A conservative spectral method for the Boltzmann equation with - PowerPoint PPT Presentation

A conservative spectral method for the Boltzmann equation with anisotropic scattering and the grazing collisions limit Jeff Haack Department of Mathematics, and Institute for Computational Engineering Science, University of Texas at Austin Joint work with with Irene M. Gamba (UT) Issues in Solving the Boltzmann Equation for Aerospace Applications ICERM, Providence RI 7 June 2013 Work supported by NSF grants DMS-0636586 and DMS-1217254 Jeff Haack (UT Austin) Conservative Spectral Boltzmann 7 Jun 2013 1 / 21

Outline of talk Spectral method for Boltzmann-type collision operators with anisotropic 1 scattering Parallelization Grazing collisions and the Landau equation 2 Jeff Haack (UT Austin) Conservative Spectral Boltzmann 7 Jun 2013 2 / 21

The Boltzmann equation Boltzmann equation: Df v ∈ R 3 , Dt ( x , v , t ) = Q ( f , f )( v ) , Q ( f , f ) is the collision operator: � � u · σ )( f ( v ′ ∗ ) f ( v ′ ) − f ( v ∗ ) f ( v )) d σ d v ∗ Q ( f , f )( v ) = S 2 B ( | u | , ˆ R 3 Jeff Haack (UT Austin) Conservative Spectral Boltzmann 7 Jun 2013 3 / 21

Numerical methods Direct Simulation Monte Carlo ◮ Bird, Nanbu, ... ◮ conservation, positivity, overcomes dimensionality ◮ noise, transients, time dependent problems, tails, low speed flows, etc. Deterministic methods (spectral, DVM....) ◮ Advantages: no noise, accuracy in v ◮ Disadvantages: Positivity, conservation, dimensionality Previous spectral works ◮ Bobylev (75), Bobylev-Rjasanow (97, 99, 00), Ibragimov-Rjasanow (02), Pareschi-Russo (00), Mouhot-Pareschi (04), Pareschi-Russo-Toscani (00) (Landau), Pareschi-Toscani-Villani (03) (Grazing collisions consistency) ◮ Weak form: Gamba-Tharkabhushanam (09, 10) - conservation, inelastic Jeff Haack (UT Austin) Conservative Spectral Boltzmann 7 Jun 2013 4 / 21

Spectral formulation of collision operator Weak (Maxwell) form of collision operator: � � f ( v ) f ( v ∗ )[ φ ( v ′ ) − φ ( v )] B ( | u | , cos θ ) d σ d v ∗ d v , Q ( f , f ) φ ( v ) d v = Let √ φ ( v ) = e − i ζ · v / ( 2 π ) 3 , then we have that the Fourier transform of the collision integral is � � Q ( ζ ) = R 3 F{ f ( v ) f ( v − u ) } ( ζ ) G ( ζ, u ) d u � � 1 f ( ζ − ξ )ˆ ˆ f ( ξ ) e − i ξ · u d ξ d u = u ∈ R 3 G ( u , ζ ) √ 2 π ) 3 ( ξ ∈ R 3 � 1 f ( ζ − ξ )ˆ ˆ f ( ξ ) ˆ √ = G ( ξ, ζ ) d ξ 2 π ) 3 ( ξ ∈ R 3 Jeff Haack (UT Austin) Conservative Spectral Boltzmann 7 Jun 2013 5 / 21

Spectral formulation of collision operator The convolution weights G ( ζ, u ) , ˆ G ( ξ, ζ ) are given by � σ ∈ S 2 B ( | u | , cos θ )( e − i ζ 2 · ( u ′ − u ) − 1) d σ. G ( ζ, u ) = � � σ ∈ S 2 B ( | u | , cos θ )( e − i ζ 2 · ( u ′ − u ) − 1) d σ d u . ˆ u ∈ R 3 e − i ξ · u G ( ξ, ζ ) = For B = | u | λ / 4 π (isotropic), this can be reduced to � ∞ r λ +2 [sinc( r | ζ | 2 )sinc( r | ξ − ζ ˆ G ( ξ, ζ ) = 2 | ) − sinc( r | ξ | )] dr 0 no time dependence Jeff Haack (UT Austin) Conservative Spectral Boltzmann 7 Jun 2013 6 / 21

