Lectures Notes on Boltzmann's Equation

Lectures notes on Boltzmann’s equation

Simone Calogero

1 Introduction

Kinetic theory describes the statistical evolution in phase-space1 of systems composed by a large number 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 which depend only on the local average dynamics of the particles. These are called macroscopic quantities. The most important examples of physical systems to which kinetic theory applies are dilute gases (where the molecules play the role of the particles) and in this case examples of macroscopic quantities are the temperature and the pressure of the gas. In fact, the temperature of a gas is a measure of the mean kinetic energy of the molecules in small regions of the gas, while the pressure is a macroscopic manifestation of the exchange of momentum between the particles and the walls of the gas container (of with any body inserted into the gas). The relation between kinetic and macroscopic 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

3 3 f = f(t, x, v), t ≥ 0 , x ∈ R , v ∈ R . 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 Z Z f(t, x, v) dv dx V Ω gives the probability to find a particle in the region V at time t with velocities v ∈ Ω. Since the number of particles is so large, the above quantity can also be interpreted as the relative number of particles in the region V at time t with velocities v ∈ Ω. Moreover, denoting by M the total mass of the system, the above integral multiplied by M can be interpreted as the total mass of the system in the region V . We shall use freely all these interpretations of the function f, which lead naturally to require that 1 3 3 0 ≤ f(t, ·, ·) ∈ L (R × R ) and that kf(t)k1 = const = 1. The Boltzmann equation is a (integro-)partial differential equation for the one particle distribution function f. Let us “derive” it in the most simple case, that is for a system of particles moving with constant velocities. Suppose (for notational simplicity) that all particles are identical (otherwise we should introduce a particles distribution for each species). If x(t) denotes the position vector at time t and v the (constant) velocity vector of a particle, the position at time t + will be given by

x(t + ) = x(t) + v .

1That is the manifold of all possible positions and velocities 2In 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 probability f(t, x(t), v) to ﬁnd the particle there, it is natural to assume that

f(t + , x(t + ), v) = f(t, x(t), v) , since we know with certainty that the particle has moved with constant velocity v in the -interval of time. Provided f is a suﬃciently smooth function, the above identity is equivalent to the free transport 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 kf(t)k1 = kf0k1. This is consistent with the interpretation 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 + ] the particle undergoes a collision with the other particles of the system. Here the word “collision” is used to refer to a general interaction which takes place in such a short time interval and small region of space that we can safely say that it occurs at time t in the position x. If the result of the collision is 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 new particle with velocity v is gained after the collision. Likewise, if a particle with velocity v is lost after 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 makes f decrease (obviously, if a particle looses velocity v, the probability to have a particle with this velocity 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 f changes as a consequence of the collision. In order to give an explicit form to G and L we have to give more information on how collisions occur. We shall consider only the case of the Boltzmann equation for binary elastic collisions.

Exercise 1. Consider a system of particles moving under the inﬂuence of an external force ﬁeld F = 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 the solution 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 with negligible probability. Moreover the collisions are assumed to be elastic, meaning that, besides the total 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 0 0 Let v, v∗ denote the velocities before the collision of two particles and v , v∗ their velocities after the collision (see Fig. 1). We also assume the particles to have the same mass (which is also conserved in the collision). Then the conservation laws of momentum and kinetic energy take the form

0 0 v + v∗ = v + v∗ (cons. of momentum) , (3) 2 2 0 2 0 2 |v| + |v∗| = |v | + |v∗| (cons. of kinetic energy) . (4)

Suppose the pre-collisional velocities (v, v∗) are given and we want to derive the post-collisional 0 0 velocities (v , v∗) from (3) and (4). Since we have 6 unknowns but only 4 equations, the above system is underdetermined. However one can prove that the manifold of the solution is a 2-sphere. Precisely we have the following

0 0 Lemma 1. A quadruple (v, v∗, v , v∗) solves (3)-(4) if and only if 0 v = v − [(v − v∗) · ω]ω, (5) 0 v∗ = v∗ + [(v − v∗) · ω]ω, (6) for some ω ∈ S2.

0 0 Proof. The claim is true for the trivial solution (v = v, v∗ = v∗) (since it is obtained for ω 0 0 orthogonal to v − v∗). Otherwise, set v∗ = v∗ + a ω, v = v + b ω; from the conservation of 0 0 momentum, equation (3), we have a = −b. Substituting v∗ = v∗ + a ω and v = v − a ω in (4) and solving for a we obtained the desired result with

0 0 v∗ − v∗ v − v ω = 0 = − 0 . |v∗ − v∗| |v − v|

3 The triple (v, v∗, ω) is called a collision conﬁguration. The direction of ω is the scattering direction of the collision. Some important geometrical properties of binary elastic collisions are collected in the following lemma, whose proof is left as exercise. Lemma 2. The following holds:

0 0 0 0 (i) |v − v∗| = |v − v∗|, |(v − v∗) · ω| = |(v − v∗) · ω|; 0 0 0 0 0 0 (ii) v = v − [(v − v∗) · ω]ω, v∗ = v∗ + [(v − v∗) · ω]ω; 0 0 (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 with velocity v results from the collision of two particles. By (ii) of Lemma 2, this is the case if the 0 0 particles collide with velocities v = v − [(v − v∗) · ω]ω and v∗ = v∗ + [(v − v∗) · ω]ω (see Fig. 2 0 0 again). The probability to ﬁnd a particle with velocity v (resp. v∗) in the point x at time t is 0 0 given by f(t, x, v ) (resp. f(t, x, v∗)). The probability to have both particles at the same time in the same position (so that they may collide) is given by the product

0 0 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 with given 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 from their velocities and scattering angle, then the product above would already give the probability density to obtain a new particle with velocity v from the collision of two particles with velocities 0 0 v and v∗. This means the all the infinitely possible collision configurations which are compatible with (3) and (4) have the same probability to occur. However this is in general not the case. The probability of collision of two particles, that is the strength of the interaction, will depend in general on the velocities of the two particles as well as on the scattering angle. In order to measure such a “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 collision of two particles with velocities v, w and scattering angle ω (that is, how probable is the collision configuration (v, w, ω)). The specific form of B depends on the model that we use to represent the particles and the forces which act during the collision. An example will be given in the next section. Having introduced the collision kernel, we can now state that the probability density of gaining a 0 0 particle v from the collision of two particles with velocities v and v∗ is given by 0 0 0 0 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 configuration in Fig. 2. But since we are not interested in the velocity v∗ of the second particle nor on the scattering angle ω, the gain term is obtained by integrating over all possible (v∗, ω): Z Z 0 0 0 0 G(t, x, v) = B(ω, v , v∗)f(t, x, v )f(t, x, v∗)dω dv∗. (7) 3 2 R S

4 The form of the loss term is now easily derived. In fact it is clear from Lemma 1 that a particle with velocity v will be lost whenever one of the two particles had velocity v before the collision4. Thus for the loss term we obtain Z Z L(t, x, v) = B(ω, v, v∗)f(t, x, v)f(t, x, v∗)dω dv∗. (8) 3 2 R S

We have almost finished to derive the Boltzmann equation in its final form. To this end we need to make a further assumption on the collision kernel: The function B = B(ω, v, w) depends on the scattering 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 the 0 0 Boltzmann equation (see Exercise 3). It implies that B(ω, v , v∗) = B(ω, v, v∗) in (7) and so we obtain Z Z 0 0 G(t, x, v) = B(ω, v, v∗)f(t, x, v )f(t, x, v∗)dω dv∗. (9) 3 2 R S Replacing (8) and (9) in (2) we obtain finally the Boltzmann equation Z Z 0 0 ∂tf + v · ∇xf = B [f(t, x, v )f(t, x, v∗) − f(t, x, v)f(t, x, v∗)] dω dv∗ (10) 3 2 R S

where B = B(|v −v∗|, |(v −v∗)·ω|). The right hand side is the collision integral. It will be denoted by 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.,

0 0 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∗)] , R 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 whenever f(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

p 2 p 2 p 0 2 p 0 2 1 + |v| + 1 + |v∗| = 1 + |v | + 1 + |v∗| where now v is the “momentum” rather than the velocity of the particles (the latter being deﬁned asv ˆ = v/p1 + |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 forces which act during the collision. In this section we present an argument to justify the form of B in

4 0 0 The exception is the solution (v = v, v∗ = v∗). However this solution is a set of zero measure on the set of all possible 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 impact between two spheres, an inﬁnite force acts instantaneously on the direction which connects the centers 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 change of momentum of any of the two single particles will also occur along this direction. Thus recalling the deﬁnition of ω in Lemma 1, we have

0 v − v x − x∗ ω = 0 = . |v − v | |x − x∗|

It is also clear by symmetry that the collision of Fig. 3 is equivalent to the collision of a sphere with diameter 2 at rest with a point particle having velocity v − v∗ (see Fig. 4). Now let dω the infinitesimal surface element on the sphere 2. The particles with velocity v − v∗ which collide with the sphere 2 on the surface element dω in the infinitesimal interval of time (t, t + dt) are contained in 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 probability of collisions of two particles, one of which has velocity v, in the same infinitesimal time interval. 3 In the hard sphere model, these are the particles contained in the cylinder C(v, v∗), ∀v∗ ∈ R . 5More rigorously, the Boltzmann equation for hard spheres is derived in the limit when the diameter of the particles tends to zero.

