Hydrodynamic Limit with Geometric Correction in Kinetic Equations

Transcription

1 Hydrodynamic Limit with Geometric Correction in Kinetic Equations Lei Wu and Yan Guo KI-Net Workshop, CSCAMM University of Maryland, College Park

2 Simple Model - Neutron Transport Equation We consider the steady homogeneous isotropic one-speed neutron transport equation in a two-dimensional unit plate. We denote the space variables as x = (x 1, x 2 ) and the velocity variables as v = (v 1, v 2 ). In the space domain Ω = { x : x 1} and the velocity domain Σ = { v : v S 1 }, the neutron density u ɛ ( x, v) satisfies { ɛ v x u ɛ + u ɛ ū ɛ = 0 for x Ω, u ɛ ( x 0, v) = g( x 0, v) for v n < 0 and x 0 Ω, (1) where ū ɛ ( x) = 1 2π S 1 u ɛ ( x, v)d v, with the Knudsen number 0 < ɛ << 1 as a parameter. We want to study the behavior of u ɛ as ɛ 0. 2

3 Features of the Equation Half boundary condition: v x u = h(x, v) for x [0, 1], u(0, v) = g 1 (v) for v (0, 1], u(1, v) = g 2 (v) for v [ 1, 0). Non-local operator: K[u]( v) = u( v )k( v, v )d v, S 1 with S 1 k( v, v )d v = 1. 3

4 Complex Model - Boltzmann Equation near Maxwellian We consider stationary Boltzmann equation for probability density F ɛ ( x, v) in a two-dimensional unit plate Ω = { x = (x 1, x 2 ) : x 1} with velocity Σ = { v = (v 1, v 2 ) R 2 } as { ɛ v x F ɛ = Q[F ɛ, F ɛ ] in Ω R 2, F ɛ ( x 0, v) = B ɛ ( x 0, v) for x 0 Ω and n( x 0 ) v < 0, where n( x 0 ) is the outward normal vector at x 0 and the Knudsen number ɛ satisfies 0 < ɛ << 1. Here we have Q[F, G] = q( ω, ( ) u v ) F( u )G( v ) F( u)g( v) d ωd u, R 2 S 1 ( ) ( ) with u = u + ω ( v u) ω, v = v ω ( v u) ω, and the hard-sphere collision kernel q( ω, u v ) = q0 u v cos φ, for positive constant q0 related to the size of ball, ω ( v u) = v u cos φ and 0 φ π/2. We intend to study the behavior of F ɛ as ɛ 0. 4

5 Complex Model - Boltzmann Equation (Cont.) We assume that the boundary data is as B ɛ ( x 0, v) = µ + ɛµ 1 2 b( x 0, v), v where µ( v) is the standard Maxwellian µ( v) = 1 2 2π exp 2. Then we havef ɛ ( x, v) = µ + ɛµ 1 2 f ɛ ( x, v), where f ɛ satisfies the equation { ɛ v x f ɛ + L[f ɛ ] = Γ[f ɛ, f ɛ ], f ɛ ( x 0, v) = b( x 0, v) for n v < 0 and x 0 Ω, for Γ[f ɛ, f ɛ ] = µ 1 2 Q[µ 1 2 f ɛ, µ 1 2 f ɛ ], L[f ɛ ] = 2µ Q[µ, µ 2 f ɛ ] = ν( v)f ɛ K[f ɛ ], ν( v) = q( v u, ω)µ( u)d ωd u R 2 S 1 K[f ɛ ]( v) = k( u, v)f ɛ ( u)d u R 2 5

6 Background Spacial Domain: R n or periodic, bounded; Temporal Domain: steady, unsteady Solution: strong(smooth), weak(renormalized), weighted. 1950s s: Case K. M., Zweifel P. F. and Larsen, E; 1D explicit solution; spectral analysis of kinetic operators; formal asymptotic expansion; 1979: Bensoussan, Alain, Lions, Jacques-L. and Papanicolaou, George C.; Boundary layers and homogenization of transport processes. Publ. Res. Inst. Math. Sci. 15 (1979), no. 1, : Bardos, C., Santos, R. and Sentis, R.; Diffusion approximation and computation of the critical size. Trans. Amer. Math. Soc. 284 (1984), no. 2, : Sone, Y; Kinetic theory and fluid dynamics. Birkhäuser Boston, Inc., Boston, MA. 6

