Regularity of Milne Problem with Geometric Correction in 3D
aa r X i v : . [ m a t h . A P ] O c t REGULARITY OF MILNE PROBLEM WITH GEOMETRIC CORRECTION IN 3D
YAN GUO AND LEI WU
Abstract.
Consider the Milne problem with geometric correction in a 3D convex domain. Via bootstrap-ping arguments, we establish W , ∞ regularity for its solutions. Combined with a uniform L estimate,such regularity leads to the validity of diffusive expansion for the neutron transport equation with diffusiveboundary conditions. Keywords:
Geometric correction; Bootstrapping; W , ∞ estimates. Contents
1. Introduction 21.1. Motivation and Formulation 21.2. Main Result 31.3. Background and Methods 41.4. Notation and Structure 62. Asymptotic Analysis 72.1. Interior Expansion 72.2. Local Coordinate System 72.3. Boundary Layer Expansion with Geometric Correction 102.4. Matching of Interior Solution and Boundary Layer 103. Well-Posedness and Decay of ǫ -Milne Problem 133.1. L Estimates 133.2. L ∞ Estimates 213.3. Exponential Decay 223.4. Diffusive Boundary 234. Regularity of ǫ -Milne Problem 254.1. Preliminaries 254.2. Mild Formulation of Normal Derivative 264.3. Mild Formulation of Velocity Derivative 374.4. Estimate of Derivatives 38Appendix A. Remainder Estimate 40A.1. L m Estimate 41A.2. L ∞ Estimate 44Appendix B. Diffusive Limit 49References 52
Mathematics Subject Classification. . Introduction
Motivation and Formulation.
Milne problem is the main tool to study boundary layer effect inkinetic equations. Here to motivate 3D ǫ -Milne problem with geometric correction, we consider the steadyneutron transport equation in a three-dimensional convex domain with diffusive boundary. In the spacedomain ~x = ( x , x , x ) ∈ Ω where ∂ Ω ∈ C and the velocity domain ~w = ( w , w , w ) ∈ S , the neutrondensity u ǫ ( ~x, ~w ) satisfies ( ǫ ~w · ∇ x u ǫ + u ǫ − ¯ u ǫ = 0 in Ω ,u ǫ ( ~x , ~w ) = P [ u ǫ ]( ~x ) + ǫg ( ~x , ~w ) for ~w · ~ν < ~x ∈ ∂ Ω , (1.1)where ¯ u ǫ ( ~x ) = 14 π Z S u ǫ ( ~x, ~w )d ~w, (1.2) P [ u ǫ ]( ~x ) = 14 π Z ~w · ~ν> u ǫ ( ~x , ~w )( ~w · ~ν )d ~w, (1.3) ~ν is the outward unit normal vector, with the Knudsen number 0 < ǫ <<
1. Also, u ǫ satisfies the normal-ization condition Z Ω ×S u ǫ ( ~x, ~w )d ~w d ~x = 0 , (1.4)and g satisfies the compatibility condition Z ∂ Ω Z ~w · ~ν< g ( ~x , ~w )( ~w · ~ν )d ~w d ~x = 0 . (1.5)A classical problem is to study the diffusive limit of (1.1) as ǫ →
0. Generally speaking, the solution u ǫ varies smoothly and slowly in the interior of Ω, and behaves like u ǫ − ¯ u ǫ = 0 which ignores ǫ ~w · ∇ x u ǫ .However, its value changes severely when approaching boundary ∂ Ω and ǫ ~w · ∇ x u ǫ becomes non-negligible.The smaller ǫ is, the more severe u ǫ changes. This indicates that u ǫ actually contains two separate parts withdifferent scalings, i.e. interior solution and boundary layer. In particular, the boundary layer is a functionwith scaled variable defined in a thin layer of thickness O ( ǫ ) close to boundary in Ω.In the region of boundary layer, assume µ denotes the distance to the boundary in the inward normaldirection and ( τ , τ ) denote a local orthogonal curvilinear coordinate system for ∂ Ω. Then we have ǫ ~w · ∇ x = − ǫ ( ~w · ~ν ) ∂∂µ + ǫR − µ ( ~w · ~t ) ∂∂τ + ǫR − µ ( ~w · ~t ) ∂∂τ , (1.6)where ( ~t , ~t ) are orthogonal tangential vectors associated with ( τ , τ ), and (cid:16) R ( τ , τ ) , R ( τ , τ ) (cid:17) denotetwo radium of principle curvature. Since ~w ∈ S , we define the spherical velocity substitution as − ~w · ~ν = sin φ,~w · ~t = cos φ sin ψ,~w · ~t = cos φ cos ψ. (1.7)With the rescaled distance η = µǫ , we may represent ǫ ~w · ∇ x = sin φ ∂∂η − ǫ (cid:18) sin ψR − ǫη + cos ψR − ǫη (cid:19) cos φ ∂∂φ + higher-order terms . (1.8)Therefore, in order to construct boundary layer, it suffices to study the 3D ǫ -Milne problem with geometriccorrection for f ǫ ( η, φ, ψ ) in ( η, φ, ψ ) ∈ [0 , L ] × [ − π/ , π/ × [ − π, π ] as sin φ ∂f ǫ ∂η − ǫ (cid:18) sin ψR − ǫη + cos ψR − ǫη (cid:19) cos φ ∂f ǫ ∂φ + f ǫ − ¯ f ǫ = S ǫ ( η, φ, ψ ) ,f ǫ (0 , φ, ψ ) = h ǫ ( φ, ψ ) for sin φ > ,f ǫ ( L, φ, ψ ) = f ǫ ( L, − φ, ψ ) , (1.9) where h ǫ and S ǫ are two functions given a priori and¯ f ǫ ( η ) = 14 π Z π − π Z π/ − π/ f ǫ ( η, φ, ψ ) cos φ d φ d ψ (1.10)in which cos φ shows up as the Jacobian of spherical coordinates in integration, and L = ǫ − n for some n tobe specified later. Note that actually f ǫ , S ǫ , h ǫ and R , R are all related to ( τ , τ ). Since boundary layeris defined locally on ∂ Ω and our analysis focuses on the case for fixed ( τ , τ ), we do not need to specify suchdependence explicitly unless necessary.1.2. Main Result.
Define norms k f k L ∞ L ∞ = sup ( η,φ,ψ ) | f ( η, φ, ψ ) | , (1.11) k f k L ∞ ( η ) = sup ( φ,ψ ) with sin φ> | f ( η, φ, ψ ) | , (1.12) k f k L L = (cid:18) Z L Z π − π Z π/ − π/ | f ( η, φ, ψ ) | cos φ d φ d ψ d η (cid:19) / , (1.13) k f k L ( η ) = (cid:18) Z π − π Z π/ − π/ | f ( η, φ, ψ ) | cos φ d φ d ψ (cid:19) / , (1.14)and the inner product as h f, g i ( η ) = Z π − π Z π/ − π/ f ( η, φ, ψ ) g ( η, φ, ψ ) cos φ d φ d ψ. (1.15) Theorem 1.1. (Well-Posedness and Decay) Assume < n < and there exist some constants M, K > uniform in ǫ , such that k h ǫ k L ∞ ≤ M, (1.16) and (cid:13)(cid:13) e Kη S ǫ (cid:13)(cid:13) L ∞ L ∞ ≤ M. Then for K > sufficiently small, there exists a constant f ǫL and the solution f ǫ ( η, φ, ψ ) to the ǫ -Milneproblem (1.9) satisfies (cid:13)(cid:13) e K η ( f ǫ − f ǫL ) (cid:13)(cid:13) L ∞ L ∞ ≤ C. (1.17) Here C ≥ denotes a universal constant independent of ǫ . Theorem 1.2. (Weighted Regularity) Assume < n < and there exist some constants M, K > uniformin ǫ , such that k h ǫ k L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂h ǫ ∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ ≤ M, (1.18) and (cid:13)(cid:13) e Kη S ǫ (cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) e Kη ∂S ǫ ∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) e Kη ∂S ǫ ∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ ≤ M. Then for K > sufficiently small, we have (cid:13)(cid:13)(cid:13)(cid:13) e K η ζ ∂ ( f ǫ − f ǫL ) ∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) e K η ζ ∂ ( f ǫ − f ǫL ) ∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ ≤ C | ln( ǫ ) | , (1.19) where the weight function ζ ( η, φ, ψ ) = − (cid:18) R − ǫηR (cid:19) ψ (cid:18) R − ǫηR (cid:19) ψ cos φ ! / . (1.20) YAN GUO AND LEI WU
If further for i = 1 , , (cid:13)(cid:13)(cid:13)(cid:13) ∂h ǫ ∂ψ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂h ǫ ∂τ i (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) e Kη ∂S ǫ ∂ψ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) e Kη ∂S ǫ ∂τ i (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ ≤ M, (1.21) we have (cid:13)(cid:13)(cid:13)(cid:13) e K η ∂ ( f ǫ − f ǫL ) ∂ψ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) e K η ∂ ( f ǫ − f ǫL ) ∂τ i (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ ≤ C | ln( ǫ ) | . (1.22) Remark 1.3.
It is easy to see ζ ≥ | sin φ | . Then Theorem 1.2 naturally implies (cid:13)(cid:13)(cid:13)(cid:13) e K η sin φ ∂ ( f ǫ − f ǫL ) ∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ ≤ C | ln( ǫ ) | , (1.23) which is bounded away from the grazing set. More importantly, due to the half-line geometry of the Milneproblem, the tangential derivatives are bounded in (1.22) up to the grazing set. This is a sharp contrast to [17] in a bounded domain. As an application, thanks to the uniform bounds for tangential derivatives (1.22), we finally establish thediffusive limit of neutron transport equation.
Corollary 1.4.
Assume g ( ~x , ~w ) ∈ C (Γ − ) satisfying (1.5). Then for the steady neutron transport equation(1.1), there exists a unique solution u ǫ ( ~x, ~w ) ∈ L ∞ (Ω × S ) satisfying (1.4). Moreover, for any < δ << ,the solution obeys the estimate k u ǫ ( ~x, ~w ) − U ǫ ( ~x ) k L ∞ (Ω ×S ) ≤ C ( δ, Ω) ǫ − δ , (1.24) where U ǫ ( ~x ) satisfies ∆ x U ǫ = 0 in Ω ,∂U ǫ ∂~ν = 1 π Z ~w · ~ν< g ( ~x, ~w ) | ~w · ~ν | d ~w on ∂ Ω , Z Ω U ǫ ( ~x )d ~x = 0 , (1.25) in which C ( δ, Ω) > denotes a constant that depends on δ and Ω . Background and Methods.
At the core of boundary layer analysis, the study of Milne problem isconsistent with the development of asymptotic analysis of kinetic equations in bounded domains. Since1960s, people have discovered several methods to study the well-posedness of Milne problem, and applythem to asymptotic expansion. We refer to the references [15], [3], [4], [29], [20], [25], [13], [12], [1], [8], [11],[16], [19], [26], [5], [2], [10], [21], [22], [23], and [24] for more details. In 1979, diffusive limit of steady neutrontransport equation was systematically investigated in [9] (see also [6] and [7]).The key idea of [9], [6] and [7] is to study the classical Milne problem as sin φ ∂f ǫ ∂η + f ǫ − ¯ f ǫ = S ǫ ( η, φ, ψ ) ,f ǫ (0 , φ, ψ ) = h ǫ ( φ, ψ ) for sin φ > , lim η →∞ f ǫ ( η, φ, ψ ) = f ǫ ∞ . (1.26)In [9], the authors proved that f ǫ is well-posed and decays exponentially fast to some constant f ǫ ∞ in L ∞ .Unfortunately, as discovered recently in [27], the lack of regularity of such classical Milne problem (1.26)has been overlooked for non-flat bounded domains. The solutions of (1.26) are singular in the normaldirection, which leads to singularity in the tangential directions, resulting in break-down of diffusive expansionwith classical Milne boundary layers.The regularity of the Milne problem is the central issue. In [27] and [28], a new approach with geometriccorrection from the next-order diffusive expansion has been introduced to ensure regularity in the cases of EGULARITY OF MILNE PROBLEM IN 3D 5
2D plate and annulus, i.e. to solve for f ǫ ( η, φ ) satisfying sin φ ∂f ǫ ∂η − ǫR − ǫη cos φ ∂f ǫ ∂φ + f ǫ − ¯ f ǫ = S ǫ ( η, φ ) ,f ǫ (0 , φ ) = h ǫ ( φ ) for sin φ > ,f ǫ ( L, φ ) = f ǫ ( L, − φ ) , (1.27)where R denotes the radius of curvature. Also, in [18], weighted W , ∞ estimates was proved to treat moregeneral 2D convex domains.There are three main ingredients to generalize our previous results to 3D convex domains.The first difficulty is the lack of conserved energy. Consider the simplest case that S = 0 and we omit ǫ temporarily. Assume F ( η, ψ ) = − ǫ (cid:18) sin ψR − ǫη + cos ψR − ǫη (cid:19) . (1.28)Taking inner product with f on both sides of (1.9), we obtain14 ∂∂η h f, f sin(2 φ ) i + 12 (cid:28) F ( η, ψ ) , ∂ ( f ) ∂φ cos φ (cid:29) + (cid:13)(cid:13) f − ¯ f (cid:13)(cid:13) L = 0 . (1.29)We may integrate by parts to get12 (cid:28) F ( η, ψ ) , ∂ ( f ) ∂φ cos φ (cid:29) = 12 h F ( η, ψ ) f, f sin(2 φ ) i . (1.30)Since F ( η, ψ ) depends on ψ when R = R in 3D, we cannot further pull F out of the integral, i.e.12 h F ( η, ψ ) f, f sin(2 φ ) i ? = 12 F ( η, ψ ) h f, f sin(2 φ ) i . (1.31)This important equality is true in (1.27) for 2D domains and yields an ordinary differential equation for h f, f sin(2 φ ) i , leading to the closure of the L estimate of the microscopic part f − ¯ f . Similarly, taking innerproduct with 1 on both sides of (1.27) and integrating by parts, we easily get the orthogonality relation h f, sin φ i = 0 , (1.32)which plays a crucial role in estimating hydrodynamical part ¯ f . Unfortunately, both (1.31) and (1.32) breakdown in 3D domains.To circumvent these two major difficulties, as Lemma 3.1 reveals, we decompose F ( η, ψ ) cos φ ∂f∂φ = ˜ F ( η ) cos φ ∂f∂φ + G ( η ) cos ψ cos φ ∂f∂φ , (1.33)where ˜ F ( η ) = − ǫR − ǫη , (1.34) G ( η ) = − ǫ ( R − R )( R − ǫη )( R − ǫη ) , (1.35)in which ˜ F behaves like 2D force (independent of ψ ) and G can be regarded as a source term. Roughlyspeaking, taking inner product with f in (1.9), we obtain (cid:13)(cid:13) f − ¯ f (cid:13)(cid:13) L L . C (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L G ( y ) (cid:10) f cos ψ, f sin φ (cid:11) ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + lower-order terms (1.36) . C + C k G k L ∞ L ∞ k f k L L , which means we cannot close the estimate for f − ¯ f alone without invoking (cid:13)(cid:13) ¯ f (cid:13)(cid:13) L L . On the other hand,taking inner product with sin φ in (1.9) indicates (cid:13)(cid:13) ¯ f (cid:13)(cid:13) L L . C Z L (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z Ls Z Lz G ( y ) (cid:10) cos ψ, ( f − ¯ f ) sin φ (cid:11) ( y )d y d z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d s + lower-order terms (1.37) . C + CL k G k L L (cid:13)(cid:13) f − ¯ f (cid:13)(cid:13) L L . YAN GUO AND LEI WU (1.36) and (1.37) form a coupled system for f − ¯ f and ¯ f and require careful analysis of the interplay betweenmicroscopic and hydrodynamic parts. Also, we have to delicately choose L = ǫ − n with 0 < n <
25 to createa small constant such that an intricate bootstrapping argument can finally close the L estimates.The second key ingredient in our analysis is to establish the regularity estimate of ǫ -Milne problem.Proving diffusive limit in transport equations requires boundary layer expansion higher than leading-orderterm, which means we need L ∞ estimate of the tangential derivatives ∂f∂τ , ∂f∂τ , and ∂f∂ψ . (1.38)In the case when R = R are constant independent of τ and τ , as in a perfect ball n | ~x | = 1 o , ∂f∂τ i for i = 1 , R and R are functions of τ i , then ∂f∂τ i relates to the normal derivative ∂f∂η , and ∂f∂ψ relates to the velocityderivative ∂f∂φ .Our main contribution is to show ∂f∂τ i and ∂f∂ψ are bounded even if R and R are not identical constantfor a general convex domain(see Theorem 4.12). Our proof is intricate and delicate, which relies on theweighted L ∞ estimates for the normal derivative with detailed analysis along the characteristic curves in thepresence of non-local operator ¯ f . The convexity and invariant kinetic distance ζ defined in (1.20) play thekey role.The third ingredient is a new L − L ∞ framework developed to improve remainder estimates in ( ǫ ~w · ∇ x u + u − ¯ u = S ( ~x, ~w ) in Ω ,u ( ~x , ~w ) = P [ u ]( ~x ) + h ( ~x , ~w ) for ~w · ~ν < ~x ∈ ∂ Ω , (1.39)The main idea is to introduce special test functions in the weak formulation to treat kernel and non-kernelparts separately. In principle, we get L estimate1 ǫ k (1 − P )[ u ] k L (Γ + ) + k ¯ u k L (Ω ×S ) + 1 ǫ k u − ¯ u k L (Ω ×S ) (1.40) ≤ C (cid:18) o (1) ǫ k u k L ∞ (Γ + ) + 1 ǫ k S k L (Ω ×S ) + 1 ǫ k S k L (Ω ×S ) + 1 ǫ k h k L (Γ − ) + k h k L (Γ − ) (cid:19) , where o (1) denotes a sufficiently small constant (see Theorem A.3). The proof relies on a careful analysisusing sharp interpolation and Young’s inequality. Finally, the utilization of modified double Duhamel’sprinciple and a bootstrapping argument yield the L ∞ estimate as k u k L ∞ (Ω ×S ) ≤ C (cid:18) ǫ k ¯ u k L (Ω ×S ) + k S k L ∞ (Ω ×S ) + k g k L ∞ (Γ − ) (cid:19) . (1.41)Our methods are currently being applied to the study of hydrodynamic limit of Boltzmann equation inthe bounded domains with boundary layer corrections.1.4. Notation and Structure.
Throughout this paper, unless specified,
C > C ( z ), itmeans a positive constant depending on the quantity z .Our paper is organized as follows: in Section 2, we present the asymptotic analysis of the equation (1.1);in Section 3, we prove the well-posedness and decay of ǫ -Milne problem, i.e. Theorem 1.1; in Section 4, weprove the weighted W , ∞ estimates in ǫ -Milne problem, i.e. Theorem 1.2; finally, in appendix, we prove theimproved L ∞ estimate of remainder equation and the diffusive limit, i.e. Corollary 1.4. EGULARITY OF MILNE PROBLEM IN 3D 7 Asymptotic Analysis
Interior Expansion.
We first try to approximate the solution of neutron transport equation (1.1).We define the interior expansion as follows: U ǫ ( ~x, ~w ) ∼ U ǫ ( ~x, ~w ) + ǫU ǫ ( ~x, ~w ) + ǫ U ǫ ( ~x, ~w ) , (2.1)where U ǫk can be determined by comparing the order of ǫ by plugging (2.1) into the equation (1.1). Thus wehave U ǫ − ¯ U ǫ = 0 , (2.2) U ǫ − ¯ U ǫ = − ~w · ∇ x U ǫ , (2.3) U ǫ − ¯ U ǫ = − ~w · ∇ x U ǫ . (2.4)Plugging (2.2) into (2.3), we obtain U ǫ = ¯ U ǫ − ~w · ∇ x ¯ U ǫ . (2.5)Plugging (2.5) into (2.4), we get U ǫ − ¯ U ǫ = − ~w · ∇ x ( ¯ U ǫ − ~w · ∇ x ¯ U ǫ ) = − ~w · ∇ x ¯ U ǫ + w ∂ x x ¯ U ǫ + w ∂ x x ¯ U ǫ + 2 w w ∂ x x ¯ U ǫ . (2.6)Integrating (2.6) over ~w ∈ S , we achieve the final form∆ x ¯ U ǫ = 0 . (2.7)which further implies U ǫ ( ~x, ~w ) satisfies the equation ( U ǫ = ¯ U ǫ , ∆ x ¯ U ǫ = 0 . (2.8)In a similar fashion, for k = 1 , U ǫk satisfies U ǫk = ¯ U ǫk − ~w · ∇ x U ǫk − , ∆ x ¯ U ǫk = − Z S ~w · ∇ x U ǫk − d ~w. (2.9)It is easy to see ¯ U ǫk satisfies an elliptic equation. However, the boundary condition of ¯ U ǫk is unknown at thisstage, since generally U ǫk does not necessarily satisfy the diffusive boundary condition of (1.1). Therefore,we have to resort to boundary layer.2.2. Local Coordinate System.