Difficulties in simulating the Boltzmann equation � � u · σ )( f ( v ′ ∗ ) f ( v ′ ) − f ( v ∗ ) f ( v )) d σ d v ∗ S 2 B ( | u | , ˆ Q ( f , f )( v ) = R 3 Issues: Dimensionality : requires O ( N 6 ) operations for a single evaluation of weighted convolution Jeff Haack (UT Austin) Conservative Spectral Boltzmann 7 Jun 2013 7 / 21

Computation is embarassingly parallel! Test problem: 1280 spatial points, 24 3 velocity mesh points Time listed: wall time using Stampede (NSF XSEDE resource) for one timestep ( ∼ 244 billion operations) nodes cores time (s) 1 16 456.313 2 32 235.315 4 64 120.762 8 128 61.345 16 256 30.943 32 512 15.252 64 1024 7.813 128 2048 4.042 [H. submitted 2013, also arXiv] Jeff Haack (UT Austin) Conservative Spectral Boltzmann 7 Jun 2013 8 / 21

Sudden Heating problem x 1 /x 0 =0.3 t/t 0 =0.5 0.5 x 1 /x 0 =0.5 x 1 /x 0 =0.1 x 1 /x 0 =0.9 0.4 g(v 1 ) 0.3 0.2 0.1 0 -2 0 2 v 1 /v 0 Jeff Haack (UT Austin) Conservative Spectral Boltzmann 7 Jun 2013 9 / 21

Difficulties in simulating the Boltzmann equation � � u · σ )( f ( v ′ ∗ ) f ( v ′ ) − f ( v ∗ ) f ( v )) d σ d v ∗ S 2 B ( | u | , ˆ Q ( f , f )( v ) = R 3 Issues: Dimensionality : requires O ( N 6 ) operations for a single evaluation of weighted convolution Conservation : collision invariants need to be preserved.   � 1   = 0 . v ∈ R 3 Q ( f , f ) v | v | 2 Solve the constrained minimization problem in O ( N 3 ) (Gamba, Tharkabhushanam) � � min 1 2 � ˜ Q − Q � 2 2 | CQ = 0 Jeff Haack (UT Austin) Conservative Spectral Boltzmann 7 Jun 2013 10 / 21

Anisotropic scattering Potentials interactions (e.g. Coulomb) include an angular component. Write B = | u | λ b (cos θ ). In this case, the untransformed weight G ( u , ζ ) is given by �� � 2 · u e − i ζ | u | 2 · σ − 1) d σ u · σ )( e i ζ G ( u , ζ ) = | u | λ S d − 1 b (ˆ � � | u | sin θ | ζ ⊥ | � � � π e i (1 − cos θ ) ζ = 2 π | u | λ · u J 0 b (cos θ ) sin θ − 1 d θ, 2 2 0 ζ ⊥ = ζ − ( ζ · u / | u | ) u / | u | . � L � π � π � � � G ( ζ, ξ ) = 4 π 2 r λ +2 r | ξ ⊥ | sin φ b (cos θ ) sin θ sin φ J 0 × 0 0 0 � � � � 1 � r ( ξ − ζ 2(1 − cos θ )) · ζ cos | ζ | cos φ J 0 2 r | ζ | sin φ sin θ � � � r ξ · ζ − cos | ζ | cos φ d θ d φ dr Jeff Haack (UT Austin) Conservative Spectral Boltzmann 7 Jun 2013 11 / 21

The Coulombic (grazing) limit It is well known that when the underlying potential is Coulombic, long distance grazing collisions dominate the collision term (Landau ’36). �� � | u | λ +2 ( δ ij − u i u j Q L = ∇ v · | u | 2 )( f ( v ∗ ) ∇ v f ( v ) − f ( v ) ∇ v f ( v ∗ )) d v ∗ Similar weak formulation gives G L ( u , ζ ) = | u | λ � 4 i ( ζ · u ) − | u | 2 | ζ ⊥ | 2 � Pareschi, Toscani, and Villani (2003): convergence to Landau for collocation. However, no computations were done in this limiting regime to our knowledge. Jeff Haack (UT Austin) Conservative Spectral Boltzmann 7 Jun 2013 12 / 21

