Lectures notes on Boltzmann’s equation

Simone Calogero

1 Introduction

Kinetic theory describes the statistical evolution in phase-space1 of systems composed by a largenumber of particles (of order 1020). The main goal of kinetic theory, as far as the physical applica-tions are concerned, is to predict the evolution of those quantities associated to the system whichdepend only on the local average dynamics of the particles. These are called macroscopic quanti-ties. The most important examples of physical systems to which kinetic theory applies are dilutegases (where the molecules play the role of the particles) and in this case examples of macroscopicquantities are the temperature and the pressure of the gas. In fact, the temperature of a gas is ameasure of the mean kinetic energy of the molecules in small regions of the gas, while the pressureis a macroscopic manifestation of the exchange of momentum between the particles and the wallsof the gas container (of with any body inserted into the gas). The relation between kinetic andmacroscopic quantities will be discussed in Section 7.

The statistical description of the dynamics is given in terms of the one particle distribution function,denoted by f , which is a function of time, position and velocity (or momentum2), that is

f = f(t, x, v), t ≥ 0 , x ∈ R3 , v ∈ R3 .

By definition, f(t, x, v) is the probability density to find a particle at time t, in the position x,with velocity v. Thus the integral ∫

V

∫Ω

f(t, x, v) dv dx

gives the probability to find a particle in the region V at time t with velocities v ∈ Ω. Since thenumber of particles is so large, the above quantity can also be interpreted as the relative numberof particles in the region V at time t with velocities v ∈ Ω. Moreover, denoting by M the totalmass of the system, the above integral multiplied by M can be interpreted as the total mass of thesystem in the region V . We shall use freely all these interpretations of the function f , which leadnaturally to require that

0 ≤ f(t, ·, ·) ∈ L1(R3 × R3)

and that ‖f(t)‖1 = const = 1. The Boltzmann equation is a (integro-)partial differential equationfor the one particle distribution function f . Let us “derive” it in the most simple case, that is fora system of particles moving with constant velocities. Suppose (for notational simplicity) that allparticles are identical (otherwise we should introduce a particles distribution for each species). Ifx(t) denotes the position vector at time t and v the (constant) velocity vector of a particle, theposition at time t+ ε will be given by

x(t+ ε) = x(t) + εv .

1That is the manifold of all possible positions and velocities2In non-relativistic mechanics, the momenum and velocity of a particle differ only by a multiplicative constant—

the mass of the particle.

1

Now if instead of knowing the exact position of the particle at time t we only know the probabilityf(t, x(t), v) to find the particle there, it is natural to assume that

f(t+ ε, x(t+ ε), v) = f(t, x(t), v) ,

since we know with certainty that the particle has moved with constant velocity v in the ε-intervalof time. Provided f is a sufficiently smooth function, the above identity is equivalent to the freetransport equation

∂tf + v · ∇xf = 0 . (1)

The solution of (1) with initial data f(0, x, v) = fin(x, v) is given by f(t, x, v) = fin(x − vt, v).Note that fin ≥ 0 implies f(t, x, v) ≥ 0 and that ‖f(t)‖1 = ‖f0‖1. This is consistent with theinterpretation of f as a probability density (i.e., of f(t, x, v)dvdx as a probability measure).

Next we make the assumption that in the (infinitesimal) interval of time [t, t + ε] the particleundergoes a collision with the other particles of the system. Here the word “collision” is used torefer to a general interaction which takes place in such a short time interval and small region ofspace that we can safely say that it occurs at time t in the position x. If the result of the collisionis to increase the number of particles with velocity v, equation (1) modifies to

∂tf + v · ∇xf = G .

Here G = G(t, x, v) ≥ 0 is called the gain term and gives the probability density that a newparticle with velocity v is gained after the collision. Likewise, if a particle with velocity v is lostafter the collision with probability L, we have

∂tf + v · ∇xf = −L .

The function L = L(t, x, v) ≥ 0 is called loss term and the minus sign indicates that it makesf decrease (obviously, if a particle looses velocity v, the probability to have a particle with thisvelocity after the collision will be smaller). In general we have

∂tf + v · ∇xf = G− L . (2)

Hence the Boltzmann equation is a balance equation: It says us how the particle distribution fchanges as a consequence of the collision. In order to give an explicit form to G and L we have togive more information on how collisions occur. We shall consider only the case of the Boltzmannequation for binary elastic collisions.

Exercise 1. Consider a system of particles moving under the influence of an external force fieldF = F (t, x). What should the equation for the one particle distribution function be in this case?Prove that the resulting equation (Vlasov equation) is consistent with the interpretation of thesolution as a probability density (i.e., the solution is non-negative and its L1 norm is constant).

2 The Boltzmann equation for binary elastic collisions

Binary elastic collisions means that we take into account only collisions between pair of particles(binary), i.e. the simultaneous collisions between three or more particles are assumed to occur withnegligible probability. Moreover the collisions are assumed to be elastic, meaning that, besides thetotal momentum of the particles, also the total kinetic energy is conserved during the collision3.

3If the kinetic energy is not conserved, the collision is called inelastic.

2

Let v, v∗ denote the velocities before the collision of two particles and v′, v′∗ their velocities afterthe collision (see Fig. 1). We also assume the particles to have the same mass (which is alsoconserved in the collision). Then the conservation laws of momentum and kinetic energy take theform

v + v∗ = v′ + v′∗ (cons. of momentum) , (3)

|v|2 + |v∗|2 = |v′|2 + |v′∗|2 (cons. of kinetic energy) . (4)

Suppose the pre-collisional velocities (v, v∗) are given and we want to derive the post-collisionalvelocities (v′, v′∗) from (3) and (4). Since we have 6 unknowns but only 4 equations, the abovesystem is underdetermined. However one can prove that the manifold of the solution is a 2-sphere.Precisely we have the following

Lemma 1. A quadruple (v, v∗, v′, v′∗) solves (3)-(4) if and only if

v′ = v − [(v − v∗) · ω]ω, (5)

v′∗ = v∗ + [(v − v∗) · ω]ω, (6)

for some ω ∈ S2.

Proof. The claim is true for the trivial solution (v′ = v, v′∗ = v∗) (since it is obtained for ωorthogonal to v − v∗). Otherwise, set v′∗ = v∗ + aω, v′ = v + b ω; from the conservation ofmomentum, equation (3), we have a = −b. Substituting v′∗ = v∗ + aω and v′ = v − aω in (4) andsolving for a we obtained the desired result with

ω =v′∗ − v∗|v′∗ − v∗|

= − v′ − v|v′ − v|

.

3

The triple (v, v∗, ω) is called a collision configuration. The direction of ω is the scattering directionof the collision. Some important geometrical properties of binary elastic collisions are collected inthe following lemma, whose proof is left as exercise.

