The Linear Boltzmann Equation - · PDF fileThe Linear Boltzmann Equation 1. ... Boltzmann...

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CHAPTER 4 The Linear Boltzmann Equation 1. Introduction One must distinguish between the “linear Boltzmann equation” and the “linearized Boltzmann equation.” The former has no self interaction, just scattering with the medium, whereas the latter is the linearization of the fully nonlinear Boltzmann equation. We will deal with the linear equation here, and will thus be interested in scattering oof a medium. The linear Boltzmann equation is the other main kinetic linear equation, after the free transport equation. It combines free transport with scattering oof a medium. The linear Boltzmann equation is used to model many systems, including neutronic dynamics, radiation transfer, cometary flow and dust particles. The linear Boltzmann equation is (1.1) 8 > > > > > > > < > > > > > > > : @ t f | (t,x,v) + v · r x f | (t,x,v) = ˆ v2V k(t, x, v, v )f (t, x, v )dv | {z } particles changing from vto v velocities - a(t, x, v)f (t, x, v) | {z } particles no longer at velocity v f | t=0 = f 0 . Here, x 2 T d or R d and v 2 V = R d or S d-1 . We also assume that a, k are at least in C 0 b , i.e. bounded continuous functions. With this, we have entered the world of nonreversible equations. Ob- serve that the linear Boltzmann equation involves “exchange phenomena” in velocity, thanks to the integral kernel k. It is thus not possible to only consider v as a parameter. As a consequence of this, there is no longer an explicit solution formula based on characteristics. Characteristic lines are now replaced by stochastic functions from a Markov process. This is because the right hand side of the equation can not be interpreted as some- thing producing characteristics. Instead, we will often regard the right hand 47

Transcript of The Linear Boltzmann Equation - · PDF fileThe Linear Boltzmann Equation 1. ... Boltzmann...

CHAPTER 4

The Linear Boltzmann Equation

1. Introduction

One must distinguish between the “linear Boltzmann equation” and the“linearized Boltzmann equation.” The former has no self interaction, justscattering with the medium, whereas the latter is the linearization of thefully nonlinear Boltzmann equation. We will deal with the linear equationhere, and will thus be interested in scattering o↵ of a medium. The linearBoltzmann equation is the other main kinetic linear equation, after thefree transport equation. It combines free transport with scattering o↵ of amedium. The linear Boltzmann equation is used to model many systems,including neutronic dynamics, radiation transfer, cometary flow and dustparticles.

The linear Boltzmann equation is

(1.1)

8>>>>>>><

>>>>>>>:

@tf |(t,x,v) + v ·rxf |(t,x,v) =

ˆv⇤2V

k(t, x, v, v⇤)f(t, x, v⇤) dv⇤| {z }particles changing from v⇤ to v velocities

� a(t, x, v)f(t, x, v)| {z }particles no longer at velocity v

f |t=0

= f0

.

Here, x 2 Td or Rd and v 2 V = Rd or Sd�1. We also assume that a, k areat least in C0

b , i.e. bounded continuous functions.With this, we have entered the world of nonreversible equations. Ob-

serve that the linear Boltzmann equation involves “exchange phenomena”in velocity, thanks to the integral kernel k. It is thus not possible to onlyconsider v as a parameter. As a consequence of this, there is no longeran explicit solution formula based on characteristics. Characteristic linesare now replaced by stochastic functions from a Markov process. This isbecause the right hand side of the equation can not be interpreted as some-thing producing characteristics. Instead, we will often regard the right hand

47

48 4. THE LINEAR BOLTZMANN EQUATION

side as a source term. We can rewrite the equation as

f(t, x, v) = f0

(x� vt, v) +

ˆ t

0

(Lf)(x� v(t� s), v)ds

where L denotes the linear operator on the right hand side of the equation.

Proof. This is just Duhamel’s principle, and can be checked by dif-ferentiation. Note that the operator @t + v · rx applied to the first termand the term inside the integral gives zero, so the only nonzero term is @tapplied to the integral, which gives Lf , the right hand side of the equation,as desired. ⇤

Observe that this yields a nice integral equation for f , but not an explicitformula.

2. The stochastic interpretation

As remarked above, we no longer have the method of characteristicsreadily available to us, but we can use some stochastic methods to arriveat a method of solution. We will take a simple version of (1.1) by taking

V = Sd�1

a = � > 0 where � is a constant

k =�

|Sd�1|This turns the equation into

(2.1) @tf + v ·rxf = �[hfi � f ]

where

hfi = 1

|Sd�1|ˆv⇤2Sd�1

f(t, x, v⇤) dv⇤

To construct the stochastic process, we let (⌧n)n�0

be a sequence of inde-pendent random variables in R

+

with exponential law (with parameter �),i.e. for t > 0

P[⌧n > t] = e��t

We thus define for n � 1, the total time after n collisions to be

Tn =n�1X

j=0

⌧j

2. THE STOCHASTIC INTERPRETATION 49

We similarly let (Vn)n�1

be a sequence of independent random variablesvalued in Sd�1 with uniform law, i.e.

P[Vn 2 A ⇢ Sd�1] =1

|Sd�1|ˆA

ds

Furthermore, we require that (⌧n)n�0

and (Vn)n�1

are independent. Giventhis set up, we will determine what the stochastic trajectories are: We define(Xt, Vt) as follows: for 0 t < T

1

Xt = x� tv Vt = v

For T1

t < T2

Xt = XT1 � (t� T1

)V1

Vt = V1

and so on until Tn t < Tn+1

Xt = XTn

� (t� Tn)Vn Vt = Vn.

