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Averaging lemmas
P-E Jabin (University of Nice)Presentation of the course
Kinetic equations : Transport equation in phase space, i.e. on
f(x,v) of x and v
d∂ f +v·∇ f = g, t ≥ 0, x,v ∈ R .t x
As for hyperbolic equation, the solution cannot be more regular
than the initial data or the right hand-side. But averages in
velocity like
∞ dρ(t,x) = f(t,x,v)φ(v)dv, φ∈ C (R ),c
usually are, the question being of course how much?Plan of the course
1. Introduction
22. L framework
p3. General L framework
4. One limit case : Averaging lemma with a full derivative
5. An example of application : Scalar Conservation LawsPlan of the first course
1. Well posedness of the basic equation
2. The 1d case
3. Local equilibrium
4. Application to the Vlasov-Maxwell systemWell posedness of the basic equation
During most of this course, we deal with the simplest
d∂ f +α(v)·∇ f = g(t,x,v), t ∈R , x ∈R , v ∈ ω, (1)t x +
dwhere ω =R or a subdomain; Or with the stationary
α(v)·∇ f = g(x,v), t ∈R , x ∈ O, v ∈ ω, (2)x +
dwhere O is open, regular in R and ω is usually rather the sphere
d−1S .
Of course (1) is really a particular case of (2) with
d −→ d +1, x −→ (t,x), α(v)−→ (1,α(v)).The fundamental relation for solutions to (1) is
f(t ,x,v) =f(t ,x−α(v)(t −t ),v)2 1 2 1
Z t −t2 1
+ g(t −s,x−α(v)s,v)ds,∀t , t2 1 2
or for solutions to (2)
Z t
f(x,v) = f(x−α(v)t,v)+ g(x−α(v)s,v)ds,∀t.
Those two formulas may be used to solve the equation but these
are not unique so an initial data must be given
0f(t = 0,x,v) = f (x,v), (3)
and for (2) the incoming value of f on the boundary must be
inf(x,v) = f (x,v), x ∈ ∂O,α(v)·ν(x)≤ 0, (4)
where ν(x) is the outward normal to O at x.With that the equation is solvable as per
0 0 d 1 0 dLet f ∈D (R ×ω) and g ∈ L (R , D (R ×ω)). Then there+loc
1 0 dis a unique solution in L (R , D (R ×ω)) to (1) with (3) in+loc
the sense of distribution given by
Z t
0f(t,x,v) = f (x−α(v)t,v)+ g(t−s,x−α(v)s,v)ds. (5)
∞ dNote that if f solves (1) then for φ∈ C (R ×ω)c
d 1f(t,x,v)φ(x,v)∈ L (R ),+locdt dR ×ω
so f has a trace at t = 0 in the weak sense and (3) perfectly
makes sense.On the other hand, the modified equation, which we will frequently
dα(v)·∇ f +f = g, x ∈R , v ∈ ω, (6)x
dis well posed in the whole R without the need for any boundary
0 d 0 dLet g ∈S (R ×ω), there exists a unique f inS (R ×ω) solution
to (6). It is given by

−tf(x,v) = g(x−α(v)t,v)e dt. (7)
Notice that taking the Fourier transform in x of (6)
ˆ(i α(v)·ξ +1)f = gˆ,
and of course 1+iα(v)·ξ never vanishes contrary to iα(v)·ξ.The 1d case
Let us study the easiest case, namely
v∂ f = g.x
p 2 p 2Of course away from v = 0, if g ∈ L (R ) then ∂ f ∈ L (R ).x
But what if f or g do not vanish around v = 0. For instance
f(x,v) = ρ(x)δ(v),
then of course, whatever ρ, in the sense of distribution
v∂ f = 0.x
So clearly concentrations have to be avoided. Let us be more
∞precise. Take φ∈ C (R), definec
ρ(x) = φ(v)f(x,v)dv.
6And compute for a bounded interval I
Z Z p|ρ(x)−ρ(y)|
kρk k,p = dxdy.W (I) 1+kp|x−y|I I
Using the equation
|ρ(x)−ρ(y)|≤ |f(x,v)−f(y,v)|φ(v)dv
≤ |f(x,v)−f(y,v)|φ(v)dv + ...
|v|<R |v|>R
and this last term is bounded by
Z Z1
|x−y||∂ f(θx +(1−θ)y,v)|φ(v)dv dθx
0 |v|>R
Z Z1 |x−y|
≤ |g(θx +(1−θ)y,v)|φ(v)dθdv
|v|0 |v|>R
Z Z 1/p1|x−y| p≤ C |g(θx +(1−θ)y,v)| φ(v)dv dθ .
1/p|R| 0 R