6 That is to say: two particles, one with velocity v and the other one with velocity v∗, which are in the cylinder C(v, v∗) must collide. Thus all the particles with velocity v and v∗ contained in C loose their velocities. By the interpretation of f as a particle density (or as a relative number of particles), 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 of collision 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 then 3 2 integrated over all possible v∗ ∈ R and ω ∈ S (because we are only interested in the probability of loosing a velocity v, regardless the velocity of the other particle and the scattering angle). If the 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 appears in the r.h.s. of the Boltzmann equation. Let T > 0 and φ : (0,T ) × R3 × R3 be a measurable function with values in a ﬁnite dimensional vector space (typically R, or R3). We denote

0 0 0 0 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 necessarily non-negative). We shall also write

0 0 0 0 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 Z Z Z Z 1 0 0 QB(f, f)φdv = qB(f, f)(φ + φ∗ − φ − φ∗)dωdv∗dv, (14) 3 4 3 3 2 R R R S for all (t, x) ∈ (0,T ) × R3 and for all functions φ and f as above such that Z Z Z 3 |qB(f, f)φ| dωdv∗dv < ∞, ∀(t, x) ∈ (0,T ) × R . (15) 3 3 2 R R S

Proof. We have Z Z Z Z 0 0 QB(f, f)φ dv = B(f f∗ − ff∗)φ dω dv∗ dv. (16) 3 3 3 2 R R R S Note that the assumption (15) allows, by Fubini’s theorem, to exchange the order of the integrals. 0 0 0 0 Thus applying the change of variable (v, v∗) → (v∗, v) (which implies (v , v∗) → (v∗, v )) to the right hand side of (16) we obtain Z Z Z Z 0 0 QB(f, f)φ dv = B(f f∗ − ff∗)φ∗ dω dv∗ dv. (17) 3 3 3 2 R R R S

7 0 0 In the right hand side of (17) we make the change of variables (v, v∗) → (v , v∗). Since this transformation leaves invariant the Lebesgue measure (see Lemma 2(iii)), we obtain Z Z Z Z 0 0 0 QB(f, f)φ dv = − B(f f∗ − ff∗)φ∗ dω dv∗ dv. (18) 3 3 3 2 R R R S

Finally, doing again the change of variable (v, v∗) → (v∗, v) we get Z Z Z Z 0 0 0 QB(f, f)φ dv = − B(f f∗ − ff∗)φ dω dv∗ dv. (19) 3 3 3 2 R R R S 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 Z QB(f, f)φ dv = 0, 3 R for almost all (t, x) ∈ (0,T )×R3 and for all measurable functions f such that the bound (15) holds.

0 0 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

0 0 φ(t, x, v) + φ(t, x, v∗) = φ(t, x, v ) + φ(t, x, v∗), (20) 0 0 i.e., φ + φ∗ = φ + φ∗, is veriﬁed if and only there exist a = a(t, x) ∈ R, b = b(t, x) ∈ R and c = 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 of 0 0 the form (21). First we notice that, since the relation between (v , v∗) and (v, v∗) are equivalent to the conservation of momentum and kinetic energy (see Lemma 1), equation (20) is veriﬁed if and only if there exists a function ψ on [0, ∞) × R3 such that 2 2 φ(v) + φ(v∗) = ψ(|v| + |v∗| , 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 that the function ψ is linear in both its variables. Replacing (22) into (20) and setting v∗ = 0 we obtain

2 0 2 0 0 2 0 ψ(|v| , v) + ψ(0, 0) = ψ(|v | , v ) + ψ(|v∗| , v∗), (23) 0 0 where v∗ = (v · ω)ω, v = v − (v · ω)ω and therefore 0 0 0 2 0 2 ∂vi ∂v∗i ∂|v | ∂|v∗| = δij − ωiωj, = ωiωj, = 2vi − 2(v · ω)ωi, = 2(v · ω)ωi. (24) ∂vj ∂vj ∂vi ∂vi

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

2 2 0 2 0 0 2 0 2∂uψ(|v| , v)vi + ∂vi ψ(|v| , v) =2∂uψ(|v | , v )vi − 2∂uψ(|v | , v )(ω · v)ωi 0 2 0 0 2 0 + ∂vi ψ(|v | , v ) − ω · ∂vψ(|v | , v )ωi 0 2 0 0 2 0 + 2∂uψ(|v∗| , v∗)(ω · v)ωi + ω · ∂vψ(|v∗| , 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| 0 0 in (25). Since for such ω there holds (v , v∗) = (0, v), we obtain the identity

2 Aij∂vi ψ(|v| , v) = Aij∂vi ψ(0, 0), (26) where A is the matrix v v A = δ − i j . (27) ij ij |v|2 2 Since the matrix A is invertible, (26) entails ∂vi ψ(|v| , v) = ∂vi ψ(0, 0), whence ψ is linear in the second variable. To prove linearity in the first variable, we proceed likewise, differentiating (25) in vj (note that the terms with ∂vi ∂vj ψ vanish because ψ is linear in the second variable) and evaluating 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 2 φ0(v) = 1, φ1(v) = v, φ2(v) = |v| 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. A measurable function f = f(t, x, v) is said to be a mild solution of the Boltzmann equation in the 3 3 interval [0,T ) with initial datum f0 if f(t, x, v) ≥ 0 for almost all (t, x, v) ∈ (0,T ) × R × R , Z t 3 QB(f, f)(s, x + v(s − t), v) ds is bounded for all (t, x) ∈ (0,T ) × R 0 and the following identity holds for almost all (t, x, v) ∈ (0,T ) × R3 × R3: Z t f(t, x, v) = f0(x − vt, v) + QB(f, f)(s, x + v(s − t), v) ds. (28) 0 The mild solution is said to be global if T = ∞.

