From Kinetic Theory to Fluid Mechanics: Viscous Surface Wave and Hydrodynamic Limit by Lei Wu B.S., Peking University, Beijing, PRC, 2010 M.S., Brown University, Providence, RI, USA, 2011 A dissertation submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy in The Division of Applied Mathematics at Brown University PROVIDENCE, RHODE ISLAND May 2015 ⃝ c Copyright 2015 by Lei Wu This dissertation by Lei Wu is accepted in its present form by The Division of Applied Mathematics as satisfying the dissertation requirement for the degree of Doctor of Philosophy. Date Yan Guo, Professor of Applied Mathematics, Advisor Date Chi-Wang Shu, Professor of Applied Mathematics, Co-Advisor Recommended to the Graduate Council Date Hongjie Dong, Associate Professor of Applied Mathematics, Reader Approved by the Graduate Council Date Peter M. Weber, Dean of the Graduate School iii Vitae Biographical Data Born on Aug 4th, 1987 in Zhangjiakou, Hebei, PRC. Education Brown University, Providence, RI, USA Doctor of Philosophy in Applied Mathematics, May 2015(expected) Adviser: Yan Guo and Chi-Wang Shu Brown University, Providence, RI, USA Master of Science in Applied Mathematics, May 2011 Peking University, Beijing, PRC Bachelor of Science in Mathematics, July 2010 Publications • Lei Wu and Yan Guo; Geometric Correction for Diffusive Expansion of Steady Neutron Transport Equation. Comm. Math. Phys. 336 (2015), no. 3, iv 1473-1553. • Lei Wu and Chi-Wang Shu; Numerical Solution of the Viscous Surface Wave with Discontinuous Galerkin Method. (2014) Accepted by ESAIM Math. Model. Numer. Anal. • Lei Wu; Well-posedness and Decay of the Viscous Surface Wave. (2014) SIAM J. Math. Anal. 46 (2014), no. 3, 2084-2135. Presentations • 2014-11-17 University of Wisconsin-Madison, Madison, WI, USA Geometric Correction of diffusive expansion for neutron transport equation. • 2014-09-19 Brown University, Providence, RI, USA Geometric Correction of diffusive expansion for neutron transport equation. v Acknowledgements First of all, I would like to show deepest gratitude to my advisers, Prof. Yan Guo and Prof. Chi-Wang Shu. Their encouragement, guidance and support help me to understand mathematics in its full depth, to think creatively and critically, to work in a strict and meticulous manner, and to enjoy the beauty of science. Their enthusiasm and inspiration will always drive me to explore the world of mathematics in future. Secondly, I would like to express thanks to my thesis reader Prof. Hongjie Dong for sparing time and providing me with detailed and constructive comments to improve this thesis. I am thankful to Prof. Ian Tice and Prof. Chanwoo Kim for inspiring discussion and providing valuable comments on my work. Further, I would like to thank my friends in Brown, Dr. Wei Wu, Dr. Yang Yang, Wei-Yang Wang, Hong Zhang, and Dr. Zheng Chen, for their support and caring both in academics and life. I would also like to thank all the members in our group. Last but not least, I owe my loving thanks to my parents. Their encouragement and caring are always the strongest support in my life. vi Abstract of “From Kinetic Theory to Fluid Mechanics: Viscous Surface Wave and Hydrodynamic Limit” by Lei Wu, Ph.D., Brown University, May 2015 In this dissertation, we mainly discuss two topics of partial differential equations in fluid dynamics and kinetic theory: viscous surface wave and diffusive limit. With respect to viscous surface wave, we consider an incompressible viscous flow without surface tension in a finite-depth domain of three dimensions, with a free top boundary and a fixed bottom boundary. The system is governed by the Navier-Stokes equations in this moving domain and the transport equation on the moving boundary. Following the framework of geometric mapping by Y. Guo and I. Tice, we further prove the local well-posedness with general smooth data and give a simpler proof of global well-posedness. Also, we construct a stable numerical scheme to simulate the evolution of this system by discontinuous Galerkin method. With respect to diffusive limit, we revisit the asymptotic analysis of a steady neutron transport equation in a two-dimensional unit disk with one-speed velocity. We disprove the classical boundary layer theory by a concrete counterexample with a different boundary layer expansion with geometric correction. Also, we provide the correct boundary layer construction with both in-flow and diffusive boundary. Contents Vitae iv Acknowledgments vi 1 Introduction 1 2 Well-Posedness and Decay of Viscous Surface Wave 3 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2.1.1 Problem Presentation . . . . . . . . . . . . . . . . . . . . . . . 4 2.1.2 Previous Results . . . . . . . . . . . . . . . . . . . . . . . . . 5 2.1.3 Geometric Formulation . . . . . . . . . . . . . . . . . . . . . . 8 2.1.4 Main Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.1.5 Convention and Terminology . . . . . . . . . . . . . . . . . . . 14 2.2 Local Well-posedness for General Data . . . . . . . . . . . . . . . . . 16 2.2.1 ϵ-Poisson Integral . . . . . . . . . . . . . . . . . . . . . . . . . 16 vii 2.2.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.2.3 Elliptic Equations . . . . . . . . . . . . . . . . . . . . . . . . . 29 2.2.4 Linear Navier-Stokes Equations . . . . . . . . . . . . . . . . . 42 2.2.5 Transport Equation . . . . . . . . . . . . . . . . . . . . . . . . 46 2.2.6 Navier-Stokes-Transport System . . . . . . . . . . . . . . . . . 51 2.3 Global Well-posedness for Horizontally Infinite Domain . . . . . . . . 71 2.3.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 2.3.2 Definitions of Energy and Dissipation . . . . . . . . . . . . . . 74 2.3.3 Interpolation Estimates . . . . . . . . . . . . . . . . . . . . . . 77 2.3.4 Nonlinear Estimates . . . . . . . . . . . . . . . . . . . . . . . 86 2.3.5 Energy Estimates . . . . . . . . . . . . . . . . . . . . . . . . . 88 2.3.6 Comparison Estimates . . . . . . . . . . . . . . . . . . . . . . 90 2.3.7 A Priori Estimates . . . . . . . . . . . . . . . . . . . . . . . . 102 2.3.8 Global Well-posedness . . . . . . . . . . . . . . . . . . . . . . 105 3 Numerical Solution to Viscous Surface Wave 106 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 3.2 Numerical Scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 3.2.1 Basic Settings . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 3.2.2 Discretization . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 3.2.3 Properties and Estimates of Discretization . . . . . . . . . . . 121 viii 3.3 Stability Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 3.4 Discussion on the Error Analysis . . . . . . . . . . . . . . . . . . . . 144 3.4.1 Velocity Error Estimates . . . . . . . . . . . . . . . . . . . . . 144 3.4.2 Pressure Error Estimates . . . . . . . . . . . . . . . . . . . . . 158 3.4.3 Error Analysis of the Fluid . . . . . . . . . . . . . . . . . . . . 165 3.5 Numerical Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 3.5.1 Accuracy Tests with SIPG . . . . . . . . . . . . . . . . . . . . 169 3.5.2 Accuracy Tests with NIPG . . . . . . . . . . . . . . . . . . . . 170 3.5.3 Discussion on the Numerical Tests . . . . . . . . . . . . . . . . 172 3.6 Conclusions and Remarks . . . . . . . . . . . . . . . . . . . . . . . . 172 4 Asymptotic Analysis of Neutron Transport Equation 174 4.1 Introduction and Notation . . . . . . . . . . . . . . . . . . . . . . . . 175 4.1.1 Problem Formulation . . . . . . . . . . . . . . . . . . . . . . . 175 4.1.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 4.1.3 Notation and Conventions . . . . . . . . . . . . . . . . . . . . 177 4.2 Asymptotic Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 4.2.1 Interior Expansion . . . . . . . . . . . . . . . . . . . . . . . . 178 4.2.2 Milne Expansion . . . . . . . . . . . . . . . . . . . . . . . . . 179 4.2.3 ϵ-Milne Expansion with Geometric Correction . . . . . . . . . 185 4.3 Well-posedness of Steady Neutron Transport Equation . . . . . . . . 190 ix 4.3.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 4.3.2 L2 Estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 4.3.3 L∞ Estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 4.3.4 Well-posedness of Transport Equation . . . . . . . . . . . . . 205 4.4 ϵ-Milne Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 4.4.1 L2 Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 4.4.2 L∞ Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . 234 4.4.3 Exponential Decay . . . . . . . . . . . . . . . . . . . . . . . . 249 4.4.4 Maximum Principle . . . . . . . . . . . . . . . . . . . . . . . . 253 4.5 Proof of Theorem 4.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . 255 4.6 Neutron Transport Equation with Diffusive Boundary . . . . . . . . . 268 4.6.1 Problem Settings . . . . . . . . . . . . . . . . . . . . . . . . . 268 4.6.2 Interior Expansion . . . . . . . . . . . . . . . . . . . . . . . . 269 4.6.3 Milne Expansion . . . . . . . . . . . . . . . . . . . . . . . . . 270 4.6.4 ϵ-Milne Expansion with Geometric Correction . . . . . . . . . 275 4.6.5 Well-Posedness of Steady Neutron Transport Equation . . . . 284 4.6.6 ϵ-Milne Problem . . . . . . . . . . . . . . . . . . . . . . . . . 287 4.6.7 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . 293 5 Future Work 301 x A Related Estimates in Viscous Surface Wave 304 A.1 Analytic Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305 A.1.1 Products in Sobolev Space . . . . . . . . . . . . . . . . . . . . 305 A.1.2 Poincare-Type Inequality . . . . . . . . . . . . . . . . . . . . . 306 A.1.3 Poisson Integral . . . . . . . . . . . . . . . . . . . . . . . . . . 306 A.1.4 Riesz Potential . . . . . . . . . . . . . . . . . . . . . . . . . . 307 A.1.5 Interpolation Estimates . . . . . . . . . . . . . . . . . . . . . . 308 A.1.6 Continuity and Temporal Derivative . . . . . . . . . . . . . . 310 A.1.7 Extension Theorem . . . . . . . . . . . . . . . . . . . . . . . . 312 A.2 Estimates for Fundamental Equations . . . . . . . . . . . . . . . . . . 313 A.2.1 Transport Estimates . . . . . . . . . . . . . . . . . . . . . . . 313 A.2.2 Elliptic Estimates . . . . . . . . . . . . . . . . . . . . . . . . . 314 B Counterexamples in Neutron Transport Equation 317 B.1 Construction of the Counterexample with In-Flow Boundary . . . . . 318 B.2 Construction of the Counterexample with Diffusive Boundary . . . . 322 xi List of Tables 3.1 L2 Error Table for Xh1 − Mh0 − Sh1 formulation with SIPG . . . . . . 169 3.2 L2 Error Table for Xh2 − Mh1 − Sh2 formulation with SIPG . . . . . . 170 3.3 L2 Error Table for Xh1 − Mh0 − Sh1 formulation with NIPG . . . . . . 171 3.4 L2 Error Table for Xh2 − Mh1 − Sh2 formulation with NIPG . . . . . . 172 xii List of Figures 3.1 Mesh Distribution in N = 4 . . . . . . . . . . . . . . . . . . . . . . . 112 3.2 Strip-Shape Mesh Distribution in N = 4 . . . . . . . . . . . . . . . . 112 xiii Chapter One Introduction 2 In classical mechanics, matter can be described under different scales: for microscopic scale, Newton’s laws dominate the movement of atoms and molecules; for macroscopic objects, fluid mechanics and solid mechanics provide effective tools to predict the behavior; in between, from mesoscopic viewpoint, kinetic theory states the statistical rules of the pattern of particles. Hilbert’s Six Problem in axiomatization of physics attempts to connect the funda- mental laws behind these theories, i.e. to derive the mesoscopic and macroscopic rules from basic Newton’s laws in microscopic scale. Following the path Hilbert himself proposed, the first step from microscope to mesoscope is the justification of kinetic theory, which involves the unfinished project of rigorous derivation of Boltzmann e- quation. The second step from mesoscope to macroscope is the hydrodynamic limit, which is the bridge between fluid equations and kinetic theory. Our research focuses on the second step and we will mainly discuss the decay of viscous surface wave and diffusive limit of neutron transport equation. In Chapter 2, we will first discuss the local and global well-posedness and decay of viscous surface wave under the framework of geometric mapping. In Chapter 3, we will provide a stable numerical scheme to viscous surface wave with discontinuous Galerkin method and analyze its convergence. In Chapter 4, we turn to the boundary layer effect in steady neutron transport equation and provide a detailed analysis of Milne problem with geometric correction. Chapter Two Well-Posedness and Decay of Viscous Surface Wave 4 2.1 Introduction 2.1.1 Problem Presentation We consider an incompressible viscous flow in the moving domain Ω(t) = {y ∈ Σ × R : −b(y1 , y2 ) < y3 < η(y1 , y2 , t)}. (2.1.1) Here we can take either Σ = R2 or Σ = (L1 T) × (L2 T) for which T denotes the 1-torus and L1 , L2 > 0 the periodicity lengths. The lower boundary b is fixed and given satisfying 0 < b(y1 , y2 ) ≤ ¯b = constant and b ∈ C ∞ (Σ). (2.1.2) When Σ = R2 , we further require b(y1 , y2 ) → b0 = positive constant as |y1 |+|y2 | → ∞ and b − b0 ∈ H s (Σ) for any s ≥ 0. When Σ = (L1 T) × (L2 T), we just denote b0 = 1/2(max{b(y1 , y2 )} + min{b(y1 , y2 )}). It is easy to see this also implies b − b0 ∈ H s (Σ) for any s ≥ 0 since Σ is bounded. We denote the initial domain Ω(0) = Ω0 . For each t, the flow is described by its velocity and pressure (u, p) : Ω(t) 7→ R3 × R which satisfies the incompressible Navier-Stokes equations    ∂t u + u · ∇u + ∇p = µ∆u   in Ω(t),       ∇·u=0 in Ω(t),       (pI − µD(u))ν = gην on {y3 = η(y1 , y2 , t)},   u=0 on {y3 = −b(y1 , y2 )}, (2.1.3)       ∂t η = u3 − u1 ∂y1 η − u2 ∂y2 η on {y3 = η(y1 , y2 , t)},         u(t = 0) = u0 in Ω0 ,     η(t = 0) = η 0 on Σ, 5 for ν the outward-pointing unit normal vector on {y3 = η}, I the 3×3 identity matrix, (Du)ij = ∂i uj + ∂j ui the symmetric gradient of u, g the gravitational constant and µ > 0 the viscosity. The fifth equation in (2.1.3) implies the free surface is convected with the fluid. Note that in (2.1.3), we have shifted the actual pressure p¯ by the constant atmosphere pressure patm according to p = p¯ + gy3 − patm . We always assume a natural condition that there exists a positive real number ρ such that η0 +b ≥ ρ > 0 on Σ, which means the initial free surface is strictly separated from the bottom. Also without loss of generality, we may assume µ = g = 1. In the following, we will use the term “infinite case” when Σ = R2 and “periodic case” when Σ = (L1 T) × (L2 T). The problem (2.1.3) possesses a natural energy structure that for the solution triple (u, p, η) which is sufficiently regular, we have (2.1.4) ∫ ∫ ∫ t∫ ∫ ∫ 1 1 1 1 1 |u(t)|2 + |η(t)|2 + |Du(s)|2 ds = |u0 |2 + |η0 |2 . 2 Ω(t) 2 Σ 2 0 Ω(s) 2 Ω0 2 Σ The first two integrals represent the kinetic energy and potential energy, while the third represents the dissipation. 2.1.2 Previous Results Historically, this problem has been studied in several different settings according to whether we consider the viscosity and surface tension. In the inviscid case, traditionally we replace the non-slip condition u = 0 with 6 non-penetration condition u · n = 0 on the bottom boundary. It is often assumed that the initial fluid is irrotational and then this curl-free property is preserved in the later time. This allows to reformulate the problem to one only on the free surface. Local well-posedness in this framework was proved by Wu [51, 52] and Lannes [31]. On the other hand, local well-posedness without this irrotational assumption was proved by Zhang-Zhang [55], Christodoulou-Lindblad [12], Lindblad [42], Coutand-Shkoller [16] and Shatah-Zheng [44]. In the viscous case, vorticity is naturally introduced at the free surface, so the surface formulation is not valid. Also, in the inviscid irrotational case, global well-posedness was shown by Wu [53] and Germain-Masmoudi-Shatah [20]. In the viscous case without surface tension, local well-posedness was proved by Solonnikov and Beale. Solonnikov [46] employed the framework of H¨older spaces to study the problem in a bounded domain, all of whose boundaries are free. Beale [8] utilized the framework of L2 spaces to study the domain like ours. Both of them took this system as a perturbation of parabolic equations and made use of the regularity results for linear equations. Abel [1] extended this result to the framework of Lp spaces. Also, Hataya [29] proved global well-posedness in the periodic case. Many authors have also considered the effects of surface tension, which can help stabilize the problem and gain regularity. e.g. global well-posedness in the small data case was shown by Bae [7]. The idea of mapping the moving domain into a fixed domain can date back to Beale’s work in [8] for graph-like domains and was also utilized by Lannes in [31]. A recent progress in this field is by Shatah-Zeng [45], which can deal with more general domains by maps with harmonic extensions. Most of above results are built in the Lagrangian coordinates and do not utilize 7 the natural energy structure of the problem in a direct manner. Hence, showing decay becomes very difficult. In [26], [25] and [24], Y. Guo and I. Tice introduced a geometric energy framework and proved the solutions with small data are globally well-posed and decay in time. The main idea in their proof was that, instead of considering the perturbation of parabolic equations, they showed the regularity under a time- dependent basis. In our paper, we follow this path and employ a similar argument. In [26], it is assumed the initial data u0 and η0 are sufficiently small to guarantee that: (1) the transform between the fixed domain and the moving domain is a dif- feomorphism; (2) the elliptic estimates in the linear Navier-Stokes problem achieve the optimal regularity; (3) in proving boundedness and contraction of the iteration sequence, the constant is small enough to obtain desired estimates. In Section 2.2 of this dissertation, we drop this smallness assumption and prove local well-posedness for general initial data. Basically, we should use different techniques to recover above three properties. Also, in proving global well-posedness and decay in [25], the authors introduced a quite complicated interpolation argument to improve the decaying rate of energy with minimal counts. In Section 2.3 of this dissertation, through redefining the en- ergy and dissipation, we greatly simplify the proof in: (1) avoiding the complicated bootstrapping argument in the interpolation estimates; (2) avoiding the bootstrap- ping argument in the comparison estimates; (3) handling the energy and dissipation with 1-minimal count and 2-minimal count simultaneously instead of separately. 8 2.1.3 Geometric Formulation In order to work in a fixed domain, we need to flatten the free surface via a coordinates transform. We follow the idea in [8] and [26], and use a slightly modified version as the geometric transform. Define a fixed domain Ω = {x ∈ Σ × R | − b0 < x3 < 0}, (2.1.5) for which we write the coordinates x ∈ Ω. In this slab, we take Σ : {x3 = 0} as the upper boundary and Σb : {x3 = −b0 } as the lower boundary. Simply denote x′ = (x1 , x2 ) and x = (x1 , x2 , x3 ). We need to utilize slightly different techniques to construct the extension of free surface in the local and global cases. Geometric Formulation in the Local Case In the local case, we cannot apply the idea in [26] directly. We introduce a parame- terized transform instead. Define the ϵ-Poisson integral for η:   ∫   ′ ηˆ(ξ)eϵ|ξ|x3 e2πix ·ξ dξ in the infinite case,   R2 η¯ϵ = P ϵ η = (2.1.6)     ∑  ′ e2πin·x eϵ|n|x3 ηˆ(n) in the periodic case, n∈(L−1 −1 1 Z)×(L2 Z) where ηˆ(ξ) and ηˆ(n) denote the Fourier transform of η(x′ ) in either continuous or discrete form and 0 < ϵ < 1 the parameter. The detailed definitions are shown in (2.2.1) and (2.2.16). Consider the geometric transform from Ω to Ω(t): b x3 Φϵ : (x1 , x2 , x3 ) 7→ (x1 , x2 , x3 + η¯ϵ (1 + )) = (y1 , y2 , y3 ). (2.1.7) b0 b0 9 This transform maps Ω into Ω(t) with the Jacobi matrix    1 0 0    ∇Φϵ =   0 1 0   (2.1.8)   Aϵ B ϵ J ϵ and the transform matrix    1 0 −A K  ϵ ϵ   A ϵ = ((∇Φϵ )−1 )T =   0 1 −B ϵ ϵ  K  (2.1.9)   0 0 Kϵ where ˜b = 1 + x3 , b0 ∂1 b ∂2 b Aϵ = x3 + ∂1 η¯ϵ˜b, Bϵ = x3 + ∂2 η¯ϵ˜b, (2.1.10) b0 b0 b η¯ϵ 1 Jϵ = + + ∂3 η¯ϵ˜b, Kϵ = ϵ . b0 b0 J Notice that this transform is not necessarily a homeomorphism for arbitrary ϵ > 0. By our assumption η0 (x′ ) + b(x′ ) ≥ ρ > 0 and Theorem 2.2.7, we can always choose a sufficiently small ϵ depending on ∥η0 ∥H 5/2 such that there exists a δ > 0 satisfying J ϵ (0) > δ > 0. Then this positive Jacobi matrix implies the transform Φϵ (t) is a homeomorphism at t = 0. Furthermore, based on Lemma 2.2.2, Lemma 2.2.3, Lemma 2.2.5 and Lemma 2.2.6, we can deduce Φϵ (0) is actually a C 1 diffeomorphism. In the following, we just write η¯ instead of η¯ϵ for simplicity, and the same convention applies to A , Φ, A, B, J and K. 10 Define the transformed operators as follows: (∇A f )i = Aij ∂j f, ∇A · ⃗g = Aij ∂j gi , ∆A f = ∇A · ∇A f, (2.1.11) N = (−∂1 η, −∂2 η, 1), (DA u)ij = Aik ∂k uj + Ajk ∂k ui , SA (p, u) = pI − DA u, where the summation should be understood in the Einstein convention. If we extend the divergence ∇A · to act on symmetric tensors in the natural way, then a straight- forward computation reveals ∇A · SA (p, u) = ∇A p − ∆A u for vector fields satisfying ∇A · u = 0. In our new coordinates, the original system (2.1.3) becomes    ∂t u − ∂t η¯˜bK∂3 u + u · ∇A u − ∆A u + ∇A p = 0 in Ω,         ∇A · u = 0 in Ω,       SA (p, u)N = ηN on Σ,      u=0 on Σb , (2.1.12)     u(x, 0) = u0 (x) in Ω,             ∂t η + u1 ∂1 η + u2 ∂2 η = u3 on Σ,      η(x′ , 0) = η (x′ ) on Σ. 0 Since A is determined by η through the transform, all the quantities above are related to η, i.e. the geometric structure of the free surface. This is the central idea of our proof. It is noticeable that in proving local well-posedness, we need to verify Φ(t) is a C 1 diffeomorphism for any t ∈ [0, T ], where the theorem holds. 11 Geometric Formulation in the Global Case In the global case, we focus on the infinite domain which always possess a flat bottom, i.e. b = constant. Since our theorem only holds for sufficiently small data, we may directly define η¯ as the Poisson integral of η which is shown in [26, (A-8)] and [26, (A-14)], i.e. η¯ = Pη. Then the map can be taken as x3 Φ : (x1 , x2 , x3 ) 7→ (x1 , x2 , x3 + η¯(1 + )) = (y1 , y2 , y3 ), (2.1.13) b which also transforms Ω to Ω(t). It is obvious that the Jacobi matrix and the trans- form matrix are identical to those in the local case except that we need to redefine the quantities ˜b = 1 + x3 /b, A = ∂1 η¯˜b, B = ∂2 η¯˜b, (2.1.14) J = 1 + η¯/b + ∂3 η¯˜b, K = 1/J. Naturally, the definitions of the weighted operators in (3.1.7) and the transformed equations (2.1.12) are the same as those in the local case with replacing the related quantities with above definitions. We can easily see as long as ∥η0 ∥H 5/2 (Σ) is sufficiently small, Φ is a C 1 diffeomorphism. Hence, the smallness itself can guarantee the validity of the transform. 2.1.4 Main Theorems Theorem 2.1.1. Let N ≥ 2 be an integer. Assume the initial data η0 + b ≥ 2b0 δ > 0 for some δ > 0. Suppose (u0 , η0 ) satisfies the estimate K0 = ∥u0 ∥2H 2N + 12 ∥η0 ∥2H 2N +1/2 (Σ) < ∞ as well as the N th compatibility conditions (2.2.159). Then there exists a 0 < T0 < 1 such that for any 0 < T < T0 , there exists a solution (u, p, η) to the system (2.1.12) on [0, T ] that satisfies the initial data. Furthermore, the solution obeys the estimate (∑ ∑ ∫ ) N j 2 N T j 2 sup ∂t u H 2N −2j + ∂t u 2N −2j+1 (2.1.15) H 0≤t≤T 0 j=0 j=0 (N ∑ −1 ∑ −1 ∫ ) j 2 N T j 2 + sup ∂t p H 2N −2j−1 + ∂t p 2N −2j H 0≤t≤T 0 j=0 j=0 ( ∑ N 2 + sup ∥η∥H 2N +1/2 (Σ) + 2 sup ∂tj η H 2N −2j+3/2 (Σ) 0≤t≤T 0≤t≤T j=1 ∑ +1 ∫ T ) N j 2 + ∂t η 2N −2j+5/2 ≤ C(Ω0 , δ)P (K0 ), H (Σ) j=2 0 where C(Ω0 , δ) is a positive constant depending on the initial domain Ω0 and the separation quantity δ, and P (·) is a polynomial in one variable satisfying P (0) = 0. The solution is unique among functions that satisfy the initial data and for which the left-hand side of the estimate (2.1.15) is finite. Moreover, η is such that the mapping Φ(t) defined by (2.1.7) is a C 2N −1 diffeomorphism for each t ∈ [0, T ]. Remark 2.1.2. Since the map Φ(t) is a C 2N −1 diffeomorphism, we may change the variables to y ∈ Ω(t) to produce the solution of (2.1.3). 13 Remark 2.1.3. Define quantities E2N , F2N and GN as follows: (2.1.16) 2 ∑ n j 2 ∑ n−1 j 2 En,2 = D2 u H 2n−2 + ∂t u 2n−2j + D2 p 2 2n−3 + H H ∂t p 2n−2j−1 H j=1 j=1 2 ∑ n j 2 + D2 η H 2n−2 (Σ) + ∂t η 2n−2j , H (Σ) j=1 ∑ n j 2 ∑ n−1 j 2 En,1 = ∥Du∥2H 2n−1 + ∂t u 2n−2j + ∥Dp∥2 2n−2 + ∂t p 2n−2j−1 (2.1.17) H H H j=1 j=1 ∑ n j 2 + ∥Dη∥2H 2n−1 (Σ) + ∂t η 2n−2j , H (Σ) j=1 ∑ n j 2 ∑ n−1 j 2 En = ∥Iλ u∥2H 0 + ∥Iλ η∥2H 0 (Σ) + ∂t u H 2n−2j + ∂t p 2n−2j−1 (2.1.18) H j=0 j=0 ∑ n j 2 + ∂t η 2n−2j , H (Σ) j=0 where Iλ u and Iλ η denote the Riesz potential of u and η as defined in (A.1.10) and (A.1.11). F2N = ∥η∥2H 4N +1/2 (Σ) . (2.1.19) The total energy is defined as follows: ∑ 2 F2N (r) GN (t) = sup E2N (r) + sup (1 + r)m+λ EN +2,m (r) + sup . (2.1.20) 0≤r≤t 0≤r≤t 0≤r≤t (1 + r) m=1 Theorem 2.1.4. Let N ≥ 4 be an integer. Suppose the initial data (u0 , η0 ) satisfies the compatibility conditions (2.2.159). Then there exists a κ > 0 such that if E2N (0) + F2N (0) ≤ κ, (2.1.21) 14 there exists a unique solution (u, p, η) to the system (2.1.12) on [0, ∞) that satisfies the initial data. Also, the solution obeys the estimate GN (∞) ≤ C(E2N (0) + F2N (0)) ≤ Cκ, (2.1.22) where C > 0 is a universal constant. Remark 2.1.5. Based on above theorem, naturally global well-posedness implies decay of En,2 and En,1 . 2.1.5 Convention and Terminology We now mention some of the definitions, notation and conventions we use throughout this chapter. 1. Unless specified, we use exactly the same notation as in [26] and [25]. 2. Throughout this chapter, C > 0 denotes a constant that only depends on the parameters N and Ω of the problem, but does not depend on the data. It is referred as universal and can change from one inequality to another. When we write C(z), it means a certain positive constant depending on quantity z. There are two exceptions to above rule in Section 2.2. The first one is in the elliptic estimates and Korn’s inequality, there are constants depending on the initial domain Ω0 . The second one is the constant C(δ) depending on the separation quantity δ. Though they are related to the initial domain Ω0 , we still call them universal since they are given implicitly. Apart from these, all the other constants related to the initial data Ω0 , u0 and η0 should be specified in detail. 15 a . b denotes a ≤ Cb, where C is a universal constant as defined above. 3. We write P (·) to denote polynomials in one variable. The polynomials always satisfy P (0) = 1 unless specified as exceptions. They may change from one inequality or equality to another. 4. We write the multi-indices N1+m = {α = (α0 , α1 , . . . , αm )} to emphasize the 0-index term is related to the temporal variable. For just spatial variables, we write Nm without the 0-index. We define the parabolic counting of such multi-indices as |α| = 2α0 + α1 + . . . + αm . We always write Df to denote the horizontal derivatives of f and ∇f for the full derivatives. For a given norm ∥·∥ and integers k, m ≥ 0, we introduce the following notation for the sums of derivatives: k 2 ∑ k 2 ∑ Dm f = ∥∂ α f ∥2 , ∇m f = ∥∂ α f ∥2 , α∈N2 m≤|α|≤k α∈N3 m≤|α|≤k k 2 ∑ k 2 ∑ D ¯ mf = ∥∂ α f ∥2 , ∇ ¯ mf = ∥∂ α f ∥2 , α∈N1+2 m≤|α|≤k α∈N1+3 m≤|α|≤k where D and ∇ denote spacial derivatives and D ¯ and ∇ ¯ denote full derivatives. Also we define k 2 k 2 k 2 k 2 D f = Dk f , ∇ f = ∇k f , k 2 k 2 k 2 k 2 D ¯ f = D ¯ f , k ∇ ¯ f = ∇ ¯ f . k 16 2.2 Local Well-posedness for General Data 2.2.1 ϵ-Poisson Integral In this section, we define the ϵ-Poisson integral which is utilized to construct the geometric transform (2.1.7), and discuss its fundamental properties. Poisson Integral in the Infinite Case For a function f defined on Σ = R2 , the ϵ-Poisson integral is defined by ∫ ′ ′ P f (x , x3 ) = ϵ fˆ(ξ)eϵ|ξ|x3 e2πix ·ξ dξ (2.2.1) R2 where fˆ(ξ) denotes the Fourier transform of f (x′ ) in R2 and 0 < ϵ < 1 is a parameter. Although P ϵ f is defined on R2 × (−∞, 0), we only consider the part R2 × (−b0 , 0) here. Lemma 2.2.1. Let P ϵ f be the ϵ-Poisson integral of function f which is in the ho- mogeneous Sobolev space H˙ q−1/2 (Σ) for q ∈ N. Then we have C ∥∇q P ϵ f ∥2H 0 ≤ ∥f ∥2H˙ q−1/2 , (2.2.2) ϵ where C > 0 is a constant independent of ϵ. In particular, we have C ∥P ϵ f ∥2H q ≤ ∥f ∥2H q−1/2 (Σ) . (2.2.3) ϵ Proof. By the definition of the Fourier transform, Fubini’s theorem and Parseval’s 17 identity, we may bound ∫ ∫ 0 2 ∥∇ Pq ϵ f ∥2H 0 ≤ (2π) 2q |ξ|2q fˆ(ξ) e2ϵ|ξ|x3 dξdx3 (2.2.4) R2 −b0 ∫ 2 ∫ 0 2q ˆ ≤ (2π) 2q |ξ| f (ξ) dξ e2ϵ|ξ|x3 dx3 R 2 −b0 ∫ 2 ( 1 − e−2ϵb0 |ξ| ) = (2π)2q |ξ|2q fˆ(ξ) dξ R2 2ϵ |ξ| ∫ 2 (2π)2q 2q−1 ˆ ≤ |ξ| f (ξ) dξ 2ϵ R2 π ≤ ∥f ∥H˙ q−1/2 (Σ) . 2 ϵ We can simply take C = π to achieve the estimate. Furthermore, considering { } 1 − e−2ϵb0 |ξ| 1 ≤ min , b0 (2.2.5) 2ϵ |ξ| 2ϵ |ξ| and 0 < ϵ < 1, we have πb0 ∥P ϵ f ∥2H 0 ≤ ∥f ∥2H 0 (Σ) (2.2.6) ϵ Hence, (2.2.3) easily follows. Lemma 2.2.2. Let P ϵ f be the ϵ-Poisson integral of function f . Then for q ∈ N and s > 1, we have C ∥∇q P ϵ f ∥2L∞ ≤ ∥f ∥2H q+s (Σ) . (2.2.7) ϵ Proof. A simple application of the Sobolev embedding theorem and Lemma 2.2.1 reveals C ∥∇q P ϵ f ∥2L∞ ≤ C ∥∇q P ϵ f ∥2H s+1/2 ≤ ∥f ∥2H q+s (Σ) . (2.2.8) ϵ 18 The following lemma illustrates the special properties of the vertical derivative. Lemma 2.2.3. Let P ϵ f be the ϵ-Poisson integral of function f . Then we have ∥∂3 P ϵ f ∥2L∞ ≤ Cϵ ∥f ∥2H 5/2 (Σ) . (2.2.9) where C > 0 is a constant independent of ϵ. Proof. We can simply bound ∥∂3 P ϵ f ∥2L∞ ≤ C ∥∂3 P ϵ f ∥2H 2 ≤ Cϵ2 ∥P ϵ f ∥2H 3 ≤ Cϵ ∥f ∥2H 5/2 (Σ) . (2.2.10) Note the first inequality is based on the Sobolev embedding theorem and the third one is a straightforward application of Lemma 2.2.1, so we only need to verify the second inequality in detail. By definition, it is easy to see ∫ ′ ∂3 P f = ϵ ϵ |ξ| fˆ(ξ)eϵ|ξ|x3 e2πix ·ξ dξ. (2.2.11) R2 Hence, we have ∫ ∫ 0 2 ∥∂3 P ϵ f ∥2H 0 = ϵ 2 |ξ|2 fˆ(ξ) e2ϵ|ξ|x3 dξdx3 (2.2.12) R2 −b0 ( ∫ ∫ 0 2 ) ϵ2 2 ˆ 2ϵ|ξ|x3 = (2π) 2 |ξ| f (ξ) e dξdx3 (2π)2 R2 −b0 ϵ2 ≤ ∥∂1 P ϵ f ∥2H 0 . (2π)2 19 Similarly, we can show for i, j = 1, 2, 3 ϵ2 ∥∂3i P ϵ f ∥2H 0 ≤ ∥∂11 P ϵ f ∥2H 0 , (2.2.13) (2π)4 ϵ2 ∥∂3ij P ϵ f ∥2H 0 ≤ 6 ∥∂111 P ϵ f ∥2H 0 . (2.2.14) (2π) Thus we have ∥∂3 P ϵ f ∥2H 2 ≤ Cϵ2 ∥P ϵ f ∥2H 3 . (2.2.15) Poisson Integral in the Periodic Case Suppose Σ = (L1 T) × (L2 T), we define the ϵ-Poisson integral as follows: ∑ ′ P ϵ f (x′ , x3 ) = e2πin·x eϵ|n|x3 fˆ(n) (2.2.16) n∈(L−1 −1 1 Z)×(L2 Z) where ∫ ′ ′ e−2πin·x ′ fˆ(n) = f (x ) dx . (2.2.17) Σ L1 L2 Now we need to show the same types of estimates as in the infinite case. Since the main ideas and the basic techniques are quite similar, we omit the detailed proofs here and only present the results. Lemma 2.2.4. Let P ϵ f be the ϵ-Poisson integral of function f which is in the ho- mogeneous Sobolev space H˙ q−1/2 (Σ) for q ∈ N. Then we have C ∥∇q P ϵ f ∥2H 0 ≤ ∥f ∥2H˙ q−1/2 , (2.2.18) ϵ 20 where C > 0 is a constant independent of ϵ. In particular, we have C ∥P ϵ f ∥2H q ≤ ∥f ∥2H q−1/2 (Σ) . (2.2.19) ϵ Lemma 2.2.5. Let P ϵ f be the ϵ-Poisson integral of function f . Then for q ∈ N and s > 1, we have C ∥∇q P ϵ f ∥2L∞ ≤ ∥f ∥2H q+s (Σ) . (2.2.20) ϵ Lemma 2.2.6. Let P ϵ f be the ϵ-Poisson integral of function f . Then we have ∥∂3 P ϵ f ∥2L∞ ≤ Cϵ ∥f ∥2H 5/2 (Σ) , (2.2.21) where C > 0 is a constant independent of ϵ. Homeomorphism of the Geometric Transform Theorem 2.2.7. If the initial data satisfies η0 + b > ρ > 0 for all x′ ∈ Σ, then we can choose a sufficiently small ϵ > 0 such that there exists a δ > 0 satisfying J ϵ (0) ≥ δ > 0. Proof. Naturally there always exists a δ > 0, such that η0 (x′ ) + b(x′ ) ≥ 2b0 δ > 0 for any x′ ∈ Σ. Based on Lemma 2.2.3 and Lemma 2.2.6, if taking ϵ ≤ δ 2 /(4C 2 ∥η0 ∥2H 5/2 ), we have ∥∂3 η¯0ϵ ∥L∞ ≤ δ/2, (2.2.22) where η¯0ϵ denotes η¯ϵ at t = 0. Furthermore, by the fundamental theorem of calculus, 21 we can estimate for any x3 ∈ [−b0 , 0] and x′ ∈ Σ ∫ 0 η0ϵ (x′ , x3 ) |¯ − η¯0ϵ (x′ , 0)| ≤ |∂3 η¯0ϵ (x′ , z)| dz ≤ b0 ∥∂3 η¯0ϵ ∥L∞ ≤ b0 δ/2. (2.2.23) −b0 Since η¯0ϵ (x′ , 0) = η0 (x′ ), we have b η¯ϵ b + η0 η¯0ϵ − η0 J ϵ (0) = + 0 + ∂3 η¯0ϵ ˜b = + + ∂3 η¯0ϵ ˜b (2.2.24) b0 b0 b0 b0 ≥ 2δ − δ/2 − δ/2 = δ > 0. Therefore, for given initial data η0 , by taking ϵ = δ 2 /(4C 2 ∥η0 ∥2H 5/2 ), we can guarantee J ϵ ≥ δ > 0. 2.2.2 Preliminaries Preliminary Lemmas Throughout this chapter, we employ the functional spaces defined in [26, Section 2]. We write H k (Ω) and H k (Σ) for usual Sobolev space of either scalar or vector functions. Define Hk (t) = {u(t) ∈ H k (Ω) : u(t)|Σb = 0}, (2.2.25) X (t) = {u(t) ∈ H 1 (Ω) : u(t)|Σb = 0, ∇A(t) · u(t) = 0}. Moreover, let L2 ([0, T ]; H k (Ω)) and L2 ([0, T ]; H k (Σ)) denotes the usual time-involved Sobolev space. Define HTk = {u ∈ L2 ([0, T ]; H k (Ω)) : u|Σb = 0}, (2.2.26) XT = {u ∈ L2 ([0, T ]; H 1 (Ω)) : u|Σb = 0, ∇A · u = 0}. 22 Define the inner product ⟨·, ·⟩Hk to denote the normalized inner product in Hk as ∫ ⟨u, v⟩H0 = Ju · v, (2.2.27) Ω ∫ ⟨u, v⟩H1 = JDA u : DA v, (2.2.28) Ω and ⟨·, ·⟩Htk to denote the normalized inner product in L2 ([0, T ]; HTk ), i.e. ∫ t ⟨u, v⟩Htk = ⟨u, v⟩Hk . (2.2.29) 0 It is easy to see Hk (t), X (t), HTk and XT are all Hilbert spaces with above inner products. Also we will need an orthogonal decomposition H0 = Y (t) ⊕ Y ⊥ (t), where Y ⊥ (t) = {∇A(t) φ(t) : φ(t) ∈ H 1 (t), φ(t)|Σ = 0}. (2.2.30) In our use of these spaces, we will often drop the (t) when there is no potential of confusion. Finally we will define the regular divergence-free space in H 1 (Ω). 1 0 Hσ = {u ∈ H 1 (Ω) : u|Σb = 0, ∇ · u = 0}. (2.2.31) Basically, following the path of [26], we need to recover all the necessary lemmas in the general data case. Since most of the results can be generalized to our case with obvious modifications, we omit the details and only present some important propositions. Only when key different techniques are employed, are all the details of the lemmas and proofs given. The following result gives a version of Korn’s type inequality for the initial domain 23 Ω0 . Here we apply Beale’s idea in [8]. Lemma 2.2.8. Suppose Ω0 is the initial domain and η0 ∈ H 5/2 (Σ). Then ∥v∥2H 1 (Ω0 ) . ∥Dv∥2H 0 (Ω0 ) , (2.2.32) for any v ∈ H 1 (Ω0 ) and v = 0 on {y3 = −b}. Proof. In the periodic case, Σ is bounded and v|Σb = 0, so naturally Korn’s inequality is valid (see the proof of [8, Lemma 2.7]). Then we consider the infinite case. First we prove decay of the initial surface η0 , i.e. for |α| ≤ 1, |∂ α η¯0 | → 0 as |x| → ∞ in the slab. For the horizontal derivative ∂ α , we have ∫ ′ α ∂ η¯0 = (2πiξ)α eϵx3 |ξ| e2πix ·ξ ηˆ0 (ξ)dξ, (2.2.33) R2 where ηˆ0 is the Fourier transform of η0 in R2 . Then ∫ (2πiξ)α eϵx3 |ξ| ηˆ0 (ξ) dξ (2.2.34) R2 ∫ ( ) α ϵx3 |ξ| 1 . |ξ| e |ηˆ0 (ξ)| (1 + |ξ| ) 5/4 dξ R2 1 + |ξ|5/4 (∫ )1/2 ( ∫ )1/2 e2ϵx3 |ξ| dξ ≤ |ξ| |ηˆ0 (ξ)| (1 + |ξ| ) dξ 2α 2 5/4 2 R2 (1 + |ξ| 5/4 2 R2 ) (∫ ) e2ϵx3 |ξ| dξ 1/2 . ∥η0 ∥H α+5/4 < ∞. R2 (1 + |ξ| 5/4 2 ) The last inequality is valid since x3 < 0. Hence, (2πiξ)α eϵx3 |ξ| ηˆ0 (ξ) ∈ L1 (R2 ). A similar proof can justify the result if ∂ α = ∂3 the vertical derivative. Since |x| → ∞ naturally implies |x′ | → ∞ in the slab, the Riemann-Lebesgue lemma implies |∂ α η¯0 | → 0 as |x| → ∞. ¯ = Φ(0) as defined in (2.1.7), which maps Ω = {R3 : We construct a mapping σ 24 −b0 < x3 < 0} to Ω0 . Denote σ ¯ (x) = x + σ(x). The above decaying property and our assumption on b lead to ∂ α σ(x) → 0 as |x| → ∞ for |α| ≤ 1. We partition the slab Ω into cubes Qj,k = {x : j < x1 < j + 1, k < x2 < k + 1, −b0 < x3 < 0}. (2.2.35) In each cube, we have Korn’s inequality ∥v∥2H 1 (Q) ≤ C(∥Dv∥2H 0 (Q) + ∥v∥2H 0 (Q) ). (2.2.36) Employing the compactness argument as Beale [8] did, under the condition v = 0 on {x3 = −b0 }, we can strengthen this result to ∥v∥2H 1 (Q) ≤ C ∥Dv∥2H 0 (Q) . (2.2.37) This argument relies on the fact that for such v ∥v∥2H 0 (Q) ≤ C ∥Dv∥2H 0 (Q) . (2.2.38) Now suppose D ⊆ Ω0 is the image of the union of such cubes contained in {|x| ≥ R} for some R. Applying last estimate to v = u ◦ σ ¯ and then transforming to Ω0 , we have ∥u∥H 1 (D) ≤ C ∥Du∥H 0 (D) + θ ∥u∥H 1 (D) , (2.2.39) where θ < 1 when R is large enough based on the decaying property. Then we use Korn’s inequality in the bounded domain Ω0 − D and combine with above estimates to obtain the desired result. 25 Next, we show the weights introduced by J and A do not affect the Sobolev norms. Although this result is quite obvious for the small data (see [26, Lemma 2.1]), now we need to utilize above Korn’s inequality to show it for the general data. Lemma 2.2.9. There exists a 0 < ϵ0 < 1, such that if ∥η − η0 ∥H 5/2 < ϵ0 , then the following relations ∫ ∥u∥2H 0 (Ω) . J |u|2 . (1 + ∥η0 ∥H 5/2 (Σ) ) ∥u∥2H 0 (Ω) , (2.2.40) Ω (2.2.41) ∫ 1 ∥u∥2H 1 (Ω) . J |DA u|2 . (1 + ∥η0 ∥H 5/2 (Σ) )3 ∥u∥2H 1 (Ω) , (1 + ∥η0 ∥H 5/2 (Σ) )3 Ω hold for all u ∈ H1 . Proof. (2.2.40) is just a natural corollary of the relation δ . ∥J∥L∞ . (1+∥∇¯ η ∥ L∞ ) . (1 + ∥η∥H 5/2 ) . (1 + ∥η0 ∥H 5/2 ) based on our assumptions and the Sobolev embedding theorem. To derive (2.2.41), notice ∫ ∫ ∫ J |DA u| = 2 J |DA0 u| + 2 J(DA u + DA0 u) : (DA u − DA0 u). (2.2.42) Ω Ω Ω For the first term on the right-hand side of (2.2.42), based on the integral substitution and Korn’s inequality we proved above, we have ∫ J |DA0 u|2 dx (2.2.43) Ω ∫ ∫ 1 1 & J0 |DA0 u| dx = 2 |Dv|2 dy 1 + ∥η0 ∥H 5/2 (Σ) Ω 1 + ∥η0 ∥H 5/2 (Σ) Ω0 1 1 & ∥v∥2H 1 (Ω0 ) & ∥u∥2H 1 (Ω) . 1 + ∥η0 ∥H 5/2 (Σ) (1 + ∥η0 ∥H 5/2 (Σ) )3 26 For the second term on the right-hand side of (2.2.42), using Korn’s inequality in slab Ω, we can naturally estimate ∫ J(DA u + DA0 u) : (DA u − DA0 u) (2.2.44) Ω ∫ . ∥J∥L∞ ∥A + A0 ∥L∞ ∥A − A0 ∥L∞ |∇u|2 ∫ Ω . ϵ0 (1 + ∥η0 ∥H 5/2 (Σ) )3 |Du|2 Ω . ϵ0 (1 + ∥η0 ∥H 5/2 (Σ) ) ∥u∥2H 1 (Ω) . 3 For given initial data, we can always take ϵ0 sufficiently small to absorb (2.2.44) into (2.2.43). Then we have ∫ 1 J |DA u|2 & ∥u∥2H 1 (Ω) . (2.2.45) Ω (1 + ∥η0 ∥H 5/2 (Σ) )3 The first part of (2.2.41) is verified. For the second part, notice ∫ ∫ J |DA u| 2 . (1 + ∥η∥H 5/2 ) |DA u|2 (2.2.46) Ω Ω . (1 + ∥η0 ∥H 5/2 ) max{1, ∥AK∥2L∞ , ∥BK∥2L∞ , ∥K∥2L∞ } ∥u∥2H 1 (Ω) . Since it is easy to see max{1, ∥AK∥2L∞ , ∥BK∥2L∞ , ∥K∥2L∞ } . 1 + (1 + ∥∇¯ η ∥2L∞ ) ∥K∥2L∞ (2.2.47) . (1 + ∥η0 ∥H 5/2 (Σ) )2 , then the second part of (2.2.41) follows. Remark 2.2.10. Throughout this section, we can always assume the restriction of η depending on ϵ0 is justified, and finally we verify this closeness condition for t ∈ [0, T ] 27 in the iteration. It is easy to see [26, Lemma 2.4] still holds in our settings. In the following, we give an updated version of [26, Lemma 2.5]. Define    K 0 0    M = M (t) = K∇Φ =   0 K 0  . (2.2.48)   AK BK 1 Then M induces a linear operator Mt : u → Mt (u) = M (t)u. Lemma 2.2.11. For each t ∈ [0, T ], k = 0, 1, 2, Mt is a bounded linear iso- morphism from H k (Ω) to H k (Ω) and from 0 Hσ1 (Ω) to X (t). The constant is giv- en by P (∥η(t)∥H 7/2 (Σ) ). If we further define M by Mu(t) = Mt u(t), then it is a bounded linear isomorphism from L2 ([0, T ]; H k (Ω)) to L2 ([0, T ]; H k (Ω)) and from L2 ([0, T ];0 Hσ1 (Ω)) to XT . The constant is given by P (sup0≤t≤T ∥η(t)∥H 7/2 (Σ) ). Proof. It is easy to see for each t ∈ [0, T ], ∥Mt u∥H 0 . ∥Mt ∥C 0 ∥u∥H 0 . P (∥η(t)∥H 5/2 (Σ) ) ∥u∥H 0 , (2.2.49) ∥Mt u∥H 1 . ∥Mt ∥C 1 ∥u∥H 1 . P (∥η(t)∥H 7/2 (Σ) ) ∥u∥H 1 , (2.2.50) (2.2.51) ∥Mt u∥H 2 . ∥Mt ∥C 1 ∥u∥H 2 + ∥u∥C 0 ∥Mt ∥H 2 . P (∥η(t)∥H 7/2 (Σ) ) ∥u∥H 2 . This implies Mt is a bounded operator from H k to H k . Since Mt is invertible, we can estimate ∥M−1 v∥H k . P (∥η(t)∥H 7/2 (Σ) ) ∥v∥H k . Hence, Mt is an isomor- phism between H k . Also [26, (2-18)] implies Mt maps divergence-free functions to divergence-A -free functions. Then it is also an isomorphism from 0 Hσ1 (Ω) to X (t). 