The Cauchy problem for the Boltzmann equation

Alexander Bastounis Thomas Holding Vittoria Silvestri

January 30, 2013

1 Introduction

New mathematical results do not emerge often from existing tools and techniques. The Boltzmannequation is one of the core equations of mathematical physics, and its Cauchy problem was and still isone of the most important open problems in the field. The new concepts that are key to the 1989 resultby DiPerna and Lions [1] are the notions of renormalised solutions of transport equations and velocityaveraging. This allows for a proof of the global existence of weak solutions via compactness argumentswithout any a priori estimates on the derivatives. The regularity and uniqueness of these solutions is stillan open problem. As is common when a new method is developed, further results came quickly. In thisproject we will present the proof of weak solutions to the Boltzmann equation and a later applicationof the same tools by DiPerna and Lions[3] to ODEs to extend the Cauchy-Lipschitz theorem to somenon-Lipschitz forcefields in a slightly weaker sense.

Contents

1 Introduction 1

2 Linear transport equations 22.1 Basic overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2.1.1 Characteristics and ‘transportation’ . . . . . . . . . . . . . . . . . . . . . . . . . . 22.1.2 Mild solutions and Duhamel’s principle . . . . . . . . . . . . . . . . . . . . . . . . 32.1.3 Algebraic properties and conservation laws . . . . . . . . . . . . . . . . . . . . . . 42.1.4 Renormalised solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.2 The setting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2.1 Function Spaces and topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2.2 Distributional solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.3 Existence and uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.3.1 Smooth coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.3.2 Rough coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.4 Velocity averaging results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.4.1 Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.4.2 Velocity Averaging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3 The Cauchy Problem for the Boltzmann Equation 133.1 The equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3.1.1 Solution concepts for the Boltzmann equation . . . . . . . . . . . . . . . . . . . . . 143.1.2 Main theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.2 Short sketch of the proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.3 A priori estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.3.1 Entropy identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.4 Truncated problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.5 Weak compactness results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.5.1 Weak compactness of (fn) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.5.2 Weak compactness of the collision term . . . . . . . . . . . . . . . . . . . . . . . . 25

1

3.5.3 Temporal regularity of f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.6 Velocity averaging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.7 Exponential formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4 Rough coefficients 354.1 Another use for renormalised solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

4.1.1 Cauchy-Lipschitz with rough coefficients . . . . . . . . . . . . . . . . . . . . . . . . 354.1.2 Strategy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364.1.3 Techniques and ideas used . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4.2 A proof using the stability result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364.3 Proof of stability result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4.3.1 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394.3.2 Pointwise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394.3.3 On compact time sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394.3.4 Existence and uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4.4 Further remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

5 Conclusions 40

A Appendix 40A.1 Weak* compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40A.2 Weak compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41A.3 Strong compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41A.4 Continuity in Banach spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

2 Linear transport equations

This section will give a refresher on basic linear transport theory, some notions of solution, and thensome existence results. To begin with the presentation will be unrigourous, with no spaces or regularityspecified. The existence and uniqueness results are sketched from [3], and the equivalence result iscombined from [3] and [2].

2.1 Basic overview

The equation we will look at is (with the Einstein summation convention),

∂u

∂t+ bi(t, x)

∂u

∂xi+ c(t, x)u = f(t, x, u)

u(0, x) = u0(x) (1)

u(t, x) : (0, T )× RN → R

To begin with we will look at when c = 0 and f = f(t, x).

2.1.1 Characteristics and ‘transportation’

Locally we can view b as constant, and we see that ∂u∂t + bi∂u∂xi

is the directional derivative along (1, b).The equation therefore states that u remains constant along this line. We can then solve the equationby finding the value of u along this line backwards in time. Globally we need to stitch up all the locallyconstant lines to form the characteristic curves.

The characteristic equations are

X(s; t, x) : R1+N → RN

Ẋ = b(t,X)

X(s; s, x) = x

2

where Ẋ = ∂X∂t . If X(t, x) solves this and u solves (1), then we observe that for each fixed s, u(t,X(t, x))obeys

d

dt[u(t,X(t, x))] =

∂u

∂t

∣∣∣∣(t,X(t,x))

+∂u

∂Xi

∂Xi∂t

=

(∂u

∂t+ bi

∂u

∂xi

)∣∣∣∣(t,X(t,x))

= 0

Noting that at t = s = 0 we have u(0, X(0; 0, x)) = u(0, x) = u0(x), we see that the value of uat (t,X(0; t, x)) is u0(x). Thus by the group property (X(s; t + τ, x) = X(s; t,X(t; τ, x))) and thateach X(s; t, ·) is a diffeomorphism, we have that if y = X(0; t, x) then x = X(t;−t, y) and henceu(t, y) = u0(X(t;−t, y)). This gives us the solution,

u(t, x) = u0(X(t;−t, x)) (2)

X(t;−t, x) means we look back along the characteristic curve that starts at (t, x) to find where itcrosses the plane t = 0. In this way, the transport equation rearranges the function u(t, ·) over time,‘transporting’ mass along the characteristics.

2.1.2 Mild solutions and Duhamel’s principle

(2) has meaning for any measurable function u0, so we can extend our notion of solution to a far moregeneral space. To do this we introduce the solution operator St.

Definition 1 (Solution Operator). The solution operator to (1) with c = f = 0 is an operator onmeasurable functions RN → RN defined by

[Stg](x) = g(X(t;−t, x))

If b depends only on x, then S obeys the group property StSs = St+s.

From now on we will write g](t, x) to mean Stg, moving our equation along the characteristics. Ingeneral this is not a group or even a semigroup due to the time dependence of b.

Definition 2 (Mild solution). The mild solution to ∂u∂t + bi∂u∂xi

= 0, u(0, x) = u0 with u0 measurable is

u]0.

All classical solutions are mild solutions.We now look at when c is not identically zero. Along the characteristics we have

d

dtu] + c]u] = 0

we can solve this via the integrating factor method to give

u](t, x) = u]0(x) exp

[−∫ t0

c](s, x) ds

]which we can use to define a new solution operator.

Definition 3 (Solution Operator). The solution operator to (1) with f = 0 is defined by

[Stg](x) = g](x) exp

[−∫ t0

c](s, x) ds

]We will now define g[ using this. This is an sort of ‘exponential’ form of the solution. Note that if

c = 0 then g[ = g].For the complete transport equation (1), we can use Duhamel’s principle. Formally, along the char-

acteristics we haved

dtu[ = f [

so integrating we obtain

u[(t, x)− u[0(x) =∫ t0

d

dtu[(s, x) dt =

∫ t0

f(s, x, u)[ dt

and we can use this for the general rigourous definition

3

Definition 4 (Mild Solutions). Let u0 be measurable, then u is a exponentially mild solution to (1) iff

• For almost all x, c], f(s, x, u)[ ∈ L1[0, T )

• For all t ∈ [0, T ) and for almost all x,

u[(t, x) = u[0(x) +

∫ t0

f(s, x, u)[ ds

If c = 0 we simply call this a mild solution. We can always take c = 0 by merging the cu term into f ,(which can depend on u).

2.1.3 Algebraic properties and conservation laws

We now look at the properties of solutions to the transport equation with f = c = 0. Suppose u is asolution to the transport equation. A renormalisation of u is given by β(u) where β : R→ R. Formally,using the chain rule,

∂β(u)

∂t+ bi

∂β(u)

∂xi= β′(u)

(∂u

∂t+ bi

∂u

∂xi

)= 0

so that β(u) also solves the transport equation with initial condition β(u(0)) = β(u0). This worksbecause the equation is linear, first order and has no feedback term as c = 0 and f = 0. Even when thisis not the case, we still obtain that β(u) satisfies

∂β(u)

∂t+ bi

∂β(u)

∂xi+ cuβ′(u) = fβ′(u)

This makes the next property useful. We look at the evolution of the mass of the solution when c = f = 0.

d

dt

∫u dx =

∫∂u

∂tdx = −

∫bi∂u

∂xidx

= −∫∂[biu]

∂xi− u ∂bi

∂xidx =

∫udiv b dx

In particular if div b = 0 then the mass is conserved, and we can approximate the Lp norms to showtheir conservation (for u0 ≥ 0). As β(u) is also a solution, we have that

d

dt

∫β(u) dx =

∫β(u) div b dx

These conservation properties can be extended to the full equation where c, f are included. We obtain

d

dt

∫β(u) dx =

∫β(u) div b− cuβ′(u)− fβ′(u) dx

2.1.4 Renormalised solutions

We will now transition the intuition discussed above into rigorous mathematical theorems. Again, werestrict to when c = f = 0. When u is very irregular, i.e. not even L1loc, we can take β ∈ C1(R) boundedand then look at the equation for β(u). Of course this equation cannot be solved in the classical sense.Instead we will view it in the sense of distributions. This allows us to define a notion of another solutionfor very general u.

Definition 5 (Renormalised Solution). u is a renormalised solution to (1) (with c = f = 0) if and onlyif for all continuously differentiable β : R→ R, β(u) obeys

∂β(u)

∂t+ bi

∂β(u)

∂xi= 0 in D′([0, T )× RN )

Depending on the space u lies in, we will require different additional restrictions on β. For example,we may demand that β ∈ L∞. We will discuss this later.

For the general case, we use the derived equation for β(u) and obtain the following definition:

4

Definition 6 (Renormalised Solution). We say that u is a renormalised solution to (1) if and only iffor all continuously differentiable β : R→ R, with |β′(z)|(1 + |z|) bounded, β(u) obeys

∂β(u)

∂t+ bi

∂β(u)

∂xi+ cuβ′(u) = fβ′(u) in D′([0, T )× RN )

The condition |β′(z)|(1 + |z|) bounded means that the extra term uβ′(u) is bounded also. To definethis with a non-zero f , we require a restriction on β to make fβ′(u) regular. This will depend heavilyon the particular f . Later we will look at the particular case of the Boltzmann equation.

The notion of renormalised solutions is particular to the form of the transport equation. It requiresthe equation to be first order and quasilinear. It allows us to deal with growth of u and f by makingthe terms in the renormalised equation of lower order.

2.2 The setting

2.2.1 Function Spaces and topologies

We saw before that the transport equation (1) can be viewed as rearranging u under the forcefield b.This means that we can make our notions of solution and spaces very general, even as general as allmeasurable functions. To do this we need to say what topology we want on these spaces.

Definition 7. We define L as the space consisting of measurable functions from RN to R that are finitealmost everywhere. We give it the topology of local convergence in measure. This has a metric

d(φ, ψ) =∑n≥1

1

2n‖|φ− ψ| ∧ 1‖L1(Bn)

This topology is equivalent to the topology induced by the notion of convergence given by un → u if andonly if β(un)→ β(u) in L1loc for all β ∈ C(R) ∩ L∞(R).

We have two variables for most of our functions, so we need to define our regularity on each and theway they depend on each other.

Definition 8. We define L1([0, T ];L∞(RN )) by the norm ‖f‖ =∫[0,T ]‖u(t, ·)‖∞ dt and similarly for

any other composed spaces with norms.

For spaces without norms, we have to be more careful.

Definition 9. The space C(R;L) consists of all continuous functions from R to L where we use thetopological definition of continuity. Similarly for other spaces.

Definition 10. f ∈ L∞(0, T ;L) means that for all β ∈ C(R) ∩ L∞(R), β(f) ∈ L∞(0, T ;L1loc).

Finally we need to define what we mean by the sum of spaces.

Definition 11. By f ∈ L1 +L∞ we merely mean that f = f1 + f2 for some f1 ∈ L1 and f2 ∈ L∞. Thissum will not be unique.

This notation is long winded and distracting. For this reason we will abbreviate. Unless otherwisespecified

• Our time interval will be [0, T ], and our spacial domain RN .