Equation (28) deﬁnes the so-called mild formulation of the Boltzmann equation. Note that smooth solutions 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 Z f0|φ0| dv dx < ∞,

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

∞ 1 6 fφ ∈ L ((0,T ); L (R )) and Z Z fφ dv dx = f0φ dv dx, for almost all t ∈ (0,T ). 6 6 R R

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 Z t f(t, x, v)φ(t, x, v) = f0(x − vt, v)φ0(x − vt, v) + QB(f, f)(s, x + v(s − t), v)φ(s, x + v(s − t), v)ds. 0 (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 the identity Z Z Z Z Z t Z Z fφdvdx = f0φ0dvdx + QB(s, x, v)φ(s, x, v)dvdxds. 3 3 3 3 3 3 R R R R 0 R R 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 total momentum (i = 1) and conservation of the total kinetic energy (i = 2). Taking φ = x ∧ v we get the conservation of the total angular momentum. See Section 7 for the deﬁnition of the macroscopic observable quantities.

6 The Entropy identity and the Maxwellian distributions

Lemma 4. Let f be measurable, strictly positive and ﬁnite a.e. such that Z Z Z 3 |qB(f, f) log f| dωdv∗dv < ∞, ∀(t, x) ∈ (0,T ) × R . 3 3 2 R R S Then Z log fQB(f, f) dv ≤ 0 3 R for almost all (t, x) and Z log fQB(f, f) dv = 0 3 R if and only if 2 f(t, x, v) = MA,β,u = A exp(−β|v − u| ), (30) a.e., where A, β, u are functions of (t, x).

Proof. Choose φ = log f in the identity (14). We obtain Z Z Z Z 0 0 ff∗ log fQB(f, f)dv = B(f f∗ − ff∗) log 0 0 dωdv∗dv. 3 3 2 f f R R S ∗ Since Y (X − Y ) log ≤ 0, for all X, Y > 0 X R with equality iﬀ X = Y , we obtain that 3 log fQB(f, f)dv ≤ 0, and the equality holds iﬀ R 0 0 0 0 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 Maxwellian distribution.

At this point it is important to emphasize that MA,β,u is not always a solution of the Boltzmann equation. 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 free transport 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) is called 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 decays rapidly at inﬁnity. Let Z Z H[f](t) = f log fdvdx (31) 3 3 R R the entropy functional. Taking the time derivative and using (11) we obtain

dH Z Z = log fQB(f, f) dvdx. (32) dt 3 3 R R 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 a solution 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, are given by

∂tρ + ∇x · (ρu) = 0, (33a) Du ρ + ∇ · σ = 0, (33b) dt x where ρ is the mass density, v is the velocity field, σ is the stress tensor, and D = ∂ + v · ∇ dt t x is the convective derivative operator. To close the system, one has to add a constitute law which permits to express σ in terms of ρ and v. For instance, for a perfect fluid σ = p(t, x)I, where p is the pressure, and (33) reduce to the Euler equations. For isentropic perfect fluids, the system is closed by assigning an equation of state, i.e., by prescribing p as a function of ρ. For viscous fluids we have σij = p δij + ∇ · v δij + ∂xi vj + ∂xj vi 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 gas can also be approximated by a continuous body. The purpose of this section is to derive the connection between the macroscopic description of the gas, based on the system (33), and the mesoscopic 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 that decays rapidly at infinity in the variable (x, v). Given a collision invariant φ that depends only on v, define Z Z Cφ[f] = fφ dv, Jφ[f] = φfvdv. (34) 3 3 R R 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 a constant, because we assume that each particle preserves its mass). The mass density ρ = ρ(t, x) is defined by Z ρ(t, x) = M f dv. (36) 3 R 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 very small, we can say with good approximation that it has its own velocity. On the other hand, uρdx must equal the total momentum of the particles in the region dx, which is given by Z M dx vf dv. 3 R Thus we obtain the following formula for the bulk velocity: R 3 vfdv u = RR . (37) 3 fdv R Lemma 5. For all sufficiently regular solutions of the Boltzmann equation, the local conservation law 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 Z τij = M vivjfdv. 3 R 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 velocity and is called internal velocity. Notice that even if a given inﬁnitesimal region dx of the gas is at rest (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 which the latter is at rest.

Exercise 7. Prove that the average of the internal velocity is zero, i.e., Z cfdv = 0. 3 R Replacing v = c + u in the deﬁnition of the tensor τ we obtain Z τij = M (ci + cj + uiuj + ciuj + ujci)fdv. 3 R

Using the result of the previous exercise, the integrals containing the mixed terms ciuj vanish. Thus we can write τij = ρuiuj + σij, where Z σij = M cicjfdv (40) 3 R is the stress tensor of the gas. Replacing in (39) we obtain, in components

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

Lemma 6. For all suﬃciently regular solutions of the Boltzmann equation, the local conservation law of momentum is satisﬁed: Du ρ + ∇ · σ = 0. (42) dt x

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 in terms of the particle distribution function f. 1 2 Finally, let us take φ = 2 M|v| ; equation (35) becomes Z Z 1 2 1 2 ∂t M |v| fdv dv + ∇x M |v| vfdv dv = 0. 2 3 2 3 R R using the decomposition v = c + u, the previous equation becomes

1 1 ∂ ρ|u|2 + ρe + ∇ · ( |u|2 + e)ρu + q + σ · u = 0, (43) t 2 x 2 where 1 Z e(t, x) = ρ−1 M |c|2fdv (internal energy) (44) 2 3 R 1 Z q(t, x) = M c|c|2fdv (heat ﬂux) (45) 2 3 R We have proved Lemma 7. For all suﬃciently regular solutions of the Boltzmann equation, the local conservation law of energy (43) holds.

