The Stationary Boltzmann equation for a two component gas in the slab.

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1 The Stationary oltzmann equation for a two component gas in the slab. Stéphane rull* Abstract The stationary oltzmann equation for hard forces in the context of a two component gas is considered in the slab. An L 1 existence theorem is proved when one component satisfies a given indata profile and the other component satisfies diffuse reflection at the boundaries. Weak L 1 compactness is extracted from the control of the entropy production term. Trace at the boundaries are also controled. *CMI, University of Aix-Marseille I, 39 rue Joliot-Curie, Marseille Cedex 13, France. 1 Introduction. Consider the stationary oltzmann problem in a slab for a two component gas ξ x f A(x, v) = Q(f A, f A + f )(x, v), (1.1) ξ x f (x, v) = Q(f, f A + f )(x, v), (1.2) x [, 1], v R 3. The collision operator Q is the oltzmann operator Q(f, g)(x, v) = (v v, ω)[f g fg ]dωdv, R 3 S = Q + (f, g)(x, v) Q (f, g)(x, v), 1

2 where Q + (f, g) Q (f, g) is the splitting into gain and loss term, f = f(x, v ), f = f(x, v ), f = f (x, v ) v = v (v v, ω)ω, v = v + (v v, ω)ω. For a more general introduction to the oltzmann equation for multicomponent gases see ([6]). The velocity component in the x-direction is denoted by ξ, and (v v, ω) denotes the Euclidean inner product in R 3. Let ω be represented by the polar angle (with polar axis along v v ) and the azimutal angle φ. The function (v v, ω) is the kernel of the collision operator Q taken for hard forces as v v β b(θ), with 0 β < 2, b L 1 +([0, 2π]), b(θ) c > 0 a.e. The boundary condition for the A component is the given indata profile f A (, v) = km (v), ξ > 0, f A (1, v) = km + (v), ξ < 0, (1.3) for some positive k. The boundary condition for the component is of diffuse reflection type f (, v) = ( ξ f (, v )dv )M (v), ξ > 0, (1.4) ξ <0 f (1, v) = ( ξ f (1, v )dv )M + (v), ξ < 0. ξ >0 M + and M are given normalized Maxwellians M (v) = 1 2πT 2 e v 2 2T and M + (v) = 1 2πT+ 2 Denote the collision frequency by ν(x, v) = (v v, ω)f(x, v )dv dω. S e v 2 2T +. In the case of two component gases, for the GK equation, some results are obtained in ([9],[10]) from a numerical point of view, when k = 1. The physical context is described in those papers. f A represents the density of a 2

3 vapor and f the density of a noncondensable gas. The case of multicomponent gases for the oltzmann operator in the slab is investigated in ([11], [12], [13]), when the noncondensable gas becomes negligible. It is proved that the noncondensable gas accumulates in a thin Knudsen layer at the boundary. In this paper, weak solutions (f A, f ) to the stationary problem in the sense of Definition 1.1 will be considered. Definition 1.1. Let M A and M be given nonnegative real numbers. (f A, f ) is a weak solution to the stationary oltzmann problem with the β-norms M A and M, if f A and f L 1 loc ((, 1) R3 ), ν L 1 loc ((, 1) R3 ), (1 + v ) β f A (x, v)dxdv = M A, (1 + v ) β f (x, v)dxdv = M, and there is a constant k > 0 such that for every test function ϕ Cc 1 ([, 1] R 3 ) such that ϕ vanishes in a neiborhood of ξ = 0, and on {(, v); ξ < 0} {(1, v); ξ > 0}, 1 = k ξm + (v)ϕ(1, v)dv k R 3, 1 ϕ (ξf A R 3 x + Q(f A, f A + f )ϕ)(x, v)dxdv ξm (v)ϕ(, v)dv, R 3, ϕ (ξf R 3 x + Q(f, f A + f )ϕ)(x, v)dxdv, = ξ M + (v)ϕ(1, v)dv( ξ f (1, v )dv ) ξ <0 ξ >0 ξm (v)ϕ(, v)dv( ξ f (, v )dv ). ξ >0 ξ <0 The main result of this paper is the following Theorem 1.1. Given β with 0 β < 2, there is a weak solution to the stationary problem with β-norms equal to one. In the case of one component, an analogous theorem with boundary conditions of type (1.3) is proved in [1]. In the case of boundary conditions of type (1.4), an analogous theorem is also shown in [2]. The case of the Povzner equation for a one component gas with diffuse-reflection boundary in the case of hard and soft forces is investigated in ([14]). For diffuse-reflection boundary conditions in n dimensions, the biting lemma is used in [8] to obtain (1.4). In the simpler one-dimensional frame of this paper, (1.4) is obtained by a weak L 1 compactness argument from a control of the entropy outflow. 3

4 The second section of this paper is devoted to a construction of approximate solutions to the problem with a modified non symmetric collision operator. The proofs are performed by monotonicity arguments which give the uniqueness of the successive approximate solutions. In the third section, the symmetry of the collision operator is re-introduced. The weak compactness in L 1 is used in this step by noticing that the sum of the two components satisfies the stationary oltzmann equation in the slab. Section 3 performs the passage to the limit in the traces. In section 4, some extensions of Theorem 1.1 are made, in particular, the case where M A and M have any positive values. 2 Approximations with fixed total masses Let r > 0, m N, µ > 0, α > 0, N. Here χ r,m is a C0 function with range [0, 1] invariant under the collision transformation J, where J(v, v, ω) = (v, v, ω), with χ r,m also invariant under the exchange of v and v and such that χ r,m (v, v, ω) = 1, min( ξ, ξ, ξ, ξ ) r, and χ r,m (v, v, ω) = 0, max( ξ, ξ, ξ, ξ ) r 1 m. The modified collision kernel is a positive C function approximating min(, µ), when n v 2 + v 2 < 2, and v v v v ω > 1 m, and v v v v ω < 1 1 m (v, v, ω) = 0, if v 2 + v 2 > n or v v v v ω < 1 2m, or v v v v ω > 1 1 2m. The functions ϕ l are mollifiers in the x-variable defined by ϕ l (x) := lϕ(lx), where ϕ C 0 (), support(ϕ) (, 1), ϕ 0, 1 ϕ(x)dx = 1. In order to use a fixed point theorem, consider the closed and convex subset of L 1 +([, 1] ), K = {f L 1 +([, 1] ), min(µ, (1 + v ) β )f(x, v)dxdv = 2}. [,1] 4

