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University of Southern California Dissertations and Theses
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Mach limits and free boundary problems in fluid dynamics
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Mach limits and free boundary problems in fluid dynamics
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Mach limits and free boundary problems in uid dynamics by Linfeng Li A Dissertation Presented to the FACULTY OF THE USC GRADUATE SCHOOL UNIVERSITY OF SOUTHERN CALIFORNIA In Partial Fulllment of the Requirements for the Degree DOCTOR OF PHILOSOPHY (MATHEMATICS) May 2023 Copyright 2023 Linfeng Li Dedication To my family. ii Acknowledgements I would like to express my sincere appreciation to my advisors Professors Juhi Jang and Igor Kukavica, whose patient guidance, encouragement, and wisdom have profoundly shaped me. I feel incredibly fortu- nate to have had the chance to work with both of them. This dissertation would not have been possible without their support. My sincere gratitude also goes to my committee members, Professors Susan Friedlander, Roger Ghanem, and Nabil Ziane for insightful suggestions and fruitful discussions. Additionally, I would like to thank Pro- fessors Mahir Hadžić and Paul Tokorcheck for their unwavering support of my professional career. I am also grateful to my collaborator, Professor Amjad Tuaha, for guiding my work towards very interesting research topics. Furthermore, I would like to thank Professors Jin Ma and Jianfeng Zhang for their warmth, kindness, and generosity. I am grateful to our math department, faculty, sta members, graduate students, and the analysis group. Over the years, I have been fortunate to have had supportive friends in the math department - Jiajun Luo, Man Luo, Bixing Qiao, Jie Ruan, Fang Sun, Gaozhan Wang, Wenqian Wu, Yusheng Wu, and Jian Zhou. I am thankful for the memories we have shared outside of our mathematical pursuits. Lastly, I want to express my deepest love and gratitude to my big family - my parents, brother, sister, brother-in-law, sister-in-law, as well as the three new additions to our family, my nephews and niece. Despite the physical distance, they have provided me with unconditional love and support. A special thanks to my dearest love, Ying Tan, for being my muse, my partner, and my best friend. iii TableofContents Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi Chapter 1: Mach Limits in Analytic Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Set-up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.3 The main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.4 Analytic estimate of the entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 1.5 Analytic estimate of@ t E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 1.6 Estimates on the velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 1.7 The Mach limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 1.8 Analyticity assumptions on the initial data . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 1.9 The Mach limit in a Gevrey norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 Chapter 2: Mach limits in analytic spaces on exterior domains . . . . . . . . . . . . . . . . . . . . . 70 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 2.2 The setting and notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 2.3 Analytic vector eld in an exterior domain and the main results . . . . . . . . . . . . . . . 75 2.4 Derivative reductions for the velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 2.5 Analytic estimate of the entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 2.6 Velocity estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 2.7 Analyticity assumptions on the initial data . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 2.8 Proof of the convergence theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 Chapter 3: Mach limits of the isentropic uids in the analytic spaces . . . . . . . . . . . . . . . . . 120 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 3.2 Reformulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 3.3 The main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 3.4 Proof of Lemma 3.3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 3.5 Commutator estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 Chapter 4: A regularity result for the free boundary compressible Euler equations of a liquid . . . 144 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 4.2 Lagrangian formulation and the main result . . . . . . . . . . . . . . . . . . . . . . . . . . 148 iv 4.3 Curl estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 4.4 Energy estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 4.5 Normal derivative estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 4.6 The improved Lagrangian ow map and Jacobian estimates . . . . . . . . . . . . . . . . . . 204 4.7 The improved curl estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 4.8 Concluding the proof of Proposition 4.2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 4.9 The general case > 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 4.10 Appendix: Hardy-Sobolev embeddings, trace estimates, and Hodge-type bounds . . . . . . 210 Chapter 5: On the local existence of solutions to the Navier-Stokes-wave system with a free interface 212 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 5.2 The model and main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 5.3 Space-time trace, interpolation, and hidden regularity inequalities . . . . . . . . . . . . . . 220 5.4 The nonhomogeneous parabolic problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 5.5 Solution to a parabolic-wave system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 5.6 Solution to the Navier-Stokes-wave system . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 Chapter 6: Maximal regularity for the Neumann-Stokes problem inH r=2;r spaces . . . . . . . . . . 277 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 6.2 The main result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278 6.3 Proof of Theorem 6.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286 v Abstract This dissertation consists of six chapters among three topics - Mach limit problem, the compressible Euler equations, and uid-structure interaction. In chapter one, we address the Mach limit problem for the Euler equations in the analytic spaces. We prove that, given analytic data, the solutions to the compressible Euler equations are uniformly bounded in a suitable analytic norm and then show that the convergence toward the incompressible Euler solution holds in the analytic norm. We also show that the same results hold more generally for Gevrey data with the convergence in the Gevrey norms. In chapter two, we extend the Mach limit problem for the Euler equations to exterior domains with analytic boundary. We prove the existence of tangential analytic vector elds for the exterior domain with constant analyticity radii, and introduce an analytic norm in which we distinguish derivatives taken from dierent directions. In chapter three, we consider the Mach limit problem of the Euler equations for isentropic uids in the analytic spaces. The results extend the isentropic uids in [77] to more general pressure laws. In chapter four, we derive a priori estimates for the compressible free boundary Euler equations in the case of a liquid without surface tension. We provide a new weighted functional framework which leads to the improved regularity of the ow map by using the Hardy inequality. One of main ideas is to decompose the initial density function. It is worth mentioning that in our analysis we do not need the higher order wave equation for the density. vi In chapter ve, we address a system of equations modeling a compressible uid interacting with an elastic body in dimension three. We prove the local existence and uniqueness of a strong solution when the initial velocity belongs to the spaceH 2+ and the initial structure velocity is inH 1:5+ , for some constant 2 (0; 1=2). In chapter six, we provide a maximal regularity theorem for the linear Stokes equation with a non-homogeneous divergence condition in a bounded domain R 3 and with the Neumann boundary conditions. We prove the existence and uniqueness of solutions such that the velocity belongs to the space H (s+1)=2;s+1 ((0;T ) ), wheres 1. vii Chapter1 MachLimitsinAnalyticSpaces 1.1 Introduction The low Mach number limit problem, which concerns the passage from slightly compressible ows to incompressible ows, is a classical singular limit problem in mathematical uid dynamics. The problem has both physical and mathematical importance. There have been many signicant works on the subject and a great deal of progress made in recent decades [2, 4, 5, 55, 59, 70, 71, 72, 73, 84, 85, 121, 131, 132, 141]. The main diculty of the problem is the presence of dierent wave speeds, which play a signicant role in the limit process. In particular, one has to address the vanishing of the acoustic waves in the limit. A study of the low Mach number limit involves two parts: the uniform bounds and existence of slightly compressible ows for a time-independent of Mach numbers and convergence to solutions of the limiting equations. Interestingly, the analysis of such a singular limit problem signicantly changes depending whether compressible uids are isentropic or non-isentropic, if compressible uids are inviscid or viscous, if initial data are well-prepared or not, if the problem is set in the whole space or domains with boundaries, or which regularity space of data is considered. In this paper, we address the low Mach number limit of the non-isentropic compressible Euler ows inR 3 in analytic and, more generally, in Gevrey spaces. 1 Before describing the results, we briey review prior relevant works (cf. [2, 4, 121] for more extensive reviews). For isentropic ows or well-prepared initial data, it is well-known that solutions of the com- pressible Euler equations with low Mach numbers exist in Sobolev spaces for a time interval independent of the Mach numbers [84, 85, 131]. When initial data are well-prepared, solutions converge to the solutions of the corresponding incompressible Euler equations with the limiting initial data [84, 85, 131]. For the isentropic ows with general initial data, the convergence is not uniform for times close to zero and initial layers are present [5, 70, 71, 72, 73, 141]. On the other hand, the non-isentropic problem with general initial data is much more involved. In this case, the pressure depends not only on the density but also on the entropy that enters into the coecients of the linearized equations, and the convergence is more subtle because the acoustic waves are governed by a wave equation with variable coecients. The rst existence and convergence of the non-isentropic problem were given in [121] and the existence result for general domains with boundary and the convergence result for exterior domains were obtained in [2]. The results above were obtained in Sobolev spaces. Recently the low Mach number limit was studied in [59] starting from dissipative measure-valued solutions of the isentropic Euler equations. Also, the Mach limit in the domains with evolving boundary was addressed in [46, 53], while for the dissipative case, see [3, 44, 45, 49, 54, 57, 60, 69, 110, 117]. For other works on analyticity for the equations involving uids, see [10, 14, 23, 35, 111], while for dierent approaches to analyticity, cf. [11, 16, 22, 61, 62, 66, 90, 125]. This paper concerns the non-isentropic equations with general analytic or Gevrey initial data inR 3 and convergence holding in these strong norms. The rst result provides a uniform in bound of the analytic solution, where> 0 represents the Mach number, while the second result asserts the convergence of the solution to the limiting equation as tends to zero. The main diculty is in obtaining the uniform analytic bound. The Mach limit in an analytic norm is then proven by interpolating the uniform boundedness result and the convergence in the Sobolev space due to Métivier and Schochet in [121]. 2 For the isentropic case, the standard energy estimate method can be applied to the velocity equation to obtain analytic estimates. However, for the non-isentropic case, the problem is more dicult since the matrix,E(u ;S ) (cf. the formulation (1.2.14)–(1.2.15)), also depends onS , and thus spatial derivative bounds cannot be obtained solely by the fundamental energy estimates. Moreover, the non-isentropic Eu- ler ows feature intriguing wave-transport structure: The divergence component of the modied velocity is governed by nonlinear acoustic equations, while the curl component and entropy are transported, and their interactions are coupled. Thus a careful analysis that captures the coupled structure of the modied velocity and the entropy is required. To accomplish these, we use the elliptic regularity for the velocity to reduce the spatial derivative to divergence and curl components. The key to the former is that the diver- gence equation for the velocity is properly balanced with the analytic energy solution, which motivates us to include time derivatives using@ t to our analytic norm; for the latter, we appeal to the transport equation of the curl component, which can be treated in a similar way as the entropy. Thus, the pure time analytic norm needs to be treated dierently than the one which also involves the spatial derivatives (cf. Sections 1.6.2 and 1.6.3 respectively). It is important to include the analytic weight in (1.3.3), which ultimately balances the time and the spatial derivatives. The main diculty in our approach is the handling of the vorticity!, which can not be treated directly. Instead, as in [2], we need to consider the equation for the modied vorticity curl(r 0 v), wherer 0 is a certain function of the entropy (cf. Section 1.6.1 below). The product and chain rules then lead to complicated analytic coupling among the entropy, divergence, vorticity, and curl(r 0 v). The paper is organized as follows. In Section 1.2, we introduce the Mach number limit problem and then formulate the symmetrized version of the compressible Euler equations. In Section 1.3, we dene the analytic norm and state the main results. The rst theorem relies on Lemma 1.3.3, the proof of which is given at the end of Section 1.6. We present the energy estimate for the transport equation in Section 1.4. Product rule and chain rules in analytic spaces are provided in Section 1.5. In Section 1.6, we estimate the 3 curl, divergence, and time-derivative components of the velocity. In Section 1.7, we prove the convergence theorem. In Section 1.8, we establish the niteness of the space-time analytic norm at the initial time under the assumption that the initial data is real-analytic in the spatial variable. In Section 1.9, we provide the Mach limit theorem in any Gevrey space. 1.2 Set-up We consider the compressible Euler equations describing the motion of an inviscid, non-isentropic gaseous uid inR 3 @ t +vr +rv = 0 (1.2.1) (@ t v +vrv) +rP = 0 (1.2.2) @ t S +vrS = 0; (1.2.3) where =(x;t)2R + is the density,v =v(x;t)2R 3 is the velocity,P =P (x;t)2R + is the pressure, andS = S(x;t)2R is the entropy of the uid. The system (1.2.1)–(1.2.3) is closed with the equation of state P =P (;S): (1.2.4) For instance, the equation of state for an ideal gas takes the form P (;S) = e S ; (1.2.5) where > 1 is the adiabatic exponent. 4 To address the low Mach number limit, we introduce the rescalings ~ t =t; ~ x =x; ~ =; ~ v = v ; ~ P =P; ~ S =S; where > 0 represents the Mach number, the ratio of the typical uid speed to the typical sound speed. We assume that the typical sound speed isO(1). For simplicity of notation, we omit tilde, and obtain the rescaled system @ t +vr +rv = 0 (1.2.6) (@ t v +vrv) + 1 2 rP = 0 (1.2.7) @ t S +vrS = 0: (1.2.8) The goal of this paper is to obtain the low Mach number limit of (1.2.6)–(1.2.8) in analytic spaces. 1.2.1 Reformulation Now, considerP , instead of, as an independent variable, we may write (1.2.4) as =(P;S); and (1.2.6) is then replaced by A 0 (@ t P +vrP ) +rv = 0; (1.2.9) where A 0 =A 0 (S;P ) = 1 (S;P ) @(S;P ) @P : 5 The equation of state for an ideal gas in (1.2.5) then reads as (P;S) =P 1 e S : To symmetrize the Euler equations, we set P = Pe p ; for a positive constant P which represents the reference state at the spatial innity so thatP = P +O(). Using@ t P =P@ t p andrP =Prp, we rewrite (1.2.9) and (1.2.7) as a (@ t p +vrp) + 1 rv = 0 (1.2.10) r (@ t v +vrv) + 1 rp = 0; (1.2.11) respectively, where a =a(S;p) =A 0 (S; Pe p ) Pe p (1.2.12) and r =r(S;p) = (S; Pe p ) Pe p : (1.2.13) In the case of an ideal gas, from(P;S) =P 1 e S , we have the expression a = 1 fora, and r = ( Pe p ) 1 1 e S 6 for r. Thus we have obtained the symmetrized version of the compressible Euler equation for non- isentropic uids inR 3 , which reads E(S;u)(@ t u +vru) + 1 L(@ x )u = 0; (1.2.14) @ t S +vrS = 0; (1.2.15) whereu = (p;v) and E(S;u) = 0 B B @ a(S;u) 0 0 r(S;u)I 3 1 C C A ; L(@ x ) = 0 B B @ 0 div r 0 1 C C A : (1.2.16) After transforming (1.2.1)–(1.2.3) to the symmetrized form (1.2.14)–(1.2.15), we now focus on the formula- tion (1.2.14)–(1.2.15). In view of (1.2.12) and (1.2.13), we assume a(S;u) =f 1 (S)g 1 (u) (1.2.17) and r(S;u) =f 2 (S)g 2 (u); (1.2.18) wheref 1 ,f 2 ,g 1 , andg 2 are positive entire real-analytic functions. 1.3 Themainresults We assume that the initial data (p 0 ;v 0 ;S 0 ) satises k(p 0 ;v 0 ;S 0 )k H 5M 0 (1.3.1) 7 and 1 X m=0 X jj=m k@ (p 0 ;v 0 ;S 0 )k L 2 (m3) + 0 (m 3)! M 0 ; (1.3.2) where 0 ;M 0 > 0 are xed constants. For > 0, dene the mixed weighted analytic space A() =fu2C 1 (R 3 ) :kuk A() <1g; where kuk A() = 1 X m=1 m X j=0 X jj=j k@ (@ t ) mj uk L 2 (j3) + (t) (m3) + (m 3)! ; (1.3.3) here,2 (0; 1] represents the mixed space-time analyticity radius and where> 0. It is convenient that the term withkuk L 2 is not included in the norm. In (1.3.2) and below we use the conventionn! = 1 when n2N. As shown in Section 1.8 below, (1.3.2) implies that with = 1 k(p 0 ;v 0 ;S 0 )k A(~ 0 ) Q(M 0 ) (1.3.4) for some functionQ, where ~ 0 = 0 =Q(M 0 ) is a suciently small constant. Note that the time derivatives of the initial data are dened iteratively by dierentiating the equations (1.2.14)–(1.2.15) and evaluating at t = 0 (cf. Section 1.8 below for details). Also observe that the norm in (1.3.3) is an increasing function of , and thus (1.3.4) holds for any2 (0; 1]. We dene the analyticity radius function (t) =(0)Kt; (1.3.5) 8 where(0) minf1; ~ 0 g is a suciently small parameter (dierent from ~ 0 ), andK 1 is a suciently large parameter, both to be determined below. The rst theorem provides a uniform in boundedness of the analytic norm on a time interval, which is independent of. Theorem 1.3.1. Assume that the initial data (p 0 ;v 0 ;S 0 ) satises (1.3.1)–(1.3.2), whereM 0 ; 0 > 0. There exist suciently small constants;(0); 0 ;T 0 > 0, depending onM 0 , such that k(p ;v ;S )(t)k A() M; 0< 0 ; t2 [0;T 0 ]; (1.3.6) where is as in (1.3.5) andK andM are suciently large constants depending onM 0 . We now turn to the Mach limit for solutions of (1.2.14)–(1.2.15) inR 3 as! 0. Denote = (0) C 0 ; (1.3.7) where(0);2 (0; 1] are xed constants chosen in the proof of Theorem 1.3.1, andC 0 > 1 is a suciently large constant to be chosen in Section 1.7. We introduce the spatial analytic norm kuk X = 1 X m=1 X jj=m k@ uk L 2 (m3) + (m 3)! ; (1.3.8) where> 0 is as in (1.3.7). Note that this is a part of our main analyticA-norm, (1.3.2). 9 By Theorem 1.3.1, for a givenM 0 and 0 > 0, the solutions (p ;v ;S ) are uniformly bounded byM in the norm ofC 0 ([0;T 0 ];X ) for xed parameters,T 0 , and2 (0; 0 ]. The second main theorem shows that solutions of (1.2.14)–(1.2.15) converge to the solution of the stratied incompressible Euler equations r(S; 0)(@ t v +vrv) +r = 0; (1.3.9) divv = 0; (1.3.10) @ t S +vrS = 0; (1.3.11) as! 0. Theorem1.3.2. Let > 0, and assume that the initial data (v 0 ;S 0 ) converges to (v 0 ;S 0 ) inX and inL 2 as! 0, andS 0 decays suciently rapidly at innity in the sense jS 0 (x)jCjxj 1 ; jrS 0 (x)jCjxj 2 ; for 0 < 0 and some constants C and > 0. Then (v ;p ;S ) converges to (v (inc) ; 0;S (inc) ) in C 0 ([0;T 0 ];X ), where (v (inc) ;S (inc) ) is the solution to (1.3.9)–(1.3.11) with the initial data (w 0 ;S 0 ), and w 0 is the unique solution of divw 0 = 0; curl(r 0 w 0 ) = curl(r 0 v 0 ); withr 0 =r(S 0 ; 0). In the rest of the paper, the constantC depends only onM 0 and 0 , and it may vary from relation to relation; we omit the superscript, and we writeS,u forS ,u . 10 Theorem 1.3.2 is proven in Section 1.7 below as a consequence of Theorem 1.3.1. The proof of Theo- rem 1.3.1 consists of a priori estimates performed on the solutions. The a priori estimates are easily justied by simply restricting the sum (1.3.2) tomm 0 wherem 0 2f6; 7;:::g is arbitrary. The estimates on the nite sums are justied since boundedness of solutions in any Sobolev norm is known by [2]. The proof of Theorem 1.3.1 relies on analytic a priori estimates on the entropyS and the (modied) velocityu. The a priori estimate needed to prove Theorem 1.3.1 is the following. Lemma1.3.3. LetM 0 > 0. For any 1, there exist constantsC, 1 ; 0 ;T 0 and a nonnegative continuous functionQ such that for all2 (0; 0 ], the norm M ; (T ) = sup t2[0;T ] (kS(t)k A((t)) +ku(t)k A((t)) ) (1.3.12) satises the estimate M ; (t)C + (t + + +(0))Q(M ; (t)); (1.3.13) fort2 [0;T 0 ] and(0)2 (0; 1 ], provided KQ(M ; (T 0 )) (1.3.14) holds. With = (t) as in (1.3.5), we use the notation (1.3.12). The constantK depends onM (and thus ultimately onM 0 ), i.e.,K =Q(M). We shall work on an interval of time such that T 0 (0) 2K : (1.3.15) 11 Thus we have(0)=2(t)(0) fort2 [0;T 0 ]. From here on, we denote by Q a positive increasing continuous function, which may change from inequality to inequality; importantly, the functionQ does not depend on, , andt. The estimates are performed on an interval of time [0;T ] where (1.3.5) holds and is such that T (0) 2K : In the rest of the paper, we allow all the constants to depend on 0 . Proof of Theorem 1.3.1 given Lemma 1.3.3. LetM 0 > 0 be as in (1.3.1)–(1.3.2). Also, xC 0 andQ 0 as the constantC and the functionQ appearing in the statement of Lemma 1.3.3, respectively. Now, choose and x M 1 > maxfC 0 ;Q 0 (M 0 )g: Then select 1 suciently small,(0) minf1; ~ 0 ; 1 g,T 1 2 (0;T 0 ], and2 (0; 0 ] suciently small, so that C 0 + (T 1 + + +(0))Q 0 (M 1 )<M 1 : Next, set T 2 = min T 1 ; (0) 2Q 0 (M 1 ) : (1.3.16) In view of (1.3.14), this last condition ensures (0) 2 (t)(0); t2 [0;T 2 ]: 12 Note thatM ; (0)M 0 . By (1.3.13)–(1.3.16) and the continuation principle, we get M ; (t)M 1 ; t2 [0;T 2 ]; and Theorem 1.3.1 is proven. Sections 1.4–1.6 are devoted to the proof of Lemma 1.3.3, thus completing the proof of Theorem 1.3.1. Remark1.3.4 (Boundedness of Sobolev norms). By [2, Theorem 1.1] theH 5 norm of (p ;v ;S ) can be estimated by a constant on a time interval [0;T 0 ], whereT 0 only depends on theH 5 norm of the initial data. More precisely, for given initial data satisfying (1.3.1), there existsT 0 > 0 and a constantC such that sup 0m5;0jm;jj=j k@ (@ t ) mj (p ;v ;S )(t)k L 2C; t2 [0;T 0 ]; 2 (0; 1]: In the rest of the paper, we always work on an interval of time [0;T ] such that 0 0. For any2 (0; 1], there exists 1 2 (0; 1] such that if 0<(0) 1 , then kS(t)k A((t)) C +tQ(M ; (t)); t2 (0;T 0 ]; (1.4.1) 13 for all2 (0; 1], providedK in (1.3.5) satises KQ(M ; (T 0 )); whereT 0 > 0 is a suciently small constant depending onM 0 . Proof of Lemma 1.4.1. Fixm2 N andjj = j where 0 j m. We apply@ (@ t ) mj to the equation (1.2.15) and take theL 2 -inner product with@ (@ t ) mj S obtaining 1 2 d dt k@ (@ t ) mj Sk 2 L 2 + vr@ (@ t ) mj S;@ (@ t ) mj S = [vr;@ (@ t ) mj ]S;@ (@ t ) mj S ; whereh;i denotes the scalar product inL 2 . Using the Cauchy-Schwarz inequality and summing over jj =j, we obtain d dt X jj=j k@ (@ t ) mj Sk L 2Ckrvk L 1 x X jj=j k@ (@ t ) mj Sk L 2 +C X jj=j k[vr;@ (@ t ) mj ]Sk L 2: With the notation (1.3.2), the above estimate implies d dt kSk A() = _ (t)kSk ~ A() + 1 X m=1 m X j=0 X jj=j (j3) + (m3) + (m 3)! d dt k@ (@ t ) mj Sk L 2 _ (t)kSk ~ A() +Ckrvk L 1 x kSk A() +C 1 X m=1 m X j=0 j X l=0 X jj=j X jj=l mj X k=0 1l+k C m;j;l;;;k ; (1.4.2) where C m;j;l;;;k = (j3) + (m3) + (m 3)! mj k k@ (@ t ) k v@ (@ t ) mjk rSk L 2 14 with kuk ~ A() = 1 X m=4 m X j=0 X jj=j k@ (@ t ) mj uk L 2 (j3) + (m 3)(t) m4 (m 3)! denoting the dissipative analytic norm corresponding to (1.3.2). In the above sums as well as below, the multiindexes;;::: are assumed to belong toN 3 0 . The third term on the far right side of (1.4.2) equals C =C 4 X m=1 m X j=0 j X l=0 X jj=j X jj=l mj X k=0 1l+k C m;j;l;;;k +C 1 X m=5 m X j=0 j X l=0 X jj=j X jj=l mj X k=0 1l+k[m=2] C m;j;l;;;k +C 1 X m=7 m X j=0 j X l=0 X jj=j X jj=l mj X k=0 [m=2]+1l+km3 C m;j;l;;;k +C 1 X m=5 m X j=0 j X l=0 X jj=j X jj=l mj X k=0 C m;j;l;;;k 1 fm2l+kmg =C 1 +C 2 +C 3 +C 4 ; where we split the sum according to the low and high values ofl +k andm. We claim that there exists T 0 > 0, such that for any2 (0; 1], there is 1 2 (0; 1] such that if 0<(0) 1 , then C 1 C; (1.4.3) C 2 Ckvk A() kSk ~ A() ; (1.4.4) C 3 Ckvk A() kSk ~ A() ; (1.4.5) C 4 Ckvk A() : (1.4.6) 15 Proof of (1.4.3): Using Hölder’s and the Sobolev inequalities,C 1 may be estimated by low-order mixed space-time derivatives, and (1.4.3) follows by appealing to Remark 1.3.4. Proof of (1.4.4): Using Hölder’s and the Sobolev inequalities we arrive at C 2 C 1 X m=5 m X j=0 j X l=0 X jj=j X jj=l mj X k=0 1l+k[m=2] (j3) + m3 (m 3)! mj k k@ (@ t ) k vk 1=4 L 2 kD 2 @ (@ t ) k vk 3=4 L 2 k@ (@ t ) mjk rSk L 2; and thus C 2 C 1 X m=5 m X j=0 j X l=0 X jj=j X jj=l mj X k=0 1l+k[m=2] a b k@ (@ t ) k vk L 2 (l3) + (l+k3) + (l +k 3)! ! 1=4 kD 2 @ (@ t ) k vk L 2 (l1) + (l+k1) + (l +k 1)! ! 3=4 k@ (@ t ) mjk rSk L 2 (jl2) + (mkl 2) mkl3 (mkl 2)! ! A m;j;l;;;k ; (1.4.7) where A m;j;l;;;k = mj k (l +k 3)! 1=4 (l +k 1)! 3=4 (mkl 2)! (mkl 2)(m 3)! (1.4.8) and a = (j 3) + l 3 4 + 3l 3 4 + (jl 2) + ; b =m 3 l +k 3 4 + 3l + 3k 3 4 + (mkl 3): (1.4.9) 16 For simplicity, we omitted indicating the dependence of a and b on j, k, and l. Since l +k 1 and 0lj, one can readily check that3=2a 3=2 and 1b 3=2, which implies a b C (1.4.10) if (0) 3 : (1.4.11) Recall the combinatorial inequality jj jj ; (1.4.12) which may also be written as j l mj k m l +k ; (1.4.13) from where we obtain A m;j;l;;;k Cm! (l +k)!(mlk)! (l +k 3)!(l +k) 3=2 (mkl 3)! (m 3)! Cm 3 (mlk) 3 C; (1.4.14) sincel +k [m=2]. Using X jj=j X jj=l x y = 0 @ X jj=l x 1 A 0 @ X j j=jl y 1 A (1.4.15) 17 from [97, Lemma 4.2], together with (1.4.7), (1.4.8)–(1.4.10), (1.4.14) and the discrete Hölder inequality, we obtain C 2 C 1 X m=5 m X j=0 j X l=0 mj X k=0 1l+k[m=2] 0 @ X jj=l k@ (@ t ) k vk L 2 (l3) + (l+k3) + (l +k 3)! 1 A 1=4 0 @ X jj=l kD 2 @ (@ t ) k vk L 2 (l1) + (l+k1) + (l +k 1)! 1 A 3=4 0 @ X j j=jl k@ (@ t ) mjk rSk L 2 (jl2) + (mkl 2) mkl3 (mkl 2)! 1 A Ckvk A() kSk ~ A() ; (1.4.16) where the last inequality follows from the discrete Young inequality. Proof of (1.4.5): We reverse the roles ofl +k andmlk and proceed as above, arriving at C 3 C 1 X m=7 m X j=0 j X l=0 X jj=j X jj=l mj X k=0 [m=2]+1l+km3 a b k@ (@ t ) k vk L 2 (l3) + l+k3 (l +k 3)! ! k@ (@ t ) mjk rSk L 2 (jl2) + (mlk 2) mlk3 (mlk 2)! ! 1=4 kD 2 @ (@ t ) mjk rSk L 2 (jl) + (mlk) mlk1 (mlk)! ! 3=4 B m;j;l;;;k ; (1.4.17) where we denote B m;j;l;;;k = mj k (l +k 3)!(mlk 2)! 1=4 (mlk)! 3=4 (mlk 2) 1=4 (mlk) 3=4 (m 3)! 18 and a = (j 3) + (l 3) + jl 2 4 + 3j 3l 4 + ; b =m 3 (l +k 3) (mlk 3) 4 3(mlk 1) 4 : Since 0lj, it is readily seen that5=2a 1=2 andb = 3=2, which implies a b C (1.4.18) if (1.4.11) holds. Using (1.4.12)–(1.4.13), we obtain B m;j;l;;;k Cm! (l +k)!(mlk)! (l +k 3)!(mlk 2)!(mlk 1) 1=2 (m 3)! Cm 3 (l +k) 3 C; (1.4.19) since [m=2] + 1l +k. Combining (1.4.17)–(1.4.19) and proceeding as in (1.4.16), we obtain C 3 Ckvk A() kSk ~ A() : Proof of (1.4.6): We splitC 4 into three sums according to the value ofl +k being equal tom2,m1, orm, and denote them byC 41 ,C 42 , andC 43 , respectively. 19 ForC 41 , we use Hölder’s and the Sobolev inequalities and obtain C 41 C 1 X m=5 m X j=0 j X l=0 X jj=j X jj=l mj X k=0 (j3) + m3 (m 3)! mj k 1 fl+k=m2g k@ (@ t ) k vk L 2k@ (@ t ) mjk rSk L 1 C 1 X m=5 m X j=0 j X l=0 mj X k=0 0 @ X jj=l k@ (@ t ) k vk L 2 (l3) + m5 (m 5)! 1 A 0 @ X j j=jl kD 2 @ (@ t ) mjk rSk L 2 1 A 3=4 0 @ X j j=jl k@ (@ t ) mjk rSk L 2 1 A 1=4 m! (m 2)! (m 5)! (m 3)! 1 fl+k=m2g Ckvk A() ; where in the second inequality we applied (1.4.12)–(1.4.13), (1.4.15) and we used ; C; in the last inequality, we estimated the low-order mixed space-time Sobolev norm ofS byC using Remark 1.3.4. ForC 42 andC 43 , we proceed as in above, by writing C 42 C 1 X m=5 m X j=0 j X l=0 X jj=j X jj=l mj X k=0 (j3) + m3 (m 3)! mj k 1 fl+k=m1g k@ (@ t ) k vk L 2k@ (@ t ) mjk rSk L 1 C 1 X m=5 m X j=0 j X l=0 mj X k=0 0 @ X jj=l k@ (@ t ) k vk L 2 (l3) + m4 (m 4)! 1 A 0 @ X j j=jl kD 2 @ (@ t ) mjk rSk L 2 1 A 3=4 0 @ X j j=jl k@ (@ t ) mjk rSk L 2 1 A 1=4 m! (m 1)! (m 4)! (m 3)! 1 fl+k=m1g Ckvk A() 20 and C 43 C 1 X m=5 m X j=0 j X l=0 X jj=j X jj=l mj X k=0 (j3) + m3 (m 3)! mj k 1 fl+k=mg k@ (@ t ) k vk L 2k@ (@ t ) mjk rSk L 1 C 1 X m=5 m X j=0 X jj=j k@ (@ t ) mj vk L 2 (j3) + m3 (m 3)! ! kD 2 rSk 3=4 L 2 krSk 1=4 L 2 Ckvk A() : Combining (1.4.2)–(1.4.6) and Remark 1.3.4 to boundkrvk L 1 x , we get d dt kSk A() kSk ~ A() ( _ +Ckvk A() ) +CkSk A() +Ckvk A() +C: (1.4.20) Now, determineK in (1.3.5) to be suciently large so that _ (t) +Ckvk A() 0; 0tT 0 ; (1.4.21) whereT 0 > 0 satises (1.3.15). The lemma is then proven by integrating (1.4.20) on [0;T 0 ], using (1.4.21), and applying the Gronwall lemma. After Section 1.5, we work with derivatives of the solution and thus instead of the norms (1.3.2) we use kuk B() = 1 X m=1 m X j=0 X jj=j k@ (@ t ) mj uk L 2 (j2) + (t) (m2) + (m 2)! (1.4.22) 21 and the corresponding dissipative analytic norm kuk ~ B() = 1 X m=3 m X j=0 X jj=j k@ (@ t ) mj uk L 2 (j2) + (m 2)(t) m3 (m 2)! : (1.4.23) It turns out that the curl component of the velocity satises an equation similar to the one for the entropy, but with the nonzero right-hand side. Thus we now consider the inhomogeneous transport equation @ t ~ S +vr ~ S =G; where ~ S = ~ S(x;t),v =v(x;t), andG =G(x;t). Lemma1.4.2. For any2 (0; 1], there exists 1 2 (0; 1] such that if 0<(0) 1 , then k ~ Sk A() k ~ S(0)k A() +C Z t 0 kG(s)k A() +kv(s)k A() ds +Ct; t2 [0;T 0 ]; (1.4.24) for some constantC and suciently smallT 0 > 0, providedK in (1.3.5) satises KCkv(t)k A() ; t2 [0;T 0 ]; (1.4.25) whereT 0 is chosen suciently small so that (1.3.15) holds. Similarly, for any 1, there exists(0) > 0 such that k ~ Sk B() k ~ S(0)k B() +C Z t 0 kG(s)k B() +kv(s)k B() ds +Ct; t2 [0;T 0 ]; (1.4.26) for some constantC and suciently smallT 0 > 0, providedK satises KCkv(t)k B() ; t2 [0;T 0 ]; (1.4.27) 22 whereT 0 is chosen suciently small so that (1.3.15) holds. Note that from denitions (1.3.2) and (1.4.22), we have kvk B() kvk A() for allv, and thus (1.4.25) implies (1.4.27). Proof. We proceed exactly as in the proof of Lemma 1.4.1. Using the Cauchy-Schwarz inequality with the inhomogeneous partG, we obtain d dt k ~ Sk A() k ~ Sk ~ A() ( _ +Ckvk A() ) +C(k ~ Sk A() +kvk A() +kGk A() ) +C: The estimate (1.4.24) then follows by using (1.4.21) and the Gronwall inequality. Analogously, we use the analytic shift (m 2)! instead of (m 3)! and proceed as in the proof of Lemma 1.4.1, we conclude d dt k ~ Sk B() k ~ Sk ~ B() ( _ +Ckvk B() ) +C(k ~ Sk B() +kvk B() +kGk B() ) +C: The assertion (1.4.26) may then be obtained by setting _ (t) +Ckvk B() 0 withC suciently large and using the Gronwall inequality. 23 1.5 Analyticestimateof@ t E In order to bound the velocity, we rst need to obtain an analytic estimate for@ t E, which in turn requires the bound on the entropy. We rst provide a product rule for the typeB norm. Lemma 1.5.1. Let k2f2; 3;:::g and > 0. For f 1 ;:::;f k 2 B(), and any 2 (0; 1], there exists 1 2 (0; 1] andT 0 > 0 such that if 0<(0) 1 , then k Y i=1 f i B() C k k X i=1 0 @ kf i k B() Y 1jk;j6=i (kf j k B() +kf j k L 2) 1 A ; fork 2, where the constant is independent ofk. proof of Lemma 1.5.1. By induction, it is sucient to prove the inequality kfgk B() Ckfk B() (kgk B() +kgk L 2) +C(kfk B() +kfk L 2)kgk B() ; (1.5.1) forf andg such that the respective right hand sides are nite. To prove the estimate (1.5.1), we use the Leibniz rule and write kfgk B() = 1 X m=1 m X j=0 X jj=j (j2) + (m2) + (m 2)! k@ (@ t ) mj (fg)k L 2 1 X m=1 m X j=0 j X l=0 X jj=j X jj=l; mj X k=0 H m;j;l;;;k ; (1.5.2) where H m;j;l;;;k = mj k (j2) + (m2) + (m 2)! k@ (@ t ) k f@ (@ t ) mjk gk L 2: 24 We split the sum on the right side of (1.5.2) according to the low and high values ofl +k andm, and we claim 2 X m=1 m X j=0 j X l=0 X jj=j X jj=l; mj X k=0 H m;j;l;;;k C(kfk B() +kfk L 2)kgk B() +C(kgk B() +kgk L 2)kfk B() ; (1.5.3) 1 X m=3 m X j=0 j X l=0 X jj=j X jj=l; mj X k=0 H m;j;l;;;k 1 fl+k=0g C(kfk B() +kfk L 2)kgk B() ; (1.5.4) 1 X m=3 m X j=0 j X l=0 X jj=j X jj=l; mj X k=0 1l+k[m=2] H m;j;l;;;k C(kfk B() +kfk L 2)kgk B() ; (1.5.5) 1 X m=3 m X j=0 j X l=0 X jj=j X jj=l; mj X k=0 [m=2]+1l+km1 H m;j;l;;;k Ckfk B() (kgk B() +kgk L 2): (1.5.6) 1 X m=3 m X j=0 j X l=0 X jj=j X jj=l; mj X k=0 H m;j;l;;;k 1 fl+k=mg Ckfk B() (kgk B() +kgk L 2); (1.5.7) Proof of (1.5.3): Form = 1, we use Hölder’s and the Sobolev inequalities and arrive at 1 X j=0 j X l=0 X jj=j X jj=l; 1j X k=0 H 1;j;l;;;k Ckf@ t gk L 2 +CkfDgk L 2 +Ckg@ t fk L 2 +CkgDfk L 2 CkD 2 fk 3=4 L 2 kfk 1=4 L 2 k@ t gk L 2 +CkD 2 fk 3=4 L 2 kfk 1=4 L 2 kDgk L 2 +CkD 2 gk 3=4 L 2 kgk 1=4 L 2 k@ t fk L 2 +CkD 2 gk 3=4 L 2 kgk 1=4 L 2 kDfk L 2 C(kfk B() +kfk L 2)kgk B() +C(kgk B() +kgk L 2)kfk B() : (1.5.8) 25 Form = 2, by Leibniz rule we write 2 X j=0 j X l=0 X jj=j X jj=l; 2j X k=0 H 2;j;l;;;k Ckf(@ t ) 2 gk L 2 +CkfD@ t gk L 2 +CkfD 2 gk L 2 +CkDf@ t gk L 2 +CkDfDgk L 2 +Ck@ t f(@ t )gk L 2 +Ck@ t fDgk L 2 +Ckg(@ t ) 2 fk L 2 +CkgD@ t fk L 2 +CkgD 2 fk L 2: (1.5.9) All the terms in (1.5.9) are estimated using Hölder and Sobolev inequalities. For illustration, we treat the fth term, for which we write kDfDgk L 2kDfk L 4kDgk L 4CkD 2 fk 3=4 L 2 kDfk 1=4 L 2 kD 2 gk 3=4 L 2 kDgk 1=4 L 2 Ckfk B() kgk B() : (1.5.10) Collecting the estimates (1.5.8)–(1.5.10), we obtain (1.5.3). Proof of (1.5.4): Using Hölder and Sobolev inequalities, we obtain 1 X m=3 m X j=0 j X l=0 X jj=j X jj=l; mj X k=0 H m;j;l;;;k 1 fl+k=0g C 1 X m=3 m X j=0 X jj=j kfk L 1k@ (@ t ) mj gk L 2 (j2) + (m2) + (m 2)! C(kfk B() +kfk L 2)kgk B() : (1.5.11) 26 Proof of (1.5.5): Using Hölder and Sobolev inequalities, we obtain 1 X m=3 m X j=0 j X l=0 X jj=j X jj=l; mj X k=0 0l+k[m=2] H m;j;l;;;k C 1 X m=3 m X j=0 j X l=0 X jj=j X jj=l; mj X k=0 0l+k[m=2] k@ (@ t ) k fk L 2 (l2) + (l+k2) + (l +k 2)! ! 1=4 kD 2 @ (@ t ) k fk L 2 l + (l+k) + (l +k)! ! 3=4 k@ (@ t ) mjk gk L 2 (jl2) + (mlk2) + (mlk 2)! ! ; where we bound the constant coecient by C analogously as in (1.4.14), and bound the and term byC analogously as in (1.4.9)–(1.4.10). Therefore, using the discrete Hölder and Young inequalities, we obtain (1.5.5). Proof of (1.5.6): We reverse the roles ofl andmlk and proceed as in the above argument, obtaining 1 X m=3 m X j=0 j X l=0 X jj=j X jj=l; mj X k=0 [m=2]+1l+km H m;j;l;;;k C 1 X m=3 m X j=0 j X l=0 X jj=j X jj=l; mj X k=0 [m=2]+1l+km k@ (@ t ) k fk L 2 (l2) + (l+k2) + (l +k 2)! ! k@ (@ t ) mjk gk L 2 (jl2) + (mlk2) + (mlk 2)! ! 1=4 kD 2 @ (@ t ) mjk gk L 2 (jl) + (mlk) + (mlk)! ! 3=4 : Therefore, using the discrete Hölder and Young inequalities, we obtain (1.5.6). 27 Proof of (1.5.7): We proceed as in (1.5.11), obtaining 1 X m=3 m X j=0 j X l=0 X jj=j X jj=l; mj X k=0 H m;j;l;;;k 1 fl+k=mg C 1 X m=3 m X j=0 X jj=j kgk L 1 k@ (@ t ) mj fk L 2 (j2) + (m2) + (m 2)! ! Ckfk B() (kgk B() +kgk L 2): Combining (1.5.3)–(1.5.7), we obtain (1.5.2). Similarly to (1.5.1) and Lemma 1.5.1, with analytic shift (m 3)! rather than (m 2)!, we also have kfgk A() Ckfk A() (kgk A() +kgk L 2) +C(kfk A() +kfk L 2)kgk A() : (1.5.12) In the case whenf belongs toL 1 but is not square integrable, we have variant formulas kfgk A() Ckfk A() (kgk A() +kgk L 2) +Ckfk L 1kgk A() ; (1.5.13) and kfgk B() Ckfk B() (kgk B() +kgk L 2) +Ckfk L 1kgk B() : (1.5.14) The proofs are similar to (1.5.1), where the modication of the proof for the variant formula (1.5.13) is to treat the termkf@ (@ t ) mj gk L 2 by Hölder’s inequality with exponents1 and 2. The next lemma provides an analytic estimate for composition of functions. 