {ST : T ∈ T , S ∈ S} is Rbounded, and R(ST ) ≤ R(S)R(T ). Another useful criteria for R-boundedness of a given family of operators is given in the following lemma. Lemma 1.6.4 Let G ⊂ C be open, K ⊂ G compact, and suppose that H : G → L(X, Y ) is a holomorphic function. Then H (K) ⊂ L(X, Y ) is R-bounded. In order to prove this assertion, fix z0 ∈ K. Since H is holomorphic in G there exists a ball Br (z0 ) such that the power series representation H (z) =

∞  k=0

H (k)(z0 )

(z − z0 )k , k!

|z − z0 | ≤ r,

∞ (k) k of H is absolutely convergent and ρ0 := k=0 H (z0 )L(X,Y ) r /k! < ∞. It follows from Remark 1.6.2(e) that R(H (Br (z0 ))) ≤ 2ρ0 . Covering K by a finite set of such balls, the assertion follows.

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

39

Operator-Valued H ∞ -Calculus and the Kalton–Weis Theorem We now turn our attention to the operator-valued H ∞ -functional calculus. Indeed, for a sectorial operator A ∈ S(X) consider an operator-valued function F ∈ H ∞ (φ ; L(X)) satisfying F (λ)(μ − A)−1 = (μ − A)−1 F (λ),

μ ∈ (A), λ ∈ φ

for some φ > φA . In this case we write F ∈ H ∞ (φ , A), where A denotes the subalgebra of all bounded operators acting on X, which commute with resolvents of A. Furthermore, we set RH ∞ (φ , A) := {F ∈ H ∞ (φ , A) : {F (z) : z ∈ φ } is R-bounded }. As in the scalar-valued case, we denote by RH0∞ (φ ; A) those elements of RH ∞ (φ ; A) for which there exist constants C, ε > 0 such that  F (z)L(X) ≤ C

|z| 1 + |z|2

ε ,

z ∈ φ .

For F ∈ RH0∞ (φ ; L(X)), the integral 1 F (A) = 2πi



F (λ)(λ − A)−1 dλ

(1.6.5)

∂φ 

is well defined as a Bochner integral where φA < φ  < φ. We now show that formula (1.6.5) give rise to a bounded functional calculus on RH ∞ (φ , A) provided A admits a bounded H ∞ -calculus. More precisely, the following theorem holds true. Theorem 1.6.5 (Kalton–Weis) Let A ∈ H∞ (X). Assume that F ∈ H ∞ (φ , A) ∞ and such that R{F (z) : z ∈  } ≤ M for some M > 0. Then for some φ > φA φ there exists a constant CA > 0, depending only on A, such that F (A) ∈ L(X) and F (A)L(X) ≤ CA M. Remark 1.6.6 It should be noted that in the situation of Theorem 1.6.5 there exists a bounded algebra homomorphism A : RH ∞ (φ ; A) → L(X) with A (F ) = F (A) as in (1.6.5) for F ∈ RH0∞ (φ ; A) satisfying the following convergence property: if (Fn ) is a bounded sequence in RH ∞ (φ ; A) such that Fn (λ)x → F (λ)x for some F ∈ RH ∞ (φ ; A) and all λ ∈ φ , x ∈ X, then A (Fn )x → A (F )x,

x ∈ X.

40

M. Hieber

The proof of Theorem 1.6.5 is based on the following key lemma. ∞ . Then there exists a Lemma 1.6.7 Let A ∈ H∞ (X) and h ∈ H0∞ (φ ) for φ > φA constant C > 0 such that  αk h(2k tA)||L(X) ≤ C sup |αk |, αk ∈ C, t > 0. || k∈Z

k∈Z

Proof The assumption h ∈ H0∞ (φ ) implies that there exist constants c, β > 0 such that |h(z)| ≤ c Setting f (z) :=



k∈Z αk h(2



k tz)

|z|β , 1 + |z|2β

z ∈ φ .

and observing that for r = t|z|

|h(2k tz)| ≤ c



k∈Z

k∈Z

(r2k )β 2c ≤ , 1 + (r2k )2β 1 − 2−β

we see that |f (z)| ≤ sup |αk | k∈Z



|h(2k tz)| ≤ C sup |αk | k∈Z

k∈Z

for some constant C > 0. Hence, the above series is absolutely converging and f ∞ by defines a bounded and analytic function on φ . Since A ∈ H∞ (X) and φ > φA assumption, we obtain 



αk h(2k tA)L(X) = f (A)L(X) ≤ CA |f |H ∞ ≤ C sup |αk |. k∈Z

k∈Z



Proof of Theorem 1.6.5 We assume first that F ∈ 1 F (A) = 2πi

 F (λ)(λ − A) 

−1

1 dλ = 2πi



H0∞ (φ ; L(X)).

Then

F (λ)λ−1/2 A1/2(λ − A)−1 dλ, 

∞ , φ). Without loss of where  is defined by  = {re±iθ : r ≥ 0} for some θ ∈ (φA + generality we may assume in the following that λ ∈  :=  ∩ {λ ∈ C : Imλ ≥ 0}. Due to our assumption that F ∈ H0∞ (φ ; L(X)), the above integrals are absolutely + convergent and for N := {λ ∈  + : 2−N ≤ |λ| ≤ 2N } we may thus write

1 N→∞ 2πi

F (A) = lim

 + N

F (λ)λ−1/2 A1/2 (λ − A)−1 dλ = lim FN (A), N→∞

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

where FN (A) =

eiθ/2 2πi

eiθ/2 = 2πi =

 

2N 2−N 2N 2−N

F (reiθ )A1/2(reiθ − A)−1 r −1/2 dr F (reiθ )h(r −1 A)r −1 dr

N−1  2 eiθ/2  F (2k teiθ )h(2−k t −1 A)t −1 dt 2πi 1 k=−N

=

eiθ/2



2πi

2

HN (t)t −1 dt,

1

and h ∈ H0∞ is defined by h(z) = z1/2(eiθ − z)−1 and HN is given by N−1 

HN (t) :=

F (2k teiθ )h(2−k t −1 A),

t ∈ [1, 2].

k=−N

For x ∈ X and x  ∈ X we may estimate HN by Lemma 1.6.7 as | < HN (t)x, x  > | = |

N−1 

< F (2k teiθ )h(2−k t −1 A)x, x  > |

k=−N

 =

|

N−1 

 k=−N

 =

|< 

≤ ||

εk2 < F (2k teiθ )h1/2 (2−k t −1 A)x, x  > |

N−1 

εk F (2k teiθ )h1/2 (2−k t −1 A)x,

k=−N

N−1 

N−1  k=−N

εk F (2k teiθ )h1/2 (2−k t −1 A)x||L2(;X)

k=−N

×||

N−1 

εk h1/2 (2−k t −1 A∗ )x  ||L2 (;X )

k=−N

≤ R(F (φ ))

N−1 

εk h1/2 (2−k t −1 A)x||L2 (;X)

k=−N

×||

N−1 

εk h1/2 (2−k t −1 A∗ )x  ||L2 (;X )

k=−N

≤ C R(F (φ ))xx  . 2

εk h1/2 (2−k t −1 A∗ )x  > |

41

42

M. Hieber

Hence, HN (t) is uniformly bounded in t ∈ [1, 2] and N ∈ N, which shows that ||F (A)||L(X) ≤ C 2 R(F (φ )). Finally, in order to prove the assertion for general F ∈ H ∞ (φ ; L(X)), we replace F by Fn , where Fn is given by Fn (z) = F (z)n and n is the approximate identity introduced in Example 1.4.4. 

R-Bounded H ∞ -Calculus and Property (α) It is now a natural question to ask whether the set of operators produced by a bounded H ∞ -calculus from a uniformly bounded set of functions is R-bounded. We then say that A ∈ H∞ (X) admits an R-bounded H ∞ -calculus provided the set {h(A) : h ∈ H ∞ (θ ), |h|θ∞ ≤ 1} is R-bounded for some θ > 0. The class of such operators is denoted by RH∞ (X) R∞ of A is defined to be the infimum of such angles θ . and the RH∞ -angle φA Note that we have the relations RH(X) ⊂ H∞ (X) ⊂ S(X). It is a very remarkable fact that, roughly speaking, the classes H∞ (X) and RH∞ (X) coincide for Banach spaces satisfying the so called property (α). A Banach space X is said to have property (α) if the contraction principle described in Remark 1.6.2(e) holds for random sequences (ri rj )i,j ∈N , i.e. more precisely if there exists a constant C > 0 such that for each n ∈ N, each subset (xij )i,j =1,...,n of X and each subset (αij )i,j =1,...,n of C with |αij | = 1 

1 1 0

0



n 



1 1

ri (s)rj (t)αij xij X dsdt ≤ C 0

i,j =1

0



n 

ri (s)rj (t)X dsdt.

i,j =1

We remark first that due to Khintchine’s inequality, see Remark 1.6.2(d), C has property (α). Secondly, Fubini’s theorem implies that Lp (; X) has property (α) for 1 ≤ p < ∞ provided X enjoys this property and  is a σ -finite measure space. Thus, closed subspaces of Lp (, C) have property (α) for the above values of p. The precise statement about the relation between the classes H∞ and RH∞ reads as follows. Proposition 1.6.8 Let X be a Banach space with property (α), A ∈ H∞ (φ ), φ  > φ and R > 0. Then the set φ

{f (A) : f ∞ ≤ R} is R-bounded. For a proof of this fact, we refer to [95].

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

43

Maximal Lp -Regularity We finally consider the problem of maximal Lp -regularity for parabolic evolution equations. By this we mean the following. Consider the inhomogeneous Cauchy problem  u (t) = Au(t) + f (t), t ≥ 0 (ACP0 ) (1.6.6) u(0) = u0 , where A is a sectorial operator on a Banach space X of spectral angle φA < π/2. Let 1 < p < ∞ and 0 < T ≤ ∞. We say that A admits maximal Lp -regularity on [0, T ) if for u0 = 0 and all f ∈ Lp ([0, T ); X) the solution of (1.6.6), given by  t e(t −s)Af (s)ds, u(t) = 0

is differentiable a.e., takes values in D(A) a.e., and u and Au belong to Lp ([0, T ); X). It follows from the closed graph theorem that in this case there exists a constant Cp such that u Lp ([0,T );X) + AuLp ([0,T );X) ≤ Cp f Lp ([0,T );X).

(1.6.7)

Observe that by (1.6.6), f = u − Au, and hence u as well as Au cannot lie in a “better” function space as f . This explains the word “maximal”. Our aim is to show that such an estimate holds true in a certain class of Banach spaces if and only if A is a so called R-sectorial operator . Here we call a sectorial operator A R-sectorial of angle φ ∈ (0, π) provided σ (A) ⊂ φ and for all φ  > φ the set {λR(λ, A) : φ  ≤ | arg(λ)| ≤ π} is R-bounded. The infimum of all φ for which the above set is R-bounded is called R. the R-angle of A, denoted by φA The class of all R-sectorial operators on X will be denoted by RS(X) . We are now in the position to state and prove the following result on the closedness of two resolvent commuting operators. Theorem 1.6.9 (Kalton–Weis Sum Theorem) Let A and B be two resolvent commuting sectorial operators on a Banach space X. Assume that B ∈ H∞ (X) and R < π. Then A + B is closed on D(A) ∩ D(B) that A is R-sectorial with φB∞ + φA and there exists a constant C > 0 such that Ax + Bx ≤ CAx + Bx,

x ∈ D(A) ∩ D(B).

(1.6.8)

Proof The idea of the proof is to show that A(A + B)−1 and B(A + B)−1 can be defined to be bounded operators on X by means of the operator-valued functional calculus developed in Theorem 1.6.5.

44

M. Hieber

More precisely, let B be the subalgebra of L(X) consisting of all operators R , π) such commuting with resolvents of B and choose φ1 ∈ (φB∞ , π), φ2 ∈ (φA that φ1 + φ2 < π. Consider the functions F and G defined by F (λ) := A(λ + A)−1 ,

G(λ) := (λ + A)n (λ)2 n (A)2

with (n ) as in Example 1.4.4. Then F ∈ RH ∞ (φ1 ; B) and G ∈ H0∞ (φ1 ; B) and G(λ)F (λ) = n (A)2 An (λ)2 . By the multiplicativity of the functional calculus, (B + A)n (B)2 n (A)2 F (B) = G(B)F (B) = An (B)2 (A)2 . By Theorem 1.6.5, F (B) ∈ L(X) and there thus exists a constant C > 0 such that n (B)2 n (A)2 Ax ≤ Cn (B)2 n (A)2 (B + A)x,

x ∈ D(A) ∩ D(B).

Letting n → ∞ gives Ax ≤ C(B + A)x,

x ∈ D(A) ∩ D(B).

Since Bx ≤ Ax + (A + B)x ≤ (C + 1)(A + B)x for x ∈ D(A) ∩ D(B), the estimate (1.6.8) follows. In order to prove the closedness of A+B on D(A)∩D(B), let xn ∈ D(A)∩D(B) with xn → x and Axn + Bxn → y in X. By (1.6.8), (Axn ) as well as (Bxn ) are Cauchy sequences so that x ∈ D(A)∩D(B) and limn→∞ (Axn +Bxn ) = Ax+Bx = y by the closedness of A and B. The proof is complete. 

Let us come back to the maximal regularity estimate (1.6.7), i.e. u Lp ([0,T );X) + AuLp ([0,T );X) ≤ Cp f Lp ([0,T );X), which we now rewrite as an inequality for operators on Lp ([0, T ); X) for 1 < p < d ∞. Indeed, let B˜ be the derivative operator dt on X˜ = Lp ([0, T ); X) and let A˜ be ˜ ˜ the extension of A to X given by (Af )(t) = −A(f (t)). Then estimate (1.6.7) is equivalent to ˜ ˜ + By ˜ ˜ ≤ C(A˜ + B)y ˜ Ay X X X˜ ,

˜ ∩ D(B). ˜ y ∈ D(A)

(1.6.9)

Of course, A˜ and B˜ are sectorial operators with commuting resolvents. We will show in the following that B˜ has a bounded H ∞ -calculus on X˜ = Lp ([0, T ); X) for all angles φ > π/2 provided X is a U MD-space. Recall that this class of spaces has been introduced in Definition 1.2.7 in Sect. 1.2. Taking this for granted for the time being and assuming that A is R-sectorial of R < π/2, we obtain the main assertion of the following characterization angle φA theorem of maximal Lp -regularity.

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

45

Corollary 1.6.10 Let A be the generator a bounded holomorphic semigroup on a UMD Banach space X. Then A has maximal Lp -regularity for one (all) p ∈ (1, ∞) R < π/2. on R+ if and only if −A is R-sectorial of angle φA For the proof of the fact that R-sectoriality is necessary for maximal Lp regularity, we refer to Proposition 3.17 of [36]. It remains to show that the derivative operator B˜ defined above has a bounded H ∞ -calculus on Lp (R+ ; X) for all angles φ > π/2 provided X is a U MD-space. We will do this by applying an operator-valued Fourier multiplier theorem. Operator-Valued Fourier Multipliers Consider the Fourier transform F on S(Rn ; X), X, Y being arbitrary Banach spaces, as described in Sect. 1.2 and let m : Rn → L(X, Y ) be a bounded and measurable function. It induces a map Tm : S(Rn ; X) → L∞ (Rn ; Y ) by Tm :

f → F −1 (m(·)[f(·)]).

(1.6.10)

Given 1 < p < ∞, we then call m an Lp -Fourier multiplier if there exists a constant Cp such that Tm f Lp (Rn ;Y ) ≤ Cp f Lp (Rn ;X) ,

f ∈ S(Rn ; X).

(1.6.11)

In this case, Tm extends uniquely to an operator Tm ∈ L(Lp (Rm ; X), Lp (Rn ; Y )) whose norm is the smallest constant Cp for which (1.6.11) holds. The following theorem is the extension of the classical Mikhlin’s theorem to the operator-valued situation. Theorem 1.6.11 (Weis) Assume that X and Y are U MD-Banach spaces and let 1 < p < ∞. Suppose that m ∈ C 1 (R \ {0}; L(X, Y )) satisfies the following conditions (i) R{m(ξ ) : ξ ∈ R \ {0}} =: M0 < ∞, (ii) R{ξ m (ξ ) : ξ ∈ R \ {0}} =: M1 < ∞. Then the operator Tm defined by (1.6.10) is bounded from Lp (R; X) into Lp (R; Y ) and there exists a constant Cp > 0 such that Tm L(Lp (R;X),Lp (R;Y )) ≤ Cp (M0 + M1 ). As discussed above we are in particular interested in the following example. d on Lp (R; X). Example 1.6.12 Let X be a U MD-space, 1 < p < ∞, and B˜ = dt ∞ Then B˜ admits a bounded H (φ )-calculus for all φ > π/2. This follows exactly in the same way as described in Example 1.4.9, now, however, with the classical Mikhlin’s theorem being replaced by Theorem 1.6.11.

It is very interesting to note that we may strengthen Theorem 1.6.11 in the case where X and Y possess in addition property (α).

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Theorem 1.6.13 Assume that X and Y are UMD-Banach spaces with property (α), 1 < p < ∞ and that M ⊂ C 1 (Rn \ {0}; L(X, Y )) satisfies the conditions   (i) R{m(ξ ) : ξ ∈ R \ {0}, m ∈ M} =:  K0 < ∞, (ii) R {ξ m (ξ ) : ξ ∈ R \ {0}, m ∈ M} =: K1 < ∞. Then the family of operators T := {Tm : m ∈ M} ⊂ L(Lp (R; X), Lp (R; Y )) defined by (1.6.10) is R-bounded with R-bound R{T } ≤ C(K0 + K1 ), where the constant C > 0 depends only on p, X, Y . Turning our attention to the multi-dimensional case, we note that an operatorvalued Fourier multiplier theorem of Lizorkin type can be deduced from Theorem 1.6.13 by induction. Theorem 1.6.14 Assume that X and Y are U MD-Banach spaces with property (α). Let 1 < p < ∞ and n ∈ N. Suppose that the family M ⊂ C n (Rn \ {0}; L(X, Y )) satisfies   R {ξ α Dξα m(ξ ) : ξ ∈ Rn \ {0}, α ∈ 0, 1n , m ∈ M} =: K < ∞. Then the family of operator T := {Tm : m ∈ M} ⊂ L(Lp (Rn ; X), Lp (Rn ; Y )) defined by (1.6.10) is R-bounded with R-bound R{T } ≤ CK, where the constant C > 0 depends only on p, X, Y . Further Results Finally for J = [0, T ] with T > 0 consider the Cauchy problem u (t) − Au(t) = f,

t ∈ J,

(1.6.12)

u(0) = u0 ,

with initial data u0 = 0. We would like to characterize those initial data u0 for which the solution u of (1.6.12) has the maximal Lp -regularity property. To this end, we recall that the real interpolation space (X, (D(A)))θ,p was introduced in Definition 1.3.16 and Remark 1.3.17 as (X, D(A))θ,p = {x ∈ X : φθ,p (x, ·) ∈ Lp (0, T )} with norm xθ,p := x + φθ,p (x, ·)Lp (0,T ) . Thus, a solution of (1.6.12) with f = 0 satisfies u ∈ Lp ((0, T ), D(A)) if and only if u0 ∈ (X, D(A))1−1/p,p . In fact, Au ∈ Lp ((0, T ); X) if and only if φ1−1/p,p (u0 , ·)Lp (0,T ) = =

 

T 0 T 0

p

t p(1−(1−1/p))−1Aet A u0 X dt p

Aet A u0 X dt

1/p

1/p

< ∞.

Combining this observation with Corollary 1.6.10 we obtain the following theorem.

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Theorem 1.6.15 Let X be a UMD space, 1 < p < ∞, and assume that −A R < π . Then (1.6.12) has precisely one solution u ∈ is R-sectorial of angle φA 2 1,p p W (J ; X) ∩ L (J ; D(A)) if and only if f ∈ Lp (J ; X) and u0 ∈ DA (1 − 1/p, p) The proof of characterization of the maximal Lp -regularity for solutions of the Cauchy problem (1.6.12) given in Corollary 1.6.10 and Theorem 1.6.15 was based on the Kalton–Weis sum Theorem 1.6.9. The latter theorem has many further consequences. We state here only the Mixed-Derivative theorem. Consider in the situation of Theorem 1.6.9 the function F given F (λ) = λα B 1−α (λ + B)−1 . The representation 1 F (λ) = 2πi

 

z−α λz(λz − B)−1 dz 1+z

for a suitable contour  shows that F (ϕ ) is R-bounded provided B is R-sectorial. We hence obtain the following result. Proposition 1.6.16 (Mixed Derivative Theorem) Let A and B be two resolvent commuting operators on a Banach space X. Assume that A ∈ H∞ (X) and B is ∞ + φ R < π. Then Aα B 1−α (A + B)−1 is a bounded operator R-sectorial with φA B on X for each α ∈ (0, 1). In particular, for each α ∈ (0, 1) D(A) ∩ D(B) = D(A + B) → D(Aα B 1−α ). A typical situation is the following: Let  ⊂ Rn be a bounded domain with smooth boundary, X0 = Lq (), X1 = H 2,q () for q ∈ (1, ∞) and for T > 0 and p ∈ (1, ∞) set E1 (T ) := Lp (0, T ; X1 ) ∩ H 1,p (0, T ; X0 ). Corollary 1.6.17 If θ ∈ [0, 1], then E1 (T ) → H θ,p (0, T ; H 2−2θ,q ()). At this point one might ask the question whether all generators of analytic semigroups on Lp (, μ) where (, μ) denotes a σ -finite measure space and 1 < p < ∞ are R-sectorial. Kalton and Lancien [90] gave a complete answer to this question. Theorem 1.6.18 (Kalton–Lancien) Assume that X has an unconditional basis and that all generators of analytic semigroups on X are R-sectorial. Then X is isomorphic to a Hilbert space. We finish this section with a result saying that R-sectorial operators are well behaved under perturbations. This fact will be used later on in various sections.

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For A ∈ S(X) and θ > φA , we set M(A)θ := sup{AR(λ, A) : λ = 0, π > | arg(λ)| > θ }; R , we set if A ∈ RS(X) and θ > φA

R(A)θ := R{AR(λ, A) : λ = 0, π > | arg(λ)| > θ }. R . Assume that B is an operator Proposition 1.6.19 Let A ∈ RS(X) and θ > φA with D(A) ⊂ D(B) and

Bx ≤ aAx + bx,

x ∈ D(A),

for some a, b ≥ 0. If a < (M(A)θ R(A)θ )−1 , then there exists ν ≥ 0 such that R A + B + ν is R-sectorial. Moreover, φA+B+ν ≤ θ for this ν. The proof follows the lines of the perturbation Theorem 1.3.15 for sectorial operators, where now the arguments for the norm of (A + B + μ)−1 have to be replaced by R-bounds. Details are left to the reader; see e.g. [36] or [95].

1.7 Quasilinear Evolution Equations In this section we present several results on the local and global well-posedness for abstract quasilinear parabolic equations as well as on the dynamics of their solutions. These results will be employed in our investigations in various sections of Part B. They are mainly due to Prüss [120], Prüss and Simonett [121], Köhne et al. [94], and Prüss et al. [126]. A convenient reference for these results is the monograph by Prüss and Simonett [124], Chapter 5. Consider the quasilinear problem v˙ + A(v)v = F (v),

t > 0,

v(0) = v0 ,

(1.7.1)

Assume that (A, F ) : Vμ → B(X1 , X0 ) × X0 and v0 ∈ Vμ . Here X1 and X0 are Banach spaces such that X1 → X0 with dense embedding and Vμ is an open subset of the real interpolation space Xγ ,μ := (X0 , X1 )μ−1/p,p ,

μ ∈ (1/p, 1].

We are mainly interested in solutions v of (1.7.1) having maximal Lp -regularity, i.e. v ∈ Hp1 (J ; X0 ) ∩ Lp (J ; X1 ) =: E1 (J ), where J = (0, T ). The trace space of this class of functions is given by Xγ := Xγ ,1 . However, to see and exploit the effect of parabolic regularization in the Lp -framework it is also useful to consider solutions in the class of weighted spaces 1 (J ; X0 ) ∩ Lp,μ (J ; X1) =: E1,μ (J ), v ∈ Hp,μ

which means t 1−μ v ∈ E1 (J ).

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Here, for μ ∈ (1/p, 1] and a time interval J ⊂ [0, ∞) we set Lpμ (J ; X) = {u ∈ L1loc (J ; X) : t 1−μ u ∈ Lp (J ; X)}, Hμ1,p (J ; X) = {u ∈ L1loc (J ; X) ∩ H 1,1(J ; X) : t 1−μ u ∈ Lp (J ; X)}, The trace space for this class of weighted spaces is given by Xγ ,μ . In our approach it is crucial to know that the operators A(v) have the property of maximal Lp regularity. We recall from Sect. 1.6 that an operator A0 in X0 with domain X1 has maximal Lp -regularity, if the linear problem v˙ + A0 v = f,

t ∈ J, v(0) = 0,

(1.7.2)

admits a unique solution v ∈ E1 (J ), for any given f ∈ Lp (J ; X0 ) =: E0 (J ). In this case we write A0 ∈ MRp (X0 ). It has been proved in [121] that in this case maximal regularity also holds in the weighted case. By this we mean that for each f ∈ Lp,μ (J ; X0 ) there exists a unique u ∈ Lp,μ (J ; X0 ) such that u , Au ∈ Lp,μ (J ; X0 ) and such that u solves (1.7.2). In this case we write A0 ∈ MRp,μ (X0 ). Proposition 1.7.1 Let p ∈ (1, ∞) and 1/p < μ ≤ 1. Then A0 ∈ MRp (X0 ) ⇐⇒ A0 ∈ MRp,μ (X0 ) Local Solutions The local existence result for Eq. (1.7.1) reads as follows. Theorem 1.7.2 Let p ∈ (1, ∞), v0 ∈ Vμ be given and suppose that (A, F ) satisfies (A, F ) ∈ C 1 (Vμ ; B(X1 , X0 ) × X0 ),

(1.7.3)

for some μ ∈ (1/p, 1]. Assume in addition that A(v0 ) has maximal Lp -regularity. Then there exist a = a(v0 ) > 0 and r = r(v0 ) > 0 with B¯ Xγ ,μ (v0 , r) ⊂ Vμ such that problem (1.7.1) has a unique solution v = v(·, v1 ) ∈ E1,μ (0, a) ∩ C([0, a]; Vμ), on [0, a], for any initial value v1 ∈ B¯ Xγ ,μ (v0 , r). In addition, t∂t v ∈ E1,μ (0, a), in particular, for each δ ∈ (0, a) we have v ∈ Hp2 ((δ, a); X0 ) ∩ Hp1 ((δ, a); X1 ) → C 1 ([δ, a]; Xγ ) ∩ C 1−1/p ([δ, a]; X1), i.e. the solution regularizes instantly.

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The next result provides information about the continuation of local solutions. Corollary 1.7.3 Let the assumptions of Theorem 1.7.2 be satisfied and assume that A(v) has maximal Lp -regularity for all v ∈ Vμ . Then the solution v of (1.7.1) has a maximal interval of existence J (v0 ) = [0, t+ (v0 )), which is characterized by the following alternatives: (i) Global existence: t+ (v0 ) = ∞; (ii) lim inft →t+ (v0 ) distXγ ,μ (v(t), ∂Vμ ) = 0; (iii) limt →t+ (v0 ) v(t) does not exist in Xγ ,μ . Stability of Equilibria and Global Solutions Next we assume that there is an open set V ⊂ Xγ such that (A, F ) ∈ C 1 (V , B(X1 , X0 ) × X0 ).

(1.7.4)

Let E ⊂ V ∩ X1 denote the set of equilibrium solutions of (1.7.1), which means that v∈E

if and only if

v ∈ V ∩ X1 and A(v)v = F (v).

Given an element v∗ ∈ E, we assume that v∗ is contained in an m-dimensional manifold of equilibria. This means that there is an open subset U ⊂ Rm , 0 ∈ U , and a C 1 -function  : U → X1 , such that • (U ) ⊂ E and (0) = v∗ , • the rank of   (0) equals m, and • A((ζ ))(ζ ) = F ((ζ )),

(1.7.5)

ζ ∈ U.

We suppose that the operator A(v∗ ) has the property of maximal Lp -regularity, and define the full linearization of (1.7.1) at v∗ by A0 w = A(v∗ )w + (A (v∗ )w)v∗ − F  (v∗ )w

for w ∈ X1 .

(1.7.6)

After these preparations we can state the following result on convergence of solutions starting near v∗ which is called the generalized principle of linearized stability. Theorem 1.7.4 Let 1 < p < ∞. Suppose v∗ ∈ V ∩ X1 is an equilibrium of (1.7.1), and suppose that the functions (A, F ) satisfy (1.7.4). Suppose further that A(v∗ ) has the property of maximal Lp -regularity and let A0 be defined in (1.7.6). Suppose that v∗ is normally stable, which means (i) (ii) (iii) (iv)

near v∗ the set of equilibria E is a C 1 -manifold in X1 of dimension m ∈ N, the tangent space for E at v∗ is isomorphic to N(A0 ), 0 is a semi-simple eigenvalue of A0 , i.e. N(A0 ) ⊕ R(A0 ) = X0 , σ (A0 ) \ {0} ⊂ C+ = {z ∈ C : Re z > 0}.

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Then v∗ is stable in Xγ , and there exists δ > 0 such that the unique solution v of (1.7.1) with initial value v0 ∈ Xγ satisfying |v0 − v∗ |γ ≤ δ exists on R+ and converges at an exponential rate in Xγ to some v∞ ∈ E as t → ∞. The next result contains information on bounded solutions in the presence of compact embeddings and of a strict Lyapunov functional. Theorem 1.7.5 Let p ∈ (1, ∞), μ ∈ (1/p, 1), μ¯ ∈ (μ, 1], with Vμ ⊂ Xγ ,μ open. Assume that (A, F ) ∈ C 1 (Vμ ; B(X1, X0 ) × X0 ), and that the embedding Xγ ,μ¯ → Xγ ,μ is compact. Suppose furthermore that v is a maximal solution which is bounded in Xγ ,μ¯ and satisfies distXγ ,μ (v(t), ∂Vμ ) ≥ η > 0, for all t ≥ 0.

(1.7.7)

Suppose that  ∈ C(Vμ ∩ Xγ ; R) is a strict Lyapunov functional for (1.7.1), which means that  is strictly decreasing along non-constant solutions. Then t+ (v0 ) = ∞, i.e. v is a global solution of (1.7.1). Its ω-limit set ω+ (v0 ) ⊂ E in Xγ is nonempty, compact and connected. If, in addition, there exists v∗ ∈ ω+ (v0 ) which is normally stable, then limt →∞ v(t) = v∗ in Xγ . The Semilinear Case We state here also a simplified version of the general result concerning semilinear evolution equations. It will then be applied to the primitive equations in Sect. 1.15. As above let X0 , X1 be Banach spaces such that X1 → X0 is densely embedded, and let A : X1 → X0 be bounded. For 0 < T ≤ ∞ consider the semi-linear problem u + Au = F (u) + f,

0 < t < T,

u(0) = u0 .

(1.7.8)

The subsequent existence and uniqueness result is based on the following assumptions. Here, for β ∈ [0, 1] define Xβ to be the complex interpolation space [X0 , X1 ]β . (H1) A has maximal Lq -regularity for q ∈ (1, ∞). (H2) F : Xβ → X0 satisfied the estimate F (u1 ) − F (u2 )X0 ≤ C(u1 Xβ + u1 Xβ )(u1 − u2 Xβ ) for some C > 0 independent of u1 , u2 . (H3) β − (μ − 1/q) ≤ 12 (1 − (μ − 1/q)), that is 2β − 1 + 1/q ≤ μ. (S) X0 is of class UMD, and the embedding H 1,q (R; X0 ) ∩ Lq (R; X1 ) → H 1−β,q (R; Xβ ) is valid for each β ∈ (0, 1) and q ∈ (1, ∞).

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Proposition 1.7.6 ([125]) Assume that the assumptions (H 1), (H 2), (H 3) and (S) hold and let u0 ∈ Xγ ,μ

and

f ∈ Lq (0, T ; X0 ).

Then there exists a time T  = T  (u0 ) with 0 < T  ≤ T such that problem (1.7.8) admits a unique solution u ∈ Hμ1,q (0, T  ; X0 ) ∩ Lqμ (0, T  ; X1 ). Furthermore, the solution u depends continuously on the data. Remark 1.7.7 Let us remark that (S) holds true whenever X0 is of class UMD and ∞ < π/2, there is an operator A# ∈ H∞ (X0 ) with domain D(A# ) = X1 satisfying φA # see Remark 1.1 of [125]. When investigating the question of a global solution, we consider t+ (u0 ) := sup{T  > 0 : Eq. (1.7.8) admits a solution on (0, T  )}. By the above Proposition 1.7.6, this set is non-empty, and we say that (1.7.8) has a global solution if for f ∈ Lq (0, T ; X0 ) one has t+ (u0 ) = T , where 0 < T ≤ ∞. Global existence results can be derived from suitable a priori bounds following [124, Theorem 5.7.1]. Proposition 1.7.8 ([124]) Assume in addition to the assumptions of Proposition 1.7.6 that for μ < μ ≤ 1 the embedding Xγ ,μ → Xγ ,μ is compact, and that for some τ ∈ (0, t+ (u0 )) the solution of (1.7.8) satisfies u ∈ Cb ([τ, t+ (u0 )); Xγ ,μ ), then there is a global solution to (1.7.8), i.e. T  = T .

Part B: Viscous Fluid Flows

In the second part of these lecture notes we investigate several problems from the theory of viscous fluid flows such as boundary layers, fluid structure interaction problems, free boundary value problems, liquid crystal flow and the primitive equations of geophysical flows. The methods and techniques established in Chapter I will play an important role in our investigation of these problems.

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We start with a discussion of conservation laws for general heat conducting fluids and deduce, assuming certain constitutive laws, the fundamental equations of viscous fluid flows, the incompressible and compressible Navier–Stokes equations. A natural step from here is the analysis of the Stokes equation in a half space Rn+ within the Lp -setting. The associated operator, the Stokes operator, is identified to be a sectorial operator on Lp (Rn+ ). It generates in particular a bounded, holomorphic p semigroup on Lσ (Rn+ ) for all p satisfying 1 < p ≤ ∞. For 1 < p < ∞, the space p n Lσ (R+ ) denotes the subspace consisting of all solenoidal vector fields in Lp (Rn+ ). Whereas the case 1 < p < ∞ is quite classical, it is not surprising that the proof for the case p = ∞ requires different methods. In this case the Helmholtz projection does not act as a bounded operator on L∞ (Rn+ ). We even show that the negative p Stokes operator admits an R-bounded H ∞ -calculus on Lσ (Rn ) for 1 < p < ∞ and obtain as a consequence of the results in Sect. 1.6 the property of maximal Lp – Lq -regularity for the Stokes equation. Section 1.10 deals with the H ∞ -calculus as well as with the maximal Lp regularity property for the classical Stokes as well as for the hydrostatic Stokes operator. The proof of these results are fairly involved and we thus present here only the key ideas of the proofs. In the Sect. 1.11 we consider stability questions for Ekman boundary layers. These boundary layers are explicit stationary solutions of the Navier–Stokes equations in the rotational framework. Considering perturbations of the Ekman layer, we are interested in the question whether there exists a global, weak solution of the perturbed systems converging to zero as t → ∞. This, of course, would imply the asymptotic stability of the Ekman layer. The methods developed in the Chapter I allow us then to prove that this is the case, provided the Reynolds number involved is small enough. Sections 1.12 and 1.13 investigate moving and free boundary value problems. In fact, the fluid-rigid body interaction problem considered in Sect. 1.12 is a typical example of a moving domain problem, whereas the two-phase problem discussed in Sect. 1.13 represents a free boundary value problem. The fluid-rigid body interaction problem will be discussed for the situation of compressible fluids. Maximal regularity of the linearized equation in Lagragian coordinates allows us to prove a local well-posedness result for strong solutions. Strong solutions for the two-phase problem for generalized Newtonian fluids are again obtained by a fixed point argument in the associated space of maximal regularity. The final two sections concern the primitive equations. These equations are a model for oceanic and atmospheric dynamics and are derived from the Navier– Stokes equations by assuming a hydrostatic balance for the pressure term. We show in Sect. 1.15 that these equations are globally strongly well-posedness for arbitrary large initial data lying in critical spaces, which in the given situation are given by μ the Besov spaces Bpq for p, q ∈ (1, ∞) with 1/p + 1/q ≤ μ ≤ 1. Finally, in Sect. 1.16 we show that the primitive equations may be obtained as the limit of anisotropically scaled Navier–Stokes equations. The proof relies on a maximal Lp regularity estimate for the differences (vε − v, ε(wε − w)) by the aspect ration ε,

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where (vε , wε ) and (v, w) are solutions of the anisotropic Navier–Stokes equations and the primitive equations, respectively.

1.8 Balance Laws In this section let  ⊂ Rn be a domain with C 1 -boundary. First Principles We begin with the balance laws of mass, momentum, and energy. They read as ∂t ρ + div(ρu) = 0

in ,

ρ(∂t + u · ∇)u + ∇π = div S

in ,

ρ(∂t + u · ∇) + div q = S : ∇u − πdiv u u = 0,

(1.8.1)

in ,

q ·ν =0

on ∂.

Here ρ means density, u velocity, π pressure, internal energy, S extra stress and q heat flux. This immediately gives conservation of the total energy. In fact, we have ρ(∂t + u · ∇)e + div(q + πu − Su) = 0

in ,

where e := |u|2 /2 + means the total mass specific energy density (kinetic plus internal). The energy flux e is given by e := q + πu − Su. Integrating over  yields  ∂t E(t) = 0,

E(t) = Ekin (t) + Eint (t) =

ρ(t, x)e(t, x)dx, 

provided q·ν =u=0

on ∂.

(1.8.2)

Hence, if (1.8.2) holds, total energy is preserved, independent of the particular choice of S and q. Thermodynamics Assume a given free energy ψ of the form ψ = ψ(ρ, θ, τ ), where θ denotes the (absolute) temperature and τ will be specified later. We then have the following thermodynamical relations: = ψ + θη η = −∂θ ψ

internal energy, entropy,

κ = ∂θ = −θ ∂θ2 ψ

heat capacity.

(1.8.3)

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Later on, for well-posedness of the heat problem, one requires κ > 0, i.e. ψ to be strictly concave with respect to θ ∈ (0, ∞). In the classical case, where ψ depends only on ρ and θ , one has the ClausiusDuhem equation ρ(∂t + u · ∇)η + div(q/θ ) = S : ∇u/θ − q · ∇θ/θ 2 + (ρ 2 ∂ρ ψ − π)(div u)/θ

in .

Hence, in this case the entropy flux η is given by η := q/θ and the entropy production by θ r := S : ∇u − q · ∇θ/θ + (ρ 2 ∂ρ ψ − π)(div u). Employing the boundary conditions (1.8.2), we obtain for the total entropy N by integration over    ∂t N(t) = r(t, x)dx ≥ 0, N(t) = ρ(t, x)η(t, x)dx, 



provided r ≥ 0 in . As div u has no sign we require π = ρ 2 ∂ρ ψ,

(1.8.4)

which is the famous Maxwell relation. Further, as S and q are independent, this requirement leads to the classical conditions S : ∇u ≥ 0

and

q · ∇θ ≤ 0.

(1.8.5)

Summarizing, we see that whatever we choose for S and q, we always have conservation of energy and the total entropy is non-decreasing provided (1.8.5), (1.8.4) and (1.8.2) are satisfied. Thus, these conditions ensure the thermodynamic consistency of the model. As an example for S and q consider the classical laws due to Newton and Fourier which are given by S := SN := 2μs D + μb div u I,

2D := (∇u + [∇u]T ),

q := −α0 ∇θ.

In this case, (1.8.5) is satisfied as soon as μs ≥ 0, 2μs + nμb ≥ 0 and α0 ≥ 0 hold. Note that it does not matter at all whether μs , μb , α0 are constants or whether they depend on ρ, θ , or on other variables.

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Model Equations for the Isothermal, Incompressible Situation In the following, we consider the isothermal situation, neglecting thus the influence of temperature variations in the flow. A fluid is said to be incompressible if div u = 0

for all t > 0, x ∈ .

In this case, the density  is constant along the trajectories associated with the velocity u. Given an incompressible fluid with homogeneous density  at initial time, i.e. x → (0, x) = 0 is constant, the density  remains constant, i.e. (t, x) = 0 for all t > 0 and all x ∈ . For nonhomogeneous and incompressible fluids, the above balance laws yield the nonhomogeneous, incompressible Navier–Stokes equations. They read as ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

∂t  + div (u) = 0

in ,

∂t (u) + div (u ⊗ u) − 2 div (μ()D(u)) + ∇π = 0

in ,

div u = 0

in .

(1.8.6)

Concerning boundary conditions, the homogeneous Dirichlet boundary condition u = 0 on ∂ is widely used. In case of nonhomogeneous fluids, we have to add side and boundary conditions for the density. In particular, in order to obtain physically meaningful solutions, we need to add the condition (t, x) ≥ 0

for all t > 0 and all x ∈ .

If the density  of the fluid is homogeneous, then the mass balance equals the divergence free condition and we obtain the homogeneous, incompressible Navier– Stokes equations. Assuming (0, x) = 1 for all x ∈  yields the equations ⎧ ⎪ ⎪ ⎨∂t u + (u · ∇)u − μu + ∇π = 0 div u = 0 ⎪ ⎪ ⎩ u=0

in , in ,

(1.8.7)

on ∂.

Of course, the above sets of equations have to be completed by imposing initial conditions for u(0) and (0), respectively. Model Equations for the Isothermal, Compressible Situation In the following, consider again the isothermal situation. A fluid is called barotropic if its pressure is a function of the density, i.e. π = π(). The law π = π() serves

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57

as an constitutive equation for the barotropic fluid model. This is governed by the following system 

∂t  + div (u) = 0

(∂t u + u · ∇u) − (μb + μs )∇( div u) − μs u + ∇π() = 0

in , in , (1.8.8)

Again, the above set of equations has to be completed by boundary and initial conditions as well as by compatibility conditions. The above system is called the compressible Navier–Stokes system. A typical boundary condition is of the form u = ub and (ub |ν) = 0 on ∂, which guarantees no inflow or outflow.

1.9 The Stokes Equation in a Half Space In this section, we consider the Stokes equation in a half space H := Rn+1 + , i.e., we consider the set of equations ut − u + p = f

in (0, ∞) × H,

div u = 0

in (0, ∞) × H,

u=0

on (0, ∞) × ∂H,

(1.9.1)

u(0) = u0 , where u = (u1 , . . . , un , un+1 )T is interpreted as the velocity field and p as the pressure of the fluid, respectively. Our aim is to show that the Stokes operator A associated in a natural way with the Stokes equation generates a bounded analytic semigroup on the solenoidal free p subspace Lσ (H ) of Lp (H ) as well as on other function spaces related to bounded and continuous functions, respectively. Moreover, we prove that −A admits an R-bounded H ∞ -calculus on these spaces. This implies in particular maximal Lp –Lq -estimates for the solution of the above Stokes equation by the results of the previous section. Let us start with the Lp -theory of the solution of (1.9.1) for 1 ≤ p ≤ ∞. If p 1 < p < ∞, one often considers the subspace Lσ (H ) of Lp (H ) consisting of all p solenoidal functions f in L (H ). This space is defined by ·Lp (H )

Lpσ (H ) := {u ∈ Cc∞ (H ) : divu = 0 in H }

.

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Then Lp (H ) can be decomposed into Lp (H ) = Lpσ (H ) ⊕ Gp (H ), 1 (H )} and there exists where Gp (H ) := {u ∈ Lp (H ) : u = ∇π for some π ∈ Hloc p p a unique projection P : L (H ) → Lσ (H ) having Gp (H ) as its null space. P is called the Helmholtz projection in Lp (H ). In fact, in the case of the whole space Rn+1 , PRn+1 is given by PRn+1 = (δij + Ri Rj )1≤i,j,≤n+1 , where Ri denotes the i-th Riesz transform introduced in Proposition 1.1.7. For the half space H = Rn+1 + , PH can be defined to be

PH := RP E,

(1.9.2)

where the extension operator E : Lp (H )n+1 → Lp (Rn+1 )n+1 is defined to be the even extension of fk to Rn+1 for k = 1, . . . , n and (Ef )n+1 is set to be the odd extension of fn+1 to Rn+1 . Moreover, the restriction operator R : Lp (Rn+1 )n+1 → Lp (H )n+1 is defined as R := 12 E ∗ , with E ∗ being the adjoint of E. It can be shown that Lpσ (H ) = {u ∈ Lp (H ) : div u = 0, γfn+1 = 0},

(1.9.3)

where γ is the trace operator and γfn+1 is interpreted for f ∈ Lp (H ) in the sence of traces. p We then associate with (1.9.1) the so-called Stokes operator A in (Lσ (H ))n+1 defined as Au : = PH u

(1.9.4) 1,p

D(A) : = (W 2,p (H ) ∩ W0 (H ) ∩ Lpσ (H ))n+1 .

(1.9.5)

Observe that the Helmholtz projection is neither bounded in L1 (H ) or L∞ (H ). p Thus the usual decomposition of Lp (H ) in Lσ (H ) and its orthogonal complement, which is true for 1 < p < ∞, is no longer possible if p = 1 or p = ∞. In order to treat also the case p = ∞, we will follow a different strategy and will solve the Stokes system by taking Fourier transforms and subsequent explicit calculations. Our strategy to solve the Stokes system is the following: Taking Fourier transforms we obtain a representation of the solution of the corresponding resolvent equation as a sum of two terms, the first being the resolvent of the Dirichlet Laplacian on Lp (Rn+1 + ), the second one being a remainder term. We then derive pointwise upper bounds on the remainder term which allow to prove Lp -estimates for the solution of the corresponding resolvent problem for 1 < p ≤ ∞, i.e., we obtain estimates of the form uλ p ≤

M f p |λ|

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

59

for the resolvent equation defined below in (1.9.6) and λ belonging to a suitable sector of the complex plane. The pointwise upper bound on the remainder term allows us also to show that for 1 < p < ∞ the negative Stokes operator admits a bounded H ∞ -calculus on p Lσ (H )n+1 . In fact, even a stronger result is true: we prove that the Laplacian on Rn and the remainder term even admit an R-bounded H ∞ -calculus on Lp (H )n+1 for 1 < p < ∞. Thus the Stokes operator admits an R-bounded H ∞ -calculus on p Lσ (H )n+1 . As a consequence, we obtain maximal Lp –Lq -regularity for the solution of the Stokes equation (1.9.1). We recall that for 0 < θ ≤ π the sector θ in the complex plane is defined by θ := {z ∈ C\{0}; |argz| < θ }. A Representation Formula and the Stokes Operator in the Lp -Setting for 1 0 we write u = (v, w)T with v = (v1 , . . . , vn )T and f = (fv , fw )T with fv = ((fv )1 , . . . , (fv )n )T . Assume that fw (ξ, 0) = 0 for all ξ ∈ Rn . Applying the Fourier transform with respect to x we obtain (λ + |ξ |2 )v(ξ, ˆ y) − ∂y2 v(ξ, ˆ y) = fˆv (ξ, y) − iξ · p(ξ, ˆ y), ˆ y) − ∂y2 w(ξ, ˆ y) = fˆw (ξ, y) − ∂y p(ξ, ˆ y), (λ + |ξ 2 |)w(ξ, iξ · v(ξ, ˆ y) + ∂y w(ξ, ˆ y) = 0,

ξ ∈ Rn , y > 0 (1.9.7) ξ ∈ Rn , y > 0 (1.9.8)

ξ ∈ Rn , y > 0

(1.9.9)

v(ξ, ˆ 0) = 0,

ξ ∈ Rn

(1.9.10)

w(ξ, ˆ 0) = 0,

ξ ∈R .

(1.9.11)

n

Multiplying Eq. (1.9.7) by iξ , applying ∂y to (1.9.8) and adding them yields |ξ |2 p(ξ, ˆ y) − ∂y2 p(ξ, ˆ y) = −∂y fˆw (ξ, y) − iξ · fˆv (ξ, y) = −( div f )ˆ(ξ, y) = 0 for ξ ∈ Rn and y > 0, where we already took into account Eq. (1.9.9). Hence p(ξ, ˆ y) = e−|ξ |y p0 (ξ ),

ξ ∈ Rn ,

y>0

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M. Hieber

for some function pˆ 0 . We thus obtain for vˆ and wˆ the following representations 1 2ω(ξ )

v(ξ, ˆ y) =

∞ [e−ω(ξ )|y−s| − e−ω(ξ )(y+s)][fˆv (ξ, s) − iξ e−|ξ |s pˆ 0 (ξ )] ds, 0

(1.9.12) ∞ [e−ω(ξ )|y−s| + e−ω(ξ )(y+s)][fˆw (ξ, s) + |ξ |e−|ξ |s pˆ 0 (ξ )] ds,

1 2ω(ξ )

w(ξ, ˆ y) =

0

(1.9.13) 1

for ξ ∈ Rn , y > 0 and where ω(ξ ) := (|λ| + |ξ |2 ) 2 for ξ ∈ Rn . In order to determine pˆ0 , consider ∂y w(ξ, ˆ y) at y = 0, i.e.  ∞ ˆ 0) = e−ω(ξ )s [fˆw (ξ, s) + |ξ |e−|ξ |s pˆ0 (ξ )] ds = 0, ∂y w(ξ,

ξ ∈ Rn .

0

This implies pˆ 0 (ξ ) = −

(ω(ξ ) + |ξ |) |ξ |



∞

e−ω(ξ )s fˆw (ξ, s) ds,

ξ = 0.

(1.9.14)

0

By assumption, iξ · fˆv (ξ, y) + ∂y fˆw (ξ, y) = 0 for ξ ∈ Rn and y > 0. Integrating by parts yields  ∞  ∞ iξ e−ω(ξ )s fˆw (ξ, s) ds = − e−ω(ξ )s fˆv (ξ, s) ds fˆwL (ξ ) := ω(ξ ) 0 0 =:

−iξ ˆL f (ξ ) ω(ξ ) v

(1.9.15)

for all ξ ∈ Rn . Inserting (1.9.14) and (1.9.15) into (1.9.12) and (1.9.13) we obtain vˆ = vˆ1 + vˆ2

(1.9.16)

wˆ = wˆ 1 + wˆ 2 with vˆ1 , vˆ2 and wˆ 1 , wˆ 2 given for ξ ∈ Rn and y > 0 by vˆ1 (ξ, y) =

1 2ω(ξ )

1 vˆ2 (ξ, y) = 2ω(ξ )



∞

[e−ω(ξ )|y−s| − e−ω(ξ )(y+s)]fˆv (ξ, s) ds

0



0

∞

[e−ω(ξ )|y−s| − e−ω(ξ )(y+s)]

(iξ ) −|ξ |s e (ω(ξ ) + |ξ |) ds fˆwL (ξ ) |ξ |

1 |ξ | [e−y|ξ | − e−ω(ξ )y ]fˆvL (ξ ) = ω(ξ ) ω(ξ ) − |ξ |

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

1 wˆ 1 (ξ, y) = 2ω(ξ ) wˆ 2 (ξ, y) = =



∞

61

[e−ω(ξ )|y−s| − e−ω(ξ )(y+s)]fˆw (ξ, s) ds

0

−1 [e−y|ξ | − e−ω(ξ )y ]fˆwL (ξ ) ω(ξ ) − |ξ | 1 iξ · [e−y|ξ | − e−ω(ξ )y ]fˆvL (ξ ). ω(ξ ) ω(ξ ) − |ξ |

Observe that v1 = (λ − D )−1 fv ,

w1 = (λ − D )−1 fw ,

where D denotes the Laplacian in Rn+1 + subject to homgeneous Dirichlet boundary conditions. It follows from Examples 1.3.13(c), (d) and Theorem 1.3.10 that for 1 < p ≤ ∞ and λ ∈ θ with θ < π there exists a constant M > 0 such that v1 Lp (Rn+1 )n ≤

M fv Lp (Rn+1 )n + |λ|

w1 Lp (Rn+1 ) ≤

M fw Lp (Rn+1 ) . + |λ|

+

+

Hence, in order to obtain Lp -estimates for v and w, we may restrict ourselves in the following to v2 and w2 . Let θ < π and define rˆv : Rn × R+ × R+ × θ → C by rˆv (ξ, y, y  , λ) := where ω(ξ ) =



e−|ξ |y − e−ω(ξ )y |ξ | −ω(ξ )y  e , ω(ξ ) − |ξ | ω(ξ )

(1.9.17)

|λ| + |ξ |2 and set

rv (x, y, y  , λ) :=

1 (2π)n

 Rn

eix·ξ rˆv (ξ, y, y  , λ)dξ.

(1.9.18)

Note that rv is well defined since the above integral is absolutely convergent for (y, y  ) = (0, 0). Observe first that by a scaling argument it suffices to consider the case |λ| = 1 and | arg λ| ≤ θ < π. In fact, rv (x, y, y  , λ) = |λ|

n−1 2

rv (|λ| 2 x, |λ| 2 y, |λ| 2 y  , 1

1

1

λ ). |λ|

Next, for z ∈ C consider the function φ given by φ(z) = (1−e−z )z−1 for z ∈ C\{0} C and note that |φ(z)| ≤ 1+|z| for Re z ≥ 0 and some suitable constant C > 0. Thus rˆv (ξ, y, y  , λ) = ye−|ξ |y e−ω(ξ )y



|ξ | φ((ω(ξ ) − |ξ |)y). ω(ξ )

62

M. Hieber

Choose now a rotation Q in Rn such that Qx = (|x|, 0, . . . , 0) and write Qξ = (a, rb),

a ∈ R,

r > 0,

b ∈ Rn−1 ,

|b| = 1.

By this coordinate transformation and by a shift of the path of integration for a to the contour a → s + iε(r + |s|), s ∈ R, without changing the value of the integral thanks to Cauchy’s theorem, for ε small enough, we obtain for a multiindex α |(∂x )α rv (x, y, y  , λ)|  ∞  ∞ 1+|α|   (r + s) n−2 dsdr r e−c(r+s)(|x|+y+y ) ye−cy ≤M 1+r +s+y 0 0  ∞  e−cs(|x|+y+y )  = Mye−cy s n+|α| ds 1+y+s 0 for some constant M > 0 independent of x ∈ Rn , y, y  > 0 and all λ ∈ C with |λ| = 1 and |arg λ| ≤ θ < π. We thus have proved the following result. Lemma 1.9.1 Let θ ∈ (0, π) and α be a multiindex. Then there exist constants M, c > 0 such that  ∞ s n+|α| −cs(|x|+y+y )  e |(∂x )α rv (x, y, y  , λ)| ≤ Mye−cy ds, 1+y+s 0 where x ∈ Rn , y, y  > 0 and λ ∈ C with |λ| = 1 and |arg λ| ≤ θ < π. Remark 1.9.2 For θ < π we define rˆw : Rn × R+ × R+ × θ → Cn by rˆw (ξ, y, y  , λ) :=

e−|ξ |y − e−ω(ξ )y iξ −ω(ξ )y  e . ω(ξ ) − |ξ | ω(ξ )

(1.9.19)

Copying the above proof we see that the assertion of Proposition 1.9.1 remains true if rv is replaced by rw , where  1 eix·ξ rˆw (ξ, y, y  , λ)dξ, x ∈ Rn , y, y  > 0. rw (x, y, y  , λ) = (2π)n Rn (1.9.20) The kernel estimates given in Lemma 1.9.1 and Remark 1.9.2 allow us to derive Lp -estimates for v2 and w2 via the following simple lemma on Lp -continuity of integral operators acting in half spaces. Lemma 1.9.3 Suppose that 1 < p ≤ ∞ and let operator in

1 p

+

1 p

= 1. Let T be an integral

Lp (Rn+1 + )

of the form  ∞ (Tf )(x, y) = k(x − x  , y, y  )f (x  , y  )dx  dy  , 0

Rn

where k : Rn × R+ × R+ → C is a measurable function.

x ∈ Rn , y > 0,

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

63

(a) Let 1 < p < ∞. If 



∞

∞

( 0

0

p

p

k(·, y, y  )1 dy  ) p dy

1

p

=: M1 < ∞,

then T ∈ L(Lp (Rn+1 ≤ M1 . + )) and T L(Lp (Rn+1 + )) (b) Let p = ∞. If  ∞ sup k(·, y, y  )1 dy  =: M2 < ∞, y>0 0

then T ∈ L(L∞ (Rn+1 + )) and T L(L∞ (Rn+1 )) ≤ M2 . +

This follows immediately by applying Young’s and Hölder’s inequalities. Combining the estimates obtained in Proposition 1.9.1 and Remark 1.9.2 with Lemma 1.9.3 we obtain the following estimates for v2 and w2 . Lemma 1.9.4 Let 1 < p ≤ ∞ and θ ∈ (0, π). Let v2 and w2 be defined as above. Then there exists a constant M > 0 such that v2 Lp (Rn+1 )n ≤

M fv Lp (Rn+1 )n , + |λ|

w2 Lp (Rn+1 ) ≤

M fv Lp (Rn+1 )n + |λ|

+

+

for all λ ∈ θ . Proof Observe that for λ ∈ θ ,  ∞ v2 (x, y) = rv (x − x  , y, y  , λ)fv (x  , y  )dx  dy  , Rn

0

n−1

1

1

x ∈ Rn , y > 0,

1

λ where rv (x, y, y  , λ) = |λ| 2 rv (|λ| 2 x, |λ| 2 y, |λ| 2 y  , |λ| ), x ∈ Rn , y, y  > 0 and rv satisfies the estimate given in Lemma 1.9.1. Combining these bounds with Lemma 1.9.3, we obtain  1 1 y  · |(∂x )α rv (x, y, y  , λ)|dx ≤ Cn e−c|λ| 2 y . 1 1 Rn 1 + |λ| 2 y (|λ| 2 (y + y  ))1+|α|

Hence, if 1 < p < ∞, then  ∞ p 1 1  p rv (·, y, y  , λ)1 dy  ≤ Cn,p |λ|− 2 |λ|− 2 (1 + |λ| 2 y)−p 0

and





∞

∞

( 0

0

p

rv (·, y, y  , λ)1 dy  ) p dy ≤ Cnp |λ|−p . p

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M. Hieber

If p = ∞, then

∞ 0

1

Rn

|rv (·, y, y  , λ)|dxdy  ≤ Cn |λ|−1 (1 + |λ| 2 y)−1 and hence 

sup

∞

rv (·, y, y  , λ)1 dy  ≤

y>0 0

Cn . |λ| 

The estimate for rw follows in exactly the same way. Summing up, we proved the following result:

n+1 Proposition 1.9.5 Let 1 < p ≤ ∞, 0 < θ < π and λ ∈ θ . Let f ∈ Lp (Rn+1 + ) T such that div f = 0 and fn+1 |∂H = 0. Let u = (v, w) be defined as in (1.9.16). Then there exists a constant M > 0 such that

uLp (Rn+1 )n+1 ≤ +

M f Lp (Rn+1 )n+1 . + |λ|

For 1 < p < ∞ and λ > 0 consider the mapping R(λ) : Lpσ (H ) → Lpσ (H ),

f → uλ , p

where uλ is defined as in (1.9.16). Let A be the Stokes operator in Lσ (H ) defined as in (1.9.4). Then R(λ)(λ − A)f = f for all f ∈ D(A) and (λ − A)R(λ)f = f p for all f ∈ Lσ (H ). Thus R(λ) = (λ − A)−1 ,

λ > 0.

Proposition 1.9.5 combined with Theorem 1.3.10 implies now the following fundamental result. Theorem 1.9.6 Let 1 < p < ∞. Then the Stokes operator A defined as in (1.9.4) p generates a bounded holomorphic semigroup on Lσ (H )n+1 . For different approaches to this basic result, we refer to [51, 62, 112, 135] and [151]. The Stokes Operator on Spaces of Bounded Functions We now consider the Stokes equation in spaces of bounded functions, more precisely in BU Cσ (H ), C0,σ (H ) and L∞ σ (H ). It is our aim to show that the Stokes operator, defined in the manner described below, is the generator of an holomorphic semigroup on these spaces (which is not strongly continuous in the case of L∞ σ (H )). To this end, define BU Cσ (H ) := {f ∈ BU C(H ); divf = 0, f (x1 , . . . , xn , 0) = 0 for all x1 , . . . , xn ∈ R}

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

65

and ||·||L∞

C0,σ (H ) := {f ∈ Cc∞ (H ); divf = 0}

.

Let Xσ (H ) be one of the spaces BU Cσ (H ) or C0,σ (H ). For θ ∈ (0, π) and λ ∈ θ consider the mapping R(λ) : Xσ (H )n+1 → L∞ (H )n+1 ,

f → uλ ,

where uλ is the solution of the Stokes equation given in (1.9.16). Theorem 1.9.6 and a direct calculation show that {R(λ); λ > 0} is a pseudo-resolvent in Xσ (H )n+1 . By this we mean a function R : U → L(Xσ (H )n+1 ), U being a subset of C, satisfying the resolvent equation given in (1.3.2). Lemma 1.9.7 Let f ∈ Xσ (H )n+1 . Then lim λR(λ)f = f.

λ→∞

Proof Notice first that R(λ)fv = (λ − D )−1 fv + v2 (λ), R(λ)fw = (λ − D )−1 fw + w2 (λ), where D denotes the Dirichlet Laplacian and v2 , w2 are defined as above. By Example 1.3.13, the Dirichlet Laplacian D generates a C0 -semigroup on BU C(H ) or C0 (H ), respectively. It hence follows that lim λ(λ − D )−1 f = f for all f ∈ Xσ (H ).

λ→∞

It thus remains to prove that limλ→∞ λv2 (λ) = 0 in Xσ (H )n and limλ→∞ λw2 (λ) = 0 in Xσ (H ). In order to do so, note first that limλ→∞ λv2 (λ) = 0 in BU Cσ (H )n if and only if  lim sup

λ→∞ y>0 0

∞ Rn

λ(n+1)/2 rv (λ1/2 x  , λ1/2 y, λ1/2 y  , 1)fv (x − x  , y  )dx  dy  = 0. (1.9.21)

Since the above inner integral equals rv (0, y, y  , 1) = 0, it follows that (1.9.21) is satisfied provided  lim sup

λ→∞ y>0 0

∞ Rn

|λ(n+1)/2 rv (λ1/2 x  , λ1/2 y, λ1/2 y  , 1)|

× |fv (x − x  , y  ) − fv (x, 0)|dx dy  = 0.

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M. Hieber

Notice that the above double integral is dominated by sup

M |x  |≤

R λ1/2

,|y  |≤

S λ1/2

|fv (x − x  , y  ) − fv (x, 0)| + 2|fv |∞ M(

1 1 + ) S R

for some M > 0 and all S, R > 0 by the kernel estimates established above. This implies the assertion if Xσ (H ) = BU Cσ (H ). The case where of Xσ (H ) = C0,σ (H ) is proved in a similar way. 

The above lemma shows that kerR(λ) = 0 for all λ > 0. Hence, by the following Lemma 1.9.8, (see e.g. [13], Prop. B.6, for a proof), there exists a closed, densely defined operator AXσ in Xσ (H )n+1 such that R(λ) = (λ − AXσ )−1 , λ > 0. We call this operator the Stokes operator AXσ on Xσ (H )n+1 . Lemma 1.9.8 Let X be a Banach space, U ⊂ C be open and R : U → L(X) be a pseudo-resolvent. Then there is an operator A on X such that R(λ) = R(λ, A) for all λ ∈ U if and only if ker R(λ) = {0} for λ ∈ U . Proposition 1.9.5 implies now the following result, which was proved first in [40]. Proposition 1.9.9 The Stokes operator AXσ generates a strongly continuous holomorphic semigroup on Xσ (H )n+1 . Finally, we consider the solution of the Stokes equation in L∞ σ (H ). This space is defined as follows: note that ∇ acts as a bounded operator from Wˆ 1,1 (H ) into L1 (H )n+1 , where Wˆ 1,1 (H ) = {f ∈ L1loc (H ); ∇f ∈ L1 (H )}. Hence Div := −∇ ∗ is a bounded operator from L∞ (H )n+1 into Wˆ 1,1 (H )∗ . We define n+1 L∞ := kerDiv. σ (H )

 ∞ Thus f ∈ L∞ σ (H ) if and only if f ∈ L (H ) and H ∇ϕf = 0 for all n+1 → L∞ (H )n+1 defined ϕ ∈ W 1,1 (H ). Consider the mapping R(λ) : L∞ σ (H ) as before. Proposition 1.9.5 and a direct calculation implies that {R(λ); λ > 0} is a pseudo-resolvent. In contrast to the situation of Xσ (H ) we do not have that limλ→∞ λv2 (λ) = 0 in L∞ σ (H ). However, the representation of the remainder term given above allows us to show that kerR(λ) = 0 in L∞ σ (H ) for all λ > 0. Thus there n+1 such that exists a closed operator AL∞ in L∞ σ (H ) σ )−1 , λ > 0. R(λ) = (λ − AL∞ σ n+1 . Note that A ∞ is not the Stokes operator in L∞ We call the operator AL∞ Lσ σ (H ) σ densely defined. However, the arguments given in the proof of Theorem 1.3.10 yield the following result, which we state with some abuse of language as follows. It was proved first in [40].

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

67

Theorem 1.9.10 The Stokes operator AL∞ generates a holomorphic semigroup on σ n+1 , which is not strongly continuous in 0. (H ) L∞ σ Remarks 1.9.11 It was shown recently by Abe and Giga [1] that the Stokes operator n generates also a holomorphic semigroup on L∞ σ () provided  ⊂ R is a bounded or exterior domain with smooth boundary ∂. Their approach is very different from the one given above and relies on certain blow up arguments. H ∞ -Calculus and R-Bounded H ∞ -Calculus for the Stokes Operator Theorem 1.9.6 can be rephrased by saying that the negative Stokes operator −A p on Lσ (H )n+1 is a sectorial operator for all p ∈ (1, ∞). It is hence natural to ask whether −A admits a bounded H ∞ -calculus on Lp (H )n+1 for every sector θ with 0 < θ < π. In order to answer this question, recall that the resolvent R(λ, A) can be represented as R(λ, A)f = (λ − D )−1 f + S(λ), where f = (fv , fw )T and S(λ)f = (v2 (λ), w2 (λ))T . The following lemma on Lp -boundedness of integral operators will be useful in the following. Lemma 1.9.12 Let T be an integral operator of the form 

∞

(Tf )(y) =

k(y, y  )f (y  )dy  ,

y > 0,

(1.9.22)

0

where k : R+ × R+ → C is a measurable function such that the above integral is well defined. Suppose that for some p ∈ (1, ∞) there exists a constant M > 0 such that |(Tf )(y)| ≤

M 1

yp

f Lp (R+ ) ,

y > 0.

If T ∈ L(Lq0 (R+ )) for some q0 ∈ (p, ∞], then T ∈ L(Lq (R+ )) for all q ∈ (p, q0 ]. Proof By assumption, Tf is dominated pointwise by a function belonging to p p the weak Lp -space Lw (R+ ). Thus T : Lp (R+ ) → Lw (R+ ) is a bounded operator. The assumption and the Marcienkiewicz interpolation theorem imply that T ∈ L(Lq (R+ )) for all q ∈ (p, q0 ]. 

Corollary 1.9.13 Let k : R+ × R+ → C be a measurable function. Suppose that there exists M > 0 such that |k(y, y  )| ≤

M y , log 1 + y + y y

y, y  > 0.

Let T be defined as in (1.9.22) and let 1 < p ≤ ∞. Then T ∈ L(Lp (R+ )).

68

M. Hieber

Proof Note first that 

∞







|k(y, y )|dy ≤ M

0

log (1 +

∞

1+

0

y y )

y y

dy  =M y



∞

log (1 + s) ds < ∞ (1 + s)s

0

which implies that T ∈ L(L∞ (R+ )). If p > 1 let p1 + p1 = 1. For f ∈ Lp (R+ ) we obtain by Hölder’s inequality |Tf (y)| ≤ M





y  1 y  )dy p f Lp (R+ )  + y  )p

log p (1 +

∞

(y

0

≤

M 1

yp

f Lp (R+ ) ,

y > 0. 

Thus the assertion follows from Lemma 1.9.12. Let now h ∈ H0∞ (θ ), where 0 < θ < π is fixed. Consider the function kh,v (x, y, y  ) =

1 2πi



h(λ)rv (x, y, y  , −λ)dλ,

x ∈ Rn , y, y  > 0,



where rv is defined as in (1.9.18) and  := {ρe±iϕ , ρ ≥ 0} with 0 < ϕ < θ . The estimate for rv given in Lemma 1.9.1 yields |kh,v (x, y, y  )| ≤ ChH ∞

 

∞

|r(x, y, y  , ρe±i(π−ϕ) )|dρ

0 ∞

≤ Ch

1

ye

H∞

−cρ 2 y 

0



∞

σ 0

ne

−cσ (|x|+y+y ) 1

ρ 2 + ρy + σ

dσ dρ

=: ChH ∞ k1 (x, y, y  ).

(1.9.23)

Now 

y |k1 (x, y, y )|dx ≤ C n y + y R 

Splitting the latter integral at s =  Rn

|k1 (x, y, y  )|dx ≤

1 y ,



∞ 0



e−csy ds, 1 + sy

y, y  > 0.

we obtain

C y log (1 +  ), y + y y

y, y  > 0.

(1.9.24)

Define now the operator Th,v in Lp (Rn+1 + ) by  (Th,v f )(x, y) := 0

∞ Rn

kh,v (x − x  , y, y  )f (x  , y  )dx  dy  ,

x ∈ Rn , y > 0. (1.9.25)

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Then, by Young’s inequality, (1.9.23), (1.9.24) and Corollary 1.9.13 we have  ∞  ∞  ∞

p |(Th,v f )(x, y)|p dxdy ≤ kh,v (·, y, y  )1 f (·, y  )p dy  dy 0

Rn

0

0 p

≤ ChH ∞



∞  ∞ 0

0

× f (·, y  )p dy  p

log(1 +

p

y ) y

dy

p . Lp (Rn+1 + )

≤ ChH ∞ f 

(1.9.26)

Moreover, by Remark 1.9.2 the function rw defined as in (1.9.20) satisfies also an estimate of the form given in Lemma 1.9.1. Thus, the function kh,w defined by  1 kh,w (x, y, y  ) := h(λ)rw (x, y, y  , −λ)dλ, x ∈ Rn , y, y  > 0 2πi  also satisfies |kh,w (x, y, y  )| ≤ ChH ∞ k1 (x, y, y  ),

x ∈ Rn , y, y  > 0.

Define the operator Th,w as in (1.9.25) with kh,v replaced by kh,w . We conclude that Th,w satisfies estimate (1.9.26). Finally note that by Example 1.4.10 the operator −D admits a bounded H ∞ (θ )-calculus on Lp (Rn+1 + ) for every θ ∈ (0, π). Summing up, we thus proved the following result. Theorem 1.9.14 Let 1 < p < ∞ and let A be the Stokes operator in p n+1 defined as in (1.9.4). Then −A admits a bounded H ∞ ( )-calculus Lσ (Rn+1 θ + ) p n+1 n+1 on Lσ (R+ ) for each θ ∈ (0, π). We finish this section by showing that −A even admits a R-bounded p n+1 for each θ ∈ (0, π). Note that this follows H ∞ -calculus on Lσ (Rn+1 + ) p immediately by combing Theorem 1.9.14 with Proposition 1.6.8 since Lσ (H ) as a closed subspace of Lp (H ) has property (α). In order to be selfcontained we, however, aim to give here a different proof which is based on the techniques developed above. To this end, let us consider first the Laplacian  on Lp (Rn ). Lemma 1.9.15 Let 1 < p < ∞. Then − admits an R-bounded H ∞ -calculus on Lp (Rn ) on the sector θ for 0 < θ < π. Proof Let h ∈ H0∞ (θ ), where 0 < θ < π. Then the Fourier transform of h(−) is given by h(|ξ |2 ) for ξ ∈ Rn . The kernel kh (·) corresponding to h(| · |2 ) is given by  1 eixξ h(|ξ |2 )dξ, x ∈ Rn . kh (x) = (2π)n Rn

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Choosing a rotation Q such that Qx = (|x|, 0, . . . , 0) and writing Qξ = (a, rb), with a ∈ R, r > 0, b ∈ Rn−1 , |b| = 1 we obtain 

∞

kh (x) = cn

 r n−2

∞ −∞

0

 h( r 2 + a 2 )ei|x|a dadr,

x ∈ Rn .

Next, deform the contour of integration via Cauchy’s theorem to a = s + i (r + |s|) for r > 0 and s ∈ R and obtain |D α kh (x)| ≤ Cα hH ∞

1 |x|n+|α|

x ∈ Rn \{0}

,

for each multiindex α. If |x| ≥ 2|y| we obtain 

1

|kh (x − y) − kh (x)| = | 0



d kh (x − ty)dt| dt

≤ |y| 0

1

dt |y| hH ∞ ≤ C n+1 hH ∞ . n+1 |x − ty| |x|

This implies that for R > 0 the uniform Hörmander condition is satisfied, i.e.: 

 sup

|x|>2|y| hH ∞ ≤R

|kh (x − y) − kh (x)|dx ≤ C|y|R

|x|>2|y|

 = C|y|R

∞ 2|y|

dx |x|n+1

dr = CR. r2

(1.9.27)

By Example 1.4.10a), − ∈ H∞ (L2 (Rn )), and thus {h(A) : h ∈ H0∞ (θ ), hH ∞ (θ ) ≤ R} ⊂ L(L2 (Rn )) is uniformly bounded. By Remarks 1.6.2(c) and (d), there is a constant C > 0 such that (

N 

2

1 2

|Hj fj | ) L2 (Rn ) ≤ CR(

j =1

N 

1

|fj |2 ) 2 L2 (Rn )

j =1

for N ∈ N, Hj := hj (A), hj ∈ H0∞ (θ ), hj H ∞ (θ ) ≤ R and fj ∈ L2 (Rn ). Set now X := RN and define K : Lp (Rn , X) → Lp (Rn , X) by (Kf )i := Hi fi for i = 1, . . . , N. The uniform Hörmander condition (1.9.27) implies that  |x|>2|y|

K(x − y) − K(x)dx ≤ CR.

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Since K acts as a bounded operator on L2 (Rn ; X), the Benedek-Calderon-Panzone Theorem 1.2.5 implies that K is Lp -bounded for 1 < p < ∞. This means that there is a constant C > 0, depending only C given in (1.9.27), R and p such that (

N 

1

|Hj fj |2 ) 2 Lp (Rn ) ≤ C(

j =1

N 

1

|fj |2 ) 2 Lp (Rn ) .

j =1

Remark 1.6.2(d) implies that the set {h(A) : h ∈ H0∞ (θ ), hH ∞ (θ ) ≤ R} ⊂ L(Lp (Rn )) is R-bounded for all R > 0 and all 0 < θ < π.



Following the proof given in Example 1.4.10(b), it is now easy to deduce that −D , where D denotes the Dirichlet Laplacian, admits an R-bounded H ∞ (θ )calculus on Lp (Rn+1 + ) for each θ ∈ (0, π). Finally, combining (1.9.25) with (1.9.26), Corollary 1.9.13 and Lemma 1.6.3 we see that the sets {Th,v : h ∈ H0∞ (θ ), hH ∞ (θ ) ≤ R} ⊂ L(Lp (Rn+1 + )), {Th,w : h ∈ H0∞ (θ ), hH ∞ (θ ) ≤ R} ⊂ L(Lp (Rn+1 + )) are also R-bounded. Summarizing, we thus proved the following result. p

Theorem 1.9.16 Let 1 < p < ∞ and let A be the Stokes operator in Lσ (Rn+1 + ). p n+1 ∞ Then −A admits an R-bounded H (θ )-calculus on Lσ (R+ ) for each θ ∈ (0, π).

1.10 The Stokes and Hydrostatic Stokes Equations on Domains Given an open set  ⊂ Rn , n ≥ 2, and a time interval J = (0, T ) for 0 < T ≤ ∞ the Stokes equation is given by the following set of equations ⎧ ∂t u − u + ∇p = f ⎪ ⎪ ⎨ div u = 0 ⎪ u=0 ⎪ ⎩ u(0) = u0

in J × , in J × , on J × ∂, in .

(1.10.1)

Here f as well as u0 are given data and u = 0 on ∂ describes the classical Dirichlet boundary condition. An important tool in order to handle the incompressibility

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condition div u = 0 is the Helmholtz projection P , allowing to decompose the p function space Lp () for 1 < p < ∞ into the solenoidal space Lσ () and gradient n fields for a large class of domains  ⊂ R . The Helmholtz Decomposition Given an open set  ⊂ Rn , the Helmholtz decomposition deals with the question whether Lp () can be decomposed into a direct sum of the space of solenoidal vector fields and the space of gradient fields. If such a decomposition holds true, the Stokes equation (1.10.1) can then be reformulated as an evolution equation in the Lp -setting. For 1 < p < ∞ and  ⊂ Rn being an arbitrary open set we set 1,p

Gp () := {u ∈ Lp () : u = ∇π for some π ∈ Wloc ()}, Lpσ () := {u ∈ Cc∞ () : div u = 0 in }

·p

Definition 1.10.1 Let 1 < p < ∞ and  ⊂ Rn be an open set. We say that the Helmholtz decomposition for Lp () exists whenever Lp () can be decomposed into Lp () = Lpσ () ⊕ Gp (). p

The unique projection operator Pp : Lp () → Lσ () having Gp () as its null space is called the Helmholtz projection. If  ⊂ Rn is an open set and 1 < p < ∞, then the Helmholtz decomposition  exists for Lp () if and only if it exists for Lp () and we have Pp = Pp , i.e., P2 is orthogonal. The existence of the Helmholtz projection for Lp () is very strongly linked  1,p () with the following weak Neumann problem: given f ∈ Lp (), find π ∈ W satisfying  (∇π − f ) · ∇ϕ = 0,

 1,p (). ϕ∈W

(1.10.2)



Note that if ∂ as well as f and π are smooth, then integration by parts yields π = div f in , ∂ν π = f · ν on ∂, where ν denotes the outer normal on ∂, which explains that we call (1.10.2) the weak Neumann problem. The existence of the Helmholtz projection can be characterized in terms of the weak Neumann problem, see e.g. [57, 60, 133] Lemma 1.10.2 Let  ⊂ Rn be an open set and 1 < p < ∞. Then the Helmholtz decomposition exists for Lp () if and only if the weak Neumann problem (1.10.2)  1,p (). admits for all f ∈ Lp ()n a unique solution π ∈ W

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Remarks 1.10.3 Let us remark that the Helmholtz decomposition exists for L2 () for any open set  ⊂ Rn and that following Maslennikova and Bogovskii [110], there exist domains  ⊂ Rn such that the Helmholtz decomposition does not hold for some p = 2. In the following proposition we consider domains for which the Helmholtz decomposition is known to exist. Note that the bent half space Hh is defined by Hh = {x = (x  , xn ) ∈ Rn : xn > h(x  )} for some bending function h : Rn−1 → R. We also say that  is a perturbed half space, if there exists an R > 0 such that  \ BR = Rn+ \ BR , where BR denotes the open ball centered at x with radius R. Proposition 1.10.4 Let  ⊂ Rn be a domain and 1 < p < ∞. Then the Helmholtz decomposition exists for Lp () provided (a)  = Rn or  = Rn+ , (b)  is a bent half space, i.e. for  = Hh provided h ∈ C 1 (Rn−1 ), ∇h∞ ≤ δ and δ > 0 is small enough, (c)  ⊂ Rn is bounded, exterior, or a perturbed half space of class C 1 , (d)  ⊂ Rn , n ≥ 2, is a bounded and convex domain, (e)  ⊂ Rn is an aperture C 1 -domain. (f)  = Rn−1 × (0, δ), n ≥ 2, δ > 0, is a layer domain. Moreover, there exists ε > 0 such that the Helmholtz decomposition for Lp () exists provided  ⊂ R3 is bounded Lipschitz domain and 32 − ε < p < 3 + ε. The range of those p is sharp. The Stokes Equation in Domains with Compact Boundaries We consider the Stokes equation (1.10.1) on domains  ⊂ Rn with smooth boundaries by means of a localization procedure. The procedure is explained for the resolvent problem ⎧ ⎨ λu − u + ∇p = f in , (1.10.3) div u = 0 in , ⎩ u = 0 on ∂, where λ is assumed to lie in a suitable sector of the complex plane. For simplicity, we consider only domains with compact boundaries. By employing finite coverings, we then may include domains of the following type: given n ≥ 2, a domain  ⊂ Rn is called a standard domain, if  coincides with Rn , Rn+ , a bounded domain, an exterior domain, or a perturbed half-space. The first step in the localization procedure consists of choosing a finite covering m n 2 (Uj )m j =1 of  ⊂ R with boundary of class C and a partition of unity (ϕj )j =1 m subordinate to (Uj )j =1 . Multiplying (1.10.3) with ϕj leads to a localized perturbed version in Uj . Since the Stokes equations are invariant under rotations and translations, and by choosing Uj sufficiently small, one may assume that the localized version of Eq. (1.10.3) is either an equation on Rn or on a bent half-space Hj := {x ∈ Rn : xn > hj (x  )},

(1.10.4)

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M. Hieber

with a certain bending function hj : Rn−1 → R. We further transforms the localized system on Hj by v(x  , xn ) := (u ◦ φ)(x  , xn ) := u(x  , xn + hj (x  )),

(x  , xn ) ∈ Rn+ .

(1.10.5)

The resulting system for v then is an equation on Rn+ . Summarizing, by this procedure the Stokes resolvent problem on a domain is reduced to finitely many equations on Rn+ or Rn . A fundamental problem arising in this approach is the fact that the condition div u = 0 is not preserved, neither under multiplication with a cut-off function nor by transformation (1.10.5). In order to overcome this difficulty, several strategies have been developed. Strategy 1 We replace the transformation (1.10.5) by   v = T u := u ◦ φ − 0, . . . , 0, (∇  hj , 0) · (u ◦ φ) .

(1.10.6)

This transformation leaves the outer normal at the boundary invariant and implies div v = 0. The price one has to pay for this is a lift of the boundary smoothness from C 2 to C 3 . This is due to the fact that ∇  hj appears in the transformation and since the Stokes system is a second order system, we require existence of third order derivatives of the bending functions hj . The transformation (1.10.6) is utilized e.g., in [59, 115, 135]. Strategy 2 Suppose the divergence problem div w = g, w|xn =0 = 0 admits a solution given by a Bogovskii operator B : g → w. Then we may correct the lacking solenoidality by the term Qv := v − B div v.

(1.10.7)

The price to pay here is that we need to prove suitable mapping properties of Bogovskii type operators. This approach to the Stokes system is performed e.g., in [59] and [60]. Strategy 3 It is also possible to work with an inhomogeneous divergence condition right from the very beginning. In this case, the inhomogeneous Stokes systems in Rn and in Rn+ needs to be solved. Localizing the inhomogeneous equations produces then perturbation terms not only in the first n lines of (1.10.3), but also in the divergence condition. We explain here strategy 1 in more detail. We consider first the Stokes resolvent problem on  = H , where H is a bent half-space with h being a suitable bending function defined in (1.10.4). Since  is a C 3 domain, we may assume h ∈ BU C 3 (Rn−1 ). This implies that the transformation T defined in (1.10.6) is an isomorphism between the Sobolev spaces involved up to order two. In particular, it can be shown that T ∈ Lis (Lpσ (H ), Lpσ (Rn+ )) ∩ Lis (D(AH,p ), D(ARn+ ,p )),

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where D(A,p ) denotes the domain of the Stokes operator A,p on . Hence, we may define AT := T AH,p T −1 , p

which is an operator in Lσ (Rn+ ) with domain D(AT ) = D(ARn+ ,p ). The smallness of ∇  h implies that B := AT − ARn+ ,p is a relatively bounded perturbation of the Stokes operator ARn+ ,p . In the next step, we set u :=

m 

ϕ j uj ,

j =1

p :=

m 

ϕj pj ,

j =1

where (uj , pj ) is the restricted bent half-space solution to data fj = φj f that corresponds to Uj . To be precise, modulo rotation and translation, one has Uj ∩ Hj = Uj ∩ ,

Uj ∩ ∂Hj = Uj ∩ ∂.

Then u solves the perturbed Stokes resolvent problem ⎧  ⎪ f+ m ⎨ λu − u + ∇p =  j =1 (−uj ϕj − ∇uj ∇ϕj + pj ∇ϕj ) in , u in , div u = m j =1 j ∇ϕj ⎪ ⎩ u=0 on ∂. (1.10.8) It remains to prove that the remainder terms are of lower order so that they may be absorbed into the terms on the left hand side. Whereas the terms uj ϕj and ∇uj ∇ϕj are standard, a further difficulty is represented by the terms pj ∇ϕj . This relates to the fact that, a priori, only an estimate on ∇p of the form ∇pp ≤ Cf p uniformly in the resolvent parameter λ is available, but no suitable estimate for the pressure itself. The following Lemma provides decay estimates for the pressure, see [115, Lemma 13]. Lemma 1.10.5 ([115]) Let θ ∈ (0, π/2), 1 < p < ∞ and (u, p) ∈ D(A,p ) ×  1,p () the unique solution of the Stokes resolvent problem (1.10.3) on Lpσ (). W  p 1 For a bounded C 2 -domain G ⊂  set pG := p − |G| G p dx ∈ L0 (G) =  {g ∈ Lp (G); G g dx = 0}. Then for each α ∈ (0, 1/2p ) and every bounded C 2 -domain G ⊂  there exists a constant C > 0, independent of λ and f , such that pG p ≤ C|λ|−α f p ,

λ ∈ π−θ , |λ| ≥ 1.

(1.10.9)

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M. Hieber

A sketch of the proof of the above lemma is as follows. Using ∇pG = ∇p and the fact that ∇pG (x) = (I − P )u(x) for x ∈ , we obtain    pG η dx = − ∇pG · ψ dx = (−D u)(I − P )ψ dx 







[(−D )1−α u](−D )α (I − P )ψ dx.

= 



By Proposition 1.4.12, −D admits a bounded H ∞ -calculus on Lp () and it  thus follows from Theorem 1.5.6 that D((−D )α ) = H 2α,p (). Note that for   2α,p α ∈ (0, 1/2p ) one has H 2α,p () = H0 (), which implies that we may shift the part (−D )α of −D onto (I − P )ψ without getting boundary terms. Summarizing, we obtain the following result. Theorem 1.10.6 Let n ≥ 2, 1 < p, q < ∞, J = (0, T ) for some T > 0 and assume that  ⊂ Rn is a standard domain of class C 3 . Then the Stokes operator defined by A,p u := P u,

1,p

D(A,p ) := W 2,p () ∩ W0 () ∩ Lpσ ()

(1.10.10)

p

admits maximal Lq -regularity on Lσ (). In particular, the solution u to the Cauchy problem u (t) − Ap u(t) = f (t), t > 0,

u(0) = u0 ,

satisfies the estimate   u Lq (J ;Lp ()) + Ap uLq (J ;Lp ()) ≤ C f Lq (J ;Lp ()) + u0 Xγ , for  p some C > 0 independent of f Lσ (), D(Ap ) 1−1/q,q .

p

∈ Lq (J ; Lσ ()) and u0 ∈ Xγ

:=

p

Moreover, Ap generates a bounded analytic C0 -semigroup on Lσ () and (a) σ (A,p ) = (−∞, 0] if  is Rn , Rn+ or an exterior domain, (b) σ (A,p ) = (−∞, −κ] for some κ = κ() > 0 provided  is bounded, 2−2/q p (c) u0 ∈ Xγ if and only if u0 ∈ Bp,q () ∩ Lσ () and u = 0 on ∂. Setting ∇π = (I d − P )(λ − Ap )−1 , one obtains the following results for the Stokes equation (1.10.1) and its corresponding resolvent Eq. (1.10.3). Corollary 1.10.7 Given the assumptions of Theorem 1.10.6, the Stokes equation (1.10.1) admits a unique solution (u, π) ∈ W 1,q (J ; Lp ())∩Lq (J ; W 2,p ()∩ 1,p p  1,p ()) and there exists a constant C > 0 such that W0 () ∩ Lσ ()) × Lq (W ut Lq (J ;Lp ()) + uLq (J ;Lp ()) + ∇ 2 uLq (J ;Lp ()) + ∇πLq (J ;Lp ())   ≤ C f Lq (J ;Lp ()) + u0 Xγ .

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First results on maximal Lp -regularity estimates for the instationary Stokes system (1.10.1) go back to the pioneering work of Solonnikov, see [135]. For a modern approach to his results, then also in the mixed Lq − Lp -context, based on the characterization of maximal Lp -regularity by the R-boundedness property of the resolvent, see the work of Geissert et al. [59] and [60]. For the halfspace Rn+ , results on the existence and analyticity of the Stokes semigroup go back to [40, 112, 151]. Giga and Sohr [64] proved for the first time global-in-time mixed Lq − Lp maximal regularity estimates for smooth exterior domains by combining a result on the boundedness of the imaginary powers of the Stokes operator with the Dore–Venni theorem [44]. A different approach to maximal Lp -regularity of the Stokes equation (1.10.1) based on pseudo-differential methods was developed by Grubb and Solonnikov in [70]. An important tool in the investigation of nonlinear problems is the representation of the domain of the fractional powers of a sectorial operator in terms of suitable function spaces. Assuming that a sectorial operator A admits a bounded H ∞ calculus on a Banach space X, Theorem 1.5.6 tells us that in case [X, D(A)]α = D(Aα ),

α ∈ (0, 1).

(1.10.11)

The following result deals with the H ∞ -calculus for the Stokes operator on

p Lσ ().

Theorem 1.10.8 ([115]) Let n ≥ 2, 1 < p < ∞, and assume that  ⊂ Rn is a standard domain of class C 3 . Then −A,p admits an R-bounded H ∞ -calculus on p p Lσ (). In particular, relation (1.10.11) holds for A = −A,p and X = Lσ (). The first proof of the boundedness of the imaginary powers of the Stokes operator on bounded domains with smooth boundaries goes back to Giga [63]. His proof is again based on Seeley’s theorem. The case of an exterior domain, due to Giga and Sohr, was treated in [64]. A first proof of the fact that −ARn+ ,p admits an R-bounded p H ∞ -calculus on Lσ (Rn+ ) was given by Desch, Hieber and Prüss in [40]. Moreover, the proof of the existence of an R-bounded H ∞ -calculus on standard domains in p Lσ () is due to Noll and Saal [115]. For the case n = 2, see the work of Abels [3]. Remark 1.10.9 Let  ⊂ Rn for n ≥ 3 be a bounded Lipschitz domains. It was p shown by Kunstmann and Weis [96] that the negative Stokes operator on Lσ () p ∞ admits a bounded H -calculus on Lσ () provided |1/p − 1/2| ≤ 1/2n. Finally, we consider the Stokes equation with inhomogeneous data of the form ⎧ ⎪ ⎪ (∂t + ω)u − u + ∇π = f ⎨ div u = g ⎪ u=h ⎪ ⎩ u(0) = u0

in R+ × , in R+ × , on R+ × ∂, in ,

(1.10.12)

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where  ⊂ Rn is a domain with compact boundary of class 3 and ω ∈ R, In the following theorem, the maximal Lq −Lp regularity estimates for the solution (u, π) of (1.10.12) are characterized by conditions on the data (f, g, h, u0 ). To this end, the set of conditions (D) is introduced: Condition (D): 2−2/q

(a) f ∈ Lq (R+ ; Lp ()), u0 ∈ Bp,q (), (b) g ∈ H 1,q (R+ ; H˙ −1,p ()) ∩ Lq (R+ ; H 1,p ()), div u0 = g(0), 1−1/2p 2−1/p (c) h ∈ Fq,p (R+ ; Lp (∂)) ∩ Lq (R+ ; Bp,p (∂)) and h(0) = u0 on ∂ if q > 3/2, (d) (g|hν ) ∈ H 1,q (R+ ; H˙ −1,p ()) and hν (0) = (ν|u0 ) on ∂. Then the following theorem holds true. Theorem 1.10.10 ([124]) Let  ⊂ Rn be a domain with compact boundary ∂ of class 3, let 1 < p, q < ∞ and q = 3, 3/2. Then there exists ω0 ∈ R such that for each ω > ω0 there exists a unique solution (u, π) of Eq. (1.10.12) within the class u ∈ H 1,q (R+ ; Lp ()) ∩ Lq (R+ , H 2,p ()) and π ∈ Lq (R+ ; H˙ 1,p ()) if and only if the data (u0 , f, g, h) satisfy the above condition (D). The proof of Theorem 1.10.10 is rather involved, see Section 7 of [124] for a very thorough study of the Stokes equation with inhomogeneous data. In addition, other types of boundary conditions as pure slip, outflow and free boundary conditions are studied there. Primitive Equations Consider the isothermal primitive equations of the form ∂t v + u · ∇v − v + ∇H π = f

in  × (0, T ),

∂z π = 0

in  × (0, T ),

div u = 0

in  × (0, T ),

(1.10.13)

v(0) = v0 . Here  = G × (−h, 0), where G = (0, 1)2 and h > 0. The velocity u of the fluid is described by u = (v, w) with v = (v1 , v2 ), and where v and w denote the horizontal and vertical components of u, respectively. Furthermore, π denotes the pressure of the fluid and f a given external force. The symbol ∇H = (∂x , ∂y )T denotes the horizontal gradient,  the three dimensional Laplacian and ∇ and div the three dimensional gradient and divergence operators. The above system is complemented by the set of boundary conditions ∂z v = 0, w = 0 on u × (0, T ), v = 0, w = 0 on b × (0, T ), u, π are periodic on l × (0, T ).

(1.10.14)

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79

Here u := G × {0}, b := G × {−h}, l := ∂G × [−h, 0] denote the upper, bottom and lateral parts of the boundary ∂, respectively. Note that w(x, y, z) = 0 z divH v(x, y, ζ ) dζ for (x, y) ∈ G, −h < z < 0, and thus divH v¯ = 0 in G, where v¯ stands for the average of v in the vertical direction, i.e., 1 v(x, ¯ y) := h



0 −h

v(x, y, z) dz,

(x, y) ∈ G.

Therefore, problem (1.10.13)–(1.10.14) is equivalent to finding a function v :  → R2 and a function π : G → R satisfying the set of equations ∂t v + v · ∇H v + w∂z v − v + ∇H π w divH v¯ v(0)

=f in  × (0, T ), 0 = z divH v dζ in  × (0, T ), =0 in G × (0, T ), = v0 , (1.10.15)

as well as the boundary conditions ∂z v = 0 on u × (0, T ), v=0 on b × (0, T ), v and π are periodic on l × (0, T ).

(1.10.16)

The Sobolev spaces equipped with periodic boundary conditions in the horizontal directions are defined by m,p

Wper () := {f ∈ W m,p () | f is periodic of order m − 1 on l }, m,p

Wper (G) := {f ∈ W m,p (G) | f is periodic of order m − 1 on ∂G}. We consider first the resolvent problem associated with the linearization of (1.10.14) within the Lp -setting. More precisely, let λ ∈ π−ε = {λ ∈ C : | arg λ| < π − ε} for some ε ∈ (0, π/2) and f ∈ Lp ()2 for some 1 < p < ∞ and consider the equation λv − v + ∇H π = f in , divH v¯ = 0 in G,

(1.10.17)

subject to the boundary conditions ∂z v = 0 on u ,

v = 0 on b ,

v and π are periodic on l .

(1.10.18)

The following resolvent estimate was deduced in [73]. Proposition 1.10.11 Let λ ∈ π−ε ∪ {0} for some ε ∈ (0, π/2). Moreover, let p ∈ (1, ∞) and f ∈ Lp (). Then Eqs. (1.10.17) and (1.10.18) admit a unique

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M. Hieber 2,p

1,p

p

solution (v, π) ∈ Wper () × Wper (G) ∩ L0 (G) and there exists a constant C > 0 such that |λ| vLp () + vW 2,p () + πW 1,p (G) ≤ Cf Lp () , λ ∈ π−ε ∪ {0}, f ∈ Lp (). (1.10.19) As in the case of the classical Helmholtz projection, the existence of the hydrostatic Helmholtz projection is closely related to the unique solvability of the Poisson problem in the weak sense. In the given situation, the equation H π = divH f in G, subject to periodic boundary conditions, plays an essential role. Lemma 1.10.12 Let p ∈ (1, ∞) and f ∈ Lp (G). Then there exists a unique 1,p p π ∈ Wper (G) ∩ L0 (G) satisfying ∇H π, ∇H φ"Lp (G) = f, ∇H φ"Lp (G) ,

1,p 

p

φ ∈ Wper (G) ∩ L0 (G).

(1.10.20)

Furthermore, there exists a constant C > 0 such that πW 1,p (G) ≤ Cf Lp (G) ,

f ∈ Lp (G).

(1.10.21)

The above Lemma 1.10.12 allows to define the hydrostatic Helmholtz projection 1,p p Pp : Lp () → Lp () as follows: given v ∈ Lp (), let π ∈ Wper (G) ∩ L0 (G) be the unique solution of Eq. (1.10.20) with f = v. ¯ One then sets Pp v := v − ∇H π,

(1.10.22)

and calls Pp the hydrostatic Helmholtz projection. It follows from Lemma 1.10.12 that Pp2 = Pp and that thus Pp is indeed a projection. We define the closed subspace Xp of Lp () as Xp := RgPp . This space plays the analogous role in p the investigations of the primitive equations as the solenoidal space Lσ () plays in the theory of the Navier–Stokes equations. The hydrostatic Helmholtz projection Pp defined as in (1.10.22) allows then to define the hydrostatic Stokes operator as follows. In fact, let 1 < p < ∞ and Xp be defined as above. Then the hydrostatic Stokes operator Ap on Xp is defined as 

Ap v := Pp v, 2,p

D(Ap ) := {v ∈ Wper ()2 : divH v¯ = 0 in G, ∂z v = 0 on u , v = 0 on b }. (1.10.23) The resolvent estimates for Eqs. (1.10.17) and (1.10.18) given in Proposition 1.10.11 yield that Ap generates a bounded analytic semigroup on Xp . More precisely, one has the following result.

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Proposition 1.10.13 Let 1 < p < ∞. Then the hydrostatic Stokes operator Ap generates a bounded analytic C0 -semigroup Tp on Xp . Moreover, there exist constants C, β > 0 such that Tp (t)f Xp ≤ Ce−βt f Xp ,

t > 0.

It was shown by Giga et al. [68] that −Ap even admits an R-bounded H ∞ -calculus on Xp of angle 0, which implies in particular maximal Lq − Lp estimates for the solution of the hydrostatic Stokes equation and allows further to characterize the domains D(−Aθp ) of the fractional powers −Aθp for 0 < θ < 1 in terms of Sobolev spaces subject to the boundary conditions given. Theorem 1.10.14 Let p ∈ (1, ∞). Then the operator −Ap admits a bounded R,∞ RH ∞ -calculus on Xp with φA = 0. p Combining this result with a characterization of the complex interpolation spaces [Xp , D(Ap )]θ proved in [83] allows then to characterize the domains D(Aθp ) of the fractional powers −Aθp for 0 < θ < 1 as follows. For p ∈ (1, ∞) and s ∈ [0, ∞) s,p

s,p

∞ () the spaces Hper () are defined as Hper () := Cper

·H s,p ()

.

Corollary 1.10.15 ([68]) Let 1 < p < ∞ and θ ∈ [0, 1] with θ ∈ / {1/2p, 1/2 + 1/2p}. Then ⎧   2θ,p ⎪ ⎪ {v ∈ Hper () ∩ Xp : ∂z v  = 0, v  = 0}, ⎪ ⎨ N D  2θ,p D((Ap )θ ) = {v ∈ Hper () ∩ Xp : v  = 0}, ⎪ D ⎪ ⎪ ⎩{v ∈ H 2θ,p ()2 ∩ X , p per

1/2 + 1/2p < θ ≤ 1, 1/2p < θ < 1/2 + 1/2p, θ < 1/2p.

1.11 Nonlinear Stability of Ekman Boundary Layers In this section we are considering the nonlinear Navier–Stokes equations in the rotational setting in the half-space R3+ subject to homogeneous Dirichlet boundary conditions. More precisely, consider the set of equations ⎧ ∂t u − νu + e3 × u + (u · ∇)u + ∇p = 0, t > 0, x ∈ R3+ , ⎪ ⎪ ⎨ div u = 0, t > 0, x ∈ R3+ , ⎪ u(t, x1 , x2 , 0) = 0, t > 0, x1 , x2 ∈ R, ⎪ ⎩ u(0, x) = u0 , x ∈ R3+ ,

(1.11.1)

where u = (u1 , u2 , u3 ) denotes the velocity field and p the pressure of an incompressible, viscous fluid. Here, e3 denotes the unit vector in x3 -direction, ν > 0 the viscosity of the fluid, and the constant  ∈ R is called the Coriolis parameter, which is equal to twice of the frequency of the rotation around the x3 axis.

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M. Hieber

It has been known for a long time that the above system has a stationary solution, which can be expressed explicitly as uE (x3 ) = u∞ (1 − e−x3 /δ cos(x3 /δ), e−x3 /δ sin(x3 /δ), 0)T ,

(1.11.2)

pE (x2 ) = −u∞ x2 ,

(1.11.3)

1/2 and u where δ is defined by δ := ( 2ν ∞ ≥ 0 is a constant. This stationary ) solution of Eq. (1.11.1) is called in honour of the swedish oceanograph Ekman, the Ekman spiral; see also [47]. It describes mathematically rotating boundary layers in geophysical fluid dynamics between a geostrophic flow and a solid boundary at which the no slip boundary condition applies. Here, δ denotes the thickness of the layer, see also [66]. In the geostrophic flow region corresponding to large x3 , there is a uniform flow with velocity u∞ in the x1 direction. Associated with u∞ , there is a pressure gradient in the x2 -direction. The Ekman spiral in R3+ matches this uniform velocity for large x3 with the no slip boundary condition at x3 = 0, i.e. we have uE (0) = 0 and

uE (x3 ) → (u∞ , 0, 0) provided x3 → ∞. In this section we are interested in stability questions for the Ekman spiral. The more general situation of stratified flows is described in detail in [93]. We consider hence perturbations of the Ekman spiral by functions u solving the above Eq. (1.11.1). To this end, set w := u − uE ,

and q := p − pE .

Then, since (uE , pE ) is a stationary solution of (1.11.1), the pair (w, q) satisfies the equations ⎧ 3 ⎪ ⎪ ∂t w − νw + e3 × w + (uE · ∇)w + w3 ∂3 uE + (w · ∇)w + ∇q = 0, t > 0, x ∈ R+ , ⎪ ⎪ ⎨ divw = 0, t > 0, x ∈ R3+ , ⎪ w(x1 , x2 , 0) = 0, t > 0, x1 , x2 ∈ R, ⎪ ⎪ ⎪ ⎩ w(0, x) = w0 , x ∈ R3+ ,

(1.11.4) where w0 = u0 − uE . We are now interested in the following questions: (a) does there exist a suitable notion of solutions (weak, mild, strong, etc.) such that for any initial data w0 ∈ L2σ (R3+ ) the above set of Eq. (1.11.4) admits a solution for all t ∈ (0, T ], where T > 0 is arbitrary or for all t ∈ [0, ∞)? (b) is the solution stable or asymptotically stable, i.e. does such a solution w satisfy w(t)2 ≤ w0 2 for all t > 0 or lim w(t)2 = 0? t →∞

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In the following, we give partial answers to the above questions. Indeed, we will show that there exists a global weak solution to the above Eq. (1.11.4) provided the Reynolds number Re given by Re = u∞ δν −1 is small enough. Secondly, assuming this condition, for every initial data w0 ∈ L2σ (R3+ ), there exists at least one global weak solution w to (1.11.4) such that  lim

T →∞ T

T +1

w(t)H 1 dt = 0,

which shows in particular that the Ekman spiral is nonlinearly stable with respect to L2 -perturbations. The Stokes-Coriolis-Ekman Operator p Rewriting the Eq. (1.11.4) as an evolution equation in Lσ (R3+ ) for 1 < p < ∞ yields 

w − ASCE w + P (w · ∇)w = 0, t > 0, w(0) = w0 .

(1.11.5) p

Here P denotes the Helmholtz projection from Lp (R3+ ) to Lσ (R3+ ) defined as p in (1.9.2) and ASCE denotes the Stokes-Coriolis-Ekman operator on Lσ (R3+ ) defined by 

ASCE w := P (νw − e3 × w − [(uE · ∇)w + w3 ∂3 uE ]) =: (AS + AC + AE )w 1,p p D(ASCE ) := W 2,p (R3+ ) ∩ W0 (R3+ ) ∩ Lσ (R3+ ).

(1.11.6) It now follows from Theorem 1.9.6 that the Stokes operator AS := P  generates p a bounded analytic semigroup et AS on Lσ (R3+ ) for all p ∈ (1, ∞). The perturbation Theorem 1.3.15 combined the standard interpolation theory implies that the StokesCoriolis-Ekman operator also generates also an holomorphic semigroup et ASCE p on Lσ (R3+ ). Moreover, combing the Perturbation Theorem 1.6.19 for R-sectorial operators with the fact that −AS is R-sectorial, see Theorem 1.9.16, it follows from Corollary 1.6.10 that there exists a constant ν ≥ 0 such that ASCE + ν has maximal Lp -regularity for all p ∈ (1, ∞). We summarize our considerations in the following proposition. Proposition 1.11.1 Let 1 < p < ∞. Then the operator ASCE generates an p holomorphic semigroup on Lσ (R3+ ) and there exists ν ≥ 0 such that −ASCE + ν p has maximal Lp -regularity on Lσ (R3+ ). The definition of a weak solution w for Eq. (1.11.4) given below requires that at least w ∈ L∞ ((0, T ); L2σ (R3+ )) for all T > 0. We are hence interested in conditions implying that et ASCE is a bounded or a contraction semigroup on L2σ (R3+ ).

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M. Hieber

For this it is useful to note that, if α > 0, then e−(·)/α vL2 (R+ ) ≤

α  v L2 (R+ ) 2

(1.11.7)

for all v ∈ H01 (R+ ). For smooth functions this follows easily from the fundamental theorem of calculus and Jensen’s inequality; the general case is then implied by approximation. Now, for v0 ∈ L2σ (R3+ ) set v(t) := et ASCE v0 . Then v satisfies v  (t) − ASCE v(t) = 0

for t > 0 and v(0) = v0 .

(1.11.8)

Multiplying with v(t) and taking into account the skew symmetry of the second and third term of ASCE yields 1 d 2 dt



 |v(t)| dx + ν



2

R3+

|∇v(t)| dx + 2

R3+

R3+

v(t) · (v3 (t) · ∂3 uE )dx = 0,

t > 0.

