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Research Article Received 4 June 2010 Published online 5 October 2010 in Wiley Online Library (wileyonlinelibrary.com) DOI: 10.1002/mma.1385 MOS subject classification: 76 P 05; 82 C 40; 82 D 05 The Boltzmann equation with potential force in the whole space Jie Sun a,b,c Communicated by I. Stratis In this paper, we consider the Cauchy problem of the Boltzmann equation with potential force in the whole space. When some more natural assumptions compared with those of the previous works are made on the potential force, we can still obtain a unique global solution to the Boltzmann equation even for the hard potential cases by energy method, if the initial data are sufficiently close to the steady state. Moreover, the solution is uniformly stable. Copyright © 2010 John Wiley & Sons, Ltd. Keywords: Boltzmann equation; potential force; whole space 1. Introduction We consider the Cauchy problem of the Boltzmann equation in the whole space as follows: t f + ·∇ x f −∇ x ·∇ f = Q(f,f ), f (0,x, ) = f 0 (x, ), (1) where f (t, x, ) is the distribution function of gas particles at time t0, with position x = (x 1 ,x 2 ,x 3 ) R 3 and velocity = ( 1 , 2 , 3 ) R 3 , while the potential function and the initial data f 0 (x, ) are given. The collision operator Q is defined by Q(f,f ) = R 3 ×S 2 B(| |, cos )(f ( )f ( ) f ()f ( ))d d, where = ( ) · , = +( ) · , S 2 , and cos = ( ) · | | . Here, under Grad’s angular cutoff assumption [1], the collision kernel B is supposed to be of the form B(| |, cos ) =| | | cos |, 01. All through this paper, the potential function is assumed to satisfy that L <, x H N , N4, (2) where >0 is sufficiently small. It is easy to check that e M is a stationary solution to Equation (1) 1 , where M is the standard Maxwellian (1 / (2) 3/ 2 ) exp(−|| 2 / 2). Our goal is to study the solution near this steady state f s = e M. a Department of Mathematics, University of Science and Technology of China, Hefei, Anhui 230026, People’s Republic of China b Department of Mathematics, City University of Hong Kong, Kowloon, Hong Kong c Joint Advanced Research Center of University of Science and Technology of China and City University of Hong Kong, Suzhou, Jiangsu 215123, People’s Republic of China Correspondence to: Jie Sun, Department of Mathematics, University of Science and Technology of China, Hefei, Anhui 230026, People’s Republic of China. E-mail: [email protected] Copyright © 2010 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2011, 34 621–632 621

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Page 1: The Boltzmann equation with potential force in the whole space

Research Article

Received 4 June 2010 Published online 5 October 2010 in Wiley Online Library

(wileyonlinelibrary.com) DOI: 10.1002/mma.1385MOS subject classification: 76 P 05; 82 C 40; 82 D 05

The Boltzmann equation with potential forcein the whole space

Jie Suna,b,c∗†

Communicated by I. Stratis

In this paper, we consider the Cauchy problem of the Boltzmann equation with potential force in the whole space. Whensome more natural assumptions compared with those of the previous works are made on the potential force, we canstill obtain a unique global solution to the Boltzmann equation even for the hard potential cases by energy method, ifthe initial data are sufficiently close to the steady state. Moreover, the solution is uniformly stable. Copyright © 2010John Wiley & Sons, Ltd.

Keywords: Boltzmann equation; potential force; whole space

1. Introduction

We consider the Cauchy problem of the Boltzmann equation in the whole space as follows:{�tf +� ·∇xf −∇x� ·∇�f =Q(f, f ),

f (0, x,�)= f0(x,�),(1)

where f (t, x,�) is the distribution function of gas particles at time t�0, with position x = (x1, x2, x3)∈R3 and velocity �= (�1,�2,�3)∈R3,while the potential function � and the initial data f0(x,�) are given. The collision operator Q is defined by

Q(f, f )=∫

R3×S2B(|�−�∗|, cos�)(f (�′)f (�′∗)−f (�)f (�∗)) d�∗ d�,

where

�′ =�−�(�−�∗) ·�, �′∗ =�∗+�(�−�∗) ·�, �∈S2,

and

cos�= (�−�∗) ·�|�−�∗| .

Here, under Grad’s angular cutoff assumption [1], the collision kernel B is supposed to be of the form

B(|�−�∗|, cos�)=|�−�∗|�|cos�|, 0���1.

