Di↵erential and Integral Equations, Volume 8, Number 3, March 1995, pp. 487 – 514.
ON LONG TIME ASYMPTOTICS OF
THE VLASOV-FOKKER-PLANCK EQUATION AND OF
THE VLASOV-POISSON-FOKKER-PLANCK SYSTEM WITH
COULOMBIC AND NEWTONIAN POTENTIALS
F. Bouchut
Université d’Orléans et CNRS, URA D1803
Département de Mathématiques, BP 6759, 450067 Orléans Cédex 2, France
J. Dolbeault*
CEREMADE, U.R.A. 749, Université Paris IX-Dauphine
Place du Maréchal de Lattre de Tassigny, 75775 Paris Cédex 16, France
(Submitted by: P.L. Lions)
Abstract. We prove that the solution of the Vlasov-Fokker-Planck equation converges to the
unique stationary solution with same mass as time tends to infinity. The same result holds in the
repulsive coulombic case for the Vlasov-Poisson-Fokker-Planck system; the newtonian attractive
case is also studied. We establish positive and negative answers to the question of existence of a
stationary solution for the last problem by examining the Poisson-Boltzmann equation.
1. Introduction. A low density gas of noninteracting particles with a velocity
dispersion is well described by the Vlasov equation
@t f + v · rx f + div v (F f ) = 0
(1.1)
in the limit of a large number of particles. Here f (t, x, v) represents the distribution
function; i.e., the probability density of finding a particle with velocity v at time t and
position x. The distribution function is usually normalized by the number of particles,
or equivalently by the total mass
Z Z
M =m
f (t, x, v) dx dv,
(1.2)
assuming that particle is of mass m. The field mF (t, x) is defined to be the macroscopic
force acting on the particles. This field is created by the particles themselves through
collective e↵ects (and modeled by a mean field approximation) or by an external source.
Vlasov’s equation is easily derived by assuming that all particles obey Newton’s equations (we assume that the velocities are small enough to neglect relativistic e↵ects)
dx
= v,
dt
m
dv
= mF (t, x).
dt
(1.3)
Vlasov’s equation does not take interactions into account either between particles or
with any background. For low densities and particles interacting at microscopic level
Received for publication December 1993.
*and Groupe de physique théorique, Laboratoire de physique quantique, IRSAMC, Université Paul
Sabatier, 118, route de Narbonne, 31062 Toulouse Cédex.
AMS Subject Classifications: 35Q20, 82A45, 82A05, 82A25, 31A30.
487
488
F. BOUCHUT AND J. DOLBEAULT
through long range two-body potentials, the first corrections one can take into account
are grazing collisions. This gives a collision term known as Fokker-Planck-Landau kernel.
The collision terms taking into account hard encounters (like Boltzmann’s collision
term for two-particle encounters) become relevant only for high densities. Here we
shall consider a simplified Fokker-Planck model containing dynamical friction forces
and a di↵usion term characteristic of Brownian motion. Newton’s equations have to be
replaced by a stochastic di↵usion equation
p
dx = v dt, m dv = m(F (t, x)
v)dt + m 2 dW,
(1.4)
where W denotes the standard Wiener process,
is a friction parameter, and
a
thermal di↵usion coefficient. According to Stokes’ law, for spherical particles of radius
a, is given by
m = 6⇡a⌘,
(1.5)
where ⌘ is the viscosity coefficient of a surrounding fluid, which has temperature T
determined by the relations
KT = m✓, ✓ = / .
(1.6)
The corresponding distribution function obeys the Vlasov-Fokker-Planck equation
@t f + v · rx f + F (t, x) · rv f = div v ( vf + rv f ).
(1.7)
In this paper, we shall consider the following case: the configuration space is RN (N
1), and the velocity-space is also RN (non-relativistic case). The macroscopic force field
takes the form
F (t, x) =
rx u0 (x)
(1.8)
F (t, x) =
rx u0 (x) + E(t, x).
(1.9)
or
It is the sum of a force deriving from an external potential u0 , and possibly of a selfconsistent force deriving from the Poisson potential through
E(t, x) =
|S N 1 |
x
⇤ ⇢,
|x|N x
with ⇢ being the spatial macroscopic density
Z
⇢(t, x) =
f (t, x, v) dv
RN
(1.10)
(1.11)
(this is a mean field approximation). The potential u0 prevents the gas of particles to
vanish in the space of positions by means of a dilution e↵ect (for large times). It is
responsible for the confinement of the particles and has the same e↵ect as a boundary.
But this formulation is much simpler because there is no need of any local boundary
condition, and therefore we avoid technical difficulties. From a physical point of view,
it is relevant for charged-particles plasmas as well as for self-gravitating particles.
The case without self-consistent forces will be called “Neutral plasma case” and denoted by (VFP). In the case with self-consistent forces (1.9), the system will be denoted
by (VPFP). Two di↵erent cases arise depending on the sign of . The coulombic case
ON LONG TIME ASYMPTOTICS
489
(VPFPc) corresponds to > 0. This model describes a gas of charged particles interacting together through a mean electrostatic field created by their spatial distribution
(the velocities are assumed to be small enough to allow us to neglect magnetic forces).
Here the collision term represents the interaction with a background medium as a thermal bath. The newtonian case (VPFPn) corresponds to < 0. It describes a system
of gravitating particles modeling a stellar system, plunged in a gravitational potential
u0 . The Fokker-Planck approximation is especially interesting here because there is no
shield e↵ect in such physical systems (see [7, 8]). It is a good approximation because all
particles of the system undergo interactions with many other particles, but each of these
interactions is small according to the fact that the corresponding impact parameters are
very large.
For introductory papers to the Vlasov-Poisson-Fokker-Planck system in an astrophysical context and critical derivation of this system, one can refer to [7, 8] and [27],
for instance. A more general introduction to kinetic theory and to the problem of grazing collisions can be found in [6]. A general review on the equilibrium solutions of the
Vlasov equation and its physical implications is given in [1]. More general considerations on the foundations of equilibrium statistical mechanics (for gravitating systems)
can be found in [22], for instance.
