On the spatial decay of the Boltzmann equation with hard potentials - - PowerPoint PPT Presentation

On the spatial decay of the Boltzmann equation with hard potentials

Haitao WANG (王海涛)

Joint with Yu-Chu Lin and Kung-Chien Wu

Swedish Summer PDEs, KTH, Stockholm August 26-28, 2019

1 / 25

Outline

Boltzmann equation Two decompositions Regularization estimate Conclusion and nonlinear theory

2 / 25

Outline

Boltzmann equation Two decompositions Regularization estimate Conclusion and nonlinear theory

3 / 25

Boltzmann with cutoff (0 < γ < 1)

• ∂tF + ξ · ∇xF = Q(F, F)

Q(F, F)(ξ) =

• R3
• S2

+

[F(ξ′)F(ξ′

∗) − F(ξ)F(ξ∗)] B(|ξ − ξ∗|, θ)dΩdξ∗,

• Angular cutoff: B(V, θ) = |V |γb(θ), 0 < b(θ) ≤ C |cos θ|
• Equilibrium: M(ξ) =

1 (2π)3/2 exp

• −|ξ|2

2

• .
• Around equilibrium M: F = M +

√ Mf.

• ∂tf + ξ · ∇xf = Lf + Γ(f, f) ,

f(0, x, ξ) = f0

• f0 cpt. supp. in x and polynomial ξ weight ξβ (β = (3/2)+)
• Lf = −ν(ξ)f + Kf.
• ν(ξ) ∼ (1 + |ξ|)γ, K: integral operator.
• KerL = span

√ M{1, ξi, |ξ|2}: 5 dimensional

4 / 25

Boltzmann with cutoff (0 < γ < 1)

• ∂tF + ξ · ∇xF = Q(F, F)

Q(F, F)(ξ) =

• R3
• S2

+

[F(ξ′)F(ξ′

∗) − F(ξ)F(ξ∗)] B(|ξ − ξ∗|, θ)dΩdξ∗,

• Angular cutoff: B(V, θ) = |V |γb(θ), 0 < b(θ) ≤ C |cos θ|
• Equilibrium: M(ξ) =

1 (2π)3/2 exp

• −|ξ|2

2

• .
• Around equilibrium M: F = M +

√ Mf.

• ∂tf + ξ · ∇xf = Lf + Γ(f, f) ,

f(0, x, ξ) = f0

• f0 cpt. supp. in x and polynomial ξ weight ξβ (β = (3/2)+)
• Lf = −ν(ξ)f + Kf.
• ν(ξ) ∼ (1 + |ξ|)γ, K: integral operator.
• KerL = span

√ M{1, ξi, |ξ|2}: 5 dimensional

4 / 25

Boltzmann with cutoff (0 < γ < 1)

• ∂tF + ξ · ∇xF = Q(F, F)

Q(F, F)(ξ) =

• R3
• S2

+

[F(ξ′)F(ξ′

∗) − F(ξ)F(ξ∗)] B(|ξ − ξ∗|, θ)dΩdξ∗,

• Angular cutoff: B(V, θ) = |V |γb(θ), 0 < b(θ) ≤ C |cos θ|
• Equilibrium: M(ξ) =

1 (2π)3/2 exp

• −|ξ|2

2

• .
• Around equilibrium M: F = M +

√ Mf.

• ∂tf + ξ · ∇xf = Lf + Γ(f, f) ,

f(0, x, ξ) = f0

• f0 cpt. supp. in x and polynomial ξ weight ξβ (β = (3/2)+)
• Lf = −ν(ξ)f + Kf.
• ν(ξ) ∼ (1 + |ξ|)γ, K: integral operator.
• KerL = span

√ M{1, ξi, |ξ|2}: 5 dimensional

4 / 25

Boltzmann with cutoff (0 < γ < 1)

• ∂tF + ξ · ∇xF = Q(F, F)

Q(F, F)(ξ) =

• R3
• S2

+

[F(ξ′)F(ξ′

∗) − F(ξ)F(ξ∗)] B(|ξ − ξ∗|, θ)dΩdξ∗,

• Angular cutoff: B(V, θ) = |V |γb(θ), 0 < b(θ) ≤ C |cos θ|
• Equilibrium: M(ξ) =

1 (2π)3/2 exp

• −|ξ|2

2

• .
• Around equilibrium M: F = M +

√ Mf.

• ∂tf + ξ · ∇xf = Lf + Γ(f, f) ,

f(0, x, ξ) = f0

• f0 cpt. supp. in x and polynomial ξ weight ξβ (β = (3/2)+)
• Lf = −ν(ξ)f + Kf.
• ν(ξ) ∼ (1 + |ξ|)γ, K: integral operator.
• KerL = span

√ M{1, ξi, |ξ|2}: 5 dimensional

4 / 25

Boltzmann with cutoff (0 < γ < 1)

• ∂tF + ξ · ∇xF = Q(F, F)

Q(F, F)(ξ) =

• R3
• S2

+

[F(ξ′)F(ξ′

∗) − F(ξ)F(ξ∗)] B(|ξ − ξ∗|, θ)dΩdξ∗,

• Angular cutoff: B(V, θ) = |V |γb(θ), 0 < b(θ) ≤ C |cos θ|
• Equilibrium: M(ξ) =

1 (2π)3/2 exp

• −|ξ|2

2

• .
• Around equilibrium M: F = M +

√ Mf.

• ∂tf + ξ · ∇xf = Lf + Γ(f, f) ,

f(0, x, ξ) = f0

• f0 cpt. supp. in x and polynomial ξ weight ξβ (β = (3/2)+)
• Lf = −ν(ξ)f + Kf.
• ν(ξ) ∼ (1 + |ξ|)γ, K: integral operator.
• KerL = span

√ M{1, ξi, |ξ|2}: 5 dimensional

4 / 25

Boltzmann with cutoff (0 < γ < 1)

• ∂tF + ξ · ∇xF = Q(F, F)

Q(F, F)(ξ) =

• R3
• S2

+

[F(ξ′)F(ξ′

∗) − F(ξ)F(ξ∗)] B(|ξ − ξ∗|, θ)dΩdξ∗,

• Angular cutoff: B(V, θ) = |V |γb(θ), 0 < b(θ) ≤ C |cos θ|
• Equilibrium: M(ξ) =

1 (2π)3/2 exp

• −|ξ|2

2

• .
• Around equilibrium M: F = M +

√ Mf.

• ∂tf + ξ · ∇xf = Lf + Γ(f, f) ,

f(0, x, ξ) = f0

• f0 cpt. supp. in x and polynomial ξ weight ξβ (β = (3/2)+)
• Lf = −ν(ξ)f + Kf.
• ν(ξ) ∼ (1 + |ξ|)γ, K: integral operator.
• KerL = span

√ M{1, ξi, |ξ|2}: 5 dimensional

4 / 25

Boltzmann with cutoff (0 < γ < 1)

• ∂tF + ξ · ∇xF = Q(F, F)

Q(F, F)(ξ) =

• R3
• S2

+

[F(ξ′)F(ξ′

∗) − F(ξ)F(ξ∗)] B(|ξ − ξ∗|, θ)dΩdξ∗,

• Angular cutoff: B(V, θ) = |V |γb(θ), 0 < b(θ) ≤ C |cos θ|
• Equilibrium: M(ξ) =

1 (2π)3/2 exp

• −|ξ|2

2

• .
• Around equilibrium M: F = M +

√ Mf.