Lemma 2. The following holds:

(i) |v − v∗| = |v′ − v′∗|, |(v − v∗) · ω| = |(v′ − v′∗) · ω|;

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

(iii) The Jacobian of the transformation (v, v∗)→ (v′, v′∗) is equal to 1.

Exercise 2. Prove Lemma 2.

The meaning of (ii) is that a binary elastic collision is a reversible process (see Fig. 2).

Recall from the Introduction that the gain term measures the probability that a new particle withvelocity v results from the collision of two particles. By (ii) of Lemma 2, this is the case if theparticles collide with velocities v′ = v − [(v − v∗) · ω]ω and v′∗ = v∗ + [(v − v∗) · ω]ω (see Fig. 2again). The probability to find a particle with velocity v′ (resp. v′∗) in the point x at time t isgiven by f(t, x, v′) (resp. f(t, x, v′∗)). The probability to have both particles at the same time inthe same position (so that they may collide) is given by the product

f(t, x, v′)f(t, x, v′∗)

provided we assume that the occurrence of two particles in the same point at the same time withgiven velocities are two independent events. This is called the molecular chaos assumption.

Now if the two particles had always the same probability (say one) to collide independently fromtheir velocities and scattering angle, then the product above would already give the probabilitydensity to obtain a new particle with velocity v from the collision of two particles with velocitiesv′ and v′∗. This means the all the infinitely possible collision configurations which are compatiblewith (3) and (4) have the same probability to occur. However this is in general not the case. Theprobability of collision of two particles, that is the strength of the interaction, will depend in generalon the velocities of the two particles as well as on the scattering angle. In order to measure sucha “collision probability” we introduce a function B = B(ω, v, w) ≥ 0, called the collision kernel,whose physical meaning is roughly speaking the following: It says us how strong is the collisionof two particles with velocities v, w and scattering angle ω (that is, how probable is the collisionconfiguration (v, w, ω)). The specific form of B depends on the model that we use to representthe particles and the forces which act during the collision. An example will be given in the nextsection.

Having introduced the collision kernel, we can now state that the probability density of gaining aparticle v from the collision of two particles with velocities v′ and v′∗ is given by

B(ω, v′, v′∗)f(t, x, v′)f(t, x, v′∗)

Equivalently, we can interpret the above expression as the probability associated to the configura-tion in Fig. 2. But since we are not interested in the velocity v∗ of the second particle nor on thescattering angle ω, the gain term is obtained by integrating over all possible (v∗, ω):

G(t, x, v) =

∫R3

∫S2

B(ω, v′, v′∗)f(t, x, v′)f(t, x, v′∗)dω dv∗. (7)

4

The form of the loss term is now easily derived. In fact it is clear from Lemma 1 that a particlewith velocity v will be lost whenever one of the two particles had velocity v before the collision4.Thus for the loss term we obtain

L(t, x, v) =

∫R3

∫S2

B(ω, v, v∗)f(t, x, v)f(t, x, v∗)dω dv∗. (8)

We have almost finished to derive the Boltzmann equation in its final form. To this end we needto make a further assumption on the collision kernel: The function B = B(ω, v, w) depends on thescattering direction ω and the velocities v, w only through the collision invariants |v−w|, |(v−w)·ω|(see (i) of Lemma 2). The role of this assumption is to assure the Galilean invariance of theBoltzmann equation (see Exercise 3). It implies that B(ω, v′, v′∗) = B(ω, v, v∗) in (7) and so weobtain

G(t, x, v) =

∫R3

∫S2

B(ω, v, v∗)f(t, x, v′)f(t, x, v′∗)dω dv∗. (9)

Replacing (8) and (9) in (2) we obtain finally the Boltzmann equation

∂tf + v · ∇xf =

∫R3

∫S2

B [f(t, x, v′)f(t, x, v′∗)− f(t, x, v)f(t, x, v∗)] dω dv∗ (10)

where B = B(|v−v∗|, |(v−v∗) ·ω|). The right hand side is the collision integral. It will be denotedby QB(f, f, )(t, x, v) for short, i.e., we write (10) as

∂tf + v · ∇xf = QB(f, f). (11)

Moreover we denote qB(f, f)(t, x, v, v∗, ω) the integrand function in the collision integral, i.e.,

qB(f, f)(t, x, v, v∗, ω) = B [f(t, x, v′)f(t, x, v′∗)− f(t, x, v)f(t, x, v∗)] ,

whence QB(f, f) =∫R3×S2 qB(f, f)dωdv∗. Throughout these notes we assume

B(a, b) ≥ 0 and B(a, b) > 0 for almost all a, b ≥ 0. (12)

This means that the probability that the particles do not collide is negligible.

Exercise 3. Prove that (10) is Galilean invariant, that is, f(t, x−ut, v−u) is a solution wheneverf(t, x, v) is a solution, for all u ∈ R3.

Exercise 4. Prove the analogue of Lemma 1 for the binary elastic collisions of relativistic particles.In this case, the conservation of energy (4) is to be replaced with√

1 + |v|2 +√

1 + |v∗|2 =√

1 + |v′|2 +√

1 + |v′∗|2

where now v is the “momentum” rather than the velocity of the particles (the latter being definedas v = v/

√1 + |v|2, where we set the speed of light and the rest mass of the particles equal to

one).

3 Example of collision kernel: The hard spheres model

The collision kernel B depends on the model that we use to represent the particles and the forceswhich act during the collision. In this section we present an argument to justify the form of B in

4The exception is the solution (v′ = v, v′∗ = v∗). However this solution is a set of zero measure on the set of allpossible collision configurations

5

the case of hard spheres collisions. This means two things: (1) particles are modelled as spheres(of diameter one5) and (2) the collision is rigid, in the sense that at the point and time of impactbetween two spheres, an infinite force acts instantaneously on the direction which connects thecenters of the two spheres (Fig. 3). The latter means that there is no friction during the impact.

Let x, x∗ denote the centers of the two spheres. Since the force is directed along x−x∗, the changeof momentum of any of the two single particles will also occur along this direction. Thus recallingthe definition of ω in Lemma 1, we have

ω =v − v′

|v − v′|=

x− x∗|x− x∗|

.

It is also clear by symmetry that the collision of Fig. 3 is equivalent to the collision of a spherewith diameter 2 at rest with a point particle having velocity v − v∗ (see Fig. 4). Now let dω theinfinitesimal surface element on the sphere 2. The particles with velocity v− v∗ which collide withthe sphere 2 on the surface element dω in the infinitesimal interval of time (t, t+ dt) are containedin the cylinder with weight |(v − v∗) · ω|dt and base dω (Fig. 4). Denote this cylinder by C(v, v∗).The volume of C is given by dω|(v − v∗)ω|dt.