28 A similar argument can justify the case of L2 ([0, T ]; H k ) and XT . Pressure as a Multiplier It is well-known that in the Navier-Stokes equations, pressure can be taken as a multiplier. With the free surface involved, [26, Proposition 2.9] gives a construction of pressure from the transformed equation (2.1.12), which is valid for small data. With obvious modifications, we can achieve the following result. Proposition 2.2.12. If Λt ∈ (H1 (t))∗ such that Λt (v) = 0 for all v ∈ X (t), then there exists a unique p(t) ∈ H0 (t) such that ⟨p(t), ∇A · v⟩H0 = Λt (v) f or all v ∈ H1 (t) (2.2.52) and ∥p(t)∥H0 . P (∥η(t)∥H 7/2 (Σ) )∥Λt ∥(H1 (t))∗ . If Λ ∈ (HT1 )∗ such that Λ(v) = 0 for all v ∈ XT , then there exists a unique p ∈ HT0 such that ⟨p, ∇A · v⟩HT0 = Λ(v) f or all v ∈ HT1 (2.2.53) and ∥p∥H0 . P (sup0≤t≤T ∥η(t)∥H 7/2 (Σ) )∥Λ∥(H1T )∗ . T Proof. Similar to the proof of Lemma 2.2.11, we can easily recover [26, Lemma 2.5 - Lemma 2.8] with only replacing the universal constant C by P (∥η(t)∥H 7/2 (Σ) ). Hence, our results naturally follow. 29 2.2.3 Elliptic Equations In this section, we study two types of elliptic problems, which are employed in deriving the linear estimates. For both equations, we prove well-posedness to the higher order regularity. A -Stokes Equation Let us consider the stationary Navier-Stokes problem.    ∇A · SA (p, u) = F   in Ω,     ∇A · u = G in Ω, (2.2.54)     SA (p, u)N = V on Σ,     u=0 on Σb . Since this problem is stationary, we temporarily ignore the time dependence of η, A , etc. The existence and uniqueness are shown in [26, Lemma 3.5 - Lemma 3.6]. Notice the estimate in [26, Lemma 3.6] does not achieve the optimal regularity of (u, p). Hence, we employ an approximation argument to improve this estimate. This is a similar result as in [26, Lemma 3.7], but with different techniques. For clarity, we divide it into two steps. In the next lemma, we first prove that the constant can actually only depend on the initial free surface. Lemma 2.2.13. Let k ≥ 3 be an integer and suppose η ∈ H k+1/2 (Σ) and η0 ∈ H k+1/2 (Σ). Then there exists ϵ0 > 0 such that if ∥η − η0 ∥H k−3/2 (Σ) ≤ ϵ0 , the solution (u, p) to the system (2.2.54) satisfies ( ) ∥u∥H r + ∥p∥H r−1 . C(η0 ) ∥F ∥H r−2 + ∥G∥H r−1 + ∥V ∥H r−3/2 (Σ) , (2.2.55) 30 for r = 2, . . . , k − 1, whenever the right-hand side is finite, where C(η0 ) is a constant depending on ∥η0 ∥H k+1/2 (Σ) . Proof. Based on [26, Lemma 3.6], we have the estimate ( ) ∥u∥H r + ∥p∥H r−1 . C(η) ∥F ∥H r−2 + ∥G∥H r−1 + ∥V ∥H r−3/2 (Σ) , (2.2.56) for r = 2, . . . , k − 1, whenever the right-hand side is finite, where C(η) is a constant depending on ∥η∥H k+1/2 (Σ) . Define ξ = η − η0 , then ξ ∈ H k+1/2 (Σ). Let us denote A0 and N0 as quantities in terms of η0 . We rewrite the system (2.2.54) as a perturbation of the initial status    ∇A0 · SA0 (p, u) = F + F 0 in Ω,       ∇A · u = G + G0 0 in Ω, (2.2.57)   SA0 (p, u)N0 = V + V 0   on Σ,     u=0 on Σb , where F 0 = ∇A0 −A · SA (p, u) + ∇A0 · SA0 −A (p, u), (2.2.58) G0 = ∇A0 −A · u, (2.2.59) V 0 = SA0 (p, u)(N0 − N ) + SA0 −A (p, u). (2.2.60) Suppose ∥ξ∥H k−3/2 (Σ) ≤ 1, which implies ∥ξ∥lH k−3/2 (Σ) ≤ ∥ξ∥H k−3/2 (Σ) < 1 for any l > 1. 31 A straightforward calculation reveals 0 F r−2 ≤ C(1 + ∥η0 ∥ k+1/2 )4 ∥ξ∥ k−3/2 (∥u∥ r + ∥p∥ r−1 ),(2.2.61) H H (Σ) H (Σ) H H 0 G r−1 ≤ C(1 + ∥η0 ∥ k+1/2 )2 ∥ξ∥ k−3/2 ∥u∥ r , (2.2.62) H H (Σ) H (Σ) H ∥V ∥H r−3/2 (Σ) ≤ C(1 + ∥η0 ∥H k+1/2 (Σ) )2 ∥ξ∥H k−3/2 (Σ) (∥u∥H r + ∥p∥H r−1 ),(2.2.63) for r = 2, . . . , k − 1. Since the initial surface function η0 satisfies all the requirements of [26, Lemma 3.6], we arrive at the estimate that for r = 2, . . . , k − 1 ( ) ∥u∥H r +∥p∥H r−1 . C(η0 ) F + F H r−2 + G + G H r−1 + V + V H r−3/2 (Σ) , 0 0 0 (2.2.64) where C(η0 ) is a constant depending on ∥η0 ∥H k+1/2 (Σ) . Combining all above, we obtain ∥u∥H r + ∥p∥H r−1 (2.2.65) ( ) . C(η0 ) ∥F ∥H r−2 + ∥G∥H r−1 + ∥V ∥H r−3/2 (Σ) +C(η0 )(1 + ∥η0 ∥H k+1/2 (Σ) )4 ∥ξ∥H k−3/2 (Σ) (∥u∥H r + ∥p∥H r−1 ). So if { } 1 1 ∥ξ∥H k−3/2 (Σ) ≤ min , , (2.2.66) 2 4CC(η0 )(1 + ∥η0 ∥H k+1/2 (Σ) )4 we can absorb the extra terms on the right-hand side of (2.2.65) into the left-hand side and get a succinct form ( ) ∥u∥H r + ∥p∥H r−1 . C(η0 ) ∥F ∥H r−2 + ∥G∥H r−1 + ∥V ∥H r−3/2 (Σ) , (2.2.67) 32 for r = 2, . . . , k − 1. The next result allows us to achieve the optimal regularity. Proposition 2.2.14. Let k ≥ 3 be an integer. Suppose η ∈ H k+1/2 (Σ) and η0 ∈ H k+1/2 (Σ) satisfying ∥η − η0 ∥H k+1/2 (Σ) ≤ ϵ0 . Then the solution (u, p) to the system (2.2.54) satisfies ( ) ∥u∥H r + ∥p∥H r−1 . C(η0 ) ∥F ∥H r−2 + ∥G∥H r−1 + ∥V ∥H r−3/2 (Σ) , (2.2.68) for r = 2, . . . , k + 1, whenever the right-hand side is finite, where C(η0 ) is a constant depending on ∥η0 ∥H k+1/2 (Σ) . Proof. If r ≤ k − 1, then this is just the conclusion of Lemma 2.2.13, so our main goal is to gain more regularity for r = k and r = k + 1. The main ideas here is utilizing the equations to control the higher order vertical derivatives by the horizontal derivatives. The weight A makes this procedure quite complicated. In the following, we first define an approximating sequence for η. In the infinite case, we let ρ ∈ C0∞ (R2 ) be such that supp(ρ) ⊂ B(0, 2) and ρ = 1 for B(0, 1). For m ∈ N, define η m by ηˆm (ξ) = ρ(ξ/m)ˆ η (ξ) whereˆdenotes the Fourier transform. For each m, η m ∈ H j (Σ) for arbitrary j ≥ 0 and also η m → η in H k+1/2 (Σ) as m → ∞. In the periodic case, we define ηˆm by throwing away the higher frequencies: ηˆm (n) = 0 for |n| ≥ m. Then η m has the same convergence property as above. Let A m and N m be defined in terms of η m . Consider the problem (2.2.54) with A and N replaced by A m and N m . Since η m ∈ H k+5/2 , we can apply [26, Lemma 3.6] to deduce the existence of (um , pm ) that 33 solves    ∇A m · SA m (pm , um ) = F in Ω,       ∇A · um = G in Ω, (2.2.69)   SA m (pm , um )N0 = V   on Σ,     um = 0 on Σb , and such that (2.2.70) ( ) ∥um ∥H r + ∥pm ∥H r−1 . C(∥η ∥H k+5/2 (Σ) ) ∥F ∥H r−2 + ∥G∥H r−1 + ∥V ∥H r−3/2 (Σ) , m for r = 2, . . . , k + 1. We can rewrite above equations in the following shape:    −∆A m um + ∇A m pm = F + ∇A m G   in Ω,     ∇A m · um = G in Ω, (2.2.71)     (pm I − DA m um )N m = V on Σ,     um = 0 on Σb . In the following, we prove an improved estimate for (um , pm ) in terms of ∥η m ∥H k+1/2 (Σ) . We divide the proof into several steps. Step 1: Preliminaries. To abuse the notation, within the following procedure, we always use (u, p, η) instead of (um , pm , η m ) to make the expressions succinct. Also it is easy to see the term ∇A m G does not affect the estimates because ∥∇A m G∥H k−1 . ∥η m ∥H k+1/2 (Σ) ∥G∥H k . Hence, we still write F to indicate the forcing terms. 34 We write explicitly each term in above equations, ∂11 u1 + ∂22 u1 + (1 + A2 + B 2 )K 2 ∂33 u1 − 2AK∂13 u1 − 2BK∂23 u1 +(AK∂3 (AK) + BK∂3 (BK) − ∂1 (AK) − ∂2 (BK) + K∂3 K)∂3 u1 + ∂1 p − AK∂3 p = F1 , (2.2.72) ∂11 u2 + ∂22 u2 + (1 + A2 + B 2 )K 2 ∂33 u2 − 2AK∂13 u2 − 2BK∂23 u2 +(AK∂3 (AK) + BK∂3 (BK) − ∂1 (AK) − ∂2 (BK) + K∂3 K)∂3 u2 + ∂2 p − BK∂3 p = F2 , (2.2.73) ∂11 u3 + ∂22 u3 + (1 + A2 + B 2 )K 2 ∂33 u3 − 2AK∂13 u3 − 2BK∂23 u3 +(AK∂3 (AK) + BK∂3 (BK) − ∂1 (AK) − ∂2 (BK) + K∂3 K)∂3 u3 + K∂3 p = F3 , (2.2.74) ∂1 u1 − AK∂3 u1 + ∂2 u2 − BK∂3 u2 + K∂3 u3 = G, (2.2.75) For convenience, we define ( ) Z = C(η0 )P (η) ∥F ∥H k−1 + ∥G∥H k + ∥V ∥H k−1/2 (Σ) , 2 2 2 (2.2.76) where C(η0 ) is a constant depending on ∥η0 ∥H k+1/2 (Σ) and P (η) is a polynomial of ∥η∥H k+1/2 (Σ) . Step 2: r = k case. With the (k − 1)th order elliptic estimate, we have ( ) ∥u∥2H k−1 + ∥p∥2H k−2 ≤ C(η0 ) ∥F ∥2H k−3 + ∥G∥2H k−2 + ∥V ∥2H k−5/2 (Σ) . Z. (2.2.77) By Lemma 2.2.13, the constant C(η0 ) only depends on ∥η0 ∥H k+1/2 . For i = 1, 2, 35 (∂i u, ∂i p) satisfies the equations    −∆A (∂i u) + ∇A (∂i p) = F¯   in Ω,     ∇A · (∂i u) = G¯ in Ω, (2.2.78)     ((∂i p)I − DA (∂i u))N = V¯ on Σ,     ∂u=0 i on Σb , where F¯ = ∂i F + ∇∂i A · ∇A u + ∇A · ∇∂i A u − ∇∂i A p, (2.2.79) ¯ = ∂i G − ∇∂ A · u, G (2.2.80) i V¯ = ∂i V − (pI − DA u)∂i N + D∂i A uN . (2.2.81) Employing the (k − 1)th order elliptic estimate, we have ( ) 2 2 2 ∥∂i u∥2H k−1 + ∥∂i p∥2H k−2 . C(η0 ) F H k−3 + G H k−2 + V H k−5/2 (Σ) . ¯ ¯ ¯ (2.2.82) Except for the derivatives of F , G and V , all the other terms on the right-hand side have the form ∥A · B∥H r , in which A = ∂ α η¯ and B = ∂ β u or ∂ β p. These kinds of estimates can be achieved by [26, Lemma A.1 - Lemma A.2]. Hence, we have the forcing estimate 2 2 ¯ k−2 + V¯ 2 k−5/2 F¯ k−3 + G (2.2.83) H H H ( ) . ∥F ∥2H k−2 + ∥G∥2H k−1 + ∥V ∥H k−3/2 (Σ) + C(η0 )P (η) ∥u∥H k−1 + ∥p∥H k−2 2 2 2 . Z. 36 In detail, this means ∥∂1 u∥2H k−1 + ∥∂1 p∥2H k−2 + ∥∂2 u∥2H k−1 + ∥∂2 p∥2H k−2 . Z. (2.2.84) Hence, most parts in ∥u∥H k + ∥p∥H k−1 have been covered by this estimate, except those with the highest order derivative of ∂3 . Multiplying (2.2.74) by A and adding it to (2.2.72) eliminate the ∂3 p term and lead to (∂11 u1 + A∂11 u3 ) + (∂22 u1 + A∂22 u3 ) + (1 + A2 + B 2 )K 2 (∂33 u1 + A∂33 u3 ) −2AK(∂13 u1 + A∂13 u3 ) − 2BK(∂23 u1 + A∂23 u3 ) +(AK∂3 (AK) + BK∂3 (BK) − ∂1 (AK) − ∂2 (BK) + K∂3 K)(∂1 u1 + A∂1 u3 ) +∂1 p = F1 + AF3 . (2.2.85) Then taking derivative ∂3k−2 on both sides, we can focus on the term (1 + A2 + B 2 )K 2 (∂3k u1 + A∂3k u3 ). The estimates of all the other terms in H 0 norm imply k ∂3 u1 + A∂3k u3 2 0 . Z. (2.2.86) H Similarly, we have k ∂3 u2 + B∂3k u3 2 0 . Z. (2.2.87) H Rearranging the terms in (2.2.75), we get K(1 + A2 + B 2 )∂3 u3 = G − ∂1 u1 − ∂2 u2 + AK(∂3 u1 + A∂3 u3 ) + BK(∂3 u2 + B∂3 u3 ). (2.2.88) Taking derivative ∂3k−1 on both sides, we obtain k 2 ∂3 u3 0 . Z. (2.2.89) H 37 Combining (2.2.86), (2.2.87) and (2.2.89), it is easy to see k 2 ∂3 u1 0 + ∂3k u2 2 0 . Z. (2.2.90) H H Plugging this into (2.2.74) and taking derivative ∂3k−2 on both sides, we get k−1 2 ∂3 p 0 . Z. (2.2.91) H Combining this with all above estimates, we have proved ∥u∥2H k + ∥p∥2H k−1 . Z. (2.2.92) This is the result for the case r = k. Step 3: r = k + 1 case: horizontal derivatives. We can take ∂j derivative for i, j = 1, 2 on both sides of (2.2.78). Then by a similar argument as in Step 2, we obtain ∥∂11 u∥2H k−1 + ∥∂12 u∥2H k−1 + ∥∂22 u∥2H k−1 + ∥∂11 p∥2H k−2 + ∥∂12 p∥2H k−2 + ∥∂22 p∥2H k−2 . Z. (2.2.93) Step 4: r = k + 1 case: mixed derivatives. Similar to the r = k case argument, for i = 1, 2, taking derivative ∂3k−2 ∂i on both sides of (2.2.85) and focusing on the term ∂3k ∂i u1 + A∂3k ∂i u3 , we get k ∂3 ∂i u1 + A∂3k ∂i u3 2 0 . Z. (2.2.94) H 38 Similarly, we have k ∂3 ∂i u2 + B∂3k ∂i u3 2 0 . Z. (2.2.95) H Plugging this result into (2.2.85) and taking derivative ∂3k−2 ∂i on both sides, we get k ∂3 ∂i u3 2 0 . Z. (2.2.96) H The direct estimates for these three terms imply k 2 ∂3 ∂i u 0 . Z. (2.2.97) H Plugging this into (2.2.74) and taking derivative ∂3k−2 ∂i on both sides, we get k−1 2 ∂3 ∂i p 0 . Z. (2.2.98) H Combining all above estimates, we have shown ∥∂3i u∥2H k−1 + ∥∂3i p∥2H k−2 . Z. (2.2.99) This is actually ∥∂13 u∥2H k−1 + ∥∂23 u∥2H k−1 + ∥∂13 p∥2H k−2 + ∥∂23 p∥2H k−2 . Z. (2.2.100) Step 5: r = k + 1 case: vertical derivatives. Again we use the same trick as above. Taking derivative ∂3k−1 on both sides of (2.2.85) and focusing on the term ∂3k+1 u1 +A∂3k+1 u3 , we can bound ∂3k+1 u1 −A∂3k+1 u3 . Similarly, we control ∂3k+1 u2 − B∂3k+1 u3 . Plugging this result into (2.2.75) and taking derivative 39 ∂3k−1 on both sides, we can estimate ∂3k+1 u3 . Then we have k+1 2 ∂3 u 0 . Z. (2.2.101) H Plugging this into (2.2.74) and taking derivative ∂3k−1 on both sides, we get k 2 ∂3 p 0 . Z. (2.2.102) H Combining all above estimate, we get ∥∂33 u∥2H k−1 + ∥∂33 p∥2H k−2 . Z. (2.2.103) Step 6: r = k + 1 case: conclusion. To synthesize, (2.2.93), (2.2.100) and (2.2.103) imply all the second order derivatives of u in the (k − 1)th order Sobolev norm and p in the (k − 2)th order Sobolev norm are bounded, so we naturally have the estimate ( ) ∥u∥2H k+1 + ∥p∥2H k . C(η0 )P (η) ∥F ∥H k−1 + ∥G∥H k + ∥V ∥H k−1/2 . 2 2 2 (2.2.104) Now let us go back to the original notation. Since P (·) is a fixed polynomial, this 40 gives the estimate ∥um ∥2H k+1 + ∥pm ∥2H k (2.2.105) ( ) . C(η0 )P (η ) ∥F ∥H k−1 + ∥G∥H k + ∥V ∥H k−1/2 (Σ) m 2 2 2 ( ) . C(η0 )P (η) ∥F ∥H k−1 + ∥G∥H k + ∥V ∥H k−1/2 (Σ) 2 2 2 ( ) . C(η0 )P (η0 ) ∥F ∥H k−1 + ∥G∥H k + ∥V ∥H k−1/2 (Σ) 2 2 2 ( ) . C(η0 ) ∥F ∥H k−1 + ∥G∥H k + ∥V ∥H k−1/2 (Σ) , 2 2 2 where C(η0 ) depends on ∥η0 ∥H k+1/2 (Σ) . This bound implies the sequence (um , pm ) is uniformly bounded in H k+1 × H k , so we can extract weakly convergent subsequence um ⇀ u0 and pm ⇀ p0 . In the second equation of (2.2.69), we multiply both sides by J m w for w ∈ C0∞ to achieve (2.2.106) ∫ ∫ ∫ ∫ GwJ m = (∇A m · um )wJ m = − um · (∇A m w)J m → − u0 · (∇A w)J Ω ∫Ω Ω Ω = (∇A · u0 )wJ, Ω which implies ∇A · u0 = G. Then we multiply the first equation of (2.2.69) by wJ m for w ∈ H1 and integrate by parts to get ∫ ∫ ∫ ∫ 1 DA m u : DA m wJ − m m p ∇A m · wJ m m = F · wJ − m V · w. (2.2.107) 2 Ω Ω Ω Σ Passing to the limit m → ∞, we deduce ∫ ∫ ∫ ∫ 1 DA u : DA wJ − 0 p ∇A · wJ = 0 F · wJ − V · w, (2.2.108) 2 Ω Ω Ω Σ 41 which reveals, upon integrating by parts again, (u0 , p0 ) satisfies (2.2.54). Since (u, p) is the unique strong solution to (2.2.54), we have u0 = u and p0 = p. The weak lower semi-continuity and the estimate (2.2.105) imply (2.2.68). A -Poisson Equation We consider the elliptic problem      ∆A p = f in Ω,  p=g on Σ, (2.2.109)      ∇A p · ν = v on Σb . The existence and uniqueness of this equation can be found in [26, Lemma 3.8]. A similar argument as in Proposition 2.2.14 shows the following result. Proposition 2.2.15. Let k ≥ 3 be an integer. Suppose η ∈ H k+1/2 (Σ) and η0 ∈ H k+1/2 (Σ). Then there exists a ϵ0 > 0 such that if ∥η − η0 ∥H k−3/2 ≤ ϵ0 , solution p to the system (2.2.109) satisfies ( ) ∥p∥H r . C(η0 ) ∥f ∥H r−2 + ∥g∥H r−1/2 (Σ) + ∥v∥H r−3/2 (Σb ) , (2.2.110) for r = 2, . . . , k − 1, whenever the right-hand side is finite, where C(η0 ) is a constant depending on ∥η0 ∥H k+1/2 (Σ) . 42 2.2.4 Linear Navier-Stokes Equations In this section, we show the well-posedness of the linear Navier-Stokes problem    ∂t u − ∆A u + ∇A u = F   in Ω,     ∇A · u = 0 in Ω, (2.2.111)     (pI − DA u)N = V on Σ,     u=0 on Σb . Following the path of I. Tice and Y. Guo in [26], the result is quite obvious and we only present the main propositions. Initial Data and Compatibility Conditions We need to define the energy and dissipation in our estimates. It is noticeable that our definitions are slightly different from those in [26]. Here are some quantities related to the initial data. ∑ N j 2 ∑ N −1 j H0 (u, p) = ∂t u(0) H 2N −2j + ∂t p(0) 2 2N −2j−1 , (2.2.112) H j=0 j=0 K0 (u0 ) = ∥u0 ∥2H 2N , (2.2.113) ∑ N j H0 (η) = ∥η(0)∥2H 2N +1/2 (Σ) + ∂t η(0) 2 2N −2j+3/2 , (2.2.114) H (Σ) j=1 K0 (η0 ) = ∥η0 ∥2H 2N +1/2 (Σ) , (2.2.115) ∑ N −1 j ∑ N −1 j H0 (F, V ) = ∂t F (0) 2 2N −2j−2 + ∂t V (0) 2 2N −2j−3/2 . (2.2.116) H H (Σ) j=0 j=0 43 Furthermore, we need to define mappings G 1 on Ω and G 2 on Σ. G 1 (v, q) = −(R + ∂t JK)∆A v − ∂t Rv + (∂t JK + R + RT )∇A q (2.2.117) +∇A · (DA (Rv) − RDA v + D∂t A v), G 2 (v, q) = DA (Rv)N − (qI − DA v)∂t N + D∂t A vN . (2.2.118) The mappings above allow us to define the higher order forcing terms. For j = 1, . . . , N , we define F 0 = F, (2.2.119) V 0 = V, (2.2.120) F j = Dt F j−1 + G 1 (Dtj−1 u, ∂tj−1 p) (2.2.121) ∑j−1 j = Dt F + Dtl G1 (Dtj−l−1 u, ∂tj−l−1 p), l=0 V j = Dt V + G 2 (Dtj−1 u, ∂tj−1 p) j−1 (2.2.122) ∑ j−1 j = Dt V + Dtl G2 (Dtj−l−1 u, ∂tj−l−1 p). l=0 Finally, we define the general quantities we need to estimate as follows: ∑ N j 2 ∑ N −1 j 2 E(u, p) = ∂t u L∞ H 2N −2j + ∂t p ∞ 2N −2j−1 , (2.2.123) L H j=0 j=0 ∑ N j 2 ∑ N −1 j 2 D(u, p) = ∂t u 2 2N −2j+1 + ∂t p 2 2N −2j , (2.2.124) L H L H j=0 j=0 K(u, p) = E(u, p) + D(u, p), (2.2.125) 44 ∑ N j 2 E(η) = ∥η∥2L∞ H 2N +1/2 (Σ) + ∂t η ∞ 2N −2j+3/2 , (2.2.126) L H (Σ) j=1 ∑ N +1 j 2 D(η) = ∂t η 2 2N −2j+5/2 , (2.2.127) L H (Σ) j=2 K(η) = E(η) + D(η), (2.2.128) (2.2.129) ∑ N −1 j 2 ∑ N j 2 K(F, V ) = ∂t F 2 2N −2j−1 + ∂tN F 2 ∗ + ∂t V 2 2N −2j−1/2 , L H X T L H (Σ) j=0 j=0 ∑ N −1 j 2 ∑ N −1 j 2 + ∂t F ∞ 2N −2j−2 + ∂t V ∞ 2N −2j−3/2 . L H L H (Σ) j=0 j=0 For j = 0, . . . , N − 1, we say that the j th compatibility conditions are satisfied if    Dj u(0) ∈ H 2 (Ω) ∩ X (0), t (2.2.130)   Π0 (V j (0) + DA0 Dtj u(0)N0 ) = 0, where Dt u = ∂t u − Ru f or R = ∂t M M −1 , (2.2.131) M is defined in (2.2.48) and Π0 (v) = v − (v · N0 )N0 |N0 |−2 f or N0 = (−∂1 η0 , −∂2 η0 , 1). (2.2.132) The construction of the higher order initial data satisfying these conditions are dis- cussed carefully in [26, pp.327-328]. Under this construction, we have the estimate H0 (u, p) . P (H0 (η)) + P (K0 (u0 )) + P (H0 (F, V )). (2.2.133) 45 Local Well-posedness of Linear System Theorem 2.2.16. Suppose K0 (u0 ) < ∞, K(η) < ∞ and the initial conditions are constructed to satisfy the j th compatibility conditions (2.2.130) for j = 0, 1, . . . , N −1, with the forcing functions satisfying K(F, V ) + H0 (F, V ) < ∞. Then there exists a universal constant T0 (η) > 0, which depends on L(η) as defined in the following, such that if 0 < T < T0 (η), then there exists a unique strong solution (u, p) to the system (2.2.111) on [0, T ] such that ∂tj u ∈ L2 ([0, T ]; H 2N −2j+1 ) ∩ C 0 ([0, T ]; H 2N −2j ) f or j = 0, . . . , N, (2.2.134) ∂tj p ∈ L2 ([0, T ]; H 2N −2j ) ∩ C 0 ([0, T ]; H 2N −2j−1 ) f or j = 0, . . . , N − 1, ∂tN +1 u ∈ XT∗ . Also, the pair (u, p) satisfies the estimate ( ) K(u, p) . L(η) P (K0 (u0 )) + P (H0 (F, V )) + P (K(F, V )) , (2.2.135) where ( ) L(η) = P (K(η)) exp T P (K(η)) . (2.2.136) Furthermore, the pair (Dtj u, ∂tj p) satisfies the system    ∂t (Dtj u) − ∆A (Dtj u) + ∇A (∂tj p) = F j in   Ω,     ∇A · (Dj u) = 0 t in Ω, (2.2.137)   SA (∂tj p, Dtj u)N = V j   on Σ,     Dj u = 0 t on Σb , in the strong sense for j = 0, . . . , N − 1 and in the weak sense for j = N . 46 Proof. This theorem is almost identical to [26, Theorem 4.8]. Since our free surface η can be general, the only difference here is we keep the polynomials of K(η) and H0 (η) in the final estimate instead of further simplifying them. 2.2.5 Transport Equation In this section, we prove the well-posedness of the transport equation    ∂t η + u1 ∂1 η + u2 ∂2 η = u3 on Σ, (2.2.138)   η(t = 0) = η0 on Σ. Local Well-posedness of Transport Equation Define the quantity ∑ N j 2 ∑ N −1 j 2 Q(u) = ∂t u L2 H 2N −2j+1 + ∂t u ∞ 2N −2j . (2.2.139) L H j=0 j=0 Obviously, we have the relation Q(u) ≤ K(u, p). Simply taking temporal derivative on both sides of (4.1.1), we can iteratively define the higher order initial data, which satisfy the estimate H0 (η0 ) . P (K0 (η0 )) + P (H0 (u, p)). (2.2.140) Theorem 2.2.17. Suppose K0 (η0 ) < ∞ and Q(u) < ∞. Then the problem (4.1.1) admits a unique solution η obeying K(η) < ∞ and satisfying the initial data ∂tj η(0) for j = 0, . . . , N . Moreover, there exists a 0 < T¯(u) < 1, depending on Q(u), such 47 that if 0 < T < T¯, then we have the estimate K(η) . P (K0 (η0 )) + P (Q(u)), (2.2.141) where P (·) is a polynomial satisfying P (0) = 0. Proof. This can be done following a similar path as in [26, Lemma 5.4], so we omit it here. Next, we introduce a lemma to describe the difference between η and η0 in a small time period. Lemma 2.2.18. If Q(u) + K0 (η0 ) < ∞ and H0 (u, p) < ∞, then for any θ > 0, there exists a T˜(θ) > 0 depending on Q(u) and K0 (η0 ), such that for any 0 < T < T˜, we have the estimate ∥η − η0 ∥2L∞ H 2N +1/2 (Σ) ≤ θ. (2.2.142) Proof. ξ = η − η0 satisfies the transport equation    ∂t ξ + u1 ∂1 ξ + u2 ∂2 ξ = u3 − u1 ∂1 η0 − u2 ∂2 η0 on Σ, (2.2.143)   ξ(0) = 0. 48 By the transport estimate in [26, Lemma A.11], we have ∥ξ∥L∞ H 2N +1/2 (Σ) (2.2.144) ( ∫ T ) ≤ exp C ∥u(t)∥H 2N +1/2 (Σ) dt 0 (∫ T ) × ∥u3 (t) − u1 (t)∂1 η0 − u2 (t)∂2 η0 ∥H 2N +1/2 (Σ) dt 0 ( )( ) √ √ . T exp T Q(u) P (K0 (η0 )) + P (H0 (u, p)) + P (Q(u)) . Hence, when T is small enough, our result naturally follows. Forcing Estimates Now we need to estimate the forcing terms which appear on the right-hand sides of the linear Navier-Stokes equations. Compared to the system (2.1.12), the forcing terms are as follows: F = ∂t η¯˜bK∂3 u − u · ∇A u, (2.2.145) V = ηN . (2.2.146) Recall we define the forcing quantities as follows: (2.2.147) ∑ N −1 j 2 2 ∑ N j 2 K(F, V ) = ∂t F 2 + ∂tN F X ∗ + ∂t V 2 2N −2j−1/2 L H 2N −2j−1 T L H (Σ) j=0 j=0 ∑ N −1 j 2 ∑ N −1 j 2 + ∂t F ∞ 2N −2j−2 + ∂t V ∞ , L H L H 2N −2j−3/2 (Σ) j=0 j=0 ∑ N −1 j ∑ N −1 j H0 (F, V ) = ∂t F (0) 2 2N −2j−2 + ∂t V (0) 2 2N −2j−3/2 . (2.2.148) H H (Σ) j=0 j=0 49 In the estimates of the Navier-Stokes-transport system, we also need some other forcing quantities. (2.2.149) ∑ N −1 j 2 N 2 ∑ N j 2 F(F, V ) = ∂t F 2 + ∂t F L2 H 0 + ∂t V ∞ 2N −2j−1/2 , L H 2N −2j−1 L H (Σ) j=0 j=0 ∑ N −1 j 2 ∑ N −1 j 2 H(F, V ) = ∂t F 2 + ∂t V 2 . (2.2.150) L H 2N −2j−1 L H 2N −2j−1/2 (Σ) j=0 j=0 Theorem 2.2.19. The forcing terms satisfy the estimates K(F, V ) . P (K(η)) + P (Q(u)), (2.2.151) H0 (F, V ) . P (K0 (η0 )) + P (K0 (u0 ), ) (2.2.152) F(F, V ) . P (K(η)) + P (Q(u)), (2.2.153) ( ) H(F, V ) . T P (K(η)) + P (Q(u)) , (2.2.154) where P (·) is a polynomial with P (0) = 0. Proof. The estimates follow from simple but lengthy computations, involving the standard argument, so we only present a sketch of the proof here. After we expand all the terms with the Leibniz rule and plug into the definitions of F j and V j , it is easy to see the basic form of estimate is ∥XY ∥2L2 H k or ∥XY ∥2L∞ H k where X includes the terms only involving η¯ and Y includes the terms involving u, p, F or V . Employing [26, Lemma A.1 - Lemma A.3], we can always bound ∥XY ∥2L2 H k . ∥X∥2L2 H k1 ∥Y ∥2L∞ H k2 or ∥XY ∥2L2 H k . ∥X∥2L∞ H k1 ∥Y ∥2L2 H k2 , and also ∥XY ∥2L∞ H k . ∥X∥2L∞ H k1 ∥Y ∥2L∞ H k2 . The principle of choice in these two options for L2 H k is not to exceed the highest order of derivatives in the right-hand sides of the estimates. A detailed analysis shows this goal can always be achieved. In (2.2.151), 50 note the trivial bound N 2 ∂t F ∗ . ∂tN F 2 2 0 . (2.2.155) X T L H Also in (2.2.154), the key part is the appearance of the constant T . We first trivially bound it as follows: ∑ N −1 j 2 ∑ N −1 j 2 ∂t F 2 2N −2j−1 + ∂t V 2 2N −2j−1/2 (2.2.156) L H L H (Σ) j=0 j=0 (N ∑ −1 ∑ −1 ) j 2 N j 2 ≤T ∂t F ∞ + ∂t V ∞ 2N −2j−1/2 . L H 2N −2j−1 L H (Σ) j=0 j=0 Then an application of the inequality 2ab ≤ a2 + b2 implies the desired estimate. Remark 2.2.20. The reason why we can get the constant T lies in that, in the nonlinear Navier-Stokes equations, velocity in the nonlinear terms actually has one less derivative than the linear terms, which makes it possible to use Lemma A.1.15. The appearance of T in (2.2.154) plays a key role in the following nonlinear iteration argument. 51 2.2.6 Navier-Stokes-Transport System In this section, we study the nonlinear system    ∂t u − ∂t η¯˜bK∂3 u + u · ∇A u − ∆A u + ∇A p = 0 in Ω,         ∇A · u = 0 in Ω,       SA (p, u)N = η¯N on Σ,      u=0 on Σb , (2.2.157)     u(x, 0) = u0 (x),             ∂t η + u1 ∂1 η + u2 ∂2 η = u3 on Σ,      η(x′ , 0) = η (x′ ). 0 We first show the compatibility conditions and then construct an iteration to show the well-posedness. Initial Data and Compatibility Conditions Define K0 = K0 (u0 ) + K0 (η0 ). (2.2.158) We say (u0 , η0 ) satisfies the N th order compatibility conditions if      ∇A0 · (Dtj u(0)) = 0 in Ω,  Dtj u(0) = 0 on Σb , (2.2.159)      Π0 (V j (0) + DA Dj u(0)N0 ) = 0 on Σ, 0 t 52 for j = 0, . . . , N − 1. The details can be found in [26, pp.338-339]. The following is a direct estimate of this construction. Theorem 2.2.21. Suppose (u0 , η0 ) satisfies K0 < ∞. Let ∂tj u(0), ∂tj η(0) for j = 0, . . . , N and ∂tj p(0) for j = 0, . . . , N − 1 be defined to satisfy the N th compatibility conditions. Then K0 ≤ H0 (u, p) + H0 (η) . P (K0 ) (2.2.160) for a given polynomial P (·) with P (0) = 0. Construction of Iteration For given initial data (u0 , η0 ), by the Sobolev extension theorem in [26, Lemma A.5], there exists a u0 defined in Ω × [0, ∞) such that Q(u0 ) . K0 (u0 ) coinciding with the initial data to the N th order. By solving the transport equation with respect to u0 , Theorem 2.2.17 implies there exists a η 0 defined in Ω × [0, T0 ) such that K(η 0 ) . (1 + K0 (η0 ))P (Q(u0 )) . P (K0 ) < ∞. This is our starting point. For any integer m ≥ 1, define the approximating sequence (um , pm , η m ) on the existence interval [0, Tm ) by solving the system      ∂t um − ∆A m−1 um + ∇A m−1 pm = ∂t η¯m−1˜bK m−1 ∂3 um−1 in Ω,       ∇A m−1 · um = 0 in Ω,      (pm I − DA m−1 um )N m−1 = η m−1 N m−1 on Σ, (2.2.161)     um = 0 on Σb ,            ∂t η m + um ∂1 η m + um ∂2 η m = um on Σ, 1 2 3 53 where (um , η m ) satisfies the common initial data (u0 , η0 ), and A m , N m , K m are given in terms of η m . This is only a formal definition of iteration. In the following theorems, we finally prove this approximating sequence can be defined for any m ∈ N and the existence intervals Tm do not shrink to 0 as m → ∞. Boundedness Theorem Before we start to prove the boundedness result, it is useful to notice that, based on the linear estimate (2.2.135) and the forcing estimates in Lemma 2.2.19, this sequence can always be constructed and we can directly derive an estimate K(um+1 , pm+1 ) + K(η m+1 ) . P (Q(um ) + K(η m ) + K0 ), (2.2.162) when T is sufficiently small, where P (·) is a polynomial satisfying P (0) = 0. Since the initial data may not be small, this estimate cannot meet our requirements. Hence, we have to go back to the energy structure and derive a stronger estimate. Theorem 2.2.22. Assume J 0 > δ > 0 and the initial data (u0 , η0 ) satisfies the N th compatibility condition. Then there exists a constant 0 < Z < ∞ and 0 < T¯ < 1 depending on K0 , such that if 0 < T ≤ T¯ and K0 < ∞, then there exists an infinite sequence (um , pm , η m )∞ m=0 satisfying the iteration equation (2.2.161) within the existence interval [0, T ) and the following properties. 1. The iteration sequence satisfies the estimate Q(um ) + K(η m ) ≤ Z (2.2.163) 54 for arbitrary m, where the temporal norm is taken with respect to [0, T ). 2. For any m, J m (t) > δ/2 for 0 ≤ t ≤ T . Proof. This theorem is basically an updated version of [26, Theorem 6.1]. However, this proof is much more complicated than the original one. The original proof directly applies the estimates in the Navier-Stokes equations and the transport equation. In our case, without the smallness assumption, the iteration will not be bounded with the same method. Hence, we have to go back to the natural energy structure. Let’s denote the above two assertions related to m as statement Pm . We use an induction to prove this theorem. Step 1: P0 case. This is only related to the initial data. Obviously, the construction of u0 leads to Q(u0 ) . K0 . By the transport estimate in Lemma 2.2.17, we have K(η 0 ) . P (K0 ) for P (0) = 0. We can choose Z ≥ P (K0 ), so the first assertion is verified. Define ξ 0 = η 0 − η0 the difference between the free surface at later time and its initial data. Then the estimate in Lemma 2.2.18 implies ∥ξ 0 ∥L∞ H 5/2 . T Z. Naturally, supt∈[0,T ] |J 0 (t) − J 0 (0)| . ∥ξ 0 ∥L∞ H 5/2 . T Z. Thus when we take T ≤ δ/(2Z), then J 0 (t) ≥ δ/2. So the second assertion is also verified. In a similar fashion, we can verify the closeness assumption in Remark 2.2.10. Hence, P0 is true. In the following, we assume Pm−1 is true for m ≥ 1 and prove Pm is also true. As long as we can show this, by induction, Pn is valid for arbitrary n ∈ N. Cer- tainly, the induction hypothesis and (2.2.162) imply Q(um−1 ) + K(η m−1 ) ≤ Z and K(um , pm ) + K(η m ) ≤ CP (K0 + Z). 55 Step 2: Pm case: estimates of um via energy structure. By Theorem 2.2.16, the pair (DtN um , ∂tN um ) satisfies the equations (2.2.137) in the weak sense, i.e.    ∂t (DtN um ) − ∆A m−1 (DtN um ) + ∇A m−1 (∂tN pm ) = F N   in Ω,     ∇A m−1 · (DN um ) = 0 t in Ω, (2.2.164)  S m−1 (DN um , ∂ N pm )N = V N    A on Σ,   t t   DN um = 0 t on Σb , where F N and V N are given in terms of um and η m−1 . Hence, we have the standard weak formulation, i.e. for any test function ψ ∈ XTm−1 , the following holds 1 ⟨∂t DtN um , ψ⟩HT0 + ⟨DA m−1 DtN um , DA m−1 ψ⟩HT0 = ⟨DtN um , F N ⟩HT0 − ⟨DtN um , V N ⟩0,Σ,T . 2 (2.2.165) Therefore, when plugging into the test function ψ = DtN um , we have the natural energy structure ∫ ∫ ∫ 1 N m 2 1 t 2 J Dt u + J DA m−1 DtN um = (2.2.166) 2 2 0 Ω ∫ Ω ∫ ∫ 1 N m 2 1 t 2 J(0) Dt u (0) + ∂t J DtN um 2 Ω 2 0 Ω ∫ t∫ ∫ t∫ + JF · Dt u − N N m V N · DtN um . 0 Ω 0 Ω We can easily give the preliminary estimates for the left-hand side (LHS) and the 56 right-hand side (RHS). 2 2 LHS & DtN um L∞ H 0 + DtN um L2 H 1 , (2.2.167) 2 RHS . P (K0 ) + T Z DtN um L∞ H 0 (2.2.168) √ √ + T Z F N L2 H 0 DtN um L∞ H 0 + T V N L∞ H −1/2 (Σ) DtN um L2 H 1/2 (Σ) 2 √ 2 . P (K0 ) + T Z DtN um L∞ H 0 + T F N L2 H 0 √ 2 √ 2 √ 2 + T Z 2 DtN um ∞ 0 + T V N ∞ −1/2 + T DtN um 2 1 , L H L H (Σ) L H for a polynomial P (0) = 0. Taking T ≤ 1/(16Z 4 ) and absorbing the extra terms on RHS into LHS imply (2.2.169) N m 2 √ √ Dt u ∞ 0 + DtN um 2 2 1 . P (K0 ) + T F N 2 2 0 + T V N 2 ∞ −1/2 . L H L H L H L H (Σ) 2 Dropping the DtN um L∞ H 0 term, we derive the further estimate for ∂tN um N m 2 √ 2 √ 2 ∂t u 2 . P (K0 ) + T F N L2 H 0 + T V N L∞ H −1/2 (Σ) (2.2.170) L H1 2 + DtN um − ∂tN um L2 H 1 . Then we need to estimate each term on RHS. For the middle two terms, it suffices to show they are bounded, However, for the last term, we need a temporal constant T within the estimate, which can be done by Lemma A.1.15. (2.2.171) (N ∑ −1 ∑ −1 ) N 2 j m 2 N j m 2 F 2 0 . P (K(η m−1 )) ∂t u 2 + ∂t p 2 1 + F(F, V ) L H L H2 L H j=0 j=0 . P (K0 + Z) + F(F, V ), 57 (2.2.172) (N ∑ −1 ∑ −1 ) N 2 j m 2 N j m 2 V ∞ −1/2 . P (K(η m−1 )) ∂t u L ∞ H 2 + ∂t p ∞ 1 + F(F, V ) L H (Σ) L H j=0 j=0 . P (K0 + Z) + F(F, V ), (2.2.173) N m √ N m Dt u − ∂tN um 2 1 . T Dt u − ∂tN um ∞ 1 L H L H (N ∑ −1 ) √ j m 2 . T P (K(η m−1 )) ∂t u L ∞ H 1 j=0 √ . T P (K0 + Z). Therefore, to sum up, we have N m 2 √ √ ∂t u 2 1 . P (K0 ) + T P (K0 + Z) + T F(F, V ). (2.2.174) L H Step 3: Pm case: estimates of um via elliptic estimates. For 0 ≤ n ≤ N − 1, the nth order Navier-Stokes system is as follows.    ∂t (Dtn um ) − ∆A (Dtn um ) + ∇A (∂tn pm ) = F n   in Ω,     ∇A · (Dn um ) = 0 t in Ω, (2.2.175)     SA (Dtn um , ∂tn pm )N = V n on Σ,     D n um = 0 t on Σb , where F n and V n are given in terms of um and η m−1 . 58 A straightforward application of the elliptic estimates reveals (2.2.176) ∥Dtn um ∥2L2 H 2N −2n+1 . ∥∂t Dtn um ∥2L2 H 2N −2n−1 + ∥F n ∥2L2 H 2N −2n−1 + ∥V n ∥2L2 H 2N −2n−1/2 . By triangle’s inequality, we have (2.2.177) 2 ∥∂tn um ∥2L2 H 2N −2n+1 . ∥F n ∥2L2 H 2N −2n−1 + ∥V n ∥2L2 H 2N −2n−1/2 + ∂tn+1 um L2 H 2N −2n−1 + ∥∂t (Dtn um − ∂tn um )∥2L2 H 2N −2n−1 + ∥Dtn um − ∂tn um ∥2L2 H 2N −2n+1 . Then we give a detailed estimates for the terms on RHS. These estimates can be easily obtained as what we did before, so we omit the details here. ∥F n ∥2L2 H 2N −2n−1 (2.2.178) (N ∑ −2 ∑ −2 ) j m 2 N j m 2 . T P (K(η m−1 )) ∂t u L∞ H 2N −2j−1 + ∂t p ∞ 2N −2j−2 + H(F, V ) L H j=0 j=0 . T P (K0 + Z) + H(F, V ), ∥V n ∥2L2 H 2N −2n−1/2 (Σ) (2.2.179) (N ∑ −2 ∑ −2 ) j m 2 N j m 2 . T P (K(η m−1 )) ∂t u ∞ 2N −2j−1 + ∂t p ∞ 2N −2j−2 L H L H j=0 j=0 +H(F, V ) . T P (K0 + Z) + H(F, V ), 59 ∥∂t (Dtn um − ∂tn um )∥2L2 H 2N −2n−1 (2.2.180) (N ∑−1 ) j m 2 . T P (K(η m−1 )) ∂t u L∞ H 2N −2j−1 j=0 . T P (K0 + Z), ∥Dtn um − ∂tn um ∥2L2 H 2N −2n+1 (2.2.181) (N ∑ −2 ) j m 2 . T P (K(η m−1 )) ∂t u ∞ 2N −2j+1 L H j=0 . T P (K0 + Z). Therefore, to sum up, we have ∑ N −1 ∥∂tn um ∥2L2 H 2N −2n+1 . T P (K0 + Z) + H(F, V ). (2.2.182) n=0 Step 4: Pm case: synthesis of estimates for um . Combining (2.2.174), (2.2.182) and Lemma A.1.14, we can deduce √ √ Q(um ) . P (K0 ) + T P (1 + K0 + Z) + T F(F, V ) + H(F, V ). (2.2.183) Then, by the forcing estimates in Lemma 2.2.19, we have F(F, V ) . P (K(η m−1 )) + P (Q(um−1 )) . P (Z) (2.2.184) ( ) H(F, V ) . T P (K(η m−1 )) + P (Q(u m−1 )) . T P (Z). (2.2.185) 60 Hence, to sum up, we achieve the estimate √ Q(um ) . P (K0 ) + T P (K0 + Z). (2.2.186) This is actually √ Q(um ) ≤ CP (K0 ) + T CP (K0 + Z) (2.2.187) for some universal constant C > 0. So we can take Z ≥ 2CP (K0 ). Taking T suffi- ciently small depending on Z, we can bound Q(um ) . 2CP (K0 ) ≤ Z. Step 5: Pm case: estimate of η m via transport estimate. The transport estimate in Lemma 2.2.17 leads to K(η m ) . P (K0 ) + P (Q(um )) (2.2.188) for P (0) = 0. Hence, we can take Z ≥ P (K0 ) + P (2CP (K0 )) to obtain K(η m ) . Z. (2.2.189) Step 6: Pm case: estimate of J m (t). For ξ m = η m − η0 , the transport estimate in Lemma 2.2.18 implies ∥ξ m ∥L∞ H 5/2 . T Z. Naturally, supt∈[0,T ] |J m (t) − J m (0)| . ∥ξ m ∥L∞ H 5/2 . T Z. Thus taking T ≤ δ/(2Z), we have J m (t) ≥ δ/2. A similar argument can justify the closeness assumption in Remark 2.2.10. 61 Synthesis. Above estimates show we can take Z = CP (K0 ) and T small enough depending on Z to obtain Q(um ) + K(η m ) ≤ Z, (2.2.190) and J m (t) ≥ δ/2 f or t ∈ [0, T ]. (2.2.191) Therefore, Pm is verified. By induction, we conclude Pn is valid for any n ∈ N. It is noticeable we can take Z = P (K0 ) where P (·) satisfies P (0) = 0. Theorem 2.2.23. Assume exactly the same conditions as Theorem 2.2.22. Then we have the estimate K(um , pm ) + K(η m ) . P (K0 ) (2.2.192) for a polynomial satisfying P (0) = 0. Proof. This is a natural corollary of above theorem. Since we know Q(um−1 ) . Z, by relation (2.2.162), it implies K(um , pm ) is bounded for any m ∈ N. Contraction Theorem Define N(v, q; T ) = ∥v∥2L∞ H 2 + ∥v∥2L2 H 3 + ∥∂t v∥2L∞ H 0 + ∥∂t v∥2L2 H 1 + ∥q∥2L∞ H 1 + ∥q∥2L2 H 2 , 2 M(ζ; T ) = ∥ζ∥2L∞ H 5/2 (Σ) + ∥∂t ζ∥2L∞ H 3/2 (Σ) + ∥∂t2 ζ∥L2 H 1/2 (Σ) . (2.2.193) Theorem 2.2.24. For j = 1, 2, suppose v j , q j , wj and ζ j satisfy the same initial 62 conditions for different j and the system      ∂t v j − ∆A j v j + ∇A j q j = ∂t ζ¯j ˜bK j ∂3 wj − wj · ∇A j wj in Ω,       ∇A j · v j = 0 in Ω,      SA j (q j , v j )N j = ζ j N j on Σ, (2.2.194)     vj = 0 on Σb ,            ∂t ζ j = w j · N j in Ω, where A j , N j and K j are in terms of ζ j . Assume K(wj , 0), K(v j , q j ) and K(ζ j ) is bounded by Z. Then there exists 0 < T˜ < 1 such that for any 0 < T < T˜, we have the contraction relations 1 N(v 1 − v 2 , q 1 − q 2 ; T ) ≤ N(w1 − w2 , 0; T ), (2.2.195) 2 M(ζ 1 − ζ 2 ; T ) . N(w1 − w2 , 0; T ). (2.2.196) Proof. We divide this proof into several steps. Step 1: Lower order equations. Define v = v 1 − v 2 , q = q 1 − q 2 , w = w1 − w2 and ζ = ζ 1 − ζ 2 , which has trivial initial condition. Then they satisfy the equation as follows:    ∂t v + ∇A1 · SA1 (q, v) = H 1 + ∇A 1 · (D(A 1 −A 2 ) v 2 )   in Ω,     ∇A · v = H 2 1 in Ω, (2.2.197)     SA1 (q, v)N 1 = H 3 + D(A 1 −A 2 ) v 2 N 1 on Σ,     v=0 on Σb , 63 where (2.2.198) H 1 = ∇(A 1 −A 2 ) (DA 2 v 2 ) − (A 1 − A 2 )∇p2 + ∂t ζ 1˜bK 1 ∂3 w + ∂t ζ ˜bK 1 ∂3 w2 +∂t ζ 1˜b(K 1 − K 2 )∂3 w2 − w · ∇A 1 w1 − w2 · ∇A 1 w − w2 · ∇(A 1 −A 2 ) w2 , H 2 = −∇(A 1 −A 2 ) · v 2 , (2.2.199) H 3 = −q 2 (N 1 − N 2 ) + DA 1 v 2 (N 1 − N 2 ) − D(A 1 −A 2 ) v 2 (N 1 − N 2 )(2.2.200) +ζN 1 + ζ 2 (N 1 − N 2 ). The solutions are sufficiently regular for us to differentiate in time, which are the following:    ∂t (∂t v) + ∇A1 · SA1 (∂t q, ∂t v) = H ˜ 1 + ∇A 1 · (D(∂ A 1 −∂ A 2 ) v 2 ) in Ω,    t t    ∇A · ∂t v = H ˜2 in Ω, 1 (2.2.201)   SA1 (∂t q, ∂t v)N ˜ 3 + D(∂ A 1 −∂ A 2 ) v 2 N   1 =H 1 on Σ,   t t   ∂v=0 t on Σb , where (2.2.202) ˜ 1 = ∂t H 1 + ∇∂ A 1 · (D(A 1 −A 2 ) v 2 ) + ∇A 1 · (D(A 1 −A 2 ) ∂t v 2 ) + ∇∂ A 1 · (DA 1 v) H t t +∇A 1 (D∂t A 1 v) − ∇∂t A 1 q, ˜ 2 = ∂t H 2 − ∇∂ A 1 · v, H (2.2.203) t (2.2.204) ˜ 3 = ∂t H 3 + D(A 1 −A 2 ) ∂t v 2 N 1 + D(A 1 −A 2 ) v 2 ∂t N 1 − SA 1 (q, v)∂t N 1 + D∂ A 1 vN 1 . H t Step 2: Energy evolution for ∂t v. 64 Multiply J 1 ∂t v on both sides to get the natural energy structure: ∫ ∫ ∫ 1 1 t |∂t v| J + 2 1 |DA 1 ∂t v|2 J 1 (2.2.205) 2 Ω 2 0 Ω ∫ t∫ ∫ t∫ ∫ t∫ = J H · ∂t v + 1 ˜1 J H ∂t q − 1 ˜2 H ˜ 3 · ∂t v 0 ∫ tΩ∫ 0 ∫Ω t ∫ 0 Σ 1 1 + |∂t v|2 ∂t J 1 − J 1 D(∂t A 1 −∂t A 2 ) v 2 : DA 1 ∂t v. 2 0 Ω 2 0 Ω LHS is simply the energy and dissipation term, so we focus on estimate of RHS. (2.2.206) ∫ t∫ ∫ t √ ˜ 1 ˜ 1 J H · ∂t v . Z 1 ˜1 H 0 ∥∂t v∥H 0 . Z ∥∂t v∥L∞ H 0 T H H L2 H 0 √ 0 Ω 0 √ ˜ 1 . T Z N H v , 2 0 L H ∫ t∫ ∫ ∫ t∫ ˜ 2 ∂t q . J H 1 ˜2 − J qH1 ˜ 2 + J 1 q∂t H (∂t J 1 q H ˜ 2) (2.2.207) 0 Ω ∫Ω 0 Ω √ √ . ˜ 2 + ∥q∥ ∞ 0 T Z J 1qH H˜ 2 + ∥q∥ ∞ 0 T Z ˜ 2 ∂ t H 2 0, L H L H L2 H 0 L H ∫Ω ( ) √ √ . ˜ 2 + T Z Nv J 1qH H˜ 2 ˜ 2 2 0 + ∂t H 2 0 Ω L H L H (2.2.208) ∫ t∫ ∫ t √ 3 ˜ 3 · ∂t v . ˜ 3 H H −1/2 ∥∂t v∥H 1/2 (Σ) . T H˜ ∥∂t v∥L2 H 1/2 (Σ) H (Σ) L∞ H −1/2 (Σ) 0 Σ 0 √ √ 3 . T Nv H ˜ , ∞ −1/2 L H (Σ) ∫ t∫ |∂t v|2 ∂t J 1 . ∂t J 1 L∞ H 2 T ∥∂t v∥2L∞ H 0 (2.2.209) 0 Ω . T ZNv . 65 ˜ i . For the following terms, it suffices to show they So we need several estimates on H are bounded. We use the usual way to estimate these quadratic terms. Since we have repeatedly used this method, we will not give the details here. (2.2.210) ( ˜ 1 H . Z ∥ζ∥L2 H 3/2 + ∥∂t ζ∥L2 H 1/2 + ∂t2 ζ L2 H −1/2 + ∥w∥L2 H 1 + ∥∂t w∥L2 H 1 L2 H 0 ) + ∥v∥L2 H 2 + ∥q∥L2 H 1 ( ) √ √ √ . Z Nv + Nw + M , (2.2.211) ( ) ( ) ˜ 2 √ √ H . Z ∥∂t ζ∥L2 H 1/2 + ∥ζ∥L2 H 1/2 + ∥v∥L2 H 1 .Z v N + M , L2 H 0 (2.2.212) ( ) ˜ 2 ∂t H . Z ∂t2 ζ L2 H 1/2 + ∥∂t ζ∥L2 H 1/2 + ∥ζ∥L2 H 1/2 + ∥∂t v∥L2 H 1 + ∥v∥L2 H 1 L2 H 0 ( ) √ √ . Z v N + M , (2.2.213) ( ) ˜ 3 H . Z ∥ζ∥L∞ H 1/2 + ∥∂t ζ∥L∞ H 1/2 + ∥q∥L∞ H 1 + ∥v∥L∞ H 2 L∞ H −1/2 (Σ) ( ) √ √ . Z Nv + M . 66 Also we have the following estimate. ∫ ˜ 2 . Z ∥q∥ ∞ 0 ˜ 2 J 1qH L H H ∞ 0 (2.2.214) L H Ω ( ) . Z ∥q∥L∞ H 0 ∥ζ∥L∞ H 1/2 + ∥∂t ζ∥L∞ H 1/2 + ∥v∥L∞ H 1 ( ) √ . Z Nv ∥ζ∥L∞ H 1/2 + ∥∂t ζ∥L∞ H 1/2 + ∥v∥L∞ H 1 . Since ζ satisfies the equation    ∂t ζ + w1 ∂1 ζ + w1 ∂2 ζ = −N 2 · w, 1 2   ζ(0) = 0, employing transport estimate and the boundedness of higher order norms, we have √ √ √ ∥ζ∥L∞ H 5/2 . T Z ∥w∥L2 H 5/2 . T Z Nw . (2.2.215) ∂t ζ satisfies the equation (2.2.216)    ∂t (∂t ζ) + w1 ∂1 (∂t ζ) + w1 ∂2 (∂t ζ) = −N 2 · ∂t w − ∂t N 2 · w − ∂t w1 ∂1 ζ − ∂t w1 ∂2 ζ, 1 2 1 2   ∂t ζ(0) = 0. Similar argument as above shows that ( ) √ √ √ ∥∂t ζ∥L∞ H 1/2 . T Z ∥∂t w∥L2 H 1/2 + ∥w∥L2 H 1/2 . T Z Nw . (2.2.217) 67 Combining above transport estimate and lemma A.1.14, we have the final version ∫ √ √ √ √ ˜2 . J 1qH T Z Nv Nw + Z Nv ∥v∥L∞ H 1 (2.2.218) Ω ( ) √ √ √ √ . T Z Nv Nw + Z Nv ∥v∥L2 H 2 + ∥∂t v∥L2 H 0 ( ) √ √ √ √ √ . T Z Nv Nw + T Z Nv ∥v∥L∞ H 2 + ∥∂t v∥L∞ H 0 ( ) √ √ √ . TZ v w N N +N .v Then we consider simplify the last term in RHS of energy structure: (2.2.219) ∫ t∫ 1 J 1 D(∂ A 1 −∂ A 2 ) v 2 2 2 0 + ϵ ∥DA 1 ∂t v∥2 2 0 , J 1 D(∂t A 1 −∂t A 2 ) v 2 : DA 1 ∂t v . t t L H L H 0 Ω 4ϵ where ϵ should be determined in order for the second term in RHS to be absorbed in LHS. 1 √ √ J D(∂ A 1 −∂ A 2 ) v 2 2 2 0 . T J 1 D(∂ A 1 −∂ A 2 ) v 2 2 ∞ 0 . T ZM. (2.2.220) t t L H t t L H To summarize, using Cauchy inequality, we have √ ∥∂t v∥2L∞ H 0 + ∥∂t v∥2L2 H 1 . T Z(Nw + Nv + M). (2.2.221) Step 3: Elliptic estimate for v. Based on standard elliptic regularity theory, we have the estimate 2 2 2 ∥v∥2H r+2 + ∥q∥2H r+1 . ∥∂t v∥2H r + H 1 H r + H 2 H r+1 + H 3 H r+ 12 (Σ) (2.2.222) 2 2 + ∇A 1 · (D(A 1 −A 2 ) v 2 ) H r + D(A 1 −A 2 ) v 2 N 1 H r+ 21 (Σ) . 