Bulletin of the Institute of Mathematics
Academia Sinica (New Series)
Vol. 5 (2010), No. 1, pp. 1-10
MIXTURE LEMMA
BY
HUNG-WEN KUO, TAI-PING LIU AND SE EUN NOH
Abstract
The Mixture Lemma plays a central role in the study of
the singularity of solutions of the Boltzmann equation. We offer
a detailed proof of the lemma. The proof depends on the proper
switching of the differentiations with respect to the space variables
to those of the microscopic velocities, and depends on the precise
regularity properties of the collision operator.
1. Introduction
Consider the Boltzmann equation
Ft + ξ · ∇x F = Q(F, F ), F = F (x, t, ξ) ≥ 0,
x, ξ ∈ R3 , t ∈ R+ ,
for the hard sphere model
Z Z
1
′
′
′
′
Q(F, G) ≡
[F (ξ )G(ξ∗ ) + F (ξ∗ )G(ξ ) − F (ξ)G(ξ∗ )
2 R3 S2+
−F (ξ∗ )G(ξ)]|(ξ − ξ∗ ) · Ω|dΩdξ∗ .
In this paper we offer a detailed proof of the Mixture Lemma of [5], [6].
Received February 16, 2010.
AMS Subject Classification: 76P05, 35L65.
Key words and phrases: Boltzmann equation, Mixture Lemma.
The authors would like to thank Professor Shih-Hsien Yu for his interest in the present
work. The first author is supported by Institute of Mathematics, Academia Sinica, the
research of the second author is supported in part by National Science Council Grant
96-2628-M-001-011 and National Science Foundation Grant DMS 0709248, and the third
author is supported by Institute of Mathematics, Academia Sinica.
1
2
HUNG-WEN KUO, TAI-PING LIU AND SE EUN NOH
[March
The Lemma is to study the singularity of the solutions for the Boltzmann
equation and is essential for the construction of the Green’s function for
the linearized Boltzmann equation. The Green’s function approach, e.g. [4],
[7], [8], [9], [10], offers more quantitative informations for the solutions of
the Boltzmann equation and is useful for the studies of interesting physical
phenomena. There is the celebrated Velocity Averaging Lemma, [3], which
is often used for the study of the combined effect of transport and collision.
It serves a different purpose of gaining space compactness for macroscopic
quantities, e.g. [2]. The Boltzmann equation for the hard sphere models is
semilinear hyperbolic and therefore the propagation of the singularities of
the solution can be studied on the level of linearized Boltzmann equation.
Consider the linearization about the normalized Maxwellian
F =M+
√
M g,
M ≡ (2π)−3/2 e−|ξ|
2 /2
.
and the resulting linearized Boltzmann equation
∂t g + ξ · ∇x g = Lg.
(1.1)
For the hard sphere model under consideration, the linearized collision operator
√
1
Lg ≡ 2 √ Q(M, M g)
M
has explicit form consisting of a multiplicative operator ν(ξ) and an integral
operator K, [1]:
Kg(ξ) ≡
Z
R3
Lg(ξ) = −ν(ξ)g(ξ) + Kg(ξ),
K(ξ, ξ∗ )g(ξ∗ )dξ∗ ,
n (|ξ|2 − |ξ |2 )2 |ξ − ξ |2 o
2
∗
∗
exp −
−
2
8|ξ
−
ξ
|
8
2π|ξ − ξ∗ |
∗
n (|ξ|2 + |ξ |2 ) o
|ξ − ξ∗ |
∗
−
exp −
,
2
4
Z
1 h − |ξ|2 1 |ξ| − u2 i
ν(ξ) ≡ √
2e 2 + |ξ| +
e 2 du .
|ξ| 0
2π
K(ξ, ξ∗ ) ≡ √
2010]
MIXTURE LEMMA
3
Rewrite the linearized Boltzmann equation as follows:
∂t g + ξ · ∇x g + ν(ξ)g = Kg,
and view this as the coupling of the integral operator K with the damped
transport equation
∂t h + ξ · ∇x h + ν(ξ)h = 0,
(1.2)
h(x, 0, ξ) = h0 (x, ξ).