Basically, we use two types of coordinate systems: Cartesian coordinatesystem for interior solution, which is stated above, and a local coordinate system in a neighborhood of theboundary for boundary layer. We need several substitution to describe solution near boundary.Substitution 1: spacial substitution:We consider the three-dimensional transport operator ~w · ∇ x . In the boundary surface, locally we can alwaysdefine an orthogonal curvilinear coordinates system ( τ , τ ) and the surface is described as ~r ( τ , τ ). Fromthe differential geometry, we know ∂ ~r and ∂ ~r denote two orthogonal tangential vectors. Then assume theoutward unit normal vector is ~ν = ∂ ~r × ∂ ~r | ∂ ~r × ∂ ~r | . (2.10)Here |·| denotes the length and ∂ i denotes derivative with respect to τ i . Let P = | ∂ ~r × ∂ ~r | = | ∂ ~r | | ∂ ~r | = P P , (2.11)with the unit tangential vectors are ~t = ∂ ~rP , ~t = ∂ ~rP . (2.12)Then in the new coordinates ( µ, τ , τ ) where µ denotes the normal distance to boundary surface, we have ~x = ~r − µ~ν. (2.13) YAN GUO AND LEI WU which further implies the operator becomes ~w · ∇ x = − (cid:18) ( ∂ ~r − µ∂ ~ν ) × ( ∂ ~r − µ∂ ~ν ) (cid:19) · ~w (cid:18) ( ∂ ~r − µ∂ ~ν ) × ( ∂ ~r − µ∂ ~ν ) (cid:19) · ~ν ∂f∂µ (2.14)+ (cid:18) ( ∂ ~r − µ∂ ~ν ) × ~ν (cid:19) · ~w (cid:18) ( ∂ ~r − µ∂ ~ν ) × ( ∂ ~r − µ∂ ~ν ) (cid:19) · ~ν ∂f∂τ − (cid:18) ( ∂ ~r − µ∂ ~ν ) × ~ν (cid:19) · ~w (cid:18) ( ∂ ~r − µ∂ ~ν ) × ( ∂ ~r − µ∂ ~ν ) (cid:19) · ~ν ∂f∂τ . Based on differential geometry, we define the first fundamental form as (
E, F, G ) and second fundamentalform as (
L, M, N ), then we have F = M = 0 and the principal curvatures are given by κ = LE , κ = NG , (2.15)and also ∂ ~ν = κ ∂ ~r, ∂ ~ν = κ ∂ ~r. (2.16)Hence, we know (cid:18) ( ∂ ~r − µ∂ ~ν ) × ( ∂ ~r − µ∂ ~ν ) (cid:19) · ~ν = (cid:18) κ κ µ − ( κ + κ ) µ + 1 (cid:19) P (2.17)= (cid:18) ( κ µ − κ µ − (cid:19) P. (cid:18) ( ∂ ~r − µ∂ ~ν ) × ( ∂ ~r − µ∂ ~ν ) (cid:19) · ~w = (cid:18) κ κ µ − ( κ + κ ) µ + 1 (cid:19) P ( ~ν · ~w ) (2.18)= (cid:18) ( κ µ − κ µ − (cid:19) P ( ~w · ~ν ) . (cid:18) ( ∂ ~r − µ∂ ~ν ) × ~ν (cid:19) · ~w = (1 − κ µ ) P P ( ~w · ∂ ~r ) . (2.19) (cid:18) ( ∂ ~r − µ∂ ~ν ) × ~ν (cid:19) · ~w = − (1 − κ µ ) P P ( ~w · ∂ ~r ) . (2.20)Hence, we have the transport operator as ~w · ∇ x = − ( ~w · ~ν ) ∂∂µ − ~w · ~t P ( κ µ − ∂∂τ − ~w · ~t P ( κ µ − ∂∂τ . (2.21)Therefore, under substitution ( x , x , x ) → ( µ, τ , τ ), the equation (1.1) is transformed into ǫ (cid:18) − ( ~w · ~ν ) ∂u ǫ ∂µ − ~w · ~t P ( κ µ − ∂u ǫ ∂τ − ~w · ~t P ( κ µ − ∂u ǫ ∂τ (cid:19) + u ǫ − ¯ u ǫ = 0 in Ω ,u ǫ (0 , τ , τ , ~w ) = P [ u ǫ ](0 , τ , τ ) + ǫg ( τ , τ , ~w ) for ~w · ~ν < , (2.22)where P [ u ǫ ](0 , τ , τ ) = 12 π Z ~w · ~ν> u ǫ (0 , τ , τ , ~w )( ~w · ~ν )d ~w, (2.23)Substitution 2: velocity substitution.Define the orthogonal velocity substitution − ~w · ~ν = sin φ,~w · ~t = cos φ sin ψ,~w · ~t = cos φ cos ψ. (2.24) EGULARITY OF MILNE PROBLEM IN 3D 9
Then we have ∂∂τ → ∂∂τ − κ P sin ψ ∂∂φ + (cid:18) ( ∂ ~r × ( ∂ ~r × ∂ ~r )) · ~t P − κ P tan φ cos ψ (cid:19) ∂∂ψ , (2.25) ∂∂τ → ∂∂τ − κ P cos ψ ∂∂φ + (cid:18) ( ∂ ~r × ( ∂ ~r × ∂ ~r )) · ~t P + κ P tan φ sin ψ (cid:19) ∂∂ψ . (2.26)Then the transport operator is as ~w · ∇ x = sin φ ∂∂µ − (cid:18) sin ψR − µ + cos ψR − µ (cid:19) cos φ ∂∂φ (2.27)+ (cid:18) cos φ sin ψP (1 − κ µ ) ∂∂τ + cos φ cos ψP (1 − κ µ ) ∂∂τ (cid:19) + (cid:18) sin ψ − κ µ ( ~t × ( ∂ ~r × ~t )) · ~t + cos ψ − κ µ ( ~t × ( ∂ ~r × ~t )) · ~t (cid:19) cos φP P ∂∂ψ , where R = 1 κ and R = 1 κ . Hence, under substitution ( w , w , w ) → ( φ, ψ ), the equation (1.1) istransformed into ǫ sin φ ∂u ǫ ∂µ − ǫ (cid:18) sin ψR − µ + cos ψR − µ (cid:19) cos φ ∂u ǫ ∂φ + ǫ (cid:18) cos φ sin ψP (1 − κ µ ) ∂u ǫ ∂τ + cos φ cos ψP (1 − κ µ ) ∂u ǫ ∂τ (cid:19) + ǫ (cid:18) sin ψ − κ µ ( ~t × ( ∂ ~r × ~t )) · ~t + cos ψ − κ µ ( ~t × ( ∂ ~r × ~t )) · ~t (cid:19) cos φP P ∂u ǫ ∂ψ + u ǫ − ¯ u ǫ = 0 ,u ǫ (0 , τ , τ , φ, ψ ) = P [ u ǫ ](0 , τ , τ ) + ǫg ( τ , τ , φ, ψ ) for sin φ > , (2.28)where P [ u ǫ ](0 , τ , τ ) = 14 π Z Z sin φ> u ǫ (0 , τ , τ , φ, ψ ) sin φ cos φ d φ d ψ, (2.29)due to Jacobian J = cos φ , in a neighborhood of the boundary.Substitution 3: scaling substitution.We define the scaled variable η = µǫ , which implies ∂∂µ = 1 ǫ ∂∂η . Then, under the substitution µ → η , theequation (1.1) is transformed into sin φ ∂u ǫ ∂η − ǫ (cid:18) sin ψR − ǫη + cos ψR − ǫη (cid:19) cos φ ∂u ǫ ∂φ + ǫ (cid:18) cos φ sin ψP (1 − ǫκ η ) ∂u ǫ ∂τ + cos φ cos ψP (1 − ǫκ η ) ∂u ǫ ∂τ (cid:19) + ǫ (cid:18) sin ψ − ǫκ η ( ~t × ( ∂ ~r × ~t )) · ~t + cos ψ − ǫκ η ( ~t × ( ∂ ~r × ~t )) · ~t (cid:19) cos φP P ∂u ǫ ∂ψ + u ǫ − ¯ u ǫ = 0 ,u ǫ (0 , τ , τ , φ, ψ ) = P [ u ǫ ](0 , τ , τ ) + ǫg ( τ , τ , φ, ψ ) for sin φ > , (2.30)where P [ u ǫ ](0 , τ , τ ) = 14 π Z Z sin φ> u ǫ (0 , τ , τ , φ, ψ ) sin φ cos φ d φ d ψ, (2.31)in a neighborhood of the boundary. Boundary Layer Expansion with Geometric Correction.
We define the boundary layer expansionas follows: U ǫ ( η, τ , τ , φ, ψ ) ∼ U ǫ ( η, τ , τ , φ, ψ ) + ǫ U ǫ ( η, τ , τ , φ, ψ ) , (2.32)where U ǫk can be defined by comparing the order of ǫ via plugging (2.32) into the equation (2.30). Thus, ina neighborhood of the boundary, we havesin φ ∂ U ǫ ∂η − ǫ (cid:18) sin ψR − ǫη + cos ψR − ǫη (cid:19) cos φ ∂ U ǫ ∂φ + U ǫ − ¯ U ǫ = 0 , (2.33)sin φ ∂ U ǫ ∂η − ǫ (cid:18) sin ψR − ǫη + cos ψR − ǫη (cid:19) cos φ ∂ U ǫ ∂φ + U ǫ − ¯ U ǫ = − G , (2.34)where G = (cid:18) cos φ sin ψP (1 − ǫκ η ) ∂ U ǫ ∂τ + cos φ cos ψP (1 − ǫκ η ) ∂ U ǫ ∂τ (cid:19) (2.35)+ (cid:18) sin ψ − ǫκ η ( ~t × ( ∂ ~r × ~t )) · ~t + cos ψ − ǫκ η ( ~t × ( ∂ ~r × ~t )) · ~t (cid:19) cos φP P ∂ U ǫ ∂ψ , and ¯ U ǫk ( η, τ , τ ) = 14 π Z π − π Z π/ − π/ U ǫk ( η, τ , τ , φ, ψ ) cos φ d φ d ψ. (2.36)Note that this formulation is always valid locally on the boundary. By open covering theorem, we can findfinite open domains to cover the whole surface. For convenience, we will not change our notation in eachopen domain.2.4. Matching of Interior Solution and Boundary Layer.
The bridge between the interior solution andthe boundary layer solution is the boundary condition of (1.1), so we first consider the boundary conditionexpansion: ( U ǫ + U ǫ ) = P [ U ǫ + U ǫ ] , (2.37)( U ǫ + U ǫ ) = P [ U ǫ + U ǫ ] + g. (2.38)Note the fact that ¯ U ǫk = P [ ¯ U ǫk ], we can simplify above conditions as follows: U ǫ = P [ U ǫ ] , (2.39) U ǫ = P [ U ǫ ] + ( ~w · U ǫ − P [ ~w · U ǫ ]) + g. (2.40)The construction of U ǫk and U ǫk are as follows:Step 0: Preliminaries.Assume the cut-off function Υ ∈ C ∞ [0 , ∞ ) are defined asΥ ( µ ) = ≤ µ ≤ R min , R min ≤ µ ≤ ∞ . (2.41)where R min = min τ ,τ { R ( τ , τ ) , R ( τ , τ ) } . (2.42)Define the length of boundary layer L = ǫ − n for 0 < n <
25 and the force as F ( ǫ ; η, ψ ) = − ǫ (cid:18) sin ψR − ǫη + cos ψR − ǫη (cid:19) . (2.43)Also, denote R φ = − φ . EGULARITY OF MILNE PROBLEM IN 3D 11
Step 1: Construction of U ǫ .Define the zeroth-order boundary layer as U ǫ ( η, τ , τ , φ, ψ ) = Υ ( ǫ n η ) (cid:18) f ǫ ( η, τ , τ , φ, ψ ) − f ǫ ,L ( τ , τ ) (cid:19) , sin φ ∂f ǫ ∂η + F ( ǫ ; η, ψ ) cos φ ∂f ǫ ∂φ + f ǫ − ¯ f ǫ = 0 ,f ǫ (0 , τ , τ , φ, ψ ) = P [ f ǫ ](0 , τ , τ ) for sin φ > ,f ǫ ( L, τ , τ , φ, ψ ) = f ǫ ( L, τ , τ , R φ, ψ ) , (2.44)with P [ f ǫ ](0 , τ , τ ) = 0 . (2.45)By Theorem 3.11, U ǫ is well-defined. It is obvious to see f ǫ = f ǫ ,L = 0 is the only solution.Step 2: Construction of U ǫ and U ǫ .Define the first-order boundary layer as U ǫ ( η, τ , τ , φ, ψ ) = Υ ( ǫ n η ) (cid:18) f ǫ ( η, τ , τ , φ, ψ ) − f ǫ ,L ( τ , τ ) (cid:19) , sin φ ∂f ǫ ∂η + F ( ǫ ; η, ψ ) cos φ ∂f ǫ ∂φ + f ǫ − ¯ f ǫ = − G ,f ǫ (0 , τ , τ , φ, ψ ) = P [ f ǫ ](0 , τ , τ ) + g ( τ , τ , φ, ψ ) for sin φ > ,f ǫ ( L, τ , τ , φ, ψ ) = f ǫ ( L, τ , τ , R φ, ψ ) , (2.46)with P [ f ǫ ](0 , τ , τ ) = 0 , (2.47)where g = ( ~w · ∇ x U ǫ ( ~x ) − P [ ~w · ∇ x U ǫ ( ~x )]) + g, (2.48)with ~x is the same boundary point as (0 , τ , τ ). To solve (2.46), we require the compatibility condition(3.102) for the boundary data (2.49) Z Z sin φ> (cid:18) g + ~w · ∇ x U ǫ ( ~x ) − P [ ~w · ∇ x U ǫ ( ~x )] (cid:19) sin φ cos φ d φ d ψ − Z L Z π − π Z π/ e − V ( s ) G cos φ d φ d ψ d s = 0 , where V (0) = 0 and ∂V∂η = − F . Note the fact ~w = (sin φ ) ~ν + (cos φ sin ψ ) ~t + (cos φ cos ψ ) ~t and Z Z sin φ> (cid:18) ~w · ∇ x U ǫ ( ~x ) − P [ ~w · ∇ x U ǫ ( ~x )] (cid:19) sin φ cos φ d φ d ψ (2.50)= Z Z sin φ> ( ~w · ∇ x U ǫ ( ~x )) sin φ cos φ d φ d ψ − π P [ ~w · ∇ x U ǫ ( ~x )]= Z Z sin φ> ( ~w · ∇ x U ǫ ( ~x )) sin φ cos φ d φ d ψ + Z Z sin φ< ( ~w · ∇ x U ǫ ( ~x )) sin φ cos φ d φ d ψ = Z π − π Z π/ − π/ ( ~w · ∇ x U ǫ ( ~x )) sin φ cos φ d φ d ψ = Z π − π Z π/ − π/ ( ~ν · ∇ x U ǫ ( ~x )) sin φ cos φ d φ d ψ = − π ∇ x ¯ U ǫ ( ~x ) · ~ν = − π ∂ ¯ U ǫ ( ~x ) ∂~ν . We can simplify the compatibility condition as follows:
Z Z sin φ> g ( τ , τ , φ, ψ ) sin φ cos φ d φ d ψ − π ∂ ¯ U ǫ ( ~x ) ∂~ν − Z L Z π − π Z π/ − π/ e − V ( s ) G cos φ d φ d ψ d s = 0 . (2.51)Then we have ∂ ¯ U ǫ ( ~x ) ∂~n = 1 π Z Z sin φ> g ( τ , τ , φ, ψ ) sin φ cos φ d φ d ψ + 1 π Z L Z π − π Z π/ − π/ e − V ( s ) G cos φ d φ d ψ d s (2.52)= 1 π Z Z sin φ> g ( τ , τ , φ, ψ ) sin φ cos φ d φ d ψ. Hence, we define the zeroth order interior solution U ǫ ( ~x, ~w ) as U ǫ = ¯ U ǫ , ∆ x ¯ U ǫ = 0 in Ω ,∂ ¯ U ǫ ∂~ν = 1 π Z Z sin φ> g ( τ , τ , φ, ψ ) sin φ cos φ d φ d ψ on ∂ Ω , Z Ω ¯ U ǫ ( ~x )d ~x = 0 . (2.53)Step 3: Construction of U ǫ .We do not expand the boundary layer to U ǫ and just terminate at U ǫ . Then we define the first orderinterior solution U ǫ ( ~x ) as U ǫ = ¯ U ǫ − ~w · ∇ x U ǫ , ∆ x ¯ U ǫ = − Z S ( ~w · ∇ x U ǫ )d ~w in Ω ,∂ ¯ U ǫ ∂~ν = − Z Ω Z S ( ~w · ∇ x U ǫ )d ~w d ~x on ∂ Ω , Z Ω ¯ U ǫ ( ~x )d ~x = Z Ω Z S ( ~w · ∇ x U ǫ − U ǫ )d ~w d ~x. (2.54)Note that here we only require the trivial boundary condition since we cannot resort to the compatibilitycondition in ǫ -Milne problem with geometric correction. Based on [27], this might lead to O ( ǫ ) error to theboundary approximation. Thanks to the improved remainder estimate, this error is acceptable.Step 4: Construction of U ǫ .By a similar fashion, we define the second order interior solution as U ǫ = ¯ U ǫ − ~w · ∇ x U ǫ , ∆ x ¯ U ǫ = − Z S ( ~w · ∇ x U ǫ )d ~w in Ω ,∂ ¯ U ǫ ∂~ν = − Z Ω Z S ( ~w · ∇ x U ǫ )d ~w d ~x on ∂ Ω , Z Ω ¯ U ǫ ( ~x )d ~x = Z Ω Z S ~w · ∇ x U ǫ d ~w d ~x. (2.55)As the case of U ǫ , we might have O ( ǫ ) error in this step due to the trivial boundary data. However, it willnot affect the diffusive limit. EGULARITY OF MILNE PROBLEM IN 3D 13 Well-Posedness and Decay of ǫ -Milne Problem We consider the ǫ -Milne problem for f ǫ ( η, φ, ψ ) in the domain ( η, φ, ψ ) ∈ [0 , L ] × [ − π/ , π/ × [ − π, π ] as sin φ ∂f ǫ ∂η + F ( ǫ ; η, ψ ) cos φ ∂f ǫ ∂φ + f ǫ − ¯ f ǫ = S ǫ ( η, φ, ψ ) ,f ǫ (0 , φ, ψ ) = h ǫ ( φ, ψ ) for sin φ > ,f ǫ ( L, φ, ψ ) = f ǫ ( L, R φ, ψ ) , (3.1)where ¯ f ǫ ( η ) = 14 π Z π − π Z π/ − π/ f ǫ ( η, φ, ψ ) cos φ d φ d ψ (3.2)in which cos φ shows up as the Jacobian of spherical coordinates in integration. Note that for φ ∈ [ − π/ , π/ φ ≥
0, which means this will not destroy the positivity of integral. Also, we have F ( ǫ ; η, ψ ) = − ǫ (cid:18) sin ψR − ǫη + cos ψR − ǫη (cid:19) . (3.3)for R and R radium of two principle curvature, and L = ǫ − n for some n > ǫ . Note that all the estimates weget will be uniform in ǫ . We define the norms in the space ( η, φ, ψ ) ∈ [0 , L ] × [ − π/ , π/ × [ − π, π ) as follows: k f k L L = (cid:18) Z L Z π − π Z π/ − π/ | f ( η, φ, ψ ) | cos φ d φ d ψ d η (cid:19) / , (3.4) k f k L ∞ L ∞ = sup ( η,φ,ψ ) | f ( η, φ, ψ ) | . (3.5)Also, we define the inner product as h f, g i ( η ) = Z π − π Z π/ − π/ f ( η, φ, ψ ) g ( η, τ , τ , φ, ψ ) cos φ d φ d ψ. (3.6)Similarly, we can define the norm at in-flow boundary as k f k L ( η ) = (cid:18) Z Z sin φ> | f ( η, φ, ψ ) | cos φ d φ d ψ (cid:19) / , (3.7) k f k L ∞ ( η ) = sup ( φ,ψ ) with sin φ> | f ( η, φ, ψ ) | , (3.8)We further assume k h k L ∞ ≤ M, (3.9)and (cid:13)(cid:13) e Kη S (cid:13)(cid:13) L ∞ L ∞ ≤ M, for M >
K > ǫ .3.1. L Estimates. L Estimates when ¯ S = 0 . Consider the equation sin φ ∂f∂η + F ( η, ψ ) cos φ ∂f∂φ + f − ¯ f = S ( η, φ, ψ ) ,f (0 , φ, ψ ) = h ( φ, ψ ) for sin φ > ,f ( L, φ, ψ ) = f ( L, R φ, ψ ) . (3.10)where F ( η, ψ ) cos φ ∂f∂φ = ˜ F ( η ) cos φ ∂f∂φ + G ( η ) cos ψ cos φ ∂f∂φ , (3.11) for ˜ F ( η ) = − ǫR − ǫη , (3.12) G ( η ) = − ǫ ( R − R )( R − ǫη )( R − ǫη ) . (3.13)Also, R φ = − φ . We may decompose the solution f ( η, φ, ψ ) = q ( η ) + r ( η, φ, ψ ) , (3.14)where the hydrodynamical part q is in the null space of the operator f − ¯ f , and the microscopic part r isthe orthogonal complement, i.e. q ( η ) = 14 π Z π − π Z π/ − π/ f ( η, φ, ψ ) cos φ d φ d ψ, r ( η, φ, ψ ) = f ( η, φ, ψ ) − q ( η ) . (3.15)Furthermore, we define a potential function ˜ V ( η ) satisfying ˜ V (0) = 0 and ∂ ˜ V∂η = − ˜ F ( η ). It is easy tocompute ˜ V ( η ) = ln (cid:18) R R − ǫη (cid:19) . Lemma 3.1.