68 Setting r = 0 and taking L∞ on both sides implies (2.2.223) 2 2 2 ∥v∥2L∞ H 2 + ∥q∥2L∞ H 1 . ∥∂t v∥2L∞ H 0 + H 1 L∞ H 0 + H 2 L∞ H 1 + H 3 L∞ H 12 (Σ) 2 2 + ∇A 1 · (D(A 1 −A 2 ) v 2 ) L∞ H 0 + D(A 1 −A 2 ) v 2 N 1 L∞ H 12 (Σ) . Setting r = 1 and taking L2 on both sides yields (2.2.224) 2 2 2 ∥v∥2L2 H 3 + ∥q∥2L2 H 2 . ∥∂t v∥2L2 H 1 + H 1 L2 H 1 + H 2 L2 H 2 + H 3 L2 H 32 (Σ) 2 2 + ∇A 1 · (D(A 1 −A 2 ) v 2 ) L2 H 1 + D(A 1 −A 2 ) v 2 N 1 L2 H 32 (Σ) . We need to estimate all the RHS terms. We can employ the usual way to estimate quadratic terms in L2 H k norm, however, for L∞ H k norm, we use lemma A.1.14. Then we have the estimate ( ) RHS . ∥∂t v∥2L2 H 1 + ∥∂t v∥2L∞ H 0 + TZ N + M . w (2.2.225) To summarize, we have ∥v∥2L∞ H 2 + ∥q∥2L∞ H 1 + ∥v∥2L2 H 3 + ∥q∥2L2 H 2 (2.2.226) ( ) . ∥∂t v∥L2 H 1 + ∥∂t v∥L∞ H 0 + T Z N + M . 2 2 w In total of (2.2.221) and (2.2.226), we get the succinct form of estimate √ Nv . T Z(Nw + Nv + M). (2.2.227) 69 Step 4: Transport estimate for ζ. ζ satisfies the equation    ∂t ζ + w1 ∂1 ζ + w1 ∂2 ζ = −N 2 · w 1 2 (2.2.228)   ζ(0) = 0 A straightforward application of transport estimate can show ( ∫ t )∫ t 2 ∥ζ∥2L∞ H 5/2 (Σ) . exp C ∥w1 ∥H 3 N · w 5/2 (2.2.229) H (Σ) √ 0 √0 . T Z ∥w∥2L2 H 5/2 (Σ) . T ZNw . Similar to the proof of transport estimate in theorem 2.2.17, we can use the transport equation to estimate the higher order derivatives. (2.2.230) 2 ∥∂t ζ∥2L∞ H 3/2 (Σ) . ζ 2 L∞ H 5/2 (Σ) ∥w∥2L∞ H 3/2 (Σ) + ∥ζ∥2L∞ H 5/2 (Σ) ∥w1 ∥2L∞ H 3/2 (Σ) . ZNw . In the same fashion, we can easily show 2 2 ∂t ζ 2 1/2 . ZNw . (2.2.231) L H To sum up, we have the estimate M . Nw . (2.2.232) Step 5: Synthesis: In (2.2.227), for T sufficiently small, we can easily absorbed all Nv term from RHS 70 to LHS and replace all M with Nw to achieve √ Nv . T ZNw . (2.2.233) Certainly, the smallness of T can guarantee the contraction. Local Well-posedness Theorem Now we can combine Theorem 2.2.23 and Theorem 2.2.24 to produce a unique strong solution to the system (2.1.12). Theorem 2.2.25. Assume (u0 , η 0 ) satisfies K0 < ∞. Then there exists a 0 < T0 < 1 such that for any 0 < T < T0 , there exists a solution triple (u, p, η) to the system (2.1.12) on [0, T ] that satisfies the initial data and obeys K(u, p) + K(η) ≤ CP (K0 ) (2.2.234) for a universal constant C > 0 and a polynomial P (·) satisfying P (0) = 0. The solution is unique among functions that satisfy the initial data and K(u, p) + K(η) < ∞. Moreover, η is such that the mapping Φ is a C 2N −1 diffeomorphism for each t ∈ [0, T ]. Proof. See [26, Theorem 6.3]. 71 2.3 Global Well-posedness for Horizontally Infi- nite Domain 2.3.1 Preliminaries In this section, we present two forms of system (2.1.12) and describe the corresponding energy structures. More details can be found in [25, Section 2]. Geometric Structure Form Suppose η, u are known and A , N , etc. are given in terms of η as usual. Then we consider the linear equations for (v, q, ξ)    ∂t v − ∂t η¯˜bK∂3 v + u · ∇A v + ∇A · SA (q, v) = F 1 in Ω,         ∇A · v = F 2 in Ω,  SA (q, v)N = ξN + F 3 on Σ, (2.3.1)       ∂t ξ − v · N = F 4 on Σ,      v=0 on Σb . Lemma 2.3.1. Suppose (u, p, η) satisfies the system (2.1.12) and (v, q, ξ) is the so- lution of the system (2.3.1). Then ( ∫ ∫ ) ∫ 1 1 1 ∂t J |v| + 2 |ξ|2 + J |DA v|2 (2.3.2) 2 Ω 2 Σ 2 Ω ∫ ∫ = J(v · F + qF ) + −v · F 3 + ξF 4 . 2 2 Ω Σ The geometric structure form is utilized to estimate the temporal derivatives. 72 Hence, we may apply differential operator ∂ α = ∂tα0 to the system (2.1.12) with the resulting equations being (2.3.1) for v = ∂ α u, q = ∂ α p and ξ = ∂ α η, where F 1 = F 1,1 + F 1,2 + F 1,3 + F 1,4 + F 1,5 + F 1,6 , (2.3.3) ∑ ∑ Fi1,1 = Cα,β ∂ β (∂t η¯˜bK)∂ α−β ∂3 ui + Cα,β ∂ α−β ∂t η¯∂ β (˜bK)∂3 ui ,(2.3.4) 0<β<α 0<β≤α ∑ ( ) Fi1,2 = − Cα,β ∂ (uj Ajk )∂ β α−β ∂k ui + ∂ Ajk ∂ β α−β ∂k p , (2.3.5) 0<β≤α ∑ Fi1,3 = Cα,β ∂ β Ajl ∂ α−β ∂l (Aim ∂m uj + Ajm ∂m ui ), (2.3.6) 0<β≤α ∑ Fi1,4 = Cα,β Ajk ∂k (∂ β Ail ∂ α−β ∂l uj + ∂ β Ajl ∂ α−β ∂l ui ), (2.3.7) 0<β<α Fi1,5 = ∂ α ∂t η¯˜bK∂3 ui , (2.3.8) Fi1,6 = Ajk ∂k (∂ α Ail ∂l uj + ∂ α Ajl ∂l ui ), (2.3.9) F 2 = F 2,1 + F 2,2 , (2.3.10) ∑ F 2,1 = − Cα,β ∂ β Aij ∂ α−β ∂j ui , (2.3.11) 0<β<α F 2,2 = −∂ α Aij ∂j ui , (2.3.12) F 3 = F 3,1 + F 3,2 , (2.3.13) ∑ F 3,1 = Cα,β ∂ β Dη(∂ α−β η − ∂ α−β p), (2.3.14) 0<β≤α ∑ ( ) F 3,2 = Cα,β ∂ (Nj Aim )∂ β α−β ∂m uj + ∂ (Nj Ajm )∂ β α−β ∂m ui , (2.3.15) 0<β≤α ∑ F4 = Cα,β ∂ β Dη · ∂ α−β u. (2.3.16) 0<β≤α In all above, Cα,β represents constants depending on α and β. 73 Perturbed Linear Form We may also take the system (2.1.12) as a perturbation of the linear system    ∂t u − ∆u + ∇p = G1 in Ω,         ∇ · u = G2 in Ω,  (pI − Du − ηI)e3 = G3 on Σ, (2.3.17)       ∂t η − u3 = G4 on Σ,      u=0 on Σb . Lemma 2.3.2. Suppose (u, p, η) satisfies the system (2.3.17). Then (2.3.18) ( ∫ ∫ ) ∫ ∫ ∫ 1 1 1 ∂t |u|2 + |η|2 + J |Du|2 = u · G2 + pG2 + −u · G3 + ηG4 . 2 Ω 2 Σ 2 Ω Ω Σ Compared to the system (2.1.12), we can write the detailed forms of the pertur- bation terms. G1 = G1,1 + G1,2 + G1,3 + G1,4 + G1,5 , (2.3.19) Gi1,1 = (δij − Aij )∂j p, (2.3.20) Gi1,2 = uj Ajk ∂k ui , (2.3.21) Gi1,3 = [K 2 (1 + A2 + B 2 ) − 1]∂33 ui − 2AK∂13 ui − 2BK∂23 ui , (2.3.22) Gi1,4 = [−K 3 (1 + A2 + B 2 )∂3 J + AK 2 (∂1 J + ∂3 A) (2.3.23) +BK 2 (∂2 J + ∂3 B) − K(∂1 A + ∂2 B)]∂3 ui , Gi1,5 = ∂t η¯(1 + x3 /b)K∂3 u3 , (2.3.24) G2 = (1 − J)(∂1 u1 + ∂2 u2 ) + A∂2 u1 + B∂3 u2 , (2.3.25) 74 ( p − η − 2(∂1 u1 − AK∂3 u1 ) ) 3 G = ∂1 η −∂2 u1 − ∂1 u2 + BK∂3 u1 + AK∂3 u2 −∂1 u3 − K∂3 u1 + AK∂3 u3 ( −∂2 u1 − ∂1 u2 + BK∂3 u1 + AK∂3 u2 ) +∂2 η p − η − 2(∂2 u2 − BK∂3 u2 ) (2.3.26) −∂2 u3 − K∂3 u2 + BK∂3 u3 ( (K − 1)∂3 u1 + AK∂3 u3 ) + (K − 1)∂3 u2 + BK∂3 u3 , 2(K − 1)∂3 u3 where based on the equations we have p − η = ∂1 η(−∂1 u3 − K∂3 u1 + AK∂3 u3 ) (2.3.27) +∂2 η(−∂2 u3 − K∂3 u2 + BK∂3 u3 ) + 2K∂3 u3 , G4 = −Dη · u. (2.3.28) 2.3.2 Definitions of Energy and Dissipation In this section we give the definitions of energy and dissipation with or without minimal counts. Compared to those in [25, (2.46) - (2.55)], the main differences here lie in the dissipation definitions, where we consider at most two vertical derivatives for velocity and one vertical derivative for pressure. It is this key improvement that significantly simplifies the whole proof. 75 We define the energy and dissipation with 2-minimal count as follows: 2n−1 2 √ 2 2n 2 E¯n,2 = D ¯ 2n−1 2 0 + ¯ 2 0 + DD J∂ n t u + D ¯ 2 η 0 , (2.3.29) H H 0 H H (Σ) 2n 2 ¯ n,2 = D D ¯ Du 0 . (2.3.30) 2 H (2.3.31) 2 2 ∑ n j 2 2 2 ∑ n−1 j 2 En,2 = D u H 2n−2 + ∂t u H 2n−2j + D p H 2n−3 + ∂t p 2n−2j−1 H j=1 j=1 2 ∑ n j 2 + D2 η H 2n−2 (Σ) + ∂t η 2n−2j , H (Σ) j=1 2n−1 2 Dn,2 = D ¯ 2 u 2 + ∇D H ¯ 22n−1 p 2 0 H (2.3.32) 2 2 ∑ n+1 j 2n−2j+2 2 + D32n−1 η H 1/2 (Σ) + ∂t D12n−1 η H 1/2 (Σ) + ∂t D0 η H 1/2 (Σ) . j=2 We define the energy and dissipation without minimal count as follows, where Iλ u and Iλ η denote the Riesz potential of u and η as defined in [25, (A.7) - (A.8)]: 2n 2 2n 2 E¯n = ∥Iλ u∥2H 0 + ∥Iλ η∥2H 0 (Σ) + D ¯ u 0 + D 0 H ¯ η 0 , 0 H (Σ) (2.3.33) 2n 2 ¯ n = ∥DIλ u∥2 0 + D D ¯ Du 0 . (2.3.34) H 0 H 76 ∑ n j 2 ∑ n−1 j 2 En = ∥Iλ u∥2H 0 + ∥Iλ η∥2H 0 (Σ) + ∂t u H 2n−2j + ∂t p 2n−2j−1 (2.3.35) H j=0 j=0 ∑ n j 2 + ∂t η 2n−2j , H (Σ) j=0 2n−1 2 Dn = ∥Iλ u∥2H 1 + D ¯ 0 u H2 + ∇D¯ 2n−1 2 0 p H0 (2.3.36) 2n−1 2 ∑ n+1 j 2n−2j+1 2 2n−2 2 + D0 η H 1/2 (Σ) + ∂t D0 η H 1/2 (Σ) + ∂t D0 η H 1/2 (Σ) . j=2 Finally, we define the energy with 1-minimal count as follows: (2.3.37) ∑ n j 2 ∑ n−1 j 2 En,1 = ∥Du∥2H 2n−1 + ∂t u 2n−2j + ∥Dp∥2 2n−2 + ∂t p 2n−2j−1 H H H j=1 j=1 ∑ n j 2 + ∥Dη∥2H 2n−1 (Σ) + ∂t η 2n−2j . H (Σ) j=1 Since we only consider the decay of energy and utilize the interpolation relation between En,2 and En,1 to estimate it, it is not necessary to define the dissipation with 1-minimum count and other horizontal quantities. Some other useful quantities are as follows: 2 2 ∑ 2 K= ∥∇u∥2L∞ + ∇ u L∞ + ∥Dui ∥2H 2 (Σ) . (2.3.38) i=1 F2N = ∥η∥2H 4N +1/2 (Σ) . (2.3.39) The total energy is defined as follows: ∑ 2 F2N (r) GN (t) = sup E2N (r) + sup (1 + r)m+λ EN +2,m (r) + sup . (2.3.40) 0≤r≤t 0≤r≤t 0≤r≤t (1 + r) m=1 77 Note that our definition of total energy is different from [25, (2.58)], since we do not include the integral of dissipation and merely concentrate on the instantaneous quantities. 2.3.3 Interpolation Estimates In the following, we provide several interpolation estimates in the form of ∥X∥2 . (EN +2,2 )θ (E2N )1−θ , (2.3.41) ∥X∥2 . (DN +2,2 )θ (E2N )1−θ , (2.3.42) where θ ∈ [0, 1], X is some quantity, and ∥·∥ is some norm. For brevity, we employ the notation in [25] to write the interpolation power in the tables below. For example, the following part of the table L2 EN +2,2 DN +2,2 (2.3.43) X θ κ should be understood as ∥X∥2L2 . (EN +2,2 )θ (E2N )1−θ , (2.3.44) ∥X∥2L2 . (DN +2,2 )κ (E2N )1−κ . (2.3.45) When we write EN +2,2 ∼ DN +2,2 in a table, it means θ is the same when interpolating between EN +2,2 and E2N , and between DN +2,2 and E2N . When we write multiple en- tries, it means the same interpolation estimates hold for each item listed. Sometimes, an interpolation index r appears. Then we can always take arbitrary r ∈ (0, 1). More 78 detailed properties of these tables can be found in [25, pp.20]. Lemma 2.3.3. The following tables encode the power in the L∞ or L2 interpolation estimates for u with its derivatives either in Ω or Σ. L∞ EN +2,2 ∼ DN +2,2 L2 EN +2,2 ∼ DN +2,2 u 1/2 u λ/(λ + 2) Du 2/(2 + r) Du (λ + 1)/(λ + 2) ∇u 1/2 ∇u λ/(λ + 2) (2.3.46) D∇u 2/(2 + r) D∇u (λ + 1)/(λ + 2) ∇2 u 1/2 ∇2 u λ/(λ + 2) D∇2 u 2/(2 + r) D∇2 u (λ + 1)/(λ + 2) The following tables encode the power in the L∞ or L2 interpolation estimates for η and η¯ with their derivatives either in Ω or Σ. L∞ EN +2,2 DN +2,2 η, η¯ (λ + 1)/(λ + 2) (λ + 1)/(λ + 3) Dη, ∇¯ η (λ + 2)/(λ + 2 + r) (λ + 2)/(λ + 3) (2.3.47) D2 η, ∇2 η¯ 1 (λ + 3)/(λ + 3 + r) ∂t η, ∂t η¯ 1 2/(2 + r) L2 EN +2,2 DN +2,2 η, η¯ λ/(λ + 2) λ/(λ + 3) Dη, ∇¯ η (λ + 1)/(λ + 2) (λ + 1)/(λ + 3) (2.3.48) D2 η, ∇2 η¯ 1 (λ + 2)/(λ + 3) ∂t η, ∂t η¯ 1 1/2 The following tables encode the power in the L∞ or L2 interpolation estimates for p 79 with its derivatives either in Ω or Σ. L∞ EN +2,2 DN +2,2 L2 EN +2,2 DN +2,2 ∇p 1/2 1/2 ∇p 0 0 Dp 2/(2 + r) 2/3 Dp 1/2 1/3 (2.3.49) D∇p 2/(2 + r) 2/(2 + r) D∇p 1/2 1/2 D2 p 1 2/(2 + r) D2 p 1 2/3 The following tables encode the power in the L∞ or L2 interpolation estimates for Gi ,. L∞ EN +2,2 DN +2,2 G1 1 (3λ + 5)/(2λ + 6) DG1 1 1 G2 1 1 DG2 1 1 (2.3.50) ∇G2 1 1 G3 1 (3λ + 5)/(2λ + 6) DG3 1 1 G4 1 1 80 L2 EN +2,2 DN +2,2 G1 (3λ + 2)/(2λ + 4)) (3λ + 3)/(2λ + 6) DG1 1 (3λ + 5)/(2λ + 6) G2 1 m2 DG2 1 1 (2.3.51) ∇G2 1 m2 G3 (2λ + 1)/(λ + 2) m1 DG3 1 m2 D2 G3 1 1 G4 1 m2 for m1 = min{(2λ2 + 6λ + 2)/(λ2 + 5λ + 6), (3λ + 3)/(2λ + 6)}, (2.3.52) m2 = min{(2λ2 + 7λ + 4)/(λ2 + 5λ + 6), (3λ + 5)/(2λ + 6)}. (2.3.53) Proof. The estimates follow directly from the Sobolev embedding theorem and [25, Lemma A.6 - Lemma A.8], using the bounds ∥Iλ η∥2H 0 . E2N and ∥Iλ ∂t η∥2H 0 . E2N , which can be directly deduced from [25, Lemma 2.8]. For Gi , we may utilize the product rules ∥XY ∥2L∞ . ∥X∥2L∞ ∥Y ∥2L∞ , (2.3.54) ∥XY ∥2L2 . ∥X∥2L∞ ∥Y ∥2L2 , (2.3.55) ∥XY ∥2L2 . ∥X∥2L2 ∥Y ∥2L∞ , (2.3.56) where in the L2 bound, we may take the larger value of θ produced by the two options. 81 Then we can show several important lemmas to improve above estimates. Lemma 2.3.4. We have the estimate r/(2+r) 2/(2+r) K . E2N EN +2 . (2.3.57) Proof. 2 2 ∑ 2 K= ∥∇u∥2L∞ + ∇ u L∞ + ∥Dui ∥2H 2 (Σ) . (2.3.58) i=1 We need to estimate each term above. Step 1: ∥∇u∥2L∞ . By definition, we have ∥∇u∥2L∞ . ∥Du∥2L∞ + ∥∂3 u∥2L∞ . (2.3.59) By the divergence equation in (2.3.17), we have 2 ∥∂3 u3 ∥2L∞ . ∥Du∥2L∞ + G2 L∞ . (2.3.60) For i = 1, 2, by the Poincar´e inequality we can naturally estimate 2 ∥∂3 ui ∥2L∞ . ∂32 ui L∞ + ∥∂3 ui ∥2L∞ (Σ) . (2.3.61) The upper boundary condition in (2.3.17) implies 2 2 ∥∂3 ui ∥2L∞ (Σ) . ∥Du3 ∥2L∞ (Σ) + G3 L∞ (Σ) . ∥D∇u3 ∥2L∞ + G3 L∞ (Σ) . (2.3.62) 82 Considering the Navier-Stokes equations in (2.3.17), we have 2 2 ∂3 ui ∞ . ∥∂t ui ∥2 ∞ + D2 ui 2 ∞ + ∥Dp∥2 ∞ + G1 2 ∞ . (2.3.63) L L L L L Therefore, we obtain (2.3.64) 2 2 2 ∥∇u∥2L∞ . ∥Du∥2L∞ + G2 L∞ + ∥D∇u3 ∥2L∞ + G3 L∞ (Σ) + ∥∂t ui ∥2L∞ + D2 ui L∞ 2 + G1 L∞ + ∥Dp∥2L∞ . The estimate of each term above with Lemma 2.3.3 satisfies our requirement. 2 Step 2: ∥∇2 u∥L∞ . By definition, we have 2 2 ∇ u ∞ . ∥D∇u∥2 ∞ + ∂32 ui 2 ∞ + ∂32 u3 2 ∞ . (2.3.65) L L L L The first term naturally satisfies our requirement while the second term has been successfully estimated in the first step above. Hence, we only need to estimate the third one. Taking ∂3 in the divergence equation of (2.3.17) reveals 2 2 ∂3 u3 ∞ . ∥D∇ui ∥2 ∞ + ∇G2 2 ∞ . (2.3.66) L L L where all the terms can be easily estimated via Lemma 2.3.3. ∑2 Step 3: i=1 ∥Dui ∥2H 2 (Σ) . 83 By the trace theorem and the Poincar´e inequality, we have ∑ 2 2 2 ∥Dui ∥2H 2 (Σ) . D3 ∇ui H 0 + D2 ∇ui H 0 + ∥Dui ∥2H 1 . (2.3.67) i=1 The first and second term are naturally estimated, so we only need to focus on the last one. By the Poincar´e inequality, we have 2 ∥Dui ∥2H 1 . ∥D∂3 ui ∥2H 0 + D2 ui H 0 , (2.3.68) and also 2 ∥D∂3 ui ∥2H 0 . D∂32 ui H 0 + ∥D∂3 ui ∥2H 0 (Σ) . (2.3.69) The boundary condition in (2.3.17) shows 2 2 2 2 ∥D∂3 ui ∥2H 0 (Σ) . D2 u3 H 0 (Σ) + DG3 H 0 (Σ) . D2 ∂3 u3 H 0 + DG3 H 0 (Σ)(2.3.70) . By the Navier-Stokes equations in (2.3.17), we have 2 2 D∂3 ui 0 . ∥D∂t ui ∥2 0 + D3 ui 2 0 + D2 p 2 0 + DG1 2 0 . (2.3.71) H H H H H Hence, in total, we achieve ∑ 2 ∥Dui ∥2H 2 (2.3.72) i=1 2 2 2 2 . D2 ∇ui H 0 + D2 u H 0 + D2 ∂3 u3 H 0 + DG3 H 0 (Σ) + ∥D∂t ui ∥2H 0 2 2 + D2 p H 0 + DG1 H 0 . The estimate of each term above with Lemma 2.3.3 satisfies our requirement, so we finish this proof. 84 Lemma 2.3.5. We have the estimate 1/(λ+3) (λ+2)/(λ+3) E¯N +2,2 . E2N DN +2,2 . (2.3.73) Proof. The dissipation DN +2,2 can control all the terms in E¯N +2,2 except ∂t η, D2 η and D2(N +2) η. The standard Sobolev interpolation estimate reveals 2(N +2) 2 D η H 0 . E2N 1/(4N −7) (4N −8)/(4N −7) DN +2,2 . (2.3.74) Since N ≥ 3, above estimate satisfies our requirement. Also Lemma 2.3.3 reveals the estimate of D2 η is not an obstacle. Hence, the dominant term is ∂t η. By the transport equation in (2.3.17), we have 2 ∥∂t η∥2H 0 (Σ) . ∥u3 ∥2H 0 (Σ) + G4 H 0 (Σ) . (2.3.75) By the Poincar´e-type inequality in Lemma A.1.4, we can estimate ∥u3 ∥2H 0 (Σ) . ∥∂3 u3 ∥2H 0 , (2.3.76) Then the divergence equation in (2.3.17) and the Poincar´e inequality imply 2 2 ∥∂3 u3 ∥2H 0 . ∥Dui ∥2H 0 + G2 H 0 . ∥D∂3 ui ∥2H 0 + G2 H 0 . (2.3.77) Similar to the third step in the proof of Lemma 2.3.4, by the Poincar´e inequality, we have 2 ∥D∂3 ui ∥2H 0 . D∂32 ui H 0 + ∥D∂3 ui ∥2H 0 (Σ) . (2.3.78) 85 The upper boundary condition in (2.3.17) shows (2.3.79) 2 2 2 2 ∥D∂3 ui ∥2H 0 (Σ) . D2 u3 H 0 (Σ) + DG3 H 0 (Σ) . D2 ∂3 u3 H 0 + DG3 H 0 (Σ) . By the Navier-Stokes equations in (2.3.17), we have 2 2 D∂3 ui 0 . ∥D∂t ui ∥2 0 + D3 ui 2 0 + D2 p 2 0 + DG1 2 0 , (2.3.80) H H H H H where D2 p dominates. Utilizing the Poincar´e inequality, we have 2 2 D p 0 . D2 ∂3 p 2 0 + D2 p 2 0 , (2.3.81) H H H (Σ) in which we may again use the upper boundary condition in (2.3.17) to estimate 2 2 D p 0 . D2 η 2 0 + D2 ∂3 u3 2 0 + D2 G3 2 0 . (2.3.82) H (Σ) H (Σ) H (Σ) H (Σ) Hence, in total, we achieve ∥∂t η∥2H 0 (2.3.83) 2 2 2 2 . D2 ∂3 u3 H 0 + DG3 H 0 (Σ) + ∥D∂t ui ∥2H 0 + D3 ui H 0 + DG1 H 0 2 2 2 2 2 + D2 ∂3 p H 0 + D2 η H 0 (Σ) + D2 ∂3 u3 H 0 (Σ) + G2 H 0 + G4 H 0 (Σ) 2 + D2 G3 H 0 (Σ) . We may easily check via Lemma 2.3.3 that each term above satisfies our requirement, so we finish this proof. 86 Lemma 2.3.6. We have the estimate (λ+1)/(λ+2) 1/(λ+2) EN +2,1 . EN +2,2 E2N . (2.3.84) Proof. We only need to estimate the lowest order terms that only appear in EN +2,1 , which are Du, Dp and Dη. Based on Lemma 2.3.3, the estimates of Du and Dη satisfy our requirement, so we focus on Dp. By a similar argument as in the proof of Lemma 2.3.5, we have ∥Dp∥2H 0 . ∥Dp∥2H 0 (Σ) + ∥D∂3 p∥2H 0 (2.3.85) 2 2 2 . ∥Dη∥2H 0 (Σ) + D∂32 u H 0 + G3 H 0 (Σ) + ∥D∆u∥2H 0 + DG1 H 0 , where each term can be estimated directly through Lemma 2.3.3. Then our result easily follows. 2.3.4 Nonlinear Estimates Now we give several estimates for the nonlinear terms in the perturbed linear form. Similar to [25, Theorem 4.1 - Theorem 4.2], the methods are quite standard, so we simply give the basic rules in estimating: 1. Since all the terms are quadratic or of higher order, we may expand the differen- tial operators with the Leibniz rule and estimate the resulting quadratic terms by the basic interpolation estimates in Lemma 2.3.3 either in L∞ or L2 norm, combining with the definitions of energy and dissipation, the trace theorem and the Sobolev embedding theorem. For the options of form (2.3.55) or (2.3.56), we should always take the higher interpolation index for EN +2,2 and DN +2,2 . 87 2. For the 2N level, there is no minimal count for the derivatives, so we only need to refer to the definitions of energy and dissipation; however, in the N + 2 level, we have to resort to Lemma 2.3.3 for terms that cannot be estimated directly. 3. Note the most important difference between our proofs and those of [25, Theo- rem 4.1 - Theorem 4.2] is we concentrate on the horizontal derivatives instead of the full derivatives. Thereafter, in the estimates related to DN +2,2 , we should try to avoid the introduction of vertical derivatives via the Sobolev embedding theorem, especially for pressure p. 4. For the term in the form of ∥X∥H 2 (Σ) , instead of using the trace theorem directly, we should first write it into ∥X∥H 0 (Σ) +∥DX∥H 0 (Σ) +∥D2 X∥H 0 (Σ) and then apply the trace theorem, which only produces one vertical derivative rather than three. Lemma 2.3.7. There exists a θ > 0 such that ¯ 2 ¯ 2(N +2)−4 1 2 ¯ 2 ¯ 2(N +2)−3 2 2 D ∇0 G + D ∇0 G (2.3.86) H0 0 2 H ¯ 2 ¯ 2(N +2)−4 3 + D ∇0 G 1/2 . E2N θ EN +2,2 , H (Σ) and ¯ 2(N +2)−1 1 2 ¯ 2(N +2)−1 2 2 2 D G 0 1 + D G 1 (2.3.87) H 2 2 H ¯ 2(N +2)−1 3 ¯ 2(N +2)−1 4 + D 2 G 1/2 + D 1 G 1/2 . E2N θ DN +2,2 . H (Σ) H (Σ) Lemma 2.3.8. There exists a θ > 0 such that 4N −2 1 2 4N −1 2 2 4N −2 3 2 ∇ ¯ 0 G H 0 + ∇ ¯ 0 G H 0 + ∇ ¯ 0 G H 1/2 (Σ) . E2N 1+θ , (2.3.88) 88 and 4N −1 1 2 4N −1 2 2 4N −1 3 2 D ¯0 G H 0 + D ¯0 G H 1 + D ¯0 G H 1/2 (Σ) (2.3.89) 4N −1 4 2 + D ¯ 0 G H 1/2 (Σ) . E2N θ D2N + KF2N . The other nonlinear terms have been proved in [25, Proposition 4.3 - Proposition 4.5]. Although our definitions of energy and dissipation are a little different, we do not even need to revise the proofs there. 2.3.5 Energy Estimates In this section, we present the energy estimates for the horizontal derivatives. Since this part has been carefully studied in [25, Section 5] and is not the main improvement of our new proof, it is not necessary to show all the details. The following is an example to demonstrate how we can obtain the same types of results as before. Lemma 2.3.9. Let ∂ α = ∂tN +2 and let F i be defined as in the geometric structure form. Then 1 2 F 0 + ∂t (JF 2 ) 2 0 + F 3 2 0 + F 4 2 0 . E2N DN +2,2 . (2.3.90) H H H (Σ) H (Σ) Also, if N ≥ 3, then there exists a θ > 0 such that 2 2 F 0 . E2N θ EN +2,2 . (2.3.91) H Proof. This is a revised version of [25, Theorem 5.2]. The proof of the first part is exactly the same as that of the original theorem. However, we utilize different 89 techniques in the second part. By definition, F 2 = F 2,1 +F 2,2 . By the same argument as [25, (5.5)], we have 2,1 2 F 0 . E2N θ EN +2,2 . (2.3.92) H For F 2,2 , a calculation reveals F 2,2 = ∂tN +2 (∂1 η¯˜bK)∂3 u1 + ∂tN +2 (∂2 η¯˜bK)∂3 u2 − ∂tN +2 K∂3 u3 . (2.3.93) We can estimate 2 2 N +2 ∂t (∂1 η¯˜bK)∂3 u1 . ∂tN +2 (∂1 η¯˜bK) ∥∂3 u1 ∥L∞ . 2 (2.3.94) H0 H0 Lemma 2.3.4 implies r/(2+r) 2/(2+r) ∥∂3 u1 ∥2L∞ . ∥∇u∥2L∞ . E2N EN +2,2 . (2.3.95) Also, for 0 ≤ |α| ≤ N + 2, there exists a θ > 0 such that ∥∂tα ∂1 η¯∥2H 0 . ∥∂tα η∥2H 1/2 (Σ) . ∥∂tα η∥H 0 (Σ) ∥∂tα η∥H 1 (Σ) . E2N EN +2,2 . 1−θ θ (2.3.96) Hence we have 2 ∑ N +2 2 N +2 ˜ 2 N +2−α ˜ ∂t (∂1 η¯bK) . ∥∂t ∂1 η¯∥H 0 ∂t α (bK) . E2N EN +2,2 . 1−θ θ (2.3.97) H0 L∞ α=0 Then when r is sufficiently small, we can naturally estimate 2 N +2 ˜ t ∂ (∂ 1 η ¯ bK)∂ 3 1 u . E2N θ EN +2,2 . (2.3.98) 0H 90 Similarly, we have 2 N +2 ˜ t ∂ (∂ 2 η ¯ bK)∂ 3 2 u . E2N θ EN +2,2 . (2.3.99) H 0 The same argument as [25, (5.11)] shows N +2 ∂t K∂3 u3 2 0 . E2N θ EN +2,2 . (2.3.100) H Therefore, our result easily follows. In a similar fashion, we can recover all the other lemmas in [25, Section 5]. Finally, we can show the same energy estimates. Proposition 2.3.10. Let F 2 be given with ∂tN +2 . Then there exists a θ > 0 such that ( ∫ ) ∂t E¯N +2,2 − 2 K∂t pF + D N +1 2 ¯ N +2,2 . E θ DN +2,2 . 2N (2.3.101) Ω Proposition 2.3.11. There exists a θ > 0 such that ∫ t ∫ t E¯2N (t) + D2N (r)dr . E2N (0) + (E2N (t)) + ¯ 3/2 (E2N (r))θ D2N dr (2.3.102) 0 ∫ t√ 0 + D2N (r)K(r)F2N (r)dr. 0 2.3.6 Comparison Estimates In this section, we prove several comparison estimates between the horizontal deriva- tives and the full derivatives, both in energy and dissipation. This is the key improve- ment in our proof, so we present all the necessary details. 91 Dissipation Comparison Estimates Proposition 2.3.12. There exists a θ > 0 such that DN +2,2 . D ¯ N +2,2 + E θ DN +2,2 . 2N (2.3.103) Proof. We define ¯ 2(N +2)−1 1 2 ¯ 2(N +2)−1 2 2 YN +2,2 = D 2 G 0 2 + D G 1 (2.3.104) H H 2 ¯ 2(N +2)−1 3 ¯ 2(N +2)−1 4 2 + D2 G 1/2 + D1 G 1/2 H (Σ) H (Σ) and Z=D ¯ N +2,2 + YN +2,2 . (2.3.105) Then the proof is divided into several steps. Step 1: Application of Korn’s inequality. Since any horizontal derivatives of velocity vanishes on the lower boundary, we can directly apply Korn’s inequality to achieve ¯ 2(N +2) 2 ¯ 2(N +2) 2 D2 u . D 2 Du =D ¯ N +2,2 . Z. (2.3.106) H1 H0 Step 2: Initial estimates of velocity and pressure. Let 0 ≤ j ≤ (N + 2) − 1 and α ∈ N2 be such that 2 ≤ 2j + |α| ≤ 2(N + 2) − 1. Hence, 92 we naturally have α j+1 2 2 ∂ ∂t u 0 ≤ ¯ 2 D 2(N +2) u 1 . Z. (2.3.107) H H We apply the operator ∂ α ∂tj ∂3 to the divergence equation in (2.3.17) and rearrange the terms as follows: ∂ α ∂tj ∂3 (∂3 u3 ) = ∂ α ∂tj ∂3 (−∂1 u1 − ∂2 u2 ) + ∂ α ∂tj ∂3 G2 . (2.3.108) Thus we have α j 2 2 2 2 ∂ ∂t ∂3 u3 0 . D¯ 22(N +2) u + ∂ α ∂t G2 H 1 . Z. j (2.3.109) H H1 Hence, it easily implies α j 2 ∆∂ ∂t u3 0 . Z. (2.3.110) H Applying the operator ∂ α ∂tj to the third equation in (2.3.17), we have α j 2 ∂ ∂t ∂3 p 0 . ∂ α ∂tj ∆u 2 0 + ∂ α ∂tj G1 2 0 . Z. (2.3.111) H H H So we only need to estimate the terms related to ∂32 ui and ∂i p for i = 1, 2. Step 3: Estimates of velocity and pressure in the upper domain. In order to achieve above estimates, we need to employ a localization argument. Define a cutoff function χ1 = χ1 (x3 ) given by χ1 ∈ Cc∞ (R) with χ(x3 ) = 1 for x3 ∈ Ω1 = [−2b/3, 0] and χ1 (x3 ) = 0 for x3 ∈ / (−3b/4, 1/2). 93 Define the curl of the velocity w = ∇ × u, which means w1 = ∂2 u3 − ∂3 u2 and w2 = ∂3 u1 − ∂1 u3 . Therefore, χ1 wi for i = 1, 2 satisfies the equations ∆∂ α ∂tj (χ1 wi ) = χ1 ∂ α ∂tj+1 wi + 2(∂3 χ1 )(∂ α ∂tj ∂3 wi ) (2.3.112) +(∂32 χ1 )(∂ α ∂tj wi ) − χ1 ∇ × (∂ α ∂tj G1 ) in Ω as well as the boundary conditions      ∂ α ∂tj (χ1 w1 ) = 2∂2 ∂ α ∂tj u3 + ∂ α ∂tj G3 · e2 on Σ,  ∂ α ∂tj (χ1 w2 ) = −2∂1 ∂ α ∂tj u3 − ∂ α ∂tj G3 · e1 on Σ, (2.3.113)      ∂ α ∂ j (χ1 w1 ) = ∂ α ∂ j (χ1 w2 ) = 0 on Σb . t t Based on the standard elliptic estimate of the Poisson equation, we have α j ∂ ∂t (χ1 wi ) 2 1 . ∆∂ α ∂tj (χ1 wi ) 2 −1 + ∂ α ∂tj (χ1 wi ) 2 1/2 . (2.3.114) H H H (Σ) Hence, we have to estimate each term on the right-hand side. Let ϕ ∈ H01 (Ω). For α ̸= 0 we may write α = β + (α − β) with |β| = 1. Then an integration by parts reveals ∫ ∫ ϕχ1 ∂ α ∂t wi = ∂ β ϕχ1 ∂ α−β ∂t wi ≤ ∥ϕ∥ 1 j+1 j+1 ¯ 2(N +2) H χ1 D2 wi .(2.3.115) Ω Ω H0 Since 2(j + 1) + |α − β| ∈ [3, 2(N + 2)], we naturally have ¯ 2(N +2) 2 ¯ 2(N +2) 2 χ1 D2 wi . D2 u . Z. (2.3.116) H0 H1 Then if we take the supremum over all ϕ such that ∥ϕ∥H 1 ≤ 1, then we may get χ1 ∂ α ∂tj+1 wi 2 −1 . Z. (2.3.117) H 94 A similar argument without integration by parts shows above result is also true for α = 0 since in this case j ≤ (N + 2) − 1 implies 4 ≤ 2(j + 1) ≤ 2(N + 2). In a similar fashion, we may apply integration by parts for ∂3 for the other terms in H −1 norm and achieve (2.3.118) 2 2(∂3 χ1 )(∂ α ∂tj ∂3 wi ) 2 −1 . (∥∂3 χ1 ∥2 ∞ + ∂32 χ1 2 ∞ ) D¯ 22(N +2) wi 0 H L L H 2(N +2) 2 . D2 u . Z, H1 2 2(N +2) (∂3 χ1 )(∂ α ∂tj wi ) 2 −1 . ∂32 χ1 2 ∞ D¯2 2 wi (2.3.119) H L H0 2 2(N +2) . D2 u . Z, H1 2 χ1 ∇ × (∂ α ∂tj G1 ) 2 −1 . (∥χ1 ∥2 ∞ + ∥∂3 χ1 ∥2 ∞ ) D¯ 22(N +2)−1 G1 . Z.(2.3.120) H L L H0 In total, above estimates together yield α j ∆∂ ∂t (χ1 wi ) 2 −1 . Z. (2.3.121) H Then for the boundary terms, a direct application of the trace theorem shows α j 2 ∂ ∂t ∂1 u3 2 1/2 + ∂ α ∂tj ∂2 u3 2 1/2 . D¯ 22(N +2) u . Z, (2.3.122) H (Σ) H (Σ) H1 α j 3 2 ∂ ∂t G 1/2 ¯ 2(N +2)−1 3 2 H (Σ) . D 2 G 1/2 . Z. (2.3.123) H (Σ) 95 Then above implies α j ∂ ∂t (χ1 wi ) 2 1/2 . Z. (2.3.124) H (Σ) Hence, the elliptic estimate gives us the final form α j 2 2 ∂ ∂t wi 1 . ∂ α ∂tj (χ1 wi ) H 1 . Z. (2.3.125) H (Ω1 ) Since ∂32 ∂ α ∂tj u1 = ∂3 ∂ α ∂tj (w2 + ∂1 u3 ) and ∂32 ∂ α ∂tj u2 = ∂3 ∂ α ∂tj (∂2 u3 − w1 ), we can deduce from above that for i = 1, 2, 2 α j 2 ∂3 ∂ ∂t ui 0 . Z, (2.3.126) H (Ω1 ) which, by considering the Navier Stokes equations in (2.3.17), further implies α j 2 ∂i ∂ ∂t p 0 . Z. (2.3.127) H (Ω1 ) Combining all above, we have achieved ¯ 2(N +2)−1 2 ¯ 2(N +2)−1 2 D2 u + D2 ∇p . Z. (2.3.128) 2 H (Ω1 ) H 0 (Ω1 ) Step 4: Estimates of velocity and pressure in the lower domain. Now we extend above estimates to the lower domain Ω2 = [−b, −b/3]. Define a cutoff function χ2 ∈ Cc∞ (R) such that χ2 (x3 ) = 1 for x3 ∈ Ω2 and χ2 (x3 ) = 0 for x3 ∈ / (−2b, −b/6). Notice this cutoff function satisfies supp(∇χ2 ) ∈ Ω1 , which will play a key role in the following. 96 Then the localized variables satisfy the equations    −∆(χ2 u) + ∇(χ2 p) = −∂t (χ2 u) + χ2 G1 + H 1 in Ω,       ∇ · (χ2 u) = χ2 G2 + H 2 in Ω, (2.3.129)     χ2 u = 0 on Σ,     χ u=0 2 on Σb , where H 1 = ∂3 χ2 (pe3 − 2∂3 u) − (∂32 χ2 )u H 2 = ∂3 χ2 u3 . (2.3.130) Let 0 ≤ j ≤ (N + 2) − 1 and α ∈ N2 be such that 2 ≤ 2j + |α| ≤ 2(N + 2) − 1. Then we may apply the differential operator ∂ α ∂tj on (2.3.129) and the elliptic estimates of the steady Navier-Stokes equations imply α j ∂ ∂t ∂i (χ2 u) 2 1 + ∂ α ∂tj ∂i (χ2 p) 2 0 . ∂ α ∂tj+1 ∂i (χ2 u) 2 −1 (2.3.131) H H H α j + ∂ ∂t ∂i (χ2 G1 + H 1 ) H −1 + ∂ α ∂tj ∂i (χ2 G2 + H 2 ) H 0 . 2 2 Similar to the argument in the upper domain, via integration by parts, it is easy to deduce the H −1 estimate that α j+1 ∂ ∂t ∂i (χ2 u) 2 −1 + ∂ α ∂tj ∂i (χ2 G1 ) 2 −1 + ∂ α ∂tj ∂i (χ2 G2 ) 2 0 . Z. (2.3.132) H H H For the remaining part, we may directly estimate α j ∂ ∂t ∂i H 1 2 −1 + ∂ α ∂tj ∂i H 2 2 0 . ∂ α ∂tj ∂i H 1 2 0 + ∂ α ∂tj ∂i H 2 2 0 (2.3.133) H H H H 2 2 ¯ 2(N +2)−1 ¯ 2(N +2)−1 . D 2 u + D 2 ∇p . Z, H 2 (Ω1 ) H 0 (Ω1 ) 97 where we utilize the property for ∇χ2 stated above. Thus we have α j 2 2 ∂i ∂ ∂t p 0 . ∂ α ∂tj ∂i (χ2 p) H 0 . Z, (2.3.134) H (Ω2 ) which, based on the Navier-Stokes equations in (2.3.17), further implies 2 α j 2 ∂3 ∂ ∂t u 0 . Z. (2.3.135) H (Ω2 ) Combining all above, we have ¯ 2(N +2)−1 2 ¯ 2(N +2)−1 2 D2 u + D 2 ∇p . Z. (2.3.136) 2 H (Ω2 ) H 0 (Ω2 ) Step 5: Estimates of free surface. With all above estimates in hand, we may derive the estimates for η by employing the boundary condition of (2.3.17), i.e. η = p − 3∂3 u3 − G33 . (2.3.137) We may take horizontal derivatives on both sides of above equation, which shows for α ∈ N2 and 3 ≤ |α| ≤ 2(N + 2) − 1, we have 2 ∥∂ α η∥2H 1/2 (Σ) . ∥∂ α p∥2H 1/2 (Σ) + ∥∂ α ∂3 u3 ∥2H 1/2 (Σ) + ∂ α G33 H 1/2 (Σ) (2.3.138) 2 2 α 3 2 ¯ 2(N +2)−1 ¯ 2(N +2)−1 ∂ G3 1/2 . ∇D 2 p + D 2 u + H (Σ) H0 H2 . Z. For the temporal derivatives of η, we need to resort to the transport equation in 98 (2.3.17), i.e. ∂t η = u3 + G4 . (2.3.139) For α ∈ N2 and 1 ≤ |α| ≤ 2(N + 2) − 1, by the trace theorem, the Poincar´e inequality and the divergence equation ∇ · u = G2 , we have 2 ∥∂ α ∂t η∥2H 1/2 (Σ) . ∥∂ α u3 ∥2H 1/2 (Σ) + ∂ α G4 H 1/2 (Σ) (2.3.140) . ∥∂ α u3 ∥2H 1 + Z . ∥∂ α u3 ∥2H 0 + ∥∂ α Du3 ∥2H 0 + ∥∂ α ∂3 u3 ∥2H 0 + Z . ∥∂ α ∂3 u3 ∥2H 0 + Z 2 . ∥∂ α Du∥2H 0 + ∂ α G2 H 0 + Z . Z. Similarly, for j = 2, . . . , (N + 2) + 1, we may apply ∂tj−1 to the transport equation to obtain that for α ∈ N2 with 0 ≤ |α| ≤ 2(N + 2) − 2j + 2, we have α j 2 2 2 ∂ ∂t η 1/2 . ∂ α ∂tj−1 u3 H 1/2 (Σ) + ∂ α ∂tj−1 G4 H 1/2 (Σ) (2.3.141) H (Σ) 2 . ∂ α ∂tj−1 u3 H 1 + Z . Z. Step 6: Synthesis. To summarize all above, since Ω = Ω1 ∪ Ω2 we have the estimate DN +2,2 . D ¯ N +2,2 + YN +2,2 . (2.3.142) 99 By Lemma 2.3.7, it is obvious that there exists a θ > 0 such that YN +2,2 . E2N θ DN +2,2 . (2.3.143) Therefore, our result naturally follows. Proposition 2.3.13. There exists a θ > 0 such that D2N . D ¯ 2N + E θ D2N + KF2N . 2N (2.3.144) Proof. It is easy to see most of the proof is identical to that of Proposition 2.3.12, except that we do not need the minimal counts and interpolation now. Also we utilize Lemma 2.3.8 instead to estimate the nonlinear terms in the perturbed linear form. Finally, the direct estimate for Riesz potential term ∥Iλ u∥2H 1 . ∥DIλ u∥2H 0 (2.3.145) can close the proof. Energy Comparison Estimates Proposition 2.3.14. There exists a θ > 0 such that EN +2,2 . E¯N +2,2 + E2N θ EN +2,2 . (2.3.146) 100 Proof. We define ¯ 2 ¯ 2(N +2)−4 1 2 ¯ 2 ¯ 2(N +2)−3 2 2 WN +2,2 = D ∇0 G + D ∇0 G (2.3.147) H0 H0 2 ¯ 2 ¯ 2(N +2)−4 3 + D ∇0 G 1/2 H (Σ) and Z = E¯N +2,2 + WN +2,2 . (2.3.148) Then we can directly apply the elliptic estimate in [26, Lemma A.15] to the perturbed linear form. For α ∈ N2 and |α| = 2, we have 2 ∥∂ α u∥2H 2(N +2)−2 + ∥∂ α p∥2H 2(N +2)−3 . ∥∂ α ∂t u∥2H 2(N +2)−4 + ∂ α G1 H 2(N +2)−4 (2.3.149) 2 2 + ∂ α G2 H 2(N +2)−3 + ∥∂ α η∥2H 2(N +2)−7/2 (Σ) + ∂ α G3 H 2(N +2)−7/2 (Σ) . Naturally, we have (2.3.150) α 1 2 ∂ G 2(N +2)−4 + ∂ α G2 2 2(N +2)−3 + ∂ α G3 2 2(N +2)−7/2 . WN +2,2 . Z, H H H (Σ) and ∥∂ α η∥2H 2(N +2)−7/2 (Σ) . E¯N +2,2 . Z. (2.3.151) So the only remaining term is ∥∂ α ∂t u∥2H 2(N +2)−4 . ∥∂t u∥2H 2(N +2)−2 , (2.3.152) which should be estimated with a finite induction in the following. 101 For j = 1, . . . , (N + 2) − 1, we have the elliptic estimate (2.3.153) j 2 ∂t u 2(N +2)−2j + ∂tj p 2 2(N +2)−2j−1 . ∂tj+1 u 2 2(N +2)−2j−2 + ∂tj G1 2 2(N +2)−2j−2 H H H H j 2 2 j 2 j 3 2 + ∂t G H 2(N +2)−2j−1 + ∂t η H 2(N +2)−2j−3/2 (Σ) + ∂t G H 2(N +2)−2j−3/2 (Σ) . By definition, it is easy to see (2.3.154) j 1 2 ∂t G 2(N +2)−2j−2 + ∂tj G2 2 2(N +2)−2j−1 + ∂tj G3 2 2(N +2)−2j−3/2 . WN +2,2 . Z, H H H (Σ) and j 2 ∂t η 2(N +2)−2j−3/2 . E¯N +2,2 . Z. (2.3.155) H (Σ) When j = (N + 2) − 1, we have j+1 2 ∂t u 2(N +2)−2j−2 = ∂tN +2 u 2 0 . E¯N +2,2 . Z. (2.3.156) H H Hence, the elliptic estimate shows N +1 2 ∂t u 2 + ∂tN +1 p 2 1 . Z. (2.3.157) H H Inductively, we can estimate all the temporal derivatives of velocity and pressure for j = 1, . . . , N to close the proof. Therefore, we have shown EN +2,2 . E¯N +2,2 + WN +2,2 . (2.3.158) 102 Based on Lemma 2.3.7, there exists a θ > 0 such that WN +2,2 . E2N θ EN +2,2 . (2.3.159) Hence, our result naturally follows. Proposition 2.3.15. There exists a θ > 0 such that E2N . E¯2N + E2N 1+θ . (2.3.160) Proof. It is easy to see the proof is almost the same as that of Proposition 2.3.14 except that there is no minimal counts now and we should utilize Lemma 2.3.8 instead. So we omit the proof here. 2.3.7 A Priori Estimates We may check [25, Lemma 9.1 - Corollary 9.3] can be recovered with obvious modi- fications, which is the following result. Proposition 2.3.16. There exists a universal constant 0 < δ < 1 such that if GN (T ) . δ, then ∫ t sup F2N (r) . F2N (0) + t D2N (r)dr, (2.3.161) 0≤r≤t 0 and ∫ t ∫ t K(r)F2N (r)dr . δ (8+2λ)/(8+4λ) F2N (0) + δ (8+2λ)/(8+4λ) D2N (r)dr, (2.3.162) 0 0 103 ∫ t√ ∫ t D2N (r)K(r)F2N (r)dr . F2N (0) + δ (8+2λ)/(16+6λ) D2N (r)dr (2.3.163) 0 0 for all 0 ≤ t ≤ T . Hence, we naturally have the a priori estimate for the 2N level. Theorem 2.3.17. There exists a universal constant 0 < δ < 1 such that if GN (T ) . δ, then ∫ t F2N (r) sup E2N (r) + D2N + sup . E2N (0) + F2N (0) (2.3.164) 0≤r≤t 0 0≤r≤t 1 + r for all 0 ≤ t ≤ T . Proof. This is identical to [25, Theorem 9.4]. Notice that in the section of energy estimates and comparison estimates, we have recovered all the necessary results for the 2N level a priori estimate here. Then we consider the a priori estimate for the N + 2 level. Theorem 2.3.18. There exists a universal constant 0 < δ < 1 such that if GN (T ) . δ, then sup (1 + r)2+λ EN +2,2 (r) . E2N (0) (2.3.165) 0≤r≤t for all 0 ≤ t ≤ T . Proof. Utilizing Lemma 2.3.5, we may employ the same argument as in [25, theorem 9.7] to show this. 104 Next, we give an interpolation argument to achieve the decay of energy with different minimal counts. Theorem 2.3.19. There exists a universal constant 0 < δ < 1 such that if GN (T ) . δ, then sup (1 + r)1+λ EN +2,1 (r) . E2N (0) (2.3.166) 0≤r≤t for all 0 ≤ t ≤ T . Proof. By Lemma 2.3.6, we have (1+λ)/(2+λ) 1/(2+λ) EN +2,1 . EN +2,2 E2N . (2.3.167) Combining with Theorem 2.3.18 and Theorem 2.3.17, we can easily see (2.3.168) ( )(1+λ)/(2+λ) ( )1/(2+λ) (1 + r)1+λ EN +2,1 (r) . (1 + r)2+λ EN +2,2 (r) E2N (r) ( )(1+λ)/(2+λ) ( )1/(2+λ) . E2N (0) E2N (0) = E2N (0). Hence, our result easily follows. Finally, we can sum up all above to achieve the a priori estimate. Theorem 2.3.20. There exists a universal constant 0 < δ < 1 such that if GN (T ) . δ, then GN (t) . E2N (0) + F2N (0) (2.3.169) 105 for all 0 ≤ t ≤ T . 2.3.8 Global Well-posedness Finally, we come to the conclusion of global well-posedness. With the a priori estimate in hand, we may directly refer to [25, Theorem 11.2], which completes the whole proof. Theorem 2.3.21. Suppose the initial data (u0 , η0 ) satisfies the 2N th compatibility condition. There exists a κ > 0 such that if ∥u0 ∥2H 4N + ∥η0 ∥2H 4N +1/2 (Σ) ≤ κ, then there exists a unique solution triple (u, p, η) on the interval [0, ∞) that satisfies the initial data. The solution obeys the estimate GN (∞) ≤ C(∥u0 ∥2H 4N + ∥η0 ∥2H 4N +1/2 (Σ) ) < Cκ. (2.3.170) Chapter Three Numerical Solution to Viscous Surface Wave 107 3.1 Introduction We consider a two-dimensional incompressible viscous flow in the moving domain Ω(t) = {y = (y1 , y2 ) ∈ Σ × R : −1 < y2 < η(y1 , t)}, (3.1.1) where Σ = T for which T denotes the 1-torus. We denote the initial domain Ω(0) = Ω0 . For each t, the flow is described by its velocity and pressure (u, p) : Ω(t) 7→ R2 ×R which satisfies the incompressible Navier-Stokes equations    ∂t u + u · ∇u + ∇p = ν∆u + S   in Ω(t),       ∇·u=0 in Ω(t),       (pI − νD(u))µ = gηµ on {y2 = η(y1 , t)},    u=0 on {y2 = −1}, (3.1.2)       ∂t η = u2 − u1 ∂y1 η on {y2 = η(y1 , t)},       u(t = 0) = u0 in Ω0 ,      η(t = 0) = η on Σ, 0 for µ the outward-pointing unit normal vector on {y2 = η}, I the 2×2 identity matrix, (Du)ij = ∂i uj + ∂j ui the symmetric gradient of u, g the gravitational constant, ν > 0 the viscosity and S = S(t, y1 , y2 ) an external source term. The fifth equation in (3.1.2) implies the free surface is convected with the fluid. Note in (3.1.2), we have shifted the actual pressure p¯ by the constant atmosphere pressure patm according to p = p¯ + gy2 − patm . We always assume the natural condition that there exists a positive number δ such that η0 + 1 ≥ δ > 0 on Σ, which means the initial free surface is strictly separated from the bottom. 108 Traditionally, based on the handling of the free surface, this type of problems can be solved via moving-grid technique as in [47], marker-and-cell method as in [28], volume-of-fluid method as in [30] and level-set method as in [23]. In each case, finite difference method, finite volume method and finite element method can be applied to solve the Navier-Stokes equation in Ω(t). However, to the best of authors’ knowl- edge, there is very little in the literature on solving this problem with discontinuous Galerkin method other than [23]. On the other hand, in spite of many computational tests presented in the literature, there is very little discussion on the stability and convergence of the numerical scheme. In this chapter, we employ the idea from [26], [24] and [50], to construct a stable numerical scheme and give detailed analysis. Our central idea is to flatten the free surface via a coordinates transform. First, we define a fixed domain Ω = {x = (x1 , x2 ) ∈ Σ × R | − 1 < x2 < 0}, (3.1.3) for which we write the coordinates x ∈ Ω. In this slab, we take Σ : {x2 = 0} as the upper boundary and Σb : {x2 = −1} as the lower boundary. Consider the geometric transform from Ω to Ω(t), which is first introduced in [8] and further extended in [26] and [50]: Φ : (x1 , x2 ) 7→ (x1 , x2 + η(1 + x2 )) = (y1 , y2 ). (3.1.4) We may directly verify this transform maps Ω into Ω(t) with the Jacobi matrix    1 0  ∇Φ =  , (3.1.5) A J 109 and the transform matrix    1 −AK  A = ((∇Φ)−1 )T =  , (3.1.6) 0 K where ˜b = 1 + x2 , A = ∂1 η˜b, J = 1 + η, K = 1/J. Here we denote the derivative with respect to x1 as ∂1 and with respect to x2 as ∂2 . Define the transformed operators as follows: (∇A f )i = Aij ∂j f, ∇A · ⃗g = Aij ∂j gi , ∆A f = ∇A · ∇A f, N = (−∂1 η, 1), (3.1.7) χ = ∂1 η, (DA u)ij = Aik ∂k uj + Ajk ∂k ui , SA (p, u) = pI − DA u, where the summation index should be understood in the Einstein convention. If we extend the divergence ∇A · to act on symmetric tensors in the natural way, then a straightforward computation reveals ∇A · SA (p, u) = ∇A p − ∆A u for vector fields satisfying ∇A · u = 0. 110 In our new coordinates, the original system (3.1.2) becomes    ∂t u − ∂t η˜bK∂2 u + u · ∇A u − ν∆A u + ∇A p = S in Ω,         ∇A · u = 0 in Ω,       SA (p, u)N = gηN on Σ,      u=0 on Σb , (3.1.8)     u(x, 0) = u0 (x) in Ω,             ∂t η = u · N on Σ,      η(x′ , 0) = η (x′ ) on Σ. 0 Since A depends on η through the transform, most of the operators in the Navier- Stokes equations are related to the free surface η. Hence, the Navier-Stokes equations and the transport equation are essentially coupled. Based on [26], the equation (3.1.8) with S = 0 possesses a natural energy equality as follows: ∫ ∫ ∫ t∫ J(t) |u(t)| + g 2 |η(t)| + ν 2 J(s) |DA u(s)|2 ds (3.1.9) ∫Ω ∫ Σ 0 Ω = J(0) |u(0)| + g 2 |η(0)|2 . Ω Σ Hence, we try to construct a numerical scheme to recover this energy stability for the numerical solution. In the following, we refer to the term “continuous case” when we consider the exact solution triple (u, p, η) which is sufficiently smooth. On the other hand, we refer to the term “discrete case” when we consider the numerical solution triple (uh , ph , ηh ). Throughout this chapter, C > 0 denotes a positive constant that only depends 111 on the parameters Ω, g and ν of the problem, and the exact solution (u, p, η). It is referred as universal and can change from one inequality to another. When we write C(z), it means a positive constant depending on the quantity z. a . b denotes a ≤ Cb, where C is a universal constant as defined above. The method we discuss in this paper belongs to the class of discontinuous Galerkin (DG) methods. DG methods are finite element methods which use completely discon- tinuous piecewise polynomial solution and test spaces. They are particularly useful for convection dominated wave problems, and have the advantage of flexibility in adap- tivity and efficient parallel implementation. We refer to the references [14, 13, 15, 49] for more details. 