The existence of solutions to the Cauchy problem for the Vlasov-Poisson-FokkerPlanck system has been proved by H.D. Victory and B.P. O’Dwyer in [30] (in the
case of strong solutions; i.e., solutions such that E(t, x) belongs to L1 ([0, +1[loc ⇥RN )
locally in time; the existence globally in time for weak solutions can be deduced from
the methods introduced by E. Hörst in [21], and R.J. DiPerna and P-L. Lions in [13, 14]
for the Vlasov-Poisson system in three dimensions. P. Degond proved in [9] the global
existence of strong solutions in dimension N = 1 or 2, G. Rein and J. Weckler gave
in [29] sufficient conditions to get a global existence result, and F. Bouchut (see [3])
gave an existence and uniqueness result for strong and global in time solutions of the
Vlasov-Poisson-Fokker-Planck system in 3 dimensions (in the attractive case as well as
in the repulsive one). Let us also mention the basic work of P-L. Lions and B. Perthame
(see [26]) concerning regularity of the Vlasov-Poisson system in 3 dimensions (see also
K. Pfa↵elmöser [28]).
For the stationary solutions, which correspond to the large time asymptotic solutions
as we prove it, we have to mention (for the repulsive case) K. Dressler’s papers [16, 17],
those of D. Gogny and P-L. Lions in [20], F. Bouchut in [2] and J. Dolbeault in [15],
and a paper by A. Krzywicki and T. Nadzieja (see [23]) for the newtonian case (in a
bounded domain). We give here the natural extension of this result to RN in the case of a
confining potential. We also mention the paper [5] which deals with the two-dimensional
problem.
The compactness lemma is directly inspired by the R.J. DiPerna and P-L. Lions
paper [11] on the Fokker-Planck-Boltzmann equation, and the renormalized solutions
are those defined by these authors in their papers on kinetic equations (see [11–14] for
example). One can also refer to P.L. Lions [25] for related compactness results.
In this paper, we shall give results on the problem of convergence to equilibrium
and on equilibrium solutions. Our equation is a simple model compared to the VlasovPoisson-Boltzmann system, for which the long time asymptotics has been studied by L.
Desvillettes and J. Dolbeault in [10]. However, we obtain a very strong convergence in
our case, and especially the convergence of the kinetic energy, which is not known at all
in the Boltzmann case.
Section II is devoted to general estimates for the Cauchy problem related to the
F. BOUCHUT AND J. DOLBEAULT
490
Vlasov-Fokker-Planck system and convergence results for long time asymptotics in the
case of a neutral plasma ((VFP): Section II.1) and in the case of self-consistent interactions ((VPFP): Section II.2). The relation between the behavior of the solutions for
asymptotically small mass (m ! 0) and for asymptotically large times (t ! 1) is analyzed in Section II.3. The case of attractive forces (VPFPn) is not known as well as the
case of repulsive forces (VPFPc), because the total energy is not bounded from below,
and because the situation for the stationary solutions is much more complicated.
Section III is devoted to the study of stationary solutions of the Vlasov-PoissonFokker-Planck system, which are Maxwellians whose densities are related to a potential
obeying an elliptic equation known as the Poisson-Boltzmann equation, or Emden equation, according to the sign of . Already known results are given in Section III.1 for
repulsive forces; III.2 is essentially concerned with existence results for the newtonian
case.
Finally, two useful results are given in the appendix: a compactness lemma for VlasovFokker-Planck equations, and a symmetry result for nonlinear elliptic equations in RN .
In conclusion, let us mention the two important points of this paper.
1) To deal with long time asymptotics, one has to establish a priori estimates which
are independent of time. According to the fact that in the model we consider here, the
long time temperature is determined because the particles are in a thermal bath, the
natural thermodynamical ensemble is the canonical ensemble. The associated thermodynamical function is the free energy function
A(t) = U (t)
where
U (t) =
ZZ
✓S(t),
(1.12)
|v|2
1
+ u0 (x) f (t, x, v) dx dv +
2
Z
|E(t, x)|2
dx
2
(1.13)
is the total energy, and ✓ = / is the temperature of the thermal bath (and of the
system in the large time limit), and where
ZZ
S(t) =
f (t, x, v) ln f (t, x, v) dx dv
(1.14)
is the physical entropy of the system.
2) The a priori estimates are not sufficient to pass to the limit. That is the reason
why we have to use the renormalization techniques introduced by R.J. DiPerna and P-L.
Lions to overcome the difficulties induced by the nonlinearity coming from the Poisson
equation, and a compactness lemma given in the first appendix.
II. Convergence to equilibrium.
II.0. Notations and main results. In this section, we consider a solution
f 2 C([0, 1[, L1 (RN ⇥ RN )),
f
0
vf
= 0,
(2.1)
of the Vlasov-Fokker-Planck equation
@t f + v · rx f + divv [(E
ru0
v)f ]
f (0, ·) = f0 ,
(2.2)
(t, x, v) 2]0, 1[⇥RN ⇥ RN , with either equation
E=0
(VFP case)(2.3)
ON LONG TIME ASYMPTOTICS
for a neutral plasma, or
E(t, x) =
|S N
x
⇤ ⇢,
1 | |x|N
⇢(t, x) =
Z
RN
491
f dv.
(VPFP case)(2.4)
The case > 0 denoted by (VPFPc) corresponds to coulombic interaction, and the case
< 0 (VPFPn) to Newtonian interactions. The parameters > 0 (di↵usion coefficient)
and > 0 (viscosity parameter) are given, and we will refer to
✓= /
(2.5)
as the temperature of the system. The external potential u0 (x) satisfies
u0 (x) 2 Lip (RN ),
lim inf
|x|!1
u0
0,
(2.6)
u0 (x)
> N ✓.
ln |x|
(2.7)
For the initial datum we assume
ZZ
f0 0,
1 + |v|2 + u0 (x) + ln+ f0 (x, v) f0 (x, v)dx dv < 1,
R2N
Z
1
|E(t = 0, x)|2 dx < 1,
2| | RN
(2.8)
which is no more than nonnegativity of the distribution function, finiteness of initial
mass, kinetic energy, external potential energy and entropy, and finiteness of internal
potential energy. For technical reasons, we assume in the (VPFP) case that
N
3,
E 2 L1 (]0, T [⇥RN ),
8 T > 0.
(2.9)
The assumptions (2.6) on u0 and (2.9) on E could be relaxed a bit, but it would be
much more difficult to justify some computations, and we will not do so.