• ∂tf + ξ · ∇xf = Lf + Γ(f, f) ,

f(0, x, ξ) = f0

• f0 cpt. supp. in x and polynomial ξ weight ξβ (β = (3/2)+)
• Lf = −ν(ξ)f + Kf.
• ν(ξ) ∼ (1 + |ξ|)γ, K: integral operator.
• KerL = span

√ M{1, ξi, |ξ|2}: 5 dimensional

4 / 25

Boltzmann with cutoff (0 < γ < 1)

• ∂tF + ξ · ∇xF = Q(F, F)

Q(F, F)(ξ) =

• R3
• S2

+

[F(ξ′)F(ξ′

∗) − F(ξ)F(ξ∗)] B(|ξ − ξ∗|, θ)dΩdξ∗,

• Angular cutoff: B(V, θ) = |V |γb(θ), 0 < b(θ) ≤ C |cos θ|
• Equilibrium: M(ξ) =

1 (2π)3/2 exp

• −|ξ|2

2

• .
• Around equilibrium M: F = M +

√ Mf.

• ∂tf + ξ · ∇xf = Lf + Γ(f, f) ,

f(0, x, ξ) = f0

• f0 cpt. supp. in x and polynomial ξ weight ξβ (β = (3/2)+)
• Lf = −ν(ξ)f + Kf.
• ν(ξ) ∼ (1 + |ξ|)γ, K: integral operator.
• KerL = span

√ M{1, ξi, |ξ|2}: 5 dimensional

4 / 25

Heuristic: Chapman-Enskog Expansion

∂tf + ξ1∂xf = Lf Micro-Macro decomposition:

Macroscopic: P0 : orthogonal projection L2

ξ → Ker L

Microscopic: P1 = Id − P0 P0L = LP0 = 0, P1L = LP1 = L, P0P1 = 0

Apply P0: ∂tP0f + ∂xP0ξ1P0f + ∂xP0ξ1P1f = 0 Apply P1: ∂tP1f + ∂xP1ξ1P0f + ∂xP1ξ1P1f = LP1f L invertible on Range P1 P1f = L−1 ∂tP1f + ∂xP1ξ1P0f + ∂xP1ξ1P1f

• ∼

P1f≪P0f L−1

∂xP1ξ1P0f

• Therefore we obtain the equation for P0f (3 dimensional)

∂t(P0f) + ∂x(P0ξ1P0f) = −∂x(P0ξ1P1f) = −∂2

x

• P0ξ1L−1P1ξ1P0f
• 5 / 25

Heuristic: Chapman-Enskog Expansion

∂tf + ξ1∂xf = Lf Micro-Macro decomposition:

Macroscopic: P0 : orthogonal projection L2

ξ → Ker L

Microscopic: P1 = Id − P0 P0L = LP0 = 0, P1L = LP1 = L, P0P1 = 0

Apply P0: ∂tP0f + ∂xP0ξ1P0f + ∂xP0ξ1P1f = 0 Apply P1: ∂tP1f + ∂xP1ξ1P0f + ∂xP1ξ1P1f = LP1f L invertible on Range P1 P1f = L−1 ∂tP1f + ∂xP1ξ1P0f + ∂xP1ξ1P1f

• ∼

P1f≪P0f L−1

∂xP1ξ1P0f

• Therefore we obtain the equation for P0f (3 dimensional)

∂t(P0f) + ∂x(P0ξ1P0f) = −∂x(P0ξ1P1f) = −∂2

x

• P0ξ1L−1P1ξ1P0f
• 5 / 25

Heuristic: Chapman-Enskog Expansion

∂tf + ξ1∂xf = Lf Micro-Macro decomposition:

Macroscopic: P0 : orthogonal projection L2

ξ → Ker L

Microscopic: P1 = Id − P0 P0L = LP0 = 0, P1L = LP1 = L, P0P1 = 0

Apply P0: ∂tP0f + ∂xP0ξ1P0f + ∂xP0ξ1P1f = 0 Apply P1: ∂tP1f + ∂xP1ξ1P0f + ∂xP1ξ1P1f = LP1f L invertible on Range P1 P1f = L−1 ∂tP1f + ∂xP1ξ1P0f + ∂xP1ξ1P1f

• ∼

P1f≪P0f L−1

∂xP1ξ1P0f

• Therefore we obtain the equation for P0f (3 dimensional)

∂t(P0f) + ∂x(P0ξ1P0f) = −∂x(P0ξ1P1f) = −∂2

x

• P0ξ1L−1P1ξ1P0f
• 5 / 25

Heuristic: Chapman-Enskog Expansion

∂tf + ξ1∂xf = Lf Micro-Macro decomposition:

Macroscopic: P0 : orthogonal projection L2

ξ → Ker L

Microscopic: P1 = Id − P0 P0L = LP0 = 0, P1L = LP1 = L, P0P1 = 0

Apply P0: ∂tP0f + ∂xP0ξ1P0f + ∂xP0ξ1P1f = 0 Apply P1: ∂tP1f + ∂xP1ξ1P0f + ∂xP1ξ1P1f = LP1f L invertible on Range P1 P1f = L−1 ∂tP1f + ∂xP1ξ1P0f + ∂xP1ξ1P1f

• ∼

P1f≪P0f L−1

∂xP1ξ1P0f

• Therefore we obtain the equation for P0f (3 dimensional)

∂t(P0f) + ∂x(P0ξ1P0f) = −∂x(P0ξ1P1f) = −∂2

x

• P0ξ1L−1P1ξ1P0f
• 5 / 25

Heuristic: Chapman-Enskog Expansion

∂tf + ξ1∂xf = Lf Micro-Macro decomposition:

Macroscopic: P0 : orthogonal projection L2

ξ → Ker L

Microscopic: P1 = Id − P0 P0L = LP0 = 0, P1L = LP1 = L, P0P1 = 0

Apply P0: ∂tP0f + ∂xP0ξ1P0f + ∂xP0ξ1P1f = 0 Apply P1: ∂tP1f + ∂xP1ξ1P0f + ∂xP1ξ1P1f = LP1f L invertible on Range P1 P1f = L−1 ∂tP1f + ∂xP1ξ1P0f + ∂xP1ξ1P1f

• ∼

P1f≪P0f L−1

∂xP1ξ1P0f

• Therefore we obtain the equation for P0f (3 dimensional)

∂t(P0f) + ∂x(P0ξ1P0f) = −∂x(P0ξ1P1f) = −∂2

x

• P0ξ1L−1P1ξ1P0f
• 5 / 25

Heuristic: Chapman-Enskog Expansion

∂tf + ξ1∂xf = Lf Micro-Macro decomposition:

Macroscopic: P0 : orthogonal projection L2

ξ → Ker L

Microscopic: P1 = Id − P0 P0L = LP0 = 0, P1L = LP1 = L, P0P1 = 0

Apply P0: ∂tP0f + ∂xP0ξ1P0f + ∂xP0ξ1P1f = 0 Apply P1: ∂tP1f + ∂xP1ξ1P0f + ∂xP1ξ1P1f = LP1f L invertible on Range P1 P1f = L−1 ∂tP1f + ∂xP1ξ1P0f + ∂xP1ξ1P1f

