Dissipative effects of the diffuse reflection boundary condition in the kinetic theory

Transcription

1 Dissipative effects of the diffuse reflection boundary condition in the kinetic theory (joint work with Tai-Ping Liu and Li-Cheng Tsai Institute of Mathematics, Academia Sinica 2012 International Conference on Nonlinear Analysis Evolutionary P.D.E. and Kinetic Theory Taipei, November

2 Introduction Boltzmann equation F t + d i=1 F ζ i = 1 Q(F, F, F = F(, t, t > 0. x i k Free molecular equation F t + d i=1 ζ i F x i = 0, F = F(, t, t > 0. Initial data and domain F(, 0 = F in (, x D R d, ζ R 3, { } D x R d : x < 1, d = 1, 2, 3.

3 Introduction Diffuse reflection boundary condition ( 1 2π 2 F(y,ζ, t = j(y, tmtw(y(ζ for y D,ξ n(y > 0, RT w (y j(y, t = ξ n(yf(y,ζ, tdζ, ξ n(y<0 1 M Tw (ζ = (2πRT w 3 2 e ζ 2 2RTw, (1 T w (y : boundary temperature. R : Boltzmann constant. ξ = (ζ 1,...,ζ d. n(y : unit normal vector to the boundary, pointed to the gas.

4 Introduction Conservation of total mass ρ 0 D = F in ( dxdζ = D R 3 D R 3 F(, t dxdζ, t 0. The equilibrating effect of the diffuse reflection boundary condition and intermolecular collsion process = the solution converges to the steady state.

5 Introduction Conservation of total mass ρ 0 D = F in ( dxdζ = D R 3 D R 3 F(, t dxdζ, t 0. The equilibrating effect of the diffuse reflection boundary condition and intermolecular collsion process = the solution converges to the steady state. First, we focus on the boundary effect without intermolecualr collision and therefore consider the free molecular equation.

6 Introduction Conservation of total mass ρ 0 D = F in ( dxdζ = D R 3 D R 3 F(, t dxdζ, t 0. The equilibrating effect of the diffuse reflection boundary condition and intermolecular collsion process = the solution converges to the steady state. First, we focus on the boundary effect without intermolecualr collision and therefore consider the free molecular equation. Assume the simplest case : the boundary temperature T w (y = T 0 is constant. In this case, we expect the solution to approach the global Maxwellian ρ 0 M T0.

7 Free molecular flow with constant T w By subtracting ρ 0 M T0, integrating out the extra microscopic velocity (for d = 1, 2 and taking 2RT 0 = 1, [ f(x,ξ, t F(, t ρ0 M T0 (ζ ] { dζ2 dζ 3, for d = 1 R 3 d dζ 3, for d = 2 f(x,ξ, tdxdξ = 0, D R d M(ξ = (π d 2 e ξ 2. We can reduce the problem to: },

8 Free molecular flow with constant T w f d t + f ξ i = 0, f = f(x,ξ, t, t > 0, x i i=1 f(x,ξ, 0 = f in (x,ξ, x D R d, ξ R d. f(y,ξ, t = (4π 1 2 j(y, tm(ξ for y D,ξ n > 0, j(y, t = ξ nf(y,ξ, tdξ. ξ n<0 (2 (3

9 Free molecular flow with constant T w PDE Approach (Characteristic Method Given x and ξ, let y B = y B (x,ξ/ ξ be the point on the boundary obtained by tracing back from x along ξ/ ξ : f(x,ξ, t = (4π 1 2 j ( y B ( x, ξ ξ, t y B x ξ M(ξ, for y B x < t, ξ f in (x ξt,ξ, for y B x ξ > t. (4 ξ x y B

10 Free molecular flow with constant T w By (4, one can derive the integral equation for boundary flux j. The integral equation of boundary flux function j for d = 1: where j + (t = j (t = 2 t t t ξf in (1 ξt,ξ dξ + H(sj (t s ds 0 ξf in ( 1 ξt,ξ dξ + H(s 0 t ( 2 3 e ( 2 s 2 1l s {s>0}, H(sds = 1. 0 H(sj + (t s ds