The damped transport equation has the solution operator St :
St h0 (x, ξ) = e−ν(ξ)t h0 (x − ξt, ξ).
(1.3)
The above coupling leads to the Mixture operators Mtk , k = 1, 2, . . . ,
when the Picard-type iterations are used in extracting parts of a Boltzmann
solution with varying degree of regularity, [5] and [6]:
Mtk g0
≡
Z tZ
0
0
s1
···
Z
s2k−1
0
St−s1 KSs1 −s2 K · · · Ss2k−1 −s2k KSs2k g0 ds2k · · · ds1 .
The function ν(ξ) behaves like 1 + |ξ|. Let ν0 be a positive lower bound
of ν(ξ):
ν(ξ) > ν0 , ξ ∈ R3 .
The main theorem as stated in [6] is the following:
Theorem 1.1. For each given k ≥ 0, there exists a positive constant Ck
such that, for any
β ∈ N30 , |β| = β 1 + β 2 + β 3 = k,
kDxβ Mtk g0 kL2x (L2 ) ≤ Ck e
ξ
−ν0 t
2
X
γ∈N30 ,γ i ≤β i
kDξγ g0 kL2x (L2 ) .
ξ
The details of its proof for k = 1 has been carried out in [5, 6] using the
combination of Fourier transform and characteristic method. In this paper,
we make use of switching the differentiations along the characteristic curves
to prove the Mixture Lemma using characteristic method only. Shih-Hsien
4
HUNG-WEN KUO, TAI-PING LIU AND SE EUN NOH
[March
Yu has recently told us that he has also carried out the proof for the case of
k = 1 using characteristic method only.
There is a delicate regularity property for the kernel K(ξ, ξ∗ ) that we
need in order for the switching of differentiations method to work. This is
done in Section 2, Lemma 2.2. The proof of the Mixture Lemma is done
in Section 3 and Section 4. For easier reading, in Section 2 and Section 3,
we present the proofs for the case of one space dimension, and in Section 4
indicate the generalization to the case of three space dimensions.
2. Regularity of Collision Kernels
In this and next sections we consider the case of one space dimention
x ∈ R1 ; the microscopic velocity is still of three dimensions ξ ∈ R3 . The
linearized Boltzmann equation becomes
∂t g + ξ 1 ∂x g = Lg.
(2.1)
The Mixture Lemma takes the following form:
Theorem 2.1. For each given k ≥ 0, there exists a positive constant Ck
such that
k
ν0 t X
k∂xk Mtk g0 kL2x (L2 ) ≤ Ck e− 2
k∂ξl 1 g0 kL2x (L2 ) .
ξ
ξ
l=0
In preparation for its proof, we now study the regularity property of the
functions ν(ξ) and K(ξ, ξ∗ ).
Lemma 2.1. For any l ≥ 1, l-th derivatives of ν(ξ) is bounded, i.e., for
some constant Cl ,
l
(2.2)
∂ξ i ν(ξ) ≤ Cl , i = 1, 2, and 3.
1
|ξ|
Proof. It suffices to consider the term
have by Taylor expansion
1
|ξ|
Z
0
|ξ|
2
− u2
e
1
du =
|ξ|
Z
0
R |ξ|
0
∞
|ξ| X
n=0
e−
u2
2
du for 0 < |ξ| < 1. We
1
u2
(− )n du
n!
2
2010]
MIXTURE LEMMA
∞
1 X 1
=
|ξ|
n!
=
1
|ξ|
n=0
∞
X
n=0
Z
5
|ξ|
(−
0
u2 n
) du
2
(−1)n |ξ|2n+1
n!2n 2n + 1
∞
X
(−1)n |ξ|2n
=
n!2n 2n + 1
n=0
which is absolutely convergent as the function of |ξ|2 and therefore is analytic
in (ξ 1 , ξ 2 , ξ 3 ). This completes the proof.