3.2 Numerical Scheme 3.2.1 Basic Settings In our construction and analysis of the numerical scheme, we focus on the semi- discrete form of the system. The time discretization is based on the Runge-Kutta method for differential-algebraic equations of index 2, as presented in [27]. We do not discuss the fully-discrete scheme in this paper. We choose the bulk domain as Ω : (x1 , x2 ) ∈ [0, 1] × [−1, 0] and the surface domain as Σ : x1 ∈ [0, 1], which are 1-periodic in the x1 direction. The surface mesh is defined by dividing Σ into N uniform elements with length h = 1/N . The bulk mesh construction can be divided into two steps: First we divide Ω into N × N uniform squares, where each element is sized h × h. Then, we divide each square 112 into two right triangles by cutting along the diagonal from the left-up corner to the right-down corner. The bulk mesh is shown in Figure 3.1. Remark 3.2.1. Since the discretization of the weighted differential operators ∇A , ∇A · and ∆A depends on η, i.e. the surface variable, we need the bulk discretization to match the surface discretization in the vertical direction. Hence, we need to first choose the strip-shape mesh as in Figure 3.2, whose projection on the upper boundary is exactly the surface mesh. Then in each strip, any triangulation is permitted. For convenience, we made the choice as in Figure 3.1. Figure 3.1: Mesh Distribution in N = 4 Figure 3.2: Strip-Shape Mesh Distribution in N =4 Let Eh be the set consisting of all the triangular elements in the bulk mesh and ∂Eh be the set of all the sides of the triangular cells. Let Fh be the set consisting of all the interval elements in the surface mesh and ∂Fh be the set of all the boundary points of intervals. Define the usual Sobolev space in Ω as H k (Ω) and in Σ as H k (Σ). We define the space Phk (Ω) where f ∈ Phk (Ω) if and only if f |E ∈ P k (E) for all E ∈ Eh (P k (E) denotes the set of polynomials of degree at most k defined on E). Similarly, we can define the space Phk (Σ). 113 We define the solution spaces as Xhk = {uh ∈ (L2 (Ω))2 : uh ∈ (Phk (Ω))2 }, (3.2.1) Mhk = {ph ∈ L2 (Ω) : ph ∈ Phk (Ω)}, (3.2.2) Shk = {ηh ∈ L2 (Σ) : ηh ∈ Phk (Σ)}, (3.2.3) for k ≥ 1. For discrete functions fh ∈ Xhk and gh ∈ Shk , we may define the Hh norms as follows: ∑ ∥fh ∥2Hh = ∥fh ∥2H 1 (E) , (3.2.4) E∈Eh ∑ ∥gh ∥2Hh = ∥gh ∥2H 1 (F ) . (3.2.5) F ∈Fh Define the Xh norm for vh ∈ Xhk as ∑( ∫ 1 ) ∥vh ∥2Xh = ∥∇vh ∥2L2 (E) + 2 [vh ] . (3.2.6) E∈Eh ∂E h Before stating the scheme, we announce the notation used below. For a given bound- ary belonging to a reference cell E ∈ Eh and a function f defined in Ω, f ext(E) denotes the value of f read from the exterior direction on the boundary and f int(E) denotes that read from the interior direction. Also for the boundary shared by two cells, we define 1 ext(E) {f } = (f + f int(E) ), (3.2.7) 2 [f ] = f int(E) − f ext(E) . (3.2.8) For the side on the boundary of Ω, we make a special definition that on Σb , f int(E) = 114 {f } = [f ] and f ext(E) = 0, while on Σ, f int(E) = f ext(E) = {f } and [f ] = 0. The definitions are reasonable, since the system (3.1.8) for u gives Dirichlet-type condition on Σb and Neumann-type condition on Σ. The numerical scheme is based on the weak formulation of the system (3.1.8). Taking test functions vh ∈ Xhk and qh ∈ Mhk−1 . We multiply Jvh and Jqh respectively on both sides of the Navier-Stokes equations and integrate over E ∈ Eh to obtain ∫ ∫ J∂t u · vh − J(∂t η˜bK∂2 u) · vh (3.2.9) E∫ E ∫ ∫ ∫ + J(u · ∇A u) · vh − ν J∆A u · vh + J∇A p · vh = J(S · vh ), ∫ E E E E J(∇A · u)qh = 0. (3.2.10) E Considering the transport equation in (3.1.8), we make an addition and subtraction to modify two terms in above formulation as (∫ ∫ ∫ ) ˜ 1 J∂t u · vh − J(∂t η bK∂2 u) · vh + ∂t η(u · vh ) (3.2.11) 2 ∂E∩Σ E (∫ E ∫ ) 1 + J(u · ∇A u) · vh − (u · N )(u · vh ) 2 ∂E∩Σ E ∫ ∫ ∫ −ν J∆A u · vh + J∇A p · vh = J(S · vh ), E ∫ E E J(∇A · u)qh = 0, (3.2.12) E where ∂E denotes the boundary of the triangular element E. This trick is to enforce the transport relation and shows its power in the stability proof. Multiplying a test functions ϕh ∈ Shk on both sides of the transport equation and integrating over F ∈ Fh , we obtain ∫ ∫ ∫ ∂t ηϕh + u¯1 ∂1 ηϕh − u¯2 ϕh = 0, (3.2.13) F F F 115 where u¯1 and u¯2 denote the traces of u1 and u2 on Σ. The weak formulations of the system (3.1.8) are based on the integration by parts of (3.2.11), (3.2.12) and (4.1.1). In the following, we define and analyze the numerical scheme term by term. 3.2.2 Discretization We define the semi-discrete scheme as follows:      κF (ηh , uh , ϕh ) = 0,       λF (χh , ψh ) = 0,      (3.2.14)     ζE (ηh , uh , vh ) + γE (uh , ηh , uh , vh ) + ναE (ηh , uh , ηh , vh )       +βE (ph , ηh , vh ) + gµE (ηh , ηh , vh ) = ωE (ηh , vh ),      ρE (uh , ηh , qh ) = 0, for any test functions ϕh ∈ Shk , ψh ∈ Shk , vh ∈ Xhk and qh ∈ Mhk−1 , where the first two equations denote the transport discretization and the last two equations denote the fluid discretization. This forms a complete differential-algebraic system for ηh (t) ∈ Shk , χh (t) ∈ Shk , uh (t) ∈ Xhk and ph (t) ∈ Mhk−1 . The detailed definitions of above multi-linear terms κ, λ, ζ, γ, α, β, µ, ω, and ρ are in the following. 116 Discretization of Transport Terms as κF and λF We define κF (ηh , uh , ϕh ) (3.2.15) ∫ ∫ ∫ ∫ = ∂t ηh ϕh − (¯ u2 )h ϕh − u1 )h ηh ϕh − (¯ ∂1 (¯ u1 )h ηh ∂1 ϕh F F F ( F ) − 1 − − +(ˆ u1 )h ηˆh ϕh |F +1/2 − (ˆu1 )h ηˆh ϕh |F −1/2 + + η ϕ [uh ]|F +1/2 − ηh ϕh [uh ]|F −1/2 + + 2 h h λF (χh , ψh ) (3.2.16) ∫ ∫ = χh ψh + ηh ∂1 ψh − ηh+ ψh− |F +1/2 + ηh+ ψh+ |F −1/2 , F F where F − 1/2 and F + 1/2 denote the left and right boundary points of F . Here, for each boundary point, “−” means the value read from the left and “+” means the value read from the right. Also, (ˆ u1 )h ηˆh denotes the numerical flux on the boundary. We always take u1 )h = {(¯ (ˆ u1 )h }, (3.2.17) and ηˆh should be determined from the sign of {(¯ u1 )h } following the upwinding rule as    η − if {(¯ h u1 )h } ≥ 0, ηˆh = (3.2.18)   ηh+ if {(¯ u1 )h } < 0, ( ) 1 − − where (¯ u1 )h is the trace of (u1 )h on Σ. Note η ϕ [uh ]|F +1/2 − ηh ϕh [uh ]|F −1/2 are + + 2 h h extra penalty terms, which helps to show the stability. Since uh is also discontinuous across the boundary point, these penalty terms are used to transform uh into {uh } on the boundary after integrating by parts in (3.2.15). 117 In all the applications below, we use χh to discretize ∂1 η, and ηh to discretize η. Hence, we denote Nh for the vector (−χh , 1). Also, Ah , Jh , Kh and Ah can be defined in the same convention. Spacial Discretization of Temporal Terms as ζE We define ζE (ηh , uh , vh ) (3.2.19) ∫ ∫ ∫ int(E) = ∂t (Jh uh ) · vh + ∂t ηh˜buh · ∂2 vh − ∂t ηh˜b(ˆ uh · vh )n2 E ∫ E ∂E 1 + ∂t ηh (uh · vh ), 2 ∂E∩Σ where nE = (n1 , n2 ) denotes the outward normal vector on ∂E. The last term in (3.2.19) is the newly-added penalty term in the formulation (3.2.11). The argument ηh in ζE (ηh , uh , vh ) denotes ηh and its derivatives. We utilize the upwinding flux in this scheme, i.e.    u+ if ∂t ηh˜b ≥ 0, h uˆh = (3.2.20)  −  uh if ∂t ηh˜b < 0, where “+” denotes the value read from the up direction and “−” denote that from the down direction. If a side is vertical, then its contribution is zero. 118 Discretization of Convection Terms as γE We define γE (uh , ηh , uh , vh ) (3.2.21) ∫ ∫ 1 = − (uh · Nh )(uh · vh ) + Jh (uh · ∇Ah uh ) · vh 2 ∂E∩Σ E ∫ ∫ int(E) int(E) 1 1 u · vh + Jh (∇Ah · uh )(uh · vh ) − [uh · Jh Ah ] · nE h 2 E 2 ∂E\∂Ω 2 ∫ int(E) ext(E) int(E) + |{uh · Jh Ah } · nE | (uh − uh ) · vh , ∂E − where ∇Ah is understood as in (3.1.7) with A replaced by Ah and ∂E − = {x ∈ ∂E : {uh · Jh Ah } · nE < 0}. (3.2.22) The first term in (3.2.21) is the newly added penalty term in the formulation (3.2.11). Since    Jh −Ah  Jh Ah =  , (3.2.23) 0 1 we only need one ηh argument in γE (uh , ηh , uh , vh ). 119 Discretization of Diffusion Terms as αE We utilize the Symmetric Internal Penalty Galerkin Method (SIPG) or the Nonsym- metric Internal Penalty Galerkin Method (NIPG) to define αE (ηh , uh , ηh , vh ) (3.2.24) ∫ ∫ 1 int(E) = Jh DAh uh : DAh vh − {DAh uh Jh Ah } · nE · vh 2 E ∂E\Σ ∫ ∫ 1 int(E) int(E) int(E) σ int(E) ± DA int(E) vh Jh Ah · nE · [uh ] + [uh ] · vh . 2 ∂E\Σ h h ∂E For the third term, if we take +, it is NIPG and if we take −, it is SIPG. The penalty constant σ can always be taken as 1 for NIPG and a sufficiently large number for SIPG. Note that we need two ηh arguments in αE (ηh , uh , ηh , vh ) to denote the dependence of Ah . Discretization of Pressure Terms as βE We define ∫ ∫ int(E) βE (ph , ηh , vh ) = − Jh ph ∇Ah · vh + vh · {ph Jh Ah } · nE . (3.2.25) E ∂E\Σ Discretization of Forcing Terms as µE We define µE (ηh , ηh , vh ) (3.2.26) ∫ ∫ 1 1 = ηh (v2 )h − vh− |∂E∩Σ+1/2 − [ηh2 ]¯ v1 )h + [ηh2 ]¯ ηh ∂1 ηh (¯ vh+ |∂E∩Σ−1/2 , ∂E∩Σ ∂E∩Σ 4 4 120 where ∂E ∩ Σ + 1/2 and ∂E ∩ Σ − 1/2 denote the left and right boundary points of ∂E ∩ Σ respectively, i.e. similar to F − 1/2 and F + 1/2. Note that the forcing term is nontrivial only for the top cells and for all other cells it is zero, µE (ηh , ηh , vh ) = 0. 1 2 − 1 vh |∂E∩Σ+1/2 − [ηh2 ]¯ [ηh ]¯ vh+ |∂E∩Σ−1/2 are extra penalty terms, which help to show the 4 4 stability. Since ηh is discontinuous across the boundary, these penalty terms are used to transform ηh2 into {ηh2 } on the boundary after integrating by parts in (3.4.60). Discretization of Source Terms as ωE We define ∫ ωE (ηh , vh ) = Jh S · vh . (3.2.27) E Discretization of Divergence Terms as ρE We define ρE (uh , ηh , qh ) (3.2.28) ∫ ∫ 1 int(E) int(E) int(E) = − Jh qh ∇Ah · uh + [uh ] · qh Jh Ah · nE . E 2 ∂E\Σ 121 3.2.3 Properties and Estimates of Discretization Estimate of Temporal Terms We consider the temporal terms, i.e. (∫ ∫ ∫ ) 1 J∂t u · vh − J(∂t η˜bK∂2 u) · vh + ∂t η(u · vh ) . (3.2.29) E E 2 ∂E∩Σ In the continuous case, a direct integration by parts reveals ∫ ∫ ∫ J∂t u · vh = ∂t (Ju) · vh − ∂t J(u · vh ), (3.2.30) ∫ E E∫ E − J(∂t η˜bK∂2 u) · vh = − ∂t η˜b∂2 u · vh (3.2.31) E ∫ E ∫ ∫ = ˜ ∂t η bu · ∂2 vh + ˜ ∂2 (∂t η b)(u · vh ) − ∂t η˜b(u · vh )n2 . E E ∂E Since ∂t J = ∂2 (∂t η˜b), we can simplify above terms into ∫ ∫ J∂t u · vh − J(∂t η˜bK∂2 u) · vh (3.2.32) ∫E E∫ ∫ = ∂t (Ju) · vh + ˜ ∂t η bu · ∂2 vh − ∂t η˜b(u · vh )n2 . E E ∂E Hence, we may define the discretization of the temporal term as (3.2.19). Since in our bulk mesh, there are only two effective boundary integrals for each triangle, we may call them the upper boundary ∂E + 1/2 and lower boundary ∂E − 1/2 without ambiguity. If we denote f (uh ) = −∂t ηh˜buh , then (3.2.19) is actually ∫ 1 ζE (ηh , uh , vh ) − ∂t ηh (uh · vh ) (3.2.33) 2 ∂E∩Σ ∫ ∫ ∫ ∫ = ∂t (Jh uh ) · vh − f (uh ) · ∂2 vh + fˆ(uh ) · vh− − fˆ(uh ) · vh+ , E E ∂E+1/2 ∂E−1/2 122 where the flux reduces to    f (u+ ) if ∂t ηh˜b ≥ 0, ˆ h f (uh ) = f (ˆ uh ) = (3.2.34)   f (u− ˜ h ) if ∂t ηh b < 0. Lemma 3.2.2. If we take vh = uh , the discretized temporal term satisfies ∑ ∫ 1 ζE (ηh , uh , uh ) ≥ ∂t Jh |uh |2 . (3.2.35) E∈Eh 2 Ω Proof. Note the fact that ∂t Jh = ∂2 (∂t ηh˜b). We can directly compute ∫ 1 ζE (ηh , uh , uh ) − ∂t ηh |uh |2 (3.2.36) 2 ∂E∩Σ ∫ ∫ ∫ ∫ ˆ − = ∂t (Jh uh ) · uh − f (uh ) · ∂2 uh + f (uh ) · uh − fˆ(uh ) · u+ h E E ∂E+1/2 ∂E−1/2 ∫ ∫ ∫ 1 1 1 = ∂t Jh |uh |2 + ∂t Jh |uh |2 − ∂2 (∂t ηh˜b) |uh |2 2 2 2 ∫E E ∫ E 1 1 − f (u− − h ) · uh + f (u+h ) · uh + 2 ∂E+1/2 2 ∂E−1/2 ∫ ∫ − + fˆ(uh ) · uh − fˆ(uh ) · u+h ∂E+1/2 ∂E−1/2 ∫ 1 = ∂t Jh |uh |2 2 ∫E ∫ 1 − − 1 − f (uh ) · uh + f (u+h ) · uh + 2 ∂E+1/2 2 ∂E−1/2 ∫ ∫ ˆ − + f (uh ) · uh − fˆ(uh ) · u+h ∂E+1/2 ∂E−1/2 ∫ 1 = ∂t Jh |uh |2 + Fˆ∂E+1/2 − Fˆ∂E−1/2 + ΘE , 2 E where ∫ ∫ 1 − − F∂E+1/2 = − ˆ f (uh ) · uh + fˆ(uh ) · u− h, (3.2.37) 2 ∂E+1/2 ∂E+1/2 ∫ ∫ 1 − − F∂E−1/2 = − ˆ f (uh ) · uh + fˆ(uh ) · u− h, (3.2.38) 2 ∂E−1/2 ∂E−1/2 123 and ∫ ( ) 1 − − ΘE = f (uh ) · uh − f (uh ) · uh + + (3.2.39) 2 ∂E−1/2 ∫ ( ) ˆ − ˆ + f (uh ) · uh − f (uh ) · uh . + ∂E−1/2 Based on the flux definition (3.2.34), we have the estimate ∫ ( − ) ˜ − u+ h + uh ΘE = ∂t ηh b(uh − uh ) · (ˆ + uh − ) ≥ 0. (3.2.40) ∂E−1/2 2 Hence, combining with (3.2.36), we obtain ∫ ∫ 1 1 ζE (ηh , uh , uh ) − ∂t ηh |uh | ≥ ∂t 2 Jh |uh |2 + Fˆ∂E+1/2 − Fˆ∂E−1/2 . (3.2.41) 2 ∂E∩Σ 2 E When summing up over all E ∈ Eh , we can easily see when ∂E ⊂ Ω, all the terms involving Fˆ are canceled out. Therefore, only the terms on ∂Ω remain, i.e. ∑ ∫ 1 ζE (ηh , uh , uh ) − ∂t ηh |uh |2 (3.2.42) E∈Eh 2 Σ ∑ 1 ∫ ∑ ∑ ≥ ∂t Jh |uh |2 + Fˆ∂E+1/2 − Fˆ∂E−1/2 . 2 E E∈E h ∂E∩Σ̸=∅ ∂E∩Σb ̸=∅ On Σb we have u− h = 0, so −F∂E−1/2 = 0, i.e. ˆ ∑ Fˆ∂E−1/2 = 0. (3.2.43) ∂E∩Σb ̸=∅ − h = uh = uh |Σ and we always take u On Σ we have u+ ˆh = uh , so it implies ∫ ∫ ∫ 1 1 Fˆ∂E+1/2 = − f (u− − h )uh + fˆ(uh )u− h = − ∂t ηh |uh |2 .(3.2.44) 2 ∂E+1/2 ∂E+1/2 2 ∂E∩Σ 124 Then summing up over E ∈ Eh gives a full integration over Σ, i.e. ∑ ∫ 1 Fˆ∂E+1/2 =− ∂t ηh |uh |2 . (3.2.45) 2 Σ ∂E∩Σ̸=∅ Therefore, combining (3.2.42), (3.2.43) and (3.2.45), we deduce ∑ ∫ ∫ ∫ 1 1 1 ζE (ηh , uh , uh ) − ∂t ηh |uh | ≥ ∂t 2 Jh |uh | − 2 ∂t ηh |uh |2 . (3.2.46) E∈Eh 2 Σ 2 E 2 Σ Hence, our result easily follows. Remark 3.2.3. This proof is based on the mesh as in Figure 1. For general trian- gulation, it is possible to have three effective boundaries for each element. However, it is easy to see Lemma 3.2.2 still holds. Estimate of Convection Terms We consider the convection term ∫ ∫ 1 J(u · ∇A u) · vh − (u · N )(u · vh ). (3.2.47) E 2 F This is the key nonlinear term in the Navier-Stokes equations. Our discretization is inspired by the idea in [22]. In the continuous case, when summing up over all E ∈ Eh , we can see ∑ γE (u, η, u, vh ) (3.2.48) E∈Eh ∫ ∫ ∫ 1 1 = J(u · ∇A u) · vh + J(∇A · u)(u · vh ) − (u · N )(u · vh ), Ω 2 Ω 2 Σ 125 where all the other boundary terms vanish. Hence, considering the continuous A - divergence-free condition for u, this discretization is consistent. Lemma 3.2.4. If we take vh = uh , the discretized convection term satisfies ∑ γE (uh , ηh , uh , uh ) ≥ 0. (3.2.49) E∈Eh Proof. We divide the proof into several steps: Step 1: Direct integration by parts. We plug the test function vh = uh into (3.2.21) to obtain ∫ 1 γE (uh , ηh , uh , uh ) + (uh · Nh ) |uh |2 (3.2.50) 2 F ∫ ∫ 1 = Jh (uh · ∇Ah uh ) · uh + Jh (∇Ah · uh )(uh · uh ) E 2 E ∫ int(E) int(E) 1 u · uh − [uh · Jh Ah ] · nE h 2 ∂E\∂Ω 2 ∫ int(E) ext(E) int(E) + |{uh · Jh Ah } · nE | (uh − uh ) · uh , ∂E − A direct integration by parts yields ∫ Jh (uh · ∇Ah uh ) · uh (3.2.51) E∫ ∫ = − Jh (uh · ∇Ah uh ) · uh − Jh (∇Ah · uh )(uh · uh ) ∫E E int(E) int(E) int(E) int(E) int(E) + (uh · Jh Ah · nE )(uh · uh ). ∂E 126 Therefore, (3.2.50) can be simplified as ∫ 1 γE (uh , ηh , uh , uh ) + (uh · Nh ) |uh |2 (3.2.52) 2 ∂E∩Σ (∫ ∫ ) 1 = − Jh (uh · ∇Ah uh ) · uh + Jh (∇Ah · uh )(uh · uh ) E 2 E ∫ int(E) int(E) 1 u · uh − [uh · Jh Ah ] · nE h 2 ∂E\∂Ω 2 ∫ int(E) ext(E) int(E) + |{uh · Jh Ah } · nE | (uh − uh ) · uh − ∫∂E int(E) int(E) int(E) int(E) int(E) + (uh · Jh Ah · nE )(uh · uh ) ∂E = I + II + III + IV, where (∫ ∫ ) 1 I=− Jh (uh · ∇Ah uh ) · uh + Jh (∇Ah · uh )(uh · uh ) , (3.2.53) E 2 E and II, III and IV can be understood respectively. Step 2: Estimates of II and IV . In IV , for e ∈ ∂E\∂Ω, we have the decomposition int(E) int(E) int(E) uh · Jh Ah (3.2.54) 1 int(E) int(E) int(E) ext(E) ext(E) ext(E) = {uh · Jh Ah } + (uh · Jh Ah − uh · Jh Ah ). 2 For e ⊂ ∂E\∂Ω, we can sum up the second term on the right-hand side of (3.2.54) over E and its neighboring cells, which have the opposition outward normal vectors, 127 to show (3.2.55) ∫ 1 int(E) int(E) int(E) ext(E) ext(E) ext(E) int(E) int(E) (uh · Jh Ah − uh · Jh Ah ) · nE (uh · uh ) e∫2 1 ext(E) ext(E) ext(E) int(E) int(E) int(E) ext(E) ext(E) + (uh · Jh Ah − uh · Jh Ah ) · (−nE )(uh · uh ) 2 ∫ e = [uh · Jh Ah ] · ne {uh · uh }, e where ne denotes the normal vector read from the same reference cell as [uh · Jh Ah ]. int(E) int(E) int(E) Note for e ⊂ ∂E ∩ ∂Ω, uh · Jh Ah = {uh · Jh Ah }. Therefore, we obtain ∑ ∑ ∫ int(E) int(E) IV = {uh · Jh Ah } · nE (uh · uh ) (3.2.56) E∈Eh e∈∂Eh e ∑ ∫ + [uh · Jh Ah ] · ne {uh · uh }. e∈∂Eh \∂Ω e Also, in II, for e ⊂ ∂E\∂Ω, we can sum up over E and its neighboring cells to achieve (3.2.57) ∫ int(E) int(E) ∫ ext(E) ext(E) 1 uh · uh 1 uh · uh − [uh · Jh Ah ] · nE − (−[uh · Jh Ah ]) · (−nE ) 2 e 2 2 2 ∫ e 1 = − [uh · Jh Ah ] · ne {uh · uh }, 2 e which further leads to ∑ ∑ ∫ 1 II = − [uh · Jh Ah ] · ne {uh · uh }. (3.2.58) 2 e E∈Eh e∈∂Eh \∂Ω Step 3: Further estimates in (3.2.52). Hence, summing over all E ∈ Eh in (3.2.52) and combining (3.2.56) and (3.2.58), we 128 deduce ∑ ∫ 1 γE (uh , ηh , uh , uh ) + (uh · Nh ) |uh |2 (3.2.59) E∈Eh 2 Σ (∫ ∫ 1 = − Jh (uh · ∇Ah uh ) · uh + Jh (∇Ah · uh )(uh · uh ) 2 Ω ∑ 1∫ Ω ) − [uh · Jh Ah ] · nE {uh · uh } 2 e e∈∂Eh \∂Ω ∑ (∫ int(E) ext(E) int(E) + |{uh · Jh Ah } · nE | (uh − uh ) · uh E∈Eh ∂E − ∫ ) int(E) int(E) + {uh · Jh Ah } · nE (uh · uh ). ∂E We can further estimate the last two terms in (3.2.59) as follows: ∑ (∫ int(E) ext(E) int(E) |{uh · Jh Ah } · nE | (uh − uh ) · uh (3.2.60) E∈Eh ∂E − ∫ ) int(E) int(E) + {uh · Jh Ah } · · nE (uh uh ) ∑ ∫ ∂E ext(E) ext(E) int(E) = |{uh · Jh Ah } · nE | uh · (uh − uh ) E∈Eh ∂E − ∑∫ int(E) int(E) + {uh · Jh Ah } · nE (uh · uh ) E∈Eh ∂E + ∩∂Ω ∑∫ int(E) int(E) + {uh · Jh Ah } · nE (uh · uh ). E∈Eh ∂E − ∩Σ 129 Then combining (3.2.59) and (3.2.60), we have the complete form ∑ ∫ 1 γE (uh , ηh , uh , uh ) + (uh · Nh ) |uh |2 (3.2.61) E∈Eh 2 Σ (∫ ∫ 1 = − Jh (uh · ∇Ah uh ) · uh + Jh (∇Ah · uh )(uh · uh ) 2 Ω ∑ 1∫ Ω ) − [uh · Jh Ah ] · nE {uh · uh } e∈∂Eh 2 e*∂Ω ∑∫ ext(E) ext(E) int(E) + |{uh · Jh Ah } · nE | uh · (uh − uh ) E∈Eh ∂E − ∑∫ int(E) int(E) + {uh · Jh Ah } · nE (uh · uh ) E∈Eh ∂E + ∩∂Ω ∑∫ int(E) int(E) + {uh · Jh Ah } · nE (uh · uh ). E∈Eh ∂E − ∩Σ Step 4: Synthesis. Summing up (3.2.50) over E ∈ Eh and adding it to (3.2.61) imply ∑ ∫ 1 γE (uh , ηh , uh , uh ) + (uh · Nh ) |uh |2 (3.2.62) E∈Eh 2 Σ ∫ 2 1 ∑ ext(E) int(E) = |{uh · Jh Ah } · nE | uh − uh 2 E∈E ∂E − ∑ 1∫ h int(E) int(E) + {uh · Jh Ah } · nE (uh · uh ) E∈Eh 2 ∂E + ∩∂Ω ∑ 1∫ int(E) int(E) + {uh · Jh Ah } · nE (uh · uh ) E∈Eh 2 ∂E − ∩Σ ∑ 1∫ int(E) int(E) ≥ {uh · Jh Ah } · nE (uh · uh ) E∈Eh 2 ∂E + ∩Σ ∑ 1∫ int(E) int(E) + {uh · Jh Ah } · nE (uh · uh ) E∈Eh 2 ∂E − ∩Σ ∫ 1 int(E) int(E) = {uh · Jh Ah } · nE (uh · uh ) 2 Σ ∫ 1 = (uh · Nh ) |uh |2 , 2 Σ 130 where the last equality can be directly verified by the definitions of Jh , Ah and Nh . Then our result naturally follows. Estimate of Diffusion Terms In order to show the coercivity of the diffusion term, we need the discrete form of Korn’s inequality. The proof here is based on [48]. Lemma 3.2.5. Assume fh ∈ Xhk satisfies for e ⊂ ∂Eh ∩ Σb , it holds that fhext |e = 0. Then for sufficiently large σ0 > 0, we have ∑∫ ∑ σ0 ∫ |Dfh | +2 [fh ]2 & ∥fh ∥2Xh & ∥fh ∥2Hh . (3.2.63) E∈Eh E e∈∂E h e h Proof. The second inequality has been shown in [22], so we turn to the first one. It is easy to see the key part is to show the derivatives of fh can be controlled. Hence, we only need to show ∑∫ ∑ σ0 ∫ |Dfh | + 2 [fh ]2 & ∥fh ∥2Hh . (3.2.64) E∈Eh E e∈∂E h e h If this is not true, then we can construct a sequence fhn ∈ Xhk satisfying ∥fhn ∥Hh = 1, (3.2.65) and ∑ ∑ σ0 ∫ 1 ∥Dfhn ∥L2 (E) + [fhn ]2 ≤ . (3.2.66) E∈Eh e∈∂E h e n h 131 To abuse the notation, we can extract the weakly convergent subsequence fhn ⇀ fh in H 1 (E) for ∀E ∈ Eh . (3.2.67) By the compact embedding theorem in each cell E and the weak lower semi-continuity of H 1 (E) norm, we have the strongly convergent subsequence fhn → fh in L2 (Ω). (3.2.68) By (3.2.66), we also have ∑ ∥Dfhn ∥L2 (Ω) → 0. (3.2.69) E∈Eh Notice the fact that in each cell E, we still have the continuous Korn’s inequality ∥fhm − fhn ∥H 1 (E) . ∥Dfhm − Dfhn )∥L2 (E) + ∥fhm − fhn ∥L2 (E) for ∀m, n ∈ N.(3.2.70) Hence, combining all above, we know fhn is a Cauchy’s sequence under the norm H 1 (E), which means it is a Cauchy’s sequence under the norm Hh . Then fhn → fh in Hh . (3.2.71) Thus this means ∥fh ∥Hh = 1, (3.2.72) which further implies Dfh = 0 in ∀E ∈ Eh , (3.2.73) 132 and ∑ σ0 ∫ [fh ]2 = 0. (3.2.74) e∈∂E h e h This means fh = (a, b) + c(−x1 , x2 ), (3.2.75) ¯ Certainly, the zero bottom implies for some constant a, b, c and fh is continuous in Ω. fh = 0, which contradicts ∥fh ∥Hh = 1. Therefore, our hypothesis is invalid and (3.2.64) holds. Lemma 3.2.6. Assume fh satisfies for e ⊂ ∂Eh ∩ Σb , it holds that fhext |e = 0. Also, ηh satisfies ηh + 1 ≥ δ > 0 and Q = sup ∥ηh ∥W 1,∞ (F ) < ∞. (3.2.76) F ∈Fh Then for sufficiently large σ0 > 0, we have ∑∫ ∑ σ0 ∫ Jh |DAh fh | + 2 [fh ]2 & C(Q) ∥fh ∥2Xh & C(Q) ∥fh ∥2Hh . (3.2.77) E∈Eh E e∈∂E h e h Proof. Note that in each cell E, the free surface ηh is smooth. Hence, we can change it back by the transform Φ−1 −1 h to a curved cell Φh E which satisfies ∫ ∫ 2 Jh (x) DAh (x) fh (x) dx = |Dfh (y)|2 dy. (3.2.78) E Φ−1 h E Hence, in this curved cell, we have the Korn’s inequality ∥fh ∥H 1 (Φ−1 E) . ∥Dfh ∥L2 (Φ−1 E) + ∥fh ∥L2 (Φ−1 E) . (3.2.79) h h h 133 Since the transform Φ−1 −1 h is a diffeomorphism between E and Φh (E), then the H 1 norms in these two spaces are comparable. Hence, we have (∫ ) ∥fh ∥H 1 (E) . C(Q) Jh |DAh fh | + ∥fh ∥L2 (E) . 2 (3.2.80) E Then a similar proof as that of Lemma 3.2.5 naturally yields the desired result. Lemma 3.2.7. Assume fh satisfies for e ⊂ ∂Eh ∩ Σb , it holds that fhext |e = 0. Also, ηh satisfies ηh + 1 ≥ δ > 0 and Q = sup ∥ηh ∥W 1,∞ (F ) < ∞. (3.2.81) F ∈Fh Then there exists a sufficiently small constant δ ′ > 0 such that for η ′ satisfying sup ∥ηh′ − ηh ∥W 1,∞ (F ) ≤ δ ′ , (3.2.82) F ∈Fh and for sufficiently large σ0′ > 0, we have ∑∫ ∑ σ′ ∫ Jh′ DA ′ fh 2 + 0 [fh ]2 & C(Q) ∥fh ∥2Xh & C(Q) ∥fh ∥2Hh . (3.2.83) E∈Eh E h e∈∂E h e h Proof. By Lemma 3.2.6, we know ∑∫ ∑ σ0 ∫ Jh |DAh fh | + 2 [fh ]2 & C(Q) ∥fh ∥2Xh & C(Q) ∥fh ∥2Hh . (3.2.84) E∈Eh E e∈∂E h e h We can rewrite our formula in a perturbed form as ∑∫ ∑∫ Jh′ DA ′ fh 2 − Jh |DAh fh |2 (3.2.85) h E∈Eh E E∈E E ∑∫ ( )( h ) ∑∫ ′ = Jh DAh′ −Ah fh 2DAh fh + DAh′ −Ah fh + (Jh′ − Jh ) |DAh fh |2 E∈Eh E E∈Eh E . δ ′ ∥fh ∥2Hh . 134 Naturally, when taking δ ′ sufficiently small, we can absorb the perturbation into the principle part. Then our result easily follows. For the diffusion term ∫ − J∆A u · vh , (3.2.86) E a direct integration by parts implies ∑∫ − J∆A u · vh (3.2.87) E∈Eh E ∫ ∑ ∫ ∫ 1 int(E) = JDA u : DA vh − {DA uJA · nE } · [vh ] − (DA u)N · vh . 2 Ω e Σ e∈∂Eh \Σ Also in the continuous case, our discretization yields ∑ αE (η, u, η, vh ) (3.2.88) E∈Eh ∫ ∑ ∫ 1 = JDA u : DA vh − {DA uJA · nE } · [vh ]. 2 Ω e e∈∂Eh \Σ ∫ We can notice the difference is the extra physical boundary term − (DA u)N · Σ int(E) vh , which contributes to the boundary condition on Σ. This is separately added to the scheme as a forcing term later combined with the pressure contribution. Hence, our scheme is consistent. Lemma 3.2.8. Assume ηh satisfies ηh + 1 ≥ δ > 0. If we take vh = uh and choose the penalty constant σ properly, the discretized diffusion term with NIPG satisfies ∑ ∫ αE (ηh , uh , ηh , uh ) ≥ Jh |DAh uh |2 . (3.2.89) E∈Eh Ω 135 If we further assume ηh satisfies Q = sup ∥ηh ∥W 1,∞ (F ) < ∞, (3.2.90) F ∈Fh then the discretized diffusion term with SIPG or NIPG satisfies ∑ αE (ηh , uh , ηh , uh ) & C(Q) ∥uh ∥2Xh . (3.2.91) E∈Eh Proof. We can directly compute ∑ αE (ηh , uh , ηh , vh ) (3.2.92) E∈Eh ∫ ∑ ∫ 1 ∑ = Jh DAh uh : DAh vh − {DAh uh Jh Ah · nE } · [vh ] 2 E∈E E e∈∂Eh \Σ e ∑ ∫ h σ ∑ ± {DAh vh Jh Ah · nE } · [uh ] + [uh ][vh ]. e h e∈∂E e∈∂Eh \Σ h For NIPG, when vh = uh , (3.2.92) reduces to ∑ ∫ 1 σ∑ αE (ηh , uh , ηh , uh ) = Jh |DAh uh |2 + [uh ]2 . (3.2.93) E∈Eh 2 Ω h e Hence, we may simply take the penalty σ = 1 and by the discrete Korn’s inequality in Lemma 3.2.6, our result naturally follows. For SIPG, when vh = uh , (3.2.92) reduces to ∑ αE (ηh , uh , ηh , uh ) (3.2.94) E∈Eh ∫ ∑ ∫ 1 σ ∑ = Jh |DAh uh | − 2 2 {DAh uh Jh Ah · nE } · [uh ] + [uh ]2 . 2 Ω e h e∈∂E e∈∂Eh \Σ h 136 By the discrete Korn’s inequality in Lemma 3.2.6, we have ∑∫ ∑ σ0 ∫ Jh |DAh uh | + 2 [uh ]2 & C(Q) ∥uh ∥2Xh & C(Q) ∥uh ∥2Hh . (3.2.95) E∈Eh E e∈∂E h e h Then we utilize H¨older’s inequality, the trace theorem and Cauchy’s inequality to estimate ∑ ∫ {DAh uh Jh Ah · nE } · [uh ] (3.2.96) e∈∂Eh \∂Ω e ∑ ≤ C(Q) ∥∇uh ∥L2 (e) ∥[uh ]∥L2 (e) e∈∂Eh \∂Ω ∑ ≤ C(Q) ∥uh ∥H 1 (e) ∥[uh ]∥L2 (e) e∈∂Eh \∂Ω ∑ C(Q) 1 ∑ ≤ CC(Q) h ∥uh ∥2H 1 (e) + ∥[uh ]∥2L2 (e) 4C h e∈∂Eh \∂Ω e∈∂Eh \∂Ω ∑ C(Q) 1 ∑ ≤ CC(Q) h ∥uh ∥2H 3/2 (E) + ∥[uh ]∥2L2 (e) 4C h E∈Eh e∈∂Eh \∂Ω C(Q) 1 ∑ ≤ CC(Q) ∥uh ∥2Hh + ∥[uh ]∥2L2 (e) . 4C h e∈∂Eh \∂Ω The last inequality is valid since uh ∈ Phk . Then we take C sufficiently small, and σ ≥ σ + C(Q)/(4C) to absorb (3.2.96) into (3.2.95). Hence, our result easily follows. Estimate of Pressure Terms For the pressure term ∫ J∇A p · vh , (3.2.97) E 137 a direct integration by parts reveals the following equality in the continuous case ∑∫ J∇A p · vh (3.2.98) E∈Eh E ∫ ∑ ∫ ∫ int(E) = − Jp∇A · vh + [vh ] · (pJA ) · ne ) + pN · vh . Ω e∈∂Eh \Σ e Σ It is easy to see in the continuous case, our discretization reduces to ∑ ∫ ∑ ∫ βE (p, η, vh ) = − Jp∇A · vh + [vh ] · (pJA ) · ne . (3.2.99) E∈Eh Ω e∈∂Eh \Σ e ∫ We can notice the difference is the extra physical boundary term pN · v int(E) , Σ which contributes to the boundary condition on Σ. This is separately added to the scheme as a forcing term later combined with the diffusion contribution. Hence, our scheme is consistent. Lemma 3.2.9. When we take vh = uh , the discretized pressure term satisfies ∑ ∫ ∑ ∫ βE (ph , ηh , uh ) = − Jh ph ∇Ah · uh + [uh ] · {ph Jh Ah } · ne ,(3.2.100) E∈Eh Ω e∈∂Eh \Σ e where ne denotes the outward normal vector read from the same direction as [uh ]. Proof. This is a direct corollary of the definition, so we omit the proof here. 138 Estimate of Forcing Terms Considering the physical boundary condition on Σ of the system (3.1.8), we need to take the upper boundary condition as a forcing term, i.e. (pI − DA u)N = ηN on Σ. (3.2.101) Multiplying the test function vh and integrating over F yield ∫ ∫ ∫ ηN vh = v2 )h − η(¯ η∂1 η(¯ v1 )h . (3.2.102) F F F In the continuous case, our discretization reduces to ∑ ∫ ∫ µE (η, η, vh ) = v2 )h − η(¯ η∂1 η(¯ v1 ) h , (3.2.103) E∈Eh Σ Σ and the penalty terms vanish. Hence, this discretization is consistent. Lemma 3.2.10. When we take vh = uh , the discretized forcing term satisfies ∑ ∫ 1 µE (ηh , ηh , uh ) ≥ ∂t |ηh |2 . (3.2.104) E∈Eh 2 Σ Proof. We may sum up over E ∈ Eh to obtain ∑ µE (ηh , ηh , vh ) (3.2.105) E∈Eh ∫ ∫ = ηh (¯ v2 )h + ηh ∂1 ηh (¯ v1 )h Σ ( Σ ) 1 ∑ − 2 + u1 )h }(ηh ) |F −1/2 . u1 )h }(ηh ) |F +1/2 − {(¯ {(¯ + 2 2 F ∈F h 139 Taking ϕh = ηh in the transport discretization (3.2.15) and integrating by parts imply (3.2.106) ∫ ∫ ∫ 1 ∂t |ηh |2 − (¯ u1 )h ηh ∂1 ηh − (¯ u2 )h ηh + (¯ u1 )− − 2 h (ηh ) |F +1/2 + (¯ h (ηh ) |F −1/2 u1 )+ + 2 2 F F F ( ) − 1 − 2 u1 )h ηˆh ηh |F +1/2 − (ˆ +(ˆ u1 )h ηˆh ηh |F −1/2 + + (ηh ) [uh ]|F +1/2 − (ηh ) [uh ]|F −1/2 = 0, + 2 2 which further leads to ∫ ∫ ∫ 1 ∂t |ηh | − 2 (¯ u2 )h ηh + (¯ u1 )h ηh ∂1 ηh (3.2.107) 2 F F F u1 )h }(ηh− )2 |F +1/2 + {(¯ −{(¯ u1 )h }(ηh+ )2 |F −1/2 u1 )h ηˆh ηh− |F +1/2 − (ˆ +(ˆ u1 )h ηˆh ηh+ |F −1/2 = 0. Summing over F ∈ Fh implies ∫ ∫ ∫ ∑ 1 ∂t |ηh | − 2 (¯ u2 )h ηh + u1 )h ηh ∂1 ηh − (¯ u1 )h }(ηh− )2 |F +1/2 {(¯ (3.2.108) 2 Σ Σ Σ F ∈Fh ∑ ∑ ∑ + {(¯ u1 )h }(ηh+ )2 |F −1/2 − u1 )h ηˆh ηh− |F +1/2 (ˆ + u1 )h ηˆh ηh+ |F −1/2 = 0. (ˆ F ∈Fh F ∈Fh F ∈Fh Since we always take uˆ1 = {(¯ u1 )h }, combining (3.2.105) and (3.2.108) yields ∑ µE (ηh , ηh , vh ) (3.2.109) E∈Eh 1 ∫ ∑ ( 1 1 = ∂t |ηh | + 2 − {(¯ u1 )h }(ηh− )2 |F +1/2 + {(¯ u1 )h }(ηh+ )2 |F −1/2 2 Σ F ∈Fh 2 2 ) +{(¯ ηh ηh− |F +1/2 − {(¯ u1 )h }ˆ ηh ηh+ |F −1/2 . u1 )h }ˆ 140 We can decompose the boundary terms in (3.2.109) to obtain 1 1 u1 )h }(ηh− )2 |F +1/2 + {(¯ − {(¯ u1 )h }(ηh+ )2 |F −1/2 (3.2.110) 2 2 +{(¯ ηh ηh− |F +1/2 − {(¯ u1 )h }ˆ u1 )h }ˆ ηh ηh+ |F −1/2 = ΨF +1/2 − ΨF −1/2 + ΘF , where 1 u1 )h }(ηh− )2 |F +1/2 + {(¯ ΨF +1/2 = − {(¯ ηh ηh− |F +1/2 , u1 )h }ˆ (3.2.111) 2 1 u1 )h }(ηh− )2 |F −1/2 + {(¯ ΨF −1/2 = − {(¯ ηh ηh− |F −1/2 , u1 )h }ˆ (3.2.112) 2 and 1 1 ΘF = {(¯ u1 )h }(ηh− )2 |F −1/2 u1 )h }(ηh+ )2 |F −1/2 − {(¯ (3.2.113) 2 2 −{(¯ u1 )h }ˆ ηh ηh+ |F −1/2 + {(¯ ηh ηh− |F −1/2 . u1 )h }ˆ By the flux definition (3.2.18), we can derive ( )( + ) − ηh + ηh− ΘF = {(¯ u1 )h } ηh − ηh + − ηˆh ≥ 0. (3.2.114) 2 When we sum up (3.2.110) over F ∈ Fh , all ΨF +1/2 are canceled out due to the periodicity. Then we have ∑ ( 1 1 u1 )h }(ηh− )2 |F +1/2 + {(¯ − {(¯ u1 )h }(ηh+ )2 |F −1/2 (3.2.115) F ∈Fh 2 2 ) +{(¯ ηh ηh− |F +1/2 u1 )h }ˆ − {(¯ ηh ηh+ |F −1/2 u1 )h }ˆ ≥ 0. Hence, combining (3.2.109) and (3.2.115), we can obtain the desired result. 141 Estimate of Divergence Terms In the continuous case, our discretization reduces to ∑ ∫ ρE (u, η, qh ) = − J(∇A · u)qh . (3.2.116) E∈Eh Ω Hence, this discretization is consistent. Lemma 3.2.11. When we take qh = ph , the discretized divergence term satisfies (3.2.117) ∑ ∫ ∑ ∫ ρE (uh , ηh , ph ) = − Jh ph ∇Ah · uh + [uh ] · {ph Jh Ah } · ne . E∈Eh Ω e∈∂Eh \Σ e Proof. This is a natural corollary of the definition, so we omit the proof here. 3.3 Stability Analysis Condition 3.3.1. The free surface ηh (t) satisfies 1 + ηh (t) ≥ δ > 0 for some δ > 0 independent of h and t. Condition 3.3.2. The free surface ηh (t) satisfies supF ∈Fh ∥ηh (t)∥W 1,∞ (F ) ≤ Q for some Q > 0 independent of h and t. Theorem 3.3.3. If S = 0, suppose Condition 3.3.1 is valid in t ∈ [0, T ] for some T > 0. Then there exists a unique numerical solution triple (uh , ph , ηh ) ∈ Xhk × Mhk−1 × Shk 142 to the scheme (3.2.14) with NIPG, which satisfies the estimate √ 2 ∫ t∫ 2 Jh (t)uh (t) 2 +g ∥ηh (t)∥2L2 (Σ) +ν Jh (s) DAh (s) uh (s) ds (3.3.1) L (Ω) 0 Ω √ 2 ≤ Jh (0)uh (0) + g ∥ηh (0)∥2L2 (Σ) , L2 (Ω) for any t ∈ [0, T ]. For general S, suppose Condition 3.3.1 and Condition 3.3.2 are valid in t ∈ [0, T ] for some T > 0. Then there exists a unique numerical solution triple (uh , ph , ηh ) ∈ Xhk × Mhk−1 × Shk to the scheme (3.2.14) with SIPG or NIPG, which satisfies the estimate 2 ∫ t √ Jh (t)uh (t) 2 + ∥ηh (t)∥L2 (Σ) + ∥uh (s)∥2Xh ds 2 (3.3.2) L (Ω) ( 2 0 ∫ t∫ ) √ . C(Q) Jh (0)uh (0) + ∥ηh (0)∥L2 (Σ) + 2 |S(r)| dr , 2 2 L (Ω) 0 Ω for any t ∈ [0, T ]. Proof. The existence and uniqueness follow from a standard argument for the differential- algebraic equations, so we omit it here and focus on the energy estimate. In the system (3.2.14), we take the test function vh = uh and qh = ph , and sum up over E ∈ Eh . Then it yields ∑ ∑ ζE (ηh , uh , uh ) + γE (uh , ηh , uh , uh ) (3.3.3) E∈Eh E∈Eh ∑ ∑ +ν αE (ηh , uh , ηh , uh ) + βE (ph , ηh , uh ) E∈Eh E∈Eh ∑ ∑ +g µE (ηh , ηh , uh ) = ωE (ηh , uh ), E∈Eh E∈Eh ∑ ρE (uh , ηh , ph ) = 0. (3.3.4) E∈Eh 143 By Lemma 3.2.9 and Lemma 3.2.11, we have ∑ ∑ βE (ph , ηh , uh ) = ρE (uh , ηh , ph ) = 0. (3.3.5) E∈Eh E∈Eh Hence, we can simplify (3.3.3) into ∑ ∑ ζE (ηh , uh , uh ) + γE (uh , ηh , uh , uh ) (3.3.6) E∈Eh E∈Eh ∑ ∑ ∑ +ν αE (ηh , uh , ηh , uh ) + g µE (ηh , ηh , uh ) = ωE (ηh , uh ). E∈Eh E∈Eh E∈Eh Then by Lemma 3.2.2, Lemma 3.2.4, Lemma 3.2.8 and Lemma 3.2.10, we have ∫ ∫ ∑ 1 g ∂t Jh |uh | + ∂t2 |ηh |2 + CC(Q) ∥uh ∥2Xh ≤ ωE (uh ). (3.3.7) 2 Ω 2 Σ E∈Eh Note Condition 3.3.1 and Condition 3.3.2 guarantee the existence of C(Q) > 0 which is independent of t. An application of Cauchy’s inequality implies ∑ ∫ ∫ ∫ ′ 1 ′ 1 ωE (uh ) ≤ C Jh2 |uh | + 2 |S| ≤ Q C 2 2 ∥uh ∥2Xh + |S|2 . (3.3.8) E∈Eh Ω 4C ′ Ω 4C ′ Ω In (3.3.8), when taking C ′ sufficiently small, we can always absorb it into CC(Q) ∥uh ∥2Xh in (3.3.7). Hence, we have ∫ ∫ ∫ 1 g CC(Q) ∂t Jh |uh | + ∂t 2 |ηh | +2 ∥uh ∥2Xh ≤ C(Q) |S|2 . (3.3.9) 2 Ω 2 Σ 2 Ω Then integrating over [0, t] leads to the desired result. The S = 0 case is easily derived from the general case without discussion of the source term. Remark 3.3.4. Condition 3.3.1 and Condition 3.3.2 are not always satisfied a priori. In [41], [3] and [43], this type of assumptions were also introduced in the numerical analysis. 144 3.4 Discussion on the Error Analysis For the continuous solution (u, η), the analysis in [50] reveals in the Navier-Stokes equations of (3.1.8), we need H 1 norm of η(t) to bound L2 norm of u(t). However, the result in [17] implies in the transport equation of (3.1.8), we need H 2 norm of u(t) to control H 1 norm of η(t). This type of inconsistent coupling cannot be improved even if we go to higher order derivatives. Hence, this implies the coupled system in (3.1.8) is not closed in the usual Sobolev norms, which means we cannot expect to obtain the error estimates for the whole system (3.1.8). In the following, we mainly analyze the error in the Navier-Stokes equations pro- vided we have the error estimates of the free surface. Condition 3.4.1. The free surface ηh satisfies the error estimates ∥ηh − η∥L2 . hk+1 , (3.4.1) ∥ηh − η∥L∞ . hk , (3.4.2) ∥χh − χ∥L2 . hk , (3.4.3) ∥χh − χ∥L∞ . hk−1 , (3.4.4) ∥∂t (ηh − η)∥L2 . hk , (3.4.5) ∥∂t (ηh − η)∥L∞ . hk−1 . (3.4.6) 3.4.1 Velocity Error Estimates In this section, we give the complete error estimates of the Navier-Stokes equations. Although the expressions look rather complicated, the basic ideas behind are rela- tively simple. Since most of the terms have been clearly estimated in [22], here we 145 mainly illustrate the basic estimating rules. Rule 1: For the estimates of the product terms as ∫ F1 F2 . . . F k , (3.4.7) we usually apply the H¨older’s inequality to obtain two terms in the L2 norm and several terms in the L∞ norm. The priority of taking the L∞ norm is as follows: 1. Fi only containing the exact solution (u, p, η) has the highest priority. 2. Fi containing the free surface error ηh − η, χh − χ or ∂t (ηh − η), or the numerical solution ηh , χh or ∂t ηh has the second priority. 3. Fi containing the projection error u − Pu or p − Pp for some P has the third priority. 4. Fi only containing the velocity error uh − Pu or the pressure error ph − Pp for some projection P, or the numerical solution uh or ph , has the lowest priority. In this fashion, we can make the best use of the regularity of the exact solutions and the known error estimates in the free surface to avoid the estimates for numerical solution in undesired norm due to the embedding theorem. Rule 2: For the boundary term, we always apply the trace theorem ∥F ∥L2 (∂E) . ∥F ∥H r (E) , (3.4.8) 146 for r > 1/2. This introduces more regularity in the estimates, so we should always try to avoid to use it directly and apply certain projection property to eliminate the boundary terms. Rule 3: Note the simple fact that for r > 0, we have hr ∥Fh ∥H r (E) . ∥Fh ∥L2 (E) , (3.4.9) where h is the mesh size and Fh ∈ Phk . This can bound the higher order Sobolev norm by the lower norm at the price of sacrificing some orders of h. Since in the following estimates, above rules can be applied in a very direct fashion, we do not give all the details, but just remark when necessary. We decompose the velocity error and the pressure error as follows: uh − u = (uh − Pu) + (Pu − u) = ϵu + δu , (3.4.10) ph − p = (ph − Pp) + (Pp − p) = ϵp + δp , (3.4.11) where P is some projection such that Pu ∈ Phk and Pp ∈ Phk−1 achieving the optimal accuracy, i.e. ∥δu ∥L2 . hk+1 , (3.4.12) ∥δp ∥L2 . hk . (3.4.13) Hence, the key part of the the error estimates is ϵu and ϵp . 147 Lemma 3.4.2. Suppose Condition 3.3.1, Condition 3.3.2 and Conditions 3.4.1 are valid. Assume the exact solution satisfies u ∈ C k+1 (Ω), p ∈ C k (Ω) and η ∈ C k+1 (Σ). Then for h sufficiently small, the numerical solution to the scheme (3.2.14) satisfies the estimate ∫ Jh |ϵu |2 + ∥ϵu ∥2Hh ∂t (3.4.14) Ω ( ) . C(Q) h k−1−r ∥ϵu ∥Hh + h ∥ϵp ∥L2 + ∥ϵu ∥L2 k−r 2 for 1/2 < r < 1. Proof. The discretization of the Navier-Stokes equation is ζE (ηh , uh , vh ) + γE (uh , ηh , uh , vh ) + ναE (ηh , uh , ηh , vh ) (3.4.15) +βE (ph , ηh , vh ) + µE (ηh , ηh , vh ) = ωE (ηh , vh ), ρE (uh , ηh , qh ) = 0. (3.4.16) The consistency implies the exact solution also satisfies this scheme, i.e. ζE (η, u, vh ) + γE (u, η, u, vh ) + ναE (η, u, η, vh ) (3.4.17) +βE (p, η, vh ) + µE (η, η, vh ) = ωE (η, vh ), ρE (u, η, qh ) = 0. (3.4.18) Therefore, taking the difference of above two sets of equations, we obtain the error 148 equations ( ) ζE (ηh , uh , vh ) − ζE (η, u, vh ) (3.4.19) ( ) + γE (uh , ηh , uh , vh ) − γE (u, η, u, vh ) ( ) +ν αE (ηh , uh , ηh , vh ) − αE (η, u, η, vh ) ( ) + βE (ph , ηh , vh ) − βE (p, η, vh ) ( ) ( ) + µE (ηh , ηh , vh ) − µE (η, η, vh ) = ωE (ηh , vh ) − ωE (η, vh ) , ( ) ρE (uh , ηh , qh ) − ρE (u, η, qh ) = 0. (3.4.20) Now we need to analyze each term in the error equations (3.4.19) and (3.4.20). We always decompose the difference of the bilinear forms into two parts: the energy part E∗ and the remaining part R∗ , such that Difference = E∗ + R∗ , (3.4.21) where E∗ builds the main body of the error equations and is put in the left-hand side (LHS), and R∗ is moved into the right-hand side (RHS) and can be taken as the perturbed source term. The ∗ can be ζ, γ or α, etc. In the following, we give details on how to make this decomposition in our settings and define each E∗ and R∗ . In all the forms, we take the test functions vh = ϵu and qh = ϵp . Step 1: Pressure Term and Divergence Term. We define the extended pressure form as (3.4.22) ∑∫ ∑ ∫ β(ph , ηh , vh ) = − Jh ph ∇Ah · vh + [vh ] · {ph Jh Ah } · ne , E∈Eh E e∈∂Eh \Σ e 149 and define the extended divergence form as (3.4.23) ∑∫ ∑ ∫ ρ(qh , ηh , uh ) = − Jh qh ∇Ah · uh + [uh ] · {qh Jh Ah } · ne . E∈Eh E e∈∂Eh \Σ e The pressure error term can be decomposed as follows: β(ph , ηh , vh ) − β(p, η, vh ) = β(ph − p, ηh , vh ) + β(p, ηh − η, vh ) (3.4.24) = β(ph − p, ηh , ϵu ) + β(p, ηh − η, ϵu ) ( ) = β(ϵp , ηh , ϵu ) + β(δp , ηh , ϵu ) + β(p, ηh − η, ϵu ) = Eβ + Rβ . The divergence error term can be decomposed as follows: 0 = ρ(qh , ηh , uh ) − ρ(qh , η, u) = ρ(qh , ηh , uh − u) + ρ(qh , ηh − η, u) (3.4.25) = ρ(ϵp , ηh , uh − u) + ρ(ϵp , ηh − η, u) ( ) = ρ(ϵp , ηh , ϵu ) + ρ(ϵp , ηh , δu ) + ρ(ϵp , ηh − η, u) = Eρ + Rρ . As in the proof of stability, the energy part satisfies β(ϵp , ηh , ϵu ) = ρ(ϵp , ηh , ϵu ), (3.4.26) which means they can be canceled out, so we only need to estimate the remaining part. We can directly obtain β(δp , ηh , ϵu ) . hk−1 ∥ϵu ∥Hh , (3.4.27) 150 β(p, ηh − η, ϵu ) . hk−1 ∥ϵu ∥Hh . (3.4.28) By applying the trace theorem for 1/2 < r < 1, we can obtain ρ(ϵp , ηh , δu ) . hk−r ∥ϵp ∥L2 . (3.4.29) Note the boundary terms vanish in the following term, so we have ρ(ϵp , ηh − η, u) . hk−r ∥ϵp ∥L2 . (3.4.30) Hence, in total we achieve Eβ = Eρ , (3.4.31) |Rβ | . hk−1 ∥ϵu ∥Hh , (3.4.32) |Rρ | . hk−r ∥ϵp ∥L2 for 1/2 < r < 1. (3.4.33) Step 2: Temporal Term. We define the extended temporal form as ζ(ηh , uh , vh ) (3.4.34) ∑∫ ∑∫ ∑∫ = ∂t (Jh uh ) · vh + ˜ ∂t ηh buh · ∂2 vh − ∂t ηh˜b(ˆ uh · vhint )n2 E∈Eh E E∈Eh E E∈Eh ∂E ∑ 1∫ + ∂t ηh (uh · vh ). E∈E 2 ∂E∩Σ h 151 The temporal error term can be decomposed as follows: ζ(ηh , uh , vh ) − ζ(η, u, vh ) = ζ(ηh , uh − u, vh ) + ζ(ηh − η, u, vh ) (3.4.35) = ζ(ηh , ϵu , vh ) + ζ(ηh , δu , vh ) + ζ(ηh − η, u, vh ) ( ) = ζ(ηh , ϵu , ϵu ) + ζ(ηh , δu , ϵu ) + ζ(ηh − η, u, ϵu ) = Eζ + Rζ . In the stability proof, we have shown ∫ 1 Eζ = ζ(ηh , ϵu , ϵu ) ≥ ∂t Jh |ϵu |2 . (3.4.36) 2 Ω Hence, we only need to estimate the remaining part. A direct estimate shows ζ(ηh , δu , ϵu ) . hk ∥ϵu ∥Hh . (3.4.37) Applying the trace theorem for 1/2 < r < 1, we can obtain ζ(ηh − η, u, ϵu ) . hk−1−r ∥ϵu ∥Hh . (3.4.38) Hence, in total we achieve ∫ 1 Eζ ≥ ∂t Jh |ϵu |2 , (3.4.39) 2 Ω |Rζ | . h k−1−r ∥ϵu ∥Hh for 1/2 < r < 1. (3.4.40) Step 3: Convection Term. 152 We define the extended convection form as ∑ γ(uh , ηh , uh , vh ) (3.4.41) E∈Eh ∑ 1∫ = − (uh · Nh )(uh · vh ) E∈Eh 2 ∂IE ∩Σ ∑∫ ∑ 1∫ + Jh (uh · ∇Ah uh ) · vh + Jh (∇Ah · uh )(uh · vh ) E∈Eh E E∈Eh 2 E ∑ 1∫ uint · vhint − [uh · Jh Ah ] · nE h E∈Eh 2 ∂E\∂Ω 2 ∑∫ + h − uh ) · vh . |{uh · Jh Ah } · nE | (uint ext int E∈Eh ∂E − The convection error term can be decomposed as follows: γ(uh , ηh , uh , vh ) − γ(u, η, u, vh ) (3.4.42) = γ(uh , ηh , uh − u, vh ) + γ(uh , ηh − η, u, vh ) + γ(uh − u, η, u, vh ) = γ(uh , ηh , ϵu , vh ) + γ(uh , ηh , δu , vh ) + γ(uh , ηh − η, u, vh ) + γ(uh − u, η, u, vh ) ( ) = γ(uh , ηh , ϵu , ϵu ) + γ(uh , ηh , δu , ϵu ) + γ(uh , ηh − η, u, ϵu ) + γ(uh − u, η, u, ϵu ) = Eγ + Rγ . Note the last term in (3.4.41) vanishes for the exact solution, so we do not need to worry about E − definition. In the stability proof, we have shown Eγ = γ(uh , ηh , ϵu , ϵu ) ≥ 0 (3.4.43) Hence, we only need to estimate the remaining part. Here we have more tools to utilize other than the rules listed before. Consider the results from [22, Proposition 4.1]. Although the norm there is ∥·∥Xh , based on the proof, it is easy to see we can further extend it to the norm ∥·∥Hh , i.e. for r > 2, w sufficiently small and 153 uh , wh , vh ∈ Phk , we have γ(uh , ηh , wh , vh ) . ∥ηh ∥W 1,∞ ∥uh ∥Hh ∥wh ∥Hh ∥vh ∥Hh , (3.4.44) γ(uh , ηh , w, vh ) . ∥ηh ∥W 1,∞ ∥uh ∥L2 ∥w∥H 2 ∥vh ∥Hh , (3.4.45) (3.4.46) (∑ ) γ(uh , ηh , w − Pw, vh ) . ∥ηh ∥W 1,∞ ∥w − Pw∥H 2 (E) ∥uh ∥L2 ∥vh ∥Hh . E∈Eh Combining this with the rules before, we can get the estimate γ(uh , ηh , δu , ϵu ) . hk−1 (1 + ∥ϵu ∥L2 ) ∥ϵu ∥Hh , (3.4.47) γ(uh , ηh − η, u, ϵu ) . hk−1 ∥ϵu ∥Hh , (3.4.48) γ(uh − u, η, u, ϵu ) . hk+1 ∥ϵu ∥Hh + ∥ϵu ∥L2 ∥ϵu ∥Hh . (3.4.49) Hence, in total we achieve Eγ ≥ 0, (3.4.50) |Rγ | . hk−1 (1 + ∥ϵu ∥L2 ) ∥ϵu ∥Hh + ∥ϵu ∥L2 ∥ϵu ∥Hh . (3.4.51) Step 4: Diffusion Term. 154 We define the extended diffusion form as α(ηh , uh , ηh , vh ) (3.4.52) ∫ ∫ 1 ∑ ∑ = Jh DAh uh : DAh vh − {DAh uh Jh Ah · nE } · [vh ] 2 E∈E E e e∈∂Eh \Σ ∑ ∫ h σ ∑ ± {DAh vh Jh Ah · nE } · [uh ] + [uh ][vh ]. e h e∈∂E e∈∂Eh \Σ h The diffusion error term can be decomposed as follows: α(ηh , uh , ηh , vh ) − α(η, u, η, vh ) (3.4.53) = α(ηh , uh − u, ηh , vh ) + α(ηh − η, u, ηh , vh ) + α(η, u, ηh − η, vh ) = α(ηh , ϵu , ηh , vh ) + α(ηh , δu , ηh , vh ) + α(ηh − η, u, ηh , vh ) + α(η, u, ηh − η, vh ) ( ) = α(ηh , ϵu , ηh , ϵu ) + α(ηh , δu , ηh , ϵu ) + α(ηh − η, u, ηh , ϵu ) + α(η, u, ηh − η, ϵu ) = Eα + Rα . In the stability proof, we have shown Eα = α(ηh , ϵu , ηh , ϵu ) ≥ K ∥ϵu ∥2Hh . (3.4.54) Hence, we only need to estimate the remaining part. It can be directly shown that α(ηh , δu , ηh , ϵu ) . hk ∥ϵu ∥Hh , (3.4.55) α(ηh − η, u, ηh , ϵu ) . hk−1 ∥ϵu ∥Hh , (3.4.56) α(η, u, ηh − η, ϵu ) . hk−1 ∥ϵu ∥Hh . (3.4.57) 155 Hence, in total we achieve Eα ≥ K ∥ϵu ∥2Hh , (3.4.58) |Rα | . hk−1 ∥ϵu ∥Hh . (3.4.59) Step 5: Forcing Term. We define the extended forcing form as ∑∫ ∑∫ µ(ηh , ηh , vh ) = ηh · (v2 )h − ηh ∂1 ηh · (v1 )h (3.4.60) E∈Eh ∂E∩Σ E∈Eh ∂E∩Σ ∑ 1 ∑ 1 + [ηh2 ] · vh− |∂E∩Σ+1/2 − [ηh2 ] · vh+ |∂E∩Σ−1/2 . E∈E 4 E∈E 4 h h The forcing error term can be decomposed as follows: µ(ηh , ηh , vh ) − µ(η, η, vh ) = µ(ηh − η, ηh , vh ) + µ(η, ηh − η, vh ) (3.4.61) = µ(ηh − η, ηh , ϵu ) + µ(η, ηh − η, ϵu ) = Rµ . There is no energy part here, so we only need to estimate the remaining part. µ(ηh − η, ηh , ϵu ) . hk ∥ϵu ∥Hh , (3.4.62) µ(η, ηh − η, ϵu ) . hk ∥ϵu ∥Hh . (3.4.63) Hence, in total we achieve |Rµ | . hk ∥ϵu ∥Hh . (3.4.64) 156 Step 6: Source Term. We define the extended source form as ∑∫ ωE (ηh , vh ) = Jh S · vh . (3.4.65) E∈Eh E Then we can directly estimate ω(ηh , S, vh ) − ϵ(η, S, vh ) = ω(ηh − η, S, vh ) (3.4.66) = ω(ηh − η, S, ϵu ) = Rω . There is no energy part here, so we only need to estimate the remaining part. ω(ηh − η, S, ϵu ) . hk+1 ∥ϵu ∥Hh . (3.4.67) Hence, it is easy to see that Hence, in total we achieve |Rω | . hk+1 ∥ϵu ∥Hh . (3.4.68) Step 7: Synthesis: We can summarize all above to simplify the error equations (3.4.19) as Eζ + Eγ + Eα + Eβ = −Rζ − Rγ − Rα − Rβ − Rµ − Rω , (3.4.69) Eρ = −Rρ . (3.4.70) 157 Then we can utilize the known facts about the energy part to simplify it as ∫ 1 ∂t Jh |ϵu |2 + K ∥ϵu ∥2Hh . Rρ − Rζ − Rγ − Rα − Rβ − Rµ − Rω . (3.4.71) 2 Ω Utilizing the known facts about the remaining part, we can further obtain ∫ 1 ∂t Jh |ϵu |2 + K ∥ϵu ∥2Hh . hk−1−r ∥ϵu ∥Hh + hk−r ∥ϵp ∥L2 (3.4.72) 2 Ω +hk−1 (1 + ∥ϵu ∥L2 ) ∥ϵu ∥Hh + ∥ϵu ∥L2 ∥ϵu ∥Hh for 1/2 < r < 1. The stability shows both uh and u are bounded in L2 , so ϵu is also bounded in L2 . Then we get ∫ 1 ∂t Jh |ϵu |2 + K ∥ϵu ∥2Hh (3.4.73) 2 Ω . hk−1−r ∥ϵu ∥Hh + hk−r ∥ϵp ∥L2 + ∥ϵu ∥L2 ∥ϵu ∥Hh for 1/2 < r < 1. Applying Cauchy’s inequality to the last term implies ∫ 1 ∂t Jh |ϵu |2 + K ∥ϵu ∥2Hh (3.4.74) 2 Ω 1 . hk−1−r ∥ϵu ∥Hh + hk−r ∥ϵp ∥L2 + ∥ϵu ∥2L2 + C ∥ϵu ∥2Hh for 1/2 < r < 1. 