Acoustic Limit for the Boltzmann equation in Optimal Scaling

Transcript

1 Acoustic Limit for the Boltzmann equation in Optimal Scaling Joint work with Yan Guo and Ning Jiang Courant Institute March 4, 2009 Kinetic FRG Young Researchers Workshop

2 The Boltzmann equation t F + v x F = Q(F, F )

3 The Boltzmann equation t F + v x F = Q(F, F ) Q(F 1, F 2 )(v) = b(v u, ω){f 1 (v )F 2 (u ) F 1 (v)f 2 (u)}dω du R 3 S 2 where v = v [(v u) ω]ω and u = u + [(v u) ω]ω

4 The Boltzmann equation t F + v x F = Q(F, F ) Q(F 1, F 2 )(v) = b(v u, ω){f 1 (v )F 2 (u ) F 1 (v)f 2 (u)}dω du R 3 S 2 where v = v [(v u) ω]ω and u = u + [(v u) ω]ω Here, we take b(v u, ω) v u γ for 3 < γ 1 0 γ 1 : hard potentials (γ = 0 for Maxwell molecules, γ = 1 for hard spheres) 3 < γ < 0 : soft potentials

5 The Boltzmann equation t F + v x F = Q(F, F ) Q(F 1, F 2 )(v) = b(v u, ω){f 1 (v )F 2 (u ) F 1 (v)f 2 (u)}dω du R 3 S 2 where v = v [(v u) ω]ω and u = u + [(v u) ω]ω Here, we take b(v u, ω) v u γ for 3 < γ 1 0 γ 1 : hard potentials (γ = 0 for Maxwell molecules, γ = 1 for hard spheres) 3 < γ < 0 : soft potentials Collision invariants {1, v k, v 2 } Q(F, F )dv = Q(F, F )v k dv = Q(F, F ) v 2 dv = 0 R 3 R 3 R 3

6 Dimensionless Form and Scalings

7 Dimensionless Form and Scalings St t F ε + v x F ε = 1 Kn Q(F ε, F ε ) ; St = ε s, Kn = ε q

8 Dimensionless Form and Scalings St t F ε + v x F ε = 1 Kn Q(F ε, F ε ) ; St = ε s, Kn = ε q F ε = µ + δ ε G ε ; Ma = δ ε (ε m ), Re Ma Kn

9 Dimensionless Form and Scalings St t F ε + v x F ε = 1 Kn Q(F ε, F ε ) ; St = ε s, Kn = ε q F ε = µ + δ ε G ε ; Ma = δ ε (ε m ), Re Ma Kn q = 1, m = 0, s = 0 : Compressible Euler (F ε F ) q = 1, m > 0, s = 0 : Acoustic Waves (G ε G)

10 Dimensionless Form and Scalings St t F ε + v x F ε = 1 Kn Q(F ε, F ε ) ; St = ε s, Kn = ε q F ε = µ + δ ε G ε ; Ma = δ ε (ε m ), Re Ma Kn q = 1, m = 0, s = 0 : Compressible Euler (F ε F ) q = 1, m > 0, s = 0 : Acoustic Waves (G ε G) Longer time scale (G ε G) q = 1, m = 1, s = 1 : Incompressible Navier-Stokes-Fourier q = 1, m > 1, s = 1 : Stokes-Fourier q > 1, m = 1, s = 1 : Incompressible Euler-Fourier

11 Two Modern Frameworks

12 Two Modern Frameworks Start from DiPerna-Lions global renormalized solutions to Boltzmann equations Justify the convergence to global weak solutions of incompressible Navier-Stokes, Stokes, and Euler flows and compressible acoustic system

13 Two Modern Frameworks Start from DiPerna-Lions global renormalized solutions to Boltzmann equations Justify the convergence to global weak solutions of incompressible Navier-Stokes, Stokes, and Euler flows and compressible acoustic system (Bardos, Golse, Levermore, Lions, Masmoudi, Saint-Raymond)