Given this definition, we have

Proposition 2.14. Defining f(t, x, v) = Ex,v[f0

(Xt, Vt)], for a fixed x, v, f0

,for Xt, Vt defined by the above process, f(t, x, v) solves (2.1) with initial dataf0

.1

Sketch of proof. For some idea of why this is true, notice that (thesecond line can be justified rigorously by conditioning)

f(t, x, v) = Ex,v [f0

(Xt, Vt)1t<T1 ] + Ex,v [f0

(Xt, Vt)1t�T1 ]

= f0

(x� tv, v)e��t + Ex,v [f(t� T1

, X1

, V1

)1t�T1 ]

Then using the law

(T1

, V1

) =�e��t1

|Sd�1| dt1 ds(v1)we have that

f(t, x, v) = f0

(x� vt, v)e��t +

ˆ t

0

� hfi (t� t1

, x� vt1

)e��t1 dt1

To see this, we just calculate the last expectation above

Ex,v[f(t� T1

, X1

, V1

)1t�T1 ] =

ˆ t

0

ˆSd�1

f(t� t1

, x� vt1

, v1

)�e��t1

|Sd�1| ds(v1) dt1

=

ˆ t

0

� hfi (t� t1

, x� vt1

)e��t1 dt1

.

1This is similar to the sense in which the heat equation has Brownian motion as“characteristics.”

50 4. THE LINEAR BOLTZMANN EQUATION

3. Stochastic trajectories, Ito calculus and PDEs (S)

4. The Cauchy Theory

4.1. The notion of solution. As a consequence of the lack of explicitformula anymore, we have to be more careful about our notion of definingsolutions, especially weak solutions. In the free transport equation, weindeed had a clear definition of weak solution, coming from the explicitsolution. In fact we could show uniqueness as distributional solutions ofour notion of weak solutions.

To simplify notation in the linear Boltzmann equation, we define

Kf :=

ˆv⇤2V

k(t, x, v, v⇤)f(t, x, v⇤) dv⇤

and we will assume that we have an additional source term

Q 2 C0([0, T ]⇥ Rd ⇥ V ).

Definition 4.15. We say that f(t, x, v) is a “generalized” (or “weak”)solution to

@tf + v ·rxf = Kf � af +Q

if for all (t, x, v) 2 (0, T )⇥ Rd ⇥ V and s such that t + s 2 (0, T ) then themap s 7! f(t+ s, x+ sv, v) is C1 and

d

ds[f(t+ s, x+ sv, v)] + a(t+ s, x+ sv, v)f(t+ s, x+ sv, v)

= (Kf +Q)(t+ s, x+ sv, v).

We have a general existence theorem for weak solutions:

Theorem 4.16. Under the assumptions

f0

2 Cb(Rd ⇥ V )

Q 2 Cb([0, T ]⇥ Rd ⇥ V )

0 a 2 Cb([0, T ]⇥ Rd ⇥ V )

0 k 2 Cb([0, T ]⇥ Rd ⇥ V ⇥ V )

supt2[0,T ],x2Rd,v2V

����ˆv⇤2V

k(t, x, v, v⇤) dv⇤

���� < M

there is a unique weak solution to the linear Boltzmann equation (with sourceQ) with initial data f

0

.

4. THE CAUCHY THEORY 51

4.2. The Duhamel principle. To prove this, we will find an heuris-tic integral version of the equation through Duhamel’s principle, and thenapply a fixed point argument to find a candidate for weak solution, thenprove existence and uniqueness.

Recall that Duhamel’s principle is basically the idea of solving a linearPDE with a source term and initial data, by solving the PDE without thesource term but with the initial data, and then solving the PDE with thesource term but zero initial data, and then adding the two solutions toobtain a total solution. For example, if the PDE is

8><

>:

@tf + Tf|{z}lin. op.

= U|{z}source

f |t=0

= f0

then we can solve the homogeneous part(@tf 1 + Tf 1 = 0

f 1|t=0

= f0

by

f 1

t = e�tTf0

and we can solve (@tf 2 + Tf 2 = U

f 2|t=0

= 0

by

f 2

t =

ˆ t

0

e�(t�s)TUs ds.

Then, the solution to the original PDE is given by

f 1

t + f 2

t = e�tTf0

+

ˆ t

0

e�(t�s)TUs ds.

Notice that U could depend on f , in which case we have just rewrittenthe PDE as an integral equation. In general, there is an arbitrary choiceof what to include in the source term and what to include in T , and thereis a trade o↵ between obtaining a simpler source term and a simpler T toexponentiate.

Applying this to the linear Boltzmann equation, the first choice we makeis T = v ·rx and Uf = Kf � af +Q. In this case, we can easily write

e�tTg(x, v) = g(x� vt, v)

52 4. THE LINEAR BOLTZMANN EQUATION

so we have

f(t, x, v) = f0

(x� tv, v) +

ˆ t

0

(Kf � af +Q)(s, x� v(t� s), v) ds.