Now we repeat the derivation of the loss term made in Section 2, using the hard spheres model.The probability of loosing a particle with velocity v in the interval (t, t+dt) equals the probabilityof collisions of two particles, one of which has velocity v, in the same infinitesimal time interval.In the hard sphere model, these are the particles contained in the cylinder C(v, v∗), ∀v∗ ∈ R3.

5More rigorously, the Boltzmann equation for hard spheres is derived in the limit when the diameter of theparticles tends to zero.

6

That is to say: two particles, one with velocity v and the other one with velocity v∗, which arein the cylinder C(v, v∗) must collide. Thus all the particles with velocity v and v∗ contained in Cloose their velocities. By the interpretation of f as a particle density (or as a relative number ofparticles), the (relative) number of particles with velocity v in C are given by

V ol(C)× f(v)dv = |(v − v∗)ω|f(v)dω dt dv

and since all the particles with velocity v∗ collide with these particles, then the probability ofcollision of two particles, one with velocity v and one with velocity v∗ is given by

|(v − v∗)ω|f(v)f(v∗)dω dt dv dv∗.

This expression is then divided by dt (to obtain the number of collisions per unit of time) and thenintegrated over all possible v∗ ∈ R3 and ω ∈ S2 (because we are only interested in the probabilityof loosing a velocity v, regardless the velocity of the other particle and the scattering angle). Ifthe resulting expression is compared with (8) we obtain

B(ω, v, v∗) = |(v − v∗) · ω|, (13)

i.e., the collision kernel for hard spheres interaction.

4 Collision invariants

In this section we study some fundamental properties of the collision integral QB(f, f) that appearsin the r.h.s. of the Boltzmann equation. Let T > 0 and φ : (0, T ) × R3 × R3 be a measurablefunction with values in a finite dimensional vector space (typically R, or R3). We denote

φ = φ(t, x, v), φ∗ = φ(t, x, v∗), φ′ = φ(t, x, v′), φ′∗ = φ(t, x, v′∗).

Moreover we denote by f any real-valued measurable function on (0, T )×R3×R3 (not necessarilynon-negative). We shall also write

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

Proposition 1. The following identity holds∫R3

QB(f, f)φdv =1

4

∫R3

∫R3

∫S2

qB(f, f)(φ+ φ∗ − φ′ − φ′∗)dωdv∗dv, (14)

for all (t, x) ∈ (0, T )× R3 and for all functions φ and f as above such that∫R3

∫R3

∫S2

|qB(f, f)φ| dωdv∗dv <∞, ∀(t, x) ∈ (0, T )× R3. (15)

Proof. We have ∫R3

QB(f, f)φdv =

∫R3

∫R3

∫S2

B(f ′f ′∗ − ff∗)φdω dv∗ dv. (16)

Note that the assumption (15) allows, by Fubini’s theorem, to exchange the order of the integrals.Thus applying the change of variable (v, v∗) → (v∗, v) (which implies (v′, v′∗) → (v′∗, v

′)) to theright hand side of (16) we obtain∫

R3

QB(f, f)φdv =

∫R3

∫R3

∫S2

B(f ′f ′∗ − ff∗)φ∗ dω dv∗ dv. (17)

7

In the right hand side of (17) we make the change of variables (v, v∗) → (v′, v′∗). Since thistransformation leaves invariant the Lebesgue measure (see Lemma 2(iii)), we obtain∫

R3

QB(f, f)φdv = −∫R3

∫R3

∫S2

B(f ′f ′∗ − ff∗)φ′∗ dω dv∗ dv. (18)

Finally, doing again the change of variable (v, v∗)→ (v∗, v) we get∫R3

QB(f, f)φdv = −∫R3

∫R3

∫S2

B(f ′f ′∗ − ff∗)φ′ dω dv∗ dv. (19)

Summing up (16)-(18) and dividing by 4 yields the claim.

Definition 1. A function φ = φ(t, x, v) is called a collision invariant if∫R3

QB(f, f)φdv = 0,

for almost all (t, x) ∈ (0, T )×R3 and for all measurable functions f such that the bound (15) holds.

By Proposition 1, it follows that φ is a collision invariant if it verifies the identity φ+φ∗ = φ′+φ′∗a.e.

Lemma 3. Let φ be measurable and finite a.e. The identity

φ(t, x, v) + φ(t, x, v∗) = φ(t, x, v′) + φ(t, x, v′∗), (20)

i.e., φ + φ∗ = φ′ + φ′∗, is verified if and only there exist a = a(t, x) ∈ R, b = b(t, x) ∈ R andc = c(t, x) ∈ R3, measurable and a.e. finite, such that

φ(t, x, v) = a(t, x) + b(t, x)|v|2 + c(t, x) · v. (21)

The proof of “if” is trivial. The proof of the “only if” part when φ is just measurable and a.e.finite is quite long. Let us restrict ourselves to show that if φ ∈ C2 solves (20), then it must be ofthe form (21). First we notice that, since the relation between (v′, v′∗) and (v, v∗) are equivalent tothe conservation of momentum and kinetic energy (see Lemma 1), equation (20) is verified if andonly if there exists a function ψ on [0,∞)× R3 such that

φ(v) + φ(v∗) = ψ(|v|2 + |v∗|2, v + v∗)

and therefore, setting v∗ = 0,φ(v) + φ(0) = ψ(|v|2, v). (22)

Note that we suppress the dependence on (t, x). The claim has now been reduced to prove thatthe function ψ is linear in both its variables. Replacing (22) into (20) and setting v∗ = 0 we obtain

ψ(|v|2, v) + ψ(0, 0) = ψ(|v′|2, v′) + ψ(|v′∗|2, v′∗), (23)

where v′∗ = (v · ω)ω, v′ = v − (v · ω)ω and therefore

∂v′i∂vj

= δij − ωiωj ,∂v′∗i∂vj

= ωiωj ,∂|v′|2

∂vi= 2vi − 2(v · ω)ωi,

∂|v′∗|2

∂vi= 2(v · ω)ωi. (24)

Thus differentiating (23) w.r.t. vi we get, denoting by u the first argument of ψ,

2∂uψ(|v|2, v)vi + ∂viψ(|v|2, v) =2∂uψ(|v′|2, v′)vi − 2∂uψ(|v′|2, v′)(ω · v)ωi

+ ∂viψ(|v′|2, v′)− ω · ∂vψ(|v′|2, v′)ωi+ 2∂uψ(|v′∗|2, v′∗)(ω · v)ωi + ω · ∂vψ(|v′∗|2, v′∗)ωi (25)

8

The previous identiy has to be satisfied for all ω ∈ S2. We may assume v 6= 0 and replace ω = v/|v|in (25). Since for such ω there holds (v′, v′∗) = (0, v), we obtain the identity

Aij∂viψ(|v|2, v) = Aij∂viψ(0, 0), (26)

where A is the matrixAij = δij −

vivj|v|2

. (27)

Since the matrix A is invertible, (26) entails ∂viψ(|v|2, v) = ∂viψ(0, 0), whence ψ is linear in thesecond variable. To prove linearity in the first variable, we proceed likewise, differentiating (25)in vj (note that the terms with ∂vi∂vjψ vanish because ψ is linear in the second variable) andevaluating the resulting identity on ω = v/|v|. The details are left as exercise.

