Stationary solutions for the ellipsoidal BGK model in a slab

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ScienceDirect J. Differential Equations 261 (2016) 5803–5828 www.elsevier.com/locate/jde

Stationary solutions for the ellipsoidal BGK model in a slab Jeaheang Bang a , Seok-Bae Yun b,∗ a Department of Mathematics, Rutgers University, Hill Center - Busch Campus, 110 Frelinghuysen Road,

Piscataway, NJ 08854-8019, USA b Department of Mathematics, Sungkyunkwan University, Suwon 440-746, Republic of Korea

Received 27 April 2016; revised 20 July 2016 Available online 31 August 2016

Abstract We address the boundary value problem for the ellipsoidal BGK model of the Boltzmann equation posed in a bounded interval. The existence of a unique mild solution is established under the assumption that the inflow boundary data does not concentrate too much around the zero velocity, and the gas is sufficiently rarefied. © 2016 Elsevier Inc. All rights reserved. MSC: 35Q20; 82C40; 35A01; 35A02; 35F30 Keywords: Ellipsoidal BGK model; Boltzmann equation; Kinetic theory of gases; Boundary value problem in a slab; Inflow boundary conditions

1. Introduction In this paper, we are interested in the boundary value problem of stationary ellipsoidal BGK model: v1

 ∂f ρ = Mν (f ) − f , ∂x τ

* Corresponding author.

E-mail addresses: [email protected] (J. Bang), [email protected] (S.-B. Yun). http://dx.doi.org/10.1016/j.jde.2016.08.022 0022-0396/© 2016 Elsevier Inc. All rights reserved.

(1.1)

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on a finite interval [0, 1] associated with the boundary condition: f (0, v) = fL (v)

for v1 > 0,

f (1, v) = fR (v)

for v1 < 0.

(1.2)

The velocity distribution function f (x, v) is the number density at x ∈ [0, 1] with velocity v = (v1 , v2 , v3 ) ∈ R3 . We have normalized the spatial domain for simplicity. τ is defined by τ = κ(1 − ν), where κ denotes the Knudsen number defined as the ratio between the mean free path and the characteristic length, and ν ∈ (−1/2, 1) is the relaxation parameter. The ellipsoidal Gaussian Mν (f ) reads   1 ρ  −1 exp − (v − U ) Tν (v − U ) , Mν (f ) = √ 2 det(2πTν ) where the local density ρ, momentum U , temperature T and the stress tensor  are given by  ρ(x) =

f (x, v)dv, R3



ρ(x)U (x) =

f (x, v)vdv, R3



(1.3) f (x, v)|v − U |2 dv,

3ρ(x)T (x) = R3



f (x, v)(v − U ) ⊗ (v − U )dv,

ρ(x)(x) = R3

and the temperature tensor Tν is defined as a linear combination of T and : ⎛ ⎜ Tν = ⎝

(1 − ν)T + ν11

ν12

ν21 ν31

(1 − ν)T + ν22 ν32

ν13

⎟ ν23 ⎠ (1 − ν)T + ν33

= (1 − ν)T I d + ν. A direct calculation gives the following cancellation property  R3

⎞ ⎛ 1   Mν (f ) − f ⎝ v ⎠ dv = 0, |v|2

leading to the conservation of following quantities along x ∈ [0, 1]: 

 f v1 dv, R3

 f v1 |v|2 dv.

f v1 vdv, R3

R3

⎞

J. Bang, S.-B. Yun / J. Differential Equations 261 (2016) 5803–5828

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Andries et al. [2] derived 



 Mν (f ) − f ln f dv ≤ 0,

R3

which gives the H -theorem for the time dependent problem (see also [10,31]). The ellipsoidal BGK model is a generalized version of the original BGK model [7,28] which has been widely used as a model equation of the Boltzmann equation. It was introduced by Holway [19] to overcome the well-known short-coming of the original BGK model: the incorrect Prandtl number in the Navier–Stokes limit. He introduced the relaxation parameter ν and generalized the local Maxwellian in the original BGK model into the ellipsoidal Gaussian by replacing the macroscopic temperature with a temperature tensor Tν parametrized by ν ∈ (−1/2, 1) (for the discussion of the range of ν, see [30]). Through a Chapmann–Enskog expansion, it can be shown that the Prandtl number of the ES–BGK model is given by 1/(1 − ν), and the desired physical Prandtl is obtained by choosing the proper relaxation parameter, which is ν = 1 − 1/P r ≈ −1/2. Note that, in the case ν = 0, the ES–BGK model reduces back to the original BGK model. Therefore, any results for the ES–BGK model automatically holds for the original BGK model either. The ES–BGK model, however, has been somewhat neglected in the literature, due mainly to the fact that the H -theorem was not verified. This was done recently by Andries et al. [2] and revived the interest on this model [1,8,15,16,21,20,23,29–33]. In this paper, we consider the ES–BGK model posed in a bounded interval with fixed inflow boundary conditions at both ends. Similar problem was considered by Ukai in [25] for the original BGK model (the case of ν = 0) using a version of the Schauder fixed point theorem to the macroscopic variables. No smallness assumption was imposed, but the uniqueness was not guaranteed. We develop here a Banach fixed point type approach for the ES–BGK model which works for the whole range of relaxation parameter (−1/2 < ν < 1), under the assumptions that the gas is sufficiently rarefied and the boundary inflow data does not concentrate too much near the zero velocity. The first assumption is a kind of smallness condition, which is typical for Banach fixed point type arguments. It is, however, not clear whether the second condition is of intrinsic nature, or mere a technicality that can be overcome by developing finer analysis. (See Remark (3) in Section 2.) Brief reference check for related works is in order. In [3], 1d stationary problem for the Boltzmann equation with Maxwellian molecules was studied in the frame work of measure valued solutions. In a series of paper [4–6], Arkeryd and Nouri studied the existence of weak solutions in L1 . Extensions of these arguments into the case of two component-gases were made in [8,9]. Gomeshi obtained the existence and uniqueness for the Boltzmann equation when the gas is sufficiently rarefied [18], which largely motivated our work. For nice survey of mathematical and physical aspects of the Boltzmann equation and BGK models, see [11,13,14,17,22,24,26,27]. 1.1. Notations To prevent confusion, we fix some notational conventions which we will keep throughout this paper. • Every constant, usually denoted by C or Ca,b,··· will be generically defined. The values of C may differ line by line, and even when the same C appears more than once in a line, they are not necessarily of the same value. But all the constants are explicitly computable in principle.

