Fractional diffusion limit for collisional kinetic equations - UMD MATH
Fractional diffusion limit for collisional kinetic equations: a Hilbert expansion approach N. Ben Abdallah1 , A. Mellet2∗and M. Puel1† 1
Institut de Math´ematiques de Toulouse, Universit´e Paul Sabatier, 118 route de Narbonne, 31062 Toulouse Cedex 9 France. 2 Department of Mathematics, University of Maryland, College Park MD 20742 USA.
Abstract We develop a Hilbert expansion approach for the derivation of fractional diffusion equations from the linear Boltzmann equation with heavy tail equilibria.
1
Setting of the result
1.1
Introduction
The linear Boltzmann equation is a simple kinetic equation describing the evolution of a particle distribution function. It couples free transport and scattering phenomena (due to collisions with the background). The properties of the scattering process determines the long time behavior of the particle distribution function. It fixes in particular the profile of the equilibrium velocity distribution. Asymptotic analysis of such an equation is a very classical problem. Typically, assuming that the mean free path (i.e. distance between two collisions) is very small and the time scale is very large, it is possible to derive a hydrodynamic type equation describing the evolution of the density of particles. Often, the thermodynamical equilibrium is given by a Maxwellian distribution function, and the mean square displacement of the particle is a linear function of time. The correct time/space scaling (x = εx0 and t = ε2 t0 ) then leads to a diffusion equation for the density of particles. This type of diffusion approximation for kinetic equations has been widely studied in several papers, see for instance [1, 4, 10, 7] (and references therein). In recent works (see [14], [13], [3]) similar asymptotic analysis are performed when the equilibrium function is not a Maxwellian distribution, but rather a heavy tail function. In this case, the diffusion coefficient appearing in the classical diffusion limit is no longer well defined and one has to modify the time scale to recover an equation for the particle density. This is known in the literature as anomalous diffusion phenomena (the mean square displacement of the particle is not a linear function of time), and it leads to fractional diffusion equations. ∗ †
Partially supported by NSF Grant DMS-0901340 N. Ben Abdallah and M. Puel were partially supported by by ANR project QUATRAIN BLAN07-2 212988.
1
Denoting by ε the ratio of the microscopic length scale (mean free path) and macroscopic length scale, we scale the space variable as x = εx0 and the time variable as t = εα t0 . The starting point of this paper is then the following rescaled Boltzmann equation: εα ∂t f ε + εv · ∇x f ε = Q(f ε )
(1)
where the particle distribution function f ε (t, x, v) depends on the time t > 0, the space variable x ∈ Rn and the velocity variable v ∈ Rn . The collision operator appearing in the right hand side is a linear integral operator of the form: Z Q(f ) = σ(x, v, v 0 ) f (v 0 )F (v) − f (v)F (v 0 ) dv 0 (2) Rn
with σ(x, v, v 0 ) = σ(x, v 0 , v). Note that the operator Q is the sum of a gain term Z Q+ (f ) = σ(x, v, v 0 )f (v 0 ) dv 0 F (v) Rn
and a loss term
Z
−
Q (f ) = −ν(v)f (v)
with ν(v) =
σ(x, v, v 0 )F (v 0 )dv 0 .
Rn
This operator has a one dimensional kernel spanned by the function F (v) and it preserves the total mass: Z Q(f ) dv = 0 for all f . (3) Rn
The usual diffusion limit corresponds to α = 2. It can be studied using the so-called Hilbert expansion method, which is based on a formal expansion of the solution in the form f ε = f 0 + εf 1 + ε2 f 2 + rε . Inserting this expansion in (1) and identifying the terms of same order in ε yields: Q(f 0 ) = 0
(4)
1
Q(f ) = v · ∇x f 2
0
0
Q(f ) = ∂t f + v · ∇x f
(5) 1
(6)
and the remainder rε solves ∂t rε + ε−1 v · ∇x rε = ε−2 Q(rε ) + ε∂t f 1 + ε2 ∂t f 2 + εv · ∇x f 2 . Equation (4) implies that f 0 (t, x, v) = ρ(t, x)F (v) and (5) then yields f 1 = Q−1 (v · ∇x f 0 ) = Q−1 (vF (v)) · ∇x ρ (assuming that such an inverse exists, see Proposition 2.2). Finally, in view of (3), the equation (6) for f 2 gives (integrating with respect to v): ∂t ρ + ∇x · j = 0 where
Z j=
1
Z
vf dv = Rn
v ⊗ Q−1 (vF (v)) dv∇x ρ.
Rn
2
We deduce that for f 2 to exist, ρ must solve ∂t ρ − ∇x · D∇x ρ = 0 with diffusion matrix
Z D=−
v ⊗ Q−1 (vF (v)) dv.
Rn
Now, assuming that ρ is a (smooth solution) of this diffusion equation, we can define f 0 , f 1 and f 2 solutions of (4), (5) and (6), and show that the remainder rε converges strongly to 0 in some L2 space. The main drawback of this method, compared with the moment method, is that it requires stronger regularity assumptions on the initial data in order to get the appropriate bounds on f 0 , f 1 and f 2 . On the other hand the method yields strong convergence of the solution (rather than weak convergence for the moment method). Hilbert expansion type methods have also proved very useful to study diffusion limits for some non linear collision operators, see [12, 9, 11]. In this paper we are considering a situation in which this limit fails because the diffusion matrix D above is infinite. This is clearly the case when the equilibrium function F , rather than a Maxwellian distribution function is a heavy tail distribution function, satisfying F (v) ∼
κ0 |v|n+α
as |v| → ∞
for some α ∈ (0, 2) (note that this α will be the same as the time scale in (1)). Such a profile has been mentioned in several context such as astrophysical plasmas ([17, 15] or in the study of granular plasmas ([5, 8] for inelastic Maxwell models, [16] for inelastic Kac model and [6] for mixture of gases with Maxwellian collision kernel). A similar study is undertaken in [14] using a relatively simple method based on the Fourier transform. This method does not generalize easily to more complicated situation involving space dependent collision kernel σ or non linear operator. In [13], an alternative method is developed which is based on the weak formulation of the Boltzmann equation (1) and the choice of particular test functions defined via an auxiliary equation. This method is somewhat analogous to the well known moment method for the usual diffusion limit. Both approaches provide the weak convergence of f ε to ρF . The aim of the present paper is to develop a Hilbert expansion approach for this anomalous diffusion regime. One of the advantages of this new method (compared with [14, 13]) will be to give strong convergence results. We also believe that this method can be used to address some nonlinear problems, such as models involving the Boltzmann-Pauli collision operator. The first difficulty in developing such an expansion method is that since α ∈ (0, 2), it is not obvious how one should identify the terms with same order (some terms have a fractional order). The originality of this expansion is thus the splitting of the different terms of equation (1) and their distribution in the different levels of the expansion. This splitting follows a heuristic based on the choice of the auxiliary function in [13]. Indeed, we notice that the advection term and the gain loss of the collision operator are of the same order whereas the gain term is of higher order (in term of ε). This leads to a rearrangement in the Hilbert expansion as explained in section 3. We are now going to list our assumptions on σ and F and give the main result of this paper (Theorem 1.1 below). In Section 2, we will present a proof of this result in the simpler case σ = 1. In this case, the expansion is relatively simple, and we will explain why in the more general case, a more complicated expansion needs to be introduced. The main theorem is then proved in Section 3.
3
1.2
Main result
In order to simplify the computation, we will assume that there exist some constants σ0 and σ1 such that 0 < σ0 ≤ σ(x, v, v 0 ) = σ(x, v 0 , v) ≤ σ1 < ∞ for all (x, v, v 0 ) ∈ R3n . (7) In particular, this implies that the collision frequency Z σ(x, v, v 0 )F (v 0 )dv 0 ν(x, v) = Rn
satisfies for all (x, v) ∈ R2n .
σ0 ≤ ν(x, v) ≤ σ1
We also need to assume that the collision frequency is regular enough and has a nice asymptotic behavior for large v. For simplicity, we will assume: ∂xk ν ∈ L∞ (Rn × Rn ) for all multi-indice k such that |k| ≤ 4 ν(x, v) = ν0 (x) ∀|v| > 1
(8)
(this last assumption can of course be replaced with lim|v|→∞ ν(x, v) = ν0 (x)). The most important assumption concerns the behavior of the equilibrium F (v) for large v. We assume: κ for |v| ≥ 1 (9) F (v) = n+α |v| for some α ∈ (0, 2) and for all v ∈ Rn .