• We will assume that the coefficients b, c,div b will be L1 in time, so that when we say b ∈ L∞ wemean b ∈ L1(0, T ;L∞(RN )). Similarly for convergence of these.

5

2.2.2 Distributional solutions

Earlier we assumed that it was obvious what is meant for u to satisfy the transport equation in thedistributional sense. We will now make this rigorous.

Definition 12 (Test functions). We define D = D([0, T ) × RN ) as the space of all C∞([0, T ] × RN )functions with compact support in [0, T )× RN , where by saying that the support is compact in [0, T ) wemean that the support can include 0, but must not contain T .

Definition 13 (Distributional Solution). By saying that u solves (1) in D′ with f = f(t, x) we meanthat for all φ ∈ D,∫ T

0

∫RN

u

{−∂φ∂t− div(bφ) + cφ

}dx dt−

∫ T0

u0φ(0, x) dx =

∫ T0

∫RN

f(t, x)φdx dt

This makes sense for u ∈ Lp if we have the conditions on the coefficients

c− div b ∈ Lqloc, b ∈ Lqloc, f(t, x) ∈ L

1loc

where p, q are Holder conjugates.In particular for u ∈ L∞, we require that c− div b, b and f are in L1loc.

In a broad sense these notions of solutions are equivalent. In particular:

Theorem 1 (Equivalence). Let the ] operator be invertable and measure preserving (for this to hold itis sufficient that div b = 0, b is independent of t and satisfies the upper bound |b(z)| ≤ C(1 + |z|)). Then:

• If u, f ∈ L1loc([0, T )× RN ), then mild and distributional solutions are equivalent.

• If fβ′(u) ∈ L1loc([0, T )× RN ) then mild solutions are renormalised solutions.

• Renormalised solutions are mild solutions, and we need only check β(z) = log(1 + z).

• If in addition c ≥ 0 and for almost all x, (cu)] ∈ L1[0, T ), then exponentially mild solutions aremild solutions.

Proof. We start with the equivalence of mild and distributional solutions. Here we will merge c into f ,so wlog we may assume that c = 0.Let u be a distributional solution, and fix a test function ϕ ∈ D(RN ). Let ρn ∈ D([0, T )) satisfyρn → 1[t1,t2] in D′([0, T )). By measure preservation and invertability the adjoint of ] is given by ](−t, x)and both operate on all measurable functions, so that when we test the equation with ϕ](−t, x)ρn(t) weobtain:

0 =

〈∂u

∂t+ bi

∂u

∂xi− f, ϕ](−t, x)ρn(t)

〉= −

〈u,

[∂

∂t+ bi

∂

∂xi

](ϕ](−t, x)ρn(t))

〉−〈f, ϕ](−t, x)ρn(t)

〉= −

〈u, ρn(t)

[∂

∂t+ bi

∂

∂xi

]ϕ](−t, x) + ϕ](−t, x)ρ′n(t)

〉−〈f, ϕ](−t, x)ρn(t)

〉= −

〈u, 0 + ϕ](−t, x)ρ′n(t)

〉−〈f, ϕ](−t, x)ρn(t)

〉= −

〈u], ϕρ′n

〉−〈f ], ϕρn

〉Where the inner product is the action of distributions in D′([0, T )× RN ) and we note that the equality[∂∂t + bi

∂∂xi

]ϕ](−t, x) = 0 follows from the chain rule (as ϕ is smooth) and the definition of ]. As ϕ was

arbitrary, we have∫ T0ρ′n(t)f

] − ρn(t)u] dt = 0 for almost all x. We then let n → ∞ to obtain that foralmost all x,

u](t2, x)− u](t1, x) =∫ t2t1

f ](s, x) ds

6

So u is a mild solution. The converse comes from reversing this argument.Now let u be a renormalised solution. Define β(z) = log(1 + z). By the definition of renormalised

solutions [∂

∂t+ bi

∂

∂xi

]β(u) = β′(u)f =

f

1 + uin D′

Next, define βδ(z) =1δβ(δ(e

z− 1)) for δ > 0, which, when δ → 0, will approximate the identity function.Note that for fixed δ each βδ is Lipschitz. We then have[

∂

∂t+ bi

∂

∂xi

]βδ(β(u)) = β

′δ(β(u))

[∂

∂t+ bi

∂

∂xi

]β(u) in D′

But we see that β′δ(β(z)) = eβ(z)β′(δ(eβ(z) − 1)) = (1 + z)β′(δz) = 1+z1+δz , so that for all δ > 0,[

∂

∂t+ bi

∂

∂xi

]βδ(β(u)) =

f

1 + δuin D′

By computation we see that vδ = βδ(β(u)) =1δ log(1 + δu), and by the equivalence of mild and distri-

butional solutions proved above, we have that for almost all x, f ] ∈ L1[0, T ). We now go back to u bynoting that u] = ev

]1 − 1, so that u] is also the time integral of some function, and hence f ], (which may

depend on u), is in L1[0, T ) as required. Then by the equivalence again we obtain that for all t > s ≥ 0,δ > 0 and almost all x,

v]δ(t, x)− v]δ(s, x) =

∫ ts

f ]

1 + δu]dτ

When we take δ → 0 we see that u is a mild solution.The converse follows from putting observing that if u is a mild solution then v]1 obeys the equation

above, and we then apply the equivalence of mild and distributional solutions.Finally we have to show that exponentially mild solutions are equivalent to mild solutions. Suppose

u is an exponentially mild solution and the given conditions on c are satisfied. Then we have

• For almost all x, c], f [ ∈ L1[0, T ).

• For 0 ≤ s < t < T and almost all x, u[(t, x)− u[(s, x) =∫ tsf(τ, x, u)[ dτ .

• c ≥ 0

• For almost all x, (cu)] ∈ L1[0, T )

and we need to show that

• For almost all x, (f − cu)] ∈ L1[0, T ).

• For 0 ≤ s < t < T and almost all x, u](t, x)− u](s, x) =∫ ts(f − cu)](τ, x) dτ

Recall that f [ = f ] exp(−∫ t0c](s, x) ds

), and so because c ≥ 0 and f [ ∈ L1[0, t) we see that f ] ∈

L1[0, T ), and hence (f − cu)] ∈ L1[0, T ). Under these conditions we can go from u[ to u] and obtain amild solution.

2.3 Existence and uniqueness

We will show existence and uniqueness with initial conditions

u0 ∈ Lp, p ∈ [1,∞] (3)

and with coefficients that satisfy

c,div b ∈ L1(L∞)b ∈ L1(W 1,qloc ) (4)

b

1 + |x|∈ L1(L1) + L1(L∞)

(5)

7

2.3.1 Smooth coefficients

Lemma 1. Assume (3) and (4), and let b(t, x), c(t, x), f(t, x) ∈ C∞. Then there is a classical smoothsolution.

Proof. By Cauchy-Lipschitz there is a unique and smooth (in both x and t) solution to the characteristicequations, and by the growth condition on b it is global. We can then define our solution by the solutionoperators defined previously, all of which is valid for in the smooth case.

2.3.2 Rough coefficients

The estimates given in 2.1.3 together with the previous smooth result and standard mollification argu-ments yield:

Proposition 2 (Existence in Lp). Assume (3) and (4) are satisfied. Then there is a unique distributionalsolution in C(0, T ;Lp) or if p =∞ merely L∞(0, T ;L∞).

Proof. Mollification allows the application of the smooth result. Using the ideas of 2.1.3 and approxi-mation we obtain the estimates

‖u(t)‖∞ ≤ ‖u0‖∞ +∫ t0

‖cu‖∞ ds ≤ C ‖u0‖∞

d

dt

∫|u(t)|p dx ≤

(∫|u(t)|p dx

)‖pc+ div b‖∞

‖u(t)‖p ≤ C′ ‖u0‖p a.e. on (0, T ) by Gronwall

We can now apply relative weak compactness criterion to see that our sequence of solutions (ũ) isrelatively weakly compact if 1 < p < ∞ and relatively weak* compact for p = ∞. A computation thatwe skip then shows that a limit of a convergent subsequence is a solution.

This leaves p = 1, for which we need to approximate u0 with elements of D(RN ) converging in L1,giving similar results.

Finally we must show uniqueness. To do this we will make the conservation laws developed aboverigorous. This requires a technical lemma:

Lemma 2. Under the conditions of the theorem, the mollification ũ = ũ(�; t, x) of the actual solution(not the solution to the equation with mollified coefficients) satisfies

∂ũ

∂t+ bi

∂ũ

∂xi+ cũ = r̃

where as �→ 0 r̃ → 0 in L1(Lp) if p

• Integrate the equation for β(u) against a test function φR = φ(·/R) where φ is 1 in |x| ≤ 1 and 0outside |x| ≥ 2. The purpose is to look at ddt

∫β(u)φR dx and close by Gronwall.

d

dt

∫β(u)φR dx = −

∫(cuβ′(u)− div bβ(u))φR dx+

∫β(u)bi

∂φR∂xi

dx

• The first integral is ok because of the boundedness of β′ and the conditions on the coefficients.

• The only problem is the second integral, for which we apply the growth condition b/(1 + |x|) ∈L1 + L∞. The problem integral is only non-zero in |x| ∈ (R, 2R) and we get 1R decay from thederivative on φR. In this region we bound

|b|R 1{R≤|x|≤2R} ≤

|b|1+|x|1{R≤|x|} which we split this into

L1 and L∞ parts. The L1 part is integrable when we bound β(u) ≤ M , while the L∞ part justadds a term to the bound for the first integral.

• Sending R→∞ we can close by Gronwall in the case p 1 then u ∈ C(0, T ;Lp) follows from the form of the transportequation. p = 1 is harder, using approximation arguments we omit here.

2.4 Velocity averaging results

The aim of this section is to establish results for the later existence proof for the Boltzmann equation,which has the form

∂f

∂t+ vi

∂f

∂xi= g

f(t, x, v) : [0, T )× RN × RN → R

We will later see the exact form of the right hand side g = Q(f, f), but for now it doesn’t matter.The equation is often written using the transport operator T = ∂∂t + vi

∂∂xi

as Tf = g. This is a

time independent inhomogeneous transport equation with b(x, v) = [v, 0]T . This means that div b =∂vi∂xi

+ ∂∂vj 0 = 0, and the homogeneous equation preserves phase-space. In this section we will look at the

properties of equations of the formTf = g

2.4.1 Characteristics

The characteristics split into two vectors and the equations have a simple form:

Ẋ(t, x, v) = −V (t, x, v)V̇ (t, x, v) = 0

X(0, x, v) = x

V (0, x, v) = −v

This has a solution given by X(t, x, v) = x− vt, V (t, x, v) = v. The characteristic map is then

f ](t, x, v) = f(t, x+ vt, v)

from which we gain our notion of mild solution.

9

2.4.2 Velocity Averaging

Here we will present one of a general class of results called velocity averaging lemmas. The transportoperator T gives no gain in regularity, but if we integrate through we obtain conservation of mass we geta very large gain in regularity. What happens if we only partially integrate through?

Definition 14 (Velocity average). We define the velocity average of f(t, x, v) by F (t, x) =∫RN f(t, x, v) dv.

We will look at the Fourier transform of Tf in the t → τ and x → y coordinates. In Fourierspace the transport operator takes the form T̂ = iτ + ivjyj , which is very similar to the Sobolev norm

‖f‖2Hs =∫R×RN (1 + |τ |

2 + |y|2)s|f̂(τ, y, v)|2 dy dτ . We can use this to show a gain in regularity from fto F .