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• We use C ,u to denote a positive constant that can be explicitly computed using only the quantities in (2.1), γ in the Theorem 2.2 and the relaxation parameter ν. C ,u is generic too in the above mentioned sense. • We fix au , a , as , cu , c , cs and γ appearing in (2.1) and Theorem 2.2 for this use only. • When there’s no risk of confusion, we write Mi (vi ) = e−C ,u vi

2

(i = 1, 2, 3)

without explicitly showing the dependence on C ,u for simplicity of notations. • When there’s no risk of confusion, we use v1 > 0 to denote either {v1 > 0} ⊂ R or {v1 > 0} × R2 according to the context. by • We define supx · L1 and · L∞ 2 2 

 sup f L1 = sup |f (x, v)|(1 + |v|2 )dv , 2

x

x

R3

= sup |f (x, v)|(1 + |v|2 ).

f L∞ 2 x,v

This paper is organized as follows. The main result is stated in the following section 2. Some relevant issues are also discussed. In section 3, we reformulate the problem in the fixed point set up. Some useful technical lemmas are recorded. Then section 4 and 5 is devoted respectively to showing that the solution map is invariant and contractive in the solution space. 2. Main result Before we state our main result, we need to define the following quantities (recall that τ = κ(1 − ν)):    1 − au au = 2 fLR dv, a = e τ |v1 | fLR dv, as = fLR dv, |v1 | R3

R3



cu = 2

R3



fLR |v| dv, c = 2

R3

e

− τ a|vu | 1



fLR |v| dv, cs =

R3

2

R3

1 fLR |v|2 dv, |v1 |

(2.1)

where we used abbreviated notation: fLR (v) = fL (v)1v1 >0 + fR (v)1v1 <0 . We define the mild solution of (1.1) as follows: Definition 2.1. f ∈ L12 ([0, 1]x × R3v ) is said to be a mild solution for (1.1) if it satisfies f (x, v) = e

− τ |v1

1|

x 0

1 + τ |v1 |

ρf (y)dy

x e 0

fL (v)

− τ |v1

1|

x y

ρf (z)dz

ρf (y)Mν (f )dy

if v1 > 0

(2.2)

J. Bang, S.-B. Yun / J. Differential Equations 261 (2016) 5803–5828

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and f (x, v) = e

− τ |v1

+

1|

1 x

1 τ |v1 |

ρf (y)dy

1 e

fR (v)

− τ |v1

1|

y x

ρf (z)dz

ρf (y)Mν (f )dy

if v1 < 0.

(2.3)

x

The main result of this paper is as follows: Theorem 2.2. Suppose that the mass of the inflow boundary data fLR ≥ 0 (not identically zero) is finite and does not concentrate too much around the zero velocity in the following sense: 1 fLR ∈ L12 , |v1 |

fLR ,

(2.4)

so that the quantities defined in (2.1) are well-defined. Suppose further that 

 fL vi dv2 dv3 =

R2

fR vi dv2 dv3 = 0

(i = 2, 3)

(2.5)

R2

and there exists a constant γ > 0 such that ⎛ ⎞⎛ ⎞   − au − au ⎜ ⎟⎜ ⎟ e τ |v1 | fL (v)|v1 |dv ⎠ ⎝ e τ |v1 | fR (v)|v1 |dv ⎠ > γ . ⎝ v1 >0

(2.6)

v1 <0

Then there exists a constant K > 0 depending only on the quantities defined in (2.1) and γ such that, if τ > K, then there exists a unique mild solution f ≥ 0 for (1.1) satisfying   a ≤ f (x, v)dv ≤ au , c ≤ f (x, v)|v|2 dv ≤ cu , R3

R3

and ⎛ ⎜ ⎝



⎞⎛ ⎟⎜ f dv ⎠ ⎝

R3



R3

⎞

⎛

⎟ ⎜ f |v|2 dv ⎠ − ⎝



⎞2 ⎟ f v1 dv ⎠ ≥ γ .

R3

Remark 2.3. (1) The last assertion of the above theorem guarantees the strict positivity of the temperature tensor for k ∈ R3 . This is important since, otherwise, the ellipsoidal Gaussian is not well-defined. (See Lemma 3.2.) (2) The well-posedness of bulk velocity follows from the above estimates since  a +c  u u   .  f (x, v)vdv  ≤ 2 R3

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J. Bang, S.-B. Yun / J. Differential Equations 261 (2016) 5803–5828

(3) The condition (2.4) is used to prove the contractiveness of the solution operator. (See the proof of Proposition 5.2.) In the literature on the stationary problem of the Boltzmann equation, truncation of the collision kernel near origin is often employed to overcome the technical difficulties arising in the small velocity region. (See, for example, [3,4,12].) Our non-concentration condition (2.4) can be understood in some sense as a weak truncation of the boundary data near zero. This is good in that we are not imposing any restriction on the equation, but bad at the same time since it excludes Maxwellian boundary data, which is the most representative distribution function in the kinetic theory. (4) Let us provide an explicit example of the boundary data which satisfies all the conditions above. Define fL and fR by fL (v) = CL 1r1 ≤v1 ≤r2 e−

|v2 |2 2

e−

|v3 |2 2

fR (v) = CR 1−r2 ≤v1 ≤−r1 e−

,

|v2 |2 2

e−

|v3 |2 2

,

with CL , CR > 0 and r2 > r1 > 0. Here 1A is the characteristic function on A. Clearly, fLR satisfies (2.5). Moreover, since fLR decays sufficiently fast and vanishes near v1 = 0, all quantities in (2.1) are well-defined. To check (2.6), we compute au = π(CL + CR )(r2 − r1 ) so that

 e

− τ a|vu | 1

fL (v)v1 dv ≥ e

−

π(CL +CR )(r2 −r1 ) τ r1

 fL (v)v1 dv R3

v1 >0

= CL e

−

π(CL +CR )(r2 −r1 ) τ r1

 

2 r

v1 dv1 r1

=

π CL e 4

π(CL +CR )(r2 −r1 ) − τ r1

e−

|v2 |2 2

e−

|v3 |2 2

 dv2 dv3

R2

(r22 − r12 ).