All through this paper, the potential function � is assumed to satisfy that

‖�‖L∞<∞, ‖∇x�‖HN ��, N�4, (2)

where �>0 is sufficiently small. It is easy to check that e−�M is a stationary solution to Equation (1)1, where M is the standardMaxwellian (1 / (2�)3/2) exp(−|�|2 / 2). Our goal is to study the solution near this steady state fs =e−�M.

aDepartment of Mathematics, University of Science and Technology of China, Hefei, Anhui 230026, People’s Republic of ChinabDepartment of Mathematics, City University of Hong Kong, Kowloon, Hong KongcJoint Advanced Research Center of University of Science and Technology of China and City University of Hong Kong, Suzhou, Jiangsu 215123, People’s Republic

of China∗Correspondence to: Jie Sun, Department of Mathematics, University of Science and Technology of China, Hefei, Anhui 230026, People’s Republic of China.†E-mail: [email protected]

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Set the perturbation

f = fs +f 1/2s u. (3)

Then the problem (1) turns into the Cauchy problem of the perturbation u,⎧⎨⎩

�tu+�·∇xu−∇x� ·∇�u=e−�Lu+e−�/2�(u, u),

u(0, x,�)= f−1/2s (f0(x,�)−fs),

(4)

where Lu and �(u, u) are defined by

Lu = M−1/2[Q(M, M1/2u)+Q(M1/2u, M)],

�(u, u) = M−1/2Q(M1/2u, M1/2u).

We denote the kernel of the linearized collision operator L by N. N⊥ denotes the orthogonal complement of N in L2� . It is

known that N is spanned by the collision invariants,

N=span{M1/2, �iM1/2, i=1, 2, 3, |�|2M1/2}.

See [2]. We denote the projection from L2� onto N by P. Following the arguments in [3] for the hard sphere case, the linearized

collision operator L can be decomposed as L=−+K , where

=∫

R3×S2B(|�−�∗|, cos�)M(�∗) d�∗ d�

and

Ku=M−1/2∫

R3×S2B(|�−�∗|, cos�)(M1/2(�′)u(�′)M(�′∗)+M1/2(�′∗)u(�′∗)M(�′)−M1/2(�∗)u(�∗)M(�)) d�∗ d�.

is called as the collision frequency.With respect to the collision linearized operator L, following the arguments in [3] for the hard sphere case, we can have

(1) There exist positive constants 0>0 and 1>0, such that

0(1+|�|)��(�)�1(1+|�|)�;

(2) K is a self-adjoint compact operator in L2� ;

(3) L is non-positive and there is a constant �0>0 such that∫R3

uLu d��−�0

∫R3

|(I−P)u|2 d� for any u∈L2(R3,d�),

which is the so-called microscopic H-theorem.

Before introducing the main result, we make some notations.Notation: We use 〈, 〉 to denote the L2 inner product in the Hilbert space L2(R3

� ) or L2(R3x ) or L2(R3

x ×R3� ), and ‖·‖ to denote the

corresponding L2 norm. When the norms need to be separated from each other, we write ‖·‖L2(R3� ), ‖·‖L2(R3

x ) and ‖·‖L2(R3x ×R3

� ),

respectively. We also define

〈u, v〉 =〈u,(�)v〉to be the weighted L2 inner product in L2(R3

� ) or L2(R3x ×R3

� ) for a pair of functions u(�) and v(�), or u(x,�) and v(x,�), respectively,

and use ‖·‖ to denote the corresponding weighted L2 norm where (�) is the collision frequency.For the multiple indices �= (�1,�2,�3) and = ( 1, 2, 3), as usual we denote

��x�

� =��1x1

��2x2

��3x3

� 1�1

� 2�2

� 3�3

.

The length of � is |�|=�1 +�2 +�3. In addition, C denotes a generic positive (generally large) constant and � denotes a genericpositive (generally small) constant. They may be different from line to line.

For later use, we define the energy functional as

[[u(t)]]2 = ∑|�+ |�N

‖��x�

�u(t)‖2,

the weighed energy functional as

‖|u(t)‖|2 = ∑|�+ |�N

‖��x�

�u(t)‖2

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and the dissipation rate as

[[u(t)]]2 = ∑

|�+ |�N‖��

x� � (I−P)u(t)‖2

+ ∑0<|�|�N

‖��xu(t)‖2

,

where N�4 is an integer.After the above preparations, we will present the key result in the following.

Theorem 1.1Let f0 = fs +f 1/2

s u0�0. Under the above assumption (2), there exist constants �0>0 and �>0 such that, if [[u(0)]]��0, then thereexists a unique global solution u(t, x,�) to Equation (4) such that f (t, x,�)= fs +f 1/2

s u�0, and satisfying

[[u(t)]]2 +�∫ t

0[[u(s)]]2

ds�C[[u(0)]]2 ∀t�0. (5)

Theorem 1.2Let all conditions in the above theorem hold. Set

f0(x,�)= fs +f 1/2s u0�0, g0(x,�)= fs +f 1/2

s v0�0.

There exist �1 ∈ (0,�0) and �1 such that, if

max [[u(0)]], [[v(0)]]��1,

then the solutions u and v corresponding to the initial data u0 and v0 obtained in the above theorem satisfy

[[u(t)−v(t)]]2 +�1

∫ t

0[[(u−v)(�)]]2

d��C[[u(0)−v(0)]]2 ∀t�0. (6)

Remark 1.3The above results also hold for the spatial dimension n�3 since the whole proof is based on the energy method only and thereare similar Sobolev inequalities to the ones in Lemma 2.1 in Rn(n�3).