28 Lemma1.5.2. Assume thatf is an entire real-analytic function. Then there exists a functionQ such that kf(S(t))k B() Q(kS(t)k A() +kS(t)k L 2); (1.5.15) and kf(S(t))k A() Q(kS(t)k A() +kS(t)k L 2); (1.5.16) whereQ also depends onf. Proof. First we prove (1.5.15). Sincef is entire, for everyR> 0 there existsN(R)> 0 such that jf (k) (x)j Nk! R k ; x2R; k2N 0 and f(S(t)) = 1 X k=0 f (k) (0)S(t) k k! : (1.5.17) By Lemma 1.5.1, we obtain f (k) (0) k! S k B() NC k k R k kSk B() (kSk B() +kSk L 2) k1 : (1.5.18) Summing (1.5.18) ink2N and using the Taylor expansion (1.5.17), we arrive at kf(S(t))k B() 1 X k=1 f (k) (0) k! S(t) k B() CN kS(t)k B() kSk B() +kSk L 2 1 X k=1 C(kSk B() +kSk L 2) R k CN 1 X k=1 C(kSk B() +kSk L 2) R k : 29 Choosing R = 2CkS(t)k B() + 2CkS(t)k L 2, we obtainkf(S(t))k B() CN, where N depends on kS(t)k B() +kS(t)k L 2. Finally, observe thatkS(t)k B() kS(t)k A() , by the denition of the norms, concluding the proof of (1.5.15). The estimate (1.5.16) is proven analogously by using (1.5.12), and we omit the details here. For the next two lemmas, assume that ~ e is one of components of the matrixE in (1.2.16), i.e., eitherr or one of the components ofa. By the assumptions (1.2.17) and (1.2.18), we have ~ e(S;u) =f(S)g(u); (1.5.19) wheref andg are positive entire real-analytic functions. The rst lemma gives the estimate of the derivative of the component of the matrixE. Lemma1.5.3. GivenM 0 > 0, and (1.5.19), wheref andg are as above. Then k@ t ~ ek B() Q(kuk A() +kuk L 2;kSk A() +kSk L 2) (1.5.20) for some functionQ. Proof. By (1.2.15), the chain rule, and product rule, we obtain @ t ~ e =f 0 (S)@ t Sg(u) +f(S)rg(u)@ t u =f 0 (S)vrSg(u) +f(S)rg(u)@ t u: Therefore, k@ t ~ ek B() kf 0 (S)vrSg(u)k B() +kf(S)rg(u)@ t uk B() =G 1 +G 2 :: (1.5.21) 30 By repeated use of (1.5.1) and (1.5.14) and Remark 1.3.5, we arrive at G 1 kf 0 (S)k B() (kvrSg(u)k B() +kvrSg(u)k L 2) +kf 0 (S)k L 1kvrSg(u)k B() kf 0 (S)k B() kg(u)k B() (kvrSk B() +kvrSk L 2) +CkvrSk B() +kvrSk L 2 +Ckg(u)k B() kvrSk B() +kvrSk L 2 +CkvrSk B() : (1.5.22) For the termkvrSk B() , we again appeal to (1.5.1), obtaining kvrSk B() Ckvk B() krSk B() +krSk L 2 +CkrSk B() kvk B() +kvk L 2 : By the denition of the analytic norms in (1.3.2) and (1.4.22), we have krSk B() = 1 X m=1 m X j=0 X jj=j k@ (@ t ) mj rSk L 2 (j2) + (m2) + (m 2)! C 1 X m=1 m X j=0 X jj=j+1 k@ (@ t ) mj Sk L 2 (j2) + (m2) + (m 2)! CkSk A() : (1.5.23) Collecting estimates (1.5.22)–(1.5.23), we obtain G 1 Q(kuk A() +kuk L 2;kSk A() +kSk L 2): (1.5.24) Using analogous arguments, we also get G 2 Q(kuk A() +kuk L 2;kSk A() +kSk L 2) (1.5.25) 31 since by denition k@ t uk B() = 1 X m=1 m X j=0 X jj=j k@ (@ t ) mj+1 uk L 2 (j2) + (m2) + (m 2)! C 1 X m=1 m X j=0 X jj=j k@ (@ t ) mj k L 2 (j3) + (m3) + (m 3)! Ckuk A() : Therefore, (1.5.20) is proven by combining (1.5.21), (1.5.24), and (1.5.25). The second lemma gives the analytic estimates for the component of the matrixE. Lemma1.5.4. Assume (1.5.19), wheref andg are as above. Then k~ e(t)k A() Q(kuk A() +kuk L 2;kSk A() +kSk L 2); (1.5.26) for some functionQ. Proof. Since ~ e(t) =f(S)g(u), the proof of the estimate (1.5.26) may be carried out by appealing to Lem- mas 1.5.1 and 1.5.2 in theA-norm. 1.6 Estimatesonthevelocity Recall that E(S;u) = 0 B B @ a(S;u) 0 0 r(S;u)I 3 1 C C A ; wherea(S;u) andr(S;u) are as in Section 1.2.1. 32 1.6.1 Estimateonthecurl We rst need to rewrite the equation (1.2.11) so to be able to estimate the curl of the velocityv. Introduce r 0 (x) =r(x; 0) (i.e.,r 0 (S) =r(S; 0)), and note that, by our assumptions, r 0 (S) =f 2 (S)g 2 (0): Dene ~ f(x;y) = 1 r 0 (x) r(x;y) = 1 g 2 (0) g 2 (y) : Since ~ f is a function ofy only and vanishes aty = 0, there exists a bounded entire functionh such that ~ f(x;y) =yh(y): Denoting ~ h(x) =xh(x); we then have ~ f(S;u) = ~ h(u): Since@ t S +vrS = 0, the equation (1.2.11) forv is equivalent to the nonlinear transport equation (@ t +vr)(r 0 v) + 1 rp = ~ hrp: Applying curl to the above equation and using curlrp = 0, we arrive at (@ t +vr) curl(r 0 v) = [vr; curl](r 0 v) + [curl; ~ h]rp: (1.6.1) 33 To treat (1.6.1), we would like to use (1.4.26) from Lemma 1.4.2 and thus we need to estimate the forcing term G = [vr; curl](r 0 v) + [curl; ~ h]rp =G 1 +G 2 in the analytic norm (1.4.22). Since ([v j @ j ; curl]w) i = ikm @ m v j @ j w k , where ikm is the permutation symbol, we may apply Lemma 1.5.1 and obtain kG 1 k B() C(kr(r 0 v)k B() +kr(r 0 v)k L 2)(krvk B() +krvk L 2): (1.6.2) From (1.5.13), (1.5.23), (1.5.26) and (1.6.2), we getkG 1 k B() Q(M ; (t)). The termG 2 may be estimated in an analogous way since ([curl; ~ h]rp) i = ijk @ j ~ h@ k p, leading to kG 2 k B() C(krpk B() +krpk L 2)(kr ~ hk B() +kr ~ hk L 2)Q(M ; (t)): Proceeding as in Lemma 1.4.2, we obtain d dt k curl(r 0 v)k B() = _ k curl(r 0 v)k ~ B() + 1 X m=1 m X j=0 X jj=j (j2) + (m2) + (m 2)! d dt k@ (@ t ) mj curl(r 0 v)k L 2; (1.6.3) where we used the notation (1.4.22)–(1.4.23). By (1.4.26) from Lemma 1.4.2, we get k curl(r 0 v)(t)k B() k curl(r 0 v)(0)k B() +Ct sup s2(0;t) kG(s)k B() +Ct sup s2(0;t) kv(s)k B() +Ct C +tQ(M ; (t)): (1.6.4) 34 Next, we estimate curlv in the analytic normB(). Denoting R 0 = 1 r 0 ; we rewrite k curlvk B() kR 0 curl(r 0 v)k B() +k[curl;R 0 ]r 0 vk B() = 1 + 2 : (1.6.5) For the term 1 , we use (1.5.14) and the curl estimate (1.6.4), obtaining 1 CkR 0 k B() (k curl(r 0 v)k B() +k curl(r 0 v)k L 2) +CkR 0 k L 1k curl(r 0 v)k B() : (1.6.6) SinceR 0 satises the homogeneous transport equation@ t R 0 +vrR 0 = 0, the inequality (1.4.26) from Lemma 1.4.2 implies kR 0 (S(t))k B() kR 0 (S(0))k B() +Ct sup s2(0;t) kv(s)k B() +CtC +tQ(M ; (t)): (1.6.7) Combining (1.6.4), (1.6.6), and (1.6.7), we obtain 1 C +tQ(M ; (t)): (1.6.8) For 2 , we rst rewrite it as [curl;R 0 ]r 0 v =R 1 rSv; 35 whereR 1 =r 0 0 =r 0 . Applying Lemma 1.5.1 and (1.5.14), we get 2 =k[curl;R 0 ]r 0 vk B() CkR 1 k B() (krSk B() +krSk L 2)(kvk B() +kvk L 2) +CkrSk B() (kR 1 k B() +kR 1 k L 1)(kvk B() +kvk L 2) +Ckvk B() (kR 1 k B() +kR 1 k L 1)(krSk B() +krSk L 2): To bound the right hand side, it suces to estimatekR 1 k B() ,krSk B() , andkvk B() , as the rest are bounded byC (cf. Remark 1.3.4). ForkR 1 k B() , sinceR 1 =r 0 0 =r 0 depends only on the entropyS, it satises the homogeneous transport equation@ t R 1 +vrR 1 = 0 and thus by Lemma 1.4.2, kR 1 (S(t))k B() kR 1 (S(0))k B() +Ct +Ct sup s2(0;t) kv(s)k B() C +tQ(M ; (t)): ForkrSk B() , by (1.5.23) and Lemma 1.4.1, we obtain krSk B() CkSk A() +tQ(M ; (t)): Forkvk B() , by the norm relation we have kvk B() 3 X m=1 m X j=0 X jj=j k@ j (@ t ) mj vk L 2 + 1 X m=4 m X j=0 X jj=j k@ j (@ t ) mj vk L 2 (j2) + (t) m2 (m 2)! C +Ckvk A() C +Q(M ; (t)): By combining the above estimates, we deduce that 2 C + (t +)Q(M ; (t)): 36 Therefore, together with (1.6.5) and (1.6.8) we arrive at k curlvk B() C + (t +)Q(M ; (t)): (1.6.9) 1.6.2 Energyequationforthepuretimederivatives(A 1 norm) In this section, we estimate the pure time-analytic norm kuk A 1 () = 1 X m=1 k(@ t ) m uk L 2 (t) (m3) + (m 3)! with the corresponding dissipative analytic norm kuk ~ A 1 () = 1 X m=4 k(@ t ) m uk L 2 (m 3)(t) m4 (m 3)! : Consider the partially linearized equation E(@ t _ u +vr _ u) + 1 L(@ x ) _ u =F; (1.6.10) where _ u = ( _ p; _ v) andE =E(S;u). The next lemma provides a dierential inequality that is used for pure time derivatives ofu. Lemma1.6.1. For all ( _ u;F ) satisfying (1.6.10), we have d dt kE 1=2 _ uk L 2C(k _ uk L 2 +kFk L 2); for a constantC 1. 37 Proof. We multiply the equation (1.6.10) by _ u and integrate inR 3 . Since L(@ x ) is skew-symmetric, we have 1 hL(@ x ) _ u; _ ui = 0; i.e., the term with 1= cancels out. Using also the Cauchy-Schwarz inequality, we get hE@ t _ u; _ uiCkr(Ev)k L 1 x k _ uk 2 L 2 +CkFk L 2k _ uk L 2; (1.6.11) and thus by Hölder’s inequality and sinceE is a positive denite symmetric matrix, we obtain from (1.6.11) d dt kE 1=2 _ uk 2 L 2 = d dt hE _ u; _ ui =h@ t E _ u; _ ui + 2hE@ t _ u; _ ui k@ t Ek L 1 x k _ uk 2 L 2 +Ckr(Ev)k L 1 x k _ uk 2 L 2 +CkFk L 2k _ uk L 2: (1.6.12) On the other hand, d dt kE 1=2 _ uk 2 L 2 = 2kE 1=2 _ uk L 2 d dt kE 1=2 _ uk L 2: (1.6.13) Now, we combine (1.6.12)–(1.6.13), and using that the low-order Sobolev norms of @ t E, @ x E, @ x v, and E 1=2 may be estimated byC (cf. Remark 1.3.4 and 1.3.5). We arrive at d dt kE 1=2 _ uk L 2 Ck _ uk 2 L 2 kE 1=2 _ uk L 2 + CkFk L 2k _ uk L 2 kE 1=2 _ uk L 2 C(k _ uk L 2 +kFk L 2); where we appealed to k _ uk L 2 =kE 1=2 E 1=2 _ uk L 2CkE 1=2 k L 1kE 1=2 _ uk L 2CkE 1=2 _ uk L 2; 38 and the lemma is proven. Using the previous lemma, the next statement provides a pure time derivative analytic estimate for the solutionu in theA 1 () norm. Lemma 1.6.2. There existt 0 > 0 suciently small depending only onM 0 and 1 > 0 suciently small depending onM ; (T ) such that fort2 (0;t 0 ) and2 (0; 1 ), we have ku(t)k A 1 C +tQ(M ; (t)); (1.6.14) for a functionQ. Proof. Form2N, we apply (@ t ) m to the equation (1.2.14). Then _ u = (@ t ) m u satises (1.6.10) with F = [E; (@ t ) m ]@ t u + [Ev; (@ t ) m ]ru: (1.6.15) Denote kuk A E = 1 X m=1 kE 1=2 (@ t ) m uk L 2 (m3) + (m 3)! (1.6.16) with the corresponding dissipative norm kuk ~ A E = 1 X m=4 kE 1=2 (@ t ) m uk L 2 (m 3) m4 (m 3)! : (1.6.17) 39 By Lemma 1.6.1 and using the notation (1.6.16)–(1.6.17), we obtain d dt kuk A E = _ kuk ~ A E + 1 X m=1 (m3) + (m 3)! d dt kE 1=2 (@ t ) m uk L 2 _ kuk ~ A E +Ckuk A 1 +C 1 X m=1 (m3) + (m 3)! kFk L 2; (1.6.18) whereF is given in (1.6.15). Note that kFk L 2 m X j=1 m j k(@ t ) j1 @ t E(@ t ) mj+1 uk L 2 + m X j=1 m j k(@ t ) j (Ev)(@ t ) mj ruk L 2 =F 1;m +F 2;m : (1.6.19) For the rst sum in (1.6.19), we have 1 X m=1 (m3) + (m 3)! F 1;m = 4 X m=1 m X j=1 (m3) + (m 3)! m j k(@ t ) j1 @ t E(@ t ) mj+1 uk L 2 + 1 X m=5 [m=2] X j=1 (m3) + (m 3)! m j k(@ t ) j1 @ t E(@ t ) mj+1 uk L 2 + 1 X m=5 m X j=[m=2]+1 (m3) + (m 3)! m j k(@ t ) j1 @ t E(@ t ) mj+1 uk L 2 =D 1 +D 2 +D 3 ; (1.6.20) where we split the sum according to the low and high values ofj. We claim D 1 C; (1.6.21) D 2 Ck@ t Ek B() kuk ~ A 1 () +Ckuk ~ A 1 () ; (1.6.22) D 3 Ck@ t Ek B() kuk A() : (1.6.23) Proof of (1.6.21): Using Hölder’s and the Sobolev inequalities, we may estimateD 1 using low order mixed space time derivative ofu andS, and from Remark 1.3.4, we obtain (1.6.21). 40 Proof of (1.6.22): Using the approach as in the estimate forS, we have D 1 = 1 X m=5 [m=2] X j=1 m3 (m 3)! m j k(@ t ) j1 @ t E(@ t ) mj+1 uk L 2 C a 1 X m=5 [m=2] X j=1 (j3) + (j 3)! k(@ t ) j1 @ t Ek L 2 ! 1=4 (j1) + (j 1)! kD 2 (@ t ) j1 @ t Ek L 2 ! 3=4 (mj 2) mj3 (mj 2)! k(@ t ) mj+1 uk L 2 A j;m ; (1.6.24) where A j;m = m! j!(mj)!(m 3)! (j 3)! 1=4 (j 1)! 3=4 (mj 2)! mj 2 Cm 3 (mj) 3 C; and a =m 3 j 3 4 + 3j 3 4 + (mj 3) 1; (1.6.25) since 1j [m=2]. By (1.6.24)–(1.6.25) and the discrete Young inequality, we obtain D 2 C 1 X m=5 [m=2] X j=1 (j3) + (j 3)! k(@ t ) j1 @ t Ek L 2 ! (mj 2) mj3 (mj 2)! k(@ t ) mj+1 uk L 2 +C 1 X m=5 [m=2] X j=1 (j1) + (j 1)! kD 2 (@ t ) j1 @ t Ek L 2 ! (mj 2) mj3 (mj 2)! k(@ t ) mj+1 uk L 2 C k@ t Ek B() +k@ t Ek L 2 kuk ~ A 1 () Ck@ t Ek B() kuk ~ A 1 () +Ckuk ~ A 1 () : 41 Proof of (1.6.23): Reversing the roles ofj andmj and proceeding as in the above argument, we may write D 3 C b 1 X m=5 m X j=[m=2]+1 (j3) + (j 3)! k(@ t ) j1 @ t Ek L 2 ! (mj2) + (mj 2)! k(@ t ) mj+1 uk L 2 ! 1=4 (mj) + (mj)! kD 2 (@ t ) mj+1 uk L 2 ! 3=4 B j;m ; (1.6.26) where B j;m = m! j!(mj)! (j 3)!(mj 2)! 1=4 (mj)! 3=4 (m 3)! Cm 3 j 3 C; and b =m 3 (j 3) + mj 2 4 + 3m 3j 4 + 0; (1.6.27) sincemj [m=2] + 1. From (1.6.26)–(1.6.27), we obtain D 3 Ck@ t Ek B() kuk A() : (1.6.28) 42 Analogously, the second sum in (1.6.19) may be separated according to low and high values ofj, ob- taining 1 X m=1 (m3) + (m 3)! F 2;m = 4 X m=1 [m=2] X j=1 (m3) + (m 3)! m j k(@ t ) j (Ev)(@ t ) mj ruk L 2 + 1 X m=5 [m=2] X j=1 (m3) + (m 3)! m j k(@ t ) j (Ev)(@ t ) mj ruk L 2 + 1 X m=5 m X j=[m=2]+1 (m3) + (m 3)! m j k(@ t ) j (Ev)(@ t ) mj ruk L 2 =D 4 +D 5 +D 6 : (1.6.29) We claim D 4 C; (1.6.30) D 5 CkEvk A() kruk A 1 () ; (1.6.31) D 6 CkEvk A() kuk A() : (1.6.32) Proof of (1.6.30): Proceeding as in the proof of (1.6.30), we obtain that the low-order mixed space-time derivatives may be estimated byC. Proof of (1.6.31): As in (1.6.24), we have D 5 C 1 X m=5 [m=2] X j=1 (m3) + (m 3)! m j k(@ t ) j (Ev)k 1=4 L 2 kD 2 (@ t ) j (Ev)k 3=4 L 2 k(@ t ) mj ruk L 2 C 1 X m=5 [m=2] X j=1 k(@ t ) j (Ev)k L 2 (j3) + (j 3)! ! 1=4 kD 2 (@ t ) j (Ev)k L 2 (j1) + (j 1)! ! 3=4 k(@ t ) mj ruk L 2 (mj3) + (mj 3)! ! CkEvk A() kruk A 1 () : 43 Proof of (1.6.32): As in (1.6.26), we arrive at D 6 C 1 X m=5 m X j=[m=2]+1 (m3) + (m 3)! m j k(@ t ) j (Ev)k L 2k(@ t ) mj ruk 1=4 L 2 k L 2kD 2 (@ t ) mj ruk 3=4 L 2 C 1 X m=5 m X j=[m=2]+1 k(@ t ) j (Ev)k L 2 (j3) + (j 3)! ! k(@ t ) mj ruk L 2 (mj2) + (mj 2)! ! 1=4 kD 2 (@ t ) mj ruk L 2 (mj) + (mj)! ! 3=4 CkEvk A() kuk A() : Collecting the above estimates (1.6.20)–(1.6.23) and (1.6.29)–(1.6.32), we obtain from (1.6.19), 1 X m=1 (m3) + (m 3)! F 1;m + 1 X m=1 (m3) + (m 3)! F 2;m D 1 +D 2 +D 3 +D 4 +D 5 +D 6 C +Ck@ t Ek B() kuk ~ A 1 () +Ckuk ~ A 1 +Ck@ t Ek B() kuk A() +CkEvk A() kruk A 1 () +CkEvk A() kuk A() ; (1.6.33) where we estimatek@ t Ek B() using Lemma 1.5.3, andkEvk A() with (1.5.13) and Lemma 1.5.4. In order to estimate the dissipative termkruk A 1 () , we recall the elliptic regularity for the div-curl system krvk L 2Ck divvk L 2 +Ck curlvk L 2; (1.6.34) which, together with the denition of theA 1 () norm, leads to kruk A 1 () CkL(@ x )uk A 1 () +Ck curlvk A 1 () : (1.6.35) 44 To treat the divergence part of the dissipative term, we rewrite the equation (1.2.14) as L(@ x )u =E(S;u)(@ t u +vru): (1.6.36) Form2N, we apply (@ t ) m to the equation (1.6.36), obtaining k(@ t ) m L(@ x )uk L 2C m X j=0 m j k(@ t ) j E(@ t ) mj+1 uk L 2 +CkEvk L 1k(@ t ) m ruk L 2 +C m X j=1 k(@ t ) j (Ev)(@ t ) mj ruk L 2: From here we arrive at kL(@ x )uk A 1 () = 1 X m=1 k(@ t ) m L(@ x )uk L 2 (m3) + (m 3)! 2 X m=1 kE(@ t ) m+1 uk L 2 (m3) + (m 3)! + 1 X m=3 kEk L 1k(@ t ) m+1 uk L 2 m3 (m 3)! +C 1 X m=1 m X j=1 m j k(@ t ) j1 @ t E(@ t ) mj+1 uk L 2 (m3) + (m 3)! +C 1 X m=1 k(@ t ) m rvk L 2 (m3) + (m 3)! +C 1 X m=1 m X j=1 m j k(@ t ) j (Ev)(@ t ) mj ruk L 2 (m3) + (m 3)! C +CkEk L 1kuk ~ A 1 +Ckrvk A 1 () +C 1 X m=1 (m3) + (m 3)! F 1;m +C 1 X m=1 (m3) + (m 3)! F 2;m ; (1.6.37) where we used the notation from (1.6.19) in the last inequality. The third term on the far right side of (1.6.37) may be absorbed in the left side of (1.6.35) when is suciently small, and the fourth and fth terms may be absorbed into the left side of (1.6.33) when is suciently small depending onM ; (T ). 45 To treat the curl part of the dissipative term, using the similar technique for the curl estimate above, we have k(@ t ) m curlvk L 2 =k(@ t ) m curl(R 0 r 0 v)k L 2 m X j=0 m j k(@ t ) j R 0 (@ t ) mj curl(r 0 v)k L 2 + m X j=0 m j k(@ t ) j rR 0 (@ t ) mj r 0 vk L 2: We use a similar technique as in the proofs of (1.4.4)–(1.4.5), obtaining k curlvk A 1 () 1 X m=1 m X j=0 m j k(@ t ) j R 0 (@ t ) mj curl(r 0 v)k L 2 (m3) + (m 3)! + 1 X m=1 m X j=0 m j k(@ t ) j rR 0 (@ t ) mj r 0 vk L 2 (m3) + (m 3)! CkR 0 k A() k curl(r 0 v)k ~ B() +CkR 0 k A() k curl(r 0 v)k B() +CkR 0 k A() kr 0 vk A() +CkR 0 k ~ A() kr 0 vk A() : (1.6.38) SinceR 0 is a function ofS, it satises the inhomogeneous transport equation @ t R 0 +vrR 0 = 0: Then Lemma 1.4.1 and (1.4.20) imply that d dt kR 0 k A() kR 0 k ~ A() ( _ +Ckvk A() ) +CkR 0 k A() +Ckvk A() +C: (1.6.39) 46 Coupling (1.6.3), (1.6.18), and (1.6.39), we arrive at d dt kuk A E +kR 0 k A() +k curl(r 0 v)k B() _ kuk ~ A E +Ckuk A 1 () +C 1 X m=1 (m3) + (m 3)! kFk L 2 +kR 0 k ~ A() ( _ +Ckvk A() ) +CkR 0 k A() +Ckvk A() + _ k curl(r 0 v)k ~ B() +CkGk B() +Ckvk B() +C: (1.6.40) Collecting the estimates (1.6.18), (1.6.33), (1.6.35), (1.6.37), (1.6.38), and (1.6.40), we arrive at d dt kuk A E +kR 0 k A() +k curl(r 0 v)k B() kuk ~ A E _ +Ck@ t Ek B() +C +CkEvk A() +Ck@ t Ek B() kuk A() +CkEvk A() kuk A() +CkEvk A() +CkEvk A() kR 0 k A() k curl(r 0 v)k B() +CkEvk A() kR 0 k A() kr 0 vk A() +kR 0 k ~ A() _ +Ckvk A() +CkEvk A() kr 0 vk A() +CkR 0 k A() +Ckvk A() +k curl(r 0 v)k ~ B() _ +CkEvk A() kR 0 k A() +CkGk B() +Ckvk B() +C; where we appealed to kuk ~ A 1 () Ckuk ~ A E by the boundedness ofkE 1=2 k L 1. Now, assume that the radius(t) decreases suciently fast so that the factors next tokuk ~ A E ,kR 0 k ~ A() , andk curl(r 0 v)k ~ B() are less than or equal to 0. Integrating the resulting inequality on [0;t], we get kuk A E ku(0)k A E +kR 0 (0)k A() +k curl(r 0 v)(0)k B() +tQ(M ; (t))C +tQ(M ; (t)); 47 and sincekuk A 1 Ckuk A E , the proof is concluded. 1.6.3 Energyequationforthemixedderivatives(A 2 norm) Here we estimate the mixed space-time analytic norm. For this purpose, denote kuk A 2 () = 1 X m=1 m X j=1 X jj=j k@ (@ t ) mj uk L 2 (j3) + (t) (m3) + (m 3)! ; and let kuk ~ A 2 () = 1 X m=4 m X j=1 X jj=j k@ (@ t ) mj uk L 2 (j3) + (m 3)(t) m4 (m 3)! be the corresponding dissipative analytic norm. Note that ku(t)k A =ku(t)k A 1 +ku(t)k A 2 : Lemma1.6.3. GivenM 0 > 0, there exist a functionQ and 0 such that for2 (0; 0 ),< 1, the solution of (1.2.14) satises ku(t)k A 2 C + (t + + +)Q(M ; (t)): (1.6.41) 48 Proof of Lemma 1.6.3. By the denition of theA 2 () norm, we have kuk A 2 () = 1 X m=1 m X j=1 X jj=j k@ (@ t ) mj uk L 2 (j3) + (m3) + (m 3)! = 3 X m=1 m X j=1 X jj=j k@ (@ t ) mj uk L 2 1 (m 3)! + 1 X m=4 m X j=4 X jj=j k@ (@ t ) mj uk L 2 j3 m3 (m 3)! + 1 X m=4 3 X j=1 X jj=j k@ (@ t ) mj uk L 2 m3 (m 3)! =P 1 +P 2 +P 3 ; where we split the sum according to the high and low values ofj andm. We claim P 1 C; (1.6.42) P 2 C + (t + + +)Q(M ; (t)); (1.6.43) P 3 C + (t + +)Q(M ; (t)): (1.6.44) Firstly, (1.6.42) follows by using Sobolev inequalities and Remark 1.3.4. Proof of (1.6.43): We rewrite equation (1.2.14) as L(@ x )u =E(S;u)(@ t u +vru): (1.6.45) 49 Form 3, andjj = j where 3 j m, we commute@ (@ t ) mj with (1.6.45), and using div-curl regularity (1.6.34), we obtain kr@ (@ t ) mj uk L 2CkL(@ x )@ (@ t ) mj uk L 2 +Ck curl(@ (@ t ) mj v)k L 2 C mj X k=0 j X l=0 X jj=l; mj k k@ (@ t ) k E@ (@ t ) mjk+1 uk L 2 +CkEvk L 1k@ (@ t ) mj ruk L 2 +C j X l=0 mj X k=0 l+k1 X jj=l; mj k k@ (@ t ) k (Ev)@ (@ t ) mjk ruk L 2 +Ck@ (@ t ) mj curlvk L 2: (1.6.46) 50 The second term on the far right side of (1.6.46) can be absorbed into the left side when is suciently small. Multiply the above estimate with appropriate weights and then sum, with change of variables we obtain P 2 = 1 X m=4 m X j=4 X jj=j k@ (@ t ) mj uk L 2 j3 m3 (m 3)! C 1 X m=3 m X j=3 X jj=j kr@ (@ t ) mj uk L 2 j2 m2 (m 2)! C 1 X m=4 m1 X j=3 X jj=j mj1 X k=0 j X l=0 X jj=l; j3 m3 (m 3)! mj 1 k k@ (@ t ) k E@ (@ t ) mjk uk L 2 +C 1 X m=3 m X j=3 X jj=j j X l=0 mj X k=0 l+k1 X jj=l; j2 m2 (m 2)! mj k k@ (@ t ) k (Ev)@ (@ t ) mjk ruk L 2 +C 1 X m=3 m X j=3 X jj=j j2 m2 (m 2)! k@ (@ t ) mj curlvk L 2 CkEuk A() +C(kEvk A() +kEvk L 2)kuk A() +Ck curlvk B() C + (t + + +)Q(M ; (t)); (1.6.47) where the last inequality follows from the estimates (1.5.13) and (1.6.9). 51 Proof of (1.6.44): Form 4 andjj = j where 1 j 3, we proceed as in (1.6.46)–(1.6.47) and obtain P 3 = 1 X m=4 3 X j=1 X jj=j k@ (@ t ) mj uk L 2 m3 (m 3)! C 1 X m=3 2 X j=0 X jj=j kr@ (@ t ) mj uk L 2 m2 (m 2)! C 1 X m=3 2 X j=0 X jj=j kL(@ x )@ (@ t ) mj uk L 2 m2 (m 2)! +C 1 X m=3 2 X j=0 X jj=j k curl(@ (@ t ) mj u)k L 2 m2 (m 2)! : (1.6.48) Therefore, P 3 C 1 X m=3 2 X j=0 X jj=j j X l=0 X jj=l; m2 (m 2)! k@ E@ (@ t ) mj+1 uk L 2 +C 1 X m=3 2 X j=0 X jj=j mj X k=1 j X l=0 X jj=l; m2 (m 2)! mj k k@ (@ t ) k1 @ t E@ (@ t ) mjk+1 uk L 2 +C 1 X m=3 2 X j=0 X jj=j j X l=0 mj X k=0 l+k1 X jj=l; m2 (m 2)! mj k k@ (@ t ) k (Ev)@ (@ t ) mjk ruk L 2 +C 1 X m=3 2 X j=0 X jj=j m2 (m 2)! k@ (@ t ) mj curlvk L 2: (1.6.49) 52 The second term on the far right side is estimated by using Lemmas 1.5.1 and 1.5.3, while the third and fourth term can be estimated analogously as in (1.6.47). For the rst term, denoted byP 31 , we use the div-curl regularity to reduce the spatial derivative. We split it according to the values ofj, obtaining P 31 = 1 X m=3 X jj=0 m2 (m 2)! kE(@ t ) m+1 uk L 2 + 1 X m=3 X jj=1 1 X l=0 X jj=l; m2 (m 2)! k@ E@ (@ t ) m uk L 2 + 1 X m=3 X jj=2 2 X l=0 X jj=l; m2 (m 2)! k@ E@ (@ t ) m1 uk L 2 Ckuk A 1 () +C 1 X m=2 kr(@ t ) m uk L 2 m2 (m 2)! +C 1 X m=3 kr 2 (@ t ) m1 uk L 2 m2 (m 2)! : (1.6.50) The rst term on the far side is bounded byC +tQ(M ; (t)) by Lemma 1.6.2, while the second and third terms can be estimated analogously to (1.6.48)–(1.6.50). Combining the resulting inequalities, we obtain P 3 C + (t + +)Q(M ; (t)); and the lemma then follows by (1.6.42)–(1.6.44). Proof of Lemma 1.3.3. The inequality (1.3.13) follows by using (1.4.1), (1.6.14), and (1.6.41). 1.7 TheMachlimit In this section, we prove the second main theorem on the Mach limit in the spaceX. Proof of Theorem 1.3.2. Let> 0 be a small constant, which is to be determined below. For the sake of con- tradiction, we assume that (v ;p ;S ) does not converge to (v (inc) ; 0;S (inc) ) inC([0;T ];X ). Then there 53 exists a sequence (v n ;p n ;S n ) which does not converge to (v (inc) ; 0;S (inc) ) inC([0;T ];X ) as n ! 0. Recall from [121, Theorem 1.4] that (v n ;p n ;S n ) converges to (v (inc) ; 0;S (inc) ) inL 1 ([0;T ];L 2 (R 3 )) as n ! 0. Fork;n2N, we denev kn (t) =v k (t)v n (t). Form2N and2N 3 0 , using integration by parts and the Cauchy-Schwarz inequality leads to k@ v kn k 2 L 2 = @ v kn ;@ v kn = (1) jj v kn ;@ 2 v kn kv kn k L 2k@ 2 v kn k L 2: Summing overjj =m withm2N such thatm 4, we obtain 1 X m=4 X jj=m k@ v kn k L 2 (m3) + (m 3)! kv kn k 1=2 L 2 1 X m=4 X jj=m k@ 2 v kn k 1=2 L 2 (m3) + (m 3)! =kv kn k 1=2 L 2 1 X m=4 X jj=m k@ 2 v kn k L 2 (2m3) + (2m3) + (2m 3)! ! 1=2 (2m 3)! 1=2 (m3) + (m 3)! (2m3) + =2 (2m3) + =2 CM 1=2 kv kn k 1=2 L 2 1 X m=4 X jj=m (2m 3)! 1=2 (m3) + (m 3)! (2m3) + =2 (2m3) + =2 M 1=2 kv kn k 1=2 L 2 1 X m=4 C m (2m 3)! 1=2 (m3) + (m 3)! (2m3) + =2 (2m3) + =2 ; (1.7.1) whereC > 0 is a xed universal constant andM is as in (1.3.6). Now choose > 0 suciently small so that= 1=CC 0 on the whole time interval [0;T 0 ], whereC 0 is suciently large, and obtain 1 X m=4 X jj=m k@ v kn k L 2 (m3) + (m 3)! M 1=2 3=2 kv kn k 1=2 L 2 1 X m=4 (2m 3)! 1=2 C m 0 (m 3)! M 1=2 3=2 kv kn k 1=2 L 2 ; (1.7.2) where we used Stirling’s formula and assumedC 0 to be suciently large so the sum converges. Analo- gously, we setp kn (t) =p k (t)p n (t) andS kn (t) =S k (t)S n (t) and proceed as in above obtaining 1 X m=4 X jj=m k@ p kn k L 2 m3 (m 3)! M 1=2 3=2 kv kn k 1=2 L 2 ; (1.7.3) 54 and 1 X m=4 X jj=m k@ S kn k L 2 m3 (m 3)! M 1=2 3=2 kS kn k 1=2 L 2 : (1.7.4) Note also thatkv kn k H 3 Ckv kn k 3=4 H 4 kv kn k 1=4 L 2 Ckv kn k 1=4 L 2 , by Remark 1.3.4, with analogous inequal- ities forp kn andS kn . SinceM and are xed constants, we infer from (1.7.2)–(1.7.4) that the sequence f(v n ;p n ;S n )g is Cauchy inC([0;T 0 ];X ) which implies that it converges inC([0;T 0 ];X ), which is a contradiction. Therefore,f(v ;p ;S )g is convergent and converges to (v (inc) ; 0;S (inc) ) inC([0;T 0 ];X ) as! 0. 1.8 Analyticityassumptionsontheinitialdata In this section, we assume that the initial data satises (1.3.2), and intend to prove that for smallern we have 3 X n=0 1 X j=0 X jj=j k@ (@ t ) n u(0)k L 2 (j+n3) + 0 (j +n 3)! ; (1.8.1) and 3 X n=0 1 X j=0 X jj=j k@ (@ t ) n S(0)k L 2 (j+n3) + 0 (j +n 3)! ; (1.8.2) 55 where > 0 is a suciently large constant depending onM 0 ; for larger values ofn, we claim that there exists a suciently small parameter> 0 depending onM 0 , such that for allk 4 we have k X n=4 1 X j=0 X jj=j k@ (@ t ) n u(0)k L 2 n3 (j+n3) + 0 (j +n 3)! 1 (1.8.3) and k X n=4 1 X j=0 X jj=j k@ (@ t ) n S(0)k L 2 n3 (j+n3) + 0 (j +n 3)! 1: (1.8.4) In (1.8.3) and (1.8.4) we then choose ~ 0 = 0 =2 and using (1.8.1)–(1.8.4), we get k(p 0 ;v 0 ;S 0 )k A(~ 0 ) = 1 X n=0 1 X j=0;n+j1 X jj=j k@ (@ t ) n (u;S)(0)k L 2 ~ (j+n3) + 0 (j +n 3)! 3 X n=0 1 X j=0 X jj=j k@ (@ t ) n (u;S)(0)k L 2 (j+n3) + 0 (j +n 3)! + 1 X n=4 1 2 n3 1 X j=0 X jj=j k@ (@ t ) n (u;S)(0)k L 2 n3 (j+n3) + 0 (j +n 3)! + 1 X n=4 1 2 n3 C; obtaining (1.3.4). In the remainder of this section, we prove (1.8.1)–(1.8.4). Forn = 0, we use the assumption (1.3.2) on the initial data to obtain 1 X j=0 X jj=j k@ (u;S)(0)k L 2 (j3) + 0 (j 3)! 0 ; (1.8.5) for some constant 0 > 0. Next, forn = 1, we apply@ to (1.2.15) wherejj =j2N 0 , which leads to @ @ t S = X @ v@ rS: 56 Therefore, 1 X j=0 X jj=j k@ @ t Sk L 2 (j2) + 0 (j 2)! C 1 X j=0 X jj=j j X l=0 X ;jj=l k@ v@ rSk L 2 (j2) + 0 (j 2)! : (1.8.6) We split the right side of (1.8.6) according to low and high valuesl. By Hölder and Sobolev inequalities, we have 1 X j=0 X jj=j k@ @ t Sk L 2 (j2) + 0 (j 2)! C 1 X j=0 X jj=j X 0l[j=2] X ;jj=l kD 2 @ vk L 2 (l1) + 0 (l 1)! ! 3=4 k@ vk L 2 (l3) + 0 (l 3)! ! 1=4 k@ rSk L 2 (jl2) + 0 (jl 2)! ! (l 1)! 3=4 (l 3)! 1=4 (jl 2)!j! (j 2)!(jl)!l! +C 1 X j=0 X jj=j X [j=2]+1lj X ;jj=l k@ vk L 2 (l3) + 0 (l 3)! ! kD 2 @ rSk L 2 (jl) + 0 (jl)! ! 3=4 k@ rSk L 2 (jl2) + 0 (jl 2)! ! 1=4 (jl)! 3=4 (jl 2)! 1=4 (l 3)!j! (j 2)!(jl)!l! : (1.8.7) One may check that (l 1)! 3=4 (l 3)! 1=4 (jl 2)!j! (j 2)!(jl)!l! C; forl [j=2], while (jl)! 3=4 (jl 2)! 1=4 (l 3)!j! (j 2)!(jl)!l! C; (1.8.8) 57 forl [j=2] + 1. Collecting the estimates (1.8.5) and (1.8.7)–(1.8.8), we obtain 1 X j=0 X jj=j k@ @ t Sk L 2 (j2) + 0 (j 2)! C(kvk A 0 ( 0 ) +kvk L 2)kSk A 0 ( 0 ) 1 ; where 1 =Q( 0 ) and kuk A 0 ( 0 ) = 1 X j=1 X jj=j k@ uk L 2 (j3) + 0 (j 3)! : As for (1.8.1), we rewrite the equation (1.2.14) as @ t u =vru ~ EL(@ x )u; (1.8.9) where we denoted ~ E(S;u) =E 1 (S;u). Applying@ to (1.8.9), wherejj =j 0, we get k@ @ t uk L 2C j X l=0 X ;jj=l k@ v@ ruk L 2 +C j X l=0 X ;jj=l k@ ~ E@ ruk L 2; from where 1 X j=0 X jj=j k@ @ t uk L 2 (j2) + 0 (j 2)! C 1 X j=0 X jj=j j X l=0 X ;jj=l k@ v@ ruk L 2 (j2) + 0 (j 2)! +C 1 X j=0 X jj=j j X l=0 X ;jj=l k@ ~ E@ ruk L 2 (j2) + 0 (j 2)! =I 1 +I 2 : (1.8.10) 58 The termI 1 can be estimated analogously as in (1.8.6)–(1.8.8), obtaining 1 Q( 0 ). For the termI 2 , we proceed as in (1.8.6)–(1.8.8), obtaining 2 Ck ~ Ek A 0 ( 0 ) kuk A 0 ( 0 ) +Ck ~ Ek L 1kuk A 0 ( 0 ) . One may easily check that the product rules in Lemmas 1.5.1 and 1.5.2 hold for the normA 0 ( 0 ). Thus we have k ~ Ek A 0 ( 0 ) Q(kuk A 0 ( 0 ) +kuk L 2;kSk A 0 ( 0 ) +kSk L 2)Q( 0 ): (1.8.11) Combining (1.8.10)–(1.8.11), we may write 1 X j=0 X jj=j k@ @ t uk L 2 (j2) + 0 (j 2)! 1 ; where 1 =Q( 0 ). Forn = 2 andn = 3, the proof is completely analogous and we obtain 1 X j=0 X jj=j k@ (@ t ) n Sk L 2 (j+n3) + 0 (j +n 3)! n and 1 X j=0 X jj=j k@ (@ t ) n uk L 2 (j+n3) + 0 (j +n 3)! n ; for suciently large n depending on 0 . Summing overn from 0 to 3, we obtain (1.8.1) and (1.8.2) for suciently large =Q( 0 ). We x for the rest of the proof. Next, we prove (1.8.3) and (1.8.4) for allk 4 using induction and starting with the casek = 4. First, we apply@ (@ t ) 3 to (1.2.15), wherejj =j 0, obtaining @ (@ t ) 3 @ t S = X 3 X n=0 3 n @ (@ t ) n v@ (@ t ) 3n rS: 59 Using the splitting argument as in (1.8.6)–(1.8.7), 1 X j=0 X jj=j k@ (@ t ) 4 Sk L 2 j+1 0 (j + 1)! C 1 X j=0 X jj=j X 0l[j=2] X ;jj=l 3 X n=0 k@ (@ t ) 3n rSk L 2 (jnl+1) + 0 (jnl + 1)! ! k@ (@ t ) n vk L 2 (l+n3) + 0 (l +n 3)! ! 1=4 kD 2 @ (@ t ) n vk L 2 (l+n1) + 0 (l +n 1)! ! 3=4 +C 1 X j=0 X jj=j X [j=2]+1lj X ;jj=l 3 X n=0 k@ (@ t ) n vk L 2 (l+n3) + 0 (l +n 3)! ! kD 2 @ (@ t ) 3n rSk L 2 (jnl+3) + 0 (jnl + 3)! ! 3=4 k@ (@ t ) 3n rSk L 2 (jnl+1) + 0 (jnl + 1)! ! 1=4 : (1.8.12) Appealing to (1.8.1) and (1.8.2), we arrive at 1 X j=0 X jj=j k@ (@ t ) 4 Sk L 2 j+1 0 (j + 1)! C 3 X n=0 0 @ 1 X j=0 X jj=j k@ (@ t ) n uk L 2 (j+n3) + 0 (j +n 3)! 1 A 0 @ 1 X j=0 X jj=j k@ (@ t ) 3n Sk L 2 (jn) + 0 (jn)! 1 A 1 2 ; (1.8.13) where we set = 1=Q(), concluding the proof of (1.8.4) fork = 4. As for (1.8.3), we apply@ (@ t ) 3 to (1.8.9), wherejj =j 0, obtaining k@ (@ t ) 4 uk L 2C j X l=0 X ;jj=l 3 n k@ (@ t ) n v@ (@ t ) 3n ruk L 2 +C j X l=0 X ;jj=l 3 n k@ (@ t ) n ~ E@ (@ t ) 3n ruk L 2: 60 Therefore, we get 1 X j=0 X jj=j k@ (@ t ) 4 uk L 2 j+1 0 (j + 1)! C 1 X j=0 X jj=j j X l=0 X ;jj=l 3 X n=0 3 n k@ (@ t ) n v@ (@ t ) 3n ruk L 2 j+1 0 (j + 1)! +C 1 X j=0 X jj=j j X l=0 X ;jj=l 3 X n=0 3 n k@ (@ t ) n ~ E@ (@ t ) 3n ruk L 2 j+1 0 (j + 1)! =I 41 +I 42 : (1.8.14) The termI 41 can be estimated as in (1.8.12)–(1.8.13), obtainingI 41 1=2, whileI 42 can be treated in a similar fashion as in (1.8.11), arriving atI 42 Ck ~ Ek A 3 ( 0 ) kuk A 3 ( 0 ) +Ck ~ Ek L 1kuk A 3 ( 0 ) , where for eachk 3, we denote kuk A k ( 0 ) = k X n=0 1 X j=0;j+n1 X jj=j k@ (@ t ) n uk L 2 (n3) + (j+n3) + 0 (j +n 3)! : (1.8.15) One can easily check that Lemma 1.5.1 and 1.5.2 hold for theA k ( 0 )-norm for eachk 3. Therefore, I 42 1=2 by choosing = 1=Q(). There, we obtain (1.8.3) fork = 4. Now we assume that we have (1.8.3) and (1.8.4) for somek 4, and prove them fork + 1. Forn 3, we apply@ (@ t ) n to (1.2.15), wherejj =j 0, obtaining @ (@ t ) n @ t S = X n X m=0 @ (@ t ) m v@ (@ t ) nm rS; 61 from where 1 X j=0 X jj=j k@ (@ t ) n+1 Sk L 2 n2 (j+n2) + 0 (j +n 2)! C 1 X j=0 X jj=j j X l=0 X ;jj=l n X m=0 n m k@ (@ t ) m v@ (@ t ) nm rSk L 2 n2 (j+n2) + 0 (j +n 2)! : We split the above sum according to the low and high values ofl +n. Using a similar argument as in (1.8.12), we get 1 X j=0 X jj=j k@ (@ t ) n+1 Sk L 2 n2 (j+n2) + 0 (j +n 2)! C 1 X j=0 X jj=j j X l=0 X ;jj=l n X m=0 kD 2 @ (@ t ) m vk L 2 (m3) + (l+m1) + 0 (l +m 1)! ! 3=4 k@ (@ t ) m vk L 2 (m3) + (l+m3) + 0 (l +m 3)! ! 1=4 k@ (@ t ) nm rSk L 2 (nm3) + (j+nlm2) + 0 (j +nlm 2)! ! 1 f0l+m[(j+n)=2]g +C 1 X j=0 X jj=j j X l=0 X ;jj=l n X m=0 kD 2 @ (@ t ) nm rSk L 2 (nm3) + (j+nlm) + 0 (j +nlm)! ! 3=4 k@ (@ t ) nm rSk L 2 (nm3) + (j+nlm2) + 0 (j +nlm 2)! ! 1=4 k@ (@ t ) m vk L 2 (m3) + (l+m3) + 0 (l +m 3)! ! 1 f[(j+n)=2]+1l+mj+ng ; 62 from where 1 X j=0 X jj=j k@ (@ t ) n+1 Sk L 2 n2 (j+n2) + 0 (j +n 2)! C n X m=0 0 @ 1 X j=0 X jj=j k@ (@ t ) m vk L 2 (m3) + (j+m3) + 0 (j +m 3)! 1 A 0 @ 1 X j=0 X jj=j k@ (@ t ) nm Sk L 2 (nm3) + (j+nm3) + 0 (j +nm 3)! 1 A : (1.8.16) Summing the above estimate inn from 3 tok, we get k+1 X n=4 1 X j=0 X jj=j k@ (@ t ) n Sk L 2 n3 (j+n3) + 0 (j +n 3)! = k X n=3 1 X j=0 X jj=j k@ (@ t ) n+1 Sk L 2 n2 (j+n2) + 0 (j +n 2)! ; which is bounded from above by C k X n=3 n X m=0 0 @ 1 X j=0 X jj=j k@ (@ t ) m vk L 2 (m3) + (j+m3) + 0 (j +m 3)! 1 A 0 @ 1 X j=0 X jj=j k@ (@ t ) nm Sk L 2 (nm3) + (j+nm3) + 0 (j +nm 3)! 1 A C 0 @ k X m=0 1 X j=0 X jj=j k@ (@ t ) m vk L 2 (m3) + (j+m3) + 0 (j +m 3)! 1 A 0 @ k X m=0 1 X j=0 X jj=j k@ (@ t ) m Sk L 2 (m3) + (j+m3) + 0 (j +m 3)! 1 A : (1.8.17) By (1.8.1) and (1.8.2), and the inductive hypothesis (1.8.3)–(1.8.4) fork, we arrive at k+1 X n=4 1 X j=0 X jj=j k@ (@ t ) n Sk L 2 n3 (j+n3) + 0 (j +n 3)! 1 2 ; where we choose = 1=Q(), which leads to (1.8.4) fork + 1. 63 As for (1.8.3), we apply@ (@ t ) n to (1.8.9) wherejj = j 0 andn 3. Similarly to (1.8.10), we obtain 1 X j=0 X jj=j k@ (@ t ) n+1 uk L 2 n2 j+n2 0 (j +n 2)! C 1 X j=0 X jj=j j X l=0 X ;jj=l n X m=0 n m k@ (@ t ) m v@ (@ t ) nm ruk L 2 n3 j+n2 0 (j +n 2)! +C 1 X j=0 X jj=j j X l=0 X ;jj=l n X m=0 n m k@ (@ t ) m ~ E@ (@ t ) nm ruk L 2 n3 j+n2 0 (j +n 2)! =J 1n +J 2n : (1.8.18) For the termJ 1n , we proceed as in (1.8.16)–(1.8.17), obtaining k X n=3 J 1n 1 2 : (1.8.19) 64 For the termJ 2n , we split the sum according to the low and high values ofl +n. Proceeding as in (1.8.16), we arrive at J 2n C 1 X j=0 X jj=j k ~ Ek L 1k@ (@ t ) n ruk L 2 n3 j+n2 0 (j +n 2)! +C 1 X j=0 X jj=j j X l=0 X ;jj=l n X m=0 kD 2 @ (@ t ) m ~ Ek L 2 (m3) + (l+m1) + 0 (l +m 1)! ! 3=4 k@ (@ t ) m ~ Ek L 2 (m3) + (l+m3) + 0 (l +m 3)! ! 1=4 k@ (@ t ) nm ruk L 2 (nm3) + (j+nlm2) + 0 (j +nlm 2)! ! 1 f1l+m[(j+n)=2]g +C 1 X j=0 X jj=j j X l=0 X ;jj=l n X m=0 kD 2 @ (@ t ) nm ruk L 2 (nm3) + (j+nlm) + 0 (j +nlm)! ! 3=4 k@ (@ t ) nm ruk L 2 (nm3) + (j+nlm2) + 0 (j +nlm 2)! ! 1=4 k@ (@ t ) m ~ Ek L 2 (m3) + (l+m3) + 0 (l +m 3)! ! 1 f[(j+n)=2]+1l+mj+ng ; and thus J 2n C 1 X j=0 X jj=j k@ (@ t ) n uk L 2 (n3) + (j+n3) + 0 (j +n 3)! +C n X m=0 0 @ 1 X j=0 X jj=j;m+j1 k@ (@ t ) m ~ Ek L 2 (m3) + (j+m3) + 0 (j +m 3)! 1 A 0 @ 1 X j=0 X jj=j k@ (@ t ) nm uk L 2 (nm3) + (j+nm3) + 0 (j +nm 3)! 1 A : 65 Summing the above estimate inn from 3 tok , we obtain k X n=3 J 2n C k X n=3 1 X j=0 X jj=j k@ (@ t ) n uk L 2 (n3) + (j+n3) + 0 (j +n 3)! +C k X n=3 n X m=0 0 @ 1 X j=0 X jj=j;m+j1 k@ (@ t ) m ~ Ek L 2 (m3) + (j+m3) + 0 (j +m 3)! 1 A 0 @ 1 X j=0 X jj=j k@ (@ t ) nm uk L 2 (nm3) + (j+nm3) + 0 (j +nm 3)! 1 A Ckuk A k ( 0 ) +Ck ~ Ek A k ( 0 ) (kuk A k ( 0 ) +kuk L 2); (1.8.20) where we used theA k ( 0 ) norm in (1.8.15). The rst term on the right side of above can be estimated by 1=4, for suciently small = 1=Q(). For the second term of the right-hand side of (1.8.20), it is easy to check that the product rules in Lemmas 1.5.1 and 1.5.2 hold for the normA k ( 0 ), and the functionQ in Lemma 1.5.2 is independent ofk. Therefore, from (1.8.20) and the inductive hypothesis (1.8.3)–(1.8.4) for k, we obtain k X n=3 J 2n 1 4 +Ck ~ Ek A k ( 0 ) (kuk A k ( 0 ) +kuk L 2) 1 4 +Q(kuk A k ( 0 ) +kuk L 2;kSk A k ( 0 ) +kSk L 2) 1 2 : (1.8.21) Finally, combining (1.8.18), (1.8.19), and (1.8.21), k+1 X n=4 1 X j=0 X jj=j k@ (@ t ) n uk L 2 n3 j+n3 0 (j +n 3)! = k X n=3 1 X j=0 X jj=j k@ (@ t ) n+1 uk L 2 n2 j+n2 0 (j +n 2)! k X n=3 J 1n + k X n=3 J 2n 1; concluding the proof of (1.8.3) fork + 1. 66 1.9 TheMachlimitinaGevreynorm Theorem 1.3.1 shows that if the initial data is analytic, then the Mach limit holds in an analytic norm. In this section, we show that if, more generally, the initial data is Gevrey, then the Mach limit holds in the Gevrey norm. Thus, assume the initial data is Gevrey regular that satises 1 X m=0 X jj=m k@ (p 0 ;v 0 ;S 0 )k L 2 (m3) + 0 (m 3)! s M 0 ; (1.9.1) wheres 1 is the Gevrey index and , 0 ;M 0 > 0 are xed constants. Note that whens = 1 we recover the class of real-analytic functions. Also, for the Sobolev regularity, we assume that we have (1.3.1). Similarly to (1.3.3), we dene the mixed weighted Gevrey norm kuk G() = 1 X m=1 m X j=0 X jj=j k@ (@ t ) mj uk L 2 (j3) + (t) (m3) + (m 3)! s ; where 2 (0; 1] represents the mixed space-time Gevrey radius and 2 (0; 1] is a xed parameter depending onM 0 . Proceeding as in Section 1.8, we can prove that with = 1 we have k(p 0 ;v 0 ;S 0 )k G(~ 0 ) Q(M 0 ) (1.9.2) for some ~ 0 > 0 depending on 0 and M 0 . Thus (1.9.2) holds for any 2 (0; 1] as it is an increasing function of. We also dene the analyticity radius function as (t) =(0)Kt; (1.9.3) 67 where(0) minf~ 0 ; 1g is a suciently small parameter, andK 1 is a suciently large parameter de- pending onM 0 . We shall work on the time interval [0;T 0 ] whereT 0 > 0 respects (1.3.15) and Remark 1.3.4. The rst theorem generalizes Theorem 1.3.1 by showing uniform boundedness in the Gevrey norms. Theorem 1.9.1. Assume that the initial data (p 0 ;v 0 ;S 0 ) satises (1.3.1) and (1.9.1), where s 1 and 0 ;M 0 > 0. There exist suciently small constants;(0); 0 ;T 0 > 0, depending on 0 ,s, andM 0 , such that k(p ;v ;S )(t)k G() M; 0< 0 ; t2 [0;T 0 ]; (1.9.4) where is as in (1.9.3) andK andM are suciently large constants depending ons andM 0 . Proof of Theorem 1.9.1. We proceed exactly as in Sections 1.4–1.6, obtaining the a priori estimates analo- gous to (1.3.13). Then we use a similar argument as in Section 1.3 to prove (1.9.4) We omit further details. Similarly to (1.3.8), we introduce the spatial Gevrey norm kuk Y = 1 X m=1 X jj=m k@ uk L 2 (m3) + (m 3)! s ; where> 0 is as in (1.3.7). The next theorem provides convergence of the solution in (1.9.1) to the corresponding incompressible Euler equation in the Gevrey space. Theorem1.9.2. Let> 0beasin (1.3.7),andassumethattheinitialdata (v 0 ;S 0 )convergesto (v 0 ;S 0 )in Y and inL 2 as! 0, andS 0 decays suciently rapidly at innity in the sense jS 0 (x)jCjxj 1 ; jrS 0 (x)jCjxj 2 ; 68 for 0 < 0 and some constants C and > 0. Then (v ;p ;S ) converges to (v (inc) ; 0;S (inc) ) in C([0;T 0 ];Y ), where (v (inc) ;S (inc) ) is the solution to (1.3.9)–(1.3.11) with the initial data (w 0 ;S 0 ), andw 0 is the unique solution of divw 0 = 0; curl(r 0 w 0 ) = curl(r 0 v 0 ); withr 0 =r(S 0 ; 0). Proof of Theorem 1.9.2. Theorem 1.9.2 follows by using arguments analogous to those in Section 1.7. 