CHAPTER 3
Linear Transport Equations
The fundamental transport operator is that of free transport
(0.1)
@
+ v · rx
@t
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 force
F (t, x, v) then we get the linear transport equation
(0.2)
@f
+ v · rx f + F · rv f = 0
@t
Still remaining in the context of linear kinetic equations, We could also
add 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 often
called the “linear Boltzmann equation”, and we shall study it in the next
chapter.
1. Construction of solutions: Lagrangian and Eulerian
viewpoints
1.1. The Lagrangian viewpoint. We will take advantage of the underlying particle trajectories to gain information about the linear transport
equation (0.2). This is the Lagrangian viewpoint of this PDE.
1
Definition 1.2. Given a force field F (t, x, v) 2 Ct0 Cx,v
, we call the characteristics of the PDE (0.2) the solutions of the following ODE system for
x, v 2 Rd
8
dX
>
>
(t, x, v) = V (t, x, v)
>
>
dt
>
<
dV
(1.1)
(t, x, v) = F (t, X(t, x, v), V (t, x, v))
>
>
dt
>
>
>
:
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 construction, taking the original ODE and ending up with a PDE. The characteristic
curves are the same information as in (0.2) but in that case we were considering things statistically. Note that if we take d = 3N , for example,
then we can describe N particles moving in spacial 3 dimensions with these
equations, and these are just the regular Newtonian equations of motion.
Proposition 1.3. Assume that F 2 C 1 (R ⇥ Rd ⇥ Rd )1 and we have some
C > 0 so that F satisfies the following growth condition
8t 2 R, x 2 Rd , v 2 Rd ,
(G)
|F (t, x, v)|  C(1 + |x| + |v|).
Then, for any x, v 2 R , there exists a unique solution to the characteristic
ODE’s (1.1) for all t 2 R. Moreover for any t 2 R the “characteristics
map”
S0,t (x, v) := (X(t, x, v), V (t, x, v))
1
is a C -di↵eomorphisme on Rd ⇥ Rd .
d
Exercise 10. Prove this proposition using the Picard-Lindelöf ODE existence theorem and the “Lemma of leaving compact sets”.
Now, with this ODE tool at hand, let us solve the Cauchy problem for
the transport equation, with even needing the compact support from the
last section.
1
1
Proposition 1.4. For f0 2 Cx,v
, F 2 Ct,x,v
satisfying (G), (0.2) admits a
1
unique C solution with initial data f0 , 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 , because since S0,t is a C 1 -di↵eomorphism, if we call the inverse function St,0
(which corresponds to solving backward the characteristics ODE’s) we have
the 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 C 1 by composition (using again the solutions X and V to the characteristics ODE’s are
C 1 ). Moreover the solution constructed thus constructed satisfies f (t, X, V ) =
f0 (x, v) and since the composed function is C 1 , so taking the total derivative
in time, we have that
d
0=
[f (t, X(t, x, v), V (t, x, v))]
dt
1For
simplicity, we are assuming more regularity than is necessary: C 0 in all variables
and locally Lipschitz in the variables x and v would be sufficiant.
1. CONSTRUCTION OF SOLUTIONS: LAGRANGIAN AND EULERIAN VIEWPOINTS
33
@f
@X
@V
+ rx f
·
+ rv f
·
@t t,X,V
@t
@t
t,X,V
t,X,V
@f
=
+ V · rx f
+ F · rv f
@t t,X,V
t,X,V
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 C 1 -di↵eomorphism, and in particular bijective, this implies that
f satisfies the transport equation at any point (t, x, v) which concludes the
proof.
⇤
Remark 1.5. We still need to study the characteristic mapping and its
jacobian. One important thing to keep in mind is that if F depends on
time (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 is
S0,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 C 1 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
= (rv · F )J
@t
and we can solve this as
✓ˆ t
◆
J(t, x, v) = J(0, x, v) exp
(rv · F )(s, x, v)ds
0
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 form
where (X, V ) solves a Hamiltonian ODE
8
Ẋ = rv H|t,X,V
>
>
>
<V̇ = r H|
x
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 obtained in the last section, using the explicit solution and Jacobian computation above. Moreover, prove that kft k1 = kf0 k1 , not just an inequality.