Assume ¯ S = 0 satisfying (3.9) and (3.10) and < n < . There exists a solution f ( η, φ, ψ ) to the equation (3.10), satisfying for some constant | f L | ≤ C , k f − f L k L L ≤ C. (3.16) The solution is unique among functions such that (A.1) holds.Proof.
As in [27, Section 6], the existence can be proved using a standard approximation argument, so wewill only focus on the a priori estimates. We divide the proof into several steps:Step 1: r Estimates.Multiplying f cos φ on both sides of (3.10) and integrating over ( φ, ψ ) ∈ [ − π/ , π/ × [ − π, π ), we get theenergy estimate 12 dd η h f, f sin φ i ( η ) = − k r ( η ) k L − ˜ F ( η ) (cid:28) ∂f∂φ , f cos φ (cid:29) ( η ) (3.17) − G ( η ) (cid:28) ∂f∂φ cos ψ, f cos φ (cid:29) ( η ) + h S, f i ( η ) . A further integration by parts in φ reveals − ˜ F ( η ) (cid:28) ∂f∂φ , f cos φ (cid:29) ( η ) = − ˜ F ( η ) h f, f sin φ i ( η ) , (3.18) − G ( η ) (cid:28) ∂f∂φ cos ψ, f cos φ (cid:29) ( η ) = − G ( η ) (cid:10) f cos ψ, f sin φ (cid:11) ( η ) . (3.19)Hence, we can simplify (3.17) as12 dd η h f, f sin φ i ( η ) = − k r ( η ) k L − ˜ F ( η ) h f, f sin φ i ( η ) (3.20) − G ( η ) (cid:10) f cos ψ, f sin φ (cid:11) ( η ) + h S, f i ( η ) . Define α ( η ) = 12 h f, f sin φ i ( η ) . (3.21)Then (3.20) can be rewritten asd α d η = − k r ( η ) k L − F ( η ) α ( η ) − G ( η ) (cid:10) f cos ψ, f sin φ (cid:11) ( η ) + h S, f i ( η ) . (3.22) EGULARITY OF MILNE PROBLEM IN 3D 15
We can solve above in [ η, L ] and [0 , η ] respectively to obtain (3.23) α ( η ) = e V ( η ) − V ( L ) α ( L ) + Z Lη e V ( η ) − V ( y ) (cid:18) k r ( y ) k L + G ( y ) (cid:10) f cos ψ, f sin φ (cid:11) ( y ) − h S, f i ( y ) (cid:19) d y, (3.24) α ( η ) = e V ( η ) α (0) + Z η e V ( η ) − V ( y ) (cid:18) − k r ( y ) k L − G ( y ) (cid:10) f cos ψ, f sin φ (cid:11) ( y ) + h S, f i ( y ) (cid:19) d y. The specular reflexive boundary f ( L, φ ) = f ( L, R φ ) ensures α ( L ) = 0. Hence, based on (3.23), we have α ( η ) ≥ Z Lη e V ( η ) − V ( y ) (cid:18) G ( y ) (cid:10) f cos ψ, f sin φ (cid:11) ( y ) − h S, f i ( y ) (cid:19) d y. (3.25)Also, (3.24) implies α ( η ) ≤ Cα (0) + Z η e V ( η ) − V ( y ) (cid:18) − G ( y ) (cid:10) f cos ψ, f sin φ (cid:11) ( y ) + h S, f i ( y ) (cid:19) d y (3.26) ≤ C k h k L + Z η e V ( η ) − V ( y ) (cid:18) − G ( y ) (cid:10) f cos ψ, f sin φ (cid:11) ( y ) + h S, f i ( y ) (cid:19) d y, due to the fact α (0) = 12 h f sin φ, f i (0) ≤ (cid:18) Z sin φ> h ( φ ) sin φ cos φ d φ (cid:19) ≤ k h k L . (3.27)Then in (3.24) taking η = L , from α ( L ) = 0, we have Z L e − V ( y ) k r ( y ) k L d y (3.28) ≤ α (0) + Z L e − V ( y ) (cid:18) − G ( y ) (cid:10) f cos ψ, f sin φ (cid:11) ( y ) + h S, f i ( y ) (cid:19) d y ≤ C k h k L + Z L e − V ( y ) (cid:18) − G ( y ) (cid:10) f cos ψ, f sin φ (cid:11) ( y ) + h S, f i ( y ) (cid:19) d y. On the other hand, we can directly estimate Z L e − V ( y ) k r ( y ) k L d y ≥ C k r k L L . (3.29)Combining above yields k r k L L ≤ C k h k L + Z L e − V ( y ) (cid:18) − G ( y ) (cid:10) f cos ψ, f sin φ (cid:11) ( y ) + h S, f i ( y ) (cid:19) d y ! . (3.30)Since h S, f i = h S, r i due to ¯ S = 0, by Cauchy’s inequality, we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L e − V ( y ) h S, r i ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C ′ k r k L L + C k S k L L , (3.31)for C ′ > C ′ k r k L L term, we deduce k r k L L ≤ C k h k L + k S k L L + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L G ( y ) (cid:10) f cos ψ, f sin φ (cid:11) ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)! (3.32) ≤ C (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L G ( y ) (cid:10) f cos ψ, f sin φ (cid:11) ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)! ≤ C (cid:16) k G k L ∞ L ∞ k f k L L (cid:17) ≤ C (cid:16) ǫ k f k L L (cid:17) . Note that this estimate is not closed since it depends on f .Step 2: Quasi-Orthogonality relation.Multiplying cos φ on both sides of (3.10) and integrating over ( φ, ψ ) ∈ [ − π/ , π/ × [ − π, π ) implydd η h sin φ, f i ( η ) = − ˜ F (cid:28) cos φ, d f d φ (cid:29) ( η ) − G (cid:28) cos φ cos ψ, d f d φ (cid:29) ( η ) + ¯ S ( η ) (3.33)= − F h sin φ, f i ( η ) − G (cid:10) sin φ cos ψ, f (cid:11) ( η ) . The specular reflexive boundary f ( L, φ ) = f ( L, R φ ) implies h sin φ, f i ( L ) = 0. Then we have h sin φ, f i ( η ) = − Z Lη e V ( η ) − ˜ V ( y ) G ( y ) (cid:10) sin φ cos ψ, f (cid:11) ( y )d y. (3.34)It is easy to see h sin φ, q i ( η ) = 0 . (3.35)Hence, we may derive h sin φ, r i ( η ) = − Z Lη e V ( η ) − ˜ V ( y ) G ( y ) (cid:10) sin φ cos ψ, f (cid:11) ( y )d y, (3.36)= − Z Lη e V ( η ) − ˜ V ( y ) G ( y ) (cid:10) sin φ cos ψ, r (cid:11) ( y )d y. Step 3: q Estimates.Multiplying sin φ cos φ on both sides of (3.10) and integrating over ( φ, ψ ) ∈ [ − π/ , π/ × [ − π, π ) lead todd η (cid:10) sin φ, f (cid:11) ( η ) = − h sin φ, r i ( η ) − ˜ F ( η ) (cid:28) sin φ cos φ, ∂f∂φ (cid:29) ( η ) (3.37) − G ( η ) (cid:28) sin φ cos φ cos ψ, ∂f∂φ (cid:29) ( η ) + h sin φ, S i ( η ) . We can further integrate by parts in φ as − ˜ F ( η ) (cid:28) sin φ cos φ, ∂f∂φ (cid:29) ( η ) = ˜ F ( η ) (cid:10) − φ, f (cid:11) ( η ) = ˜ F ( η ) (cid:10) − φ, r (cid:11) ( η ) , (3.38)(3.39) − G ( η ) (cid:28) sin φ cos φ cos ψ, ∂f∂φ (cid:29) ( η ) = G ( η ) (cid:10) − φ, f cos ψ (cid:11) ( η ) = G ( η ) (cid:10) − φ, r cos ψ (cid:11) ( η ) , to obtain dd η (cid:10) sin φ, f (cid:11) ( η ) = − h sin φ, r i ( η ) + ˜ F ( η ) (cid:10) − φ, r (cid:11) ( η ) (3.40)+ G ( η ) (cid:10) − φ, r cos ψ (cid:11) ( η ) + h sin φ, S i ( η ) . Define β ( η ) = (cid:10) sin φ, f (cid:11) ( η ) . (3.41)Then we can simplify (3.37) as d β d η = D ( η ) , (3.42)where D ( η ) = − h sin φ, r i ( η ) + ˜ F ( η ) (cid:10) − φ, r (cid:11) ( η ) (3.43)+ G ( η ) (cid:10) − φ, r cos ψ (cid:11) ( η ) + h sin φ, S i ( η ) . We can integrate over [0 , η ] in (3.42) to obtain β ( η ) = β (0) + Z η D ( y )d y. (3.44) EGULARITY OF MILNE PROBLEM IN 3D 17
The quasi-orthogonal relation implies D ( η ) = 2 Z Lη e V ( η ) − ˜ V ( y ) G ( y ) (cid:10) sin φ cos ψ, f (cid:11) ( y )d y + ˜ F ( η ) (cid:10) − φ, r (cid:11) ( η ) (3.45)+ G ( η ) (cid:10) − φ, r cos ψ (cid:11) ( η ) + h sin φ, S i ( η ) . Hence, we deduce β ( η ) − β (0) = 2 Z η Z Lz e V ( z ) − V ( y ) G ( y ) (cid:10) sin φ cos ψ, r (cid:11) d y d z + Z η ˜ F ( y ) (cid:10) − φ, r (cid:11) ( y )d y (3.46)+ Z η G ( y ) (cid:10) − φ, r cos ψ (cid:11) ( y )d y + Z η h sin φ, S i ( y )d y. For the boundary data, β (0) = (cid:10) sin φ, f (cid:11) (0) ≤ (cid:18) h f, f | sin φ |i (0) (cid:19) / k sin φ k / L ≤ C (cid:18) h f, f | sin φ |i (0) (cid:19) / . (3.47)Obviously, we have h f, f | sin φ |i (0) = Z sin φ> h ( φ ) sin φ cos φ d φ − Z sin φ< (cid:18) f (0 , φ ) (cid:19) sin φ cos φ d φ. (3.48)However, based on the definition of α ( η ), we can obtain Z sin φ> h ( φ ) sin φ cos φ d φ + Z sin φ< (cid:18) f (0 , φ ) (cid:19) sin φ cos φ d φ = 2 α (0) (3.49) ≥ Z L e − V ( y ) (cid:18) G ( y ) (cid:10) f cos ψ, f sin φ (cid:11) ( y ) − h S, r i ( y ) (cid:19) d y. Hence, we can deduce − Z sin φ< (cid:18) f (0 , φ ) (cid:19) sin φ cos φ d φ (3.50) ≤ Z sin φ> h ( φ ) sin φ cos φ d φ − Z L e − V ( y ) (cid:18) G ( y ) (cid:10) f cos ψ, f sin φ (cid:11) ( y ) − h S, r i ( y ) (cid:19) d y ≤ k h k L + C (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L (cid:18) G ( y ) (cid:10) f cos ψ, f sin φ (cid:11) ( y ) − h S, r i ( y ) (cid:19) d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Hence, we obtain (cid:16) β (0) (cid:17) ≤ h f, f | sin φ |i (0) ≤ C k h k L + C (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L (cid:18) G ( y ) (cid:10) f cos ψ, f sin φ (cid:11) ( y ) − h S, r i ( y ) (cid:19) d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . (3.51) Note that k G k L ∞ L ∞ ≤ Cǫ . Since ˜ F ∈ L [0 , L ] ∩ L [0 , L ], r ∈ L ([0 , L ] × [ − π, π )), and S exponentiallydecays, using Cauchy’s inequality, we have | β ( L ) | ≤ C k h k L + C (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L G ( y ) (cid:10) f cos ψ, f sin φ (cid:11) ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / + C (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L h S, r i ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / (3.52)+ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L ˜ F ( y ) (cid:10) − φ, r (cid:11) ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L Z Lz e V ( z ) − V ( y ) G ( y ) (cid:10) sin φ cos ψ, r (cid:11) d y d z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L G ( y ) (cid:10) − φ, r cos ψ (cid:11) ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L h sin φ, S i ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L G ( y ) (cid:10) f cos ψ, f sin φ (cid:11) ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / + C k h k L + C k r k L L + C (cid:13)(cid:13)(cid:13) ˜ F (cid:13)(cid:13)(cid:13) L L k r k L L + k G k L L k r k L L + L k G k L L k r k L L + k S k L L ≤ C + C (1 + ǫ − n ) k r k L L + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L G ( y ) (cid:10) f cos ψ, f sin φ (cid:11) ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / ≤ C (1 + ǫ − n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L G ( y ) (cid:10) f cos ψ, f sin φ (cid:11) ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / ≤ C (1 + ǫ − n ) (cid:16) ǫ k f k L L (cid:17) ≤ C (1 + ǫ − n ) (cid:16) ǫ k f − f L k L L + ǫ k f L k L L (cid:17) ≤ C (1 + ǫ − n ) (cid:16) ǫ k f − f L k L L + ǫ − n | f L | (cid:17) , where we define f L = q L = β ( L ) k sin φ k L . (3.53)Therefore, for 0 < n <
23 and ǫ sufficiently small, absorbing | f L | , we have | f L | ≤ C + Cǫ k f − f L k L L . (3.54)Thus, we naturally have k r k L L ≤ C (cid:16) ǫ k f k L L (cid:17) (3.55) ≤ C (cid:16) ǫ k f − f L k L L + ǫ k f L k L L (cid:17) ≤ C (cid:16) ǫ k f − f L k L L (cid:17) , Furthermore, we have β ( L ) − β ( η ) = Z Lη D ( y )d y (3.56)= (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z Lη ˜ F ( y ) (cid:10) − φ, r (cid:11) ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z Lη Z Lz e V ( z ) − V y G ( y ) (cid:10) sin φ cos ψ, r (cid:11) d y d z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z Lη G ( y ) (cid:10) − φ, r cos ψ (cid:11) ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z Lη h sin φ, S i ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Note β ( η ) = (cid:10) sin φ, f (cid:11) ( η ) = (cid:10) sin φ, q (cid:11) ( η ) + (cid:10) sin φ, r (cid:11) ( η ) = q ( η ) k sin φ k L + (cid:10) sin φ, r (cid:11) ( η ) . (3.57) EGULARITY OF MILNE PROBLEM IN 3D 19
Thus considering S decays exponentially, we can estimate k q − q L k L L ≤ C k r k L L + C k β ( η ) − β ( L ) k L L (3.58) ≤ C k r k L L + Z L (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z Lη ˜ F ( y ) (cid:10) − φ, r (cid:11) ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d η + Z L (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z Lη Z Lz e V ( z ) − V y G ( y ) (cid:10) sin φ cos ψ, r (cid:11) d y d z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d η + Z L (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z Lη G ( y ) (cid:10) − φ, r cos ψ (cid:11) ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d η + Z L (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z Lη h sin φ, S i ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d η ≤ C k r k L L + k r k L L Z L Z Lη (cid:12)(cid:12)(cid:12) ˜ F ( y ) (cid:12)(cid:12)(cid:12) d y d η + L k G k L L k r k L L + L k G k L L k r k L L + Z L (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z Lη S ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d η ≤ C (1 + ǫ − n k r k L L ) ≤ C (1 + ǫ − n ) (cid:16) ǫ k f − f L k L L (cid:17) . Therefore, for 0 < n <
25 , we have k q − q L k L L ≤ C (cid:16) ǫ k f − f L k L L (cid:17) . (3.59)Step 4: Synthesis.For 0 < n <
25 , we have k r k L L ≤ C (cid:16) ǫ k f − f L k L L (cid:17) , (3.60) k q − q L k L L ≤ C (cid:16) ǫ k f − f L k L L (cid:17) , (3.61)which further implies k f − f L k L L ≤ k r k L L + k q − q L k L L ≤ C (cid:16) ǫ k f − f L k L L (cid:17) . (3.62)Hence, for ǫ sufficiently small, we have | f L | ≤ C and k f − f L k L L ≤ C. (3.63)In order to show the uniqueness of the solution, we assume there are two solutions f and f to the equation(3.10) satisfying above estimates. Then f ′ = f − f satisfies the equation sin φ ∂f ′ ∂η + F ( η ) cos φ ∂f ′ ∂φ + f ′ − ¯ f ′ = 0 ,f ′ (0 , φ, ψ ) = 0 for sin φ > ,f ′ ( L, φ, ψ ) = f ′ ( L, R φ, ψ ) . (3.64)Assume | f ′ L | ≤ C and k f ′ − f ′ L k L L ≤ C. (3.65)Then we can repeat the proof procedure and obtain k f ′ − f ′ L k L L ≤ C + Cǫ k f ′ − f ′ L k L L . (3.66)Note that in this proof, O (1) term C purely comes from the boundary data and source term. Since all dataare zero in f ′ equation, we have k f ′ − f ′ L k L L ≤ Cǫ k f ′ − f ′ L k L L , (3.67) which implies f ′ = f ′ L is a constant. Then based on zero boundary data, we must have f ′ = 0. (cid:3) S = 0 Case.
Consider the ǫ -Milne problem for f ( η, φ ) in ( η, φ, ψ ) ∈ [0 , L ] × [ − π/ , π/ × [ − π, π ) witha general source term sin φ ∂f∂η + F ( η ) cos φ ∂f∂φ + f − ¯ f = S ( η, φ, ψ ) ,f (0 , φ, ψ ) = h ( φ, ψ ) for sin φ > ,f ( L, φ, ψ ) = f ( L, R φ, ψ ) , (3.68)where F = ˜ F + G . Lemma 3.2.
Assume S = 0 satisfying (3.9) and (3.10) and < n < . There exists a solution f ( η, φ, ψ ) of the problem (3.10), satisfying for some constant | f L | ≤ C , k f − f L k L L ≤ C. (3.69) The solution is unique among functions such that (3.69) holdsProof.
We can utilize superposition property for this linear problem, i.e. write S = ¯ S + ( S − ¯ S ) = S Q + S R .Then we solve the problem by the following steps.Step 1: Construction of auxiliary function f .We first solve f as the solution to sin φ ∂f ∂η + F ( η ) cos φ ∂f ∂φ + f − ¯ f = S R ( η, φ, ψ ) ,f (0 , φ, ψ ) = h ( φ, ψ ) for sin φ > ,f ( L, φ, ψ ) = f ( L, R φ, ψ ) . (3.70)Since ¯ S R = 0, by Lemma 3.1, we know there exists a unique solution f satisfying the L estimate.Step 2: Construction of auxiliary function f .We seek a function f satisfying − π Z π − π Z π/ − π/ (cid:18) sin φ ∂f ∂η + F ( η ) cos φ ∂f ∂φ (cid:19) cos φ d φ d ψ + S Q = 0 . (3.71)The following analysis shows this type of function can always be found. An integration by parts transformsthe equation (3.71) into − Z π − π Z π/ − π/ ∂f ∂η sin φ cos φ d φ d ψ − Z π − π Z π/ − π/ F ( η ) f sin φ cos φ d φ d ψ + 4 πS Q = 0 . (3.72)Setting f ( φ, η ) = a ( η ) sin φ. (3.73)and plugging this ansatz into (3.72), we have − d a d η Z π − π Z π/ − π/ sin φ cos φ d φ d ψ − a ( η ) Z π − π Z π/ − π/ F ( η ) sin φ cos φ d φ d ψ + 4 πS Q = 0 . (3.74)Hence, we have − d a d η − ¯ F ( η ) a ( η ) + 2 S Q = 0 , (3.75)where ¯ F ( η ) = Z π − π Z π/ − π/ F ( η ) sin φ cos φ d φ d ψ ∼ (cid:18) ǫR − ǫη + ǫR − ǫη (cid:19) (3.76) EGULARITY OF MILNE PROBLEM IN 3D 21
This is a first order linear ordinary differential equation, which possesses infinite solutions. We can directlysolve it to obtain a ( η ) = e − R η ¯ F ( y )d y (cid:18) a (0) + Z η e R y ¯ F ( z )d z S Q ( y )d y (cid:19) . (3.77)We may take a (0) = − Z L e R y ¯ F ( z )d z S Q ( y )d y. (3.78)Based on the exponential decay of S Q , we can directly verify a ( η ) decays exponentially to zero as η → L and f satisfies the L estimate.Step 3: Construction of auxiliary function f .Based on above construction, we can directly verify Z π − π Z π/ − π/ (cid:18) − sin φ ∂f ∂η − F ( η ) cos φ ∂f ∂φ − f + ¯ f + S Q (cid:19) cos φ d φ d ψ = 0 . (3.79)Then we can solve f as the solution to sin φ ∂f ∂η + F ( η ) cos φ ∂f ∂φ + f − ¯ f = − sin φ ∂f ∂η − F ( η ) cos φ ∂f ∂φ − f + ¯ f + S Q ,f (0 , φ, ψ ) = − a (0) sin φ for sin φ > ,f ( L, φ, ψ ) = f ( L, R φ, ψ ) . (3.80)By (3.79), we can apply Lemma 3.1 to obtain a unique solution f satisfying the L estimate.Step 4: Construction of auxiliary function f .We now define f = f + f and an explicit verification shows sin φ ∂f ∂η + F ( η ) cos φ ∂f ∂φ + f − ¯ f = S Q ( η, φ, ψ ) ,f (0 , φ, ψ ) = 0 for sin φ > ,f ( L, φ, ψ ) = f ( L, R φ, ψ ) , (3.81)and f satisfies the L estimate.In summary, we deduce that f + f is the solution of (3.68) and satisfies the L estimate. (cid:3) Combining all above, we have the following theorem.