7 Asymptotic Analysis - Perturbation Theory The main goal is to study the behavior of parameterized problems as the parameter goes to a limit. Algebraic Equations: Regular: (x ɛ ) 2 + ɛx ɛ 1 = 0. Singular: ɛ(x ɛ ) 2 + x ɛ 1 = 0. Differential Equations: Regular: (y ɛ ) + ɛ(y ɛ ) + y ɛ = 1 with y ɛ (0) = 0 and y ɛ (1) = 1. Singular: ɛ(y ɛ ) + (y ɛ ) + y ɛ = 1 with y ɛ (0) = 0 and y ɛ (1) = 1. Ingredients: interior solution; boundary layer; decay; cut-off function in 1D and 2D. 7

8 Hilbert Expansion The classical method is to introduce a power series in ɛ: 1 Define the formal expansion x ɛ ɛ k x k, k=0 y ɛ (t) ɛ k y k (t), k=0 where x k and y k (t) are independent of ɛ. 2 Then plugging this expansion into the original equations, we obtain a series of relations for x k and y k (t), which can be solved or estimated directly. 3 Finally, we can estimate the remainder N R N [x] = x ɛ ɛ k x k, k=0 N R N [y] = y ɛ (t) ɛ k y k (t). k=0 8

9 Hilbert Expansion(Cont.) This method can be used to analyzed both the interior solution and boundary(initial) layer. The convergence here is different from that of power series. Not all asymptotic relations can be expressed in power series with respect to the parameter. Hilbert expansion is not the only expansion to analyze asymptotic behaviors. This procedure is ideal. We may encounter difficulties in each step. Sometimes a trade-off is inevitable. 9

10 10 Interior Solution We define the interior expansion as follows: U( x, v) ɛ k U k ( x, v), k=0 where U k can be defined by comparing the order of ɛ via plugging this expansion into the neutron transport equation. Thus, we have U 0 Ū 0 = 0, U 1 Ū 1 = v x U 0, U 2 Ū 2 = v x U 1,... U k Ū k = v x U k 1.

11 11 Interior Solution (Cont.) We can show U 0 ( x, v) satisfies the equation { U0 ( x, v) = Ū 0 ( x), x Ū 0 = 0. Similarly, we can derive U k ( x, v) for k 1 satisfies { U k = Ū k v x U k 1, x Ū k = 0. We need to determine the boundary data of U k.

12 12 Boundary Layer The boundary layer can be constructed as follows: Polar coordinates: (x 1, x 2 ) (r, θ). Boundary layer scaling: η = (1 r)/ɛ. We define the boundary layer expansion as follows: U (η, θ, v) which satisfies ɛ k U k (η, θ, v), k=0 ( v n) U η ɛ U + ( v τ) 1 ɛη θ + U U = 0, where n is the outer normal vector and τ is the tangential vector.

13 13 Boundary Layer(cont.) By comparing the order of ɛ, we have the relation ( v n) U 0 η + U 0 U 0 = 0, ( v n) U 1 η + U 1 U 1 1 = ( v τ) 1 ɛη... ( v n) U k η + U k U 1 k = ( v τ) 1 ɛη in a neighborhood of the boundary. U 0 θ, U k 1, θ

14 14 Matching of Interior Solution and Boundary Layer We define the boundary layer U 0 as U 0 = f 0 (η, θ, v) f 0 (, θ) ( v n) f 0 η + f 0 f0 = 0, f 0 (0, θ, v) = g(θ, v) for n v < 0, lim η f 0 (η, θ, v) = f 0 (, θ), and the interior solution U 0 as U 0 ( x, v) = Ū 0 ( x), x Ū 0 = 0, Ū 0 ( x 0 ) = f 0 (, θ).