13 The identities derived in this section apply to solutions of the Boltzmann equation. In particular they 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 macroscopic quantities for a generic (not necessarily global) Maxwellian we ﬁnd that:

3 3 3/2 β(t, x) = ,A(t, x) = , (46a) 4e(t, x) 4πe(t, x) 2 σ (t, x) = ρe(t, x)δ , q = 0. (46b) ij 3 ij The function (of (t, x)) 2 1 Z 1 p = ρe = |c|2fdv = Tr(σ), 3 3 3 3 R is the isotropic pressure.

Exercise 8. Proof the identities (46). We conclude this section by proving that the Maxwellians minimize the entropy functional. Let Z h[f](t, x) = f log fdv 3 R 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 function with 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 Z Z Z (f log f − M log M)dv = (f log f − f log M)dv + log M(f − M)dv. 3 3 3 R R R Moreover Z Z Z log M(f − M) = log A (f − M)dv − β |v − u|2(f − M)dv. 3 3 3 R R R Both terms in the r.h.s. of the previous identity vanish, because f and M have the same ρ, u, e. Whence Z Z (f log f − M log M)dv = (f log f − M log M)dv. 3 3 R R Using the elementary identity

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

we obtain Z h[f] − h[M] ≥ (f − M)dv = 0. 3 R Since equality in (47) occurs only for z = y, the proof is complete.

14 8 Global existence and uniqueness of mild solutions for small data

In this section we prove the existence of global, mild, continuous solutions of the Boltzmann equation 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 Boltzmann equation is Z t f(t, x + vt, v) = f0(x, v) + QB(f, f)(s, x + vs, v) ds. 0 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

Z t ] ] f (t, x, v) = f0(x, v) + QB(f, f) (s, x, v), 0

or, in terms of f ] alone,

Z t Z Z ] ] 0 0 ] 0 0 f (t, x, v) = f0(x, v) + B[f (s, x + (v − v )s, v )f (s, x + (v − v∗)s, v∗) 3 2 0 R S ] ] − 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 > 0 and deﬁne the Banach space

0 3 3 −β(|x|2+|v|2) Mβ = {f ∈ C ([0, ∞) × R × R ): |f(t, x, v)| ≤ αe , for some α > 0}, with norm 2 2 kfk = sup eβ(|x| +|v| )|f(t, x, v)|. t,x,v

R Denote by Mβ = {f ∈ Mβ : kfk ≤ R}, the ball of radius R and centered on f = 0 in Mβ. Then R for all β > 0 there exists R0 = R0(β, b) such that for all R ≤ R0 and f0 ∈ Mβ , there exists a ] 2R unique f ∈ Mβ solution of (48).

Note that the solutions of Theorem 2 are NOT yet mild solutions, because it is not claimed that the are non-negative! We divide the proof in several lemmata. Let us begin with a calculus lemma.

Lemma 8. The integral ∞ Z 2 I = e−β|x+τ(v−v∗)| dτ 0 is bounded as rπ 1 I ≤ . β |v − v∗|

15 Proof. We write

∞ 2 Z 2 2 I = e−β|x| e−β[τ |v−v∗| +2τx·(v−v∗)]dτ 0 −β|x|2 Z ∞ e −β[s2+2sx· (v−v∗) ] = e |v−v∗| ds, |v − v∗| 0

v−v∗ 2 where the second equality follows by the change of variables s = τ|v − v∗|. Let η = ∈ S |v−v∗| and rewrite the exponent of the integrand function as

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

Then

−β(|x|2−(x·η)2) ∞ e Z 2 I = e−β(s+x·η) ds |v − v∗| 0 ∞ 1 Z 2 rπ 1 ≤ e−βy dy = . |v − v∗| −∞ β |v − v∗|

+ Next recall that the collision integral QB(f, f) consists of a gain term QB(f, f) and a loss term − QB(f, f), i.e., + − QB(f, f) = QB(f, f) − QB(f, f), where Z Z + 0 0 QB(f, f)(t, x, v) = Bf(t, x, v )f(t, x, v∗)dωdv∗, (49) 3 2 R S Z Z − QB(f, f)(t, x, v) = f(t, x, v) Bf(t, x, v∗)dωdv∗. (50) 3 2 R S + − We remark that QB and QB 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