Theorem 3.3.
For the ǫ -Milne problem (3.1), there exists a unique solution f ( η, φ, ψ ) satisfying the esti-mates k f − f L k L L ≤ C (3.82) for some constant f L satisfying | f L | ≤ C. (3.83)3.2. L ∞ Estimates.
This section is similar to Section 3 of [18] with obvious modifications, so we omit theproof here and only present the main results.
Theorem 3.4.
The solution f ( η, φ, ψ ) to the Milne problem (3.1) satisfies k f − f L k L ∞ L ∞ ≤ C (cid:18) k f − f L k L L (cid:19) . (3.84) Theorem 3.5.
There exists a unique solution f ( η, φ, ψ ) to the ǫ -Milne problem (3.1) satisfying k f − f L k L ∞ L ∞ ≤ C. (3.85) Exponential Decay.
In this section, we prove the spatial decay of the solution to the Milne problem.
Theorem 3.6.
Assume (3.9) and (3.10) hold and < n < . For K > sufficiently small, the solution f ( η, φ, ψ ) to the ǫ -Milne problem (3.1) satisfies (cid:13)(cid:13) e K η ( f − f L ) (cid:13)(cid:13) L ∞ L ∞ ≤ C. (3.86) Proof.
Let V = f − f L . Then V satisfies sin φ ∂ V ∂η + F ( η, ψ ) cos φ ∂ V ∂φ + V − ¯ V = S, V (0 , φ, ψ ) = p ( φ, ψ ) = h ( φ, ψ ) − f L for sin φ > , V ( L, φ, ψ ) = V ( L, R φ, ψ ) . (3.87)We divide the analysis into several steps:Step 1: L Estimates.Assume ¯ S = 0. We continue using the notation F = ˜ F + G and the decomposition V = r V + q V . Now wenaturally have ( q V ) L = 0. The quasi-orthogonal property reveals h V , V sin φ i φ ( η ) = h r V , r V sin φ i φ ( η ) + 2 h r V , q V sin φ i φ ( η ) + h q V , q V sin φ i φ ( η ) (3.88)= h r V , r V sin φ i φ ( η ) − q V ( η ) Z Lη e V ( η ) − ˜ V ( y ) G ( y ) (cid:10) sin φ cos ψ, r V (cid:11) ( y )d y. Multiplying e K η V cos φ on both sides of equation (3.87) and integrating over ( φ, ψ ) ∈ [ − π/ , π/ × [ − π, π ),we obtain 12 dd η (cid:18) e K η h V , V sin φ i φ ( η ) (cid:19) + F ( η ) (cid:18) e K η h V , V sin φ i φ ( η ) (cid:19) (3.89)= e K η K h V , V sin φ i φ ( η ) − h r V , r V i φ ( η ) − G ( η ) (cid:18) e K η (cid:10) V cos ψ, V sin φ (cid:11) φ ( η ) (cid:19) + e K η h S, r V i φ ( η )= e K η (cid:18) K h r V , r V sin φ i φ ( η ) − h r V , r V i φ ( η ) (cid:19) + 4e K η K q V ( η ) Z Lη e V ( η ) − ˜ V ( y ) G ( y ) (cid:10) sin φ cos ψ, r V (cid:11) ( y )d y − G ( η ) (cid:18) e K η (cid:10) V cos ψ, V sin φ (cid:11) φ ( η ) (cid:19) + e K η h S, r V i φ ( η ) . For K < min { / , K } , we have32 k r V ( η ) k L ≥ − K h r V , r V sin φ i φ ( η ) + h r V , r V i φ ( η ) ≥ k r V ( η ) k L . (3.90)Similar to the proof of Lemma 3.1, formula as (3.89) and (3.90) imply (cid:13)(cid:13) e K η r V (cid:13)(cid:13) L L ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L K η K q V ( η ) Z Lη e V ( η ) − ˜ V ( y ) G ( y ) (cid:10) sin φ cos ψ, r V (cid:11) ( y )d y d η (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (3.91)+ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L G ( η ) (cid:18) e K η (cid:10) V cos ψ, V sin φ (cid:11) φ ( η ) (cid:19) d η (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L e K η h S, r V i φ ( η )d η (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ CL k G k L L (cid:13)(cid:13) e K η r V (cid:13)(cid:13) L L (cid:13)(cid:13) e K η q V (cid:13)(cid:13) L L + Cǫ (cid:13)(cid:13) e K η V (cid:13)(cid:13) L L + C (cid:13)(cid:13) e K η S (cid:13)(cid:13) L L (cid:13)(cid:13) e K η r V (cid:13)(cid:13) L L ≤ Cǫ (cid:13)(cid:13) e K η q V (cid:13)(cid:13) L L + Cǫ − n (cid:13)(cid:13) e K η r V (cid:13)(cid:13) L L + ǫ (cid:13)(cid:13) e K η V (cid:13)(cid:13) L L + C k S k L L ≤ C + C (cid:13)(cid:13) e K η q V (cid:13)(cid:13) L L + Cǫ − n (cid:13)(cid:13) e K η r V (cid:13)(cid:13) L L + ǫ (cid:13)(cid:13) e K η V (cid:13)(cid:13) L L . EGULARITY OF MILNE PROBLEM IN 3D 23
Hence, for ǫ sufficiently small, we know (cid:13)(cid:13) e K η r V (cid:13)(cid:13) L L ≤ C + Cǫ (cid:13)(cid:13) e K η q V (cid:13)(cid:13) L L + Cǫ k V k L L . (3.92)Then similar to the proof of Lemma 3.1, we deduce (cid:13)(cid:13) e K η q V (cid:13)(cid:13) L L (3.93) ≤ (cid:13)(cid:13) e K η r V (cid:13)(cid:13) L L + Z L e K η (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z Lη ˜ F ( y ) (cid:10) − φ, r V (cid:11) ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d η + Z L e K η (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z Lη Z Lz e V ( z ) − V y G ( y ) (cid:10) sin φ cos ψ, r V (cid:11) d y d z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d η + Z L e K η (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z Lη G ( y ) (cid:10) − φ, r V cos ψ (cid:11) ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d η + Z L e K η (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z Lη h sin φ, S i ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d η ≤ C + C (cid:13)(cid:13) e K η r V (cid:13)(cid:13) L L + C (cid:13)(cid:13) e K η r V (cid:13)(cid:13) L L (cid:18) Z L Z Lη e K ( η − y ) F ( y )d y d η (cid:19) + L k G k L L (cid:13)(cid:13) e K η r V (cid:13)(cid:13) L L + L k G k L L (cid:13)(cid:13) e K η r V (cid:13)(cid:13) L L + Z L e K η (cid:18) Z Lη k S ( y ) k L ∞ d y (cid:19) d η ≤ C + C (1 + ǫ − n ) (cid:13)(cid:13) e K η r V (cid:13)(cid:13) L L ≤ C + C (cid:13)(cid:13) e K η r V (cid:13)(cid:13) L L ≤ C + Cǫ (cid:13)(cid:13) e K η q V (cid:13)(cid:13) L L + Cǫ (cid:13)(cid:13) e K η V (cid:13)(cid:13) L L , (3.94)which implies (cid:13)(cid:13) e K η q V (cid:13)(cid:13) L L ≤ C + Cǫ (cid:13)(cid:13) e K η V (cid:13)(cid:13) L L . (3.95)In summary, we have (cid:13)(cid:13) e K η V (cid:13)(cid:13) L L ≤ (cid:13)(cid:13) e K η q V (cid:13)(cid:13) L L + (cid:13)(cid:13) e K η r V (cid:13)(cid:13) L L ≤ C + Cǫ (cid:13)(cid:13) e K η V (cid:13)(cid:13) L L , (3.96)which yields (cid:13)(cid:13) e K η V (cid:13)(cid:13) L L ≤ C. (3.97)This is the desired result when ¯ S = 0. By the method introduced in Lemma 3.2, we can extend above L estimates to the general S case. Note all the auxiliary functions constructed in Lemma 3.2 satisfy the desiredestimates.Step 2: L ∞ Estimates.This is similar to the proof of exponential decay in [18], so we omit the details here. We have (cid:13)(cid:13) e K η V (cid:13)(cid:13) L ∞ L ∞ ≤ C + C (cid:13)(cid:13) e K η V (cid:13)(cid:13) L L . (3.98)Combining (3.97) and (3.98), we deduce the desired result (cid:13)(cid:13) e K η ( f − f L ) (cid:13)(cid:13) L ∞ L ∞ ≤ C. (3.99) (cid:3) Diffusive Boundary.
In this subsection, we consider the ǫ -Milne problem with diffusive boundary as sin φ ∂f∂η + F ( η, ψ ) cos φ ∂f∂φ + f − ¯ f = S ( η, φ, ψ ) ,f (0 , φ, ψ ) = h ( φ, ψ ) + P [ f ](0) for sin φ > ,f ( L, φ, ψ ) = f ( L, R φ, ψ ) , (3.100)where P [ f ](0) = − π Z Z sin φ< f (0 , φ, ψ ) sin φ cos φ d φ d ψ, (3.101) Similar to [27, Section 6], we can easily prove that
Lemma 3.7.
In order for the equation (3.100) to have a solution f ( η, φ, ψ ) ∈ L ∞ ([0 , L ] × [ − π, π ) × [ − π, π ) × [ − π/ , π/ , the boundary data h and the source term S must satisfy the compatibility condition Z Z sin φ> h ( φ, ψ ) sin φ cos φ d φ d ψ + Z L Z π − π Z π/ − π/ e − V ( s ) S ( s, φ, ψ ) cos φ d φ d ψ d s = 0 . (3.102) In particular, if S = 0 , then the compatibility condition reduces to Z Z sin φ> h ( φ, ψ ) sin φ cos φ d φ d ψ = 0 . (3.103)It is easy to see if f is a solution to (3.100), then f + C is also a solution for any constant C . Hence, inorder to obtain a unique solution, we need a normalization condition P [ f ](0) = 0 . (3.104)The following lemma in [27, Section 6] tells us the problem (3.100) can be reduced to the ǫ -Milne problemwith in-flow boundary (3.1). Lemma 3.8.
If the boundary data h and S satisfy the compatibility condition (3.102), then the solution f to the ǫ -Milne problem (3.1) with in-flow boundary as f = h on sin φ > is also a solution to the ǫ -Milneproblem (3.100) with diffusive boundary, which satisfies the normalization condition (3.104). Furthermore,this is the unique solution to (3.100) among the functions satisfying (3.104) and k f ( η, φ, ψ ) − f L k L L ≤ C . In summary, based on above analysis, we can utilize the known result for ǫ -Milne problem (3.1) to obtainthe desired results of the solution to the ǫ -Milne problem (3.100). Theorem 3.9.
There exists a unique solution f ( η, φ, ψ ) to the ǫ -Milne problem (3.100) with the normaliza-tion condition (3.104) satisfying for some constant | f L | < C , k f ( η, φ, ψ ) − f L k L L ≤ C. (3.105) Theorem 3.10.
The unique solution f ( η, φ, ψ ) to the ǫ -Milne problem (3.100) with the normalization con-dition (3.104) satisfying for some constant | f L | < C , k f ( η, φ, ψ ) − f L k L ∞ L ∞ ≤ C. (3.106) Theorem 3.11.
There exists K > such that the solution f ( η, φ, ψ ) to the ǫ -Milne problem (3.100) withthe normalization condition (3.104) satisfies (cid:13)(cid:13)(cid:13)(cid:13) e K η (cid:18) f ( η, φ, ψ ) − f L (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ ≤ C. (3.107) EGULARITY OF MILNE PROBLEM IN 3D 25 Regularity of ǫ -Milne Problem We continue studying the ǫ -Milne problem with in-flow boundary as sin φ ∂f∂η + F ( η, ψ ) cos φ ∂f∂φ + f − ¯ f = S ( η, φ, ψ ) ,f (0 , φ, ψ ) = h ( φ, ψ ) for sin φ > ,f ( L, φ, ψ ) = f ( L, R φ, ψ ) . (4.1)Here we already omit the superscript ǫ and dependence on ( τ , τ ). Besides (3.9) and (3.10), we furtherassume (cid:13)(cid:13)(cid:13)(cid:13) ∂h∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂h∂ψ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂h∂τ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂h∂τ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ ≤ M, (4.2)and (cid:13)(cid:13)(cid:13)(cid:13) e Kη ∂S∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) e Kη ∂S∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) e Kη ∂S∂ψ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) e Kη ∂S∂τ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) e Kη ∂S∂τ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ ≤ M, (4.3)for some M, K >
0. Define a potential function V ( η, ψ ) satisfying V (0 , ψ ) = 0 and ∂V∂η = − F ( η, ψ ). Also,we know L = ǫ − n for 0 < n <
25 .
Lemma 4.1.
We have e − V (0 ,ψ ) = 1 and e − V ( L,ψ ) = (cid:18) − ǫ − n R (cid:19) sin ψ (cid:18) − ǫ − n R (cid:19) cos ψ . (4.4) Also, for R = max { R , R } and R ′ = min { R , R } which are the maximum and minimum of R and R ,we have R ′ − ǫηR ′ ≤ e − V ( η,ψ ) ≤ R − ǫηR . (4.5) Proof.
We directly compute V ( η, ψ ) = sin ψ ln (cid:18) R R − ǫη (cid:19) + cos ψ ln (cid:18) R R − ǫη (cid:19) , (4.6)and e − V ( η,ψ ) = (cid:18) R − ǫηR (cid:19) sin ψ (cid:18) R − ǫηR (cid:19) cos ψ . (4.7)Hence, our result naturally follows. (cid:3) Preliminaries.
It is easy to see V ( η, φ, ψ ) = f ( η, φ, ψ ) − f L satisfies the equation sin φ ∂ V ∂η + F ( η, ψ ) cos φ ∂ V ∂φ + V − ¯ V = S ( η, φ, ψ ) , V (0 , φ, ψ ) = p ( φ, ψ ) for sin φ > , V ( L, φ, ψ ) = V ( L, R φ, ψ ) . (4.8)where p ( φ, ψ ) = h ( φ, ψ ) − f L . (4.9)We intend to estimate the normal, tangential and velocity derivative. This idea is motivated by [17] and[18]. Define a distance function ζ ( η, φ, ψ ) as ζ ( η, φ, ψ ) = − (cid:18) e − V ( η,ψ ) cos φ (cid:19) ! / . (4.10)Note that the closer ( η, φ, ψ ) is to the grazing set which satisfies η = 0 and sin φ = 0, the smaller ζ is. Inparticular, at grazing set, ζ = 0. Also, we have 0 ≤ ζ ≤ Lemma 4.2.
We have sin φ ∂ζ∂η + F ( η, ψ ) cos φ ∂ζ∂φ = 0 . (4.11) Proof.
We may directly compute ∂ζ∂η = 12 − (cid:18) e − V ( η,ψ ) cos φ (cid:19) ! − / (cid:18) − − V ( η,ψ ) cos φ (cid:19) F ( η, ψ ) = − e − V ( η,ψ ) F ( η, ψ ) cos φζ , (4.12) ∂ζ∂φ = 12 − (cid:18) e − V ( η,ψ ) cos φ (cid:19) ! − / (cid:18) − − V ( η,ψ ) cos φ (cid:19) ( − sin φ ) = e − V ( η,ψ ) cos φ sin φζ . (4.13)Hence, we know (4.14)sin φ ∂ζ∂η + F ( η, ψ ) cos φ ∂ζ∂φ = − sin φ (cid:18) e − V ( η,ψ ) F ( η, ψ ) cos φ (cid:19) + F ( η, ψ ) cos φ (cid:18) e − V ( η,ψ ) cos φ sin φ (cid:19) ζ = 0 . (cid:3) As a matter of fact, we are able to prove some preliminary estimates that are based on the characteristicsof V itself instead of the derivative. In the following, let 0 < δ << Lemma 4.3.
Assume (3.9), (3.10), (4.2) and (4.3). For sin φ > δ , we have (cid:12)(cid:12)(cid:12)(cid:12) sin φ ∂ V ∂η ( η, φ ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:18) δ (cid:19) . (4.15) Lemma 4.4.
Assume (3.9), (3.10), (4.2) and (4.3). For sin φ < with | E ( η, φ ) | ≤ e − V ( L ) , if it satisfies min φ ′ sin φ ′ ≥ δ where ( η ′ , φ ′ ) are on the same characteristics as ( η, φ ) with sin φ ′ ≥ , then we have (cid:12)(cid:12)(cid:12)(cid:12) sin φ ∂ V ∂η ( η, φ ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:18) δ (cid:19) . (4.16) Lemma 4.5.
Assume (3.9), (3.10), (4.2), (4.3) and (cid:12)(cid:12)(cid:12)(cid:12) ∂ ¯ V ∂η (cid:12)(cid:12)(cid:12)(cid:12) ≤ C (1 + | ln( ǫ ) | + | ln( η ) | ) . For sin φ ≤ and | E ( η, φ ) | ≥ e − V ( L ) , we have (cid:12)(cid:12)(cid:12)(cid:12) sin φ ∂ V ∂η ( η, φ ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C (1 + | ln( ǫ ) | ) . (4.17)The proofs of Lemma 4.3, Lemma 4.4 and Lemma 4.5 are similar to those in [18] with obvious modifica-tions, so we omit the details here. Remark 4.6.
Estimates in Lemma 4.3, Lemma 4.4 and Lemma 4.5 can provide pointwise bounds of deriva-tives. However, they are not uniform estimates due to presence of δ and ln( ǫ ) . We need weighted L ∞ estimates of derivatives to close the proof. Also, the estimate (cid:12)(cid:12)(cid:12)(cid:12) ∂ ¯ V ∂η (cid:12)(cid:12)(cid:12)(cid:12) ≤ C (1 + | ln( ǫ ) | + | ln( η ) | ) are not knowna priori, so we need an iteration argument. Mild Formulation of Normal Derivative.
In this and next subsection, we will prove stronger apriori estimates of derivatives. Consider the ǫ -transport problem for A = ζ ∂ V ∂η as sin φ ∂ A ∂η + F ( η, ψ ) cos φ ∂ A ∂φ + A = ˜ A + S A , A (0 , φ, ψ ) = p A ( φ, ψ ) for sin φ > , A ( L, φ, ψ ) = A ( L, R φ, ψ ) , (4.18)where p A and S A will be specified later with˜ A ( η, φ, ψ ) = 14 π Z π − π Z π/ − π/ ζ ( η, φ, ψ ) ζ ( η, φ ∗ , ψ ) A ( η, φ ∗ , ψ ) cos φ ∗ d φ ∗ d ψ. (4.19) EGULARITY OF MILNE PROBLEM IN 3D 27
Lemma 4.7.