F (v) ≤ κ
(10)
Finally, we assume that the following symmetry conditions are satisfied: ν(x, −v) = ν(x, v) for all (x v) ∈ R2n . (11) R This implies in particular that the flux Rn vF (v) dv = 0 (when it is defined, that is when α > 1), which we know is a necessary condition for the derivation of diffusion type equation. F (−v) = F (v) ,
We can now state our main result: Theorem 1.1. Assume that (7), (8), (9), (10), (11) hold and let f ε be the solution of (1) with initial condition fin (x, v) = ρin (x)F (v) such that ρin ∈ H 4 (Rn ), then ||f ε − ρ0 F ||L∞ (0,∞;L2 −1 (Rn ×Rn )) −→ 0,
as ε → 0,
(12)
F
where ρ0 satisfies
in [0, ∞) × Rn , in Rn
∂t ρ0 + L(ρ0 ) = 0 ρ0 (0, x) = ρin (x)
(13)
and L is an elliptic operator defined by the singular integral: Z ρ(x) − ρ(y) L(ρ) = P.V. γ(x, y) dy. |y − x|n+α Rn with
Z γ(x, y) = ν0 (x) ν0 (y)
∞
z α e−z
0
4
R1 0
ν0 ((1−s)x+sy) ds
(14)
dz
Note that γ(x, y) = γ(y, x) and that condition (7) yields 0 < γ1 < γ(x, y) ≤ γ2 . In particular the operator L is a self-adjoint elliptic operator of order α (comparable to (−∆)α/2 ). Before giving the proof of this theorem in the general case (Section 3), we will present two simpler cases that can be handled with a more direct Hilbert method based on the Fourier Transform and for which the result is obtained in a L1 framework as well as in a L2 framework with less regularity assumptions on the initial data.
2
Space homogeneous cross sections
2.1
Constant cross section
In this section, we prove Theorem 1.1 in the simpler framework where the cross section σ is constant (for the sake of simplicity we take σ = 1). We then have Z f (v 0 ) dv 0 F (v) − f (v). (15) Q(f ) = Rn
In that case, the use of the Fourier transform (as in [14]) allows for explicit computations and a much simpler proof. We have the following proposition: Proposition 2.1. Let f ε be the solution of (1) with Q given by (15) where F satisfies (9-11) and with initial condition fin (x, v) = ρin (x)F (v) such that ρin ∈ W 3,1 (Rn ). Then ||f ε − ρ0 F ||L∞ (0,∞;L1 (Rn ×Rn )) −→ 0, where ρ0 satisfies
(
α
as ε → 0
∂t ρ0 + κ(−∆) 2 ρ0 = 0
in [0, ∞) × Rn
ρ0 (0, x) = ρin (x)
in Rn
with Z κ= Rn
(w · e)2 1 dw. 2 1 + (w · e) |w|n+α
(16)
(17)
(18)
If instead we have ρin ∈ H 3 (Rn ), then ||f ε − ρ0 F ||L∞ (0,∞;L2 −1 (Rn ×Rn )) −→ 0,
as ε → 0.
(19)
F
The proof relies on the following Hilbert expansion for f ε : f ε = f0ε + g1ε + g2ε + rε , with Q(f0ε ) = 0
(20)
(1 + εv · ∇x )g1ε = −εv · ∇x f0ε
(21)
Q(g2ε ) = εα ∂t f0ε − Q+ (g1ε ) 5
(22)
and rε solution of ∂t rε + ε1−α v · ∇x rε = ε−α Q(rε ) − ∂t g1ε − ∂t g2ε − ε1−α v · ∇g2ε .
(23)
In order to investigate these equations, we recall the following result, which holds for general operators of the form (2), whenever (7) holds (see [7]): Proposition 2.2. i) The operators Q+ and Q are bounded operator on L2F −1 (Rn ). ii) The kernel of Q has dimension 1 and is spanned by F . iii) There exists a constant c > 0 such that Z Z 1 1 − Q(f )f dv ≥ c |f − ρF |2 dv. F F Rn Rn iv) For all h ∈ L2F −1 (Rn ), the equation Q(g) = h has aRsolution if and only if is a unique such solution (denoted Q−1 (h)) satisfying Rn g(v) dv = 0.
R
h(v) dv = 0, and there
v) The operator Q−1 is bounded in L2F −1 (Rn ): kQ−1 (h)kL2 −1 ≤ CkhkL2 −1 F
for all h such that
F
R
Rn
h(v) dv = 0.
(24)
Z vi)
Q(f )sgn(f ) dv ≤ 0 for all f . Equation (20) then yields (using (ii)): f0ε (x, v, t) = ρε (x, t)F (v).
and so equation (21) becomes Tε g1ε = −εv · ∇x ρε F where we introduced the operator Tε = 1 + εv · ∇x . Assuming (for the time being) that this operator can be inverted, we can then write g1ε = −Tε−1 (εv · ∇x ρε )F.
(25)
Finally, the solvability condition for (22) (using (iv)) is Z α ε ε ∂t ρ F − Q+ (g1ε ) dv = 0 which reads εα ∂t ρε −
Z
g1ε dv = 0.
Using (25), we see that this is a condition on ρε , which we can write as Z α ε ε ∂t ρ + Tε−1 (εv · ∇x ρε )F dv = 0.
(26)
Because of the simple form of the operator Q, this condition also implies that the right hand side of (22) is actually 0 (rather than only having zero integral), and thus g2ε = 0. 6
Remarks 2.3. The usual Hilbert expansion consists in writing f ε = ρF + f1ε + f2ε + rε where ν(x, v)f1ε = Q+ (f1ε ) − εv · ∇x ρF, Q(f2ε ) + εv · ∇x f1ε = εα ∂t ρF. We can see that the scheme that we use here consists in reshuffling the terms Q+ (f1ε ) and εv · ∇x f1ε in the first order and second order equation. Note also that while in the usual Hilbert expansion, the first term is solution of the asymptotic equation, here we have ρε which is solution of an approximation of the limiting model (26). In order to prove Proposition 2.1, we thus have to show that ρε , solution of (26), exists and that g1ε and rε go to zero in some appropriate norm. This is best done using Fourier transform (with respect to x): We denote by b h(ξ) the Fourier transform, with respect to x, of a function h. Taking the Fourier transform in (21), we get: (1 + iεv · ξ)gb1ε = −iεv · ξ fb0 and so
iεv · ξ gb1ε = − F (v) ρbε . 1 + iεv · ξ
(27)
∂t ρbε + aε (ξ) ρbε = 0.
(28)
We can thus rewrite (26) as with ε
a (ξ) = ε
−α
Z Rn
iεv · ξ F (v) dv. 1 + iεv · ξ
We now need the following lemma (see also [14]): Lemma 2.4. For all ξ, aε (ξ) ≥ 0 and |aε (ξ) − κ|ξ|α | ≤ Cε2−α |ξ|2 −→ 0,
(29)
with κ given by (18). Furthermore, there exist positive constants c1 and c2 such that, c1 min(ε−α , |ξ|α ) ≤ aε (ξ) ≤ c2 (|ξ|α + ε(2−α) |ξ|2 )
(30)
Proof of Lemma 2.4. A simple computation (using the properties of F ) yields Z Z iεv · ξ (εv · ξ)2 F (v) dv = F (v) dv 2 Rn 1 + iεv · ξ Rn 1 + (εv · ξ) Z (εv · ξ)2 = F (v) dv 2 |v|≤1 1 + (εv · ξ) Z (εv · ξ)2 1 + dv. 2 n+α |v|≥1 1 + (εv · ξ) |v| The first integral is clearly bounded by Cε2 |ξ|2 , and using the change of variable w = εv|ξ| we can write the second integral as Z Z 1 (w · e)2 1 (εv · ξ)2 α α dv = ε |ξ| dw 2 n+α 2 n+α |v|≥1 1 + (εv · ξ) |v| |w|≥ε|ξ| 1 + (w · e) |w| 7
We deduce α ε
α
α
2
2
α
|ε a (ξ) − ε κ|ξ| | ≤ Cε |ξ| + ε |ξ|
α
Z |w|≤ε|ξ|
2
(w · e)2 1 dw 2 1 + (w · e) |w|n+α
2
≤ Cε |ξ|
This implies (29) and the second inequality in (30). The first inequality in (30) is obtained by distinguishing two cases: R (εv·ξ)2 1 α α and the result If ε|ξ| ≤ 1, then the computation above yields |v|≥1 1+(εv·ξ) 2 |v|n+α dv ≥ cε |ξ| follows. If ε|ξ| ≥ 1, then we have Z Z 1 1 1 (εv · ξ)2 α α dv ≥ ε |ξ| dw 2 n+α n+α |w|≥ε|ξ| 2 |w| |v|≥1 1 + (εv · ξ) |v| ≥ 1 and so aε ≥ ε−α .