4C Taking C sufficiently small, we can absorb it into the left-hand side to achieve ∫ 1 K ∂t Jh |ϵu |2 + ∥ϵu ∥2Hh . (3.4.75) 2 Ω 2 hk−1−r ∥ϵu ∥Hh + hk−r ∥ϵp ∥L2 + ∥ϵu ∥2L2 for 1/2 < r < 1. This is the desired result. 158 3.4.2 Pressure Error Estimates In order for further analysis, we first need to show the inf-sup condition for our pressure discretization. ∑ Lemma 3.4.3. In the pressure discretization β(ph , ηh , vh ) = E∈Eh βE (ph , ηh , vh ), suppose Condition 3.3.1, Condition 3.3.2 and Conditions 3.4.1 are valid. Assume the exact solution satisfies η(t) ∈ C k+1 (Σ). Then for sufficiently small time T which is independent of h, there exists a constant Ξ > 0 such that β(ph , ηh , vh ) inf sup ≥ Ξ, (3.4.76) ph ∈Mhk−1 vh ∈X k h ∥ph ∥L2 ∥vh ∥Hh for t ∈ [0, T ], where Ξ is independent of h. Proof. The pressure discretization is as follows. (3.4.77) ∑∫ ∑ ∫ β(ph , ηh , vh ) = − Jh ph (∇Ah · vh ) + [vh ] · {ph Jh Ah } · ne . E∈Eh E e∈∂Eh \Σ e We divide our proof into several steps: Step 1: Classical Inf-sup condition. In [21], it is shown that for any v ∈ H 1 (Ω), there exists a projection Rh v ∈ Xhk such that ∫ rh ∇ · (Rh v − v) = 0 for ∀rh ∈ Phk−1 , (3.4.78) E ∫ rh [Rh v] = 0 for ∀rh ∈ Phk−1 , (3.4.79) ∂E ∥Rh v − v∥Hh ≤ C ∥v∥H 1 . (3.4.80) 159 Also, it is a classical result that for qh ∈ Mhk−1 (Ω) and v ∈ H01 (Ω), there exists Ξ1 > 0 such that ∫ q (∇ Ω h · v) sup ≥ Ξ1 ∥qh ∥L2 . (3.4.81) v∈H01 ∥v∥H 1 Combining above results, taking vh = Rh v, we can naturally show ∑ ∫ E∈Eh qh (∇ · vh ) sup E ≥ Ξ2 ∥qh ∥L2 . (3.4.82) vh ∈Xhk ∥vh ∥Hh Step 2: Inf-sup condition without perturbations. Let J0 A0 denote the values of JA at t = 0, and J¯0 A¯0 denote the average of J0 A0 in each cell E which is a piece-wise constant. Then considering the mapping between Ω0 and Ω, we can obtain ∫ J0 qh (∇A0 · v) sup Ω ≥ Ξ3 ∥qh ∥L2 . (3.4.83) v∈H01 ∥v∥H 1 Since we can naturally obtain J0 A0 − J¯0 A¯0 ∞ ≤ C0 h, (3.4.84) L which implies ∫ J¯0 qh (∇A¯0 · v) sup Ω ≥ (Ξ3 − C0 h) ∥qh ∥L2 . (3.4.85) v∈H01 ∥v∥H 1 Furthermore, from a lengthy but not difficult computation, we can easily see property (3.4.78), (3.4.79) and (3.4.80) are still valid if we replace the usual gradient operator ∇ by weighted gradient J¯0 ∇A¯0 This means we can still find the projection Rh v for 160 each v ∈ H01 such that ∫ rh J¯0 ∇A¯0 · (Rh v − v) = 0 for ∀rh ∈ Phk−1 , (3.4.86) E ∫ rh [Rh v] = 0 for ∀rh ∈ Phk−1 , (3.4.87) ∂E ∥Rh v − v∥Hh ≤ C ∥v∥H 1 . (3.4.88) Hence, by taking vh = Rh v, based on the property (3.4.86) and (3.4.88), we can show ∑ ∫ E∈Eh J¯0 qh (∇A¯0 · vh ) Ξ3 − C0 h sup E ≥ ∥qh ∥L2 . (3.4.89) vh ∈Xhk ∥vh ∥Hh 1+C In the following, we consider a perturbation of this inf-sup condition. Step 3: Time perturbations. We turn to our pressure discretization. Since J∇A v = J∂1 v1 − A∂2 v1 + ∂2 v2 . (3.4.90) We can separate the bulk integral as follows. (3.4.91) ∑∫ ∑∫ ∑∫ − Jh ph ∇Ah · vh = − Jph ∇A · vh − ph (Jh ∇Ah − J∇A ) · vh . E∈Eh E E∈Eh E E∈Eh E 161 we can directly estimate ∑∫ − Jph ∇A · vh (3.4.92) E∈Eh E ∑∫ ∑∫ = − J0 ph ∇A0 · vh − ph (J∇A − J0 ∇A0 ) · vh E∈Eh E E∈Eh E ∑∫ ∑∫ = − ph J¯0 ∇A¯0 · vh − ph (J0 ∇A0 − J¯0 ∇A¯0 ) · vh E∈Eh E E∈Eh E ∑∫ − ph (J∇A − J0 ∇A0 ) · vh . E∈Eh E By the analysis in Step 2, we have ∑ ∫ E∈Eh ph J¯0 ∇A¯0 · vh Ξ3 − C0 h sup E ≥ ∥ph ∥L2 . (3.4.93) vh ∈Xhk ∥vh ∥Hh 1+C By Taylor’s expansion around the average and continuity of exact solution, we have ∑∫ ph (J0 ∇A0 − J¯0 ∇A¯0 ) · vh ≤ C1 h ∥ph ∥L2 ∥vh ∥Hh , (3.4.94) E∈Eh E for some C1 > 0. Furthermore, for sufficiently small time T , [2] shows ∥η(t) − η0 ∥H 1 can be arbitrarily small. Then we can obtain ∑∫ − ph (J∇A − J0 ∇A0 ) · vh ≤ C2 (T ) ∥ph ∥L2 ∥vh ∥Hh . (3.4.95) E∈Eh E Also, by the error estimates for the free surface, there exists constants C3 > 0 satis- fying ∑∫ − ph (Jh ∇Ah − J∇A ) · vh ≤ C3 hk−1 ∥ph ∥L2 ∥vh ∥Hh . (3.4.96) E∈Eh E 162 In summary, we have (3.4.97) ∑∫ ( ) Ξ3 − C0 h − Jh ph ∇Ah · vh ≥ − C1 h − C2 (T ) − C3 hk−1 ∥ph ∥L2 ∥vh ∥Hh . E∈Eh E 1+C Step 4: Boundary perturbations. On each boundary, taking E as a reference cell, denoting p∗ as the value read from outside and p read from inside, we have 1 1 {ph Jh Ah } = ph Jh Ah + p∗h Jh∗ Ah∗ (3.4.98) 2 2 ( ) 1 ∗ ∗ ∗ ∗ = (Cph + C ph ) + {ph Jh Ah } − (Cph + C ph ) . 2 where J ∗ A ∗ is taken in the same convention of p∗ , and matrix C is the average of Jh Ah and C ∗ of J ∗ A ∗ on that boundary. By Taylor’s expansion around the average and the continuity of the numerical solution in each cell, we have the natural estimate 1 {ph Jh Ah } − (Cph + C ∗ p∗h ) (3.4.99) ( 2) ( ) ( ) 1 1 ∗ ∗ ∗ ∗ ∗ = Jh Ah − C ph + J A − C ph . h |ph | + |ph | . 2 2 h h Then we can decompose the boundary integral in the same fashion as ∑ ∫ ∑ ∫ 1 [vh ] · {ph Jh Ah } · ne = [vh ] · (Cph + C ∗ p∗h ) · ne (3.4.100) e e 2 e∈∂Eh \Σ e∈∂Eh \Σ ∑ ∫ ( ) 1 ∗ ∗ + [vh ] · {ph Jh Ah } − (Cph + C ph ) · ne . e 2 e∈∂Eh \Σ Since we define vh = Rh v, due to ph , p∗h ∈ Phk−1 , we can utilize the orthogonal property 163 (3.4.87) to show ∑ ∫ [vh ] · (Cph + C ∗ p∗h ) · ne = 0. (3.4.101) e∈∂Eh \Σ e For the remaining term in (3.4.100), we can apply H¨older’s inequality and the trace theorem to obtain ∑ ∫ ( ) ∗ ∗ [vh ] · {ph Jh Ah } − (Cph + C ph ) · ne (3.4.102) e∈∂Eh \Σ e ∑ . h ∥ph ∥L2 (e) ∥vh ∥L2 (e) e∈∂Eh \Σ ∑ . h ∥ph ∥H 3/4 (E) ∥vh ∥Hh E∈Eh . h 1/4 ∥ph ∥L2 ∥vh ∥Hh . In summary, we have ∑ ∫ [vh ] · {ph Jh Ah } · ne ≤ C4 h1/4 ∥ph ∥L2 ∥vh ∥Hh . (3.4.103) e∈∂Eh \Σ e Step 5: Synthesis. In total, summarizing (3.4.97) and (3.4.103), we can obtain (3.4.104) ( ) β(ph , ηh , vh ) Ξ3 − C0 h sup ≥ − C1 h − C2 (T ) − C3 hk−1 − C4 h1/4 ∥ph ∥L2 . vh ∈Xhk ∥vh ∥Hh 1+C C2 can be arbitrarily small as long as we take T sufficiently small. Hence, if h is also sufficiently small, we can obtain β(ph , ηh , vh ) Ξ3 sup ≥ ∥ph ∥L2 . (3.4.105) vh ∈Xhk ∥vh ∥Hh 2C 164 Hence, taking Ξ = Ξ3 /(2C), we have shown β(ph , ηh , vh ) inf sup ≥ Ξ, (3.4.106) ph ∈Mhk−1 vh ∈X k h ∥ph ∥L2 ∥vh ∥H 1 which is the inf-sup condition desired. Lemma 3.4.4. Suppose Condition 3.3.1, Condition 3.3.2 and Conditions 3.4.1 are valid. Assume the exact solution satisfies u ∈ C k+1 (Ω), p ∈ C K (Ω) and η ∈ C k+1 (Σ). Then for sufficiently small T and h, the numerical solution to the scheme (3.2.14) satisfies the estimate ∫ t ∥ϵp ∥L2 (3.4.107) 0 ( ∫ t ∫ ) 1 t . C(Q) h k−1−r + ∥ϵu ∥L2 + ∥ϵu (t)∥L2 + ∥ϵu ∥Hh for 1/2 < r < 1, 0 h 0 within t ∈ [0, T ]. Proof. We can take any test function vh ∈ Xhk in the error equation (3.4.19) and integrate over time [0, t]. Here we do not distinguish between the energy part and remaining part, but only concentrate on ∫ T β(ph , ηh , vh ). (3.4.108) 0 All the other terms can be moved into the right-hand side and estimated in terms of ∥vh ∥Hh . Basically, the estimates are similar to the velocity error estimates and the only exception is the energy part, where we cannot obtain the perfect estimates as in the stability proof now. Hence, we need to give some explanations for their estimates. For the temporal term, since vh is independent of time, we can integrate by part for 165 the temporal derivative and obtain ∫ t ∫ t ζ(ηh , ϵu , vh ) . ∥ϵu (t)∥L2 ∥vh ∥Hh + ∥vh ∥Hh ∥ϵu ∥Hh . (3.4.109) 0 0 We may directly estimate the convection term as ∫ t γ(uh , ηh , ϵu , vh ) (3.4.110) ∫ t 0 ∫ t 1 . ∥vh ∥Hh ∥uh ∥Hh ∥ϵu ∥Hh . ∥vh ∥Hh ∥uh ∥L2 ∥ϵu ∥Hh 0 h 0 ∫ t 1 . ∥vh ∥Hh ∥ϵu ∥Hh . h 0 Also, we can estimate the diffusion term as ∫ t ∫ t α(ηh , ϵu , ηh , vh ) . ∥vh ∥Hh ∥ϵu ∥Hh . (3.4.111) 0 0 Based on the inf-sup condition in Lemma 3.4.3, we can deduce the desired result. 3.4.3 Error Analysis of the Fluid Theorem 3.4.5. Suppose Condition 3.3.1, Condition 3.3.2 and Conditions 3.4.1 are valid. Assume the exact solution satisfies u ∈ C k+1 (Ω), p ∈ C k (Ω) and η ∈ C k+1 (Σ). Then for sufficiently small T and h, the numerical solution to the scheme (3.2.14) satisfies the estimates ∥uh − u∥L2 . C(Q)hk−1−r , (3.4.112) (∫ t )1/2 ∥uh − u∥Hh2 . C(Q)hk−1−r , (3.4.113) 0 ∫ t ∥ph − p∥L2 . C(Q)hk−1−r f or 1/2 < r < 1, (3.4.114) 0 166 in t ∈ [0, T ]. Proof. In Lemma 3.4.2 and Lemma 3.4.4, we have shown ∫ Jh |ϵu |2 + ∥ϵu ∥2Hh ∂t (3.4.115) Ω ( ) . C(Q) h k−1−r ∥ϵu ∥Hh + h ∥ϵp ∥L2 + ∥ϵu ∥L2 k−r 2 for 1/2 < r < 1, and ∫ t ∥ϵp ∥L2 (3.4.116) ( 0 ∫ t ∫ ) 1 t . C(Q) h k−1−r + ∥ϵu ∥L2 + ∥ϵu (t)∥L2 + ∥ϵu ∥Hh for 1/2 < r < 1. 0 h 0 Integrating over [0, t] for any t ∈ [0, T ] in (3.4.115), we obtain ∫ ∫ t Jh (t) |ϵu (t)| + 2 ∥ϵu ∥2Hh (3.4.117) Ω ( 0 ∫ t ∫ t ∫ t ) . C(Q) h k+2 +h k−1−r ∥ϵu ∥Hh + h k−r ∥ϵp ∥L2 + ∥ϵu ∥L2 . 2 0 0 0 Then we plug (3.4.116) into (3.4.117) to eliminate ϵp and obtain ∫ ∫ t Jh (t) |ϵu (t)| + 2 ∥ϵu ∥2Hh (3.4.118) Ω ( 0 ∫ t ∫ t ) . C(Q) h 2k−1−2r +h k−1−r ∥ϵu ∥Hh + h ∥ϵu (t)∥L2 + k−r ∥ϵu ∥L2 2 0 0 ( (∫ t )2 ∫ t ) . C(Q) h 2k−1−2r +h 2k−2−2r + ∥ϵu ∥Hh + h ∥ϵu (t)∥L2 + k−r ∥ϵu ∥L2 2 ( 0 ) 0 √ ∫ t ∫ t . C(Q) h 2k−2−2r + t ∥ϵu ∥Hh + h ∥ϵu (t)∥L2 + 2 k−r ∥ϵu ∥L2 . 2 0 0 √ ∫t When T is sufficiently small, we can absorb C(Q) t 0 ∥ϵu ∥2Hh into the left-hand side 167 of (3.4.118) to achieve (3.4.119) ∫ ∫ t ( ∫ t ) Jh (t) |ϵu (t)|2 + ∥ϵu ∥2Hh . C(Q) h2k−2−2r + h ∥ϵu (t)∥L2 + k−r ∥ϵu ∥L2 . 2 Ω 0 0 Since (3.4.119) holds for any t ∈ [0, T ], we can take the maximum to get (3.4.120) ( ) max ∥ϵu (t)∥2L2 . C(Q) h 2k−2−2r +h k−r max ∥ϵu (t)∥L2 + T max ∥ϵu (t)∥L2 . 2 t∈[0,T ] t∈[0,T ] t∈[0,T ] When T and h are sufficiently small, we can absorb C(Q)(hk−r + T ) maxt∈[0,T ] ∥ϵu ∥2L2 into the left-hand side of (3.4.120) to obtain max ∥ϵu (t)∥2L2 . C(Q)h2k−2−2r . (3.4.121) t∈[0,T ] Then this leads to max ∥ϵu (t)∥L2 . C(Q)hk−1−r . (3.4.122) t∈[0,T ] Hence, we plug (3.4.122) into (3.4.119) and (3.4.116) to achieve ∫ t ∥ϵu ∥2Hh . C(Q)h2k−2−2r , (3.4.123) ∫ 0 t ∥ϵp ∥L2 . C(Q)hk−1−r . (3.4.124) 0 Combining with the projection errors δu and δp , we show the desired result. Remark 3.4.6. The convergent rate k − 1 − r is not optimal. If we can obtain a nicer error estimate of the temporal term, then this rate can be further improved. 168 3.5 Numerical Tests We perform the accuracy tests for our numerical scheme. Our tests are based on a set of exact solutions η = Z sin(2π(x1 − t)), (3.5.1) u1 = Z(6x52 + 15x42 + 12x32 + 3x22 ) sin(2π(x1 − t)) + (6x52 + 5x42 + 1), (3.5.2) u2 = πZ 2 (4x62 + 15x52 + 21x42 + 13x32 + 3x22 ) sin(4π(x1 − t)) (3.5.3) +2πZ(4x62 + 7x52 + 2x42 − x32 ) cos(2π(x1 − t)), p = Z sin(2π(x1 − t)), (3.5.4) with the source term ( ) S1 = (u1 )t − ηt (1 + x2 )K∂2 u1 + u1 (∂1 u1 − AK∂2 u1 ) + Ku2 ∂2 u1 (3.5.5) ( −ν ∂11 u1 + (1 + A2 )K 2 ∂22 u1 − 2AK∂12 u1 ) +(AK ∂2 A − A∂1 K − K∂1 A)∂2 u1 + (∂1 p − AK∂2 p), 2 ( ) S2 = (u2 )t − ηt (1 + x2 )K∂2 u2 + u1 (∂1 u2 − AK∂2 u2 ) + Ku2 ∂2 u2 (3.5.6) ( −ν ∂11 u2 + (1 + A2 )K 2 ∂22 u2 − 2AK∂12 u2 ) +(AK ∂2 A − A∂1 K − K∂1 A)∂2 u2 + K∂2 p, 2 where 0 < Z < 1 can be any fixed constant, A and K are defined as in (3.1.7), and differential operators ∂t , ∂i and ∂ij are defined in the usual sense. In the following test, we always take Z = 0.1, ν = 0.05 and T = 1/8. 169 3.5.1 Accuracy Tests with SIPG The following are the error tables for our numerical scheme with SIPG: Table 3.1: L2 Error Table for Xh1 − Mh0 − Sh1 formulation with SIPG (a) Free Surface Error η − ηh (b) Velocity Error u − uh N Error Order N Error Order 4 9.4028E-4 - 4 4.8233E-2 - 8 2.5493E-4 1.8830 8 1.3863E-2 1.7988 16 6.6612E-5 1.9362 16 3.6182E-3 1.9378 32 1.7409E-5 1.9359 32 9.4871E-4 1.9312 64 4.9397E-6 1.8173 64 2.4593E-4 1.9477 (c) Pressure Error p − ph N Error Order 4 3.1028E-3 - 8 1.6255E-3 0.9327 16 8.2709E-4 0.9748 32 4.1570E-4 0.9925 64 2.0818E-4 0.9977 170 Table 3.2: L2 Error Table for Xh2 − Mh1 − Sh2 formulation with SIPG (a) Free Surface Error η − ηh (b) Velocity Error u − uh N Error Order N Error Order 4 1.4435E-4 - 4 5.8590E-3 - 8 1.6283E-5 3.1481 8 7.7408E-4 2.9201 16 2.1495E-6 2.9213 16 9.8674E-5 2.9717 32 2.9961E-7 2.8428 32 1.2586E-5 2.9709 (c) Pressure Error p − ph N Error Order 4 9.6076E-4 - 8 2.3182E-4 2.0512 16 5.4527E-5 2.0880 32 1.2942E-5 2.0749 3.5.2 Accuracy Tests with NIPG The following are the error tables for our numerical scheme with NIPG: 171 Table 3.3: L2 Error Table for Xh1 − Mh0 − Sh1 formulation with NIPG (a) Free Surface Error η − ηh (b) Velocity Error u − uh N Error Order N Error Order 4 9.3809E-4 - 4 4.5428E-2 - 8 2.5179E-4 1.8975 8 1.8218E-2 1.3182 16 6.5861E-5 1.9347 16 6.7565E-3 1.4310 32 1.7258E-5 1.9322 32 2.1758E-3 1.6347 64 4.9108E-6 1.8132 64 7.5074E-4 1.5352 (c) Pressure Error p − ph N Error Order 4 3.2035E-3 - 8 1.6481E-3 0.9588 16 8.4487E-4 0.9640 32 4.2453E-4 0.9929 64 2.1239E-4 0.9992 172 Table 3.4: L2 Error Table for Xh2 − Mh1 − Sh2 formulation with NIPG (a) Free Surface Error η − ηh (b) Velocity Error u − uh N Error Order N Error Order 4 1.4810E-4 - 4 6.0580E-3 - 8 1.7895E-5 3.0489 8 1.2693E-3 2.2548 16 3.3218E-6 2.4295 16 1.9305E-4 2.7170 32 8.5392E-7 1.9598 32 2.7534E-5 2.8097 (c) Pressure Error p − ph N Error Order 4 9.2305E-4 - 8 2.8213E-4 1.7100 16 7.5701E-5 1.8980 32 2.0531E-5 1.8825 3.5.3 Discussion on the Numerical Tests In above accuracy tests, we can obtain the optimal order of convergence when applying scheme (3.2.14) with SIPG, and the sub-optimal order with NIPG. However, in both cases, our numerical results are much better than the analytical result obtained in Theorem 3.4.5. 3.6 Conclusions and Remarks In this chapter, we construct a stable numerical scheme to solve the system (3.1.8) with discontinuous Galerkin method, and discuss the error analysis in the fluid with 173 certain assumptions. Although we focus on the 2-D fluid throughout this chapter, it is easy to see this scheme can be naturally extended to the 3-D case with periodic settings for two horizontal directions. The main restriction to our scheme is that we require the free surface to be a single-valued function of the horizontal variables, which is not always true in practice, especially when the topological structure of the free surface varies during the evolution. Our scheme is non-conservative. Hence, it might not give the qualitatively correct simulation if the exact solution possesses singularities. However, for smooth exact solutions, our scheme can give a quite good approximation. Here we use the piecewise polynomial P k due to the connection between the bulk discretization and surface discretization. Note that Qk , which denotes the piecewise polynomials of degree at most k for each variables, is acceptable in the scheme, but not suitable in analysis since it cannot satisfy the desired inf-sup condition for pressure term in the error analysis. Chapter Four Asymptotic Analysis of Neutron Transport Equation 175 4.1 Introduction and Notation 4.1.1 Problem Formulation We consider a homogeneous isotropic steady neutron transport equation in a two- dimensional unit disk Ω = {⃗x = (x1 , x2 ) : |⃗x| ≤ 1} with one-speed velocity Σ = {w ⃗= ⃗ ∈ S 1 } as (w1 , w2 ) : w    ϵw ⃗ · ∇x uϵ + uϵ − u¯ϵ = 0 in Ω, (4.1.1)   uϵ (⃗x0 , w) ⃗ = g(⃗x0 , w) ⃗ · ⃗n < 0 and ⃗x0 ∈ ∂Ω, ⃗ for w where ∫ ϵ 1 u¯ (⃗x) = uϵ (⃗x, w)d ⃗ w.˜ (4.1.2) 2π S1 and ⃗n is the outward normal vector on ∂Ω, with the Knudsen number 0 < ϵ << 1. We will study the diffusive limit of uϵ as ϵ → 0. Based on the flow direction, we can divide the boundary Γ = {(⃗x, w) ⃗ : ⃗x ∈ ∂Ω} into the in-flow boundary Γ− , the out-flow boundary Γ+ , and the grazing set Γ0 as Γ− = {(⃗x, w) ⃗ : ⃗x ∈ ∂Ω, w ⃗ · ⃗n < 0}, (4.1.3) Γ+ = {(⃗x, w) ⃗ : ⃗x ∈ ∂Ω, w ⃗ · ⃗n > 0}, (4.1.4) Γ0 = {(⃗x, w) ⃗ : ⃗x ∈ ∂Ω, w ⃗ · ⃗n = 0}. (4.1.5) It is easy to see Γ = Γ+ ∪ Γ− ∪ Γ0 . Hence, the boundary condition is only given on Γ− . 176 The study of neutron transport equation dates back to 1950s. The main methods include the explicit formula and spectral analysis of the transport operators(see [33, 32, 35, 34, 36, 37, 38, 40, 39]). A classical result in [9] states that ∥uϵ − U0 − U0 ∥L∞ = O(ϵ) (4.1.6) where U0 is the Knudsen layer to the Milne problem (4.2.25) while U0 is the corre- sponding interior solution to the Laplace equation (4.2.26). The goal of this paper is to construct a counterexample to such a result in a disk. 4.1.2 Main Results ⃗ ∈ C 3 (Γ− ). Then for the steady neutron transport Theorem 4.1.1. Assume g(⃗x0 , w) ⃗ ∈ L∞ (Ω × S 1 ) satisfies equation (4.1.1), the unique solution uϵ (⃗x, w) ∥uϵ − U0ϵ − U0ϵ ∥L∞ = O(ϵ) (4.1.7) where the interior solution U0ϵ and boundary layer U0ϵ are defined in (4.2.51) and (4.2.50). Moreover, if g(θ, ϕ) = cos ϕ, then there exists a C > 0 such that ∥uϵ − U0 − U0 ∥L∞ ≥ C > 0 (4.1.8) when ϵ is sufficiently small, where U0 and U0 are defined in (4.2.26) and (4.2.25). Remark 4.1.2. θ and ϕ are defined in (4.2.13) and (4.2.37). For the diffusive boundary case, the zeroth order classical Knudsen layer is always absent. Our method leads to a counterexample to the classical Knudsen layer expansion at first order(see Theorem 4.6.9). 177 Our results demonstrates that the classical Knudsen layer expansion in Theorem 4.2.1 breaks down in a unit disk in L∞ . Even though the ϵ-Milne equation with geometric correction (4.2.50) only modifies the Milne equation (4.2.25) with an ϵ order term, the difference between the solutions can be order 1 via a contradiction argument in Step 5 of Section 5. We remark that L∞ is a natural space to characterize boundary layer contributions, whose Lp norm is O(ϵ1/p ) for 1 ≤ p < ∞. Unfortunately, our new expansion can only be established in a disk. A new mathematical theory is needed to characterize such a diffusive limit in a general domain. Our analysis is based on a careful study of the ϵ-Milne problem with geometric correction. Our proof is self-contained and without any use of probabilistic techniques as in [9]. Some recent results on stationary Boltzmann equation with a similar type of techniques can be found in [5, 4, 6]. 4.1.3 Notation and Conventions Throughout this chapter, C > 0 denotes a constant that only depends on the param- eter Ω, but does not depend on the data. It is referred as universal and can change from one inequality to another. When we write C(z), it means a certain positive constant depending on the quantity z. 178 4.2 Asymptotic Analysis 4.2.1 Interior Expansion We define the interior expansion as follows: ∑ ∞ ⃗ ∼ U (⃗x, w) ϵk Uk (⃗x, w), ⃗ (4.2.1) k=0 where Uk can be defined by comparing the order of ϵ via plugging (4.2.1) into the equation (4.1.1). Thus, we have U0 − U¯0 = 0, (4.2.2) U1 − U¯1 = −w ⃗ · ∇x U0 , (4.2.3) U2 − U¯2 = −w ⃗ · ∇x U 1 , (4.2.4) ... Uk − U¯k = −w ⃗ · ∇x Uk−1 . (4.2.5) The following analysis reveals the equation satisfied by Uk : Plugging (4.2.2) into (4.2.3), we obtain U1 = U¯1 − w ⃗ · ∇x U¯0 . (4.2.6) Plugging (4.2.6) into (4.2.4), we get U2 − U¯2 = −w ⃗ · ∇x (U¯1 − w ⃗ · ∇x U¯0 ) = −w ⃗ · ∇x U¯1 + w ⃗ 2 ∆x U¯0 + 2w1 w2 ∂x1 x2 U¯0 .(4.2.7) 179 ⃗ ∈ S 1 , we achieve the final form Integrating (4.2.7) over w ∆x U¯0 = 0, (4.2.8) which further implies U0 (⃗x, w) ⃗ satisfies the equation    U0 = U¯0 (4.2.9)   ∆x U0 = 0 ⃗ for k ≥ 1 satisfies Similarly, we can derive Uk (⃗x, w)    Uk = U¯k − w ⃗ · ∇x Uk−1 (4.2.10)   ∆x U¯k = 0 4.2.2 Milne Expansion In order to determine the boundary condition fir Uk , it is well known that we need to define the boundary layer expansion. Hence, we need several substitutions: Substitution 1: We consider the substitution into quasi-polar coordinates uϵ (x1 , x2 , w1 , w2 ) → uϵ (µ, θ, w1 , w2 ) with (µ, θ, w1 , w2 ) ∈ [0, 1) × [−π, π) × S 1 defined as    x1 = (1 − µ) cos θ, (4.2.11)   x2 = (1 − µ) sin θ. 180 Here µ denotes the distance to the boundary ∂Ω and θ is the space angular variable. In these new variables, equation (4.1.1) can be rewritten as  ( ) ϵ ( ) ϵ   −ϵ w1 cos θ + w2 sin θ ∂u − ϵ w1 sin θ − w2 cos θ ∂u     ∫ ∂µ 1−µ ∂θ 1  +uϵ − uϵ dw ˜ = 0, (4.2.12)   2π S 1    uϵ (0, θ, w , w ) = g(θ, w , w ) for w cos θ + w sin θ < 0. 1 2 1 2 1 2 Substitution 2: We further define the stretched variable η by making the scaling transform for uϵ (µ, θ, w1 , w2 ) → uϵ (η, θ, w1 , w2 ) with (η, θ, w1 , w2 ) ∈ [0, 1/ϵ) × [−π, π) × S 1 as    η = µ/ϵ, (4.2.13)   θ = θ, which implies ∂uϵ 1 ∂uϵ = . (4.2.14) ∂µ ϵ ∂η Then equation (4.1.1) is transformed into  ( ) ϵ ( ) ϵ   − ∂u − ϵ − ∂u   w 1 cos θ + w 2 sin θ w1 sin θ w 2 cos θ   ∫ ∂η 1 − ϵη ∂θ 1  +u −ϵ ϵ u dw ˜ = 0, (4.2.15)   2π S 1    uϵ (0, θ, w , w ) = g(θ, w , w ) for w cos θ + w sin θ < 0. 1 2 1 2 1 2 Substitution 3: Define the velocity substitution for uϵ (η, θ, w1 , w2 ) → uϵ (η, θ, ξ) with (η, θ, ξ) ∈ [0, 1/ϵ)× 181 [−π, π) × [−π, π) as    w1 = − sin ξ, (4.2.16)   w2 = − cos ξ. Here ξ denotes the velocity angular variable. We have the succinct form for (4.1.1) as  ∫ π  ∂uϵ ϵ ∂uϵ 1  sin(θ + ξ) − cos(θ + ξ) +u − ϵ uϵ dξ = 0, ∂η 1 − ϵη ∂θ 2π −π (4.2.17)   uϵ (0, θ, ξ) = g(θ, ξ) for sin(θ + ξ) > 0. We now define the Milne expansion of boundary layer as follows: ∑ ∞ U (η, θ, ϕ) ∼ ϵk Uk (η, θ, ϕ), (4.2.18) k=0 where Uk can be determined by comparing the order of ϵ via plugging (4.2.18) into the equation (4.2.17). Thus, in a neighborhood of the boundary, we have ∂U0 sin(θ + ξ) + U0 − U¯0 = 0, (4.2.19) ∂η ∂U1 1 ∂U0 sin(θ + ξ) + U1 − U¯1 = cos(θ + ξ) , (4.2.20) ∂η 1 − ϵη ∂θ ... ∂Uk 1 ∂Uk−1 sin(θ + ξ) + Uk − U¯k = cos(θ + ξ) , (4.2.21) ∂η 1 − ϵη ∂θ where ∫ π 1 U¯k (η, θ) = Uk (η, θ, ξ)dξ. (4.2.22) 2π −π 182 The construction of Uk and Uk in [9] can be summarized as follows: Step 1: Construction of U0 and U0 . Assume the cut-off function ψ and ψ0 are defined as    1 0 ≤ µ ≤ 1/2, ψ(µ) = (4.2.23)   0 3/4 ≤ µ ≤ ∞.    1 0 ≤ µ ≤ 1/4, ψ0 (µ) = (4.2.24)   0 3/8 ≤ µ ≤ ∞. Then the zeroth order boundary layer solution is defined as  ( )   U0 (η, θ, ξ) = ψ0 (ϵη) f0 (η, θ, ξ) − f0 (∞, θ) ,        sin(θ + ξ) ∂f0 + f − f¯ = 0, 0 0 ∂η (4.2.25)     f0 (0, θ, ξ) = g(θ, ξ) for sin(θ + ξ) > 0,      lim f (η, θ, ξ) = f0 (∞, θ). η→∞ 0 Assuming g ∈ L∞ (Γ− ), by Theorem 4.4.12, we can show there exists a unique solution f0 (η, θ, ξ) ∈ L∞ ([0, ∞) × [−π, π) × [−π, π)). Hence, U0 is well-defined. Then we can define the zeroth order interior solution as      U0 = U¯0 ,  ∆x U¯0 = 0 in Ω, (4.2.26)      U¯0 = f0 (∞, θ) on ∂Ω. Step 2: Construction of U1 and U1 . 183 Define the first order boundary layer solution as  ( )   U1 (η, θ, ξ) = ψ0 (ϵη) f1 (η, θ, ξ) − f1 (∞, θ) ,        sin(θ + ξ) ∂f1 + f − f¯ = cos(θ + ξ) ψ(ϵη) ∂U0 , 1 1 ∂η 1 − ϵη ∂θ (4.2.27)     f1 (0, θ, ξ) ⃗ · ∇x U0 (⃗x0 , w) = w ⃗ for sin(θ + ξ) > 0,      limη→∞ f1 (η, θ, ξ) = f1 (∞, θ). where (⃗x0 , w) ⃗ is the same point as (0, θ, ξ). Define the first order interior solution as      U1 = U¯1 − w ⃗ · ∇x U0 ,   ∆x U¯1 = 0 in Ω, (4.2.28)     U¯1 = f1 (∞, θ) on ∂Ω. Step 3: Generalization to arbitrary k. Similar to above procedure, we can define the k th order boundary layer solution as  ( )   Uk (η, θ, ξ) = ψ0 (ϵη) fk (η, θ, ξ) − fk (∞, θ) ,       ∂fk ψ(ϵη) ∂Uk−1  sin(θ + ξ) + fk − f¯k = cos(θ + ξ) , ∂η 1 − ϵη ∂θ (4.2.29)     ⃗ · ∇x Uk−1 (⃗x0 , w) fk (0, θ, ξ) = w ⃗ for sin(θ + ξ) > 0,      limη→∞ fk (η, θ, ξ) = fk (∞, θ). Define the k th order interior solution as      Uk = U¯k − w ⃗ · ∇x Uk−1 ,  ∆x U¯k = 0 in Ω, (4.2.30)      U¯k = fk (∞, θ) on ∂Ω. In [9, pp.136], the author proved the following result: Theorem 4.2.1. Assume g(⃗x0 , w) ⃗ is sufficiently smooth. Then for the steady neutron 184 ⃗ ∈ L∞ (Ω × S 1 ) satisfies transport equation (4.1.1), the unique solution uϵ (⃗x, w) ∥uϵ − U0 − U0 ∥L∞ = O(ϵ). (4.2.31) Our work begins with a crucial observation that based on Remark 4.4.15, the existence of solution f1 requires the source term ψ ∂U0 cos(θ + ξ) ∈ L∞ ([0, ∞) × [−π, π) × [−π, π)). (4.2.32) 1 − ϵη ∂θ Since the support of ψ(ϵη) depends on ϵ, by (4.2.25), this in turn requires ( ) ∂ f0 (η, θ, ξ) − f0 (∞, θ) ∈ L∞ ([0, ∞) × [−π, π) × [−π, π)) (4.2.33) ∂θ Note that Z = ∂θ (f0 − f0 (∞, θ)) satisfies the equation   ∂Z ∂f0   sin(θ + ξ) + Z − Z¯ = − cos(θ + ξ) ,   ∂η ∂η  ∂g(θ, ξ) ∂f0 (∞, θ)   Z(0, θ, ξ) = − for sin(θ + ξ) > 0, (4.2.34)   ∂θ ∂θ   limη→∞ Z(η, θ, ξ) = Z(∞, θ). In order for Z ∈ L∞ ([0, ∞) × [−π, π) × [−π, π)), assuming the boundary data ∂θ g ∈ L∞ (Γ− ), we require the source term ∂f0 − cos(θ + ξ) ∈ L∞ ([0, ∞) × [−π, π) × [−π, π)). (4.2.35) ∂η On the other hand, as shown by Lemma B.1.1, we can show for specific g, it holds / L∞ ([0, ∞) × [−π, π) × [−π, π)). Due to intrinsic singularity for (4.2.25), that ∂η f0 ∈ the construction in [9] breaks down. 185 In fact, in general geometry with curved boundary, we need to control the normal derivative of the boundary layer solution for the Milne expansion. 4.2.3 ϵ-Milne Expansion with Geometric Correction Our main goal is to overcome the difficulty in estimating ψ ∂Uk cos(θ + ξ) . (4.2.36) 1 − ϵη ∂θ We introduce one more substitution to decompose the term (4.2.36). Substitution 4: We make the rotation substitution for uϵ (η, θ, ξ) → uϵ (η, θ, ϕ) with (η, θ, ϕ) ∈ [0, 1/ϵ)× [−π, π) × [−π, π) as      η = η,  θ = θ, (4.2.37)      ϕ = θ + ξ, and transform the equation (4.1.1) into  ( ϵ ) ∫ π  ∂uϵ ϵ ∂u ∂uϵ 1  sin ϕ − cos ϕ + +u − ϵ uϵ dϕ = 0, ∂η 1 − ϵη ∂ϕ ∂θ 2π −π (4.2.38)   uϵ (0, θ, ϕ) = g(θ, ϕ) for sin ϕ > 0. Inspired by [11], [54] and [5], the most important idea is to include the most singular term ϵ ∂uϵ − cos ϕ (4.2.39) 1 − ϵη ∂ϕ 186 in the Milne problem. We define the ϵ-Milne expansion with geometric correction of boundary layer as follows: ∑ ∞ U (η, θ, ϕ) ∼ ϵ ϵk Ukϵ (η, θ, ϕ), (4.2.40) k=0 where Ukϵ can be determined by comparing the order of ϵ via plugging (4.2.40) into the equation (4.2.38). Thus, in a neighborhood of the boundary, we have ∂U0ϵ ϵ ∂U ϵ sin ϕ − cos ϕ 0 + U0ϵ − U¯0ϵ = 0, (4.2.41) ∂η 1 − ϵη ∂ϕ ϵ ∂U ϵ ∂U ϵ 1 ∂U ϵ sin ϕ 1 − cos ϕ 1 + U1ϵ − U¯1ϵ = cos ϕ 0 , (4.2.42) ∂η 1 − ϵη ∂ϕ 1 − ϵη ∂θ ... ∂Ukϵ ϵ ∂U ϵ 1 ∂U ϵ sin ϕ − cos ϕ k + Ukϵ − U¯kϵ = cos ϕ k−1 . (4.2.43) ∂η 1 − ϵη ∂ϕ 1 − ϵη ∂θ where ∫ π 1 U¯kϵ (η, θ) = Ukϵ (η, θ, ϕ)dϕ. (4.2.44) 2π −π It is important to note the solution Ukϵ depends on ϵ. We refer to the cut-off function ψ and ψ0 as (4.2.23) and (4.2.24), and define the force as ϵψ(ϵη) F (ϵ; η) = − , (4.2.45) 1 − ϵη 187 Define the interior expansion as follows: ∑ ∞ ⃗ ∼ U ϵ (⃗x, w) ϵk Ukϵ (⃗x, w) ⃗ (4.2.46) k=0 where Ukϵ satisfies the same equations as Uk in (4.2.9) and (4.2.10). Here, to high- light its dependence on ϵ via the ϵ-Milne problem and boundary data, we add the superscript ϵ. The bridge between the interior solution and the boundary layer solution is the boundary condition of (4.1.1), so we first consider the boundary condition expansion U0ϵ + U0ϵ = g, (4.2.47) U1ϵ + U1ϵ = 0, (4.2.48) ... Ukϵ + Ukϵ = 0. (4.2.49) The construction of Ukϵ and Ukϵ are as follows: Step 1: Construction of U0ϵ and U0ϵ . Define the zeroth order boundary layer solution as  ( )   U0ϵ (η, θ, ϕ) = ψ0 (ϵη) f0 (η, θ, ϕ) − f0 (∞, θ) , ϵ ϵ       ϵ ϵ  sin ϕ ∂f0 + F (ϵ; η) cos ϕ ∂f0 + f ϵ − f¯ϵ = 0, 0 0 ∂η ∂ϕ (4.2.50)     f0ϵ (0, θ, ϕ) = g(θ, ϕ) for sin ϕ > 0,      lim f ϵ (η, θ, ϕ) = f0ϵ (∞, θ). η→∞ 0 In contrast to the classical Milne problem (4.2.25), the key advantage is, due to the ∂F (ϵ; η) geometry, = 0, such that (4.2.50) is invariant in θ. ∂θ 188 Then we define the zeroth order interior solution U0ϵ (⃗x) as      U0ϵ = U¯0ϵ ,  ∆x U¯0ϵ = 0 in Ω, (4.2.51)      U¯0ϵ = f0ϵ (∞, θ) on ∂Ω. ∂U0ϵ Step 2: Estimates of . ∂θ By Theorem 4.4.13, we can easily see f0ϵ is well-defined in L∞ (Ω ×S 1 ) and approaches f0ϵ (∞) exponentially fast as η → ∞. Then we can naturally derive Z = ∂θ (f0ϵ −f0ϵ (∞)) also satisfies the same type of ϵ-Milne problem   ∂Z ∂Z   sin ϕ + F (ϵ; η) cos ϕ + Z − Z¯ = 0,   ∂η ∂ϕ ∂g ∂f0ϵ (∞)  Z(0, θ, ϕ) = − for sin ϕ > 0, (4.2.52)   ∂θ ∂θ   limη→∞ Z(η, ϕ) = C. By Theorem 4.4.13, we can see Z → C exponentially fast as η → ∞. It is natural to obtain this constant C must be zero. Hence, if g ∈ C r (Γ− ), it is obvious to check f0ϵ (∞) ∈ C r (∂Ω). By the standard elliptic estimate in (4.2.51), there exists a unique solution U¯0ϵ ∈ W r,p (Ω) for arbitrary p ≥ 2 satisfying ϵ U¯ 0 W r,p (Ω) ≤ C(Ω) ∥f0ϵ (∞)∥W r−1/p,p (∂Ω) , (4.2.53) which implies ∇x U¯0ϵ ∈ W r−1,p (Ω), ∇x U¯0ϵ ∈ W r−1−1/p,p (∂Ω) and U¯0ϵ ∈ C r−1,1−2/p (Ω). Step 3: Construction of U1ϵ and U1ϵ . 189 Define the first order boundary layer solution as  ( )   U1ϵ (η, θ, ϕ) = ψ0 (ϵη) f1ϵ (η, θ, ϕ) − f1ϵ (∞, θ) ,       ϵ ϵ ∂U ϵ  sin ϕ ∂f1 + F (ϵ; η) cos ϕ ∂f1 + f ϵ − f¯ϵ = ψ(ϵη) cos ϕ 0 , ∂η ∂ϕ 1 1 1 − ϵη ∂θ (4.2.54)     f1ϵ (0, θ, ϕ) ⃗· = w ∇x U0ϵ (⃗x0 , w) ⃗ for sin ϕ > 0,      lim f ϵ (η, θ, ϕ) = f1ϵ (∞, θ). η→∞ 1 where (⃗x0 , w) ⃗ is the same point as (0, θ, ϕ). Then we define the first order interior solution U1ϵ (⃗x) as      U1ϵ = U¯1ϵ − w ⃗ · ∇x U0ϵ ,  ∆x U¯1ϵ = 0 in Ω, (4.2.55)      U¯1ϵ = f1ϵ (∞, θ) on ∂Ω. ∂U1ϵ Step 4: Estimates of . ∂θ By Theorem 4.4.13, we can easily see f1ϵ is well-defined in L∞ (Ω ×S 1 ) and approaches f1ϵ (∞) exponentially fast as η → ∞. Also, since w ⃗ · ∇x U0ϵ ∈ W r−1−1/p,p (∂Ω), ∂θ f1ϵ is well-defined and decays exponentially fast. Hence, f1ϵ (∞, θ) ∈ W r−1−1/p,p (∂Ω). By the standard elliptic estimate in (4.2.55), there exists a unique solution U¯1ϵ ∈ W r−1,p (Ω) and satisfies ϵ U¯1 r−1,p ≤ C(Ω) ∥f1ϵ (∞)∥ r−1−1/p,p W (Ω) W (∂Ω) , (4.2.56) which implies ∇x U¯1ϵ ∈ W r−2,p (Ω), ∇x U¯1ϵ ∈ W r−2−1/p,p (∂Ω) and U¯1ϵ ∈ C r−2,1−2/p (Ω). Step 5: Generalization to arbitrary k. In a similar fashion, as long as g is sufficiently smooth, above process can go on. We 190 construct the k th order boundary layer solution as  ( )   Ukϵ (η, θ, ϕ) = ψ0 (ϵη) fkϵ (η, θ, ϕ) − fkϵ (∞, θ) ,       ϵ ϵ ϵ ∂Uk−1  sin ϕ ∂fk + F (ϵ; η) cos ϕ ∂fk + f ϵ − f¯ϵ = ψ(ϵη) cos ϕ , ∂η ∂ϕ k k 1 − ϵη ∂θ (4.2.57)     fkϵ (0, θ, ϕ) ⃗· = w ∇x Uk−1 ϵ (⃗x0 , w) ⃗ for sin ϕ > 0,      lim f ϵ (η, θ, ϕ) = fkϵ (∞, θ). η→∞ k Then we define the k th order interior solution as      Ukϵ = U¯kϵ − w ⃗ · ∇x Uk−1 ϵ ,  ∆x U¯kϵ = 0 in Ω, (4.2.58)      U¯kϵ = fkϵ (∞, θ) on ∂Ω. For g ∈ C k+1 (Γ− ), the interior solution and boundary layer solution can be well- defined up to k th order, i.e. up to Ukϵ and Ukϵ . 4.3 Well-posedness of Steady Neutron Transport Equation In this section, we consider the well-posedness of the steady neutron transport equa- tion    ϵw ⃗ · ∇x u + u − u¯ = f (⃗x, w) ⃗ in Ω, (4.3.1)   u(⃗x0 , w) ⃗ for ⃗x0 ∈ ∂Ω and w ⃗ = g(⃗x0 , w) ⃗ · ⃗n < 0. 191 We define the L2 and L∞ norms in Ω × S 1 as usual: (∫ ∫ )1/2 ∥f ∥L2 (Ω×S 1 ) = |f (⃗x, w)| 2 ⃗ dwd˜ ˜ x , (4.3.2) Ω S1 ∥f ∥L∞ (Ω×S 1 ) = sup |f (⃗x, w)| ⃗ . (4.3.3) (⃗ x,w)∈Ω×S ⃗ 1 Define the L2 and L∞ norms on the boundary as follows: ( ∫∫ )1/2 ∥f ∥L2 (Γ) = |f (⃗x, w)| ⃗ |w⃗ · ⃗n| dwd˜ 2 ˜ x , (4.3.4) Γ ( ∫∫ )1/2 ∥f ∥L2 (Γ± ) = |f (⃗x, w)| ⃗ |w⃗ · ⃗n| dwd˜ 2 ˜ x , (4.3.5) Γ± ∥f ∥L∞ (Γ) = sup |f (⃗x, w)| ⃗ , (4.3.6) (⃗ x,w)∈Γ ⃗ ∥f ∥L∞ (Γ± ) = sup |f (⃗x, w)| ⃗ . (4.3.7) (⃗ x,w)∈Γ ⃗ ± 4.3.1 Preliminaries In order to show the L2 and L∞ well-posedness of the equation (4.3.1), we start with some preparations with the penalized neutron transport equation. ⃗ ∈ L∞ (Ω × S 1 ) and g(x0 , w) Lemma 4.3.1. Assume f (⃗x, w) ⃗ ∈ L∞ (Γ− ). Then for the penalized transport equation    λuλ + ϵw ⃗ · ∇x uλ + uλ = f (⃗x, w)⃗ in Ω, (4.3.8)   uλ (⃗x0 , w) ⃗ for ⃗x0 ∈ ∂Ω and w ⃗ = g(⃗x0 , w) ⃗ · ⃗n < 0. ⃗ ∈ L∞ (Ω × S 1 ) with λ > 0 as a penalty parameter, there exists a solution uλ (⃗x, w) satisfying ∥uλ ∥L∞ (Ω×S 1 ) ≤ ∥f ∥L∞ (Ω×S 1 ) + ∥g∥L∞ (Γ− ) . (4.3.9) 192 Proof. The characteristics (X(s), W (s)) of the equation (4.3.8) which goes through (⃗x, w) ⃗ is defined by      (X(0), W (0)) = (⃗x, w) ⃗    dX(s) = ϵW (s), (4.3.10)   ds     dW(s)  = 0. ds which implies    X(s) = ⃗x + (ϵw)s ⃗ (4.3.11)   W (s) = w ⃗ Hence, we can rewrite the equation (4.3.8) along the characteristics as (4.3.12) ∫ tb ⃗ = g(⃗x − ϵtb w, uλ (⃗x, w) ⃗ −(1+λ)tb + ⃗ w)e f (⃗x − ϵ(tb − s)w, ⃗ −(1+λ)(tb −s) ds, ⃗ w)e 0 where the backward exit time tb is defined as ⃗ = inf{t ≥ 0 : (⃗x − ϵtw, tb (⃗x, w) ⃗ ∈ Γ− }. ⃗ w) (4.3.13) Then we can naturally estimate −(1+λ)tb 1 − e−(1+λ)tb ∥uλ ∥L∞ (Ω×S 1 ) ≤ e ∥g∥L∞ (Γ− ) + ∥f ∥L∞ (Ω×S 1 ) (4.3.14) 1+λ ≤ ∥g∥L∞ (Γ− ) + ∥f ∥L∞ (Ω×S 1 ) . Since uλ can be explicitly traced back to the boundary data, the existence naturally follows from above estimate. ⃗ ∈ L∞ (Ω × S 1 ) and g(x0 , w) Lemma 4.3.2. Assume f (⃗x, w) ⃗ ∈ L∞ (Γ− ). Then for the 193 penalized neutron transport equation    λuλ + ϵw ⃗ · ∇x uλ + uλ − u¯λ = f (⃗x, w) ⃗ in Ω, (4.3.15)   uλ (⃗x0 , w) ⃗ for ⃗x0 ∈ ∂Ω and w ⃗ = g(⃗x0 , w) ⃗ · ⃗n < 0. ⃗ ∈ L∞ (Ω × S 1 ) satisfying with λ > 0, there exists a solution uλ (⃗x, w) ( ) 1+λ ∥uλ ∥L∞ (Ω×S 1 ) ≤ ∥f ∥L∞ (Ω×S 1 ) + ∥g∥L∞ (Γ− ) . (4.3.16) λ Proof. We define an approximating sequence {ukλ }∞ 0 k=0 , where uλ = 0 and    λuk + ϵw λ ⃗ · ∇x ukλ + ukλ − u¯k−1 λ = f (⃗x, w) ⃗ in Ω, (4.3.17)   ukλ (⃗x0 , w) ⃗ for ⃗x0 ∈ ∂Ω and w ⃗ = g(⃗x0 , w) ⃗ · ⃗n < 0. By Lemma 4.3.1, this sequence is well-defined and ukλ L∞ (Ω×S 1 ) < ∞. The characteristics and the backward exit time are defined as (4.3.10) and (4.3.13), so we rewrite equation (4.3.17) along the characteristics as ukλ (⃗x, w) ⃗ (4.3.18) ∫ tb = g(⃗x − ϵtb w, ⃗ −(1+λ)tb + ⃗ w)e (f + u¯k−1 x − ϵ(tb − s)w, λ )(⃗ ⃗ −(1+λ)(tb −s) ds. ⃗ w)e 0 We define the difference v k = ukλ − uk−1 λ for k ≥ 1. Recall (4.1.2) for v¯k , then v k satisfies ∫ tb v k+1 (⃗x, w) ⃗ = v¯k (⃗x − ϵ(tb − s)w, ⃗ −(1+λ)(tb −s) ds. ⃗ w)e (4.3.19) 0 194 Since v¯k L∞ (Ω×S 1 ) ≤ v k L∞ (Ω×S 1 ) , we can directly estimate (4.3.20) ∫ k+1 tb 1 − e−(1+λ)tb v ∞ ≤ v k L∞ (Ω×S 1 ) e−(1+λ)(tb −s) ds ≤ v k ∞ . L (Ω×S 1 ) 1+λ L (Ω×S 1 ) 0 Hence, we naturally have k+1 1 v ∞ ≤ v k ∞ . (4.3.21) L (Ω×S 1 ) 1+λ L (Ω×S 1 ) Thus, this is a contraction sequence for λ > 0. Considering v 1 = u1λ , we have ( )k−1 k 1 1 v ∞ ≤ uλ ∞ , (4.3.22) L (Ω×S 1 ) 1+λ L (Ω×S 1 ) for k ≥ 1. Therefore, ukλ converges strongly in L∞ to a limit solution uλ satisfying ∑∞ k 1+λ ∥uλ ∥L∞ (Ω×S 1 ) ≤ v ∞ 1 ≤ u1λ ∞ . (4.3.23) L (Ω×S ) λ L (Ω×S 1 ) k=1 Since u1λ can be rewritten along the characteristics as (4.3.24) ∫ tb ⃗ = g(⃗x − ϵtb w, u1λ (⃗x, w) ⃗ −(1+λ)tb + ⃗ w)e f (⃗x − ϵ(tb − s)w, ⃗ −(1+λ)(tb −s) ds, ⃗ w)e 0 based on Lemma 4.3.1, we can directly estimate 1 uλ ∞ ≤ ∥f ∥L∞ (Ω×S 1 ) + ∥g∥L∞ (Γ− ) . (4.3.25) L (Ω×S 1 ) Combining (4.3.23) and (4.3.25), we can easily deduce the lemma. 195 4.3.2 L2 Estimate It is easy to see when λ → 0, the estimate in Lemma 4.3.2 blows up. Hence, we need to show a uniform estimate of the solution to the penalized neutron transport equation (4.3.15). Lemma 4.3.3. (Green’s Identity) Assume f (⃗x, w), ⃗ ∈ L2 (Ω × S 1 ) and w ⃗ g(⃗x, w) ⃗· ∇x f, w ⃗ · ∇x g ∈ L2 (Ω × S 1 ) with f, g ∈ L2 (Γ). Then ∫∫ ( ) ∫ ⃗ · ∇x f )g + (w (w ⃗ · ∇x g)f d˜ xdw ˜ = f gdγ, (4.3.26) Ω×S 1 Γ ⃗ · ⃗n)ds on the boundary. where dγ = (w Proof. See [10, Chapter 9] and [19]. Lemma 4.3.4. The solution uλ to the equation (4.3.15) satisfies the uniform estimate ϵ ∥¯ uλ ∥L2 (Ω×S 1 ) (4.3.27) ( ) ≤ C(Ω) ∥uλ − u¯λ ∥L2 (Ω×S 1 ) + ∥f ∥L2 (Ω×S 1 ) + ϵ ∥uλ ∥L2 (Γ+ ) + ϵ ∥g∥L2 (Γ− ) , for 0 ≤ λ << 1 and 0 < ϵ << 1. Proof. Applying Lemma 4.3.3 to the solution of the equation (4.3.15). Then for any ϕ ∈ L2 (Ω × S 1 ) satisfying w ⃗ · ∇x ϕ ∈ L2 (Ω × S 1 ) and ϕ ∈ L2 (Γ), we have (4.3.28) ∫∫ ∫ ∫∫ ∫∫ λ uλ ϕ + ϵ uλ ϕdγ − ϵ ⃗ · ∇x ϕ)uλ + (w (uλ − u¯λ )ϕ ∫∫ Ω×S 1 Γ Ω×S 1 Ω×S 1 = f ϕ. Ω×S 1 196 Our goal is to choose a particular test function ϕ. We first construct an auxiliary function ζ. Since uλ ∈ L∞ (Ω × S 1 ), it naturally implies u¯λ ∈ L∞ (Ω) which further leads to u¯λ ∈ L2 (Ω). We define ζ(⃗x) on Ω satisfying    ∆ζ = u¯λ in Ω, (4.3.29)   ζ = 0 on ∂Ω. In the bounded domain Ω, based on the standard elliptic estimate, we have ∥ζ∥H 2 (Ω) ≤ C(Ω) ∥¯ uλ ∥L2 (Ω) . (4.3.30) We plug the test function ϕ = −w ⃗ · ∇x ζ (4.3.31) into the weak formulation (4.3.28) and estimate each term there. Naturally, we have ∥ϕ∥L2 (Ω) ≤ C ∥ζ∥H 1 (Ω) ≤ C(Ω) ∥¯ uλ ∥L2 (Ω) . (4.3.32) Easily we can decompose (4.3.33) ∫∫ ∫∫ ∫∫ −ϵ ⃗ · ∇x ϕ)uλ = −ϵ (w ⃗ · ∇x ϕ)¯ (w uλ − ϵ ⃗ · ∇x ϕ)(uλ − u¯λ ). (w Ω×S 1 Ω×S 1 Ω×S 1 We estimate the two term on the right-hand side separately. By (4.3.29) and (4.3.31), 197 we have ∫∫ −ϵ ⃗ · ∇x ϕ)¯ (w uλ (4.3.34) Ω×S 1 ∫∫ ( ) = ϵ u¯λ w1 (w1 ∂11 ζ + w2 ∂12 ζ) + w2 (w1 ∂12 ζ + w2 ∂22 ζ) Ω×S 1 ∫∫ ( ) 2 2 = ϵ u¯λ w1 ∂11 ζ + w2 ∂22 ζ ∫ Ω×S 1 = ϵπ u¯λ (∂11 ζ + ∂22 ζ) Ω = ϵπ ∥¯ uλ ∥2L2 (Ω) 1 = ϵ ∥¯ uλ ∥2L2 (Ω×S 1 ) . 