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Stability analysis of nonlinear fluid models around affine motions
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Stability analysis of nonlinear fluid models around affine motions
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STABILITY ANALYSIS OF NONLINEAR FLUID MODELS AROUND AFFINE MOTIONS by Calum Rickard 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 APPLIED MATHEMATICS August 2021 Copyright 2021 Calum Rickard Logic is the foundation of the certainty of all the knowledge we acquire. -Leonhard Euler ii Acknowledgments First and foremost, I would like to thank my advisor Professor Juhi Jang. Without her support, this dissertation would not have been possible. From her excellent teaching, to introducing me to mathematical uid dynamics, to guiding my work on very interesting research problems and to providing invaluable career guidance, she has provided consistent enthusiasm, patience, insight and mentorship. Many thanks to Professor Igor Kukavica for always being very supportive of my work and for cultivating a culture of research excellence among analysis graduate students. His support has given me condence throughout my graduate research which I am highly grate- ful for. I also owe great thanks to Professor Susan Friedlander for being highly encouraging of my research and providing me with many professional development opportunities throughout my graduate studies. These opportunities have prepared me well as I pursue an academic career and I very much appreciate her support. I would also like to thank my collaborator Professor Mahir Had zi c who I am very fortunate to have had the pleasure to work with and he has been extremely supportive of the research in this dissertation. Further thanks to Professor Sergey Lototsky for guiding me through the early stages of my PhD and later supporting my teaching. In addition, I will always remember fondly the many discussions about triathlon we would have. I am grateful to Professor Mohammed Ziane and Professor Paul Newton for acting as committee members, and providing valued support and comments. iii I would like to thank the Department of Mathematics at USC, in particular Professor Susan Montgomery, Professor Eric Friendlander and Amy Yung, for supporting me as a graduate student in numerous ways throughout my PhD. The support of the Fulbright Foreign Student Program and Fulbright New Zealand has been immense in helping me to pursue graduate studies at USC as a New Zealander and I am very grateful for their assistance. I am fortunate to have had Nicolle Sandoval Gonz alez and Fanhui Xu as graduate student mentors during my time at USC who have helped me greatly as I have progressed through the program. The USC Triathlon team has provided me with a means of personal growth and fulll- ment outside of mathematics and I have made lifelong friendships through the club. My father Dr. Graham Rickard and mother Sandra Rickard have always been there for me from New Zealand and I would like to thank them for their continued support of my education and career. Finally, I would like to thank my partner Abegail for her unwavering support of my studies: I am very lucky to have met someone so encouraging and kind. iv Table of Contents Epigraph ii Acknowledgments iii Abstract ix Part I: Global Existence of the Nonisentropic Compressible Euler Equa- tions with Vacuum Boundary Surrounding a Variable Entropy State 1 Chapter 1: Introduction 1 1.1 Nonisentropic Ane Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Chapter 2: Formulation and Main Global Existence Result 14 2.1 Lagrangian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.2 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.3 High-order Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.4 Main Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Chapter 3: Curl Estimates 26 Chapter 4: Energy Estimates 46 Chapter 5: Large Adiabatic Constant 62 5.1 Curl Estimates > 5 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 5.2 Energy Estimates > 5 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 v Chapter 6: Appendices 74 6.1 based inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 6.2 Derivative Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 6.3 Weighted Sobolev-Hardy Inequality . . . . . . . . . . . . . . . . . . . . . . . 77 6.4 Scaling Analysis and Ane Motion . . . . . . . . . . . . . . . . . . . . . . . 77 Part II: Global Solutions to the Compressible Euler Equations with Heat Transport by Convection around Dyson's Isothermal Ane Solutions 82 Chapter 7: Introduction 82 7.1 Existence Theories for the Euler System . . . . . . . . . . . . . . . . . . . . 84 7.2 Isothermal Ane Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 Chapter 8: Formulation and Main Global Existence Result 89 8.1 Lagrangian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 8.2 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 8.3 High-order Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 8.4 Main Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 Chapter 9: Finite Propagation 102 Chapter 10: High Order Energy Estimates 119 Chapter 11: High Order Curl Estimates 131 Chapter 12: Energy Inequality and Proof of the Main Theorem 145 Chapter 13: Appendices 148 13.1 Energy Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 13.2 Time Based Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 13.3 Local Well Posedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 13.4 Curl Equations Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 vi 13.5 Coercivity Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 Part III: The Vacuum Boundary Problem for the Spherically Symmetric Compressible Euler Equations with Positive Density and Unbounded Entropy 161 Chapter 14: Introduction 161 14.1 Spherically Symmetric Lagrangian Coordinates . . . . . . . . . . . . . . . . 165 14.2 Nonisentropic Spherically Symmetric Ane Motion . . . . . . . . . . . . . . 166 Chapter 15: Formulation and Main Global Existence Result 170 15.1 Perturbation of Ane Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 170 15.2 The H Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 15.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 15.4 High-order Norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 15.5 Main Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 Chapter 16: Energy Estimates 180 16.1 Dierentiated Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180 16.2 Time Weight Manipulation . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 16.3 Energy Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 16.4 High-order Quantities from Energy Identity . . . . . . . . . . . . . . . . . . 185 16.5 Main Energy Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 Chapter 17: Proof of the Main Theorem 203 Chapter 18: Appendices 206 18.1 Dierential Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 18.2 Hardy-Sobolev Embedding . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 18.3 Time Based Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 18.4 Local Well-Posedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 vii 18.5 Coercivity Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 Bibliography 212 viii Abstract This dissertation consists of three parts. In Part I, global existence for the nonisentropic compressible Euler equations with vacuum boundary for all adiabatic constants > 1 is shown through perturbations around a rich class of background nonisentropic ane motions. The notable feature of the nonisentropic motion lies in the presence of non- constant entropies, and it brings a new mathematical challenge to the stability analysis of nonisentropic ane motions. In particular, the estimation of the curl terms requires a careful use of algebraic, nonlinear structure of the pressure. With suitable regularity of the underlying ane entropy, we are able to adapt the weighted energy method developed for the isentropic Euler [20] to the nonisentropic problem. For large values, inspired by [48], we use time-dependent weights that allow some of the top-order norms to potentially grow as the time variable tends to innity. We also exploit coercivity estimates here via the fundamental theorem of calculus in time variable for norms which are not top-order. In Part II, global solutions to the compressible Euler equations with heat transport by convection in the whole space are shown to exist through perturbations of Dyson's isother- mal ane solutions [12]. This setting presents new diculties because of the vacuum at innity behavior of the density. In particular, the perturbation of isothermal motion intro- duces a Gaussian function into our stability analysis and a novel nite propagation result is proven to handle potentially unbounded terms arising from the presence of the Gaussian. Crucial stabilization-in-time eects of the background motion are mitigated through the use of this nite propagation result however and a careful use of the heat transport for- mulation in conjunction with new time weight manipulations are used to establish global ix existence. The heat transport by convection oers unique physical insights into the model and mathematically, we use a controlled spatial perturbation in the analysis of this feature of our system which leads us to exploit source term estimates as part of our techniques. In Part III, global stability of the spherically symmetric nonisentropic compressible Euler equations with positive density around global-in-time background ane solutions is shown in the presence of free vacuum boundaries. Vacuum is achieved despite a non-vanishing density by considering a negatively unbounded entropy and we use a novel weighted energy method whereby the exponential of the entropy will act as a changing weight to handle the degeneracy of the vacuum boundary. Spherical symmetry introduces a coordinate singular- ity near the origin for which we adapt a method developed for the Euler-Poisson system [17] to our problem. x Part I Global Existence of the Nonisentropic Compressible Euler Equations with Vacuum Boundary Surrounding a Variable Entropy State Chapter 1 Introduction We consider compressible Euler equations for ideal gases in three space dimensions (@ t u + uru) +rp = 0; (1.0.1) @ t + div(u) = 0; (1.0.2) @ t + ur + ( 1) div(u) = 0; (1.0.3) where u is the uid velocity vector eld, is the density, is the internal energy, p is the pressure and > 1 is the adiabatic constant. Coupled with the equation of state for an ideal gas p(;) = ( 1); (1.0.4) 1 equations (1.0.1)-(1.0.4) describe the compressible ow of an inviscid, non-conducting and adiabatic gas. It is often convenient to use another unknown - the entropy S - instead of the internal energy. Equation (1.0.3) is then equivalently replaced by @ t S + urS = 0; (1.0.5) and the equation of state reads p(;S) = e S : (1.0.6) Note that the entropy is just transported by the ow, and therefore the entropy formulation will be in particular useful in Lagrangian coordinates. See Section 2.1. A special case where the entropy S remains constant represents the isentropic process and in that case, the equation of state relates the pressure of the gas to the density only: p = and the unknown variables for the system are the density and velocity. In this part, we are interested in the dynamics of the nonisentropic gas whose entropy changes in both time and space but is nite. We study the vacuum free boundary problem: that is, we consider the Euler equations (1.0.1)-(1.0.3) and (1.0.5) in the following time dependent open bounded domain (t)R 3 with boundary @ (t) where t2 [0;T ] for some T > 0: The boundary conditions are then the physical vacuum boundary condition coupled with kinematic boundary condition p = 0 on @ (t); (1.0.7) 1< @ @n < 0 on @ (t); (1.0.8) V(@ (t)) = u n(t) on @ (t); (1.0.9) 2 with n the outward unit normal vector to @ (t), @ @n the outward normal derivative, and V(@ (t)) the normal velocity of @ (t). The condition (1.0.8) is most convenient for us to express in terms of the internal energy . When the entropy is bounded from below and from above, the physical vacuum condition (1.0.8) can be written as 1< @ 1 @n < 0: (1.0.10) Physically, this condition implies a nontrivial acceleration of the gas at the boundary in the normal direction and mathematically, it implies an inherent lack of regularity of the enthalpy at the vacuum boundary. Moreover it is clear from (1.0.10) that the solution is not even C 1 at the boundary. This is not just a mathematical curiosity, but it is in fact fundamental to give up the requirement of smoothness in order to track the evolution of compactly supported gases; we are led to study the free boundary problem and it turns out that (1.0.10) is a key condition to guarantee well-posedness of the free boundary. The physical vacuum condition is not merely a technical condition but it is realized for a wide range of physical systems for gaseous uids. In particular a key impetus for the physical vacuum condition (1.0.10) comes from astrophysics. A famous class of equilibria of the gravitational Euler-Poisson system, known as the Lane-Emden stars, satisfy (1.0.10) and any rigorous theory that attempts to study the nonlinear dynamics in the vicinity of Lane- Emden stars must contend with (1.0.10) [5,14,23]. It also arises naturally in the context of Euler equations with damping for the gas ow through a porous medium [33, 34, 53]. For more detail on the physical vacuum, we refer to [9,24{27,36]. Finally, we recall we are working on the time interval [0;T ] and we consider the initial conditions ((0;); u(0;);S(0;); (0)) = ( 0 ; u 0 ;S 0 ; 0 ): (1.0.11) Collectively, we will study the vacuum free boundary nonisentropic Euler system (1.0.1)- (1.0.11). 3 Before we move on, we brie y discuss some known results for the Euler equations. Due to vast literature, we will only mention the works relevant to the present part. We refer to [9,20,27,35] and reference therein for more thorough review. We begin with the Cauchy problem in the whole space. It is well-known that the Euler equations are hyperbolic and the existence of C 1 local-in-time positive density solutions follows from the theory of symmetric hyperbolic systems [30, 37]. Serre [47] and Grassin [15] proved global existence for a special class of initial data in the whole space by the perturbation of expansive wave solutions to the vectorial Burgers equation with linearly growing velocities at innity - a related idea was used in the work of Rozanova [45]. On the other hand, Sideris [49] showed that singularities must form if the density is a strictly positive constant outside of a bounded set. A detailed description of shock formation starting with smooth initial data for irrotational relativistic uids around is given by Christodoulou [7] for special-relativistic uids and Christodoulou-Miao [8] in the nonrelativistic case. For a more general framework covering a wider class of equations leading to shock formation see the works of Speck and Luk-Speck [35, 52]. Makino-Ukai- Kawashima [40] proved that singularities form starting from compactly supported smooth solutions. We remark that these singularity and shock formation results do not apply to the physical vacuum free boundary problem. In the vacuum free boundary framework, a lot of important progress has been made in the past decade. Local well-posedness for compressible Euler equations with physical vacuum has been established by Coutand-Shkoller [9] and Jang-Masmoudi [27]. First examples of global-in-time solutions surrounded by vacuum and satisfying the physical vacuum condition were given by Sideris [51]. These are the so-called ane motions found by a separation-of-variables ansatz for the Lagrangian ow map (t;y) (t;y) =A(t)y; (1.0.12) where t7!A(t) is an unknown 3 3 matrix. Such an ansatz severely reduces the dynamic degrees of freedom resulting in an ordinary dierential equation (ODE) for the matrix 4 A(t). We will therefore refer to such solutions as ODE-type solutions. It should be noted, that the idea of considering the ansatz (1.0.12) was considered before in the context of nonisentropic ows and goes back to the works of Ovsyannikov [41] and Dyson [12], wherein the ODE satised byA(t) was already discovered. The ane motions constructed by Sideris importantly satisfy the physical vacuum condition (1.0.10), which is a critical assumption in the general well-posedness framework developed in [9,27]. We remark that the ansatz (1.0.12) has been widely exploited in various physical appli- cations beyond the solutions considered in this dissertation. For instance, it is explicitly seen to hold for special physical solutions to the Navier-Stokes equations; specically, as described in [38], a jet which compresses in one coordinate direction and stretches in another, a uid which drains purely along one direction from planes parallel to the alternate coor- dinate plane, and the pure rotation of a rigid body, all obey (1.0.12). Another physical application of the ane ansatz (1.0.12) can be found in the dynamics of gaseous stars gov- erned by the Euler-Poisson system; in particular, in [14,39], the star collapse and expansion were demonstrated by using (1.0.12) in the radially symmetric setting. In the isentropic case, the nonlinear stability of the Sideris solutions was shown by Had zi c-Jang [20] for 2 (1; 5 3 ] and then extended to the full range > 1 by Shkoller- Sideris [48]. In a recent work [42], Parmeshwar-Had zi c-Jang showed the global existence of expanding solutions with small densities without relying on the background ane solutions, again in the class of isentropic ows. Also very recently in the isentropic setting with damping [53], Zeng has established the existence of almost global solutions by perturbing background approximate solutions known as Barenblatt solutions with sub-linear expansion which contrasts with the linear expansion considered otherwise. By contrast to the isentropic case, it has remained an open question to construct open sets of initial data in the physically important nonisentropic case that lead to global exis- tence in the presence of free vacuum boundaries. This is the main goal of this part. Indeed, the expansion of gas into vacuum is an important physical phenomenon [16]. Our result 5 demonstrates the linear expansion of nonisentropic gas into vacuum is a stable mechanism that avoids shock formation. 1.1 Nonisentropic Ane Motion As mentioned above, Sideris [51] constructed a family of ane motions satisfying the phys- ical vacuum boundary condition. Such solutions are blobs of gas initially occupyingB 1 (0) - the unit ball inR 3 . Their evolving support is given as the image ofB 1 (0) under the matrix A(t) i.e. (t) =A(t)B 1 (0), wheret7!A(t) is an a priori unknown matrix. This generically gives us a gas supported on an evolving ellipsoid. At the level of Lagrangian coordinates, this translates into separating variables and writing the ow map in the form (1.0.12), see [12, 41, 51]. After plugging this back into the Lagrangian formulation of the problem, an algebraic manipulation leads to the following fundamental system of ODEs satised by A(t) A 00 (t) =(detA(t)) 1 A(t) > ; (1.1.1) (A(0);A 0 (0);)2 GL + (3)M 33 R + ; (1.1.2) for xed > 0. In the aboveM 33 denotes the set of 3 3 matrices overR and GL + (3) = fA2 M 33 : detA > 0g. With A2 C(R; GL + (3))\C 1 (R;M 3 ) solving this system of ODEs, the associated solution of the Euler equations is given by u A (t;x) =A 0 (t)A(t) 1 x; (1.1.3) A (t;x) = (jA(t) 1 xj)=(detA(t)); (1.1.4) A (t;x) = (jA(t) 1 xj)=(detA(t)) 1 : (1.1.5) We should think of the density A and the internal energy A as the basic proles = (r) and = (r), modulated by the matrix A(t) through (1.1.4){(1.1.5) respectively. 6 Following Sideris [51], to see why we have the fundamental system of ODEs (1.1.1) rst write x = A(t)y with y2 B 1 (0) which follows from (t) = A(t)B 1 (0). Then using the chain rule d dt u A (t;x(t;y)) =@ t u A +@ t xru A =@ t u A +A 0 (t)yru A =@ t u A + u A ru A : (1.1.6) On the other hand with (1.1.3), d dt u A (t;x(t;y)) = d dt (A 0 (t)y) =A 00 (t)y: (1.1.7) Also using the chain rule, with r =jyj as above, r x =A > r y =A > y@ r : (1.1.8) Then substituting these formulas as well as (1.1.4)-(1.1.5) and the equation of state (1.0.4) into the momentum equation (1.0.1) we nd (detA(t)) 1 (r)A 00 (t)y + (detA(t)) A(t) > y p 0 (r) = 0; (1.1.9) where p = ( 1) . Separate time and space variables using the common constant > 0 and eliminate y to obtain A 00 (t) =(detA(t)) 1 A(t) > ; (1.1.10) p 0 (r) =r (r): (1.1.11) Hence we have derived the fundamental system ODEs for A(t) (1.1.1) and also the funda- mental ODE in space, see (1.1.13) below. Next we require the density prole has the following properties (Lemma 1 [51]) 2C 0 [0; 1]\C 1 [0; 1); (1.1.12a) 7 (r)> 0 for r2 [0; 1); (1.1.12b) 0 (0) = (1) = 0; (1.1.12c) 0< lim r!1 (1r) (r)<1 for some > 0: (1.1.12d) Now the corresponding internal energy prole is found by solving the fundamental ODE in space ( p) 0 (r) =r (r); (1.1.13) where we recall p = ( 1) and > 0. From (1.0.5) the entropy of the Sideris ane solution satises, using the same argument as (1.1.6), d dt S A (t;x(t;y)) =@ t S A + u A rS A = 0: (1.1.14) Hence S A (t;x) depends on y only. Moreover using (1.1.4)-(1.1.5) and the equivalent equa- tions of state (1.0.4)-(1.0.6), ( 1) (r) (r) (detA(t)) = (r) (detA(t)) e S A (y) : (1.1.15) Therefore canceling (detA(t)) we see thatS A (y) must be a function ofr =jyj =jA(t) 1 xj, that is, the entropy of the Sideris ane solutions is given by S A (t;x) = S(jA(t) 1 xj): (1.1.16) Here, S = S(r) is dened through the relationship p = ( ) e S . Through (1.1.13) we can obtain an explicit formula for S in terms of which together with (1.1.12a)-(1.1.12d) denes the nonisentropic entropy prole associated with the Sideris ane solutions S(r) = ln 1 r ` (`)d` ( (r)) ! : (1.1.17) 8 This formula highlights an important feature of the nonisentropic ane setting: the solu- tions are dened as a class of functions through the choice of which becomes an additional parameter in the solution scheme. In particular, the space which \parametrizes" the non- isentropic ane motions is innite dimensional, by contrast to the isentropic case: Remark 1.1.1. In the isentropic case, is xed as follows [51] (r) = ( 1) 2 (1r 2 ) 1 1 : (1.1.18) On the other hand, in the nonisentropic case, the choice of is essentially arbitrary as long as the constraints (1.1.12a)-(1.1.12d) are met. Remark 1.1.2. One can in fact provide a dierent derivation of the ane motions using only the Eulerian formulation of the problem. This approach exploits heavily the rich sym- metries of the problem. The details are given in Appendix 6.4. Motivated by physical considerations [2], it is important to isolate the ane motions with uniformly bounded entropies up to the vacuum boundary. Using the formula (1.1.17), we obtain the following simple, but important characterization of ane entropy behavior. Lemma 1.1.3. For r2 [0; 1); 0< (e S )(r)<1. Furthermore (e S )(1) = lim r!1 (e S )(r) = 8 > > > > > > < > > > > > > : 1 ( 1)> 1 (L 1 )= ( 1) = 1 0 ( 1)< 1 ; where > 0 from (1.1.12d) is the particular value such that 0< lim r!1 (1r) (r) :=L<1: 9 Proof. First by (1.1.12a)-(1.1.12b), 0< (r)<1 for r2 [0; 1) and hence 0< (e S )(r) = 1 r ` (`)d` (r) <1 for r2 [0; 1): Now lim r!1 (e S )(r) = lim r!1 1 r ` (`)d` (r) = lim r!1 (1r) (r) 1 r ` (`)d` (1r) = L lim r!1 1 r ` (`)d` (1r) = L lim r!1 r (r) (1r) 1 (L'Hospital's Rule) = L lim r!1 (r) (1r) r (1r) ( 1)1 = L 1 lim r!1 r (1r) ( 1)1 ; wherefrom the claim follows. Lemma 1.1.3 shows that the only value of allowing a uniformly bounded entropy is = 1 1 . In this part, we restrict our attention to the class of nonisentropic ane solutions with nite entropies: namely, we demand = 1 1 > 0 in (1.1.12d). Then with = 1 1 and in view of the condition (1.1.12a)-(1.1.12c), we shall from now on consider the proles of the form (r) = (1r) 1 1 (r); (1.1.19) where 2 C k [0; 1]; > 0 satisfying 0 (0) = 1 1 (0) with k2 N to be specied. Here we demand the condition 0 (0) = 1 1 (0) to ensure the regularity of at the center as in (1.1.12c). In the following, we show that the ane entropy enjoys the same regularity as . Lemma 1.1.4. Let 2C k [0; 1]; > 0 be given for some k2Z 0 . With (1.1.19), we have that e S 2C k [0; 1]. Proof. The case k = 0 follows from Lemma 1.1.3. For k 1, rst notice that by (1.1.17) and (1.1.19) (e S )(r) = 1 r `(`)(1`) 1 1 d` (1r) 1 ((r)) : (1.1.20) 10 To prove higher-order regularity for S we rst consider general functions of the form z(r) := 1 r a(`)(1`) X d` (1r) Y b(r) ; (1.1.21) wherea;b2C k [0; 1],b6= 0 andX;Y 2R >0 withY X +1. Now use the product rule, the fundamental theorem of calculus to dierentiate z, and then integrate-by-parts to obtain z 0 (r) =a(r)(1r) XY [b(r)] 1 + 1 r a(`)(1`) X d` Y (1r) Y1 (b(r)) 1 [b(r)] 2 b 0 (r)(1r) Y = Y X + 1 1 (1r) XY a(r)[b(r)] 1 + Y X + 1 1 r a 0 (`)(1`) X+1 d` (1r) Y +1 b(r) 1 X + 1 a(r)[b(r)] 2 b 0 (r)(1r) X+1Y 1 X + 1 b 0 (r) 1 r a 0 (`)(1`) X+1 d` (1r) Y [b(r)] 2 =: (i) + (ii) + (iii) + (iv): (1.1.22) The key step in the above calculation was to use integration by parts to obtain a desirable form for (i) by combining terms to avoid potentially unbounded negative powers of (1r). On this note, the term (i) is either 0 (when Y =X + 1) or C k (when Y X). Also term (iii) is C k1 . For term (ii) notice that j(ii)j = Y X + 1 1 r a 0 (`)(1`) X+1 d` (1r) Y +1 b(r) (sup 0r1 ja 0 (r)j) 1 r (1`) X+1 d` (1r) Y +1 jb(r)j = (sup 0r1 ja 0 (r)j) 1 X+2 (1r) X+2 (1r) Y +1 jb(r)j : Hence (ii) is bounded since Y + 1X + 1 + 1 =X + 2. Similarly, term (iv) is bounded. Furthermore, we note that terms (ii) and (iv) are of the same general form as z(r) except with C k1 functions replacing the C k functions a;b inside the expression for z. Therefore we can repeat the same procedure as for the rst derivative abovek1 times to obtain that z2 C k [0; 1]. Since e S can be realized as a special instance of the function z with a(r) = 11 r(r), b(r) =(r) , X = 1 1 , and Y =X + 1, see (1.1.20), we conclude e S 2C k [0; 1], as claimed. We collect these results into an important consequence that will be used throughout this part. Corollary 1.1.5. Let 2C k [0; 1]; > 0 be given for some k2Z 0 . With (1.1.19), then there exist constants 0<cC such that for all r2 [0; 1] 0<c (e S )(r)C; (1.1.23) d dr j (e S )(r) C; 1jk: (1.1.24) Proof. The positive lower bound in (1.1.23) follows from Lemma 1.1.3 and the upper bounds in (1.1.23)-(1.1.24) follow from Lemma 1.1.4 since C k functions are bounded. To conclude our characterization of nonisentropic ane motion, we nally provide pre- cise asymptotics-in-time for A(t). Lemma 1.1.6. Consider the initial value problem (1.1.1){(1.1.2) with> 0. For 2 (1; 5 3 ], the unique solution A(t) to the fundamental system (1.1.1){(1.1.2) has the property detA(t) 1 +t 3 ; t 0 (1.1.25) Furthermore in this case, there exist matrices A 0 ;A 1 ;M(t) such that A(t) =A 0 +tA 1 +M(t); t 0: (1.1.26) where A 0 ;A 1 are time-independent and M(t) satises the bounds kM(t)k =o t!1 (1 +t); k@ t M(t)k. (1 +t) 33 : (1.1.27) 12 For > 5 3 , given matrices A 0 ;A 1 with A 1 positive denite, there exists a unique solution A(t) to the fundamental system (1.1.1){(1.1.2) such that (1.1.25), (1.1.26) and (1.1.27) hold. Proof. For all > 1, we use Theorem 3 and Lemma 6 from [51] to obtain the results. For 2 (1; 5 3 ], we additionally use Lemma A.1 from [20]. For > 5 3 , we additionally use Lemma 1 from [48]. In this part we restrict our attention to the class of nonisentropic ane solutions expand- ing linearly in each coordinate direction: namely we require detA(t) 1 +t 3 ; t 0: (1.1.28) By Lemma 1.1.6, for 2 (1; 5 3 ] this is not a restriction at all in fact since A(t) will imme- diately satisfy (1.1.28). For > 5 3 , Lemma 1.1.6 shows there exists a rich class of A(t) satisfying (1.1.28). We denote the set of ane motions under consideration byS . To recap, the setS is parametrized by the quadruple (A(0);A 0 (0);;)2 GL + (3)M 33 R + Z k ; (1.1.29) where Z k := 2C k [0; 1] :> 0; 0 (0) = 1 1 (0) (1.1.30) and we take k2 N suciently large (to be specied later in Theorems 2.4.1 and 2.4.2). We recall (A(0);A 0 (0);) are parameters for the fundamental system (1.1.1){(1.1.2) and appears in the formula for (1.1.19). The choice of 2C k [0; 1] highlights a new freedom in the specication of ane motions with respect to the isentropic setting. With our set of nonisentropic ane motions S in hand, the goal of this part is to establish the global-in-time stability of the nonisentropic Euler system (1.0.1)-(1.0.11) for all > 1 by perturbing around the expanding ane motions. 13 Chapter 2 Formulation and Main Global Existence Result 2.1 Lagrangian Coordinates In order to analyze the stability problem for ane motions, we will use the Lagrangian formulation that brings the problem onto the xed domain. We rst dene the ow map as follows @ t (t;y) = u(t;(t;y)); (2.1.1) (0;y) = 0 (y); (2.1.2) where 0 is a suciently smooth dieomorphism to be specied. We introduce the notation A := [D] 1 (Inverse of the Jacobian matrix) (2.1.3) J := det[D] (Jacobian determinant) (2.1.4) f := (Lagrangian density) (2.1.5) g :=S (Langrangian entropy) (2.1.6) a :=J A (Cofactor matrix): (2.1.7) 14 In this framework material derivatives reduce to pure time derivatives and in particular, the entropy equation (1.0.5) is simply reformulated as @ t g(t;y) = 0 which implies g(t;y) =S 0 ( 0 (y)): (2.1.8) In other words, the entropy remains constant along uid particle worldline given by ow maps. Furthermore, it is well-known [9, 27] that the conservation of mass equation (1.0.2) gives f(t;y) = (J (t;y)) 1 0 ( 0 (y))J (0;y): (2.1.9) Finally using the nonisentropic equation of state p = e S the momentum equation (1.0.1) is reformulated as f@ tt i + [A ] k i (f e g ) ;k = 0: (2.1.10) Here we use coordinatesi = 1; 2; 3 with the Einstein summation convention and the notation F; k to denote the k th partial derivative of F . Next introduce the following notations S(y) :=S 0 ( 0 (y)); (2.1.11) w(y) := [ 0 ( 0 (y))J (0;y)] 1 : (2.1.12) Then using [A ] k i =J 1 [a ] k i , we obtain w @ tt i + [a ] k i (w 1+ J ( 1 1) e S ) ;k = 0; (2.1.13) where we set := 1 1 : Using the Piola identity ([a ] k i ) ;k = 0; (2.1.14) 15 we rewrite (2.1.13) as w @ tt i + (w 1+ e S [A ] k i J 1 ) ;k = 0: (2.1.15) Ane motions described in Section 1.1 can be realized as special solutions of (2.1.15) of the form (t;y) =A(t)y. In this caseA > = [D] > =A(t) > andJ = detA. Hence the ansatz transforms (2.1.15) into w A tt y + (detA) 1 A > r(w 1+ e S ) = 0: (2.1.16) We have that w 1+ e S is independent of t and hence (2.1.16) will hold if we require A tt =(detA) 1 A > ; (2.1.17) w y =r(w 1+ e S ); (2.1.18) for > 0. At this stage, we demand [w(y)] = (1jyj) (jyj); (2.1.19) S(y) = S(jyj); (2.1.20) where we require by recalling (1.1.30) 2Z k for k2N taken suciently large to be specied later by Theorems 2.4.1 and 2.4.2: (2.1.21) Note with Corollary 1.1.5 and (2.1.20)-(2.1.21), e S 2C k [0; 1] with e S c> 0. We observe that (2.1.17)-(2.1.21) are nothing but the ane solutions described in Section 1.1, and produce the set of ane motionsS under consideration. Fix an element ofS . Remark 2.1.1. Through (2.1.11)-(2.1.12), the initial data 0 , S 0 for our problem are chosen such that (2.1.19)-(2.1.21), which include the regularity demands on , are satised. 