34
3. LINEAR TRANSPORT EQUATIONS
1.2. The Eulerian viewpoint. We will study solutions to (0.2) with
x, v 2 Rd . We will further assume that the force is divergence free in
velocity, i.e. that rv · F = 0. Under this general assumption, we have the
following a priori estimates:
Proposition 1.7. Consider f is a C 1 solution on [0, T ] to the linear transport equation (0.2) with initial data f (t = 0, x, v) = f0 (x, v) such that
9R > 0,
Then for all t
(1.2)
8t 2 [0, T ],
0 and p 2 [1, 1)
supportft ⇢ B(0, R).
kf (t, x, v)kLp (Rdx ⇥Rdv ) = kf0 (x, v)kLp (Rdx ⇥Rdv ) .
Also, for p = 1, we have
(1.3)
kf (t, x, v)kL1 (Rdx ⇥Rdv )  kf0 (x, v)kL1 (Rdx ⇥Rdv ) .
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 C 1 (R; R) then
ˆ
ˆ
(1.4)
(f (t, x, v)) dx dv =
(f0 (x, v)) dx dv
Rdx ⇥Rdv
Rdx ⇥Rdv
Proof. We will prove (1.4) and then show that this implies the rest of
the proposition. Letting 2 C 1 (R; R), we claim that if f is a C 1 solution
to the transport equation satisfying the support condition above, then so is
(f ). This is a simple application of the chain rule:
✓
◆
@
@f
0
(f )+rx ·(v (f ))+F ·rv ( (f )) = (f )
+ v · rx f + F · rv f = 0.
@t
@t
This shows that (f ) is a solution to the same transport with initial
conditions (f0 ). Integrating this over Rdx ⇥ Rdv , the divergence terms are
zero by Green’s theorem (note that we have used that rv · F = 0 to move
F inside and outside of the v-derivative term), so we thus have shown that
ˆ
ˆ
@
@
(f ) dx dv =
(f ) dx dv = 0
@t
@t
which establishes (1.4). We readily deduce the statement of conservation
of 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 first
instance of a regularisation argument. Let ✏ (s) = |s| ✏ (|s|), with ✏ 2
[0, 1], ✏ (s) % 1 as ✏ ! 0, and ✏ 2 C 1 ((0, 1)) (notice that this means
that ✏ (0) = 0✏ (0) = 0, so we may extend it to a C 1 function on all of R).
1. CONSTRUCTION OF SOLUTIONS: LAGRANGIAN AND EULERIAN VIEWPOINTS
35
For example, to construct such a ✏ , let ' be a C 1 function, with compact
support 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), notice
that because f0 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
0 for t
0. Now let us generalize the
✏ (f (t, x, v)) = 0, and thus ft
example: consider some initial data f0 with L1 bound kf0 kL1 , i.e.
8x, v 2 Rd ,
kf0 kL1  f0 (x, v)  kf0 kL1 .
We then consider a C 1 mollified cuto↵ function ✏ on R+ which vanishes on
[0, kf0 kL1 , and is identically one on [kf0 kL1 + ✏, +1). Then we check that
1
✏ (|f ) is C solution with the compact support property, and therefore
ˆ
ˆ
8t 2 [0, T ],
✏ (|ft |) =
✏ (|f0 |) = 0
which implies that |ft | does not values in [kf0 kL1 + ✏, +1). Since it is true
for any ✏ > 0 it concludes the proof.
⇤
Remark 1.8. Another very natural argument, which cannot be performed
here only due to artificial obstacles since we restrict ourselves to functions
with compact, is the following: if f0 2 L1 , then letting
f˜0 = f0
kf0 k1
we obtain a new solution
f˜t = ft = kf0 k1
for t 2 [0, T ]. Then since f˜0  0, we deduce that f˜  0 for all t
0.
Thus, f  kf0 k1. Applying the same argument to f gives the desired
inequality:
kf k1  kf0 k1 .
36
3. LINEAR TRANSPORT EQUATIONS
This argument can be made rigorous by showing that any solution can
be approximated by compactly supported solutions. This is another instance
of a regularisation argument, used here to justify this a priori estimate. If
used for the Lp norms we would obtain that (1.2) holds in the sense that
there is equality if the RHS is finite, and if the RHS is infinite, then the
LHS is as well.
We shall in any case revisit the proof of these L1 bound estimates in a
more general setting, using the so-called characteristics method, and show
that there is indeed equality.
