on the spatial decay of the boltzmann equation with hard
play

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 ( ) Joint with Yu-Chu Lin and Kung-Chien Wu Swedish Summer PDEs, KTH, Stockholm August 26-28, 2019 1 / 25 Outline Boltzmann equation Two


  1. 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

  2. Outline Boltzmann equation Two decompositions Regularization estimate Conclusion and nonlinear theory 2 / 25

  3. Outline Boltzmann equation Two decompositions Regularization estimate Conclusion and nonlinear theory 3 / 25

  4. Boltzmann with cutoff (0 < γ < 1) • ∂ t F + ξ · ∇ x F = Q ( F, F ) � � [ F ( ξ ′ ) F ( ξ ′ Q ( F, F )( ξ ) = ∗ ) − F ( ξ ) F ( ξ ∗ )] B ( | ξ − ξ ∗ | , θ ) d Ω dξ ∗ , R 3 S 2 + • Angular cutoff: B ( V, θ ) = | V | γ b ( θ ) , 0 < b ( θ ) ≤ C | cos θ | � � −| ξ | 2 1 • Equilibrium: M ( ξ ) = (2 π ) 3 / 2 exp . 2 √ • Around equilibrium M : F = M + M f . • ∂ t f + ξ · ∇ x f = Lf + Γ( f, f ) , f (0 , x, ξ ) = f 0 • f 0 cpt. supp. in x and polynomial ξ weight � ξ � β ( β = (3 / 2) + ) • Lf = − ν ( ξ ) f + Kf . • ν ( ξ ) ∼ (1 + | ξ | ) γ , K : integral operator. √ M{ 1 , ξ i , | ξ | 2 } : 5 dimensional • Ker L = span 4 / 25

  5. Boltzmann with cutoff (0 < γ < 1) • ∂ t F + ξ · ∇ x F = Q ( F, F ) � � [ F ( ξ ′ ) F ( ξ ′ Q ( F, F )( ξ ) = ∗ ) − F ( ξ ) F ( ξ ∗ )] B ( | ξ − ξ ∗ | , θ ) d Ω dξ ∗ , R 3 S 2 + • Angular cutoff: B ( V, θ ) = | V | γ b ( θ ) , 0 < b ( θ ) ≤ C | cos θ | � � −| ξ | 2 1 • Equilibrium: M ( ξ ) = (2 π ) 3 / 2 exp . 2 √ • Around equilibrium M : F = M + M f . • ∂ t f + ξ · ∇ x f = Lf + Γ( f, f ) , f (0 , x, ξ ) = f 0 • f 0 cpt. supp. in x and polynomial ξ weight � ξ � β ( β = (3 / 2) + ) • Lf = − ν ( ξ ) f + Kf . • ν ( ξ ) ∼ (1 + | ξ | ) γ , K : integral operator. √ M{ 1 , ξ i , | ξ | 2 } : 5 dimensional • Ker L = span 4 / 25

  6. Boltzmann with cutoff (0 < γ < 1) • ∂ t F + ξ · ∇ x F = Q ( F, F ) � � [ F ( ξ ′ ) F ( ξ ′ Q ( F, F )( ξ ) = ∗ ) − F ( ξ ) F ( ξ ∗ )] B ( | ξ − ξ ∗ | , θ ) d Ω dξ ∗ , R 3 S 2 + • Angular cutoff: B ( V, θ ) = | V | γ b ( θ ) , 0 < b ( θ ) ≤ C | cos θ | � � −| ξ | 2 1 • Equilibrium: M ( ξ ) = (2 π ) 3 / 2 exp . 2 √ • Around equilibrium M : F = M + M f . • ∂ t f + ξ · ∇ x f = Lf + Γ( f, f ) , f (0 , x, ξ ) = f 0 • f 0 cpt. supp. in x and polynomial ξ weight � ξ � β ( β = (3 / 2) + ) • Lf = − ν ( ξ ) f + Kf . • ν ( ξ ) ∼ (1 + | ξ | ) γ , K : integral operator. √ M{ 1 , ξ i , | ξ | 2 } : 5 dimensional • Ker L = span 4 / 25