14 Two Modern Frameworks Start from DiPerna-Lions global renormalized solutions to Boltzmann equations Justify the convergence to global weak solutions of incompressible Navier-Stokes, Stokes, and Euler flows and compressible acoustic system (Bardos, Golse, Levermore, Lions, Masmoudi, Saint-Raymond) Start from smooth solutions to fluid equations Construct solutions to Boltzmann equations via Hilbert or Chapman-Enskog expansions

15 Two Modern Frameworks Start from DiPerna-Lions global renormalized solutions to Boltzmann equations Justify the convergence to global weak solutions of incompressible Navier-Stokes, Stokes, and Euler flows and compressible acoustic system (Bardos, Golse, Levermore, Lions, Masmoudi, Saint-Raymond) Start from smooth solutions to fluid equations Construct solutions to Boltzmann equations via Hilbert or Chapman-Enskog expansions (Caflisch, De Masi, Esposito, Lebowitz, Guo)

16 Acoustic Limit Acoustic scaling t F ε + v x F ε = 1 ε Q(F ε, F ε ), F ε = µ 0 + δg ε µ 0 1 (2π) 3/2 exp( v 2 2 ), δ 0 as ε 0 (δ = εm )

17 Acoustic Limit Acoustic scaling t F ε + v x F ε = 1 ε Q(F ε, F ε ), F ε = µ 0 + δg ε µ 0 1 (2π) 3/2 exp( v 2 2 ), δ 0 as ε 0 (δ = εm ) Formal Acoustic limit ε(g ε + v x G ε ) + LG ε = δq(g ε, G ε ) { σ + v u + By letting ε 0, LG = 0 G = ( v ) } θ µ 0

18 Acoustic Limit Acoustic scaling t F ε + v x F ε = 1 ε Q(F ε, F ε ), F ε = µ 0 + δg ε µ 0 1 (2π) 3/2 exp( v 2 2 ), δ 0 as ε 0 (δ = εm ) Formal Acoustic limit ε(g ε + v x G ε ) + LG ε = δq(g ε, G ε ) { σ + v u + By letting ε 0, LG = 0 G = ( v ) } θ µ 0 G ε formally satisfy the local conservation laws and thus σ, u, θ satisfy the acoustic system t σ + x u = 0, t u + x (σ + θ) = 0, σ(x, 0) = σ 0 (x) u(x, 0) = u 0 (x) 3 2 tθ + x u = 0, θ(x, 0) = θ 0 (x)

19 Rigorous Justification In the framework of renormalized solutions

20 Rigorous Justification In the framework of renormalized solutions Golse-Levermore (CPAM, 02) : the convergence for m > 1 2 Jiang-Levermore-Masmoudi (08) : the case m = 1 2 Due to the lack of local conservation laws and regularity of renormalized solutions, the case m < 1 2 remains an open question

21 Rigorous Justification In the framework of renormalized solutions Golse-Levermore (CPAM, 02) : the convergence for m > 1 2 Jiang-Levermore-Masmoudi (08) : the case m = 1 2 Due to the lack of local conservation laws and regularity of renormalized solutions, the case m < 1 2 remains an open question In the framework of classical solutions J-Jiang (08) : global-in-time uniform convergence for the case m = 1

22 Rigorous Justification In the framework of renormalized solutions Golse-Levermore (CPAM, 02) : the convergence for m > 1 2 Jiang-Levermore-Masmoudi (08) : the case m = 1 2 Due to the lack of local conservation laws and regularity of renormalized solutions, the case m < 1 2 remains an open question In the framework of classical solutions J-Jiang (08) : global-in-time uniform convergence for the case m = 1 The purpose of this work is to establish the acoustic limit in optimal scaling (0 < m < 1) via a recent L 2 L setting.