Similarly,  e

− τ a|vu | 1

fR (v)v1 dv ≥

π(CL +CR )(r2 −r1 ) π − τ r1 (r22 − r12 ). CR e 4

v1 >0

Therefore, ⎛ ⎜ ⎝

⎞⎛

 e

v1 >0

≥

− τ a|vu | 1

⎟⎜ fL (v)|v1 |dv ⎠ ⎝

⎞

 e

− τ a|vu | 1

⎟ fR (v)|v1 |dv ⎠

v1 <0

4(C +C )(R−r) π2 − R τ rL 1 (r22 − r12 )2 , C L CR e 4

which is strictly positive. (5) At first sight, the definition of a , c and γ seems a little dangerous in that they contain τ inside the integral, which may lead to a kind of circular reasoning in the choice of τ . But, when

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τ is very large (whose size is determined only by fLR ), we can treat a , c and γ as if they are independent of τ . We only consider c : Take r > 0 small enough such that   1 fLR (v)|v|2 dv ≥ fLR (v)|v|2 dv. 2 |v1 |≥r

R3

− τ a|vu |

fLR (v)|v|2 dv ≥ e− τ r

Then, we observe  e

1



au

fLR (v)|v|2 dv |v1 |≥r

R3

1 au ≥ e− τ r 2



fLR (v)|v|2 dv. R3

Since we are going to take τ sufficiently large, and au and r does not depend on τ , we can assume −

au

that τ is large enough such that e 2τ r1 ≥ 1/2 which yields   1 − τ a|vu | 2 1 e fLR (v)|v| dv ≥ fLR (v)|v|2 dv. 4 R3

R3

That is, 1 c ≥ cu . 8 a and γ can be treated similarly. 3. Fixed point set-up We will find the solution for (1.1) as a fixed point of a solution map defined on the following function space:

  = f ∈ L12 ([0, 1]x × R3v ) | f satisfies (A), (B), (C) endowed with the metric d(f, g) = sup f − g L1 , where (A), (B) and (C) denote x∈[0,1]

2

• (A) f is non-negative: f (x, v) ≥ 0 for x, v ∈ [0, 1] × R3 . • (B) The macroscopic field is well-defined:  a ≤ f (x, v)dv ≤ au , R3

 c ≤

f (x, v)|v|2 dv ≤ cu . R3

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• (C) The following lower bound holds: ⎛ ⎜ ⎝



⎞⎛ ⎟⎜ f dv ⎠ ⎝

R3



R3

 2     ⎟   2 f |v| dv ⎠ −  f vdv  ≥ γ .   R3  ⎞

In view of (2.2) and (2.3), we define our solution map by (f ) = + (f )1v1 >0 + − (f )1v1 <0 , where + (f ) and − (f ) are + (f )(x, v) = e

− τ |v1

1|

x

ρf (y)dy

0

x

1 + τ |v1 |

e

fL (v) x

− τ |v1

1|

y

ρf (z)dz

(3.1) ρf (y)Mν (f )dy

if v1 > 0

ρf (y)Mν (f )dy

if v1 < 0.

0

and − (f )(x, v) = e

− τ |v1

1|

1 x

1 + τ |v1 |

ρf (y)dy

1 e

fR (v)

− τ |v1

1|

y x

ρf (z)dz

(3.2)

x

Our main Theorem 2.2 then follows directly once we show that the solution map is invariant and contractive in . The remaining of this paper is devoted to the proof of these two properties, which is stated in Proposition 4.1 and Proposition 5.2 respectively. Before we move on to the existence proof, we record some technical lemmas that will be useful throughout this paper. Lemma 3.1. [29] Suppose ρ(x) > 0. Then Tν satisfies Cν1 T I d ≤ Tν ≤ Cν2 T I d, where Cν1 = min{1 − ν, 1 + 2ν} and Cν2 = max{1 − ν, 1 + 2ν}. Lemma 3.2. Let f ∈ . Then the macroscopic fields U , T constructed from f satisfy |U | ≤

a u + cu , 2a

γ cu ≤T ≤ 2 3a 3au

and Cν1 for all k ∈ R3 .

γ cu 2 |k|2 ≤ k  Tν k ≤ Cν2 |k| , 2 3a 3au

J. Bang, S.-B. Yun / J. Differential Equations 261 (2016) 5803–5828

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Proof. For the first inequality, we compute     f vdv   a u + cu 3 R |ρU | |U | = . =  ≤ ρ 2a R3 f dv Besides, we observe that T is represented as (3ρT + ρ|U |2 ) − |ρU |2 ρ −1 3ρ  2   −1   2 dv −  f |v| f vdv f dv   3 3 3 R R R  = . 3 R3 f dv

T=

We then ignore the last term on the numerator to get  2 cu 3 f |v| dv ≤ T ≤ R . 3a 3 R3 f dv For the lower bound, we recall the definition of γ in (2.6) to obtain  T=

R3

f dv

 

2     f |v|2 dv −  R3 f vdv  γ ≥ 2.  2 3au 3 R3 f dv R3

Now, the last assertion follows directly from this estimates on T and Lemma 3.1.