16 With an ane motion fromS xed, we dene the modied ow map :=A 1 . Then A > = A > A > ,J = (detA)J whereA > ,J are the equivalents of (2.1.3), (2.1.4) respectively. Now from (2.1.15) we have w (@ tt i + 2[A 1 ] i` @ t A `j @ t j + [A 1 ] i` @ tt A `j j ) + (detA) 1 [A 1 ] i` [A 1 ] j` (w 1+ e S [A ] k j J 1 ) ;k = 0: (2.1.22) Via (2:1:17) we rewrite the above equation as w ((detA) 1 3 @ tt i +2(detA) 1 3 [A 1 ] i` @ t A `j @ t j )+w i` ` +(w 1+ e S ij [A ] k j J 1 ) ;k = 0; (2.1.23) where the notation := (detA) 2 3 A 1 A > has been introduced. Next make a change of time variable by setting d dt = (detA) 1 3 : Then we can formulate (2.1.23) as w ((detA) 1 @ i 1 3 (detA) 2 (detA) @ i + 2(detA) 1 [A 1 ] i` @ A `j @ j ) +w i` ` + (w 1+ e S ij [A ] k j J 1 ) ;k = 0: (2.1.24) Writing A =O where := (detA) 1 3 and O2 SL(3), we have A 1 A = 1 I +O 1 O . The equation is then w ( 3 3 @ i + 3 4 @ i + 2 3 3 ij @ j ) +w i` ` + (w 1+ e S ij [A ] k j J 1 ) ;k = 0: (2.1.25) where we have dened := O 1 O . Note (y) y corresponds to ane motion. Intro- ducing the perturbation (;y) :=(;y)y; (2.1.26) 17 and using (2.1.18), equation (2.1.25) can be written in terms of w 3 3 @ i + @ i + 2 ij @ j +w i` ` + w 1+ e S ij [A ] k j J 1 k j ; k = 0; (2.1.27) with the initial conditions (0;y) = 0 (y); (0;y) = V(0;y) = V 0 (y); (y2 =B 1 (0)): (2.1.28) Above we have introduced the notation V :=@ which will be used interchangeably. Remark 2.1.2 (Initial entropy S 0 and initial density 0 ). The initial entropy S 0 and the initial density 0 are connected to the background ane motion via S 0 (x) =S A (( 0 A (0)) 1 (x)); 0 (x) = A (( 0 A (0)) 1 (x)) det[D( 1 0 (x))] 1 ; where the composed maps are dened by 0 (y) :=A 1 (0) 0 (y) and A (0)(y) :=A(0)y. 2.2 Notation For ease of notation rst set A :=A ; J :=J : (2.2.1) UsingA [D] = Id, we have the dierentiation formulae forA andJ @A k i =A k ` @ ` ; s A s i ; @J =JA s ` @ ` ; s (2.2.2) for @ =@ or @ =@ i , i = 1; 2; 3. 18 Let F : !R 3 and f : !R be an arbitrary vector eld and function respectively. First dene the gradient and divergence along the ow map respectively [r F] i r :=A s r F i ;s ; div F :=A s ` F ` ;s : (2.2.3) For curl estimates, introduce the anti-symmetric curl and cross product matrices respec- tively [Curl A F] i j := jm A s m F i ; s im A s m F j ; s ; [Arf F] i j := jm A s m f ;s F i im A s m f ;s F j : (2.2.4) For any k2Z 0 and any continuous, non-negative function ' : !R + , we consider the weighted L 2 norm kfk 2 k;' := 'w k jf(y)j 2 dy: (2.2.5) We generalize this denition to vector elds F and 2-tensors T : !M 33 kFk 2 k;' := 3 X i=1 kF i k 2 k;' ; kTk 2 k;' := 3 X i;j=1 kT ij k 2 k;' : (2.2.6) The weight function ' will often include the smooth cut-o function : B 1 (0)! [0; 1] which satises = 0 on B 1 4 (0) and = 1 on B 1 (0)nB 3 4 (0): (2.2.7) The following derivative operators will be used near the boundary = @ ji :=y j @ i y i @ j ; i;j = 1; 2; 3; X r :=r@ r where @ r := y r r and r =jyj; (2.2.8) which represent angular derivatives tangent to the boundary and the radial derivative nor- mal to the boundary. 19 2.3 High-order Quantities Our time weights will dier depending on whether 2 (1; 5 3 ] or > 5 3 . This is because we take a slightly dierent approach for > 5 3 by an adaptation of [48], applied to our nonisentropic setting. On this note, introduce the following dependent exponents d( ) := 8 > > < > > : 3 3 if 1< 5 3 2 if > 5 3 ; b( ) :=d( ) + 3 3 = 8 > > < > > : 0 if 1< 5 3 5 3 if > 5 3 : (2.3.1) Let N 2 N. To measure the size of the deviation , we dene the high-order weighted Sobolev norm as follows S N (; V)() := sup 0 0 n X a+jjN d( ) kX a r = @ Vk 2 a+; e S +kX a r = @ k 2 a+; e S + X a+jjN1 kr X a r = @ k 2 a++1; e S +kdiv X a r = @ k 2 a++1; e S + X a+jj=N b( ) kr X a r = @ k 2 a++1; e S + b( ) kdiv X a r = @ k 2 a++1; e S o + sup 0 0 n X jjN d( ) k@ Vk 2 ;(1 )e S +k@ k 2 ;(1 )e S + X jjN1 kr @ k 2 +1;(1 )e S +kdiv @ k 2 +1;(1 )e S + X jj=N b( ) kr @ k 2 +1;(1 )e S + b( ) kdiv @ k 2 +1;(1 )e S o : (2.3.2) Modied curl terms arise during energy estimates which are not a priori controlled by the normS N (). These are measured via the following high-order quantity 20 B N [V]() := sup 0 0 n X a+jjN1 kCurl A X a r = @ Vk 2 a++1; e S + X a+jj=N b( ) kCurl A X a r = @ Vk 2 a++1; e S o + sup 0 0 n X jjN1 kCurl A @ Vk 2 +1;(1 )e S + X jj=N b( ) kCurl A @ Vk 2 +1;(1 )e S o ; (2.3.3) withB N [] dened in the same way: replaces V in (2.3.3). 2.4 Main Theorem Before giving our main theorem, rst dene the important related quantities 1 := lim !1 () () ; 0 := d( ) 2 1 ; (2.4.1) where we recall the quantity d( ) = 8 > > < > > : 3 3 if 1< 5 3 2 if > 5 3 : Local Well-Posedness. Next, we give the local well-posedness of our system. Theorem 2.4.1. Suppose > 1. Fix N 2de + 12. Let k N + 1 in (2.1.21). Then suppose we have initial data for (2.1.27) ( 0 ; V 0 ) that satisesS N ( 0 ; V 0 ) +B N (V 0 )<1. In this case there exists a unique solution ((); V()) : !R 3 R 3 to (2.1.27)-(2.1.28) for all2 [0;T ], for someT > 0. The solution has the propertyS N (; V)()+B N [V]() 2(S N ( 0 ; V 0 ) +B N (V 0 )) for each 2 [0;T ]. Furthermore, the map [0;T ]37!S N ()2 R + is continuous. Proof. The proof follows by adapting the argument in [27]. Notably, [27] proves local well- posedness for the isentropic Euler equations using spatial weights to handle the physical vacuum condition and we are considering the nonisentropic physical vacuum setting that 21 also includes time weights. However using the regularity of S andw, which are independent of the solutions, as well as the solution independent time weights, we can obtain bounds on S N andB N through the estimates in Chapter 3 and Chapter 4. These crucial esimates let us useS N andB N in the techniques of [27]. A priori assumptions. Finally before our main theorem, make the following a priori assumptions on our local solutions from Theorem 2.4.1 S N ()< 1 3 ; kA Idk W 1;1 ( ) < 1 3 ; kDk W 1;1 ( ) < 1 3 ; kJ Idk W 1;1 ( ) < 1 3 ; (2.4.2) for all 2 [0;T ]. We are now ready to give our main theorem. Theorem 2.4.2. Suppose > 1. Fix N 2de + 12. Let k N + 1. Consider a xed quadruple (A(0);A 0 (0);;)2 GL + (3)M 33 R + Z k ; (2.4.3) parametrizing a nonisentropic ane motion from the setS so that detA(t) 1+t 3 ; t 0. Then there is an " 0 > 0 such that for every "2 [0;" 0 ] and pair of initial data for (2.1.27) ( 0 ; V 0 ) satisfyingS N ( 0 ; V 0 ) +B N (V 0 )", there exists a global-in-time solution, (; V), to the initial value problem (2.1.27)-(2.1.28). Proof. With our a priori assumptions, our high-order quantities for our local solution from Theorem 2.4.1 will be shown to satisfy the curl and energy inequalities (2.4.5)-(2.4.7) stated below in our curl estimates and energy estimate Propositions. Then via a similar continuity argument to that presented in [20] we can rstly show that our a priori assumptions are in fact improved thus justifying making them originally. For instance, by the fundamental theorem of calculus and our modied weighted Sobolev-Hardy inequality Lemma 6.3.1 kDk W 1;1 =k 0 DVk W 1;1 0 e 0 0 S N ( 0 )d 0 ."< 1 6 ; 2 [0;T ) (2.4.4) 22 for " > 0 small enough andT T is from the continuity argument. Note the continuity argument will give thatT =1. Similar arguments apply to the remaining a priori assump- tions. Secondly from this continuity argument we can deduce we have a global-in-time solution to the initial value problem (2.1.27)-(2.1.28). Therefore the rest of the treatment will be devoted to proving the following curl and energy estimate Propositions respectively which will establish the curl and energy inequal- ities needed for the proof of Theorem 2.4.2. Proposition (Curl Estimates). Suppose > 1. Let (; V) : !R 3 R 3 be a unique local solution to (2.1.27)-(2.1.28) on [0;T ] with T > 0 xed and assume (; V) satises the a priori assumptions (2.4.2). Fix N 2de + 12. Let k N + 1 in (2.1.21). Then for all 2 [0;T ], the following inequalities hold for some 0< 1 B N [V](). 8 > > < > > : e 2 0 S N (0) +B N [V](0) +e 2 0 S N () if 1< < 5 3 e 2 0 S N (0) +B N [V](0) + (1 + 2 )e 2 0 S N () if 5 3 ; (2.4.5) B N []().S N (0) +B N [V](0) +S N () + 0 e 0 0 S N ( 0 )d 0 : (2.4.6) Proposition (Energy Estimate). Suppose > 1. Let (; V) : !R 3 R 3 be a unique local solution to (2.1.27)-(2.1.28) on [0;T ] with T > 0 xed and assume (; V) satises the a priori assumptions (2.4.2). Fix N 2de + 12. Let kN + 1 in (2.1.21). Then for all 2 [0;T ], we have the following inequality for some 0< 1 S N ().S N (0) +B N []() + 0 (S N ( 0 )) 1 2 (B N [V]( 0 )) 1 2 d 0 +S N () + 0 e 0 0 S N ( 0 )d 0 : (2.4.7) We henceforth assume we are working with a unique local solution (; V) : !R 3 R 3 to (2.1.27)-(2.1.28) such thatS N (; V) +B N [V]<1 on [0;T ] with T > 0 xed: Theorem 23 2.4.1 ensures the existence of such a solution, and furthermore we assume this local solution satises the a priori assumptions (2.4.2). To prove the above key results, we apply weighted energy estimates developed in [20] which enable us to handle exponentially growing-in-time coecients and the vacuum bound- ary. The exponentially growing time weights capture the stabilizing eect of the expanding background ane motion and were used crucially in [20] to close the estimates. As seen in our high order quantities, only spatial derivative operators are used so as to keep intact the exponential structure of time weights which function as stabilizers allowing us to close estimates. Further, the increase in spatial weightw in accordance with an increase in radial derivatives seen inS N andB N will be essential in closing energy estimates by avoiding potentially dangerous negative powers of w near the boundary. Among others, in the nonisentropic setting, we have to contend with the presence of e S when proceeding with our estimates. The regularity of e S given by Corollary 1.1.5 will be important for us, used throughout our analysis. However this will only be useful at the stage of isolated derivatives of e S , and it will be seen both in the curl estimates and energy estimates that we will need to be careful when taking derivatives of expressions involving e S . In the curl estimates we almost immediately have to deal with the presence of e S as the usual derivation of the curl does not reveal a desired structure. We will derive the curl equation carefully by exploiting the algebraic nonlinear structure of the pressure in (1.0.6) such that it is decoupled from the main equation and that it still exhibits a good structure without loss of derivatives. The fact that e S is a radial function from our ane development will be crucial in obtaining favorable terms with respect to our high order quantities and weight structure. In the energy estimates, new commutator formulae are obtained for dierentiating the pressure term which are carefully acquired to match the weight structure of our high order norm. Furthermore to establish the results for all > 1, the proofs of the curl and energy inequalities will dier for 2 (1; 5 3 ] and > 5 3 . This is because we need to eliminate the 24 anti-damping eect encountered in [20] for > 5 3 . Hence we will need to consider dierent high-order quantities for > 5 3 , which we rst saw in Section 2.3. Moreover, in this setting directly from the new equation structure we consider, many of our terms will contain time weights with negative powers. Thus to control such terms without this decay, we apply a coercivity estimate technique which is not used in the 2 (1; 5 3 ] case. This technique employs the fundamental theorem of calculus to express in terms of V and initial data with the coercivity estimates given in Lemma 5.0.1. Due to this dierence in methodologies for dierent , in Chapters 3-4 we give the analysis for 2 (1; 5 3 ]. In Chapter 5 we give the analysis for > 5 3 . Thus henceforth we x 2 (1; 5 3 ] in Chapters 3-4 and > 5 3 in Chapter 5, with the corresponding implications for S N andB N given in Section 2.3. 25 Chapter 3 Curl Estimates To control the modied curl in the nonisentropic setting, we crucially have to contend with the presence of e S in our formulation. There is no natural decoupling to exploit because of this term: instead we need to articially decouple the system through multiplication by an appropriate power of e S . To then analyze the modied curl we are led to introduce novel cross product terms: rst recall the cross product matrix introduced in (2.2.4) [Arf F] i j := jm A s m f ;s F i im A s m f ;s F j : (3.0.1) The related cross product commutator will also be needed [@ ; Arf]F i j :=@ ( jm A s m )f; s F i @ ( im A s m )f; s F j : (3.0.2) These two cross product quantities dened above are unique to the nonisentropic setting. Finally before we derive desirable forms for our curl matrices we introduce a term that will let us commute the time derivative outside of the curl operator [@ ; Curl A ]F i j :=@ ( jm A s m ) F; i s @ ( im A s m ) F; j s : (3.0.3) We rst derive the equations for Curl A V and Curl A . 26 Lemma 3.0.1. Suppose 2 (1; 5 3 ]. Let (; V) : ! R 3 R 3 be a unique local solution to (2.1.27)-(2.1.28) on [0;T ] with T > 0 xed. Then for all 2 [0;T ] the curl matrices Curl A V and Curl A can be written in the following desirable forms Curl A V = 1 Ar( S) V + (0)Curl A (V(0)) (0)Ar( S) V(0) + 1 0 [@ ; Curl A ]Vd 0 1 0 [@ ; Ar( S)]Vd 0 2 0 Curl A ( V)d 0 + 2 0 Ar( S) ( V)d 0 + 0 43 r( S) d 0 0 43 A [D]r( S) d 0 ; (3.0.4) and Curl A = 1 Ar( S) + Curl A ([(0)]) 1 Ar( S)(0) +(0)Curl A (V(0)) 0 1 ( 0 ) d 0 (0)Ar( S) V(0) 0 1 ( 0 ) d 0 + 0 [@ ; Curl A ]d 0 1 0 [@ ; Ar( S)]d 0 + 0 1 ( 0 ) 0 0 ( 00 )[@ ; Curl A ]Vd 00 d 0 1 0 1 ( 0 ) 0 0 ( 00 )[@ ; Ar( S)]Vd 00 d 0 0 2 ( 0 ) 0 0 ( 00 ) Curl A ( V)d 00 d 0 + 1 0 2 ( 0 ) 0 0 ( 00 ) Ar( S) ( V)d 00 d 0 + 0 2 ( 0 ) 0 0 ( 00 ) 43 r( S) d 00 d 0 0 1 ( 0 ) 0 0 ( 00 ) 43 A [D]r( S) d 00 d 0 : (3.0.5) Proof. First divide (2.1.25) by w and use V = = to obtain 3 3 V + V + 2 V ++w A > r(w 1+ e S J 1 ) +w 1+ e S r(J 1 ) = 0; (3.0.6) 27 where we have split the gradient term and moved away from coordinates at this stage. Next, multiply (3.0.6) by e S to obtain a r f term 3 3 e S V + V + 2 V +e S + ( + 1)r we ( 1 +1 ) S J 1 = 0: (3.0.7) Since Curl A (r f) = 0 apply Curl A to (3.0.7) 3 3 Curl A e S V + Curl A e S V + 2Curl A e S V +Curl A e S = 0: (3.0.8) Then Curl A e S V + Curl A e S V +2Curl A e S V + 33 Curl A e S = 0: (3.0.9) Reformulate this as @ Curl A e S V =[@ ; Curl A ] e S V 2 Curl A e S V 43 Curl A e S ; (3.0.10) where we have used Curl A e S V =@ Curl A e S V [@ ; Curl A ] e S V ; (3.0.11) Integrate (3.0.10) from 0 to Curl A e S V = (0)Curl A e S [V(0)] + 1 0 [@ ; Curl A ] e S V d 0 2 0 Curl A e S V d 0 0 43 Curl A e S d 0 : (3.0.12) 28 Via (3.0.3) we obtain @ Curl A e S [] = (0)Curl A e S [V(0)] + [@ ; Curl A ] e S + 1 0 [@ ; Curl A ] e S V d 0 2 0 Curl A e S V d 0 0 43 Curl A e S d 0 : (3.0.13) Again integrating from 0 to Curl A e S [] = Curl A e S [0] +(0)Curl A e S [V(0)] 0 1 ( 0 ) d 0 + 0 [@ ; Curl A ] e S [ 0 ] d 0 + 0 1 ( 0 ) 0 0 [@ ; Curl A ] e S V[ 00 ] d 00 d 0 0 2 ( 0 ) 0 0 ( 00 ) Curl A e S V[ 00 ] d 00 d 0 (3.0.14) 0 ( 0 ) 0 0 ( 00 ) 43 Curl A e S [ 00 ] d 00 d 0 : (3.0.15) Now note Curl A e S F =e S Curl A F + Ar(e S ) F; (3.0.16) where we recall the denitions introduced in (2.2.4). Also via (3.0.3) we have [@ ; Curl A ] e S F =e S [@ ; Curl A ]F + [@ ; Ar(e S )]F; (3.0.17) Then using Curl A () = 0; (3.0.18) we can rewrite (3.0.12) as e S Curl A V =Ar(e S ) V +e S (0)Curl A (V(0)) + (0)Ar(e S ) V(0) 29 + 1 0 e S [@ ; Curl A ]Vd 0 + 1 0 [@ ; Ar(e S )]Vd 0 2 0 e S Curl A ( V)d 0 2 0 Ar(e S ) ( V)d 0 0 43 Ar(e S ) ()d 0 : (3.0.19) Now multiply (3.0.19) by e S and note e S ; s = 1 ( S); s e S ; (3.0.20) to obtain Curl A V = 1 Ar( S) V + (0)Curl A (V(0)) (0)Ar( S) V(0) + 1 0 [@ ; Curl A ]Vd 0 1 0 [@ ; Ar( S)]Vd 0 2 0 Curl A ( V)d 0 + 2 0 Ar( S) ( V)d 0 + 0 43 Ar( S) ()d 0 : (3.0.21) Now since =O(1) for our estimates, it is not a priori clear if the last term of (3.0.21) can be controlled. To this end, rst noteA = [D] 1 and =y +. We then have A k j k j =A k l l j A k l [D] l j =A k l ( l j [Dy] l j [D] l j ) =A k l [D] l j ; (3.0.22) Second notice that, since S is a radial function from (2.1.20) and hence S; s = (y s =jyj) S 0 (jyj), js ( S); s ik y k is ( S); s jk y k = S 0 (jyj) jyj js y s ik y k is y s jk y k = 0: (3.0.23) Then [Ar( S) ()] i j = jm ( s m A s ` [D] ` m )( S); s ik k im ( s m A s ` [D] ` m )( S); s jk k 30 = js ( S); s ik k is ( S); s jk k + js ( S); s ik y k is ( S); s jk y k ( jm A s ` [D] ` m ( S); s ik k im A s ` [D] ` m ( S); s jk k ) = ( js ( S); s ik k is ( S); s jk k ) ( jm A s ` [D] ` m ( S); s ik k im A s ` [D] ` m ( S); s jk k ) := [r( S) ] i j [A [D]r( S) ] i j : (3.0.24) Hence using (3.0.24), we rewrite (3.0.21) and obtain our desirable form for Curl A V, (3.0.4). Following the same approach as for obtaining (3.0.4), notably using (3.0.16), (3.0.17), mul- tiplying by e S and nally using (3.0.24), from (3.0.15) we obtain our desirable form for Curl A , (3.0.5). Next we prove some preliminary bounds for several of the terms unique to our nonisen- tropic setting. These bounds on terms arising from the derivation above will be used in our main curl estimate. Lemma 3.0.2. Suppose 2 (1; 5 3 ]. Let (; V) : ! R 3 R 3 be a unique local solution to (2.1.27)-(2.1.28) on [0;T ] with T > 0 xed and assume (; V) satises the a priori assumptions (2.4.2). FixN 2de+12. LetkN +1 in (2.1.21). Then for all2 [0;T ], the following bounds hold for a +jjN,jjN k 0 43 X a r = @ (r( S) )d 0 k 2 +a+1; e S +k 0 43 @ (r( S) )d 0 k 2 1+;(1 )e S .e 2 0 S N (); (3.0.25) k 0 43 X a r = @ (A [D]r( S) )d 0 k 2 +a+1; e S +k 0 43 @ (A [D]r( S) )d 0 k 2 1+;(1 )e S .e 2 0 S N (); (3.0.26) k 0 2 0 0 43 X a r = @ (r( S) )d 00 d 0 k 2 +a+1; e S +k 0 2 0 0 43 @ (r( S) )d 00 d 0 k 2 1+;(1 )e S . 0 e 0 0 S N ( 0 )d 0 ; (3.0.27) k 0 1 0 0 43 X a r = @ (A [D]r( S) )d 00 d 0 k 2 +a+1; e S 31 +k 0 1 0 0 43 @ (A [D]r( S) )d 00 d 0 k 2 1+;(1 )e S . 0 e 0 0 S N ( 0 )d 0 : (3.0.28) Proof. Proof of (3.0.25)-(3.0.26). By the denition of r( S) introduced in (3.0.24), we have X a r = @ [(r( S) )] i j =X a r = @ ( js S; s ik k is S; s jk k ): (3.0.29) We restrict our focus to the left term only X a r = @ ( js S; s ik k ) = js ik ((X a r = @ S; s ) k + S; s (X a r = @ k )) + X 1a 0 +j 0 jN1 C a 0 ; 0( js (X a 0 r = @ 0 S; s ) ik (X aa 0 r = @ 0 k )): (3.0.30) Schematically consider the rst two terms on the right hand side of (3.0.30) (X a r = @ (D S)) | {z } =:C 1 + (D S)(X a r = @ ) | {z } =:C 2 (3.0.31) Notice that for C 1 , since X a r = @ (D S) is bounded by Corollary 1.1.5 and Lemma 6.2.2, k 0 43 C 1 d 0 k 2 +a+1; e S = w 1++a e S 2 2 2 0 43 (X a r = @ (D S))d 0 2 dy . 1 2 0 43 d 0 2 sup 0 0 kk 2 ; e S .U()S N (); (3.0.32) where we also have used (2.1.19) and (6.1.6), and we dene U() = 1 2 j 0 43 d 0 j 2 . Now for U, using (6.1.4) and (6.1.3), we have U.e 2 1 0 e (43 ) 1 0 d 0 2 32 . 8 > > > > > > < > > > > > > : e 2 1 e 2(43 ) 1 if 1< < 4 3 e 2 1 2 if = 4 3 e 2 1 (e 2(43 ) 1 + 1) if 4 3 < 5 3 : (3.0.33) By the denition of 1 (2.4.1), we have2 1 + 2(4 3 ) 1 =4 0 and also, 1 0 . Furthermore, for = 4 3 , 0 = 1 2 1 , and so in this case 2 e 2 1 = 2 e 1 e 2 0 .e 2 0 : (3.0.34) Therefore by (3.0.33) U.e 2 0 : (3.0.35) Hence nally for C 1 by (3.0.32), we have k 0 43 C 1 d 0 k 2 +a+1; e S .e 2 0 S N (): (3.0.36) For C 2 , by employing Fubini's theorem to interchange the spatial and time integrals, k 0 43 C 1 d 0 k 2 +a+1; e S = w 1++a e S 2 2 2 0 43 (D S)(X a r = @ )d 0 2 dy .U() sup 0 0 kX a r = @ k 2 a+; e S .e 2 0 S N (); (3.0.37) where we have again used Corollary 1.1.5 and Lemma 6.2.2 to bound the entropy term. We have also used our bound for U (3.0.35) as well as (2.1.19) and (6.1.6). The low order commutator terms on the right hand side of (3.0.30) can be estimated in the same way as C 1 and C 2 k 0 43 C a 0 ; 0( js (X a 0 r = @ 0 S; s ) ik (X aa 0 r = @ 0 k ))d 0 k 2 1++a; e S .e 2 0 S N (); (3.0.38) 33 for a 0 +j 0 jN 1. Hence by (3.0.36), (3.0.37) and (3.0.38), and an analogous argument for thekk 1+;(1 )e S norms, we obtain (3.0.25). By an analogous argument additionally using the a priori assumptions (2.4.2) in the case and also Lemma 6.3.1 for the low order commutator terms, we get (3.0.26). Proof of (3.0.27)-(3.0.28). With 0 < < 1 xed to be specied later and using the Cauchy-Schwarz inequality in conjunction with Fubini's theorem to take advantage of the fact that time integrals of negative powers are bounded by (6.1.4) k 0 2 ( 0 ) 0 0 ( 00 ) 43 (X a r = @ r( S) )d 00 d 0 k 2 +a+1; e S . 0 1 (( 0 )) 2(1) d 0 0 1 ( 0 ) 2 0 0 ( 00 ) 43 (X a r = @ r( S) )d 00 ! 2 d 0 w 1++a e S dy . 0 1 ( 0 ) 2 k 0 0 ( 00 ) 43 (X a r = @ r( S) )d 00 k 2 1++a; e S d 0 . 0 e 0 0 S N ( 0 )d 0 ; (3.0.39) where we conclude the nal bound by a similar argument to that giving (3.0.25) if we can show 1 ( 0 ) 2 0 0 ( 00 ) 43 d 00 2 .e 0 : (3.0.40) Now if 6= 4 3 1 ( 0 ) 2 0 0 ( 00 ) 43 d 00 2 .e 2 1 0 0 0 e (43 ) 1 00 d 00 2 . 8 > > < > > : e 2 1 0 e 2(43 ) 1 0 if 1< < 4 3 e 2 1 0 (e 2(43 ) 1 0 + 1) if 4 3 < 5 3 = 8 > > < > > : e (86 2) 1 0 if 1< < 4 3 e (86 2) 1 0 +e 2 1 0 if 4 3 < 5 3 : (3.0.41) 34 Here, specify as follows = 8 > > < > > : 1 2 if 11 9 < 4 3 or 4 3 < 5 3 (i) r where r2 [1 9 9 4 ; 1) if 1< < 11 9 (ii) (3.0.42) In case (i) (8 6 2) 1 = 2 3 3 0 4 0 0 ; (3.0.43) since 2 3 3 3 for 11 9 . Also in case (i),2 1 = 1 0 . In case (ii) (8 6 2) 1 =4 0 + 2(2 2) 3 3 0 0 (3.0.44) since 2(22) 3 3 3 for 1 9 9 4 . Finally for = 4 3 , set = 3 4 , and then 1 ( 0 ) 2 0 0 ( 00 ) 43 d 00 2 .e 3 2 1 0 0 0 e (43 ) 1 00 d 00 2 .e 3 2 1 0 ( 0 ) 2 =e 0 0 ( 0 ) 2 e 1 0 .e 0 0 : (3.0.45) In all cases, we have the same result and hence 1 ( 0 ) 2 0 0 ( 00 ) 43 d 00 2 .e 0 : (3.0.46) So with an analogous argument for thekk 1+;(1+ )e S norms, we have (3.0.27). By an analogous argument, with (3.0.26) replacing (3.0.25), we can get (3.0.28). Before proving our main curl estimate result, we also prove bounds on commutator terms that will immediately arise when applying our derivative operators. Lemma 3.0.3. Suppose 2 (1; 5 3 ]. Let (; V) : ! R 3 R 3 be a unique local solution to (2.1.27)-(2.1.28) on [0;T ] with T > 0 xed and assume (; V) satises the a priori 35 assumptions (2.4.2). FixN 2de+12. LetkN +1 in (2.1.21). Then for all2 [0;T ], the following bounds hold for some 0< 1, X 1a+jjN h X a r = @ ; Curl A i V() 2 1++a; e S + X jjN [@ ; Curl A ] V() 2 1+;(1 )e S .e 2 0 S N () 1 +P (S N ()) ; (3.0.47) X 1a+jjN h X a r = @ ; Curl A i () 2 1++a; e S + X jjN [@ ; Curl A ]() 2 1+;(1 )e S .S N (0) +S N () + (1 +P (S N ())) 0 e 0 0 S N ( 0 )d 0 ; (3.0.48) with P a polynomial of degree at least 1. Proof. Proof of (3.0.47). Note h X a r = @ ; Curl A i V k i =X a r = @ Curl A V k i Curl A (X a r = @ V) k i =X a r = @ im A s m V k ; s km A s m V i ; s im A s m (X a r = @ V) k ; s km A s m (X a r = @ V) i ; s = km A s m (X a r = @ V) i ; s X a r = @ (A s m V) i ; s ) im A s m (X a r = @ V) k ; s X a r = @ (A s m V k ; s ) : (3.0.49) Since the two terms in the last line above are estimated similarly noting derivative count, we restrict our focus to the second term only. Via the Leibniz rule A s m (X a r = @ V) k ; s X a r = @ (A s m V k ; s ) = (X a r = @ A s m )V k ; s | {z } =:A X 1a 0 +j 0 jN1 a 0 a; 0 X a 0 r = @ 0 A s m X aa 0 r = @ 0 V k ; s A s m [X a r = @ ;@ s ]V k | {z } =:B (3.0.50) Now for A rst note ifjj = 0 then X a r = @ A s m =X a r A s m =A s i (X a r s ); k A k m A s i [X a1 r ;@ m ]X r i A k i 36 X 1a 0 a c a 0X a 0 r (A s i A k m )X aa 0 r X r i ; k +X a1 r A s i [@ k ;X r ] i A k m ; (3.0.51) and if jj > 0 then for some e ` given by (e ` ) ` = 1; (e ` ) i = 0 for i 6= ` and ` 1 ;` 2 2 f1; 2; 3g;` 1 6=` 2 X a r = @ A s m =A s i (X a r = @ i ); k A k m A s i [X a r = @ e ` ;@ k ] = @ ` 1 ` 2 i A k m X 0 0 because using (3.0.51), thejj = 0 case follows analogously. Using (3.0.52) A =A k j (X a r = @ j ); m A m i V k ; s | {z } =:A 1 A k i [X a r = @ e ` ;@ m ] = @ ` 1 ` 2 s A m i V k ; s X 0 0 xed and assume (; V) satises the a priori assumptions (2.4.2). Fix N 2de + 12. Let k N + 1 in (2.1.21). Then for all 2 [0;T ], the following inequalities hold for some 0< 1 B N [V](). 8 > > < > > : e 2 0 S N (0) +B N [V](0) +e 2 0 S N () if 1< < 5 3 e 2 0 S N (0) +B N [V](0) + (1 + 2 )e 2 0 S N () if = 5 3 ; (3.0.61) B N []().S N (0) +B N [V](0) +S N () + 0 e 0 0 S N ( 0 )d 0 : (3.0.62) Proof. Proof of (3.0.61). Apply X a r = @ to (3.0.4) Curl A X a r = @ V =[X a r = @ ; Curl A ]V + 1 X a r = @ (Ar( S) V) 39 + (0)X a r = @ Curl A (V(0)) (0)X a r = @ (Ar( S) V(0)) + 1 0 X a r = @ [@ ; Curl A ]Vd 0 1 0 X a r = @ [@ ; Ar( S)]Vd 0 2 0 X a r = @ Curl A ( V)d 0 + 2 0 X a r = @ (Ar( S) ( V))d 0 + 0 43 X a r = @ (r( S) )d 0 0 43 X a r = @ (A [D]r( S) )d 0 : (3.0.63) The bound on the rst term on the right hand side of (3.0.63) follows from Lemma 3.0.3. The second term is similar to the rst term but lower order and hence is also bounded. For the third term we have (0)X a r = @ Curl A V(0) 2 1++a; e S .e 2 1 (S N (0) +B N [V](0)); (3.0.64) where we apply (6.1.4). The fourth term is similar to the third term but lower order and hence is also bounded. We now consider the fth term on the right hand side of (3.0.63) X a r = @ [@ ; Curl A ]V k j =X a r = @ @ ( jm A s m ) V; k s @ ( km A s m ) V; j s : (3.0.65) As the other tern can be estimated in the same way, we restrict our focus to the rst term only X a r = @ @ ( jm A s m ) V; k s =X a r = @ h (@ jm A s m + jm @ A s m ) V; k s i =@ jm X a r = @ A s m V; k s +A s m X a r = @ V; k s + jm X a r = @ @ A s m V; k s +@ A s m X a r = @ V; k s + X 1a 0 +j 0 jN1 C a 0 ; 0 @ jm X a 0 r = @ 0 A s m + jm X a 0 r = @ 0 @ A s m X aa 0 r = @ 0 V k ; s : (3.0.66) 40 Schematically consider the rst two terms on the right hand side of (3.0.66) @ X a r = @ DDV | {z } =:D 1 +@ DX a r = @ DV | {z } =:D 2 + DVX a r = @ DV | {z } =:D 3 (3.0.67) For D 1 , using the exponential bounds on and given by (6.1.4) and (6.1.6) respectively w 1++a e S 1 2 0 @ X a r = @ DDVd 0 2 dy .e 2 1 w 1++a e S 0 e 1 0 @ X a r = @ DDV d 0 2 dy .e 2 1 sup 0 0 w 1++a e S jX a r = @ Dj 2 0 je 1 0 j@ DVj d 0 2 dy .e 2 1 sup 0 0 kX a r = @ Dk 2 1++a; e S 0 e 1 0 j@ je 0 0 (S N ( 0 )) 1 2 d 0 2 .e 2 1 (S N ()) 2 ; (3.0.68) where we applied Lemma 6.3.1. For D 2 , one must rst integrate by parts in 1 0 @ DX a r = @ DVd 0 = 1 @ DX a r = @ D 0 (3.0.69) 1 0 ( @ D +@ D +@ DV)X a r = @ Dd 0 : By the same argument as for D 1 k 1 0 @ DX a r = @ DVd 0 k 2 1++a; e S .e 2 1 S N (0) + (1 + 2 )e 2 1 +Q() S N (); (3.0.70) with Q() = 8 > > > > > > < > > > > > > : e 4 0 if 1< < 4 3 2 e 4 0 if = 4 3 e 2 1 if 4 3 < 5 3 : (3.0.71) 41 For D 3 , we also rst integrate by parts in 1 0 DVX a r = @ DVd 0 = 1 h DVX a r = @ D i 0 1 0 ( DV +@ DV + DV )X a r = @ Dd 0 : (3.0.72) The rst term on the right-hand side of (3.0.72) is bounded by 1 S N (0) +kDVk L 1 ( ) (S N ()) 1 2 .e 1 S N (0) +e 0 S N (); (3.0.73) where we applied Lemma 6.3.1. For the second term using the expontential boundedness of (6.1.4) 1 0 DVX a r = @ Dd 0 . sup 0 0 jX a r = @ Dje 1 0 e 1 0 kDVk L 1 ( ) d 0 . sup 0 0 jX a r = @ Dj(S N ()) 1 2 e 1 0 e ( 1 0 ) 0 d 0 | {z } =:L : (3.0.74) Notice that after integration and using 0 1 L. 8 > > < > > : e 0 if 1< < 5 3 e 0 if = 5 3 : (3.0.75) Hence k 1 0 DX a r = @ Dd 0 k 2 1++a; e S . 8 > > < > > : e 2 0 (S N ()) 2 if 1< < 5 3 2 e 2 0 (S N ()) 2 if = 5 3 : (3.0.76) 42 The remaining terms on the right-hand side of (3.0.72) are estimated in a similar way except when considering the last term which includesDV : there we use (2.1.27) to rewriteDV , and then an analogous proof completes the argument. Therefore k 1 0 ( 0 )D 3 ( 0 )d 0 k 2 1++a; e S . 8 > > < > > : e 2 0 S N () if 1< < 5 3 (1 + 2 )e 2 0 S N () if = 5 3 : (3.0.77) Finally using a similar approach X 1a 0 +j 0 jN1 C a 0 ; 0 @ jm X a 0 r = @ 0 A s m + jm X a 0 r = @ 0 @ A s m X aa 0 r = @ 0 V k ; s 2 1++a; e S .e 2 0 S N () 1 +P (S N ()) .e 2 0 S N (): (3.0.78) where we have used Lemma 6.3.1 and (2.4.2). Thus we have k 1 0 X a r = @ [@ ; Curl A ]Vd 0 k 2 1++a; e S . 8 > > < > > : e 2 0 S N () if 1< < 5 3 (1 + 2 )e 2 0 S N () if = 5 3 : (3.0.79) The sixth term on the right hand side of (3.0.63) is similar to the fth term but lower order and hence is also bounded. Proceeding in an analogous way to the fth term estimate, we have for the seventh term on the right hand side of (3.0.63) k 2 0 X a r = @ Curl A ( V)d 0 k 2 1++a; e S . 8 > > < > > : e 2 0 S N () if 1< < 5 3 (1 + 2 )e 2 0 S N () if = 5 3 : (3.0.80) The eighth term is similar to the seventh term but lower order and hence is also bounded. Bounds on the last two terms on the right hand side of (3.0.63) follow from Lemma 3.0.2. 43 Combining the above analysis, and an analogous argument for thekk 1+;(1 )e S norms, we obtain (3.0.61). Proof of (3.0.62). Apply X a r = @ to (3.0.5) Curl A X a r = @ =[X a r = @ ; Curl A ] + 1 X a r = @ Ar( S) +X a r = @ Curl A ([(0)]) 1 X a r = @ (Ar( S))(0) +(0)X a r = @ Curl A (V(0)) 0 1 ( 0 ) d 0 (0)X a r = @ Ar( S) V(0) 0 1 ( 0 ) d 0 + 0 X a r = @ [@ ; Curl A ]d 0 1 0 X a r = @ [@ ; Ar( S)]d 0 + 0 1 ( 0 ) 0 0 ( 00 )X a r = @ [@ ; Curl A ]Vd 00 d 0 1 0 1 ( 0 ) 0 0 ( 00 )X a r = @ [@ ; Ar( S)]Vd 00 d 0 0 2 ( 0 ) 0 0 ( 00 )X a r = @ Curl A ( V)d 00 d 0 + 1 0 2 ( 0 ) 0 0 ( 00 )X a r = @ Ar( S) ( V)d 00 d 0 + 0 2 ( 0 ) 0 0 ( 00 ) 43 X a r = @ r( S) d 00 d 0 0 1 ( 0 ) 0 0 ( 00 ) 43 X a r = @ A [D]r( S) d 00 d 0 : (3.0.81) The bound on the rst term on the right hand side of (3.0.81) follows from Lemma 3.0.3. The second term is similar to the rst term but lower order and hence is also bounded. The third term is bounded byS N (0). Noting the fact thatkX a r = @ Ak 2 1++a; e S .S N established in the proof of Lemma 3.0.3, the fourth term is similar to the third term but lower order and hence is also bounded in the same way as the third term. The fth term is bounded byS N (0) +B N [V](0) using (6.1.4). Using again the bound on X a r = @ A , the sixth term is similar to the fth term but lower order and hence is also bounded in the same way as the third term. For the seventh term estimate, a similar argument to that used to prove the bound (3.0.79) gives us kX a r = @ [@ ; Curl A ]k.e 0 (S N ()) 1 2 : (3.0.82) 44 Then via the Cauchy-Schwarz inequality and Fubini's Theorem, we have for the seventh term 0 X a r = @ [@ ; Curl A ]d 0 2 1++a; e S . 