Exercise 5. Prove the linearity of ψ in the first variable.

It follows by Proposition 1 and Lemma 3 that any function of the form (21) is a collision invariant.We denote

φ0(v) = 1, φ1(v) = v, φ2(v) = |v|2

and call them the fundamental collision invariants.

5 Mild solutions and conservation laws

Definition 2. Let f0 = f0(x, v) be a measurable, almost everywhere non-negative function. Ameasurable function f = f(t, x, v) is said to be a mild solution of the Boltzmann equation in theinterval [0, T ) with initial datum f0 if f(t, x, v) ≥ 0 for almost all (t, x, v) ∈ (0, T )× R3 × R3,∫ t

0

QB(f, f)(s, x+ v(s− t), v) ds is bounded for all (t, x) ∈ (0, T )× R3

and the following identity holds for almost all (t, x, v) ∈ (0, T )× R3 × R3:

f(t, x, v) = f0(x− vt, v) +

∫ t

0

QB(f, f)(s, x+ v(s− t), v) ds. (28)

The mild solution is said to be global if T =∞.

Equation (28) defines the so-called mild formulation of the Boltzmann equation. Note that smoothsolutions verify the identity (28).

Proposition 2. Let φ = φ(t, x, v) be a C1 collision invariant that solves the free transport equation∂tφ+ v · ∇xφ = 0 with initial data φ0(x, v) = φ(0, x, v). Let f0 ≥ 0 a.e., such that∫

f0|φ0| dv dx <∞,

i.e., f0φ0 ∈ L1. Let f be a mild solution in the interval [0, T ) such that (15) holds. Then

fφ ∈ L∞((0, T );L1(R6))

and ∫R6

fφ dv dx =

∫R6

f0φdv dx, for almost all t ∈ (0, T ).

9

Proof. Notice first that since φ solves the free transport equation, then

φ(t, x, v) = φ(s, x+ v(s− t), v)

holds, for all (s, t, x, v). Then, multiplying (28) by φ(t, x, v) we get

f(t, x, v)φ(t, x, v) = f0(x− vt, v)φ0(x− vt, v) +

∫ t

0

QB(f, f)(s, x+ v(s− t), v)φ(s, x+ v(s− t), v)ds.

(29)We now integrate in the variables (x, v) and exchange the order of the integrals (thanks to (15))in such a way that the first integral is over the variable x. After a translation in x, we obtain theidentity ∫

R3

∫R3

fφdvdx =

∫R3

∫R3

f0φ0dvdx+

∫ t

0

∫R3

∫R3

QB(s, x, v)φ(s, x, v)dvdxds.

The last term vanishes, because φ is a collision invariant. The result follows.

Choosing φ = φi(v) we get: conservation of the total mass (i = 0), conservation of the totalmomentum (i = 1) and conservation of the total kinetic energy (i = 2). Taking φ = x ∧ v we getthe conservation of the total angular momentum. See Section 7 for the definition of the macroscopicobservable quantities.

6 The Entropy identity and the Maxwellian distributions

Lemma 4. Let f be measurable, strictly positive and finite a.e. such that∫R3

∫R3

∫S2

|qB(f, f) log f | dωdv∗dv <∞, ∀(t, x) ∈ (0, T )× R3.

Then ∫R3

log fQB(f, f) dv ≤ 0

for almost all (t, x) and ∫R3

log fQB(f, f) dv = 0

if and only iff(t, x, v) = MA,β,u = A exp(−β|v − u|2), (30)

a.e., where A, β, u are functions of (t, x).

Proof. Choose φ = log f in the identity (14). We obtain∫log fQB(f, f)dv =

∫R3

∫R3

∫S2

B(f ′f ′∗ − ff∗) logff∗f ′f ′∗

dωdv∗dv.

Since

(X − Y ) logY

X≤ 0, for all X,Y > 0

with equality iff X = Y , we obtain that∫R3 log fQB(f, f)dv ≤ 0, and the equality holds iff

ff∗ = f ′f ′∗, i.e., log f + log f∗ = log f ′ + log f ′∗.

The claim follows by Lemma 3.

10

Definition 3. A distribution of the form (30), with A > 0 and β > 0, is called a Maxwelliandistribution.

At this point it is important to emphasize that MA,β,u is not always a solution of the Boltzmannequation. Since

qB(MA,β,u,MA,β,u) = 0,

the condition for MA,β,u to be a solution of the Boltzmann equation is that it solves the freetransport equation:

∂tMA,β,u + v · ∇xMA,β,u = 0.

A simple and important case is when A, β, u are constants. In this case, the function MA,β,u(v) iscalled a global Maxwellian. Note however that a global Maxwellian is not integrable over all space(since it is independent of x).

We now use the previous lemma to establish an important identity, known as the entropy identity.Let us assume that f is a positive classical solution of the Boltzmann equation, which decaysrapidly at infinity. Let

H[f ](t) =

∫R3

∫R3

f log fdvdx (31)

the entropy functional. Taking the time derivative and using (11) we obtain

dH

dt=

∫R3

∫R3

log fQB(f, f) dvdx. (32)

It follows by Lemma 4 that the entropy is non-increasing, and that at the stationary points of H,the solution must be a Maxwellian, i.e., f(t, x, v) = MA,β,u.

Exercise 6. Find the general conditions on the functions A, β, u of (t, x) such that MA,β,u is asolution of the Boltzmann equation and give an example.

7 Macroscopic balance equations

The fundamental laws of the mechanics of continuous bodies, in the absence of external forces, aregiven by

∂tρ+∇x · (ρu) = 0, (33a)

ρDu

dt+∇x · σ = 0, (33b)

where ρ is the mass density, v is the velocity field, σ is the stress tensor, and

D

dt= ∂t + v · ∇x

is the convective derivative operator. To close the system, one has to add a constitute law whichpermits to express σ in terms of ρ and v. For instance, for a perfect fluid σ = p(t, x)I, where p isthe pressure, and (33) reduce to the Euler equations. For isentropic perfect fluids, the system isclosed by assigning an equation of state, i.e., by prescribing p as a function of ρ. For viscous fluidswe have

σij = p δij +∇ · v δij + ∂xivj + ∂xjvi

and the second equation of the system (33) reduces to the Navier-Stokes equation.

11

On a sufficiently large scale, in which the discrete, molecular structure can be neglected, a gascan also be approximated by a continuous body. The purpose of this section is to derive theconnection between the macroscopic description of the gas, based on the system (33), and themesoscopic description, which is based on the Boltzmann equation.