69 Chapter2 Machlimitsinanalyticspacesonexteriordomains 2.1 Introduction The incompressible limit concerns the passage from compressible uids to incompressible uids as the Mach number tends to zero. The rst rigorous result on this singular limit problem can be traced back to Klainerman and Majda [84] (see also Ebin [55]), with a great deal of activity and progress in recent decades [2, 3, 4, 5, 44, 45, 49, 54, 57, 59, 60, 69, 70, 71, 72, 73, 85, 110, 117, 121, 131, 132, 141]. A general approach to this problem is to rst prove the existence of solutions on some time interval independent of the Mach number and then to show the convergence to solutions of the limiting equations when the Mach number tends to zero. A key ingredient in establishing a uniform time interval is a uniform upper bound, while in showing the convergence, the most critical issue is the vanishing of the acoustic waves. It is well-known that the analysis depends on various settings such as isentropic vs. non-isentropic, inviscid vs. viscous, well-prepared data vs. general data, or the whole space vs. domain with a boundary. For the non-isentropic problem with general data, the non-isentropic Euler ows feature intriguing wave-transport structure, and the coecients of governing equations (for instance, seeE in (2.2.4) and (2.2.5)) depend on the dependent variables, which makes the low Mach number limit a dicult problem. Métivier and Schochet in [121] gave the rst satisfactory answer by making use of the microlocal defect measure, and Alazard in [2] extended the existence result to the case of domains with boundary and the 70 convergence result for exterior domains. Both results were obtained in Sobolev spaces. In a recent work [77], we studied the non-isentropic problem with general analytic or Gevrey initial data inR 3 and proved that the convergence holds in these strong norms. The goal of this paper is to extend our previous results on analytic/Gevrey convergence in the zero Mach limit from [77] to the more dicult case of the domain with boundary. The natural setting for the Mach limit is the case of an exterior domain since there the convergence was established in Sobolev spaces. One of the main diculties in our approach is the existence of a complete family of analytic vector elds which respect to the boundary conditionv j @ = 0. In Theorem 2.3.1 below, we construct a family of tangential analytic vector elds for the exterior domain with constant analyticity radii. The existence of a family of complete tangential analytic vector elds was established by Komatsu in [81] for the case of a bounded domain. The completeness refers to the fact that the vector elds span the tangential space in the closure of the domain. In our proof of analytic boundedness, we require for the tangential elds to span only in a neighborhood of the boundary. In order to construct the necessary tangential elds, we use interior analytic hypoellipticity for the Dirichlet problem (cf. [108]) and analytic regularization by the heat kernel. As an additional benet, we obtain a low number of tangential vector elds (three) needed for the construction. In order to obtain uniform boundedness in analytic spaces, we rst establish boundedness of the entropy, divergence, followed by then normal and tangential derivative reduction schemes for the velocity. The normal and tangential derivatives can be reduced by using elliptic regularity, which leads to the estimates of divergence component, curl component, pure time derivatives, commutators, and lower- order terms associated to the boundary. In order to obtain that the Mach limit holds, we need to change the approach from [77, Section 7] since the interpolation inequality used there results in boundary terms which can not be handled. Instead, we use a simpler interpolation inequality (2.8.1) and a discrete dominated convergence theorem. As a byproduct of the Mach convergence, we obtain the analyticity of solutions 71 of the stratied incompressible Euler equation in an exterior (or bounded) domain. For the unstratied version of the Euler equations, the analyticity was proven in [98] using dierent methods. The paper is organized as follows. In Section 2.2, we recall from [2, 121] symmetrization of the com- pressible Euler equations. In Section 2.3, we construct the tangential vector elds for the exterior domain and state the main results. The a priori estimate needed for the uniform analytic boundedness, stated in Lemma 2.3.5, is given at the end of Section 2.6 and relies on the bounds on entropy and the velocity given in previous sections. The proof of the second main theorem is given in Section 2.8. We emphasize that the construction of the system of analytic tangential elds also applies to the case of a bounded domain; only when proving the Mach limit convergence, we rely on the fact that the domain is external. All con- siderations also apply in the Gevrey norm, or more generally to the spaces used in the book by Lions and Magenes [108]. The approach in this paper benets from ideas in [35]. For other approaches to analyticity, we refer to [10, 11, 14, 16, 22, 23, 46, 53, 61, 62, 66, 90, 97, 111, 125]. 2.2 Thesettingandnotation We address the incompressible limit for classical solutions of the compressible Euler equations for non- isentropic uids in an exterior domain R 3 with analytic boundary@ . More specically, we consider the compressible Euler equations for an inviscid, non-isentropic uid @ t +vr +rv = 0; (2.2.1) (@ t v +vrv) +rP = 0; (2.2.2) @ t S +vrS = 0; (2.2.3) 72 where: [0;T )!R + is the density,v : [0;T )!R 3 is the velocity,P : [0;T )!R + is the pressure, andS : [0;T )!R is the entropy of the uid. To close the system (2.2.1)–(2.2.3), we assume the equation of state P =P (;S): For instance, in the case of ideal gas the equation of state reads P (;S) = e S ; where > 1 is the adiabatic exponent. After some rescalings and a change of variables (cf. [121, 2]), we consider the symmetrized version of the compressible Euler equations for non-isentropic uids E(S;u)(@ t u +vru) + 1 L(@ x )u = 0; (2.2.4) @ t S +vrS = 0; (2.2.5) whereu = (p;v) T and E(S;u) = 0 B B @ a(S;u) 0 0 r(S;u)I 3 1 C C A ; L(@ x ) = 0 B B @ 0 div r 0 1 C C A : The parameter > 0 represents the Mach number. In agreement with the equation of state of the ideal gas, we adopt the assumption a(S;u) =f 1 (S)g 1 (u) 73 and r(S;u) =f 2 (S)g 2 (u); wheref 1 ,f 2 ,g 1 , andg 2 are positive entire real-analytic functions. We impose the impermeability boundary condition vj @ = 0: (2.2.6) Due to presence of the boundary, we also require some compatibility conditions. Since the matrixE(S;u) is invertible, we obtain @ t v =vrv + 1 r rp: (2.2.7) Dierentiating (2.2.7) with respect to time recursively, we get @ k t v(0) =A k (u(0);S(0)); for some functionsA k , wherek2 N 0 . We say that the initial data satisfy the compatibility condition of all orders if A k (u(0);S(0))j @ = 0; k2N 0 : One may readily check that the above condition is satised in the smooth or non-analytic Gevrey case for initial data (p 0 ;v 0 ;S 0 ) vanishing in a neighborhood of the boundary@ . 74 2.3 Analyticvectoreldinanexteriordomainandthemainresults Assume that R 3 is an exterior domain, located on one side of its nonempty, compact, and analytic boundary@ . Denote byd =d(x) the signed distance function to the boundary@ , taking positive values inside and negative values outside . Since@ is smooth, we have rd =(x) on@ ; (2.3.1) where is the unit outward normal vector. Moreover, the signed distance function is a real-analytic func- tion in a neighborhood of the boundary @ . Namely, we may extend with the formula (2.3.1) to a neighborhood 0 =fx2 : d(x)< 0 g of the boundary@ such that 1 X l=0 X jj=l l (l 3)! k@ k L 1 ( 0 ) . 1; (2.3.2) for some constants 0 ;> 0. 2.3.1 Analyticvectorelds A vector eldX istangential to@ ifXd = 0 on@ . SuchX may be restricted to@ byXf = (X ~ f)j @ forf 2 C 1 (@ ), where ~ f 2 C 1 ( ) is an arbitrary extension off. In fact, if ~ f vanishes on@ , then there exists ~ g2C 1 ( ) such that ~ f = ~ gd near@ . Existence of global analytic vector elds in a bounded domain inR 3 has been proven in Komatsu [81]. Here we construct a family of tangential elds for an exterior domain R 3 . An additional benet from 75 the construction is that we only need three tangential vector elds (which is the minimum possible by the hairy ball theorem). Theorem 2.3.1. There exist analytic vector eldsX 0 ;T 1 ;T 2 ;T 3 dened globally on with the following properties. 1. The eldsT 1 ;T 2 ;T 3 are tangential to@ , with T j = 3 X i=1 b ij (x)@ i ; 1j 3; (2.3.3) where the coecientsb ij (x) are real-analytic functions with constant analyticity radii, i.e., j@ b ij (x)j.C jj jj!; x2 ; 2N 3 0 ; for 1i;j 3. 2. There exists 0 > 0, such that the partial derivatives may be expressed as @ @x k = k (x)X 0 + 3 X j=1 jk (x)T j ; 1k 3 (2.3.4) on 0 , for some analytic coecients k (x) and jk (x), with 1j;k 3. The same proof works in all space dimensions, with a change in the number of vector elds. If the dimension isn, then the number of tangential elds becomesn(n 1)=2. In the proof of the above proposition, we also need the following statement, which follows from [108, Theorem 8.1.3]. 76 Lemma 2.3.2. (Local analytic hypoellipticity) Let be an exterior domain inR 3 with analytic boundary, and letR 0 > 0 be such that B R 0 =2 @ : Assume thatf satises j@ f(x)jMR jj jj!; 2N 3 0 ; x2B 2R 0 \ ; (2.3.5) for some constantsM;R> 0, and suppose that solves ( + 1) =f in (2.3.6) = 0 on@ : (2.3.7) Then we have j@ (x)jC(M +k k L 2 (B 2R 0 \ ) )(CR) jj jj!; 2N 3 0 ; (2.3.8) forx2B R 0 \ , whereC depends on@ andR 0 . This statement also applies to the case =R 3 , in which case we refer to it as the interior hypoellip- ticity. It follows directly from [108, Theorem 8.1.3]. We say that : !R is a globally dening function if 2C 1 and j @ = 0 and r j @ 6= 0: 77 Proof of Theorem 2.3.1. First we construct globally dening functions which have constant analyticity radii. Let R 0 4 be a radius such that c b B R 0 =2 . We pick an arbitrary globally dening function d 0 2 C 1 (R 3 ) which is compactly supported inB R 0 and it satises rd 0 j @ 6= 0: (2.3.9) For2 (0; 1], we dened :R 3 !R by d (x) = Z R 3 H(xy;)d 0 (y)dy; x2R 3 ; (2.3.10) whereH : R 3 R!R is the heat kernel. Consider the Dirichlet problem (2.3.6)–(2.3.7) withf = f = ( + 1)d . Sincef satises (2.3.5), it follows from Lemma 2.3.2 that the solution of (2.3.6)–(2.3.7) satises (2.3.8) forx2B R 0 \ ; note thatk k L 2 ( ) is uniformly bounded in2 (0; 1]. In order to obtain (2.3.8) for remainingx, we consider (2.3.6)–(2.3.7) withf = ( + 1)d and the version of Lemma 2.3.2 with = R 3 . Fix anyy2 R 3 nB R 0 . Sincef satises (2.3.5) inB 2 (y), Lemma 2.3.2 implies that the solution of (2.3.6) satises (2.3.8) for allx2B 1 (y). Thus, we obtain j@ (x)jCM(CR) jj jj!; 2N 3 0 ; x2 : (2.3.11) By (2.3.9)–(2.3.10) and continuity, we infer that r j @ 6= 0; for suciently small> 0, which implies that is a globally dening function. 78 Next, we pick an arbitrary globally dening function satisfying (2.3.11). Choose 0 > 0 so that r 6= 0 on 0 . For 1j;k 3, we set X 0 = 3 X i=1 @ @x i @ @x i and T jk = @ @x j @ @x k @ @x k @ @x j : (2.3.12) Since 2 C 1 ( ) vanishes on @ , there exists some ~ g 2 C 1 ( ) such that = ~ gd near @ . For 1j;k 3, we compute T jk d = @(~ gd) @x j @d @x k @(~ gd) @x k @d @x j =d @~ g @x j @d @x k @~ g @x k @d @x j + ~ g @d @x j @d @x k @d @x k @d @x j = 0; on@ ; which implies thatT jk are tangential vector elds, and thus (2.3.3) is proven. From (2.3.12), we have @ @x k X 0 + 3 X j=1 @ @x j T jk =jr j 2 @ @x k ; for 1k 3. Thus (2.3.4) follows sincer 6= 0 on 0 . To show that one may choose only three rather than nine tangential elds, note thatT jk =T kj for 1j;k 3, and thus alsoT 11 =T 22 =T 33 = 0. 2.3.2 Mainresults Let R 3 be an exterior domain with an analytic boundary @ . We assume that the initial data (p 0 ;v 0 ;S 0 ) satises k(p 0 ;v 0 ;S 0 )k H 5 ( ) M 0 (2.3.13) 79 and 1 X j=0 (j3) + 0 (j 3)! k@ j x (p 0 ;v 0 ;S 0 )k L 2 ( ) M 0 ; (2.3.14) for some xed constants 0 ;M 0 > 0. It can be shown, analogously to Lemma 2.4.7 below, that (2.3.14) implies 1 X j=0 1 X k=0 (j+k3) + 0 (j +k 3)! k@ j x T k (p 0 ;v 0 ;S 0 )k L 2 ( ) Q(M 0 ); (2.3.15) for some functionQ and some constant 0 > 0. In (2.3.14)–(2.3.15) and below, we useT = (T 1 ;T 2 ;T 3 ) to denote the tangential vector elds from Theorem 2.3.1, and use@ x = (@ 1 ;@ 2 ;@ 3 ) to denote the gradient. We adopt the following agreement for the iterative derivativesT k and@ j x . The symbolT k is understood in the tensorial sense, i.e.,T k stands for the list of all possible operatorsT 1 T k , where i 2f1; 2; 3g for 1ik. We also adopt the same agreement for@ j x . For > 0, we dene the mixed weighted analytic space A() =fu2C 1 ( ):kuk A() <1g; where kuk A() = X (j;k;i)2N 3 0 (j1) + k (t) (j+k+i3) + (j +k +i 3)! k@ j x T k (@ t ) i uk L 2 ( ) ; (2.3.16) here2 (0; 1] represents the mixed space-time analyticity radius, while and , where 0< 1 are parameters which represent the balances of radii in dierent directions. In (2.3.16) and below we use 80 the conventionn! = 1 forn2N. We show in Section 2.7 that (2.3.15) implies, regardless of the choices of2 (0; 1] and 2 (0; 1], that we have k(p 0 ;v 0 ;S 0 )k A(~ 0 ) Q(M 0 ); for some functionQ, where ~ 0 = 0 =Q(M 0 ) represents the mixed space-time analyticity radius. Note that the time derivatives of the initial data are dened iteratively by dierentiating the equations (2.2.4)–(2.2.5) and then evaluating att = 0. The analyticity radius function is dened as (t) =(0)Kt; (2.3.17) where(0) minf1; ~ 0 g is a suciently small parameter (dierent from ~ 0 ), andK 1 is a suciently large parameter, both to be determined below. Our rst main theorem provides a uniform in boundedness of the analytic norm on a time interval independent of. Theorem 2.3.3. Let R 3 be an exterior domain with analytic boundary @ . We assume that the initial data (p 0 ;v 0 ;S 0 ) satises (2.3.13)–(2.3.14) for some xed constantsM 0 ; 0 > 0. Also, suppose that (p 0 ;v 0 ;S 0 ) satises compatibility condition of all orders. Then there exist suciently small constants, ,(0), 0 ,T 0 > 0, depending onM 0 , such that k(p ;v ;S )(t)k A() M; 0< 0 ; t2 [0;T 0 ]; where is as in (2.3.17) andK andM are suciently large constants depending onM 0 . We note that the same statement and the proof apply the case of a bounded domain. 81 The second main result states that the solutions of (2.2.4)–(2.2.5) converge to the solution of the strat- ied incompressible Euler equations r(S; 0)(@ t v +vrv) +r = 0; (2.3.18) divv = 0; (2.3.19) @ t S +vrS = 0; (2.3.20) as! 0. We dene the spatial analytic space X() =fu2C 1 ( ):kuk X() <1g; where kuk X() = 1 X j=0 (j3) + (j 3)! k@ j x uk L 2 ( ) ; for some constant> 0. Theorem2.3.4. Let R 3 beanexteriordomainwithananalyticboundary. Assumethattheinitialdata (p 0 ;v 0 ;S 0 ) satisfy (2.3.13)–(2.3.14) uniformly for xed 0 ;M 0 > 0 and the compatibility condition of all orders. Also, suppose that (v 0 ;S 0 ) converge to (v 0 ;S 0 ) inH 3 ( ), withS 0 decaying in the sense of jS 0 (x)j.jxj 1 ; and jrS 0 (x)j.jxj 2 ; 82 for 0< 0 andsomeconstant > 0. Then (v ;p ;S )convergesto (v (inc) ; 0;S (inc) )2L 1 ([0;T 0 ];X()) inL 2 ([0;T 0 ];X()), where2 (0; 0 ] is a suciently small constant and (v (inc) ;S (inc) ) is the solution to (2.3.18)–(2.3.20) with the initial data (w 0 ;S 0 ), andw 0 is the unique solution of divw 0 = 0; curl(r 0 w 0 ) = curl(r 0 v 0 ); withr 0 =r(S 0 ; 0). In the rest of the paper, we omit the superscript, and writeS,u, andv forS ,u , andv . The constant C and the functionQ denote a generic constant and a positive increasing function, respectively, which depend only onM 0 , 0 , and the exterior domain ; they may vary from an inequality to an inequality. The domain of dependence in the norms is understood to be unless stated otherwise. We writea.b if there exists a constantC > 0 such thataCb. In order to prove Theorem 2.3.3, we establish analytic a priori estimates of the entropy S and the (modied) velocityu. The a priori estimate needed to prove Theorem 2.3.3 is the following. Lemma2.3.5. LetM 0 ; 0 > 0. For suciently small parameters and satisfying 0< 1, there existconstants 1 , 0 ,T 0 > 0,andanonnegativecontinuousfunctionQsuchthatforall2 (0; 0 ],thenorm M ;; (T ) = sup t2[0;T ] (kS(t)k A((t)) +ku(t)k A((t)) ) satises the estimate M ;; (t). 1 + (t + + + +(0))Q(M ;; (t)); 83 fort2 (0;T 0 ] and(0)2 (0; 1 ], provided KQ(M ;; (T 0 )) (2.3.21) holds, where andK are as in (2.3.17). The constantK in (2.3.21) depends onM, which implies thatK eventually depends onM 0 . In the rest of the paper, we work on an interval of time such that T 0 (0) 2K : (2.3.22) Thus from (2.3.17) we have(0)=2(t)(0) fort2 [0;T 0 ]. For the proof of Theorem 2.3.3 given Lemma 2.3.5, cf. [77]. Sections 2.4–2.6 are devoted to the proof of Lemma 2.3.5, thus completing the proof of Theorem 2.3.3. By [2, Theorem 1.1] theH 5 norm of (p;v;S) can be estimated by a constant on a time interval [0;T 0 ], whereT 0 only depends on theH 5 norm of the initial data. Thus we may assume sup (j;k;i)2N 3 0 ;0j+k+i5 k@ j x T k (@ t ) i (p;v;S)(t)k L 2. 1; t2 [0;T 0 ]; 2 (0; 1]: (2.3.23) In particular, ifF is a smooth function ofu andS, then there exists a constantC > 0 depending on the functionF such that kF (u(t);S(t))k L 1. 1; t2 [0;T 0 ]; 2 (0; 1]: In the rest of the paper, we work on the time interval [0;T ] where 0 0 such that krvk L 2 ( ) .k divvk L 2 ( ) +k curlvk L 2 ( ) +kvk H 1=2 (@ ) +kvk L 2 ( ) ; (2.4.5) for allv2H 1 ( ). Proof of Lemma 2.4.2. Using theH 2 regularity for the Laplace equation kvk H 2 ( ) .kvk L 2 ( ) +kvk H 3=2 (@ ) ; combined with the trace theorem, we arrive at kvk H 2 ( ) .kvk L 2 ( ) +kTvk H 1 ( ) +kvk L 2 ( ) ; concluding the proof of (2.4.4). Proof of Lemma 2.4.1. Letj 2. Appealing to theH 2 regularity (2.4.4), we obtain k@ j x T k (@ t ) i vk L 2 ( ) .k@ j2 x T k (@ t ) i vk L 2 ( ) +kT@ j2 x T k (@ t ) i vk H 1 ( ) +k@ j2 x T k (@ t ) i vk L 2 ( ) ; 86 from where we obtain by using the vector calculus identity v =r divv curl curlv the inequality k@ j x T k (@ t ) i vk L 2 ( ) .k div@ j2 x T k (@ t ) i vk _ H 1 ( ) +k curl@ j2 x T k (@ t ) i vk _ H 1 ( ) +kT@ j2 x T k (@ t ) i vk H 1 ( ) +k@ j2 x T k (@ t ) i vk L 2 ( ) .k@ j2 x T k (@ t ) i divvk _ H 1 ( ) +k@ j2 x T k (@ t ) i curlvk _ H 1 ( ) +k@ j2 x T k+1 (@ t ) i vk H 1 ( ) +k@ j2 x T k (@ t ) i vk L 2 ( ) +k@ j1 x [T k ; div](@ t ) i vk L 2 ( ) +k@ j1 x [T k ; curl](@ t ) i vk L 2 ( ) +k[T;@ j2 x ]T k (@ t ) i vk H 1 ( ) ; and (2.4.2) follows. On the other hand, the inequality (2.4.3) follows from the elliptic regularity (2.4.5) and the boundary condition (2.4.1). 2.4.2 Tangentialderivativereduction The following lemma allows us to reduce the number of tangential derivatives. Lemma2.4.4. Forj = 1 andk 1, we have k@ x T k (@ t ) i vk L 2 ( ) .kT k (@ t ) i divvk L 2 ( ) +kT k (@ t ) i curlvk L 2 ( ) +kT k (@ t ) i vk L 2 ( ) + k X l=1 k l k@ x T kl (@ t ) i vk L 2 ( ) kT l k L 1 ( 0 ) +kT kl (@ t ) i vk L 2 ( ) k@ x T l k L 1 ( 0 ) +kT kl (@ t ) i vk L 2 ( ) kT l k L 1 ( 0 ) +k[T k ; div](@ t ) i vk L 2 ( ) +k[T k ; curl](@ t ) i vk L 2 ( ) : (2.4.6) 87 Note that whenj = 0 andk 1, we may simply use kT k (@ t ) i vk L 2 ( ) .k@ x T k1 (@ t ) i vk L 2 ( ) (2.4.7) and apply (2.4.6) with the reducedk. Proof of Lemma 2.4.4. Forj = 1 andk 1, we appeal to (2.4.5) and obtain k@ x T k (@ t ) i vk L 2 ( ) .k divT k (@ t ) i vk L 2 ( ) +k curlT k (@ t ) i vk L 2 ( ) +kT k (@ t ) i vk H 1=2 (@ ) +kT k (@ t ) i vk L 2 ( ) : (2.4.8) Now, note that T k (@ t ) i v =T k (@ t ) i (v) k X l=1 k l T kl (@ t ) i vT l ; with the rst term on the right side vanishing by (2.4.1). Thus we get kT k (@ t ) i vk H 1=2 (@ ) k X l=1 k l kT kl (@ t ) i vT l k H 1=2 (@ ) : Extending the unit normal vector to 0 as in Section 2.3, and using the trace theorem, we obtain kT kl (@ t ) i vT l k H 1=2 (@ ) .kT kl (@ t ) i vT l k H 1 ( 0 ) : 88 By the Leibniz rule and Hölder’s inequality, we conclude that kT kj (@ t ) i vT j k H 1 ( 0 ) k@ x T kj (@ t ) i vk L 2 ( ) kT j k L 1 ( 0 ) +kT kj (@ t ) i vk L 2 ( ) k@ x T j k L 1 ( 0 ) +kT kj (@ t ) i vk L 2 ( ) kT j k L 1 ( 0 ) : (2.4.9) Combining (2.4.8)–(2.4.9), we then obtain (2.4.6). 2.4.3 Commutatorestimates Here we recall a Leibniz formula fork-folded commutators. Given two linear dierential operatorsY ,Z, the adjoint operator adY (Z) is dened as adY (Z) = [Y;Z] =YZZY: Recall a Leibniz-type formula [T k ;Z] = k X m=1 k m ((adT) m (Z))T km ; k2N (2.4.10) from [35, Lemma 3.4], which holds for any dierential operatorZ. From [81, Lemma 5.3], we also recall an analytic estimate for the adjoint operator. Lemma2.4.5 ([81]). LetY 1 ;:::;Y m andY 0 be analytic vector elds dened on R 3 such that Y n = 3 X i=1 a i n @ i ; n = 0; 1;:::;m; 89 where max jj=k j@ a i n j.k!K k 1 ; i = 1; 2; 3; n = 0; 1;:::;m; k2N 0 ; for someK 1 1. Then there exists K 1 ; K 2 1 such that (adY m ::: adY 1 )(Y 0 ) = 3 X i=1 b i m @ i ; where max jj=k j@ b i m j. (k +m)! K k 1 K m 2 ; fori = 1; 2; 3 andk2N 0 . In the following lemma, we derive commutator estimates for various spatial derivative operators. Lemma2.4.6. There exists a constant K 1 such that k@ j x [T k ; div](@ t ) i vk L 2. k X k 0 =1 j X j 0 =0 k k 0 j j 0 (j 0 +k 0 )! K j 0 +k 0 k@ jj 0 +1 x T kk 0 (@ t ) i vk L 2; (2.4.11) k@ j x [T k ;r](@ t ) i vk L 2. k X k 0 =1 j X j 0 =0 k k 0 j j 0 (j 0 +k 0 )! K j 0 +k 0 k@ jj 0 +1 x T kk 0 (@ t ) i vk L 2; (2.4.12) k@ j x [T k ; curl](@ t ) i vk L 2. k X k 0 =1 j X j 0 =0 k k 0 j j 0 (j 0 +k 0 )! K j 0 +k 0 k@ jj 0 +1 x T kk 0 (@ t ) i vk L 2; (2.4.13) k@ j x [@ k x ;T]T l (@ t ) i vk L 2. k X k 0 =1 j X j 0 =0 k k 0 j j 0 (j 0 +k 0 )! K j 0 +k 0 k@ kk 0 +jj 0 +1 x T l (@ t ) i vk L 2; (2.4.14) fori;j;l2N 0 andk2N. 90 Proof of Lemma 2.4.6. By (2.4.10), we have the expansion [T k ; div]v = 3 X s=1 [T k ;@ s ]v s = 3 X s=1 k X k 0 =1 k k 0 ((adT) k 0 (@ s ))T kk 0 v s : Using Theorem 2.3.1 and Lemma 2.4.5, we obtain (adT) k 0 (@ s ) = 3 X l=1 b l k 0 ;s @ l ; k 0 2N; s2f1; 2; 3g; (2.4.15) where max jj=k j@ b l k 0 ;s j. (k +k 0 )! K k 1 K k 0 2 ; k2N 0 ; k 0 2N; s;l2f1; 2; 3g; (2.4.16) for some constants K 1 ; K 2 1. From the expression (2.4.15) and the Leibniz rule, we arrive at @ j x [T k ; div](@ t ) i v =@ j x 3 X s=1 k X k 0 =1 3 X l=1 k k 0 b l k 0 ;s @ l T kk 0 (@ t ) i v s = 3 X s=1 k X k 0 =1 3 X l=1 j X j 0 =0 k k 0 j j 0 @ j 0 x b l k 0 ;s @ jj 0 x @ l T kk 0 (@ t ) i v s : Using (2.4.16), we obtain k@ j x [T k ; div](@ t ) i vk L 2. 3 X s=1 k X k 0 =1 j X j 0 =0 3 X l=1 k k 0 j j 0 k@ j 0 x b l k 0 ;s @ jj 0 +1 x T kk 0 (@ t ) i v s k L 2 . k X k 0 =1 j X j 0 =0 k k 0 j j 0 (j 0 +k 0 )! K j 0 1 K k 0 2 k@ jj 0 +1 x T kk 0 (@ t ) i vk L 2; and (2.4.11) follows by setting K max( K 1 ; K 2 ). The proofs of (2.4.12)–(2.4.14) are analogous. 91 The next lemma provides an analytic estimate for the unit normal vector to the boundary. Lemma2.4.7. There exists a constant ~ > 0 such that 1 X j=0 1 X k=0 ~ j+k (j +k 3)! k@ j x T k k L 1 ( 0 ) . 1; (2.4.17) where is the unit normal vector to the boundary satisfying (2.3.2). Proof of Lemma 2.4.7. We claim that there exists a constant > 0, such that for allk2N 0 , we have 1 X j=0 j+k (j +k 3)! k@ j x T k k L 1 ( 0 ) C k+1 0 ; (2.4.18) for some constantC 0 1. In (2.4.18), we then choose ~ = =2C 0 to get 1 X j=0 1 X k=0 ~ j+k (j +k 3)! k@ j x T k k L 1 ( 0 ) 1 X j=0 1 X k=0 j+k 2 k C k 0 (j +k 3)! k@ j x T k k L 1 ( 0 ) . 1; obtaining (2.4.17). In the remainder of the proof, we proceed by induction to prove (2.4.18) for allk2N 0 . Firstly, we use (2.3.2) to obtain (2.4.18) fork = 0 by taking =. Now we assume that (2.4.18) holds for somek2N 0 and aim to show that it also holds fork + 1. Using (2.3.3), the Leibniz rule, and Hölder’s inequality, we arrive at 1 X j=0 j+k+1 (j +k 2)! k@ j x T k+1 k L 1 ( 0 ) C 3 X i=1 3 X l=1 1 X j=0 j X j 0 =0 j j 0 j+k+1 (j +k 2)! k@ j 0 x b il k L 1 ( 0 ) k@ jj 0 +1 x T k k L 1 ( 0 ) C 1 X j=0 j X j 0 =0 j 0 C j 0 j!(jj 0 +k 2)! (jj 0 )!(j +k 2)! jj 0 +k+1 (jj 0 +k 2)! k@ jj 0 +1 x T k k L 1 ( 0 ) ! C 1 X j 00 =1 j 00 +k (j 00 +k 3)! k@ j 00 x T k k L 1 ( 0 ) 1 X j=j 00 1 (C 1 ) j+1j 00 j!(j 00 +k 3)! (j 00 1)!(j +k 2)! ; (2.4.19) 92 where C 1 1 is a constant. Note that the sum in j is dominated by a constant C uniformly for all j 00 ;k2N 0 , by taking 1=2C 1 . Thus from (2.4.19) and the induction hypothesis fork, we have 1 X j=0 j+k+1 (j +k 2)! k@ j x T k+1 k L 1 ( 0 ) CC k+1 0 ; concluding the proof of (2.4.18) fork + 1 by simply takingC 0 C. 2.5 Analyticestimateoftheentropy The following lemma provides an analytic estimate for the entropyS. Lemma2.5.1. LetM 0 > 0. For any, 2 (0; 1], there exists 1 2 (0; 1] such that if 0<(0) 1 , then kS(t)k A((t)) . 1 +tQ(M ;; (t)); t2 (0;T 0 ]; (2.5.1) for all2 (0; 1], providedK in (2.3.17) satises KQ(M ;; (T 0 )); whereT 0 > 0 is a suciently small constant depending onM 0 . Proof of Lemma 2.5.1. Let 1 = 4 : (2.5.2) 93 Now, we x (j;k;i)2 N 3 0 , apply@ j x T k (@ t ) i to the equation (2.2.5), and take theL 2 -inner product with @ j x T k (@ t ) i S, obtaining 1 2 d dt k@ j x T k (@ t ) i Sk 2 L 2 + vr@ j x T k (@ t ) i S;@ j x T k (@ t ) i S = [vr;@ j x T k (@ t ) i ]S;@ j x T k (@ t ) i S ; whereh;i denotes the scalar product in L 2 . Using the Cauchy-Schwarz inequality and the boundary condition (2.2.6), we obtain d dt k@ j x T k (@ t ) i Sk L 2.krvk L 1 x k@ j x T k (@ t ) i Sk L 2 +k[vr;@ j x T k (@ t ) i ]Sk L 2: (2.5.3) Using the notation (2.3.16) and kuk ~ A() = X j+k+i4 k@ j x T k (@ t ) i uk L 2 (j1) + k (j +k +i 3)(t) j+k+i4 (j +k +i 3)! ; the estimate (2.5.3) implies d dt kSk A() = _ (t)kSk ~ A() + X (j;k;i)2N 3 0 (j1) + k (j+k+i3) + (j +k +i 3)! d dt k@ j x T k (@ t ) i Sk L 2 . _ (t)kSk ~ A() +krvk L 1 x kSk A() + X (j;k;i)2N 3 0 X (j 0 ;k 0 ;i 0 )(j;k;i) j 0 +k 0 +i 0 1 C j;k;i;j 0 ;k 0 ;i 0 +kvk L 1 x X (j;k;i)2N 3 0 (j1) + k (j+k+i3) + (j +k +i 3)! k@ j x [T k ;r](@ t ) i Sk L 2; (2.5.4) where C j;k;i;j 0 ;k 0 ;i 0 = (j1) + k (j+k+i3) + (j +k +i 3)! j j 0 k k 0 i i 0 k@ j 0 x T k 0 (@ t ) i 0 v@ jj 0 x T kk 0 (@ t ) ii 0 rSk L 2: 94 We split the third term on the far right side of (2.5.4) according to the low and high values ofj 0 +k 0 +i 0 . We claim that there exists a constantC > 0 such that I 1 = X 0j+k+i4 X (j 0 ;k 0 ;i 0 )(j;k;i) j 0 +k 0 +i 0 1 C j;k;i;j 0 ;k 0 ;i 0. 1; (2.5.5) I 2 = X j+k+i5 X (j 0 ;k 0 ;i 0 )(j;k;i) 1j 0 +k 0 +i 0 [(j+k+i)=2] C j;k;i;j 0 ;k 0 ;i 0.kvk A() kSk ~ A() +kvk A() ; (2.5.6) I 3 = X j+k+i5 X (j 0 ;k 0 ;i 0 )(j;k;i) [(j+k+i)=2]+1j 0 +k 0 +i 0 j+k+i3 C j;k;i;j 0 ;k 0 ;i 0.kvk A() kSk ~ A() +kvk A() ; (2.5.7) I 4 = X j+k+i5 X (j 0 ;k 0 ;i 0 )(j;k;i) j+k+i2j 0 +k 0 +i 0 C j;k;i;j 0 ;k 0 ;i 0.kvk A() ; (2.5.8) I 5 = X (j;k;i)2N 3 0 (j1) + k (j+k+i3) + (j +k +i 3)! k@ j x [T k ;r](@ t ) i Sk L 2. 1 +kSk ~ A() : (2.5.9) The proofs of (2.5.5)–(2.5.8) are analogous to those in Section 4 in [77]. Here we only outline necessary modications. Proofof (2.5.5): We apply Hölder and Sobolev inequalities on the factork@ j 0 x T k 0 (@ t ) i 0 v@ jj 0 x T kk 0 (@ t ) ii 0 rSk L 2 and then use (2.4.12) to estimatek@ j x T k (@ t ) i rSk L 2. Then (2.5.5) follows by appealing to (2.3.23). Proof of (2.5.6): Using Hölder and Sobolev inequalities, we obtain C j;k;i;j 0 ;k 0 ;i 0 1 fj+k+i5g 1 f1j 0 +k 0 +i 0 [(j+k+i)=2]g . j 0 +1 k 0 (j 0 +k 0 +i 0 1) + (j 0 +k 0 +i 0 1)! k@ j 0 +2 x T k 0 (@ t ) i 0 vk L 2 ! 3=4 (j 0 1) + k 0 (j 0 +k 0 +i 0 3) + (j 0 +k 0 +i 0 3)! k@ j 0 x T k 0 (@ t ) i 0 vk L 2 ! 1=4 jj 0 kk 0 (j+k+ij 0 k 0 i 0 3) + (j +k +ij 0 k 0 i 0 3)! k@ jj 0 x T kk 0 (@ t ) ii 0 rSk L 2 ! : (2.5.10) 95 In (2.5.10), we bounded the rest of the powers of, , and byC, which is possible by (2.5.2), and bounded the rest of the factors involving combinatorial symbols byC sincej 0 +k 0 +i 0 [(j +k +i)=2]. From (2.5.10), we use the discrete Hölder and Young inequalities to get I 2 .kvk A() kSk ~ A() +kvk A() X j+k+i3 j k (j+k+i3) + (j +k +i 3)! k@ j x [T k ;r](@ t ) i Sk L 2: (2.5.11) Appealing to (2.4.12), the sum in the second term on the right side of above may be estimated by X j+k+i3 k X k 0 =1 j X j 0 =0 k k 0 j j 0 j k (j+k+i3) + (j +k +i 3)! (j 0 +k 0 )! K j 0 +k 0 k@ jj 0 +1 x T kk 0 (@ t ) i Sk L 2 . X j+k+i3 k X k 0 =1 j X j 0 =0 ( K) j 0 ( K) k 0 1 fi3g jj 0 kk 0 (j+k+ij 0 k 0 3) + (j +k +ij 0 k 0 3)! k@ jj 0 +1 x T kk 0 (@ t ) i Sk L 2 ! A j;k;i;j 0 ;k 0 + X j+k+i3 k X k 0 =1 j X j 0 =0 ( K) j 0 ( K) k 0 1 f0i2g jj 0 kk 0 (j+k+ij 0 k 0 3) + (j +k +ij 0 k 0 3)! k@ jj 0 +1 x T kk 0 (@ t ) i Sk L 2 ! B j;k;i;j 0 ;k 0: (2.5.12) In (2.5.12), we denote A j;k;i;j 0 ;k 0 = k k 0 j j 0 (j +k +ij 0 k 0 3)!(j 0 +k 0 )! (j +k +i 3)! 1 fi3g (2.5.13) and B j;k;i;j 0 ;k 0 = k k 0 j j 0 (j +k +ij 0 k 0 3)!(j 0 +k 0 )! (j +k +i 3)! 1 f0i2g : 96 Applying the combinatorial inequality k k 0 j j 0 k +j k 0 +j 0 to (2.5.13), we obtain A j;k;i;j 0 ;k 0. (j +k +ij 0 k 0 3)!(j +k)! (j +k +i 3)!(j +kj 0 k 0 )! . 1; (2.5.14) sincei 3. Similarly, B j;k;i;j 0 ;k 0. (j +k +ij 0 k 0 3)!(j +k)! (j +k +i 3)!(j +kj 0 k 0 )! 1 f0i2g : Ifj 0 +k 0 [(j +k)=2] + 1, then there exists a constantC 1 such that B j;k;i;j 0 ;k 0.C j 0 +k 0 ; uniformly for alli2f0; 1; 2g; if 1j 0 +k 0 [(j +k)=2], then we have B j;k;i;j 0 ;k 0. 1; (2.5.15) 97 uniformly for alli2f0; 1; 2g. Combining (2.5.12), (2.5.14)–(2.5.15), and switching indices fromjj 0 to j 00 andkk 0 tok 00 , we obtain X j+k+i3 j k j+k+i3 (j +k +i 3)! k@ j x [T k ;r](@ t ) i Sk L 2 . 1 X j 00 =0 1 X k 00 =0 1 X i=3 j 00 k 00 (j 00 +k 00 +i3) + (j 00 +k 00 +i 3)! k@ j 00 +1 x T k 00 (@ t ) i Sk L 2 ! 1 X j=j 00 1 X k=k 00 +1 ( K) jj 00 ( K) kk 00 + 1 X j 00 =0 1 X k 00 =0 2 X i=0 j 00 k 00 (j 00 +k 00 +i3) + (j 00 +k 00 +i 3)! k@ j 00 +1 x T k 00 (@ t ) i Sk L 2 ! 1 X j=j 00 1 X k=k 00 +1 (C K) jj 00 (C K) kk 00 : (2.5.16) Note that the sums inj andk in (2.5.16) are bounded byC for 1=C K and 1=C K, uniformly for allj 0 ;k 0 ;i2N 0 . Therefore, by (2.5.11) and (2.5.16) we conclude I 2 .kvk A() kSk ~ A() +kvk A() X 1j+k+i3 k@ j x T k (@ t ) i Sk L 2.kvk A() kSk ~ A() +kvk A() ; where the last inequality follows from (2.3.23). Proof of (2.5.7): Using Hölder and Sobolev inequalities, we obtain C j;k;i;j 0 ;k 0 ;i 0 1 fj+k+i5g 1 f[(j+k+i)=2]+1j 0 +k 0 +i 0 j+k+i3g . (j 0 1) + k 0 (j 0 +k 0 +i 0 3) + (j 0 +k 0 +i 0 3)! k@ j 0 x T k 0 (@ t ) i 0 vk L 2 ! jj 0 +2 kk 0 (j+k+ij 0 k 0 i 0 1) + (j +k +ij 0 k 0 i 0 )! k@ jj 0 +2 x T kk 0 (@ t ) ii 0 rSk L 2 ! 3=4 jj 0 kk 0 (j+k+ij 0 k 0 i 0 3) + (j +k +ij 0 k 0 i 0 3)! k@ jj 0 x T kk 0 (@ t ) ii 0 rSk L 2 ! 1=4 : (2.5.17) 98 Similarly to (2.5.10), by (2.5.2), the remaining factors of, , and are bounded byC; the product of factors involving combinatorial symbols may be bounded byC sincej 0 +k 0 +i 0 [(j +k +i)=2] + 1. By (2.5.17), using the discrete Hölder and the discrete Young inequalities we arrive at I 3 .kvk A() kSk ~ A() +kvk A() X j+k+i3 j k (j+k+i3) + (j +k +i 3)! k@ j x [T k ;r](@ t ) i Sk L 2: Proceeding as in (2.5.12)–(2.5.16), we obtain (2.5.7). Proof of (2.5.8): Using Hölder and Sobolev inequality, we obtain I 4 . X j+k+i5 X (j 0 ;k 0 ;i 0 )(j;k;i) j+k+i2j 0 +k 0 +i 0 (j 0 1) + k 0 j 0 +k 0 +i 0 3 (j 0 +k 0 +i 0 3)! k@ j 0 x T k 0 (@ t ) i 0 vk L 2 ! k@ jj 0 x T kk 0 (@ t ) ii 0 rSk H 2 .kvk A() ; where the last inequality follows from (2.3.23). Proof of (2.5.9): For low values ofj +k +i, from (2.3.23) and (2.4.12), we obtain X 0j+k+i4 (j1) + k (j+k+i3) + (j +k +i 3)! k@ j x [T k ;r](@ t ) i Sk L 2. 1: (2.5.18) 99 For high values ofj +k +i, by (2.4.12), we have X j+k+i5 (j1) + k (j+k+i3) + (j +k +i 3)! k@ j x [T k ;r](@ t ) i Sk L 2 . X j+k+i5 k X k 0 =1 j X j 0 =0 (j1) + k (j+k+i3) + (j +k +i 3)! k k 0 j j 0 (j 0 +k 0 )! K j 0 +k 0 k@ jj 0 +1 x T kk 0 (@ t ) i Sk L 2 . X j+k+i5 k X k 0 =1 j X j 0 =0 (j +k)!(j +k +ij 0 k 0 3)! (j +kj 0 k 0 )!(j +k +i 3)! ( K) j 0 +k 0 jj 0 kk 0 (j+k+ij 0 k 0 3) + (j +k +ij 0 k 0 3)! k@ jj 0 +1 x T kk 0 (@ t ) i Sk L 2 ! ; where the last inequality follows from (2.5.2). We then proceed as in (2.5.12)–(2.5.16) to obtain X j+k+i5 (j1) + k (j+k+i3) + (j +k +i 3)! k@ j x [T k ;r](@ t ) i Sk L 2. 1 +kSk ~ A() : (2.5.19) Thus (2.5.9) follows from (2.5.18) and (2.5.19). Combining (2.5.4) and (2.5.5)–(2.5.9), we obtain d dt kSk A() .kSk ~ A() _ (t) +kvk A() + 1 +kSk A() +kvk A() + 1; (2.5.20) where we used (2.3.23) to boundkrvk L 1 x andkvk L 1 x . Now, we chooseK in (2.3.17) to be suciently large so that the term next tokSk ~ A() is less than or equal to 0. The lemma then follows by integrating (2.5.20) on [0;T 0 ] and using the Gronwall lemma. 100 2.6 Velocityestimates In this section, we use derivative reductions for the divergence, curl, and pure time derivative components of the velocity. First, we splitkvk A() as kvk A() = 5 X l=1 X J l (j1) + k (i+j+k3) + (i +j +k 3)! k@ j x T k (@ t ) i vk L 2 = 5 X l=1 U l ; (2.6.1) where U l = X J l (j1) + k (i+j+k3) + (i +j +k 3)! k@ j x T k (@ t ) i vk L 2; l = 1;:::; 5; and J 1 =f(j;k;i)2N 3 0 : j 2g; J 2 =f(j;k;i)2N 3 0 : j = 1;k 1g; J 3 =f(j;k;i)2N 3 0 : j = 1;k = 0g; J 4 =f(j;k;i)2N 3 0 : j = 0;k 1g; J 5 =f(j;k;i)2N 3 0 : j = 0;k = 0g: For simplicity of notation, we abbreviate, for (j;k;i)2N 3 0 , U j;k;i = (j1) + k (i+j+k3) + (i +j +k 3)! k@ j x T k (@ t ) i vk L 2 for the velocity coecients arising in expansions below. Similarly, we denoteE =f(j;k;i)2N 3 0 : j 1g, and we write D j;k;i = (j1) + k (i+j+k3) + (i +j +k 3)! k@ j1 x T k (@ t ) i divvk L 2 101 for the coecients containing the divergence and C j;k;i = (j1) + k (i+j+k3) + (i +j +k 3)! k@ j1 x T k (@ t ) i curlvk L 2; for those with the curl. The termU 1 : First we estimate the sumU 1 . By (2.4.2), we have U 1 . U 1 +U 1 + X E D j;k;i + X E C j;k;i + X J 1 (j1) + k (i+j+k3) + (i +j +k 3)! k@ j1 x [T k ; div](@ t ) i vk L 2 + X J 1 (j1) + k (i+j+k3) + (i +j +k 3)! k@ j1 x [T k ; curl](@ t ) i vk L 2 + X J 1 (j1) + k (i+j+k3) + (i +j +k 3)! k[T;@ j2 x ]T k (@ t ) i vk L 2 + X J 1 (j1) + k (i+j+k3) + (i +j +k 3)! k@ x [T;@ j2 x ]T k (@ t ) i vk L 2 =I 11 +I 12 +I 13 +I 14 +I 15 +I 16 +I 17 +I 18 : (2.6.2) For the termI 15 , we use (2.4.11) to write I 15 . X J 1 k X k 0 =1 j1 X j 0 =0 k k 0 j 1 j 0 j1 k (i+j+k3) + (i +j +k 3)! (j 0 +k 0 )! K j 0 +k 0 k@ jj 0 x T kk 0 (@ t ) i vk L 2 . 1 X i=0 1 X k 00 =0 1 X j 00 =0 j 00 k 00 (i+j 00 +k 00 2) + (i +j 00 +k 00 2)! k@ j 00 +1 x T k 00 (@ t ) i vk L 2 1 X j=j 00 +1 1 X k=k 00 +1 (k +j 1)!(i +j 00 +k 00 2)! (j 00 +k 00 )!(i +j +k 3)! jj 00 1 kk 00 K kk 00 +jj 00 1 : (2.6.3) 102 Note that the sum inj;k is dominated by K 1 X j=j 00 +1 1 X k=k 00 +1 (k +j 1)!(i +j 00 +k 00 2)! (j 00 +k 00 )!(i +j +k 3)! jj 00 1 kk 00 1 K kk 00 +jj 00 2 . ; (2.6.4) uniformly for alli;k 00 ;j 00 2N 0 , by taking 1=CK. Thus, from (2.6.3)–(2.6.4) we obtain I 15 . U 1 + U 2 + U 3 : (2.6.5) The termI 16 can be treated analogously asI 15 . Namely, using (2.4.13), we get I 16 . U 1 + U 2 + U 3 : (2.6.6) For the termI 17 , we appeal to (2.4.14) obtaining I 17 . X J 1 j2 X j 0 =1 j 2 j 0 j 0 ! K j 0 j1 k (i+j+k3) + (i +j +k 3)! k@ jj 0 1 x T k (@ t ) i vk L 2 . 1 X i=0 1 X k=0 1 X j 00 =0 j 00 k (i+j 00 +k2) + (i +j 00 +k 2)! k@ j 00 +1 x T k (@ t ) i vk L 2 ! 1 X j=j 00 +3 (i +j 00 +k 2)!(j 2)! (i +j +k 3)!j 00 ! jj 00 1 K jj 00 2 : (2.6.7) Note that the sum inj is dominated by 1 X j=j 00 +3 (i +j 00 +k 2)!(j 2)! (i +j +k 3)!j 00 ! jj 00 2 K jj 00 2 .; (2.6.8) uniformly for alli;k;j 00 2N 0 , by taking 1=C K. From (2.6.7)–(2.6.8), we obtain I 17 .U 1 +U 2 +U 3 : (2.6.9) 103 The termI 18 can be estimated analogously asI 17 , and we arrive at I 18 .U 1 +U 2 +U 3 : (2.6.10) Collecting the estimates (2.6.2), (2.6.5)–(2.6.6), and (2.6.9)–(2.6.10), we obtain U 1 . + + U 1 + ( + )U 2 + ( + )U 3 + X E D j;k;i + X E C j;k;i : (2.6.11) The termU 2 : Next, we estimate the sumU 2 . By (2.4.6), we have U 2 . X E D j;k;i + X E C j;k;i +U 4 + 1 X k=1 1 X i=0 k X l=1 k l k (k+i2) + (k +i 2)! k@ x T kl (@ t ) i vk L 2kT l k L 1 ( 0 ) + 1 X k=1 1 X i=0 k X l=1 k l k (k+i2) + (k +i 2)! kT kl (@ t ) i vk L 2k@ x T l k L 1 ( 0 ) + 1 X k=1 1 X i=0 k X l=1 k l k (k+i2) + (k +i 2)! kT kl (@ t ) i vk L 2kT l k L 1 ( 0 ) + 1 X k=1 1 X i=0 k (k+i2) + (k +i 2)! k[T k ; div](@ t ) i vk L 2 + 1 X k=1 1 X i=0 k (k+i2) + (k +i 2)! k[T k ; curl](@ t ) i vk L 2 =I 21 +I 22 +I 23 +I 24 +I 25 +I 26 +I 27 +I 28 : (2.6.12) 104 We split the sumI 24 according to the values ofk andi, obtaining I 24 = 1 X k=1 1 X i=2 k X l=1 k l k (k+i2) + (k +i 2)! k@ x T kl (@ t ) i vk L 2kT l k L 1 ( 0 ) + 1 X k=2 k X l=1 k l k (k2) + (k 2)! k@ x T kl vk L 2kT l k L 1 ( 0 ) + 1 X k=2 k X l=1 k l k (k1) + (k 1)! k@ x T kl (@ t )vk L 2kT l k L 1 ( 0 ) + 1 X i=0 (i1) + (i 1)! k@ x (@ t ) i vk L 2kTk L 1 ( 0 ) =I 241 +I 242 +I 243 +I 244 : For the termI 241 , we estimate I 241 . 1 X k=1 1 X i=2 k X l=1 l l ~ l kl (k+il2) + (k +il 2)! k@ x T kl (@ t ) i vk L 2 ! ~ l (l 3)! kT l k L 1 ( 0 ) k!(k +il 2)!(l 3)! l!(kl)!(k +i 2)! . 1 X k=1 1 X i=2 k X l=1 l1 ~ l1 kl (k+il2) + (k +il 2)! k@ x T kl (@ t ) i vk L 2 ! ~ l (l 3)! kT l k L 1 ( 0 ) : We choose ~ and use (2.4.17) to get I 241 . U 2 + U 3 : The termsI 242 ,I 243 , andI 244 can be estimated analogously toI 241 . Thus we arrive at I 24 . U 2 + U 3 : (2.6.13) 105 Similarly, the termsI 25 andI 26 are treated analogously toI 24 , obtaining I 25 +I 26 . U 2 + U 3 : (2.6.14) For the termI 27 , we use (2.4.11) and obtain I 27 . 1 X k=1 1 X i=0 k X k 0 =1 k k 0 k (i+k2) + (k +i 2)! k 0 ! K k 0 k@ x T kk 0 (@ t ) i vk L 2 . 1 X i=0 1 X k 00 =0 k 00 (i+k 00 2) + (i +k 00 2)! k@ x T k 00 (@ t ) i vk L 2 1 X k=k 00 +1 k!(i +k 00 2)! k 00 !(k +i 2)! kk 00 K kk 00 : (2.6.15) Note that the sum ink is bounded by K 1 X k=k 00 +1 k!(i +k 00 2)! k 00 !(k +i 2)! kk 00 1 K kk 00 1 . ; (2.6.16) uniformly for allk 00 ;i2N, by taking 1=CK. Thus from (2.6.15)–(2.6.16) we estimate I 27 . U 2 + U 3 : (2.6.17) The termI 28 can be estimated analogously asI 27 by using (2.4.13), and we arrive at I 28 . U 2 + U 3 : (2.6.18) Collecting the estimates (2.6.12), (2.6.13)–(2.6.14), and (2.6.17)–(2.6.18), we conclude U 2 . U 2 + U 3 +U 4 + X E D j;k;i + X E C j;k;i ; 106 from where we arrive at U 2 . U 3 +U 4 + X E D j;k;i + X E C j;k;i ; (2.6.19) by taking 1=C. The termsU 3 andU 4 : ForU 3 , we have by (2.4.3), U 3 . X E D j;k;i + X E C j;k;i +U 5 : (2.6.20) ForU 4 , we have by (2.4.7), U 4 . 1 X k=1 1 X i=0 k1 (k+i3) + (k +i 3)! k@ x T k1 (@ t ) i vk L 2 = U 2 + U 3 : The termU 5 : We claim that there existst 0 > 0 suciently small depending only onM 0 and 1 > 0 suciently small depending onM ;; (T ), such that 1 X i=0 (i3) + (i 3)! k(@ t ) i (p;v)k L 2. 1 +tQ(M ;; (t)); (2.6.21) for all2 (0; 1 ). The proof of (2.6.21) relies on the energy estimate of the partially linearized equation as in the case = R 3 , cf. [77, Lemma 6.2]. Here we only outline the modication needed for the presence of boundary. Instead of using the elliptic regularity, we appeal to Lemma 2.4.3 to estimate the dissipative term. Namely, we have k(@ t ) i rvk L 2.k(@ t ) i divvk L 2 +k(@ t ) i curlvk L 2 +k(@ t ) i vk H 1=2 (@ ) +k(@ t ) i vk L 2 .k(@ t ) i divvk L 2 +k(@ t ) i curlvk L 2 +k(@ t ) i vk L 2; 107 sincev = 0 on@ . Then we may proceed as in [77, Lemma 6.2] since the termk(@ t ) i vk L 2 can be absorbed in theA norm. It then follows from (2.6.21) that U 5 . 