Since  |

R3+

v(t) · (v3 (t) · ∂3 uE )dx| ≤

2 

e(·)/2δ (∂3 uE )j v3 (t)2 e−(·)/2δ vj (t)2 ,

j =1

and since ⎛ ⎞ cos(x3 /δ) + sin(x3 /δ) u∞ −x3 /δ ⎝ ∂3 uE (x3 ) = e cos(x3 /δ) − sin(x3 /δ) ⎠ , δ 0 we see that e(·)/2δ (∂3 uE )j v3 (t)2 ≤

√ u∞ −(·)/2δ e 2 v3 (t)2 , δ

j = 1, 2.

(1.11.9)

Estimate (1.11.7) implies now  |

R3+

v(t) · (v3 (t) · ∂3 uE )dx| ≤

√ 2u∞ δ∇v(t)22 .

√ d Thus, dt v(t)22 ≤ 0 for all t > 0 provided 2u∞ δ ≤ ν. Therefore et ASCE v0 2 = v(t)2 ≤ v0 2 for all t > 0 provided this condition is fulfilled. Setting Re := u∞ δν −1 ,

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we thus proved the following result t ASCE on Lemma 1.11.2 The √ operator ASCE generates a contraction semigroup e 3 2 Lσ (R+ ) provided 2Re ≤ 1. Moreover, if

√ 2Re < 1,

(1.11.10)

then there exists a constant C > 0 such that the solution v of (1.11.8) satisfies  v(t)22 + C

t

∇v(s)22 ds ≤ v(0)22 ,

0

t ≥ 0.

(1.11.11)

Remarks 1.11.3 (a) One easily sees that there exists also a constant C > 0 such that the solution v of (1.11.8) satisfies  v(t)22 + C

s

t

∇v(τ )22 ds ≤ v(s)22 ,

t ≥ s.

(1.11.12)

for all s ≥ 0. (b) Combining Lemma 1.11.2 with Proposition 1.4.11, it follows that ASCE admits a bounded H ∞ (θ )-calculus on L2σ (R3+ ) for any angle θ > π/2 provided condition (1.11.10) is satisfied. Furthermore, applying Theorem 1.5.6 yields 1/2

D(ASCE ) = H01 (R3+ ) ∩ L2σ (R3+ ).

(1.11.13) 1/2

It thus follows that there exists a constant C > 0 such that for u ∈ D(ASCE ) 1/2

ASCE u2 ≤ C(u2 + ∇u2 ), 1/2

∇u2 ≤ C(ASCE u2 + u2 ).

(1.11.14)

(c) Assume (1.11.10). For α > 0, let AαS and AαSCE be defined according to Theorem 1.5.3 and define q by 1/q = 1/2 − 2α/3. Then uq ≤ CAαS u2 for u ∈ D(Aα ) and one can show that uq ≤ C(η, α)(ASCE + η)α u2 , (A∗SCE + η)−α u2 ≤ C(η, α)u2 ,

u ∈ L2σ (R3+ )

u ∈ D(AαSCE ) (1.11.15) (1.11.16)

86

M. Hieber ∗

(d) For t ≥ 0 and v0 ∈ L2σ (R3+ ) set v ∗ (t) := et ASCE v0 . Then one can show as above that there exists a constant C > 0 such that  t v ∗ (t)22 + C ∇v ∗ (s)22 ds ≤ v ∗ (0)22 , t ≥ 0. (1.11.17) 0

The following result concerning the strong stability of et ASCE on L2σ R3+ will be one of the main ingredients for our stability estimates of weak solutions of system (1.11.5). Proposition 1.11.4 Assume that condition (1.11.10) is satisfied. Then lim et ASCE v2 = 0,

t →∞

for every v ∈ L2σ (R3+ ), i.e. the Stokes-Coriolis-Ekman semigroup on L2σ (R3+ ) is strongly stable. Proof Consider the operator ∂1 := ∂1 I d3 on X := L2σ (R3+ )3 with domain D(∂1 ) = X ∩ H 1 (R; L2 (R2+ ))3 . It can be shown that ∂1∗ = −∂1 and that ∂1 has dense range in X. First let f ∈ Rg(∂1 ) and g ∈ D(∂1 ) with ∂1 g = f . By Lemma 1.11.2, et ASCE is a contraction semigroup on X and thus by Hölder’s inequality e

t ASCE

1 f 2 ≤ t



t

e

sASCE

0

1 f 2 ds ≤ √ ( t



t 0

esASCE ∂1 g22 ds)1/2 .

Since the lower order terms of ASCE do not depend on the x1 -variable, one can show that et ASCE ∂1 g2 = ∂1 et ASCE g2 ,

g ∈ D(∂1 ).

Hence, by the above energy inequality (1.11.12)  te

t ASCE

f 22

t

≤ 

0 t

≤ 0

 e

sASCE

∂1 g22 ds

t

= 0

∂1 esASCE g22 ds

∇esASCE g22 ds ≤ Cg22 .

For general f ∈ X, note that since Rg ∂1 is dense in X we may approximate f by h ∈ Rg(∂1 ) and obtain thus the assertion. 

Stability of Weak Solutions We now state the definition of a weak solution to Eq. (1.11.4).

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Definition 1.11.5 Let w0 ∈ L2σ (R3+ ) and T > 0. A function w is called a weak 1/2 solution of Eq. (1.11.4) if w ∈ L∞ ((0, T ); L2σ (R3+ )) ∩ L2 ((0, T ); D(ASCE )) and 

T

− 

w, φ"h (t)dt + ν

0

T

 

w3 · ∂3 uE , φ"h(t)dt + 

0

T

∇w, ∇φ"h(t)dt +

0 T

+



(uE · ∇)w, φ"h(t)dt

0 T



T

e3 × w, φ"h(t)dt +

0

w · ∇w, φ"h(t)dt

0

= w0 , φ"h(0), 1/2

holds for all φ ∈ D(ASCE ) and all h ∈ C 1 ([0, T ], R) with h(T ) = 0. In the following we prove the existence of a global weak solution to the problem (1.11.4) provided the Reynolds number Re = u∞ δν −1 is small enough. More precisely, we assume from now on that condition (1.11.10) is satisfied. Note that our proof is inspired by a technique developed by Miyakawa and Sohr in [114]. We subdivide our considerations into three steps. Step 1: Approximate Local Solutions We first introduce smoothing operators Jk given by Jk := k(k − ASCE )−1 ,

k ∈ N,

p

on Xp := Lσ (R3+ ) for 1 < p < ∞. They fulfill the following mapping properties. Lemma 1.11.6 (a) (b) (c) (d)

Jk L(X2 ) ≤ 1 for all k ∈ N, Jk L(Xp ) ≤ C for some C > 0 and all k ≥ k0 and some k0 ≥ 0, Jk uL∞ ≤ C(k)uX2 for all k ∈ N, 1/2 ∇Jk uX2 ≤ C(uX2 + ∇uX2 ) for some C > 0 and all u ∈ D(ASCE ) and all k ∈ N.

In fact, since ASCE generates a contraction semigroup on X2 , see Lemma 1.11.2, the Hille-Yoshida Theorem 1.3.6 implies (a). Assertion (b) follows from Proposition 1.11.1. The Gagliardo-Nirenberg inequality implies Jk u∞ ≤ 1/4 3/4 Cu2 (ASCE + 1)Jk u2 for u ∈ X2 and (c) follows e.g. from the fact that ∞ ASCE ∈ H (X2 ); see Remark 1.11.3. Finally, (d) follows by combining the above assertions. We now set w0k := Jk w0 and Fk w := −P (Jk w · ∇)w,

k ∈ N,

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M. Hieber

and construct approximate solutions wk to Eq. (1.11.5) by solving the integral equations 

t

wk (t) = et ASCE w0k +

e(t −s)ASCE Fk wk (s)ds,

k ∈ N.

(1.11.18)

0 1/2

To this end, consider for T > 0 the Banach space X := C([0, T ]; D(ASCE )) equipped with the norm 1/2

uT := sup (u(t)2 + ASCE u(t)2 ). 0≤t ≤T

For M > 0 and k ∈ N consider the closed set S(k, M, T ) := {u ∈ X, u(0) = w0k , uT ≤ M}, as well as the nonlinear operator k defined on S(k, M, T ) given by 

t

k u(t) := et ASCE w0k +

e(t −s)ASCE Fk u(s)ds.

0

Note that by Lemma 1.11.6(c) and Remark 1.11.14 1/2

Fk u2 ≤ C(k)(u22 + ASCE u22 ),

k ∈ N.

We thus may estimate k u as follows 

1/2

k uT ≤ w0k 2 + ASCE w0k 2 + sup {C(k)  + sup {C(k) 0≤t ≤T

0

0≤t ≤T

t

t

0

1/2

u22 + ASCE u22 ds}

1

(t − s)− 2 eβ(t −s)(u22 + ASCE u22 )ds} 1/2

1

1/2

≤ w0k 2 + ASCE w0k 2 + C(k)M 2 (T + eβT T 2 ) for some β ≥ 0. Similarly, 1

k u1 − k u2 T ≤ C(k)M(T + eβT T 2 )u1 − u2 T . 1/2

Fix now M in such a way that w0k 2 + ASCE w0k 2 ≤ 1

1

M 2

and then T ∗ < T such

βT 2 that C(k)M 2 (T + eβT T 2 ) ≤ M 2 and C(k)M(T + e T ) < 1. Then k is a strict ∗ contraction in S(k, M, T ) and by Banach fixed point theorem, there exists a unique wk in S(k, M, T ∗ ) satisfying (1.11.18) for t ∈ (0, T ∗ ).

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Step 2: A Priori Bounds for Fixed T 1/2 In the following we prove a priori bounds for wk (T ) and ASCE wk (T ) for all T > 0. ∗ To this end, note that Fk wk ∈ C([0, T ]; X2 ) and that wk solves wk (t) + ASCE wk = Fk wk ,

t ∈ (0, T ∗ ).

(1.11.19)

Multiplying (1.11.19) with wk and integrating by parts yields as above  wk (T )22

T

+C 0

∇wk (s)22 ds ≤ w0 22 .

(1.11.20)

1/2

In order to show that ASCE wk (T )2 is bounded for all fixed T > 0, note that by (1.11.14) it suffices to show this for ∇wk (T )2 . As above we see that  ∇wk 22 ≤ C(k)w0 22 + C

T 0

 wk (s)22 ds + C(k)

T 0

(1 + wk (s)22 )∇wk (s)22 ds.

Hence, by Gronwall’s inequality there exists C = C(w0 2 , T , k) such that ∇wk (T )2 ≤ C < ∞. Combining this with (1.11.20), we see that 1/2

sup {wk (t)2 + ASCE wk (t)2 } ≤ M < ∞.

0≤t ≤T

Step 3: Weak Convergence In this final step we show that the approximate global solutions (wk ) constructed above, converge weakly to some weak solution w of system (1.11.4). Fix T > 0. The above inequality (1.11.20) and (1.11.14) implies that 

T 0



1/2

ASCE wk (s)22 ds ≤ C

T 0

(wk (s)22 + ∇wk (s)22 )ds

≤ C(T + C)w(0)22 , and hence that wk ∈ L2 (0, T ; D(ASCE )) ∩ L∞ (0, T ; L2σ (R3+ ) =: E =: E1 ∩ E2 , 1/2

k ∈ N.

Thus (wk ) is a bounded sequence in E. Since E1 is reflexive, there exists a subsequence of (wk ) converging weakly in E1 . Further, (wk ) possesses a weak-star convergent subsequence in E2 . In order to investigate the strong convergence of (wk ) we write wk = wk1 + wk2 with  t wk(1) (t) := et ASCE w0k , and wk(2) (t) := e(t −s)ASCE Fk wk (s)ds. 0

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M. Hieber

Performing the same calculations which led to (1.11.20), we obtain wk(1) (T ) − wl(1) (T )22 +



T 0

1/2

ASCE (wk(1) (s) − wl(1) (s))22 ds

≤ C(T )(Jk − Jl )w0 22 → 0 (1)

(2)

as k, l → ∞, since Jk w0 → w0 in X2 . Hence (wk ) as well as (wk ) are bounded sequences in E. Consider next the term Fk wk in Lrσ (R3+ ) for r = 5/4. The Gagliardo–Nirenberg inequality yields 2/5

3/5

Fk wk r ≤ wk 2 ∇Jk wk 2 ∇wk 2 . 1/2

Since ∇Jk wk 2 ≤ C(ASCE wk 2 + wk 2 ), we see that by (1.11.20) that 

T 0

Fk wk rr dt ≤ Cw0 2 . (2)

Hence, (Fk wk ) is a bounded sequence in Lr (0, T ; Lrσ (R3+ )). By construction, wk is the solution of the Cauchy problem wk (t) + ASCE wk (t) = Fk wk (t), w(0) = 0.

t > 0,

By Proposition 1.11.1 there exists a constant νr > 0 such that −ASCE + νr admits maximal Lr -regularity on Lrσ (R3+ ). A scaling argument shows not only e−νr t wk (t) but also (wk2 ) ∈ Lr (0, T ; D(ASCE,r )) ∩ W 1,r (0, T ; Lrσ (R3+ )) is a bounded sequence in this space. By Lemma 1.11.6(b), (Jk wk(2) )k≥k0 is as well a bounded sequence in this space.l It thus follows from Theorem III.2.1 in [147] that (wk(2) ) and (Jk wk(2) )k≥k0 are relatively compact in L2 (K × (0, T )) for any fixed (2) (2) compact set K ⊂ R3+ . Hence, (wk ) and (Jk wk ) converge in L2 (R3+ × (0, T )). Therefore, wk (s) → w(s) and Jk wk (s) → w(s) for a.a. s ∈ (0, T ) for some function w ∈ E. Finally, we need to verify that the function w constructed above is in fact a weak solution of our problem (1.11.4). We refrain, however, from giving details at this point. We thus proved that for any initial value w0 ∈ L2σ (R3+ ) there exists a weak solution w to (1.11.4) satisfying the energy inequality  w(t)22

t

+C s

∇w(τ )22 dτ ≤ w(s)22

(1.11.21)

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√ for almost all s ≥ 0, all t > s and where w(0) = w0 provided 2Re < 1. The advantage of our quite lengthy procedure is that we now have a representation formula for w, which allows us to deduce asymptotic properties of w. 3/4 In the following let T0 > 0 and for u ∈ D(ASCE ) set w := (ASCE + 1)1/4u. Then, by the momentum inequality (1.5.1) 1/3

2/3

(ASCE + 1)1/4u2 ≤ C(ASCE + 1)3/4u2 u2 for some C > 0. Hence, by (1.5.2) and (1.11.14)

w(t)2 ≤ Cw(t)H 1 (ASCE + 1)−1/4 w(t)2 . 2/3

Hölder’s inequality combined with the energy inequality (1.11.21) yields 

T0 +1 T0

w(t)22 dt

≤C



T0 +1 T0

(ASCE + 1)−1/4 w(t)22 dt

2/3

with C independent of T0 and w. Next, for h > 0 and s, t satisfying 0 ≤ s ≤ τ ≤ t < T0 , consider  h (τ ) := U (τ )

t

h (τ − σ )U (σ )w(σ )dσ,

s ∗

where U (τ ) = (ASCE + 1)−1/4 e(t −τ )ASCE and h (τ ) = h−1 (h−1 τ ) for a mollifier ρ ∈ C ∞ (R). Since w is a weak solution of (1.11.4), integration by parts yields < w(t), h (t) > − < w(s), h (s) >  t =− < (w(τ ) · ∇)w(τ ), h (τ ) > dτ 

s



t

+

t

< w(τ ), U (τ ) s

s

d h (τ − σ )U (σ )w(σ )dσ > dτ. dτ

(1.11.22)

Now, Fubini’s theorem as well as standard properties of mollifiers imply that 2 < w(s), h (s) > → e(t −s)ASCE (ASCE + 1)−1/4 w(s)22 , 2 < w(t), h (t) > → (ASCE + 1)−1/4w(t)22 , as h → 0. Notice next that by symmetry of h , the second term on the right hand side of (1.11.22) vanishes, whereas the first term on the right hand side of (1.11.22) can be estimated as  t   t   < (w(τ ) · ∇)w(τ ), h (τ ) > dτ  ≤ Cw0 2 ∇w(τ )22 dτ.  s

s

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Indeed,   

s

t

 t   < (w(τ ) · ∇)w(τ ), h (τ ) > dτ  ≤ C w(τ )6 ∇w(τ )2 h 3 dτ s



≤ C sup {h 3 } s≤τ ≤t

t

s

∇w(τ )22 dτ

and the estimates given in Remark 1.11.3(c) imply that sups≤τ ≤t h (τ )3 ≤ Cw0 2 . Hence, letting h → 0 in (1.11.22) yields (ASCE +1)

−1/4

w(t)22

≤ e

(t −s)ASCE

(ASCE +1)

−1/4

 w(s)22 +C

t s

∇w(τ )22 dτ.

The energy inequality (1.11.21) implies that for ε > 0 there exists m0 ∈ N such that w(tm2 )22 − wm1 22 ≤ ε. Thus, for given ε > 0 there exists T1 > 0 such that 

T

T0

∇w(τ )22 dτ < ε

for all T > T1 . For fixed T2 > 0, Proposition 1.11.4 implies that e(t −T2 )ASCE (ASCE+1 )−1/4 w(T2 )22 ≤ ε0 provided t > T3 for some T3 > 0. Hence, 

T3 +1 T3

(ASCE + 1)−1/4 w(t)22 ≤



T3 +1

T3

+C



e(t −s)ASCE (ASCE + 1)−1/4w(s)22 T3 +1  t

T3

T2

∇w(τ )22 dτ dt < Cε0 .

Thus, there exists T > 0 such that 

T +1

T

w(τ )H 1 dτ < ε.

Summing up, we proved the following result. √ Theorem 1.11.7 Assume that 2Re < 1. Then, given w0 ∈ L2σ (R3+ ), there exists a least one weak solution w to Eq. (1.11.4) with w(0) = w0 satisfying  lim

T →∞ T

T +1

w(τ )H 1 dτ = 0.

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For generalizations of this result to the situation of stratified flows we refer to the work of Koba [93]. We refer to the same reference for the existence of strong solutions for data being small in suitable norms.

1.12 Fluid-Rigid Body Interaction Problems for Compressible Fluids The analysis of the movement of rigid or elastic bodies immersed in a fluid is a classical problem in fluid mechanics. In order to describe the problem, denote the bounded domain occupied by the body by B(t) and let D(t) be the exterior domain filled by the fluid, i.e. D(t) := R3 \B(t). The interface between body and fluid is denoted by (t). In this section we are interested in the case where the body is a rigid body and the fluid is a compressible fluid within the barotropic regime. In this case, the motion of the fluid is given by the equations t + div (v) = 0 (vt + (v · ∇)v) − div T (v, p) = 0 v = vB v(0) = v0 (0) = 0

in JT × D(t) in JT × D(t), in JT × (t), in D(0), in D(0),

(1.12.1)

Here JT = (0, T ) for some T > 0 and , v and p denote the density, velocity and pressure of the fluid, respectively. We assume that the fluid is of barotropic type, i.e. that the pressure p = p() satisfies the relation p ∈ C ∞ (R+ ) and p () > 0 for all  > 0. The stress tensor T (v, p) is given by T (v, p) = 2μD(v) + (μ − μ) div vI − pI. Here D(v) = 12 (∇v) + (∇v)T denotes the deformation tensor, I is the 3 × 3 identity matrix, μ > 0 and μ are constants satisfying the relation μ + μ > 0. The fluid equations are coupled to the balance equations for the momentum and the angular momentum of the rigid body which read as  mη (t) − (t ) T (v, p)n(t, x)dσ  (J ω) (t) − (t )(x − xc ) × T (v, p)n(t, x)dσ η(0) ω(0)

= F (t), t ∈ JT , = M(t), t ∈ JT , = η0 , = ω0 ,

(1.12.2)

and which contain the drag force and the torque extended by the fluid onto the body. The constants m and J denote the body’s mass and inertia tensor. Moreover, xc is the position of its center of gravity and η = η(t) and ω = ω(t) denote its translational

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and angular velocity. Hence vB (t, x) := η(t) + ω(t) × (x − xc (t)),

x ∈ (t).

The functions F and M are external forces and torques. Since the domain of the fluid D(t) depends on the motion of the rigid body, the problem is a so called moving domain problem. We start by rewriting Eqs. (1.12.1) and (1.12.2) into the following system of equations for the unknowns , v and η, ω ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

t + div (v) = 0

in JT × D (t)

(vt + (v · ∇)v) − div T (v, p) = 0 mη (t) −



in JT × D (t),

v(t, x) = η(t) + ω(t) × (x − xc (t))

(t) T (v, p)n(t, x)dσ = F (t) ⎪ ⎪  ⎪ ⎪  ⎪ (J ω) (t) − (t) (x − xc ) × T (v, p)n(t, x)dσ = M(t) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ v(0) = v0 , (0) = 0 ⎪ ⎪ ⎪ ⎪ ⎩ η(0) = η0 , ω(0) = ω0 .

on JT × (t), t ∈ JT , t ∈ JT , in D (0),

(1.12.3) We note state our main result on the existence of unique, local strong solution to system (1.12.3). Let us denote by ¯ the mean value of 0 in D(0). Theorem 1.12.1 Let 1 < p < ∞ and 3 < q < ∞. Let T0 > 0 and F, M ∈ Lp (JT0 ; R3 ) and D(0) be an exterior domain of class C 2,1 . Assume that 2−2/p

0 − ¯ ∈ W 1,q (D(0)), v0 ∈ Bq,p

(D(0)) and η0 , ω0 ∈ R3

are satisfying the compatibility condition v0 = η0 + ω0 × x. Then there exists T ∈ (0, T0 ] such that the system (1.12.3) admits a unique, strong solution (u, , η, ω) on JT = (0, T ) within the class  ∈ W 1,p (JT ; Lq (D(·))) ∩ Lp (JT ; W 1,q (D(·))), v ∈ W 1,p (JT ; Lq (D(·))3 ) ∩ Lp (JT ; W 2,q (D(·))3 ), (η, ω) ∈ W 1,p (JT ; R6 ). The above regularity properties for  and v in Sobolev spaces on domains D(t) depending on t are understood in the following way: by v ∈ W 1,p (JT ; Lq (D(·))3 )∩ Lp (JT ; W 2,q (D(·))3 ) we mean that X∗ v ∈ W 1,p (JT ;Lq (D)3 ) ∩ Lp (JT ;W 2,q (D)3 ), where X∗ v(t, y) := v(t, X(t, y)) and X is being defined as the solution of Eq. (1.12.4) below. The regularity space for  is defined in an analogous way. In order to prove Theorem 1.12.1 it is natural to transform the original problem on D(t) to a problem on the fixed domain D(0). We will use a nonlinear, local change of coordinates for this, which only acts on a neighborhood of the rigid body.

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This transformation goes back to Inoue and Wakimoto [89]. For compressible fluids it is also natural to switch, after this first change of coordinates, from Eulerian to Lagrangian coordinates. After these two transformations, our approach relies on maximal regularity estimates for the linearized transformed problem. Finally, we rewrite the nonlinear problem as a fixed point problem in the space of maximal regularity. Let us start by introducing a diffeomorphism X between the time dependent and the fixed domain by considering the equation 

∂t X(t, y) = b(t, X(t, y)), (0, T ) × R3 , X(0, y) = y, y ∈ R3 ,

(1.12.4)

where b determines the modified velocity due to this change of coordinates. More precisely, we choose open balls B1 , B2 ⊂ R3 such that B ⊂ B1 ⊂ B1 ⊂ B2 and define a cut-off function χ ∈ C ∞ (R3 ; [0, 1]) by  χ(y) :=

1 if y ∈ B1 , 0 if y ∈ D \ B2 ,

(1.12.5)

as well as a time-dependent vector field b : [0, T ] × R3 → R3 by b(t, x) := χ(x − xc (t))[m(t)(x − xc (t)) + η(t)].

(1.12.6)

Then b ∈ W 1,p (0, T ; Cc∞ (R3 )) and that b| = m(x − xc ) + η. Given η, ω ∈ W 1,p (0, T ), the Eq. (1.12.4) admits a unique solution X ∈ C 1 ((0, T ); C ∞ (Rn )) by the Picard–Lindelöf theorem. Denoting by JX the Jacobian matrix of X, one can show that JX is invertible provided T > 0 is small enough. Performing this change of coordinates and switching also to Lagrangian coordinates in a second step, we obtain the system ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

mκ 

⎪ I − ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩



θt + ¯ div u = f0 (u, θ, κ, )

in JT × D,

ut − div T (u, θ ) = f1 (u, θ, κ, )

in JT × D,

−





u=0

on JT × ,

T (u, θ )νdσ = g0 (u, θ, κ, )

t ∈ JT ,

ξ × T (u, θ )νdσ = g1 (u, θ, κ, )

t ∈ JT ,



κ(0) = η0 ,

u(0) = w0

in

D,

θ (0) = 0 − ¯

in

D,

(1.12.7)

(0) = ω0 ,

where the right hand sides f0 , f1 , g0 and g1 collect all the nonlinear terms. Here T , ν, κ,  and I denote the transformed stress tensor, normal, translational, angular velocities and inertia tensor.

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We now aim to prove a maximal regularity theorem for the linearized problem of (1.12.7), which reads as ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

mκ  I

−



t + γ1 div v = f0

in JT × D,

vt − div T (v, ) = f1

in JT × D,

−





v=0

on JT × ,

T (v, )νdσ = g0

t ∈ JT ,

ξ × T (v, )νdσ = g1

t ∈ JT ,



v(0) = v0

in D,

(0) = 0 − ¯

in D,

(1.12.8)

η(0) = η0 , ω(0) = ω0 .

Here, f = (f0 , f1 ) and g = (g0 , g1 ) are given functions, T (v, ) is given by T (v, ) = 2αD(v) + (β − α) div vI − γ2 I , where α, β, γ1 and γ2 are constants satisfying α > 0, α + β > 0 and γ1 , γ2 > 0. Step 1: Maximal Regularity for the Fluid Equation Consider the set of equations describing the compressible fluid ⎧ ⎪ t + γ1 div v = f0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨vt − div T (v, ) = f1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

in JT × D, in JT × D, on JT × ,

v=0

v(0) = v0 ,

in D,

(0) = 0 − ¯

in D.

(1.12.9)

We rewrite the above system (1.12.9) as an evolution equation on W 1,q (D) × i.e. for w = (, v), f = (f0 , f1 ) and w0 = (0 − , ¯ v0 ) we consider the equation

Lq (D)3 ,

wt + Aw = f, w(0) = w0 where  A :=

0 γ1 div γ2 ∇ AD



with domain D(A) := W 1,q (D) × D(AD ) and where 1,q

AD v := αv + (β − α)∇ div v for v ∈ D(AD ) := (W 2,q (D) ∩ W0 (D))3 .

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97

Let us decompose A as  A=

0 γ1 div γ2 ∇ AD



 =

0 γ1 div 0 AD



 +

0 0 γ2 ∇ 0

 =: A0 + A1 .

Note that AD is a parameter elliptic operator in sense of [36] subject to Dirichlet boundary conditions. Hence, by Theorem 8.2 of [36], AD admits maximal regularity on Lq (D)3 . Moreover, since div is a bounded operator from D(AD ) to W 1,q (D) it follows that A0 admits maximal regularity on W 1,q (D) × Lq (D)3 . Furthermore, since ∇ maps W 1,q (D) boundedly into Lq (D)3 , it follows that A is a bounded perturbation of A0 and that thus A admits maximal regularity on W 1,q (D)×Lq (D)3 . Step 2: Maximal Regularity for the Rigid Body Equations Let   mI 0 I= 0 I be the constant momentum matrix of our problem. For 0 < ε ≤ 1 − 1/q we define the operator    hνdσ ε+1/q,q 3×3 6   (D, R ) → R , h → . Jξ,ν : Wloc  ξ × hνdσ ε/2+1/q,q

The boundedness of the trace operator γ : Hloc

(D) → Lq (∂D) implies that

|Jξ,ν h| ≤ Cγ hLq () ≤ ChW ε+1/q,q (D) , h ∈ W ε+1/q,q (D).

(1.12.10)

Hence, the fourth and fifth line of the system (1.12.8) may be rewritten as  I

κ 



 − Jξ,ν T (v, ) =

g0 g1

 (1.12.11)

κ(0) = η0 , (0) = ω0 . We then obtain 1/p 0 −  ¯ W 1,q (D)). Jξ,ν I ]Lp ≤ C(T Yp,q T +T

By interpolation, H α,p (JT ; H 2−2α,q (D)) → L2p (JT ; W 1+1/q+ε/2,q (D)) provided 1/2p < α < 1/2 − 1/2q. Thus, by (1.12.10) and the mixed derivative Theorem 1.6.16 Jξ,ν D(v)Lp ≤ CvLp (JT ;W 1+ε/2+1/q,q (D)) ≤ CT 1/2p vL2p (JT ;W 1+ε/2+1/q,q (D)) ≤ CT 1/2p vH α,p (JT ;H 2−2α,q (D)) ≤ CT 1/2p vXp,q T

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provided 0 < ε < 2 − 2/q. Similarly, Jξ,ν div vI p ≤ CT β vXp,q T for some constant β > 0. We thus proved the following result. Proposition 1.12.2 Let D be an exterior domain with boundary class of C 2,1 and 2−2/p let p, q ∈ (1, ∞) as well as T > 0. Let η0 , ω0 ∈ R3 , v0 ∈ Bp,q (D) and 1,q p 1,q p 0 − ¯ ∈ W (D). Assume that f0 ∈ L (JT ; W (D)), f1 ∈ L (JT ; Lq (D)3 ) and g0 , g1 ∈ Lp (JT ; R3 ). Then the system (1.12.8) admits a unique solution T := W 1,p (JT ; Lq (D)3 ) ∩ Lp (JT ; W 2,q (D)3 ), v ∈ Xp,q T  ∈ Yp,q := W 1,p (JT ; W 1,q (D)),

κ ∈ W 1,p (JT ; R6 ),  ∈ W 1,p (JT ; R6 ). and there exists a constant C > 0 such that vXp,q + Yp,q T T + κW 1,p (J ) + W 1,p (J ) T T ≤ C(f0 Lp (JT ;W 1,q (D)) + f1 Lp (JT ;Lq (D)) + g0 Lp (JT ) + g1 Lp (JT ) + v0 B 2−2/p (D) + 0 −  ¯ W 1,q (D) + |η0 | + |ω0 |). p,q

We finally solve (1.12.7) by an application of the contraction principle. To this end, we define the set ZR,T by 1,p

T T × Yp,q,0 × W0 (JT ; R6 ) : uXp,q ZR,T := {(u, θ, κ, ) ∈ Xp,q,0 T

+ θ Yp,q T + (κ, )Lp (JT ) ≤ L}, T T T T where Xp,q,0 = {u ∈ Xp,q : u(0) = 0}, Yp,q,0 = {θ ∈ Yp,q : θ (0) = 0} 1,p

and W0 (JT ; R6 ) = {(κ, ) ∈ W 1,p (JT ; R6 ) : η(0) = ω(0) = 0}. Given ˜ ∈ ZR,T , there exists a unique solution (u, θ, κ, ) to the linear (u, ˜ θ˜ , κ, ˜ ) problem (1.12.8) with initial data u0 = 0, 0 − ¯ = 0, η0 = ω0 = 0 and right ˜ gˆi (u, ˜ for i = 0, 1. Set uˆ := u − u∗ , θˆ := θ − θ ∗ , hand sides fˆi (u, ˜ θ˜ , κ, ˜ ), ˜ θ˜ , κ, ˜ ) ∗ ∗ ˆ κˆ := κ − κ and  :=  −  , where (u∗ , θ ∗ , κ ∗ , ∗ ) is the unique solution to ˜ ∈ ZR,T system (1.12.8) with f0 = f1 = g0 = g1 = 0 and γ1 = . For (u, ˜ θ˜ , κ, ˜ ) we define the map  by ˜ := (u, ˆ (u, ˜ θ˜ , κ, ˜ ) ˆ θˆ , κ, ˆ ), and show that  is a contraction from ZR,T into itself. For this, we need to estimate ˜ and gˆi (u, ˜ We will not do this here in ˜ θ˜ , κ, ˜ ) ˜ θ˜ , κ, ˜ ). the nonlinear terms fˆi (u,

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99

detail, but refer to [74]. The unique solution of Eq. (1.12.7) is then transferred back to the unique solution of the original system (1.12.3) within the regularity class stated in Theorem 1.12.1. The proof of Theorem 1.12.1 is complete.

1.13 Two-Phase Free Boundary Value Problems for a Class of Non-Newtonian Fluids In this section we consider a two-phase free boundary value problem for generalized Newtonian fluids. The problem reads as follows: let n ≥ 2 and 0 ⊂ Rn be a hypersurface that separates a region 1 (0) filled with a viscous, incompressible fluid from 2 (0), the complement of 1 (0) in Rn . The region 2 (0) is also occupied with a second incompressible, viscous fluid and it is assumed that the two fluids are immiscible. Denoting by (t) the position of 0 at time t, (t) is then the interface separating the two fluids occupying the regions 1 (t) and 2 (t). We recall from Sect. 1.8 that an incompressible fluid is subject to the set of equations ρ(∂t u + u · ∇u) = div T , div u = 0, where ρ denotes the density of the fluid and where the stress tensor T can be decomposed as T = τ − qI . Here q denotes the pressure and τ the tangential part of the stress tensor of the fluid. For a Newtonian fluid, τ is given by τ = 2μD(u), where D(u) = [∇u + (∇u)T ]/2 denotes the deformation tensor and μ the viscosity coefficient of the fluid. A Two-Phase Problem for Generalized Newtonian Fluids In this section, we consider a class of non-Newtonian fluids, where τ as above is replaced by τ = 2μ(|D(u)|2 )D(u) for some function μ satisfying μ ∈ C 3 ([0, ∞))

and

μ(0) > 0.

(1.13.1)

In the special case of power law fluids, one has μ(|D(u)|2 ) = ν + β|D(u)|d−2 for some d ≥ 1 and constants ν, β ≥ 0. If d < 2, the fluid is then called a shear thinning fluid, if d > 2 it is called a shear thickening fluid. Fluids of this type are special cases of so called Stokesian fluids, which were investigated analytically for fixed domains by Amann in [7] and [9].

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The motion of the two immiscible, incompressible and viscous fluids is then governed by the following set of equations ⎧ ρ(∂t v + v · ∇v) = div T − ργa eN , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ div = 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −[[T n ]] = σ H n ⎪ ⎪ ⎨ [[v]] = 0 ⎪ ⎪ ⎪ ⎪ V = v · n ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ v|t =0 = v0 ⎪ ⎪ ⎪ ⎪ ⎩ |t =0 = 0 ,

in (t), in (t), on (t), on (t),

(1.13.2)

on (t), in 0 ,

where t > 0 and where we assumed that the unknown (t) is determined by an unknown scalar function h = h(t, x  ) for x  ∈ Rn−1 , that is (t) = {(x  , xn ) : x  ∈ Rn−1 , xn = h(t, x  )} and (t) = 1 (t) ∪ 2 (t) with n (t) = {(x  , xn ) : x  ∈ Rn−1 , (−1)n (xn − h(t, x  )) > 0},

n = 1, 2.

Here, en = (0, . . . , 0, 1)T and the normal field on (t), pointing from 1 (t) into 2 (t), is denoted by n (t, ·). Moreover, V (t, ·) and H (t, ·) denote the normal velocity and mean curvature of (t), respectively. Furthermore, γa denotes the gravitational acceleration and σ the coefficient of the surface tension. We suppose that the stress tensor T is given by the generalized Newtonian type described above, that is, for given scalar functions μ1 , μ2 : [0, ∞) → R, T = χ1 (t )T1 (v, q) + χ2 (t )T2 (v, q), Tj (v, π) = −qI + 2μj (|D(v)|2 )D(v),

j = 1, 2,

N 2 and where |D(u)|2 = i,j =1 (Dij (u)) . The function χD denotes the indicator n function of a set D ⊂ R , and the density ρ is defined by ρ = χ1 (t )ρ1 + χ2 (t ) ρ2 for the densities ρj > 0 of the j -th fluid. The system is complemented by the initial fluid velocity v0 and the given initial height function h0 , as well as 0 = 1 (0) ∪ 2 (0) and 0 given by j (0) = {(x  , xn ) : x  ∈ Rn−1 , (−1)j (xn − h0 (x  )) > 0}, 0 = {(x  , xn ) : x  ∈ Rn−1 , xn = h0 (x  )}.

j = 1, 2,

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In addition, we denote the unit normal field, pointing from 1 (0) into 2 (0), on 0 by n0 . The quantity [[f ]] = [[f ]](t, x) is the jump of the quantity f , which is defined on (t) across the interface (t) as [[f ]](t, x) = lim {f (t, x + εn ) − f (t, x − εn )} ε→0+

for x ∈ (t).

The problem then is to find functions v, q, and h solving the Eq. (1.13.2). Well-posedness results for the above system (1.13.2) in the case of Newtonian fluids and in the special case of one-phase flows with or without surface tension were first obtained by Solonnikov [136–138], Shibata and Shimizu [130, 132]. The case of an ocean of infinite extend and which is bounded below by a solid surface and bounded above by a free surface was treated first by Beale [15] and Tani and Tanaka [146] Besides the articles cited already above, the two-phase problem for Newtonian fluids was studied by Denisova in [34] and [35], and by Tanaka in [145] using Lagrangian coordinates. Indeed, Denisova proved local wellposedness in the r,r/2 Newtonian case in W2 for r ∈ (5/2, 3) for the case where one of the domains is bounded and Tanaka obtained wellposedness (including thermo-capillary convecr,r/2 tion) in W2 for r ∈ (7/2, 4). Prüss and Simonett were using in [124] a different approach by transforming problem (1.13.2) to a problem on a fixed domain via the Hanzawa transform and applying then an optimal regularity approach for the linearized equations. Like this they proved wellposedness of the above problem in the case of Newtonian fluids. Problems of the above kind for non-Newtonian fluids were treated by Abels in [4] in the context of measure-valued varifold solutions. His result covers in particular the situation where μj (s) = νj s (d−2)/2 for j = 1, 2 and d ∈ (1, ∞). Note, however, that his approach does not give the uniqueness of a solution. We refer also to [5]. In the main result of this section we show that system (1.13.2) admits a unique, strong solution on (0, T ) for arbitrary T > 0 provided the viscosity functions μn fulfill (1.13.1) and the initial data are sufficiently small in their natural norms. More precisely, we have the following result. Theorem 1.13.1 Let p ∈ (n + 2, ∞) and J = (0, T ) for some T > 0. Suppose that ρ1 > 0, ρ2 > 0, γa ≥ 0, σ > 0 and that μj ∈ C 3 ([0, ∞)) and μj (0) > 0,

j = 1, 2.

2−2/p

Then there exists ε0 > 0 such that for (v0 , h0 ) ∈ Wp satisfying the compatibility conditions

3−2/p

(0 )n × Wp

(Rn−1 )

[[μ(|D(v0 )|2 )D(v0 )n0 − {n0 · μ(|D(v0 )|2 )D(v0 )n0 }n0 ]] = 0 on 0 , div0 = 0

in 0 ,

[[v0 ]] = 0 on 0 ,

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M. Hieber

with μ(|D(v0 )|2 ) = χ1 (0) μ1 (|D(v0 )|2 ) + χ2 (0) μ2 (|D(v0 )|2 ) as well as the smallness condition v0 W 2−2/p ( ) + h0 W 3−2/p (Rn−1 ) < ε0 , 0

p

p

the system (1.13.2) admits a unique solution (v, q, h) within the class v ∈ (Hp1 (J, Lp ((t)) ∩ Lp (J, Hp2 ((t))))n , q ∈ Lp (J, H˙ p1 ((t))), 2−1/(2p)

h ∈ Wp

1/2−1/(2p)

∩ Wp

2−1/p

(J, Lp (Rn−1 )) ∩ Hp1 (J, Wp

(Rn−1 ))

3−1/p

(J, Hp2 (Rn−1 )) ∩ Lp (J, Wp

(Rn−1 )).

Remarks 1.13.2 (a) Some remarks on our notation are in order at this point. Setting ˙ n = Rn \ Rn , R 0

Rn0 = {(x  , xn ) : x  ∈ Rn−1 , xn = 0},

we mean by v ∈ Hp1 (J, Lp ((t))) ∩ Lp (J, Hp2 ((t)))N that ˙ n )) ∩ Lp (J, Hp2 (R ˙ n )))n , "∗ v = v ◦ " ∈ (Hp1 (J, Lp (R where " and "∗ are defined below by (1.13.4) and (1.13.5), respectively. The regularity statement for q is understood in the same way. (b) The assumption p > n + 2 implies that h ∈ BU C(J, BU C 2 (Rn−1 )),

∂t h ∈ BU C(J, BU C 1 (Rn−1 )),

which means that the condition on the free interface can be understood in the classical sense. (c) Typical examples of viscosity functions μ satisfying our conditions are given by  d−2  with d = 2, 4, 6, or d ≥ 8, μ(s) = ν 1 + s 2 μ(s) = ν(1 + s)

d−2 2

with 1 ≤ d < ∞

for ν > 0. For more information and details we refer e.g. to the work of [43] and [108]. Obviously, if d = 2, the above viscosity functions correspond to the Newtonian situation. Let us remark at this point that our proof of Theorem 1.13.1 is inspired by the work by Pr¨uss and Simonett in [122] and [123]. Our strategy may be described as follows: we first transform the system (1.13.2) to a problem on a fixed domain. Then maximal regularity properties of the associated linearized problem due to Prüss and Simonett [123] enable us prove Theorem 1.13.1 by the contraction principle.

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

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Transformation to a Fixed Domain Let us start the proof of Theorem 1.13.1 by calculating the divergence of the stress tensor, i.e. by calculating explicitly div{μN (|D(u)|2 )D(u)}

for N = 1, 2.

Let us remark first that given a vector u of length m for m ≥ 2, we denote by ui its i-th component and by u its tangential component, i.e. u = (u1 , . . . , um )T and u = (u1 , . . . , um−1 )T . We then obtain (div{μN (|D(u)|2 )D(u)})i =

n 1  {2μ˙ N (|D(u)|2 )Dij (u)Dkl (u) 2 j,k,l=1

+ μN (|D(u)|2 )δik δj l }(∂j ∂k ul + ∂j ∂l uk ). For vectors u, v we set AN (u)v := (An,1 (u)v, . . . , An,N (u)v)T , where AN,i (u)v := −

n 

j,k,l

AN,i (D(u))(∂j ∂k vl + ∂j ∂l vk )

and

j,k,l=1 j,k,l

AN,i (D(u)) :=

1 2μ˙ N (|D(u)|2 )Dij (u)Dkl (u) + μN (|D(u)|2 )δik δj l 2

for N = 1, 2 and i = 1, . . . , n. We then have AN (u)u = − div{μN (|D(u)|2 )D(u)}

and AN (0)u = −μN (0)(u + ∇ div u).

In addition, we set A(u)v := χ1 (t )A1 (u)v + χ2 (t )A2 (u)v

and ! q := q + ργa xn .

Then system (1.13.2) may be rewritten as ⎧ ρ(∂t v + v · ∇v) − μ(0)v + ∇! q = −(A(v) − A(0))v ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ div = 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −[[T!n ]] = σ H n + [[ρ]]γa xN n ⎪ ⎨ [[v]] = 0 ⎪ ⎪ ⎪ ⎪ ⎪ V = v · n ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ v|t =0 = v0 ⎪ ⎪ ⎪ ⎩ |t =0 = 0 ,

in (t), in (t), on (t), on (t), on (t), in 0 , (1.13.3)

q ) + χ2 (t )T2 (v, ! q ) and μ(0) = χ1 (t ) μ1 (0) + χ2 (t ) μ2 (0). where T! = χ1 (t )T1 (v, !

104

M. Hieber

Our first aim is to transform the problem (1.13.3) to a problem on the fixed ˙ n . To this end, we define a transformation " on J × R ˙ n for J = (0, T ) domain R with T > 0 as " ˙ n $ (τ, ξ  , ξn ) → (t, x  , xn ) ∈ {s} × (s), " :J ×R s∈J 



with t = τ, x = ξ , xn = ξn + h(τ, ξ  )

(1.13.4)

for some scalar-valued function h. Note that det J " = 1, where J " denotes the Jacobian matrix of ". We now define u(τ, ξ ) := "∗ v(t, x) := v("(τ, ξ )),

π(τ, ξ ) := "∗! q (t, x),

(1.13.5)

as well as ˙ n → Rn , "∗ f (τ, ξ ) := f ("−1 (t, x)) for f : R

(1.13.6)

where "−1 is given by "−1 (t, x) = (t, x  , xn −h(t, x  )). Hence, the system (1.13.3) ˙n is reduced to the following problem on R ⎧ ˙ n, ⎪ ρ∂τ u − μ(0)u + ∇π = F (u, π, h) in R ⎪ ⎪ ⎪ ⎪ ⎪ ˙ n, ⎪ div u = Fd (u, h) in R ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −[[μ(0)(DN uj + Dj un )]] = Gj (u, [[π]], h) on Rn0 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ [[π]] − 2[[μ(0)DN uN ]] − ([[ρ]]γa + σ  )h = GN (u, h) on Rn0 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

[[u]] = 0 ∂τ h − un = Gh (u , h)

on Rn0 , on Rn0 ,

u|t =0 = u0

˙ n, on R

h|t =0 = h0

on Rn−1 , (1.13.7)

where j = 1, . . . , n − 1 and F = (F1 , . . . , Fn )T . The terms on the right hand side of (1.13.7) are given by Fi (u, π, h) := ρ{(∂τ h)Dn ui − (u · ∇)ui + (u · ∇  h)Dn ui } −μ(0)

n 

Fjj (h)ui + (Di h)DN π + Ai (u, h)

j =1

Gj (u, [[π]], h) := σ H(h)Dj h − {([[ρ]]γa + σ  )h}Dj h + [[π]]Dj h + Bj (u, h) Gn (u, h) := −σ H(h) + Bn (u, h), Fd (u, h) := (Dn u ) · ∇  h = Dn (u · ∇  h), Gh (u, h) := −u · ∇  h.

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

105

Here Ai (u, h), Bj (u, h) and Bn (u, h) are given by Ai (u, h) :=

n 

j,k,#

Ai

j,k,#

(E(u, h)) − Ai

(0) (Dj Dk u# + Dj D# uk )

j,k,#=1

−

N 

j,k,#

Ai

j,k,#

(E(u, h)) − Ai

(0)

j,k,#=1

× (Fj k (h)u# + Fj # (h)uk ),

i = 1, . . . , n,

Bj (u, h) := − [[μ(|E(u, h)|2)Dn un ]]Dj h + [[{μ(|E(u, h)|2) − μ(0)}(Dn uj + Dj un )]] −

n−1  [[μ(|E(u, h)|2 )(Dj uk + Dk uj )]]Dk h k=1

+

n−1  [[μ(|E(u, h)|2 )(DN uj Dk h k=1

+ Dn uk Dj h)]]Dk h,

j = 1, . . . , n − 1,

Bn (u, h) :=2[[{μ(|E(u, h)|2) − μ(0)}Dn un ]] + [[μ(|E(u, h)|2)Dn un ]]|∇  h|2 −

n−1  [[μ(|E(u, h)|2 )(Dn uk + Dk un )]]Dk h, k=1

j,k,l

where Ai

j,k,l

j,k,l

(E(u, h)) := χRn− Ai,1 (E(u, h)) + χRn+ Ai,2 (E(u, h)). In particular,

μ(|E(u, h)|2 ) = χRn− μ1 (|E(u, h)|2 ) + χRn+ μ2 (|E(u, h)|2 ), ρ = χRn− ρ1 + χRn+ ρ2 , μ(0) = χRn− μ1 (0) + χRn+ μ2 (0). Finally, in order to simplify our notation, we set G(u, [[π]], h) := (G1 (u, [[π]], h), . . . , Gn−1 (u, [[π]], h), Gn (u, h))T A(u, h) := (A1 (u, h), . . . , An (u, h))T , B(u, h) := (B1 (u, h), . . . , Bn (u, h))T .

106

M. Hieber

The above set of Eq. (1.13.7) leads to the following associated linear problem ⎧ ˙ n, ⎪ ρ∂t u − νu + ∇π = f in R ⎪ ⎪ ⎪ ⎪ ⎪ ˙ n, ⎪ div u = fd in R ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −[[ν(Dn uj + Dj un )]] = gj on Rn0 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ [[π]] − 2[[μDn un ]] − ([[ρ]]γa + σ  )h = gn on Rn0 , (1.13.8) ⎪ [[u]] = 0 on Rn0 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂t h − un = gh on Rn0 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ˙ n, ⎪ u|t =0 = u0 in R ⎪ ⎪ ⎪ ⎪ ⎩ h|t =0 = h0 on Rn−1 , where j = 1, . . . , n − 1 and set g = (g1 , . . . , gn )T . Here, ρ = ρ1 χRn− + ρ2 χRn+ ,

ν = ν1 χRn− + ν2 χRn+

with positive constants ρN , νN for N = 1, 2. Characterization of Maximal Regularity by Data The optimal regularity property of the solution of the above problem (1.13.8) will be of central importance in the following. The following result due to Prüss and Simonett [122] characterizes the set of right-hand sides of Eq. (1.13.8) which yield a unique solution of (1.13.8) in the space of maximal regularity. Proposition 1.13.3 (Prüss–Simonett [122, 123]) Let 1 < p < ∞, p = 3/2, 3, a > 0 and J = (0, a). Suppose that ρN > 0, νN > 0, γa ≥ 0 and σ > 0,

N = 1, 2.

Then, Eq. (1.13.8) admits a unique solution (u, π, h) with regularity ˙ n )) ∩ Lp (J, Hp2 (R ˙ n )))n , u ∈ (Hp1 (J, Lp (R ˙ n )), π ∈ Lp (J, H˙ p1 (R 1/2−1/(2p)

[[π]] ∈ Wp

2−1/(2p)

h ∈ Wp

1−1/p

(J, Lp (Rn−1 )) ∩ Lp (J, Wp

2−1/p

(J, Lp (Rn−1 )) ∩ Hp1 (J, Wp

(Rn−1 )), 3−1/p

(Rn−1 )) ∩ Lp (J, Wp

(Rn−1 ))

if and only if the data (f, fd , g, gh , u0 , h0 ) satisfy the following regularity and compatibility conditions: ˙ n ))n , f ∈ Lp (J, Lp (R ˙ n )), fd ∈ Hp1 (J, H˙ p−1 (Rn )) ∩ Lp (J, Hp1 (R

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations 1/2−1/(2p)

g ∈ (Wp

1−1/(2p)

gh ∈ Wp

1−1/p

(J, Lp (Rn−1 )) ∩ Lp (J, Wp

2−1/p

(J, Lp (Rn−1 )) ∩ Lp (J, Wp

2−2/p

˙ n )n , (R

h0 ∈ Wp

3−2/p

(Rn−1 ),

fd |t =0 = div u0

˙n in R

u0 ∈ Wp

and [[u0 ]] = 0

gj |t =0 = −[[ν(Dn u0j + Dj u0n )]]

107

(Rn−1 )))n ,

(Rn−1 )),

on RN−1

if p > 3/2,

on Rn−1 if p > 3

for all j = 1, . . . , n − 1. Moreover, the solution map [(f, fd , g, gh , u0 , h0 ) → (u, π, [[π]], h)] is continuous between the corresponding spaces. The Nonlinear Problem For a > 0 let J = (0, a) and set ˙ n )) ∩ Lp (J, Hp2 (R ˙ n )))N | [[u]] = 0}, E1 (a) := {u ∈ (Hp1 (J, Lp (R ˙ n )), E2 (a) := Lp (J, H˙ p1 (R 1/2−1/(2p)

E3 (a) := Wp

2−1/(2p)

E4 (a) := Wp

1−1/p

(J, Lp (Rn−1 )) ∩ Lp (J, Wp

2−1/p

(J, Lp (Rn−1 )) ∩ Hp1 (J, Wp

1/2−1/(2p)

∩ Wp

(Rn−1 )),

(Rn−1 )) 3−1/p

(J, Hp2 (Rn−1 )) ∩ Lp (J, Wp

(Rn−1 ))

as well as ˙ n ))N , F1 (a) := Lp (J, Lp (R ˙ n )), F2 (a) := Hp1 (J, H˙ p−1 (Rn )) ∩ Lp (J, Hp1 (R 1/2−1/(2p)

F3 (a) := (Wp

1−1/(2p)

F4 (a) := Wp

1−1/p

(J, Lp (Rn−1 )) ∩ Lp (J, Wp

2−1/p

(J, Lp (Rn−1 )) ∩ Lp (J, Wp

(Rn−1 )))N ,

(Rn−1 )).

The solution space E(a) and the data space F(a) are defined for a > 0 by E(a) := {(u, π, q, h) ∈ E1 (a) × E2 (a) × E3 (a) × E4 (a) | [[π]] = q}, F(a) := F1 (a) × F2 (a) × F3 (a) × F4 (a). and are endowed with their natural norms. Finally, for (u, π, q, h) ∈ E(a) consider the nonlinear mapping N defined by N (u, π, q, h) := (F (u, π, h), Fd (u, h), G(u, q, h), Gh (u, h)). The mapping N possesses the following properties.

(1.13.9)

108

M. Hieber

Lemma 1.13.4 Let n + 2 < p < ∞, a > 0 and r > 0. Suppose that μN (s) ∈ C 3 ([0, ∞)) for N = 1, 2 and in addition that ρ1 > 0, ρ2 > 0, γa ≥ 0 and σ > 0 are constants. Then N ∈ C 1 (BE(a) (r), F(a)), N (0) = 0 and DN (0) = 0. Finally, let us return to the nonlinear problem (1.13.7). Let us define the space of initial data I by 2−2/p

I := Wp

˙ n )n × Wp (R

3−2/p

(Rn−1 ).

The following result shows that problem (1.13.7) on the fixed domain admits a unique strong solution provided the data u0 and h0 are sufficiently small in their corresponding norms. Proposition 1.13.5 Let n+2 < p < ∞ and a > 0. Suppose that μN ∈ C 3 ([0, ∞)) and that ρN > 0, μN (0) > 0, γa ≥ 0 and σ > 0 for N = 1, 2. Then there exist positive constants ε0 and δ0 (depending on a and p), such that system (1.13.7) admits a unique solution (u, π, [[π]], h) in BE(a) (δ0 ) provided the initial data (u0 , h0 ) ∈ I satisfy the smallness condition (u0 , h0 )I < ε0 , as well as the compatibility conditions [[μ(|E(u0, h0 )|2 )E(u0 , h0 )n0 − {n0 · μ(|E(u0, h0 )|2 )E(u0 , h0 )n0 }n0 ]] = 0 on Rn0 , (1.13.10) div u0 = Fd (u0 , h0 )

˙ n, in R

[[u0 ]] = 0

on Rn0 .

For z = (u, π, q, h) ∈ E(a), the nonlinear problem (1.13.7) can be restated as Lz = N (z),

(u, h)|t =0 = (u0 , h0 ),

(1.13.11)

where L denotes the linear operator on the left side of (1.13.8) with ν = μ(0) and N is defined in (1.13.9). We divide the proof into several steps. Step 1 We introduce an auxiliary function z∗ ∈ E(a) that satisfies Lz∗ = (0, fd∗ , g ∗ , gh∗ ),

(u∗ , h∗ )|t =0 = (u0 , h0 ),

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

109

where fd∗ ∈ F2 (a) and fd∗ |t =0 = Fd (u0 , h0 ), g ∗ ∈ F3 (a)

and g ∗ |t =0 = G(u0 , [[π0 ]], h0 ),

gh∗ ∈ F4 (a)

and gh∗ |t =0 = Gh (u0 , h0 ),

(1.13.12)

and where z∗ E(a) ≤ C0 (u0 , h0 )I with a positive constant C0 independent of (u0 , h0 ). The last inequality then yields Lz∗ F(a) ≤ C1 C0 (u0 , h0 )I

(1.13.13)

for some positive constant C1 . Step 2 In order to proceed with our proof, we define closed subspaces 0 E(a), 0 F(a) of E(a), F(a) as 0 E(a)

:= {z = (u, π, q, h) ∈ E(a) : (u, q, h)|t =0 = (0, 0, 0)},

0 F(a)

:= {(f, fd , g, gh ) ∈ F(a) : (fd , g, gh )|t =0 = (0, 0, 0)},

respectively. We then replace z by z + z∗ in (1.13.11) in order to obtain Lz = N (z + z∗ ) − Lz∗ =: K0 (z),

z ∈ 0 E(a).

This leads to the fixed point equation z = L−1 0 K0 (z),

z ∈ 0 E(a),

(1.13.14)

where L0 denotes the restriction of L to 0 E(a). By (1.13.12), we have K0 (z) ∈ 0 F(a) for any z ∈ 0 E(a). In addition, it follows from Lemma 1.13.4 and (1.13.13) that K0 ∈ C 1 (B0 E(a) (r), 0 F(a)) for r > 0. Consequently, L−1 0 K0 : 0 E(a) → 0 E(a) is well-defined. Step 3 In this last step, we prove that  := L−1 0 K0 is a contraction mapping in B0 E(a) (r0 ) for some positive number r0 . By Lemma 1.13.4, we may choose r0 > 0 small enough such that sup z∈BE(a) (2r0 )

DN (z)L(E(a),F(a)) ≤

1 4L−1 0 L(0 F(a),0 E(a))

In addition, we choose ε0 sufficiently small and satisfying  r r0 0 0 < ε0 < min ( , −1 C0 2C1 C0 L0 L(

 . 0 F(a),0 E(a))

For z ∈ B0 E(a) (r0 ), the mean value theorem and (1.13.13) imply that

.

110

M. Hieber

(z)0 E(a) ≤ L−1 0 L(0 F(a),0 E(a)) K0 (z)0 F(a)   ∗ ∗ ≤ L−1 0 L(0 F(a),0 E(a)) N (z + z ) − N (0)F(a) + Lz F(a)  ≤ L−1 sup DN (¯z)L(E(a),F(a))z + z∗ E(a) 0 L(0 F(a),0 E(a)) z¯ ∈BE(a) (2r0 )

 r0 r0 + = r0 , + C1 C0 ε0 ≤ 2 2 which yields that  is a mapping from B0 E(a) (r0 ) into itself. Given z1 , z2 ∈ B0 E(a) (r0 ) and noting that  −1  N (z1 + z∗ ) − N (z2 + z∗ ) , (z1 ) − (z2 ) = L−1 0 (K0 (z1 ) − K0 (z2 )) = L0 we obtain by the mean value theorem ∗ ∗ (z1 ) − (z2 )0 E(a) ≤ L−1 0 L(0 F(a),0 E(a)) N (z1 + z ) − N (z2 + z )0 F(a)   sup ≤ L−1 DN (¯z)L(E(a),F(a)) z1 − z2 0 E(a) 0 L(0 F(a),0 E(a)) z¯ ∈BE(a) (2r0 )

≤

1 z1 − z2 0 E(a) . 4

This implies that  is a contraction on B0 E(a) (r0 ) and we thus obtain the existence of a unique solution z ∈ B0 E(a)(r0 ) of (1.13.14). The proof of Proposition 1.13.5 is complete. 

Proof of Theorem 1.13.1 Observe that the compatibility conditions stated in of Theorem 1.13.1 are satisfied if and only if (1.13.10) is satisfied. The mapping "h0 given by "h0 (ξ  , ξn ) := (ξ  , ξn + h0 (ξ  ))

˙n for (ξ  , ξn ) ∈ R

3−2/p ˙ n onto 0 with defines for h0 ∈ Wp (Rn−1 ) a C 2 -diffeomorphism from R    inverse "−1 h0 (x , xn ) = (x , xn − h0 (x )). Thus there exists a constant C(h0 ) such that

C(h0 )−1 v0 W 2−2/p ( ) ≤ u0 W 2−2/p (R˙ n ) ≤ C(h0 )v0 W 2−2/p ( ) . p

0

p

p

0

Hence, the smallness condition in Theorem 1.13.1 implies the smallness condition in Proposition 1.13.5 and the latter then yields a unique solution (u, π, [[π]], h) ∈ BE(a) (δ0 ) of (1.13.7). Finally, setting (v, q) = ("∗ u, "∗ π) = (u ◦ "−1 , π ◦ "−1 ),

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

111

where "∗ is defined as in (1.13.6), we obtain a unique solution (v, q, h) of the original problem (1.13.2) with the regularities stated in Theorem 1.13.1. The proof is complete. 

1.14 Nematic Liquid Crystals The Ericksen-Leslie model describing the hydrodynamic flow of nematic liquid crystals was derived by Ericksen [49] from the classical balance laws for mass, linear and angular momentum and by considering it as an anisotropic fluid. The system was completed by Leslie [97] with the addition of constitutive relations. For a derivation of this model based on thermodynamical principles we refer to [77]. The non-isothermal Ericksen-Leslie model for incompressible fluids subject to general Leslie stress SL and isotropic free energy ψ(θ, τ ) developed and described in [77] reads as ⎧ ⎪ ρDt u + ∇π = div S in , ⎪ ⎪ ⎪ ⎪ div u = 0 in , ⎪ ⎪ ⎨ ρDt + div q = S : ∇u + div(λ∇dDt d) in , ⎪ γ Dt d − μV V d − div[λ∇]d = λ|∇d|2 d + μD Pd Dd in , ⎪ ⎪ ⎪ ⎪ on ∂, u = 0, q · ν = 0, ∂ν d = 0 ⎪ ⎪ ⎩ in . ρ(0) = ρ0 , u(0) = u0 , θ (0) = θ0 , d(0) = d0 (1.14.1) The unknown variables u, π, θ, d denote velocity, pressure, (absolute) temperature, and director, respectively, Dt = ∂t + u · ∇ the Lagrangian derivative, and Pd := I − d ⊗ d. We also impose the condition |d| = 1

in (0, T ) × .

(1.14.2)

Here incompressible means that the density ρ is constant and isotropic means that the free energy ψ is a function only of θ and τ = |∇d|22 /2. These equations have to be supplemented by the thermodynamical laws for the internal energy , entropy η, heat capacity κ and Ericksen tension λ, according to = ψ + θ η,

η = −∂θ ψ,

κ = ∂θ ,

λ = ρ∂τ ψ,

(1.14.3)

and by the constitutive laws ⎧ ⎪ ⎪ ⎪ ⎪ ⎨

S = SN + SE + SLstretch + SLdiss , SN = 2μs D,

⎪ SLstretch = ⎪ ⎪ ⎪ ⎩ SLdiss =

D = (∇u + [∇u]T )/2,

q = −α∇θ,

SE = −λ∇d[∇d]T

μD +μV −μV n ⊗ d + μD2γ d ⊗ n, 2γ γ μL +μ2P μP (n ⊗ d + d ⊗ n) + γ 2γ

n = μV V d + μD Pd Dd − γ Dt d, (Pd Dd ⊗ d + d ⊗ Pd Dd) + μ0 (Dd|d)d ⊗ d.

(1.14.4)

112

M. Hieber

All coefficients μj , αj and γ are functions of θ, τ , in accordance with the principle of equi-presence. For thermodynamic consistency and well-posedness we require μs > 0,

α > 0,

μ0 , μL ≥ 0,

κ, γ > 0,

λ, λ + 2τ ∂τ λ > 0.

(1.14.5)

It is convenient to write the equation for the internal energy as an equation for the temperature θ . It reads as ρκDt θ +div q = (S −(1−θ ∂θ λ/λ)SE ) : ∇u+div(λ∇)d ·Dt d +(θ ∂θ λ)∇d : ∇Dt d. The Simplified System Due to the complexity of the system (1.14.1), (1.14.3)–(1.14.5) we analyze first a simplified version, which reads as ⎧ ⎪ ⎪ ∂t u + (u · ∇)u − νu + ∇π = ⎪ ⎪ ⎪ ∂t d + (u · ∇)d = ⎨ div u = ⎪ ⎪ ⎪ (u, ∂ν d) = ⎪ ⎪ ⎩ (u, d)|t =0 =

−λdiv([∇d]T∇d) γ (d + |∇d|2 d) 0 (0, 0) (u0 , d0 )

in (0, T ) × , in (0, T ) × , in (0, T ) × , on (0, T ) × ∂, in . (1.14.6)

As above, the function u : (0, ∞) ×  → Rn describes the velocity field, π : (0, ∞)× → R the pressure, and d : (0, ∞)× → Rn represents the macroscopic molecular orientation of the liquid crystal. We impose again the condition |d| = 1

in (0, T ) × .

(1.14.7)

We will show in the following that this condition is indeed preserved by the above system. In the simplified model the parameters ν > 0, λ > 0 and γ > 0 are constants and for simplicity we set ν = λ = γ = 1. Our main idea is to consider (1.14.6) not as a semilinear equation as done in other approaches but as a quasilinear evolution equation of the form z (t) + A(z(t))z(t) = F (z(t)),

t ∈ J,

z(0) = z0 .

(1.14.8)

We thus incorporate the term div([∇d]T∇d) into the quasilinear operator A given by #

$ A PB(d) A(d) = , 0 D where A denotes the Stokes operator, D the Neumann Laplacian, and B is given by [B(d)h]i := ∂i dl hl + ∂k dl ∂k ∂i hl ,

(1.14.9)

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for which we employ the sum convention. Note that B(d)d = div([∇d]T ∇d). We now reformulate (1.14.6) equivalently as a quasilinear parabolic evolution equation for the unknown z = (u, d). To this end, for 1 < q < ∞ define the Banach space X0 by X0 := Lq,σ () × Lq ()n , where  ⊂ Rn is a bounded domain with boundary ∂ ∈ C 2 . The NeumannLaplacian Dq in Lq () is defined by Dq = − with domain D(Dq ) := {d ∈ Hq2 ()n : ∂ν d = 0 on ∂}. It follows from Proposition 1.4.12 that Dq has the property of Lp -maximal regularity. Denoting by P : Lq ()n → Lq,σ () the Helmholtz projection defined in Sect. 1.10, we consider the Stokes Operator Aq = −P in Lq,σ () with domain D(Aq ) = {u ∈ Hq2 ()n : div u = 0 in , u = 0 on ∂}. By Theorem 1.10.6, Aq has the property of Lp -maximal regularity. Next, we define the space X1 by X1 := D(Aq ) × D(Dq ), equipped with d

its canonical norms. Then X1 → X0 densely. The quasilinear part A(z) of (1.14.8) is given by the tri-diagonal matrix # $ Aq PBq (d) A(z) = , 0 Dq where the operator Bq is given as in (1.14.9). By the tri-diagonal structure of A(z) and by the regularity of Bq we see that A(z) also has the property of Lp maximal regularity, for each z ∈ C 1 ()2n . Note that the conditions (A) and (F) of Theorem 1.7.2 hold, as soon as we have the embedding Xγ ,μ → C 1 ()2n . The space Xγ is given by Xγ = (X0 , X1 )1−1/p,p = DAq (1 − 1/p, p) × DDq (1 − 1/p, p); 1 (J ; X ) ∩ L see [8, 38]. The trace space of the class z ∈ Hp,μ 0 p,μ (J ; X1 ) reads as

Xγ ,μ = (X0 , X1 )μ−1/p,p = DAq (μ − 1/p, p) × DDq (μ − 1/p, p), provided p ∈ (1, ∞) and μ ∈ (1/p, 1]. In order to obtain the embeddings Xγ → C 1 ()2n and more generally Xγ ,μ → 1 C ()2n we impose on p, q ∈ (1, ∞) the conditions n 2 + < 1, p q

1 1 n + + < μ ≤ 1. 2 p 2q

(1.14.10)

Standard Sobolev embedding theorems can then be applied. Furthermore, we recall from [148, Theorem 4.3.3] and [10, Theorem 3.4], respectively, the following characterizations of the interpolation spaces involved, d ∈ DDq (μ − 1/p, p)

⇔

2μ−2/p

d ∈ Bqp

()n , ∂ν d = 0 on ∂,

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and u ∈ DAq (μ − 1/p, p)

⇔

2μ−2/p

u ∈ Bqp

()n ∩ Lq,σ (), u = 0 on ∂.

Observe that both of these characterizations make sense, since the condition (1.14.10) guarantees the existence of the trace. Applying Theorem 1.7.2 we obtain the following result on local well-posedness of (1.14.6). Theorem 1.14.6 Let p, q, μ be subject to (1.14.10), and assume z0 = (u0 , d0 ) ∈ 2μ−2/p Xγ ,μ , which means that u0 , d0 ∈ Bqp ()n satisfy the compatibility conditions div u0 = 0 in ,

u0 = ∂ν d0 = 0 on ∂.

Then for some a = a(z0 ) > 0, there is a unique solution 1 z ∈ Hp,μ (J, X0 ) ∩ Lp,μ (J ; X1 ),

J = [0, a],

of (1.14.6) on J . Moreover, z ∈ C([0, a]; Xγ ,μ) ∩ C((0, a]; Xγ ), i.e. the solution regularizes instantly in time. It depends continuously on z0 and exists on a maximal time interval J (z0 ) = [0, t + (z0 )). Therefore problem (1.14.6), i.e. (1.14.8), generates a local semi-flow in its natural state space Xγ ,μ . The following lemma tells that the condition (1.14.7) is preserved by (1.14.6). Lemma 1.14.7 Suppose that μ, p, q are satisfying (1.14.10) and let z0 = 1 (J ; X ) ∩ L (u0 , d0 ) ∈ Xγ ,μ with |d0 | ≡ 1, a > 0. Let z ∈ Hp,μ 0 p,μ (J ; X1 ) be a solution of (1.14.6) on the interval J = [0, a]. Then |d(t)| ≡ 1 holds for all t ∈ [0, a]. We consider next the linearization of (1.14.6) at z∗ ∈ E0 , which is given by the linear evolution equation z˙ + A∗ z = f,

z(0) = z0 ,

in X0 , where A∗ = diag(Aq , Dq ),

D(A∗ ) = X1 .

As  is bounded, the spectrum σ (Aq ) consists only of positive eigenvalues and 0 ∈ σ (Aq ). On the other hand, Dq has 0 as an eigenvalue, which is semi-simple and the remaining part of σ (Dq ) consist only of positive eigenvalues. Thus σ (A∗ )\{0} ⊂ [δ, ∞) for some δ > 0 and the kernel of A∗ is given by N(A∗ ) = {0} × Rn ,

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which equals the tangent space. Verifying that the equilibrium is normally stable, we obtain by Theorem 1.7.4 the following result on global existence for data close to an equilibrium point. Theorem 1.14.8 Let p, q satisfy the first inequality in (1.14.10). Then each equilibrium z∗ ∈ {0} × Rn is stable in Xγ , i.e. there exists > 0 such that a solution z(t) of (1.14.6) with initial value z0 ∈ Xγ , |z0 − z∗ |Xγ ≤ , exists globally and converges exponentially to some z∞ ∈ {0} × Rn in Xγ , as t → ∞. The Ericksen-Leslie System with General Leslie Stress The Ericksen-Leslie equations with general Lesie stress lead to a mixed order system. For this reason, in order to formulate our main well-posedness result in this situation, we introduce a functional analytic setting as follows. Denote the principal variable by v = (u, θ, d). Then v belongs to the base space X0 defined by X0 := Lq,σ () × Lq (; R) × Hq1 (; Rn ), where 1 < p, q < ∞ and σ indicates solenoidal vector fields. Following Theorem 1.10.6, Proposition 1.4.12 and Remark 1.4.13, the regularity space will be X1 := {u ∈ Hq2 (; Rn ) ∩ Lq,σ () : u = 0 on ∂} × Y1 , with Y1 := {(θ, d) ∈ Hq2 () × Hq3 (; Rn ) : ∂ν θ = ∂ν d = 0 on ∂}. We consider solutions within the class 1 (J ; X0) ∩ Lp,μ (J ; X1 ), v ∈ Hp,μ

where J = (0, a) with 0 < a ≤ ∞ is an interval and μ ∈ (1/p, 1] indicates a time weight, as before. The time trace space of this class is given by 2(μ−1/p)

Xγ ,μ = {u ∈ Bqp

()n ∩ Lq,σ () : u = 0 on ∂} × Yγ ,μ ,

where 2(μ−1/p)

Yγ ,μ = {(θ, d) ∈ Bqp

1+2(μ−1/p)

() × Bqp

(; Rn ) : ∂ν θ = ∂ν d = 0 on ∂},

whenever the boundary traces exist. It satisfies 2(μ−1/p)

Xγ ,μ → Bqp

1+2(μ−1/p)

()n+1 × Bqp

()n → C()n+1 × C 1 ()n ,

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provided n 1 + < μ ≤ 1. p 2q

(1.14.11)

For brevity we set Xγ := Xγ ,1 , as before. The state manifold of the problem is defined by SM = {v ∈ Xγ : θ (x) > 0, |d(x)|2 = 1 in }. The fundamental well-posedness result regarding the general Ericksen-Leslie system reads then as follows. Theorem 1.14.9 Let J = (0, a), 1 < p, q < ∞, 1 ≥ μ > 1/2 + 1/p + n/2q, and assume that ψ ∈ C 4 ((0, ∞) × [0, ∞)) as well as α, μj , γ ∈ C 2 ((0, ∞) × [0, ∞)), j = S, V , D, P , L, 0, and the positivity conditions (1.14.5). Then the following assertions are valid: (i) (Local Well-Posedness) Let v0 ∈ Xγ ,μ . Then for some a = a(v0 ) > 0, there is a unique solution 1 v ∈ Hp,μ (J ; X0 ) ∩ Lp,μ (J ; X1 )

of (1.14.1)–(1.14.4) on J . Moreover, v ∈ C([0, a]; Xγ ,μ) ∩ C((0, a]; Xγ ), i.e., the solution regularizes instantly in time. It depends continuously on v0 and exists on a maximal time interval J (v0 ) = [0, t + (v0 )). Moreover, 1 t∂t v ∈ Hp,μ (J ; X0 ) ∩ Lp,μ (J ; X1 ),

a < t + (v0 ),

and E(t) ≡ E0 and −N is a strict Lyapunov functional. Furthermore, the problem (1.14.1), (1.14.3), (1.14.4) generates a local semi-flow in its natural state manifold SM. (ii) (Stability of Equilibria) Any equilibrium v∗ ∈ E of (1.14.1)–(1.14.4) is stable in Xγ . Moreover, for each v∗ ∈ E there is ε > 0 such that if v0 ∈ SM with |v0 − v∗ |Xγ ,μ ≤ ε, then the solution v of (1.14.1)–(1.14.4) with initial value v0 exists globally in time and converges at an exponential rate in Xγ to some v∞ ∈ E. (iii) (Long-Time Behaviour) (a) Suppose that sup

t ∈(0,t +(v0 ))



v(t)Xγ ,μ + 1/θ (t)L∞ < ∞.

Then t + (v0 ) = ∞ and v is a global solution.

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(b) If v is a global solution, bounded in Xγ ,μ and with 1/θ bounded, then v converges exponentially in SM to an equilibrium v∞ ∈ E of (1.14.1)– (1.14.4), as t → ∞. It is remarkable that the above theorem holds true without any structural assumptions on the Leslie coefficients, except for Condition 1.14.5. In particular, the above well-posedness results hold true without assuming Parodi’s relation and no conditions for μV , μD , μP are needed. Remarks 1.14.10 (a) A related class of models also dealing with the non-isothermal situation was presented by Feireisl et al. [53] as well as by Feireisl, Frémond, Rocca and Schimperna in [54]. Their model includes stretching as well as rotational terms and is consistent with the fundamental laws of Thermodynamics. The equation for the director d, however, is given in the penalized form. They show that the presence of the term |∇d|22 in the internal energy as well as the stretching term d · ∇u give rise, in order to respect the laws of Thermodynamics, to two new non dissipative contributions in the stress tensor S and in the flux q. (b) Wu, Xu and Liu reconsidered in [155] the isothermal penalized Ericksen-Leslie model. Their main result says that under certain assumptions on the data and the Leslie coefficients, the penalized Ericksen-Leslie system admits a unique, global solution provided the viscosity is large enough. Wang et al. [153] proved local well-posedness of the isothermal general Ericksen-Leslie system as well as global well-posedness for small initial data under various conditions on the Leslie coefficients, which ensure that the energy of the system is dissipated. (c) Let us mention here also the well-posedness results concerning the nonisothermal situation obtained in [78] and the very recent work of De Anna and Liu [33], in which the non-isothermal compressible Ericksen-Leslie system is investigated, too. In contrast to the approach in [33], the approach described here does not impose any structural assumptions on the Leslie-coefficients. Key ideas of the proof of Theorem 1.14.9 We recall that the parameter functions are having the regularity properties μj , α, γ ∈ C 2 ((0, ∞) × [0, ∞)),

ψ ∈ C 4 ((0, ∞) × [0, ∞)).

(1.14.12)

and that furthermore the positivity conditions (1.14.5) are assumed to be fulfilled. Step 1: Linearization We linearize Eq. (1.14.1) at an initial value v0 = [u0 , θ0 , d0 ]T and drop all terms of lower order. This yields the principal linearization ⎧ ⎨

Lπ (∂t , ∇)vπ = f in J × , u = ∂ν θ = ∂ν d = 0 on J × ∂, ⎩ u = θ = d = 0 on {0} × .

(1.14.13)

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Here J = (0, a), vπ = [u, π, θ, d]T is the unknown, and f = [fu , fπ , fθ , fd ]T are the given data. The differential operator Lπ (∂t , ∇) is defined via its symbol Lπ (z, iξ ), which is given by ⎡

Mu (z, ξ ) ⎢ iξ T Lπ (z, iξ ) = ⎢ ⎣ 0 −iR0 (ξ )

⎤ iξ 0 izR1 (ξ )T ⎥ 0 0 0 ⎥, 0 mθ (z, ξ ) −izθ0 ba(ξ ) ⎦ 0 −iba(ξ ) Md (z, ξ )

(1.14.14)

with b = ∂θ λ, and λ1 = ∂τ λ. We also introduce the parabolic part of this symbol by dropping pressure gradient and divergence, i.e. ⎡

⎤ Mu (z, ξ ) 0 izR1 (ξ )T L(z, iξ ) = ⎣ 0 mθ (z, ξ ) izθ0 ba(ξ ) ⎦ . −iR0 (ξ ) iba(ξ ) Md (z, ξ )

(1.14.15)

The entries of these matrices are given by mθ = ρκz + α|ξ |2 ,

a(ξ ) = ξ · ∇d0 ,

Md = γ z + λ|ξ |2 + λ1 a(ξ ) ⊗ a(ξ ) = md (z, ξ ) + λ1 a(ξ ) ⊗ a(ξ ), μD + μV μD − μV P0 ξ ⊗ d0 + (ξ |d0 )P0 , 2 2 μD + μV μD − μV + μP )P0 ξ ⊗ d0 + ( + μp )(ξ |d0 )P0 , R1 = ( 2 2 R0 =

Mu = ρz + μs |ξ |2 + μ0 (ξ |d0 )2 d0 ⊗ d0 + a1 (ξ |d0 )P0 ξ ⊗ d0 , + a2 (ξ |d0 )2 P0 + a3 |P0 ξ |2 d0 ⊗ d0 + a4 (ξ |d0 )d0 ⊗ P0 ξ. Here P0 = Pd0 = I − d0 ⊗ d0 , and aj are certain coefficients. Step 2: Maximal Lp -Regularity Let 1 < p, q < ∞ and assume that (1.14.12) holds. Then (1.14.13) admits a unique solution vπ = [u, π, θ, d]T satisfying (u, θ ) ∈ 0 H 1p (J ; Lq ())n+1 ∩ Lp (J ; Hq2())n+1 , π ∈ Lp (J ; H˙ q1()), d ∈ 0 H 1p (J ; Hq1())n ∩ Lp (J ; Hq3())n , if and only if (fu , fθ ) ∈ Lp (J ; Lq ())n+1 ,

fd ∈ Lp (J ; Hq1 ())n ,

1 fπ ∈ 0 H 1p (J ; 0 H −1 q ()) ∩ Lp (J ; Hq ()).

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In order to prove this, we set J := diag(I, 1/θ0 , zI ) and show first that the symbol J¯L is accretive for Re z > 0, i.e. the associated system is strongly elliptic. Observe that for proving this we do not need any conditions for the coefficients μD , μV , μP , ∂θ λ. Next, we perform a Schur reduction to reduce the above symbol to a symbol depending only for u. This implies that the resulting generalized Stokes symbol for (u, π) is strongly elliptic. Then we may apply a result due to Bothe and Prüss [18] to prove maximal Lp regularity in the case of Rn . In a second step we verify the Lopatinskii-Shapiroo condition to obtain the corresponding result on the half space. A localization procedure finishes the proof of the above assertion. Step 3: Local Existence We rewrite Problem (1.14.1) as an abstract quasi-linear evolution equation of the form v˙ + A(v)v = F (v),

t > 0, v(0) = v0 ,

(1.14.16)

replacing Dt d appearing in the equations for u and θ by the equation for d. Here v = (u, θ, d) and the Helmholtz projection P is applied to the equation for u. The base space will be X0 := Lq,σ () × Lq () × Hq1 (; Rn ). Then, by Theorem 1.7.2 for some a = a(z0 ) > 0 there is a unique solution 1 v ∈ Hp,μ (J, X0 ) ∩ Lp,μ (J ; X1 ),

J = [0, a],

of (1.14.16), i.e., (1.14.1) on J . Moreover, t[

d 1 ]v ∈ Hp,μ (J ; X0 ) ∩ Lp,μ (J ; X1 ), dt

and it can be shown that |d(t, x)|2 ≡ 1, E(t) ≡ E0 , and −N is a strict Lyapunov functional; see [76, 77] for details. Furthermore, the problem (1.14.16) generates a local semi-flow in its natural state manifold SM. Step 4: Dynamics The linearization of (1.14.1) at an equilibrium v∗ = (0, θ∗ , d∗ ) is given by the operator A∗ = A(v∗ ) in X0 . This operator has maximal Lp -regularity, it is the negative generator of a compact analytic C0 -semigroup, and it has compact resolvent. So its spectrum consists only of countably many eigenvalues of finite multiplicity, which have all positive real parts, hence are stable, except for 0. The eigenvalue 0 is semi-simple. Its eigenspace is given by N(A∗ ) = {(0, ϑ, d) : ϑ ∈ R, d ∈ Rn }, ¯ when ignoring the constraint hence it coincides with the set of constant equilibria E, |d|2 = 1 and conservation of energy. Therefore each such equilibrium is normally

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stable and Theorem 1.7.4 implies assertion (ii). Finally, assertion (iii) follows from Theorem 1.7.8. 

1.15 Global Strong Well-Posedness of the Primitive Equations The primitive equations for the ocean and atmosphere are considered to be a fundamental model for geophysical flows which is derived from Navier–Stokes equations assuming a hydrostatic balance for the pressure term in the vertical direction. The mathematical analysis of the primitive equations was pioneered by Lions, Teman and Wang in a series of articles [103, 104]; for a survey of known results and further references, we refer to an article by Li and Titi [99]. In contrast to Navier–Stokes equations, the 3D primitive equations admit a unique, global, strong solution for arbitrary large data in H 1 . This breakthrough result was proved by Cao and Titi [25] in 2007 using energy methods. A different approach to the primitive equations, based on methods of evolution equations, has been presented in [73]. There a Fujita-Kato type iteration scheme was developed in addition to H 2 -a priori bounds for the solution. In this section we present an alternative approach to the primitive equations based on techniques from maximal Lq -regularity. This approach has several advantages compared to the two other approaches. In fact, the regularizing effect of the solution plays an important role when extending local solutions to global ones by means of certain a priori bounds. We mention here the H 2 -a priori bounds used in [73, 83]. In the following, we show that a priori bounds in the maximal regularity space L2 (0, T ; H 2 ) ∩ H 1 (0, T ; L2 ) are already sufficient to prove the global existence of a solution within the Lq –Lp -class. Secondly, our approach allows to prove the existence and uniqueness of a global, strong solution for initial values lying in critical spaces, which in the given situation μ are the Besov spaces Bpq for p, q ∈ (1, ∞) and with 1/p + 1/q ≤ μ ≤ 1. Here, we use in an essential way the concept of time weights for maximal Lp -regularity; see Sect. 1.7 and [124] as well as [125] for details and proofs. Choosing in particular 1 = H 1 , we rediscover in particular the p = q = 2 and μ = 1 and noting that B22 celebrated result by Cao and Titi [25]. The precise formulation of the primitive equations reads as follows: Consider a cylindrical domain  = G × (−h, 0) ⊂ R3 with G = (0, 1) × (0, 1), h > 0 and denote by v :  → R2 the vertical velocity of the fluid and by πs : G → R its surface pressure. The primitive equations are then given by ⎧ ⎨ ∂t v + v · ∇H v + w(v) · ∂z v − v + ∇H πs = f, in  × (0, T ), divH v = 0, in  × (0, T ), ⎩ v(0) = v0 , in ,

(1.15.1)

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121

where x, y ∈ G denote the horizontal coordinates and z ∈ (−h, 0) the vertical one. We use the notations  = ∂x2 + ∂y2 + ∂z2 , divH v = ∂x v1 + ∂y v2

∇H = (∂x , ∂y )T ,  1 0 and v := v(·, ·, ξ )dξ. h −h

Here the horizontal velocity w = w(v) is given by  w(v)(x, y, z) = −

z −h

divH v(x, y, ξ )dξ,

where w(x, y, −h) = w(x, y, 0) = 0. The Eq. (1.15.1) are supplemented by mixed boundary conditions on u = G × {0},

b = G × {−h}

and l = ∂G × (−h, 0),

i.e., the upper, bottom and lateral parts of the boundary ∂, respectively, defined by v, πs are periodic on l × (0, ∞), v = 0 on D × (0, ∞)

and ∂z v = 0 on N × (0, ∞).

(1.15.2)

The Dirichlet, Neumann and mixed boundary conditions are comprised by the notation D ∈ {∅, u , b , u ∪ b }

and N = (u ∪ b ) \ D .

p

Given p ∈ (1, ∞), the space Lσ () of hydrostatically solenoidal vector fields is given as in Sect. 1.10 as the subspace of Lp ()2 defined by p

∞ ()2 : div v = 0} Lσ () := {v ∈ Cper H

·Lp ()2

.

(1.15.3)

∞ () and C ∞ (G) Here horizontal periodicity is modeled by the function spaces Cper per consists of all smooth functions, which are periodic with respect to x, y coordinates but not necessarily in the z coordinate. Furthermore, following Sect. 1.10 there exists a continuous projection Pp , called the hydrostatic Helmholtz projection, from p Lp ()2 onto Lσ (). In particular, Pp annihilates the pressure term ∇H πs .

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The Hydrostatic Stokes Operator For p ∈ (1, ∞) and s ∈ [0, ∞) define the spaces s,p

∞ () Hper () := Cper

·H s,p ()

s,p

∞ (G) and Hper (G) := Cper

·H s,p (G)

,

0,p

where Hper := Lp . Here H s,p () denotes the Bessel potential spaces, which are defined as restrictions of Bessel potential spaces on the whole space to , compare e.g. [148, Definition 3.2.2.]. For p, q ∈ (1, ∞) and s ∈ [0, ∞) we also define the periodic Besov spaces s ∞ () Bpq,per () := Cper

·B s

pq ()

s ∞ (G) and Bpq,per (G) := Cper

·B s

pq (G)

,

s denotes Besov spaces, which are defined as restrictions of Besov spaces where Bpq s (R3 ), compare e.g. [148, Definitions 3.2.2]. defined on the whole space Bp,q p

Following Sect. 1.10, the hydrostatic Stokes operator Ap in Lσ () is defined as Ap v := Pp v,

  2,p p D(Ap ) := {v ∈ Hper ()2 : ∂z v  = 0, v  = 0} ∩ Lσ (). N

D

By Theorem 1.10.14, Ap ∈ H ∞ (Lσ ()) with ∞ A = 0. In particular, Ap admits the property of maximal Lq -regularity. Following Proposition 1.7.1 this is equivalent to maximal Lq -regularity of Ap in time-weighted spaces. Recall that these spaces were defined in Sect. 1.7 for μ ∈ (1/q, 1] by p

Lqμ (J ; D(Ap )) = {v ∈ L1loc (J ; D(Ap )) : t 1−μ v ∈ Lq (J ; D(Ap ))}, p

p

p

p

Hμ1,q (J ; Lσ ()) = {v ∈ Lqμ (J ; Lσ ()) ∩ H 1,1 (J ; Lσ ()) : t 1−μ vt ∈ Lq (J ; Lσ ())}.

The natural trace spaces of these spaces are determined by real interpolation (·, ·)θ,q for θ ∈ (0, 1) and p, q ∈ (1, ∞). Thanks to Theorem 1.10.14 these spaces can be identified explicitly as described in the following lemma. Lemma 1.15.1 Let θ ∈ (0, 1), p, q Then ⎧ p 2θ ⎪ ⎪ ⎨{v ∈ Bpq,per () ∩ Lσ () : p 2θ Xθ,q = {v ∈ Bpq,per () ∩ Lσ () : ⎪ ⎪ p ⎩ 2θ () ∩ L (), B pq,per

σ

p

∈ (1, ∞) and Xθ,q := (Lσ (), D(Ap ))θ,q .   ∂z v  = 0, v  = 0}, N D  v  = 0}, D

1 1 2 + 2p 1 2p < θ

< θ < 1, 1 1 2 + 2p , 1 2p .


0 (depending only on  and p, s) such that Fp (v, v  )H s,p ()2 ≤ CvH s+1+1/p,p v  H s+1+1/p,p , for all v, v  ∈ H s+1+1/p,p ()2 , i.e., the mapping Fp (·, ·) : H s+1+1/p,p ()2 × H s+1+1/p,p ()2 → H s,p ()2 is a continuous bilinear map. p

Denoting by Xβ = [Lσ (), D(Ap )]β the complex interpolation space of order p β ∈ [0, 1], we verify that [Lσ (), D(Ap )](1+1/p)/2 ⊂ H 1+1/p,p ()2 , which for β = 12 (1 + 1/p) yields the existence a constant C > 0, independent of v, v  , such that   Fp (v) − Fp (v  )Lp () ≤ C vXβ + v  Xβ v − v  Xβ . σ

(1.15.4)

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The local well-posedness of the primitive equations follows now from Proposition 1.7.6 by using Theorem 1.10.14 and the remark after Proposition 1.7.6 for condition (S) and (H 1) and estimate (1.15.4) for the conditions (H 2) and (H 3). More precisely, the following result holds true. Proposition 1.15.4 (Local Well-Posedness) Let p, q ∈ (1, ∞) with 1/p + 1/q ≤ 1, μ ∈ [1/p + 1/q, 1] and T > 0. Assume that v0 ∈ Xμ−1/q,q

and

p

Pp f ∈ Lqμ (0, T ; Lσ ()).

Then there exists T  = T  (v0 ) with T  ∈ (0, T ] and a unique, strong solution v ∈ Hμ1,q (0, T  ; Lσ ()) ∩ Lqμ (0, T  ; D(Ap )) p

to (1.15.1) on (0, T  ). The proof of global well-posedness of the primitive equations is based on the following a priori bound in H 1 (0, T ; L2 ) ∩ L2 (0, T ; H 2 ). Theorem 1.15.5 (A Priori Bounds) There exists a continuous function B satisfying the following property: any solution of (1.15.1) fulfilling for 0 < T < ∞ the conditions v ∈ H 1 (0, T ; L2σ ())) ∩ L2 (0, T ; D(A2 )),  v0 ∈ {H 1 ∩ L2σ () : v  = 0}, P2 f ∈ L2 (0, T ; L2σ ()) D

satisfies   vH 1 (0,T ;L2 ()))∩L2(0,T ;D(A2 )) ≤ B v0 H 1 () , P2 f L2 (0,T ;L2 ()), T . σ

The proof of Theorem 1.15.2 is based on Proposition 1.15.4, the above a priori bound and the following result within the L2 -setting.  Proposition 1.15.6 Let 0 < T < ∞ and v0 ∈ {H 1 ∩ L2σ () : v  = 0}. D

L2 (0, T ; L2σ ()),

then there exists a unique, strong solution v to the (a) If P2 f ∈ primitive Eq. (1.15.1) within the class v ∈ H 1 (0, T ; L2σ ())) ∩ L2 (0, T ; D(A2 )). (b) If in addition t → t · P2 ft (t) ∈ L2 (0, T ; L2σ ()), then t · vt ∈ H 1 (0, T ; L2σ ())) ∩ L2 (0, T ; D(A2 )). Before giving the proof of Proposition 1.15.6 we note the following elementary fact about extending regularity of solutions from (0, T  ) for any 0 < T  < T to (0, T ).

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

Lemma 1.15.7 Let v ∈ E1,μ (0, T  ) for any 0 < T  < T , and < C for some constant C > 0. Then v ∈ E1,μ (0, T ).

125

sup vE1,μ (0,T  )

0 0 and that the solutions in E1,1 (0, T  ) are unique. Assume now that t+ (v0 ) < T . By Theorem 1.15.5, vE1,1 (0,T  ) ≤   B v0 H 1 () , P2 f L2 (0,T ;L2 ()) , t+ (v0 ) for any 0 < T  < t+ (v0 ). Hence, by Lemma 1.15.7 we have v ∈ E1,1 (0, t+ (v0 )). Since the trace in E1,1 (0, t+ (v0 )) is well-defined, v(t+ (v0 )) can be taken as new initial value, thus extending the solution beyond t+ (v0 ) and contradicting the assumption. Hence, t+ (v0 ) = T and combing again Theorem 1.15.5 with Lemma 1.15.7 yields v ∈ E1,1 (0, T ). This proves part (a). Assertion (b) is proved by Angenent’s method. For details see [65]. 

Proof of Theorem 1.15.2 Using Angenent’s method we verify that the local solution v obtained in Proposition 1.15.4 enjoys additional time regularity. In particular, we deduce that v ∈ H 1,q (δ, T ; D(Ap )) → C 0 (δ, T ; D(Ap )) for some 0 < δ ≤ T  and 0 < T  < T . Now, using v(T  ) as new initial value, and taking advantage of the embedding D(Aq ) ⊂ (L2σ (), D(A2 ))1/2,q for q ∈ [6/5, ∞) and the additional assumption P2 f ∈ W 1,2 (δ, T ; L2σ ()), we see that v is also an L2 solution, at least for δ > 0. This holds for q ∈ [6/5, ∞), and for q ∈ (1, 6/5) by a bootstrapping argument; see [83]. By Proposition 1.15.6 there exists now a global L2 solution v ∈ Cb (δ, D(A2 )). Lemma 1.15.1 and classical embedding results yield D(A2 ) → Xμ,q

for 0 ≤ μ − μ < 2 − p2 ,

and the compactness of the embedding Xμ,q → Xμ,q for 1/p < μ < μ < 1. Since vCb (δ,T ;Xμ,q ) ≤ CvCb (δ,T ;D(A2 )) , Proposition 1.7.8 applies and the solution v exists hence globally, that is for any T > 0. 

1.16 Justification of the Hydrostatic Approximation for the Primitive Equations by Scaled Navier–Stokes Equations In this section we show that the primitive equations can be obtained as the limit of anisotropically scaled Navier–Stokes equations. The scaling parameter ε > 0

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represents the ratio of the depth to the horizontal width. Such an approximation is motivated by the fact that for large-scale oceanic dynamics, this aspect ratio ε is rather small and implies anisotropic viscosity coefficients. For an aspect ratio ε, i.e., in the case where the spacial domain can be represented as ε = G × (−ε, +ε) for some G ⊂ R2 , and a horizontal and vertical eddy viscosity 1 and ε2 , respectively, the system can be rescaled into the form ⎧ ⎨

∂t vε + uε · ∇vε − vε + ∇H pε = 0, ε(∂t wε + uε · ∇wε − wε ) + 1ε ∂z pε = 0, ⎩ div uε = 0,

(1.16.1)

in the time-space domain (0, T ) × 1 , which is independent of the aspect ratio. We refer to [98] for more details on this rescaling procedure. Here the horizontal and vertical velocities vε and wε describe the three-dimensional velocity uε , while pε denotes the pressure of the fluid. In the following, we show convergence results of the above system in the strong sense within the Lp –Lq -setting hereby extending a previous result due to Li and Titi [98] to a more general setting. Our method is very different from the one introduced by [98]. Whereas they rely on second order energy estimates, our approach is based on maximal Lp –Lq -regularity estimates for the heat or Stokes equation. Consider the cylindrical domain  := (0, 1)2 × (−1, 1). Let u = (v, w) be the solution of the primitive equations ⎧ ⎪ ∂t v + u · ∇v − v + ∇H p = ⎪ ⎪ ⎪ ⎪ ∂z p = ⎪ ⎪ ⎨ div u = ⎪ p periodic in x, y ⎪ ⎪ ⎪ ⎪ v, w periodic in x, y, z, ⎪ ⎪ ⎩ u(0) =

0 0 0

in (0, T ) × , in (0, T ) × , in (0, T ) × ,

(PE)

even and odd in z, in , u0

and uε = (vε , wε ) be the solution of the anisotropic Navier–Stokes equations ⎧ ∂t vε + uε · ∇vε − vε + ∇H pε = ⎪ ⎪ ⎪ ⎪ ⎪ ∂t wε + uε · ∇wε − wε + ε12 ∂z pε = ⎪ ⎪ ⎨ div uε = ⎪ pε periodic in x, y, z, ⎪ ⎪ ⎪ ⎪ , w v ⎪ ε ε periodic in x, y, z, ⎪ ⎩ uε (0) =

0 0 0 even even and odd u0

in (0, T ) × , in (0, T ) × , in (0, T ) × , in z, in z, in .

(NSε )

Here v and vε denote the (two-dimensional) horizontal velocities, w and wε the vertical velocities, and p and pε denote the pressure term for the primitive equations as well as the Navier–Stokes equations, respectively. These are functions of three space variables x, y ∈ (0, 1), z ∈ (−1, 1). Since w is odd, the divergence free condition for the primitive equation translates into divH v = 0, where

1 Analysis of Viscous Fluid Flows: An Approach by Evolution Equations

v(x, y) =

 1 1 2 −1 v(x, y, z) dz,

127

and 

w(·, ·, z) = −

z −1

divH v(·, ·, ζ ) dζ.

The Setting s,p For p, q ∈ (1, ∞) and s ∈ [0, ∞) let the Bessel potential spaces Hper () = ·H s,p

·B s

p,q s ∞ () ∞ () Cper and the Besov spaces Bp,q,per () = Cper be defined s,p as in Sect. 1.15. The space H () denotes the Bessel potential space of order s, with norm  · H s,p defined via the restriction of the corresponding space defined on the whole space to , see e.g. [148, Definition 3.2.2.]. Similarly, s () denotes a Besov space on , which is defined again by restrictions of Bp,q functions on the whole space to , see again [148, Definition 3.2.2.]. Note that 0,p s s Lp () = Hper () and Bp,2,per () = Hp,per (). The anisotropic structure of the s,p primitive equations motivates the definition of the Bessel potential spaces Hxy := s,p s,p 2 s,p H ((0, 1) ) and Hz := H (−1, 1) for the horizontal and vertical variables, p 0,p p 0,p respectively. Similarly as above we write Lxy := Hxy and Lz := Hz and set s,p r,q r,q Hxy Hz := H s,p ((0, 1)2 ; Hz ). The divergence free conditions in the above sets of equations can be encoded into the space of solenoidal functions

·Lp

∞ ()3 : div u = 0} Lpσ () = {u ∈ Cper

·Lp

p

∞ ()2 : div v = 0} Lσ () = {v ∈ Cper H

and .

For given p, q ∈ (1, ∞) we set X0 := Lq (),

2,q

X1 := Hper (),

q

X0v := {v ∈ Lσ () : v even in z}, 2,q

q

X1v := {v ∈ Hper ()2 ∩ Lσ () : v even in z}, X0u := {(v1 , v2 , w) ∈ Lqσ () : v1 , v2 even w odd in z}, 2,q

X1u := {(v1 , v2 , w) ∈ Hper ()3 ∩ Lqσ () : v1 , v2 even w odd in z}, and consider the trace space Xγ defined by Xγ = (X0u , X1u )1−1/p,p . Given p, q ∈ (1, ∞) and following [68], the trace space Xγ can be characterized as ⎧ 2−2/p q 3 ⎪ ⎪ {(v1 , v2 , w) ∈ Bp,q,per () ∩ Lσ () : v = (v1 , v2 ) even, w odd in z, ⎪ ⎪ ⎨ (∂z v, w) = 0 at z = −1, 0, 1}, 1 > Xγ = 2−2/p q 3 ⎪ {(v1 , v2 , w) ∈ Bp,q,per () ∩ Lσ () : v = (v1 , v2 ) even, w odd in z, ⎪ ⎪ ⎪ ⎩ w = 0 at z = −1, 0, 1}, 1
0 and (v, w) and (vε , wε ) be solutions of (PE) and (NSε ), respectively. Then there exists a

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129

constant C > 0, independent of ε, such that (Vε , εWε )E1 (T ) ≤ Cε for sufficiently small ε. In particular, (vε , εwε ) → (v, 0) in Lp (0, T ; H 2,q ()) ∩ H 1,p (0, T ; Lq ()) as ε → 0 with convergence rate O(ε). Remarks 1.16.2 (a) If the solution u = (v, w) of the primitive equations exists globally in time, the convergence rate is uniform for all T ∈ (0, ∞], see Remark 4.10.b) in [55]. For example, if p = q = 2 and the initial data are mean value free, one can show that the solution to the primitive equations exists globally in E1 (T ) with T = ∞. (b) Let us note that the above result covers in particular the case p = q = 2, which was investigated before in [98]. (c) The scaled Navier–Stokes equations are locally well-posed in the maximal regularity space of the torus and the parity conditions are preserved. The above Theorem 1.16.1 yields that for each T > 0 there exists an ε > 0 such that the solution exists on (0, T ). (d) The primitive equations are well-posed for all times T > 0 provided u0 ∈ Xγ , see Sect. 1.15. We start the proof of Theorem 1.16.1 by considering the linearization of (1.16.2). More specifically, given F ∈ E0 (T ) and U0 ∈ Xγ , we consider the linear problem ⎧ ∂t U − U = ⎪ ⎪ ⎨ ∇ε · U = ⎪ U, P ⎪ ⎩ U (0) =

F − ∇ε P 0 periodic U0

in (0, T ) × , in (0, T ) × , in x, y, z, in ,

(1.16.3)

where ∇ε := (∂x , ∂y , ε−1 ∂z )T . The following proposition gives a maximal regularity estimate for U , where the constants are independent of the aspect ratio and the pressure gradient. Proposition 1.16.3 Let p, q ∈ (1, ∞), T > 0, F ∈ E0 (T ), U0 ∈ Xγ and ε > 0. Then there is a unique solution U, P to the Eq. (1.16.3) with U ∈ E1 (T ) and ∇ε P ∈ E0 (T ), where P is unique up to a constant. Moreover, there exist constants C > 0 and CT > 0, independent of ε, such that U E1 (T ) ≤ CF E0 (T ) + CT U0 Xγ .

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In order to prove Proposition 1.16.3 we define the ε-dependent Helmholtz projection Pε by Pε := Id − ∇ε −1 ε divε ,

where ε = ∇ε · ∇ε .

The following lemma shows that Pε is a bounded projection with uniform norm bound independent of ε. Lemma 1.16.4 Let ε > 0, q ∈ (1, ∞) and assume that F = (fH , fz ) ∈ Lq () 1,q and P ∈ Hper () are satisfying the equation −(H + ε−2 ∂z2 )P = div(fH , ε−1 fz ) for ε > 0. Then there exists a constant C > 0, independent of ε, such that (∇H P , ε−1 ∂z P )q ≤ CF q . Proof For kε = (k1 , k2 , ε−1 k3 )T and m(k) = − k⊗k ∈ R3×3 set mε (k) = m(kε ). |k|2 Then (∇H , ε−1 ∂z )P = F −1 mε F F and k∇mε (k) = kε ∇m(kε ). Hence, sup sup |k γ D γ mε (k)| =

γ ∈{0,1}3 k =0

sup

sup |kεγ D γ m(kε )| = 1,

γ ∈{0,1}3 kε =0

and Markincinkiewicz’s theorem, see Corollary 1.1.9, implies that mε is an Lp Fourier multiplier satisfying F −1 mε F L(Lq ()) ≤ C for some C = C(q) > 0 independent of ε. 

Applying Lemma 1.16.4 to (1.16.3) and taking into account the equalities Pε  = Pε and Pε ∇ε P = 0, Eq. (1.16.3) reduces to the heat equation with right hand side Pε F . Maximal Lp -regularity estimates of the three-dimensional Laplacian in the periodic setting yield U E1 (T ) ≤ CPε F E0 (T ) + CT U0 Xγ ≤ CF E0 (T ) + CT U0 Xγ .



Finally, we state that the solution u = (v, w) of the primitive equations belongs to the maximal regularity class Eu1 (T ). For a detailed proof, see [55], Prop. 4.8. Proposition 1.16.5 Let p, q fulfill Assumption (A) and let v be the strong solution of the primitive equations associated to v0 satisfying (v0 , w0 ) ∈ Xγ . Then u = (v, w) ∈ Eu1 (T ) for all T > 0. Corollary 1.16.6 Let T > 0 and p, q ∈ (1, ∞) such that 1/p + 1/q ≤ 1. Let (Vε , Wε ) ∈ Eu1 (T ) denote the solution of Eq. (1.16.2) for some u = (v, w) ∈ E1 (T ) and initial data U0 ∈ Xγ and set Xε (T ) := (Vε , εWε )E1 (T ) . Then, for any

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131

η ∈ [0, 1 − 1/p − 1/q] there exists a constant C > 0, independent of ε, such that , + Xε (T ) ≤ CT η Xε (T )uE1 (T ) + Xε2 (T ) , + + εC uE1 (T ) + T η u2E1 (T ) + CU0 Xγ , for all T ∈ [0, T ]. Proof of Theorem 1.16.1 Fix T > 0 and let u be the solution of Eq. (PE). Then Proposition 1.16.5 implies u ∈ E1 (T ). We show that Xε (T ) = (Vε , εWε )E1 (T ) ≤ εC((v, w)E1 (·) , T , p, q) for all T ∈ [0, T ] and ε > 0 small enough. To this end, by the uniform continuity of the mapping T → uE1 (T ) on [0, T ], there is a T ∗ ∈ [0, T ] such that p p uE1 (T +T ∗ ) − uE1 (T ) ≤ (2CT η )−p for all T ∈ [0, T − T ∗ ] and where C denotes the constant given in Corollary 1.16.6. The latter with U0 = 0 implies CXε2 (T ) − 12 Xε (T ) + ε ≥ 0,

T ∈ [0, T ∗ ].