Section II is subdivided in three subsections, in which we prove that a limit exists
when t ! 1, successively in the cases of a neutral plasma (VFP) in II.1., and of
repulsive interactions in II.2. We also consider the case of attractive interactions, but it
is of greater difficulty because the stationary problem is not well solved (the stationary
problem is the subject of the third section). In II.3 we show how to relate the behaviors
of f when t ! 1 and when the particle mass m ! 0 for (VPFPc). The main results of
this section are the following.
Theorem A. (VFP case). Given any f0 (x, v) satisfying (2.8), we consider the solution
f (with E = 0) of (VFP). Then
with
N
in L1 (RN
x ⇥ Rv ) as t ! 1,
f (t, x, v) ! fs (x, v)
fs (x, v) =
⇣ ZZ
Moreover, for any T > 0,
ZZ Z
|v|2 |f (t, x, v)
⌘ e (|v2 |/2+u0 (x))/✓
R
f0
.
(2⇡✓)N/2 e u0 /✓
fs (x, v)| dt dx dv ! 0,
(2.11)
⌧ ! 1,
]⌧,⌧ +T [⇥R2N
ZZ Z
]⌧,⌧ +T [⇥R2N
|rv
p
f
rv
p 2
fs | dt dx dv ! 0,
(2.10)
⌧ ! 1.
(2.12)
F. BOUCHUT AND J. DOLBEAULT
492
Theorem B. (VPFPc case). Given any solution f of (VPFPc) satisfying (2.1), (2.8)
and (2.9), we have
N
in L1 (RN
x ⇥ Rv ) as t ! 1,
f (t, x, v) ! fs (x, v)
with
fs (x, v) =
⇣ ZZ
f0
⌘e
(|v 2 |/2+u0 (x)+us (x))/✓
(2⇡✓)N/2
and where us is the unique solution of ( > 0)
us =
in LN/(N
1),1
⇣ ZZ
f0
⌘e
R
e
(2.13)
(2.14)
(u0 +us )/✓
(u0 (x)+us (x))/✓
R
e
(2.15)
(u0 +us )/✓
(RN ). Moreover, for any T > 0,
ZZ Z
|v|2 |f (t, x, v) fs (x, v)| dt dx dv ! 0,
⌧ ! 1,
(2.16)
]⌧,⌧ +T [⇥R2N
ZZ Z
]⌧,⌧ +T [⇥R2N
|rv
p
f
rv
p 2
fs | dt dx dv ! 0,
⌧ ! 1.
Theorem C. (VPFPn case). Given any solution f of (VPFPn) satisfying (2.1), (2.8)
and (2.9), two situations may occur. Either
Z
lim sup
|E(t, x)|2 dx = +1,
(2.17)
t!1
RN
or for any sequence (tn ) such that tn ! 1, there exists a stationary solution
fs (x, v) =
⇣ ZZ
f0
⌘e
⇣ ZZ
f0
(|v 2 |/2+u0 (x)+us (x))/✓
(2⇡✓)N/2
where us is a solution of ( < 0)
us =
in LN/(N
1),1
⌘e
R
e
(u0 +us )/✓
,
(2.18)
(u0 (x)+us (x))/✓
R
e
(2.19)
(u0 +us )/✓
(RN ), such that (after extraction of a subsequence)
f (tn , x, v) ! fs (x, v)
N
in L1 (RN
x ⇥ Rv )
The proofs of these Theorems rely on the estimate
ZZ
p
p
dA
=
|v f + 2✓rv f |2 dx dv
dt
R2N
where A(t) is defined by (1.12).
as n ! 1.
(2.20)
in ]0, 1[,
(2.21)
ON LONG TIME ASYMPTOTICS
493
Remarks. 1. For a given field E(t, x) 2 L1 (]0, T [⇥RN ) and a given datum f0 satisfying (2.8), f is uniquely determined by (2.2) and the regularity condition (2.1). The
problem (VFP) is consequently well-posed. However, for (VPFP), we do not know a priori if such a solution exists, and we therefore assume it does. Particularly, for attractive
forces, a solution may not exist for all times if N 4 (see E. Hörst [21]). Nevertheless,
the condition (2.9) imposed on E is at least acceptable in N = 3 dimensions, because
in this case, such a solution exists and is unique under the additional assumptions
ZZ
1
6
f0 2 L (R ),
|v|k f0 (x, v) dx dv < 1 for some k > 6,
R6
in the coulombic case (VPFPc) as well as in the newtonian one (VPFPn); a proof of
this claim can be found in F. Bouchut [3].
2. Each solution conserves the total mass
ZZ
ZZ
f (t, x, v) dx dv =
f0 (x, v) dx dv ⌘ M.
Particularly, the only solution with null mass is f = 0.
3. Two simple choices of u0 are u0 (x) = |x| or u0 (x) = C ln(1 + |x|), C > N ✓.
II.1. The linear problem. We consider the problem (VFP). Before proving Theorem A, we have to state some preliminary and quite technical results. We first write a
lemma, which states the possibility of integrating equation (2.2) with respect to (x, v).
Then we state two basic estimates, one for L2 initial datum (Proposition 2.2), and one
for L ln L initial datum (Proposition 2.3), proving in some sense that it is possible to
“multiply” the equation for f by some nonlinear functionals of f . This allows us to
study the decrease of the free energy, which is the key of the long time behavior, and
to obtain Theorem A.
Lemma 2.1. (Mass conservation). Let T > 0, F (t, x) 2 L1 (]0, T [⇥RN ), f0 2 L1 (R2N
x,v ),
H 2 L1 (]0, T [⇥R2N ), and consider the solution f 2 C([0, T ], L1 (R2N )) of
@t f + v · rx f + divv (F
Then we have
d
dt
ZZ
R2N
v)f
f (t, x, v) dx dv =
vf
in ]0, T [⇥R2N , f (0, .) = f0 .
=H
ZZ
R2N
H(t, x, v) dx dv
in ]0, T [.
