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- right hand-side
- averaging lemmas
- vlasov- maxwell system
- basic equation
- general lp
- kinetic equations

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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

Z

∞ dρ(t,x) = f(t,x,v)φ(v)dv, φ∈ C (R ),c

dR

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

0

or for solutions to (2)

Z t

f(x,v) = f(x−α(v)t,v)+ g(x−α(v)s,v)ds,∀t.

0

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

speciﬁed

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

Theorem

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)

0

∞ dNote that if f solves (1) then for φ∈ C (R ×ω)c

Z

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

use,

dα(v)·∇ f +f = g, x ∈R , v ∈ ω, (6)x

dis well posed in the whole R without the need for any boundary

condition

Theorem

0 d 0 dLet g ∈S (R ×ω), there exists a unique f inS (R ×ω) solution

to (6). It is given by

Z

∞

−tf(x,v) = g(x−α(v)t,v)e dt. (7)

0

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

Z

ρ(x) = φ(v)f(x,v)dv.

R

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

Z

|ρ(x)−ρ(y)|≤ |f(x,v)−f(y,v)|φ(v)dv

RZ Z

≤ |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

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