Unlike ν(ξ), the differentiations of the kernel K(ξ, ξ∗ ) are not all smooth,
even integrable. However, it is the smoothness of certain combinations of
the differentiations of K(ξ, ξ∗ ) that are needed. To express this, we have the
following notations:
n
1
2
(1) A ≡ α ∈ N6k
0 α = (α1 , α2 , . . . , α2k ) α2j = 0, α2j−1 = 0, 1, α2j−1 =
o
α32j−1 = 0, for j = 1, . . . , k
(2) H(ξ, ξ∗ ) ≡ |K(εξ, εξ∗ )| + |∂ξ∗1 K(εξ, εξ∗ )|, for any fixed ε < 1.
(3) using the change of variables T : Ξ = (ξ1 , ξ2 , . . . , ξ2k ) → V = (V1 , V2 , . . .,
V2k ) where V1 ≡ ξ − ξ1 , Vi ≡ ξi−1 − ξi for i = 2, 3, . . . , 2k, to rewrite the
kernel:
P
Pi
Pi−1
Pi
(4) Ki ≡ K(ξ − i−1
j=1 Vj , ξ −
j=1 Vj ), Hi ≡ H(ξ −
j=1 Vj , ξ −
j=1 Vj ).
Lemma 2.2. For 1 < l < 2k and α ∈ A, we have, for some constant Cα ,
|DVα Kl | ≤ Cα Hl .
Proof. We focus only on the most interesting term in Kl :
n (|ξ −
1 Gl
1
Fl ≡
e ≡
exp −
|Vl |
|Vl |
Pl−1
2
j=1 Vj |
− |ξ −
|Vl |2
Pl
2 2o
j=1 Vj | )
.
Note that we can find m such that αi = 0 for i > m and α1m = 1 for the
given α in A, α 6= 0. And it is easy to see that DVα Kl = 0 in case of m > l,
so we consider the following two cases.
6
HUNG-WEN KUO, TAI-PING LIU AND SE EUN NOH
[March
Case A (m < l) Note that
P
Pl
2
2
1
−2(|ξ − l−1
∂Gl
∂Gl
j=1 Vj | − |ξ −
j=1 Vj | )(−2Vl )
=
=
, for n < l, (2.3)
1
2
∂Vn
|Vl |
∂V11
1 2
Vl
∂ 2 Gl
∂ 2 Gl
=
=
, for n1 and n2 < l and l > 2,
(2.4)
∂Vn11 ∂Vn12
|Vl |
∂[V11 ]2
we obtain multi index α′ by replacing αm with 0. Direct calculation then
gives us
′
DVα Fl = DVα Fl ·
=
X
β<α′
≤ Cm Fl
≤ Cm Fl
∂Gl 1
∂V
!1
α′
β
[|α|/2] ′
∂Gl DVβ Fl DVα −β
∂V11
∂Gl |α|−2n ∂ 2 Gl n
∂V11
∂[V11 ]2
n=1
P
Pl
2
2 |V 1 | |α| [|α|/2]
X 4||ξ − l−1
j=1 Vj | − |ξ −
j=1 Vj | | |α|−2n
l
X
|Vl |
|Vl |
n=1
.
Here we have used the fact that the third derivatives of Gl vanishes, (2.3)
and (2.4). It is easy to see that, for any fixed ε < 1,, there is a constant Cm
such that |DVα Fl | is bounded by
|DVα Fl |
n
(|ξ −
1
≤ Cm
exp − ε2
|Vl |
Pl−1
2
j=1 Vj |
− |ξ −
|Vl |2
Pl
2 2o
j=1 Vj | )
.
Case B (m = l) In this case, we observe that
∂V 1 Kl = −∂ξ∗ K(ξ −
l
l−1
X
j=1
Vj , ξ −
l
X
Vj ),
j=1
and use the same argument as in Case A to complete the proof.