11 Free molecular flow with constant T w The integral equation of boundary flux function j for d = 2, 3 : j(y, t = + e n<0 s<t ξ n<0 ξ < y y B t ( ξ n f in (x ξt,ξdξ G(α, s j (y B (y, e, t s ds where G(α, s 1 Put π d 1 2 H(s = ( 2 cosα cos α>0 s { dα, for d = 2 dω(e, for d = 3 }, d+2 ( 2 e 2 cosα s 1l {s>0,cosα>0}. { dα G(α, s dω(e }.

12 Free molecular flow with constant T w For the equations (2 and (3, we can construct the density function f from j by (4. Thus our focus will be on the boundary flux j instead of the solution f. For f to decay, we need to show the decay of j. Definition The apriori norm of j is a function of t defined by N(t = sup 0 s t where j ± (t = j(±1, t. ( (s+1 d j L y (s for d = 2, 3, ( (s+1 j + (s + j (s for d = 1,

13 Free molecular flow with constant T w Theorem (Deacy of boundary flux Fix any µ > 4. For all F in ( L,µ, the boundary flux j(y, t, (3, exists globally and decays to zero uniformly in y at a rate of (t + 1 d. More precisely, N(t = O(1 F in L,µ, t 0, where F in L,µ ess sup x D,ζ R 3 (1+ ζ µ F in (, The choice of the weighted norm is for f in L, f in x,ξ L 1, sup ξ f in (x,ξ dξ = O(1 F in x,ξ L,µ x D R d.

14 Free molecular flow with constant T w Corollary (Pointwise estimate of solution F(, t ρ 0 M(ζ = O(1 F in L,µ M(ζ (t + 1 d for 4 < ξ, t 1 for ξ < 4 t.

15 Free molecular flow with constant T w Corollary (Pointwise estimate of solution F(, t ρ 0 M(ζ = O(1 F in L,µ M(ζ (t + 1 d for 4 < ξ, t 1 for ξ < 4 t. Corollary (L p estimate of the solution F(,, t ρ 0 M L p = O(1 F in L,µ (t + 1 d p. Tetsuro Tsuji, Kazuo Aoki, and François Golse: J. Stat. Phys. (2010, numerical result for p = 1.

16 Free molecular flow with constant T w Idea of Proofs The global existence for boundary flux j with a crude estimate: j(y, t = O(1 F in L,µ e Ct. To obtain boundedness and the optimal decay rate of j, we need a crucial property, namely conservation law. It turns out that we need the fluctuation estimate. To estimate the flux fluctuation, we use the Law of Large Numbers and more detailed analysis on the derivative of the kernel G(α, s and H(s.

17 Free molecular flow with constant T w Idea of Proofs The global existence for boundary flux j with a crude estimate: j(y, t = O(1 F in L,µ e Ct. To obtain boundedness and the optimal decay rate of j, we need a crucial property, namely conservation law. It turns out that we need the fluctuation estimate. To estimate the flux fluctuation, we use the Law of Large Numbers and more detailed analysis on the derivative of the kernel G(α, s and H(s. Due to diffuse reflection boundary condition and symmetric geometry, we have the explicit stochastic formulation and thereby the explicit pointwise estimates of the solutions.

18 Free molecular flow with constant T w Idea of Proofs The global existence for boundary flux j with a crude estimate: j(y, t = O(1 F in L,µ e Ct. To obtain boundedness and the optimal decay rate of j, we need a crucial property, namely conservation law. It turns out that we need the fluctuation estimate. To estimate the flux fluctuation, we use the Law of Large Numbers and more detailed analysis on the derivative of the kernel G(α, s and H(s. Due to diffuse reflection boundary condition and symmetric geometry, we have the explicit stochastic formulation and thereby the explicit pointwise estimates of the solutions. Shih-Hsien Yu, Arch. Ration. Mech. Anal. 192, (2009