(1.16.4)

Since Xε (0) = 0 we may solve the above quadratic inequality for ε < (16C)−1 and obtain Xε ≤ 2ε on [0, T ∗ ]. Observe next that inequality (1.16.4) holds on a time interval independent of ε. More specifically, if one replaces T ∗ by Tε < T ∗ , where Tε is the maximal existence time of (NSε ) with initial data u0 , then similarly as above we obtain Xε ≤ 2ε on [0, Tε ], which yields a contradiction to the maximality of the existence time. Assume now that there exists some m ∈ N such that mT ∗ < T and Xε ≤ ε2Km in [0, mT ∗ ], where K1 = 1 and Km = 21/p [(2CcmT ∗ + 1) Km−1 + 1] and cT denotes the embedding constant of E1 (T ) → L∞ (0, T ; Xγ ). Let (V˜ε , εW˜ ε )(T ) = (Vε , εWε )(T + mT ∗ ) be the unique solution of problem (1.16.2) with respect to u(T ˜ ) = u(T + mT ∗ ) and initial data U0 = (Vε , εWε )(mT ∗ ). Setting p X˜ εp (T ) := (V˜ε , εW˜ ε )E1 (T ) = Xεp (T + mT ∗ ) − Xεp (mT ∗ ),

Corollary 1.16.6 and the argument about the ε-independency of the time interval given above imply C X˜ ε2 (T ) − 12 X˜ ε (T ) + ε + CU0 Xγ ≥ 0,

T ∈ [0, min{T ∗ ; T − mT ∗ }].

By assumption U0 Xγ ≤ cmT ∗ Xε (mT ∗ ) ≤ cmT ∗ ε2Km . Since X˜ ε (0) = 0 and X˜ ε is continuous in [0, min{T ∗ ; T − mT ∗ }], we may solve the quadratic inequality for ε < (16C(1 + 2CcmT ∗ Km ))−1 and obtain X˜ ε ≤ ε2(1 + 2CcmT ∗ Km ) in

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[0, min{T ∗ ; T − mT ∗ }]. Hence, by the assumption on m Xεp (T ) ≤ (ε2(1 + 2CcmT ∗ Km ))p + Xεp (mT ∗ ) ≤ 2 [2ε(1 + 2CcmT ∗ Km ) + ε2Km ]p = (ε2Km+1 )p , for all T ∈ [mT ∗ , min{(m+1)T ∗ ; T }]. The assumption on m implies Xε ≤ ε2Km+1 in [0, min{(m + 1)T ∗ ; T }]. By induction we get Xε ≤ ε2KM in [0, T ] with M = [ TT∗ ]. The proof of Theorem 1.16.1 is complete. 

1.17 Notes Sections 1.1 and 1.2 The content of Sect. 1.1 is very standard and can be found in many books, for example [86] or [141]. Excellent sources for more information on Fourier transforms, Fourier multipliers and the Hilbert transform are [45, 69, 140] and [14]. The corresponding vector-valued setting is also well studied and documented. In fact, basic properties on Bochner integrals and Banach space valued Lp -spaces can be found e.g. in Section 1.1 of [13]. Properties of Banach space valued distributions and Fourier transforms are described in detail e.g. in [8, 12, 95] or [87]. It is remarkable that the fundamental theorem on scalar valued singular integrals based on Calderon-Zygmund theory extends without difficulties to the Banach space valued setting described in Theorem 1.2.3. A very thorough study of UMD-spaces can be found in the monograph by Hytönen et al. [87]. Their significant role in the theory of scalar- or vector-valed Lp Fourier multipliers was recognized by Bourgain in [22] and is described in detail in Chapter I of [95], see also [88]. We only note here that UMD spaces are necessarily reflexive spaces and that the converse is not true. Due to the work of Burkholder [24] it is known that a Banach space X is a UMD-space if and only if X is ζ -convex in the sense that there exists a symmetric, biconvex function ζ : X × X → R satisfying ζ (0, 0) > 0 and ζ (x, y) ≤ x + y for x, y ∈ X with x = y = 1. The abbreviation UMD stands for the property of unconditional martingale differences and it was shown by Bourgain [20] that spaces possessing this property coincide with UMD-spaces as defined in Sect. 1.2. There are many assertions in vector-valued harmonic analysis and probability theory which are equivalent to the UMD property, see e.g. [23] and [87]. Section 1.3 The theory of semigroups is well described in various books. The approach described in Sect. 1.3 based on Laplace transforms is described in detail in [13]. For other approaches see e.g. [48] or [117]. The books [8] and [106] concentrate on analytic semigroups and their properties. A short introduction to the theory of semigroups can be found e.g. in [50].

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133

Section 1.4 The concept of sectorial operators is again a classical one and was already used many authors such as Hille and Phillips, Triebel [148] and Tanabe [144] and Kato [92]. Operators admitting a bounded H ∞ -functional calculus have been introduced by McIntosh [113], where also the convergence lemma, Proposition 1.4.5, is proved. Excellent sources for further properties of the H ∞ -calculus are [71, 95] and [88]. The examples presented in Example 1.4.9 and 1.4.10 are standard and are based on Mikhlin’s theorem. More complicated examples of operators admitting a bounded H ∞ -calculus are discussed in Proposition 1.4.11. Its proof is based on the following vector-valued version of the transference principle from harmonic analysis. Note that for an operator-valued kernel k ∈ L1 (R, L(X)) the convolution operator Sk : Lp (R; X) → Lp (R; X) is defined for 1 ≤ q ≤ ∞ by  Sk f (t) =

∞ −∞

k(s)[f (t − s)]ds,

f ∈ Lp (R; X).

Let U be a C0 -group of bounded operators on a Banach space X with sup U (t)L(X) ≤ CU < ∞, t ∈R

k(t)U (s) = U (s)k(t) for all s, t ∈ and assume that k ∈ L1 (R; L(X)) is satisfying ∞ R. Then, given x ∈ X, the mapping Tk : x → ∞ k(t)[U (t)]xdt defines a bounded operator on X such that Tk L(X) ≤ CU2 Sk L(Lp (X)) . For details see [77] and [88]. Another important class of operators admitting a bounded H ∞ -calculus is closely related to boundary value problems of order 2m subject to general boundary conditions. They are of the form λu + A(x, D)u = f in , Bj (x, D)u = gj on ∂,

j = 1, . . . , m.

Here  ⊂ Rn+1 is a domain with compact, smooth boundary, f and gj are given functions, A is a differential operator of order 2m of the form   A(x, D) = |α|≤2m aα (x)D α and Bj (x, D) = |β|≤mj bj,β (x)D β with mj ≤ 2m for j = 1, . . . m. We denote by AB the realization of A(x, D) in Lp () with domain D(AB ) = {u ∈ Hp2m () : Bj (x, D)u = 0 on ∂, j = 1, . . . , m}. Assume that the top-order coefficients of A(x, D) are Hölder continuous and that the above boundary value problem satifies certain ellipticity, smoothness as well as the Lopatinskii-Shapiro condition. It was then shown in [37] that, roughly speaking,

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M. Hieber

there exists μ > 0 such that AB + μ admits a bounded H ∞ -calculus on Lp (). For the precise assumptions and the precise assertion, we refer to Theorem 2.3 of [37]. At this point we refer further to a result by Duong and Li [46] on bounded H ∞ -calculus for elliptic operators on Rn satisfying only very weak smoothness assumption on the coefficients, more precisely only a smallness condition on the BM0-norm of the top-order coefficients. Section 1.5 For general expositions concerning fractional powers of operators we refer e.g. to the books of Triebel [148] and Amann [8]. The approach described in Sect. 1.5 uses an extended H ∞ -calculus which is described in more detail e.g. in Section I.2 of [36] or in Appendix B of [95]. We follow here closely the presentation in [95] in which one finds also additional information on representation formulas for Aα and on interpolation and extrapolation scales. The characterization of the domain D(Aα ) of a sectorial operators A in terms of complex interpolation spaces is important in many applications. It should be noted that Theorem 1.5.6 holds true under the weaker assumption that A admits bounded imaginary powers. Observe that a sectorial operator is said to admit bounded imaginary powers if Ais ∈ L(X) for all s ∈ R and there exists a constant C > 0 such that Ais  ≤ C for all s with |s| ≤ 1. It seems that Theorem 1.5.6 goes back to [148]. The reiteration theorem in interpolation theory yields in particular [Xα , Xβ ]θ = Xα(1−θ)+θβ ,

0 ≤ α < β ≤ 1, θ ∈ (0, 1),

for the fractional power spaces associated to operators A having bounded imaginary powers. The latter class of operators plays a central role in the Dore–Venni approach to maximal Lp -regularity of evolution equations. Section 1.6 The operator-valued H ∞ -calculus in its general form as presented in Sect. 1.6 was developed in a fundamental article by Kalton and Weis [91] in 2001. Their approach uses in an essential way the notation of R-boundedness of a set of bounded operators. It is shown that the functional calculus for scalar-valued bounded holomorphic functions described in Sect. 1.4 can be extended to operatorvalued functions F provided the range {F (λ) : λ ∈ θ } of the function F is not only bounded but R-bounded. The notion of R-boundedness is implicitly already contained in the article [21]. In [17], this property was called “Riesz-property”; the name “randomized boundedness” appears in [28] and it this property is also often also known as “Rademacher boundedness”. For detailed proofs of the assertions given in Remark 6.2, Lemmas 6.3 and 6.4, we refer to [36, 95] or [88]. The Key Lemma 1.6.7 is also being used in a crucial way in the proof of the boundedness of H ∞ -calculus for operators associated to general boundary value problems subject to general boundary conditions describred above. Proposition 1.6.8 asserts that the classes H∞ (X) and RH∞ (X) coincide for Banach spaces satisfying the so-called property (α). Banach spaces having this property were introduced by Pisier in 1978, see [119]. The property (α) also plays a crucial role in the theory of vector-valued Fourier multipliers. This was already observed by Zimmermann in [156] in 1989. In particular, the vector-valued Fourier

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multiplier theorems stated in Theorem 1.6.13 and 1.6.14 require the additional assumption that X possesses property (α). For more information on this and on Fourier multipliers we refer to [11, 36, 124, 154] and [88]. The definition of an R-bounded H ∞ -calculus is a natural extension of the concept of the bounded H ∞ -calculus. It appeared in an article by Desch, Hieber and Prüss [40] on the Stokes problem in the half space Rn+ and in the abstract setting by Kalton and Weis in [91]. The fact that the Laplacian or more generally, parameter elliptic operators of angle ϕA < π, admit an R-bounded H ∞ -calculus on Lp (Rn ) was shown in [40] by combining kernel estimates with Theorem 1.2.3. Of course, taking into account the property (α) of Lp (Rn ), this follows immediately also from Proposition 1.6.8. Our approach to the problem of maximal Lp -regularity for evolution equations of the form u (t) + Au(t) = f (t), t > 0, u(0) = 0 is based on the beautiful Kalton–Weis sum theorem which was proved in [91]. The problem whether the sum of two resolvent commuting operators is closed was already investigated in 1975 by Da Prato and Grisvard in [32]. They proved a maximal Lp -regularity result for real interpolation spaces between X and D(A) via the sum method. By the same method, Dore and Venni [44] succeeded in 1987 in proving that A + B is closed provided X is a UMD space and A as well as B admit bounded imaginary powers with power angles θA and θB such that θA + θB < π. The Kalton–Weis theorem requires thus less for A, namely R-Boundedness instead of bounded imaginary powers for A, but more on B, namely bounded H ∞ -calculus instead of bounded imaginary powers. In order to deduce a maximal Lp -regularity estimate of the form u Lp ([0,T );X) + AuLp ([0,T );X) ≤ Cp f Lp ([0,T );X), from the sum theorem, we need to show that the derivative operator B = d/dt has a bounded H ∞ -calculus on Lp (R+ ; X). Assuming that X has the UMD property, it follows easily from the vector-valued Fourier multiplier Theorem 1.6.11 that B has a bounded H ∞ -calculus on Lp (R+ ; X) for all angles φ > π/2. It is interesting to note that the UMD property of X is not only sufficient for this assertion but also necessary. Thus, the UMD property may also be characterized in terms of spectral theory. The notion of R-sectorial operators goes back to Clément and Prüss [29] and Weis [154]. The operator-valued version of Mikhlin’s theorem was proved first by Weis in [154]. For detailed treatments also of the n-dimensional situation, we refer to Chapter I.3 and I.4 of [36], Chapters 1.3 and 1.4 of [95] and [124]. s (Rn ))-regularity result for the Let us finally mention a maximal L1 (0, T ; B˙ p,1 n solution of the heat equation on R for certain homogeneous Besov spaces which is due to Danchin and Mucha, see [30, 31]. More precisely, letϕ : Rn → [0, 1] be a function with support in {ξ ∈ Rn : 1/2 ≤ |ξ | ≤ 2} such that j ∈Z ϕ(2−j ξ ) = 1 for ˙ j )j ∈Z over Rn is given by all ξ = 0. Then the Littlewood-Paley decomposition ((Δ) ˙ j u := F −1 (ϕ(2−j ·)F u), 

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where F denotes the Fourier transform. Moreover, we denote by Sh (Rn ) the set of all tempered distributions u over Rn such that for all functions θ ∈ Cc∞ (Rn ) one has limλ→∞ θ (λD)u = 0 in L∞ (Rn ). For u ∈ Sh (Rn ), s ∈ R and 1 ≤ p, r ≤ ∞ set uB˙ s

p,r (R

n)

˙ j uLp (Rn ) l r (Z) := 2sj 

s (Rn ) is defined as Then the homogeneous Besov space B˙ p,r s (Rn ) := {u ∈ Sh (Rn ) : uB˙ s B˙ p,r

p,r (R

n)

< ∞}.

Consider the Gaussian semigroup T (t) = et  introduced in Example 1.3.13b). Then it is not difficult to show that there exist two constants c, C > 0 such that for all j ∈Z ˙ j hLp (Rn ) . ˙ j hLp (Rn ) ≤ Ce−c2  et   2j

Hence, by the variation of constant formula, the solution u of the inhomogeneous heat equation ut − u = f,

t ∈ J := (0, T ), x ∈ Rn ,

u(0) = u0

(1.17.1)

satisfies   ˙ j uL∞ (J ;Lp (Rn )) +22j  ˙ j uL1 (J ;Lp (Rn )) ≤ C  ˙ j u0 Lp (Rn )) + ˙ j f L1 (J ;Lp (Rn )) 

Multiplying with 2j s and summing over j yields the following maximal regularity estimate uL∞ (J ;B˙ s

p,1 (R

n ))

+ u L1 (J ;B˙ s (Rn )) + uL1 (J ;B˙ s (Rn )) p,1 p,1   ≤ C f L1 (J ;B˙ s (Rn )) + u0 B˙ s (Rn ) . p,1

p,1

(1.17.2)

We thus proved the following result: Let s ∈ R, p ∈ [1, ∞] and assume that f ∈ s (Rn )) and u ∈ B ˙ s (Rn ). Then the heat Eq. (1.17.1) admits a unique L1 (J ; B˙ p,1 0 p,1 s (Rn )) satisfying solution in C([0, T ); B˙ p,1 s (Rn )) ut , u ∈ L1 (J ; B˙ p,1

and there is a constant C > 0 such that (1.17.2) is satisfied. We finally note that s (Rn ) is replaced an estimate of the form (1.17.2) cannot hold true if the space B˙ p,1 p by any reflexive space and in particular by an L space for 1 < p < ∞. Let us emphasize that estimates of the form (1.17.2) can be deduced from a more abstract point of view also from the classical Da Prato-Grisvard theorem [32], which is valid also for the case p = 1.

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Section 1.7 There is a huge amount of articles dealing with abstract quasilinear evolution equations. Early work by the Japanese school founded by Kato and Tanaba and the Russian school founded by Sobolevskii was continued by Amann, von Wahl, Da Prato, Lundardi and many others. We refer here e.g. to the monographs [8, 106] and [144] and to a series of articles by Amann started with [6]. The results described in this section go back to Clément and Li [27] and [120]. For a very thorough study of quasilinear parabolic problems (including the theory of time weights and the principle of generalized linearized stability due to Prüss, Simonett and Zacher [126]) we refer to the excellent monograph by Prüss and Simonett [124]. Section 1.8 The content of this section is rather standard and can be found in many textbooks, for example [116]. For a recent comprehensive investigation, see [109]. Section 1.9 The Stokes equation in the halfspace Rn+ often serves as a model problem for this equation in bounded, exterior or more complicated types of domains. Indeed, the localization procedure described in Sect. 1.10 is based on a good understanding of this or the corresponding resolvent equation. Following [40], we describe an approach which differs from the ones given e.g. by Solonnikov [135], McCracken [112] and Ukai [151]. The advantage of our approach is twofold: first, the representation formula for the solution of the Stokes resolvent problem developed in Sect. 1.9 allows to show by fairly easy means that −A admits an p R-bounded H ∞ -calculus on Lσ (Rn+ ), where A denotes the Stokes operator in p Lσ (Rn+ ) defined as in (1.9.4). Secondly, our representation allows to deduce L∞ -estimates for the solution of n the Stokes resolvent problem which imply that the Stokes operator in L∞ σ (R+ ) generates a holomorphic semigroup on this space (which, of course, is not strongly continuous in 0). Note that the usual localization procedure does not work in this setting, which means that generator results for the Stokes semigroup on L∞ σ (),  a bounded or exterior domain with smooth boundary, are much more difficult to obtain. Recently this problem was solved by Abe and Giga [1]. Their proof is based on a contradiction argument. For an approach based on resolvent estimates which is inspired by the Masuda-Stewart technique for elliptic operator, see [2]. Section 1.10 There is a vast amount of literature concerning the linear Stokes equations. An excellent sourse of information are the monographs by Galdi [57], Sohr [134], Robinson et al. [128] and Tsai [150]. Pioneering results in the context of Lp -spaces are due to Solonnikov [135], Giga [62, 63], Giga and Sohr [64], Farwig and Sohr [51] and Shibata and Shimizu [130] and Abels [3]. For a recent survey on this topic we refer to the work of Hieber and Saal [79]. Section 1.11 An excellent source for mathematical models and their analysis arising in geophysics is the book by Chemin et al. [26]. They proved the following very remarkable fact concerning the Navier–Stokes equations in the rotational setting in

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R3 : given u(0) ∈ L2 (R2 )3 + H 1/2(R3 )3 , there exists 0 > 0 such that for every  ∈ R with || ≥ 0 , the equation ⎧ ⎨ ∂t u − νu + e3 × u + (u · ∇)u + ∇p = 0, t > 0, x ∈ R3 , divu = 0, t > 0, x ∈ R3 , ⎩ u(0, x) = u0 , x ∈ R3 ,

(1.17.3)

admits a unique, global solution. Their proof is based on dispersive estimates for certain terms in the solution operator of the linear part due to the Coriolis force. Global well-posedness results for the above equation for arbitrary  but initial data being small with respect to H 1/2 or other function spaces go back to Hieber and Shibata [81] and Giga et al. [67]. For more information on geophysical fluids we refer also to the book by Pedlosky [118]. It is a remarkable fact that the Navier–Stokes equations in the rotational seeting, i.e. with an additional term representing the Coriolis force, admit an explicit stationary solution. This stationary solution is called the Ekman spiral and the study of its stability properties is an important task, see e.g. [26, 42]. In this context, the Stokes-Coriolis-Ekman semigroup arises naturally. It seems to be unknown whether the Stokes-Coriolis-Ekman semigroup defined as in (1.11.2) remains a bounded semigroup on L2σ (Rn+ ) for large Reynolds numbers. An answer to this question would be helpful for clarifying questions related to the stability/instability of the Ekman spiral for large Reynolds numbers. Our presentation here follows [72] and [93]. The latter booklet studies more generally stationary solutions of the rotating Navier–Stokes-Boussinesq equations with stratification effects, i.e. ⎧ ⎪ ∂t u − νu + e3 × u + (u · ∇)u + ∇p = Ge3 , t > 0, x ∈ R3+ , ⎪ ⎪ ⎪ 2 ⎪ ∂t θ − κθ + (u · ∇)θ = −N u3 , t > 0, x ∈ R3+ , ⎨ divu = 0, t > 0, x ∈ R3+ , ⎪ ⎪ ⎪ u(t, x1 , x2 , 0) = (a1 , b1 , 0), t > 0, x1 , x2 ∈ R, ⎪ ⎪ ⎩ t > 0, x1 , x2 ∈ R, θ (t, x1 , x2 , 0) = c1 , (1.17.4) subject to initial data. Here G and N represent the gravitational force and the BruntVäisälä frequency. For further results concerning Ekman layers in rotating fluids we refer to the work of Masmoudi [111] and Rousset [129]. Hieber and Stannat considered in [82] also the situation of the stochastic Navier– Stokes-Coriolis equation in T2 × (0, b). They proved, as a stochastic analogue of the deterministic stability result described in Theorem 1.11.7, stochastic stability of the Ekman spiral by considering stationary martingale solutions.

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Section 1.12 The analysis of the motion of a body immersed in a liquid is a classical problem in fluid mechanics. The methods used depend on the assumptions whether the body is considered to be rigid or elastic or whether the fluid is incompressible, compressible, viscolelastic or has a complex stress tensor. For a survey of known results and methods concerning the stationary and instationary problem for Newtonian fluids and for prescribed, steady, self-propelled and free movements, see e.g. the articles by Galdi [56] and Galdi and Neustupa [58]. There are several possibilities to transform this moving domain problem to a probem on a fixed domain. One possible transformation was introduced by Galdi [56]: it is linear and the whole space is rotated and shifted back to its original position at every time t > 0. This transformation generates in the fluid equations a drift term with unbounded coefficients, i.e. the new fluid operator is of the form Lu := P (u + (ω × x · ∇)u − ω × u) in the purely rotational case. Here P denotes the Helmholtz projection and ω the rotational velocity. One of the fundamental difficulties of this approach is that the transformed problem is no longer parabolic. A second approach is characterized by a non-linear, “local” change of coordinates which only acts in a suitable bounded neighborhood of the obstacle. Tucsnak, Cumsille and Takahashi used this transform going back to Inoue and Wakimoto [89] and showed the existence of a unique, local strong L2 -solution to the fluid rigid body problem in two and three space dimensions, see [142, 143]. The Lp -theory of this approach was developed in [61] for Newtonian and also for generalized Newtonian fluids. For weak solutions, see e.g. [41] and [52]. Section 1.12 follows the approach based on this second change of variables, however now in the situation of a compressibe fluid, see [74]. For a different approach, see [19]. The methods for investigating the case of an elastic body are again different. We refer here e.g. to the fundamental article by Raymond and Vanninathan [127]. Section 1.13 There is an enormous amount of literature concerning free boundary value problems for viscous fluids. The monograph [124] is an excellent source of information about the state of the art concerning strong solutions and sharp interfaces. Pioneering results for Newtonian fluids in the one phase setting go back to Solonnikov [136, 137], Beale [15], Tani and Tanaka [146] and Shibata and Shimizu [131]; for the two-phase situation we refer to Denisova [34] and Prüss and Simonett [122]. For the spin-coating problem, see [39]. Problems of this kind for non-Newtonian fluids were treated by Abels [4] in the context of measure-valued varifold solutions. The approach presented here for generalized non-Newtonian fluids follows [80]. Section 1.14 The Ericksen-Leslie models describing the hydrodynamic flow of nematic liquid crystal flow was pioneered by Ericksen [49]. Starting from the classical balance laws for mass, linear and angular momentum he introduced the a set of equation describing the evolution of liquid crystals. The system was completed by Leslie [97] with the addition of constitutive equations. For more

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information see the book by Virga [152] and the survey article [102] by Lin and Wang. For a derivation of this model from thermodynamical principles we refer to the survey paper [77]. The latter and [152] contain many further results and references. The mathematical analysis of the simplified system started with the work of Lin and Liu, see [100] and [101]. The approach to the simplified system presented here is taken form [84]. The non-isothermal Ericksen-Leslie model for incompressible fluids subject to general Leslie stress SL and general Ericksen tensor can be found e.g. in [77]. It is shown to be thermodynamically consistent. The well-posedness results concerning the situation of general Leslie stress go back to [76] and [78], see also [77]. Observe that these results do not make any structural assumptions on the Leslie coefficients and that Parodi’s condition is not assumed in contrast to previous work, see e.g. [105] and [33]. For results dealing with anisotropic elasticity, i.e. with the general Ericksen tensor SE but with vanishing Leslie tensor SL , in particular without stretching, see the recent work of Hong et al. [85] and Ma, Li and Gong [107]. Let us mention here also that in the case of  = Rn , the local existence result given in Theorem 1.14.9 extends to the situation of general Ericksen stress tensor, see [78]. Sections 1.15 and 1.16 The mathematical analysis of the primitive equations was initiated by Lions, Teman and Wang in a series of articles [103, 104]. For a survey of known results and further references, we refer to an survey article by Li and Titi [99]. A breakthrough result in this contex was proved by Cao and Titi [25] in 2007. They proved that the primitive equations admit a unique, global, strong solution for arbitrary large data in H 1 . A different approach to the primitive equations based on methods of evolution equations has been initiated in [73]. The results presented in Sects. 1.15 and 1.16 go back to the work of Giga et al. [65] and Furukawa et al. [55].

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Chapter 2

Partial Regularity for the 3D Navier–Stokes Equations James C. Robinson

2.1 Introduction These notes give a relatively quick introduction to some of the main results for the three-dimensional Navier–Stokes equations, concentrating in particular on ‘partial regularity’ results that limit the size of the set of (potential) singularities, both in time (Sect. 2.2.5) and in space-time (Sect. 2.4). We will consider the Navier–Stokes equations ∂t u − u + (u · ∇)u + ∇p = 0,

∇ · u = 0,

(2.1)

on the three-dimensional torus T3 , i.e. [0, 2π)3 with ‘periodic boundary conditions’. We choose this domain for simplicity, since it is both bounded and boundaryless. Often we will consider the initial-value problem, i.e. we solve (2.1) for t > 0 given u(x, 0) = u0 (x). The existence of weak solutions (solutions with finite kinetic energy) on the whole space has been known since the work of Leray [22], but whether or not smooth solutions exist for all time is still unresolved, and is one of the Clay Foundation’s Million-Dollar Millennium Problems [12]. The first ‘partial regularity’ results were due to Scheffer who showed that the 1-dimensional Hausdorff measure of the set of spatial singularities is finite at the first singular time [33], that the Hausdorff dimension of the set of space-time singularities is no larger than 5/3 [35], and that the 1/2-dimensional Hausdorff measure of the set of singular times is zero ([32]; cf. our Proposition 2.12).

J. C. Robinson () Department of Mathematics, University of Warwick, Coventry, UK e-mail: [email protected] © Springer Nature Switzerland AG 2020 G. P. Galdi, Y. Shibata (eds.), Mathematical Analysis of the Navier-Stokes Equations, Lecture Notes in Mathematics 2254, https://doi.org/10.1007/978-3-030-36226-3_2

147

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The ingredients that go into known partial regularity results are usually (i) a conditional regularity result, i.e. a guarantee of local regularity of solutions under a local smallness condition and (ii) a global bound on solutions that involves the same quantity as the conditional regularity result. Then (ii) can be used to show that the condition in (i) cannot be violated at too many points, which serves to limit the size of the singular set. A space-time point (x, t) is termed ‘regular’ if u ∈ L∞ (U ) for some space-time neighbourhood U of (x, t). That this is a reasonable definition of a regular point is a consequence of a local conditional regularity result due to Serrin [37]: local boundedness is sufficient for spatial smoothness. We given a sketch of this result in Sect. 2.3 (in fact we show that if u ∈ Lα (U ) for any α > 5 then u is spatially smooth within U ; and we also indicate how to extend this result to the case u ∈ L5 (U )). The most well known partial regularity result for the Navier–Stokes equations is due to Caffarelli et al. [3] [hereafter CKN] and guarantees that the 1-dimensional Hausdorff measure of the set of space-time singularities is zero; in particular the Hausdorff dimension of this set is no larger than one. The hard part of their proof is a local conditional regularity result. In fact they proved two such results; we only prove the first in these notes (and then in a simplified form, neglecting the pressure), but this is perhaps the harder of the two: from what we prove it is relatively straightforward to show that the box-counting (Minkowski) dimension of the set of space-time singularities is bounded by 5/3. We then present a result to due Robinson and Sadowski [30] that makes crucial use of this bound on the dimension of the singular set: we show that given any weak solution u, the ‘particle trajectories’, i.e. solutions of X˙ = u(X, t), exist and are unique for almost every choice of initial condition, so that it makes sense to consider even such an irregular solution from a Lagrangian point of view. The point of including this result is two-fold: it provides an application of the partial regularity theory, and demonstrates that, despite the lack of a complete theory of existence and uniqueness, one can prove interesting results about solutions of the 3D Navier– Stokes equations that are valid without invoking any unproved assumptions.

2.1.1 Notation and Preliminaries We will use AB to mean that A ≤ cB for some absolute constant c; and A r B to mean that A ≤ c(r)B. Similarly we use A ≈ B to mean that c1 A ≤ B ≤ c2 B for some constants c1 and c2 .

2 Partial Regularity for the 3D Navier–Stokes Equations

We write ∇ = (∂1 , ∂2 , ∂3 ), where ∂j = ∇ · u = div u,

∂ ∂xj ,

∇ × u = curl u,

149

so that

and

[(u · ∇)u]i =

3  (uj ∂j )ui . j =1

For functions on the torus we can easily define the norms in the Sobolev spaces H s (T3 ) using Fourier series: if u=



uˆ k eik·x

k∈Z3

then u2H s :=



(1 + |k|2 )s |uˆ k |2

k

and u2H˙ s :=



|k|2s |uˆ k |2 .

k

Note that if u has zero average then uˆ 0 = 0, and so uH s ≈ uH˙ s . Also note that from these expressions it is easy to obtain the generalised Poincaré inequality uH s1  uH s2 for all s2 ≥ s1 . It is useful to recall that H s is an algebra if s > 3/2, i.e. fgH s  f H s gH s

f, g ∈ H s , s > 3/2.

(2.2)

Let   Cσ∞ (T3 ) = ϕ ∈ [C ∞ (T3 )]3 : ∇ · ϕ = 0,

T3

ϕ=0 ,

i.e. divergence-free smooth periodic functions with zero average; the subscript σ is used to indicate that the functions are divergence free. We set H := completion of Cσ∞ (T3 ) in the norm of L2 (T3 ), denoting the L2 norm by  · , and V := completion of Cσ∞ (T3 ) in the norm of H 1 (T3 ) = H ∩ H 1 (T3 ). When we omit the domain from a spatial integral it should be understood to be T3 .

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We will often use ‘Lebesgue interpolation’ (which follows from Hölder’s inequality), in particular the inequality 1/4

3/4

uL4 ≤ uL2 uL6 . Coupled with the Sobolev embedding H 1 (T3 ) ⊂ L6 (T3 ) this yields (when the Ladyzhenskaya inequality 1/4

3/4

uL4  uL2 ∇uL2 .



u = 0)

(2.3)

2.2 Basic Existence and Uniqueness Results We will not discuss in detail here the proof of the existence of solutions, referring instead to the relevant sections of Robinson et al. [31], hereafter RRS. However, we show how the regularity of weak and strong solutions (and their existence) is to be expected, by using a priori estimates obtained by manipulating the equations as if all the terms were smooth.

2.2.1 Weak Solutions Both the definition of weak solutions and the proof of their existence are based on the following a priori energy estimate for smooth solutions. If we take the inner product of (2.1) with u and integrate then two terms vanish: for the nonlinear term we have1   ui (∂i uj )uj = − ui uj (∂i uj ) and so this term vanishes, and for the pressure term 

 (∂i p)ui = −

p(∂i ui ) = 0,

since ∂i ui = div u = 0.

1 This

is a particular case of the very useful anti-symmetry property (u · ∇)v, w" = − (u · ∇)w, v",

which we will use from time to time in what follows.

(2.4)

2 Partial Regularity for the 3D Navier–Stokes Equations

151

Therefore 1 d u2 + ∇u2 = 0 2 dt from which it follows that 1 u(t)2 + 2



t

∇u(s)2 ds =

0

1 u0 2 ; 2

this estimate is at the basis of the proof that for any u0 ∈ H there exists at least one weak solution u ∈ L∞ (0, T ; L2 ) ∩ L2 (0, T ; H 1) for every T > 0. [One can perform the same estimates for the Euler equations (when ν = 0) and show that the kinetic energy is constant. The required level of regularity for solutions to ensure that this holds is the subject of the Onsager Conjecture, resolved by work of Constantin et al. [6] and Isett [18]; see also Buckmaster et al. [2].] We now define the notion of a weak solution more precisely. Let Dσ = {ϕ ∈ Cc∞ (T3 × [0, ∞))3 : div ϕ(t) = 0 for all t ∈ [0, ∞)}. Definition 2.1 A weak solution of (2.1) is a function u such that u ∈ L∞ (0, T ; H ) ∩ L2 (0, T ; V ) and 

s 0



s

− u, ∂t ϕ" + 0



s

∇u, ∇ϕ" +

for every T > 0

(u · ∇)u, ϕ" = u0 , ϕ(0)" − u(s), ϕ(s)"

0

(2.5) for all ϕ ∈ Dσ and almost every s > 0. [Note that this definition is distinct from requiring that u satisfies the equations ‘in the sense of distributions’. However, one can prove that given a weak solution u there exists a corresponding pressure such that the original equations hold in this sense, see Chapter 5 of RRS (Section 5.2).] Exercise Derive (2.5) as a consequence of (2.1) when u is smooth by taking the inner product with ϕ ∈ Dσ , integrating in both space and time, and integrating by parts. The existence of weak solutions on T3 can be proved rigorously via a Galerkin procedure,2 see Chapter 4 of RRS, for example, which also yields a solution that 2 Leray’s 1934 paper treats the equations on the whole space and does not use the Galerkin approach (see O˙za´nski and Pooley [25], for a modern treatment of the methods in his paper). A ‘Galerkinlike’ argument for the equations on the whole space can be found in the book by Chemin et al. [4]. The first proof of existence of solutions on bounded domains was due to Hopf [17].

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satisfies the strong energy inequality [(2.6), below]. These are known as Leray– Hopf weak solutions. Theorem 2.2 Given u0 ∈ H there exists at least one weak solution that satisfies the strong energy inequality 1 u(t)2 + 2



t

∇u(τ )2 dτ ≤

s

1 u(s)2 2

(2.6)

for every t ≥ s, for a set of s of full measure that includes s = 0. The ‘energy inequality’ is (2.6) allowing only s = 0. It is not known whether Leray–Hopf weak solutions are unique. Very recently, Buckmaster and Vicol [1] have shown that if one does not require the energy inequality then distributional solutions are not unique, and there are reasons to suspect that this non-uniqueness extends to Leray–Hopf weak solutions [19].

2.2.2 Strong Solutions We can show that if u0 ∈ V then smoother solutions exist with u ∈ L∞ (0, T ; V ) ∩ L2 (0, T ; H 2 ),

(2.7)

at least for some T > 0, and if the norm of u0 in V is sufficiently small then the solutions exist for all t ≥ 0. A weak solution with the regularity in (2.7) is called a strong solution on [0, T ]. Taking the L2 inner product of (2.1) with u we obtain 1 d ∇u2 + u2 ≤ 2 dt

 [(u · ∇)u] · u

≤ u∞ ∇uu  ∇u

3/2

≤

(2.8)

u

3/2

1 u2 + C∇u6 . 2

Exercise Use Fourier series to show that uL∞  ∇u1/2 u1/2 whenever u ∈ H 2 (T3 ).

(2.9)

2 Partial Regularity for the 3D Navier–Stokes Equations

153

Therefore d ∇u2 + u2  ∇u6 . dt

(2.10)

In particular d ∇u2 ≤ c∇u6 , dt from which it follows that ∇u(t)2 ≤ 

∇u0 2 1 − 2ct∇u0 4

(2.11)

.

These calculations suggest that if u0 ∈ H 1 there exists T = T (∇u0 ) [for example, T = ∇u0 −4 /4c] such that u ∈ L∞ (0, T ; H 1). Once we have u ∈ L∞ (0, T ; H 1 ) we can return to (2.10) and integrate from 0 to T to show that u ∈ L2 (0, T ; H 2 ): 

T

∇u(T )2 +



T

u(s)2 ds  ∇u(0)2 +

0

∇u(s)6 ds.

0

Again, to make this rigorous one can argue using Galerkin approximations (Chapter 6 in RRS). Since u ∈ L∞ (0, T ; V ) ∩ L2 (0, T ; H 2) it follows that ∂t u ∈ L2 (0, T ; H ). Indeed, choosing ϕ ∈ H we have, using (2.9), |(∂t u, ϕ)| ≤ |(u, ϕ)| + |(u · ∇u, ϕ)| ≤ uϕ + u∞ ∇uϕ  [u + ∇u3/2 u1/2 ]ϕ,

(2.12)

and the time integrability of ∂t u follows from the regularity of u. This implies, in particular, that u ∈ C 0 ([0, T ]; V ) (see Theorem 4 in Section 5.9.2 in Evans [9] or Theorem 7.2 in Robinson [27] for details). Furthermore, strong solutions are unique (actually, something stronger is true, as we will soon see in the next section). If u and v are strong solutions and we consider their difference w = v − u then ∂t w − w + (v · ∇)v − (u · ∇)u +∇(p − q) = 0, /0 1 . (v·∇)w+(w·∇)u

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and if we take the inner product with w then 1 d w2 + ∇w2 = − 2 dt

 (w · ∇)u · w

(2.13)

≤ w2L4 ∇u  w1/2 ∇w3/2 ∇u, 1/4

3/4

using the Ladyzhenskaya inequality wL4  wL2 ∇wL2 (see (2.3)). Now we use Young’s inequality to split the right-hand side, and absorb the ∇w2 using the left-hand side, to obtain 1 d w2  ∇u4 w2 ; 2 dt

(2.14)

since ∇u ∈ L∞ (0, T ; V ) we can use Gronwall’s inequality to deduce that w = 0 for all t ∈ [0, T ], i.e. that u = v. The inequality (2.10) can also be used to prove the global-in-time existence of strong solutions when the initial condition is sufficiently small, if we use the Poincaré inequality ∇u  u. Indeed, we then have d ∇u2 ≤ c∇u6 − λ∇u2 ; dt

(2.15)

if we take ∇u0 4 < 04 := λ/c then ∇u(t) is non-increasing and so the solution remains bounded in H 1 for all positive times. Putting all these facts together we have the following result on the existence and uniqueness of strong solutions. Theorem 2.3 If u0 ∈ V then there exists a time T = T (∇u0 ) such that there exists a unique strong solution u ∈ L∞ (0, T ; V ) ∩ L2 (0, T ; H 2). Furthermore there exists 0 > 0 such that whenever ∇u0  < 0 this strong solution exists for all t ≥ 0. It turns out that strong solutions are also smooth in space (while they exist): so if u is a strong solution on (0, T ) it is smooth on (ε, T ] for any ε > 0. We can prove this inductively, following Constantin and Foias [5]. Lemma 2.4 If u0 ∈ H k and u is a strong solution on (0, T ) then u ∈ L∞ (0, T ; H k ) ∩ L2 (0, T ; H k+1 ). Proof We will show that if s ≥ 1 then u ∈ L∞ (0, T ; H s ) ∩ L2 (0, T ; H s+1) and u0 ∈ H s+1

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implies that u ∈ L∞ (0, T ; H s+1) ∩ L2 (0, T ; H s+2). If we take the inner product of the equation with u in H s+1 we obtain  2 1 d u2H s+1 + ∇u2H s+1 ≤ | (u · ∇)u, u"H s+1  2 dt  u2H s+1 ∇uH s+1 ≤

1 ∇u2H s+1 + cu4H s+1 , 2

where we have used the fact that H s+1 is an algebra (see (2.2)) and Young’s inequality to split the right-hand side. Therefore d u2H s+1 + ∇u2H s+1  u4H s+1 = u2H s+1 u2H s+1 . dt Since by assumption u ∈ L2 (0, T ; H s+1) and u0 ∈ H s+1 it follows—dropping the ∇u2H s+1 term and integrating in time—that in fact u ∈ L∞ (0, T ; H s+1). Now we can integrate again, but this time retaining the ∇u term, to see that u ∈ L2 (0, T ; H s+2). We can start the induction with s = 1 since for any strong solution we have u ∈ L∞ (0, T ; H 1 ) ∩ L2 (0, T ; H 2). 

With this result it is easy to prove the smoothness of strong solutions. Proposition 2.5 If u0 ∈ V and u ∈ L∞ (0, T ; H 1) ∩ L2 (0, T ; H 2 ) is a strong solution then u ∈ L∞ (ε, T ; H k ) for every k ∈ N and ε > 0. Proof Fix ε > 0. Since u ∈ L∞ (0, T ; V ) ∩ L2 (0, T ; H 2 ), there exists a time t1 ∈ (0, ε) such that u(t1 ) ∈ H 2 . Since strong solutions are unique, the strong solution v(t) with initial condition u(t1 ) coincides with u(t1 + t), so it follows from Lemma 2.4 that u ∈ L∞ (t1 , T ; H 2) ∩ L2 (t1 , T ; H 3). Now we can find t2 ∈ (t1 , ε) such that u(t2 ) ∈ H 3 ; it follows from Lemma 2.4 (again using the uniqueness of strong solutions) that u ∈ L∞ (t2 , T ; H 3 ) ∩ L2 (t2 , T ; H 4). Continuing in this way we obtain the result as stated. 

Corollary 2.6 If u0 ∈ V and u ∈ L∞ (0, T ; H 1)∩L2 (0, T ; H 2 ) is a strong solution then u ∈ C 0 ((0, T ]; H k ) for every k ≥ 0. In particular the solution is smooth in space: u(t) ∈ C ∞ (T3 ) for every t ∈ (0, T ]. Proof It follows from an estimate similar to that in (2.12) that we have ∂t u ∈ L∞ (0, T ; H k ) for every k ≥ 0, and so u ∈ C 0 ((0, T ]; H k ) for every k ≥ 0 [see Theorem 4 in Section 5.9.2 in Evans [9] or Theorem 7.2 in Robinson [27]]. Thus for every t ∈ (0, T ] we have u(t) ∈ H k for every k ≥ 0, and hence u(t) ∈ C ∞ . 

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2.2.3 Weak-Strong Uniqueness One very useful property of strong solutions is that they are unique in the class of weak solutions that satisfy the energy inequality (‘weak-strong uniqueness’). The following argument gives an indication of why this is true, but it is not rigorous since it requires more smoothness of the two solutions than we actually have. Suppose that u is a strong solution and v is a weak solution, and consider, as above, the difference w = u − v. If we neglect any smoothness considerations and follow our previous calculations that led to (2.14) then we obtain 1 d w2  ∇u4 w2 ; 2 dt noticing that the smoothness of v does not play a role here, we can again use the fact that ∇u ∈ L∞ (0, T ; V ) to deduce that w = 0 for all t ∈ [0, T ], i.e. that u = v. However, this argument is not valid, since the time derivative of a weak solution does not have sufficient regularity to ensure that ∂t w, w" makes sense. Indeed, if v is a weak solution then the regularity of ∂t v is essentially determined by that of (v · ∇)v, and 1/2

| (v · ∇)v, ϕ"| = | (v · ∇)ϕ, v"| ≤ u2L4 ∇ϕ ≤ vL2 ∇v3/2 ∇ϕ, which shows that ∂t v ∈ L4/3 (0, T ; H −1 ) [we used the anti-symmetry property of the nonlinear term (2.4)]. This is not enough to allow us to pair ∂t w (whose regularity is limited by that of ∂t v) with w, which is only in L2 (0, T ; H 1 ). A rigorous proof of the weak-strong uniqueness property essentially proceeds as follows: (i) a strong solution u is sufficiently regular that the NSE hold as an equality in L2 (0, T ; H ), so one can take the inner product with a weak solution v ∈ L2 (0, T ; H ); (ii) a strong solution u has sufficient regularity to be used as a test function in the definition of a weak solution v; (iii) add the two equations from (i) and (ii) and use the fact that v satisfies the energy inequality to deduce that w = u − v satisfies  t   t   1 2 2  w(t) + ∇w ≤  (w · ∇)w, u" , (2.16) 2 0 0 (this is the rigorous version of (2.13) in this context) and from here one can argue to show that w ≡ 0. Indeed, it follows that 1 w(t)2 + 2



t



t

∇w  2

0

w1/2 ∇w3/2 ∇u

0



t

≤c 0

∇u4 w2 +

1 2



t 0

∇w2

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and so we obtain 

t

w(t) + 2



t

∇w(s) ds  2

0

∇u(s)4 w(s)2 ds.

0

Since u is a strong solution we have u ∈ L∞ (0, T ; H 1 ), and since u(0) = v(0) we also have w(0) = 0, from which it follows that w(t) ≡ 0 for all t ∈ (0, T ). See Section 4 of Galdi [15] or Chapter 6 in RRS for details. Now let us consider what implications this has for weak solutions. One is that all Leray–Hopf weak solutions are eventually regular. We know that such solutions satisfy the energy inequality 1 u(t)2 + 2



t

∇u(s)2 ds ≤

0

1 u0 2 , 2

and we have seen that if ∇u0  < 0 then the solution is strong for all t ≥ 0 (see (2.15)). From the energy inequality it follows that if T ≥ 1 2 u0 2 then 20

there exists a t0 ∈ [0, T ] such that ∇u(t0 ) < 0 ; weak-strong uniqueness now guarantees that u coincides with the strong solution through u(t0 ) for all t ≥ t0 . Since any weak solution satisfies u ∈ L2 (0, T ; H 1 ), we have u(t1 ) ∈ H 1 for almost every t1 ; it follows from weak-strong uniqueness that for t ≥ t1 the weak solution u coincides with the strong solution with initial data u(t1 ); it therefore follows from (2.11) that ∇u(t2 )2 ≤ 

∇u(t1 )2 1 − 2c(t2 − t1 )∇u(t1 )4

t1 ≤ t2 ≤ T .

(2.17)

while expression within the square root in the denominator remains positive.

2.2.4 Regular and Singular Times Let us call a time t ∈ (0, ∞) a regular time if ∇u ∈ L∞ (U ) for some neighbourhood U of {t}. Let R denote the set of all regular times. This is clearly open, so we can write R as a disjoint union of open intervals: R=

∞ "

(ai , bi );

(2.18)

i=1

since weak (Leray–Hopf) solutions are eventually strong (and so bounded) we can take b1 = ∞. We now show that u(t) is regular if t ∈ R.

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Lemma 2.7 If t ∈ R then u(t) ∈ C ∞ . Proof If t ∈ R then u ∈ L∞ ((t − δ, t + δ), V ) for some δ > 0. In particular there exists t0 ∈ (t − δ, t − δ/2) such that u(t0 ) ∈ V ; the regularity result of Corollary 2.6 now implies that u(t) ∈ C ∞ . 

We will call a time a singular time if it is not regular. Thus, if T is a singular time then there exists a sequence of times tj → T such that ∇u(tj ) ≥ j , and so, by rearranging3 (2.17) we obtain, for any t < T , ∇u(t)4 ≥

∇u(tj )4 ; 1 + 2c(tj − t)∇u(tj )4

(2.19)

hence, on letting j → ∞, it follows that 1 ∇u(t)2 ≥ √ . 2c(T − t)

(2.20)

In particular, any singular time T must actually be a ‘blowup time’ with ∇u(t) → ∞ as t → T − . [While the H 1 norm must blow up, it is not known whether any weak solution v that coincides with u on (0, T ) must satisfy lim supt →T + ∇v(t) = ∞.] Alternatively we can interpret (2.20) as a ‘local regularity’ result. If we start from (2.20) and integrate from T − r to T then we obtain 

T T −r

3 ∇u(t) dt ≥ 2

1 2c



T T −r

1

3

√ dt = T −t

2 1/2 r . c

√ If we set ε∗ = 2/c then we have the following conditional ‘ε-regularity’ result. Lemma 2.8 There exists an absolute constant ε∗ such that if r −1/2



T T −r

∇u(s)2 ds < ε∗

then u is regular at time T .

argument proceeds by contradiction: if (2.19) does not hold then ∇u(t)4 < RHS of (2.19). In this case it follows from (2.17) that a strong solution v with v(t) = u(t) exists on the time interval [t, tj ] and ∇v(tj ) < ∇u(tj ). But by weak-strong uniqueness we must have u = v on [t, tj ], which yields a contradiction.

3 A more careful

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2.2.5 ‘Partial Regularity’: The Set of Singular Times All the partial regularity results that we will discuss rely on two components: an ε-regularity result, like Lemma 2.8, and a global bound on the same quantity, such as ∇u ∈ L2 ( × (0, T )). These two ingredients can be combined to limit the size of ‘singular’ points of u: there cannot be too many of these, otherwise the integrability would be violated.

2.2.5.1 Dimensions To describe the ‘size’ of a set we will use its box-counting dimension (also called the Minkowski dimension). There are a number of equivalent definitions of this dimension; we focus on two here (which also work for subsets of infinitedimensional spaces) and will use both in what follows. For more details see Falconer [11] or Robinson [28]. Definition in Terms of Coverings Let X be compact set. We let N(X, ε) be the minimum number of balls of radius ε whose centres lie in X needed to cover X, and define the box-counting dimension of X to be dimB (X) := lim sup ε→0

log N(X, ε) . − log ε

Essentially this extracts the exponent d from N(X, ε) ∼ ε−d . If d > dimB (X) then N(X, ε) ≤ ε−d for ε sufficiently small; while for each d < dimB (X) there is a sequence εj → 0 such that N(X, εj ) > εj−d . Definition in Terms of ε-Separated Points Let M(X, ε) be the maximal number of ε-separated points in X, and use the same definition as above with N(X, ε) replaced by M(X, ε). This gives the same quantity. [To show that these definitions are equivalent: (1) M(X, ε) ≤ N(X, ε) since none of the ε-separated balls cover the centres of the others (2) We have N(X, ε) ≤ M(X, ε/3)—if some x ∈ X does not belong to an ε-ball about an ε/3-separated point then B(x, ε/3) and B(y, ε/3) are disjoint for every ε/3-separated y, giving a new ε/3-separated point x.] As above, if d < dimB (X) then there is a sequence εj → 0 for which it is possible to find a set of at least εj−d points that are εj separated. It can be useful to mix these definitions, as the following simple result shows. Lemma 2.9 The set Aα := {n−α : n ∈ N} ∪ {0} has box-counting dimension 1/(1 + α).

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Proof Points in Aα are a distance ε apart for n−α − (n + 1)−α ∼ αn−(1+α) ≥ ε, i.e. for n  ε−1/(1+α). It follows that M(X, ε/2)  ε−1/(1+α). However, n intervals of length ε will certainly cover these points, and the remaining points are contained in the interval [0, εα/(1+α)], which requires no more than ε−1/(1+α) intervals of length ε to cover: N(Aα , ε)  ε−1/(1+α). These upper and lower bounds imply that dimB (Aα ) = 1/(1+α) as claimed.  Exercise Suppose that (en ) is an orthonormal subset of a Hilbert space H , and define Hα := {n−α en : n ∈ N} ∪ {0}. Show that dimB (Hα ) = 1/α. The following consequence of the covering definition will be useful later (and can itself serve as the basis for yet another equivalent definition of the box-counting dimension in Euclidean spaces). Lemma 2.10 If X is a compact subset of Rn and d > dimB (X) then there exists C > 0 such that μ(O(X, ε)) ≤ Cεn−d , where O(X, ε) = {x + y : x ∈ X, |y| < ε}

(2.21)

is the ε-neighbourhood of X. Proof By definition, for any d > dimB (X), we have N(X, ε) < Cε−d for some C > 0. Since X can be covered by Cε−d balls of radius ε, it follows that the εneighbourhood of X can be covered by Cε−d balls of radius 2ε, of total measure no more than 2n Cεn−d . 

Exercise Show that if X ⊂ Rn then dimB (X) = n − sup{s : μ(O(X, ε)) ≤ Cεs for some C > 0}. Corollary 2.11 If X ⊂ Rn and dimB (X) < n then μ(X) = 0. Proof Choose d with dimB (X) < d < n. Since X ⊂ O(X, ε) for every ε > 0, it follows that μ(X) ≤ μ(O(X, ε)) ≤ Cεn−d and so μ(X) = 0.



We can now use this definition to deduce a ‘partial regularity’ result concerning the singular times of a Navier–Stokes solution.

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2.2.5.2 Dimension of the Set of Singular Times We now bound the box-counting dimension of the set of singular times. The result in this form is due to Robinson and Sadowski [29], but the proof we give is based on Kukavica [21]. Proposition 2.12 If T = {t ≥ 0 : ∇u(t) is unbounded in a neighbourhood of t} then dimB (T ) ≤ 1/2. Proof It follows from Lemma 2.8 that if t is a singular time then 

t t −r

∇u(s)2 ds ≥ ε0 r 1/2

for every r > 0. Suppose now, for a contradiction, that the set of all singular times T satisfies dimB (T ) > d > 1/2. Then there is a sequence εj → 0 for which one can find a collection Tj of at least εj−d points that are εj separated. Then the intervals (t − εj /2, t) are disjoint for distinct points t in Tj , and so 

T 0

−(d−1/2)

∇u(s)2 ds  εj−d (ε0 εj ) = ε0 εj 1/2

;

letting j → ∞ contradicts the integrability of ∇u2 .



Using Corollary 2.11 it follows that the set of singular times has zero measure, i.e. almost every time is regular. Exercise While the fact that any weak solution satisfies u ∈ L2 (0, T ; H 1) implies that u(t) ∈ H 1 for almost every t, this is not the same as saying that almost every time is regular; to see this construct a function f ∈ L2 (0, 1) for which the set of singular times, sing(f ) := {t ∈ [0, 1] : f is unbounded in a neighbourhood of t}, has full measure. The proof of Proposition 2.12 relies only on two ingredients, namely that the quantity X(t) := ∇u(t)2 satisfies X˙  X3



T

and 0

Using only these two facts we can do no better.

X(s) ds < ∞.

(2.22)

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J. C. Robinson

Lemma 2.13 For each 0 < d < 1/2 there exists a function Xd : [0, 1] → R that satisfies (2.22) and for which dimB (sing(Xd )) = d. Proof Let Aα be the set from Lemma 2.9; if α = (1 − d)/d then dimB (Aα ) = d, and d < 1/2 corresponds to choosing some α > 1. Now let  −1/2 Xd = n−α − t

for

(n + 1)−α < t < n−α

with Xd (n−α ) chosen arbitrarily. Then −3/2 1 1  −α n −t X˙ d (t) = = Xd (t)3 2 2 for almost every t ∈ [0, 1], and 

1

Xd (t) dt =

0

∞  

n−α

−α n=1 (n+1)

(n−α − t)−1/2 dt

 1/2 ∞  1 1 1 − = 2 nα (n + 1)α n=1

≤

∞  α 1/2 1.



The above result shows that it is not possible to improve the bound on the boxcounting dimension using only the ingredients from (2.22) that went into the proof of Proposition 2.12. But with a new definition one can prove something a little stronger. The following argument is due to Kukavica [21]. Given r > 0, let F s,r (X) := N(X, r)r s = inf{nr s : covers of X by n balls of radius r, n ∈ N} and set F s (X) := lim supr→0 F s,r (X). This quantity has the following properties: (i) dimB (X) = inf{s ≥ 0 : F s (X) = 0}; (ii) Hs (X) ≤ F s (X), where Hs is the s-dimensional Hausdorff measure. Since the inequality in (ii) can be strict, the following result improves on the bound in Lemma 2.12 and on the result of Scheffer [34] that H1/2 (T ) = 0.

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163

Proposition 2.14 F 1/2 (T ) = 0. Proof Fix ε > 0 and choose ti ∈ T such that (ti − r/3, ti + r/3) form a maximal family of disjoint intervals with centres in T . The expanded intervals (ti − r, ti + r) then form a cover O all of T . Since dimB (T ) ≤ 1/2 we can choose r small enough that the measure of O is small enough that  O

∇u2 < ε.

Since ti ∈ T we have 

ti ti −r/3

∇u(s)2 ds ≥ ε∗ (r/3)1/2 ;

since the intervals (ti − r/3, ti ) are disjoint we have ε∗ N(T , r/3)(r/3)

1/2

≤

 

≤

ti −r/3

i

O

ti

∇u2 ds

∇u(s)2 ds < ε 

and the result follows.

2.3 Serrin’s Local Regularity Result for u ∈ L5+ (Q∗ ) In this section we prove a local conditional regularity result, due to Serrin [37]. We take a solution that solves the Navier–Stokes equations only on some region U of space-time, and show that if u ∈ Lα (U ) for some α > 5 then in fact u is smooth in the spatial variables within U . We then give a sketch of the proof of the same result when u ∈ L5 (U ). [Two comments are in order here. First, Serrin’s result is more general, allowing for different integrability in space and time: if u ∈ Lr (a, b; Ls ()) with 2 3 + 0. The means that in space-time the ‘natural’ set to consider is not a ball in R4 , but rather the ‘cylinder’ Q∗r (x, t) = Br (x) × (t − r 2 /2, t + r 2 /2)

(2.25)

(we choose the factor 1/2 in the time direction for later convenience). By ‘solving the Navier–Stokes equations on U ’ we mean that u is a weakly divergence-free distributional solution of the equations on U , i.e. that4  u, ∂t ϕ" + u, ϕ" + u ⊗ u : ∇ϕ" dx dt = 0,

(2.26)

U

for every ϕ ∈ Dσ (U ), where Dσ (U ) := {ϕ ∈ [Cc∞ (U )]3 : ∇ · ϕ(x, t) = 0} and (u ⊗ v)ij = ui vj . In terms of ‘local solutions’ within parabolic cylinders, it is easy to see that if (u, p) solve the Navier–Stokes equations in Q∗1 (0), then the rescaled (uλ , pλ ) solve the equations on Q∗1/λ (0). In this section we give a (sketch) proof of the following local conditional regularity theorem. Theorem 3.1 Suppose that u solves the Navier–Stokes equations on Q∗r (x, t) and u ∈ Lα (Q∗r (x, t)) for some α > 5. Then in fact u is smooth in the spatial variables in Q∗r/2 (x, t). The following is an almost immediate corollary. Note that in particular this shows that the above theorem is true with the conclusion holding in Q∗r (x, t). Corollary 3.2 Let U be an open set in space-time, and u a solution of the Navier– Stokes equations on U such that u ∈ Lα (U ) with α > 5. Then u is smooth in the spatial variables within U . Proof Take (x, t) ∈ U , choose r > 0 such that Q∗r (x, t) ∈ U , and apply Theorem 3.1 to show that u is smooth in the spatial variables within Q∗r/2 (x, t).  We will show that if u ∈ Lα (Q∗1 ) then u is spatially smooth in Q∗1/2 ; Theorem 3.1 then follows using the rescaling in (2.24).

4 Here

we use a colon for the matrix product, i.e. A : B =

3

i,j =1 Aij Bij .

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165

The key idea is to work with the vorticity form of the Navier–Stokes equations: if u is smooth then we can define ω = ∇ × u (the vorticity), and then ω satisfies the vorticty equation5 ωt − ω + (u · ∇)ω − (ω · ∇)u = 0. If we rewrite this as ωt − ω = (ω · ∇)u − (u · ∇)ω = div(ω ⊗ u − u ⊗ ω)

(2.27)

then we have recast the equation as a heat equation for ω, although admittedly the right-hand side depends on ω and on u (which itself also depends on ω—we will discuss recovering u from ω later). [With a little care, it is possible to show something similar based on the formulation of what it means for u to be a ‘local weak solution’ in (2.26): for any test function φ ∈ [Cc∞ (U )]3 we take ϕ = curl φ, which is divergence free and so an element of Dσ (U ). We can therefore use this ϕ as a test function in (2.26) to give  u, ∂t (∇ × φ)" + u, (∇ × φ)" + u ⊗ u, ∇(∇ × φ)" dx dt = 0, U

which after manipulations similar to those leading from the Navier–Stokes equations to (2.27) yields the weak form of the vorticity equation  ω, ∂t φ" + ω, φ" + (u ⊗ ω) − (ω ⊗ u), ∇φ" dx dt = 0, U

see RRS for details.] The formulation in (2.27) is convenient, since it does not contain the pressure and properties of solutions of the heat equation are well understood. In particular if we set  2 ct −3/2 e−|x| /4t t > 0 K(x, t) = 0 otherwise

5 The vorticity equation follows on taking the curl of the Navier–Stokes equations and using the two vector identities

1 ∇|u|2 = (u · ∇)u + u × ω 2

and

∇ × (a × b) = a(∇ · b) − b(∇ · a) + (b · ∇)a − (a · ∇)b

along with the fact that both u and ω are divergence free.

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J. C. Robinson

(K is the ‘heat kernel’) and the support of f is contained in Q∗r (a, s) then K & f , defined by setting  K & f (x, t) :=

s+r 2 /2  s−r 2 /2

K(x − ξ, t − τ )f (ξ, τ ) dξ dτ, B(a,r)

is the solution for (x, t) ∈ Q∗r (a, s) of ωt − ω = f (x, t)

with ω(x, s − r 2 /2) = 0.

We can use this to write down a representation formula for ω within Q∗r , r < 1, in terms of g = ω ⊗ u − u ⊗ ω: ω = K & (div g) + H (x, t) = ∇K & g + H (x, t),

(2.28)

where H (x, t) is a solution of the heat equation (and hence smooth in both space and time). The term H (x, t) therefore plays only a minimal role in what follows. [Note that there are also some subtleties here that we have ignored. For example, are u and ω (and so g) actually smooth enough to use the representation in (2.28)? Serrin’s ‘rigorous’ derivation is very short on details, but the idea is relatively standard: find a way to mollify Eq. (2.26) and take limits. One method is to start with a test function φ ∈ Dσ and then use ϕ = ψε ∗ φ in (2.26), where ψε is a standard mollifying function.] In order to exploit this representation formula, we will use Young’s inequality for convolutions, which we recall here. Note that r = ∞ is included: we obtain f ∗ g ∈ L∞ if p−1 + q −1 ≤ 1, where ∗ denotes convolution over the whole space. Lemma 3.3 (Young’s Inequality) Let 1 ≤ p, q, r ≤ ∞ satisfy 1 1 1 +1= + . r p q Then for all f ∈ Lp , g ∈ Lq , we have f ∗ g ∈ Lr with f ∗ gLr ≤ f Lp gLq .

(2.29)

2.3.1 Step 1: Show that ω ∈ L∞ (Q∗s ), s < 1 If we apply Young’s inequality using ∇K as the kernel, then we obtain the following, which is fundamental to the first part of Serrin’s argument. In the statement we use Q∗ρ to denote any space-time cylinder.

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Lemma 3.4 If g ∈ Lq (Q∗ρ ) for some q ≤ 5 then ∇K & gLr (Q∗ρ ) ≤ cgLq (Q∗ρ ) for any r such that 1 1 1 > − ; r q 5

(2.30)

if g ∈ Lq (Q∗ρ ) for some q > 5 then ∇K & g ∈ L∞ (Q∗ρ ). Proof We have ∇K ∈ Lp ((0, ρ 2 ) × B2r (0)) for any 1 ≤ p < 5/4. To see this, first observe that |∇K(x, t)| ≤ c|x|t −5/2e−|x|

2 /4t

.

Now one can calculate p ∇KLp ((0,ρ 2)×B (0) 2ρ



ρ2

≤



0

c|x|p t −5p/2 e−p|x|

2 /4t

B2ρ (0)



ρ2

=c

t

−5p/2 p/2+3/2

4

t

0

≤ c



ρ2

dx dt 5 p −|y|2

Bρ √2p/t (0)

|y| e

dy

dt

t −2p+(3/2) dt < ∞,

0

provided that 1 ≤ p < 5/4. The result now follows from Lemma 3.3.



Corollary 3.5 If ω ∈ Lρ (Q∗r ) and u ∈ Lα (Q∗r ), then if 1 1 1 + ≥ α ρ 5

(2.31)

1 1 1 1 > + − . σ α ρ 5

(2.32)

1 1 1 + < α ρ 5

(2.33)

we have ω ∈ Lσ (Q∗r ) provided that

If

then ω ∈ L∞ (Q∗r ).

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Proof Since u ∈ Lα (Q∗r ) and ω ∈ Lρ (Q∗r ) we have 

g = u ⊗ ω + ω ⊗ u ∈ Lρ (Q∗r ) with 1 1 1 = + .  ρ α ρ Now we can use Lemma 3.4 to show that if ρ  ≤ 5, i.e. under condition (2.31), then ω = (∇K & g) + H ∈ Lσ (Q∗r ) for any σ satisfying (2.32). Similarly, if ρ  > 5, i.e. under condition (2.33), it follows that ω ∈ L∞ (Q∗r ). 

If u is a weak solution we know that u ∈ L2 (0, T ; H 1); it follows that ∇u ∈ so in particular ω ∈ L2 (Q∗1 ). We now iterate the above argument to improve the regularity of ω and show that ω ∈ L∞ (Q∗r ) for every r < 1. L2 (0, T ; L2 ),

Proposition 3.6 If ω ∈ L2 (Q∗r ) and u ∈ Lα (Q∗r ) for some α > 5 then in fact ω ∈ L∞ (Q∗r ). Proof We apply Corollary 3.5 repeatedly: it is enough to show that ω ∈ Lρ (Q∗r ) with ρ satisfying (2.33), since the next step of the iteration then shows that ω ∈ L∞ (Q∗r ). If ω ∈ Lρn (Q∗r ) then Corollary 3.5 implies that while α −1 + ρn−1 ≥ 5−1 we have ω ∈ Lρn+1 (Q∗r ) if # $ # $ 1 1 1 1 1 1 + + − > − . α ρn+1 α ρn 5 α Since α > 5 this shows that α −1 + ρn−1 decreases by a constant factor (anything less than 1/5 − 1/α) with each iteration. It follows that after a finite number of iterations condition (2.33) will be satisfied, and then u ∈ L∞ (Q∗r ) as claimed. 

[As an example, if we assume that u ∈ L∞ (Q∗r ) and start with ω ∈ L2 (Q∗r ) then on the first iteration we obtain ω ∈ Lp for any p < 10/3; on the second ω ∈ Lp for any p < 10; and then on the third ω ∈ L∞ .]

2.3.2 Show that u ∈ Cx∞ (Q∗s ) for any s < 1 Proving smoothness of u is based on an iterative process consisting of two stages. It turns out that we have to keep track of the smoothness in time and in space separately. To do this we adopt the notation Lt Lx (Q∗r ) p

q

and

Lt Cxα (Q∗r ) p

2 Partial Regularity for the 3D Navier–Stokes Equations

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to consist of functions f in Lp ((−r 2 /2, r 2 /2); Lq (Br ))

Lp ((−r 2 /2, r 2 /2); C α (Br )),

and

respectively. The first stage in this iteration is the reconstruction of u from ω via the Biot– Savart Law: if we take the curl of the relation ω = ∇ × u then we obtain − u = ∇ × ω,

(2.34)

where we have used the vector identity ∇ × (∇ × u) = ∇(∇ · u) − u and the fact that u is divergence free. On the whole space it follows, by inverting the Laplacian in (2.34), that u=−

1 4π

 R3

(x − y) × ω(y) dy; |x − y|3

if we only know ω on a subset U of R3 then instead we have u(x) = −

1 4π

 U

(x − y) × ω(y) dy + H (x), |x − y|3

(2.35)

where H is harmonic (H = 0). For proofs of these facts see Chapter 12 in RRS. As a consequence of this representation formula, it is possible to show (in line with what one would guess from (2.34)) that u is ‘one derivative better’ than ω, i.e. ω ∈ L∞ (t1 , t2 ; C k,α (BR ))

⇒



u ∈ L∞ (t1 , t2 ; C k+1,α (BR  ))

(2.36)

for R  < R and for any 0 < α  < α (see Theorem 12.6 in RRS). In order to improve the regularity of ω, since we know that ω satisfies ωt − ω = div(ωu − uω) we can appeal to ‘standard’ results for the heat equation: if in Q∗R

(2.37)

0,α ∗ ω ∈ L∞ t Cx (QR )

(2.38)

ωt − ω = divg then for any R  < R and any 0 < α < 1 we have ∞ ∗ g ∈ L∞ t Lx (QR )

⇒

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J. C. Robinson

and (by considering derivatives of the Eq. (2.37)) ∞ ∗ ∂ k g ∈ L∞ t Lt (QR )

⇒

k,α ∗ ω ∈ L∞ t Cx (QR  ),

(2.39)

∞ where by ∂ k g ∈ L∞ t Lx we mean that all derivatives of order up to k are bounded k (note that this is a little weaker than ω ∈ L∞ t Cx ); we also have k,α ∗ g ∈ L∞ t Cx (QR )

⇒

∞ ∗ ∂ k+1 ω ∈ L∞ t Lx (QR  ).

(2.40)

[Full proofs can be found in Appendix D of RRS.] We now use these to reach a degree of regularity at which we can establish a more systematic induction (throughout what follows we take α, α  with 0 < α  < α < 1) and R > R  > R  > R  > R (4) > · · · > R/2: ∞ ∗ u, ω ∈ L∞ t Lx (QR )

⇒

∞ ∗ ωu − uω ∈ L∞ t Lx (QR )

∞ ∗ ωu − uω ∈ L∞ t Lx (QR )

⇒

0,α ∗ ω ∈ L∞ t Cx (QR ) 1,α

by (2.38)

0,α ∗ ω ∈ L∞ t Cx (QR )

⇒

u ∈ L∞ t Cx

(Q∗R )

0,α ∗ u, ω ∈ L∞ t Cx (QR )

⇒

0,α ∗ ωu − uω ∈ L∞ t Cx (QR )

0,α ∗ ωu − uω ∈ L∞ t Cx (QR )

⇒

∞ ∗ ∂ω ∈ L∞ t Lx (QR  )

by (2.36)

by (2.39).



1,α ∗ ∗ ∞ ∞ We now have u ∈ L∞ t Cx (QR  ) and ∂ω ∈ Lt Lx (QR  ) and from here the k,α k ∞ ∞ ∗ induction process to improve from u ∈ L∞ t Cx , ∂ ω ∈ Lt Lx on QR (2k+1) to 

k+1,α ∗ ∞ , ∂ k+1 ω ∈ L∞ u ∈ L∞ t Cx t Lx on QR (2k+3) is more systematic:



k ∗ u ∈ L∞ t Cx (QR  ) k ∞ ∞ ∂ ω ∈ Lt Lx (Q∗R  )

-

∞ ∗ ∂ k (ωu − uω) ∈ L∞ t Lx (QR  ) k,α 

u, ω ∈ L∞ t Cx

⇒

∞ ∗ ∂ k (ωu − uω) ∈ L∞ t Lx (QR  )

⇒

k,α ∗ ω ∈ L∞ t Cx (QR  ) k,α 

by (2.39)

(Q∗R  )

⇒

ωu − uω ∈ L∞ t Cx

k,α ∗ ωu − uω ∈ L∞ t Cx (QR  )

⇒

∞ ∗ ∂ k+1 ω ∈ L∞ t Lx (QR  )

k,α ∗ ω ∈ L∞ t Cx (QR  )

⇒

k+1,α u ∈ L∞ (Q∗R  ) t Cx





(Q∗R  ) by (2.40) by (2.36).

k It is clear that continuing in this way leads to u ∈ L∞ t Cx for every k ∈ N, and so u is spatially smooth as claimed. Note, in particular, that the bounds on the spatial derivatives of u are uniform with respect to time within Q∗R/2 . This will be useful in the final proof in these notes (of the almost-everywhere uniqueness of particle trajectories), where this will ensure that u is Lipschitz continuous in space with a Lipschitz constant that is uniform in time.

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2.3.3 The Case u ∈ L5 (Q∗) The following argument6 gives an indication of how one can prove regularity for the boundary case u ∈ L5 (and more generally when u ∈ Lr (a, b; Ls ()) with equality in the condition (2.23)). The first observation is that it is sufficient to prove the result under a smallness condition on uL5 (Q∗r ) , since if u ∈ L5 (U ) then it follows that for any ε > 0, for each point (x, t) there is an r > 0 such that uL5 (Q∗r ) < ε. Starting from the equality ω = ∇K & g + H (x, t),

(2.41)

choose any r < ∞. Then, recalling that g = ωu − uω, we have gLm  uL5 ωLr ,

where

1 1 1 = + , m 5 r

and so using Lemma 3.4 we obtain7 ωLr ≤ CuL5 ωLr + H Lr .

(2.42)

If uL5 < ε := 1/2C then this inequality yields ωLr ≤ 2H Lr .

(2.43)

The Biot–Savart Law in (2.35) allows us to recover u from ω (at each fixed time t); since this is essentially a convolution with a kernel of the order of |x|−2 at infinity, the kernel is in the weak Lebesgue space L3/2,∞ . Away from the endpoint values (1 < p, q, r < ∞) Young’s convolution inequality (2.29) still holds when the Lp norm on the right-hand side replaced by the norm in Lp,∞ , f ∗ gLr  f Lp,∞ gLq ,

6 This is the basis of the proof given in the paper by Takahashi [39], although rather than following exactly the argument here he works with the equation for φω, where φ is a cutoff function. 7 Here our argument has what is potentially a fatal flaw: in the inequality (2.42) we have assumed / Lr since that ω ∈ Lr , which is in fact what we want to prove (the inequality is trivially true if ω ∈ then both sides are infinite). However, this can be circumvented by considering estimates for the equation

∂t W ε − W ε = div(W ε uε − uε W ε ), where uε is a mollified version of the original function u, and then taking limits as ε → 0.

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J. C. Robinson

and so it follows that if ω ∈ Lx for some q > 3 then u ∈ L∞ x with q

uL∞  ωLq . In particular if we use (2.43) to bound ω in L6 (U ) (for example) then for each t we have uL∞  ωL6 and hence 6 6 u ∈ L∞ t Lx ⊂ L (U ).

This finishes our sketch proof, since we have already shown that if u ∈ L6 (U ) then u is spatially smooth in U .

2.4 Space-Time Partial Regularity We will now develop—following Caffarelli et al. [3]—a theory that to some extent parallels that for regular times, but now for space-time points. We will call a spacetime point (x, t) is regular if it has a space-time neighbourhood U such that u ∈ L∞ (U ). It then follows from the results of Serrin sketched in the previous section that u is smooth (in space, at least) in U . If (x, t) is not regular then it is singular, and we set S := {(x, t) : T3 × (0, ∞) : (x, t) is a singular point}. We again respect the ‘parabolic scaling’ of the Eq. (2.24); we now take our basic domain to be the ‘non-anticipating’ parabolic cylinder Qr (a, s) := {(x, t) : |x − a| < r, s − r 2 < t < s}. Note that (a, s) is the ‘centre-right’ point of Qr (a, s) and is not contained in Qr (a, s). We have Q∗r (a, s) = Qr (a, s + r 2 /2), where Q∗r is the centred cylinder we defined earlier in (2.25). We will show that there exists an absolute constant ε0 > 0 such that if for some r > 0 we have  1 |u|3 + |p|3/2 < ε0 r 2 Qr (a,s) then u is bounded on Qr/2 (a, s).

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2.4.1 The Navier–Stokes Inequality The Navier–Stokes equations themselves require two equalities: one expressing conservation of momentum and the other the incompressibility condition. We now replace the Navier–Stokes ‘equality’ by the following inequality: u · (∂t u − u + (u · ∇)u + ∇p) ≤ 0. This is essentially the pointwise form of the energy equality, changed to an inequality. If we require this equation to hold weakly, then we obtain (iii) of the following definition. We retain the incompressibility condition in (i) and the relation between the pressure and u is given in (ii). Definition 4.1 A pair (u, p) is a weak solution of the Navier–Stokes inequality (NSI) on U × (t1 , t2 ) if (i) u ∈ L∞ (t1 , t2 ; L2 (U )), ∇u ∈ L2 (U × (t1 , t2 )) with ∇ · u = 0 almost everywhere, and p ∈ L5/3 (U × (t1 , t2 )); (ii) −p = ∂i ∂j (ui uj ) for a.e. t ∈ (t1 , t2 ); and (iii) the local energy inequality  |u(t)| ϕ + 2 2

U

 t t1

|∇u| ϕ ≤ 2

U

 t t1

|u| (∂t ϕ + ϕ) + 2

U

 t t1

(|u|2 + 2p)u · ∇ϕ U

is valid for every t ∈ (t1 , t2 ) for every choice of non-negative test function ϕ ∈ Cc∞ (U × (t1 , t2 )). Exercise Derive the local energy equality under the assumption that u is smooth. CKN prove something very like the following theorem, although their condition on the pressure is a little different since they only assumed the regularity p ∈ L5/4 (U × (t1 , t2 )), which was the best known to hold in bounded domains when they wrote their paper. Note that the theorem does not require (u, p) to be a weak solution of the Navier–Stokes equations, only of the Navier–Stokes inequality. Theorem 4.2 There exists an absolute constant ε0 > 0 such that if (u, p) is a weak solution of the Navier–Stokes inequality on Q1 (0, 0) and  |u|3 + |p|3/2 ≤ ε0 Q1 (0,0)

then u ∈ L∞ (Q1/2 (0, 0)).

(2.44)

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2.4.2 Partial Regularity It is easy to show that if (u, p) solves the NSI on Qr (0, 0) then (ur , pr ), where ur (x, t) = ru(rx, r 2 t)

and

pr (x, t) = r 2 p(rx, r 2 t),

satisfies the NSI on Q1 and   1 |ur |3 + |pr |3/2 = 2 |u|3 + |p|3/2 . r Qr Q1 This observation yields a more useful version of Theorem 4.2. Theorem 4.3 If (u, p) is a weak solution of the Navier–Stokes inequality on Qr and  1 |u|3 + |p|3/2 < ε0 r 2 Qr then u ∈ L∞ (Qr/2). From here we can easily obtain an upper bound on the dimension of the singular set; but it is useful first to recast Theorem 4.3 in what is actually a slightly weaker form. Note that since (x, t) ∈ / Qr (x, t), Theorem 4.3 does not actually give a condition to guarantee that (x, t) is a regular point, but this is easily remedied. Recall that previously we defined the ‘centred’ cylinder Q∗r (x, t) about (x, t) to be Q∗r (x, t) = Br (x) × (t − r 2 /2, t + r 2 /2); we can deduce the following from Theorem 4.3. Corollary 4.4 If (u, p) is a weak solution of the Navier–Stokes inequality on Q∗r (x, t) and 1 r2

 Q∗r (x,t )

|u|3 + |p|3/2 < ε0 /4

then u ∈ L∞ (Q∗r/4(x, t)); in particular, (x, t) is a regular point. Proof Since Q∗r (x, t) = Qr (x, t + r 2 /2) ⊃ Qr/2 (x, t + r 2 /32) it follows from (2.45) that 1 (r/2)2

 Qr/2 (x,t +r 2 /32)

|u|3 + |p|3/2 < ε0 ,

(2.45)

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and we can use Theorem 4.3 to guarantee that u ∈ L∞ (Qr/4(x, t + r 2 /32)). Since Qr/4 (x, t + r 2 /32) = Q∗r/4 (x, t) 

the theorem follows as stated.

In order to use this result to bound the dimension of the singular set we first obtain some space-time integrability of u and p from the regularity we know for weak solutions, namely u ∈ L∞ (0, T ; L2 ) ∩ L2 (0, T ; L6 ) (since H 1 ⊂ L6 ). We use Lebesgue interpolation to show that 2/5

3/5

uL10/3 (U ) ≤ uL2 (U ) uL6 (U ) and hence 10/3

4/3

uL10/3 (U ) ≤ uL2 (U ) u2L6 (U ) , from which it follows that 

T 0

 |u|10/3 < ∞, U

i.e. that u ∈ L10/3(U × (0, T )). Note that p ∈ L5/3 (U × (0, T )) by assumption. Now, using Hölder’s inequality we have 

 |u| ≤ Qr

-1/10

-9/10  |u|

3

10/3

1

Qr

Qr

which yields 

|u|10/3  r −5/9



-10/9 |u|3

Qr

.

Qr

At an irregular point we cannot have (2.45), and so either 

 |u|10/3  r 5/3

Qr

|p|5/3  r 5/3 .

or

(2.46)

Qr

[The second lower bound on the pressure term follows in a similar way, again using Hölder’s inequality and (2.45).] We now combine these for a dimension estimate; one can find the basis of the following argument in CKN, although they treat only the Hausdorff dimension.

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Proposition 4.5 ([30]) If (u, p) is a weak solution of the Navier–Stokes inequality on U × (a, b) and S denotes the set of space-time singularities then dimB (S ∩ K) ≤ 5/3 for any compact subset K of U × (a, b). Proof Suppose that d = dimB (S ∩ K) > 5/3 and take δ with 5/3 < δ < d. Then (j ) there exists a sequence j → 0, so that for each j there is a maximal collection {zn } of at least j−δ points in S ∩ K that are 2 j separated. For each j it follows that the j -balls centred at the zn are disjoint, and from Corollary 4.4 and (2.46)—assuming that the lower bound on the u integral holds for infinitely many points—we have 

 |u|10/3 ≥ K

j

Q∗

|u|10/3 

5/3 −δ j j ,

j

which on letting j → ∞ contradicts the fact that u ∈ L10/3 (U × (a, b)). A similar contradiction occurs if the lower bound on the p integral holds for infinitely many points, since p ∈ L5/3 (U × (a, b)). 

Lemma 2.11 now implies that almost-every space-time point is regular. Exercise Suppose that (u, p) is a weak solution of the Navier–Stokes inequality on R3 . Show that the set of space-time singularities is bounded. (Use a contradiction argument again.)

2.4.3 Idea of the Proof of Theorem 4.2 In order to prove Theorem 4.2 we will use an inductive argument to show that for every (a, s) ∈ Q1/2 (0, 0) and every rn := 2−n we have 1 rn

 sup s−rn2 0 and let  × (0, T ) = ∞ n=1 Kn with Kn compact. If P denotes the projection onto the spatial component then dimB (P [S ∩ Kn ]) ≤ dimB (S ∩ Kn ) ≤ 5/3; we have used the fact that dimB is non-increasing under the Lipschitz mapping P (Lemma 5.5) and the bound on the box-counting dimension of S from Theorem 4.5. Theorem 5.4 now guarantees that almost every trajectory avoids the set P (S∩Kn ) for all t ≥ 0, and so in particular avoids S∩Kn for every n, so almost every trajectory avoids S. 

Finally, we can deduce that the Lagrangian trajectories are unique for almost every initial condition. To do this we need to use the fact that u is smooth away

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J. C. Robinson

from the space-time singular set, which follows from the local regularity result due to Serrin from Sect. 2.3: if u ∈ L∞ (U ×(a, b)) then in fact u is smooth in the spatial variables on U × (a, b). By a suitable weak solution we mean a Leray–Hopf weak solution that is also a weak solution of the Navier–Stokes inequality. Corollary 5.7 If u is a suitable weak solution corresponding to an initial condition u0 ∈ V then almost every initial condition a ∈ T3 gives rise to a unique particle trajectory. Proof Taking u0 ∈ H 1 (T3 ) implies that trajectories are unique on some [0, T ), so the issue is uniqueness for t ≥ T , in which range of times the partial regularity results of Sect. 2.4 serve to limit the dimension of the set of space-time singularities. Let Xa (·) be a trajectory that avoids the singular set for all t ≥ 0, and suppose (for a contradiction) that there are two trajectories that pass through the space-time point (Xa (t), t) for some particular t > 0. Since (Xa (t), t) ∈ / S it is a regular point; by definition this means that u ∈ L∞ (U ) for some space-time neighbourhood U = W × (t − ε, t + ε) of (Xa (t), t), and then the result of Serrin in Theorem 3.1 guarantees that u is (spatially) smooth in U , and in particular Lipschitz continuous on W with a constant that is uniform over (t − ε, t + ε). It follows (see the discussion at the beginning of this section) that the solution of X˙ = u(X, t) is unique at (Xa (t), t), so there cannot be two trajectories passing through this point. 

References 1. T. Buckmaster, V. Vicol, Nonuniquenness of weak solutions to the Navier–Stokes equation. arXiv:1709.10033 (2017) 2. T. Buckmaster, C. De Lellis, L. Székelyhidi Jr., V. Vicol, Onsager’s conjecture for admissible weak solutions. arXiv:1701.08678 (2017) 3. L. Caffarelli, R. Kohn, L. Nirenberg, Partial regularity of suitable weak solutions of the Navier– Stokes equations. Comm. Pure. Appl. Math. 35, 771–931 (1982) 4. J.-Y. Chemin, B. Desjardins, I. Gallagher, E. Grenier, Mathematical Geophysics (Oxford University Press, Oxford, 2006) 5. P. Constantin, C. Foias, Navier–Stokes Equations (University of Chicago Press, Chicago, 1988) 6. P. Constantin, E. Weinan, E.S. Titi, Onsager’s conjecture on the energy conservation for solutions of Euler’s equation. Commun. Math. Phys. 165, 207–209 (1994) 7. M. Dashti, J.C. Robinson, A simple proof of uniqueness of the particle trajectories for solutions of the Navier–Stokes equations. Nonlinearity 22, 735–746 (2009) 8. L. Escauriaza, G. Seregin, V. Šverák, L3,∞ -solutions of Navier–Stokes equations and backward uniqueness. Russ. Math. Surv. 58, 211–250 (2003) 9. L.C. Evans, Partial Differential Equations (American Mathematical Society, Providence, 2010) 10. E.B. Fabes, B.F. Jones, N.M. Rivière, The initial value problem for the Navier–Stokes equations with data in Lp . Arch. Ration. Mech. Anal. 45, 222–240 (1972) 11. K.J. Falconer, Fractal Geometry (Wiley, Chichester, 1990) 12. C.L. Fefferman, Existence and smoothness of the Navier–Stokes equation, in The Millennium Prize Problems (Clay Mathematics Institute, Cambridge, 2000), pp. 57–67

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13. C. Foias, C. Guillopé, R. Temam, New a priori estimates for Navier–Stokes equations in dimension 3. Comm. Partial Diff. Equ. 6, 329–359 (1981) 14. C. Foias, C. Guillopé, R. Temam, Lagrangian representation of a flow. J. Diff. Equ. 57, 440– 449 (1985) 15. G.P. Galdi, An introduction to the Navier–Stokes initial-boundary value problem, in ed. by G.P. Galdi, J.G. Heywood, R. Rannacher. Fundamental Directions in Mathematical Fluid Dynamics (Birkhäuser, Basel, 2000), pp. 1–70 16. P. Hartman, Ordinary Differential Equations (Wiley, Baltimore, 1973) 17. E. Hopf, Über die Aufgangswertaufgave für die hydrodynamischen Grundliechungen. Math. Nachr. 4, 213–231 (1951) 18. P. Isett, A proof of Onsager’s conjecture. Ann. Math. 188, 871–963 (2018) 19. H. Jia, V. Šverák, Are the incompressible 3d Navier–Stokes equations locally ill-posed in the natural energy space? J. Func. Anal. 268, 3734–3766 (2015) 20. I. Kukavica, Partial regularity results for solutions of the Navier–Stokes system, in ed. by J.C. Robinson, J.L. Rodrigo, Partial Differential Equations and Fluid Mechanics (Cambridge University Press, Cambridge, 2009), pp. 121–145 21. I. Kukavica, The fractal dimension of the singular set for solutions of the Navier–Stokes system. Nonlinearity 22, 2889–2900 (2009) 22. J. Leray, Essai sur le mouvement d’un liquide visqueux emplissant l’espace. Acta Math. 63, 193–248 (1934) 23. F. Lin, A new proof of the Caffarelli–Kohn–Nirenberg theorem. Comm. Pure Appl. Math. 51, 241–257 (1998) 24. W. O˙za´nski, The Partial Regularity Theory of Caffarelli, Kohn, and Nirenberg and Its Sharpness. Lecture Notes in Mathematical Fluid Mechanics (Birkhäuser/Springer, 2019) 25. W. O˙za´nski, B. Pooley, Leray’s fundamental work on the Navier–Stokes equations: a modern review of “Sur le mouvement d’un liquid visqueux emplissant l’espace”. in ed. by C.L. Fefferman, J.L. Rodrigo, J.C. Robinson. Partial Differential Equations in Fluid Mechanics. LMS Lecture Notes (Cambridge University Press, Cambridge, 2018) 26. A.P. Robertson, On measurable selections. Proc. Roy. Soc. Edinburgh Sect. A 73, 1–7 (1974) 27. J.C. Robinson, Infinite-Dimensional Dynamical Systems (Cambridge University Press, Cambridge, 2001) 28. J.C. Robinson, Dimensions, Embeddings, and Attractors (Cambridge University Press, Cambridge, 2011) 29. J.C. Robinson, W. Sadowski, Decay of weak solutions and the singular set of the threedimensional Navier–Stokes equations. Nonlinearity 20, 1185–1191 (2007) 30. J.C. Robinson, W. Sadowski, Almost everywhere uniqueness of Lagrangian trajectories for suitable weak solutions of the three-dimensional Navier–Stokes equations. Nonlinearity 22, 2093–2099 (2009) 31. J.C. Robinson, J.L. Rodrigo, W. Sadowski, The Three-Dimensional Navier–Stokes Equations (Cambridge University Press, Cambridge, 2016) 32. V. Scheffer, Turbulence and Hausdorff dimension, in Turbulence and Navier–Stokes Equation, Orsay 1975. Springer Lecture Notes in Mathematics, vol. 565 (Springer, Berlin, 1976), pp. 174–183 33. V. Scheffer, Partial regularity of solutions to the Navier–Stokes equations. Pacific J. Math. 66, 535–552 (1976) 34. V. Scheffer, Hausdorff measure and the Navier–Stokes equations. Comm. Math. Phys. 55, 97– 112 (1977) 35. V. Scheffer, The Navier–Stokes equations on a bounded domain. Comm. Math. Phys. 73, 1–42 (1980) 36. V. Scheffer, Nearly one-dimensional singularities of solutions to the Navier–Stokes inequality. Comm. Math. Phys. 110, 525–551 (1987)

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37. J. Serrin, On the interior regularity of weak solutions of the Navier–Stokes equations. Arch. Rat. Mech. Anal. 9, 187–195 (1962) 38. M. Struwe, On partial regularity result for the Navier–Stokes equations. Comm. Pure Appl. Math. 41, 437–458 (1988) 39. S. Takahashi, On interior regularity criteria for weak solutions of the Navier–Stokes equations. Manuscripta Math. 69, 237–254 (1990)

Chapter 3

R Boundedness, Maximal Regularity and Free Boundary Problems for the Navier Stokes Equations Yoshihiro Shibata

3.1 Introduction 3.1.1 Free Boundary Problem for the Navier–Stokes Equations In this chapter, we study free boundary problem for the Navier–Stokes equations, and R bounded solution operators and Lp –Lq maximal regularity theorem for the Stokes equations with free boundary conditions. Typical problem for the free boundary problem of the Navier–Stokes equations are (P1) the motion of an isolated liquid mass and (P2) the motion of a viscous incompressible fluid contained in an ocean of infinite extent. The mathematical problem for the free boundary problem of the Navier– Stokes equations is to find a time dependent domain Ωt , t being time variable, in the N-dimensional Euclidean space RN , the velocity vector field, v(x, t) = ) (v (x, t), . . . , v (x, t)), where ) M denotes the transposed M, and the pressure 1 N

Y. Shibata () Department of Mathematics, Waseda University, Tokyo, Japan © Springer Nature Switzerland AG 2020 G. P. Galdi, Y. Shibata (eds.), Mathematical Analysis of the Navier-Stokes Equations, Lecture Notes in Mathematics 2254, https://doi.org/10.1007/978-3-030-36226-3_3

193

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Y. Shibata

field p = p(x, t) satisfying the Navier–Stokes equations in Ωt : ⎧ ∂t v + (v · ∇)v − Div (μD(v) − pI) = 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ div v = 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ (μD(v) − pI)nt = σ H (Γt )nt − p0 nt ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ Vn = v · nt ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ v|t =0 = v0 in Ω0 , Ωt |t =0 = Ω0 ,

"

in

Ωt × {t},

0 0 and by Abels [1] in the Lp SobolevSlobodetskii space when σ = 0. The global in time unique existence theorem for small initial velocities was proved in the L2 Sobolev-Slobodetskii space by Beale [5, 6] and Tani and Tanaka [72] in the case that σ > 0 and by Saito [38] in the Lp in time and Lq in space setting in the case that σ = 0. The decay rate was studied by Beale and Nishida [7] (cf. also [24] for the detailed proof), Lynn and Sylvester [27], Hataya [23] and Hataya-Kawashima [25] in the L2 framework with polynomial decay, and by Saito [38] with exponential decay. Recently, the local well-posedness in general unbounded domains was proved by Shibata [44] and [55] under the assumption that the initial domain is uniformly C 3 and the weak Dirichlet problem is uniquely solvable in the initial domain. To transform Eq. (3.1) to the problem in the reference domain, in [44] the Lagrange transform was used and in [55] the Hanzawa transform was used. Moreover, in the case that Ωt = {x ∈ RN | xN < η(x  , t) x  = (x1 , . . . , xN−1 ) ∈ RN−1 }, which is corresponding to the ocean problem without bottom physically, the global in time unique existence theorem for Eq. (3.1) was proved by Saito and Shibata [39, 40], and in the case that Ω is an exterior domain in RN (N ≥ 3), the local and global in time unique existence theorems were proved by Shibata [46, 48, 49]. In such unbounded domains case, we can only show that the Lq space norm of solutions of the Stokes equations with free boundary conditions decay polynomially, and so we have to choose different exponents p and q to guarantee the Lp integrability on time interval (0, ∞). This is one of the reasons why the maximal Lp –Lq regularity theory with freedom of choice of p and q is necessary in the study of the free boundary problems. In Sect. 3.8 below, it is explained how to prove the global in time unique existence theorem by combining the decay properties of the lower order terms and the maximal Lp –Lq regularity of the highest order terms. Finally, the author is honored to refer the readers to the lecture note written by Solonnikov, who

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203

is a pioneer of the study of free boundary problems of the Navier- Stokes equations, and Denisova [67], where it is given an overview of studies achieved by Solonnikov and his followers about the free boundary problems of the Navier–Stokes equations in a bounded domain and related topics. We remark that the two phase fluid flows separated by sharp interface problem has been studied by Abels [2], Denisova and Solonnikov [12–15], Giga and Takahashi [20, 69], Nouri and Poupaud [32], Pruess, Simonett et al. [26, 35–37, 58], Shibata and Shimizu [51], Shimizu [56, 57], Tanaka [70] and references therein.

3.1.4 Further Notation This section is ended by explaining further notation used in this lecture note. We denote the sets of all complex numbers, real numbers, integers, and natural numbers by C, R, Z, and N, respectively. Let N0 = N ∪ {0}. For any multi-index α = αN α1 α (α1 , . . . , αN ) ∈ NN 0 we set ∂x h = ∂1 · · · ∂N h with ∂i = ∂/∂xi . For any scalor function f , we write ∇f = (∂1 f, . . . , ∂N f ), ∇ n f = (∂xα f | |α| = n),

¯ = (f, ∂1 f, . . . , ∂N f ), ∇f ∇¯ n f = (∂xα f | |α| ≤ n)

(n ≥ 2),

where ∂x0 f = f . For any m-vector of functions, f = ) (f1 , . . . , fm ), we write ∇f = (∇f1 , . . . , ∇fm ),

¯ = (∇f ¯ 1 , . . . , ∇f ¯ m ), ∇f

∇ n f = (∇ n f1 , . . . , ∇ n fm ),

∇¯ n f = (∇¯ n f1 , . . . , ∇¯ n fm ).

For any N-vector of functions, u = ) (u1 , . . . , uN ), sometime ∇u is regarded as an N × N matrix of functions whose (i, j )th component is ∂i uj , that is ⎛

⎞ ∂1 u1 ∂2 u1 · · · ∂N u1 ⎜ ∂1 u2 ∂2 u2 · · · ∂N u2 ⎟ ⎜ ⎟ ∇u = ⎜ . .. . . .. ⎟ . . ⎝ . . . ⎠ . ∂1 uN ∂2 uN · · · ∂N uN For any m-vector V = (v1 , . . . , vm ) and n-vector W = (w1 , . . . , wn ), V ⊗ W denotes an (m, n) matrix whose (i, j )th component is Vi Wj . For any (mn, N) matrix A = (Aij,k | i = 1, . . . , m, j = 1, . . . , n, k = 1, . . . , N),  AV ⊗ W denotes n an N column vector whose ith component is the quantity: m j =1 j =1 Aj k,i vj wk . For any N vector a, ai denotes the ith component of a and for any N × N matrix A, Aij denotes the (i, j )th component of A, and moreover, the N × N matrix whose

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(i, j )th component is Kij is written as (Kij ). Let δij be the Kronecker delta symbol, that is δii = 1 and δij = 0 for i = j . In particular, I = (δij ) is the N × N identity N matrix. Let a · b =< a, b >= j =1 aj bj for any N-vectors a and b. For any N-vector a, let 0 a = aτ := a− < a, n > n. For any two N × N matrices A and B, the quantity A : B is defined by A : B = trAB = N i,j =1 Aij Bj i . Given  1 < q < ∞, let q = q/(q − 1). For L > 0, let BL = {x ∈ RN | |x| < L} and SL = {x ∈ RN | |x| = L}. Moreover, for L < M, we set DL,M = {x ∈ RN | L < |x| < M}. For any domain G in RN , let C0∞ (G) be the set of all C ∞ functions whose supports are compact and contained in G. Let (u, v)G = G u · v dx and (u, v)∂G =  u · v ds, where ds denotes the surface element on ∂G and ∂G is the boundary of ∂G G. For T > 0, G × (0, T ) = {(x, t) | x ∈ G, t ∈ (0, T )} is written simply by GT . s (G) be the standard Lebesgue, For 1 ≤ q ≤ ∞, let Lq (G), Hqm (G), and Bq,p s (G) Sobolev, and Besov spaces on G, and let  · Lq (G) ,  · Hqm (G) , and  · Bq,p 0 s denote their respective norms. We write Lq (G) as Hq (G), and Bq,q (G) simply as Wqs (G). For any Banach space X with norm  · X , let Xd = {(f1 , . . . , fd ) | fi ∈ d X (i = 1, as  · X , which is defined by . d. . , d)}, and write the norm of X simply f X = j =1 fj X for f = (f1 , . . . , fd ) ∈ Xd . For 1 ≤ p ≤ ∞, Lp ((a, b), X) and Hpm ((a, b), X) denote the standard Lebesgue and Sobolev spaces of X-valued functions defined on an interval (a, b), and their respective norms are denoted by  · Lp ((a,b),X) and  · Hpm ((a,b),X). For θ ∈ (0, 1), Hpθ (R, X) denotes the standard X-valued Bessel potential space defined by Hpθ (R, X) = {f ∈ Lp (R, X) | f Hpθ (R,X) < ∞}, 

1/p p F −1 [(1 + τ 2 )θ/2 F [f ](τ )](t)X dt , f Hpθ (R,X) = R

where F and F −1 denote the Fourier transform and the inverse Fourier transform, respectively. 1 (G) by setting For 1 ≤ q < ∞, we define homogeneous spaces Hˆ q1 (G) and Hˆ q,0 Hˆ q1 (G) = {u ∈ Lq,loc (Ω) | ∇u ∈ Lq (G)N }, 1 (G) = {u ∈ Lq,loc (Ω) | ∇u ∈ Lq (G)N , Hˆ q,0

u|∂G = 0},

For 1 < q < ∞, the solenoidal space Jq (G) on G of this chapter is defined by setting Jq (G) = {u ∈ Lq (G)N | (u, ∇ϕ)G = 0 for any ϕ ∈ Hˆ q1 ,0 (G)}.

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For two Banach spaces X and Y , X + Y = {x + y | x ∈ X, y ∈ Y }, L(X, Y ) denotes the set of all bounded linear operators from X into Y and L(X, X) is written simply as L(X). For a domain U in C, Hol (U, L(X, Y )) denotes the set of all L(X, Y )-valued holomorphic functions defined on U . Let RL(X,Y ) ({T (λ) | λ ∈ U }) be the R norm of the operator family T (λ) ∈ Hol (U, L(X, Y )). Let Σ

0

= {λ ∈ C \ {0} | | arg λ| ≤ π −

0 },

Σ

0 ,λ0

= {λ ∈ Σ

0

| |λ| ≥ λ0 },

C+,λ0 = {λ ∈ C | Re λ ≥ max(0, λ0 )}. N Let RN + = {x = (x1 , . . . , xN ) | xN > 0} and R0 = {x = (x1 , . . . , xN ) | xN = 0}. The letter C denotes a generic constant and Ca,b,c,··· denotes that the constant Ca,b,c,··· depends on a, b, c, · · · . The value of C and Ca,b,c,··· may change from line to line.

3.2 Preliminaries In this section, we study a uniform C k domain, a uniform C k domain whose inside has a finite covering, the weak Dirichlet problem, Besov spaces on the boundary, and the Laplace- Beltrami operator on the boundary.

3.2.1 Definitions of Domains We first introduce the definition of a uniform C k (k = 2 or 3) domain. Definition 3.2.1 Let k ∈ N. We say that Ω is a uniform C k domain, if there exist positive constants a1 , a2 , and A such that the following assertion holds: For any x0 = (x01, . . . , x0N ) ∈ Γ there exist a coordinate number j and a C k function h(x  ) defined on Ba 1 (x0 ) such that hH∞ k (B  (x  )) ≤ A and a1

0

Ω ∩ Ba2 (x0 ) = {x ∈ RN | xj > h(x  ) (x  ∈ Ba1 (x0 ))} ∩ Ba2 (x0 ), Γ ∩ Ba2 (x0 ) = {x ∈ RN | xj = h(x  ) (x  ∈ Ba 1 (x0 ))} ∩ Ba2 (x0 ). Here, we have set y  = (y1 , . . . , yj −1 , yj +1 , . . . , yN ) (y ∈ {x, x0}), Ba 1 (x0 ) = {x  ∈ RN−1 | |x  − x0 | < a1 }, Ba2 (x0 ) = {x ∈ RN | |x − x0 | < a2 }. The uniform C k domains are characterized as follows.

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Proposition 3.2.2 Let k = 2 or k = 3. Let Ω be a uniform C k domain in RN . Then, for any M1 ∈ (0, 1), there exist constants M2 > 0 and 0 < r0 < 1, at most countably many N-vector of functions Φj ∈ C k (RN )N and points xj0 ∈ Ω and xj1 ∈ Γ such that the following assertions hold: (i) The maps: RN $ x → Φj (x) ∈ RN are bijections satisfying the following conditions: ∇Φj = Aj + Bj , ∇(Φj )−1 = Aj,− + Bj,− , where Aj and Aj,− are N × N constant orthogonal matrices, and Bj and Bj,− are N × N matrices of C k−1 (RN ) functions defined on RN satisfying the conditions: (Bj , Bj,− )L∞ (RN ) ≤ M1 , and ∇(Bj , Bj,− )L∞ (RN ) ≤ CA , where CA is a constant depending on constants A, a1 and a2 appearing in Definition 3.2.1. Moreover, if k = 3, then ∇ 2 (Bj , Bj − )L∞ (RN ) ≤ M2 . 9

9

∞ ∞ 0 N 1 )) , B (x 0 ) ⊂ Ω, (ii) Ω = B (x ) ∪ (Φ (R ) ∩ B (x r j r r0 j + 0 0 j =1 j =1 j j 1 1 Φj (RN 0 ) ∩ Br0 (xj ) = Γ ∩ Br0 (xj ). (iii) There exist C ∞ functions ζji and ζ˜ji (i = 0, 1, j ∈ N) such that

0 ≤ ζji , ζ˜ji ≤ 1, supp ζji ⊂ supp ζ˜ji ⊂ Br0 (xji ), ζ˜ji = 1 on supp ζji , 1  ∞ 

ζji = 1 on Ω,

i=0 j =1

∞  j =1

ζj1 = 1 on Γ , ∇ζji H k−1 (RN ) , ∇ ζ˜ji H k−1 (RN ) ≤ M2 . ∞

∞

ij

(iv) Below, for the notational simplicity, we set Bji = Br0 (xji ). For each j , let #k i = ∅ for (k = 1, . . . , mij ) be numbers for which Bji ∩ B i ij = ∅ and Bji ∩ Bm ij

#k

m ∈ {#k | k = 1, . . . , mij }. Then, there exists an L ≥ 2 independent of M1 such that mij ≤ L. Proof Proposition 3.2.2 was essentially proved by Enomoto-Shibata [18, i Appendix], except for ∇(Bji , Bj,− )L∞ (RN ) ≤ CA . There, it was proved that i )L∞ (RN ) ≤ CM2 , ∇(Bji , Bj,− i ) depends on M2 . Since the proof of Proposithat is the estimate of ∇(Bji , Bj,− tion 3.2.2 is almost the same as in [18, Appendix], we shall give an idea how to improve this point below. Let x0 = (x0 , x0N ) ∈ Γ and we assume that

Ω ∩ Bβ (x0 ) = {x ∈ RN | xN > h(x  ) (x  ∈ Bα (x0 ))} ∩ Bβ (x0 ), Γ ∩ Bβ (x0 ) = {x ∈ RN | xN = h(x  ) (x  ∈ Bα (x0 ))} ∩ Bβ (x0 ). We only consider the case where k = 3. In fact, by the same argument, we can improve the estimate in the case where k = 2. We assume that h ∈ C 3 (Ba 1 (x0 )),

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

hH∞ 3 (B 

 a1 (x0 ))

207

≤ A, and x0N = h(x0 ). Below, C denotes a generic constant

depending on A, a1 and a2 but independent of . Let ρ(y) be a function in C0∞ (RN ) such that ρ(y) = 1 for |y  | ≤ 1/2 and |yN | ≤ 1/2 and ρ(y) = 0 for |y  | ≥ 1 or |yN | ≥ 1. Let ρ (y) = ρ(y/ ). We consider a C ∞ diffeomorphism: xj = Φj (y) = x0j +

N 

N 

tj,k yk +

k=1

sj,k# yk y# ρ (y).

k,#=1

Here, tj,k and sj,k# are some constants satisfying the conditions (3.22), (3.20), and (3.21), below. Let G (y) = ΦN (y) − h(Φ1 (y), . . . , ΦN−1 (y)). Notice that G (0) = x0N − h(x0 ) = 0. We choose tj,k and s#,mn in such a way that N−1  ∂h ∂G (0) = tN,N − (x  )tk,N = 0, ∂yN ∂xk 0 k=1

N−1  ∂h ∂G (0) = tN,j − (x  )tk,j = 0, ∂yj ∂xk 0

(3.20)

k=1

N−1  ∂h ∂ 2G (0) = sN,#m + sN,m# − (x  )(sk,#m + sk,m# ) ∂y# ∂ym ∂xk 0 k=1

−

N−1  j,k=1

(3.21)

∂ 2G (x  )tj,# tk,m = 0. ∂xj ∂xk 0

Moreover, setting ⎛

t1,1 t2,1 ⎜ t1,2 t2,2 ⎜ T =⎜ . .. ⎝ .. . t1,N t2,N

⎞ · · · tN,1 · · · tN,2 ⎟ ⎟ . ⎟, .. . .. ⎠ · · · tN,N

we assume that T is an orthogonal matrix, that is N  #=1

t#,m t#,n = δmn =

 1

for m = n,

0

for m = n.