2

Lemma 3.3. Let f ∈ . Then there exist positive constants C ,u depending only on the quantities (2.1) and γ such that Mν (f )(1 + |v|2 ) ≤ C ,u e−C ,u |v| . 2

Remark 3.4. The two C ,u do not necessarily represent the same constant. Proof. We only consider Mν |v|2 since the estimate for Mν is similar and simpler. Since Tν is symmetric, it is diagonalizable. Let λi (i = 1, 2, 3) denote the eigenvalues. Then, from Lemma 3.1, we can deduce that det Tν = λ1 λ2 λ3 ≥ {Cν1 }3 T 3 . On the other hand, Lemma 3.1 also implies −(v − U ) Tν−1 (v − U ) ≤ − Therefore, we have

|v − U |2 . Cν2 T

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| − |v−U ρ 2Cν2 T |v|2 Mν (f )|v| ≤ e {2πCν1 }3/2 T 3/2

2

≤

2

2

2

2

|U | − |v| ρ 2Cν2 T e 2Cν2 T |v|2 e {2πCν1 }3/2 T 3/2

|U | − |v| CCν2 ρ 2Cν2 T e 4Cν2 T ≤ e {2πCν1 }3/2 T 1/2

where we have used the boundedness: x 2 e−x < C for some C > 0. We then apply the lower and upper bounds of Lemma 3.2 to get the desired result. 2 2

4.  maps  into itself The main result of this section is the following proposition, which says that the elements of  are mapped into  by our solution map : Proposition 4.1. Let f ∈ . Then, under the assumption of Theorem 2.2, we have (f ) ∈ . We divide the proof into Lemma 4.1, 4.2, 4.3, and Lemma 4.5. Lemma 4.1. Let f ∈ . Then (f )(x, v) ≥ 0. Proof. From the proof of Lemma 3.3, we see that ρ ρ ≥ , √ 2 3/2 {2πC det 2πTν νT } which, in view of Lemma 3.2, implies   ρ 3a 3/2 ≥ a > 0. √ 2πcu det 2πTν Hence, we have Mν (f ) > 0. Therefore, we can ignore the second term in (3.1) to conclude + (f )(x, v) ≥ e Similarly, we have

− τ |v1

1|

x 0

ρf (y)dy

fL (v) ≥ 0.

J. Bang, S.-B. Yun / J. Differential Equations 261 (2016) 5803–5828

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− (f )(x, v) ≥ 0. 2

This completes the proof.

Lemma 4.2. Assume f ∈ . Then we have  (f )dv ≥ a ,

 (f )|v|2 dv ≥ c .

R3

R3

Proof. We only prove the second one. Recall from the previous proof that (f ) ≥ e

− τ |v1

1|

x 0

ρf (y)dy

fL (v)1v1 >0 + e

1

− τ |v1

1|

x

ρf (y)dy

fR (v)1v1 <0 .

Then, since ρf ≤ au , we have (f ) ≥ e ≥e

u − τxa |v | 1

− τ a|vu | 1

=e

− τ a|vu | 1

fL (v)1v1 >0 + e fL (v)1v1 >0 + e

u − (1−x)a τ |v | 1

− τ a|vu | 1

fR (v)1v1 <0

fR (v)1v1 <0

fLR .

Integrating with respect to |v|2 dv, we obtain the desired lower bound:   − au (f )|v|2 dv ≥ e τ |v1 | fLR |v|2 dv ≥ c . 2 R3

R3

Lemma 4.3. Let f ∈ . Then we have  (f )dv ≤ au ,

 (f )|v|2 dv ≤ cu .

R3

(4.1)

R3

Proof. We only prove the second one. Consider   x − 1 ρ (y)dy + (f )|v|2 dv = e τ |v1 | 0 f fL (v)|v|2 dv R3

v1 >0

 x + v1 >0 0

1 − e τ |v1 |

Using ρf ≥ a , we estimate the first term of  e v1 >0

− τ |v1

1|

x 0

ρf (y)dy

 R3

v1 >0

ρf (y)Mν (f )|v|2 dydv.

+ |v|2 dv as

 fL (v)|v|2 dv ≤

x y ρf (z)dz τ |v1 |

e

a x 1|

− τ |v

 fL (v)|v|2 dv ≤ v1 >0

fL (v)|v|2 dv.

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By Lemma 3.3, we compute  x v1 >0 0

1 − e τ |v1 |

≤ C ,u

x y ρf (z)dz τ |v1 |

⎧ ⎪ ⎨ x   ⎪ ⎩

0

v1 >0

x  ≤ C ,u au 0 v1 >0

x  ≤ C ,u au 0 v1 >0

ρf (y)Mν (f )|v|2 dydv

1 e τ |v1 |

1 − e τ |v1 |

x ρf (z)dz − y τ |v | 1



M1 (v1 )dv1 ρf (y)dy

⎫⎧ ⎪ ⎬⎪ ⎨ ⎪ ⎭⎪ ⎩

M2 M3 dv2 dv3

R2

x y a dz τ |v1 |

M1 (v1 )dv1 dy

1 − a τ(x−y) |v1 | M (v )dv dy e 1 1 1 τ |v1 |

≡ C ,u au I. Then divide the domain of integration into the following two regions

I=

⎧ ⎪ ⎨ x  ⎪ ⎩

x  +

0 |v1 |<τ

0 |v1 |>τ

⎫ ⎪ ⎬

1 − a τ(x−y) |v1 | M1 (v1 )dv1 dy e ⎪ ⎭ τ |v1 |

≡ I1 + I2 . For I1 , we carry out the integration on x first: I1 =

⎩

|v1 |<τ

=

1 a

1 ≤ a

⎫ ⎬ 1 − a τ(x−y) |v1 | dy M1 (v1 )dv1 e ⎭ τ |v1 |

⎧ x ⎨





0

 1−e

|v1 |<τ



 1−e



a x 1|

− τ |v

a

− τ |v |

M1 (v1 )dv1  M1 (v1 )dv1 ,

1

|v1 |<τ

and decompose the integration further as ⎧ ⎪ ⎪ ⎨ 

1 a ⎪ ⎪ ⎩

|v1 |< τ1

 + 1 τ <|v1 |<τ

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭

1−e

a

− τ |v | 1

 M1 (v1 )dv1 ≡ I11 + I12 .