Remark 1.4In this paper, we use the perturbation f = fs +f 1/2

s u not the perturbation f = fs +M1/2u in order to avoid the term 12 �·∇x�u. If it

exists, we can not get rid of the zeroth-order term of the macroscopic part of u under the assumption (2) with respect to �. Onemore thing, if one uses the perturbation f = fs +M1/2u, then from the property (1), the dissipation rate from L is not enough tocontrol the term that increases in velocity with the rate |�| when 0��<1, which is mentioned in [4, 5].

Remark 1.5For the Boltzmann equation in the whole space, so far there are two kinds of L2 energy methods available. One is initiated by Liuand Yu [6] and developed by Yang and coworkers [7, 8] in terms of the macro-micro decomposition around a local Maxwellian.The other method is founded by Guo [9, 10], and developed by Strain [11] and Strain and Guo [12, 13] due to the decompositionaround a global Maxwellian.

There have been a lot of amazing works about the Boltzmann equation with external force in the whole space such as [5, 14, 15]and so on. We discuss the solution near the steady state with a more natural potential force (2) compared with the previous works.In [14], the potential � is restricted as follows:

‖�‖L2 + ∑1�|�|�N+1

‖��x�‖L3<�0,

where �0 is sufficiently small. In [4, 5], the potential � is assumed to satisfy

‖�‖L∞<∞, ‖(1+|x|)2∇x�‖L∞ + ∑2�|�|�N

‖(1+|x|)��x�‖L∞<�,

where � is sufficiently small.

2. Global existence

For further analysis, we need the following Sobolev’s inequalities from [16].

Lemma 2.1Let u∈H2(R3). Then

(1) ‖u‖L6�C‖∇u‖,(2) ‖u‖L∞�C‖∇u‖H1 .

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We also need to state some properties about the linearized collision operator L and the collision operator �, which can beobtained by the similar arguments in [9].

Lemma 2.2The derivatives of are uniformly bounded. Let | |=k, then for any small �>0, there exists a positive constant Ck,� such that, for

any u∈Hk(R3 ×R3),

‖� �Ku‖2��

∑| ′|=k

‖� ′� u‖2 +Ck,�‖u‖2.

Lemma 2.3Let u, v and w be proper functions. We can have

|〈� ��(u, v), w〉| � C

∑ 1+ 2�

∫R3

(‖� 1 u‖‖� 2 v‖‖w‖+‖� 1 u‖‖� 2 v‖‖w‖) dx,

‖〈�(u, v), w〉‖L2x+‖〈�(v, u), w〉‖L2

x� C sup

x‖w‖L2

�‖sup

x‖u‖L2

�‖v‖.

(7)

2.1. Local existence

We will construct a local solution to the Boltzmann equation (4) in this subsection. The construction is based on the uniform energyestimate for a sequence of iterating approximate solutions.

Theorem 2.4For any sufficiently small �>0 and �>0, there exists a constant T∗>0 such that if [[u(0)]]2 is small enough, then there exists a uniqueclassical solution u(t, x,�) to the Boltzmann equation (4) in [0, T∗]×R3 ×R3 such that

sup0�t�T∗

[[u(t)]]2 +∫ T∗

0‖|u(t)‖|2 dt��2

and [[u(t)]]2 is continuous over [0, T∗]. If f0 = fs +f 1/2s u0�0, then

f (t, x,�)= fs +f 1/2s u(t, x,�)�0.

ProofIn order to solve (4), we construct the following iterating sequence {un}(n�0):⎧⎪⎪⎪⎨

⎪⎪⎪⎩u0(t, x,�)=u0(x,�),

un+1t +� ·∇xun+1 −∇x�·∇�un+1 +e−�un+1 =e−�Kun +e−�/2�(un, un),

un+1(0, x,�)=u0(x,�).

(8)

Continually, we will prove that there exists T∗>0 such that for sufficiently small �, if

sup0�t�T∗

[[un(t)]]2 +∫ T∗

0‖|un(t)‖|2 dt��2,

then

sup0�t�T∗

[[un+1(t)]]2 +∫ T∗

0‖|un+1(t)‖|2 dt��2. (9)

We estimate un+1 step by step. Multiplying equation (8) by un+1 and integrating the result with respect to (x,�) over R3 ×R3, wecan get

d

dt‖un+1‖2 +�‖un+1‖2

�C‖un‖2 +C[[un(t)]]2‖|un(t)‖|2 (10)

where � is a positive constant. Applying ��x(|�|>0) to Equation (8), then multiplying it by ��

xun+1 and integrating the result withrespect to (x,�) over R3 ×R3 yield that

d

dt‖��

xun+1‖2 +��‖��xun+1‖2

� C�∑

|�′|�|�|‖��′

x un+1‖2 +C�

∑|�′|�|�|−1

‖∇���′x un+1‖2

+C‖��xun‖2 +C�

∑|�′|�|�|

‖��′x un‖2 +C[[un(t)]]2‖|un(t)‖|2, (11)

where �� is a positive constant.