We have k A k L ∞ L ∞ ≤ C (cid:18) k p A k L ∞ + k S A k L ∞ L ∞ (cid:19) (4.20)+ C | ln( ǫ ) | (cid:18) k V k L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂p∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + k S k L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ (cid:19) . The rest of this subsection will be devoted to the proof of this lemma. We first introduce some notation.Define the energy as before E ( η, φ, ψ ) = e − V ( η,ψ ) cos φ. (4.21)Along the characteristics, where this energy is conserved and ζ is a constant, the equation can be simplifiedas follows: sin φ d A d η + A = ˜ A + S A . (4.22)An implicit function η + ( η, φ, ψ ) can be determined through | E ( η, φ, ψ ) | = e − V ( η + ,ψ ) . (4.23)which means ( η + , φ , ψ ) with sin φ = 0 is on the same characteristics as ( η, φ, ψ ). Define the quantities for0 ≤ η ′ ≤ η + as follows: φ ′ ( φ, η, η ′ , ψ ) = cos − (e V ( η ′ ,ψ ) − V ( η,ψ ) cos φ ) , (4.24) R φ ′ ( φ, η, η ′ , ψ ) = − cos − (e V ( η ′ ,ψ ) − V ( η,ψ ) cos φ ) = − φ ′ ( φ, η, η ′ , ψ ) , (4.25)where the inverse trigonometric function can be defined single-valued in the domain [0 , π/
2] and the quantitiesare always well-defined due to the monotonicity of V . Note that sin φ ′ ≥
0, even if sin φ <
0. Finally we put G η,η ′ ,ψ ( φ ) = Z ηη ′ φ ′ ( φ, η, ξ, ψ )) d ξ. (4.26)Similar to ǫ -Milne problem, we can define the solution along the characteristics as follows: A ( η, φ, ψ ) = K [ p A ] + T [ ˜ A + S A ] , (4.27)whereRegion I:For sin φ > K [ p A ] = p A ( φ ′ (0) , ψ ) exp( − G η, ,ψ ) (4.28) T [ ˜ A + S A ] = Z η ( ˜ A + S A )( η ′ , φ ′ ( η ′ ) , ψ )sin( φ ′ ( η ′ )) exp( − G η,η ′ ,ψ )d η ′ . (4.29)Region II:For sin φ < | E ( η, φ, ψ ) | ≤ e − V ( L,ψ ) , K [ p A ] = p A ( φ ′ (0) , ψ ) exp( − G L, ,ψ − G L,η,ψ ) (4.30) T [ ˜ A + S A ] = Z L ( ˜ A + S )( η ′ , φ ′ ( η ′ ) , ψ )sin( φ ′ ( η ′ )) exp( − G L,η ′ ,ψ − G L,η,ψ )d η ′ (4.31)+ Z Lη ( ˜ A + S )( η ′ , R φ ′ ( η ′ ) , ψ )sin( φ ′ ( η ′ )) exp( − G η ′ ,η,ψ )d η ′ . Region III:
For sin φ < | E ( η, φ, ψ ) | ≥ e − V ( L,ψ ) , K [ p A ] = p A ( φ ′ ( φ, η, , ψ ) exp( − G η + , ,ψ − G η + ,η,ψ ) (4.32) T [ ˜ A + S A ] = Z η + ( ˜ A + S A )( η ′ , φ ′ ( η ′ ) , ψ )sin( φ ′ ( η ′ )) exp( − G η + ,η ′ ,ψ − G η + ,η,ψ )d η ′ (4.33)+ Z η + η ( ˜ A + S A )( η ′ , R φ ′ ( η ′ ) , ψ )sin( φ ′ ( η ′ )) exp( − G η ′ ,η,ψ )d η ′ . Then we need to estimate K [ p A ] and T [ ˜ A + S A ] in each region. We assume 0 < δ << < δ << Region I: sin φ > . A direct computation reveals |K [ p A ] | ≤ k p A k L ∞ , (4.34) |T [ S A ] | ≤ k S A k L ∞ L ∞ . (4.35)Hence, we only need to estimate I = T [ ˜ A ]. We divide it into several steps:Step 0: Preliminaries.We have E ( η ′ , φ ′ ) = (cid:18) R − ǫη ′ R (cid:19) sin ψ (cid:18) R − ǫη ′ R (cid:19) cos ψ cos φ ′ . (4.36)We can directly obtain ζ ( η ′ , φ ′ , ψ ) ≤ R ′ s R ′ − (cid:18) ( R ′ − ǫη ′ ) cos φ ′ (cid:19) = 1 R ′ q R ′ − ( R ′ − ǫη ′ ) + ( R ′ − ǫη ′ ) sin φ ′ , (4.37) ≤ p R ′ − ( R ′ − ǫη ′ ) + q ( R ′ − ǫη ′ ) sin φ ′ R ′ ≤ C (cid:18)p ǫη ′ + sin φ ′ (cid:19) , and ζ ( η ′ , φ ′ , ψ ) ≥ R p R − ( R − ǫη ′ ) ≥ C p ǫη ′ . (4.38)Also, we know for 0 ≤ η ′ ≤ η ,sin φ ′ = p − cos φ ′ ≤ s − (cid:18) R ′ − ǫηR ′ − ǫη ′ (cid:19) cos φ (4.39)= q ( R ′ − ǫη ′ ) sin φ + (2 R ′ − ǫη − ǫη ′ )( ǫη − ǫη ′ ) cos φR ′ − ǫη ′ . (4.40)Since 0 ≤ (2 R ′ − ǫη − ǫη ′ )( ǫη − ǫη ′ ) cos φ ≤ R ′ ǫ ( η − η ′ ) , (4.41)we have sin φ ≤ sin φ ′ ≤ q sin φ + ǫ ( η − η ′ ) , (4.42)which means 12 q sin φ + ǫ ( η − η ′ ) ≤ φ ′ ≤ φ . (4.43) EGULARITY OF MILNE PROBLEM IN 3D 29
Therefore, − Z ηη ′ φ ′ ( y ) d y ≤ − Z ηη ′ q sin φ + ǫ ( η − y ) d y (4.44)= 1 ǫ (cid:18) sin φ − q sin φ + ǫ ( η − η ′ ) (cid:19) = − η − η ′ sin φ + q sin φ + ǫ ( η − η ′ ) ≤ − η − η ′ q sin φ + ǫ ( η − η ′ ) . Define a cut-off function χ ∈ C ∞ [ − π, π ] satisfying χ ( φ ) = ( | sin φ | ≤ δ, | sin φ | ≥ δ, (4.45)In the following, we will divide the estimate of I into several cases based on the value of sin φ , sin φ ′ , ǫη ′ and ǫ ( η − η ′ ). Let denote the indicator function. We write I = Z η { sin φ ≥ δ } + Z η { ≤ sin φ ≤ δ } { χ ( φ ∗ ) < } + Z η { ≤ sin φ ≤ δ } { χ ( φ ∗ )=1 } {√ ǫη ′ ≥ sin φ ′ } (4.46)+ Z η { ≤ sin φ ≤ δ } { χ ( φ ∗ )=1 } {√ ǫη ′ ≤ sin φ ′ } { sin φ ≤ ǫ ( η − η ′ ) } + Z η { ≤ sin φ ≤ δ } { χ ( φ ∗ )=1 } {√ ǫη ′ ≤ sin φ ′ } { sin φ ≥ ǫ ( η − η ′ ) } = I + I + I + I + I . Step 1: Estimate of I for sin φ ≥ δ .Based on Lemma 4.3, we know (cid:12)(cid:12)(cid:12)(cid:12) sin φ ∂ V ∂η (cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:18) δ (cid:19)(cid:18) k V k L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂p∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ (cid:19) . (4.47)Hence, we have | I | ≤ C (cid:12)(cid:12)(cid:12)(cid:12) ∂ V ∂η (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cδ (cid:18) k V k L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂p∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ (cid:19) . (4.48)Step 2: Estimate of I for 0 ≤ sin φ ≤ δ and χ ( φ ∗ ) < I = 14 π Z η (cid:18) Z π − π Z π/ − π/ ζ ( η ′ , φ ′ , ψ ) ζ ( η ′ , φ ∗ , ψ ) (1 − χ ( φ ∗ )) A ( η ′ , φ ∗ , ψ ) cos φ ∗ d φ ∗ d ψ (cid:19) φ ′ exp( − G η,η ′ ,ψ )d η ′ = 14 π Z η (cid:18) Z π − π Z π/ − π/ ζ ( η ′ , φ ′ , ψ )(1 − χ ( φ ∗ )) V ( η ′ , φ ∗ , ψ ) ∂η ′ cos φ ∗ d φ ∗ d ψ (cid:19) φ ′ exp( − G η,η ′ ,ψ )d η ′ . Based on the ǫ -Milne problem of V assin φ ∗ ∂ V ( η ′ , φ ∗ , ψ ) ∂η ′ + F ( η ′ , ψ ) cos φ ∗ ∂ V ( η ′ , φ ∗ , ψ ) ∂φ ∗ + V ( η ′ , φ ∗ , ψ ) − ¯ V ( η ′ ) = S ( η ′ , φ ∗ , ψ ) , (4.50)we have ∂ V ( η ′ , φ ∗ , ψ ) ∂η ′ = − φ ∗ (cid:18) F ( η ′ , ψ ) cos φ ∗ ∂ V ( η ′ , φ ∗ , ψ ) ∂φ ∗ + V ( η ′ , φ ∗ , ψ ) − ¯ V ( η ′ ) − S ( η ′ , φ ∗ , ψ ) (cid:19) . (4.51) Hence, we have˜ A = Z π − π Z π/ − π/ ζ ( η ′ , φ ′ , ψ )(1 − χ ( φ ∗ )) ∂ V ( η ′ , φ ∗ ) ∂η ′ cos φ ∗ d φ ∗ d ψ (4.52)= − Z π − π Z π/ − π/ ζ ( η ′ , φ ′ , ψ )(1 − χ ( φ ∗ )) 1sin φ ∗ (cid:18) V ( η ′ , φ ∗ , ψ ) − ¯ V ( η ′ ) − S ( η ′ , φ ∗ , ψ ) (cid:19) cos φ ∗ d φ ∗ d ψ − Z π − π Z π/ − π/ ζ ( η ′ , φ ′ , ψ )(1 − χ ( φ ∗ )) 1sin φ ∗ F ( η ′ , ψ ) cos φ ∗ ∂ V ( η ′ , φ ∗ , ψ ) ∂φ ∗ cos φ ∗ d φ ∗ d ψ = ˜ A + ˜ A . We may directly obtain (4.53) (cid:12)(cid:12)(cid:12) ˜ A (cid:12)(cid:12)(cid:12) ≤ Z π − π Z π/ − π/ ζ ( η ′ , φ ′ , ψ )(1 − χ ( φ ∗ )) 1sin φ ∗ (cid:18) V ( η ′ , φ ∗ , ψ ) − ¯ V ( η ′ ) − S ( η ′ , φ ∗ , ψ ) (cid:19) cos φ ∗ d φ ∗ d ψ ≤ Rδ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z π − π Z π/ − π/ (cid:18) V ( η ′ , φ ∗ , ψ ) − ¯ V ( η ′ ) − S ( η ′ , φ ∗ , ψ ) (cid:19) cos φ ∗ d φ ∗ d ψ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C ( δ ) (cid:18) k V k L ∞ L ∞ + k S k L ∞ L ∞ (cid:19) . On the other hand, an integration by parts yields˜ A = Z π − π Z π/ − π/ ∂∂φ ∗ (cid:18) ζ ( η ′ , φ ′ , ψ )(1 − χ ( φ ∗ )) 1sin φ ∗ F ( η ′ , ψ ) cos φ ∗ (cid:19) V ( η ′ , φ ∗ , ψ ) cos φ ∗ d φ ∗ d ψ, (4.54)which further implies (cid:12)(cid:12)(cid:12) ˜ A (cid:12)(cid:12)(cid:12) ≤ Cǫδ k V k L ∞ L ∞ ≤ C ( δ ) k V k L ∞ L ∞ . (4.55)Since we can use substitution to show Z η φ ′ exp( − G η,η ′ ,ψ )d η ′ ≤ , (4.56)we have | I | ≤ C ( δ ) (cid:18) k V k L ∞ L ∞ + k S k L ∞ L ∞ (cid:19) Z η φ ′ exp( − G η,η ′ ,ψ )d η ′ (4.57) ≤ C ( δ ) (cid:18) k V k L ∞ L ∞ + k S k L ∞ L ∞ (cid:19) . Step 3: Estimate of I for 0 ≤ sin φ ≤ δ , χ ( φ ∗ ) = 1 and √ ǫη ′ ≥ sin φ ′ .Based on (4.37), this implies ζ ( η ′ , φ ′ , ψ ) ≤ C p ǫη ′ . Then combining this with (4.38), we can directly obtain (4.58) Z π − π Z π/ − π/ ζ ( η ′ , φ ′ , ψ ) ζ ( η ′ , φ ∗ , ψ ) χ ( φ ∗ ) A ( η ′ , φ ∗ , ψ ) cos φ ∗ d φ ∗ d ψ ≤ C Z π − π Z δ − δ A ( η ′ , φ ∗ , ψ ) cos φ ∗ d φ ∗ d ψ ≤ Cδ k A k L ∞ L ∞ . Hence, we have | I | ≤ Cδ k A k L ∞ L ∞ Z η φ ′ exp( − G η,η ′ ,ψ )d η ′ ≤ Cδ k A k L ∞ L ∞ . (4.59)Step 4: Estimate of I for 0 ≤ sin φ ≤ δ , χ ( φ ∗ ) = 1, √ ǫη ′ ≤ sin φ ′ and sin φ ≤ ǫ ( η − η ′ ).Based on (4.37), this implies ζ ( η ′ , φ ′ , ψ ) ≤ C sin φ ′ . (4.60) EGULARITY OF MILNE PROBLEM IN 3D 31
Based on (4.44), we have − G η,η ′ ,ψ = − Z ηη ′ φ ′ ( y ) d y ≤ − η − η ′ p ǫ ( η − η ′ ) ≤ − C r η − η ′ ǫ . (4.61)Hence, we know | I | ≤ C Z η (cid:18) Z π − π Z π/ − π/ ζ ( η ′ , φ ′ , ψ ) ζ ( η ′ , φ ∗ , ψ ) χ ( φ ∗ ) A ( η ′ , φ ∗ , ψ ) cos φ ∗ d φ ∗ d ψ (cid:19) φ ′ exp( − G η,η ′ ,ψ )d η ′ (4.62) ≤ C Z η (cid:18) Z π − π Z δ − δ ζ ( η ′ , φ ∗ , ψ ) A ( η ′ , φ ∗ , ψ ) cos φ ∗ d φ ∗ d ψ (cid:19) ζ ( η ′ , φ ′ , ψ )sin φ ′ exp( − G η,η ′ ,ψ )d η ′ ≤ Cδ k A k L ∞ L ∞ Z η √ ǫη ′ exp( − G η,η ′ ,ψ )d η ′ ≤ Cδ k A k L ∞ L ∞ Z η √ ǫη ′ exp (cid:18) − C r η − η ′ ǫ (cid:19) d η ′ Define z = η ′ ǫ , which implies d η ′ = ǫ d z . Substituting this into above integral, we have | I | ≤ Cδ k A k L ∞ L ∞ Z η/ǫ √ z exp (cid:18) − C r ηǫ − z (cid:19) d z (4.63)= Cδ k A k L ∞ L ∞ Z √ z exp (cid:18) − C r ηǫ − z (cid:19) d z + Z η/ǫ √ z exp (cid:18) − C r ηǫ − z (cid:19) d z ! . We can estimate these two terms separately. Z √ z exp (cid:18) − C r ηǫ − z (cid:19) d z ≤ Z √ z d z = 2 . (4.64) Z η/ǫ √ z exp (cid:18) − C r ηǫ − z (cid:19) d z ≤ Z η/ǫ exp (cid:18) − C r ηǫ − z (cid:19) d z t = ηǫ − z ≤ Z ∞ t e − Ct d t < ∞ . (4.65)Hence, we know | I | ≤ Cδ k A k L ∞ L ∞ . (4.66)Step 5: Estimate of I for 0 ≤ sin φ ≤ δ , χ ( φ ∗ ) = 1, √ ǫη ′ ≤ sin φ ′ and sin φ ≥ ǫ ( η − η ′ ).Based on (4.37), this implies ζ ( η ′ , φ ′ , ψ ) ≤ C sin φ ′ . Based on (4.44), we have − G η,η ′ ,ψ = − Z ηη ′ φ ′ ( y ) d y ≤ − C ( η − η ′ )sin φ . (4.67)Hence, we have | I | ≤ C k A k L ∞ L ∞ Z η (cid:18) Z π − π Z δ − δ ζ ( η ′ , φ ∗ , ψ ) cos φ ∗ d φ ∗ d ψ (cid:19) exp (cid:18) − C ( η − η ′ )sin φ (cid:19) d η ′ (4.68) Here, we use a different way to estimate the inner integral. We use substitution to find Z δ − δ ζ ( η ′ , φ ∗ , ψ ) cos φ ∗ d φ ∗ ≤ Z δ − δ (cid:18) R − ( R − ǫη ′ ) cos φ ∗ (cid:19) / d φ ∗ (4.69) sin φ ∗ small ≤ C Z δ − δ cos φ ∗ (cid:18) R − ( R − ǫη ′ ) cos φ ∗ (cid:19) / d φ ∗ = C Z δ − δ cos φ ∗ (cid:18) R − ( R − ǫη ′ ) + ( R − ǫη ′ ) sin φ ∗ (cid:19) / d φ ∗ y =sin φ ∗ = C Z δ − δ (cid:18) R − ( R − ǫη ′ ) + ( R − ǫη ′ ) y (cid:19) / d y. Define p = p R − ( R − ǫη ′ ) = p Rǫη ′ − ǫ η ′ ≤ C p ǫη ′ , (4.70) q = R − ǫη ′ ≥ C, (4.71) r = pq ≤ C p ǫη ′ . (4.72)Then we have Z δ − δ ζ ( η ′ , φ ∗ , ψ ) d φ ∗ ≤ C Z δ − δ p + q y ) / d y (4.73) ≤ C Z − p + q y ) / d y ≤ C Z − r + y ) / d y ≤ C Z r + y ) / d y = (cid:18) ln( y + p r + y ) − ln( r ) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:18) ln(2 + p r + 4) − ln r (cid:19) ≤ C (cid:18) r ) (cid:19) ≤ C (cid:18) | ln( ǫ ) | + | ln( η ′ ) | (cid:19) . Hence, we know | I | ≤ C k A k L ∞ L ∞ Z η (cid:18) | ln( ǫ ) | + | ln( η ′ ) | (cid:19) exp (cid:18) − C ( η − η ′ )sin φ (cid:19) d η ′ (4.74)We may directly compute (cid:12)(cid:12)(cid:12)(cid:12)Z η (cid:18) | ln( ǫ ) | (cid:19) exp (cid:18) − C ( η − η ′ )sin φ (cid:19) d η ′ (cid:12)(cid:12)(cid:12)(cid:12) ≤ C sin φ (1 + | ln( ǫ ) | ) . (4.75)Hence, we only need to estimate (cid:12)(cid:12)(cid:12)(cid:12)Z η | ln( η ′ ) | exp (cid:18) − C ( η − η ′ )sin φ (cid:19) d η ′ (cid:12)(cid:12)(cid:12)(cid:12) . (4.76)If η ≤
2, using Cauchy’s inequality, we have (cid:12)(cid:12)(cid:12)(cid:12)Z η | ln( η ′ ) | exp (cid:18) − C ( η − η ′ )sin φ (cid:19) d η ′ (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:18) Z η ln ( η ′ )d η ′ (cid:19) / (cid:18) Z η exp (cid:18) − C ( η − η ′ )sin φ (cid:19) d η ′ (cid:19) / (4.77) ≤ (cid:18) Z ln ( η ′ )d η ′ (cid:19) / (cid:18) Z η exp (cid:18) − C ( η − η ′ )sin φ (cid:19) d η ′ (cid:19) / ≤ p sin φ. EGULARITY OF MILNE PROBLEM IN 3D 33 If η ≥
2, we decompose and apply Cauchy’s inequality to obtain (cid:12)(cid:12)(cid:12)(cid:12)Z η | ln( η ′ ) | exp (cid:18) − C ( η − η ′ )sin φ (cid:19) d η ′ (cid:12)(cid:12)(cid:12)(cid:12) (4.78) ≤ (cid:12)(cid:12)(cid:12)(cid:12)Z | ln( η ′ ) | exp (cid:18) − C ( η − η ′ )sin φ (cid:19) d η ′ (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)Z η ln( η ′ ) exp (cid:18) − C ( η − η ′ )sin φ (cid:19) d η ′ (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:18) Z ln ( η ′ )d η ′ (cid:19) / (cid:18) Z exp (cid:18) − C ( η − η ′ )sin φ (cid:19) d η ′ (cid:19) / + ln(2) (cid:12)(cid:12)(cid:12)(cid:12)Z η exp (cid:18) − C ( η − η ′ )sin φ (cid:19) d η ′ (cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:18)p sin φ + sin φ (cid:19) ≤ C p sin φ. Hence, we have | I | ≤ C (1 + | ln( ǫ ) | ) p δ k A k L ∞ L ∞ . (4.79)Step 6: Synthesis.Collecting all the terms in previous steps, we have proved | I | ≤ C (1 + | ln( ǫ ) | ) p δ k A k L ∞ L ∞ + Cδ k A k L ∞ L ∞ (4.80)+ Cδ (cid:18) k V k L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂p∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ (cid:19) + C ( δ ) (cid:18) k V k L ∞ L ∞ + k S k L ∞ L ∞ (cid:19) . Therefore, we know | A | I ≤ k p A k L ∞ + k S A k L ∞ L ∞ + C (1 + | ln( ǫ ) | ) p δ k A k L ∞ L ∞ + Cδ k A k L ∞ L ∞ (4.81)+ Cδ (cid:18) k V k L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂p∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ (cid:19) + C ( δ ) (cid:18) k V k L ∞ L ∞ + k S k L ∞ L ∞ (cid:19) . Region II: sin φ < and | E ( η, φ, ψ ) | ≤ e − V ( L ) . K [ p A ] = p A ( φ ′ (0) , ψ ) exp( − G L, ,ψ − G L,η,ψ ) (4.82) T [ ˜ A + S A ] = Z L ( ˜ A + S )( η ′ , φ ′ ( η ′ ) , ψ )sin( φ ′ ( η ′ )) exp( − G L,η ′ ,ψ − G L,η,ψ )d η ′ (4.83)+ Z Lη ( ˜ A + S )( η ′ , R φ ′ ( η ′ ) , ψ )sin( φ ′ ( η ′ )) exp( − G η ′ ,η,ψ )d η ′ . A direct computation reveals |K [ p A ] | ≤ k p A k L ∞ , (4.84) |T [ S A ] | ≤ k S A k L ∞ L ∞ . (4.85)Hence, we only need to estimate II = T [ ˜ A ]. In particular, we can decompose T [ ˜ A ] = Z L ˜ A ( η ′ , φ ′ ( η ′ ) , ψ )sin( φ ′ ( η ′ )) exp( − G L,η ′ ,ψ − G L,η,ψ )d η ′ + Z Lη ˜ A ( η ′ , R φ ′ ( η ′ ) , ψ )sin( φ ′ ( η ′ )) exp( − G η ′ ,η,ψ )d η ′ (4.86)= Z η ˜ A ( η ′ , φ ′ ( η ′ ) , ψ )sin( φ ′ ( η ′ )) exp( − G L,η ′ ,ψ − G L,η,ψ )d η ′ + Z Lη ˜ A ( η ′ , φ ′ ( η ′ ) , ψ )sin( φ ′ ( η ′ )) exp( − G L,η ′ ,ψ − G L,η,ψ )d η ′ + Z Lη ˜ A ( η ′ , R φ ′ ( η ′ ) , ψ )sin( φ ′ ( η ′ )) exp( − G η ′ ,η,ψ )d η ′ . The integral Z η · · · can be estimated as in Region I, so we only need to estimate the integral Z Lη · · · . Also,noting that fact that exp( − G L,η ′ ,ψ − G L,η,ψ ) ≤ exp( − G η ′ ,η,ψ ) , (4.87) we only need to estimate Z Lη ˜ A ( η ′ , R φ ′ ( η ′ ) , ψ )sin( φ ′ ( η ′ )) exp( − G η ′ ,η,ψ )d η ′ . (4.88)Here the proof is almost identical to that in Region I, so we only point out the key differences.Step 0: Preliminaries.We need to update one key result. For 0 ≤ η ≤ η ′ ,sin φ ′ = p − cos φ ′ ≤ s − (cid:18) R − ǫηR − ǫη ′ (cid:19) cos φ (4.89)= q ( R − ǫη ′ ) sin φ + (2 R − ǫη − ǫη ′ )( ǫη ′ − ǫη ) cos φR − ǫη ′ ≤ | sin φ | . (4.90)Then we have − Z η ′ η φ ′ ( y ) d y ≤ − η ′ − η | sin φ | . (4.91)In the following, we will divide the estimate of II into several cases based on the value of sin φ , sin φ ′ and ǫη ′ . We write II = Z Lη { sin φ ≤− δ } + Z Lη {− δ ≤ sin φ ≤ } { χ ( φ ∗ ) < } (4.92)+ Z Lη {− δ ≤ sin φ ≤ } { χ ( φ ∗ )=1 } {√ ǫη ′ ≥ sin φ ′ } + Z Lη {− δ ≤ sin φ ≤ } { χ ( φ ∗ )=1 } {√ ǫη ′ ≤ sin φ ′ } = II + II + II + II . Step 1: Estimate of II for sin φ ≤ − δ .We first estimate sin φ ′ . Along the characteristics, we knowe − V ( η ′ ,ψ ) cos φ ′ = e − V ( η,ψ ) cos φ, (4.93)which implies cos φ ′ = e V ( η ′ ,ψ ) − V ( η,ψ ) cos φ ≤ e V ( L,ψ ) − V (0 ,ψ ) cos φ = e V ( L,ψ ) − V (0 ,ψ ) q − δ . (4.94)Based on Lemma 4.1, we can further deduce thatcos φ ′ ≤ (cid:18) − ǫ − n R ′ (cid:19) − q − δ . (4.95)Then we have sin φ ′ ≥ s − (cid:18) − ǫ − n R ′ (cid:19) − (1 − δ ) ≥ δ − ǫ − n > δ , (4.96)when ǫ is sufficiently small. Based on Lemma 4.4, we know (cid:12)(cid:12)(cid:12)(cid:12) sin φ ∂ V ∂η (cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:18) δ (cid:19)(cid:18) k V k L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ (cid:19) . (4.97)Hence, we have | II | ≤ | sin φ | (cid:12)(cid:12)(cid:12)(cid:12) ∂ V ∂η (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cδ (cid:18) k V k L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ (cid:19) . (4.98) EGULARITY OF MILNE PROBLEM IN 3D 35
Step 2: Estimate of II for − δ ≤ sin φ ≤ χ ( φ ∗ ) < I based on the integral Z Lη φ ′ exp( − G η ′ ,η,ψ )d η ′ ≤ . (4.99)Then we have | II | ≤ C ( δ ) (cid:18) k V k L ∞ L ∞ + k S k L ∞ L ∞ (cid:19) . (4.100)Step 3: Estimate of II for − δ ≤ sin φ ≤ χ ( φ ∗ ) = 1 and √ ǫη ′ ≥ sin φ ′ .This is identical to the estimate of I , we have | II | ≤ Cδ k A k L ∞ L ∞ . (4.101)Step 4: Estimate of II for − δ ≤ sin φ ≤ χ ( φ ∗ ) = 1 and √ ǫη ′ ≤ sin φ ′ .This step is different. We do not need to further decompose the cases. Based on (4.91), we have, − G η,η ′ ≤ − η ′ − η | sin φ | . (4.102)Then following the same argument in estimating I , we obtain | II | ≤ C k A k L ∞ L ∞ Z Lη (cid:18) | ln( ǫ ) | + | ln( η ′ ) | (cid:19) exp (cid:18) − η ′ − η | sin φ | (cid:19) d η ′ (4.103)If η ≥
2, we directly obtain (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z Lη | ln( η ′ ) | exp (cid:18) − η ′ − η | sin φ | (cid:19) d η ′ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L ln( η ′ ) exp (cid:18) − η ′ − η | sin φ | (cid:19) d η ′ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (4.104) ≤ ln(2) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L exp (cid:18) − η ′ − η | sin φ | (cid:19) d η ′ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C p | sin φ | . If η ≤
2, we decompose as (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z Lη | ln( η ′ ) | exp (cid:18) − η ′ − η | sin φ | (cid:19) d η ′ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (4.105) ≤ (cid:12)(cid:12)(cid:12)(cid:12)Z η | ln( η ′ ) | exp (cid:18) − η ′ − η | sin φ | (cid:19) d η ′ (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L | ln( η ′ ) | exp (cid:18) − η ′ − η | sin φ | (cid:19) d η ′ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . The second term is identical to the estimate in η ≥
2. We apply Cauchy’s inequality to the first term (cid:12)(cid:12)(cid:12)(cid:12)Z η | ln( η ′ ) | exp (cid:18) − η ′ − η | sin φ | (cid:19) d η ′ (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:18) Z η ln ( η ′ )d η ′ (cid:19) / (cid:18) Z η exp (cid:18) − η ′ − η ) | sin φ | (cid:19) d η ′ (cid:19) / (4.106) ≤ (cid:18) Z ln ( η ′ )d η ′ (cid:19) / (cid:18) Z η exp (cid:18) − η ′ − η ) | sin φ | (cid:19) d η ′ (cid:19) / ≤ C p | sin φ | . Hence, we have | II | ≤ C (1 + | ln( ǫ ) | ) p δ k A k L ∞ L ∞ . (4.107)Step 5: Synthesis. Collecting all the terms in previous steps, we have proved | II | ≤ C (1 + | ln( ǫ ) | ) p δ k A k L ∞ L ∞ + Cδ k A k L ∞ L ∞ (4.108)+ Cδ (cid:18) k V k L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂p∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ (cid:19) + C ( δ ) (cid:18) k V k L ∞ L ∞ + k S k L ∞ L ∞ (cid:19) . Therefore, we know | A | II ≤ k S A k L ∞ L ∞ + k p A k L ∞ + C (1 + | ln( ǫ ) | ) p δ k A k L ∞ L ∞ + Cδ k A k L ∞ L ∞ (4.109)+ Cδ (cid:18) k V k L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂p∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ (cid:19) + C ( δ ) (cid:18) k V k L ∞ L ∞ + k S k L ∞ L ∞ (cid:19) . Region III: sin φ < and | E ( η, φ, ψ ) | ≥ e − V ( L ) . We still ignore ψ dependence. Based on [27, Lemma4.7, Lemma 4.8], we still have |K [ p A ] | ≤ k p A k L ∞ , (4.110) |T [ S A ] | ≤ k S A k L ∞ L ∞ . (4.111)Hence, we only need to estimate III = T [ ˜ A ]. Note that | E ( η, φ, ψ ) | ≥ e − V ( L,ψ ) impliese − V ( η,ψ ) cos φ ≥ e − V ( L,ψ ) . (4.112)Hence, based on Lemma 4.1, we can further deduce thatcos φ ≥ e V ( η,ψ ) − V ( L,ψ ) ≥ e V (0 ,ψ ) − V ( L,ψ ) ≥ (cid:18) − ǫ − n R ′ (cid:19) . (4.113)Hence, we know | sin φ | ≤ s − (cid:18) − ǫ − n R ′ (cid:19) ≤ ǫ − n . (4.114)Hence, when ǫ is sufficiently small, we always have | sin φ | ≤ ǫ − n ≤ δ . (4.115)This means we do not need to bother with the estimate of sin φ ≤ − δ as Step 1 in estimating I and II .Since we can decompose T [ ˜ A ] = Z η ˜ A ( η ′ , φ ′ ( η ′ ) , ψ )sin( φ ′ ( η ′ )) exp( − G η + ,η ′ ,ψ − G η + ,η,ψ )d η ′ (4.116) (cid:18) Z η + η ˜ A ( η ′ , φ ′ ( η ′ ) , ψ )sin( φ ′ ( η ′ )) exp( − G η + ,η ′ ,ψ − G η + ,η,ψ )d η ′ + Z η + η ( ˜ A + S A )( η ′ , R φ ′ ( η ′ ) , ψ )sin( φ ′ ( η ′ )) exp( − G η ′ ,η,ψ )d η ′ (cid:19) . Then the integral Z η ( · · · ) is similar to the argument in Region I, and the integral Z η + η ( · · · ) is similar to theargument in Region II. Hence, combining the methods in Region I and Region II, we can show the desiredresult, i.e. | A | III ≤ k p A k L ∞ + k S A k L ∞ L ∞ + C (1 + | ln( ǫ ) | ) p δ k A k L ∞ L ∞ + Cδ k A k L ∞ L ∞ (4.117)+ C ( δ ) (cid:18) k V k L ∞ L ∞ + k S k L ∞ L ∞ (cid:19) . EGULARITY OF MILNE PROBLEM IN 3D 37
Estimate of Normal Derivative.
Combining the analysis in these three regions, we have | A | ≤ k p A k L ∞ + k S A k L ∞ L ∞ + C (1 + | ln( ǫ ) | ) p δ k A k L ∞ L ∞ + Cδ k A k L ∞ L ∞ (4.118)+ Cδ (cid:18) k V k L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂p∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ (cid:19) + C ( δ ) (cid:18) k V k L ∞ L ∞ + k S k L ∞ L ∞ (cid:19) . Taking supremum over all ( η, φ, ψ ), we have k A k L ∞ L ∞ ≤ k p A k L ∞ + k S A k L ∞ L ∞ + C (1 + | ln( ǫ ) | ) p δ k A k L ∞ L ∞ + Cδ k A k L ∞ L ∞ (4.119)+ Cδ (cid:18) k V k L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂p∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ (cid:19) + C ( δ ) (cid:18) k V k L ∞ L ∞ + k S k L ∞ L ∞ (cid:19) . Then we choose these constants to perform absorbing argument. First we choose 0 < δ << Cδ ≤ . (4.120)Then we take δ = δ | ln( ǫ ) | − such that C (1 + | ln( ǫ ) | ) p δ ≤ Cδ ≤ . (4.121)for ǫ sufficiently small. Note that this mild decay of δ with respect to ǫ also justifies the assumption in CaseIII and the proof of Lemma 4.4 that ǫ − n ≤ δ , (4.122)for ǫ sufficiently small. Here since δ and C are independent of ǫ , there is no circulant argument. Hence, wecan absorb all the term related to k A k L ∞ L ∞ on the right-hand side of (4.119) to the left-hand side to obtain k A k L ∞ L ∞ ≤ C (cid:18) k p A k L ∞ + k S A k L ∞ L ∞ (cid:19) (4.123)+ C | ln( ǫ ) | (cid:18) k V k L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂p∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + k S k L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ (cid:19) . Mild Formulation of Velocity Derivative.
Consider the general ǫ -Milne problem for B = ζ ∂ V ∂φ as sin φ ∂ B ∂η + F ( η, ψ ) cos φ ∂ B ∂φ + B = S B , B (0 , φ, ψ ) = p B ( φ, ψ ) for sin φ > , B ( L, φ, ψ ) = B ( L, R φ, ψ ) , (4.124)where p B and S B will be specified later. This is much simpler than normal derivative, since we do not have˜ B . Then by a direct argument that |K [ p B ] | ≤ k p B k L ∞ , (4.125) |T [ S B ] | ≤ k S B k L ∞ L ∞ . (4.126)we can get the desired result. Lemma 4.8.
We have k B k L ∞ L ∞ ≤ k p B k L ∞ + k S B k L ∞ L ∞ . (4.127) Estimate of Derivatives.
In this subsection, we combine above a priori estimates of normal andvelocity derivatives.
Theorem 4.9.
Assume (3.9), (3.10), (4.2) and (4.3). The normal and velocity derivatives of V are well-defined a.e. and satisfy (cid:13)(cid:13)(cid:13)(cid:13) ζ ∂ V ∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ζ ∂ V ∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ ≤ C | ln( ǫ ) | . (4.128) Proof.
Based on the analysis in [18], derivatives are a.e. well-defined. Collecting the estimates for A and B in Lemma 4.7 and Lemma 4.8, we have k A k L ∞ L ∞ ≤ C (cid:18) k p A k L ∞ + k S A k L ∞ L ∞ (cid:19) (4.129)+ C | ln( ǫ ) | (cid:18) k V k L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂p∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + k S k L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂S∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ (cid:19) , k B k L ∞ L ∞ ≤ k p B k L ∞ + k S B k L ∞ L ∞ . (4.130)Taking derivatives on both sides of (4.8) and multiplying ζ , based on Lemma 4.2, we have p A = ǫ cos φ ∂p∂φ + p − ¯ V (0) , (4.131) p B = sin φ ∂p∂φ , (4.132) S A = ∂F∂η B cos φ + ζ ∂S∂η , (4.133) S B = A cos φ + F B sin φ + ζ ∂S∂φ . (4.134)Since | F ( η ) | + (cid:12)(cid:12)(cid:12)(cid:12) ∂F∂η (cid:12)(cid:12)(cid:12)(cid:12) ≤ ǫ , by absorbing A and B on the right-hand side of (4.129) and (4.130), we derive A ≤ C | ln( ǫ ) | , (4.135) B ≤ C | ln( ǫ ) | . (4.136) (cid:3) Theorem 4.10.
Assume (3.9), (3.10), (4.2) and (4.3). The normal and velocity derivatives of V arewell-defined a.e. and satisfy (cid:13)(cid:13)(cid:13)(cid:13) e K η ζ ∂ V ∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) e K η ζ ∂ V ∂φ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ ≤ C | ln( ǫ ) | . (4.137) Proof.
This proof is almost identical to Theorem 4.9. The only difference is that S A is added by K A sin φ and S B added by K B sin φ . When K is sufficiently small, we can also absorb them into the left-hand side.Hence, this is obvious. (cid:3) Corollary 4.11.
Assume (3.9), (3.10), (4.2) and (4.3). We have (cid:13)(cid:13)(cid:13)(cid:13) e K η sin φ ∂ V ∂η (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ ≤ C | ln( ǫ ) | . (4.138) Proof.
This is a natural result of Theorem 4.10 since ζ ( η, φ, ψ ) ≥ | sin φ | . (cid:3) Now we pull τ i for i = 1 , ψ dependence back and study the tangential derivatives and velocityderivative. Theorem 4.12.
Assume (3.9), (3.10), (4.2) and (4.3). We have for i = 1 , , (cid:13)(cid:13)(cid:13)(cid:13) e K η ∂ V ∂τ i (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ ≤ C | ln( ǫ ) | , (4.139) (cid:13)(cid:13)(cid:13)(cid:13) e K η ∂ V ∂ψ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ ≤ C | ln( ǫ ) | (4.140) EGULARITY OF MILNE PROBLEM IN 3D 39
Proof.
Following a similar fashion in proof of Theorem 4.10, using iteration and characteristics, we can show ∂ V ∂τ i is a.e. well-defined, so here we focus on the a priori estimate. Let W = ∂ V ∂τ i . Taking τ i derivative onboth sides of (4.8), we have W satisfies the equation sin φ ∂ W ∂η + F ( η, ψ ) cos φ ∂ W ∂φ + W − ¯ W = ∂S∂τ i + ǫ (cid:18) ∂ τ i R sin ψ ( R − ǫη ) + ∂ τ i R cos ψ ( R − ǫη ) (cid:19)(cid:18) cos φ ∂ V ∂φ (cid:19) , W (0 , φ, ψ ) = ∂p∂τ i ( φ, ψ ) for sin φ > , W ( L, φ, ψ ) = W ( L, R φ, ψ ) . (4.141)Our assumptions on S verify (cid:13)(cid:13)(cid:13)(cid:13) e K η ∂S∂τ i (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ ≤ C. (4.142)For η ∈ [0 , L ], we have (cid:13)(cid:13)(cid:13)(cid:13) ∂ τ i R sin ψ ( R − ǫη ) + ∂ τ i R cos ψ ( R − ǫη ) (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ ≤ C, (4.143)and C ǫ ≤ F ( η, ψ ) ≤ C ǫ. (4.144)Based on Corollary 4.11 and the equation (4.8), we know (cid:13)(cid:13)(cid:13)(cid:13) e K η (cid:18) ǫ cos φ ∂ V ∂φ (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ ≤ C | ln( ǫ ) | . (4.145)Therefore, the source term in the equation (4.141) is in L ∞ and decays exponentially. By Theorem 3.6, wehave (cid:13)(cid:13) e K η ( W − W L ) (cid:13)(cid:13) L ∞ L ∞ ≤ C | ln( ǫ ) | , (4.146)for some constant W L . It is easy to see this W L must be zero due to decay of V . Similarly, let W ′ = ∂ V ∂ψ .Taking ψ derivative on both sides of (4.8), we have W ′ satisfies the equation sin φ ∂ W ′ ∂η + F ( η, ψ ) cos φ ∂ W ′ ∂φ + W ′ = ∂S∂ψ + ǫ (cid:18) sin(2 ψ ) R − ǫη − sin(2 ψ ) R − ǫη (cid:19)(cid:18) cos φ ∂ V ∂φ (cid:19) , W ′ (0 , φ, ψ ) = ∂p∂ψ ( φ, ψ ) for sin φ > , W ′ ( L, φ, ψ ) = W ′ ( L, R φ, ψ ) . (4.147)We may directly estimate (cid:13)(cid:13)(cid:13)(cid:13) sin(2 ψ ) R − ǫη − sin(2 ψ ) R − ǫη (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ ≤ C, (4.148)which means the source term is in L ∞ and decays exponentially. This equation does not involve ¯ W ′ term,which makes it even simpler. Hence, we get (cid:13)(cid:13) e K η W ′ (cid:13)(cid:13) L ∞ L ∞ ≤ C | ln( ǫ ) | , (4.149)for some constant W ′ L . Naturally we have W ′ L must be zero. (cid:3) We finally come to the ǫ -Milne problem with diffusive boundary. Theorem 4.13.