Alternatively, we could let T = v ·rx + a. It is easy to check that

e�tT g(x, v) = exp

✓�ˆ t

0

a(s, x� (t� s)v, v) ds

◆g(x� tv, v)

so we alternatively obtain

f(t, x, v) = f0

(x� tv, v) exp

✓�ˆ t

0

a(s, x� (t� s)v, v) ds

+

ˆ t

0

(Kf +Q)(s, x� (t� s)v, v) exp

✓�ˆ t

s

a(⌧, x� (t� ⌧)v, v)d⌧

◆ds

4.3. Proof of the main result. Now, taking the Duhamel principleas a heuristic, we will prove existence and uniqueness of weak solutions.

Proof. We define

F [f0

, Q] := f0

(x� tv, v) +

ˆ t

0

Q(s, x� (t� s)v, v) ds

⌧f :=

ˆ t

0

(Kf � af)(s, x� (t� s)v, v) ds

Duhamel’s principle suggests that we should look for solutions to

f = F [f0

, Q] + ⌧f

As such, we consider the seriesX

n�0

⌧n[F [f0

, Q]]

and we will show that it converges in Cb([0, T ]⇥Rd⇥V ) (i.e. in L1 norm).First, note that

kF [f0

, Q]kL1(Rd⇥V )

kf0

kL1(Rd⇥V )

+ tkQtkL1(Rd⇥V )

sokF [f

0

, Q]kL1([0,T ]⇥Rd⇥V )

kf0

kL1(Rd⇥V )

+ TkQtkL1(Rd⇥V )

Now, we will prove by induction that

(Hn) k(⌧ng)tkL1(Rd⇥V )

tn(

:=Mz }| {M + kakL1)n

n!kgkL1

([0,T ]⇥Rd⇥V )

4. THE CAUCHY THEORY 53

H0

is trivial, so we will prove that Hn ) Hn+1

.

⌧n+1g(t, x, v) =

ˆ t

0

[K(⌧ng)� a(⌧ng)](s, x� (t� s)v, v) ds

|⌧n+1g|(t, x, v) ˆ t

0

|K⌧ng|+ a|⌧ng|)(s, x� (t� s)v, v) ds.

With the assumptions of the Theorem, we have that

|K(⌧ng)|(s, x� (t� s)v, v) M suph2[0,s]

k(⌧ng)hkL1(Rd⇥V )

.

Further, for 0 s t

|a⌧ng|(s, x� (t� s)v, v) kakL1k(⌧ng)skL1(Rd⇥V )

.

Combining these and Hn gives

|⌧n+1g|(t, x, v) ˆ t

0

MM

nsn

n!kgkL1 ds

and thus

k⌧n+1gkL1(Rd⇥V )

(M)n+1

tn+1

(n+ 1)!kgkL1

completing the induction step. These estimates combine to give

k(⌧n[F [f0

, Q]])tkL1(Rd⇥V )

(M)ntn

n![kf

0

kL1 + TkQkL1 ]

for all n � 0. These combine to show that the original series converges inL1, and because each term is in Cb, it in fact converges in Cb. We denotethe limit

f :=X

n�0

⌧n[F [f0

, Q]].

Finally, by using the bounds on ⌧ it is continuous on Cb, so we have that

f = F [f0

, Q] +X

n�1

⌧n[F [f0

, Q]] = F [f0

, Q] + ⌧

X

n�0

⌧n[F [f0

, Q]]

!.

This implies that f satisfies the Duhamel principle integral equation, i.e.

f = F [f0

, Q] + ⌧f.

Thus, we have f 2 Cb solving

f(t, x, v) = f0

(x� tv, v) +

ˆ t

0

(Kf � af +Q)(s, x� (t� s)v, v) ds.

54 4. THE LINEAR BOLTZMANN EQUATION

Thus, for t 2 (0, T ) and s with t+ s 2 (0, T )

f(t+ s, x+ sv, v) = f(x� vt, v)+

ˆ t+s

0

(Kf � af +Q)(u, x� (t�u)v, v) du.

The right hand side of the equation is C1 in s, and

d

ds[f(t+ s, x+ sv, v)] = (Kf � af +Q)(t+ s, x+ sv, v).

This shows that f is a generalized solution2, and hence we have shownexistence. If we have two solutions f 1, f 2, then their di↵erence g = f 1 � f 2

is a weak solution with f0

= Q = 0. Thus

g = ⌧g.

However,

k(⌧ng)tkL1(Rd⇥V )

(M)ntn

n!kgkL1

but the coe�cient eventually is less than 1 for large enough n, even though⌧ng = g, which implies that g = 0 a.e., and because it is continuous g = 0everywhere. This establishes uniqueness. ⇤

5. Construction of semigroups and Hille-Yosida Theorem (S)

6. Maximum Principle and L1 bounds

As for the free transport equation it is useful to obtain estimates on thesolutions. In fact, it is even more important here, because we do not havean explicit solution formula. We have the following “maximum principle”

6.1. The main result.

Proposition 6.17. If we have f0

, Q, a, k � 0 and all in Cb (on the appro-priate spaces), and furthermore have the usual finite mass assumption onk

supt,x,v

ˆv⇤2V

k(t, x, v, v⇤) dv⇤ < 1

then, the (we have existence and uniqueness by the previous proposition)generalized solution ft is non-negative, i.e. ft � 0 for t � 0.