19 Free molecular flow with constant T w Conservation of total mass j(y, t = 1 ( j(y, tm(ξ 1 f(x,ξ, t dxdξ D (4π 1 2 D R d = 1 ( j(y, tm(ξ 1 f D (4π 1 in (x ξt,ξ dxdξ D ξ < y B x t ξ > y B x t ( j(y, t j(y B, t y B x M(ξdxdξ ξ

20 Free molecular flow with constant T w Split the integration domain D R d into four parts: { I = ξ < y B x }, initial value contribution,slow particles; t { yb x II = < ξ < y B x }, slow particles; t t/2 { } yb x y III = < ξ < B x, relatively slow particles; t/2 (log(t + 1/2 { } yb x IV = (log(t + 1/2 < ξ, residual componet, fluctuation of j.

21 Free molecular flow with constant T w Lemma I III IV = O(1 ( (t + 1 d F in L,µ +J(t, = O(1 log(t + 2 sup t 2 <s<t ( j L y (s, II = O(1 (t + 1 dj(t, = sup t (t log(t+1,t 2 y,y D j(y, t j(y, t. ( j L (s for d = 2, 3, y J(t sup ( 0<s<t j + (s + j (s for d = 1.

22 Free molecular flow with constant T w Then we have j(y, t = O(1 (t + 1 d F in L,µ + O(1 log(t + 2 J(t + sup j(y, t j(y, t t (t log(t+1,t 2 y,y D Need to estimate the boundary flux fluctuation. Since ( ( j(y, t j(y, t = j(y, t j(y, t + j(y, t j(y, t, we may consider temporal and spacial fluctuations separately.

23 Free molecular flow with constant T w Theorem (Temporal Fluctuation Estimate Let t < t, t /n 1, j(y, t j(y, t = O(1 nd+1 log(t + 2 ( (t + 1 d+1 F in L,µ +J(t ( log n ( + O(1 sup j L t y (s 2 <s<t n 1 2 (t t, for d = 2, 3, j ± (t j ± (t = O(1 n2 log(t + 2 ( (t F in L,µ +J(t ( + O(1 sup j + (s + j (s t 2 <s<t t t n.

24 Free molecular flow with constant T w Theorem (Spacial Fluctuation Estimate Let y, y D. Provided t/n 1, j(y, t j(y, t = O(1 nd+1 log(t + 2 ( (t + 1 d+1 F in L,µ +J(t ( log n + O(1 sup t 2 <s<t ( j L y (s n 1 2, for d = 2, 3, j ± (t j (t = O(1 n2 log(t + 2 ( (t F in L,µ ( ( log(t O(1 sup j + (s + j (s + t n 2 <s<t +J(t 1 log(t + 2.

25 Free molecular flow with constant T w By conservation law and the fluctuation estimate, we have ( j(y, t = O(1 [ + O(1 1 log(t nd+1 log(t + 2 (t + 1 d+1 1 log(t nd+1 log(t + 2 (t + 1 d+1 ( log n + n log(t n log(t + 2, for d = 2, 3 1 log(t + 2, for d = 1 F in L,µ ] J(t.

26 Free molecular flow with constant T w We now choose any r (0,(d + 1 1, say r = (d + 2 1, and set n = n(t = t r. Under this choice, there exists t > 0 such that ( j L (t, for d = 2, 3 ( y j + (t + j (t, for d = 1 O(1 F in L,µ This implies 1 2 J(t = O(1 F in L,µ J(t, for all t > t.

27 Free molecular flow with constant T w Now we have the boundedness of j. By fluctuation estimate and conservation law again, we have ( j(y, t = O(1 F in L,µ 1 (t + 1 d + nd+1 log(t + 2 (t + 1 d+1 ( 1 log n 2 +O(1 1 log(t + 2, for d = 2, 3 log(t n log(t log(t + 2, for d = 1 n 1 2 sup t 2 <s<t ( j L y (s.