Lemma 3. Let f ∈ L2(R × RN × RN ) with compact support, and Tf ∈ L2(R × RN × RN ), thenF (t, x) ∈ H1/2(R× RN ).Proof. We wish to bound ‖F‖H1/2 . We compute

F̂ (τ, y) =

∫RN

f̂(τ, y, v) dv

So as f ∈ L2(R×RN ×RN ) it is sufficient to show that∫R×RN (|τ |

2 + |y|2)1/2|F̂ (τ, y)|2 dy dτ is bounded.

Note that∫R×RN |F̂ (τ, y)|

2 dy dτ =∥∥∥f̂∥∥∥2

L2< ∞. Let ρ = (|τ |2 + |y|2)1/2. We will show that |F̂ |2 ≤

Cρ

(∥∥∥f̂∥∥∥L2v

+∥∥∥T̂ f∥∥∥

L2v

)and the result will follow as by Jensen’s inequality∫

R×RN‖u‖2L2v dx dt ≤

∫R×RN×RN

|u|2 dv dx dt

Note that |T̂ f |2 = |τ + vjyj |2|f̂ |2. We want to use this to obtain 1ρ . To do this, we split the integralaccording to whether the velocity makes this small or large. Define τ0 = τρ and y

0j =

yjρ . Then

|T̂ | = |τ + vjyj | = ρ|τ0 + vjy0j |. For |T̂ | ≥ 1, if K is the compact support of f in v, we have

|F̂large|2 =

(∫|T̂ |≥1

f̂ dv

)2

≤

(∫|T̂ |≥1;v∈K

|T̂ |−2 dv

)(∫RN|T̂ |2|f̂ |2 dv

)by Cauchy-Schwarz

≤ 1ρ2

(∫|T̂ |≥1;v∈K

|τ0 + vjy0j |−2 dv

)·∥∥∥T̂ f∥∥∥2

L2v

≤ 1ρ2

(∫|τ0+vjy0j |≥1/ρ;v∈K

|τ0 + vjy0j |−2 dv

)·∥∥∥T̂ f∥∥∥2

L2v

≤ Cρ

∥∥∥T̂ f∥∥∥2L2v

by direct integration

While for |T̂ | < 1 we obtain:

|F̂small|2 =

(∫|T̂ |

The compactness of H1/2 in L2 allows the conversion of weak compactness into strong compactness.The following lemma and corollary will later be used on the Boltzmann equation.

Lemma 4. Let the following assumptions hold:

(fn) ⊂ L1((0, T )× RN × RN ) be weakly relatively compact. (6)(ψn) be bounded in L

∞((0, T )× RN × RN ) and converge a.e. to ψ

Additionally, assume that (Tfn) be weakly relatively compact in L1loc((0, T )×RN×RN ). Then the velocity

average∫RN fnψn dv is strongly compact in L

1((0, T )× RN ).

Corollary 1. Furthermore, if in addition fn → f weakly in L1((0, T ) × RN × RN ) then the velocityaverage converges strongly to

∫RN fψ dv in L

1((0, T )× RN ).

Proof. By Egorov’s theorem have ψn → ψ uniformly except on a set of arbitrarily small measure, aswe only care about L1 convergence, we can take ψn → ψ uniformly everywhere and in fact we can takeψn = ψ. In a similar way, by uniform integrability (as (fn) is relatively weakly compact), for any � > 0we can take a fixed compact set K on whose complement all the integrals are < � in magnitude. Forthis reason we may assume that (fn) and f have fixed compact support K. We can also approximateψ ∈ L∞ by smooth functions ψk uniformly bounded in L∞, so that fnψn satisfies the same conditionsas fn. This means we may take ψ = 1, as the error is∫

R×RN

∣∣∣∣∫ ψkfn dv − ∫ ψfn dv∣∣∣∣ dx dt ≤ ∫R×RN×RN

|ψk − ψ||fn| dv dx dt→ 0

because (gn) is uniformly integrable. Finally by the compact support of (fn), we have that Tfn is weaklyrelatively compact in the whole space L1 rather than simply L1loc.

In summary, we now have (fn) compactly supported and weakly relatively compact in L1, with Tfn

weakly relatively compact in L1, and we want to show (Fn) strongly compact in L2. The result would

then follow from the velocity averaging lemma as H1/2 ⊂⊂ H0 = L2 by Rellich-Kondrachov, and becauseof the compact support, the compactness in L1.

To do this we need to approximate with L2 functions. We define the large and small parts of fn bythe solutions to:

Tfsmalln = (Tfn)1{|Tfn| ≤M}Tf largen = (Tfn)1{|Tfn| > M}

f largen∣∣t=0

= fsmalln∣∣t=0

= 0

By linearity of T and uniqueness of solutions, fn = fsmalln +f

largen . We can then use the explicit solution

operator to express f largen as

f large]n =

∫ t0

(Tg)]1{|Tfn| > M}] ds

Then integrating and using measure preservation of the solution operator, we see that uniformly in n∥∥f largen ∥∥L1 → 0 as M →∞However, by measure preservation again, both (fsmalln ) and (Tf

smalln ) are bounded in L

2, which is whatwe needed for the velocity averaging lemma.

Lemma 5. Assume that the following hold

fn → f weakly in L1+((0, T )× RN × RN )For each fixed δ > 0, let (Tβδ(f

n)) be weakly relatively compact in L1loc((0, T )× RN × RN ) (7)where (βδ)δ>0 is a uniformly Lipschitz family in C(R;R) with βδ(0) = 0and βδ(z)→ z as δ → 0 uniformly on compact sets.

Additionally, assume that (6) holds. Then∫RN fnψn dv →

∫RN fψ dv.

11

Proof. First note that:

‖fn − βδ(fn)‖L1 =∫

1{|fn| < a}|fn − βδ(fn)|+∫

1{|fn| ≥ a}|fn − βδ(fn)|

The first term tends to 0 by the uniform convergence on compact sets. Using the bound βδ(fn) ≤M |fn|

where M is the uniform Lipschitz constant, we see that the second integral is less than (1+M)∫

1{|fn| ≥a}|fn| which tends to zero by the dominated convergence theorem. Hence we have that fn−βδ(fn)→ 0in L1.

Now we apply lemma 4 to (βδ(fn)) for each fixed δ > 0, and obtain the result using the triangle

inequality, taking δ → 0.

We can also deduce vector valued versions of these results, but first we need a technical lemma.

Lemma 6. Let

• (E, ν) be an arbitrary measure space.

• U ⊂ RN .

• (un) be weakly compact in L1(K) for any compact K ⊂ U with suppψ ⊂ K × E.

• (ψk) be uniformly bounded in L∞(U ;L1(E)) and converge to 0 in L1(U × E).

Then supn ‖unψk‖L1(U×E) → 0 as k →∞.

Proof. First it should be noted that the reason (E, ν) is so general is that we need nothing from it.Let Km be a chain of compact sets in U that eventually surround any point. Then we note that by

the convergence of ψk we have the k-independent estimate

supn‖ψkun‖L1(Km×E) ≤ C sup

n‖un‖L1(Km)

We then bound the large part of un,

‖ψkun1{|un| ≥M}‖L1(Km×E) ≤ C ‖un1{|un| ≥M}‖L1(Km) ≤ γm(M)

where by the weak compactness of (un), γm(M)→ 0 as M →∞.The small part of un can then be bounded by

‖unψk1{|un| < M}‖L1(Km×E) ≤M ‖ψk‖L1(Km×E)

We then take k →∞ and then M →∞ to obtain the result.

Corollary 2. Let (6) and (7) hold but with ψn, ψ and their convergence in L∞((0, T )×RN×RN ;L1(RN )).

Then the velocity average converges,∫RN

fnψn dv →∫RN

fψ dv strongly in L1((0, T )× RN × RN )

Proof. By arguments given in prior proofs we can reduce to when ψn = ψ. Now ψ(t, x, v) is valued inL1(RN ). To keep this space distinct we will label it E. If ψ separates into a sum

ψ(t, x, v) =

m∑i=1

φi(t, x, v)ϕi where φi ∈ L1((0, T )× RN × RN ) and ϕi ∈ L1(E)

then the result is immediate by lemma 5. Otherwise we can approximate ψ with functions of this formand apply the technical lemma.

12

3 The Cauchy Problem for the Boltzmann Equation

3.1 The equation

In this section, we turn to the discussion of the Cauchy problem for the Boltzmann equation, namely∂f

∂t(t, x, v) + v · ∇xf(t, x, v) = Q(f, f)(t, x, v)

f(0, x, v) = f0(x, v)(B)

where (t, x, v) ∈ R+ × Rd × Rd. This equation was first formulated in 1872 by Ludwig Boltzmann tomodel the time evolution of the molecular density of a rarefied gas. Note that if the gas is so rarifiedthat the molecules can be assumed not to interact with each other, then we expect their density to followthe linear transport equation

∂f

∂t+ v · ∇xf = 0 ,

and hence to remain constant along each molecular path. In this case the problem can be solved explicitlyjust by evolving the initial datum along characteristics. If, on the other hand, the molecules interactwith each other, this conservation no more holds, and a term to describe the change of the moleculardensity due to these interactions needs to be added to the equation. This corresponds to Q(f, f) in (B).In order to explain the structure of Q(f, f) we need to choose a model for the molecular interaction. Inthis notes, we focus on the so called hard spheres model, according to which the molecules are describedas perfectly elastic spheres moving independently (following the laws of classical mechanics) until theycollide. Call v and v∗ the velocities of two molecules colliding at time t in position x, and v

′, v′∗ their

Figure 1: Schematic representation of a collision between two molecules.

velocities after the collision respectively. Since we are assuming the collisions to be perfectly elastic, themomentum and kinetic energy must be conserved, and we obtain the first relations

v + v∗ = v′ + v′∗

|v|2 + |v∗|2 = |v′|2 + |v′∗|2(8)

From this, if we let ω be the unit vector, directed along the line joining the centres of the spheres, wededuce {

v′ = v − ω[(v − v∗) · ω]v′∗ = v∗ + ω[(v − v∗) · ω]

(9)

and this easily implies the relations

v · v∗ = v′ · v′∗|v − v∗| = |v′ − v′∗|

(v − v∗) · ω = −(v′ − v′∗) · ω(10)

that will be useful later on. Equations (9) allow us to explicitly define the collision operator Q(f, f) asthe average change in the molecular densities of colliding particles, weighted with the collision kernel.

Q(f, f) =

∫Rd

∫Sd−1

q(v − v∗, ω)(f ′f ′∗ − ff∗)dv∗dω

13

where f = f(t, x, v), f∗ = f(t, x, v∗), f′ = f(t, x, v′), f ′∗ = f(t, x, v

′∗), and the hard spheres collision kernel

q is given byq(v − v∗, ω) = |(v − v∗) · ω| .

Observe that we can interpret q as the probability of a collision happening between two molecules havingvelocities v, v∗ and such that the unit vector joining their centres is ω. The collision operator Q is themain problem with trying to solve the Boltzmann equation. It is non-linear and also non-local in thesense that its value at a point depends on the values of f everywhere. Despite this, Q splits nicely intopositive and negative parts, and the negative part can be split into local and non-local parts.

Q+(f, f) =

∫ ∫f ′∗f′q(v − v∗, ω)dv∗dω

Q−(f, f) = f · L(f)

L(f) =

∫ ∫Sd−1

q(v − v∗, ω)f∗dωdv∗ =∫A(v − v∗)f∗dv∗ = A ∗ f

A(z) =

∫Sd−1

q(z, ω)dω .

Observe that Q− has a local part f and a non-local part A ∗ f . The non-local part will be the mainangle of attack, once we have a bound on it, we can use an entropy inequality to bound Q+. In this waywe control the nasty non-local part of Q+ with the nicer but still non-local part of Q−.We remark here that more general collision kernels can be considered, and in fact Assumptions 1 beloware enough to prove the main existence result, namely Theorem 5.