2Note that this last argument can be reversed (i.e. integrated) to show that anygeneralized solution obeys Duhamel’s principle

6. MAXIMUM PRINCIPLE AND L1 BOUNDS 55

Remark 6.18. For a linear equation, in general, using superposition, wecan drop any sort of sign requirements. In fact, less important than thesign here is the explicit lower bound. In fact, as a general principle, if f1is a stationary (independent of time) in Cb, then f

0

Cf1 implies thatft Cf1 (and similarly with �).

To prove this last part of the remark, in our case, defining gt = ft�Cf1,we have that g

0

0, so by the above proposition, gt 0 for all t � 0, whichis what we claimed.

6.2. Heuristic. One can give a “heuristic proof” of the proposition,by writing down Duhamel’s principle for the equation, and then “inductingon time” to say that the terms in the integrals, which are at time less thant are positive by “induction” and this implies that it is true for t as well.

More precisely we have

f(t, x, v) = f0

(x� tv, v) exp

✓�ˆ t

0

a(s, x� (t� s)v, v) ds

+

ˆ t

0

(Kf +Q)(s, x� (t� s)v, v) exp

✓�ˆ t

s

a(⌧, x� (t� ⌧)v, v)d⌧

◆ds.

In this formula, as soon as the initial data f0

is non-negative, and fs isnon-negative for s < t, we “deduce” that ft is non-negative, and it seemsclear intuitively that the non-negativity is propagated along time.

6.3. Proof of the main result. Let us now give a more rigorous proof,which justifies the previous intuition through a rigorous iterative scheme:

Proof. We know that the solution ft is given by

ft =X

n�0

⌧nt [F [f0

, Q]]

In an exercise above, we checked that the second integral formulation is alsovalid, in which we define

⌧g =

ˆ t

0

Kg(s, x� (t� s)v, v)e�´t

s

a(u,x�v(t�u),v)du ds

F [f0

, Q] = f0

(x� vt, v)e�´t

0 a(s,x�v(t�s),v) ds

+

ˆ t

0

Q(s, x� v(t� s), v)e�´t

s

a(u,x�(t�u)v,v)du ds.

Furthermore, we showed that we have (Cb, k·kL1) bounds on these operatorsmaking the above sum converge in Cb. From their explicit formula, it is not

56 4. THE LINEAR BOLTZMANN EQUATION

hard to see that because f0

, Q � 0, F [f0

, Q] � 0, and if g � 0, Kg � 0,implying that ⌧g � 0 (using k � 0). ⇤

7. The Problem of Relaxation to Equilibrium

We would like to show “entropic convergence” which is a method ofproving long time convergence to some steady state. This is di↵erent thanthe phase mixing and decay phenomena from the previous sections. Wewould like to show that ft ! f1 strongly as t ! +1, where f1 is somesort of thermodynamical equilibrium (this is not a reversible process, so itonly works for positive time).

We will introduce a rigorous form of Boltzmann’s H-theorem3, which issometimes referred to as a spectral gap. We will also discuss hypercoercivity,which is a method for overcoming the lack of entropy production at localequilibrium.

We’ll study the simplest equation possible which still retains the desiredphenomena. As such, we’ll take x 2 Td and v 2 Rd and study8>>>>>><

>>>>>>:

@tf + v ·rxf = ⇢[f ]M � f| {z }:=Lf

M := 1

(2⇡)d/2e�v2/2 (M is often called the “Maxwellian distribution”)

⇢[f ] :=´v2Rd

f(t, x, v) dv

ft=0

= f0

2 L1

x,v \ C1

b

where C1

b is the space of C1 functions such that kfkC1

b

:= kfkL1+krx,vfkL1 <1 Given this setting, the tools we have just proven can be combined in thefollowing exercise

We shall need the following Cauchy theorem, which is proved in thecorresponding example sheet:

Theorem 7.19. Assuming that the initial data is L1

x,v, there exists a uniqueglobal solution in L1

x,v. If furthermore the initial data is Ck with boundedderivates with k > d + 1, then the solution is di↵erentiable in t and x andis a classical solution, and preserves the mass (L1 norm).

7.1. Relaxation to Equilibrium in the Spatially HomogeneousCase. Notice that if f(t, v) does not depend on x for all time (for that itis enough that the initial does not depend on x), then our equation reduces

3This was first proved by Carleman, following a proof in the linear case by Hilbert,relying on techniques of Weyl.

7. THE PROBLEM OF RELAXATION TO EQUILIBRIUM 57

to (@tf = Lf

ft=0

= f0

Note that at equilibrium, f = cM , and the space of such functions, we willdenote by RM . Also note that ⇢[ft] = ⇢[f

0

] and that f0

� 0 implies thatft � 0. Hence it is straightforward to write the following explicit solution

f(t, v) = fin(v)e�t + ⇢M

�1� e�t

�.

The convergence to the equilibrium ⇢M is obvious from this formula, how-ever we shall in this section study this convergence from a purely functionalinequality viewpoint, independent from the flow of the equation. This shallprove useful for the spatially inhomogeneous case.

In order to study convergence to the equilibrium, we introduce a weightedL2 space by the norm

kfk2L 2 :=

ˆf 2M�1 dv

We have an obvious Hilbert space structure on L 2, the space of L2 functionswith finite L 2-norm. This allows us to state the following proposition whichcan be regarded either as a H-theorem or a spectral gap theorem

Proposition 7.20.

(1) The operator L is bounded on L 2 and in fact is symmetric andnon-positive. Its null space is exactly RM .