0 e 0 S N ( 0 )d 0 : (3.0.83) The eighth term is similar to the seventh term but lower order and hence is also bounded. For the ninth term 0 1 ( 0 ) 0 0 ( 00 )X a r = @ [@ ; Curl A ]Vd 00 d 0 2 1++a; e S . 2 4 0 (1 + 0 ) 2 ( 0 ) d 0 0 1 (1 + 0 ) 2 ( 0 ) 0 0 ( 00 )X a r = @ [@ ; Curl A ]Vd 00 ! 2 d 0 3 5 w 1++a e S dx . 0 1 (1 + 0 ) 2 ( 0 ) 0 0 ( 00 )X a r = @ [@ ; Curl A ]Vd 00 2 1++a; e S d 0 . 0 e 0 S N ( 0 )d 0 ; where the nal bound follows from similar arguments to those giving the bound (3.0.79). The tenth term is similar to the ninth term but lower order and hence is also bounded. Similarly to the ninth term proof but via (3.0.80), for the eleventh term we have 0 2 ( 0 ) 0 0 ( 00 )X a r = @ Curl A ( V)d 00 d 0 2 1++a; e S . 0 e 0 0 S N ( 0 )d 0 : The twelfth term is similar to the eleventh term but lower order and hence is also bounded. Bounds on the last two terms on the right hand side of (3.0.81) follow from Lemma 3.0.2. Combining the above analysis, and an analogous argument for thekk 1+;(1 )e S norms, we obtain (3.0.62). 45 Chapter 4 Energy Estimates Before proving our main energy inequality, we rst introduce the two energy based high order quantities which arise directly from the problem. Begin by diagonalizing then positive symmetric matrix = (detA) 2 3 A 1 A > 2 SL(3) as follows =P > QP; P2 SO(3); Q = diag(d 1 ;d 2 ;d 3 ); d i > 0 eigenvalues of ; (4.0.1) and then dene M a; :=Pr X a r = @ P > andN :=Pr @ P > : (4.0.2) Denoting the usual dot product onR 3 byh;i, introduce the high-order energy functional E N (; V)() =E N () = 1 2 X a+jjN h 3 3 D 1 X a r = @ V; X a r = @ V E + D 1 X a r = @ ; X a r = @ Ei w a+ + J 1 h 3 X i;j=1 d i d 1 j (M a; ) j i 2 + 1 div X a r = @ 2 i w a++1 e S dy + 1 2 X jjN (1 ) h 3 3 1 @ V; @ V + 1 @ ; @ i w + (1 )J 1 h 3 X i;j=1 d i d 1 j (N ) j i 2 + 1 (div @ ) 2 i w 1+ e S dy; (4.0.3) 46 and the dissipation functional D N (V) =D N () = 5 3 2 3 3 h X a+jjN D 1 X a r = @ V; X a r = @ V E w a+ + (1 ) X jjN 1 @ V; @ V w i dy: (4.0.4) We are requiring 5 3 in this formulation, which givesD N (V) 0. Next, we give key identities which will be used in our estimates. First from Lemma 4.3 [20] we have the following modied energy identity: Lemma 4.0.1. Recalling the matrix quantities introduced in (4.0.1) and (4.0.2), the fol- lowing identities hold `j (r X a r = @ ) i j 1 im (r X a r = @ ) m `; = 1 2 d d 0 @ 3 X i;j=1 d i d 1 j (M a; ) j i ) 2 1 A +T a; ; (4.0.5) `j (r @ ) i j 1 im (r @ ) m `; = 1 2 d d 0 @ 3 X i;j=1 d i d 1 j (N ) j i ) 2 1 A +T ; (4.0.6) where the error termsT a; andT are given as follows T a; = 1 2 3 X i;j=1 d d d i d 1 j ((M a; ) j i ) 2 Tr QM a; Q 1 @ PP > M > a; +M > a; P@ P > ; (4.0.7) T = 1 2 3 X i;j=1 d d d i d 1 j ((N ) j i ) 2 Tr QN Q 1 @ PP > N > +N > P@ P > : (4.0.8) Next we have two important commutation results which will be used crucially when dierentiating the nonlinear pressure term in our equation. 47 Lemma 4.0.2. Forq2R + , T : !M 33 and anyi;j;`2f1; 2; 3g the following identities hold X r 1 w q (w 1+q e S T k i ); k = 1 w 1+q (w 2+q e S X r T k i ); k +C q+1 i [T]; (4.0.9) = @ j` 1 w q (w 1+q e S T k i ); k = 1 w q (w 1+q e S = @ j` T k i ); k +C q ij` [T]; (4.0.10) whereC q+1 i [T] is C q+1 i [T] = (X r (we S )we S ) y j r 2 = @ jk T k i +w(X r (e S )e S ) y k r 2 X r T k i + h (1 +q)X r (w; k e S ) +X r ((e S ); k w) i T k i ; (4.0.11) andC q ij` [T] is C q ij` [T] =we S y m k` = @ mj kj = @ m` r 2 + k` y j kj y ` r 2 X r T k i + (1 +q)( = @ j` w; k )e S T k i +w = @ j` (e S ); k T k i for j;` = 1; 2; 3: (4.0.12) Proof. Compute X r 1 w q (w 1+q e S T k i ); k =X r h we S T k i ; k +(1 +q)w; k e S T k i + (e S ); k wT k i i =we S X r T k i ; k +X r (we S )T k i ; k +(1 +q)w; k e S X r T k i + (1 +q)T k i X r (w; k e S ) + (e S ); k wX r T k i + T k i X r ((e S ); k w) =we S (X r T k i ); k +we S [X r ;@ k ]T k i + (2 +q)w; k e S X r T k i w; k e S X r T k i + (e S ); k wX r T k i +X r (we S )T k i ; k +(1 +q)T k i X r (w; k e S ) + T k i X r ((e S ); k w) = 1 w 1+q (w 2+q e S X r T k i ); k +we S [X r ;@ k ]T k i w; k e S X r T k i +X r (we S )T k i ; k + (1 +q)T k i X r (w; k e S ) + T k i X r ((e S ); k w): 48 We have now obtained favorable terms except for the following three terms which we rewrite using our commutator identity (6.2.2), [X r ;@ k ] =@ k and (6.2.1) to express @ k in terms of X r and = @ jk , we S [X r ;@ k ]T k i w; k e S X r T k i +X r (we S )T k i ; k = (X r (we S )we S )T k i ; k w; k e S X r T k i = (X r (we S )we S ) y j r 2 = @ jk T k i + y k r 2 X r T k i w; k e S X r T k i = (X r (we S )we S ) y j r 2 = @ jk T k i +w(X r (e S )e S ) y k r 2 X r T k i ; where to obtain the last line we have used the fact that w is a radial function, say f(jyj), by (2.1.19) and hence w; k = y k r f 0 (jyj) which implies (X r w) y k r 2 w; k =y ` y ` r f 0 (jyj) y k r 2 y k r f 0 (jyj) = 0: Combining the above calculations we obtain (4.0.9). Now sincew and S are radial functions by (2.1.20) and (2.1.19), = @ j` w = = @ j` (e S ) = 0 and so we have = @ j` 1 w q (w 1+q e S T k i ); k = = @ j` h we S T k i ; k +(1 +q)w; k e S T k i + (e S ); k wT k i i =we S = @ j` T k i ; k +(1 +q)w; k e S = @ j` T k i + (1 +q)( = @ j` w; k )e S T k i +w(e S ); k = @ j` T k i ; k +w = @ j` (e S ); k T k i =we S ( = @ j` T k i ); k +we S [ = @ j` ;@ k ]T k i + (1 +q)w; k e S = @ j` T k i + (1 +q)( = @ j` w; k )e S T k i +w(e S ); k = @ j` T k i ; k +w = @ j` (e S ); k T k i = 1 w q (w 1+q e S = @ j` T k i ); k +we S ( k` @ j kj @ ` )T k i + (1 +q)( = @ j` w; k )e S T k i +w = @ j` (e S ); k T k i = 1 w q (w 1+q e S = @ j` T k i ); k +we S y m k` = @ mj kj = @ m` r 2 + k` y j kj y ` r 2 X r T k i + (1 +q)( = @ j` w; k )e S T k i +w = @ j` (e S ); k T k i ; where [ = @ j` ;@ k ]T k i has been rewritten using (6.2.2) and (6.2.1). 49 In the next Lemma, we give some useful results concerning our quantitiesA ,J and and also our derivative operators, Lemma 4.0.3. ForA ,J and , the following identities hold A k j J 1 k j = (A k j k j )J 1 + k j (J 1 1); (4.0.13) A k j k j =A k l [D] l j ; (4.0.14) ij = ip (A j p +A j l l ; p ): (4.0.15) For any derivative operator L2f = @ ij ;X r g i;j=1;2;3 L(A k j J 1 ) =J 1 A k ` A s j (L ` ); s 1 J 1 A k j A s ` (L ` ); s J 1 A k ` A s j [L;@ s ] ` 1 J 1 A k j A s ` [L;@ s ] ` : (4.0.16) Proof. First (4.0.13) is straightforward to verify by expanding the right-hand side. Sec- ond, (4.0.14) was proven in Chapter 3, (3.0.22). Next, (4.0.15) is proven in the following calculation where we recall A = [D] 1 and use =y +, ij = ip j p = ip A j l l ; p = ip (A j p +A j l l ; p ): (4.0.17) Lastly, using our dierentiation formulae forA andJ (2.2.2) which hold when generalized to X r and = @, and also again the formula =y +, L(A k j J 1 ) =J 1 A k ` A s j L( ` ; s ) 1 J 1 A k j A s ` L( ` ; s ) =J 1 A k ` A s j L(y ` ; s + ` ; s ) 1 J 1 A k j A s ` L(y ` ; s + ` ; s ) =J 1 A k ` A s j L( ` ; s ) 1 J 1 A k j A s ` L( ` ; s ) =J 1 A k ` A s j (L ` ); s 1 J 1 A k j A s ` (L ` ); s J 1 A k ` A s j [L;@ s ] ` 1 J 1 A k j A s ` [L;@ s ] ` : (4.0.18) 50 Finally, before we prove our main energy inequality, it is worth formally stating the equivalence of our high order normS N and high order energy functionalE N . Lemma 4.0.4. Let (; V) : ! R 3 R 3 be a unique local solution to (2.1.27)-(2.1.28) on [0;T ] with T > 0 xed and assume (; V) satises the a priori assumptions (2.4.2). Fix N 2de + 12. Let kN + 1 in (2.1.21). Then there are constants C 1 ;C 2 > 0 so that C 1 S N () sup 0 0 E N ( 0 )C 2 S N (): (4.0.19) Proof. Recall the quantities introduced in Section 2.3: notably the denitionsS N (2.3.2) andE N (4.0.3), and forE N the associated decomposition of (4.0.1) and denition of the conjugatesM a; ,N (4.0.2). Then the equivalence ofS N andE N is a straightforward application of Lemma 6.1.1 to give bounds on and associated matrix quantities, and importantly for the nonisentropic setting, we also use the positive lower and upper bounds of e S given by Corollary 1.1.5 to conclude the result. We are now ready to prove our central energy inequality which will be essential in the proof of our main result Theorem 2.4.2. Proposition 4.0.5. Suppose 2 (1; 5 3 ]. Let (; V) : ! R 3 R 3 be a unique local solution to (2.1.27)-(2.1.28) on [0;T ] with T > 0 xed and assume (; V) satises the a priori assumptions (2.4.2). Fix N 2de + 12. Let k N + 1 in (2.1.21). Then for all 2 [0;T ], we have the following inequality for some 0< 1 E N () + 0 D N ( 0 )d 0 .S N (0) +B N []() + 0 (S N ( 0 )) 1 2 (B N [V]( 0 )) 1 2 d 0 +S N () + 0 e 0 0 S N ( 0 )d 0 : (4.0.20) 51 Proof. Zeroth order estimate. Multiplying (2.1.27) by 1 im @ m and integrating over w ( 3 3 @ i + 3 4 @ i + 2 3 3 ij @ j +w i` ` ) 1 im @ m dy + w 1+ e S ij A k j J 1 k j ; k 1 im @ m dy = 0: (4.0.21) Recognizing the perfect time derivative structure of the rst integral 1 2 d d 3 3 w h 1 @ ;@ idy + w jj 2 dy + 5 3 2 3 4 w h 1 @ ;@ idy 3 3 2 w h@ 1 @ ;@ idy + 2 3 3 w h 1 @ ; @ idy: (4.0.22) After integrating from 0 to , we see the rst three terms in (4.0.22) will contribute to the left hand side of (4.0.20), and also toS N (0) on the right hand side of (4.0.20) by the fundamental theorem of calculus and the equivalence of our norm and energy functional, given by Lemma 4.0.4. For the last two terms in (4.0.22), before time integration, we rewrite them as 3 3 2 w h[@ 1 4 1 ]@ ;@ idy: (4.0.23) Now with A =O, = (detA) 1 3 ;O2 SL(3), we have @ 1 4 1 =@ (O > O) 4O > OO 1 O =@ 1 + 2(O > OO > O ): (4.0.24) Since O > OO > O is anti-symmetric, we can reduce (4.0.23) to 3 3 2 w h@ 1 @ ;@ idy; (4.0.25) which is then bounded bye 0 S N () via@ 1 = 1 (@ ) 1 and (6.1.6). So the last two terms in (4.0.22) contribute to 0 e 0 0 S N ( 0 )d 0 in (4.0.20) after time integration. 52 Returning to the second integral in (4.0.21), we integrate by parts and apply the identities (4.0.13)-(4.0.15) proven in Lemma 4.0.3 to obtain w 1+ e S ij A k j J 1 k j 1 im @ m ; k dy = w 1+ e S J 1 ij A k l l ; j 1 im @ m ; k dy w 1+ e S (J 1 1)@ k ; k dy = w 1+ e S J 1 ip A j p ` ; j 1 im A k ` @ m ; k dy + w 1+ e S J 1 ip A j l l ; p A k ` ` ; j 1 im @ m ; k dy w 1+ e S (J 1 1)@ k ; k dy = w 1+ e S J 1 `j [r ] i j 1 im [r @ ] m ` dy + w 1+ e S J 1 [Curl A ] ` i 1 im [r @ ] m ` dy + w 1+ e S J 1 A j l l ; m A k ` ` ; j @ m ; k dy w 1+ e S (J 1 1)@ k ; k dy =: (i) + (ii) + (iii) + (iv): (4.0.26) By Lemma 4.0.1 (i) = 1 2 d d w 1+ e S J 1 3 X i;j=1 d i d 1 j ((M 0;0 ) j i ) 2 dy + w 1+ e S J 1 T 0;0 dy + 1 2 w 1+ e S J 1 1 J 3 X i;j=1 d i d 1 j ((M 0;0 ) j i ) 2 dy + w 1+ e S J 1 `j [r ] j i 1 im [r ] m p [r @ ] p ` dy; (4.0.27) with the last term above arising from commuting @ andr which is needed for Lemma 4.0.1. Now after time integration, the rst term in (4.0.27) contributes toE N () in (4.0.20), and also toS N (0) by the fundamental theorem of calculus and Lemma 4.0.4. Before time integration, the second term in (4.0.27) is bounded by e 0 S N () due to the desirable form ofT 0;0 (4.0.7). In particularT 0;0 contains derivatives ofP andd i which are exponentially bounded using (6.1.8). Also before time integration, the third term in (4.0.27) is bounded by e 0 S N () due to Lemma 6.3.1 which we use to bound the D 53 term arising fromJ . Hence the second and third terms contribute to 0 e 0 0 S N ( 0 )d 0 in (4.0.20) after time integration. For the last term in (4.0.27), note schematically we can write, J 1 `j [r ] j i 1 im [r ] m p [r @ ] p ` asJ 1 AD 1 ADAD : Then using the a priori assumptions (2.4.2) to boundJ 1 andA , the boundedness of and 1 (6.1.6), and the exponential bound on D from Lemma 6.3.1, we obtain 0 w 1+ e S J 1 AD 1 ADAD dyd 0 . 0 e 0 0 (S N ( 0 )) 3 2 d 0 . 0 e 0 0 S N ( 0 )d 0 ; (4.0.28) where we have also used the a priori assumptionS N ()< 1 3 . Now (iii) is similar to the last term in (4.0.27) and so an analogous argument used to bound that term above will give us that (iii) also contributes to 0 e 0 0 S N ( 0 )d 0 . We consider (ii). Schematically [Curl A ]AD = 1 2 d d ([Curl A ] 2 ) (@ )ADAD (@ A )DAD: (4.0.29) Then using the boundedness ofJ 1 and 1 we have that after time integration, the rst term on the right hand side of (4.0.29) will lead to (ii) contributing toB N [] in (4.0.20). Now using the exponential boundedness of@ and applying Lemma 6.3.1 to@ A , an analogous argument to that used to bound the last term in (4.0.27) will give us that the last two terms on the right hand side of (4.0.29) will lead to contributions to 0 e 0 0 S N ( 0 )d 0 after time integration. For (iv), write 1J 1 = 1 Tr[D]+O(jDj 2 ) usingJ = det[D] = det[Id+D] = 1+Tr[D]+O(jDj 2 ); (4.0.30) 54 to bound (iv) by e 0 S N () where we again apply Lemma 6.3.1. To complete the zeroth order estimate, we obtain the full expression forE N in (4.0.20) by adding the following formula 1 2 d d 1 w 1+ e S J 1 jdiv j 2 dy = 1 2 w 1+ e S @ (J 1 )jdiv j 2 +J 1 @ (jdiv j 2 ) dy; (4.0.31) with the right-hand side in turn bounded bye 0 S N () by Lemma 6.3.1. Hence after time integration, the left-hand side of (4.0.31) completesE N in (4.0.20), contributes toS N (0) by the fundamental theorem of calculus and the right-hand side of (4.0.31) contributes to 0 e 0 0 S N ( 0 )d 0 . High order estimates. Fix (a;) with a +jj 1. First, rearrange (2.1.27) as 3 4 (@ i + @ i + 2 ij @ j ) + i` ` + 1 w w 1+ e S ij A k j J 1 k j ; k = 0: (4.0.32) Apply X a r = @ to (4.0.32) and multiply by 3 4 (@ X a r = @ i + @ X a r = @ i + 2 ij @ X a r = @ j ) + i` X a r = @ ` + 1 w +a w 1++a e S ij X a r = @ A k j J 1 k j ; k = R a; i ; (4.0.33) where we have used Lemma 4.0.2 to compute X a r = @ h 1 w w 1+ e S ij A k j J 1 k j ; k i withR a; i dened as the lower order terms arising from multiple applications of (4.0.11) and (4.0.10) in Lemma 4.0.2, and applying our derivative operators to the resultant expressions from (4.0.11) and (4.0.10). More specically, using the favorable expressionsC q+1 i andC q ij` , (4.0.11) and (4.0.12) respectively, derived in Lemma 4.0.2, the regularity of our entropy term given by Corollary 1.1.5, the dierentiation formulae forA andJ (2.2.2), and also the decomposition of the 55 spatial derivative intoX r and = @ (6.2.1), we have the following favorable schematic form for R a; i R a; i = a+jj X `=0 0 B B @ 2 6 6 4 X a 0 +j 0 j=` a 0 a;j 0 jjj+1 c 1 a 0 ; 0 ;` X a 0 r = @ 0 +c 2 a 0 ; 0 ;` wX a 0 r = @ 0 D 3 7 7 5 Y a 00 +j 00 jNmax(`;N`) c a 00 ; 00 ;` X a 00 r = @ 00 +w 2 4 Y a 000 +j 000 jNmax(`;N`) c a 000 ; 000 ;` X a 000 r = @ 000 D 3 5 1 A ; (4.0.34) for all i2f1; 2; 3g, where c 1 a 0 ; 0 ;` , c 2 a 0 ; 0 ;` , c a 00 ; 00 ;` and c a 000 ; 000 ;` are bounded coecients on B 1 (0) away from the origin. Note here for simplicity we are including the terms A and J without derivatives in our bounded coecients since they are bounded by our a priori assumptions (2.4.2). Since is also bounded by (6.1.6), we in addition include in our smooth coecients. The important feature of the above expression is that any time we have a potentially top orderD term, we have a weightw multiplying such terms: this is seen in the second sum- mation in the expression. This is a direct result of the commutator forms (4.0.11)-(4.0.12) we carefully obtained in Lemma 4.0.2 to have this property. Then with the expression (4.0.34), Lemma 6.3.1 and the desirable form of [X a r = @ ;@ s ] (3.0.57) derived in the curl estimates, kR a; i k 2 a+; e S . (1 +P (S N ))S N ; (4.0.35) with P a polynomial of degree at least 1. Multiply (4.0.33) by w +a 1 im @ X a r = @ m and integrate over 1 2 d d 3 3 w +a h 1 X a r = @ V; X a r = @ Vidy + w +a jX a r = @ j 2 dy + 5 3 2 3 3 w +a h 1 X a r = @ V; X a r = @ Vidy 1 2 3 3 w +a D @ 1 4 1 X a r = @ V;X a r = @ V E dy 56 + w 1++a e S ij X a r = @ A k j J 1 k j ; k 1 im @ X a r = @ m dy = w +a R i a; 1 im X a r = @ m dy: (4.0.36) By the same reasoning as the zeroth order case, the rst four integrals on the left hand side of (4.0.36) contribute to the energy inequality (4.0.20) in the same way. Also, after time integration, the right hand side of (4.0.36) in addition contributes to 0 e 0 0 S N ( 0 )d 0 using (4.0.35) and (2.4.2). Now compute the last integral on the left hand side of (4.0.36) using the derivative formula forA k j J 1 (4.0.16) and integration by parts w 1++a e S ij X a r = @ A k j J 1 k j ; k 1 im @ X a r = @ m dy = w 1++a e S ij J 1 A k ` A s j X a r = @ ` ; s 1 J 1 A k j A s ` X a r = @ ` ; s +C a;;k j () ; k 1 im @ X a r = @ m dy = w 1++a e S J 1 A k ` ij A s j X a r = @ ` ; s `j A s j X a r = @ i ; s +J 1 A k ` `j A s j X a r = @ i ; s + 1 J 1 ij A k j A s ` X a r = @ ` ; s ; k 1 im @ X a r = @ m dy + ij w 1++a e S C a;;k j ; k 1 im @ X a r = @ m dy = w 1++a e S J 1 A k ` [Curl A X a r = @ ] ` i +A k ` `j [r X a r = @ ] i j + 1 ij A k j div (X a r = @ ) 1 im @ X a r = @ m ; k dy + ; k w 1++a e S J 1 A k ` [Curl A X a r = @ ] ` i +A k ` `j [r X a r = @ ] i j + 1 ij A k j div (X a r = @ ) 1 im @ X a r = @ m dy + ij w 1++a e S C a;;k j ; k 1 im @ X a r = @ m dy =:I 1 +I 2 +I 3 ; (4.0.37) where we deneC a;;k j () to be the lower order terms arising from successive iterations of (4.0.16) in Lemma 4.0.3, and dierentiating the result of (4.0.16). In particular, using the desirable form of [X a r = @ ;@ s ] (3.0.57) obtained in the curl estimates, the Leibniz rule, and 57 also the decomposition of the spatial derivative intoX r and = @ (6.2.1), we have the following desirable schematic form forC a;;k j () C a;;k j () = a+jj X `= N 2 0 B B @ 2 6 6 4 X a 0 +j 0 j=` a 0 a+1;j 0 jjj+1 c a 0 ; 0 ;` X a 0 r = @ 0 3 7 7 5 Y a 00 +j 00 jN` c a 00 ; 00 ;` X a 00 r = @ 00 1 C C A ; (4.0.38) for every j;k2f1; 2; 3g, where c a 0 ; 0 ;` and c a 00 ; 00 ;` are bounded coecients on B 1 (0) away from the origin. Note here for simplicity we are including the terms A andJ without derivatives in our bounded coecients since they are bounded by our a priori assumptions (2.4.2). With this formula (4.0.38), we can see using Lemma 6.3.1 to handle the lower order terms in the product Q a 00 +j 00 j N 2 that kC a;;k j ()k 2 +a; e S .S N (): (4.0.39) Furthermore, again using the desirable form of [X a r = @ ;@ s ] (3.0.57) and the decomposition of the spatial derivative into X r and = @ (6.2.1), we can obtain kDC a;;k j ()k 2 1++a; e S .S N (): (4.0.40) Estimation of I 3 : 0 I 3 d 0 = 0 w 1++a e S C a;;k j ; k 1 im @ X a r = @ m dyd 0 . 0 kX a r = @ V m k +a; e S kC a;;k j ()k +a; e S +kX a r = @ V m k 1++a; e S kDC a;;k j k 1++a; e S d 0 . 0 e 0 0 S N ( 0 )d 0 ; (4.0.41) 58 where we have used (6.1.6), (2.1.19), Lemma 1.1.4 and (4.0.39)-(4.0.40). Estimation of I 2 : 0 I 2 d 0 = 0 B 3 4 (0)nB 1 4 (0) ; k w 1++a e S J 1 A k ` [Curl A X a r = @ ] ` i +A k ` `j [r X a r = @ ] i j + 1 ij A k j div (X a r = @ ) 1 im @ X a r = @ m dyd 0 . 0 kD k L 1 ( ) kX a r = @ V m k +a; e S kr X a r = @ k 1++a; e S + X jja+jj k@ Vk ;(1 )e S X jja+jj kr @ k 1+;(1 )e S d 0 . 0 e 0 0 S N ( 0 )d 0 ; (4.0.42) where we have used that ; k has support in B 3 4 (0)nB 1 4 (0) and the fact that we can write X a r = @ in terms of @ in B 3 4 (0)nB 1 4 (0). Estimation of I 1 : 0 I 1 d 0 = 0 w 1++a e S J 1 A k ` [Curl A X a r = @ ] ` i 1 im @ X a r = @ m ; k dyd 0 + 0 w 1++a e S J 1 A k ` `j [r X a r = @ ] i j + 1 ij A k j div (X a r = @ ) 1 im @ X a r = @ m ; k dyd 0 = 0 w 1++a e S J 1 A k ` 1 im [Curl A X a r = @ ] ` i @ X a r = @ m ; k dyd 0 + 0 w 1++a e S J 1 `j [r X a r = @ ] i j 1 im [r @ X a r = @ ] m ` + 1 div (X a r = @ )div (@ X a r = @ ) dyd 0 : Integrating by parts in the curl term 0 w 1++a e S J 1 A k ` 1 im [Curl A X a r = @ ] ` i @ X a r = @ m ; k dyd 0 = w 1++a e S J 1 A k ` 1 im [Curl A X a r = @ ] ` i X a r = @ m ; k dy 0 0 w 1++a e S J 1 A k ` 1 im [Curl A X a r = @ V] ` i X a r = @ m ; k dyd 0 59 0 w 1++a e S @ J 1 A k ` 1 im [Curl A X a r = @ ] ` i X a r = @ m ; k dyd 0 0 w 1++a e S J 1 A k ` 1 im [Curl A X a r = @ ] ` i X a r = @ m ; k dyd 0 0 w 1++a e S J 1 A k ` 1 im [Curl A X a r = @ ] ` i X a r = @ m ; k dyd 0 :=B 1 +B 2 +R 1 +R 2 +R 3 : (4.0.43) Now rewriting the gradient and divergence terms 0 w 1++a e S J 1 `j [r X a r = @ ] i j 1 im [r @ X a r = @ ] m ` + 1 div (X a r = @ )div (@ X a r = @ ) dyd 0 = 0 w 1++a e S J 1 `j [r X a r = @ ] i j 1 im @ [r X a r = @ ] m ` + 1 div (X a r = @ )@ div (X a r = @ ) dyd 0 0 J 1 `j [r X a r = @ ] i j 1 im @ A k ` X a r = @ m ; k + 1 div (X a r = @ )@ A k j X a r = @ j ; k dyd 0 =E 1 +R 4 : (4.0.44) For B 1 B 1 = w 1++a e S J 1 A k ` 1 im X a+jjN [Curl A X a r = @ ] ` i X a r = @ m ; k dy 0 .S N (0) +kr X a r = @ k 2 1++a; e S + 1 kCurl A X a r = @ k 2 1++a; e S .S N (0) +S N () +B N [](); (4.0.45) where we have introduced through the Young inequality. For B 2 jB 2 j = 0 w 1++a e S J 1 A k ` 1 im [Curl A X a r = @ V] ` i X a r = @ m ; k dxd 0 . 0 kr X a r = @ k 1++a; e S kCurl A X a r = @ Vk 1++a; e S d 0 . 0 (S N ( 0 )) 1 2 (B N [V]( 0 )) 1 2 d 0 : (4.0.46) 60 For E 1 E 1 = 1 2 0 d d 8 < : w 1++a e S J 1 0 @ 3 X i;j=1 d i d 1 j (M a; ) j i ) 2 + 1 div X a r = @ 2 1 A dy 9 = ; d 0 + 0 w 1++a e S J 1 T a; dyd 0 1 2 0 w 1++a e S @ J 1 0 @ 3 X i;j=1 d i d 1 j (M a; ) j i ) 2 + 1 div X a r = @ 2 1 A dyd 0 :=E 2 +R 5 +R 6 ; (4.0.47) where we have used Lemma 4.0.1. Then E 2 contributes toE N () in (4.0.20), and also to S N (0) by the fundamental theorem of calculus. Finally, by the Young inequality, Lemma 6.1.1, a priori bounds (2.4.2) and Lemma 6.3.1 jR i j. 0 e 0 0 S N ( 0 )d 0 fori = 1; 2; 3; 4; 5; 6: (4.0.48) We have thus obtained (4.0.20) except for@ terms inE N ();D N ();S N () andB N (). To estimate such@ terms, we note that by (2.1.19),w> 0 inB 3 4 (0) where the 1 estimates will be obtained. Therefore a similar calculation leads to the @ contribution to (4.0.20). Hence we have obtained the full energy inequality (4.0.20). 61 Chapter 5 Large Adiabatic Constant We now prove the case for > 5 3 by modifying the above analysis. We use the strategy from [48], applied to our nonisentropic setting. The rst step is to eliminate the anti- damping eect encountered in [20]. We divide (2.1.27) by 3 5 to obtain w 2 @ i + @ i + 2 ij @ j + 53 w i` ` + 53 w 1+ e S ij A k j J 1 k j ; k = 0: (5.0.1) Then with > 5 3 ,S N andB N are dening according to the expressions given in Section 2.3. Also as dened in Section 2.4, in this case we do not distinguish between 0 and 1 0 = 1 : (5.0.2) Next, the energy based high order quantities which arise directly from the problem are modied as follows since we are considering (5.0.1). ~ E N (; V)() = ~ E N () = 1 2 X a+jjN h 2 D 1 X a r = @ V; X a r = @ V E + 53 D 1 X a r = @ ; X a r = @ Ei w a+ + 53 J 1 h 3 X i;j=1 d i d 1 j (M a; ) j i 2 + 1 div X a r = @ 2 i w a++1 e S dy 62 + 1 2 X jjN (1 ) h 2 1 @ V; @ V + 53 1 @ ; @ i w + 53 (1 )J 1 h 3 X i;j=1 d i d 1 j (N ) j i 2 + 1 (div @ ) 2 i w 1+ e S dy; (5.0.3) ~ D N () = 3 5 2 53 X a+jjN jX a r = @ j 2 w a+ + J 1 h 3 X i;j=1 d i d 1 j (M a; ) j i 2 + 1 div X a r = @ 2 i w a++1 e S + X jjN (1 )j@ j 2 w + (1 )J 1 h 3 X i;j=1 d i d 1 j (N ) j i 2 + 1 (div @ ) 2 i w 1+ e S dy: (5.0.4) We are requiring > 5 3 in this formulation, which gives ~ D N () 0. Unique to the > 5 3 case, we introduce a similar term to ~ E N which does not include top order quantities but will be controlled through our coercivity Lemma 5.0.1 below C N1 (; V)() =C N1 () = 1 2 X a+jjN1 D 1 X a r = @ ; X a r = @ E w a+ + J 1 h 3 X i;j=1 d i d 1 j (M a; ) j i 2 + 1 div X a r = @ 2 i w a++1 e S dy + 1 2 X jjN1 (1 ) 1 @ ; @ w + (1 )J 1 h 3 X i;j=1 d i d 1 j (N ) j i 2 + 1 (div @ ) 2 i w 1+ e S dy: (5.0.5) Before outlining the proofs of our main curl and energy inequalities for the large > 5 3 case, we give the most important and useful result in this setting which will allow us to 63 overcome the time weights with negative powers which arise from the new equation structure (5.0.1). Lemma 5.0.1 (Coercivity Estimates). Suppose > 5 3 . Let (; V) : ! R 3 R 3 be a unique local solution to (2.1.27)-(2.1.28) on [0;T ] with T > 0 xed and assume (; V) satises the a priori assumptions (2.4.2). Fix N 2de + 12. Let kN + 1 in (2.1.21). Fix (a;) with 0a +jjN 1 and with 0jjN 1. Then for all 2 [0;T ], we have the following inequalities kX a r = @ k 2 a+; e S . sup 0 0 f 2 kX a r = @ Vk 2 a+; e S g +kX a r = @ (0)k 2 a+; e S ; (5.0.6a) k@ k 2 ;(1 ) . sup 0 0 f 2 k@ Vk 2 ;(1 )e S g +k@ (0)k 2 ;(1 )e S ; (5.0.6b) kr X a r = @ k 2 a++1; e S . sup 0 0 f X a 0 +j 0 j=a+jj+1 2 kX a 0 r = @ 0 Vk 2 a++1; e S g +kr X a r = @ (0)k 2 a++1; e S ; (5.0.7a) kr @ k 2 +1; e S . sup 0 0 f X j 0 j=jj+1 2 k@ 0 Vk 2 +1; e S g +kr @ (0)k 2 +1; e S ; (5.0.7b) kdiv X a r = @ k 2 a++1; e S . sup 0 0 f X a 0 +j 0 j=a+jj+1 2 kVk 2 a++1; e S g +kdiv X a r = @ (0)k 2 a++1; e S ; (5.0.8a) kdiv @ k 2 +1; e S . sup 0 0 f X j 0 j=jj+1 2 k@ 0 Vk 2 +1; e S g +kdiv @ (0)k 2 +1; e S : (5.0.8b) 64 Proof. Proof of (5.0.6a)-(5.0.6b). By the fundamental theorem of calculus, and the expo- nential boundedness of (6.1.4) and therefore time integrability of negative powers of , X a r = @ = 0 X a r = @ Vd 0 +X a r = @ (0) = 0 1 X a r = @ Vd 0 +X a r = @ (0) . sup 0 0 fX a r = @ Vg +X a r = @ (0): (5.0.9) Therefore applying Cauchy's inequality (ab.a 2 +b 2 ; a;b2R) kX a r = @ k 2 a+; e S . sup 0 0 f 2 kX a r = @ Vk 2 a+; e S g +kX a r = @ (0)k 2 a+; e S : (5.0.10) This proves (5.0.6a), and (5.0.6b) is similar. Proof of (5.0.6b)-(5.0.7a). By a similar coercivity estimate to (5.0.9)-(5.0.10) kr X a r = @ k 2 a++1; e S . sup 0 0 f 2 kr X a r = @ Vk 2 a++1; e S g +kr X a r = @ (0)k 2 a++1; e S : (5.0.11) Now using the decomposition of spatial derivatives into angular derivatives and radial deriva- tive (6.2.1), and also our a priori bounds (2.4.2), we have sup 0 0 2 kr X a r = @ Vk 2 a++1; e S . sup 0 0 f X a 0 +j 0 j=a+jj+1 2 kX a 0 r = @ 0 Vk 2 a++1; e S g: (5.0.12) Then (5.0.11)-(5.0.12) imply (5.0.7a). The proof of (5.0.7b) is similar but simpler since we do not need the decomposition formula. Proof of (5.0.8a)-(5.0.8b). Finally the proofs of (5.0.8a)-(5.0.8b) are similar to the proofs of (5.0.7a)-(5.0.7b). Remark 5.0.2. Lemma 5.0.1 allows us to control terms without time weights with negative powers using our normS N () and initial dataS N (0). In the > 5 3 case, such terms do 65 not appear immediately from our equation because of the new structure (5.0.1). Notably, Lemma 5.0.1 will let us include the new quantityC N1 in our energy inequality. However we cannot use Lemma 5.0.1 to control top order quantities since that would require control of N + 1 derivatives of V which we do not have. This is why rstly we have the particular structure ofS N in this case where we separate top order terms from lower order terms, and secondly why we only include N 1 derivatives inC N1 . 5.1 Curl Estimates > 5 3 For the curl estimates for > 5 3 we note that we will start from the same equation (2.1.25) as for the 2 (1; 5 3 ] case. So when controlling the curl quantities, we will articially include the time weight with negative power 53 . Therefore this will not pose an issue in most of the curl estimates since we either simply use the boundedness of 53 if we do wish to include it in our high order quantities, or at the top order, include it in our high order quantities. The only time we will have to consider it more carefully is when controlling the Curl A terms when we integrate this time weight. We will give the method for this situation in Lemma 5.1.1 below. Otherwise, we note in the > 5 3 case that from the structure ofS N allowable by the coercivity Lemma 5.0.1, the fact that 0 = 1 here, and our asymptotic behavior ofA which still holds by Lemma 1.1.6 because of our requirement that detA 1 +t 3 , the methods of the curl estimates for 2 (1; 5 3 ] in Chapter 3 will hold for > 5 3 and we will obtain similar bounds on our curl quantities. Lemma 5.1.1. Suppose > 5 3 . Let (; V) : ! R 3 R 3 be a unique local solution to (2.1.27)-(2.1.28) on [0;T ] with T > 0 xed and assume (; V) satises the a priori assumptions (2.4.2). Fix N 2de + 12. Let k N + 1 in (2.1.21). Let a +jj = N. Assume the following bound is known 0 0 ( 00 )X a r = @ [@ ; Curl A ]Vd 00 2 1++a; e S .S N ( 0 ): (5.1.1) 66 Then 53 0 1 ( 0 ) 0 0 ( 00 )X a r = @ [@ ; Curl A ]Vd 00 d 0 2 1++a; e S . 0 e 0 S N ( 0 )d 0 : (5.1.2) Proof. Applying the boundedness of 53 53 0 1 ( 0 ) 0 0 ( 00 )X a r = @ [@ ; Curl A ]Vd 00 d 0 2 1++a; e S . 53 2 4 0 (1 + 0 ) 2 ( 0 ) d 0 0 1 (1 + 0 ) 2 ( 0 ) 0 0 ( 00 )X a r = @ [@ ; Curl A ]Vd 00 ! 2 d 0 3 5 w 1++a e S dy . 53 0 1 (1 + 0 ) 2 ( 0 ) 0 0 ( 00 )X a r = @ [@ ; Curl A ]Vd 00 2 1++a; e S d 0 . sup 0 0 [( 0 ) 53 ] 0 1 (1 + 0 ) 2 ( 0 ) 0 0 ( 00 )X a r = @ [@ ; Curl A ]Vd 00 2 1++a; e S d 0 = sup 0 0 [( 0 ) 53 ] 0 ( 0 ) 53 ( 0 ) 53 (1 + 0 ) 2 ( 0 ) 0 0 ( 00 )X a r = @ [@ ; Curl A ]Vd 00 2 1++a; e S d 0 . sup 0 0 [( 0 ) 53 ] sup 0 0 1 ( 0 ) 53 0 ( 0 ) 53 (1 + 0 ) 2 ( 0 ) 0 0 ( 00 )X a r = @ [@ ; Curl A ]Vd 00 2 1++a; e S d 0 = 0 ( 0 ) 53 (1 + 0 ) 2 ( 0 ) 0 0 ( 00 )X a r = @ [@ ; Curl A ]Vd 00 2 1++a; e S d 0 . 