In this section we assume that f is a classical solution (i.e., C1) of the Boltzmann equation thatdecays rapidly at infinity in the variable (x, v). Given a collision invariant φ that depends only onv, define

Cφ[f ] =

∫R3

fφ dv, Jφ[f ] =

∫R3

φfvdv. (34)

It follows that Cφ, Jφ verify the equation

∂tCφ +∇x · Jφ = 0. (35)

Let M > 0 be the total mass of the system, i.e., the sum of the mass of each particle (thus M is aconstant, because we assume that each particle preserves its mass). The mass density ρ = ρ(t, x)is defined by

ρ(t, x) = M

∫R3

f dv. (36)

Consider an infinitesimal volume dx of the gas. The moment of this gas region is given by uρ dx,where u is the bulk velocity of the region dx, i.e., the velocity of dx as a whole. Since dx is verysmall, we can say with good approximation that it has its own velocity. On the other hand, uρdxmust equal the total momentum of the particles in the region dx, which is given by

M dx

∫R3

vf dv.

Thus we obtain the following formula for the bulk velocity:

u =

∫R3 vfdv∫R3 fdv

. (37)

Lemma 5. For all sufficiently regular solutions of the Boltzmann equation, the local conservationlaw of mass is satisfied:

∂tρ+∇x · (ρu) = 0. (38)

Proof. Use φ = M in (35).

Now, replacing φ = Mv in (35) we obtain

∂t(ρu) +∇x · τ = 0, (39)

where τ is the second order tensor with components

τij = M

∫R3

vivjfdv.

The velocity of each single molecule can be decomposed as

v = c+ u,

where c = v − u measure the deviation of the velocity of each molecule from the bulk velocityand is called internal velocity. Notice that even if a given infinitesimal region dx of the gas is atrest (u = 0), the internal velocity is not zero; in fact in this case it coincides with the molecules

12

velocity v. Thus c is the velocity of the molecules in the region dx in the reference frame in whichthe latter is at rest.

Exercise 7. Prove that the average of the internal velocity is zero, i.e.,∫R3

cfdv = 0.

Replacing v = c+ u in the definition of the tensor τ we obtain

τij = M

∫R3

(ci + cj + uiuj + ciuj + ujci)fdv.

Using the result of the previous exercise, the integrals containing the mixed terms ciuj vanish.Thus we can write

τij = ρuiuj + σij ,

where

σij = M

∫R3

cicjfdv (40)

is the stress tensor of the gas. Replacing in (39) we obtain, in components

∂t(ρui) + ∂xj (ρuiuj) + ∂xjσij = 0. (41)

Lemma 6. For all sufficiently regular solutions of the Boltzmann equation, the local conservationlaw of momentum is satisfied:

ρDu

dt+∇x · σ = 0. (42)

Proof. Use (41) and the local conservation of mass.

Note that in this case the constitute law needs not to be given: all quantities are expressed interms of the particle distribution function f .

Finally, let us take φ = 12M |v|

2; equation (35) becomes

∂t

(1

2M

∫R3

|v|2fdv)dv +∇x

(1

2M

∫R3

|v|2vfdv)dv = 0.

using the decomposition v = c+ u, the previous equation becomes

∂t

(1

2ρ|u|2 + ρe

)+∇x ·

((1

2|u|2 + e)ρu+ q + σ · u

)= 0, (43)

where

e(t, x) = ρ−1 1

2M

∫R3

|c|2fdv (internal energy) (44)

q(t, x) =1

2M

∫R3

c|c|2fdv (heat flux) (45)

We have proved

Lemma 7. For all sufficiently regular solutions of the Boltzmann equation, the local conservationlaw of energy (43) holds.

13

The identities derived in this section apply to solutions of the Boltzmann equation. In particularthey apply to the Maxwellian distributions. In the simplest case, i.e., for the global Maxwellian,they are trivially satisfied, because MA,β,u(v) is independent of (t, x). Computing the macroscopicquantities for a generic (not necessarily global) Maxwellian we find that:

β(t, x) =3

4e(t, x), A(t, x) =

(3

4πe(t, x)

)3/2

, (46a)

σij(t, x) =2

3ρe(t, x)δij , q = 0. (46b)

The function (of (t, x))

p =2

3ρe =

1

3

∫R3

|c|2fdv =1

3Tr(σ),

is the isotropic pressure.

Exercise 8. Proof the identities (46).

We conclude this section by proving that the Maxwellians minimize the entropy functional. Let

h[f ](t, x) =

∫R3

f log fdv

the entropy density.

Theorem 1. Let f ≥ 0—not necessarily solution of the Boltzmann equation—, with finite entropy,mass density ρ, bulk velocity u and internal energy e. Let Mρ,u,e the Maxwellian associated to(ρ, u, e) (i.e., the coefficients A, β are given by (46a). Then

h[f ] ≥ h[Mρ,u,e]

and equality holds iff f = Mρ,u,e, that is to say, the Maxwellian Mρ,u,e is the distribution functionwith the least entropy among those with density ρ, bulk velocity u and internal energy e.

Proof. Let us denote M = Mρ,u,e for notational simplicity. Then∫R3

(f log f −M logM)dv =

∫R3

(f log f − f logM)dv +

∫R3

logM(f −M)dv.

Moreover ∫R3

logM(f −M) = logA

∫R3

(f −M)dv − β∫R3

|v − u|2(f −M)dv.

Both terms in the r.h.s. of the previous identity vanish, because f and M have the same ρ, u, e.Whence ∫

R3

(f log f −M logM)dv =

∫R3

(f log f −M logM)dv.

Using the elementary identity

z log z − z log y + y − z ≥ 0, ∀y, z > 0, (47)

we obtain

h[f ]− h[M ] ≥∫R3

(f −M)dv = 0.

Since equality in (47) occurs only for z = y, the proof is complete.

14

8 Global existence and uniqueness of mild solutions for smalldata

In this section we prove the existence of global, mild, continuous solutions of the Boltzmannequation for small data. Moreover, the solution verifies the standard global conservation laws(mass, momentum and energy). Recall that the mild (or integral) formulation of the Boltzmannequation is

f(t, x+ vt, v) = f0(x, v) +

∫ t

0

QB(f, f)(s, x+ vs, v) ds.

Now, given a function g = g(t, x, v), we denote

g](t, x, v) = g(t, x+ vt, v).

We may rewrite the mild formulation of the Boltzmann equation in the form

f ](t, x, v) = f0(x, v) +

∫ t

0

QB(f, f)](s, x, v),

or, in terms of f ] alone,

f ](t, x, v) = f0(x, v) +

∫ t

0

∫R3

∫S2

B[f ](s, x+ (v − v′)s, v′)f ](s, x+ (v − v′∗)s, v′∗)

− f ](s, x, v)f ](s, x+ (v − v∗)s, v∗)]dωdv∗. (48)

Our first goal is to prove the following

Theorem 2. Assume that the collision kernel satisfies B ≤ b|(v− v∗)|, for some b > 0. Let β > 0and define the Banach space

Mβ = f ∈ C0([0,∞)× R3 × R3) : |f(t, x, v)| ≤ αe−β(|x|2+|v|2), for some α > 0,

with norm‖f‖ = sup

t,x,veβ(|x|2+|v|2)|f(t, x, v)|.