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Finally, applying ��x�

� (| |>0) to Equation (8), then multiplying it by ��x�

�un+1 and integrating the result with respect to (x,�) over

R3 ×R3, we can obtain

d

dt‖��

x� �un+1‖2 +�

�‖��x�

�un+1‖2 � C�‖|un+1(t)‖|2 +C

∑ ′<

(‖∇x��x�

′un+1‖2 +‖��

x� ′un+1‖2)

+C[[��xun]]2 +C[[un(t)]]2‖|un(t)‖|2, (12)

where � � is a positive constant.

A suitable linear combination of the inequalities (10)–(12) can yield that for 0�t�T∗,

[[un+1(t)]]2 +

∫ t

0‖|un+1(�)‖|2 d��C[[u(0)]]2 +C

∫ t

0[[un(�)]]2 d�+C

∫ t

0[[un(�)]]2‖|un(�)‖|2 d�. (13)

If [[u(0)]]2, � and T∗ are small enough, then we can have that statement (9) holds. By a similar discussion like above, we can provethat the sequence {un} is a Cauchy sequence. We take n→∞ and obtain a classical local solution u to Equation (4). The proof forthe uniqueness of the local solution u and the positivity of f = fs +f 1/2

s u is omitted. �

2.2. A priori estimate

In this subsection, we will try to prove the following theorem.

Theorem 2.5Suppose that u(t, x,�) is a solution in [0, T]×R3 ×R3 to the Cauchy problem (4). Under assumption (2) about �, there exist positiveconstants � and � that are independent of T , such that if

sup0�t�T

[[u(t)]]2��2, (14)

then

[[u(t)]]2 +�∫ t

0[[u(s)]]2

ds�C[[u(0)]]2 (15)

for ∀t ∈ [0, T].

The proof of the above theorem is divided into two parts: one part is to obtain the macroscopic dissipation rate, and the otherpart is to deal with the microscopic dissipation rate.

2.2.1. Macroscopic dissipation. We will discuss the macroscopic dissipation rate in this part. For fixed (t, x), u can be decomposed as

u = u1 +u2, u1 =Pu, u2 = (I−P)u,

Pu = (a(t, x)+b(t, x) ·�+c(t, x)|�|2)M1/2.(16)

For later use, we show the local conservation laws for the macroscopic parts below. Multiplying Equation (4) by M1/2, �M1/2 and(|�|2 / 2)M1/2, respectively, and integrating the results with respect to � over R3, we can have⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

�t(a+3c)+∇x ·b− 12 ∇x� ·b=0,

�tb+∇x(a+5c)+∇x ·〈�⊗�M1/2, u2〉+ 12 ∇x�(a+c)

− 12 ∇x�·〈�⊗�M1/2, u2〉=0,

�t(3a+15c)+5∇x ·b+∇x ·〈�|�|2M1/2, u2〉− 12 ∇x� ·b

− 12 ∇x�·〈�|�|2M1/2, u2〉=0.

(17)

From the above conservation of mass and energy, we can get

�tc+ 13 ∇x ·b+ 1

6 ∇x ·〈�|�|2M1/2, u2〉+ 16 ∇x�·b− 1

12 ∇x� ·〈�|�|2M1/2, u2〉=0. (18)

It is easy to see that the above local conservation laws are the linearized Euler-type equations that are not closed due to theappearance of the microscopic part u2. Therefore, we need to study the higher moments of u2,

〈�⊗�M1/2, u2〉, 〈�|�|2M1/2, u2〉.

In order to study the higher moments of u2, we rewrite Equation (4) as follows:

�tu1 +� ·∇xu1 −∇x�·∇�u1 =−�tu2 + l+n, (19)

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where

l =e−�Lu+∇x� ·∇�u2 −�·∇xu2, n=e−�/2�(u, u).

Then, putting u1 = (a+b ·�+c|�|2)M1/2 into Equation (19), we can obtain

(�ta−∑

i�xi �bi

)M1/2 +∑

i

(�tbi +�xi a−2c�xi �+ 1

2a�xi �

)�iM

1/2 +∑i

(�tc+�xi bi + bi

2�xi �

)|�i|2M1/2

+∑i<j

(�xi bj +�xj bi + 1

2�xi �bj + 1

2�xj �bi

)�i�jM

1/2 +∑i

(�xi c+ c

2�xi �

)�i|�|2M1/2 =−�tu2 + l+n. (20)

In the following, we define two kinds of bounded linear operators: Aij and Bi , i=1, 2, 3; j=1, 2, 3.