3. One Space Dimension
Equipped with the above lemmas, we are now ready to prove the Mixture
2010]
MIXTURE LEMMA
7
Lemma.
The Proof of Theorem 2.1.
With the explicit form of the solution operator (1.3) for the damped
transport equation, the Mixture operator Mtk g0 is
Mtk g0
Z Z
=
T
−ν(ξ)(t−s1 )
e
R6k
2k
hY
i
e−ν(ξi )(si −si+1 ) K(ξi−1 , ξi )
i=1
×g0 (x −
2k
X
i=0
ξi1 (si − si+1 ), ξ2k )dΞdS,
where we have set ξ0 ≡ ξ, , s0 ≡ t, , s2k+1 ≡ 0 and used the notations
S ≡ (s1 , s2 , . . . , s2k ),
T ≡ [0, t] × [0, s1 ] × · · · × [0, s2k−1 ].
We now change variables T as introduced in previous section, put ξ as V0 .
The key observation is that one should do integrate by parts with respect to
V instead of with respect to ξ. This way the differentiations can be evenly
distributed to the differentiation with respect to the components of V :
∂xk Mtk g0
Z Z
=
T
e−ν(ξ)(t−s1 )
R6k
i=1
×∂xk g0 (x −
=
Z Z
T
2k
X
i=0
e−ν(ξ)(t−s1 )
R6k
×
=
2k
hY
k
X
ξi1 (si − si+1 ), ξ2k )dΞdS
2k
hY
e−ν(ξ−
Pi
j=1
Vj )(si −si+1 )
Ki
i=1
k
X
|α|=0,α∈A
|α|
(−1)
|α|=0,α∈A
×
i
e−ν(ξi )(si −si+1 ) K(ξi−1 , ξi )
k−|α|
Dα ∂ξ 1 g0 (x
2k
Z Z
T
R6k
D
α
n
−
2k
X
i=0
Vi si , ξ −
−ν(ξ)(t−s1 )
e
2k
hY
i
1
s1 · s3 · · · s2k−1
2k
X
Vi )dV dS
i=1
e−ν(ξ−
Pi
j=1
Vj )(si −si+1 )
i=1
2k
io
2k
X
X
1
k−|α|
∂ξ 1 g0 (x −
Vi si , ξ −
Vi )dV dS.
s1 · s3 · · · s2k−1 2k
i=0
Ki
i=1
(3.1)
8
HUNG-WEN KUO, TAI-PING LIU AND SE EUN NOH
[March
With (2.2) and Lemma 2.2, we can easily see that
2k
n
hY
io
Pi
Dα e−ν(ξ)(t−s1 )
e−ν(ξ− j=1 Vj )(si −si+1 ) Ki
−ν0 t
≤ Ck e
i=1
2k
hY i
Hi
i=1
2k
hY
i
×
(si − si+1 )k ,
i=1
where Ck is a generic constant depending only on k. Thus from the Hölder
inequality,
|∂xk Mtk g0 |
k
X
−ν0 t
≤ Ck e
×
α∈A,|α|=0
Z Z
dV dS
≤ Ck e−
R6k
T
1
2
ν0 t
2
dV dS
k
X
|α|=0
1
2
Z Z
T
R6k
2k
hY
i=1
i Q2k (s − s )2k
1
2
i+1
i=0 i
Hi
dV dS
s1 · s3 · · · s2k−1
hY i
X
X
1
k−|α|
Hi |∂ξ 1 g0 (x −
Vi si , ξ −
Vi )|2
s1 · s3 · · · s2k−1
Z Z
T
R6k
2k
2k
2k
i=1
i=0
i=1
hQ
2k
i=1 Hi
i
k−|α|
s1 · s3 · · · s2k−1
|∂ξ 1
g0 (x −
2k
X
i=0
Vi si , ξ −
2k
X
Vi )|2
i=1
(3.2)
,
Here we have used the integrability of kernel K and ∂ξ K and the fact that
the time integral is of the polynomial order. We now integrate the square of
(3.2) over R × R3 to get
k∂xk Mtk g0 kL2x (L2 )
ξ
≤ Ck e
−ν0 t
2
k
X
m=0
k∂ξk−m
g0 kL2x (L2 ) .