Z t 3 − ] 4π b ] ] −β(|x|2+|v|2) |QB(f1, f2) (s, x, v)|ds ≤ 2 kf1kkf2ke . 0 β

Proof. By the deﬁnition of the loss term, Z Z − ] ] ] |QB(f1, f2) (t, x, v)| ≤ b|f1(t, x, v)| |v − v∗|f2(t, x + t(v − v∗), v∗)dωdv∗ 3 2 R S 2 2 Z Z 2 2 ] ] −β(|x| +|v| ) −β|x+t(v−v∗)| −β|v∗| ≤ bkf1kkf2ke |v − v∗|e dωdv∗. 3 2 R S Therefore

t ∞ Z 2 2 Z 2 Z 2 − ] ] −β(|x| +|v| ) −β|v∗| −β|x+t(v−v∗)| |QB(f1, f2)ds ≤ 4πbkf1kkf2ke e |v − v∗| e dtdv∗ 3 0 R 0 2 2 rπ Z 2 ] ] −β(|x| +|v| ) −β|v∗| ≤ 4πbkf1kkf2ke e dv∗. β 3 R 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

Z t 3 + ] 4π b ] ] −β(|x|2+|v|2) |QB(f1, f2) (s, x, v)|ds ≤ 2 kf1kkf2ke . 0 β

Proof. To begin with we have the estimate

Z Z 0 2 0 2 0 2 0 2 + ] ] ] −β|x+(v−v )t| −β|v | −β|x+(v−v∗)t| −β|v∗| |QB(f1, f2) | ≤ bkf1kkf2k |v − v∗|e e dωdv∗ 3 2 R S Z Z 2 2 0 2 0 2 ] ] −β|v| −β|v∗| −β[|x+(v−v )t| +|x+(v−v∗)t| ] = bkf1kkf2k |v − v∗|e e dωdv∗, 3 2 R S 2 2 0 2 0 2 where the equality follows from the conservation of energy: |v| + |v∗| = |v | + |v∗| . Now, a direct computation, using the conservation of momentum and energy, shows that

0 2 0 2 2 2 |x + (v − v )t| + |x + (v − v∗)t| = |x| + |x + t(v − v∗)| , whence we obtain

Z Z 2 2 2 2 + ] ] ] −β|v| −β|v∗| −β|x| −β|x+t(v−v∗)| |QB(f1, f2) | ≤ bkf1kkf2k |v − v∗|e e e e dωdv∗ 3 2 R S 2 2 Z 2 2 ] ] −β|v| −β|x| −β|x+t(v−v∗)| −β|v∗| = 4πbkf1kkf2ke e e |v − v∗|e dv∗. 3 R Integrating in time we get

t ∞ Z 2 2 Z 2 Z 2 + ] ] ] −β(|v| +|x| ) −β|v∗| −β|x+t(v−v∗)| |QB(f1, f2) |ds ≤ 4πkf1kkf2ke |v − v∗|e e dtdv∗ 3 0 R 0 3 4π b 2 2 ≤ kf ]kkf ]ke−β(|x| +|v| ), β2 1 2 where the last inequality follows as in Lemma 9. Lemma 11. Let T ± be the operators f0

f Z t Z Z T +[g] = 0 + Bg(s, x + (v − v0)s, v0)g(s, x + (v − v0 )s, v0 )dωdv ds, f0 ∗ ∗ ∗ 2 3 2 0 R S f Z t Z Z T −[g] = 0 − g(s, x, v) Bg(s, x + (v − v )s, v )dωdv ds. f0 ∗ ∗ ∗ 2 3 2 0 R S There exists R (β, b) such that, for all R ≤ R and f ∈ M R, T ± maps M R into M R. 0 0 0 β f0 β β

Proof. By Lemma 9 and Lemma 10 we have

3 3 ± 1 4π b 2 R 4π b 2 kT [g]k ≤ kf0k + kgk ≤ + kgk . f0 2 β2 2 β2

R Now, when g ∈ Mβ we have R 4π3b kT ±[g]k ≤ + R2. f0 2 β2

± β2 ± Whence kT [g]k ≤ R for R ≤ 3 = R . The fact that T [g] is continuous when g is continuous f0 8π b 0 f0 is straightforward.

17 To conclude the proof of Theorem 2 it is enough to prove that, for a properly small R, 1 kT ±[g] − T ±[¯g]k < kg − g¯k. f0 f0 2 For this implies, by the ﬁxed point theorem, that the equation(s) T ±[g] = g have only one solution, f0 which we denote g . Obviously, g = g + g ∈ M 2R solves T [g] = g, where T = T + + T −. ± − − β f0 f0 f0 f0 But this is just our Boltzmann equation in the mild form (for g = f ]), so the existence part of Theorem 2 follows. The uniqueness follows from the fact that kT [g] − T [¯g]k ≤ kT +[g] − T +[¯g]k + kT −[g] − T −[¯g]k < kg − g¯k. f0 f0 f0 f0 f0 f0 Thus we conclude if we prove the following

Lemma 12. There exists R0 = R0(β, b) such that for R < R0 1 kT ±[g] − T ±[¯g]k < kg − g¯k. f0 f0 2