(2) We have

hLf, fiL 2 :=

ˆ(Lf)(f)M�1 �kf � ⇢[f

0

]Mk2L 2

(3) This implies that

d

dtkf � ⇢Mk2L 2 �kf � ⇢Mk2L 2

(4) By Gronwall’s inequality, this implies that

fte�t/2���! ⇢M

in L 2.

Remark 7.21. Letting ⇧ be the orthogonal projection onto RM , notethat L is the complement (in L2) of ⇧, because it is easy to show that⇧(f) = ⇢[f ]M .

58 4. THE LINEAR BOLTZMANN EQUATION

This proposition gives Liapunov functions which quantify the distance toequilibrium, i.e. entropy kfk2L2 , the H-function �kfk2L2 , or relative entropykf � ⇢Mk2L2 .

Proof. To see the L2 bounds, note thatˆ

⇢[f ]2M2M�1 =

ˆ⇢[f ]2M = ⇢[f ]2 =

✓ˆf dv

◆2

✓ˆ

f 2M�1

◆✓ˆM

Thus k⇢[f ]Mk2L 2 kfk2L 2 .To show non-positivity, we compute the Dirichlet form

hLf, fiL 2 =

ˆ(Lf)(f)M�1

=

ˆv2Rd

ˆv⇤2Rd

f(v)f(v⇤)M(v)M(v)�1 dv⇤ dv �ˆv2Rd

f(v)2

M(v)dv

=

ˆv,v⇤

ff⇤ �ˆv,v⇤

f(v)2

M(v)2M(v⇤)M(v) dv⇤ dv

=

ˆ f

M

f⇤M⇤

� f 2

M

�MM⇤

=1

2

ˆ 2f

M

f⇤M⇤

� f 2

M2

� f 2

⇤M2⇤

�MM⇤

= �1

2

ˆv,v⇤

f

M� f⇤

M⇤

�2

MM⇤ 0

It is not hard to extend this to show that L is symmetric, by showing that

hLf, giL 2 = �1

2

ˆ f

M� f⇤

M⇤

� g

M� g⇤

M⇤

�MM⇤

Using this, we have that

d

dtkftk2L 2 = 2 hLf, fi = �1

2

ˆv,v⇤

f

M� f⇤

M⇤

�2

MM⇤

If we define the “entropy production functional”

D(f) = �1

2

ˆv,v⇤

f

M� f⇤

M⇤

�2

MM⇤

Note that if D(f) = 0, then fM = f⇤

M⇤, which immediately implies that

f = cM , so in fact f = ⇢M .

7. THE PROBLEM OF RELAXATION TO EQUILIBRIUM 59

Using Jensen’s inequality for ' = x2, we have that

D(f) = �1

2

ˆv,v⇤

f

M� f⇤

M⇤

�2

MM⇤

�ˆv

ˆv⇤

f

M� f⇤

M⇤

�M⇤ dv⇤

�2

M dv

=

ˆv

f

M� ⇢[f ]

�2

M dv

= kf � ⇢[f ]Mk2L 2

We have just showed that

d

dtkf � ⇢Mk2L 2 �kf � ⇢Mk2L 2

and using this, we can conclude the remaining claims in the propositionusing Gronwall’s inequality. ⇤Remark 7.22. In terms of spectral theory, the operator L is self adjoint,non-positive, with null space RM , and it has a “spectral gap” of size 1/2.That is, because L is self adjoint, the spectrum is entirely real, and becauseit is nonpositive, it is contained in (�1, 0]. Finally from the spectral gapthe spectrum is included in (�1,�1/2] which follows from from the factthat

kf � ⇢Mk2L 2 e�tkf0

� ⇢Mk2L 2

.Let us prove this last claim and show that if 0 6= � 2 �(L), is in the

spectrum (i.e. (L � �) is non invertible) then � 2 (1,�1/2]. Indeed forany � 2 (�1/2, 0), one has

(L� �)�1 =

ˆ+1

0

e(L��)t dt

exists and provides a bounded inverse to L.

Thus, we’ve found an L2-Hilbert setting in which to describe entropy.However, for nonlinear problems these methods are not in general applica-ble. Instead, what is more commonly used is the Boltzmann entropy andrelative entropy

H(f) :=

ˆf log f dv

H(f |⇢M) :=

ˆf log

f

⇢Mdv

60 4. THE LINEAR BOLTZMANN EQUATION

As a sketch of their use, we give the following computation

d

dt

ˆf log

f

⇢M=

ˆ@tf +

ˆ(@tf) log

f

⇢M

=d

dt

ˆv

f| {z }

=0

+

ˆLf log

f

⇢M

=

ˆv,v⇤

f⇤M logf

⇢M�ˆv

f log f⇢M

=

ˆv,v⇤

0

BB@f⇤M⇤

logf

⇢M| {z }swap v, v⇤

� f

Mlog

f

⇢M

1

CCAMM⇤

=

ˆv,v⇤

f

M

✓log

f⇤⇢M⇤

� logf

⇢M

◆MM⇤

=

ˆv,v⇤

f

M

✓log

f⇤M⇤

� logf

M

◆MM⇤

=1

2

ˆv,v⇤

✓f

M� f⇤

M⇤

◆✓log

f⇤M⇤

� logf

M

◆MM⇤

=1

2

ˆv,v⇤

✓f

M� f⇤

M⇤

◆✓log

f/M

f⇤/M⇤

◆MM⇤

Using the convexity X � Y log XY � 0, this expression is always positive,

and we have equality if and only if f = ⇢M .

