Boltzmann Equation and Hydrodynamics beyond Navier-Stokes

Transcription

1 Boltzmann Equation and Hydrodynamics beyond Navier-Stokes Alexander Bobylev Keldysh Institute for Applied Mathematics, RAS, Moscow

2 Chapman-Enskog method and Burnett equations Notation: f (x, v, t) - distribution function (x R 3, v R 3, t ) Boltzmann equation: Df = 1 ε Q(f, f ), D = t + v x, ε - Knudsen number, ε + Macroscopic variables: (in the notation f, g = R 3 dvf (v)g(v)) density ρ = f, 1 R + bulk velocity u = 1 ρ f, v R3 temperature T = 1 3ρ f, c 2 R +, c = v u (thermal velocity) 7 / 3

3 Equations of hydrodynamics (based on: Ψ, Q(f, f ) = if Ψ = 1, v, v 2 ) ρ t + divρu = ρd u α + p + ε π αβ = x α x( β ) 3 2 ρd u α T + pdivu + ε π αβ + divq x β = where Fluxes: D = t + u x, p = ρt (pressure) επ αβ = f, c α c β c 2 3 δ αβ, c = v u, εq α = 1 2 f, c α c 2, α, β = 1, 2, 3 Main problem: How to express fluxes π(f ) and q(f ) in terms of macroscopic variables (ρ, u, T )? 8 / 3

4 Chapman-Enskog Expansion We use the famous Chapman-Enskog expansion (Hilbert 1912, Enskog 1917, Chapman 1917, Burnett 1935, Chapman & Cowling 1939) and obtain Equations of hydrodynamics in symbolic form ρ ρ ( E N S 2 u = A B ) + εa1 + ε A u, t T T L A, k =,1,... where all are nonlinear operators k

5 Chapman-Enskog Expansion Truncation problem The standard truncation rule ρ ρ ( E N S 2 u = A B ) + εa1 + ε A u, t T T ( n 1) "Neglect all terms of order O ε + " does not work for n=2: Burnett equations are ill-posed. What shall we do? Is the way of truncation unique?

6 General Approach (Poincaré,...) Take an abstract evolution equation (or simply a vector ODE) dy dt = T (y; ε) = A(y) + εb(y) + ε2 C(y) +..., y Y ; A(y), B(y),... are differentiable non-linear operators: A(y + h) A(y) = A (y)h + O( h 2 ) 1. Let the natural truncation Y t = A(y) + εb(y) + ε 2 C(y) + O(ε 3 ) does not work. What then? 2. Use the change of variables (all operators are time-independent): z = y + ε 2 R(y) Then y z ε 2 R(z) and simple calculation yields Z t = A(z) + εb(z) + ε 2 {C(z) + [R, A] (z)} + O(ɛ 3 ), where [R, A] (z) = R (z)a(z) A (z)r(z). 18 / 3

7 Proposition Any truncated equation y t = A(y) + εb(y) + ε 2 C(y) +... (2) is formally equaivalent to a family of equations z t = A(z) + εb(z) + ε 2 C(z) +..., (3) where and z = y + ε 2 R(y) y = z ε 2 R(z) +..., C(z) = C(z) + R (z)a(z) A (z)r(z) with any differentiable operator R. Remark: Note that even classical Navier-Stokes Eqs. are not defined uniquely (only Euler Eqs. are unique). Remaining problem: Find the operator R(z) that makes Eqs. (3) good. 19 / 3

8 Hierarchy of problems (with increasing difficulty) Linearized BE in R or Tor 3 3 Nonlinear BE in R or Tor 3 3 Linearized BE in other domains + boundary conditions Nonlinear BE in other domains + boundary conditions Separate question: stationary BE Main goal is to understand how to improve the Navier-Stokes results

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11 Sketch of proof (in R) Step1 Fourier transform ikx ˆ (k,t) dxe (x,t), ρ = F( ρ)= ρ û= F( u), Tˆ = F( T). ˆ ρ Then ẑ = uˆ and we obtain ODEs ˆT ˆz + ikm ˆz = εk M ˆz + ε ( ik ) M ˆz t Note that has three distinct eigenvalues M λ =, λ ± =± 5/ 3 =± c.