5 A non negative function f K and θ [0, 1] being given, let us construct the solutions F A and F of the boundary value problem αf A + ξ x F A = χ r,m F A S 1 + F (x, v ) f ϕ A+F 1 + f ϕ (x, v )dv dω F A χ r,m f ϕ S 1 + f ϕ (x, v )dv dω, (x, v) (, 1), (2.1) F A (, v) = λm (v), ξ > 0, F A (1, v) = λm + (v), ξ < 0, and αf + ξ x F = χ r,m F S 1 + F (x, v ) f ϕ A+F 1 + f ϕ (x, v )dv dω F χ r,m f ϕ S 1 + f ϕ (x, v )dv dω, (x, v) (, 1), (2.2) F (, v) = θλm (v), ξ > 0, F (1, v) = (1 θ)λm + (v), ξ < 0. The number λ > 0 will be fixed in (2.13). Let us prove that the solution to (2.1, 2.2) is the monotone L 1 limit of the sequences (F l A ) l N and (F l ) l N defined by αf l+1 A + ξ x F l+1 A = F l+1 A and αf l+1 S χ r,m F l A 1 + F l A +F l F 0 A = 0, (y, v ) f ϕ 1 + f ϕ (x, v )dv dω χ r,m f ϕ S 1 + f ϕ (x, v )dv dω, (x, v) [, 1], (2.3) + ξ x F l+1 = F l A(, v) = λm (v), ξ > 0, F l A(1, v) = λm + (v), ξ < 0, S2 χ r,m F l 1 + F l A +F l F 0 = 0, (x, v ) f ϕ 1 + f ϕ (x, v )dv dω F l+1 χ r,m f ϕ S2 1 + f ϕ (x, v )dv dω, (x, v) [, 1], (2.4) F l+1 (, v) = θ lλm (v), ξ > 0, F l+1 (1, v) = (1 θ l)λm + (v), ξ < 0. 5

6 The sequences (FA l ) l N and (F l ) l N are well defined, since they solve linear equations. They are nonnegative, for l 1. Let F l = FA l + F l. It satisfies F 0 = 0, αf l+1 + ξ R3v S x F l+1 = χ r,m F l (x, v ) f ϕ 1 + F l 1 + f ϕ (x, v )dv dω F l+1 χ r,m f ϕ S 1 + f ϕ (x, v )dvdω, (x, v) [, 1], (2.5) F l+1 (, v) = (θ l + 1)λM (v), ξ > 0, F l+1 (1, v) = (2 θ l )λm + (v), ξ < 0. First F 1 0. Write the equation (2.5) in the exponential form, F l+1 α 1+x ξ R 0 = (1 + θ l )λm (v)e 1+x ξ R3v S χ r,m F l And so, F l 0 1+x ξ e R S χr,m f ϕ 1+ f ϕ αs R 0 s R S χr,m f ϕ 1+ f ϕ (x+τξ,v )dv dωdτ (x+τξ,v )dv dωdτ (2.6) (x + sξ, v ) f ϕ 1 + f ϕ (x + sξ, v )dv dvdωds, ξ > 0. F l+1 (x, v) F l (x, v) = 0 1+x ξ R3v S χ r,m ( F l e αs R 0 s R S χr,m f ϕ 1+ f ϕ 1 + F l (x+τξ,v )dv dωdτ (x + sξ, v ) F l (x + sξ, v ) ) (2.7) 1 + F l f ϕ 1 + f ϕ (x + sξ, v )dv dωds, ξ > 0. It follows that (F l ) l N is a nondecreasing sequence for ξ > 0. Moreover, being compactly supported because of the truncation for v 2 + v 2 > n, (F l ) l N is a bounded sequence. Hence, (F l ) l N converges a.e to some F in a nondecreasing way. Passing to the limit in the equation (2.7) when 6

7 l +, α 1+x ξ R 0 F (x, v) = (1 + θ)λm (v)e 1+x ξ x ξ e R S χr,m f ϕ 1+ f ϕ αs R 0 s R S χr,m f ϕ 1+ f ϕ (x+τξ,v )dv dωdτ (x+τξ,v )dv dωdτ χ r,m F S 1 + F (x + sξ, v ) f ϕ 1 + f ϕ (x + sξ, v )dv dvdωds, ξ > 0. For the same reasons, (2.8) F (x, v) = (2 θ)λm + (v)e x ξ e αs R 0 s α 1 x ξ R 0 1 x RR 3 ξ v S χr,m f ϕ 1+ f ϕ R S χr,m f ϕ 1+ f ϕ (x+τξ,v )dv dωdτ (x+τξ,v )dv dωdτ χ r,m F S 1 + F (x + sξ, v ) f ϕ 1 + f ϕ (x + sξ, v )dv dvdωds, ξ < 0. Let us show that FA l is a converging sequence. Consider QA+ l defined by Q A+ l = χ r,m F S 1 + F (x + sξ, v ) f ϕ 1 + f ϕ (x + sξ, v )dv dvdω. y using that F A F and a convolution with respect to the v variable ([2], [7]), it holds that Q A+ l is strongly compact in L 1. So, (FA l ) l N converges to some F A in L 1. For the same reasons, (F l ) l N converges to some F. Passing to the limit in (??), we find that there are F A and F solutions to (2.1) and (2.2). Lemma 2.1. The equations (2.1) and (2.2) each has a unique solution which is strictly positive. Proof of Lemma 2.1. Let F A and G A be two solutions to the equation (2.1). Consider ψ ε, the approximation of the sign function, and φ ε defined by φ ε (x) = ε + x 2 a primitive of ψ ε. Subtract the equation satisfied by G A to the equation satisfied by F A, multiply by ψ ε (F A (x, v) G A (x, v)) and (2.9) 7