28 Free molecular flow with constant T w Note that N(t sup ( j L (s = O(1 y t 2 <s<t (t + 1 d. As before, choose any r (0,(d and set n = n(t = t r. It follows that, for some t > 0, ( j L (t, for d = 2, 3 ( y j + (t + j (t, for d = 1 F in L,µ O(1 (t + 1 d This implies 1 2 N(t = O(1 F in L,µ. N(t (t + 1 d, for all t > t.

29 Free molecular flow with constant T w Optimal decay rate [j(y, tm(ξ f(x,ξ, t] dxdξ I [J(tM(ξ+ f in (x ξt,ξ ] dxdξ ξ < 2 t = O(1 (t + 1 d ( J(t+ F in L,µ Therefore, the obtained convergence rate, (t + 1 d, is optimal..

30 Free molecular flow with constant T w For d = 2, 3, j(y, t = j (0 (y, t+j (1 (y, t+j (2 (y, t+..., j (k (y, t = y (k B ξ(t s 1... s k D s s k <t ( ( ξ n k f in y (k B ξ(t s 1... s k,ξ k l=1 ( G(α l, s l ds l { dαl dω(e l } dξ,

31 Free molecular flow with constant T w ( j (k (y, t = + A 1 k l=1 A 2 ( ( ξ n k f in y (k B ξ(t s 1... s k,ξ ( G(α l, s l ds l { dαl dω(e l A 1 + A 2, where { A 1 0 < s s k < t } { 2 { } { t A 2 2 < s s k < t y (k B y (k B } dξ } ξ(t s 1... s k D, } ξ(t s 1... s k D.

32 Free molecular flow with constant T w A 1 F in L,µ ξ < 4 t F in L,µ =O(1 (t + 1 d+1. A 2 F in L,µ t 2 <s s k <t ξ dζ (1+ ζ µ ( k l=1 G(α l, s l ds l { dαl dω(e l ξ dζ (1+ ζ µ ( k { } dαl G(α l, s l ds l dω(e l l=1 t t 2 } H k (sds.

33 Free molecular flow with constant T w Theorem (Law of Large Numbers There exists some constant C > 0 such that, for any γ with γ/n 1 d+1 > C, H n (sds = P{ X X n n E[X 1 ] > γ} s n E[X 1 ] >γ = O(1 nd logγ γ d+1. For k n t, t 2 <s s k <t k l=1 G(α l, s l ds l { dαl dω(e l } = O(1 k d log(t + 2 (t + 1 d+1.

34 Free molecular flow with constant T w Theorem Assume t/n 1, for d = 2, 3, j(y, t = O(1 nd+1 log(t + 2 ( (t + 1 d+1 F in L,µ +J(t +Λ n (y, t. Λ n (y, t 0<s s n< t 2 j ( y (n B, t s 1... s n ( n l=1 G(α l, s l ds l { dαl dω(e l }.

35 Free molecular flow with constant T w Lemma For t > t n, y D, Λ n (y, t Λ n (y, t = O(1 sup For all t n, y, y D, t 2 <s<t ( j L y (s ( n d+1 log(t + 2 (t + 1 d+1 + ( log n Λ n (y, t Λ n (y, t = O(1 sup ( j L (s y t 2 <s<t n 1 2 (t t. ( log n n 1 2.

36 Free molecular flow with constant T w The Fundamental Theorem of Calculus For temporal fluctuation, Λ n (y, t Λ n (y, t = t Λ n t s (y, sds For spacial fluctuation, Λ n (y, t Λ n (y, t d = 1:the boundary is discrete = estimate directly. d = 2:use the polar coordinates = Λ n (θ, t Λ n (0, t d = 3:choose the z axis such that y, y lie on the equator and use cylindrical coordinates = Λ n (θ, 0, t Λ n (0, 0, t Λ n (θ, t Λ n (0, t = θ 0 Λ n θ (θ, tdθ

37 Free molecular flow with constant T w A direct computation yields ( Λ n (y, t t sup ( ( j L (s H t y n t 2 <s<t n G(α n 1, s 1 G s (α l, s l G(α n, s n d n s l=1 { d n α d n Ω(e It turns out that we need to estimate n G(α 1, s 1 G { s (α l, s l G(α n, s n d n d s n } α d n Ω(e l=1 ( = O (n log n 1 2. }.