We deduce Corollary 2.5. For any ρin and ε > 0, there exists a unique ρε solution of (28), given by ε ρbε (t, ξ) = e−ta (ξ) ρc in (ξ)
Furthermore, ρε satisfies || |ξ|k ρbε (t)||L∞ (Rn ) ≤ || |ξ|k ρc in ||L∞ (Rn ) and || |ξ|k ∂t ρbε (t)||L∞ (Rn ) ≤ || |ξ|k+2 ρc in ||L∞ (Rn ) for all k ≥ 0. Finally we have ||ρε − ρ0 ||L∞ (0,T ;L1 (Rn )) ≤ CT ||ρin ||W 2,1 (Rn ) and ||ρε − ρ0 ||L∞ (0,T ;L1 (Rn )) ≤ CT ||ρin ||H 2 (Rn ) where ρ0 is the unique solution of (17). Proof. We only prove the last assertion: We note that ∂t (ρbε − ρb0 ) + aε (ξ)(ρbε − ρb0 ) = (κ|ξ|α − aε (ξ))ρb0 and so Lemma 2.4 yields d ε |ρb − ρb0 | ≤ Cε2−α |ξ|2 ρb0 . dt We deduce
d ε ||ρ − ρ0 ||L1 (Rn ) ≤ Cε2−α ||ρ0 ||W 2,1 (Rn ) dt which gives the L1 (Rn ) convergence of ρε . A similar computation (multiplying by ρbε − ρb0 ) gives the L2 (Rn ) convergence.
8
Proof of Proposition 2.1. We now notice that (27) yields |εv · ξ| |gb1ε | ≤ p F (v)ρbε 1 + (εv · ξ)2 and thus ||gb1ε ||L∞ ≤ sup n 1 N ξ (R ;Lv (R )))
(Z
)
|εv · ξ|
p F (v) dv ρbε (ξ) 1 + (εv · ξ)2 ≤ sup C(εα |ξ|α + ε|ξ|)ρbε ξ
ξ
≤ C (εα + ε) ||(1 + |ξ|2 )ρbε ||L∞ (Rn ) , and similarly ≤ sup C(εα |ξ|α + ε|ξ|)∂t ρbε ||∂t gb1ε ||L∞ N n 1 ξ (R ;Lv (R ))) ξ
≤ sup C(εα |ξ|α + ε|ξ|)aε (ξ)ρbε ξ
so, using Lemma 2.4, we deduce ||∂t gb1ε ||L∞ ≤ (εα + ε)||(1 + |ξ|3 )ρbε ||L∞ (Rn ) . n 1 N ξ (R ;Lv (R ))) It follows that g1ε and ∂t g1ε converge to 0 strongly in L∞ (0, ∞; L1 (Rn × Rn )). Finally, multiplying the equation for the remainder rε , (23), by sign(rε ) (and using Proposition 2.2 (vi)), we get ∂t ||rε ||L1 (Rn ) ≤ ||∂t g1ε ||L1 (Rn ) . Using the fact that rε (t = 0) = −g1ε (t = 0), we deduce ||rε ||L∞ (0,T ;L1 (R2n )) −→ 0 and so ||f ε − ρε F ||L∞ (0,T ;L1 (Rn ×Rn )) −→ 0. Similarly, we have (using the computations of Lemma 2.4): Z Z |εv · ξ|2 2 2 1 ε b |g1 | dv ≤ F (v) dv ρbε 2 F (v) 1 + (εv · ξ) n n R R ≤ C(εα |ξ|α + ε2 |ξ|2 )ρbε
2
and so ||gb1ε ||L2 −1 (Rn ×Rn ) ≤ Cεα ||ρε ||H 1 (Rn ) F
and in a same way ||∂t gb1ε ||L2 −1 (Rn ×Rn ) ≤ Cεα ||ρε ||H 3 (Rn ) . F
A similar argument (multiplying (23) by rε /F ) gives the strong convergence in L∞ (0, ∞; L2F −1 (Rn × Rn )).
9
2.2
Case of space homogeneous cross section
We now show, in a still relatively simple case, why the general case is more complicated. In this section, we assume that Q is given by: Z Q(f ) = σ(v, v 0 )[f (v 0 )F (v) − f (v)F (v 0 )] dv 0 (31) Rn
with σ depending only on v and v 0 (and not on x), and satisfying 0 < σ0 ≤ σ(v, v 0 ) = σ(v 0 , v) ≤ σ1 . We might try to proceed as in the previous section, writing f ε = f0ε + g1ε + g2ε + rε , with f0ε (x, v, t) = ρε (x, t)F (v), g1ε = −Tε−1 (εv · ∇x ρε ), F Q(g2ε ) = εα ∂t ρε F − Q+ (g1ε ). where the operator Tε is now defined by Tε (f ) = νf + εv · ∇x f . The solvability condition for g2ε (obtained by integrating this last equation with respect to v) reads Z ∂t ρε = ε−α Q+ (g1ε ) dv n ZR −α = ε νg1ε dv n R Z −α = −ε ν(v)Tε−1 (εv · ∇x ρε )F dv. Rn
We thus obtain that ρε must solve ∂t ρε + Lε (ρε ) = 0 with Lε (ρ) = ε−α
Z Rn
ν(v)Tε−1 (εv · ∇x ρε )F dv.
Once again, we can identify the nature of this operator using Fourier transform: We get Z iεv · ξ −α ε \ L (ρ) = ε ν(v)F (v) dv ρbε Rn ν(v) + iεv · ξ = aε (ξ)ρbε . However, unlike the previous section, we do not have Q(g2ε ) = 0. Instead, we have Q(g2ε ) = −εα Lε (ρε )F − Q+ (g1ε ) which, in Fourier, reads: Q(gb2ε ) = −εα
Z Rn
iεv 0 · ξ [ν(v 0 ) − σ(v, v 0 )]F (v 0 ) dv 0 ρbε F (v) ν(v 0 ) + iεv 0 · ξ
10
and so |εv 0 · ξ| p F (v 0 ) dv 0 |ρbε |F (v) 2 0 2 n σ0 + (εv · ξ) R α α α ≤ Cε (ε |ξ| + ε|ξ|)|ρbε |F (v).
|Q(gb2ε )| ≤ 2σ1 εα
Z
This clearly implies that g2ε goes to zero as ε goes to 0, but it also means that it will be difficult to control the term ε1−α v · ∇g2ε because of the lack of integrability of vF (at least when α < 1). For this reason, we need to add a term in the expansion, as explained in the next section.
3
Proof of Theorem 1.1
3.1
A reshuffled Hilbert expansion
In order to prove Theorem 1.1, we use a Hilbert type expansion and some techniques similar to those first introduced in [13]. We recall that f ε is a solution of εα ∂t f ε + εv · ∇x f ε = Q(f ε ) where Q(f ) = Q+ (f ) − νf
with Q+ (f ) =
Z
(32)
σ(x, v, v 0 )F (v)f (v 0 )dv 0 ,
Rn
and we write the following expansion f ε = ρε (t, x)F (v) + g1ε (t, x, v) + g2ε (t, x, v) + g3ε (t, x, v) + rε (t, x, v)
(33)
where the terms g1ε , g2ε , g3ε satisfy (ν(x, v) + εv · ∇x )g1ε = −εv · ∇x (ρε F ) Q(g2ε ) = −Q+ (g1ε ) − εα ∂t ρε F (ν(x, v) + εv · ∇x )g3ε = −εv · ∇x g2ε .
(34) (35) (36)
The remainder solves ∂t rε + ε1−α v · ∇x rε =
Q+ (g3ε ) Q(rε ) ε ε ε − ∂ g − ∂ g − ∂ g + . t t t 1 2 3 εα εα
(37)
We recall that we defined the operator Tε (g) = νg + εv · ∇x g. A simple computation (see [13]) shows that this operator is invertible with Z z Z +∞ − ν(x − εvs, v) ds 0 Tε−1 (f ) = e f (x − εvz, v) dz
(38)
(39)
0
for any function f (x, v). In particular, equations (34) and (36) are solvable without additional conditions, but (35) requires the following solvability condition (obtained by integrating (35) with respect to v): ∂t ρε + Lε (ρε ) = 0 (40) 11
with Lε (ρ) = −ε−α
Z Rn
= −ε−α Z −α = ε
Z Rn
Rn
Q+ (g1ε )dv ν(x, v)g1ε dv
ν(x, v) Tε−1 (εv · ∇x ρε ) F dv.