• ∼

P1f≪P0f L−1

∂xP1ξ1P0f

• Therefore we obtain the equation for P0f (3 dimensional)

∂t(P0f) + ∂x(P0ξ1P0f) = −∂x(P0ξ1P1f) = −∂2

x

• P0ξ1L−1P1ξ1P0f
• 5 / 25

Heuristic: Chapman-Enskog Expansion

∂tf + ξ1∂xf = Lf Micro-Macro decomposition:

Macroscopic: P0 : orthogonal projection L2

ξ → Ker L

Microscopic: P1 = Id − P0 P0L = LP0 = 0, P1L = LP1 = L, P0P1 = 0

Apply P0: ∂tP0f + ∂xP0ξ1P0f + ∂xP0ξ1P1f = 0 Apply P1: ∂tP1f + ∂xP1ξ1P0f + ∂xP1ξ1P1f = LP1f L invertible on Range P1 P1f = L−1 ∂tP1f + ∂xP1ξ1P0f + ∂xP1ξ1P1f

• ∼

P1f≪P0f L−1

∂xP1ξ1P0f

• Therefore we obtain the equation for P0f (3 dimensional)

∂t(P0f) + ∂x(P0ξ1P0f) = −∂x(P0ξ1P1f) = −∂2

x

• P0ξ1L−1P1ξ1P0f
• 5 / 25

Heuristic: Chapman-Enskog Expansion

∂t(P0f) + ∂x(P0ξ1P0f) = ∂2

x

• P0ξ1(−L)−1P1ξ1
• viscosity matrix

P0f

• P0f is 3 dimensional, in terms of the basis, the above equation is a

viscous conservation law system. Diagonalize the system to give 3 eigenvalues

• λ1 = −
• 5/3, λ2 = 0, λ3 =
• 5/3
• 5/3 sound speed, 0 is the background velocity. The numerical

values are due to our linearization M = (2π)−3/2 exp(−|ξ|2/2). In general {λ1 = u − c, λ2 = u, λ3 = u + c}

6 / 25

Heuristic: Chapman-Enskog Expansion

∂t(P0f) + ∂x(P0ξ1P0f) = ∂2

x

• P0ξ1(−L)−1P1ξ1
• viscosity matrix

P0f

• P0f is 3 dimensional, in terms of the basis, the above equation is a

viscous conservation law system. Diagonalize the system to give 3 eigenvalues

• λ1 = −
• 5/3, λ2 = 0, λ3 =
• 5/3
• 5/3 sound speed, 0 is the background velocity. The numerical

values are due to our linearization M = (2π)−3/2 exp(−|ξ|2/2). In general {λ1 = u − c, λ2 = u, λ3 = u + c}

6 / 25

Heuristic: Chapman-Enskog Expansion

∂t(P0f) + ∂x(P0ξ1P0f) = ∂2

x

• P0ξ1(−L)−1P1ξ1
• viscosity matrix

P0f

• P0f is 3 dimensional, in terms of the basis, the above equation is a

viscous conservation law system. Diagonalize the system to give 3 eigenvalues

• λ1 = −
• 5/3, λ2 = 0, λ3 =
• 5/3
• 5/3 sound speed, 0 is the background velocity. The numerical

values are due to our linearization M = (2π)−3/2 exp(−|ξ|2/2). In general {λ1 = u − c, λ2 = u, λ3 = u + c}

6 / 25

Outline

Boltzmann equation Two decompositions Regularization estimate Conclusion and nonlinear theory

7 / 25

Long wave-short wave decomposition

• ∂tf + ξ · ∇xf = Lf ,

(t, x, ξ) ∈ R+ × R3 × R3 , f(0, x, ξ) = f0(x, ξ) f(t, x, ξ) =

• R3 eiηx+(−iξ·η+L)t ˆ

f0 (η, ξ) dη fL(t, x, ξ) =

• |η|≤1

eiηx+(−iξ·η+L)t ˆ f0 (η, ξ) dη fS(t, x, ξ) =

• |η|>1

eiηx+(−iξ·η+L)t ˆ f0 (η, ξ) dη

8 / 25

Fluid structure 0 ≤ γ < 1 (Liu-Yu)

Let c =

• 5/3 be the sound speed associated with normalized global

Maxwellian |fL|L2

ξ ≤ C

 (1 + t)−2

• 1 + (|x| − ct)2

1 + t −N + (1 + t)−3/2

• 1 + |x|2

1 + t −N +1{|x|≤ct} (1 + t)−3/2

• 1 + |x|2

1 + t −3/2 + e−ct   f0L1

xL2 ξ .

fSL2

x,ξ e−ctf0L2 x,ξ 9 / 25

Singular-Regular decomposition

∂tf + ξ · ∇xf + ν(ξ)f = Kf.

• ∂th(0) + ξ · ∇xh(0) + ν(ξ)h(0) = 0 ,

h(0)(0, x, ξ) = f0(x, ξ) The jth order approximation h(j), j ≥ 1

• ∂th(j) + ξ · ∇xh(j) + ν(ξ)h(j) = Kh(j−1) ,

h(j)(0, x, ξ) = 0 Intuition: h(j) becomes more and more regular. We can define the singular part and the regular part: W (9) =

9

• j=0

h(j) , R(9) = f − W (9)

• ∂tR(9) + ξ · ∇xR(9) = LR(9) + Kh(9) ,

R(9)(0, x, ξ) = 0

10 / 25

Singular-Regular decomposition

∂tf + ξ · ∇xf + ν(ξ)f = Kf.

• ∂th(0) + ξ · ∇xh(0) + ν(ξ)h(0) = 0 ,

h(0)(0, x, ξ) = f0(x, ξ) The jth order approximation h(j), j ≥ 1

• ∂th(j) + ξ · ∇xh(j) + ν(ξ)h(j) = Kh(j−1) ,

h(j)(0, x, ξ) = 0 Intuition: h(j) becomes more and more regular. We can define the singular part and the regular part: W (9) =

9

• j=0

h(j) , R(9) = f − W (9) ∂tR(9) + ξ · ∇xR(9) = LR(9) + Kh(9) , R(9)(0, x, ξ) = 0

10 / 25

Singular-Regular decomposition

∂tf + ξ · ∇xf + ν(ξ)f = Kf.

• ∂th(0) + ξ · ∇xh(0) + ν(ξ)h(0) = 0 ,

h(0)(0, x, ξ) = f0(x, ξ) The jth order approximation h(j), j ≥ 1

• ∂th(j) + ξ · ∇xh(j) + ν(ξ)h(j) = Kh(j−1) ,

h(j)(0, x, ξ) = 0 Intuition: h(j) becomes more and more regular. We can define the singular part and the regular part: W (9) =

9

• j=0

h(j) , R(9) = f − W (9) ∂tR(9) + ξ · ∇xR(9) = LR(9) + Kh(9) , R(9)(0, x, ξ) = 0

10 / 25

Singular-Regular decomposition

∂tf + ξ · ∇xf + ν(ξ)f = Kf.