Proof. We prove the claim for T − and leave the statement for T + as exercise. We have f0 f0 Z t h Z Z T −[g] − T −[¯g] = − g(s, x, v) Bg(s, x + (v − v )s, v ) f0 f0 ∗ ∗ 3 2 0 R S Z Z i − g¯(s, x, v) Bg¯(s, x + (v − v∗)s, v∗) ds 3 2 R S Z t Z Z = − (g − g¯)(s, x, v) Bg(s, x + (v − v∗)s, v∗)dωdv∗ds 3 2 0 R S Z t Z Z + g¯(s, x, v) B(g − g¯)(s, x + (v − v∗)s, v∗)dωdv∗ds. 3 2 0 R S To each of the last two integrals we apply the argument of Lemma 9, and so doing we obtain 4π3b 8π3b kT −[g] − T −[g]k ≤ (kgk + kg¯k)kg − g¯k ≤ Rkg − g¯k. f0 f0 β2 β2 The claim for T − follows. f0

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 is non-negative. Let RB denote the linear operator Z Z RB(g)(t, x, v) = Bg(t, x, v∗)dωdv∗, 3 2 R S using which we can rewrite the loss term as − QB(f, f) = fRB(f). + In the following we shall make use of the monotonicity property of R and QB, 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, ] ] ] + ] ∂tlk+1 + lk+1RB(uk) = QB(lk, lk) , lk+1(0) = f0, ] ] ] + ] ∂tuk+1 + uk+1RB(lk) = QB(uk, uk) , uk+1(0) = f0,

18 Lemma 13. Assume that the beginning condition holds:

3 3 u1 ≤ u0, for all (t, x, v) ∈ [0, ∞) × R × R (BC).

Then the sequences {uk}, {lk} verify

3 3 lk−1 ≤ lk ≤ uk ≤ uk−1, for all (t, x, v) ∈ [0, ∞) × R × R .

3 3 In particular, lk ↑ l, uk ↓ u and l ≤ u, for all (t, x, v) ∈ [0, ∞) × R × R .

Proof. We may write lk as Z t ] R t ] R t ] − 0 RB (uk) ds − τ RB (uk) ds + ] lk+1 = f0e + e QB(lk, lk) dτ 0

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

] ] h R t ] R t ] i − 0 RB (uk) ds − 0 RB (uk−1) ds lk+1 − lk =f0 e − e t Z h R t ] R t ] i − τ RB (uk) ds − τ RB (uk−1) ds + ] + e − e QB(lk, lk) dτ 0 t Z h R t ] i − τ R(uk−1) ds + ] + ] + e QB(lk, lk) − QB(lk−1, lk−1) . 0

] ] Using (51), it follows that lk+1 ≥ lk Similarly we may write uk as Z t ] R t ] R t ] − 0 RB (lk) ds − τ RB (lk) ds + ] uk+1 = f0e + e QB(uk, uk) dτ 0

] ] ] by which it follows that uk+1 ≥ 0 and, arguing as before, uk+1 ≤ uk. Moreover, since uk ≥ lk, the ] ] ] ] ] ] previous equation also gives uk+1 ≥ lk+1. Thus we have proved that lk−1 ≤ lk ≤ uk ≤ uk−1 and since 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 QB(l0, l0) = 0, ] R t ] − 0 R(u0) ds l1 = f0e . ] This implies from one hand that l1 ≥ 0 and, on the other hand, that l1 ≤ f0. Moreover, since RB(l0) = 0, Z t ] + ] ] u1 = f0 + QB(u0, u0) ds ≥ f0 ≥ l1. 0

R ] R Proposition 3. There exists R0 = R0(b, β) such that for all R ≤ R0, if f0 ∈ Mβ and u0 ∈ Mβ , ] ] ] ] 2R then uk, lk, u , l ∈ Mβ .

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

Z t ] + ] lk ≤ f0 + QB(lk, lk) ds 0 Z t ] + ] uk ≤ f0 + QB(uk, uk) ds 0 and the estimates of Lemma 9 and Lemma 10.

] 2R R Corollary 3. Assume that the beginning condition holds for u0 ∈ Mβ and f0 ∈ Mβ . Then for a properly small R (depending on b, β), u = l and u] coincides with the solution f ] of (48) from Theorem 2, which is therefore a mild solution of the Boltzmann equation.

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

Z t ] + ] − ] l = f0 + [QB(l, l) − QB(l, u) ]ds 0 Z t ] + ] − ] u = f0 + [QB(u, u) − QB(u, l) ]ds. 0

Thus the diﬀerence u] − l] is Z t ] ] + ] + ] − − u − l = [QB(u, u − l) + QB(u − l, l) + QB(l, u − l) − QB(u − l, l)]dτ 0

and reasoning as in the proof of Lemma 12 we get, for a proper small R, ku]−l]k < ku]−l]k ⇒ 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 of the Boltzmann equation for small data) has been reduced to construct a function u0 that satisﬁes ] ] ] 2R the beginning condition u1 ≤ u0 or, equivalently, u1 ≤ u0, and u0 ∈ Mβ . Observe that

Z t ] ] u1 = f0 + Q+(u0, u0) (s, x, v)ds 0 Z t Z Z ] 0 0 ] 0 0 = Bu0(s, x + (v − v )s, v )u0(s, x + (v − v∗)s, v∗)dωdv∗ds. (52) 3 2 0 R S