⎫ ⎪ ⎬ ⎪ ⎭

J. Bang, S.-B. Yun / J. Differential Equations 261 (2016) 5803–5828

We use rough estimates 1 − e

a

− τ |v |

5815

≤ 1 and M1 (v1 ) ≤ 1 to control I11 by

1



1 a

1 1 . a τ

dv1 ≤ |v1 |< τ1

a

−

On the other hand, we expand 1 − e τ |v1 | by Taylor series and use M1 (v1 ) ≤ 1 to find         a 1 a 2 1 a 3 1 − I12 = + + · · · M1 (v1 )dv1 a τ |v1 | 2! τ |v1 | 3! τ |v1 | 1 τ <|v1 |<τ

1  ≤  a  = ≤

τ

τ  2   τ  3  a   1 a 1 a    dr  +  dr  +  dr  + · · · τr 2! τ r 3! τ r

1 τ

1 τ

1 τ

1 1 a 2 τ 4 − 1 1 a τ 2 − 1 1 a 3 τ 6 − 1 ln τ 2 + + + ··· τ 2! τ 2 τ 2 · 3! τ 3 τ 2 3 · 4! τ 4 τ 3



1 ea 1 ln τ 2 + . τ a τ

In the last line, we used 1 a 2 τ 4 − 1 1 a τ 2 − 1 1 a 3 τ 6 − 1 + + ··· 2! τ 2 τ 2 · 3! τ 3 τ 2 3 · 4! τ 4 τ 3 1 a τ 2 − 1 1 a 2 τ 4 − 1 1 a 3 τ 6 − 1 + + ··· 2! τ τ 2 2 · 3! τ τ 4 3 · 4! τ τ 6   1 1 a 2 a 3 a 4 + + ··· ≤ a 2! 3! 4! τ

=

≤

ea 1 , a τ

which again is the consequence of (τ n − 1)/τ n < 1. Finally, I2 is estimated as follows: ⎧ 1 ⎨

 I2 ≤ |v1 |>τ

 ≤ |v1 |>τ

≤

1 τ2

⎩ 0

⎧ 1 ⎨ ⎩ 

0

⎫ ⎬ 1 − a τ(x−y) |v1 | dy e M1 (v1 )dv1 ⎭ τ |v1 | ⎫ ⎬

1 dy M1 (v1 )dv1 τ |v1 | ⎭

M1 (v1 )dv1 |v1 |>τ

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1 ≤ 2 τ

M1 (v1 )dv1 R3

≤ C ,u

1 . τ2

The above decomposition of velocity domain is largely motivated from [18]. In conclusion, we obtain the following estimate for I :  I ≤ C ,u

   1 1 1 ln τ + 1 ln τ 2 + + 2 ≤ C ,u , τ τ τ τ

(4.2)

where C ,u > 0 depends only on ν, quantities in (2.1) and γ . Finally, we gather these estimates to obtain     ln τ + 1 + (f )|v|2 dv ≤ fL (v)|v|2 dv + C ,u . τ v1 >0

v1 >0

By an identical argument, similar estimate can be derived for −(f ): 

− (f )dv ≤

v1 <0



 fR (v)|v|2 dv + C ,u

 ln τ + 1 . τ

v1 <0

Summing up these estimates and recalling the definition of cu , we get   1 ln τ + 1 (f )|v| dv ≤ cu + C ,u , 2 τ



2

R3

2

which gives the second estimate in (4.1) for sufficiently large τ . Lemma 4.4. For i = 2, 3, we have     ln τ + 1   .  (f )vi dv  ≤ C ,u τ R3

Proof. We only prove for i = 2. We integrate (3.1) with respect to v2 dv2 dv3 to get F+ (x, v1 ) = e where

− τ |v1

1|

x 0

ρ(y)dy

1 FL,+ (v1 ) + τ |v1 |



F+ =

e

− τ |v1

1|

x y

ρ(z)dz

ρ(y)G(y, v1 )dy,

0

 + (f )v2 dv2 dv3 , G =

R2

x

 Mν (f )v2 dv2 dv3 , FL,+ (v1 ) =

R2

fL (v)v2 dv2 dv3 . R2

J. Bang, S.-B. Yun / J. Differential Equations 261 (2016) 5803–5828

5817

By our assumption on fL , we have FL,+ (v1 ) = 0, so that 1 F+ (x, v1 ) = τ |v1 |

x e

x

− τ |v1

1|

y

ρ(z)dz

ρ(y)G(y, v1 )dy.

0

Using Lemma 3.3, we compute G(x, v1 ) ≤ C ,u M1



 M2 M3 |v2 |dv2 dv3 ≤ C ,u M1 (v1 ).

R2

Substituting this into (4.3), 1 F+ (x, v1 ) ≤ C ,u τ |v1 | ≤ C ,u

au τ |v1 |

x e

− τ |v1

1|

x y

ρ(z)dz

ρ(y)M1 (v1 )dy

0

x e

−

a (x−y) τ |v1 |

M1 (v1 )dy.

0

Now, integrating on v1 > 0 and recalling (4.2), we get 

x  F+ (x, v1 )dv1 = C ,u 0 v1 >0

v1 >0

 ≤ C ,u

1 − a τ(x−y) |v1 | M1 (v1 )dv1 dy e τ |v1 |

 ln τ + 1 . τ

That is, 

 + (f )v2 dv ≤ C ,u

 ln τ + 1 . τ

v1 >0

Applying the same type of argument to − (f ), we can derive      ln τ + 1   − (f )v2 dv  ≤ C ,u .  τ v1 <0

We sum these up to obtain the desired result. 2

(4.3)

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J. Bang, S.-B. Yun / J. Differential Equations 261 (2016) 5803–5828