Aij(u)=〈(�i�j −1)M1/2, u〉, Bi(u)=〈(|�|2 −5)�iM1/2, u〉

where u∈L2� . Then, applying the operators to Equation (20), we can have the following equations for the higher moments of u2:

⎧⎪⎪⎪⎨⎪⎪⎪⎩

�t(Aii(u2)+2c)+2�xi bi +bi�xi �=Aii(l+n), i=1, 2, 3,

�tAij(u2)+�xi bj +�xj bi + 12 �xi �bj + 1

2 �xj �bi =Aij(l+n), i �= j,

�tBi(u2)+10�xi c+5c�xi �=Bi(l+n), i=1, 2, 3.

(21)

Using the similar analysis in [9], it is easy to see that the above system yields the following equations that contain parabolicterms.

Lemma 2.6For fixed j,

−�t

[∑i

�xi Aij(u2)+ 1

2�xj Ajj(u2)

]−�xbj −�xj �xj bj = 1

2

∑i

�xi (�xi �bj +�xj �bi)− 1

2

∑i �=j

�xj (bi�xi �)+ 1

2

∑i �=j

�xj Aii(l+n)−∑i

�xi Aij(l+n) (22)

holds.

Before estimating the dissipation of the macroscopic part of u, we state the following lemma first.

Lemma 2.7Let |�|�N−1. We have

‖Aij(��xl)‖+‖Bi(�

�xl)‖ � C

∑|�|�N

‖��xu2‖,

‖Aij(��xn)‖+‖Bi(�

�xn)‖ � C[[u(t)]][[u(t)]].

From Lemma 2.3, we know that the proof of the above lemma can follow the arguments in [17]. In the following, we use theidea of [18] to estimate the macroscopic dissipation rate by introducing a free energy functional.

Applying ��x(|�|�N−1) to Equation (22), multiplying it by ��

xbj and integrating the final equation with respect to x over R3, wecan obtain

d

dt

⟨∑i

�xi Aij(��xu2)+ 1

2�xj Ajj(�

�xu2),−��

xbj

⟩+‖��

x∇xbj‖2 +‖��x�xj bj‖2

=−⟨∑

i�xi Aij(�

�xu2)+ 1

2�xj Ajj(�

�xu2),�t�

�xbj

⟩− 1

2

⟨∑i �=j

Aii(��x(l+n)),��

x�xj bj

⟩+∑

i〈Aij(�

�x(l+n)),��

x�xi bj〉

−1

2

∑i〈��

x(�xi �bj +�xj �bi)),��x�xi bj〉+ 1

2

∑i �=j

〈��x(bi�xi �),��

x�xj bj〉. (23)

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From the local conservation law (17)2, we can get

d

dt

(∑i〈Aij(�

�xu2),�xi �

�xbj〉+ 1

2〈Ajj(�

�xu2),��

x�xj bj〉)

+ 1

2‖��

x∇xbj‖2

��‖��x∇x(a+3c, c)‖2 + C

�‖∇x�

�xu2‖2 +C

∑|�′|�N

‖��′x u2‖2

+C�∑

0<|�′|�N−1‖��′

x (a+3c, b, c)‖2 +C[[u(t)]]2[[u(t)]]2 , (24)

where � is any positive constant.Applying ��

x(|�|�N−1) to Equation (17)2, multiplying it by ��x∇x(a+3c) and integrating the final equation with respect to x over

R3, we can have

d

dt〈��

xbj,�xj ��x(a+3c)〉+‖�xj �

�x(a+3c)‖2

=−〈�xj ��xbj,��

x�t(a+3c)〉−2〈�xj ��xc,�xj �

�x(a+3c)〉

−⟨��

x

(�xj �

a+c

2

),�xj �

�x(a+3c)

⟩+ 1

2

∑i〈��

x(�xi �Aij(u2)),

��x�xj (a+3c)〉−∑

i〈�xi Aij(�

�xu2),��

x�xj (a+3c)〉. (25)

From the local conservation law (17)1, we can get the following estimate:

d

dt〈��

xbj,�xj ��x(a+3c)〉+ 1

2‖�xj �

�x(a+3c)‖2

�C‖�xj ��xc‖2 +C

∑i

‖�xi ��xb‖2 +C‖∇x��

xu2‖2

+C�∑

0<|�′|�N−1‖��′

x (a+3c, b, c)‖2 +C�∑

|�′|�N−1‖��′

x u2‖2. (26)

Applying ��x(|�|�N−1) to Equation (21)3, multiplying it by ��

x�xi c and integrating the final equation with respect to x over R3, wecan get

d

dt〈Bi(�

�xu2),��

x�xi c〉+10‖��x�xi c‖2 =〈Bi(�

�x�xi u2),−��

x�tc〉+〈Bi(��x(l+n)),��

x�xi c〉−5〈��x(c�xi �),��

x�xi c〉. (27)

From the local conservation law (18), we can obtain

d

dt〈Bi�

�xu2),�xi �

�xc〉+5‖�xi �

�xc‖2�C�

∑0<|�′|�N−1

‖��′(b, c)‖2 +C‖∇x ·��

xb‖2 +C∑

|�′|�N‖��′

x u2‖2 +C[[u(t)]]2[[u(t)]]2 . (28)

From (24), (26) and (28), we can get the following lemma.