  7. Boltzmann with cutoff (0 < γ < 1) • ∂ t F + ξ · ∇ x F = Q ( F, F ) � � [ F ( ξ ′ ) F ( ξ ′ Q ( F, F )( ξ ) = ∗ ) − F ( ξ ) F ( ξ ∗ )] B ( | ξ − ξ ∗ | , θ ) d Ω dξ ∗ , R 3 S 2 + • Angular cutoff: B ( V, θ ) = | V | γ b ( θ ) , 0 < b ( θ ) ≤ C | cos θ | � � −| ξ | 2 1 • Equilibrium: M ( ξ ) = (2 π ) 3 / 2 exp . 2 √ • Around equilibrium M : F = M + M f . • ∂ t f + ξ · ∇ x f = Lf + Γ( f, f ) , f (0 , x, ξ ) = f 0 • f 0 cpt. supp. in x and polynomial ξ weight � ξ � β ( β = (3 / 2) + ) • Lf = − ν ( ξ ) f + Kf . • ν ( ξ ) ∼ (1 + | ξ | ) γ , K : integral operator. √ M{ 1 , ξ i , | ξ | 2 } : 5 dimensional • Ker L = span 4 / 25

  8. Boltzmann with cutoff (0 < γ < 1) • ∂ t F + ξ · ∇ x F = Q ( F, F ) � � [ F ( ξ ′ ) F ( ξ ′ Q ( F, F )( ξ ) = ∗ ) − F ( ξ ) F ( ξ ∗ )] B ( | ξ − ξ ∗ | , θ ) d Ω dξ ∗ , R 3 S 2 + • Angular cutoff: B ( V, θ ) = | V | γ b ( θ ) , 0 < b ( θ ) ≤ C | cos θ | � � −| ξ | 2 1 • Equilibrium: M ( ξ ) = (2 π ) 3 / 2 exp . 2 √ • Around equilibrium M : F = M + M f . • ∂ t f + ξ · ∇ x f = Lf + Γ( f, f ) , f (0 , x, ξ ) = f 0 • f 0 cpt. supp. in x and polynomial ξ weight � ξ � β ( β = (3 / 2) + ) • Lf = − ν ( ξ ) f + Kf . • ν ( ξ ) ∼ (1 + | ξ | ) γ , K : integral operator. √ M{ 1 , ξ i , | ξ | 2 } : 5 dimensional • Ker L = span 4 / 25

  9. Boltzmann with cutoff (0 < γ < 1) • ∂ t F + ξ · ∇ x F = Q ( F, F ) � � [ F ( ξ ′ ) F ( ξ ′ Q ( F, F )( ξ ) = ∗ ) − F ( ξ ) F ( ξ ∗ )] B ( | ξ − ξ ∗ | , θ ) d Ω dξ ∗ , R 3 S 2 + • Angular cutoff: B ( V, θ ) = | V | γ b ( θ ) , 0 < b ( θ ) ≤ C | cos θ | � � −| ξ | 2 1 • Equilibrium: M ( ξ ) = (2 π ) 3 / 2 exp . 2 √ • Around equilibrium M : F = M + M f . • ∂ t f + ξ · ∇ x f = Lf + Γ( f, f ) , f (0 , x, ξ ) = f 0 • f 0 cpt. supp. in x and polynomial ξ weight � ξ � β ( β = (3 / 2) + ) • Lf = − ν ( ξ ) f + Kf . • ν ( ξ ) ∼ (1 + | ξ | ) γ , K : integral operator. √ M{ 1 , ξ i , | ξ | 2 } : 5 dimensional • Ker L = span 4 / 25