(3.22)

208

We write

Y. Shibata

∂h  (x ) simply by hj and set ∂xj 0 < = N−1 = ?  = Hj = >1 + h2# = 1 + h2j + h2j +1 + · · · + h2N−1 . #=j

Let tN,N−j =

hN−j , HN−j HN+1−j

tN−k,N−j = −

hN−k hN−j HN−j HN+1−j

for k = 1, . . . , j − 1, and tN−j,N−j =

HN+1−j , HN−1

tk,N−j = 0,

for k = 1, . . . , N − j − 1 and j = 1, . . . , N − 1, and ti,N = −

hi , H1

tN,N =

1 H1

for i = 1, . . . , N − 1. Then, we see that such tj,k satisfy (3.20) and (3.22). In particular, ∂G 1 (0) = . ∂yN H1

(3.23)

Moreover, assuming the symmetry: s#,j k = s#,kj , we have sN,j k

N−1 ∂ 2h 1  = (x  )tm,j tn,k , 2H2 ∂xm ∂xn 0 m,n=1

si,j k

N−1 ∂ 2h hi  =− (x0 )tm,j tn,k . 2H12 m,n=1 ∂xm ∂xn

By successive approximation, we see that there exists a constant 0 > 0 such that for any ∈ (0, 0 ) there exists a function ψ ∈ C 3 (B  (0)) satisfying the following

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

209

conditions: ψ (0) = ∂i ψ (0) = ∂i ∂j ψ (0) = 0, ψ L∞ (B  (0)) ≤ C 2 , ∂i ∂j ψ L∞ (B  (0)) ≤ C,

∂i ψ L∞ (B  (0)) ≤ C , ∂i ∂j ∂k ψ L∞ (B  (0)) ≤ C

−1

,

G (y  , ψ (y  )) = 0 for y  ∈ B  (0),

(3.24)

where i, j and k run from 1 through N − 1. Notice that xN − h (x  ) = G (y) = G (y  , ψ (y  ))  1 (∂N G )(y  , ψ (y  ) + θ (yN − ψ (y  ))) dθ (yN − ψ (y  )) + 0

˜ (y))(yN − ψ (y  )), = ((∂N G )(0) + G (3.25) where we have used G (y  , ψ (y  )) = 0 and ˜ (y) = G



1  1 N−1 

{

0

0

(∂# ∂N G )(τy  , τ (ψ (y  ) + θ (yN − ψ (y  ))))y#

#=1

+ (∂N G )(τy  , τ (ψ (y  ) + θ (yN − ψ (y  ))))(ψ (y  ) + θ (yN − ψ (y  )))} dθ dτ. ˜ (y)| ≤ 1/(2H1) for Since (∂N G )(0) = 1/H1 , choosing 0 > 0 so small that |G   |y| ≤ 0 , we see that xN − h(x ) ≥ 0 and yN − ψ (y ) ≥ 0 are equivalent. Let ω be a function in C0∞ (RN−1 ) such that ω(y  ) = 1 for |y  | ≤ 1/2 and ω(y  ) = 0 for |y  | ≥ 1 and set ω (y  ) = ψ (y  )ω(y  / ). Then, by (3.24) we have ω L∞ (RN−1 ) ≤ C 2 , ∂i ∂j ω L∞ (RN−1 ) ≤ C,

∂i ω L∞ (RN−1 ) ≤ C , ∂i ∂j ∂k ω L∞ (RN−1 ) ≤ C

−1

(3.26) .

where i, j , and k run from 1 through N − 1. Setting Ψ (z) = Φ (z , zN + ω (z )), that is yN = zN + ω (z ), and yj = zj for j = 1, . . . , N − 1, we see that there

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Y. Shibata

exists an 0 > 0 such that for any ∈ (0, 0 ), the map: z → x = Ψ (z) is a diffeomorphism of C 3 class from RN onto RN . Since ∂xm = tm,k + bm,k , ∂zk

∂xm = tm,N + bn,N ∂zN

where we have set bm,k =

 ∂ (sm,ij yi yj ρ (y)) ∂zk

i,j =1

+ {tm,N +

N  ∂ ∂ω  (sm,ij yi yj ρ (y))} (z ), ∂zk ∂zk

i,j =1

bn,N =

 i,j =1

∂ (sm,ij yi yj ρ (y)), ∂zN

let A and B be the N × N matrices whose (m, n)th components are tm,n and bm,n , respectively. Then, by (3.26), A is an orthogonal matrix and B satisfies the estimates: BL∞ (RN ) ≤ C ,

∇BL∞ (RN ) ≤ C,

∇ 2 BL∞ (RN ) ≤ C

−1

.

Moreover, by (3.25) we have ˜ (z , zN + ω (z )))(zN + (ω(z / ) − 1)ψ (z )), xN − h(x  ) = (1/H1 + G which shows that when |z | ≤ /2, xN ≥ h(x  ) and zN ≥ 0 are equivalent. We can construct the sequences of C0∞ (RN ) functions, {ζji }, {ζ˜ji }, by standard manner (cf. Enomoto-Shibata [18, Appendix]). This completes the proof of Proposition 3.2.2. 

To show the a priori estimates for the Stokes equations with free boundary condition in a general domain, we need some restriction of the domains. In this chapter note, we adopt the following conditions. Definition 3.2.3 Let k = 2 or 3 and let Ω be a domain in RN . We say that Ω is a uniformly C k domain whose inside has a finite covering if Ω is a uniformly C k domain in the sense of Definition 3.2.1 and the following assertion hold: (v) Let ζji be the partition of unity given in Proposition 3.2.2 (iii) and set ψ 0 =   9∞ ∞ 0 1 0 there exists a finite j =1 ζj . Let O = supp ∇ψ ∪ j =1 supp ∇ζj . Then, 9ι number of subdomains Oj (j = 1, . . . , ι) such that O ⊂ j =1 Oj and each Oj

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

211

satisfies one of the following conditions: (a) There exists an R > 0 such that Oj ⊂ ΩR , where ΩR = {x ∈ Ω | |x| < R}. (b) There exist a translation τ , a rotation A, a domain D ⊂ RN−1 , a coordinate function a(x  ) defined for x  ∈ D, and a positive constant b such that 0 ≤ a(x  ) < b for x ∈ D, A ◦ τ (Oj ) ⊂ {x = (x  , xN ) | x  ∈ D, a(x  ) ≤ xN ≤ b} ⊂ A ◦ τ (Ω), {x = (x  , xN ) ∈ RN | x  ∈ D, xN = a(x  )} ⊂ A ◦ τ (Γ ). Here, for any subset E of RN , A(E) = {Ax | x ∈ E} with some orthogonal matrix A and τ (E) = {x + y | x ∈ E} with some y ∈ RN . Example 3.2.4 Let Ω be a domain whose boundary Γ is a C k hypersurface. If Ω satisfies one of the following conditions, then Ω is a uniform C k domain whose inside has a finite covering. (1) Ω is bounded, or Ω is an exterior domain, that is, Ω = RN \ O with some bounded domain O. (2) Ω = RN + (half space), or Ω is a perturbed half space, that is, there exists an R R N R > 0 such that Ω ∩ B R = RN + ∩ B , where B = {x ∈ R | |x| > R}. (3) Ω is a layer L or a perturbed layer, that is, there exists an R > 0 such that Ω ∩ B R = L ∩ B R . Here L = {x = (x  , xN ) ∈ RN | x  = (x1 , . . . , xN−1 ) ∈ RN−1 , a < xN < b} for some constants a and b for which a < b. (4) Ω is a tube, that is, there exists a bounded domain D in RN−1 such that Ω = D × R. (5) There exist an R > 0 and several orthogonal transforms, Ri (i = 1, . . . , M),

9M N R. ∩ B R R such that Γ ∩ B R = i=1 i 0 (6) There exist an R > 0, half tubes, Ti (i = 1, . . . , M), and orthogonal transforms, 9M R Ri (i = 1, . . . , M), such that Ω ∩ B R = i=1 Ri Ti ∩ B , where what Ti is a half tube means that Ti = Di × [0, ∞) with some bounded domain Di of RN−1 . N i In the following, we write Br0 (xji ), Φj (RN + ), and Φj (R0 ) simply by Bj , Ωj and 1 Γj , respectively. In view of Proposition 3.2.2 (ii), we have Ωj ∩ Bj = Ω ∩ Bj1 and Γj ∩ Bj1 = Γ ∩ Bj1 . By the finite intersection property stated in Proposition 3.2.2 (iv), for any r ∈ [1, ∞) there exists a constant Cr,L such that ∞ + j =1

f rL

,1 r

i r (Ω∩Bj )

≤ Cr,L f Lr (Ω)

for any f ∈ Lr (Ω).

(3.27)

212

Y. Shibata

n (RN ) ≤ c0 Let n ∈ N0 , f ∈ Hqn (Ω), and let ηji be functions in C0∞ (Bji ) with ηji H∞ 1 1 for some constant c0 independent of j ∈ N. Since Ω ∩ Bj = Ωj ∩ Bj , by (3.27)

∞  j =1

q ηj0 f H n (RN ) q

+

∞  j =1

q

q

ηj1 f H n (Ωj ) ≤ Cq f H n (Ω) . q

(3.28)

q

3.2.2 Besov Spaces on Γ We now define Besov spaces on Γ . Before turning to it, we recall some basic facts. Let f be a function defined on Γ such that supp f ⊂ Γ ∩Bj1 ∩Bk1 . Let fj = f ◦Φj−1 , and then by Proposition 3.2.2 (iv), we see that for any s ∈ [−1, 3] and j , k ∈ N, fj W s (RN ) ≤ Cs,q fk W s (RN ) . q

q

0

(3.29)

0

In fact, in the case that s = 0, 1, 2, 3, noting Wqs = Hqs , we see that the inequality (3.29) follows from the direct calculations. When s = −1, it follows from duality argument. Finally, in the case that s ∈ Z, the inequality (3.29) follows from real interpolation. Let Γj = Φj (RN 0 ) be spaces given in Proposition 3.2.2 and the space Wqs (Γj ) and its norm  · Wqs (Γj ) are defined by Wqs (Γj ) = {f | f ◦ Φj ∈ Wqs (RN 0 )},

f Wqs (Γj ) = f ◦ Φj−1 W s (RN ) . q

0

In view of (3.29), if supp f ⊂ Γ ∩ Bj1 ∩ Bk1 , then f Wqs (Γj ) ≤ Cs,q f Wqs (Γk ) .

(3.30)

For s ∈ [−1, 3], we now define Wqs (Γ ) by Wqs (Γ )

= {f =

∞ 

fj | supp fj ⊂ Γ ∩ Bj1 ,

fj ∈ Wqs (Γj ),

j =1

f Wqs (Γ ) =

∞  j =1

q

fj W s (Γj ) q

1/q

< ∞}.

(3.31)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

213

Since each Wqs (Γj ) is a Banach space, so is Wqs (Γ ). Given f ∈ Wqs (Γ ), we have ζ˜j f ∈ Wqs (Γj ),

∞  j =1

ζ˜j f W s (Γj ) ≤ (c0 Cs,q 3N )q f W s (Γ ) . q

q

q

(3.32)

q

 1j s In fact, let f = ∞ j =1 fj ∈ Wq (Γ ), where fj satisfy (3.31). For each j , let #k (k = 1, . . . , m1j ) be the numbers given in Proposition 3.2.2 (iv) for which Bj1 ∩ B 11j = ∅ and

Bj1

1 ∩ Bm

= ∅ for m ∈

1j {#k

#k

|k=

ζ˜j f =

1, . . . , m1j }, 

and then

ζ˜j f#1j . k

k=1,...,m1j

Since supp ζ˜j f#1j ⊂ Γ ∩ B 11j ∩ Bj1 , by (3.30) #k

k



ζ˜j f Wqs (Γj ) ≤



ζ˜j f#1j Wqs (Γj ) ≤ Cs,q

k=1,2,...,m1j

≤ c0 Cs,q

k



ζ˜j f#1j Wqs (Γ 1j ) #k

k

k=1,2,...,m1j

f#1j Wqs (Γ 1j ) < ∞,

k=1,2,...,m1j

k

#k

where c0 is the number appearing in Proposition 3.2.2. Thus, we have ζ˜j f ∈ Wqs (Γj ). Since m1j ≤ L with some constant L independent of M1 as follows from Proposition 3.2.2, we have ∞  j =1

ζ˜j f W s (Γj ) ≤ q

q

∞  (c0 Cs,q )q (m1j )q−1 j =1

≤ (c0 Cs,q L)q

 k=1,...,m1j

∞ 

q

f#1j W s (Γ 1j ) k

q

#k

q

fn W s (Γn ) , q

n=1

which yields (3.32).

3.2.3 The Weak Dirichlet Problem 1 (Ω) be the homogeneous Sobolev space defined by letting Let Hˆ q,0 1 (Ω) = {ϕ ∈ Lq,loc (Ω) | ∇ϕ ∈ Lq (Ω)N , Hˆ q,0

ϕ|Γ = 0}.

(3.33)

214

Y. Shibata

Let 1 < q < ∞. The variational equation: (∇u, ∇ϕ)Ω = (f, ∇ϕ)Ω

for all ϕ ∈ Hˆ q1 ,0 (Ω)

(3.34)

is called the weak Dirichlet problem, where q  = q/(q − 1). Definition 3.2.5 We say that the weak Dirichlet problem (3.34) is uniquely solvable 1 (Ω) if for any f ∈ L (Ω)N , problem (3.34) admits a unique solution u ∈ in Hˆ q,0 q Hˆ 1 (Ω) possessing the estimate: ∇uLq (Ω) ≤ CfLq (Ω) . q,0

We define an operator K acting on f ∈ Lq (Ω)N by u = K(f). Remark 3.2.6 (1) When q = 2, the weak Dirichlet problem is uniquely solvable for any domain Ω, which is easily proved by using the Hilbert space structure of the space 1 (Ω). But, for any q = 2, speaking generally, we do not know whether the Hˆ 2,0 weak Dirichlet problem is uniquely solvable. 1−1/q (2) Given f ∈ Lq (Ω)N and g ∈ Wq (Γ ), we consider the weak Dirichlet problem: (∇u, ∇ϕ)Ω = (f, ∇ϕ)Ω

for every ϕ ∈ Hˆ q1 ,0 (Ω),

(3.35)

subject to u = g on Γ . Let G be an extension of g to Ω such that G = g on Γ 1 (Ω) and GHq1 (Ω) ≤ CgW 1−1/q (Γ ) for some constant C > 0. Let v ∈ Hˆ q,0 q be a solution of the weak Dirichlet problem: (∇v, ∇ϕ)Ω = (f − ∇G, ∇ϕ)Ω

for every ϕ ∈ Hˆ q1 ,0 (Ω).

1 (Ω) is a unique solution of Eq. (3.35) Then, u = G + v ∈ Hq1 (Ω) + Hˆ q,0 possessing the estimate:

∇uLq (Ω) ≤ C(gW 1−1/q (Γ ) + fLq (Ω) ) q

for some constant C > 0. Example 3.2.7 When Ω is a bounded domain, an exterior domain, half space, a perturbed half space, layer, a perturbed layer, and a tube, then the weak Dirichlet problem is uniquely solvable for q ∈ (1, ∞). Theorem 3.2.8 Let 1 < q < ∞. Let Ω be a uniform C 2 domain. Given f ∈ 1 (Ω) be a unique solution of the weak Dirichlet problem (3.34) Lq (Ω)N , let u ∈ Hˆ q,0 possessing the estimate: ∇uLq (Ω) ≤ CfLq (Ω) . If we assume that div f ∈ Lq (Ω) in addition, then ∇ 2 u ∈ Lq (Ω) and ∇ 2 uLq (Ω) ≤ CM2 ,q (div fLq (Ω) + fLq (Ω) ).

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

215

Proof This theorem will be proved in Sect. 3.2.6, after some preparations for weak Laplace problem in RN in Sect.3.2.4 and weak Dirichlet problem in the half space in Sect. 3.2.5 below. 

Remark 3.2.9 From Theorem 3.2.8, we know the existence of solutions of the strong Dirichlet problem: Δu = div f in Ω,

u|Γ = 0.

(3.36)

But, the uniqueness of solutions of Eq. (3.36) does not hold generally. For example, let B1 = {x ∈ RN | |x| > 1} for N ≥ 2, and let f (x) be a function defined by f (x) =

 log |x|

when N = 2,

|x|−(N−2)

− 1 when N ≥ 3.

Then, f (x) satisfies the homogeneous Dirichlet problem: Δf = 0 in B1 and f |∂B1 = 0, where ∂B1 = {x ∈ RN | |x| = 1}.

3.2.4 The Weak Laplace Problem in RN In this subsection, we consider the following weak Laplace problem in RN : (∇u, ∇ϕ)RN = (f, ∇ϕ)RN

for any ϕ ∈ Hˆ q1 (RN ).

(3.37)

We shall prove the following theorem. Theorem 3.2.10 Let 1 < q < ∞. Then, for any f ∈ Lq (RN )N , the weak Laplace problem (3.37) admits a unique solution u ∈ Hˆ q1 (RN ) possessing the estimate: ∇uLq (RN ) ≤ CfLq (RN ) . Moreover, if we assume that div f ∈ Lq (RN ) in addition, then ∇ 2 u ∈ Lq (RN )N and

2

∇ 2 uLq (RN ) ≤ Cdiv fLq (RN ) . Proof To prove the theorem, we consider the strong Laplace equation: Δu = div f in RN . Let 1 (D) = {f ∈ Lq (D)N | div f ∈ Lq (D)}, Hq,div

(3.38)

216

Y. Shibata

where D is any domain in RN . Since C0∞ (RN )N is dense both in Lq (RN )N and 1 Hq,div (RN ), we may assume that f ∈ C0∞ (RN )N . Let F [f ] = fˆ and F −1 denote respective the Fourier transform of f and the Fourier inverse transform. We then set u = −F

−1

+ F [div f](ξ ) , |ξ 2 |

= −F

−1

+ N

j =1 iξj F [fj ](ξ ) |ξ |2

,

for f = ) (f1 , . . . , fN ). By the Fourier multiplier theorem, we have ∇uLq (RN ) ≤ CfLq (RN ) ,

(3.39)

∇ 2 uLq (RN ) ≤ Cdiv fLq (RN ) .

Of course, u satisfies Eq. (3.38). We now prove that u satisfies the weak Laplace equation (3.38). For this purpose, we use the following lemma. Lemma 3.2.11 Let 1 < q < ∞ and let dq (x) =

 (1 + |x|2 )1/2

for N = q,

(1 + |x|2 )1/2 log(2 + |x|2 )1/2

for N = q.

Then, for any ϕ ∈ Hˆ q1 (RN ), there exists a constant c for which ϕ − c   ≤ C∇ϕLq (RN )   dq Lq (RN ) with some constant independent of ϕ and c. Proof For a proof, see Galdi [19, Chapter II].



To use Lemma 3.2.11, we use a cut-off function, ψR , of Sobolev’s type defined as follows: Let ψ be a function in C ∞ (R) such that ψ(t) = 1 for |t| ≤ 1/2 and ψ(t) = 0 for |t| ≥ 1, and set ψR (x) = ψ

ln ln |x| . ln ln R

Notice that |∇ψR (x)| ≤

1 c , ln ln R |x| ln |x|

supp ∇ψR ⊂ DR ,

(3.40)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

217

√

where we have set DR = {x ∈ RN | e ln R ≤ |x| ≤ R}. Noting that f ∈ C0∞ (RN )N , by (3.38) for large R > 0 and ϕ ∈ Hˆ q1 (RN ) we have (f, ∇ϕ)RN = (f, ∇(ϕ − c))RN = −(div f, ϕ − c)RN = −(ψR div f, ϕ − c)RN = −(ψR Δu, ϕ − c)RN

(3.41)

= ((∇ψR ) · (∇u), ϕ − c)RN + (ψR ∇u, ∇ϕ)RN , where c is a constant for which ϕ − c   ≤ C∇ϕLq  (RN )   dq  Lq  (RN )

(3.42)

with some constant C > 0. By (3.40) and (3.42), we have ϕ − c   |((∇ψR ) · (∇u), ϕ − c)RN | ≤ dq  (∇ψR ) · (∇u)Lq (RN )   dq  Lq  (RN ) ≤

C ∇uLq (DR ) ∇ϕLq  (RN ) → 0 ln ln R (3.43)

as R → ∞. By (3.41) and (3.43) we see that u satisfies the weak Dirichlet problem (3.37). The uniqueness follows from the existence theorem just proved for the dual 2 problem. Moreover, if div f ∈ Lq (RN ) in addition, then ∇ 2 u ∈ Lq (RN )N , and so by (3.39) we complete the proof of Theorem 3.2.10. 

3.2.5 The Weak Dirichlet Problem in the Half Space Case In this subsection, we consider the following weak Dirichlet problem in RN +: (∇u, ∇ϕ)RN = (f, ∇ϕ)RN +

+

for any ϕ ∈ Hˆ q1 ,0 (RN + ).

(3.44)

We shall prove the following theorem. N Theorem 3.2.12 Let 1 < q < ∞. Then, for any f ∈ Lq (RN + ) , the weak Dirichlet N 1 problem (3.44) admits a unique solution u ∈ Hˆ q,0 (R+ ) possessing the estimate: ∇uLq (RN ) ≤ CfLq (RN ) . +

+

N N 2 Moreover, if we assume that div f ∈ Lq (RN + ) in addition, then ∇ u ∈ Lq (R+ ) and

∇ 2 uLq (RN ) ≤ Cdiv fLq (RN ) . +

+

2

218

Y. Shibata

N Proof We may assume that f = ) (f1 , . . . , fN ) ∈ C0∞ (RN + ) in the following, N N N 1 because C0∞ (RN + ) is dense both in Lq (R+ ) and Hq,div (R+ ). We first consider the strong Dirichlet problem:

Δu = div f

in RN +,

u|xN =0 = 0.

(3.45)

e o For any function, f (x), defined in RN + , let f and f be the even extension and the odd extension of f defined by letting

 f (x) = e

f (x  , xN ) f (x  , −x

xN > 0,

N)

f (x) = o

xN < 0,

 f (x  , xN ) −f (x  , −x

xN > 0, N)

xN < 0, (3.46)

where x  = (x1 , . . . , xN−1 ) ∈ RN−1 and x = (x  , xN ) ∈ RN . o o e Noting that (div f) = N−1 j =1 ∂j (fj ) + ∂N (fN ) , we define u by letting u = −F −1 = −F

−1

+ F [(div f)o ](ξ ) , |ξ |2

+

N−1 j =1

iξj F [(fj )o ](ξ ) + iξN F [(fN )e ](ξ ) , |ξ |2

.

We then have ∇uLq (RN ) ≤ CfLq (RN + ),

∇ 2 uLq (RN ) ≤ Cdiv fLq (RN ) , +

(3.47)

and moreover u satisfies Eq. (3.45). We next prove that u satisfies the weak Dirichlet problem Eq. (3.44). For this purpose, instead of Lemma 3.2.11, we use the Hardy type inequality: 

∞  x

p f (y) dy

0

x

−r−1

0

1/p dy

≤ (p/r)



∞

(yf (y))p y −r−1 dy

1/p ,

0

(3.48) where f ≥ 0, p ≥ 1 and r > 0 (cf. Stein [68, A.4 p.272]). Of course, using zero extension of f suitably, we can replace the interval (0, ∞) by (a, b) for any 0 ≤ a < b < ∞ in (3.48). Let DR,2R = {x ∈ RN + | R ≤ |x| ≤ 2R}. Using (3.48), ) we see that for any ϕ ∈ Hˆ q1 ,0 (RN + lim R −1 ϕLq  (DR,2R ) = 0.

R→∞

(3.49)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .





In fact, using ϕ|xN =0 = 0, we write ϕ(x , xN ) =

b



|ϕ(x  , xN )|q dxN ≤

a

(∂s ϕ)(x  , s) ds. Thus,

0

by (3.48) we have 

xN

219

bq  q   b  |(∂N ϕ)(x  , xN )|q dxN q − 1 a

for any 0 < a < b. Let ER1 = {x ∈ RN | |x  | ≤ 2R, R/2 ≤ xN < 2R}, ER2 = {x ∈ RN | 0 ≤ xN ≤ 2R, R/2 ≤ |x  | ≤ 2R}, and then DR,2R ⊂ ER1 ∪ ER2 . Thus, by (3.48), 



|ϕ(x)|q dx DR,2R

1/q 

Rq    ≤  q −1 



|x  |≤R





|∂N ϕ(x)|q dxN dx 

R/2 2R

+ R/2≤|x |≤2R

2R



|∂N ϕ(x)|q dxN dx 

1/q  ,

0

which leads to (3.49). Let ω be a function in C0∞ (RN ) such that ω(x) = 1 for |x| ≤ 1 and ϕ(x) = 0 for |x| ≥ 2, and we set ωR (x) = ω(x/R). For any ϕ ∈ Hˆ q1 ,0 (RN + ) and for large R > 0, we have (div f, ϕ)RN = (ωR div f, ϕ)RN = (ωR Δu, ϕ)RN +

+

+

= −((∇ωR ) · (∇u), ϕ)RN − (ωR ∇u, ∇ϕ)RN . +

(3.50)

+

By (3.49) |((∇ωR ) · (∇u), ϕ)RN | ≤ R −1 ∇uLq (DR,2R ) ϕLq  (DR,2R ) → 0 +

as R → ∞. On the other hand, (div f, ϕ)RN = −(f, ∇ϕ)RN , where we have used N f ∈ C0∞ (RN + ) . Thus, by (3.50) we have

+

+

(∇u, ∇ϕ)RN = (f, ∇ϕ)RN +

+

for any ϕ ∈ Hˆ q1 ,0 (RN + ). This shows that u is a solution of the weak Dirichlet problem. The uniqueness follows from the existence of solutions for the dual problem, which completes the proof of Theorem 3.2.12. 

220

Y. Shibata

3.2.6 Regularity of the Weak Dirichlet Problem In this subsection, we shall prove Theorem 3.2.8 in Sect. 3.2.3. Let ζji (i = 0, 1, j ∈ N) be cut-off functions given in Proposition 3.2.2. We first consider the regularity of ζj0 u. For this purpose, we use the following lemma. Lemma 3.2.13 Let Ω be a uniformly C 2 domain in RN . Then, there exists a constant c1 > 0 independent of j ∈ N such that ϕH 1 (Ωj ∩B 1 ) ≤ c1 ∇ϕLq (Ωj ∩B 1 )

1 for any ϕ ∈ Hˆ q,0 (Ωj ),

ψH 1 (Ω∩B 1 ) ≤ c1 ∇ψLq (Ω∩B 1 )

1 for any ψ ∈ Hˆ q,0 (Ω),

q

j

q

j

j

j

ϕ − cj0 (ϕ)H 1 (B 0 ) ≤ c1 ∇ϕLq (B 0 )

for any ϕ ∈ Hˆ q1 (RN ),

ψ − cj0 (ψ)H 1 (B 0 ) ≤ c1 ∇ψLq (B 0 )

for any ψ ∈ Hˆ q1 (Ω).

q

q

j

j

j

j

Here, cj0 (ϕ) and cj0 (ψ) are suitable constants depending on ϕ and ψ, respectively. Proof For a proof, see Shibata [42, in the proofs of Lemma 3.4 and Lemma 3.5]. 

Continuation of Proof of Theorem 3.2.8 Let Lemma 3.2.13 such that

cj0

=

cj0 (ϕ)

u − cj0 Lq (B 0 ) ≤ c1 ∇uLq (B 0 ) . j

j

be a constant in

(3.51)

For any ϕ ∈ Hˆ q1 (RN ), we have (∇(ζj0 (u − cj0 )), ∇ϕ)RN = ((∇ζj0 )(u − cj0 ), ∇ϕ)RN + (∇u, ∇(ζj0 ϕ))RN − ((∇u) · (∇ζj0 ), ϕ)RN = −((Δζj0 )(u − cj0 ) + 2(∇ζj0 ) · (∇u) + ζj0 div f, ϕ)RN , where we have used (∇u, ∇(ζj0 ϕ))RN = (f, ∇(ζj0 ϕ))RN = −(ζj0 div f, ϕ)RN . Let f = (Δζ 0 )(u − c0 ) + 2(∇ζ 0) · (∇u) + ζ 0 div f. Since C ∞ (RN ) ⊂ Hˆ q1 (RN ), for any j

j

ϕ ∈ C0∞ (RN ) we have

j

j

0

(Δ(ζj0 (u − cj0 )), ϕ)RN = (f, ϕ)RN , which yields that Δ(ζj0 (u − cj0 )) = f

in RN

(3.52)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

221

in the sense of distribution. By Lemma 3.2.13, f ∈ Lq (RN ) and f Lq (RN ) ≤ C(div fLq (B 0 ) + ∇uLq (B 0 ) ). j

j

(3.53)

From (3.52) it follows that ∂k ∂# Δ(ζj0 (u − cj0 )) = ∂k ∂# f for any k, # ∈ N. Since both sides are compactly supported distributions, we can apply the Fourier transform and the inverse Fourier transform. We then have ∂k ∂# (ζj0 (u − cj0 )) = F −1

+ξ ξ , k # F [f ](ξ ) . |ξ |2

By the Fourier multiplier theorem, we have ∂k ∂# (ζj0 (u − cj0 ))Lq (RN ) ≤ Cf Lq (RN ) . Since ∂k ∂# (ζj0 (u − cj0 )) = ζj0 ∂k ∂# u + (∂k ζj0 )∂# u + (∂# ζj0 )∂k u + (∂k ∂# ζj0 )(u − cj0 ), by (3.51) and (3.53) we have ζj0 ∇ 2 u ∈ Lq (RN )N and ζj0 ∇ 2 uLq (Ω) ≤ CM2 (div fLq (B 0 ) + ∇uLq (B 0 ) ). j

j

(3.54)

We next consider ζj1 u. For any ϕ ∈ Hˆ q1 ,0 (Ωj ), we have (∇(ζj1 u), ∇ϕ)ΩJ = (g, ϕ)Ωj ,

(3.55)

where we have set g = −(ζj1div f + 2(∇u) · (∇ζj1 ) + (Δζj1 )u). By Lemma 3.2.13, g ∈ Lq (Ωj ) and gLq (Ωj ) ≤ C(div fLq (Ω∩B 1 ) + ∇uLq (Ω∩B 1 ) ). j

j

(3.56)

We use the symbols given in Proposition 3.2.2. Let ak# and bk# be the (k, #)th component of N × N matrices Aj and Bj given in Proposition 3.2.2. By the change of variables: y = Φj (x), the variational Eq. (3.55) is transformed to N 

((δk# + Ak# )∂k v, ∂# ϕ)RN = (h, ϕ)RN .

k,#=1

+

+

(3.57)

222

Y. Shibata

Here, we have set v = ζj1 u ◦ Φj , Ak# =

N 

J = det(Aj + Bj ) = 1 + J 0 ,

h = g ◦ Φj ,

{a#m bkm + a#m J0 (ak# + bk# ) + b#m J (akm + bkm )}.

m=1

By Proposition 3.2.2 and (3.56), we have Ak# L∞ (RN ) ≤ CM1 ,

∇Ak# L∞ (RN ) ≤ CA ,

(3.58)

hLq (RN ) ≤ C(div fLq (B 1 ∩Ω) + ∇uLq (B 1 ∩Ω) ), +

j

j

where CA is a constant depending a1 , a2 and A appearing in Definition 3.2.1. Since supp g ⊂ Φ −1 (Bj1 ) ∩ Ω, by Lemma 3.2.13 |(h, ϕ)RN | ≤ hLq (B 1 ∩RN ) ϕL +

j

+

1 N q  (Bj ∩R+ )

≤ CgLq (B 1 ∩Ω) ∇ϕL j

N q  (R+ )

for any ϕ ∈ Hˆ q1 ,0 (RN + ), where C is a constant independent of j ∈ N. Thus, by N ≤ the Hahn-Banach theorem, there exists an h ∈ Lq (RN + ) such that hLq (RN +) N 1 ˆ  (R+ ). In particular, Ch N and (h, ∇ϕ) N = (h, ϕ) N for any ϕ ∈ H 1 Lq (Bj ∩R+ )

R+

R+

q ,0

div h = −h ∈ Lq (RN + ). Thus, the variational problem (3.57) reads N 

((δk# + Ak# )∂k v, ∂# ϕ)RN = (h, ∇ϕ)RN +

k,#=1

+

for any ϕ ∈ Hˆ q1 ,0 (RN + ).

N We now prove that if M1 ∈ (0, 1) is small enough, then for any g ∈ Lq (RN +) , 1 (RN ) of the variational problem: there exists a unique solution w ∈ Hˆ q,0 + N 

((δk# + Ak# )∂k w, ∂# ϕ)RN = (g, ∇ϕ)RN +

k,#=1

+

for any ϕ ∈ Hˆ q1 ,0 (RN + ),

(3.59)

possessing the estimate: ∇wLq (RN ) ≤ CgLq (RN ) . +

(3.60)

+

1 N N Moreover, if div g ∈ Lq (RN + ), then ∇w ∈ Hq (R+ ) and

∇ 2 wLq (RN ) ≤ Cdiv gLq (RN ) + CA gLq (RN ) . +

+

+

(3.61)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

223

In fact, we prove the existence of w by the successive approximation. Let w1 ∈ 1 (RN ) be a solution of the weak Dirichlet problem: Hˆ q,0 + for any ϕ ∈ Hˆ q1 ,0 (RN + ).

(∇w1 , ∇ϕ)RN = (g, ∇ϕ)RN +

+

(3.62)

By Theorem 3.2.12, w1 uniquely exists and satisfies the estimate: ∇w1 Lq (RN ) ≤ CgLq (RN ) . +

(3.63)

+

2 1 N N Additionally, we assume that div g ∈ Lq (RN + ), and then ∇ w1 ∈ Hq (R+ ) and

∇ 2 w1 Lq (RN ) ≤ Cdiv gLq (RN ) . +

(3.64)

+

N 1 (RN ), let w 1 Given wj ∈ Hˆ q,0 j +1 ∈ Hˆ q,0 (R+ ) be a solution of the weak Dirichlet + problem:

(∇wj +1 , ∇ϕ)RN = (g, ∇ϕ)RN − +

+

N 

(Ak# ∂k wj , ∂# ϕ)RN

+

k,#=1

(3.65)

for any ϕ ∈ Hˆ q1 ,0 (RN + ). By Theorem 3.2.12 and (3.58), wj +1 exists and satisfies the estimate: ∇wj +1 Lq (RN ) ≤ C(gLq (Ω+ ) + M1 ∇wj Lq (RN ) ). +

+

(3.66)

Applying Theorem 3.2.12 and (3.58) to the difference wj +1 − wj , we have ∇(wj +1 − wj )Lq (RN ) ≤ CM1 ∇(wj − wj −1 )Lq (RN ) . +

+

(3.67)

Choosing CM1 ≤ 1/2 in (3.67), we see that {wj }∞ j =1 is a Cauchy sequence in N N 1 1 ˆ Hq,0 (R+ ), and so the limit w ∈ Hq,0 (R+ ) exists and satisfies the weak Dirichlet problem (3.59). Moreover, taking the limit in (3.66), we have ∇wLq (RN ) ≤ CgLq (Ω+ ) + CM1 ∇wLq (RN ) . +

+

Since CM1 ≤ 1/2, we have ∇wLq (RN ) ≤ 2CgLq (Ω+ ) . Thus, we have proved + 1 (Ω ) that the weak Dirichlet problem (3.59) admits at least one solution w ∈ Hˆ q,0 + possessing the estimate (3.60). The uniqueness follows from the existence of solutions to the dual problem. Thus, we have proved the unique existence of solutions of Eq. (3.59).

224

Y. Shibata

N provided that div g ∈ L (RN ). By We now prove that ∇w ∈ Hq1 (RN q +) + N Theorem 3.2.12, ∇w1 ∈ Hq1 (RN + ) and w1 satisfies the estimate:

∇ 2 w1 Lq (RN ) ≤ Cdiv gHq1 (Ω+ ) . +

N Additionally, we assume that ∇ 2 wj ∈ Lq (RN + ) , and then applying Theorem 3.2.12 N 2 and to (3.65) and using (3.58) give that ∇ 2 wj +1 ∈ Lq (RN +)

∇ 2 wj +1 Lq (RN ) ≤ Cdiv gLq (RN ) + CM1 ∇ 2 wj Lq (RN ) + CA ∇wj Lq (RN ) . +

+

+

+

(3.68) And also, applying Theorem 3.2.12 to the difference wj +1 − wj and using (3.58), we have ∇ 2 (wj +1 − wj )Lq (RN ) +

≤ CM1 ∇ (wj − wj −1 )Lq (RN ) + CA ∇(wj − wj −1 )Lq (RN ) , 2

+

+

which, combined with (3.67), leads to ∇ 2 (wj +1 − wj )Lq (RN ) + ∇(wj − wj −1 )Lq (RN ) +

+

≤ CM1 ∇ (wj − wj −1 )Lq (RN ) + C(CA + 1)M1 ∇(wj −1 − wj −2 )Lq (RN ) . 2

+

+

Choosing M1 > 0 so small that CM1 ≤ 1/2 and (CA + 1)M1 ≤ 1/2, then we have ∇ 2 (wj +1 − wj )Lq (RN ) + ∇(wj − wj −1 )Lq (RN ) ≤ (1/2)j −1 L +

+

with L = ∇ 2 (w3 − w2 )Lq (RN ) + ∇(w2 − w1 )Lq (RN ) . From this it follows that +

+

N N 2 {∇ 2 wj }∞ j =1 is a Cauchy sequence in Lq (Ω), which yields that ∇ w ∈ Lq (R+ ) . Moreover, taking the limit in (3.68) and using (3.60) gives that 2

∇ 2 wLq (RN ) ≤ Cdiv gLq (RN ) + (1/2)∇ 2 wLq (RN ) + CA gLq (RN ) . +

+

+

+

which leads to (3.61). Applying what we have proved and using the estimate: div hLq (RN ) + hLq (RN ) ≤ ChLq (RN ) ≤ C(div fLq (B 1 ) + ∇uLq (B 1 ) ), +

+

+

j

j

N which follows from (3.58), we have ∇v = ∇(ζj1 u ◦ Φj ) ∈ Hq1 (RN + ) and

∇(ζj1 u ◦ Φj )H 1 (RN ) ≤ C(div fLq (B 1 ∩Ω) + ∇uLq (B 1 ∩Ω) ). q

+

j

j

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

225

Since uLq (B 1 ∩Ω) ≤ c1 ∇uLq (B 1 ∩Ω) as follows from Lemma 3.2.13, we have j

j

ζj1 ∇ 2 uLq (Ω) ≤ C(div fLq (B 1 ∩Ω) + ∇uLq (B 1 ∩Ω) ). j

j

(3.69)

Combining (3.54), (3.69) and (3.28) gives ∇ 2 uLq (Ω) ≤ C(div fLq (Ω) + ∇uLq (Ω) ) ≤ C(div fLq (Ω) + fLq (Ω) ), which completes the proof of Theorem 3.2.8.

3.2.7 Laplace–Beltrami Operator In this subsection, we introduce the Laplace–Beltrami operators and some important formulas from differential geometry. Let Γ be a hypersurface of class C 3 in RN . Let Γ be parametrized as p = φ(θ ) = ) (φ1 (θ ), . . . , φN (θ )) locally at p ∈ Γ , where θ = (θ1 , . . . , θN−1 ) runs through a domain Θ ⊂ RN−1 . Let τi = τi (p) =

∂ φ(θ ) = ∂i φ ∂θi

(i = 1, . . . , N − 1),

(3.70)

which forms a basis of the tangent space Tp Γ of Γ at p. Let n = n(p) denote the outer unit normal of Γ at p. Notice that < τi , n >= 0.

(3.71)

Here and in the following, < ·, · > denotes a standard inner product in RN . To introduce the formula of n, we notice that ⎛ ∂φ

··· ⎜ . . det ⎜ ⎝ .. . . ∂φN ∂θ1 · · · 1

∂θ1

∂φ1 ∂φ1 ∂θN−1 ∂θk

.. .

⎞

.. ⎟ ⎟ . ⎠=0

∂φN ∂φN ∂θN−1 ∂θk

for any k = 1, . . . , N − 1. Thus, to satisfy (3.71), n is defined by n=

) (h , . . . , h ) 1 N

H

(3.72)

226

Y. Shibata

with H =

?

N 2 j =1 hi

and ⎛

∂φ1 ∂θ1

··· .. . ···

⎜ . ⎜ . ⎜ . ⎜ ∂φi−1 ˆ ⎜ ∂θ ∂(φ1 , . . . , φi , . . . , φN ) N+i 1 hi = = (−1) det ⎜ ⎜ ∂φi+1 · · · ∂(θ1 , . . . , θN−1 ) ⎜ ∂θ1 ⎜ . ⎜ . .. . ⎝ . ∂φN ∂θ1 · · ·

∂φ1 ⎞ ∂θN−1

⎟ ⎟ ⎟ ∂φi−1 ⎟ ⎟ ∂θN−1 ⎟ ∂φi+1 ⎟ . ∂θN−1 ⎟ .. ⎟ ⎟ . ⎠ .. .

∂φN ∂θN−1

For example, when N = 3, ∂φ ∂θ1 ×  n=  ∂φ  ∂θ1 ×

h2 =

∂φ ∂θ2  ∂φ  ∂θ2 

= H −1) (h1 , h2 , h3 )

h1 =

∂φ2 ∂φ3 ∂φ3 ∂φ2 − , ∂θ1 ∂θ2 ∂θ1 ∂θ2

∂φ3 ∂φ1 ∂φ1 ∂φ3 ∂φ1 ∂φ2 ∂φ2 ∂φ1 − , h3 = − , H = ∂θ1 ∂θ2 ∂θ1 ∂θ2 ∂θ1 ∂θ2 ∂θ1 ∂θ2

?

h21 + h22 + h23 .

Let gij = gij (p) =< τi , τj >

(i, j = 1, . . . , N − 1),

and let G be an (N − 1) × (N − 1) matrix whose (i, j )th components are gij . The matrix G is called the first fundamental form of Γ . In the following, we employ Einstein’s summation convention, which means that equal lower and upper indices are to be summed. Since for any ξ ∈ RN−1 with ξ = 0, < Gξ, ξ >= gij ξ i ξ j =< ξ i τi , ξ j τj >= |ξ i τi |2 > 0, G is a positive symmetric matrix, and therefore G−1 exists. Let g ij be the (i, j )th component of G−1 and let τ i = g ij τj . Using g ik gkj = gik g kj = δji , where δji are the Kronecker delta symbols defined by δii = 1 and δji = 0 for i = j , we have < τi , τ j >=< τ i , τj >= δji .

(3.73) j

In fact, < τi , τ j >=< τi , g j k τk >= g j k < τi , τk >= g j k gki = δi . Thus, {τ j }N−1 j =1

is a dual basis of {τi }N−1 i=1 . In particular, we have τ i = g ij τj ,

τi = gij τ j .

(3.74)

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227

For any a ∈ Tp Γ , we write a = a i τi = a i τ i . By (3.73), we have < a, τ i >=< a j τj , τ i >= a j < τj , τ i >= a i and < a, τi >= aj < τ j , τi >= ai , and so a =< a, τ i > τi =< a, τi > τ i .

(3.75)

In particular, by (3.73) and (3.74), we have ai = gij a j ,

a i = g ij aj

with ai =< a, τi > and a i =< a, τ i >. Notice that {τ1 , . . . , τN−1 , n} forms a basis of RN . Namely, for any N-vector b ∈ RN we have b = bi τi + < b, n > n = bi τ i + < b, n > n with bi =< b, τ i > and bi =< b, τi >. We next consider τij = ∂i ∂j φ = ∂j τi . Notice that τij = τj i . Let Λkij =< τij , τ k >,

#ij =< τij , n >,

(3.76)

and then, we have τij = Λkij τk + #ij n.

(3.77)

Let L be an N − 1 × N − 1 matrix whose (i, j )th component is #ij , which is called the second fundamental form of Γ . Let H(Γ ) =

1 1 tr (G−1 L) = g ij #ij . N −1 N −1

(3.78)

The H(Γ ) is called the mean curvature of H(Γ ). Let g = det G. One of the most important formulas in this section is √ √ ∂i ( gg ij τj ) = gg ij #ij n.

(3.79)

This formula will be proved below. The Λkij is called Christoffel symbols. We know the formula: Λrij =

1 rk g (∂i gj k + ∂j gki − ∂k gij ). 2

(3.80)

228

Y. Shibata

In fact, ∂i gj k = ∂i < τj , τk >=< τij , τk > + < τj , τki >, ∂j gki = ∂j < τk , τi >=< τj k , τi > + < τk , τij >, ∂k gij = ∂k < τi , τj >=< τki , τj > + < τi , τj k >, and so we have ∂i gj k + ∂j gki − ∂k gij = 2 < τij , τk >, which implies (3.80). We now prove (3.79) by studying several steps. We first prove ∂k gij = gj r Λrki + gir Λrkj .

(3.81)

In fact, by (3.77) and < τi , n >= 0, ∂k gij =< τki , τj > + < τi , τkj >=< Λrki τr , τj > + < τi , Λrkj τr > = gj r Λrki + gir Λrkj . We next prove j

∂k g ij = −g ir Λrk − g j r Λirk .

(3.82)

In fact, by g ij gj # = δ#i , we have 0 = ∂k (g ij gj k ) = (∂k g ij )gj # + g ij ∂k gj # . Using (3.81), we have (∂k g ij )gj # = −g ij ∂k gj # = −g ij (g#r Λrkj + gj r Λrk# ) = −g ij g#r Λrkj − δri Λrk# = −g ij g#r Λrkj − Λik# , and therefore ∂k g im = (∂k g ij )δjm = (∂k g ij )gj # g #m = −g #m (g ij g#r Λrkj + Λik# ) m# i i# m m# i = −δrm g ij Λrkj − g #m Λik# = −g ij Λm kj − g Λk# = −g Λk# − g Λk# . m Setting m = j and # = r in the above formula, we have (3.82), because Λm k# = Λ#k i i and Λk# = Λ#k .

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229

We next prove j

∂i g = 2Λij g.

(3.83)

Recall that g = det G and the (i, j )th component of G is < τi , τj >. From the definition of differentiation, we have det G(x + ei Δxi ) − det G(x) Δxi =

det(G(x) + ∂i G(x)Δxi ) − det G(x) + O(Δxi ) Δxi

=

det G(x)(det(I + ∂i G(x)G−1 (x)Δxi ) − 1) + O(Δxi ) Δxi

= det G(x) tr (∂i G(x)G−1 (x)) + O(Δxi ). Thus, we have ∂i g = det G tr (∂i GG−1 ). Using (3.77) and < τi , n >= 0, we have tr (∂i GG−1 ) = ∂i (< τj , τk >)g kj = (< τij , τk > + < τj , τik >)g kj = ((Λrij τr , τk > + < Λrik τr , τj >)g kj = (grk Λrij + grj Λrik )g kj j

= 2Λij Putting these two formulas gives (3.83). We next prove √ √ j ∂i ( gg ij ) = − gg ik Λik .

(3.84)

In fact, by (3.82) and (3.83), 1 1 1 √ j (∂i g)g ij + ∂i g ij = 2Λ# gg ij − g ir Λri − g j r Λiri √ ∂i ( gg ij ) = g 2g 2g i# j

j

= g j r Λ#r# − g ir Λri − g j r Λ##r = −g ik Λik . Thus, we have (3.84).

230

Y. Shibata

We now prove (3.79). By (3.77) and (3.84), √ √ √ ∂i ( gg ij τj ) = ∂i ( gg ij )τj + gg ij ∂i τij √ √ j = − gg ik Λik τj + gg ij (Λkij τk + #ij n) √ √ √ = − gg ij Λkij τk + gg ij Λkij τk + gg ij #ij n √ = gg ij #ij n. Thus, we have (3.79). We now introduce the Laplace–Beltrami operator ΔΓ on Γ , which is defined by 1 √ ΔΓ f = √ ∂i ( gg ij ∂j f ). g By (3.84), we have j

ΔΓ f = g ij ∂i ∂j f − g ik Λik ∂j f.

(3.85)

By (3.78) and (3.79), we have ΔΓ φ = (N − 1)H(Γ )n. Usually, we put H (Γ ) = (N − 1)H(Γ ), and so we have ΔΓ x = H (Γ )n for x ∈ Γ .

(3.86)

One fundamental result for the Laplace–Beltrami operator is the following. Lemma 3.2.14 Let 1 < q < ∞. Assume that Ω is a uniform C 3 domain and let Γ be the boundary of Ω. Then, there exists a large number m > 0 such that for −1/q 2−1/q any f ∈ Wq (Γ ), there exists a unique v ∈ Wq (Γ ) such that v satisfies the equation (m − ΔΓ )v = f

on Γ

and the estimate vW 2−1/q (Γ ) ≤ Cf W −1/q (Γ ) q

for some constant C > 0.

q

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

231

Remark 3.2.15 (1) Lemma 3.2.14 was proved by Amann–Hieber–Simonett [4, Theorem 10.3] in the case where Γ has finite covering, and by Shibata [44, Theorem 2.2] in the case where Γ is the boundary of a uniform C 2 domain. (2) Let (m − ΔΓ )−1 be defined by (m − ΔΓ )−1 f = v.

3.2.8 Parametrized Surface Let φt : Ω → RN be an injection map with suitable regularity for each time t ≥ 0. Let Γ be a C 3 hypersurface ⊂ Ω and set Γt = {x = φt (y) | y ∈ Γ }. First we prove Theorem 3.1.1. Let J (t) be the Jacobian of the map x = φ(y). Let w(x, t) = (∂t φt )(φ −1 (x)). We shall prove that ∂ J (t) = (div x w(x, t))J (t). ∂t

(3.87)

This formula is called a Reynolds transport theorem. In fact, in view of the definition of differentiation, we consider J (t + Δt) − J (t) . Δt Writing φt = ) (x1 (y, t), . . . , xN (y, t)), we have J (t) = det X(t) with ⎞ ∂x1 . . . ∂y N ⎜ . . .. ⎟ ⎟ . . X(t) = ⎜ . . ⎠. ⎝ . ∂xN ∂xN ∂y1 . . . ∂yN ⎛ ∂x

1

∂y1

By mean value theorem, we write X(t + Δt) = X(t) + X (t)Δt + O((Δt)2 ) with ⎛ ∂2x

1

∂y1 ∂t

⎜ . X (t) = ⎜ ⎝ ..

∂ 2 xN ∂y1 ∂t

... .. . ...

∂ 2 x1 ∂yN ∂t

⎞

.. ⎟ ⎟ . ⎠.

∂ 2 xN ∂yN ∂t

232

Y. Shibata

Using this symbol, we write J (t + Δt) − J (t) det X(t + Δt) − det X(t) = Δt Δt =

det(X(t) + ΔtX (t) + O((Δt)2 )) − det X(t) Δt

=

det X(t)(det(I + ΔtX (t)X−1 (t) + O((Δt)2 )) − 1) Δt

= (det X(t))tr(X (t)X−1 (t)) + O(Δt). Since the (i, j )th component of X (t)X(t)−1 is N  ∂wi ∂ 2 xi ∂yk = , ∂yk ∂t ∂xj ∂xj k=1

where w(x) = (∂t φ)(φ −1 (x)) = div x w, which shows (3.87). We also prove that ∂ |Γt | = − ∂t

) (w , . . . , w ), 1 N



we have tr (X (t)X−1 )(t) =

H (Γt ) < nt , φ˙ > dσ,

(3.88)

Γt

where |Γt | is the area of Γt , φ˙ = ∂t φ, and dσ is the surface element of Γt . The formula < nt , φ˙ > denotes the velocity of the evolution of Γt with respect to nt . To prove (3.88), we have to parametrize Γt . Let Γ be parametrized as y = y(θ ) for θ ∈ Θ ⊂ RN−1 , and then the first fundamental form of Γt is given by Gt = (gij (t)) with gij (t) =< τi (t), τj (t) >, where τi (t) =

∂φt (y(θ )) . ∂θi

ij Let g(t) = det Gt and G−1 t = (g (t)). Since the surface element of Γt is given by √ dσ = gdθ , we have

 |Γ (t)| =

dσ = Γt

 

g(t) dθ.

Θ

Thus, d |Γ (t)| = dt

 Θ

d g(t) dθ = dt

 Θ

g˙ √ dθ 2 g

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233

To find g, ˙ we calculate (g(t + Δt) − g(t))/Δt as follows: g(t + Δt) − g(t) Δt det G(t + Δt) − det G(t) = Δt =

det(G(t) + ∂t G(t)Δt + O((Δt)2 ) − det G(t) Δt

=

det(G(t)(det(I + ∂t G(t)G(t)−1 Δt + O((Δt)2 )) − 1) Δt

= det G(t))tr(∂t G(t)G(t)−1 ) + O(Δt). Thus, we have g˙ = g tr (∂t G(t)G(t)−1 ) = g(< τ˙i , τj > + < τi , τ˙j >)g j i = g(< τ˙i , τj > g j i + < τ˙i , τj > g ij ) = 2gg ij < τ˙i , τj >, where we have used g ij = g j i . Putting these formulas together gives d |Γ (t)| = dt

 

√ ij gg < τ˙i , τj > dθ Θ

√ ˙ τj >) dθ − ∂i ( gg ij < φ,

= Θ



√ < ∂i ( gg ij τj ), φ˙ > dθ

Θ



=− Θ



1 √ √ √ < ∂i ( gg ij ∂j φ), φ˙ > g dθ = − g



< ΔΓt φ, φ˙ > dσ

Γt

H (Γt ) < nt , φ˙ > dσ,

=− Γt

where we have used (3.86). This shows (3.88).

3.2.9 Example of Mean Curvature Let Γ be a closed hypersurface in R3 defined by |x| = r(ω) for ω ∈ S1 = {ω ∈ R3 | |ω| = 1}. We introduce the polar coordinates: ω1 = cos ϕ sin θ, ω2 = sin ϕ sin θ, ω3 = cos θ

234

Y. Shibata

for ϕ ∈ [0, 2π) and θ ∈ [0, π). And then, Γ is represented by x1 = r(ϕ, θ ) cos ϕ sin θ, x2 = r(ϕ, θ ) sin ϕ sin θ, x3 = r(ϕ, θ ) cos θ. Let H (Γ ) be the doubled mean curvature of Γ . In the sequel, we will show the following well-known formula (cf. [59]): H (Γ ) =

∂ sin θ rθ  1  ∂ rϕ   + r sin θ ∂ϕ sin θ r 2 + |∇r|2 ∂θ r 2 + |∇r|2 2

(3.89)

, − r 2 + |∇r|2 where ∇r = (rθ , r/ sin θ ), and so |∇r|2 = rθ2 + (rϕ / sin θ )2 . From the definition, we have H (Γ ) =

2 

g ij #ij ,

ij =1

where g ij is the (i, j )th component of the inverse matrix of the first fundamental form and #ij is the (i, j )th component of the second fundamental form. We have to find g ij and #ij . We first find gij . Since ⎛ ⎞ ⎛ ⎞ cos ϕ sin θ − sin ϕ sin θ ∂x = rϕ ⎝ sin ϕ sin θ ⎠ + r ⎝ cos ϕ sin θ ⎠ , ∂ϕ cos θ 0 ⎛ ⎞ ⎛ ⎞ cos ϕ sin θ cos ϕ cos θ ∂x = rθ ⎝ sin ϕ sin θ ⎠ + r ⎝ sin ϕ cos θ ⎠ , ∂θ cos θ − sin θ the first fundamental form G is given by G =

  g11 g12 with g12 g22

g11 =

∂x ∂x · = rϕ2 + r 2 sin2 θ, ∂ϕ ∂ϕ

g12 =

∂x ∂x · = rϕ rθ , ∂ϕ ∂θ

g22 =

∂x ∂x · = rθ2 + r 2 . ∂θ ∂θ

(3.90)

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And then, g = det G = r 2 sin2 θ (r 2 + rθ2 + (rϕ / sin θ )2 ) = r 2 sin2 θ (r 2 + |∇r|2 ), where |∇r|2 = rθ2 + (rϕ / sin θ )2 . Thus, G

−1

  1 rθ2 + r 2 −rϕ rθ = . g −rϕ rθ rϕ2 + r 2 sin2 θ

and so g 11 =

rϕ2 + r 2 sin2 θ rθ2 + r 2 rϕ rθ , g 12 = g 21 = − , g 22 = . g g g

(3.91)

We next calculate the second fundamental form of Γ . For this purpose, we first calculate the unit outer normal nΓ to Γ , which is given by ⎞ ⎛ rϕ r sin ϕ − rθ r cos ϕ sin θ cos θ + r 2 cos ϕ sin2 θ 1 nΓ = √ ⎝−rϕ r cos ϕ − rθ r sin ϕ sin θ cos θ + r 2 sin ϕ sin2 θ ⎠ . g rθ r sin2 θ + r 2 sin θ cos θ

(3.92)

We next find the second fundamental form. For this, first we calculate the second derivatives of x as follows: ⎛

∂ 2x ∂ϕ 2 ∂ 2x ∂ϕ∂θ

∂ 2x ∂θ 2

⎞ ⎛ ⎞ ⎛ ⎞ cos ϕ sin θ − sin ϕ sin θ − cos ϕ sin θ = rϕϕ ⎝ sin ϕ sin θ ⎠ + 2rϕ ⎝ cos ϕ sin θ ⎠ + r ⎝ − sin ϕ sin θ ⎠ ; cos θ 0 0 ⎛ ⎛ ⎛ ⎞ ⎞ ⎞ cos ϕ sin θ cos ϕ cos θ − sin ϕ sin θ = rϕθ ⎝ sin ϕ sin θ ⎠ + rϕ ⎝ sin ϕ cos θ ⎠ + rθ ⎝ cos ϕ sin θ ⎠ (3.93) cos θ − sin θ 0 ⎛ ⎞ − sin ϕ cos θ + r ⎝ cos ϕ cos θ ⎠ ; 0 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ cos ϕ sin θ cos ϕ cos θ − cos ϕ sin θ = rθθ ⎝ sin ϕ sin θ ⎠ + 2rθ ⎝ sin ϕ cos θ ⎠ + r ⎝ − sin ϕ sin θ ⎠ . (3.94) cos θ − sin θ − cos θ

236

Y. Shibata

Thus, we have #11 =
∂ϕ 2 Γ

1 = √ (r 2 rϕϕ sin θ − 2rrϕ2 sin θ + r 2 rθ sin2 θ cos θ − r 3 sin3 θ ), g #12 =
∂ϕ∂θ Γ

(3.95)

1 = √ (r 2 rϕθ sin θ − 2rrϕ rθ sin θ − r 2 rϕ cos θ ), g ∂ 2x ,n > ∂θ 2 Γ 1 = √ (r 2 rθθ sin θ − 2rθ2 r sin θ − r 3 sin θ ). g

#22 =
0 for k = 1, 2, 3 and # = 1, 2. We then define Ωt by Ωt = {x = y + ω(y)Hρ (y, t)n(y) + ξ(t) | y ∈ Ω} (t ∈ (0, T )), where ω(y) is a C ∞ function which equals 1 near Γ and zero far from Γ . For the notational simplicity, we set Ψρ (y, t) = ω(y)Hρ (y, t)n(y). The transformation: x = y + Ψρ (y, t) + ξ(t)

(3.99)

is called the Hanzawa transform, which was originally introduced by Hanzawa [22] to treat classical solutions of the Stefan problem. Usually, ξ(t) is also unknown functions and to prove the local well-posedness, we set ξ(t) = 0. In the following, we study how to change the equations and boundary conditions under such transformation. Assume that sup Ψρ (·, t)H∞ 1 (RN ) ≤ δ,

t ∈(0,T )

(3.100)

where δ is a small positive number determined in such a way that several conditions stated below will be satisfied. We first choose 0 < δ < 1, and then the Hanzawa transform (3.99) is injective for each t ∈ (0, T ). In fact, let xi = yi +Ψρ (yi , t)+ξ(t) (i = 1, 2). We then have |x1 − x2 | = |y1 + Ψρ (y1 , t) − (y2 + Ψρ (y2 , t)| ≥ |y1 − y2 | − ∇Ψρ (·, t)L∞ (RN ) |y1 − y2 | ≥ (1 − δ)|y1 − y2 |. Thus, the condition 0 < δ < 1 implies if y1 = y2 then x1 = x2 , that is, the Hanzawa transform is injective. We now set Ωt = {x = y + Ψρ (y, t) + ξ(t) | y ∈ Ω} (t ∈ (0, T ). Notice that the Hanzawa transform maps Ω onto Ωt injectively.

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3.3.2 Transformation of Equations and the Divergence Free Condition In this subsection, for the latter use we consider more general transformation: x = y + Ψ (y, t), where Ψ (y, t) satisfies the condition: sup Ψ (·, t)H∞ 1 (RN ) ≤ δ

t ∈(0,T )

with some small δ > 0. Moreover, we assume that ∂t Ψ exists. Let v and p be solutions of Eq. (3.1), and let u(y, t) = v(y + Ψ (y, t), t),

q(y, t) = p(y + Ψ (y, t), t).

We now show that the first equation in (3.1) is transformed to ∂t u − Div (μD(u) − qI) = f(u, Ψ )

in Ω T ,

(3.101)

and the divergence free condition: div v = 0 in (3.1) is transformed to div u = g(u, Ψ ) = div g(u, Ψ )

in Ω T .

(3.102)

Here, f(u, Ψ ), g(u, Ψ ) and g(u, Ψ ) are suitable non-linear functions with respect to u and ∇Ψ given in (3.113) and (3.108), below. Let ∂x/∂y be the Jacobi matrix of the transformation (3.99), that is, ⎞ ∂1 Ψ1 ∂2 Ψ1 . . . ∂N Ψ1 ⎜ ∂1 Ψ2 ∂2 Ψ2 . . . ∂N Ψ2 ⎟ ⎟ ⎜ ∇Ψ = ⎜ . . ⎟ .. . . ⎝ .. . .. ⎠ . ∂1 ΨN ∂2 ΨN . . . ∂N ΨN ⎛

∂Ψj ∂x = I + ∇Ψ (y, t) = (δij + ), ∂y ∂yi

where Ψ (y, t) = ) (Ψ1 (y, t), . . . , ΨN (y, t)), and ∂i Ψj = ∂x −1 ∂y

= I+

∂Ψj . If 0 < δ < 1, then ∂yi

∞  (−∇Ψ (y, t))k k=1

exists, and therefore there exists an N × N matrix V0 (k) of C ∞ functions defined on |k| < δ such that V0 (0) = 0 and ∂x −1 ∂y

= I + V0 (∇Ψ (y, t)).

(3.103)

Here and in the following, k = (kij ) and kij are the variables corresponding to ∂i Ψj .

240

Y. Shibata

Let V0ij (k) be the (i, j )th component of V0 (k). We then have  ∂ ∂ = (δij + V0ij (k)) , ∂xi ∂yj N

∇x = (I + V0 (k))∇y ,

(3.104)

j =1

where ∇z = ) (∂/∂z, . . . , ∂/∂zN ) for z = x and y. By (3.104), we can write D(v) as D(v) = D(u) + DD (k)∇u with D(u)ij =

∂uj ∂ui + , ∂yj ∂yi

N  ∂uj ∂ui (DD (k)∇u)ij = V0j k (k) + V0ik (k) . ∂yk ∂yk

(3.105)

k=1

We next consider div v. By (3.104), we have div x v =

N  ∂vj j =1

∂xj

=

N 

(δj k + V0j k (k))

j,k=1

∂uj = div y u + V0 (k) : ∇u. ∂yk

(3.106)

Let J be the Jacobian of the transformation (3.99). Choosing δ > 0 small enough, we may assume that J = J (k) = 1 + J0 (k), where J0 (k) is a C ∞ function defined for |k| < σ such that J0 (0) = 0. To obtain another representation formula of div x v, we use the inner product (·, ·)Ωt . For any test function ϕ ∈ C0∞ (Ωt ), we set ψ(y) = ϕ(x). We then have (div x v, ϕ)Ωt = −(v, ∇ϕ)Ωt = −(J u, (I + V0 )∇y ψ)Ω = (div ((I + ) V0 )J u), ψ)Ω = (J −1 div ((I + ) V0 )J u), ϕ)Ωt , which, combined with (3.106), leads to div x v = div y u + V0 (k) : ∇u = J −1 (div y u + div y (J ) V0 (k)u)).

(3.107)

Recalling that J = J (k) = 1 + J0 (k), we define g(u, Ψ ) and g(u, Ψ ) by letting g(u, Ψ ) = −(J0 (k)div u + (1 + J0 (k))V0 (k) : ∇u), g(u, Ψ ) = −(1 + J0 (k))) V0 (k)u,

(3.108)

and then by (3.107) we see that the divergence free condition: div v = 0 is transformed to Eq. (3.102). In particular, it follows from (3.107) that J0 (k)div u + J (k)V0 (k) : ∇u = div (J (k)) V0 (k)u).

(3.109)

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241

To derive Eq. (3.101), we first observe that N  ∂ (μD(v)ij − pδij ) ∂xj j =1

=

N 

 ∂ ∂q (D(u)ij + (DD (k)∇u)ij ) − (δij + V0ij ) , ∂yk ∂yj N

μ(δj k + V0j k )

j =1

j,k=1

(3.110) where we have used (3.105). Since  ∂Ψj ∂vi ∂ ∂vi [vi (y + Ψ (y, t), t)] = (x, t) + (x, t), ∂t ∂t ∂t ∂xj N

j =1

we have N  ∂Ψj ∂ui ∂ui ∂vi , = − (δj k + V0j k ) ∂t ∂t ∂t ∂yk j,k=1

and therefore, N N  ∂Ψj ∂ui ∂ui ∂vi  ∂vi + + )(δj k + V0j k (k)) vj = (uj − . ∂t ∂xj ∂t ∂t ∂yk j =1

(3.111)

j,k=1

Putting (3.110) and (3.111) together gives 0=

∂u

i

∂t −μ

N 

+

j,k=1 N 

(uj −

∂Ψj ∂ui )(δj k + V0j k (k)) ∂t ∂yk

(δj k + V0j k (k))

j,k=1

−

N 

(δij + V0ij (k))

j =1

∂ (D(u)ij + (DD (k)∇u)ij ) ∂yk

∂q . ∂yj

Since (I + ∇Ψ )(I + V0 ) = (∂x/∂y)(∂y/∂x) = I, N  i=1

(δmi + ∂m Ψi )(δij + V0ij (k)) = δmj ,

(3.112)

242

Y. Shibata

and so we have 0=

N N ∂u   ∂Ψi ∂ui i + )(δj k + V0j k (k)) (δmi + ∂m Ψi ) (uj − ∂t ∂t ∂yk i=1

j,k=1

N 

−μ

(δmi + ∂m Ψi )(δj k + V0j k (k))

i,j,k=1

−

∂ (D(u)ij + (DD (k)∇u)ij ) ∂yk

∂q . ∂ym

Thus, changing i to # and m to i in the formula above, we define an N-vector of functions f(u, Ψ ) by letting f(u, Ψ )|i = −

N 

(uj −

j,k=1

−

N 

∂i Ψ#

#=1

∂Ψj ∂ui )(δj k + V0j k (k)) ∂t ∂yk

∂u

#

∂t

+

N 

(uj −

j,k=1

∂Ψj ∂u# )(δj k + V0j k (k)) ∂t ∂yk

N N   ∂ ∂ +μ (DD (k)∇u)ij + V0j k (k) (D(u)ij + (DD (k)∇u)ij ) ∂yj ∂yk j =1

+

N  j,k,#=1

∂i Ψ# (δj k + V0j k (k))

j,k=1

∂ (D(u)#j + (DD (k)∇u)#j ) , ∂yk

(3.113)

where f(u, Ψ )|i denotes the ith complonent of f(u, Ψ ). We then see that Eq. (3.6) is transformed to Eq. (3.101).

3.3.3 Transformation of the Boundary Conditions In this subsection, we consider the Hanzawa transform given in Sect.3.3.1. To represent Γ locally, we use the local coordinates near x#1 ∈ Γ such that 1 Ω ∩ B#1 = {y = Φ# (p) | p ∈ RN + } ∩ B# , 1 Γ ∩ B#1 = {y = Φ# (p , 0) | (p , 0) ∈ RN 0 } ∩ B# .

(3.114)

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243

Let {ζ#1}#∈N be a partition of unity given in Proposition 3.2.2. In the following we use the formula: f =

∞ 

ζ#1 f

in Γ

#=1

for any function, f , defined on Γ . We write ρ = ρ(y(p1 , . . . , pN−1 , 0), t) in the following. By the chain rule, we have N  ∂ρ ∂Ψρ ∂Φ#,m ∂ = Ψρ (Φ# (p1 , . . . , pN−1 , 0), t) = |pN =0 , ∂pi ∂pi ∂ym ∂pi

(3.115)

m=1

where we have set Φ# = letting

) (Φ , . . . , Φ #,1 #,N ),

and so, ∂ρ/∂pi is defined in B#1 by

N  ∂Ψρ ∂Φ#,m ∂ρ = ◦ Φ# . ∂pi ∂ym ∂pi

(3.116)

m=1

We first represent nt . Recall that Γt is given by x = y +ρ(y, t)n+ξ(t) for y ∈ Γ (cf. (3.97)). Let nt = a(n +

N−1 

b i τi )

with τi =

i=1

∂ ∂ y= Φ# (p , 0). ∂pi ∂pi

These vectors τi (i = 1, . . . , N − 1) form a basis of the tangent space of Γ at y = y(p1 , . . . , yN−1 ). Since |nt |2 = 1, we have 1 = a 2 (1 +

N−1 

gij bi bj )

with gij = τi · τj

(3.117)

i,j =1

because τi · n = 0. The vectors space of Γt , and so

∂x (i = 1, . . . , N − 1) form a basis of the tangent ∂pi

∂x · nt = 0. Thus, we have ∂pi

0 = a(n +

N−1  j =1

b j τj ) · (

∂y ∂ρ ∂n + n+ρ ). ∂pi ∂pi ∂pi

(3.118)

244

Y. Shibata

∂y ∂n ∂y ∂y = n · τi = 0, · n = 0 (because of |n|2 = 1), and · = ∂pi ∂pi ∂pi ∂pj τi · τj = gij , by (3.118) we have

Since n ·

N−1  ∂ρ ∂n + (gij + ρ · τj )bj = 0. ∂pi ∂pi

(3.119)

i=1

∂n · τj . Since ∂pi N 2 n is defined in R as an N-vector of C functions with nH∞ 2 (RN ) < ∞, we can write Let H be an (N − 1) × (N − 1) matrix whose (i, j )th component is

N  ∂n ∂Φ#,m ∂Φ# ∂n · τj = ◦ Φ# · , ∂pi ∂ym ∂pi ∂pj

(3.120)

m=1

and then H is defined in RN and H H∞ 2 (RN ) ≤ C with some constant C independent of # ∈ N. Under the assumption (3.100), we may assume that the ∂ρ ∂ρ ∂ρ inverse of I + ρH G−1 exists. Let ∇Γ ρ = ( ,..., ) with = ∂p1 ∂pN−1 ∂pi ∂ ρ(Φ# (p , 0)), and then by (3.119) we have ∂pi b = −(G + ρH )−1 ∇Γ ρ = −G−1 (I + ρH G−1 )−1 ∇Γ ρ.

(3.121)

Putting (3.121) and (3.117) together gives a = (1+ < Gb, b >)−1/2 = (1+ < (I + ρH G−1 )−1 ∇Γ ρ, G−1 (I + ρH G−1 )−1 ∇Γ ρ >)−1/2 . Thus, we have nt = n −

N−1  i,j =1

g ij τi

∂ρ + VΓ (ρ, ∇Γ ρ) ∂pj

where we have set VΓ (ρ, ∇Γ ρ) = − < G−1 ((I + ρH G−1 )−1 − I)∇Γ ρ, τ > + {(1+ < (I + ρH G−1 )−1 ∇Γ ρ, G−1 (I + ρH G−1 )−1 ∇Γ ρ >)−1/2 − 1} × (n− < G−1 (I + ρH G−1 )−1 ∇Γ ρ, τ >).

(3.122)

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From (3.116), ∇Γ ρ is extended to RN by letting ∇Γ ρ = (∇Φ# )∇Ψρ ◦ Φ# := Ξρ,# , and so VΓ (ρ, ∇Γ ρ) can be extended to B#1 by VΓ (ρ, ∇Γ ρ) = − < G−1 ((I + Ψρ H G−1 )−1 − I)Ξρ,# , τ > + {(1+ < (I + Ψρ H G−1 )−1 Ξρ,# , G−1 (I + ρH G−1 )−1 Ξρ,# >)−1/2 − 1}

(3.123)

× (n− < G−1 (I + ρH G−1 )−1 Ξρ,# , τ >), where H is an (N − 1) × (N − 1) matrix whose (i, j )th component, Hij , is given by Hij =

N  ∂n ∂Φ#,m ∂Φ# ◦ Φ# · . ∂ym ∂pi ∂pj

m=1

Thus, we may write ¯ ∇Ψ ¯ ρ ⊗ ∇Ψ ¯ ρ VΓ (ρ, ∇Γ ρ) = VΓ,# (k) ¯ = VΓ,# (y, k) ¯ defined on B 1 × {k¯ | |k| ¯ ≤ δ} with on B#1 with some function VΓ,# (k) # VΓ,# (0) = 0 possessing the estimate ¯ ∂ ¯ VΓ,# (·, k)) ¯ (VΓ,# (k), H 1 (B 1 ) ≤ C k ∞

#

with some constant C independent of #. Here and in the following k¯ are the ¯ ρ = (Ψρ , ∇Ψρ ). In view of (3.122), we have corresponding variables to ∇Ψ nt = n −

N−1  i,j =1

g ij τi

∂ρ ¯ ∇Ψ ¯ ρ ⊗ ∇Ψ ¯ ρ + VΓ,# (k) ∂yj

on B#1 ∩ Γ .

(3.124)

Let ¯ = VΓ (k)

∞ 

¯ ζ#1 VΓ,# (k),

(3.125)

#=1

and then we have ¯ ∂ ¯ VΓ (·, k)) ¯ H 1 (Ω) ≤ C (VΓ (k), k ∞ ¯ ≤ δ with some constant C > 0. for |k|

(3.126)

246

Y. Shibata

In view of (3.116) and (3.122), the unit outer normal nt is also represented by nt = n −

N−1 

N 

g ij τi

i,j =1 m=1

∂Ψρ ∂Φ#,m ¯ ∇Ψ ¯ ρ ⊗ ∇Ψ ¯ ρ ◦ Φ# + VΓ,# (k) ∂ym ∂pi

on B#1 , and so we may write ˜ Γ,# (∇Ψ ¯ ρ )∇Ψ ¯ ρ nt = n + V ˜ Γ,# (k) ¯ =V ˜ Γ,# (y, k) ¯ defined on B 1 × {k¯ | |k| ¯ ≤ δ} possessing for some functions V # the estimate: ¯ ˜ Γ,# , ∂ ¯ V ˜ (V k Γ,# )(·, k))H 1 (B 1 ) ≤ C ∞

#

¯ ≤δ for |k|

(3.127)

with some constant C independent of # ∈ N. Thus, setting ˜ Γ (k) ¯ = V

∞ 

¯ ˜ Γ,# (k), ζ#1 V

(3.128)

˜ Γ (∇Ψ ¯ ρ )∇Ψ ¯ ρ nt = n + V

(3.129)

#=1

we have

and ¯ ∂ ¯ VΓ (·, k)) ¯ H 1 (Ω) ≤ C ˜ Γ (·, k), (V k ∞

¯ ≤ δ. for |k|

(3.130)

We now consider the kinematic equation: VΓt = v · nt in (3.1). Since x = y + ρ(y, t)n + ξ(t), by (3.124) we have VΓt = =

∂x · nt ∂t ∞ 

ζ#1 < (∂t ρ)n + ξ  (t), n −

N−1  i,j =1

#=1

g ij τi

∂ρ ¯ ∇Ψ ¯ ρ ⊗ ∇Ψ ¯ ρ> + VΓ,# (k) ∂pj

¯ ∇Ψ ¯ ρ ⊗ ∇Ψ ¯ ρ> = ∂t ρ + ξ (t) · n + ∂t ρ < n, VΓ (k) 

¯ ∇Ψ ¯ ρ ⊗ ∇Ψ ¯ ρ >. − < ξ  (t) | ∇¯ Γ ρ > + < ξ  (t), VΓ (k) Here, for any N-vector of functions, d, we have set =

N−1  i,j =1

g ij < τi , d >

∂ρ . ∂pj

(3.131)

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247

On the other hand, v · nt =

∞  #=1

ζ#1 u · (n −

N−1  i,j =1

g ij τi

∂ρ ¯ ∇Ψ ¯ ρ ⊗ ∇Ψ ¯ ρ) + VΓ,# (k) ∂pj

¯ ∇Ψ ¯ ρ ⊗ ∇Ψ ¯ ρ. = n · u− < u | ∇Γ ρ > +u · VΓ (k) From the consideration above, the kinematic equation is transformed to the following equation: ∂t ρ + ξ  (t) · n − n · u+ < u | ∇Γ ρ >= d(u, ρ)

on Γ × (0, T )

(3.132)

with ¯ ∇Ψ ¯ ∇Ψ ¯ ρ ⊗ ∇Ψ ¯ ρ − ∂t ρ < n, VΓ (k) ¯ ρ ⊗ ∇Ψ ¯ ρ> d(u, ρ) = u · VΓ (k) ¯ ∇Ψ ¯ ρ ⊗ ∇Ψ ¯ ρ >. + < ξ  (t) | ∇Γ ρ > − < ξ  (t), VΓ (k)

(3.133)

We next consider the boundary condition: (μD(v) − pI)nt = σ H (Γt )nt − p0 nt

(3.134)

in Eq. (3.1). It is convenient to divide the formula in (3.134) into the tangential part and normal part on Γt as follows: t μD(v)nt = 0, < μD(v)nt , nt > −p = σ < H (Γt )nt , nt > −p0 .

(3.135) (3.136)

Here, t is defined by t d = d− < d, nt > nt for any N-vector of functions, d. By (3.129) we can write ˜ Γ (∇Ψ ˜ Γ (∇Ψ ¯ ρ )∇Ψ ¯ ρ , d > n+ < n, d > V ¯ ρ )∇Ψ ¯ ρ t d = 0 d+ < V ˜ Γ (∇Ψ ˜ Γ (∇Ψ ¯ ρ )∇Ψ ¯ ρ, d > V ¯ ρ )∇Ψ ¯ ρ, + n h (u, Ψρ ) = −μ{0 DD (k)∇u+ < V ˜ Γ (k) ¯ k¯ + < n, D(u) + DD (k)∇u > V ˜ Γ (k) ¯ k, ¯ D(u) + DD (k)∇u > V ˜ Γ,# (k) ¯ k} ¯ + n. To consider the transformation of the boundary condition (3.136), we first consider the Laplace–Beltrami operator on Γt . From (3.85), we have ij

ij

j

ΔΓt f = gt ∂i ∂j f − gt Λt ik ∂j f,

(3.140)

j

where Λt ik are Christoffel symbols of Γt defined by (3.76). Recall x = y+ρn+ξ(t). The first fundamental form of Γt = (gt ij ) is given by gt ij =
= gij + g˜ij ρ + ∂xi ∂xj ∂pi ∂pj

(3.141)

with g˜ ij =< τi ,

∂n ∂n ∂n ∂n > + < τj , >+< , >. ∂pj ∂pi ∂pi ∂pj

In view of (3.114), we can write τi and ∂j n as τi =

N−1  ∂Φ# (p) ∂y = , ∂pi ∂pj j =1

 ∂n ∂Φ#k (p) ∂n = . ∂pj ∂yk ∂pj N

(3.142)

k=1

Let H be an (N − 1) × (N − 1) matrix whose (i, j )th components are g˜ij , and then Gt = G + ρH + ∇p ρ ⊗ ∇p ρ = G(I + ρG−1 H + G−1 ∇p ρ ⊗ ∇p ρ). ∂ρ Here, ∇p ρ = ( ∂p , . . . , ∂p∂ρ ). Choosing δ > 0 small enough in (3.100), we know 1 N−1

that the inverse of I + ρG−1 H + ∇p ρ ⊗ ∇p ρ exists, and so

−1 −1 −1 −1 = G−1 − ρG−1 H G−1 + O2 , G−1 t = (I + ρG H + G ∇p ρ ⊗ ∇p ρ)) G

that is, ij

gt = g ij + ρhij + O2 , where hij denotes the (i, j )th component of −G−1 H G−1 . Here and in the following, O2 denotes some nonlinear function with respect to ρ and ∇p ρ of the

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249

form: O2 = a0 Ψρ2 +

N 

a j Ψρ

j =1

N  ∂Ψρ ∂Ψρ ∂Ψρ + bj k ∂yj ∂yj ∂yk

(3.143)

j,k=1

with suitable functions aj , and bj k possessing the estimates |(aj , bj k )| ≤ C,

|∇(aj , bj k )| ≤ C(|∇Ψρ | + |∇ 2 Ψρ |),

(3.144)

provided that (3.100) holds. Moreover, the Christoffel symbols are given by Λktij = gtk# < τt ij , τt # >, where τt i =

∂x ∂i

and τt ij =

∂2x ∂pi ∂pJ

. Since

τt ij = τij + ∂i ∂j (ρn) = τij + ρ∂i ∂j n + (∂i ρ)∂j n + (∂j ρ)∂i n + (∂i ∂j ρ)n, we have < τt ij , τt # > =< τij , τ# > +ρ(< ∂i ∂j n, τ# > + < τij , ∂# n >) + ∂j ρ < ∂i n, τ# > +∂i ρ < ∂j n, τ# > . Thus, Λktij = (g k# + hk# ρ + O2 )(< τij , τ# > +ρ(< ∂i ∂j n, τ# > + < τij , ∂# n >) + ∂j ρ < ∂i n, τ# > +∂i ρ < ∂j n, τ# >) = Λkij + ρg k# (< ∂i ∂j n, τ# > + < τij , ∂# n >) + ∂j ρ g k# < ∂i n, τ# > +∂i ρ g k# < ∂j n, τ# > +O2 , and so we may write Λktij = Λkij + ρAkij + ∂i ρBjk + ∂j ρBik + O2

(3.145)

with Akij = g k# (< ∂i ∂j n, τ# > + < τij , ∂# n >), Bjk = g k# < ∂i n, τ# >,

Bik = g k# < ∂j n, τ# > .

Combining these formulas with (3.140) gives ΔΓt f = ΔΓ f +

N−1  i,j =1

(h ρ + O2 )∂i ∂j f + ij

N−1  k=1

(hk + O2 )∂k f

(3.146)

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Y. Shibata

with hk = −

N−1 

(g ij Akij ρ + g ij Bjk ∂i ρ + g ij Bik ∂j ρ + Λkij hij ρ).

i,j =1

Thus, we have H (Γt )nt = ΔΓt (y + ρn) = ΔΓ y + ρΔΓ n + nΔΓ ρ + g ij (∂i ρ∂j n + ∂j ρ∂i n) + ρhij ∂i ∂j y + (ρhij ∂i ∂j ρ)n + hk ∂k y + O2 ∂i ∂j ρ + O2 , which, combined with (3.122) and < n, ∂j n >= 0, leads to < H (Γt )nt , nt > =< ΔΓ y, n > +ρ < ΔΓ n, n > +ΔΓ ρ + ρhij < ∂i ∂j y, n > + (hij ρ + O2 )∂i ∂j ρ − g ij < ΔΓ y, τi > ∂j ρ + O2 . Noting that ΔΓ y = H (Γ )n, we have < H (Γt )nt , nt > = ΔΓ ρ + H (Γ ) + ρ < ΔΓ n, n > + ρhij < ∂i ∂j y, n > +(hij ρ + O2 )∂i ∂j ρ + O2 . ¯ defined on Recalling (3.143) and (3.144), we see that there exists a function VΓ (k) ¯ ≤ C} satisfying the estimate: Ω × {k¯ | |k| ¯ ∂ ¯ V (·, k)) ¯ H 1 (Ω) ≤ C sup (VΓ (·, k), k Γ ∞

¯ |k|≤δ

with some constant C, where ∂k¯ denotes the partial derivatives with respect to ¯ for which variables k, ∞ 

¯¯ ¯ k¯ ⊗ k, ζ#1 ((hij ρ + O2 )∂i ∂j ρ + O2 ) = VΓ (k)

(3.147)

#=1

where k¯ = (Ψρ , ∇Ψρ ) and k¯¯ = (Ψρ , ∇Ψρ , ∇ 2 Ψρ ). Therefore, we have ¯ ρ )∇Ψ ¯ ρ ⊗ ∇¯ 2 Ψρ , < H (Γt )nt , nt > = Δγ ρ + H (Γ ) + Bρ + VΓ (∇Ψ where ∇¯ 2 Ψρ = (Ψρ , ∇Ψρ , ∇ 2 Ψρ ) and we have set Bρ = {< ΔΓ n, n > +

∞  #=1

ζ#1 (

N−1  i,j =1

hij < ∂i ∂j y, n >)}ρ.

(3.148)

(3.149)

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To turn to Eq. (3.136), in view of (3.105), (3.129), and (3.128), we write ¯ k¯ > < n, μD(v)nt > =< n, μD(u)n > + < n, μD(u)VΓ (k) ¯ k) ¯ >, + < n, μ(DD (k)∇u)(n + VΓ (k)

(3.150)

where k = ∇Ψρ and k¯ = (Ψρ , ∇Ψρ ). Thus, from (3.148), (3.150) and (3.2) it follows that the boundary condition (3.136) is transformed to < n, μD(u)n > −(q + c0 ) − σ (ΔΓ ρ + Bρ) = hN (u, Ψρ ),

(3.151)

where c0 is a constant and we have set ¯ k¯ > hN (u, Ψρ ) = − < n, μD(u)VΓ (k) ¯¯ ¯ k) ¯ > +σ VΓ (k)( ¯ k, ¯ k), − < n, μ(DD (k)∇u)(n + VΓ (k)

(3.152)

¯ ρ , and k¯¯ = ∇¯ 2 Ψρ . Setting q+c0 = q , we have ∇q = ∇q , where k = ∇Ψρ , k¯ = ∇Ψ and so we may assume that c0 = 0.

3.3.4 Linearization Principle In this subsection, we give a linearization principle1 of Eq. (3.1) for the local wellposedness and the global well-posedness. Putting (3.101), (3.102), (3.132), (3.138), and (3.151) together, we see that Eq. (3.1) is transformed to the equations: ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

∂t u − Div (μD(u) − qI) = f(u, Ψρ )

in Ω T ,

div u = g(u, Ψρ ) = div g(u, Ψρ )

in Ω T ,

∂t ρ + ξ  (t) · n − u · n+ < u | ∇Γ ρ >= d(u, ρ)

on Γ T ,

⎪ ⎪ (μD(u)n)τ = h (u, Ψρ ) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ < μD(u)n, n > −(q + σ H (Γ )) − σ (ΔΓ + B)ρ = hN (u, Ψρ ) ⎪ ⎪ ⎪ ⎩ u|t =0 = u0 in Ω, ρ|t =0 = ρ0 on Γ .

on Γ T , on Γ T , (3.153)

1 The linearization principle means how to divide a nonlinear equation into a linear part and a non-linear part.

252

Y. Shibata

Local Well-Posedness To prove the local well-posedness, the unique existence of local in time solutions, we take ξ(t) = 0. For the reference body Ω, we choose the boundary Γ0 of the initial domain in such a way that Γ0 = {x = y + h0 (y)n | y ∈ Γ } and h0 B 3−1/p−1/q (Γ ) is small enough. But, we do not want to have any restriction q,p on the size of the initial velocity, u0 . Thus, we approximate u0 by uκ defined by letting  1 κ T (s)u˜ 0 ds, uκ = κ 0 where u˜ 0 is a suitable extension of u0 to RN satisfying the condition: u˜ 0 B 2(1−1/p) (RN ) ≤ Cu0 B 2(1−1/p) (Ω) , q,p

(3.154)

q,p

and T (s) is some analytic semigroup defined on RN satisfying the estimate: T (·)u˜ 0 L

2(1−1/p) N (R )) ∞ ((0,∞),Bq,p

≤ Cu0 B 2(1−1/p) (Ω) , q,p

T (·)u˜ 0 Lp ((0,∞),Hq2 (RN )) + T (·)u˜ 0 Hp1 ((0,∞),Lq (RN )) ≤ Cu0 B 2(1−1/p) (Ω) . q,p

(3.155) By (3.155) uκ B 2(1−1/p) (Ω) ≤ Cu0 B 2(1−1/p) (Ω) , q,p

q,p

uκ Hq2 (Ω) ≤ Cu0 B 2(1−1/p) (Ω) κ −1/p .

(3.156)

q,p

The original idea to use uκ goes back to Padula and Solonnikov [34]. From Eq. (3.153), the linearization principle of Eq. (3.1) is the following equations: ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

∂t u − Div (μD(u) − qI) = f(u, Ψρ )

in Ω T

div u = g(u, Ψρ ) = div g(u, Ψρ )

in Ω T

∂t ρ+ < uκ | ∇Γ ρ > −u · n = d(u, Ψρ )+ < uκ − u | ∇Γ ρ >

in Γ T

(μD(u)n)τ = h (u, Ψρ )

in Γ T

< μD(u)n, n > −q − σ (ΔΓ + B)ρ = hN (u, Ψρ )

in Γ T

u|t =0 = u0 in Ω,

ρ|t =0 = ρ0 on Γ . (3.157)

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Global Well-Posedness In this lecture note, we only treat the global well-posedness in the case where the boundary is compact, that is Ω is a bounded domain or an exterior domain. The case where the boundary is unbounded was treated, for example, by Saito and Shibata [39, 40] in the Lp –Lq framework. When Ω is a bounded domain, roughly speaking, the global well-posedess holds in the maximal Lp –Lq regularity class provided that initial data are small and orthogonal to the rigid motion and the reference domain Ω is very closed to a ball, and moreover the solutions decay exponentially, which will be proved in Sect. 3.7 below. When Ω is an exterior domain, we consider (3.1) with σ = 0, that is without surface tension. In this case, instead of the Hanzawa transform, we use the partial Lagrange transform. Because, if we use the Hanzawa transform, we can not obtain necessary regularity of ρ representing Γt . Roughly speaking, we have the global wellposedness provided that initial data are small, and moreover the solutions decay with polynomial order. For detailed, see Sect. 3.8 below.

3.4 Maximal Lp –Lq Regularity To prove the local and global well-posedness for Eq. (3.1), in this lecture note the maximal regularity theorem for the linear part (the left hand side of Eq. (3.157)) plays an essential role. In this section, we study the maximal Lp –Lq regularity theorem and the generation of C0 analytic semigroup associated with the Stokes equations with free boundary conditions.

3.4.1 Statement of Maximal Regularity Theorems In this subsection, we shall state the maximal Lp –Lq regularity for the following three problems: ⎧ ∂t u − Div (μD(u) − pI) = f, ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

div u = g = div g

(μD(u) − pI)n = h u = u0

in Ω T , on Γ T , in Ω;

(3.158)

254

Y. Shibata

⎧ ∂t v − Div (μD(v) − qI) = 0, div v = 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ∂t h − n · v + F1 v = d ⎪ ⎪ (μD(v) − qI)n − (F2 h + σ ΔΓ h)n = 0 ⎪ ⎪ ⎪ ⎩ (v, h)|t =0 = (0, h0 ) ⎧ ∂t w − Div (μD(w) − rI) = 0, div w = 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ∂t ρ + Aκ · ∇  ρ − w · n + F1 w = d Γ

⎪ ⎪ ⎪ ⎪ ⎪ ⎩

(μD(w) − rI)n − (F2 ρ + σ ΔΓ ρ)n = 0 (w, ρ)|t =0 = (0, 0)

in Ω T , on Γ T ,

(3.159)

on Γ T , in Ω × Γ ; in Ω T , on Γ T , on Γ T ,

(3.160)

in Ω × Γ .

Here, F1 and F2 are linear operators such that F1 vW 2−1/q (Γ ) ≤ M0 vHq1 (Ω) , q

F2 hHq1 (Ω) ≤ C0 hHq2 (Ω)

(3.161)

with some constant M0 . If we consider the total problem: ⎧ ∂ U − Div (μD(U ) − P I) = f, div U = g = div g ⎪ ⎪ t ⎪ ⎪ ⎪ ⎨ ∂t H + Aκ · ∇Γ H − U · n + F1 U = d ⎪ ⎪ (μD(U ) − P I)n − (F2 H + σ ΔΓ H )n = h ⎪ ⎪ ⎪ ⎩ (U, H )|t =0 = (u0 , h0 )

in Ω T , on Γ T , on Γ T , in Ω × Γ , (3.162)

then, U , P and H are given by U = u + v + w and P = p + q + r and H = h + ρ, where (u, p) is a solution of Eq. (3.158), (v, q, h) a solution of Eq. (3.159) with d = 0, and (w, r, ρ) a solution of Eq. (3.160) by replacing d by d +n·u−F1 u−Aκ ·∇Γ h. We now introduce a solenoidal space Jq (Ω) defined by Jq (Ω) = {f ∈ Lq (Ω)N | (f, ∇ϕ)Ω = 0

for any ϕ ∈ Hˆ q1 ,0 (Ω)}.

(3.163)

Before stating the maximal Lp –Lq regularity theorems for the three equations given above, we introduce the assumptions on μ, σ , and Aκ .

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Assumption on μ, σ and Aκ There exist positive constants m0 , m1 , m2 , m3 , a and b for which, m0 ≤ μ(x), σ (x) ≤ m1 , |Aκ (x)| ≤ m2 ,

|∇(μ(x), σ (x))| ≤ m1

|Aκ (x) − Aκ (y)| ≤ m2 |x − y|a Aκ W 2−1/r (Γ ) ≤ m3 κ r

−b

for any x ∈ Ω, for any x, y ∈ Γ ,

(κ ∈ (0, 1)),

(3.164)

where r is an exponent in (N, ∞). Moreover, in view of (3.164), for Bji = Br0 (xji ) given in Proposition 3.2.2 we assume that |μ(x) − μ(xj0 )| ≤ M1 ,

for xj0 ∈ Bji ,

(3.165)

|μ(x) − μ(xj1 )| ≤ M1 , |σ (x) − σ (xj1 )| ≤ M1 , |Aκ (x) − Aκ (xj1 )| ≤ M1 (3.166) for any x ∈ Bj1 . Since Hq1 (Ω) is usually not dense in Hˆ q1 (Ω), it does not hold that div u = div g implies (u, ∇ϕ)Ω = (g, ∇ϕ)Ω for all ϕ ∈ Hˆ q1 (Ω). Of course, the opposite direction holds. Thus, finally we introduce the following definition. Definition 3.4.1 For u, g ∈ Lq (Ω)N , we say that div u = div g in Ω if u − g ∈ Jq (Ω). To solve the divergence equation div u = g in Ω, it is necessary to assume that g is given by g = div g for some g, and so we define the space DIq (G) by DIq (G) = {(g, g) | g ∈ Hq1 (G), g ∈ Lq (G), g = div g in G}, where G is any domain in RN . Before stating the main results in this section, we give a definition of the uniqueness. For Eq. (3.158), the uniqueness is defined as follows: •

If v and p with v ∈ Lp ((0, T ), Hq2 (Ω)N ) ∩ Hp1 ((0, T ), Lq (Ω)N ), 1 (Ω)) p ∈ Lp ((0, T ), Hq1 (Ω) + Hˆ q,0

satisfy the homogeneous equations: ⎧ ∂t v − Div (μD(v) − pI) = 0, ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ then, v = 0 and p = 0.

div v = 0

in Ω T ,

(μD(v) − pI)n = 0

on Γ T ,

v|t =0 = 0

on Ω,

(3.167)

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Y. Shibata

For Eq. (3.159) and (3.160), the uniqueness is defined as follows: •

If v, p and ρ with v ∈ Lp ((0, T ), Hq2 (Ω)N ) ∩ Hp1 ((0, T ), Lq (Ω)N ), 1 (Ω)), p ∈ Lp ((0, T ), Hq1 (Ω) + Hˆ q,0 3−1/q

ρ ∈ Lp ((0, T ), Wq

2−1/q

(Γ )) ∩ Hp1 ((0, T ), Wq

(Γ ))

satisfy the homogeneous equations: ⎧ ∂t v − Div (μD(v) − pI) = 0, div v = 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ∂t ρ + Aκ · ∇Γ ρ − v · n + F1 v = 0

in Ω T , on Γ T ,

⎪ ⎪ (μD(v) − pI − ((F2 + σ ΔΓ )ρ)I)n = 0 ⎪ ⎪ ⎪ ⎩ (v, ρ)|t =0 = (0, 0)

on Γ T ,

(3.168)

on Ω × Γ ,

then, v = 0, p = 0, and ρ = 0. Notice that when Aκ = 0, the uniqueness is stated in the same manner as in (3.168). We now state the maximal Lp –Lq regularity theorem. Theorem 3.4.2 Let 1 < p, q < ∞ with 2/p + 1/q = 1. Assume that Ω is a uniform C 2 domain and that the weak Dirichlet problem is uniquely solvable for index q. Then, there exists a γ0 > 0 such that the following assertion holds: Let 2(1−1/p) u0 ∈ Bq,p (Ω)N be initial data for Eq. (3.158) and let f, g, g, and h be functions in the right side of Eq. (3.158) such that f ∈ Lp ((0, T ), Lq (Ω)N ), and e−γ t g ∈ Hp1 (R, Lq (Ω)N ),

e−γ t g ∈ Lp (R, Hq1 (Ω)) ∩ Hp (R, Lq (Ω)), 1/2

e−γ t h ∈ Lp (R, Hq1 (Ω)N ) ∩ Hp (R, Lq (Ω)N ) 1/2

(3.169)

for any γ ≥ γ0 . Assume that the compatibility condition: u0 − g|t =0 ∈ Jq (Ω)

and div u0 = g|t =0

in Ω

(3.170)

holds. In addition, we assume that the compatibility condition: (μD(u0 )n)τ = (h|t =0 )τ

on Γ

(3.171)

holds for 2/p + 1/q < 1. Then, problem (3.158) admits solutions u and p with u ∈ Lp ((0, T ), Hq2 (Ω)N ) ∩ Hp1 ((0, T ), Lq (Ω)N ), 1 (Ω)) p ∈ Lp ((0, T ), Hq1 (Ω) + Hˆ q,0

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possessing the estimate: uLp ((0,T ),Hq2 (Ω)) + ∂t uLp ((0,T ),Lq (Ω)) ≤ Ceγ T (u0 B 2(1−1/p) (Ω) + fLp ((0,T ),Lq (Ω)) + e−γ t gHp1 (R,Lq (Ω)) q,p

+ e

−γ t

(g, h)Lp (R,Hq1 (Ω)) + e−γ t (g, h)H 1/2 (R,L p

q (Ω))

)

for some constant C > 0 and any γ ≥ γ0 , where C is independent of γ . Moreover, if we assume that the weak Dirichlet problem is uniquely solvable for q  = q/(q − 1) in addition, then the uniqueness for Eq. (3.158) holds. Theorem 3.4.3 Let 1 < p < ∞, 1 < q ≤ r and 2/p + 1/q = 1. Assume that Ω is a uniform C 3 domain and that the weak Dirichlet problem is uniquely solvable for 3−1/p−1/q (Γ ) be initial data for Eq. (3.159) and let d be a function in q. Let h0 ∈ Bq,p the right side of Eq. (3.159) such that 2−1/q

d ∈ Lp ((0, T ), Wq

(3.172)

(Γ )).

Then, problem (3.159) admits solutions v, q and h with v ∈ Lp ((0, T ), Hq2 (Ω)N ) ∩ Hp1 ((0, T ), Lq (Ω)N ), 1 (Ω)), q ∈ Lp ((0, T ), Hq1 (Ω) + Hˆ q,0 3−1/q

h ∈ Lp ((0, T ), Wq

2−1/q

(Γ )) ∩ Hp1 ((0, T ), Wq

(Γ ))

possessing the estimate: vLp ((0,T ),Hq2 (Ω)) + ∂t vLp ((0,T ),Lq (Ω)) + hL

3−1/q (Γ )) p ((0,T ),Wq

+ ∂t hL

2−1/q (Γ )) p ((0,T ),Wq

≤ Ceγ T (h0 B 3−1/p−1/q (Γ ) + dL q,p

2−1/q (Γ )) p ((0,T ),Wq

)

for some constants C > 0 and any γ ≥ γ0 , where C is independent of γ . Moreover, if we assme that the weak Dirichlet problem is uniquely solvable for q  = q/(q − 1) in addition, then the uniqueness for Eq. (3.159) holds. Theorem 3.4.4 Let 1 < p, q < ∞. Assume that Ω is a uniform C 3 domain and that the weak Dirichlet problem is uniquely solvable for q. Let d be a function in the right side of Eq. (3.160) such that 2−1/q

d ∈ Lp ((0, T ), Wq

(Γ )).

(3.173)

258

Y. Shibata

Then, problem (3.160) admits unique solutions w, r and ρ with w ∈ Lp ((0, T ), Hq2 (Ω)N ) ∩ Hq1 ((0, T ), Lq (Ω)N ), 1 r ∈ Lp ((0, T ), Hq1 (Ω) + Hˆ q,0 (Ω)), 3−1/q

ρ ∈ Lp ((0, T ), Wq

2−1/q

(Γ )) ∩ Hp1 ((0, T ), Wq

(Γ ))

possessing the estimate: wLp ((0,T ),Hq2 (Ω)) + ∂t wLp ((0,T ),Lq (Ω)) + ρL

3−1/q (Γ )) p ((0,T ),Wq

≤ Ceγ κ

−b T

dL

+ ∂t ρL

2−1/q (Γ )) p ((0,T ),Wq

2−1/q (Γ )) p ((0,T ),Wq

for some constants C > 0 and any γ ≥ γ0 with some γ0 , where C is independent of γ and κ ∈ (0, 1), where κ ∈ (0, 1) and b is the constant appearing in (3.164). Moreover, if we assume that Ω is a uniform C 3 domain whose inside has a finite covering and that the weak Dirichlet problem is uniquely solvable for q  = q/(q−1) in addition, then the uniqueness for Eq. (3.160) holds. Remark 3.4.5 The uniqueness follows from the existence of solutions to the dual problem in the case of Eqs. (3.158) and (3.159). But, the uniqueness for the Eq. (3.160) follows from the a priori estimates, because we can not find a suitable dual problem for Eq. (3.160). Thus, in addition, we need the assumption that the inside of Ω has a finite covering. Applying Theorems 3.4.2–3.4.4, we have the following corollary. Corollary 3.4.6 Let 1 < p < ∞, 1 < q ≤ r and 2/p + 1/q = 1. Assume that Ω is a uniform C 3 domain and that the weak Dirichlet problem is uniquely solvable for q. Then, there exists a γ0 for which the following assertion holds: Let 2(1−1/p) 3−1/p−1/q u0 ∈ Bq,p (Ω)N and h0 ∈ Bq,p (Γ ) be initial data for Eq. (3.162) and let f, g, g, d, and h be functions appearing in the right side of Eq. (3.162) and satisfying the conditions: 2−1/q

f ∈ Lp ((0, T ), Lq (Ω)N ),

d ∈ Lp ((0, T ), Wq

e−γ t g ∈ Lp (R, Hq1 (Ω)) ∩ Hp (R, Lq (Ω)), 1/2

e−γ t g ∈ Hp1 (R, Lq (Ω)N ),

e−γ t h ∈ Lp (R, Hq1 (Ω)N ) ∩ Hp (R, Lq (Ω)N ). 1/2

(Γ )),

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259

for any γ ≥ γ0 . Assume that the compatibility conditions (3.170) and (3.171) are satisfied. Then, problem (3.162) admits solutions U P , and H with U ∈ Lp ((0, T ), Hq2 (Ω)N ) ∩ Hp1 ((0, T ), Lq (Ω)N ), 1 (Ω)), P ∈ Lp ((0, T ), Hq1 (Ω) + Hq,0 3−1/q

H ∈ Lp ((0, T ), Wq

2−1/q

(Γ )) ∩ Hp1 ((0, T ), Wq

(Γ ))

possessing the estimate: U Lp ((0,T ),Hq2 (Ω)) + ∂t U Lp ((0,T ),Lq (Ω)) + H L + ∂t H L ≤ Ce2γ κ

−b T

3−1/q (Γ ) p ((0,T ),Wq

2−1/q (Γ )) p ((0,T ),Wq

{u0 B 2(1−1/p) (Ω) + κ −b h0 B 3−1/p−1/q (Γ ) q,p

q,p

+ fLp ((0,T ),Lq (Ω)) + e−γ t ∂t gLp (R,Lq (Ω)) + e−γ t (g, h)Lp (R,Hq1 (Ω)) + e−γ t (g, h)H 1/2 (R,L p

q (Ω))

+ dL

2−1/q (Γ )) p ((0,T ),Wq

}

for any γ ≥ γ0 with some constant C independent of γ and κ, where κ ∈ (0, 1), and b is the constant appearing in (3.164). Proof As was mentioned after Eq. (3.162), U , P and H are given by U = u+v+w, P = p+q+r and H = h+ρ. Since u and p are solutions of Eq. (3.158), by Theorem 3.4.2 we have uLp ((0,T ),Hq2 (Ω)) + ∂t uLp ((0,T ),Lq (Ω)) ≤ Ceγ T {u0 B 2(1−1/p) (Ω) + fLp ((0,T ),Lq (Ω)) + e−γ t ∂t gLp (R,Lq (Ω)) q,p

+ e−γ t (g, h)Lp (R,Hq1 (Ω)) + e−γ t (g, h)H 1/2 (R,L p

q (Ω))

}. (3.174)

Since v, q and h are solutions of Eq. (3.159) with d = 0, applying Theorem 3.4.3 with d = 0 yields that vLp ((0,T ),Hq2 (Ω)) + ∂t vLp ((0,T ),Lq (Ω)) + hL + ∂t hL

2−1/q

p ((0,T ),Wq

(Γ ))

3−1/q (Γ )) p ((0,T ),Wq

≤ Ceγ T h0 B 3−1/p−1/q (Γ ) .

(3.175)

q,p

Finally, recalling that w, r and ρ are solutions of Eq. (3.160) with d replaced by d + n · u − F1 u − Aκ · ∇Γ h, applying Theorem 3.4.4 and using the estimate Aκ · ∇Γ hW 2−1/q (Γ ) ≤ CAκ Hr2 (Ω) hW 3−1/q (Γ ) , q

q

260

Y. Shibata

which follows from the Sobolev imbedding theorem and the assumption: N < r < ∞, we have wLp ((0,T ),Hq2 (Ω)) + ∂t wLp ((0,T ),Lq (Ω)) + ρL ≤ Ceγ κ

3−1/q (Γ )) p ((0,T ),Wq

−b T

+ ∂t ρL

2−1/q (Γ )) p ((0,T ),Wq

{h0 B 3−1/p−1/q (Γ ) q,p

+ d + n · u − F1 u − Aκ · ∇Γ hL ≤ Ceγ κ

−b T

{κ −b hL

3−1/q (Γ ) p ((0,T ),Wq

2−1/q (Γ )) p ((0,T ),Wq

+ dL

}

2−1/q (Γ )) p ((0,T ),Wq

+ uLp ((0,T ),Hq2 (Ω)) }, which, combined with (3.174) and (3.175), leads to the required estimate. This completes the proof of Corollary 3.4.6. 

