A new regularization possibility for the Boltzmann equation with soft potentials

We consider a simplified Boltzmann equation: spatially homogeneous, two-dimensional, radially symmetric, with Grad's angular cutoff, and linearized around its initial condition. We prove that for a sufficiently singular velocity cross section, the solution may become instantaneously a function, even if the initial condition is a singular measure. To our knowledge, this is the first regularization result in the case with cutoff: all the previous results were relying on the non-integrability of the angular cross section. Furthermore, our result is quite surprising: the regularization occurs for initial conditions that are not too singular, but also not too regular. The objective of the present work is to explain that the singularity of the velocity cross section, which is often considered as a (technical) obstacle to regularization, seems on the contrary to help the regularization.


Introduction
Let f t (dv) be the velocity distribution in a 2d spatially homogeneous dilute gas at time t ≥ 0. Then under some physical assumptions, f solves the Boltzmann equation: for some γ ∈ (−2, 1], some angular cross section β (a nonnegative symmetric measure on (−π, π)\{0}), for all sufficiently regular functions ϕ : where, for R θ the rotation of angle θ, We refer to Villani [13] and Desvillettes [4] for reviews on this equation. When γ < 0, we speak of soft potentials. The suject of the present paper is regularization: can f t be more regular than f 0 , as soon as t > 0 ?
• finally, the most general result (but also weaker) is the one of Alexandre-Desvillettes-Villani-Wennberg [2]: in any dimension, for any γ, any ν ∈ (0, 2), any initial condition with finite mass, energy, entropy, √ f t instantaneously belongs to H ν/2 loc . Let us observe that when tracking the constants in [2], we realize that the result is weaker and weaker when γ becomes more and more negative. These results are based on lowerbounds of the entropy dissipation functionnal. Such an idea was initiated by Lions [10], see also Villani [12].
Our goal in the present paper is to show that the explosion of |v − v * | γ near v = v * , when γ < 0, is not an obstacle to regularization: ont the contrary, even in the case with cutoff, such a singular interaction kernel may provide some regularization.
We start with a simplified equation, namely we linearize the Boltzmann equation around its initial condition. We show that for a specific singular initial condition (an uniform distribution on a circle), the solution instantaneously becomes a function, in the case with cutoff, even if the angular cross section is not a function.
The present result is completely new, since all the previous results were relying on the explosion near 0 of the angular cross section β. It is furthermore very surprising, since, as we will show, no regularization may hold if f 0 is too singular nor too regular: if f 0 is a measure that is not a function but is too smooth in some sense, then it will never become a function. This kind of phenomenon is fully nonlinear. However, our result is very weak, since it is qualitative (we only prove that the solution immediately becomes a function), and since we consider a simplified equation.
Regularization is motivated by many other subjects for which regularity estimates are needed: convergence to equilibrium, uniqueness,... For example, it is shown in [8] that uniqueness holds (with γ < 0 and a possibly non-cutoff angular cross-section) for sufficiently smooth solutions (in some L p , with p large enough). We are far from such a quantitative regularization.
Let us finally mention that a similar result should hold for the (nonlinear) Boltzmann equation.
Formal proof. We assume here that one may apply (1) with any bounded measurable ϕ. Since f 0 is a radially symmetric probability measure, so is f t for all t ≥ 0.
Step 1. First, applying (1) with ϕ = 1 {0} , one easily deduces that f t ({0}) = 0 for all t ≥ 0. Indeed, f 0 ({0}) = 0, and simple considerations using that f t is radially symetric show that for all v, all Step 2. Next, for any Lebesgue-null A ⊂ R 2 , one gets convinced that for all θ = 0, Here, one has to use that f t is radially symmetric and does not give weight to 0. As a consequence, Thus if f 0 (A) = 0, we deduce that f t (A) = 0 for all t ≥ 0. This implies that f t (dv) has a density, except maybe on C = {|v| = 1}.
Step 3. It remains to check that This implies (see Falconer [6,Theorem 4.13 p 64]) that either f t (C) = 0 or that the Haussdorff dimension of C is greater than |γ|, the latter being excluded since dim H (C) = 1 < |γ|.
Unfortunately, we are not able to justify the use of such test functions in the nonlinear case.
We state our result in Section 2, we prove it in Section 3. An appendix lies at the end of the paper.