(2.22)
(2.23)
Proof. The solution f of (2.22) is obtained as the unique fixed point of the equation
ZZ
f (t, x, v) =
G(t, x, v, ⇠, ⌫)f0 (⇠, ⌫) d⇠ d⌫
⇠,⌫
ZZZ
+
r⌫ G(s, x, v, ⇠, ⌫)(F f )(t s, ⇠, ⌫) d⇠ d⌫ ds
(2.24)
0<s<t,⇠,⌫
ZZZ
+
G(s, x, v, ⇠, ⌫)H(t s, ⇠, ⌫) d⇠ d⌫ ds,
0<s<t,⇠,⌫
where G(t, x, v, ⇠, ⌫) is the Green function, solution of
@t G + v · rx G
divv vG
G(0, x, v, ⇠, ⌫) = (x
vG
⇠, v
⌫)
= 0 in ]0, 1[⇥R2N
x,v ,
F. BOUCHUT AND J. DOLBEAULT
494
(see F. Bouchut [3]). Hence we may integrate (2.24) to obtain
ZZ
f (t, x, v) dx dv =
ZZ
f0 (⇠, ⌫) d⇠ d⌫ +
ZZZ
H(t
s, ⇠, ⌫) d⇠ d⌫ ds,
0<s<t,⇠,⌫
which is equivalent to (2.23).
Proposition 2.2. (L2 initial datum) Let F (t, x) 2 L1 (]0, T [⇥RN ) for any T > 0,
f0 2 L2 (R2N ), and consider the solution f 2 C ([0, 1[, L2 (R2N )) of
@t f + v · rx f + divv (F
v)f
vf
in ]0, 1[⇥R2N , f (0, .) = f0 .
=0
(2.25)
Then for each T > 0, r⌫ f 2 L2 (]0, T [⇥R2N ) and
kf (T, .)k2  eN
ZZZ
]0,T [⇥R2N
For each
2 C 2 (R) such that
00
T /2
kf0 k2 ,
(2.26)
|rv f |2 dt dx dv 
2 L1 ,
@t (f ) + v · rx (f ) + divv F (f )
1
kf0 k22 eN
2
T
.
(2.27)
(f ) solves
v · rv (f )
(f ) = N f
0
00
(f )
(f )|rv f |2.
(2.28)
Proof. Let ⇢n (t, x, v) be a sequence of smooth functions which tends to the Dirac mass
when n ! 1. We assume ⇢n to be of the form
⇢n (t, x, v) = ⇢tn (t)⇢xn (x)⇢vn (v),
and ⇢tn is supposed to have its support in the negative real axis. We define
fn = ⇢n ⇤ f 2 C 1 ([0, 1[, H m (R2N )).
t,x,v
We easily obtain from (2.25) that
hn ⌘ @t fn + v · rx fn + F · rv fn
= [(divx
divv (vfn )
divv )(v⇢n )] ⇤ f + (F [(rv ⇢n ) ⇤ f ]
v fn
(2.29)
(rv ⇢n ) ⇤ (F f )) ⌘
h(1)
n
+
h(2)
n ,
so that hn 2 L1 (]0, T [, L2 (R2N )). We have
t
h(1)
n (t, X) = ⇢n ⇤
t
Z
R2N
kh(1)
n kL1 (]0,T [, L2 (R2N ))  C
By (2.29) we get vrx fn
ZZ
divv )(v⇢x,v
n )(X
(divx
sup
0tT +1/n,|h|1/n
Y )(f (t, Y )
kf (t, X + h)
f (t, X)) dY,
f (t, X)kL2X ! 0.
n!1
vrv fn 2 L1 (]0, T [, L2 ), and therefore
fn (v · rx fn
v · rv fn ) =
N
2
ZZ
fn2
a.e. t > 0
ON LONG TIME ASYMPTOTICS
495
(consider the space H 1 (R2N ) of all u 2 L2 such that rx u and rv u 2 L2 . The above
identity is true with u 2 H 1 such that vrx u
vrv u 2 L2 instead of fn , which can be
seen by truncating u and using the dominated convergence theorem). Taking the scalar
product of (2.29) and fn for almost every t > 0, we get
d
dt
ZZ
fn2
2
ZZ
N
2
fn2 +
ZZ
|rv fn |2 =
ZZ
|rv fn |2 =
ZZZ
fn hn ,
hence
⇥
ZZ
fn2 ⇤T
2 0
ZZZ
N
2
]0,T [⇥R2N
We know that
]0,T [⇥R2N
ZZZ
and we have
ZZZ
fn h(2)
n
=
ZZZ
fn2 +
ZZZ
fn hn .
(2.30)
]0,T [⇥R2N
fn h(1)
n ! 0,
rv fn ⇢n ⇤ (F f )  C
ZZZ
|rv fv |2
1/2
,
so that by (2.30), rv fv is bounded in L2 (]0, T [⇥R2N ) independently of n. We hence
obtain
2
rv f 2 L2 , rv fn ! rv f in L2 , h(2)
n ! 0 in L .
We get from (2.30) that
ZZ
⇥
f 2 ⇤T
2 0
0
Multiplying (2.29) by
0
ZZZ
f2 +
]0,T [⇥R2N
which gives (2.26) and (2.27).
Now let 2 C 2 (R) such that
rv (fn ) =
ZZZ
N
2
00
]0,T [⇥R2N
2 L1 . We have
(fn )rv fn ,
v
(fn ) =
|rv f |2 = 0
0
(fn )
(fn ) 2 C 2 ,
v fn
00
+
(fn )|rv fn |2 .
(fn ), we get
@t (fn )+v · rx (fn ) + F · rv (fn )
= N fn
0
(fn )
00
v · rv (fn )
(fn )|rv fn |2 + hn
0
v
(fn )
(fn ).
Since is at most quadratic at infinity, we have (fn ) ! (f ) in L1 locally,
0
(f ) in L2 locally, and we may pass to the limit to obtain (2.28). ⇤
In the same spirit we have
0
(fn ) !
Proposition 2.3. (L ln L initial datum) Let F (t, x) 2 L1 (]0, T [⇥R2N ) (8T > 0),
f0 2 L1 (R2N ) such that
f0
0,
ZZ
R2N
(1 + |v|2 + u0 (x) + ln+ f0 (x, v))f0 (x, v) dx dv < 1,
(2.31)
F. BOUCHUT AND J. DOLBEAULT
496
where u0 satisfies (2.6) and (2.7), and consider the solution f 2 C([0, 1[, L1 (R2N )) of
Lf ⌘ @t f + v · rx f + divv (F
v)f
vf
Then
f ln f 2 C([0, 1[, L1 (R2N )),
p we have
2
rv f 2 L (]0, T [⇥R2N ), and
= 0 in ]0, 1[⇥R2N , f (0, .) = f0 . (2.32)
p
f 2 C([0, 1[, L2 (R2N )),
p
L(f ln f ) = N f 4 |rv f |2 in ]0, 1[⇥R2N ,
p
p
rv f = rv f /2 f , rv f 2 L1 (]0, T [⇥R2N ), |v||rv f | 2 L1 (]0, T [⇥R2N ),
ZZ
ZZ
ZZ
p
d
f ln f = N
f 4
|rv f |2 in ]0, 1[.
dt
p
p
Moreover, for each ✏ > 0, (✏ + f )
✏ 2 C([0, 1[, L2 (R2N )) solves
p
p
p
p
f
|rv f |2
p
L( "+f
") = N
( "+f
") +
,
4 ("+f )3/2
2 "+f
p
p
p
rv ( " + f
") = rv f /2 " + f 2 L2 (]0, T [⇥R2N ).
