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首页无截断Boltzmann方程的Gevrey正则性扩展:软位势与温和/临界奇异性
无截断Boltzmann方程的Gevrey正则性扩展:软位势与温和/临界奇异性
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更新于2024-07-16
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本文探讨了无截断空间齐次Boltzmann方程在软位势下的Gevrey正则性,这是在碰撞理论中的一个重要概念,它衡量了解的光滑程度。作者张腾飞和殷朝阳,来自中山大学数学与计算科学学院,针对两个关键的奇异性情况进行了深入研究:一是温和奇异性($s < \frac{1}{2}$),这代表势函数的局部行为相对较弱;二是临界奇异性($s = \frac{1}{2}$),在这种情况下,位势的特殊性可能导致方程行为更为复杂。 与之前的文献相比,该研究的主要创新在于扩展了对参数γ的处理范围。在软位势下,作者以前的工作通常考虑γ在$(-\frac{5}{2}, 0)$内,而在这篇新论文中,这个范围被进一步拓宽。此外,研究不再将γ与$s$的整体效应一并考虑,而是分别对待这两个参数,从而提供了一个更精确的分析框架。 核心部分是作者们通过建立了一个新的强制估计,即改进后的柯西型估计,来证明Gevrey正则性的传播。Gevrey正则性意味着方程的解在高阶导数上具有更强的衰减性质,这对于理解粒子碰撞过程中的长期行为至关重要。这种正则性结果对于理解和控制无截断Boltzmann方程的稳定性以及其在软散射情况下的行为具有重要意义。 这篇首发论文在无截断空间齐次Boltzmann方程的分析上取得了显著进展,不仅深化了我们对软位势下非平凡奇异性下解的了解,也为未来的数值模拟和理论研究提供了有价值的理论基础。
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0.3 Main results
Now we give our main result of propagation of Gevrey regularity in spatially homogeneous
case as follows:
Theorem 1. Let ν > 1(which is independent of s) and assume that 0 < s ≤ 1/2, −5/2 < γ < 0.
Let f(t, v) be a smooth Maxwellian decay solution to the Cauchy problem (0.2). If there exist
ρ
0
, δ
0
such that
sup
α
ρ
0|α|
ke
δ
0
hvi
2
∂
α
v
f(0)k
L
2
{α!}
ν
< +∞, (0.5)
then there exist ρ > 0 and δ, κ > 0 with δ > κT such that
sup
t∈(0,T ]
sup
α
ρ
|α|
ke
(δ−κt)hvi
2
∂
α
v
f(t)k
L
2
{α!}
ν
< +∞. (0.6)
Remark 1. It should be noted that the above theorem is similar as Theorem 1.2 in [17] and
Theorem 1.3 in [18], but we consider here Φ = |v|
γ
and γ ∈ (−5/2, 0).
Arguing as in Section 4 of [17], and thanks to the interpolation inequality with weight
(Lemma 3.7 of [21]), for γ ∈ [−2, 0),
kW
l
F
(α,β)
k
2
H
s/2
. kW
l+γ/2
F
(α,β)
k
H
s
kW
l−γ/2
F
(α,β)
k
L
2
. kW
l+γ/2
F
(α,β)
k
H
s
kW
l+1
F
(α,β)
k
L
2
,
we would obtain the Gevrey smoothing effect of order 1/s as follows:
Theorem 2. Assume that 0 < s ≤ 1/2, γ ∈ [−2, 0). Let ν = 1/s and let f(t, v) be a smooth
Maxwellian decay solution to the Cauchy problem (0.2), then for any t
0
∈ (0, T ), there exist
ρ > 0 and δ, κ > 0 with δ > κT such that
sup
t∈[t
0
,T ]
sup
α
ρ
|α|
ke
(δ−κt)hvi
2
∂
α
v
f(t)k
L
2
{α!}
ν
< +∞. (0.7)
Remark 2. From the above interpolation inequality, we know that the additional requirement
γ ≥ −2 is natural, which is different from the assumption γ ∈ (−5/2, 0) in Theorem 1.
Remark 3. We believe that these results are also valid for spatially inhomogeneous case, as
we did in [19].
0.4 The structure of the paper
The remainder of the paper proceeds as follows. In the next section we give some prelimi-
naries and a main lemma, by using which we can complete the proof of Theorem 1 immediately.
In Section 3 we prove a coercivity estimate for collision operator which is different from what
we used before. The proof of the main lemma will be given in Section 4.
- 6 -
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http://www.paper.edu.cn
1 Preliminaries
First of all, we introduce some basic definitions (see [17] for details).
Let l, r ∈ Z
+
which will be chosen later. For δ, ρ > 0 we set:
kfk
δ,l,ρ,α,r
,
ρ
|α|
khvi
l
e
δhvi
2
∂
α
v
fk
L
2
{(α − r)!}
ν
,
where α = (α
1
, α
2
, . . . , α
n
) ∈ Z
n
+
, and we denote
(α − r)! = (α
1
− r)! · · · (α
n
− r)!.
Now we give the following definition:
kfk
l,ρ,r,N
(t) , sup
rn≤|α|≤N
kfk
δ−κt,l,ρ,α,r
, (1.1)
with fixed δ, κ > 0 such that δ > κT . Here N is a fixed large number satisfying rn ≤ |α| ≤ N.
Then for h > 1 we can obtain
kfk
l,ρ(1+h)
−ν
,r,N
(t) ≤
(r!)
n
h
r
ν
kfk
l,ρ,0,N
(t). (1.2)
Now let ρ = ρ
0
in the above inequality and take a large enough h, then it follows from the
initial assumption (0.5) that kfk
l,ρ
0
(1+h)
−ν
,r,N
(0) is as small as we want, where δ can be chosen
any positive less than δ
0
> 0 in (0.5).
Thus, to prove (0.6), it suffices to prove, under the assumption that kfk
l,ρ,r,N
(0) is suffi-
ciently small,
sup
t∈(0,T ]
kfk
l,ρ,r,N
(t) < ∞. (1.3)
Above, ρ = ρ
0
(1 + h)
−ν
.
We point out that we will consider the Cauchy problem in R
3
in the paper.
Lemma 1. If l ≥ 4 and r > 1 + ν/(ν − 1) then for any α satisfying 3r ≤ |α| ≤ N we have
kf(t)k
2
δ−κt,l,ρ,α,r
+ 2κ
Z
t
0
kf(τ)k
2
δ−κτ,l+1,ρ,α,r
dτ
≤kf(0)k
2
δ,l,ρ,α,r
+ C
κ
Z
t
0
kfk
2
l,ρ,r,N
(τ) + kfk
2(i+β)/β
l,ρ,r,N
(τ)
dτ
+
κ
10
sup
3r≤|α|≤N
Z
t
0
kf(τ)k
2
δ−κτ,l+1,ρ,α,r
dτ, (1.4)
where i = 1
s<1/2
+ 2 · 1
s=1/2
, and β = −γ · 1
γ∈[−1,0)
+ 1
γ∈(−5/2,−1)
.
Mimicking the scheme in Section 2 in [17] (It is only a matter of using the Bernoulli
equation in ODE theory with a different order), we can prove Theorem 1 by virtue of this
lemma. We omit the details. The proof of this lemma will be given in Section 4.
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