Lemma 2.8There exists a positive constant � that is independent of T such that for t ∈ [0, T] and any �>0,

d

dt

∑|�|�N−1

[∑ij

〈Aij(��xu2),��

x(�xi bj +�xj bi)〉+∑j〈Ajj(�

�xu2),�xj �

�xbj〉

]

+�∑

|�|�N−1

∑j

‖��x∇xbj‖2�(C�+�)

∑|�|�N−1

‖∇x��x(a+3c, c)‖2 + C

∑0<|�|�N

‖��xu2‖2

+C∑

|�|�N‖��

xu2‖2 +C[[u(t)]]2[[u(t)]]2 , (29)

d

dt

∑|�|�N−1

∑j〈��

xbj,�xj ��x(a+3c)〉+�

∑|�|�N−1

‖∇x��x(a+3c)‖2

�C∑

|�|�N−1

(∑j

‖�xj ��x(b, c)‖2 +‖∇x��

xu2‖2

)+C�

∑|�|�N−1

‖��xu2‖2 (30)

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and

d

dt

∑|�|�N−1

∑i〈Bi(�

�xu2),�xi �

�xc〉+�

∑|�|�N−1

‖∇x��xc‖2

�C�∑

0<|�|�N−1‖��

x(b, c)‖2 +C∑

|�|�N‖��

xu2‖2 +C∑

|�|�N−1‖∇x ·��

xb‖2

+C[[u(t)]]2[[u(t)]]2 (31)

hold when �>0 is small enough.

For further discussion, we define a free energy functional Efree(u(t)) by

Efree(u(t)) = ∑|�|�N−1

ij

〈Aij(��xu2),��

x(�xi bj +�xj bi)〉

+ ∑�|�N−1

j

〈Ajj(��xu2),�xj �

�xbj〉+�0

∑|�|�N−1

i

〈Bi(��xu2),�xi �

�xc〉

+�1∑

|�|�N−1j

〈��xbj,�xj �

�x(a+3c)〉, (32)

where �0 and �1 will be decided in the following Corollary 2.1. From Lemma 2.8 and the definitions of operators Aij and Bi , we canhave the following corollary.

Corollary 2.1There exist positive constants �, �0 and �1 that are independent of T , such that

d

dtEfree(u(t))+�

∑0<|�|�N

‖��x(a+3c, b, c)‖2�C‖u2‖2

L2� (HN

x )+C[[u(t)]]2[[u(t)]]2

(33)

and

|Efree(u(t))|�C‖u‖2L2

� (HNx )

(34)

for sufficiently small � in (31) and �.

2.2.2. Microscopic dissipation. We will present the microscopic dissipation rate by the standard energy method in this part. MultiplyingEquation (4) by u and integrating the result with respect to (x,�) over R3 ×R3, we can get

d

dt‖u‖2 +�‖u2‖2

�C[[u(t)]]2[[u(t)]]2 , (35)

where � is a positive constant. Applying ��x(|�|>0) to Equation (8), then multiplying it by ��

xu and integrating the result with respectto (x,�) over R3 ×R3 yield that

d

dt‖��

xu‖2 +��‖��xu2‖2

�C�∑

|�′|<|�|‖��′

x u2‖2 +C�

∑0<|�′|<|�|

‖∇���′x u‖2 +C�

∑0<|�′|�2

‖∇���′x u‖2 +C�‖��

xu‖2 +C[[u(t)]]2[[u(t)]]2 , (36)

where �� is a positive constant.Applying operator I−P to Equation (4), then we get the following equation:

�tu2 +�·∇xu2 −∇x� ·∇�u2 =−(I−P)(� ·∇xu1 −∇x� ·∇�u1)+P(�·∇xu2 −∇x� ·∇�u2)+e−�Lu2 +e−�/2�(u, u). (37)

Applying ��x�

� (| |>0) to the above equation, then multiplying it by ��

x� �u and integrating the final equation with respect to (x,�)

over R3 ×R3, we can obtain, for any �>0, there exists a positive constant � � such that

d

dt‖��

x� �u2‖2 +�

�‖��x�

�u2‖2

�C�∑

|�′+ ′|�N‖��′

x � ′� u2‖2

+�∑

| ′|=| |‖��

x� ′� u2‖2 +C�

∑0<|�′|�N−1

‖��′x (a+3c, b, c)‖2

+C∑

′< ‖��

x� ′

u2‖2 +C∑

| ′|=1‖∇x��

x� − ′u2‖2 +C�‖��

xu2‖2 +C‖∇x��xu2‖2

+C‖∇x��x(a+3c, b, c)‖2 +C[[u(t)]]2[[u(t)]]2

. (38)

The estimate about the microscopic dissipation rate is obtained now.

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From the above estimates (35), (36) and (38), we can have the following lemma.