(2.33)
(2.34)
(2.35)
(2.36)
(2.37)
Estimate (2.36) will be useful in treating the problem (VPFP). However, it is not
necessary for (VFP).
Proof. We have |v|2 f , u0 f 2 C([0, 1[, L1 ). This yields a bound on f ln f (ln f =
max( ln f, 0)); namely, for any K > 0,
f ln f  f +
1
(|v|2 /2+u0 (x))f +(1+(|v|2 /2+u0 (x))/K)e
K
1
e
(|v|2 /2+u0 (x))/K
, (2.38)
which can be seen by examining this inequality in the domains where f > e 1 ,
e 1 exp (|v|2 /2 + u0 (x))/K < f  e 1 , and f  e 1 exp (|v|2 /2 + u0 (x))/K respectively. We decompose the proof into two steps.
First case. Assume f0 2 L2 . Define
" (f )
= (" + f ) ln(1 + f /") + f ln " 2 C([0, 1[, L1 ).
0
" (f )
= ln(1 + f /") + 1 + ln ",
We have
00
" (f )
= 1/(" + f ).
By Proposition 2.2. and Lemma 2.1. we get
d
dt
L
ZZ
= N (f " ln(1 + f /"))
|rv f |2 /(" + f ),
ZZ
ZZ
|rv f |2
(f
)
=
N
(f
"
ln(1
+
f
/"))
.
"
"+f
" (f )
Since f 2 C([0, 1[, L2 ), we have f ln f 2 L1 (]0, T [, L1 ) and
" (f )
f ln f = " ln(1 + f /") + f ln(1 + "/f ),
so by dominated convergence,
ZZ
|
" (f )
f ln f | ! 0,
" ! 0.
(2.39)
ON LONG TIME ASYMPTOTICS
But we have
ZZ
⇥
⇤T
" (f ) 0 = N T
ZZ
f0
ZZZ
N
497
ZZZ
" ln(1 + f /")
]0,T [⇥R2N
|rv f |2
,
"+f
]0,T [⇥R2N
so |rv f |2 /(✏ + f ) is bounded in L1 (]0, T [⇥R2N ) independently of ✏, and consequently
|rv f |2 /f 2 L1 (]0, T [⇥R2N ),
Hence
" (f )
|rv f |2
|rv f |2
!
"+f
f
in L1 , " ! 0.
! g in C([0, T ], L1 ) solution of
|rv f |2
,
f
Lg = N f
g(0) = f0 ln f0 .
We deduce (2.33) since g = f ln f . Equation (2.36) is valid by Proposition 2.2, and
letting ✏ ! 0, since
p
p
rv f
rv ( " + f
") = p
2 "+f
is bounded in L2 (]0, T [⇥R2N ), we get
p
rv f 2 L2 (]0, T [⇥R2N ).
All remaining assertions are quite obvious.
Second case. General case. We have f 2 C([0, 1[, L1 ) \ C(]0, 1[, L2 ). The real
difficulty is for small time. We prove precisely why f ln f 2 C([0, 1[, L1 ). Let us fix
t > 0. The function " defined in (2.39) is convex, hence we can write for X, Y 2 R2N ,
N t
f0 (X))
" (e
Since for x
" (f (t, Y
0,
x ln " 
(eN t f0 (X)
))
" (x)
f (t, Y ))
0
" (f (t, Y
))
 2x + x ln+ x + x ln+ ",
we have " (eN t f0 (X)) 2 L1X . Denoting by k the solution of Lk = 0, k(0) =
we have for all Y that
eN
k(t, .)
t
" (f (t, Y
[eN t f (t, .)
))
eN t f (t, Y )]
0
" (f (t, Y
))
0.
N t
f0 ),
" (e
0.
Choose . = Y . We get
|
Hence
Z
|
" (f (t, .))
" (f (t, Y
Z
f (t, Y ) ln " 
" (f (t, Y
2 L1 ,
))|  e
N t
Z
" (f (t, Y
))|  e
|
N t
" (e
N t
))  e
N t
k(t, Y ),
|k(t, Y )| + | ln "|f (t, Y ).
f0 )| + | ln "|
Z
f0
(2eN t f0 + eN t f0 ln+ (eN t f0 ) + eN t f0 ln+ " + | ln "|eN t f0 ) + | ln "|
Z
Z
Z
 2 f0 + f0 ln+ (eN t f0 ) + (ln+ " + 2| ln "|) f0
Z
Z
 (2 + 3| ln "| + N t) f0 + f0 ln+ f0 .
e
N t
Z
f0
F. BOUCHUT AND J. DOLBEAULT
498
But we have, for x
0,
x ln+ x 
so
Z
1 (x)
 2x + x ln+ x,
f ln+ f  (2 + N t)
Z
f0 +
Z
f0 ln+ f0 .
Repeating the same computation with the convex function x ln+ x we get in fact
ZZ
ZZ
ZZ
+
+
f ln f 
f0 ln f0 + N t
f0 .
(2.40)
Define f0n = f0 1f0 n 2 L1 \ L1 , and let fn be the solution of Lfn = 0, fn (0, .) = f0n .
We have fn ! f in C([0, T ], L1 ), and by (2.40),
(f
fn ) ln+ (f
fn ) ! 0 in L1 (]0, T [, L1 ).
We have
|
" (f )
" (fn )|
 (f
fn )(|1 + ln "| + ln(1 + f /")),
and, for ⌘ > 0,
(f
fn ) ln(1 + f /")  ⌘(1 + f /")1f >" + (f
 2⌘f /" + (f
fn ) ln+
fn )(1 + ln+ (f
f
fn
⌘
+ (f
fn )) + (f
fn ) ln 2
1
fn ) ln+ .