38 Free molecular flow with constant T w For spacial fluctuation estimate, the symmetry of D is used. y (i B α i θ i y (i 1 B n Λ n (y, t = Λ n (θ, t = G(α l, s l s s n< t l=1 2 ( n j (θ +θ θ n, t s 1... s n dα l ds l l=1

39 Free molecular flow with constant T w Λ n (θ, t = Consequently, s s n< t 2 n l=1 G ( α l + θ 2n, s l ( n j (θ θ n, t s 1... s n dα l ds l. Λ n (θ, t θ ( j sup L (s y t 2 <s<t 1 n G(α 2n 1, s 1 G α (α l, s l G(α n, s n d n αd n s. l=1 l=1

40 Stochastic Process Forward stochastic process diffuse reflection x (2 B x (1 B diffuse reflection free transport β 1 V 1 (x in,ξ in β2 free transport ξ x in in V 2 (x in,ξ in x (3 B P{V l ξ + dξ} = (4π 1/2 ξ n(x (l B M(ξ1l{ ξ n(x (l B >0} dξ.

41 Stochastic Process The joint probability density function: { { }} [β,β + dβ] P S l [τ,τ + dτ], w l = w + dω(w { dβ G(β,τdτ dω(w The domain is spherical symmetric and the boundary temperature is uniform. Thus we can integrate out the dependence on reflected direction w l and obtain i.i.d. random variables X l by : ( P{X l [τ,τ + dτ]} = { dβ G(β,τ dω(w }. } dτ H(τdτ.

42 Boltzmann equation with variable T w Assume that T w (y is a measurable function with 0 < inf T w(y T T sup T w (y <. D D W.L.O.G, we may set T = 1, 1 F D in (dxdζ = 1, D R 3 M(ζ M T = (π 3 2 exp( ζ 2.

43 Boltzmann equation with variable T w Stationary free molecular equation d S ζ i = 0, x i i=1 ( 1 2π 2 S(y,ζ = j(y, tmtw(y(ζ, y D, ξ n > 0, RT w (y j(y, t = ξ ns(y,ζ dζ, ξ n<0 1 S(dxdζ = 1. D D R 3 S( = 1 ( 1 2π 2 MTw(y C S RT w (y B B (ζ.

44 Boltzmann equation with variable T w We expand F around S instead of M, F = S + Mφ: φ d t + φ ζ i 1 x i k Lφ = 1 ( k L S M M i=1 φ(y,ζ M(ζ = ( + 1 ( k M Q S M + Mφ, S M + Mφ. ( 1 2π 2 RT w (y ξ nφ(y,ζ, t M(ζ dζ M Tw(y(ζ, ξ n<0 (7 y D, ξ n > 0. (8 φ in Mdxdζ = 0. (9

45 Boltzmann equation with variable T w, S DFr t, S Fr t are the solution operators defined as the following: φ d t + φ ζ i + 1 x i k Lφ = 0, φ(, 0 = φ in( i=1 diffuse reflection boundary condition (8 φ d t + φ ζ i + 1 x i k νφ = 0, φ(, 0 = φ in( i=1 diffuse reflection boundary condition (8 d + S LB t F F ζ t i = 0, F(, 0 = F x in ( i i=1 diffuse reflection boundary condition (1

46 Boltzmann equation with variable T w Theorem (Free molecular flow If F in dxdζ = 0, S Fr t (F in ( = O(1 F in L,5 The main difficulties are: { M (t + 1 d 1l { ξ > 4 t} + 1l { ξ < 4 t} }.

47 Boltzmann equation with variable T w Theorem (Free molecular flow If F in dxdζ = 0, S Fr t (F in ( = O(1 F in L,5 The main difficulties are: { M (t + 1 d 1l { ξ > 4 t} + 1l { ξ < 4 t} The probability density functions G (α i, s i 2RT w (y (i B 2RT w (y (i B depend on the boundary temperature T w (y so that we don t have i.i.d. random variables. }.