• ∂th(0) + ξ · ∇xh(0) + ν(ξ)h(0) = 0 ,

h(0)(0, x, ξ) = f0(x, ξ) The jth order approximation h(j), j ≥ 1

• ∂th(j) + ξ · ∇xh(j) + ν(ξ)h(j) = Kh(j−1) ,

h(j)(0, x, ξ) = 0 Intuition: h(j) becomes more and more regular. We can define the singular part and the regular part: W (9) =

9

• j=0

h(j) , R(9) = f − W (9) ∂tR(9) + ξ · ∇xR(9) = LR(9) + Kh(9) , R(9)(0, x, ξ) = 0

10 / 25

Singular-Regular decomposition

∂tf + ξ · ∇xf + ν(ξ)f = Kf.

• ∂th(0) + ξ · ∇xh(0) + ν(ξ)h(0) = 0 ,

h(0)(0, x, ξ) = f0(x, ξ) The jth order approximation h(j), j ≥ 1

• ∂th(j) + ξ · ∇xh(j) + ν(ξ)h(j) = Kh(j−1) ,

h(j)(0, x, ξ) = 0 Intuition: h(j) becomes more and more regular. We can define the singular part and the regular part: W (9) =

9

• j=0

h(j) , R(9) = f − W (9) ∂tR(9) + ξ · ∇xR(9) = LR(9) + Kh(9) , R(9)(0, x, ξ) = 0

10 / 25

Singular-Regular decomposition

∂tf + ξ · ∇xf + ν(ξ)f = Kf.

• ∂th(0) + ξ · ∇xh(0) + ν(ξ)h(0) = 0 ,

h(0)(0, x, ξ) = f0(x, ξ) The jth order approximation h(j), j ≥ 1

• ∂th(j) + ξ · ∇xh(j) + ν(ξ)h(j) = Kh(j−1) ,

h(j)(0, x, ξ) = 0 Intuition: h(j) becomes more and more regular. We can define the singular part and the regular part: W (9) =

9

• j=0

h(j) , R(9) = f − W (9) ∂tR(9) + ξ · ∇xR(9) = LR(9) + Kh(9) , R(9)(0, x, ξ) = 0

10 / 25

Singular-Regular decomposition

• Damped transport operator: h(t) = Sth0

∂th + ξ · ∇xh + ν(ξ)h = 0, with h(0, x, ξ) = h0 Let α ≥ 0. Then for 0 ≤ γ < 1 and any positive number σ,

• St ξ−α h0
• L∞

ξ

≤ C sup

y e− ν0

3 t (1 + |x − y|)− α 1−γ

t−σ t−σ + (1 + |x − y|)

σγ 1−γ |h0 (y, ·)|L∞ ξ . 11 / 25

Singular-Regular decomposition

• Damped transport operator: h(t) = Sth0

∂th + ξ · ∇xh + ν(ξ)h = 0, with h(0, x, ξ) = h0 Let α ≥ 0. Then for 0 ≤ γ < 1 and any positive number σ,

• St ξ−α h0
• L∞

ξ

≤ C sup

y e− ν0

3 t (1 + |x − y|)− α 1−γ

t−σ t−σ + (1 + |x − y|)

σγ 1−γ |h0 (y, ·)|L∞ ξ . 11 / 25

Singular-Regular decomposition

• Damped transport operator: h(t) = Sth0

∂th + ξ · ∇xh + ν(ξ)h = 0, with h(0, x, ξ) = h0 Let α ≥ 0. Then for 0 ≤ γ < 1 and any positive number σ,

• St ξ−α h0
• L∞

ξ

≤ C sup

y e− ν0

3 t (1 + |x − y|)− α 1−γ

t−σ t−σ + (1 + |x − y|)

σγ 1−γ |h0 (y, ·)|L∞ ξ . 11 / 25

Singular-Regular decomposition

• 0 < γ < 1, with polynomial weight ξα
• Sth0
• L∞

ξ ≤ C sup

y e− ν0

3 t (1 + |x − y|)− α 1−γ

t−σ t−σ + (1 + |x − y|)

σγ 1−γ |h0 (y, ·)|L∞ ξ (ξα) .

• 0 < γ < 1, with exponential weight eα|ξ|p (Lin-Wang-Wu, JSP 2018)
• Sth0
• L∞

ξ ≤ sup

y e −c0

• t+α

1−γ p+1−γ |x−y| p p+1−γ

• |h0 (y, ·)|L∞

ξ (eα|ξ|p)

• γ = 1, (Liu-Yu, CPAM 2004)
• Sth0
• L∞

ξ ≤ sup

y e−c0(t+|x−y|) |h0 (y, ·)|L∞

ξ 12 / 25

Singular-Regular decomposition

• 0 < γ < 1, with polynomial weight ξα
• Sth0
• L∞

ξ ≤ C sup

y e− ν0

3 t (1 + |x − y|)− α 1−γ

t−σ t−σ + (1 + |x − y|)

σγ 1−γ |h0 (y, ·)|L∞ ξ (ξα) .

• 0 < γ < 1, with exponential weight eα|ξ|p (Lin-Wang-Wu, JSP 2018)
• Sth0
• L∞

ξ ≤ sup

y e −c0

• t+α

1−γ p+1−γ |x−y| p p+1−γ

• |h0 (y, ·)|L∞

ξ (eα|ξ|p)

• γ = 1, (Liu-Yu, CPAM 2004)
• Sth0
• L∞

ξ ≤ sup

y e−c0(t+|x−y|) |h0 (y, ·)|L∞

ξ 12 / 25

Singular-Regular decomposition

• 0 < γ < 1, with polynomial weight ξα
• Sth0
• L∞

ξ ≤ C sup

y e− ν0

3 t (1 + |x − y|)− α 1−γ

t−σ t−σ + (1 + |x − y|)

σγ 1−γ |h0 (y, ·)|L∞ ξ (ξα) .

• 0 < γ < 1, with exponential weight eα|ξ|p (Lin-Wang-Wu, JSP 2018)
• Sth0
• L∞

ξ ≤ sup

y e −c0

• t+α

1−γ p+1−γ |x−y| p p+1−γ

• |h0 (y, ·)|L∞

ξ (eα|ξ|p)

• γ = 1, (Liu-Yu, CPAM 2004)
• Sth0
• L∞

ξ ≤ sup

y e−c0(t+|x−y|) |h0 (y, ·)|L∞

ξ 12 / 25

Singular-Regular decomposition

• Let j ≥ 1, β = (3/2)+, 0 ≤ γ < 1,
• h(j)
• L∞

ξ

≤ e−c0t x−

γ+αj +βj 1−γ

f0L∞

x L∞ ξ,β

where αj = j(2−γ)

j+1 , βj = β j+1.

• Key fact: |Kg(ξ)| ≤ ξ−(2−γ) |g|L∞

ξ .

• Note: γ + αj + βj > 3/2.
• Why W (9) ?
• How to estimate R(9)

13 / 25

Singular-Regular decomposition

• Let j ≥ 1, β = (3/2)+, 0 ≤ γ < 1,
• h(j)
• L∞

ξ

≤ e−c0t x−

γ+αj +βj 1−γ

f0L∞

x L∞ ξ,β

where αj = j(2−γ)

j+1 , βj = β j+1.

