xxxx Unique moment set from the order of magnitude method Henning Struchtrup University of Victoria, Canada
¡ ε λ ¢ Goal: Find approximation to kinetic equation for Knudsen order O Chapman-Enskog expansion: • gives classical transport laws for λ = 1 • stability problems for λ ≥ 2 • full summation not feasible Grad moment method: • no relation to Knudsen number • (mostly) hyperbolic equations • unphysical subshocks order of magnitude method [HS2004] : • combines ideas of Grad and CE methods • moments and their equations directly related to Knudsen order • regularizing terms remove subshocks, damp high frequencies
Moment method in kinetic theory kinetic equation: Knudsen number ε ∂ f ∂ f = 1 ∂ t + c k ε S ( f ) ∂ x k moment method: replace kinetic equation with equations for moments Z u A = ψ A ( c i ) fdc A = 1 , . . . , N moment equations: multiply kinetic equation with ψ A ( c i ) and integrate ∂ u A ∂ t + ∂ F Ak = 1 ε P A ∂ x k with F Ak = R ψ A ( c i ) c k fd c , P A = R ψ A ( c i ) S ( f ) d c Question 1: which moments, ψ A ( c i ) = ?? Question 2: how many moments, N = ?? Question 3: how to close the equations, F Ak ( u B ) = ?? , P A ( u B ) = ?? Q3 answered by Grad [1949] , Q1 and Q2 today
Order of magnitude method [HS 2004] Step 1: • set up moment system for arbitrary number of moments N • use in fi nite system (if possible), or close with Grad method • must use arbitrary complete function system for moments Step 2: • Chapman-Enskog expansion to fi nd leading ε − order of moments • linear combination of moments such that number of moments at given ε − order is minimal • repeat for next order of magnitude Step 3: • use ε − orders to rescale equations for new moments • use scaling for model reduction to a given order of accuracy
Step 0: A simple kinetic model and its properties kinetic model for 1-D heat transfer: simpli fi ed phonon/photon model with scattering R κ ( μ ) fd μ ∙ ¸ ∂ f ∂ t + μ∂ f ∂ x = − 1 R κ ( μ ) d μ ετ κ ( μ ) f − f ( x, t, μ ) - distribution function, ε - Knudsen number, μ = cos ϑ - direction cosine anisotropic scattering probability ( γ = 0 for isotropic scattering) κ ( μ ) = 1 + γμ 2 energy density and heat fl ux are moments Z 1 Z 1 u 0 ( x, t ) = f ( x, t, μ ) d μ , w 1 ( x, t ) = μ f ( x, t, μ ) d μ − 1 − 1 energy is conserved ∂ u 0 ∂ t + ∂ w 1 ∂ x = 0 H-theorem ∂η ∂ t + ∂φ ∂ x = σ ≥ 0 entropy density, fl ux, generation Z 1 Z 1 Z μ f 2 d μ , σ = 1 κ ( μ ) ( f − f 0 ) 2 d μ ≥ 0 f 2 d μ , φ = − η = − τε − 1 − 1
Step 1: Grad closure for monomials (2 N + 1) monomials: Z 1 μ 2 α fd μ u α = , α = 0 , 1 , . . . , N − 1 Z 1 μ 2 α − 1 fd μ w α = , α = 1 , . . . , N − 1 γ 1 2 α +1 + nested moment equations: 2 α +3 φ α = 1+ γ 3 ∂ u 0 ∂ t + ∂ w 1 ∂ x = 0 ∂ u α ∂ t + ∂ w α +1 = − 1 τε [ u α + γ u α +1 − φ α ( u 0 + γ u 1 )] ∂ x ∂ w α ∂ t + ∂ u α ∂ x = − 1 τε [ w α + γ w α +1 ] needs closure for u N +1 and w N +1