48 Boltzmann equation with variable T w Theorem (Free molecular flow If F in dxdζ = 0, S Fr t (F in ( = O(1 F in L,5 The main difficulties are: { M (t + 1 d 1l { ξ > 4 t} + 1l { ξ < 4 t} The probability density functions G (α i, s i 2RT w (y (i B 2RT w (y (i B depend on the boundary temperature T w (y so that we don t have i.i.d. random variables. The spacial differentiations on G (α i, s i 2RT w (y (i B 2RT w (y (i B break the spacial fluctuation estimate. }.

49 Boltzmann equation with variable T w φ d t + φ ζ i = 1 x i k νφ i=1 With estimates of free molecular flow, comparison method + Duhamel s principle + characteristic method =

50 Boltzmann equation with variable T w φ d t + φ ζ i = 1 x i k νφ i=1 With estimates of free molecular flow, comparison method + Duhamel s principle + characteristic method = Theorem (Free molecular flow with damping Consider φ in L, a, 0 a 1, and φ in Mdxdξ = 0, S DFr ( t (φ in (1+ ζ a = O(1 φ in L, a 1 (t + 1 d 1l { ξ > 4 t} + 1l { ξ < 4 t} + t k We have the term t k because there is no conservation law for damped transport equation..

51 Boltzmann equation with variable T w For the linearized Boltzmann equation, iteration and t k 1 = local existence and estimates.

52 Boltzmann equation with variable T w For the linearized Boltzmann equation, iteration and t k 1 = local existence and estimates. Theorem (Local Existence for linearized Boltzmann equation Consider φ in L, a, 0 a 1. c > 0, for t k < c, S LB ( t (φ in (1+ ζ a = O(1 φ in L, a 1 (t + 1 d 1l { ξ > 4 t} + 1l { ξ < 4 t} + t k.

53 Boltzmann equation with variable T w For the linearized Boltzmann equation, iteration and t k 1 = local existence and estimates. Theorem (Local Existence for linearized Boltzmann equation Consider φ in L, a, 0 a 1. c > 0, for t k < c, S LB ( t (φ in (1+ ζ a = O(1 φ in L, a 1 (t + 1 d 1l { ξ > 4 t} + 1l { ξ < 4 t} + t k From the local estimate to the global estimate: Define F(t = sup 0 s t e ν 1 s k φ(s L, a, for 0 < ν 1 < infν(ζ Show F(t F(t ck, for all t 2ck. Then this implies F(t = O(1..

54 Boltzmann equation with variable T w Using characteristic method: φ(, t = e ν(ζc φ(x ckξ,ζ, t ck1l {τ>ck} ( 1 + j(y B, t τe ν(ζ τ 2π 2 k MT(yB RT(y B (ζ1l {τ<ck} + 1 k min{τ,ck} 0 e ν(ζ k s K(φ(s(x sξ,ζ, t sds For j(y B, t τ and K(φ(s(x sξ,ζ, t s, we trace back to the time t 2ck and use the local estimate to obtain the smallness in R.H.S..

55 Boltzmann equation with variable T w Recall the local estimate for linearized Boltzmann equation: ξ <ε 1 (t + 1 d 1l { ξ > 4 1 for t 1. t} For the part O(11l { ξ < 4, t 1, using t} K(ζ,ζ (1+ ζ a dζ = (1+ ζ a And we always require t k 1. O(ε, for d = 1 O(ε 2 logε, for d = 2 O(ε 2, for d = 3

56 Boltzmann equation with variable T w e ν 1 t k φ(t F(t ck (1+ ζ a 1,d = 1 e (ν 0 ν 1 c + Ce 2cν 1 log(ck + 2 1,d = 2 k ck c 2 1 1l {τ>ck},d = 3 ck + 1 1,d = 1 +Ce 2cν 1 1 (ck + 1 d + 1 log(ck + 2,d = 2 k ck c 1 1l {τ<ck},d = 3 ck + 1 for some positive constant C.