  10. Boltzmann with cutoff (0 < γ < 1) • ∂ t F + ξ · ∇ x F = Q ( F, F ) � � [ F ( ξ ′ ) F ( ξ ′ Q ( F, F )( ξ ) = ∗ ) − F ( ξ ) F ( ξ ∗ )] B ( | ξ − ξ ∗ | , θ ) d Ω dξ ∗ , R 3 S 2 + • Angular cutoff: B ( V, θ ) = | V | γ b ( θ ) , 0 < b ( θ ) ≤ C | cos θ | � � −| ξ | 2 1 • Equilibrium: M ( ξ ) = (2 π ) 3 / 2 exp . 2 √ • Around equilibrium M : F = M + M f . • ∂ t f + ξ · ∇ x f = Lf + Γ( f, f ) , f (0 , x, ξ ) = f 0 • f 0 cpt. supp. in x and polynomial ξ weight � ξ � β ( β = (3 / 2) + ) • Lf = − ν ( ξ ) f + Kf . • ν ( ξ ) ∼ (1 + | ξ | ) γ , K : integral operator. √ M{ 1 , ξ i , | ξ | 2 } : 5 dimensional • Ker L = span 4 / 25

  11. Boltzmann with cutoff (0 < γ < 1) • ∂ t F + ξ · ∇ x F = Q ( F, F ) � � [ F ( ξ ′ ) F ( ξ ′ Q ( F, F )( ξ ) = ∗ ) − F ( ξ ) F ( ξ ∗ )] B ( | ξ − ξ ∗ | , θ ) d Ω dξ ∗ , R 3 S 2 + • Angular cutoff: B ( V, θ ) = | V | γ b ( θ ) , 0 < b ( θ ) ≤ C | cos θ | � � −| ξ | 2 1 • Equilibrium: M ( ξ ) = (2 π ) 3 / 2 exp . 2 √ • Around equilibrium M : F = M + M f . • ∂ t f + ξ · ∇ x f = Lf + Γ( f, f ) , f (0 , x, ξ ) = f 0 • f 0 cpt. supp. in x and polynomial ξ weight � ξ � β ( β = (3 / 2) + ) • Lf = − ν ( ξ ) f + Kf . • ν ( ξ ) ∼ (1 + | ξ | ) γ , K : integral operator. √ M{ 1 , ξ i , | ξ | 2 } : 5 dimensional • Ker L = span 4 / 25

  12. Heuristic: Chapman-Enskog Expansion ∂ t f + ξ 1 ∂ x f = Lf Micro-Macro decomposition: Macroscopic: P 0 : orthogonal projection L 2 ξ → Ker L Microscopic: P 1 = Id − P 0 P 0 L = LP 0 = 0 , P 1 L = LP 1 = L, P 0 P 1 = 0 Apply P 0 : ∂ t P 0 f + ∂ x P 0 ξ 1 P 0 f + ∂ x P 0 ξ 1 P 1 f = 0 Apply P 1 : ∂ t P 1 f + ∂ x P 1 ξ 1 P 0 f + ∂ x P 1 ξ 1 P 1 f = LP 1 f L invertible on Range P 1 P 1 f = L − 1 � � ∂ t P 1 f + ∂ x P 1 ξ 1 P 0 f + ∂ x P 1 ξ 1 P 1 f P 1 f ≪ P 0 f L − 1 � � ∼ ∂ x P 1 ξ 1 P 0 f Therefore we obtain the equation for P 0 f (3 dimensional) � � ∂ t ( P 0 f ) + ∂ x ( P 0 ξ 1 P 0 f ) = − ∂ x ( P 0 ξ 1 P 1 f ) = − ∂ 2 P 0 ξ 1 L − 1 P 1 ξ 1 P 0 f x 5 / 25