3.4.2 R Bounded Solution Operators To prove Theorems 3.4.2–3.4.4, we use R bounded solution operators associated with the following generalized resolvent problems: 

λu − Div (μD(u) − pI) = f,

div u = g = div g

in Ω,

(μD(u) − pI)n = h

on Γ ;

⎧ λv − Div (μD(v) − qI) = f, div v = g = div g ⎪ ⎪ ⎨ λρ − v · n + F1 v = d ⎪ ⎪ ⎩ (μD(v) − qI)n − (F2 ρ + σ ΔΓ ρ)n = h ⎧ ⎪ ⎪ λv − Div (μD(v) − qI) = f, div v = g = div g ⎨ λρ + Aκ · ∇Γ ρ − v · n + F1 v = d ⎪ ⎪ ⎩ (μD(v) − qI)n − (F2 ρ + σ ΔΓ ρ)n = h

(3.176)

in Ω, on Γ ,

(3.177)

on Γ ;

in Ω, on Γ ,

(3.178)

on Γ .

In the following, we consider Eq. (3.177) and (3.178) at the same time. For this, we set A0 = 0, and then Eq. (3.177) is represented by Eq. (3.178) with κ = 0.

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

261

We make a definition. Definition 3.4.7 Let X and Y be two Banach spaces. A family of operators T ⊂ L(X, Y ) is called R bounded on L(X, Y ), if there exist constants C > 0 and p ∈ [1, ∞) such that for each n ∈ N, {Tj }nj=1 ⊂ T , and {fj }nj=1 ⊂ X, we have 

n 

rk Tk fk Lp ((0,1),Y ) ≤ C

k=1

n 

rk fk Lp ((0,1),X).

k=1

Here, the Rademacher functions rk , k ∈ N, are given by rk : [0, 1] → {−1, 1} t → sign (sin 2k πt). The smallest such C is called R-bound of T on L(X, Y ), which is denoted by RL(X,Y ) T . We introduce the definition of the uniqueness of solutions. For Eq. (3.176), the uniqueness is defined as follows: •

Let λ ∈ U ⊂ C. If u and p with u ∈ Hq2 (Ω)N ,

1 p ∈ Hq1 (Ω) + Hˆ q,0 (Ω)

satisfy the homogeneous equations: 

λu − Div (μD(u) − pI) = 0,

div u = 0

in Ω,

(μD(u) − pI)n = 0

on Γ ,

(3.179)

then u = 0 and p = 0. For Eq. (3.177) and (3.178), the uniqueness is defined as follows: •

Let λ ∈ U ⊂ C. If v, q and ρ with v ∈ Hq2 (Ω)N ,

1 q ∈ Hq1 (Ω) + Hˆ q,0 (Ω),

3−1/q

ρ ∈ Wq

(Γ )

satisfy the homogeneous equations: ⎧ λv − Div (μD(v) − qI) = 0, div v = 0 ⎪ ⎪ ⎨ λρ + Aκ · ∇Γ ρ − v · n + F1 v = 0 ⎪ ⎪ ⎩ (μD(v) − qI − ((F2 + σ ΔΓ )ρ)I)n = 0 then u = 0, q = 0 and ρ = 0. We have the following theorems.

in Ω, on Γ , on Γ ,

(3.180)

262

Y. Shibata

Theorem 3.4.8 Let 1 < q < ∞ and 0 < 0 < π/2. Assume that Ω is a uniform C 3 domain and that the weak Dirichlet problem is uniquely solvable for q. Let A0 = 0 and let Aκ (κ ∈ (0, 1)) be an N − 1 vector of real valued functions satisfying (3.164). Assume that 1 < q ≤ r when κ ∈ (0, 1). (1) (Existence) Let Xq (Ω) = {F = (f, d, h, g, g) | f ∈ Lq (Ω)N , (g, g) ∈ DIq (Ω), 2−1/q

h ∈ Hq1 (Ω)N , d ∈ Wq

(Γ )},

Xq (Ω) = {F = (F1 , F2 , . . . , F7 ) | F1 , F3 , F7 ∈ Lq (Ω)N , F4 ∈ Hq1 (Ω)N , 2−1/q

Λκ,λ0

(Γ )}, F5 ∈ Lq (Ω), F6 ∈ Hq1 (Ω), F2 ∈ Wq   Σ 0 ,λ0 for κ = 0, 1 for κ = 0, γκ = = −b for κ ∈ (0, 1). C+,λ0 for κ ∈ (0, 1), κ

Then, there exist a constant λ0 > 0 and operator families A1 (λ), P1 (λ) and H1 (λ) with A1 (λ) ∈ Hol (Λκ,λ0 γκ , L(Xq (Ω), Hq2(Ω)N )), 1 (Ω))), P1 (λ) ∈ Hol (Λκ,λ0 γκ , L(Xq (Ω), Hq1(Ω) + Hˆ q,0

H1 (λ) ∈ Hol (Λκ,λ0 γκ , L(Xq (Ω), Hq3(Ω))) such that for every λ = γ + iτ ∈ Λκ,λ0 γκ and (f, d, h, g, g) ∈ Xq (Ω), v = A1 Fλ and q = P1 (λ)Fλ , and ρ = H1 (λ)Fλ are solutions of Eq. (3.178), where Fλ = (f, d, λ1/2 h, h, λ1/2 g, g, λg). Moreover, we have RL(X

2−j (Ω)N ) q (Ω),Hq

({(τ ∂τ )# (λj/2 A1 (λ)) | λ ∈ Λκ,λ0 γκ }) ≤ rb ;

RL(Xq (Ω),Lq (Ω)N ) ({(τ ∂τ )# (∇P1 (λ)) | λ ∈ Λκ,λ0 γκ }) ≤ rb ; RL(Xq (Ω),H 3−k (Ω)) ({(τ ∂τ )# (λk H1 (λ)) | λ ∈ Λκ,λ0 γκ }) ≤ rb q

for # = 0, 1, j = 0, 1, 2, and k = 0, 1 with some constant rb > 0. (2) (Uniqueness) (i) When κ = 0, if we assume that the weak Dirichlet problem is uniquely solvable for q  = q/(q − 1) in addition, then the uniqueness for Eq. (3.177) holds.

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

263

(ii) When κ ∈ (0, 1), if we assume that Ω is a uniformly C 3 domain whose inside has a finite covering and that the weak Dirichlet problem is uniquely solvable for q  = q/(q − 1) in addition, then the uniqueness for Eq. (3.178) holds. Remark 3.4.9 (1) F1 , F2 , F3 , F4 , F5 , F6 , and F7 are variables corresponding to f, d, λ1/2 h, h, λ1/2 g, g, and, λg, respectively. (2) We define the norms  · Xq (Ω) and  · Xq (Ω) by (f, d, h, g, g)Xq (Ω) = (f, g)Lq (Ω) + (g, h)Hq1 (Ω) + dW 2−1/q (Γ ) ; q

(F1 , . . . , F7 )Xq (Ω) = (F1 , F3 , F7 )Lq (Ω) + (F4 , F6 )Hq1 (Ω) + F2 W 2−1/q (Γ ) . q

Theorem 3.4.10 Let 1 < q < ∞ and 0
0 and operator families A2 (λ) and P2 (λ) with A2 (λ) ∈ Hol (Σ

0 ,λ0

, L(X˜q (Ω), Hq2 (Ω)N )),

P2 (λ) ∈ Hol (Σ

0 ,λ0

1 , L(X˜q (Ω), Hq1 (Ω) + Hˆ q,0 (Ω)))

such that for every λ ∈ Σ 0 ,λ0 and (f, h, g, g) ∈ X˜ q (Ω), u = A2 (λ)F˜ λ and p = P2 (λ)F˜ λ are solutions of Eq. (3.176), where we have set F˜ λ = (f, λ1/2 h, h, g, λ1/2 g, λg).

Moreover, we have RL(X˜

({(τ ∂τ )# (λj/2 A2 (λ)) | λ ∈ Σ

0 ,λ0

}) ≤ rb ;

RL(X˜q (Ω),Lq (Ω)N ) ({(τ ∂τ )# (∇P2 (λ)) | λ ∈ Σ

0 ,λ0

}) ≤ rb

2−j (Ω)N ) q (Ω),Hq

for # = 0, 1, j = 0, 1, 2 with some constant rb > 0.

264

Y. Shibata

(Uniqueness) If we assume that the weak Dirichlet problem is uniquely solvable for q  = q/(q − 1) in addition, then the uniqueness for Eq. (3.176) holds.

3.4.3 Stokes Operator and Reduced Stokes Operator Since the pressure term has no time evolution, sometimes it is convenient to eliminate the pressure terms, for example to formulate the problem in the semigroup setting. For u ∈ Hq2 (Ω)N and h ∈ Hq3 (Ω), we introduce functionals K0 (u) and 1 (Ω) be a unique solution of the weak Dirichlet K(u, h). Let K0 (u) ∈ Hq1 (Ω) + Hˆ q,0 problems: (∇K0 (u), ∇ϕ)Ω = (Div (μD(u)) − ∇div u, ∇ϕ)Ω

(3.181)

for any ϕ ∈ Hˆ q1 ,0 (Ω), subject to K0 (u) = μ < D(u)n, n > −div u on Γ . 1 (Ω) be a unique solution of the weak Dirichlet And, let K(u, h) ∈ Hq1 (Ω) + Hˆ q,0 problem:

(∇K(u, h), ∇ϕ)Ω = (Div (μD(u)) − ∇div u, ∇ϕ)Ω

(3.182)

for any ϕ ∈ Hˆ q1 ,0 (Ω), subject to K(u, h) = μ < D(u)n, n > −(F2 h + σ ΔΓ h) − div u

on Γ .

By Remark 3.2.6, we know the unique existence of K0 (u) and K(u, h) satisfying the estimates: ∇K0 (u)Lq (Ω) ≤ C∇uHq1 (Ω) , ∇K(u, h)Lq (Ω) ≤ C(∇uHq1 (Ω) + hW 3−1/q (Γ ) )

(3.183)

q

for some constant C depending on q. We consider the reduced Stokes equations: 

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

λu − Div (μD(u) − K0 (u)I) = f

in Ω,

(μD(u) − K0 (u)I)n = h

on Γ ;

(3.184)

λu − Div (μD(u, h) − K(u, h)I) = f

in Ω,

λh + Aκ · ∇Γ h − n · u + F1 u = d

on Γ ,

(μD(u) − K(u, h)I)n − (F2 h + σ ΔΓ h)n = h

on Γ .

(3.185)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

265

Notice that both of the boundary conditions in Eq. (3.184) and (3.185) are equivalent to (μD(u)n)τ = hτ

and div u = n · h on Γ .

(3.186)

We now study the equivalence between Eqs. (3.176) and (3.184). The equivalence between Eqs. (3.178) and (3.185) are similarly studied. We first assume that Eq. (3.176) is uniquely solvable. Given f ∈ Lq (Ω)N in the right side of Eq. (3.184), let g ∈ Hq1 (Ω) be a unique solution of the variational equation: λ(g, ϕ)Ω + (∇g, ∇ϕ)Ω = (−f, ∇ϕ)Ω

for any ϕ ∈ Hq1 ,0 (Ω)

(3.187)

subject to g = n · h on Γ . The unique existence of g is guaranteed for λ ∈ Σ with some large λ0 > 0. From (3.187) it follows that (g, ϕ)Ω = (−λ−1 (f + ∇g), ∇ϕ)Ω ,

0 ,λ0

(3.188)

and so (g, g) ∈ DIq with g = λ−1 (f+∇g). Thus, from the assumption we know that 1 (Ω). In Eq. (3.176) admits unique solutions u ∈ Hq2 (Ω)N and p ∈ Hq1 (Ω) + Hˆ q,0 view of the first equation in Eq. (3.176) and Definition 3.4.1, for any ϕ ∈ Hˆ q1 ,0 (Ω) we have (f, ∇ϕ)Ω = (λu − Div (μD(u) − pI), ∇ϕ)Ω = (λu, ∇ϕ)Ω − (∇div u, ∇ϕ)Ω − (Div (μD(u)) − ∇div u, ∇ϕ)Ω + (∇p, ∇ϕ)Ω = λ(λ−1 (f + ∇g), ∇ϕ)Ω − (∇g, ∇ϕ)Ω + (∇(p − K0 (u)), ∇ϕ)Ω , and so, (∇(p − K0 (u)), ∇ϕ)Ω = 0

for any ϕ ∈ Hˆ q1 ,0 (Ω).

Moreover, by the second equation in Eq. (3.176) and (3.187), we have p − K0 (u) = −h · n + div u = −g + g = 0

on Γ .

Thus, the uniqueness implies that p = K0 (u), which, combined with Eq. (3.176), shows that u satisfies Eq. (3.184). Conversely, we assume that Eq. (3.184) is uniquely solvable. Let f ∈ Lq (Ω)N and h ∈ Hq1 (Ω). Let θ ∈ Hq1 (Ω) + Hˆ q1 (Ω) be a unique solution of the weak Dirichlet problem: (∇θ, ∇ϕ)Ω = (f, ∇ϕ)Ω

for any ϕ ∈ Hˆ q1 ,0 (Ω),

266

Y. Shibata

subject to θ = n · h on Γ . Setting p = θ + q in (3.176), we then have λu − Div (μD(u) − qI) = f − ∇θ,

div u = g = div g in Ω,

(μD(u) − qI)n = h− < h, n > n on Γ . Let f = f − ∇θ and h = h− < h, n > n, and then h · n = 0 on Γ

for any ϕ ∈ Hˆ q1 ,0 (Ω).

and (f , ∇ϕ)Ω = 0

(3.189)

1 (Ω) be a solution to the weak Given (g, g) ∈ DIq (Ω)N , let K ∈ Hq1 (Ω) + Hˆ q,0 Dirichlet problem:

(∇K, ∇ϕ)Ω = (λg − ∇g, ∇ϕ)Ω

for any ϕ ∈ Hˆ q1 ,0 (Ω),

(3.190)

subject to K = −g on Γ . Let u be a solution of the equations: 

λu − Div (μD(u) − K0 (u)I) = f + ∇K 

(μD(u) − K0 (u)I)n = h + gn

in Ω, on Γ .

(3.191)

1 (Ω) we have By (3.181), and (3.189)–(3.191), for any ϕ ∈ Hˆ q,0

(∇K, ∇ϕ)Ω = (f + ∇K, ∇ϕ)Ω = (λu − Div (μD(u) − K0 (u)I), ∇ϕ)Ω

(3.192)

= (λu, ∇ϕ)Ω − (∇div u, ∇ϕ)Ω . Since Hq1 ,0 (Ω) ⊂ Hˆ q1 ,0 (Ω), by (3.190) and (3.192) we have λ(div u − g, ϕ)Ω + (∇(div u − g), ∇ϕ)Ω = 0 for any ϕ ∈ Hq1 ,0 (Ω). Here, we have used the fact that div g = g ∈ Hq1 (Ω). Recalling that h · n = 0 and putting (3.191) and the boundary condition of (3.181) together gives g = (gn + h ) · n =< μD(u)n, n > −K0 (u) = div u on Γ . Thus, the uniqueness implies that div u = g in Hq1 (Ω), which, combined with (3.190) and (3.192), leads to (g, ∇ϕ)Ω = (u, ∇ϕ)Ω

for any ϕ ∈ Hˆ q1 ,0 (Ω),

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

267

because we may assume that λ = 0. Since K = −g on Γ , by (3.192) we have λu − Div (μD(u) − (K0 (u) − K)I) = f ,

div u = g = div g

in Ω,



on Γ .

(μD(u) − (K0 (u) − K)I)n = h

Thus, u and p = K0 (u) − K − θ are required solutions of Eq. (3.176).

3.4.4 R-Bounded Solution Operators for the Reduced Stokes Equations In the following theorem, we state the existence of R bounded solution operators for the reduced Stokes equations (3.185). Theorem 3.4.11 Let 1 < q < ∞ and 0 < < π/2. Let Λκ,λ0 be the set defined in Theorem 3.4.8. Assume that Ω is a uniform C 3 domain and the weak Dirichlet problem is uniquely solvable for q. Let A0 = 0 and Aκ (κ ∈ (0, 1)) be an N − 1 vector of real valued functions satisfying (3.164). Assume that 1 < q ≤ r when κ ∈ (0, 1). Set Yq (Ω) = {(f, d, h) | f ∈ Lq (Ω)N ,

2−1/q

d ∈ Wq

(Γ ), h ∈ Hq1 (Ω)N }, 2−1/q

Yq (Ω) = {(F1 , . . . , F4 ) | F1 , F3 ∈ Lq (Ω)N , F2 ∈ Wq

(Γ )

F4 ∈ Hq1 (Ω)N }. (3.193) Then, there exist a constant λ∗ ≥ 1 and operator families: Ar (λ) ∈ Hol (Λκ,λ∗ γκ , L(Yq (Ω), Hq2 (Ω)N )), Hr (λ) ∈ Hol (Λκ,λ∗ γκ , L(Yq (Ω), Hq3 (Ω))) such that for any λ ∈ Λκ,λ∗ γκ and (f, d, h) ∈ Yq (Ω), u = Ar (λ)(f, d, λ1/2 h, h),

h = Hr (λ)(f, d, λ1/2 h, h),

are solutions of (3.185), and RL(Y

2−j (Ω)N ) q (Ω),Hq

({(τ ∂τ )# (λj/2 Ar (λ)) | λ ∈ Λκ,λ∗ γκ }) ≤ rb ,

RL(Yq (Ω),Hq3−k (Ω)) ({(τ ∂τ )# (λk Hr (λ)) | λ ∈ Λκ,λ∗ γκ }) ≤ rb

268

Y. Shibata

for # = 0, 1, j = 0, 1, 2 and k = 0, 1. Here, rb is a constant depending on m1 , m2 , m3 , λ∗ , p, q, and N, but independent of κ ∈ (0, 1), and γκ is the number defined in Theorem 3.4.8. If we assume that Ω is a uniformly C 3 domain whose inside has a finite covering and that the weak Dirichlet problem is uniquely solvable for q  = q/(q − 1) in addition, then the uniqueness for Eq. (3.185) holds. Remark 3.4.12 The norm of space Yq (Ω) is defined by (f, d, h)Yq (Ω) = fLq (Ω) + dW 2−1/q (Γ ) + hHq1 (Ω) ; q

and the norm of space Yq (Ω) is defined by (F1 , F2 , F3 , F4 )Yq (Ω) = (F1 , F3 )Lq (Ω) + F2 W 2−1/q (Γ ) + F4 Wq1 (Ω) . q

Remark 3.4.13 By the equivalence of Eqs. (3.178) and (3.185) that was pointed out in Sect. 3.4.3, we see easily that Theorem 3.4.8 follows immediately from Theorem 3.4.11. Concerning the existence of R bounded solution operator for Eq. (3.184), we have the following theorem. Theorem 3.4.14 Let 1 < q < ∞ and 0 < < π/2. Assume that Ω is a uniform C 2 domain and the weak Dirichlet problem is uniquely solvable for q. Set Y˜q (Ω) = {(f, h) | f ∈ Lq (Ω)N , h ∈ Hq1 (Ω)N }, Y˜ q (Ω) = {(F1 , F3 , F4 ) | F1 , F3 ∈ Lq (Ω)N , F4 ∈ Hq1 (Ω)N }.

(3.194)

Then, there exist a constant λ∗∗ ≥ 1 and operator family A0r (λ) with A0r (λ) ∈ Hol (Σσ,λ∗∗ , L(Y˜q (Ω), Hq2(Ω)N )) such that for any λ ∈ Σσ,λ∗∗ and (f, h) ∈ Y˜q (Ω), u = A0r (λ)(f, λ1/2 h, h) is a solution of (3.184), and RL(Y˜

2−j (Ω)N ) q (Ω),Hq

({(τ ∂τ )# (λj/2 A0r (λ)) | λ ∈ Σσ,λ∗∗ }) ≤ rb

for # = 0, 1 and j = 0, 1, 2. Here, rb is a constant depending on m1 , λ0 , p, q, and N. If we assume that the weak Dirichlet problem is uniquely solvable for q  = q/(q − 1) in addition, then the uniqueness for Eq. (3.184) holds. Remark 3.4.15 Theorem 3.4.14 was proved by Shibata [43] and can be proved by the same argument as in the proof of Theorem 3.4.11. Thus, we may omit its proof.

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269

3.4.5 Generation of C 0 Analytic Semigroup In this subsection, we consider the following two initial-boundary value problems for the reduced Stokes operator: ⎧ ∂t u − Div (μD(u) − K0 (u)I) = 0 ⎪ ⎪ ⎨ (μD(u) − K0 (u)I)n = 0 ⎪ ⎪ ⎩ u|t =0 = u0 ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

in Ω ∞ , on Γ ∞ ,

(3.195)

in Ω;

∂t v − Div (μD(v) − K(v, h)I) = 0

in Ω ∞ ,

∂t h − n · v + F1 v = 0

on Γ ∞ ,

⎪ (μD(v) − K(v, h)I)n − (F2 h + σ ΔΓ h)n = 0 ⎪ ⎪ ⎪ ⎪ ⎩ v|t =0 = v0 in Ω, h|t =0 = h0 on Γ ,

on Γ ∞ ,

(3.196)

where we have set Ω ∞ = Ω × (0, ∞) and Γ ∞ = Γ × (0, ∞). Let u and (v, h) be solutions of Eqs. (3.195) and (3.196), respectively. Roughly speaking, if u ∈ Jq (Ω) for any t > 0, then, u and p = K0 (u) are unique solutions of Eq. (3.158) with g = g = h = 0. And, if v ∈ Jq (Ω) for any t > 0, then v, q = K(v, h), and h are unique solutions of (3.159) with d = 0 and (v, h)|t =0 = (v0 , h0 ). Let us introduce spaces and operators to describe (3.195) and (3.196) in the semigroup setting. Let Dq1 (Ω) = {u ∈ Jq (Ω) ∩ Hq2 (Ω)N | (μD(u)n)τ = 0 on Γ }, A1q u = Div (μ(D(u) − K0 (u)I)

for u ∈ Dq1 (Ω); 2−1/q

Hq (Ω) = {(v, h) | v ∈ Jq (Ω), h ∈ Wq

(Γ )}, 3−1/q

Dq2 (Ω) = {(v, h) ∈ Hq (Ω) | v ∈ Hq2 (Ω)N , h ∈ Wq

(3.197) (Γ ),

(μD(v)n)τ = 0 on Γ }, A2q (v, h) = (Div (μD(v) − K(v, h)I), n · v|Γ ) for (v, h) ∈ Dq2 (Ω). Since div u = 0 for u ∈ Dq1 (Ω), by (3.186) (μD(u)n)τ = 0 is equivalent to (μD(u) − K0 (u)I)n = 0. And also, for (v, h) ∈ Dq2 (Ω), (μD(v)n)τ = 0 is equivalent to (μD(v) − K(v, h)I)n − (F2 h + σ ΔΓ h) = 0.

270

Y. Shibata

Using the symbols defined in (3.197), we see that Eqs. (3.195) and (3.196) are written as ∂t u − A1q u = 0

(t > 0),

u|t =0 = u0 ;

(3.198)

∂t U (t) − A2q U (t) = 0

(t > 0),

U (t)|t =0 = U0 ,

(3.199)

where u(t) ∈ Dq1 (Ω) for t > 0 and u0 ∈ Jq (Ω), and U (t) = (v, h) ∈ Dq2 (Ω) for t > 0 and U0 = (v0 , h0 ) ∈ Hq (Ω). The corresponding resolvent problem to (3.198) is that for any f ∈ Jq (Ω) and λ ∈ Σ 0 ,λ0 we find u ∈ Dq1 (Ω) uniquely solving the equation: λu − A1q u = f

in Ω

(3.200)

possessing the estimate: |λ|uLq (Ω) + uHq2 (Ω) ≤ CfLq (Ω) .

(3.201)

And also, the corresponding resolvent problem to (3.199) is that for any F ∈ Hq (Ω) and λ ∈ Σ 0 ,λ0 we find U ∈ Dq2 (Ω) uniquely solving the equation: λU − A2q U = F

in Ω × Γ

(3.202)

possessing the estimate: |λ|U Hq (Ω) + U Dq2 (Ω) ≤ CF Hq (Ω) ,

(3.203)

where for U = (v, h) we have set U Hq (Ω) = vLq (Ω) + hW 2−1/q (Γ ) , U Dq2 (Ω) = vHq2 (Ω) + hW 3−1/q (Γ ) . q

q

Since R boundedness implies boundedness as follows from Definition 3.4.7 with n = 1, by Theorem 3.4.14, we know the unique existence of u ∈ Dq1 (Ω) satisfying (3.200) and (3.201). And also, by Theorem 3.4.11, we know the unique existence of U ∈ Dq2 (Ω) satisfying (3.202) and (3.203). Thus, using standard semigroup theory, we have the following theorem. Theorem 3.4.16 Let 1 < q < ∞. Assume that Ω is a uniform C 2 domain in RN and that the weak Dirichlet problem is uniquely solvable for q and q  = q/ (q − 1). Then, problem (3.198) generates a C 0 analytic semigroup {T1 (t)}t ≥0 on

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

271

Jq (Ω) satisfying the estimates: T1 (t)u0 Lq (Ω) + t (∂t T1 (t)u0 Lq (Ω) + T1 (t)u0 Hq2 (Ω) ) ≤ Ceγ t u0 Lq (Ω) , (3.204) ∂t T1 (t)u0 Lq (Ω) + T1 (t)u0 Hq2 (Ω) ≤ Ce u0 Hq2 (Ω) γt

(3.205)

for any t > 0 with some constants C > 0 and γ > 0. Theorem 3.4.17 Let 1 < q < ∞. Assume that Ω is a uniform C 3 domain in RN and that the weak Dirichlet problem is uniquely solvable for q and q  = q/(q − 1). Then, problem (3.199) generates a C 0 analytic semigroup {T2 (t)}t ≥0 on Hq (Ω) satisfying the estimates: T2 (t)U0 Hq (Ω) + t (∂t T2 (t)U0 Hq (Ω) + T2 (t)U0 Dq2 (Ω) ) ≤ Ceγ t U0 Hq (Ω) , (3.206) ∂t T2 (t)U0 Hq (Ω) + T2 (t)U0 Dq2 (Ω) ≤ Ceγ t U0 Dq2 (Ω)

(3.207)

for any t > 0 with some constants C > 0 and γ > 0. We now show the following maximal Lp –Lq regularity theorem for Eqs. (3.195) and (3.196). Theorem 3.4.18 Let 1 < p, q < ∞. Assume that Ω is a uniform C 2 domain in RN and that the weak Dirichlet problem is uniquely solvable for q and q  = q/(q − 1). 1 (Ω) be a subspace of B 2(1−1/p) (Ω)N defined by Let Dq,p q,p 1 Dq,p (Ω) = (Jq (Ω), Dq1 (Ω))1−1/p,p ,

where (·, ·)1−1/p,p denotes a real interpolation functor. Then, there exists a γ > 1 (Ω), problem (3.195) admits a unique 0 such that for any initial data u0 ∈ Dq,p solution u with e−γ t u ∈ Hp1 ((0, ∞), Lq (Ω)N ) ∩ Lp ((0, ∞), Hq2(Ω)N ), possessing the estimate: e−γ t ∂t uLp ((0,∞),Lq (Ω)) + e−γ t uLp ((0,∞),Hq2 (Ω)) ≤ Cu0 B 2(1−1/p) (Ω) . q,p (3.208) Here, for any Banach space X with norm  · X we have set e−γ t f Lp ((a,b),X) =



b a

(e−γ t f (t)X )p dt

1/p .

272

Y. Shibata 2(1−1/p)

1 (Ω) ⊂ B Remark 3.4.19 Since Dq,p (Ω), in view of a boundary trace q,p 1 (Ω), we have theorem, we see that for u0 ∈ Dq,p

⎧ ⎪ ⎪ ⎨ u0 ∈ Jq (Ω),

(μD(u0 )n)τ = 0

on Γ

⎪ ⎪ ⎩ u0 ∈ Jq (Ω) 1−2/p

because D(u0 ) ∈ Bq,p exist for

2 p

+

1 q

(Ω), and so D(u0 )|Γ exists for

for

1 2 + < 1, p q

for

1 2 + > 1, p q

2 p

+

1 q

< 1, but it does not

> 1.

Theorem 3.4.20 Let 1 < p, q < ∞. Assume that Ω is a uniform C 3 domain in RN and that the weak Dirichlet problem is uniquely solvable for q and q  = q/(q − 1). 2 (Ω) be a subspace of B 2(1−1/p) (Ω)N × B 3−1/p−1/q (Γ ) defined by Let Dq,p q,p q,p 2 Dq,p (Ω) = (Hq (Ω), Dq2 (Ω))1−1/p,p . 2 (Ω), Then, there exists a γ > 0 such that for any initial data (u0 , h0 ) ∈ Dq,p problem (3.196) admits a unique solution U = (v, h) with

e−γ t U ∈ Hp1 ((0, ∞), Hq (Ω)) ∩ Lp ((0, ∞), Dq2 (Ω)) possessing the estimate: e−γ t ∂t U Lp ((0,∞),Hq (Ω)) + e−γ t U Lp ((0,∞),Dq2 (Ω))

(3.209)

≤ C(u0 B 2(1−1/p) (Ω) + h0 B 3−1/p−1/q (Γ ) ). q,p

q,p

Proof of Theorem 3.4.20 We only prove Theorem 3.4.20, because Theorem 3.4.18 can be proved by the same argument. To prove Theorem 3.4.20, we observe that 

∞ 0

=

≤

(e−γ t ∂t T2 (t)U0 Hq (Ω) )p dt

∞  

2j+1

j j =−∞ 2

∞  j =−∞

1/p

(e−γ t ∂t T2 (t)U0 Hq (Ω) )p dt

(2j +1 − 2j )(

sup t ∈(2j ,2j+1 )

1/p

(e−γ t ∂t T2 (t)U0 Hq (Ω) )p

1/p .

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

273

We now introduce Banach spaces #sp for s ∈ R and 1 ≤ p ≤ ∞ which are sets of all sequences, (aj )j ∈Z , such that (aj )j ∈Z #sp < ∞, where we have set (aj )j ∈Z #sp

⎧

⎨ ∞ (2j s |a |)p 1/p j j =1 = ⎩sup{2j s |a | | j ∈ Z} j

for 1 ≤ p < ∞, for p = ∞.

Let aj =

sup t ∈(2j ,2j+1 )

e−γ t ∂t T2 (t)U0 Hq (Ω) ,

and then 

∞ 0

≤

(e−γ t ∂t T2 (t)u0 Hq (Ω) )p dt

∞ 

(2j/p aj )p

1/p

1/p (3.210)

= (aj )j ∈Z #1/p . p

j =−∞

By real interpolation theory (cf. Bergh and Löfström [8, 5.6.Theorem]), we have 1/p #p = (#1∞ , #0∞ )1−1/p,p . Moreover, by (3.206) and (3.207), we have (aj )j ∈Z #1∞ ≤ CU0 Hq (Ω) ,

(aj )j ∈Z #0∞ ≤ CU0 Dq (Ω) ,

and therefore, by real interpolation (aj )j ∈Z #1/p = (aj )j ∈Z (#1

0 ∞ ,#∞ )1−1/p,p

p

≤ CU0 (Hq (Ω),Dq (Ω))1−1/p,p .

Putting this and (3.210) together gives e−γ t ∂t T2 (t)U0 Lp ((0,∞),Hq (Ω)) ≤ CU0 Dq,p (Ω) . Analogously, we have e−γ t T2 (t)U0 Lp ((0,∞),Dq (Ω)) ≤ CU0 Dq,p (Ω) . We now prove the uniqueness. Let U satisfy the homogeneous equation: ∂t U − A2q U = 0

(t > 0),

U |t =0 = 0,

(3.211)

and the condition: e−γ t U ∈ Hp1 ((0, ∞), Hq (Ω)) ∩ Lp ((0, ∞), Dq2 (Ω)).

(3.212)

274

Y. Shibata

Let U0 be the zero extension of U to t < 0. In particular, from (3.211) it follows that ∂t U0 − A2q U0 = 0

(t ∈ R).

(3.213)

For any λ ∈ C with Re λ > γ , we set Uˆ (λ) =



∞ −∞

e−λt U0 (t) dt =



∞

e−λt U (t) dt.

0

By (3.212) and Hölder’s inequality, we have Uˆ (λ)Dq (Ω) ≤



∞



e−(Re λ−γ )tp dt

1/p

0

e−γ t U Lp ((0,∞),Dq2 (Ω))



≤ ((Re λ − γ )p )−1/p e−γ t U Lp ((0,∞),Dq2 (Ω)) . Since λUˆ (λ) =

∞ 0

e−λt ∂t U (t) dt, we also have 

λUˆ (λ)Hq (Ω) ≤ ((Re λ − γ )p )−1/p e−γ t ∂t U Lp ((0,∞),Hq (Ω)) . Thus, by (3.213), we have Uˆ (λ) ∈ Dq2 (Ω) satisfies the homogeneous equation: λUˆ (λ) − A2q Uˆ (λ) = 0

in Ω × Γ .

Since the uniqueness of the resolvent problem holds for λ ∈ Σ 0 ,λ0 , we have Uˆ (λ) = 0 for any λ ∈ C with Re λ > max(λ0 , γ ). By the Laplace inverse transform, we have U0 (t) = 0 for t ∈ R, that is U (t) = 0 for t > 0, which shows the uniqueness. This completes the proof of Theorem 3.4.20. 

3.4.6 Proof of Maximal Regularity Theorem We first prove Theorem 3.4.2 with the help of Theorem 3.4.10. Proof of Theorem 3.4.2 The key tool in the proof of Theorem 3.4.2 is the Weis operator valued Fourier multiplier theorem. To state it we need to make a few definitions. For a Banach space, X, D(R, X) denotes the space of X-valued C ∞ (R) functions with compact support D (R, X) = L(D(R), X) the space of X-valued distributions. And also, S(R, X) denotes the space of X-valued rapidly decreasing functions and S  (R, X) = L(S(R), X) the space of X-valued tempered distributions. Let Y be another Banach space. Then, given m ∈ L1,loc (R, L(X, Y )),

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

275

we define an operator Tm : F −1 D(R, X) → S  (R, Y ) by letting Tm φ = F −1 [mF [φ]] for all F φ ∈ D(R, X),

(3.214)

where F and F −1 denote the Fourier transform and its inversion formula, respectively. Definition 3.4.21 A Banach space X is said to be a UMD Banach space, if the Hilbert transform is bounded on Lp (R, X) for some (and then all) p ∈ (1, ∞). Here, the Hilbert transform H operating on f ∈ S(R, X) is defined by [Hf ](t) =

1 lim π →0+



f (s) ds t −s

|t −s|>

(t ∈ R).

Theorem 3.4.22 (Weis [74]) Let X and Y be two UMD Banach spaces and 1 < p < ∞. Let m be a function in C 1 (R \ {0}, L(X, Y )) such that RL(X,Y )({m(τ ) | τ ∈ R \ {0}}) = κ0 < ∞, RL(X,Y ) ({τ m (τ ) | τ ∈ R \ {0}}) = κ1 < ∞. Then, the operator Tm defined in (3.214) is extended to a bounded linear operator from Lp (R, X) into Lp (R, Y ). Moreover, denoting this extension by Tm , we have Tm f Lp (R,Y ) ≤ C(κ0 + κ1 )f Lp (R,X)

for all f ∈ Lp (R, X)

with some positive constant C depending on p. We now construct a solution of Eq. (3.158). Let f0 be the zero extension of f outside of (0, T ), that is f0 (t) = f(t) for t ∈ (0, T ) and f0 (t) = 0 for t ∈ (0, T ). Notice that f0 , g, g, and h are defined on the whole line R. Thus, we first consider the equations: ⎧ ∂t u1 − Div (μD(u1 ) − q1 I) = f0 ⎪ ⎪ ⎨ div u1 = g = div g ⎪ ⎪ ⎩ (μD(u1 ) − q1 I)n = h

in Ω × R, in Ω × R,

(3.215)

on Γ × R.

Let FL be the Laplace transform with respect to the time variable t defined by fˆ(λ) = FL [f ](λ) = for λ = γ + iτ ∈ C. Obviously,

 R

e−λt f (t) dt

276

Y. Shibata

 FL [f ](λ) =

e−iτ t e−γ t f (t) dt = F [e−γ t f ](τ ).

R

Applying the Laplace transform to Eq. (3.215) gives ⎧ λuˆ 1 − Div (μD(uˆ 1 ) − qˆ 1 I) = ˆf0 ⎪ ⎪ ⎨ div uˆ 1 = gˆ = div gˆ ⎪ ⎪ ⎩ (μD(uˆ 1 ) − qˆ 1 I)n = hˆ

in Ω, in Ω,

(3.216)

on Γ .

Applying Theorem 3.4.10, we have uˆ 1 = A2 (λ)Fλ and qˆ 1 = P2 (λ)Fλ for λ ∈ Σ 0 ,λ0 , where 1/2 ˆ ˆ Fλ = (ˆf0 (λ), λ1/2 h(λ), g(λ), ˆ g(λ), ˆ λˆg(λ)). h(λ)λ

Let FL−1 be the inverse Laplace transform defined by FL−1 [g](t)

1 = 2π



1 e g(τ ) dτ = e 2π R λt



γt

R

eiτ t g(τ ) dτ

for λ = γ + iτ ∈ C. Obviously, FL−1 [g](t) = eγ t F −1 [g](t),

FL FL−1 = FL−1 FL = I.

Setting Λγ1/2 f = FL−1 [λ1/2 FL [f ]] = eγ t F −1 [λ1/2 F [e−γ t f ]], and using the facts that λˆg(λ) = FL [∂t g](λ),

λ1/2 fˆ(λ) = FL [Λγ1/2 f ] = F [e−γ t Λγ1/2 f ]

for f ∈ {g, h}, we define u1 and q1 by u1 (·, t) = FL [A2 (λ)Fλ ] = eγ t F −1 [A2 (λ)F [e−γ t F  (t)](τ )], q1 (·, t) = FL [P2 (λ)Fλ ] = eγ t F −1 [P2 (λ)F [e−γ t F  (t)](τ )], with F  (t) = (f0 , Λγ h, h, Λγ g, g, ∂t g), where γ is chosen as γ > λ0 , and so γ + iτ ∈ Σ 0 ,λ0 for any τ ∈ R. By Cauchy’s theorem in the theory of one complex variable, u1 and q1 are independent of choice of γ whenever γ > λ0 and the condition (3.169) is satisfied for γ > λ0 . Noting that 1/2

1/2

∂t u1 = FL−1 [λA1 (λ)Fλ ] = eγ t F −1 [λA1 (λ)F [e−γ t F  (t)](τ )],

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

277

and applying Theorem 3.4.22, we have e−γ t ∂t u1 Lp (R,Lq (Ω)) + e−γ t u1 Lp (R,Hq2 (Ω)) + e−γ t ∇q1 Lp ((R,Lq (Ω)) ≤ Crb e−γ t F  Lp (R,Hq (Ω)) ≤ Crb {e−γ t fLp ((0,T ),Lq (Ω)) + e−γ t gLp (R,Hq1 (Ω)) + e−γ t Λγ1/2 gLp (R,Lq (Ω)) + e−γ t ∂t gLp (R,Lq (Ω)) + e−γ t hLp (R,Hq1 (Ω)) + e−γ t Λγ1/2 hLp (R,Lq (Ω)) }. (3.217) We now write solutions u and q of Eq. (3.158) by u = u1 + u2 and q = q1 + q2 , where u2 and q2 are solutions of the following equations: ⎧ ∂t u2 − Div (μD(u2 ) − q2 I) = 0, div u2 = 0 ⎪ ⎪ ⎨ (μD(u2 ) − q2 I)n = 0 ⎪ ⎪ ⎩ u2 = u0 − u1 |t =0

in Ω × (0, ∞), on Γ × (0, ∞),

(3.218)

in Ω.

Notice that div u2 = 0 in Ω × (0, ∞) means that u2 ∈ Jq (Ω) for any t > 0. By real interpolation theory, we know that sup e−γ t u1 (t)B 2(1−1/p) (Ω)

t ∈(0,∞)

q,p

≤ C(e−γ t u1 Lp ((0,∞),Hq2(Ω)) + e−γ t ∂t u1 Lp ((0,∞),Lq (Ω)) ).

(3.219)

In fact, this inequality follows from the following theory (cf. Tanabe[73, p.1]): Let X1 and X2 be two Banach spaces such that X2 is a dense subset of X1 , and then Lp ((0, ∞), X2 ) ∩ Hp1 ((0, ∞), X1 ) ⊂ C([0, ∞), (X1 , X2 )1−1/p,p ),

(3.220)

and sup u(t)(X1 ,X2 )1−1/p,p ≤ C(uLp ((0,∞),X2) + ∂t uLp ((0,∞),X1) ).

t ∈(0,∞)

2(1−1/p)

Since Bq,p

(Ω) = (Lp (Ω), Hq2 (Ω))1−1/p,p , we have (3.219). Thus, 2(1−1/p)

u0 − u1 |t =0 ∈ Bq,p

(Ω).

By the compatibility condition (3.170) and (3.215), we have (u0 − u1 |t =0 , ∇ϕ) = (u0 − g|t =0, ∇ϕ) = 0

for any ϕ ∈ Hˆ q1 ,0 (Ω).

(3.221)

278

Y. Shibata

Moreover, if 2/p+1/q < 1, then by the compatibility condition (3.171) and (3.215), we have (μD(u0 − u1 |t =0 )n)τ = (μD(u0 )n)τ − (h|t =0 )τ = 0 on Γ . 1 (Ω). Applying Theorem 3.4.18, we Thus, if 2/p + /q = 1, then u0 − u1 |t =0 ∈ Dq,p  see that there exists a γ > 0 such that Eq. (3.218) admits unique solutions u2 with q2 = K0 (u2 ) and 

e−γ t u2 ∈ Hp1 ((0, ∞), Lq (Ω)N ) ∩ Lp ((0, ∞), Hq2 (Ω)N )

(3.222)

possessing the estimate: 



e−γ t ∂t u2 Lp ((0,∞),Lq (Ω)) + e−γ t u2 Lp ((0,∞),Hq2 (Ω)) ≤ Cu0 − u1 |t =0 B 2(1−1/p) (Ω) .

(3.223)

q,p

Thus, setting u = u1 + u2 and q = q1 + K0 (u2 ) and choosing γ0 in such a way that γ0 > max(λ0 , γ  ), by (3.215), (3.217)–(3.219), (3.222) and (3.223), we see that u and q are required solutions of Eq. (3.158). Employing the same argument as in the proof of the uniqueness of Theorem 3.4.20, we can show the uniqueness. This completes the proof of Theorem 3.4.2. 

Employing the same argument as above, we can show Theorem 3.4.3, and so we may omit the proof of Theorem 3.4.3. Thus, we finally give a Proof of Theorem 3.4.4 Let d0 be the zero extension of d outside of (0, T ), that is d0 (t) = d(t) for t ∈ (0, T ) and d0 (t) = 0. Employing the same argument as in the proof of Theorem 3.4.2 above, we can show the existence of solutions, w, r and ρ, of the equations: ⎧ ∂t w − Div (μD(w) − rI) = 0, div u = 0 ⎪ ⎪ ⎨ ∂t ρ + Aκ · ∇Γ ρ − w · n + F1 w = d0 ⎪ ⎪ ⎩ (μD(w) − rI)n − (F2 ρ + σ ΔΓ ρ)n = 0

in Ω × R, on Γ × R,

(3.224)

on Γ × R,

possessing the estimate: e−γ t wLp (R,Hq2 (Ω)) + e−γ t ∂t wLp (R,Lq (Ω)) + e−γ t ρL

3−1/q (Γ )) p (R,Wq

≤ Ce−γ t d0 L

+ e−γ t ∂t ρL

2−1/q (Γ )) p (R,Wq

2−1/q (Γ )) p (R,Wq

≤ CdL

2−1/q (Γ )) q ((0,T ),Wq

(3.225)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

279

for any γ ≥ λ0 κ −b , where we have used (3.220) and (3.221). In particular, for any > 0 and γ ≥ λ0 κ −b , we have e−γ t wLp ((−∞,−

),Hq2 (Ω))

+ e−γ t ρL

≤ e−γ t wLp (R,Hq2 (Ω)) + e−γ t ρL

p ((−∞,−

3−1/q

),Wq

(Γ ))

≤ CdL

3−1/q (Γ )) p (R,Wq

2−1/q (Γ )) p ((0,T ),Wq

By the monotonicity of e−γ t , we have e−γ t wLp ((−∞,−

),Hq2 (Ω))

≥ e (wLp ((−∞,− γ

+ e−γ t ρL

),Hq2 (Ω))

3−1/q

p ((−∞,−

+ ρL

),Wq

p ((−∞,−

(Γ ))

3−1/q

),Wq

(Γ ))

).

Putting these inequalities together gives wLp ((−∞,−

),Hq2 (Ω))

≤ Ce−γ dL

+ ρL

p ((−∞,−

3−1/q

),Wq

(Γ ))

2−1/q (Γ )) p ((0,T ),Wq

for any γ ≥ λ0 κ −b . Thus, letting γ → ∞, we have wLp ((−∞,− Since

),Hq2 (Ω))

+ ρL

p ((−∞,−

3−1/q

),Wq

(Γ ))

= 0.

> 0 is chosen arbitrarily, we have wLp ((−∞,0),Hq2 (Ω)) + ρL

3−1/q (Γ )) p ((−∞,0),Wq

= 0,

which shows that w = 0 and ρ = 0 for t < 0, because 2(1−1/p)

w ∈ C(R, Bq,p

(Ω)N ),

3−1/p−1/q

ρ ∈ C(R, Wq,p

(Γ )).

By the monotonicity of e−γ t , we have e−γ t wLp (R,Hq2 (Ω)) + e−γ t ∂t wLp (R,Lq (Ω)) ≥ e−γ t wLp ((0,T ),Hq2 (Ω)) + e−γ t ∂t wLp ((0,T ),Lq (Ω)) ≥ e−γ T (wLp ((0,T ),Hq2 (Ω)) + ∂t wLp ((0,T ),Lq (Ω)) ). Similarly, we have e−γ t ρL

3−1/q (Γ )) p (R,Wq

≥ e−γ T (ρL

+ e−γ t ∂t ρL

3−1/q (Γ )) p ((0,T ),Wq

2−1/q (Γ )) p (R,Wq

+ ∂t ρL

2−1/q (Γ )) p ((0,T ),Wq

).

.

280

Y. Shibata

Thus, by (3.225), we have wLp ((0,T ),Hq2 (Ω)) + ∂t wLp ((0,T ),Lq (Ω)) + ρL

3−1/q (Γ )) p ((0,T ),Wq

≤ Ceγ T dL

+ ∂t ρL

2−1/q (Γ )) p ((0,T ),Wq

2−1/q (Γ )) p ((0,T ),Wq

for any γ ≥ λ0 κ −b . This completes the proof of the existence part of Theorem 3.4.4. Employing the same argument as in the proof of the uniqueness of Theorem 3.4.20, we can show the uniqueness. This completes the proof of Theorem 3.4.4. 

3.5 R Bounded Solution Operators In this section, we mainly prove Theorem 3.4.11. The operators F1 and F2 can be treated by perturbation method, and so the Sects. 3.5.1–3.5.8 below devote to proving the existence part for Eq. (3.185) for the following equations: ⎧ λu − Div (μD(u, h) − K(u, h)I) = f ⎪ ⎪ ⎨ λh + Aκ · ∇Γ h − n · u = d ⎪ ⎪ ⎩ (μD(u) − K(u, h)I)n − (σ ΔΓ h)n = h

in Ω, on Γ ,

(3.226)

on Γ .

And then, in Sect. 3.5.9, we prove the existence part of Theorem 3.4.11 for Eq. (3.185) by using a perturbation method. The existence part of Theorem 3.4.14 can be proved in the same manner as in the proof of the existence part of Theorem 3.4.11 and also has been proved by Shibata [43], and so we may omit its proof. Concerning the uniqueness part, we first prove Theorem 3.4.14 in Sect. 3.5.10. Finally, the uniqueness part of Theorem 3.4.11 will be proved in Sect. 3.5.11 by showing apriori estimates for Eq. (3.186) under the assumption that Ω is a uniform C 3 domain whose inside has a finite covering.

3.5.1 Model Problem in RN ; Constant μ Case In this subsection, we assume that μ is a constant satisfying the assumption (3.164), that is m0 ≤ μ ≤ m1 . Given u ∈ Hq2 (RN )N , let u = K0 (u) be a unique solution of the weak Laplace problem: (∇u, ∇ϕ)RN = (Div (μD(u)) − ∇div u, ∇ϕ)RN

(3.227)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

281

for any ϕ ∈ Hˆ q1 (RN ). In this subsection, we consider the resolvent problem: λu − Div (μD(u) − K0 (u)I) = f

in RN ,

(3.228)

and prove the following theorem. Theorem 3.5.1 Let 1 < q < ∞, 0 < < π/2, and λ0 > 0. Then, there exists an operator family A0 (λ) ∈ Hol (Σ ,λ0 , L(Lq (RN )N , Hq2 (RN )N )) such that for any λ = γ + iτ ∈ Σ ,λ0 and f ∈ Lq (RN )N , u = A0 (λ)f is a unique solution of Eq. (3.228) and RL(L

q (R

N )N ,H 2−j (RN )N ) q

({(τ ∂τ )# (λj/2 A0 (λ)) | λ ∈ Σ

,λ0 })

≤ rb (λ0 )

(3.229)

for # = 0, 1 and j = 0, 1, 2, where rb (λ0 ) is a constant depending on , λ0 , m0 , m1 , q and N, but independent of μ ∈ [m0 , m1 ]. Proof We first consider the Stokes equations: λu − Div (μ(D(u) − qI) = f,

div u = g = div g in RN .

(3.230)

Since Div (μD(u) − qI) = μΔu + μ∇div u − ∇q, applying div to (3.230), we have λdiv g − 2μΔg + Δq = div f, and so, q = 2μg + Δ−1 (div f − λdiv g). Combining this with (3.230) gives λu − μΔu = f − ∇Δ−1 div f − μ∇g + λ∇Δ−1 div g.

(3.231)

We now look for a solution formula for Eq. (3.228). Let g be a solution of the variational problem: (λg, ϕ)RN + (∇g, ∇ϕ)RN = (−f, ∇ϕ)RN

for any ϕ ∈ Hˆ q1 (RN ),

and then this g is given by g = (λ − Δ)−1 div f. According to (3.188), we set g = λ−1 (f + ∇g). Inserting these formulas into (3.231) gives λu − μΔu = f − (μ − 1)∇g = f − (μ − 1)(λ − Δ)−1 ∇div f.

282

Y. Shibata

Thus, we have u = Fξ−1

+ F [f](ξ ) , + , ξ ξ · F [f](ξ ) −1 + (μ − 1)F , ξ λ + μ|ξ |2 (λ + μ|ξ |2 )(λ + |ξ |2 )

where F and Fξ−1 denote the Fourier transform and its inversion formula defined by  F [f ](ξ ) =

RN

e

−ix·ξ

f (x) dx,

Fξ−1 [g(ξ )](x)

1 = (2π)N

 RN

eix·ξ g(ξ ) dξ.

Thus, we define an operator family A0 (λ) acting on f ∈ Lq (RN )N by A0 (λ)f = Fξ−1

+ F [f](ξ ) , + , ξ ξ · F [f](ξ ) −1 + (μ − 1)F . ξ λ + μ|ξ |2 (λ + μ|ξ |2 )(λ + |ξ |2 )

To prove the R-boundedness of A0 (λ), we use the following lemma. Lemma 3.5.2 Let 0
0 for any τ ∈ R \ {0}. Let Tn be an operator-valued Fourier multiplier defined by Tn f = F −1 [nF [f ]] for any f with F [f ] ∈ D(R, Lq (D)). Then, Tn is extended to a bounded linear operator from Lp (R, Lq (D)) into itself. Moreover, denoting this extension also by Tn , we have Tn L(Lp (R,Lq (D))) ≤ Cp,q,D γ .

284

Y. Shibata

Proof The assertions (a) and (b) follow from [17, p.28, Proposition 3.4], and the assertions (c) and (d) follow from [17, p.27, Remarks 3.2] (see also Bourgain [11]). 

3.5.2 Perturbed Problem in RN In this subsection, we consider the case where μ(x) is a real valued function satisfying (3.164). Let x0 be any point in RN and let d0 be a positive number such that Bd0 (x0 ) ⊂ RN . In view of (3.164), we assume that |μ(x) − μ(x0 )| ≤ m1 M1

for x ∈ Bd0 (x0 ),

(3.234)

where we have set M1 = d0 . We assume that M1 ∈ (0, 1) below. Let ϕ be a function in C0∞ (RN ) which equals 1 for x ∈ Bd0 /2 (x0 ) and 0 outside of Bd0 (x0 ). Let μ(x) ˜ = ϕ(x)μ(x) + (1 − ϕ(x))μ(x0).

(3.235)

Let K˜ 0 (u) ∈ Hˆ q1 (RN ) be a unique solution of the weak Laplace problem: (∇u, ∇ϕ)RN = (Div (μD(u)) ˜ − ∇div u, ∇ϕ)RN

for any ϕ ∈ Hˆ q1 (RN ). (3.236)

We consider the resolvent problem: λu − Div (μD(u) ˜ − K˜ 0 (u)I) = f in RN .

(3.237)

We shall prove the following theorem. Theorem 3.5.5 Let 1 < q < ∞ and 0 < λ0 ≥ 1 and an operator family A˜0 (λ) with A˜ 0 (λ) ∈ Hol (Σ such that for any λ ∈ Σ Eq. (3.237), and RL(L

q (R

,λ0

< π/2. Then, there exist M1 ∈ (0, 1),

,λ0 , L(Lq (R

) , Hq2 (RN )N ))

N N

˜ and f ∈ Lq (RN )N , u = A(λ)f is a unique solution of

N )N ,H 2−j (RN )N ) q

({(τ ∂τ )# (λj/2 A˜0 (λ)) | λ ∈ Σ

,λ0 })

≤ r˜b

for # = 0, 1 and j = 0, 1, 2. Here, r˜b is a constant independent of M1 and λ0 . Proof Let u = Kx0 (u) ∈ Hˆ q1 (RN ) be a unique solution of the weak Laplace equation: (∇u, ∇ϕ)RN = (Div (μ(x0 )D(u) − ∇div u, ∇ϕ)RN

(3.238)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

285

for any ϕ ∈ Hˆ q1 (RN ). We consider the resolvent problem: λu − Div (μ(x0 )D(u) − Kx0 (u)I) = f

in RN .

(3.239)

Let Bx0 (λ) ∈ Hol (Σ ,1 , L(Lp (RN )N , Hq2 (RN )N )) be a solution operator of Eq. (3.239) such that for any λ ∈ Σ ,1 and f ∈ Lq (RN )N , u = Bx0 (λ)f is a unique solution of Eq. (3.239) and RL(L

q (R

N )N ,H 2−j (RN )N ) q

({(τ ∂τ )# (λj/2 Bx0 (λ)) | λ ∈ Σ

,1 })

≤ γ0

(3.240)

for # = 0, 1 and j = 0, 1, 2, where γ0 is a constant independent of M1 and ∇ϕ. Such an operator is given in Theorem 3.5.1 with μ = μ(x0) and λ0 = 1. Inserting the formula: u = Bx0 (λ)f into (3.237) gives λu − Div (μ(x)D(u) ˜ − K˜ 0 (u)I) = f − R(λ)f

in RN ,

(3.241)

where we have set R(λ)f = Div (μ(x)D(B ˜ x0 (λ)f) − μ(x0 )D(Bx0 (λ)f)) − ∇(K˜ 0 (Bx0 (λ)f) − Kx0 (Bx0 (λ)f)).

(3.242)

We shall estimate R(λ)f. For any ϕ ∈ Hˆ q1 (RN ), by (3.236) and (3.238), we have (∇(K˜ 0 (Bx0 (λ)f) − Kx0 (Bx0 (λ)f)), ∇ϕ)RN = ((Div ((μ(x) ˜ − μ(x0))D(Bx0 (λ)f)), ∇ϕ)RN . Since μ(x) ˜ − μ(x0 ) = ϕ(x)(μ(x) − μ(x0 )), by (3.234) and (3.164), we have Div ((μ(x) ˜ − μ(x0))D(Bx0 (λ)f)Lq (RN ) ≤ M1 ∇ 2 Bx0 (λ)fLq (RN ) + Cm1 ,∇ϕ ∇Bx0 (λ)fLq (RN ) . Here and in the following, Cm1 ,∇ϕ denotes a generic constant depending on m1 and ∇ϕL∞ (RN ) . Thus, we have R(λ)fLq (RN ) ≤ CM1 ∇ 2 Bx0 (λ)fLq (RN ) + Cm1 ,∇ϕ ∇Bx0 (λ)fLq (RN ) . (3.243) Here and in the following, C denotes a generic constants independent of M1 , m1 , and ∇ϕL∞ (RN ) . Let λ0 be any number ≥1 and let n ∈ N, {λk }nk=1 ⊂ (Σ ,λ0 )n ,

286

Y. Shibata

and {Fk }nk=1 ⊂ (Lq (RN )N )n . By (3.243), (3.240) and Proposition 3.5.4, we have 

1 0



n 

q

rk (u)R(λk )fk L

q (R

N)

du

k=1 q



≤ 2q−1 M1

1



0



q

q



q

rk (u)∇ 2 Bx0 (λk )fk L

1



0

1



0 n 

≤2

n 

du

q

rk (u)∇Bx0 (λk )fk L

q (R

q

rk (u)∇ 2 Bx0 (λk )fk L

N)

du

q (R

N)

du

k=1 −q/2

q

q (M1

N)

k=1

+ 2q−1 Cm1 ,∇ϕ λ0 q−1

q (R

k=1

+ 2q−1 Cm1 ,∇ϕ ≤ 2q−1 M1

n 



1



0

n 

q

q (R

N)

du

k=1

q −q/2 q + Cm1 ,∇ ϕ λ0 )γ0

q

1/2

rk (u)λk ∇Bx0 (λk )fk L



1



0

n 

q

rk (u)fk L

q (R

N)

du.

k=1

q

Choosing M1 so small that 2q−1 M1 γ0 ≤ (1/4)(1/2)q−1 and λ0 ≥ 1 so large that q q −q/2 ≤ (1/4)(1/2)q−1, we have 2q−1 Cm1 ,∇ϕ γ0 λ0 RL(Lq (RN ) ({R(λ) | λ ∈ Σ

,λ0 })

≤ 1/2.

Analogously, we have RL(Lq (RN ) ({τ ∂τ R(λ) | λ ∈ Σ Thus, (I − R(λ))−1 = I +

∞

j =1 R(λ)

j

,λ0 })

≤ 1/2.

exists and

RL(Lq (RN ) ({(τ ∂τ )# (I − R(λ))−1 | λ ∈ Σ

,λ0 })

≤4

for # = 0, 1.

(3.244)

Setting A˜0 (λ) = Bx0 (λ)(I − R(λ))−1 , by (3.240), (3.244) and Proposition 3.5.4, we see that A˜ 0 (λ) is a solution operator satisfying the required properties with r˜b = 4γ0 . To prove the uniqueness of solutions of Eq. (3.237), let u ∈ Hq2 (RN )N be a solution of the homogeneous equation: λu − Div (μD(u) ˜ − K˜ 0 (u)I) = 0 in RN .

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

287

And then, u satisfies the non-homogeneous equation: λu − Div (μ(x0 )D(u) − Kx0 (u)I) = Ru

in RN ,

(3.245)

where we have set Ru = −Div ((μ(x) ˜ − μ(x0 ))D(u)) + ∇(K˜ 0 (u) − Kx0 (u)). Analogously to the proof of (3.234), we have RuLq (RN ) ≤ CM1 ∇ 2 uLq (RN ) + Cm1 ,∇ϕ ∇uLq (RN ) . On the other hand, applying Theorem 3.5.1 to (3.245) for λ ∈ Σ

,1 ,

(3.246) we have

|λ|uLq (RN ) + |λ|1/2uHq1 (RN ) + uHq2 (RN ) ≤ CRuLq (RN ) .

(3.247)

Combining (3.246) and (3.247) gives 1/2

(λ0

− CCm1 ,∇ϕ )uHq1 (RN ) + (1 − CM1 )uHq2 (RN ) ≤ 0.

Choosing M1 ∈ (0, 1) so small that 1 − CM1 > 0 and λ0 ≥ 1 so large that 1/2 λ0 − CCm1 ,∇ϕ > 0, we have u = 0. This proves the uniqueness, and therefore we have proved Theorem 3.5.5 

3.5.3 Model Problem in RN + In this section, we assume that μ, δ, and Aκ (κ ∈ [0, 1)) are constants and an N − 1 constant vector satisfying the conditions: m0 ≤ μ, σ ≤ m1 ,

A0 = 0,

|Aκ | ≤ m2 (κ ∈ (0, 1)).

Let N RN + = {(x1 , . . . , xN ) ∈ R | xN > 0}, N RN 0 = {(x1 , . . . , xN ) ∈ R | xN = 0},

n0 = ) (0, . . . , 0, −1).

(3.248)

288

Y. Shibata

3−1/q N N 1 N ˆ1 Given u ∈ Hq2 (RN (RN + ) and h ∈ Wq 0 ), let K(u, h) ∈ Hq (R+ ) + Hq,0 (R+ ) be a unique solution of the weak Dirichlet problem:

(∇K(u, h), ∇ϕ)RN = (Div (μD(u)) − ∇div u, ∇ϕ)RN +

+

(3.249)

for any ϕ ∈ Hˆ q1 ,0 (RN ), subject to K(u, h) =< μD(u)n0 , n0 > −σ Δ h − div u N−1 2  2 on RN j =1 ∂ h/∂xj . In this section, we consider the half space 0 , where Δ h = problem: ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

λu − Div (μD(u) − K(u, h)I) = f

in RN +,

λh + Aκ · ∇  h − u · n0 = d

on RN 0 ,

(μD(u) − K(u, h)I)n0 − σ (Δ h)n0 = h

on RN 0 ,

(3.250)

where ∇  = (∂1 , . . . , ∂N−1 ). The last equations in (3.250) are equivalent to (μD(u)n0 )τ = hτ

and div u = h · n0

on RN 0 .

Here, we have set hτ = h− < h, n0 > n0 . We shall show the following theorem. Theorem 3.5.6 Let 1 < q < ∞, let μ, σ , and Aκ are constants and an N − 1 constant vector satisfying the conditions in (3.248). Let Λκ,λ0 be the set defined in N Theorem 3.4.8. Let Yq (RN + ) and Yq (R+ ) be spaces defined by replacing Ω and Γ N N by R+ and R0 in Theorem 3.4.11. Then, there exist a constant λ0 ≥ 1 and operator families: 2 N N A0 (λ) ∈ Hol (Λκ,λ0 , L(Yq (RN + ), Hq (R+ ) )), 3 N H0 (λ) ∈ Hol (Λκ,λ0 , L(Yq (RN + ), Hq (R+ )))

(3.251)

such that for any λ = γ + iτ ∈ Λκ,λ0 and (f, d, h) ∈ Yq (RN + ), u = A0 (λ)(f, d, λ1/2 h, h),

h = H0 (λ)(f, d, λ1/2 h, h),

are unique solutions of (3.250), and RL(Y

2−j N N (RN q (R+ ),Hq +) )

({(τ ∂τ )# (λj/2 A0 (λ)) | λ ∈ Λκ,λ0 }) ≤ rb ,

RL(Yq (RN ),Hq3−k (RN )) ({(τ ∂τ )# (λk H0 (λ)) | λ ∈ Λκ,λ0 }) ≤ rb , +

(3.252)

+

for # = 0, 1, j = 0, 1, 2 and k = 0, 1. Here, rb is a constant depending on m0 , m1 , m2 , λ0 , q, and N.

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

289

Remark 3.5.7 In this section, what the constant c depends on m0 , m1 , m2 means that the constant c depends on m0 , m1 , m2 , but is independent of μ, σ and Aκ whenever μ ∈ [m0 , m1 ], σ ∈ [m0 , m1 ], and |Aκ | ≤ m2 for κ ∈ (0, 1). To prove Theorem 3.5.6, as an auxiliary problem, we first consider the following equations: 

λv − Div (μD(v) − θ I) = 0,

div v = 0

in RN +,

(μD(v) − θ I)n0 = h

on RN 0 ,

(3.253)

and we shall prove the following theorem, which was essentially proved by Shibata and Shimizu [54]. Theorem 3.5.8 Let 1 < q < ∞,

∈ (0, π/2), and λ0 > 0. Let

N N Yq (RN + ) = {(G1 , G2 ) | G1 ∈ Lq (R+ ) ,

G2 ∈ Hq1 (RN )N },

N N Hˆ q1 (RN + ) = {θ ∈ Lq,loc (R+ ) | ∇θ ∈ Lq (R+ )}.

Then, there exists a solution operator V(λ) with V(λ) ∈ Hol (Σ

,λ0 , L(Y



2 N N (RN + ), Hq (R+ ) ))

N 1/2 h, h) are such that for any λ = γ + iτ ∈ Σ ,λ0 and h ∈ Hq1 (RN + ) , v = V(λ)(λ unique solutions of Eq. (3.253) with some θ ∈ Hˆ q1 (RN + ) and

RL(Y  (RN ),H 2−j (RN )N ) ({(τ ∂τ )# (λj/2 V(λ)) | λ ∈ Σ q

+

q

+

,λ0 })

≤ rb (λ0 )

for # = 0, 1, and j = 0, 1, 2. Here, rb (λ0 ) is a constant depending on m0 , m1 , m2 , , λ0 , N, and q. Proof To prove Theorem 3.5.8, we start with the solution formulas of Eq. (3.253), which were obtained in Shibata and Shimizu [54] essentially, but for the sake of the completeness of the paper as much as possible and also for the later use, we will derive them in the following. Applying the partial Fourier transform with respect to x  = (x1 , . . . , xN−1 ) to Eq. (3.253), we have ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

λvˆj + μ|ξ  |2 − μ∂N2 vˆj + iξj θˆ = 0, λvˆN + μ|ξ  |2 − μ∂N2 vˆN + ∂N θˆ = 0 N−1 

iξj vˆj + ∂N vˆN = 0

(xN > 0), (3.254) (xN > 0),

j =1

μ(∂N vˆj + iξj vˆN ) = gj ,

2μ∂N vˆN − θˆ = gN

for xN = 0.

290

Y. Shibata

Here, for f = f (x  , xN ), x  = (x1 , . . . , xN−1 ) ∈ RN−1 , xN ∈ (a, b), fˆ denotes the partial Fourier transform of f with respect to x  defined by fˆ(ξ  , xN ) = F  [f (·, xN )](ξ  ) =





RN−1

with ξ  = (ξ1 , . . . , ξN−1 ) ∈ RN−1 and x  · ξ  = gj = −hˆ j (ξ  , 0). To obtain solution formula, we set vˆj = αj e−AxN + βj e−BxN , with A = |ξ  | and B =



iξk αk − AαN = 0,

k=1

N−1 j =1

xj ξj , and we have set

θˆ = ωe−AxN

λμ−1 + |ξ  |2 , and then from (3.254) we have

μαj (B 2 − A2 ) + iξj ω = 0, N−1 



e−ix ·ξ f (x  , xN ) dx 

N−1 

μαN (B 2 − A2 ) − Aω = 0,

(3.255)

iξk βk − BβN = 0,

(3.256)

k=1

μ{(Aαj + Bβj ) − iξj (αN + βN )} = gj ,

(3.257)

2μ(AαN + BβN ) + ω = gN .

(3.258)

The solution formula of Eq. (3.250) was given in Shibata and Shimizu [54], but there is an error in the formula in [54, (4.17)] such as μ{(Aαj + Bβj ) + iξj (αN + βN )} = hˆ j (ξ  , 0), which should read μ{(Aαj + Bβj ) − iξj (αN + βN )} = −hˆ j (ξ  , 0) as (3.257) above. The formulas obtained in [54] are correct, but we repeat here how to obtain αj , βj and ω, because this error confuses readers. We first drive 2 × 2 system of equations with respect to αN and βN . Multiplying (3.257) with iξj , summing up the resultant formulas from j = 1 through N − 1  and writing iξ  · m = N−1 j =1 iξj mj for mj ∈ {αj , βj , gj } give Aiξ  · α  + Biξ  · β  + A2 (αN + βN ) = μ−1 iξ  · g  .

(3.259)

By (3.256), iξ  · α  = AαN ,

iξ  · β  = BβN ,

(3.260)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

291

which, combined with (3.259), leads to 2A2 αN + (A2 + B 2 )βN = μ−1 iξ  · g  .

(3.261)

By (3.255), ω=

μ(B 2 − A2 ) αN , A

(3.262)

which, combined with (3.258), leads to (A2 + B 2 )αN + 2ABβN = μ−1 AgN .

(3.263)

Thus, setting   2 A + B 2 2A2 L= 2AB A2 + B 2

(Lopatinski matrix),

we have L

   −1    βN μ iξ · g = . αN μ−1 AgN

Since det L = (A2 + B 2 )2 − 4A3 B = A4 − 4A3 B + 2A2 B 2 + B 4 = (B − A)D(A, B) with D(A, B) = B 3 + AB 2 + 3A2 B − A3 , we have L−1 =

 2  1 A + B 2 −2A2 . (B − A)D(A, B) −2AB A2 + B 2

Thus, we have βN =

1 ((A2 + B 2 )iξ  · g  − 2A3gN ), μ(B − A)D(A, B)

−1 (2ABiξ  · g  − (A2 + B 2 )AgN ). αN = μ(B − A)D(A, B)

(3.264)

292

Y. Shibata

In particular, vˆN = αN e−AxN + βN e−BxN = αN (e−AxN − e−BxN ) + (αN + βN )e−BxN . We have αN + βN =

1 ((A2 + B 2 − 2AB)iξ  · g  + ((A2 + B 2 )A − 2A3)gN ) (B − A)D(A, B)

=

1 ((B − A)2 iξ  · g  + A(B 2 − A2 )gN ) μ(B − A)D(A, B)

=

1 ((B − A)2 iξ  · g  + A(B − A)(A + B)gN ) μ(B − A)D(A, B)

=

1 ((B − A)iξ  · g  + A(A + B)gN ). μD(A, B) (3.265)

Setting M(xN ) =

e−BxN − e−AxN , B−A

we have vˆN =

A M(xN )(2Biξ  · g  − (A2 + B 2 )gN ) μD(A, B) e−BxN + ((B − A)iξ  · g  + A(A + B)gN ). μD(A, B)

(3.266)

By (3.262) and (3.264), ω= =

μ(B 2 − A2 ) αN A

μ(B 2 − A2 ) −1 (2ABiξ  · g  − (A2 + B 2 )AgN ) A μ(B − A)D(A, B)

=−

(A + B) (2Biξ  · g  − (A2 + B 2 )gN ) D(A, B)

and so θˆ = −

(A + B)e−AxN (2Biξ  · g  − (A2 + B 2 )gN ). D(A, B)

(3.267)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

293

By (3.255), αj = −

iξj ω μ(B 2 − A2 )

=

iξj A+B (2Biξ  · g  − (A2 + B 2 )gN ) μ(B 2 − A2 ) D(A, B)

=

iξj (2Biξ  · g  − (A2 + B 2 )gN ). μ(B − A)D(A, B)

(3.268)

By (3.257) βj =

1 1 gj + (iξj (αN + βN ) − Aαj ). μB B

By (3.265) and (3.268) iξj (αN + βN ) − Aαj =

iξj {(B − A)2 iξ  · g  + A(B − A)(A + B)gN μ(B − A)D(A, B) − A(2Biξ  · g  − (A2 + B 2 )gN )}

=

iξj {(A2 − 4AB + B 2 )iξ  · g  + 2AB 2 gN )}, μ(B − A)D(A, B)

and therefore βj =

iξj 1 gj + {(A2 − 4AB + B 2 )iξ  · g  + 2AB 2 gN )}. μB μ(B − A)D(A, B)B (3.269)

Combining (3.268) and (3.269) gives vˆj =

iξj e−AxN e−BxN gj + {2Biξ  · g  − (A2 + B 2 )gN } μB μ(B − A)D(A, B) +

=

iξj e−BxN {(A2 − 4AB + B 2 )iξ  · g  + 2AB 2 gN )} μ(B − A)D(A, B)B

1 gj e−BxN + I iξ  · g  + I IgN , μB

with I=

iξj e−AxN iξj e−BxN 2B + (A2 − 4AB + B 2 ), μ(B − A)D(A, B) μ(B − A)D(A, B)B

II = −

iξj e−BxN iξj e−AxN (A2 + B 2 ) + 2AB μ(B − A)D(A, B) μ(B − A)D(A, B)

294

Y. Shibata

We proceed as follows: I =

iξj e−BxN iξj (e−AxN − e−BxN ) 2B + (A2 − 4AB + 3B 2 ) μ(B − A)D(A, B) μ(B − A)D(A, B)B

=− II = − =

2iξj BM(xN ) iξj (3B − A)e−BxN + ; μD(A, B) μD(A, B)B iξj e−BxN (A2 − 2AB + B 2 ) iξj (e−AxN − e−BxN ) 2 (A + B 2 ) − μ(B − A)D(A, B) μ(B − A)D(A, B)

iξj (A2 + B 2 )M(xN ) iξj e−BxN (B − A) − . μD(A, B) μD(A, B)

Therefore, we have vˆj =

iξj M(xN ) e−BxN gj − (2Biξ  · g  − (A2 + B 2 )gN ) μB μD(A, B) iξj e−BxN ((3B − A)iξ  · g  − B(B − A)gN ). + μD(A, B)B

(3.270)

To define solution operators for Eq. (3.250), we make preparations. Lemma 3.5.9 Let s ∈ R and 0 < < π/2. Then, there exists a positive constant c depending on , m1 and m2 such that c(|λ|1/2 + A) ≤ Re B ≤ |B| ≤ (μ−1 |λ|)1/2 + A,

(3.271)

c(|λ|1/2 + A)3 ≤ |D(A, B)| ≤ 6((μ−1 |λ|)1/2 + A)3

(3.272)

for any λ ∈ Σ and μ ∈ [m1 , m2 ]. Proof The inequality in the left side of (3.271) follows immediately from Lemma 3.5.2. Notice that D(A, B) = B 3 + 3A2 B + AB 2 − A3 = B(B 2 + 2A2 ) + A(A2 + μ−1 λ) − A3 = B(μ−1 λ + 4A2 ) + μ−1 Aλ. If we consider the angle of B(μ−1 λ + 4A2 ) and −μ−1 Aλ, then we see easily that D(A, B) = 0. Thus, studying the following three cases: R1 |λ|1/2 ≤ A, R1 A ≤ |λ|1/2 and R1−1 A ≤ |λ|1/2 ≤ R1 A for sufficient large R1 > 0, we can prove the inequality in the left side of (3.272). The detailed proof was given in Shibata and Shimizu [50]. The independence  of the constant c of λ ∈Σ and μ ∈ [m0 , m1 ] follows from the homogeneity: μ−1 (m2 λ) + (mA)2 = m μ−1 λ + A2 and D(mA, mB) = m3 D(A, B) for any m > 0 and the compactness of the interval [m0 , m1 ]. 

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

295

To introduce the key tool of proving the R boundedness in the half space, we make a definition. Definition 3.5.10 Let V be a domain in C, let Ξ = V × (RN−1 \ {0}), and let m : Ξ → C; (λ, ξ  ) → m(λ, ξ  ) be C 1 with respect to τ , where λ = γ + iτ ∈ V , and C ∞ with respect to ξ  ∈ RN−1 \ {0}. (1) m(λ, ξ  ) is called a multiplier of order s with type 1 on Ξ , if the estimates: 



|∂ξκ m(λ, ξ  )| ≤ Cκ  (|λ|1/2 + |ξ  |)s−|κ | , 



|∂ξκ (τ ∂τ m(λ, ξ  ))| ≤ Cκ  (|λ|1/2 + |ξ  |)s−|κ |  hold for any multi-index κ ∈ NN 0 and (λ, ξ ) ∈ Ξ with some constant Cκ   depending solely on κ and V . (2) m(λ, ξ  ) is called a multiplier of order s with type 2 on Ξ , if the estimates: 



|∂ξκ m(λ, ξ  )| ≤ Cκ  (|λ|1/2 + |ξ  |)s |ξ  |−|κ | , 



|∂ξκ (τ ∂τ m(λ, ξ  ))| ≤ Cκ  (|λ|1/2 + |ξ  |)s |ξ  |−|κ |  hold for any multi-index κ ∈ NN 0 and (λ, ξ ) ∈ Ξ with some constant Cκ  depending solely on κ  and V .

Let Ms,i (V ) be the set of all multipliers of order s with type i on Ξ for i = 1, 2. For m ∈ Ms,i (V ), we set M(m, V ) = max|κ  |≤N Cκ  . Let Fξ−1  be the inverse partial Fourier transform defined by   Fξ−1  [f (ξ , xN )](x )

1 = (2π)N−1





RN−1



eix ·ξ f (ξ  , xN ) dξ  .

Then, we have the following two lemmata which have been proved essentially by Shibata and Shimizu [54, Lemma 5.4 and Lemma 5.6]. Lemma 3.5.11 Let 0 < < π/2, 1 < q < ∞, and λ0 > 0. Given m ∈ M−2,1 (Λκ,λ0 ), we define an operator L(λ) by 

∞

[L(λ)g](x) = 0

 1/2 −B(xN +yN ) Fξ−1 e g(ξ ˆ  , yN )](x  ) dyN .  [m(λ, ξ )λ

Then, we have RL(L

2−j N N (RN q (R+ ),Hq +) )

({(τ ∂τ )# (λj/2 ∂xα L(λ)) | λ ∈ Λκ,λ0 }) ≤ rb (λ0 )

for # = 0, 1 and j = 0, 1, 2, where τ denotes the imaginary part of λ, and rb (λ0 ) is a constant depending on M(m, Λκ,λ0 ), , λ0 , N, and q.

296

Y. Shibata

Lemma 3.5.12 Let 0 < < π/2, 1 < q < ∞, and λ0 > 0. Given m ∈ M−2,2 (Λκ,λ0 ), we define operators Li (λ) (i = 1, . . . , 4) by 

∞

[L1 (λ)g](x) = 

0



0



0

∞

[L2 (λ)g](x) =

∞

[L3 (λ)g](x) =

∞

[L4 (λ)g](x) = 0

 −B(xN +yN ) Fξ−1 g(ξ ˆ  , yN )](x  ) dyN ,  [m(λ, ξ )Ae  −A(xN +yN ) Fξ−1 g(ξ ˆ  , yN )](x  ) dyN ,  [m(λ, ξ )Ae  2 Fξ−1 ˆ  , yN )](x  ) dyN ,  [m(λ, ξ )A M(xN + yN )g(ξ  1/2 Fξ−1 AM(xN + yN )g(ξ ˆ  , yN )](x  ) dyN .  [m(λ, ξ )λ

Then, we have RL(L

2−j N N (RN q (R+ ),Hq +) )

({(τ ∂τ )# (λj/2 ∂xα Li (λ)) | λ ∈ Λκ,λ0 }) ≤ rb (λ0 )

for # = 0, 1 and j = 0, 1, 2, where τ denotes the imaginary part of λ, and rb (λ0 ) is a constant depending on M(m, Λκ,λ0 ), , λ0 , N, and q. To construct solution operators, we use the following lemma. Lemma 3.5.13 Let 0 < < π/2, 1 < q < ∞ and λ0 > 0. Given multipliers, n1 ∈ M−2,1 (Λκ,λ0 ), n2 ∈ M−2,2 (Λκ,λ0 ), and n3 ∈ M−1,2 (Λκ,λ0 ), we define operators Ti (λ) (i = 1, 2, 3) by 1/2 −BxN ˆ  , 0)](x  ), T1 (λ)h = Fξ−1 e n1 (λ, ξ  )h(ξ  [λ −BxN ˆ  , 0)](x  ), n2 (λ, ξ  )h(ξ T2 (λ)h = Fξ−1  [Ae  ˆ   T3 (λ)h = Fξ−1  [AM(xN )n3 (λ, ξ )h(ξ , 0)](x ).

Let N 1 N Zq (RN + ) = {(G3 , G4 ) | G3 ∈ Lq (R+ ), G4 ∈ Hq (R+ )}. N 2 Then, there exist operator families Ti (λ) ∈ Hol (Λκ,λ0 , L(Yq (RN + ), Hq (R+ ))) such 1/2 h, h) and that for any λ = γ + iτ ∈ Λκ,λ0 and h ∈ Hq1 (RN + ), Ti (λ)h = Ti (λ)(λ

RL(Y  (RN ),H 2−j (RN )) ({(τ ∂τ )# (λj/2 Ti (λ)) | λ ∈ Λκ,λ0 }) ≤ rb (λ0 ) q

+

q

+

(3.273)

for # = 0, 1, j = 0, 1, 2, where rb (λ0 ) is a constant depending on M(ni , Λκ,λ0 ) (i = 1, 2, 3), , λ0 , N, and q.

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

297

Proof By Volevich’s trick we write T1 (λ)h



∞

=− 0



∞

=− 

0 ∞

+ 0

−

Fξ−1  [

1/2 −B(xN +yN ) ˆ  , yN )](x  ) dyN Fξ−1 e n1 (λ, ξ  )∂N h(ξ  [λ 1/2 −B(xN +yN ) Fξ−1 e  [λ

N−1  ∞ j =1

∂ ˆ  , yN ))](x  ) dyN (λ1/2 e−B(xN +yN ) n1 (λ, ξ  )h(ξ ∂yN

0

λ1/2 ˆ  , yN )](x  ) dyN n1 (λ, ξ  )λ1/2 h(ξ μB

−B(xN +yN ) Fξ−1  [Ae

λ1/2 iξj n1 (λ, ξ  )F [∂j h(·, yN )]](x  ) dyN , B A

where we have used the formula: B=

N−1  A iξj μ−1 λ + A2 λ = − iξj . μB μB B A j =1

Let T1 (λ)(G3 , G4 )  ∞ 1/2 −B(xN +yN ) Fξ−1 e n1 (λ, ξ  )F [∂N G4 (·, yN )]](x  ) dyN =−  [λ 

0

∞

+ 0

−

1/2 −B(xN +yN ) Fξ−1 e  [λ

N−1  ∞ j =1

0

λ1/2 n1 (λ, ξ  )F [G3 (·, yN )]](x  ) dyN μB

−B(xN +yN ) Fξ−1  [Ae

λ1/2 iξj n1 (λ, ξ  )F [∂j G4 (·, yN )]](x  ) dyN , B A

and then, T1 (λ)h = T1 (λ)(λ1/2 h, h). Moreover, Lemmas 3.5.11 and 3.5.12 yield (3.273) with j = 1, because n1 (λ, ξ  ) ∈ M−2,1 (Λκ,λ0 ),

λ1/2 n1 (λ, ξ  ) ∈ M−2,1 (Λκ,λ0 ), μB

λ1/2 iξj n1 (λ, ξ  ) ∈ M−2,2 (Λκ,λ0 ). B A Analogously, we can prove the existence of T2 (λ).

298

Y. Shibata

To construct T3 (λ), we use the formula: ∂ M(xN ) = −e−BxN − AM(xN ), ∂xN and then, by Volevich’s trick we have T3 (λ)h



∞

=−

Fξ−1  [

0

∂ ˆ  , yN ))](x  ) dyN = −I + I I (AM(xN + yN )n3 (λ, ξ  )h(ξ ∂yN

with 

∞

I = 

0 ∞

II = 0

  ˆ  Fξ−1  [AM(xN + yN )n3 (λ, ξ )∂N h(ξ , yN )](x ) dyN ; −B(xN +yN ) ˆ  , yN )](x  ) dyN . Fξ−1 + A2 M(xN + yN ))n3 (λ, ξ  )h(ξ  [(Ae

Using the formula: 1=

N−1 B2 λ1/2 1/2 A λ1/2 1/2  iξj = λ + A = λ − iξj , B2 μB 2 B2 μB 2 B2 j =1

we have 

∞

I = 0



+  II = 0

0 ∞

1/2 Fξ−1 AM(xN + yN )  [λ ∞

2 Fξ−1  [A M(xN + yN )

λ1/2 ˆ  , yN )](x  ) dyN n3 (λ, ξ  )∂N h(ξ μB 2

A ˆ  , yN )](x  ) dyN ; n3 (λ, ξ  )∂N h(ξ B2

−B(xN +yN ) Fξ−1 + A2 M(xN + yN ))  [(Ae

λ1/2 ˆ  , yN )](x  ) dyN n3 (λ, ξ  )λ1/2 h(ξ μB 2 N−1  ∞ −B(xN +yN ) − Fξ−1 + A2 M(xN + yN ))  [(Ae ×

j =1

×

0

iξj n3 (λ, ξ  )F [∂j h(·, yN )]](x  ) dyN . B2

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

299

Let T3 (λ)(G3 , G4 )  ∞ λ1/2 1/2 Fξ−1 AM(xN + yN ) n (λ, ξ  )F [∂N G4 (·, yN )]](x  ) dyN =−  [λ 2 3 μB 0  ∞ A 2   − Fξ−1  [A M(xN + yN ) 2 n3 (λ, ξ )F [∂N G4 (·, yN )]](x ) dyN B 0  ∞ −B(xN +yN ) + Fξ−1 + A2 M(xN + yN ))  [(Ae 0

λ1/2 n3 (λ, ξ  )F [G3 (·, yN )]](x  ) dyN μB 2 N−1  ∞ −B(xN +yN ) − Fξ−1 + A2 M(xN + yN ))  [(Ae ×

j =1

×

0

iξj n3 (λ, ξ  )F [∂j G4 (·, yN )]](x  ) dyN , B2

and then T3 (λ)h = T3 (λ)(λ1/2 h, h). Moreover, Lemma 3.5.12 yields (3.273) for j = 3, because n3 (λ) ∈ M−2,2 (Λκ,λ0 ),

λ1/2 n3 (λ, ξ  ) ∈ M−2,2 (Λκ,λ0 ), μB 2

iξj n3 (λ, ξ  ) ∈ M−2,2 (Λκ,λ0 ). B2 

This completes the proof of Lemma 3.5.13. Continuation of Proof of Theorem 3.5.8 Let vj (x) = Fξ−1 ˆj (ξ  , xN )](x  ),  [v and then by (3.266) and (3.270) we have vN = Fξ−1 

+

N−1 ,  A M(xN )((A2 + B 2 )hˆ N (ξ  , 0) − 2B iξ# hˆ # (ξ  , 0)) (x  ) μD(A, B) #=1

− Fξ−1 

N−1 , + Ae−BxN  iξ# ((B − A) hˆ # (ξ  , 0) + (A + B)hˆ N (ξ  , 0)) (x  ); μD(A, B) A #=1

, , + λ1/2 + A −1 1/2 −BxN ˆ   −BxN ˆ  vk = −Fξ−1 λ e (ξ , 0) (x ) − F Ae (ξ , 0) (x  ) h h  k  k ξ μ2 B 3 μB 3

300

Y. Shibata N−1 +  1 iξk (2B AM(xN ) + Fξ−1 iξ# hˆ # (ξ  , 0)  A μD(A, B) #=1 , − (A2 + B 2 )hˆ N (ξ  , 0)) (x  ) N−1 +  iξk 1 Ae−BxN ((3B − A) − Fξ−1 iξ# hˆ # (ξ  , 0)  A μD(A, B)B #=1 , − B(B − A)hˆ N (ξ  , 0)) (x  ),

for k = 1, . . . , N − 1, where we have used the formula treat the first term of vˆj in (3.270). Since

λ 1 A2 = 2 3 + to μB μ B μB 3

A2 + B 2 iξk Biξ# iξk A2 + B 2 Biξ# , , , ∈ M−1,2 (Σ μD(A, B) μD(A, B) A μD(A, B) A μD(A, B)

and

λ1/2 ∈ M−2,1 (Σ μ2 B 3

,λ0 ),

A+B A B − A iξ# , , ∈ M−2,2 (Σ μD(A, B) A μD(A, B) μB 3

,λ0 )

iξk (3B − A)iξ# iξk B(B − A) , ∈ M−2,2 (Σ A μD(A, B)B A μD(A, B)B

,λ0 ),

,λ0 ),

by Lemma 3.5.13 we have Theorem 3.5.8.



We next consider the equations: ⎧ λw − Div (μD(w) − qI) = 0, div w = 0 ⎪ ⎪ ⎨ λh + Aκ · ∇  h − w · n0 = d ⎪ ⎪ ⎩ (μD(w) − qI)n0 − σ (Δ h)n0 = 0

in RN +, on RN 0 ,

(3.274)

on RN 0 .

We shall prove the following theorem. Theorem 3.5.14 Let 1 < q < ∞ and ∈ (0, π/2). Then, there exist a λ1 > 0 and solution operators Wκ (λ) and Hκ (λ) with 2 N N Wκ (λ) ∈ Hol (Λκ,λ1 , L(Hq2 (RN + ), Hq (R+ ) )), 3 N Hκ (λ) ∈ Hol (Λκ,λ1 , L(Hq2 (RN + ), Hq (R+ ))),

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

301

such that for any λ = γ + iτ ∈ Λκ,λ1 and d ∈ Hq2 (RN + ), w = Wκ (λ)d and h = Hκ (λ)d are unique solutions of Eq. (3.274) with some q ∈ Hˆ q1 (Ω), and RL(H 2 (RN ),Hq2−k (RN )N ) ({(τ ∂τ )# (λk/2 Wκ (λ)) | λ ∈ Λκ,λ1 }) ≤ rb (λ1 ), q

+

+

RL(H 2 (RN ),Hq3−m (RN )) ({(τ ∂τ )# (λm Hκ (λ)) | λ ∈ Λκ,λ1 }) ≤ rb (λ1 ) q

+

+

for # = 0, 1, k = 0, 1, 2, and m = 0, 1, where rb (λ1 ) is a constant depending on m0 , m1 , m2 , , λ1 , N, and q. Proof We start with solution formulas. Applying the partial Fourier transform to Eq. (3.274), we have the following generalized resolvent problem: λwˆ j + μ|ξ  |2 wˆ j − μ∂N2 wˆ j + iξj qˆ = 0

(xN > 0),

λwˆ N + μ|ξ  |2 wˆ N − μ∂N2 wˆ N + ∂N qˆ = 0

(xN > 0),

N−1 

iξj wˆ j + ∂N wˆ N = 0

(xN > 0),

j =1

μ(∂N wˆ j (0) + iξj wˆ N (0)) = 0,

2μ∂N wˆ N − q = −σ A2 hˆ

λhˆ +

N−1 

iξj Aκj hˆ + wˆ N = dˆ

for xN = 0, for xN = 0.

j =1

(3.275) Here, we have set Aκ = (Aκ1 , . . . , AκN−1 ). Using the solution formulas given ˆ in (3.266), (3.267) and (3.270) with gj = 0 (j = 1, . . . , N − 1), and gN = σ A2 h, we have wˆ j =

iξj M(xN ) iξj e−BxN ˆ σ A2 (A2 + B 2 )hˆ − σ A2 (B − A)h, μD(A, B) μD(A, B)

wˆ N = − qˆ = −

AM(xN ) e−BxN ˆ σ A2 (A2 + B 2 )hˆ + σ A3 (A + B)h, μD(A, B) μD(A, B) (A + B)A2 (A2 + B 2 )e−AxN ˆ h. D(A, B)

Inserting the formula of wˆ N |xN =0 into the last equation in (3.275), we have (λ + iξ  · Aκ )hˆ +

σ A3 (A + B) ˆ ˆ h = d, μD(A, B)

(3.276)

302

Y. Shibata

where we have set iξ  · Aκ =

N−1 j =1

iξj Aκj , which implies that

μD(A, B) ˆ d hˆ = Eκ

(3.277)

with Eκ = μ(λ + iξ  · Aκ )D(A, B) + σ A3 (A + B). Thus, we have the following solution formulas: wˆ j = iξj M(xN )

σ A2 (B − A) ˆ σ A2 (A2 + B 2 ) ˆ d − iξj e−BxN d, Eκ Eκ

wˆ N = −AM(xN ) qˆ =

σ A3 (A + B) ˆ σ A2 (A2 + B 2 ) ˆ d + e−BxN d, Eκ Eκ

(3.278)

σ (A + B)A2 (A2 + B 2 )e−AxN ˆ d. Eκ

Concerning the estimation for Eκ , we have the following lemma. Lemma 3.5.15 (1) Let 0 < < π/2 and let E0 be the function defined in (3.277) with A0 = 0. Then, there exists a λ1 > 0 and c > 0 such that the estimate: |E0 | ≥ c(|λ| + A)(|λ|1/2 + A)3

(3.279)

holds for (λ, ξ  ) ∈ Σ ,λ1 × (RN−1 \ {0}). (2) Let κ ∈ (0, 1) and let Eκ be the function defined in (3.277). Then, there exists a λ1 > 0 and c > 0 such that |Eκ | ≥ c(|λ| + A)(|λ|1/2 + A)3

(3.280)

holds for (λ, ξ  ) ∈ C+,λ1 × (RN−1 \ {0}). Here, the constant c in (1) and (2) depends on λ1 , m0 , m1 , and m2 . Proof We first study the case where |λ| ≥ R1 A for large R1 > 0. Note that |λ| ≥ λ1 . Since |B| ≤ A + μ−1/2 |λ|1/2 and since Λκ,λ1 ⊂ Σ , by Lemma 3.5.9 we have |Eκ | ≥ μ|λ||D(A, B)| − μ|Aκ ||A||D(A, B)| − σ A3 (A + μ−1/2 |λ|1/2 ) ≥ cμ|λ|(|λ|1/2 + A)3 − μm2 CR1−1 |λ|(|λ|1/2 + A)3 − σ R1−1 |λ|(|λ|1/2 + A)3 − μ−1/2 σ |λ|1/2(|λ|1/2 + A)3 ≥ (cμ/2)|λ|(|λ|1/2 + A)3 + ((cμ/2) − μm2 CR1−1 − σ R1−1 − σ/(μ|λ|)1/2)|λ|(|λ|1/2 + A)3 .

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

303

Thus, choosing R1 > 0 and λ1 > 0 so large that (cμ/4) − μm2 CR1−1 − σ R1−1 ≥ 0 and (cμ/4) − σ/(μλ1 )1/2 ≥ 0, we have |E˜ κ | ≥ (cμ/2)|λ|(|λ|1/2 + A)3 ≥ (cμ/4)(|λ| + R1 A)(|λ|1/2 + A)3

(3.281)

provided that |λ| ≥ R1 A and λ ∈ Λκ,λ1 . When κ = 0, we may assume that m2 = 0 above. We now consider the case where |λ| ≤ R1 A. We assume that λ ∈ Σ ,λ1 . In 1/2 this case, we have A ≥ R1−1 |λ|1/2λ1 , and so, choosing λ1 large enough, we have −1/2 −1/2 B = A(1 + O(λ1 )). In particular, D(A, B) = 4A3(1 + O(λ1 )). Thus, we have −1/2

Eκ = 4μ(λ + iξ  · Aκ )A3 (1 + O(λ1

−1/2

)) + 2σ A4 (1 + O(λ1

)).

We first consider the case where κ = 0. Using Lemma 3.5.2, we have −1/2

|E0 | ≥ |4μλA3 + 2σ A4 | − 4μ|λ|A3O(λ1

−1/2

≥ (sin )(4μ|λ|A3 + 2σ A4 ) − O(λ1

−1/2

) − 2σ A4 O(λ1

)

)(4μ|λ|A3 + 2σ A4).

−1/2

Thus, choosing λ1 > 0 so large that (sin /2) − O(λ1

) ≥ 0, we have

|E0 | ≥ (sin /2)(4μ|λ|A3 + 2σ A4 ) ≥ c(|λ| + A)A3 ≥ c/23(|λ| + A)(A + R1−1 λ1 |λ|1/2 )3 . 1/2

This completes the proof of (1). We next consider the case of κ ∈ (0, 1). Taking the real part gives −1/2

Re Eκ = 4μ(Re λ)A3 (1 + O(λ1 −1/2

+ 2σ A4 (1 + O(λ1

−1/2

)) + O(λ1

)(Im λ + Aκ · ξ  )A3

)).

Since Re λ ≥ λ1 > 0 and |λ| ≤ R1 A, we have −1/2

Re Eκ ≥ 2σ A4 − (4μ(m2 + R1 ) + 2σ )O(λ1

)A4 , −1/2

and so, choosing λ1 > 0 so large that σ − (4μ(m2 + R1 ) + 2σ )O(λ1 have

) ≥ 0, we

|Eκ | ≥ Re Eκ ≥ σ A4 ≥ (σ/24 )(A + R1−1 |λ|)(A + R1−1 λ1 |λ|1/2 )3 . 1/2

This completes the proof of Lemma 3.5.15.



304

Y. Shibata

Continuation of Proof of Theorem 3.5.14 Let wj = Fξ−1 ˆ j ], q = Fξ−1 q] and  [w  [ˆ −1 −AxN ˆ ∞ η = ϕ(xN )Fξ  [e h], where ϕ(xN ) ∈ C0 (R) equals to 1 for xN ∈ (−1, 1) and 0 for xN ∈ [−2, 2]. Notice that η|xN =0 = h. In view of (3.278) and Volevich’s trick, we define Wκj (λ) by 

∞

Wκj (λ)d = 0

+ − (Ae−B(xN +yN ) + A2 M(xN + yN )) Fξ−1 

iξj σ (A2 + B 2 ) F [Δ d](ξ  , yN ) A Eκ

×

+ Ae−B(xN +yN ) 

, iξj σ B(B − A)   F [Δ d](ξ  , yN ) (x  ) dyN A Eκ

+ σ (A2 + B 2 )  − A2 M(xN + yN ) Fξ−1 F [∂j ∂N d](ξ  , yN )  Eκ 0 , σ A(B − A)  + Ae−B(xN +yN ) F [∂j ∂N d](ξ  , yN ) (x  ) dyN , Eκ

+

∞

ˆ  , yN ). We have Wκj (λ)d = wj . where we have used F  [Δ d](ξ  , yN ) = −A2 d(ξ By Lemma 3.5.15, we see that A2 + B 2 , Eκ

A2 + B 2 ξj , Eκ A

A(B − A) , Eκ

B(B − A) ξj Eκ A

belong to M−2,2 (Λκ,λ1 ), and so by Lemma 3.5.12, we have RL(H 2 (RN ),H 2−k (RN )) ({(τ ∂τ )# (λk/2 Wκj (λ)) | λ ∈ Λκ,λ1 }) ≤ rb (λ1 ) q

+

q

+

for # = 0, 1 and k = 0, 1, 2, where rb (λ1 ) is a constant depending on m0 , m1 , m2 and λ1 . Analogously, we have RL(H 2 (RN ),H 2−k (RN )) ({(τ ∂τ )# (λk/2 WκN (λ)) | λ ∈ Λκ,λ1 }) ≤ rb (λ1 ) q

+

q

+

for # = 0, 1 and k = 0, 1, 2. Thus, our final task is to construct Hκ (λ). In view of (3.277), we define Hκ (λ) acting on d ∈ Hq2 (RN + ) by + μD(A, B) ˆ  ,  e−AxN Hκ (λ)d = ϕ(xN )Fξ−1 d(ξ , 0) (x ).  Eκ Since ϕ(xN ) equals one for xN ∈ (−1, 1), we have Hκ (λ)d|xN =0 = h. Recalling the definition of hˆ given in (3.277) and using Volevich’s trick, we have Hκ (λ)d =

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

305

ϕ(xN ){Ωκ (λ)d + Hκ2 (λ)d} with 

+ , μD(A, B) ˆ  , yN ) (x  ) dyN , Ae−A(xN +yN ) Fξ−1 ϕ(yN )d(ξ  Eκ 0  ∞ + , μD(A, B) ˆ  , yN )) (x  ) dyN . Hκ2 (λ)d = − Fξ−1 e−A(xN +yN ) ∂N (ϕ(yN )d(ξ  Eκ 0 Ωκ (λ)d =

∞

We use the following lemma. Lemma 3.5.16 Let Λ be a domain in C and let 1 < q < ∞. Let ϕ and ψ be two C0∞ ((−2, 2)) functions. Given m ∈ M0,2 (Λ), we define operators L6 (λ) and L7 (λ) acting on g ∈ Lq (RN + ) by  [L6 (λ)g](x) = ϕ(xN ) 

∞

+ , e−A(xN +yN ) m(λ, ξ  )g(ξ Fξ−1 ˆ  , yN )ψ(yN ) dyN , 

∞

+ , Ae−A(xN +yN ) m(λ, ξ  )g(ξ Fξ−1 ˆ  , yN )ψ(yN ) dyN . 

0

[L7 (λ)g](x) = ϕ(xN ) 0

Then, RL(Lq (RN )) ({(τ ∂τ )# Lk (λ) | λ ∈ Λ}) ≤ rb

(3.282)

+

for # = 0, 1 and k = 6, 7, where rb is a constant depending on M(m, Λ). Here, M(m, Λ) is the number defined in Definition 3.5.10. Proof Using the assertion for L2 (λ) in Lemma 3.5.12, we can show (3.282) immediately for k = 7, and so we show (3.282) only in the case that k = 6 below. In view of Definition 3.4.7, for any n ∈ N, we take {λj }nj=1 ⊂ Λ, {gj }nj=1 ⊂ Lq (RN + ), and rj (u) (j = 1, . . . , n) are Rademacher functions. For the notational simplicity, we set |||L6 (λ)g||| = 

n  j =1

=



1

rj (u)L6 (λj )gj Lq ((0,1),Lq (RN )) +



0

n 

1/q q du . Lq (RN +)

rj (u)L6 (λj )gj 

j =1

By the Fubini–Tonelli theorem, we have  |||L6 (λ)g|||q = 0

 = 0

1 ∞

|

n 

RN−1 j =1

0 ∞  1 0



n  j =1

rj (u)L6 (λj )gj |q dy  dxN du

q

rj (u)L6 (λj )gj L

q

(RN−1 )

du dxN .

306

Y. Shibata

Since 



|∂ξα (e−A(xN +yN ) m(λ, ξ  ))| ≤ Cα  |ξ  |−|α | for any xN ≥ 0, yN ≥ 0, (λ, ξ  ) ∈ Λ × (RN−1 \ {0}), and α  ∈ NN−1 , by Theorem 3.5.1 we have 

1



0

n  j =1

+ , q e−A(xN +yN ) m(λj , ξ  )gˆj (ξ  , yN ) (y  )L rj (u)Fξ−1  

1

≤ CN,q M(m, Λ)

n 



0

q

rj (u)gj (·, yN )L

j =1

q (R

N−1 )

q (R

N−1 )

du

(3.283)

du.

For any xN ≥ 0, by Minkowski’s integral inequality, Lemma 3.5.3, and Hölder’s inequality, we have 

1 0



n  j =1

= |ϕ(xN )|



1



∞



0

0

q

 1  0

q

Fξ−1  [



rj (u)e−A(xN +yN ) m(λj , ξ  )gˆj (ξ  , yN )](y  )

du (RN−1 )

∞ 0

∞  1

0

n  j =1

Fξ−1  [

0

n 

rj (u)e−A(xN +yN ) m(λj , ξ  )gˆj (ξ  , yN )](y  )

j =1

× ψ(yN )Lq (RN−1 ) dyN ≤ |ϕ(xN )|

du

(RN−1 )

1/q

q

× ψ(yN ) dyN L ≤ |ϕ(xN )|

1/q

q

rj (u)L6 (λj )gj L

q

1/q du

Fξ−1  [

n 

rj (u)e−A(xN +yN ) m(λj , ξ  )

j =1

× gˆj (ξ  , yN )](y  )L

1/q

q

q

|ψ(yN )| dyN

du (RN−1 ) 

≤ CN,q M(m, Λ)|ϕ(xN )|

∞  1

0



0

n  j =1

1/q

q

rj (u)gj (·, yN )L

q

(RN−1 )

du

× |ψ(yN )| dyN ≤ CN,q M(m, Λ)|ϕ(xN )|

 0

∞ 1 0



n  j =1

1/q

q

rj (u)gj (·, yN )L

q

(RN−1 )

du dyN

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

×



∞



|ψ(yN )|q dyN

0

= CN,q M(m, Λ)|ϕ(xN )|

1/q 



1



0

×



∞



|ψ(yN )|q dyN

307

n 

1/q q du Lq (RN +)

rj (u)gj (·, yN )

j =1

1/q 

0

Putting these inequalities together and using Hölder’s inequality gives 

1 0



n 

q du Lq (RN +)

rj (u)L6 (λj )gj 

j =1



∞

≤ (CN,q M(m, Λ))q



0

×



∞



1

|ϕ(xN )|q dxN

|ψ(yN )|q dyN



0

n 

q du Lq (RN +)

rj (u)gj 

j =1

q/q  ,

0

and so, we have 

n  j =1

rj L6 (λj )gj Lq ((0,1),Lq (RN )) +

≤ CN,q M(m, Λ)ϕLq (R) ψLq  (R)

n  j =1

rj gj Lq ((0,1),Lq (RN )) . +



This shows Lemma 3.5.16. Continuation of Proof of Theorem 3.5.14 For j + |α  | + k ≤ 3 and j = 0, 1, we write  λj ∂xα ∂Nk Hκ (λ)d

=

k 

(j, α  , k)

∈

k−n j α n k Cn (∂N ϕ(xN ))[λ ∂x  ∂N Ωκ (λ)d

N0 × NN−1 0

× N0 with



+ λj ∂xα ∂Nn Hκ2 (λ)d],

n=0

and then 

λj ∂xα ∂Nn Ωκ (λ)d  ∞ + j  α n −1 −A(xN +yN ) μλ (iξ ) (−A) D(A, B) Fξ  Ae ϕ(yN ) = (1 + A2 )Eκ 0 , F  [(1 − Δ )d](ξ  , yN ) (x  ) dyN ;

308

Y. Shibata

 λj Hκ2 (λ)d =





λj ∂xα ∂Nn Hκ2 (λ)d =

∞

0

+ μλj D(A, B) e−A(xN +yN ) Fξ−1  Eκ 0 , ˆ  , yN )) (x  ) dyN ; ∂N (ϕ(yN )d(ξ ∞

+ j  α n −A(xN +yN ) μλ (iξ ) (−A) D(A, B) Fξ−1 e  2 (1 + A )Eκ

ˆ  , yN ))](x  ) dyN ∂N (ϕ(yN )d(ξ −

N−1  ∞

 + μλj (iξ  )α (−A)n D(A, B) iξj Ae−A(xN +yN ) Fξ−1  (1 + A2 )Eκ A k=1 0 , ∂N (ϕ(yN )F [∂j d(·, yN )](ξ  )) (x  ) dyN

for |α  | + n ≥ 1. Here, we have used the formula: 1=

N−1  A iξj 1 1 + A2 = − iξj 2 2 1+A 1+A 1 + A2 A j =1

in the third equality. By Lemmas 3.5.9 and 3.5.15, we see that multipliers: 

λj (iξ  )α An D(A, B) , (1 + A2 )Eκ

λj D(A, B) , Eκ





λj (iξ  )α An D(A, B) λj (iξ  )α An D(A, B) ξj , (1 + A2 )Eκ (1 + A2 )Eκ A belong to M0,2 (Λκ,λ1 ), because j + |α  | + n ≤ 3 and j = 0, 1. Thus, using Lemma 3.5.16, we see that for any n ∈ N, {λj }nj=1 ⊂ Λκ,λ1 , and {dj }nj=1 ⊂ Hq2 (RN + ), the inequality: 

n 



r# (·)(∂Nk−n ϕ)(λ# )j ∂xα ∂Nn (Ωκ (λ# ), Hκ2 (λ# ))d# Lq ((0,1),Lq (RN ))

#=1

≤ C

n 

r# (·)d# Lq ((0,1),H 2(RN )) q

#=1

+

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

309

holds, which leads to 

n 



r# (·)(λ# )j ∂xα ∂Nk Hκ (λ# )d# Lq ((0,1),Lq (RN )) +

#=1

≤ C

n 

r# (·)d# Lq ((0,1),H 2(RN )) . q

#=1

+

Here, C is a constant depending on N, q, m0 , m1 , and m2 . This shows that RL(H 2 (RN ),Hq3−k (RN )) ({λk Hκ (λ) | λ ∈ Λκ,λ1 }) ≤ rb (λ1 ) +

q

+

for k = 0, 1. Here, rb (λ1 ) is a constant depending on N, q, m0 , m1 , and m2 , but independent of μ, σ ∈ [m0 , m1 ] and |Aκ | ≤ m2 for κ ∈ [0, 1). Analogously, we have RL(H 2 (RN ),H 3−k (RN )) ({τ ∂τ (λk Hκ (λ)) | λ ∈ Λκ,λ1 }) ≤ rb (λ1 ) q

+

+

q

for k = 0, 1. This completes the proof of Theorem 3.5.14.