3.1.1 Solution concepts for the Boltzmann equation

The concepts of mild and renormalised solutions for the Boltzmann equation read as follows:

Definition 15 (Mild solution of the Boltzmann equation). A mild solution to the Boltzmann equation(B) is a measurable function f(t, x, v) : [0,∞)× Rd × Rd for which

• Q±(f, f)] ∈ L1loc[0,∞) for almost all x, v, (so the integral below makes sense).

• For all t ≥ 0 and for almost all x, v, we have f ](t, x, v) = f0(x, v) +∫ t0Q(f, f)](s, x, v) ds.

Definition 16 (Renormalised solution of the Boltzmann equation). A renormalised solution to theBoltzmann equation is a measurable function f(t, x, v) ∈ L1+([0,∞)loc × Rd × Rd), for which

•Q±(f, f)

1 + f∈ L1loc([0,∞)× Rd × Rd) (11)

(so that it exists as a distribution).

• For all Lipschitz continuous functions β(z) : [0,∞)→ R for which |β′(z)|(1+ |z|) ∈ L∞, f satisfies

T [β(f)] = β′(f)Q(f, f) in D′

Note how the bound on the derivative of β combined with the condition on Q acts to put the righthand side of the equation in L1loc.

Lemma 7 (Equivalence for Boltzmann equation). The two notions of solutions are equivalent in thefollowing sense.

• If f is a renormalised solution, but for only β(z) = log(1 + z), then f is a mild solution.

• If f is a mild solution, and (11) holds, then f is a renormalised solution.

Proof. We just apply Theorem 1.

14

3.1.2 Main theorem

In order to state the main existence result for (B) we need to make the following assumptions.

Assumptions 1. Let z = v − v∗. Assume for the collision kernel q(z, ω) the following properties:

• q depends only on |z| and |z · ω|

•1

1 + |v|2

∫|v∗|≤R

A(v − v∗)dv∗ −→ 0 as |v| → ∞ (12)

• A ∈ L∞loc(Rd).

Note that the hard spheres kernel q(v−v∗, ω) = |(v−v∗) ·ω| trivially satisfies all the above properties.We also make the following assumptions on the initial datum.

Assumptions 2. Assume the initial datum f0 of (B) belongs to L1+(Rd × Rd) and is such that∫ ∫

f0(1 + |x|2 + |v|2)dxdv

• From this, we obtain the compactness of Qn+(f

n,fn)

1+fn via the entropy inequality

Qn+(fn, fn) ≤ KQn−(fn, fn) +

1

logKen(f

n) .

• Hence, for now, fn L1

⇀ f , andQn±(f

n,fn)

1+fn is weakly compact. The problem is that we don’t know

whetherQn±(f

n,fn)

1+fn goes toQ±(f,f)

1+f , because of the non linearity.

• On the other hand, with a careful use of the velocity averaging results, we can show that the aboveconvergence holds, even strongly, for the respective velocity averages, i.e.∫

Qn±(fn, fn)φdv

1 +∫fndv

−→∫Q±(f, f)φdv

1 +∫fdv

in L1((0, T )× Rd), for each compactly supported function φ ∈ L∞((0, T )× Rd × Rd).

• To conclude, we introduce an equivalent form of (B), and show that it is satisfied by f . This,together with the right integrability conditions on Q±(f,f)1+f , implies that f is a renormalised solution

to (B), as claimed.

3.3 A priori estimates

Let f be a solution to (B) with initial datum f0 satisfying Assumptions 2. Take ψ ∈ C∞(Rd) and assumewe have all the necessary decay for the calculations below to make sense.We start by integrating ψ against the collision operator:∫

ψ(v)Q(f, f)dv =

∫ ∫ ∫q(v − v∗, ω)ψ(v)(f ′∗f ′ − f∗f)dv∗dvdω (13)

Recall that q(v∗ − v, ω) = q(v − v∗, ω) and exchange v and v∗, to get∫ψ(v)Q(f, f)dv =

∫ ∫ ∫q(v − v∗, ω)ψ(v∗)(f ′∗f ′ − f∗f)dv∗dvdω (14)

Now operate in (13) the change of variables (v, v∗) 7→ (v, v′∗), or more explicitly{v′ = v − ω[ω · (v − v∗)]v′∗ = v∗ + ω[ω · (v − v∗)]

and note that the Jacobian associated with this transformation is 1. We get∫ψ(v)Q(f, f)dv =

∫ ∫ ∫q(v′ − v′∗, ω)ψ(v′)(f∗f − f ′∗f ′)dv∗dvdω

= −∫ ∫ ∫

q(v − v∗, ω)ψ(v′)(f ′∗f ′ − f∗f)dv∗dvdω(15)

where we have used the second relation in (10) for the second equality. Finally we once again exchangev and v∗ in (15) to get∫

ψ(v)Q(f, f)dv = −∫ ∫ ∫

q(v − v∗, ω)ψ(v′∗)(f ′∗f ′ − f∗f)dv∗dvdω (16)

Adding up equations (13)-(16) we obtain the identity:∫ψ(v)Q(f, f)dv =

1

4

∫ ∫ ∫q(v − v∗, ω)(f ′∗f ′ − f∗f)[ψ + ψ∗ − ψ′ − ψ′∗]dv∗dvdω (17)

This motivates the following definition.

16

Definition 17. A function φ : Rd −→ R is said to be collision invariant if

φ+ φ∗ = φ′ + φ′∗

This definition has a simple interpretation. Observe that, by (17), if ψ is collision invariant then∫Q(f, f)φ(v)dv = 0 .

Since this integral represents the rate of change in the average value of φ due to collisions, the aboverelation tells us that this average value stays unchanged.

Then (8) immediately implies that any function of the form

ψ(v) = a+ b · v + c|v|2 (18)

is collision invariant1, and hence gives 0 when integrated against the collision operator. To make use ofthat, we go back to equation (B), multiply it by (18) and integrate with respect to v to get

0 =

∫Q(f, f)ψ(v)dv =

∫ψ

(∂f

∂t+ v · ∇xf

)dv

=∂

∂t

∫ψfdv + divx

(∫ψvfdv

)Next, we integrate over x ∈ Rd and use the total time-derivative to see that:

d

dt

∫ψfdv = 0 .

This gives the following conservation laws:

• Conservation of mass ∫ ∫f(t, x, v)dxdv =

∫ ∫f0(x, v)dxdv (19)

• Conservation of momentum∫ ∫f(t, x, v)v dxdv =

∫ ∫f0(x, v)v dxdv (20)

• Conservation of kinetic energy∫ ∫f(t, x, v)|v|2dxdv =

∫ ∫f0(x, v)|v|2dxdv (21)

Another conservation law is obtained integrating the Boltzmann equation against ψ(t, x, v) = |x−tv|2with respect to x and v. By (8) and (10), in fact, we deduce that

|x− tv′|2 + |x− tv′∗|2 = 2|x|2 + t2(|v′|2 + |v′∗|2)− 2t x · (v′ + v′∗)= 2|x|2 + t2(|v|2 + |v∗|2)− 2t x · (v + v∗)= |x− tv|2 + |x− tv∗|2

so that∫Q(f, f)|x− tv|2dv = 0. Apply the same reasoning as before to get

d

dt

∫ ∫f(t, x, v)|x− tv|2dxdv = 0

and hence the conservation law∫ ∫f(t, x, v)|x− tv|2dxdv =

∫ ∫f0(x, v)|x|2dxdv . (22)

1In fact, it can be shown that all collision invariants have this form, see [2] for details.

17

3.3.1 Entropy identity

Multiply the Boltzmann equation by (1 + log f) and integrate over v:∫(1 + log f)(∂tf + v · ∇xf)dv =

∫Q(f, f)(1 + log f)dv

So ∫Tfdv + ∂t

[∫f log fdv

]+ divx

[∫v · f log f

]=

∫Q(f, f)dv +

∫Q(f, f) log fdv

and using that f solves the Boltzmann equation we get

∂t

[∫f log fdv

]+ divx

[∫v · f log f

]=

∫Q(f, f) log fdv

= −14

∫ ∫ ∫q(v − v∗, ω)(f ′∗f ′ − f∗f) log

f ′f ′∗ff∗

dv∗dvdω

where in the last equality we have used relation (17) with ψ = log f . Now integrate over x ∈ Rd, toobtain the identity

d

dt

∫ ∫f log fdxdv +

∫ ∫e(f)dxdv = 0 (23)

with

e(f) =1

4

∫ ∫q(v − v∗, ω)(f ′∗f ′ − f∗f) log

f ′f ′∗ff∗

dv∗dω . (24)

Note that the non-negativity of e(f) implies that∫ ∫

f log f is non-increasing in time.

Lemma 8. Let g be any non negative function on Rd × Rd such that∫ ∫

g| log g|dxdv

Apply the above result with g = f(t, x+ vt, v) = f ](t, x, v), to get∫ ∫f ]| log f ]|dxdv ≤

∫ ∫f ] log f ]dxdv + 2

∫ ∫f ](|x|2 + |v|2)dxdv + Cd

and the change of variables (x, v) 7→ (x− vt, v) gives us∫ ∫f | log f |dxdv ≤

∫ ∫f log fdxdv + 2

∫ ∫f(|x− vt|2 + |v|2)dxdv + Cd . (26)

Lemma 9. (A priori estimates)Suppose that q is a non negative measurable function in L∞loc(Rd×Rd×Sd−1), q = q(x, |v−v∗|, |(v−v∗)·ω|)and q grows at most polynomially in x and v − v∗. Assume further that f is a positive solution to theCauchy problem (B) with initial datum satisfying Assumptions 2, such that f ∈ C1((0,∞),S(Rd×Rd)),where S denotes the Schwartz space. If, moreover, | log f | grows at most polynomially in (x, v) uniformlyin compact time intervals, then the following hold:∫ ∫

f(t, x, v)dxdv =

∫ ∫f0(x, v)dxdv∫ ∫

f(t, x, v)|v|2dxdv =∫ ∫

f0(x, v)|v|2dxdv∫ ∫f(t, x, v)|x− tv|2dxdv =

∫ ∫f0(x, v)|x|2dxdv∫ ∫

f(t, x, v) log f(t, x, v)dxdv +

∫ t0

∫ ∫e(f)(s)dxdvds =

∫ ∫f0 log f0dxdv .

These identities in turn imply:∫ ∫f(t, x, v)(1 + |x|2 + |v|2)dxdv ≤

∫ ∫f0(x, v)(1 + 2|x|2 + (2t2 + 1)|v|2)dxdv (27)∫ ∫

f(t, x, v)| log f(t, x, v)|dxdv +∫ ∞0

∫ ∫e(f)(s, x, v)dxdvds ≤

∫ ∫f0(| log f0|+ 2|x|2 + 2|v|2) +Cd

(28)

Note that the assumptions of the lemma are enough to carry out all the a priori computations made inthe previous section, and these computations are not affected by allowing the collision kernel to dependon the position. The assumptions on the growth of the logarithm is needed to make sure that integralssuch as

∫ ∫f log f are convergent, as f being in S(Rd ×Rd) in (x, v) potentially allows for a decay that

is too rapid for the integral to converge.Let us now turn to the proof.