2. Dispersion Estimates
What is dispersion? In wave mechanics, it is the phenomenon in which
waves of di↵erent frequencies travel at di↵erent velocities, e.g. in optics light
traveling in a dispersive medium. This leads to wave packets spreading out
and dispersing at infinity, and other phenomenon. In particle dynamics,
particles disperse because they have all di↵erent velocities, which to their
spatial density to“spread out” then “escape at infinity,” which is really the
phenomenon we are interested in. This means that there are two aspects in
the dispersion phenomenon: a local one and a global one in the all space.
As an aside, notice that from a quantum mechanical viewpoint, these
become the same phenomenon, when the (quantized) velocity is associated
with the frequency of waves.
We will study this phenomenon in the simple setting of x, v 2 Rd for
the free transport equation, which we recall is
(2.1)
@ t f + v · rx f = 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 mechanism is the fact that in phase space the volume is conserved, but it tends
to “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 and
decay in time. The first dispersive estimate that we have is:
2In
general, one must assume that the characteristics are not “trapped,” escape at
infinity 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 direction
under the flow of the linear transport equation, while maintaining the same volume in phase space.
Lemma 2.9. If f is a solution to the free transport equation
(
@ t f + v · rx f = 0
f |t=0 = f0
and f 2 L1 \ L1
x,v for all p 2 [1, 1], then we have for t
0 that
d
1  |t|
kft kL1
kf0 kL1x (L1
.
x (Lv )
v )
⇣
⌘
Note that d p1 1q  0, so this gives us information about the decay
of the norms as t ! 1.
(2.2)
Proof. With no force, we have the exact representation
f (t, x, v) = f0 (x
tv, v)
and thus for t > 0
1 = sup
kft kL1
x (Lv )
x2Rd
ˆ
 sup
ˆ
x2Rd
=
Rd
|f0 (x
sup |f0 (x
| {zvt}, y)| dv
Rd y2Rd
1
sup
td x2Rd
ˆ
vt, v)| dv
:=V
sup |f0 (V, y)|dV
Rd y2Rd
38
3. LINEAR TRANSPORT EQUATIONS
ˆ
1
= d
sup |f0 (V, y)|dV
t Rd y2Rd
1
= d kf0 kL1x (L1
v )
t
⇤
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
(
@ t f + v · rx f = 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
that
1
1
(2.3)
kft kLqx (Lpv )  |t| d( p q ) kf0 kLpx (Lqv ) .
0
Proof. By conservation of mass, e.g. (1.2), we have that
kft kLpx (Lpv ) = kf0 kLpx (Lpv )
and thus, combining these two bounds with Riesz-Thorin Interpolation gives
the desired bounds.
⇤
We also in general expect transfer of regularity, from v to x. This can
be made precise in the following lemma
Lemma 2.11. Again, if f is a solution to the free transport equation
(
@ t f + v · rx f = 0
f |t=0 = f0
and for f0 is regular enough, then for s 2 N
k(trx + rv )s ft kL1x,v = k(rv )s f kL1x,v .
Note that the operator on the left commutes with itself, so raising it to s is
well 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 being
zero is trivial, and the second is a short computation. As a consequence of
this, because f is a solution of the free transport equation, then so is rx f
and (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
" s
#
s
Y
X
Ai , C =
A1 · · · Ai 1 [Ai , C] Ai+1 · · · As
| {z }
| {z }
i=1
i=1
I if i=1
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 equation, then
@t [(trx + rv )s f ] + v · rx [(trx + rv )s f ] = 0
and by conservation of L1 norm, we have that
k((trv + rx )s f )|t kL1v,x = krsv f0 kL1x,v
as desired.
⇤
One must be careful with this result. Because there might be cancellation, we cannot draw direct information about x-derivatives of f . For
example, because f (t, x, v) = f0 (x tv, v), it is clear that rsv f = O(ts ),
which makes the lemma hard to use.
Note that it also gives the simpler result that
krsx ft kL1x,v = krsx f0 kL1x,v
Exercise 14. Show that the previous two results extend to f a solution of
the more general equation
@t f + a(v) · rx f = 0
where a(v) 2 C (R ; R ) is a “uniform di↵eomorphism,” i.e. we have the
two sided bound
0 < C1  | Jac a(v)|  C2
d
for all v 2 R .