• Key fact: |Kg(ξ)| ≤ ξ−(2−γ) |g|L∞

ξ .

• Note: γ + αj + βj > 3/2.
• Why W (9) ?
• How to estimate R(9)

13 / 25

Singular-Regular decomposition

• Let j ≥ 1, β = (3/2)+, 0 ≤ γ < 1,
• h(j)
• L∞

ξ

≤ e−c0t x−

γ+αj +βj 1−γ

f0L∞

x L∞ ξ,β

where αj = j(2−γ)

j+1 , βj = β j+1.

• Key fact: |Kg(ξ)| ≤ ξ−(2−γ) |g|L∞

ξ .

• Note: γ + αj + βj > 3/2.
• Why W (9) ?
• How to estimate R(9)

13 / 25

Singular-Regular decomposition

• Let j ≥ 1, β = (3/2)+, 0 ≤ γ < 1,
• h(j)
• L∞

ξ

≤ e−c0t x−

γ+αj +βj 1−γ

f0L∞

x L∞ ξ,β

where αj = j(2−γ)

j+1 , βj = β j+1.

• Key fact: |Kg(ξ)| ≤ ξ−(2−γ) |g|L∞

ξ .

• Note: γ + αj + βj > 3/2.
• Why W (9) ?
• How to estimate R(9)

13 / 25

Singular-Regular decomposition

• Let j ≥ 1, β = (3/2)+, 0 ≤ γ < 1,
• h(j)
• L∞

ξ

≤ e−c0t x−

γ+αj +βj 1−γ

f0L∞

x L∞ ξ,β

where αj = j(2−γ)

j+1 , βj = β j+1.

• Key fact: |Kg(ξ)| ≤ ξ−(2−γ) |g|L∞

ξ .

• Note: γ + αj + βj > 3/2.
• Why W (9) ?
• How to estimate R(9)

13 / 25

Time-like region: |x| < Mt

f = fL

• long wave

+ fS

• short wave

= W (9) singular + R(9)

• regular

f = fL + W (9) + fR Tail part fR = R(9) − fL = fS − W (9) fRL2 = fS − W (9)L2 ≤ e−Ctf0L2 (0 ≤ γ < 1) ∇2

xR(9)L2 ≤

t

• Gt−s∇2

xh(9)(s)

• L2 ds

Goal: ∇2

xh(9)L2

14 / 25

Time-like region: |x| < Mt

f = fL

• long wave

+ fS

• short wave

= W (9) singular + R(9)

• regular

f = fL + W (9) + fR Tail part fR = R(9) − fL = fS − W (9) fRL2 = fS − W (9)L2 ≤ e−Ctf0L2 (0 ≤ γ < 1) ∇2

xR(9)L2 ≤

t

• Gt−s∇2

xh(9)(s)

• L2 ds

Goal: ∇2

xh(9)L2

14 / 25

Time-like region: |x| < Mt

f = fL

• long wave

+ fS

• short wave

= W (9) singular + R(9)

• regular

f = fL + W (9) + fR Tail part fR = R(9) − fL = fS − W (9) fRL2 = fS − W (9)L2 ≤ e−Ctf0L2 (0 ≤ γ < 1) ∇2

xR(9)L2 ≤

t

• Gt−s∇2

xh(9)(s)

• L2 ds

Goal: ∇2

xh(9)L2

14 / 25

Time-like region: |x| < Mt

f = fL

• long wave

+ fS

• short wave

= W (9) singular + R(9)

• regular

f = fL + W (9) + fR Tail part fR = R(9) − fL = fS − W (9) fRL2 = fS − W (9)L2 ≤ e−Ctf0L2 (0 ≤ γ < 1) ∇2

xR(9)L2 ≤

t

• Gt−s∇2

xh(9)(s)

• L2 ds

Goal: ∇2

xh(9)L2

14 / 25

Time-like region: |x| < Mt

f = fL

• long wave

+ fS

• short wave

= W (9) singular + R(9)

• regular

f = fL + W (9) + fR Tail part fR = R(9) − fL = fS − W (9) fRL2 = fS − W (9)L2 ≤ e−Ctf0L2 (0 ≤ γ < 1) ∇2

xR(9)L2 ≤

t

• Gt−s∇2

xh(9)(s)

• L2 ds

Goal: ∇2

xh(9)L2

14 / 25

Time-like region: |x| < Mt

f = fL

• long wave

+ fS

• short wave

= W (9) singular + R(9)

• regular

f = fL + W (9) + fR Tail part fR = R(9) − fL = fS − W (9) fRL2 = fS − W (9)L2 ≤ e−Ctf0L2 (0 ≤ γ < 1) ∇2

xR(9)L2 ≤

t

• Gt−s∇2

xh(9)(s)

• L2 ds

Goal: ∇2

xh(9)L2

14 / 25

Space-like region: |x| > Mt

• f = W (9) + R(9), we have accurate description of W (9).
• (Weighted energy estimate for R(9), 0 < γ < 1)

w (t, x, ξ) = 5 (δ (x − Mt))

2 1−γ (1 − χ)

+

• (1 − χ) [δ (x − Mt)] ξγ+1

D

+ 3 ξ2

D

• χ.

where ξD = (D + |ξ|2)1/2, χ = χ

• δ(x−Mt)

ξ1−γ

D

• and µ(x, ξ) = w(0, x, ξ).

Let u = wR(9), then u solves the equation ∂tu + ξ · ∇xu − (∂tw + ξ · ∇xw)w−1u − wL

• w−1u
• = wKh(9) .

15 / 25

Space-like region: |x| > Mt

• f = W (9) + R(9), we have accurate description of W (9).
• (Weighted energy estimate for R(9), 0 < γ < 1)

w (t, x, ξ) = 5 (δ (x − Mt))

2 1−γ (1 − χ)

+

• (1 − χ) [δ (x − Mt)] ξγ+1

D

+ 3 ξ2

D

• χ.

where ξD = (D + |ξ|2)1/2, χ = χ

• δ(x−Mt)

ξ1−γ

D

• and µ(x, ξ) = w(0, x, ξ).

Let u = wR(9), then u solves the equation ∂tu + ξ · ∇xu − (∂tw + ξ · ∇xw)w−1u − wL

• w−1u
• = wKh(9) .

15 / 25

Space-like region: |x| > Mt

• f = W (9) + R(9), we have accurate description of W (9).
• (Weighted energy estimate for R(9), 0 < γ < 1)

w (t, x, ξ) = 5 (δ (x − Mt))

2 1−γ (1 − χ)

+

• (1 − χ) [δ (x − Mt)] ξγ+1

D

+ 3 ξ2

D

• χ.

where ξD = (D + |ξ|2)1/2, χ = χ

• δ(x−Mt)

ξ1−γ

D

• and µ(x, ξ) = w(0, x, ξ).

Let u = wR(9), then u solves the equation ∂tu + ξ · ∇xu − (∂tw + ξ · ∇xw)w−1u − wL

• w−1u
• = wKh(9) .

