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CHAPTER 3

Linear Transport Equations

The fundamental transport operator is that of free transport

(0.1)@

@t+ v ·rx

We can think of this as evolving x ! x + v dt and v ! v in time dt, i.e.the motion of a particle which feels no force. If there is an external forceF (t, x, v) then we get the linear transport equation

(0.2)@f

@t+ v ·rxf + F ·rvf = 0

Still remaining in the context of linear kinetic equations, We could alsoadd a scattering term on the right, representing scattering o↵ of a medium,but we will not worry about this for now. With scattering, (0.2) is oftencalled the “linear Boltzmann equation”, and we shall study it in the nextchapter.

1. Construction of solutions: Lagrangian and Eulerianviewpoints

1.1. The Lagrangian viewpoint. We will take advantage of the un-derlying particle trajectories to gain information about the linear transportequation (0.2). This is the Lagrangian viewpoint of this PDE.

Definition 1.2. Given a force field F (t, x, v) 2 C0

t C1

x,v, we call the char-acteristics of the PDE (0.2) the solutions of the following ODE system forx, v 2 Rd

(1.1)

8>>>>><

>>>>>:

dX

dt(t, x, v) = V (t, x, v)

dV

dt(t, x, v) = F (t,X(t, x, v), V (t, x, v))

X(0, x, v) = x, V (0, x, v) = v.

31

32 3. LINEAR TRANSPORT EQUATIONS

This is linked to (0.2) in the opposite sense of the Liouville construc-tion, taking the original ODE and ending up with a PDE. The characteristiccurves are the same information as in (0.2) but in that case we were con-sidering things statistically. Note that if we take d = 3N , for example,then we can describe N particles moving in spacial 3 dimensions with theseequations, and these are just the regular Newtonian equations of motion.

Proposition 1.3. Assume that F 2 C1(R ⇥ Rd ⇥ Rd)1 and we have someC > 0 so that F satisfies the following growth condition

(G) 8t 2 R, x 2 Rd, v 2 Rd, |F (t, x, v)| C(1 + |x|+ |v|).Then, for any x, v 2 Rd, there exists a unique solution to the characteristicODE’s (1.1) for all t 2 R. Moreover for any t 2 R the “characteristicsmap”

S0,t(x, v) := (X(t, x, v), V (t, x, v))

is a C1-di↵eomorphisme on Rd ⇥ Rd.

Exercise 10. Prove this proposition using the Picard-Lindelof ODE exis-tence theorem and the “Lemma of leaving compact sets”.

Now, with this ODE tool at hand, let us solve the Cauchy problem forthe transport equation, with even needing the compact support from thelast section.

Proposition 1.4. For f0

2 C1

x,v, F 2 C1

t,x,v satisfying (G), (0.2) admits aunique C1 solution with initial data f

0

, given implicitly by

8t 2 R, x 2 Rd, v 2 Rd, f(t,X(t,X(t, x, v), V (t, x, v)) = f0

(x, v).

Note that this actually gives a well defined implicit definition of f , be-cause since S

0,t is a C1-di↵eomorphism, if we call the inverse function St,0

(which corresponds to solving backward the characteristics ODE’s) we havethe explicit formula

f(t, x, v) = f0

(St,0(x, v)).

Proof. By definition, f(t, x, v) = f0

(St,0(x, v)) which is C1 by compo-sition (using again the solutions X and V to the characteristics ODE’s areC1). Moreover the solution constructed thus constructed satisfies f(t,X, V ) =f0

(x, v) and since the composed function is C1, so taking the total derivativein time, we have that

0 =d

dt[f(t,X(t, x, v), V (t, x, v))]

1For simplicity, we are assuming more regularity than is necessary: C0 in all variablesand locally Lipschitz in the variables x and v would be su�ciant.

1. CONSTRUCTION OF SOLUTIONS: LAGRANGIAN AND EULERIAN VIEWPOINTS33

=@f

@t

��t,X,V

+rxf��t,X,V

· @X@t

+rvf��t,X,V

· @V@t

=@f

@t

��t,X,V

+ V ·rxf��t,X,V

+ F ·rvf��t,X,V

This shows that f satisfies the transport equation at any point

(t,X(t, x, v), V (t, x, v)) = (t, S0,t(x, v)).