Let us choose u0 of the form −β|x−tv|2 u0(t, x, v) = e w(v). From (52) we obtain

t 2 Z Z Z 0 2 0 2 ] ] −β|x| 0 0 −β(|x+s(v −v)| +|x+s(v−v∗)| ) u1 − u0 = f0 − e + Bw(v )w(v∗)e dωdv∗ds 3 2 0 R S Z t Z Z −β|x|2 −β|x+s(v−v∗)|2 0 0 = e Be w(v )w(v∗)dωdv∗ds, 3 2 0 R S where we used that

0 2 0 2 2 2 |x + s(v − v )| + |x + s(v − v∗)| = |x| + |x + s(v − v∗)| ,

20 which we already used in the proof of Lemma 10. Thus by Lemma 8 we obtain s 2 Z Z ] ] β|x|2 β|x|2 πb 0 0 (u1 − u0)e ≤ sup f0e − w(v) + w(v )w(v∗)dωdv∗. 3 β 3 2 x∈R R S Let β|x|2 ψ(v) = sup f0e . 3 x∈R ] ] Thus a suﬃcient condition for u1 ≤ u0 is that T (w)(v) = w(v), (53) where T is the operator s 2 Z Z πb 0 0 T (w)(v) = ψ(v) + w(v )w(v∗)dωdv∗. β 3 2 R S Let G denote the space

0 3 −β|v|2 G = {g ∈ C (R ): |g(v)| ≤ αe , for some α > 0}

β|v|2 with norm kgkG = supv |e g(v)|. Let GR denotes the ball in G with center on g = 0. R Lemma 14. Let f0 ∈ Mβ . 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

0 3 3 Theorem 4. Consider an initial data 0 ≤ f0 ∈ C (R × R ) for the Boltzmann equation such that

−β(|x|2+|v|2) f0(x, v) ≤ ce ,

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

2 2 f(t, x, v) ≤ 2ce−β(|x−vt| +|v| ).

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)

∞ 3 3 Multiplying (54) by a test function φ ∈ Cc ([0,T ) × R × R ) and integrating by parts we obtain the identity Z T Z Z Z Z Z T Z Z − f(∂tφ+v·∇xφ)dvdxdt+ φ(0, x, v)f0(x, v)dvdx = φQB(f, f)dxdvdt. 3 3 3 3 3 3 0 R R R R 0 R R (55)

21 1 3 3 Deﬁnition 4. Let f0 ∈ Lloc(R × R ) a.e. non-negative. An a.e. non-negative function f ∈ 1 3 3 Lloc([0,T ) × R × R ) 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

1 3 3 • QB(f, f) ∈ Lloc([0,T ) × R × R ); ∞ 3 3 • The identity (55) is veriﬁed for all φ ∈ Cc ([0,T ) × R × R .

The solution is global if T = ∞.

Now let β(f) = log(1 + f). Substituting in the Boltzmann equation we obtain 1 ∂ β(f) + v · ∇ β(f) = Q (f, f). (56) t x 1 + f B

1 3 3 Deﬁnition 5. Let f0 ∈ Lloc(R × R ) a.e. non-negative. An a.e. non-negative function f ∈ 1 3 3 Lloc([0,T )×R ×R ) 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 3 3 • 1+f ∈ Lloc([0,T ) × R × R ); • 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

Z T Z Z Z Z − log(1 + f)[∂t + v · ∇xφ]dvdxdt + φ(0, x, v) log(1 + f0)dvdx 3 3 3 3 0 R R R R Z T Z Z Q (f, f) = φ B dvdxdt 3 3 1 + f 0 R R ∞ 3 3 holds for all φ ∈ Cc ([0,T ) × R × R ). 1 3 3 Theorem 5. Assume that f ∈ Lloc([0, ∞) × R × R ) is a.e. non-negative. If

± 1 3 3 QB(f, f) ∈ Lloc([0, ∞) × R × R ), (57) then the concept of mild, renormalized and distributional solution are equivalent. Under the milder condition Q±(f, f) B ∈ L1 ([0, ∞) × 3 × 3), (58) 1 + f loc R R 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 the Boltzmann equation: Theorem 6. Under suitable conditions on the collision kernel B, which includes the hard sphere 1 3 3 model B = |ω · (v − v∗)|, and for f0 ∈ L (R × R ) such that Z Z Z Z 2 2 f0(1 + |x| + |v| )dvdx and f0| log f0|dvdx are bounded, 3 3 3 3 R R R R

22 there exists f ∈ C([0, ∞),L1(R3×R3)), renormalized solution of the Boltzmann equation. Moreover Q−(f, f) Q+(f, f) B ∈ L∞([0, ∞),L1( 3 × B )), B ∈ L1([0, ∞),L1( 3 × B )), 1 + f R R 1 + f R R for all balls BR, thus in particular f is also a mild solution. The solution preserves mass and satisﬁes Z Z sup f(1 + |x − vt|2 + |v|2 + | log f|)dvdx < ∞. 3 3 t≥0 R R