15 / 25

Space-like region: |x| > Mt

Define wLw−1 = Lw and H+ = {(x, ξ) : δ (x − Mt) > 2 ξ1−γ

D

}, H0 = {(x, ξ) : ξ1−γ

D

≤ δ (x − Mt) ≤ 2 ξ1−γ

D

}, H− = {(x, ξ) : δ (x − Mt) < ξ1−γ

D

}.

• u, Lwuξ dx ≤
• u, Luξ dx + CD−2

ξγ |P1u|2 dξdx +CD− 3

2 − γ 2

• H+ [δ (x − Mt)]−1 |P0u|2 dξdx +
• H0∪H− |P0u|2 dξdx
• .
• The red part can be controlled by
• R3
• u, (∂tw)w−1u
• ξ dx.
• d

dtu2 H2

xL2 ξ uH2 xL2 ξwKh(9)H2 xL2 ξ + R(9)2

H2

xL2 ξ

• Goal: Estimate Kh(9)H2

xL2 ξ(µ) 16 / 25

Space-like region: |x| > Mt

Define wLw−1 = Lw and H+ = {(x, ξ) : δ (x − Mt) > 2 ξ1−γ

D

}, H0 = {(x, ξ) : ξ1−γ

D

≤ δ (x − Mt) ≤ 2 ξ1−γ

D

}, H− = {(x, ξ) : δ (x − Mt) < ξ1−γ

D

}.

• u, Lwuξ dx ≤
• u, Luξ dx + CD−2

ξγ |P1u|2 dξdx +CD− 3

2 − γ 2

• H+ [δ (x − Mt)]−1 |P0u|2 dξdx +
• H0∪H− |P0u|2 dξdx
• .
• The red part can be controlled by
• R3
• u, (∂tw)w−1u
• ξ dx.
• d

dtu2 H2

xL2 ξ uH2 xL2 ξwKh(9)H2 xL2 ξ + R(9)2

H2

xL2 ξ

• Goal: Estimate Kh(9)H2

xL2 ξ(µ) 16 / 25

Space-like region: |x| > Mt

Define wLw−1 = Lw and H+ = {(x, ξ) : δ (x − Mt) > 2 ξ1−γ

D

}, H0 = {(x, ξ) : ξ1−γ

D

≤ δ (x − Mt) ≤ 2 ξ1−γ

D

}, H− = {(x, ξ) : δ (x − Mt) < ξ1−γ

D

}.

• u, Lwuξ dx ≤
• u, Luξ dx + CD−2

ξγ |P1u|2 dξdx +CD− 3

2 − γ 2

• H+ [δ (x − Mt)]−1 |P0u|2 dξdx +
• H0∪H− |P0u|2 dξdx
• .
• The red part can be controlled by
• R3
• u, (∂tw)w−1u
• ξ dx.
• d

dtu2 H2

xL2 ξ uH2 xL2 ξwKh(9)H2 xL2 ξ + R(9)2

H2

xL2 ξ

• Goal: Estimate Kh(9)H2

xL2 ξ(µ) 16 / 25

Space-like region: |x| > Mt

Define wLw−1 = Lw and H+ = {(x, ξ) : δ (x − Mt) > 2 ξ1−γ

D

}, H0 = {(x, ξ) : ξ1−γ

D

≤ δ (x − Mt) ≤ 2 ξ1−γ

D

}, H− = {(x, ξ) : δ (x − Mt) < ξ1−γ

D

}.

• u, Lwuξ dx ≤
• u, Luξ dx + CD−2

ξγ |P1u|2 dξdx +CD− 3

2 − γ 2

• H+ [δ (x − Mt)]−1 |P0u|2 dξdx +
• H0∪H− |P0u|2 dξdx
• .
• The red part can be controlled by
• R3
• u, (∂tw)w−1u
• ξ dx.
• d

dtu2 H2

xL2 ξ uH2 xL2 ξwKh(9)H2 xL2 ξ + R(9)2

H2

xL2 ξ

• Goal: Estimate Kh(9)H2

xL2 ξ(µ) 16 / 25

Space-like region: |x| > Mt

Define wLw−1 = Lw and H+ = {(x, ξ) : δ (x − Mt) > 2 ξ1−γ

D

}, H0 = {(x, ξ) : ξ1−γ

D

≤ δ (x − Mt) ≤ 2 ξ1−γ

D

}, H− = {(x, ξ) : δ (x − Mt) < ξ1−γ

D

}.

• u, Lwuξ dx ≤
• u, Luξ dx + CD−2

ξγ |P1u|2 dξdx +CD− 3

2 − γ 2

• H+ [δ (x − Mt)]−1 |P0u|2 dξdx +
• H0∪H− |P0u|2 dξdx
• .
• The red part can be controlled by
• R3
• u, (∂tw)w−1u
• ξ dx.
• d

dtu2 H2

xL2 ξ uH2 xL2 ξwKh(9)H2 xL2 ξ + R(9)2

H2

xL2 ξ

• Goal: Estimate Kh(9)H2

xL2 ξ(µ) 16 / 25

Outline

Boltzmann equation Two decompositions Regularization estimate Conclusion and nonlinear theory

17 / 25

Regularization: Estimate of Kh(9)

Mixture operator: T4 = [0, t] × [0, s1] × [0, s2] × [0, s3] × [0, s4] dS4 = ds1ds2ds3ds4ds5. Mtg =

• T4

KSt−s1KSs1−s2KSs2−s3KSs3−s4KSs4−s5Kg(·, s5)dS4 .

• Kh(9) = Mth(4).
• Key observation: Good properties of K.
• Key observation: Dt = t∇x + ∇ξ, we have [Dt, ∂t + ξ · ∇x] = 0
• Key observation: DtStg0L2(µ) te−c0tg0L2

xH1 ξ (µ)

• Mt :regularization. h(4) :cancel the weight µ.
• Kh(9)H2

xL2 ξ(µ) t7 (1 + t)2 e−c0t

f0L∞

ξ L2 x + f0L2

• Gualdani, Mischler and Mouhot.

18 / 25

Regularization: Estimate of Kh(9)

Mixture operator: T4 = [0, t] × [0, s1] × [0, s2] × [0, s3] × [0, s4] dS4 = ds1ds2ds3ds4ds5. Mtg =

• T4

KSt−s1KSs1−s2KSs2−s3KSs3−s4KSs4−s5Kg(·, s5)dS4 .

• Kh(9) = Mth(4).
• Key observation: Good properties of K.
• Key observation: Dt = t∇x + ∇ξ, we have [Dt, ∂t + ξ · ∇x] = 0
• Key observation: DtStg0L2(µ) te−c0tg0L2

xH1 ξ (µ)

• Mt :regularization. h(4) :cancel the weight µ.
• Kh(9)H2

xL2 ξ(µ) t7 (1 + t)2 e−c0t

f0L∞

ξ L2 x + f0L2

• Gualdani, Mischler and Mouhot.

18 / 25

Regularization: Estimate of Kh(9)

Mixture operator: T4 = [0, t] × [0, s1] × [0, s2] × [0, s3] × [0, s4] dS4 = ds1ds2ds3ds4ds5. Mtg =

• T4

KSt−s1KSs1−s2KSs2−s3KSs3−s4KSs4−s5Kg(·, s5)dS4 .