23 Main Theorem Theorem (Guo-J-Jiang) Let τ > 0 be any given time and let σ(0, x) = σ 0 (x), u(0, x) = u 0 (x), θ(0, x) = θ 0 (x) H s, s 4 be any given initial data to the acoustic system. Then,

24 Main Theorem Theorem (Guo-J-Jiang) Let τ > 0 be any given time and let σ(0, x) = σ 0 (x), u(0, x) = u 0 (x), θ(0, x) = θ 0 (x) H s, s 4 be any given initial data to the acoustic system. Then, there exist an ε 0 > 0 and a δ 0 > 0 such that for each 0 < ε ε 0 and 0 < δ δ 0, there exists a constant C > 0 so that sup G ε (t) G(t) + sup G ε (t) G(t) 2 C{ ε 0 t τ 0 t τ δ + δ} where ε δ 0 as ε 0, and C depends only on τ and the initial data σ 0, u 0, θ 0.

25 Idea of Proof

26 Idea of Proof The acoustic system is the linearization about the homogeneous state [1, 0, 1] of the compressible Euler system t ρ + (u )ρ + ρ u = 0 ρ t u + ρ(u )u + ρ T + T ρ = 0 t T + (u )T T u = 0 (1)

27 Idea of Proof The acoustic system is the linearization about the homogeneous state [1, 0, 1] of the compressible Euler system t ρ + (u )ρ + ρ u = 0 ρ t u + ρ(u )u + ρ T + T ρ = 0 t T + (u )T T u = 0 (1) Instead of estimating G ε G directly, make a detour in two steps

28 Idea of Proof The acoustic system is the linearization about the homogeneous state [1, 0, 1] of the compressible Euler system t ρ + (u )ρ + ρ u = 0 ρ t u + ρ(u )u + ρ T + T ρ = 0 t T + (u )T T u = 0 (1) Instead of estimating G ε G directly, make a detour in two steps The first step is to show for µ δ = ρ δ (t,x) exp [2πT δ (t,x)] 3/2 F ε µ δ = O(ε) } { [v uδ (t,x)] 2 local Maxwellians 2T δ (t,x) defined from the solutions of (1) constructed from the acoustic initial data ρ 0 = 1 + δσ 0, u 0 = δu 0, T 0 = 1 + δθ 0

29 Idea of Proof The acoustic system is the linearization about the homogeneous state [1, 0, 1] of the compressible Euler system t ρ + (u )ρ + ρ u = 0 ρ t u + ρ(u )u + ρ T + T ρ = 0 t T + (u )T T u = 0 (1) Instead of estimating G ε G directly, make a detour in two steps The first step is to show for µ δ = ρ δ (t,x) exp [2πT δ (t,x)] 3/2 F ε µ δ = O(ε) } { [v uδ (t,x)] 2 local Maxwellians 2T δ (t,x) defined from the solutions of (1) constructed from the acoustic initial data ρ 0 = 1 + δσ 0, u 0 = δu 0, T 0 = 1 + δθ 0 before the time of possible shock formation, which is of the order of 1 δ in the acoustic scaling longer than any fixed time τ

30 Idea of Proof The second step is to show µ δ µ 0 = δg + O(δ 2 ) within the time scale of 1 δ

31 Idea of Proof The second step is to show µ δ µ 0 = δg + O(δ 2 ) within the time scale of 1 δ Such an estimate confirms that the solution G of the acoustic equation is the first order (linear) approximation of the Euler equations

32 Idea of Proof The second step is to show µ δ µ 0 = δg + O(δ 2 ) within the time scale of 1 δ Such an estimate confirms that the solution G of the acoustic equation is the first order (linear) approximation of the Euler equations The proof of Theorem : G ε (t) G(t) = F ε (t) µ 0 G(t) δ

33 Idea of Proof The second step is to show µ δ µ 0 = δg + O(δ 2 ) within the time scale of 1 δ Such an estimate confirms that the solution G of the acoustic equation is the first order (linear) approximation of the Euler equations The proof of Theorem : Rewrite F ε (t) µ 0 δ G ε (t) G(t) = F ε (t) µ 0 G(t) δ G(t) = F ε (t) µ δ (t) δ + µδ (t) µ 0 δg(t) δ Therefore, sup G ε (t) G(t) + sup G ε (t) G(t) 2 C{ ε 0 t τ 0 t τ δ + δ}