Denote by MRβ = f ∈ Mβ : ‖f‖ ≤ R, the ball of radius R and centered on f = 0 in Mβ. Then

for all β > 0 there exists R0 = R0(β, b) such that for all R ≤ R0 and f0 ∈ MRβ , there exists a

unique f ] ∈M2Rβ solution of (48).

Note that the solutions of Theorem 2 are NOT yet mild solutions, because it is not claimed thatthe are non-negative!

We divide the proof in several lemmata. Let us begin with a calculus lemma.

Lemma 8. The integral

I =

∫ ∞0

e−β|x+τ(v−v∗)|2dτ

is bounded as

I ≤√π

β

1

|v − v∗|.

15

Proof. We write

I = e−β|x|2

∫ ∞0

e−β[τ2|v−v∗|2+2τx·(v−v∗)]dτ

=e−β|x|

2

|v − v∗|

∫ ∞0

e−β[s2+2sx· (v−v∗)|v−v∗|]ds,

where the second equality follows by the change of variables s = τ |v − v∗|. Let η = v−v∗|v−v∗| ∈ S

2

and rewrite the exponent of the integrand function as

s2 + 2sx · η = (s+ x · η)2 − (x · η)2.

Then

I =e−β(|x|2−(x·η)2)

|v − v∗|

∫ ∞0

e−β(s+x·η)2ds

≤ 1

|v − v∗|

∫ ∞−∞

e−βy2

dy =

√π

β

1

|v − v∗|.

Next recall that the collision integral QB(f, f) consists of a gain term Q+B(f, f) and a loss term

Q−B(f, f), i.e.,QB(f, f) = Q+

B(f, f)−Q−B(f, f),

where

Q+B(f, f)(t, x, v) =

∫R3

∫S2

Bf(t, x, v′)f(t, x, v′∗)dωdv∗, (49)

Q−B(f, f)(t, x, v) = f(t, x, v)

∫R3

∫S2

Bf(t, x, v∗)dωdv∗. (50)

We remark that Q+B and Q−B might not be well defined individually, even though QB is. However

for the solutions of Theorem 2 the gain and loss term are both well defined.

Lemma 9 (Estimate for the loss term). The following estimate holds∫ t

0

|Q−B(f1, f2)](s, x, v)|ds ≤ 4π3b

β2‖f ]1‖‖f

]2‖e−β(|x|2+|v|2).

Proof. By the definition of the loss term,

|Q−B(f1, f2)](t, x, v)| ≤ b|f ]1(t, x, v)|∫R3

∫S2

|v − v∗|f ]2(t, x+ t(v − v∗), v∗)dωdv∗

≤ b‖f ]1‖‖f]2‖e−β(|x|2+|v|2)

∫R3

∫S2

|v − v∗|e−β|x+t(v−v∗)|2−β|v∗|2dωdv∗.

Therefore∫ t

0

|Q−B(f1, f2)ds ≤ 4πb‖f ]1‖‖f]2‖e−β(|x|2+|v|2)

∫R3

e−β|v∗|2

|v − v∗|∫ ∞

0

e−β|x+t(v−v∗)|2dtdv∗

≤ 4πb‖f ]1‖‖f]2‖e−β(|x|2+|v|2)

√π

β

∫R3

e−β|v∗|2

dv∗.

Since the last integral is equal to π3/2/β, the proof is complete.

16

Lemma 10 (Estimate for the gain term). The following estimate holds∫ t

0

|Q+B(f1, f2)](s, x, v)|ds ≤ 4π3b

β2‖f ]1‖‖f

]2‖e−β(|x|2+|v|2).

Proof. To begin with we have the estimate

|Q+B(f1, f2)]| ≤ b‖f ]1‖‖f

]2‖∫R3

∫S2

|v − v∗|e−β|x+(v−v′)t|2−β|v′|2e−β|x+(v−v′∗)t|2−β|v′∗|

2

dωdv∗

= b‖f ]1‖‖f]2‖∫R3

∫S2

|v − v∗|e−β|v|2−β|v∗|2e−β[|x+(v−v′)t|2+|x+(v−v′∗)t|

2]dωdv∗,

where the equality follows from the conservation of energy: |v|2 + |v∗|2 = |v′|2 + |v′∗|2. Now, adirect computation, using the conservation of momentum and energy, shows that

|x+ (v − v′)t|2 + |x+ (v − v′∗)t|2 = |x|2 + |x+ t(v − v∗)|2,

whence we obtain

|Q+B(f1, f2)]| ≤ b‖f ]1‖‖f

]2‖∫R3

∫S2

|v − v∗|e−β|v|2

e−β|v∗|2

e−β|x|2

e−β|x+t(v−v∗)|2dωdv∗

= 4πb‖f ]1‖‖f]2‖e−β|v|

2

e−β|x|2

∫R3

e−β|x+t(v−v∗)|2 |v − v∗|e−β|v∗|2

dv∗.

Integrating in time we get∫ t

0

|Q+B(f1, f2)]|ds ≤ 4π‖f ]1‖‖f

]2‖e−β(|v|2+|x|2)

∫R3

|v − v∗|e−β|v∗|2

∫ ∞0

e−β|x+t(v−v∗)|2dtdv∗

≤ 4π3b

β2‖f ]1‖‖f

]2‖e−β(|x|2+|v|2),

where the last inequality follows as in Lemma 9.

Lemma 11. Let T±f0 be the operators

T+f0

[g] =f0

2+

∫ t

0

∫R3

∫S2

Bg(s, x+ (v − v′)s, v′)g(s, x+ (v − v′∗)s, v′∗)dωdv∗ds,

T−f0 [g] =f0

2−∫ t

0

g(s, x, v)

∫R3

∫S2

Bg(s, x+ (v − v∗)s, v∗)dωdv∗ds.

There exists R0(β, b) such that, for all R ≤ R0 and f0 ∈MRβ , T±f0 maps MR

β into MRβ .

Proof. By Lemma 9 and Lemma 10 we have

‖T±f0 [g]‖ ≤ 1

2‖f0‖+

4π3b

β2‖g‖2 ≤ R

2+

4π3b

β2‖g‖2.

Now, when g ∈MRβ we have

‖T±f0 [g]‖ ≤ R

2+

4π3b

β2R2.

Whence ‖T±f0 [g]‖ ≤ R for R ≤ β2

8π3b = R0. The fact that T±f0 [g] is continuous when g is continuousis straightforward.