Since St is a C1-di↵eomorphism, and in particular bijective, this implies thatf satisfies the transport equation at any point (t, x, v) which concludes theproof. ⇤Remark 1.5. We still need to study the characteristic mapping and itsjacobian. One important thing to keep in mind is that if F depends ontime (as we are allowing it to) then we do not get a semigroup! That is,it is not necessarily true that St+s = St � Ss. What we have of course isS0,t = Ss,t �S0,s. However, we have that as long as rv ·F = 0, the Jacobian

of S0,t is 1.

Proposition 1.6. For F 2 C1 with rv · F = 0, defining J(t, x, v) =det(DX,DV ), we have that J(t, x, v) = 1 for all t.

The proof of this is exactly the same as solution to (2) in Exercise 2.Note that the proof shows that, even without assuming that rv · F = 0

@J

@t= (rv · F )J

and we can solve this as

J(t, x, v) = J(0, x, v) exp

✓ˆ t

0

(rv · F )(s, x, v)ds

◆

which shows that becase J(0, x, v) = 1, J(t, x, v) is never zero.

Exercise 11. Redo the proof of this proposition in the more abstract formwhere (X, V ) solves a Hamiltonian ODE

8>>><

>>>:

X = rvH|t,X,V

V = �rxH|t,X,V

X(0, x, v) = x

V (0, x, v) = v

Exercise 12. For rv · F = 0, give a new proof of the Lp, L1 bounds ob-tained in the last section, using the explicit solution and Jacobian compu-tation above. Moreover, prove that kftk1 = kf

0

k1, not just an inequality.


1.2. The Eulerian viewpoint. We will study solutions to (0.2) withx, v 2 Rd. We will further assume that the force is divergence free invelocity, i.e. that rv · F = 0. Under this general assumption, we have thefollowing a priori estimates:

Proposition 1.7. Consider f is a C1 solution on [0, T ] to the linear trans-port equation (0.2) with initial data f(t = 0, x, v) = f

0

(x, v) such that

9R > 0, 8t 2 [0, T ], supportft ⇢ B(0, R).

Then for all t � 0 and p 2 [1,1)

(1.2) kf(t, x, v)kLp

(Rd

x

⇥Rd

v

)

= kf0

(x, v)kLp

(Rd

x

⇥Rd

v

)

.

Also, for p = 1, we have

(1.3) kf(t, x, v)kL1(Rd

x

⇥Rd

v

)

kf0

(x, v)kL1(Rd

x

⇥Rd

v

)

.

Furthermore, if f0

� 0, then f � 0 for all t � 0.We will in fact prove (what will turn out to be stronger than (1.2) and

(1.3)) that for � 2 C1(R;R) then

(1.4)

ˆRd

x

⇥Rd

v

�(f(t, x, v)) dx dv =

ˆRd

x

⇥Rd

v

�(f0

(x, v)) dx dv

Proof. We will prove (1.4) and then show that this implies the rest ofthe proposition. Letting � 2 C1(R;R), we claim that if f is a C1 solutionto the transport equation satisfying the support condition above, then so is�(f). This is a simple application of the chain rule:

@

@t�(f)+rx·(v�(f))+F ·rv(�(f)) = �0(f)

✓@f

@t+ v ·rxf + F ·rvf

◆= 0.

This shows that �(f) is a solution to the same transport with initialconditions �(f

0

). Integrating this over Rdx ⇥ Rd

v, the divergence terms arezero by Green’s theorem (note that we have used that rv · F = 0 to moveF inside and outside of the v-derivative term), so we thus have shown that

@

@t

ˆ�(f) dx dv =

ˆ@

@t�(f) dx dv = 0

which establishes (1.4). We readily deduce the statement of conservationof Lp norms for p 2 (1,+1), i.e. (1.2) for p 2 (1,+1).

Now, to prove the rest of the proposition, we shall perform the firstinstance of a regularisation argument. Let �✏(s) = |s|�✏(|s|), with �✏ 2[0, 1], �✏(s) % 1 as ✏ ! 0, and �✏ 2 C1((0,1)) (notice that this meansthat �✏(0) = �0

✏(0) = 0, so we may extend it to a C1 function on all of R).