34 Refined Estimates of Acoustic and Euler solutions

35 Refined Estimates of Acoustic and Euler solutions Introduce difference variables (σd δ, uδ d, θδ d ) given by the second order perturbation in δ of Euler solutions ρ δ = 1 + δσ + δ 2 σ δ d, uδ = δu + δ 2 u δ d, T δ = 1 + δθ + δ 2 θ δ d

36 Refined Estimates of Acoustic and Euler solutions Introduce difference variables (σd δ, uδ d, θδ d ) given by the second order perturbation in δ of Euler solutions ρ δ = 1 + δσ + δ 2 σ δ d, uδ = δu + δ 2 u δ d, T δ = 1 + δθ + δ 2 θ δ d Indeed, (σd δ, uδ d, θδ d ) satisfy the linear system of equations t σ δ d + (u δ )σ δ d + ρ δ u δ d + δ[ σ u δ d + ( u)σ δ d] = (σu) ρ δ t u δ d + ρ δ (u δ )u δ d + ρ δ θ δ d + T δ σ δ d + δ[( t u)σ δ d + ρ δ (u δ d )u + θ δ d σ + σ δ d θ] = σ t u ρ δ (u )u (σθ) t θ δ d + (u δ )θ δ d T δ u δ d + δ[ θ u δ d ( u)θδ d] = u θ 2 3 θ u

37 Refined Estimates of Acoustic and Euler solutions Introduce difference variables (σd δ, uδ d, θδ d ) given by the second order perturbation in δ of Euler solutions ρ δ = 1 + δσ + δ 2 σ δ d, uδ = δu + δ 2 u δ d, T δ = 1 + δθ + δ 2 θ δ d Indeed, (σd δ, uδ d, θδ d ) satisfy the linear system of equations t σ δ d + (u δ )σ δ d + ρ δ u δ d + δ[ σ u δ d + ( u)σ δ d] = (σu) ρ δ t u δ d + ρ δ (u δ )u δ d + ρ δ θ δ d + T δ σ δ d + δ[( t u)σ δ d + ρ δ (u δ d )u + θ δ d σ + σ δ d θ] = σ t u ρ δ (u )u (σθ) t θ δ d + (u δ )θ δ d T δ u δ d + δ[ θ u δ d ( u)θδ d] = u θ 2 3 θ u The coefficients are smooth and have the uniform bounds up to time τ. The above system can be symmetrized.

38 Refined Estimates of Acoustic and Euler solutions Standard energy method of linear symmetric hyperbolic system (σd δ, uδ d, θδ d ) H s C (2) and thus (ρ δ 1 δσ, u δ δu, T δ 1 δθ) H s Cδ 2

39 Refined Estimates of Acoustic and Euler solutions Standard energy method of linear symmetric hyperbolic system and thus (σ δ d, uδ d, θδ d ) H s C (2) (ρ δ 1 δσ, u δ δu, T δ 1 δθ) H s Cδ 2 Consider µ δ as a function of δ and expand it around δ = 0 to derive µ δ (t) = µ 0 + Gδ + µ (δ ) δ 2 2 The uniform boundedness of µ (δ ) 2 + µ (δ ) follows from the uniform estimate (2)

40 Hilbert Expansion

41 Hilbert Expansion Following Caflisch, take the form F ε = 5 n=0 εn F n + ε 3 F ε R where F 0,..., F 5 solve the equations (F 0 = µ δ µ) 0 = Q(F 0, F 0 ) { t + v x }F 0 = Q(F 0, F 1 ) + Q(F 1, F 0 )... { t + v x }F 5 = Q(F 0, F 6 ) + Q(F 6, F 0 ) + i+j=6 1 i 5,1 j 5 Q(F i, F j )