Assume (3.9), (3.10), (4.2) and (4.3). There exists K > such that the unique solution f ( η, φ, ψ ) to the ǫ -Milne problem (3.100) with the normalization condition (3.104) satisfies for i = 1 , , (cid:13)(cid:13)(cid:13)(cid:13) e K η ∂ ( f − f L ) ∂τ i (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ ≤ C | ln( ǫ ) | , (4.150) (cid:13)(cid:13)(cid:13)(cid:13) e K η ∂ ( f − f L ) ∂ψ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ L ∞ ≤ C | ln( ǫ ) | (4.151) Appendix A. Remainder Estimate
In this section, we consider the remainder equation for u ( ~x, ~w ) as ( ǫ ~w · ∇ x u + u − ¯ u = f ( ~x, ~w ) in Ω ,u ( ~x , ~w ) = P [ u ]( ~x ) + h ( ~x , ~w ) for ~w · ~ν < ~x ∈ ∂ Ω , (A.1)where ¯ u ( ~x ) = 14 π Z S u ( ~x, ~w )d ~w, (A.2) P [ u ]( ~x ) = 14 π Z ~w · ~ν> u ( ~x , ~w )( ~w · ~ν )d ~w, (A.3) ~ν is the outward unit normal vector, with the Knudsen number 0 < ǫ <<
1. To guarantee uniqueness, weneed the normalization condition Z Ω ×S u ( ~x, ~w )d ~w d ~x = 0 . (A.4)Also, the data f and h satisfy the compatibility condition Z Ω ×S f ( ~x, ~w )d ~w d ~x + ǫ Z ∂ Ω Z ~w · ~ν< h ( ~x , ~w )( ~w · ~ν )d ~w d ~x = 0 . (A.5)Based on the flow direction, we can divide the boundary Γ = { ( ~x, ~w ) : ~x ∈ ∂ Ω } into the in-flow boundaryΓ − , the out-flow boundary Γ + and the grazing set Γ asΓ − = { ( ~x, ~w ) : ~x ∈ ∂ Ω , ~w · ~ν < } (A.6)Γ + = { ( ~x, ~w ) : ~x ∈ ∂ Ω , ~w · ~ν > } (A.7)Γ = { ( ~x, ~w ) : ~x ∈ ∂ Ω , ~w · ~ν = 0 } (A.8)It is easy to see Γ = Γ + ∪ Γ − ∪ Γ . Hence, the boundary condition is only given for Γ − . We define the L p norm with 1 ≤ p < ∞ and L ∞ norms in Ω × S as usual: k f k L p (Ω ×S ) = (cid:18) Z Ω Z S | f ( ~x, ~w ) | p d ~w d ~x (cid:19) /p , (A.9) k f k L ∞ (Ω ×S ) = sup ( ~x, ~w ) ∈ Ω ×S | f ( ~x, ~w ) | . (A.10)Define the L p norm with 1 ≤ p < ∞ and L ∞ norms on the boundary as follows: k f k L p (Γ) = (cid:18) Z Z Γ | f ( ~x, ~w ) | p | ~w · ~ν | d ~w d ~x (cid:19) /p , (A.11) k f k L p (Γ ± ) = (cid:18) Z Z Γ ± | f ( ~x, ~w ) | p | ~w · ~ν | d ~w d ~x (cid:19) /p , (A.12) k f k L ∞ (Γ) = sup ( ~x, ~w ) ∈ Γ | f ( ~x, ~w ) | , (A.13) k f k L ∞ (Γ ± ) = sup ( ~x, ~w ) ∈ Γ ± | f ( ~x, ~w ) | . (A.14)The direct application of energy method as in [27] and [28], we may obtain Lemma A.1.
Assume f ( ~x, ~w ) ∈ L ∞ (Ω × S ) and h ( x , ~w ) ∈ L ∞ (Γ − ) . Then for the transport equation(A.1), there exists a unique solution u ( ~x, ~w ) ∈ L (Ω × S ) satisfying ǫ k (1 − P )[ u ] k L (Γ + ) + k u k L (Ω ×S ) ≤ C (cid:18) ǫ k f k L (Ω ×S ) + 1 ǫ k h k L (Γ − ) (cid:19) , (A.15)Based on classical L − L ∞ framework, we are able to show k u k L ∞ (Ω ×S ) ≤ C (Ω) (cid:18) ǫ k u k L (Ω ×S ) + k f k L ∞ (Ω ×S ) + k h k L ∞ (Γ − ) (cid:19) . (A.16)Therefore, it is natural to deduce L ∞ estimate. EGULARITY OF MILNE PROBLEM IN 3D 41
Theorem A.2.
Assume f ( ~x, ~w ) ∈ L ∞ (Ω × S ) and h ( x , ~w ) ∈ L ∞ (Γ − ) . Then the solution u ( ~x, ~w ) to thetransport equation (A.1) satisfies k u k L ∞ (Ω ×S ) ≤ C (Ω) (cid:18) ǫ k f k L (Ω ×S ) + 1 ǫ k h k L (Γ − ) + k f k L ∞ (Ω ×S ) + k h k L ∞ (Γ − ) (cid:19) . However, the estimates here is not strong enough to close the diffusive limit, so we must further improvethem.A.1. L m Estimate.
In this subsection, we try to improve previous estimates. In the following, we assume m is an integer and let o (1) denote a sufficiently small constant. Theorem A.3.
Assume f ( ~x, ~w ) ∈ L ∞ (Ω × S ) and h ( x , ~w ) ∈ L ∞ (Γ − ) . Then for ≤ m ≤ , u ( ~x, ~w ) satisfies ǫ k (1 − P )[ u ] k L (Γ + ) + k ¯ u k L m (Ω ×S ) + 1 ǫ k u − ¯ u k L (Ω ×S ) (A.1.1) ≤ C (cid:18) o (1) ǫ m k u k L ∞ (Γ + ) + 1 ǫ k f k L (Ω ×S ) + 1 ǫ k f k L m m − (Ω ×S ) + 1 ǫ k h k L (Γ − ) + k h k L m (Γ − ) (cid:19) . Proof.
We divide the proof into several steps:Step 1: Kernel Estimate.Applying Green’s identity to the equation (A.1). Then for any φ ∈ L (Ω ×S ) satisfying ~w ·∇ x φ ∈ L (Ω ×S )and φ ∈ L (Γ), we have ǫ Z Γ uφ d γ − ǫ Z Z Ω ×S ( ~w · ∇ x φ ) u + Z Z Ω ×S ( u − ¯ u ) φ = Z Z Ω ×S f φ. (A.1.2)Our goal is to choose a particular test function φ . We first construct an auxiliary function ζ . Naturally u ∈ L ∞ (Ω × S ) implies ¯ u ∈ L m (Ω) which further leads to (¯ u ) m − ∈ L m m − (Ω). We define ζ ( ~x ) on Ωsatisfying ∆ ζ = (¯ u ) m − − | Ω | Z Ω (¯ u ) m − d ~x in Ω ,∂ζ∂~ν = 0 on ∂ Ω . (A.1.3)In the bounded domain Ω, based on the standard elliptic estimate, we have a unique ζ satisfying k ζ k W , m m − (Ω) ≤ C (cid:13)(cid:13) (¯ u ) m − (cid:13)(cid:13) L m m − (Ω) = C k ¯ u k m − L m (Ω) , (A.1.4)and Z Ω ζ ( ~x )d ~x = 0 . (A.1.5)We plug the test function φ = − ~w · ∇ x ζ (A.1.6)into the weak formulation (A.1.2) and estimate each term there. By Sobolev embedding theorem, we havefor 1 ≤ m ≤ k φ k L (Ω) ≤ C k ζ k H (Ω) ≤ C k ζ k W , m m − (Ω) ≤ C k ¯ u k m − L m (Ω) , (A.1.7) k φ k L m m − (Ω) ≤ C k ζ k W , m m − (Ω) ≤ C k ¯ u k m − L m (Ω) . (A.1.8)Easily we can decompose − ǫ ZZ Ω ×S ( ~w · ∇ x φ ) u λ = − ǫ Z Z Ω ×S ( ~w · ∇ x φ )¯ u λ − ǫ Z Z Ω ×S ( ~w · ∇ x φ )( u λ − ¯ u λ ) . (A.1.9) We estimate the two term on the right-hand side of (A.1.9) separately. By (A.1.3) and (A.1.6), we have − ǫ Z Z Ω ×S ( ~w · ∇ x φ )¯ u = ǫ Z Z Ω ×S ¯ u (cid:18) w ( w ∂ ζ + w ∂ ζ ) + w ( w ∂ ζ + w ∂ ζ ) (cid:19) (A.1.10)= ǫ Z Z Ω ×S ¯ u (cid:18) w ∂ ζ + w ∂ ζ (cid:19) = 2 ǫπ Z Ω ¯ u ( ∂ ζ + ∂ ζ )= ǫ k ¯ u k mL m (Ω) . In the second equality, above cross terms vanish due to the symmetry of the integral over S . On the otherhand, for the second term in (A.1.9), H¨older’s inequality and the elliptic estimate imply − ǫ Z Z Ω ×S ( ~w · ∇ x φ )( u − ¯ u ) ≤ Cǫ k u − ¯ u k L m (Ω ×S ) k∇ x φ k L m m − (Ω) (A.1.11) ≤ Cǫ k u − ¯ u k L m (Ω ×S ) k ζ k W , m m − (Ω) ≤ Cǫ k u − ¯ u k L m (Ω ×S ) k ¯ u k m − L m (Ω) . Based on (A.1.4), (A.1.7), (A.1.8), Sobolev embedding theorem and the trace theorem, we have (A.1.12) k∇ x ζ k L m m − (Γ) ≤ C k∇ x ζ k W m , m m − (Γ) ≤ C k∇ x ζ k W , m m − (Ω) ≤ C k ζ k W , m m − (Ω) ≤ C k ¯ u k m − L m (Ω) . We may also decompose ~w = ( ~w · ~ν ) ~ν + ~w ⊥ to obtain ǫ Z Γ uφ d γ = ǫ Z Γ u ( ~w · ∇ x ζ )d γ (A.1.13)= ǫ Z Γ u ( ~ν · ∇ x ζ )( ~w · ~ν )d γ + ǫ Z Γ u ( ~w ⊥ · ∇ x ζ )d γ = ǫ Z Γ u ( ~w ⊥ · ∇ x ζ )d γ. Based on (A.1.4), (A.1.8) and H¨older’s inequality, we have ǫ Z Γ uφ d γ = ǫ Z Γ u ( ~w ⊥ · ∇ x ζ )d γ (A.1.14)= ǫ Z Γ P [ u ]( ~w ⊥ · ∇ x ζ )d γ + ǫ Z Γ + (1 − P )[ u ]( ~w ⊥ · ∇ x ζ )d γ + ǫ Z Γ − h ( ~w ⊥ · ∇ x ζ )d γ = ǫ Z Γ + (1 − P )[ u ]( ~w ⊥ · ∇ x ζ )d γ + ǫ Z Γ − h ( ~w ⊥ · ∇ x ζ )d γ ≤ Cǫ k∇ x ζ k L m m − (Γ) (cid:18) k (1 − P )[ u ] k L m (Γ + ) + k h k L m (Γ − ) (cid:19) ≤ Cǫ k ¯ u k m − L m (Ω ×S ) (cid:18) k (1 − P )[ u ] k L m (Γ + ) + k h k L m (Γ − ) (cid:19) . Hence, we know ǫ Z Γ uφ d γ ≤ Cǫ k ¯ u k m − L m (Ω ×S ) (cid:18) k (1 − P )[ u ] k L m (Γ + ) + k h k L m (Γ − ) (cid:19) . (A.1.15)Also, we have Z Z Ω ×S ( u − ¯ u ) φ ≤ C k φ k L (Ω ×S ) k u − ¯ u k L (Ω ×S ) ≤ C k ¯ u k m − L m (Ω) k u − ¯ u k L (Ω ×S ) , (A.1.16) Z Z Ω ×S f φ ≤ C k φ k L (Ω ×S ) k f k L (Ω ×S ) ≤ C k ¯ u k m − L m (Ω) k f k L (Ω ×S ) . (A.1.17) EGULARITY OF MILNE PROBLEM IN 3D 43
Collecting terms in (A.1.10), (A.1.11), (A.1.15), (A.1.16) and (A.1.17), we obtain ǫ k ¯ u k L m (Ω ×S ) ≤ C (cid:18) ǫ k u − ¯ u k L m (Ω ×S ) + k u − ¯ u k L (Ω ×S ) + k f k L (Ω ×S ) (A.1.18)+ ǫ k (1 − P )[ u ] k L m (Γ + ) + ǫ k h k L m (Γ − ) (cid:19) , Step 2: Energy Estimate.In the weak formulation (A.1.2), we may take the test function φ = u to get the energy estimate12 ǫ Z Γ | u | d γ + k u − ¯ u k L (Ω ×S ) = Z Z Ω ×S f u. (A.1.19)Hence, by L estimate as in Theorem A.1 and [18], this naturally implies ǫ k (1 − P )[ u ] k L (Γ + ) + k u − ¯ u k L (Ω ×S ) ≤ k f k L (Ω ×S ) + Z Z Ω ×S f u + k h k L (Γ − ) . (A.1.20)On the other hand, we can square on both sides of (A.1.18) to obtain ǫ k ¯ u k L m (Ω ×S ) ≤ C (cid:18) ǫ k u − ¯ u k L m (Ω ×S ) + k u − ¯ u k L (Ω ×S ) + k f k L (Ω ×S ) (A.1.21)+ ǫ k (1 − P )[ u ] k L m (Γ + ) + ǫ k h k L m (Γ − ) (cid:19) , Multiplying a sufficiently small constant on both sides of (A.1.21) and adding it to (A.1.20) to absorb k u − ¯ u k L (Ω ×S ) and ǫ k ¯ u k L (Ω ×S ) , we deduce ǫ k (1 − P )[ u ] k L (Γ + ) + ǫ k ¯ u k L m (Ω ×S ) + k u − ¯ u k L (Ω ×S ) (A.1.22) ≤ C (cid:18) ǫ k u − ¯ u k L m (Ω ×S ) + ǫ k (1 − P )[ u ] k L m (Γ + ) + k f k L (Ω ×S ) + Z Z Ω ×S f u + k h k L (Γ − ) + ǫ k h k L m (Γ − ) (cid:19) . By interpolation estimate and Young’s inequality, for 32 ≤ m ≤
3, we have k (1 − P )[ u ] k L m (Γ + ) ≤ k (1 − P )[ u ] k m L (Γ + ) k (1 − P )[ u ] k m − m L ∞ (Γ + ) (A.1.23)= (cid:18) ǫ m − m k (1 − P )[ u ] k m L (Γ + ) (cid:19)(cid:18) ǫ m − m k (1 − P )[ u ] k m − m L ∞ (Γ + ) (cid:19) ≤ C (cid:18) ǫ m − m k (1 − P )[ u ] k m L (Γ + ) (cid:19) m + o (1) (cid:18) ǫ m − m k (1 − P )[ u ] k m − m L ∞ (Γ + ) (cid:19) m m − ≤ Cǫ m − m k (1 − P )[ u ] k L (Γ + ) + o (1) ǫ m k (1 − P )[ u ] k L ∞ (Γ + ) ≤ Cǫ m − m k (1 − P )[ u ] k L (Γ + ) + o (1) ǫ m k u k L ∞ (Ω ×S ) . Similarly, we have k u − ¯ u k L m (Ω ×S ) ≤ k u − ¯ u k m L (Ω ×S ) k u − ¯ u k m − m L ∞ (Ω ×S ) (A.1.24)= (cid:18) ǫ m − m k u − ¯ u k m L (Ω ×S ) (cid:19)(cid:18) ǫ m − m k u − ¯ u k m − m L ∞ (Ω ×S ) (cid:19) ≤ C (cid:18) ǫ m − m k u − ¯ u k m L (Ω ×S ) (cid:19) m + o (1) (cid:18) ǫ m − m k u − ¯ u k m − m L ∞ (Ω ×S ) (cid:19) mm − ≤ Cǫ m − m k u − ¯ u k L (Ω ×S ) + o (1) ǫ m k u − ¯ u k L ∞ (Ω ×S ) . We need this extra ǫ m for the convenience of L ∞ estimate. Then we know for sufficiently small ǫ and32 ≤ m ≤
3, , ǫ k (1 − P )[ u ] k L m (Γ + ) ≤ Cǫ − m − m k (1 − P )[ u ] k L (Γ + ) + o (1) ǫ m k u k L ∞ (Γ + ) (A.1.25) ≤ ǫ m k (1 − P )[ u ] k L (Γ + ) + o (1) ǫ m k u k L ∞ (Γ + ) , (A.1.26) ≤ o (1) ǫ k (1 − P )[ u ] k L (Γ + ) + o (1) ǫ m k u k L ∞ (Γ + ) . Similarly, we have ǫ k u − ¯ u k L m (Ω ×S ) ≤ ǫ − m − m k u − ¯ u k L (Ω ×S ) + o (1) ǫ m k u k L ∞ (Ω ×S ) (A.1.27) ≤ ǫ m − k u − ¯ u k L (Ω ×S ) + o (1) ǫ m k u k L ∞ (Ω ×S ) , (A.1.28) ≤ o (1) k u − ¯ u k L (Ω ×S ) + o (1) ǫ m k u k L ∞ (Ω ×S ) . By (A.1.20), we can absorb k u − ¯ u k L (Ω ×S ) and ǫ k (1 − P )[ u ] k L (Γ + ) into left-hand side to obtain ǫ k (1 − P )[ u ] k L (Γ + ) + ǫ k ¯ u k L m (Ω ×S ) + k u − ¯ u k L (Ω ×S ) (A.1.29) ≤ C (cid:18) o (1) ǫ m k u k L ∞ (Ω ×S ) + k f k L (Ω ×S ) + Z Z Ω ×S f u + k h k L (Γ − ) + ǫ k h k L m (Γ − ) (cid:19) . We can decompose
Z Z Ω ×S f u = Z Z Ω ×S f ¯ u + Z Z Ω ×S f ( u − ¯ u ) . (A.1.30)H¨older’s inequality and Cauchy’s inequality imply Z Z Ω ×S f ¯ u ≤ k f k L m m − (Ω ×S ) k ¯ u k L m (Ω ×S ) ≤ Cǫ k f k L m m − (Ω ×S ) + o (1) ǫ k ¯ u k L m (Ω ×S ) , (A.1.31)and Z Z Ω ×S f ( u − ¯ u ) ≤ C k f k L (Ω ×S ) + o (1) k u − ¯ u k L (Ω ×S ) . (A.1.32)Hence, absorbing ǫ k ¯ u k L m (Ω ×S ) and k u − ¯ u k L (Ω ×S ) into left-hand side of (A.1.29), we get ǫ k (1 − P )[ u ] k L (Γ + ) + ǫ k ¯ u k L m (Ω ×S ) + k u − ¯ u k L (Ω ×S ) (A.1.33) ≤ C (cid:18) o (1) ǫ m k u k L ∞ (Γ + ) + k f k L (Ω ×S ) + 1 ǫ k f k L m m − (Ω ×S ) + k h k L (Γ − ) + ǫ k h k L m (Γ − ) (cid:19) , which implies1 ǫ k (1 − P )[ u ] k L (Γ + ) + k ¯ u k L m (Ω ×S ) + 1 ǫ k u − ¯ u k L (Ω ×S ) (A.1.34) ≤ C (cid:18) o (1) ǫ m k u k L ∞ (Γ + ) + 1 ǫ k f k L (Ω ×S ) + 1 ǫ k f k L m m − (Ω ×S ) + 1 ǫ k h k L (Γ − ) + k h k L m (Γ − ) (cid:19) . (cid:3) A.2. L ∞ Estimate.