8 integrate on [, 1], 1 1 α[f A G A ](y, v)ψ ε (y, v)dy + φ ε (F A (1, v) G A (1, v)) χ r,m F A G A S (1 + F A+F )(1 + G A+G ) (y, v ) f ϕ 1 + f ϕ (y, v )dv dωdy 1 ψ ε [F A (y, v) G A (y, v)] χ r,m f ϕ S 1 + f ϕ (y, v )dv dωdy. (2.10) Passing to the limit when ε tends to 0, in (2.10), 1 1 α [F A G A ](y, v) dy + ξ [F A G A ](1, v) χ r,m F A G A S2 (1 + F A+F )(1 + G A+G ) (y, v ) f ϕ 1 + f ϕ (y, v )dv dωdy 1 [F A G A ](y, v) χ r,m f ϕ R3 v S f ϕ (y, v )dv dωdy. (2.11) Integrating (2.11) on, As, 1 1 α [F A G A ](y, v) dydv + ξ [F A G A ](1, v) dv χ r,m F A G A R3 v S2 (1 + F A+F )(1 + G A+G ) (y, v ) f ϕ 1 + f ϕ (y, v )dv dωdydv 1 [F A G A ](y, v) χ r,m f ϕ R3 v S 1 + f ϕ (y, v )dv dωdydv. (2.12) [F A G A ](y, v) (1 + F A+F F A G A )(1 + G A+G ) (y, v), the right-hand side of the equation (2.12) is negative. Then, for ξ > 0, 1 α [F A G A ](y, v) dydvdydv 0. Analogously, the same result holds for ξ < 0. So, F A = G A. Similarly, the equation (2.2) has a unique solution. This proves Lemma

9 Let f A = f = F A min(µ, (1 + v ) β )F A (x, v)dxdv, F min(µ, (1 + v ) β )F (x, v)dxdv. The functions f A and f are well defined since F A and F strictly positive. Indeed, writing the equations (2.1) and (2.2) in the exponential form, α 1+x ξ R 0 F A (x, v) λm (v)e 1+x ξ F A (x, v) λm + (v)e α 1 x ξ R 0 1 x ξ So using that 1 (α + ν(x, v))dx 2 + 2µ, R S2 χ r,m f ϕ 1+ f ϕ R S2 χ r,m f ϕ 1+ f ϕ (x+τξ,v )dv dωdτ, ξ > 0, (x+τξ,v )dv dωdτ, ξ < 0. Analogously, F A (x, v) λm (v)e 2+2µ ξ, ξ > 0, F A (x, v) λm + (v)e 2+2µ ξ, ξ < 0. Let λ be defined by F (x, v) θλm (v)e 2+2µ ξ, ξ > 0, F (x, v) (1 θ)λm + (v)e 2+2µ ξ, ξ < 0. 1 λ = min( ; M (v)min(µ, (1 + v ) β )e 2+2µ ξ dv And so, 1 ). M +(v)min(µ, (1 + v ) β )e 2+2µ ξ dv min(µ, (1 + v ) β )F A (x, v)dxdv 1 9

10 and min(µ, (1 + v ) β )F (x, v)dxdv 1. The functions f A and f are solutions to αf A + ξ x f A = χ r,m S2 f ϕ f A and S2 χ r,m f A (, v) = f A (1, v) = 1 + f ϕ f A 1 + F (x, v ) f ϕ 1 + f ϕ (x, v )dv dω (x, v )dv dω, (x, v) (, 1), (2.13) λ min(µ, (1 + v ) β )F A (x, v)dxdv M (v), ξ > 0, λ min(µ, (1 + v ) β )F A (x, v)dxdv M +(v), ξ < 0, αf + ξ x f = χ r,m f (v, v, ω) R F (x, v ) f ϕ v S2 1 + f ϕ (x, v )dv dω f (x, v) χ r,m f ϕ S2 1 + f ϕ (x, v )dv dω, (x, v) (, 1), (2.14) λ f (, v) = min(µ, (1 + v ) β )F (x, v)dxdv θm (v), ξ > 0, f (1, v) = Recall that F is solution to λ min(µ, (1 + v ) β )F (x, v)dxdv (1 θ)m +(v), ξ < 0. αf + ξ x F = χ r,m F R F (x, v ) f ϕ v S2 1 + f ϕ (x, v )dv dω F χ r,m f ϕ S2 1 + f ϕ (x, v )dv dω, (x, v) (, 1), (2.15) F (, v) = (θ + 1)λM (v), ξ > 0, F (1, v) = (2 θ)λm + (v), ξ < 0. A fixed point argument will now be used in order to solve (2.13, 2.14) with f = f A + f. Define T on K [0, 1] by T (f, θ) = (f A + f, θ) with θ = ξ f (, v)dv ξ f (, v)dv + ξf (2.16) (1, v)dv 10