57 Boltzmann equation with variable T w Theorem (Linear Stability Consider φ in L, a, 0 a 1, φ in Mdxdζ = 0. There exists a C 1 > 0, such that for each 0 < ν 1 < ν 0 infν(ζ, S LB t (φ in L, a C φ in L, a e ν 1 k t, provided k C 1 (ν 0 ν 1 2 for d = 1, k log(k + 2 C 1(ν 0 ν for d = 2, k C 1 (ν 0 ν for d = 3, (10 where C is a constant independent of k, ν 1, and T.

58 Boltzmann equation with variable T w In terms of S LB t, the full Boltzmann equation (7 is equivalent to φ(t = S LB t 1 t S LB t s k 0 (φ in + 1 k t 0 ( S LB t s L( S M ds+ M ( Q(S M + Mφ, S M + Mφ M ds. (11 In view of (11, the equation for the steady state solution Φ = Φ( is Φ = 1 k 1 k 0 0 S LB s ( S LB s L( S M ds+ M ( Q(S M + MΦ, S M + MΦ ds. (12 M

59 Boltzmann equation with variable T w Using Picard iteration to solve stationary full Boltzmann equation: ( Q φ M,ψ M ν M = O(1 φ L ψ ζ L, ζ L ζ S( M(ζ = O(1 T M(ζ. k 1 = the linear global estimates 1 T 1 = (Φ (i Φ (i 1 = [O(1(1 T ] i+1

60 Boltzmann equation with variable T w Theorem (Nonlinear Stationary Solution Assume 1 T 1 and k 1, the steady state solution Φ of (7 exists, with Φ L = O(1 T, (13 D R 3 Φ( M(ζdxdζ = 0. (14

61 Boltzmann equation with variable T w We have already obtain the steady state solution F 0 S + MΦ for full Boltzmann equation. Moreover, from (14, F 0 dxdζ = 1. For general initial boundary value problem, we expand F around F 0 : F = F 0 + Mψ. The equation for ψ is ψ d ψ + ζ t i 1 x i k Lψ = 2 ( k M Q Mψ, F0 M i=1 + 1 k M Q(ψ M,ψ M, (15 ψ(, 0 = ψ in ( L, given, ψ in ( M(ζdxdζ = 0. D R 3

62 Boltzmann equation with variable T w Theorem (Nonlinear Stability For 0 < ν 2 < ν 0, assume (10 and (1 T + ψ in (ν 0 ν 2 2 1, the solution ψ of (15 exists with ψ(t L C ψ in L e ν 2 k t, where C is a constant independent of k, ν 2, T. S LB t (φ in L, a C φ in L, a e ν 1 k t, with ν 1 = ν 2 +ν 0 2 [ ψ (i ψ (i 1 = ψ in O(1 (1 T ] + ψ in i (ν 0 ν 1 2 e ν 1 k t, i 1.

63 Free molecular equation with variable T w = P { s < Z Z n < s } = s<s s n<s n l=1 { G (α l, s l 2RT w (y (l B dαl dω(e l σ s< σn <s 2RTw(y (1 B 2RTw(y (n B 2RT s<σ σ n< 2RT s n l=1 n l=1 ( G(α l,σ l 2RT w (y (l B } ds l { dαl dω(e l ( { dαl G(α l,σ l dω(e l } dσ l } dσ l = P{ 2RT s < X X n < 2RT s }.