Lemma 4.5. Let f ∈ . Then, for sufficiently large τ , we have 2 ⎞⎛ ⎞  ⎛       ⎟⎜ ⎟  ⎜  2 ⎝ (f )dv ⎠ ⎝ (f )|v| dv ⎠ −  (f )vdv  ≥ γ .   R3  R3 R3 Proof. Since we have shown (f ) ≥ 0 in Lemma 4.1, we can apply the Cauchy–Schwarz inequality as 2 ⎛ ⎞⎛ ⎞        ⎜ ⎟⎜ ⎟   2 ⎝ (f )dv ⎠ ⎝ (f )|v| dv ⎠ −  (f )vdv    R3  R3 R3 2 ⎛ ⎞2       ⎜ ⎟   ≥ ⎝ (f )|v|dv ⎠ −  (f )vdv    R3  R3 2 ⎛ ⎞2       ⎜ ⎟   ≥ ⎝ (f )|v1 |dv ⎠ −  (f )vdv  .   R3  R3 In the last line, we have used |v| ≥ |v1 |. Then we decompose the last term as ⎛ ⎜ ⎝



R3

⎞2

 2 ⎟   (f )|v1 |dv ⎠ −  (f )vdv  ⎛

⎜ ≥⎝

R3



⎞2

⎟ (f )|v1 |dv ⎠ −

1≤i≤3

R3

⎛ ⎜ =⎝



2

     (f )vi dv 

⎞2 ⎟ (f )|v1 |dv ⎠ −

R3

R3

 (f )v1 dv

2

−R

R3

≡ I − R, where R = M22 + M32 + 2M1 M2 + 2M2 M3 + 2M3 M1 ,     for Mi =  R3 (f )vi dv . Since M1 is bounded: M1 ≤ au + cu , we see from Lemma 4.4 that R can be made arbitrarily small by taking τ sufficiently large:  R ≤ C ,u

 ln τ + 1 . τ

J. Bang, S.-B. Yun / J. Differential Equations 261 (2016) 5803–5828

5819

For the estimate of I , we use the simple identity: a 2 − b2 = (a − b)(a + b) to bound I from below by

I≥

⎧ ⎪ ⎨

(f )(|v1 | + v1 )dv

⎪ ⎩

R3

=4

⎧ ⎪ ⎨ 

(f )|v1 |dv

⎪ ⎩

v1 >0

⎫⎧ ⎪ ⎬⎪ ⎨

(f )(|v1 | − v1 )dv

⎪ ⎭⎪ ⎩

⎫⎧ ⎪ ⎬⎪ ⎨  ⎪ ⎭⎪ ⎩

R3

v1 <0

⎫ ⎪ ⎬ ⎪ ⎭

⎫ ⎪ ⎬

(f )|v1 |dv . ⎪ ⎭

We then recall from (3.1) that (f ) ≥ e

− τ |v1

1|

x

ρf (y)dy

0

fL 1v1 >0 + e

− τ |v1

1|

1 x

ρf (y)dy

fR 1v1 <0

to obtain ⎧ ⎪ ⎨  4

⎪ ⎩

(f )|v1 |dv

v1 >0

⎛

⎜ ≥ 4⎝

⎫⎧ ⎪ ⎬⎪ ⎨  ⎪ ⎭⎪ ⎩

v1 <0

 e

− τ |v1

1|

x 0

⎜ ≥ 4⎝

e

− τ a|vu | 1

⎟⎜ fL |v1 |dv ⎠ ⎝

v1 >0

e

v1 <0

 e

⎞



⎟⎜ fL |v1 |dv ⎠ ⎝

⎞⎛



⎪ ⎭

⎞⎛

ρf (y)dy

v1 >0

⎛

(f )|v1 |dv

⎫ ⎪ ⎬

− τ a|vu | 1

− τ |v1

1|

1 x

ρf (y)dy

⎟ fR |v1 |dv ⎠

⎞

⎟ fR |v1 |dv ⎠ .

v1 <0

In view of (2.6), we see that the last term is bounded from below by 4γ . In summary, we have derived the following estimate:  2     ⎟⎜ ⎟  ⎜  2 ⎝ (f )dv ⎠ ⎝ (f )|v| dv ⎠ −  (f )vdv    R3  R3 R3   ln τ + 1 ≥ 4γ − C ,u . τ ⎛



⎞⎛



⎞

Therefore, upon choosing sufficiently large τ , we can get the desired result. 2 5.  is contractive in  The goal of this section is to show that the solution map is contractive in . First, we consider the continuity property of the ellipsoidal Gaussian.

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J. Bang, S.-B. Yun / J. Differential Equations 261 (2016) 5803–5828

Proposition 5.1. Let f , g be elements of . Then the non-isotropic Gaussian Mν satisfies the following continuity property: |Mν (f ) − Mν (g)| ≤ C ,u sup f − g L1 e−C ,u |v| . 2

2

x

Proof. (1) We expand Mν (f ) − Mν (g) as 1

∂Mν (θ ) dθ ∂ρ

Mν (f ) − Mν (g) = (ρf − ρg ) 0

1 + (Uf − Ug )

∂Mν (θ ) dθ ∂U

0

1 + (Tf − Tg )

(5.1)

∂Mν (θ ) dθ ∂Tν

0

≡ I 1 + I2 + I 3 , where ∂Mν (θ ) ∂Mν = (ρθ , Uθ , Tθ ) ∂X ∂X     for (ρθ , Uθ , Tθ ) = (1 − θ ) ρf , Uf , Tf + θ ρg , Ug , Tg . Since (ρθ , Uθ , Tθ ) is a linear combination of macroscopic fields of f and g, all lemmas in the previous sections hold the same. Therefore, instead of restating the corresponding lemmas, we refer to them whenever such estimates are needed for (ρθ , Uθ , Tθ ). (a) Estimate for I1 : Since we have ∂Mν (θ ) 1 = Mν (θ ), ∂ρ ρθ it follows directly from ρθ ≥ a and Lemma 3.3 that    ∂Mν (θ )  −C ,u |v|2   .  ∂ρ  ≤ C ,u e (b) Estimate for I2 : An explicit computation gives  ∂Mν (θ ) 1 = − (v − Uθ ) Tθ−1 + Tθ−1 (v − Uθ ) Mν (θ ). ∂U 2 Let X = v − Uθ and observe

(5.2)