11 where (f A, f ) is solution to (2.13, 2.14). Lemma 2.2. T is a continuous and compact map from K [0, 1] into itself. Proof of Lemma 2.2. It is clear that T maps K [0, 1] into itself. Let T A, T, and T respectively map f K into f A, f, F solutions to (2.13), (2.14, (2.17). First, T is compact in L 1. Indeed, let f l be a bounded sequence in L 1 ([, 1] ). Using the following exponential form of F l and F l = T (f l ), α 1+x F l ξ R 0 1+x ξ (x, v) = λ(1 + θ l )M (v)e + 0 with 1+x ξ α 1+x ξ R 0 s e Q + l (x, v) = R S2 χ r,m f l ϕ m,n,µ 1+ fl ϕ S2 χ r,m F l 1 + F l R S2 χ r,m f l ϕ m,n,µ 1+ fl ϕ (x+τξ,v )dτdv dω Q + l (x, v ) f l ϕ (x, v 1 + f l ϕ )dv dω. ecause of the convolution of f l by ϕ in the x variable, ( S2 χ r,m f l ϕ 1 + f l ϕ (x + τξ, v )dτdv dω) (x+τξ,v )dτdv dω (x + sξ, v)ds, (2.17) ξ > 0, is strongly compact in L 1. Using a convolution by a mollifier in the v variable([7], [2]), the gain term Q + l is strongly compact in L 1. Hence, F l is strongly compact in L 1 and so T is a compact map. Let us show that T is a continuous map. Let f l be a converging sequence in L 1 ([, 1] ). Then, f l is a bounded sequence in L 1. y compactness of T, F l has a subsequence still denoted F l converging to some F. F l satisfies αf l + ξ R3v S2 x F l = χ r,m F l (x, v ) f l ϕ (x, v 1 + F l 1 + f l ϕ )dv dω F l χ r,m f l ϕ (x, v S2 1 + f l ϕ )dv dω, (x, v) (, 1), (2.18) F φ(l) (, v) = (θ l + 1)λM (v), ξ > 0, F φ(l) (1, v) = (2 θ l )λm + (v), ξ < 0. 11

12 Writing (2.18) in the exponential form, α 1+x F l ξ R 0 1+x ξ = (1 + θ l )λm (v)e + 0 R3v S2 χ r,m F l 1+x ξ 1 + F l e R S2 χ r,m f l ϕ m,n,µ 1+ fl ϕ αs R 0 s R S2 χ r,m f l ϕ m,n,µ 1+ fl ϕ (x+τξ,v )dv dωdτ (x+τξ,v )dv dωdτ (x + sξ, v ) f l ϕ (x + sξ, v 1 + f l ϕ )dv dvdωds, ξ > 0. Furthermore, F φ(l) F and f φ(l) ϕ f ϕ in L 1 ([, 1] ). So, there is a subsequence to F φ(l) still denoted by F φ(l) converging a.e. to F. Furthermore, χ r,m (v, v, ω) F l 1 + F l (y, v ) f l ϕ (y, v 1 + f l ϕ ) 2 χ r,m (v, v, ω). So, by the Lebegue theorem applied to (2.19), F is solution to αf + ξ x F = χ r,m F R F (x, v ) f ϕ v S2 1 + f ϕ (x, v )dv dω, F χ r,m f ϕ S2 1 + f ϕ (x, v )dv dω, (x, v) (, 1), (2.20) F (, v) = (θ + 1)λM (v), ξ > 0, F (1, v) = (2 θ)λm + (v), ξ < 0. Reasoning as in the proof of Lemma 2.1, (2.20) is proved to have a unique solution. Then, the whole sequence (F l ) l N converges to the solution to the equation (2.20). This proves that T is continuous. For analogous reasons, T A and T are continuous and compact maps. Therefore, the first component of T, T A + T, is a compact and continuous map. It is clear that the second component of T is compact. So, it remains to show its continuity. Consider f l converging to f in L 1 and let us show that θ l converges to θ. Recall that θ is defined in (2.16). Thanks to the truncation in the ξ variable ( R 3 ξ2 f l (x, v)dv) is uniformly bounded in x. And so, Q l (f l v, f l ) converges to Q(f, f) in L 1. Consider a continuously differentiable function ϕ defined [, 1] with values in [0, 1] such that ϕ() = 0 and ϕ(1) = 1. f l satisfies (2.19) 12

13 the equation (2.14), 1 ξ x [(f l (x, v) f (x, v))ϕ(x)]dx = Since ϕ and ϕ are bounded, it follows that ξ(f(1, l v) f (1, v)) dv c + c + c α (Q(f l, f l ) Q(f, f))ϕ(x)dx (f l f )(x, v)ξ x ϕ(x)dx (f l f )(x, v)ϕ(x)dx. (Q(f l, f l ) Q(f, f)) dxdv (f l f )(x, v) dxdv (f l f )(x, v) dxdv. Q(f l, f l ) and f l converge respectively to Q(f, f) and f in L 1 ([, 1] ). So, f l (1,.) tends to f (1,.) in L 1. Analogously, f l (,.) tends to f (,.) in L 1. Then, ξf l (1, v)dv tends to ξf (1, v)dv. In the same way, ξ f l (, v)dv tends to ξ f (, v)dv. Hence, the sequence ( θ l ) converges to θ. Finally, T is a compact and continuous map from the convex closed set K [0, 1] into itself. It follows from the Schauder fixed point theorem that there is (f, θ) with f = f A + f θ = ξ f (, v)dv ξf (1, v)dv + ξ f (, v)dv that satisfy αf A + ξ x f A = χ r,m f A R F (x, v ) f ϕ l v S2 1 + f ϕ (x, v )dv dω l f A χ r,m f ϕ l S2 1 + f ϕ (x, v )dv dω, (x, v) (, 1), (2.21) l f A (, v) = k A M (v), ξ > 0, f A (1, v) = k A M + (v), ξ < 0 with 13