1
d
d
3. Dispersion estimates for wave operators (S)
State and prove basic dispersion estimates for the wave equation and
for the Schrödinger equation in the whole space.
Explain the link between transport and wave operators, through the
Wigner’s transform formalism.
40
3. LINEAR TRANSPORT EQUATIONS
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 (by
approximating by sequence). Finish with the abstract general notion of
interpolation space. Emphasize the intuition to deduce from it for linear
problems: “it is enough to prove functional inequalities at extremal values of the coefficients” (remark it can take several interpolation steps for
reconstructing the whole convex hull of parameters).
5. Averaging Lemma
As remarked above, the regularity transfer estimates of the previous
section are quite difficult to use. Motivated to find better regularity statements, we will prove that if we average solutions to the free transport equation in velocity, then we have improved regularity estimates. As such we
prove
Lemma 5.12. For f 2 L2 (Rt ⇥ Rdx ⇥ Rdv ), a solution to the inhomogeneous
free transport equation
(
@ t f + v · rx f = s
f (t = 0) = f0
d
for f0 (x, v) 2 L2x,v and s(t, x, v) 2 L2t,x,v then for ' 2 L1
c (R ), a compactly
supported L1 “test function” we have that there is C > 0 such that
ˆ
h
i
f (·, ·, v)'(v) dv
 C kf kL2t,x,v + kskL2t,x,v
1/2
L2t (Hx
)
Recall that we define the fractional-1/2 Sobolev norm
✓ˆ
◆2
2
kgkHx1/2 :=
|ĝ(⇠)| (1 + |⇠|)d⇠
´
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 not
generalize well. Also, with more work we could prove more regularity of the
averaged 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
ˆ
fˆ(⌧, ⇠, v) := e 2⇡i⌧ t 2⇡i⇠·x f (t, x, v) dt dx
5. AVERAGING LEMMA
41
Notice that taking the Fourier transform changes the inhomogeneous
transport equation to
i(⌧ + v · ⇠)fˆ = ŝ
For now, we assume |⇠|
1. We bound the Fourier transform of the
velocity averaged distribution as follows. Fixing ↵ > 0, we split the integral
into two terms, and estimate both individually.
ˆ
ˆ
'(v)
ˆ
f (⌧, ⇠, v)'(v) dv 
i(⌧ + v · ⇠)fˆ(⌧, ⇠, v)
dv
i(⌧ + v · ⇠)
|⌧ +v·⇠|>↵
ˆ
+
fˆ(⌧, ⇠, v)'(v) dv

|⌧ +v·⇠|↵
ˆ
ŝ(⌧, ⇠, v)
|⌧ +v·⇠|>↵
+
✓ˆ
'(v)
dv
i(⌧ + v · ⇠)
◆1/2 ✓ˆ
2
'(v) dv
|⌧ +v·⇠|↵,|v|M
Rd
ˆ2
|f | dv
◆1/2
We can bound the second term, by bounding ' and then estimating the
size of the set being integrated over. Thus the second term is bounded as

C
kfˆ(⌧, ⇠, ·)kL2v
(1 + |⇠|)1/2
Now, we bound the first term, starting with an application of CauchySchwartz, and then decomposing v into a vector parallel to ⇠ and one perpendicular, i.e. v = v k + v ?
✓ˆ
◆1/2
'(v)2
 kŝkL2v
dv
2
|⌧ +v·⇠|>↵,|v|M |⌧ + v · ⇠|
✓ˆ
◆1/2
ˆ
'(v)2
?
k
 kŝkL2v
dv
dv
k
2
|v ? |M
|⌧ +v k ·⇠|>↵,|v k |M |⌧ + v · ⇠|
0
11/2
ˆ
B
C
'(v)2
kC
B
 CkŝkL2v @
dv
k
A
2
|⌧ +v k ·⇠|>↵,|v k |M | ⌧ + v · ⇠ |
| {z }
 CkŝkL2v
=
✓ˆ
C
kŝkL2v
|⇠|1/2
↵<|V|⌧ +M |⇠|
:=V
dV
|V|2 |⇠|
◆1/2
42
3. LINEAR TRANSPORT EQUATIONS
Thus, we have that
ˆ
ˆ
ˆ
2
ˆ
(1 + |⇠|)
f (⌧, ⇠, v)'(v) dv d⇠d⌧  C
|⇠| 1
|⇠| 1
h
i
kŝk2L2v + kfˆk2L2v d⇠d⌧
For |⇠|  1, note that because |⇠| is bounded from above, we can just apply
Hölder’s inequality to get that
ˆ
ˆ
2
ˆ
(1 + |⇠|)
f ' dv d⇠d⌧  Ckfˆk2L2
⌧,⇠,v
|⇠|1
Combining these two bounds gives the desired result.