• Kh(9) = Mth(4).
• Key observation: Good properties of K.
• Key observation: Dt = t∇x + ∇ξ, we have [Dt, ∂t + ξ · ∇x] = 0
• Key observation: DtStg0L2(µ) te−c0tg0L2

xH1 ξ (µ)

• Mt :regularization. h(4) :cancel the weight µ.
• Kh(9)H2

xL2 ξ(µ) t7 (1 + t)2 e−c0t

f0L∞

ξ L2 x + f0L2

• Gualdani, Mischler and Mouhot.

18 / 25

Regularization: Estimate of Kh(9)

Mixture operator: T4 = [0, t] × [0, s1] × [0, s2] × [0, s3] × [0, s4] dS4 = ds1ds2ds3ds4ds5. Mtg =

• T4

KSt−s1KSs1−s2KSs2−s3KSs3−s4KSs4−s5Kg(·, s5)dS4 .

• Kh(9) = Mth(4).
• Key observation: Good properties of K.
• Key observation: Dt = t∇x + ∇ξ, we have [Dt, ∂t + ξ · ∇x] = 0
• Key observation: DtStg0L2(µ) te−c0tg0L2

xH1 ξ (µ)

• Mt :regularization. h(4) :cancel the weight µ.
• Kh(9)H2

xL2 ξ(µ) t7 (1 + t)2 e−c0t

f0L∞

ξ L2 x + f0L2

• Gualdani, Mischler and Mouhot.

18 / 25

Regularization: Estimate of Kh(9)

Mixture operator: T4 = [0, t] × [0, s1] × [0, s2] × [0, s3] × [0, s4] dS4 = ds1ds2ds3ds4ds5. Mtg =

• T4

KSt−s1KSs1−s2KSs2−s3KSs3−s4KSs4−s5Kg(·, s5)dS4 .

• Kh(9) = Mth(4).
• Key observation: Good properties of K.
• Key observation: Dt = t∇x + ∇ξ, we have [Dt, ∂t + ξ · ∇x] = 0
• Key observation: DtStg0L2(µ) te−c0tg0L2

xH1 ξ (µ)

• Mt :regularization. h(4) :cancel the weight µ.
• Kh(9)H2

xL2 ξ(µ) t7 (1 + t)2 e−c0t

f0L∞

ξ L2 x + f0L2

• Gualdani, Mischler and Mouhot.

18 / 25

Regularization: Estimate of Kh(9)

Mixture operator: T4 = [0, t] × [0, s1] × [0, s2] × [0, s3] × [0, s4] dS4 = ds1ds2ds3ds4ds5. Mtg =

• T4

KSt−s1KSs1−s2KSs2−s3KSs3−s4KSs4−s5Kg(·, s5)dS4 .

• Kh(9) = Mth(4).
• Key observation: Good properties of K.
• Key observation: Dt = t∇x + ∇ξ, we have [Dt, ∂t + ξ · ∇x] = 0
• Key observation: DtStg0L2(µ) te−c0tg0L2

xH1 ξ (µ)

• Mt :regularization. h(4) :cancel the weight µ.
• Kh(9)H2

xL2 ξ(µ) t7 (1 + t)2 e−c0t

f0L∞

ξ L2 x + f0L2

• Gualdani, Mischler and Mouhot.

18 / 25

Regularization: Estimate of Kh(9)

Mixture operator: T4 = [0, t] × [0, s1] × [0, s2] × [0, s3] × [0, s4] dS4 = ds1ds2ds3ds4ds5. Mtg =

• T4

KSt−s1KSs1−s2KSs2−s3KSs3−s4KSs4−s5Kg(·, s5)dS4 .

• Kh(9) = Mth(4).
• Key observation: Good properties of K.
• Key observation: Dt = t∇x + ∇ξ, we have [Dt, ∂t + ξ · ∇x] = 0
• Key observation: DtStg0L2(µ) te−c0tg0L2

xH1 ξ (µ)

• Mt :regularization. h(4) :cancel the weight µ.
• Kh(9)H2

xL2 ξ(µ) t7 (1 + t)2 e−c0t

f0L∞

ξ L2 x + f0L2

• Gualdani, Mischler and Mouhot.

18 / 25

Outline

Boltzmann equation Two decompositions Regularization estimate Conclusion and nonlinear theory

19 / 25

Conclusion

Theorem

Let f be a solution to the linearized Boltzmann equation with initial data compactly supported in the x-variable and bounded in L∞

ξ,β space,

β = (3/2)+, then

• W (9)
• L∞

ξ ≤ e−c0t x− 3/2 1−γ f0L∞ x L∞ ξ,β.

(a) |x| < Mt

• R(9)
• L2

ξ

≤ CN        (1 + t)−2 1 + (|x|−ct)2

1+t

−N + (1 + t)−3/2 1 + |x|2

1+t

−N +1{|x|≤ct} (1 + t)−3/2 1 + |x|2

1+t

−3/2        f0L∞

x L∞ ξ,β .

(b) |x| > Mt

• R(9)
• L2

ξ

≤ (1 + t)−N/4(t + |x|)− 3/2

1−γ f0L∞ x L∞ ξ,β . 20 / 25

Conclusion

Bootstrap from L2

ξ to L∞ ξ :

f = Stf0+ t St−sK(W (9)+R(9))(s)ds = W (10)+ t St−sKR(9)(s)ds .

Theorem (L∞

ξ estimate)

Let f be a solution to the linearized Boltzmann equation with initial data compactly supported in the x-variable and bounded in L∞

ξ,β space,

β = (3/2)+, then |f|L∞

ξ ≤ CN

          (1 + t)−2 1 + (|x|−ct)2

1+t

− 3/2

1−γ

+ (1 + t)−3/2 1 + |x|2

1+t

−N +1{|x|≤ct} (1 + t)−3/2 1 + |x|2

1+t

−3/2 +(1 + t)−3/2(t + |x|)− 3/2

1−γ

          f0L∞

x L∞ ξ,β . 21 / 25

Conclusion

Bootstrap from L2

ξ to L∞ ξ :

f = Stf0+ t St−sK(W (9)+R(9))(s)ds = W (10)+ t St−sKR(9)(s)ds .

Theorem (L∞

ξ estimate)

Let f be a solution to the linearized Boltzmann equation with initial data compactly supported in the x-variable and bounded in L∞

ξ,β space,

β = (3/2)+, then |f|L∞

ξ ≤ CN

          (1 + t)−2 1 + (|x|−ct)2

1+t

− 3/2

1−γ

+ (1 + t)−3/2 1 + |x|2

1+t

−N +1{|x|≤ct} (1 + t)−3/2 1 + |x|2

1+t

−3/2 +(1 + t)−3/2(t + |x|)− 3/2

1−γ

          f0L∞

x L∞ ξ,β . 21 / 25

Nonlinear theorem

• ∂tf + ξ · ∇xf = Lf + Γ(f, f) ,

f(0, x, ξ) = f0 small

• f0 ∈ HN

x L2 ξ(ξp), N ≥ 4, p ≥ 2.

• Existence, uniqueness and optimal time decay:

Ukai, Kawashima, Y. Guo, Strain, T. Yang, R-J. Duan, etc. Define E(f)(t) =

|α|≤N ∂α x f2 L2 + |α|≤N ξp ∂α x f2 L2 .