Proof of Theorem 3.5.6 To prove Theorem 3.5.6, in view of the consideration in Sect. 3.4.3, we first consider the equation: div v = g

in RN +.

(3.284)

Here, g is a solution of the variational equation: λ(g, ϕ)RN + (∇g, ∇ϕ)RN = (−f, ∇ϕ)RN +

+

+

for any ϕ ∈ Hq1 ,0 (RN + ),

(3.285)

subject to g = ρ on Γ . We have the following theorem. Lemma 3.5.17 Let 1 < q < ∞, 0
0. Let

N N 1 N Yq (RN + ) = {(f, ρ) | f ∈ Lq (R+ ) , ρ ∈ Hq (R+ )}, N N N 1 N Yq (RN + ) = {(F1 , G3 , G4 ) | F1 ∈ Lq (R+ ) , G3 ∈ Lq (R+ ), G4 ∈ Hq (R+ )}.

Let g be a solution of the variational problem (3.285). Then, there exists an operator 2 N N family B0 (λ) ∈ Hol (Σ ,λ0 , L(Yq (RN + ), Hq (R+ ) )) such that for any λ ∈ Σ ,λ0 1/2 ρ, ρ), and (f, ρ) ∈ Yq (RN + ), problem (3.284) admits a solution v = B0 (λ)(f, λ and RL(Y  (RN ),H 2−j (RN )N ) ({(τ ∂τ )# (λj/2 B0 (λ)) | λ ∈ Σ q

+

q

+

,λ0 })

≤ rb (λ0 )

for # = 0, 1 and j = 0, 1, 2, where rb (λ0 ) is a constant depending on , λ0 , N, and q.

310

Y. Shibata

Proof This lemma was proved in Shibata [45, Lemma 9.3.10], but for the sake of completeness of this lecture note as much as possible, we give a proof. Let g1 be a solution of the equation: (λ − Δ)g1 = div f

in RN +,

g1 |xN =0 = 0,

and let g2 be a solution of the equation: (λ − Δ)g2 = 0

in RN +,

g2 |xN =0 = ρ.

And then, g = g1 + g2 is a solution of Eq. (3.285). To construct g1 and g2 , we use the even extension, f e , and odd extension, f o , of a function, f , which  has been o introduced in (3.46). Let f = ) (f1 , . . . , fN ). Notice that (div f)o = N−1 j =1 ∂j fj + e ∂N fN . We define g1 by letting g1 =

Fξ−1

+ F [(div f)o ](ξ ) , λ + |ξ |2

=

Fξ−1

+ N−1 iξ F [f o ](ξ ) + iξ F [f e ](ξ ) , k N k=1 k N . λ + |ξ |2

And also, the g2 is defined by −B0 xN g2 (x) = Fξ−1 ρ(ξ ˆ  , 0)](x  ) =  [e

∂h , ∂xN

(3.286)

 −1 −B0 xN where we have set B0 = λ + |ξ  |2 and h(x) = −Fξ−1 ρ(ξ ˆ  , 0)](x  ).  [B0 e By Volevich’s trick, we have 

∞

h(x) = 0



−1 −B0 (xN +yN )   (∂N ρ)(ξ , yN )](x  ) dyN Fξ−1  [B0 e ∞

− 0



∞

= 0

 

Fξ−1  [ ∞

+ 0

∞

− 0

+

−B0 (xN +yN ) Fξ−1 ρ(ξ ˆ  , yN )](x  ) dyN  [e

λ1/2 1/2 −B0 (xN +yN )   λ e (∂N ρ)(ξ , yN )](x  ) dyN B03

Fξ−1  [ Fξ−1  [

N−1  ∞ j =1

0

A B03

   Ae−B0 (xN +yN ) (∂ N ρ)(ξ , yN )](x ) dyN

1 1/2 −B0 (xN +yN )  λ e (λ1/2 ρ)(ξ  , yN )](x  ) dyN B02

Fξ−1  [

1 iξj −B0 (xN +yN )   Ae (∂j ρ)(ξ , yN )](x  ) dyN . B02 A

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

311

Let Zq (RN + ) be the same space as in Lemma 3.5.13. We then define an operator H (λ) acting on (G3 , G4 ) ∈ Zq (RN + ) by setting 

∞

H (λ)(G3 , G4 ) = 0



∞

+ 

Fξ−1  [

0 ∞

− 0

+

λ1/2 1/2 −B0 (xN +yN )   λ e ∂N G4 (ξ , yN )](x  ) dyN B03

Fξ−1  [ Fξ−1  [

N−1  ∞ j =1

0

A B03

  Ae−B0 (xN +yN ) ∂ N G4 (ξ , yN )](x ) dyN

1 1/2 −B0 (xN +yN ) @  λ e G3 (ξ , yN )](x  ) dyN B02

Fξ−1  [

1 iξj −B0 (xN +yN )   Ae ∂j G4 (ξ , yN )](x  ) dyN . B02 A

By Lemmas 3.5.11 and 3.5.12, we see that 2 N H (λ) ∈ Hol (Σ , L(Yq (RN + ), Hq (R+ ))),

h = H (λ)(λ1/2 ρ, ρ),

RL(Y  (RN ),H 2−j (RN )) ({(τ ∂τ )# (λj/2 H (λ)) | λ ∈ Σ q

+

q

+

,λ0 })

≤ rb (λ)

(3.287)

for # = 0, 1 and j = 0, 1, 2. Moreover, we have (λ − Δ)H (λ)(G3 , G4 ) = 0 in RN +.

(3.288)

Let v1 be an N-vector of functions defined by v1 =

−Fξ−1

o e + iξ F [g ](ξ ) , + N−1 , 1 −1 ξ( k=1 ξk F [fk ](ξ ) + ξN F [fN ](ξ )) = F . ξ |ξ |2 (λ + |ξ |2 )|ξ |2

We see that div v1 = g1 in RN + . Moreover, by Lemmas 3.5.2 and 3.5.3, there N N 1 2 exists an operator family B0 (λ) ∈ Hol (Σ , L(Lq (RN + ), Hq (R+ ) )) such that 1 v1 = B0 (λ)f and RL(L

2−j N N N (RN q (R+ ) ,Hq +) )

({(τ ∂τ )# (λj/2 B01 (λ)) | λ ∈ Σ

,λ0 })

≤ rb (λ0 )

for # = 0, 1, j = 0, 1, 2, and λ0 > 0, where rb (λ0 ) is a constant depending on , λ0 , N and q. Let v2j = Fξ−1

o + ξ ξ F [he ](ξ ) , + , j N −1 iξj F [g2 ](ξ ) = −F ξ |ξ |2 |ξ |2

(j = 1, . . . , N)

312

Y. Shibata

and let v2 = ) (v21 , . . . , v2N ), and then by (3.286) we have div v2 = g2 in RN +. 2 (λ), . . . , B 2 (λ)) acting on (G , G ) ∈ Thus, we define an operator B02 (λ) = (B01 3 4 0N Zq (RN + ) by 2 (λ)(G3 , G4 ) = Fξ−1 B0j

+ ξ ξ F [H (λ)(G , G )e ](ξ ) , j N 3 4 . |ξ |2

By (3.288), we have v2 = B02 (λ)(λ1/2 ρ, ρ). Noting that ∂N f e = (∂N f )o and ∂k ∂N f e = (∂k ∂N f )0 (k = 1, . . . , N − 1), we have 2 λB0j (λ)(G3 , G4 ) = Fξ−1 2 λ1/2 ∇B0j (λ)(G3 , G4 ) = Fξ−1 2 ∂k ∇B0j (λ)(G3 , G4 ) = Fξ−1

+ ξ ξ F [λH (λ)(G , G )e ](ξ ) , j N 3 4 , |ξ |2 + ξ ξ F [λ1/2 (∂ H (λ)(G , G ))o ](ξ ) , j

4

|ξ |2

,

+ ξ ξ F [λ1/2 (∂ ∂ H (λ)(G , G ))o ](ξ ) , j k N 3 4 , |ξ |2

Moreover, since ∂N2 H (λ)(G3 , G4 ) = −(λ − from (3.288), we have ∂N2 B02 (λ)(G3 , G4 ) = Fξ−1

3

N

N−1 j =1

∂j2 )H (λ)(G3 , G4 ) as follows

, + ξ2 N  e F [((λ − Δ )H (λ)(G , G )) ](ξ ) . 3 4 |ξ |2

Thus, by (3.287) and the Fourier multiplier theorem, we see that RL(Y  (RN ),H 2−j (RN )N ) ({(τ ∂τ )# (λj/2 B02(λ)) | λ ∈ Σ q

+

q

+

,λ0 })

≤ rb (λ0 )

for # = 0, 1, j = 0, 1, 2, and λ0 > 0, where rb (λ0 ) is a constant depending on , λ0 , N, and q. Since v = v1 + v2 is a solution of Eq. (3.284), setting B0 (λ)(F1 , G3 , G4 ) = B01 (λ)F1 + B02 (λ)(G3 , G4 ), we see that B0 (λ) is the required operator, which completes the proof of Lemma 3.5.17. 

We now prove Theorem 3.5.6. Let (f, d, h) ∈ Yq (RN + ). Let g be a solution of Eq. (3.285) with ρ = n0 · h, and let u, q and h be solutions of the equations: ⎧ λu − Div (μD(u) − qI) = f, div u = g = div g ⎪ ⎪ ⎨ λh + Aκ · ∇Γ h − u · n0 = d ⎪ ⎪ ⎩ (μD(u) − qI − σ (Δ h)I)n0 = h

in RN +, on RN 0 ,

(3.289)

on RN 0 .

Then, according to what pointed out in Sect. 3.4.3, u and h are solutions of Eq. (3.250). Thus, we shall look for u, q and h below. In view of (3.187) and (3.188), applying Lemma 3.5.17 with ρ = n0 ·h, we define u0 by u0 = B0 (f, λ1/2 n0 ·h, n0 ·h).

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

313

Notice that div u0 = g = div g0 with g = λ−1 (f + ∇g). We then look for w0 , q, and h satisfying the equations: ⎧ λw − Div (μD(w0 ) − qI) = f − f0 , div w0 = 0 ⎪ ⎪ 0 ⎨ λh + Aκ · ∇  h − w0 · n0 = d + d0 ⎪ ⎪ ⎩ (μD(w0 ) − qI)n0 − σ (Δ h)n0 = h − h0

in RN +, on RN 0 ,

(3.290)

on RN 0 ,

where we have set f0 = λu0 − Div (μD(u0 )),

d0 = u0 · n0 ,

h0 = μD(u0 )n0 .

To solve Eq. (3.290), we first consider the equations: 

λU1 − Div (μD(U1 ) − P1 I) = F,

div U1 = 0

∂N (U1 · n0 ) = 0,

P1 = 0

in RN +, on RN 0 .

(3.291)

e e o N ) ˜ For F = ) (F1 , . . . , FN ) ∈ Lq (RN + ) , let F = (F1 , . . . , FN−1 , FN ). Let B1 (λ) N N and P1 (λ) be operators acting on F ∈ Lq (R+ ) defined by

+ F [F](ξ ˜ ) − ξ ξ · F [F](ξ ˜ )|ξ |−2 , , λ + μ|ξ |2 + ξ · F [F](ξ ˜ ), . P1 (λ)F = Fξ−1 |ξ |2 B1 (λ)F = Fξ−1

As was seen in Shibata and Shimizu [54, p.587] or [50, Proof of Theorem 4.3], U1 = B1 (λ)F and P1 = P1 (λ)F satisfy Eq. (3.291). Moreover, employing the same argument as in Sect. 3.5.1, by Lemmas 3.5.2 and 3.5.3, we see that

for any

B1 (λ) ∈ Hol (Σ

N N 2 N N ,λ0 , L(Lq (R+ ) , Hq (R+ ) )),

P1 (λ) ∈ Hol (Σ

N N 1 N ,λ0 , L(Lq (R+ ) , Hˆ q (R+ )))

∈ (0, π/2) and λ0 > 0, and moreover RL(L

2−j N N N (RN q (R+ ) ,Hq +) )

({(τ ∂τ )# (λj/2 B1 (λ)) | λ ∈ Σ

,λ0 })

≤ rb (λ0 )

for # = 0, 1 and j = 0, 1, 2, where rb (λ0 ) is a constant depending on , λ0 , m0 , and m1 . In particular, we set u1 = B1 (λ)(f − f0 ),

q1 = P1 (λ)(f − f0 ).

(3.292)

314

Y. Shibata

We now let u = u0 + u1 + U2 and q = q1 + P2 , and then ⎧ λU2 − Div (μD(U2 ) − P2 I) = 0, div U2 = 0 ⎪ ⎪ ⎨ λh + Aκ · ∇  h − U2 · n0 = d + d2 ⎪ ⎪ ⎩ (μD(U2 ) − P2 I)n0 − σ (Δ h)n0 = h − h2

in RN +, on RN 0 ,

(3.293)

on RN 0 ,

where we have set d2 = n0 · (u0 + u1 ),

h2 = μD(u0 + u1 ).

N Thus, for H ∈ Hq1 (RN + ) we consider the equations:



λU2 − Div (μD(U2 ) − P2 I) = 0,

div U2 = 0

in RN +,

(μD(U2 ) − P2 I)n0 = H

on RN 0 ,

(3.294)

and then by Theorem 3.5.8, we see that U2 = V(λ)(λ1/2 H, H) is a unique solutions 1/2 (h − of Eq. (3.294) with some P2 ∈ Hˆ q1 (RN + ). In particular, we set u2 = V(λ)(λ h2 ), (h − h2 )). We finally let u = u0 + u1 + u2 + u3 and q = q1 + q2 + q3 , and then u3 , q3 and h are solutions of the equations: ⎧ λu − Div (μD(u3 ) − q3 I) = 0, div u3 = 0 ⎪ ⎪ 3 ⎨ λh + Aκ · ∇  h − u3 · n0 = d + d3 ⎪ ⎪ ⎩ (μD(u3 ) − q3 I)n0 − σ (Δ h)n0 = 0

in RN +, on RN 0 ,

(3.295)

on RN 0 ,

where d3 = n0 · (u0 + u1 + u2 ). Setting W(λ) = ) (W1 (λ), . . . , WN (λ)), by Theorem 3.5.14, we see that u3 = W(λ)(d + d3 ) and h = Hκ (λ)(d + d3 ) are unique solutions of Eq. (3.295) with some q3 ∈ Hˆ q1 (RN + ). Since the composition of two R-bounded operators is also R bounded as follows from Proposition 3.5.4, we see easily that given ∈ (0, π/2), there exist λ1 > 0 and operator families A0 (λ) and H0 (λ) satisfying (3.251) such that u = A0 (λ)(f, d, λ1/2 h, h) and h = H0 (λ)(f, d, λ1/2 h, h) are unique solutions of Eq. (3.250), and moreover the estimate (3.252) holds. This completes the proof of Theorem 3.5.6.

3.5.4 Problem in a Bent Half Space Let Φ : RN → RN : x → y = Φ(x) be a bijection of C 1 class and let Φ −1 be its inverse map. We assume that ∇Φ and ∇Φ −1 have the forms: ∇Φ = A + B(x) and ∇Φ −1 = A−1 + B−1 (y), where A and A−1 are N × N orthogonal matrices with

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

315

constant coefficients and B(x) and B−1 (y) are matrices of functions in C 2 (RN ) such that (B, B−1 )L∞ (RN ) ≤ M1 ,

∇(B, B−1 )L∞ (RN ) ≤ CA ,

∇ 2 (B, B−1 )L∞ (RN ) ≤ M2 .

(3.296)

Here, CA is a constant depending on constants A, a1 , a2 appearing in Definition 3.2.1. We choose M1 > 0 small enough and M2 large enough eventually, and so we may assume that 0 < M1 ≤ 1 ≤ CA ≤ M2 . Let Ω+ = Φ(RN + ) and Γ+ = Φ(RN ). In the sequel, a and b (x) denote the (i, j )th element of A ij ij −1 and 0 (B−1 ◦ Φ)(x), respectively. Let n+ be the unit outer normal to Γ+ . Setting Φ −1 = ) (Φ−1,1 , . . . , Φ−1.N ), we see that Γ+ is represented by Φ−,N (y) = 0, which yields that n+ (x) = −

) (a (∇Φ−1,N ) ◦ Φ(x) N1 + bN1 (x), . . . , aNN + bNN (x)) =− .  2 1/2 |∇Φ−1,N ) ◦ Φ(x)| ( N j =1 (aNj + bNj (x)) )

(3.297) Obviously, n+ is defined on RN and n+ denotes the unit outer normal to Γ+ for y = Φ(x  , 0) ∈ Γ+ . By (3.296), writing n+ = −) (aN1 , . . . , aNN ) + b+ (x),

(3.298)

we see that b+ is an N-vector defined on RN , which satisfies the estimates: b+ L∞ (RN ) ≤ CN M1 , ∇b+ L∞ (RN ) ≤ CN CA , ∇ 2 b+ L∞ (RN ) ≤ CM2 . (3.299) We next give the Laplace–Beltrami operator on Γ+ . Let  ∂Φ ∂Φ (x) · (x) = (aik + bik (x))(aj k + bj k (x)) = δij + g˜ +ij (x) ∂xi ∂xj N

g+ij (x) =

k=1

 with g˜+ij = N k=1 (aik bj k (x) + aj k bik (x) + bik (x)bj k (x)). Since Γ+ is given by yN = Φ(x  , 0), letting G(x) be an N ×N matrix whose (i, j )th element √ are g+ij (x), we see that G(x  , 0) is the 1st fundamental matrix of Γ+ . Let g+ := det G and let ij g+ (x) denote the (i, j )th component of the inverse matrix, G−1 , of G. By (3.296), we can write g+ = 1 + g˜+ ,

ij

ij

g+ (x) = δij + g˜+ (x)

with ij

(g˜+ , g˜+ )L∞ (RN ) ≤ CN M1 , ij

∇ 2 (g˜+ , g˜+ )L∞ (RN ) ≤ CM2 .

ij

∇(g˜+ , g˜+ )L∞ (RN ) ≤ CN CA ,

(3.300)

316

Y. Shibata

The Laplace–Beltrami operator ΔΓ+ is given by N−1 

(ΔΓ+ f )(y) =

i,j =1

∂ 1 ∂ ij {g+ (x  , 0)g+ (x  , 0) f (Φ(x  , 0))} g+ (x  , 0) ∂xi ∂xj

= Δ f (Φ(x  , 0)) + D+ f (3.301) for y = Φ(x  , 0) ∈ Γ+ . Here, (D+ f )(y) =

N−1  i,j =1

ij

g˜ + (x)

N−1  j ∂ 2 (f ◦ Φ) ∂(f ◦ Φ) (x) + g+ (x) (x) ∂xi ∂xj ∂xj

for y = Φ(x)

j =1

with j g+ (x)

N−1 1  ∂ ij = (g+ (x)g+ (x)). g+ (x) ∂xi i=1

By (3.300) D+ f H 1 (RN ) ≤ CN M1 ∇ 3 f Lq (RN ) + CM2 f H 2 (RN ) . q

+

+

q

+

(3.302)

We now formulate problem treated in this section. Let y0 be any point of Γ+ and let d0 be a positive number such that |μ(y) − μ(y0)|, |σ (y) − σ (y0 )| ≤ m1 M1 , |Aκ (y) − Aκ (y0 )| ≤ m2 M1

for any y ∈ Ω+ ∩ Bd0 (y0 ); for any y ∈ Γ+ ∩ Bd0 (y0 ). (3.303)

In addition, μ, σ , and Aκ satisfy the following conditions: m0 ≤ μ(y), σ (y) ≤ m1 , |Aκ (y)| ≤ m2

|∇μ(y)|, |∇σ (y)| ≤ m1 for any y ∈ Γ+ ,

for any y ∈ Ω+ ,

Aκ W 2−1/q (Ω r

+)

≤ m3 κ −b (3.304)

for any κ ∈ (0, 1). In view of (3.164) and (3.304), to have (3.303) for given M1 ∈ (0, 1) it suffices to choose d0 > 0 in such a way that d0 ≤ M1 and d0a ≤ M1 . We assume that N < r < ∞ according to (3.164) and let A0 = 0. Let ϕ(y) be

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

317

a function in C0∞ (RN ) which equals 1 for y ∈ Bd0 /2 (y0 ) and 0 in the outside of Bd0 (y0 ). We assume that ∇ϕH∞ 1 (RN ) ≤ M2 . Let μy0 (y) = ϕ(y)μ(y) + (1 − ϕ(y))μ(y0), σy0 (y) = ϕ(y)σ (y) + (1 − ϕ(y))σ (y0), Aκ,y0 (y) = ϕ(y)Aκ (y) + (1 − ϕ(y))Aκ (y0 ). In the following, C denotes a generic constant depending on m0 , m1 , m2 , m3 , N, and q, and CM2 denotes a generic constant depending on M2 , m0 , m1 , m2 , m3 , N, and q. Given v ∈ Hq2 (Ω+ )N and h ∈ Hq3 (Ω+ ), let Kb (v, h) be a unique solution of the weak Dirichlet problem: (∇Kb (v, h), ∇ϕ)Ω+ = (Div (μy0 D(v)) − ∇div v, ∇ϕ)Ω+

(3.305)

for any ϕ ∈ Hˆ q1 ,0 (Ω+ ) subject to Kb (v, h) =< μy0 D(v)n+ , n+ > −σy0 ΔΓ+ h − div v

on Γ+ .

We then consider the following equations: ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

λv − Div (μy0 D(v) − Kb (v, h)I) = g

in Ω+ ,

λh + Aκ,y0 · ∇Γ+ h − v · n+ = gd

on Γ+ ,

(μy0 D(v) − Kb (v, h)I)n+ − σy0 (ΔΓ+ h)n+ = gb

on Γ+ .

(3.306)

The following theorem is a main result in this section. Theorem 3.5.18 Let 1 < q < ∞ and 0 < < π/2. Let γκ be the number defined in Theorem 3.4.8. Then, there exist M1 ∈ (0, 1), λ˜ 0 ≥ 1 and operator families Ab (λ) and Hb (λ) with Ab (λ) ∈ Hol (Λκ,λ˜ 0 γκ , L(Yq (Ω+ ), Hq2 (Ω+ )N )), Hb (λ) ∈ Hol (Λκ,λ˜ 0 γκ , L(Yq (Ω+ ), Hq3 (Ω+ ))) such that for any λ = γ + iτ ∈ Λκ,λ˜ 0 γκ and (g, gd , gb ) ∈ Yq (Ω+ ), u = Ab (λ)(g, gd , λ1/2 gb , gb ),

h = Hb (λ)(g, gd , λ1/2 gb , gb )

318

Y. Shibata

are unique solutions of Eq. (3.306), and RL(Y

2−j (Ω+ )N ) q (Ω+ ),Hq

({(τ ∂τ )# (λj/2 Ab (λ)) | λ ∈ Λκ,λ˜ 0 γκ }) ≤ rb ,

RL(Yq (Ω+ ),H 3−k (Ω+ )) ({(τ ∂τ )# (λk Hb (λ)) | λ ∈ Λκ,λ˜ 0 γκ }) ≤ rb ,

(3.307)

q

for # = 0, 1, j = 0, 1, 2, and k = 0, 1. Here, rb is a constant depending on m0 , m1 , m2 , N, q, and , but independent of M1 and M2 , and moreover, λ˜ 0 is a constant depending on M2 . Below, we shall prove Theorem 3.5.18. By the change of variables y = Φ(x), we transform Eq. (3.306) to a problem in the half-space. Let y0 = Φ(x0 ),

μ(x) ˜ = ϕ(Φ(x))(μ(Φ(x)) − μ(Φ(x0 ))),

σ˜ (x) = ϕ(Φ(x))(σ (Φ(x)) − σ (Φ(x0 ))), A˜ κ (x) = ϕ(Φ(x))(Aκ (Φ(x)) − Aκ (Φ(x0 ))). Notice that μy0 (Φ(x)) = μ(y0 ) + μ(x), ˜

σy0 (Φ(x)) = σ (y0 ) + σ˜ (x),

Aκ (Φ(x  , 0)) = Aκ (y0 ) + A˜ κ (x). We may assume that m1 , m2 , m3 ≤ M2 . Recalling that ∇ϕH∞ 1 (RN ) ≤ M2 , by (3.303) and (3.304) we have (μ, ˜ σ˜ )L∞ (RN ) ≤ m1 M1 ,

∇(μ, ˜ σ˜ )L∞ (RN ) ≤ CM2 ,

A˜ κ L∞ (RN ) ≤ m2 M1 ,

(3.308)

0

∇ A˜ κ W 1−1/q (RN ) ≤ CM2 κ q

−b

0

for κ ∈ (0, 1).

Since x = Φ −1 (y), we have  ∂ ∂ = (akj + bkj (x)) ∂yj ∂xk N

k=1

where (∇Φ −1 )(Φ(x)) = (aij + bij (x)). Let g := det ∇Φ,

g˜ = g − 1.

(3.309)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

319

By (3.296), ˜gL∞ (RN ) ≤ CN M1 ,

∇ 2 g˜ L∞ (RN ) ≤ CM2 . (3.310)

∇ g˜ L∞ (RN ) ≤ CN CA ,

By the change of variables: y = Φ(x), the weak Dirichlet problem: for any ϕ ∈ Hˆ q1 ,0 (Ω+ ),

(∇u, ∇ϕ)Ω+ = (k, ∇ϕ)Ω+

subject to u = k on Γ+ , is transformed to the following variational problem: for any ψ ∈ Hˆ q1 ,0 (RN + ),

(∇v, ∇ψ)RN + (B 0 ∇v, ∇ψ)RN = (h, ∇ψ)RN +

+

+

(3.311) subject to v = h, where h = g(A−1 + B−1 ◦ Φ)k ◦ Φ and h = k ◦ Φ. Moreover, B 0 0 , given by is an N × N matrix whose (#, m)th component, B#m 0 B#m = g˜ δ#m + g

N  (a#j bmj (x) + amj b#j (x) + b#j bmj (x)). j =1

By (3.296), we have 0 B#m L∞ (RN ) ≤ CN M1 ,

0 ∇B#m L∞ (RN ) ≤ CN CA ,

0 ∇ 2 B#m L∞ (RN ) ≤ CM2 .

(3.312)

Lemma 3.5.19 Let 1 < q < ∞. Then, there exist an M1 ∈ (0, 1) and an operator K1 with N 1 N N ˆ1 K1 ∈ L(Lq (RN + ) , Hq (R+ ) + Hq,0 (R+ )) N 1 N such that for any f ∈ Lq (RN + ) and f ∈ Hq (R+ ), v = K1 (f, f ) is a unique solution of the variational problem:

for any ψ ∈ Hˆ q1 ,0 (RN + ),

(∇v, ∇ψ)RN + (B 0 ∇v, ∇ψ)RN = (f, ∇ψ)RN +

+

+

(3.313) subject to v = f on RN 0 , which possesses the estimate: ∇vLq (RN ) ≤ CM2 (fLq (RN ) + f H 1 (RN ) ). +

+

q

+

Proof We know the unique existence theorem of the variational problem: (∇v, ∇ψ)RN = (f, ∇ψ)RN +

+

for any ψ ∈ Hˆ q1 ,0 (RN + ),

(3.314)

320

Y. Shibata

subject to v = f on RN + . Thus, choosing M1 > 0 small enough in (3.312) and using the Banach fixed point theorem, we can easily prove the lemma. 

Using the change of the unknown functions: u = A−1 v ◦ Φ as well as the change of variable: y = Φ(x), we will derive the problem in RN + from (3.306). Noting that A = ) A−1 , by (3.309) we have Dij (v) =

N 

aki a#j Dk# (u) + bijd : ∇u

(3.315)

k,#=1

N with bijd : ∇u = k,#=1 akj b#i Dk# (u). Setting b+ (x) = in (3.298), by (3.298) we have

) (b

+1 , . . . , b+N )

< D(v)n+ , n+ >=< D(u)n0 , n0 > +B 1 : ∇u

(3.316)

where we have set B 1 : ∇u = −2

N  i,j =1

+

N 

N 

aj i b+i Dj N (u) +

aki a#j b+i b+j Dk# (u)

i,j,k,#=1

(bijd : ∇u)(aNi + b+i )(aNj + b+j ).

i,j =1

By (3.296), we have B 1 : ∇uLq (RN ) ≤ CN M1 ∇uLq (RN ) , +

+

B : ∇uH 1 (RN ) ≤ CN {M1 ∇ uLq (RN ) + CA uH 1 (RN ) }. 1

2

q

+

+

q

(3.317)

+

And also, div v = div u + B 2 : ∇u

with

B 2 : ∇u =

M  N  ∂u# ( bkj a#j ) . ∂xk

(3.318)

#,k=1 j =1

By (3.296), we have B 2 : ∇uLq (RN ) ≤ CN M1 ∇uLq (RN ) , +

+

B : ∇uH 1 (RN ) ≤ CN {M1 ∇ uLq (RN ) + CA uH 1 (RN ) }. 2

2

q

+

+

q

+

(3.319)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

321

By (3.315), we have A−1 Div (μy0 D(v)) = Div (μ(y0 )D(u)) + R1 : u

(3.320)

with R1 : u = ) (R1 : u|1 , . . . , R1 : u|N ), and N N   ∂ ∂ d (μ(x)D ˜ asi akj ((μ(y0) + μ(x))b ˜ sk (u)) + ij : ∇u) ∂xk ∂xk

R1 : u|s =

k=1

+

N 

i,j,k=1

a#j bkj

j,k,#,m=1

N  ∂ ∂ d {(μ(y0 ) + μ(x))D ˜ asi bkj {(μ(y0 ) + μ(x))b ˜ s# (u)} + ij :∇u}. ∂xk ∂xk i,j,k=1

By (3.296) and (3.308), R1 : uLq (RN ) ≤ CN m1 M1 ∇ 2 uLq (RN ) + CM2 uH 1 (RN ) . +

+

q

+

(3.321)

And also, by (3.315) (A−1 + B−1 ◦ Φ −1 )Div (μy0 D(v)) = Div (μ(y0)D(u)) + R2 : u with R2 : u = (R2 : u|1 , . . . , R2 : u|N ) and R2 : u|s = R1 : u|s

+

N 

bsi (akj +bkj )

i,j,k=1

N  ∂ d [(μ(x) ˜ + μ(y0 )){ a#i amj D#m (u)+bij :∇u}]. ∂xk #,m=1

By (3.296) R2 : uLq (RN ) ≤ CN m1 M1 ∇ 2 uLq (RN ) + CM2 uH 1 (RN ) . +

+

q

+

(3.322)

And also, we have (A−1 + B−1 ◦ Φ −1 )(∇div v) ◦ Φ = ∇div u + R3 : u with R3 : u = (R3 : u|1 , . . . , R3 : u|N ) and R3 : u|s =

∂ (B 2 : ∇u) ∂xs +

N  N  ∂ { (asi bki + bsi (aki + bki ))} (div u + B 2 : ∇u). ∂xk k=1 i=1

322

Y. Shibata

By (3.296) R3 : uLq (RN ) ≤ CN m1 (M1 ∇ 2 uLq (RN ) + CA uH 1 (RN ) ). +

+

(3.323)

+

q

Let f(u) := g(A−1 + B−1 ◦ Φ)(Div (μy0 D(v)) − ∇div v) ◦ Φ, and then f(u) = Div (μ(y0)D(u)) − ∇div u + R4 : u with R4 : u = (R4 : u|1 , . . . R4 : u|N ) and R4 : u|s = g˜ (Div (μ(y0)D(u)) − ∇div u) + gR2 : u − gR3 : u. By (3.296), (3.308), (3.310)–(3.322), and (3.323), R4 : uLq (RN ) ≤ CN (m1 + 1)M1 ∇ 2 uLq (RN ) + CM2 uH 1 (RN ) . +

+

+

q

(3.324)

In view of (3.301), (3.316) and (3.318), setting ρ = h ◦ Φ, 1 f (u, ρ) =< μ(x)D(u)n ˜ ˜ : ∇u 0 , n0 > +(μ(y0 ) + μ(x))B

− σ˜ (x)Δ ρ − (σ (y0 ) + σ˜ (x))D+ ρ − B 2 : ∇u, we have < μy0 D(v)n+ , n+ > −σy0 ΔΓ+ h − div v =< μ(y0 )D(u)n0 , n0 > −σ (y0 )Δ ρ − div u + f (u, ρ). Thus, K1 (u, ρ) = Kb (v, h) ◦ Φ satisfies the variational equation: (∇K1 (u, ρ), ∇ψ)RN + (B 0 ∇K1 (u, ρ), ∇ψ)RN +

+

= (Div (μ(y0 )D(u)) − ∇div u + R4 : u, ∇ψ)RN

+

for any ψ ∈ Hˆ q1 ,0 (RN + ), subject to K1 (u, ρ) =< μ(y0 )D(u)n0 , n0 > −σ (y0 )Δ ρ − div u + f (u, ρ)

on RN 0 .

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

323

N ˜ ˆ1 Let K(u, ρ) ∈ Hq1 (RN + ) + Hq,0 (R+ ) be a unique solution of the weak Dirichlet problem:

˜ (∇ K(u, ρ), ∇ψ)RN = (Div (μ(y0 )D(u)) − ∇div u, ∇ψ)RN +

+

for any ψ ∈ Hˆ q1 ,0 (RN + ) subject to ˜ K(u, ρ) =< μ(y0 )D(u)n0 , n0 > −σ (y0 )Δ ρ − div u

on RN 0 .

˜ ρ) + K2 (u, ρ), we then see that K2 (u, ρ) satisfies the Setting K1 (u, ρ) = K(u, variational equation: ˜ (∇K2 (u, ρ), ∇ψ)RN + (B 0 ∇K2 (u, ρ), ∇ψ)RN = (R4 : u − B 0 ∇ K(u, ρ), ∇ψ)RN +

+

+

N for any ψ ∈ Hˆ q1 ,0 (RN + ), subject to K2 (u, ρ) = f (u, ρ) on R0 . In view of Lemma 3.5.19, we have

˜ ρ), f (u, ρ)). K2 (u, ρ) = K1 (R4 : u − B 0 ∇ K(u, By Lemma 3.5.19, (3.312), (3.317), (3.319), (3.302), and (3.324), we have ∇K2 (u, ρ)Lq (RN ) +

≤ CN (1 + m1 )M1 (∇ 2 uLq (RN ) + ∇ 3 ρLq (RN ) ) +

+

+ CM2 (uH 1 (RN ) + ρH 2 (RN ) ). +

q

q

+

Since A−1 ∇Kb (v, h)|s =

N 

asi (aki + bki )

i,k=1

 ∂ ˜ ∂ ˜ K(u, ρ) + K(u, ρ) ( asi bki ) ∂xs ∂xk N

=

∂ K1 (u, ρ) ∂xk N

k=1 i=1

+

N  k=1

(δks +

N  i=1

asi bki )

∂ K2 (u, ρ), ∂xk

by (3.320) we see that the first equation of Eq. (3.306) is transformed to ˜ ρ)I) + R5 (u, ρ) = h in RN λu − Div (μ(y0 )D(u) − K(u, +,

(3.325)

324

Y. Shibata

where h = A−1 g ◦ Φ, R 5 (u, ρ) = (R5 (u, ρ)|1 , . . . , R5 (u, ρ)|N ), and R5 (u, ρ)|s = −R1 : u|s +

N  N  ∂ ˜ ( asi bki ) K(u, ρ) ∂xk k=1 i=1

+

N 

N 

k=1

i=1

(δks +

asi bki )

∂ K2 (u, ρ). ∂xk

By (3.298), we have v · n+ = −() A−1 u) · ) (aN1 , . . . , aNN ) + () A−1 u) · b+ = u · n0 + u · (A−1 b+ ), and so the second equation of Eq. (3.306) is transformed to λρ + Aκ (y0 ) · ∇  ρ − u · n0 + R6κ (u, ρ) = hd with hd = gd ◦ Φ and R60 (u, ρ) = −u · (A−1 b+ )

for κ = 0,

R6κ (u, ρ) = A˜ κ (x)∇  ρ − u · (A−1 b+ )

for κ ∈ (0, 1).

By (3.298) and (3.315), we have A−1 μy0 D(v)n+ = μ(y0 )D(u)n0 + R71 (u), where R71 (u) is an N-vector of functions whose sth component, R71 (u)|s , is defined by R71 (u)|s = −μ(x)D ˜ ˜ sN (u) + (μ(y0 ) + μ(x)) ×

N 

(aij b+j Dsi (u) + asi bijd : ∇u(−aNj + b+j )).

i,j =1

By (3.298), ˜ ˜ A−1 Kb (v, h)n+ = K(u, ρ)n0 + K(u, ρ)A−1 b+ + K2 (u, ρ)(n0 + A−1 b+ ). By (3.301), A−1 σy0 (ΔΓ+ h)n+ = σ (y0 )(Δ ρ)n0 + σ˜ (x)(Δ ρ)n0 + (σ˜ (x) + σ (y0 )){(Δ ρ)(A−1 b+ ) + (D+ ρ)(n0 + A−1 b+ )}.

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

325

Putting formulas above together yields that the third equation of Eq. (3.306) is transformed to the equation: ˜ (μ(y0)D(u) − K(u, ρ)I)n0 − σ (y0 )(Δ ρ)n0 + R7 (u, ρ) = hb

on RN 0 ,

where hb = A−1 gb ◦ Φ, and ˜ R7 (u, ρ) = R71 (u, ρ) − K(u, ρ)(A−1 b+ ) − K2 (u, ρ)(n0 + A−1 b+ ) − σ˜ (x)(Δ ρ)n0 − (σ˜ (x) + σ (y0 )){(Δ ρ)(A−1 b+ ) + (D+ ρ)(n0 + A−1 b+ )}. Summing up, we have seen that Eq. (3.306) is transformed to the following equations: ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

˜ ρ)I) + R5 (u, ρ) = h λu − Div (μ(y0 )D(u) − K(u, λρ + Aκ (y0 ) · ∇  ρ − u · n0 + Rκ6 (u, ρ) = hd ˜ ρ)I)n0 − σ (y0 )(Δ ρ)n0 + R7 (u, ρ) = hb (μ(y0 )D(u) − K(u,

in RN +, on RN 0 , on RN 0 , (3.326)

where h = A−1 g ◦ Φ, hd = gd ◦ Φ, hd = A−1 gd ◦ Φ, and R5 (u, ρ), R6κ (u, ρ) and R7 (u, ρ) are linear in u and ρ and satisfy the estimates: R5 (u, ρ)Lq (RN ) ≤ CM1 (∇ 2 uLq (RN ) + ∇ 3 ρLq (RN ) ) +

+

+

+ CM2 (uH 1 (RN ) + ρH 2 (RN ) ), q

+

q

+

R60 (u, ρ)W 2−1/q (RN ) q

≤ CM1 ∇ uLq (RN ) + CM2 uH 1 (RN ) ,

R6κ (u, ρ)W 2−1/q (RN ) q

≤ CM1 (∇ uLq (RN ) + ∇ ρLq (RN ) )

0

0

2

+

+

q

2

3

+

+ CM2 (uH 1 (RN ) + κ q

+

+

−b

ρH 2 (RN ) ), +

q

(3.327)

R7 (u, ρ)Lq (RN ) ≤ CM1 (∇uLq (RN ) + ∇ 2 ρLq (RN ) ) +

+

+

+ CM2 (uLq (RN ) + ρH 1 (RN ) ), +

q

+

R7 (u, ρ)H 1 (RN ) ≤ CM1 (∇ 2 uLq (RN ) + ∇ 3 ρLq (RN ) ) q

+

+

+

+ CM2 (uH 1 (RN ) + ρH 2 (RN ) ). q

+

q

+

Here and in the following, C denotes a generic constant depending on N, q, m1 , and m2 and CM2 a generic constant depending on N, q, m1 , m2 , m3 and M2 . By Theorem 3.5.6, there exists a large number λ0 and operator families A0 (λ) and

326

Y. Shibata

H0 (λ) with 2 N N A0 (λ) ∈ Hol (Λκ,λ0 , L(Y(RN + ), Hq (R+ ) )), 3 N H0 (λ) ∈ Hol (Λκ,λ0 , L(Y(RN + ), Hq (R+ )))

such that for any λ ∈ Λκ,λ0 and (f, d, h) ∈ Yq (RN + ), u and ρ with u = A0 (λ)Fλ (f, d, h),

ρ = H0 (λ)Fλ (f, d, h),

where Fλ (f, d, h) = (f, d, λ1/2 h, h), are unique solutions of the equations: ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

˜ λu − Div (μ(y0 )D(u) − K(u, ρ)I) = f

in RN +,

λρ + Aκ (y0 ) · ∇  ρ − u · n0 = d

on RN 0 ,

˜ (μ(y0 )D(u) − K(u, ρ)I)n0 − σ (y0 )(Δ ρ)n0 = h,

on RN 0 ,

and RL(Y (RN ),H 2−j (RN )N ) ({(τ ∂τ )s (λj/2 A0 (λ)) | λ ∈ Λκ,λ0 }) ≤ rb , +

+

q

RL(Y (RN ),H 3−k (RN )) ({(τ ∂τ )s (λk H0 (λ)) | λ ∈ Λκ,λ0 }) ≤ rb +

q

+

for s = 0, 1, j = 0, 1, 2, and k = 0, 1. Here, rb is a constant depending on , N, m1 , and m2 . Let u = A0 (λ)Fλ (h, hd , hb ) and ρ = H0 (λ)Fλ (h, hd , hb ) in (3.326). Then, Eq. (3.326) is rewritten as ⎧ ˜ λu − Div (μ(y0 )D(u) − K(u, ρ)I) + R5 (u, ρ) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ = h + R8 (λ)Fλ (h, hd , hb ) ⎪ ⎪ ⎪ ⎪ ⎪ λρ + A · ∇  ρ − u · n + R 6 (u, ρ) ⎨ κ 0 κ ⎪ ⎪ = hd + R8d (λ)Fλ (h, hd , hb ) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ˜ (μ(y0 )D(u) − K(u, ρ)I)n0 − σ (y0 )(Δ ρ)n0 + R7 (u, ρ) ⎪ ⎪ ⎪ ⎪ ⎩ = hb + R8b (λ)Fλ (h, hd , hb )

in RN +, in RN +, on RN 0 , (3.328)

where we have set R8 (λ)(F1 , F2 , F3 , F4 ) = R5 (A0 (λ)(F1 , F2 , F3 , F4 ), H0 (λ)(F1 , F2 , F3 , F4 )),

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

327

R8d (λ)(F1 , F2 , F3 , F4 ) = R6κ (A0 (λ)(F1 , F2 , F3 , F4 ), H0 (λ)(F1 , F2 , F3 , F4 )), R8b (λ)(F1 , F2 , F3 , F4 ) = R7 (A0 (λ)(F1 , F2 , F3 , F4 ), H0 (λ)(F1 , F2 , F3 , F4 )). Let R9 (λ)F = (R8 (λ)F, R8d (λ)F, R8b (λ)F ) for F = (F1 , F2 , F3 , F4 ) ∈ Yq (RN + ). Notice that R9 (λ)F = (R8 (λ)F, R8d (λ)F, λ1/2 R8b (λ)F, R8b (λ)F ) ∈ Yq (RN + ), for F = (F1 , F2 , F3 , F4 ) ∈ Yq (RN + ) and that the right side of Eq. (3.328) is written as (h, hd , hb ) + F 9 (λ)Fλ (h, hd , hb ). By (3.327), (3.299), Proposition 3.5.4, and Theorem 3.5.6, we have −1/2

RL(Yq (RN )) ({(τ ∂τ )# (Fλ R9 (λ)) | λ ∈ Λκ,λ1 }) ≤ CM1 + CM2 (λ1 +

+ λ−1 1 γκ ) (3.329)

for any λ1 ≥ λ0 . Here and in the following, C denotes a generic constant depending on N, , m1 , m2 , and CA , and CM2 denotes a generic constant depending on N, , m1 , m2 , m3 , CA , and M2 . Choosing M1 so small that CM1 ≤ 1/4 and choosing −1/2 λ1 > 0 so large that CM2 λ1 ≤ 1/8 and CM2 λ−1 1 γκ ≤ 1/8, by (3.329) we have RL(Yq (RN )) ({(τ ∂τ )# (Fλ R9 (λ)) | λ ∈ Λκ,λ1 }) ≤ 1/2

(3.330)

+

2 γ , for # = 0, 1. Since γκ ≥ 1 and we may assume that CM2 ≥ 1, if λ1 ≥ 64CM κ 2 −1/2

then CM2 λ1 ≤ 1/8 and CM2 λ−1 1 γκ ≤ 1/8. N Recall that for F = (F1 , F2 , F3 , F4 ) ∈ Yq (RN + ) and (h, hd , hb ) ∈ Yq (R+ ),

(F1 , F2 , F3 , F4 )Yq (RN ) = (F1 , F3 )Lq (RN ) + F2 W 2−1/q (RN ) + F4 H 1 (RN ) , +

+

q

q

0

+

(h, hd , hb )Xq (RN ) = hLq (RN ) + hd W 2−1/q (RN ) + hb H 1 (RN ) +

+

q

q

0

+

(3.331) (cf. Remark 3.4.12, where Ω should be replaced by RN + ). By (3.330) we have Fλ (R9 (λ)Fλ (h, hd , hb ))Yq (RN ) ≤ (1/2)Fλ (h, hd , hb )Yq (RN ) . +

+

(3.332)

328

Y. Shibata

In view of (3.331), when λ = 0, Fλ (h, hd , hb )Yq (RN ) is an equivalent norm to + ∞ 9 j (h, hd , hb )Xq (RN ) . Thus, by (3.332) (I + R9 (λ)Fλ )−1 = j =1 (−R (λ)Fλ ) +

exists in L(Xq (RN + )). Setting u = A0 (λ)Fλ (I + R9 (λ)Fλ )−1 (h, hd , hb ), ρ = H0 (λ)Fλ (I + R9 (λ)Fλ )−1 (h, hd , hb ),

(3.333)

by (3.328) we see that solutions of Eq. (3.326). In view of (3.328),  u and ρ are N 9 j (I + Fλ R9 (λ))−1 = ∞ j =0 (−Fλ R (λ)) exists in L(Yq (R+ )), and RL(Yq (RN )) ({(τ ∂τ )# (I + Fλ R9 (λ))−1 | λ ∈ Λκ,λ1 }) ≤ 4 +

(3.334)

for # = 0, 1. Since Fλ (I + R (λ)Fλ ) 9

−1

∞ ∞   9 j = Fλ (−R (λ)Fλ ) = ( (−Fλ R9 (λ))j )Fλ j =0

j =0

= (I + Fλ R (λ)) 9

−1

Fλ ,

defining operators A1 (λ) and H1 (λ) acting on F = (F1 , F2 , F3 , F4 ) ∈ Yq (RN + ) by A1 (λ)F = A0 (λ)(I + Fλ R9 (λ))−1 F,

H1 (λ)F1 = H0 (λ)(I + Fλ R9 (λ))−1 F,

by (3.333) u = A1 (λ)Fλ (h, hd , hb ) and ρ = H1 (λ)Fλ (h, hd , hb ) are solutions of Eq. (3.326). Moreover, by (3.334) and Theorem 3.5.6 RL(Y

2−j N N (RN q (R+ ),Hq +) )

({(τ ∂τ )# (λj/2 A1 (λ)) | λ ∈ Λκ,λ1 γκ }) ≤ 4rb ,

RL(Yq (RN ),H 3−k (RN )) ({(τ ∂τ )# (λk H1 (λ)) | λ ∈ Λκ,λ1 γκ }) ≤ 4rb , +

q

(3.335)

+

for # = 0, 1, j = 0, 1, 2 and k = 0, 1. Recalling that v = () A−1 u) ◦ Φ −1 , h = A−1 g ◦ Φ,

h = ρ ◦ Φ −1 ,

hd = gd ◦ Φ,

hd = A−1 gd ◦ Φ,

we define operators Ab (λ) and Hb (λ) acting on F = (F1 , F2 , F3 , F4 ) ∈ Yq (Ω+ ) by Ab (F1 , F2 , F3 , F4 ) = ) A−1 [A1 (λ)(A−1 F1 ◦ Φ, F2 ◦ Φ, A−1 F3 ◦ Φ, F4 ◦ Φ)] ◦ Φ −1 , Hb (F1 , F2 , F3 , F4 ) = [H1 (λ)(A−1 F1 ◦ Φ, F2 ◦ Φ, A−1 F3 ◦ Φ, F4 ◦ Φ)] ◦ Φ −1 .

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

329

Obviously, given any (g, gd , gb ) ∈ Yq (Ω+ ), u = Ab (λ)Fλ (g, gd , gb ) and h = Hb (λ)Fλ (g, gd , gb ) are solutions of Eq. (3.306). From (3.296) we have g ◦ Φ −1 Hq# (Ω+ ) ≤ CA gH # (RN ) q

for # = 0, 1, 2,

+

∇ 3 (g ◦ Φ −1 )Lq (Ω+ ) ≤ CA ∇ 2 gH 1 (RN ) + CM2 ∇gLq (RN ) , q

h ◦ ΦH # (RN ) ≤ CA hHq# (Ω+ ) q

+

+

+

for # = 0, 1, 2,

and so, in view of (3.335) we can choose λ˜ 0 ≥ λ1 suitably large such that Ab (λ) and Hb (λ) satisfy the estimates: RL(Y

2−j (Ω+ )N ) q (Ω+ ),Hq

({(τ ∂τ )# (λj/2 Ab (λ)) | λ ∈ Λκ,λ1 γκ }) ≤ Crb ,

RL(Yq (RN ),Hq3−k (RN )) ({(τ ∂τ )# (λk Hb (λ)) | λ ∈ Λκ,λ1 γκ }) ≤ Crb , +

+

for # = 0, 1, j = 0, 1, 2 and k = 0, 1, where C and rb are constants independent of M2 . This completes the existence part of Theorem 3.5.18. The uniqueness can be proved by showing a priori estimates of solutions of Eq. (3.326) with (h, hd , hb ) = (0, 0, 0) in the same manner as in the proof of Theorem 3.5.5. This completes the proof of Theorem 3.5.18 without the operators L1 and L2 .

3.5.5 Some Preparation for the Proof of Theorem 3.4.11 In the following, we use the symbols given in Proposition 3.2.2 in Sect. 3.2.1 and N we write Ωj = Φj (RN + ), and Γj = Φj (R0 ) for the sake of simplicity. Recall that Bji = Br0 (xji ). In view of the assumptions (3.164)–(3.166), we may assume that |μ(x) − μ(xji )| ≤ M1

for any x ∈ Bji ;

|σ (x) − σ (xj1 )| ≤ M1

for any x ∈ Γj ∩ Bji ;

|Aκ (x) − Aκ (xj1 )| ≤ M1

for any x ∈ Γj ∩ Bji ;

m0 ≤ μ(x), σ (x) ≤ m1 , |Aκ (x)| ≤ m2

(3.336)

|∇μ(x)|, |∇σ (x)| ≤ m1 for any x ∈ Γ ,

for any x ∈ Ω,

Aκ W 2−1/q (Γ ) ≤ m3 κ −b r

(3.337)

for any κ ∈ (0, 1). Here, m0 , m1 , m2 , m3 , b and r are constants given in (3.164). We next prepare some propositions used to construct a parametrix. Proposition 3.5.20 Let X be a Banach space and X∗ its dual space, while  · X ,  · X∗ , and < ·, · > denote the norm of X, the norm of X∗ , and the duality pairing

330

Y. Shibata

between of X and X∗ , respectively. Let n ∈ N, l = 1, . . . , n, and {al }nl=1 ⊂ C, l ∞ ∗ ∞ and let {fjl }∞ j =1 be sequences in X and {gj }j =1 , {hj }j =1 be sequences of positive numbers. Assume that there exist maps Nj : X → [0, ∞) such that | < fjl , ϕ > | ≤ M3 gjl Nj (ϕ)

n A  B   al fjl , ϕ  ≤ M3 hj Nj (ϕ) 

(l = 1, . . . , n),

l=1

for any ϕ ∈ X with some positive constant M3 independent of j ∈ N and l = 1, . . . , n. If ∞  q gjl < ∞,

∞  

j =1

j =1

hj

q

∞  

< ∞,

Nj (ϕ)

q 

≤ (M4 ϕX )q



j =1

with 1 < q < ∞ and q  = q/(q −1) for some positive constant M4 , then the infinite l l ∗ sum f = ∞ j =1 fj exists in the strong topology of X and f l X∗ ≤ M3 M4

∞   l q 1/q gj ,

n     al f l  

j =1

l=1

X∗

≤ M3 M4

∞   q 1/q hj . j =1

(3.338) Proof For a proof, see Proposition 9.5.2 in Shibata [45].



The following propositions are used to define the infinite sum of R-bounded operator families defined on RN and Ωj . Proposition 3.5.21 Let 1 < q < ∞, i = 0, 1, and n ∈ N0 . Set Hj0 = RN and n (RN ) ≤ c1 for any Hj1 = Ωj . Let ηji be a function in C0∞ (Bji ) such that ηji H∞ j ∈ N with some constant c independent of j ∈ N. Let f (j 1 j  ∞ i ∈ N) be elements q in Hqn (Hji ) such that ∞ j =1 fj  n i < ∞. Then, j =1 ηj fj converges some Hq (Hj )

f ∈ Hqn (Ω) strongly in Hqn (Ω), and

f Hqn (Ω) ≤ Cq {

∞ 

q }1/q . Hqn (Hij )

fj 

j =1

Proof For a proof, see Proposition 9.5.3 in Shibata [45].



Proposition 3.5.22 Let 1 < q < ∞ and n = 2, 3. Then we have the following assertions. n−1/q

(1) There exist extension maps Tnj : Wq n−1/q

(Γj ) → Hqn (Ωj ) such that for any

h ∈ Wq (Γj ), Tnj h = h on Γj and Tnj hHqn (Ωj ) ≤ ChW n−1/q (Γ ) with j q some constant C > 0 independent of j ∈ N.

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . . n−1/q

(2) There exists an extension map TnΓ : Wq

331

(Γ ) → Hqn (Ω) such that for h ∈

n−1/q Wq (Γ ),

TnΓ h = h on Γ and TnΓ hHqn (Ω) ≤ ChW n−1/q (Γ ) with some q constant C > 0. 

Proof For a proof, see Proposition 9.5.4 in Shibata [45]. C0∞ (Bj1 )

(j ∈ N) Proposition 3.5.23 Let 1 < q < ∞ and n = 2, 3 and let ηj ∈ n (RN ) ≤ c2 for some constant c2 independent of j ∈ N. Then, we have with ηj H∞ the following two assertions: n−1/q

(1) Let (Γj ) satisfying the condition: ∞ fj (j q ∈ N) be functions in Wq ∞ f  < ∞, and then the infinite sum η f j j j converges to n−1/q j =1 j =1 Wq (Γj ) n−1/q

some f ∈ Wq

n−1/q

(Γ ) strongly in Wq

f W n−1/q (Γ ) ≤ Cq { q

n−1/q

(2) For any h ∈ Wq

∞ 

(Γ ) and q

fj 

j =1

n−1/q

Wq

(Γj )

}1/q .

(Γ ),

∞ 

q

ηj h

j =1

q

n−1/q

Wq

(Γj )

≤ Ch

n−1/q

Wq

(Γ )

.

Proof For a proof, see Proposition 9.5.5 in Shibata [45].



3.5.6 Parametrix of Solutions of Eq. (3.184) In this subsection, we construct a parametrix for Eq. (3.226). Let {ζji }j ∈N and {ζ˜ji }j ∈N (i = 0, 1) be sequences of C0∞ functions given in Proposition 3.2.2, and let N (f, d, h) ∈ Yq (Ω) (cf. (3.193)). Recall that Ωj = Φj (RN + ) and Γj = Φj (R0 ). Let μij (x) = ζ˜ji (x)μ(x) + (1 − ζ˜ji (x))μ(xji ), σj (x) = ζ˜j1 (x)σ (x) + (1 − ζ˜j1 (x))σ (xj1 ), Aκ,j (x) = ζ˜j1 (x)Aκ (x) + (1 − ζ˜j1 (x))Aκ (xj1 ). Notice that ζji μ = ζji μij ,

ζj1 σ = ζj1 σj ,

ζj1 Aκ = ζj1 Aκ,j ,

332

Y. Shibata

because ζ˜j1 = 1 on supp ζj1 . We consider the equations: λu0j − Div (μ0j D(u0j ) − K0j (u0j )I) = ζ˜j0 f

in RN ;

λu1j − Div (μ1j D(u1j ) − K1j (u1j , hj )I) = ζ˜j1 f

in Ωj ,

λhj + Aκ,j · ∇Γj hj − nj · uj = ζ˜j1 d

on Γj ,

(μ1j D(u1j ) − K1j (u1j , hj )I)nj − σj (ΔΓj hj )nj = ζ˜j1 h

on Γj .

⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

(3.339)

(3.340)

Here, for u ∈ Hq2 (RN )N , K0j (u) ∈ Hˆ q1 (RN ) denotes a unique solution of the weak Laplace equation: (∇K0j (u), ∇ϕ)RN = (Div (μ0j D(u)) − ∇div u, ∇ϕ)RN

(3.341)

for any ϕ ∈ Hˆ q1 (RN ). And, for u ∈ Hq2 (Ωj ) and h ∈ Hq3 (Ωj ), K1j (u, hj ) ∈ Hq1 (Ωj ) + Hˆ 1 (Ωj ) denotes a unique solution of the weak Dirichlet problem: q,0

(∇K1j (u, h), ∇ϕ)Ωj = (Div (μ1j D(u)) − ∇div u, ∇ϕ)Ωj

(3.342)

for any ϕ ∈ Hˆ q1 ,0 (Ωj ), subject to K1j (u, h) =< μ1j D(u)nj , nj > −div u − σj ΔΓj h on Γj . Moreover, we denote the unit outer normal to Γj by nj , which is defined on RN and satisfies the estimate: nj L∞ (RN ) ≤ C,

∇nj L∞ (RN ) ≤ CA ,

∇ 2 nj L∞ (RN ) ≤ CM2 .

Let ∇Γj = (∂1 , . . . , ∂N−1 ) with ∂j = ∂/∂xj for y = Φj (x  , 0) ∈ Γj and let ΔΓj be the Laplace–Beltrami operator on Γj , which have the form: ΔΓj f = Δ f + DΓj f where Δ f = j

N−1 j =1

∂j2 f and DΓj f =

on Φj−1 (Γj ),

N−1

j k.#=1 ak# ∂k ∂# f

+

N−1 k=1

j

and ak satisfy the following estimates: j

ak# L∞ (RN ) ≤ CM1 , j

j

j

j

j

(∂1 ak# , . . . , ∂N−1 ak# , ak )L∞ (RN ) ≤ CA , j

(∂1 ak# , . . . , ∂N−1 ak# , ak )H∞ 1 (RN ) ≤ CM2 .

j

ak ∂k f , and ak#

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

333

Notice that nj = n and ΔΓj = ΔΓ on Γj ∩ Bj1 = Γ ∩ Bj1 . We know the existence of K0j (u0 ) ∈ Hˆ q1 (RN ) possessing the estimate: j

∇K0j (u0j )Lq (RN ) ≤ C∇u0j Hq1 (RN ) .

(3.343)

 Let ρ be a function in C0∞ (Br0 ) such that RN ρ dx = 1. Below, this ρ is fixed. Since K0j (u0j ) + c also satisfy the variational Eq. (3.341) for any constant c, adjusting constants we may assume that  Bj0

K0j (u0j )ρ(x − xj0 ) dx = 0.

(3.344)

Moreover, choosing M1 ∈ (0, 1) suitably small, we have the unique existence 1 (Ω ) of Eq. (3.342) possessing the of solutions K1j (u1j , hj ) ∈ Hq1 (Ωj ) + Hˆ q,0 j estimates: ∇K1j (u1j , hj )Lq (Ωj ) ≤ C(∇u1j Hq1 (Ωj ) + hj W 3−1/q (Γ ) ). j

q

(3.345)

Let Yq (Ωj ) and Yq (Ωj ) be the spaces defined in (3.193) replacing Ω by Ωj . By Theorem 3.5.1 and Theorem 3.5.18, there exist constants M1 ∈ (0, 1) and λ0 ≥ 1, which are independent of j ∈ N, and operator families S0j (λ) ∈ Hol (Σ

,λ0 , L(Lq (R

) , Hq2 (RN )N )),

N N

S1j (λ) ∈ Hol (Λκ,λ0 γκ , L(Yq (Ωj ), Hq2 (Ωj )N )), Hj (λ) ∈ Hol (Λκ,λ0 γκ , L(Yq (Ωj ), Hq3 (Ωj ))) such that for each j ∈ N, Eq. (3.339) admits a unique solution u0j = S0j (λ)ζ˜j0 f and Eq. (3.340) admits unique solutions u1j = S1j (λ)ζ˜j1 Fλ (f, d, h) and hj = Hj (λ)ζ˜j1 Fλ (f, d, h), where Fλ (f, d, h) = (f, d, λ1/2 h, h), and ζ˜j1 Fλ (f, d, h) = (ζ˜j1 f, ζ˜j1 d, λ1/2 ζ˜j1 h, ζ˜j1 h). Moreover, there exists a number rb > 0 independent of M1 , M2 , and j ∈ N such that RL(Lq (RN )N ,H 2−k (RN )N ) ({(τ ∂τ )# (λk/2 S0j (λ)) | λ ∈ Σ q

RL(Yq (Ωj ),Hq2−k (Ωj )N ) ({(τ ∂τ )# (λk/2 S1j (λ)) | λ ∈ Λ

,λ0 })

≤ rb ,

,λ0 γκ })

≤ rb ,

RL(Yq (Ωj ),H 3−n (Ωj )) ({(τ ∂τ )# (λn Hj (λ)) | λ ∈ Λκ,λ0 γκ }) ≤ rb , q

for # = 0, 1, j ∈ N, k = 0, 1, 2, and n = 0, 1. Notice that λ0 γκ ≥ λ0 .

(3.346)

334

Y. Shibata

By (3.346), we have |λ|u0j Lq (RN ) + |λ|1/2 u0j Hq1 (RN ) + u0j Hq2 (RN ) ≤ rb ζj0 fLq (RN ) , |λ|u1j Lq (Ωj ) + |λ|1/2 u1j Hq1 (Ωj ) + u1j Hq2 (Ωj ) + |λ|hj Hq2 (Ωj ) + hj Hq3 (Ωj ) ≤ rb (ζ˜j1 fLq (Ωj ) + ζ˜j1 dW 2−1/q (Γ ) + |λ|1/2 hLq (Ωj ) + hHq1 (Ωj ) ) j

q

(3.347) for λ ∈ Σκ,λ0 γκ . Let u=

1  ∞ 

ζji uij ,

i=0 j =1

h=

∞ 

ζj1 hj .

(3.348)

j =1

Then, by (3.339), (3.340), (3.347), Proposition 3.5.21, and Proposition 3.5.23, we have u ∈ Hq2 (Ω)N , h ∈ Hq3 (Ω), and |λ|uLq (Ω) + |λ|1/2uHq1 (Ω) + uHq2 (Ω) + |λ|hHq2 (Ω) + hHq3 (Ω) ≤ Cq rb (fLq (Ω) + dW 2−1/q (Γ ) + |λ|1/2hLq (Ω) + hHq1 (Ω) ) q

for λ ∈ Λκ,λ0 γκ . Moreover, we have ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

λu − Div (μD(u) − K(u, h)I) = f − V 1 (λ)(f, d, h)

in Ω,

λh + Aκ · ∇Γ h − u · n = d − Vκ2 (λ)(f, d, h)

on Γ ,

(μD(u) − K(u, h)I − (σ ΔΓ h)I)n = h − V 3 (λ)(f, d, h)

on Γ ,

(3.349)

where we have set V 1 (λ)(f, d, h) = V11 (λ)(f, d, h) + V21 (λ)(f, d, h), V11 (λ)(f, d, h) =

1  ∞ 

[Div (μ(D(ζji uij ) − ζji D(uij )))

i=0 j =1

+ Div (ζji μij D(uij )) − ζji Div (μij D(uij ))], V21 (λ)(f, d, h) = ∇K(u, h) −

∞  j =1

ζj0 ∇K0j (u0j ) −

∞  j =1

ζj1 ∇K1j (u1j , hj ),

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

Vκ2 (λ)(f, d, h) =

∞ 

335

Aκ (x) · ((∇Γ ζj1 )hj ),

j =1

V 3 (λ)(f, d, h) = V13 (λ)(f, d, h) − V23 (λ)(f, d, h) − V33 (λ)(f, d, h), V13 (λ)(f, d, h) =

∞ 

μ(D(ζj1 u1j ) − ζj1 D(u1j ))n,

j =1

V23 (λ)(f, d, h) = {

∞ 

ζj1 K1j (u1j , hj ) − K(u, h)}n,

j =1

V33 (λ)(f, d, h) =

∞ 

σ (ΔΓj (ζj1 h1j ) − ζj1 ΔΓj h1j ).

j =1

For F = (F1 , F2 , F3 , F4 ) ∈ Yq (Ω), we define operators Ap (λ) and Bp (λ) acting on F by Ap (λ)F =

∞ 

ζj0 Sj0 (λ)ζ˜j0 F1 +

j =1

Bp (λ)F =

∞ 

∞ 

ζj1 S1j (λ)ζ˜j1 F,

j =1

(3.350)

ζj1 Hj (λ)ζ˜j1 F.

j =1

Then, by Proposition 3.5.21 and (3.346), we have u = Ap (λ)Fλ (f, d, h), h = Hp (λ)Fλ (f, d, h), and Ap (λ) ∈ Hol (Λκ,λ1 , L(Yq (Ω), Hq2(Ω)N )), Bp (λ) ∈ Hol (Λκ,λ1 , L(Yq (Ω), Hq3 (Ω))), RL(Y

2−j (Ω)N ) q (Ω),Hq

({(τ ∂τ )# (λj/2 Ap (λ)) | λ ∈ Λκ,λ1 })

−1/2

≤ (C + CM2 λ1

(3.351) )rb ,

RL(Yq (Ω),Hq3−k (Ω)) ({(τ ∂τ )# (λk Bp (λ)) | λ ∈ Λκ,λ1 }) ≤ (C + CM2 λ−1 1 )rb for # = 0, 1, j = 0, 1, 2, and k = 0, 1 for any λ1 ≥ λ0 γκ .

336

Y. Shibata

3.5.7 Estimates of the Remainder Terms For F = (F1 , F2 , F3 , F4 ) ∈ Yq (Ω), let V 1 (λ)F = V11 (λ)F + V21 (λ)F, V11 (λ)F =

∞  [Div (μ(D(ζj0 S0j (λ)ζ˜j0 F1 ) − ζj0 D(S0j (λ)ζ˜j0 F1 ))) j =1

+ Div (ζj0 μ0j D(S0j (λ)ζ˜j0 F1 )) − ζj0 Div (μ0j D(S0j (λ)ζ˜j0 F1 ))] +

∞  [Div (μ(D(ζj1 S1j (λ)ζ˜j1 F ) − ζj1 D(S1j (λ)ζ˜j1 F ))) j =1

+ Div (ζj1 μ1j D(S1j (λ)ζ˜j1 F )) − ζj1 Div (μ1j D(S1j (λ)ζ˜j1 F ))], V21 (λ)F = ∇K(Ap (λ)F, Bp (λ)F ) −

∞ 

ζj0 ∇K0j (S0j (λ)ζ˜j0 F1 )

j =1

−

∞ 

ζj1 ∇K1j (S1j (λ)ζ˜j1 F, Hj (λ)ζ˜j1 F ),

j =1

Vκ2 (λ)F =

∞ 

Aκ (x) · ((∇Γ ζj1 )Hj (λ)ζ˜j1 F ),

j =1

V 3 (λ)F = V13 (λ)F + V23 (λ)F + V33 (λ)F, V13 (λ)F

=

∞ 

μ(D(ζj1 S1j (λ)ζ˜j1 F ) − ζj1 D(ζj1 S1j (λ)ζ˜j1 F ))n

j =1

V23 (λ)F = {

∞ 

ζj1 K1j (S1j (λ)ζ˜j1 F, Hj (λ)ζ˜j1 F ) − K(Ap (λ)F, Bp (λ)F )}n,

j =1

V33 (λ)F =

∞ 

σ (ΔΓj (ζj1 Hj (λ)ζ˜j1 F ) − ζj1 ΔΓj (Hj (λ)ζ˜j1 F )).

j =1

Notice that Vκ2 (λ)F = 0 for κ = 0. Let V (λ)(f, d, h) = (V 1 (λ)(f, d, h), V 2 (λ)(f, d, h), V 3 (λ)(f, d, h)), V(λ)F = (V 1 (λ)F, Vκ2 (λ)F, V 3 (λ)F ).

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

337

˜ d, h), Since u0j = S0j (λ)ζ˜j0 f, u1j = S1j (λ)ζ˜j1 Fλ (f, d, h), and hj = Hj (λ)ζj1 Fλ (f, we have V (λ)(f, d, h) = V(λ)Fλ (f, d, h).

(3.352)

In what follows, we shall prove that RL(Yq (Ω)) ({(τ ∂τ )# (Fλ V(λ)) | λ ∈ Λκ,λ˜ 0 })

(3.353)

˜ −1/2 )) ≤ Cq rb ( + CM2 , (λ˜ −1 0 γκ + λ0

for # = 0, 1 and λ˜ 0 ≥ λ0 γκ , where γκ is the number given in Theorem 3.4.8. To prove (3.353), we use Propositions 3.2.2, 3.5.4, and 3.5.20–3.5.23, (3.301), (3.302), (3.336), (3.337), (3.28) and (3.346). In the following, λ˜ 0 is any number such that λ˜ 0 ≥ λ0 γκ . We start with the following estimate of V11 (λ): −1/2

RL(Yq (Ω),Lq (Ω)N ) ({(τ ∂τ )# V11 (λ) | λ ∈ Λκ,λ˜ 0 }) ≤ CM2 rb λ˜ 0

(3.354)

for # = 0, 1. In fact, since D#,m (ζji u) − ζji D#m (u) = (∂# ζji )um + (∂m ζji )u# , and N i n n div (ζji u) − ζji div u = k=1 (∂k ζj )uk , for any n ∈ N, {λ# }#=1 ⊂ (Λκ,λ˜ 0 ) , and n n {F# = (F1# , F2# , F3# , F4# )}#=1 ⊂ Yq (Ω) , we have 

1



0

n 

q

r# (u)V11 (λ# )F# Lq (Ω) du

#=1 q

q

≤ Cq M2

∞   {



1



0

n 

n 

1



0

≤

n 

∞   {

1

≤

n 

r# (u)λ# S0j (λ# )ζ˜j0 F1# H 1 (RN ) du 1/2



n 

q

q

#=1

r# (u)λ# S1j (λ# )ζ˜j1 F# H 1 (Ω ) du} 1/2

q

q

#=1

1 0



0

∞   { j =1



j

q

q q −q/2 q Cq M2 λ˜ 0 rb

+

q

#=1

j =1



q

q r# (u)S1j (λ# )ζ˜j1 F# H 1 (Ω ) du}

q −q/2

+

r# (u)S0j (λ# )ζ˜j0 F1# H 1 (RN ) du

#=1

≤ Cq M2 λ˜ 0 q



0

j =1

+

1

1



n 

0

j

q r# (u)ζ˜j0 F1# L

q (R

#=1

r# (u)ζ˜j1 F# Yq (Ωj ) du} q

#=1

2q q −q/2 q Cq M2 λ˜ 0 rb



1 0



n  #=1

q

r# (u)F# Lq (Ω) du.

N)

du

338

Y. Shibata

This shows that −1/2 RL(Yq (Ω),Lq (Ω)N ) ({V11 (λ) | λ ∈ Λκ,λ˜ 0 }) ≤ CM2 rb λ˜ 0 .

Analogously, we can show that −1/2

RL(Yq (Ω),Lq (Ω)N ) ({τ ∂τ V11 (λ) | λ ∈ Λκ,λ˜ 0 }) ≤ CM2 rb λ˜ 0

,

and therefore we have (3.354). For r ∈ (N, ∞) and q ∈ (1, r], by the extension of functions defined on Γj to Ωj and Sobolev’s imbedding theorem, we have abW 2−1/q (Γ ) ≤ Cq,r,K aHq2 (Ωj ) bW 2−1/q (Γ q

j

q

2−1/q

for any a ∈ Hr2 (Ω) and b ∈ Wq

j)

(Γj ). Applying this inequality, we have

Aκ · ((∇Γ ζj1 )Hj (λ)ζ˜j1 F )W 2−1/q (Γ ) ≤ Cq,r M2 m3 κ −b Hj (λ)ζ˜j1 F Hq2 (Ωj ) q

j

for κ ∈ (0, 1). Thus, employing the same argument as in the proof of (3.354), we have RL(Y

2−1/q (Γ )) q (Ω),Wq

−b ({(τ ∂τ )# Vκ2 (λ) | λ ∈ Λκ,λ˜ 0 }) ≤ CM2 rb λ˜ −1 0 κ

for # = 0, 1 and κ ∈ (0, 1). Employing the same argument as in the proof of (3.354), we also have −1/2

3 RL(Yq (Ω),Lq (Γ )N ) ({(τ ∂τ )# (λ1/2 Vm (λ)) | λ ∈ Λκ,λ˜ 0 }) ≤ CM2 rb λ˜ 0

,

−1/2 3 RL(Yq (Ω),Hq1 (Γ )N ) ({(τ ∂τ )# Vm (λ) | λ ∈ Λκ,λ˜ 0 }) ≤ CM2 rb λ˜ 0 ,

for # = 0, 1, m = 1 and 3. Noting that μζj1 = μ1j ζj1 and σ ζj1 = σj1 ζj1 , we have ∞ 

ζj1 K1j (S1j (λ)ζ˜j1 F, Hj (λ)ζ˜j1 F ) − K(Ap (λ)F, Bp (λ)F )

j =1

=

∞ 

μ < ζj1 D(S1j (λ)ζ˜j1 F ) − D(ζj1 S1j (λ)ζ˜j1 F ), n >

j =1 ∞  {ζj1 div S1j (λ)ζ˜j1 F − div (ζj1 S1j (λ)ζ˜j1 F )} − j =1

−

∞  j =1

σ {ζj1 ΔΓj (Hj (λ)ζ˜j1 F ) − ΔΓj (ζj1 Hj (λ)ζ˜j1 F )}

(3.355)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

339

on Γ , where we have used ΔΓ = ΔΓj and n = nj on Γj ∩ Bj1 . Employing the same argument as in the proof of (3.354), we have −1/2 RL(Yq (Ω),Lq (Γ )N ) ({(τ ∂τ )# (λ1/2 V23 (λ)) | λ ∈ Λκ,λ˜ 0 }) ≤ CM2 rb λ˜ 0 ; −1/2

RL(Yq (Ω),Hq1 (Γ )N ) ({(τ ∂τ )# V23 (λ) | λ ∈ Λκ,λ˜ 0 }) ≤ CM2 rb λ˜ 0 for # = 0, 1. The final task is to prove that

−1/2 RL(Yq (Ω),Lq (Γ )N ) ({(τ ∂τ )# V21 (λ) | λ ∈ Λκ,λ˜ 0 }) ≤ Cq,r ( + CM2 , λ˜ 0 )rb (3.356)

for # = 0, 1. For this purpose, we use Lemma 3.2.13 and the following lemma. Lemma 3.5.24 Let 1 < q < ∞. For u ∈ Hq2 (RN ), let K0j (u) be a unique solution of the weak Laplace equation (3.341) satisfying (3.344). Then, we have K0j (u)Lq (B 0 ) ≤ C∇uLq (RN ) .

(3.357)

j

Proof Let ρ be the same function in (3.344). Let ψ be any function in C0∞ (Bj0 ) and we set  0 ˜ ψ(x) = ψ(x) − ρ(x − xj ) ψ(y) dy. RN

Then, ψ˜ ∈ C0∞ (Bj0 ),

 RN

ψ˜ dx = 0,

˜ ψ L

≤ Cq  ψ

.

(3.358)

˜ ˆ −1 N ≤ Cq  ψL  (RN ) . ψ H (R ) q

(3.359)

0 q  (Bj )

0 q  (Bj )

Moreover, N ψ˜ ∈ Hˆ q1 (RN )∗ = Hˆ q−1  (R ),

q

In fact, by Lemma 3.2.13, for any ϕ ∈ Hˆ q1 (RN ), there exists a constant ej for which ϕ − ej Lq (B 0 ) ≤ cq ∇ϕLq (B 0 ) . j

j

Thus, by (3.358), we have ˜ ϕ)RN | = |(ψ, ˜ ϕ − ej )RN | ≤ ψ ˜ |(ψ, L

0 q  (Bj )

˜ ≤ Cq ψ

ϕ − ej Lq (B 0 )

0 q  (Bj )

j

∇ϕLq (B 0 ) , j

340

Y. Shibata

which yields (3.359). Let Ψ be a function in Hˆ q1 (RN ) such that ∇Ψ ∈ Hq1 (RN )N , ˜ θ )R N (∇Ψ, ∇θ )RN = (ψ,

for any θ ∈ Hˆ q1 (RN ), (3.360)

˜ L  (RN ) + ψ ˜ ˆ −1 N ). ∇Ψ H 1 (RN ) ≤ C(ψ H (R ) q q

q

By (3.358) and (3.359), we have ∇Ψ H 1 (RN ) ≤ Cq  ψLq  (RN ) .

(3.361)

q

By (3.344), (3.360), and the divergence theorem of Gauß, we have (K0j (u), ψ)RN = (K0j (u), ψ˜ )RN = (∇K0j (u), ∇Ψ )RN = (Div (μ0j D(u)) − ∇div u, ∇Ψ )RN = −(μ0j D(u), ∇ 2 Ψ )RN + (div u, ΔΨ )RN , and therefore by (3.361) |(K0j (u), ψ)RN | ≤ C∇uLq (RN ) ψLq  (RN ) , 

which proves (3.357). This completes the proof of Lemma 3.5.24.

and h ∈ let Lemma 3.5.25 Let 1 < q < ∞. For u ∈ 1 1 ˆ K1j (u, h) ∈ Hq (Ωq ) + Hq,0 (Ωj ) be a unique solution of the weak Dirichlet problem (3.342). Then, we have Hq2 (Ωj )

Hq3 (Ωj ),

K1j (u, h)Lq (Ωj ∩B 1 ) ≤ C(∇uLq (Ωj ) + hHq2 (Ωj ) j

1/q

1−1/q

1/q

1−1/q

+ ∇ 2 uLq (Ωj ) ∇uLq (Ωj ) + hH 3 (Ω ) hH 2 (Ω ) ). q

j

j

Here, the constant C depends on q and CA . Remark 3.5.26 By Young’s inequality, we have K1j (u, h)Lq (Ωj ∩B 1 ) ≤ (∇ 2 uLq (Ωj ) + hHq3 (Ωj ) ) j

+ C (∇uLq (Ωj ) + hHq2 (Ωj ) ) for any

∈ (0, 1) with some constant C

,q

(3.362)

depending on and q.

Proof For a proof, see Lemma 3.4 in Shibata [42].



3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

341

1 (λ) + V 1 (λ), To prove (3.356), we divide V21 (λ) into two parts as V21 (λ) = ∇V21 22 where 1 V21 (λ)F = K(Ap (λ)F, Bp (λ)F ) −

∞ 

ζj0 K0j (S0j (λ)ζ˜j0 F1 )

j =1

−

∞ 

ζj1 K1j (S1j (λ)ζ˜j0 F, Hj (λ)ζ˜j1 F ),

j =1 ∞ 

(∇ζj0 )K0j (S0j (λ)ζ˜j0 F1 )

1 (λ)F = V22

j =1

+

∞  (∇ζj1 )K1j (S1j (λ)ζ˜j0 F, Hj (λ)ζ˜j1 F ). j =1

1 F, ∇ϕ) = By (3.182), (3.341), and (3.342), for any ϕ ∈ Hˆ q1 0 (Ω) we have (∇V21 Ω I − I I , where

I = (Div (μD(Ap (λ)F )) − ∇div (Ap (λ)F ), ∇ϕ)Ω , II =

∞ 

((∇ζj0 )K0j (S0j (λ)ζ˜j0 F1 ), ∇(ϕ − ej ))Ω

j =1

+

∞  ((∇ζj1 )K1j (S1j (λ)ζ˜j1 F, Hj (λ)ζ˜j1 F ), ∇ϕ)Ω j =1

+

∞  (∇K0j (S0j (λ)ζ˜j0 F1 ), ∇(ζj0 (ϕ − ej )))Ω j =1

+

∞  (∇K1j (S1j (λ)ζ˜j1 F, Hj (λ)ζ˜j1 F ), ∇(ζj1 ϕ))Ω j =1

−

∞  ((∇ζj0 )∇K0j (S0j (λ)ζ˜j0 F1 ), ϕ − ej )Ω j =1

∞  − ((∇ζj1 )∇K1j (S1j (λ)ζ˜j1 F, Hj (λ)ζ˜j1 F ), ϕ)Ω . j =1

342

Y. Shibata

Here and in the following, ej = cj0 (ϕ) are constants given in Lemma 3.2.13. By the definition (3.350), we have I =

∞  (Div (μD(ζj0 S0j (λ)ζ˜j0 F1 )) − ∇div (ζj0 S0j (λ)ζ˜j0 F1 ), ∇ϕ)RN j =1

+

∞ 

(Div (μD(ζj1 S1j (λ)ζ˜j1 F )) − ∇div (ζj1 S1j (λ)ζ˜j1 F ), ∇ϕ)Ω

j =1

=

∞  (ζj0 Div (μ0j D(S0j (λ)ζ˜j0 F1 )) − ζj0 ∇div (S0j (λ)ζ˜j0 F1 ), ∇ϕ)RN j =1

+

∞ 

(ζj1 Div (μ1j D(S1j (λ)ζ˜j1 F )) − ζj1 ∇div (S1j (λ)ζ˜j1 F ), ∇ϕ)Ω + I I I,

j =1

where III =

∞  (Div (μD(ζj0 S0j (λ)ζ˜j0 F1 )) − ζj0 Div (μD(S0j (λ)ζ˜j0 F1 )), ∇ϕ)RN j =1

−

∞  (∇div (ζj0 S0j (λ)ζ˜j0 F1 ) − ζj0 ∇div (S0j (λ)ζ˜j0 F1 ), ∇ϕ)Ω j =1

+

∞  (Div (μD(ζj1 S1j (λ)ζ˜j1 F )) − ζj1 Div (μD(S1j (λ)ζ˜j1 F )), ∇ϕ)RN j =1

−

∞  (∇div (ζj1 S1j (λ)ζ˜j1 F ) − ζj1 ∇div (S1j (λ)ζ˜j1 F ), ∇ϕ)Ω .

(3.363)

j =1

Since ζj0 (ϕ − ej ) ∈ Hˆ q1 ,0 (RN ), and ζj1 ϕ ∈ Hˆ q1 ,0 (Ωj ), by (3.341) and (3.342), we have II =

∞  (ζj0 Div (μ0j D(S0j (λ)ζ˜j0 F1 )) − ζj0 ∇div (S0j (λ)ζ˜j0 F1 ), ∇ϕ)RN j =1

+

∞  j =1

(ζj1 Div (μ1j D(S1j (λ)ζ˜j1 F )) − ζj1 ∇div (S1j (λ)ζ˜j1 F ), ∇ϕ)Ω + I V ,

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

343

where we have set IV =

∞   2(K0j (S0j (λ)ζ˜j0 F1 )(∇ζj0 ), ∇ϕ)Ω + (K0j (S0j (λ)ζ˜j0 F1 )(Δζj0 ), ϕ − ej )Ω j =1

− (μ0j D(S0j (λ)ζ˜j0 F1 ) : (∇ 2 ζj0 ), ϕ − ej )Ω − (μ0j D(S0j (λ)ζ˜j0 F1 )(∇ζj0 ), ∇ϕ)Ω + (div (S0j (λ)ζ˜j0 F1 )(Δζj0 ), ϕ − ej )Ω + (div (S0j (λ)ζ˜j0 F1 )(∇ζj0 ), ∇ϕ)Ω + 2(K1j (S1j (λ)ζ˜j1 F, Hj (λ)ζ˜j1 F )(∇ζj1 ), ∇ϕ)Ω + (K1j (S1j (λ)ζ˜j1 F, Hj (λ)ζ˜j1 F )(Δζj1 ), ϕ)Ω − (μ1j D(S1j (λ)ζ˜j1 F1 ) : (∇ 2 ζj1 ), ϕ)Ω − (μ1j D(S1j (λ)ζ˜j1 F1 )(∇ζj1 ), ∇ϕ)Ω  + (div (S1j (λ)ζ˜j1 F1 )(Δζj1 ), ϕ)Ω + (div (S1j (λ)ζ˜j1 F1 )(∇ζj1 ), ∇ϕ)Ω .

Thus, we have 1 (∇V21 (λ)F, ∇ϕ)Ω = I I I + I V .

(3.364)

We let define operators L(λ) and M(λ) acting on F ∈ Yq (Ω) by the following formulas: L(λ)F =

∞ 

(Div (μD(ζj0 S0j (λ)ζ˜j0 F1 )) − ζj0 Div (μD(S0j (λ)ζ˜j0 F1 ))

j =1

−

∞ 

(∇div (ζj0 S0j (λ)ζ˜j0 F1 ) − ζj0 ∇div (S0j (λ)ζ˜j0 F1 ))

j =1

+

∞ 

(Div (μD(ζj1 S1j (λ)ζ˜j1 F )) − ζj1 Div (μD(S1j (λ)ζ˜j1 F ))

j =1

−

∞ 

(∇div (ζj1 S1j (λ)ζ˜j1 F ) − ζj1 ∇div (S1j (λ)ζ˜j1 F ))

j =1

+2

∞  (∇ζj0 )K0j (S0j (λ)ζ˜j0 F1 ) j =1

+2

∞  (∇ζj1 )K1j (S1j (λ)ζ˜j1 F, Hj (λ)ζ˜j1 F ) j =1

344

Y. Shibata

−

∞ 

μ0j D(S0j (λ)ζ˜j0 F1 )(∇ζj0 ) +

j =1

−

∞ 

∞ 

div (S0j (λ)ζ˜j0 F1 )(∇ζj0 )

j =1

μ1j D(S1j (λ)ζ˜j1 F )(∇ζj1 ) +

j =1

∞ 

div (S1j (λ)ζ˜j1 F1 )(∇ζj1 );

j =1

> = −

∞ 

(μ0j D(S0j (λ)ζ˜j0 F1 ) : (∇ 2 ζj0 ), ϕ − ej )Ω

j =1

+

∞  (div (S0j (λ)ζ˜j0 F1 )(Δζj0), ϕ − ej )Ω j =1

+

∞  (K0j (S0j (λ)ζ˜j0 F1 )(Δζj0 ), ϕ − ej )Ω j =1

−

∞  (μ1j D(S1j (λ)ζ˜j1 F ) : (∇ 2 ζj1 ), ϕ)Ω j =1

∞  + (div (S1j (λ)ζ˜j1 F )(Δζj1 ), ϕ)Ω j =1

+

∞  (K1j (S1j (λ)ζ˜j1 F, Hj (λ)ζ˜j1 F )(Δζj1 ), ϕ)Ω . j =1

−1 Here and in the following, Wˆ q,0 (Ω) denotes the dual space of Hˆ q1 ,0 (Ω), and −1 > denotes the duality between Wˆ q,0 (Ω) and Hˆ q1 ,0 (Ω). Moreover, by (3.182) and (3.342), for x ∈ Γ we have 1 V21 (λ)F =< μD(Ap (λ)F )n, n > −σ ΔΓ Bp (λ)F − div Ap (λ)F

−

∞ 

ζj1 {< (μ(xj1 )D(S1j (λ)ζ˜j1 F )nj , nj > −σ (xj1 )ΔΓj Hj (λ)ζ˜j1 F

j =1

− div S1j (λ)ζ˜j1 F } =

∞  j =1


−

∞  j =1

σ ΔΓ (ζj1 Hj (λ)ζ˜j1 F )

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

−

∞ 

345

div (ζj1 S1j (λ)ζ˜j1 F )

j =1

−

∞ 

ζj1 {< μ(xj1 )D(S1j (λ)ζ˜j1 F )nj , nj > −σ (xj1 )ΔΓj Hj (λ)ζ˜j1 F

j =1

− div S1j (λ)ζ˜j1 F }. Thus, we define an operator Lb (λ) acting on F ∈ Yq (Ω) by letting ∞  [< μ(x)(D(ζj1S1j (λ)ζ˜j1 F ) − ζj1 D(S1j (λ)ζ˜j1 F ))n, n >

Lb (λ)F =

j =1

− σ (ΔΓ (ζj1 Hj (λ)ζ˜j1 F ) − ζj1 ΔΓ Hj (λ)ζ˜j1 F ) − (∇ζj1 )S1j (λ)ζ˜j1 F ], 1 (λ)F = L (λ)F on Γ . and then, V21 b We now prove the R boundedness of operator families L(λ), M(λ) and Lb (λ). We first prove that −1/2 RL(Yq (Ω),Wˆ −1 (Ω)) ({(τ ∂τ )# M(λ) | λ ∈ Λκ,λ˜ 0 }) ≤ ( + Cq, λ˜ 0 )rb

(3.365)

q,0

for # = 0, 1. In fact, if we set > = −(μ(xj0 )D(S0j (λ)ζ˜j0 F1 ) : (∇ 2 ζj0 ), ϕ − ej )Ω + (div (S0j (λ)ζ˜j0 F1 )(Δζj0), ϕ − ej )Ω + (K0j (S0j (λ)ζ˜j0 F1 )(Δζj0 ), ϕ − ej )Ω , > = −(μ(xj1 )D(S1j (λ)ζ˜j1 F ) : (∇ 2 ζj1 ), ϕ)Ω + (div (S1j (λ)ζ˜j1 F )(Δζj1 ), ϕ)Ω + (K1j (S1j (λ)ζ˜j1 F, Hj (λ)ζ˜j1 F )(Δζj1 ), ϕ)Ω , then, by Lemmas 3.2.13, 3.5.24 and (3.362), we have | > | ≤ CM2 ∇S0j (λ)ζ˜j0 F Lq (RN ) ∇ϕL

0 q  (Bj )

,

| > | ≤ (S1j ζ˜j1 F Hq2 (Ωj ) + Hj (λ)ζ˜j1 F Hq3 (Ωj ) ) +C

1 ,M2 (S1j ζ˜j F Hq1 (Ωj )

+ Hj (λ)ζ˜j1 F Hq2 (Ωj ) )∇ϕLq (B 1 ∩Ω) . j

346

Y. Shibata

By (3.27), we have ∞ 

q Lq  (Bj0 )

∇ϕ

j =1

+

∞ 

q Lq  (Bj1 ∩Ω)

∇ϕ

j =1

q

≤ Cq  ∇ϕL

q  (Ω)

for any ϕ ∈ Hˆ q1 ,0 (Ω). By (3.346), (3.27), and Proposition 3.5.23, we have ∞  j =1

∇S0j (λ)ζ˜j0 F1 L q

q

≤ rb (

∞  j =1

ζ˜j0 F1 L

q

+ (RN )

q

(RN )

q

+

∞  j =1 ∞  j =1

(S1j (λ)ζ˜j1 F H 2 (Ω ) + Hj (λ)ζ˜j1 F H 3 (Ω ) ) q

q

j

q

j

q

ζ˜j1 F Yq (Ωj ) ) ≤ rb Cq F Yq (Ω) < ∞. q

Thus, by Proposition 3.5.20, M(λ)F = in Wˆ −1 (Ω) for any F ∈ Yq (Ω) and

q

∞

0 j =1 Mj (λ)F

q

+

∞

1 j =1 Mj (λ)F

exists

q,0

q Wq,0 (Ω)

M(λ)F  ˆ −1 q

≤ CM2 +

q

∞  j =1 ∞  j =1

+C

q ∇S0j (λ)ζ˜j0 F1 L

q ,M2

q

(RN )

+

q

∞  j =1

q q Hj (λ)ζ˜j1 F H 3 (Ω ) + C ,M2 j

q

∞  j =1

q S1j (λ)ζ˜j1 F H 2 (Ω

∞  j =1

j)

q

q S1j (λ)ζ˜j1 F H 1 (Ω q

j)

Hj (λ)ζ˜j1 F H 1 (Ω ) . q

j

q

Analogously, by Proposition 3.5.20 we have 

n  #=1

q Wq,0 (Ω)

r# (u)M(λ# )F#  ˆ −1

q

≤ CM2 +

q

∞ n   q  r# (u)∇S0j (λ# )ζ˜j0 F1# L j =1

n ∞   q { r# (u)S1j (λ# )ζ˜j1 F# H 2 (Ω j =1

+

q (R

n  #=1

N)

#=1

q

#=1

q r# (u)Hj (λ# )ζ˜j1 F# H 3 (Ω ) } q

j

j)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

+C

+

q ,M2

n 

∞ 

{

j =1

n 

347

r# (u)S1j (λ# )ζ˜j1 F# H 1 (Ω q

j)

q

#=1

r# (u)Hj (λ# )ζ˜j1 F# H 1 (Ω ) }. q

q

#=1

j

Noting that Ω ∩ Bj1 = Ωj ∩ Bj1 , by (3.346), (3.28), Proposition 3.5.23, and Proposition 3.5.4, we have 

1 0



n 

q du Wq,0 (Ω)

r# (u)M(λ# )F#  ˆ −1

#=1 −q/2 q rb

≤ CM2 λ˜ 0 q

+

q q rb

+C

≤ Cq (





∞ 1 0 j =1

∞ 1



0 j =1

q ˜ −q/2 r q ,M2 λ0 b

q

+C



n 

r# (u)ζ˜j0 F1# L q

q (R

du

N)

#=1

n 

r# (u)ζ˜j1 F# Yq (Ωj ) du q

#=1



∞ 1



0 j =1

q ˜ −q/2 )r q ,M2 λ0 b

n 

r# (u)ζ˜j1 F# Yq (Ωj ) du q

#=1



1 0



n 

q

r# (u)F# Yq (Ω) du,

#=1

which shows (3.365). Analogously, we can prove −1/2

RL(Yq (Ω),Lq (Ω)N ) ({(τ ∂τ )# L(λ) | λ ∈ Λκ,λ˜ 0 }) ≤ CM2 rb λ˜ 0

,

−1/2

RL(Yq (Ω),Hq1 (Ω)N ) ({(τ ∂τ )# Lb (λ) | λ ∈ Λκ,λ˜ 0 }) ≤ CM2 rb λ˜ 0

(3.366)

for # = 0, 1. We now use the following lemma. −1 Lemma 3.5.27 Let 1 < q < ∞. Then, there exists a linear map E from Wˆ q,0 (Ω) −1 N ˆ and into Lq (Ω) such that for any F ∈ W (Ω), E(F )Lq (Ω) ≤ CF  ˆ −1 q,0

< F, ϕ >= (E(F ), ∇ϕ)Ω

Wq,0 (Ω)

for all ϕ ∈ Hˆ q1 ,0 (Ω).

Proof The lemma follows from the Hahn-Banach theorem by indentifying Hˆ q1 ,0 (Ω) with a closed subspace of Lq  (Ω)N via the mapping: ϕ → ∇ϕ. 

348

Y. Shibata

Applying Lemma 3.5.27 and using (3.364) and (3.365), we have 1 (∇V21 (λ)F, ∇ϕ)Ω = (L(λ)F + E(M(λ)F ), ∇ϕ)Ω

for all ϕ ∈ Hˆ q1 ,0 (Ω), (3.367)

1 (λ)F = L (λ)F on Γ , and subject to V21 b −1/2 RL(Yq (Ω),Lq (Ω)N ) ({(τ ∂τ )# E ◦ M(λ) | λ ∈ Σσ,λ˜ 0 }) ≤ C( + Cq, λ˜ 0 )rb , (3.368)

where E ◦ M(λ) denotes a bounded linear operator family acting on F by 1 (λ)F = L (λ)F + E ◦ M(λ)F = E(M(λ)F ). By Remark 3.2.6, we have V21 b K(L(λ)F + E(M(λ)F ) − ∇Lb (λ)F ), and so by (3.366) and (3.368), we see that 1 (λ) ∈ Hol (Σ N ∇V21 σ,λ˜ 0 , L(Yq (Ω), Lq (Ω) )) and 1 RL(Yq (Ω),Lq (Ω)N ) ({(τ ∂τ )# ∇V21 (λ) | λ ∈ Σσ,λ˜ 0 }) −1/2 ≤ Cq ( + CM2 , λ˜ 0 )rb

for # = 0, 1.

(3.369)

Finally, by Lemma 3.5.24, (3.362), (3.28), and Proposition 3.5.20, we have 1 V22 (λ) ∈ Hol (Σσ,λ˜ 0 , L(Yq (Ω), Lq (Ω)N )), −1/2 1 (λ) | λ ∈ Σσ,λ˜ 0 }) ≤ Cq ( + C λ˜ 0 )rb RL(Yq (Ω),Lq (Ω)N ) ({(τ ∂τ )# V21 1 (λ) + for # = 0, 1, which, combined with (3.369) and the formula: V21 (λ) = ∇V21 2 V22 (λ), leads to (3.356).

3.5.8 Proof of Theorem 3.4.11, Existence Part for Eq. (3.226) Choosing so small that Cq,r rb ≤ 1/4, and λ˜ 0 so large that −1/2

˜ Cq rb CM2 , (λ˜ −1 0 γκ + λ0

) ≤ 1/4

(3.370)

in (3.353), we have RL(Yq (Ω)) ({(τ ∂τ )# Fλ V(λ) | λ ∈ Λκ,λ˜ 0 }) ≤ 1/2

(3.371)

for # = 0, 1. Let λ∗ be a large number for which λ∗ ≥ (8Cq rb CM2 , )2 , and then  j setting λ˜ 0 = λ∗ γκ , we have (3.370). By (3.370), (I−Fλ V(λ))−1 = ∞ j =1 (Fλ V(λ)) exists in Hol (Λκ,λ˜ 0 , L(Yq (Ω))) and RL(Yq (Ω)) ({(τ ∂τ )# (I − Fλ V(λ))−1 | λ ∈ Λκ,λ˜ 0 }) ≤ 4

(3.372)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

349

for # = 0, 1. Moreover, by (3.352) and (3.371) Fλ V (λ)(f, d, h)Yq (Ω) ≤ (1/2)Fλ(f, d, h)Yq (Ω) .

(3.373)

Since Fλ (f, d, h)Yq (Ω) gives an equivalent norm in Yq (Ω) for λ = 0, by (3.373)  j (I − V (λ))−1 = ∞ j =0 V (λ) exists in L(Yq (Ω)). Since u = Ap (λ)Fλ (f, d, h) and h = Bp (λ)Fλ (f, d, h) satisfy Eq. (3.349), setting v = Ap (λ)Fλ (I − V (λ))−1 (f, d, h),

ρ = Bp (λ)(λ)Fλ (I − V (λ))−1 (f, d, h),

we see that v ∈ Hq2 (Ω)N , ρ ∈ Hq3 (Ω) and v and ρ satisfy the equations: ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

λv − Div (μD(v) − K(v, ρ)I) = f

in Ω,

+ Aκ · ∇Γ ρ

−v·n=d

on Γ ,

(μD(v) − K(v, ρ)I − (σ ΔΓ ρ)I)n = h

on Γ .

λρ

(3.374)

Moreover, by (3.352) we have Fλ (I − V (λ))−1 = (I − Fλ V(λ))−1 Fλ . Thus, setting Ar (λ) = Ap (λ)(I − Fλ V(λ))−1 ,

Hr (λ) = Bp (λ)(I − Fλ V(λ))−1

we see that v = Ar (λ)Fλ (f, d, h) and ρ = Hr (λ)Fλ (f, d, h) are solutions of Eq. (3.374). Since we may assume that λ∗ γκ ≥ λ1 in (3.351), by (3.351) and (3.372), we have RL(Y

2−j (Ω)N ) q (Ω),Hq

({(τ ∂τ )# (λj/2 Ar (λ)) | λ ∈ Λκ,λ∗ γκ }) ≤ Cq rb ,

RL(Yq (Ω),Hq3−k (Ω)) ({(τ ∂τ )# (λk Hr (λ)) | λ ∈ Λκ,λ∗ γκ }) ≤ Cq rb ,

(3.375)

for # = 0, 1, j = 0, 1, 2 and k = 0, 1. This completes the proof of the existence part of Theorem 3.4.11 for Eq. (3.226).

3.5.9 A Proof of the Existence Part of Theorem 3.4.11 We now prove the existence of R-bounded solution operators of Eq. (3.185). Let Ar (λ) and Hr (λ) be the operators constructed in the previous subsection. Let (f, d, h) ∈ Yq (Ω). Let u = Ar (λ)Fλ (f, d, h) and h = Hr (λ)Fλ (f, d, h), where,

350

Y. Shibata

F(f, d, h) = (f, d, λ1/2 h, h) ∈ Yq (Ω) for (f, d, h) ∈ Yq (Ω). Then u and h satisfy the equations: ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

λu − Div (μD(u, h) − K(u, h)I) = f

in Ω,

λh + Aκ · ∇Γ h − n · u + F1 u = d + F1 u (μD(u) − K(u, h)I)n − (F2 h + σ ΔΓ h)n = h − (F2 h)n

on Γ , on Γ . (3.376)

Let VR1 (λ)(f, d, h) = −F1 Ar (λ)Fλ (f, d, h), VR2 (λ)(f, d, h) = F2 Hr (λ)Fλ (f, d, h),

VR1 (λ)F = −F1 Ar (λ)F, VR2 (λ)F = F2 Hr (λ)F

with Fλ (f, d, h) = (f, d, λ1/2 h, h) and F = (F1 , F2 , F3 , F4 ) ∈ Yq (Ω). Notice that F1 u = −VR1 (λ)(f, d, h) = −VR1 (λ)Fλ (f, d, h), F2 h = VR2 (λ)(f, d, h) = VR2 (λ)Fλ (f, d, h). Let VR (λ) = (0, VR1 (λ), VR2 (λ)) and VR (λ) = (0, VR1 (λ), VR2 (λ)), and then VR (λ) = VR (λ)Fλ .

(3.377)

By (3.377) and (3.161), we have −1/2

RL(Yq (Ω)) ({(τ ∂τ )# Fλ VR (λ) | λ ∈ Λκ,λ2 }) ≤ M0 (λ2

+ λ−1 2 )rb

(# = 0, 1)

for any λ2 > λ∗ γκ . Choosing λ1 so large, we have RL(Yq (Ω)) ({(τ ∂τ )# Fλ VR (λ) | λ ∈ Λκ,λ2 }) ≤ 1/2 (# = 0, 1).

(3.378)

By (3.377) and (3.378), Fλ VR (λ)(f, d, h)Yq (Ω) ≤ (1/2)Fλ (f, d, h)Yq (Ω) . Since Fλ (f, d, h)Yq (Ω) gives an equivalent norm in Yq (Ω) for λ = 0, by (3.373)  j (I − VR (λ))−1 = ∞ j =0 VR (λ) exists in L(Yq (Ω)). Since u = Ar (λ)Fλ (f, d, h) and h = Hr (λ)Fλ (f, d, h) satisfy Eq. (3.376), setting v = Ar (λ)Fλ (I − VR (λ))−1 (f, d, h),

ρ = Hr (λ)Fλ (I − VR (λ))−1 (f, d, h),

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

351

we see that v ∈ Hq2 (Ω)N , ρ ∈ Hq3 (Ω) and v and ρ satisfy the equations: ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

λv − Div (μD(v) − K(v, ρ)I) = f

in Ω,

λρ + Aκ · ∇Γ ρ − v · n + F1 v = d

on Γ ,

(μD(v) − K(v, ρ)I − ((F2 ρ + σ ΔΓ )ρ)I)n = h

on Γ .

(3.379)

Moreover, by (3.377) we have Fλ (I−VR (λ))−1 = (I−Fλ VR (λ))−1 Fλ . Thus, setting A˜ r (λ) = Ar (λ)(I − Fλ V(λ))−1 ,

H˜ r (λ) = Hr (λ)(I − Fλ V(λ))−1

we see that v = A˜ r (λ)Fλ (f, d, h) and ρ = H˜ r (λ)Fλ (f, d, h) are solutions of Eq. (3.379). And, by (3.377) and (3.378), there exists a large number λ∗∗ ≥ λ∗ for which RL(Y

2−j (Ω)N ) q (Ω),Hq

({(τ ∂τ )# (λj/2 A˜ r (λ)) | λ ∈ Λκ,λ∗∗ γκ }) ≤ Cq rb ,

RL(Yq (Ω),Hq3−k (Ω)) ({(τ ∂τ )# (λk H˜ r (λ)) | λ ∈ Λκ,λ∗∗ γκ }) ≤ Cq rb , for # = 0, 1, j = 0, 1, 2 and k = 0, 1. This completes the proof of the existence part of Theorem 3.4.11.