17

To conclude the proof of Theorem 2 it is enough to prove that, for a properly small R,

‖T±f0 [g]− T±f0 [g]‖ < 1

2‖g − g‖.

For this implies, by the fixed point theorem, that the equation(s) T±f0 [g] = g have only one solution,

which we denote g±. Obviously, g = g− + g− ∈ M2Rβ solves Tf0 [g] = g, where Tf0 = T+

f0+ T−f0 .

But this is just our Boltzmann equation in the mild form (for g = f ]), so the existence part ofTheorem 2 follows. The uniqueness follows from the fact that

‖Tf0 [g]− Tf0 [g]‖ ≤ ‖T+f0

[g]− T+f0

[g]‖+ ‖T−f0 [g]− T−f0 [g]‖ < ‖g − g‖.

Thus we conclude if we prove the following

Lemma 12. There exists R0 = R0(β, b) such that for R < R0

‖T±f0 [g]− T±f0 [g]‖ < 1

2‖g − g‖.

Proof. We prove the claim for T−f0 and leave the statement for T+f0

as exercise. We have

T−f0 [g]− T−f0 [g] =−∫ t

0

[g(s, x, v)

∫R3

∫S2

Bg(s, x+ (v − v∗)s, v∗)

− g(s, x, v)

∫R3

∫S2

Bg(s, x+ (v − v∗)s, v∗)]ds

=−∫ t

0

(g − g)(s, x, v)

∫R3

∫S2

Bg(s, x+ (v − v∗)s, v∗)dωdv∗ds

+

∫ t

0

g(s, x, v)

∫R3

∫S2

B(g − g)(s, x+ (v − v∗)s, v∗)dωdv∗ds.

To each of the last two integrals we apply the argument of Lemma 9, and so doing we obtain

‖T−f0 [g]− T−f0 [g]‖ ≤ 4π3b

β2(‖g‖+ ‖g‖)‖g − g‖ ≤ 8π3b

β2R‖g − g‖.

The claim for T−f0 follows.

Exercise 9. Prove the claim of Lemma 12 for T+f0

.

To show that f is indeed a mild solution of the Boltzmann equation, it remains to prove that it isnon-negative. Let RB denote the linear operator

RB(g)(t, x, v) =

∫R3

∫S2

Bg(t, x, v∗)dωdv∗,

using which we can rewrite the loss term as

Q−B(f, f) = fRB(f).

In the following we shall make use of the monotonicity property of R and Q+B , namely

u(t, x, v) ≤ w(t, x, v), ∀(t, x, v)⇒ R(u) ≤ R(w), Q(u, u) ≤ Q(w,w), ∀(t, x, v). (51)

Let us define the sequences lk, uk inductively by

l0 ≡ 0, 0 ≤ u0,

∂tl]k+1 + l]k+1RB(uk)] = Q+

B(lk, lk)], lk+1(0) = f0,

∂tu]k+1 + u]k+1RB(lk)] = Q+

B(uk, uk)], uk+1(0) = f0,

18

Lemma 13. Assume that the beginning condition holds:

u1 ≤ u0, for all (t, x, v) ∈ [0,∞)× R3 × R3 (BC).

Then the sequences uk, lk verify

lk−1 ≤ lk ≤ uk ≤ uk−1, for all (t, x, v) ∈ [0,∞)× R3 × R3.

In particular, lk ↑ l, uk ↓ u and l ≤ u, for all (t, x, v) ∈ [0,∞)× R3 × R3.

Proof. We may write lk as

l]k+1 = f0e−

∫ t0RB(uk)]ds +

∫ t

0

e−∫ tτRB(uk)]dsQ+

B(lk, lk)]dτ

and it follows by induction that lk ≥ 0. We prove the claim of the lemma by induction. Assume

lk−1 ≤ lk ≤ uk ≤ uk−1

holds for some k ≥ 1. Then

l]k+1 − l]k =f0

[e−

∫ t0RB(uk)]ds − e−

∫ t0RB(uk−1)]ds

]+

∫ t

0

[e−

∫ tτRB(uk)]ds − e−

∫ tτRB(uk−1)]ds

]Q+B(lk, lk)]dτ

+

∫ t

0

[e−

∫ tτR(uk−1)]ds

(Q+B(lk, lk)] −Q+

B(lk−1, lk−1)])].

Using (51), it follows that l]k+1 ≥ l]k Similarly we may write uk as

u]k+1 = f0e−

∫ t0RB(lk)]ds +

∫ t

0

e−∫ tτRB(lk)]dsQ+

B(uk, uk)]dτ

by which it follows that u]k+1 ≥ 0 and, arguing as before, u]k+1 ≤ u]k. Moreover, since uk ≥ lk, the

previous equation also gives u]k+1 ≥ l]k+1. Thus we have proved that l]k−1 ≤ l]k ≤ u]k ≤ u]k−1 andsince this is true for all (t, x, v), we may remove the symbol ] and conclude that

lk ≤ lk+1 ≤ uk+1 ≤ uk, for all (t, x, v).

To complete the induction argument, it remains to show that

0 ≤ l1 ≤ u1 ≤ u0.

The inequality u1 ≤ u0 is the beginning condition assumption. To prove 0 ≤ l1 ≤ u1 we write,since Q+

B(l0, l0) = 0,

l]1 = f0e−

∫ t0R(u0)]ds.

This implies from one hand that l1 ≥ 0 and, on the other hand, that l]1 ≤ f0. Moreover, sinceRB(l0) = 0,

u]1 = f0 +

∫ t

0

Q+B(u0, u0)]ds ≥ f0 ≥ l]1.

Proposition 3. There exists R0 = R0(b, β) such that for all R ≤ R0, if f0 ∈ MRβ and u]0 ∈ MR

β ,

then u]k, l]k, u

], l] ∈M2Rβ .

19

Proof. The proof follow by induction as a consequence of the inequalities

l]k ≤ f0 +

∫ t

0

Q+B(lk, lk)]ds

u]k ≤ f0 +

∫ t

0

Q+B(uk, uk)]ds

and the estimates of Lemma 9 and Lemma 10.

Corollary 3. Assume that the beginning condition holds for u]0 ∈ M2Rβ and f0 ∈ MR

β . Then for

a properly small R (depending on b, β), u = l and u] coincides with the solution f ] of (48) fromTheorem 2, which is therefore a mild solution of the Boltzmann equation.

Proof. By the monotone convergence theorem, u, l satisfy

l] = f0 +

∫ t

0

[Q+B(l, l)] −Q−B(l, u)]]ds

u] = f0 +

∫ t

0

[Q+B(u, u)] −Q−B(u, l)]]ds.