42 Hilbert Expansion Following Caflisch, take the form F ε = 5 n=0 εn F n + ε 3 F ε R where F 0,..., F 5 solve the equations (F 0 = µ δ µ) 0 = Q(F 0, F 0 ) { t + v x }F 0 = Q(F 0, F 1 ) + Q(F 1, F 0 )... { t + v x }F 5 = Q(F 0, F 6 ) + Q(F 6, F 0 ) + Remainder equation for F ε R i+j=6 1 i 5,1 j 5 t F ε R + v xf ε R 1 ε {Q(µ, F ε R ) + Q(F ε R, µ)} Q(F i, F j ) 5 = ε 2 Q(FR ε, F R ε ) + ε i 1 {Q(F i, FR ε ) + Q(F R ε, F i)} + ε 2 A i=1

43 Compressible Euler Limit Theorem Assume that the solution to the Euler equations [ρ, u, T ] is smooth and ρ(t, x) has a positive lower bound for 0 t τ. Furthermore, assume that the temperature T (t, x) satisfies the condition T M < max T (t, x) < 2T M. Let t [0,τ],x Ω F ε (0, x, v) = µ(0, x, v) + 5 n=1 εn F n (0, x, v) + ε 3 FR ε (0, x, v) 0. Then

44 Compressible Euler Limit Theorem Assume that the solution to the Euler equations [ρ, u, T ] is smooth and ρ(t, x) has a positive lower bound for 0 t τ. Furthermore, assume that the temperature T (t, x) satisfies the condition T M < max T (t, x) < 2T M. Let t [0,τ],x Ω F ε (0, x, v) = µ(0, x, v) + 5 n=1 εn F n (0, x, v) + ε 3 FR ε (0, x, v) 0. Then there is an ε 0 > 0 such that for 0 < ε ε 0, and for any β 9 2γ 2, there exists a constant C τ (µ, F 1,..F 5 ) such that sup ε 3 2 (1 + v 2 ) β F R ε(t) + sup FR ε(t) 2 0 t τ µ 0 t τ µ C τ {ε 3 2 (1 + v 2 ) β F R ε (0) + FR ε(0) } 2 µ(0) + 1. µ(0)

45 Compressible Euler Limit Theorem Assume that the solution to the Euler equations [ρ, u, T ] is smooth and ρ(t, x) has a positive lower bound for 0 t τ. Furthermore, assume that the temperature T (t, x) satisfies the condition T M < max T (t, x) < 2T M. Let t [0,τ],x Ω F ε (0, x, v) = µ(0, x, v) + 5 n=1 εn F n (0, x, v) + ε 3 FR ε (0, x, v) 0. Then there is an ε 0 > 0 such that for 0 < ε ε 0, and for any β 9 2γ 2, there exists a constant C τ (µ, F 1,..F 5 ) such that sup ε 3 2 (1 + v 2 ) β F R ε(t) + sup FR ε(t) 2 0 t τ µ 0 t τ µ C τ {ε 3 2 (1 + v 2 ) β F R ε (0) + FR ε(0) } 2 µ(0) + 1. µ(0) Remark F ε = 5 n=0 εn F n + ε 3 F ε R = F ε µ δ = O(ε)

46 Proof of Euler Limit Introduce f ε F R ε and h ε {1 + v 2 } β µ F R ε µm } where β 9 2γ 1 2 and µ M = exp { v 2 (2πT M ) 3/2 2T M so that c 1 µ M µ c 2 µ α M for some 1/2 < α < 1

47 Proof of Euler Limit Introduce f ε F R ε and h ε {1 + v 2 } β µ F R ε µm } where β 9 2γ 1 2 and µ M = exp { v 2 (2πT M ) 3/2 2T M so that c 1 µ M µ c 2 µ α M for some 1/2 < α < 1 L 2 estimates for f ε t f ε + v x f ε + 1 ε Lf ε = { t + v x } µ µ f ε + ε 2 Γ(f ε, f ε ) + 5 i=1 Difficulty is coming from the boxed term v 3 f ε ε i 1 {Γ( F i, f ε ) + Γ(f ε F i, )} + ε 2 Ā µ µ