(41)
In view of (37), it is clear that in order to prove Theorem 1.1, we need to show that g1ε , g2ε and g3ε Q+ (g ε )
can be defined and that the terms ∂t g1ε , ∂t g2ε ∂t g3ε and εα 3 in the right hand side of (37) go to 0 in L1 (0, ∞; L2F −1 (Rn × Rn )). The first step, detailed in the next section, will be to study the properties of this operator Lε and show that the equation (40) has a smooth solution ρε . In section 3.3, we will derive some important estimate on Q−1 (h) that will be needed to get some estimates on g2ε . In Section 3.4, we derive estimates on g1ε , g2ε and g3ε and their derivatives and show that f ε − ρε F converges to zero strongly. The final step (Section 3.5) is to show that ρε , solution of (40), converges strongly to ρ0 solution of (13).
3.2
Properties of the operator Lε and the equation (40)
First, of all, we note that for any function ρ(x), (39) and (41) imply Z ε −α ν(x, v) Tε−1 (εv · ∇x ρε ) F dv L (ρ) = ε n R Z Z +∞ R z = ε−α e− 0 ν(x−εvs,v) ds εv · ∇x ρ(x − εvz) F (v) dz dv. Rn
0
Alternatively, we can also write Tε−1 (εv · ∇x ρ) Z +∞ R z e− 0 ν(x−εvs,v) ds εv · ∇x ρ(x − εvz) dz = 0 Z +∞ R z d ρ(x − εvz) dz =− e− 0 ν(x−εvs,v) ds dz 0 Z +∞ R z d =− e− 0 ν(x−εvs,v) ds ρ(x − εvz) − ρ(x) dz dz 0 Z +∞ R z =− e− 0 ν(x−εvs,v) ds ν(x − εvz, v) ρ(x − εvz) − ρ(x) dz 0
which leads to ε
L (ρ) = −ε
−α
Z Rn
Z
+∞
e−
Rz 0
ν(x−εvs,v) ds
ν(x − εvz, v) ρ(x − εvz) − ρ(x) F (v) dz dv
(42)
0
which is the formula found in [13]. We can now prove the following lemma: Lemma 3.1. For all k integer, k ≥ 2, the operator Lε is a bounded operator from H k (Rn ) to H k−2 (Rn ): ||Lε (ρ)||H k−2 (Rn ) ≤ C||ρ||H k (Rn ) for all ρ ∈ H k (Rn ). (43) 12
Proof. We first prove that ||Lε (ρ)||L2 (Rn ) ≤ C||ρ||H 2 (Rn )
for all ρ ∈ H 2 (Rn ).
A similar computation is performed in [13], and recalled here for the reader’s convenience. We write Z ε −α ν(x, v) Tε−1 (εv · ∇x ρ) F dv L (ρ) = ε I1ε
= where I1ε
=ε
−α
Rn I2ε + I3ε
+
Z |v|≤1
I2ε = ε−α
ν(x, v) Tε−1 (εv · ∇x ρ) F dv,
Z
ν(x, v) Tε−1 (εv · ∇x ρ) F dv,
1≤|v|≤1/ε
I3ε = ε−α
Z |v|≥1/ε
ν(x, v) Tε−1 (εv · ∇x ρ) F dv.
We have Tε−1 (εv
Z
+∞
· ∇x ρ) =
e−
Rz 0
ν(x−εvs,v) ds
εv · ∇x ρ(x − εvz) dz
0
Z =
+∞ h
e−
Rz 0
ν(x−εvs,v) ds
i − e−ν(x,v)z εv · ∇x ρ(x − εvz) dz
0
Z
+∞
+
e−ν(x,v)z [εv · ∇x ρ(x − εvz) − εv · ∇x ρ(x)] dz
0
Z +
+∞
e−ν(x,v)z εv · ∇x ρ(x) dz
0
and so an integration by parts yields Z +∞ h R i z −1 Tε (εv · ∇x ρ) = e− 0 ν(x−εvs,v) ds − e−ν(x,v)z εv · ∇x ρ(x − εvz) dz 0 Z +∞ 1 + e−ν(x,v)z ε2 v · Dx2 ρ(x − εvz) · v dz ν(x, v) 0 Z +∞ + e−ν(x,v)z εv · ∇x ρ(x) dz. 0
Using the symmetry assumption on F and ν, we deduce: Z Z +∞ h R i z ε −α I1 = ε ν(x, v) e− 0 ν(x−εvs,v) ds − e−ν(x,v)z εv · ∇x ρ(x − εvz) F (v) dz dv |v|≤1
+ε−α
Z |v|≤1
0
Z
+∞
e−ν(x,v)z ε2 v · Dx2 ρ(x − εvz) · v dzF (v) dv
0
and since F is bounded, and Rz − 0 ν(x−sεv,v)ds − e−zν(x,v) ≤ Cz 2 e−σ1 z ε|v| e 13
(44)
we deduce ||I1ε ||L2 (Rn ) ≤ Cε2−α ||ρ||H 2 (Rn ) Next, Assumption (9) and the change of variable w = εv yields: Z Z +∞ R z ε −α e− 0 ν(x−εvs,v) ds ν(x, v) ε · ∇x ρ(x − εvz) I2 = ε 1≤|v|≤1/ε 0 Z +∞ R − 0z ν(x−ws,w/ε) ds
Z
e
=
ε≤|w|≤1
ν(x, w/ε) w · ∇x ρ(x − wz)
0
1 |v|n+α
dz dv
1 dz dw |w|n+α
and proceeding as with I1ε , we deduce I2ε
Z
Z ν(x, w/ε)
=
+∞ h
e−
Rz 0
ν(x−ws,w/ε) ds
i − e−ν(x,w/ε)z w · ∇x ρ(x − wz)
0
ε≤|w|≤1
Z
Z
+ ε≤|w|≤1
+∞
e−ν(x,w/ε)z w · Dx2 ρ(x − wz) · w dz
0
1 dz dw |w|n+α
1 dw |w|n+α
and so ||I2ε ||L2 (Rn ) ≤ C||ρ||H 2 (Rn ) . Finally, to estimate I3ε , we use the formula Tε−1 (εv · ∇x ρ) = −
Z
+∞
e−
Rz 0
ν(x−εvs,v) ds
ν(x − εvz, v) ρ(x − εvz) − ρ(x) dz.
0
Assumption (9) and the change of variable w = εv then yields: Z Z +∞ R z ε I3 = − e− 0 ν(x−ws,w/ε) ds ν(x, w/ε) ν(x − wz, v) ρ(x − wz) − ρ(x) |w|≥1
0
1 dz dw |w|n+α
and so ||I3ε ||L2 (Rn ) ≤ C||ρ||L2 (Rn ) .
We now show how to obtain higher order estimates: We have Z ε ε −α ∂xi L (ρ ) = ε ∂xi ν(x, v)Tε−1 (εv · ∇x ρ)F (v)dv Rn
−ε
−α
Z Rn
+ε
−α
Z Rn
ν(x, v)Tε−1 (∂xi νTε−1 (εv · ∇x ρ))F (v) dv ν(x, v)Tε−1 (εv · ∇x ∂xi ρ)F (v)dv.
The last term is just Lε (∂xi ρ) which can thus be bounded by the previous step. The first term is ∂ ν treated in a same way as Lε (ρ) except that we need to use the fact that xνi is even and bounded. Therefore, we will only detail the computations concerning the second term. We split it into three integrals corresponding respectively to |v| < 1, 1 < |v| < 1ε and |v| > 1ε .
14
1 ε
First of all, we handle the |v| > variable. It leads to I1ε
= −ε−α
Z |v|> 1ε
part. For that purpose, we use (9) and the usual change of
ν(x, v)Tε−1 (∂xi νTε−1 (εv · ∇x ρ))F (v) dv z
Z Z
w = − ν(x, ) ε |w|>1
ν(x − ws,
+∞ −
Z
e
0
0
w Z +∞− ) ds ε ∂x ν(x − wz, w ) e i ε 0
ν(x − w(z + z),
z
Z
ν(x − w(s + z), 0
w ) ds ε
w 1 )[ρ(x − w(z + z)) − ρ(x − εvz)] n+α dvdzdz. ε |w|
Thus, ||I1ε ||L2 (Rn ) ≤ C||ρ||L2 (Rn ) . Next, for |v| < 1, we write Z ε −α I2 = −ε ν(x, v)Tε−1 (∂xi νTε−1 (εv · ∇x ρ))F (v) dv |v|
Institut de Math´ematiques de Toulouse, Universit´e Paul Sabatier, 118 route de Narbonne, 31062 Toulouse Cedex 9 France. 2 Department of Mathematics, University of Maryland, College Park MD 20742 USA.