2 In the second equality, above cross terms vanish due to the symmetry of the integral over S 1 . On the other hand, for the second term in (4.3.33), H¨older’s inequality and the elliptic estimate imply ∫∫ −ϵ ⃗ · ∇x ϕ)(uλ − u¯λ ) ≤ C(Ω)ϵ ∥uλ − u¯λ ∥L2 (Ω×S 1 ) ∥ζ∥H 2 (Ω) (w (4.3.35) Ω×S 1 ≤ C(Ω)ϵ ∥uλ − u¯λ ∥L2 (Ω×S 1 ) ∥¯ uλ ∥L2 (Ω×S 1 ) . Based on (4.3.30), (4.3.32), the boundary condition of the penalized neutron transport equation (4.3.15), the trace theorem, H¨older’s inequality and the elliptic estimate, we have ∫ ∫ ∫ ϵ uλ ϕdγ = ϵ uλ ϕdγ + ϵ uλ ϕdγ (4.3.36) Γ− Γ ( Γ+ ) ≤ C(Ω) ϵ ∥uλ ∥L2 (Γ+ ) ∥¯ uλ ∥L2 (Ω×S 1 ) + ϵ ∥g∥L2 (Γ− ) ∥¯ uλ ∥L2 (Ω×S 1 ) , 198 ∫∫ ∫∫ ∫∫ λ uλ ϕ = λ u¯λ ϕ + λ (uλ − u¯λ )ϕ (4.3.37) Ω×S 1 ∫ ∫Ω×S 1 Ω×S 1 = λ (uλ − u¯λ )ϕ Ω×S 1 ≤ C(Ω)λ ∥¯ uλ ∥L2 (Ω×S 1 ) ∥uλ − u¯λ ∥L2 (Ω×S 1 ) , ∫∫ (uλ − u¯λ )ϕ ≤ C(Ω) ∥¯ uλ ∥L2 (Ω×S 1 ) ∥uλ − u¯λ ∥L2 (Ω×S 1 ) , (4.3.38) Ω×S 1 ∫∫ f ϕ ≤ C(Ω) ∥¯ uλ ∥L2 (Ω×S 1 ) ∥f ∥L2 (Ω×S 1 ) . (4.3.39) Ω×S 1 Collecting terms in (4.3.34), (4.3.35), (4.3.36), (4.3.37), (4.3.38) and (4.3.39), we obtain ϵ ∥¯ uλ ∥L2 (Ω×S 1 ) (4.3.40) ( ) ≤ C(Ω) (1 + ϵ + λ) ∥uλ − u¯λ ∥L2 (Ω×S 1 ) + ϵ ∥uλ ∥L2 (Γ+ ) + ∥f ∥L2 (Ω×S 1 ) + ϵ ∥g∥L2 (Γ− ) , When 0 ≤ λ < 1 and 0 < ϵ < 1, we get the desired uniform estimate with respect to λ. ⃗ ∈ L∞ (Ω × S 1 ) and g(x0 , w) Theorem 4.3.5. Assume f (⃗x, w) ⃗ ∈ L∞ (Γ− ). Then for ⃗ ∈ the steady neutron transport equation (4.3.1), there exists a unique solution u(⃗x, w) L2 (Ω × S 1 ) satisfying ( ) 1 1 ∥u∥L2 (Ω×S 1 ) ≤ C(Ω) 2 ∥f ∥L2 (Ω×S 1 ) + 1/2 ∥g∥L2 (Γ− ) . (4.3.41) ϵ ϵ Proof. In the weak formulation (4.3.28), we may take the test function ϕ = uλ to get 199 the energy estimate ∫ ∫∫ 1 λ ∥uλ ∥2L2 (Ω×S 1 ) + ϵ |uλ | dγ + ∥uλ − 2 u¯λ ∥2L2 (Ω×S 1 ) = f uλ . (4.3.42) 2 Γ Ω×S 1 Hence, this naturally implies ∫∫ 1 1 ϵ ∥uλ ∥2L2 (Γ+ ) + ∥uλ − u¯λ ∥2L2 (Ω×S 1 ) ≤ f uλ + ϵ ∥g∥2L2 (Γ− ) . (4.3.43) 2 Ω×S 1 2 On the other hand, we can square on both sides of (4.3.27) to obtain ϵ2 ∥¯ uλ ∥2L2 (Ω×S 1 ) (4.3.44) ( ) ≤ C(Ω) ∥uλ − u¯λ ∥L2 (Ω×S 1 ) + ∥f ∥L2 (Ω×S 1 ) + ϵ ∥uλ ∥L2 (Γ+ ) + ϵ ∥g∥L2 (Γ− ) . 2 2 2 2 2 2 Multiplying a sufficiently small constant on both sides of (4.3.44) and adding it to (4.3.43) to absorb ∥uλ ∥2L2 (Γ+ ) and ∥uλ − u¯λ ∥2L2 (Ω×S 1 ) , we deduce ϵ ∥uλ ∥2L2 (Γ+ ) + ϵ2 ∥¯ uλ ∥2L2 (Ω×S 1 ) + ∥uλ − u¯λ ∥2L2 (Ω×S 1 ) (4.3.45) ( ∫∫ ) ≤ C(Ω) ∥f ∥2L2 (Ω×S 1 ) + f uλ + ϵ ∥g∥2L2 (Γ− ) . Ω×S 1 Hence, we have (4.3.46) ( ∫∫ ) ϵ ∥uλ ∥2L2 (Γ+ ) + ϵ2 ∥uλ ∥2L2 (Ω×S 1 ) ≤ C(Ω) ∥f ∥L2 (Ω×S 1 ) + 2 f uλ + ϵ ∥g∥L2 (Γ− ) . 2 Ω×S 1 A simple application of Cauchy’s inequality leads to ∫∫ 1 f uλ ≤ 2 ∥f ∥2L2 (Ω×S 1 ) + Cϵ2 ∥uλ ∥2L2 (Ω×S 1 ) . (4.3.47) Ω×S 1 4Cϵ 200 Taking C sufficiently small, we can divide (4.3.46) by ϵ2 to obtain ( ) 1 1 1 ∥uλ ∥L2 (Γ+ ) + ∥uλ ∥L2 (Ω×S 2 ) ≤ C(Ω) 4 ∥f ∥L2 (Ω×S 2 ) + ∥g∥L2 (Γ− ) . 2 2 2 2 (4.3.48) ϵ ϵ ϵ Since above estimate does not depend on λ, it gives a uniform estimate for the penal- ized neutron transport equation (4.3.15). Thus, we can extract a weakly convergent subsequence uλ → u as λ → 0. The weak lower semi-continuity of norms ∥·∥L2 (Ω×S 2 ) and ∥·∥L2 (Γ+ ) implies u also satisfies the estimate (4.3.48). Hence, in the weak formu- lation (4.3.28), we can take λ → 0 to deduce that u satisfies equation (4.3.1). Also uλ − u satisfies the equation    ϵw ⃗ · ∇x (uλ − u) + (uλ − u) − (¯ uλ − u¯) = −λuλ in Ω, (4.3.49)   (uλ − u)(⃗x0 , w) ⃗ = 0 for ⃗x0 ∈ ∂Ω and w ⃗ · ⃗n < 0. By a similar argument as above, we can achieve ( ) λ ∥uλ − u∥2L2 (Ω×S 2 ) ≤ C(Ω) 4 ∥uλ ∥L2 (Ω×S 2 ) . 2 (4.3.50) ϵ When λ → 0, the right-hand side approaches zero, which implies the convergence is actually in the strong sense. The uniqueness easily follows from the energy estimates. 4.3.3 L∞ Estimate ⃗ ∈ L∞ (Ω × S 1 ) and g(x0 , w) Theorem 4.3.6. Assume f (⃗x, w) ⃗ ∈ L∞ (Γ− ). Then ⃗ ∈ for the neutron transport equation (4.3.1), there exists a unique solution u(⃗x, w) 201 L∞ (Ω × S 1 ) satisfying ( ) 1 1 ∥u∥L∞ (Ω×S 1 ) ≤ C(Ω) 3 ∥f ∥L∞ (Ω×S 1 ) + 3/2 ∥g∥L∞ (Γ− ) . (4.3.51) ϵ ϵ Proof. We divide the proof into several steps to bootstrap an L2 solution to an L∞ solution: Step 1: Double Duhamel iterations. The characteristics of the equation (4.3.1) is given by (4.3.10). Hence, we can rewrite the equation (4.3.1) along the characteristics as ∫ tb −tb ⃗ = g(⃗x − ϵtb w, u(⃗x, w) ⃗ w)e ⃗ + f (⃗x − ϵ(tb − s)w, ⃗ −(tb −s) ds (4.3.52) ⃗ w)e ∫ tb ( ∫ 0 ) 1 + u(⃗x − ϵ(tb − s)w,⃗ w ˜ t e−(tb −s) ds. ⃗ t )dw 2π 0 S 1 where the backward exit time tb is defined as (4.3.13). Note we have replaced u¯ by the integral of u over the dummy velocity variable w ⃗ t . For the last term in this formulation, we apply the Duhamel’s principle again to u(⃗x − ϵ(tb − s)w, ⃗ w⃗ t ) and obtain u(⃗x, w) ⃗ (4.3.53) ∫ tb = g(⃗x − ϵtb w, ⃗ −tb + ⃗ w)e f (⃗x − ϵ(tb − s)w, ⃗ −(tb −s) ds ⃗ w)e ∫ tb ∫ 0 1 + g(⃗x − ϵ(tb − s)w ⃗ − ϵsb w ⃗ t )e−sb dw ⃗ t, w ˜ t e−(tb −s) ds 2π 0 S 1 ∫ tb ∫ ( ∫ s b ) 1 −(sb −r) + f (⃗x − ϵ(tb − s)w ⃗ − ϵ(sb − r)w ⃗ t, w ⃗ t )e dr dw ˜ t e−(tb −s) ds 2π 0 S 1 0 ( )2 ∫ tb ∫ 1 + e−(tb −s) 2π ( ∫ sb ∫ 0 S 1 ) −(sb −r) × e u(⃗x − ϵ(tb − s)w ⃗ − ϵ(sb − r)w ⃗ t, w⃗ s )dw˜ s dr dw ˜ t ds, 0 S1 202 where we introduce another dummy velocity variable w ⃗ s and sb (⃗x, w, ⃗ t ) = inf{r ≥ 0 : (⃗x − ϵ(tb − s)w ⃗ s, w ⃗ − ϵrw ⃗ t ) ∈ Γ− }. ⃗ t, w (4.3.54) Step 2: Estimates of all but the last term in (4.3.53). We can directly estimate as follows: g(⃗x − ϵtb w, ⃗ −tb ≤ ∥g∥L∞ (Γ− ) , ⃗ w)e (4.3.55) ∫ ∫ 1 tb g(⃗x − ϵ(tb − s)w ⃗ − ϵsb w ⃗ t, w ⃗ t )e −sb dw ˜ te −(tb −s) ds ≤ ∥g∥L∞ (Γ− ) , (4.3.56) 2π 0 S1 ∫ tb f (⃗x − ϵ(tb − s)w, ⃗ w)e ⃗ −(tb −s) ds ≤ ∥f ∥L∞ (Ω×S 1 ) , (4.3.57) 0 (4.3.58) ∫ tb ∫ (∫ ) 1 sb f (⃗x − ϵ(tb − s)w ⃗ − ϵ(sb − r)w ⃗ t, w ⃗ t )e−(sb −r) dr dw ˜ te −(tb −s) ds 2π 0 S1 0 ≤ ∥f ∥L∞ (Ω×S 1 ) . Step 3: Estimates of the last term in (4.3.53). Now we decompose the last term in (4.3.53) as ∫ tb ∫ ∫ sb ∫ ∫ tb ∫ ∫ ∫ ∫ tb ∫ ∫ ∫ = + = I1 + I2 , (4.3.59) 0 S1 0 S1 0 S1 sb −r≤δ S1 0 S1 sb −r≥δ S1 203 for some δ > 0. We can estimate I1 directly as ∫ tb (∫ sb ) −(tb −s) I1 ≤ e ∥u∥L∞ (Ω×S 1 ) dr ds ≤ δ ∥u∥L∞ (Ω×S 1 ) . (4.3.60) 0 max(0,sb −δ) Then we can bound I2 as ∫ tb ∫ ∫ max(0,sb −δ) I2 ≤ C (4.3.61) ∫0 S1 0 × |u(⃗x − ϵ(tb − s)w ⃗ − ϵ(sb − r)w ⃗ s )| e−(tb −s) dw ⃗ t, w ˜ s drdw ˜ t ds. S1 By the definition of tb and sb , we always have ⃗x − ϵ(tb − s)w ⃗ − ϵ(sb − r)w ⃗ t ∈ Ω. ¯ Hence, we may interchange the order of integration and apply H¨older’s inequality to obtain ∫ tb ∫ max(0,sb −δ) ∫ ∫ I2 ≤ C 1Ω (⃗x − ϵ(tb − s)w ⃗ − ϵ(sb − r)w ⃗ t) (4.3.62) 0 0 S1 S1 |u(⃗x − ϵ(tb − s)w ⃗ − ϵ(sb − r)w ⃗ s )| e−(tb −s) dw ⃗ t, w ˜ t dw ˜ s drds ∫ tb ∫ ( ∫ max(0,sb −δ) ∫ ≤ C 1Ω (⃗x − ϵ(tb − s)w ⃗ − ϵ(sb − r)w ⃗ t) 0 S1 0 S1 )1/2 |u(⃗x − ϵ(tb − s)w ⃗ − ϵ(sb − r)w ⃗ t, w ⃗ s )| dw ˜ t dr 2 e−(tb −s) dw ˜ s ds. ⃗ t ∈ S 1 , which is essentially a one-dimensional variable. Thus, we may write Note w it in a new variable ψ as w ⃗ t = (cos ψ, sin ψ). Then we define the change of variable [−π, π) × R → Ω : (ψ, r) → (y1 , y2 ) = ⃗y = ⃗x − ϵ(tb − s)w ⃗ − ϵ(sb − r)w ⃗ t , i.e.    y1 = x1 − ϵ(tb − s)w1 − ϵ(sb − r) cos ψ, (4.3.63)   y2 = x2 − ϵ(tb − s)w2 − ϵ(sb − r) sin ψ. Therefore, for sb − r ≥ δ, we can directly compute the Jacobian ∂(y1 , y2 ) −ϵ(sb − r) sin ψ ϵ cos ψ = ϵ2 (sb − r) ≥ ϵ2 δ. ∂(ψ, r) = (4.3.64) ϵ(sb − r) cos ψ ϵ sin ψ 204 Hence, we may simplify (4.3.62) as ∫ tb ∫ (∫ )1/2 C I2 ≤ √ |u(⃗y , w 2 ⃗ s )| d˜ y e−(tb −s) dw ˜ s ds. (4.3.65) ϵ δ 0 S1 Ω Then we may further utilize Cauchy’s inequality and the L2 estimate of u in Theorem 4.3.5 to obtain ∫ tb ( ∫ ∫ )1/2 C I2 ≤ √ |u(⃗y , w 2 ⃗ s )| d˜ y dw ˜s e−(tb −s) ds (4.3.66) ϵ δ 0 S 1 Ω ∫ tb C = √ e−(tb −s) ∥u∥L2 (Ω×S 1 ) ds ϵ δ 0 C ≤ √ ∥u∥L2 (Ω×S 1 ) ϵ δ ( ) C(Ω) 1 1 ≤ √ ∥f ∥L2 (Ω×S 1 ) + 3/2 ∥g∥L2 (Γ− ) δ ϵ3 ϵ ( ) C(Ω) 1 1 ≤ √ ∥f ∥L∞ (Ω×S 1 ) + 3/2 ∥g∥L∞ (Γ− ) . δ ϵ3 ϵ In summary, collecting (4.3.55), (4.3.56), (4.3.57), (4.3.58), (4.3.60) and (4.3.66), for fixed 0 < δ < 1, we have ( ) C(Ω) 1 1 ⃗ ≤ δ ∥u∥L∞ (Ω×S 1 ) + √ |u(⃗x, w)| ∥f ∥L∞ (Ω×S 1 ) + 3/2 ∥g∥L∞ (Γ− ) . (4.3.67) δ ϵ3 ϵ Then we may take 0 < δ ≤ 1/2 to obtain ( ) 1 C(Ω) 1 1 ⃗ ≤ ∥u∥L∞ (Ω×S 1 ) + √ |u(⃗x, w)| ∥f ∥L∞ (Ω×S 1 ) + 3/2 ∥g∥L∞ (Γ− ) . (4.3.68) 2 δ ϵ3 ϵ Taking supremum of u over all (⃗x, w), ⃗ we have ( ) 1 C(Ω) 1 1 ∥u∥L∞ (Ω×S 1 ) ≤ ∥u∥L∞ (Ω×S 1 ) + √ ∥f ∥L∞ (Ω×S 1 ) + 3/2 ∥g∥L∞ (Γ− ) .(4.3.69) 2 δ ϵ3 ϵ 205 Finally, absorbing ∥u∥L∞ (Ω×S 1 ) , for fixed 0 < δ ≤ 1/2, we get ( ) 1 1 ∥u∥L∞ (Ω×S 1 ) ≤ C(Ω) 3 ∥f ∥L∞ (Ω×S 1 ) + 3/2 ∥g∥L∞ (Γ− ) . (4.3.70) ϵ ϵ 4.3.4 Well-posedness of Transport Equation ⃗ ∈ L∞ (Γ− ). Then for the steady neutron transport Theorem 4.3.7. Assume g(x0 , w) ⃗ ∈ L∞ (Ω × S 1 ) satisfying equation (4.1.1), there exists a unique solution uϵ (⃗x, w) 1 ∥uϵ ∥L∞ (Ω×S 1 ) ≤ C(Ω) ∥g∥L∞ (Γ− ) . (4.3.71) ϵ3/2 Proof. We can apply Theorem 4.3.6 to the equation (4.1.1). The result naturally follows. 4.4 ϵ-Milne Problem We consider the ϵ-Milne problem for f ϵ (η, θ, ϕ) in the domain (η, θ, ϕ) ∈ [0, ∞) × [−π, π) × [−π, π)   ∂f ϵ ∂f ϵ   sin ϕ + F (ϵ; η) cos ϕ + f ϵ − f¯ϵ = S ϵ (η, θ, ϕ),   ∂η ∂ϕ f ϵ (0, θ, ϕ) = hϵ (θ, ϕ) for sin ϕ > 0, (4.4.1)      limη→∞ f ϵ (η, θ, ϕ) = f∞ ϵ (θ), 206 where ∫ π ¯ϵ 1 f (η, θ) = f ϵ (η, θ, ϕ)dϕ, (4.4.2) 2π −π F (ϵ; η) is defined as (4.2.45), |hϵ (θ, ϕ)| ≤ M, (4.4.3) and |S ϵ (η, θ, ϕ)| ≤ M e−Kη , (4.4.4) for M > 0 and K > 0 uniform in ϵ and θ. We may further define a potential function V (ϵ; η) satisfying V (ϵ; 0) = 0 and F (ϵ; η) = −∂η V (ϵ, η). The following lemma illustrates the main properties of F and V. Lemma 4.4.1. V (ϵ; η) is monotonically increasing with respect to η and satisfies 0 ≤ V (ϵ; η) ≤ ln 4. (4.4.5) Define V∞ (ϵ) = lim V (ϵ; η). (4.4.6) η→∞ Then V∞ (ϵ) = V∞ independent of ϵ. For any 0 ≤ y ≤ z ≤ ∞ ∫ z − ln 4 ≤ F (ϵ; η)dη ≤ 0. (4.4.7) y 207 For any σ > 0 and η ≥ 0, we have eV (η+σ)−V (η) ≤ 1 + 4ϵσ. (4.4.8) Moreover, we have ∫ ∞ ∫ ∞ |F (y)|2 dydη ≤ 3 − ln 4. (4.4.9) 0 η ∫ ∞ |F (ϵ; η)|2 dη ≤ 3ϵ. (4.4.10) 0 ∥F (ϵ; η)∥L∞ ≤ 4ϵ. (4.4.11) Proof. Since F (ϵ; η) ≤ 0, by definition, we know V (ϵ; η) is monotonically increasing with respect to η. Since V (ϵ; 0) = 0, V (ϵ; η) ≥ 0. Then based on (4.2.23), we can direct estimate ∫ η ∫ η ϵψ(ϵy) V (ϵ; η) = V (ϵ; 0) − F (ϵ; y)dy = dy (4.4.12) 0 0 1 − ϵy ∫ y= 3 4ϵ 3 ϵ dy = − ln(1 − ϵy) 4ϵ ≤ = ln 4. 0 1 − ϵy y=0 This verifies (4.4.5). Also, letting µ = ϵη, we have (4.4.13) ∫ ∞ ∫ ∞ ∫ ∞ ϵψ(ϵη) ψ(ϵη) ψ(µ) V∞ (ϵ) = − dη = − d(ϵη) = − dµ ≤ ln 4. 0 1 − ϵη 0 1 − ϵη 0 1−µ 208 Hence, we have V∞ (ϵ) = V∞ independent of ϵ. Since ∫ z F (ϵ; η)dη = V (ϵ; y) − V (ϵ; z), (4.4.14) y we can naturally obtain (4.4.7). Moreover, we have ∫ η+σ ∫ η+σ ϵψ(ϵy) V (η + σ) − V (η) = − F (ϵ; y)dy = dy (4.4.15) η η 1 − ϵy ∫ y=min{(η+σ), 3 } } 3 min{(η+σ), 4ϵ ϵψ(ϵy) 4ϵ ≤ dy ≤ − ln(1 − ϵy) (4.4.16) . η 1 − ϵy y=η If η + σ ≤ 3 4ϵ , we have 1 − ϵη ϵσ eV (η+σ)−V (η) ≤ =1+ ≤ 1 + 4ϵσ. (4.4.17) 1 − ϵ(η + σ) 1 − ϵ(η + σ) On the other hand, if η + σ ≥ 3 4ϵ and η ≤ 3 4ϵ , we define σ ′ = 3 4ϵ − η to obtain ′ eV (η+σ)−V (η) ≤ eV (η+σ )−V (η) ≤ 1 + 4ϵσ ′ ≤ 1 + 4ϵσ. (4.4.18) Finally, if η + σ ≥ 3 4ϵ and η ≥ 3 4ϵ , we have eV (η+σ)−V (η) = e0 = 1. (4.4.19) Therefore, (4.4.8) follows. 209 Furthermore, considering the cut-off function (4.2.23), we have ∫ ∞ ∫ ∞ |F (y)|2 dydη (4.4.20) 0 η ∫ ∫ ∫ 3 ∫ 3 ϵψ(ϵy) 2 ∞ ∞ ϵ2 4ϵ 4ϵ ≤ 1 − ϵy dydη ≤ dydη 0 η 0 η (1 − ϵy)2 ∫ 3 ( y= 3 ) ∫ 3 ( ) 4ϵ ϵ 4ϵ 4ϵ ϵ = dη = 4ϵ − dη 0 1 − ϵy y=η 0 1 − ϵη ( ) η= 4ϵ3 = 4ϵη + ln(1 − ϵη) = 3 − ln 4. η=0 Also, we can directly estimate ∫ ∞ ∫ ∞ ∫ 3 ϵ2 ψ 2 (ϵη) 4ϵ ϵ2 |F (ϵ; η)| dη =2 dη ≤ dη (4.4.21) 0 0 (1 − ϵη)2 0 (1 − ϵη)2 η= 3 ϵ 4ϵ = = 3ϵ. 1 − ϵη η=0 Finally, based on the definition of F and ψ, we obtain ϵψ(ϵη) ϵ |F (ϵ; η)| = ≤ = 4ϵ. (4.4.22) 1 − ϵη 1 − ϵ 4ϵ3 For notational simplicity, we omit ϵ and θ dependence in f ϵ in this section. The same convention also applies to F (ϵ; η), V (ϵ; η), S ϵ (η, θ, ϕ) and hϵ (θ, ϕ). However, our estimates are independent of ϵ and θ. In this section, we introduce some special notation to describe the norms in the 210 space (η, ϕ) ∈ [0, ∞) × [−π, π). Define the L2 norm as follows: (∫ π )1/2 ∥f (η)∥L2 = |f (η, ϕ)| dϕ 2 , (4.4.23) −π (∫ ∞ ∫ π )1/2 ∥f ∥L2 L2 = |f (η, ϕ)| dϕdη2 . (4.4.24) 0 −π Define the inner product in ϕ space ∫ π ⟨f, g⟩ϕ (η) = f (η, ϕ)g(η, ϕ)dϕ. (4.4.25) −π Define the L∞ norm as follows: ∥f (η)∥L∞ = sup |f (η, ϕ)| , (4.4.26) ϕ∈[−π,π) ∥f ∥L∞ L∞ = sup |f (η, ϕ)| , (4.4.27) (η,ϕ)∈[0,∞)×[−π,π) (∫ π )1/2 ∥f ∥L∞ L2 = sup |f (η, ϕ)| dϕ 2 . (4.4.28) η∈[0,∞) −π Since the boundary data h(ϕ) is only defined on sin ϕ > 0, we naturally extend above definitions on this half-domain as follows: (∫ )1/2 ∥h∥L2 = |h(ϕ)| dϕ 2 , (4.4.29) sin ϕ>0 ∥h∥L∞ = sup |h(ϕ)| . (4.4.30) sin ϕ>0 Lemma 4.4.2. We have ∥h∥L2 ≤ C ∥h∥L∞ ≤ CM (4.4.31) M ∥S∥L2 L2 ≤ C (4.4.32) K ∥S∥L∞ L2 ≤ C∥S∥L∞ L∞ ≤ CM (4.4.33) 211 Proof. They can be verified via direct computation, so we omit the proofs here. 4.4.1 L2 Estimates Finite Slab with S¯ = 0 Consider the ϵ-Milne problem for f L (η, ϕ) in a finite slab (η, ϕ) ∈ [0, L] × [−π, π)   ∂f L ∂f L   sin ϕ + F (η) cos ϕ + f L − f¯L = S(η, ϕ),   ∂η ∂ϕ f L (0, ϕ) = h(ϕ) for sin ϕ < 0, (4.4.34)      f L (L, ϕ) = f L (L, Rϕ), where Rϕ = −ϕ and S satisfies S(η) ¯ = 0 for any η. We may decompose the solution f L (η, ϕ) = qfL (η) + rfL (η, ϕ), (4.4.35) where the hydrodynamical part qfL is in the null space of the operator f L − f¯L , and the microscopic part rfL is the orthogonal complement, i.e. ∫ π 1 qfL (η) = f L (η, ϕ)dϕ rfL (η, ϕ) = f L (η, ϕ) − qfL (η). (4.4.36) 2π −π In the following, when there is no confusion, we simply write f L = q L + rL . ¯ Lemma 4.4.3. Assume S(η) = 0 for any η ∈ [0, L] with (4.4.3) and (4.4.4). Then 212 there exists a solution f (η, ϕ) to the finite slab problem (4.4.34) satisfying ∫ ( )2 L L 2 M r (η) 2 dη ≤ C M + < ∞, (4.4.37) L K 0 ( )2 ( ) L 2 L q (η) 2 ≤ C 1 + M + M 1/2 1 + η + r (η) L2 , (4.4.38) L K ⟨sin ϕ, rL ⟩ϕ (η) = 0, (4.4.39) for arbitrary η ∈ [0, L]. Proof. We divide the proof into several steps: Step 1: Assume ∥H∥L∞ L∞ < ∞ and ∥h∥L∞ < ∞, then the solution fλ (η, ϕ) to the penalized ϵ-transport equation   ∂fλ ∂fλ   λfλ + sin ϕ + F (η) cos ϕ + fλ = H(η, ϕ),   ∂η ∂ϕ fλ (0, ϕ) = h(ϕ) for sin ϕ < 0, (4.4.40)      fλ (L, ϕ) = fλ (L, Rϕ). satisfies ∥fλ ∥L∞ L∞ ≤ ∥h∥L∞ + ∥H∥L∞ L∞ . (4.4.41) The proof of (4.4.41): To construct the solution of (4.4.40), we define the energy as E(η, ϕ) = cos ϕe−V (η) . (4.4.42) This curve with constant energy is the characteristics of the equation (4.4.40). Hence, 213 on this curve the equation can be simplified as follows: dfλ λfλ + sin ϕ + fλ = H. (4.4.43) dη An implicit function η + (η, ϕ) can be determined through |E(η, ϕ)| = e−V (η ) . + (4.4.44) which means (η + , ϕ0 ) with sin ϕ0 = 0 is on the same characteristics as (η, ϕ). Define the quantities for 0 ≤ η ′ ≤ η + as follows: ′ ϕ′ (ϕ, η, η ′ ) = cos−1 (cos ϕeV (η )−V (η) ), (4.4.45) ′ Rϕ′ (ϕ, η, η ′ ) = − cos−1 (cos ϕeV (η )−V (η) ) = −ϕ′ (ϕ, η, η ′ ), (4.4.46) where the inverse trigonometric function can be defined single-valued in the domain [0, π) and the quantities are always well-defined due to the monotonicity of V . Finally we put ∫ η 1+λ Gλη,η′ (ϕ) = dξ. (4.4.47) η′ sin(ϕ′ (ϕ, η, ξ)) We can define the solution to (4.4.40) along the characteristics as follows: Case I: For sin ϕ > 0, (4.4.48) ∫ η H(η ′ , ϕ′ (ϕ, η, η ′ )) fλ (η, ϕ) = h(ϕ′ (ϕ, η, 0)) exp(−Gλη,0 ) + exp(−Gλη,η′ )dη ′ . 0 sin(ϕ′ (ϕ, η, η ′ )) Case II: 214 For sin ϕ < 0 and |E(η, ϕ)| ≤ e−V (L) , fλ (η, ϕ) = h(ϕ′ (ϕ, η, 0)) exp(−GλL,0 − GλL,η ) (4.4.49) (∫ L H(η ′ , ϕ′ (ϕ, η, η ′ )) + ′ (ϕ, η, η ′ )) exp(−GλL,η′ − GλL,η )dη ′ 0 sin(ϕ ∫ L ) H(η ′ , Rϕ′ (ϕ, η, η ′ )) λ ′ + exp(Gη,η′ )dη . η sin(ϕ′ (ϕ, η, η ′ )) Case III: For sin ϕ < 0 and |E(η, ϕ)| ≥ e−V (L) , fλ (η, ϕ) = h(ϕ′ (ϕ, η, 0)) exp(−Gλη+ ,0 − Gλη+ ,η ) (4.4.50) ( ∫ η+ H(η ′ , ϕ′ (ϕ, η, η ′ )) + ′ ′ exp(−Gλη+ ,η′ − Gλη+ ,η )dη ′ 0 sin(ϕ (ϕ, η, η )) ∫ η+ ) H(η ′ , Rϕ′ (ϕ, η, η ′ )) λ ′ + exp(Gη,η′ )dη . η sin(ϕ′ (ϕ, η, η ′ )) Note the fact d λ 1+λ ′ Gη,η′ (ϕ) = − . (4.4.51) dη sin(ϕ′ (ϕ, η, η ′ )) Hence, we can directly estimate as follows: In Case I: |fλ (η, ϕ)| (4.4.52) ∫ η 1 ≤ exp(−Gλη,0 ) ∥h∥L∞ + ∥H∥L∞ L∞ exp(−Gλη,η′ )dη ′ sin(ϕ′ (ϕ, η, η ′ )) 0 η 1 = exp(−Gλη,0 ) ∥h∥L∞ + ∥H∥L∞ L∞ exp(−Gη,η′ ) λ 1+λ ( ) 0 1 = exp(−Gλη,0 ) ∥h∥L∞ + 1 − exp(−Gλη,0 ) ∥H∥L∞ L∞ ≤ ∥h∥L∞ + ∥H∥L∞ L∞ . 1+λ 215 In Case II: |fλ (η, ϕ)| ≤ exp(−GλL,0 − GλL,η ) ∥h∥L∞ (4.4.53) { ( ) 1 + exp(−GL,η ) 1 − exp(−GL,0 ) ∥H∥L∞ L∞ λ λ 1+λ ( ) } + 1 − exp(−GL,η ) ∥H∥L∞ L∞ λ ≤ ∥h∥L∞ + ∥H∥L∞ L∞ . In Case III: |fλ (η, ϕ)| ≤ exp(−Gλη+ ,0 − Gλη+ ,η ) ∥h∥L∞ { ( ) 1 + exp(−Gη+ ,η ) 1 − exp(−Gη+ ,0 ) ∥H∥L∞ L∞ λ λ 1+λ ( ) } + 1 − exp(−Gη+ ,η ) ∥H∥L∞ L∞ λ ≤ ∥h∥L∞ + ∥H∥L∞ L∞ . This completes the proof of (4.4.41). Step 2: Assume ∥S∥L∞ L∞ < ∞ and ∥h∥L∞ < ∞, then the solution fλL (η, ϕ) to the penalized ϵ-Milne equation   ∂fλL ∂fλL   L λfλ + sin ϕ + F (η) cos ϕ + fλL − f¯λL = S(η, ϕ),   ∂η ∂ϕ  fλL (0, ϕ) = h(ϕ) for sin ϕ < 0, (4.4.54)     fλL (L, ϕ) = fλL (L, Rϕ). where f¯λL is defined as (4.4.2), satisfies ( ) L fλ ∞ ∞ ≤ 1 + λ ∥h∥L∞ + ∥S∥L∞ L∞ . (4.4.55) L L λ 216 The proof of (4.4.55): In order to construct the solution of (4.4.54), we iteratively L ∞ define the sequence {fm }m=0 as f0L = 0 and  L L  ∂fm ∂fm   λf L + sin ϕ + F (η) cos ϕ L + fm − f¯m−1 L = S(η, ϕ),   m ∂η ∂ϕ  L fm (0, ϕ) = h(ϕ) for sin ϕ < 0,(4.4.56)     L L fm (L, ϕ) = fm (L, Rϕ). Based on the analysis in Step 1 with H = S + f¯m−1 L L , we know fm is well-defined and L fm ∞ ∞ < ∞. We further define gm L L = fm − fm−1 L for m ≥ 1. Then gm L can be L L rewritten along the characteristics as follows: Case I: For sin ϕ > 0, ∫ η L g¯m (η ′ ) L gm+1 (η, ϕ) = exp(−Gη,η′ )dη ′ . (4.4.57) 0 sin(ϕ′ (ϕ, η, η ′ )) Case II: For sin ϕ < 0 and |E(η, ϕ)| ≤ e−V (L) , (∫ L g¯m L (η ′ ) L gm+1 (η, ϕ) = ′ (ϕ, η, η ′ )) exp(−GL,η′ − GL,η )dη ′ (4.4.58) 0 sin(ϕ ∫ L ) g¯mL (η ′ ) ′ + ′ ′ exp(Gη,η′ )dη . η sin(ϕ (ϕ, η, η )) Case III: For sin ϕ < 0 and |E(η, ϕ)| ≥ e−V (L) , (∫ η+ g¯mL (η ′ ) L gm+1 (η, ϕ) = exp(−Gη+ ,η′ − Gη+ ,η )dη ′ (4.4.59) 0 sin(ϕ′ (ϕ, η, η ′ )) ∫ η+ ) L g¯m (η ′ ) ′ + exp(Gη,η′ )dη . η sin(ϕ′ (ϕ, η, η ′ )) 217 In all three cases, we can always obtain L 1 gm+1 ≤ gm L . (4.4.60) L∞ L∞ 1+λ L∞ L∞ For λ > 0, this is a contraction sequence. Also, we have L g1 ∞ ∞ = f1L ∞ ∞ ≤ ∥h∥ ∞ + ∥H∥ ∞ ∞ . (4.4.61) L L L L L L L L Hence, fm converges strongly in L∞ ([0, L] × [−π, π)) to fλL which satisfies (4.4.54). Also, fλL satisfies ( ) L fλ ∞ ∞ ≤ 1 + λ ∥h∥L∞ + ∥S∥L∞ L∞ . (4.4.62) L L λ Naturally, we obtain fλL ∈ L2 ([0, L] × [−π, π)). However, we can see the estimate blows up when λ → 0. Therefore, we need to show the uniform estimate of fλL with respect to λ. Similarly, we can define rλL and qλL for fλL as in (4.4.36). Step 3: rλL satisfies ∫ ∫ L L 2 L rλ (η) 2 dη ≤ 4 ∥h∥2 2 + 8 ∥S(η)∥2L2 dη. (4.4.63) L L 0 0 The proof of (4.4.63): Multiplying fλL on both sides of (4.4.54) and integrating over ϕ ∈ [−π, π), we get the energy estimate 1 d L L ⟨f , f sin ϕ⟩ϕ (η) (4.4.64) 2 dη λ λ 2 2 ∂f L = −λ fλL (η) L2 − rλL (η) L2 − F (η)⟨ λ , fλL cos ϕ⟩ϕ (η) + ⟨S, fλL ⟩ϕ (η). ∂ϕ 218 A further integration by parts reveals ∂fλL L 1 −F (η)⟨ , fλ cos ϕ⟩ϕ (η) = − F (η)⟨fλL , fλL sin ϕ⟩ϕ (η). (4.4.65) ∂ϕ 2 ¯ Also, the assumption S(η) = 0 leads to ⟨S, fλL ⟩ϕ (η) = ⟨S, qλL ⟩ϕ (η) + ⟨S, rλL ⟩ϕ (η) = ⟨S, rλL ⟩ϕ (η). (4.4.66) Hence, we have the simplified form of (4.4.64) as follows: 1 d L L ⟨f , f sin ϕ⟩ϕ (η) (4.4.67) 2 dη λ λ 2 2 1 = −λ fλL (η) L2 − rλL (η) L2 − F (η)⟨fλL , fλL sin ϕ⟩ϕ (η) + ⟨S, rλL ⟩ϕ (η). 2 Define 1 α(η) = ⟨fλL , fλL sin ϕ⟩ϕ (η). (4.4.68) 2 Then (4.4.67) can be rewritten as follows: dα 2 2 = −λ fλL (η) L2 − rλL (η) L2 − F (η)α(η) + ⟨S, rλL ⟩ϕ (η). (4.4.69) dη We can integrate above on [η, L] and [0, η] respectively to obtain (∫ L ) α(η) = α(L) exp F (y)d(y) (4.4.70) η ∫ (∫ )( ) L y L 2 L 2 + exp F (z)dz λ fλ (y) L2 + rλ (y) L2 − ⟨S, rλ ⟩ϕ (y) dy, L η η ( ∫ η ) α(η) = α(0) exp − F (y)d(y) (4.4.71) ∫ η ( ∫ η 0 )( ) L 2 L 2 + exp − F (z)dz − λ fλ (y) L2 − rλ (y) L2 + ⟨S, rλ ⟩ϕ (y) dy. L 0 y 219 The specular reflexive boundary fλL (L, ϕ) = fλL (L, Rϕ) ensures α(L) = 0. Hence, based on (4.4.70), we have ∫ L (∫ y )( ) α(η) ≥ exp F (z)dz − ⟨S, rλL ⟩ϕ (y) dy. (4.4.72) η η Also, (4.4.71) implies )( ∫ η ) ( ∫ η α(η) ≤ α(0) exp[V (η)] + exp − F (z)dz ⟨S, rλ ⟩ϕ (y) dy (4.4.73) L 0 y ∫ η ( ∫ η )( ) ≤ 2 ∥h∥L2 + 2 exp − F (z)dz ⟨S, rλ ⟩ϕ (y) dy, L 0 y due to the fact (∫ ) 1 1 1 α(0) = ⟨sin ϕfλL , fλL ⟩ϕ (0) ≤ h(ϕ) sin ϕdϕ ≤ ∥h∥2L2 . 2 (4.4.74) 2 2 sin ϕ>0 2 Then in (4.4.71) taking η = L, from α(L) = 0, we have ∫ (∫ ) L y 2 exp F (z)dz rλL (y) L2 dy (4.4.75) 0 0 ∫ L (∫ y ) ≤ α(0) + exp F (z)dz ⟨S, rλL ⟩ϕ (y)dy 0 0 ∫ L (∫ y ) 1 ≤ ∥h∥L2 + 2 exp F (z)dz ⟨S, rλL ⟩ϕ (y)dy. 2 0 0 On the other hand, by (4.4.7), we can directly estimate as follows: ∫ (∫ ) ∫ L y 2 1 L L 2 exp F (z)dz rλL (y) L2 dy ≥ rλ (y) 2 dy. L (4.4.76) 0 0 4 0 Combining (4.4.75) and (4.4.76) yields ∫ ∫ (∫ ) L L 2 L y rλ (η) 2 dη ≤ 2 ∥h∥2 2 + 4 exp F (z)dz ⟨S, rλL ⟩ϕ (y)dy.(4.4.77) L L 0 0 0 220 By Cauchy’s inequality and (4.4.7), we have ∫ (∫ ) ∫ L L y exp F (z)dz ⟨S, rλ ⟩ϕ (y)dy ≤ L ⟨S, rλ ⟩ϕ (y)dy L (4.4.78) 0 0 0 ∫ ∫ L 1 L 2 ≤ L rλ (η) L2 dη + 2 ∥S(η)∥2L2 dη. 8 0 0 Therefore, summarizing (4.4.77) and (4.4.78), we deduce (4.4.63). Step 4: qλL satisfies ( ) L 2 L qλ (η) 2 ≤ 256π (1 + λ) 1 + η + rλ (η) 2 2 1/2 (4.4.79) L L ( ∫ L ∫ η ) × 1 + ∥h∥L2 + 2 ∥S(η)∥L2 dη + 2 ∥S(y)∥L∞ dy . 0 0 The proof of (4.4.79): Multiplying sin ϕ on both sides of (4.4.54) and integrating over ϕ ∈ [−π, π) lead to d ⟨sin2 ϕ, fλL ⟩ϕ (η) = −λ⟨sin ϕ, fλL ⟩ϕ (η) − ⟨sin ϕ, rλL ⟩ϕ (η) (4.4.80) dη ∂f L −F (η)⟨sin ϕ cos ϕ, λ ⟩ϕ (η) + ⟨sin ϕ, S⟩ϕ (η). ∂ϕ We can further integrate by parts as follows: (4.4.81) ∂fλL −F (η)⟨sin ϕ cos ϕ, ⟩ϕ (η) = F (η)⟨cos(2ϕ), fλL ⟩ϕ (η) = F (η)⟨cos(2ϕ), rλL ⟩ϕ (η), ∂ϕ to obtain d ⟨sin2 ϕ, fλL ⟩ϕ (η) = −λ⟨sin ϕ, fλL ⟩ϕ (η) − ⟨sin ϕ, rλL ⟩ϕ (η) (4.4.82) dη +F (η)⟨cos(2ϕ), rλL ⟩ϕ (η) + ⟨sin ϕ, S⟩ϕ (η). 221 Define βλL (η) = ⟨sin2 ϕ, fλL ⟩ϕ (η). (4.4.83) Then we can simplify (4.4.80) as follows: dβλL = DλL (η, ϕ), (4.4.84) dη where DλL (η, ϕ) (4.4.85) = −λ⟨sin ϕ, fλL ⟩ϕ (η) − ⟨sin ϕ, rλL ⟩ϕ (η) + F (η)⟨cos(2ϕ), rλL ⟩ϕ (η) + ⟨sin ϕ, S⟩ϕ (η). Since (4.4.86) −λ⟨sin ϕ, fλL ⟩ϕ (η) = −λ⟨sin ϕ, rλL ⟩ϕ (η) − λ⟨sin ϕ, qλL ⟩ϕ (η) = −λ⟨sin ϕ, rλL ⟩ϕ (η). we can further get DλL (η, ϕ) (4.4.87) = −λ⟨sin ϕ, rλL ⟩ϕ (η) − ⟨sin ϕ, rλL ⟩ϕ (η) + F (η)⟨cos(2ϕ), rλL ⟩ϕ (η) + ⟨sin ϕ, S⟩ϕ (η). We can integrate over [0, η] in (4.4.84) to obtain ∫ η βλL (η) = βλL (0) + DλL (y)dy. (4.4.88) 0 It is important to note that DλL only depends on rλL and is independent of qλL . Then 222 we can directly estimate L Dλ (η) ∞ (4.4.89) ( L ) ≤ 2π 1 + λ + |F (η)| rλL (η) L2 + ∥S(η)∥L∞ ≤ 4π(1 + λ) rλL (η) L2 + ∥S(η)∥L∞ . Also, for the initial data βλL (0) = ⟨sin2 ϕ, fλL ⟩ϕ (0) (4.4.90) ( )1/2 ( )1/2 3/2 ≤ ⟨fλL , fλL |sin ϕ|⟩ϕ (0) ∥sin ϕ∥L2 ≤ 8 ⟨fλL , fλL |sin ϕ|⟩ϕ (0) . Obviously, we have ∫ ∫ ( )2 ⟨fλL , fλL |sin ϕ|⟩ϕ (0) = h (ϕ) sin ϕdϕ − 2 L fλ (0, ϕ) sin ϕdϕ. (4.4.91) sin ϕ>0 sin ϕ<0 However, based on the definition of α(η) and (4.4.72), we can obtain ∫ ∫ ( )2 2 L h (ϕ) sin ϕdϕ + fλ (0, ϕ) sin ϕdϕ (4.4.92) sin ϕ>0 sin ϕ<0 = 2α(0) ∫ L (∫ y )( ) ≥ 2 exp F (z)dz − ⟨S, rλ ⟩ϕ (y) dy L 0 0 ∫ 1 L ≥ − ⟨S, rλL ⟩ϕ (y)dy. 2 0 Hence, we can deduce ∫ ( )2 − L fλ (0, ϕ) sin ϕdϕ (4.4.93) sin ϕ<0 ∫ ∫ 1 L ≤ 2 h (ϕ) sin ϕdϕ + ⟨S, rλL ⟩ϕ (y)dy sin ϕ>0 2 0 ∫ L ∫ 1 1 L ≤ ∥h∥L2 + 2 L 2 rλ (η) L2 dη + ∥S(η)∥2L2 dη. 4 0 4 0 223 From (4.4.63), we can deduce βλL (0)2 (4.4.94) ∫ ∫ L L 2 L ≤ 64⟨fλL , fλL |sin ϕ|⟩ϕ (0) ≤ 128 ∥h∥2L2 + 16 rλ (η) 2 dη + 16 ∥S(η)∥2L2 dη L 0 0 ∫ L ≤ 192 ∥h∥L2 + 192 2 ∥S(η)∥2L2 dη. 0 From (4.4.63), (4.4.88), (4.4.89) and (4.4.94), we have L βλ (η) ∞ (4.4.95) L ∫ L ≤ 8 + 192 ∥h∥2L2 + 192 ∥S(η)∥2L2 dη ∫ η 0 ∫ η L +2π(1 + λ) rλ (y) L2 dy + ∥S(y)∥L∞ dy 0 0 ∫ L ∫ η ≤ 8 + 192 ∥h∥L2 + 192 2 ∥S(η)∥L2 dη + 2 ∥S(y)∥L∞ dy 0 0 (∫ )1/2 L 2 η +2π(1 + λ)η 1/2 rλ (y) L2 dy 0 ( ∫ L ∫ η ) ≤ 64π(1 + λ)(1 + η ) ∥h∥L2 + 1/2 2 ∥S(η)∥L2 dη + 2 ∥S(y)∥L∞ dy . 0 0 By (4.4.83) this implies (4.4.96) L 2 qλ (η) 2 ≤ 2 βλL (η) 2 + 2 rλL (η) 2 ≤ 4π βλL (η) ∞ + 2 rλL (η) 2 , L L L L L which completes the proof of (4.4.79). Step 5: Passing to the limit. Since estimates (4.4.63) and (4.4.79) are uniform in λ, we can take weakly convergent subsequence fλL → f L ∈ L2 ([0, L] × [−π, π)) as λ → 0. Hence, f L is the solution of (4.4.34) and satisfies the estimates (4.4.37) and (4.4.38). 224 Step 6: Orthogonality relation (4.4.39). A direct integration over ϕ ∈ [−π, π) in (4.4.34) implies d df L ⟨sin ϕ, f L ⟩ϕ (η) = −F ⟨cos ϕ, ⟩ϕ (η) + S(η) ¯ = −F ⟨sin ϕ, f L ⟩ϕ (η). (4.4.97) dη dϕ thanks to S¯ = 0. The specular reflexive boundary f L (L, ϕ) = f L (L, Rϕ) implies ⟨sin ϕ, f L ⟩ϕ (L) = 0. Then we have ⟨sin ϕ, f L ⟩ϕ (η) = 0. (4.4.98) It is easy to see ⟨sin ϕ, q L ⟩ϕ (η) = 0. (4.4.99) Hence, we may derive ⟨sin ϕ, rL ⟩ϕ (η) = 0. (4.4.100) This leads (4.4.39) and completes the proof of (4.4.3). Infinite Slab with S¯ = 0 We turn to the ϵ-Milne problem for f (η, ϕ) in the infinite slab (η, ϕ) ∈ [0, ∞)×[−π, π)   ∂f ∂f   sin ϕ + F (η) cos ϕ + f − f¯ = S(η, ϕ),   ∂η ∂ϕ f (0, ϕ) = h(ϕ) for sin ϕ > 0, (4.4.101)      limη→∞ f (η, ϕ) = f∞ . 225 Define r and q for f as rL and q L for f L . ¯ Lemma 4.4.4. Assume S(η) = 0 for any η ∈ [0, ∞) with (4.4.3) and (4.4.4). Then there exists a solution f (η, ϕ) of the infinite slab problem (4.4.101), satisfying ( )2 M ∥r∥L2 L2 ≤ C M+ < ∞, (4.4.102) K ⟨sin ϕ, r⟩ϕ (η) = 0, (4.4.103) for any η ∈ [0, ∞). Also there exists a constant q∞ = f∞ ∈ R such that the following estimates hold, ( )2 M |q∞ | ≤ C 1 + M + < ∞, (4.4.104) K (4.4.105) ( ∫ ∞ ∫ ∞ ) ∥q(η) − q∞ ∥L2 ≤ C ∥r(η)∥L2 + |F (y)| ∥r(y)∥L2 dy + ∥S(y)∥L∞ dy , η η ( )2 M ∥q − q∞ ∥L2 L2 ≤ C M+ < ∞. (4.4.106) K The solution is unique among functions such that (4.4.102), (4.4.104)and (4.4.106) hold. Proof. Step 1: Existence and estimates (4.4.102) and (4.4.103). By the uniform estimates from Lemma 4.4.4, the solution f L of the finite problem (4.4.34) in the slab [0, L] is uniformly bounded in L2loc ([0, ∞); L2 [−π, π)). Then there exists a subsequence such that q L ⇀ q, (4.4.107) rL ⇀ r, (4.4.108) 226 weakly in L2loc ([0, ∞); L2 [−π, π)). Also, f = q + r satisfies the boundary condition at η = 0. This shows the existence of the solution. Then property (4.4.102) naturally holds due to the weak lower semi-continuity of norm ∥·∥L2 L2 . Also, the orthogonal relation (4.4.103) is preserved. Step 2: Estimates (4.4.104), (4.4.105) and (4.4.106). We continue using the notation in Step 5 of the proof of Lemma 4.4.3. Recall (4.4.83) to (4.4.85) with λ = 0 and L = ∞. We have β(η) = ⟨sin2 ϕ, f ⟩ϕ (η) (4.4.109) and dβ = D(η, ϕ), (4.4.110) dη where D(η, ϕ) = −⟨sin ϕ, r⟩ϕ + F (η)⟨cos(2ϕ), r⟩ϕ + ⟨sin ϕ, S⟩ϕ (η). (4.4.111) The orthogonal relation (4.4.103) implies D(η, ϕ) = F (η)⟨cos(2ϕ), r⟩ϕ + ⟨sin ϕ, S⟩ϕ (η). (4.4.112) Hence, we can integrate (4.4.110) over [0, η] to show ∫ η ∫ η β(η) − β(0) = F (y)⟨cos(2ϕ), r⟩ϕ (y)dy + ⟨sin ϕ, S⟩ϕ (y)dy. (4.4.113) 0 0 Based on Lemma 4.4.1, since F ∈ L1 [0, ∞) ∩ L2 [0, ∞), r ∈ L2 ([0, ∞) × [−π, π)), and S exponentially decays, by (4.4.94) and (4.4.102), there exists some constant β∞ such 227 that β∞ = limη→∞ β(η) satisfying (4.4.114) ∫ ∫ ∞ ∞ |β∞ | ≤ |β(0)| + F (y)⟨cos(2ϕ), r⟩ϕ (y)dy + ⟨sin ϕ, S⟩ϕ (y)dy 0 ∫ ∞ 0 ≤ 8 + 192 ∥h∥2L2 + 200 ∥S(η)∥2L2 dη + 2π∥F ∥L2 L2 ∥r∥L2 L2 0 ( )2 M ≤ C 1+M + . K We define q∞ = β∞ / ∥sin ϕ∥2L2 . Hence, the estimate of |q∞ | in (4.4.104) is valid. Moreover, (4.4.115) ∫ ∞ ∫ ∞ ∫ ∞ β∞ − β(η) = D(y)dy = F (y)⟨cos(2ϕ), r⟩ϕ (y)dy + ⟨sin ϕ, S⟩ϕ (y)dy. η η η Note β(η) = ⟨sin2 ϕ, f ⟩ϕ (η) = ⟨sin2 ϕ, q⟩ϕ (η) + ⟨sin2 ϕ, r⟩ϕ (η) (4.4.116) = q(η) ∥sin ϕ∥2L2 + ⟨sin2 ϕ, r⟩ϕ (η). Thus we can estimate ∥sin ϕ∥2L2 ∥q(η) − q∞ ∥L2 (4.4.117) ( ) √ √ = 2π ∥sin ϕ∥L2 ∥q(η) − q∞ ∥L∞ = 2π β(η) − ⟨sin ϕ, r⟩ϕ (η) − β∞ 2 2 ( ∫ ∞ √ ≤ 2π ⟨sin ϕ, r⟩ϕ (η) + 2 |F (y)⟨cos(2ϕ), r(y)⟩ϕ dy| dη η ∫ ∞ ) + |⟨sin ϕ, S⟩ϕ (y)| dy η ( ∫ ∞ ∫ ∞ ) ≤ 2π ∥r(η)∥L2 + 2 |F (y)| ∥r(y)∥L2 dy + ∥S(y)∥L∞ dy . η η 228 This implies (4.4.105). Furthermore, we integrate (4.4.117) over η ∈ [0, ∞). Cauchy’s inequality and (4.4.9) imply (4.4.118) ∫ ∞(∫ ∞ )2 ∫ ∞∫ ∞ |F (y)| ∥r(y)∥L2 dy dη ≤ ∥r∥2L2 L2 |F (y)|2 dydη 0 η 0 η ≤ C. The exponential decays shows ∫ ∞ (∫ ∞ )2 ∥S(y)∥L∞ dy dη ≤ C. (4.4.119) 0 η Hence, the estimate of ∥q − q∞ ∥L2 L2 in (4.4.106) naturally follows. Step 3: Uniqueness In order to show the uniqueness of the solution, we assume there are two solutions f1 and f2 to the equation (4.4.101) satisfying (4.4.102) and (4.4.103). Then f ′ = f1 − f2 satisfies the equation   ∂f ′ ∂f ′   sin ϕ + F (η) cos ϕ + f ′ − f¯′ = 0,   ∂η ∂ϕ  f ′ (0, ϕ) = 0 for sin ϕ > 0, (4.4.120)     limη→∞ f ′ (η, ϕ) = f∞ ′ . Similarly, we can define r′ and q ′ . Multiplying e−V (η) f ′ on both sides of (4.4.120) and integrating over ϕ ∈ [−π, π) yields ( ) ( ) 1 d ′ ′ −V (η) ′ 2 −V (η) ⟨f , f sin ϕ⟩ϕ (η)e = − ∥r (η)∥L2 e ≤ 0. (4.4.121) 2 dη 229 This is due to the fact ( ) 1 d ′ ′ −V (η) ⟨f , f sin ϕ⟩ϕ (η)e (4.4.122) 2 dη ( ′ ) ( ) ′ df −V (η) 1 ′ ′ −V (η) = ⟨f , sin ϕ⟩ϕ (η)e + F (η)⟨f , f sin ϕ⟩ϕ (η)e dη 2 ( ′ ) ( ′ ) ′ df −V (η) ′ df −V (η) = ⟨f , sin ϕ⟩ϕ (η)e + F (η)⟨f , cos ϕ⟩ϕ (η)e . dη dϕ Thus, we have 1 γ(η) = ⟨f ′ , f ′ sin ϕ⟩ϕ (η)e−V (η) . (4.4.123) 2 is decreasing. Since r′ ∈ L2 ([0, ∞)×[−π, π)) and q ′ −q∞ ′ ∈ L2 ([0, ∞)×[−π, π)), there exists a convergent subsequence ηk → ∞ satisfying ∥r′ (ηk )∥L2 → 0 and q ′ (ηk ) − q∞ ′ → 0. Hence, this implies 1 ′ ′ ⟨r , r sin ϕ⟩ϕ (ηk )e−V (ηk ) → 0. (4.4.124) 2 Also, due to the fact that q ′ (ηk ) is independent of ϕ and it is finite, we have γ(ηk ) → 0. (4.4.125) By the monotonicity, γ(η) decreases to zero and γ(η) ≥ 0. Then we can integrate (4.4.121) over η ∈ [0, ∞) to obtain ∫ ∞ ∥r′ (y)∥L2 e−V (y) dy, 2 γ(∞) − γ(0) = −2 (4.4.126) 0 which implies ∫ ∞ ′ ′ −V (0) ∥r′ (y)∥L2 e−V (y) dy. 2 γ(0) = ⟨f , f sin ϕ⟩ϕ (0)e =2 (4.4.127) 0 230 Also, we know (4.4.128) ∫ 1 1 ′ ′ 0 ≤ ⟨f ′ , f ′ sin ϕ⟩ϕ (0)e−V (0) = ⟨f , f sin ϕ⟩ϕ (0) ≤ (f ′ )2 (ϕ) sin ϕdϕ = 0. 2 2 sin ϕ>0 Naturally, we have ∫ ∞ ′ ′ −V (0) ∥r′ (y)∥L2 e−V (y) dy = 0. 2 ⟨f , f sin ϕ⟩ϕ (0)e =2 (4.4.129) 0 Hence, we have r′ = 0 and f ′ (η, ϕ) = q ′ (η). Plugging this into the equation (4.4.120) reveals ∂η q ′ = 0. Therefore, f ′ (η, ϕ) = C for some constant C. Naturally the bound- ary data leads to C = 0. In conclusion, we must have f ′ = 0, which means f1 = f2 , and the uniqueness follows. S¯ ̸= 0 Case Consider the ϵ-Milne problem for f (η, ϕ) in (η, ϕ) ∈ [0, ∞) × [−π, π) with a general source term   ∂f ∂f   sin ϕ + F (η) cos ϕ + f − f¯ = S(η, ϕ),   ∂η ∂ϕ f (0, ϕ) = h(ϕ) for sin ϕ > 0, (4.4.130)      limη→∞ f (η, ϕ) = f∞ . Lemma 4.4.5. Assume (4.4.3) and (4.4.4) hold. Then there exists a solution f (η, ϕ) of the problem (4.4.130), satisfying ( )2 M ∥r∥L2 L2 < C 1 + M + ≤ ∞, (4.4.131) K ∫ ∞ ⟨sin ϕ, r⟩ϕ (η) = − ¯ eV (η)−V (y) S(y)dy. (4.4.132) η 231 Also there exists a constant q∞ = f∞ ∈ R such that the following estimates hold, ( )2 M |q∞ | ≤ C 1 + M + < ∞, (4.4.133) K (4.4.134) ( ∫ ∞ ∫ ∞ ) ∥q(η) − q∞ ∥L2 ≤ C ∥r(η)∥L2 + |F (y)| ∥r(y)∥L2 dy + ∥S(y)∥L∞ dy , η η ( )2 M ∥q − q∞ ∥L2 L2 ≤ C 1+M + < ∞. (4.4.135) K The solution is unique among functions satisfying ∥f − f∞ ∥L2 L2 < ∞. Proof. We can apply superposition property for this linear problem, i.e. write S = S¯ + (S − S) ¯ = SQ + SR . Then we solve the problem by the following steps. For simplicity, we just call the estimates (4.4.131), (4.4.133), (4.4.134) and (4.4.135) as the L2 estimates. Step 1: Construction of auxiliary function f 1 . We first solve f 1 as the solution to   ∂f 1 ∂f 1   sin ϕ + F (η) cos ϕ + f 1 − f¯1 = SR (η, ϕ),   ∂η ∂ϕ f 1 (0, ϕ) = h(ϕ) for sin ϕ > 0, (4.4.136)      limη→∞ f 1 (η, ϕ) = f∞ 1 . Since S¯R = 0, by Lemma 4.4.4, we know there exists a unique solution f 1 satisfying the L2 estimate. Step 2: Construction of auxiliary function f 2 . 232 We seek a function f 2 satisfying ∫ π ( ) 1 ∂f 2 ∂f 2 − sin ϕ + F (η) cos ϕ dϕ + SQ = 0. (4.4.137) 2π −π ∂η ∂ϕ The following analysis shows this type of function can always be found. An integration by parts transforms the equation (4.4.137) into ∫ π ∫ π ∂f 2 − sin ϕ dϕ − F (η) sin ϕf 2 dϕ + 2πSQ = 0. (4.4.138) −π ∂η −π Setting f 2 (ϕ, η) = a(η) sin ϕ. (4.4.139) and plugging this ansatz into (4.4.138), we have ∫ π ∫ π da − sin ϕdϕ − F (η)a(η) 2 sin2 ϕdϕ + 2πSQ = 0. (4.4.140) dη −π −π Hence, we have da − − F (η)a(η) + 2SQ = 0. (4.4.141) dη This is a first order linear ordinary differential equation, which possesses infinitely many solutions. We can directly solve it to obtain ∫η ( ∫ η ∫y ) − F (y)dy F (z)dz a(η) = e 0 a(0) + e 0 2SQ (y)dy . (4.4.142) 0 We may take ∫ ∞ ∫ y a(0) = − e 0 F (z)dz 2SQ (y)dy. (4.4.143) 0 233 Based on the exponential decay of SQ , we can directly verify a(η) decays exponen- tially to zero as η → ∞ and f 2 satisfies the L2 estimate. Step 3: Construction of auxiliary function f 3 . Based on above construction, we can directly verify ∫ π ( ) ∂f 2 ∂f 2 − sin ϕ − F (η) cos ϕ − f + f¯ + SQ dϕ = 0. 2 2 (4.4.144) −π ∂η ∂ϕ Then we can solve f 3 as the solution to (4.4.145)    ∂f 3 ∂f 3 2 ¯3 = − sin ϕ ∂f − F (η) cos ϕ ∂f 2   sin ϕ + F (η) cos ϕ + f 3 − f   ∂η ∂ϕ ∂η ∂ϕ    −f 2 + f¯2 + SQ ,     f 3 (0, ϕ) = −a(0) sin ϕ for sin ϕ > 0,      limη→∞ f 3 (η, ϕ) = f∞ 3 . By (4.4.144), we can apply Lemma 4.4.4 to obtain a unique solution f 3 satisfying the L2 estimate. Step 4: Construction of auxiliary function f 4 . We now define f 4 = f 2 + f 3 and an explicit verification shows   ∂f 4 ∂f 4   sin ϕ + F (η) cos ϕ + f 4 − f¯4 = SQ (η, ϕ),   ∂η ∂ϕ f 4 (0, ϕ) = 0 for sin ϕ > 0, (4.4.146)      limη→∞ f 4 (η, ϕ) = f∞ 4 , and f 4 satisfies the L2 estimate. 234 In summary, we deduce that f 1 + f 4 is the solution of (4.4.130) and satisfies the L2 estimate. A direct computation of ⟨sin ϕ, f i ⟩ϕ (η) for i = 1, 2, 3, 4 leads to (4.4.132). From ∥f − f∞ ∥L2 L2 < ∞ , we deduce f¯ − f∞ L2 L2 < ∞, a similar argument as in Lemma 4.4.4 shows the uniqueness of solution. Combining all above, we have the following theorem. Theorem 4.4.6. For the ϵ-Milne problem (4.4.1), there exists a unique solution f (η, ϕ) satisfying the estimates ( ) M ∥f − f∞ ∥L2 L2 ≤C 1+M + < ∞, (4.4.147) K for some real number f∞ satisfying ( )2 M |f∞ | ≤ C 1 + M + < ∞. (4.4.148) K 4.4.2 L∞ Estimates Finite Slab Consider the ϵ-transport problem for f (η, ϕ) in a finite slab (η, ϕ) ∈ [0, L] × [−π, π)   ∂f ∂f   sin ϕ + F (η) cos ϕ + f = H(η, ϕ),   ∂η ∂ϕ f (0, ϕ) = h(ϕ) for sin ϕ > 0, (4.4.149)      f (L, ϕ) = f (L, Rϕ). 235 Define the energy as follows: E(η, ϕ) = cos ϕe−V (η) . (4.4.150) In the plane (η, ϕ) ∈ [0, ∞) × [−π, π), on the curve ϕ = ϕ(η) with constant energy, we can see dE ∂E ∂E ∂ϕ ∂ϕ = + = cos ϕF (η)e−V (η) − sin ϕe−V (η) = 0, (4.4.151) dη ∂η ∂ϕ ∂η ∂η which further implies ∂ϕ cos ϕF (η) = . (4.4.152) ∂η sin ϕ Plugging this into the equation (4.4.149), on this curve, we deduce ( ) df ∂f ∂f ∂ϕ 1 ∂f ∂f = + = sin ϕ + cos ϕF (η) . (4.4.153) dη ∂η ∂ϕ ∂η sin ϕ ∂η ∂ϕ Hence, this curve with constant energy is exactly the characteristics of the equation (4.4.149). Also, on this curve the equation can be simplified as follows: df sin ϕ + f = H. (4.4.154) dη An implicit function η + (η, ϕ) can be determined through |E(η, ϕ)| = e−V (η ) . + (4.4.155) 236 which means (η + , ϕ0 ) with sin ϕ0 = 0 is on the same characteristics as (η, ϕ). Define the quantities for 0 ≤ η ′ ≤ η + as follows: ′ ϕ′ (ϕ, η, η ′ ) = cos−1 (cos ϕeV (η )−V (η) ), (4.4.156) ′ Rϕ′ (ϕ, η, η ′ ) = − cos−1 (cos ϕeV (η )−V (η) ) = −ϕ′ (ϕ, η, η ′ ), (4.4.157) where the inverse trigonometric function can be defined single-valued in the domain [0, π) and the quantities are always well-defined due to the monotonicity of V . We can see (η ′ , ϕ′ (ϕ, η, η ′ )) and (η, ϕ) are on the same characteristics. Finally we put ∫ η 1 Gη,η′ (ϕ) = dξ. (4.4.158) η′ sin(ϕ′ (ϕ, η, ξ)) We can rewrite the solution to the equation (4.4.149) along the characteristics as follows: Case I: For sin ϕ > 0, (4.4.159) ∫ η ′ ′ ′ H(η , ϕ (ϕ, η, η )) f (η, ϕ) = h(ϕ′ (ϕ, η, 0)) exp(−Gη,0 ) + exp(−Gη,η′ )dη ′ . 0 sin(ϕ′ (ϕ, η, η ′ )) Case II: 237 For sin ϕ < 0 and |E(η, ϕ)| ≤ e−V (L) , f (η, ϕ) = h(ϕ′ (ϕ, η, 0)) exp(−GL,0 − GL,η ) (4.4.160) (∫ L H(η ′ , ϕ′ (ϕ, η, η ′ )) + ′ (ϕ, η, η ′ )) exp(−GL,η′ − GL,η )dη ′ 0 sin(ϕ ∫ L ) H(η ′ , Rϕ′ (ϕ, η, η ′ )) ′ + exp(Gη,η′ )dη . η sin(ϕ′ (ϕ, η, η ′ )) Case III: For sin ϕ < 0 and |E(η, ϕ)| ≥ e−V (L) , f (η, ϕ) = h(ϕ′ (ϕ, η, 0)) exp(−Gη+ ,0 − Gη+ ,η ) (4.4.161) ( ∫ η+ H(η ′ , ϕ′ (ϕ, η, η ′ )) + ′ (ϕ, η, η ′ )) exp(−Gη+ ,η′ − Gη+ ,η )dη ′ 0 sin(ϕ ∫ η+ ) H(η ′ , Rϕ′ (ϕ, η, η ′ )) ′ + exp(Gη,η′ )dη . η sin(ϕ′ (ϕ, η, η ′ )) Infinite Slab Consider the ϵ-transport problem for f (η, ϕ) in an infinite slab (η, ϕ) ∈ [0, ∞)×[−π, π)   ∂f ∂f   sin ϕ + F (η) cos ϕ + f = H(η, ϕ),   ∂η ∂ϕ f (0, ϕ) = h(ϕ) for sin ϕ > 0, (4.4.162)      limη→∞ f (η, ϕ) = f∞ . We can define the solution via taking limit L → ∞ in (4.4.159), (4.4.160) and (4.4.161) as follows: f (η, ϕ) = Ah(ϕ) + T H(η, ϕ), (4.4.163) 238 where Case I: For sin ϕ > 0, Ah(ϕ) = h(ϕ′ (ϕ, η, 0)) exp(−Gη,0 ) (4.4.164) ∫ η H(η ′ , ϕ′ (ϕ, η, η ′ )) T H(η, ϕ) = ′ (ϕ, η, η ′ )) exp(−Gη,η′ )dη ′ . (4.4.165) 0 sin(ϕ Case II: For sin ϕ < 0 and |E(η, ϕ)| ≤ e−V∞ , Ah(ϕ) = 0 (4.4.166) ∫ ∞ ′ ′ ′ H(η , Rϕ (ϕ, η, η )) T H(η, ϕ) = ′ ′ exp(Gη,η′ )dη ′ . (4.4.167) η sin(ϕ (ϕ, η, η )) Case III: For sin ϕ < 0 and |E(η, ϕ)| ≥ e−V∞ , Ah(ϕ) = h(ϕ′ (ϕ, η, 0)) exp(−Gη+ ,0 − Gη+ ,η ) (4.4.168) ( ∫ η+ H(η ′ , ϕ′ (ϕ, η, η ′ )) T H(η, ϕ) = ′ (ϕ, η, η ′ )) exp(−Gη+ ,η′ − Gη+ ,η )dη ′ (4.4.169) 0 sin(ϕ ∫ η+ ) H(η ′ , Rϕ′ (ϕ, η, η ′ )) ′ + exp(Gη,η′ )dη . η sin(ϕ′ (ϕ, η, η ′ )) Notice that lim exp(−GL,η ) = 0, (4.4.170) L→∞ 239 for sin ϕ < 0 and |E(η, ϕ)| ≤ e−V∞ . Hence, above derivation is valid. In order to achieve the estimate of f , we need to control T H and Ah. Preliminaries We first give several technical lemmas to be used for proving L∞ estimates of f . Lemma 4.4.7. For any 0 ≤ β ≤ 1, we have βη e Ah ∞ ≤ ∥h∥ ∞ . (4.4.171) L L In particular, ∥Ah∥L∞ ≤ ∥h∥L∞ . (4.4.172) Proof. Since ϕ′ is always in the domain [0, π), we naturally have 0 ≤ sin(ϕ′ (ϕ, η, ξ)) ≤ 1, (4.4.173) which further implies 1 ≥ 1. (4.4.174) sin(ϕ′ (ϕ, η, ξ)) Combined with the fact η + ≥ η, we deduce exp(−Gη,0 ) ≤ e−η (4.4.175) exp(−Gη+ ,0 − Gη+ ,η ) ≤ exp(−Gη+ ,0 ) ≤ exp(−Gη,0 ) ≤ e−η . (4.4.176) Hence, our result easily follows. 240 Lemma 4.4.8. The integral operator T satisfies ∥T H∥L∞ L∞ ≤ C∥H∥L∞ L∞ , (4.4.177) and for any 0 ≤ β ≤ 1/2 βη e T H ∞ ∞ ≤ C eβη H ∞ ∞ , (4.4.178) L L L L where C is a universal constant independent of data. Proof. For (4.4.177), when sin ϕ > 0 ∫ η 1 |T H| ≤ |H(η ′ , ϕ′ (ϕ, η, η ′ ))| ′ ′ )) exp(−Gη,η′ )dη ′ (4.4.179) sin(ϕ (ϕ, η, η 0 ∫ η 1 ≤ ∥H∥L∞ L∞ ′ ′ exp(−Gη,η′ )dη ′ . 0 sin(ϕ (ϕ, η, η )) We can directly estimate ∫ η ∫ ∞ 1 ′ ′ exp(−Gη,η′ )dη ′ ≤ e−z dz = 1, (4.4.180) 0 sin(ϕ (ϕ, η, η )) 0 and (4.4.177) naturally follows. For sin ϕ < 0 and |E(η, ϕ)| ≤ e−V∞ , ∫ ∞ 1 |T H| ≤ |H(η ′ , ϕ′ (ϕ, η, η ′ ))| ′ ′ exp(Gη,η′ )dη ′ (4.4.181) η sin(ϕ (ϕ, η, η )) ∫ ∞ 1 ≤ ∥H∥L∞ L∞ ′ ′ exp(Gη,η′ )dη ′ . η sin(ϕ (ϕ, η, η )) we have ∫ ∞ ∫ 0 1 ′ ′ exp(Gη,η′ )dη ′ ≤ ez dz = 1, (4.4.182) η sin(ϕ (ϕ, η, η )) −∞ 241 and (4.4.177) easily follows. The case sin ϕ < 0 and |E(η, ϕ)| ≥ e−V∞ can be proved combining above two techniques, so we omit it here. For (4.4.178), when sin ϕ > 0, η ≥ η ′ and β < 1/2, since Gη,η′ ≥ η − η ′ , we have 1 1 1 β(η − η ′ ) − Gη,η′ ≤ β(η − η ′ ) − (η − η ′ ) − Gη,η′ ≤ − Gη,η′ . (4.4.183) 2 2 2 Then it is natural that ∫ η 1 exp(β(η − η ′ ) − Gη,η′ )dη ′ (4.4.184) sin(ϕ′ (ϕ, η, η ′ )) ∫0 η 1 ≤ exp(−Gη,η′ /2)dη ′ sin(ϕ′ (ϕ, η, η ′ )) ∫0 ∞ ≤ e−z/2 dz = 2. 