48 For any κ > 0 and a = 1/(3 γ) { t + v x } µ f ε, f ε = + µ v κ ε a v κ ε a { x ρ 2 + x u 2 + x T 2 } {1 + v 2 } 3/2 f ε 1 v κ ε a f ε 2 + { x ρ + x u + x T } {1 + v 2 } 3/4 f ε 1 v κ ε a 2 2 C κ ε 2 h ε f ε 2 + C {1 + v 2 } 3/4 Pf ε 1 v κ ε a C {1 + v 2 } 3/4 {I P}f ε 1 v κ ε a 2 2 C κ ε 2 h ε f ε 2 + C f ε Cκ3 γ {I P}f ε 2 ν ε Due to the fact {1 + v 2 } 3/2 f ε {1 + v 2 } γ 3 h ε for β 3/2 + (3 γ) and the fact µ M < Cµ

49 For any κ > 0 and a = 1/(3 γ) { t + v x } µ f ε, f ε = + µ v κ ε a v κ ε a { x ρ 2 + x u 2 + x T 2 } {1 + v 2 } 3/2 f ε 1 v κ ε a f ε 2 + { x ρ + x u + x T } {1 + v 2 } 3/4 f ε 1 v κ ε a 2 2 C κ ε 2 h ε f ε 2 + C {1 + v 2 } 3/4 Pf ε 1 v κ ε a C {1 + v 2 } 3/4 {I P}f ε 1 v κ ε a 2 2 C κ ε 2 h ε f ε 2 + C f ε Cκ3 γ {I P}f ε 2 ν ε Due to the fact {1 + v 2 } 3/2 f ε {1 + v 2 } γ 3 h ε for β 3/2 + (3 γ) and the fact µ M < Cµ d dt f ε δ 0 2ε {I P}f ε 2 ν C{ ε ε 3/2 h ε +1}( f ε f ε 2 )

50 L estimates for h ε sup {ε 3/2 h ε (s) } C{ ε 3/2 h 0 + sup f ε (s) 2 + ε 7/2 } 0 s τ 0 s τ

51 L estimates for h ε sup {ε 3/2 h ε (s) } C{ ε 3/2 h 0 + sup f ε (s) 2 + ε 7/2 } 0 s τ 0 s τ Following Caflisch, define L M g = 1 µm {Q(µ, µ M g) + Q( µ M g, µ)} = {ν(µ) + K}g where ν(µ) = R 3 S 2 b(v v, ω)µ(v )dv dω (1 + v ) γ

52 L estimates for h ε sup {ε 3/2 h ε (s) } C{ ε 3/2 h 0 + sup f ε (s) 2 + ε 7/2 } 0 s τ 0 s τ Following Caflisch, define L M g = 1 µm {Q(µ, µ M g) + Q( µ M g, µ)} = {ν(µ) + K}g where ν(µ) = R 3 S 2 b(v v, ω)µ(v )dv dω (1 + v ) γ To treat soft potentials, employ a cutoff trick of Guo and Strain

53 L estimates for h ε sup {ε 3/2 h ε (s) } C{ ε 3/2 h 0 + sup f ε (s) 2 + ε 7/2 } 0 s τ 0 s τ Following Caflisch, define L M g = 1 µm {Q(µ, µ M g) + Q( µ M g, µ)} = {ν(µ) + K}g where ν(µ) = R 3 S 2 b(v v, ω)µ(v )dv dω (1 + v ) γ To treat soft potentials, employ a cutoff trick of Guo and Strain Smooth cutoff function 0 χ m 1 such that for any m > 0, χ m (s) 1 for s m and χ m (s) 0 for s 2m. Split K = K m + K c. K m g(v) Cm 3+γ ν(µ) g and K c g(v) = R 3 l(v, v )g(v )dv where the kernel l satisfies l(v, v ) C m exp{ c v v 2 } v v (1 + v + v ) 1 γ.