Abstract We develop a Hilbert expansion approach for the derivation of fractional diffusion equations from the linear Boltzmann equation with heavy tail equilibria.
1
Setting of the result
1.1
Introduction
The linear Boltzmann equation is a simple kinetic equation describing the evolution of a particle distribution function. It couples free transport and scattering phenomena (due to collisions with the background). The properties of the scattering process determines the long time behavior of the particle distribution function. It fixes in particular the profile of the equilibrium velocity distribution. Asymptotic analysis of such an equation is a very classical problem. Typically, assuming that the mean free path (i.e. distance between two collisions) is very small and the time scale is very large, it is possible to derive a hydrodynamic type equation describing the evolution of the density of particles. Often, the thermodynamical equilibrium is given by a Maxwellian distribution function, and the mean square displacement of the particle is a linear function of time. The correct time/space scaling (x = εx0 and t = ε2 t0 ) then leads to a diffusion equation for the density of particles. This type of diffusion approximation for kinetic equations has been widely studied in several papers, see for instance [1, 4, 10, 7] (and references therein). In recent works (see [14], [13], [3]) similar asymptotic analysis are performed when the equilibrium function is not a Maxwellian distribution, but rather a heavy tail function. In this case, the diffusion coefficient appearing in the classical diffusion limit is no longer well defined and one has to modify the time scale to recover an equation for the particle density. This is known in the literature as anomalous diffusion phenomena (the mean square displacement of the particle is not a linear function of time), and it leads to fractional diffusion equations. ∗ †
Partially supported by NSF Grant DMS-0901340 N. Ben Abdallah and M. Puel were partially supported by by ANR project QUATRAIN BLAN07-2 212988.
1
Denoting by ε the ratio of the microscopic length scale (mean free path) and macroscopic length scale, we scale the space variable as x = εx0 and the time variable as t = εα t0 . The starting point of this paper is then the following rescaled Boltzmann equation: εα ∂t f ε + εv · ∇x f ε = Q(f ε )
(1)
where the particle distribution function f ε (t, x, v) depends on the time t > 0, the space variable x ∈ Rn and the velocity variable v ∈ Rn . The collision operator appearing in the right hand side is a linear integral operator of the form: Z Q(f ) = σ(x, v, v 0 ) f (v 0 )F (v) − f (v)F (v 0 ) dv 0 (2) Rn
with σ(x, v, v 0 ) = σ(x, v 0 , v). Note that the operator Q is the sum of a gain term Z Q+ (f ) = σ(x, v, v 0 )f (v 0 ) dv 0 F (v) Rn
and a loss term
Z
−
Q (f ) = −ν(v)f (v)
with ν(v) =
σ(x, v, v 0 )F (v 0 )dv 0 .
Rn
This operator has a one dimensional kernel spanned by the function F (v) and it preserves the total mass: Z Q(f ) dv = 0 for all f . (3) Rn
The usual diffusion limit corresponds to α = 2. It can be studied using the so-called Hilbert expansion method, which is based on a formal expansion of the solution in the form f ε = f 0 + εf 1 + ε2 f 2 + rε . Inserting this expansion in (1) and identifying the terms of same order in ε yields: Q(f 0 ) = 0
(4)
1
Q(f ) = v · ∇x f 2
0
0
Q(f ) = ∂t f + v · ∇x f
(5) 1
(6)
and the remainder rε solves ∂t rε + ε−1 v · ∇x rε = ε−2 Q(rε ) + ε∂t f 1 + ε2 ∂t f 2 + εv · ∇x f 2 . Equation (4) implies that f 0 (t, x, v) = ρ(t, x)F (v) and (5) then yields f 1 = Q−1 (v · ∇x f 0 ) = Q−1 (vF (v)) · ∇x ρ (assuming that such an inverse exists, see Proposition 2.2). Finally, in view of (3), the equation (6) for f 2 gives (integrating with respect to v): ∂t ρ + ∇x · j = 0 where
Z j=
1
Z
vf dv = Rn
v ⊗ Q−1 (vF (v)) dv∇x ρ.
Rn
2
We deduce that for f 2 to exist, ρ must solve ∂t ρ − ∇x · D∇x ρ = 0 with diffusion matrix
Z D=−
v ⊗ Q−1 (vF (v)) dv.
Rn
Now, assuming that ρ is a (smooth solution) of this diffusion equation, we can define f 0 , f 1 and f 2 solutions of (4), (5) and (6), and show that the remainder rε converges strongly to 0 in some L2 space. The main drawback of this method, compared with the moment method, is that it requires stronger regularity assumptions on the initial data in order to get the appropriate bounds on f 0 , f 1 and f 2 . On the other hand the method yields strong convergence of the solution (rather than weak convergence for the moment method). Hilbert expansion type methods have also proved very useful to study diffusion limits for some non linear collision operators, see [12, 9, 11]. In this paper we are considering a situation in which this limit fails because the diffusion matrix D above is infinite. This is clearly the case when the equilibrium function F , rather than a Maxwellian distribution function is a heavy tail distribution function, satisfying F (v) ∼
κ0 |v|n+α
as |v| → ∞
for some α ∈ (0, 2) (note that this α will be the same as the time scale in (1)). Such a profile has been mentioned in several context such as astrophysical plasmas ([17, 15] or in the study of granular plasmas ([5, 8] for inelastic Maxwell models, [16] for inelastic Kac model and [6] for mixture of gases with Maxwellian collision kernel). A similar study is undertaken in [14] using a relatively simple method based on the Fourier transform. This method does not generalize easily to more complicated situation involving space dependent collision kernel σ or non linear operator. In [13], an alternative method is developed which is based on the weak formulation of the Boltzmann equation (1) and the choice of particular test functions defined via an auxiliary equation. This method is somewhat analogous to the well known moment method for the usual diffusion limit. Both approaches provide the weak convergence of f ε to ρF . The aim of the present paper is to develop a Hilbert expansion approach for this anomalous diffusion regime. One of the advantages of this new method (compared with [14, 13]) will be to give strong convergence results. We also believe that this method can be used to address some nonlinear problems, such as models involving the Boltzmann-Pauli collision operator. The first difficulty in developing such an expansion method is that since α ∈ (0, 2), it is not obvious how one should identify the terms with same order (some terms have a fractional order). The originality of this expansion is thus the splitting of the different terms of equation (1) and their distribution in the different levels of the expansion. This splitting follows a heuristic based on the choice of the auxiliary function in [13]. Indeed, we notice that the advection term and the gain loss of the collision operator are of the same order whereas the gain term is of higher order (in term of ε). This leads to a rearrangement in the Hilbert expansion as explained in section 3. We are now going to list our assumptions on σ and F and give the main result of this paper (Theorem 1.1 below). In Section 2, we will present a proof of this result in the simpler case σ = 1. In this case, the expansion is relatively simple, and we will explain why in the more general case, a more complicated expansion needs to be introduced. The main theorem is then proved in Section 3.
3
1.2
Main result
In order to simplify the computation, we will assume that there exist some constants σ0 and σ1 such that 0 < σ0 ≤ σ(x, v, v 0 ) = σ(x, v 0 , v) ≤ σ1 < ∞ for all (x, v, v 0 ) ∈ R3n . (7) In particular, this implies that the collision frequency Z σ(x, v, v 0 )F (v 0 )dv 0 ν(x, v) = Rn
satisfies for all (x, v) ∈ R2n .
σ0 ≤ ν(x, v) ≤ σ1
We also need to assume that the collision frequency is regular enough and has a nice asymptotic behavior for large v. For simplicity, we will assume: ∂xk ν ∈ L∞ (Rn × Rn ) for all multi-indice k such that |k| ≤ 4 ν(x, v) = ν0 (x) ∀|v| > 1
(8)
(this last assumption can of course be replaced with lim|v|→∞ ν(x, v) = ν0 (x)). The most important assumption concerns the behavior of the equilibrium F (v) for large v. We assume: κ for |v| ≥ 1 (9) F (v) = n+α |v| for some α ∈ (0, 2) and for all v ∈ Rn .