1. CONSTRUCTION OF SOLUTIONS: LAGRANGIAN AND EULERIAN VIEWPOINTS35

For example, to construct such a �✏, let ' be a C1 function, with compactsupport in (0, 3), increasing and equal to 1 on [1, 2]. Then, letting

�✏(s) =

('(s/✏) s ✏

1 ✏ s,

it is not hard to see that this has the desired properties. Then applying(1.4) to �✏(f), and letting ✏ ! 0, we have obtained conservation of the L1

norm, i.e. (1.2).Now for p = 1, the proof shall make use of an interesting classical

techniques. First suppose that f0

� 0. Then, using �✏(s) = �✏(�s), noticethat because f

0

� 0, �✏(f0) ⌘ 0 (because it vanishes on [0,1)). Thus, by(1.4) ˆ

�✏(f(t, x, v)) dv dx = 0

but because the integrand is positive and continuous, this implies that�✏(f(t, x, v)) = 0, and thus ft � 0 for t � 0. Now let us generalize theexample: consider some initial data f

0

with L1 bound kf0

kL1 , i.e.

8x, v 2 Rd, �kf0

kL1 f0

(x, v) kf0

kL1 .

We then consider a C1 mollified cuto↵ function �✏ on R+ which vanishes on[0, kf

0

kL1 , and is identically one on [kf0

kL1 + ✏,+1). Then we check that�✏(|f) is C1 solution with the compact support property, and therefore

8t 2 [0, T ],

ˆ�✏(|ft|) =

ˆ�✏(|f0|) = 0

which implies that |ft| does not values in [kf0

kL1 + ✏,+1). Since it is truefor any ✏ > 0 it concludes the proof. ⇤Remark 1.8. Another very natural argument, which cannot be performedhere only due to artificial obstacles since we restrict ourselves to functionswith compact, is the following: if f

0

2 L1, then letting

f0

= f0

� kf0

k1we obtain a new solution

ft = ft = kf0

k1for t 2 [0, T ]. Then since f

0

0, we deduce that f 0 for all t � 0.Thus, f kf

0

k1. Applying the same argument to �f gives the desiredinequality:

kfk1 kf0

k1.


This argument can be made rigorous by showing that any solution canbe approximated by compactly supported solutions. This is another instanceof a regularisation argument, used here to justify this a priori estimate. Ifused for the Lp norms we would obtain that (1.2) holds in the sense thatthere is equality if the RHS is finite, and if the RHS is infinite, then theLHS is as well.

We shall in any case revisit the proof of these L1 bound estimates in amore general setting, using the so-called characteristics method, and showthat there is indeed equality.

2. Dispersion Estimates

What is dispersion? In wave mechanics, it is the phenomenon in whichwaves of di↵erent frequencies travel at di↵erent velocities, e.g. in optics lighttraveling in a dispersive medium. This leads to wave packets spreading outand dispersing at infinity, and other phenomenon. In particle dynamics,particles disperse because they have all di↵erent velocities, which to theirspatial density to“spread out” then “escape at infinity,” which is really thephenomenon we are interested in. This means that there are two aspects inthe dispersion phenomenon: a local one and a global one in the all space.

As an aside, notice that from a quantum mechanical viewpoint, thesebecome the same phenomenon, when the (quantized) velocity is associatedwith the frequency of waves.

We will study this phenomenon in the simple setting of x, v 2 Rd forthe free transport equation, which we recall is

(2.1) @tf + v ·rxf = 0

but this could be studied in more general settings.2

The results below can be found in a 1996 paper by Castella-Perthame,however this is just an paradigmatic example. The key underlying mecha-nism is the fact that in phase space the volume is conserved, but it tendsto “stretch in the x direction,” as can be seen in Figure 1.

This is a nice heuristic, how do we take advantage of it mathematically?The answer is to translate it into an interplay between integrability anddecay in time. The first dispersive estimate that we have is:

2In general, one must assume that the characteristics are not “trapped,” escape atinfinity and are “flat enough” which is requiring some bound on the gradient of the force,or on the curvature of the manifold.

2. DISPERSION ESTIMATES 37

Figure 1. Over time, a domain stretches in the x directionunder the flow of the linear transport equation, while main-taining the same volume in phase space.

Lemma 2.9. If f is a solution to the free transport equation(@tf + v ·rxf = 0

f |t=0

= f0

and f 2 L1 \ L1x,v for all p 2 [1,1], then we have for t � 0 that

(2.2) kftkL1x

(L1v

)

|t|�dkf0

kL1x

(L1v

)

.