64 Free molecular equation with variable T w Spacial fluctuation for the two dimensional case. Given two boundary points, y, y, let y be point of degree zero, and denote the polar angle of y by θ. Denote the relative polar angle of y (l B with respect to y(l 1 B by θ l, i.e. θ +θ θ l stands for the absolute polar angle of y (l B. Also we have θ l = π 2α l. Put T l = T w (θ +θ θ l, and T l = T w (θ θ l. y (l B θ l α l y (l 1 B

65 Free molecular equation with variable T w Λ m (y, t = Λ m (θ, t = j σ 2RT σ m T 2RT m< T t 2 m G(α l,σ l l=1 ( θ+θ θ m, t σ 1 2RT1... σm 2RTm d m σd m α j = σ σ m 2R T 1 2R Tm < T T t 2 ( θ θ m, t σ 1 m l=1 G ( α l + θ 2m,σ l 2R T 1... σm where T l = T w (θ+θ 1 + +θ m lθ m. 2R T m d m σd m α,

66 Free molecular equation with variable T w = Λ m (y, t Λ m (y, t = Λ m (θ, t Λ m (0, t σ σ m < T T 2R T 1 2R Tm t 2 m l=1 ( j θ θ m, t σ 1 j σ 1 2RT 1 G ( α l + θ 2m,σ l 2R T 1... σm σ m < T 2RT m T t 2 2R T m m G(α l,σ l l=1 d m σd m α ( θ θ m, t σ 1 2RT 1... σm 2RT m d m σd m α

67 Free molecular equation with variable T w U 1 = j σ σ m 2R T 1 2R Tm < T T t 2 m l=1 ( θ θ m, t σ 1 σ 1 2RT σ m < T 2RT m T t 2 G ( α l + θ 2m,σ l 2R T 1... σm can be estimated by Law of Large Numbers. 2R T m d m σd m α

68 Free molecular equation with variable T w { σ σ m < } T t 2RT T 2 σ σ m < 2R T 1 2R T m { T t T 2 σ 1 2RT σ m 2RT m < } T t T 2

69 Free molecular equation with variable T w U 2 = j σ 1 2RT 1 [ j σ m < T 2RT m T t 2 m l=1 ( θ θ m, t σ 1 G ( α l + θ 2m,σ l 2R T 1... σm 2R T m ( θ θ m, t σ 1 2RT 1... σm 2RT m ] d m σd m α σ σ m m E[X 1 ] < m = temporal fluctuation σ σ m m E[X 1 ] > m = Law of Large Numbers

70 Free molecular equation with variable T w U 3 = σ 1 2RT 1 j σ m < T 2RT m T t 2 ( m G ( α l + θ 2m,σ m l G(α l,σ l l=1 l=1 ( θ θ m, t σ 1 2RT 1... σm 2RT m d m σd m α just like the spacial fluctuation in the case of constant temperature.

71 Maxwell-type boundary condition Maxwell-type boundary condition: ( 1 2π 2 g(y,ξ, t = α j(y, tmt(y (ξ RT(y +(1 αg(y,ξ 2(ξ nn, t, y D,ξ n > 0, 0 < α < 1. For free molecular flow, we have g(x,ξ, t = g in (x ξt,ξ1l {t<τ1 }+ m α (1 α k 1 j k=1 ( y (k, t τ 1 kτ 2 ( 2π RT(y (k 1 2 M T(y(k (ξ k 1 } +(1 α m g in (y (m ξ m (t τ 1 (m 1τ 2,ξ m 1l {t>τ1 }, where m = ξ t x y 1 +1 y 1 y 2

72 Maxwell-type boundary condition j(y, t = n k n k=0 l=1 k 1 +k 2 + +k l =l { (1 α k j (k in (y, t (1 α k k 1 k l j (k k 1 k l in (y (k1,,k l, t s 1 s l l 2RT(y (1 α k (k1,,k i i αg φ s i 2RT(y (k1,,k i, i ds i dφ i i=1 k i k i +

73 Maxwell-type boundary condition j(y, t = n k n k=0 l=1 k 1 +k 2 + +k l =l { (1 α k j (k in (y, t (1 α k k 1 k l j (k k 1 k l in (y (k1,,k l, t s 1 s l l 2RT(y (1 α k (k1,,k i i αg φ s i 2RT(y (k1,,k i, i ds i dφ i i=1 k i k i + j(y, t = O(1 1 α e 1 α 1 (1+αt d, for d = 1, 2.

74 Thank you for your attention!!