15 15 Good Results In 1979 s and 1984 s papers, the author showed that both the interior and boundary layer expansion can be constructed to higher order and then proved the following theorem: Theorem Assume g( x 0, v) is sufficiently smooth. Then for the steady neutron transport equation, the unique solution u ɛ ( x, v) L (Ω S 1 ) satisfies u ɛ U 0 U L 0 = O(ɛ). This is a remarkable result!

16 16 Think about it The proof is based on the following key theorem: Theorem Consider the Milne problem ( v n) f η + f f = S(η, θ, v), f(0, θ, v) = h(θ, v) for n v < 0, lim η f(η, θ, v) = f (θ), with e β 0η S L L C, h L C. Then for β > 0 sufficiently small, there exists a unique solution f(η, θ, v) L satisfying e βη (f f ) L C. L

17 17 Think about it(cont.) In the proof of 1979 s and 1984 s papers, we have to go to U 1 and U 1 at least. In all the known results, in order to show the L well-posedness of Milne problem, we need the source term is in L and exponentially decays. Thus in order to show the well-posedness of U 1, we need 1 ( v τ) 1 ɛη which further needs U 0 θ L ([0, ) [ π, π) S 1 ), ( v τ θ ) U 0 η L ([0, ) [ π, π) S 1 ). This is not always true. We have counterexamples to illustrate this fact.

18 18 Counterexample Lemma For the Milne problem sin(θ + ξ) f η + f f = 0, f(0, θ, ξ) = g(θ, ξ) for sin(θ + ξ) > 0, lim η f(η, θ, ξ) = f(, θ), if g(θ, ξ) = cos(3(θ + ξ)), then we have f η L ([0, ) [ π, π) [ π, π)).

19 Counterexample (Cont.) The central idea of the proof is by contradiction: 1 By maximum principle, we have f(0, θ, ξ) 1 for sin(θ + ξ) < 0. Then this implies f(0, θ) We can obtain η f(0, θ, ξ) L [ π, π) [ π, π) is a.e. well-defined and satisfies the formula 3 Finally, we can directly estimate η f(0, θ, ξ) = f(0, θ) f(0, θ, ξ). sin(θ + ξ) f lim (0, θ, ξ) =. ξ θ + η which is a contradiction. 19

20 20 Boundary Layer with Geometric Correction Polar coordinates: (x 1, x 2 ) (r, θ). Boundary layer scaling: η = (1 r)/ɛ. Change of Variables: v n = v n and v τ = v τ. We define the boundary layer expansion as follows: U ɛ (η, θ, v n, v τ ) k=0 ɛ k U ɛ k (η, θ, v n, v τ ), which satisfies U ɛ v n η + ɛ ( U ɛ v τ 1 ɛη θ + v2 τ U ɛ v n U ɛ ) v n v τ + U ɛ U ɛ = 0 v τ

21 21 Where is the Singularity? The singular term is decomposed into three terms v τ U ɛ 0 θ v2 τ U ɛ 0 v n + v n v τ U ɛ 0 v τ. By comparing the order of ɛ, we have the relation U ɛ 0 v n η U ɛ 1 v n η U ɛ k v n η + U ɛ 0 U ɛ 0 = 0, + U ɛ 1 + U ɛ U ɛ 1 =... k U ɛ k = 1 U (v ɛ 0 τ 1 ɛη θ v2 τ U ɛ 0 v n 1 U (v ɛ U ɛ k 1 τ v 2 k 1 τ 1 ɛη θ v n U ɛ ) 0 + v n v τ, v τ + v n v τ U ɛ k 1 v τ ), in a neighborhood of the boundary.

22 22 Boundary Layer with Geometric Correction(cont.) Putting the singular terms together, we have the relation U ɛ 0 v n η + ɛ ( U ɛ U ɛ ) vτ v n v τ + U ɛ 0 1 ɛη v n v U ɛ 0 = 0, τ U ɛ 1 v n η + ɛ ( U ɛ U ɛ ) vτ v n v τ + U ɛ 1 1 ɛη v n v U ɛ 1 = τ... U ɛ k v n η + ɛ ( ) vτ 2 + U ɛ k 1 ɛη U ɛ k = U ɛ k v n U ɛ k v n v τ v τ in a neighborhood of the boundary. 1 1 ɛη v U ɛ 0 τ θ, 1 1 ɛη v U ɛ τ k 1 θ,