Thus the difference u] − l] is

u] − l] =

∫ t

0

[Q+B(u, u− l)] +Q+

B(u− l, l)] +Q−B(l, u− l)−Q−B(u− l, l)]dτ

and reasoning as in the proof of Lemma 12 we get, for a proper smallR, ‖u]−l]‖ < ‖u]−l]‖ ⇒ u = l.Substituting in the equation satisfied by u (or l) we obtain that u is a solution of (48) and therefore,by the uniqueness statement of Theorem 2, must coincide with f ].

Thus the main purpose of this section (proving the existence and uniqueness of mild solutions ofthe Boltzmann equation for small data) has been reduced to construct a function u0 that satisfies

the beginning condition u1 ≤ u0 or, equivalently, u]1 ≤ u]0, and u]0 ∈M2R

β . Observe that

u]1 = f0 +

∫ t

0

Q+(u0, u0)](s, x, v)ds

=

∫ t

0

∫R3

∫S2

Bu]0(s, x+ (v − v′)s, v′)u]0(s, x+ (v − v′∗)s, v′∗)dωdv∗ds. (52)

Let us choose u0 of the formu0(t, x, v) = e−β|x−tv|

2

w(v).

From (52) we obtain

u]1 − u]0 = f0 − e−β|x|

2

+

∫ t

0

∫R3

∫S2

Bw(v′)w(v′∗)e−β(|x+s(v′−v)|2+|x+s(v−v′∗)|

2)dωdv∗ds

= e−β|x|2

∫ t

0

∫R3

∫S2

Be−β|x+s(v−v∗)|2w(v′)w(v′∗)dωdv∗ds,

where we used that

|x+ s(v − v′)|2 + |x+ s(v − v′∗)|2 = |x|2 + |x+ s(v − v∗)|2,

20

which we already used in the proof of Lemma 10. Thus by Lemma 8 we obtain

(u]1 − u]0)eβ|x|

2

≤ supx∈R3

(f0e

β|x|2)− w(v) +

√πb2

β

∫R3

∫S2

w(v′)w(v′∗)dωdv∗.

Letψ(v) = sup

x∈R3

(f0e

β|x|2).Thus a sufficient condition for u]1 ≤ u

]0 is that

T (w)(v) = w(v), (53)

where T is the operator

T (w)(v) = ψ(v) +

√πb2

β

∫R3

∫S2

w(v′)w(v′∗)dωdv∗.

Let G denote the space

G = g ∈ C0(R3) : |g(v)| ≤ αe−β|v|2

, for some α > 0

with norm ‖g‖G = supv |eβ|v|2

g(v)|. Let GR denotes the ball in G with center on g = 0.

Lemma 14. Let f0 ∈ MRβ . There exists R0 = R=0(b, β) such that for all R ≤ R0, T maps G2R

into G2R and it is a contraction.

The proof of this result is left as exercise.

Exercise 10. Prove the previous lemma.

In conclusion we have proved the following

Theorem 4. Consider an initial data 0 ≤ f0 ∈ C0(R3×R3) for the Boltzmann equation such that

f0(x, v) ≤ ce−β(|x|2+|v|2),

for some c, β > 0. Then there exists c0 > 0 such that for all c < c0, the Boltzmann equation has aunique mild solution f ∈ C0(R3 × R3) such that

f(t, x, v) ≤ 2ce−β(|x−vt|2+|v|2).

Exercise 11. Prove that the mild solutions in Theorem 4 satisfy the conservation laws of mass,momentum and energy.

9 Other concepts of solutions

In this section we introduce two more concepts of solution to the Boltzmann equation:

∂tf + v · ∇xf = QB(f, f) (54)

Multiplying (54) by a test function φ ∈ C∞c ([0, T )× R3 × R3) and integrating by parts we obtainthe identity

−∫ T

0

∫R3

∫R3

f(∂tφ+v·∇xφ)dvdxdt+

∫R3

∫R3

φ(0, x, v)f0(x, v)dvdx =

∫ T

0

∫R3

∫R3

φQB(f, f)dxdvdt.

(55)

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Definition 4. Let f0 ∈ L1loc(R3 × R3) a.e. non-negative. An a.e. non-negative function f ∈

L1loc([0, T )×R3 ×R3) is said to be a local solution of the Boltzmann equation (54) in the sense of

distributions with initial datum f0 in the interval [0, T ) iff

• QB(f, f) ∈ L1loc([0, T )× R3 × R3);

• The identity (55) is verified for all φ ∈ C∞c ([0, T )× R3 × R3.

The solution is global if T =∞.

Now let β(f) = log(1 + f). Substituting in the Boltzmann equation we obtain

∂tβ(f) + v · ∇xβ(f) =1

1 + fQB(f, f). (56)

Definition 5. Let f0 ∈ L1loc(R3 × R3) a.e. non-negative. An a.e. non-negative function f ∈

L1loc([0, T )×R3×R3) is said to be a local solution of the Boltzmann equation (54) in the renormalized

sense with initial datum f0 in the interval [0, T ) iff

• QB(f,f)1+f ∈ L1

loc([0, T )× R3 × R3);

• The equation (56) is verified in the sense of distributions.

The renormalized solution is said to be global if T =∞.

The condition that (56) holds in the sense of distributions means that

−∫ T

0

∫R3

∫R3

log(1 + f)[∂t + v · ∇xφ]dvdxdt+

∫R3

∫R3

φ(0, x, v) log(1 + f0)dvdx

=

∫ T

0

∫R3

∫R3

φQB(f, f)

1 + fdvdxdt

holds for all φ ∈ C∞c ([0, T )× R3 × R3).

Theorem 5. Assume that f ∈ L1loc([0,∞)× R3 × R3) is a.e. non-negative. If

Q±B(f, f) ∈ L1loc([0,∞)× R3 × R3), (57)

then the concept of mild, renormalized and distributional solution are equivalent. Under the mildercondition

Q±B(f, f)

1 + f∈ L1

loc([0,∞)× R3 × R3), (58)

the concept of mild and renormalized solution are equivalent.

The celebrated Theorem by Di Perna and Lions prove the existence of global mild solutions of theBoltzmann equation:

Theorem 6. Under suitable conditions on the collision kernel B, which includes the hard spheremodel B = |ω · (v − v∗)|, and for f0 ∈ L1(R3 × R3) such that∫

R3

∫R3

f0(1 + |x|2 + |v|2)dvdx and

∫R3

∫R3

f0| log f0|dvdx are bounded,

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there exists f ∈ C([0,∞), L1(R3×R3)), renormalized solution of the Boltzmann equation. Moreover

Q−B(f, f)

1 + f∈ L∞([0,∞), L1(R3 ×BR)),

Q+B(f, f)

1 + f∈ L1([0,∞), L1(R3 ×BR)),

for all balls BR, thus in particular f is also a mild solution. The solution preserves mass andsatisfies

supt≥0

∫R3

∫R3

f(1 + |x− vt|2 + |v|2 + | log f |)dvdx <∞.

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