0 e 0 S N ( 0 )d 0 : (5.1.3) Using the technique of Lemma 5.1.1 for similar terms and then otherwise an analogous argument to the 2 (1; 5 3 ] case as described above, we obtain our curl estimates in this case. Proposition 5.1.2. Suppose > 5 3 . Let (; V) : !R 3 R 3 be a unique local solution to (2.1.27)-(2.1.28) on [0;T ] with T > 0 xed and assume (; V) satises the a priori 67 assumptions (2.4.2). FixN 2de+12. LetkN +1 in (2.1.21). Then for all2 [0;T ], the following inequalities hold for some 0< 1 B N [V]().e 2 0 S N (0) + ~ B N (0) + (1 + 2 )e 2~ 0 S N (); (5.1.4) ~ B N []().S N (0) +B N (0) +S N () + 0 e 0 0 S N ( 0 )d 0 : (5.1.5) 5.2 Energy Estimates > 5 3 Energy estimates for > 5 3 depend on the new equation structure (5.0.1). This gives rise to the forms of ~ E N and ~ D N . However with the coercivity Lemma 5.0.1 and the norm-energy equivalence result in this case, term by term estimates will use the same techniques as in the 2 (1; 5 3 ] case since as in the curl estimates, we have the desirable asymptotic behavior ofA which still holds by Lemma 1.1.6 because of our requirement that detA 1 +t 3 . Also we have 0 = 1 . Furthermore, as in the curl estimates, when estimating terms which include the time weight 53 we either simply use the boundedness of 53 if we do wish to include it in our high order quantities, or at the top order, include it in our high order quantities. The only time we will have to consider it more carefully is when we integrate this time weight after integration by parts in. We will give the details for the unique term that arises from this in the outline of our proof of the energy inequality below. Therefore we give the norm-energy equivalence, and then outline the proof for the energy inequality in this case, placing emphasis on the new energy identity structure while omitting the details which are similar to the 2 (1; 5 3 ] case. Lemma 5.2.1. Let (; V) : ! R 3 R 3 be a unique local solution to (2.1.27)-(2.1.28) on [0;T ] with T > 0 xed and assume (; V) satises the a priori assumptions (2.4.2). Fix N 2de + 12. Let kN + 1 in (2.1.21). Then there are constants C 1 ;C 2 > 0 so that C 1 S N () sup 0 0 fE N ( 0 ) +C N1 ( 0 )gC 2 (S N () +S N (0)): (5.2.1) 68 Proof. The proof follows from an analogous argument to the proof of Lemma 4.0.4 from the 2 (1; 5 3 ] case, in conjunction with Lemma 5.0.1 to control terms without time weights with negative powers which are included inC N1 ( 0 ) byS N () +S N (0). We are now ready to give the energy inequality in this case. Proposition 5.2.2. Suppose > 1. Let (; V) : !R 3 R 3 be a unique local solution to (2.1.27)-(2.1.28) on [0;T ] with T > 0 xed and assume (; V) satises the a priori assumptions (2.4.2). FixN 2de+12. LetkN +1 in (2.1.21). Then for all2 [0;T ], we have the following inequality for some 0< 1, ~ E N () +C N1 () + 0 ~ D N ( 0 )d 0 .S N (0) +B N []() + 0 (S N ( 0 )) 1 2 (B N [V]( 0 )) 1 2 d 0 +S N () + 0 e 0 0 S N ( 0 )d 0 : (5.2.2) Outline of Proof. Zeroth order estimate. Multiplying (5.0.1) by 1 im @ m and inte- grating over w ( 2 @ i + @ i + 2 2 ij @ j + 53 w i` ` ) 1 im @ m dy + 53 w 1+ e S ij A k j J 1 k j ; k 1 im @ m dy = 0: (5.2.3) Recognizing the perfect time derivative structure of the rst integral 1 2 d d 2 w h 1 @ ;@ idy + 53 w jj 2 dy + 3 5 2 43 w jj 2 dy 2 2 w h@ 1 @ ;@ idy + 2 2 w h 1 @ ; @ idy: (5.2.4) Returning to the second integral in (5.2.3), we integrate by parts to obtain 53 w 1+ e S ij A k j J 1 k j 1 im @ m ; k dy 69 = 53 w 1+ e S J 1 ij A k l l ; j 1 im @ m ; k dy 53 w 1+ e S (J 1 1)@ k ; k dy = 53 w 1+ e S J 1 ip A j p ` ; j 1 im A k ` @ m ; k dy + 53 w 1+ e S J 1 ip A j l l ; p A k ` ` ; j 1 im @ m ; k dy 53 w 1+ e S (J 1 1)@ k ; k dy = 53 w 1+ e S J 1 `j [r ] i j 1 im [r @ ] m ` dy + 53 w 1+ e S J 1 [Curl A ] ` i 1 im [r @ ] m ` dy + 53 w 1+ e S J 1 A j l l ; m A k ` ` ; j @ m ; k dy 53 w 1+ e S (J 1 1)@ k ; k dy =: (i) + (ii) + (iii) + (iv); (5.2.5) By Lemma 4.0.1 (i) = 1 2 d d 53 w 1+ e S J 1 3 X i;j=1 d i d 1 j ((M 0;0 ) j i ) 2 dy + 3 5 2 53 w 1+ e S J 1 3 X i;j=1 d i d 1 j ((M 0;0 ) j i ) 2 dy + 53 w 1+ e S J 1 T 0;0 dy + 53 2 w 1+ e S J 1 1 J 3 X i;j=1 d i d 1 j ((M 0;0 ) j i ) 2 dy + 53 w 1+ e S J 1 `j [r ] j i 1 im [r ] m p [r @ ] p ` dy: (5.2.6) Now after time integration, the second term in (5.2.6) contributes to 0 ~ D N ( 0 )d 0 in (5.2.2). High order estimate Fix (a;) with a +jj 1. First, rearrange (5.0.1) as 2 @ i + @ i + 2 ij @ j + 53 i` ` + 53 1 w w 1+ e S ij A k j J 1 k j ; k = 0: (5.2.7) Apply X a r = @ to (5.2.7) and multiply by 2 @ X a r = @ i + @ X a r = @ i + 2 ij @ X a r = @ j + 53 i` X a r = @ ` + 53 1 w +a w 1++a e S ij X a r = @ A k j J 1 k j ; k = 53 R a; i : (5.2.8) 70 Multiply (5.2.8) by w +a 1 im @ X a r = @ m and integrate over 1 2 d d 2 w +a h 1 X a r = @ V; X a r = @ Vidy + 53 w +a jX a r = @ j 2 dy + 3 5 2 53 w +a jX a r = @ j 2 dy 2 2 w +a D @ 1 4 1 X a r = @ V;X a r = @ V E dy + 53 w 1++a e S ij X a r = @ A k j J 1 k j ; k 1 im @ X a r = @ m dy = 53 w +a R i a; 1 im X a r = @ m dy: (5.2.9) Now compute the last integral on the left hand side of (5.2.9) using integration by parts 53 w 1++a e S ij X a r = @ A k j J 1 k j ; k 1 im @ X a r = @ m dy = 53 w 1++a e S ij J 1 A k ` A s j X a r = @ ` ; s 1 J 1 A k j A s ` = @ ` ; s +C a;;k j () ; k 1 im @ X a r = @ m dy = 53 w 1++a e S J 1 A k ` ij A s j X a r = @ ` ; s `j A s j X a r = @ i ; s +J 1 A k ` `j A s j X a r = @ i ; s + 1 J 1 ij A k j A s ` X a r = @ ` ; s ; k 1 im @ X a r = @ m dy + 53 ij w 1++a e S C a;;k j ; k 1 im @ X a r = @ m dy = 53 w 1++a e S J 1 A k ` [Curl A X a r = @ ] ` i +A k ` `j [r X a r = @ ] i j + 1 ij A k j div (X a r = @ ) 1 im @ X a r = @ m ; k dy + 53 ; k w 1++a e S J 1 A k ` [Curl A X a r = @ ] ` i +A k ` `j [r X a r = @ ] i j + 1 ij A k j div (X a r = @ ) 1 im @ X a r = @ m dy + 53 ij w 1++a e S C a;;k j ; k 1 im @ X a r = @ m dy =:I 1 +I 2 +I 3 : (5.2.10) 71 Estimation of I 1 : 0 I 1 d 0 = 0 53 w 1++a e S J 1 A k ` [Curl A X a r = @ ] ` i 1 im @ X a r = @ m ; k dyd 0 + 0 53 w 1++a e S J 1 A k ` `j [r X a r = @ ] i j + 1 ij A k j div (X a r = @ ) 1 im @ X a r = @ m ; k dyd 0 = 0 53 w 1++a e S J 1 A k ` 1 im [Curl A X a r = @ ] ` i @ X a r = @ m ; k dyd 0 + 0 53 w 1++a e S J 1 `j [r X a r = @ ] i j 1 im [r @ X a r = @ ] m ` + 1 div (X a r = @ )div (@ X a r = @ ) dyd 0 = 53 w 1++a e S J 1 A k ` 1 im [Curl A X a r = @ ] ` i X a r = @ m ; k dx 0 0 53 w 1++a e S J 1 A k ` 1 im [Curl A X a r = @ V] ` i X a r = @ m ; k dxd 0 + (3 5) 0 43 w 1++a e S J 1 A k ` 1 im [Curl A X a r = @ ] ` i X a r = @ m ; k dxd 0 0 53 w 1++a e S @ J 1 A k ` 1 im [Curl A X a r = @ ] ` i X a r = @ m ; k dxd 0 0 53 w 1++a e S J 1 A k ` 1 im [Curl A X a r = @ ] ` i X a r = @ m ; k dxd 0 0 53 w 1++a e S J 1 A k ` 1 im [Curl A X a r = @ ] ` i X a r = @ m ; k dxd 0 + 0 53 w 1++a e S J 1 `j [r X a r = @ ] i j 1 im @ [r X a r = @ ] m ` + 1 div (X a r = @ )@ div (X a r = @ ) dyd 0 0 53 J 1 `j [r X a r = @ ] i j 1 im @ A k ` X a r = @ m ; k + 1 div (X a r = @ )@ A k j X a r = @ j ; k d 0 :=B 1 +B 2 +B 3 +R 1 +R 2 +R 3 +E 1 +R 4 ; (5.2.11) where we have used integration by parts. ForB 3 , which is a term unique to the > 5 3 case, (3 5) 0 43 w 1++a e S J 1 A k ` 1 im [Curl A X a r = @ ] ` i X a r = @ m ; k dxd 0 = (3 5) 0 43 1 53 2 53 2 w 1++a e S J 1 A k ` 1 im [Curl A X a r = @ ] ` i X a r = @ m ; k dxd 0 . sup 0 0 53 kr X a r = @ k 2 1++a; e S + 1 sup 0 0 53 kCurl A X a r = @ k 2 1++a; e S ! 0 33 2 d 0 . ~ S N () + ~ B N [](); (5.2.12) 72 where we have used the Young inequality with " and (6.1.4). For E 1 E 1 = 1 2 0 d d 8 < : 53 w 1++a e S J 1 0 @ 3 X i;j=1 d i d 1 j (M a; ) j i ) 2 + 1 div X a r = @ 2 1 A dy 9 = ; d 0 + 0 3 5 2 53 J 1 h 3 X i;j=1 d i d 1 j (M a; ) j i 2 + 1 div X a r = @ 2 i w a++1 e S dyd 0 + 0 53 w 1++a e S J 1 T a; dyd 0 1 2 0 53 w 1++a e S @ J 1 0 @ 3 X i;j=1 d i d 1 j (M a; ) j i ) 2 + 1 div X a r = @ 2 1 A dyd 0 :=E 2 +D 1 +R 5 +R 6 ; (5.2.13) where we have used Lemma 4.0.1. ThenD 1 contributes to 0 ~ D N ( 0 )d 0 in (5.2.2). This concludes our energy inequality proof outline in the > 5 3 case. 73 Chapter 6 Appendices 6.1 based inequalities We have the following useful based inequalities, summarized by the Lemma below. Lemma 6.1.1. Suppose > 1. Fix an ane motion A(t) from the setS under consider- ation, namely require detA(t) 1 +t 3 ; t 0: (6.1.1) Let 1 := lim !1 () () ; 0 := d( ) 2 1 ; (6.1.2) where () = detA() and d( ) = 8 > > < > > : 3 3 if 1< 5 3 2 if > 5 3 : Then we have the following properties 0< 0 = 0 ( ) 1 ; (6.1.3) e 1 . ().e 1 ; 0; (6.1.4) X a+jjN kX a r = @ Vk 1++a; e S + X jjN k@ Vk 1+;(1 )e S .e 0 S N () 1 2 (6.1.5) k k.e 1 ; kk +k 1 kC; (6.1.6) 74 3 X i=1 d i + 1 d i C (6.1.7) 3 X i=1 j@ d i j +k@ Pk.e 1 (6.1.8) jwj 2 .h 1 w; wi.jwj 2 ; w2R 3 ; (6.1.9) for C > 0. Proof. The result (6.1.3) is clear from the denition of 0 . For inequalities (6.1.4) and (6.1.6) through (6.1.9) we rst note that by Lemma 1.1.6, there exist matrices A 0 ;A 1 ;M(t) such that A(t) =A 0 +tA 1 +M(t); t 0: (6.1.10) where A 0 ;A 1 are time-independent and M(t) satises the bounds kM(t)k =o t!1 (1 +t); k@ t M(t)k. (1 +t) 33 : (6.1.11) We also note detA(t) 1 +t 3 . Then inequalities (6.1.4) and (6.1.6) through (6.1.9) follow from Lemma A.1 [20]. Finally, (6.1.5) follows from the denition ofS N (2.3.2) and properties (6.1.3)-(6.1.4) above. 6.2 Derivative Operators Our derivative operators = @ ji and X r satisfy the following identities: Lemma 6.2.1. For i2f1; 2; 3g we have the decomposition @ i = y j r 2 = @ ji + y i r 2 X r : (6.2.1) 75 For i;j;k;m2f1; 2; 3g; we have the commutator identities [ = @ ji ;X r ] = 0; [ = @ ji ; = @ ik ] = = @ jk ; [@ m ;X r ] =@ m ; [@ m ; = @ ji ] = mj @ i mi @ j : (6.2.2) Proof. These properties are straightforward consequences of the denitions introduced in (2.2.8) [19]. The following Lemma will help when dierentiating radial functions. Lemma 6.2.2. Given a radial function f : !R, say f(jyj) with derivatives with respect tojyj denoted by the standard prime notation, and a +jjN, k2f1; 2; 3g, we have X r y k jyj = 0; (6.2.3) X r y k jyj 2 = y k jyj 2 ; (6.2.4) X a r f(jyj) = a X j=1 p j (jyj)f (j) (jyj); (6.2.5) where p j is some polynomial with max degree j; = @ ji f(jyj) = 0; (6.2.6) = @ y k jyj = 8 > > < > > : y ` =jyj 0; (6.2.7) for some `2f1; 2; 3g; = @ y k jyj 2 = 8 > > < > > : y ` =jyj 2 0 ; (6.2.8) for some `2f1; 2; 3g: 76 Proof. Both (6.2.3) and (6.2.6) are straightforward applications of the denitions introduced in (2.2.8). Now (6.2.5) follows from induction using the following results X r f(jyj) =jyjf 0 (jyj); X r (jyj n ) =njyj n for n2Z 0 : Next (6.2.7) follows easily from the following facts, the rst of which is a straightforward calculation, = @ ji y k jyj = jk y i ik y j jyj ; = @ ii = 0 for i;j2 1; 2; 3: Similar arguments give (6.2.4) and (6.2.8) to conclude the proof. 6.3 Weighted Sobolev-Hardy Inequality By (2.1.19), w behaves like a distance function. Then from Proposition C.2 [20], and also using Corollary 1.1.5, we have the following modied weighted Sobolev-Hardy inequality: Lemma 6.3.1. For any u2C 1 (B 1 (0)), we have sup B 1 (0)nB 1 4 (0) w a 1 2 e S X a 1 r = @ 1 u . X a+jja 1 +j 1 j+de+6 kX a r = @ uk a+; e S + X jja 1 +j 1 j+2 k@ uk ;(1 )e S ; (6.3.1) sup B 1 (0)nB 1 4 (0) w a 1 2 e S DX a 1 r = @ 1 u . X a+jja 1 +j 1 j+de+6 kr X a r = @ uk a++1; e S + X jja 1 +j 1 j+2 kr @ uk +1;(1 )e S : (6.3.2) 6.4 Scaling Analysis and Ane Motion An alternative derivation of the nonisentropic ane motion discussed in Section 1.1 is available directly from the structure of the original Eulerian equations. 77 First for any given (; u;S;p) and A2 GL + (3), consider the following mass critical transformation (t;x) = (detA) 1 ^ (s;y); (6.4.1) u(t;x) = (detA) 13 6 A ^ u(s;y); (6.4.2) S(t;x) = ^ S(s;y); (6.4.3) p(t;x) = (detA) ^ p(s;y); (6.4.4) ds dt = (detA) 13 6 ; (6.4.5) y =A 1 x: (6.4.6) If (; u;S;p) solve (1.0.1)-(1.0.2) and (1.0.4)-(1.0.5) in (t), then (^ ; ^ u; ^ S; ^ p) solve @ s ^ + div (^ ^ u) = 0 (6.4.7) ^ (@ s ^ u + ^ ur^ u) + r(^ p) = 0 (6.4.8) @ s ^ S + ^ ur ^ S = 0 (6.4.9) ^ p = ^ e ^ S ; (6.4.10) in ^ (s) =A 1 (s) with = (detA) 2 3 A 1 A > . Motivated by the nearly invariant transformation (6.4.1)-(6.4.6), we are looking for a path R + 3t7!A(t)2 GL + (3); and consider the transformation (t;x) = (detA(s)) 1 ~ (s;y); (6.4.11) u(t;x) = (detA(s)) 13 6 A(s) ~ u(s;y); (6.4.12) S(t;x) = ~ S(s;y); (6.4.13) p(t;x) = (detA(s)) ~ p(s;y); (6.4.14) 78 ds dt = (detA(t)) 13 6 ; (6.4.15) y =A(t) 1 x: (6.4.16) We seekA(t) such that this transformation solves the vacuum free boundary nonisentropic Euler system (1.0.1)-(1.0.11). Introduce (s) := detA(s) 1 3 ; B(s) :=A 1 A s : Then (1.0.1)-(1.0.2) and (1.0.4)-(1.0.5) can be written as ~ @ s ~ u + 1 3 2 s ~ uB~ u +B(yr)~ u + (~ ur)~ u + r(~ p) = 0; (6.4.17) @ s ~ 3 s ~ +Byr~ + div(~ ~ u) = 0; (6.4.18) @ s ~ S + ~ ur ~ S +Byr ~ S = 0; (6.4.19) ~ p = ~ e ~ S : (6.4.20) Next make the important change of variables U(s;y) := ~ u(s;y) +B(s)y: (6.4.21) Notice that UrU = ~ ur~ u +Byr~ u +B~ u +B 2 y; @ s U =@ s ~ u +B s y; div(By) =Tr(A 1 A s ) = @ s detA(s) detA(s) =3 s : 79 Then (6.4.17)-(6.4.19) can be expressed as @ s U + (Ur)U + 1 3 2 s Id 2B U B s B 2 +B 1 3 2 s y + 1 ~ r(~ p) = 0; (6.4.22) @ s ~ + div (~ U) = 0; (6.4.23) @ s ~ S + Ur ~ S = 0: (6.4.24) We now seek a special solution of (6.4.22)-(6.4.24) and (6.4.20) in = B 1 (0) by setting U = 0. Then (6.4.23)-(6.4.24) reduce to @ s ~ = @ s ~ S = 0 which implies ~ and ~ S are only y-dependent. Furthermore by (6.4.20), ~ p is also only y-dependent. Now for (6.4.22) to hold, we need to solve B s B 2 + 1 3 2 s B y = 1 ~ r~ p: (6.4.25) Let 1 ~ r~ p =y in : (6.4.26) By a simple argument, this equation implies ~ p and ~ are radial functions. Thus ~ S is also radial. Now (6.4.25) will be satised if B solves the ODE system B s B 2 + 1 3 2 s B =: (6.4.27) Now using the denition of B and dt ds = (detA(t)) 13 6 , we have B s = detA 3 1 6 @ t (A 1 A t detA 3 1 6 ) = detA 3 1 3 A 1 A t A 1 A t detA 3 1 3 A 1 A tt 3 1 6 detA 3 1 3 A 1 A t @ t detA detA =B 2 + 3 1 2 s B detA 3 1 3 A 1 A tt : 80 Thus recalling the denition of , (6.4.25) is equivalent to the following ODE system for A detA 3 1 3 A 1 A tt = detA 2 3 A 1 A > : which can be written exactly as the ane fundamental system A tt = detA 1 A > : (6.4.28) Finally since ~ and ~ p are radial, (6.4.26) gives the fundamental ane ODE ~ p 0 (r) =r~ (r): (6.4.29) Then using (6.4.20) and our vacuum condition, from (6.4.29) we can obtain the explicit ane entropy formula for e ~ S in terms of ~ e ~ S (r) = 1 r `~ (`)d` (~ (r)) : (6.4.30) 81 Part II Global Solutions to the Compressible Euler Equations with Heat Transport by Convection around Dyson's Isothermal Ane Solutions Chapter 7 Introduction We consider compressible Euler equations with heat transport by convection for ideal gases in three space dimensions @ t + div(u) = 0; (7.0.1) (@ t u + uru) +rp = 0; (7.0.2) (@ t T + urT ) +T div(u) = 0; (7.0.3) where u : R + R 3 ! R 3 is the uid velocity vector eld, : R + R 3 ! R + is the density, T : R + R 3 ! R + is the temperature, p : R + R 3 ! R + is the pressure and > 0 is the heat capacity at constant volume [31], a physical constant. Equations (7.0.1) and (7.0.2) express the usual conservation of mass and momentum respectively. Equation 82 (7.0.3) expresses the conservation of energy in terms of the temperature T in the form of heat transport by convection. This formulation follows from the ideal gas assumption which lets us express the internal energy e as a linear function of temperature: e =T [31]. Our equation of state is the usual equation of state for an ideal gas given in terms of independent unknowns and T [29,31] p(;T ) =T: (7.0.4) Together, equations (7.0.1)-(7.0.4) describe the compressible ow of an inviscid, non- conducting and calorically perfect [1], ideal gas. The equations (7.0.1)-(7.0.3) under con- sideration can also be derived in a kinetic theory framework from the Boltzmann equa- tion [18, 46] where the temperature occurs naturally as an unknown and the specic heat capacity = 3 2 appears modeling the monoatomic gas. Collectively, we will study the Cauchy problem in the whole space for the Euler system with heat transport (7.0.1)-(7.0.4). Notably we are interested in a regime that is almost isothermal. Following Dyson [12], we consider a system isothermal if the temperature function is space independent. We will consider a wider class of potentially space dependent temperature proles by perturbing around a background space independent temperature. Our model admits isothermal solu- tions in the sense of Dyson and allows the temperature to vary in time. Mathematically, we are deviating slightly from the notion of the isothermal Euler models which do not consider temperature but instead focus on the equation of statep =, see [11,28] for example. How- ever with the equation of state (7.0.4)p =T , andT close to being space independent, our framework shares some mathematical similarities with such previously considered models. Beyond the special Dyson solution [12] discussed below in Section 7.2, to the best of our knowledge there are no known previous global existence results for almost isothermal or isothermal Euler, with or without heat convection. The main goal of this part is to 83 construct open sets of initial data that lead to global solutions to the physically important almost isothermal Euler system with heat transport. 7.1 Existence Theories for the Euler System Before we introduce the special Dyson solution, we brie y review some known results for the Euler equations relevant to us. Firstly the Euler equations are hyperbolic and the existence ofC 1 local-in-time positive density solutions follows from the theory of symmetric hyperbolic systems [30,37]. However smooth solutions are expected to breakdown in nite time: the classical result of Sideris [49] shows that singularities must form if the density is a strictly positive constant outside of a bounded set. Makino-Ukai- Kawashima [40] proved that singularities form for compactly supported smooth solutions moving into vacuum. A detailed description of shock formation for irrotational uids starting with smooth initial data is given by Christodoulou-Miao [8]. For a general framework in this direction see the works of Speck and Luk-Speck [35, 52]. Very recently Buckmaster-Shkoller-Vicol [4] give a constructive proof of shock formation from an open set of initial conditions leading to vorticity formation. Further it is only known in one space dimension that the isentropic (p = for a constant > 1) Euler system allows for a globally dened notion of a unique weak solution [6,10,32]. While compression contributes to the breakdown of solutions, expansion provides a mechanism that can produce global solutions. The so-called ane motions, found across dierent works by Ovsyannikov [41], Dyson [12] and Sideris [51], are special expanding global solutions obtained using a separation-of-variables ansatz for the Lagrangian ow map, see Section 7.2. In particular, Dyson [12] obtained special ane solutions to the isothermal system with heat transport, where a space independent initial temperature was considered. A lot of progress has been made in the vacuum free boundary problem in this direction: notably the Sideris ane solutions [51] are special solutions in this setting. Then in the isentropic setting, global stability of background ane solutions has been proven by Had zi c- Jang [20] for 2 (1; 5 3 ] and then extended to the full range > 1 by Shkoller-Sideris [48]. 84 For the nonisentropic (variable entropy) vacuum free boundary problem, Rickard-Had zi c- Jang [44] recently established global existence through perturbations around a rich class of background nonisentropic ane motions. Parmeshwar-Had zi c-Jang [42] showed the global existence of expanding solutions in the vacuum setting with small densities without relying on a perturbation. Small density global solutions in the whole space were obtained by Serre [47] and Grassin [15] for a special class of initial data by the perturbation of expansive wave solutions to the vectorial Burgers equation with linearly growing velocities at innity - a related idea was used in the work of Rozanova [45]. We remark that our result does not require smallness of the density, considers a dierent pressure law to nonisentropic models and allows for isothermal solutions in the sense of Dyson [12]. Finally, there is limited work on the mathematical isothermal Euler models which do not consider temperature but instead focus on the equation of state p =. Akin to Sideris' work on nonisentropic Euler [49], Dong [11] proved a blow up result and a nite propagation result when p =. Jenssen-Tsikkou [28] provide a construction of blow-up solutions in the radial setting, and these are shown to be weak solutions to the original p = system. 7.2 Isothermal Ane Motion A special global-in-time solution to the isothermal Euler system with heat transport in the whole space was given by Dyson [12]. Crucially, Dyson makes following the space independence assumption on the initial ane temperature prole T A (0;x)T > 0; (7.2.1) With this assumption in hand, start by writing x(y;t) =A(t)y: (7.2.2) 85 Then by computing d dt x = u A , u A (x;t) =A 0 (t)A(t) 1 x: (7.2.3) Then div(u A ) = tr(A 0 (t)A(t) 1 ). Hence by Jacobi's formula (detA(t)) 0 = detA(t)div(u A ), we obtain from (7.0.3) @ t T A +T A (detA(t)) 0 detA(t) = 0 (7.2.4) Then we have @ t (ln [T A detA(t)]) = 0: (7.2.5) Then using (7.2.1) T A =T [detA(0)] 1 (detA(t)) 1 : (7.2.6) Hence T A does not depend on space for all time. Then, after plugging (7.2.2) into the momentum and continuity equations, one can eectively separate variables and discover the associated density solution of the Euler equations A (t;x) = e 1 2 jA(t) 1 xj 2 detA(t) ; (7.2.7) and fundamental system of ODEs for A(t) A 00 (t) =T [detA(0)] 1 (detA(t)) 1 A(t) > ; (7.2.8) (A(0);A 0 (0))2 GL + (3)M 33 : (7.2.9) In the above M 33 denotes the set of 3 3 matrices over R and GL + (3) =fA2 M 33 : detA > 0g. Notably (7.2.8) is the same ODE system discovered by Sideris [51]. We let A2C(R; GL + (3))\C 1 (R;M 3 ) be the global solution of this system of ODEs. Remark 7.2.1. The density A is a Gaussian function modulated by the matrixA(t) (7.2.7). This is unique to the isothermal whole space problem and in particular we note A dispereses 86 to 0 asjxj!1, that is, exhibits vacuum at innity behavior. This is dierent to the form of A considered in the vacuum free boundary perturbation problem [20, 44, 48] where A achieves a vacuum state locally through the modulation of a distance function 1jyj 2 . Specically, in the isentropic problem [20, 48], A is xed to a particular distance function whereas in the nonisentropic problem [44], A has innite dimensional freedom associated with its distance function. The isothermal momentum A u A decays exponentially in space because of the Gaussian function while the ane velocity is linearly growing in space. To conclude our characterization of isothermal ane motion, we provide precise asymptotics-in-time for A(t). Lemma 7.2.2. Consider the initial value problem (7.2.8){(7.2.9) and note T [detA(0)] 1 > 0. For 3 2 , the unique solution A(t) to the fundamental system (7.2.8){(7.2.9) has the property detA(t) 1 +t 3 ; t 0 (7.2.10) Furthermore in this case, there exist matrices A 0 ;A 1 ;M(t) such that A(t) =A 0 +tA 1 +M(t); t 0: (7.2.11) where A 0 ;A 1 are time-independent and M(t) satises the bounds kM(t)k =o t!1 (1 +t); k@ t M(t)k. (1 +t) 3 : (7.2.12) For 0 < < 3 2 , given matrices A 0 ;A 1 with A 1 positive denite, there exists a unique solution A(t) to the fundamental system (7.2.8){(7.2.9) such that (7.2.10), (7.2.11) and (7.2.12) hold. 87 Proof. For all> 0, we use Theorem 3 and Lemma 6 from [51] to obtain the results. Now note 8 > > < > > : 0<< 3 2 ) 1 + 1> 5 3 3 2 ) 1< 1 + 1 5 3 ; (7.2.13) and 1 ( 1 + 1) = 1 . Then for 3 2 , we additionally use Lemma A.1 from [20]. For 0<< 3 2 , we additionally use Lemma 1 from [48]. In this part we restrict our attention to the class of isothermal ane solutions expanding linearly in each coordinate direction: namely we require detA(t) 1 +t 3 ; t 0: (7.2.14) By Lemma 7.2.2, for 3 2 this is not a restriction at all in fact sinceA(t) will immediately satisfy (7.2.14). For 0<< 3 2 , Lemma 7.2.2 shows there exists a rich class ofA(t) satisfying (7.2.14). Remark 7.2.3. With expanding ane solutions in hand, we give a physical interpretation of our ane temperature prole T A = T [detA(0)] 1 (detA(t)) 1 : Since detA(t) 1 +t 3 , our space independent ane temperature T A ! 0 as t!1, that is, the isothermal gas becomes cooler for large time. This is a consequence of the heat convection equation (7.0.3). We denote the set of ane motions under consideration byS . To recap, the setS is parametrized by the quadruple (A(0);A 0 (0); T )2 GL + (3)M 33 R + : (7.2.15) With our set of isothermal ane motionsS in hand, the goal of this part is to establish the global-in-time stability of the isothermal Euler system with heat transport (7.0.1)-(7.0.4) by perturbing around the expanding ane motions. 88 Chapter 8 Formulation and Main Global Existence Result 8.1 Lagrangian Coordinates In order to analyze the stability problem for ane motions, we will use the Lagrangian formulation that elucidates the perturbation of the background ane motion compared to the Eulerian formulation. In particular, the Lagrangian formulation allows us to separate variables in a sense which crucially allows us to take full advantage of the time stabilizing mechanism provided by the expanding background motion and on the other hand, isolate the spatial Gaussian prole which will require careful treatment. Notably the heat transport equation is necessary in providing sucient decay to close estimates and establish global stability. We rst dene the ow map as follows @ t (t;y) = u(t;(t;y)); (8.1.1) (0;y) = 0 (y); (8.1.2) where 0 is a suciently smooth dieomorphism to be specied. We introduce the notation A := [D] 1 (Inverse of the Jacobian matrix), (8.1.3) J := det[D] (Jacobian determinant), (8.1.4) 89 v := u (Lagrangian velocity); (8.1.5) f := (Lagrangian density); (8.1.6) T :=T (Langrangian temperature); (8.1.7) a :=J A (Cofactor matrix): (8.1.8) In this framework material derivatives reduce to pure time derivatives and in particular, the temperature equation (7.0.3) is reformulated as @ t T +T [A ] j i v i ; j = 0 (8.1.9) By the standard calculation @ t J =J [A ] j i v i ; j @ t T T + @ t J J = 0: (8.1.10) Then we have @ t (ln [T J ]) = 0; (8.1.11) which implies T =T 0 ( 0 (y)) J (0;y) J 1 : (8.1.12) Furthermore is well-known [9,27] that the conservation of mass equation (7.0.1) gives f(t;y) = (J (t;y)) 1 0 ( 0 (y))J (0;y): (8.1.13) Finally using the equation of state for an ideal gas p =T the momentum equation (7.0.2) is reformulated as f@ tt i + [A ] k i (fT ) ;k = 0: (8.1.14) Here we use coordinatesi = 1; 2; 3 with the Einstein summation convention and the notation F; k to denote the k th partial derivative of F . 90 Next introduce the following notation w(y) := 0 ( 0 (y))J (0;y): (8.1.15) Then using [A ] k i =J 1 [a ] k i and the formula forT (8.1.12) we obtain w@ tt i + [a ] k i wJ 1 T 0 ( 0 (y)) J (0;y) J 1 ! ;k = 0; (8.1.16) Using the Piola identity ([a ] k i ) ;k = 0 we rewrite (8.1.16) as w@ tt i + w[A ] k i T 0 ( 0 (y)) J (0;y) J 1 ! ;k = 0: (8.1.17) Ane motions described in Section 7.2 can be realized as special solutions of (8.1.17) of the form(t;y) =A(t)y if we make the space independence assumption on our initial ane temperature T A (0;x)T > 0; (8.1.18) Now in this case, (t;y) = A(t)y, A > = A(t) > and J = detA. Hence the ansatz transforms (8.1.17) into wA tt y +T [detA(0)] 1 (detA) 1 A > r(w) = 0; (8.1.19) We have that w is independent of t and hence (8.1.19) will hold if we require A tt =T [detA(0)] 1 (detA) 1 A > A > ; (8.1.20) wy =r(w): (8.1.21) At this stage, we demand w(y) =w(jyj) =e jyj 2 2 : (8.1.22) 91 We observe that (8.1.20)-(8.1.22) are nothing but the ane solutions described in Section 7.2, and produce the set of ane motionsS under consideration. Fix an element ofS . Remark 8.1.1. Through (8.1.15), the initial data 0 for our problem is chosen such that (8.1.22) is satised. With an ane motion fromS xed, we dene the modied ow map :=A 1 . Then A > = A > A > ,J = (detA)J whereA > ,J are the equivalents of (8.1.3), (8.1.4) respectively. Now from (8.1.17) we have w(@ tt i + 2[A 1 ] i` @ t A `j @ t j + [A 1 ] i` @ tt A `j j ) + [detA(0)] 1 (detA) 1 [A 1 ] i` [A 1 ] j` w[A ] k j T 0 ( 0 (y)) J (0;y) J 1 ! ;k = 0: (8.1.23) Via (8.1.20) we rewrite the above equation as w(@ tt i + 2[A 1 ] i` @ t A `j @ t j ) +wT [detA(0)] 1 (detA) 2 3 1 i` ` + [detA(0)] 1 (detA) 2 3 1 w ij [A ] k j T 0 ( 0 (y)) J (0;y) J 1 ! ;k = 0; (8.1.24) where the notation := (detA) 2 3 A 1 A > has been introduced. WritingA =O where := (detA) 1 3 andO2 SL(3), we haveA 1 A t = 1 t I+O 1 O t . For ease of notation set C := T [detA(0)] 1 > 0. Taking out of a factor of T from the last term of (8.1.24) the equation is w(@ tt i + 2 t @ t i + 2 ij @ t j ) +Cw 2 3 i` ` + +C 2 3 w ij [A ] k j T 0 ( 0 (y)) T J (0;y) J 1 ! ;k = 0; (8.1.25) where we have dened :=O 1 O t : Note(y)y corresponds to ane motion. Introducing the perturbation (;y) :=(;y)y; (8.1.26) 92 and using (8.1.21), equation (8.1.25) can be written in terms of w(@ tt i + 2 t @ t i + 2 ij @ t j ) +Cw 2 3 i` ` +C 2 3 w ij [A ] k j T 0 ( 0 (y)) T J (0;y) J 1 k j !! ;k = 0: (8.1.27) Now we discuss how we choose the class of general initial data T 0 for our original problem. In particular we suppose T 0 ( 0 (y)) = [J (0;y)] 1 T (1 +(y)); (8.1.28) where : R 3 ! R is a smooth compactly supported function in the unit ball in H k (R 3 ) with the following smallness condition kk 2 H k (R 3 ) ; (8.1.29) for k2Z 0 taken suciently large and > 0 taken suciently small, to be specied later by Theorems 8.4.1 and 8.4.2. Remark 8.1.2. Through (8.1.28), we are allowing our initial temperature to vary spatially in a controlled way from the space independent ane temperature. As we will see, 1 + will appear in our high order energy and itself will contribute as a source term in our estimates. This motivates the smallness assumption (8.1.29). Now from (8.1.28) T 0 ( 0 (y))[J (0;y)] 1 T = 1 +(y): (8.1.30) Then the last term of (8.1.27) is C 2 3 (w ij ([A ] k j J 1 +[A ] k j J 1 k j )) ;k =C 2 3 (w ij (1 +)([A ] k j J 1 k j )) ;k +C 2 3 ik (w) ;k ; (8.1.31) 93 where the second term on the right hand side will act as a source term in our estimates. Thus from (8.1.27) we have @ tt i + 2 t @ t i + 2 ij @ t j +C 2 3 i` ` + C 2 3 w (w ij (1 +)([A ] k j J 1 k j )) ;k + C 2 3 w ik (w) ;k = 0: (8.1.32) Make the time of change variable d dt = 1 : (8.1.33) Then we can formulate (8.1.32) as 1 2 @ i + 3 @ i + 2 2 ij @ j +C 2 3 i` ` + C 2 3 w (w ij (1 +)([A ] k j J 1 k j )) ;k + C 2 3 w ik (w) ;k = 0: (8.1.34) where we dene =O 1 O which implies = 1 . To take advantage of additional time decay arising from the heat transport, let 0<< min 3 ; 2 ; := 3 > 0: (8.1.35) Multiply (8.1.34) by 2+ to nally obtain @ i + 1+ @ i + 2 ij @ j +C i` ` + C w (w ij (1 +)([A ] k j J 1 k j )) ;k + C w ik (w) ;k = 0: (8.1.36) We consider (8.1.36) with the initial conditions (0;y) = 0 (y); (0;y) = V(0;y) = V 0 (y): (8.1.37) Above we have introduced the notation V :=@ which will be used interchangeably. 94 Remark 8.1.3 (Eulerian initial density 0 and temperatureT 0 ). The Eulerian initial den- sity 0 and temperature T 0 are connected to the background ane motion via 0 (x) =w(( 0 A (0)) 1 (x)) det[D( 1 0 (x))] 1 ; T 0 (x) = det[D( 1 0 (x))] 1 T [1 +(( 0 A (0)) 1 (x))] where the composed maps are dened by 0 (y) :=A 1 (0) 0 (y) and A (0)(y) :=A(0)y. 8.2 Notation For ease of notation rst set A :=A ; J :=J : (8.2.1) UsingA [D] = Id, we have the dierentiation formulae forA andJ @A k i =A k ` @ ` ; s A s i ; @J =JA s ` @ ` ; s (8.2.2) for @ =@ or @ =@ i , i = 1; 2; 3. Let F : !R 3 and f : !R be an arbitrary vector eld and function respectively. First dene the gradient and divergence along the ow map respectively [r F] i r :=A s r F i ;s ; div F :=A s ` F ` ;s : (8.2.3) For curl estimates, introduce the anti-symmetric curl and cross product matrices respec- tively [Curl A F] i j := jm A s m F i ; s im A s m F j ; s ; [Arf F] i j := jm A s m f ;s F i im A s m f ;s F j : (8.2.4) 95 We work with L 2 based norms. Dene kk :=kk L 2 (R 3 ) : (8.2.5) Cartesian derivative operators will be used. For 2Z 3 0 let @ :=@ 1 y 1 @ 2 y 2 @ 3 y 3 : (8.2.6) For our gradient energy contribution which arises directly from our problem, we need to diagonalize then positive symmetric matrix = (detA) 2 3 A 1 A > 2 SL(3) as follows =P > QP; P2 SO(3); Q = diag(d 1 ;d 2 ;d 3 ); d i > 0 eigenvalues of : (8.2.7) Then the following quantity will appear in the energy which arises directly from the problem N :=Pr @ P > : (8.2.8) Finally dene the important related quantities 1 := lim !1 () () ; 0 := 2 1 ; (8.2.9) where we recall 0<< min 3 ; 2 . 