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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

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