⌘
We conclude that " (fn ) ! " (f ) in L1 ((0, T ), L1 ) and that " (f ) 2 C([0, 1), L1 ).
For ⌘ > 0, f 2 C([⌘, 1), L2 ) so we may apply the results of the first case. By using
the same ideas as in the first case, we conclude by letting ✏ ! 0 that " (f ) ! f ln f
in L1 (]0, T [, L1 ), so that f ln f 2 C([0, T ], L1 ). The remaining details are left to the
reader.
Proof of Theorem A. We first prove the decrease of the free energy. Consider a
solution of (VFP) (neutral plasma case). We have |v|2 f 2 C([0, 1[, L1 ),
ZZ
ZZ
ZZ
ZZ
d
|v|2
f=
v · ru0 f
|v|2 f + N
f
(2.41)
dt
2
(obtained formally by multiplying (2.2) by |v|2 ), u0 f 2 C([0, 1[, L1 ),
ZZ
ZZ
d
u0 f =
v · ru0 f
dt
(obtained formally by multiplying
(2.2) by u0 ),pand by Proposition 2.3
p
f ln f 2 C([0, 1), L1 ), f 2 C([0, 1[, L2 ), rv f 2 L2 (]0, T [⇥R2N ),
p
L(f ln f ) = N f 4 |rv f |2
(obtained formally by multiplying (2.2) by 1 + ln f ), and by Lemma 2.1,
ZZ
ZZ
ZZ
p
d
f ln f = N
f 4
|rv f |2 in ]0, 1[.
dt
(2.42)
(2.43)
(2.44)
ON LONG TIME ASYMPTOTICS
Since
we have
499
p |rv f |
p
p
|v||rv f | = |v| f p
= 2|v| f |rv f | 2 L1 (]0, T [⇥R2N ),
f
ZZ
v · rv f =
N
ZZ
f
for a.e. t > 0,
and by (2.41), (2.42) and (2.44), setting
ZZ
ZZ
ZZ
|v|2
A(t) ⌘
f+
u0 f + ✓
f ln f,
2
(2.45)
we obtain
d
A(t) =
dt
=
=
ZZ
ZZ
ZZ
|v|2 f + N
|v|2 f
|v
2
ZZ
ZZ
ZZ
f + ✓(N
2
v · rv f
p
rv f
f + ✓ p |2 =
f
ZZ
f
ZZ
|v
4
ZZ
|rv
|rv f |2
f
p 2
f| )
(2.46)
p
p
f + 2✓rv f |2 .
Now since dA/dt  0 we have A(t)  A(0). But using (2.38), we can bound A(t) from
below:
ZZ
ZZ
ZZ
ZZ
|v|2
+
A(t)
f+
u0 f + ✓
f ln f ✓
f ln f
2
ZZ
ZZ
ZZ
(2.47)
(1 ✓/K)
(|v|2 /2 + u0 )f + ✓
f ln+ f ✓
f0
ZZ
2
✓e 1
(1 + (|v|2 /2 + u0 (x))/K)e (|v| /2+u0 (x))/K .
Assumption (2.7) ensures that we can find some K > ✓ such that the last integral is
finite. Hence we obtain
ZZ
ZZ
|v|2
(1 ✓/K)
(
+ u0 )f + ✓
f ln+ f  A(0) + C,
2
and conclude that
ZZ
f,
ZZ
|v|2
f,
2
ZZ
u0 f,
ZZ
+
f ln f,
ZZ
f ln f
are bounded independently of t
0. According to Dunford-Pettis’ Theorem, these
estimates imply that the functions (f (t))t 0 are weakly compact in L1 (R2N ). In order
to get the strong convergence result stated in Theorem A, we need a more precise result,
which is that the family
f⌧ (t, x, v) ⌘ f (t + ⌧, x, v),
⌧
0,
is strongly compact in C([0, T ], L1 (R2N )) for any T > 0.
To obtain this result, we read (2.2) as
@t f + v · rx f
divv (vf )
vf
= ru0 · rv f, f (0, .) = f0 .
(2.48)
(2.49)
F. BOUCHUT AND J. DOLBEAULT
500
Obviously,
f⌧ is also a solution of this equation, with initial datum f (⌧, .), and by (2.44),
p
rv f⌧ is bounded in L2 (]0, T [⇥R2N )). Since
p
p
rv f⌧ = 2 f⌧ rv f⌧
is bounded in L2 (]0, T [, L1 (R2N )), we may apply to f⌧ a new compactness lemma
explicitly stated in Appendix 1, and conclude that for any ⌘ > 0, and ! bounded
open subset of R2N , f⌧ is compact in C([⌘, T ], L1 (!)). The above stated compactness
result (2.49) easily follows, taking into account the weak compactness. We are now
in a position to prove the strong convergence (2.10), just by considering converging
subsequences. Let ⌧n ! 1 such that f⌧ n ! f1 in C([0, T ], L1 ) for any T > 0. We just
have to prove that f1 does not depend on t and is indeed fs .
Obviously, f1 2 C([0, 1[, L1 ) is nonnegative, and solves (VFP). But (2.46) can be
written
ZZZ
p
p
A(⌧ + T ) A(⌧ ) =
|v f⌧ + 2✓rv f⌧ |2 dx dv dt,
]0,T [⇥R2N
and since A(t) is non-increasing and bounded, it must have a limit when t ! 1, and
we conclude that
p
p
v f⌧ + 2✓rv f⌧ ! 0, ⌧ ! 1, in L2 (]0, T [, ⇥R2N ),
(2.50)
so that
2✓rv
Denoting
p
f1 =
2
g1 ⌘ e|v|
v
p
f1 .
(2.51)
/2✓
f1 ,
(2.52)
p
equation (2.51) reads rv g1 = 0, hence g1 only depends on t and x, and the equation
satisfied by f1 can be rewritten on g1 as
1
@t g1 + v · rx g1 =
ru0 · vg1 .
✓
We consequently obtain, identifying the coefficients of the v-polynomial with 0,
1
@t g1 = 0, rx g1 =
ru0 g1 ,
(2.53)
✓
and g1 (x) 2 L1 (RN ). Equation (2.53) is easily solved by
g1 = C e
u0 /✓
,
f1 = C e
(|v|2 /2+u0 (x))/✓
.