J. Bang, S.-B. Yun / J. Differential Equations 261 (2016) 5803–5828

5821

|X  Tθ−1 | = sup X  {Tθ }−1 Y |Y |=1



1 sup (X + Y ) {Tθ }−1 (X + Y ) − X  {Tθ }−1 X − Y  {Tθ }−1 Y 2 |Y |=1   |X + Y |2 + |X|2 + 1 ≤C Tθ   1 + |v − Uθ |2 ≤C , Tθ =

which is, by Lemma 3.2, bounded by C ,u (1 + |v|2 ). Similarly, we can derive |{Tθ }−1 (v − Uθ )| ≤ C ,u (1 + |v|2 ). With these computations and Lemma 3.2 and Lemma 3.3, we have  ∂M (θ )  2 ν    ≤ C ,u Mν (θ )(1 + |v|2 ) ≤ C ,u e−C ,u |v| .  ∂U (c) Estimate for I3 : We first observe    ∂Mν (θ ) 1 1 ∂ det Tθ ∂Tθ = + (v − Uθ ) Tθ−1 − Tθ−1 (v − Uθ ) Mν (θ ). ∂Tij 2 det Tθ ∂Tθij ∂Tij θ Since each entry of ∂∂TTθij is either 1 or 0, we have

         (v − Uθ ) T −1 ∂Tθ T −1 (v − Uθ ) ≤ (v − Uθ ) T −1  T −1 (v − Uθ ) θ θ θ θ   ∂T θij

≤ C ,u (1 + |v| ). 2

Since det Tθ is a homogeneous polynomial of entries of Tθ : 

Cij k mn Tθij Tθk Tθmn ,

i,j,k, ,m,n

for some constants Cij k mn ,

∂ det Tθ ∂ Tθij

is written in the following form. 

Cij mn Tθij Tθmn

i,j,m,n

for some constants Cij mn . Therefore, in view of Lemma 3.1 and Lemma 3.2, we have    ∂ det Tθ    ≤ CT 2 ≤ C ,u . θ  ∂T  θij Hence, Lemma 3.3 yields

(5.3)

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J. Bang, S.-B. Yun / J. Differential Equations 261 (2016) 5803–5828

 ∂M (θ )  2 ν     ≤ C ,u (1 + |v|2 )Mν (θ ) ≤ C ,u e−C ,u |v| . ∂Tij Plugging all these estimates into (5.1) gives |Mν (f ) − Mν (g)| 

2 ≤ C ,u |ρf − ρg | + |Uf − Ug | + |Tf − Tg | e−C ,u |v| . It remains to estimate the macroscopic fields. The first term is estimated straightforwardly:  |ρf − ρg | = |f − g|dv ≤ C sup f − g L1 . 2

x

R3

We divide the second term into two parts and estimate separately as 1 1 |ρf Uf − ρg Ug | + |ρf − ρg ||Ug | ρf ρf   |Ug | 1 ≤ |f − g||v|dv + |f − g|dv ρf ρf

|Uf − Ug | ≤

R3

R3

≤ C ,u sup f − g L1 . 2

x

The last term is decomposed similarly: |Tf − Tg | ≤

1 1 |ρf Tf − ρg Tg | + |ρf − ρg ||Tg | = J1 + J2 , ρf ρf

where J1 and J2 are computed as    1   J1 ≤ |f − g| 3−1 (1 − ν)|v − U |2 I d + ν(v − U ) ⊗ (v − U ) dv a R3

≤

1 a



|f − g|(1 + |v|2 )dv R3

≤ C ,u sup f − g L1 , 2

x

and  J2 ≤ C ,u

|f − g|dv ≤ C ,u sup f − g L1 . 2

x

R3

We now substitute these estimates into (5.4) to obtain |Mν (f ) − Mν (g)| ≤ C ,u sup f − g L1 e−C ,u |v| . 2

x

2

2

(5.4)

J. Bang, S.-B. Yun / J. Differential Equations 261 (2016) 5803–5828

Proposition 5.2. Suppose f, g ∈ . Then, under the assumption of Theorem 2.2, satisfies sup (f ) − (g) L1 ≤ α sup f − g L1 2

x∈[0,1]

2

x∈[0,1]

for some constant α < 1 depending on the quantities in (2.1), γ , ν and κ. Proof. We first consider + (f ). We write + (f ) = I (f ) + II(f, f, f ), where I (f ) and II(f, g, h) are defined by I (f ) = e

− τ |v1

1|

x 0

ρf (y)dy

fL (v),

and 1 II(f, g, h) = τ |v1 |

x e

− τ |v1

1|

x y

ρf (z)dz

ρg (y)Mν (h)dy.

0

(i) The estimate for I (f ) − I (g): Consider x x

 − 1 ρ (y)dy − 1 ρ (y)dy I (f ) − I (g) = e τ |v1 | 0 f − e τ |v1 | 0 g fL (v).

By the mean value theorem, there exists 0 < θ < 1 such that e

− τ |v1

1|

x 0

= −e

ρf (y)dy

−e

− τ |v1

1|

x 0

ρg (y)dy

⎫ ⎧ ⎬ ⎨ 1 x 0 (1−θ)ρf (y)+θρg (y)dy ρf (y) − ρg (y)dy . ⎭ ⎩ τ |v1 |

x

− τ |v1 | 1

0

Then, since |ρf (y) − ρg (y)| ≤ sup f − g L1 , 2

x∈[0,1]

and ρf , ρg ≥ a , we have     − 1  x ρf (y)dy − τ |v1 | 0x ρg (y)dy   e τ |v1 | 0 1 − e   ≤e

≤e

− τ |v1

1|

− τ |v1

1|

x 0

x 0

(1−θ)ρf (y)+θρg (y)dy

 (1−θ)ρf +θρg dy

1 τ |v1 |

x |ρf (y) − ρg (y)|dy 0

x sup f − g L1 2 τ |v1 | x∈[0,1]



5823

5824

J. Bang, S.-B. Yun / J. Differential Equations 261 (2016) 5803–5828

≤

1 − τ |v1 | 0x (1−θ)a +θa dy 1 e sup f − g L1 2 τ |v1 | x∈[0,1]