⇤
Note that in general, we can show that we gain a p1 derivative in x if
working with Lp data. This is due to the fact that we have regularization
in x only in the v direction for each fixed v.
Exercise 15. Assume that we have a more regular test function, ' 2
W 1,1 (Rdv )3, but less regular source term, i.e.
s = rv · u
and prove a similar result.
u 2 L2t,x,v
Exercise 16. In this exercise, we try to understand why the averaging
lemma “degenerates” when working in L1 . We define
T ' : L1t,x,v ! L1t,x
where
T '(g)(t, x) :=
for f solving
(
ˆ
g 7! T '(g)
f (t, x, v)'(v) dv
@ t f + v · rx f + f = g
f (t = 0) = f0
(the additional f makes the solution more regular, and we could remove it
by multiplying by an exponential term). We would like to show that T ' is
not compact (even locally). More precisely, construct gn 2 L1t,x,v such that
gn (v) = 0 for |v| 1 and gn ! x0 ,v0 in the sense of (Cb )⇤ (i.e. weakly in
the sense of measures, or equivalently in the dual of continuous bounded
functions). Compute the associated fn . Then, take ' a regularized version
of B(0,1) , the characteristic function of the unit ball. Compute T '(gn ) and
compute 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 Lp 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 more
general geometry. State the key inequalities (and maybe prove GagliardoNirenberg, or do the dimension 1 case).
Revisit the interpolation section by applying it to this new scale of space
H k.
7. Phase Mixing
This is another fundamental aspect of the transport equation (for simplicity we restrict to the free transport equation), when x is confined. We
have “mixing” (i.e. ergodization) of the characteristic curves in phase space.
In physics this is related to plasmas (e.g. aurora borealis) and galactic dynamics. This is also somehow related to Landau Damping.
Here, we will study the free transport equation on Td
(
@ t f + v · rx f = 0
f (t = 0) = f0
(more interesting, but still tractable is the linearized Vlasov-Poisson equation, which we will not discuss here).
HERE INCLUDE A PICTURE OF THE ERGODIC PHASE MIXING
Lemma 7.13. 4
(1) For f0 2 L2x Hv1 , then for all ' 2 L2x Hv1 , as t ! ±1
ˆ
ˆ ✓ˆ ◆ ✓ˆ ◆
1
O( 1+|t|
)
ft '
!
f0
'
x,v
v
x
x
That is, in the appropriate sense, we have the weak convergence
✓ˆ ◆
ft !
f0 (v) := F (v)
x
(2) For f0 2
2 N⇤ then defining
ˆ
⇢t := ft
v
ˆ
ˆ
⇢1 := ⇢t =
f0
L1x Wvd+↵,1 ↵
x
4Note
x,v
that we are not worrying (and have not been worrying) about solutions having
enough regularity in order to be “classical.” This is because we have a simple explicit
formula that immediately gives us the notion of weak solution, and everything we have
said holds in the classical sense if the initial data is smooth enough.
44
3. LINEAR TRANSPORT EQUATIONS
we have that
k⇢t
⇢ 1 kL 1

x
C
(1 + |t|)d+↵
Note that we have no dispersion so we don’t expect particles to escape to
infinity. Instead ergodization yields weak convergence. Also note that, weak
convergence can be compatible with reversibility. [For example, u(t, x) =
eitx u0 (x) is a reversible system, but the Riemann Lebesgue lemma says that
this weakly converges to zero]
Furthermore, we are obtaining strong convergence in the partial average
of the solution. We have that ⇢ (which is an observable) is exhibiting
irreversiblity, but the information is just being put into hidden degrees of
freedom.
Proof of (1). We Fourier transform our equation in x, v, taking x !
k, v ! ⌘, i.e.