54 Letting K w g wk( g w ) where w(v) = {1 + v 2 } β t h ε + v x h ε + ν(µ) ε hε + 1 ε K w h ε = ε2 w + 5 i=1 ε i 1 w {Q(F i, hε µ M µm w Q( hε µ M µm w, hε µ M ) w ) + Q( hε µ M, F i )} + ε 2 Ã, w

55 Letting K w g wk( g w ) where w(v) = {1 + v 2 } β t h ε + v x h ε + ν(µ) ε hε + 1 ε K w h ε = ε2 w + 5 i=1 ε i 1 By Duhamel s principle 0 w {Q(F i, hε µ M µm w Q( hε µ M µm w, hε µ M ) w ) + Q( hε µ M, F i )} + ε 2 Ã, w h ε (t, x, v) = exp{ νt ε }hε (0, x vt, v) t ( ) ν(t s) 1 exp{ } 0 ε ε K w m h ε (s, x v(t s), v)ds t ( ) ν(t s) 1 exp{ } 0 ε ε K w c h ε (s, x v(t s), v)ds t ( ν(t s) ε 2 w + exp{ } Q( hε µ M, hε ) µ M ) (s, x v(t s), v)ds ε µm w w +...

56 t ( ) ν(t s) 1 exp{ } 0 ε ε K w m h ε (s, x v(t s), v)ds t Cm 3+γ exp{ 0 ν(t s) } ν ds sup h ε (t) Cm 3+γ ε ε 0 t τ sup h ε (t) 0 t τ

57 t ( ) ν(t s) 1 exp{ } 0 ε ε K w m h ε (s, x v(t s), v)ds t Cm 3+γ exp{ 0 And since w µm Q( hε µm t 0 exp{ 0 ν(t s) } ν ds sup h ε (t) Cm 3+γ ε ε 0 t τ w, hε µm w ( ν(t s) ε 2 w } ε µm Q( hε µ M w ) Cν(µ) h ε 2 sup h ε (t) 0 t τ, hε ) µ M ) (s, x v(t s), v)ds w t Cε 2 ν(µ)(t s) exp{ }ν(µ) h ε (s) 2 ε ds Cε 3 sup h ε (s) 2 0 s t

58 t ( ) ν(t s) 1 exp{ } 0 ε ε K w m h ε (s, x v(t s), v)ds t Cm 3+γ exp{ 0 And since w µm Q( hε µm t 0 exp{ 0 ν(t s) } ν ds sup h ε (t) Cm 3+γ ε ε 0 t τ w, hε µm w ( ν(t s) ε 2 w } ε µm Q( hε µ M w ) Cν(µ) h ε 2 sup h ε (t) 0 t τ, hε ) µ M ) (s, x v(t s), v)ds w t Cε 2 ν(µ)(t s) exp{ }ν(µ) h ε (s) 2 ε ds Cε 3 sup h ε (s) 2 0 s t Let l w (v, v ) be the corresponding kernel associated with Kw c. t ( ) ν(t s) 1 exp{ } 0 ε ε K w c h ε (s, x v(t s), v)ds 1 t ν(t s) exp{ } l w (v, v )h ε (s, x v(t s), v ) dv ds ε 0 ε R 3

59 Use the Duhamel expression again 1 t ε exp{ 0 t ν(t s) } l w (v, v )h ε (s, x v(t s), v ) dv ds ε R 3 1 ν(t s) exp{ } sup l w (v, v ) dv ε 0 ε v R 3 exp{ νs ε }hε (0, x v(t s) v s, v )ds + 1 t ν(t s) ε 2 exp{ } l w (v, v ) ε R 3 s ε 2 t s exp{ ν(v )(s s 1 ) } {K m h ε }(s 1, x v(t s) v (s s 1 ), v ) dv ds 1 ds ε ν(t s) exp{ } ε 0 R 3 R 3 l w (v, v )l w (v, v ) exp{ ν(v )(s s 1 ) }h ε (s 1, x v(t s) v (s s 1 ), v ) dv dv ds 1 ds ε