14 k A = λ min(µ, (1 + v ) β )F A (x, v)dxdv and αf + ξ x f = χ r,m f R F (x, v ) f ϕ l v S2 1 + f ϕ (x, v )dv dω l f χ r,m f ϕ l S2 1 + f ϕ (x, v )dv dω, (x, v) (, 1), l f (, v) = λ ( ξ f (, v)dv ξf (1, v)dv + ξ f (, v)dv )M (v), ξ > 0, f (1, v) = λ ( (1, v)dv ξf (1, v)dv + ξ f (, v)dv )M +(v), ξ < 0, with λ = λ min(µ, (1 + v ) β )F (x, v)dxdv. Keeping, l,, r, m, µ fixed, denote f,α,l,r,m,µ by f α and study the passage to the limit when α tends to 0. Writing the equations (2.21, 2.20) in exponential form and using the averaging lemmas together with a convolution with a mollifier ([2],[7]) give that fa α and F α are strongly compact in L 1 ([, 1] ). Denote by f A and F the limits of fa α and F α. The passage to the limit when α tends to 0 in the equation (2.21) yields ξ x f A = χ r,m f A S 1 + F (x, v ) f ϕ l 1 + f ϕ (x, v )dv dω l f A χ r,m f ϕ l S 1 + f ϕ (x, v )dv dω, (x, v) (, 1), l λ f A (, v) = min(µ, (1 + v ) β )F A (x, v)dxdv M (v), ξ > 0, (2.22) f A (1, v) = λ min(µ, (1 + v ) β )F A (x, v)dxdv M +(v), ξ < 0, with min(µ, (1 + v ) β )f A (x, v)dxdv = 1. 14

15 For the same reasons, the limit f of f α satisfies ξ x f = χ r,m f R F (x, v ) f ϕ l v S2 1 + f ϕ (x, v )dv dω l f χ r,m f ϕ l S2 1 + f ϕ (x, v )dv dω, (x, v) (, 1), (2.23) l f (, v) = σ()λ M (v), ξ > 0, f (1, v) = σ(1)λ M + (v), ξ < 0, with min(µ, (1 + v ) β )f (x, v)dxdv = 1, where and σ() = σ(1) = λ = ξ f (, v)dv ξf (1, v)dv + ξ f (, v)dv, ξf (1, v)dv ξf (1, v)dv + ξ f (, v)dv λ min(µ, (1 + v ) β )F (x, v)dxdv. The passage to the limit in (2.22) and in (2.23), when l tends to is similar and implies that the limits f A and f of fa l and f l are solutions to ξ x f A = χ r,m f A R F (x, v f ) v S2 1 + f (x, v )dv dω f A χ r,m f S2 1 + f (x, v )dv dω, (x, v) (, 1), (2.24) f A (, v) = k A M (v), ξ > 0, f A (1, v) = k A M + (v), ξ < 0, with min(µ, (1 + v) β )f A (x, v)dxdv = 1, where k A is defined in (2.21) before passing to the limit and ξ x f = χ r,m f R F (x, v f ) v S2 1 + f (x, v )dv dω f χ r,m f S2 1 + f (x, v )dv dω, (x, v) (, 1), (2.25) f (, v) = σ()λ M (v), ξ > 0, f (1, v) = σ(1)λ M + (v), ξ < 0, 15

16 min(µ, (1 + v ) β )f (x, v)dxdv = 1, with, σ() = σ(1) = ξ f (, v)dv ξf (1, v)dv + ξ f (, v)dv, ξf (1, v)dv ξf (1, v)dv + ξ f (, v)dv, and F = F A +F, with F A and F defined by (2.1, 2.2) after passing to the limit. We are going to pass to the limit when tends to the infinity (2.24, 2.25). The sequences of solutions (f A ) and (f ) to (2.24) and (2.25) are weakly compact in L 1 ([, 1] ). Indeed, f = f A + f satisfies the oltzmann equation for a one component gas. y using the entropy production term ([2]), we find that f is weakly compact in L 1 ([, 1] ). The weak L 1 compactness of f A and f follows from 0 f A f and 0 f f. Denote by f A and f the limits of f A and f, up to subsequences. Furthermore, Q + (f, f ) and Q (f, f ) are weakly compact in L 1 ([, 1] ) ([2]). So, by the inequalities Q + (f A, f ) Q + (f, f ), Q + (f, f ) Q + (f, f ), Q (f A, f ) Q (f, f ) and Q (f, f ) Q (f, f ), the collision terms Q + (f, f ), Q (f A, f ), Q (f A, f ) and Q (f, f ) are weakly compact in L 1 ([, 1] ). So, f A and f satisfying (2.24, 2.25), ξ x f A and ξ x f are weakly compact in L1 ([, 1] ). Pass to the limit when + in the weak formulation of (2.25), [,1] ξ x ϕ(x, v)f (x, v)dxdv + χ r,m ϕ(x, v) [,1] S2 f ( (x, v ) 1 + F A +F = σ ()λ f 1 + f (x, v ) f f (x, v) 1 + f (x, v ))dv dωdxdv (2.26) ξf (1, v)ϕ(1, v)dv ξf (, v)ϕ(, v)dv ξm (v)ϕ(, v)dv + σ (1)λ ξm + (v)ϕ(1, v)dv, 16