=

1 − τa|v x | 1 sup f − g 1 . e L2 τ |v1 | x

Using this, we integrate  |I (f ) − I (g)|(1 + |v|2 )dv R3

≤

   x  − 1 ρ (y)dy   − τ |v11 | 0x ρf (y)dy − e τ |v1 | 0 g e fL (v)(1 + |v|2 )dv

v1 >0

≤

⎧ ⎪ ⎨  ⎪ ⎩

v1 >0

≤

1 e τ |v1 |

a x − τ |v | 1

⎫ ⎪ ⎬

fL (v)(1 + |v|2 )dv sup f − g L1 2 ⎪ ⎭ x

1 (as + cs ) sup f − g L1 . 2 τ x

Taking supreme in x, we have sup I (f ) − I (g) L1 ≤ 2

x

1 (as + cs ) sup f − g L1 . 2 τ x

(ii) The estimate for II(f ) − II(g): We divide it into three parts as II(f, f, f ) − II(g, g, g) ! " ! " = II(f, f, f ) − II(g, f, f ) + II(g, f, f ) − II(g, g, f ) ! " + II(g, g, f ) − II(g, g, g) = II 1 + II 2 + II 3 . By a similar manner as for I (f ), we first compute  x   − 1 ρ (z)dz   − τ |v11 | yx ρf (z)dz − e τ |v1 | y g e 

≤e

≤e ≤

− τ |v1

1|

− τ |v1

1|

x y

x y

(1−θ)ρf (z)+θρg (z)dz

(1−θ)a +θa dz

⎧ x ⎨ ⎩

y

1 τ |v1 |

x |ρg (z) − ρg (z)|dz y

⎫ ⎬ 1 dz ρf − ρg L∞ x τ |v1 | ⎭

x − y − a τ(x−y) |v1 | e sup f − g L1 2 τ |v1 | x

J. Bang, S.-B. Yun / J. Differential Equations 261 (2016) 5803–5828

≤

5825

C − a 2τ(x−y) |v1 | sup f − g

e L12 , a x

where we boundedness of xe−x (x > 0). With this, (4.2) and Lemma 3.3, we  used the uniform 2 bound R3 |II 1 |(1 + |v| )dv by  |II 1 |(1 + |v|2 )dv R3

 x ≤ R3 0

x  1  − τ |v1 | yx ρf (z)dz − 1 ρ (z)dz  1 − e τ |v1 | y g  ρf (y)Mν (f )(1 + |v|2 )dydv e τ |v1 |

⎧ ⎫ ⎪ ⎪  x ⎨ ⎬ a (x−y) 1 − 2τ |v | au 1 M1 (v1 )dydv1 sup f − g L1 e = 2 ⎪ a ⎪ τ |v1 | ⎩ ⎭ x v1 >0 0

 ≤ C ,u

 ln τ + 1 sup f − g L1 . 2 τ x

We can treat II 2 similarly:  |II 2 |(1 + |v|2 )dv R3

 = R3

1 τ |v1 |

≤ C ,u

≤ C ,u

− τ |v1

1|

⎪ ⎩

y

ρg (z)dz

⎪ ⎩

R3

1 − τ |v1 | 1 e τ |v1 |

0 v1 >0

⎧ ⎪ ⎨ x  ⎪ ⎩

x

|ρf (y) − ρg (y)|Mν (f )(1 + |v|2 )dydv

0

⎧ ⎪ ⎨ x 

 ≤ C ,u

e

⎧ x ⎪ ⎨ 0

≤ C ,u

x

0 v1 >0

x y

ρg (z)dz

x

1 e τ |v1 |

− τ |v1 | 1

1 e τ |v1 |

a (x−y) − τ |v | 1

y

⎫ ⎪ ⎬

M(f )(1 + |v|2 )dv |ρf (y) − ρg (y)|dy ⎪ ⎭

ρg (z)dz

⎫ ⎪ ⎬

M1 (v1 )dv1 dy sup f − g L1 2 ⎪ ⎭ x ⎫ ⎪ ⎬

M1 (v1 )dv1 dy sup f − g L1 2 ⎪ ⎭ x

 ln τ + 1 sup f − g L1 . 2 τ x

For the estimate of II 3 , we use Proposition 5.1 as

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J. Bang, S.-B. Yun / J. Differential Equations 261 (2016) 5803–5828

 |II 3 |(1 + |v|2 )dv R3

 = R3

1 τ |v1 |

≤ au C ,u

x e

− τ |v1

⎪ ⎩

⎧ ⎪ ⎨ x  ⎪ ⎩

x y

ρ(z)dz

ρg (y)|Mν (f ) − Mν (g)|(1 + |v|2 )dydv

0

⎧ ⎪ ⎨  v1 >0

≤ C ,u

1|

0 v1 >0

1 τ |v1 |

x e

− τ |v1

1|

x y

ρ(z)dz −C ,u |v|2

e

0

1 e τ |v1 |

a (x−y) − τ |v | 1

⎫ ⎪ ⎬

(1 + |v|2 )dvdy sup f − g L1 2 ⎪ ⎭ x

⎫ ⎪ ⎬

e−C ,u |v| dv1 dy sup f − g L1 . 2 ⎪ ⎭ x 2

Therefore, in view of (4.2), we have  II 3 ≤ C ,u

 ln τ + 1 sup f − g L1 . 2 τ x

We now gather all these estimates to obtain sup + (f ) − + (g) L1 ≤ C ,u



2

x

 1 (as + cs ) + Cτ sup f − g L1 , 2 τ x

where Cτ =

ln τ + 1 . τ

In a similar fashion, we can derive the corresponding estimate for −(f ): −



−

sup (f ) − (g) L1 ≤ C ,u 2

x

 1 (as + cs ) + Cτ sup f − g L1 . 2 τ x

Therefore, we conclude that  sup (f ) − (g) L1 ≤ C ,u x

2

  1 as + cs + Cτ sup f − g L1 . 2 τ x

This gives the desired result for sufficiently large τ > 0.

2

Acknowledgments The work of S.-B. Yun was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Science, ICT and Future Planning (NRF-2014R1A1A1006432).

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