Note that �d⇣

1

p � 1

q

⌘ 0, so this gives us information about the decay

of the norms as t ! 1.

Proof. With no force, we have the exact representation

f(t, x, v) = f0

(x� tv, v)

and thus for t > 0

kftkL1x

(L1v

)

= supx2Rd

��ˆRd

|f0

(x� vt, v)| dv��

supx2Rd

��

ˆRd

supy2Rd

|f0

(x� vt| {z }:=V

, y)| dv��

=1

tdsupx2Rd

��

ˆRd

supy2Rd

|f0

(V , y)|dV��


=1

td

ˆRd

supy2Rd

|f0

(V , y)|dV

=1

tdkf

0

kL1x

(L1v

)

⇤By an instance of an interpolation argument idea we can deduce the

more general following Corollary

Corollary 2.10. If f is a solution to the free transport equation(@tf + v ·rxf = 0

f |t=0

= f0

and f 2 Lp for all p 2 [1,1], then for 1 p q 1 we have for t � 0that

(2.3) kftkLq

x

(Lp

v

)

|t|�d( 1p

� 1q

)kf0

kLp

x

(Lq

v

)

.

Proof. By conservation of mass, e.g. (1.2), we have that

kftkLp

x

(Lp

v

)

= kf0

kLp

x

(Lp

v

)

and thus, combining these two bounds with Riesz-Thorin Interpolation givesthe desired bounds. ⇤

We also in general expect transfer of regularity, from v to x. This canbe made precise in the following lemma

Lemma 2.11. Again, if f is a solution to the free transport equation(@tf + v ·rxf = 0

f |t=0

= f0

and for f0

is regular enough, then for s 2 Nk(trx +rv)

sftkL1x,v

= k(rv)sfkL1

x,v

.

Note that the operator on the left commutes with itself, so raising it to s iswell defined, without any ambiguity.

Proof. Observe that

[rx, @t + v ·rx] = [trx +rv, @t + v ·rx] = 0

where [A,B] = AB � BA is the commutator. The first commutator beingzero is trivial, and the second is a short computation. As a consequence ofthis, because f is a solution of the free transport equation, then so is rxfand (trx +rv)f .

3. DISPERSION ESTIMATES FOR WAVE OPERATORS (S) 39

Exercise 13. Show that

[AB,C] = A[B,C] + [A,C]B

and more generally that"

sY

i=1

Ai, C

#=

sX

i=1

A1

· · ·Ai�1| {z }I if i=1

[Ai, C]Ai+1

· · ·As| {z }I if i=s

(even if the Ai’s do not commute.As an application prove that

[(trx +rv)s, @t + vrx] = 0

for s 2 N.Given this exercise, it is clear that if f solves the linear transport equa-

tion, then@t [(trx +rv)

sf ] + v ·rx [(trx +rv)sf ] = 0

and by conservation of L1 norm, we have that

k((trv +rx)sf)|tkL1

v,x

= krsvf0kL1

x,v

as desired. ⇤One must be careful with this result. Because there might be cancel-

lation, we cannot draw direct information about x-derivatives of f . Forexample, because f(t, x, v) = f

0

(x � tv, v), it is clear that rsvf = O(ts),

which makes the lemma hard to use.Note that it also gives the simpler result that

krsxftkL1

x,v

= krsxf0kL1

x,v

Exercise 14. Show that the previous two results extend to f a solution ofthe more general equation

@tf + a(v) ·rxf = 0

where a(v) 2 C1(Rd;Rd) is a “uniform di↵eomorphism,” i.e. we have thetwo sided bound

0 < C1

| Jac a(v)| C2

for all v 2 Rd.

3. Dispersion estimates for wave operators (S)

State and prove basic dispersion estimates for the wave equation andfor the Schrodinger equation in the whole space.

Explain the link between transport and wave operators, through theWigner’s transform formalism.


4. A crash-course on interpolation theory (S)

State the Riesz-Thorin interpolation theorem and give the two proofs(real and complex interpolation), maybe also sketch the original proof (byapproximating by sequence). Finish with the abstract general notion ofinterpolation space. Emphasize the intuition to deduce from it for linearproblems: “it is enough to prove functional inequalities at extremal val-ues of the coe�cients” (remark it can take several interpolation steps forreconstructing the whole convex hull of parameters).