In this subsection, we prove the L ∞ estimate. We consider the characteristics thatreflect several times on the boundary. Definition A.4. (Stochastic Cycle) For fixed point ( t, ~x, ~w ) with ( ~x, ~w ) / ∈ Γ , let ( t , ~x , ~w ) = (0 , ~x, ~w ) . For ~w k +1 such that ~w k +1 · ~ν ( ~x k +1 ) > , define the ( k + 1) -component of the back-time cycle as ( t k +1 , ~x k +1 , ~w k +1 ) = ( t k + t b ( ~x k , ~w k ) , ~x b ( ~x k , ~w k ) , ~w k +1 ) (A.2.1) where t b ( ~x, ~w ) = inf { t > ~x − ǫt ~w / ∈ Ω } (A.2.2) x b ( ~x, ~w ) = ~x − ǫt b ( ~x, ~w ) ~w / ∈ Ω (A.2.3)
EGULARITY OF MILNE PROBLEM IN 3D 45
Set X cl ( s ; t, ~x, ~w ) = X k t k +1 ≤ s
For T > sufficiently large, there exists constants C , C > independent of T , such thatfor k = C T / , Z Q k − j =1 µ j t k ( t,~x, ~w, ~w ,..., ~w k − ) See [14, Lemma 4.1]. (cid:3) Theorem A.6. Assume f ( ~x, ~w ) ∈ L ∞ (Ω × S ) and h ( x , ~w ) ∈ L ∞ (Γ − ) . Then for the steady neutrontransport equation (A.1), there exists a unique solution u ( ~x, ~w ) ∈ L ∞ (Ω × S ) satisfying k u k L ∞ (Ω ×S ) ≤ C (cid:18) ǫ m k f k L (Ω ×S ) + 1 ǫ m k f k L m m − (Ω ×S ) + k f k L ∞ (Ω ×S ) (A.2.8)+ 1 ǫ m k h k L (Γ − ) + 1 ǫ m k h k L m (Γ − ) + k h k L ∞ (Γ − ) (cid:19) . Proof. We divide the proof into several steps:Step 1: Mild formulation.We rewrite the equation (A.1) along the characteristics as u ( ~x, ~w ) = h ( ~x − ǫt ~w, ~w )e − t + P [ u ]( ~x − ǫt ~w, ~w )e − t (A.2.9)+ Z t f ( ~x − ǫ ( t − s ) ~w, ~w )e − ( t − s ) d s + Z t ¯ u ( ~x − ǫ ( t − s ) ~w )e − ( t − s ) d s . Note that here P [ u ] is an integral over µ at ~x , using stochastic cycle, we may rewrite it again along thecharacteristics to ~x . This process can continue to arbitrary ~x k . Then we get u ( ~x, ~w ) = e − t H + k − X l =1 (cid:18) Z Q lj =1 e − t l +1 G l Y j =1 d σ j (cid:19) + k − X l =1 (cid:18) Z Q lj =1 e − t l +1 P [ u ]( ~x k , ~w k − ) l Y j =1 d σ j (cid:19) (A.2.10)= I + II + III. where H = h ( ~x − ǫt ~w, ~w ) (A.2.11)+ Z t f ( ~x − ǫ ( t − s ) ~w, ~w )e s d s + Z t ¯ u ( ~x − ǫ ( t − s ) ~w )e s d s ,G = h ( ~x l − ǫt l +1 ~w l , ~w l ) (A.2.12)+ Z t l f ( ~x l − ǫ ( t l +1 − s l +1 ) ~w l , ~w l )e s l +1 d s l +1 + Z t l ¯ u ( ~x l − ǫ ( t l +1 − s l +1 ) ~w l )e s l +1 d s l +1 . We need to estimate each term on the right-hand side of (A.2.10). Step 2: Estimate of mild formulation.We first consider III . We may decompose it as III = k − X l =1 Z Q lj =1 P [ u ]( ~x k , ~w k − )e − t l +1 l Y j =1 d σ j (A.2.13)= k − X l =1 Z Q lj =1 t k ≤ T P [ u ]( ~x k , ~w k − )e − t l +1 l Y j =1 d σ j + k − X l =1 Z Q lj =1 t k ≥ T P [ u ]( ~x k , ~w k − )e − t l +1 l Y j =1 d σ j , = III + III , where T > k = C T / . By Lemma A.5, we deduce | III | ≤ C (cid:18) (cid:19) C T / k u k L ∞ (Ω ×S ) . (A.2.14)Also, we may directly estimate | III | ≤ C e − T k u k L ∞ (Ω ×S ) . (A.2.15)Then taking T sufficiently large, we know | III | ≤ δ k u k L ∞ (Ω ×S ) , (A.2.16)for δ > I and II related to h and f ,which we denote as I and II . For fixed T , it is easy to see | I | + | II | ≤ k f k L ∞ (Ω ×S ) + k h k L ∞ (Γ − ) . (A.2.17)Step 3: Estimate of ¯ u term.The most troubling terms are related to ¯ u . Here, we use the trick as in [14] and [27]. Collecting the resultsin (A.2.16) and (A.2.17), we obtain | u | ≤ A + (cid:12)(cid:12)(cid:12)(cid:12)Z t ¯ u ( ~x − ǫ ( t − s ) ~w )e − ( t − s ) d s (cid:12)(cid:12)(cid:12)(cid:12) (A.2.18)+ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k − X l =1 Z Q lj =1 (cid:18) Z t l ¯ u ( ~x l − ǫ ( t l +1 − s l +1 ) ~w l )e − ( t l +1 − s l +1 ) d s l +1 (cid:19) l Y j =1 d σ j !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , = A + I + II , where A = k f k L ∞ (Ω ×S ) + k h k L ∞ (Γ − ) + δ k u k L ∞ (Ω ×S ) . (A.2.19)By definition, we know | I | = (cid:12)(cid:12)(cid:12)(cid:12)Z t (cid:18) Z S u ( ~x − ǫ ( t − s ) ~w, ~w s )d ~w s (cid:19) e − ( t − s ) d s (cid:12)(cid:12)(cid:12)(cid:12) , (A.2.20)where ~w s ∈ S is a dummy variable. Then we can utilize the mild formulation (A.2.10) to rewrite u ( ~x − ǫ ( t − s ) ~w, ~w s ) along the characteristics. We denote the stochastic cycle as ( t ′ k , ~x ′ k , ~w ′ k ) correspondingly EGULARITY OF MILNE PROBLEM IN 3D 47 and ( t ′ , ~x ′ , ~w ′ ) = (0 , ~x − ǫ ( t − s ) ~w, ~w s ). Then | I | ≤ (cid:12)(cid:12)(cid:12)(cid:12)Z t (cid:18) Z S A d ~w s (cid:19) e − ( t − s ) d s (cid:12)(cid:12)(cid:12)(cid:12) (A.2.21)+ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z t (cid:18) Z S Z t ′ ¯ u ( ~x ′ − ǫ ( t ′ − s ′ ) ~w s )e − ( t ′ − s ′ ) d s ′ d ~w s (cid:19) e − ( t − s ) d s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Z t (cid:18) Z S k − X l ′ =1 Z Q l ′ j ′ =1 (cid:18) Z t ′ l ′ ¯ u ( ~x l ′ − ǫ ( t ′ l ′ +1 − s ′ l ′ +1 ) ~w l ′ )e − ( t ′ l ′ +1 − s ′ l ′ +1 ) d s ′ l ′ +1 (cid:19) l ′ Y j ′ =1 d σ j ′ d ~w s (cid:19) e − ( t − s ) d s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , = | I , | + | I , | + | I , | . It is obvious that | I , | = (cid:12)(cid:12)(cid:12)(cid:12)Z t (cid:18) Z S A d ~w s (cid:19) e − ( t − s ) d s (cid:12)(cid:12)(cid:12)(cid:12) ≤ A (A.2.22) ≤ k f k L ∞ (Ω ×S ) + k h k L ∞ (Γ − ) + δ k u k L ∞ (Ω ×S ) . Then by definition, we know | I , | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z t (cid:18) Z S Z t ′ ¯ u ( ~x ′ − ǫ ( t ′ − s ′ ) ~w s )e − ( t ′ − s ′ ) d s ′ d ~w s (cid:19) e − ( t − s ) d s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . (A.2.23)We may decompose this integral Z t Z S Z t ′ = Z t Z S Z t ′ − s ′ ≤ δ + Z t Z S Z t ′ − s ′ ≥ δ = I , , + I , , . (A.2.24)For I , , , since the integral is defined in the small domain [ t ′ − δ, t ′ ], it is easy to see | I , , | ≤ δ k u k L ∞ (Ω ×S ) . (A.2.25)For I , , , applying H¨older’s inequality, we get | I , , | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z t Z S Z t ′ − s ′ ≥ δ ¯ u ( ~x ′ − ǫ ( t ′ − s ′ ) ~w s )e − ( t ′ − s ′ ) e − ( t − s ) d s ′ d ~w s d s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (A.2.26) ≤ (cid:18) Z t Z S Z t ′ − s ′ ≥ δ e − ( t ′ − s ′ ) e − ( t − s ) d s ′ d ~w s d s (cid:19) m − m (cid:18) Z t Z S Z t ′ − s ′ ≥ δ ~x ′ − ǫ ( t ′ − s ′ ) ~w s ∈ Ω | ¯ u | m ( ~x ′ − ǫ ( t ′ − s ′ ) ~w s )e − ( t ′ − s ′ ) e − ( t − s ) d s ′ d ~w s d s (cid:19) m ≤ (cid:18) Z t Z S Z t ′ − s ′ ≥ δ ~x ′ − ǫ ( t ′ − s ′ ) ~w s ∈ Ω | ¯ u | m ( ~x ′ − ǫ ( t ′ − s ′ ) ~w s )e − ( t ′ − s ′ ) e − ( t − s ) d s ′ d ~w s d s (cid:19) m . Since ~w s ∈ S , we can express it as (sin φ cos ψ, sin φ sin ψ, cos φ ). Then considering ~x ′ − ǫ ( t ′ − s ′ ) ~w s ∈ ¯Ω,we apply the substitution ( φ, ψ, s ′ ) → ( y , y , y ) as ~y = ~x ′ − ǫ ( t ′ − s ′ ) ~w s , (A.2.27)whose Jacobian is (cid:12)(cid:12)(cid:12)(cid:12) ∂ ( y , y , y ) ∂ ( φ, ψ, r ′ ) (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − ǫ ( t ′ − s ′ ) cos φ cos ψ ǫ ( t ′ − s ′ ) sin φ sin ψ ǫ sin φ cos ψ − ǫ ( t ′ − s ′ ) cos φ sin ψ − ǫ ( t ′ − s ′ ) sin φ cos ψ ǫ sin φ sin ψǫ ( t ′ − s ′ ) sin φ ǫ cos φ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (A.2.28)= ǫ ( t ′ − s ′ ) sin φ. (A.2.29) Hence, we can further decompose I , , into | sin φ | ≤ δ and | sin φ | ≥ δ . In the first part, the integral can bebounded by δ k u k L ∞ (Ω ×S ) . Then for the second part, we have (cid:12)(cid:12)(cid:12)(cid:12) ∂ ( y , y , y ) ∂ ( φ, ψ, r ′ ) (cid:12)(cid:12)(cid:12)(cid:12) ≥ ǫ δ . (A.2.30)Therefore, we know | I , , | ≤ ǫ m δ m k ¯ u k L m (Ω ×S ) . (A.2.31)Hence, we have shown | I , | ≤ δ k u k L ∞ (Ω ×S ) + 1 ǫ m δ m k ¯ u k L m (Ω ×S ) . (A.2.32)After a similar but tedious computation, we can show | I , | ≤ δ k u k L ∞ (Ω ×S ) + 1 ǫ m δ m k ¯ u k L m (Ω ×S ) . (A.2.33)Hence, we have proved | I | ≤ δ k u k L ∞ (Ω ×S ) + 1 ǫ m δ m k ¯ u k L m (Ω ×S ) + k f k L ∞ (Ω ×S ) + k h k L ∞ (Γ − ) . (A.2.34)In a similar fashion, we can show | II | ≤ δ k u k L ∞ (Ω ×S ) + 1 ǫ m δ m k ¯ u k L m (Ω ×S ) + k f k L ∞ (Ω ×S ) + k h k L ∞ (Γ − ) . (A.2.35)Step 4: Synthesis.Summarizing all above, we have shown | u | ≤ δ k u k L ∞ (Ω ×S ) + 1 δ m ǫ m k ¯ u k L (Ω ×S ) + k f k L ∞ (Ω ×S ) + k h k L ∞ (Γ − ) (A.2.36) ≤ δ k u k L ∞ (Ω ×S ) + 1 δ m ǫ m k u k L (Ω ×S ) + k f k L ∞ (Ω ×S ) + k h k L ∞ (Γ − ) . Since ( ~x, ~w ) are arbitrary and δ is small, taking supremum on both sides and applying Lemma A.1, we have k u k L ∞ (Ω ×S ) ≤ C (cid:18) ǫ m k ¯ u k L m (Ω ×S ) + k f k L ∞ (Ω ×S ) + k g k L ∞ (Γ − ) (cid:19) . (A.2.37)Considering Theorem A.3, we obtain k u k L ∞ (Ω ×S ) ≤ C (cid:18) ǫ m k f k L (Ω ×S ) + 1 ǫ m k f k L m m − (Ω ×S ) + k f k L ∞ (Ω ×S ) (A.2.38)+ 1 ǫ m k h k L (Γ − ) + 1 ǫ m k h k L m (Γ − ) + k h k L ∞ (Γ − ) (cid:19) + o (1) k u k L ∞ (Γ + ) . Absorbing k u k L ∞ (Ω ×S ) into the left-hand side, we obtain k u k L ∞ (Ω ×S ) ≤ C (cid:18) ǫ m k f k L (Ω ×S ) + 1 ǫ m k f k L m m − (Ω ×S ) + k f k L ∞ (Ω ×S ) (A.2.39)+ 1 ǫ m k h k L (Γ − ) + 1 ǫ m k h k L m (Γ − ) + k h k L ∞ (Γ − ) (cid:19) . This is the desired estimate. (cid:3) EGULARITY OF MILNE PROBLEM IN 3D 49 Appendix B. Diffusive Limit Corollary B.1. Assume g ( ~x , ~w ) ∈ C (Γ − ) satisfying (1.5). Then for the steady neutron transport equation(1.1), there exists a unique solution u ǫ ( ~x, ~w ) ∈ L ∞ (Ω × S ) satisfying (1.4). Moreover, for any < δ << ,the solution obeys the estimate k u ǫ − U ǫ k L ∞ (Ω ×S ) ≤ C ( δ, Ω) ǫ − δ , (B.1) where U ǫ is defined in (2.53).Proof. Based on Theorem A.6, we know there exists a unique u ǫ ( ~x, ~w ) ∈ L ∞ (Ω × S ), so we focus on thediffusive limit. We can divide the proof into several steps:Step 1: Remainder definitions.We define the remainder as R = u ǫ − X k =0 ǫ k U ǫk − X k =0 ǫ k U ǫk = u ǫ − Q − Q , (B.2)where Q = U ǫ + ǫU ǫ + ǫ U ǫ , (B.3) Q = U ǫ + ǫ U ǫ . (B.4)Noting the equation (2.30) is equivalent to the equation (1.1), we write L to denote the neutron transportoperator as follows: L [ u ] = ǫ ~w · ∇ x u + u − ¯ u (B.5)= sin φ ∂u∂η − ǫ (cid:18) sin ψR − ǫη + cos ψR − ǫη (cid:19) cos φ ∂u∂φ + u − ¯ u + G [ u ] , where G [ u ] = ǫ (cid:18) cos φ sin ψP (1 − ǫκ η ) ∂u∂τ + cos φ cos ψP (1 − ǫκ η ) ∂u∂τ (cid:19) (B.6)+ ǫ (cid:18) sin ψ − ǫκ η ( ~t × ( ∂ ~r × ~t )) · ~t + cos ψ − ǫκ η ( ~t × ( ∂ ~r × ~t )) · ~t (cid:19) cos φP P ∂u∂ψ . Step 2: Estimates of L [ Q ].The interior contribution can be estimated as L [ Q ] = ǫ ~w · ∇ x Q + Q − ¯ Q = ǫ ~w · ∇ x U ǫ . (B.7)Based on classical elliptic estimates, we have kL [ Q ] k L ∞ (Ω ×S ) ≤ (cid:13)(cid:13) ǫ ~w · ∇ x U ǫ (cid:13)(cid:13) L ∞ (Ω ×S ) ≤ Cǫ k∇ x U ǫ k L ∞ (Ω ×S ) ≤ Cǫ . (B.8)This implies kL [ Q ] k L (Ω ×S ) ≤ Cǫ , (B.9) kL [ Q ] k L m m − (Ω ×S ) ≤ Cǫ , (B.10) kL [ Q ] k L ∞ (Ω ×S ) ≤ Cǫ . (B.11)Step 3: Estimates of L Q .Since U ǫ = 0, we only need to estimate U ǫ = ( f ǫ − f ǫ ,L ) · ψ = V ψ where f ǫ ( η, τ , τ , φ, ψ ) solves the ǫ -Milne problem and V = f ǫ − f ǫ ,L . The boundary layer contribution can be estimated as L [ ǫ U ǫ ] = sin φ ∂ ( ǫ U ǫ ) ∂η − ǫ (cid:18) sin ψR − ǫη + cos ψR − ǫη (cid:19) cos φ ∂ ( ǫ U ǫ ) ∂φ + ( ǫ U ǫ ) − ( ǫ ¯ U ǫ ) + G [ ǫ U ǫ ] (B.12)= ǫ sin φ (cid:18) ψ ∂ V ∂η + V ∂ψ ∂η (cid:19) − ǫψ (cid:18) sin ψR − ǫη + cos ψR − ǫη (cid:19) cos φ ∂ V ∂φ + ∂ V ∂τ + ψ V − ψ ¯ V + ψ G [ V ] ! = ǫψ (cid:18) sin φ ∂ V ∂η − ǫ (cid:18) sin ψR − ǫη + cos ψR − ǫη (cid:19) cos φ ∂ V ∂φ + V − ¯ V (cid:19) + ǫ sin φ ∂ψ ∂η V − ǫψ G [ V ] ! = ǫ sin φ ∂ψ ∂η V − ǫψ G [ V ] ! . Since ψ = 1 when η ≤ R min / (4 ǫ n ), the effective region of ∂ η ψ is η ≥ R min / (4 ǫ n ) which is further andfurther from the origin as ǫ → 0. By Theorem 3.11, the first term in (B.12) can be bounded as (cid:13)(cid:13)(cid:13)(cid:13) ǫ sin φ ∂ψ ∂η V (cid:13)(cid:13)(cid:13)(cid:13) L ∞ (Ω ×S ) ≤ Cǫ e − K ǫn ≤ Cǫ . (B.13)Then we turn to the crucial estimate in the second term of (B.12), by Theorem 4.13, we have k ǫψ G [ V ] k L ∞ (Ω ×S ) ≤ Cǫ (cid:18) (cid:13)(cid:13)(cid:13)(cid:13) ∂ V ∂τ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ (Ω ×S ) + (cid:13)(cid:13)(cid:13)(cid:13) ∂ V ∂τ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ (Ω ×S ) + (cid:13)(cid:13)(cid:13)(cid:13) ∂ V ∂ψ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ (Ω ×S ) (cid:19) (B.14) ≤ Cǫ | ln( ǫ ) | . Also, the exponential decay of ∂ V ∂τ by Theorem 4.13 and the rescaling η = µ/ǫ implies k ǫψ G [ V ] k L (Ω ×S ) ≤ ǫ (cid:18) (cid:13)(cid:13)(cid:13)(cid:13) ∂ V ∂τ (cid:13)(cid:13)(cid:13)(cid:13) L (Ω ×S ) + (cid:13)(cid:13)(cid:13)(cid:13) ∂ V ∂τ (cid:13)(cid:13)(cid:13)(cid:13) L (Ω ×S ) + (cid:13)(cid:13)(cid:13)(cid:13) ∂ V ∂ψ (cid:13)(cid:13)(cid:13)(cid:13) L (Ω ×S ) (cid:19) (B.15) ≤ ǫ Z π − π Z (1 − µ ) (cid:16) (cid:13)(cid:13)(cid:13)(cid:13) ∂ V ∂τ ( µ, τ ) (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂ V ∂τ ( µ, τ ) (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂ V ∂ψ ( µ, τ ) (cid:13)(cid:13)(cid:13)(cid:13) L ∞ (cid:17) d µ d τ ! / ≤ ǫ Z π − π Z /ǫ (1 − ǫη ) (cid:16) (cid:13)(cid:13)(cid:13)(cid:13) ∂ V ∂τ ( η, τ ) (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂ V ∂τ ( η, τ ) (cid:13)(cid:13)(cid:13)(cid:13) L ∞ + (cid:13)(cid:13)(cid:13)(cid:13) ∂ V ∂ψ ( η, τ ) (cid:13)(cid:13)(cid:13)(cid:13) L ∞ (cid:17) d η d τ ! / ≤ Cǫ Z π − π Z /ǫ e − K η | ln( ǫ ) | d η d τ ! / ≤ Cǫ | ln( ǫ ) | . Similarly, we have k ǫψ G [ V ] k L m m − (Ω ×S ) ≤ Cǫ − m | ln( ǫ ) | . (B.16)In total, we have kL [ Q ] k L (Ω ×S ) ≤ Cǫ | ln( ǫ ) | , (B.17) kL [ Q ] k L m m − (Ω ×S ) ≤ Cǫ − m | ln( ǫ ) | , (B.18) kL [ Q ] k L ∞ (Ω ×S ) ≤ Cǫ | ln( ǫ ) | . (B.19)Step 4: Diffusive Limit. EGULARITY OF MILNE PROBLEM IN 3D 51 In summary, since L [ u ǫ ] = 0, collecting estimates in Step 2 and Step 3, we can prove kL [ R ] k L (Ω ×S ) ≤ Cǫ | ln( ǫ ) | , (B.20) kL [ R ] k L m m − (Ω ×S ) ≤ Cǫ − m | ln( ǫ ) | , (B.21) kL [ R ] k L ∞ (Ω ×S ) ≤ Cǫ | ln( ǫ ) | . (B.22)Also, based on our construction, it is easy to see R − P [ R ] = − ǫ ( ~w · ∇ x U ǫ − P [ ~w · ∇ x U ǫ ]) , (B.23)which further implies k R − P [ R ] k L (Γ − ) ≤ Cǫ , (B.24) k R − P [ R ] k L m (Γ − ) ≤ Cǫ , (B.25) k R − P [ R ] k L ∞ (Γ − ) ≤ Cǫ (B.26)Hence, the remainder R satisfies the equation ( ǫ ~w · ∇ x R + R − ¯ R = L [ R ] for ~x ∈ Ω ,R − P [ R ] = R − P [ R ] for ~w · ~ν < ~x ∈ ∂ Ω . (B.27)It is easy to verify R satisfies the normalization condition (A.4) and the data satisfies the compatibilitycondition (A.5). By Theorem A.6, we have for 2 < m ≤ k R k L ∞ (Ω ×S ) ≤ C (cid:18) ǫ m kL [ R ] k L (Ω ×S ) + 1 ǫ m kL [ R ] k L m m − (Ω ×S ) + kL [ R ] k L ∞ (Ω ×S ) (B.28)+ 1 ǫ m k R − P [ R ] k L (Γ − ) + 1 ǫ m k R − P [ R ] k L m (Γ − ) + k R − P [ R ] k L ∞ (Γ − ) (cid:19) ≤ C ǫ m (cid:18) ǫ | ln( ǫ ) | (cid:19) + 1 ǫ m (cid:18) ǫ − m | ln( ǫ ) | (cid:19) + (cid:18) ǫ | ln( ǫ ) | (cid:19) + 1 ǫ m ( ǫ ) + 1 ǫ m ( ǫ ) + ( ǫ ) ! ≤ Cǫ − m | ln( ǫ ) | ≤ Cǫ − δ (B.29)Note that the constant C might depend on m and thus depend on δ . Since it is easy to see (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X k =1 ǫ k U ǫk + X k =0 ǫ k U ǫk (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L ∞ (Ω ×S ) ≤ Cǫ, (B.30)our result naturally follows. This completes the proof of diffusive limit. (cid:3) References [1] K. Aoki, C. Bardos, and S. Takata , Knudsen layer for gas mixtures , J. Statist. 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