Main result
In the whole paper, the angular cross section is supposed to be finite, and to vanish on {|θ| ≥ π/2}. For the nonlinear Boltzmann equation, this last condition is not restrictive, for symmetrical reasons, see the introduction of [2]. We also impose that β vanishes near 0 for simplicity.
We now define the notion weak solutions we will use. We denote by Lip(R 2 ) the set of globally Lipschitz functions from R 2 to R. where with We will check later that in our situation, all the terms make sense in (4).
Our main result writes as follows.
Let us comment on this result. Consider γ ∈ (−2, 0), and an initial condition f 0 satisfying Then we consider a solution (f t ) t≥0 to LB(f 0 , γ, β). Applying (4) with some nonnegative Lipschitz function ϕ and using (A1), we immediately get This result is fully nonlinear and quite surprising: if f 0 is regular enough to satisfy (6)  greater than |γ|, we may find a probability measure f 0 on R 2 with f 0 (A) = 1 and such that such Let us insist on the fact that initial conditions satisfying (6) are more regular than the uniform distribution on {|v| = 1}: the latter gives positive weight to some sets with lower dimension. Note also that on the contrary, f 0 has to be sufficiently regular. If for example we assume that f 0 = 1 2 (δ v0 + δ v1 ), no regularization may hold (due to the indicator function in (6)). The same argument applies to f 0 = 1 2 (δ v0 + g 0 ), for some bounded probability density g 0 . Finally, let us mention that our result holds even if β = δ θ0 + δ −θ0 , for some fixed θ 0 ∈ (0, π/2): the regularization does really not follow from the regularity of the angular cross section.

Proof
The aim of this section is to prove Theorem 2. We assume in the whole section that (A1 − A2) hold, that r 0 = 1 (for simplicity), and that γ ∈ (−2, −1) is fixed. Thus our initial condition f 0 is defined by for all measurable ϕ : R 2 → R + , where e α := (cos α, sin α).
Finally, we prove that our solution leaves immediately the unit circle. Proof. We divide the proof into two steps.
Step 2. We now conclude. Consider the measure µ T (dr) = T 0 dsλ s (dr)1 {r =1} . Since µ T is finite and µ T ({1}) = 0, the de la Vallée Poussin Lemma 7 ensures us that there exists a function g : R + → R + , with g(∞) = ∞, such that x → xg(1/x) is nondecreasing on R + , and ∞ 0 µ T (dr)g(1/|r 2 −1|) < ∞. Since |γ| ≥ 1 by assumption, we deduce that Letting ε tend to 0, we get T 0 dsλ s ({1}) = 0. Since T is arbitrarily large, this ends the proof. We may now conclude the Proof of Theorem 2. We consider the solution (f t ) t≥0 built in Proposition 3, and the associated radius distribution (λ t ) t≥0 . Owing to the Radon-Nikodym Theorem, to Lemmas 5 and 6, we deduce that for a.e. t ≥ 0, λ t (dr) has a density λ t (r) with respect to the Lebesgue measure on R + . Then we deduce from (8)  We conclude the section with the Proof of Proposition 3. We split the proof into 4 steps.
Step 1. We first check the existence of a solution. We introduce, for n ≥ 1, the operator A n , of which the expression is the same as (5) with min(|v − v * | γ , n) instead of |v − v * | γ . Then we observe that with our choice for f 0 , as shown in the appendix, we have for all ϕ ∈ Lip(R 2 ), In particular, (15) implies that all the terms make sense in (4). One easily checks, by classical methods (Gronwall Lemma and Picard iteration using the total variation norm), that there exists a unique solution (f n t ) t≥0 to LB n (f 0 , γ, β), where A is replaced by A n . The obtained solution f n is clearly radially symmetric. Next, using (4) and (17), we deduce that (a) using ϕ(v) = |v|, gives some equicontinuity estimates. It is then standard that up to extraction of a (not relabelled) subsequence, (f n t ) t≥0 tends to some family of (radially symmetric) probability measures (f t ) t≥0 , in the sense that for all ϕ ∈ Lip(R 2 ), for all T ≥ 0, lim n sup [0,T ] | R 2 (f n t − f t )(dv)ϕ(v)| = 0. This also implies that for all t ≥ 0, all ϕ : R 2 → R continuous and bounded, lim n R 2 f n t (dv)ϕ(v) = R 2 f t (dv)ϕ(v). We deduce that (f t ) t≥0 solves LB(f 0 , γ, β), by passing to the limit in LB n (f 0 , γ, β), using the convergence properties of f n to f , the Lebesgue dominated convergence Theorem, as well as (15-16-17-18).
Step 2. Next, point (ii) of the statement is a simple consequence of the radial symmetry of (f t ) t≥0 .