Proof. The first three relations have already been shown. The fourth is obtained integrating (23) in thetime interval [0, t]. It only remains to prove (27) and (28).For the first one we split the integral, and use the conservation of mass and kinetic energy to get∫ ∫

f(t, x, v)(1 + |v|2)dxdv =∫ ∫

f0(x, v)(1 + |v|2)dxdv

To bound the remaining term we use the inequality |x|2 ≤ (|x − tv| + t|v|)2 ≤ 2(|x − tv|2 + t2|v|2) and(21)-(22), to get ∫ ∫

f(t, x, v)|x|2dxdv ≤ 2∫ ∫

f0(x, v)(|x|2 + t2|v|2)dxdv

which yields the required result. Finally, for the last inequality combine (26) with (23) and use theconservation laws to get∫ ∫

f(t, x, v)| log f(t, x, v)|dxdv ≤∫ ∫

f log fdxdv + 2

∫ ∫f(|x− vt|2 + |v|2)dxdv + Cd

= −∫ t0

∫ ∫e(f)(s)dxdvds+

∫ ∫f0 log f0dxdv + 2

∫ ∫f0(|x|2 + |v|2)dxdv + Cd

≤ −∫ ∞0

∫ ∫e(f)(s)dxdvds+

∫ ∫f0| log f0|dxdv + 2

∫ ∫f0(|x|2 + |v|2)dxdv + Cd

19

where in the last inequality we have used the fact that∫intf log f is non increasing in time, and the

lemma is proved.

3.4 Truncated problems

We now want to make use of the a priori estimates. The idea is to define some nicer problems, for whichwe can prove existence and uniqueness of solutions in the class of functions satisfying all the assumptionsof Lemma 9. For these solutions the conclusions of the above lemma will still hold.To implement this strategy, we have to identify the problems in proving the existence of a solution to(B), and find a way to avoid them. These are the following:

• Large relative velocity collisions.Note that Assumptions 1 on the collision kernel do not control its behaviour for large velocities.

• Non-linearity of the collision operator.In order to deal with the convergence of Q(f, f) we will introduce a suitable additional renormal-ization, aimed to slow down the growth of the collision operator.

• Regularity of the initial datum.Lemma 9 takes the initial datum in the Schwartz space, but the theorem claims the existence of asolution when f0 is merely L

1+(Rd × Rd). We will have to use some approximation argument.

We take care of these problems as follows:Assume that the collision kernel q satisfies Assumptions 1, and consider a sequence of truncated kernelsqn such that:

• qn ∈ C∞0,+(Rd × Sd−1) for all n ≥ 1,

• the (qn) satisfy the second and third assumptions in (1) uniformly in n,

• each qn vanishes for |(v− v∗) ·ω| < δn, where (δn) is a sequence of positive real numbers such thatδn ↘ 0,

• qn −→ q a.e.

For each n we then define the truncated (and renormalized) collision operator

Qn(f, f) =

∫ ∫qn(f

′f ′∗ − ff∗)dωdv∗1 + δn

∫|f |dv

.

Finally, if f0 satisfies Assumptions 2 we approximate it by a sequence (fn0 ) such that:

• fn0 ∈ S(Rd × Rd) for all n ≥ 1,

• for each fn, there exists a positive real number µn such that fn0 ≥ µne−(|x|2+|v|2),

• ∫ ∫fn0 (1 + |x|2 + |v|2)dxdv −→

∫ ∫f0(1 + |x|2 + |v|2)dxdv as n→∞ (29)

• ∫ ∫f0| log f0|dxdv −→

∫ ∫f0| log f0|dxdv as n→∞ (30)

We have then obtained the sequence of truncated problems∂fn

∂t+ v · ∇xfn = Qn(fn, fn)

fn(0, ·) = fn0(Bn)

Recall that A(z) =∫Sd−1

q(z, ω)dω and set

An(z) =

∫Sd−1

qn(z, ω)dω

20

so that

Qn(fn, fn) = Qn+(fn, fn)−Qn−(fn, fn) where Qn−(fn, fn) =

fn(An ∗ fn)1 + δn

∫fndv

.

Proposition 4. For each n ≥ 1, the Cauchy problem (Bn) has a unique global solutionfn ∈ C1((0,∞),S(Rd × Rd)). Moreover, fn satisfies all the assumptions of Lemma 9, and hence theconservation laws (19)-(22) and inequalities (27)-(28).

To prove this result, we will establish the following lemma which we will use to bound∫Qn(f, f)dv.

Lemma 10. Suppose a, b ∈ L1v(Rd). Then∫ ∫ ∫qn|a′||b′∗|dωdv∗dv1 =

∫ ∫ ∫qn |a(v1 − ω[ω · (v1 − v∗)])| |b (v∗ + ω [ω · (v1 − v∗)])| dωdv∗dv1

≤ cn||a||L1v ||b||L1v

for some constant cn.

Proof. To prove this result, we simply make the change of variables u(ω) = ω[ω · (v1 − v∗)] for fixed v1and v∗. The Jacobian associated with this transformation is given by (ω · (v1 − v∗))d and so∫ ∫ ∫

qn|a||b|dωdv∗dv1 =∫ ∫ ∫

qn |a(v1 − u)| |b (v∗ + u)|1

|ω · (v1 − v∗)|ddudv∗dv1

Note that the integrand is defined everywhere, as qn vanishes if |ω · (v1 − v∗)| < δn. Thus∫ ∫ ∫qn|a||b|dωdv∗dv1 ≤

1

δdn

∫ ∫ ∫qn |a(v1 − u)| |b (v∗ + u)| dudv∗dv1

Moreover, qn is supported only on a compact set of u, which we will denote by Kn. Additionally, qn isbounded on this set by some constant αn. Thus∫ ∫ ∫

qn|a||b|dωdv∗dv1 ≤αnδdn

∫ ∫ ∫Kn

|a(v1 − u)| |b (v∗ + u)| dudv∗dv1

Applying Tonelli’s theorem yields∫ ∫ ∫qn|a||b|dωdv∗dv1 ≤

αnδdn

∫ ∫ ∫Kn

|a(v1 − u)| |b (v∗ + u)| dv∗dv1du

=αnδdn

∫Kn

(∫|a(v1 − u)| dv1

)(∫|b (v∗ + u)| dv∗

)du

≤ αnδdn||a||L1v ||b||L1vλ(Kn)

and setting cn =αnδdnλ(Kn) yields the desired inequality.

We will use this lemma to establish L1 bounds on the truncated collision kernel.

Lemma 11. There are constants Cn such that for any f, g ∈ L1 ∩ L∞(Rd), we have∫|Qn(f, f)(v)−Qn(g, g)(v)|dv ≤ Cn||f − g||L1v

and ∫|Qn(f, f)(v)|dv ≤ Cn||f − g||L1v

21

Proof. To show this, we split the collision kernel into Qn+ and Qn− where

Qn+(f, f) =

∫ ∫qn(f

′f ′∗)dωdv∗1 + δn

∫|f |dv

.

and

Qn−(f, f) =

∫ ∫qn(ff∗)dωdv∗

1 + δn∫|f |dv

.

It is simple to see that

Qn+(f, f)−Qn+(g, g) =∫ ∫

qn[(1 + δn

∫|g|dv)(f ′f ′∗)− (1 + δn

∫|f |dv)(g′g′∗)

]dωdv∗(

1 + δn∫|f |dv

) (1 + δn

∫|g|dv

)=

∫ ∫qn[(f ′f ′∗ − g′g′∗) + δn(f ′f ′∗

∫|g|dv − g′g′∗

∫|f |dv)

]dωdv∗(

1 + δn∫|f |dv

) (1 + δn

∫|g|dv

)We will bound the term ∣∣∣∣δn(f ′f ′∗ ∫ |g|dv − g′g′∗ ∫ |f |dv)∣∣∣∣To do this, we first perform the following simple calculation∣∣∣∣δn(f ′f ′∗ ∫ |g|dv − g′g′∗ ∫ |f |dv)∣∣∣∣ = ∣∣∣∣δn ∫ f ′f ′∗|g| − g′g′∗|f |dv∣∣∣∣

=

∣∣∣∣δn ∫ f ′|g| (f ′∗ − g′∗) dv + ∫ f ′g′∗ (|g| − |f |) dv + ∫ |f |g′∗ (f ′ − g′) dv∣∣∣∣≤ δn

(||g||L1v |f

′| (|f ′∗ − g′∗|) + ||g − f ||L1v |f′||g′∗|+ ||f ||L1v |g

′∗| |f ′ − g′|

)Each of our terms involves exactly one of f ′, g′ or (f ′− g′), exactly one of f ′∗, g′∗ or (f ′∗− g′∗) and exactlyone of ||f ||L1v , ||g||L1v or ||f

′ − g′||L1v . This was done deliberately so that we can apply Lemma 10 to seethat ∫ ∫ ∫

qn

∣∣∣∣δn(f ′f ′∗ ∫ |g|dv − g′g′∗ ∫ |f |dv)∣∣∣∣ dωdv∗dv1 ≤ 3cnδn (||g||L1v ||f ||L1v ||f − g||L1v)Similar arguments yield∫ ∫ ∫

qn |(f ′f ′∗ − g′g′∗)| dωdv∗dv1 ≤ βn[(||f ||L1v + ||g||L1v )||f − g||L1v

]and so∫

|Qn+(f, f)(v)−Qn+(g, g)(v)|dv ≤||f − g||L1v

[βn(||f ||L1v + ||g||L1v ) + 3cnδn

(||g||L1v ||f ||L1v

)]1 + δn(||f ||L1v + ||g||L1v ) + δ2n||f ||L1v ||g||L1v

≤ Cn||f − g||L1v

A slightly easier argument (there is no need to change coordinates) can be used to show that∫|Qn−(f, f)(v)−Qn−(g, g)(v)|dv ≤ Dn||f − g||L1v

Combining these two results leaves us∫|Qn(f, f)(v)−Qn(g, g)(v)|dv ≤ Cn||f − g||L1v

The second bound follows by taking g = 0.

We can now prove Proposition 4.

22

Proof. We construct the following scheme to approximate the required fn. We define the sequence (fm)recursively by

Tf0 = 0

f0(0, x, v) = fn0 (x, v)

Tfm+1 = Qn(fm, fm)

fm+1(0, x, v) = fn0 (x, v)

We can explicitly compute these as

f ]0(t, x, v) = fn0 (x, v)

f ]m+1(t, x, v) = fn0 (x, v) +

∫ t0

Qn(f ]m, f]m) ds

We will show that the operator that maps fm → fm+1 is a contraction on C([0, τ);L1x,v) for sufficientlysmall τ . Then we have a solution to the integral form of the equation. In other words a mild solution.

‖fm+1 − fm‖C([0,τ);L1x,v) =∥∥∥f ]m+1 − f ]m∥∥∥

C([0,τ);L1x,v)

=

∥∥∥∥∫ t0

Qn(f ]m, f]m)(s, x, v)−Qn(f

]m−1, f

]m−1)(s, x, v) ds

∥∥∥∥C([0,τ);L1x,v)

≤∫ τ0

∥∥∥Qn(f ]m, f ]m)−Qn(f ]m−1, f ]m−1)∥∥∥L1x,v

dt

=

∫ τ0

‖Qn(fm, fm)−Qn(fm−1, fm−1)‖L1x,v dt

≤ τCn ‖fm − fm−1‖C([0,τ);L1v,x)

So we may take τ small enough for this to be a contraction. Then by the fixed point theorem we have asolution in [0, τ). As none of our estimates depended on the norm of the initial condition we may thenrestart the process at τ/2 say, and continue to get existence for all time.

In fact we can prove that the limit is in C([0, τ),S(Rd × Rd)). To do this we bound the x and vderivatives of fm+1 by moving them under the integral. When the derivatives hit qn they produce afunction that obeys all the same bounds as qn does, and when they hit fm they produce its derivatives. Inthis way the same argument gives that any derivative of the solution is also in C([0, τ),S(Rd×Rd)). Thisargument also works for multiplication by polynomials: we see that a polynomial in v can be absorbedinto the collision kernel which still obeys the same assumptions as before, and a polynomial in x is takencare of by the fact that the initial data is Schwartz. This is enough to give the Schwartz space result.