8.3 High-order Quantities Let N 2 N. To measure the size of the deviation , we dene the high-order weighted Sobolev norm as follows S N () := sup 0 0 n X jjN k@ Vk 2 +k@ k 2 + X jjN1 kr @ k 2 +kdiv @ k 2 96 + X jj=N kr @ k 2 + kdiv @ k 2 o : (8.3.1) Modied curl terms arise during energy estimates which are not a priori controlled by the normS N (). These are measured via the following high-order quantity B N [V]() = sup 0 0 n X jjN1 kCurl A @ Vk 2 + X jj=N kCurl A @ Vk 2 o ; (8.3.2) withB N [] dened in the same way: replaces V in (8.3.2). 8.4 Main Theorem Local Well-Posedness. Before giving our main theorem, we give the local well-posedness of our system. Theorem 8.4.1. Fix N 4. Then there are " 0 > 0, 0 > 0 and T > 0 such that for every "2 (0;" 0 ], 2 (0; 0 ] , pair of compactly supported initial data for (8.1.36) ( 0 ; V 0 ) satisfying S N ( 0 ; V 0 ) +B N (V 0 )"; supp 0 B 1 (0); supp V 0 B 1 (0); (8.4.1) spatial temperature perturbation appearing in (8.1.36) satisfying kk 2 H N+1 (R 3 ) ; suppB 1 (0); (8.4.2) there exists a unique solution ((); V()) : ! R 3 R 3 to (8.1.36)-(8.1.37) for all 2 [0;T ]. The solution has the propertyS N (; V)() +B N [V]()." for each 2 [0;T ]. Furthermore, the map [0;T ]37!S N ()2R + is continuous. Sketch of proof. The construction of a local Eulerian solution from generic initial data is given in Appendix 13.3. With an initial ow map specied by our Lagrangian initial data satisfying (8.4.1), we dene the Eulerian initial data 0 andT 0 , see Remark 8.1.3, and choose 97 u 0 to have the same regularity. From the theory of symmetric hyperbolic systems [30], the associated local Eulerian solution then preserves the regularity of its initial data. Then using the Picard Iteration Theorem for ODEs to solve for the ow map, we obtain a perturbation , and associated V, which solves (8.1.36)-(8.1.37). TheS N (; V)() +B N [V]() bound is established through regularity obtained from the Eulerian solution and ODE theory in conjunction with elliptic regularity, desirable curl equations, and nally a standard div-curl estimate. A priori assumptions. Finally before our main theorem, make the following a priori assumptions on our local solutions from Theorem 8.4.1 kA Idk L 1 (R 3 ) < 1 3 ; kDk L 1 (R 3 ) < 1 3 ; kJ Idk L 1 (R 3 ) < 1 3 ; S N ()< 1=3; kDVk L 1 (R 3 ) C; kDV k L 1 (R 3 ) C; (8.4.3) for all 2 [0;T ]. We are now ready to give our main theorem. Theorem 8.4.2. Fix N 4. Consider a xed triple (A(0);A 0 (0);T )2 GL + (3)M 33 R + ; (8.4.4) parametrizing an isothermal ane motion from the setS so that detA(t) 1 +t 3 ;t 0. Then there are " 0 > 0 and 0 > 0 such that for every "2 (0;" 0 ], 2 (0; 0 ] , pair of compactly supported initial data for (8.1.36) ( 0 ; V 0 ) satisfying S N ( 0 ; V 0 ) +B N (V 0 )"; supp 0 B 1 (0); supp V 0 B 1 (0); (8.4.5) spatial temperature perturbation appearing in (8.1.36) satisfying kk 2 H N+1 (R 3 ) ; suppB 1 (0); (8.4.6) 98 there exists a global-in-time solution, (; V), to the initial value problem (8.1.36)-(8.1.37) and a constant C > 0 such that S N (; V)()C(" +); B N [V]()C(" +)(1 + 2 )e 2 0 ; 0 <1; (8.4.7) where we recall 1 = lim !1 () () ; 0 = 2 1 and 0<< min 3 ; 2 . We believe Theorem 8.4.2 is the rst global existence result for the almost isothermal Euler system with heat transport. Henceforth we assume we are working with a unique local solution (; V) :R 3 !R 3 R 3 to (8.1.36)-(8.1.37) such thatS N (; V) +B N [V] <1 and supp 0 B 1 (0), supp V 0 B 1 (0) on [0;T ] with T > 0 xed: Theorem 8.4.1 ensures the existence of such a solution, and furthermore we assume this local solution satises the a priori assumptions (8.4.3). To prove our main result, we apply high order energy estimates. A similar methodology to [44] enables us to handle exponentially growing-in-time coecients. The exponentially growing time weights take advantage of the stabilizing eect of the expanding background ane motion and were used crucially in [44] to close estimates. As seen in our high order quantities, only spatial derivative operators are used so as to keep intact the exponential structure of time weights which function as stabilizers allowing us to close estimates. The primary diculty in the isothermal setting is the presence of the Gaussian function w =e jyj 2 2 : (8.4.8) In particular, as seen in our high order quantities, we cannot includew as a weight function since then we will have no hope to control unavoidable lower order terms using any kind of weighted Sobolev embedding. However if we do not use w as a weight function, we instead must contend with potentially unbounded y terms arising from the following unavoidable calculation resulting from the Gaussian form of w w; k w =y k : (8.4.9) 99 Therefore without any further analysis, since we are working in the whole space, there is no chance to close estimates containing clearly unbounded y terms. Furthermore since such y terms arise from the nonlinear pressure term, they persist throughout the levels of higher order derivatives so we must deal with them at each stage. There is no such diculty in the vacuum free boundary problem since in that setting w behaves like a distance function on a bounded domain instead. The key to overcoming this issue is that we are able to establish a nite propagation result for our equation. This will be proven immediately below in Theorem 9.0.1. We will show that starting from compactly supported initial data as specied in Theorems (8.4.1)- (8.4.2), the support of our solution grows at most linearly in from B 1 (0), the support of our initial data. Therefore y terms can be estimated by a function linear in . This nite propagation result is applied in conjunction with the crucial exponentially growing time weights and hence it will be seen both in the energy estimates and curl esti- mates that we will need to be careful with any expression involving y to establish sucient time decay for our norm. In particular in the curl estimates, novel algebraic manipulations of time weights are required but fortunately, we are still able to close estimates because we have sucient time decay from our equation. On this note, it is in particular the temperature formulation which provides us with slightly more time decay than the isothermal setting without temperature and this allows us to close estimates. Our spatial perturbation of the temperature through (8.1.28) on the other hand means we have to deal with source terms in our estimates which arise from the last term of (8.1.36). Through the compact support and smallness of , we can use similar methods to [42] to overcome these source terms and still establish global existence. Furthermore, from the time decay provided by the temperature formulation, many of our terms will contain time weights with negative powers. Thus to control such terms without this decay, we apply a coercivity estimate technique from [44]. This technique employs the fundamental theorem of calculus to express in terms of V and initial data with the coercivity estimates given in Lemma 13.5.1. 100 As mentioned above, we prove the fundamental nite propagation result immediately below in Chapter 9. Then in Chapter 10 we prove our higher order energy estimates and in Chaper 11 we establish high order curl estimates. Finally in Chapter 12, we prove our Main Theorem 8.4.2 using a continuity argument. 101 Chapter 9 Finite Propagation We rst state our nite propagation theorem and then prove the local energy and curl estimate results required to prove the theorem, with the nal proof of the theorem given at the end of the Chapter. Theorem 9.0.1 (Finite Propagation). Let (; V) : !R 3 R 3 be a unique local solution to (8.1.36)-(8.1.37) on [0;T ] for T > 0 xed with supp 0 B 1 (0), supp V 0 B 1 (0) and assume (; V) satises the a priori assumptions (8.4.3). Fix N 4. Suppose in (8.1.36) satiseskk 2 H N+1 (R 3 ) and suppB 1 (0) where> 0 is xed. Then there existsK > 0 such that ((;y); V(;y)) = (0; 0); (9.0.1) for 2 [0;T ] andjyj> 1 +K. Local energy estimates are used to prove this theorem. To this end, we rst dene the required local energy and curl quantities. For all of the following denitions, we x y2R 3 , 2 [0;T ] and K > 0, and with xed, then let s2 [0;]. First dene the cone C s C s :=f(x;s 0 ) :jxyjK(s 0 ); 0s 0 sg: (9.0.2) 102 Cross-sections of C s are U(s 0 ) :=fx :jxyjK(s 0 )g for s 0 2 [0;s]: (9.0.3) We then dene our local energy and boundary energy, where we recall the denitions of d i (8.2.7) andN (8.2.8), e(s) := 1 2 U(s) 0 @ w 1 V; V +Cw jj 2 +wC (1 +)J 1 h 3 X i;j=1 d i d 1 j ([N 0 ] j i ) 2 i + + C w (1 +)(div) 2 (s;x)dx e @ (s) := 1 2 @U(s) 0 @ w 1 V; V +Cw jj 2 +wC (1 +)J 1 h 3 X i;j=1 d i d 1 j ([N 0 ] j i ) 2 i + C w (1 +)(div) 2 (s;x)dS(x): (9.0.4) As is the case in high order estimates, modied curl energy arise during energy estimates which are not a priori controlled by the normS N (). The quantities we will need to control are the local and V curl energies as well as the boundary curl energy b (s) := U(s) w(1 +)jCurl A j 2 (s;x)dx b V (s) := U(s) w(1 +)jCurl A Vj 2 (s;x)dx b @ (s) := @U(s) w(1 +)jCurl A j 2 (s;x)dS(x) (9.0.5) Remark 9.0.2. We include w at this level unlike our higher order quantities which is allowable because we do not need high order embeddings here. In fact it is necessary because we do not have higher order derivative control for our local quantities. We now prove our local energy estimate result. Proposition 9.0.3. Let (; V) : ! R 3 R 3 be a unique local solution to (8.1.36)- (8.1.37) on [0;T ] for T > 0 xed with supp 0 B 1 (0), supp V 0 B 1 (0) and assume 103 (; V) satises the a priori assumptions (8.4.3). Fix N 4. Suppose in (8.1.36) satises kk 2 H N+1 (R 3 ) and supp B 1 (0) where > 0 is xed. Fix y2 R 3 and 2 [0;T ]. Then there exists K > 0 in (9.0.3) such that for s2 [0;], e(s).e(0) +b (0) + U(0) jj 2 +jDj 2 dx +b (s) + s 0 b V (s 0 )ds 0 + s 0 e(s 0 )ds 0 : (9.0.6) Proof. Energy estimates First, since we include w in our local energy, rewrite our equation (8.1.36) as follows w( @ i + 1+ @ i + 2 ij @ j +C i` ` ) +C (w ij (1 +)(A k j J 1 k j )) ;k +C ik (w) ;k = 0: (9.0.7) Multiply (9.0.7) by 1 im @ m , and integrate over C s s 0 U(s 0 ) w( @ i + 1+ @ i + 2 ij @ j +C i` ` ) 1 im @ m dxds 0 + s 0 U(s 0 ) C (w ij (1 +)(A k j J 1 k j )); k 1 im @ m dxds 0 + s 0 U(s 0 ) C ik (w) ;k 1 im @ m dxds 0 = 0: (9.0.8) Recognizing the perfect time derivative structure of the rst integral in (9.0.8), we rewrite it as s 0 U(s 0 ) 1 2 d d (w 1 V; V +Cw jj 2 )dxds 0 + s 0 U(s 0 ) 1 2 + 1+ w 1 V; V dxds 0 + s 0 U(s 0 ) 2 w @ 1 V; V + 2 w 1 V; V dxds 0 + s 0 U(s 0 ) 1 wjj 2 dxds 0 : (9.0.9) 104 For the rst integral in (9.0.9), apply the Dierentiation Formula for Moving Regions, which interchanges the derivative and integrals, and then the Fundamental Theorem of Calculus to write it as, U(s) 1 2 w 1 V; V +Cw jj 2 (s;x)dx U(0) 1 2 w 1 V; V +Cw jj 2 (0;x)dx + s 0 @U(s 0 ) K 1 2 w 1 V; V +Cw jj 2 dSds 0 : (9.0.10) The second integral in (9.0.9) is positive, 1 2 + 1+ = (2) 1 2 and 0<< 2; so we leave it as is. For the third integral in (9.0.9), note with A =O, = (detA) 1 3 ;O2 SL(3), we have @ 1 4 1 =@ (O > O) 4O > OO 1 O =@ 1 + 2(O > OO > O ): (9.0.11) Since O > OO > O is anti-symmetric, we can reduce the third integral in (9.0.9) to s 0 U(s 0 ) 2 w @ 1 V; V dxds 0 (9.0.12) which is then bounded by s 0 e(s 0 )ds via @ 1 = 1 (@ ) 1 and (13.2.6). The fourth integral in (9.0.9) is positive so we leave it as is. For the second integral in (9.0.8), integrate by parts in x and apply the identities (13.1.3)-(13.1.5) s 0 U(s 0 ) C w ij (1 +)(A k j J 1 k j ) 1 im @ m ; k dxds 0 + s 0 @U(s 0 ) C w ij (1 +)(A k j J 1 k j ) 1 im @ m n k dSds 0 = s 0 U(s 0 ) C w(1 +)J 1 `j [r ] i j 1 im [r @ ] m ` dxds 0 + s 0 U(s 0 ) C w(1 +)J 1 [Curl A ] ` i 1 im [r @ ] m ` dxds 0 105 + s 0 U(s 0 ) C w(1 +)J 1 A j l l ; m A k ` ` ; j @ m ; k dxds 0 s 0 U(s 0 ) C w(1 +)(J 1 1)@ k ; k dxds 0 + s 0 @U(s 0 ) C w ij (1 +)(A k j J 1 k j ) 1 im @ m n k dSds 0 ; (9.0.13) where n is the outward facing unit normal toU(s 0 ). Using Lemma 13.1.1, the Dierentiation Formula for Moving Regions and the Fundamental Theorem of Calculus, the rst integral on the right hand side of (9.0.13) gives a gradient energy contribution with some remainder terms U(s) C w 2 (1 +)J 1 3 X i;j=1 d i d 1 j ([N ] j i ) 2 (s;x)dx U(0) C w 2 (1 +)J 1 3 X i;j=1 d i d 1 j ([N ] j i ) 2 (0;x)dx + s 0 @U(s 0 ) K C w 2 (1 +)J 1 3 X i;j=1 d i d 1 j ([N ] j i ) 2 dSds 0 + s 0 U(s 0 ) C w(1 +)J 1 T 0;0 dxds 0 + s 0 U(s 0 ) C 1 w(1 +)J 1 3 X i;j=1 d i d 1 j ([N ] j i ) 2 dxds 0 + 1 2 s 0 U(s 0 ) C w(1 +)J 1 1 J 3 X i;j=1 d i d 1 j ([N ] j i ) 2 dxds 0 + s 0 U(s 0 ) C w(1 +)J 1 `j [r ] j i 1 im [r ] m p [r @ s 0] p ` dxds 0 : (9.0.14) The same tools, and in addition (13.1.6): 1J 1 = 1 Tr[D] +O(jDj 2 ), applied to the second to last integral on the right hand side of (9.0.13) gives a divergence energy contribution with some remainder terms 1 s 0 U(s 0 ) C w(1 +) ` ; ` @ k ; k dxds 0 + s 0 U(s 0 ) C w(1 +)O(jDj 2 )@ k ; k dxds 0 = 1 2 s 0 U(s 0 ) Cw d d (1 +)(div) 2 dxds 0 + 2 s 0 U(s 0 ) Cw 1 (1 +)(div) 2 dxds 0 106 + s 0 U(s 0 ) C w(1 +)O(jDj 2 )@ k ; k dxds 0 = U(s) C 2 w (1 +)(div) 2 (s;x)dx U(0) C 2 w (1 +)(div) 2 (0;x)dx + s 0 @U(s 0 ) K 1 2 Cw (1 +)(div) 2 dSds 0 + 2 s 0 U(s 0 ) Cw 1 (1 +)(div) 2 dxds 0 + s 0 U(s 0 ) C w(1 +)O(jDj 2 )@ k ; k dxds 0 : (9.0.15) For the boundary term on the right hand side of (9.0.13), apply (13.1.3): A k j J 1 k j =A k ` [D] ` j J 1 k j ( 1 Tr[D] +O(jDj 2 ); and then using = 3 = 2 2 3 . 2 3= 2 = 2 2 , we have that there exists C 1 > 0 such that C w ij (1 +)(A k j J 1 k j ) 1 im @ m n k C 1 0 @ w 2 1 V; V + wC 2 (1 +)J 1 h 3 X i;j=1 d i d 1 j [N ] j i 2 i 1 A ; (9.0.16) where we also use the a priori assumptions (8.4.3) and the boundedness of . Therefore at this stage we take K >C 1 . For the curl term on the right hand side of (9.0.13), rst write as a perfect time derivative and then integrate by parts in s 0 U(s 0 ) C d d h w(1 +)J 1 [Curl A ] ` i 1 im [r ] m ` i dxds 0 s 0 U(s 0 ) C d d h w(1 +)J 1 [Curl A ] ` i 1 im A m p i p ; ` dxds 0 = U(s) C w(1 +)J 1 [Curl A ] ` i 1 im [r ] m ` (x;s)dx U(0) C w(1 +)J 1 [Curl A ] ` i 1 im [r ] m ` (0;s)dx + s 0 @U(s 0 ) KC w(1 +)J 1 [Curl A ] ` i 1 im [r ] m ` dSds 0 107 s 0 U(s 0 ) C d d h w(1 +)J 1 [Curl A ] ` i 1 im A m p i p ; ` dxds 0 = (i) + (ii) + (iii) + (iv): (9.0.17) Integral (i) on the right hand side of (9.0.17) is estimated using Young's inequality which gives for some xed 0< 1 1 j(i)j U(s) 2 w 1 2 (1 +) 1=2 [Curl A ] ` i 2 w 1 2 (1 +) 1=2 CJ 1 1 im [r ] m ` (x;s)dx . 1 1 U(s) w(1 +)jCurl A j 2 (x;s)dx + 1 U(s) w(1 +)jDj 2 (x;s)dx . 1 1 b (s) + 1 e(s): (9.0.18) Then 1 e(s) is absorbed by the left hand side of our energy inequality. Integral (ii) is bounded by initial data e(0) +b (0): The boundary integral (iii) on the right hand side of (9.0.17) is also estimated using Young's inequality which gives for some xed 0< 2 1 j(iii)j. K 2 s 0 @U(s 0 ) w(1 +)jCurl A j 2 dSds 0 + 2 K s 0 @U(s 0 ) w C 2 (1 +)J 1 h 3 X i;j=1 d i d 1 j Pr P > j i 2i dSds 0 .K s 0 b @ (s 0 )ds 0 + 2 K s 0 e @ (s 0 )ds 0 : (9.0.19) The term 2 K s 0 e @ (s 0 )ds 0 can then be absorbed by the positive boundary energy contribu- tion from above. For K s 0 b @ (s 0 )ds 0 , we note we can recover this term as follows s 0 U(s 0 ) w(1 +) d d h jCurl A j 2 i dxds 0 = U(s) w (1 +)jCurl A j 2 dx U(0) w (1 +)jCurl A j 2 dx +K s 0 @U(s 0 ) w (1 +)jCurl A j 2 dSds 0 =b (s)b (0) +K s 0 b @ (s 0 )ds 0 : (9.0.20) 108 On the other hand note for the left hand side of (9.0.20) s 0 U(s 0 ) w(1 +) d d h jCurl A j 2 i dxds 0 = s 0 U(s 0 ) w 1 (1 +)jCurl A j 2 dxds 0 + 2 s 0 U(s 0 ) w (1 +)Curl A d d [Curl A ] dxds 0 = s 0 U(s 0 ) w 1 (1 +)jCurl A j 2 dxds 0 + 2 s 0 U(s 0 ) w (1 +)Curl A Curl @ (A ) dxds 0 + 2 s 0 U(s 0 ) w (1 +)Curl A Curl A Vdxds 0 : (9.0.21) The rst two terms on the right hand side above are remainder terms and the last term is estimated bounded by a remainder term and s 0 b V (s 0 )ds 0 contribution s 0 U(s 0 ) w (1 +)jCurl A j 2 dxds 0 + s 0 U(s 0 ) w (1 +)jCurl A Vj 2 dxds 0 : (9.0.22) For integral (iv) (iv) = s 0 U(s 0 ) C d d h w(1 +)J 1 [Curl A ] ` i 1 im A m p i p ; ` dxds 0 = s 0 U(s 0 ) C 1 w(1 +)J 1 [Curl A ] ` i 1 im A m p p ; ` dxds 0 s 0 U(s 0 ) C w(1 +) d d h J 1 1 im A m p i [Curl A ] ` i p ; ` dxds 0 s 0 U(s 0 ) C w(1 +)J 1 1 im A m p [Curl @ (A ) ] ` i p ; ` dxds 0 s 0 U(s 0 ) C w(1 +)J 1 [Curl A V] ` i 1 im A m p p ; ` dxds 0 ; (9.0.23) where the last term on the right hand side of above is estimated by s 0 b V (s 0 )ds 0 + s 0 e(s 0 )ds 0 : (9.0.24) 109 All remainder integrals either appear as positive contributions on the left hand side of our inequality or they are bounded by s 0 e(s 0 )ds 0 : (9.0.25) It remains to estimate the third integral in (9.0.8) which will contribute source terms s 0 U(s 0 ) C ik (w) ;k 1 im @ m dxds 0 = s 0 U(s 0 ) (C w ik x k +C w ik ;k ) 1 im @ m dxds 0 : (9.0.26) where we have used w; k =x k w; k . For the rst term on the right hand side of (9.0.26), rst recall y2 R 3 and 2 [0;T ] are xed. Then for x2 U(s 0 ), rst note w(x)jxj 2 . 1 because w is a Gaussian function. Then s 0 U(s 0 ) C w ik x k 1 im @ m dxds 0 . s 0 U(s 0 ) =2 w ik x k 1 im V m dx ds 0 . s 0 U(s 0 ) wjVj 2 dxds 0 ! 1=2 U(s 0 ) 2 wjxj 2 jj 2 dx ! 1=2 ds 0 . s 0 [e(s 0 )] 1=2 U(s 0 ) 2 jj 2 dx ! 1=2 ds 0 . s 0 e(s 0 )ds 0 + s 0 U(s 0 ) 2 jj 2 dxds 0 . s 0 e(s 0 )ds 0 + s 0 U(0) 2 jj 2 dxds 0 . s 0 e(s 0 )ds 0 + U(0) jj 2 dx s 0 2 ds 0 . s 0 e(s 0 )ds 0 + U(0) jj 2 dx (9.0.27) 110 where we have used U(s)U(s 0 ) for s 0 s and the integrability of negative powers of . A similar argument for the second term on the left hand side of (9.0.26), this time simply using w. 1, gives the bound s 0 U(s 0 ) C w ik ;k 1 im @ m dxds 0 . s 0 e(s 0 )ds 0 + U(0) jDj 2 dx: (9.0.28) This concludes the proof (9.0.6). Next we prove the local curl estimates required to control the curl quantities on the right hand side of (9.0.6). Before proving the result, note that we say C(s 0 ) is integrable in s 0 if 1 0 C(s 0 )ds 0 . 1 and we say C(s 0 ;;jyj) is integrable in s 0 if 1 0 C(s 0 ;;jyj)ds 0 .C(;jyj): Proposition 9.0.4. Let (; V) : ! R 3 R 3 be a unique local solution to (8.1.36)- (8.1.37) on [0;T ] for T > 0 xed with supp 0 B 1 (0), supp V 0 B 1 (0) and assume (; V) satises the a priori assumptions (8.4.3). Fix N 4. Suppose in (8.1.36) satises kk 2 H N+1 (R 3 ) and supp B 1 (0) where > 0 is xed. Fix y2 R 3 and 2 [0;T ]. Take K > 0 from (9.0.3). Let s2 [0;]. Let s2 [0;s 0 ]. Then b V (s 0 ).C(s 0 )b V (0) +C(s 0 ;;jyj)e(0) +C(;jyj) sup 0s 00 s 0 e(s 00 ) +C(s 0 ;;jyj) U(s 0 ) jDj 2 dx; (9.0.29) b (s).b (0) +b V (0) +C(;jyj)e(0) +C(;jyj) s 0 sup 0s 00 s 0 e(s 00 )ds 0 +C(;jyj) U(0) jDj 2 dx; (9.0.30) where C(s 0 ) and C(s 0 ;;jyj) are integrable in s 0 . Proof. Proof of (9.0.29). From our velocity curl equation derivation Lemma 13.4.1 (13.4.1) we have the desirable form for our local curl estimate Curl A V = 1 1 + Ax V + (1 +)(1 +) Ar V + (0)Curl A (V(0)) (0)Ax V(0) (1 +) (0)Ar V(0) (1 +)(1 +) 111 + 1 s 0 0 [@ ; Curl A ]Vds 00 1 (1 +) s 0 0 [@ ; Ax]Vds 00 (1 +)(1 +) s 0 0 [@ ; Ar]Vds 00 2 s 0 0 Curl A ( V)ds 00 + 2 (1 +) s 0 0 Ax ( V)ds 00 + 2 (1 +)(1 +) s 0 0 Ar ( V)ds 00 + C (1 +) s 0 0 1 x ds 00 C (1 +) s 0 0 1 A [D]x ds 00 + C (1 +)(1 +) s 0 0 1 Ar ds 00 C (1 +)(1 +) s 0 0 1 Ar xds 00 (9.0.31) For the purpose of our local curl estimates dene thekk 2 (w(1+);L 2 (U(s)) norm as follows kfk 2 w(1+);L 2 (U(s)) := U(s) w(1 +)jfj 2 (s;x)dx (9.0.32) We take thekk 2 w(1+);L 2 (U(s 0 )) norm, wheres 0 2 [0;s] , of (9.0.31), multiply by [(s 0 )] and estimate the right hand side. As can be seen from (9.0.31), many terms and hence estimates are similar. Therefore we give the key estimates below and remark similar arguments will hold for the other terms. For the rst term on the right hand side of (9.0.31), 1 1+ Ax V, rst recall y2R 3 and 2 [0;T ] are xed. Then for x2U(s 0 ) where s 0 2 [0;s], rst note jxj 2 .jxyj 2 +jyj 2 K 2 (s 0 ) 2 +jyj 2 K 2 2 +jyj 2 : (9.0.33) Then using the boundedness of [(s 0 )] k 1 1 + Ax Vk 2 w(1+);L 2 (U(s 0 )) .C(;jyj)e(s 0 ): (9.0.34) 112 For the sixth term on the right hand side of (9.0.31), 1 (1+) s 0 0 [@ ; Ax]Vds 00 , [@ ; Curl A ]V k j =@ ( jm A s m ) V; k s @ ( km A s m ) V; j s : (9.0.35) As the other term can be estimated in the same way, we restrict our focus to the rst term only @ ( jm A s m ) V; k s = (@ jm A s m + jm @ A s m ) V; k s =@ jm A s m V; k s + jm @ A s m V; k s : (9.0.36) Schematically consider the right hand side of (9.0.36) @ DDV + DVDV (9.0.37) For the rst term in (9.0.37), one must rst integrate by parts in s 00 1 s 0 0 @ DDVds 00 = 1 (@ DD) s 0 0 (9.0.38) 1 s 0 0 ( @ D +@ D +@ DV)Dds 00 : Then k 1 (@ DD) s 0 0 k 2 w(1+);L 2 (U(s 0 )) .e(s 0 ) +C(s 0 )e(0); (9.0.39) where we include 1 + in our energy here, and C(s 0 ) is integrable in s 0 . Next k 1 s 0 0 @ DDds 00 k 2 w(1+);L 2 (U(s 0 )) . 1 [(s 0 )] 2 sup 0s 00 s 0 U(s 0 ) w(1 +)jDj 2 s 0 0 @ Dds 00 2 dx . 1 [(s 0 )] 2 sup 0s 00 s 0 U(s 0 ) w(1 +)jDj 2 dx s 0 0 ds 00 2 . sup 0s 00 s 0 U(s 0 ) w(1 +)jDj 2 dx 113 . sup 0s 00 s 0 U(s 00 ) w(1 +)jDj 2 dx . sup 0s 00 s 0 e(s 00 ): (9.0.40) Similar arguments apply to the remaining terms in (9.0.38). For the second term in (9.0.37), we also rst integrate by parts in s 00 1 s 0 0 DVDVds 00 = 1 (DVD) s 0 0 (9.0.41) 1 s 0 0 ( DV +@ DV + DV )Dds 00 : Then, noting our a priori assumptions (8.4.3) used to control DV , a similar argument to the rst term in (9.0.37) give the same bounds here. Therefore [(s 0 )] k 1 (1 +) s 0 0 [@ ; Ax]Vds 00 k 2 w(1+);L 2 (U(s 0 )) . sup 0s 00 s 0 e(s 00 ) +C(s 0 )e(0): (9.0.42) For the last term on the right hand side of (9.0.31), C (1+)(1+) s 0 0 1 Arxds 00 , rst note using Lemma 13.2.1 1 2 j s 0 0 1 ds 00 j 2 .e 2 1 s 0 s 0 0 e (1) 1 s 00 ds 00 2 .e 2 1 s 0 (e 2(1) 1 s 0 + 1) .e 2 0 s 0 ; (9.0.43) which is integrable in s 0 . Then with this and using that U(s 0 )U(s) for ss 0 as well as the boundedness of 1 1+ and (9.0.33) [(s 0 )] k C (1 +)(1 +) s 0 0 1 Ar xds 00 k 2 w(1+);L 2 (U(s 0 )) .C(s 0 ;;jyj) U(s 0 ) jDj 2 dx .C(s 0 ;;jyj) U(0) jDj 2 dx: (9.0.44) 114 Proof of (9.0.30). From our perturbation curl equation derivation Lemma 13.4.1 (13.4.2) we have the desirable form for our local curl estimate Curl A = Curl A ([(0)]) +(0) Curl A (V(0)) s 0 1 (s 0 ) ds 0 (0)Ax V(0) 1 + s 0 1 (s 0 ) ds 0 (0)Ar V(0) (1 +)(1 +) s 0 1 (s 0 ) ds 0 + s 0 [@ ; Curl A ]ds 0 + 1 1 + s 0 Ax Vds 0 + (1 +)(1 +) s 0 Ar Vds 0 + s 0 1 (s 0 ) s 0 0 (s 00 )[@ ; Curl A ]Vds 00 ds 0 1 1 + s 0 1 (s 0 ) s 0 0 (s 00 )[@ ; Ax]Vds 00 ds 0 (1 +)(1 +) s 0 1 (s 0 ) s 0 0 (s 00 )[@ ; Ar]Vds 00 s 0 2 (s 0 ) s 0 0 (s 00 ) Curl A ( V)ds 00 ds 0 + 2 1 + s 0 1 (s 0 ) s 0 0 (s 00 ) Ax ( V)ds 00 ds 0 + 2 (1 +)(1 +) s 0 1 (s 0 ) s 0 0 (s 00 ) Ar ( V)ds 00 ds 0 + C 1 + s 0 1 (s 0 ) s 0 0 (s 00 ) 1 x ds 00 ds 0 C 1 + s 0 1 (s 0 ) s 0 0 (s 00 ) 1 A [D]x ds 00 ds 0 + C (1 +)(1 +) s 0 1 s 0 0 1 Ar ds 00 ds 0 C (1 +)(1 +) s 0 1 s 0 0 1 Ar xds 00 ds 0 : (9.0.45) We take thekk 2 w(1+);L 2 (U(s)) norm of (9.0.45), multiply by [(s)] and estimate the right hand side. As can be seen from (9.0.45), many terms and hence estimates are similar. Therefore we give the key estimates below and remark similar arguments will hold for the other terms. For the fth term on the right hand side of (9.0.45), s 0 [@ ; Curl A ]ds 0 , notingU(s) U(s 0 ) for s 0 s, [(s)] k s 0 [@ ; Curl A ]ds 0 k 2 w(1+);L 2 (U(s)) 115 . [(s)] U(s) w(1 +) s 0 [@ ; Curl A ]ds 0 2 dx . [(s)] U(s) w(1 +)( s 0 1ds 0 )( s 0 j[@ ; Curl A ]j 2 ds 0 )dx] . s 0 U(s) w(1 +)j[@ ; Curl A ]j 2 dxds 0 . s 0 U(s 0 ) w(1 +)j[@ ; Curl A ]j 2 dxds 0 . s 0 e(s 0 )ds 0 : (9.0.46) For the eighth term on the right hand side of (9.0.45), s 0 1 (s 0 ) s 0 0 (s 00 )[@ ; Curl A ]Vds 00 ds 0 , rst note k s 0 1 (s 0 ) s 0 0 (s 00 )[@ ; Curl A ]Vds 00 ds 0 k 2 w(1+);L 2 (U(s)) = U(s) w(1 +) s 0 1 [(s 0 )] 1=2 1 [(s 0 )] 1=2 s 0 0 (s 00 )[@ ; Curl A ]Vds 00 ds 0 2 dx . U(s) w(1 +)[( s 0 1 (s 0 ) ds 0 ) s 0 1 (s 0 ) ( s 0 0 (s 00 )[@ ; Curl A ]Vds 00 ) 2 ds 0 ]dx . s 0 U(s) w(1 +) 1 (s 0 ) ( s 0 0 (s 00 )[@ ; Curl A ]Vds 00 ) 2 dxds 0 . s 0 U(s 0 ) w(1 +) 1 (s 0 ) ( s 0 0 (s 00 )[@ ; Curl A ]Vds 00 ) 2 dxds 0 (9.0.47) where we note U(s) U(s 0 ) for s 0 s. Hence applying the argument above used for the sixth term on the right hand side of (9.0.31) [(s)] k s 0 1 (s 0 ) s 0 0 (s 00 )[@ ; Curl A ]Vds 00 ds 0 k 2 w(1+);L 2 (U(s)) . s 0 C(s 0 ) sup 0s 00 s 0 e(s 00 ) +C(s 0 )(s 0 )e(0)ds 0 .e(0) + s 0 sup 0s 00 s 0 e(s 00 )ds 0 : (9.0.48) 116 Finally, using (9.0.43), for the last term on the right hand side of (9.0.45) we have [(s)] k C (1 +)(1 +) s 0 1 s 0 0 1 Ar xds 00 ds 0 k 2 w(1+);L 2 (U(s)) .C(;jyj) U(s) jDj 2 dx .C(;jyj) U(0) jDj 2 dx: (9.0.49) This concludes the proof of our local curl estimate Proposition 9.0.4. We are now ready to prove our Finite Propagation Theorem 9.0.1. Proof of Theorem 9.0.1. Combining the results of the energy estimate and curl estimate Propositions above, namely the inequalities (9.0.6), (9.0.29) and (9.0.30), we have estab- lished the following energy inequality e(s).C(;jyj)e(0) +b (0) +b V (0) + U(0) jj 2 dx +C(;jyj) U(0) jDj 2 dx +C(;jyj) s 0 sup 0s 00 s 0 fe(s 00 )gds 0 : (9.0.50) Then taking the supremum over 0s 0 s, sup 0s 0 s fe(s 0 )g.C(;jyj)e(0) +b (0) +b V (0) + U(0) jj 2 dx +C(;jyj) U(0) jDj 2 dx +C(;jyj) s 0 sup 0s 00 s 0 fe(s 00 )gds 0 : (9.0.51) SinceC(;jyj)e(0)+b (0)+b V (0)+ U(0) jj 2 +jDj 2 dx is constant ins, applying Gronwall's inequality, we have for some C > 0, 0 sup 0s 0 s fe(s 0 )g)C " C(;jyj)e(0) +b (0) +b V (0) + U(0) jj 2 dx +C(;jyj) U(0) jDj 2 dx # exp s 0 C ds 0 : (9.0.52) 117 Hence 0 sup 0s 0 s fe(s 0 )g)C " C(;jyj)e(0) +b (0) +b V (0) + U(0) jj 2 dx +C(;jyj) U(0) jDj 2 dx # e C s : (9.0.53) Ifjyj> 1 +K, thenjxj> 1 for x2U(0). Thus e(0) =b (0) =b V (0) = 0 since supp 0 B 1 (0); supp V 0 B 1 (0): (9.0.54) Furthermore U(0) jj 2 dx = U(0) jDj 2 dx = 0 since suppB 1 (0): (9.0.55) Hence e(s) = 0 forjyj> 1 +K. Thus (s;y) = V(s;y) = 0 forjyj> 1 +K. As s2 [0;] is arbitrary, ((;y); V(;y)) = (0; 0); (9.0.56) for 2 [0;T ] andjyj> 1 +K. 118 Chapter 10 High Order Energy Estimates With our nite propagation theorem in hand, we are now ready to proceed with high order energy estimates. To this end, we rst introduce the two energy based high order quantities which arise directly from the problem. Denoting the usual dot product on R 3 byh;i, introduce the high-order energy functional E N () = 1 2 X jjN R 3 h 1 @ V; @ V +C 1 @ ; @ i +C (1 +)J 1 h 3 X i;j=1 d i d 1 j ([N ] j i ) 2 + 1 (div @ ) 2 i dy; (10.0.1) and the dissipation functional D N () =C 2 R 3 X jjN j@ j 2 + (1 +)J 1 h 3 X i;j=1 d i d 1 j ([N ] j i ) 2 + 1 (div @ ) 2 i dy: (10.0.2) We also introduce a similar term toE N which does not include top order quantities but will be controlled through our coercivity Lemma 13.5.1 C N1 () = 1 2 X jjN1 C 1 @ ; @ 119 +C(1 +)J 1 h 3 X i;j=1 d i d 1 j (N ) j i 2 + 1 (div @ ) 2 i dy: (10.0.3) Remark 10.0.1. Our coercivity Lemma 13.5.1 given in Appendix 13.5 allows us to control terms without time weights with negative powers using our normS N () and initial data S N (0). Such terms do not appear immediately from our equation (8.1.36). Notably, Lemma 13.5.1 will let us includeC N1 in our energy inequality. However we cannot use Lemma 13.5.1 to control top order quantities since that would require control of N + 1 derivatives of V which we do not have. This is why we have the particular structure ofS N where we separate top order terms from lower order terms, and secondly why we only include N 1 derivatives inC N1 . Finally, before we prove our main energy inequality, we give the norm-energy equivalence as follows. Lemma 10.0.2. Let (; V) : !R 3 R 3 be a unique local solution to (8.1.36)-(8.1.37) on [0;T ] for T > 0 xed with supp 0 B 1 (0), supp V 0 B 1 (0) and assume (; V) satises the a priori assumptions (8.4.3). Fix N 4. Suppose in (8.1.36) satises kk 2 H N+1 (R 3 ) and supp B 1 (0) where > 0 is xed. Then there are constants C 1 ;C 2 > 0 so that C 1 S N () sup 0 0 fE N ( 0 ) +C N1 ( 0 )gC 2 (S N () +S N (0)): (10.0.4) Proof. Recall the denitionS N (8.3.1), the decomposition of (8.2.7) and the denition of the conjugateN (8.2.8). Then (10.0.4) is a straightforward application of Lemma 13.2.1 to give bounds on and associated matrix quantities and the boundedness of given by (8.1.29), in conjunction with Lemma 13.5.1 to control terms without time weights with negative powers, which are included inC N1 ( 0 ), byS N () +S N (0). We are now ready to prove our central high order energy inequality which will be essential in the proof of our main result Theorem 8.4.2. 120 Proposition 10.0.3. Let (; V) : ! R 3 R 3 be a unique local solution to (8.1.36)- (8.1.37) on [0;T ] for T > 0 xed with supp 0 B 1 (0), supp V 0 B 1 (0) and assume (; V) satises the a priori assumptions (8.4.3). Fix N 4. Suppose in (8.1.36) satises kk 2 H N+1 (R 3 ) and supp B 1 (0) where > 0 is xed. Then for all 2 [0;T ], we have the following inequality for some 0< 1, E N () +C N1 () + 0 D N ( 0 )d 0 .S N (0) + +B N []() +S N () + 0 (1 + 0 )(S N ( 0 )) 1 2 (B N [V]( 0 )) 1 2 d 0 + 0 (1 + 0 )e 0 0 S N ( 0 )d 0 : (10.0.5) Proof. First write (8.1.36) as follows @ i + 1+ @ i + 2 ij @ j +C i` ` +C ( ij (1 +)(A k j J 1 k j )) ;k C y k (1 +) ij (A k j J 1 k j ) +C ik ;k C ik y k = 0: (10.0.6) Zeroth order estimate Multiply (10.0.6) by 1 im @ m and integrate over 0 R 3 dyd 0 , 0 R 3 @ i + 1+ @ i + 2 ij @ j +C i` ` 1 im @ m dyd 0 + 0 R 3 C ( ij (1 +)(A k j J 1 k j )) ;k 1 im @ m dyd 0 0 R 3 C y k (1 +) ij (A k j J 1 k j ) 1 im @ m dyd 0 + 0 R 3 C ik ;k C ik y k 1 im @ m dyd 0 = 0: (10.0.7) The zeroth order estimate for the rst and second integrals in (10.0.7) follows by a similar argument to the nite propagation energy estimate argument without boundary contribu- tions and we do not isolate a divergence contribution: instead we control 1J 1 using (13.1.6). For the third integral in (10.0.7) apply (13.1.3) 0 R 3 C y k (1 +) ij (A k j J 1 k j ) 1 im @ m dyd 0 121 = 0 R 3 C y k (1 +) ij A k ` ` ; j J 1 1 im @ m dyd 0 + 0 R 3 C (1 +) ik y k ( 1 Tr[D] +O(jDj 2 )) 1 im @ m dyd 0 (10.0.8) Applying the Finite Propagation Theorem 9.0.1 to controly, and the boundedness of, the rst integral on the right hand side of (10.0.8) is bounded by 0 R 3 ((1 +K 0 ) =2 ) =2 i ; j @ m dyd 0 . 