7.2. Convergence to Equilibrium in a Confined Setting. Now,we will study 8

><

>:

@tf + v ·rxf = Lf

f |t=0

= f0

x 2 Td, v 2 Rd

Similarly to the homogeneous case, notice that M is still an equilibrium,because rxM = 0. So, as a heuristic, when looking for stationary solutions,we will look for Lf = 0 and v · rxf = 0. We have showed that Lf = 0implies that f = ⇢(x)M , and then v ·rxf = 0 implies that rx⇢ = 0, as vis arbitrary. Thus ⇢ = ⇢1, a constant, which is prescribed by preservationof total mass, because

⇢1 =

ˆx,v

⇢1M =

ˆft =

ˆf0

dx dv

7. THE PROBLEM OF RELAXATION TO EQUILIBRIUM 61

We call ⇢1M the “global equilibrium.”We will now redefine L 2 by the new norm

kfk2L 2 :=

ˆTd⇥Rd

f 2M�1 dx dv

Now, a naive guess would be to repeat the computation from the previoussection. If we do this, we get

d

dtkft � ⇢1Mk2L 2 =

d

dtkftk2L 2

= �2

ˆx,v

(v ·rxf)f| {z }=v·r

x

(f2)

M�1 � (Lf)fM�1

= �2

ˆx,v

(Lf)fM�1

�ˆx,v

(ft � ⇢[f ]M)2M�1

In the second line, we used Green’s theorem to show that the first termvanishes, and after that, we used the computation from the previous section(which did not rely on the fact that f was spatially homogeneous).

Examining the results of this computation, we see that we did not reallyget something useful, as ⇢[f ] 6= ⇢1 in general now, because

´v ft is not

preserved, and can change with time. Thus, we can not use Duhamel’sprinciple, and are not able to show any sort of convergence.

Part of the problem is the degeneracy of the coercivity property of L,namely that in L 2

x,v

NullL = {⇢(x)M}

which does not let us apply the techniques from before. However, we knowthat v · rx “opposes” having ⇢(x) which is not constant. So, intuitively,it won’t “allow” a solution to stop at ⇢(x)M for ⇢(x) not constant. Soour heuristic picture is that L pushes the solution towards ⇢(x)M , a localequilibrium, and then v ·rx pushes the solution towards global equilibrium,both forces ensuring that the other does not push the solution too far inthe wrong direction.

This intuition is the basis for hypocoercivity theorem. It is related tohypoelliptic theory by is instead focused on long term behavior. An example

62 4. THE LINEAR BOLTZMANN EQUATION

of a hypoelliptic4 PDE is(@tf + v ·rxf = �vf

x 2 Td, v 2 Rd

From elliptic regularity, we expect regularization in v, but it is a surpris-ing result that we also have regularization in x, due to similar e↵ects asdescribed above.

We will give some simplified results about hypocoercivity.

Proposition 7.23. There is a functional E which is equivalent to the squareof the L 2 norm, i.e.

C1

kfk2L 2 E(f) C2

kfk2L 2

such thatd

dtE(f � ⇢1M) ��E(f � ⇢1M)

for some � > 0. This implies that E ! 0 exponentially, and that

ftCe��t���! f1 = ⇢1M

in L 2.

We will prove this below, after proving a preliminary lemma.Taking the Fourier transform in x (our results will hold for x 2 Td and

x 2 Rd, in which case ⇠ 2 Zd or Rd respectively), we have that

@tf + i(v · ⇠)f = ⇢(t, ⇠)M � f

We’ll establish the following lemma, which will also be useful in the uncon-fined case.

Lemma 7.24. 5 There are > 0 and K > 0 such that, defining

e[f ](t, ⇠) :=

✓ˆ|f |2M�1 dv

◆(t, ⇠) + Re

✓ˆA⇠[f ]fM

�1 dv

where

A⇠[f ] =�i⇠

1 + ⇠2· ⇢[vf ]M

4This theory is due originally to Hormander.5This is known as an auxilary operator (the term coming from hypoelliptic theory),

and this is a simplification of the work [23] (also inspired by the work [32]). It is alsorelated to the multiplier operator for dispersive PDE and even more to the Kawashimacompensating function, see [38] and also the book [25].

7. THE PROBLEM OF RELAXATION TO EQUILIBRIUM 63

(we are abusing notation slightly, defining ⇢[vf ] =´vf dv), then we have

that

C 01

⇣|f |2M�1 dv

⌘ e(t, ⇠) C 0

2

✓ˆ|f |2M�1 dv

and

@te(t, ⇠) � K⇠2

1 + ⇠2e(t, ⇠).

Proof. We have that

kA⇠[f ]k2L 2v

=

ˆ|A⇠[f ]|2M�1 dv C

✓ˆ|f |2M�1 dv

because

|⇢[vf ]| =����ˆ

vf dv

����

=

����ˆ

vfM1/2

M1/2

����

✓ˆ

|f |2M�1

◆1/2✓ˆ

v2M dv

◆1/2

| {z }<1

.