23 ɛ-milne Problem with Geometric Correction Consider the substitution v n = sin φ and v τ = cos φ. The construction of the boundary layer depends on the properties of the Milne problem for f ɛ (η, θ, φ) in the domain (η, θ, φ) [0, ) [ π, π) [ π, π) where sin φ f ɛ η + F(ɛ; η) cos φ f ɛ φ + f ɛ ɛ f = S ɛ (η, θ, φ), f ɛ (0, θ, φ) = h ɛ (θ, φ) for sin φ > 0, lim η f ɛ (η, θ, φ) = f (θ). ɛ F(ɛ; η) = ɛψ(ɛη) { 1 0 µ 1/2, 1 ɛη, ψ(µ) = 0 3/4 µ, and h ɛ (θ, φ) C, S ɛ (η, θ, φ) Ce β 0 η, for C and β 0 uniform in ɛ and θ. 23

24 24 ɛ-milne Problem with Geometric Correction (Cont.) Theorem For β > 0 sufficiently small, there exists a unique solution f ɛ (η, θ, φ) L to the ɛ-milne problem satisfying e βη (f ɛ f ) ɛ L C, L where C depends on the data h ɛ and S ɛ. Theorem The solution f ɛ (η, θ, φ) to the ɛ-milne problem with S ɛ = 0 satisfies the maximum principle, i.e. min hɛ (θ, φ) f ɛ (η, θ, φ) max hɛ (θ, φ). sin φ>0 sin φ>0

25 25 ɛ-milne Problem with Geometric Correction (Cont.) Basic ideas: penalized finite slab finite slab infinite slab; homogeneous inhomogeneous. 1 Using energy estimate to define f ɛ and show f ɛ f ) ɛ L C. 2 L 2 2 Using the characteristics to get f ɛ f ) ɛ L C + C L f ɛ f ) ɛ L. 2 L 2 3 Applying the similar techniques to the equation satisfied by F ɛ = e βη f ɛ.

26 26 Remainder Estimate Theorem Assume f( x, v) L (Ω S 1 ) and g(x 0, v) L (Γ ). Then for the remainder equation { ɛ w x R + R R = f( x, v) in Ω, R( x 0, w) = g( x 0, w) for x 0 Ω and v n < 0, there exists a unique solution R( x, v) L (Ω S 1 ) satisfying ( ) 1 R L (Ω S 1 ) C(Ω) ɛ f 5/2 L (Ω S 1 ) + g L (Γ ).

27 27 Main Theorem Theorem Assume g( x 0, v) C 2 (Γ ). Then for the steady neutron transport equation (1), the unique solution u ɛ ( x, v) L (Ω S 1 ) satisfies u ɛ U ɛ 0 U ɛ L = O(ɛ) Moreover, if g(θ, v n, v τ ) = v τ, then there exists a C > 0 such that u ɛ U 0 U L 0 C > 0 when ɛ is sufficiently small. Remark The comparison of L p and L result. 0

28 28 Main Theorem (Cont.) Proof of u ɛ U 0 U 0 L C > 0 is as follows: 1 The problem can be simplified into the estimate of solutions u in Milne problem and U in ɛ-milne problems with exactly the same boundary data v τ Rewriting the solution along the characteristics, we can obtain the estimate at point (η, φ) = (nɛ, ɛ) as u(nɛ, ɛ) = ū(0) + e n ( ū(0) + 3) + o(ɛ), U(nɛ, ɛ) = Ū(0) + e 1 1+2n ( Ū(0) + 3) + o(ɛ). 3 We can derive lim ( ū(0) ɛ 0 + 3) ( Ū(0) + 3) L = 0. and ū(0) + 3 = O(1) with Ū(0) + 3 = O(1). Due to the smallness of ɛ, we can obtain U(nɛ, ɛ) u(nɛ, ɛ) = O(1).