12 Step 2 Diagonalization of M New variables w w+ = w s 3 s = ˆ ρ T, ˆ w ˆ Tˆ c ˆ ± = ρ + ± u 2 Then and we obtain w + ikc Λ w= εk M w + ε (ik) M w, t where Λ = diag( 1, 1, )

13 Step 3 Follows from Lemma n If w + Λ w = ε Aw, w R, Λ = diag( λ,..., λ ), λ λ for i j, t 1 n i j then there is a unique matrix B such that (a) diag B = and (b) the substitution w= y+ε B y leads to equation t ( ) ( 2 diag ε ) y +Λ y = ε A y + O. Proof y 1 ε B w Let ( ) 1 = + and therefore ε [ ] t ( ) ( 2 ε ) y +Λ y = A+ B, Λ y + O. { ij} { ij} then [ ] ( ) ij ( λ j λi ) A = a, B = b, B, Λ = B Λ Λ B = b. a ij Hence B = b ij =, i j;bii = completes the proof. λi λj ij ij

14 Structure of Chapman-Enskog expansion for linearized Boltzmann equation Linearization of BE : = + = ( 2 ) g(x,v,t): 1 ε 3/ 2 v / 2 π f m g m, m e 3 3 gt + v gx = Lg, g t= = g ; x R,v R ; * g,g dv g (v)g (v), g g,g, dx g. = = < R Fourier transform: ˆ = ˆ + ˆ = ˆ ˆ = 3 ε ik x 3 g(k,v,t ) dx e g(...), gt ik vg Lg, g t = g, k R. R R 2

15 Null-space of L: { } 2 ( ) ( 3 R ) N L = g L, Lg = { 2 } N = Span m, v m, v m, dim N = 5. Basis in N(L): k ei v,, i 1,..., 5 such that e i,e j ij, k = = δ 5 k v e,e = k, = = 1, = = =. 3 ( ) i j γi δij γ1 γ2 γ3 γ4 γ5 w1 Hydrodynamic vector: w = M, wi = e i,g( k,v,t ), w 5 t k g( k,v,t ) = exp A iε k, g ( k,v ), where ε k ( ) ( ) μω = μ ω μ ω 2 A, L v, C, S.

16 Lemma (Arsen ev, 1965). 12 / For hard spheres with diameter d = π there is a number r > such that the eigenvalue problem ( ) λg = A μω, g, μ r, has exactly 5 linearly independent solutions { λ,g ; j = 1,..., 5} j j having the following form: n j = j( ) = n n= 1 λ λ μ μ λ ( j), n ( j ( ) ( ) ) j μω j ω μ n ( ω) j n n= 1 g v;, = e v; + g v;, e,g =. ( j)

17 Connection with Chapman-Enskog expansion Chapman-Enskog method leads to equation where n w( k,t ) : wt = Uw = ε U n w, n= 5 2 N S ( ) ( ) 3 B U = k diag 1, 1,,,, U1 = k U, U2 = i k U,... 3 U What can be said about the structure of?

18 Connection with Chapman-Enskog expansion Proposition { λ ( ) ( )} 1 ε λ5 ε 1 U = BΛB, Λ = i k,..., i k, where 1 diag ε B = { b 1 5} ij ;i,j =,...,, bij = k e i,g j iε k, k. Obviously the series ( j ) ( ) n ( ij ) ( ) λ = λ iε k, b = δ + b iε k ; i, j = 1,..., 5; j n ij ij n j= 1 n= 1 converge for such k R that 3 k n r. ε

19 Asymptotic expansion of solutions Chapman - Enskog equations of hydrodynamics : ( as) 1 wt = BΛ B w, w t= = w, where ( ε) ( ) ( as) Λ ε are analytic at ε =. B,,w "Diagonal" equations: (the connection with "changes of variables" is obvious). Then and we can use two independent truncations: w= By y = Λ y, y = y = B w wt = Be y () Λt t 1 t= ( as)

20 () () ( n l ) Λ t () l w t = B e y, n, l, where nl B=B () l ( l+ 1) () l ( l+ 1) ( n) ( n+ 1 + ε ) = + ε Λ =Λ + ε O, y y O, O. Final estimates for n =, 1, 2 : F ( ) wk,t w k,t = O + O ( ) ( ) ( n) ( l ε ε ) k x ne uniformly in time. Note that n=1 for Navier-Stokes and n=2 for Burnett equations.

21 Conclusions 1 We began with the problem regularization of classical Burnett equations and show that this can be done by transformation to new hydrodynamic variables. It was also shown that the way of truncation of the Chapman-Enskog expansion is not unique (even at the Navier-Stokes level).

22 2 What is the meaning of these transformations? How to find the optimal one? Does it exist? These questions were considered in detail for the linearized Boltzmann equation. It was shown that the Chapman-Enskog expansion has a special structure which allows to pass to "diagonal" equations of hydrodynamics (5 independent equations). This is the meaning of the optimal transformation in the linear case.

23 3 The "diagonal" Navier-Stokes equations are: three heat equations for two viscous and one thermal modes respectively. Plus two linearized Burgers equations for two sound modes. The corresponding solutions are valid with error O ( ε ) uniformly in time.

24 4 The "diagonal" Burnett equations are: the same three heat equations plus two linearized Burgers-KdV equations for sound modes. The uniform in time error is quadratic in in this case provided that the diagonalizing transformation and the asymptotic initial data are also computed with quadratic error. ε

25 . THANK YOU!...