This proves the first claim, by taking small enough.Now, note that

d

dt

ˆ|f |2M�1 dv =

ˆv

(@tf)fM�1 +

ˆv

(@tf)fM�1

= �ˆ

i(v · ⇠)f fM�1 dv �ˆ

f(iv · ⇠)fM�1 dv +

ˆ(Lf)f + Lf f

= 2

ˆL(Re f) Re fM�1 dv + 2

ˆL(Im f) Im fM�1 dv

�kf � ⇢[f ]Mk2L 2v

.

To show the last line, note that the first two terms cancel, and the finalterm we can bound by using the results in the homogeneous case (regarding⇠ as fixed). Now we write

d

dt

ˆA⇠[f ]fM

�1 dv = �ˆ

A⇠[iv · ⇠f ]fM�1 +

ˆA⇠[Lf ]fM

�1

ˆA⇠[f ]i(v · ⇠)fM�1 +

ˆA⇠[f ]LfM

�1

:= I1

+ I2

+ I3

+ I4

.

64 4. THE LINEAR BOLTZMANN EQUATION

Note that I4

= 0 becauseˆ �i⇠

1 + ⇠2· ⇢[vf ]ML[f ]M�1 dv =

�i⇠

1 + ⇠2· ⇢[vf ]

ˆL[f ]

| {z }=0

dv = 0.

Now, we compute

I1

= �

ˆv

⇠ ⌦ ⇠

1 + ⇠2: ⇢[v ⌦ vf ]fMM�1

= �⇠ ⌦ ⇠

1 + ⇠2: ⇢[v ⌦ vf ]⇢[f ]

= �|⇢[f ]|2 ⇠ ⌦ ⇠

1 + ⇠2: ⇢[v ⌦ vM ]| {z }

=Id

�⇠ ⌦ ⇠

1 + ⇠2⇢[v ⌦ v(f � ⇢[f ]M ]⇢[f ]

= I1,1 + I

1,2.

In the above, we used the following notation for matrix contraction:

(aij) : (bij) =X

ij

aijbij.

Now, note that

I1,1 = �⇠ ⌦ ⇠

1 + ⇠2: Id |⇢[f ]|2

=�⇠2

1 + ⇠2|⇢[f ]|2

and that

|I1,2| |⇠|

1 + ⇠2

������⇢[v (f � ⇢[f ]M)| {z }

:=g

⇢[f ]

������

C|⇠|

1 + ⇠2kgkL 2

v

|⇢[f ]|.This combines to give

I1

� ⇠2

1 + ⇠2+

⇠2

1 + ⇠2kgkL 2 |⇢[f ]|.

Now, we bound I3

:

I3

=

✓ �i⇠

1 + ⇠2· ⇢[vf ]

◆ˆv

(iv · ⇠)fMM�1 dv

=

✓ �i⇠

1 + ⇠2· ⇢[vf ]

◆i⇠ · ⇢[vf ]

7. THE PROBLEM OF RELAXATION TO EQUILIBRIUM 65

=|⇠ · ⇢[vf ]|21 + ⇠2

=|⇠ · ⇢[vg]|21 + ⇠2

.

In the last line, we used that

⇢[v⇢[f ]M ] = ⇢[f ]

ˆv

vM = 0.

By Cauchy-Schwartz, this gives that

I3

C 0|⇠|21 + ⇠2

kgk2L 2v

.

Finally, we have that

I2

=�i⇠

1 + ⇠2· ⇢[vL[f ]]⇢[f ] = i⇠

1 + ⇠2⇢[vL[g]]⇢[f ]

where we used that L[⇢[f ]M ] = L[M ]⇢[f ] = 0 because L[M ] = 0. Thus, wehave that

|I2

| C 00|⇠|1 + ⇠2

kgkL 2v

|⇢[f ]|.Now, gathering all of these terms, and using the inequality (for ✏ > 0)

|ab| 1

2✏a2 +

2b2

with a = kgkL 2v

and b = |⇠|1+⇠2 |⇢[f ]|, we get that

@te �1� C 0⇠2

1 + ⇠2� C 00

2✏� C

2✏

�kgk2L 2

v

⇠2

1 + ⇠2� C 00✏⇠2

2(1 + ⇠2)2� C✏⇠2

2(1 + ⇠2)2

�|⇢[f ]|2.

We can take ✏ small enough so that

C 00⇠2

2(1 + ⇠2)2+

C⇠2

2(1 + ⇠2)2

� 1

2

⇠2

1 + ⇠2

Note that we can choose ✏ = ✏(C,C 00), independently of ⇠. Secondly, set small enough so that

C 0⇠2

1 + ⇠2+

C 00

2✏� C

2✏

� 1

2

and notice that = (✏, C, C 0, C 00) can also be chosen independently of ⇠.

66 4. THE LINEAR BOLTZMANN EQUATION

Now, with these choices, we have that

@te �1

2kgk2L 2

v

2

✓⇠2

1 + ⇠2

◆|⇢[f ]|2.

We may as well assume that 1, and using

kf � ⇢[f ]Mk2L 2v

+ |⇢[f ]|2 =ˆv

[(f � ⇢[(f ]M)2 + |⇢[f ]M |2]M�1 dv � kfk2L 2v

we have that

@te �

2

⇠2

1 + ⇠2kfk2L 2

v

2C1

⇠2

1 + ⇠2e(t, ⇠).