Finally we argue that as the limit obeys f ](t, x, v) = fn]0 +∫ t0Qn(f, f)] ds, and the integrand is smooth

and compactly supported we can deduce that f is time differentiable, i.e. f ∈ C1([0, τ),S(Rd × Rd)).

To prove that | log fn| grows at most polynomially in (x, v), observe that fm ∈ C[0, T ] has initialcondition fm(0, x, v) = f

n0 (x, v) > 0. Thus fm is non-negative for some time interval. Hence (on this

time interval)

Tfm+1 = Qn(fm, fm)

= Qn+(fm, fm)−Qn−(fm, fm)

≥ −fm(An ∗ fm)1 + δn

∫fm

Now because (An∗fm)1+δn

∫fm≥ Cn for some Cn > 0, we have

Tfm+1 ≥ −Cnfm

23

Send m→∞ to get the inequalityTfn ≥ Cnfn

This implies ∂t(fn]) = (Tfn)] ≥ Cnfn] and hence, by Gronwall’s inequality, we obtain

fn] ≥ fn0 (t, x, v)e−Cnt .

Going back along the characteristic, we finally get

fn ≥ fn0 (t, x− vt, v)e−Cnt ≥ µne−(|x−vt|2+|v|2)e−Cnt

as claimed.

3.5 Weak compactness results

For now, we have a sequence (fn) of solutions to the problems (Bn) for which all the a priori estimates,together with relations (29)-(30), hold. In this section we want to show that the two sequences

fn,Qn±(f

n, fn)

1 + fnn ≥ 1

are contained is weakly compact sets: this will allow to apply the velocity averaging results in a later step.The main tool we will use to prove weak compactness is the Dunford-Pettis (DP) criterion (Theorem15). In particular, we will make use of it in the following form:

Corollary 3. Let h : R+ → R+ be a continuous function, w : U → R+ be in L∞loc(U), and assumeh(s)/s → ∞ as s → ∞ and w(y) → ∞ as |y| → ∞. If, then, (fn) is a sequence which is uniformlybounded in L1(U) and such that

supn

∫Rd

[h(|fn|) + |fn|(1 + w)] dy 0. The claim then follows by applying Corollary 3 with U = (0, T )× Rd × Rd and

h(fn) = fn| log fn|w(x, v) = |x|2 + |v|2 .

Up to the extraction of a subsequence (that we still call (fn)), hence

fn ⇀ f in L1((0, T )× Rd × Rd) .

24

3.5.2 Weak compactness of the collision term

Theorem 5. For any T > 0, R > 0, the sequences

Qn−(fn, fn)

1 + fnQn+(f

n, fn)

1 + fnn ≥ 1

are relatively weakly compact in L1((0, T )× Rd ×BR), where BR = {v ∈ Rd : ‖v‖ ≤ R}.

To prove this, we start from the negative part Qn−(fn, fn), and show that it is a uniformly integrable

sequence, so that weak compactness follows from DP criterion. We then obtain control on the positivepart via the following inequality.

Lemma 13 (Entropy inequality). For all K ≥ 1 we have

Qn±(fn, fn) ≤ KQn∓(fn, fn) +

1

logKen(f

n) ∀K ≥ 1 , (31)

where en is defined as in (24) replacing q with qn.

Note that this remarkable inequality gives us control on the global part of the collision operator interms of the local one, and vice versa. This is, as will become clear from the proof, due to the particularform of the collision kernel, and is therefore characteristic of the Boltzmann equation.

Proof. Since n is fixed, we drop it in the notation. Define the events

AK = {(v, v∗, ω) : f ′f ′∗ ≥ Kff∗}BK = A

cK .

Then

Q+(f, f) =

∫ ∫q(v − v∗, ω)f ′f ′∗ [ 1AK + 1BK ] dv∗dω

≤ K∫ ∫

q(ff∗) 1BKdv∗dω +

∫ ∫q(f ′f ′∗ − ff∗) 1AKdv∗dω +

∫ ∫q(ff∗) 1AKdv∗dω

≤ K∫ ∫

q(ff∗) 1BKdv∗dω +1

logK

∫ ∫q(f ′f ′∗ − ff∗) log

f ′f ′∗ff∗

1AKdv∗dω +K

∫ ∫q(ff∗) 1AKdv∗dω

= KQ−(f, f) +1

logK

∫ ∫q(f ′f ′∗ − ff∗) log

f ′f ′∗ff∗

1AKdv∗dω

as claimed.

We are now ready to prove Theorem 5.

Proof. Let us start by proving that the sequenceQn−(f

n, fn)

1 + fnsatisfies the DP criterion in L1((0, T ) ×

Rd ×BR), and is therefore weakly compact. Note that, since

Qn−(fn, fn)

1 + fn≤ f

n

1 + fn· An ∗ f

n

1 + δn∫fndv

≤ An ∗ fn

it suffices to prove that the sequence An ∗fn is uniformly integrable, i.e. it satisfies properties (a)-(b)-(c)of Definition 18.To show (a) observe that∫Rd

∫BR

(An ∗ fn)dvdx =∫Rd

∫BR

∫RdAn(v − z)fn(z)dzdvdx =

∫Rd

∫Rdfn(x, z)

(∫BR

An(z − v)dv)dzdx

≤ C∫Rd

∫BR

fn(x, z)(1 + |z|2)dzdx = C∫Rd

∫BR

fn0 (x, z)(1 + |z|2)dzdx

25

by (12), and the last term is uniformly bounded in n by (29). Let us now move to conditions (b) and(c). We split the proof in several steps.

Step 1Assume ‖A‖L1(Rd) K

(An ∗ fn)dvdx −→ 0 as K →∞

uniformly in n (and t). To this end, split the integral as follows:∫Rd

∫|v|>K

(An ∗ fn)dvdx ≤∫Rd

∫Rd

∫RdAn(v − v∗)fn(x, v∗) 1(|v∗|≥K/2)dv∗dvdx

+

∫Rd

∫Rd

∫Rd

1(|v|≥K/2) · 1(|v∗|≤K/2)An(v − v∗)fn(x, v∗)dv∗dvdx

= I + II

For the first term, integrating over v and using the a priori bound (21), we get

I ≤ an∫Rd

∫Rdfn(x, v∗) 1(|v∗|≥K/2)dv∗dx ≤

an(K/2)2

∫Rd

∫Rd|v∗|2fn(x, v∗)dv∗dx

=4anK2

∫Rd

∫Rd|v∗|2fn0 (x, v∗)dv∗dx ≤

C

K2−→ 0 as K →∞

where, by (29), the constant C is independent of n. For the second one:

II ≤

(∫|z|≥K/2

An(z)dz

)‖fn‖L1(Rd×Rd) =

(∫|z|≥K/2

An(z)dz

)‖fn0 ‖L1(Rd×Rd)

≤ C ′(∫|z|≥K/2

An(z)dz

)2Note that, as pointed out in [2], we could have set R = ∞ in this argument. We will need, though, R to be finite in

the next step, where we remove the integrability assumption on A.

26

where C ′ is a uniform bound on ‖fn0 ‖L1(Rd×Rd) again given by (29). From the integrability assumptionon A, then, we deduce that the last term then becomes arbitrarily small as K →∞, and (c) follows.

Step 2We now remove the integrability condition on A. Note that if the An’s are all supported in a compactset then all the previous calculations continue to hold, and we are done. If not, take K > 0 and truncatethe An’s by defining

An,K(z) := An(z) · 1(|z|≤K) .Then weak compactness will follow from

supn‖An,K ∗ fn −An ∗ fn‖L∞((0,T );L1(Rd×BR)) −→ 0 as K →∞ (32)

To see that, we calculate explicitly that

‖An,K ∗ fn −An ∗ fn‖L1(Rd×BR) =∫Rd

∫BR

∫RdAn(v − v∗) 1|v−v∗|≥Kf

n(x, v∗)dv∗dvdx

and observe that if K > R then

|v∗| = |v∗ − v + v| ≥ |v∗ − v| − |v| ≥ K −R

so that1(|v∗−v|≥K) ≤ 1(|v∗|≥K−R) .

Using this inequality and integrating then An with respect to v we get

‖An,K ∗ fn −An ∗ fn‖L1(Rd×BR) ≤∫Rd

∫Rd

(∫BR

An(v − v∗)dv)

1(|v∗|≥K−R)fn(x, v∗)dv∗dx

≤ ε∫Rd

∫Rd

(1 + |v∗|2)fndv∗dx+Cε

(K −R)2

∫Rd

∫Rd|v∗|fn(x, v∗)dv∗dx

≤ ε∫Rd

∫Rd

(1 + |v∗|2)fn0 dv∗dx+Cε

(K −R)2

∫Rd

∫Rd|v∗|fn0 (x, v∗)dv∗dx

and (32) follows by letting K →∞ and then ε→ 0, and observing that the convergence is uniform in n,again by (29).

Step 3 We deduce the weak compactness of the (renormalized) positive part of the collision operatorfrom that of the negative part via entropy inequality, by noting that for each R > 0 and A Borel set ofLebesgue measure λ(A), (31) implies

supn

∫ T0

∫Rd

∫Rd

Qn+(fn, fn)

1 + fn[

1A + 1(|x|+|v|≥R)]dvdxdt −→ 0 as λ(A)→ 0, R→∞.

and uniform integrability follows.

3.5.3 Temporal regularity of f

Combining the weak compactness of the above sequences with the a priori estimates, we can obtain someinformation about the regularity in time of the limit f . Our goal is to show that

f ∈ C((0, T ), L1(Rd × Rd)) .

We start by introducing, for any fixed δ > 0, the new sequence

gnδ =1

δlog(1 + δfn) (33)

and observing that each gnδ solves

Tgnδ =1

1 + δfnQn(fn, fn) ,

where T denotes, as usual, the transport operator.

27

Lemma 14. If the sequence (gnδ ) is defined as in (33), it holds:

supn

supt∈(0,T )

∫Rd

∫Rd|gnδ − fn|dxdv −→ 0 as δ ↘ 0 (34)

Proof. Consider βδ(s) =1δ (1 + δs), so that g

nδ = βδ(f

n). Note thatd

dtβδ(s) =

1

1 + δs, so that for each

s > 0, βδ(s) ≥s

1 + δs. For any R > 0, therefore, we have

(s− βδ(s)) = (s− βδ(s)) 1(s

as h↘ 0. Moreover,

supt∈(0,T )

∥∥fn](t+ h)− fn](t)∥∥L1(Rd×Rd) ≤ sup

t∈(0,T )

∫ ∫|fn](t+ h)− fn](t)|dxdv

≤ supt∈(0,T )

[∫ ∫|fn](t+ h)− gn]δ (t+ h)|+ |g

n]δ (t+ h)− g

n]δ (t)|( 1BR + 1BcR) + |f

n](t)− gn]δ (t)|dxdv]

≤ φ′(h) + supt

∫Rd

∫BR

|gn]δ (t+ h)− gn]δ (t)|

and by taking the supremum over n and using (35) we conclude that

supn

supt∈(0,T )

∥∥fn](t+ h)− fn](t)∥∥L1(Rd×Rd) −→ 0

as h↘ 0. Being the above convergence uniform in n, the weak limit f ] of fn] also satisfies

supt∈(0,T )

∥∥f ](t+ h)− f ](t)∥∥L1(Rd×Rd) −→ 0

as h↘ 0, and therefore belongs to C((0,∞), L1(Rd × Rd)).From this information we deduce that f ∈ C((0,∞), L1(Rd×Rd)) by the change of variables (x+vh, v) 7→(x, v):

0←− supt∈(0,T )

∫ ∫|f ](t+ h)− f ](t)|dxdv = sup

t∈(0,T )

∫ ∫|f(t+ h, x+ vh, v)− f(t, x+ vh, v)|dxdv

= supt∈(0,T )

∫ ∫|f(t+ h, x, v)− f(t, x, v)|dxdv = sup

t∈(0,T )‖f(t+ h)− f(t)‖L1(Rd×Rd)

and we conclude thatf ∈ C((0,∞), L1(Rd × Rd))

as claimed.