Approximating Boltzmann by Landau Using Screened Coulomb: −3 x 10 C 10 B = | u | − 3 sin 4 ( θ/ 2)1 θ ≥ ε , 8 ε ∼ r 0 /λ 3 D . 6 f(v 1 ) Isotropic initial condition: 4 � | v |− 1 . 5 � � 2 � 2 1 − 10 1 . 5 f ( v , 0) = 100 e . 0 −1 −0.5 0 0.5 1 v 1 ε = 10 − 4 , N = 16 Solid lines: Landau solution. Dashed lines: Boltzmann solution. t = 0 , 1 , 2 , 5 , 10. Jeff Haack (UT Austin) Conservative Spectral Boltzmann 7 Jun 2013 13 / 21

Scattering kernel Make the following assumptions on the scattering kernel. Let ε > 0 be the small parameter associated with the grazing collision limit. A family of kernels b ε are grazing if (Villani, Bobylev): � π 0 b ε (cos θ ) sin 2 ( θ/ 2) sin θ d θ → Λ 0 < ∞ Λ ε = 2 π ∀ θ 0 > 0 , b ε (cos θ ) → 0 uniformly on θ ≥ θ 0 � π 0 b ε (cos θ ) sin θ θ k d θ → 0 , This corresponds to k > 2 Some examples sin θ Coulomb: b ε (cos θ ) sin θ = C sin 4 ( θ/ 2) log sin( ε/ 2)1 θ ≥ ε ε -linear: b ε (cos θ ) sin θ = 8 ε πθ 4 1 θ ≥ ε , Note for ε -linear: � π � b ε (cos θ ) sin θ ( θ 2 ) d θ ≈ C ε ε = C (1 − ε � π � π ) θ 0 Jeff Haack (UT Austin) Conservative Spectral Boltzmann 7 Jun 2013 14 / 21

Convergence of Boltzmann to Landau � � Q ε ( ζ ) = R 3 F{ f ε ( v ) f ε ( v − u ) } ( ζ ) G ε ( ζ, u ) d u Theorem (H., Gamba.) Assume that f ε satisfies A ( ζ ) |F{ f ε ( v ) f ε ( v − u ) } ( ζ ) | ≤ 1 + | u | 3 , (1) with A uniformly bounded in ζ , and that b (cos θ ) is the screened Rutherford cross section. Then the rate of convergence of the Boltzmann collision operator with grazing collisions to the Landau collision operator is given by � � � � 1 � � � � Q L [ f ε ]( ζ ) − � Q b ε [ f ε ]( ζ ) � ≤ O → 0 as ε → 0 . (2) | log sin( ε/ 2) | Jeff Haack (UT Austin) Conservative Spectral Boltzmann 7 Jun 2013 15 / 21

sketch of proof �� � S d − 1 b ε (cos θ )( e i ζ 2 · u e − i ζ | u | 2 · σ − 1) d σ G ε ( u , ζ ) = | u | λ � π � 2 π = | u | λ b ε (cos θ ) sin θ 0 0 � � 2 ((1 − cos θ ) ζ · u + | u | ζ · j sin θ sin φ + | u | ζ · k sin θ cos φ ) − 1 i × e d φ d θ, First two terms of the expansion: � π 2 | u | − 3 cos( θ/ 2) sin( θ/ 2) i ( u · ζ ) − 1 2 sin( θ/ 2) cos( θ/ 2)( u · ζ ) 2 d θ = − log sin( ε/ 2) ε = | u | − 3 � 4 i ( u · ζ ) − | u | 2 | ζ ⊥ | 2 � � 1 � (1 + cos ε ) 2( u · ζ ) 2 + 1 − | u | − 3 4 | ζ ⊥ | 2 | u | 2 log sin( ε/ 2) . Jeff Haack (UT Austin) Conservative Spectral Boltzmann 7 Jun 2013 16 / 21

sketch of proof � Q b ε [ f ε ]( ζ ) = � � Q L [ f ε ]( ζ ) + R n F{ f ε ( v ) f ε ( v − u ) } ( ζ ) R ( ζ, u ) du (3) � = � Q L [ f ε ]( ζ ) + R n F{ f ε ( v ) f ε ( v − u ) } ( ζ ) � � ( u · ζ ) 2 � | u | − 3 + 1 4 | ζ ⊥ | 2 | u | 2 × (1 + cos ε ) | log sin( ε/ 2) | 2 � � ∞ | u | − 3 + G b ε , n ( ζ, u ) du . | log sin( ε/ 2) | n =3 Jeff Haack (UT Austin) Conservative Spectral Boltzmann 7 Jun 2013 17 / 21