1 +tQ(M ;; (t)); t2 (0;t 0 ); (2.6.22) 2.6.1 Nonhomogeneoustransportequation As shown in [77] and [2], the curl component of the velocity satises the non-homogeneous transport equation. Thus, we consider @ t ~ S +vr ~ S =G; where ~ S = ~ S(t;x),v =v(t;x) andG =G(t;x). Denote kuk B() = X (j;k;i)2N 3 0 j k (j+k+i2) + (j +k +i 2)! k@ j x T k (@ t ) i uk L 2; with the corresponding dissipative analytic norm kuk ~ B() = X j+k+i3 j k (j +k +i 2) j+k+i3 (j +k +i 2)! k@ j x T k (@ t ) i uk L 2: Lemma2.6.1. For any; 2 (0; 1], there exists 1 2 (0; 1] such that if 0<(0) 1 , then k ~ S(t)k A() .k ~ S(0)k A() + Z t 0 kG(s)k A() +kv(s)k A() +t; 108 for some constantC > 0 and suciently smallT 0 > 0, providedK in (2.3.17) satises KCkvk A() ; t2 [0;T 0 ]; (2.6.23) where C 1 is a suciently large constant, and T 0 is chosen so that (2.3.22) holds. Similarly, for any ; 2 (0; 1], there exists 1 2 (0; 1] such that if 0<(0) 1 , then k ~ S(t)k B() .k ~ S(0)k B() + Z t 0 kG(s)k B() +kv(s)k B() +t; for some constantC > 0 and suciently smallT 0 > 0, providedK in (2.3.17) satises KCkvk B() ; t2 [0;T 0 ]; (2.6.24) forC 1 suciently large and whereT 0 is chosen so that (2.3.22) holds. Note thatkvk A() kvk B() for allv, and thus (2.6.23) implies (2.6.24). 2.6.2 Productrule In this section, we introduce the product rule which is needed in the estimates of divergence, curl, and pure time derivative components of the velocity. We denote kuk Y = sup (j;k;i)2N 3 0 ;0j+k+i4 k@ j x T k (@ t ) i uk L 2: The following lemma provides a product rule for the analyticB-norm. 109 Lemma 2.6.2. Letk2f2; 3;:::g and > 0. Forf 1 ;:::;f k 2 B(), and any, 2 (0; 1], there exists 1 2 (0; 1] andT 0 > 0 such that if 0<(0) 1 , then k k Y i=1 f i k B() C k k X i=1 kf i k B() Y 1jk;j6=i kf j k B() +kf j k Y : The next statement provides an analytic estimate for composition of functions. Lemma2.6.3. Assume thatf is an entire real-analytic function. Then for > 0 andw2A(), we have kf(w)k A() .Q(kwk A() +kwk Y ) and kf(w)k B() .Q(kwk B() +kwk Y ); for some functionQ. Recall thatE is assumed to be the product of two entire real-analytic functions. Suppose that ~ e is one of the components of the matrixE. Thus, ~ e(S;u) =f(S)g(u): (2.6.25) In the next statement, we provide an analytic estimates of ~ e. Lemma2.6.4. LetM 0 > 0 and assume that ~ e satises (2.6.25). Then k@ t ~ ek B() .Q(kuk A() +kuk Y ;kSk A() +kSk Y ); 2 (0; 1] 110 and k~ ek A() .Q(kuk A() +kuk Y ;kSk A() +kSk Y ); 2 (0; 1]; whereQ is a function. The proofs of Lemmas 2.6.1, 2.6.2, 2.6.3, and 2.6.4 are analogous to Lemmas 4.2, 5.1, 5.2, and 5.3 in [77]. Thus we simply state them and refer to [77] for details. 2.6.3 Thedivergencecomponents In this section, we estimate the sum of terms involving the divergence of the velocity. First, we rewrite the equation (2.2.4) as L(@ x )u =E(S;u)(@ t u +vru): (2.6.26) Forj 2 andk;i2N 0 , we commute@ j1 x T k (@ t ) i with (2.6.26), obtaining k@ j1 x T k (@ t ) i L(@ x )uk L 2 . j1 X j 0 =0 k X k 0 =0 i X i 0 =0 j 1 j 0 k k 0 i i 0 k@ j 0 x T k 0 (@ t ) i 0 E@ j1j 0 x T kk 0 (@ t ) ii 0 +1 uk L 2 +kEvk L 1k@ j1 x T k (@ t ) i ruk L 2 + j1 X j 0 =0 k X k 0 =0 i X i 0 =0 j 1 j 0 k k 0 i i 0 k@ j 0 x T k 0 (@ t ) i 0 (Ev)@ j1j 0 x T kk 0 (@ t ) ii 0 ruk L 2: (2.6.27) 111 Multiplying the above estimate with appropriate weights and following the arguments from [77, Lemma 6.3], which is justied since we have Lemmas 2.6.2, 2.6.3, and 2.6.4, we obtain 1 X j=2 1 X k=0 1 X i=0 D j;k;i . ( +)Q(M ;; (t)): (2.6.28) Forj = 1,k = 0, andi2N 0 , we proceed as in [77, Lemma 6.3] to get 1 X i=0 D 1;0;i . 1 + (t +)Q(M ;; (t)): (2.6.29) Forj = 1,k2N, andi2N 0 , from (2.4.7), we have kT k (@ t ) i L(@ x )uk L 2.k@ x T k1 (@ t ) i L(@ x )uk L 2; from where we proceed as in (2.6.27)–(2.6.28) to obtain 1 X k=1 1 X i=0 D 1;k;i . ( +)Q(M ;; (t)): (2.6.30) Combining (2.6.28)–(2.6.29) and (2.6.30), we arrive at X E D j;k;i; . 1 + ( + +t +)Q(M ;; (t)): (2.6.31) 2.6.4 Thecurlcomponents As in [77, Section 6.1], we use Lemmas 2.6.2, 2.6.3, and 2.6.4 to obtain X E C j;k;i . 1 + (t +)Q(M ;; (t)); (2.6.32) 112 for allt2 [0;T 0 ], 2.6.5 Thepressureestimates The analytic norm of the pressure can be recovered by the mixed space-time derivatives and pure time derivatives. Namely, forj2N andk2N 0 , we have k@ j x T k (@ t ) i pk L 2.k@ j1 x T k (@ t ) i L(@ x )uk L 2 +k@ j1 x [T k ;r](@ t ) i pk L 2: (2.6.33) The rst term on the right side is estimated in Section 2.6.3, while the second term is estimated analogously to (2.5.9). Thus, we arrive at 1 X j=1 1 X k=0 1 X i=0 (j1) + k (j+k+i3) + (j +k +i 3)! k@ j x T k (@ t ) i pk L 2. 1 + ( + +t +)Q(M ;; (t)): (2.6.34) Forj = 0 andk2N, we may use (2.4.7) and (2.6.33)–(2.6.34) to get 1 X k=1 1 X i=0 k (k+i3) + (k +i 3)! kT k (@ t ) i pk L 2. 1 X k=1 1 X i=0 k (k+i3) + (k +i 3)! k@ x T k1 (@ t ) i pk L 2 . 1 + ( + +t +)Q(M ;; (t)): (2.6.35) Combining (2.6.34)–(2.6.35) with the pure time derivative estimates of the pressure obtained in (2.6.21), we arrive at kpk A() . 1 + ( + +t +)Q(M ;; (t)): (2.6.36) 2.6.6 Proofofthemainlemma Here we combine the results of the previous section to prove Lemma 2.3.5. 113 Proof of Lemma 2.3.5. Combining the estimates (2.6.1), (2.6.11), (2.6.19), (2.6.20)–(2.6.22), and (2.6.31)–(2.6.32), we arrive at kvk A() . 1 + (t + + +(0) + )Q(M ;; (t)); (2.6.37) by taking 1=C and =C. Thus Lemma 2.3.5 follows by combining (2.6.36)–(2.6.37) with (2.5.1). 2.7 Analyticityassumptionsontheinitialdata In this section, we assume the initial data satises (2.3.14), and show that for low values ofi, 3 X i=0 1 X j;k=0 k@ j x T k (@ t ) i u(0)k L 2 (j+k+i3) + 0 (j +k +i 3)! ; (2.7.1) and 3 X i=0 1 X j;k=0 k@ j x T k (@ t ) i S(0)k L 2 (j+k+i3) + 0 (j +k +i 3)! ; (2.7.2) where > 0 is a suciently large constant depending onM 0 ; for high values ofi, there exists a suciently large constantC > 1 such that for alln 4 we have n X i=4 1 X j;k=0 k@ j x T k (@ t ) i u(0)k L 2 j+k+i3 0 C i3 (j +k +i 3)! 1; (2.7.3) and n X i=4 1 X j;k=0 k@ j x T k (@ t ) i S(0)k L 2 j+k+i3 0 C i3 (j +k +i 3)! 1: (2.7.4) 114 In (2.7.3) and (2.7.4) we then choose ~ 0 = 0 =C and using (2.7.1)–(2.7.2), we obtain k(p 0 ;v 0 ;S 0 )k A(~ 0 ) X (j;k;i)2N 3 0 k@ j x T k (@ t ) i (u;S)(0)k L 2 ~ (j+k+i3) + 0 (j +k +i 3)! 3 X i=0 1 X j;k=0 k@ j x T k (@ t ) i S(0)k L 2 (j+k+i3) + 0 (j +k +i 3)! + 1 X i=4 1 X j;k=0 k@ j x T k (@ t ) i S(0)k L 2 j+k+i3 0 C i3 (j +k +i 3)! + 1; and we conclude as in [77, Section 8]. The proofs of (2.7.1)–(2.7.4) are analogous to those in [77, Section 8] by using the commutator estimate (2.4.12), and thus we omit the details. 2.8 Proofoftheconvergencetheorem The following lemma is needed in the proof of the Theorem 2.3.4. Lemma2.8.1. ThereexistsaconstantC 1 > 1suchthatforany2N 3 0 withjj =j wherej2N 0 ,wehave k@ j x uk L 2C j 1 kuk 1=(j+1) L 2 k@ j+1 x uk j=(j+1) L 2 ; (2.8.1) for allu2H j+1 ( ). We emphasize thatC 1 isj-independent. Proof of Lemma 2.8.1. We proceed by induction onj2N 0 . The casej = 0 is clear. Now we assume that (2.8.1) holds for somej2 N 0 and aim to prove it for the casej + 1. Let2 N 3 0 be any multiindex with 115 jj =j + 1. We choose a multiindex 2N 3 0 such that there exists2N 3 0 for which + = and jj = 1. Let + = 2 . Using integration by parts, we obtain k@ uk 2 L 2 = Z @ @ u@ u i d Z @ u@ + udx =I 1 +I 2 ; (2.8.2) for somei2f1; 2; 3g. For the termI 1 , we use the Cauchy-Schwarz inequality to get I 1 Z @ j@ uj 2 d 1=2 Z @ j@ uj 2 d 1=2 Ck@ uk 1=2 L 2 kr@ uk 1=2 L 2 k@ uk 1=2 L 2 kr@ uk 1=2 L 2 ; (2.8.3) where the last inequality follows from the trace theorem. For the term I 2 , using the Cauchy-Schwarz inequality, we arrive at I 2 k@ uk L 2k@ + uk L 2: (2.8.4) Summing (2.8.2) overjj =j + 1 and using (2.8.3)–(2.8.4), we obtain k@ j+1 x uk 2 L 2 C X jj=j+1 k@ uk 1=2 L 2 kr@ uk 1=2 L 2 k@ uk 1=2 L 2 kr@ uk 1=2 L 2 + X jj=j+1 k@ uk L 2k@ + uk L 2 Ck@ j x uk 1=2 L 2 k@ j+1 x uk L 2k@ j+2 x uk 1=2 L 2 +Ck@ j x uk L 2k@ j+2 x uk L 2: Using the Cauchy-Schwarz inequality to absorbk@ j+1 x uk L 2 from the rst factor and then using the induc- tion hypothesis for the casej, we obtain k@ j+1 x uk 2 L 2 Ck@ j x uk L 2k@ j+2 x uk L 2CC j 1 kuk 1=(j+1) L 2 k@ j+1 x uk j=(j+1) L 2 k@ j+2 x uk L 2; 116 from where we arrive at k@ j+1 x uk L 2C (j+1)=(j+2) C j(j+1)=(j+2) 1 kuk 1=(j+2) L 2 k@ j+2 x uk (j+1)=(j+2) L 2 C j+1 1 kuk 1=(j+2) L 2 k@ j+2 x uk (j+1)=(j+2) L 2 ; by takingC 1 C. Therefore, (2.8.1) is proven forj + 1 for some suciently large constantC 1 > 1. Next we prove the second main result on convergence of the Mach limit. Proof of Theorem 2.3.4. Denote the spatial analyticity radius = (0) C 0 ; where(0),2 (0; 1] are xed constants as in Theorem 2.3.3 andC 0 2 is a suciently large constant to be determined below. It is proved in Theorem 2.3.3 that if the initial data (p 0 ;v 0 ;S 0 ) satises (2.3.13)– (2.3.14) for some constantsM 0 , 0 > 0, as well as the compatibility condition of all orders, then k(p ;v ;S )(t)k X() .M 2 ; 2 (0; 0 ]; t2 [0;T 0 ]; for some parameters,(0), 0 , andT 0 > 0. We proceed dierently than in [77, Section 7]. For the sake of contradiction, we assume that (v ;p ;S ) does not converge to (v (inc) ; 0;S (inc) ) inL 2 ([0;T ];X()). Then there exists a sequencef n g! 0 such that f(v n ;p n ;S n )g does not converge to (v (inc) ; 0;S (inc) ) inL 2 ([0;T ];X()). Recall from [2, Theorem 1.2] 117 that (v n ;p n ;S n ) converges to (v (inc) ; 0;S (inc) ) in L 2 ([0;T ];L 2 ( )) as n ! 0. We dene v kn (t) = v k (t)v n (t), fork;n2N. Using Lemma 2.8.1, we obtain kn (t) = 1 X m=4 X jj=m m3 (m 3)! k@ v kn k L 2 x 1 X m=4 C m 1 m3 (m 3)! kv kn k 1=(m+1) L 2 x k@ m+1 x v kn k m=(m+1) L 2 x ; (2.8.5) and thus, using the Minkowski and Hölder inequalities k kn (t)k L 2 t . 1 X m=4 C m 1 m3 (m 3)! kv kn k 1=(m+1) L 2 x;t k@ m+1 x v kn k m=(m+1) L 2 x;t ; (2.8.6) where the space and time domains are understood to be and [0;T ], respectively. To show convergence to zero, we shall apply the discrete dominated convergence theorem. To get a uniform bound, we use the discrete Young inequality, we get 1 X m=4 (C 1 ) m3 (m 3)! k@ m+1 x v kn k m=(m+1) L 2 x;t = 1 X m=4 (C 1 ) m3 (m 3)! k@ m+1 x v kn k L 2 x;t m=(m+1) (C 1 ) m3 (m 3)! 1=(m+1) 1 X m=4 m(C 1 ) m3 (m + 1)(m 3)! k@ m+1 x v kn k L 2 x;t + 1 X m=4 (C 1 ) m3 (m + 1)(m 3)! : Now, choose =(0)=C 1 C 2 , whereC 1 is the constant from Lemma 2.8.1 andC 2 1. We obtain 1 X m=4 (C 1 ) m3 (m 3)! k@ m+1 x v kn k m=(m+1) L 2 x;t CM 2 1 C 1 1 +C: (2.8.7) Thus the discrete dominated convergence theorem applies, and it thus follows from (2.8.5)–(2.8.7) that 1 X m=4 X jj=m (m3) + (m 3)! k@ v kn k L 2 x;t ! 0 ask;n!1 (2.8.8) 118 sincekv kn (t)k L 2 x;t ! 0 ask;n!1. Note that from (2.3.23) we have kv kn k 2 L 2 t H 3 Ckv kn k 3=2 H 4 kv kn k 1=2 L 2 x;t +Ckv kn k 2 L 2 x;t Ckv kn k 1=2 L 2 x;t +Ckv kn k 2 L 2 x;t ! 0 ask;n!1: (2.8.9) From (2.8.8)–(2.8.9) and analogous inequalities forp kn andS kn , we infer that the sequencef(v n ;p n ;S n )g is Cauchy in L 2 ([0;T 0 ];X()), which along with (v ;p ;S ) ! (v (inc) ; 0;S (inc) ) in L 2 ([0;T ];L 2 ( )) leads to a contradiction. Therefore,f(v ;p ;S )g is convergent and thus goes to (v (inc) ; 0;S (inc) ) in L 2 ([0;T 0 ];X()) as! 0. 119 Chapter3 Machlimitsoftheisentropicuidsintheanalyticspaces 3.1 Introduction In this paper we consider the incompressible limit of the compressible Euler equations for isentropic uids inR 3 . First, the study of Mach limit problem involves uniform bounds and existence of solution for a time interval independent of the Mach number> 0. Second, one can justify the convergence to the solution of the limiting equations as tends to zero. For the non-isentropic ows with general initial data, the rst existence and convergence results were obtained by Métivier and Schochet in [121] for the whole space and the torus. Their approach relies on the fact that one can obtain uniform bounds by applying some suitable operators to the equations. For domains with boundaries, the existence results and convergence in exterior domains were given by Alazard in [2]. The above-mentioned results hold in the Sobolev spaces. In a recent work [77], we studied the Mach limit in the analytic spaces in the whole space. The proof relies on the elliptic regularity for the velocity and the energy estimates for the entropy, curl, and the time derivative components of the velocity. In another recent work [78], the Mach limit in the analytic spaces for the exterior domains with analytic boundaries was addressed. For the isentropic ows with well-prepared initial data, it is well-established that the solution of the compressible isentropic Euler equations exists for a time interval independent of the Mach 120 number, and converges to the corresponding incompressible Euler equations as Mach number tends to zero (cf. [84, 85]). For other works on the isentropic ows, we refer the reader to [44, 49, 85, 110, 117]. The main goal of this paper is to provide a simple and new approach to the Mach limit of the isentropic Euler equations in the analytic spaces for more general pressure laws. Our rst main result shows that, given analytic initial data, the solution is uniformly bounded on a time interval, which is independent of the Mach number. Our second main theorem shows that the solution of a slightly compressible Euler equations converges to the limiting equations as Mach number tends to zero. As the equation of state in [77] is assumed to be entire, we extend the results for isentropic uids to the case where the equation of state is more general. The proof diers to [77] and the major simplication is that the analytic energy estimates are performed using only spatial derivatives of the velocity. As a result, the main diculty is in obtaining the explicit expressions for an arbitrary partial dierential derivatives of the matrixE(u ) (cf. the formulation (3.2.7)) in terms ofu . We overcome this by using the multivariate Faa di Bruno formula (cf. [13, 38]), which generalizes chain rule to higher-order derivatives. Moreover, we introduce a power series version of the multivariate Faa di Bruno formula which in turn estimate the commutator terms. The paper is organized as follows. In Section 3.2, we introduce the Mach limit problem for the isen- tropic Euler equations and then formulate the symmetrized hyperbolic system. In Section 3.3, we dene the analytic spaces and state our main results. The proof of the rst theorem relies on Lemma 3.3.3, which is proved in Section 3.4 assuming two key estimates on the convection term and the singular term, Lemma 3.4.1 and 3.4.2. Section 3.5 introduces the multivariate Faa di Bruno formula and and contains the proof of Lemma 3.4.1 and 3.4.2. 121 3.2 Reformulation We consider the compressible Euler equations describing the motion of an inviscid, isentropic uid inR 3 @ t +vr +rv = 0; (3.2.1) (@ t v +vrv) +rP = 0; (3.2.2) where = (t;x)2 R + is the density,v = v(t;x)2 R 3 is the velocity, andP = P (t;x)2 R + is the pressure. To close the system (3.2.1)–(3.2.2), we assume that the uid is barotropic and the equation of state is given in the form =R(P ); whereR: R + ! R + is a smooth function satisfying@R=@P > 0. For instance, in the case of an ideal gaseous uid, =KP 1= ; whereK > 0 is some constant and > 1 is the adiabatic exponent. After some rescalings and change of variables (cf. [121, 2, 77]), we are led to the symmetric hyperbolic system a(@ t p +v rp ) + 1 rv = 0; (3.2.3) r(@ t v +v rv ) + 1 rp = 0; (3.2.4) 122 wherev =v (t;x)2R 3 is the velocity,p =p (t;x)2R is the pressure variation,2R + is the Mach number; in agreement with the equation of state for ideal gas, we adopt the assumption thata andr are positive real-analytic functions of the form a =a(p ); r =r(p ): (3.2.5) Thus we have obtained the symmetrized version of the compressible Euler equations for isentropic uids inR 3 , which reads E(@ t u +v ru ) + 1 L(@ x )u = 0; (3.2.6) whereu = (p ;v ) T and E(u ) = 0 B B @ a(p ) 0 0 r(p )I 3 1 C C A ; L(@ x ) = 0 B B @ 0 div r 0 1 C C A : (3.2.7) From here on, we focus on the formulation (3.2.6) inR 3 with initial datau 0 = (p 0 ;v 0 ) T . 3.3 Themainresults We assume that the initial data (p 0 ;v 0 ) satises 1 X m=0 X jj=m m 0 (m 3)! k@ (p 0 ;v 0 )k L 2 (R 3 ) M 0 ; (3.3.1) where 0 ,M 0 > 0 are xed constants. For > 0, we dene the analytic space A() =fu2C 1 (R 3 ):kuk A() <1g; 123 where kuk A() = 1 X m=0 X jj=m (t) m (m 3)! k@ uk L 2 (R 3 ) : (3.3.2) We emphasize that, as opposed to [77, 78], the norm does not contain time derivative of the velocity. In (3.3.2) and below, we use the convention thatn! = 1 forn2N. We dene the analyticity radius function as (t) = 0 Kt; (3.3.3) whereK 1 is a suciently large parameter to be determined below. We shall work on the time interval [0;T ] such that T 0 2K : (3.3.4) In view of (3.3.3)–(3.3.4), we have 0 =2(t) 0 fort2 [0;T ]. The rst main theorem provides a uniform in boundedness of the solutions in the analytic norm on a time interval, which is independent of. Theorem 3.3.1. Assume that the initial data (p 0 ;v 0 ) satisfy (3.3.1), where 0 , M 0 > 0. Then there exist suciently small constants 0 , T 0 > 0, depending on 0 andM 0 , such that the solution (p ;v ) of (3.2.6) satisfy the estimate k(p ;v )(t)k A() M; 0< 0 ; t2 [0;T 0 ]; (3.3.5) where is as in (3.3.3) andK andM are suciently large constants depending onM 0 . 124 Our second main theorem states that the solution of (3.2.6) converge to the solution of the stratied incompressible Euler equations r 0 (@ t v +vrv) +r = 0; (3.3.6) divv = 0; (3.3.7) as! 0, wherer 0 =r(0) is as in (3.2.5). Theorem 3.3.2. Assume that the initial data (p 0 ;v 0 ) satisfy (3.3.1) uniformly for xed 0 ,M 0 > 0. Also, suppose that v 0 converges to v 0 in H 3 (R 3 ). Then (v ;p ) converge to (v (inc) ; 0)2 L 1 ([0;T 0 ];A()) in L 2 ([0;T ];A()), where2 (0; 0 ] is a suciently small constant andv (inc) is the solution to (3.3.6)–(3.3.7) with initial dataw 0 , andw 0 is the unique solution of divw 0 = 0; curl(r 0 w 0 ) = curl(r 0 v 0 ): In order to prove Theorem 3.3.1, we establish the following a priori estimate of the (modied) velocity. Lemma3.3.3. Assume the initial data (p 0 ;v 0 ) satisfy (3.3.1), whereM 0 ; 0 > 0. Then there exist constants 0 ;T 0 > 0, such that for all2 (0; 0 ], the norm M (T ) := sup t2[0;T ] k(p (t);v (t))k A((t)) (3.3.8) satises the estimate M (t)C +CtM (t); (3.3.9) 125 fort2 [0;T 0 ], provided KQ(M (T )); (3.3.10) whereK is as in (3.3.3) andQ is a nonnegative continuous function. For the proof of Theorem 3.3.1 given Lemma 3.3.3, we refer the reader to [2, 77]. As a consequence of Theorem 3.3.1, the proof of Theorem 3.3.2 is analogous to the one in [77, Section 7] by using the interpo- lation inequality and the Stirling’s formula, and thus we omit further details. In the rest of the paper, the symbolC denotes a generic constant depending onM 0 and 0 , which may vary from inequality to inequality. For simplicity of the notation, we omit the superscript, and writep andv forp andv . Remark 3.3.4. (Boundedness of Sobolev norm) By (3.3.1) and [2, Theorem 1.1], theH 3 norm of (p ;v ) can be estimated by a constant on a time interval [0;T 0 ], whereT 0 > 0 depends only on 0 andM 0 . In particular, ifF is a smooth function ofu = (p ;v ), then there exists constantC > 0 depending onF , such that kF (u )k L 1 x C; t2 [0;T 0 ]; 2 (0; 1]: From here on, we shall work on the time interval [0;T ] whereT2 [0;T 0 ]. Remark 3.3.5. (Mach limit in a Gevrey norm) Assume that the initial data is Gevrey regular, we may generalize Theorem 3.3.1 and 3.3.2 to the Gevrey norms by showing uniform boundedness of the solution and convergence in Gevrey spaces. The proofs of Remark 3.3.5 are analogous to those in [77, Section 9], and thus we omit the details. 126 3.4 ProofofLemma3.3.3 In this section we prove Lemma 3.3.3 and thus complete the proof of Theorem 3.3.1. Since the matrixE is symmetric positive denite, we may rewrite the equation (3.2.6) as @ t u +vru + 1 E 1 L(@ x )u = 0: (3.4.1) Fixm2N 0 and2N 3 0 withjj =m. We apply@ to the equation (3.4.1) to obtain @ t @ u +vr@ u + 1 E 1 L(@ x )@ u =F ; (3.4.2) where F = [vr;@ ]u + 1 [E 1 L(@ x );@ ]u: Taking theL 2 -inner product of (3.4.2) with@ u and using the Cauchy-Schwarz and Hölder inequalities, we obtain 1 2 d dt k@ uk 2 L 2 + 1 hE 1 L(@ x )@ u;@ uiCkrvk L 1 x k@ uk 2 L 2 +CkF k L 2k@ uk L 2; (3.4.3) whereh;i denotes the scalar product inL 2 . The second term on the left side can be rewritten as 1 hE 1 L(@ x )@ u;@ ui = 1 hL(@ x )E 1=2 @ u;E 1=2 @ ui 1 h[L(@ x );E 1=2 ]@ u;E 1=2 @ ui: (3.4.4) Since the operatorL(@ x ) is formally skew-adjoint, the rst term on the right side of (3.4.4) is cancelled out. For the second term on the right side of (3.4.4), using the product rule and chain rule, it suces to 127 estimate the lower-order terms. Note that the matrixE depends onu, the factor 1= is cancelled out. From (3.4.3)–(3.4.4), we use the Hölder inequality and Remark 3.3.4 to get d dt k@ uk L 2Ck@ uk L 2 +CkF k L 2: Summing overjj =m, we arrive at d dt X jj=m k@ uk L 2C X jj=m kF k L 2 +C X jj=m k@ uk L 2: (3.4.5) Denote the dissipative analytic norm by kuk ~ A() = 1 X m=1 X jj=m m m1 (m 3)! k@ uk L 2: Using the above notation and (3.3.2), the estimate (3.4.5) implies that d dt kuk A() = _ kuk ~ A() + 1 X m=0 X jj=m m (m 3)! d dt k@ uk L 2 _ kuk ~ A() +Ckuk A() +I 1 +I 2 ; (3.4.6) where I 1 =C 1 X m=1 X jj=m m (m 3)! k[vr;@ ]uk L 2; (3.4.7) and I 2 = C 1 X m=1 X jj=m m (m 3)! k[E 1 L(@ x );@ ]uk L 2: (3.4.8) 128 In order to estimateI 1 , we use the following lemma, the proof of which is given in Section 3.5 below. Lemma3.4.1. There exists a constantT 0 > 0, depending on 0 andM 0 , such that I 1 Ckuk A() kuk ~ A() ; 2 (0; 1]; t2 [0;T 0 ]; (3.4.9) for some constantC > 0. The following lemma, the proof of which is given in Section 3.5 below, shall be used to estimateI 2 . Lemma3.4.2. There exist suciently small constants 0 ;T 0 > 0, depending on 0 andM 0 such that I 2 Ckuk ~ A() +Ckuk A() kuk ~ A() ; 2 (0; 0 ]; t2 [0;T 0 ]; (3.4.10) for some constantC > 0. It follows from (3.4.6)–(3.4.8) and Lemmas 3.4.1 and 3.4.2 that d dt kuk A() kuk ~ A() ( _ +Ckuk A() +C) +Ckuk A() ; 2 (0; 0 ]; t2 [0;T 0 ]; (3.4.11) for some constantC > 0. Now, assume that the radius(t) decreases suciently fast so that the factor next tokuk ~ A() is less than or equal to 0, namely, KQ(M (T )); for some nonnegative continuous functionQ. Integrating (3.4.11) in time from 0 tot, we get ku(t)k A() C +CtM (t): 129 Thus, the proof of Lemma 3.3.3 is concluded. 3.5 Commutatorestimates Before we prove Lemma 3.4.1 and 3.4.2, we introduce the multivariate Faa di Bruno formula which shall be used throughout this section. For multi-indices;2N 3 0 where = ( 1 ; 2 ; 3 ) and = ( 1 ; 2 ; 3 ), we introduce a linear order onN 3 0 by writing provided one of the following holds: (i) jj<jj; (ii)jj =jj and 1 < 1 ; (iii) jj =jj, 1 = 1 ;:::; k = k and k+1 < k+1 for some 1k< 3. We write x =x 1 1 x 2 2 x 3 3 ; wherex = (x 1 ;x 2 ;x 3 )2R 3 . With the above notations, we recall [38, Theorem 2.1]. Lemma 3.5.1. (Multivariate Faa di Bruno) Letg : R 3 !R be a scalar function, that is smooth in a neigh- borhoodofsomepointx 0 2R 3 . Leth:R!Rbeascalarfunction,smoothinaneighborhoodofy 0 =g(x 0 ). Denef(x) =h(g(x)). Then for any2N 3 0 withjj 1, we have @ f(x 0 ) = jj X i=1 h (i) (y 0 ) jj X s=1 X Ps(i;) ! s Y l=1 (@ l g(x 0 )) k l k l ! l ! k l ; (3.5.1) 130 where P s (i;) = (k 1 ;:::;k s ; 1 ;:::; s )2N:::NN 3 0 :::N 3 0 : 0 1 s ; s X l=1 k l =i and s X l=1 k l l = : (3.5.2) The next lemma provides a multivariate Faa di Bruno formula for power series, which is needed in the proof of Lemma 3.4.2. Lemma3.5.2. (Power series Faa di Bruno) Suppose (x) = 1 X n=0 a n x n ; x2R; wherea n 2R for alln2N 0 , and (x) = X 2N 3 0 ;jj1 b x ; x2R 3 ; whereb 2Rfor2N 3 0 withjj 1. Thenthecomposition( (x))isapowerseriesandcanbewrittenas ( (x)) = X 2N 3 0 c x ; x2R 3 ; wherec (0;0;0) =a 0 and c = jj X i=1 jj X s=1 X Ps(i;) i k 1 ;:::;k s a i s Y l=1 b k l l ; (3.5.3) for2N 3 0 withjj 1. 131 Proof of Lemma 3.5.2. The composition( (x)) is a formal power series and thus we assume that ( (x)) = X 2N 3 0 d x ; x2R 3 ; (3.5.4) whered 2 R with2 N 3 0 . For the constant term of (3.5.4) we haved (0;0;0) = a 0 . Fix2 N 3 0 where jj 1. We apply@ to (3.5.4) on both sides and evaluate atx =0, obtaining @ (( (x)))j x=0 =!d : (3.5.5) On the other hand, using Lemma 3.5.1, we arrive at @ (( (x)))j x=0 = jj X i=1 jj X s=1 X Ps(i;) (i) ( (0))! s Y l=1 (@ l (0)) k l k l ! l ! k l = jj X i=1 jj X s=1 X Ps(i;) i!a i ! s Y l=1 l ! k l b k l l k l ! l ! k l : (3.5.6) Combining (3.5.4)–(3.5.6), we obtain (3.5.3) and thus conclude the proof of Lemma 3.5.2. Proof of Lemma 3.4.1. Using the Leibniz rule, we obtain I 1 C 1 X m=1 m X j=1 X jj=m X jj=j; m (m 3)! k@ vr@ uk L 2 = 1 X m=1 m X j=1 I m;j ; (3.5.7) where I m;j =C X jj=m X jj=j; m (m 3)! k@ vr@ uk L 2: 132 We split the right side of (3.5.7) according to low and high values ofj. We claim that there exists a constant C > 0 such that 1 X m=1 [m=2] X j=1 I m;j Ckuk A() kuk ~ A() (3.5.8) and 1 X m=1 m X j=[m=2]+1 I m;j Ckuk A() kuk ~ A() : (3.5.9) Proof of (3.5.8): Using Hölder and Sobolev inequalities we get 1 X m=1 [m=2] X j=1 I m;j C 1 X m=1 [m=2] X j=1 X jj=m X jj=j; m (m 3)! k@ vk 1=4 L 2 kD 2 @ vk 3=4 L 2 kr@ uk L 2 C 3=2 1 X m=1 [m=2] X j=1 X jj=m X jj=j; (mj + 1) mj (mj 2)! kr@ uk L 2 j (j 3)! k@ vk L 2 1=4 j+2 (j 1)! kD 2 @ vk L 2 3=4 A m;j;; ; (3.5.10) where A m;j;; = (j 3)! 1=4 (j 1)! 3=4 (mj 2)! (mj + 1)(m 3)! : (3.5.11) Recall the combinatorial inequality jj jj ; (3.5.12) 133 which implies A m;j;; m! j!(mj)! (j 3)! 1=4 (j 1)! 3=4 (mj 2)! (mj + 1)(m 3)! Cm 3 (mj) 3 C; (3.5.13) sincej [m=2]. Using X jj=m X ;jj=j x y = 0 @ X jj=j x 1 A 0 @ X j j=mj y 1 A : (3.5.14) from [97, Lemma 4.2], together with (3.5.10)–(3.5.13) and the discrete Hölder inequality, we obtain 1 X m=1 [m=2] X j=1 I m;j C 1 X m=1 [m=2] X j=1 0 @ X jj=j j (j 3)! k@ vk L 2 1 A 1=4 0 @ X jj=j j+2 (j 1)! kD 2 @ vk L 2 1 A 3=4 0 @ X j j=mj (mj + 1) mj (mj 2)! k@ ruk L 2 1 A Ckuk A() kuk ~ A() ; (3.5.15) where the last inequality follows from the discrete Young inequality. Proof of (3.5.9): We reverse the roles ofj andmj and proceed as in (3.5.10), obtaining 1 X m=1 m X j=[m=2]+1 I m;j C 1 X m=1 m X j=[m=2]+1 X jj=m X jj=j; m (m 3)! kD 2 r@ uk 3=4 L 2 kr@ uk 1=4 L 2 k@ vk L 2 C 3=2 1 X m=0 m X j=[m=2]+1 X jj=m X jj=j; (mj + 3) mj+2 (mj)! kD 2 r@ uk L 2 3=4 (mj + 1) mj (mj 2)! kr@ uk L 2 1=4 j (j 3)! k@ vk L 2 B m;j;; ; (3.5.16) 134 where B m;j;; = (mj 2)! 1=4 (mj)! 3=4 (j 3)! (mj + 3) 3=4 (mj + 1) 1=4 (m 3)! : Using (3.5.12), we get B m;j;; m! j!(mj)! (mj 2)! 1=4 (mj)! 3=4 (j 3)! (mj + 3) 3=4 (mj + 1) 1=4 (m 3)! Cm 3 j 3 C; (3.5.17) sincej [m=2] + 1. We combine (3.5.16)–(3.5.17) and proceed as in (3.5.15), obtaining 1 X m=1 m X j=[m=2]+1 I m;j Ckuk A() kuk ~ A() : Combining (3.5.7)–(3.5.9), we conclude the proof of Lemma 3.4.1. Proof of Lemma 3.4.2. Using the Leibniz rule, we obtain I 2 C 1 X m=1 m X j=1 X jj=m X jj=j; m (m 3)! k@ E 1 r@ uk L 2: (3.5.18) We denote the non-trivial element of the matrixE 1 (x) byh(x). Using Lemma 3.5.1, we arrive at I 2 C 1 X m=1 m X j=1 j X i=1 j X s=1 X jj=m X jj=j; X Ps(i;) i1 m ! (m 3)! kh (i) (u) s Y l=1 (@ l p) k l k l ! l ! k l r@ uk L 2 = 1 X m=1 m X j=1 J m;j ; (3.5.19) 135 where J m;j =C j X i=1 j X s=1 X jj=m X jj=j; X Ps(i;) i1 m ! (m 3)! kh (i) (u) s Y l=1 (@ l p) k l k l ! l ! k l r@ uk L 2: We split the far right side of (3.5.19) according to the low and high values ofj. We claim that there exists a constantC > 0 such that 1 X m=1 [m=2] X j=1 J m;j Ckuk ~ A() +Ckuk A() kuk ~ A() (3.5.20) and 1 X m=1 m X j=[m=2]+1 J m;j Ckuk A() kuk ~ A() : (3.5.21) Proof of (3.5.20): Using Hölder and Sobolev inequalities, we arrive at J m;j C j X i=1 j X s=1 X jj=m X jj=j; X Ps(i;) C i i1 m ! (m 3)! kr@ uk L 2kh (i) (u)k L 1 s Y l=1 1 k l ! l ! k l kD 2 @ l pk 3k l =4 L 2 k@ l pk k l =4 L 2 C j X i=1 j X s=1 X jj=m X jj=j; X Ps(i;) C i i1 i 3i=2 (mj + 1) mj (mj 2)! kr@ uk L 2 i (i 3)! kh (i) (u)k L 1 C m;j;i;s s Y l=1 j l j+2 (j l j 1)! kD 2 @ l pk L 2 ! 3k l =4 j l j (j l j 3)! k@ l pk L 2 ! k l =4 ; (3.5.22) where2 (0; 1] is some constant to be determined below and C m;j;i;s = (i 3)!(mj 2)!! (m 3)!(mj + 1) s Y l=1 (j l j 1)! 3k l =4 (j l j 3)! k l =4 k l ! l ! k l : (3.5.23) 136 For; 1 ; ; k 2N 3 0 andk2N, using (3.5.12) and induction, we obtain 1 ; 2 ; ; k jj j 1 j;j 2 j; ;j k j ; (3.5.24) where = k X i=1 i : From (3.5.23)–(3.5.24) we arrive at C m;j;i;s Cm! j!(mj)! (mj 2)! (m 3)!(mj + 1) i k 1 ;:::;k s jj! Q s l=1 j l j! k l s Y l=1 (j l j 1)! 3k l =4 (j l j 3)! k l =4 Cm 3 (mj) 3 i k 1 ;:::;k s C i k 1 ;:::;k s ; (3.5.25) sincej [m=2]. Recall thath(x) is an analytic function. Thus, for any xed > 0, there exists a constant 2 (0; 1] such that kD i h(x)k L 1 C(i 3)! i ; jxj; i2N 0 ; (3.5.26) for some constantC > 0. It follows from Remark 3.3.4 that kD i h(u)k L 1 C(i 3)! i ; x2R 3 ; i2N 0 ; (3.5.27) 137 by choosing=CM (T ). Combining (3.5.22), (3.5.25), and (3.5.27), together with splitting the low and high values ofi, we obtain 1 X m=1 [m=2] X j=1 J m;j C 1 X m=1 [m=2] X j=1 j X i=2 j X s=1 X jj=m X jj=j; X Ps(i;) i k 1 ;:::;k s (mj + 1) mj (mj 2)! kr@ uk L 2 C i i=2 i 3i=2 s Y l=1 j l j+2 (j l j 1)! kD 2 @ l pk L 2 ! 3k l =4 j l j (j l j 3)! k@ l pk L 2 ! k l =4 + C 3=2 1 X m=1 [m=2] X j=1 X jj=m X jj=j; (mj + 1) mj (mj 2)! kr@ uk L 2 j+2 (j 1)! kD 2 @ pk L 2 3=4 j (j 3)! k@ pk L 2 1=4 =J 1 +J 2 ; (3.5.28) where the inequality follows from P s (1;) = 8 > > > > < > > > > : f(1;)g; s = 1; ;; s 2. For the termJ 1 , we set the power series (x) = 1 X n=2 (C 1=2 1 3=2 ) n x n ; x2R and (x) = X 2N 3 0 ;jj1 b x ; x2R 3 ; 138 where b = jj+2 (jj 1)! kD 2 @ pk L 2 ! 3=4 jj (jj 3)! k@ pk L 2 ! 1=4 (3.5.29) for2N 3 0 withjj 1. It follows from Lemma 3.5.2 that 1 X n=2 (C 1=2 1 3=2 ) n 0 @ X 2N 3 0 ;jj1 b x 1 A n = X 2N 3 0 ;jj1 e x ; x2R 3 ; (3.5.30) where e = jj X i=2 jj X s=1 X Ps(i;) (C 1=2 1 3=2 ) i i k 1 ;:::;k s s Y l=1 j l j+2 (j l j 1)! kD 2 @ l pk L 2 ! 3k l =4 j l j (j l j 3)! k@ l pk L 2 ! k l =4 ; (3.5.31) for2N 3 0 withjj 1. From (3.5.14), (3.5.28), and (3.5.31), we arrive at J 1 C 1 X m=1 [m=2] X j=1 X jj=m X jj=j; (mj + 1) mj (mj 2)! kr@ uk L 2 e C 1 X m=1 [m=2] X j=1 X jj=j e 0 @ X j j=mj (mj + 1) mj (mj 2)! kr@ uk L 2 1 A Ckuk ~ A() 1 X n=2 C 1=2 1 3=2 X 2N 3 0 ;jj1 b n ; (3.5.32) where the last inequality follows from (3.5.30) withx = (1; 1; 1). Note that from (3.5.29) and the discrete Young inequality, we have X 2N 3 0 ;jj1 b Ckuk A() : (3.5.33) 139 Choosing 2 3 =CM (T ) 2 and combining (3.5.32)–(3.5.33), we arrive at J 1 Ckuk ~ A() : (3.5.34) For the termJ 2 , we proceed as in (3.5.10)–(3.5.15) to get J 2 Ckuk A() kuk ~ A() : (3.5.35) Combining (3.5.28) and (3.5.34)–(3.5.35), we conclude the proof of (3.5.20). Proof of (3.5.21): Using Hölder and Sobolev inequalities, together with (3.5.24), we arrive at J m;j C j X i=1 j X s=1 X jj=m X jj=j; X Ps(i;) C i i1 m jj! (m 3)!k s !j s j! ks kh (i) (u)k L 1 s1 Y l=1 1 k l !j l j! k l kD 2 @ l pk 3k l =4 L 2 k@ l pk k l =4 L 2 ! k@ s pk L 2kD 2 @ s pk 3(ks1)=4 L 2 k@ s pk (ks1)=4 L 2 kD 2 r@ uk 3=4 L 2 kr@ uk 1=4 L 2 C j X i=1 j X s=1 X jj=m X jj=j; X Ps(i;) C i i1 i 3i=2 i (i 3)! kh (i) (u)k L 1 s1 Y l=1 j l j+2 (j l j 1)! kD 2 @ l pk L 2 ! 3k l =4 j l j (j l j 3)! k@ l pk L 2 ! k l =4 jsj (j s j 3)! k@ s pk L 2 ! jsj+2 (j s j 1)! kD 2 @ s pk L 2 ! 3(ks1)=4 jsj (j s j 3)! k@ s pk L 2 ! (ks1)=4 (mj + 3) mj+2 (mj)! kD 2 r@ uk L 2 3=4 (mj + 1) mj (mj 2)! kr@ uk L 2 1=4 D m;j;i;s ; (3.5.36) 140 where D m;j;i;s = (i 3)!(mj 2)! 1=4 (mj)! 3=4 jj! (m 3)!(mj + 1) 1=4 (mj + 3) 3=4 (j s j 1)! 3(ks1)=4 (j s j 3)! (ks1)=4 (j s j 3)! k s !j s j! ks s1 Y l=1 (j l j 1)! 3k l =4 (j l j 3)! k l =4 k l !j l j! k l : Using (3.5.12), we arrive at D m;j;i;s C i k 1 ;:::;k s m!(j s j 3)! (m 3)!j s j! C i k 1 ;:::;k s m 3 j s j 3 : Note that from (3.5.2) we get jj = s X l=1 k l j l jj s j s X l=1 k l =ij s j; from where D m;j;i;s C i k 1 ;:::;k s m 3 i 3 jj 3 C i k 1 ;:::;k s i 3 ; (3.5.37) 141 sincejj =j [m=2] + 1. From (3.5.27) and (3.5.36)–(3.5.37), we obtain 1 X m=1 m X j=[m=2]+1 J m;j C 1 X m=1 m X j=[m=2]+1 j X i=2 j X s=1 X jj=m X jj=j; X Ps(i;) C i i=2 i 3i=2 i k 1 ;:::;k s jsj (j s j 3)! k@ s pk L 2 ! (mj + 1) mj (mj 2)! kr@ uk L 2 1=4 (mj + 3) mj+2 (mj)! kD 2 r@ uk L 2 3=4 s1 Y l=1 j l j+2 (j l j 1)! kD 2 @ l pk L 2 ! 3k l =4 j l j (j l j 3)! k@ l pk L 2 ! k l =4 jsj+2 (j s j 1)! kD 2 @ s pk L 2 ! 3(ks1)=4 jsj (j s j 3)! k@ s pk L 2 ! (ks1)=4 +C 1 X m=1 m X j=[m=2]+1 X jj=m X jj=j; j (j 3)! k@ pk L 2 (mj + 1) mj (mj 2)! kr@ uk L 2 1=4 (mj + 3) mj+2 (mj)! kD 2 r@ uk L 2 3=4 =J 3 +J 4 : (3.5.38) For the termJ 3 , we use the change of variablei 0 =i 1 to get J 3 C 1 X m=1 m X j=[m=2]+1 X jj=m X jj=j; X !;j!j1 j!j X i 0 =1 j!j X s=1 X Ps(i 0 ;!) C i 0 (i 0 +1)=2 i 0 +1 3(i 0 +1)=2 i 0 k 0 1 ;:::;k 0 s j!j (j!j 3)! k@ ! pk L 2 ! (mj + 1) mj (mj 2)! kr@ uk L 2 1=4 (mj + 3) mj+2 (mj)! kD 2 r@ uk L 2 3=4 s Y l=1 j 0 l j+2 (j 0 l j 1)! kD 2 @ 0 l pk L 2 ! 3k 0 l =4 j 0 l j (j 0 l j 3)! k@ 0 l pk L 2 ! k 0 l =4 ; (3.5.39) 142 where the element ofP s (i 0 ;!) is denoted by (k 0 1 :::;k 0 s ; 0 1 ;::: 0 s ). Fixm;j;k2N 0 withkjm. Using (3.5.14) we get X jj=m X ;jj=j X !;j!j=k x ! y z ! = 0 @ X j j=k x 1 A 0 @ X j j=mj y 1 A 0 @ X j j=jk z 1 A : (3.5.40) From (3.5.39)–(3.5.40), we proceed as in (3.5.28)–(3.5.34) to get J 3 C 1 X m=1 m X j=[m=2]+1 j X k=1 X j j=jk e X j j=k k (k 3)! k@ pk L 2 0 @ X j j=mj (mj + 1) mj (mj 2)! kr@ uk L 2 1 A +Ckuk 2 A() kuk ~ A() Ckuk ~ A() kuk A() 1 X n=2 C 1=2 1 3=2 X 2N 3 0 ;jj1 b n +Ckuk 2 A() kuk ~ A() Ckuk ~ A() kuk A() ; (3.5.41) by choosing 2 3 =CM (T ) 2 , where e and b are as in (3.5.31) and (3.5.29). For the term J 4 , we proceed as in (3.5.10)–(3.5.15) to get J 4 Ckuk A() kuk ~ A() : (3.5.42) Combining (3.5.38)–(3.5.42), we conclude the proof of (3.5.21). The proof of Lemma 3.4.2 is thus completed by combining (3.5.19)–(3.5.21). 