Step 1: Grad closure for monomials Grad-type distribution: just a polynomial [e.g., from maximizing entropy...] X N X N υ β μ 2 β + ω β μ 2 β − 1 , f G = β =0 β =1 where X N X N A − 1 B − 1 υ α = αβ u β , ω α = αβ w β β =0 β =1 2 2 A αβ = , B αβ = 2 ( α + β ) + 1 2 ( α + β ) − 1 closure for higher moments: N N X X u N +1 = ξ α u α , w N +1 = ζ α w α α =0 α =1 with coe ffi cients X N X N 2 2 2 ( N + β ) + 3 A − 1 ξ α = , ζ α = 2 ( N + β ) + 1 B βα β =0 β =1
Step 1: Grad closure for monomials Closed equations for (2 N + 1) monomials ∂ u 0 ∂ t + ∂ w 1 ∂ x = 0 X N X N ∂ u α ∂ w β ∂ x = − 1 U αβ u β + 1 ∂ t + R αβ τε ( φ α − γξ 0 δ α N ) u 0 τε β =1 β =1 X N ∂ w α ∂ t + ∂ u α ∂ x = − 1 W αβ w β τε β =1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 0 1 1 − γφ 1 γ 1 γ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 0 1 − γφ 2 1 γ 1 γ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ . ... ... ... ... ... ... ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ . . ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ R αβ = , U αβ = , W αβ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ... ... ... ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 1 γ γ ζ 1 ζ 2 ζ 3 · · · · · · ζ N γξ 1 − γφ N γξ 2 γξ 3 · · · 1 + γξ N γζ 1 γζ 2 γζ 3 · · · · · · 1 + γζ N Order of Magnitude Method, next steps: ¡ ε λ ¢ • Chapman-Enskog expansions to determine leading order of moments as O • reconstruct moments so that minimum number of variables at given order λ
Step 2: O of M method, isotropic scattering ( γ = 0 ) moment equations for γ = 0 : ∂ u 0 ∂ t + ∂ w 1 ∂ x = 0 µ ¶ ∂ u α ∂ t + ∂ w α +1 = − 1 u 0 u α − ∂ x τε 2 α + 1 ∂ w α ∂ t + ∂ u α ∂ x = − 1 τε w α equilibrium: limit ε → 0 u 0 u α | E = , w α | E = 0 2 α + 1 ¡ ε 0 ¢ = ⇒ all u α are O ¡ ε 1 ¢ = ⇒ all w α are at least O
Step 2: O of M method, isotropic scattering ( γ = 0 ) fi rst non-equilibrium moments: subtract equilibrium values from higher moments u 0 u (1) α = u α − u α | E = u α − ( α =1 , 2 ,... ) 2 α + 1 w (1) α = w α − w α | E = w α ( α =1 , 2 ,... ) ¡ ε 0 ¢ = ⇒ u 0 is O ¡ ε 1 ¢ ⇒ u (1) α , w (1) = α are at least O new moment equations: variables u 0 , u (1) α , w (1) α ∂ t + ∂ w (1) ∂ u 0 1 = 0 ∂ x ∂ u (1) ∂ w (1) + ∂ w (1) 1 = − 1 α 1 α +1 τε u (1) − ( α =1 , 2 ,... ) α ∂ t 2 α + 1 ∂ x ∂ x ∂ w (1) ∂ x + ∂ u (1) 1 ∂ u 0 ∂ x = − 1 α α τε w (1) + ( α =1 , 2 ,... ) α ∂ t 2 α + 1 Remark: balance laws for new variables, i.e., only the variable in time derivative
Step 2: O of M method, isotropic scattering ( γ = 0 ) expand high order variables in Chapman-Enskog series: α = ε u (1) α , 1 + ε 2 u (1) α = ε w (1) α , 1 + ε 2 w (1) u (1) w (1) α , 2 + . . . , α , 2 + . . . leading terms only: 2 α + 1 τ ∂ u 0 1 0 = u (1) ∂ x = w (1) , − ( α =1 , 2 ,... ) α , 1 α , 1 the w (1) α are linearly dependent 3 w (1) 2 α + 1 w (1) α , 1 = ( α =2 , 3 ,... ) 1 , 1 second non-equilibrium moments: use above u (2) α = u (1) ( α =1 , 2 ,... ) α 3 2 α + 1 w (1) w (2) α = w (1) α − ( α =2 , 3 ,... ) 1 ¡ ε 0 ¢ = ⇒ u 0 is O ¡ ε 1 ¢ w (1) = ⇒ is O 1 ¡ ε 2 ¢ u (2) α , w (2) = ⇒ α are at least O