Defining the constant in front to be K, we have finished the lemma. ⇤It is important to note that this lemma does not depend on whether

x 2 Td or Rd, as it is a “mode-by-mode” study of the problem. Now, thislemma gives the following in the confined setting:

Proposition 7.25. There is a function E[f ], equivalent to the square ofthe L 2 norm, i.e.

C1

ˆx2Td

ˆv2Rd

f 2M�1 dx dv E[f ] C2

ˆx2Td

ˆv2Rd

f 2M�1 dx dv

such that there is � > 0 with6

d

dtE[ft � f1] ��E[ft � f1]

where f1 = ⇢1M with ⇢1 =´x,v f0. As a consequence, we have that

kft � f1k2L 2x,v

Ce��tkf0

� f1k2L 2x,v

Proof. Defining, for ⇠ 2 Zd

e(t, ⇠) := e[ \ft � f1](⇠)

where e is as defined in the above Lemma 7.24, we have that

@te �K⇠2

1 + ⇠2e

For ⇠ 6= 0, we have that

@te �K

2e

6It is important to note that � does not depend on f0.

7. THE PROBLEM OF RELAXATION TO EQUILIBRIUM 67

and when ⇠ = 0, note that A0

(g) = 0 for any g, so

e(t, 0) =

ˆ ���ˆx

(f(t, x, v)� f1(v) dx| {z }

:=hfi

���2

M�1 dv = k hfi k2L 2v

.

Because f1 is a stationary solution, we have that

@t(f � f1) + v ·rx(f � f1) = L(f � f1).

Integrating this, the x-divergence term vanishes by Green’s theorem, so wehave that

@t hfi = L hfi .By our spatially homogeneous results, we have that

@tk hfi k2L 2v

�k hfi k2L 2v

which implies that

@te(t, 0) �e(t, 0).

Now, we may assume that K 1, and as such, combining the above resultsgives

@te �K

2e

so, defining � = K2

, and letting

E(t) =X

e(t, ⇠)

we have just shown that

d

dtE(t) ��E(t).

To see that E is equivalent to the square of the L 2 norm, we sum theequivalence for e and then use Plancherel’s theorem

C1

ˆ X⇠

|f |2M�1 dv E(t) C2

ˆ X⇠

|f |2M�1 dv

which implies that

C 01

ˆx,v

|f |2M�1 dv E[f ] C 02

ˆx,v

|f |2M�1 dv.

This finishes the proof. ⇤

68 4. THE LINEAR BOLTZMANN EQUATION

7.3. Relaxation to Zero in an Unconfined Setting. As a heuristic,a stationary solution, f1, to

(7.1)

8><

>:

@tf + v ·rxf = Lf

f |t=0

= f0

x 2 Rd, v 2 Rd

should havev ·rxf1 = 0

andLf1 = 0.

These combine to imply that f1 = cM for some constant c. By integrabilityin x, we have that c = 0. Thus, we expect dispersion. Hence in this casethe transport operator “wins” over the collision operator in the long time,and imposes the same algebraic decay rate as for the heat equation. Assuch, we have the following proposition, which follows from Lemma 7.24.

Proposition 7.26. For ft a solution to (7.1), we have the bounds

kftk2L 2v

C

(1 + t)d/2

hkf

0

k2L 2 + kf0

k2L 2v

(L1x

)

i

where

kf0

k2L 2v

(L1x

)

=

ˆRd

����ˆTd

|f0

| dx����2

M�1 dv.

Proof. Lemma 7.24 gives that for ⇠ 2 Rd

@te �K⇠2

1 + ⇠2e

so by Gronwall, we have that

e(t, ⇠) e�Kt⇠

2

1+⇠

2 e(0, ⇠)

so if we define

E(t) =

ˆ⇠

e(t, ⇠) d⇠ =

ˆ|⇠|1

e(t, ⇠) d⇠ +

ˆ|⇠|>1

e(t, ⇠) d⇠ := I1

+ I2

then we have the bounds

I2

ˆ|⇠|>1

e�Kt⇠

2

1+⇠

2 e(0, ⇠) d⇠

e�Kt/2

ˆ⇠

e(0, ⇠)

10. EXERCISES 69

e�Kt/2

ˆ⇠,v

|f |2M�1 dv d⇠

Ce�Kt/2kfk2L 2

and

I1

ˆ|⇠|1

e�Kt⇠

2

1+⇠

2 e(0, ⇠) d⇠

sup|⇠|1

|e(0, ⇠)|! ˆ

|⇠|1

e�Kt⇠

2

2 d⇠.

For t � 1, under the substitution ⌘ =pt⇠ˆ

|⇠|1

e�Kt⇠

2

2 d⇠ =1

td/2

ˆ|⌘|p

t

e�K⌘/2d⌘ C

td/2.

Furthermore, we can bound the supremum by

sup|⇠|1

|e(0, ⇠)| ˆv

sup|⇠|1

|f0

(⇠, v)|2M�1 dv = C

ˆv

✓ˆx

|f0

| dx◆

2

M�1 dv.

For t < 1 we can use the bounds from the fixed point construction, i.e.kftkL 2 Ckf

0

kL 2 , and the above bound holds for t � 1. Combining all ofthis proves the proposition.

8. Hypocoercivity and decay of semigroups (S)

Present the construction of the norm of coercivity in general in a Banachfor a decaying semigroup.

9. Bibliographical and historical notes

10. Exercises