143 Chapter4 AregularityresultforthefreeboundarycompressibleEulerequations ofaliquid 4.1 Introduction In this paper, we study the vacuum free boundary problem for compressible uids inR 3 described by the compressible Euler equations (@ t u +uru) +rp = 0 in (t); (4.1.1) @ t +ur + divu = 0 in (t); (4.1.2) where (t) represents an open subset ofR 3 which is occupied by the uid at the timet, whose boundary is denoted by@ (t). In (4.1.1)–(4.1.2),u = u(t;x), = (t;x), andp = p(t;x) denote the velocity eld, density, and pressure, respectively. To close the system, we consider the equation of state p() =K 1 K 2 (4.1.3) 144 for polytropic uids, whereK 1 ;K 2 > 0 are constants and > 1 is the adiabatic constant (see [33]). The system (4.1.1)–(4.1.3) is supplemented with the initial, kinematic, and vacuum boundary conditions (0) = (4.1.4) (u(0;x);(0;x)) = (u 0 (x); 0 (x)); x2 (4.1.5) V((t)) =u(t;x)n(t); x2 (t) (4.1.6) p(t;x) = 0; x2 (t); (4.1.7) where (t) @ (t) is the moving free boundary, n(t) is the outward unit normal vector to (t), and V(@ (t)) is the normal velocity of (t). In order to model a compressible liquid, we assume that the initial density is strictly positive everywhere in the domain. This problem is known to be ill-posed (see [56]) unless the Rayleigh-Taylor sign condition @p @n > 0 on (t); (4.1.8) holds for some constant > 0. The condition (4.1.8) is a natural physical condition which indicates that the pressure, and thus also the density, is greater in the interior than on the boundary. In recent years, there have been a lot of eort made toward understanding the well-posedness of this free-boundary problem. For the incompressible Euler equations, the divergence of the velocity eld is zero and the density is a xed constant. Additionally, the pressure is not determined by the equation of state but rather a Lagrange multiplier enforcing the divergence free condition. For the existence theory, we refer the reader to [25, 31, 88, 124, 127, 128, 133, 142, 143, 144, 145] in various settings including irrotationality (i.e., curlu = 0) and [36, 39, 40, 47, 101, 136, 137] for rotational uids. 145 Concerning the compressible uids, we distinguish the cases between a liquid and a gas. The gas prob- lem refers to the equation of state (4.1.3) with the constantK 2 = 0, which combined with the boundary condition (4.1.7) leads to the vanishing of the density on the moving boundary. As a consequence, the system becomes degenerate along the vacuum boundary ([37, 80]) and the standard method of symmetriz- able hyperbolic equations cannot be applied. We refer to [41, 76, 79, 80, 115] and reference therein for the well-posedness results. In the case of a liquid, a commonly used equation of state is (4.1.3) with a positive constantK 2 > 0, which is the pressure law we treat in this paper. As opposed to the vanishing of the den- sity on the boundary in the case of a gas, the density remains strictly positive everywhere in the domain and thus the Euler equations are uniformly hyperbolic. The existence theory was established by Lind- blad in [100] using the Nash-Moser construction. Trakhinin provided a dierent proof for the local well- posedness in [140], using the theory of symmetric hyperbolic systems. Then, Coutand, Hole, and Shkoller in [34] proved the well-posedness using the vanishing viscosity method and the time-dierentiated a priori estimates. A recent work [116] by Luo and Zhang established the local well-posedness of the compressible gravity water wave problem using the Alinhac’s good unknowns. For other local well-posedness results in various settings, we refer to [51, 52, 53, 67, 99, 106]. In this paper, we derive a priori estimates for the compressible free boundary Euler equations without surface tension in the case of a liquid. Using a decomposition of the initial density (see (4.2.14)–(4.2.15)), we provide a new weighted functional framework for the liquid problem. An application of the Hardy inequality then leads to the improved regularity of the ow map and hence Jacobian, which in turn closes the energy estimates. Note that in [34], the higher-order regularity of the Jacobian is obtained using the wave equation of the density, which requires a high time derivative regularity of the initial data. Here we provide a direct method and a dierent functional framework with a low time-dierentiated regularity. Under the Rayleigh-Taylor sign condition, (4.2.12) below, the initial density grows as the distance func- tion in the inward normal direction. Also, the initial density is strictly bounded from below. In light of 146 this, we decompose the initial density into two parts (see (4.2.14)–(4.2.15)): one is degenerate along the boundary with the growth rate of the distance function and the other is uniformly bounded from below. A careful analysis of each part leads to weighted and homogeneous structure of the functional space. For the degenerate part, we deploy the Hardy inequality, Lemma 4.10.1, and obtain an improvedH 0:5 regularity of the ow map which in turn leads to the improved regularity of the Jacobian (see Section 4.6). The part of the initial density which is uniformly bounded from below produces a positive energy contribution on the boundary, allowing us to close the estimate for the improved regularity using the elliptic estimate. In the pure tangential energy estimates, the Jacobian and cofactor matrix enter as a highest order term which requires improved regularity of the Jacobian and ow map. To overcome this diculty, we uncover the improved regularity by using the Euler momentum equation and a div-curl elliptic estimate. Namely, we introduce a new fractionalH 1:5 Sobolev space so that the highest order term involving the Jacobian and the cofactor matrix can be controlled at the expense of two time derivatives and the above-mentioned improved regularity of the ow map. As a result, the scaling between space and time is necessarily one to two at the pure tangential level. For the energy estimates with at least two time derivatives, the top order term involving the Jacobian and cofactor matrix can be handled by using integration by parts in space and time. The fact thatDJ and@ t v are of the same regularity is essential to close the estimates. Consequently, the scaling between space and time becomes one to one when at least two time derivatives are involved. The normal derivative estimate of the solution is obtained by the elliptic regularity, which requires divergence, curl, time, and tangential components of the solution. Another major diculty lies in the energy estimate with pure time derivatives. The boundary integrals of the tangential component of the ow map cannot be controlled directly by the normal component or the trace lemma. Instead, we resort to the momentum equation restricted to the boundary (see (4.4.34)) and show that the normal component dominates the tangential component on the free boundary, at least for a short time interval due to the closeness of the cofactor matrix to its initial state. 147 The paper is organized as follows. In Section 4.2, we introduce the Lagrangian formulation of the free boundary problem and the assumptions on the domain and initial data. Also, we state the main result, Theorem 4.2.1. The a priori estimate needed for the boundedness of the solution, stated in Proposition 4.2.2, is proven in Section 4.8. In Sections 4.3–4.8, we restrict ourselves to the case = 2 in (4.1.3) for the equation of state, while in Section 4.9 we explain the modications needed for the general case > 1. In Section 4.3, we provide curl estimates which shall be needed when using the div-curl lemma. In Section 4.4, we derive energy estimates of the tangential, time, and divergence components of the solution, while in Section 4.5, we provide the normal derivative estimates of the solution using the elliptic regularity. In order to close the estimate, we establish the improved regularity of the Lagrangian ow map and the Jacobian in Section 4.6 and the improved regularity of the curl in Section 4.7. At last, in Appendix 4.10, we recall some auxiliary lemmas used in the proof. 4.2 Lagrangianformulationandthemainresult 4.2.1 Lagrangianformulation The ow map(t;): x7!(t;x)2 (t) associated with the uid velocity, dened as a solution to @ t (t;x) =u(t;(t;x)); t> 0; x2 ; (0;x) =x; x2 ; 148 denotes the location of a particle at timet that is initially placed at a Lagrangian labelx. We dene the following Lagrangian quantities: v(t;x) =u(t;(t;x)) (Lagrangian velocity); f(t;x) =(t;(t;x)) (Lagrangian density); A = [D] 1 (inverse of deformation tensor); J = detD (Jacobian determinant); a =JA (transpose of cofactor matrix): The notation F; k is used to represent @F=@x k , i.e., the partial derivative of F with respect to the La- grangian variable x k . We adopt the Einstein’s summation convention for repeated indices. The Latin indices i;j;k;l;m;r;s are summed from 1 to 3, while the Greek indices ; are summed from 1 to 2. Without loss of generality, we set the constantsK 1 = K 2 = 1 in (4.1.3). Namely, the equation of state reads p() = 1: (4.2.1) Using the Lagrangian quantities, the equations (4.1.1)–(4.1.7) can be rewritten in a xed domain as f t +fA j i v i ; j = 0 in [0;T ] ; (4.2.2) fv i t +A k i f ; k = 0 in [0;T ] ; (4.2.3) f = 1 on [0;T ] ; (4.2.4) (;v;f) = (Id;u 0 ; 0 ) inft = 0g ; (4.2.5) 149 where = (0) denote the initial vacuum free boundary. SinceJ t = JA j i v i ; j andJ(0) = 1, it follows that f = 0 J 1 ; (4.2.6) which indicates that the initial density can be viewed as a parameter in the Euler equations. Thus, using (4.2.6), the system (4.2.2)–(4.2.5) becomes 0 v i t +a k i ( 0 J ); k = 0 in [0;T ] ; (4.2.7) 0 J 1 = 1 on [0;T ] ; (4.2.8) (;v) = (Id;u 0 ) inft = 0g : (4.2.9) Since 0 = 1 on [0;T ] , it follows that J = 1 on [0;T ] : From (4.2.7) we write equivalently v i t + ( 0 J 1 ) 2 A k i ( 0 J 1 ); k = 0 in [0;T ] (4.2.10) and v i t ( 0 J 1 ) 2 0 J 3 a k i J; k + ( 0 J 1 ) 2 0 ; k a k i J 2 = 0 in [0;T ] : (4.2.11) 150 The three equivalent equations (4.2.7) and (4.2.10)–(4.2.11) are being used for dierent purposes: (4.2.7) is used for the energy estimates, (4.2.10) is used for estimates of the vorticity, while (4.2.11) is used for the Jacobian and the normal derivative estimates. 4.2.2 Thereferencedomain To avoid the use of local coordinates charts, we assume the initial domain R 3 at timet = 0 is given by =f(x 1 ;x 2 ;x 3 ) : (x 1 ;x 2 )2T 2 ;x 3 2 (0; 1)g; whereT 2 denotes the 2-dimensional torus with length 1. At timet = 0, the reference vacuum boundary is the top boundary =fx 3 = 1g, whileN = (0; 0; 1) denotes the outward unit normal vector to . The reference bottom boundaryfx 3 = 0g is xed with the boundary condition u 3 = 0 on [0;T ]fx 3 = 0g. The moving vacuum free boundary is given by (t) =f(t;x 1 ;x 2 ; 1) : (x 1 ;x 2 )2T 2 g. 4.2.3 Notations Given a vector eldF over , we useDF , divF , and curlF to denote its full gradient, divergence, and curl. Namely, we denote by [DF ] i j =F i ; j divF =F r ; r [curlF ] i = ijk F k ; j ; 151 where ijk represents the Levi-Civita symbol. We denote the Lie derivatives along the trajectory map by [D F ] i r =A s r F i ; s div F =A s r F r ; s [curl F ] i = ijk A s j F k ; s ; which correspond to Eulerian full gradient, divergence, and curl represented in Lagrangian coordinates. The symbol @ is used for tangential derivative @ = (@ 1 ;@ 2 ), whileD = (@ 1 ;@ 2 ;@ 3 ) stands for full spatial derivative. For integersk 0, we denote byH k ( ) the standard Sobolev spaces of orderk with corresponding normskk k . For real number s 0, the Sobolev spaces H s ( ) and the normskk s are dened by interpolation. TheL p norms on are denoted bykk L p ( ) or simplykk L p when no confusion can arise. The negative-order Sobolev spacesH s ( ) are dened via duality, namely, H s ( ) := (H s ( )) 0 : Note that in our conguration the boundary@ has two parts: the reference free boundary =fx 3 = 1g and the xed bottom boundaryfx 3 = 0g. We shall work on the Sobolev spaces on the reference free boundary and use the notationsH s () andjj s for the Sobolev spaces and norms, withs 0. The negative-order Sobolev spaces associated to the boundary are dened analogously as the domain , while theL p norms on are denoted bykk L p () . 152 4.2.4 Theinitialdensity Under (4.1.8), we assume that the initial pressure satises the physical sign condition @p 0 @N < 0 on : From (4.2.1) the above condition is equivalent to 1 0 @ 0 @N < 0 on : (4.2.12) Since > 1 and 0 = 1 on , the condition (4.2.12) implies that 0 1 grows like the distance function to the reference boundary . Throughout this paper, we assume that 0 (x) = 1 +w(x); x2 ; (4.2.13) wherew(x) = 1x 3 is the distance function to the reference free boundary. It is clear that the Rayleigh- Taylor sign condition (4.2.12) is satised for any > 1. We introduce the decomposition 0 (x) = d (x) + u (x), where d (x) = ((1 +w(x)) (1 + w(x) 2 )) 1 ; x2 (4.2.14) and u (x) = (1 + w(x) 2 ) 1 ; x2 : (4.2.15) 153 It is readily checked that d (x): ! [0;1) is well-dened and d = 0 on . Moreover, we have that u 1 for x2 . The d part of the density is degenerate along the boundary, while the u part is uniformly bounded from below. 4.2.5 TheJacobianandcofactormatrix We recall the following dierentiation identities: @ t J =a s r v r ; s ; (4.2.16) @ t a k i =J 1 (a s r a k i a s i a k r )v r ; s : (4.2.17) Using (4.2.16)–(4.2.17) and the identitya =JA, we obtain @ t A k i =A s i A k r v r ; s : (4.2.18) Note that the identities (4.2.16)–(4.2.18) become spatial-dierentiation formulas by replacing v r ; s with D r ; s on the right hand side, whereD =@ l withl = 1; 2; 3. Namely, we have DJ =a s r D r ; s ; (4.2.19) Da k i =J 1 (a s r a k i a s i a k r )D r ; s ; (4.2.20) DA k i =A s i A k r D r ; s : (4.2.21) The cofactor matrix satises the Piola identity which reads a k i ; k = 0; (4.2.22) 154 fori = 1; 2; 3. By the denition of the cofactor matrix, we have that a 3 1 a 3 2 a 3 3 = 2 ; 1 3 ; 2 3 ; 1 2 ; 2 3 ; 1 1 ; 2 1 ; 1 3 ; 2 1 ; 1 2 ; 2 1 ; 2 2 ; 1 : (4.2.23) Note that only tangential derivatives appear in each entry on the right hand side, which shall play an essential role in our analysis. 4.2.6 Mainresult The system (4.2.7)–(4.2.9) admits a conserved physical energy E(t) = Z 1 2 0 jv(t)j 2 + 1 1 0 J 1 (t) +J(t) dx: A direct computation shows that formally d dt E(t) = 0, assuming sucient dierentiability of the solution. However, it is too weak to control the regularity of the evolving free boundary. Instead, we introduce the higher order energy functional E(t) =kvk 2 4 +k@ t vk 2 3 +k@ t Jk 2 4 +k @ 3 Jk 2 1:5 +k @ 3 k 2 1:5 +k @ 3 curl vk 2 0:5 + 4 X l=2 k @ 4l @ l t k 2 1:5 + 5 X l=2 (k 2 d @ 5l @ l t Dk 2 0 +k@ l t vk 2 5l +k@ l t Jk 2 5l +j@ l t j 2 5l ) + 1: (4.2.24) Note that the time derivatives att = 0 are dened iteratively by dierentiating (4.2.10) and evaluating at t = 0. The following theorem is our main result. 155 Theorem 4.2.1. Let > 1. Suppose that ((t);v(t)) is a smooth solution of (4.2.7)–(4.2.9) on some time interval [0;T 0 ] and the initial data satisesE(0) <1. Then there exists a suciently small constantT2 (0;T 0 ) such that(t) 1=2 in (t) fort2 [0;T ] and sup t2[0;T ] E(t)M; whereM > 0 is a constant depending on the initial data. In this paper we restrict ourselves to derive a priori estimates and thus a smooth solution is assumed to be given. As a result, there is no need to state the compatibility conditions for the initial data. Nevertheless, we refer the reader to [34] where the solution is constructed using the tangentially smoothed approximate system. We denote byM 0 =P (E(0)) some constant depending on the initial data, which may vary from line to line. Throughout this paper, the notationA.B means thatACB for some constantC > 0 which depends onM 0 and . The notationc l;m 2R stands for some constant that depends onl;m2N, which may vary from line to line. In fact, the true value ofc l;m is not essential in our analysis. The a priori estimate needed to prove Theorem 4.2.1 is the following. Proposition4.2.2. Let2 (0; 1). There exists a constantT > 0 and a nonnegative continuous functionP such that E(t). 1 + sup t2[0;T ] E(t) +C tP ( sup t2[0;T ] E(t)); (4.2.25) forallt2 [0;T ],whereC > 0isaconstantthatdependson. Weemphasizethattheimplicitconstantused in (4.2.25) is independent of. 156 For the proof of Theorem 4.2.1 given Proposition 4.2.2, we take > 0 suciently small and use the Gronwall inequality and the standard continuity argument. Throughout Sections 4.3–4.8, we set = 2 in (4.2.7), (4.2.10)–(4.2.11), (4.2.14)–(4.2.15), and (4.2.24). We shall prove that the a priori bound (4.2.25) in Proposition 4.2.2, thus completing the proof of Theorem 4.2.1. In proving Proposition 4.2.2, we assume the following: (i) 1 2 J 3 2 , (ii)kk 4 C +Tk@ t k 4 C, (iii) kJk 4 C +Tk@ t Jk 4 C, (iv)k@ t Jk 3 C +Tk@ 2 t Jk 3 C, (v)k@ 2 t vk 2 C +Tk@ 3 t vk 2 C, (vi)kAI 3 k H 2 +kaI 3 k H 2CT , on [0;T ] for some constantC > 0, whereI 3 is the three-dimensional identity matrix. The above assumptions can be justied by using the fundamental theorem of calculus and takingT > 0 suciently small, after we establish the a priori bounds. 4.3 Curlestimates The following lemma provides the curl estimates of the solutions andv. Lemma4.3.1. We have sup t2[0;T ] 5 X l=2 k d @ 5l @ l t curl(t)k 2 0 + sup t2[0;T ] 4 X l=2 k@ l t D 4l curlv(t)k 2 0 +k @ 4l @ l t curl(t)k 2 0:5 + sup t2[0;T ] (k @ 3 curl(t)k 2 0:5 +kD 3 curlv(t)k 2 0 ). 1 +TP ( sup t2[0;T ] E(t)): (4.3.1) 157 Proof of Lemma 4.3.1. We start with the estimates ofk@ l t D 4l curlvk 2 0 forl = 2; 3; 4. Applying the La- grangian curl to (4.2.10), we get curl v t = 0: (4.3.2) From (4.2.18) and (4.3.2) it follows that [curl v] k t = kji (A s j ) t v i ; s = kji A m j A s r v r ; m v i ; s =:Q 0 (A;Dv); (4.3.3) where Q 0 is a quadratic function of A and Dv. In (4.3.3) and below, we use [F ] k to denote the k-th component of a vector eldF . Using the fundamental theorem of calculus, we obtain [curl v(t)] k = [curlu 0 ] k + Z t 0 Q 0 (A(t 0 );Dv(t 0 ))dt 0 : (4.3.4) Fixl2f2; 3; 4g. Applying@ l t D 4l to the above equation, we obtain [curl @ l t D 4l v(t)] k = 4l X m=0 l X n=0 m+n1 c m;n kji @ n t D m A s j @ ln t D 4lm v i ; s +@ l1 t D 4l Q 0 (A;Dv): Inserting the identity [curl @ l t D 4l v] k = [@ l t D 4l curlv] k + kji @ l t D 4l v i ; s R t 0 (A s j ) t dt 0 to the above equa- tion, we obtain [@ l t D 4l curlv] k = jki @ l t D 4l v i ; s Z t 0 (A s j ) t dt 0 + 4l X m=0 l X n=0 m+n1 c m;n kji @ n t D m A s j @ ln t D 4lm v i ; s +@ l1 t D 4l Q 0 (A;Dv); 158 from where k@ l t D 4l curlvk 2 0 .k@ l t D 5l v(AI 3 )k 2 0 | {z } K 1 +k@ l1 t D 4l Q 0 (A;Dv)k 2 0 | {z } K 2 +k@ l t D 4l ADvk 2 0 | {z } K 3 + 4l X m=0 l X n=0 3m+n1 k@ n t D m A@ ln t D 4lm Dvk 2 0 | {z } K 4 : The termK 1 is estimated using the Hölder and Sobolev inequalities as K 1 .k@ l t D 5l vk 2 0 kAI 3 k 2 L 1.TP ( sup t2[0;T ] E(t)); sincek@ l t vk 2 5l E(t) forl = 2; 3; 4 andkAI 3 k 2 .T . The highest order term (in terms of derivative counts on) inK 2 can be written as kji (A m j A s r v r ; m @ l1 t D 4l v i ; s +A m j A s r v i ; s @ l1 t D 4l v r ; m ): Using the fundamental theorem of calculus, we obtain k@ l1 t D 5l vk 2 0 .k@ l1 t D 5l v(0)k 2 0 + Z T 0 k@ l t D 5l vk 2 0 . 1 +TP ( sup t2[0;T ] E(t)); where we used the Jensen’s inequality. From the Hölder and Sobolev inequalities it follows that kAADv@ l1 t D 5l vk 2 0 . 1 +TP ( sup t2[0;T ] E(t)): 159 where we drop the indices for simplicity. The rest of the terms inK 2 are of lower order which can be treated in a similar fashion using the Hölder and Sobolev inequalities and the fundamental theorem of calculus. Thus, we have K 2 . 1 +TP ( sup t2[0;T ] E(t)): Similarly, the termK 3 is estimated as K 3 .k@ l t D 4l Ak 2 0 kDvk 2 L 1. 1 +TP ( sup t2[0;T ] E(t)): The termK 4 consists of essentially lower order terms which can be estimated analogously using the Hölder and Sobolev inequalities and the fundamental theorem of calculus, and we obtain K 4 . 1 +TP ( sup t2[0;T ] E(t)): Consequently, we conclude 4 X l=2 k@ l t D 4l curlv(t)k 2 0 . 1 +TP ( sup t2[0;T ] E(t)): (4.3.5) The termkD 3 curlvk 2 0 is treated in a similar fashion using the above arguments, and we arrive at kD 3 curlv(t)k 2 0 . 1 +TP ( sup t2[0;T ] E(t)): (4.3.6) Now, we derive the estimates ofk d @ 5l @ l t curlk 2 0 , wherel = 2; 3; 4; 5. We rewrite (4.3.4) as [curl@ t (t)] k = [curlu 0 ] k + jki v i ; s Z t 0 (A s j ) t dt 0 + Z t 0 Q 0 (A;Dv)dt 0 : (4.3.7) 160 Fixl2f2; 3; 4; 5g. Applying d @ 5l @ l1 t to the above equation, we arrive at k d @ 5l @ l t curlk 2 0 .k d @ 5l @ l2 t (DvA t )k 2 0 | {z } J 1 +k d @ 5l @ l2 t (@ t Dv(AI 3 ))k 2 0 | {z } J 2 +k d @ 5l @ l2 t Q 0 (A;Dv)k 2 0 | {z } J 3 : The highest order term inJ 1 scales like d @ t A @ 5l @ l1 t D which can be estimated using the fundamental theorem of calculus as k d @ t A @ 5l @ l1 t Dk 2 0 .k d @ 5l @ l1 t D(0)k 2 0 + Z T 0 k d @ 5l @ l t Dk 2 0 . 1 +TP ( sup t2[0;T ] E(t)); sincek d @ 5l @ l t Dk 2 0 E(t), forl = 2; 3; 4; 5. The rest of the terms inJ 1 are of lower order which can be treated in a similar fashion using the Hölder and Sobolev inequalities and the fundamental theorem of calculus. Thus, we have J 1 . 1 +TP ( sup t2[0;T ] E(t)): The highest order term inJ 2 is estimated using the Hölder and Sobolev inequalities as k d @ 5l @ l t D(AI 3 )k 2 0 .k d @ 5l @ l t Dk 2 0 kAI 3 k 2 L 1.TP ( sup t2[0;T ] E(t)); sincekAI 3 k 2 E(t). The rest of the terms inJ 2 , as well as the termJ 3 , are treated analogously using the above arguments. As a result, we get J 2 +J 3 . 1 +TP ( sup t2[0;T ] E(t)): 161 Combining the above estimates, we conclude k d @ 5l @ l t curl(t)k 2 0 . 1 +TP ( sup t2[0;T ] E(t)): (4.3.8) Next we derive the estimates ofk @ 4l @ l t curlk 2 0:5 forl = 2; 3; 4. Fixl2f2; 3; 4g. Applying @ 4l @ l1 t to (4.3.7), we obtain k @ 4l @ l t curlk 2 0:5 .k @ 4l @ l2 t (DvA t )k 2 0:5 | {z } I 1 +k @ 4l @ l2 t (@ t Dv(AI 3 )k 2 0:5 | {z } I 2 +k @ 4l @ l2 t Q 0 (A;Dv)k 2 0:5 | {z } I 3 : For the termI 1 , from the Leibniz rule it follows that I 1 .k @ 4l @ l2 t DvA t k 2 0:5 +kDv @ 4l @ l2 t A t k 2 0:5 + 4l X m=0 l2 X n=0 m+n=1 k @ m @ n t Dv @ 4lm @ l2n t A t k 2 0:5 =:I 11 +I 12 +I 13 : (4.3.9) Recall the multiplicative Sobolev inequality kfgk r .kfk 1:5+ kgk r ; (4.3.10) for 0r 1:5 and> 0. From (4.3.10) it follows that I 11 .k @ 4l @ l1 t Dk 2 0:5 kA t k 2 1:5+ . 1 +TP ( sup t2[0;T ] E(t)) 162 where we used the fundamental theorem of calculus andk @ 4l @ l t k 2 1:5 +k@ 2 t vk 2 3 E(t), forl = 2; 3; 4. Similarly, we have I 12 .k @ 4l @ l1 t Ak 2 0:5 kDvk 2 1:5+ . 1 +TP ( sup t2[0;T ] E(t)): In addition to (4.3.10), we recall the multiplicative Sobolev inequality kfgk 0:5 .kfk 1 kgk 1 : (4.3.11) For the termI 13 , using (4.3.11), we get I 13 . 4l X m=0 l2 X n=0 m+n=1 k @ m @ n t Dvk 2 1 k @ 4lm @ l1n t Ak 2 1 . 1 +TP ( sup t2[0;T ] E(t)): where we used the fundamental theorem of calculus. For the termI 2 , from the Leibniz rule it follows that I 2 .k @ 4l @ l t D(AI 3 )k 2 0:5 +k@ t Dv @ 4l @ l2 t (AI 3 )k 2 0:5 + 4l X m=0 l2 X n=0 m+n=1 k @ m @ n+1 t Dv @ 4lm @ l2n t (AI 3 )k 2 0:5 =:I 21 +I 22 +I 23 : We bound the termI 21 using (4.3.10) as I 21 .k @ 4l @ l t Dk 2 0:5 kAI 3 k 2 1:5+ .TP ( sup t2[0;T ] E(t)); (4.3.12) sincekAI 3 k. T . The termsI 22 ,I 23 , andI 3 are estimated analogously using the above arguments. Thus, we have I 22 +I 23 +I 3 . 1 +TP ( sup t2[0;T ] E(t)): 163 Consequently, we conclude 4 X l=2 k @ 4l @ l t curlk 2 0:5 . 1 +TP ( sup t2[0;T ] E(t)): (4.3.13) Finally, we derive the estimate ofk @ 3 curlk 2 0:5 . An application of the fundamental theorem of calculus to (4.3.7) leads to [curl(t)] k = [t curlu 0 ] k + Z t 0 jki v i ; s Z t 0 0 (A s j ) t dt 00 dt 0 + Z t 0 Z t 0 0 Q 0 (A;Dv)dt 00 dt 0 : Applying @ 3 to the above equation, we arrive at k @ 3 curlk 2 0:5 .Tk @ 3 curlu 0 k 2 0:5 | {z } L 1 +k @ 3 Z t 0 jki v i ; s (A(t 0 )I 3 )dt 0 k 2 0:5 | {z } L 2 +k @ 3 Z t 0 Z t 0 0 Q 0 (A;Dv)dt 00 dt 0 k 2 0:5 | {z } L 3 : The termL 1 is bounded byTM 0 sincek @ 3 curlu 0 k 0:5 =k @ 3 curl v(0)k 0:5 . For the termL 2 , we use the Leibniz rule, obtaining L 2 .k Z t 0 @ 3 Dv(AI 3 )dt 0 k 2 0:5 +k Z t 0 Dv @ 3 Adt 0 k 2 0:5 + 2 X l=1 k Z t 0 @ l Dv @ 3l Adt 0 k 2 0:5 =:L 21 +L 22 +L 23 : For the termL 21 , we integrate by parts in time, leading to L 21 .k Z t 0 @ 3 D@ t Adt 0 k 2 0:5 +k @ 3 D(AI 3 ) t 0 =t t 0 =0 k 2 0:5 =:L 211 +L 212 : (4.3.14) 164 For the termL 211 , we appeal to (4.3.10), obtaining L 211 . Z T 0 k @ 3 Dk 2 0:5 k@ t Ak 2 1:5+ dt 0 .TP ( sup t2[0;T ] E(t)): (4.3.15) The termL 212 is treated in a similar fashion as in (4.3.12), and we arrive at L 212 .TP ( sup t2[0;T ] E(t)): Similarly, the termL 22 is estimated as L 22 . Z t 0 kDv @ 3 Ak 2 0:5 dt 0 . Z t 0 kDvk 2 2 k @ 3 Ak 2 0:5 dt 0 .TP ( sup t2[0;T ] E(t)): (4.3.16) The termsL 23 andL 3 are treated using similar arguments as in (4.3.14)–(4.3.16). Therefore, we conclude the estimate k @ 3 curlk 2 0:5 . 1 +TP ( sup t2[0;T ] E(t)): (4.3.17) Collecting the estimates in (4.3.5)–(4.3.6), (4.3.8), (4.3.13), and (4.3.17), we complete the proof of the lemma. 4.4 Energyestimates The following lemma provides @ 4 energy estimates of the solutionsv,J, and. 165 Lemma4.4.1. We have sup t2[0;T ] Z j @ 4 J(t)j 2 +j @ 4 v(t)j 2 + sup t2[0;T ] Z j @ 4 3 (t)j 2 . 1 +TP ( sup t2[0;T ] E(t)): (4.4.1) Proof of Lemma 4.4.1. Applying @ 4 to (4.2.7) and taking the inner product with @ 4 v, we obtain 1 2 d dt Z 0 j @ 4 vj 2 + Z @ 4 a k i ( 2 0 J 2 ); k @ 4 v i | {z } J 1 + Z a k i ( 2 0 @ 4 J 2 ); k @ 4 v i | {z } J 2 = 3 X l=1 c l Z @ 4l a k i ( 2 0 @ l J 2 ); k @ 4 v i | {z } J 3 : (4.4.2) Estimate ofJ 1 in (4.4.2): From (4.2.20) and the Leibniz rule it follows that J 1 = Z J 1 a s i a k r @ 4 r ; s ( 2 0 J 2 ); k @ 4 v i | {z } J 11 + Z J 1 a s r a k i @ 4 r ; s ( 2 0 J 2 ); k @ 4 v i | {z } J 12 + 3 X l=1 c l Z @ l (J 1 (a s r a k i a s i a k r )) @ 4l r ; s ( 2 0 J 2 ); k @ 4 v i | {z } J 13 : For the termJ 11 , we integrate by parts in@ s , obtaining Z T 0 J 11 = Z T 0 Z fx 3 =0g J 1 a 3 i a k r @ 4 r ( 2 0 J 2 ); k @ 4 v i | {z } =0 Z T 0 Z J 1 a 3 i a 3 r @ 4 r ( 2 0 J 2 ); 3 @ 4 v i | {z } J 111 + Z T 0 Z J 1 a s i a k r @ 4 r ( 2 0 J 2 ); k @ 4 v i ; s | {z } J 112 + Z T 0 Z (J 1 a s i a k r ( 2 0 J 2 ); k ); s @ 4 r @ 4 v i | {z } J 113 ; (4.4.3) 166 where we note thata k r ( 2 0 J 2 ); k = a 3 r ( 2 0 J 2 ); 3 since 2 0 J 2 = 1 on [0;T ]. In (4.4.3), we have also used that 3 = 0 onfx 3 = 0g [0;T ], which leads toa 3 i @ 4 v i = 0 onfx 3 = 0g [0;T ]. We rewrite the termJ 111 as J 111 = 1 2 Z T 0 d dt Z J 1 ( 2 0 J 2 ); 3 j @ 4 r a 3 r j 2 | {z } J 1111 + 1 2 Z T 0 Z (J 1 ( 2 0 J 2 ); 3 ) t j @ 4 r a 3 r j 2 | {z } J 1112 + Z T 0 Z J 1 ( 2 0 J 2 ); 3 @ 4 r @ t a 3 r @ 4 i a 3 i | {z } J 1113 : Note that from (4.2.13) we may assume that J 1 ( 2 0 J 2 ); 3 =J 1 ( 2 0 ; 3 J 2 2J 3 J; 3 2 0 ) 1 2 ; (4.4.4) on [0;T ], which can be justied by using the fundamental theorem of calculus and taking T > 0 suciently small after we establish the a priori bounds. From (4.4.4) it follows that J 1111 = 1 2 Z J 1 ( 2 0 J 2 ); 3 j @ 4 r a 3 r j 2 (T ) +M 0 1 4 Z j @ 4 r a 3 r j 2 (T ) +M 0 : (4.4.5) For the termJ 1112 , using the Hölder and Sobolev inequalities and Lemma 4.10.3, we get J 1112 . Z T 0 (kDJ t k L 1 + 1)j @ 4 j 2 0 . Z T 0 (kJ t k H 3 + 1)k @ 3 k 2 1:5 .TP ( sup t2[0;T ] E(t)); sincek@ 3 t k 2 1:5 E(t). Similarly, the termJ 1113 is estimated as J 1113 .TP ( sup t2[0;T ] E(t)): 167 For the termJ 112 , we integrate by parts in time, obtaining J 112 = Z T 0 Z J 1 a s i a k r @ 4 @ t r ( 2 0 J 2 ); k @ 4 i ; s | {z } J 1121 + Z J 1 a s i a k r @ 4 r ( 2 0 J 2 ); k @ 4 i ; s T 0 | {z } J 1122 Z T 0 Z (J 1 a s i a k r ( 2 0 J 2 ); k ) t @ 4 r @ 4 i ; s | {z } J 1123 : It is clear thatJ 1121 + R T 0 J 12 = 0. Note that @ 4 J =a s i @ 4 i ; s + P 3 l=1 c l @ l a s i @ 4l i ; s . For the pointwise in time termJ 1122 , using the Hölder and Sobolev inequalities and the fundamental theorem of calculus, we arrive at J 1122 .k @ 4 J(T )k 2 0 +k @ 4 (T )k 2 0 + 3 X l=1 k @ l a(T ) @ 4l D(T )k 2 0 + 1. 1 +TP ( sup t2[0;T ] E(t)); sincek@ t Jk 2 4 +kvk 2 4 E(t). For the termJ 1123 , we appeal to Lemma 4.10.5, obtaining J 1123 . Z T 0 (1 +kDJ t k L 1)k @ 4 k 0:5 k @ 4 Dk 0:5 . Z T 0 (1 +kJ t k 3 )k @ 3 k 2 1:5 .TP ( sup t2[0;T ] E(t)); where we used the Hölder and Sobolev inequalities. The termJ 113 is estimated using the Hölder and Sobolev inequalities as J 113 . Z T 0 (1 +kD 2 Jk L 1)k @ 4 k 0 k @ 4 vk 0 .TP ( sup t2[0;T ] E(t)): The termJ 13 consists of essentially lower-order terms which can be estimated using the Hölder and Sobolev inequalities and the fundamental theorem of calculus as Z T 0 J 13 .TP ( sup t2[0;T ] E(t)): 168 Collecting (4.4.5) and the above estimates, we conclude that, after integrating in time of (4.4.2), 1 2 Z 0 j @ 4 v(T )j 2 + 1 4 Z j @ 4 r a 3 r j 2 (T ) + Z T 0 J 2 . 1 +TP ( sup t2[0;T ] E(t)) + Z T 0 J 3 : (4.4.6) Usingj @ 4 r a 3 r jj @ 4 3 a 3 3 jj @ 4 a 3 j andkaI 3 k 2 .T on [0;T ] , we obtain Z j @ 4 3 a 3 3 j 2 (T )CT Z j @ 4 (T )j 2 + Z j @ 4 r a 3 r j 2 (T )CTk @ 3 (T )k 2 1:5 + Z j @ 4 r a 3 r j 2 (T ); (4.4.7) where we used the Hölder and Sobolev inequalities and Lemma 4.10.3. Combining (4.4.6)–(4.4.7), we arrive at 1 2 Z 0 j @ 4 v(T )j 2 + 1 8 Z j @ 4 3 (T )j 2 + Z T 0 J 2 . 1 +TP ( sup t2[0;T ] E(t)) + Z T 0 J 3 : (4.4.8) Estimate ofJ 2 in (4.4.2): We integrate by parts in@ k and use the Piola identity (4.2.22) to get J 2 = Z a 3 i 2 0 @ 4 J 2 @ 4 v i Z fx 3 =0g a 3 i 2 0 @ 4 J 2 @ 4 v i Z a k i 2 0 @ 4 J 2 @ 4 v i ; k = Z a k i 2 0 @ 4 J 2 @ 4 v i ; k ; (4.4.9) since @ 4 J 2 = 0 on [0;T ] anda 3 i @ 4 v i = 0 onfx 3 = 0g [0;T ]. Note that from (4.2.16) and the Leibniz rule it follows @ 4 @ t J =a k i @ 4 v i ; k + @ 4 a k i v i ; k + 3 X l=1 c l @ l a k i @ 4l v i ; k : (4.4.10) 169 Inserting (4.4.10) to (4.4.9), we arrive at J 2 = 2 Z 2 0 a k i J 3 @ 4 J @ 4 v i ; k + 3 X l=1 c l Z 2 0 a k i @ l J 3 @ 4l J @ 4 v i ; k | {z } J 24 = 2 Z 2 0 J 3 @ 4 J @ 4 @ t J | {z } J 21 Z 2 0 J 3 @ 4 J @ 4 a k i v i ; k | {z } J 22 + 3 X l=1 c l Z 2 0 J 3 @ 4 J @ l a k i @ 4l v i ; k | {z } J 23 +J 24 : The termJ 21 can be rewritten as J 21 = d dt Z 2 0 J 3 j @ 4 Jj 2 Z ( 2 0 J 3 ) t j @ 4 Jj 2 | {z } J 0 21 : (4.4.11) Using the Hölder and Sobolev inequalities, we obtain Z T 0 J 0 21 .TP ( sup t2[0;T ] E(t)): The highest order term inJ 22 is of the form R 2 0 J 3 @ 4 J @ 4 DDv, which can be treated using the Hölder and Sobolev inequalities and Lemma 4.10.5 as Z T 0 Z 2 0 J 3 @ 4 J @ 4 DDv. Z T 0 k @ 4 Jk 0:5 k @ 4 Dk 0:5 . Z T 0 k @ 3 Jk 1:5 k @ 3 k 1:5 .TP ( sup t2[0;T ] E(t)); sincek @ 3 Jk 2 1:5 +k @ 3 k 2 1:5 E(t). The rest of the terms inJ 22 are of lower order which can be estimated using the Hölder and Sobolev inequalities and the fundamental theorem of calculus. Therefore, we obtain Z T 0 J 22 . 1 +TP ( sup t2[0;T ] E(t)): 170 The termJ 23 consists of lower-order terms which can be estimated using the Hölder and Sobolev inequal- ities and the fundamental theorem of calculus as Z T 0 J 23 .TP ( sup t2[0;T ] E(t)); For the termJ 24 , we integrate by parts in@ k and use the Piola identity (4.2.22), obtaining J 24 = 3 X l=1 c l Z 2 0 a 3 i @ l J 3 @ 4l J @ 4 v i + 3 X l=1 c l Z fx 3 =0g 2 0 a 3 i @ l J 3 @ 4l J @ 4 v i + 3 X l=1 c l Z a k i ( 2 0 @ l J 3 @ 4l J); k @ 4 v i = 3 X l=1 c l Z a k i ( 2 0 @ l J 3 @ 4l J); k @ 4 v i ; (4.4.12) since @ l J 3 = 0 on [0;T ] forl2N anda 3 i @ 4 v i = 0 onfx 3 = 0g [0;T ]. Thus, using the Hölder and Sobolev inequalities, the termJ 24 is estimated as Z T 0 J 24 .TP ( sup t2[0;T ] E(t)): (4.4.13) Combining (4.4.8), (4.4.11), and the above estimates, we conclude 1 2 Z 0 j @ 4 v(T )j 2 + 2 0 J 3 (T )j @ 4 J(T )j 2 + 1 8 Z j @ 4 3 (T )j 2 . 1 +TP ( sup t2[0;T ] E(t)) + Z T 0 J 3 : (4.4.14) Estimate ofJ 3 in (4.4.2): We proceed analogously as in (4.4.12)–(4.4.13), obtaining Z T 0 J 3 .TP ( sup t2[0;T ] E(t)): (4.4.15) 171 Concluding the proof : Combining (4.4.14)–(4.4.15), we arrive at 1 2 Z 0 j @ 4 v(T )j 2 + 2 0 J 3 j @ 4 J(T )j 2 + 1 8 Z j @ 4 3 (T )j 2 . 1 +TP ( sup t2[0;T ] E(t)): The proof of the lemma is thus completed. The following lemma provides @ 3 @ 2 t energy estimates of the solutionsv,J, and. Lemma4.4.2. For2 (0; 1), we have sup t2[0;T ] Z j d D @ 3 @ 2 t (t)j 2 +j @ 3 @ 2 t v(t)j 2 +j @ 3 @ 2 t J(t)j 2 + sup t2[0;T ] Z j @ 3 @ 2 t (t)j 2 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); (4.4.16) whereC > 0 is a constant depending on. Proof of Lemma 4.4.2. Applying @ 3 @ 2 t to (4.2.7) and taking the inner product with @ 3 @ 2 t v i , we arrive at 1 2 d dt Z 0 j @ 3 @ 2 t vj 2 + Z @ 3 @ 2 t a k i ( 2 0 J 2 ); k @ 3 @ 2 t v i | {z } J 1 + Z a k i ( 2 0 @ 3 @ 2 t J 2 ); k @ 3 @ 2 t v i | {z } J 2 = 3 X l=0 2 X m=0 1m+l4 c l;m Z @ 3l @ 2m t a k i ( 2 0 @ l @ m t J 2 ); k @ 3 @ 2 t v i | {z } J 3 : (4.4.17) Estimate ofJ 1 in (4.4.17): Using the decomposition 2 0 = 2 d + 2 u (see (4.2.14)–(4.2.15)), we split the termJ 1 as J 1 = Z @ 3 @ 2 t a k i ( 2 d J 2 ); k @ 3 @ 2 t v i | {z } G + Z @ 3 @ 2 t a k i ( 2 u J 2 ); k @ 3 @ 2 t v i | {z } L : (4.4.18) 172 First we estimate the termG in (4.4.18). Using integration by parts in@ k , we arrive at G = Z @ 3 @ 2 t a 3 i 2 d J 2 @ 3 @ 2 t v i Z fx 3 =0g @ 3 @ 2 t a 3 i 2 d J 2 @ 3 @ 2 t v i Z @ 3 @ 2 t a k i 2 d J 2 @ 3 @ 2 t v i ; k ; (4.4.19) where we used the Piola identity (4.2.22). The rst term on the right hand side vanishes since d = 0 on [0;T ]. The second term on the right hand side of (4.4.19) also vanishes since @ 3 @ 2 t a 3 i @ 3 @ 2 t v i = 0 on fx 3 = 0g [0;T ]. Therefore, from (4.2.17) and the Leibniz rule it follows that G = Z A s i A k r @ 3 @ 2 t r ; s 2 d J 1 @ 3 @ 2 t v i ; k | {z } G 1 Z A s r A k i @ 3 @ 2 t r ; s 2 d J 1 @ 3 @ 2 t v i ; k | {z } G 2 + 3 X l=0 1 X m=0 m+l1 c l;m Z @ l @ m t (J 1 (a s r a k i a s i a k r )) @ 3l @ 2m t r ; s 2 d J 2 @ 3 @ 2 t v i ; k | {z } G 3 : (4.4.20) It is clear that A s i A k r @ 3 @ 2 t r ; s @ 3 @ 2 t v i ; k =A s r @ 3 @ 2 t i ; s A k r @ 3 @ 2 t @ t i ; k + (A s i @ 3 @ 2 t r ; s A s r @ 3 @ 2 t i ; s )A k r @ 3 @ 2 t @ t i ; k : (4.4.21) The rst term on the right hand side can be rewritten as A s r @ 3 @ 2 t i ; s A k r @ 3 @ 2 t @ t i ; k = [D @ 3 @ 2 t ] i r [D @ 3 @ 2 t @ t ] i r = 1 2 d dt jD @ 3 @ 2 t j 2 1 2 @ 3 @ 2 t i ; s @ 3 @ 2 t i ; k (A s r A k r ) t ; (4.4.22) 173 while the second term on the right hand side of (4.4.21) can be rewritten as (A s i @ 3 @ 2 t r ; s A s r @ 3 @ 2 t i ; s )A k r @ 3 @ 2 t @ t i ; k = ijk @ 3 @ 2 t k ; r A r j iml @ 3 @ 2 t @ t l ; s A s m = curl @ 3 @ 2 t curl @ 3 @ 2 t @ t = 1 2 d dt j curl @ 3 @ 2 t j 2 + 1 2 @ 3 @ 2 t k ; r @ 3 @ 2 t k ; s (A r j A s j ) t 1 2 @ 3 @ 2 t k ; r @ 3 @ 2 t j ; s (A r j A s k ) t : (4.4.23) Combining (4.4.21)–(4.4.23), we arrive at G 1 = 1 2 Z 2 d J 1 d dt jD @ 3 @ 2 t j 2 j curl @ 3 @ 2 t j 2 + Z Q( d @ 3 @ 2 t D) = 1 2 d dt Z 2 d J 1 jD @ 3 @ 2 t j 2 j curl @ 3 @ 2 t j 2 + Z Q( d @ 3 @ 2 t D); whereQ is a quadratic function withL 1 ([0;T ] ) coecients which may vary from line to line. Inte- gratingG 1 in time from 0 toT , we get Z T 0 G 1 = 1 2 Z 2 d J 1 jD @ 3 @ 2 t (T )j 2 j curl @ 3 @ 2 t (T )j 2 + Z T 0 Z Q( d @ 3 @ 2 t D) +M 0 : For the termG 2 , we have G 2 = Z 2 d J 1 div @ 3 @ 2 t div @ 3 @ 2 t @ t = 1 2 d dt Z 2 d J 1 j div @ 3 @ 2 t j 2 + Z Q( d @ 3 @ 2 t D); which leads to Z T 0 G 2 = 1 2 Z 2 d J 1 j div @ 3 @ 2 t (T )j 2 + Z T 0 Z Q( d @ 3 @ 2 t D) +M 0 : (4.4.24) 174 From (4.2.16) and the Leibniz rule it follows that div @ 3 @ 2 t =A s r @ 3 @ 2 t r ; s =J 1 @ 3 @ 2 t J + 3 X l=0 1 X m=0 l+m1 c l;m J 1 @ l @ m t a s r @ 3l @ 2m t r ; s : (4.4.25) Inserting (4.4.25) to the rst term on the right hand side of (4.4.24), we arrive at 1 2 Z 2 d J 1 j div @ 3 @ 2 t (T )j 2 = 1 2 Z 2 d J 3 j @ 3 @ 2 t J(T )j 2 + 3 X l=0 1 X m=0 l+m1 c l;m Z 2 d J 3 j @ l @ m t a @ 3l @ 2m t Dj 2 (T ) + 3 X l=0 1 X m=0 l+m1 c l;m Z 2 d J 3 @ 3 @ 2 t J @ l @ m t a @ 3l @ 2m t D(T ) =: 1 2 Z 2 d J 3 j @ 3 @ 2 t J(T )j 2 +G 21 +G 22 ; where we drop the indices for simplicity. The highest order terms inG 21 can be treated using the Hölder and Sobolev inequalities and the fundamental theorem of calculus as Z 2 d J 3 j@ t D @ 3 @ t Dj 2 (T ). Z j d @ 3 @ t Dj 2 (T ). 1 +TP ( sup t2[0;T ] E(t)) and Z 2 d J 3 j @D @ 2 @ 2 t Dj 2 (T ). Z j @ 2 @ 2 t Dj 2 (T ). 1 +TP ( sup t2[0;T ] E(t)); 175 sincek d @ 3 @ 2 t Dk 2 0 +kvk 2 4 +k@ 2 t vk 2 3 E(t). The rest of the terms inG 21 are of lower order which can be treated in a similar fashion using the Hölder and Sobolev inequalities and the fundamental theorem of calculus. Thus, we arrive at G 21 . 1 +TP ( sup t2[0;T ] E(t)): (4.4.26) For the termG 22 , using the Young inequality, we obtain G 22 .k d @ 3 @ 2 t J(T )k 2 0 +C 3 X l=0 1 X m=0 l+m1 k d J 3 @ l @ m t a @ 3l @ 2m t Dk 2 0 (T ) .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); where the last inequality follows from (4.4.26). For the termG 3 in (4.4.20), we integrate by parts in time, leading to Z T 0 G 3 = 3 X l=0 1 X m=0 l+m1 c l;m Z @ l @ m t (J 1 (a s r a k i a s i a k r )) @ 3l @ 2m t r ; s 2 d J 2 @ 3 @ 2 t i ; k t=T t=0 | {z } G 31 + 3 X l=0 1 X m=0 l+m1 c l;m Z T 0 Z ( @ l @ m t (J 1 (a s r a k i a s i a k r )) @ 3l @ 2m t r ; s 2 d J 2 ) t @ 3 @ 2 t i ; k | {z } G 32 : We bound the termG 31 using the Young and Sobolev inequalities and the fundamental theorem of calculus as G 31 .k d @ 3 @ 2 t D(T )k 2 0 +C 3 X l=0 1 X m=0 l+m1 k d @ l @ m t (J 1 (a s r a k i a s i a k r )) @ 3l @ 2m t r ; s k 2 0 (T ) + 1 .C + sup t2[0;T ] E(t) +TP ( sup t2[0;T ] E(t)); 176 sincek d @ 3 @ 2 t Dk 2 0 +k@ 2 t Jk 2 3 +k@ 2 t vk 2 3 E(t). For the termG 32 , we use the Hölder and Sobolev inequalities and the fundamental theorem of calculus, obtaining G 32 . 3 X l=0 1 X m=0 l+m1 Z T 0 k d ( @ l @ m t (J 1 (a s r a k i a s i a k r )) @ 3l @ 2m t r ; s J 2 ) t k 0 k d @ 3 @ 2 t Dk 0 .TP ( sup t2[0;T ] E(t)): Collecting the above estimates, we conclude Z T 0 G = 1 2 Z 2 d J 1 jD @ 3 @ 2 t (T )j 2 2 d J 1 j curl @ 3 @ 2 t (T )j 2 2 d J 3 j @ 3 @ 2 t J(T )j 2 +G 0 ; (4.4.27) whereG 0 consists of the terms satisfying G 0 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)): (4.4.28) Next we estimate the termL in (4.4.18). From (4.2.17) and the Leibniz rule it follows that L = Z J 1 a s i a k r @ 3 @ 2 t r ; s ( 2 u J 2 ); k @ 3 @ 2 t v i | {z } L 1 + Z J 1 a s r a k i @ 3 @ 2 t r ; s ( 2 u J 2 ); k @ 3 @ 2 t v i | {z } L 2 + 3 X l=0 1 X m=0 l+m1 c l;m Z @ l @ m t (J 1 (a s r a k i a s i a k r )) @ 3l @ 2m t r ; s ( 2 u J 2 ); k @ 3 @ 2 t v i | {z } L 3 : (4.4.29) 177 For the termL 1 , we integrate by parts in@ s , obtaining Z T 0 L 1 = Z T 0 Z J 1 a 3 i a 3 r @ 3 @ 2 t r ( 2 u J 2 ); 3 @ 3 @ 2 t v i | {z } L 11 + Z T 0 Z J 1 a s i a k r @ 3 @ 2 t r ( 2 u J 2 ); k @ 3 @ 2 t v i ; s | {z } L 12 + Z T 0 Z (J 1 a s i a k r ( 2 u J 2 ); k ); s @ 3 @ 2 t r @ 3 @ 2 t v i | {z } L 13 ; (4.4.30) where we note thata k r ( 2 u J 2 ); k = a 3 r ( 2 u J 2 ); 3 since 2 u J 2 = 1 on [0;T ]. The termL 11 can be rewritten as L 11 = 1 2 Z T 0 d dt Z J 1 ( 2 u J 2 ); 3 j @ 3 @ 2 t r a 3 r j 2 | {z } L 111 + 1 2 Z T 0 Z (J 1 ( 2 u J 2 ); 3 ) t j @ 3 @ 2 t r a 3 r j 2 | {z } L 112 + Z T 0 Z J 1 ( 2 u J 2 ); 3 @ 3 @ 2 t r @ t a 3 r @ 3 @ 2 t i a 3 i | {z } L 113 : Note that from (4.2.15) we may assume that J 1 ( 2 u J 2 ); 3 =J 1 ( 2 u ; 3 J 2 2J 3 J; 3 2 u ) 1 2 on [0;T ] ; (4.4.31) from where L 111 = 1 2 Z J 1 ( 2 u J 2 ); 3 j @ 3 @ 2 t r a 3 r j 2 (T ) +M 0 1 4 Z j @ 3 @ 2 t r a 3 r j 2 (T ) +M 0 : (4.4.32) Using the Hölder and Sobolev inequalities, we get L 112 +L 113 . Z T 0 (kDJ t k L 1 + 1)j @ 3 @ 2 t j 2 0 .TP ( sup t2[0;T ] E(t)); 178 sincej @ 3 @ 2 t j 2 0 E(t). Now we claim that sup t2[0;T ] j @ 3 @ 2 t (t)j 2 0 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); (4.4.33) where = 1; 2. To prove the claim, we resort to the boundary dynamics for the ow map. In fact, from the momentum equation (4.2.11), we see that satises @ 2 t i 2a 3 i J; 3 2a 3 i = 0 on [0;T ] ; i = 1; 2; 3; since 0 ; 3 =1, 0 ; = 0, andJ = 0 = 1 on [0;T ] . Then from a 3 3 @ 2 t =a 3 3 (2a 3 J; 3 +2a 3 ) =a 3 (2a 3 3 J; 3 +2a 3 3 ) =a 3 @ 2 t 3 ; = 1; 2; we deduce the following algebraic relation between@ 2 t and@ 2 t 3 : @ 2 t = (a 3 3 ) 1 a 3 @ 2 t 3 on [0;T ] ; = 1; 2: (4.4.34) Applying @ 3 to the above equation, we get@ 2 t @ 3 = @ 3 (a 3 3 ) 1 a 3 @ 2 t 3 , which leads to j@ 2 t @ 3 j 2 0 .j(a 3 3 ) 1 a 3 @ 3 @ 2 t 3 j 2 0 | {z } I 1 +j @ 3 ((a 3 3 ) 1 a 3 )@ 2 t 3 j 2 0 | {z } I 2 +j @((a 3 3 ) 1 a 3 ) @ 2 @ 2 t 3 j 2 0 | {z } I 3 +j @ 2 ((a 3 3 ) 1 a 3 ) @@ 2 t 3 j 2 0 | {z } I 4 : (4.4.35) 179 The termI 1 is estimated using the Hölder and Sobolev inequalities as I 1 .ka 3 k L 1j @ 3 @ 2 t 3 j 2 0 .TP ( sup t2[0;T ] E(t)); (4.4.36) where we usedkaI 3 k 2 .T . For the termI 2 , using the Leibniz rule, we get I 2 .j @ 3 (a 3 3 ) 1 a 3 j 2 0 | {z } I 21 +j(a 3 3 ) 1 @ 3 a 3 j 2 0 | {z } I 22 + 2 X l=1 j @ l (a 3 3 ) 1 @ 3l a 3 j 2 0 | {z } I 23 : (4.4.37) Using the Hölder and Sobolev inequalities and the fundamental theorem of calculus, we arrive at I 21 .j @ 3 (a 3 3 ) 1 j 2 0 ka 3 k 2 L 1.TP ( sup t2[0;T ] E(t)); (4.4.38) where we usedj @ 4 j 2 0 .k @ 3 k 2 1:5 and (4.2.23) in the last inequality. For the termI 22 , from (4.2.23) and the Leibniz rule it follows that I 22 .j @ 3 ( @ @ 3 )j 2 0 .j @ @ 4 3 j 2 0 | {z } I 221 +j @ 4 @ 3 j 2 0 | {z } I 222 + 2 X l=1 j @ l+1 @ 4l 3 j 2 0 | {z } I 223 : (4.4.39) The termI 221 is estimated using the Hölder and Sobolev inequalities as I 221 .j @ 4 3 j 2 0 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); (4.4.40) where the last inequality follows from Lemma 4.4.1. For the termI 222 , we use the Hölder inequality to get I 222 .j @ 4 j 2 0 j @ 3 j 2 L 1 () .TP ( sup t2[0;T ] E(t)); (4.4.41) 180 sincek @ 3 k L 1 () . T which holds by the fundamental theorem of calculus. The termI 223 is estimated using the Hölder and Sobolev inequalities and the fundamental theorem of calculus as I 223 .k @ 2 k 2 L 1 ( ) j @ 3 j 2 0 . 