0 R 3 j i ; j j 2 dy 1=2 R 3 j@ m j 2 dy 1=2 d 0 . 0 e 0 0 S N d 0 (10.0.9) A similar argument holds for the second integral on the right hand side of the rst term on the right hand side of (10.0.8). For the fourth integral in 10.0.7) 0 R 3 C ik ;k C ik y k 1 im @ m dyd 0 = 0 R 3 C ik ;k 1 im @ m dyd 0 0 R 3 C ik y k 1 im @ m dyd 0 : (10.0.10) Using the compact support of to boundy k and then the smallness ofkk L 2 (R 3 ) , the second integral on the right hand side of (10.0.8) is bounded by 0 R 3 C@ m dyd 0 (10.0.11) . 0 R 3 2 dx 1=2 R 3 j@ m j 2 dx 1=2 . 0 1=2 e 0 0 (S N ) 1=2 d 0 . 0 e 0 0 d 0 + 0 e 0 0 S N d 0 . + 0 e 0 0 S N d 0 : (10.0.12) 122 A similar argument directly using the smallness ofkDk L 2 (R 3 ) gives that the third integral on the right hand side (10.0.8) is bounded by + 0 e 0 0 S N d 0 : (10.0.13) To complete the zeroth order estimate, we obtain the full expression forE N in (10.0.5) by adding the following formula 0 1 2 d d 1 R 3 J 1 jdiv j 2 dyd 0 = 0 1 2 R 3 @ (J 1 )jdiv j 2 +J 1 @ (jdiv j 2 ) dyd 0 ; (10.0.14) with the right-hand side in turn bounded by e 0 S N () by the H 2 (R 3 ) ,! L 1 (R 3 ) embedding. Then the left-hand side of (10.0.14) completesE N in (10.0.5) and contributes toS N (0) by the fundamental theorem of calculus. High order estimates Fix 2Z 3 0 withjj 1. Apply @ to (10.0.6) @ @ i + 1+ @ @ i + 2 ij @ @ j +C i` @ ` +C ij (@ [(1 +)(A k j J 1 k j )]) ;k C ij @ [y k (1 +)(A k j J 1 k j )] +C ik (@ ) ;k C ik @ (y k ) = 0: (10.0.15) Multiply (10.0.15) by 1 im @ @ m and integrate overR 3 and then from 0 to , 0 1 2 d d R 3 h 1 @ V; @ Vidy +C R 3 j@ j 2 dy d 0 + 0 C 2 R 3 j@ j 2 dyd 0 0 2 R 3 @ 1 4 1 @ V;@ V dyd 0 +C 0 R 3 ij (@ [(1 +)(A k j J 1 k j )]) ;k 1 im @ @ m dyd 0 C 0 R 3 ij @ [y k (1 +)(A k j J 1 k j )] 1 im @ m dyd 0 123 +C 0 R 3 ik (@ ) ;k 1 im @ m dyd 0 C 0 R 3 ik @ (y k ) 1 im @ m dyd 0 = 0: (10.0.16) By the same reasoning as the zeroth order case, the rst four integrals on the left hand side of (10.0.16) contribute to the energy inequality (10.0.5) in the same way. For the fourth to last integral on the left hand side of (10.0.16), rst note via the Leibniz rule @ [(1 +)(A k j J 1 k j )] = (1 +)@ (A k j J 1 ) + X 0j 0 jjj1 c 0@ 0 ()@ 0 (A k j J 1 ): (10.0.17) Then the fourth to last integral on the left hand side of (10.0.16) is C 0 R 3 ij (@ [(1 +)(A k j J 1 k j )]) ;k 1 im @ @ m dyd 0 =C 0 R 3 ij [(1 +)@ (A k j J 1 )] ;k 1 im @ @ m dyd 0 +C 0 R 3 ij 0 @ X 0j 0 jjj1 c 0@ 0 ()@ 0 (A k j J 1 ) 1 A ;k 1 im @ @ m dyd 0 : (10.0.18) For the rst integral on the right hand side of (10.0.18), rst note using our dierentiation formulae forA andJ (8.2.2) @ y i (A k j J 1 ) =J 1 A k ` A s j (@ y i ` ); s 1 J 1 A k j A s ` (@ y i ` ); s : (10.0.19) Then using this and applying integration by parts which holds due to nite propagation, the rst integral on the right hand side of (10.0.18) is C 0 R 3 ij [(1 +)@ (A k j J 1 )]; k 1 im @ @ m dyd 0 =C 0 R 3 ij [(1 +)(J 1 A k ` A s j @ ` ; s 1 J 1 A k j A s ` @ ` ; s +C ;k j ())]; k 1 im @ @ m dyd 0 124 =C 0 R 3 [(1 +)(J 1 A k ` ( ij A s j @ ` ; s `j A s j @ i ; s ) +J 1 A k ` `j A s j @ i ; s + 1 J 1 ij A k j A s ` @ ` ; s )]; k 1 im @ @ m dyd 0 +C 0 R 3 ij [(1 +)C ;k j ]; k 1 im @ @ m dyd 0 =C 0 R 3 (1 +)J 1 (A k ` [Curl A @ ] ` i +A k ` `j [r @ ] i j + 1 ij A k j div (@ )) 1 im @ (@ m ); k dyd 0 +C 0 R 3 ij [(1 +)C ;k j ]; k 1 im @ @ m dyd 0 =:I 1 +I 2 ; (10.0.20) where we deneC ;k j to be the lower order terms resulting from dierentiating the result of (10.0.19). Note kC ;k j k 2 .S N ; kD(C ;k j )k 2 .S N : (10.0.21) Then using the boundedness of and D jI 2 j. 0 k@ @ kkC ;k j k +k@ @ k kD(C ;k j )kd 0 . 0 e 0 0 S N d 0 : (10.0.22) ForI 1 rewrite the gradient and divergence terms, and integrate by parts in the curl term as follows I 1 = 0 R 3 (1 +)J 1 A k ` [Curl A @ ] ` i 1 im @ (@ m ); k dyd 0 + 0 R 3 (1 +)J 1 A k ` `j [r @ ] i j + 1 ij A k j div (@ ) 1 im @ (@ m ); k dyd 0 = 0 R 3 (1 +)J 1 A k ` 1 im [Curl A @ ] ` i @ (@ m ); k dyd 0 + 0 R 3 (1 +)J 1 `j [r @ ] i j 1 im [r @ @ ] m ` + 1 div (@ )div (@ @ ) dyd 0 = R 3 (1 +)J 1 A k ` 1 im [Curl A @ ] ` i (@ m ); k dy 0 0 R 3 (1 +)J 1 A k ` 1 im [Curl A @ V] ` i (@ m ); k dyd 0 + 0 1 R 3 (1 +)J 1 A k ` 1 im [Curl A @ ] ` i (@ m ); k dyd 0 125 0 R 3 (1 +)@ J 1 A k ` 1 im [Curl A @ ] ` i (@ m ); k dyd 0 0 R 3 (1 +)J 1 A k ` 1 im [Curl A @ ] ` i (@ m ); k dyd 0 0 R 3 (1 +)J 1 A k ` 1 im [Curl A @ ] ` i (@ m ); k dyd 0 + 0 R 3 (1 +)J 1 `j [r @ ] i j 1 im @ [r @ ] m ` + 1 div (@ )@ div (@ ) dyd 0 0 R 3 (1 +)J 1 `j [r @ ] i j 1 im @ A k ` (@ m ); k + 1 div (@ )@ A k j @ j ; k dyd 0 (10.0.23) :=B 1 +B 2 +B 3 +R 1 +R 2 +R 3 +E 1 +R 4 : (10.0.24) For B 1 , for 0< 1, B 1 .S N (0) +S N () +B N [](); (10.0.25) where we have introduced through the Young inequality. For B 2 , jB 2 j. 0 (S N ) 1=2 (B N [V]) 1=2 d 0 : (10.0.26) For B 3 , rst apply the Fundamental Theorem of Calculus as follows Curl A @ ( 0 ;y) = 0 0 @ Curl A @ ( 00 ;y)d 00 + Curl A @ (0;y) = 0 0 Curl A @ V( 00 ;y)d 00 + 0 0 Curl @ (A ) @ ( 00 ;y)d 00 + Curl A @ (0;y) . 0 sup 0 00 0 fCurl A @ Vg + 0 sup 0 00 0 fCurl @ (A ) @ g + Curl A @ (0;y): (10.0.27) Then using the boundedness of jB 3 j. 0 R 3 ( 0 sup 0 00 0 fj[Curl A @ V] ` i jg + 0 sup 0 00 0 fj[Curl @ (A ) @ ] ` i jg +j[Curl A @ (0)] ` i j) (@ m ); k dyd 0 . 0 0 (S N ) 1=2 (B N [V]) 1=2 + 0 e 0 0 S N d 0 126 + sup 0 0 kr @ k 2 + 1 kCurl A @ (0)k 2 ! 0 d 0 . 0 0 (S N ) 1=2 (B N [V]) 1=2 + 0 e 0 0 S N d 0 +S N +S N (0): (10.0.28) For E 1 E 1 = 1 2 0 d d 8 < : R 3 (1 +)J 1 0 @ 3 X i;j=1 d i d 1 j (N ) j i ) 2 + 1 (div @ ) 2 1 A dy 9 = ; d 0 + 0 2 R 3 (1 +)J 1 h 3 X i;j=1 d i d 1 j (N ) j i 2 + 1 div X a r = @ 2 i w a++1 e S dyd 0 + 0 R 3 (1 +)J 1 T dyd 0 1 2 0 R 3 (1 +)@ J 1 0 @ 3 X i;j=1 d i d 1 j (N ) j i ) 2 + 1 (div @ ) 2 1 A dyd 0 :=E 2 +D 1 +R 5 +R 6 ; (10.0.29) where we have used Lemma 13.1.1. Then E 2 contributes toE N () in (10.0.5), and also to S N (0) by the fundamental theorem of calculus. AlsoD 1 contributes to 0 D N ( 0 )d 0 in (10.0.5). Now jR i j. 0 e 0 0 S N d 0 ; (10.0.30) for i = 1; 2; 3; 4; 5; 6. For the second integral on the right hand side of (10.0.18) rst note 0 @ X 0j 0 jjj1 c 0@ 0 ()@ 0 (A k j J 1 ) 1 A ;k =@ ( ;k )A k j J 1 +@ ()(A k j J 1 ) ;k + 0 @ X 1j 0 jjj1 c 0@ 0 ()@ 0 (A k j J 1 ) 1 A ;k : (10.0.31) 127 For the integral resulting from the rst term on the right hand side of (10.0.31) C 0 R 3 ij @ ( ;k )A k j J 1 1 im @ @ m dyd 0 . 0 R 3 (@ ) ;k @ @ m dyd 0 . 0 R 3 (@ ) 2 ;k dx 1=2 R 3 j@ m j 2 dx 1=2 . 0 1=2 e 0 0 (S N ) 1=2 d 0 . 0 e 0 0 d 0 + 0 e 0 0 S N d 0 . + 0 e 0 0 S N d 0 : (10.0.32) For the remaining integrals from the last two terms on the right hand side of (10.0.31) with both the H 2 (R 3 ) ,! L 1 (R 3 ) embedding and boundedness ofkk H N+1 (R 3 ) , the resulting integrals are bounded by 0 e 0 0 S N d 0 : (10.0.33) For the third to last integral on the left hand side of (10.0.16), rst note: @ [y k (1 +)(A k j J 1 k j )] =y k @ [(1 +)(A k j J 1 k j )] + X j 0 j=jj1 c 0 ;k @ 0 [(1 +)(A k j J 1 k j )]; (10.0.34) where c 0 ;k are non-negative constants depending on 0 ;k. Then the third to last integral on the left hand side of (10.0.16) is C 0 R 3 ij @ [y k (1 +)(A k j J 1 k j )] 1 im @ m dyd 0 =C 0 R 3 ij y k @ [(1 +)(A k j J 1 k j )] 1 im @ m dyd 0 C 0 R 3 ij X j 0 j=jj1 c 0 ;k @ 0 [(1 +)(A k j J 1 k j )]] 1 im @ m dyd 0 (10.0.35) 128 For the rst integral on the right hand side of (10.0.35), use nite propagation and then a combination of the H 2 (R 3 ) ,! L 1 (R 3 ) embedding, boundedness ofkk H N+1 (R 3 ) and (13.1.7) to estimate C 0 R 3 ij y k @ [(1 +)(A k j J 1 k j )] 1 im @ m dyd 0 0 R 3 C 1 +K ij @ [(1 +)(A k j J 1 k j )] 1 im @ m dyd 0 . 0 e 0 0 S N d 0 : (10.0.36) Using a combination of the H 2 (R 3 ),!L 1 (R 3 ) embedding, boundedness ofkk H N+1;1 (R 3 ) and (13.1.7), the second integral on the right hand side of (10.0.35) is straightforward to bound by 0 e 0 0 S N d 0 : (10.0.37) For the second to last integral on the right hand side (10.0.16), a similar argument to (10.0.32) gives C 0 R 3 ik (@ ) ;k 1 im @ m dyd 0 . + 0 e 0 0 S N d 0 : (10.0.38) For the last integral on the left hand side of (10.0.16) on the rst note @ (y k ) =y k @ + X j 0 j=jj1 c 0 ;k @ 0 : (10.0.39) Then the last integral on the left hand side of (10.0.16) is C 0 R 3 ik @ (y k ) 1 im @ m dyd 0 =C 0 R 3 ik y k @ 1 im @ m dyd 0 C 0 R 3 ik X j 0 j=jj1 c 0 ;k @ 0 1 im @ m dyd 0 : (10.0.40) 129 Use nite propagation in the same way as in (10.0.36) and then a similar argument to (10.0.32) to bound the rst integral on the right hand side of (10.0.40) by + 0 e 0 0 S N d 0 : (10.0.41) Finally use a similar argument to (10.0.32) to bound the second integral on the right hand side of (10.0.40) by + 0 e 0 0 S N d 0 : (10.0.42) This concludes the energy estimate. 130 Chapter 11 High Order Curl Estimates We now prove the high order curl estimates necessary to control the curl contributions on the right hand side of our energy inequality (10.0.5). Proposition 11.0.1. Let (; V) : ! R 3 R 3 be a unique local solution to (8.1.36)- (8.1.37) on [0;T ] for T > 0 xed with supp 0 B 1 (0), supp V 0 B 1 (0) and assume (; V) satises the a priori assumptions (8.4.3). Fix N 4. Suppose in (8.1.36) satises kk 2 H N+1 (R 3 ) and supp B 1 (0) where > 0 is xed. Then for all 2 [0;T ], we have the following inequalities for some 0< 1 B N [V]().e 2 0 S N (0) +B N [V](0) +e 2 0 + (1 + 2 )e 2 0 S N (); (11.0.1) B N []().S N (0) +B N [V](0) + +S N () + 0 e 0 0 S N ( 0 )d 0 : (11.0.2) Proof. Proof of (11.0.1). We have derived the equation (13.4.1) for Curl A V in Lemma 13.4.1. Apply @ to (13.4.1) Curl A @ V =[@ ; Curl A ]V + 1 1 + @ (Ay V) + 1 + @ ((1 +) 1 (Ar V)) + (0)@ (Curl A (V(0))) (0)@ (Ay V(0)) (1 +) + (0)@ ((1 +) 1 (Ar V(0)) (1 +) + 1 0 @ [@ ; Curl A ]Vd 0 1 (1 +) 0 @ [@ ; Ay]Vd 0 + (1 +) 0 @ ((1 +) 1 [@ ; Ar]V)d 0 131 2 0 @ (Curl A ( V))d 0 + 2 (1 +) 0 @ (Ay ( V))d 0 + 2 (1 +) 0 @ ((1 +) 1 Ar ( V))d 0 + C 2 0 1 @ (y )d 0 C 2 0 1 @ (A [D]y )d 0 + C (1 +) 0 1 @ ((1 +) 1 (Ar ))d 0 C (1 +) 0 1 @ ((1 +) 1 (Ar y))d 0 : (11.0.3) We take thekk 2 norm of (11.0.3), and ifjj = N then multiply by [()] , and then estimate the right hand side. As can be seen from (11.0.3), many terms and hence estimates are similar. Therefore we give the key estimates below and remark similar arguments will hold for the other terms. For the second term on the right hand side of (11.0.3), 1 1+ @ (AyV), forjjN1 use nite propagation, R 3 1 (1 +) 2 j@ (Ay V)j 2 . (1 + 2 )e 2 0 S N (11.0.4) Forjj =N, similarly k 1 1 + @ (Ay V)k 2 . (1 + 2 )e 2 0 S N ; (11.0.5) including inS N if needed. For the third term on the right hand side of (11.0.3), use a combination of theH 2 (R 3 ),! L 1 (R 3 ) embedding and boundedness ofkk H N+1;1 (R 3 ) e 2 0 S N & 8 > > < > > : k 1 + @ ((1 +) 1 (Ar V))k 2 if jjN 1; k 1 + @ ((1 +) 1 (Ar V))k 2 if jj =N: (11.0.6) 132 For the fth term on the right hand side of (11.0.3), we use the compact support of@ @ (0) for N from nite propagation to in this case boundjyj. 1, and we conclude e 2 0 (S N (0))& 8 > > < > > : k (0)@ (Ay V(0)) (1 +) k 2 if jjN 1; k (0)@ (Ay V(0)) (1 +) k 2 if jj =N: (11.0.7) For eighth term on the right hand side of (11.0.3), (1+) 0 @ ((1 + ) 1 [@ ; Ar]V)d 0 , rst compute @ [@ ; Ay]@ k j =@ (@ ( jm A s m )y s @ k @ ( km A s m )y s @ j ): (11.0.8) For the rst term on the right hand side, @ (@ jm A s m y s + jm @ A s m y s )@ k =@ jm ((@ A s m )y s @ k +A s m y s (@ @ k )) + jm (@ (@ A s m )y s @ k +@ A s m y s @ @ k ) +R: (11.0.9) We are denoting byR above favorable remainder terms. Schematically consider the rst two terms on the right hand side of above @ @ DyV | {z } =:D 1 +@ Ay@ V | {z } =:D 2 + (@ DV)yV | {z } =:D 3 + DVy@ V | {z } =:D 4 (11.0.10) For D 1 R 3 1 2 0 @ @ DyVd 0 2 dy .e 2 1 R 3 0 e 1 0 (1 +K 0 )@ @ DV d 0 2 dy .e 2 1 sup 0 0 R 3 j@ Dj 2 0 je 1 0 (1 + 0 )@ Vjd 0 2 dy .e 2 1 sup 0 0 k@ Dk 2 0 e 1 0 (1 + 0 )e 1 0 e 0 0 (S N ) 1=2 d 0 2 .e 2 0 S N (); (11.0.11) 133 where we use the H 2 (R 3 ),!L 1 (R 3 ) embedding onkVk L 1. For D 2 , R 3 1 2 j 0 @ Ay@ @ d 0 j 2 . R 3 1 2 (1 +K) 2 sup 0 0 j@ @ j 2 j 0 1d 0 j 2 .e 2 1 0 (1 +K) 2 e 2 0 S N 2 .e 2 0 0 S N (11.0.12) For D 3 , rst integrate by parts in , 1 0 (@ @ D)y@ d 0 = 1 (V (@ D)y) 0 1 0 ( @ y +@ @ y + @ y)@ Dd 0 =A +B +C +D: (11.0.13) Then kAk. (1 +K)k@ k L 1k@ Dk + 1 S N (0). (1 +)e 0 (S N ) 1=2 k@ Dk +e 0 S N (0); kBk. sup 0 0 k@ Dke 1 0 (1 + 0 )e 1 0 0 0 k@ k L 1d 0 ; . sup 0 0 k@ Dk(S N ) 1=2 e 0 (1 +) kCk. sup 0 0 k@ Dk(S N ) 1=2 e 0 : (11.0.14) Now for D in the same way as for B rst note e 1 0 e 1 0 0 0 d 0 .e 0 : (11.0.15) Then using our equation for (8.1.36) to rewrite @ and noting each term gains a factor from @ i in (8.1.36) and our a priori assumptionS N < 1=3, we have kDk. sup 0 0 k@ Dke 0 (1 +); (11.0.16) 134 For D 4 , R 3 1 2 j 0 DVy@ @ d 0 j 2 . R 3 1 2 (1 +K) 2 kDVk 2 L 1 sup 0 0 j@ @ j 2 j 0 1d 0 j 2 .e 2 1 0 (1 +K) 2 e 2 0 S N 2 .e 2 0 0 S N (11.0.17) Hence (1 + 2 )e 2 0 S N & 8 > > < > > : k 1 (1 +) 0 @ [@ ; Ay]Vd 0 k 2 if jjN 1; k 1 (1 +) 0 @ [@ ; Ay]Vd 0 k 2 if jj =N: (11.0.18) For the thirteenth term on the right hand side of (11.0.3), C 2 0 1 @ (y)d, rst compute @ [(y )] i j =@ ( js y s ik k is y s jk k ): (11.0.19) For the left term @ ( js y s ik k ) = js ik y s @ + X j 0 j=jj1 c v 0 ;s js ik @ 0 k (11.0.20) Schematically consider the rst term y@ :=C 1 (11.0.21) Then: k C 2 0 1 C 1 d 0 k 2 = R 3 C 2 4 2 0 1 y@ d 0 2 dy . (1 + 2 ) 1 2 0 1 d 0 2 sup 0 0 k@ k 2 . (1 + 2 )U()S N (); (11.0.22) 135 where we dene U() = 1 2 j 0 1 d 0 j 2 . Now for U, U.e 2 1 0 e (1) 1 0 d 0 2 .e 2 1 (e 2(1) 1 + 1) .e 2 0 (11.0.23) Hence k C 2 0 1 C 1 d 0 k 2 . (1 + 2 )e 2 0 S N (11.0.24) Similar for the sum in (11.0.20) except bound is just e 2 0 S N there. Hence (1 + 2 )e 2 0 S N & 8 > > > < > > > : k C 2 0 1 @ (y )d 0 k 2 if jjN 1 k C 2 0 1 @ (y )d 0 k 2 if jj =N (11.0.25) For the fourteenth term on the right hand side of (11.0.3), C 2 0 1 @ (A [D]y )d 0 , rst rewrite [A [D]y ] i j = jm A s ` ` ; m y s ik k im A s ` ` ; m y s jk k = ( jm A s ` ` ; m y s ik k im A s ` ` ; m y s jk k ) + ( jm A s ` ` ; m y s ik y k im A s ` ` ; m y s jk y k ) = [A [D]y ] i j + [A [D]y y] i j : (11.0.26) The A [D]y term is similar to other terms and analogous arguments apply to estimate this term. For the A [D]y y term, rst compute: @ ( jm A s ` ` ; m y s ik y k ) =c(@ (DD))yy | {z } =:G 1 + X j 0 j=jj2 c(@ 0 (DD)) + X j 0 j=jj1 c(@ 0 (DD))y: (11.0.27) 136 The two sums above are straightforward to handle using methods from above andH 2 (R 3 ),! L 1 (R 3 ) embedding. For G 1 , for 0jjN 1, k C 2 0 1 G 1 d 0 k 2 = R 3 C 2 4 2 0 c 1 @ (DD)yyd 0 2 dy . 1 2 0 (1 +K 0 ) 2 1 d 0 2 sup 0 0 k@ k 2 . 1 2 0 (1 +K 0 ) 2 1 d 0 2 sup 0 0 k@ k 2 . 1 2 0 1 d 0 2 sup 0 0 k@ k 2 .e 2 1 (e 2 1 1 + 1)S N () .e 2 0 S N (): (11.0.28) For = N a similar argument holds except there we include the additional, articially introduced, time weight inS N . Hence (1 + 2 )e 2 0 S N & 8 > > > < > > > : k C 2 0 1 @ (A [D]y )d 0 k 2 if jjN 1 k C 2 0 1 @ (A [D]y )d 0 k 2 if jj =N (11.0.29) For the last term on the right hand side of (11.0.3), use the compact support of and it's derivatives to boundy, and then apply (11.0.23) in combination with theH 2 (R 3 ),!L 1 (R 3 ) embedding andkk 2 H N+1;1 (R 3 ) to obtain e 2 0 & 8 > > > < > > > : k C (1 +) 0 1 @ ((1 +) 1 (Ar y))d 0 k 2 if jjN 1 k C (1 +) 0 1 @ ((1 +) 1 (Ar y))d 0 k 2 if jj =N (11.0.30) This concludes the proof of (11.0.1). 137 Proof of (11.0.2). Using the Fundamental Theorem of Calculus we rewrite two terms in the equation (13.4.2) for Curl A in Lemma 13.4.1. Apply @ to the resulting equation Curl A @ =[@ ; Curl A ] + 1 1 + @ (Ay) + 1 + @ ((1 +) 1 Ar) +@ (Curl A ([(0)])) 1 1 + @ (Ay(0)) 1 + @ ((1 +) 1 Ar(0)) +(0)@ (Curl A (V(0))) 0 1 ( 0 ) d 0 (0)@ (Ay V(0)) 1 + 0 1 ( 0 ) d 0 (0)@ ((1 +) 1 Ar V(0)) 1 + 0 1 ( 0 ) d 0 + 0 @ ([@ ; Curl A ])d 0 + 1 1 + 0 @ ([@ ; Ay]) + (1 +) 0 @ ((1 +) 1 [@ ; Ar])d 0 + 0 1 ( 0 ) 0 0 ( 00 )@ ([@ ; Curl A ]V)d 00 d 0 1 1 + 0 1 ( 0 ) 0 0 ( 00 )@ ([@ ; Ay]V)d 00 d 0 (1 +) 0 1 ( 0 ) 0 0 ( 00 )@ ((1 +) 1 [@ ; Ar]V)d 00 d 0 0 2 ( 0 ) 0 0 ( 00 )@ (Curl A ( V))d 00 d 0 + 2 1 + 0 1 ( 0 ) 0 0 ( 00 )@ (Ay ( V))d 00 d 0 + 2 1 + 0 1 ( 0 ) 0 0 ( 00 )@ ((1 +) 1 Ar ( V))d 00 d 0 + C 1 + 0 1 0 0 1 @ (y )d 00 d 0 C 1 + 0 1 0 0 1 @ (A [D]y )d 00 d 0 + C 1 + 0 1 0 0 1 @ ((1 +) 1 (r ))d 00 d 0 C 1 + 0 1 0 0 1 @ ((1 +) 1 (Ar y))d 00 d 0 (11.0.31) We take thekk 2 norm of (11.0.31), and ifjj = N then multiply by [()] , and then estimate the right hand side. As can be seen from (11.0.31) and (11.0.3), many terms and 138 hence estimates are similar. Therefore we give the key estimates below and remark similar arguments to these and the proof of (11.0.1) will hold for the other terms. For the second term on the right hand side of (11.0.31), 1 1+ @ (Ay), @ (Ay) = y@ (A) +R; (11.0.32) whereR are favorable remainder terms. Note @ (A) =A@ +@ 1 ((D 2 )) (11.0.33) Then kyA@ k 2 .k 0 y@ @ d 0 +y@ (0)k 2 . R 3 j 0 (1 +K 0 )@ @ d 0 j 2 +j@ (0)j 2 dy . R 3 j 0 (1 +K 0 ) =4 =4 =2 @ @ d 0 j 2 dy +S N (0) . R 3 ( 0 (1 +K 0 ) 2 e 0 0 d 0 )( 0 e 0 0 j@ @ j 2 d 0 )dy +S N (0) . 0 e 0 0 S N d 0 +S N (0): (11.0.34) Next, ky@ 1 ((D 2 ))k 2 .S N ky@ ` k 2 L 1 (R 3 ) wherej ` j N 2 . (ky@ ` (0)k L 1 + 0 ky@ ` @ k L 1) 2 .S N (0) + ( 0 (1 +K 0 )e 0 0 p S N ) 2 .S N (0) + ( 0 (1 +K 0 )e 0 0 d 0 )( 0 e 0 0 S N d 0 ) .S N (0) + 0 e 0 0 S N d 0 (11.0.35) 139 Thus, since top order here with is simple, S N (0) + 0 e 0 0 S N d 0 & 8 > > < > > : k 1 1 + @ (Ay)k 2 if jjN 1 k 1 1 + @ (Ay)k 2 if jj =N (11.0.36) For the eleventh term on the right hand side of (11.0.31), 1 1+ 0 @ ([@ ; Ay]), note @ ([@ ; Ay]) =y@ (@ (A )) +R; (11.0.37) where R are favorable remainder terms. Note @ (A ) = (@ )A + (@ A ). Also @ (A) =@ 1 ((D 2 )) +A@ (). Then k 0 y@ @ 1 ((D 2 ))d 0 k 2 . R 3 j 0 (1 +K 0 )e 0 2 0 e 0 2 0 @ h @ ` d 0 j 2 dy wherej h j N 2 + 1;j ` j N 2 .k@ ` k 2 L 1 R 3 ( 0 e 0 0 j@ h j 2 d 0 )( 0 e 0 0 (1 +K 0 ) 2 d 0 )dy . 0 e 0 0 S N d 0 : (11.0.38) Similar for y@ A@ , except just boundA . Next, @ (@ A) = (@ (@ D)) +@ 1 ([@ D][D]): (11.0.39) ForjjN 1, k 0 y@ (@ D)d 0 k 2 . R 3 j 0 (1 +K 0 )e 0 2 0 e 0 2 0 e 0 0 (@ +1 @ )d 0 j 2 dy .kk 2 L 1 R 3 ( 0 e 0 0 e 2 0 0 j@ +1 @ j 2 d 0 )( 0 e 0 0 (1 +K 0 ) 2 d 0 )dy . 0 e 0 0 S N d 0 : (11.0.40) 140 For =N we integrate by parts in 0 y@ (@ +1 )d 0 = y@ +1 j 0 0 y(@ )@ +1 d 0 0 (@ )y@ +1 d 0 = (I) + (II) + (III): (11.0.41) Note k(I)k 2 .S N (0) + 2 R 3 jyj 2 j@ +1 j 2 dy .S N (0) +S N kyk 2 L 1 .S N (0) +S N (ky(0)k L 1 + 0 ky@ k L 1d 0 ) 2 .S N (0) +S N (S N (0) + ( 0 e 0 0 p S N (1 +K 0 )d 0 ) 2 ) .S N (0) +S N (S N (0) + ( 0 e 0 0 (1 +K 0 ) 2 d 0 )( 0 e 0 0 S N d 0 )) .S N (0) + 0 e 0 0 S N d 0 : (11.0.42) For (II), let & > 0 be such that 2& < 0 : Then, k(II)k 2 = R 3 j 0 y@ @ +1 d 0 j 2 . R 3 ( 0 e 2& 0 jy@ j 2 d 0 )( 0 e 2& 0 j@ +1 j 2 d 0 )dy . 0 e 2& 0 ky@ k 2 L 1d 0 S N 0 e 2& 0 d 0 . 0 e 2& 0 (1 +K 0 ) 2 e 0 0 e 0 0 S N d 0 . 0 e 0 0 S N d 0 : (11.0.43) 141 Now (III) is similar to other estimates as well as top order inside integral argument. Also @ 1 ([@ D][D]) estimate is similar to other estimates. Hence S N (0) + 0 e 0 0 S N d 0 & 8 > > < > > : k 1 1 + 0 @ ([@ ; Ay])k 2 if jjN 1 k 1 1 + 0 @ ([@ ; Ay])k 2 if jj =N (11.0.44) For the fourteenth term on the right hand side of (11.0.31) which is as follows 1 1 + 0 1 ( 0 ) 0 0 ( 00 )@ ([@ ; Ay]V)d 00 d 0 we rst note @ ([@ ; Ay]@ ) =y@ (@ (A )@ ) +R; (11.0.45) whereR are favorable remainder terms. Then k 0 1 0 0 y@ (@ (A )@ )d 00 d 0 k 2 . R 3 j 0 (1 +K 0 )(1 + 0 ) 1=2 1 (1 + 0 ) 1=2 0 0 @ (@ (A )@ )d 00 )d 0 jdy . R 3 [( 0 (1 +K 0 ) 2 (1 + 0 ) 2 d 0 ) 0 1 (1 + 0 ) 2 ( 0 0 @ (@ (A )@ )d 00 ) 2 ]dy . 0 1 (1 + 0 ) 2 k 0 @ (@ (A )@ )d 00 k 2 d 0 : (11.0.46) Then by a similar argument to other estimates, 0 e 0 0 S N d 0 & 8 > > > < > > > : k 1 1 + 0 1 ( 0 ) 0 0 ( 00 )@ ([@ ; Ay]V)d 00 d 0 k 2 if jjN 1; k 1 1 + 0 1 ( 0 ) 0 0 ( 00 )@ ([@ ; Ay]V)d 00 d 0 k 2 if jj =N: (11.0.47) 142 For the nineteenth term on the right hand side of (11.0.31), C 1+ 0 1 0 0 1 @ (y )d 00 d 0 , we x 0 < X < 1 4 suciently small so that 4 + 1 2 + 2 X > 0. Then x 0<Y < 4 + 1 2 + 2 X. Then k 0 C 2 0 0 1 (@ (y ))d 00 d 0 k 2 . R 3 ( 0 1 2X d 0 ) 0 1 22X ( 0 0 1 (@ (y ))d 00 ) 2 d 0 dy . 0 1 22X k 0 0 1 @ (y )d 00 k 2 d 0 . 0 e 0 0 S N ; (11.0.48) where we conclude the nal bound by a similar argument to other estimates since 1 22X j 0 1 (1 +K) 2 dj 2 .e (2X2) 1 0 e (1+Y ) 1 0 d 0 2 .e (2X2) 1 (e 2(1+Y ) 1 + 1) .e 0 : (11.0.49) Hence 0 e 0 0 S N d 0 & 8 > > > < > > > : k C 1 + 0 1 0 0 1 @ (y )d 00 d 0 k 2 if jjN 1 k C 1 + 0 1 0 0 1 @ (y )d 00 d 0 k 2 if jj =N (11.0.50) For the twentieth term on the right hand side of (11.0.31), C 1+ 0 1 0 0 1 @ (A [D]y )d 00 d 0 , rst A [D]y = A [D]y + A [D]y y: (11.0.51) 143 Then a similar argument to above holds except at top orderjj =N. At top order order, jj =N, note, with X dened as above, k 0 C 2 0 0 1 (@ (y ))d 00 d 0 k 2 . R 3 [ 0 1 ( 0 ) 2X d 0 0 1 ( 0 ) 22X ( 0 0 1 (@ (y ))d 00 ) 2 d 0 dy . sup 0 0 [( 0 ) ] 0 22X k 0 0 1 (@ (y ))d 00 k 2 d 0 . sup 0 0 [( 0 ) ] sup 0 0 [ 1 ( 0 ) ] 0 1 22X k 0 0 1 (@ (y ))d 00 k 2 d 0 = 0 1 22X k 0 0 1 (@ (y ))d 00 k 2 d 0 . 0 e 0 0 S N d 0 ; (11.0.52) where the last bound follows from a similar argument to above, where we include in S N if needed. Hence 0 e 0 0 S N d 0 & 8 > > > < > > > : k C 1 + 0 1 0 0 1 @ (A [D]y )d 00 d 0 k 2 if jjN 1; k C 1 + 0 1 0 0 1 @ (A [D]y )d 00 d 0 k 2 if jj =N: (11.0.53) Finally for the last term on the right hand side of (11.0.31), using a combination of the H 2 (R 3 ),!L 1 (R 3 ) embedding andkk 2 H N+1;1 (R 3 ) , a similar argument, without apply- ing nite propagation, to other estimates and then (10.0.32) gives + 0 e 0 0 S N d 0 & 8 > > > < > > > : k C 1 + 0 1 0 0 1 @ ((1 +) 1 (Ar y))d 00 d 0 k 2 if jjN 1; k C 1 + 0 1 0 0 1 @ ((1 +) 1 (Ar y))d 00 d 0 k 2 if jj =N: (11.0.54) This concludes the proof of (11.0.2). 144 Chapter 12 Energy Inequality and Proof of the Main Theorem By the Local Well-Posedness Theorem 8.4.1 there exists a unique solution to (8.1.36) on some time interval [0;T ];T > 0: From Propositions 10.0.3 and 11.0.1 with > 0 chosen small enough and by using the equivalence of the normS N and the modied energyE N + C N1 as well as a similar argument to (10.0.32), we conclude that there exist universal constants c 1 ;c 2 ;c 3 ;c 4 1 such that for any 0 T S N (; )c 1 S N ( ) +c 2 +c 3 (S N (0) +B N [V](0)) +c 4 e 0 2 0 S N ( 0 ; )d 0 : (12.0.1) HereS N (; ) denotes the sliced norm ofS N from to with sup 0 0 replaced by sup 0 . By a standard well-posedness estimate, we deduce that the time of existence T is inversely proportional to the size of the initial data, i.e.: T (S N (0) +B N [V](0)) 1 . Choose "> 0 so small that the time of existence T satises e 0 T=4 0 c 4 ; sup T S N ()c S N (0) +B N [V](0) + (12.0.2) wherec> 0 is a universal constant provided by the local-in-time well-posedness theory. Let C = 3(c 1 c +c 2 +c 3 ): (12.0.3) 145 Dene T := sup 0 f solution to (8.1.36) exists on [0;) andS N ()C S N (0) +B N [V](0) + g: (12.0.4) Observe thatT T since cC . Letting = T 2 in (12.0.1) for any 2 [ T 2 ;T ] we obtain S N (; T 2 )c 1 S N ( T 2 )+c 2 +c 3 S N (0) +B N [V](0) +c 4 T 2 e 0 2 0 S N ( 0 ; T 2 )d 0 : (12.0.5) Therefore, using (12.0.2) we conclude that for any 2 [ T 2 ;T ] S N (; T 2 )c 1 S N ( T 2 ) +c 2 +c 3 S N (0) +B N [V](0) + c 4 0 e 0 T=4 S N (; T 2 ) c 1 S N ( T 2 ) +c 2 +c 3 S N (0) +B N [V](0) +S N (; T 2 ): (12.0.6) Since by (12.0.2),S N ( T 2 )c S N (0) +B N [V](0) + we conclude from (12.0.6) that S N (; T 2 )c 1 c S N (0) +B N [V](0) + +c 2 +c 3 S N (0) +B N [V](0) +S N (; T 2 ) (12.0.7) which for suciently small gives S N (; T 2 ) 2(c 1 c +c 2 +c 3 ) S N (0) +B N [V](0) + <C S N (0) +B N [V](0) + ; (12.0.8) and hence S N ()<C S N (0) +B N [V](0) + : (12.0.9) It is now easy to check the a priori bounds in (8.4.3) are in fact improved. For instance, by the fundamental theorem of calculus kDk W 1;1 =k 0 DVk W 1;1 0 e 0 0 S N ( 0 )d 0 ."< 1 6 ; 2 [0;T ) (12.0.10) for "> 0 small enough. Similar arguments apply to the remaining a priori assumptions. 146 From the continuity of the map7!S N ( 0 ) and the denition ofT we conclude thatT = 1 and the solution to (8.1.36) exists globally-in-time. Furthermore, also using Proposition 11.0.1, the global bound (8.4.7) follows. 147 Chapter 13 Appendices 13.1 Energy Identities We give key identities which will be used in our estimates. First from Lemma 4.3 [20] we have the following modied energy identity: Lemma 13.1.1. Recalling the matrix quantities introduced in (8.2.7) and (8.2.8), the fol- lowing identity holds `j (r @ ) i j 1 im (r @ ) m `; = 1 2 d d 0 @ 3 X i;j=1 d i d 1 j ([N ] j i ) 2 1 A +T : (13.1.1) where the error termT is given as follows T = 1 2 3 X i;j=1 d d d i d 1 j ([N ] j i ) 2 Tr QN Q 1 @ PP > N > +N > P@ P > : (13.1.2) In the next Lemma, we give some useful results concerning our quantitiesA ,J and Lemma 13.1.2. ForA ,J and , the following identities hold A k j J 1 k j = (A k j k j )J 1 + k j (J 1 1); (13.1.3) A k j k j =A k l [D] l j ; (13.1.4) ij = ip (A j p +A j l l ; p ); (13.1.5) 148 1J 1 = 1 Tr[D] +O(jDj 2 ); (13.1.6) A k j J 1 k j =A k ` [D] ` j J 1 k j ( 1 Tr[D] +O(jDj 2 )): (13.1.7) Proof. First (13.1.3) is straightforward to verify by expanding the right-hand side. For (13.1.4), rst noteA = [D] 1 and =y +. We then have A k j k j =A k l l j A k l [D] l j =A k l ( l j [Dy] l j [D] l j ) =A k l [D] l j : (13.1.8) Next, (13.1.5) is proven in the following calculation where we recall A = [D] 1 and use =y +, ij = ip j p = ip A j l l ; p = ip (A j p +A j l l ; p ): (13.1.9) Now (13.1.6) follows from the calculation J = det[D] = det[Id +D] = 1 + Tr[D] +O(jDj 2 ): (13.1.10) Finally (13.1.7) follows from (13.1.3)-(13.1.4) and (13.1.6). 13.2 Time Based Inequalities We have the following useful based inequalities, summarized by the Lemma below. Lemma 13.2.1. Fix an ane motion A(t) from the setS under consideration, namely require detA(t) 1 +t 3 ; t 0: (13.2.1) Let 1 := lim !1 () () ; 0 := 2 1 : (13.2.2) 149 Then we have the following properties 0< 0 = 0 ( ) 1 ; (13.2.3) e 1 . ().e 1 ; 0; (13.2.4) X a+jjN kX a r = @ Vk 1++a; e S + X jjN k@ Vk 1+;(1 )e S .e 0 S N () 1 2 (13.2.5) k k.e 1 ; kk +k 1 kC; (13.2.6) 3 X i=1 d i + 1 d i C (13.2.7) 3 X i=1 j@ d i j +k@ Pk.e 1 (13.2.8) jwj 2 .h 1 w; wi.jwj 2 ; w2R 3 ; (13.2.9) for C > 0. Proof. The result (13.2.3) is clear from the denition of 0 . For inequalities (13.2.4) and (13.2.6) through (13.2.9) we rst note that by Lemma 7.2.2, there exist matrices A 0 ;A 1 ;M(t) such that A(t) =A 0 +tA 1 +M(t); t 0: (13.2.10) where A 0 ;A 1 are time-independent and M(t) satises the bounds kM(t)k =o t!1 (1 +t); k@ t M(t)k. (1 +t) 33 : (13.2.11) We also note detA(t) 1 +t 3 . Then inequalities (13.2.4) and (13.2.6) through (13.2.9) follow from Lemma A.1 [20]. Finally, (13.2.5) follows from the denition ofS N (8.3.1) and properties (13.2.3)-(13.2.4) above. 150 13.3 Local Well Posedness We construct a local solution for our original Euler system (7.0.1)-(7.0.4) from generic initial data. First write the equations (7.0.1)-(7.0.3) in terms of (; u;T ) as follows where we use the material derivative D Dt = @ t + ur and we have multiplied by diag(T 2 ; 2 TI 3 ; 2 ) where I 3 is the 3 3 identity matrix, T 2 D Dt +T 2 div(u) = 0 (13.3.1) 2 T Du Dt +T 2 r + 2 TrT = 0 (13.3.2) 2 DT Dt + 2 T div(u) = 0 (13.3.3) Let () denote the symmetric 5 5 system of equations (13.3.1)-(13.3.3). We rst note a xed ane solution ( A ; u A ;T A ) solves () with initial data (( A ) 0 ; (u A ) 0 ; (T A ) 0 ) = ( A (0;); u A (0;);T A (0;)): (13.3.4) Next we construct initial data, using generic initial data ( 0 ; u 0 ;T 0 ) and our ane initial data (( A ) 0 ; (u A ) 0 ; (T A ) 0 ), for a modied system which allows to us to avoid the fact that we want to have 0 ! 