17 where σ () = σ (1) = ξ f (, v)dv ξ f (, v)dv + ξf, (1, v)dv ξf (1, v)dv ξ f (, v)dv + ξf, (1, v)dv and ϕ is a test function belonging to C 1 c ([, 1] ). First, lim [,1] = lim χ r,m ϕ(x, v) S2 [,1] f 1 + F A +F (x, v ) f 1 + f (x, v )dxdvdv dω S χ r,m ϕ(x, v )f (x, v)f (x, v )dxdvdv dω. χ r,m being bounded, f and ξ x f being weakly compact in L 1 ([, 1] S), an averaging lemma applies, so that ( R χ r,m 3 m,n,µ f (x, v )ϕ(x, v )dv dω) N converges to v S2 R χ r,m 3 m,n,µ f(x, v )ϕ(x, v )dv dω in L 1 ([, 1] ). Furthermore, v S2 because of the truncation in the ξ variable for small velocities, this sequence is bounded. f and ξ x f being weakly compact in L1 ([, 1] ), using once more an averaging lemma ( χ r,m f S (x, v)f (x, v )ϕ(x, v )dv dωdxdv) N [,1] converges to [,1] S2 χ r,m f (x, v)f(x, v )ϕ(x, v )dv dωdxdv. So, Q (f, f ) converges to Q(f, f) in L 1. Hence, reasoning as in the proof of Lemma 2.2, we find that σ () (resp σ (1)) tends to σ() ( resp σ(1)), when tends to infinity. y passing to the limit when tends to infinity in (2.26), it holds that there is f r,µ solution to f r,µ ξ x f r,µ = χ r,m f r,µ S (x, v )f r,µ (x, v )dv dω S2 χ r,m f r,µ (x, v )dv dω, (x, v) (, 1), (2.27) f r,µ (, v) = σ()λ M (v), ξ > 0, f r,µ (1, v) = σ(1)λ M + (v), ξ < 0, 17

18 with Here, σ() = min(µ, (1 + v ) β )f r,µ (x, v)dxdv = 1. ξf r,µ r,µ ξ f (, v)dv r,µ ξ f (, v)dv (1, v)dv + and σ(1) = ξf r,µ r,µ ξf (1, v)dv r,µ. ξ f (, v)dv (1, v)dv + Moreover, so that, ξf r,µ (1, v)dv ξf r,µ (, v)dv = 0, ξ f r,µ (1, v)dv + ξ f r,µ (, v)dv = λ. Hence, the boundary conditions in (2.27) write f r,µ (, v) = ξ f r,µ (, v)dvm (v), ξ > 0, f r,µ (1, v) = ξ f r,µ (1, v)dvm +(v), ξ < 0. The passage to the limit when m + and n + in the equation (2.27) is performed as before and implies that there is f r,µ satisfying, ξ x f r,µ = χ r µ (v v, ω)f r,µ (x, v )f r,µ (x, v )dv dω S2 f r,µ with χ r µ (v v, ω)f r,µ (x, v )dv dω, (x, v) (, 1), S2 f r,µ (, v) = M (v) ξ f r,µ (, v)dv, ξ > 0, (2.28) f r,µ (1, v) = M +(v) ξf r,µ (1, v)dv, ξ < 0, 18

19 min(µ, (1 + v ) β )f r,µ (x, v)dxdv = 1. For the same reasons, there is f r,µ A satisfying, ξ x f r,µ A = χ r µ (v v, ω)f r,µ A (x, v )f r,µ (x, v )dv dω S 2 f r,µ A S2 χ r µ (v v, ω)f r,µ (x, v )dv dω, (x, v) (, 1), f r,µ A (, v) = k AM (v), ξ > 0, f r,µ A (1, v) = k AM + (v), ξ < 0, (2.29) with min(µ, (1 + v ) β )f r,µ A (x, v)dxdv = 1, where k A is defined in the equation (2.21) before passing to the limit. 3 End of the proof of the main theorem Let (r ) N with lim + r = 0 and (µ ) N with lim + µ = +, f A = f r,µ A and f = f r,µ. The passage to the limit when + is now performed in the weak formulations satisfied by f A and f. A positive number δ being fixed, let ϕ be a test function vanishing for δ δ and for v 1 δ. Since f = f A + f satisfies the oltzmann equation for a one component gas, using the entropy production term ([1]), f, Q (f, f ) and Q + (f, f ) are weakly compact in L 1 ([, 1] {v R 3 ; ξ δ, v 1 δ })). The weak compactness of f A, f, Q (f A, f ), Q + (f A, f ), Q (f, f ) and Q + (f, f ) follows from the inequalities 0 f A f, 0 f f, Q (f A, f ) Q (f, f ), Q + (f A, f ) Q + (f, f ), Q (f, f ) Q (f, f ) and Q + (f, f ) Q + (f, f ). Denote by f A and f the limits of f A and f, up to subsequences. As in ([1],[2]), lim [Q (f A, f ) Q (f A, f)]ϕdxdv = 0. (3.1) 19