1 +TP ( sup t2[0;T ] E(t)): (4.4.42) Similarly, we bound the termI 23 as I 23 . 1 +TP ( sup t2[0;T ] E(t)): (4.4.43) For the termI 3 , using the Hölder and Sobolev inequalities, we obtain I 3 .j @ 2 @ 2 t j 2 0 . 1 +TP ( sup t2[0;T ] E(t)); (4.4.44) sincej @ 2 @ 3 t j 2 0 E(t), where we used the fundamental theorem of calculus in the last inequality. The highest order term inI 4 is scales likej @ 3 @@ 2 t 3 j 2 0 which is treated using the Hölder and Sobolev inequal- ities as j @ 3 @@ 2 t 3 j 2 0 .j @ 3 j 2 L 4 () j @@ 2 t 3 j 2 L 4 () .k @ 3 k 2 1 k@ 2 t k 2 2 . 1 +TP ( sup t2[0;T ] E(t)); where we used the Sobolev embeddingH 0:5 () ,! L 4 (), Lemma 4.10.3, and the fundamental theorem of calculus. The rest of the terms inI 4 are of lower order which can treated in a similar fashion as above. Thus, we have I 4 . 1 +TP ( sup t2[0;T ] E(t)): (4.4.45) Collecting the estimates (4.4.35)–(4.4.45), we complete the proof of the claim (4.4.33). 181 Next we estimate the termL 12 in (4.4.30). Using integration by parts in time, we obtain L 12 = Z T 0 Z J 1 a s i a k r @ 3 @ 2 t v r ( 2 u J 2 ); k @ 3 @ 2 t i ; s | {z } L 121 + Z J 1 a s i a k r @ 3 @ 2 t r ( 2 u J 2 ); k @ 3 @ 2 t i ; s T 0 | {z } L 122 Z T 0 Z (J 1 a s i a k r ( 2 u J 2 ); k ) t @ 3 @ 2 t r @ 3 @ 2 t i ; s | {z } L 123 : It is clear thatL 121 + R T 0 L 2 = 0. Note that from (4.2.16) and the Leibniz rule it follows that @ 3 @ 2 t J =a s i @ 3 @ 2 t i ; s + @ 3 @ t a s i @ t i ; s +@ t a s i @ 3 @ t i ; s + 3 X l=0 1 X m=0 1l+m3 (l;m)6=(0;1) c l;m @ l @ m t a s i @ 3l @ 2m t i ; s : (4.4.46) Inserting (4.4.46) into the termL 122 and using the Young, Hölder, Sobolev inequalities and the fundamental theorem of calculus, we arrive at L 122 . 1 +k @ 3 @ 2 t J(T )k 2 0 +k @ 3 @ t a@ t Dk 2 0 (T ) +C k @ 3 @ 2 t (T )k 2 0 + 3 X l=0 1 X m=0 1l+m3; (l;m)6=(0;1) k @ l @ m t a @ 3l @ 2m t Dk 2 0 (T ).C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); where we recall thatkvk 2 4 +k@ 2 t vk 2 3 +k@ 2 t Jk 2 3 E(t). For the termL 123 , we appeal to the Hölder and Sobolev inequalities and Lemma 4.10.5, obtaining L 123 . Z T 0 k @ 3 @ 2 t k 0:5 k @ 3 @ 2 t Dk 0:5 . Z T 0 k @ 2 @ 2 t k 2 1:5 .TP ( sup t2[0;T ] E(t)); 182 sincek @ 2 @ 2 t k 2 1:5 E(t). The termL 13 in (4.4.30) is estimated using the Hölder and Sobolev inequalities as L 13 . Z T 0 (1 +kD 2 Jk L 1)k @ 3 @ 2 t k 0 k @ 3 @ 2 t vk 0 .TP ( sup t2[0;T ] E(t)); sincekJk 2 4 E(t). The termL 3 in (4.4.29) consists of essentially lower order terms which can be estimated using the Hölder and Sobolev inequalities and the fundamental theorem of calculus as Z T 0 L 3 .TP ( sup t2[0;T ] E(t)): Collecting (4.4.17), (4.4.27)–(4.4.28), and (4.4.32) and the above estimates, we conclude that 1 2 Z 0 j @ 3 @ 2 t vj 2 (T ) + 2 d J 1 jD @ 3 @ 2 t (T )j 2 2 d J 1 j curl @ 3 @ 2 t (T )j 2 2 d J 3 j @ 3 @ 2 t J(T )j 2 + 1 4 Z j @ 3 @ 2 t r a 3 r j 2 (T ) + Z T 0 J 2 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)) + Z T 0 J 3 : Usingj @ 3 @ 2 t r a 3 r jj @ 3 @ 2 t 3 a 3 3 jj @ 3 @ 2 t a 3 j andkaI 3 k 2 .T , we obtain 1 2 Z 0 j @ 3 @ 2 t vj 2 (T ) + 2 d J 1 jD @ 3 @ 2 t (T )j 2 2 d J 1 j curl @ 3 @ 2 t (T )j 2 2 d J 3 j @ 3 @ 2 t J(T )j 2 + 1 8 Z j @ 3 @ 2 t j 2 (T ) + Z T 0 J 2 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)) + Z T 0 J 3 : (4.4.47) where we used (4.4.33) in the last inequality. 183 Estimate ofJ 2 in (4.4.17): We integrate by parts in@ k and use the Piola identity (4.2.22) to get J 2 = Z a 3 i 2 0 @ 3 @ 2 t J 2 @ 3 @ 2 t v i Z fx 3 =0g a 3 i 2 0 @ 3 @ 2 t J 2 @ 3 @ 2 t v i Z a k i 2 0 @ 3 @ 2 t J 2 @ 3 @ 2 t v i ; k = Z a k i 2 0 @ 3 @ 2 t J 2 @ 3 @ 2 t v i ; k ; (4.4.48) since @ 3 @ 2 t J 2 = 0 on [0;T ] anda 3 i @ 3 @ 2 t v i = 0 onfx 3 = 0g [0;T ]. Note that from (4.2.16) and the Leibniz rule it follows @ 3 @ 3 t J =a k i @ 3 @ 2 t v i ; k +v i ; k @ 3 @ 2 t a k i + 2@ t a k i @ 3 @ t v i ; k + 3 X l=0 2 X m=0 1m+l4 (l;m)6=(0;1) c l;m @ l @ m t a k i @ 3l @ 2m t v i ; k : (4.4.49) Combing (4.4.48)–(4.4.49) , we arrive at J 2 = 2 Z 2 0 a k i J 3 @ 3 @ 2 t J @ 3 @ 2 t v i ; k + 3 X l=0 1 X m=0 l+m1 c l;m Z 2 0 a k i @ l @ m t J 3 @ 3l @ 2m t J @ 3 @ 2 t v i ; k | {z } J 25 = 2 Z 2 0 J 3 @ 3 @ 2 t J @ 3 @ 3 t J | {z } J 21 Z 2 0 J 3 @ 3 @ 2 t Jv i ; k @ 3 @ 2 t a k i | {z } J 22 2 Z 2 0 J 3 @ 3 @ 2 t J@ t a k i @ 3 @ t v i ; k | {z } J 23 + 3 X l=0 2 X m=0 1m+l4 (l;m)6=(0;1) c l;m Z 2 0 J 3 @ 3 @ 2 t J @ l @ m t a k i @ 3l @ 2m t v i ; k | {z } J 24 +J 25 : The termJ 21 can be rewritten as J 21 = d dt Z 2 0 J 3 j @ 3 @ 2 t Jj 2 Z ( 2 0 J 3 ) t j @ 3 @ 2 t Jj 2 | {z } J 0 21 ; (4.4.50) 184 where the termJ 0 21 satises Z T 0 J 0 21 .TP ( sup t2[0;T ] E(t)): For the termJ 22 , we integrate by parts in time, obtaining Z T 0 J 22 = Z T 0 Z @ 3 @ t J( 2 0 J 3 v i ; k @ 3 @ 2 t a k i ) t | {z } J 221 + Z @ 3 @ t J 2 0 J 3 v i ; k @ 3 @ 2 t a k i j t=0 t=T | {z } J 222 : (4.4.51) For the termJ 221 , we integrate by parts in @ to get J 221 = Z T 0 Z @ 3 @ t J( 2 0 J 3 v i ; k ); t @ 3 @ 2 t a k i + Z T 0 Z @ 3 @ t J 2 0 J 3 v i ; k @ 3 @ 3 t a k i = Z T 0 Z @( @ 3 @ t J( 2 0 J 3 v i ; k ); t ) @ 2 @ 2 t a k i Z T 0 Z @( @ 3 @ t J 2 0 J 3 v i ; k ) @ 2 @ 3 t a k i ; (4.4.52) since integration by parts in @ does not produce any boundary terms. Therefore, using the Hölder and Sobolev inequalities and the fundamental theorem of calculus, we obtain J 221 .TP ( sup t2[0;T ] E(t)); (4.4.53) sincek@ t Jk 2 4 +k@ 2 t vk 2 3 E(t). For the termJ 222 , we integrate by parts in @, obtaining J 222 = Z @( @ 3 @ t J 2 0 J 3 v i ; k ) @ 2 @ 2 t a k i j t=T t=0 = Z @ 4 @ t J 2 0 J 3 v i ; k @ 2 @ 2 t a k i j t=T t=0 + Z @ 3 @ t J @( 2 0 J 3 v i ; k ) @ 2 @ 2 t a k i j t=T t=0 : (4.4.54) From the Young, Hölder, and Sobolev inequalities and the fundamental theorem of calculus it follows that J 222 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); (4.4.55) 185 sincek@ 2 t Jk 2 3 +k@ 2 t vk 2 3 E(t). For the termJ 23 , we proceed analogously as in (4.4.51)–(4.4.55), obtaining Z T 0 J 23 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)): The termJ 24 consists of essentially lower order terms which can be estimated using the Hölder and Sobolev inequalities and the fundamental theorem of calculus as Z T 0 J 24 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)): For the termJ 25 , we integrate by parts in@ k , obtaining J 25 = 3 X l=0 1 X m=0 l+m1 c l;m Z ( 2 0 a k i @ l @ m t J 3 @ 3l @ 2m t J); k @ 3 @ 2 t v i ; (4.4.56) from where Z T 0 J 25 . 1 +TP ( sup t2[0;T ] E(t)): Combining (4.4.47), (4.4.50), and the above estimates, we arrive at 1 2 Z ( 0 j @ 3 @ 2 t vj 2 (T ) + 2 d J 1 jD @ 3 @ 2 t (T )j 2 2 d J 1 j curl @ 3 @ 2 t (T )j 2 + 2 0 J 3 j @ 3 @ 2 t J(T )j 2 ) + 1 8 Z j @ 3 @ 2 t j 2 (T ) .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)) + Z T 0 J 3 ; (4.4.57) since 2 d 2 0 . 186 Estimate ofJ 3 in (4.4.17): The termJ 3 consists of essentially lower order terms which can be treated in a similar fashion as in (4.4.56), and we obtain J 3 . 1 +TP ( sup t2[0;T ] E(t)): (4.4.58) Concluding the proof : Combining (4.4.57)–(4.4.58), we get 1 2 Z 0 j @ 3 @ 2 t vj 2 (T ) + 2 d J 1 jD @ 3 @ 2 t (T )j 2 2 d J 1 j curl @ 3 @ 2 t (T )j 2 + 2 0 J 3 j @ 3 @ 2 t J(T )j 2 + 1 8 Z j @ 3 @ 2 t j 2 (T ).C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)): The proof of the lemma is thus completed by the curl estimate in Lemma 4.3.1. The following lemma provides @ 2 @ 3 t , @@ 4 t , and@ 5 t energy estimates of the solutions,J, andv. Lemma4.4.3. For2 (0; 1), we have sup t2[0;T ] 5 X l=3 Z j d @ 5l @ l t D(t)j 2 +j @ 5l @ l t v(t)j 2 +j @ 5l @ l t J(t)j 2 + sup t2[0;T ] 5 X l=3 Z j @ 5l @ l t (t)j 2 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); (4.4.59) whereC > 0 is a constant depending on. Proof of Lemma 4.4.3. Similar arguments as in Lemma 4.4.2 lead to sup t2[0;T ] 4 X l=3 Z j d @ 5l @ l t D(t)j 2 +j @ 5l @ l t v(t)j 2 +j @ 5l @ l t J(t)j 2 + sup t2[0;T ] 4 X l=3 Z j @ 5l @ l t (t)j 2 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); (4.4.60) 187 It remains to establish the@ 5 t energy estimates. Applying@ 5 t to (4.2.7) and taking the inner product with @ 5 t v, we obtain 1 2 d dt Z 0 j@ 5 t vj 2 + Z @ 5 t a k i ( 2 0 J 2 ); k @ 5 t v i | {z } J 1 + Z a k i ( 2 0 @ 5 t J 2 ); k @ 5 t v i | {z } J 2 = 4 X l=1 c l Z @ l t a k i ( 2 0 @ 5l t J 2 ); k @ 5 t v i | {z } J 3 : (4.4.61) Estimate ofJ 1 in (4.4.61): Using the decomposition 2 0 = 2 d + 2 u (see (4.2.14)–(4.2.15)), we split the termJ 1 as J 1 = Z @ 5 t a k i ( 2 d J 2 ); k @ 3 @ 2 t v i | {z } G + Z @ 5 t a k i ( 2 u J 2 ); k @ 3 @ 2 t v i | {z } L : (4.4.62) For the termG, we proceed in a similar fashion as in Lemma 4.4.2, obtaining Z T 0 G = 1 2 Z 2 d J 1 jD @ 5 t (T )j 2 2 d J 1 j curl @ 5 t (T )j 2 2 d J 3 j@ 5 t J(T )j 2 +G 0 ; (4.4.63) whereG 0 consists of the terms satisfying Z T 0 G 0 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)): (4.4.64) Next we estimate the termL in (4.4.62). Using (4.2.17) and the Leibniz rule, we get L = Z @ 5 t r ; s J 1 a s i a k r ( 2 u J 2 ); k @ 5 t v i | {z } L 1 + Z @ 5 t r ; s J 1 a s r a k i ( 2 u J 2 ); k @ 5 t v i | {z } L 2 + 4 X l=1 c l Z @ l t (J 1 (a s r a k i a s i a k r ))@ 5l t r ; s ( 2 u J 2 ); k @ 5 t v i | {z } L 3 : 188 For the termL 1 , we integrate by parts in@ s , leading to Z T 0 L 1 = Z T 0 Z fx 3 =0g @ 5 t r J 1 a 3 i a k r ( 2 u J 2 ); k @ 5 t v i | {z } =0 Z T 0 Z @ 5 t r J 1 a 3 i a 3 r ( 2 u J 2 ); 3 @ 5 t v i | {z } L 11 + Z T 0 Z @ 5 t r J 1 a s i a k r ( 2 u J 2 ); k @ 5 t v i ; s | {z } L 12 + Z T 0 Z @ 5 t r (J 1 a s i a k r ( 2 u J 2 ); k ); s @ 5 t v i | {z } L 13 ; sincea 3 i @ 5 t v i = 0 onfx 3 = 0g [0;T ] and 2 u J 2 = 1 on [0;T ]. The termL 11 can be rewritten as L 11 = 1 2 Z T 0 d dt Z J 1 ( 2 u J 2 ); 3 j@ 5 t r a 3 r j 2 | {z } L 111 + 1 2 Z T 0 Z (J 1 ( 2 u J 2 ); 3 ) t j@ 5 t r a 3 r j 2 | {z } L 112 + Z T 0 Z J 1 ( 2 u J 2 ); 3 @ 5 t i a 3 i @ 5 t r @ t a 3 r | {z } L 113 : From (4.4.31) it follows that L 111 = 1 2 Z J 1 ( 2 u J 2 ); 3 j @ 5 r a 3 r j 2 (T ) +M 0 1 4 Z j @ 5 r a 3 r j 2 (T ) +M 0 : (4.4.65) Using the Hölder and Sobolev inequalities, we get L 112 +L 113 . Z T 0 (1 +kDJ t k L 1)j@ 5 t j 2 0 .TP ( sup t2[0;T ] E(t)); sincek@ t Jk 2 4 +j@ 5 t j 2 0 E(t). For the termL 12 , we integrate by parts in time, obtaining L 12 = Z @ 5 t r J 1 a s i a k r ( 2 u J 2 ); k @ 5 t i ; s T 0 | {z } L 121 Z T 0 Z @ 6 t r J 1 a s i a k r ( 2 u J 2 ); k @ 5 t i ; s | {z } L 122 Z T 0 Z (J 1 a s i a k r ( 2 u J 2 ); k ) t @ 5 t r @ 5 t i ; s | {z } L 123 ; 189 Note that from (4.2.16) and the Leibniz rule it follows that @ 5 t J =a s i @ 5 t i ; s + 4 X l=1 c l @ l t a s i @ 5l t i ; s : (4.4.66) Inserting (4.4.66) to the termL 121 and using the Young, Hölder, and Sobolev inequalities and the funda- mental theorem of calculus, we obtain L 121 .k@ 5 t J(T )k 2 0 +C k@ 5 t (T )k 2 0 + 4 X l=1 k@ l t a@ 5l t Dk 2 0 (T ) + 1 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); where we recall thatk@ 5 t Jk 2 0 +k@ 3 t vk 2 2 +k@ 4 t vk 2 1 +k@ 5 t vk 2 0 E(t). It is clear thatL 122 + R T 0 L 2 = 0. For the termL 123 , we use the Hölder and Sobolev inequalities, obtaining L 123 . Z T 0 (1 +kDJ t k L 1)k@ 5 t k 0 k@ 5 t Dk 0 .TP ( sup t2[0;T ] E(t)); sincek@ 4 t vk 2 1 E(t). Similarly, the termsL 13 andL 3 are estimated as L 13 + Z T 0 L 3 .TP ( sup t2[0;T ] E(t)): Combining (4.4.61), (4.4.63)–(4.4.65), and the above estimates, we conclude 1 2 Z 0 j@ 5 t v(T )j 2 + 2 d J 1 jD @ 5 t (T )j 2 2 d J 1 j curl @ 5 t (T )j 2 2 d J 3 j@ 5 t J(T )j 2 + 1 4 Z j@ 5 t r a 3 r j 2 (T ) + Z T 0 J 2 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)) + Z T 0 J 3 : (4.4.67) 190 Now we claim that sup t2[0;T ] j@ 5 t (t)j 2 0 . 1 +TP ( sup t2[0;T ] E(t)); (4.4.68) where = 1; 2. Applying@ 3 t to (4.4.34), we get j@ 5 t j 2 0 .j(a 3 3 ) 1 a 3 @ 5 t 3 j 2 0 | {z } I 1 +j@ 3 t ((a 3 3 ) 1 a 3 )@ 2 t 3 j 2 0 | {z } I 2 +j@ t ((a 3 3 ) 1 a 3 )@ 4 t 3 j 2 0 | {z } I 3 +j@ 2 t ((a 3 3 ) 1 a 3 )@ 3 t 3 j 2 0 | {z } I 4 (4.4.69) The termI 1 is estimated using the Hölder and Sobolev inequalities as I 1 .ka 3 k L 1 ( ) j@ 5 t 3 j 2 0 .TP ( sup t2[0;T ] E(t)); (4.4.70) where we usedkaI 3 k 2 .T andj@ 5 t j 2 0 E(t). For the termI 2 , using the Leibniz rule, we get I 2 .j@ 3 t (a 3 3 ) 1 a 3 j 2 0 | {z } I 21 +j(a 3 3 ) 1 @ 3 t a 3 j 2 0 | {z } I 22 + 2 X l=1 j@ l t (a 3 3 ) 1 @ 3l t a 3 j 2 0 | {z } I 23 : (4.4.71) We bound the termI 21 using the Hölder and Sobolev inequalities and the fundamental theorem of calculus as I 21 .j@ 3 t (a 3 3 ) 1 j 2 0 ka 3 k 2 L 1.TP ( sup t2[0;T ] E(t)): (4.4.72) Similarly, we estimate the termI 22 as I 22 .j@ 3 t ( @ @)j 2 0 . 1 +TP ( sup t2[0;T ] E(t)); (4.4.73) 191 sincej@ 4 t @j 2 0 E(t). The termI 23 consists of lower-order terms which can be estimated in a similar fashion using the Hölder and Sobolev inequalities and the fundamental theorem of calculus as I 23 . 1 +TP ( sup t2[0;T ] E(t)): (4.4.74) For the termI 3 , we have I 3 .j@ 4 t j 2 0 . 1 +TP ( sup t2[0;T ] E(t)); (4.4.75) sincej@ 5 t j 2 0 E(t). For the termI 4 , we proceed analogously as in (4.4.73), obtaining I 4 . 1 +TP ( sup t2[0;T ] E(t)): (4.4.76) Collecting (4.4.69)–(4.4.76), we complete the proof of the claim (4.4.68). From (4.4.67)–(4.4.68) and using j@ 5 t r a 3 r jj@ 5 t 3 a 3 3 jj@ 5 t a 3 j andkaI 3 k 2 .T , we conclude 1 2 Z 0 j@ 5 t v(T )j 2 + 2 d J 1 jD @ 5 t (T )j 2 2 d J 1 j curl @ 5 t (T )j 2 2 d J 3 j@ 5 t J(T )j 2 + 1 8 Z j@ 5 t (T )j 2 + Z T 0 J 2 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)) + Z T 0 J 3 : (4.4.77) Estimate ofJ 2 in (4.4.61): We integrate by parts in@ k and use the Piola identity (4.2.22) to get J 2 = Z a 3 i 2 0 @ 5 t J 2 @ 5 t v i Z fx 3 =0g a 3 i 2 0 @ 5 t J 2 @ 5 t v i Z a k i 2 0 @ 5 t J 2 @ 5 t v i ; k = Z a k i 2 0 @ 5 t J 2 @ 5 t v i ; k ; 192 since@ 5 t J 2 = 0 on [0;T ] anda 3 i @ 5 t v i = 0 onfx 3 = 0g [0;T ]. Using @ 6 t J =a k i @ 5 t v i ; k +v i ; k @ 5 t a k i + 4 X l=1 c l @ l t a k i @ 5l t v i ; k ; we obtain J 2 = 2 Z a k i 2 0 J 3 @ 5 t J@ 5 t v i ; k + 4 X l=1 c l Z 2 0 a k i @ l t J 3 @ 5l t J@ 5 t v i ; k | {z } J 24 = 2 Z 2 0 J 3 @ 5 t J@ 6 t J | {z } J 21 2 Z 2 0 J 3 @ 5 t Jv i ; k @ 5 t a k i | {z } J 22 + 4 X l=1 c l Z 2 0 J 3 @ 5 t J@ l t a k i @ 5l t v i ; k | {z } J 23 +J 24 : The termJ 21 can be rewritten as Z T 0 J 21 = Z T 0 d dt Z 2 0 J 3 j@ 5 t Jj 2 Z T 0 Z ( 2 0 J 3 ) t j@ 5 t Jj 2 | {z } J 0 21 ; (4.4.78) where the termJ 0 21 satises J 0 21 .TP ( sup t2[0;T ] E(t)): The highest order term inJ 22 is of the form R 2 0 J 3 @ 5 t JDv@ 5 t D, which can be treated using the Hölder and Sobolev inequalities as Z T 0 Z 2 0 J 3 @ 5 t JDv@ 5 t D. Z T 0 k@ 5 t Jk 0 k@ 5 t Dk 0 .TP ( sup t2[0;T ] E(t)); 193 sincek@ 4 t vk 2 1 +k@ 5 t Jk 2 0 E(t). The rest of the terms inJ 22 are of lower order which can be treated in a similar fashion using the Hölder and Sobolev inequalities and the fundamental theorem of calculus. Thus, we obtain Z T 0 J 22 .TP ( sup t2[0;T ] E(t)): Similarly, the termJ 23 is estimated as Z T 0 J 23 . 1 +TP ( sup t2[0;T ] E(t)): For the termJ 24 , we integrate by parts in@ k , obtaining J 24 = 4 X l=1 c l Z ( 2 0 a k i @ l t J 3 @ 5l t J); k @ 5 t v i : (4.4.79) From the Hölder and Sobolev inequalities and the fundamental theorem of calculus it follows that Z T 0 J 24 .TP ( sup t2[0;T ] E(t)); (4.4.80) sincek@ 4 t Jk 2 1 +k@ 3 t Jk 2 2 +k@ 5 t vk 2 0 E(t). Combining (4.4.77)–(4.4.78) and the above estimates, we conclude 1 2 Z 0 j@ 5 t v(T )j 2 + 2 d J 1 jD @ 5 t (T )j 2 2 d J 1 j curl @ 5 t (T )j 2 + 2 0 J 3 j@ 5 t J(T )j 2 + 1 8 Z j@ 5 t (T )j 2 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)) + Z T 0 J 3 ; (4.4.81) since 2 0 2 d . 194 Estimate ofJ 3 in (4.4.61): For the termJ 3 , we proceed analogously as in (4.4.79)–(4.4.80), obtaining Z T 0 J 3 .TP ( sup t2[0;T ] E(t)): (4.4.82) Concluding the proof : Combining (4.4.81)–(4.4.82), we arrive at 1 2 Z 0 j@ 5 t v(T )j 2 + 2 d J 1 jD @ 5 t (T )j 2 2 d J 1 j curl @ 5 t (T )j 2 + 2 0 J 3 j@ 5 t J(T )j 2 + 1 8 Z j@ 5 t (T )j 2 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)): (4.4.83) From the curl estimates in Lemma 4.3.1 we complete the @ 5 t energy estimates. Combining (4.4.60) and (4.4.83), we conclude the proof of the lemma. 4.5 Normalderivativeestimates In the following three lemmas, we derive the normal derivative estimates ofJ andv. Lemma4.5.1. For2 (0; 1), we have sup t2[0;T ] k@ 4 t v(t)k 2 1 +k@ 4 t J(t)k 2 1 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); (4.5.1) whereC > 0 is a constant depending on. Proof of Lemma 4.5.1. We start with the termk@ 4 t vk 2 1 . Using Lemma 4.10.2, we have k@ 4 t vk 2 1 .k@ 5 t divk 2 0 | {z } K 1 +k@ 5 t curlk 2 0 | {z } K 2 +k@ 5 t @Nk 2 H 0:5 (@ ) | {z } K 3 +k@ 5 t k 2 0 | {z } K 4 : 195 From the Leibniz rule it follows that div@ 5 t = @ 5 t J + ( s r a s r )@ 5 t r ; s + P 4 l=1 c l @ l t a s r @ 5l t r ; s . Using Lemma 4.4.3, the Hölder and Sobolev inequalities, and the fundamental theorem of calculus, we obtain K 1 .k@ 5 t Jk 2 0 +k(aI 3 )@ 5 t Dk 2 0 + 4 X l=1 k@ l t a@ 5l t Dk 2 0 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); (4.5.2) sincekaI 3 k 2 .T andk@ 4 t vk 2 1 E(t). The termK 2 is already estimated in Lemma 4.3.1. For the term K 3 , we appeal to Lemma 4.10.4, obtaining K 3 .k@ 5 t @k 2 0 +k div@ 5 t k 2 0 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); where the last inequality follows from Lemma 4.4.3 and (4.5.2). We bound the termK 4 using the funda- mental theorem of calculus as K 4 . 1 +P ( sup t2[0;T ] E(t)); sincek@ 5 t vk 2 0 E(t). Consequently, we conclude the estimate k@ 4 t vk 2 1 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)): (4.5.3) Next we establish the estimate ofk@ 4 t Jk 2 1 . We rewrite (4.2.11) as v i t 2 0 J 3 a k i J; k 2a 3 i J 2 = 0; (4.5.4) 196 since 0 ; 3 =1 and 0 ; = 0. Lettingi = 3 in the above equation and applying@ 4 t to the resulting identity, we obtain 2 0 a 3 3 @ 4 t J; 3 = 4a 3 3 @ 4 t J +J 3 @ 6 t 3 + 0 @ 4 t (a 3 J 2 ; ) + 3 X l=0 c l @ l t J 2 @ 4l t a 3 3 + 4 X l=1 c l 0 @ l t a 3 3 @ 4l t J 2 ; 3 + 4 X l=1 c l 0 a 3 3 @ l t J 3 @ 4l t J; 3 + 3 X l=1 c l a 3 3 @ l t J 3 @ 4l t J : (4.5.5) Now we estimate the terms on the right hand side of (4.5.5) inL 2 . Using the Hölder and Sobolev inequalities and the fundamental theorem of calculus, we obtain ka 3 3 @ 4 t Jk 2 0 .ka 3 3 k 2 L 1k@ 4 t Jk 2 0 . 1 +TP ( sup t2[0;T ] E(t)); (4.5.6) sincek@ 5 t Jk 2 0 E(t). The termkJ 3 @ 6 t k 2 0 is treated using the Hölder and Sobolev inequalities and Lemma 4.4.3 as kJ 3 @ 6 t k 2 0 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)): From the Leibniz rule it follows that kJ 3 0 @ 4 t (a 3 J 2 ; )k 2 0 .k@ 4 t @J 2 k 2 0 + 4 X l=1 k@ l t a@ 4l t @J 2 k 2 0 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); (4.5.7) 197 where we appealed to the Hölder and Sobolev inequalities and Lemma 4.4.3. The rest of the terms on the right hand side of (4.5.5) are of lower order which can estimated in a similar fashion as in (4.5.5)–(4.5.7). Collecting the above estimates, we infer from (4.5.5) that k@ 4 t J; 3 k 2 0 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)): Combining the already-established estimates ofk@ 4 t @Jk 2 0 in Lemma 4.4.3, we conclude the estimate of k@ 4 t Jk 2 1 . The proof of the lemma is thus completed by combining (4.5.3). Lemma4.5.2. For2 (0; 1), we have sup t2[0;T ] 3 X l=2 k@ l t v(t)k 2 5l +k@ l t J(t)k 2 5l .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); (4.5.8) whereC > 0 is a constant depending on. Proof of Lemma 4.5.2. We start with the termk@ 3 t vk 2 2 . Using Lemma 4.10.2, we have k@ 3 t vk 2 2 .k@ 4 t divk 2 1 | {z } K 1 +k@ 4 t curlk 2 1 | {z } K 2 +k@ 4 t @Nk 2 H 0:5 (@ ) | {z } K 3 +k@ 4 t k 2 1 | {z } K 4 : From the Leibniz rule it follows that div@ 4 t =@ 4 t J + ( s r a s r )@ 4 t r ; s + P 3 l=1 c l @ l t a s r @ 4l t r ; s . Thus, we have K 1 .k@ 4 t Jk 2 1 | {z } K 11 +k(aI 3 )@ 4 t Dk 2 1 | {z } K 12 + 3 X l=1 k@ l t a@ 4l t Dk 2 1 | {z } K 13 : (4.5.9) 198 The termK 11 is already estimated in Lemma 4.5.1. For the termK 12 , using the Leibniz rule, we obtain K 12 .kaI 3 k 2 L 1k@ 4 t Dk 2 1 +kD 2 k 2 L 1k@ 4 t Dk 2 0 . 1 +TP ( sup t2[0;T ] E(t)); (4.5.10) sincekaI 3 k 2 .T andk@ 3 t vk 2 2 +k@ 4 t vk 2 1 E(t). In the last inequality of (4.5.10), we also used the Hölder and Sobolev inequalities and the fundamental theorem of calculus. Similarly, the termK 13 is estimated as K 13 . 1 +TP ( sup t2[0;T ] E(t)): The termK 2 is already estimated in Lemma 4.3.1. For the termK 3 , we appeal to Lemmas 4.10.3–4.10.4 and 4.4.3, obtaining K 3 .k@ 4 t @ 2 Nk H 0:5 (@ ) +k@ 4 t @Nk L 2 (@ ) .k@ 4 t @ 2 k 2 0 +k div@ 4 t @k 2 0 +k@ 4 t k 2 2 +C k@ 4 t k 2 0 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); where we also used the Sobolev interpolation in the second inequality and (4.5.9) in the last inequality. We bound the termK 4 using the fundamental theorem of calculus as K 4 . 1 +TP ( sup t2[0;T ] E(t)); sincek@ 4 t vk 2 1 E(t). Consequently, we conclude the estimate k@ 3 t vk 2 2 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)): (4.5.11) 199 Next we derive the estimate ofk@ 3 t Jk 2 2 . The estimate ofk@ 3 t @ 2 Jk 2 0 is already established in Lemma 4.4.3. For the estimate ofk@ 3 t @@ 3 Jk 2 0 , we proceed analogously as in Lemma 4.5.1 by applying @ 3 t @ to (4.5.4), obtaining k@ 3 t @@ 3 Jk 2 0 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)): (4.5.12) It remains to estimatek@ 3 t @ 33 Jk 2 0 . Applying@ 3 t to (4.5.4), we obtain 2 0 a 3 3 @ 3 t J; 3 = 4a 3 3 @ 3 t J +J 3 @ 5 t 3 + 0 @ 3 t (a 3 J 2 ; ) + 2 X l=0 c l @ l t J 2 @ 3l t a 3 3 + 3 X l=1 c l 0 @ l t a 3 3 @ 3l t J 2 ; 3 + 3 X l=1 c l 0 a 3 3 @ l t J 3 @ 3l t J; 3 + 2 X l=1 c l a 3 3 @ l t J 3 @ 3l t J : Applying@ 3 to the above equation, while noting that 0 ; 3 =1, we obtain 2 0 a 3 3 @ 3 t J; 33 = 6a 3 3 @ 3 t J; 3 2J 3 0 (J 3 a 3 3 ); 3 @ 3 t J; 3 +4J 3 (J 3 a 3 3 ); 3 @ 3 t J +J 3 @ 5 t 3 + 0 @ 3 t (a 3 J 2 ; ) + 2 X l=0 c l @ l t J 2 @ 3l t a 3 3 + 3 X l=1 c l 0 @ l t a 3 3 @ 3l t J 2 ; 3 + 3 X l=1 c l 0 a 3 3 @ l t J 3 @ 3l t J; 3 + 2 X l=1 c l a 3 3 @ l t J 3 @ 3l t J ; 3 : (4.5.13) Now we estimate the terms on the right hand side of (4.5.13) inL 2 . Using the Hölder and Sobolev inequal- ities and the fundamental theorem of calculus, we obtain ka 3 3 @ 3 t J; 3 k 2 0 .ka 3 3 k 2 L 1k@ 3 t J; 3 k 2 0 . 1 +TP ( sup t2[0;T ] E(t)): Similarly, we have kJ 3 0 (J 3 a 3 3 ); 3 @ 3 t J; 3 k 2 0 . 1 +TP ( sup t2[0;T ] E(t)) 200 and kJ 3 (J 3 a 3 3 ); 3 @ 3 t Jk 2 0 . 1 +TP ( sup t2[0;T ] E(t)) The termkJ 3 @ 5 t ; 3 k 2 0 is already estimated in Lemma 4.5.1. The highest order term inJ 3 ( 0 @ 3 t (a 3 J 2 ; )); 3 scales like@ 3 t @J; 3 which is estimated in (4.5.12). The rest of the terms inJ 3 ( 0 @ 3 t (a 3 J 2 ; )); 3 , as well as the terms on the right hand side of (4.5.13) are of lower order which can estimated using the Hölder and Sobolev inequalities and the fundamental theorem of calculus. Collecting the above estimates, we infer from (4.5.13) that k@ 3 t J; 33 k 2 0 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)): (4.5.14) The estimates ofk@ 2 t vk 2 3 andk@ 2 t Jk 2 3 follow in a similar fashion using the above arguments. Therefore, the proof of the lemma is completed by combining (4.5.11) and (4.5.14). Lemma4.5.3. For2 (0; 1), we have sup t2[0;T ] kv(t)k 2 4 +k@ t v(t)k 2 3 +k@ t J(t)k 2 4 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); (4.5.15) whereC > 0 is a constant depending on. Proof of Lemma 4.5.3. Using the fundamental theorem of calculus, we obtain k@ t vk 2 3 . 1 +TP ( sup t2[0;T ] E(t)); 201 sincek@ 2 t vk 2 3 E(t). Next we estimate the termkvk 2 4 . From Lemma 4.10.2 it follows that kvk 2 4 .k@ t divk 2 3 | {z } K 1 +k@ t curlk 2 3 | {z } K 2 +k@ t @Nk 2 H 2:5 (@ ) | {z } K 3 +k@ t k 2 0 | {z } K 4 : Note that div@ t =@ t J +( s r a s r )@ t r ; s . From the Hölder and Sobolev inequalities and the fundamental theorem of calculus it follows that K 1 .k@ t Jk 2 3 +k(aI 3 )@ t Dk 2 3 .k@ t Jk 2 3 +kaI 3 k 2 L 1k@ t Dk 2 3 +kaI 3 k 2 3 k@ t Dk 2 L 1 . 1 +TP ( sup t2[0;T ] E(t)); (4.5.16) sincekaI 3 k 2 .T andk@ 2 t Jk 2 3 +kvk 2 4 E(t). The termK 2 is already estimated in Lemma 4.3.1. For the termK 3 , we appeal to Lemmas 4.10.3–4.10.4 and 4.4.1, we obtain K 3 .k@ t @ 4 Nk H 0:5 (@ ) +k@ t @Nk L 2 (@ ) .k@ t @ 4 k 2 0 +k@ t @ 3 divk 2 0 +k@ t k 2 2 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); where we used (4.5.16) in the last inequality. We bound the termK 4 using the fundamental theorem of calculus as K 4 . 1 +TP ( sup t2[0;T ] E(t)): Consequently, we conclude the estimate kvk 2 4 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)): (4.5.17) 202 Finally, we derive the estimate ofk@ t Jk 2 4 . Applying@ t D 3 to (4.5.4), whereD = (@ 1 ;@ 2 ;@ 3 ), we obtain 2 0 J 3 a 3 3 @ t D 3 J; 3 = 2@ t D 3 ( 0 J 3 a 3 3 )J; 3 2J 3 0 a 3 @ t D 3 J; 2@ t D 3 ( 0 J 3 a 3 )J; +D 3 @ 3 t 3 2J 2 @ t D 3 a 3 3 2a 3 3 @ t D 3 J 2 + 1 X l=0 3 X m=0 1l+m3 c l;m @ l t D m ( 0 J 3 a 3 3 )@ 1l t D 3m J; 3 + 1 X l=0 3 X m=0 1l+m3 c l;m @ l t D m ( 0 J 3 a 3 )@ 1l t D 3m J; + 1 X l=0 3 X m=0 1l+m3 c l;m @ l t D m a 3 3 @ 1l t D 3m J 2 =:I 1 +I 2 +I 3 +I 4 +I 5 +I 6 +I 7 +I 8 +I 9 : The highest order term inI 1 scales like 0 J; 3 D 3 @v, which is estimated using Hölder and Sobolev inequal- ities and (4.5.17). The rest of the terms inI 1 are of lower order which can be estimated using the Hölder and Sobolev inequalities and the fundamental theorem of calculus. Thus, we have kI 1 k 2 0 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)): For the termI 2 , we have kI 2 k 2 0 .ka 3 k 2 L 1k@ t Jk 2 4 . 1 +TP ( sup t2[0;T ] E(t)); sincekaI 3 k 2 .T . The highest order term inI 3 is of the form 0 a 3 J; @ t D 3 J which can be estimated using the fundamental theorem of calculus, sincek@ 2 t Jk 2 3 E(t). The rest of the terms inI 3 are of lower order which can be estimated using the Hölder and Sobolev inequalities and the fundamental theorem of calculus. Thus, we have kI 3 k 2 0 . 1 +TP ( sup t2[0;T ] E(t)): 203 The termI 4 is already estimated in Lemma 4.5.2, while the termI 5 is estimated using Hölder and Sobolev inequalities and (4.5.17). The termI 6 is estimated using the fundamental theorem of calculus, since k@ 2 t Jk 2 3 E(t). The termsI 7 ,I 8 , andI 9 are of lower order which can be estimated using the Hölder and Sobolev inequalities and the fundamental theorem of calculus as kI 7 k 2 0 +kI 8 k 2 0 +kI 9 k 2 0 . 1 +TP ( sup t2[0;T ] E(t)): Consequently, we conclude k@ t J; 3 k 2 3 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)): The estimate ofk@ t @Jk 2 3 follows analogously using the above arguments. The proof of the lemma is thus completed by combining (4.5.17). 4.6 TheimprovedLagrangianowmapandJacobianestimates The next two lemmas provide the estimates of the improved regularity of solutionsJ and. Lemma4.6.1. For2 (0; 1), we have sup t2[0;T ] k @ 3 J(t)k 2 1:5 +k @ 3 (t)k 2 1:5 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); (4.6.1) whereC > 0 is a constant depending on. 204 Proof of Lemma 4.6.1. From Lemma 4.10.2 it follows that k @ 3 k 2 1:5 .k div @ 3 k 2 0:5 | {z } J 1 +k curl @ 3 k 2 0:5 | {z } J 2 +j @ 4 Nj 2 0 | {z } J 3 +k @ 3 k 2 0 | {z } J 4 : Note that from the Leibniz rule it follows that @ 3 div = @ 3 J + ( s r a s r ) @ 3 r ; s + P 2 l=1 c l @ l a s r @ 3l r ; s . Thus, we have J 1 .k @ 3 Jk 2 0:5 | {z } J 11 +k(aI 3 ) @ 3 Dk 2 0:5 | {z } J 12 +k @a @ 2 Dk 2 0:5 | {z } J 13 +k @ 2 a @Dk 2 0:5 | {z } J 14 : The termJ 11 is estimated using the fundamental theorem of calculus as J 11 . 1 +TP ( sup t2[0;T ] E(t)); sincek@ t Jk 2 4 E(t). We bound the termJ 12 using the multiplicative Sobolev inequality (4.3.10) as J 12 .kaI 3 k 2 1:5+ k @ 3 Dk 2 0:5 .TP ( sup t2[0;T ] E(t)); sincekaI 3 k 2 .T . For the termJ 13 , we proceed analogously asJ 12 , leading to J 13 .k @ak 2 1:5+ k @ 2 Dk 2 0:5 . 1 +TP ( sup t2[0;T ] E(t)); sincekvk 2 4 E(t). The termJ 14 is estimated using (4.3.11) as J 14 .k @ 2 ak 2 1 k @Dk 2 1 . 1 +TP ( sup t2[0;T ] E(t)): 205 The termJ 2 is already established in Lemma 4.3.1, while the termJ 3 is already estimated in Lemma 4.4.1. We bound the termJ 4 using the fundamental theorem of calculus as J 4 . 1 +TP ( sup t2[0;T ] E(t)): Consequently, we conclude the estimate k @ 3 k 2 1:5 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)): (4.6.2) It remains to estimate the termk @ 3 Jk 2 1:5 . Applying @ 3 to (4.5.4), we obtain 2 @ 3 J; 3 = 2(a 3 3 ) 1 a 3 @ 3 J; + 1 0 (a 3 3 ) 1 J 3 @ 2 t @ 3 3 + 3 X l=0 @ l t a 3 3 @ 3l J 2 + 3 X l=1 c l @ l ( 0 J 3 a 3 3 ) @ 3l J; 3 + 3 X l=1 c l @ l ( 0 J 3 a 3 ) @ 3l J; : (4.6.3) Now we estimate the terms on the right hand side of (4.6.3) inH 0:5 . Using (4.3.10), we obtain k(a 3 3 ) 1 a 3 @ 3 J; k 2 0:5 .k(a 3 3 ) 1 a 3 k 2 2 k @ 3 @Jk 2 0:5 .ka 3 k 2 2 k @ 3 Jk 2 1:5 .TP ( sup t2[0;T ] E(t)); sincekaI 3 k 2 .T , where we used the fact thatH 2 is an algebra. Similarly, we have k 1 0 (a 3 3 ) 1 J 3 @ 2 t @ 3 3 k 2 0:5 .k 1 0 (a 3 3 ) 1 J 3 k 2 2 k@ 2 t @ 3 k 2 0:5 .k@ 2 t @ 3 k 2 0:5 : From Lemmas 4.10.1 and 4.4.2 it follows that k@ 2 t @ 3 3 k 2 0:5 . Z w(j@ 2 t @ 3 Dj 2 +j@ 2 t @ 3 j 2 ).k d @ 2 t @ 3 Dk 2 0 +k@ t vk 2 3 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); 206 where we have crucially used that 2 d = (1 +w) 2 (1 +w) =w +w 2 w: The rest of the terms on the right hand side of (4.6.3) are of lower order which can estimated analogously using the above arguments. Thus, we infer that k @ 3 J; 3 k 2 0:5 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)): Similar arguments show that k @ 3 @Jk 2 0:5 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)): Therefore, we complete the proof of the Lemma by combining (4.6.2). Lemma4.6.2. For2 (0; 1), we have sup t2[0;T ] 4 X l=2 k @ 4l @ l t (t)k 2 1:5 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)) (4.6.4) whereC > 0 is a constant depending on. The proof of Lemma 4.6.2 is analogously to the proof of Lemma 4.6.1 by using Lemma 4.4.3 and thus we omit the details. 4.7 Theimprovedcurlestimates 207 Lemma4.7.1. For2 (0; 1), we have sup t2[0;T ] k @ 3 curl v(t)k 2 0:5 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); (4.7.1) whereC > 0 is a constant depending on. Proof of Lemma 4.7.1. Applying @ 3 to (4.3.4), we obtain k @ 3 curl vk 2 0:5 .k @ 3 curlu 0 k 2 0:5 | {z } I 1 +k Z t 0 @ 3 Q 0 (A(t 0 );Dv(t 0 ))dt 0 k 2 0:5 | {z } I 2 : The termI 1 is bounded byM 0 sincek @ 3 curlv(0)k 0:5 =k @ 3 curlu 0 k 0:5 . From (4.3.3), we infer that the highest order term inI 2 can be written as kji [ @ 3 v i ; s A m j A s r v r ; m + @ 3 v r ; m A m j A s r v i ; s ]: Using integration by parts in time, where we drop the indices for simplicity, we obtain k Z t 0 @ 3 DvDvAAdt 0 k 2 0:5 .k Z t 0 @ 3 D(DvAA) t dt 0 k 2 0:5 | {z } I 21 +k @ 3 DDvAAk 2 0:5 (t) | {z } I 22 : For the termI 21 , we appeal to the multiplicative Sobolev inequality (4.3.10), obtaining I 21 . Z T 0 k @ 3 Dk 2 0:5 k(DvAA) t k 2 2 .TP ( sup t2[0;T ] E(t)): Similarly, we bound the termI 32 as I 22 .k @ 3 Dk 2 0:5 kDvAAk 2 2 .C + sup t2[0;T ] E(t) +C TP ( sup t2[0;T ] E(t)); 208 where we appealed to Lemma 4.6.1 in the last inequality. The rest of the terms inI 2 are of lower order which can be estimated using the Hölder and Sobolev inequalities and the fundamental theorem of calcu- lus. Consequently, we conclude the proof of the lemma. 4.8 ConcludingtheproofofProposition4.2.2 We sum the estimates in Lemmas 4.3.1, 4.4.1– 4.4.3, 4.5.1–4.5.3, 4.6.1–4.6.2, and 4.7.1, thus completing the proof of Proposition 4.2.2. The proof of Theorem 4.2.1 is thus concluded. 4.9 Thegeneralcase > 1 In this section we prove Theorem 4.2.1 for general > 1. The proof is similar to those in Sections 4.3–4.8, thus we only outline the necessary modications and omit further details. The Lagrangian curl of (4.2.10) equals zero for any > 1. Consequently, similar estimates in Section 4.3 can be carried out. From (4.2.15), we may assume that J 1 ( u J ); 3 =J 1 ( u ; 3 J u J 1 J; 3 ) 4 ; on [0;T ] . Thus, using the decomposition 0 = d + u (see (4.2.14)–(4.2.15)), similar energy estimates in Section 4.4 hold for > 1. The normal derivative estimates in Section 4.5 relies on the equation (4.2.11). With the additional term ( 0 J 1 ) 2 , we get the same normal derivative estimates using the energy estimates. From (4.2.14) it is readily checked that d (x) w(x) 2 ; x2 ; 209 from where we obtain kFk 2 0:5 . Z w(jDFj 2 +jFj 2 ).k 2 d DFk 2 0 +kFk 2 0 : Therefore, the improved regularity in Section 4.6 can also be carried out using the above Hardy-type inequality. In Section 4.7, the improved curl estimates hold for > 1. Consequently, we conclude the proof of Proposition 4.2.2 and thus completing the proof of Theorem 4.2.1. 4.10 Appendix:Hardy-Sobolevembeddings,traceestimates,andHodge- typebounds In this section we provide some auxiliary results. We rst recall the following Hardy type inequality (cf. [86]). Lemma4.10.1 (Hardy inequality). For any givena> 0 and nonnegative integerba=2, we have kFk 2 b a 2 C b X k=0 Z w(x) a jD k F (x)j 2 dx; for some constantC > 0 depending ona,b, and . In particular, letp = 1; 2, then kFk 2 1 p 2 C Z w(x) p jDF (x)j 2 +jF (x)j 2 dx; for some constantC > 0 depending on andp. The following lemma provides a Hodge type elliptic estimates (cf. [139]). 210 Lemma4.10.2 (Hodge type estimate). LetF be a vector eld over ands 1. Then we have kFk s C kFk 0 +k divFk s1 +k curlFk s1 +k @FNk H s1:5 (@ ) ; (4.10.1) for some constantC > 0, whereN denotes the outward unit normal to@ . The estimate (4.10.1) follows from the well-known identityF = curl curlFr divF and the analysis of second order elliptic operators. We recall from [1] the following Sobolev trace theorem. Lemma4.10.3. LetF2H s ( ) wheres> 1 2 . Then we have kFk 2 H s 1 2 (@ ) CkFk 2 s ; for some constantC > 0. The next lemma provide the normal trace estimate which is needed in the elliptic estimates (cf. [1, 32]). Lemma4.10.4 (Normal trace). LetF2L 2 ( ) and divF2L 2 ( ). Then k @FNk 2 H 0:5 (@ ) C k @Fk 2 0 +k divFk 2 0 ; for some constantC > 0, whereN denotes the outward unit normal vector to@ . We recall from [39] the following duality inequality. Lemma4.10.5 (Duality estimate). LetF2H 0:5 ( ). Then we have k @Fk 2 0:5 CkFk 2 0:5 ; for some constantC > 0. 211 Chapter5 OnthelocalexistenceofsolutionstotheNavier-Stokes-wavesystem withafreeinterface 5.1 Introduction The purpose of this paper is to establish the local-in-time existence of solutions for the free boundary uid- structure interaction model with low regularity assumptions on the initial data. The model describes the interaction problem between an elastic structure and a viscous compressible uid in which the structure is immersed. Mathematically, the dynamics of the uid are captured by the compressible Navier-Stokes equations in the velocity and density variables (u;), while the elastic dynamics are described by a second- order elasticity equation (we replace it with a wave equation to simplify the presentation) in the vector variables (w;w t ) representing the displacement and velocity of the structure. The interaction between the structure and the uid is mathematically described by velocity and stress matching boundary conditions at the moving interface separating the solid and uid regions. Since the interface position evolves with time and is not known a priori, this is a free-boundary problem. Important features of the problem are a mis- match between parabolic and hyperbolic regularity and the complexity of the stress-matching condition on the free boundary. 