3.5.10 Uniqueness In this subsection, we shall prove the uniqueness part of Theorem 3.4.14. The uniqueness part of Theorem 3.4.11 will be proved in the next subsection. Let u ∈ Hq2 (Ω)N satisfy the homogeneous equations: λu − Div (μD(u) − K0 (u)I) = 0 in Ω,

(μD(u) − K0 (u)I)n = 0 on Γ . (3.380)

We shall prove that u = 0 below. Let λ0 be a large positive number such that for any λ ∈ Σ ,λ0 the existence part of Theorem 3.4.14 holds for q  = q/(q − 1). Let Jq (Ω) be a solenoidal space defined in (3.163) and let g be any element in Jq  (Ω). Let v ∈ Hq2 (Ω)N be a solution of the equations: λ¯ v − Div (μD(v) − K0 (v)I) = g in Ω,

(μD(v) − K0 (v)I)n = 0 on Γ . (3.381)

1 (Ω), we have We first observe that v ∈ Jq  (Ω). In fact, for any ϕ ∈ Hˆ q,0

¯ ∇ϕ)Ω − (Div (μD(v)), ∇ϕ)Ω + (∇K0 (v), ∇ϕ)Ω 0 = (g, ∇ϕ)Ω = λ(v, ¯ = λ(v, ∇ϕ)Ω − (∇div v, ∇ϕ)Ω .

(3.382)

352

Y. Shibata

1 (Ω) ⊂ Hˆ 1 (Ω), for any ϕ ∈ H 1 (Ω), we have Since Hq,0 q,0 q,0

0 = λ¯ (div v, ϕ)Ω + (∇div v, ∇ϕ)Ω .

(3.383)

Choosing λ0 > 0 larger if necessary, we have the uniquness of the resolvent problem (3.383) for the weak Dirichlet operator, and so div v = 0. Putting this and (3.382) 1 (Ω), that is v ∈ J  (Ω). Analogously, together gives (v, ∇ϕ)Ω = 0 for any ϕ ∈ Hˆ q,0 q we see that u ∈ Jq (Ω), because u satisifies Eq. (3.380). Since K0 (v) ∈ Hq1 (Ω) + Hˆ q1 ,0 (Ω), we write K0 (v) = A1 + A2 with A1 ∈ 1 Hq  (Ω) and A2 ∈ Hˆ q1 ,0 (Ω). Since u ∈ Jq (Ω), we see that div u = 0 in Ω. Thus, by the definition of the solenoidal space Jq (Ω) and the divergence theorem of Gauss (u, ∇K0 (v))Ω = (u, ∇A1 )Ω + (u, ∇A2 )Ω = (u · n, A1 )Γ − (div u, A1 )Ω = (u · n, K0 (v))Γ , where we have used K0 (v) = A1 on Γ . Analogously, we have (∇K0 (u), v)Ω = (K0 (u), v · n)Γ . Thus, by the divergence theorem of Gauss we have (u, g)Ω = (u, λ¯ v) − (u, Div (μD(v) − K0 (v)I))Ω μ = λ(u, v) − (u, (μD(v) − K0 (v))n)Γ + ( D(u), D(v))Ω 2 1 = λ(u, v) + (μD(u), D(v))Ω . 2 Analogously, we have 0 = (λu − Div (μD(u) − K0 (u)I), v)Ω − (u, Div (μD(v) − K0 (v)I))Ω 1 = λ(u, v) + (μD(u), D(v))Ω . 2 Combining these two equalities yields that (u, g)Ω = 0

for any g ∈ Jq  (Ω).

(3.384)

For any f ∈ C0∞ (Ω)N , let ψ ∈ Hˆ q1 ,0 (Ω) be a solution to the variational equation 1 (Ω). Let g = f − ∇ψ, and then (f, ∇ϕ)Ω = (∇ψ, ∇ϕ)Ω for any ϕ ∈ Hˆ q,0 g ∈ Jq  (Ω) and (u, ∇ψ)Ω = 0, because u ∈ Jq (Ω). Thus, by (3.384), (u, f)Ω = (u, g)Ω = 0, which, combined with the arbitrariness of the choice of f, leads to u = 0. This completes the proof of the uniqueness part of Theorem 3.4.14.

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

353

3.5.11 A Priori Estimate In this subsection, we prove a priori estimates of solutions of Eq. (3.185), from which the uniqueness part of Theorem 3.4.11 follows immediately. In the previous subsection, we used the dual problem for Eq. (3.184) to prove the uniqueness, but in the present case, it is not clear what is a suitable dual problem for Eq. (3.185) to prove the uniqueness. This is the reason why we consider the a priori estimates. To prove the a priori estimates, we need a slight restriction of the domain Ω. Namely, in this subsection, we assume that Ω is a uniform C 3 domain whose inside has a finite covering (cf. Definition 3.2.3), which is used to estimate the local norm of K(u, h). The following theorem is the main result of this section. Theorem 3.5.28 Let 1 < q < ∞. Let Ω be a uniformly C 3 domain whose inside has a finite covering. Then, there exists a λ0 > 0 such that for any λ ∈ Λσ,λ0 γκ and (u, h) ∈ Hq2 (Ω)N × Hq3 (Ω) satisfying Eq. (3.185), we have |λ|uLq (Ω) + |λ|1/2uHq1 (Ω) + uHq2 (Ω) + |λ|hHq2 (Ω) + hHq3 (Ω) ≤ C{fLq (Ω) + dW 2−1/q (Γ ) + |λ|1/2 hLq (Ω) + hHq1 (Ω) }. q

(3.385) Corollary 3.5.29 Let 1 < q < ∞. Let Ω be a uniformly C 3 domain whose inside has a finite covering. Then, there exists a λ0 > 0 such that the uniqueness holds for Eq. (3.185) for any λ ∈ Λσ,λ0 γκ . In what follows, we shall prove Theorem 3.5.28. We first consider Eq. (3.226). Let u ∈ Hq2 (Ω)N and h ∈ Hq3 (Ω) satisfy Eq. (3.226). We use the same notation as in Sect. 3.5.6, Proposition 3.2.2 and Definition 3.2.3. Let ψ 0 be the function given ∞ 0 0 in Definition 3.2.3, that is ψ0 = j =1 ζj , where ζj are the functions given in Proposition 3.2.2. Notice that supp ψ 0 ⊂ Ω. Let u0 = ψ 0 u, and then u0 satisfies the equations: λu0 − Div (μD(u0 ) − K0 (u0 )I) = f0

in Ω,

(μD(u0 ) − K0 (u0 )I)n = 0

on Γ .

(3.386)

Here, we have set f0 = ψ 0 f + ψ 0 Div (μD(u)) − Div (μD(ψ 0 u)) − (ψ 0 ∇K(u, h) − ∇K0 (ψ 0 u)), and we have used the fact: K0 (u0 ) =< μD(u0 )n, n > −div u0 = 0

on Γ .

354

Y. Shibata

We also let u1j = ζj1 u and hj = ζj1 h, and then u1j and hj satisfy the equations: ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

λu1j − Div (μ(xj1 )D(u1j ) − K1j (u1j , h1j )I) = f1j

in Ωj ,

λhj + Aκ (xj1 ) · ∇Γj hj − nj · uj = dj

on Γj ,

(μ(xj1)D(u1j ) − K1j (u1j , h1j )I)nj − σ (xj1 )(ΔΓj hj )nj = hj

on Γj . (3.387)

Here, we have set f1j = ζj1 f + ζj1 Div (μ(x)D(u)) − Div (μ(x)D(ζj1u)) + Div ((μ(x) − μ(xj1 ))D(ζj1 u)) − (ζj1 ∇K(u, h) − ∇K1j (ζj1 u, ζj1 h)); dj = ζj1 d − ζj1 (Aκ (x) − Aκ (xj1 )) · ∇Γj hj − Aκ (xj1 ) · (ζj1 ∇Γj h − ∇Γj (ζj1 h)); hj = ζj1 h − {ζj1 (μ(x) − μ(xj1 ))D(u) + μ(xj1)(ζj1 D(u) − D(ζj1 u))}n + (ζj1 K(u, h) − K1j (ζj1 u, ζj1 h))n + ζj1 (σ (x) − σ (xj1 ))(ΔΓ h)n + σ (xj1 )(ζj1 ΔΓ h − ΔΓ (ζj1 h))n. Set q

q

q

Eλ (u, h) = |λ|q uLq (Ω) + |λ|q/2 uH 1 (Ω) + uH 2 (Ω) q

+ |λ|

q

q hH 2 (Ω) q

q

q + hH 3 (Ω) . q

Employing the similar argument to that in Sect. 3.5.7 and using Theorem 3.4.14 to estimate u0 in addition, for any positive number ω we have q

Eλ (u, h) ≤ C{f0 Lq (Ω) +

∞  j =1

q

q

(f1j Lq (Ωj ) + dj 

2−1/q

Wq

q

(Γj )

q

+ |λ|q/2 hj Lq (Ωj ) + hj H 1 (Ω ) )} j

q

≤

q C{fLq (Ω) q

q

+ d

2−1/q

Wq

(Γ )

+ |λ|q/2 hLq (Ω)

q

+ hH 1 (Ω) + γκq hH 2 (Ω) q

q

q q + (ωq + M1 )(uH 2 (Ω) q

q

+ hH 3 (Ω) + |λ|q/2 hHq2 (Ω) ) q

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

355

q

+ Cω,M2 (uHq1 (Ω) + |λ|q/2 uLq (Ω) + hH 2 (Ω) q

+ |λ|

q/2

q hH 1 (Ω) ) + q

q K(u, h)Lq (O) }.

(3.388)

Here, Cω,M2 is a constant depending on ω and M2 , and we have used the assumption that ∞ "

supp ∇ψ 0 ∪

supp ∇ζj1 = O

(3.389)

j =1

in Definition 3.2.3. To estimate K(u, h)Lq (O) , we need the following Poincarés’ type lemma. Lemma 3.5.30 Let 1 < q < ∞ and let Ω be a uniformly C 2 domain whose inside has a finite covering. Let O be a set given in (3.389). Then, we have ϕLq (O) ≤ Cq,O ∇ϕLq (Ω)

1 for any ϕ ∈ Hˆ q,0 (Ω)

with some constant Cq,O depending solely on O and q. Proof Let Oi (i = 1, . . . , ι) be the sub-domains given in Definition 3.2.3, and then it is sufficient to prove that ϕLq (Oi ) ≤ C∇ϕLq (Ω)

for any ϕ ∈ Hˆ q,0 (Ω) and i = 1, . . . , ι.

If Oi ⊂ ΩR for some R > 0, since ϕ|Γ = 0, by the usual Poincarés’ inequality we have ϕLq (Oi ) ≤ ϕLq (ΩR ) ≤ C∇ϕLq (ΩR )

1 for any ϕ ∈ Hˆ q,0 (Ω).

Let Oi be a subdomain for which the condition (b) in Definition 3.2.3 holds. Since the norms for ϕ(A ◦ τ (y)) and ϕ(y) are equivalent, without loss of generality we may assume that Oi ⊂ {x = (x  , xN ) ∈ RN | a(x  ) < xN < b,

x  ∈ D} ⊂ Ω,

{x = (x  , xN ) ∈ RN | xN = a(x  )

x  ∈ D} ⊂ Γ.

1 (Ω), we can write Since ϕ ∈ Hˆ q,0

ϕ(x  , xN ) =



xN a(x )

(∂s ϕ)(x  , s) ds.

356

Y. Shibata

because ϕ(x  , a(x  )) = 0. By Hardy inequality (3.48), we have 

b

−q

a(x  )

≤ ≤

|ϕ(x  , xN )|q xN dxN





b a(x  )

q q −1

xN

a(x  )



b

a(x  )

1/q

|(∂s ϕ)(x  , s)| ds

q

1/q

−q

xN dxN

|s∂s ϕ(x  , s)|q s −q ds

1/q ,

and so, by Fubini’s theorem we have  Oi

|ϕ(x)|q dx

1/q

≤ ≤

 

dx  D

dx  D

qb ≤ q −1



 

b a(x  ) b

1/q

−q

a(x  )

dx

|ϕ(x  , xN )|q dxN

|ϕ(x  , xN )|q xN bq dxN



D



b a(x )

|∂N ϕ(x)|q dxN

1/q

1/q



≤ bq ∇ϕLq (Ω) . 

This completes the proof of Lemma 3.5.30.

We now prove that for any ω > 0 there exists a constant Cω,M2 depending on ω and M2 such that K(u, h)Lq (O) ≤ ω(uHq2 (Ω) + hHq3 (Ω) ) + Cω,M2 (uHq1 (Ω) + hHq2 (Ω) ). For this purpose, we estimate |(K(u, h), ψ)Ω | for any ψ 1 (Ω) we have Lemma 3.5.30, for any ϕ ∈ Hˆ q,0

(3.390)

∈ C0∞ (O). By

|(ϕ, ψ)Ω | ≤ ϕLq (O) ψLq  (O) ≤ Cq,O ∇ϕLq (O) ψLq  (O) . Thus, by the Hahn-Banach theorem, there exists a g ∈ Lq  (Ω)N such that gLq  (Ω) ≤ Cq  ,O ψLq  (O) and (ϕ, ψ)Ω = (∇ϕ, g)Ω

(3.391)

1 (Ω). In particular, div g = −ψ, and therefore div g  for any ϕ ∈ Hˆ q,0 Lq (Ω) ≤ ψLq  (O) . By the assumption of the unique existence of solutions of the weak Dirichlet problem and its regularity theorem, Theorem 3.2.8 in Sect. 3.2.6 below,

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

357

there exists a Ψ ∈ Hˆ q1 ,0 (Ω) such that ∇ 2 Ψ ∈ Lq (Ω)N , Ψ satisfies the weak Dirichlet problem: (∇Ψ, ∇ϕ)Ω = (g, ∇ϕ)Ω

1 for any ϕ ∈ Hˆ q,0 (Ω)

(3.392)

and the estimate: ∇Ψ H 1 (Ω) ≤ Cq,O ψLq  (O) .

(3.393)

q

Let L = K(u, h)−{< μD(u)n, n > −σ ΔΓ h−div u}, where the Laplace–Beltrami 1 (Ω). Thus, by (3.391), operator ΔΓ is suitably extended in Ω, and then L ∈ Hˆ q,0 (3.392) with ϕ = L and the divergence theorem of Gauß, |(L, ψ)Ω | = |(∇L, g)Ω | = |(∇Ψ, ∇L)Ω | ≤ |(Div (μD(u)) − ∇div u, ∇Ψ )Ω | + |(∇{< μD(u)n, n > −σ ΔΓ h − div u}, ∇Ψ )Ω | ≤ CM2 {(∇uLq (Γ ) + (h, ∇h, ∇ 2 h)Lq (Γ ) )∇Ψ Lq (Γ ) + (∇uLq (Ω) + hHq2 (Ω) )∇ 2 Ψ Lq (Ω) }. 1/q

1−1/q

Using the interpolation inequality: vLq (Γ ) ≤ C∇vLq (Ω) vLq (Ω) and (3.393), we have |(L, ψ)Ω | ≤ {ω(∇ 2 uLq (Ω) + hHq3 (Ω) ) + Cω,M2 (∇uLq (Ω) + hHq2 (Ω) )}ψLq  (O) , which leads to LLq (Ω) ≤ ω(∇ 2 uLq (Ω) + hHq3 (Ω) ) + Cω,M2 (∇uLq (Ω) + hHq2 (Ω) ). Thus, we have (3.390). Putting (3.388) and (3.390) together and choosing ω and M1 small enough and λ0 large enough, we have (3.385). This completes the proof of Theorem 3.5.28 for Eq. (3.226). For Eq. (3.185), using the result we have proved just now, we see that Eλ (u, h) ≤ C(fLq (Ω) + dW 2−1/q (Γ ) + |λ|1/2 hLq (Ω) + hHq1 (Ω) + R) q

with R = F1 uW 2−1/q (Γ ) + |λ|1/2 F2 hLq (Ω) + F2 hHq1 (Ω) ). q

358

Y. Shibata

Since R ≤ C(uHq1 (Ω) + |λ|1/2 hHq2 (Ω) ) as follows from (3.161), choosing |λ| suitably large, R can be absorbed by Eλ (u, h), which completes the proof of Theorem 3.5.28.

3.6 Local Well-Posedness for Arbitrary Initial Velocity Fields in a Uniform C 3 Domain In this section, we shall prove the unique existence of local in time solutions of Eq. (3.157). Namely, we shall prove the following theorem. Theorem 3.6.1 Let 2 < p < ∞, N < q < ∞, B > 0 and let Ω be a uniform C 3 domain whose inside has a finite covering. Assume that 2/p + N/q < 1 and that the weak Dirichlet problem is uniquely solvable for indices q and q  = q/(q − 2(1−1/p) 3−1/p−/q 1). Let u0 ∈ Bq,p (Ω)N and ρ0 ∈ Wq,p (Γ ) be initial data such that u0 B 2(1−1/p (Ω) ≤ B holds and the compatibility conditions: q,p

u0 − g(u0 , Ψρ |t =0 ) ∈ Jq (Ω),

div u0 = g(u0 , Ψρ |t =0 )

(μD(u0 )n)τ = h (u0 , Ψρ |t =0)

in Ω,

on Γ

are satisfied. Then, there exist a time T > 0 and a small number > 0 depending on B such that if ρ0 B 3−1/p−1/q (Γ ) ≤ , then problem (3.157) admits unique solutions q,p u, q and ρ with u ∈ Lp ((0, T ), Hq2 (Ω)N ) ∩ Hp1 ((0, T ), Lq (Ω)N ), 1 (Ω)), q ∈ Lp ((0, T ), Hq1 (Ω) + Hˆ q,0 3−1/q

ρ ∈ Lp ((0, T ), Wq

2−1/q

(Γ )) ∩ Hp1 ((0, T ), Wq

(Γ ))

possessing the estimate (3.100) and Ep,q,T (u, ρ) ≤ CB for some constant C independent of B. Here, we have set Ep,q,T (u, ρ) = uLp ((0,T ),Hq2 (Ω)) + ∂t uLp ((0,T ),Lq (Ω)) + ∂t ρL + ρL

1−1/q (Γ )) ∞ ((0,T ),Wq

3−1/p (Ω)) p ((0,T ),Wq

.

+ ∂t ρL

2−1/q (Ω)) p ((0,T ),Wq

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

359

In what follows, we shall prove Theorem 3.6.1 by the Banach fixed point theorem. Since N < q < ∞, by Sobolev’s inequality we have the following estimates: f L∞ (Ω) ≤ Cf Hq1 (Ω) , fgHq1 (Ω) ≤ Cf Hq1 (Ω) gHq1 (Ω) , fgW 1−1/q (Γ ) ≤ Cf W 1−1/q (Γ ) gW 1−1/q (Γ ) . q

q

(3.394)

q

Moreover, since 2/p + N/q < 1, we have f H∞ , 1 (Ω) ≤ Cf  2(1−1/p) B (Ω) q,p

f H∞ . 2 (Γ ) ≤ Cf  3−1/p−1/q B (Γ ) q,p

(3.395)

We define an underlying space UT by letting UT = {(u, ρ) | u ∈ Hp1 ((0, T ), Lq (Ω)N ) ∩ Lp ((0, T ), Hq2 (Ω)N ), 2−1/q

ρ ∈ Hp1 ((0, T ), Wq u|t =0 = u0 Ep,q,T (u, ρ) ≤ L,

in Ω,

3−1/q

(Γ )) ∩ Lp ((0, T ), Wq

ρ|t =0 = ρ0

(Γ )),

on Γ ,

sup ρ(·, t)H∞ 1 (Γ ) ≤ δ},

t ∈(0,T )

where L is a number determined later. Since L is chosen large and small eventually, we may assume that 0 < < 1 < L in the following. Thus, for example, we will use the inequality: 1 + + L < 3L below. Let (v, h) ∈ UT and let u, q and ρ be solutions of the linear equations: ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

∂t u − Div (μD(u) − qI) = f(v, Ψh )

in Ω T ,

div u = g(u, Ψh ) = div g(v, Ψh )

in Ω T ,

∂t ρ+ < uκ | ∇Γ ρ > −u · n = dκ (v, Ψh )

on Γ T ,

⎪ ⎪ (μD(u)n)τ = h (v, Ψh ) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ < μD(u)n, n > −q − σ (ΔΓ ρ + Bρ) = hN (v, Ψh ) ⎪ ⎪ ⎩ (u, ρ)|t =0 = (u0 , ρ0 )

on Γ T , on Γ T , in Ω × Γ .

Here, the operator B is defined in (3.149) and we have set dκ (v, Ψh ) = d(v, Ψh )+ < uκ − v | ∇Γ h > .

(3.396)

360

Y. Shibata

Let Ψh = ωHh n, where Hh is an extension of h to Ω such that Hh (·, t)Hqk (Ω) ≤ Ch(·, t)W k−1/q (Γ ) , q

∂t Hh (·, t)Hq# (Ω) ≤ C∂t h(·, t)W #−1/q (Γ )

(3.397)

q

for k = 1, 2, 3 and # = 1, 2 with some constants C. In particular, for (v, h) ∈ UT , we may assume that sup Ψh (·, t)H∞ 1 (RN ) ≤ δ.

(3.398)

t ∈(0,T )

In fact, we can assume that supt ∈(0,T ) h(·, t)H∞ 1 (Γ ) ≤ δ with smaller δ. Since (v, h) ∈ UT , we have Ep,q,T (v, h) ≤ L,

(3.399)

that is ∂t hL

1−1/q (Γ )) ∞ ((0,T ),Wq

+ ∂t hL

2−1/q (Γ )) p ((0,T ),Wq

+ hL

3−1/q (Γ )) p ((0,T ),Wq

+ ∂t vLp ((0,T ),Lq (Ω)) + vLp ((0,T ),Hq2 (Ω)) ≤ L.

(3.400)

Moreover, we use the assumption: u0 B 2(1−1/p) (Ω) ≤ B, q,p

ρ0 B 3−1/p−1/q (Γ ) ≤ ,

(3.401)

q,p

where is a small constant determined later. To obtain the estimates of u and ρ, we shall use Corollary 3.4.6. Thus, we shall estimate the nonlinear functions appearing in the right hand side of Eq. (3.396). We start with proving that 

f(v, Ψh )Lp ((0,T ),Lq (Ω)) ≤ C{T 1/p (L + B)2 + ( + T 1/p L)L}.

(3.402)

The definition of f(v, Ψh ) is given by replacing ξ(t), ρ and u by 0, h and v in (3.113). Since |V0 (k)| ≤ C|k| when |k| ≤ δ, by (3.398) we have f(v, Ψh )Lq (Ω) ≤ C{vL∞ (Ω) ∇vLq (Ω) + ∂t Ψh L∞ (Ω) ∇vLq (Ω) + ∇Ψh L∞ (Ω) ∂t vLq (Ω) + ∇Ψh L∞ (Ω) ∇ 2 vLq (Ω) + ∇ 2 Ψh Lq (Ω) ∇vL∞ (Ω) }.

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

361

By (3.397) and (3.394), we have v(·, t)L∞ (Ω) ≤ Cv(·, t)Hq1 (Ω) , ∂t Ψh (·, t)L∞ (Ω) ≤ C∂t h(·, t)W 1−1/q (Γ ) , q

(3.403)

∇Ψh (·, t)L∞ (Ω) ≤ Ch(·, t)W 2−1/q (Γ ) , q

∇v(·, t)L∞ (Ω) ≤ Cv(·, t)Hq2 (Ω) , and so, f(v, Ψh )Lp ((0,T ),Lq (Ω)) ≤ C{v2L + vL∞ ((0,T ),Hq1 (Ω)) ∂t hL + hL

2−1/q (Γ )) ∞ ((0,T ),Wq

1 ∞ ((0,T ),Hq (Ω))

1−1/q (Γ )) ∞ ((0,T ),Wq

T 1/p

T 1/p

(3.404)

(∂t vLp ((0,T ),Lq (Ω)) + vLp ((0,T ),Hq2 (Ω)) )}.

Here and in the following, we use the estimate: f Lp ((0,T ),X) ≤ T 1/p f L∞ ((0,T ),X) for the lower order term. In what follows, we shall use the following inequalities: vL

2(1−1/p) (Ω)) ∞ ((0,T ),Bq,p

≤ C{u0 B 2(1−1/p) (Ω) q,p

+ vLp ((0,T ),Hq2 (Ω)) + ∂t vLp ((0,T ),Lq (Ω)) },

(3.405) hL

3−1/p−1/q (Γ )) ∞ ((0,T ),Bq,p

≤ C{ρ0 B 3−1/p−1/q (Γ ) q,p

+ hL

3−1/q (Γ )) p ((0,T ),Wq

+ ∂t hL

2−1/q (Γ )) p ((0,T ),Wq

},

(3.406)  h(·, t)L

2−1/q

∞ ((0,T ),Wq

(Γ ))

≤ ρ0 W 2−1/q (Γ ) + q

T 0

≤ ρ0 W 2−1/q (Γ ) + T q

∂s h(·, s)W 2−1/q (Γ ) ds

1/p 

q

L,

(3.407)

for some constant C independent of T . The inequalities: (3.405) and (3.406) will be proved later. Combining (3.404) with (3.405), (3.407), (3.400) and (3.401) gives (3.402). We next consider dκ (u, Ψh ). Since ξ(t) = 0, by (3.133) we have ¯ h )∇Ψ ¯ h ⊗ ∇Ψ ¯ h − ∂t h < n, VΓ (∇Ψ ¯ h )∇Ψ ¯ h ⊗ ∇Ψ ¯ h>. d(v, Ψh ) = v · VΓ (∇Ψ

362

Y. Shibata

Applying (3.126), (3.398) and (3.394), we have ¯ h )∇Ψ ¯ h ⊗ ∇Ψ ¯ h  1−1/q VΓ (∇Ψ ≤ Ch(·, t)W 2−1/q (Γ ) , W (Γ ) q

q

¯ h )∇Ψ ¯ h ⊗ ∇Ψ ¯ h  2−1/q ≤ Ch(·, t)W 2−1/q (Γ ) h(·, t)W 3−1/q (Γ ) , VΓ (∇Ψ W (Γ ) q

q

q

and so, by (3.403) and (3.394) d(v, Ψh )W 1−1/q (Γ ) ≤ C(∂t h(·, t)W 1−1/q (Γ ) + v(·, t)Hq1 (Ω) )h(·, t)W 2−1/q (Γ ) , q

q

q

d(v, Ψh )W 2−1/q (Γ ) ≤ C{(∂t h(·, t)W 2−1/q (Γ ) + v(·, t)Hq2 (Ω) )h(·, t)W 2−1/q (Γ ) q

q

q

+ (∂t h(·, t)W 1−1/q (Γ ) q

+ v(·, t)Hq1 (Ω) )h(·, t)W 2−1/q (Γ ) h(·, t)W 3−1/q (Γ ) }. q

q

Thus, by (3.400), (3.401), (3.405), and (3.407), we have 

sup d(v, Ψh )W 1−1/q (Γ ) ≤ C(L + B)( + T 1/p L), q

t ∈(0,T )

d(v, Ψh )L



2−1/q (Γ )) p ((0,T ),Wq

(3.408)

≤ C(L + B)L( + T 1/p L).

By (3.156) and (3.394), we have  < uκ − v | ∇Γ h > W 1−1/q (Γ ) q

≤ C(u0 B 2(1−1/p) (Ω) + v(·, t)Hq1 (Ω) )h(·, t)W 2−1/q (Γ ) , q

q,p

 < uκ − v | ∇Γ h > W 2−1/q (Γ )

(3.409)

q

≤ C{(uκ Hq2 (Ω) + v(·, t)Hq2 (Ω) )h(·, t)W 2−1/q (Γ ) q

+ uκ − v(·, t)Hq1 (Ω) h(·, t)W 3−1/q (Γ ) }. q

By (3.405), (3.407), (3.400), and (3.401), we have  < uκ − v | ∇Γ h > L

1−1/q (Γ )) ∞ ((0,T ),Wq 

≤ C(B + L)( + T 1/p L). By the definition of uκ , we have uκ − u0 =

1 κ



κ 0

(T (s)u˜ 0 − u0 ) ds,

(3.410)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

363

and so, we have uκ − u0 Lq (Ω) ≤

1 κ



≤ Cκ

κ



0 1/p 

s 0

∂s T (r)u˜ 0 Lq (Ω) dr ds

u0 B 2(1−1/p) (Ω) . q,p

Let s be a positive number such that 1 < s < 2(1 − 1/p), and then by interpolation theory and (3.156) we have 1−1/s

1/s

uκ − u0 Hq1 (Ω) ≤ Cuκ − u0 Lq (Ω) uκ − u0 W s (Ω) q

≤ Cκ Writing v(·, t) − u0 =

t 0

(1−1/s)/p 

u0 B 2(1−1/p) (Ω) .

(3.411)

q,p

∂r v(·, r) dr, we have 

v(·, t) − u0 Lq (Ω) ≤ LT 1/p . On the other hand, by (3.405) v(·, t) − u0 B 2(1−1/p) (Ω) ≤ L + u0 B 2(1−1/p) (Ω) , q,p

q,p

and so, 

v(·, t) − u0 Hq1 (Ω) ≤ (L + B)T (1−1/s)/p .

(3.412)

Putting (3.411) and (3.412) together gives 



sup v(·, t) − uκ Hq1 (Ω) ≤ C(κ (1−1/s)/p + T (1−1/s)/p )(L + B).

t ∈(0,T )

(3.413)

By (3.409)  < uκ − v | ∇Γ h > L

2−1/q (Γ )) p ((0,T ),Wq

≤ C{(uκ Hq2 (Ω) T 1/p + vLp ((0,T ),Hq2 (Ω)) )hL + uκ − vL∞ ((0,T ),Hq1 (Ω)) hL

2−1/q (Γ )) ∞ ((0,T ),Wq

3−1/q (Γ )) p ((0,T ),Wq

}.

364

Y. Shibata

Thus, by (3.156), (3.400), (3.401), (3.407) and (3.413), we have  < uκ − v | ∇Γ h > L

2−1/q (Γ )) p ((0,T ),Wq 

≤ C{(Bκ −1/p T 1/p + L)( + T 1/p L) 

(3.414) 

+ L(L + B)(κ (1−1/s)/p + T (1−1/s)/p )} for some constant s ∈ (1, 2(1 − 1/p)). Putting (3.408), (3.410), and (3.414) gives 

d˜κ (v, Ψh )L

1−1/q (Γ )) ∞ ((0,T ),Wq

d˜κ (v, Ψh )L

2−1/q (Γ )) p ((0,T ),Wq

≤ C(L + B)( + T 1/p L), 

≤ C{(L + Bκ −1/p T 1/p )( + T 1/p L)







+ L(L + B)( + T 1/p L + κ (1−1/s)/p + T (1−1/s)/p )},

(3.415)

where s is a constant ∈ (1, 2(1 − 1/p)). We now estimate g(v, Ψh ) and g(v, Ψh ), which are defined in (3.108). In view of Corollary 3.4.6, we have to extend g(v, Ψh ) and g(v, Ψh ) to the whole time interval R. Before turning to the extension of these functions, we make a few definitions. 2(1−1/p) N N 2(1−1/p) (R ) be an extension of u0 ∈ Bq,p (Ω)N to RN such that Let u˜ 0 ∈ Bq,p u0 = u˜ 0 2(1−1/p)

for X ∈ {Hqk , Bq,p

in Ω,

u˜ 0 X(RN ) ≤ Cu0 X(Ω)

}. Let

Tv (t)u0 = e−(2−Δ)t u˜ 0 = F −1 [e−(|ξ |

2 +2)t

F [u˜ 0 ](ξ )],

(3.416)

and then Tv (0)u0 = u0 in Ω and et Tv (t)u0 X(Ω) ≤ Cu0 X(Ω) , et Tv (·)u0 Hp1 ((0,∞),Lq (Ω)) + et Tv (·)u0 Lp ((0,∞),Hq2 (Ω)) ≤ Cu0 B 2(1−1/p) (Ω) .

(3.417)

q,p

Let W, P , and Ξ be solutions of the equations: ⎧ ∂t W + λ0 W − Div (μD(W) − P I) = 0, div W = 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ∂t Ξ + λ0 Ξ − W · n = 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

(μD(W)n − P I)n − (σ ΔΓ Ξ )n = 0 (W, Ξ )|t =0 = (0, ρ0 )

in Ω × (0, ∞), on Γ × (0, ∞), on Γ × (0, ∞), in Ω × Γ .

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

365

Since WL

2(1−1/p) (Ω)) ∞ ((0,∞),Bq,p

≤ C(WLp ((0,∞),Hq2 (Ω)) + ∂t WLp ((0,∞),Lq (Ω)) ), Ξ L

3−1/p−1/q (Γ )) ∞ ((0,∞),Bq,p

≤ C(Ξ L

3−1/q (Γ )) p ((0,∞),Wq

+ ∂t Ξ L

2−1/q (Γ )) p ((0,∞),Wq

),

as follow from real interpolation results (3.220) and (3.221), choosing λ0 large enough, by Theorem 3.4.3, we know the existence of W and Ξ with W ∈ Lp ((0, ∞), Hq2 (Ω)N ) ∩ Hp1 ((0, ∞), Lq (Ω)N ), 3−1/q

Ξ ∈ Lp ((0, ∞), Wq

2−1/q

(Γ )) ∩ Hp1 ((0, ∞), Wq

(Γ ))

possessing the estimates: et WL

2(1−1/p) (Ω)) ∞ ((0,∞),Bq,p

+ et Ξ L

3−1/p−1/q (Γ )) ∞ ((0,∞),Bq,p

+ et WLp ((0,∞),Hq2 (Ω)) + et ∂t WLp ((0,∞),Lq (Ω)) + et Ξ L

3−1/q (Γ )) p ((0,∞),Wq

+ et ∂t Ξ L

2−1/q (Γ )) p ((0,∞),Wq

≤ Cρ0 B 3−1/p−1/q (Γ ) . q,p

Moreover, by the trace theorem and the kinematic equation, we have et ∂t Ξ L

1−1/q (Γ )) ∞ ((0,∞),Wq

≤ λ0 et Ξ L

1−1/q (Γ )) ∞ ((0,∞),Wq

+ et n · WL

1−1/q (Γ )) ∞ ((0,∞),Wq

≤ Cρ0 W 3−1/p−1/q (Γ ) . q,p

Setting Th (t)ρ0 = ΨΞ , we have Th (0)ρ0 = Ψρ0 in Ω, and et Th (·)ρ0 L

3−1/p (Ω)) ∞ (0,∞),Bq,p

+ et ∂t Th (·)ρ0 L∞ ((0,∞),Hq1 (Ω))

+ et Th (·)ρ0 Lp ((0,∞),Hq3 (Ω)) + et ∂t Th (·)ρ0 Lp ((0,∞),Hq2 (Ω)) ≤ Cρ0 B 3−1/p−1/q (Γ ) . q,p

(3.418)

366

Y. Shibata

Let ψ(t) be a function of C ∞ (R) which equals one for t > −1 and zero for t < −2. Given a function, f (t), defined on (0, T ), the extension, eT [f ], of f is defined by letting ⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎨f (t) eT [f ](t) = ⎪f (2T − t) ⎪ ⎪ ⎪ ⎩ 0

for t < 0, for 0 < t < T , for T < t < 2T ,

(3.419)

for t > 2T .

Obviously, eT [f ](t) = f (t) for t ∈ (0, T ) and eT [f ](t) = 0 for t ∈ (0, 2T ). Moreover, if f |t =0 = 0, then ⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎨(∂ f )(t) t ∂t eT [f ](t) = ⎪ ⎪−(∂t f )(2T − t) ⎪ ⎪ ⎩ 0

for t < 0, for 0 < t < T , for T < t < 2T ,

(3.420)

for t > 2T .

We now define the extensions, E1 [v], and E2 [Ψh ], of v and Ψh to R by letting E1 [v] = eT [v − Tv (t)u0 ] + ψ(t)Tv (|t|)u0 , E2 [Ψh ] = eT [Ψh − Th (t)ρ0 ] + ψ(t)Th (|t|)ρ0 .

(3.421)

Since v|t =0 = Tv (0)u0 = u0 and Ψh |t =0 = Th (0)ρ0 , we can differentiate E1 [v] and E2 [Ψh ] once with respect to t and we can use the formula (3.420). Obviously, we have E1 [v] = v,

E2 [Ψh ] = Ψh

in Ω T .

(3.422)

Let g(v, ˜ Ψh ) = −{J0 (∇E2 [Ψh ])div E1 [v] + (1 + J0 (∇E2 [Ψh ]))V0 (∇E2 [Ψh ]) : ∇E1 [v]},

(3.423)

)

g˜ (v, Ψh ) = −(1 + J0 (∇E2 [Ψh ])) V0 (∇E2 [Ψh ])E1 [v], and then, applying the Hanzawa transform: x = y + E2 [Ψh ] instead of x = y + Ψh , by (3.108), (3.109), and (3.422), we have g(v, ˜ Ψh ) = g(v, Ψh ), g˜ (v, Ψh ) = g(v, Ψh ) div g˜ (v, Ψh ) = g(v, ˜ Ψh )

in Ω T , in Ω × R.

(3.424)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

367

By (3.418) and (3.407), we have sup E2 [Ψh ]H∞ 1 (Ω) ≤ C( sup Ψh H 2 (Ω) + T (·)ρ0 L ,H 2 (Ω) ) ∞ q q t ∈R

t ∈(0,T )



≤ C(ρ0 W 3−1/p1/q (Γ ) + T 1/p L). q

Thus, we choose T and ρW 3−1/p−1/q (Ω) so small that q,p

sup E2 [Ψh ]H∞ 1 (Ω) ≤ δ. t ∈R

(3.425)

Since V0 (0) = 0, we may write g˜ (v, Ψh ) as g˜ (v, Ψh ) = Vg (∇E2 [Ψh ])∇E2 [Ψh ] ⊗ E1 (v)

(3.426)

with some matrix of C 1 functions, Vg (k), defined on |k| < δ such that (Vg , Vg )L∞ (|k|≤δ) ≤ C, where Vg denotes the derivative of Vg with respect to k. We may write the time derivative of g˜ (v, Ψh ) as ∂t g˜ (v, Ψh ) = Vg (∇E2 [Ψh ])∇E2 [Ψh ] ⊗ (∂t E1 (v)) + Vg (∇E2 [Ψh ])(∂t ∇E2 [Ψh ]) ⊗ E1 (v) + (Vg (∇E2 [Ψh ])∂t ∇E2 [Ψh ])∇E2 [Ψh ] ⊗ E1 (v), and so by (3.425) we have ∂t g˜ (v, Ψh )Lq (Ω) ≤ C{∇E2 [Ψh ]Hq1 (Ω) ∂t E1 [v]Lq (Ω) + ∂t ∇E2 [Ψh ]Lq (Ω) E1 [v]Hq1 (Ω) }.

(3.427)

Thus, using (3.420), (3.394), (3.407), (3.404), (3.400), (3.401), (3.417), and (3.418), for any γ ≥ 0, we have e−γ t ∂t g˜ (v, Ψh )Lp (R,Lq (Ω)) ≤ C(Ψh L∞ ((0,T ),Hq2 (Ω)) + Th (·)Ψρ0 L∞ ((0,∞),Hq2(Ω)) ) × (∂t vLp ((0,T ),Lq (Ω)) + Tv (·)u0 Hp1 ((0,∞),Lq (Ω)) ) + T 1/p (∂t Ψh L∞ ((0,T ),Hq1 (Ω)) + ∂t Th (·)ρ0 L∞ ((0,∞),Hq1 (Ω)) ) × (vL∞ ((0,T ),Hq1 (Ω)) + Tv (·)u0 L∞ ((0,∞),Hq1 (Ω)) ) 

≤ C(LT 1/p + + (L + )T 1/p )(L + B) 

≤ C( + L(T 1/p + T 1/p ))(L + B).

(3.428)

368

Y. Shibata

We next prove that ˜ Ψh )H 1/2 (R,L e−γ t g(v, p

q (Ω))

+ e−γ t g(v, ˜ Ψh )Lp (R,Hq1 (Ω))





(3.429)

≤ C(( + T 1/p L)1/2 L1/2 + + T 1/p L)(L + B) for any γ ≥ 0. In the sequel, to estimate fgH 1/2 (R,L (Ω)) , we use the following q p two lemmas. Lemma 3.6.2 Let 1 < p < ∞ and N < q < ∞. Let 1 (R, Lq (Ω)), f ∈ L∞ (R, Hq1 (Ω)) ∩ H∞ 1/2

g ∈ Hp (R, Lq (Ω)) ∩ Lp (R, Hq1 (Ω)). Then, we have fgH 1/2 (R,L p

q (Ω))

+ fgLp (R,Hq1 (Ω))

1/2

≤ C(f L

1/2

1 ∞ (R,Hq (Ω)

× (gH 1/2 (R,L p

∂t f L∞ (R,Lq (Ω)) + f L∞ (R,Hq1 (Ω)) )

q (Ω))

+ gLp (R,Hq1 (Ω)) ).

Proof We can prove the lemma by using the complex interpolation: 1/2

1/2

Hp (R, Lq (Ω)) ∩ Lp (R, Hq (Ω)) = (Lp (R, Lq (Ω)), Hp1(R, Lq (Ω)) ∩ Lp (R, Hq1 (Ω)))[1/2]. 

To estimate ∇vH 1/2 (R,L p

q (Ω))

, we use the following lemma.

Lemma 3.6.3 Let 1 < p, q < ∞ and let Ω be a uniform C 2 domain. Then, 1/2

Hp1 (R, Lq (Ω)) ∩ Lp (R, Hq2 (Ω)) ⊂ Hp (R, Hq1 (Ω)) and uH 1/2 (R,H 1 (Ω)) ≤ C{uLp (R,Hq2 (Ω)) + ∂t uLp (R,Lq (Ω)) }. p

q

Remark 3.6.4 This lemma was mentioned in Shibata–Shimizu [53] in the case that Ω is bounded and was proved by Shibata [48] in the case that Ω is a uniform C 2 domain.

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

369

Since J0 (0) = 0 and V0 (0) = 0, by (3.425) we may write g(v, ˜ Ψh ) = Vg (∇E2 [Ψh ])∇E2 [Ψh ] ⊗ ∇E1 [v]

(3.430)

with some matrix of C 1 functions, Vg (k), defined on |k| < δ such that (Vg , Vg )L∞ (|k|≤δ) ≤ C, where Vg denotes the derivative of Vg with respect to k. By (3.394), (3.397), (3.425), (3.421), (3.418), (3.407), and (3.401), we have Vg (∇E2 [Ψh ])∇E2 [Ψh ]L∞ (R,Hq1 (Ω)) ≤ CE2 [Ψh ]L∞ (R,Hq2 (Ω)) 

≤ C(Th (·)ρ0 L∞ (R,Hq2 (Ω)) + Ψh L∞ (R,Hq2 (Ω)) ) ≤ C( + T 1/p L); ∂t (Vg (∇E2 [Ψh ])∇E2 [Ψh ])L∞ (R,Lq (Ω)) ≤ C∂t E2 [Ψh ]L∞ (R,Hq1 (Ω)) Thus, by Lemmas 3.6.2 and 3.6.3 we have e−γ t g(v, ˜ Ψh )H 1/2 (R,L p

q (Ω))

+ e−γ t g(v, ˜ Ψh )Lp (R,Hq1 (Ω))





≤ C(( + T 1/p L)1/2 L1/2 + + T 1/p L)

(3.431)

× (e−γ t E1 [v]Hp1 (R,Lq (Ω)) + e−γ t E1 [v]Lp (R,Hq2 (Ω)) ) for any γ ≥ 0, which, combined with (3.400), (3.401), and (3.417), leads to (3.429). We finally estimate h (v, Ψh ) and hN (v, h) given in (3.139) and (3.152). In view of (3.139), we may write h (v, Ψh ) as ¯ h )∇Ψ ¯ h ⊗ ∇v h (v, Ψh ) = Vh (∇Ψ

(3.432)

¯ = V (y, k), ¯ defined on Ω × {k¯ | |k| ¯ ≤ δ} with some matrix of C 1 functions, Vh (k) h possessing the estimate : ¯ ∂ ¯ V (·, k)) ¯ H 1 (Ω) ≤ C sup (Vh (·, k), k h ∞

¯ |k|≤δ

¯ with some constant C. Here, ∂k¯ Vh denotes the derivative of Vh with respect to k.  We extend h (v, Ψh ) to the whole time interval R by letting ¯ 2 [Ψh ]))∇E ¯ 2 [Ψh ] ⊗ ∇E1 [v]. h˜  (v, Ψh ) = Vh (∇E

(3.433)

Employing the same argument as in the proof of (3.429), for any γ > 0 we have e−γ t h˜  (v, Ψh )H 1/2 (R,L p



q (Ω))

+ e−γ t h˜  (v, Ψρ )Lp (R,Hq1 (Ω)) 

≤ C(( + T 1/p L)1/2 L1/2 + + T 1/p L)(L + B).

(3.434)

370

Y. Shibata

We finally consider hN (v, Ψh ). In (3.152) we may write ¯ k¯ > − < n, μ(DD (k)∇v)(n + VΓ (k) ¯ k) ¯ > − < n, μD(v)VΓ (k) ¯ = Vh,N (k)∇Ψ h ⊗ ∇v ¯ = Vh,N (y, k), ¯ defined on Ω × {k | with some matrix of C 1 functions, Vh,N (k) |k| ≤ δ} such that ¯ ∂ ¯ Vh,N (·, k)) ¯ H 1 (Ω) ≤ C. sup (Vh,N (·, k), k ∞

¯ |k|≤δ

Thus, we may write hN (v, Ψh ) as ¯ h )∇Ψh ⊗ ∇v + σ VΓ (∇Ψ ¯ h )∇Ψ ¯ h ⊗ ∇¯ 2 Ψh , hN (v, Ψh ) = Vh,N (∇Ψ

(3.435)

and so, we can define the extension of hN (v, Ψh ) by letting ¯ 2 [Ψh ])∇E ¯ 2 [Ψh ] ⊗ ∇E1 [v] h˜ N (v, Ψh ) = Vh,N (∇E

(3.436)

¯ 2 [Ψh ])∇E ¯ 2 [Ψh ] ⊗ ∇¯ 2 E2 [Ψh ]. + σ VΓ1 (∇E Using Lemmas 3.6.2, 3.6.3, (3.417), and (3.418), we have h˜ N (v, Ψh )Lp (R,Hq1 (Ω)) + h˜ N (v, Ψh )H 1/2 (R,L p

≤ C(( + T

1/p 

L)1/2 L1/2 + + T

1/p 

q (Ω))

L)

× (e−γ t E1 [v]Lp (R,Hq2 (Ω)) + e−γ t E1 [v]Hp1 (R,Lq (Ω)) + e−γ t ∇¯ 2 E2 [Ψh ]Lp (R,Hq1 (Ω)) + e−γ t ∇¯ 2 E2 [Ψh ]H 1/2 (R,L p

q (Ω))

).

1/2

By the fact that Hp1 (R, Lq (Ω)) ⊂ Hp (R, Lq (Ω)), we have e−γ t ∇¯ 2 E2 [Ψh ]H 1/2 (R,L p

q (Ω))

≤ e−γ t ∇¯ 2 E2 [Ψh ]Hp1 (R,Lq (Ω)) ≤ C( + L).

Therefore, by (3.417), (3.418), (3.400), and (3.401), we have h˜ N (v, Ψh )Lp (R,Hq1 (Ω)) + h˜ N (v, Ψh )H 1/2 (R,L p





q (Ω))

≤ C(( + T 1/p L)1/2 L1/2 + + T 1/p L)(B + L).

(3.437)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

371

Let E˜ p,q,T (u, ρ) = uLp ((0,T ),Hq2 (Ω)) + ∂t uLp ((0,T ),Lq (Ω)) + ρL

3−1/q (Γ )) p ((0,T ),Wq

+ ∂t ρL

2−1/q (Γ )) p ((0,T ),Wq

.

Notice that Ep,q,T (u, ρ) = E˜ p,q,T (u, ρ) + ∂t ρL∞ ((0,T ),Hq1 (Ω)) . Applying Corollary 3.4.6 and using the estimates (3.402), (3.415), (3.428), (3.434), (3.437), and (3.401), we have −b E˜ p,q,T (u, ρ) ≤ Ceγ κ T D

(3.438)

with D = B + κ −b + T 1/p (L + B)2 



+ ( + T 1/p L)L + (L + Bκ −1/p T 1/p )( + T 1/p L) 





+ L(L + B)( + T 1/p L + κ (1−1/s)/p + T (1−1/s)/p ) 



+ ( + L(T 1/p + T 1/p ) + ( + T 1/p L)1/2 L1/2 )(L + B)} for some positive constants γ and C independent of B, and T . Combining (3.405) and (3.406) with (3.438) yields that uL ρL

2(1−1/p) (Ω)) ∞ ((0,T ),Bq,p

3−1/p−1/q (Γ )) ∞ ((0,T ),Bq,p

≤ C(B + E˜ p,q,T (u, ρ)), (3.439)

≤ C( + E˜ p,q,T (u, ρ)).

Noting that 3 − 1/p − 1/q > 2 − 1/q, by the kinematic equation in Eq. (3.396) and (3.439) we have sup ∂t ρ(·, t)W 1−1/q (Γ ) q

t ∈(0,T )

≤ uκ Hq1 (Ω) sup ρ(·, t)W 2−1/q (Γ ) t ∈(0,T )

q

+ C sup u(·, t)Hq1 (Ω) + sup d(·, t)W 1−1/q (Γ ) t ∈(0,T )

t ∈(0,T )

q



≤ CB( + E˜ p,q,T ) + C(B + E˜ p,q,T (u, ρ)) + C(L + B)( + T 1/p L). (3.440)

372

Y. Shibata

< 1 and B ≥ 1, combining (3.438) and (3.440)

Since we may assume that 0 < yields that



Ep,q,T (u, ρ) ≤ A1 B + A2 (L + B)( + T 1/p L) + A3 Beγ κ

−b T

D

(3.441)

for some constants A1 , A2 and A3 independent of B, , κ and T . Let κ = and then we have D = B + 1−b + D  with D =

1/p

(L + B)2 ( +

1/p 

+ L(L + B)( + + ( + L( Choosing

1/p 

1/p 

+

1/p

L)L + (L + B)( +

L+2

(1−1/s)/p 

)+( +

1/p 

1/p 

= T,

L)

)

L)1/2 L1/2 )(L + B).

∈ (0, 1) so small that D  ≤ 1, (L + B)( + T

1−b 1/p 

≤ B,

γ κ −b T = γ

L) = (L + B)( +

1−b 1/p 

≤ 1,

L) ≤ 2B

by (3.441), we have Ep,q,T (u, ρ) ≤ A1 B + 2A2 B + 3A3 eB. Thus, setting L = A1 B + 2A2 B + 3A3 eB, we finally obtain Ep,q,T (u, ρ) ≤ L.

(3.442)

Let M be a map defined by letting M(v, h) = (u, ρ), and then by (3.442) M maps UT into itself. We can also prove that M is a contraction map. Namely, choosing κ = = T smaller if necessary, we can show that for any (vi , hi ) ∈ UT (i = 1, 2), Ep,q,T (M(v1 , h1 ) − M(v2, h2 )) ≤ (1/2)Ep,q,T ((v1 , ρ1 ) − (v2 , ρ2 )). Thus, by the contraction mapping principle, we have Theorem 3.6.1, which completes the proof of Theorem 3.6.1. We finally prove the inequalities (3.405) and (3.406). Let E1 [v] be the function given in (3.421). By (3.422) and (3.221), we have vL

2(1−1/p) (Ω)) ∞ ((0,T ),Bq,p

≤ E1 [v]L

2(1−1/p) (Ω)) ∞ ((0,T ),Bq,p

≤ C{E1 [v]Lp ((0,∞),Hq2 (Ω)) + ∂t E1 [v]Lp ((0,∞),Lq (Ω)) },

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

373

which, combined with (3.417), leads to the inequality (3.405). Analogously, using E2 [Ψh ] given in (3.421) and (3.221), we have Ψh L

3−1/p−1/q (Γ )) ∞ ((0,T ),Bq,p

≤ C{Ψh Lp ((0,T ),Hq3 (Ω)) + ∂t Ψh Lp ((0,T ),Hq2 (Ω)) }, which, combined with (3.418) and (3.98), leads to the inequality in (3.406).

3.7 Global Well-Posedness in a Bounded Domain Closed to a Ball In this section, we study the global well-posedness of Eq. (3.1). As was stated in Sect. 3.3.4, we consider the problem in the following setting. Let BR be the ball of radius R centered at the origin and let SR be a sphere of radius R centered at the origin. We assume that R N ωN , where |D| denotes the Lebesgue measure of a N Lebesgue measurable set D in RN and ωN is the area of S1 . 

(A.1) |Ω| = |BR | = x dx = 0.

(A.2) Ω

(A.3) Γ is a normal perturbation of SR given by Γ = {x = y + ρ0 (y)n(y) | y ∈ SR } with a given small function ρ0 (y) defined on SR . Here, n = y/|y| is the unit outer normal to SR . n is extended to RN by n = R −1 y. Let Γt be given by Γt = {x = y + ρ(y, t)n + ξ(t) | y ∈ SR } = {x = y + R −1 ρ(y, t)y + ξ(t) | y ∈ SR },

(3.443)

where ρ(y, t) is an unknown function with ρ(y, 0) = ρ0 (y) for y ∈ SR . Let ξ(t) be the barycenter point of the domain Ωt defined by ξ(t) =

1 |Ω|

 x dx. Ωt

374

Y. Shibata

Here, by (3.9), we have |Ωt | = |Ω|. Notice that ξ(t) is also unknown. It follows from the assumption (A.2) that ξ(0) = 0. Moreover, by (3.18) we have ξ  (t) =

1 |Ω|

 v(x, t) dx.

(3.444)

Ωt

Given any function ρ defined on SR , let Hρ be a suitable extension of ρ such that Hρ = ρ on SR and (3.98) holds. We define the Hanzawa transform centered at ξ(t) by letting x = hρ (y, t) = (1 + R −1 Hρ (y, t))y + ξ(t)

for y ∈ BR .

(3.445)

In the following, we set Ψρ = R −1 Hρ y and we assume that sup Ψρ (·, t)H∞ 1 (B ) ≤ δ. R

(3.446)

t ∈(0,T )

We will choose δ so small that several conditions hold. For example, since |Ψρ (y, t) − Ψρ (y  , t)| ≤ ∇Ψρ (·, t)L∞ (BR ) |y − y  |, if 0 < δ < 1/2, then |hρ (y, t) − hρ (y  , t)| ≥ (1 − δ)|y − y  | ≥ (1/2)|y − y  | for any y, y  ∈ BR , and so the Hanzawa transform is a bijective map from BR onto Ωt with Ωt = {x = hρ (y, t) | y ∈ BR }

for t ∈ (0, T ).

(3.447)

Let v0 (x) be an initial data for Eq. (3.1), and then we set u0 (y) = v0 (hρ0 (y)), where hρ0 (y) = y + R −1 Hρ0 y. Notice that hρ0 (y) = hρ (y, 0) if ρ|t =0 = ρ0 . Let v and p satisfy Eq. (3.1) and we set u(y, t) = v(hρ (y, t), t),

q(y, t) = p(hρ (y, t), t),

p0 =

σ (N − 1) . R (3.448)

We then see from the consideration in Sect. 3.3 that u, q and ρ satisfy the following equations: ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

∂t u − Div (μD(u) − qI) = f(u, Ψρ )

in BRT ,

div u = g(u, Hρ ) = div g(u, Ψρ )

in BRT ,

˜ Ψρ ) ∂t ρ − n · P u = d(u,

on SRT ,

⎪ ⎪ 0 (μD(u)n) = h (u, Ψρ ) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ < μD(u)n, n > −q − σ Bρ = hN (u, Ψρ ) ⎪ ⎪ ⎪ ⎩ (u, ρ)|t =0 = (u0 , ρ0 )

on SRT , on SRT , in BR × SR ,

(3.449)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

375

T where BRT = B R × (0, T ), SR = SR × (0, T ), n = y/|y| for y ∈ SR , 1 N −1 Pu = u − u dy, B = ΔSR + , and ΔSR is the Laplace–Beltrami |BR | BR R2 operator on SR . Here, the functions in the right side of (3.449), f(u, Ψρ ), g(u, Ψρ ), g(u, Ψρ ), and h (u, Ψρ ), are nonlinear terms given in (3.113), (3.108), and (3.139), respectively. And, hN (u, Ψρ ) is given in the formula (3.459) of Sect. 3.7.1 below. Since dx = dy + J0 (k)dy with

⎛ ∂Ψρ1 (y,t ) ⎜ J0 (k) = det ⎜ ⎝

∂y1

.. .

··· .. .

∂ΨρN (y,t ) ∂y1

···

∂Ψρ1 (y,t ) ⎞ ∂yN

.. .

∂ΨρN (y,t ) ∂yN

⎟ ⎟ ⎠

for Ψρ = ) (Ψρ1 , . . . , ΨρN ), by (3.444) we have 1 |BR |

ξ  (t) =

 u(y, t) dy + BR

1 |BR |

 u(y, t)J0 (k) dy,

(3.450)

BR

˜ Ψρ ), where k = ∇Ψρ . Thus, by (3.132) and (3.133), we have ∂t ρ − n · P u = d(u, ˜ where d(u, Ψρ ) is given by letting ˜ Ψρ ) = d(u, Ψρ )− < u | ∇Γ ρ > + 1 d(u, |Ω|

 u(y, t)J0 (k) dy,

(3.451)

BR

with d(u, Ψρ ) given in (3.133). We now state our main result of this section. For this purpose, we make several definitions. From the assumptions (A.1) and (A.2) it follows that ρ0 should satisfy the following conditions: R N ωN = N



 dx =



Ω

0=

 |ω|=1 0



xi dx = 

=

R+ρ0 (Rω)



 r N−1 dr dω =

R+ρ0 (Rω)

Ω

|ω|=1 0

|ω|

(R + ρ0 (Rω))N+1 ωi dω N +1

|ω|=1

(R + ρ0 (Rω))N dω, N

ωi r N dr dω

for i = 1, . . . , N, and so, we have the compatibility conditions for ρ0 as follows: N  k=1

 N Ck SR

(R −1 ρ0 (y))k dω = 0,

N+1  k=1

 N+1 Ck

yi (R −1 ρ0 (y))k dω = 0

SR

(3.452)

376

Y. Shibata

N! where N Ck = k!(N−k)! and dω denotes the surface element of SR , because   S1 dω = ωN and S1 ωi dω = 0. Let Rd = {u | D(u) = 0}, and then Rd is a finite dimensional vector space spanned by constant N-vectors and first order polynomials xi ej − xj ei (i, j = 1, . . . , N), where ei are N-vector whose ith component is 1 and other components are zero. Let Md be the dimension of Rd and let p# = |BR |−1/2 e# 1 (x e − x e ) with (# = 1, . . . , N), and p# (# = N + 1, . . . , Md ) be one of cR i j j i  1 cR = (N + 2)/(2R 2 |BR |). Since (p# , pm )BR = δ#,m for #, m = 1, . . . , Md , the Md forms a orthogonal basis of Rd with respect to the L2 (BR ) inner-product set {p# }#=1 (·, ·)BR . We know that Bxi = 0 on SR for i  = 1, . . . , N. Setting ϕ1 = |SR |−1/2 and ϕ# = 2 2 cR x#−1 (# = 2, . . . , N +1) with cR = N/(R N+1 ωN ), we see that (ϕi , ϕj )SR = δij (i, j = 1, . . . , N +1), and therefore the set {ϕi }N+1 i=1 forms an orthogonal basis of the space {ψ | Bψ = 0 on SR } ∪ C with respect to the L2 (SR ) inner-product (·, ·)SR . In this section, we set

Sp,q ((a, b)) = {(u, q, ρ) | u ∈ Hp1 ((a, b), Lq (BR )N ) ∩ Lp ((a, b), Hq2(BR )N ), q ∈ Lp ((a, b), Hq1(SR )), 2−1/q

ρ ∈ Hp1 ((a, b), Wq

3−1/q

(SR )) ∩ Lp ((a, b), Wq

(SR ))}.

(3.453)

Our main result of this section is the following. Theorem 3.7.1 Let p and q be real numbers such that 2 < p < ∞, N < q < ∞ and 2/p + N/q < 1. Assume that (A.1), (A.2) and (A.3) hold. Then, there exists 2(1−1/p) a small number > 0 such that if initial data u0 ∈ Bq,p (BR ) and ρ0 ∈ 3−1/p−/q Bq,p (SR ) satisfy the smallness condition: u0 B 2(1−1p) (B q,p

R)

+ ρ0 B 3−1/p−1/q (S q,p

R)

≤ ,

(3.454)

the compatibility conditions (3.452) and div u0 = g(u0 , ρ0 ) = div g(u0 , ρ0 ) 

(μD(u0 )n)τ = g (u0 , ρ0 )

in BR , on SR ,

(3.455)

and the orthogonal condition: (v0 , ei )Ω = 0,

(v0 , xi ej − xj ei )Ω = 0,

(3.456)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

377

for i, j = 1, . . . , N, then problem (3.449) with T = ∞ admits a unique solution (u, q, ρ) ∈ Sp,q ((0, ∞)) possessing the estimate: eηt ∂t uLp ((0,∞),Lq (BR )) + eηt uLp ((0,∞),Hq2(BR )) + eηt ∇qLp ((0,∞),Lq (BR )) + eηt ∂t ρL

2−1/q (SR )) p ((0,∞),Wq

+ eηt ρL

3−1/q (SR )) p ((0,∞),Wq

≤C

for some positive constants C and η independent of .

3.7.1 Derivation of Nonlinear Term hN (u, Ψρ ) in Eq. (3.449) In this subsection, we derive the nonlinear term hN (u, Ψρ ) in (3.449). Let ω ∈ S1 be represented by ω = ω(p1 , . . . , pN−1 ) with a local coordinate (p1 , . . . , pN−1 ), and then for x = (R + ρ)ω + ξ(t) ∈ Γt , we have ∂ρ ∂x = (R + ρ)τj + ω ∂pj ∂pj ∂ω . Since τj · ω = 0, the (i, j )th component of the first fundamental where τj = ∂p j form Gt of Γt is given by

gt ij =

∂x ∂x ∂ρ ∂ρ · = (R + ρ)2 gij + , ∂pi ∂pj ∂pi ∂pj

where gij = τi · τj are the (i, j )th elements of the first fundamental form, G, of S1 , and so Gt = (R + ρ)2 (G + (R + ρ)−2 ∇p ρ ⊗ ∇p ρ) = (R + ρ)2 G(I + (R + ρ)−2 (G−1 ∇p ρ) ⊗ ∇p ρ), where ∇p ρ = ) (∂ρ/∂p1 , . . . , ∂ρ/∂pN−1 ). Since det(I + a ⊗ b ) = 1 + a · b ,

(I + a ⊗ b )−1 = I −

a ⊗ b 1 + a · b

for any N − 1 vectors a and b ∈ RN−1 , we have −2 G−1 = (R + ρ) I− t

(R + ρ)−2 (G−1 ∇p ρ) ⊗ ∇p ρ G−1 1 + (R + ρ)−2 < G−1 ∇p ρ, ∇p ρ >

= (R + ρ)−2 G−1 + O2 .

(3.457)

378

Y. Shibata

Here and in the following, O2 denotes a symbol of the form: O2 = a0 Ψρ2 +

N 

b j Ψρ

j =1

N  ∂Ψρ ∂Ψρ ∂Ψρ + cij ∂yj ∂yi ∂yj i,j =1

with some coefficients a0 , bj and cij satisfying the estimate: |(a0 , bj , cij )(y, t)| ≤ C,

|∇(a, bj , aij )(y, t)| ≤ C(|∇Ψρ (y, t)| + |∇ 2 Ψρ (y, t)|)

provided that (3.446) holds. In particular, we have gt = (R + ρ)−2 g ij + O2 , ij

componentwise. We next calculate the Christoffel symbols of Γt . Since τt i = (R + ρ)τi +

∂ρ ω, ∂pi

τt ij = (R + ρ)τij +

∂ρ ∂ρ ∂ 2ρ τi + τj + ω, ∂pj ∂pi ∂pi ∂pj

we have < τt ij , τt m > = (R + ρ)2 < τij , τm > +(R + ρ)( +

∂ρ ∂ρ ∂ρ #ij + gi# + gj # ) ∂pm ∂pj ∂pi

∂ 2 ρ ∂ρ , ∂pi ∂pj ∂pm

and so Λktij = gtkm < τt ij , τt m > =< (R + ρ)−2 g km + O2 , (R + ρ)2 < τij , τm > + (R + ρ)(

∂ρ ∂ρ ∂ 2 ρ ∂ρ ∂ρ #ij + gim + gj m )+ > ∂pm ∂pj ∂pi ∂pi ∂pj ∂pm

= Λkij + (R + ρ)−1 g km ( + ((R + ρ)−2 g km

∂ρ ∂ρ ∂ρ #ij + δik + δjk ) ∂pm ∂pj ∂pi

∂ρ ∂ 2ρ + O2 ) + O2 . ∂pm ∂pi ∂pj

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

379

Thus, ij

ΔΓt f = gt (∂i ∂j f − Λktij ∂k f ) = (R + ρ)−2 g ij (∂i ∂j f − Λkij ∂k f ) + (Ak ∇p2 ρ)∂k f + O2 ⊗ (∇¯ 2 f ) where ∇¯ 2 f is an (N − 1)2 + N vector of the form: ∇¯ 2 f = (∂i ∂j f, ∂i f, f | i, j = 1, . . . , N − 1), ∂i = ∂/∂pi , and Ak ∇p2 ρ = ((R + ρ)−4 g k# g ij

∂ρ ∂ 2ρ + O2 ) , ∂p# ∂pi ∂pj

and so H (Γt )nt = ΔΓt [(R + ρ)ω + ξ(t)] = (R + ρ)−2 g ij (∂i ∂j − Λkij ∂k )((R + ρ)ω) + (Ak ∇p2 ρ)∂k ((R + ρ)ω) + O2 ⊗ ∇¯ 2 ((R + ρ)ω) = (R + ρ)−1 g ij (∂i ∂j ω − Λkij ∂k ω) + (R + ρ)−2 g ij (∂i ρ∂j ω + ∂j ρ∂i ω) + (R + ρ)−2 g ij (∂i ∂j ρ − Λkij ∂k ρ)ω + (Ak ∇p2 ρ)(∂k ρ)ω + (Ak ∇p2 ρ)(R + ρ)∂k ω + O2 ⊗ ∇¯ 2 ρ Combining this formula with (3.122), recalling n = ω in this case and using < ∂i ω, ω >= 0, < ωτ# >= 0, ΔS1 ω = −(N − 1)ω, and (3.115) gives < H (Γt )nt , nt > = −(R + ρ)−1 (N − 1) + (R + ρ)−2 ΔS1 ρ + (O1 + O2 ) ⊗ ∇p2 ρ + O2 where O1 is a symbol of the form: O1 = a 0 ρ +

N−1  j =1

bj

∂ρ ∂pj

with coefficients a0 and bj satisfying the estimates: |(a0 , bj )(y, t)| ≤ C,

|∇(a0 , bj )(y, t)| ≤ C(|∇Ψρ (y, t)| + |∇ 2 Ψρ (y, t)|)

380

Y. Shibata

provided that (3.446) holds. Since (R + ρ)−1 = R −1 − ρR −2 + O(ρ 2 ), (R + ρ)−2 ΔS1 ρ = R −2 ΔS1 ρ + 2R −3 ρΔS1 ρ + O2 ⊗ ∇p2 ρ, we have < H (Γt )nt , nt > =−

(3.458)

N −1 + Bρ + (O1 + O2 ) ⊗ ∇p2 ρ + O2 . R

Setting p0 = −(N − 1)/R, we have < μD(u)n, n > −q − σ Bρ = hN (u, Ψρ ) on SRT . In view of (3.147), (3.150), and (3.435), hN (u, Ψρ ) may be defined by letting ˜  (∇Ψ ¯ ρ )∇Ψ ¯ ρ ⊗ ∇u + σ V ¯ ρ )∇Ψ ¯ ρ ⊗ ∇¯ 2 Ψρ , hN (u, Ψρ ) = Vh,N (∇Ψ Γ

(3.459)

¯ and V ˜  (k) ¯ are functions defined on BR × {k¯ | |k| ¯ ≤ δ} possessing where Vh,N (k) Γ the estimate: ¯ ∂ ¯ Vh,N (·, k)) ¯ H 1 (B ) ≤ C, sup (Vh,N (·, k), k ∞ R

¯ |k|≤δ

˜ Γ (·, k), ¯ ∂¯ V ¯ ˜ sup (V 1 (B ) ≤ C, k Γ (·, k))H∞ R

¯ |k|≤δ

for some constant C.

3.7.2 Local Well-Posedness In this subsection, we prove the local well-posedness of Eq. (3.449). Theorem 3.7.2 Let N < q < ∞, 2 < p < ∞ and T > 0. Assume that 2/p + N/q < 1. Then, there exists a constant > 0 depending on T such that if initial 2(1−1/p) 3−1p−1q data u0 ∈ Bq,p (BR ) and ρ0 ∈ Bq,p (SR ) satisfy the smallness condition: u0 B 2(1−1/p) (B q,p

R)

+ ρ0 B 3−1/p−1/q (S q,p

R)

≤

2

(3.460)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

381

and the compatibility condition: div u0 = g(u0 , ρ0 )

in BR ,

0 (μD(u0 )n) = μh (u0 , ρ0 )

on SR ,

(3.461)

then problem (3.449) admits unique solutions (u, q, ρ) ∈ Sp,q ((0, T )) possessing the estimates: sup Ψρ (·, t)H∞ 1 (B ) ≤ δ, R

Ep,q,T (u, ρ) ≤ .

0 0 such that the following assertion holds: Let u0 ∈ Bq,p (BR )N 3−1/p−1/q and ρ0 ∈ Bq,p (SR ) be initial data for Eq. (3.462) and let f, g, g, d, h be given functions in the right side of Eq. (3.462) with f ∈ Lp ((0, T ), Lq (BR )N ), e−γ t g ∈ Hp1 (R, Lq (BR )),

e−γ t g ∈ Hp1 (R, Hq1 (BR )) ∩ Hp (R, Lq (BR )), 1/2

2−1/q

d ∈ Lp ((0, T ), Wq

e−γ t h ∈ Hp1 (R, Hq1 (BR )N ) ∩ Hp (R, Lq (BR )N ) 1/2

(SR )),

382

Y. Shibata

for all γ ≥ γ0 . Assume that the compatibility condition: div u0 = g|t =0 in BR holds. In addition, the compatibility condition: 0 (μD(u0 )) = 0 (h|t =0 ) on Γ holds provided 2/p + 1/q < 1. Then, problem (3.462) admits unique solutions (u, q, ρ) ∈ Sp,q ((0, ∞)) possessing the estimate: ∂t uLp ((0,T )Lq (BR )) + uLp ((0,T ),Hq2 (BR )) + ∂t ρL + ρL

3−1/q (SR )) p ((0,T ),Wq

≤ Ceγ T (u0 B 2(1−1/p) (B q,p

+ fLp ((0,T ),Lq (BR )) + e

−γ t

2−1/q (SR )) p ((0,T ),Wq

+ ρ0 B 3−1/p−1/q (S

R)

q,p

(g, h)Lp (R,Hq1 (BR )) + e

+ e−γ t ∂t gLp (R,Lq (BR )) + dL

2−1/q (SR ) p ((0,T ),Wq

−γ t

R)

(g, h)H 1/2 (R,L p

q (BR ))

(3.463)

)

for any γ ≥ γ0 with some constant C independent of γ . Proof of Theorem 3.7.2 In what follows, using the Banach fixed point argument, we prove Theorem 3.7.2. Let U be the underlying space defined by U = {(u, ρ) | u ∈ Hp1 ((0, T ), Lq (BR )N ) ∩ Lp ((0, T ), Hq2 (BR )N ), 2−1/q

ρ ∈ Hq1 ((0, T ), Wq u|t =0 = u0

in BR ,

3−1/q

(SR )) ∩ Lp ((0, T ), Wq

ρ|t =0 = ρ0

(SR )),

on SR ,

sup Ψρ (·, t)H∞ 1 (B ) ≤ δ, R

Ep,q,T (u, ρ) ≤ }.

(3.464)

0 L ≤ C hL

1−1/q (SR )) ∞ ((0,T ),Wq

2−1/q (SR )) ∞ ((0,T ),Wq

≤ C 2,

¯ ∇Ψ ¯ h ⊗ ∇Ψ ¯ h>  < ξ  (t) | VΓ (k) L

1−1/q (SR )) ∞ ((0,T ),Wq

384

Y. Shibata

≤ C h2

2−1/q

L∞ ((0,T ),Wq

 < ξ  (t) | ∇Γ h > L ≤ C hL

≤ C 3,

(SR ))

2−1/q (SR )) p ((0,T ),Wq

≤ C 2,

3−1/q (SR )) p ((0,T ),Wq

¯ ∇Ψ ¯ h ⊗ ∇Ψ ¯ h>  < ξ  (t) | VΓ (k) L ≤ C (1 + hL

2−1/q (SR )) p ((0,T ),Wq

2−1/q (SR )) ∞ ((0,T ),Wq

)hL

3−1/q (SR )) p ((0,T ),Wq

≤ C 2 (1 + ).

Estimating the rest of d(v, h) given in (3.133) defined by replacing u and ρ by v and h in the similar manner to the proof of (3.408), we have d(v, Ψh )L

1−1/q (SR )) ∞ ((0,T ),Wq

≤ C 2 , d(v, Ψh )L

2−1/q (SR )) p ((0,T ),Wq

≤ C 2.

Moreover, by (3.394), (3.405), (3.406), and (3.467), we have  < v | ∇Γ h > L

1−1/q (SR )) ∞ ((0,T ),Wq

≤ CvL∞ ((0,T ),Hq1 (BR )) hL  < v | ∇Γ h > L

2−1/q (SR )) ∞ ((0,T ),Wq

2−1/q (SR )) p ((0,T ),Wq

≤ C{vLp ((0,T ),Hq2 (BR )) hL + vL∞ ((0,T ),Hq1 (BR )) hL And also,  BR

≤ C 2;

2−1/q (SR )) ∞ ((0,T ),Wq

3−1/q (SR )) p ((0,T ),Wq

} ≤ C 2.



|v(y, t)J0 (k)| dy ≤ |BR |1/q v(·, t)Lq (BR ) ∇Ψh (·, t)L∞ (BR ) ,

and so by (3.467) we have   

BR

  

BR

  vJ0 dy 

1−1/q

(SR ))

2−1/q

(SR ))

L∞ ((0,T ),Wq

  vJ0 dy 

Lp ((0,T ),Wq

≤ C 2, ≤ C 2.

Combining estimates obtained above gives ˜ Ψh ) d(v, L

1−1/q (SR )) ∞ ((0,T ),Wq

˜ Ψh ) d(v, L

2−1/q (SR )) p ((0,T ),Wq

≤ C 2, ≤ C 2.

(3.470)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

385

We now consider g(v, Ψh ) and g(v, Ψh ). Let g(v, ˜ Ψh ) and g˜ (v, Ψh ) be the extension of g(v, Ψh ) and g(v, Ψh ) to the whole time line R given in (3.426) and (3.430), respectively. By (3.427) we have e−γ t ∂t g˜ (v, Ψh )Lp (R,Lq (BR )) ≤ C{E2 [Ψh ]L∞ (R,Hq2 (BR )) e−γ t ∂t E1 [v]Lp (R,Lq (BR )) + e−γ t E2 [Ψh ]Lp (R,Hq2 (BR )) E1 [v]L∞(R,Hq1 (BR )) }, and so by (3.417), (3.418), and (3.467), we have e−γ t ∂t g˜ (v, Ψh )Lp (R,Lq (BR )) ≤ C 2 .

(3.471)

Analogously, by (3.417), (3.418), Lemmas 3.6.2, 3.6.3, (3.464), (3.405), and (3.406), we have ˜ Ψh )H 1/2 (R,L e−γ t g(v, p

q (BR ))

+ e−γ t g(v, ˜ Ψh )Lp (R,Hq1 (BR )) ≤ C 2 .

(3.472)

We next consider h (v, Ψh ). Let h˜  (v, Ψh ) be the extension of h (v, Ψh ) to the whole time line RN given by (3.433). By Lemmas 3.6.2 and 3.6.3, we have e−γ t h˜  (v, Ψh )H 1/2 (R,L p

q (BR ))

+ e−γ t h˜  (v, Ψh )Lp (R,Hq1 (Ω))

¯ 2 [Ψh ]L∞ (R,Lq (BR )) ) ¯ 2 [Ψh ]L (R,H 1 (B )) + ∂t ∇E ≤ C(∇E ∞ R q × (e−γ t E1 [v]Lp (R,Hq2 (BR )) + e−γ t ∂t E1 [v]Lp (R,Lq (BR )) ). Thus, by (3.417), (3.418), and (3.467), we have e−γ t h˜  (v, Ψh ))H 1/2 (R,L p

q (BR ))

+ e−γ t h˜  (v, Ψh )Lp (R,Hq1 (BR )) ≤ C 2 .

(3.473)

We finally consider hN (v, Ψh ) given in (3.459). Let h˜ N (v, Ψh ) be the extension of hN (v, Ψh ) to the whole time interval R given by ¯ 2 [Ψh ])(∇E ¯ 2 [Ψh ], ∇E1 [v]) h˜ N (v, Ψh ) = Vh,N (∇E ˜  (∇E ¯ 2 [Ψh ])(∇E ¯ 2 [Ψh ], ∇¯ 2 E2 [Ψh ]). + σV Γ By Lemmas 3.6.2 and 3.6.3, we have e−γ t h˜ N (v, Ψh )H 1/2 (R,L p

q (BR ))

+ e−γ t h˜ N (v, Ψh )Lp (R,Hq1 (BR ))

¯ 2 [Ψh ]L (R,H 1 (B )) ) ¯ 2 [Ψh ]L∞ (R,Lq (BR )) + ∇E ≤ C(∂t ∇E ∞ R q

(3.474)

386

Y. Shibata

× (e−γ t ∂t E1 [v]Lp (R,Lq (BR )) + e−γ t E1 [v]Lp (R,Hq2 (BR )) + e−γ t ∂t E2 [Ψh ]Lp (R,Hq2 (BR )) + e−γ t E2 [Ψh ]Lp (R,Hq3 (BR )) ). Thus, by (3.417), (3.418), and (3.467), we have ˜ Ψh ) 1/2 e−γ t h(v, H ((R,L p

q (BR ))

˜ Ψh )L ((R,H 1 (B )) ≤ C 2 . + e−γ t h(v, p R q

(3.475)

Applying Theorem 3.7.3 to Eq. (3.466) and using (3.469)–(3.473), and (3.475), we have uLp ((0,T ),Hq2 (BR )) + ∂t uLp ((0,T ),Lq (BR )) + ρL

3−1/q (SR )) p ((0,T ),Wq

+ ∂t ρL

2−1/q (SR )) p ((0,T ),Wq

≤ Ceγ T

2

(3.476)

for some positive constants C and γ . Moreover, by the third equation in (3.466), we have ∂t ρL

1−1/q (SR )) ∞ ((0,T ),Wq

˜ Ψh ) ≤ C(uL∞ ((0,T ),Hq1 (BR )) + d(v, L

(3.477) 1−1/q (SR )) ∞ ((0,T ),Wq

),

which, combined with (3.405), (3.406), and (3.470), yields ∂t ρL

1−1/q (SR )) ∞ ((0,T ),Wq

≤ C(u0 B 2(1−1/p) (B q,p

R ))

+ uLp ((0,T ),Hq2 (BR )) + ∂t uLp ((0,T ),Lq (BR )) + C 2 ). Combining this inequality with (3.476) gives Ep,q,T (u, ρ) ≤ Ceγ T

2

(3.478)

for some positive constants C and γ . Let P be a map defined by P(v, h) = (u, ρ), and then choosing so small that Ceγ T ≤ 1, by (3.478) we see that Ep,q,T (P(v, h)) ≤ , which shows that P maps U into itself. Let (vi , hi ) (i = 1, 2) be any two elements of U , and then we have Ep,q,T (P(v1, h1 ) − P(v2 , h2 )) ≤ Ceγ T Ep,q,T ((v1 , h1 ) − (v2 , h2 )) for some positive constants C and γ , where we have used Ep,q,T ((vi , ρi )) ≤ for i = 1, 2. Choosing so small that Ceγ T ≤ 1/2, we see that P is a contraction map from U into itself, and so by the Banach fixed point theorem, there exists a unique (u, ρ) ∈ U satisfying P(u, ρ) = (u, ρ). Thus, (u, ρ) is a required unique solution of Eq. (3.449), which completes the proof of Theorem 3.7.2.

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

387

3.7.3 Decay Estimates of Solutions for the Linearized Equations To prove Theorem 3.7.1 the key tool is decay properties of solutions of the Stokes equations: ⎧ ⎪ ∂t u − Div (μD(u) − pI) = f ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ div u = g = div g ⎪ ⎪ ⎨ ∂t ρ − n · P u = d ⎪ ⎪ ⎪ ⎪ ⎪ (μD(u) − pI)n − σ (Bρ)n = h ⎪ ⎪ ⎪ ⎪ ⎩ (u, ρ)|t =0 = (u0 , ρ0 )

in BRT , in BRT , on SRT ,

(3.479)

on SRT , in BR × SR .

Here and in the following, n = y/|y| ∈ S1 . We will prove the following theorem. N+1 Theorem 3.7.4 Let 1 < p, q < ∞ and 2/p + 1/q = 0. Let {p# }M #=1 and {ϕj }j =1 be orthogonal basis of Rd = {u | D(u) = 0} and {ψ | Bψ = 0 on SR } ∪ C, which are given before Theorem 3.7.1. Then, there exists an η > 0 such that the 2(1−1/p) 3−1/p−1/q following assertion holds: Let u0 ∈ Bq,p (BR )N and ρ0 ∈ Bq,p (SR ) be initial data for Eq. (3.479) and let f, g, g, d, h be given functions in the right side of Eq. (3.479). Assume that

f ∈ Lp ((0, T ), Lq (BR )N ),

2−1/q

d ∈ Lp ((0, T ), Wq

(SR )),

and that there exist g˜η , g˜ η and h˜ η such that eηt g = g˜ η , eηt g = g˜ η , eηt h = h˜ η for t ∈ (0, T ), div g˜ η = g˜η for t ∈ R, and 1/2

g˜η ∈ Lp (R, Hq1 (BR )) ∩ Hp (R, Lq (BR )),

g˜ η ∈ Hp1 (R, Lq (BR )),

1/2 h˜ η ∈ Lp (R, Hq1 (BR )N ) ∩ Hp (R, Lq (BR )N ).

Assume that the compatibility condition: div u0 = g|t =0 in BR holds. In addition, the compatibility condition: (μD(u0 )n)τ = hτ |t =0 on Γ holds provided 2/p + 1/q < 1. Then, problem (3.479) admits unique solutions (u, p, ρ) ∈ Sp,q ((0, T )) possessing the estimate:  Ip,q,T (u, ρ; η) ≤ C Jp,q,T (u0 , ρ0 , f, g, g, d, h; η) +

M   #=1

+

T 0

(eηs |(u(·, s), p# )BR |)p ds

N+1   T j =1

0

1/p

(eηs |(ρ(·, s), ϕj )SR |)p ds

1/p 

(3.480)

388

Y. Shibata

for some constant C independent of η. Here and in the following, we set Ip,q,T (u, ρ; η) = eηt uLp ((0,T ),Hq2 (BR )) + eηt ∂t uLp ((0,T ),Lq (BR )) + eηt ρL

3−1/q (SR )) p ((0,T ),Wq

+ eηt ∂t ρL

2−1/q (SR )) p ((0,T ),Wq

Jp,q,T (u0 , ρ0 , f, g, g, d, h; η) = u0 B 2(1−1/p) (B q,p

+ e fLp ((0,T ),Lq (BR )) + e dL ηt

ηt

+ (gη , hη )H 1/2 (R,L p

q (BR ))

R)

;

+ ρ0 B 3−1/p−1/q (S q,p

2−1/q (SR )) p ((0,T ),Wq

R)

+ ∂t gη Lp (R,Lq (BR ))

+ (gη , hη )Lp (R,Hq1 (BR )) .

To prove Theorem 3.7.3, we first consider the following shifted equations: ⎧ ⎪ ∂t u1 + λ1 u1 − Div (μD(u1 ) − p1 I) = f in BRT , ⎪ ⎪ ⎪ ⎪ ⎪ in BRT , ⎪ ⎨ div u1 = g = div g ∂t ρ1 + λ1 ρ1 − n · P u1 = d ⎪ ⎪ ⎪ ⎪ (μD(u1 ) − p1 I)n − σ (Bρ1 )n = h ⎪ ⎪ ⎪ ⎩ (u , ρ )| 1 1 t =0 = (u0 , ρ0 )

on SRT , on

(3.481)

SRT ,

on BR × SR .

For the shifted equation (3.481), we have Theorem 3.7.5 Let 1 < p, q < ∞ and T > 0. Assume that 2/p + 1/q = 1. 2(1−1/p) Let λ0 be a constant given in Theorem 3.4.8. Let u0 ∈ Bq,p (BR )N and ρ0 ∈ 3−1/p−1/q Bq,p (SR ) be initial data for Eq. (3.481) and let f, g, g, d, h be given functions in the right side of Eq. (3.481) satisfying the same condition as in Theorem 3.7.4. Moreover, there exist g˜0 , g˜ 0 and h˜ 0 such that g = g˜0 , g = g˜ 0 and h = h˜ 0 for t ∈ (0, T ), div g˜ 0 = g˜0 for t ∈ R, and 1/2

g˜0 ∈ Lp (R, Hq1 (BR )) ∩ Hp (R, Lq (BR )),

g˜ 0 ∈ Hp1 (R, Lq (BR )),

1/2 h˜ 0 ∈ Lp (R, Hq1 (BR )N ) ∩ Hp (R, Lq (BR )N ).

Assume that the compatibility condition: div u0 = g|t =0 in BR holds. In addition, the compatibility condition: (μD(u0 )n)τ = hτ |t =0 on Γ holds provided 2/p + 1/q < 1. Then, for any λ1 > λ0 , problem (3.481) admits unique solutions (u1 , p1 , ρ1 ) ∈ Sp,q ((0, T )) possessing the estimate: Ip,q,T (u1 , ρ1 ; 0) ≤ CJp,q,T (u0 , ρ0 , f, g, g, d, h; 0) for some constant C independent of T . Proof Let A2 (λ), P2 (λ) and H2 (λ) be operators given in Theorem 3.4.8. Let λ1 and λ2 be numbers for which λ1 − λ0 > λ2 > 0. Then, for any λ ∈ −λ2 + Σ we

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

389

have λ + λ1 ∈ λ0 + Σ , and so by Theorem 3.4.8 we have RL(X

2−j (Ω)N ) q (Ω),Hq

({(τ ∂τ )# (λj/2 A2 (λ + λ1 )) | λ ∈ −λ2 + Σ 0 }) ≤ rb ;

RL(Xq (Ω),Lq (Ω)N ) ({(τ ∂τ )# (∇P2 (λ + λ1 )) | λ ∈ −λ2 + Σ 0 }) ≤ rb ; RL(Xq (Ω),W 3−k (Γ )) ({(τ ∂τ )# (λk H2 (λ + λ1 )) | λ ∈ −λ2 + Σ 0 }) ≤ rb . q

for # = 0, 1, j = 0, 1, 2, and k = 0, 1. Thus, we can choose γ = 0 in the argument on Sect. 3.4.6, and so we have the theorem. 

For any η > 0, eηt u1 , eηt p1 and eηt ρ1 satisfy the equations: ∂t (eηt u1 ) + (λ1 − η)eηt u1 − Div (μD(eηt u1 ) − eηt p1 I) = eηt f

8

div eηt u1 = eηt g = div eηt g 8 ∂t (eηt ρ1 ) + (λ1 − η)eηt ρ1 − n · P (eηt u1 ) = eηt d (μD(eηt u1 ) − eηt p1 I)n) − σ (B(eηt ρ1 ))n = eηt h (eηt u1 , eηt ρ1 )|t =0 = (u0 , ρ0 )

in BRT , on SRT ,

on BR × SR .

Given η > 0, we choose λ1 > 0 in such a way that λ1 − λ0 > η > 0, and then by Theorem 3.7.5, we have the following corollary. Corollary 3.7.6 Let 1 < p, q < ∞, T > 0 and η > 0. Assume that 2/p+1/q = 1. 2(1−1/p) 3−1/p−1/q Let u0 ∈ Bq,p (BR )N and ρ0 ∈ Bq,p (SR ) be initial data for Eq. (3.481) and let f, g, g, d, h be given functions in the right side of Eq. (3.481) satisfying the same conditions as in Theorem 3.7.4. Assume that the compatibility condition: div u0 = g|t =0 in BR holds. In addition, the compatibility condition: (μD(u0 )n)τ = hτ |t =0 on Γ holds provided 2/p + 1/q < 1. Then, there exists a λ1 > 0 such that problem (3.481) admits unique solutions (u1 , p1 , ρ1 ) ∈ Sp,q ((0, T )) possessing the estimate: Ip,q,T (u1 , ρ1 ; η) ≤ CJp,q,T (u0 , ρ0 , f, g, g, d, h; η)

(3.482)

for some constant C. We consider solutions u, p and ρ of problem (3.479) of the form: u = u1 + v, p = p1 + q and ρ = ρ1 + h, where u1 , p1 and ρ1 are solutions of the shifted equations (3.481), and then v, q and h should satisfy the equations: ⎧ ⎪ ⎪ ∂t v − Div (μD(v) − qI) = −λ1 u1 , ⎪ ⎪ ⎨ ∂ h − n · P v = −λ ρ t 1 1 ⎪ (μD(v) − qI)n − σ (Bh)n = 0 ⎪ ⎪ ⎪ ⎩ (v, h)|t =0 = (0, 0)

div v = 0

in BRT , on SRT , on SRT , on BR × SR .

(3.483)

390

Y. Shibata

Recall the definition of Jq (Ω) given in (3.163) in Sect. 3.4.5, that is Jq (BR ) = {f ∈ Lq (BR )N | (f, ∇ϕ)BR = 0

for any ϕ ∈ Hˆ q1 ,0 (BR )}.

Recall that 1 Hq,0 (BR ) = {ϕ ∈ Hq1 (BR ) | ϕ|SR = 0}, 1 (BR ) = {ϕ ∈ Lq,loc (BR ) | ∇ϕ ∈ Lq (BR )N , ϕ|SR = 0}. Hˆ q,0

Since C0∞ (BR ) is dense in Hˆ q1 ,0 (BR ), the necessary and sufficient condition in order 1 (B ) be a solution of the that u ∈ Jq (BR ) is that div u = 0 in BR . Let ψ ∈ Hq,0 R variational equation: (∇ψ, ∇ϕ)BR = (u1 , ∇ϕ)BR

for any ϕ ∈ Hq1 ,0 (BR ),

(3.484)

and let w = u1 − ∇ψ. Then, w ∈ Jq (BR ) and wLq (BR ) + ψHq1 (BR ) ≤ Cu1 Lq (BR ) .

(3.485)

Using w and ψ , we can rewrite the first equation in (3.483) as follows: ∂t v − Div (μD(v) − (q + λ1 ψ)I) = −λ1 w,

div v = 0 in BRT .

Thus, in what follows we may assume that u1 ∈ Hp1 ((0, T ), Jq (BR )) ∩ Lp ((0, T ), Hq2 (BR )N ).

(3.486)

According to the argument in Sect. 3.4.3, we introduce a functional P (v, h) ∈ 1 (B ) that is a unique solution of the weak Dirichlet problem Hq1 (BR ) + Hˆ q,0 R (∇P (v, h), ∇ϕ)BR = (Div (μD(v)) − ∇div v, ∇ϕ)BR

(3.487)

for any ϕ ∈ Hˆ q1 ,0 (BR ), subject to P (v, h) = μ < D(v)n, n > −σ (Bh) − div v

on SR .

(3.488)

And then, to handle problem (3.483) in the semigroup setting, we consider the initial value problem: ⎧ ⎪ ∂t v − Div (μD(v) − P (v, h)I) = 0 ⎪ ⎪ ⎪ ⎨ ∂ h − n · Pv = 0 t ⎪ (μD(v) − qI)n − σ (Bh)n = 0 ⎪ ⎪ ⎪ ⎩ (v, h)|t =0 = (v0 , ρ0 )

in BR × (0, ∞), on SR × (0, ∞), on SR × (0, ∞), on BR × SR .

(3.489)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

391

Note that (μD(v) − P (v, h)I)n − σ (Bh)n = 0 on SR × (0, ∞) is equivalent to (D(v)n)τ = 0,

div v = 0

on SR × (0, ∞).

(3.490)

Defining Hq (BR ), Dq (BR ) and Aq (v, h) by 2−1/q

Hq (BR ) = {(v, h) | v ∈ Jq (BR ),

h ∈ Wq

(SR )}, 3−1/q

Dq (BR ) = {(v, h) ∈ Hq (BR ) | v ∈ Hq2 (BR )N , h ∈ Wq

(SR ),

(D(v)n)τ = 0 on SR }, Aq (v, h) = (Div (D(v) − P (v, h)I), −n · P v)

for (v, h) ∈ Dq (BR ), (3.491)

we see that Eq. (3.489) is formulated by ∂t U = Aq U

(t > 0),

U |t =0 = U0

(3.492)

with U = (v, h) ∈ Dq (BR ) for t > 0 and U0 = (u0 , ρ0 ) ∈ Hq (BR ). According to Theorem 3.4.17, we see that Aq generates a C 0 semigroup {T (t)}t ≥0 on Hq (BR ). Moreover, if we define J˙q (BR ) = {f ∈ Jq (BR ) | (f, p# )BR = 0 (# = 1, . . . , M)}; W˙ q# (SR ) = {g ∈ Wq# (SR ) | (g, ϕj )SR = 0 (j = 1, . . . , N + 1)}; H˙ q (BR ) = {(f, g) | f ∈ J˙q (BR ),

g ∈ W˙ q

2−1/q

(SR )};

(3.493)

(f, g)Hq = fLq (BR ) + gW 2−1/q (S ) ; q

R

(v, h)Dq = vHq2 (BR ) + hW 3−1/q (S ) , q

R

then we have Theorem 3.7.7 Let 1 < q < ∞. Then, {T (t)}t ≥0 is exponentially stable on H˙ q , that is T (t)(f, g)Hq ≤ Ce−η1 t (f, g)Hq

(3.494)

for any t > 0 and (f, g) ∈ H˙ q (BR ) with some positive constants C and η1 . Postponing the proof of Theorem 3.7.7 to the next section, we continue to prove Theorem 3.7.3. Let u˜ 1 = u1 −

M  (u1 (·, t), p# )BR p# , #−1

ρ˜1 = ρ1 −

N+1  j =1

(ρ1 (·, t), ϕj )SR ϕj ,

392

Y. Shibata

and then (u˜ 1 , p# )BR = 0 (# = 1, . . . , M) and (ρ˜1 , ϕj )SR = 0 (j = 1, . . . , N + 1). Moreover, since div p# = 0, by (3.486) we have u˜ 1 ∈ Lp ((0, T ), J˙q (BR )). Let ˜ (˜v, h)(·, s) =



s

T (s − r)(−λ1 u˜ 1 (·, r), −λ1 ρ˜1 (·, r)) dr,

0

and then by the Duhamel principle, v˜ and h˜ satisfy the equations: ⎧ ˜ = −λ1 u˜ 1 , ⎪ ∂t v˜ − Div (μD(˜v) − P (˜v, h)I) ⎪ ⎪ ⎪ ⎨ ∂ h˜ − n · P v˜ = −λ ρ˜ t 1 1 ⎪ ˜ ˜ =0 (μD(˜v) − P (˜v, h)I)n − σ (B h)n ⎪ ⎪ ⎪ ⎩ ˜ t =0 = (0, 0) on BR × SR . (˜v, h)|

div v˜ = 0 in BRT , on SRT , on SRT ,

(3.495)

By (3.494), ˜ (˜v, h)(·, s)Hq ≤ C ≤C



s



s 0

e−η1 (s−r)(u˜ 1 (·, r), ρ˜1 (·, r))Hq dr

1/p 

e−η1 (s−r) dr

0

s 0

e−η1 (s−r)(u˜ 1 (·, r), ρ˜1 (·, r))Hq dr p

1/p .

Choosing η > 0 smaller if necessary, we may assume that 0 < ηp < η1 without loss of generality. Thus, by the inequality above we have 

t 0

˜ (eηs (˜v, h)(·, s)Hq )p ds

≤C

 t  0

=

0

 t  

0

= 0

s 0

t

s

p eηsp e−η1 (s−r)(u1 (·, r), ρ1 (·, r))Hq dr ds

e−(η1 −pη)(s−r)(eηr (u1 (·, r), ρ1 (·, r))Hq )p dr ds

(eηr (u1 (·, r), ρ1 (·, r))Hq )p

≤ (η1 − pη)−1

 0

T



t

e−(η1 −pη)(s−r) ds dr

r

(eηr (u1 (·, r), ρ1 (·, r))Hq )p dr,

which, combined with (3.482), leads to ˜ L ((0,t ),H ) ≤ CJp,q,T (u0 , ρ0 , f, g, g, d, h; η) eηs (˜v, h) p q

(3.496)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

393

˜ and k˜ satisfy the shifted equations: for any t ∈ (0, T ). If w ⎧ ˜ = f, ⎪ ˜ + λ1 w ˜ − Div (μD(w) ˜ − P (w, ˜ k)I) ∂t w ⎪ ⎪ ⎪ ⎨ ∂ k˜ + λ k˜ − n · P w ˜ =d t 1 ⎪ ˜ ˜ =0 ˜ − P (w, ˜ k)I)n − σ (B k)n (μD(w) ⎪ ⎪ ⎪ ⎩ ˜ t =0 = (0, 0) on BR × SR , ˜ k)| (w,

˜ =0 div w

in BRT , on SRT , on SRT ,

where we have set ˜ d = −λ1 ρ˜1 + λ1 h,

f = −λ1 u˜ 1 + λ1 v˜ , by (3.482) and (3.496), we have

˜ η) ≤ CJp,q,T (u0 , ρ0 , f, fd , fd , g, h; η). ˜ k; Ip,q,T (w, ˜ = v˜ But, noting that v˜ and h˜ satisfy Eq. (3.495), by the uniqueness we see that w and k˜ = h˜ for t ∈ (0, T ), and so we have ˜ η) ≤ CJp,q,T (u0 , ρ0 , f, fd , fd , g, h; η). Ip,q,T (˜v, h; Let v = v˜ − λ1

M  

t

#=1 0

h = h˜ − λ1

(u1 (·, s), p# )BR ds p# ,

N+1  t j =1

0

(ρ1 (·, s), ϕj )SR ds ϕj .

In this case, ˜ + λ1 (N − 1)R −2 σ P (v, h) = P (˜v, h)



t 0

(ρ1 (·, s), ϕ1 )SR ds ϕ1 .

In fact, letting C = λ1 (N − 1)R −2 σ



t 0

(ρ1 (·, s), ϕ1 )SR ds ϕ1

for notational simplicity, we have ˜ + C)), ∇ψ)BR = 0 (∇(P (v, h) − (P (˜v, h)

(3.497)

394

Y. Shibata

for any ψ ∈ Hˆ q1 ,0 (BR ), because ∇ϕ1 = 0. Moreover, on SR we have ˜ + C) = 0 P (v, h) − (P (˜v, h) because D(p# ) = 0, div p# = 0, and Bh = B h˜ − (N − 1)R −2 λ1



t 0

(ρ1 (·, s), ϕ1 )SR ds ϕ1 .

˜ + C. Thus, we have P (v, h) = P (˜v, h) By (3.495), we have ∂t v − Div (μD(v) − P (v, h)I) = −λ1 u1 ,

div v = 0

(μD(v) − P (v, h)I)n − σ (Bh)n = 0

in BRT , on SRT .

Recall that p# · n|SR = 0 for # = N + 1, . . . , M. Moreover, recalling that p# = |BR |−1 e# (# = 1, . . . , N), we have 

−1 −1 P p# = |BR | e# dy = 0, e# − |BR | BR

and therefore, ∂t h − n · P v = ∂t h˜ − n · P v˜ − λ1

N+1 

(ρ1 (·, t), ϕj )SR ϕj = −λ1 ρ1

on SRT .

j =1

Summing up, we have proved that v, q = P (v, h), and h satisfy the Eq. (3.483). By (3.497), we have eηt ∂t (v, h)Lp ((0,T ),Hq ) ≤ CJp,q,T (u0 , ρ0 , f, fd , fd , g, h; η)

(3.498)

for any t ∈ (0, T ). To estimate eηt (v, h)Lp ((0,T ),Dq ) , we use the following lemma. 3−1/q

Lemma 3.7.8 Let 1 < q < ∞. Let u ∈ Hq2 (BR )N ∩ Jq (BR ) and ρ ∈ Wq satisfy the equations: ⎧ ⎪ ⎪ ⎨ −Div (μD(u) − P (u, ρ)I) = f

n · Pu = g ⎪ ⎪ ⎩ (μD(u) − P (u, ρ)I)n − σ (Bρ)n = 0

(SR )

in BR , on SR , on SR .

(3.499)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

395

Then, there exists a constant C > 0 such that M N+1     |(u, p# )BR | + |(ρ, ϕj )SR | . (u, ρ)Dq ≤ C (f, g)Hq +

(3.500)

j =1

#=1

Postponing the proof of Lemma 3.7.8 to the next section, we continue to prove Theorem 3.7.3. By (3.483), v and h satisfy the elliptic equations: − Div (μD(v) − P (v, h)I) = −λ1 u1 − ∂t v,

div v = 0

in BR ,

n · P v = λ1 ρ1 + ∂t h

on SR ,

(μD(v − P (v, h)I)n − σ (Bh)n = 0

on SR ,

and therefore, applying Lemma 3.7.8 and using (3.498) yield that eηt vLp ((0,T ),Hq2 (BR )) + eηt hL ((0,T ),W 3−1/q (S )) p R q  ≤ C Jp,q,T (u0 , ρ0 , f, fd , fd , g, h; η) +

M   #=1

+

T 0

M   j =1

T 0

(eηs |(v(·, s), p# )BR |)p ds

(eηs |(h(·, s), ϕj )SR |)p ds

1/p

(3.501)

1/p  .

Let u = u1 + v, p = p1 + q and ρ = ρ1 + h. By (3.481) and (3.483), u, p and ρ satisfy the equations (3.479). Since 

T 0

≤

(eηs |(v(·, s), p# )BR |)p ds 

T 0



T 0

≤

(eηs |(u(·, s), p# )BR |)p ds

(e |(h(·, s), ϕj )SR |) ds ηs



p

T 0

1/p

1/p

+ CJη,T , (3.502)

1/p

(eηs |(ρ(·, s), ϕj )SR |)p ds

1/p

+ CJη,T ,

where we have set Jη,T = Jp,q,T (u0 , ρ0 , f, fd , fd , g, h; η), as follows from (3.482), by (3.498), (3.501), and (3.502), we see that u, p and ρ satisfy the inequality (3.480).

396

Y. Shibata

3.7.4 Exponential Stability of Continuous Analytic Semigroup Associated with Eq. (3.489) In this subsection, we shall prove Theorem 3.7.7 stated in the previous subsection. For this purpose, we consider the equations: (λI − Aq )U = F

(3.503)

for F = (f, g) ∈ J˙q (BR ) and U = (v, h) ∈ Dq (BR )∩J˙q (BR ), which is the resolvent problem corresponding to Eq. (3.492). Here, I is the identity operator, J˙q (BR ) is the space defined in (3.493), and Dq (BR ) and Aq are the domain and the operator defined in (3.491). Since R boundedness implies the usual boundedness of operator families, by Theorem 3.4.8 and the observation in Sect. 3.4.3, we have the following theorem. Theorem 3.7.9 Let 1 < q < ∞ and 0 < 0 < π/2. Then, there exists a λ0 > 0 such that for any λ ∈ Σ 0 ,λ0 and F ∈ Hq (BR ), Eq. (3.503) admits a unique solution U ∈ Dq (BR ) possessing the estimate: |λ|U Hq + U Dq ≤ CF Hq

(3.504)

for some constant C > 0. Here, Hq (BR ) is the space defined in (3.491). Our task in this subsection is to prove the following theorem. Theorem 3.7.10 Let 1 < q < ∞ and let C+ = {λ ∈ C | Re λ ≥ 0}. Then, for any λ ∈ C+ and F ∈ H˙ q (BR ), Eq. (3.503) admits a unique solution U ∈ Dq (BR ) ∩ H˙ q (BR ) possessing the estimate (3.504). In the following, we shall prove Theorem 3.7.10. We start with the following lemma. Lemma 3.7.11 Let 1 < q < ∞, let λ ∈ C \ (−∞, 0) and F = (f, g) ∈ H˙ q (BR ). If U = (v, h) ∈ Dq (BR ) satisfies Eq. (3.503), then U = (v, h) belongs to H˙ q (BR ). Proof First we prove that v ∈ J˙q (BR ). By (3.503), we know that v ∈ Jq (BR ) ∩ Hq2 (BR )N satisfies the equations: λv − Div (μD(v) − P (v, h)I) = f,

div v = 0

in BR ,

(μD(v) − P (v, q)I)n − σ (Bh)n = 0

on SR .

Since F = (f, g) ∈ H˙ q (BR ), we know that f ∈ Jq (BR ) and (f, p# )BR = 0 for # = 1, . . . , M, and so by the divergence theorem of Gauß we have 0 = (f, p# )BR = (λv − Div (μ(D(v) − P (v, h)I), p# )BR μ = λ(v, p# )BR − σ (Bh, n · p# )SR + (D(v), D(p# ))BR − (P (v, h), div p# )BR . 2

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

397

We see that (Bh, n · p# )SR = 0

(# = 1, . . . , M).

(3.505)

In fact, recalling that p# = |BR |−1 e# (# = 1, . . . , N) and n = y/|y| ∈ S1 , we have (Bh, n · p# )SR = R −1 |BR |−1 (h, By# )SR = R −3 |BR |−1 (h, (N − 1 + ΔS1 )y# )SR = 0 for # = 1, . . . , N. Moreover, p# · n = 0 for # = N + 1, . . . , M because p# (# = N + 1, . . . , M) are equal to cij (xi ej − xj ei ) for some i and j and constant cij , and therefore (Bh, n · p# )SR = 0 for # = N + 1, . . . , M. Since D(p# ) = 0 and div p# = 0, we have λ(v, p# )BR = 0, which, combined with λ = 0, leads to (v, p# )BR = 0, that is, v ∈ J˙q (BR ) ∩ Dq (BR ). 3−1/q 3−1/q (SR ). We know that h ∈ Wq (SR ) and g Next, we prove that h ∈ W˙ q satisfies the equation: λh − n · P v = g

on SR .