60 The estimate for double l term : Divide into Case 1 : v N Case 2 : v N, v 2N or v 2N, v 3N Case 3a : v N, v 2N, v 3N, and s s 1 κε Case 3b : v N, v 2N, v 3N, and s s 1 κε

61 The estimate for double l term : Divide into Case 1 : v N Case 2 : v N, v 2N or v 2N, v 3N Case 3a : v N, v 2N, v 3N, and s s 1 κε Case 3b : v N, v 2N, v 3N, and s s 1 κε To see how to get f ε 2, in Case 3b, integrate over v to get C N h ε (s 1, x (s s 1 )v, v ) dv v 2N { } 1/2 C N 1 Ω (x (s s 1 )v ) h ε (s 1, x (s s 1 )v, v ) 2 dv v 2N C N κ 3/2 ε 3/2 { y x 1 (s s 1)3N C N{(s s 1 ) 3/2 + 1} κ 3/2 ε 3/2 h ε (s 1, y, v ) 2 dy { h ε (s 1, y, v ) 2 dy Ω } 1/2 } 1/2 Here we have made a change of variable y = x (s s 1 )v, and for s s 1 κε, dy dv 1 κ 3 ε 3

62 Thus approximate version of Case 3b C t 0 B C N,κ ε 7/2 C N,κ ε 7/2 C N,κ ε 3/2 s εκ 0 e ν(v)(t s) ε e ν(v )(s s 1 ) ε l N (v, v )l N (v, v ) h ε (s 1, x 1 (s s 1 )v, v ) ds 1 dv dv ds t s κε 0 0 v 3N t s κε 0 0 { v 3N e ν(v)(t s) ε e ν(v )(s s 1 ) ε {(s s 1 ) 3/2 + 1} { sup f ε (s) 2 0 s t 1/2 h ε (s 1, y, v ) dy} 2 dv ds 1 ds Ω e ν(v)(t s) ε e ν(v )(s s 1 ) ε {(s s 1 ) 3/2 + 1} Ω f ε (s 1, y, v ) 2 dydv } 1/2 ds 1 ds

63 Thus approximate version of Case 3b C t 0 B C N,κ ε 7/2 C N,κ ε 7/2 C N,κ ε 3/2 s εκ 0 e ν(v)(t s) ε e ν(v )(s s 1 ) ε l N (v, v )l N (v, v ) h ε (s 1, x 1 (s s 1 )v, v ) ds 1 dv dv ds t s κε 0 0 v 3N t s κε 0 0 { v 3N e ν(v)(t s) ε e ν(v )(s s 1 ) ε {(s s 1 ) 3/2 + 1} { sup f ε (s) 2 0 s t 1/2 h ε (s 1, y, v ) dy} 2 dv ds 1 ds Ω e ν(v)(t s) ε e ν(v )(s s 1 ) ε {(s s 1 ) 3/2 + 1} Ω f ε (s 1, y, v ) 2 dydv } 1/2 ds 1 ds In summary, one can obtain for any κ > 0 and large N > 0,

64 sup {ε 3/2 h ε (s) } {Cm γ+3 + C N,m κ + C m 0 s t N } sup {ε 3/2 h ε (s) } 0 s t +Cε 7/2 + C ε,n ε 3/2 h 0 + εc sup {ε 3/2 h ε (s) } 2 0 s t +C m,n,κ sup f ε (s) 2 0 s t

65 sup {ε 3/2 h ε (s) } {Cm γ+3 + C N,m κ + C m 0 s t N } sup {ε 3/2 h ε (s) } 0 s t +Cε 7/2 + C ε,n ε 3/2 h 0 + εc sup {ε 3/2 h ε (s) } 2 0 s t +C m,n,κ sup f ε (s) 2 0 s t For sufficiently small ε > 0, first choosing m small, then N sufficiently large, and finally letting κ small so that {Cm γ+3 + C N,m κ + Cm N } < 1 2, one gets sup {ε 3/2 h ε (s) } C{ ε 3/2 h 0 + sup f ε (s) 2 + ε 7/2 } 0 s τ 0 s τ and this completes the proof.