F (v) ≤ κ
(10)
Finally, we assume that the following symmetry conditions are satisfied: ν(x, −v) = ν(x, v) for all (x v) ∈ R2n . (11) R This implies in particular that the flux Rn vF (v) dv = 0 (when it is defined, that is when α > 1), which we know is a necessary condition for the derivation of diffusion type equation. F (−v) = F (v) ,
We can now state our main result: Theorem 1.1. Assume that (7), (8), (9), (10), (11) hold and let f ε be the solution of (1) with initial condition fin (x, v) = ρin (x)F (v) such that ρin ∈ H 4 (Rn ), then ||f ε − ρ0 F ||L∞ (0,∞;L2 −1 (Rn ×Rn )) −→ 0,
as ε → 0,
(12)
F
where ρ0 satisfies
in [0, ∞) × Rn , in Rn
∂t ρ0 + L(ρ0 ) = 0 ρ0 (0, x) = ρin (x)
(13)
and L is an elliptic operator defined by the singular integral: Z ρ(x) − ρ(y) L(ρ) = P.V. γ(x, y) dy. |y − x|n+α Rn with
Z γ(x, y) = ν0 (x) ν0 (y)
∞
z α e−z
0
4
R1 0
ν0 ((1−s)x+sy) ds
(14)
dz
Note that γ(x, y) = γ(y, x) and that condition (7) yields 0 < γ1 < γ(x, y) ≤ γ2 . In particular the operator L is a self-adjoint elliptic operator of order α (comparable to (−∆)α/2 ). Before giving the proof of this theorem in the general case (Section 3), we will present two simpler cases that can be handled with a more direct Hilbert method based on the Fourier Transform and for which the result is obtained in a L1 framework as well as in a L2 framework with less regularity assumptions on the initial data.
2
Space homogeneous cross sections
2.1
Constant cross section
In this section, we prove Theorem 1.1 in the simpler framework where the cross section σ is constant (for the sake of simplicity we take σ = 1). We then have Z f (v 0 ) dv 0 F (v) − f (v). (15) Q(f ) = Rn
In that case, the use of the Fourier transform (as in [14]) allows for explicit computations and a much simpler proof. We have the following proposition: Proposition 2.1. Let f ε be the solution of (1) with Q given by (15) where F satisfies (9-11) and with initial condition fin (x, v) = ρin (x)F (v) such that ρin ∈ W 3,1 (Rn ). Then ||f ε − ρ0 F ||L∞ (0,∞;L1 (Rn ×Rn )) −→ 0, where ρ0 satisfies
(
α
as ε → 0
∂t ρ0 + κ(−∆) 2 ρ0 = 0
in [0, ∞) × Rn
ρ0 (0, x) = ρin (x)
in Rn
with Z κ= Rn
(w · e)2 1 dw. 2 1 + (w · e) |w|n+α
(16)
(17)
(18)
If instead we have ρin ∈ H 3 (Rn ), then ||f ε − ρ0 F ||L∞ (0,∞;L2 −1 (Rn ×Rn )) −→ 0,
as ε → 0.
(19)
F
The proof relies on the following Hilbert expansion for f ε : f ε = f0ε + g1ε + g2ε + rε , with Q(f0ε ) = 0
(20)
(1 + εv · ∇x )g1ε = −εv · ∇x f0ε
(21)
Q(g2ε ) = εα ∂t f0ε − Q+ (g1ε ) 5
(22)
and rε solution of ∂t rε + ε1−α v · ∇x rε = ε−α Q(rε ) − ∂t g1ε − ∂t g2ε − ε1−α v · ∇g2ε .
(23)
In order to investigate these equations, we recall the following result, which holds for general operators of the form (2), whenever (7) holds (see [7]): Proposition 2.2. i) The operators Q+ and Q are bounded operator on L2F −1 (Rn ). ii) The kernel of Q has dimension 1 and is spanned by F . iii) There exists a constant c > 0 such that Z Z 1 1 − Q(f )f dv ≥ c |f − ρF |2 dv. F F Rn Rn iv) For all h ∈ L2F −1 (Rn ), the equation Q(g) = h has aRsolution if and only if is a unique such solution (denoted Q−1 (h)) satisfying Rn g(v) dv = 0.
R
h(v) dv = 0, and there
v) The operator Q−1 is bounded in L2F −1 (Rn ): kQ−1 (h)kL2 −1 ≤ CkhkL2 −1 F
for all h such that
F
R
Rn
h(v) dv = 0.
(24)
Z vi)
Q(f )sgn(f ) dv ≤ 0 for all f . Equation (20) then yields (using (ii)): f0ε (x, v, t) = ρε (x, t)F (v).
and so equation (21) becomes Tε g1ε = −εv · ∇x ρε F where we introduced the operator Tε = 1 + εv · ∇x . Assuming (for the time being) that this operator can be inverted, we can then write g1ε = −Tε−1 (εv · ∇x ρε )F.
(25)
Finally, the solvability condition for (22) (using (iv)) is Z α ε ε ∂t ρ F − Q+ (g1ε ) dv = 0 which reads εα ∂t ρε −
Z
g1ε dv = 0.
Using (25), we see that this is a condition on ρε , which we can write as Z α ε ε ∂t ρ + Tε−1 (εv · ∇x ρε )F dv = 0.
(26)
Because of the simple form of the operator Q, this condition also implies that the right hand side of (22) is actually 0 (rather than only having zero integral), and thus g2ε = 0. 6
Remarks 2.3. The usual Hilbert expansion consists in writing f ε = ρF + f1ε + f2ε + rε where ν(x, v)f1ε = Q+ (f1ε ) − εv · ∇x ρF, Q(f2ε ) + εv · ∇x f1ε = εα ∂t ρF. We can see that the scheme that we use here consists in reshuffling the terms Q+ (f1ε ) and εv · ∇x f1ε in the first order and second order equation. Note also that while in the usual Hilbert expansion, the first term is solution of the asymptotic equation, here we have ρε which is solution of an approximation of the limiting model (26). In order to prove Proposition 2.1, we thus have to show that ρε , solution of (26), exists and that g1ε and rε go to zero in some appropriate norm. This is best done using Fourier transform (with respect to x): We denote by b h(ξ) the Fourier transform, with respect to x, of a function h. Taking the Fourier transform in (21), we get: (1 + iεv · ξ)gb1ε = −iεv · ξ fb0 and so
iεv · ξ gb1ε = − F (v) ρbε . 1 + iεv · ξ
(27)
∂t ρbε + aε (ξ) ρbε = 0.
(28)
We can thus rewrite (26) as with ε
a (ξ) = ε
−α
Z Rn
iεv · ξ F (v) dv. 1 + iεv · ξ
We now need the following lemma (see also [14]): Lemma 2.4. For all ξ, aε (ξ) ≥ 0 and |aε (ξ) − κ|ξ|α | ≤ Cε2−α |ξ|2 −→ 0,
(29)
with κ given by (18). Furthermore, there exist positive constants c1 and c2 such that, c1 min(ε−α , |ξ|α ) ≤ aε (ξ) ≤ c2 (|ξ|α + ε(2−α) |ξ|2 )
(30)
Proof of Lemma 2.4. A simple computation (using the properties of F ) yields Z Z iεv · ξ (εv · ξ)2 F (v) dv = F (v) dv 2 Rn 1 + iεv · ξ Rn 1 + (εv · ξ) Z (εv · ξ)2 = F (v) dv 2 |v|≤1 1 + (εv · ξ) Z (εv · ξ)2 1 + dv. 2 n+α |v|≥1 1 + (εv · ξ) |v| The first integral is clearly bounded by Cε2 |ξ|2 , and using the change of variable w = εv|ξ| we can write the second integral as Z Z 1 (w · e)2 1 (εv · ξ)2 α α dv = ε |ξ| dw 2 n+α 2 n+α |v|≥1 1 + (εv · ξ) |v| |w|≥ε|ξ| 1 + (w · e) |w| 7
We deduce α ε
α
α
2
2
α
|ε a (ξ) − ε κ|ξ| | ≤ Cε |ξ| + ε |ξ|
α
Z |w|≤ε|ξ|
2
(w · e)2 1 dw 2 1 + (w · e) |w|n+α
2
≤ Cε |ξ|
This implies (29) and the second inequality in (30). The first inequality in (30) is obtained by distinguishing two cases: R (εv·ξ)2 1 α α and the result If ε|ξ| ≤ 1, then the computation above yields |v|≥1 1+(εv·ξ) 2 |v|n+α dv ≥ cε |ξ| follows. If ε|ξ| ≥ 1, then we have Z Z 1 1 1 (εv · ξ)2 α α dv ≥ ε |ξ| dw 2 n+α n+α |w|≥ε|ξ| 2 |w| |v|≥1 1 + (εv · ξ) |v| ≥ 1 and so aε ≥ ε−α .