Step 2: O of M method, isotropic scattering ( γ = 0 ) u 0 , w (1) w (1) 1 , u (2) α , w (2) equations for u 0 = ˆ = ε ˆ α 1 w (1) ∂ ˆ ∂ t + ε∂ ˆ u 0 1 = 0 ∂ x ∙ ¸ w (1) + ∂ u (2) ε∂ ˆ ∂ x = − 1 1 + τ ∂ ˆ u 0 w (1) 1 1 ˆ ∂ t τ 3 ∂ x ∂ u (2) w (1) + ∂ w (2) (2 α + 3) (2 α + 1) ε∂ ˆ 4 α = − 1 α 1 α +1 τε u (2) + ( α =1 , 2 ,... ) α ∂ t ∂ x ∂ x ∂ w (2) ∂ u (2) ∂ x + ∂ u (2) 3 ∂ x = − 1 α α 1 τε w (2) − ( α =2 , 3 ,... ) α ∂ t 2 α + 1
Step 2: O of M method, isotropic scattering ( γ = 0 ) expand high order variables in Chapman-Enskog series: α = ε 2 u (2) α , 2 + ε 3 u (2) α = ε 2 w (2) α , 3 + ε 3 w (2) u (2) w (2) α , 3 + . . . , α , 3 + . . . leading terms only: w (1) (2 α + 3) (2 α + 1) τ ∂ ˆ 4 α = u (2) 1 − ( α =1 , 2 ,... ) α , 2 ∂ x 0 = w (2) ( α =2 , 3 ,... ) α , 2 the u (2) α , 2 are linearly dependent 15 α u (2) (2 α + 3) (2 α + 1) u (2) α , 2 = ( α =1 , 2 ,... ) 1 , 2 third non-equilibrium moments: use above 15 α (2 α + 3) (2 α + 1) u (2) u (3) α = u (2) α − ( α =2 , 3 ,... ) 1 w (3) α = w (2) ( α =2 , 3 ,... ) α ¡ ε 0 ¢ = ⇒ u 0 is O ¡ ε 1 ¢ w (1) = ⇒ is O 1 ¡ ε 2 ¢ u (2) = ⇒ is O 1 ¡ ε 3 ¢ u (3) α , w (3) α are at least O = ⇒
Step 2: O of M method, isotropic scattering ( γ = 0 ) u 0 , w (1) w (1) 1 , u (2) u (2) and u (3) α , w (3) = ε 2 ˆ equations for u 0 = ˆ = ε ˆ α 1 1 1 w (1) ∂ ˆ ∂ t + ε∂ ˆ u 0 1 = 0 ∂ x ∙ ¸ w (1) u (2) ε∂ ˆ + ε 2 ∂ ˆ ∂ x = − 1 1 + τ ∂ ˆ u 0 w (1) 1 1 ˆ 3 ∂ t τ ∂ x " # u (2) + ∂ w (3) w (1) ε 2 ∂ ˆ = − 1 1 − 4 15 τ ∂ ˆ u (2) 1 2 1 τ ε ˆ ∂ t ∂ x ∂ x + ∂ w (3) ∂ u (3) ∂ w (3) 15 α = − 1 α 2 α +1 τε u (3) − ( α =2 , 3 ,... ) α ∂ t (2 α + 3) (2 α + 1) ∂ x ∂ x ∂ w (3) ∂ x + ∂ u (3) u (2) (2 α + 3) (2 α + 1) ε 2 ∂ ˆ 9 ( α − 1) ∂ x = − 1 α α 1 τε w (3) + ( α =2 , 3 ,... ) α ∂ t
Step 2: O of M method, isotropic scattering ( γ = 0 ) expand high order variables in Chapman-Enskog series: α = ε 3 u (3) α , 3 + ε 4 u (3) α = ε 3 w (3) α , 3 + ε 4 w (3) u (3) w (3) α , 4 + . . . , α , 4 + . . . leading terms only: 0 = u (3) ( α =2 , 3 ,... ) α , 3 u (2) (2 α + 3) (2 α + 1) τ ∂ ˆ 9 ( α − 1) ∂ x = w (3) 1 − ( α =2 , 3 ,... ) α , 3 the w (3) α , 3 are linearly dependent 35 ( α − 1) w (3) (2 α + 3) (2 α + 1) w (3) α , 3 = ( α =3 , 4 ,... ) 2 , 3 fourth non-equilibrium moments: use above u (4) α = u (3) ( α =2 , 3 ,... ) α 35 ( α − 1) (2 α + 3) (2 α + 1) w (3) w (4) α = w (3) α − ( α =3 , 4 ,... ) 2 ¡ ε 0 ¢ ¡ ε 1 ¢ ¡ ε 2 ¢ ¡ ε 3 ¢ , w (1) , u (2) , w (3) = ⇒ u 0 is O is O is O is O 1 1 2 ¡ ε 4 ¢ u (4) α , w (4) α are at least O = ⇒
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