5. Averaging Lemma

As remarked above, the regularity transfer estimates of the previoussection are quite di�cult to use. Motivated to find better regularity state-ments, we will prove that if we average solutions to the free transport equa-tion in velocity, then we have improved regularity estimates. As such weprove

Lemma 5.12. For f 2 L2(Rt ⇥Rdx ⇥Rd

v), a solution to the inhomogeneousfree transport equation

(@tf + v ·rxf = s

f(t = 0) = f0

for f0

(x, v) 2 L2

x,v and s(t, x, v) 2 L2

t,x,v then for ' 2 L1c (Rd), a compactly

supported L1 “test function” we have that there is C > 0 such that��ˆ

f(·, ·, v)'(v) dv��L2t

(H1/2x

)

ChkfkL2

t,x,v

+ kskL2t,x,v

i

Recall that we define the fractional-1/2 Sobolev norm

kgkH1/2

x

:=

✓ˆ|g(⇠)|2(1 + |⇠|)d⇠

◆2

It is possible to generalize this result to f0

, s 2 Lp implying that´f' dv 2

W s,p, but the proof we give relies on the Fourier transform, which does notgeneralize well. Also, with more work we could prove more regularity of theaveraged function, and other more intricate versions.

Proof. We will take the Fourier transform in t and x, sending t ! ⌧ ,x ! ⇠. For our convention, we use

f(⌧, ⇠, v) :=

ˆe�2⇡i⌧ t�2⇡i⇠·xf(t, x, v) dt dx

5. AVERAGING LEMMA 41

Notice that taking the Fourier transform changes the inhomogeneoustransport equation to

i(⌧ + v · ⇠)f = s

For now, we assume |⇠| � 1. We bound the Fourier transform of thevelocity averaged distribution as follows. Fixing ↵ > 0, we split the integralinto two terms, and estimate both individually.��ˆ

f(⌧, ⇠, v)'(v) dv

�� ˆ|⌧+v·⇠|>↵

i(⌧ + v · ⇠)f(⌧, ⇠, v) '(v)

i(⌧ + v · ⇠) dv��

+

��ˆ|⌧+v·⇠|↵

f(⌧, ⇠, v)'(v) dv

��

��ˆ|⌧+v·⇠|>↵

s(⌧, ⇠, v)'(v)

i(⌧ + v · ⇠) dv��

+

✓ˆ|⌧+v·⇠|↵,|v|M

'(v)2 dv

◆1/2✓ˆ

Rd

|f |2 dv◆

1/2

We can bound the second term, by bounding ' and then estimating thesize of the set being integrated over. Thus the second term is bounded as

C

(1 + |⇠|)1/2kf(⌧, ⇠, ·)kL2v

Now, we bound the first term, starting with an application of Cauchy-Schwartz, and then decomposing v into a vector parallel to ⇠ and one per-pendicular, i.e. v = vk + v?

kskL2v

✓ˆ|⌧+v·⇠|>↵,|v|M

'(v)2

|⌧ + v · ⇠|2 dv◆

1/2

kskL2v

✓ˆ|v?|M

dv?ˆ|⌧+vk·⇠|>↵,|vk|M

'(v)2

|⌧ + vk · ⇠|2 dvk◆

1/2

CkskL2v

0

BB@

ˆ|⌧+vk·⇠|>↵,|vk|M

'(v)2

| ⌧ + vk · ⇠| {z }:=V

|2 dvk

1

CCA

1/2

CkskL2v

✓ˆ↵<|V|⌧+M |⇠|

dV|V|2|⇠|

◆1/2

=C

|⇠|1/2kskL2v


Thus, we have thatˆ|⇠|�1

(1 + |⇠|)��ˆ

f(⌧, ⇠, v)'(v) dv

��2

d⇠d⌧ C

ˆ|⇠|�1

hksk2L2

v

+ kfk2L2v

id⇠d⌧

For |⇠| 1, note that because |⇠| is bounded from above, we can just applyHolder’s inequality to get thatˆ

|⇠|1

(1 + |⇠|)��ˆ

f' dv

��2

d⇠d⌧ Ckfk2L2⌧,⇠,v

Combining these two bounds gives the desired result. ⇤Note that in general, we can show that we gain a 1

p derivative in x ifworking with Lp data. This is due to the fact that we have regularizationin x only in the v direction for each fixed v.