D(f)(t) =

1≤|α|≤N ∂α x P0f2 L2 + |α|≤N ξp ∂α x P1f2 L2

σ .

If we assume E(f)(0) ≤ η, then

• E(f)(t) +

t

0 D(f)(s)ds ≤ E(f)(0) .

• f2

L2 = P0f2 L2 + P1f2 L2 ≤ η(1 + t)−3/2 ,

• 1≤|α|≤N ∂α

x P0f2 L2 + |α|≤N ξp ∂α x P1f2 L2 ≤ η(1 + t)−5/2 .

22 / 25

Nonlinear theorem

• ∂tf + ξ · ∇xf = Lf + Γ(f, f) ,

f(0, x, ξ) = f0 small

• f0 ∈ HN

x L2 ξ(ξp), N ≥ 4, p ≥ 2.

• Existence, uniqueness and optimal time decay:

Ukai, Kawashima, Y. Guo, Strain, T. Yang, R-J. Duan, etc. Define E(f)(t) =

|α|≤N ∂α x f2 L2 + |α|≤N ξp ∂α x f2 L2 .

D(f)(t) =

1≤|α|≤N ∂α x P0f2 L2 + |α|≤N ξp ∂α x P1f2 L2

σ .

If we assume E(f)(0) ≤ η, then

• E(f)(t) +

t

0 D(f)(s)ds ≤ E(f)(0) .

• f2

L2 = P0f2 L2 + P1f2 L2 ≤ η(1 + t)−3/2 ,

• 1≤|α|≤N ∂α

x P0f2 L2 + |α|≤N ξp ∂α x P1f2 L2 ≤ η(1 + t)−5/2 .

22 / 25

Nonlinear theorem

• ∂tf + ξ · ∇xf = Lf + Γ(f, f) ,

f(0, x, ξ) = f0 small

• f0 ∈ HN

x L2 ξ(ξp), N ≥ 4, p ≥ 2.

• Existence, uniqueness and optimal time decay:

Ukai, Kawashima, Y. Guo, Strain, T. Yang, R-J. Duan, etc. Define E(f)(t) =

|α|≤N ∂α x f2 L2 + |α|≤N ξp ∂α x f2 L2 .

D(f)(t) =

1≤|α|≤N ∂α x P0f2 L2 + |α|≤N ξp ∂α x P1f2 L2

σ .

If we assume E(f)(0) ≤ η, then

• E(f)(t) +

t

0 D(f)(s)ds ≤ E(f)(0) .

• f2

L2 = P0f2 L2 + P1f2 L2 ≤ η(1 + t)−3/2 ,

• 1≤|α|≤N ∂α

x P0f2 L2 + |α|≤N ξp ∂α x P1f2 L2 ≤ η(1 + t)−5/2 .

22 / 25

Nonlinear theorem

• ∂tf + ξ · ∇xf = Lf + Γ(f, f) ,

f(0, x, ξ) = f0 small

• f0 ∈ HN

x L2 ξ(ξp), N ≥ 4, p ≥ 2.

• Existence, uniqueness and optimal time decay:

Ukai, Kawashima, Y. Guo, Strain, T. Yang, R-J. Duan, etc. Define E(f)(t) =

|α|≤N ∂α x f2 L2 + |α|≤N ξp ∂α x f2 L2 .

D(f)(t) =

1≤|α|≤N ∂α x P0f2 L2 + |α|≤N ξp ∂α x P1f2 L2

σ .

If we assume E(f)(0) ≤ η, then

• E(f)(t) +

t

0 D(f)(s)ds ≤ E(f)(0) .

• f2

L2 = P0f2 L2 + P1f2 L2 ≤ η(1 + t)−3/2 ,

• 1≤|α|≤N ∂α

x P0f2 L2 + |α|≤N ξp ∂α x P1f2 L2 ≤ η(1 + t)−5/2 .

22 / 25

Nonlinear theorem

Let u = wf, then u solves the equation ∂tu + ξ · ∇xu − (∂tw + ξ · ∇xw)w−1u − wL

• w−1u
• = wΓ(w−1u, f) .
• u, wΓ(w−1u, f)
• ξ − u, Γ(u, f)ξ
• ≤ D−2 |P1u|L2

σ

• |u|L2

σ |ξp f|L2 ξ + |u|L2 ξ |ξp f|L2 σ

• +D−2 |P0u|L2

ξ

• |u|L2

ξ |ξp f|L2 ξ + |u|L2 ξ |ξp f|L2 ξ

• 23 / 25

Nonlinear theorem

Let u = wf, then u solves the equation ∂tu + ξ · ∇xu − (∂tw + ξ · ∇xw)w−1u − wL

• w−1u
• = wΓ(w−1u, f) .
• u, wΓ(w−1u, f)
• ξ − u, Γ(u, f)ξ
• ≤ D−2 |P1u|L2

σ

• |u|L2

σ |ξp f|L2 ξ + |u|L2 ξ |ξp f|L2 σ

• +D−2 |P0u|L2

ξ

• |u|L2

ξ |ξp f|L2 ξ + |u|L2 ξ |ξp f|L2 ξ

• 23 / 25

Nonlinear theorem

2

• j=0
• R3
• R3
• ∇j

xu

• ∇j

x

• wΓ(w−1u, f)
• dξdx

≤ η1/2(1 + t)−9/8 P1u2

H2

xL2 σ + u2

H2

xL2 ξ

• + D−2η1/2(1 + t)−9/8

P1u2

H2

xL2 σ + u2

H2

xL2 ξ

• + D−2 P1u2

H2

xL2 σ + D−2 u2

H2

xL2 ξ D(f)

+ D−2η1/2(1 + t)−9/8 u2

H2

xL2 ξ .

Theorem (Energy estimate for fully nonlinear equation)

Let f be a solution to the Boltzmann equation with small initial data f0 ∈ HN

x L2 ξ(ξp), N ≥ 4, and compactly supported in x, if |x| > Mt for

some M large, then |f|L2

ξ ≤ C(1 + |x|)− p 1−γ f0H4 xL2 ξ(ξp) . 24 / 25

Nonlinear theorem

2

• j=0
• R3
• R3
• ∇j

xu

• ∇j

x

• wΓ(w−1u, f)
• dξdx

≤ η1/2(1 + t)−9/8 P1u2

H2

xL2 σ + u2

H2

xL2 ξ

• + D−2η1/2(1 + t)−9/8

P1u2

H2

xL2 σ + u2

H2

xL2 ξ

• + D−2 P1u2

H2

xL2 σ + D−2 u2

H2

xL2 ξ D(f)

+ D−2η1/2(1 + t)−9/8 u2

H2

xL2 ξ .

Theorem (Energy estimate for fully nonlinear equation)

Let f be a solution to the Boltzmann equation with small initial data f0 ∈ HN

x L2 ξ(ξp), N ≥ 4, and compactly supported in x, if |x| > Mt for

some M large, then |f|L2

ξ ≤ C(1 + |x|)− p 1−γ f0H4 xL2 ξ(ξp) . 24 / 25

End THANK YOU

25 / 25