20 Next, performing the change of variable (v, v, ω) (v, v, ω), the same property is obtained for the gain term, 1 1 Q + (f A, f )ϕdxdv = Q + (f A, f)ϕdxdv. lim + Analogously, the same result holds for f. It remains to pass to the limit in the boundary terms (1.4) i.e to prove the weak convergence in L 1 ({v, ξ > 0}) ( resp L 1 ({v, ξ < 0})) of f (1,.) ( resp. f (,.)) to f (1,.) (resp. f (,.)). First, the fluxes ξf (1, v)dv and ξ f (, v)dv are controled in the following way. From (2.28) written in the exponential form, it holds that R 0 f (x, v) f (, v)e 1+x ξ R 0 f (x, v) f (1, 1 x v)e ξ Recall that, R R S χ f (x+sξ,v )dv dωds 1, ξ >, v 2, 2 S χ f (x+sξ,v )dv dωds 1, ξ <, v 2. (3.2) 2 ν (x, v) = χ f (x, v )dv dω. S2 For v satisfying v 2 with ξ > 1 2 or ξ < 1 2, 1 ν (z,v) ξ dz is uniformly bounded from above. Hence, using the definition of the boundary conditions (1.4) in (3.2), f (x, v) cm (v) ξ f (, v)dv, ξ > 1, v 2, 2 f (x, v) cm +(v) ξf (1, v)dv, ξ <, v 2. 2 So, f being non negative, c c 1 {ξ> 1 2, v 2} {ξ< 1 2, v 2} f ξf (1, v)dv + (x, v)dxdv ξ f (, v)dv. min(µ, (1 + v ) β )f (x, v)dxdv ξf (1, v)dv + 20 ξ f (, v)dv.

21 Since 1 R min(µ, (1 + 3 v )β )f v (x, v)dxdv = 1, the fluxes ξf (1, v)dv and ξ f (, v)dv are bounded uniformly w.r.t. Furthermore, the energy fluxes are controlled. Indeed, by conservation of the energy for f, ξv 2 f (1, v)dv + ξ v 2 f (, v)dv ξv 2 f (, v)dv + ξ v 2 f (1, v)dv. y definition of the boundary conditions (2.29) and (2.28), ξv 2 f (1, v)dv + ξ v 2 f (, v)dv (k + ξ f (, v )dv ) ξv 2 M (v)dv (3.3) ξ <0 +(k + ξ f (1, v )dv ) ξ v 2 M + (v)dv. ξ >0 The right-hand side of (3.3) being bounded, ξv 2 f (1, v)dv + ξ v 2 f (, v)dv c. Finally, the entropy fluxes are controled. Indeed, f = f A + f satisfies the following equation ξ ( f (log(f ) 1) ) = Q (f, f )log(f ). (3.4) x Using a Green s formula and an entropy estimate in (3.4), leads to ξf (1, v) log f (1, v)dv + ξ f (, v) log f (, v)dv ( ξ f (1, v )dv + k ) ξ >0 ξ M + (v)log(m + (v)( ξ f (1, v )dv + k ))dv ξ >0 +( ξ f (, v )dv + k ) ξ <0 M (v) log(m (v)( ξ f (, v )dv + k ))dv. ξ <0 21

22 y the Dunford-Pettis criterion ([5]), f (1,.) is weakly compact in L 1 ({v, ξ > 0}). Let one of its subsequence still denoted by f (1,.), converging weakly to some g + in L 1 ({v, ξ > 0}). It remains to prove that g + = f (1,.). Consider a test function ϕ vanishing on { ξ δ} { v 1 δ } and satisfying ϕ(x, v) = ϕ 1(x)ϕ 2 (v) with ϕ 1 = 1 in a neighborhood of 1. Recall that the trace f (1, v) can be defined by f (1, v) = lim ɛ0 0 (ϕf ) satisfies the equation 1 ɛ 0 ɛ0 0 f (1 ɛ, v)dɛ ([4]). ξ (ϕf ) x = ξ ϕ x f + Q (f, f )ϕ. (3.5) Integrating (3.5) on [1 ε, 1] and using a Green s formula, it holds that 1 ɛ 0 1 ɛ0 ɛ 0 0 ɛ0 + 1 ɛ0 ɛ (f (1, v) f (1 ɛ, v))ϕ 2(v)dvdɛ 1 ɛ 0 Q (f, f )(x, v)ϕ(x, v) dxdvdɛ (3.6) 1 1 ɛ 0 f (x, v)ξ ϕ(x, v) dxdvdɛ. x Let η > 0 be given. y the weak compactness of f and Q (f, f ) in L 1 ([, 1] {v, ξ δ, v 1 δ }), there exists ɛ 0 > 0 such that for ɛ 0 < ɛ 0, uniformly w.r.t, 1 Q (f, f )(x, v)ϕ(x, v) dxdv < η 1 ɛ 0 2, 1 f (x, v)ξ 1 ɛ 0 x ϕ(x, v) dxdv < η 2. So the inequality (3.6) gives, for ɛ 0 < ɛ 0, uniformly w.r.t, 1 ɛ0 ξ(f ɛ (1, v) f (1 ɛ, v))ϕ 2(v)dɛdv η. (3.7) 0 0 y the weak compactness of f in L1 ([, 1] {v, ξ δ, v 1 δ }), 22

23 ɛ0 0 f (1 ɛ, v)ϕ 2(v)dvdɛ ɛ0 0 f (1 ɛ, v)ϕ 2 (v)dvdɛ when tends to infinity. Passing to the limit when tends to infinity in the inequality (3.7), implies that 1 ɛ0 ξ(g + (1, v) f (1 ɛ, v))ϕ 2 (v)dvdɛ η. ɛ 0 0 Next, passing to the limit when ɛ 0 tends to 0, ξ(g + (1, v) f (1, v))ϕ 2 (v)dv η. (3.8) The inequality (3.8) being true for all η > 0, g + and f (1,.) are equal on the sets {v, ξ δ, v 1 δ } for all δ > 0 and so a.e. The same result holds for f (,.). Therefore, we can pass to the limit in (2.28, 2.29). This ends the proof of Theorem Some extensions Theorem 1.1 can be generalized to the case of any values of the β-norms, M A and M. Corollary 4.1. Given β with 0 β < 2, M A > 0 and M > 0, there is a weak solution to the stationary problem with β-norms M A and M. Proof of Corollary 4.1. Define M = M A + M. In the first part of the proof of Theorem 1.1, choose λ satisfying, and λ min( λ min( M A M (v)min(µ, (1 + v ) β )e 2+M ξ dv ; M A ) M +(v)min(µ, (1 + v ) β )e 2+M ξ dv M M (v)min(µ, (1 + v ) β )e 2+M ξ dv ; M ). M +(v)min(µ, (1 + v ) β )e 2+M ξ dv 23