Exercise 15. Assume that we have a more regular test function, ' 2W 1,1(Rd

v)3, but less regular source term, i.e.

s = rv · u u 2 L2

t,x,v

and prove a similar result.

Exercise 16. In this exercise, we try to understand why the averaginglemma “degenerates” when working in L1. We define

T' : L1

t,x,v ! L1

t,x g 7! T'(g)

where

T'(g)(t, x) :=

ˆf(t, x, v)'(v) dv

for f solving (@tf + v ·rxf + f = g

f(t = 0) = f0

(the additional f makes the solution more regular, and we could remove itby multiplying by an exponential term). We would like to show that T' isnot compact (even locally). More precisely, construct gn 2 L1

t,x,v such thatgn(v) = 0 for |v| � 1 and gn ! �x0,v0 in the sense of (Cb)⇤ (i.e. weakly inthe sense of measures, or equivalently in the dual of continuous boundedfunctions). Compute the associated fn. Then, take ' a regularized versionof �B(0,1), the characteristic function of the unit ball. Compute T'(gn) andcompute its weak limit, which will show there is a problem if we want T'to be compact.

3Recall that we let for k 2 N and p 2 [1,1], W k,p := {g(v) : D↵g 2 L

p for |↵| < k}

7. PHASE MIXING 43

6. Some notions of Sobolev spaces (S)

Restrict the discussion to Rd and Td. Give as a remark the idea in moregeneral geometry. State the key inequalities (and maybe prove Gagliardo-Nirenberg, or do the dimension 1 case).

Revisit the interpolation section by applying it to this new scale of spaceHk.

7. Phase Mixing

This is another fundamental aspect of the transport equation (for sim-plicity we restrict to the free transport equation), when x is confined. Wehave “mixing” (i.e. ergodization) of the characteristic curves in phase space.In physics this is related to plasmas (e.g. aurora borealis) and galactic dy-namics. This is also somehow related to Landau Damping.

Here, we will study the free transport equation on Td

(@tf + v ·rxf = 0

f(t = 0) = f0

(more interesting, but still tractable is the linearized Vlasov-Poisson equa-tion, which we will not discuss here).

HERE INCLUDE A PICTURE OF THE ERGODIC PHASE MIXING

Lemma 7.13. 4

(1) For f0

2 L2

xH1

v , then for all ' 2 L2

xH1

v , as t ! ±1ˆx,v

ft'O( 1

1+|t|)��!ˆv

✓ˆx

f0

◆✓ˆx

'

◆�

That is, in the appropriate sense, we have the weak convergence

ft !✓ˆ

x

f0

◆(v) := F (v)

(2) For f0

2 L1

xWd+↵,1v ↵ 2 N⇤ then defining

⇢t :=

ˆv

ft

⇢1 :=

ˆx

⇢t =

ˆx,v

f0

4Note that we are not worrying (and have not been worrying) about solutions havingenough regularity in order to be “classical.” This is because we have a simple explicitformula that immediately gives us the notion of weak solution, and everything we havesaid holds in the classical sense if the initial data is smooth enough.


we have that

k⇢t � ⇢1kL1x

C

(1 + |t|)d+↵

Note that we have no dispersion so we don’t expect particles to escape toinfinity. Instead ergodization yields weak convergence. Also note that, weakconvergence can be compatible with reversibility. [For example, u(t, x) =eitxu

0

(x) is a reversible system, but the Riemann Lebesgue lemma says thatthis weakly converges to zero]

Furthermore, we are obtaining strong convergence in the partial averageof the solution. We have that ⇢ (which is an observable) is exhibitingirreversiblity, but the information is just being put into hidden degrees offreedom.