We deduce Corollary 2.5. For any ρin and ε > 0, there exists a unique ρε solution of (28), given by ε ρbε (t, ξ) = e−ta (ξ) ρc in (ξ)
Furthermore, ρε satisfies || |ξ|k ρbε (t)||L∞ (Rn ) ≤ || |ξ|k ρc in ||L∞ (Rn ) and || |ξ|k ∂t ρbε (t)||L∞ (Rn ) ≤ || |ξ|k+2 ρc in ||L∞ (Rn ) for all k ≥ 0. Finally we have ||ρε − ρ0 ||L∞ (0,T ;L1 (Rn )) ≤ CT ||ρin ||W 2,1 (Rn ) and ||ρε − ρ0 ||L∞ (0,T ;L1 (Rn )) ≤ CT ||ρin ||H 2 (Rn ) where ρ0 is the unique solution of (17). Proof. We only prove the last assertion: We note that ∂t (ρbε − ρb0 ) + aε (ξ)(ρbε − ρb0 ) = (κ|ξ|α − aε (ξ))ρb0 and so Lemma 2.4 yields d ε |ρb − ρb0 | ≤ Cε2−α |ξ|2 ρb0 . dt We deduce
d ε ||ρ − ρ0 ||L1 (Rn ) ≤ Cε2−α ||ρ0 ||W 2,1 (Rn ) dt which gives the L1 (Rn ) convergence of ρε . A similar computation (multiplying by ρbε − ρb0 ) gives the L2 (Rn ) convergence.
8
Proof of Proposition 2.1. We now notice that (27) yields |εv · ξ| |gb1ε | ≤ p F (v)ρbε 1 + (εv · ξ)2 and thus ||gb1ε ||L∞ ≤ sup n 1 N ξ (R ;Lv (R )))
(Z
)
|εv · ξ|
p F (v) dv ρbε (ξ) 1 + (εv · ξ)2 ≤ sup C(εα |ξ|α + ε|ξ|)ρbε ξ
ξ
≤ C (εα + ε) ||(1 + |ξ|2 )ρbε ||L∞ (Rn ) , and similarly ≤ sup C(εα |ξ|α + ε|ξ|)∂t ρbε ||∂t gb1ε ||L∞ N n 1 ξ (R ;Lv (R ))) ξ
≤ sup C(εα |ξ|α + ε|ξ|)aε (ξ)ρbε ξ
so, using Lemma 2.4, we deduce ||∂t gb1ε ||L∞ ≤ (εα + ε)||(1 + |ξ|3 )ρbε ||L∞ (Rn ) . n 1 N ξ (R ;Lv (R ))) It follows that g1ε and ∂t g1ε converge to 0 strongly in L∞ (0, ∞; L1 (Rn × Rn )). Finally, multiplying the equation for the remainder rε , (23), by sign(rε ) (and using Proposition 2.2 (vi)), we get ∂t ||rε ||L1 (Rn ) ≤ ||∂t g1ε ||L1 (Rn ) . Using the fact that rε (t = 0) = −g1ε (t = 0), we deduce ||rε ||L∞ (0,T ;L1 (R2n )) −→ 0 and so ||f ε − ρε F ||L∞ (0,T ;L1 (Rn ×Rn )) −→ 0. Similarly, we have (using the computations of Lemma 2.4): Z Z |εv · ξ|2 2 2 1 ε b |g1 | dv ≤ F (v) dv ρbε 2 F (v) 1 + (εv · ξ) n n R R ≤ C(εα |ξ|α + ε2 |ξ|2 )ρbε
2
and so ||gb1ε ||L2 −1 (Rn ×Rn ) ≤ Cεα ||ρε ||H 1 (Rn ) F
and in a same way ||∂t gb1ε ||L2 −1 (Rn ×Rn ) ≤ Cεα ||ρε ||H 3 (Rn ) . F
A similar argument (multiplying (23) by rε /F ) gives the strong convergence in L∞ (0, ∞; L2F −1 (Rn × Rn )).
9
2.2
Case of space homogeneous cross section
We now show, in a still relatively simple case, why the general case is more complicated. In this section, we assume that Q is given by: Z Q(f ) = σ(v, v 0 )[f (v 0 )F (v) − f (v)F (v 0 )] dv 0 (31) Rn
with σ depending only on v and v 0 (and not on x), and satisfying 0 < σ0 ≤ σ(v, v 0 ) = σ(v 0 , v) ≤ σ1 . We might try to proceed as in the previous section, writing f ε = f0ε + g1ε + g2ε + rε , with f0ε (x, v, t) = ρε (x, t)F (v), g1ε = −Tε−1 (εv · ∇x ρε ), F Q(g2ε ) = εα ∂t ρε F − Q+ (g1ε ). where the operator Tε is now defined by Tε (f ) = νf + εv · ∇x f . The solvability condition for g2ε (obtained by integrating this last equation with respect to v) reads Z ∂t ρε = ε−α Q+ (g1ε ) dv n ZR −α = ε νg1ε dv n R Z −α = −ε ν(v)Tε−1 (εv · ∇x ρε )F dv. Rn
We thus obtain that ρε must solve ∂t ρε + Lε (ρε ) = 0 with Lε (ρ) = ε−α
Z Rn
ν(v)Tε−1 (εv · ∇x ρε )F dv.
Once again, we can identify the nature of this operator using Fourier transform: We get Z iεv · ξ −α ε \ L (ρ) = ε ν(v)F (v) dv ρbε Rn ν(v) + iεv · ξ = aε (ξ)ρbε . However, unlike the previous section, we do not have Q(g2ε ) = 0. Instead, we have Q(g2ε ) = −εα Lε (ρε )F − Q+ (g1ε ) which, in Fourier, reads: Q(gb2ε ) = −εα
Z Rn
iεv 0 · ξ [ν(v 0 ) − σ(v, v 0 )]F (v 0 ) dv 0 ρbε F (v) ν(v 0 ) + iεv 0 · ξ
10
and so |εv 0 · ξ| p F (v 0 ) dv 0 |ρbε |F (v) 2 0 2 n σ0 + (εv · ξ) R α α α ≤ Cε (ε |ξ| + ε|ξ|)|ρbε |F (v).
|Q(gb2ε )| ≤ 2σ1 εα
Z
This clearly implies that g2ε goes to zero as ε goes to 0, but it also means that it will be difficult to control the term ε1−α v · ∇g2ε because of the lack of integrability of vF (at least when α < 1). For this reason, we need to add a term in the expansion, as explained in the next section.
3
Proof of Theorem 1.1
3.1
A reshuffled Hilbert expansion
In order to prove Theorem 1.1, we use a Hilbert type expansion and some techniques similar to those first introduced in [13]. We recall that f ε is a solution of εα ∂t f ε + εv · ∇x f ε = Q(f ε ) where Q(f ) = Q+ (f ) − νf
with Q+ (f ) =
Z
(32)
σ(x, v, v 0 )F (v)f (v 0 )dv 0 ,
Rn
and we write the following expansion f ε = ρε (t, x)F (v) + g1ε (t, x, v) + g2ε (t, x, v) + g3ε (t, x, v) + rε (t, x, v)
(33)
where the terms g1ε , g2ε , g3ε satisfy (ν(x, v) + εv · ∇x )g1ε = −εv · ∇x (ρε F ) Q(g2ε ) = −Q+ (g1ε ) − εα ∂t ρε F (ν(x, v) + εv · ∇x )g3ε = −εv · ∇x g2ε .
(34) (35) (36)
The remainder solves ∂t rε + ε1−α v · ∇x rε =
Q+ (g3ε ) Q(rε ) ε ε ε − ∂ g − ∂ g − ∂ g + . t t t 1 2 3 εα εα
(37)
We recall that we defined the operator Tε (g) = νg + εv · ∇x g. A simple computation (see [13]) shows that this operator is invertible with Z z Z +∞ − ν(x − εvs, v) ds 0 Tε−1 (f ) = e f (x − εvz, v) dz
(38)
(39)
0
for any function f (x, v). In particular, equations (34) and (36) are solvable without additional conditions, but (35) requires the following solvability condition (obtained by integrating (35) with respect to v): ∂t ρε + Lε (ρε ) = 0 (40) 11
with Lε (ρ) = −ε−α
Z Rn
= −ε−α Z −α = ε
Z Rn
Rn
Q+ (g1ε )dv ν(x, v)g1ε dv
ν(x, v) Tε−1 (εv · ∇x ρε ) F dv.