0 for large x. To this end choose ( 0 ; u 0 ; T 0 ) as follows 0 ='( 0 C 1 ) + (1') (( A ) 0 C 1 ) + (1 ) 0 ; (13.3.5) u 0 ='u 0 + (1')(u A ) 0 ; (13.3.6) T 0 ='T 0 + (1')(T A ) 0 ; (13.3.7) where '; 2 C 1 c (R 3 ) are such that, for xed R > 0 and > 0, ' = 1 on B(0; R 2 ), ' = 0 onR 3 nB(0;R), = 1 on B(0;R + 2) and = 0 onR 3 nB(0;R + 3), and C 1 > 0 is such thatk 0 k L 1 (R 3 ) C 1 2 . Note the existence of C 1 > 0 will be guaranteed through the 151 regularity of 0 . Then by construction we will have inf R 3f 0 +C 1 g> 0. Now we consider ( 0 ; u 0 ; T 0 ) as initial data for the modied system T 2 D Dt + ( +C 1 )T 2 div(u) = 0 (13.3.8) ( +C 1 ) 2 T Du Dt + ( +C 1 )T 2 r + ( +C 1 ) 2 TrT = 0 (13.3.9) ( +C 1 ) 2 DT Dt + ( +C 1 ) 2 T div(u) = 0 (13.3.10) Let (-) denote the symmetric 5 5 system of equations (13.3.8)-(13.3.10). Then, with suciently regular ( 0 ; u 0 ;T 0 ) which will be specied by the Lagrangian formulation, by Theorem II [30] there exists ^ T > 0 such that ( ; u ; T ) is a solution to (-) with initial data ( 0 ; u 0 ; T 0 ). Now let ( B ; u B ;T B ) = ( +C 1 ; u ; T ): (13.3.11) Since ( ; u ; T ) solve (-) we have that ( B ; u B ;T B ) solve () with initial data (( B ) 0 ; (u B ) 0 ; (T B ) 0 ) = ( 0 +C 1 ; u 0 ; T 0 ). Next let K =f(x;t)j 0tT 1 ;x2B(0;R + +Mt)g M = 3 c ( sup 0tT 1 fk B (T B ) 2 k L 1 (B(0;R+2)) +k( B ) 2 T B u B k L 1 (B(0;R+2)) +k( B ) 2 T B k L 1 (B(0;R+2)) +k A (T A ) 2 k L 1 (B(0;R+2)) +k( A ) 2 T A u A k L 1 (B(0;R+2)) +k( A ) 2 T A k L 1 (B(0;R+2)) g) c> 0 is such that c sup 0tT fk(T B ) 2 k L 1 (R 3 ) +k( B ) 2 T B k L 1 (R 3 ) +k( B ) 2 k L 1 (R 3 ) g; T 1 = min( ^ T; 2M ); > 0 is suciently small. (13.3.12) 152 Now we take (; u;T ) = 8 > > < > > : ( B ; u B ;T B ) in K; ( A ; u A ;T A ) outside K: (13.3.13) Then (; u;T ) is a solution of () onR 3 [0;T 1 ] with the following initial data 0 ='( B ) 0 + (1')( A ) 0 ; u 0 ='(u B ) 0 + (1')(u A ) 0 ; T 0 ='(T B ) 0 + (1')(T A ) 0 ; (13.3.14) since (; u;T ) is a solution in K and outside K, and applying a classical property of local uniqueness of solutions to () to get that (; u;T ) is continuous across @K. 13.4 Curl Equations Derivation Here we give the derivations of the equations satised by the modied curl of our velocity and perturbation which will be used for the purpose of our estimates. Lemma 13.4.1. Let (; V) : !R 3 R 3 be a unique local solution to (8.1.36)-(8.1.37) on [0;T ] forT > 0 xed. Then for all2 [0;T ] the curl matrices Curl A V and Curl A satisfy the equations Curl A V = 1 1 + Ay V + (1 +)(1 +) Ar V + (0)Curl A (V(0)) (0)Ay V(0) (1 +) (0)Ar V(0) (1 +)(1 +) + 1 0 [@ ; Curl A ]Vd 0 1 (1 +) 0 [@ ; Ay]Vd 0 (1 +)(1 +) 0 [@ ; Ar]Vd 0 2 0 Curl A ( V)d 0 + 2 (1 +) 0 Ay ( V)d 0 + 2 (1 +)(1 +) 0 Ar ( V)d 0 153 + C (1 +) 0 1 y d 0 C (1 +) 0 1 A [D]y d 0 + C (1 +)(1 +) 0 1 Ar d 0 C (1 +)(1 +) 0 1 Ar yd 0 : (13.4.1) and Curl A = Curl A ([(0)]) +(0)Curl A (V(0)) 0 1 ( 0 ) d 0 (0)Ay V(0) 1 + 0 1 ( 0 ) d 0 (0)Ar V(0) (1 +)(1 +) 0 1 ( 0 ) d 0 + 0 [@ ; Curl A ]d 0 + 1 1 + 0 Ay Vd 0 + (1 +)(1 +) 0 Ar Vd 0 + 0 1 ( 0 ) 0 0 ( 00 )[@ ; Curl A ]Vd 00 d 0 1 1 + 0 1 ( 0 ) 0 0 ( 00 )[@ ; Ay]Vd 00 d 0 (1 +)(1 +) 0 1 ( 0 ) 0 0 ( 00 )[@ ; Ar]Vd 00 0 2 ( 0 ) 0 0 ( 00 ) Curl A ( V)d 00 d 0 + 2 1 + 0 1 ( 0 ) 0 0 ( 00 ) Ax ( V)d 00 d 0 + 2 (1 +)(1 +) 0 1 ( 0 ) 0 0 ( 00 ) Ar ( V)d 00 d 0 + C 1 + 0 1 ( 0 ) 0 0 ( 00 ) 1 x d 00 d 0 C 1 + 0 1 ( 0 ) 0 0 ( 00 ) 1 A [D]y d 00 d 0 + C (1 +)(1 +) 0 1 0 0 1 Ar d 00 d 0 C (1 +)(1 +) 0 1 0 0 1 Ar yd 00 d 0 : (13.4.2) Proof. Writing (8.1.36) without the source term outside the nonlinearity, we have @ i + 1+ @ i +2 ij @ j +C i` ` + C w (w ij ((1+)A k j J 1 k j )) ;k = 0: (13.4.3) 154 Return back to via = +y, @ i + 1+ @ i + 2 ij @ j +C i` ` + C w (w ij (1 +)A k j J 1 ) ;k = 0: (13.4.4) Multiply by w 1 1+ (1 +) 1+ w 1 1+ (1 +) 1+ ( @ i + 1+ @ i + 2 ij @ j +C i` ` ) +w 1 1+ 1 (1 +) 1+ C (w ij A k j J 1 ); k = 0 (13.4.5) Note w 1 1+ 1 (1 +) 1+ C (w ij A k j J 1 ); k =C (1 +) ij A k j (w 1 1+ (1 +) 1 1+ J 1 ); k : (13.4.6) Moving away from coordinates we then have w 1 1+ (1 +) 1+ (@ + 1+ + 2 @ +C ) +C (1 +)A T r(w 1 1+ (1 +) 1 1+ J 1 ) = 0: (13.4.7) Note C (1 +)A T r(w 1 1+ (1 +) 1 1+ J 1 ) =C (1 +)r (w 1 1+ (1 +) 1 1+ J 1 ): (13.4.8) Since Curl A (r f) = 0, apply Curl A to (13.4.7), Curl A w 1 1+ (1 +) 1+ ( @ + 1+ @ + 2 @ +C ) = 0: (13.4.9) Now note Curl A (w 1 1+ (1 +) 1+ F) =w 1 1+ (1 +) 1+ Curl A F + Ar(w 1 1+ (1 +) 1+ ) F: (13.4.10) 155 Also (w 1 1+ (1 +) 1+ ); s = 1 1 + w 1 1+ 1 w; s (1 +) 1+ 1 + (1 +) 1+ 1 ; s w 1 1+ = 1 1 + w 1 1+ (1 +) 1+ y s 1 + (1 +) 1+2 1+ ; s w 1 1+ (13.4.11) since w; s =y s w. So Ar(w 1 1+ (1+) 1+ )F = w 1 1+ (1 +) 1+ 1 + AyF (1 +) 1+2 1+ w 1 1+ 1 + ArF (13.4.12) where [Ay F] i j := jm A s m x s F i im A s m y s F j : (13.4.13) So multiplying (13.4.9) by w 1 1+ (1 +) 1+ we have and using Curl A () = 0 we have Curl A (@ ) + 1+ Curl A (@ ) + 2 Curl A ( @ ) 1 1 + Ax @ + 1+ @ + 2 @ +C (1 +)(1 +) Ar @ + 1+ @ + 2 @ +C = 0: (13.4.14) Divide by Curl A (@ ) + 1 Curl A (@ ) + 2Curl A ( @ ) 1 1 + Ax @ + 1 @ + 2 @ +C (1 +)(1 +) Ar @ + 1 @ + 2 @ +C = 0: (13.4.15) Note Curl A (V ) =@ ( Curl A (V)) [@ ; Curl A ] (V); (13.4.16) 156 where [@ ; Curl A ]F i j :=@ ( jm A s m ) F; i s @ ( im A s m ) F; j s : (13.4.17) Then @ ( Curl A (V)) =[@ ; Curl A ] (V) 2 Curl A ( V) + 1 1 + Ay @ + @ + 2 @ +C 1 + (1 +)(1 +) Ar @ + @ + 2 @ +C 1 : (13.4.18) Integrate from 0 to 0 , where 0 2 [0;], Curl A (V) = (0)Curl A ([V(0)]) + 1 0 0 [@ ; Curl A ] (V)d 00 2 0 0 Curl A ( V)d 00 + 1 (1 +) 0 0 Ax @ + @ + 2 @ +C 1 d 00 + (1 +)(1 +) 0 0 Ar @ + @ + 2 @ +C 1 d 00 : (13.4.19) Note (Ax@ ) =@ (Ax@ ) (Ax@ )[@ ; Ax]@ ; (13.4.20) where [@ ; Ax]F i j :=@ ( jm A s m )x s F i @ ( im A s m )y s F j : (13.4.21) Using a similar result for (Ar@ ), we have Curl A V = 1 1 + Ay V + (1 +)(1 +) Ar V + (0)Curl A (V(0)) (0)Ax V(0) (1 +) (0)Ar V(0) (1 +)(1 +) + 1 0 0 [@ ; Curl A ]Vd 00 1 (1 +) 0 0 [@ ; Ax]Vd 00 157 (1 +)(1 +) 0 0 [@ ; Ar]Vd 00 2 0 0 Curl A ( V)d 00 + 2 (1 +) 0 0 Ax ( V)d 00 + 2 (1 +)(1 +) 0 0 Ar ( V)d 00 + C (1 +) 0 0 1 Ay ()d 00 + C (1 +)(1 +) 0 0 1 Ar ()d 00 : (13.4.22) Now [Ax ()] i j = jm ( s m A s ` [D] ` m )x s ik k im ( s m A s ` [D] ` m )y s jk k = js y s ik k is y s jk k + js y s ik y k is y s jk y k ( jm A s ` [D] ` m y s ik k im A s ` [D] ` m y s jk k ) = ( js y s ik k is y s jk k ) ( jm A s ` [D] ` m x s ik k im A s ` [D] ` m y s jk k ) := [y ] i j [A [D]y ] i j : (13.4.23) Second notice that [Ar ()] i j = jm A s m ;s ik k im A s m ;s jk k + jm A s m ;s ik y k im A s m ;s jk y k = [Ar ] i j + [Ar y] i j : (13.4.24) Hence we have (13.4.1). Now for Curl A , rst note @ (Curl A ) = Curl A (@ ) + [@ ; Curl A ]: (13.4.25) So integrating (13.4.25) from 0 to via (13.4.22) we have (13.4.2). 158 13.5 Coercivity Estimates We give a useful result which will allow us to overcome the time weights with negative powers which arise from our equation structure. Lemma 13.5.1 (Coercivity Estimates). Let (; V) : !R 3 R 3 be a unique local solution to (8.1.36)-(8.1.37) on [0;T ] for T > 0 xed with supp 0 B 1 (0), supp V 0 B 1 (0) and assume (; V) satises the a priori assumptions (8.4.3). Fix N 4. Suppose in (8.1.36) satiseskk 2 H N+1 (R 3 ) and supp B 1 (0) where > 0 is xed. Fix with 0jjN 1. Then for all 2 [0;T ], we have the following inequalities k@ k 2 . sup 0 0 f k@ Vk 2 g +k@ (0)k 2 (13.5.1) kr @ k 2 . sup 0 0 f X j 0 j=jj+1 k@ 0 Vk 2 g +kr @ (0)k 2 (13.5.2) kdiv @ k 2 . sup 0 0 f X j 0 j=jj+1 k@ 0 Vk 2 g +kdiv @ (0)k 2 : (13.5.3) Proof. Proof of (13.5.1). By the fundamental theorem of calculus, and the exponential boundedness of (13.2.4) and therefore time integrability of negative powers of , @ = 0 @ Vd 0 +@ (0) = 0 2 2 @ Vd 0 +@ (0) . sup 0 0 f 2 @ Vg +@ (0): (13.5.4) Therefore applying Cauchy's inequality (ab.a 2 +b 2 ; a;b2R) k@ k 2 . sup 0 0 f k@ Vk 2 g +k@ (0)k 2 : (13.5.5) Proof of (13.5.2). By a similar coercivity estimate to (13.5.4)-(13.5.5) kr @ k 2 . sup 0 0 f kr @ Vk 2 g +kr @ (0)k 2 : (13.5.6) 159 Now using our a priori bounds (8.4.3), we have sup 0 0 kr @ Vk 2 . sup 0 0 f X j 0 j=jj+1 k@ 0 Vk 2 g: (13.5.7) Then (13.5.6)-(13.5.7) imply (13.5.2). Proof of (13.5.3). Finally the proof of (13.5.3) is similar to the proof of (13.5.2). 160 Part III The Vacuum Boundary Problem for the Spherically Symmetric Compressible Euler Equations with Positive Density and Unbounded Entropy Chapter 14 Introduction We consider the free boundary compressible Euler equations for ideal gases in three space dimensions (@ t u + uru) +rp = 0 on (t); (14.0.1) @ t + div(u) = 0 on (t); (14.0.2) @ t S + urS = 0 on (t); (14.0.3) coupled with equation of state for an ideal gas p(;S) = e S ; (14.0.4) 161 where u is the uid velocity vector eld, is the density,S is the entropy,p is the pressure, > 1 is the adiabatic constant and (t)R 3 is a time dependent, open bounded domain with boundary @ (t) where t2 [0;T ] for some T > 0: Our boundary conditions are the physical vacuum boundary condition coupled with kinematic boundary condition p = 0 on @ (t); (14.0.5) 1< @c 2 s @n < 0 on @ (t); (14.0.6) V(@ (t)) = u n(t) on @ (t); (14.0.7) with n the outward unit normal vector to @ (t), @ @n the outward normal derivative, c s := q @p @ = p 1 e S is the speed of sound andV(@ (t)) the normal velocity of @ (t). Finally, we demand positivity of the initial density throughout the starting domain and boundary 0 > 0 in (0)[@ (0): (14.0.8) Our study of the Euler system with vacuum (14.0.5) and a positive density (14.0.8) was motivated by the work of Ovsyannikov [41] and Borisov-Kilin-Mamaev [3] in which examples of global-in-time solutions were given in this context. Considering the presence of vacuum (14.0.5) and our equation of state (14.0.4), a positive density implies formally our entropy must become unbounded and approach negative innity towards the boundary. Unbounded entropy proles have featured in work on black holes [21, 22] and an entropy diverging to negative innity was studied in the context of dynamical systems used to model particle states [13]. In our study of the vacuum free boundary nonisentropic Euler system with positive initial density (14.0.1)-(14.0.8) we will make the assumption of spherical symmetry. The novel weight structure unique to this problem whereby the entropy prole will function as our changing weight leads to a fundamental loss of weight with respect to derivatives in 162 the vorticity equation. Thus to avoid this problem, we assume radial symmetry for the Lagrangian ow map (t;y) (t;y) =(t;r)y; r =jyj; (14.0.9) where will now function as our radial ow map in Lagrangian variables. This assumption will avoid the need to have control of the problematic curl term due to loss of weight, but will introduce a coordinate singularity near the origin r = 0 for which we employ the methodolodgy developed in the recent work [17] to handle this singularity. In Eulerian coordinates, the spherically symmetric free boundary Euler equations are (@ t u +u@ r u) +@ r p = 0 in (0;R(t)); (14.0.10) @ t (r 2 ) +@ r (r 2 u) = 0 in (0;R(t)); (14.0.11) @ t S +u@ r S = 0 in (0;R(t)); (14.0.12) where u(t;x) = (x=r)u(t;r); (t;x) =(t;r); S(t;x) =S(t;r); r =jxj; (14.0.13) and the moving domain (t) =B R(t) (0). Collectively, we will study the Cauchy problem for the spherically symmetric vacuum free boundary nonisentropic Euler system with positive density. To the best of our knowledge there are no known previous global-in-time existence results for this system, with or without spherical symmetry. The main goal of this part is to construct open sets of initial data that lead to global solutions to the positive density nonisentropic Euler system with spherical symmetry in the presence of free vacuum boundaries. Before we move on, we brie y discuss some known results relevant to this part. It is well-known that the Euler equations are hyperbolic and in the whole space the existence of C 1 local-in-time positive density solutions follows from the theory of symmetric hyperbolic 163 systems [30, 37]. Serre [47] and Grassin [15] proved global existence in the whole space for a special class of initial data by perturbing solutions to the vectorial Burgers equation. Recently, Rickard [43] proved global-in-time well-posedness of the Euler equations with heat transport by the pertubation of Dyson's [12] isothermal ane solutions, see below for details on ane solutions. In the other direction, Sideris [49] showed that singularities must form if the density is a strictly positive constant outside of a bounded set. Christodoulou-Miao [8] give a thorough description of shock formation for irrotational uids starting from smooth initial data. We refer to the works of Speck and Luk-Speck [35, 52] for a more general framework with respect to shock formation. Buckmaster-Shkoller-Vicol [4] recently gave a constructive proof of shock formation leading to vorticity formation from an open set of initial data. We remark that these singularity and shock formation results do not apply to the physical vacuum free boundary problem. Important examples of global-in-time solutions are given by Ovsyannikov [41], Dyson [12] and Sideris [50, 51]. These are the so-called ane motions which are special expanding global solutions found by a separation-of-variables ansatz for the Lagrangian ow map, see Section 14.2. Notably for our current work, the ane solutions found by Ovsyannikov [41] can be used to construct positive density solutions in the presence of vacuum boundaries, see Borisov-Kilin-Mamaev [3]. For the vacuum free boundary problem, local well-posedness with physical vacuum has been proven by Coutand-Shkoller [9] and Jang-Masmoudi [27]. In particular, we will adapt the method of Jang-Masmoudi [27] to obtain local well-posedness for our system. The global ane solutions of Sideris [50, 51] were constructed in the free vacuum boundary setting and importantly satisfy the physical vacuum condition (14.0.6). In the isentropic (constant entropy) case, Had zi c-Jang [20] and Shkoller-Sideris [48] estabilished the nonlinear stability of the Sideris solutions. For the nonisentropic setting with vanishing density and bounded entropy, Rickard-Had zi c-Jang [44] proved global existence by perturbing around 164 a rich class of nonisentropic ane motions. Finally, in the isentropic setting, Parmeshwar- Had zi c-Jang [42] do not rely on background ane solutions and obtain global existence of small density expanding solutions. In the previous vacuum free boundary problem results, the density is vanishing at the boundary. This contrasts with the current part in which we consider a positive density. In particular, our entropy prole will instead behave like a distance function and will operate as a changing weight in our analysis. 14.1 Spherically Symmetric Lagrangian Coordinates To study the vacuum free boundary nonisentropic Euler system with positive initial density (14.0.1)-(14.0.8) we move to Lagrangian coordinates which brings the problem onto a xed domain. Dene the ow map as follows @ t (t;y) = u(t;(t;y)); (14.1.1) (0;y) = 0 (y); (14.1.2) where 0 is a suciently smooth dieomorphism. Introduce the notation A := [D] 1 ; (Inverse of the Jacobian matrix) (14.1.3) J := det[D]; (Jacobian determinant) (14.1.4) (y) := 0 ( 0 (y))J (0;y) (Lagrangian density prole) (14.1.5) d(y) :=e S 0 ( 0 (y)) (Langrangian entropy prole). (14.1.6) Then it is known [44] that the nonisentropic Euler equations (14.0.1)-(14.0.3) reduce to @ 2 t i + ( d [A ] k i J 1 ) ;k = 0: (14.1.7) 165 As mentioned above, we make the assumption of spherical symmetry for the remainder of this part. Therefore we make the ansatz (t;y) =(t;r)y; r =jyj: (14.1.8) ThenA andJ transform as follows [23] A = k i r y k y i ( + r r)r ; (14.1.9) J = 2 ( + r r): (14.1.10) Since we are considering the spherically symmetric case, make the assumption that and d are radial functions (y) = (r); (14.1.11) d(y) =d(r): (14.1.12) With these formulae and the fact that @ k = y k r @ r , by substituting (14.1.8) into (14.1.7) we obtain tt + 2 r @ r d( 2 ( + r r)) = 0: (14.1.13) 14.2 Nonisentropic Spherically Symmetric Ane Motion By making the further ansatz (t;r) =a(t) for scalar a(t) with a(0)> 0, the fundamental ane ODEs found in [12,41,50] are obtained. The ansatz transforms (14.1.13) into a tt + a 23 r @ r ( d) = 0: (14.2.1) 166 We have that d is independent oft and hence (14.2.1) will hold if we require the following fundamental ane ODEs to hold a tt =a 23 ; (14.2.2) r =@ r ( d): (14.2.3) Positive density motions in the presence of vacuum boundaries were obtained from the Ovsyannikov [41] ane solutions, see Borisov-Kilin-Mamaev [3]. For our positive density free boundary Euler system, special ane solutions are obtained through having the following properties 2C 0 [0; 1]\C 1 [0; 1]; (14.2.4a) (r)> 0 for r2 [0; 1]; (14.2.4b) 0 (0) = 0: (14.2.4c) Then we consider the following form for d which solves (14.2.3) d(r) = 1 r ` (`)d` ( (r)) : (14.2.5) With (14.2.5), we have the fundamental property d(1) = 0 which is crucial to permit a solution to our positive density Euler system. Our next calculation shows why the vacuum boundary condition (14.0.6) is satised by our ane solution ( 1 d) 0 (1) = lim r!1 1 r ` (`)d` (r)(r 1) = lim r!1 r (r) 0 (r)(r 1) + (r) =1; (14.2.6) 167 where we have used L'Hospital's Rule. Then with a2 C(R;R + )\C 1 (R;R) solving the ODE (14.2.2), satisfying (14.2.4a)-(14.2.4c) and nally d being given by (14.2.5), the associated solution of the positive density Euler system is u a (t;x) = a 0 (t)x a(t) ; (14.2.7) a (t;x) = (r) a(t) ; (14.2.8) S a (t;x) = lnd(r): (14.2.9) noting that r =j x a(t) j here. Remark 14.2.1 (Unbounded Entropy). From (14.2.9), the property lim r!1 d(r) =d(1) = 0 corresponds to our ane entropy approaching negative innity as it approaches the bound- ary. Now at this stage, we consider the proles of the form (r) =(r); (14.2.10) where 2C k [0; 1]; > 0 satisfying 0 (0) = 0 with k2N to be specied. Here we demand the condition 0 (0) = 0 to ensure the regularity of at the center as in (14.2.4c). Then from (14.2.5), we immediately have d2C k [0; 1] also. Finally, we note for all > 1 we have the asymptotics a(t) 1 +t; t 0: (14.2.11) This follows from Theorem 3 [51] if we consider the special case of the full 3D problem where A(t) = diag(a(t);a(t);a(t)). 168 We denote the set of ane motions under consideration byS . To recap, the setS is parametrized by the quadruple (a(0);a 0 (0);)2R + RZ k ; (14.2.12) where Z k := n 2C k [0; 1] :> 0; 0 (0) = 0 o (14.2.13) and we take k2N suciently large (to be specied later in Theorems 15.5.1 and 15.5.3). Remark 14.2.2 (Eulerian description of spherically symmetric solutions). At this stage it is worth giving the connections between the Eulerian description of spherically symmetric solutions (14.0.13), spherically symmetric Lagrangian coordinates and background ane motion. For the velocity, using (14.1.1) and (14.1.8), u(t;(t;y)) =@ t (t;y) = t (t;r)y = (y=r)(r t (r;t)) so that u(t;j(t;y)j) =r t (t;r): (14.2.14) For the density (t;j(t;y)j) = (r) 2 ( + r r) : (14.2.15) For the entropy S(t;j(t;y)j) = lnd(r): (14.2.16) Remark 14.2.3 (Eulerian initial density 0 and entropy S 0 ). The spherically symmetric Eulerian initial density 0 and entropy S 0 are connected to the background ane motion via 0 (jxj) = (j 1 0 (x)j) [ 0 (j 1 0 (x)j)] 2 ( 0 (j 1 0 (x)j) + 0 0 (j 1 0 (x)j)j 1 0 (x)j) S 0 (jxj) = lnd(j 1 0 (y)j); (14.2.17) where 0 0 (r) =@ r 0 (r). 169 Chapter 15 Formulation and Main Global Existence Result 15.1 Perturbation of Ane Motion We derive the equation for the perturbation of our ane motion. With an ane motion a xed, dene the modied ow map = a . We note considering the Lagrangian motion given by (t;r)y then theJ equivalent ofJ is given by J = 2 ( + r r) :=J (15.1.1) ThusJ =a 3 J from (14.1.10). Now from (14.1.13) we have (a tt + 2a t t +a tt ) +a 23 2 r @ r dJ ) = 0: (15.1.2) Apply (14.2.2) and multiply by a 3 2 a 3 1 tt +a 3 2 a t t + + 2 r @ r dJ ) = 0: (15.1.3) Next make a change of time variable by setting d dt = 1 a : 170 Then we can formulate (15.1.3) as a 3 3 +a 3 4 a + + 2 r @ r dJ ) = 0: (15.1.4) Note (r) 1 corresponds to ane motion. Introducing the perturbation (;r) :=(;r) 1; (15.1.5) equation (15.1.4) can be written in terms of a 3 3 +a 3 4 a + (1 +) + (1 +) 2 r @ r dJ = 0: (15.1.6) Finally apply (14.2.3) and note (1 +) (1 +) 2 =(1 +) to obtain a 3 3 +a 3 4 a (1 +) + (1 +) 2 r @ r d (J 1) = 0: (15.1.7) 15.2 The H Equation We derive the equation to be used in our estimates that will help us overcome the coordinate singularity near the origin r = 0. Divide (15.1.7) by a 3 3 +a 3 4 a (1 +) + (1 +) 2 r @ r ( d(J 1)) = 0: (15.2.1) Note that using the ane ODE r =@ r ( d) we can write 1 r@ r [( d)(J 1)] = 1 dr@ r (J )r 2 (J 1): (15.2.2) Now @ r (J ) = J 1 @ r J; (15.2.3) 171 and @ r J =@ r ( 2 ( + r r)) =@ r ( 2 (1 + + r r)) = 2 r rr + (2 2 + 2 r r) r + (2 r ) + 2 r = 2 r rr + (2 2 + 2 r r + 2) r + (2 r ) (15.2.4) where we recall = 1 (15.2.5) and hence r = r . Thus 1 dr@ r (J ) = 1 d 2 r 2 rr J 1 1 dr(2 2 + 2 r r + 2) r J 1 1 dr(2 r )J 1 : (15.2.6) Then from (15.2.2) 1 (r@ r )(( d)(J 1)) = 1 d 2 r 2 rr J 1 1 rd(2 2 + 2 r r + 2)J 1 r 1 dr(2 r )J 1 r 2 (J 1) = rJ 1 2 1 @ r ( d 1 r 2 @ r (r 3 ))r 2 ((J 1) + J 1 2 [r r + 3]) 1 rdJ 1 (2 r r + 2 2 2 ) r 1 dr2 r J 1 (15.2.7) We notice that 2 2 2 r = 2(1) r =2 r ; (15.2.8) 172 which leads to a cancellation on the last line of (15.2.7). Also J 1 = (J 1) + ((J 1) + (J 1)) (15.2.9) J 1 2 (r r + 3) = 2 (r r + 3) + (J 1 1) 2 (r@ r + 3): (15.2.10) Then (J 1) + J 1 2 (r r + 3) = (J 1) + 2 (r@ r + 3) + ((J 1) + (J 1)) + (J 1 1) 2 (r@ r + 3): (15.2.11) Now using J = 2 ( + r r) = (1 +) 2 (1 + + r r) = 1 + 1 r 2 r 3 + 2 + 3 3 r ; (15.2.12) we have J 1 = 1 r 2 r 3 + 2 + 3 3 r = 1 r 2 (3r 2 ( + 2 + 3 3 ) +r 3 ( r + 2 r + 3 2 r ) = 3 + 3 2 + 3 +r r + 2 r +r 2 r : (15.2.13) On the other hand, notice that 2 (r@ r + 3) = (1 +) 2 (r@ r + 3) =r@ r + 3 + 2r@ r + 2 r@ r + 3 3 + 6 2 : (15.2.14) Hence canceling terms (J 1) + 2 (r@ r + 3) = (3 2 2 3 ): (15.2.15) 173 Therefore we now have 2 r 2 (r@ r )( d(J 1)) = 4 J +1 r @ r ( d 1 r 2 @ r (r 3 )) 1 dJ 1 2 3 r r 2 ((J 1) + (J 1) (3 2 2 3 ) + (J 1 1) 2 (r@ r + 3)): (15.2.16) Rewrite the second line of (15.2.16) as follows 1 dJ 1 2 3 r r =2 1 d 3 J +1 r 2 (r@ r ) 2 (15.2.17) Then let R 1 [] :=2 1 d 3 J +1 r 2 (r@ r ) 2 (15.2.18) R 2 [] := 2 ((J 1) + (J 1) (3 2 2 3 ) + (J 1 1) 2 (r@ r + 3)): (15.2.19) Therefore (15.2.1) can be written as a 3 3 +a 3 4 a 4 J +1 r @ r ( d 1 r 2 @ r (r 3 ))(1+)+R 1 []+R 2 [] = 0: (15.2.20) Next let H :=r. So = H r . Then H solves a 3 3 H +a 3 4 a H 4 J +1 @ r ( d 1 r 2 @ r (r 2 H))H 1 + H r +rR 1 [ H r ] +rR 2 [ H r ] = 0; (15.2.21) with the initial conditions H(0;y) =H 0 (r); H (0;r) =@ H 0 (r); (r2 [0; 1]): (15.2.22) 174 15.3 Notation First introduce the radial equivalent of the three-dimensional divergence operator D r := 1 r 2 @ r (r 2 ) (15.3.1) To avoid singularities at r = 0 when applying high-order derivatives, dene D j := 8 > > < > > : (@ r D r ) j 2 if j is even D r (@ r D r ) j1 2 if j is odd (15.3.2) and setD 0 = 1. Also dene D i := 8 > > < > > : D 0 for i = 0 D i1 @ r for i 1 : (15.3.3) We will use the following elliptic operators to derive high-order equations L k f := 1 1+k( 1) d k @ r +k( 1) d 1+k D r f (15.3.4) L k h := 1 1+k( 1) d k D r +k( 1) d 1+k @ r h : (15.3.5) Then dene L j D j := 8 > > < > > : L j D j if j is even L j D j if j is odd : (15.3.6) Now for any k2Z 0 , we consider the weighted L 2 norm kfk 2 k := 1 0 d k (f) 2 r 2 dr (15.3.7) 175 The smooth cut-o function 0 such that = 1 on [0; 1 2 ], = 0 on [ 3 4 ; 1] and 0 0; (15.3.8) will be useful. The following vector elds will be important in obtaining high-order estimates successfully taking into account the coordinate singularity near the origin P 2j+2 := ( j+1 Y k=1 @ r V k :V k 2fD r ; 1 r g ) ; (15.3.9) P 2j+1 := ( V j+1 j Y k=1 @ r V k :V k 2fD r ; 1 r g ) ; (15.3.10) for j 0, and setP 0 =f1g. Also dene P 2j+2 :=fW@ r :W2P 2j+1 g P 2j+1 :=fW@ r :W2P 2j g; (15.3.11) for j 0, and setP 0 =f1g. 15.4 High-order Norm Our time weights will dier depending on whether 2 (1; 5 3 ] or > 5 3 . This is because we take a slightly dierent approach for > 5 3 by an adaptation of [48], applied to our spherically symmetric nonisentropic setting. On this note, introduce the following dependent exponents d( ) := 8 > > < > > : 3 3 if 1< 5 3 2 if > 5 3 ; b( ) :=d( ) + 3 3 = 8 > > < > > : 0 if 1< 5 3 5 3 if > 5 3 : (15.4.1) 176 LetN2N. To measure the size H, we dene the high-order weighted Sobolev norm as follows S N (H;H ) =S N () := sup 0 0 ( N X i=0 a d( ) kD i H k 2 i + N1 X i=0 kD i+1 Hk 2 i+1 + X i=N a b( ) kD i+1 Hk 2 i+1 ) : (15.4.2) 15.5 Main Theorem Before giving our main theorem, rst dene the important a() related quantities a 1 := lim !1 a () a() ; a 0 := d( ) 2 a 1 : (15.5.1) Local Well-Posedness. Next, we give the local well-posedness of our system. Theorem 15.5.1. Suppose > 1. Fix N 8. Let k N in (14.2.13). Then there are 0 > 0, 0 > 0 and T > 0 such that for every 2 (0; 0 ], 2 (0; 0 ] and pair of initial data for (15.2.21) (H 0 ;@ H 0 ) satisfyingS N (H 0 ;@ H 0 ) andkH(0)k 2 0 , there exists a unique solution (H();H ()) : [0; 1]!RR to (15.2.21)-(15.2.22) for all 2 [0;T ]. The solution has the propertyS N (H;H ). for each 2 [0;T ]. Furthermore, the map [0;T ]37!S N ()2R + is continuous. We give the sketch of the proof of Theorem 15.5.1 in Appendix 18.4. A priori assumptions. Finally before our main theorem, make the following a priori assumptions on our local solutions from Theorem 15.5.1 S N ()< 1 3 ; (15.5.2) jJ 1j< 1 3 ; (15.5.3) j@ r ()j =j@ r H r j< 1 3 ; (15.5.4) j@ 2 r ()j =j@ 2 r H r j< 1 3 : (15.5.5) 177 Remark 15.5.2. UsingJ = 2 ( + r r) and r = r , (15.5.3)-(15.5.4) imply 1.jj. 1: (15.5.6) Furthermore, usingJ r = ( 3 + 2 r r) r and r = r , (15.5.4)-(15.5.6) imply 1.jJ r j. 1: (15.5.7) We are now ready to give our main theorem. Theorem 15.5.3. Suppose > 1. Fix N 8. Let kN. Consider a xed quadruple (a(0);a 0 (0);)2R + RZ k ; (15.5.8) parametrizing a nonisentropic ane motion from the set S . Then there are 0 > 0 and 0 > 0 such that for every 2 (0; 0 ], 2 (0; 0 ] and pair of initial data for (15.2.21) (H 0 ;@ H 0 ) satisfyingS N (H 0 ;@ H 0 ) andkH(0)k 2 0 , there exists a global-in-time solution, (H;H ), to the initial value problem (15.2.21)-(15.2.22) and a constant C > 0 such that S N (H;H )()C(" +); 0 <1: (15.5.9) We believe Theorem 15.5.3 is the rst global existence result for the positive density vacuum boundary Euler system. Henceforth we assume we are working with a unique local solution (H;H ) : [0; 1]! RR to (15.2.21)-(15.2.22) such thatS N (H) <1 on [0;T ] with T > 0 xed: Theorem 15.5.1 ensures the existence of such a solution, and furthermore we assume this local solution satises the a priori assumptions (15.5.2)-(15.5.5). To prove our main result, we apply weighted energy estimates. A similar methodology to [17] in conjunction with a weighted energy estimate method similar to that used in [44] allows us to simultaneously handle the coordinate singularity near the origin, the exponentially 178 growing-in-time coecients and the vacuum boundary. Firstly the particular choice of derivative operators in combination with the introduction of special vector eld classes, see Section 15.3, allows us to circumvent the coordinate singularity at r = 0. Secondly the exponentially growing time weights take advantage of the stabilizing eect of the expanding background ane motion. Finally the increase in spatial weight d in accordance with an increase in derivatives will be essential in avoiding potentially dangerous negative powers of d near the boundary. It is worth noting that the number of derivatives required to close estimates and prove the main theorem here does not depend on . This is because the weight structure involving d does not depend on . This contrasts to previous free boundary works using weighted estimates with a vanishing density, see [20,44,48] for example. Coercivity estimates are employed to account for the fact that our equation structure does not include a zeroth order contribution of H which is seen in our denition ofS N . Furthermore, we also use coercivity estimates to obtain results for all > 1 because of the time weight manipulation necessary for > 5 3 , see Section 16.2. Immediately below in Chapter 16 we prove our high order energy estimates and in Chapter 17 we prove our Main Theorem 15.5.3 by means of a continuity argument. 179 Chapter 16 Energy Estimates 16.1 Dierentiated Equation Let 1in. ApplyD i to our equation for H (15.2.21) a 3 3 D i H +a 3 4 a D i H D i HD i H 2 r D i 4 J +1 @ r ( d 1 r 2 @ r (r 2 H)) +D i rR 1 [ H r ] +rR 2 [ H r ] = 0: (16.1.1) Using the product rule forD i Lemma 18.1.1, we compute D i 4 J +1 @ r ( d 1 r 2 @ r (r 2 H)) =D i 1 @ r ( d 1 r 2 @ r (r 2 H)) 4 J +1 + D i1 @ r 4 J +1 1 @ r ( d 1 r 2 @ r (r 2 H)) + D i1 4 J +1 D r 1 @ r ( d 1 r 2 @ r (r 2 H)) 4 J +1 D i1 D r 1 @ r ( d 1 r 2 @ r (r 2 H)) : (16.1.2) The last line of (16.1.2) can be written using the commutator as follows [ D i1 ; 4 J +1 ]D r 1 @ r ( d 1 r 2 @ r (r 2 H)) : (16.1.3) 180 Next we computeD i h 1 @ r ( d 1 r 2 @ r (r 2 H)) i . To this end, introduce L k f := 1 1+k( 1) d k @ r +k( 1) d 1+k D r f (16.1.4) L k h := 1 1+k( 1) d k D r +k( 1) d 1+k @ r h (16.1.5) L j D j := 8 > > < > > : L j D j if j is even L j D j if j is odd : (16.1.6) Recalling the notation L k (15.3.4) and L k (15.3.5), rst D r L k f =L 1+k D r f + [( +k( 1))D r ( 2 r d) + (1 +k)D r (d r 1 )]D r f =L 1+k D r f +Q + D r f (16.1.7) where we dene Q + := 1 r ((2( +k( 1)) 2 r d + (2(1 +k) 1 )d r )+ + (( +k( 1))(( 2) 3 2 r + 2 rr )d + (2 + 2k)( 1) 2 r d r + (1 +k) 1 d rr : (16.1.8) Also @ r L k h =L 1+k @ r h + " @ r @ r ( +k( 1) d 1+k ) 1+k( 1) d k ! 2 r @ r ( +k( 1) d 1+k ) 1+k( 1) d k # @ r h =L 1+k @ r h +Q @ r h: (16.1.9) where we dene Q := ( +k( 1))(( 2) 3 2 r d + 2 rr d + 2 r d r )+ + (1 +k)(d rr 1 + ( 1) 2 r d r ) 2 r (( +k( 1)) 2 r d + (1 +k) 1 d r ): (16.1.10) 181 Then using the commutation rule forD i L 0 Lemma 18.1.2 D i 1 @ r ( d 1 r 2 @ r (r 2 H)) =D i L 0 H =L i D i H + i1 X j=0 q ij D ij H; (16.1.11) where q ij = 2+j X k=1 P k `=0 c ijk` @ ` r d r 2+jk ; (16.1.12) and c ijk` are bounded functions on [0; 1]. Now returning to (16.1.1), we have a 3 3 D i H +a 3 4 a D i H D i HD i H 2 r 4 J +1 L i D i H 4 J +1 i1 X j=0 q ij D ij H D i1 @ r 4 J +1 L 0 H [ D i1 ; 4 J +1 ]D r L 0 H +D i rR 1 [ H r ] +rR 2 [ H r ] = 0: (16.1.13) Let C i [H] := 4 J +1 i1 X j=0 q ij D ij H + [ D i1 ; 4 J +1 ]D r L 0 H: (16.1.14) Then we write (16.1.13) as follows a 3 3 D i H +a 3 4 a D i H 4 J +1 L i D i H =D i H +D i H 2 r D i rR 1 [ H r ] +rR 2 [ H r ] + C i [H] + D i1 @ r 4 J +1 L 0 H : (16.1.15) 182 16.2 Time Weight Manipulation For > 5 3 we need to eliminate our equivalent of the anti-damping eect encountered in [20]. We use the strategy from [48], applied to our spherically symmetric positive density nonisentropic setting. First recall our dependent exponents d( ) = 8 > > < > > : 3 3 if 1< 5 3 2 if > 5 3 ; b( ) =d( ) + 3 3 = 8 > > < > > : 0 if 1< 5 3 5 3 if > 5 3 : (16.2.1) We then multiply (16.1.15) by a b( ) to obtain a d( ) D i H +a d( )1 a D i H a b( ) 4 J +1 L i D i H =a b( ) D i H +a b( ) D i H 2 r a b( ) D i rR 1 [ H r ] +rR 2 [ H r ] + a b( ) C i [H] + a b( ) D i1 @ r 4 J +1 L 0 H : (16.2.2) 16.3 Energy Identity Multiply (16.2.2) by r 2 d i D i H and integrate in r from 0 to 1 1 0 a d( ) D i H D i H d i r 2 dr + 1 0 a d( )1 a (D i H ) 2 d i r 2 dr a b( ) 8 > > < > > : 1 0 4 1+k( 1) J +1 @ r +i( 1) d 1+i D r D i H r 2 D i H if i is even 1 0 4 1+k( 1) J +1 D r +i( 1) d 1+i @ r D i H r 2 D i H if i is odd = 1 0 a b( ) D i HD i H r 2 d i dr + 1 0 a b( ) D i H 2 r D i H r 2 d i dr 1 0 a b( ) D i rR 1 [ H r ] +rR 2 [ H r ] D i H r 2 d i dr + 1 0 a b( ) C i [H]r 2 d i D i H dr + 1 0 a b( ) D i1 @ r 4 J +1 L 0 H r 2 d i D i H dr: (16.3.1) 183 We rewrite the third term on the left hand side of (16.3.1) and obtain 1 0 a d( ) D i H D i H d i r 2 dr + 1 0 a d( )1 a (D i H ) 2 d i r 2 dr + 1 0 a b( ) 4 J +1 D i+1 HD i+1 H r 2 1 d i+1 dr = 1 0 a b( ) D i HD i H r 2 d i dr + 1 0 a b( ) D i H 2 r D i H r 2 d i dr 1 0 a b( ) +i( 1) d 1+i D i+1 HD i H r 2 @ r 4 1+i( 1) J +1 dr 1 0 a b( ) D i rR 1 [ H r ] +rR 2 [ H r ] D i H r 2 d i dr + 1 0 a b( ) C i [H]r 2 d i D i H dr + 1 0 a b( ) D i1 @ r 4 J +1 L 0 H r 2 d i D i H dr: (16.3.2) Writing (16.3.2) using perfect time derivatives d d 1 2 a d( ) 1 0 (D i H ) 2 d i r 2 dr + a b( ) 2 1 0 4 J +1 (D i+1 H) 2 1 d i+1 r 2 dr ! + 2d( ) 2 a d( )1 a 1 0 (D i H ) 2 d i r 2 dr b( ) 2 a b( )1 a 1 0 4 J +1 (D i+1 H) 2 1 d i+1 r 2 dr =a b( ) 1 0 D i HD i H r 2 d i dr +a b( ) 1 0 D i H 2 r D i H r 2 d i dr + a b( ) 2 1 0 @ 4 J +1 (D i+1 H) 2 1 d i+1 r 2 dr a b( ) 1 0 +i( 1) d i+1 D i+1 HD i H r 2 @ r 4 1+i( 1) J +1 dr 1 0 a b( ) D i rR 1 [ H r ] +rR 2 [ H r ] D i H r 2 d i dr + 1 0 a b( ) C i [H]r 2 d i D i H dr + 1 0 a b( ) D i1 @ r 4 J +1 L 0 H r 2 d i D i H dr: (16.3.3) Remark 16.3.1. The second line of the left hand side of (16.3.3) is nonnegative for both 2 (1; 5 3 ] and > 5 3 where we note 2d( ) = 0 for > 5 3 and b( ) = 0 for 2 (1; 5 3 ]. 