29 29 Unsteady Neutron Transport Equation We consider a homogeneous isotropic unsteady neutron transport equation in a two-dimensional unit disk Ω = { x = (x 1, x 2 ) : x 1} with one-speed velocity Σ = { v = (v 1, v 2 ) : v S 1 } as ɛ 2 t u ɛ + ɛ v x u ɛ + u ɛ ū ɛ = 0 in [0, ) Ω, u ɛ (0, x, v) = h( x, v) in Ω u ɛ (t, x 0, v) = g(t, x 0, v) for v n < 0 and x 0 Ω, where ū ɛ (t, x) = 1 2π S 1 u ɛ (t, x, v)d v. and n is the outward normal vector on Ω, with the Knudsen number 0 < ɛ << 1. The initial and boundary data satisfy the compatibility condition h( x 0, v) = g(0, x 0, v) for v n < 0 and x 0 Ω. (2)

30 30 Remainder Estimate Theorem Assume f(t, x, v) L ([0, ) Ω S 1 ), h( x, v) L (Ω S 1 ) and g(t, x 0, v) L ([0, ) Γ ). Then for the remainder equation ɛ 2 t R + ɛ v x R + R R = f(t, x, v) in [0, ) Ω, R(0, x, v) = h( x, v) in Ω R(t, x 0, v) = g(t, x 0, v) for v n < 0 and x 0 Ω, there exists a unique solution R(t, x, v) L ([0, ) Ω S 1 ) satisfying R L ([0, ) Ω S 1 ) ( 1 C(Ω) ɛ f 5/2 L ([0, ) Ω S 1 ) + h L (Ω S 1 ) + g L ([0, ) Γ ) ).

31 31 Diffusive Limit Theorem Assume g(t, x 0, v) C 2 ([0, ) Γ ) and h( x, v) C 2 (Ω S 1 ). Then for the unsteady neutron transport equation (2), the unique solution u ɛ (t, x, v) L ([0, ) Ω S 1 ) satisfies u ɛ U ɛ 0 U ɛ I,0 U ɛ B,0 L = O(ɛ), for the interior solution U ɛ, the initial layer U ɛ, and the boundary layer 0 I,0 U ɛ B,0.

32 32 Boltzmann Equation near Maxwellian We turn back to the stationary Boltzmann equation { ɛ v x F ɛ = Q[F ɛ, F ɛ ] in Ω R 2, F ɛ ( x 0, v) = B ɛ ( x 0, v) for x 0 Ω and n( x 0 ) v < 0, and F ɛ ( x, v) = µ + ɛµ 1 2 f ɛ ( x, v), where f ɛ satisfies the equation { ɛ v x f ɛ + L[f ɛ ] = Γ[f ɛ, f ɛ ], f ɛ ( x 0, v) = b( x 0, v) for n v < 0 and x 0 Ω,

33 33 Hydrodynamic Limit of Stationary Boltzmann Equation Theorem For given b( x 0, v) sufficiently small and 0 < ɛ << 1, there exists a unique positive solution F ɛ = µ + ɛµ 1 2 f ɛ to the stationary Boltzmann equation, where ( N f ɛ = ɛ 3 R N + k=1 ) ( N ) ɛ k F ɛ k + ɛ k F ɛ k, k=1 for N 3, R N satisfies the remainder equation, F ɛ k and F ɛ are interior k solution and boundary layer. Also, there exists a C > 0 such that f ɛ satisfies v ϑ e ζ v 2 f ɛ L C, for any ϑ > 2, 0 ζ 1/4.

34 34 Steady Navier-Stokes-Fourier System In particular, the leading order interior solution satisfies F ɛ 1 = ( µ ρ ɛ 1 + uɛ 1,1 v 1 + u ɛ 1,2 v 2 + θ ɛ 1 ( v )), with x (ρ ɛ 1 + θɛ 1 ) = 0, u ɛ 1 x u ɛ 1 γ 1 x u ɛ 1 + xp ɛ 2 = 0, x u ɛ 1 = 0, u ɛ 1 xθ ɛ 1 γ 2 x θ ɛ 1 = 0, and suitable Dirichlet-type boundary conditions.

35 35 Ongoing and Future Work Steady problem in smooth domain(general smooth convex domain, annulus). Detailed structure of boundary layer(how does U ɛ depend on ɛ?). Higher dimensional problems. Boltzmann equation with time.

36 Thank you for your attention! 36