Proof of (1). We Fourier transform our equation in x, v, taking x !k, v ! ⌘, i.e.

f(t, k, ⌘) :=

ˆe�2⇡ix·k/L�2⇡iv·⌘f(t, x, v) dx dv

This gives the linear transport equation as

@tf � k ·r⌘f = 0

for which the solution is

f(t, k, ⌘) = f0

(k, ⌘ + kt)

For ' 2 L2

xH1

v we define

g(t, x, v) := f(t, x, v)� F (v)

where

F (v) =

ˆx

f(t, x, v) dx =

ˆx

f0

(x, v) dx

(the second inequality is by di↵erentiating in time under the integral sign,and then using Green’s Theorem to show that the time derivative is zero).We have that

g(t, k, ⌘) = f(t, k, ⌘)� 1k=0

F (⌘)|{z}=

˜f0(0,⌘)

Thus, using the Plancherel Theorem and Cauchy-Schwartz we have that(using the fact that g(t, 0, ⌘) = 0, so we may ignore the k = 0 term)

��ˆ

g'

�� C

X

k 6=0

ˆ⌘

|gt|2(1 + |⌘ + kt|2)!

1/2 X

k 6=0

ˆ⌘

|'(k, ⌘)|2(1 + |⌘ + kt|2)

!1/2

7. PHASE MIXING 45

C

0

@X

k 6=0

ˆ⌘

|f0

(t, k, ⌘ + kt)|2(1 + | ⌘ + kt| {z }:=⌘

)|21

A1/2

⇥ X

k 6=0

ˆ⌘

|'(k, ⌘)|2(1 + |⌘ + kt|2)

!1/2

Ckf0

kL2x

H1v

supk 6=0

✓1

(1 + |⌘|2)(1 + |⌘ + kt|2)◆�

1/2

k'kL2x

H1v

If |⌘| < |kt|/21

(1 + |⌘|2)(1 + |⌘ + kt|2) 1

1 + |⌘ + kt|2 1

1 + |kt|2/4 1

1 + |t|2/4If |⌘| � |kt|/2, then

1

(1 + |⌘|2)(1 + |⌘ + kt|2) 1

1 + |⌘|2 1

1 + |kt|2/4 1

1 + |t|2/4(In both of these we used k 6= 0). Combining these gives the desired con-vergence. ⇤

Proof of (2). Letting t > 0, we have defined

⇢(t, x) :=

ˆft(x, v) dv.

Fourier transforming Td ⇥ Rd 3 (x, v) ! (k, ⌘) 2 Zd ⇥ Rd gives that

⇢(t, k) := ft(k, 0) = f0

(k, kt)

Thus, we definer(t, x) = ⇢(t, x)� ⇢1

We assume, for simplicity that |Td| = 1, but we could leave it as a constantin our expressions if we wanted. Now, computing gives

r(t, k) = ⇢(t, k)� 1k=0

⇢1This gives that for t > 1

k⇢t � ⇢1kL1 X

k2Zd

|r(t, k)|

X

k 6=0

|r(t, k)|

X

k 6=0

|f0

(k, kt)|(1 + |kt|)d+↵

(1 + |kt|)d+↵


. kf0

kL1x

W d+↵,1v

X

k 6=0

1

(1 + |kt|)d+↵

!(*)

kf0

kL1x

W d+↵,1v

1

|t|d+↵

X

k 6=0

1

|k|d+↵

!

| {z }<1

. kf0

kL1x

W d+↵,1v

1

|t|d+↵.

For t < 1, the inequality trivially holds. In the step marked (*), we usedthe result of following exercise:

Exercise 17. Show that

|f0

(k, kt)|(1 + |kt|)d+↵ .

2

4X

|j|d+↵

(@jvf0)(k, kt)

3

5

and thus deduce that

supk 6=0

|f0

(k, kt)|(1 + |kt|)d+↵ . kf0

kL1x

W d+↵

v

.

⇤Exercise 18. Consider the following equations:

@tf + v ·rxf + (v ⇥ B) ·rxf = 0

where x = (x1

, x2

, 0) 2 R3, v = (v1

, v2

, 0) 2 R3 and B = (0, 0, 1), modelingconfinement by a magnetic field, and

@tf + v ·rxf � x ·rvf = 0

where x, v 2 Rd, modeling confinement by a harmonic potential F =�rxV , V (x) = |x|2

2

.

(1) What are the characteristic curves?(2) Do you expect dispersion and/or phase mixing? (does not need a

proof, just heuristic reasoning).

8. Complete integrability and phase mixing (S)

9. Bibliographical and historical notes

10. Exercises

linear transport equations - wordpress.com · 2012. 3. 6. · chapter 3 linear transport equations...

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