1
This completes the proof of Theorem 2.1.
ξ
Remark 3.1. This result can be generalized to the function space Lpx (Lpξ )
for 1 ≤ p ≤ ∞.
For the case of 1 ≤ p < ∞, we apply Hölder inequality to (3.1) with
2010]
MIXTURE LEMMA
9
p
Hölder conjugate ( p−1
, p) to yield
|∂xk Mtk g0 |
×
Z Z
T
k
X
−ν0 t
≤ Ck e
R6k
α∈A,|α|=0
Z Z
T
2k
hY
1
s1 · s3 · · · s2k−1
2k
hY
R6k
i=1
i
i Q2k (s − s )2k
1− 1
p
i+1
i=0 i
Hi
dV dS
s1 · s3 · · · s2k−1
2k
X
k−|α|
Hi |∂ξ 1 g0 (x− Vi si , ξ
i=1
i=0
−
2k
X
Vi )|p dV dS
i=1
1
p
.
(3.3)
We can then apply the same argument as in the above proof with the exponent p to establish the same estimate as in Theorem 2.1 for the spaces
Lpx (Lpξ ) for 1 ≤ p < ∞.
For the case of p = ∞, it is easy to see that
|∂xk Mtk g0 |
−ν0 t
≤ Ck e
k
X
α∈A,|α|=0
Z Z
T
R6k
k−|α|
∞ .
×k∂ξ 1 g0 kL∞
x (Lξ )
2k
hY
i=1
i Q2k (s − s )k
i+1
i=0 i
Hi
dV dS
s1 · s3 · · · s2k−1
4. Three Space Dimensions
For the three dimensional space case, the distribution of space differentiations to that of microscopic differentiations is also done evenly and we
have the corresponding notations: For any β ∈ N30 with |β| = k, one can
P
decompose β into β = ki=1 βi so that βi ∈ N3 and |βi | = 1. Define the set
Aβ ≡
n
α ∈ N6k
0 : α = (α1 , α2 , . . . , α2k ) , α2j = 0, α2j−1 = βj or 0,
for j = 1, . . . , k
and for α ∈ Aβ , denote α
e≡
Pk
i=1 α2i−1
∈ N30 .
A direct computation yields
Dxβ g0 (x
=
−
2k
X
i=0
ξi1 (si − si+1 ), ξ2k )
1
s1 · s3 · · · s2k−1
k
X
|α|=0,α∈Aβ
α
DVα Dξβ−e
g0 (x −
2k
2k
X
i=0
Vi si , ξ −
2k
X
i=1
Vi ).
o
10
HUNG-WEN KUO, TAI-PING LIU AND SE EUN NOH
[March
Then we can apply the same argument as for the one space dimensional case
to establish
kDxβ Mtk g0 kL2x (L2 ) ≤ Ck e
ξ
−ν0 t
2
X
γ∈N30 ,γ i ≤β i
kDξγ g0 kL2x (L2 ) .
ξ
As for the one space dimensional case, the estimate also hold for Lp
spaces.
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Institute of Mathematics, Academia Sinica, Taipei 10617, Taiwan, R.O.C. and
Department of Mathematics, National Taiwan University, Taipei 10617, Taiwan, R.O.C.
E-mail: [email protected]
Institute of Mathematics, Academia Sinica, Taipei 10617, Taiwan, R.O.C. and
Department of Mathematics, Stanford University, Standford, CA 94305, U.S.A.
E-mail: [email protected]
Institute of Mathematics, Academia Sinica, Taipei 10617, Taiwan, R.O.C.
E-mail: [email protected]