0 This leads to βη e T H (4.4.185) ∫ η 1 ≤ eβη |H(η ′ , ϕ′ (ϕ, η, η ′ ))| ′ ′ exp(−Gη,η′ )dη ′ sin(ϕ (ϕ, η, η )) 0 ∫ η βη 1 ≤ e H L∞ L∞ ′ ′ exp(β(η − η ′ ) − Gη,η′ )dη ′ 0 sin(ϕ (ϕ, η, η )) βη ≤ C e H L∞ L∞ , and (4.4.178) naturally follows. For sin ϕ < 0 and |E(η, ϕ)| ≤ e−V∞ , note for η ′ ≥ η 1 1 1 β(η − η ′ ) + Gη,η′ ≤ β(η − η ′ ) + (η − η ′ ) + Gη,η′ ≤ Gη,η′ . (4.4.186) 2 2 2 Then (4.4.178) holds by obvious modifications of sin ϕ > 0 case. The case sin ϕ < 0 and |E(η, ϕ)| ≥ e−V∞ can be shown by combining above two cases, so we omit it here. Lemma 4.4.9. For any δ > 0 there is a constant C(δ) > 0 independent of data such 242 that ∥T H∥L∞ L2 ≤ C(δ)∥H∥L2 L2 + δ∥H∥L∞ L∞ . (4.4.187) Proof. We divide the proof into several steps. Step 1: The case of sin ϕ > 0. We consider ∫ |T H(η)|2 dϕ (4.4.188) sin ϕ>0 ∫ (∫ )2 H(η ′ , ϕ′ (ϕ, η, η ′ )) η ′ = exp(−Gη,η′ )dη dϕ sin ϕ>0 0 sin(ϕ′ (ϕ, η, η ′ )) ∫ (∫ )2 ∫ ( ∫ )2 = 1{sin(ϕ′ (ϕ,η,η′ ))>m} . . . + 1{sin(ϕ′ (ϕ,η,η′ ))≤m} . . . = I1 + I2 . By Cauchy’s inequality and (4.4.180), we get ∫ (∫ η ) ′ ′ ′ 2 ′ I1 ≤ |H(η , ϕ (ϕ, η, η ))| dη (4.4.189) sin ϕ>0 0 (∫ η ) exp(−2Gη,η′ ) ′ × 1{sin(ϕ′ (ϕ,η,η′ ))>m} 2 ′ dη dϕ sin (ϕ (ϕ, η, η ′ )) 0 ∫ (∫ η ) 1 exp(−2Gη,η′ ) ′ ≤ ∥H∥L2 L2 2 1{sin(ϕ′ (ϕ,η,η′ ))>m} dη dϕ m sin ϕ>0 0 sin(ϕ′ (ϕ, η, η ′ )) π ≤ ∥H∥2L2 L2 . m On the other hand, for η ′ ≤ η, we can directly estimate ϕ′ (ϕ, η, η ′ ) ≥ ϕ. Hence, we have the relation sin ϕ ≤ sin(ϕ′ (ϕ, η, η ′ )). (4.4.190) 243 Therefore, we can directly estimate I2 as follows: (4.4.191) ∫ (∫ η )2 1 ′ I2 ≤ ∥H∥2L∞ L∞ 1{sin(ϕ′ (ϕ,η,η′ ))≤m} exp(−Gη,η′ )dη dϕ sin ϕ>0 0 sin(ϕ′ (ϕ, η, η ′ )) ∫ (∫ η )2 1 ′ ≤ ∥H∥2L∞ L∞ 1{sin ϕ≤m} exp(−Gη,η′ )dη dϕ sin ϕ>0 0 sin(ϕ′ (ϕ, η, η ′ )) ∫ (∫ η )2 1 ′ = ∥H∥L∞ L∞ 2 1{sin ϕ≤m} ′ ′ exp(−Gη,η′ )dη dϕ. sin ϕ>0 0 sin(ϕ (ϕ, η, η )) Note ∫ η ∫ ∞ 1 exp(−Gη,η′ )dη ′ ≤ e−z dz = 1. (4.4.192) 0 sin(ϕ (ϕ, η, η ′ )) ′ 0 Therefore, for m sufficiently small, we have ∫ I2 ≤ ∥H∥2L∞ L∞ 1{sin ϕ≤m} dϕ ≤ 4m∥H∥2L∞ L∞ . (4.4.193) sin ϕ>0 Summing up (4.4.189) and (4.4.193), for m sufficiently small, we deduce (4.4.187). Step 2: The case of sin ϕ < 0 and |E(η, ϕ)| ≤ e−V∞ . We can decompose ∫ 1{|E(η,ϕ)|≤e−V∞ } |T H|2 dϕ (4.4.194) sin ϕ<0 ∫ (∫ ∞ )2 H(η ′ , Rϕ′ (ϕ, η, η ′ )) ′ = 1{|E(η,ϕ)|≤e−V∞ } exp(Gη,η′ )dη dϕ sin ϕ<0 η sin(ϕ′ (ϕ, η, η ′ )) ∫ (∫ )2 ∫ ( ∫ )2 = 1{sin(ϕ′ (ϕ,η,η′ ))>m} . . . + 1{sin(ϕ′ (ϕ,η,η′ ))≤m} 1{η′ −η≥σ} . . . ∫ (∫ )2 1{sin(ϕ′ (ϕ,η,η′ ))≤m} 1{η′ −η≤σ} . . . = I1 + I2 + I3 . 244 We can directly estimate I1 as follows: ∫ (∫ ∞ ) ′ ′ ′ 2 ′ I1 ≤ |H(η , Rϕ (ϕ, η, η ))| dη (4.4.195) sin ϕ<0 η (∫ ∞ ) exp(2Gη,η′ ) ′ × 1{sin(ϕ′ (ϕ,η,η′ ))>m} 2 ′ dη dϕ η sin (ϕ (ϕ, η, η ′ )) ∫ (∫ ∞ ) 1 exp(2Gη,η′ ) ′ ≤ ∥H∥L2 L2 2 1{sin(ϕ′ (ϕ,η,η′ ))>m} dη dϕ m sin ϕ<0 η sin(ϕ′ (ϕ, η, η ′ )) π ≤ ∥H∥2L2 L2 . m On the other hand, for I2 we have ∫ I2 ≤ ∥H∥2L∞ L∞ 1{|E(η,ϕ)|≤e−V∞ } (4.4.196) sin ϕ<0 (∫ ∞ )2 exp(Gη,η′ ) × 1{sin(ϕ′ (ϕ,η,η′ ))≤m} 1{η′ −η≥σ} dη ′ dϕ. η sin(ϕ′ (ϕ, η, η ′ )) Note ∫ η 1 η′ − η σ Gη,η′ = ′ dy ≤ − =− . (4.4.197) η′ sin(ϕ (ϕ, η, y)) m m Then we can obtain ∫ (∫ −σ/m )2 dϕ = 4e− m ∥H∥2L∞ L∞ . (4.4.198) σ I2 ≤ ∥H∥2L∞ L∞ z e dz sin ϕ<0 ∞ For I3 , we can estimate as follows: ∫ I3 ≤ ∥H∥2L∞ L∞ 1{|E(η,ϕ)|≤e−V∞ } (4.4.199) sin ϕ<0 (∫ ∞ )2 exp(Gη,η′ ) × 1{sin(ϕ′ (ϕ,η,η′ ))≤m} 1{η′ −η≤σ} ′ (ϕ, η, η ′ )) dη ′ dϕ sin(ϕ η ∫ ≤ ∥H∥L∞ L∞ 2 1{|E(η,ϕ)|≤e−V∞ } sin ϕ<0 (∫ η+σ )2 exp(Gη,η′ ) × 1{sin(ϕ′ (ϕ,η,η′ ))≤m} 1{η′ −η≤σ} dη ′ dϕ. η sin(ϕ′ (ϕ, η, η ′ )) 245 Note ∫ ∞ ∫ 0 1 exp(Gη,η′ )dη ′ ≤ ez dz = 1. (4.4.200) η sin(ϕ′ (ϕ, η, η ′ )) −∞ ′ Then 1 ≤ α = eV (η )−V (η) ≤ eV (η+σ)−V (η) ≤ 1+4σ due to (4.4.8), and for η ′ ∈ [η, η +σ], ( ) sin ϕ (ϕ, η, η )) = sin cos (α cos ϕ) , sin(ϕ′ (ϕ, η, η ′ )) < m lead to ′ ′ −1 (4.4.201) √ ( ) √ α2 − 1 − sin ϕ′ (ϕ, η, η ′ ) 2 √ cos2 ϕ′ (ϕ, η, η ′ ) |sin ϕ| = 1 − cos2 ϕ = 1− 2 = √ α α √ α −1+m 2 2 (1 + 4σ) − 1 + m 2 2 √ ≤ ≤ ≤ 9σ + m2 . α α Hence, we can obtain (4.4.202) ∫ ∫ I3 ≤ ∥H∥2L∞ L∞ 1{sin(ϕ′ (ϕ,η,η′ ))≤m} dϕ ≤ ∥H∥2L∞ L∞ 1{|sin ϕ|≤√9σ+m2 } dϕ √ sin ϕ<0 sin ϕ<0 ≤ 4 9σ + m .2 Summarizing (4.4.195), (4.4.198) and (4.4.202), for sufficiently small σ, we can always choose m small enough to guarantee the relation (4.4.187). Step 3: The case of sin ϕ < 0 and |E(η, ϕ)| ≥ e−V∞ . 246 We can decompose T H as follows: T H(η, ϕ) (4.4.203) ( ∫ η+ ∫ η+ ) = . . . exp(−Gη+ ,η′ − Gη+ ,η )dη ′ + . . . exp(Gη,η′ )dη ′ ∫ η0 η = . . . exp(−Gη+ ,η′ − Gη+ ,η )dη ′ 0 ( ∫ η+ ∫ η+ ) ′ ′ + . . . exp(−Gη+ ,η′ − Gη+ ,η )dη + . . . exp(Gη,η′ )dη η η = I1 + I2 . For I1 , we can apply a similar argument as in Step 1 and for I2 , a similar argument as in Step 2 conclude the lemma. Estimates of ϵ-Milne Equation Consider the equation satisfied by z = f − f∞ as follows:   ∂z ∂z   sin ϕ + F (η) cos ϕ + z = z¯ + S,   ∂η ∂ϕ  z(0, ϕ) = p(ϕ) = h(ϕ) − f∞ for sin ϕ > 0, (4.4.204)     limη→∞ z(η, ϕ) = 0. Lemma 4.4.10. Assume (4.4.3) and (4.4.4) hold. Then there exists a constant C depending on the data such that the solution of equation (4.4.204) verifies ( ) M ∥z∥L∞ L∞ ≤C 1+M + + ∥z∥L2 L2 . (4.4.205) K 247 Proof. We first show the following important facts: ∥¯ z ∥L2 L2 ≤ ∥z∥L2 L2 , (4.4.206) 1 ∥¯ z ∥ L∞ L∞ ≤ ∥z∥L∞ L2 . (4.4.207) 2π We can directly derive them by Cauchy’s inequality as follows: ∫ ∞ ∫ π ( )2 ( ∫ π )2 1 ∥¯ z ∥2L2 L2 = z(η, ϕ)dϕ dϕdη (4.4.208) −π 2π −π ∫ ∞ ∫ π ( )( ∫ π 0 ) 1 ≤ 2 z (η, ϕ)dϕ dϕdη −π 2π −π ∫ ∞(∫ π 0 ) = z (η, ϕ)dϕ dη = ∥z∥2L2 L2 . 2 0 −π ( ∫ π )2 1 ∥¯ z ∥2L∞ L∞ 2 = sup z¯ (η) = sup z(η, ϕ)dϕ (4.4.209) η η 2π −π ()2 ( ∫ π )( ∫ π ) 1 ≤ sup 2 z (η, ϕ)dϕ 2 1 dϕ η 2π −π −π (∫ π ) 1 = sup z (η, ϕ)dϕ = ∥z∥2L∞ L2 . 2 2π η −π By (4.4.204), z = Ap + T (¯ z + S) leads to T (¯ z + S) = z − Ap, (4.4.210) Then by Lemma 4.4.9, (4.4.206) and (4.4.207), we can show ∥z − Ap∥L∞ L2 (4.4.211) ( ) ( ) ≤ C(δ) ∥¯ z ∥L2 L2 + ∥S∥L2 L2 + δ ∥¯ z ∥L∞ L∞ + ∥S∥L∞ L∞ ( ) ( ) ≤ C(δ) ∥z∥L2 L2 + ∥S∥L2 L2 + δ ∥z∥L∞ L2 + ∥S∥L∞ L∞ . 248 Therefore, based on Lemma 4.4.7 and (4.4.211), we can directly estimate ∥z∥L∞ L2 (4.4.212) ( ) ( ) ≤ ∥Ap∥L∞ + C(δ) ∥z∥L2 L2 + ∥S∥L2 L2 + δ ∥z∥L∞ L2 + ∥S∥L∞ L∞ ( ) ( ) ≤ ∥p∥L∞ + C(δ) ∥z∥L2 L2 + ∥S∥L2 L2 + δ ∥z∥L∞ L2 + ∥S∥L∞ L∞ . We can take δ = 1/2 to obtain ( ) ∥z∥L∞ L2 ≤ C ∥z∥L2 L2 + ∥S∥L2 L2 + ∥S∥L∞ L∞ + ∥p∥L∞ . (4.4.213) Therefore, based on Lemma 4.4.8, (4.4.213) and (4.4.207), we can achieve ∥z∥L∞ L∞ (4.4.214) ( ) ≤ ∥Ap∥L∞ L∞ + ∥T (¯ z + S)∥L∞ L∞ ≤ C ∥p∥L∞ + ∥¯z ∥L∞ L∞ + ∥S∥L∞ L∞ ( ) ≤ C ∥p∥L∞ + ∥S∥L∞ L∞ + ∥z∥L∞ L2 ( ) ≤ C ∥p∥L∞ + ∥S∥L∞ L∞ + ∥S∥L2 L2 + ∥z∥L2 L2 . Since ∥p∥L∞ , ∥S∥L2 L2 and ∥S∥L∞ L∞ are finite, our result easily follows. Lemma 4.4.10 naturally implies the following. Theorem 4.4.11. The solution f (η, ϕ) to the Milne problem (4.4.1) satisfies ( ) M ∥f − f∞ ∥L∞ L∞ ≤C 1+M + + ∥f − f∞ ∥L2 L2 . (4.4.215) K Combining Theorem 4.4.11 and Theorem 4.4.6, we deduce the main theorem. Theorem 4.4.12. There exists a unique solution f (η, ϕ) to the ϵ-Milne problem 249 (4.4.1) satisfying ( ) M ∥f − f∞ ∥L∞ L∞ ≤C 1+M + . (4.4.216) K 4.4.3 Exponential Decay In this section, we prove the spatial decay of the solution to the Milne problem. Theorem 4.4.13. Assume (4.4.3) and (4.4.4) hold. For K0 > 0 sufficiently small, the solution f (η, ϕ) to the ϵ-Milne problem (4.4.1) satisfies ( ) Kη e 0 (f − f∞ ) ∞ ∞ ≤ C 1 + M + M , (4.4.217) L L K Proof. Define Z = eK0 η g for z = f − f∞ . We divide the analysis into several steps: Step 1: We have ∫ ∞ (∫ π ) ( )2 M ∥Z∥2L2 L2 = e 2K0 η (f (η, ϕ) − f∞ ) dϕ dη ≤ C 1 + M + 2 .(4.4.218) 0 −π K The proof of (4.4.218): The orthogonal property (4.4.39) reveals ⟨f, f sin ϕ⟩ϕ (η) = ⟨r, r sin ϕ⟩ϕ (η). (4.4.219) Multiplying e2K0 η f on both sides of equation (4.4.1) and integrating over ϕ ∈ [−π, π), we obtain ( ) ( ) 1 d 1 e2K0 η ⟨r, r sin ϕ⟩ϕ (η) + F (η) e 2K0 η ⟨r, r sin ϕ⟩ϕ (η) (4.4.220) 2 dη 2 ( ) −e 2K0 η K0 ⟨r, r sin ϕ⟩ϕ (η) − ⟨r, r⟩ϕ (η) = e2K0 η ⟨S, f ⟩ϕ (η). 250 For K0 < min{1/2, K}, we have 3 1 ∥r(η)∥2L2 ≥ −K0 ⟨r, r sin ϕ⟩ϕ (η) + ⟨r, r⟩ϕ (η) ≥ ∥r(η)∥2L2 . (4.4.221) 2 2 Similar to the proof of Lemma 4.4.3 and Lemma 4.4.4, formula as (4.4.220) and (4.4.221) imply ∫ ( )2 K η 2 ∞ M e 0 r 2 = 2K0 η e ⟨r, r⟩ϕ (η)dη ≤ C 1 + M + . (4.4.222) L L2 K 0 From (4.4.105), Cauchy’s inequality and (4.4.9), we can deduce ∫ ∞ (∫ π ) e 2K0 η (f (η, ϕ) − f∞ ) dϕ dη 2 (4.4.223) −π 0 ∫ ∞ (∫ π ) ∫ ∞ (∫ π ) ≤ e2K0 η 2 r (η, ϕ)dϕ dη + e2K0 η (q(η) − q∞ ) dϕ dη 2 −π −π ∫ ∞ 0 0 ≤ e2K0 η ∥r(η)∥2L2 dη 0 ∫ ∞ (∫ ∞ )2 ∫ ∞ (∫ ∞ )2 + e 2K0 η |F (y)| ∥r(y)∥L2 dy dη + e 2K0 η ∥S(y)∥L∞ dy dη 0 η 0 η ( )2 M ≤ C 1+M + K (∫ ∞ )( ∫ ∞ ∫ ∞ ) +C e2K0 η ∥r(η)∥L2 dη 2 e 2K0 (η−y) 2 F (y)dydη 0 0 η ∫ ∞ (∫ ∞ )2 + e 2K0 η ∥S(y)∥L∞ dy dη 0 η ( )2 M ≤ C 1+M + K (∫ ∞ )( ∫ ∞ ∫ ∞ ) +C e2K0 η ∥r(η)∥L2 dη 2 2 F (y)dydη 0 0 η ∫ ∞ (∫ ∞ )2 + e 2K0 η ∥S(y)∥L∞ dy dη 0 η ( )2 M ≤ C 1+M + . K This completes the proof of (4.4.218) when S¯ = 0. By the method introduced in 251 Lemma 4.4.5, we can extend above L2 estimates to the general S case. Note all the auxiliary functions constructed in Lemma 4.4.5 satisfy the estimates (4.4.218). Step 2: We have ( ) M ∥Z∥L∞ L∞ ≤C 1+M + + ∥Z∥L2 L2 . (4.4.224) K Proof of (4.4.224): Z satisfies the equation    sin ϕ ∂Z + F (η) cos ϕ ∂Z + Z = Z¯ + eK0 η S + K0 sin ϕZ, ∂η ∂ϕ (4.4.225)   Z(0, ϕ) = p(ϕ) = h(ϕ) − f∞ for sin ϕ > 0. Since we know Z = Ap + T (Z¯ + eK0 η S + K0 sin ϕZ) leads to T (Z¯ + eK0 η S + K0 sin ϕZ) = Z − Ap, (4.4.226) then by Lemma 4.4.9, (4.4.206) and (4.4.207), we can show ∥Z − Ap∥L∞ L2 (4.4.227) ( ) Kη ¯ 0 ≤ C(δ) Z L2 L2 + e S L2 L2 + K0 ∥Z∥L2 L2 ( ) Kη +δ Z¯ ∞ ∞ + e S ∞ ∞ + K0 ∥Z∥ ∞ ∞ L L 0 L L L L ( ) Kη ≤ C(δ) ∥Z∥L2 L2 + e 0 S L2 L2 + K0 ∥Z∥L2 L2 ( ) Kη +δ ∥Z∥L∞ L2 + e 0 S L∞ L∞ + K0 ∥Z∥L∞ L∞ . 252 Therefore, based on Lemma 4.4.7 and (4.4.211), we can directly estimate ∥Z∥L∞ L2 (4.4.228) ( ) Kη ≤ ∥Ap∥L∞ + C(δ) ∥Z∥L2 L2 + e S L2 L2 + K0 ∥Z∥L2 L2 0 ( ) Kη +δ ∥Z∥L∞ L2 + e S L∞ L∞ + K0 ∥Z∥L∞ L∞ 0 ( ) Kη ≤ ∥p∥L∞ + 2C(δ) ∥Z∥L2 L2 + e 0 S L2 L2 ( ) Kη +δ ∥Z∥L∞ L2 + e 0 S L∞ L∞ + K0 ∥Z∥L∞ L∞ . We can take δ = 1/2 to obtain (4.4.229) ( ) ∥Z∥L∞ L2 ≤ C ∥Z∥L2 L2 + ∥S∥L2 L2 + ∥S∥L∞ L∞ + ∥p∥L∞ + K0 ∥Z∥L∞ L∞ . Then based on Lemma 4.4.7, Lemma 4.4.8 and Lemma 4.4.9, we can deduce ∥Z∥L∞ L∞ ≤ eK0 η Ap L∞ + eK0 η T S L∞ L∞ + Z¯ L∞ L∞ (4.4.230) ≤ ∥p∥L∞ + eK0 η S L∞ L∞ + Z¯ L∞ L∞ ≤ ∥p∥L∞ + eK0 η S L∞ L∞ + ∥Z∥L∞ L2 ( ) Kη Kη 0 0 ≤ C ∥Z∥L2 L2 + e S L2 L2 + e S L∞ L∞ + ∥p∥L∞ + K0 ∥Z∥L∞ L∞ . Taking K0 sufficiently small, this completes the proof of (4.4.224). Combining (4.4.218) and (4.4.224), we deduce (4.4.217). 253 4.4.4 Maximum Principle Theorem 4.4.14. The solution f (η, ϕ) to the ϵ-Milne problem (4.4.1) with S = 0 satisfies the maximum principle, i.e. min h(ϕ) ≤ f (η, ϕ) ≤ max h(ϕ). (4.4.231) sin ϕ>0 sin ϕ>0 Proof. We claim it is sufficient to show f (η, ϕ) ≤ 0 whenever h(ϕ) ≤ 0. Suppose this claim is justified. Denote m = minsin ϕ>0 h(ϕ) and M = maxsin ϕ>0 h(ϕ). Then f 1 = f − M satisfies the equation   ∂f 1 ∂f 1   sin ϕ + F (η) cos ϕ + f 1 − f¯1 = 0,   ∂η ∂ϕ  f 1 (0, ϕ) = h(ϕ) − M for sin ϕ > 0, (4.4.232)     limη→∞ f 1 (η, ϕ) = f∞ 1 . Hence, h − M ≤ 0 implies f 1 ≤ 0 which is actually f ≤ M . On the other hand, f 2 = m − f satisfies the equation   ∂f 2 ∂f 2   sin ϕ + F (η) cos ϕ + f 2 − f¯2 = 0,   ∂η ∂ϕ  f 2 (0, ϕ) = m − h(ϕ) for sin ϕ > 0, (4.4.233)     limη→∞ f 2 (η, ϕ) = f∞ 2 . Thus, m − h ≤ 0 implies f 2 ≤ 0 which further leads to f ≥ m. Therefore, the maximum principle is established. We now prove if h(ϕ) ≤ 0, we have f (η, ϕ) ≤ 0. We divide the proof into several steps: Step 1: Penalized ϵ-Milne problem in a finite slab. 254 Assuming h(ϕ) ≤ 0, we then consider the penalized Milne problem for fλL (η, ϕ) in the finite slab (η, ϕ) ∈ [0, L] × [−π, π)   ∂fλL ∂fλL   λf L + sin ϕ + F (η) cos ϕ + fλL − f¯λL = 0,   λ ∂η ∂ϕ  fλL (0, ϕ) = h(ϕ) for sin ϕ < 0,(4.4.234)     fλL (L, ϕ) = fλL (L, Rϕ). In order to construct the solution of (4.4.234), we iteratively define the sequence L ∞ {fm }m=1 as f0L = 0 and (4.4.235)  L L  ∂f ∂f   L λfm + sin ϕ m + F (η) cos ϕ m + fm L − f¯m−1 L = 0,   ∂η ∂ϕ L  fm (0, ϕ) = h(ϕ) for sin ϕ < 0,     L L fm (L, ϕ) = fm (L, Rϕ). L Along the characteristics, it is easy to see we always have fm < 0. In the proof of L Lemma 4.4.3, we have shown fm converges strongly in L∞ ([0, L] × [−π, π)) to fλL which satisfies (4.4.234). Also, fλL satisfies L fλ ∞ ∞ ≤ 1 + λ ∥h∥ ∞ . (4.4.236) L L L λ Naturally, we obtain fλL ∈ L2 ([0, L] × [−π, π)) and fλL ≤ 0. Step 2: ϵ-Milne problem in a finite slab. Consider the Milne problem for f L (η, ϕ) in a finite slab (η, ϕ) ∈ [0, L] × [−π, π)   ∂f L ∂f L   sin ϕ + F (η) cos ϕ + f L − f¯L = 0,   ∂η ∂ϕ f L (0, ϕ) = h(ϕ) for sin ϕ < 0, (4.4.237)      f L (L, ϕ) = f L (L, Rϕ). 255 In the proof of Lemma 4.4.3, we have shown fλL is uniformly bounded in L2 ([0, L) × [−π, π)) with respect to λ, which implies we can take weakly convergent subsequence fλL ⇀ f L as λ → 0 with f L ∈ L2 ([0, L] × [−π, π)). Naturally, we have f L (η, ϕ) ≤ 0. Step 3: ϵ-Milne problem in an infinite slab. Finally, in the proof of Lemma 4.4.4, by taking L → ∞, we have f L ⇀ f in L2loc ([0, L) × [−π, π), (4.4.238) where f satisfies (4.4.1). Certainly, we have f (η, ϕ) ≤ 0. This justifies the claim in Step 1. Hence, we complete the proof. Remark 4.4.15. Note that when F = 0, then all the previous proofs can be recov- ered and Theorem 4.4.12, Theorem 4.4.13 and Theorem 4.4.14 still hold. Hence, we can deduce the well-posedness, decay and maximum principle of the classical Milne problem   ∂f   sin ϕ + f − f¯ = S(η, ϕ),   ∂η f (0, ϕ) = h(ϕ) for sin ϕ > 0, (4.4.239)      lim η→∞ f (η, ϕ) = f∞ . 4.5 Proof of Theorem 4.1.1 We divide the proof into several steps: Step 1: Remainder definitions. 256 We may rewrite the asymptotic expansion as follows: ∑ ∞ ∑ ∞ uϵ ∼ ϵk Ukϵ + ϵk Ukϵ . (4.5.1) k=0 k=0 The remainder can be defined as ∑ N ∑ N RN = uϵ − ϵk Ukϵ − ϵk Ukϵ = uϵ − QN − QN , (4.5.2) k=0 k=0 where ∑ N QN = ϵk Ukϵ , (4.5.3) k=0 ∑ N QN = ϵk Ukϵ . (4.5.4) k=0 Noting the equation (4.2.38) is equivalent to the equation (4.1.1), we write L to denote the neutron transport operator as follows: Lu = ϵw ⃗ · ∇x u + u − u¯ (4.5.5) ( ) ∂u ϵ ∂u ∂u = sin ϕ − cos ϕ + + u − u¯. ∂η 1 − ϵη ∂ϕ ∂θ Step 2: Estimates of LQN . The interior contribution can be estimated as LQ0 = ϵw ⃗ · ∇x Q0 + Q0 − Q ⃗ · ∇x U0ϵ + (U0ϵ − U¯0ϵ ) = ϵw ¯ 0 = ϵw ⃗ · ∇x U0ϵ . (4.5.6) We have ⃗ · ∇x U0ϵ | ≤ Cϵ |∇x U0ϵ | ≤ Cϵ. |ϵw (4.5.7) 257 This implies |LQ0 | ≤ Cϵ. (4.5.8) Similarly, for higher order term, we can estimate LQN = ϵw ⃗ · ∇x QN + QN − Q ⃗ · ∇x UNϵ . ¯ N = ϵN +1 w (4.5.9) We have N +1 ϵ ⃗ · ∇x UNϵ ≤ CϵN +1 |∇x UNϵ | ≤ CϵN +1 . w (4.5.10) This implies |LQN | ≤ CϵN +1 . (4.5.11) Step 3: Estimates of LQN . The boundary layer solution is Ukϵ = (fkϵ − fkϵ (∞)) · ψ0 = Vk ψ0 where fkϵ (η, θ, ϕ) solves the ϵ-Milne problem and Vk = fkϵ − fkϵ (∞). Notice ψ0 ψ = ψ0 , so the boundary layer 258 contribution can be estimated as LQ0 (4.5.12) ( ) ∂Q0 ϵ ∂Q0 ∂Q0 = sin ϕ − cos ϕ + + Q0 − Q¯0 ∂η 1 − ϵη ∂ϕ ∂θ ( ) ( ) ∂V0 ∂ψ0 ψ0 ϵ ∂V0 ∂V0 = sin ϕ ψ0 + V0 − cos ϕ + + ψ0 V0 − ψ0 V¯0 ∂η ∂η 1 − ϵη ∂ϕ ∂θ ( ) ( ) ∂V0 ∂ψ0 ψ0 ψϵ ∂V0 ∂V0 = sin ϕ ψ0 + V0 − cos ϕ + + ψ0 V0 − ψ0 V¯0 ∂η ∂η 1 − ϵη ∂ϕ ∂θ ( ) ∂V0 ϵψ ∂V0 ∂ψ0 ψ0 ϵ ∂V0 = ψ0 sin ϕ − cos ϕ + V0 − V¯0 + sin ϕ V0 − cos ϕ ∂η 1 − ϵη ∂ϕ ∂η 1 − ϵη ∂θ ∂ψ0 ψ0 ϵ ∂V0 = sin ϕ V0 − cos ϕ . ∂η 1 − ϵη ∂θ Since ψ0 = 1 when η ≤ 1/(4ϵ), the effective region of ∂η ψ0 is η ≥ 1/(4ϵ) which is further and further from the origin as ϵ → 0. By Theorem 4.4.13, the first term in (4.5.12) can be controlled as ∂ψ sin ϕ V0 ≤ Ce− ϵ ≤ Cϵ. 0 K0 (4.5.13) ∂η For the second term in (4.5.12), we have ψ0 ϵ ∂V0 ∂V0 − 1 − ϵη cos ϕ ∂θ ≤ Cϵ ∂θ ≤ Cϵ. (4.5.14) This implies |LQ0 | ≤ Cϵ. (4.5.15) 259 Similarly, for higher order term, we can estimate ( ) ∂QN ϵ ∂QN ∂QN LQN = sin ϕ − cos ϕ + + QN − Q¯N (4.5.16) ∂η 1 − ϵη ∂ϕ ∂θ ∑ k ∂ψ0 ψ0 ϵk+1 ∂Vk = ϵi sin ϕ Vi − cos ϕ . i=0 ∂η 1 − ϵη ∂θ Away from the origin, the first term in (4.5.16) can be controlled as ∑k ∂ψ K0 Vi ≤ Ce− ϵ ≤ Cϵk+1 . 0 ϵi sin ϕ (4.5.17) ∂η i=0 For the second term in (4.5.16), we have ψ0 ϵk+1 ∂V ∂Vk − k 1 − ϵη cos ϕ ∂θ ≤ Cϵ ∂θ ≤ Cϵ . k+1 k+1 (4.5.18) This implies |LQN | ≤ Cϵk+1 . (4.5.19) Step 4: Proof of (4.1.7). In summary, since Luϵ = 0, collecting (4.5.2), (4.5.11) and (4.5.19), we can prove |LRN | ≤ CϵN +1 . (4.5.20) Consider the asymptotic expansion to N = 3, then the remainder R3 satisfies the equation    ϵw ⃗ · ∇x R3 + R3 − R ¯ 3 = LR3 for ⃗x ∈ Ω, (4.5.21)   R3 (⃗x0 , w) ⃗ = 0 for w ⃗ · ⃗n < 0 and ⃗x0 ∈ ∂Ω. 260 By Theorem 4.3.6, we have C(Ω) C(Ω) ∥R3 ∥L∞ (Ω×S 1 ) ≤ ∥LR3 ∥ L ∞ (Ω×S 1 ) ≤ (Cϵ4 ) = C(Ω)ϵ. (4.5.22) ϵ3 ϵ3 Hence, we have ∑ 3 ∑ 3 ϵ u − ϵk Ukϵ − ϵk Ukϵ = O(ϵ). (4.5.23) k=0 k=0 L∞ (Ω×S 1 ) Since it is easy to see ∑3 ∑ 3 ϵ k ϵ ϵ Uk + ϵ Uk k = O(ϵ), (4.5.24) k=1 k=1 L∞ (Ω×S 1 ) our result naturally follows. This completes the proof of (4.1.7). Step 5: Basic settings to show (4.1.8). By (4.2.25), the solution f0 satisfies the Milne problem   ∂f0   sin(θ + ξ) + f0 − f¯0 = 0,   ∂η f0 (0, θ, ξ) = g(θ, ξ) for sin(θ + ξ) > 0, (4.5.25)      limη→∞ f0 (η, θ, ξ) = f0 (∞, θ). For convenience of comparison, we make the substitution ϕ = θ + ξ to obtain   ∂f0   sin ϕ + f0 − f¯0 = 0,   ∂η f0 (0, θ, ϕ) = g(θ, ϕ) for sin ϕ > 0, (4.5.26)      lim η→∞ f0 (η, θ, ϕ) = f0 (∞, θ). 261 Assume (4.1.8) is incorrect, such that lim ∥(U0 + U0 ) − (U0ϵ + U0ϵ )∥L∞ = 0. (4.5.27) ϵ→0 Since the boundary g(ϕ) = cos ϕ independent of θ, by (4.2.25) and (4.2.50), it is obvious the limit of zeroth order boundary layer f0 (∞, θ) and f0ϵ (∞, θ) satisfy f0 (∞, θ) = C1 and f0ϵ (∞, θ) = C2 for some constant C1 and C2 independent of θ. By (4.2.26) and (4.2.51), we can derive the interior solutions are indeed constants U0 = C1 and U0ϵ = C2 . Hence, we may further derive lim ∥(f0 (∞) + U0 ) − (f0ϵ (∞) + U0ϵ )∥L∞ = 0. (4.5.28) ϵ→0 For 0 ≤ η ≤ 1/(2ϵ), we have ψ0 = 1, which means f0 = U0 + f0 (∞) and f0ϵ = U0ϵ + f0ϵ (∞) on [0, 1/(2ϵ)]. Define u = f0 + 2, U = f0ϵ + 2 and G = g + 2 = cos ϕ + 2, then u(η, ϕ) satisfies the equation   ∂u   sin ϕ + u − u¯ = 0,   ∂η u(0, ϕ) = G(ϕ) for sin ϕ > 0, (4.5.29)      lim η→∞ u(η, ϕ) = 2 + f0 (∞), and U (η, ϕ) satisfies the equation   ∂U ∂U   sin ϕ + F (ϵ; η) cos ϕ + U − U¯ = 0,   ∂η ∂ϕ U (0, ϕ) = G(ϕ) for sin ϕ > 0, (4.5.30)      limη→∞ U (η, ϕ) = 2 + f0ϵ (∞). Based on (4.5.28), we have lim ∥U (η, ϕ) − u(η, ϕ)∥L∞ = 0. (4.5.31) ϵ→0 262 Then it naturally implies lim U¯ (η) − u¯(η) L∞ = 0. (4.5.32) ϵ→0 Step 6: Continuity of u¯ and U¯ at η = 0. For the problem (4.5.29), we have for any r0 > 0 (4.5.33) (∫ ∫ ) 1 |¯ u(η) − u¯(0)| ≤ |u(η, ϕ) − u(0, ϕ)| dϕ + |u(η, ϕ) − u(0, ϕ)| dϕ . 2π sin ϕ≤r0 sin ϕ≥r0 Since we have shown u ∈ L∞ ([0, ∞) × [−π, π)), then for any δ > 0, we can take r0 sufficiently small such that ∫ 1 C δ |u(η, ϕ) − u(0, ϕ)| dϕ ≤ arcsin r0 ≤ . (4.5.34) 2π sin ϕ≤r0 2π 2 For fixed r0 satisfying above requirement, we estimate the integral on sin ϕ ≥ r0 . By Ukai’s trace theorem, u(0, ϕ) is well-defined in the domain sin ϕ ≥ r0 and is continuous. Also, by consider the relation ∂u u¯(0) − u(0, ϕ) (0, ϕ) = , (4.5.35) ∂η sin ϕ we can obtain in this domain ∂η u is bounded, which further implies u(η, ϕ) is uniformly continuous at η = 0. Then there exists δ0 > 0 sufficiently small, such that for any 0 ≤ η ≤ δ0 , we have ∫ ∫ 1 1 δ δ |u(η, ϕ) − u(0, ϕ)| dϕ ≤ dϕ ≤ . (4.5.36) 2π sin ϕ≥r0 2π sin ϕ≥r0 2 2 263 In summary, we have shown for any δ > 0, there exists δ0 > 0 such that for any 0 ≤ η ≤ δ0 , δ δ |¯ u(η) − u¯(0)| ≤ + = δ. (4.5.37) 2 2 Hence, u¯(η) is continuous at η = 0. By a similar argument along the characteristics, we can show U¯ (η, ϕ) is also continuous at η = 0. In the following, by the continuity, we assume for arbitrary δ > 0, there exists a δ0 > 0 such that for any 0 ≤ η ≤ δ0 , we have |¯ u(η) − u¯(0)| ≤ δ, (4.5.38) U¯ (η) − U¯ (0) ≤ δ. (4.5.39) Step 7: Milne formulation. We consider the solution at a specific point (η, ϕ) = (nϵ, ϵ) for some fixed n > 0. The solution along the characteristics can be rewritten as follows: ∫ nϵ − sin 1 e− sin ϵ (nϵ−κ) 1 1 nϵ u(nϵ, ϵ) = G(ϵ)e ϵ + u¯(κ)dκ, (4.5.40) 0 sin ϵ ∫ nϵ ∫ nϵ ∫ nϵ − 1 e− 1 1 dζ dζ U (nϵ, ϵ) = G(ϵ0 )e 0 sin ϕ(ζ) + κ sin ϕ(ζ) U¯ (κ)dκ, (4.5.41) 0 sin ϕ(κ) where we have the conserved energy along the characteristics E(η, ϕ) = cos ϕe−V (η) , (4.5.42) in which (0, ϵ0 ) and (ζ, ϕ(ζ)) are in the same characteristics of (nϵ, ϵ). 264 Step 8: Estimates of (4.5.40). We turn to the Milne problem for u. We have the natural estimate ∫ nϵ ∫ nϵ − sin 1 1 e− ϵ (nϵ−κ) dκ + o(ϵ) 1 1 (nϵ−κ) e ϵ dκ = (4.5.43) sin ϵ ϵ 0 0 ∫ nϵ κ 1 = e−n e ϵ dκ + o(ϵ) ϵ ∫0 n = e−n eζ dζ + o(ϵ) 0 −n = (1 − e ) + o(ϵ). Then for 0 < ϵ ≤ δ0 , we have |¯ u(0) − u¯(κ)| ≤ δ, which implies ∫ nϵ ∫ nϵ − sin 1 1 e− sin ϵ (nϵ−κ) 1 1 (nϵ−κ) e ϵ u¯(κ)dκ = u¯(0)dκ + O(δ) (4.5.44) 0 sin ϵ 0 sin ϵ = (1 − e−n )¯ u(0) + o(ϵ) + O(δ). For the boundary data term, it is easy to see G(ϵ)e− sin ϵ nϵ = e−n G(ϵ) + o(ϵ) 1 (4.5.45) In summary, we have u(nϵ, ϵ) = (1 − e−n )¯ u(0) + e−n G(ϵ) + o(ϵ) + O(δ). (4.5.46) Step 9: Estimates of (4.5.41). We consider the ϵ-Milne problem for U . For ϵ << 1 sufficiently small, ψ(ϵ) = 1. Then we may estimate cos ϕ(ζ)e−V (ζ) = cos ϵe−V (nϵ) , (4.5.47) 265 which implies 1 − nϵ2 cos ϕ(ζ) = cos ϵ. (4.5.48) 1 − ϵζ and hence √ √ ϵ(nϵ − ζ)(2 − ϵζ − nϵ2 ) sin ϕ(ζ) = 1 − cos2 ϕ(ζ) = cos2 ϵ + sin2 ϵ. (4.5.49) (1 − ϵζ)2 For ζ ∈ [0, ϵ] and nϵ sufficiently small, by Taylor’s expansion, we have 1 − ϵζ = 1 + o(ϵ), (4.5.50) 2 − ϵζ − nϵ2 = 2 + o(ϵ), (4.5.51) sin2 ϵ = ϵ2 + o(ϵ3 ), (4.5.52) cos2 ϵ = 1 − ϵ2 + o(ϵ3 ). (4.5.53) Hence, we have √ sin ϕ(ζ) = ϵ(ϵ + 2nϵ − 2ζ) + o(ϵ2 ). (4.5.54) √ Since ϵ(ϵ + 2nϵ − 2ζ) = O(ϵ), we can further estimate 1 1 = √ + o(1) (4.5.55) sin ϕ(ζ) ϵ(ϵ + 2nϵ − 2ζ) (4.5.56) ∫ √ nϵ √ nϵ 1 ϵ + 2nϵ − 2ζ ϵ + 2nϵ − 2κ − dζ = + o(ϵ) = 1 − + o(ϵ). κ sin ϕ(ζ) ϵ κ ϵ 266 Then we can easily derive the integral estimate ∫ nϵ ∫ nϵ 1 e− 1 dζ κ dκ sin ϕ(ζ) (4.5.57) sin ϕ(κ) 0 ∫ nϵ √ 1 e− ϵ+2nϵ−2κ = e1 ϵ √ dκ + o(ϵ) 0 ϵ(ϵ + 2nϵ − 2κ) ∫ 1 1 (1+2n)ϵ −√ σϵ 1 = e e √ dσ + o(ϵ) 2 ϵ ϵσ ∫ 1+2n 1 1 √ 1 = e e− ρ √ dρ + o(ϵ) 2 1 ρ ∫ √1+2n = e1 e−t dt + o(ϵ) 1 √ = (1 − e1− 1+2n ) + o(ϵ). Then for 0 < ϵ ≤ δ0 , we have U¯ (0) − U¯ (κ) ≤ δ, which implies ∫ nϵ ∫ nϵ 1 e− 1 dζ κ U¯ (κ)dκ sin ϕ(ζ) (4.5.58) sin ϕ(κ) ∫ nϵ ∫ 0 nϵ 1 e− κ sin ϕ(ζ) dζ 1 = U¯ (0)dκ + O(δ) 0 sin ϕ(κ) √ = (1 − e1− 1+2n )U¯ (0) + o(ϵ) + O(δ). For the boundary data term, since G(ϕ) is C 1 , a similar argument shows ∫ nϵ √ √ G(ϵ0 )e− 1 dζ 0 sin ϕ(ζ) = e1− 1+2n G( 1 + 2nϵ) + o(ϵ). (4.5.59) Therefore, we have √ √ √ U (nϵ, ϵ) = (1 − e1− 1+2n )U¯ (0) + e1− 1+2n G( 1 + 2nϵ) + o(ϵ) + O(δ). (4.5.60) Step 10: Proof of (4.1.8). 267 In summary, we have the estimate u(nϵ, ϵ) = (1 − e−n )¯ u(0) + e−n G(ϵ) + o(ϵ) + O(δ), (4.5.61) (4.5.62) √ √ √ U (nϵ, ϵ) = (1 − e1− 1+2n )U¯ (0) + e1− 1+2n G( 1 + 2nϵ) + o(ϵ) + O(δ). The boundary data is G = cos ϕ + 2. Then by the maximum principle in Theorem 4.4.14, we can achieve 1 ≤ u(0, ϕ) ≤ 3 and 1 ≤ U (0, ϕ) ≤ 3. Since (4.5.63) ∫ π ∫ ∫ 1 1 1 u¯(0) = u(0, ϕ)dϕ = u(0, ϕ)dϕ + u(0, ϕ)dϕ 2π −π 2π sin ϕ>0 2π ∫ ∫ sin ϕ<0 1 1 = (2 + cos ϕ)dϕ + u(0, ϕ)dϕ 2π sin ϕ>0 2π sin ϕ<0 ∫ 1 = 1+ u(0, ϕ)dϕ, 2π sin ϕ<0 we naturally obtain 3/2 ≤ u¯(0) ≤ 5/2. Similarly, we can obtain 3/2 ≤ U¯ (0) ≤ 5/2. Furthermore, for ϵ sufficiently small, we have √ G( 1 + 2nϵ) = 3 + o(ϵ), (4.5.64) G(ϵ) = 3 + o(ϵ). (4.5.65) Hence, we can obtain u(nϵ, ϵ) = u¯(0) + e−n (−¯ u(0) + 3) + o(ϵ) + O(δ), (4.5.66) √ U (nϵ, ϵ) = U¯ (0) + e1− 1+2n (−U¯ (0) + 3) + o(ϵ) + O(δ). (4.5.67) Then we can see limϵ→0 U¯ (0) − u¯(0) L∞ = 0 naturally leads to limϵ→0 (−¯ u(0) + 3) − (−U¯ (0) + 3) L∞ = 0. Also, we have −¯ u(0) + 3 = O(1) and 268 √ −U¯ (0) + 3 = O(1). Due to the smallness of ϵ and δ, and also e−n ̸= e1− 1+2n , we can obtain |U (nϵ, ϵ) − u(nϵ, ϵ)| = O(1). (4.5.68) However, above result contradicts our assumption that limϵ→0 ∥U (η, ϕ) − u(η, ϕ)∥L∞ = 0 for any (η, ϕ). This completes the proof of (4.1.8). 4.6 Neutron Transport Equation with Diffusive Bound- ary 4.6.1 Problem Settings We consider the steady neutron transport equation in a two-dimensional unit disk with diffusive boundary. In the space domain Ω = {⃗x : |⃗x| ≤ 1} and the velocity domain Σ = {w ⃗ ∈ S 1 }, the neutron density uϵ (⃗x, w) ⃗: w ⃗ satisfies      ⃗ · ∇x uϵ + (1 + ϵ2 )uϵ − u¯ϵ = 0 in Ω, ϵw  ⃗ = Puϵ (⃗x0 ) + ϵg(⃗x0 , w) uϵ (⃗x0 , w) ⃗ (4.6.1)      for w ⃗ · ⃗n < 0 and ⃗x0 ∈ ∂Ω, where ∫ ϵ 1 u¯ (⃗x) = uϵ (⃗x, w)d ⃗ w,˜ (4.6.2) 2π S1 269 ∫ 1 Pu (⃗x0 ) = ϵ uϵ (⃗x0 , w)( ⃗ w ⃗ · ⃗n)dw, ˜ (4.6.3) 2 w·⃗ ⃗ n>0 with the Knudsen number 0 < ϵ << 1. 4.6.2 Interior Expansion We define the interior expansion as follows: ∑ ∞ ⃗ ∼ U (⃗x, w) ϵk Uk (⃗x, w), ⃗ (4.6.4) k=0 where Uk can be defined by comparing the order of ϵ by plugging (4.6.4) into the equation (4.6.1). Thus we have U0 − U¯0 = 0, (4.6.5) U1 − U¯1 = −w ⃗ · ∇x U0 , (4.6.6) U2 − U¯2 = −w ⃗ · ∇x U1 − U0 , (4.6.7) ... Uk − U¯k = −w ⃗ · ∇x Uk−1 − Uk−2 . (4.6.8) The following analysis reveals the equation satisfied by Uk : Plugging (4.6.5) into (4.6.6), we obtain U1 = U¯1 − w ⃗ · ∇x U¯0 . (4.6.9) 270 Plugging (4.6.9) into (4.6.7), we get U2 − U¯2 = −w ⃗ · ∇x (U¯1 − w ⃗ · ∇x U¯0 ) (4.6.10) = −w ⃗ · ∇x U¯1 + w ⃗ 2 ∆x U¯0 + 2wx wy ∂xy U¯0 − U0 . ⃗ ∈ S 1 , we achieve the final form Integrating (4.6.10) over w ∆x U¯0 − U¯0 = 0. (4.6.11) which further implies U0 (⃗x, w) ⃗ satisfies the equation    U0 = U¯0 , (4.6.12)   ∆x U¯0 − U¯0 = 0. In a similar fashion, U1 satisfies    U1 = U¯k − w ⃗ · ∇x U0 , ∫ (4.6.13)   ∆x U¯1 − U¯1 = − ⃗ · ∇x U0 dw. w ˜ S1 ⃗ for k ≥ 2 satisfies Also, Uk (⃗x, w)    Uk = U¯k − w ⃗ · ∇x Uk−1 − Uk−2 , ∫ ∫ (4.6.14)   ∆x Uk − Uk = − ¯ ¯ ⃗ · ∇x Uk−1 dw w ˜− Uk−2 dw. ˜ S1 S1 4.6.3 Milne Expansion In order to determine the boundary condition for Uk , it is well known that we need to define the boundary layer expansion. We still perform the substitutions (4.2.11), 271 (4.2.13) and (4.2.16) to the equation (4.6.1) to obtain the form  ∫ π  ∂uϵ ϵ ∂uϵ 1  sin(θ + ξ) − cos(θ + ξ) + (1 + ϵ )u − 2 ϵ uϵ dξ = 0, ∂η 1 − ϵη ∂θ 2π −π (4.6.15)   uϵ (0, θ, ξ) = Puϵ (0, θ) + ϵg(θ, ξ) for sin(θ + ξ) > 0, where ∫ 1 Pu (0, θ) = − ϵ uϵ (0, θ, ξ) sin(θ + ξ)dξ. (4.6.16) 2 sin(θ+ξ)<0 Define the Milne expansion of boundary layer as follows: ∑ ∞ U (η, θ, ϕ) ∼ ϵk Uk (η, θ, ϕ) (4.6.17) k=0 where Uk is defined by comparing the order of ϵ via plugging (4.6.17) into the equation (4.6.15). Thus, in a neighborhood of the boundary, we have ∂U0 sin(θ + ξ) + U0 − U¯0 = 0, (4.6.18) ∂η ∂U1 1 ∂U0 sin(θ + ξ) + U1 − U¯1 = cos(θ + ξ) , (4.6.19) ∂η 1 − ϵη ∂θ ∂U2 1 ∂U1 sin(θ + ξ) + U2 − U¯2 = cos(θ + ξ) − U0 , (4.6.20) ∂η 1 − ϵη ∂θ ... ∂Uk 1 ∂Uk−1 sin(θ + ξ) + Uk − U¯k = cos(θ + ξ) − Uk−2 , (4.6.21) ∂η 1 − ϵη ∂θ where ∫ π 1 U¯k (η, θ) = Uk (η, θ, ξ)dξ. (4.6.22) 2π −π 272 The construction of Uk and Uk in [9] can be summarized as follows: Step 1: Construction of U0 . Define ψ and ψ0 as (4.2.23) and (4.2.24). Then the zeroth order boundary layer solution can be defined as  ( )   U0 = ψ0 (ϵη) f0 (η, θ, ξ) − f0 (∞, θ) ,        sin(θ + ξ) ∂f0 + f − f¯ = 0, 0 0 ∂η (4.6.23)     f0 (0, θ, ξ) = Pf0 (0, θ) for sin(θ + ξ) > 0,      limη→∞ f0 (η, θ, ξ) = f0 (∞, θ), with Pf0 (0, θ) = 0. (4.6.24) It is easy to see U0 = f0 = 0. Step 2: Construction of U1 and U0 . Define the first order boundary layer solution as  ( )   U1 = ψ0 (ϵη) f1 (η, θ, ξ) − f1 (∞, θ) ,        sin(θ + ξ) ∂f1 + f − f¯ = cos(θ + ξ) ψ(ϵη) ∂U0 , 1 1 ∂η 1 − ϵη ∂θ (4.6.25)     f1 (0, θ, ξ) = Pf1 (0, θ) + g1 (θ, ξ) for sin(θ + ξ) > 0,      limη→∞ f1 (η, θ, ξ) = f1 (∞, θ), with Pf1 (0, θ) = 0, (4.6.26) 273 ⃗ · ∇x U0 − P(w where g1 = w ⃗ · ∇x U0 ) + g in which (⃗x0 , w) ⃗ and (0, θ, ξ) denote the same point. Since U0 = 0, we can obtain a unique f1 (η, θ, ξ) ∈ L∞ ([0, ∞) × [−π, π) × [−π, π)). Hence, U1 is well-defined. By the compatibility condition (4.6.99), we can define the zeroth order interior solution as      U0 = U¯0 ,   ∆x U¯0 − U¯0 = 0 in Ω, (4.6.27)   ∫   ∂U0 1   = g(θ, ξ) sin(θ + ξ)dξ on ∂Ω. ∂⃗n π sin(θ+ξ)>0 Step 3: Construction of U2 and U1 . Define the second order boundary layer solution as  ( )   U2 = ψ0 (ϵη) f2 (η, θ, ξ) − f2 (∞, θ) ,        sin(θ + ξ) ∂f2 + f − f¯ = cos(θ + ξ) ψ(ϵη) ∂U1 − ψ(ϵη)U0 (η, θ, ξ), 2 2 ∂η 1 − ϵη ∂θ (4.6.28)     f2 (0, θ, ξ) = Pf2 (0, θ) + g2 (θ, ξ) for sin(θ + ξ) > 0,      limη→∞ f2 (η, θ, ξ) = f2 (∞, θ), with Pf2 (0, θ) = 0. (4.6.29) ⃗ · ∇x U1 − P(w where g2 = w ⃗ · ∇x U1 ) + U0 − PU0 . By the compatibility condition (4.6.99), we define the first order interior solution as    U1 = U¯1 − w⃗ · ∇x U0 ,     ∫  ∆x U¯1 − U¯1 = − ⃗ · ∇x U0 )dw (w ˜ in Ω, (4.6.30)   S1 ∫ ∞∫ π     ∂U1 1 ψ(ϵs) ∂U1  = cos(θ + ξ) (s, θ, ξ)dξds on ∂Ω. ∂⃗n π 0 −π 1 − ϵs ∂θ 274 Step 4: Generalization to arbitrary k. Define the k th order boundary layer solution as  ( )   Uk = ψ0 (ϵη) fk (η, θ, ξ) − fk (∞, θ) ,       ∂fk ψ(ϵη) ∂Uk−1  sin(θ + ξ) + fk − f¯k = cos(θ + ξ) − ψ(ϵη)Uk−2 (η, θ, ξ), ∂η 1 − ϵη ∂θ (4.6.31)     fk (0, θ, ξ) = Pfk (0, θ) + gk (θ, ξ) for sin(θ + ξ) > 0,      limη→∞ fk (η, θ, ξ) = fk (∞, θ), with Pfk (0, θ) = 0. (4.6.32) ⃗ · ∇x Uk−1 − P(w where gk = w ⃗ · ∇x Uk−1 ) + Uk−2 − PUk−2 . By the compatibility condition (4.6.99), we can define the (k − 1)th order interior solution as (4.6.33)    Uk−1 = U¯k−1 − w⃗ · ∇x Uk−2 − Uk−3 ,     ∫ ∫       ∆x Uk−1 − Uk−1 = − ¯ ¯ ⃗ · ∇x Uk−2 )dw (w ˜− Uk−3 dw˜ in Ω,     S1 ∫ ( S1 ) ∂ U¯k−1 1 π  = w⃗ · ∇x ( w ⃗ · Uk−2 + Uk−3 ) sin ϕdϕ   ∂⃗n π −π∫ ∫ (   1 ∞ π ψ(ϵs)   ∂Uk−1   + cos(θ + ξ)   π 0 −π ) 1 − ϵs ∂θ      −ψ(ϵs)Uk−2 (s, θ, ξ)dξds on ∂Ω. In [9, pp.143], the author proved the expansion can be applied to the second order of ϵ. Based on Remark 4.6.8 to the Milne problem, in order to show the existence of a solution f2 (η, θ, ξ) ∈ L∞ ([0, ∞) × [−π, π) × [−π, π)), we at least require the source 275 term ψ ∂U1 cos(θ + ξ) ∈ L∞ ([0, ∞) × [−π, π) × [−π, π)), (4.6.34) 1 − ϵη ∂θ since U0 = 0. We thus need to require ∂(f1 − f1 (∞, θ)) ∈ L∞ ([0, ∞) × [−π, π) × [−π, π)). (4.6.35) ∂θ Since Z = ∂θ (f1 − f1 (∞)) satisfies the equation   ∂Z ∂f1  sin(θ + ξ) + Z − Z¯ = − cos(θ + ξ) , ∂η ∂η (4.6.36)   ∂g1 Z(0, θ, ξ) = PZ(0, θ) + (θ, ξ) for sin(θ + ξ) > 0, ∂θ in order for Z ∈ L∞ ([0, ∞) × [−π, π) × [−π, π)), assuming the boundary data ∂θ g1 ∈ L∞ (Γ− ), we require the source term ∂f1 − cos(θ + ξ) ∈ L∞ ([0, ∞) × [−π, π) × [−π, π)). (4.6.37) ∂η On the other hand, by Lemma B.2.1, we can show for specific g, it holds that ∂η f1 ∈ / L∞ ([0, ∞) × [−π, π) × [−π, π)). Due to the intrinsic singularity in the solution to (4.6.25), the construction in [9] breaks down. 4.6.4 ϵ-Milne Expansion with Geometric Correction In order to overcome the difficulty in estimating ψ ∂Uk cos(θ + ξ) , (4.6.38) 1 − ϵη ∂θ 276 we introduce one more substitution (4.2.37) to decompose this term and transform the equation (4.6.1) into  ( ϵ ) ∫ π  ∂uϵ ϵ ∂u ∂uϵ 1  sin ϕ − cos ϕ + + (1 + ϵ )u − 2 ϵ uϵ dϕ = 0, ∂η 1 − ϵη ∂ϕ ∂θ 2π −π (4.6.39)   uϵ (0, θ, ϕ) = Puϵ (0, θ) + ϵg(θ, ϕ) for sin ϕ > 0, with ∫ 1 Pu (0, θ) = − ϵ uϵ (0, θ, ξ) sin ϕdϕ. (4.6.40) 2 sin ϕ<0 We define the ϵ-Milne expansion of boundary layer as follows: ∑ ∞ U (η, θ, ϕ) ∼ ϵ ϵk Ukϵ (η, θ, ϕ) (4.6.41) k=0 where Ukϵ can be defined by comparing the order of ϵ via plugging (4.6.41) into the equation (4.6.39). Thus, in a neighborhood of the boundary, we have ∂U0ϵ ϵ ∂U ϵ sin ϕ − cos ϕ 0 + U0ϵ − U¯0ϵ = 0 (4.6.42) ∂η 1 − ϵη ∂ϕ ∂U1ϵ ϵ ∂U1ϵ ¯ 1 ∂U0ϵ sin ϕ − cos ϕ + U1 − U1 = ϵ ϵ cos ϕ (4.6.43) ∂η 1 − ϵη ∂ϕ 1 − ϵη ∂θ ∂U ϵ ϵ ∂U ϵ 1 ∂U ϵ sin ϕ 2 − cos ϕ 2 + U2ϵ − U¯2ϵ = cos ϕ 1 − U0ϵ (4.6.44) ∂η 1 − ϵη ∂ϕ 1 − ϵη ∂θ ... (4.6.45) ϵ ∂Ukϵ ϵ ∂Ukϵ 1 ∂Uk−1 sin ϕ − cos ϕ + Ukϵ − U¯kϵ = cos ϕ − Uk−2 ϵ ∂η 1 − ϵη ∂ϕ 1 − ϵη ∂θ where ∫ π 1 U¯kϵ (η, θ) = Ukϵ (η, θ, ϕ)dϕ. (4.6.46) 2π −π 277 We refer to the cut-off function ψ and ψ0 as (4.2.23) and (4.2.24), and define the force as (4.2.45). Define the interior expansion as follows: ∑ ∞ ⃗ ∼ U ϵ (⃗x, w) ϵk Ukϵ (⃗x, w) ⃗ (4.6.47) k=0 where Ukϵ satisfies the same equations as Uk in (4.6.12), (4.6.13) and (4.6.14). Here, to highlight its dependence on ϵ via the ϵ-Milne problem and boundary data, we add the superscript ϵ. The bridge between the interior solution and the boundary layer solution is the boundary condition of (4.6.1), so we first consider the boundary condition expansion: (U0ϵ + U0ϵ ) = P(U0ϵ + U0ϵ ), (4.6.48) (U1ϵ + U1ϵ ) = P(U1ϵ + U1ϵ ) + g, (4.6.49) (U2ϵ + U2ϵ ) = P(U2ϵ + U2ϵ ), (4.6.50) ... (Ukϵ + Ukϵ ) = P(Ukϵ + Ukϵ ). (4.6.51) Note the fact that U¯kϵ = P U¯kϵ , we can simplify above conditions as follows: U0ϵ = PU0ϵ , (4.6.52) U1ϵ = PU1ϵ + (w ⃗ · U0ϵ − P(w ⃗ · U0ϵ )) + g, (4.6.53) ⃗ · U1ϵ )) + (U0ϵ − PU0ϵ ), ⃗ · U1ϵ − P(w U2ϵ = PU2ϵ + (w (4.6.54) ... Ukϵ = PUkϵ + (w ⃗ · Uk−1 ϵ − P(w ⃗ · Uk−1 ϵ ϵ )) + (Uk−2 − PUk−2 ϵ ). (4.6.55) The construction of Ukϵ and Ukϵ are as follows: 278 Step 1: Construction of U0ϵ . Define the zeroth order boundary layer solution as  ( )   U0ϵ (η, θ, ϕ) = ψ0 (ϵη) f0 (η, θ, ϕ) − f0 (∞, θ) , ϵ ϵ       ϵ ϵ  sin ϕ ∂f0 + F (ϵ; η) cos ϕ ∂f0 + f ϵ − f¯ϵ = 0, 0 0 ∂η ∂ϕ (4.6.56)     f0ϵ (0, θ, ϕ) = Pf0ϵ (0, θ) for sin ϕ > 0,      lim f ϵ (η, θ, ϕ) = f0ϵ (∞, θ), η→∞ 0 with Pf0ϵ (0, θ) = 0. (4.6.57) By Theorem 4.6.5, U0ϵ is well-defined. It is obvious to see f0ϵ = f0ϵ (∞) = 0 is the only solution. Step 2: Construction of U1ϵ and U0ϵ . Define the first order boundary layer solution as  ( )   U1ϵ (η, θ, ϕ) = ψ0 (ϵη) f1 (η, θ, ϕ) − f1 (∞, θ) , ϵ ϵ       ∂f1ϵ ∂f ϵ ψ(ϵη) ∂U ϵ  sin ϕ + F (ϵ; η) cos ϕ 1 + f1ϵ − f¯1ϵ = cos ϕ 0 , ∂η ∂ϕ 1 − ϵη ∂θ (4.6.58)     f1ϵ (0, θ, ϕ) = Pf1ϵ (0, θ) + g1 (θ, ϕ) for sin ϕ > 0,      limη→∞ f1ϵ (η, θ, ϕ) = f1ϵ (∞, θ), with Pf1ϵ (0, θ) = 0, (4.6.59) 279 where ⃗ · ∇x U0ϵ (⃗x0 ) − P(w g1 = ( w ⃗ · ∇x U0ϵ (⃗x0 ))) + g, (4.6.60) with ⃗x0 is the same boundary point as (0, θ) and ⃗ = (− sin(ϕ − θ), − cos(ϕ − θ)), w (4.6.61) ⃗n = (cos θ, sin θ). (4.6.62) To solve (4.6.58), we require the compatibility condition (4.6.99) for the boundary data ∫ ( ) g+w ⃗ · ∇x U0 (⃗x0 ) − P(w ϵ ⃗ · ∇x U0 (⃗x0 )) sin ϕdϕ ϵ (4.6.63) sin ϕ>0 ∫ ∞∫ π ψ ∂U ϵ + e−V (s) cos ϕ 0 (s, θ, ϕ)dϕds = 0. 0 −π 1 − ϵs ∂θ Note the fact ∫ ( ) ⃗· w ∇x U0ϵ (⃗x0 ) − P(w ⃗· ∇x U0ϵ (⃗x0 )) sin ϕdϕ (4.6.64) ∫ sin ϕ>0 = ⃗ · ∇x U0ϵ (⃗x0 )) sin ϕdϕ − 2P(w (w ⃗ · ∇x U0ϵ (⃗x0 )) ∫sin ϕ>0 ∫ = ⃗ · ∇x U0 (⃗x0 )) sin ϕdϕ + (w ϵ ⃗ · ∇x U0ϵ (⃗x0 )) sin ϕdϕ (w ∫sinπ ϕ>0 sin ϕ<0 = ⃗ · ∇x U0ϵ (⃗x0 )) sin ϕdϕ (w −π ∂ U¯ ϵ (⃗x0 ) = −π∇x U¯0ϵ (⃗x0 ) · ⃗n = −π 0 . ∂⃗n 280 We can simplify the compatibility condition as follows: ∫ ∂ U¯0ϵ (⃗x0 ) g(ϕ) sin ϕdϕ − π (4.6.65) ∂⃗n ∫ ∞ ∫ π sin ϕ>0 ψ ∂U ϵ + e−V (s) cos ϕ 0 (s, θ, ϕ)dϕds = 0. 0 −π 1 − ϵs ∂θ Then we have ∂ U¯0ϵ (⃗x0 ) (4.6.66) ∫∂⃗ n ∫ ∫ 1 1 ∞ π −V (s) ψ ∂U ϵ = g(θ, ϕ) sin ϕdϕ + e cos ϕ 0 (s, θ, ϕ)dϕds π sin ϕ>0 π 0 −π 1 − ϵs ∂θ ∫ 1 = g(θ, ϕ) sin ϕdϕ. π sin ϕ>0 Hence, we define the zeroth order interior solution U0ϵ (⃗x, w) ⃗ as      U0ϵ = U¯0ϵ ,   ∆x U¯0ϵ − U¯0ϵ = 0 in Ω, (4.6.67)   ∫   ∂ U¯0ϵ 1   = g(θ, ϕ) sin ϕdϕ on ∂Ω. ∂⃗n π sin ϕ>0 Step 3: Analysis of U0ϵ and U1ϵ . By Theorem 4.6.6, we can easily see f1ϵ is well-defined in L∞ (Ω × S 1 ) and approaches f1ϵ (∞) exponentially fast as η → ∞. Since f0ϵ = 0, then if g ∈ C r (Γ− ), it is obvious to check ∂n U0ϵ ∈ C r (∂Ω). Hence, by the standard elliptic estimate, there exists a unique solution U0ϵ ∈ W r+1,p (Ω) for arbitrary p ≥ 2 satisfying ϵ ϵ ∂ U¯0 U¯ ≤ C(Ω) , (4.6.68) 0 W r+1,p (Ω) ∂⃗n r−1/p,p W (∂Ω) which implies ∇x U0ϵ ∈ W r,p (Ω), ∇x U0ϵ ∈ W r−1/p,p (∂Ω) and U0ϵ ∈ C r,1−2/p (Ω). 281 Step 4: Construction of U2ϵ and U1ϵ . Define the second order boundary layer solution as  ( )   U2ϵ (η, θ, ϕ) = ψ(ϵη) f2 (η, θ, ϕ) − f2 (∞, θ) , ϵ ϵ       ϵ ϵ ∂U ϵ  sin ϕ ∂f2 + F (ϵ; η) cos ϕ ∂f2 + f ϵ − f¯ϵ = ψ(ϵη) cos ϕ 1 − ψ(ϵη)U0ϵ , ∂η ∂ϕ 2 2 1 − ϵη ∂θ (4.6.69)     f2ϵ (0, θ, ϕ) = Pf2ϵ (0, θ) + g2 (θ, ϕ) for sin ϕ > 0,      lim f ϵ (η, θ, ϕ) = f2ϵ (∞, θ), η→∞ 2 with Pf2ϵ (0, θ) = 0, (4.6.70) where ⃗ · ∇x U1ϵ (⃗x0 ) − P(w g2 = (w ⃗ · ∇x U1ϵ (⃗x0 ))) + U0ϵ (⃗x0 ) − PU0ϵ (⃗x0 ). (4.6.71) In order for equation (4.6.69) being well-posed, we require the compatibility condition (4.6.99) for the boundary data ∫ ( ) ⃗· w ∇x U1ϵ (⃗x0 ) − P(w ⃗· sin ϕdϕ ∇x U1ϵ (⃗x0 ))) (4.6.72) sin ϕ>0 ∫ ∞∫ π ( ) −V (s) ψ ∂U1ϵ + e cos ϕ (s, θ, ϕ) − ψU0 (s, θ, ϕ) dϕds = 0. ϵ 0 −π 1 − ϵs ∂θ Similarly, we can directly verify the relation ∫ ( ) ∂ U¯1ϵ (⃗x0 ) ⃗· w ∇x U1ϵ (⃗x0 ) − P(w ⃗· ∇x U1ϵ (⃗x0 )) sin ϕdϕ = −π . (4.6.73) sin ϕ>0 ∂⃗n 282 We can simplify the compatibility condition as follows: (4.6.74) ∫ ∞∫ ( ) ∂ U¯1ϵ (⃗x0 ) π ψ ∂U ϵ −π + e−V (s) cos ϕ 1 (s, θ, ϕ) − ψU0ϵ (s, θ, ϕ) dϕds = 0. ∂⃗n 0 −π 1 − ϵs ∂θ Then we have (4.6.75) ∫ ∞ ∫ ( ) ∂ U¯1ϵ (⃗x0 ) 1 π ψ ∂U ϵ −π = e−V (s) cos ϕ 1 (s, θ, ϕ) − ψU0ϵ (s, θ, ϕ) dϕds ∂⃗n π 1 − ϵs ∂θ ∫0 η ∫ −π π 1 ψ ∂U ϵ = e−V (s) cos ϕ 1 (s, θ, ϕ)dϕds. π 0 −π 1 − ϵs ∂θ Hence, we define the first order interior solution U1ϵ (⃗x) as    U1ϵ = U¯1ϵ − w ⃗ · ∇x U0ϵ ,   ∫   ∆x U¯1ϵ − U¯1ϵ = − ⃗ · ∇x U0ϵ )dw (w ˜ in Ω, (4.6.76)   ∫ S 1 ∫   ∂ U¯1ϵ 1 ∞ π −V (s) ψ(ϵs) ∂U ϵ   = e cos ϕ 1 (s, θ, ϕ)dϕds on ∂Ω. ∂⃗n π 0 −π 1 − ϵs ∂θ Step 5: Analysis of U1ϵ and U2ϵ . By Theorem 4.6.6, we can easily see f2ϵ is well-defined in L∞ (Ω × S 1 ) and approaches f2ϵ (∞) exponentially fast as η → ∞. By above analysis, it is obvious to check ∂n U¯1ϵ ∈ C r−1 (∂Ω). Hence, by the standard elliptic estimate, there exists a unique solution U¯1ϵ ∈ W r,p (Ω) for arbitrary p ≥ 2 satisfying ( ¯ϵ ) ϵ U¯1 r,p ≤ C(Ω) ∂ U1 + ∥∇x U0 ∥W r−2,p (Ω) ϵ (4.6.77) W (Ω) ∂⃗n r−1−1/p,p W (∂Ω) which implies ∇x U1ϵ ∈ W r−1,p (Ω), ∇x U1ϵ ∈ W r−1−1/p,p (∂Ω) and U1ϵ ∈ C r−1,1−2/p (Ω). 283 Step 6: Generalization to arbitrary k. Then this process can proceed to arbitrary k as long as g is sufficiently smooth. We can always determine Ukϵ and Uk−1 ϵ simultaneously based on the compatibility condition. Define the k th order boundary layer solution as  ( )   Ukϵ (η, θ, ϕ) = ψ(ϵη) fk (η, θ, ϕ) − fk (∞, θ) , ϵ ϵ       ∂fkϵ ∂fkϵ ψ ϵ ∂Uk−1  sin ϕ + F (ϵ; η) cos ϕ ¯ + fk − fk = ϵ ϵ cos ϕ − ψUk−2 ϵ , ∂η ∂ϕ 1 − ϵη ∂θ (4.6.78)     fkϵ (0, θ, ϕ) = Pfkϵ (0, θ) + gk (θ, ϕ) for sin ϕ > 0,      limη→∞ fkϵ (η, θ, ϕ) = fkϵ (∞, θ), with Pfkϵ (0, θ) = 0, (4.6.79) where ⃗ · ∇x Uk−1 gk = ( w ϵ (⃗x0 ) − P(w ⃗ · ∇x Uk−1 ϵ ϵ (⃗x0 ))) + Uk−2 (⃗x0 ) − PUk−2 ϵ (⃗x0 ). (4.6.80) Hence, we define the (k − 1)th order interior solution as    = U¯k−1 −w ⃗ · ∇x Uk−2 − Uk−3   ϵ Uk−1 ϵ ϵ ϵ ,   ∫ ∫       ∆x Uk−1 − Uk−1 = − ¯ ϵ ¯ ϵ ⃗ · ∇x Uk−2 )dw (w ϵ ˜− ϵ Uk−3 dw ˜ in Ω,     S ∫ ( 1 S 1 ) ∂ U¯k−1 ϵ 1 π (4.6.81)  = w⃗ · ∇x ( w⃗ · Uk−2 + Uk−3 ) sin ϕdϕ ϵ ϵ   ∂⃗n π −π∫ ∫ (   1 ∞ π −V (s) ψ(ϵs) ϵ   ∂Uk−1   + e cos ϕ   π 0 −π ) 1 − ϵs ∂θ      −ψ(ϵs)Uk−2 ϵ (s, θ, ϕ)dϕds on ∂Ω. In particular, for g ∈ C k+1 (Γ− ), we can define the interior solution up to k th order and the boundary layer solution up to (k + 1)th order, i.e. up to Ukϵ and Uk+1 ϵ . 284 4.6.5 Well-Posedness of Steady Neutron Transport Equation In this section, we consider the well-posedness of the steady neutron transport equa- tion      ⃗ · ∇x u + (1 + ϵ2 )u − u¯ = f (⃗x, w) ϵw ⃗ in Ω,  ⃗ = Pu(⃗x0 ) + ϵg(⃗x0 , w) u(⃗x0 , w) ⃗ (4.6.82)      for w ⃗ · ⃗n < 0 and ⃗x0 ∈ ∂Ω. We define the L∞ norms in Ω × S 1 as usual: ∥f ∥L∞ (Ω×S 1 ) = sup |f (⃗x, w)| ⃗ . (4.6.83) (⃗ x,w)∈Ω×S ⃗ 1 ⃗ ∈ L∞ (Ω × S 1 ) and g(x0 , w) Theorem 4.6.1. Assume f (⃗x, w) ⃗ ∈ L∞ (Γ− ). Then for the transport equation      ⃗ · ∇x u + (1 + ϵ2 )u − u¯ = f (⃗x, w) ϵw ⃗ in Ω,  ⃗ = Pu(⃗x0 ) + ϵg(⃗x0 , w) u(⃗x0 , w) ⃗ (4.6.84)      for ⃗x0 ∈ ∂Ω and w ⃗ · ⃗n < 0, ⃗ ∈ L∞ (Ω × S 1 ) satisfying there exists a unique solution u(⃗x, w) ( ) 1 1 ∥u∥L∞ (Ω×S 1 ) ≤ C 2 ∥f ∥L∞ (Ω×S 1 ) + 2 ∥g∥L∞ (Γ− ) . (4.6.85) ϵ ϵ Proof. We iteratively construct an approximating sequence {uk }∞ 0 k=0 where u = 0 and      ⃗ · ∇x ukλ + (1 + ϵ2 )uk − u¯k−1 = f (⃗x, w) ϵw ⃗ in Ω,  ⃗ = Puk−1 (⃗x0 ) + ϵg(⃗x0 , w) uk (⃗x0 , w) ⃗ (4.6.86)      for ⃗x0 ∈ ∂Ω and w ⃗ · ⃗n < 0. 