3.6 Velocity averaging

To make use of the compactness results proved in the preceding section we go back to the sequences

gnδ =1

δlog(1 + δfn)

indexed by δ > 0, and recall that each gnδ solves

Tgnδ =1

1 + δfnQn(fn, fn) . (36)

The main advantage in using this sequences is that now we know that the RHS of (36) is weakly relativelycompact in L1((0, T )× Rd ×BR) and such is, therefore, Tgnδ . Moreover, since

0 ≤ Tgnδ ≤ fn

and fn is weakly compact in L1((0, T ) × Rd × Rd), we deduce from the DP criterion that Tgnδ is alsoweakly compact in the same space. Therefore the velocity averaging results apply to (gnδ ), and we deducethe following.

Proposition 6. For each T > 0 we have:

(i)

∫fndv →

∫fdv a.e. and in L1((0, T )× Rd).

(ii) An ∗ fn → A ∗ f a.e. and in L1((0, T )× Rd ×BR), for all R > 0.

29

(iii) For each function ϕ compactly supported in L∞((0, T )× Rd × Rd),∫Qn±(f

n, fn)ϕdv

1 +∫fndv

−→∫Q±(f, f)ϕdv

1 +∫fdv

as n→∞ in L1((0, T )× Rd).

Proof. Note that the family (indexed by δ > 0) of functions βδ(s) =1δ log(1 + δs) satisfies all the

assumptions of Lemma 5, namely (βδ) is a uniformly Lipschitz family in C(R;R) with βδ(0) = 0 andβδ(z)→ z as δ → 0 uniformly on compact sets. Moreover, as already noted, the sequence Tgnδ = Tβδ(fn)is weakly relatively compact in L1((0, T )×Rd ×BR). Hence we can apply Corollary 1 with ψn ≡ ψ ≡ 1to obtain the convergence of the velocity averages∫

fndv →∫fdv

in L1((0, T ) × Rd). For (ii) we use the same truncation argument of step 2 in the proof of Theorem 5.By (12), which holds uniformly in n, for any K > 0 we have∫

|v∗|≤RAn(v − v∗)dv∗ =

(∫|v∗|≤R

An(v − v∗)dv∗

)(1(|v|≤K) + 1(|v|>K)

)≤ sup|v|≤K

(∫|v∗|≤R

An(v − v∗)dv∗

)+ C(1 + |v|2) 1(|v|>K)

(37)

where the last integral is finite since An ∈ L∞loc(Rd;L1(BR)). We write An,K(v) := An(v − v∗) 1(|v|>K).We have

‖An ∗ fn −A ∗ f‖L1((0,T )×Rd×Rd;L1(BR)) ≤ ‖An ∗ fn −An,K ∗ fn‖L1((0,T )×Rd×Rd;L1(BR))

+ ‖An,K ∗ fn −AK ∗ f‖L1((0,T )×Rd×Rd;L1(BR))+ ‖AK ∗ fn −A ∗ f‖L1((0,T )×Rd×Rd;L1(BR)) .

But by (12)

‖An ∗ fn −An,K ∗ fn‖L1((0,T )×Rd×Rd;L1(BR)) =∫ T0

∫ ∫|v|>K

An ∗ fndvdxdt

≤ C∫ T0

∫ ∫|v|>K

fn(1 + |v|2)dv −→ 0

as K →∞ (recall that supn∫fn(1 + |v|2)dv

By (i) and (ii) An ∗ fn and (1 +∫fndv) converge almost everywhere to A ∗ f and 1 +

∫fdv respectively,

and henceAn ∗ fn

1 +∫fndv

ϕ −→ A ∗ f1 +

∫fdv

ϕ a.e.

We can therefore apply again Corollary 1 to get∫Qn−(f

n, fn)ϕdv

1 +∫fndv

=

∫fn(An ∗ fn)ψndv −→

∫f(A ∗ f)ϕdv =

∫Q−(f, f)ϕdv

1 +∫fdv

in L1((0, T )× Rd), and this concludes the proof.

3.7 Exponential formulation

Another kind of solutions to the Boltzmann equation can be obtained by viewing (A ∗ f)f as a linearfeedback term with variable coefficient c(t, x, v) = A ∗ f which we think of as a given function. Theequation is then

Tf + cf = Q+(f, f) = Q+

We then have the following form of solutions:

d

dt

[exp

(∫ t0

c] ds

)f ]]

= exp

(∫ t0

c] ds

)Q+]

Duhamel’s principle then gives

exp

(∫ t0

c] ds

)f ] = f0 +

∫ t0

exp

(∫ τ0

c](s, x, v) ds

)]Q+](τ, x, v) dτ

This is unpleasant looking, but with some notation we can make it nicer.Define T−1g as the solution to Tf = g, f |t=0 = 0. Then (T−1g)] =

∫ t0g](s, x, v) ds. Also define

F = T−1c so F ] =∫ t0c](s, x, v) ds, then T−1F = e

−FT−1eF (an operator composition viewing e±F asmultiplication operators). Then we see that

exp

(∫ t0

c](s, t, v) ds

)= eF

](t,x,v)

∫ t0

(eF )]Q+](τ, x, v) dτ = (T−1eFQ+)] = (eFT−1F Q+)]

So the equation is (eF f)] = f0 + (eFT−1F Q

+)], and undoing the characteristic map ] and dividing by eF

we obtain

f = f0e−F + T−1F Q

+(f, f) where F = T−1(A ∗ f)= e−F (f0 + T

−1[eFQ+(f, f)])

This form has some advantages, in particular due to the explicit formula for T−1 we can see that itis a continuous and weakly continuous map from L1((0, T ) × Rd × Rd) to C([0, T ];L1(Rd × Rd)) whichpreserves non-negativity. Furthermore for F ∈ C((0, T );L1(Rd×Rdloc)), we see that T

−1F is strongly and

weakly continuous from L1((0, T )× Rd × Rdloc) into C((0, T );L1(Rd × Rdloc)).

Theorem 7. We havef = f0e

−F + T−1F Q+(f, f)

where F = T−1(A ∗ f).

Proof. To prove this result, we first show that f ≥ f0e−F + T−1F Q+(f, f). To this end, we once againconsider the sequence gnδ =

1δ log(1+δf

n). As shown earlier, this sequence is weakly (relatively) compactas n→∞, so we assume that

gnδ → gδ

31

for some gδ weakly as n→∞. It is easy to see, as in the proof of Lemma 5, that limδ→0 gδ = f as δ → 0strongly in L1.

Similar arguments to the above tell us that the sequence

lnδ :=fn

1 + δfn

is weakly compact. Thus we may assume that lnδ → lδ weakly as n → ∞. Moreover, lδ → f as δ → 0strongly in L1.

We will also approximate F by a sequence Fn = T−1(An ∗fn) and we see that T (eFn) = eFn(An ∗fn)

Finally, we notice thatQn+(fn, fn)

1 + δfn

is weakly compact, so we can assumeQn+(fn, fn)

1 + δfn→ Q+,δ

weakly for some Q+,δ.Now that we have built up some convergent sequences, we perform the following computations

T (eFn gnδ ) = e

FnT (gnδ ) + gnδ T (e

Fn)

=eFn

1 + δfnQn(fn, fn) + gnδ e

Fn(An ∗ fn)

=eFn

1 + δfnQn+(f

n, fn) + eFn(An ∗ fn)[gnδ −

fn1 + δfn

]By integrating, we see that

(eFngnδ )(t, x, v)− (eFngnδ )(0, x, v) =∫ t0

eFn

1 + δfnQn+(f

n, fn)(s, x− (t− s)v, v)ds

+

∫ t0

eFn(An ∗ fn)[gnδ −

fn1 + δfn

](s, x− (t− s)v, v)ds

= T−1(eFnQn+(f

n, fn)

1 + δfn

)(t, x, v)

+ T−1(eFn(An ∗ fn)

[gnδ −

fn1 + δfn

])(t, x, v)

If we define T−1Fn by T−1Fn

:= e−FnT−1eFn then the above equation reads

gnδ =1

δlog(1 + δfn0 )e

−Fn + T−1Fn

((An ∗ fn)

[gnδ −

fn1 + δfn

])+ T−1Fn

(Qn+(f

n, fn)

1 + δfn

)(38)

We will need to bound the final term on the right hand side. To do this, we see that for any non-negativeϕ ∈ L∞ with compact support∫ [(

Qn+(fn, fn)

1 + δfn

)ϕ

1 +∫fndv1

]dv ≤

∫ [Qn+(f

n, fn)ϕ

1 +∫fndv1

]dv

From the prior results on velocity averaging Proposition 6 (ii), sending n→∞ yields

lim supn→∞

∫ [(Qn+(f

n, fn)

1 + δfn

)ϕ

1 +∫fndv1

]dv ≤

∫ [Q+(f, f)

ϕ

1 +∫fdv1

]dv

But the left hand side converges to ∫Q+,δ

ϕ

1 +∫fdv1

dv

32

and so ∫Q+,δ

ϕ

1 +∫fdv1

dv ≤∫Q+(f, f)

ϕ

1 +∫fdv1

dv

Because this holds for any ϕ ≥ 0, we conclude that

Q+,δ ≤ Q+(f, f)

almost surely in v, and so because T−1 is a positive functional we have

T−1(Q+,δ) ≤ T−1(Q+(f, f))

It is easy to see that if xn → x weakly in L1 then T−1Fn (xn)(t)→ T−1F (x)(t) in L

1. We will now returnto equation (38). Sending n→∞ gives

gδ =1

δlog(1 + δf0)e

−F + T−1F ((A ∗ f) [gδ − lδ]) + T−1F (Q+,δ)

≤ 1δ

log(1 + δf0)e−F + T−1F ((A ∗ f) [gδ − lδ]) + T

−1F (Q+(f, f))

Notice that

1

δlog(1 + δf0) = log

[(1 + δf0)

1δ

]→ log(ef0) = f0

as δ → 0.Also by the continuity of T−1F ,

T−1F ((A ∗ f) [gδ − lδ])→ T−1F ((A ∗ f) [f − f ]) = 0

and sof ≤ f0e−F + T−1F Q+(f, f)

It remains to show thatf ≥ f0e−F + T−1F Q+(f, f)

Our argument to show this will be similar to the one we have just performed. This time, we considerthe sequence of cut-off functions (hnδ ) such that h

nδ = min(f

n, 1δ ). It is clear that hnδ → hδ weakly for

some hδ satisfyinghδ ↑ f

We also get Qn+(hnδ , h

nδ )→ Q+(hδ, hδ) weakly. A similar computation to the one earlier (except this

time considering T (eF fn)) gives

fn = fn0 e−Fn + T−1Fn (Q

n+(f

n, fn))

But fn ≥ hnδ , and again because T−1Fn

is positive

fn ≥ fn0 e−Fn + T−1Fn (Qn+(h

nδ , h

nδ ))

We send n→∞ in this and conclude that

f ≥ f0e−F + T−1F (Q+(hδ, hδ))

Next, by the monotone convergence theorem,

Q+(hδ, hδ) ↑ Q+(f, f)

and sof ≥ f0e−F + T−1F (Q+(f, f))

as was claimed.

Theorem 8. The f constructed above is indeed a renormalized solution.