Lemma 2.9There exists a positive constant � independent of T such that for any t ∈ [0, T], if � in (38) and � are small enough, we have

1

2

d

dt‖u‖2 +�‖u2‖2

�C[[u(t)]]2[[u(t)]]2 , (39)

1

2

d

dt

∑1�|�|�N

‖��xu‖2 +�

∑1�|�|�N

‖��xu2‖2

� C�∑

|�|�N−1‖∇x��

x(a, b, c)‖2

+C�‖u2‖2 +C�

∑|�|�N−1

‖��x∇�u2‖2

+C[[u(t)]]2[[u(t)]]2 , (40)

1

2

d

dt

∑| |=k

|�|+| |�N

‖��x�

�u2‖2 +�∑

| |=k|�|+| |�N

‖��x�

�u2‖2

�C�∑

|�|�N−1‖∇x��

x(a, b, c)‖2 +C�∑

|�′|+| ′|�N‖��′

x � ′� u2‖2

+C∑

|�|�N−k‖��

xu2‖2

+C∑

|�|�N−k‖��

x∇xu2‖2 +C

∑|�|�N−k

‖��x∇x(a+3c, b, c)‖2

+C∑

| ′|<k|�|�N−k

‖��x�

′� u2‖2 +C

∑| ′|=k−1|�|�N−k

‖∇x��x�

′� u2‖2 +C[[u(t)]]2[[u(t)]]2

, (41)

where k =1,. . . , N.

From (33), (39) and (40), we can know that there exist positive constants � and M that are independent of T , such that

M‖u‖2L2

� (HNx )

+Efree(u(t))=O(1)‖u‖2L2

� (HNx )

(42)

and

d

dt[M‖u‖2

L2� (HN

x )+Efree(u(t))]+�

[ ∑|�|�N

‖��xu2‖2

+ ∑0<|�|�N

‖��x(a+3c, b, c)‖2

]

�C�∑

|�|�N−1‖��

x∇�u2‖2 +C[[u(t)]]2[[u(t)]]2

(43)

when � is sufficiently small.A suitable linear combination of (41) and (43) yields that there exist positive constants �2 and � such that

d

dt

⎡⎢⎣M‖u‖2

L2� (HN

x )+�2

∑| |>0

|�|+| |�N

‖��x�

�u2‖2 +Efree(u(t))

⎤⎥⎦+�

[ ∑|�+ |�N

‖��x�

�u2‖2 + ∑0<|�|�N

‖��x(a+3c, b, c)‖2

]�C[[u(t)]]2[[u(t)]]2

. (44)

For simplicity, we define two energy functionals as follows:

E(u(t))=M‖u‖2L2

� (HNx )

+�2∑

| |>0|�|+| |�N

‖��x�

�u2‖2 +Efree(u(t))

and

D(u(t))= ∑|�+ |�N

‖��x�

�u2‖2 + ∑0<|�|�N

‖��x(a+3c, b, c)‖2.

Then, the inequality (44) can be rewritten as

d

dtE(u(t))+�D(u(t))�C[[u(t)]]2[[u(t)]]2

. (45)

It is easy to check that

E(u(t))=O(1)[[u(t)]]2, D(u(t))=O(1)[[u(t)]]2 .

Therefore, Theorem 2.5 holds if � is sufficiently small. As we know, Theorems 2.4 and 2.5 as well as the continuum argument implythe global existence. Hence, Theorem 1.1 is proved.

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3. Uniform stability

In this section, we will study the uniform stability of the unique solution obtained in Theorem 1.1. It is supposed that there existsolutions u(t, x,�), v(t, x,�) to the Boltzmann equation (4) corresponding to the initial data u0(x,�), v0(x,�) with

max{[[u(0)]], [[v(0)]]}��1, (46)

where �1 ∈ (0,�0) will be decided later. Since �1<�0, from Theorem 1.1, we can have

[[u(t)]]2 +�∫ t

0[[u(s)]]2

ds � C[[u(0)]]2,

[[v(t)]]2 +�∫ t

0[[v(s)]]2

ds � C[[v(0)]]2,

(47)

for any t�0.If we set w =u−v, it is easy to see that w satisfies{

�tw+� ·∇xw−∇x� ·∇�w =e−�Lw+e−�/2�(w, u)+e−�/2�(v, w),

w(0, x,�)=u0 −v0,(48)

only the collision part e−�/2�(u, u) in (4) is replaced by e−�/2(�(w, u)+�(v, w)). Hence, we can study w as in Section 2.2.For fixed (t, x), w can be decomposed as

w = w1 +w2, w1 =Pw, w2 = (I−P)w,

Pw = (aw(t, x)+bw(t, x) ·�+cw(t, x)|�|2)M1/2.(49)

For the coefficients aw , bw and cw , we also have the following equation:⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