212 The well-posedness and local-in-time existence were rst obtained in [42, 43] in 2005, where the au- thors used the Lagrangian coordinate system to x the domain and the Tychono xed point theorem to construct a local-in-time solution given an initial uid velocityu 0 2H 5 and structural velocityw 1 2H 3 . In [91, 92], the authors established a priori estimates for the local existence of solutions using direct esti- mates givenu 0 2H 3 for the initial velocity andw 1 2H 5=2+r , wherer2 (0; ( p 2 1)=2), for the initial structural velocity. A key ingredient in obtaining the result was the hidden regularity trace theorem for the wave equations established in [107, 26, 102, 103, 130, 140]. A wealth of literature on wave-heat coupled systems on a non-moving domain was instrumental in further understanding the heat-wave interaction phenomena (cf. [7, 8, 9, 50, 20, 21, 94, 95, 96, 104]). Recently, a sharp regularity result for the case when the initial domain is a at channel was obtained by Raymond and Vanninathan [129]. The authors study the system in the Lagrangian coordinate setting and obtain local-in-time solutions for the 3D model when the initial velocityu 0 2H 1+ and the initial structural velocityw 1 2H 1=2++ , where2 (1=2; 1) and > 0. More recently, Boulakia, Guerrero, and Takahashi obtained in [24] a unique local-in-time solution given initial datau 0 2H 2 andw 1 2H 9=8 for the case of a general domain. Concerning compressible uids, the global existence of weak solutions for the interaction with a rigid structure was obtained in [48] for the density lawp = with 2 and in [57] with >d=2. In [17], the authors obtained the existence and uniqueness for the initial density 0 belonging toH 3 , the velocityu 0 inH 4 , and the structure displacement and velocity (w;w t ) inH 3 H 2 . A similar result was later obtained by Kukavica and Tuaha [93] with less regular initial data ( 0 ;u 0 ;w 1 )2H 3=2+r H 3 H 3=2+r , where r2 (0; ( p 2 1)=2). In [18], the existence of a regular global solution is proved for small initial data. In a recent work [19], the authors proved the existence of a unique local-in-time strong solution of the interaction problem between a compressible uid and elastic structure, which is modeled by the Saint- Venant Kirchho system. For some other works on uid-structure models, cf. [6, 12, 27, 29, 30, 28, 48, 58, 65, 63, 64, 74, 75, 82, 87, 89, 104, 105, 112, 113, 114, 118, 119, 120, 135]. 213 In this paper, we provide a natural proof of the existence of a unique local-in-time solution to the sys- tem under a low regularity assumptionsu 0 2 H 2+ andw 1 2 H 1:5+ , for some constant2 (0; 1=2), in the case of the at initial conguration. Our proof relies on a maximal regularity type theorem for the nonhomegeneous linear parabolic problem with Neumann type conditions on the uid-structure interface, in addition to the hidden regularity theorems (cf. Lemma 5.3.3–5.3.4) for the wave equation. The time reg- ularity of the solution is obtained using the energy estimates, which, combined with the elliptic regularity, yield the spatial regularity of the solutions. An essential ingredient of the proof of the main results is a trace inequality kuk H ((0;T );L 2 ( c)) .kuk 1=(2r+1) L 2 ((0;T );H r ( f )) kuk 2r=(2r+1) H 2r=(2r1) ((0;T );L 2 ( f )) +kuk L 2 ((0;T );H r ( f )) ; for functions which are Sobolev in the time variable and square integrable on the boundary (cf. Lemma 5.3.1 and (5.3.7) below). This is used essentially in the proof of the existence for the nonlinear parabolic-wave system, Theorem 5.5.2, and in the proof of the main result, Theorem 5.2.1. The construction of a unique solution for the uid-structure problem is obtained via the Banach xed point theorem. The scheme involves solving the nonlinear parabolic-wave system with the variable coe- cients treated as given forcing perturbations. The iteration scheme then requires solving the wave equation with Dirichlet data using the integrated velocity matching conditions and appealing to the hidden trace regularity Lemma 5.3.4 for the normal derivative. Note that the conguration we adopt, (5.2.8) with the periodic boundary conditions in they 1 andy 2 directions, is needed only in Lemma 5.3.4. In addition, the solution in each iteration step is used to prescribe new variable coecients for the next iteration step. The contracting property of the Navier-Stokes-wave system is maintained by taking a suciently short time to ensure closeness of the Jacobian and the inverse matrix of the ow map to their initial states. 214 The paper is organized as follows. In Section 5.2 we introduce the uid-structure model and state our main result. In Section 5.3 we provide the trace inequality, interpolation, and hidden regularity lemmas. Section 5.4 provides the maximal regularity for the nonhomogeneous parabolic problem, which is needed in the proof of local existence to the nonlinear parabolic-wave system in Section 5.5. In Section 5.6, we prove the main result, Theorem 5.2.1, using the local existence obtained in Section 5.5 and construct a unique solution using the Banach xed point theorem. 5.2 Themodelandmainresults We consider the uid-structure problem for a free boundary system involving the motion of an elastic body immersed in a compressible uid. Let f (t) and e (t) be the domains occupied by the uid and the solid body at time t inR 3 , whose common boundary is denoted by c (t). The uid is modeled by the compressible Navier-Stokes equations, which in Eulerian coordinates reads t + div(u) = 0 in (0;T ) f (t); (5.2.1) u t +(ur)u div(ru + (ru) T )r divu +rp = 0 in (0;T ) f (t); (5.2.2) where =(t;x)2R + is the density,u =u(t;x)2R 3 is the velocity,p =p(t;x)2R + is the pressure, and; > 0 are physical constants. (We remark that the condition for and can be relaxed to > 0 and 3 + 2 > 0.) The system (5.2.1)–(5.2.2) is dened on f (t) which set to f = f (0) and evolves in time. The dynamics of the coupling between the compressible uid and the elastic body are best described 215 in the Lagrangian coordinates. Namely, we introduce the Lagrangian ow map(t;): f ! f (t) and rewrite the system (5.2.1)–(5.2.2) as R t Ra kj @ k v j = 0 in (0;T ) f ; (5.2.3) @ t v j Ra kl @ k (a ml @ m v j +a mj @ m v l )Ra kj @ k (a mi @ m v i ) +Ra kj @ k (q(R 1 )) = 0 in (0;T ) f ; (5.2.4) for j = 1; 2; 3, where R(t;x) = 1 (t;(t;x)) is the reciprocal of the Lagrangian density, v(t;x) = u(t;(t;x)) is the Lagrangian velocity,a(t;x) = (r(t;x)) 1 is the inverse matrix of the ow map andq is a given function of the density. The system (5.2.3)–(5.2.4) is expressed in terms of Lagrangian coordinates and posed in a xed domain f . On the other hand, the elastic body is modeled by the wave equation in Lagrangian coordinates, which is posed in a xed domain e as w tt w = 0 in (0;T ) e ; (5.2.5) where (w;w t ) are the displacement and the structure velocity. The interaction boundary conditions are the velocity and stress matching conditions, which are formulated in Lagrangian coordinates over the xed common boundary c = c (0) as v j =@ t w j on (0;T ) c ; (5.2.6) @ k w j k =Ja kl (a ml @ m v j +a mj @ m v l ) k +Ja kj a mi @ m v i k Ja kj q k on (0;T ) c ; (5.2.7) 216 forj = 1; 2; 3, whereJ(t;x) = det(r(t;x)) is the Jacobian and is the unit normal vector to c , which is outward with respect to e . In the present paper, we consider the reference congurations = f [ e [ c , f , and e given by =fy = (y 1 ;y 2 ;y 3 )2R 3 : (y 1 ;y 2 )2T 2 ; 0<y 3 <L 3 g; f =fy = (y 1 ;y 2 ;y 3 )2R 3 : (y 1 ;y 2 )2T 2 ; 0<y 3 <L 1 or L 2 <y 3 <L 3 g; e =fy = (y 1 ;y 2 ;y 3 )2R 3 : (y 1 ;y 2 )2T 2 ;L 1 <y 3 <L 2 g; (5.2.8) where 0 < L 1 < L 2 < L 3 andT 2 is the two-dimensional torus with the side 2. Thus, the common boundary is expressed as c =f(y 1 ;y 2 )2R 2 : (y 1 ;y 2 ;y 3 )2 ;y 3 =L 1 or y 3 =L 2 g; while the outer boundary is represented by f =fy2 :y 3 = 0g[fy2 :y 3 =L 3 g: To close the system, we impose the homogeneous Dirichlet boundary condition v = 0 on (0;T ) f ; (5.2.9) on the outer boundary f and the periodic boundary conditions forw,, andu on the lateral boundary, i.e., w(t;);(t;(t;));u(t;(t;)) periodic in they 1 andy 2 directions: (5.2.10) 217 Note that the inverse matrix of the ow mapa satises the ODE system @a @t (t;x) =a(t;x)rv(t;x)a(t;x) in (0;T ) f ; (5.2.11) a(0) =I 3 in f ; (5.2.12) while the Jacobian satises the ODE system @J @t (t;x) =J(t;x)a kj (t;x)@ k v j (t;x) in (0;T ) f ; (5.2.13) J(0) = 1 in f : (5.2.14) The initial data of the system (5.2.3)–(5.2.5) is given as (R;v;w;w t )(0) = (R 0 ;v 0 ;w 0 ;w 1 ) in f f e e ; (R 0 ;v 0 ;w 0 ;w 1 ) periodic in they 1 andy 2 directions; (5.2.15) wherew 0 = 0. Denote H r;s ((0;T ) f ) =H r ((0;T );L 2 ( f ))\L 2 ((0;T );H s ( f )); with the corresponding norm kfk 2 H r;s ((0;T ) f ) =kfk 2 H r ((0;T );L 2 ( f )) +kfk 2 L 2 ((0;T );H s ( f )) : For simplicity of notation, we frequently write K s ((0;T ) f ) =H s=2;s ((0;T ) f ) 218 and K s c =H s=2;s ( c ); where the dependence on time is understood from the context. Our main result states the local-in-time existence of solution to the system (5.2.3)–(5.2.5) with the mixed boundary conditions (5.2.6)–(5.2.10) and the initial data (5.2.15). Theorem 5.2.1. Lets2 (2; 2 + 0 ] for some constant 0 2 (0; 1=2). Assume thatR 0 2 H s ( f );R 1 0 2 H s ( f ),v 0 2H s ( f ),w 1 2H s1=2 ( e ), andw 0 = 0, with the compatibility conditions w 1j =v 0j on c ; v 0j = 0 on f ; (@ k v 0j +@ j v 0k ) k +@ i v 0i j q(R 1 0 ) j = 0 on c ; @ k (@ k v 0j +@ j v 0k ) +@ j @ k v 0k @ k (q(R 1 0 )) = 0 on f ; forj = 1; 2; 3. Then the system (5.2.3)–(5.2.5) with the boundary conditions (5.2.6)–(5.2.10) and the initial data (5.2.15) admits a unique solution v2K s+1 ((0;T ) f ) R2H 1 ((0;T );H s ( f )) w2C([0;T ];H s+1=4 0 ( e )) w t 2C([0;T ];H s3=4 0 ( e )); 219 forsomeconstantT > 0,wherethecorrespondingnormsareboundedbyafunctionofthenormsoftheinitial data. Remark 5.2.2. We assumev 0 2 H s ( f ) wheres2 (2; 2 + 0 ] for some 0 2 (0; 1=2) since the elliptic regularity forkvk L 2 t H 4 x in (5.4.28) requires thatR 1 2 L 1 H 2 ((0;T ) f ). From the density equation (5.2.3), we deduce that the regularity for the initial velocity must be at least inH 2 ( f ), showing the opti- mality of the ranges 2. It would be interesting to nd whether the statement of the theorem holds for the borderline cases = 2. The proof of the theorem is given in Section 5.6 below. For simplicity, we present the proof for the pressure lawq(R) =R, noting that the general case follows completely analogously. 5.3 Space-timetrace,interpolation,andhiddenregularityinequalities In this section, we provide four auxiliary results needed in the xed point arguments. The rst lemma provides an estimate for the trace in a space-time norm and is an essential ingredient when constructing solutions to the nonlinear parabolic-wave system in Section 5.5 below. Lemma 5.3.1. Letr > 1=2 and 0. Ifu2 L 2 ((1;1);H r ( f ))\H 2r=(2r1) ((1;1);L 2 ( f )), thenu2H ((1;1);L 2 ( c )), and for all2 (0; 1], we have the inequality kuk H ((1;1);L 2 (c)) kuk H 2r=(2r1) ((1;1);L 2 ( f )) +C 12r kuk L 2 ((1;1);H r ( f )) ; (5.3.1) whereC > 0 is a constant. The above lemma was proven in [62] 220 Proof of Lemma 5.3.1. It is sucient to prove the inequality foru2C 1 0 (RR 3 ) with the trace taken on the set = (t;x 1 ;x 2 ;x 3 )2RR 3 :x 3 = 0 : the general case is settled by the partition of unity and straightening of the boundary. Since it should be clear from the context, we usually do not distinguish in notation between a function and its trace. Denoting by ^ u the Fourier transform ofu with respect to (t;x 1 ;x 2 ;x 3 ), we have kuk 2 H ((0;T );L 2 ()) . Z 1 1 Z 1 1 Z 1 1 (1 + 2 ) Z 1 1 ^ u( 1 ; 2 ; 3 ;)d 3 2 dd 1 d 2 : Denote by = 2r 1 2 (5.3.2) the quotient between the exponentsr and 2r=(2r 1) in (5.3.1). Then, with > 0 to be determined below, we have kuk 2 H ((0;T );L 2 ()) . Z R 3 (1 + 2 ) Z 1 1 ^ u( 1 ; 2 ; 3 ;)d 3 2 dd 1 d 2 . Z R 3 (1 + 2 ) Z 1 1 (1 + ( 2 1 + 2 2 ) + 2 2 3 + 2 ) =2 (1 + ( 2 1 + 2 2 ) + 2 2 3 + 2 ) =2 j^ ujd 3 ! 2 dd 1 d 2 . Z R 3 (1 + 2 ) Z 1 1 1 + ( 2 1 + 2 2 ) + 2 2 3 + 2 j^ uj 2 d 3 Z 1 1 d 3 (1 + ( 2 1 + 2 2 ) + 2 2 3 + 2 ) ! dd 1 d 2 ; where we used the Cauchy-Schwarz inequality in 3 . Note that, using a substitution, Z 1 1 dx (A 2 + 2 x 2 ) . 1= A 1= 2 ; A;> 0; (5.3.3) 221 provided satises 2 > 1, which is by (5.3.2) equivalent to > 2r 1 : (5.3.4) Note that 2 > 1 implies 1= 2 < 0 for the exponent ofA in (5.3.3). Now we use (5.3.3) for the integral in 3 withA = (1 + ( 2 1 + 2 2 ) + 2 ) 1=2 , while noting that (1 + 2 ) A 1= 2 = (1 + 2 ) (1 + ( 2 1 + 2 2 ) + 2 ) 1=2 (1 + 2 ) +1=2 (1 + ( 2 1 + 2 2 ) + 2 2 3 + 2 ) +1=2 ; provided 1=2 , i.e., 2r 2r 1 : (5.3.5) Under the condition (5.3.5), we thus obtain kuk 2 H ((0;T );L 2 ( c)) . 1= Z RR 3 1 + ( 2 1 + 2 2 ) + 2 2 3 + 2 +1=2 j^ uj 2 d 3 d 2 d 1 d . 1= Z RR 3 1 + 2 ( 2 1 + 2 2 + 2 3 ) + 2 +1=2 j^ uj 2 d 3 d 2 d 1 d . 2 kuk 2 L 2 ((0;T );H +1=2 ( f )) + 1= kuk 2 H +1=2 ((0;T );L 2 ( f )) ; for all2 (0; 1]. Using (5.3.2), we get kuk 2 H ((0;T );L 2 ( c)) . 2 kuk 2 L 2 ((0;T );H r ( f )) + 2=(2r1) kuk 2 H 2r=(2r1) ((0;T );L 2 ( f )) ; (5.3.6) for all2 (0; 1]. Finally, note that = 2r=(2r1) satises (5.3.4)–(5.3.5) under the conditionr> 1=2. 222 Optimizing2 (0; 1] in (5.3.6) by using = kuk L 2 ((0;T );H r ( f )) kuk L 2 ((0;T );H r ( f )) +kuk H 2r=(2r1) ((0;T );L 2 ( f )) ! (2r1)=2r ; we obtain a trace inequality kuk H ((0;T );L 2 ( c)) .kuk 1=(2r+1) L 2 ((0;T );H r ( f )) kuk 2r=(2r+1) H 2r=(2r1) ((0;T );L 2 ( f )) +kuk L 2 ((0;T );H r ( f )) ; (5.3.7) which is a more explicit version of (5.3.1). The second lemma provides a space-time interpolation inequality needed in several places in Sec- tions 5.5 and 5.6 below. Lemma 5.3.2. Let; > 0. Ifu2 H ((1;1);L 2 ( f ))\L 2 ((1;1);H ( f )), then we have that u2H ((1;1);H ( f )) for all2 (0;) and2 (0;) such that + 1: In addition, for all2 (0; 1], we have the inequality kuk H ((1;1);H ( f )) kuk H ((1;1);L 2 ( f )) +C kuk L 2 ((1;1);H ( f )) ; whereC > 0 is a constant. 223 Proof of Lemma 5.3.2. Using a partition of unity, straightening of the boundary, and a Sobolev extension, it is sucient to prove the inequality in the case f =R 3 andu2C 1 0 (RR 3 ). Then, using the Parseval identity and the denition of the Sobolev, we only need to prove (1 +jj 2 )(1 +jj 2 )(1 +jj 2 ) +C (1 +jj 2 ); (5.3.8) for2R and2R n , where2 (0; 1]. Finally, (5.3.8) follows from the Young’s inequality. In the last part of this section, we address the regularity for the wave equation. We rst recall the hidden regularity result for the wave equation (cf. [107]) w tt w = 0 in (0;T ) e ; (5.3.9) w = on (0;T ) c ; (5.3.10) w periodic in they 1 andy 2 directions; (5.3.11) (w;w t )(0;) = (w 0 ;w 1 ): (5.3.12) Lemma5.3.3 ([107]). Assume that (w 0 ;w 1 )2H ( e )H 1 ( e ), where 1, and 2L 2 ((0;T );H ( c ))\H ((0;T );L 2 ( c )); 224 withthecompatibilityconditions j t=0 =w 0 j c and@ t j t=0 =w 1 j c . Thenthereexistsasolution (w;w t )2 C([0;T ];H ( e )H 1 ( e )) of (5.3.9)–(5.3.12), which satises the estimate kwk C([0;T ];H ( e)) +kw t k C([0;T ];H 1 ( e)) + @w @ H 1 ((0;T );L 2 (c)) + @w @ L 2 ((0;T );H 1 (c)) .kw 0 k H ( e) +kw 1 k H 1 ( e) +k k H ((0;T );L 2 (c)) +k k L 2 ((0;T );H (c)) ; where the implicit constant depends on e . In the nal lemma of this section, we recall an essential trace regularity result for the wave equation from [129]. Lemma5.3.4 ([129]). Assume that (w 0 ;w 1 )2H +2 ( e )H +1 ( e ), where 0< < 5=2, and 2L 2 ((0;T );H +2 ( c ))\H =2+1 ((0;T );H =2+1 ( c )); with the compatibility conditions j t=0 = w 0 j c and @ t j t=0 = w 1 j c . Then there exists a solutionw of (5.3.9)–(5.3.12) such that @w @ L 2 ((0;T );H +1 (c)) .kw 0 k H +2 ( e) +kw 1 k H +1 ( e) +k k L 2 ((0;T );H +2 (c)) +k k H =2+1 ((0;T );H =2+1 (c)) ; where the implicit constant depends on e . 5.4 Thenonhomogeneousparabolicproblem In this section, we consider the parabolic problem @ t uR div(ru + (ru) T )Rr divu =f in (0; 1) f ; (5.4.1) 225 with the nonhomogeneous boundary conditions and the initial data (@ k u j +@ j u k ) k +@ k u k j =h j on (0; 1) c ; (5.4.2) u = 0 on (0; 1) f ; (5.4.3) u periodic in they 1 andy 2 directions; (5.4.4) u(0) =u 0 in f ; (5.4.5) forj = 1; 2; 3. To state the maximal regularity for (5.4.1)–(5.4.5), we consider the homogeneous version when (5.4.2)–(5.4.5) is replaced by (@ k u j +@ j u k ) k +@ k u k j = 0 on (0; 1) c ; (5.4.6) u = 0 on (0; 1) f ; (5.4.7) u periodic in they 1 andy 2 directions; (5.4.8) u(0) = 0 in f ; (5.4.9) forj = 1; 2; 3. Lemma5.4.1. Assume thatf(0) = 0 and (R;R 1 )2 (L 1 t H 2 x ([0; 1] f )\H 1 t L 1 x ([0; 1] f )) 2 : (5.4.10) Then the parabolic problem (5.4.1) with the boundary conditions and the initial data (5.4.6)–(5.4.9) admits a solutionu satisfying kuk K 2 ((0;1) f ) .kfk K 0 ((0;1) f ) (5.4.11) 226 and kuk K 4 ((0;1) f ) .kfk K 2 ((0;1) f ) ; (5.4.12) where the implicit constants depend on the norms ofR andR 1 in (5.4.10). Proof. Analogously to [109, Theorem 3.2], the parabolic problem (5.4.1) admits a solutionu2K 2 ((0; 1) f ) iff2K 0 ((0; 1) f ) andu2K 4 ((0; 1) f ) iff2K 2 ((0; 1) f ). Below, the norm of dependence on time and space are understood as (0; 1) and f , unless stated otherwise. In the reminder of the proof we shall prove the regularity. Taking theL 2 -inner product of (5.4.1) withu, we arrive at 1 2 d dt Z f juj 2 Z f Ru j @ k (@ k u j +@ j u k ) Z f Ru j @ j @ k u k = Z f fu: (5.4.13) For the second and third terms on the left side of (5.4.13), we integrate by parts with respect to@ k and@ j to get Z f Ru j @ k (@ k u j +@ j u k ) = Z c Ru j (@ k u j +@ j u k ) k + Z f R@ k u j (@ k u j +@ j u k ) + Z f u j @ k R(@ k u j +@ j u k ) (5.4.14) and Z f Ru j @ j @ k u k = Z c Ru j @ k u k j + Z f R@ j u j @ k u k + Z f u j @ j R@ k u k : (5.4.15) 227 Inserting (5.4.14)–(5.4.15) into (5.4.13) and appealing to (5.4.6)–(5.4.7), we get 1 2 d dt Z f juj 2 + Z f R@ k u j (@ k u j +@ j u k ) + Z f R@ j u j @ k u k = Z f fu Z f u j @ k R(@ k u j +@ j u k ) Z f u j @ j R@ k u k .kfk 2 L 2 +kuk 2 L 2 +kuk L 4krRk L 4kruk L 2; (5.4.16) where the last inequality follows from the Hölder and Young inequalities. Note that for anyv2H 1 ( f ), using the Sobolev and Young’s inequalities, we have kvk L 4 ( f ) .kvk 3=4 H 1 ( f ) kvk 1=4 L 2 ( f ) .kvk H 1 ( f ) +C kvk L 2 ( f ) ; (5.4.17) for any2 (0; 1], whereC > 0 is a constant depending on. We integrate (5.4.16) in time from 0 tot and use (@ k u j +@ j u k )@ k u j = 1 2 3 X j;k=1 (@ k u j +@ j u k ) 2 ; (5.4.18) obtaining ku(t)k 2 L 2 + 3 X j;k=1 Z t 0 Z f R(@ k u j +@ j u k ) 2 + Z t 0 Z f Rj@ k u k j 2 .kfk 2 L 2 t L 2 x + Z t 0 kuk 2 L 2 + Z t 0 kuk 2 H 1 +C Z t 0 kuk L 2kuk H 1 .kfk 2 L 2 t L 2 x + ( +C ) Z t 0 kuk 2 H 1 +C ; Z t 0 kuk 2 L 2 ; (5.4.19) for any; 2 (0; 1], where we used (5.4.17) and the Young’s inequality. By Korn’s inequality, we get Z t 0 kuk 2 H 1 . 3 X j;k=1 Z t 0 Z f R(@ k u j +@ j u k ) 2 + Z t 0 kuk 2 L 2 : (5.4.20) 228 From (5.4.19)–(5.4.20) it follows that ku(t)k 2 L 2 +kuk 2 L 2 t H 1 x .kfk 2 L 2 t L 2 x + Z t 0 kuk 2 L 2 ; (5.4.21) by choosing; > 0 suciently small. By Gronwall’s inequality, we obtain ku(t)k 2 L 2 .kfk 2 L 2 t L 2 x ; (5.4.22) and then, after using (5.4.22) in (5.4.21), we arrive at kuk 2 L 2 t H 1 x .kfk 2 L 2 t L 2 x : (5.4.23) Next, we take theL 2 -inner product of (5.4.1) withu t , obtaining Z f ju t j 2 Z f Ru tj @ k (@ k u j +@ j u k ) Z f Ru tj @ j @ k u k = Z f fu t : (5.4.24) Then, proceeding as in (5.4.14)–(5.4.15), we get Z f Ru tj @ k (@ k u j +@ j u k ) = Z c Ru tj (@ k u j +@ j u k ) k + Z f R@ k u tj (@ k u j +@ j u k ) + Z f u tj @ k R(@ k u j +@ j u k ) (5.4.25) and Z f Ru tj @ j @ k u k = Z c Ru tj @ k u k j + Z f R@ j u tj @ k u k + Z f u tj @ j R@ k u k : (5.4.26) 229 Inserting (5.4.25)–(5.4.26) into (5.4.24) and appealing to (5.4.6)–(5.4.7), we arrive at Z f ju t j 2 + 2 d dt Z f R@ k u j (@ k u j +@ j u k ) + 2 d dt Z f R@ j u j @ k u k = Z f fu t + Z f u tj @ k R(@ k u j +@ j u k ) + Z f u tj @ j R@ k u k + 2 Z f R t @ k u j (@ k u j +@ j u k ) + 2 Z f R t @ j u j @ k u k .C kfk 2 L 2 +ku t k 2 L 2 +krRk L 4kruk L 4ku t k L 2 +kR t k L 1kruk 2 L 2 ; for any2 (0; 1], where we used Hölder’s and Young’s inequalities. Integrating in time from 0 tot and using the Young, Sobolev, and Korn inequalities with (5.4.17)–(5.4.18), we get ku t k 2 L 2 t L 2 x +ku(t)k 2 H 1 .C kfk 2 L 2 t L 2 x +ku t k 2 L 2 t L 2 x +ku(t)k 2 L 2 + Z t 0 ( kuk H 2 +C kuk H 1)ku t k L 2 + Z t 0 kR t k L 1kuk 2 H 1 .C kfk 2 L 2 t L 2 x + ( + + ~ C )ku t k 2 L 2 t L 2 x +ku(t)k 2 L 2 + kuk 2 L 2 t H 2 x +C ;~ kuk 2 L 2 t H 1 x + Z t 0 kR t k L 1kuk 2 H 1 ; (5.4.27) for any ; ; ~ 2 (0; 1], where we usedku(0)k H 1 = 0 in the last inequality by (5.4.9). For the space regularity, note thatu is the solution of the elliptic problem div(ru + (ru) T )r divu = @ t u R + f R in (0;T ) f ; (5.4.28) with the boundary conditions (@ k u j +@ j u k ) k +@ k u k j = 0 on (0; 1) c ; (5.4.29) u = 0 on (0; 1) f ; (5.4.30) 230 forj = 1; 2; 3. From the elliptic regularity for (5.4.28)–(5.4.30) it follows that kuk H 2. 1 R u t L 2 + 1 R f L 2 .ku t k L 2 +kfk L 2; (5.4.31) from where kuk L 2 t H 2 x .ku t k L 2 t L 2 x +kfk L 2 t L 2 x : (5.4.32) Combining (5.4.22)–(5.4.23), (5.4.27), and (5.4.32), we obtain ku t k 2 L 2 t L 2 x +ku(t)k 2 H 1 .kfk 2 L 2 t L 2 x + Z t 0 kR t k L 1kuk 2 H 1 ; (5.4.33) by taking; ; ~ > 0 suciently small. Using Gronwall’s inequality, we arrive at ku(t)k 2 H 1 Ckfk 2 L 2 t L 2 x exp C Z t kR t ()k L 1d Ckfk 2 L 2 t L 2 x ; and thus (5.4.33) implies kuk 2 H 1 t L 2 x .kfk 2 L 2 t L 2 x : (5.4.34) From (5.4.32) and (5.4.34) it follows that kuk K 2.kuk L 2 t H 2 x +kuk H 1 t L 2 x .kfk L 2 t L 2 x ; completing the proof of (5.4.11). 231 Dierentiating (5.4.1) in time and taking theL 2 -inner product withu t , we arrive at 1 2 d dt Z f ju t j 2 Z f Ru tj @ k (@ k u tj +@ j u tk ) Z f Ru tj @ j @ k u tk = Z f f t u t + Z f R t u tj @ k (@ k u j +@ j u k ) + Z f R t u tj @ j @ k u k : (5.4.35) We proceed as in (5.4.14)–(5.4.15) to obtain Z f Ru tj @ k (@ k u tj +@ j u tk ) = Z c Ru tj (@ k u tj +@ j u tk ) k + Z f u tj @ k R(@ k u tj +@ j u tk ) + Z f R@ k u tj (@ k u tj +@ j u tk ) (5.4.36) and Z f Ru tj @ j @ k u tk = Z c Ru tj @ k u tk j + Z f u tj @ j R@ k u tk + Z f R@ j u tj @ k u tk : (5.4.37) Inserting (5.4.36)–(5.4.37) into (5.4.35), we get 1 2 d dt Z f ju t j 2 + Z f R@ k u tj (@ k u tj +@ j u tk ) + Z f R@ j u tj @ k u tk .kf t k 2 L 2 +ku t k 2 L 2 +kR t k L 1ku t k L 2kuk H 2 +ku t k L 4krRk L 4ku t k H 1; where we used Young’s, Hölder’s, and Sobolev inequalities. Integrating in time from 0 tot and using the Young’s and Korn’s inequality and (5.4.17)–(5.4.18), we obtain ku t (t)k 2 L 2 +ku t k 2 L 2 t H 1 x .kfk 2 H 1 t L 2 x + Z t 0 kR t k L 1kuk 2 H 2 + Z t 0 kR t k L 1ku t k 2 L 2 + ( + C )ku t k 2 L 2 t H 1 x +C ; ku t k 2 L 2 t L 2 x ; 232 for any; 2 (0; 1], sinceku t (0)k L 2 =kf(0)k L 2 = 0. From the elliptic regularity (5.4.31) and (5.4.34) it follows that ku t (t)k 2 L 2 +ku t k 2 L 2 t H 1 x .kfk 2 H 1 t L 2 x +ku t k 2 L 2 t L 2 x + Z t 0 kR t k L 1ku t k 2 L 2 + Z t 0 kR t k L 1kfk 2 L 2 .kfk 2 H 1 t L 2 x +ku t k 2 L 2 t L 2 x + Z t 0 kR t k L 1ku t k 2 L 2 (5.4.38) by taking; > 0 suciently small, where we used kfk L 1 t L 2 x .kfk H 1 t L 2 x (5.4.39) in the last inequality. Appealing to Gronwall’s inequality, (5.4.38) implies ku t (t)k 2 L 2 .kfk 2 K 2 ; (5.4.40) and then, after using (5.4.40) in (5.4.38), we arrive at ku t k 2 L 2 t H 1 x .kfk 2 K 2 : Dierentiating (5.4.1) in time and taking theL 2 -inner product withu tt , we obtain Z f ju tt j 2 + 2 d dt Z f R@ k u tj (@ k u tj +@ j u tk ) + 2 d dt Z f R@ j u tj @ k u tk = Z f f t u tt Z f u ttj @ k R(@ k u tj +@ j u tk ) Z f u ttj @ j R@ k u tk + 2 Z f R t @ k u tj (@ k u tj +@ j u tk ) + 2 Z f R t @ j u tj @ k u tk + Z f u ttj R t @ k (@ k u j +@ j u k ) + Z f u ttj R t @ jk u k ; 233 where we integrated by parts in spatial variables. We proceed as in (5.4.35)–(5.4.38) to get ku tt k 2 L 2 t L 2 x +ku t (t)k 2 H 1 .C ~ kfk 2 H 1 t L 2 x +ku t (t)k 2 L 2 + Z t 0 ku t k H 2ku tt k L 2 +C ~ Z t 0 kR t k 2 L 1kfk 2 L 2 + ( C + ~ + ~ )ku tt k 2 L 2 t L 2 x +C ; ku t k 2 L 2 t H 1 x +C ~ Z t 0 (1 +kR t k 2 L 1)ku t k 2 H 1 ; for any; ; ~ 2 (0; 1], where we used the Young’s, Hölder, Sobolev, and Korn inequalities. Note thatu t is the solution of the elliptic problem div(ru t + (ru t ) T )r divu t =R 1 u tt +R 2 u t R t +R 1 f t R 2 R t f in (0; 1) f ; with the boundary conditions (@ k u tj +@ j u tk ) k +@ k u tk j = 0 in (0; 1) c ; u tj = 0 in (0; 1) f ; forj = 1; 2; 3. The elliptic regularity implies that ku t k H 2.ku tt k L 2 +ku t R t k L 2 +kf t k L 2 +kR t fk L 2 .ku tt k L 2 +ku t k L 2kR t k L 1 +kf t k L 2 +kR t k L 1kfk L 2; (5.4.41) where we used Hölder’s inequality. From (5.4.40)–(5.4.41), we obtain ku tt k 2 L 2 t L 2 x +ku t (t)k 2 H 1 .kfk 2 K 2 +ku t (t)k 2 L 2 + Z t 0 (1 +kR t k 2 L 1)ku t k 2 H 1 + Z t 0 kR t k 2 L 1kfk 2 L 2 .kfk 2 K 2 + Z t 0 (1 +kR t k 2 L 1)ku t k 2 H 1 ; 234 by taking; ; ~ > 0 suciently small, where we used (5.4.39). Appealing to Gronwall’s inequality, we arrive at ku t (t)k 2 H 1 .kfk 2 K 2 ; whence ku tt k 2 L 2 t L 2 x .kfk 2 K 2 : (5.4.42) From theH 4 regularity of the elliptic problem (5.4.28)–(5.4.30) and (5.4.41) it follows that kuk H 4.kR 1 u t k H 2 +kR 1 fk H 2 .ku tt k L 2 +kR t k L 1ku t k L 2 +kR t k L 1kfk L 2 +kf t k L 2 +kfk H 2; (5.4.43) sinceH 2 is an algebra. We combine (5.4.40) and (5.4.42)–(5.4.43) to get kuk K 4 =kuk L 2 t H 4 x +kuk H 2 t L 2 x .ku tt k L 2 t L 2 x +kR t k L 2 t L 1 x ku t k L 1 t L 2 x +kR t k L 2 t L 1 x kfk H 1 t L 2 x +kfk K 2.kfk K 2; completing the proof of (5.4.12). The following lemma provides a maximal regularity for the parabolic system (5.4.1)–(5.4.5). Lemma5.4.2. Lets2 (2; 2 + 0 ] for some 0 2 (0; 1=2). Assume the compatibility conditions h j (0) =(@ k u 0j +@ j u 0k ) k +@ k u 0k j on c ; (5.4.44) u 0j = 0 on f ; (5.4.45) 235 forj = 1; 2; 3. Suppose that (R;R 1 )2 (L 1 ([0; 1];H 2 ( f ))\H 1 ([0; 1];L 1 ( f ))) 2 (5.4.46) and (h;f)2K s1=2 ((0; 1) c )K s1 ((0; 1) f ): (5.4.47) Then the system (5.4.1)–(5.4.5) admits a solutionu satisfying kuk K s+1 ((0;1) f ) .khk K s1=2 ((0;1)c) +ku 0 k H s +kfk K s1 ((0;1) f ) ; (5.4.48) where the implicit constant depends on the norms ofR andR 1 in (5.4.46). Proof. From [109, Theorem 2.3] and the compatibility conditions (5.4.44)–(5.4.45) it follows that there exists v2K s+1 satisfying the boundary conditions and the initial data (5.4.2)–(5.4.5) with kvk K s+1.khk K s1=2 c +ku 0 k H s (5.4.49) and (@ t vR div(rv + (rv) T )Rr divv)j t=0 =f(0); (5.4.50) sinces> 1=2. Now we consider the homogeneous parabolic problem @ t wR div(rw + (rw) T )Rr divw =F in (0; 1) f ; (5.4.51) 236 with the homogeneous boundary conditions and the initial data (@ k w j +@ j w k ) k +@ k w k j = 0 on (0; 1) c ; (5.4.52) w = 0 on (0; 1) f ; (5.4.53) w periodic in they 1 andy 2 directions; (5.4.54) w(0) = 0 in f ; (5.4.55) forj = 1; 2; 3, where F =f +@ t vR div(rv + (rv) T )Rr divv: (5.4.56) Note that (5.4.50) implies thatF (0) = 0. By Lemma 5.4.1, there exists a solutionw to the system (5.4.51)– (5.4.56) satisfying kwk K 2.kFk K 0 (5.4.57) and kwk K 4.kFk K 2; (5.4.58) where the implicit constants depend on the norms ofR andR 1 in (5.4.46). From [109, Theorem 6.2] and (5.4.57)–(5.4.58) it follows that kwk K s+1.kFk K s1; (5.4.59) 237 sinces6= integer + 1=2 ands=26= integer. From (5.4.56), we get kFk K s1.kfk K s1 +kv t k K s1 +kRr 2 vk K s1: (5.4.60) For the second term on the right side of (5.4.60), we obtain kv t k K s1.kv t k L 2 t H s1 x +kv t k H (s1)=2 t L 2 x .kvk K s+1; where we used Lemma 5.3.2. Regarding the last term on the right side of (5.4.60), we appeal to the Hölder and Sobolev inequalities, yielding kRr 2 vk L 2 t H s1 x .kRk L 1 t H 2 x kr 2 vk L 2 t H s1 x .kvk K s+1 and kRr 2 vk H (s1)=2 t L 2 x .kRk W (s1)=2;4 t L 1 x kr 2 vk L 4 t L 2 x +kRk L 1 t H 2 x kr 2 vk H (s1)=2 t L 2 x .kRk H 1 t L 1 x kr 2 vk H (s1)=2 t L 2 x +kvk K s+1.kvk K s+1; (5.4.61) since 3=2 > > > < > > > > : 1 on [0; ~ T ); 0 on (2 ~ T; 1]: 239 We assume that there exists a constantC > 0 such thatk@ t ~ T (t)k L 1 (0;1) C ~ T 1 . We shall start with cuto solutions (R;w;v) on the time interval (0; 1) in order to apply Lemmas 5.3.1–5.3.2 with implicit con- stants independent of ~ T . Then we recover the solutions to (5.5.1)–(5.5.8) by restricting the cuto solutions to a small time interval (0; ~ T ). To provide the maximal regularity for the system (5.5.1)–(5.5.8), we state the following necessary a pri- ori density estimates. Lemma5.5.1. Lets2 (2; 2 + 0 ] where 0 2 (0; 1=2). Consider the ODE system R t R ~ T divv = 0 in (0; 1) f ; (5.5.9) R(0) =R 0 on f : (5.5.10) Assume that (R 0 ;R 1 0 ;v 0 )2 H s ( f )H s ( f )H s ( f ) andkvk K s+1 ((0;1) f ) M for some constant M 1. Then for ~ T2 (0; 1=M 4s+4 ), we have (i)kRk L 1 t L 1 x +kR 1 k L 1 t L 1 x +kRk L 1 t H s x +kR 1 k L 1 t H s x . 1 , (ii)kR 1 k H 1 t L 1 x +kRk H 1 t L 1 x +kRk H 1 t H 1 x . 1, (iii)kRk H 1 t H s x .M, wherethenormofdependenceis [0; 1] f . Weemphasizethattheimplicitconstantsintheaboveinequalities are independent ofM. Proof of Lemma 5.5.1. (i) The solution of the ODE system (5.5.9)–(5.5.10) reads R(t;x) =R 0 (x)e R t 0 ~ T () divv()d in [0; 1] f : (5.5.11) 240 From the Hölder and Sobolev inequalities it follows that kRk L 1 t L 1 x .kR 0 k H se R 2 ~ T 0 k ~ T () divv()k L 1d .C ~ T 1=2 M . 1 and kR 1 k L 1 t L 1 x .kR 1 0 k H se R 2 ~ T 0 k ~ T () divv()k L 1d .C ~ T 1=2 M . 1; for ~ T2 (0; 1=M 4s+4 ). Similarly, we have kRk L 1 t H s x .kR 0 k H ske R 2 ~ T 0 ~ T () divv()d k L 1 t H s x . 1 and kR 1 k L 1 t H s x .kR 1 0 k H ske R 2 ~ T 0 ~ T () divv()d k L 1 t H s x . 1: (ii) From (5.5.9), we use the Hölder and Sobolev inequalities to get k(R 1 ) t k L 2 t L 1 x .kR 2 R t k L 2 t L 1 x .kR ~ T k L 1 t L 1 x k divvk L 2 t L 1 x ((0;2 ~ T ) f ) .kvk L 2 t H s+1=2 x ((0;2 ~ T ) f ) ; (5.5.12) since s > 2. Recall that for constants 0 < r < r 0 and f 2 H r 0 , we have the Sobolev interpolation inequality kfk H r.kfk H r 0 + r=(rr 0 ) kfk L 2; (5.5.13) 241 for any2 (0; 1]. From (5.5.12)–(5.5.13) it follows that k(R 1 ) t k L 2 t L 1 x .kvk L 2 t H s+1 x ((0;2 ~ T ) f ) + 2s1 kvk L 2 t L 2 x ((0;2 ~ T ) f ) .M + 2s1 ~ T 1 2 M; (5.5.14) where we have used thatkfk L 1 t L 2 x .kfk H (s+1)=2 t L 2 x sinces> 2. Taking = 1=M in (5.5.14), we arrive at k(R 1 ) t k L 2 t L 1 x . 1 +M 2s+2 ~ T 1 2 . 1; for ~ T2 (0; 1=M 4s+4 ). Similarly, we have kR t k L 2 t L 1 x +kR t k L 2 t H 1 x . 1: (iii) From (5.5.9) and the Hölder inequality it follows that kR t k L 2 t H s x .kR ~ T k L 1 t H s x k divvk L 2 t H s x .kvk L 2 t H s+1 x .M: Therefore, we conclude the proof of (iii). The following theorem provides the local existence for the parabolic-wave system (5.5.1)–(5.5.8). 242 Theorem5.5.2. Suppose thats2 (2; 2 + 0 ] for some sucientlysmall 0 2 (0; 1=2). Assume the compat- ibility conditions w 1j =v 0j on c ; v 0j = 0 on f ; (@ k v 0j +@ j v 0k ) k +@ i v 0i j R 1 0 j @ k w 0j k = 0 on c ; @ k (@ k v 0j +@ j v 0k ) +@ j @ k v 0k @ k (R 1 0 ) = 0 on f : Suppose that the initial data satisfy (v 0 ;w 0 ;w 1 ;R 1 0 ;R 0 )2H s ( f )H s+1=2 ( e )H s1=2 ( e )H s ( f )H s ( f ) and the nonhomogeneous terms satisfy (f;h)2K s1 ((0; 1) f )K s1=2 ((0; 1) c ): Then there exists a unique solution to the system (5.5.1)–(5.5.8) satisfying (v;R;w;w t )2K s+1 ((0; ~ T ) f )H 1 ((0; ~ T );H s ( f )) C([0; ~ T ];H s+1=4 0 ( e ))C([0; ~ T ];H s3=4 0 ( e )); where ~ T > 0isaconstantandthecorrespondingnormsareboundedbyafunctionoftheinitialdataandthe nonhomogeneous terms. 243 We dene Z =fv2K s+1 ((0; 1) f ) :v(0) =v 0 in f ;v = 0 on (0; 1) f ; v periodic in they 1 andy 2 directions; andkvk K s+1 ((0;1) f ) Mg; (5.5.15) where M 1 is a constant to be determined below. For v2Z, we rst obtain the solution R using (5.5.11). Then, we solve the wave equation (5.5.3) forw with the boundary condition and the initial data w(t) =w(0) + Z t 0 ~ T ()v(;x)d on (0; 1) c ; (5.5.16) (w;w t )(0) = (w 0 ;w 1 ) in e : (5.5.17) With (R;w) constructed, we dene a mapping : v(2Z)7! v; where v is the solution of the nonhomogeneous parabolic problem @ t vR div(r v + (r v) T )Rr div v =fRrR 1 in (0; 1) f ; (5.5.18) with the boundary conditions and the initial data (@ k v j +@ j v k ) k +@ k v k j =@ k w j k +R 1 j +h j on (0; 1) c ; v = 0 on (0; 1) f ; v periodic in they 1 andy 2 directions; v(0) =v 0 in f ; (5.5.19) 244 forj = 1; 2; 3. We shall prove later in this section that is a contraction mapping and use the Banach xed-point theorem. 5.5.1 Uniformboundednessoftheiterativesequence In this section we show that the mapping is well-dened fromZ toZ, for some suciently large constant M 1. Let ~ T 2 (0; 1=M 4s+4 ) be a constant. We emphasize that the implicit constants below in this section are independent ofM and ~ T . From Lemmas 5.4.2 and 5.5.1 it follows that k vk K s+1 ((0;1) f ) . @w @ K s1=2 c +khk K s1=2 ((0;1) c) +kv 0 k H s +kfk K s1 ((0;1) f ) +kR 1 rRk K s1 ((0;1) f ) +kR 1 k K s1=2 ((0;1) c) : (5.5.20) Here and below, the norm of dependence on time and space are understood as (0; 1) and f , unless stated otherwise. For space component of the rst term on the right side of (5.5.20), we appeal to Lemma 5.3.4 and obtain @w @ L 2 t H s1=2 x ( c) .kw 0 k H s+1=2 ( e) +kw 1 k H s1=2 ( e) +kwk L 2 t H s+1=2 x ( c) +kwk H s=2+1=4 t H s=2+1=4 x ( c) .kwk L 2 t H s+1=2 x ( c) +k@ t wk H s=23=4 t H s=2+1=4 x ( c) + 1: (5.5.21) Regarding the rst term on the far right side, using (5.5.16) andw(0) = 0, we get kwk L 2 t H s+1=2 x ( c) . Z 2 ~ T 0 k Z t 0 ~ T vdk 2 H s+1=2 ( c) dt ! 1 2 + Z 1 2 ~ T k Z t 0 ~ T vdk 2 H s+1=2 ( c) dt 1 2 . ~ Tkvk L 2 t H s+1 x + ~ T 1 2 kvk L 2 t H s+1 x . ~ T 1 2 M; (5.5.22) 245 where we used the Cauchy-Schwarz and trace inequalities. For the second term on the far right side of (5.5.21), using the Sobolev interpolation and Young inequalities, we have k@ t wk H s=23=4 t H s=2+1=4 x ( c) .k@ t wk s=23=4 H 1 t H 1=2 x ( c) k@ t wk 7=4s=2 L 2 t H (2s+5)=2(72s) x ( c) .k@ t wk H 1 t H 1=2 x ( c) + (32s)=(72s) k@ t wk L 2 t H (2s+5)=2(72s) x ( c) :=I 1 +I 2 ; (5.5.23) for any2 (0; 1]. For the termI 1 , using the trace inequality and the Leibniz rule, we obtain I 1 .k@ t ~ T vk L 2 t H 1=2 x ( c) +k ~ T @ t vk L 2 t H 1=2 x ( c) +k ~ T vk L 2 t H 1=2 x ( c) .k@ t ~ T vk L 2 t H 1 x +k@ t vk L 2 t H 1 x +kvk L 2 t H 1 x =:I 11 +I 12 +I 13 : (5.5.24) The termI 11 is estimated using the Sobolev and Hölder inequalities as I 11 . ~ T 1 kvk L 2 t H 1 x ((0;2 ~ T ) f ) . ~ T 1=2 kvk L 1 t H 1 x ((0;2 ~ T ) f ) . ~ T 1=2 kvk H s=2 t H 1 x . ~ T 1=2 M; (5.5.25) where we used Lemma 5.3.2 in the last inequality. Let = ~ T 1=2 ~ in (5.5.25), where ~ 2 (0; 1] is a constant, we get I 11 . ~ M: (5.5.26) Similarly, the termsI 12 andI 13 are estimated as I 12 .k@ t vk L 2 t H s1 x .kvk H (s+1)=2 t L 2 x +kvk L 2 t H s+1 x . ~ M (5.5.27) 246 and I 13 .kvk L 2 t H s+1 x . ~ M: (5.5.28) For the termI 2 , we use the trace inequality and (5.5.13) to get I 2 . (32s)=(72s) k ~ T vk L 2 t H (2s+5)=2(72s)+1=2 x ((0;2 ~ T ) f ) . (32s)=(72s) kvk L 2 t H 6=(72s) x ((0;2 ~ T ) f ) . (32s)=(72s) ( 1 kvk L 2 t H s+1 x ((0;2 ~ T ) f ) + 6=(2s 2 5s1) 1 kvk L 2 t L 2 x ((0;2 ~ T ) f ) ); (5.5.29) for any 1 2 (0; 1]. Let 1 = (2s3)=(72s) ~ 1 in (5.5.29), where > 0 is as in (5.5.23) and ~ 1 2 (0; 1] is a constant. From (5.5.29) it follows that I 2 . ~ 1 M + (2s3)(2s 2 +5s+7)=(72s)(2s 2 5s1) C ~ 1 kvk L 2 t L 2 x ((0;2 ~ T ) f ) . ~ 1 M + ~ T (2s3)(2s 2 +5s+7)=2(72s)(2s 2 5s1) C ~ 1 ;~ kvk L 2 t L 2 x ((0;2 ~ T ) f ) : (5.5.30) For the termkvk L 2 t L 2 x ((0;2 ~ T ) f ) , using the fundamental theorem of calculus, we obtain kvk L 2 t L 2 x ((0;2 ~ T ) f ) .k Z t 0 @ t vdk L 2 t L 2 x ((0;2 ~ T ) f ) +kv 0 k L 2 t L 2 x ((0;2 ~ T ) f ) . ~ Tk@ t vk L 2 t L 2 x ((0;2 ~ T ) f ) + ~ T 1 2 . ~ T 3 2 M + ~ T 1 2 : (5.5.31) Inserting (5.5.31) to (5.5.30), we get I 2 . ~ 1 M + ~ T 3=2+(2s3)(2s 2 +5s+7)=2(72s)(2s 2 5s1) C ~ 1 ;~ M + ~ T 1=2+(2s3)(2s 2 +5s+7)=2(72s)(2s 2 5s1) C ~ 1 ;~ . ( ~ 1 + ~ T 1 C ~ 1 ;~ )M + ~ T 2 ; (5.5.32) 247 where 1 := 1 (s) = 3 2 + (2s 3)(2s 2 + 5s + 7) 2(7 2s)(2s 2 5s 1) (5.5.33) and 2 := 2 (s) = 1 2 (2s 3)(2s 2 + 5s + 7) 2(7 2s)(2s 2 5s 1) : (5.5.34) It is readily checked that 1 (2)> 0 and 2 (2) = 0. Combining (5.5.23)–(5.5.28) and (5.5.32), we arrive at k@ t wk H s=23=4 t H s=2+3=4 x ( c) . (~ + ~ 1 + ~ T 1 C ~ 1 ;~ )M + ~ T 2 : (5.5.35) From (5.5.21)–(5.5.22) and (5.5.35) it follows that @w @ L 2 t H s1=2 x ((0;1) c) . 1 + (~ + ~ 1 + ~ T 1 C ~ 1 ;~ + ~ T 1 2 )M + ~ T 2 : (5.5.36) For the time component of the rst term on the right side of (5.5.20), we use the trace inequality to get @w @ H s=21=4 t L 2 x ( c) .kr( ~ T v)k L 2 t L 2 x ( c) + @w @ L 2 t L 2 x ( c) .k ~ T vk L 2 t H 2 x + @w @ L 2 t H s1=2 x ( c) . 1 + (~ + ~ 1 + ~ T 1 C ~ 1 ;~ + ~ T 1=4 )M + ~ T 2 ; (5.5.37) since 2 0. Thus, we have shown that the mapping : v7! v is well-dened fromZ toZ, for some constantM 1. Now we x the constantsM 1. Using Lemma 5.3.3, we have the interior regularity estimate kwk C([0;1];H s+1=4 0 ( e)) +kw t k C([0;1];H s3=4 0 ( e)) .kw 0 k H s+1=4 0 +kw 1 k H s3=4 0 +kwk H s+1=4 0 ;s+1=4 0 ((0;1) c) : (5.5.43) For the last term on the right side, we appeal to Lemma 5.3.1 and (5.5.22), yielding kwk H s+1=4 0 ;s+1=4 0 ((0;1) c) .kvk H s3=4 0 t L 2 x ((0;1) c) +kwk L 2 t L 2 x ((0;1) c) +kwk L 2 t H s+1=4 0 x ((0;1) c) .kvk H (s+1)=2 t L 2 x +kvk L 2 t H (s+1)=(2s+5+4 0 ) x +kwk L 2 t H s+1=2 x ( c) . 1; (5.5.44) sinces 3=4 0 (s + 1)=2. From (5.5.43)–(5.5.44) it follows that kwk C([0;1];H s+1=4 0 ( e)) +kw t k C([0;1];H s3=4 0 ( e)) C; (5.5.45) for some constantC > 0. 