Since (g, ϕj )SR = 0, by the divergence theorem of Gauß we have 0 = (g, ϕj )SR = λ(h, ϕj )SR − (P v · n, ϕj )SR  = λ(h, ϕj )SR − div ((P v)ϕj ) dx. BR

Since div P v = div v = 0 and since ∂# ϕj are constants, we have  div ((P v)ϕj ) dx BR

 (div (P v))ϕj dx +

= BR

N+1  #=1

 v# − |BR |−1 BR

v# dy (∂# ϕj ) dx BR

= 0, (3.506) Thus, λ(h, ϕj )SR = 0, which, combined with λ = 0, leads to (h, ϕj )SR = 0. 3−1/q (SR ). This completes the proof of Lemma 3.7.11. Therefore, we have h ∈ W˙ q 

Combining Theorem 3.7.9 and Lemma 3.7.11, we have the following Corollary.

398

Y. Shibata

Corollary 3.7.12 Let 1 < q < ∞ and 0 < 0 < π/2. Then, there exists a positive constant λ0 such that for any λ ∈ Σ 0 ,λ0 and (f, g) ∈ H˙ q (BR ), Eq. (3.503) admits a unique solution (v, h) ∈ Dq ∩ H˙ q (BR ) possessing the estimates (3.504). In view of Corollary 3.7.12, in order to prove Theorem 3.7.10 it suffices to prove the following theorem. Theorem 3.7.13 Let 1 < q < ∞ and let λ0 be the same positive number as in Corollary 3.7.12. Let Qλ0 = {λ ∈ C | Re λ ≥ 0, |λ| ≤ λ0 }. Then, for any λ ∈ Qλ0 and (f, g) ∈ H˙ q (BR ), Eq. (3.503) admits a unique solution (v, h) ∈ Dq (BR ) ∩ H˙ q (BR ) possessing the estimate: (v, h)Dq ≤ C(f, g)Hq

(3.507)

with some constant C independent of λ ∈ Qλ0 . Proof We write D˙ q = Dq (BR )∩ H˙ q (BR ) for the sake of simplicity. We first observe that Aq D˙ q ⊂ H˙ q (BR ).

(3.508)

In fact, for (v, h) ∈ D˙ q , we set Aq (v, h) = (f, g), that is Div (μ(D(v) − P (v, h)I) = f in BR and n · P v = g on SR . For any ϕ ∈ Hˆ q1 ,0 (BR ), by (3.487) and the fact that div v = 0, we have (f, ∇ϕ)BR = (Div (μD(v) − P (v, h)I), ∇ϕ)BR = (∇div v, ∇ϕ)BR = 0, which implies f ∈ Jq (BR ). Next, we observe that (f, p# )BR = (Div (μD(v) − P (v, h)I), p# )BR = σ (Bh, n · p# )SR −

μ (D(v), D(p# ))BR + (P (v, h), div p# )BR . 2

Thus, by (3.505) and the facts that D(p# ) = 0 and div p# = 0, we have (f, p# )BR = 0, and so, f ∈ J˙q (BR ). Finally, by (3.506) we have  (g, ϕj )SR = (n · P v, ϕj )SR = which leads to g ∈ W˙ q

3−1/q

div ((P v)ϕj ) dx = 0, BR

(SR ). This completes the proof of (3.508).

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

399

In view of Corollary 3.7.12, (λ0 I − Aq )−1 exists as a bounded linear operator from H˙ q (BR ) onto D˙ q , and then, the equation (3.503) is rewritten as (f, g) = (λI − Aq )(v, h) = (λ − λ0 )(v, h) + (λ0 I − Aq )(v, h) = (I + (λ − λ0 )(λ0 I − Aq )−1 )(λ0 I − Aq )(v, h). If (I + (λ − λ0 )(λ0 I − Aq )−1 )−1 exists as a bounded linear operator from H˙ q (BR ) into itself, then we have (v, h) = (λ0 I − Aq )−1 (I + (λ − λ0 )(λ0 I − Aq )−1 )−1 (f, g).

(3.509)

Thus, our task is to prove the existence of the inverse operator (I + (λ − 3−1/q λ0 )(λ0 I − Aq )−1 )−1 . Since Hq2 (BR )N and Wq (SR ) are compactly embedded 2−1/q

into Lq (BR )N and Wq (SR ), respectively, as follows from the Rellich compact embedding theorem, (λ0 I − Aq )−1 is a compact operator from H˙ q into itself. Thus, in view of Riesz-Schauder theory, in order to prove the existence of the inverse operator (I + (λ − λ0 )(λ0 I − Aq )−1 )−1 it suffices to prove that the kernel of the map I + (λ − λ0 )(λ0 I − Aq )−1 is trivial. Thus, let (f, g) be an element in H˙ q (BR ) such that (I + (λ − λ0 )(λ0 I − Aq )−1 )(f, g) = (0, 0).

(3.510)

Our task is to prove that (f, g) = (0, 0). Since (f, g) = −(λ − λ0 )(λ0 I − Aq )−1 (f, g) ∈ D˙ q , we have (λ0 I − Aq )(f, g) = −(λ − λ0 )(f, g), and so, (f, g) ∈ D˙ q satisfies the homogeneous equation: (λI − Aq )(f, g) = (0, 0).

(3.511)

Namely, (f, g) ∈ D˙ q satisfies the homogeneous equations: ⎧ ⎪ ⎪ ⎨

λf − Div (μD(f) − P (f, g)I) = 0

in BR ,

λg − n · P f = 0 ⎪ ⎪ ⎩(μD(f) − P (f, g)I)n − σ (Bg)n = 0

on SR ,

(3.512)

on SR .

First we consider the case where 2 ≤ q < ∞. Since (f, g) ∈ D˙ q ⊂ D˙ 2 , by (3.512) and the divergence theorem of Gauß, we have 0 = (λf − Div (D(f) − P (f, g)I), f)BR μ = λf2L2 (BR ) − σ (Bg, n · f)SR + D(f)2L2 (BR ) − (P (f, g), div f)BR . 2

400

Y. Shibata

For h ∈ Hq2 (SR ) and g = ) (g1 , . . . , gN ), we have (Bh, P g · n)SR = (Bh, g · n)SR ,

(3.513)

because recalling n = y/|y| ∈ S1 , we have N 

|BR |

−1

BR

j =1

=



N 

|BR |

j =1

g# dyR −1 (Bh, y# )SR

−1

 BR

g# dy R −1 (h, R −2 (N − 1 + ΔS1 )y# )SR = 0.

Moreover, div f = 0, because f ∈ J˙q (BR ). Thus, noting that λg = P f · n on SR , we have ¯ λf2L2 (BR ) − σ λ(Bg, g)SR +

μ D(f)2L2 (BR ) = 0. 2

(3.514)

To treat (Bg, g)SR , we use the following lemma. Lemma 3.7.14 Let H˙ 22 (SR ) = {h ∈ H22 (SR ) | (h, 1)SR = 0,

(h, xj )SR = 0 (j = 1, . . . , N)}.

Then, − (Bh, h) ≥ ch2L2 (SR )

(3.515)

for any h ∈ H˙ 22 (SR ) with some constant c > 0. Postponing the proof of Lemma 3.7.14, we continue the proof of Theorem 3.7.10. 3−1/q Since g ∈ W˙ q (SR ) ⊂ H˙ 22 (SR ), taking the real part of (3.514) we have μ D(f)2L2 (BR ) 2 μ ≥ Re λ(f2L2 (BR ) + cσ g2L2 (SR ) ) + D(f)2L2 (BR ) , 2

0 = Re λ(f2L2 (BR ) − σ (Bg, g)SR ) +

which, combined with Re λ ≥ 0, leads to D(f) = 0. But, (f, p# )BR = 0 for # = 1, . . . , M, and so, f = 0. Thus, by the first equation in (3.512), ∇P (f, g) = 0, and so, P (f, g) = f0 with some constant f0 , which, combined with the third equation in (3.512), leads to Bg = −σ −1 f0 on SR . Since (g, 1)SR = |SR |(g, ϕ1 )SR = 0, we have −σ −1 |SR |f0 = (Bg, 1)SR = (g, ΔSR 1)SR + R −2 (N − 1)(g, 1)SR = 0,

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

401

and so, f0 = 0, which implies that Bg = 0 on SR . Recalling that g ∈ 3−1/q (SR ) ⊂ H˙ 22 (SR ), by Lemma 3.7.14 g = 0. Thus, we have (f, g) = (0, 0), W˙ q and so, the formula (3.509) holds. Namely, problem (3.503) admits a unique solution(v, h) ∈ D˙ q possessing the estimate: (v, h)Dq ≤ Cλ (f, g)Hq (BR )

(3.516)

with some constant Cλ depending on λ, when 2 ≤ q < ∞ and λ ∈ Qλ0 . Before considering the case where 1 < q < 2, at this point we give a Proof of Lemma 3.7.14 Let {λj }∞ j =1 be the set of all eigen-values of the Laplace– Beltrami operator ΔSR on SR . We may assume that λ1 > λ2 > λ3 > · · · > λj > · · · → −∞, and then λ1 = 0 and λ2 = −(N − 1)R −2 . Let Ej be the eigen-space corresponding to λj , and then the dimension of Ej is finite (cf. Neri [31, Chapter III, Spherical Harmonics]). Let dj = dim Ej , and then d1 = 1 and d2 = N. Especially, E1 = {a | a ∈ C} and E2 = {a1 x1 + · · · + aN xN | ai ∈ C (i = 1, . . . , N)}. Let {ϕij }dj i=1 be the orthogonal basis of Ei in L2 (SR ), and then for any h ∈ H˙ 22 (SR ) we have h=

di ∞  

aij ϕij

(aij = (h, ϕij )SR ),

i=3 j =1

because (h, ϕij )SR = 0 for i = 1, 2. Thus, we have −(Bh, h)SR =

di ∞   i=3 j =1

|aij |2 (−λi − (N − 1)R −1 )ϕij 2L2 (SR ) .

Since −λi − (N − 1)R −1 ≥ c with some positive constant c for any i ≥ 3, we have (3.515), which completes the proof of Lemma 3.7.14.

 Next, we consider the case where 1 < q < 2. Let (f, g) ∈ D˙ q satisfy the homogeneous equations (3.512). First, we prove that (f, g)BR = 0

for any g ∈ J˙q  (BR ).

(3.517)

Let (u, ρ) ∈ D˙ q  be a solution of the equations: ⎧ ⎪ ⎪ ⎨

¯ − Div (μD(u) − P (u, ρ)I) = g in BR , λu ¯ − n · P u = 0 on SR , λρ

⎪ ⎪ ⎩(μD(u) − P (u, ρ)I)n − σ (Bρ)n = 0 on S . R

(3.518)

402

Y. Shibata

Since λ ∈ Qλ0 , λ¯ ∈ Qλ0 , and moreover 2 < q  < ∞, and so, by the fact proved above we know the unique existence of (u, ρ) ∈ D˙ q  . By (3.512), (3.518) and the divergence theorem of Gauß, we have (f, g)BR = (f, λ¯ u − Div (μD(u) − P (u, ρ)I))BR = λ(f, u)BR − (f · n, σ Bρ)SR +

μ (D(f), D(u))BR − (div f, P (u, ρ))BR . 2

Noting that P f · n = λg and div f = 0 and using (3.513), we have (f, g)BR = λ(f, u)BR + σ λ{(∇SR g, ∇SR ρ)SR − R −2 (N − 1)(g, ρ)SR } +

μ (D(f), D(u))BR . 2

(3.519)

On the other hand, we have 0 = (λf − Div (μD(f) − P (f, g)I), u)BR μ = λ(f, u)BR − σ (Bg, n · u)SR + (D(f), D(u))BR − (P (f, g), div u)BR . 2 Noting that P u · n = λ¯ ρ and div u = 0 and using (3.513), we have 0 = λ(f, u)BR + σ λ{(∇SR g, ∇SR ρ)SR − R −2 (N − 1)(g, ρ)SR } +

μ (D(f), D(u))BR , 2

which, combined with (3.519), leads to (3.517). Next, we prove that (f, g)BR = 0 for any g ∈ Lq  (BR )N . Given g ∈ Lq  (BR )N , let ψ ∈ Hˆ q1 ,0 (BR ) be a solution to the variational equation: (∇ψ, ∇ϕ)BR = (g, ∇ϕ)BR

1 for any ϕ ∈ Hˆ q,0 (BR ).

Let h = g − ∇ψ and we decompose g as g = ∇ψ + h −

M M   (h, p# )BR p# + (h, p# )BR p# . j =1

j =1

 Since f ∈ J˙q (BR ), we have (f, g)BR = (f, h − M j =1 (h, p# )p# )BR . Since div p# = 0,  M h − j =1 (h, p# )p# ∈ J˙q  (BR ), and so, by (3.517) (f, h − M j =1 (h, p# )p# )BR = 0, which implies that (f, g)BR = 0 for any g ∈ Lq  (BR ). Thus, we have f = 0. By the first equation in (3.512), ∇P (f, g) = 0 in BR , which leads to P (f, g) = f0 with some constant f0 . Thus, by the third equation in (3.512), we have Bg =

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

403

−σ −1 f0 on SR . Since (g, 1)SR = 0, we have −σ −1 f0 |SR | = (Bg, 1)SR = R −2 (N − 1)(g, 1)SR = 0, which leads to f0 = 0. Thus, we have Bg = 0 on SR . By the hypoellipticity of the operator ΔSR , we see that g ∈ H22 (SR ), and so, g ∈ H˙ 22 (SR ), which, combined with Lemma 3.7.14, leads to g = 0. Thus, the formula (3.509) holds, and therefore problem (3.503) admits a unique solution (v, h) ∈ D˙ q possessing the estimate (3.516) when 1 < q < 2 and λ ∈ Qλ0 . Finally, we prove that the constant in the estimate (3.516) is independent of λ ∈ Qλ0 . Let λ ∈ Qλ0 and μ ∈ C, and we consider the equation: (μI − Aq )(v, h) = (f, g).

(3.520)

We write this equation as follows: (f, g) = ((μ − λ)I + (λI − Aq ))(v, h) = (I + (μ − λ)(λI − Aq )−1 )(λI − Aq )(v, h). Since (μ−λ)(λI−Aq )−1 L(H˙ q (BR )) ≤ |μ−λ|Cλ as follows from (3.516), where ·  ˙ denotes the operator norm of the bounded linear operator from H˙ q (BR ) L(Hq (BR ))

into itself, choosing μ ∈ C in such a way that |μ − λ|Cλ ≤ 1/2, we see that the inverse operator (I + (μ − λ)(λI − Aq )−1 )−1 exists as a bounded linear operator from H˙ q (BR ) into itself and (I + (μ − λ)(λI − Aq )−1 )−1 L(H˙ q (BR )) ≤ 2. Thus, (v, h) = (λI − Aq )−1 (I + (μ − λ)(λI − Aq )−1 )−1 (f, g) belongs to D˙ q and solves the Eq. (3.520). Moreover, (v, h)Dq ≤ (λI − Aq )−1 L(Hq ,Dq ) (I + (μ − λ)(λI − Aq )−1 L(H˙ q (BR )) (f, g)Hq ≤ 2Cλ (f, g)Hq , provided that |μ − λ| ≤ (2Cλ )−1 , where  · L(H˙ q ,Dq ) denotes the operator norm of bounded linear operators from H˙ q (BR ) into Dq (BR ). Since Qλ0 is a compact set, we have (3.507), which completes the proof of Theorem 3.7.10. 

Proof of Lemma 3.7.8 Finally we prove Lemma 3.7.8. Let u˜ = u −

M  #=1

(u, p# )BR p# ,

ρ˜ = ρ −

N+1  j =1

(ρ, ϕj )SR ϕj .

404

Y. Shibata

Since D(p# ) = 0, div p# = 0, and Bϕj = 0 (j = 2, . . . , N + 1), we have ˜ ρ), (∇P (u, ˜ ∇ψ)BR = (μDiv D(u) − ∇div u, ∇ψ)BR = (∇P (u, ρ), ∇ψ)BR for any ψ ∈ Hˆ q1 ,0 (BR ), subject to ˜ ρ) P (u, ˜ =< μD(u)n, n > −σ Bρ +

σ (N − 1) (ρ, ϕ1 )SR ϕ1 − div u R2

on Γ ,

˜ ϕ) (ρ, ϕ1 )SR ϕ1 . Since ∇ϕ1 = 0, u˜ and ρ˜ satisfy and so, P (u, ˜ = P (u, ρ) + σ (N−1) R2 ˜ ρ) Eq. (3.499). Moreover, (u, ˜ ∈ H˙ q (BR ) ∩ Dq (BR ), and therefore by (3.508) and ˜ ρ) Theorem 3.7.10 with λ = 0, we have (u, ˜ Dq (BR ) ≤ C(f, g)Hq (BR ) , which, combined with the estimate: ˜ ρ) ˜ Dq (BR ) + C (u, ρ)Dq (BR ) ≤ (u,

M 

|(u, p# )BR | +

N+1 

 |(ρ, ϕj )SR | ,

j =1

#=1

leads to (3.500). This completes the proof of Lemma 3.7.8.

3.7.5 Global Wellposedness, a Proof of Theorem 3.7.1 In this section, we prove Theorem 3.7.1. Assume that the initial data u0 ∈ 2−2/p 3−1/p−1/q Bq,p (BR )N and ρ0 ∈ Wq,p (SR ) satisfy the smallness condition: u0 Hq2 (BR ) + ρ0 W 3−1/p−1/q (S q,p

R)

≤

(3.521)

with small constant > 0 as well as the compatibility condition (3.171). For the notational simplicity, we write I = u0 Hq2 (BR ) + ρ0 W 3−1/p−1/q (S ) , R

q,p

E˜ p,q,T (u, ρ; η) = e uLp ((0,T ),Hq2 (BR )) + e ∂t uLp ((0,T ),Lq (BR )) ηt

+ eηt ρL

ηt

3−1/q (SR )) p ((0,T ),Wq

+ eηt ∂t ρL

Ep,q,T (u, ρ; η) = E˜ p,q,T (u, ρ; η) + eηt ∂t ρL

2−1/q (SR )) p ((0,T ),Wq

1−1/q (SR )) ∞ ((0,T ),Wq

.

Notice that eηt uL

2(1−1/p) (BR )) ∞ ((0,T ),Bq,p

+ eηt ρL

≤ C(I + E˜ p,q,T (u, ρ; η)),

3−1/p−1/q (SR )) ∞ ((0,T ),Bq,p

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

405

which follows from (3.405) and (3.406). Since we choose small enough eventually, we may assume that 0 < I ≤ < 1. Let T0 be a positive number > 2. In view of Theorem 3.7.2, there exists a constant 1 > 0 depending on T0 such that if I ≤ 1 , then for any T ∈ (0, T0 ] problem (3.479) admits a unique solution (u, q, ρ) ∈ Sp,q ((0, T )) satisfying the condition: sup Ψρ (·, t)H∞ 1 (B ) ≤ δ R

(3.522)

0 0 independent of , T , and T0 , where η is the same positive constant as in Theorem 3.7.3. In fact, let r0 ( ) and r± ( ) be three different solutions of the algebraic equation: x 3 + x 2 + − M3−1 x = 0 with r0 ( ) = M3 + O( 2 ), r+ ( ) = M3−1 + O(M3−2 ) + O( ), and r− ( ) = −1 − M3−1 + O(M3−2 ) + O( ) as M3 → ∞ and → 0. Since Ep,q,T (u, ρ; η) ≥ 0 > r− ( ), by (3.523) one of the following cases holds: Ep,q,T (u, ρ; η) ≤ r0 ( ),

Ep,q,T (u, ρ; η) ≥ r+ ( ).

Since we may change M3 larger for (3.523) to hold if necessary, and since we choose 1 > 0 relatively small, we may assume that e By (3.524), Ep,q,1/η (u, ρ; η) ≤ e

1

1

< M3 /2.

(3.524)

< M3 /2 < r+ ( ), and therefore,

Ep,q,T (u, ρ; η) ≤ r0 ( )

for T ∈ (0, 1/η).

But, Ep,q,T (u, ρ; η) is a continuous function with respect to T ∈ (0, T0 ), which yields that Ep,q,T (u, ρ; η) ≤ r0 ( )

for any T ∈ (0, T0 ).

(3.525)

406

Y. Shibata

Let T0 = T0 − 1/2. Thus, choosing > 0 small enough and employing the same argument as that in proving Theorem 3.7.2, we see that there exist unique solutions (v, p, h) ∈ Sp,q ((T0 , T0 + 1)) of the equations: ⎧ ⎪ ⎪ ∂t v − Div (μD(v) − pI) = f(v, Ψh ) ⎪ ⎪ ⎪ ⎪ div v = g(v, Ψh ) = div g(v, Ψh ) ⎪ ⎪ ⎪ ⎨ ∂ h − ω · P v = d(v, ˜ Ψh ) t

in BR × (T0 , T0 + 1), in BR × (T0 , T0 + 1),

⎪ (μD(v)ω)τ = h (v, Ψh ) ⎪ ⎪ ⎪ ⎪ ⎪ < μD(v)ω, ω > −p − σ Bρ = hN (v, Ψh ) ⎪ ⎪ ⎪ ⎩ (v, h)|t =T0 = (u(·, T0 ), ρ(·, T0 ))

on SR × (T0 , T0 + 1),

on SR × (T0 , T0 + 1),

(3.526)

on SR × (T0 , T0 + 1), in BR × SR ,

which satisfies the condition: sup

T0 b > , − b p > 1, b − , q1 p q1 2q2 2q1 1 N  N 2 N N p > 1, + p < 1, bp > 1, b − + < 1. b≥ , q2 2q2 2 2q2 q2 p (3.555)

1 < q1 < 2,

416

Y. Shibata

Remark 3.8.3 The exponent q2 is used to control the nonlinear terms, and so q2 is chosen in such a way that N < q2 < ∞. Let 1 1 1 + , = q1 N q2

1 1 1 = + . q3 q1 q2

(3.556)

Since we have to take q3 > 1, we have 1>

2 1 1 1 + , + = q1 q2 N q2

which leads to q2 >

2N . N −1

Thus, we assume that max(N,

2N ) < q2 < ∞. N −2

Remark 3.8.4 If we choose δ > 0 small enough in (3.545), then x = Xu (y, t) becomes a diffeomorphism with suitable regularity from Ω onto Ωt , and so the original problem (3.543) is globally well-posed.

3.8.1 Maximal Lp –Lq Regularity in an Exterior Domain In this subsection, we study the maximal Lp –Lq regularity of solutions to the Stokes equations with free boundary condition: ⎧ ∂t u − Div (μD(u) − qI) = f, div u = g = div g ⎪ ⎪ ⎨ (μD(u) − qI)n = h ⎪ ⎪ ⎩ u|t =0 = u0

in Ω T , on Γ T , in Ω.

Let 1 Hˆ q,0 (Ω) = {ϕ ∈ Lq,loc (Ω) | ∇ϕ ∈ Lq (Ω)N ,

Jq (Ω) = {u ∈ Lq (Ω)N | (u, ∇ϕ)Ω = 0 We start with the following proposition.

ϕ|Γ = 0},

for any ϕ ∈ Hˆ q1 ,0 (Ω)}.

(3.557)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

417

Proposition 3.8.5 Let 1 < q < ∞. If u ∈ Hq1 (Ω) satisfies div u = 0 in Ω, then u ∈ Jq (Ω). To prove Proposition 3.8.5, we need the following lemma. Lemma 3.8.6 Let 1 < q < ∞, m ∈ N0 and let G be a bounded domain whose 0 (G) = L (G) and for m ≥ 1 boundary ∂G is a hypersurface of C m+1 class. Let Hq,0 q let m Hq,0 (G) = {f ∈ H m (G) | ∂xα f |∂G = 0

for |α| ≤ m − 1}.

m (G) → H m+1 (G)N having the Then, there exists a linear operator B : Hq,0 q,0 following properties:  ∞ (1) There exists a ρ ∈ C0 (G) such that ρ ≥ 0, ρ dx = 1, and  G div B[f ] = f − ρ f dx. In particular, if f dx = 0, then div B[f ] = f . Ω

G

(2) We have the estimate:

B[f ]Hqk+1 (G) ≤ Cq,k,G f Hqk (G)

(k = 0, . . . , m).

m+1 (3) If f = ∂g/∂xi with some g ∈ Hq,0 (G), then

B[f ]Hqk (G) ≤ Cq,k,G gHqk (G)

(k = 0, . . . , m).

Remark 3.8.7 m+1 (1) Since B[f ] ∈ Hq,0 (G), if the 0 extension of B[f ] to RN is written simply by B[f ], then B[f ] ∈ Hqm+1 (RN )N and supp B[f ] ⊂ G. (2) Lemma 3.8.6 was proved by Bogovski [9, 10] (cf. also Galdi [19]).

To apply Lemma 3.8.6, we use the following lemma. Lemma 3.8.8 Let 1 < q < ∞ and let 2R < L1 < L2 < L3 < L4 < 5R. Let χ be a function in C ∞ (RN ) such that χ(x) = 1 for x ∈ BL2 and χ(x) = 0 for x ∈ BL3 . If v ∈ Hq2 (G)N , G ∈ {RN , Ω, Ω5R }, satisfies div v = 0 in DL1 ,L4 , then 3 (DL2 ,L3 ). Here, we have set (∇χ) · v ∈ Hq,0,a  3 Hq,0,a (DL2 ,L3 )

= {f ∈

3 Hq,0 (DL2 ,L3 )

|

f (x) dx = 0}. DL2 ,L3

Here and in the following, we set DL,M = {x ∈ RN | L < |x| < M} for 0 < L < M. To prove Lemma 3.8.8, we need the following lemma.

418

Y. Shibata

Lemma 3.8.9 Let 1 < q < ∞ and let 2R < L1 < L2 < L3 < L4 < L5 < L6 < 5R. Let χ be a function in C ∞ (RN ) such that supp ∇χ ⊂ DL3 ,L4 . If u ∈ Hq2 (G), G ∈ {RN , Ω, Ω5R }, satisfies div u = 0 in DL1 ,L6 , then there exists a v ∈ Hq2 (RN )N such that supp v ⊂ DL2 ,L5 , div v = 0 in RN and (∇χ) · v = (∇χ) · u in RN . Proof Let A0 , A1 , . . ., A5 and B0 , B1 , . . ., B5 be numbers such that L2 = A5 < A4 < A3 < A2 < A1 < A0 < L3 < L4 < B0 < B1 < B2 < B3 < B4 < B5 = L5 . Let ϕ ∈ C0∞ (RN ) such that ϕ(x) = 1 for A2 < |x| < B2 and ϕ(x) = 0 for |x| < A3 or |x| > B3 . Note that ϕ(x) = 1 on supp ∇χ. Let E = DA4 ,A1 ∪ DB1 ,B4 . Since div u = 0 in DL1 ,L6 and supp ϕ ⊂ DA3 ,B3 , we have div (ϕu) = (∇ϕ) · u in 2 (E). Moreover, we have DA4 ,B4 and div (ϕu) = (∇ϕ) · u ∈ Hq,0 





(∇ϕ) · u dx = E

(∇ϕ) · u dx = DA4 ,B4



=− SA4

div (ϕu) dx DA4 ,B4

x · (ϕu) dτ + |x|



x · (ϕu) dτ = 0, |x|

S B4

2 which leads to (∇ϕ) · u ∈ Hq,0,a (E). By Lemma 3.8.6, v = ϕu − B[(∇ϕ) · u] has 2 N the properties: v ∈ Hq (R ), supp v ⊂ DA3 ,B3 , and div v = 0 in RN . Moreover, (∇χ) · u = (∇χ) · v in RN , because ϕ = 1 on supp ∇χ and B[(∇ϕ) · u] = 0 on supp ∇χ. 

Proof of Lemma 3.8.8 By Lemma 3.8.9, there exists a w ∈ Hq2 (RN )N such that (∇χ) · v = (∇χ) · w in RN , supp w ⊂ BL4 and div w = 0 in RN . Then, the assertion follows from the following observation: 





(∇χ) · v dx = DL2 ,L3

(∇χ) · w dx = DL2 ,L3

 =

(∇χ) · w dx BL4



div (χw) dx = BL4

SL4

x · (χw) dτ = 0, |x|

because χ|SL4 = 0. This completes the proof of Lemma 3.8.8.



Proof of Proposition 3.8.5 We use the Sobolev cut off function. Let χ(t) ∈ C0∞ ((−1, 1)) equal one for |t| ≤ 1/2, and set χL (x) = χ

ln ln |x| . ln ln L

Notice that |∇χL (x)| ≤

1 c ln ln L |x| ln |x|

(3.558)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

419

√ and ∇χL (x) vanishes outside of Ω˙ L , where Ω˙ L = {x ∈ Ω | exp ln L < |x| < L}. Since u ∈ Hq1 (Ω) and div u = 0 in Ω, for any ϕ ∈ Hˆ q1 ,0 (Ω) we have (u, ∇ϕ)Ω = lim (χL u, ∇ϕ)Ω = − lim (∇χL · u, ϕ)Ω . L→∞

L→∞

Let ϕ0 be the zero extension of ϕ to RN , that is ϕ0 (x) = ϕ(x) for x ∈ Ω and ϕ0 (x) = 0 for x ∈ Ω. Since ϕ ∈ Hˆ q1 ,0 (Ω), ϕ0 ∈ Hˆ q1 (RN ), where Hˆ q1 (RN ) = {ψ ∈ Lq  ,loc (RN ) | ∇ψ ∈ Lq  (RN )N }. Then, we know (cf. [19]) that there exist constants c = 0 and C for which ϕ − c   0 ≤ C∇ϕ0 Lq  (RN ) = C∇ϕLq  (Ω) ,   Lq  (RN ) d where d(x) = (1 + |x|) log(2 + |x|). Noting that supp ∇χL ⊂ ΩR , we have  (∇χL · u, ϕ)Ω = c

Ω˙ L

(∇χL (x)) · u(x) dx

 +

Ω˙ L

(d(x)(∇χL)(x) · u(x))

ϕ0 (x) − c dx. d(x)

By Lemma 3.8.8, we have  Ω˙ L

(∇χL (x)) · u(x) dx = 0.

Moreover, by (3.558) and Hölder’s inequality, we have   

ϕ0 (x) − c  dx  d(x) Ω˙ L 

1/q  ϕ − c    0 ≤ (|d(x)(∇χL)(x)||u(x)|)q dx   Lq  (RN ) d Ω˙ L 

1/q C ≤ |u(x)|q dx ∇ϕLq (Ω) → 0 as L → ∞. ln ln L Ω˙ L (d(x)(∇χL )(x) · u(x))

Therefore, we have (u, ∇ϕ)Ω = 0 for any ϕ ∈ Hˆ q  ,0 (Ω), that is u ∈ Jq (Ω). This completes the proof of Proposition 3.8.5. 

420

Y. Shibata

We next consider the weak Dirichlet problem: (∇u, ∇ϕ)Ω = (f, ∇ϕ)Ω

for any ϕ ∈ Hˆ q1 ,0 (Ω).

(3.559)

Then, we know the following fact. Proposition 3.8.10 Let 1 < q < ∞ and let Ω be an exterior domain in RN (N ≥ 2). Then, the weak Dirichlet problem is uniquely solvable. Namely, for any f ∈ 1 (Ω) possessing the Lq (Ω)N , problem (3.559) admits a unique solution u ∈ Hˆ q,0 estimate: ∇uLq (Ω) ≤ CfLq (Ω) . Remark 3.8.11 (1) This proposition was proved by Pruess and Simonett [37, Section 7.4] and by Shibata [48, Theorem 18] independently. (2) Let Ω = RN \ S1 and Γ = S1 , where S1 denotes the unit sphere in RN . Let f (x) =

 ln |x| |x|−(N−2)

N = 2, −1

N ≥ 3.

Then, f (x) satisfies the strong Dirichlet problem: Δf = 0 in Ω and f |Γ = 0. 1 (Ω) provided that q > N/(N − 1). However, f does not Moreover, f ∈ Hˆ q,0 satisfy the weak Dirichlet problem: (∇f, ∇ϕ)Ω = 0 for any ϕ ∈ Hˆ q1 ,0 (Ω). In fact, C0∞ (Ω) is not dense in Hˆ q1 ,0 (Ω) when 1 < q  < N. The detailed has been studied in Shibata [48, Appendix A]. By Theorem 3.4.2 in Sect. 3.4, we have the following theorem. Theorem 3.8.12 Let 1 < p, q < ∞ with 2/p + 1/q = 1 and 0 < T < ∞. Let 2(1−1/p)

u0 ∈ Bq,p

(Ω)N , f ∈ Lp ((0, T ), Lq (Ω)N ), 1/2

g ∈ Lp (R, Hq1 (Ω)) ∩ Hp (R, Lq (Ω)),

g ∈ Hp1 (R, Lq (Ω)N ),

(3.560)

1/2

h ∈ Hp (R, Lq (Ω)N ) ∩ Lp (R, Hq1 (Ω)N ) which satisfy the compatibility condition: div u0 = g|t =0

in Ω

and, in addition, (μD(u0 )n − h|t =0 )τ = 0 on Γ

(3.561)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

421

provided that 2/p + 1/q < 1. Then, problem (3.557) admits unique solutions u and p with u ∈ Lp ((0, T ), Hq2 (Ω)N ) ∩ Hp1 ((0, T ), Lq (Ω)N ),

(3.562)

1 p ∈ Lp ((0, T ), Hq1 (Ω) + Hˆ q,0 (Ω))

satisfying the estimates uLp ((0,T ),Hq2 (Ω)) + ∂t uLp ((0,T ),Lq (Ω)) ≤ Cγ eγ T u0 B 2(1−1/p) (Ω) + fLp ((0,T ),Lq (Ω) + (g, h)Lp (R,Hq1 (Ω)) q,p

+ (g, h)H 1/2 (R,L (Ω)) + gHp1 (R,Lq (Ω)) (3.563) p

q

for some positive constants C and γ . 1−2/p

Remark 3.8.13 In the case where 2/p + 1/q < 1, μD(u0 ) ∈ Bq,p (Ω) and 1/2 1 − 2/p > 1/q, and so μD(u0 )|Γ exists. Since h ∈ Hp (R, Lq (Ω)N ) ∩ θ/2 Lp (R, Hq1 (Ω)N ), by complex interpolation theory, h ∈ Hp (R, Hq1−θ (Ω)N ) for any θ ∈ (0, 1). Since 2/p+1/q < 1, we can choose θ in such a way that 1−θ > 1/q and 1/p < θ/2. Thus, h|t =0 ∈ Hq1−θ (Ω)N , and so the trace of h|t =0 to Γ exists.

3.8.2 Local Well-Posedness of Eq. (3.547) In this subsection, we prove the local well-posedness of Eq. (3.547). The following theorem is the main result of this subsection. Theorem 3.8.14 Let 2 < p < ∞, N < q < ∞ and S > 0. Let Ω be an exterior domain in RN (N ≥ 2) whose boundary Γ is a C 2 compact hypersurface. Assume that 2/p + N/q < 1. Then, there exists a time T > 0 depending on S 2(1−1/p) such that if initial data u0 ∈ Bq,p (Ω)N satisfies u0 B 2(1−1/p) (Ω) ≤ S and the q,p compatibility condition: div u0 = 0 in Ω, (D(u0 )n)τ = 0 on Γ , then problem (3.543) admits a unique solution (u, q) with u ∈ Lp ((0, T ), Hq2 (Ω)N ) ∩ Hp1 ((0, T ), Lq (Ω)N ), 1 (Ω)) q ∈ Lp ((0, T ), Hq1 (Ω) + Hˆ q,0

(3.564)

422

Y. Shibata

possessing the estimate: uLp ((0,T ),Hq2 (Ω)) + ∂t uLp ((0,T ),Lq (Ω)) ≤ CS,  T κ(·)u(·, s)H∞ 1 (Ω) ds ≤ δ 0

for some constant C > 0 independent of T and S. Here, δ is the constant appearing in (3.545). Proof Let T and L be a positive numbers determined late and let IT = {u ∈ Lp ((0, T ), Hq2 (Ω)N ) ∩ Hp1 ((0, T ), Lq (Ω)N ) | u|t =0 = u0 , < u >T := uLp ((0,T ),Hq2 (Ω)) + ∂t uLp ((0,T ),Lq (Ω)) ≤ L,  T κ(·)u(·, s)H∞ 1 (Ω) ds ≤ δ}. 0

Since T > 0 is chosen small enough eventually, we may assume that 0 < T ≤ 1. Given v ∈ IT , let u be a solution of linear equations: ⎧ ⎪ ∂t u − Div (μD(u) − qI) = f(v) in Ω T , ⎪ ⎪ ⎪ ⎨ div u = g(v) = div g(v) in Ω T , ⎪ (μD(u) − qI)n = h(v) on Γ T , ⎪ ⎪ ⎪ ⎩ u|t =0 = u0 in Ω.

(3.565)

Notice that 

T

< v >T ≤ L, 0

κ(·)v(·, s)H∞ 1 (Ω) ds ≤ δ.

(3.566)

To solve (3.565), we use Theorem 3.8.12. To this end, we introduce E[v] = E1 [v], which is the functions defined in (3.421) of Sect. 3.6, and eT , which is the extension map defined by (3.419). Recall that the following formulas hold: < E1 [v] >∞ ≤ C(u0 B 2(1−1/p) (Ω) + < v >T ) ≤ C(S + L). q,p

(3.567)

For the sake of simplicity, we may write g(v), g(v) and h(v) given in Eqs. (3.549) and (3.550) as 

t

g(v) = v1 ( 0

 ∇(κv) ds) 0

t

∇(κv) ds ⊗ ∇v,

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .



t

g(v) = v2 ( 

 ∇(κv) ds)

0

∇(κv) ds ⊗ v,

0

t

h(v) = v3 (

t

423

 ∇(κv) ds)

0

t

∇(κv) ds ⊗ ∇v,

0

with some matrices of C 1 functions v1 (k), v2 (k), and v3 (k) defined for |k| < δ. Note that v3 (k) depends also on x ∈ RN with sup (v3 (·, k), ∂k v3 (·, k)H∞ 1 (RN ) ≤ C

|k|≤δ

with some constant C, because n is defined on RN with nH∞ 1 (RN ) < ∞. Let eT be the operator defined in (3.419), and set g0 = eT [g(v)],

g0 = eT [g(v)],

h0 = eT [h(v)].

Since ⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎨[g(v)](·, t) g0 (·, t) = ⎪ [g(v)](·, 2T − t) ⎪ ⎪ ⎪ ⎩ 0 ⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎨[g(v)](·, t) g0 (·, t) = ⎪ [g(v)](·, 2T − t) ⎪ ⎪ ⎪ ⎩ 0

t < 0, 0 < t < T, T < t < 2T , t > 2T , t < 0, 0 < t < T, T < t < 2T , t > 2T ,

as follows from g(v) = 0 and g(v) = 0 at t = 0, and since div g(v) = g(v) for 0 < t < T , we have div g0 = g0 for any t ∈ R. Moreover, we have 

t

g0 = eT [v1 (



0

 

 ∇(κv) ds)

0

0

t

∇(κv) ds ⊗ E1 [v]],

0

t

h0 = eT [v3 (

∇(κv) ds ⊗ ∇E1 [v]],

0

t

g0 = eT [v2 (

t

∇(κv) ds)

 ∇(κv) ds)

t

∇(κv) ds ⊗ ∇E1 [v]],

0

because E1 [v] = v in (0, T ). Here, E1 [v] has been defined in (3.421).

(3.568)

424

Y. Shibata

Let u and q be solutions of the linear equations: ⎧ ⎪ ⎪ ⎪∂t u − Div (μD(u) − qI) = f ⎪ ⎨ div u = g = div g 0

⎪ ⎪ ⎪ ⎪ ⎩

in Ω T , 0

(μD(u) − qI)n = h0

in Ω T ,

(3.569)

on Γ T ,

u|t =0 = u0

in Ω,

and then u and q are also solutions of the Eq. (3.565), because g0 = g(v), g0 = g(v) and h0 = h(v) for t ∈ (0, T ). In the following, using the Banach fixed point theorem, we prove that there exists a unique u ∈ IT such that u = v, which is a required solution of Eq. (3.547). Applying Theorem 3.8.12 gives that < u >T ≤ C{u0 B 2(1−1/p) (Ω) + f0 Lp (R,Lq (Ω)) + (g0 , h0 )Lp (R,Hq1 (Ω)) q,p

+ (g0 , h0 )H 1/2 (R,L p

q (Ω))

+ ∂t g0 Lp (R,Lq (Ω)) },

(3.570)

provided that the right hand side in (3.570) is finite. In the following, C denotes t generic constants independent of T and L. Recalling k = ∇ 0 (κv) ds and the definition of f(u)|i given in (3.548), we have  f(v)Lq (Ω) ≤ C v(·, t)L∞ (Ω) ∇v(·, t)Lq (Ω) 

T

+ 

0 T

+ 0

2 v(·, s)H∞ 1 (Ω) ds(∇ v, ∂t v)Lq (Ω)

 v(·, s)Hq2 (Ω) ds∇v(·, t)L∞ (Ω) .

(3.571)

By Hölder’s inequality, the Sobolev inequality and the assumption: 2/p +N/q < 1, 

T 0



T 0

v(·, s)H∞ 1 (Ω) ds ≤ C v(·, s)Hq2 (Ω) ds ≤ C



T

0

1/p

q

0



p

v(·, s)H 2 (Ω) ds

T

p

v(·, s)H 2 (Ω) ds q

1/p









T 1/p ≤ CT 1/p L, T 1/p ≤ CT 1/p L,

v(·, t)L∞ (Ω) ≤ Cv(·, t)Hq1 (Ω) , ∇v(·, t)L∞ (Ω) ≤ Cv(·, t)B 2(1−1/p) (Ω) . q,p

(3.572)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

425

Moreover, by (3.220) and (3.460), we have sup v(·, t)B 2(1−1/p) (Ω) ≤ sup E1 [v]B 2(1−1/p)(Ω) ≤ C(S + L). q,p

0≤t ≤T

t ∈(0,∞)

q,p

(3.573) By (3.572) and (3.573) sup (v(·, t)L∞ (Ω) v(·, t)Hq1 (Ω) ) ≤ C sup v(·, t)2H 1 (Ω)

0≤t ≤T

q

0T v(·, 2T − t)Hq2 (Ω) T ⎪ ⎪ ⎪ ⎪ ⎩0

t < 0, 0 < t < T, T < t < 2T , t > 2T ,

which, combined with (3.566), gives that 

g0 Lp (R,Hq1 (Ω)) ≤ CL2 T 1/p .

(3.576)

Analogously, 

h0 Lp (R,Hq1 (Ω)) ≤ CL2 T 1/p .

(3.577)

426

Y. Shibata

Since 

t

 0



t

∂t 0





∇(κv) dsL∞((0,T ),Hq1 (Ω)) ≤ CT 1/p < v >T ≤ CT 1/p L, ∇(κv) dsL∞ ((0,T ),Lq (Ω)) ≤ CvL∞ ((0,T ),Hq1 (Ω)) ≤ C(S + L)

as follows from (3.573), applying Lemmas 3.6.2 and 3.6.3 to g0 and h0 , and using the formula of g0 and h0 given in (3.568) and the estimates (3.566), we have 

g0 H 1/2 (R,L p

q (Ω))

h0 H 1/2 (R,L

q (Ω))

p

≤ C(L + S)LT 1/2p , 

≤ C(L + S)LT 1/2p .

(3.578) (3.579)

We finally estimate ∂t g0 . To this end, we write 

t

∂t g0 = v2 (

 ∇(κv) ds)

0

t

+ v2 ( 

∇(κv) ds)∇(κv) ⊗ E1 [v](·, t)

0 t

 ∇(κv) ds)∇(κv)

0



2T −t

∂t g0 = v2 (

 ∇(κv) ds)

0

 

t

∇(κv) ds ⊗ E1 [v](·, t) for t ∈ (0, T ),

0 2T −t

∇(κv) ds ⊗ ∂t E1 [v](·, t)

0 2T −t

+ v2 ( + v2 (

∇(κv) ds ⊗ ∂t E1 [v](·, t)

0



+ v2 (

t

∇(κv) ds)∇(κv) ⊗ E1 [v](·, t)

0 2T −t



2T −t

∇(κv) ds)∇(κv)

0

∇(κv) ds ⊗ E1 [v](·, t)

0

for t ∈ (T , 2T ), and ∂t g0 = 0 for t ∈ [0, 2T ], where v2 (k) = ∇k v2 (k). By (3.573), ∂t g0 Lp (R,Lq (Ω)) ≤ C(T

1 p

1

+ T p )(L + S)2 ,

which, combined with (3.570), (3.575)–(3.579), leads to < u >T ≤ C(L + S)2 (T

1 p

+T

1 p

+T

1 2p

Choosing T > 0 so small that C(L + S)(T

1 p

+T

1 p

+T

1 2p

) ≤ 1/2,

).

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

427

we have < u >T ≤ (C + 1/2)S + L/2. Thus, choosing L = (2C + 1)S, we have < u >T ≤ L.

(3.580)

Moreover, we have 

T 0

1/p κ(·)u(·, s)H∞ 1 (Ω) ds ≤ Cq κH 1 (Ω) T ∞

≤ Cq κH∞ 1 (Ω) LT





T

1/p

q

0

1/p 

p

u(·, s)H 2 (Ω) ds

, 

1/p ≤ δ, we have and so choosing T > 0 in such a way that Cq κH∞ 1 (Ω) LT



T 0

κ(·)u(·, s)H∞ 1 (Ω) ds ≤ δ.

Thus, u ∈ IT . Let Q be a map defined by Qv = u, and then Q maps IT into itself. Given vi ∈ IT (i = 1, 2), considering the equations satisfied by u2 − u1 = Qv2 − Qv1 and employing the same argument as that in proving (3.580), we can show that < Qv1 − Qv2 >T ≤ C(L + S)(T

1 p

+T

1 p

+T

1 2p

) < v 1 − v 2 >T

holds. Choosing T smaller if necessary, we may assume that C(L + S)(T 1 2p

1 p

1

+T p +

T ) ≤ 1/2, and so Q is a contration map on IT . By the Banach fixed point theorem, there exists a unique u ∈ IT such that Qu = u, which is a required unique solution of Eq. (3.543). This completes the proof of Theorem 3.8.14. 

Employing the similar argument to that in Sect. 3.7.2, we can prove the following theorem, which is used to prove the global well-posedness. Theorem 3.8.15 Let 2 < p < ∞, N < q < ∞ and T > 0. Let Ω be an exterior domain in RN (N ≥ 2), whose boundary Γ is a C 2 compact hypersurface. Assume that 2/p +N/q < 1. Then, there exists an 1 > 0 depending on T such that if initial 2(1−1/p) data u0 ∈ Bq,p (Ω)N satisfies u0 B 2(1−1/p) (Ω) ≤ 1 and the compatibility q,p condition (3.564), then problem (3.543) admits a unique solution (u, q) with u ∈ Lp ((0, T ), Hq2 (Ω)N ) ∩ Hp1 ((0, T ), Lq (Ω)N ), 1 q ∈ Lp ((0, T ), Hq1 (Ω) + Hˆ q,0 (Ω))

428

Y. Shibata

possessing the estimate:  uLp ((0,T ),Hq2 (Ω)) + ∂t uLp ((0,T ),Lq (Ω)) ≤

T

1, 0

κ(·)u(·, s)H∞ 1 (Ω) ≤ δ.

3.8.3 A New Formulation of Eq. (3.547) Let T > 0 and let u ∈ Hp1 ((0, T ), Lq (Ω)N ) ∩ Lp ((0, T ), Hq2 (Ω)N ), 1 q ∈ Lp ((0, T ), Hq1 (Ω) + Hˆ q,0 (Ω))

(3.581)

be solutions of Eq. (3.547) satisfying the condition (3.545). In what follows, we rewrite Eq. (3.547) in order that the nonlinear terms have suitable decay properties. In the following, we repeat the argument in Sects. 3.3.2 and 3.3.3. Let V0 (k) = ∂y = (V0ij (k)) be the N × N matrix defined in (3.103) in Sect. 3.3.2 and set ∂x I + V0 (k) := (aij (t)), where k = {kij | i, j = 1, . . . , N} are the variables t corresponding to 0 ∇(κ(y)v(y, s)) ds. And also, let nt = ) (nt 1 , . . . , nt N ) and n = ) (n1 , . . . , nN ) be respective the unit outer normals to Γt and Γ . Since 0 = nt · dx =

N  j =1

we see that

) ∂x n t

j,k=1

is parallel to n, that is

∂y

and so we have nt = c) and (3.309) we have

N 

ntj dxj =

) ∂x n t

∂y

ntj

∂xj dyk , ∂yk

= cn for some c ∈ R \ {0},

∂y n. Since |nt | = 1, we have (3.546). Thus, by (3.103) ∂x

 ∂ ∂ = aj i (t) , ∂xi ∂yj N

nt i = d(t)

j =1

N 

aj i (t)nj

(3.582)

j =1

where d(t) = |(I + ) V0 (k))n|. Let J be the Jacobian of the partial Lagrange transform (3.544) and set 

t

#ij = δij + 0

∂ (κ(y)ui (y, s)) ds ∂yj

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

429

where u = ) (u1 , . . . , uN ). Let aij (t) = δij + a˜ ij (t), J (t) = 1 + J˜(t), #ij (t) = δij + #˜ij (t)

(3.583)

with 

t

a˜ ij (t) = bij (

J˜(t) = K(

∇(κ(y)v(y, s)) ds),

0

#˜ij (t) = mij (



t

t

∇(κ(y)v(y, s)) ds),

0



t

∇(κ(y)v(y, s)) ds) :=

0



0

∂ (κ(y)ui (y, t)) ds. ∂yj (3.584)

Here, if we use the symbols defined in Sect. 3.3.2, then bij = V0ij and K = J0 , and so bij and K are smooth functions defined on {k | |k| ≤ δ} such that bij (0) = K(0) = 0. Let v(x, t) = u(y, t) and p(x, t) = q(y, t), and then v and p are solutions of Eq. (3.1) with 

t

Ωt = {x = y + 

κ(y)u(y, s) ds | y ∈ Ω},

0 t

Γt = {x = y +

κ(y)u(y, s) ds | y ∈ Γ }.

0

By (3.582), ∂vj ∂vi + = Dij,t (u) := Dij (u) + D˜ ij (t)∇u ∂xj ∂xi with ∂uj ∂ui Dij (u) = + , ∂yj ∂yj

D˜ ij (t)∇u =

N  ∂uj ∂ui (a˜ kj (t) + a˜ ki (t) ). ∂yk ∂yk

(3.585)

k=1

By (3.107) in Sect. 3.3.2 we also have an important formula: div v =

N  ∂vj j =1

∂xj

=

N 

J (t)akj (t)

j,k=1

N  ∂uj ∂ = (J (t)akj (t)uj ), ∂yk ∂yk

(3.586)

j,k=1

and so, the divergence free condition: div v = 0 leads to N  j,k=1

(a˜ kj (t) + J˜(t)akj (t))

N  ∂uj ∂ = {(a˜ kj (t) + J˜(t)akj (t))uj } = 0. ∂yk ∂yk j,k=1

(3.587)

430

Y. Shibata

And then, Eq. (3.547) is written as follows: N 

#is (t)(∂t ui + (1 − κ)

i=1

uj akj (t)

j,k=1 N 

−μ

J (t)akj (t)

j,k=1

∂ui ) ∂yk

∂ ∂q Dij,t (u) − =0 ∂yk ∂ys

in Ω T ,

N  ∂uj ∂ = (J (t)akj (t)uj ) = 0 ∂yk ∂yk

in Ω T ,

#is (t)akj (t)

i,j.k=1 N 

N 

j,k=1

N 

μ

#is (t)akj (t)Dij,t (u)nk − qns = 0

on Γ T ,

i,j,k=1

u|t =0 = u0

in Ω,

where s runs from 1 through N. Here, we have used the fact that (#ij ) = A−1 . In order to get some decay properties of the nonlinear terms, we write  T  T  t ∇(κ(y)u(y, s)) ds = ∇(κ(y)u(y, s)) ds − ∇(κ(y)u(y, s)) ds. 0

0

t

In (3.584), by the Taylor formula we write aij (t) = aij (T ) + Aij (t),

#ij (t) = #ij (T ) + Lij (t),

Dij,t (u) = Dij,T (u) + Dij (t)∇u,

J (t) = J (T ) + J (t)

(3.588)

with  Aij (t) = − 

1

0 T

×

bij (



T



T

∇(κ(y)u(y, s)) ds − θ

0

∇(κ(y)u(y, s)) ds) dθ

t

∇(κ(y)u(y, s)) ds

t

 Lij (t) = −

T

t

Dij (t)∇u =

∂ (κ(y)ui (y, s)) ds, ∂yj

N  ∂uj ∂ui (Akj (t) + Aki (t) ), ∂yk ∂yk k=1



J (t) = − 

1

0

t



T



T

∇(κ(y)u(y, s)) ds − θ

0 T

×

K (

∇(κ(y)u(y, s)) ds,

t

∇(κ(y)u(y, s)) ds) dθ

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

431

where bij and K  are derivatives of bij and K with respect to k. By the relation: N 

#is (T )asm (T ) = δsi ,

(3.589)

s=1

the first equation in (3.588) is rewritten as follows: ∂t um −

N 

akj (T )

j,k=1

∂ (μDmj,T (u) − δmj q) = f˜m (u) ∂yk

with f˜m (u) = −

N 

asm (T ){

s=1

+μ

Lis (t)∂t ui +

i=1

N 

asm (T )

s=1

+

N 

N 

N 

(1 − κ)#is (t)akj (t)uj

i,j,k=1

N  

#is (T )akj (T )

i,j,k=1

#is (T )Akj (t)

i,j,k=1

∂ (Dij (t)∇u) ∂yk

∂ui } ∂yk (3.590)

N   ∂ ∂ Dij,t (u) + Lis (t)akj (t) Dij,t (u) . ∂yk ∂yk i,j,k=1

Next, by (3.586) C div u = g(u) ˜ = div g˜ (u) with C div u =

N  j,k=1

g(u) ˜ =

N  j,k=1

g˜k (u) =

J (T )akj (T )

N  ∂uj ∂ = (J (T )akj (T )uj ), ∂yk ∂yk j,k=1

(J (T )Akj (t) + J (t)akj (t))

∂uj , ∂yk

N  (J (T )Akj (t) + J (t)akj (t))uj , j =1

(3.591)

g˜ (u) = ) (g˜1 (u), . . . , g˜N (u)).

432

Y. Shibata

Finally, we consider the boundary condition. Let n˜ be an N-vector defined on ˜ H∞ ˜ is simply written RN such that n˜ = n on Γ and n 2 (RN ) ≤ C. In what follows, n by n = ) (n1 , . . . , nN ). By (3.582) and (3.589) N 

akj (T )(μDmj,T (u) − δmj q)nk = h˜ m (u)

j,k=1

with h˜ m (u) = −μ

N 

(akj (T )Dmj (t)∇u + Akj (t)Dmj,t (u))nk

j,k=1 N 

−μ

(3.592) asm (T )Lis (t)akj (t)Dij,t (u)nk .

i,j,k,s=1

By (3.586), N 

akj (T )

j,k=1

∂ (μDmj,T (u) − δmj q) ∂yk

= J (T )−1

N N  ∂  [ {J (T )akj (T )(μDmj,T (u) − δmj q)}]. ∂yk k=1

j =1

Thus, letting Smk (u, q) =

N 

J (T )akj (T )(Dmj,T (u) − δmj q),

˜ S(u, q) = (Sij (u, q)),

j =1

˜f(u) = ) (f˜1 (u), . . . , f˜N (u)),

˜ h(u) = ) (h˜ 1 (u), . . . , h˜ N (u)),

and using (3.586), we see that u and q satisfy the following equations: ⎧ ˜ ⎪ ∂ u − J (T )−1 Div S(u, q) = ˜f(u) ⎪ ⎪ t ⎪ ⎪ ⎪ ⎨ Cu = g(u) div ˜ = div g˜ (u) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

˜ ˜ S(u, q)n = J (T )h(u) u|t =0 = u0

in Ω T , in Ω T , on Γ T , in Ω.

(3.593)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

433

This is a new formula of Eq. (3.547) which is satisfied by local in time solutions u and q of Eq. (3.547). The corresponding linear equations to Eq. (3.593) is the followings: ⎧ ˜ ⎪ ∂t u − J (T )−1 Div S(u, q) = f ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ C div u = g = div g ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

˜ S(u, q)n = h u|t =0 = u0

in Ω T , in Ω T ,

(3.594)

on Γ T , in Ω.

We call Eq. (3.594) the slightly perturbed Stokes equations.

3.8.4 Slightly Perturbed Stokes Equations In this subsection we summarize some results obtained by Shibata [49] concerning the slightly perturbed Stokes equations. Let r be an exponent such that N < r < ∞. Let aij (T ), a˜ ij (T ), J (T ) and J˜(T ) be functions defined in (3.583) with t = T . We assume that (a˜ ij (T ), J˜(T ))L∞ (Ω) + ∇(a˜ ij (T ), J˜(T ))Lr (Ω) ≤ σ

(3.595)

with some small constant σ > 0. In view of Theorem 3.8.15, we can choose σ > 0 small as much as we want if we choose the initial data small. Since supp κ ⊂ B2R , a˜ ij (T ) and J˜(T ) vanish for x ∈ B2R . In the following, we write aij (T ), a˜ ij (T ), J (T ) and J˜(T ) simply by aij , a˜ ij , J and J˜, respectively. And also, we write A = (aij (T )). To state the compatibility condition for Eq. (3.594), we modify the equation slightly. Notice that it follows from (3.586) with uj = δmj q that N  j,k=1

J −1

N N   ∂ ∂ ∂q (J akj δmj q) = akj (δmj q) = akm ∂yk ∂yk ∂yk j,k=1

k=1

= ) A∇q|m . In the following, we set ˜ ˜ = ) A∇q, D(u) ∇q = (Dij,T (u)), n˜ = dn−1 ) An, n˜ i = dn−1

N  j =1

aj i (T )nj ,

434

Y. Shibata

 1/2 , where we have set n ˜ = with dn = ( N i,j =1 aij (T )ni nj ) ) ˜ = 1. And then, we can write n = (n1 , . . . , nN ). Note that |n|

) (n ˜ 1 , . . . , n˜ N )

˜ ˜ ˜ J −1 Div S(u, q) = J −1 Div (J AμD(u)) − ∇q,

and

(3.596)

˜ ˜ ˜ − qI)n. S(u, q)n = J dn (μD(u) And then, Eq. (3.594) is rewritten as ⎧ −1 ˜ ˜ =f ⎪ + ∇q ⎪ ∂t u − J Div (J AμD(u)) ⎪ ⎪ ⎪ ⎪ ⎨ Cu = g = div g div ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

in Ω T , in Ω T ,

˜ n˜ − qn˜ = (J dn )−1 h μD(u)

(3.597)

on Γ T ,

u|t =0 = u0

in Ω.

To obtain the maximal Lp –Lq regularity and some decay properties of solutions of Eq. (3.597), we first consider the following generalized resolvent problem corresponding to Eq. (3.593): ⎧ ˜ ˜ =f ⎪ λu − J −1 Div (J AμD(u)) + ∇q ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ Cu = g = div g div ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

˜ n˜ − qn˜ = (J dn )−1 h μD(u) u|t =0 = u0

in Ω T , in Ω T ,

(3.598)

on Γ T , in Ω.

The following theorem was proved by Shibata [49]. Theorem 3.8.16 Let 1 < q ≤ r and 0 < 0 < π/2. Assume that Ω is an exterior domain whose boundary Γ is a compact C 2 hypersurface. Let Xq (Ω) = {(f, g, g, h) | f, g ∈ Lq (Ω)N ,

g ∈ Hq1 (Ω),

Xq (Ω) = {(F1 , . . . , FN ) | F1 , F4 , F6 ∈ Lq (Ω)N , F3 ∈ Lq (Ω),

h ∈ Hq1 (Ω)N },

F2 ∈ Hq1 (Ω),

F5 ∈ Hq1 (Ω)N }.

Then, there exist a constant λ0 > 0 and operator families As (λ) and Ps (λ) with As (λ) ∈ Hol (Σ

0 ,λ0

, L(Xq (Ω), Hq2(Ω)N )),

Ps (λ) ∈ Hol (Σ

0 ,λ0

1 , L(Xq (Ω), Hq1(Ω) + Hˆ q,0 (Ω)))

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

435

such that for any λ ∈ Σ 0 ,λ0 and (f, g, g, h) ∈ Xq (Ω), u = As (λ)Fλ and q = Ps (λ)Fλ are unique solutions of Eq. (3.598), where Fλ = (f, g, λ1/2 g, λg, h, λ1/2 h). Moreover, we have RL(X  (Ω),H 2−j (Ω)N ) ({(τ ∂τ )# (λj/2 As (λ)) | λ ∈ Σ

0 ,λ0

}) ≤ rb ,

RL(Xq (Ω),Lq (Ω)N ) ({(τ ∂τ )# (∇Ps (λ)) | λ ∈ Σ

0 ,λ0

}) ≤ rb

q

q

for # = 0, 1 and j = 0, 1, 2 with some constant rb . To obtain the decay properties of solutions to Eq. (3.597), we first consider the time shifted equations: ⎧ ˜ ˜ =f ⎪ ∂t u + λ0 u − J −1 Div (J AμD(u)) + ∇q ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ C div u = g = div g ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

˜ n˜ − qn˜ = (J dn )−1 h μD(u) u|t =0 = u0

in Ω T , in Ω T , on Γ T ,

(3.599)

in Ω.

Employing the same argument as in Sect. 3.4.6, we have the following maximal Lp –Lq regularity theorem for Eq. (3.599) with large λ0 > 0. Theorem 3.8.17 Let 1 < p, q < ∞ and assume that 2/p + N/q = 1. Then, there exist constants σ > 0 and λ0 > 0 such that if (3.595) holds, then the following 2(1−1/p) assertion holds: Let u0 ∈ Bq,p (Ω)N be initial data for Eq. (3.599) and let f, g, g, d, h be given functions for Eq. (3.599) with f ∈ Lp (R, Lq (Ω)N ),

1/2

g ∈ Hp1 (R, Hq1 (Ω)) ∩ Hp (R, Lq (BR )), 1/2

g ∈ Hp1 (R, Lq (Ω)N ), h ∈ Hp1 (R, Hq1 (Ω)N ) ∩ Hp (R, Lq (Ω)N ). Assume that the compatibility conditions: D u0 = g|t =0 div

in Ω.

When 2/p + 1/q < 1, in addition we assume that ˜ 0 )n) ˜ 0 (h|t =0 ) ˜ 0 (μD(u ˜ = J dn 

on Γ ,

436

Y. Shibata

˜ 0 d = d− < d, n˜ > n. ˜ Then, problem (3.599) where for any N-vector d we set  admits unique solutions u and q with u ∈ Hp1 ((0, T ), Lq (Ω)N ) ∩ Lp ((0, T ), Hq2 (Ω)N ), 1 (Ω)) q ∈ Lp ((0, T ), Hq1 (Ω) + Hˆ q,0

possessing the estimate: uLp ((0,T ),Hq2 (Ω)) + ∂t uLp ((0,T ),Lq (Ω)) ≤ C(u0 B 2(1−1/p) (Ω) + fLp (R,Lq (Ω)) + (g, h)H 1/2 (R,L q,p

p

q (Ω))

+ (g, h)Lp (R,Hq1 (Ω)) + ∂t gLp (R,Lq (Ω)) ) for some constant C. Since ∂t (< t >b u) =< t >b ∂t u + b < t >b−1 u, if u and q satisfy Eq. (3.599), then < t >b u and < t >b q satisfy the equations: ˜ ∂t (< t >b u) + λ0 (< t >b u) − J (T )−1 Div (J AμD(< t >b u)) ˜ +∇(< t >b q) =< t >b ˜f + b < t >b−1 u

in Ω T ,

C < t >b u =< t >b g˜ = div (< t >b g˜ ) div

in Ω T ,

˜ ˜ < t >b qn˜ =< t >b (J dn )−1 h˜ μD(< t >b u)n−

on Γ T ,

< t >b u|t =0 = u0

in Ω.

Thus, repeated use of Theorem 3.8.17 yields that  < t >b uLp ((0,T ),Hq2 (Ω)) +  < t >b ∂t uLp ((0,T ),Lq (Ω)) ≤ C(u0 B 2(1−1/p) (Ω) +  < t >b fLp (R,Lq (Ω)) q,p

+ (gb , hb )H 1/2 (R,L p

q (Ω))

+ ∂t gb Lp (R,Lq (Ω)) ),

+ (gb , hb )Lp (R,Hq1 (Ω)) (3.600)

provided that the right hand side is finite. Here, gb , gb and hb are suitable extension of < t >b g, < t >b g, and < t >b h to the whole time interval R such that < t > g = gb , < t >b g = gb , < t >b h = hb for t ∈ (0, T ) and div gb = gb for t ∈ R.

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

437

We next consider so called Lp –Lq decay estimate of semigroup associated with Eq. (3.597). Thus, we formulate Eq. (3.597) in the semigroup setting. For this purpose, we have to eliminate q. We start with the weak Dirichlet problem: ˜ J ∇ϕ) ˜ Ω = (f, J ∇ϕ) ˜ Ω (∇u,

for any ϕ ∈ Hˆ q1 ,0 (Ω).

(3.601)

Here, ˜ = )( ∇ϕ

N 

 ∂ϕ ∂ϕ ,..., akN ) = ) A∇ϕ. ∂xk ∂xk N

ak1

k−1

k−1

Cu = div u and ∇ϕ ˜ = ∇ϕ in RN \ B2R . Since a˜ ij and J˜ vanish outside of B2R , div Thus, by Proposition 3.8.10, we have the following result. Proposition 3.8.18 Let 1 < q ≤ r. Then, for any f ∈ Lq (Ω)N problem (3.601) 1 (Ω) possessing the estimate: admits a unique solution u ∈ Hˆ q,0 ∇uLq (Ω) ≤ CfLq (Ω) . We next consider the following slightly perturbed Dirichlet problem. Given u ∈ Hq2 (Ω)N , let K(u) be a unique solution of the weak Dirichlet problem: ˜ ˜ ˜ Ω = (μJ −1 Div (J AD(u)) C u, J ∇ϕ) ˜ Ω (∇K(u), J ∇ϕ) − ∇˜ div

(3.602)

for any ϕ ∈ Hˆ q1 ,0 (Ω), subject to ˜ n, ˜ n˜ > −C K(u) =< D(u) div u. This problem also can be treated by small perturbation of the weak Dirichlet problem. We now consider the evolution equations: ⎧ ˜ ˜ ∂t u − J −1 Div (J AμD(u)) + ∇K(u) =0 ⎪ ⎪ ⎨ ˜ n˜ − K(u)n˜ = 0 μD(u) ⎪ ⎪ ⎩ u|t =0 = u0 ,

in Ω × (0, ∞), on Γ × (0, ∞),

(3.603)

which is corresponding to Eq. (3.594) with λ0 = 0, f = g = g = h = 0, and u0 ∈ J˜q (Ω). Here, we have set ˜ Ω =0 J˜q (Ω) = {f ∈ Lq (Ω) | (f, J ∇ϕ)

for any ϕ ∈ Hˆ q1 ,0 (Ω)}.

438

Y. Shibata

1 (Ω) be solutions of the Given f ∈ J˜q (Ω), let u ∈ Hq2 (Ω)N and q ∈ Hq1 (Ω) + Hˆ q,0 resolvent equations:



˜ ˜ λu − J −1 Div (J AμD(u)) + ∇K(u) = f,

C div u = 0

in Ω,

˜ n˜ − K(u)n˜ = 0 μD(u)

on Γ .

(3.604)

Employing the same argument as in Sect. 3.4.3, we have q = K(u), and so by Theorem 3.8.16 we have u = As (λ)f and |λ|uLq (Ω) + uHq2 (Ω) ≤ rb fLq (Ω) for any λ ∈ Σ

0 ,λ0

. Let

˜ n˜ − K(u)n| ˜ Γ = 0}, Ds,q (Ω) = {u ∈ J˜q (Ω) ∩ Hq2 (Ω)N | μD(u) ˜ ˜ As u = J −1 Div (J AμD(u)) − ∇K(u)

for u ∈ Ds,q (Ω).

Then, Eq. (3.603) is written by u˙ − As u = 0

u|t =0 = u0

t > 0,

˜ n˜ − K(u)n| ˜ Γ = 0 for u ∈ Ds,q (Ω) is equivalent for u0 ∈ J˜q (Ω). Notice that μD(u) to ˜ n] ˜ 0 [μD(u) ˜ =0 

on Γ .

(3.605)

We then see that the operator As generates a C 0 analytic semigroup {Ts (t)}t ≥0 on J˜q (Ω). Moreover, we have the following theorem. Theorem 3.8.19 Assume that N ≥ 3 and that μ is a positive constant. Then, there exists a σ > 0 such that if the assumption (3.595) holds, then for any q ∈ (1, r], there exists a C 0 analytic semigroup {Ts (t)}t ≥0 associated with Eq. (3.603) such that for any u0 ∈ J˜q (Ω), u is a unique solution of Eq. (3.603) with u = Ts (t)u0 ∈ C 0 ([0, ∞), J˜q (Ω)) ∩ C 1 ((0, ∞), Lq (Ω)N ) ∩ C 0 ((0, ∞), Ds,q (Ω)). Moreover, for any p ∈ [q, ∞) or p = ∞ and for any f ∈ J˜q (Ω) and t > 0 we have the following estimates: Ts (t)fLp (Ω) ≤ Cq,p t ∇Ts (t)fLp (Ω) ≤ Cq,p t

− 12



1 1 q −p

− 12 − 12



fLq (Ω) ,

1 1 q −p

fLq (Ω) .

(3.606)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

439

Remark 3.8.20 This theorem was proved by Shibata [49]. We finally consider the following equations: ⎧ ˜ ˜ = f, C + ∇q div u = 0 ∂t u − J −1 Div (J AμD(u)) ⎪ ⎪ ⎨ ˜ n˜ − qn˜ = 0 μD(u) ⎪ ⎪ ⎩ u|t =0 = 0

in Ω T , on Γ T ,

(3.607)

in Ω.

1 (Ω) be a solution of the weak Dirichlet problem: Let ψ ∈ Hˆ q,0

˜ Ω ˜ J ∇ϕ) ˜ Ω = (f, J ∇ϕ) (∇ψ,

for any ϕ ∈ Hˆ q1 ,0 (Ω).

˜ Let g = f − ∇ψ, and then g ∈ J˜q (Ω) and gLq (Ω) + ∇ψLq (Ω) ≤ CfLq (Ω) . Using this decomposition, we can rewrite Eq. (3.607) as ⎧ ˜ Cu = 0 ˜ − ψ) = g, div ∂t u − J −1 Div (J AμD(u)) + ∇(q ⎪ ⎪ ⎨ ˜ n˜ − (q − ψ)n˜ = 0 μD(u) ⎪ ⎪ ⎩ u|t =0 = 0

in Ω T , on Γ T , in Ω,

where we have used the formula (3.596) and ψ|Γ = 0. Since g ∈ J˜q (Ω) for any t ∈ (0, T ), we see that by Duhamel’s principle, we have  u=

t



t

T (t − s)g(s) ds =

0

T (t − s)(f(s) − ∇ψ(s)) ds.

(3.608)

0

Moreover, we have q = K(u) + ψ. This is a solution formula of Eq. (3.607).

3.8.5 Estimates for the Nonlinear Terms ˜ Let ˜f(u), g(u), ˜ g˜ (u) and h(u) are functions defined in Sect. 3.8.3. In this subsection, we estimate these functions. In the following we write  < t > wLp ((0,T ),X) = α



T

(< t >α w(·, t)X )p dt

1/p

1 ≤ p < ∞,

0

 < t >α wL∞ ((0,T ),X) = esssup < t >α w(·, t)X 0b ˜fLp ((0,T ),Lq2 (Ω))

(3.609)

≤ C(I + [u]2T ),

where [u]T is the norm defined in Theorem 3.8.1 and I = u0 B 2(1−1/p) (Ω) + q2 ,p

u0 B 2(1−1/p) (Ω) . Here and in the following, C denotes generic constants indepenq1 /2,p

dent of I, [u]T , δ, and T . The value of C may change from line to line. Since we choose I small enough eventually, we may assume that 0 < I ≤ 1. Especially, we use the estimates: I 2 ≤ I, below. Since 

β

α

I[u]T ≤

∇(κu(·, s))L∞ (Ω) ds

≤ C(1 + α)

−b+ p1



β α



β α

1 2 (I + [u]2T ) ≤ I + [u]2T 2

p (< s >b u(·, s)H∞ 1 (Ω) ) ds

1/p ,

∇ 2 (κu(·, s))Lq (Ω) ds

≤ C(1 + α)

N −b+ 2q + p1



β

(< s >

2

N b− 2q

2

u(·, s)Hq2

2

α

(Ω) )

p

1/p ds

for any 0 ≤ α < β ≤ T , where q ∈ [1, q2 ], we have  

β α β

α

∇(κu(·, s))L∞ (Ω) ds ≤ C[u]T (1 + α)

−b+ p1

, (3.610)

∇ (κu(·, s))Lq (Ω) ds ≤ C[u]T 2

for any 0 ≤ α < β ≤ T , where q ∈ [1, q2 ], because b > from (3.555). By (3.405), we have sup < t >

b− 2qN

+

1 p

as follows

u(·, t)B 2(1−1/p) (Ω) ≤ C(u0 B 2(1−1/p) (Ω)

2

q2 ,p

t ∈(0,T )

+

N 2q2

N b− 2q

2

q2 ,p

uLp ((0,T ),Hq2 (Ω)) +  < t > 2

N b− 2q

2

∂t uLp ((0,T ),Lq2 (Ω)) }. (3.611)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . . 2(1−1/p)

Since 2/p + N/q2 < 1, Bq2 ,p so by (3.611) 

b− qN

2

441

1 (Ω), and (Ω) is continuously imbedded into H∞

uL∞ ((0,T ),H∞ 1 (Ω)) ≤ C(I + [u]T ).

(3.612)

Applying (3.545), (3.610) and (3.611) to the formulas in (3.583) and (3.584) and N N N and −b + 2q ≤ − 2q , which follows using the fact that −b + p1 < − 2q 2 2 2 from (3.555), give (aij (t), J (t), #ij (t), Aij (t), J (t), Lij (t))L∞ (Ω) ≤ C, (Aij (t), J (t), Lij (t))L∞ (Ω)  T −b+ p1 − N ≤C ∇(κu(·, s))L∞ (Ω) ds ≤ C[u]T < t > ≤ C[u]T < t > 2q2 , t

∇(aij (t), J (t), #ij (t), Aij (t), J (t), Lij (t))Lq (Ω)  T ≤C ∇ 2 (κu(·, s))Lq (Ω) ≤ C[u]T , 0

∂t (aij (t), J (t), #ij (t), Aij (t), J (t), Lij (t))L∞ (Ω) ≤ C∇(κu(·, t))L∞ (Ω) ≤ C(I + [u]T ) < t >

−b+ 2qN

2

≤ C(I + [u]T ) < t >

N − 2q

2

(3.613)

for any t ∈ (0, T ], where q ∈ [1, q2 ]. Moreover, we have (a˜ ij , J˜, #˜ij , Aij , J , Lij )(x, t) = 0

for x ∈ B2R and t ∈ [0, T ].

By (3.613) and (3.614), asm (T )Lis (t)∂t ui (t)Lq (Ω) ≤ C[u]T < t > for any q ∈ [1, q2 ]. Since

1 p

b asm (T )Lis ∂t ui Lp ((0,T ),Lq (Ω)) ≤ C(I + [u]2T ) for any q ∈ [1, q2 ]. Next, by Hölder’s inequality, < t >b u(·, t) · ∇u(·, t)Lq1 /2 (Ω) N

≤< t > 2q1 u(·, t)Lq1 (Ω) < t >

N b− 2q

1

∇u(·, t)Lq1 (Ω)

(3.614)

442

Y. Shibata

and so, by (3.613), we have  < t >b asm (T )#is akj uj

∂ui L ((0,T ),Lq1 /2 (Ω)) ≤ C[u]2T . ∂ξk p

Since < t >b u · ∇u(·, t)Lq2 (Ω) N

≤< t > 2q2 u(·, t)L∞ (Ω) < t >

b− 2qN

2

∇u(·, t)Lq2 (Ω) ,

by (3.613)  < t >b asm (T )#is akj uj

∂ui L ((0,T ),Lq2 (Ω)) ≤ C(I + [u]T )[u]T ≤ C(I + [u]2T ). ∂ξk p

Since N  ∂ 2 uj ∂ 2 ui ∂ (Dij (t)∇u) = (Amj (t) + Ami (t) ) ∂ξk ∂ξk ∂ξm ∂ξk ∂ξm m=1

N 

+

((

m=1

∂uj ∂ ∂ui ∂ Amj (t)) +( Ami (t)) ), ∂ξm ∂ξm ∂ξk ∂ξm

by (3.613) and (3.614), < t >b 

∂ (Dij (·)∇u)Lq (Ω) ∂ξk

≤ C[u]T {< t >

b− 2qN

2

∇ 2 u(·t)Lq2 (Ω) + < t >b ∇u(·, t)L∞ (Ω) }, (3.615)

for any q ∈ [1, q2 ], and therefore  < t >b asm (T )#is (T )akj (T )

∂ (Dij (·)∇u)Lp ((0,T ),Lq (Ω)) ≤ C[u]2T ∂ξk

for any q ∈ [1, q2 ]. Since N  ∂ 2 uj ∂ ∂ 2 ui Dij,T (u) = (amj (T ) + ami (T ) ) ∂ξk ∂ξk ∂ξm ∂ξk ∂ξm m=1

+

N  m=1

((

∂uj ∂ ∂ui ∂ amj (T )) +( ami (T )) ), ∂ξm ∂ξm ∂ξk ∂ξm

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

443

by (3.613) and (3.614), < t >b asm (T )#is (T )Akj (t) ≤ C[u]T {< t >

b− 2qN

2

∂ Dij,T (u)Lq (Ω) ∂ξk

∇ 2 u(·t)Lq2 (Ω) + < t >b ∇u(·, t)L∞ (Ω) },

and so  < t >b asm (T )#is (T )Akj

∂ Dij,T (u)Lp ((0,T ),Lq (Ω)) ≤ C[u]2T ∂ξk

for any q ∈ [1, q2 ]. Analogously, we have  < t >b asm (T )Lis akj

∂ Dij,T (u)Lp ((0,T ),Lq (Ω)) ≤ C[u]2T ∂ξk

for any q ∈ [1, q2 ]. Summing up, we have obtained (3.609). We now consider g˜ and h˜ = ) (h˜ 1 (u), . . . , h˜ N (u)), which have been defined 1

in (3.591) and (3.592), respectively. To estimate the Hp2 norm, we use the following lemma. 1

1 (R, L (Ω)) and g ∈ H 2 (R, L (Ω)). Assume that Lemma 3.8.21 Let f ∈ H∞ ∞ q2 p f (x, t) = 0 for (x, t) ∈ BR × R. Then,

fg

1

Hp2 (R,Lq (Ω))

≤ Cq f H∞ 1 (R,L (Ω)) g ∞

1

Hp2 (R,Lq2 (Ω))

(3.616)

.

for any q ∈ [1, q2] with some constant Cq depending on q and q2 . Proof To prove the lemma, we use the fact that 1

Hp2 (R, Lq (Ω)) = (Lp (R, Lq (Ω)), Hp1 (R, Lq (Ω)))[ 1 ] ,

(3.617)

2

where (·, ·)[1/2] denotes a complex interpolation functor. Let q ∈ [1, q2]. Noting that f (x, t) = 0 for (x, t) ∈ BR × R, we have ∂t (fg)Lq (Ω) ≤ ∂t f L∞ (Ω) gLq2 (Ω) + f L∞ (Ω) ∂t gLq2 (Ω) , and therefore ∂t (fg)Lp (R,Lq (Ω)) ≤ Cf H∞ 1 (R,L (Ω)) gH 1 (R,L ∞ q p

2

(Ω))

444

Y. Shibata

for any q ∈ [1, q2 ]. Moreover, we easily see that fgLp (R,Lq (Ω)) ≤ Cf L∞ (R,L∞ (Ω)) gLp (R,Lq2 (Ω)) . Thus, by (3.617), we have (3.616), which completes the proof of Lemma 3.8.21.  To use the maximal Lp –Lq estimate, we have to extend g, ˜ g˜ and h˜ to R with respect to time variable. For this purpose, we introduce an extension operator e˜T . Let f be a function defined on (0, T ) such that f |t =T = 0, and then e˜T is an operator acting on f defined by ⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎨f (t) [e˜T f ](t) = ⎪ f (−t) ⎪ ⎪ ⎪ ⎩ 0

(t > T ), (0 < t < T ),

(3.618)

(−T < t < 0), (t < −T ).

Lemma 3.8.22 Let 1 < p < ∞, 1 ≤ q ≤ q2 , and 0 ≤ a ≤ b. Let 1 ((0, T ), L (Ω)) and g ∈ H 1 ((0, T ), L (Ω)) ∩ L ((0, T ), H 2 (Ω)). f ∈ H∞ ∞ q2 p p q2 Assume that f |t =T = 0 and f = 0 for (x, t) ∈ BRT . Let < t >= (1 + t 2 )1/2 . Then, we have e˜T (< t >a f ∇g)

1

Hp2 (R,Lq (Ω))

N

≤ C < t > 2q2 f H∞ 1 ((0,T ),L (Ω)) ( < t > ∞ +

N b− 2q

2

N b− 2q

2

gLp ((0,T ),Hq2 (Ω)) 2

∂t gLp ((0,T ),Lq2 (Ω)) + g|t =0 B 2(1−1/p) (Ω) ). q2 ,p

(3.619) a−b+

N

b−

N

2q2 f (t) and g0 (t) =< t > 2q2 g(t), and then Proof Let f0 (t) =< t > 2(1−1/p) N a < t > f ∇g = f0 ∇g0 . Let h be a function in Bq2 ,p (R ) such that h = g|t =0 in Ω and hB 2(1−1/p) (Ω) ≤ Cg|t =0 B 2(1−1/p) (Ω) . Similarly to (3.416), we define q2 ,p

q2 ,p

Tv (t)h by letting Tv (t)h = e−(2−Δ) h. Recall the operator eT defined in (3.419) and note that g0 |t =0 = g|t =0 = Tv (t)h|t =0 in Ω. Let G(t) = eT [g0 − Tv (·)h](t) + Tv (t)h

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

445

for t > 0 and let

[ιg](t) =

 G(t)

⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎨f (t) 0 [ιf ](t) = ⎪ f0 (−t) ⎪ ⎪ ⎪ ⎩ 0

(t > 0), (t < 0),

G(−t)

(t > T ), (0 < t < T ), (−T < t < 0), (t < −T ).

Since G(t) = g0 (t) for 0 < t < T , we have ⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎨f (t)∇g (t) 0 0 e˜T [< t >a f ∇g](t) = ⎪ f0 (−t)∇g0 (−t) ⎪ ⎪ ⎪ ⎩ 0 ⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎨f (t)∇G(t) 0 = ⎪ f ⎪ 0 (−t)∇G(−t) ⎪ ⎪ ⎩ 0

(t > T ) (0 < t < T ) (−T < t < 0) (t < −T ) (t > T ) (0 < t < T ) (−T < t < 0)

= [ιf ](t)∇[ιg](t).

(t < −T )

By Lemma 3.8.21, e˜T [< t >a f ∇g]

1

Hp2 (R,Lq (Ω))

= [ιf ]∇[ιg]

1

Hp2 (R,Lq (Ω))

≤ Cιf H∞ 1 (R,L (Ω)) ∇(ιg) q

1

Hp2 (R,Lq2 (Ω))

.

Since f0 (t)|t =T = 0, we have N

2q ιf H∞ 1 (R,L (Ω)) = 2f0 H 1 ((0,T ),L (Ω)) ≤  < t > 2 f H 1 ((0,T ),L (Ω)) , ∞ ∞ ∞ ∞ ∞

because a − b ≤ 0. To estimate ∇(ιg)

1

Hp2 (R,Lq2 (Ω))

, we use Lemma 3.6.3. And then, by

Lemma 3.6.3 and (3.417), we have ∇(ιg)

1

Hp2 (R,Lq2 (Ω))

≤ C(ιgHp1 (R,Lq

2

(Ω))

≤ C(GHp1 ((0,∞),Lq

2

+ ιgLp (R,Hq2 (Ω)) )

(Ω))

2

+ GLp ((0,∞),Hq2 (Ω)) )

≤ C(g0 − Tv (·)hHp1 ((0,T ),Lq

2

2

(Ω))

+ g0 − Tv (·)hLp ((0,T ),Hq2

2

(Ω))

446

Y. Shibata

+ Tv (·)hHp1 ((0,T ),Lq ≤ C( < t >

N b− 2q

2

(Ω))

+ Tv (·)hLp ((0,∞),Hq2 (Ω)) ) 2

∂t gLp ((0,T ),Lq2 (Ω)) +  < t >

2

N b− 2q

2

gLp ((0,T ),Hq2 (Ω)) 2

+ g|t =0 B 2(1−1/p) (Ω) ). q2 ,p



This completes the proof of Lemma 3.8.22.

Recall the definitions of g(u) ˜ and h˜ m (u) given in (3.591) and (3.592). By Lemma 3.8.22 and (3.613) e˜T [< t >a g(u)] ˜

1

Hp2 (R,Lq (Ω))

≤

N 

N

 < t > 2q2 (J (T )Akj (·) + Tv (·)akj (·))H∞ 1 ((0,T ),L (Ω)) ∞

j,k=1

× ( < t >

N b− 2q

2

uLp ((0,T ),Hq2 (Ω)) +  < t >

N b− 2q

2

2

∂t uLp ((0,T ),Lq2 (Ω))

+ u0 B 2(1−1/p) (Ω) ) q2 ,p

≤ C(I + [u]2T )

(3.620)

for any a ∈ [0, b] and q ∈ [1, q2]. Analogously, we have ˜ e˜T [< t >a h(u)]

1

Hp2 (R,Lq (Ω))

≤ C(I + [u]2T )

(3.621)

for any a ∈ [0, b] and q ∈ [1, q2]. Next, by (3.613), (3.614) and (3.618), e˜T [< t >a g(u)] ˜ Lp (R,Hq1 (Ω)) ≤

N 

N

 < t > 2q2 (J (T )Akj (·) + J (·)akj (·))L∞ ((0,T ),L∞ (Ω))

j,k=1

× +

N 

N b− 2q

2

uLp ((0,T ),Hq2 (Ω)) 2

∇(J (T )Akj (·) + J (·)akj (·))L∞ ((0,T ),Lq (Ω))

j,k=1

×  < t >b uLp ((0,T ),H∞ 1 (Ω)) ≤ C[u]2T

(3.622)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

447

for any a ∈ [0, b] and q ∈ [1, q2]. Analogously, we have 2 ˜ e˜T [< t >b h(u)] Lp ((0,T ),Hq1 (Ω)) ≤ C[u]T

(3.623)

for any a ∈ [0, b] and q ∈ [1, q2]. We finally consider g˜ = ) (g˜1 (u), . . . , g˜N (u)). Let g˜ k (u) be functions given in (3.591). Since ∂t g˜k (u) =

N 

(J (T )∂t Akj (t) + (∂t J (t))akj (t) + J (t)(∂t akj (t))uj )

j =1

+

N 

(J (T )Akj (t) + J (t)akj (t))∂t uj

j =1

and since (J (T ), akj (t), J (t))L∞ (Ω) ≤ C as follows from (3.613), by (3.614) we have e˜T [< t >a ∂t gk (u)]Lp (R,Lq (Ω)) ≤

N  N ( < t > 2q2 ∂t (Akj , J , akj )L∞ ((0,T ),L∞ (Ω)) j =1

×

N b− 2q

2

uLp ((0,T ),Lq2 (Ω))

N

+  < t > 2q2 (Akj , J )L∞ ((0,T ),L∞ (Ω))  < t >

b− 2qN

2

∂t uLp ((0,T ),Lq2 (Ω)) ),

which, combined with (3.613), leads to e˜T [< t >a g˜ (u)]Lp (R,Lq (Ω)) ≤ C(I + [u]2T )

(3.624)

for any a ∈ [0, b] and q ∈ [1, q2].

3.8.6 A Proof of Theorem 3.8.1 The strategy of proving Theorem 3.8.1 is to prolong local in time solutions to any time interval, which is the same idea as that in Sect. 3.7.5. Let T be a positive number >2. Let u and q be solutions of Eq. (3.547), which satisfy the regularity condition (3.553) and the condition (3.545). Let [·]T be the norm defined in (3.554). To prove Theorem 3.8.1 it suffices to prove that [u]T ≤ C(I + [u]2T )

(3.625)

448

Y. Shibata

for some constant C > 0, where I = u0 B 2(1−1/p) (Ω) + u0 B 2(1−1/p) (Ω) . q2 ,p

q1 /2,p

If we show (3.625), employing the same argument as that in Sect. 3.7.5 and using Theorem 3.8.15 concerning the almost global unique existence theorem, we can show that there exists a small constant > 0 such that if I ≤ then [u]T ≤ C for some constant C > 0 independent of , and so we can prolong u to any time interval beyond (0, T ). Thus, we have Theorem 3.8.1. In view of Theorem 3.8.15, there exists an 1 > 0 such that if u0 B 2(1−1/p) (Ω) ≤ 1 , then u and q mentioned above q2 ,p

surely exist. We assume that 0 < ≤ 1 . Thus, our task below is to prove (3.625). In the following, we use the results stated in Sect. 3.8.4 with r = q2 and Sect. 3.8.5. As was seen in Sect. 3.8.2, u and q satisfy Eq. (3.593). To estimate u, we divide u and q into two parts as u = w + v, and q = r + p, where w and r are solutions of the equations: ⎧ ˜ ˜ = ˜f(u) ⎪ ∂t w + λ0 w − J (T )−1 Div (J AμD(w)) + ∇q ⎪ ⎪ ⎪ ⎨ Cw = g(u) div ˜ = div g˜ (u)

in Ω T ,

⎪ ⎪ ⎪ ⎪ ⎩

on Γ T ,

˜ ˜ ˜ = h(u) J dn (μD(w) n˜ − qn) w|t =0 = u0

in Ω T ,

(3.626)

in Ω,

and v and p are solutions of the equations: ⎧ ˜ ˜ = −λ0 w ⎪ ∂t v − J (T )−1 Div (J AμD(v)) + ∇r ⎪ ⎪ ⎪ ⎨ C div v = 0 ˜ n˜ − rn˜ = 0 μD(v)

⎪ ⎪ ⎪ ⎪ ⎩

v|t =0 = 0

in Ω T , in Ω T ,

(3.627)

on Γ T , in Ω.

Concerning the estimate of w, applying (3.600) and using estimations (3.609), (3.620)–(3.624), we have  < t >b ∂t wLp ((0,T ),Lq (Ω)) +  < t >b wLp ((0,T ),Hq2 (Ω)) ≤ C(I + [u]2T ) (3.628) for q = q1 /2, q1 , q2 . By Sobolev’s inequality, we have 

T 0

 (< s > w(·, s)H∞ 1 (Ω) ) ds ≤ b

p

0

T

(< s >b w(·, s)Hq2

2

(Ω) )

p

ds, (3.629)

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

449

because N < q2 < ∞. By (3.405), N

sup < s > 2q1 w(·, t)Lq1 (Ω) ≤ C(u0 

2(1−1/q1 )

Bq1 ,p

0 2q1 wLp ((0,T ),Hq2

1

(Ω)) ).

(3.630) Since q1 /2 < q1 < q2 , we have u0 

(Ω)

≤ I. Thus, putting (3.628)–

[w]T ≤ C(I + [u]2T ).

(3.631)

2(1−1/q1 )

Bq1 ,p

(3.630) together yields that

We next consider v. Let ψ be a solution of the weak Dirichlet problem ˜ Ω ˜ J ∇ϕ) ˜ Ω = (−λ0 w, J ∇ϕ) (∇ψ,

for any ϕ ∈ Hˆ q1 ,0 (Ω).

Let Pw = −λ0 w − ∇ψ. Since ∇ψLq (Ω) ≤ Cq wLq (Ω) for any q ∈ (1, q2 ], we have PwLq (Ω) ≤ Cq wLq (Ω)

(3.632)

for any q ∈ (1, q2 ]. Moreover, by (3.608), we have 

t

v(·, t) =

T (t − s)(Pw)(·, s) ds.

(3.633)

0

Using the estimates (3.606) and (3.632) yields that  ∇ j v(·, t)Lr (Ω) ≤ Cr,q˜1

0

+ Cr,q˜2

t −1



(t − s)

t

t −1

− j2 − N2

(t − s)



− j2 − N2

1 1 q˜1 − r



1 1 q˜2 − r

w(·, s)Lq˜1 (Ω) ds

w(·, s)Lq˜2 (Ω) ds (3.634)

for j = 0, 1, for any t > 1 and for any indices r, q˜1 and q˜2 such that 1 < q˜1 , q˜2 ≤ r ≤ ∞ and q˜1 , q˜2 ≤ q2 , where ∇ 0 v = v and ∇ 1 v = ∇v.

450

Y. Shibata

Recall that T > 2. In what follows, we prove that 

T

2

p (< t >b v(·, t)H∞ 1 (Ω) ) dt

1/p

≤ C(I + [u]2T ),

(3.635)

sup (< t > 2q1 v(·, t)Lq1 (Ω) ) ≤ C(I + [u]2T ),

(3.636)

N

2≤t ≤T



T

(< t >

N b− 2q

1

1

2



v(·, t)Hq1

T

(< t >

N b− 2q

2

2

p (Ω) ) dt

v(·, t)Lq2 (Ω) )p dt

1/p

1/p

≤ C(I + [u]2T ),

(3.637)

≤ C(I + [u]2T ).

(3.638)

By (3.634) with r = ∞, q˜1 = q1 /2 and q˜2 = q2 ,  v(·, t)H∞ 1 (Ω) ≤ C

t 0

T (t − s)Pw(·, s)H∞ 1 (Ω) ds

= C(I∞ (t) + I I∞ (t) + I I I∞ (t)) with  I∞ (t) =  I I∞ (t) =  I I I∞ (t) =

t /2

(t − s)

− qN

w(·, s)Lq1 /2 (Ω) ds,

1

0 t −1

(t − s)

− qN

1

t /2 t t −1

(t − s)

w(·, s)Lq1 /2 (Ω) ds,

N − 2q − 12 2

w(·, s)Lq2 (Ω) ds.

Since I∞ (t) ≤ (t/2)

− qN



t /2

1



< s >−bp ds

1/p 

0

≤ C(bp − 1)

−1/p 

t /2 0

(I + [u]2T )t

(< s >b w(·, s)Lq1 /2 (Ω) )p ds

1/p

− qN 1

as follows from the condition: bp > 1 in (3.555), by the condition: ( qN1 − b)p > 1 in (3.555), we have 

T 2

 (< t >b I∞ (t))p dt ≤ C

T

2

≤ C((



− qN −b p 1

dt (I + [u]2T )p

N − b)p − 1)−1 (I + [u]2T )p . q1

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

451

By Hölder’s inequality, < t >b I I∞ (t)  t −1 −N (t − s) q1 < s >b w(·, s)Lq1 /2 (Ω) ds ≤C t /2

≤C



t −1

(t − s)

− qN

1

1/p ds

t /2

×



t −1

(t − s)

t /2

≤C

N q1

−1

− qN

1

(< s >b w(·, s)Lq1 /2 (Ω) )p ds

−1/p 

t −1

(t − s)

t /2

− qN

1

1/p

(< s >b w(·, s)Lq1 /2 (Ω) )p ds

1/p ,

because N/q1 = N/q2 + 1 > 1. By the Fubini–Tonelli theorem and (3.631), 

T 2

(< t >b I I∞ (t))p dt

≤C ≤C ≤C Since

N 2q2

N q1 N q1

N

q1 +

1 2

−1

− p 



T

p

2

−1 −1

− p 

(t − s)

t /2 T −1

p

−p

t −1

dt

1

− qN

1

(< s >b w(·, s)Lq1 /2 (Ω) )p ds 

(< s > w(·, s)Lq1 /2 (Ω) ) ds b

p

2s

(t − s)

− qN

1

dt

s+1

(I + [u]2T )p .

< 1 as follows from q2 > N, by Hölder’s inequality,

< t >b I I I∞ (t)  t − N −1 ≤C (t − s) 2q2 2 < s >b w(·, s)Lq2 (Ω) ds t −1

≤C ×



t t −1



(t − s)

t t −1

N − 2q − 12

(t − s)

2

N − 2q − 12 2

1/p ds (< s >b w(·, s)Lq2 (Ω) )p ds

1/p

 N

1/p 1 −1/p t − N −1 ≤C − (t − s) 2q2 2 (< s >b w(·, s)Lq2 (Ω) )p ds . 2q2 2 t −1

452

Y. Shibata

By the Fubini–Tonelli theorem, we have  2

T

(< t >b I I I∞ (t))p dt

p N − p ≤C 1− 2q2  T  t − N −1 × dt (t − s) 2q2 2 (< s >b w(·, s)Lq2 (Ω) )p ds

2

t −1

p N − p ≤C 1− 2q2  T  × (< s >b w(·, s)Lq2 (Ω) )p ds

1

N −p (I + [u]2T )p . =C 1− 2q2

s+1

(t − s)

− 2qN − 12 2

dt

s

Summing up, we have obtained (3.635). Next, we prove (3.636). By (3.634) with r = q1 , q˜1 = q1 /2 and q˜2 = q1 , v(·, t)Lq1 (Ω) ≤ C(Iq1 ,1 (t) + I Iq1 ,1 (t) + I I Iq1 ,1 (t)) with  Iq1 ,1 (t) =  I Iq1 ,1 (t) =  I I Iq1 ,1 (t) =

t /2

(t − s)

N − 2q

1

0 t −1

(t − s)

N − 2q

1

t /2 t t −1

w(·, s)Lq1 /2 (Ω) ds, w(·, s)Lq1 /2 (Ω) ds,

w(·, s)Lq1 (Ω) ds.

By (3.631) Iq1 ,1 (t) ≤ (t/2) ×

− 2qN

t /2

1



< s >−bp ds

1/p

0



T 0

≤ Ct



N − 2q

1

(< s >b w(·, s)Lq1 /2 (Ω) )p ds

(I + [u]2T ).

1/p

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

453

Analogously, by Hölder’s inequality and (3.631),  I Iq1 ,1 (t) ≤ C

t −1

(t − s)

t /2

≤C ×

N − 2q

1



−b

t −1

< s >−b < s >b w(·, s)Lq1 /2 (Ω) ds (t − s)

− Np 2q

1/p



1

ds

t /2



T

0

(< s >b w(·, s)Lq1 /2 (Ω) )p ds

1/p

Np 1/p −b− 2qN + p1 1

(I + [u]2T ) 2q1 Np 1/p − N ≤C 1− < t > 2q1 (I + [u]2T ), 2q1

=C 1−

because b >

1 p .

Finally, by (3.631),

I I Iq1 ,1 (t) ≤ Ct −b ≤ Ct

−b

≤ Ct

−b



t

t −1



< s >b w(·, s)Lq1 /2 (Ω) ds

1/p 

t

T

ds

t −1

0

(< s >b w(·, s)Lq1 /2 (Ω) )p ds

1/p

+ [u]2T ).

(I

Summing up, we have obtained (3.636). Next, we prove (3.637). By (3.634) with r = q1 , q¯1 = q1 /2 and q¯2 = q1 , v(·, t)Hq1

1

(Ω)

≤ C(Iq1 ,2 (t) + I Iq1 ,2 (t) + I I Iq1 ,2 (t))

with  Iq1 ,2 (t) =  I Iq1 ,2 (t) =  I I Iq1 ,2 (t) =

t /2

(t − s)

N − 2q

1

0 t −1

(t − s)

N − 2q

1

t /2 t t −1

w(·, s)Lq1 /2 (Ω) ds,

1

w(·, s)Lq1 /2 (Ω) ds,

(t − s)− 2 w(·, s)Lq1 (Ω) ds.

454

Y. Shibata

By (3.631), Iq1 ,2 (t) ≤ (t/2) ×

t /2



< s >−bp ds

1

1/p

0



t /2 0

≤ Ct



N − 2q

N − 2q

1

(< s >b w(·, s)Lq1 /2 (Ω) )p ds

1/p

(I + [u]2T ),

and so, by the condition: ( qN1 − b)p > 1 in (3.555) 

T

(< t >

N b− 2q

1

2

Iq1 ,2 (t))p dt

1/p

N

−1/p ≤ C ( − b)p − 1 (I + [u]2T ). q1

By Hölder’s inequality,

N b− 2q

I Iq1 ,2 (t)  t −1 − N − N (t − s) 2q1 < s >b w(·, s)Lq1 /2 (Ω) ds ≤ C < t > 2q1 1

t /2

≤C ×

T



T

(< t > 2

(t − s)

− Np 2q

1/p



1

ds

(< s >b w(·, s)Lq1 /2 (Ω) )p ds

≤ C(1 + t) 1 p  )p

t −1

1

t /2

 0

Since ( qN1 −



− 2qN

− qN − p1 1

(I + [u]2T ).

> 1 as follows from

b− 2qN

1

1/p

I Iq1 ,2 (t))p dt

=1+

N q1

1/p

N q2

>1=

1 p

+

1 p ,

we have

N

−1/p ≤ C ( − b)p − 1 (I + [u]2T ). q1

Since

N b− 2q

1

 I I Iq1 ,2 (t) ≤ ≤

t

1

t −1



(t − s)− 2 < s >

t

1

t −1

×



(t − s)− 2 ds t

t −1

N b− 2q

1

w(·, s)Lq1 (Ω) ds

1/p

(t − s)− 2 (< s >b w(·, s)Lq1 (Ω) )p ds 1

1/p ,

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

by the Fubini–Tonelli theorem, we have 

T

(< t > 2

N b− 2q



p

≤2



T

≤ 2 p

I I Iq1 ,2 (t))p dt

1

dt 

p p

0



(< s >b w(·, s)Lq1 (Ω) )p ds

s+1

× s

(t − s)− 2 (< s >b w(·, s)Lq1 (Ω) )p ds 1

t −1

2 T

t

1

(t − s)− 2 dt = 2p  < t >b wLp ((0,T ),Lq1 (Ω)) ,

which, combined with (3.628) with q = q1 , leads to 

T

(< t >

b− 2qN

I I Iq1 ,2 (t))p dt

1

2

1/p

≤ C(I + [u]2T ).

Summing up, we have obtained (3.637). Finally, we prove (3.638). By (3.634) with r = q2 , q˜1 = q1 /2 and q˜2 = q2 , v(·, t)Lq2 (Ω) ≤ C(Iq2 (t) + I Iq2 (t) + I I Iq2 (t)) with  Iq2 (t) =  I Iq2 (t) =  I I Iq2 (t) =

t /2

(t − s)

− N2



0 t −1

(t − s)

− N2

2 1 q1 − q2



2 1 q1 − q2

w(·, s)Lq1 /2 (Ω) ds,

w(·, s)Lq1 /2 (Ω) ds,

t /2 t t −1

w(·, s)Lq2 (Ω) ds.

By Hölder’s inequality, Iq2 (t) ≤ (t/2) ×

− N2



2 1 q1 − q2

 t /2



< s >−bp ds

1/p

0



t /2 0

≤C

(< s >b w(·, s)Lq1 /2 (Ω) )p ds

− N2



2 1 q1 − q2

(I + [u]2T )

1/p

455

456

Y. Shibata

for t ≥ 2. Since 1 N N N 2 − − b− = − b, 2 q1 q2 2q2 q1 by the condition: ( qN1 − b)p > 1 in (3.555), 

T

(< t > 2

≤C



N b− 2q

Iq2 (t))p dt

2

T

t



− qN −b p 1

1/p

1/p dt

2

(I + [u]2T )

−1/p N (I + [u]2T ). ≤ C ( − b)p − 1 q1 Since N 2 N 1 N 1 2 = = − + + 1 > 1, 2 q1 q2 2 q2 N 2q2 by Hölder’s inequality

N b− 2q

2

 ≤C

I Iq2 (t)

t −1

(t − s)



− 2qN +1 2

b− 2qN

t /2

≤C



t −1

(t − s)



N − 2q +1 2

2

w(·, s)Lq1 /2 (Ω) ds

1/p ds

t /2

×



t −1

(t − s)



N − 2q +1 2

t /2

(< s >b w(·, s)Lq1 /2 (Ω) )p ds

1/p



N −1/p  t −1

1/p N − 2q +1 2 ≤C (t − s) (< s >b w(·, s)Lq1 /2 (Ω) )p ds , 2q2 t /2

and so, by the Fubini–Tonelli theorem and (3.631) 

T

(< t > 2

N b− 2q

2

I Iq2 (t))p dt



 t −1 N −p/p  T N − 2q +1 2 dt (t − s) (< s >b w(·, s)Lq1 /2 (Ω) )p ds ≤C 2q2 2 t /2

3 R Boundedness, Maximal Regularity and Free Boundary Problems for the. . .

457

N −p/p  T (< s >b w(·, s)Lq1 /2 (Ω) )p ds 2q2 0

 2s N −p N − 2q +1 2 × (t − s) dt ≤ C (I + [u]2T )p . 2q2 s+1

≤C

Analogously, by Hölder’s inequality



N b− 2q

I I Iq2 (t) ≤ C

2

≤C =C

t t −1



t t −1

w(·, s)Lq2 (Ω) ds,

2

1/p 

t

t −1



b− 2qN

ds

t t −1

(< s >b w(·, s)Lq2 (Ω) )p ds

(< s >b w(·, s)Lq2 (Ω) )p ds

1/p

1/p ,

and so, by the Fubini–Tonelli theorem and (3.631) 

T

(< t > 2



N b− 2q

T

≤C 0

2

 I I Iq2 (t))p dt ≤ C



T

dt 2



(< s >b w(·, s)Lq2 (Ω) )p ds

t t −1

s+1 s

(< s >b w(·, s)Lq2 (Ω) )p ds dt ≤ C(I + [u]2T )p .

Summing up, we have obtained (3.638). We next consider the case where t ∈ (0, 2). In view of Theorem 3.8.16, the usual maximal Lp –Lq theorem holds for Eq. (3.627), and so we have vLp ((0,2),Hq2(Ω)) + ∂t vLp ((0,2),Lq (Ω)) ≤ Cq λ0 wLp ((0,2),Lq (Ω)) ≤ C(I + [u]2T )

(3.639)

for any q ∈ [q1 /2, q2 ]. Since 2(1−1/p) > N/q2 +1 as follows from 2/p+N/q2 < 1, by (3.405), Sobolev’s inequality, and (3.639), we have 

2 0

v(·, t)H∞ 1 (Ω) dt

1/p

≤ C sup v(·, t)B 2(1−1/p) (Ω) q2 ,p

0