  13. Heuristic: Chapman-Enskog Expansion ∂ t f + ξ 1 ∂ x f = Lf Micro-Macro decomposition: Macroscopic: P 0 : orthogonal projection L 2 ξ → Ker L Microscopic: P 1 = Id − P 0 P 0 L = LP 0 = 0 , P 1 L = LP 1 = L, P 0 P 1 = 0 Apply P 0 : ∂ t P 0 f + ∂ x P 0 ξ 1 P 0 f + ∂ x P 0 ξ 1 P 1 f = 0 Apply P 1 : ∂ t P 1 f + ∂ x P 1 ξ 1 P 0 f + ∂ x P 1 ξ 1 P 1 f = LP 1 f L invertible on Range P 1 P 1 f = L − 1 � � ∂ t P 1 f + ∂ x P 1 ξ 1 P 0 f + ∂ x P 1 ξ 1 P 1 f P 1 f ≪ P 0 f L − 1 � � ∼ ∂ x P 1 ξ 1 P 0 f Therefore we obtain the equation for P 0 f (3 dimensional) � � ∂ t ( P 0 f ) + ∂ x ( P 0 ξ 1 P 0 f ) = − ∂ x ( P 0 ξ 1 P 1 f ) = − ∂ 2 P 0 ξ 1 L − 1 P 1 ξ 1 P 0 f x 5 / 25

  14. Heuristic: Chapman-Enskog Expansion ∂ t f + ξ 1 ∂ x f = Lf Micro-Macro decomposition: Macroscopic: P 0 : orthogonal projection L 2 ξ → Ker L Microscopic: P 1 = Id − P 0 P 0 L = LP 0 = 0 , P 1 L = LP 1 = L, P 0 P 1 = 0 Apply P 0 : ∂ t P 0 f + ∂ x P 0 ξ 1 P 0 f + ∂ x P 0 ξ 1 P 1 f = 0 Apply P 1 : ∂ t P 1 f + ∂ x P 1 ξ 1 P 0 f + ∂ x P 1 ξ 1 P 1 f = LP 1 f L invertible on Range P 1 P 1 f = L − 1 � � ∂ t P 1 f + ∂ x P 1 ξ 1 P 0 f + ∂ x P 1 ξ 1 P 1 f P 1 f ≪ P 0 f L − 1 � � ∼ ∂ x P 1 ξ 1 P 0 f Therefore we obtain the equation for P 0 f (3 dimensional) � � ∂ t ( P 0 f ) + ∂ x ( P 0 ξ 1 P 0 f ) = − ∂ x ( P 0 ξ 1 P 1 f ) = − ∂ 2 P 0 ξ 1 L − 1 P 1 ξ 1 P 0 f x 5 / 25

  15. Heuristic: Chapman-Enskog Expansion ∂ t f + ξ 1 ∂ x f = Lf Micro-Macro decomposition: Macroscopic: P 0 : orthogonal projection L 2 ξ → Ker L Microscopic: P 1 = Id − P 0 P 0 L = LP 0 = 0 , P 1 L = LP 1 = L, P 0 P 1 = 0 Apply P 0 : ∂ t P 0 f + ∂ x P 0 ξ 1 P 0 f + ∂ x P 0 ξ 1 P 1 f = 0 Apply P 1 : ∂ t P 1 f + ∂ x P 1 ξ 1 P 0 f + ∂ x P 1 ξ 1 P 1 f = LP 1 f L invertible on Range P 1 P 1 f = L − 1 � � ∂ t P 1 f + ∂ x P 1 ξ 1 P 0 f + ∂ x P 1 ξ 1 P 1 f P 1 f ≪ P 0 f L − 1 � � ∼ ∂ x P 1 ξ 1 P 0 f Therefore we obtain the equation for P 0 f (3 dimensional) � � ∂ t ( P 0 f ) + ∂ x ( P 0 ξ 1 P 0 f ) = − ∂ x ( P 0 ξ 1 P 1 f ) = − ∂ 2 P 0 ξ 1 L − 1 P 1 ξ 1 P 0 f x 5 / 25

Download Presentation
Download Policy: The content available on the website is offered to you 'AS IS' for your personal information and use only. It cannot be commercialized, licensed, or distributed on other websites without prior consent from the author. To download a presentation, simply click this link. If you encounter any difficulties during the download process, it's possible that the publisher has removed the file from their server.

Recommend


More recommend