250 5.5.2 Contractingproperty In this section we shall prove k(v 1 ) (v 2 )k K s+1 1 2 kv 1 v 2 k K s+1; v 1 ;v 2 2Z; (5.5.46) whereM 1 is xed as in (5.5.42) and ~ T2 (0; 1=M 4s+4 ) is a suciently small constant to be determined below. We emphasize that the implicit constants below are allowed to depend on M. Let v 1 ;v 2 2Z and (R 1 ; 1 ; 1t ; v 1 ) and (R 2 ; 2 ; 2t ; v 2 ) be the corresponding solutions of (5.5.11), (5.5.16)–(5.5.17), and (5.5.18)–(5.5.19) with the same initial data (R 0 ;w 0 ;w 1 ;v 0 ) and the same nonhomogeneous terms (f;h). We denote ~ V = v 1 v 2 , ~ v =v 1 v 2 , ~ R =R 1 R 2 , and ~ = 1 2 . The dierence ~ V satises ~ V t R 1 div(r ~ V + (r ~ V ) T )R 1 r div ~ V =g in (0; 1) f ; with the boundary conditions and the initial data (@ k ~ V j +@ j ~ V k ) k +@ k ~ V k j =@ k ~ j k R 1 1 R 1 2 ~ R j on (0; 1) c ; ~ V = 0 on (0; 1) f ; ~ V periodic in they 1 andy 2 directions; ~ V (0) = 0 in f ; forj = 1; 2; 3, where g =R 1 rR 1 1 +R 2 rR 1 2 + ~ R div(r v 2 + (r v 2 ) T ) + ~ Rr div v 2 : (5.5.47) 251 Proof of Theorem 5.5.2. We proceed as in (5.5.20) to obtain k ~ Vk K s+1 ((0;1) f ) . @ ~ @ K s1=2 c +k ~ RR 1 1 R 1 2 rR 2 k K s1 ((0;1) f ) +kR 1 1 r ~ Rk K s1 ((0;1) f ) +k ~ Rr 2 v 2 k K s1 ((0;1) f ) +kR 1 1 R 1 2 ~ Rk K s1=2 ((0;1) c) ; (5.5.48) where the last inequality follows from (5.5.47). Here and below, the norm of dependence on time and space are understood as (0; 1) and f , unless stated otherwise. The dierence ~ satises the wave equation ~ tt ~ = 0 in (0; 1) e ; with the boundary condition and the initial data ~ (t) = ~ (0) + Z t 0 ~ T ~ v()d on (0; 1) c ; ( ~ ; ~ t )(0) = (0; 0) in e : For the rst term on the right side of (5.5.48), we proceed as in (5.5.21)–(5.5.37) to obtain @ ~ @ K s1=2 c . ( + 1 + ~ T 1 C 1 ; + ~ T 1=4 )k~ vk K s+1; (5.5.49) for any; 1 2 (0; 1], where 1 is a constant as in (5.5.33). Since the dierence ~ R satises the ODE system ~ R t ~ T ~ R divv 2 =R 1 ~ T div ~ v in (0; 1) f ; (5.5.50) ~ R(0) = 0 in f ; (5.5.51) 252 the solution is given by ~ R(t;x) =e R t 0 ~ T () divv 2 ()d Z t 0 e R 0 ~ T divv 2 ~ T ()R 1 () div ~ v()d in (0; 1) f : (5.5.52) For the second term on the right side of (5.5.48), we obtain k ~ RR 1 1 R 1 2 rR 2 k L 2 t H s1 x .k ~ Rk L 2 t H s x kR 1 1 k L 1 t H s x kR 1 2 k L 1 t H s x kR 2 k L 1 t H s x .k ~ Rk L 1 t H s x ; (5.5.53) where we used the Hölder inequality and Lemma 5.5.1. From (5.5.52) it follows that k ~ RR 1 1 R 1 2 rR 2 k L 2 t H s1 x .k ~ Rk L 1 t H s x . ~ T 1=2 k~ vk L 2 t H s+1 x ; (5.5.54) for ~ T2 (0; 1=M 4s+4 ). For the time component, we have k( ~ RR 1 1 R 1 2 rR 2 ) t k L 2 t L 2 x .k ~ R t rR 2 k L 2 t L 2 x +k ~ RR 1t rR 2 k L 2 t L 2 x +k ~ RR 2t rR 2 k L 2 t L 2 x +k ~ RrR 2t k L 2 t L 2 x .k ~ Rk L 1 t L 1 x k divv 2 k L 2 t L 4 x krR 2 k L 1 t L 4 x +kR 1 k L 1 t L 1 x k div ~ vk L 2 t L 1 x ((0;2 ~ T ) f ) krR 2 k L 1 t L 2 x +k ~ Rk L 1 t L 1 x k divv 1 k L 2 t L 4 x krR 2 k L 1 t L 4 x +k ~ Rk L 1 t L 1 x k divv 2 k L 2 t L 4 x krR 2 k L 1 t L 4 x +k ~ Rk L 1 t L 1 x krR 2t k L 2 t L 2 x . ( +C ~ T 1=2 )k~ vk K s+1; (5.5.55) for any2 (0; 1], where we used the Hölder inequality, Lemma 5.5.1, (5.5.50), and (5.5.52). For the third term on the right side of (5.5.48), we get kR 1 r ~ Rk L 2 t H s1 x .kR 1 k L 1 t H s x k ~ Rk L 1 t H s x . ~ T 1 2 k~ vk K s+1 (5.5.56) 253 and kR 1 r ~ Rk H (s1)=2 t L 2 x .k(R 1 r ~ R) t k L 2 t L 2 x .kR 1t r ~ Rk L 2 t L 2 x +kR 1 r ~ R t k L 2 t L 2 x .kR 1t k L 2 t L 1 x kr ~ Rk L 1 t L 2 x +kR 1 k L 1 t L 1 x kr ~ R t k L 2 t L 2 x . ( +C ~ T 1=2 )k~ vk K s+1; (5.5.57) where we appealed to Lemma 5.5.1, (5.5.50), and (5.5.52). Regarding the fourth term on the right side of (5.5.48), it follows that k ~ Rr 2 v 2 k L 2 t H s1 x .k ~ Rk L 1 t H s x k v 2 k L 2 t H s+1 x . ~ T 1=2 k~ vk K s+1 (5.5.58) and k ~ Rr 2 v 2 k H (s1)=2 t L 2 x .k ~ Rk L 1 t L 1 x kr 2 v 2 k H (s1)=2 t L 2 x +k ~ Rk W (s1)=2;4 t L 1 x kr 2 v 2 k L 4 t L 2 x . ~ T 1=2 k~ vk K s+1kr 2 v 2 k H (s1)=2 t L 2 +k ~ Rk H 1 t L 1 x kr 2 v 2 k H (s1)=2 t L 2 . ( +C ~ T 1=2 )k~ vk K s+1; (5.5.59) where we used Lemma 5.3.2 in the last inequality. For the last term on the right side of (5.5.48), using the trace inequality, we arrive at kR 1 1 R 1 2 ~ Rk L 2 t H s1=2 x ( c) .k ~ Rk L 2 t H s x . ~ T 1=2 k~ vk K s+1 (5.5.60) and kR 1 1 R 1 2 ~ Rk H s=21=4 t L 2 x ( c) .kR 1 1 R 1 2 ~ Rk H 1 t H 1 x .k(R 1 1 R 1 2 ~ R) t k L 2 t H 1 x +kR 1 1 R 1 2 ~ Rk L 2 t H 1 x ; (5.5.61) 254 sinces=2 1=4 1. For the rst term on the right side of (5.5.61), we proceed as in (5.5.55) to obtain k(R 1 1 R 1 2 ~ R) t k L 2 t H 1 x .kR 1t ~ Rk L 2 t H 1 x +kR 2t ~ Rk L 2 t H 1 x +k ~ R t k L 2 t H 1 x . ( +C ~ T 1=2 )k~ vk K s+1: (5.5.62) The second term on the right side of (5.5.61) is estimated analogously as in (5.5.60), and we get kR 1 1 R 1 2 ~ Rk L 2 t H 1 x . ~ Tk~ vk K s+1: (5.5.63) From (5.5.48)–(5.5.49) and (5.5.53)–(5.5.63) it follows that k ~ Vk K s+1. ( + 1 + ~ T 1 C 1 ; + ~ T 1=4 )k~ vk K s+1: Taking , 1 , and ~ T > 0 suciently small, we conclude the proof of (5.5.46). Thus, the mapping is contracting fromZ toZ. From (5.5.45) and Lemma 5.5.1 it follows that there exists a unique solution (v;R;w;w t )2K s+1 ((0; ~ T ) f )H 1 ((0; ~ T );H s ( f )) C([0; ~ T ];H s+1=4 0 ( e ))C([0; ~ T ];H s3=4 0 ( e )); for some ~ T > 0. 5.6 SolutiontotheNavier-Stokes-wavesystem In this section, we provide the local existence for the coupled Navier-Stokes-wave system (5.2.3)–(5.2.5) with the boundary conditions (5.2.6)–(5.2.10) and the initial data (5.2.15). Throughout this section, let s2 (2; 2 + 0 ] where 0 2 (0; 1=2) is a suciently small constant. Letv2Z whereZ is as in (5.5.15), 255 with constantM 1 to be determined below. Let ~ T (t) be a smooth cuto function as dened in Sec- tion 5.5, where ~ T 2 (0; 1=2) is a constant to be determined below. We shall start with cuto solutions (R;w;;J;a;v) on the time interval (0; 1), in order to apply Lemmas 5.3.1–5.3.2 with implicit constants independent of ~ T . We then recover the solutions to (5.2.3)–(5.2.5) by restricting the cuto solutions to a small time interval (0; ~ T ). Let(t;x) =x+ R t 0 ~ T ()v(;x)d be the corresponding Lagrangian ow map anda(t;x) = (r(t;x)) 1 be the inverse matrix of the ow map, whileJ(t;x) = det(r(t;x)) denotes the Jacobian. Denoteb =aI 3 , whereI 3 is the three-dimensional identity matrix. First, we consider the ODE system R t R ~ T a kj @ k v j = 0 in [0; 1] f ; (5.6.1) R(0) =R 0 on f : (5.6.2) The solution is given by the formula R(t;x) =R 0 (x)e R t 0 ~ T ()a kj (;x)@ k v j (;x)d in [0; 1] f : Then we solve the wave equation (5.2.5) forw with the boundary condition and the initial data w(t;x) =w(0) + Z t 0 ~ T ()v(;x)d on [0; 1] c ; (w;w t )(0) = (w 0 ;w 1 ) in e : With (R;w;;J;a) constructed, we dene a mapping : v2Z7! v; 256 where v is the solution of the nonhomogeneous parabolic problem @ t v j R@ k (@ j v k +@ k v j )R@ j @ k v k =f j in (0; 1) f ; (@ k v j +@ j v k ) k +@ k v k j =@ k w j k +h j in (0; 1) c ; v periodic in they 1 andy 2 directions; v(0) =v 0 in f ; where we denote f j =R@ k (b mk @ m v j +b mj @ m v k ) +Rb kl @ k (b ml @ m v j +b mj @ m v l ) +Rb kl @ k (@ l v j +@ j v l ) +R@ j (b mi @ m v i ) +Rb kj @ k (b mi @ m v i ) +Rb kj @ k @ i v i Rb kj @ k R 1 R@ j R 1 =:I 1 +I 2 +I 3 +I 4 +I 5 +I 6 +I 7 +I 8 (5.6.3) and h j =(1J)(@ k v j +@ j v k ) k +(1J)@ k v k j Jb kl (b ml @ m v j +b mj @ m v l ) k +Jb kj R 1 k + (J 1)R 1 j J(b mk @ m v j +b mj @ m v k ) k Jb kl (@ l v j +@ j v l ) k Jb kj b mi @ m v i k Jb mi @ m v i j Jb kj @ i v i k +R 1 j =:K 1 +K 2 +K 3 +K 4 +K 5 +K 6 +K 7 +K 8 +K 9 +K 10 +K 11 ; (5.6.4) forj = 1; 2; 3. Before we bound the terms in (5.6.3)–(5.6.4) and construct a contraction mapping as in Section 5.5, we provide some necessary a priori estimates on the variable coecients. 5.6.1 TheLagrangianowmap,Jacobianmatrix,anddensityestimates We start with the necessary Jacobian and the inverse matrix of the ow map estimates. 257 Lemma 5.6.1. Suppose thatkvk K s+1 ((0;1) f ) M for some suciently large constantM 1. Then the following statements hold for ~ T2 (0; 1=M 80 ): (i)kbk L 1 t H s x +kbk H 1 t L 1 x +kbk H 1 t H 1 x . ~ T 1=20 , (ii)kbk H 1 t H s x .M, (iii)k1Jk L 1 t H s x . ~ T 1=20 , (iv) kJk L 1 t L 1 x +kJ 1 k L 1 t L 1 x +kJk L 1 t H s x +kJ 1 k L 1 t H s x . 1, (v)kJ 1 k H 1 t L 1 x +kJk H 1 t L 1 x +kJk H 1 t H 1 x . 1, (vi) kJk H 1 t H s x .M, wherethenormofdependenceis [0; 1] f . Weemphasizethattheimplicitconstantsintheaboveinequalities are independent ofM. Proof of Lemma 5.6.1. (i) From (5.2.11) we have b t = ~ T (b(rv)b +brv + (rv)b +rv): (5.6.5) Note thatb(0) = 0. From the fundamental theorem of calculus, it follows that fort2 (0; 2 ~ T ), we have kb(t)k H s. Z t 0 kbk 2 H skrvk H sd + Z t 0 kbk H skrvk H sd + Z t 0 krvk H sd . Z t 0 kvk H s+1(kbk 2 H s +kbk H s)d + ~ T 1=2 M: From the nonlinear Gronwall’s inequality we obtain kbk L 1 t H s x ((0;2 ~ T ) f ) . ~ T 1=2 M. ~ T 1=20 ; 258 for ~ T2 (0; 1=M 80 ). Therefore, we infer that kbk L 1 t H s x . ~ T 1=20 ; (5.6.6) sinceb t = 0 on (2 ~ T; 1). From (5.6.5)–(5.6.6), we use the Hölder and Sobolev inequalities, together with (5.5.13) to obtain kb t k L 2 t L 1 x .kvk L 2 t H s+1=2 x ((0;2 ~ T ) f ) . 1 kvk L 2 t H s+1 x ((0;2 ~ T ) f ) + 2s1 1 kvk L 2 t L 2 x ((0;2 ~ T ) f ) . 1 M + 2s1 1 ~ T 1=2 M; for any 1 2 (0; 1], sincekvk L 2 t L 2 x ((0;2 ~ T ) f ) . ~ T 1=2 kvk H (s+1)=2 t L 2 x ((0;2 ~ T ) f ) . Letting 1 = ~ T 1=20 M 1 , we have kb t k L 2 t L 1 x . ~ T 1=20 + ~ T (92s)=20 M 2s+2 . ~ T 1=20 ; (5.6.7) for ~ T2 (0; 1=M 80 ). Similarly, we have kb t k L 2 t H 1 x . ~ T 1=20 : (5.6.8) Combining (5.6.6)–(5.6.8), we conclude the proof of (i). (ii) From (5.6.5) and the Hölder inequality it follows that kb t k L 2 t H s x .krvk L 2 t H s x kbk 2 L 1 t H s x +krvk L 2 t H s x kbk L 1 t H s x +krvk L 2 t H s x .M; completing the proof of (ii). 259 (iii) Since the Jacobian satises the ODE system @J @t (t;x) = ~ T (t)J(t;x)a kj (t;x)@ k v j (t;x) in [0; 1] f ; (5.6.9) J(0) = 1 in f ; (5.6.10) the solution is given by the explicit formula J(t;x) =e R t 0 ~ T ()a kj (;x)@ k v j (;x)d in [0; 1] f Using the nonlinear Sobolev estimate, we have kJ 1k L 1 t H s x .C ~ T 1=2 M 1. ~ T 1=20 ; for some constantC > 0, where ~ T2 (0; 1=M 80 ) andM 1 is a suciently large constant. The proofs of (iv), (v), and (vi) are analogous to the proof of Lemma 5.5.1 using (i)–(iii). Thus we omit the details. The following lemma provides the necessary a priori density estimates. Lemma5.6.2. Assume that (R 0 ;R 1 0 ;b)2H s ( f )H s ( f )L 1 ([0;T ];H s ( f )); andkvk K s+1 ((0;1) f ) M whereM 1 is a constant. Then for ~ T 2 (0; 1=M 4s+4 ), the solution to the ODE system (5.6.1)–(5.6.2) satises (i)kRk L 1 t L 1 x +kR 1 k L 1 t L 1 x +kRk L 1 t H s x +kR 1 k L 1 t H s x . 1, 260 (ii)kRk H 1 t L 1 x +kR 1 k H 1 t L 1 x +kRk H 1 t H 1 x . 1, (iii)kRk H 1 t H s x .M, wherethenormofdependenceis [0; 1] f . Weemphasizethattheimplicitconstantsintheaboveinequalities are independent ofM. The proof of Lemma 5.6.2 is analogous to the proof of Lemma 5.5.1. Thus we omit the details. 5.6.2 Uniformboundednessoftheiterativesequence In this section we shall prove that the mapping is well-dened fromZ toZ, for some suciently large M 1 and suciently small ~ T2 (0; 1=M 80 ). From Lemmas 5.4.2 and 5.6.2 it follows that k vk K s+1 ((0;1) f ) . @w @ K s1=2 ( c) +kfk K s1 ((0;1) f ) +khk K s1=2 ((0;1) c) +kv 0 k H s; (5.6.11) wheref andh are as in (5.6.3)–(5.6.4). Here and below, the norm of dependence on time and space are understood as (0; 1) and f , unless stated otherwise. We emphasize that the implicit constants in this section are independent ofM. For the rst term on the right side of (5.6.11), we proceed as in (5.5.21)–(5.5.37) to obtain @w @ K s1=2 ( c) . 1 + ( + 1 + ~ T 1 C 1 ; + ~ T 1 4 )M + ~ T 2 ; (5.6.12) for any; 1 2 (0; 1], where 1 ; 2 are constants depending ons as in (5.5.33)–(5.5.34). Next, we estimate theK s1 norm of the terms on the right side of (5.6.3) forj = 1; 2; 3. For the space component of the termI 1 , we use the Hölder inequality and Lemmas 5.6.1–5.6.2 to get kI 1 k L 2 t H s1 x .kRrbr vk L 2 t H s1 x +kRbr 2 vk L 2 t H s1 x .kbk L 1 t H s x k vk L 2 t H s+1 x . ~ T 1=20 k vk K s+1: (5.6.13) 261 For the time component, we have kI 1 k H (s1)=2 t L 2 x .kRrbr vk H (s1)=2 t L 2 x +kRbr 2 vk H (s1)=2 t L 2 x .kRk L 1 t L 1 x krbk L 1 t L 4 x kr vk H (s1)=2 t L 4 x +kRk W (s1)=2;4 t L 1 x krbk L 1 t L 4 x kr vk L 4 t L 4 x +kRk L 1 t L 1 x krbk W (s1)=2;4 t L 2 x kr vk L 4 t L 1 x +kRk L 1 t L 1 x kbk L 1 t L 1 x kr 2 vk H (s1)=2 t L 2 x +kRk W (s1)=2;4 t L 1 x kbk L 1 t L 6 x kr 2 vk L 4 t L 3 x +kRk L 1 t L 1 x kbk W (s1)=2;4 t L 6 x kr 2 vk L 4 t L 3 x : (5.6.14) From Lemma 5.3.2 and Sobolev inequality it follows that kr vk H (s1)=2 t L 4 x +kr vk L 4 t L 4 x +kr vk L 4 t L 1 x +kr 2 vk H (s1)=2 t L 2 x +kr 2 vk L 4 t L 3 x .k vk H (s+1)=2 t L 2 x +k vk L 2 t H s+1 x .k vk K s+1; (5.6.15) since 2 0. Combining (5.6.17)–(5.6.19) and (5.6.22)–(5.6.23), we have kK 1 k K s1=2 c . ( +C T 1=30 M 2s1 )k vk K s+1; (5.6.24) for any2 (0; 1]. The termK 2 is estimated analogously toK 1 , and we arrive at kK 2 k K s1=2 c . ( +C T 1=30 M 2s1 )k vk K s+1; (5.6.25) for any2 (0; 1]. For the space component of the termK 3 , we use the Hölder and trace inequalities to obtain kK 3 k L 2 t H s1=2 x ( c) .kJbbr vk L 2 t H s x .kJk L 1 t H s x kbk 2 L 1 t H s x k vk L 2 t H s+1 x . ~ T 1=20 k vk K s+1; (5.6.26) where we appealed to Lemma 5.6.1. For the time component, using Lemma 5.3.1, we have kK 3 k H s=21=4 t L 2 x ( c) . 3 kJbbr vk H s=2 t L 2 x + 12s 3 kJbbr vk L 2 t H s x . 3 kJk H s=2 t L 3 x kr vk L 1 t L 6 x + 3 kbk H s=2 t L 3 x kr vk L 1 t L 6 x + 3 kr vk H s=2 t L 2 x + 12s 3 kbk L 1 t H s x k vk L 2 t H s+1 x =:K 31 +K 32 +K 33 +K 34 ; (5.6.27) for any 3 2 (0; 1]. The termK 31 is estimated analogously as in (5.6.20)–(5.6.22), and we obtain K 31 .k vk K s+1; 265 by taking 3 = ~ T (s2)=2 M 1 in (5.6.27), where2 (0; 1] is a constant. The termkbk H s=2 t L 3 x is estimated analogously as in (5.6.21) and we get kbk H s=2 t L 3 x . ~ T (2s)=2 M + 1: Therefore, we have K 32 .k vk K s+1: The termK 33 is estimated using Lemma 5.3.2 as K 33 .k vk K s+1: The termK 34 is estimated analogously to (5.6.23), and we get K 34 .C ~ T 1=30 M 2s1 k vk K s+1: (5.6.28) Combining (5.6.26)–(5.6.28), we arrive at kK 3 k K s1=2 c . ( +C T 1=30 M 2s1 )k vk K s+1: (5.6.29) Regarding the termK 4 , we proceed as in (5.5.40)–(5.5.41), obtaining kK 4 k K s1=2 c .kJbR 1 k L 2 t H s x +kJbR 1 k H 1 t H 1 x . 1 +kJ t k L 2 t H 1 x kbk L 1 t H s x +kb t k L 2 t H 1 x +kR 2 R t k L 2 t H 1 x . 1 + ~ T 1=20 M; (5.6.30) 266 where we used Lemmas 5.6.1–5.6.2. The termK 5 is estimated in a similar fashion asK 4 , and we arrive at kK 5 k K s1=2 c . 1: (5.6.31) The termsK 6 ,K 7 ,K 8 ,K 9 , andK 10 are estimated analogously asK 3 , and we have kK 6 k K s1=2 c +kK 7 k K s1=2 c +kK 7 k K s1=2 c +kK 8 k K s1=2 c +kK 9 k K s1=2 c +kK 10 k K s1=2 c . ( +C T 1=30 M 2s1 )k vk K s+1; (5.6.32) for any2 (0; 1]. For the termK 11 , we proceed as in (5.5.40)–(5.5.41) to obtain K 11 . 1: (5.6.33) Collecting the estimates (5.6.24)–(5.6.25) and (5.6.29)–(5.6.33), we conclude khk K s1=2 c . ( +C T 1=30 M 2s1 )k vk K s+1 + 1: (5.6.34) From (5.6.11)–(5.6.12), (5.6.16), and (5.6.34) it follows that k vk K s+1. ( + 1 + ~ T 1=20 +C ; 1 ~ T 1 +C T 1=30 M 2s1 )k vk K s+1 + 1 + ~ T 1=4 M + ~ T 2 ; for any; 1 2 (0; 1], where 1 ; 2 are constants as in (5.5.33)–(5.5.34). Note that 2 (2) = 0. Taking, 1 , and ~ T suciently small to get k vk K s+1M; (5.6.35) 267 by allowingM 1 suciently large, sinces2 (2; 2 + 0 ] for some suciently small 0 > 0. Thus, the mapping : v7! v is well-dened fromZ toZ, for some constantM 1 as in (5.6.35). Now we x the constantsM 1. We proceed as in (5.5.43)–(5.5.45) to obtain the interior regularity estimate kwk C([0;1];H s+1=4 0 ( e)) +kw t k C([0;1];H s3=4 0 ( e)) C; (5.6.36) for some constantC > 0. 5.6.3 Contractingproperty In this section we shall prove k(v 1 ) (v 2 )k K s+1 1 2 kv 1 v 2 k K s+1; v 1 ;v 2 2Z; (5.6.37) whereM 1 is xed as in (5.6.35) and ~ T > 0 is a suciently small constant to be determined below. Note that the implicit constants below are allowed to depend onM. Let v 1 ;v 2 2 Z and ( 1 ;J 1 ;a 1 ) and ( 2 ;J 2 ;a 2 ) be the corresponding cuto Lagrangian ow map, Jacobian, and the inverse matrix of the ow map. First we solve for (R 1 ;R 2 ) from (5.6.1)–(5.6.2) with the same initial dataR 0 . Then we solve for ( 1 ; 1t ) and ( 2 ; 2t ) from (5.2.5) with the boundary conditions (5.5.16) and the same initial data (w 0 ;w 1 ). To obtain the next iterate ( v 1 ; v 2 ), we solve (5.5.18) with the boundary conditions and the initial data (5.5.19). Denote b 1 = a 1 I 3 , b 2 = a 2 I 3 , ~ b = b 1 b 2 , 268 ~ V = v 1 v 2 , ~ v =v 1 v 2 , ~ R =R 1 R 2 , ~ = 1 2 , ~ = 1 2 , and ~ J =J 1 J 2 . The dierence ~ V satises @ t ~ V j ~ R@ k (@ j ~ V k +@ k ~ V j ) ~ R@ j @ k ~ V k = ~ f j in (0; 1) f ; (@ k ~ V j +@ j ~ V k ) k +@ k ~ V k j =@ k ~ j k + ~ h j in (0; 1) c ; ~ V periodic in they 1 andy 2 directions; ~ V (0) = 0 in f ; (5.6.38) where ~ f j = ~ R@ k (@ j v 2k +@ k v 2j ) + ~ R@ j @ k v 2k + ~ R@ k (b 1mk @ m v 1j +b 1mj @ m v 1k ) +R 2 @ k (b 1mk @ m ~ V j +b 1mj @ m ~ V k ) +R 2 @ k ( ~ b mk @ m v 2j + ~ b mj @ m v 2k ) + ~ Rb 1kl @ k (b 1ml @ m v 1j +b 1mj @ m v 1l ) +R 2 ~ b kl @ k (b 1ml @ m v 1j +b 1mj @ m v 1l ) +R 2 b 2kl @ k ( ~ b ml @ m v 1j + ~ b mj @ m v 1l ) +R 2 b 2kl @ k (b 2ml @ m ~ V j +b 2mj @ m ~ V l ) + ~ Rb 1kl @ k (@ l v 1j +@ j v 1l ) +R 2 ~ b kl @ k (@ l v 1j +@ j v 1l ) +R 2 b 2kl @ k (@ l ~ V j +@ j ~ V l ) +R@ j (b 1mi @ m v 1i ) +R 2 @ j ( ~ b mi @ m v 1i ) +R 2 @ j (b 2mi @ m ~ V ) + ~ Rb 1kj @ k (b 1mi @ m v 1i ) +R 2 ~ b kj @ k (b 1mi @ m v 1i ) +R 2 b 2kj @ k ( ~ b mi @ m v 1i ) +R 2 b 2kj @ k (b 2mi @ m ~ V i ) + ~ Rb 1kj @ k @ i v 1i +R 2 ~ b kj @ k @ i v 1i +R 2 b 2kj @ k @ i ~ V i R 1 1 R 1 2 ~ Rb 1kj @ k R 1 +R 1 2 ~ b kj @ k R 1 +R 1 2 b 2kj @ k ~ RR 1 1 R 1 2 ~ R@ j R 1 +R 1 2 @ j ~ R (5.6.39) 269 and ~ h j = ~ J(@ k v 1j +@ j v 1k ) k +(1J 2 )(@ k ~ V j +@ j ~ V k ) k ~ J@ k v 1k j +(1J 2 )@ k ~ V k j + ~ Jb 1kl (b 1ml @ m v 1j +b 1mj @ m v 1l ) k +J 2 ~ b kl (b 1ml @ m v 1j +b 1mj @ m v 1l ) k +J 2 b 2kl ( ~ b ml @ m v 1j + ~ b mj @ m v 1l ) k +J 2 b 2kl (b 2ml @ m ~ V j +b 2mj @ m ~ V l ) k + ~ Jb 1kj R 1 1 k +J 2 ~ b kj R 1 1 k J 2 b 2kj R 1 1 R 1 2 ~ R k ~ JR 1 1 j (J 2 1)R 1 1 R 1 2 ~ R j ~ J(b 1mk @ m v 1j +b 1mj @ m v 1k ) k J 2 ( ~ b mk @ m v 1j + ~ b mj @ m v 1k ) k J 2 (b 2mk @ m ~ V j +b 2mj @ m ~ V 1k ) k ~ Jb 1kl (@ l v 1j +@ j v 1l ) k J 2 ~ b kl (@ l v 1j +@ j v 1l ) k J 2 b 2kl (@ l ~ V j +@ j ~ V l ) k ~ Jb 1kj b 1mi @ m v 1i k J 2 ~ b kj b 1mi @ m v 1i k J 2 b 2kj ~ b mi @ m v 1i k J 2 b 2kj b 2mi @ m ~ V i k ~ Jb 1mi @ m v 1i j J 2 ~ b mi @ m v 1i j J 2 b 2mi @ m ~ V i j ~ Jb 1kj @ i v 1i k J 2 ~ b kj @ i v 1i k J 2 b 2kj @ i ~ V i k R 1 1 R 1 2 ~ R j ; (5.6.40) forj = 1; 2; 3. Before we bound the terms on the right sides of (5.6.39) and (5.6.40), we provide necessary a priori estimates for the dierences of the density, Jacobian, and inverse matrix of the ow map. Lemma 5.6.3. Let v 1 ;v 2 2Z. Supposekv 1 k K s+1 ((0;1) f ) M andkv 2 k K s+1 ((0;1) f ) M, where M 1 is xed as in (5.5.42). Then for ~ T2 (0; 1=M 100 ), we have (i)k ~ bk L 1 t H s x +k ~ bk H 1 t L 1 x . ~ T 1=20 k~ vk K s+1, (ii)k ~ Rk L 1 t H s x +k ~ Rk H 1 t L 1 x . ~ T 1=20 k K s+1, (iii)k ~ Jk L 1 t H s x +k ~ Jk H 1 t L 1 x . ~ T 1=20 k~ vk K s+1, (iv) k ~ Rk H 1 t H s x +k ~ bk H 1 t H s x +k ~ Jk H 1 t H s x .k~ vk K s+1, where the norm of dependence is [0; 1] f . 270 Proof of Lemma 5.6.3. (i) The dierence ~ b satises the ODE ~ b t = ~ T [ ~ b(rv 1 )b 1 +b 2 (r~ v)b 1 +b 2 (rv 2 ) ~ b + (r~ v)b 1 + (rv 2 ) ~ b + ~ b(rv 1 ) +b 2 (r~ v) +r~ v] in (0; 1) f ; (5.6.41) with the initial datab(0) = 0. From the fundamental theorem of calculus, we have k ~ b(t)k H s. Z t 0 k ~ bk H skrv 1 k H skb 1 k H s + Z t 0 kb 2 k H skr~ vk H skb 1 k H s + Z t 0 kb 2 k H skrv 2 k H sk ~ bk H s + Z t 0 kr~ vk H skb 1 k H s + Z t 0 kr~ vk H s + Z t 0 krv 2 k H sk ~ bk H s + Z t 0 k ~ bk H skrv 1 k H s + Z t 0 kb 2 k H skr~ vk H s . Z t 0 k~ vk H s+1 + Z t 0 k ~ bk H s(kv 1 k H s+1 +kv 2 k H s+1); fort2 (0; 2 ~ T ), where the last inequality follows from Lemma 5.6.1. Using Gronwall’s inequality, we arrive at k ~ bk L 1 t H s x ((0;2 ~ T ) f ) . ~ T 1=2 k~ vk K s+1: (5.6.42) Therefore, we have k ~ bk L 1 t H s x . ~ T 1=2 k~ vk K s+1; since ~ b t = 0 on (2 ~ T; 1). From (5.6.41) and the Hölder and Sobolev inequalities it follows that k ~ b t k L 2 t L 1 x .k ~ bk L 1 t L 1 x +kr~ vk L 2 t L 1 x ((0;2 ~ T ) f ) . ~ T 1=2 k~ vk K s+1 + 1 k~ vk L 2 t H s+1 x ((0;2 ~ T ) f ) + 2s1 1 k~ vk L 2 t L 2 x ((0;2 ~ T ) f ) . ~ T 1=20 k~ vk K s+1; (5.6.43) 271 where we used Lemmas 5.3.2 and 5.6.1. (ii) Since the dierence ~ R satises the ODE ~ R t ~ T ~ R(divv 2 +b 1kj @ k v 1j ) = ~ T (R 1 div ~ v +R 2 ~ b kj @ k v 1j +R 2 b 2kj @ k ~ v j ) in (0;T ) f ; (5.6.44) ~ R(0) = 0 in f ; (5.6.45) the solution is given by the formula ~ R(t;x) =e R t 0 ~ T (divv 2 +b 1kj @ k v 1j )d Z t 0 e R 0 ~ T (divv 2 +b 1kj @ k v 1j ) ~ T (R 1 div ~ v +R 2 ~ b kj @ k v 1j +R 2 b 2kj @ k ~ v j )d in [0;T ] f : From the Hölder inequality it follows that k ~ Rk L 1 t H s x . Z 2 ~ T 0 (k~ vk H s+1 +k ~ brv 1 k H s)d. ~ T 1=2 k~ vk K s+1; (5.6.46) where we used (5.6.42) in the last inequality. Using (5.6.44) and the Hölder and Sobolev inequalities, we obtain k ~ R t k L 2 t L 1 x .k ~ Rk L 1 t L 1 x krv 2 k L 2 t L 1 x +k ~ Rk L 1 t L 1 x kb 1 k L 1 t L 1 x krv 1 k L 2 t L 1 x +kR 1 k L 1 t L 1 x kr~ vk L 2 t L 1 x +kR 2 k L 1 t L 1 x k ~ bk L 1 t L 1 x krv 1 k L 2 t L 1 x +kR 2 k L 1 t L 1 x kb 2 k L 1 t L 1 x kr~ vk L 2 t L 1 x . ( 1 + 2s1 1 ~ T 1=2 )k~ vk K s+1. ~ T 1=20 k~ vk K s+1; (5.6.47) by taking 1 = ~ T 1=20 in the last inequality. Thus we conclude the proof of (ii) by combining (5.6.46)– (5.6.47). 272 (iii) The dierence ~ J satises the same ODE system (5.6.44)–(5.6.45) and from (ii) it follows that k ~ Jk L 1 t H s x . ~ T 1=20 k~ vk K s+1: (iv) The proof is analogous to (i)–(iii) using (5.6.43) and (5.6.47). Thus we omit the details. Proof of Theorem 5.2.1. From Lemmas 5.4.2 and 5.6.2 it follows that the solution ~ V of (5.6.38) satises k ~ Vk K s+1. @ ~ @ K s1=2 c +k ~ hk K s1=2 c +k ~ fk K s1; (5.6.48) where ~ f j and ~ h j are as in (5.6.39)–(5.6.40), forj = 1; 2; 3. We proceed as in (5.5.21)–(5.5.37) to bound the rst term on the right side of (5.6.48), obtaining @ ~ @ K s1=2 c . ( + 1 + ~ T 1 C ; 1 + ~ T 1=4 )k~ vk K s+1; (5.6.49) for any; 1 2 (0; 1]. Next we estimate the K s1 norm of terms on the right side of (5.6.39) for j = 1; 2; 3. The term ~ Rb 1kj @ k (b 1mi @ m v 1i ) is bounded as k ~ Rb 1kj @ k (b 1mi @ m v 1i )k L 2 t H s1 x .k ~ Rk L 1 t H s x kb 1 k 2 L 1 t H s x k v 1 k L 2 t H s+1 x . ~ T 1=20 k~ vk K s+1 273 and k ~ Rb 1kj @ k (b 1mi @ m v 1i )k H (s1)=2 t L 2 x .k ~ Rk W (s1)=2;4 t L 1 x kr 2 v 1 k L 4 t L 2 x +k ~ Rk L 1 t L 1 x kb 1 k W (s1)=2;4 t L 1 x kb 1 k L 1 t L 1 x kr 2 v 1 k L 4 t L 4 x +k ~ Rk L 1 t L 1 x kr 2 v 1 k H (s1)=2 t L 2 x +k ~ Rk W (s1)=2;4 t L 1 x krb 1 k L 1 t L 6 x kr 2 v 1 k L 4 t L 3 x +k ~ Rk L 1 t L 1 x kr 2 v 1 k H (s1)=2 t L 2 x +k ~ Rk L 1 t L 1 x krb 1 k W (s1)=2;4 t L 6 x kr 2 v 1 k L 4 t L 3 x +k ~ Rk L 1 t L 1 x kb 1 k W (s1)=2;4 t L 1 x krb 1 k L 1 t L 6 x kr 2 v 1 k L 4 t L 3 x . ~ T 1=20 k~ vk K s+1; where we used Lemma 5.6.3. The termR 2 b 2kj @ k @ i ~ V i is estimated as kR 2 b 2kj @ k @ i ~ V i k L 2 t H s1 x .kR 2 k L 1 t H s x kb 2 k L 1 t H s x kr 2 ~ Vk L 2 t H s1 x . ~ T 1=20 k ~ Vk K s+1 and kR 2 b 2kj @ k @ i ~ V i k H (s1)=2 t L 2 x .kR 2 k W (s1)=2;4 t L 1 x kr 2 ~ Vk L 4 t L 2 x +kb 2 k W (s1)=2;4 t L 1 x kr 2 ~ Vk L 4 t L 2 x +kb 2 k L 1 t L 1 x kr 2 ~ Vk H (s1)=2 t L 2 x . ~ T 1=20 k ~ Vk K s+1; where we used Lemmas 5.6.1–5.6.2. Other terms on the right side of (5.6.39) are treated analogously as in Theorem 5.5.2 using Lemmas 5.6.1–5.6.3, and we arrive at k ~ fk K s1. ~ T 1=20 k~ vk K s+1 + ~ T 1=20 k ~ Vk K s+1; (5.6.50) where ~ T2 (0; 1=M 100 ). 274 Next we estimate theK s1=2 c norm of the terms on the right side of (5.6.40) forj = 1; 2; 3. The term (1J 2 )@ k ~ V j k is estimated using trace inequality and Lemma 5.6.1 as k(1J 2 )(@ k ~ V j +@ j ~ V k ) k k L 2 t H s1=2 x ( c) .k(1J 2 )(@ k ~ V j +@ j ~ V k ) k k L 2 t H s x .k1J 2 k L 1 t H s x k ~ Vk L 2 t H s+1 x . ~ T 1=20 k ~ Vk K s+1: For the time component, we proceed analogously as in (5.6.18)–(5.6.23), obtaining k(1J 2 )(@ k ~ V j +@ j ~ V k ) k k H s=21=4 t L 2 x ( c) . ( 3 +C 3 ~ T 1=30 )k ~ Vk K s+1; for any 3 2 (0; 1]. Other terms on the right side of (5.6.40) are treated analogously as in Theorem 5.5.2 using Lemmas 5.6.1–5.6.3, and we arrive at k ~ hk K s1=2 c . ~ T 1=20 k~ vk K s+1 + ( 3 +C 3 ~ T 1=30 )k ~ Vk K s+1; (5.6.51) for any 3 2 (0; 1], where ~ T2 (0; 1=M 100 ). Since the terms involvingk ~ Vk K s+1 on the right side of (5.6.50)–(5.6.51) are absorbed to the left side (5.6.48) by taking 3 ; ~ T > 0 suciently small, we obtain from (5.6.48)–(5.6.51) that k ~ Vk K s+1 1 2 k~ vk K s+1; completing the proof of (5.6.37) by taking; 1 > 0 suciently small. From (5.6.36) and Lemma 5.6.2 it follows that the system (5.2.3)–(5.2.10) admits a unique solution (v;R;w;w t )2K s+1 ((0; ~ T ) f )H 1 ((0; ~ T );H s ( f )) C([0; ~ T ];H s+1=4 0 ( e ))C([0; ~ T ];H s3=4 0 ( e )); 275 with the corresponding norms bounded by a function of the initial data. 276 Chapter6 MaximalregularityfortheNeumann-StokesprobleminH r=2;r spaces 6.1 Introduction In this paper, we provide a maximal regularity type theorem for the linear Stokes system with the Neumann boundary conditions and nonhomogeneous divergence. Such maximal regularity theorems in dierent functional spaces have been studied early on by Grubb and Solonnikov [68]. Still they have attracted much interest recently [122, 123, 126] due to their important role in the study of free boundary problems [7, 8, 10, 17, 21, 29, 75, 83, 91, 118, 126, 129]. Our main result, Theorem 6.2.1, is a variant of maximal regularity for the non-homogeneous Stokes equation with the non-homogeneous divergence, stated similarly to [122], but in dierent functional spaces. The feature of the approach in [122] and Theorem 6.2.1 is the representation (6.2.6) for g t . An important new idea is the presence of the fourth term on the left side of (6.2.7), i.e., we provide the con- trol of the pressure tracekpk H s=21=4;s1=2 1 . The control of this quantity is obtained by a space-time trace inequality (see Lemma 6.3.1). The main diculty in applying the theorem from [68] is that the boundary term (which is the sixth term on the right-hand side of (6.2.7)) has a higher time regularity than the Grubb-Solonnikov interior pressure terms (i.e., the second and third terms on the left of (6.2.7)). To overcome this, we prove an inequality satised for a boundary pressure term, which allows us to absorb the high time regularity 277 pressure term on the right. Another important ingredient in the proof of the main theorem is a trace inequality for functions which are Sobolev in the time variable and square-integrable on the boundary (see Lemma 6.3.1 and (6.3.1) below). The trace regularity is used essentially in the proof of the existence for the nonhomogeneous Stokes equation. We believe that the proof with one simple correctorz is new (the proof in [122] uses two correctors). The main dicult term resulting from the Dirichlet condition on the outside boundary is treated by the inequality (6.3.11) which is valid due to the Neumann condition forz on 2 . 6.2 Themainresult We consider the linear Stokes system u t u +rp =f in (0;T ) divu =g in (0;T ) ; (6.2.1) where R 3 is a bounded domain with the boundary @ . The system is supplemented with two boundary conditions @u @N pN =h 1 on (0;T ) 1 (6.2.2) u =h 2 on (0;T ) 2 ; (6.2.3) where 1 and 2 are the two disjoint components of the boundary with@ = 1 [ 2 andN is the unit normal vector to the boundary 1 . We impose the initial condition u(0;) =u 0 in (6.2.4) 278 and the compatibility condition u 0 j 2 =h 2 j 2 : (6.2.5) Denote H r;s ((0;T ) ) =H r ((0;T );L 2 ( ))\L 2 ((0;T );H s ( )); with the norm kfk 2 H r;s ((0;T ) ) =kfk 2 H r ((0;T );L 2 ( )) +kfk 2 L 2 ((0;T );H s ( )) : We shall use the notationH r;s , involving functions on (0;T ) , for the analogous space corresponding to the boundary set . The following theorem, which states the maximal regularity theorem for the linear Stokes equation with a nonhomogeneous divergence (cf. [68, 122, 134]), is our main result of this paper. Theorem6.2.1. Lets 1. Then, assuming u 0 2H s ( ) and (f;g;h 1 ;h 2 )2 (H s=21=2;s1 ;H s=2;s ;H s=21=4;s1=2 1 ;H s=2+1=4;s+1=2 2 ); with the structural condition g t = + divb; (6.2.6) where ;b2H s=21=2;s1 ; 279 there exists a unique solution (u;p) to the system (6.2.1)–(6.2.5) satisfying kuk H s=2+1=2;s+1 +kpk H s=21=2;s +krpk H s=21=2;s1 +kpk H s=21=4;s1=2 1 .ku 0 k H s +kfk H s=21=2;s1 +kgk H s=2;s +kk H s=21=2;s1 +kbk H s=21=2;s1 +kh 1 k H s=21=4;s1=2 1 +kh 2 k H s=2+1=4;s+1=2 2 ; (6.2.7) where the implicit constant depends on . The proof of the theorem is given in the next section. 6.3 ProofofTheorem6.2.1 First we recall two auxiliary lemmas from [83] which are needed in the proof of the main result. The rst lemma provides an estimate for the trace in a space-time norm. Lemma6.3.1 ([83]). Lets> 1=2and 0. Ifu2L 2 ((0;T );H s ( ))\H 2s=(2s1) ((0;T );L 2 ( )),then u2H ((0;T );L 2 ()), and for all 0 2 (0; 1] we have the inequality kuk H ((0;T );L 2 ()) 0 kuk H 2s=(2s1) ((0;T );L 2 ( )) +C 0 kuk L 2 ((0;T );H s ( )) ; (6.3.1) whereC 0 > 0 is a constant depending on 0 . The second lemma provides a space-time interpolation inequality needed in several instances in the proof of the main theorem. Lemma6.3.2 ([83]). Let; > 0. Ifu2H ((0;T );L 2 ( ))\L 2 ((0;T );H ( )), then u2H ((0;T );H ( )) and for all 0 > 0, we have the inequality kuk H ((0;T );H ( )) 0 kuk H ((0;T );L 2 ( )) +C 0 kuk L 2 ((0;T );H ( )) ; 280 for all2 (0;) and2 (0;) such that + 1; whereC 0 > 0 is a constant depending on 0 . In (6.2.7), as well as in the rest of the paper, we do not indicate the domains (0;T ) or in the norms as they are understood from the context. However, we always use a complete notation for norms involving the boundary traces on 1 and 2 . Proof of Theorem 6.2.1. As in [122], letz be the solution of z =g in (0;T ) z = 0 on 1 @z @N = 0 on 2 : (6.3.2) Note that the time derivativez t satises z t = ~ g + divb in (0;T ) z t = 0 on 1 @z t @N = 0 on 2 ; (6.3.3) by (6.2.6). The dierence ~ u =urz satises ~ u t ~ u +rp =frz t + rz in (0;T ) f (6.3.4) div ~ u = 0 in (0;T ) f ; (6.3.5) 281 subject to the mixed boundary conditions @~ u @N pN =h 1 @rz @N on (0;T ) 1 ~ u =rz +h 2 on (0;T ) 2 (6.3.6) and the initial condition ~ u(0;) =u 0 rz(0) in : (6.3.7) Denote the right-hand sides of (6.3.4), (6.3.6) 1 , (6.3.6) 2 , and (6.3.7) by ~ f, ~ h 1 , ~ h 2 and ~ u 0 , respectively. Note that based on the estimates in the proof of [68, Theorem 7.5], Theorem 6.2.1 holds for the special case g = ~ g = b = 0 sinces 1, but without the fourth term on the left side of (6.2.7). Applying it to the system (6.3.4)–(6.3.7), we obtain k~ uk H s=2+1=2;s+1 +kpk H s=21=2;s +krpk H s=21=2;s1 .k~ u 0 k H s +k ~ fk H s=21=2;s1 +k ~ h 1 k H s=21=4;s1=2 1 +k ~ h 2 k H s=2+1=4;s+1=2 2 ; whence k~ uk H s=2+1=2;s+1 +kpk H s=21=2;s +krpk H s=21=2;s1 .ku 0 k H s +kz(0)k H s+1 +kfk H s=21=2;s1 +krz t k H s=21=2;s1 +krzk H s=21=2;s1 +kh 1 k H s=21=4;s1=2 1 +kh 2 k H s=2+1=4;s+1=2 2 +k@rz=@Nk H s=21=4;s1=2 1 +krzk H s=2+1=4;s+1=2 2 : (6.3.8) 282 We now need to show that each of the terms on the right-hand side of (6.3.8) is dominated by the right- hand side of (6.2.7). The rst, third, sixth, and seventh terms in (6.3.8) already appear in (6.2.7), so we only need to estimate the remaining ones. For the second term, we have kz(0)k H s+1.kg(0)k H s1 =k divu 0 k H s1.ku 0 k H s: For the fourth term in (6.3.8), we write krz t k H s=s1=2;s1.krz t k L 2 H s1 +krz t k H s=21=2 L 2 .k~ gk L 2 H s2 +kbk L 2 H s1 +k~ gk H s=21=2 H 1 +kbk H s=21=2 L 2 .kk H s=21=2;s1 +kbk H s=21=2;s1; where we used the elliptic regularity for (6.3.3) in the second inequality. For the fth term in (6.3.8), we have krzk H s=21=2;s1 =krgk H s=21=2;s1.kgk L 2 H s +kgk H s=21=2 H 1.kgk H s=2;s; where we used Lemma 6.3.2 in the last step. For the eighth term in (6.3.8), we write k@rz=@Nk H s=21=4;s1=2 1 .k@rz=@Nk L 2 H s1=2 ( 1 ) +k@rz=@Nk H s=21=4 L 2 ( 1 ) ; (6.3.9) where for simplicity of notation, we write 1 instead of (0;T ) 1 as the time interval is understood. For the rst term in (6.3.9), we use the Sobolev trace inequality and obtain k@rz=@Nk L 2 H s1=2 ( 1 ) .kD 2 zk L 2 H s.kgk L 2 H s; 283 which is dominated by the third term in (6.2.7). For the second term in (6.3.9), we use the space-time trace inequality in Lemma 6.3.1 to write k@rz=@Nk H s=21=4 L 2 ( 1 ) .kD 2 zk H s=2;s.kgk H s=2;s: Regarding the ninth term in (6.3.8), we have krzk H s=2+1=4;s+1=2 2 .krzk L 2 H s+1=2 ( 2 ) +krzk H s=2+1=4 L 2 ( 2 ) : (6.3.10) For the rst term on the right-hand side of (6.3.10), we use the trace inequality to write krzk L 2 H s+1=2 ( 2 ) .krzk L 2 H s+1.kzk L 2 H s+2.kgk L 2 H s: For the second term, we claim that (6.3.2) 3 implies krzk L 2 ( 2 ) .kzk H 1 ( 2 ) ; (6.3.11) for allt2 (0;T ). To prove this, note that, by straightening of the boundary and using a partition of unity, we may assume thatz is smooth and compactly supported inR 3 , while 2 is replaced byR 2 =R 3 \fx 3 = 0g. Then@z=@x 3 = 0 onR 2 , and the inequality (6.3.11) follows. Now, using (6.3.11), we get krzk H s=2+1=4 L 2 ( 2 ) .kzk H s=2+1=4 H 1 ( 2 ) .kzk H s=2+1=4 H 3=2 .kz t k H s=23=4 H 3=2 +kzk L 2 H 3=2.kk H s=23=4 H 1=2 +kbk H s=23=4 H 1=2 +kgk L 2 L 2 .kk H s=23=4 L 2 +kbk H s=23=4 H 1=2 +kgk L 2 L 2: (6.3.12) 284 Finally, for the second term on the far right side of (6.3.12), we apply Lemma 6.3.2 and obtainkbk H s=23=4 H 1=2. kbk H s=21=s;s1. Therefore, all the terms on the far right side of (6.3.12) are bounded by the right-hand side of (6.2.7). This concludes estimates for all the terms in (6.3.8), and (6.2.7) is proven, except for the bound of the fourth term on the left, i.e., the termkpk H s=21=4;s1=2 1 . 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Abstract (if available)
Abstract
This dissertation consists of six chapters among three topics - Mach limit problem, the compressible Euler equations, and fluid-structure interaction.
In chapter one, we address the Mach limit problem for the Euler equations in the analytic spaces. We prove that, given analytic data, the solutions to the compressible Euler equations are uniformly bounded in a suitable analytic norm and then show that the convergence toward the incompressible Euler solution holds in the analytic norm. We also show that the same results hold more generally for Gevrey data with the convergence in the Gevrey norms. In chapter two, we extend the Mach limit problem for the Euler equations to exterior domains with analytic boundary. We prove the existence of tangential analytic vector fields for the exterior domain with constant analyticity radii, and introduce an analytic norm in which we distinguish derivatives taken from different directions. In chapter three, we consider the Mach limit problem of the Euler equations for isentropic fluids in the analytic spaces. The results extend the isentropic fluids in [77] to more general pressure laws.
In chapter four, we derive a priori estimates for the compressible free boundary Euler equations in the case of a liquid without surface tension. We provide a new weighted functional framework which leads to the improved regularity of the flow map by using the Hardy inequality. One of main ideas is to decompose the initial density function. It is worth mentioning that in our analysis we do not need the higher order wave equation for the density.
In chapter five, we address a system of equations modeling a compressible fluid interacting with an elastic body in dimension three. We prove the local existence and uniqueness of a strong solution when the initial velocity belongs to the space $H^{2+\epsilon}$ and the initial structure velocity is in $H^{1.5+\epsilon}$, for some constant $\epsilon\in (0, 1/2)$. In chapter six, we provide a maximal regularity theorem for the linear Stokes equation with a non-homogeneous divergence condition in a bounded domain $\Omega\subset R^3$ and with the Neumann boundary conditions. We prove the existence and uniqueness of solutions such that the velocity belongs to the space $H^{(s+1)/2, s+1} ((0, T) \times \Omega)$, where $s ≥ 1$.
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Creator
Li, Linfeng
(author)
Core Title
Mach limits and free boundary problems in fluid dynamics
School
College of Letters, Arts and Sciences
Degree
Doctor of Philosophy
Degree Program
Mathematics
Degree Conferral Date
2023-05
Publication Date
03/31/2023
Defense Date
03/22/2023
Publisher
University of Southern California
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fluid dynamics,free boundary problem,Mach limits,OAI-PMH Harvest
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theses
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English
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Jang, Juhi (
committee chair
), Kukavica, Igor (
committee chair
), Ghanem, Roger (
committee member
), Ziane, Nabil (
committee member
)
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linfengli507@gmail.com,lli265@usc.edu
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Tags
fluid dynamics
free boundary problem
Mach limits