184 16.4 High-order Quantities from Energy Identity We now dene our high order quantities. First from (16.3.3), dene E i = 1 2 a d( ) 1 0 (D i H ) 2 d i r 2 dr + a b( ) 2 1 0 4 J +1 (D i+1 H) 2 1 d i+1 r 2 dr D i = 2d( ) 2 a d( )1 a 1 0 (D i H ) 2 d i r 2 dr b( ) 2 a b( )1 a 1 0 4 J +1 (D i+1 H) 2 1 d i+1 r 2 dr R i = 1 0 a b( ) D i HD i H r 2 d i dr + 1 0 a b( ) D i H 2 r D i H r 2 d i dr + 2 1 0 a b( ) @ 4 J +1 (D i+1 H) 2 1 d i+1 r 2 dr 1 0 a b( ) +i( 1) d 1+i D i+1 HD i H r 2 @ r 4 1+i( 1) J +1 dr 1 0 a b( ) D i rR 1 [ H r ] +rR 2 [ H r ] D i H r 2 d i dr + 1 0 a b( ) C i [H]r 2 d i D i H dr + 1 0 a b( ) D i1 @ r 4 J +1 L 0 H r 2 d i D i H dr: (16.4.1) Then let E N () = N X i=0 E i () (16.4.2) D N () = N X i=0 D i () (16.4.3) R i =Z i 1 +Z i 2 +Z i 3 +Z i 4 +Z i 5 +Z i 6 +Z i 7 : (16.4.4) where we will show in the zeroth order estimate Z 0 6 =Z 0 7 = 0. Therefore from (16.3.3), we have d d (E N ()) +D N () = N X i=0 R i = N X i=0 (Z i 1 +Z i 2 +Z i 3 +Z i 4 +Z i 5 +Z i 6 +Z i 7 ): (16.4.5) 185 16.5 Main Energy Inequality To establish our central energy inequality, we primarily need to estimate the right hand of (16.4), that is, estimate Z i j for all j = 1; 2; 3; 4; 5; 6; 7 and all 0iN. Before this, we introduce a term only present for > 5 3 which is similar toE i but does not include top order quantities or time weights with negative powers, and will be controlled through our coercivity Lemma 18.5.1 C i; = 1 >5=3 1 0 4 2J +1 (D i+1 H) 2 1 d i+1 r 2 dr; (16.5.1) where 1 >5=3 = 8 > > < > > : 1 if > 5 3 0 if 1< 5 3 . Then let C N1 () = N1 X i=0 C i; (): (16.5.2) Remark 16.5.1. Our coercivity Lemma 18.5.1 given in Appendix 18.5 will let us control C N1 . However we cannot use Lemma 18.5.1 to handle top order terms as that would require control of N + 1 derivatives of @ H. This is why we do not include top order terms inC N1 . Finally, prior to proving our main energy inequality, it is worth formally stating the equivalence of our high order normS N and high order energy functionalE N . Lemma 16.5.2. Let (H;H ) : [0; 1] ! RR be a unique local solution to (15.2.21)- (15.2.22) on [0;T ] for T > 0 xed withkH(0)k 2 0 <1 and assume (H;H ) satises the a 186 priori assumptions (15.5.2)-(15.5.5). Fix N 8. Let kN in (14.2.13). Then there are constants C 1 ;C 2 > 0 so that C 1 S N () sup 0 0 fE N ( 0 ) +C N1 ( 0 )gC 2 (S N () +S N (0)): (16.5.3) Proof. Recall the denitionS N (15.4.2). Then (16.5.3) is a straightforward application of the uniform boundedness of J (15.5.3), (15.5.6) and (14.2.10), in conjunction with Lemma 18.5.1 to control terms without time weights with negative powers, which are included inC N1 ( 0 ), byS N () +S N (0). We are now ready to prove our central high order energy inequality which will be essential in the proof of our main result Theorem 15.5.3. Proposition 16.5.3. Suppose > 1. Let (H;H ) : [0; 1] ! RR be a unique local solution to (15.2.21)-(15.2.22) on [0;T ] for T > 0 xed withkH(0)k 2 0 <1 and assume (H;H ) satises the a priori assumptions (15.5.2)-(15.5.5). Fix N 8. Let k N in (14.2.13). Then for all 2 [0;T ], we have the following inequality for some 0< 1 E N () +C N1 () + 0 D N ( 0 )d 0 .S N (0) +kH(0)k 2 0 +S N () + 0 e a 0 0 S N ( 0 )d 0 : (16.5.4) Proof. We integrate our energy identity written in terms of our high order quantities (16.4.5) from 0 to , and for > 5 3 apply Lemma 18.5.1, to obtain the left hand side of (16.5.4). Our goal is then to estimatejZ i j j for all j = 1; 2; 3; 4; 5; 6; 7 and all 0iN. j = 1: Fix i 1. Suppose i =` + 1 for 0`N 1. Then jZ i 1 j =ja b( ) 1 0 D i HD i H r 2 d i drj .a b( ) ( 1 0 (D i H) 2 r 2 d i dr) 1=2 ( 1 0 (D i H ) 2 r 2 d i dr) 1=2 =a b( ) ( 1 0 (D `+1 H) 2 r 2 d `+1 dr) 1=2 (e d( ) e d( ) 1 0 (D i H ) 2 r 2 d i dr) 1=2 . (S N ) 1=2 e a 0 (S N ) 1=2 =e a 0 S N (): (16.5.5) 187 For i = 0, we rst need to derive the energy identity (16.3.3) for i = 0. Multiplying our equation for H (15.2.21) by a b( ) we have a d( ) H +a d( )1 a H a b( ) 4 J +1 L 0 H =a b( ) H+a b( ) H 2 r a b( ) (rR 1 [ H r ]+rR 2 [ H r ]+rR 3 [ H r ]): (16.5.6) Multiply (16.5.6) by r 2 H and integrate in r from 0 to 1 1 0 a d( ) H H r 2 dr + 1 0 a d( )1 a (H ) 2 r 2 dr 1 0 a b( ) 4 J +1 @ r ( dD r H)H r 2 dr = 1 0 a b( ) HH r 2 dr + 1 0 a b( ) H 2 r H r 2 dr 1 0 a b( ) (rR 1 [ H r ] +rR 2 [ H r ] +rR 3 [ H r ])H r 2 dr: (16.5.7) Rewriting the third term of the left hand side of (16.5.7) we then have 1 0 a d( ) H H r 2 dr + 1 0 a d( )1 a (H ) 2 r 2 dr + 1 0 a b( ) 4 J +1 D 1 HD 1 H r 2 1 ddr = 1 0 a b( ) HH r 2 dr + 1 0 a b( ) H 2 r H r 2 dr 1 0 a b( ) dD 1 H@ r 4 J +1 H r 2 dr 1 0 a b( ) (rR 1 [ H r ] +rR 2 [ H r ] +rR 3 [ H r ])H r 2 dr: (16.5.8) Writing (16.5.8) using perfect time derivatives d d 1 2 a d( ) 1 0 (H ) 2 r 2 dr + 2 a b( ) 1 0 4 J +1 (D 1 H) 2 1 dr 2 dr + 2d( ) 2 a d( )1 a 1 0 (H ) 2 r 2 dr b( ) 2 a b( )1 a 1 0 4 J +1 (D 1 H) 2 1 dr 2 dr =a b( ) 1 0 HH r 2 dr +a b( ) 1 0 H 2 r H r 2 dr + 2 a b( ) 1 0 @ 4 J +1 (D 1 H) 2 1 dr 2 dr a b( ) 1 0 dD 1 H@ r 4 J +1 H r 2 dra b( ) 1 0 (rR 1 [ H r ] +rR 2 [ H r ] +rR 3 [ H r ])H r 2 dr: (16.5.9) 188 Therefore in terms of our high order quantities at the zeroth order level, we have d d (E 0 ()) +D 0 () =R 0 =Z 0 1 +Z 0 2 +Z 0 3 +Z 0 4 +Z 0 5 +Z 0 6 +Z 0 7 ; (16.5.10) where Z 0 6 =Z 0 7 = 0. Returning to our j = 1 estimate, that is ourjZ i 1 j estimate, for i = 0 using Young's inequality with 0< 1 we have 0 Z 1 0 d 0 = 0 a b( ) 1 0 HH r 2 drd 0 = 0 1 0 a b( ) a d( ) 4 Ha d( ) 4 H r 2 drd 0 . 0 1 0 a d( ) 2 H 2 r 2 drd 0 + 0 1 0 a d( ) 2 H 2 r 2 drd 0 . sup 0 0 f 1 0 H 2 r 2 drg 0 a d( ) 2 d 0 + 0 1 0 a d( ) 2 a d( ) H 2 r 2 drd 0 . sup 0 0 f 1 0 H 2 r 2 drg + 0 e a 0 0 S N ( 0 )d 0 : (16.5.11) For 1 0 H 2 r 2 dr apply a coercivity estimate as follows H = 0 H d 0 +H(0) = 0 a 3 3 2 a 3 3 2 H d 0 +H(0). sup 0 0 a 3 3 2 H +H(0); (16.5.12) and hence using Cauchy's inequality (ab.a 2 +b 2 ; a;b2R) we have 1 0 (H) 2 r 2 dr.S N () +kH(0)k 2 0 : (16.5.13) Therefore 0 Z 0 1 d 0 .S N () +kH(0)k 2 0 + 0 e a 0 0 S N ( 0 ): (16.5.14) 189 j = 2: Fix 0iN. Then with 0 a smooth cut-o function such that = 1 on [0; 1 2 ], = 0 on [ 3 4 ; 1] and 0 0, jZ i 2 j = a b( ) 1 0 D i H 2 r D i H r 2 d i dr = a b( ) 3 4 0 D i H 2 r D i H r 2 d i dr +a b( ) 1 1 2 D i H 2 r D i H r 2 d i (1 )dr . 3 4 0 D i H 2 r D i H r 2 d i dr + 1 1 2 D i H 2 r D i H r 2 d i (1 )dr : (16.5.15) Next recalling the denition of the vector eldP i given in Section 15.3, rst noteD i 2P i . Then by the product rule forP i Lemma 18.1.3 D i H 2 r = i X k=0 X A2P k B2 P ik c D i AB k A[H]B H r ; (16.5.16) for some real valued constants c D i BC k . Now note for B2 P ik , B H r = C[H] for C 2 P i+1k . Then we can write in terms of low order and high order derivatives D i H 2 r = i=2 X k=0 X A2P k C2P i+1k c D i AC k A[H]C[H]: (16.5.17) Hence applying Lemma 18.1.6 to estimateP i+1k usingD i+1k , ourL 1 embedding forP k (18.2.4) and the fact that 1.d. 1 on [0; 3 4 ], we have for the left hand integral on the last line of (16.5.15) 3 4 0 D i H r D i H r 2 d i dr . i=2 X k=0 X A2P k C2P i+1k kA(H)k L 1 3 4 0 C[H]D i H r 2 d i dr . (S N () +kH(0)k 2 ) i=2 X k=0 X C2P i+1k 3 4 0 (C[H]) 2 r 2 d i 2 dr ! 1=2 3 4 0 2 (D i H ) 2 r 2 d i dr ! 1=2 . i=2 X k=0 X C2P i+1k 3 4 0 (C[H]) 2 r 2 2 dr ! 1=2 1 0 (D i H ) 2 r 2 d i dr 1=2 190 . i=2 X k=0 3 4 0 (D i+1k H) 2 r 2 2 dr ! 1=2 e a 0 (S N ()) 1 2 . i=2 X k=0 3 4 0 (D i+1k H) 2 r 2 2 d i+1 dr ! 1=2 e a 0 (S N ()) 1 2 . (S N ()) 1 2 e a 0 (S N ()) 1 2 .e a 0 S N (): (16.5.18) For the second integral on the right hand side of (16.5.15), also use P k u = k X `=0 c ` (r)D ` (H); (16.5.19) where c ` are smooth functions of r on [ 1 4 ; 1] in this case to obtain 1 1 2 D i H 2 r D i H r 2 d i (1 )dr . i=2 X k=0 X A2P k C2P i+1k kA(H)k L 1 1 1 2 C[H]D i H r 2 d i (1 )dr ! . (S N () +kH(0)k 2 ) i=2 X k=0 i+1k X `=0 1 1 2 D ` HD i H r 2 d i (1 )dr . i=2 X k=0 i+1k X `=0 ( 1 1 2 a d( ) 2 (D ` H) 2 r 2 d i dr) + 1 1 2 a d( ) 2 (D i H ) 2 r 2 d i dr .a d( ) 2 S N () +a d( ) 2 kH(0)k 2 +e a 0 S N (): (16.5.20) Therefore combining the above analysis 0 Z i 2 d 0 .S N () +kH(0)k 2 0 + 0 e a 0 0 S N ( 0 )d 0 : (16.5.21) j = 3: For Z i 3 = 2 1 0 @ 4 J +1 (D i+1 H) 2 1 d i+1 r 2 dr, rst compute: @ ( 4 J ( +1) ) = 4 3 (@ )J ( +1) ( + 1)J ( +2) (@ J ) 4 : (16.5.22) 191 Now using = + 1 @ =@ =@ ( 1 r H) = 1 r H ; @ J =@ ( 3 + 2 r r); @ ( r ) =@ r ( ) =@ r H r ; r@ ( r ) =r@ r ( H r ) =r( 1 r 2 H + 1 r @ r H ) =@ r H 1 r H =D 1 H 3 r H : (16.5.23) Hence Z i 3 = 2 1 0 (4 3 J ( +1) ) H r ( + 1) 4 J ( +2) (3 2 H r + 2H +D 1 H 3 H r ) (D i+1 H) 2 1 d i+1 r 2 dr . 1 0 H r (D i+1 H) 2 d i+1 r 2 dr + 1 0 H (D i+1 H) 2 d i+1 r 2 dr + 1 0 (D 1 H )(D i+1 H) 2 d i+1 r 2 dr . (k H r k 1 +kH k 1 +kD 1 H k 1 ) 1 0 (D i+1 H) 2 d i+1 r 2 dr: (16.5.24) Applying the L 1 embedding (18.2.3) k H r k 1 . 3 X k=2 3 4 0 (D k H ) 2 r 2 dr + 3 X k=2 1 1 4 d 2 (D k H ) 2 dr ! 1 2 . 3 X k=2 3 4 0 (D k H ) 2 r 2 d k dr + 2 X k=0 1 1 4 r 2 d 2 (D k H ) 2 dr ! 1 2 . 3 X k=2 1 0 (D k H ) 2 r 2 d k dr + 2 X k=0 1 0 r 2 d 2 (D k H ) 2 dr ! 1 2 .e a 0 (S N ) 1=2 : (16.5.25) Applying the L 1 embedding (18.2.1) kH k 1 . 3 X k=2 3 4 0 (D k H ) 2 r 2 dr + 2 X k=0 1 1 4 d 2 (D k H ) 2 dr ! 1 2 192 . 3 X k=2 3 4 0 (D k H ) 2 r 2 d k dr + 2 X k=0 1 1 4 r 2 d 2 (D k H ) 2 dr ! 1 2 . 3 X k=2 1 0 (D k H ) 2 r 2 d k dr + 2 X k=0 1 0 r 2 d 2 (D k H ) 2 dr ! 1 2 .e a 0 (S N ) 1=2 : (16.5.26) Applying the L 1 embedding (18.2.2) and using that D k D 1 =D k+1 kD 1 H k 1 . 2 X k=1 3 4 0 ( D k D 1 H ) 2 r 2 dr + 3 X k=0 1 1 4 d 4 ( D k D 1 H ) 2 dr ! 1 2 . 2 X k=1 3 4 0 (D k+1 H ) 2 r 2 d k+1 dr + 3 X k=0 1 1 4 r 2 d k+1 (D k+1 H ) 2 dr ! 1 2 . 2 X k=1 1 0 (D k+1 H ) 2 r 2 d k+1 dr + 3 X k=0 1 0 r 2 d k+1 (D k H ) 2 dr ! 1 2 .e a 0 (S N ) 1=2 : (16.5.27) Then since 1 0 (D i+1 H) 2 d i+1 r 2 dr.S N from (16.5.24) we have jZ i 3 j.e a 0 (S N ) 3 2 .e a 0 S N : (16.5.28) j = 4: For Z i 4 = 1 0 a b( ) +i( 1) d 1+i D i+1 HD i H r 2 @ r 4 1+i( 1) J +1 dr rst note j@ r ( 4 J ( +1) (1+i( 1)) j =j4 3 r J ( +1) (1+i( 1)) ( + 1)J ( +2) J r (1+i( 1) (1 +i( 1)) (2+i( 1)) r 4 J ( +1) j . 1: (16.5.29) Then jZ i 4 j. 1 0 d 1+i D i+1 HD i H r 2 dr 193 . ( 1 0 d 1+i (D i+1 H) 2 r 2 dr) 1=2 ( 1 0 d i (D i H ) 2 r 2 dr) 1=2 .e a 0 (S N ) 1=2 (S N ) 1=2 =e a 0 S N : (16.5.30) j = 5: Recall Z i 5 = 1 0 a b( ) D i rR 1 [ H r ] +rR 2 [ H r ] D i H r 2 d i dr: (16.5.31) We start by considering 1 0 a b( ) D i rR 2 [ H r ] D i H r 2 d i dr = 1 0 D i1 D r [rR 2 [ H r ]] D i H r 2 d i dr: (16.5.32) sinceR 2 requires more care. We combine this term with the j = 7, Z i 7 , term which is as follows 1 0 a b( ) D i1 @ r 4 J +1 L 0 H r 2 d i D i H dr (16.5.33) to obtain a b( ) 1 0 D i1 @ r ( 4 J +1 )L 0 HD r [rR 2 [ H r ]] D i H r 2 d i dr :=Z i R 2 57 for i 1: (16.5.34) We note for i = 0, we have Z 0 R 2 57 := 1 0 rR 3 [ H r ]H r 2 dr. We examine the structure of Z i R 2 57 for i 1. Note @ r ( 4 J +1 )L 0 HD r [rR 2 [ H r ]] = @ r (J 1 r(D r H) 4 +@ r (rR 2 [ H r ]) + @ r ( 4 )J 1 L 0 H + 4 1 d(@ r D r H)@ r (J 1 ) 2 r (rR 2 [ H r ]): (16.5.35) Next note R 2 [] =R a 2 [] +R b 2 [] where R a 2 [] := 2 ((J 1) + (J 1) + (J 1 1) 2 (r@ r + 3)); R b 2 [] := 2 (3 2 + 2 3 ): (16.5.36) 194 Now @ r (rR a 2 []) =@ r [r 2 ((J 1) + (J 1) + (J 1 1) 2 (r@ r + 3))] =@ r (r 2 )((J 1) + (J 1) + (J 1 1) 2 (r@ r + 3)) r 2 @ r (J 1 + (J 1) + (J 1 1) 2 (r@ r + 3) : (16.5.37) Using @ r (J 1) = J 1 @ r J we have @ r (J 1 + (J 1) + (J 1 1) 2 (r@ r + 3) = J 1 @ r J + @ r J ( ( + 1)J 2 @ r J ) 2 (r@ r + 3) + (J 1 1)@ r ( 2 (r@ r + 3)) = ( + 1)J 2 (@ r J ) 2 (r@ r + 3) + (J 1 1)@ r ( 2 (r@ r + 3)) + ( J 1 )@ r J = ( + 1)J 2 (@ r J ) 2 (r@ r + 3) (J 1 1)(@ r J 2 (r@ 2 r + 4@ r )) + 2 (J 1 1) r (r@ r + 3): (16.5.38) Thus using r@ r ( H r ) + 3 H r =D r H we can write @ r (rR a 2 []) =r 2 ( + 1)J 2 (@ r J ) 2 D r H +R 3 []; (16.5.39) where we dene R 3 [] =R 3 [ H r ] := r 2 (J 1 1)(@ r J 2 (r@ 2 r + 4@ r )) 2r 2 (J 1 1) r (r@ r + 3) @ r (r 2 )[(J 1) + (J 1) + (J 1 1) 2 (r@ r + 3)]: (16.5.40) 195 Returning to (16.5.35), noting @ r (J 1) = J 1 @ r J , we can then cancel the unfavorable with respect to weight d term ( + 1)J 2 (@ r J )rD r H 4 and obtain @ r ( 4 J +1 )L 0 HD r [rR 2 [ H r ]] = r@ r ( 4 )J 1 D r H + @ r ( 4 ) 1 d(@ r D r H)J 1 + 4 1 d(@ r D r H)@ r J 1 2 r (rR 2 [ H r ])R 3 [ H r ]@ r [rR b 2 [ H r ]] = r@ r ( 4 )J 1 D r H + 1 d@ r ( 4 J +1 )(@ r D r H)R 3 [ H r ] 2 r (rR 2 [ H r ])@ r [rR b 2 [ H r ]]: (16.5.41) In terms of derivative count with respect to weight d, the only potentially concerning com- ponent above is r 2 (J 1 1)(@ r J 2 (r@ 2 r + 4@ r )) inR 3 [ H r ] =R 3 []. Noting J = 3 + 2 r r we rewrite this term in a favorable form using the following identity @ r J 2 (r@ 2 r + 4@ r ) = 2(@ r ) 2 r: (16.5.42) Therefore we have written Z i R 2 57 for i 1 in a desirable form. We now estimate Z i R 2 57 for i = 0 and then for i 1. For i = 0, note Z 0 R 2 57 =a b( ) 1 0 rR 2 []H r 2 dr =a b( ) 1 0 r( 2 ((J 1 1) + (J 1))H r 2 dr 1 0 a b( ) r( 2 ((J 1 1) 2 (r@ r + 3) +a b( ) (3 2 + 2 3 ))H r 2 dr = (i) + (ii): (16.5.43) For (i), we use Taylor series to write J 1 + (J 1) = ( + 1) 1 0 (1s)(1 +s(J 1)) 2 ds (J 1) 2 : (16.5.44) 196 Now (J 1) 2 = ((1 +) 2 (1 + + r r) 1) 2 = ((1 + H r ) 2 (1 +H r ) 1) 2 = ((1 + 2H r + H 2 r 2 )(1 +H r ) 1) 2 = (1 +H r + 2 r H + 2 r HH r + H 2 r 2 + H 2 Hr r 1) 2 = (D r H + 2 r H(D r H 2 H r ) + H 2 r 2 + H 2 r (D r H 2 r H)) 2 = (D r H +D r H 2 r H +D r H H 2 r 3 r 2 H 2 2H 3 r 2 ) 2 = (D r H) 2 + 4H(DrH) 2 r + 2H 2 (DrH) 2 r 6H 2 DrH r + 4H 2 (DrH) 2 r 2 4H 3 (DrH) 2 r 2 + + H 4 (DrH) 2 r 2 12H 3 (DrH) r 3 14H 4 DrH r 3 4H 5 DrH r 3 + 9H 4 r 4 + 12H 5 r 4 + 4H 6 r 4 : (16.5.45) First using (16.5.44) and the boundedness of 2 ( +1) 1 0 (1s)(1 +s(J 1)) 2 ds , we have for integral (i) j(i)j. 1 0 r 3 (J 1) 2 H dr : (16.5.46) As can be seen from (16.5.45) many contributions from (J 1) 2 are similar. Therefore we give the key estimates below and remark similar arguments will hold for the other terms. For the (D r H) 2 term in (16.5.45), apply the L 1 embedding (18.2.1) k 1 0 r 3 (D r H) 2 H drj.kH k 1 S N .e a 0 S N : (16.5.47) For the 2H 2 (DrH) 2 r term in (16.5.45), use theL 1 embedding (18.2.1) and a similar argument to that used for Z 1 0 1 0 r 2 H 2 (D r H) 2 H dr .kD 1 Hk 2 1 a d( ) 2 1 0 r 2 H 4 dr +a d d( ) 2 1 0 r 2 (H ) 2 dr .S N (a d( ) 2 kHk 2 1 1 0 r 2 H 2 dr +e a 0 S N ) .a d( ) 2 S N +a d( ) 2 kH(0)k 2 0 +e a 0 S N ; (16.5.48) 197 and therefore 0 1 0 r 3 H 2 (D r H) 2 H drd 0 .S N +kH(0)k 2 0 + 0 e a 0 0 S N d 0 : (16.5.49) For integral (ii) j(ii)j. 1 0 r 3 D 1 HH dr + 1 0 rH 2 H dr + 1 0 H 3 H dr . (S N ) 1=2 e a 0 (S N ) 1=2 +a d( ) 2 (k H r k 2 1 +k H r k 4 1 ) 1 0 H 2 r 2 dr +e a 0 S N : (16.5.50) Therefore 0 (ii)d 0 .e a 0 S N +S N +kH(0)k 2 0 + 0 e a 0 0 S N d 0 : (16.5.51) Thus we have 0 jZ 0 R 2 57 jd 0 .S N +kH(0)k 2 0 + 0 e a 0 0 S N d 0 : (16.5.52) Recalling R 3 [] =r 2 (J 1 1)(2( r ) 2 r) 2r 2 (J 1 1) r (r@ r + 3) @ r (r 2 )[(J 1) + (J 1) + (J 1 1) 2 (r@ r + 3)]; (16.5.53) note for i 1 Z i R 2 57 =a b( ) 1 0 D i1 h r@ r ( 4 )J 1 D r H + 1 d@ r ( 4 J +1 )(@ r D r H)R 3 [] 2 r (rR 2 [ H 4 ])@ r [rR b 2 [ H r ]] i D i H r 2 d i dr =a b( ) 1 0 D i1 [ r@ r ( 4 )J 1 D r H] + D i1 [ 1 d@ r ( 4 J +1 )(@ r D r H)] D i1 [R 3 []] D i1 (2R 2 [ H r ]) D i1 [@ r (rR b 2 ( H r )] D i H r 2 d i dr 198 = 1 0 a b( ) D i1 [ r@ r ( 4 )J 1 D r H]D i H d i r 2 dr + 1 0 a b( ) D i1 [ 1 d@ r ( 4 J +1 )(@ r D r H)] D i d i r 2 dr 1 0 a b( ) D i1 [R 3 []]DH d i r 2 dr 1 0 a b( ) D i1 (2R 2 [ H r ]) D i d i r 2 dr 1 0 a b( ) D i1 (@ r (rR b 2 [ H r ]))D i H d i r 2 dr = (a) + (b) + (c) + (d) + (e): (16.5.54) For (a) rst rewrite r@ r ( 4 ) =r4 3 r = 4 3 (r@ r ( H r )) = 4 3 (D r H 3 H r ) = 4 3 D r H 12 3H r : (16.5.55) By the product rule forP i ; P i Lemma 18.1.3 applied twice D i1 [ 3 J 1 D r HD r H] = X A 1 2P `+1 ;A 2 2P ` 2 +1 ; A 3 2 P ` 3 ;A 4 2 P ` 4 ` 1 +` 2 +` 3 +` 4 =i1 c A 1 A 2 A 3 A 4 ` 1 ` 2 ` 3 ` 4 A 1 (H)A 2 (H)A 3 ( 3 )A 4 (J 1 ): (16.5.56) Now by the chain rule for P i Lemma 18.1.4, for A 3 2 P ` 3 , A 3 ( 3 ) = ` 3 X k=1 3k X i 1 +:::+i k =` 3 (A 3 ) j 2 P i j c k;i 1 ;:::;i k k Y j=1 (A 3 ) j : (16.5.57) Next note for (A 3 ) j 2 P i j (A 3 ) j () = (A 3 ) j1 @ r () = (A 3 ) j1 @ r ( H r ) = (A 3 ) j+1 (H) (16.5.58) for some (A 3 ) j1 2P i j 1 , (A 3 ) j+1 2P i j +1 . Similarly apply Lemma 18.1.4 to obtain for A 4 2 P ` 4 A 4 (J 1 ) = ` 4 X k=1 J 1k X i 1 +:::+i k =` 4 (A 4 ) j 2 P i j c k;i 1 ;:::;i k k Y j=1 (A 4 ) j J: (16.5.59) 199 Now for (A 4 ) j 2 P i j (A 4 ) j (J ) = (A 4 ) j ( 3 + 2 r r) = (A 4 ) j ( 3 ) + (A 4 ) j ( 2 r r) (16.5.60) We already know how to handle (A 4 ) j ( 3 ) and we can apply Lemma 18.1.3 and Lemma 18.1.4 to handle (A 4 ) j ( 2 r r) noting that r r =r@ r ( H r ) =D r H 3 H r ; (16.5.61) and for some A i j 2 P i j , A i j 1 2P i j 1 , A i j +1 2P i j +1 , A i j (D r H) =A i j 1 @ r (D r H) =A i j +1 (H): (16.5.62) An analogous computation applies to D i1 [ 3 J 1H r D r H] since 1 r 2 fD r ; 1 r g, see the denition ofP k for k 1 (15.3.9)-(15.3.10). Therefore for non-empty nite sets L :=f`2N >0 :d i 4 e`ig6=; such thatjLj<1 (16.5.63) K(`) :=fk2N 0 :kd i1 2 e + 1g such that 3jK(`)j<1: (16.5.64) we have D i1 [ r@ r ( 4 )J 1 D r H] = X `2L A ` 2P ` [c ` A ` (H) Y k2K(`) A k 2P k A k (H)]; (16.5.65) wherec ` are bounded coecients on [0; 1] from above and below, 1.jc ` j. 1. We can now estimate (a) j(a)j = 1 0 D i1 [ r@ r ( 4 )J 1 D r H]D i H d i r 2 dr . X `2L A ` 2P ` Y k2K(`) A k 2P k kA k (H)k L 1 1 0 A ` (H)D i H d i r 2 dr 200 . X `2L A ` 2P ` (S N +kH(0)k 2 0 )(a d( ) 2 1 0 (A ` (H)) 2 r 2 d i dr +a d d( ) 2 1 0 (D i H ) 2 r 2 d i dr) .e a 0 S N +(S N +kH(0)k 2 0 )(S N +kH(0)k 2 0 ) .e a 0 S N +S N +kH(0)k 2 0 : (16.5.66) Therefore 0 j(a)jd 0 .S N +kH(0)k 2 0 + 0 e a 0 0 S N d 0 : (16.5.67) For (b) we have D i1 [ 1 d@ r ( 4 J +1 )@ r D r H] = X A L 2P L 2Li+1 c L A L (H)d ( Y A k (L)2P k(L) k(L)b i1 2 c+2 A k(L) (H)) + X A ` 2P ` 2`i c ` A ` (H) Y A k (`)2P k(`) k(`)b i1 2 c+2 A k(`) (H): (16.5.68) Then by a similar argument to (a), we obtain the same bound. For (c), there are no troubling terms that are beyond similar methods to the methods of (a) and (b), so we obtain the same bound for (c). For (d), an analogous argument gives the same bound. For (e), an analogous argument gives the same bound. Therefore we have estimated Z i R 2 57 for i 1. Next we estimate 1 0 a b( ) D i [rR 1 [ H r ]]D i H r 2 d i dr :=Z i R 1 5 : (16.5.69) NoteD i = D i1 D r = D i1 (@ r + 2 r ): Then there are no troubling terms that are beyond similar methods to the methods of (a) and (b) fromZ i R 2 57 fori 1, and methods similar to previous estimates, and therefore we obtain the same bound as for (a) and (b) from Z i R 2 57 for Z i R 1 5 : 201 j = 6 : Note using the denition of q ij (18.1.3) and the commutator identity Lemma 18.1.5 C i [H] = 4 J +1 q ij D ij H + [ D i1 ; 4 J +1 ]D r L 0 H = 4 J +1 i1 X j=0 2+j X k=1 P k `=0 c@ ` r d r 2+jk ! D ij H ! + (i 1)@ r ( 4 J +1 ) D i2 (D r L 0 H) + X 2ki A2 P k ;B2 P ik cA( 4 J +1 )B(D r L 0 H): (16.5.70) Then note D r L 0 H =D r ( 1 @ r ( dD r H)) =D r ( 1 (@ r ( d)D r H + d(@ r D r H)) =D r ( @r ( d) D r H + 1 d(@ r D r H)) =D r ( @r ( d) )D r H + @r ( d) (@ r D r H) +@ r ( 1 d)@ r D r H + 1 d(D r @ r D r H): (16.5.71) Therefore we can writeZ i 6 in a form to which can use analogous methods to the methods of (a) and (b) from Z i R 2 57 for i 1, and methods similar to previous estimates, and therefore we obtain the same bound as for (a) and (b) from Z i R 2 57 for Z i 6 . This completes the energy estimate. 202 Chapter 17 Proof of the Main Theorem Applying the Local Well-Posedness Theorem 15.5.1 we have that on some time interval [0;T ];T > 0 there exists a unique solution to (15.2.21). By Proposition 16.5.3 with > 0 chosen small enough and using the norm-modied energy equivalence Lemma 16.5.2, we obtain that there are universal constants c 1 ;c 2 ;c 3 ;c 4 1 such that for any 0 T S N (; )c 1 S N ( ) +c 2 +c 3 S N (0) +c 4 e a 0 0 S N ( 0 ; )d 0 : (17.0.1) AboveS N (; ) denotesS N with sup 0 instead of sup 0 0 . Applying a standard well-posedness estimate, we have that the time of existence T is inversely proportional to the size of the initial data, that is: T (S N (0)) 1 . Choose "> 0 so small that the time of existence T satises e a 0 T=4 a 0 c 4 ; sup T S N ()c S N (0) + (17.0.2) where c> 0 is a universal constant from the local well-posedness theory. Then let C = 3(c 1 c +c 2 +c 3 ): (17.0.3) 203 Now dene T := sup 0 f solution to (15.2.21) exists on [0;) andS N ()C S N (0) + g: (17.0.4) Notice thatT T since cC . Setting = T 2 in (17.0.1) for any 2 [ T 2 ;T ] we have S N (; T 2 )c 1 S N ( T 2 ) +c 2 +c 3 S N (0) +c 4 T 2 e a 0 2 0 S N ( 0 ; T 2 )d 0 : (17.0.5) Therefore, applying (17.0.2) we deduce that for any 2 [ T 2 ;T ] S N (; T 2 )c 1 S N ( T 2 ) +c 2 +c 3 S N (0) + c 4 a 0 e a 0 T=4 S N (; T 2 ) c 1 S N ( T 2 ) +c 2 +c 3 S N (0) +S N (; T 2 ): (17.0.6) By (17.0.2),S N ( T 2 )c S N (0) + and so from (17.0.6) S N (; T 2 )c 1 c S N (0) + +c 2 +c 3 S N (0) +S N (; T 2 ): (17.0.7) For suciently small this gives S N (; T 2 ) 2(c 1 c +c 2 +c 3 ) S N (0) + <C S N (0) + ; (17.0.8) and thus S N ()<C S N (0) + : (17.0.9) It is now straightforward to verify the a priori bounds (15.5.2)-(15.5.5) are in fact improved. For example, by the fundamental theorem of calculus, with A2P 2 j@ r H r j =j 0 AH j. 0 e a 0 0 S N ( 0 )d 0 ."< 1 6 ; 2 [0;T ) (17.0.10) 204 for"> 0 small enough. Analogous arguments apply to the remaining a priori assumptions. Recalling the denition ofT 17.0.4 and by the continuity of the map 7!S N ( 0 ), we conclude thatT =1 and the solution to (15.2.21) exists globally-in-time. Moreover the global bound (15.5.9) follows. 205 Chapter 18 Appendices 18.1 Dierential Operators We give a series of useful results concerning our dierential operatorsD i and vector elds P i . First we have the product rule forD i which is straightforward to prove using induction. Lemma 18.1.1. For any i 1 the following holds D i (fg) = (D i f)g + D i1 (f@ r g) + D i1 (gD r f) +g( D i1 D r f): (18.1.1) Next we have a commutation rule forD i L 0 which also follows by induction, in a similar fashion to Lemma B.1 [17] Lemma 18.1.2. For any i2Z >0 D i L 0 X =L i D i X + i1 X j=0 q ij D ij X (18.1.2) where q ij = 2+j X k=1 P k `=0 c ijk` @ ` r d r 2+jk ; (18.1.3) and c ijk` are bounded functions on [0; 1]. 206 Next for the vector eldsP i and P i , we give the product rule and chain rule from Lemma A.4 [17] and Lemma A.5 [17] respectively. Lemma 18.1.3. Let i2N be given. (a) For any A2P i the following identity holds: A (fg) = i X k=0 X B2P k C2 P ik c ABC k BfCg; (18.1.4) for some real-valued constants c ABC k . (b) For any A2 P i the following identity holds: A (fg) = i X k=0 X B2 P k C2 P ik c ABC k BfCg; (18.1.5) for some real-valued constants c ABC k . Lemma 18.1.4. Let a2 R;i2 N be given and x a vectoreld W 2 P i . Then for any suciently smooth f the following identity holds W (f a ) = i X k=1 f ak X i 1 +:::i k =i W j 2 P i j c k;i 1 ;:::;i k k Y j=1 W j f (18.1.6) for some real constants c k;i 1 ;:::;i k . Next we give a useful commutator identity from Lemma B.2 [17] Lemma 18.1.5. For any i2Z >0 [ D i ;e]X =i@ r e D i1 X + X 2ki A2 P k ;B2 P ik c (Ae)(BX): (18.1.7) Finally we give an important estimate from Lemma A.3 [17] which lets us controlP i usingD i . 207 Lemma 18.1.6. SupposeD i X is bounded in L 2 ([0; 3 4 ];r 2 dr). Then we have the following estimate: X D i 2P i 3 4 0 jD i Xj 2 r 2 2 dr. 3 4 0 jD i Xj 2 r 2 2 dr (18.1.8) where 0 is a smooth cuto function satisfying = 1 on [0; 1 2 ], = 0 on [ 3 4 ; 1], and 0 0. 18.2 Hardy-Sobolev Embedding From Lemma C.2 [17] we have the following weighted L 1 embedding Lemma 18.2.1. For any smooth u : [0; 1]!R and any m2Z >0 , we have kuk 2 1 . 2 X k=1 3 4 0 (D k u) 2 r 2 dr + m+1 X k=0 1 1 4 d 2m (D k u) 2 dr (18.2.1) kuk 2 1 . 2 X k=1 3 4 0 ( D k u) 2 r 2 dr + m+1 X k=0 1 1 4 d 2m ( D k u) 2 dr (18.2.2) k u r k 2 1 . 2 X k=1 3 4 0 (D k u) 2 r 2 dr + m+1 X k=0 1 1 4 d 2m (D k u) 2 dr: (18.2.3) As a corollary we have the following embedding result specically involving the vector eldP k Corollary 18.2.2. For any smooth u : [0; 1]!R and any m2Z >0 , we have for k2Z >0 kP k uk 2 1 . 2 X j=1 3 4 0 (P j+k u) 2 r 2 dr + m+1 X j=0 1 1 4 d 2m (P j+k u) 2 dr: (18.2.4) Proof. First note P k u = k X `=0 c ` (r)D ` (H) (18.2.5) where c ` are smooth functions of r on [ 1 4 ; 1]. Next note D j (P k u) =P j+k u for k even, (18.2.6) 208 D j (P k u) =P j+k u for k odd. (18.2.7) Then (18.2.4) follows from (18.2.1)-(18.2.2). 18.3 Time Based Inequalities We have a simple but crucial result concerning a() and the related quantities a 1 and a 0 . Lemma 18.3.1. Assume > 1. Fix an ane motion a(t) from the setS under consider- ation. Let a 1 := lim !1 a () a() ; a 0 := d( ) 2 a 1 ; (18.3.1) where d( ) = 8 > > < > > : 3 3 if 1< 5 3 2 if > 5 3 . Then 0 5 3 , using (14.2.2) and similarly consideringA(t) = diag(a(t);a(t);a(t)), we can apply Lemma 6 [51] to express a(t) in the form a(t) =a 0 +ta 1 +m(t); (18.3.4) such that a 0 ;a 1 are time-independent and m(t) satises the bounds jm(t)j =o t!1 (1 +t);j@ t m(t)j. (1 +t) 33 : (18.3.5) We also recalla(t) 1+t. Then (18.3.3) follows from Lemma A.1 [20] in this case also. 209 18.4 Local Well-Posedness Here we sketch the proof of the local well posedness Theorem 15.5.1 for our system. We rst recall Theorem 15.5.1. Theorem 2.1. Suppose > 1. Fix N 8. Let k N in (14.2.13). Then there are 0 > 0, > 0 and T > 0 such that for every 2 (0; 0 ], 2 (0; 0 ] and pair of initial data for (15.2.21) (H 0 ;@ H 0 ) satisfyingS N (H 0 ;@ H 0 ) andkH(0)k 2 0 , there exists a unique solution (H();H ()) : [0; 1]!RR to (15.2.21)-(15.2.22) for all 2 [0;T ]. The solution has the propertyS N (H;H ). for each 2 [0;T ]. Furthermore, the map [0;T ]37!S N ()2R + is continuous. Sketch of proof. The proof follows by adapting the argument in [24, 27]. Notably, [24, 27] establishes the existence theory based on a suitable approximate scheme and a priori bounds. We will design the approximate scheme for H :=D r H and H fromH from which we can apply the result of [24, 27]. The jth approximations (H j ;@ H j ) and (H j ;@ H j ) are constructed as follows. The initial data (H 0 ;@ H 0 ) such thatS N (H 0 ;@ H 0 ) is used for the rst approximation j = 1, that is, we let (H 1 ;@ H 1 ) = (D r H 0 ;D r @ H 0 ) and (H 1 ;@ H 1 ) = (H 0 ;@ H 0 ). Then we obtain the approximate (j +1) th solutions by induction: for j 1, we let (H j+1 ;@ H j+1 ) solve the linear PDE a 3 3 @ 2 H j+1 +a 3 4 a @ H j+1 4 J +1 L 1 H j+1 H j+1 =D 1 (H j ) 2 r D 1 h rR 1 [ H j r ] +rR 2 [ H j r ] +rR 3 [ H j r ] i + C 1 [H j ] + @ r 4 J +1 L 0 H j ; (18.4.1) with the initial data (H j+1 ;@ H j+1 )j =0 = (D r H 0 ;D r @ H 0 ): Equation (18.4.1) mimics (16.1.15) for i = 1. The right hand side (16.1.15) is evaluated using H j which is indi- cated by the subscript j. The boundS N (H j ;@ H j )<1 depends only on and hence we 210 can apply the duality argument in [24, 27] to obtain the existence of (H j+1 ;@ H j+1 ). We then dene H j+1 by H j+1 = 1 r 2 r 0 H j+1 (r 0 ) 2 dr 0 (18.4.2) and have thatS N (H j+1 ;@ H j+1 )<1 from a priori estimates with this bound depending only on . Finally, as j!1, we extract a subsequence and obtain the limit (H;@ H) of (H j ;@ H j ) that is a solution to (15.2.21) on [0;T ] for some T = T () > 0 with S N (H;@ H).. 18.5 Coercivity Estimates We give a useful result which will let us handle time weights with negative powers which are present in our equation structure when > 5 3 . Lemma 18.5.1 (Coercivity Estimate). Let (H;H ) : [0; 1]! RR be a unique local solution to (15.2.21)-(15.2.22) on [0;T ] for T > 0 xed withkH(0)k 2 0 <1 and assume (H;H ) satises the a priori assumptions (15.5.2)-(15.5.5). Fix N 8. Let k N in (14.2.13). Fix 0iN 1. Then for all 2 [0;T ], we have kD i Hk 2 i . sup 0 0 fa 2 kD i H k 2 i g +kD i H(0)k 2 i : (18.5.1) Proof. 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Abstract (if available)
Abstract
This dissertation consists of three parts. In Part I, global existence for the nonisentropic compressible Euler equations with vacuum boundary for all adiabatic constants ? > 1 is shown through perturbations around a rich class of background nonisentropic affine motions. The notable feature of the nonisentropic motion lies in the presence of nonconstant entropies, and it brings a new mathematical challenge to the stability analysis of nonisentropic affine motions. In particular, the estimation of the curl terms requires a careful use of algebraic, nonlinear structure of the pressure. With suitable regularity of the underlying affine entropy, we are able to adapt the weighted energy method developed for the isentropic Euler by Hadži? and Jang to the nonisentropic problem. For large ? values, inspired by Shkoller and Sideris, we use time-dependent weights that allow some of the top-order norms to potentially grow as the time variable tends to infinity. We also exploit coercivity estimates here via the fundamental theorem of calculus in time variable for norms which are not top-order. ? In Part II, global solutions to the compressible Euler equations with heat transport by convection in the whole space are shown to exist through perturbations of Dyson's isothermal affine solutions. This setting presents new difficulties because of the vacuum at infinity behavior of the density. In particular, the perturbation of isothermal motion introduces a Gaussian function into our stability analysis and a novel finite propagation result is proven to handle potentially unbounded terms arising from the presence of the Gaussian. Crucial stabilization-in-time effects of the background motion are mitigated through the use of this finite propagation result however and a careful use of the heat transport formulation in conjunction with new time weight manipulations are used to establish global existence. The heat transport by convection offers unique physical insights into the model and mathematically, we use a controlled spatial perturbation in the analysis of this feature of our system which leads us to exploit source term estimates as part of our techniques. ? In Part III, global stability of the spherically symmetric nonisentropic compressible Euler equations with positive density around global-in-time background affine solutions is shown in the presence of free vacuum boundaries. Vacuum is achieved despite a non-vanishing density by considering a negatively unbounded entropy and we use a novel weighted energy method whereby the exponential of the entropy will act as a changing weight to handle the degeneracy of the vacuum boundary. Spherical symmetry introduces a coordinate singularity near the origin for which we adapt a method developed for the Euler-Poisson system by Guo, Hadži? and Jang to our problem.
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Rickard, Calum
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Stability analysis of nonlinear fluid models around affine motions
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Applied Mathematics
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2021-08
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