(41)
In view of (37), it is clear that in order to prove Theorem 1.1, we need to show that g1ε , g2ε and g3ε Q+ (g ε )
can be defined and that the terms ∂t g1ε , ∂t g2ε ∂t g3ε and εα 3 in the right hand side of (37) go to 0 in L1 (0, ∞; L2F −1 (Rn × Rn )). The first step, detailed in the next section, will be to study the properties of this operator Lε and show that the equation (40) has a smooth solution ρε . In section 3.3, we will derive some important estimate on Q−1 (h) that will be needed to get some estimates on g2ε . In Section 3.4, we derive estimates on g1ε , g2ε and g3ε and their derivatives and show that f ε − ρε F converges to zero strongly. The final step (Section 3.5) is to show that ρε , solution of (40), converges strongly to ρ0 solution of (13).
3.2
Properties of the operator Lε and the equation (40)
First, of all, we note that for any function ρ(x), (39) and (41) imply Z ε −α ν(x, v) Tε−1 (εv · ∇x ρε ) F dv L (ρ) = ε n R Z Z +∞ R z = ε−α e− 0 ν(x−εvs,v) ds εv · ∇x ρ(x − εvz) F (v) dz dv. Rn
0
Alternatively, we can also write Tε−1 (εv · ∇x ρ) Z +∞ R z e− 0 ν(x−εvs,v) ds εv · ∇x ρ(x − εvz) dz = 0 Z +∞ R z d ρ(x − εvz) dz =− e− 0 ν(x−εvs,v) ds dz 0 Z +∞ R z d =− e− 0 ν(x−εvs,v) ds ρ(x − εvz) − ρ(x) dz dz 0 Z +∞ R z =− e− 0 ν(x−εvs,v) ds ν(x − εvz, v) ρ(x − εvz) − ρ(x) dz 0
which leads to ε
L (ρ) = −ε
−α
Z Rn
Z
+∞
e−
Rz 0
ν(x−εvs,v) ds
ν(x − εvz, v) ρ(x − εvz) − ρ(x) F (v) dz dv
(42)
0
which is the formula found in [13]. We can now prove the following lemma: Lemma 3.1. For all k integer, k ≥ 2, the operator Lε is a bounded operator from H k (Rn ) to H k−2 (Rn ): ||Lε (ρ)||H k−2 (Rn ) ≤ C||ρ||H k (Rn ) for all ρ ∈ H k (Rn ). (43) 12
Proof. We first prove that ||Lε (ρ)||L2 (Rn ) ≤ C||ρ||H 2 (Rn )
for all ρ ∈ H 2 (Rn ).
A similar computation is performed in [13], and recalled here for the reader’s convenience. We write Z ε −α ν(x, v) Tε−1 (εv · ∇x ρ) F dv L (ρ) = ε I1ε
= where I1ε
=ε
−α
Rn I2ε + I3ε
+
Z |v|≤1
I2ε = ε−α
ν(x, v) Tε−1 (εv · ∇x ρ) F dv,
Z
ν(x, v) Tε−1 (εv · ∇x ρ) F dv,
1≤|v|≤1/ε
I3ε = ε−α
Z |v|≥1/ε
ν(x, v) Tε−1 (εv · ∇x ρ) F dv.
We have Tε−1 (εv
Z
+∞
· ∇x ρ) =
e−
Rz 0
ν(x−εvs,v) ds
εv · ∇x ρ(x − εvz) dz
0
Z =
+∞ h
e−
Rz 0
ν(x−εvs,v) ds
i − e−ν(x,v)z εv · ∇x ρ(x − εvz) dz
0
Z
+∞
+
e−ν(x,v)z [εv · ∇x ρ(x − εvz) − εv · ∇x ρ(x)] dz
0
Z +
+∞
e−ν(x,v)z εv · ∇x ρ(x) dz
0
and so an integration by parts yields Z +∞ h R i z −1 Tε (εv · ∇x ρ) = e− 0 ν(x−εvs,v) ds − e−ν(x,v)z εv · ∇x ρ(x − εvz) dz 0 Z +∞ 1 + e−ν(x,v)z ε2 v · Dx2 ρ(x − εvz) · v dz ν(x, v) 0 Z +∞ + e−ν(x,v)z εv · ∇x ρ(x) dz. 0
Using the symmetry assumption on F and ν, we deduce: Z Z +∞ h R i z ε −α I1 = ε ν(x, v) e− 0 ν(x−εvs,v) ds − e−ν(x,v)z εv · ∇x ρ(x − εvz) F (v) dz dv |v|≤1
+ε−α
Z |v|≤1
0
Z
+∞
e−ν(x,v)z ε2 v · Dx2 ρ(x − εvz) · v dzF (v) dv
0
and since F is bounded, and Rz − 0 ν(x−sεv,v)ds − e−zν(x,v) ≤ Cz 2 e−σ1 z ε|v| e 13
(44)
we deduce ||I1ε ||L2 (Rn ) ≤ Cε2−α ||ρ||H 2 (Rn ) Next, Assumption (9) and the change of variable w = εv yields: Z Z +∞ R z ε −α e− 0 ν(x−εvs,v) ds ν(x, v) ε · ∇x ρ(x − εvz) I2 = ε 1≤|v|≤1/ε 0 Z +∞ R − 0z ν(x−ws,w/ε) ds
Z
e
=
ε≤|w|≤1
ν(x, w/ε) w · ∇x ρ(x − wz)
0
1 |v|n+α
dz dv
1 dz dw |w|n+α
and proceeding as with I1ε , we deduce I2ε
Z
Z ν(x, w/ε)
=
+∞ h
e−
Rz 0
ν(x−ws,w/ε) ds
i − e−ν(x,w/ε)z w · ∇x ρ(x − wz)
0
ε≤|w|≤1
Z
Z
+ ε≤|w|≤1
+∞
e−ν(x,w/ε)z w · Dx2 ρ(x − wz) · w dz
0
1 dz dw |w|n+α
1 dw |w|n+α
and so ||I2ε ||L2 (Rn ) ≤ C||ρ||H 2 (Rn ) . Finally, to estimate I3ε , we use the formula Tε−1 (εv · ∇x ρ) = −
Z
+∞
e−
Rz 0
ν(x−εvs,v) ds
ν(x − εvz, v) ρ(x − εvz) − ρ(x) dz.
0
Assumption (9) and the change of variable w = εv then yields: Z Z +∞ R z ε I3 = − e− 0 ν(x−ws,w/ε) ds ν(x, w/ε) ν(x − wz, v) ρ(x − wz) − ρ(x) |w|≥1
0
1 dz dw |w|n+α
and so ||I3ε ||L2 (Rn ) ≤ C||ρ||L2 (Rn ) .
We now show how to obtain higher order estimates: We have Z ε ε −α ∂xi L (ρ ) = ε ∂xi ν(x, v)Tε−1 (εv · ∇x ρ)F (v)dv Rn
−ε
−α
Z Rn
+ε
−α
Z Rn
ν(x, v)Tε−1 (∂xi νTε−1 (εv · ∇x ρ))F (v) dv ν(x, v)Tε−1 (εv · ∇x ∂xi ρ)F (v)dv.
The last term is just Lε (∂xi ρ) which can thus be bounded by the previous step. The first term is ∂ ν treated in a same way as Lε (ρ) except that we need to use the fact that xνi is even and bounded. Therefore, we will only detail the computations concerning the second term. We split it into three integrals corresponding respectively to |v| < 1, 1 < |v| < 1ε and |v| > 1ε .
14
1 ε
First of all, we handle the |v| > variable. It leads to I1ε
= −ε−α
Z |v|> 1ε
part. For that purpose, we use (9) and the usual change of
ν(x, v)Tε−1 (∂xi νTε−1 (εv · ∇x ρ))F (v) dv z
Z Z
w = − ν(x, ) ε |w|>1
ν(x − ws,
+∞ −
Z
e
0
0
w Z +∞− ) ds ε ∂x ν(x − wz, w ) e i ε 0
ν(x − w(z + z),
z
Z
ν(x − w(s + z), 0
w ) ds ε
w 1 )[ρ(x − w(z + z)) − ρ(x − εvz)] n+α dvdzdz. ε |w|
Thus, ||I1ε ||L2 (Rn ) ≤ C||ρ||L2 (Rn ) . Next, for |v| < 1, we write Z ε −α I2 = −ε ν(x, v)Tε−1 (∂xi νTε−1 (εv · ∇x ρ))F (v) dv |v|