Since the mass is conserved, the constant C is determined, and we obtain f1 = fs ,
which achieves the proof of (2.10). To prove (2.12), we just observe that (2.50) and the
computations (2.46) lead to
ZZZ
ZZZ
ZZ
ZZ
ZZ
p
p
|v|2 f⌧ + 4 ✓
|rv f⌧ |2 ! 2N T f0 = T |v|2 fs + 4 ✓T |rv fs |2 ,
]0,T [⇥R2N
]0,T [⇥R2N
and by lower-semi-continuity arguments that
ZZZ
ZZ
ZZZ
|v|2 f⌧ ! T
|v|2 fs ,
]0,T [⇥R2N
R2N
]0,T [⇥R2N
|rv
ZZ
p 2
f⌧ | ! T
(2.12) follows by standard measure theory arguments. ⇤
An obvious consequence follows.
R2N
|rv
p 2
fs | .
ON LONG TIME ASYMPTOTICS
501
Corollary 2.4. For any M 0, the problem (VFP) admits a unique stationary solution
of mass M , which is given by
2
fs (x, v) = M
e (|v| /2+u0 (x))/✓
R
.
(2⇡✓)N/2 e u0 /✓
(2.54)
Remarks. Theorem A deals with the solution of a linear equation, which one may
consider as quite simple, although its solution is not explicit. We may wonder whether
the convergence rate is exponential or not, which should involve a spectral analysis.
Notice that by a simple density argument, the convergence (2.10) also holds for the
solutions f of (2.2) corresponding to any initial datum f0 2 L1 , not necessarily satisfying
(2.8), and not necessarily nonnegative. We have chosen this formulation because the
proof of Theorem A involves a nonlinear analysis,pwhich is based on writing equations
on nonlinear functions of f , namely f ln f and f . It is the essential reason why
this analysis can be carried out when considering the nonlinear coupling with Poisson
equation, which is the subject of the next section.
II.2. Coupling with the Poisson equation. We now consider a solution of
the Fokker-Planck equation (2.2), where the field E is coupled with f by the Poisson
equation (2.4). From now on, the dimension N is supposed to be at least 3.
Before proving Theorems B and C, we first consider steady states for the problem
(2.1), (2.2), (2.4)–(2.9) in the repulsive case ( > 0), which means that we look for
solutions f of (VPFPc) which are independent of time.
Proposition 2.5. (Steady states of (VPFPc)). For any M
us 2 L2N/(N
2)
(RN ) such that rus 2 L2 , e
0, there exists a unique
(u0 +us )/✓
2 L1 ,
(2.55)
solution of ( > 0)
Furthermore, us
e (u0 +us )/✓
us = M R (u +u )/✓ .
e 0 s
(2.56)
0 and M 7! us is non-decreasing. The corresponding distribution
2
e (|v| /2+u0 (x)+us (x))/✓
R
fs (x, v) ⌘ M
(2⇡✓)N/2 e (u0 +us )/✓
(2.57)
is the unique steady solution of (2.1)–(2.9) of mass M .
We refer to Section III and the references therein for a discussion of this problem.
The uniqueness of the stationary solution is indeed an easy consequence of Theorem B.
Proof of Theorem B. We essentially follow the proof of Theorem A. The new difficulty is that the field E is only bounded in L2x , and f in L1x,v , hence the product Ef is unbounded. To overcome this difficulty, we use the renormalization technique introduced
by R.J. DiPerna and P.L. Lions in [11, 13]. We notice that E 2 C([0, 1[, Lp (RN )),
N
N 1 < p < 1, and
Z
ZZ
d |E|2
dx =
E · v f dx dv.
dt 2
F. BOUCHUT AND J. DOLBEAULT
502
Applying Proposition 2.3 and performing the same computations as in Theorem A, we
obtain that the free energy
A(T ) ⌘
ZZ
satisfies
|v|2
1
f+
2
Z
|E|2
+
2
ZZ
dA
=
dt
|2✓rv
and we again deduce that
ZZ
f,
ZZ
|v|2
f,
2
Z
ZZ
u0 f + ✓
ZZ
f ln f 2 C([0, 1[)
p
p
f + v f |2 dx dv
|E|2
,
2
ZZ
u0 f,
ZZ
in ]0, 1[,
ZZ
f ln+ f,
f ln f
(2.58)
(2.59)
(2.60)
are bounded independently of t
0. Hence (f (t))t 0 is weakly compact in L1 (R2N ).
We define
f⌧ (t, x, v) = f (⌧ + t, x, v).
(2.61)
We now prove that
(f⌧ )⌧
0
is strongly compact in C([0, T ], L1 (R2N )) for any T > 0.
(2.62)
We again apply the compactness lemma of Appendix 1, to
'" (f ) ⌘
p
"+f
p
" 2 C([0, 1[, L2 )
(2.63)
(for a fixed ✏). By Proposition 2.3 we have
L'" (f ) = N
f
p
2 "+f
'" (f ) +
|rv f |2
,
4 (" + f )3/2
or equivalently
f
p
2 "+f
L0 f ⌘ @t f + v · rx f
L0 '" (f ) = N
|rv f |2
4 (" + f )3/2
divv vf
v f.
'" (f ) +
(E
ru0 )rv '" (f ),
(2.64)
As before we have
p
p
f⌧ + v f⌧ ! 0
⌧ !1
p
4✓2 |rv f⌧ |2 |v|2 f⌧ ! 0
2✓rv
⌧ !1
vf⌧
rv '" (f⌧ ) + p
2✓ " + f⌧
! 0
⌧ !1
in L2 (]0, T [⇥R2N ),
in L1 (]0, T [⇥R2N ),
(2.65)
in L2 (]0, T [⇥R2N ).
Let (v) 2 Cc1 (RN ). We have
L0 [ (v)'" (f⌧ )] = L0 ['" (f⌧ )]
v · rv '" (f⌧ )
(
v
'" (f⌧ ) + 2rv rv ('" (f⌧ )));
ON LONG TIME ASYMPTOTICS
503
the second term is bounded in L1 (]0, T [, L2 ), the third in L1 (]0, 1[, L2 ) and the fourth
in L2 (]0, T [⇥R2N ). For the first we have
L0 ['" (f⌧ )] = N
E⌧
⇥
f
p ⌧
2 " + f⌧
ru0 )(rv ('" (f⌧ ))+
⇤
'" (f⌧ ) +
|rv
p 2
f⌧ |
vf
p ⌧
+ (E⌧
2✓ " + f⌧
|v|2 f⌧
f⌧
4✓2 (" + f⌧ )3/2
vf⌧
|v|2 f⌧2
ru0 ) p
+ 2
.