�t(aw +3cw)+∇x ·bw − 12 ∇x�·bw =0,

�tbw +∇x(aw +5cw)+∇x ·〈�⊗�M1/2, w2〉+ 12 ∇x�(aw +cw)− 1

2 ∇x�·〈�⊗�M1/2, w2〉=0,

�tcw + 13 ∇x ·bw + 1

6 ∇x ·〈�|�|2M1/2, w2〉+ 16 ∇x�·bw − 1

12 ∇x�·〈�|�|2M1/2, w2〉=0,

�t(Aii(w2)+2cw)+2�xi biw +bi

w�xi �=Aii(l+n), i=1, 2, 3,

�tAij(w2)+�xi bjw +�xj bi

w + 12 �xi �b

jw + 1

2 �xj �biw =Aij(l+n), i �= j,

�tBi(w2)+10�xi cw +5cw�xi �=Bi(l+n), i=1, 2, 3,

(50)

where l and n are replaced by

l =e−�Lw+∇x� ·∇�w2 −�·∇xw2, n=e−�/2�(w, u)+e−�/2�(v, w).

Instead of Lemma 2.7, we have the following lemma.

Lemma 3.1Let |�|�N−1. We have

‖Aij(��xl)‖+‖Bi(�

�xl)‖ � C

∑|�|�N

‖��xw2‖,

‖Aij(��xn)‖+‖Bi(�

�xn)‖ � C([[u(t)]]+[[v(t)]])[[w(t)]].

Thus, for the macroscopic dissipation rate of w, we can have

d

dtEfree(w(t))+�

∑0<|�|�N

‖��x(aw +3cw, bw, cw)‖2�C‖w2‖2

L2� (HN

x )+C([[u(t)]]2 +[[v(t)]]2)[[w(t)]]2

(51)

and

|Efree(w(t))|�C‖w‖2L2

� (HNx )

, (52)

where the free energy functional Efree(w(t)) is defined as (32).With respect to the microscopic part of w, we only need to reconsider such terms as

〈��x�

� (e−�/2�(w, u)+e−�/2�(v, w)),��x�

�w2〉,

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where |�+ |�N. From Lemma 2.3, we can obtain

〈��x�

� (e−�/2�(w, u)+e−�/2�(v, w)),��x�

�w2〉� C

�{([[u]]2

+[[v]]2 )[[w]]2 +([[u]]2 +[[v]]2)[[w]]2

}+�‖��x�

�w2‖2, (53)

for any �>0.Then we can have the following lemma for the microscopic dissipation rate of w like Lemma 2.9.

Lemma 3.2There exists a positive constant � independent of T such that for any t�0, if � in (53) and � are small enough, we can have

1

2

d

dt‖w‖2 +�‖w2‖2

�C{([[u]]2 +[[v]]2

)[[w]]2 +([[u]]2 +[[v]]2)[[w]]2 }, (54)

1

2

d

dt

∑1�|�|�N

‖��xw‖2 +�

∑1�|�|�N

‖��xw2‖2

�C�∑

|�|�N−1‖∇x�

�x(aw, bw, cw)‖2 +C�‖w2‖2

+C�∑

|�|�N−1‖��

x∇�w2‖2

+C{([[u]]2 +[[v]]2

)[[w]]2 +([[u]]2 +[[v]]2)[[w]]2 }, (55)

1

2

d

dt

∑| |=k

|�|+| |�N

‖��x�

�w2‖2 +�

∑| |=k

|�|+| |�N

‖��x�

�w2‖2

�C�∑

|�|�N−1‖∇x��

x(aw, bw, cw)‖2 +C�∑

|�′|+| ′|�N‖��′

x � ′� w2‖2

+C∑

|�|�N−k‖��

xw2‖2

+C∑

|�|�N−k‖��

x∇xw2‖2 +C

∑|�|�N−k

‖��x∇x(aw +3cw, bw, cw)‖2

+C∑

| ′|<k|�|�N−k

‖��x�

′� w2‖2 +C

∑| ′|=k−1|�|�N−k

‖∇x��x� ′

� w2‖2 +C{([[u]]2 +[[v]]2

)[[w]]2 +([[u]]2 +[[v]]2)[[w]]2 }, (56)

where k =1,. . . , N.

By the similar analysis to that in Section 2.2, from (51) and (54)–(56), we know that

d

dtE(w(t))+�D(w(t))�C{([[u]]2

+[[v]]2 )[[w]]2 +([[u]]2 +[[v]]2)[[w]]2

}, (57)

and

E(w(t))=O(1)[[w(t)]]2, D(w(t))=O(1)[[w(t)]]2 ,

where E(w(t)) and D(w(t)) are defined as the ones in the above section.From the assumption (46), letting �1 be sufficiently small, we can get that there exists a positive constant � such that

d

dtE(w(t))+�D(w(t))�C{([[u]]2

+[[v]]2 )E(w(t)). (58)

From Gronwall’s inequality and assumption (47), we know that

E(w(t))�CE(w(0)). (59)

Then we can figure out that there exists a positive constant �3 such that

[[u(t)−v(t)]]2 +�3

∫ t

0[[(u−v)(�)]]2

d��C[[u(0)−v(0)]]2.

Therefore, Theorem 1.2 is proved.

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