33

Proof. Firstly, we have

0 ≤ Q−(f, f)1 + f

=(A ∗ f)f

1 + f≤ A ∗ f

But we know from Lemma 6 that limn→∞An ∗fn → A∗f in L1((0, T )×Rd×BR) sense for all T,R > 0.Thus

Q−(f, f)

1 + f∈ L1((0, T )× Rd × Rdloc)

for any T > 0.

We will show the (harder) corresponding result for Q+(f,f)1+f . To do this, recall that, from the entropy

inequality (13),

Qn±(fn, fn) ≤ 2Qn∓(fn, fn) + 4

en(fn)

log(2)

Next, for any positive δ, we divide by the positive function (1 + δ∫fndv) to get

Qn±(fn, fn)

1 + δ∫fndv

≤ 2Qn∓(f

n, fn)

1 + δ∫fndv

+ 4en(f

n)

log(2)(1 + δ∫fndv)

(39)

We have shown in Proposition 6 that∫fndv →

∫fdv. We also showed in that same proposition that

for any function ϕ ∈ L∞((0, T )× Rd × Rd) with compact support∫Rd Q

n±(f

n, fn)ϕdv

1 + δ∫fndv

→∫Rd Q±(f, f)ϕdv

1 + δ∫fdv

Moreover, since en(fn) are L1(R+ × Rd × Rd) uniformly bounded, they converge vaguely to some finite

measure µ. The absolutely continuous part of µ (which we denote by e) is in L1((0, T )× Rd × Rd) andis non-negative and so, by taking limits in (39)∫

Rd Q±(f, f)ϕdv

1 + δ∫fdv

≤∫Rd Q∓(f, f)ϕdv

1 + δ∫fdv

+ 4e(ϕ)

log(2)(1 + δ∫fdv)

By sending δ → 0, we see that∫RdQ±(f, f)ϕdv ≤

∫RdQ∓(f, f)ϕdv +

4e(ϕ)

log(2)

and because e is absolutely continuous we get

Q±(f, f) ≤ Q∓(f, f) + E (40)

where E ∈ L1((0, T )× Rd × Rd) is non-negative. Thus

Q+(f, f)

1 + f≤ Q−(f, f)

1 + f+

E

1 + f≤ Q−(f, f)

1 + f+ E

and so Q+(f,f)1+f ∈ L1((0, T )× Rd × Rdloc).

Next, we show that Q+(f, f)#(t, x, v) is in L1(0, T ) for almost all (x, v). To do this, recall that by

Lemma 7 we have T−1F (Q+(f, f)(t, x, v)) ∈ L1(Rd×Rdloc) for every t > 0 and so[T−1F (Q+(f, f)(t, x, v))

]# ∈L1(Rd × Rdloc). Writing this out fully, we see that for compact sets K ⊂ Rd

∞ >∫K

∫Rd

(e−F (t))#(x, v)[T−1eF (t)(Q+(f, f)(t, x, v))

]#dxdv

>

∫K

∫Rd

(e−F (t))#(x, v)

∫ t0

eF#(s)

(Q#+(f, f)(s, x, v)

)dsdxdv

>

∫K

∫Rd

∫ t0

e−(F#(t)−F#(s))(x, v)

(Q#+(f, f)(s, x, v)

)dsdxdv

34

But F#(t, x, v) =∫ t0(A ∗ f)(s, x, v). We know that A ∗ f is non-negative, so F# increases in t and is also

non-negative. Thus

∞ >∫K

∫Rd

∫ t0

e−(F#(t))(x, v)

(Q#+(f, f)(s, x, v)

)dsdxdv (41)

It is clear that for each t > 0 we have F#(t) ∈ L1(Rd×Rdloc). Define Ω0 to be the set of (x, v) ∈ Rd×Rdsuch that F#(t)

• The equation is satisfied for t ∈ R.

• X(·, x) ∈ C1(R), b(X) ∈ C(R).

4.1.2 Strategy

The Cauchy-Lipschitz theorem is usually proven using a fixed point argument involving the Lipschitzcondition. Here we will use a completely different method; we will view Cauchy-Lipschitz as a statementabout PDEs.

Theorem 10 (Cauchy-Lipshitz (smooth case)). Let b ∈ C∞(RN ;RN ). Then the ODE stated in (42)has a unique smooth solution (in both variables) X(t, x) and

∂Xi(t, x)

∂t= bi(X)

In particular, the solutions of the ODE are the characteristics of the transport equation

∂u

∂t− bi

∂u

∂xi= 0 (43)

u(0, x) = x

Instead of looking at the original ODE we will attack the transport equation. We will derive stabilityresults using renormalised solutions, and then return to the original ODE via the characteristics.

4.1.3 Techniques and ideas used

Other than the notion of renormalised solutions, the proofs that follow use standard regularising andapproximating methods.

4.2 A proof using the stability result

We now have the machinery to state the stability result that allows us to prove the main theorem.

Theorem 11 (Stability). Let bn → b, cn → c, div bn → div b respectively in L1loc and assume that b,div band c satisfy

b ∈W 1,1loc , div b ∈ L∞, c ∈ L∞ (44)

|b|1 + |x|

∈ L1 + L∞ (45)

Furthermore, let un be the renormalized solution to (1) (with b replaced by bn etc.) with initial dataun0 ∈ L such that un0 converges in L to some u0 ∈ L. We also assume that un is a bounded sequence inL∞(L). Then un → u in C(L) where u is the renormalized solution of the original transport equation(1).

Theorem 12 (Existence and measure preservation). Let u0 ∈ L. Then there is a unique renormalizedsolution u to (1) such that u ∈ L∞(L) ∩ C(L).

We shall now restate and prove the main theorem using these results. Later, we will show that theseresults are indeed valid.

Theorem 13 (Main result). Assume that (44) and (45) are satisfied and that c, div b = 0 and b dependonly on x. Then there is a unique X(t, x) ∈ C(R;L) such that a.e. in x we have

(i) X(·, x) ∈ C1(R), b(X(·, x)) ∈ C(R)

(ii) ∂Xi∂t = bi(X) i.e. the original ODE.

Also (with no a.e. restriction) we have

36

(a) For a fixed t, X(t, x) preserves Lebesgue measure (i.e. µ ◦ X(t, ·) = µ where µ is the Lebesguemeasure)so that for all test functions φ we have

∫φ(X(t, x))dx =

∫φ(x)dx.

(b) (Group property) For all t, s and for almost all x we have X(t+ s, x) = X(t,X(s, x)).

(c) The equation

∂[βi(X)]

∂t=

∂βi∂Xj

bj(X) (46)

β(X(0, x)) = β(x)

holds in the sense of distributions for all β ∈ C1(RN ;RN ) with β, |(Dβ)(z)|(1 + |z|) ∈ L∞. This isthe notion of renormalized solutions restated in the RN → RN case.5

(d) β(X) ∈ L1loc(RN ;C(R)) for all such β.

Proof. We start with existence. The plan is as follows:

1. Mollify b and use the smooth version of Cauchy-Lipschitz and characteristics to formulate a se-quence of solutions to the transport equation.

2. Fix a β, show compactness and apply the stability result to gain existence and uniqueness of thesolution to (46).

3. Deduce measure preservation and group property from the transport equation.

4. Use properties of a particular β to deduce continuity.

5. Use similar compactness ideas to deduce that b(X) is continuous.

6. Use (46) to deduce the statements that hold for almost all x.

We will now execute this plan:

1. Apply the smooth version of Cauchy-Lipschitz to solve ˙̃X = b̃(X̃) where X̃ and b̃ are obtainedby a convolution with the function 1

�Nρ(·/�) where ρ ∈ D+(RN ) and

∫ρ = 1. Doing this yields

a unique smooth X̃(t, x) which obeys ∂X̃∂t = b̃(X̃) and X̃(0, x) = x. Furthermore, provided u0 is

smooth, we can use the chain rule, the group property and the measure preservation of X̃ to seethat ũ = u0(X̃) solves

∂ũ

∂t= b̃i

∂ũ

∂xi, ũ(0, x) = u0(x)

and so by choosing u0(x) = xi (i.e. u0 is a projection) we deduce that X̃ satisfies

∂X̃i∂t

= b̃j∂X̃i∂xj

, X̃(0, x) = x

2. As everything is smooth, we know that a function is a renormalised solution to the perturbedODE if and only if it is a solution to the perturbed ODE. We wish to extract a subsequencefrom X̃ that converges for any β that obeys the conditions on (46). By combining the equations∂X̃i∂t = b̃i(X̃) = b̃j

∂X̃i∂xj

and the chain rule we see that

∂[βi(X̃)

]∂t

= b̃j∂[βi(X̃)

]∂xj

=∂βi

∂X̃jb̃j(X̃)

which is the perturbed version of (46). Our assumptions on β and (45) tell us that for every

i, ∂βi(z)∂zj b̃j(z) ∈ L1 + L∞ . By the measure preservation property this is true for z = X̃, i.e.

∂βi∂X̃j

b̃j(X̃) ∈ L1 + L∞ uniformly in t, and we obtain

5Here and throughout, ∂βi∂Xj

means ∂βi∂xj

evaluated at X.

37

• ∂[βi(X̃)]∂t is bounded in L∞(R;L1 + L∞)

• ∂[βi(X̃)]∂t is relatively compact in the sets L∞(−T, T ;L1(BR)) for all R, T

But this is just the statement that u is a mild solution to ∂u∂t − bi∂u∂xi

= 0, and certainly u ∈ L1loc(R×RN )as it has compact support. The forcing term is 0 ∈ L1loc(R × RN ). Also X is invertible by the groupproperty and preserves measure, so we can apply theorem 1 to show that u is a distributional solutionand hence unique.

4.3 Proof of stability result

We have already seen that the transport equation has some stability in lemma 2. However, the mostgeneral result requires many separate approximation arguments, so to simplify things we will only con-sider the case where c, cn = 0 and b depends only on x (which was all that was needed above). We willlook at the equation

∂u

∂t− bi

∂u

∂xi= 0

with bn → b and div bn → div b in L1loc. To remove some of the sub/super-scripts we will no longerassume that ũ is a convolution with ρ as in the previous section: instead, we will write ũ = un and adopta similar notation for other functions. The sequence of initial conditions will then be given by ũ0.

4.3.1 Stability

We have to prove that for all admissible β, β(ũ) → β(u) in C(0, T ;L1loc). To do this we will first provea.e. convergence, then show uniformity over compact sets. We will fix an admissible β and set ṽ = β(ũ).

4.3.2 Pointwise

The key idea here is that if β(u) solves the equation then β(u) itself is a renormalized solution, so β(u)2

is a solution.By the definition of ũ we have that in D′([0, T )× RN ),

∂ṽ

∂t− b̃i

∂ṽ

∂xi= 0

and ṽ|(t=0) = ṽ0 = β(ũ0). Without loss of generality we can assume that vn converge in L∞((0, T )×RN )to some v in the weak∗ sense. By the convergence of b̃ and div b̃, v will solve the equation, and becauseof the convergence of ũ0 it has initial condition v0 = β(u0).

In the same way we can take w̃ = β(ũ)2, to obtain a w with w̃ → w such that w solves the equationwith initial condition w0 = β(u0)2. But then because v solves the equation it is a renormalised solutionby theorem 1. Hence v2 (the square of v) solves the equation with initial condition β(u0)2 = w0. Byuniqueness we then have that v2 = w. However, ṽ2 → v2 weak* in L∞((0, T ) × RN ) and hence inL2(L2loc). This gives us convergence locally in measure. As ṽ = β(ũ) where β was arbitrary (under theconditions imposed), we can vary β over strictly increasing functions to deduce that ũ → u locally inmeasure to some u. Therefore ṽ → β(u) and so u is a renormalised solution.

4.3.3 On compact time sets

We now need to show that the convergence is uniform on our time interval [0, T ]. T