285 By Lemma 4.3.1, this sequence is well-defined and uk L∞ (Ω×S 1 ) < ∞. We rewrite equation (4.6.86) along the characteristics as ⃗ −(1+ϵ )tb 2 ⃗ = (ϵg + Puk−1 )(⃗x − ϵtb w, uk (⃗x, w) ⃗ w)e (4.6.87) ∫ tb ⃗ −(1+ϵ )(tb −s) ds. 2 + (f + u¯k−1 )(⃗x − ϵ(tb − s)w, ⃗ w)e 0 where the backward exit time tb is defined as (4.3.13). We define the difference v k = uk − uk−1 for k ≥ 1. Then v k satisfies (4.6.88) ∫ tb ⃗ −(1+ϵ ⃗ −(1+ϵ 2 )t 2 )(t −s) ⃗ = Pv k (⃗x − ϵtb w, v k+1 (⃗x, w) ⃗ w)e b + v¯k (⃗x − ϵ(tb − s)w, ⃗ w)e b ds. 0 Since v¯k L∞ (Ω×S 1 ) ≤ v k L∞ (Ω×S 1 ) and Pv k L∞ (Ω×S 1 ) ≤ v k L∞ (Ω×S 1 ) , we can directly estimate (4.6.89) ∫ k+1 tb v ∞ ≤ e−(1+ϵ )tb v k+1 L∞ (Ω×S 1 ) + v k L∞ (Ω×S 1 ) 2 e−(1+ϵ 2 )(t −s) b ds L (Ω×S 1 ) 0 k+1 1 ≤ e−(1+ϵ 2 )t b v ∞ + (1 − e −(1+ϵ2 )tb k ) v L∞ (Ω×S 1 ) . L (Ω×S 1 ) 1 + ϵ2 Hence, we naturally have k+1 1 v ≤ v k ∞ . (4.6.90) L∞ (Ω×S 1 ) 1 + ϵ2 L (Ω×S 1 ) Thus, this is a contraction iteration. Considering v 1 = u1 , we have ( )k−1 k 1 1 v ∞ ≤ u ∞ . (4.6.91) L (Ω×S 1 ) 1 + ϵ2 L (Ω×S 1 ) 286 for k ≥ 1. Therefore, uk converges strongly in L∞ to the limiting solution u satisfying (4.6.92) ∑ ∞ k 1 + ϵ2 2 ∥u∥L∞ (Ω×S 1 ) ≤ v ∞ ≤ u1 ∞ 1) ≤ u1 ∞ . L (Ω×S 1 ) ϵ2 L (Ω×S ϵ2 L (Ω×S 1 ) k=1 Since u1 satisfies the equation ∫ tb −(1+ϵ2 )tb ⃗ −(1+ϵ b −s) 2 )(t 1 ⃗ = g(⃗x − ϵtb w, u (⃗x, w) ⃗ w)e ⃗ + f (⃗x − ϵ(tb − s)w, ⃗ w)e ds. 0 Based on Lemma 4.3.1, we can directly estimate 1 u ∞ ≤ ∥f ∥L∞ (Ω×S 1 ) + ∥g∥L∞ (Γ− ) . (4.6.93) L (Ω×S 1 ) Combining (4.6.92) and (4.6.93), we can naturally obtain the existence and estimates. Also, the uniqueness easily follows from the energy estimates. ⃗ ∈ L∞ (Γ− ). Then for the steady neutron transport Theorem 4.6.2. Assume g(x0 , w) ⃗ ∈ L∞ (Ω × S 1 ) satisfying equation (4.6.1), there exists a unique solution uϵ (⃗x, w) C ∥uϵ ∥L∞ (Ω×S 1 ) ≤ ∥g∥L∞ (Γ− ) . (4.6.94) ϵ2 Proof. We can apply Theorem 4.6.1 to the equation (4.6.1). The result naturally follows. 287 4.6.6 ϵ-Milne Problem We consider the ϵ-Milne problem for f ϵ (η, θ, ϕ) in the domain (η, θ, ϕ) ∈ [0, ∞) × [−π, π) × [−π, π) (4.6.95)  ϵ ϵ  ∂f ∂f   sin ϕ + F (ϵ; η) cos ϕ + f ϵ − f¯ϵ = S ϵ (η, θ, ϕ),   ∂η ∂ϕ  f ϵ (0, θ, ϕ) = hϵ (θ, ϕ) + Pf ϵ (0) f or sin ϕ > 0,     limη→∞ f ϵ (η, θ, ϕ) = f∞ ϵ (θ), where ∫ 1 Pf (η, θ) = − ϵ f ϵ (η, θ, ϕ) sin ϕdϕ, (4.6.96) 2 sin ϕ<0 F (ϵ; η) is defined as (4.2.45), |hϵ (θ, ϕ)| ≤ M, (4.6.97) and |S ϵ (η, θ, ϕ)| ≤ M e−Kη , (4.6.98) for M and K uniform in ϵ and θ. For notational simplicity, we omit ϵ and θ dependence in f ϵ in this section. The same convention also applies to F (ϵ; η), V (ϵ; η), S ϵ (η, θ, ϕ) and hϵ (θ, ϕ). However, our estimates are independent of ϵ and θ. 288 Compatibility Condition Lemma 4.6.3. In order for the equation (4.6.95) to have a solution f ∈ L∞ ([0, ∞) × [−π, π)), the boundary data h and the source term S must satisfy the compatibility condition ∫ ∫ ∞ ∫ π h(ϕ) sin ϕdϕ + e−V (s) S(s, ϕ)dϕds = 0. (4.6.99) sin ϕ>0 0 −π In particular, if S = 0, then the compatibility condition reduces to ∫ h(ϕ) sin ϕdϕ = 0. (4.6.100) sin ϕ>0 Proof. We just integrate over ϕ ∈ [−π, π) on both sides of (4.6.95) and integrate by parts to achieve ∫ π ∫ π ∫ π d f (η, ϕ) sin ϕdϕ + F (η) f (η, ϕ) sin ϕdϕ = S(η, ϕ)dϕ. (4.6.101) dη −π −π −π ∫π This is a first order ordinary differential equation for −π f (η, ϕ) sin ϕdϕ. The initial data is ∫ π ∫ ∫ f (0, ϕ) sin ϕdϕ = f (0, ϕ) sin ϕdϕ + f (0, ϕ) sin ϕdϕ (4.6.102) −π sin ϕ>0 sin ϕ<0 ∫ ( ) ∫ = h(ϕ) + Pf (0) sin ϕdϕ + f (0, ϕ) sin ϕdϕ ∫sin ϕ>0 ∫ sin ϕ<0 = h(ϕ) sin ϕdϕ + Pf (0) sin ϕdϕ − 2Pf (0) ∫sin ϕ>0 sin ϕ>0 = h(ϕ) sin ϕdϕ. sin ϕ>0 289 Hence, we can directly solve (4.6.101) as (4.6.103) ∫ π (∫ ∫ η∫ π ) −V (s) f (η, ϕ) sin ϕdϕ = eV (η) h(ϕ) sin ϕdϕ + e S(s, ϕ)dϕds . −π sin ϕ>0 0 −π The problem (4.6.95) requires f (η, ϕ) → f∞ as η → ∞. Hence, we must have ∫ π lim f (η, ϕ) sin ϕdϕ = 0. (4.6.104) η→∞ −π Therefore, the only possibility to justify above requirement is ∫ ∫ ∞ ∫ π h(ϕ) sin ϕdϕ + e−V (s) S(s, ϕ)dϕds = 0. (4.6.105) sin ϕ>0 0 −π This is the desired compatibility condition. If S = 0, then above condition reduces to ∫ h(ϕ) sin ϕdϕ = 0. (4.6.106) sin ϕ>0 Reduction to In-flow ϵ-Milne Problem It is easy to see if f is a solution to (4.6.95), then f + C is also a solution for any constant C. Hence, in order to obtain a unique solution, we need a normalization condition Pf (0) = 0. (4.6.107) 290 The following lemma tells us the problem (4.6.95) can be reduced to the ϵ-Milne problem (4.4.1) with in-flow boundary. Lemma 4.6.4. If the boundary data h and S satisfy the compatibility condition (4.6.99), then the solution f to the ϵ-Milne problem (4.4.1) with in-flow boundary as f = h on sin ϕ > 0 is also a solution to the ϵ-Milne problem (4.6.95) with dif- fusive boundary, which satisfies the normalization condition (4.6.107). Furthermore, this is the unique solution to (4.6.95) among the functions satisfying (4.6.107) and ∥f − f∞ ∥L2 L2 < ∞. Proof. Consider f satisfies the ϵ-Milne problem with in-flow boundary as follows:   ∂f ∂f   sin ϕ + F (η) cos ϕ + f − f¯ = S(η, ϕ),   ∂η ∂ϕ f (0, ϕ) = h(ϕ) for sin ϕ > 0, (4.6.108)      limη→∞ f (η, ϕ) = f∞ . Then there exists a f∞ such that ∥f − f∞ ∥L∞ L∞ < ∞ and f decays to f∞ exponen- tially. Therefore, z = f − f∞ satisfies the equation   ∂z ∂z   sin ϕ + F (η) cos ϕ + z − z¯ = S(η, ϕ),   ∂η ∂ϕ z(0, ϕ) = h(ϕ) − f∞ for sin ϕ > 0, (4.6.109)      limη→∞ z(η, ϕ) = 0. Multiplying e−V (η) on both sides of (4.6.109) and integrating over ϕ ∈ [−π, π) imply ∫ π ∫ π d −V (η) e z(η, ϕ) sin ϕdϕ = e−V (η) S(η, ϕ)dϕ. (4.6.110) dη −π −π Since z decays exponentially with respect to η, we know z ∈ L1 ([0, ∞) × [−π, π)). 291 Hence, we can integrate (4.6.110) over η ∈ [0, ∞) to obtain ∫ π ∫ ∞ ∫ π −V (0) 0− e z(0, ϕ) sin ϕdϕ = e−V (s) S(s, ϕ)dϕds, (4.6.111) −π 0 −π which further implies ∫ π ∫ ∞ ∫ π z(0, ϕ) sin ϕdϕ + e−V (s) S(s, ϕ)dϕds = 0. (4.6.112) −π 0 −π Then we may compute (4.6.113) ∫ π ∫ ∫ z(0, ϕ) sin ϕdϕ = z(0, ϕ) sin ϕdϕ + z(0, ϕ) sin ϕdϕ −π ∫sin ϕ>0 sin ϕ<0 = (h(ϕ) − f∞ ) sin ϕdϕ − 2Pz(0) sin ϕ>0 ∫ ( ) = h(ϕ) sin ϕdϕ − 2 Pz(0) + f∞ . sin ϕ>0 Combining (4.6.112), (4.6.113) and the compatibility condition (4.6.99), we have Pz(0) + f∞ = 0. (4.6.114) Since Pz(0) = Pf (0) − Pf∞ (0) = Pf (0) − f∞ . (4.6.115) naturally, we have Pf (0) = 0. Hence, f is a solution to the ϵ-Milne problem (4.6.95) with the normalization condition (4.6.107). By Cauchy’s inequality, we can deduce the fact that ∥f − f∞ ∥L2 L2 < ∞ implies f¯ − f∞ L2 L2 < ∞. Then by a similar argument as Step 3 in the proof of Lemma 4.4.4, we can show the uniqueness. 292 In summary, based on above analysis, we can utilize the known result for ϵ-Milne problem with in-flow boundary to obtain the well-posedness, decay and maximum principle of the solution to the ϵ-Milne problem (4.6.95). Theorem 4.6.5. There exists a unique solution f (η, ϕ) to the ϵ-Milne problem (4.6.95) with the normalization condition (4.6.107) satisfying ( ) M ∥f − f∞ ∥L∞ L∞ ≤C 1+M + . (4.6.116) K Theorem 4.6.6. For K0 > 0 sufficiently small, the solution f (η, ϕ) to the ϵ-Milne problem (4.6.95) with the normalization condition (4.6.107) satisfies ( ) Kη M e 0 (f − f∞ ) ∞ ∞ ≤ C 1 + M + . (4.6.117) L L K Theorem 4.6.7. The solution to the ϵ-Milne problem (4.6.95) with S = 0 and the normalization condition (4.6.107) satisfies the maximum principle, i.e. min h(ϕ) ≤ f (η, ϕ) ≤ max h(ϕ). (4.6.118) sin ϕ>0 sin ϕ>0 Remark 4.6.8. Note that when F = 0, then all the previous proofs can be recovered and Theorem 4.6.5, Theorem 4.6.6 and Theorem 4.6.7 still hold. Hence, we can obtain the well-posedness, decay and maximum principle of the classical Milne problem   ∂f   sin ϕ + f − f¯ = S(η, ϕ),   ∂η f (0, ϕ) = Pf (0) + h(ϕ) for sin ϕ > 0, (4.6.119)      lim η→∞ f (η, ϕ) = f∞ , with the normalization condition (4.6.107). Note that now we always have V (η) = 0. 293 4.6.7 Main Results ⃗ ∈ C 3 (Γ− ). Then for the steady neutron transport Theorem 4.6.9. Assume g(⃗x0 , w) ⃗ ∈ L∞ (Ω × S 1 ) satisfies equation (4.6.1), the unique solution uϵ (⃗x, w) ∥uϵ − U0ϵ − U0ϵ ∥L∞ = O(ϵ), (4.6.120) Moreover, if g(θ, ϕ) = cos ϕ, then there exists a C > 0 such that ∥uϵ − U0 − U0 − ϵU1 − ϵU1 ∥L∞ ≥ Cϵ > 0, (4.6.121) when ϵ is sufficiently small. Proof. We can divide the proof into several steps: Step 1: Remainder definitions. Note that the boundary layer solution depends on ϵ due to the force and the inte- rior solution also depends on ϵ due the boundary condition. We may rewrite the asymptotic expansion as follows: ∑ ∞ ∑ ∞ ϵ u ∼ k ϵ Ukϵ + ϵk Ukϵ . (4.6.122) k=0 k=0 The remainder can be defined as ∑ N ∑ N RN = u − ϵ ϵ k Ukϵ − ϵk Ukϵ = u − QN − QN , (4.6.123) k=0 k=0 294 where ∑ N QN = ϵk Ukϵ , (4.6.124) k=0 ∑N QN = ϵk Ukϵ . (4.6.125) k=0 Noting the equation (4.6.39) is equivalent to the equation (4.6.1), we use L to denote the neutron transport operator as follows: Lu = ϵw ⃗ · ∇x u + (1 + ϵ2 )u − u¯ (4.6.126) ( ) ∂u ϵ ∂u ∂u = sin ϕ − cos ϕ + + (1 + ϵ2 )u − u¯ ∂η 1 − ϵη ∂ϕ ∂θ Step 2: Estimates of LQN . The interior contribution can be estimated as LQ0 = ϵw ⃗ · ∇x Q0 + (1 + ϵ2 )Q0 − Q ¯0 (4.6.127) = (Q0 − Q ⃗ · ∇x U0ϵ + ϵ2 U0ϵ = ϵw ¯ 0 ) + ϵw ⃗ · ∇x U0ϵ + ϵ2 U0ϵ . We can directly estimate |ϵw ⃗ · ∇x U0ϵ | ≤ Cϵ |∇x U0ϵ | ≤ Cϵ, (4.6.128) 2 ϵ ϵ U0 ≤ Cϵ2 |U0ϵ | ≤ Cϵ2 . (4.6.129) This implies |LQ0 | ≤ Cϵ. (4.6.130) 295 For higher order term, we can estimate (4.6.131) LQN = ϵw ⃗ · ∇x QN + (1 + ϵ2 )QN − Q ⃗ · ∇x UNϵ + ϵN +2 UNϵ + ϵN +1 UNϵ −1 . ¯ N = ϵN +1 w We have N +1 ϵ ⃗ · ∇x UNϵ ≤ CϵN +1 |∇x UNϵ | ≤ CϵN +1 , w (4.6.132) N +2 ϵ ϵ UN + ϵN +1 UNϵ −1 ≤ CϵN +2 |UNϵ | + CϵN +1 UNϵ −1 ≤ CϵN +1 . (4.6.133) This implies |LQN | ≤ CϵN +1 . (4.6.134) Step 3: Estimates of LQN . The boundary layer solution is Ukϵ = (fkϵ − fkϵ (∞)) · ψ0 = Vk ψ0 where fkϵ (η, θ, ϕ) solves the ϵ-Milne problem and Vk = fkϵ − fkϵ (∞). Notice ψ0 ψ = ψ0 , so the boundary layer 296 contribution can be estimated as LQ0 (4.6.135) ( ) ∂Q0 ϵ ∂Q0 ∂Q0 = sin ϕ − cos ϕ + + (1 + ϵ2 )Q0 − Q¯0 ∂η 1 − ϵη ∂ϕ ∂θ ( ) ( ) ∂V0 ∂ψ0 ψ0 ϵ ∂V0 ∂V0 = sin ϕ ψ0 + V0 − cos ϕ + ∂η ∂η 1 − ϵη ∂ϕ ∂θ +(1 + ϵ2 )ψ0 V0 − ψ0 V¯0 ( ) ( ) ∂V0 ∂ψ0 ψ0 ψϵ ∂V0 ∂V0 = sin ϕ ψ0 + V0 − cos ϕ + ∂η ∂η 1 − ϵη ∂ϕ ∂θ +(1 + ϵ )ψ0 V0 − ψ0 V¯0 2 ( ) ∂V0 ϵψ ∂V0 ¯ ∂ψ0 = ψ0 sin ϕ − cos ϕ + V0 − V0 + sin ϕ V0 ∂η 1 − ϵη ∂ϕ ∂η ψ0 ϵ ∂V0 − cos ϕ + ϵ2 ψ0 V0 1 − ϵη ∂θ ∂ψ0 ψ0 ϵ ∂V0 = sin ϕ V0 − cos ϕ + ϵ2 ψ0 V0 . ∂η 1 − ϵη ∂θ Since ψ0 = 1 when η ≤ 1/(4ϵ), the effective region of ∂η ψ0 is η ≥ 1/(4ϵ) which is further and further from the origin as ϵ → 0. By Theorem 4.6.6, the first term in (4.6.135) can be controlled as ∂ψ sin ϕ ≤ Ce− ϵ0 ≤ Cϵ. 0 K V (4.6.136) ∂η 0 For the second term in (4.6.135), we have ψ0 ϵ ∂V ∂V0 − 0 1 − ϵη cos ϕ ∂θ ≤ Cϵ ∂θ ≤ Cϵ. (4.6.137) For the third term in (4.6.135), we have 2 ϵ ψ0 V0 ≤ Cϵ. (4.6.138) 297 This implies |LQ0 | ≤ Cϵ. (4.6.139) For higher order term, we can estimate (4.6.140) ( ) ∂QN ϵ ∂QN ∂QN LQN = sin ϕ − cos ϕ + + (1 + ϵ2 )QN − Q¯N ∂η 1 − ϵη ∂ϕ ∂θ ∑ N ∂ψ0 ψ0 ϵN +1 ∂VN = ϵi sin ϕ Vi − cos ϕ + ϵN +2 ψ0 VN + ϵN +1 ψ0 VN −1 . i=0 ∂η 1 − ϵη ∂θ Away from the origin, the first term in (4.6.140) can be controlled as N ∑ ∂ψ K0 Vi ≤ Ce− ϵ ≤ CϵN +1 . 0 i ϵ sin ϕ (4.6.141) ∂η i=0 For the second term in (4.6.140), we have ψ0 ϵN +1 ∂VN N +1 ∂VN − ≤ ∂θ ≤ Cϵ N +1 1 − ϵη cos ϕ Cϵ . (4.6.142) ∂θ For the third term in (4.6.140), we have N +2 ϵ ψ0 VN + ϵN +1 ψ0 VN −1 ≤ CϵN +1 . (4.6.143) This implies |LQN | ≤ CϵN +1 . (4.6.144) Step 4: Proof of (4.6.120). 298 In summary, since Luϵ = 0, collecting (4.6.123), (4.6.134) and (4.6.144), we can prove |LRN | ≤ CϵN +1 . (4.6.145) Consider the asymptotic expansion to N = 2, then the remainder R2 satisfies the equation    ϵw ⃗ · ∇x R2 + R2 − R ¯ 2 = LR2 for ⃗x ∈ Ω, (4.6.146)   R2 = PR2 for w ⃗ · ⃗n < 0 and ⃗x0 ∈ ∂Ω. By Theorem 4.6.1, we have C Cϵ3 ∥R2 ∥L∞ (Ω×S 1 ) ≤ ∥LR2 ∥ ∞ 1 L (Ω×S ) ≤ ≤ Cϵ. (4.6.147) ϵ2 ϵ2 Hence, we have ∑ 2 ∑ 2 ϵ ϵ u − ϵ Uk − k ϵ ϵ Uk k = O(ϵ). (4.6.148) k=0 k=0 L∞ Since it is easy to see ∑2 ∑ 2 ϵ k ϵ ϵ Uk + ϵ Uk k ≤ Cϵ, (4.6.149) k=1 k=1 L∞ our result naturally follows. This completes the proof of (4.6.120). Step 5: Proof of (4.6.121). 299 By (4.6.25), the solution f1 satisfies the Milne problem   ∂f1   sin(θ + ξ) + f1 − f¯1 = 0,   ∂η  f1 (0, θ, ξ) = Pf1 (0, θ) + g1 (θ, ξ) for sin(θ + ξ) > 0, (4.6.150)     limη→∞ f1 (η, θ, ξ) = f1 (∞, θ). For convenience of comparison, we make the substitution ϕ = θ + ξ to obtain   ∂f1   sin ϕ + f1 − f¯1 = 0,   ∂η f1 (0, ϕ) = Pf1 (0) + g1 (θ, ϕ) for sin ϕ > 0, (4.6.151)      lim η→∞ f1 (η, θ, ξ) = f1 (∞, θ). Assume (4.6.121) is incorrect. For our g(ϕ) = cos ϕ which is independent of θ, since U0 = U0ϵ = 0 and U0 = U0ϵ = 0, we have lim ∥(U1 + U1 ) − (U1ϵ + U1ϵ )∥L∞ = 0. (4.6.152) ϵ→0 Since now U1 and U1ϵ are independent of θ, by (4.6.30) and (4.6.76), we can directly estimate (4.6.153) ∫ ∞∫ ( ) ∂ U¯1ϵ π ψ(ϵs) ∂U ϵ = − e−V (s) cos ϕ 1 (s, ϕ) − ψ(ϵs)U0ϵ (s, ϕ) dϕds = 0, ∂⃗n 0 −π 1 − ϵη ∂θ and also (4.6.154) ∫ ∞ ∫ ( ) ∂ U¯1 π ψ(ϵs) ∂U1 = − e−V (s) cos ϕ (s, ϕ) − ψ(ϵs)U0 (s, ϕ) dϕds = 0. ∂⃗n 0 −π 1 − ϵη ∂θ Hence, we have U¯1ϵ = U¯1 in the domain, which further implies U1ϵ = U1 . Therefore, 300 we can obtain lim ∥U1 − U1ϵ ∥L∞ = 0. (4.6.155) ϵ→0 Then on the boundary of sin ϕ > 0, these two boundary layer solutions satisfy U1 = g − f1 (∞), (4.6.156) U1ϵ = g − f1ϵ (∞). (4.6.157) Naturally, we have the estimate limϵ→0 ∥f1ϵ (∞) − f1 (∞)∥L∞ = 0 based on above as- sumptions. Hence, we may further derive lim ∥(f1 (∞) + U1 ) − (f1ϵ (∞) + U1ϵ )∥L∞ = 0. (4.6.158) ϵ→∞ For 0 ≤ η ≤ 1/(2ϵ), we have ψ0 = 1, which means f1 = U1 + f1 (∞) and f1ϵ = U1ϵ + f1ϵ (∞) on [0, 1/(2ϵ)]. Since we have PU1ϵ (0) = −f1ϵ (∞) and PU1 (0) = −f1 (∞), we have recovered the normalization condition, i.e. Pf1 (0) = Pf1ϵ (0) = 0. Note that g satisfies the compatibility condition (4.6.100). Therefore, the ϵ-Milne problem satisfied by f1 and f1ϵ can be reduced to the ϵ-Milne problem with in-flow boundary. Hence, we can naturally obtain the desired result through the proof of Theorem 4.1.1. Chapter Five Future Work 302 In this dissertation, we mainly talk about the viscous surface flow and diffusive limit in neutron transport equation. In the future, we would like to extend these results to more general settings and get a deeper understanding of these phenomena. For the viscous surface wave, in the domain with infinite cross-section and curved bottom, although the local well-posedness has been proved in Chapter 2, the global well-posedness is still open. The main difficulty is the curved bottom will absorb the derivatives aiming at interpolation estimates with minimal counts and lower the decaying rate. This requires us to delicately redesign the energy and dissipation and improve interpolation exponent. On the other hand, we can also consider the problem in more general models, either in an infinite-depth domain, or with surface tension involved. For the hydrodynamic limit, although we disprove the classical boundary layer theory, the current technique can only tackle the two-dimensional steady problem in a unit plate. There are at least three directions to extend this result: two-dimensional steady problem in general smooth convex domain, which involves more complicated influence due to non-constant boundary curvature and requires further decomposition of the tangential component; three dimensional steady problem in a unit ball, which includes more tangential components in the Milne problem and natural singularity due to coordinate substitution; one-dimensional unsteady problem with general initial and boundary data, which requires a composite of initial layer, boundary layer, and initial-boundary layer whose analysis involves evolutionary Milne problem. Further, if we extend the result from neutron transport equation to Boltzmann equation, a new type of difficulty arising is the regularity requirement. Since we do not general know whether there exist smooth solutions to fluid equations and kinetic equations, the analysis of boundary effect has to be in a new framework with weaker assumptions. 303 As for the numerical analysis, it is a classical problem to treat multi-scale problem in simulation of kinetic equations. We intend to utilize the asymptotic-preserving scheme to construct a uniform scheme for these two domains. The main difficulty is the analyze of discrete boundary layer, especially discrete Milne problem. This requires us to recover the results for continuous Milne problem in the discrete settings with weaker assumptions. Appendix A Related Estimates in Viscous Surface Wave 305 A.1 Analytic Tools A.1.1 Products in Sobolev Space We will need some estimates of the products of functions in Sobolev spaces. Since these results have been proved in lemma A.1 and lemma A.2 of [?], we will present the statement of the lemmas here without proof. Lemma A.1.1. Let U denote either Σ or Ω. 1. Let 0 ≤ r ≤ s1 ≤ s2 be such that s1 > n/2. Let f ∈ H s1 (U ), g ∈ H s2 (U ). Then f g ∈ H r (U ) and ∥f g∥H r . ∥f ∥H s1 ∥g∥H s2 . (A.1.1) 2. Let 0 ≤ r ≤ s1 ≤ s2 be such that s2 > r + n/2. Let f ∈ H s1 (U ), g ∈ H s2 (U ). Then f g ∈ H r (U ) and ∥f g∥H r . ∥f ∥H s1 ∥g∥H s2 . (A.1.2) 3. Let 0 ≤ r ≤ s1 ≤ s2 be such that s2 > r + n/2. Let f ∈ H −r (Σ), g ∈ H s2 (Σ). Then f g ∈ H −s1 (Σ) and ∥f g∥H −s1 . ∥f ∥H −r ∥g∥H s2 . (A.1.3) Lemma A.1.2. Suppose that f ∈ C 1 (Σ) and g ∈ H 1/2 (Σ). Then f g ∈ H 1/2 (Σ) and ∥f g∥H 1/2 . ∥f ∥C 1 ∥g∥H 1/2 . (A.1.4) 306 A.1.2 Poincare-Type Inequality We need several Poincare-type inequality in Ω. Since all these lemmas have be proved in lemma A.10-A.13 in [25], we will only give the statement here without proof. Lemma A.1.3. It holds that ∥f ∥2L2 (Ω) . ∥f ∥2L2 (Σ) + ∥∂3 f ∥2L2 (Ω) , (A.1.5) for all f ∈ H 1 (Ω). Also, if f ∈ W 1,∞ (Ω), then ∥f ∥2L∞ (Ω) . ∥f ∥2L∞ (Σ) + ∥∂3 f ∥2L∞ (Ω) . (A.1.6) Lemma A.1.4. It holds that ∥f ∥H 0 (Σ) . ∥∂3 f ∥H 0 for f ∈ H 1 (Ω) such that f = 0 on Σb . It also holds that ∥f ∥L∞ (Σ) . ∥∂3 f ∥L∞ (Ω) for f ∈ W 1,∞ (Ω) such that f = 0 on Σb . Lemma A.1.5. It holds that ∥u∥H 1 . ∥Du∥H 0 for all u ∈ H 1 (Ω; R3 ) such that f = 0 on Σb . Lemma A.1.6. It holds that ∥f ∥H 1 . ∥∇f ∥H 0 for all f ∈ H 1 (Ω) such that f = 0 on Σb . Also, ∥f ∥W 1,∞ (Ω) . ∥∇f ∥L∞ (Ω) for all f ∈ W 1,∞ (Ω) such that f = 0 on Σb . A.1.3 Poisson Integral For a function f defined on Σ = R2 , the Poisson integral Pf in R2 × (−∞, 0) is defined by ∫ ′ ′ Pf (x , x3 ) = fˆ(ξ)e2π|ξ|x3 e2πix ·ξ dξ, (A.1.7) R2 307 where fˆ(ξ) is the Fourier transform of f (x′ ) on R2 . Then within the slab R2 × (−b, 0), we have the following estimate based on lemma A.5 in [25]. Lemma A.1.7. The Poisson integral satisfies that for q ∈ N, ∥∇q Pf ∥2H 0 . ∥f ∥2H˙ q−1/2 (Σ) , (A.1.8) ∥∇q Pf ∥2H 0 . ∥f ∥2H˙ q (Σ) , (A.1.9) where H˙ s denotes the usual homogeneous Sobolev space. A.1.4 Riesz Potential For a function f defined in Ω, we define the Riesz potential ∫ 0 ∫ ′ ′ Iλ f (x , x3 ) = fˆ(ξ, x3 ) |ξ|−λ e2πix ·ξ dξdx3 . (A.1.10) −b R2 Similarly, for f defined on Σ, we set ∫ ′ ′ Iλ f (x ) = fˆ(ξ) |ξ|−λ e2πix ·ξ dξ. (A.1.11) R2 We have the following lemmas to describe the product of Riesz potential and its interaction with the horizontal derivatives as lemma A.3 and A.4 in [25]. Lemma A.1.8. Let λ ∈ (0, 1). If f ∈ H 0 (Ω) and g, Dg ∈ H 1 (Ω), then ∥Iλ (f g)∥H 0 . ∥f ∥H 0 ∥g∥λH 1 ∥Dg∥1−λ H1 . (A.1.12) 308 If f ∈ H 0 (Σ) and g, Dg ∈ H 1 (Σ), then ∥Iλ (f g)∥H 0 (Σ) . ∥f ∥H 0 (Σ) ∥g∥λH 1 (Σ) ∥Dg∥1−λ H 1 (Σ) . (A.1.13) Lemma A.1.9. Let λ ∈ (0, 1). If f ∈ H k (Ω) for k ≥ 1 an integer, then Iλ Dk f 0 . Dk−1 f λ 0 Dk f 1−λ . (A.1.14) H H H0 A.1.5 Interpolation Estimates Here we record several interpolation estimate utilized in our proof. Since they have been proved in lemma A.6-A.8 and lemma 3.18 in [25], we will omit the proof now. Lemma A.1.10. For s, q > 0 and 0 ≤ r ≤ s, we have the estimate q/(r+q) r/(r+q) ∥f ∥H s (Σ) . ∥f ∥H s−r (Σ) ∥f ∥H s+q (Σ) , (A.1.15) whenever the right hand side is finite. Lemma A.1.11. Let Pf be the Poisson integral of f , defined on Σ. Let λ ≥ 0, q, s ∈ N, and r ≥ 0. Then the following estimates hold. 1. Let s q+λ θ= , 1−θ = . (A.1.16) q+s+λ q+s+λ Then ( )θ ( )1−θ q+s 2 ∥∇ q Pf ∥2H 0 . ∥Iλ f ∥2H 0 D f 0 . (A.1.17) H 309 2. Let r + s > 1, r+s−1 q+λ+1 θ= , 1−θ = . (A.1.18) q+s+r+λ q+s+r+λ Then ( )θ ( )1−θ q+s 2 ∥∇ q Pf ∥2L∞ . ∥Iλ f ∥2H 0 D f r . (A.1.19) H 3. Let s > 1. Then ∥∇q Pf ∥2L∞ . ∥Dq f ∥2H s . (A.1.20) Lemma A.1.12. Let f be defined on Σ. Let λ ≥ 0. Then we have the following estimates. 1. Let q, s ∈ (0, ∞) and s q+λ θ= , 1−θ = . (A.1.21) q+s+λ q+s+λ Then ( )θ ( )1−θ q+s 2 ∥D q f ∥2H 0 . ∥Iλ f ∥2H 0 D f 0 . (A.1.22) H 2. Let q, s ∈ N, r ≥ 0, r + s > 1, r+s−1 q+λ+1 θ= , 1−θ = . (A.1.23) q+s+r+λ q+s+r+λ Then ( )θ ( )1−θ q+s 2 ∥D q f ∥2L∞ . ∥Iλ f ∥2H 0 D f r . (A.1.24) H 310 Lemma A.1.13. Let f be defined in Ω. Let λ ≥ 0, q, s ∈ N, and r ≥ 0. Then we have the following estimates. 1. Let s q+λ θ= , 1−θ = . (A.1.25) q+s+λ q+s+λ Then ( )θ ( )1−θ q+s 2 ∥D q f ∥2H 0 . ∥Iλ f ∥2H 0 D f 0 . (A.1.26) H 2. Let r + s > 1 r+s−1 q+λ+1 θ= , 1−θ = . (A.1.27) q+s+r+λ q+s+r+λ Then ( )θ ( )1−θ q+s 2 ∥D q f ∥2L∞ . ∥Iλ f ∥2H 1 D f r+1 , (A.1.28) H and ( )θ ( )1−θ q+s 2 ∥D q f ∥2L∞ (Σ) . ∥Iλ f ∥2H 1 D f r+1 . (A.1.29) H A.1.6 Continuity and Temporal Derivative In the following, we give two important lemmas to connect L2 H k norm and L∞ H k norm. Lemma A.1.14. suppose that u ∈ L2 ([0, T ]; H s1 (Ω)) and ∂t u ∈ L2 ([0, T ]; H s2 (Ω)) 311 for s1 ≥ s2 ≥ 0 and s = (s1 + s2 )/2. Then u ∈ C 0 ([0, T ]; H s (Ω)) and satisfies the estimate ∥u∥2L∞ H s ≤ ∥u(0)∥2H s + ∥u∥2L2 H s1 + ∥∂t u∥2L2 H s2 , (A.1.30) where the L2 H k norm and L∞ H k norm are evaluated in [0, T ]. Proof. Considering the extension theorem in Sobolev space, we only need to prove this result in the Rn case. The periodic case can be derived in a similar fashion. Using Fourier transform, (∫ ) ∂t ∥u(t)∥2H s = 2R ⟨ξ⟩ uˆ(ξ, t)∂t uˆ(ξ, t)dξ 2s (A.1.31) ∫ Rn ≤ 2 ⟨ξ⟩2s |ˆ u(ξ, t)||∂t uˆ(ξ, t)|dξ ∫R n = 2 ⟨ξ⟩s1 |ˆ u(ξ, t)|⟨ξ⟩s2 |∂t uˆ(ξ, t)|dξ ∫ R ∫ n ≤ ⟨ξ⟩ |ˆ 2s1 2 u(ξ, t)| dξ + ⟨ξ⟩2s2 |∂t uˆ(ξ, t)|2 dξ Rn Rn = ∥u(t)∥2H s1 + ∥∂t u(t)∥2H s2 . So integrate with respect to time on [0, t] ∥u(t)∥2H s ≤ ∥u(0)∥2H s + ∥u∥2L2 H s1 + ∥∂t u∥2L2 H s2 . (A.1.32) The following lemma shows the estimate in another direction. Lemma A.1.15. For any u ∈ L∞ ([0, T ]; H k ) within [0, T ], we must have u ∈ L2 ([0, T ]; H k ) and satisfies the estimate ∥u∥2L2 H k ≤ T ∥u∥2L∞ H k . (A.1.33) 312 Proof. The result is simply based on the definition of these two norms. A.1.7 Extension Theorem The following are two extension theorems which will be used to construct start point of iteration from initial data in proving well-posedness of Naiver-Stokes-transport system. Since it is identical as lemma A.5 and A.6 in [26], we omit the proof here. Lemma A.1.16. Suppose that ∂tj u(0) ∈ H 2N −2j (Ω) for j = 0, . . . , N , then there exists a extension u achieving the initial data, such that ∂tj u ∈ L2 ([0, ∞); H 2N −2j+1 (Ω)) ∩ L∞ ([0, ∞); H 2N −2j (Ω)) for j = 0, . . . , N . Moreover, ∑ N j 2 2 ∑ N j ∂t u 2 + ∂tj u L∞ H 2N −2j . ∂t u(0) 2 2N −2j . (A.1.34) L H 2N −2j+1 H j=0 j=0 Lemma A.1.17. Suppose that ∂tj p(0) ∈ H 2N −2j−1 (Ω) for j = 0, . . . , N −1, then there exists a extension p achieving the initial data, such that ∂tj p ∈ L2 ([0, ∞); H 2N −2j (Ω)) ∩ L∞ ([0, ∞); H 2N −2j−1 (Ω)) for j = 0, . . . , N − 1. Moreover, ∑ N −1 j 2 ∑ N −1 j ∂t p 2 2N −2j + ∂tj p 2 ∞ 2N −2j−1 . ∂t p(0) 2 2N −2j−1 . (A.1.35) L H L H H j=0 j=0 313 A.2 Estimates for Fundamental Equations A.2.1 Transport Estimates Let Σ be either infinite or periodic. Consider the equation    ∂t η + u · Dη = g, in Σ × (0, T ) (A.2.1)   η(t = 0) = η0 . We have the following estimate of the regularity solution to this equation, which is a particular case of a more general result proved in proposition 2.1 of [17]. Note that the result in [17] is stated for Σ = R2 , but the same result holds in the periodic case as described in [18]. Lemma A.2.1. Let η be a solution to equation (A.2.1). Then there exists a universal constant C > 0 such that for any s ≥ 3 ( ∫ t )( ∫ t ) ∥η∥L∞ H s ≤ exp C ∥u(r)∥H s (Σ) dr ∥η0 ∥H s + ∥g(r)∥H s dr . (A.2.2) 0 0 Proof. Use p = p2 = 2, r = 2 and σ = s in proposition 2.1 of [18]. See the proof of Theorem 5.4 in [25]. Lemma A.2.2. Let η be a solution to (A.2.1). Then there is a universal constant C > 0 such that for any 0 ≤ s < 2 (A.2.3) ( ∫ t )( ∫ t ) ∥η∥L∞ H s ≤ exp C ∥Du(r)∥H 3/2 (Σ) dr ∥η0 ∥H s (Σ) + ∥g(r)∥H s (Σ) dr . 0 0 Proof. The same as lemma A.9 in [25]. 314 A.2.2 Elliptic Estimates We need two different type of elliptic estimates in our proof either in weak sense or strong sense. ∫ Lemma A.2.3. (weak solution) Assume Ω ψdx = 0. We call (u, p) ∈ H01 (Ω)×H 0 (Ω) a weak solution to the elliptic equation    −∆u + ∇p = ϕ ∈ H −1 (Ω),   in Ω     ∇ · u = ψ ∈ H 0 (Ω), in Ω (A.2.4)     u = 0, on Σ     u = 0, on Σb if ∇ · u = ψ in Ω and (u, p) satisfy the weak formulation that for arbitrary f ∈ H01 (Ω), it holds that ⟨Du : Df ⟩H 0 + ⟨p, ∇ · f ⟩H 0 = ⟨ϕ, f ⟩H −1 , (A.2.5) where ⟨·, ·⟩H 0 represents the inner product in H 0 (Ω) and ⟨·, ·⟩H −1 denotes the dual pairing between H01 (Ω) and H −1 (Ω). Then we have the estimate ∥u∥2H 1 + ∥p∥2H 0 . ∥ϕ∥2H −1 + ∥ψ∥2H 0 . (A.2.6) Proof. An obvious modification for the proof of lemma 3.3 in [8] implies that for any ψ ∈ H 0 (Ω), there exists a v ∈ H01 (Ω) such that ∇ · v = ψ in Ω and satisfies the estimate ∥v∥2H 1 . ∥ψ∥2H 0 . Then we may change to the unknown w = u − v, which 315 satisfies the system    −∆w + ∇p = ϕ + ∆v,   in Ω     ∇ · w = 0, in Ω (A.2.7)     w = 0, on Σ     w = 0. on Σb Then we have the new weak formulation that for arbitrary f ∈ H01 (Ω), it is valid that ⟨Dw : Df ⟩H 0 + ⟨p, ∇ · f ⟩H 0 = ⟨ϕ, f ⟩H −1 − ⟨Dv : Df ⟩H 0 . (A.2.8) We may further restrict that f satisfies ∇ · f = 0 as the pressureless weak formulation that ⟨Dw : Df ⟩H 0 = ⟨ϕ, f ⟩H −1 − ⟨Dv : Df ⟩H 0 . (A.2.9) Riesz theorem implies that there exists a pressureless weak solution w ∈ H01 (Ω) satisfying ∇ · w = 0 and the estimate ∥w∥2H 1 . ∥ϕ∥2H −1 + ∥v∥2H 1 . ∥ϕ∥2H −1 + ∥ψ∥2H 0 . (A.2.10) Then it is wellknown that pressure can be taken as the multiplier for the Navier-Stokes system. Similar to proposition 2.9 in [26], we have that there exists a p ∈ H 1 (Ω) such that the weak formulation (A.2.8) holds and it satisfies ∥p∥2H 0 . ∥ϕ∥2H −1 + ∥ψ∥2H 0 + ∥w∥2H 1 . ∥ϕ∥2H −1 + ∥ψ∥2H 0 . (A.2.11) Therefore, (u, p) satisfy the weak formulation (A.2.5) and our result easily follows. 316 Lemma A.2.4. (strong solution) Suppose (u, p) solve the equation    −∆u + ∇p = ϕ ∈ H r−2 (Ω),   in Ω     ∇ · u = ψ ∈ H r−1 (Ω), in Ω (A.2.12)     (pI − Du)e3 = φ ∈ H r−3/2 (Σ), on Σ     u = 0, on Σb in the strong sense. Then for r ≥ 2, ∥u∥2H r + ∥p∥2H r−1 . ∥ϕ∥2H r−2 + ∥ψ∥2H r−1 + ∥φ∥2H r−3/2 . (A.2.13) whenever the right hand side is finite. Proof. The same as lemma A.15 in [26]. Appendix B Counterexamples in Neutron Transport Equation 318 B.1 Construction of the Counterexample with In- Flow Boundary Lemma B.1.1. For the Milne problem   ∂f   sin(θ + ξ) + f − f¯ = 0,   ∂η f (0, θ, ξ) = g(θ, ξ) for sin(θ + ξ) > 0, (B.1.1)      limη→∞ f (η, θ, ξ) = f (∞, θ), if g(θ, ξ) = cos(3(θ + ξ)), then we have ∂f / L∞ ([0, ∞) × [−π, π) × [−π, π)). ∈ (B.1.2) ∂η Proof. We divide the proof into several steps: we first assume ∂η f ∈ L∞ ([0, ∞) × [−π, π) × [−π, π)) and then show it can lead to a contradiction. Step 1: Maximum principle Theorem 4.4.14 implies the solution f to the problem (B.1.1) satisfies the maximum principle, i.e. for any (η, θ, ϕ) min g(θ, ξ) ≤ f (η, θ, ξ) ≤ max g(θ, ξ). (B.1.3) sin(θ+ξ)>0 sin(θ+ξ)>0 We can see the data g(θ, ξ) = cos(3(θ + ξ)) satisfying g(θ, −θ) = 1 and |g| ≤ 1. Based on the maximum principle, we can derive for any (η, θ, ξ) f (η, θ, ξ) ≤ 1. (B.1.4) Hence, certainly we have f (0, θ, ξ) ≤ 1 for sin(θ + ξ) < 0. 319 Step 2: Estimates of f¯(0, θ). We can directly estimate ∫ π 1 f¯(0, θ) = f (0, θ, ξ)dξ (B.1.5) 2π −π (∫ ∫ ) 1 = f (0, θ, ξ)dξ + f (0, θ, ξ)dξ 2π sin(θ+ξ)<0 sin(θ+ξ)>0 ∫ 1 1 ≤ f (0, θ, ξ)dξ + . 2π sin(θ+ϕ)>0 2 By the choice of g, we naturally have ∫ ∫ f (0, θ, ξ)dξ = g(θ, ξ)dξ = 0. (B.1.6) sin(θ+ξ)>0 sin(θ+ξ)>0 Then this implies 1 f¯(0, θ) ≤ . (B.1.7) 2 Step 3: Definition of trace. It is easy to see ∂η f satisfies the Milne problem ∂(∂η f ) sin(θ + ξ) + ∂η f − ∂η f = 0. (B.1.8) ∂η Since we have ∂η f ∈ L∞ ([0, ∞) × [−π, π) × [−π, π)) which implies ∂η f ∈ L∞ ([0, L] × [−π, π)), by Ukai’s trace theorem, we may define the trace of ∂η f on η = 0 satisfying ∂η f (0, θ, ϕ) ∈ L∞ [−π, π) × [−π, π). However, we can define the trace of ∂η f in another fashion. For any ξ ̸= −θ and ξ ̸= π − θ, we have sin(θ + ξ) ̸= 0. Since we have f ∈ L∞ ([0, ∞) × [−π, π) × [−π, π)) 320 as well as f¯ ∈ L∞ [0, ∞) × [−π, π), by the Milne problem (B.1.1), it is naturally to define for η > 0 f¯(η, θ) − f (η, θ, ξ) ∂η f (η, θ, ξ) = . (B.1.9) sin(θ + ξ) Since ∂η f ∈ L∞ ([0, ∞) × [−π, π) × [−π, π)), we know f is continuous with respect to η for a.e. (θ, ξ). Taking η → 0 defines the trace for ∂η f at(0, θ, ξ) f¯(0, θ) − f (0, θ, ξ) ∂η f (0, θ, ξ) = . (B.1.10) sin(θ + ξ) Since the grazing set {(θ, ξ) : θ + ξ = 0 or θ + ξ = π} is zero-measured on the boundary η = 0, then we have the trace of ∂η f is a.e. well-defined. By the uniqueness of trace of ∂η f , above two types of traces must coincide with each other a.e.. Then we may combine them both and obtain ∂η f (0, θ, ξ) ∈ L∞ [−π, π) × [−π, π) is a.e. well-defined and satisfies the formula f¯(0, θ) − f (0, θ, ξ) ∂η f (0, θ, ξ) = . (B.1.11) sin(θ + ξ) Step 4: Contradiction. Therefore, we may consider the limiting process ∂f f¯(0, θ) − f (0, θ, ξ) lim + (0, θ, ξ) = lim + . (B.1.12) ξ→−θ ∂η ξ→−θ sin(θ + ξ) Since we know as ξ → −θ+ , it follows that sin(θ + ξ) → 0+ (B.1.13) f¯(0, θ) − f (0, θ, ξ) → f¯(0, θ) − g(θ, −θ) = f¯(0, θ) − 1 < 0. (B.1.14) 321 Then this leads to ∂f lim + (0, θ, ξ) = −∞. (B.1.15) ξ→−θ ∂η / L∞ [−π, π) × [−π, π). This contradicts our result in the which means ∂η f (0, θ, ξ) ∈ previous step. Hence, our assumption that ∂η f ∈ L∞ ([0, ∞) × [−π, π) × [−π, π)) cannot be true. Step 5: Another contradiction. There is another way to show this fact. Since ∂η f ∈ L∞ ([0, L] × [−π, π) × [−π, π)). Also, we have f ∈ L∞ ([0, L] × [−π, π) × [−π, π)). Then this implies f is Lipschitz continuous in [0, ∞) with respect to η for a.e. (θ, ξ). Hence, this implies f¯ is al- so Lipschitz continuous in η ∈ [0, ∞). Without loss of generality, we may assume ∥∂η f ∥∞ ≤ M . Thus we have for a.e. (θ, ξ) ∈ sin(θ + ϕ) > 0 |f (η, θ, ξ) − f (0, θ, ξ)| ≤ M η (B.1.16) f¯(η, θ) − f¯(0, θ) ≤ M η. (B.1.17) Then f (η, θ, ξ) − f¯(η, θ) (B.1.18) ≥ f (0, θ, ξ) − f¯(0, θ) − |f (η, θ, ξ) − f (0, θ, ξ)| − f¯(η, θ) − f¯(0, θ) ≥ f¯(0, θ) − g(θ, −θ) − 2M η. Since we know f¯(0, θ) − g(θ, −θ) ≥ C > 0 for some constant C. Then as long as η ≤ C/(4M ), we have f (η, θ, ξ) − f¯(η, θ) ≥ C . (B.1.19) 2 322 Since for (θ, ξ) not in the grazing set, we always have f¯(η, θ) − f (η, θ, ξ) ∂η f (η, θ, ξ) = . (B.1.20) sin(θ + ξ) then |∂η f | can be arbitrarily large as long as sin(θ + ξ) is sufficiently small, and also it possesses a positive measure. This implies ∂η f ∈ L∞ ([0, ∞) × [−π, π) × [−π, π)) cannot be true. B.2 Construction of the Counterexample with D- iffusive Boundary Lemma B.2.1. For the Milne problem   ∂f   sin(θ + ξ) + f − f¯ = 0,   ∂η f (0, θ, ξ) = Pf (0, θ) + g1 (θ, ξ) for sin(θ + ξ) > 0, (B.2.1)      limη→∞ f (η, θ, ξ) = f (∞, θ), with Pf (0, θ) = 0. (B.2.2) If g(θ, ξ) = cos(3(θ + ξ)), then we have ∂f / L∞ ([0, ∞) × [−π, π) × [−π, π)). ∈ (B.2.3) ∂η Proof. For g(θ, ξ) = cos(3(θ + ξ)), we can easily derive U0 = 0. Hence g1 = g. Note that g satisfies the compatibility condition (4.6.100). With the normalization 323 condition Pf (0, θ) = 0, we can see this problem reduces to the Milne problem with in-flow boundary   ∂f   sin(θ + ξ) + f − f¯ = 0,   ∂η f (0, θ, ξ) = g(θ, ξ) for sin(θ + ξ) > 0, (B.2.4)      limη→∞ f (η, θ, ξ) = f1 (∞, θ). Therefore, we can complete the proof by Theorem B.1.1. Bibliography [1] H. Abels, The initial-value problem for the Navier-Stokes equations with a free surface in Lq -Sobolev spaces, Adv. Differential Equations, 10 (2005), pp. 45–64. [2] S. Agmon, A. Douglis, and L. Nirenberg, Estimates near the boundary for solutions of elliptic partial differential equations satisfying general boundary conditions i., Comm. Pure Appl. Math., 12 (1959), pp. 623–727. [3] M. J. Ahn, H. Y. Lee, and M. R. Ohm, Error estimates for fully discrete approximation to a free boundary problem in polymer technology, Appl. Math. Comput., 138 (2003), pp. 227–238. [4] L. Arkeryd, R. Esposito, R. Marra, and A. Nouri, Stability for Rayleigh-Benard convective solutions of the Boltzmann equation, Arch. Ration. Mech. Anal., 198 (2010), pp. 125–187. [5] , Ghost effect by curvature in planar Couette flow, Kinet. Relat. Models, 4 (2011), pp. 109–138. [6] L. Arkeryd and A. Nouri, On a Taylor-Couette type bifurcation for the stationary nonlinear Boltzmann equation, J. Stat. Phys., 124 (2006), pp. 401– 443. [7] H. Bae, Solvability of the free boundary value problem of the Navier-Stokes equations, Discrete Contin. Dyn. Syst., 29 (2011), pp. 769–801. [8] J. T. Beale, The initial value problem for the Navier-Stokes equations with a free surface, Comm. Pure Appl. Math., 34 (1981), pp. 359–392. [9] A. Bensoussan, J.-L. Lions, and G. C. Papanicolaou, Boundary layers and homogenization of transport processes, Publ. Res. Inst. Math. Sci., 15 (1979), pp. 53–157. [10] C. Cercignani, R. Illner, and M. Pulvirenti, The mathematical theory of dilute gases, Springer-Verlag, New York, 1994. [11] C. Cercignani, R. Marra, and R. Esposito, The Milne problem with a force term, Transport Theory Statist. Phys., 27 (1998), pp. 1–33. 324 325 [12] D. Christodoulou and H. Lindblad, On the motion of the free surface of a liquid, Comm. Pure Appl. Math., 53 (2000), pp. 1536–1602. [13] B. Cockburn, S. Hou, and C.-W. Shu, The Runge-Kutta local projection discontinuous Galerkin finite element method for conservation laws IV. the mul- tidimensional case, Math. Comp., 54 (1990), pp. 545–581. [14] B. Cockburn and C.-W. Shu, TVB Runge-Kutta local projection discontinu- ous Galerkin finite element method for conservation laws II. general framework, Math. Comp., 52 (1989), pp. 411–435. [15] , Runge-Kutta discontinuous Galerkin methods for convection-dominated problems, J. Sci. Comput., 16 (2001), pp. 173–261. [16] D. Coutand and S. Shkoller, Well-posedness of the free-surface incom- pressible Euler equations with or without surface tension, J. Amer. Math. Soc., 20 (2007), pp. 829–930. [17] R. Danchin, Estimates in Besov spaces for transport and transport-diffusion e- quations with almost Lipschitz coefficients, Rev. Mat. Iberoamericana, 21 (2005), pp. 863–888. [18] , Fourier analysis methods for PDEs, Preprint, http://perso-math.univ- mlv.fr/users/danchin.raphael/recherche.html (2005). [19] R. Esposito, Y. Guo, C. Kim, and R. Marra, Non-isothermal boundary in the Boltzmann theory and Fourier law, Comm. Math. Phys., 323 (2013), pp. 177– 239. [20] P. Germain, N. Masmoudi, and J. Shatah, Global solutions for the gravity water waves equation in dimension 3, Ann. of Math., 175 (2012), pp. 691–754. [21] V. Girault, B. Riviere, and M. F. Wheeler, A discontinuous Galerkin method with nonoverlapping domain decomposition for the Stokes and Navier- Stokes problems, Math. Comp., 74 (2005), pp. 53–84. [22] , A splitting method using discontinuous Galerkin for the transient in- compressible Navier-Stokes equations, ESAIM Math. Model. Numer. Anal., 39 (2005), pp. 1115–1147. [23] J. Grooss and J. S. Hesthaven, A level set discontinuous Galerkin method for free surface flows, Comput. Methods Appl. Mech. Engrg., 195 (2006), p- p. 3406–3429. [24] Y. Guo and I. Tice, Almost exponential decay of periodic viscous surface waves without surface tension, Anal. PDE, 6 (2013), pp. 459–531. [25] , Decay of viscous surface waves without surface tension in horizontally in- finite domains, Arch. Ration. Mech. Anal., 207 (2013), pp. 1429–1533. [26] , Local well-posedness of the viscous surface wave problem without surface tension, Anal. PDE, 6 (2013), pp. 287–369. 326 [27] E. Hairer, C. Lubich, and M. Roche, The numerical solution of differential- algebraic systems by Runge-Kutta methods, Springer-Verlag, Berlin, 1989. [28] F. H. Harlow and J. E. Welch, Numerical calculation of time-dependent viscous incompressible flow of fluid with free surface, Phys. Fluids, 8 (1965), pp. 2182–2189. [29] Y. Hataya, Decaying solution of a Navier-Stokes flow without surface tension, J. Math. Kyoto Univ., 49 (2009), pp. 691–717. [30] C. Hirt and B. Nichols, Volume of fluid method (VOF) for the dynamics of free boundaries, J. Comput. Phys., 39 (1981), pp. 201–225. [31] D. Lannes, Well-posedness of the water-waves equations, J. Amer. Math. Soc., 18 (2005), pp. 605–654. [32] E. W. Larsen, A functional-analytic approach to the steady, one-speed neu- tron transport equation with anisotropic scattering, Comm. Pure Appl. Math., 27 (1974), pp. 523–545. [33] , Solutions of the steady, one-speed neutron transport equation for small mean free paths, J. Mathematical Phys., 15 (1974), pp. 299–305. [34] , Neutron transport and diffusion in inhomogeneous media I., J. Mathemat- ical Phys., 16 (1975), pp. 1421–1427. [35] , Asymptotic theory of the linear transport equation for small mean free paths II., SIAM J. Appl. Math., 33 (1977), pp. 427–445. [36] E. W. Larsen and J. D’Arruda, Asymptotic theory of the linear transport equation for small mean free paths I., Phys. Rev., 13 (1976), pp. 1933–1939. [37] E. W. Larsen and G. J. Habetler, A functional-analytic derivation of Case’s full and half-range formulas, Comm. Pure Appl. Math., 26 (1973), p- p. 525–537. [38] E. W. Larsen and J. B. Keller, Asymptotic solution of neutron transport problems for small mean free paths, J. Mathematical Phys., 15 (1974), pp. 75–81. [39] E. W. Larsen and P. F. Zweifel, On the spectrum of the linear transport operator, J. Mathematical Phys., 15 (1974), pp. 1987–1997. [40] , Steady, one-dimensional multigroup neutron transport with anisotropic s- cattering, J. Mathematical Phys., 17 (1976), pp. 1812–1820. [41] H. Y. Lee, Error analysis of finite element approximation of a Stefan prob- lem with nonlinear free boundary condition, J. Appl. Math. Comput., 22 (2006), pp. 223–235. [42] H. Lindblad, Well-posedness for the motion of an incompressible liquid with free surface boundary, Ann. of Math., 162 (2005), pp. 109–194. 327 [43] R. H. Nochetto and C. Verdi, An efficient linear scheme to approximate parabolic free boundary problems: error estimates and implementation, Math. Comp., 51 (1988), pp. 27–53. [44] J. Shatah and C. Zeng, Geometry and a priori estimates for free boundary problems of the Euler equation, Comm. Pure Appl. Math., 61 (2008), pp. 698– 744. [45] , Local well-posedness for fluid interface problems, Arch. Ration. Mech. Anal., 199 (2011), pp. 653–705. [46] V. Solonnikov, Solvability of a problem on the motion of a viscous incom- pressible fluid that is bounded by a free surface, Math. USSR-Izv., 11 (1978), pp. 1323–1358. [47] M. Sussman and M. Y. Hussaini, A discontinuous spectral element method for the level set equation, J. Sci. Comput., 19 (2003), pp. 479–500. [48] L.-h. Wang, On Korn’s inequality, J. Comput. Math., 21 (2003), pp. 321–324. [49] M. F. Wheeler, An elliptic collocation-finite element method with interior penalties, SIAM J. Numer. Anal., 15 (1978), pp. 152–161. [50] L. Wu, Well-posedness and decay of the viscous surface wave, SIAM J. Math. Anal., 46 (2014), pp. 2084–2135. [51] S. Wu, Well-posedness in Sobolev spaces of the full water wave problem in 2-D, Invent. Math., 130 (1997), pp. 39–72. [52] , Well-posedness in Sobolev spaces of the full water wave problem in 3-D, J. Amer. Math. Soc., 12 (1999), pp. 445–495. [53] , Global well-posedness of the 3-D full water wave problem, Invent. Math., 184 (2011), pp. 125–220. [54] X. Yang, Asymptotic behavior on the milne problem with a force term, J. Dif- ferential Equations, 252 (2012), pp. 4656–4678. [55] P. Zhang and Z. Zhang, On the free boundary problem of three-dimensional incompressible Euler equations, Comm. Pure Appl. Math., 61 (2008), pp. 877– 940.