ˆ
˜
f (t, k, ⌘) := e 2⇡ix·k/L 2⇡iv·⌘ f (t, x, v) dx dv
This gives the linear transport equation as
@t f˜ k · r⌘ f˜ = 0
for which the solution is
f˜(t, k, ⌘) = f˜0 (k, ⌘ + kt)
For ' 2 L2x Hv1 we define
g(t, x, v) := f (t, x, v)
where
F (v) =
ˆ
x
f (t, x, v) dx =
F (v)
ˆ
f0 (x, v) dx
x
(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
g̃(t, k, ⌘) = f˜(t, k, ⌘) 1k=0 F̃ (⌘)
| {z }
=f˜0 (0,⌘)
Thus, using the Plancherel Theorem and Cauchy-Schwartz we have that
(using the fact that g̃(t, 0, ⌘) = 0, so we may ignore the k = 0 term)
!1/2
!1/2
ˆ
2
Xˆ
Xˆ
|
'(k,
˜
⌘)|
g'  C
|g̃t |2 (1 + |⌘ + kt|2 )
(1 + |⌘ + kt|2 )
⌘
k6=0
k6=0 ⌘
7. PHASE MIXING
0
C@
Xˆ
k6=0
45
11/2
|f˜0 (t, k, ⌘ + kt)|2 (1 + | ⌘ + kt)|2 A
| {z }
⌘
:=⌘
!1/2
|'(k,
˜ ⌘)|2
⇥
(1 + |⌘ + kt|2 )
k6=0 ⌘

✓
◆ 1/2
1
 Ckf0 kL2x Hv1 sup
k'kL2x Hv1
(1 + |⌘|2 )(1 + |⌘ + kt|2 )
k6=0
If |⌘| < |kt|/2
(1 +
If |⌘|
Xˆ
1
1
1
1



2
2
2
+ |⌘ + kt| )
1 + |⌘ + kt|
1 + |kt| /4
1 + |t|2 /4
|⌘|2 )(1
|kt|/2, then
1
1
1
1



2
2
2
2
(1 + |⌘| )(1 + |⌘ + kt| )
1 + |⌘|
1 + |kt| /4
1 + |t|2 /4
(In both of these we used k 6= 0). Combining these gives the desired convergence.
⇤
Proof of (2). Letting t > 0, we have defined
ˆ
⇢(t, x) := ft (x, v) dv.
Fourier transforming Td ⇥ Rd 3 (x, v) ! (k, ⌘) 2 Zd ⇥ Rd gives that
⇢ˆ(t, k) := f˜t (k, 0) = f˜0 (k, kt)
Thus, we define
r(t, x) = ⇢(t, x) ⇢1
We assume, for simplicity that |Td | = 1, but we could leave it as a constant
in our expressions if we wanted. Now, computing gives
r̃(t, k) = ⇢ˆ(t, k)
This gives that for t > 1
k⇢t
⇢1 kL1 


X
k2Zd
X
k6=0
X
k6=0
1k=0 ⇢1
|r̂(t, k)|
|r̂(t, k)|
|f˜0 (k, kt)|
(1 + |kt|)d+↵
(1 + |kt|)d+↵
46
3. LINEAR TRANSPORT EQUATIONS
!
1
. kf0 kL1x Wvd+↵,1
(1 + |kt|)d+↵
k6=0
!
X 1
1
 kf0 kL1x Wvd+↵,1 d+↵
|t|
|k|d+↵
k6=0
|
{z
}
X
(*)
<1
. kf0 kL1x Wvd+↵,1
1
|t|d+↵
.
For t < 1, the inequality trivially holds. In the step marked (*), we used
the result of following exercise:
Exercise 17. Show that
2
|f˜0 (k, kt)|(1 + |kt|)d+↵ . 4
and thus deduce that
X
|j|d+↵
3
(@vj f0 )(k, kt)5
sup |f˜0 (k, kt)|(1 + |kt|)d+↵ . kf0 kL1x Wvd+↵ .
k6=0
⇤
Exercise 18. Consider the following equations:
@t f + v · rx f + (v ⇥ B) · rx f = 0
where x = (x1 , x2 , 0) 2 R3 , v = (v1 , v2 , 0) 2 R3 and B = (0, 0, 1), modeling
confinement by a magnetic field, and
@ t f + v · rx f
x · rv f = 0
where x, v 2 Rd , modeling confinement by a harmonic potential F =
2
rx V , V (x) = |x|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