24 In the fixed point part, choose K = {f L 1 +([, 1] R 3 ); min(µ, (1 + v ) β )f(x, v)dxdv = M}. (4.1) Reasoning as before, there are f A and f solutions to the equation (1.1) with respective β norms M A and M. In the presence of several gases of the A component and several gases of the component, Corollary 4.1 can be generalized. Consider (f A1...f ANA, f 1...f N ) with f Ai satisfying x f A i = Q(f Ai, f A1 ) Q(f Ai, f ANA ) + Q(f Ai, f 1 ) Q(f Ai, f N ), (x, v) (, 1), f Ai (, v) = k Ai M (v), ξ > 0, f Ai (1, v) = k Ai M + (v), ξ < 0, and f i satisfying x f i = Q(f i, f A1 ) Q(f i, f ANA ) + Q(f i, f 1 ) Q(f i, f N ), (x, v) (, 1), f i (, v) = ( ξ f i (, v )dv )M (v), ξ > 0, ξ <0 f i (1, v) = ( ξ f i (1, v )dv )M + (v), ξ < 0. Corollary 4.2. Given β with 0 β < 2, M A1...M ANA and M 1...M N there are weak solutions f A1...f N to the stationary problem with respective β-norms M A1,..., M N. Proof of Corollary 4.2. Let f = f A f AN + f f N. The oltzmann operator corresponding to the A i component is Q(f Ai, f) = Q(f Ai, f A1 )+...+Q(f Ai, f ANA )+Q(f Ai, f 1 )+...+Q(f Ai, f N ), ([6]). For the fixed point step, define K as in (4.1) and consider the compact and continuous map T from the closed convex set K [0, 1] N into itself, ξ >0 T : (f, θ 1,..., θ N ) (f A f ANA + f 1...f N, θ 1,..., θ N ), with θ Ni = ξ f Ni (, v)dv ξ f Ni (, v)dv + ξf Ni (1, v)dv. 24

25 The corollary follows by the same arguments as before. Remark that the case of a one component gas with one boundary condition of the type (1.3) and another of the type (1.4) can also be solved. It comes back to the diffuse-reflection problem solved in ([2]). Furthermore, this problem can be generalized to several components by reasoning as in the proof of Corollary 4.2. Theorem 1.1 can also be generalized to the case of a convex combination of boundary conditions of the type (1.3) and (1.4), f(, v) = a( f(1, v) = a( ξ x f = Q(f, f), (x, v) (, 1) R3 v, ξ f(, v)dv)m (v) + (1 a)km (v), ξ > 0, ξf(1, v)dv)m + (v) + (1 a)km + (v), ξ < 0, (4.2) a [0, 1]. Corollary 4.3. Given β with 0 β < 2, M > 0 there is a weak solution to the stationary problem (4.2) with the β-norm M. References [1] Arkeryd, L.; Nouri, A. The stationary oltzmann Equation in the Slab with Given Weighted Mass for Hard and Soft Forces. Ann.Scuola.Norm.Sup.Pisa 1998, 27, [2] Arkeryd, L.; Nouri, A. L 1 solutions to the stationary oltzmann equation in a slab. Annales de la Faculte des sciences de Toulouse 2000, 9, [3] Arkeryd, L.; Maslova, N. On diffuse reflection at the boundary for the oltzmann equation and related equations. J.Stat.Phys. 1994, 77, [4] Arkeryd, L.; Heintz, A. On the solvability and asymptotics of the oltzmann equation in irregular domains. Comm.Part.Diff.Eqs. 1997, 22, [5] Cercignani, C.; Illner, R.; Pulvirenti, M. Existence and Uniqueness Results. The mathematical theory of dilute gases. Springer: New York, 1994,

26 [6] Cercignani, C. The oltzmann Equation. The oltzmann equation and its applications. Scottish Academic Press, 1988, [7] Lions, PL. Compactness in oltzmann equation via Fourier integral operators and applications. J.Math.Kyoto Univ. 1994, 34, ( [8] Mischler, S. On the weak-weak convergences and applications to the initial boudary value problem for kinetic equations, preprint [9] Sone, Y.; Aoki, K.; Doi, T. Kinetic theory analysis of gas flows condensing on a plane condensed phase: Case of a mixture of a vapor and noncondensable gas. Transport theory and statistical physics, 1992, 21(4-6), [10] Taguchi, S.; Aoki, K.; Takata S. Vapor flows condensing at incidence onto a plane condensed phase in the presence of a noncondensable gas.i. Subsonic condensation. Physics of fluids, 2003, volume 15, number 3, [11] Aoki, K.; Takata, S.; Taguchi S. Vapor flows with evaporation and condensation in the continuum limit: effect of a trace of non condensable gas, European Journal of Mechanics /Fluids 22, [12] Aoki, K. The behaviour of a vapor-gas mixture in the continuum limit: Asymptotic analysis based on the oltzman equation. T.J. artel, M.A.Gallis eds: AIP, Melville, 2001, [13] Aoki, K.; Takata, S.; Kosuge, S. Vapor flows caused by evaporation and condensation on two parallel plane surfaces: Effect of the presence of a noncondensable gas. Physics of fluids, 1998, volume 10, number 6, [14] Panferov, V.A. On the interior boundary-value problem for the stationary Povzner equation with hard and soft interaction, preprint