4✓ (" + f⌧ )3/2
2✓ " + f⌧
The first term is bounded in L1 (]0, T [, L2 ), the second tends to 0 in L1 (]0, T [⇥R2N ),
the third multiplied by (v) tends to 0 in L1 (]0, T [⇥R2N ), the fourth multiplied by (v)
is bounded in L1 (]0, T [, L1 ), and the fifth is bounded in L1 (]0, T [, L1 ).
Applying successively the compactness lemma to all these second members, we obtain
that for any ⌘ > 0, and ! ⇢ R2N bounded,
(v)'" (f⌧ )
is compact in C([⌘, T ], L1 (!)),
hence '" (f⌧ ) is compact in C([⌘, T ], L1 (!)), and also f⌧ (we recall that ✏ is fixed). We
conclude therefore that f⌧ is compact in C([0, T ], L1 (R2N )).
Since assertion (2.16) is proved in the same way as in Theorem A, it just remains for
Theorem B to prove that if ⌧n ! 1 is such that
f⌧n ! f1 in C([0, T ], L1 ) for any T > 0,
n!1
(2.66)
then f1 is independent of t and is indeed fs .
For proving it, we pass to the limit in the renormalized equation on '" (f⌧ ), and then
let ✏ ! 0.
We have obviously f1 2 C([0, 1[, L1 ), f1 0,
rv
p
f1 =
vp
f1 .
2✓
(2.67)
We pass to the limit in (2.64), using (2.65) and obtain
L0 ['" (f1 )] = N
f
p 1
2 "+f1
'" (f1 ) +(E1 ru0 )
2
vf
|v|2 f1
p 1
+ 2
.
2✓ "+f1 4✓ ("+f1 )3/2
By Lebesgue’s theorem we obtain, letting ✏ ! 0,
L0
Let g1 ⌘ e
|v|2
4✓
p
f1 =
N p
f1 + (E1
2
ru0 )
p
vp
f1 + 2 |v|2 f1 .
2✓
4✓
(2.68)
p
f1 2 C([0, 1[, L2loc ). By (2.67) we get rv g1 = 0, and by (2.68),
L0 g1 = e
|v|2
4✓
p
N
f1 + (E1
p
v f1
ru0 )
.
2✓
Since g1 only depends on (t, x), we obtain
@t g1 + v · rx g1 = (E1
ru0 )
v
g1 .
2✓
F. BOUCHUT AND J. DOLBEAULT
504
Hence
@t g1 = 0,
rx g1 = (E1
ru0 )
g1
,
2✓
(2.69)
with g1 (x) 2 L2 (RN ). But
E1 =
u1 2 L2N/(N
ru1 ,
2)
,
u1
0,
which comes from the limit in Poisson equation. Consequently, (2.69) is solved by
g1 = C e
u0 +u1
2✓
.
(2.70)
This result is not easily proved, because we cannot multiply (2.69) by e(u0 +u1 )/2✓ which
is a priori not locally integrable. A possible way to overcome this difficulty is to consider
ln" (x) ⌘ ln x if x
", ln " if x  ".
The gradient r ln" (g1 ) is bounded in L2 + L1 , hence after extraction
ln"0 (g1 )
C"0 ! · in L1loc ,
and we deduce (2.70). The details are left to the reader. This yields
f1 (x, v) = C 2 e
|v|2
2✓
e
(u0 (x)+u1 (x))/✓
,
(x, v) 2 R2N
and the constant C is determined by mass conservation. Inserting into the Poisson
equation we get
⇣ ZZ
⌘ e (u0 +u1 )/✓
u1 =
f0 R (u +u )/✓ ,
e 0 1
which implies that u1 = us and f1 = fs . This ends the proof of Theorem B. ⇤
The proof of Theorem C is following the previous one, it only di↵ers by the fact that
the free energy is not necessarily bounded from below (the internal potential energy is
negative), and by the nonuniqueness of the solution of (PBn).
Remarks. 1. In some cases, it is possible to prove that the stationary problem (PBn)
has a unique solution. Hence, if
Z
lim sup
|E(t, x)|2 dx < 1,
t!1
RN
then f (t, x, v) ! fs (x, v) in L1 (R2N ) exactly as in the coulombic case (Theorem B).
2. If
Z
lim sup
|E(t, x)|2 dx = 1,
t!1
RN
it is possible to prove, using an estimate close to (2.38), that
ZZ
f (t, x, v)(u/2 + ✓ ln f (t, x, v))(1
1St ) dx dv,
St = {(x, v) : 1 < f (t, x, v) < exp (|v|2 /2 + u0 (x) + u(t, x)/2)/✓},
ON LONG TIME ASYMPTOTICS
505
is bounded from below independently of t, and that
Z
ZZ
1
2
|E(t, x)| dx ⇠
f (t, x, v)u(t, x) dx dv.
t!1
RN
St
II. 3. The small mass approximation (coulombic interaction). Consider the
problem stated in Section II.2. in the coulombic case (repulsive interaction case). In a
physical context, the constants , , and the potential u0 depend on the mass m. We
look for an asymptotic behavior of f when m ! 0. Following Stokes’ law, the friction
coefficient is given by
= 6⇡a⌘/m
for spherical particles of radius a, where ⌘ denotes the coefficient of viscosity of a
surrounding fluid. We refer to S. Chandrasekhar [7 ,8] for a derivation
and physical
p
comments on the Fokker-Planck model. If a is proportional to m (relevant for a
two-dimensional problem), we have
= ¯ /m,
¯
= ¯/m1/2 , ✓ = ✓/m,
u0 = ū0 /m,
= ¯ /m3/2 ,
(2.71)
where the overlines denote constants independent of m. If we write the new distribution
function in the form
f¯(t̄, x, v̄) = m
N/2
f (m1/2 t̄, x, v̄/m1/2 ),
(2.72)
then it solves
@t̄ f¯ + v̄ · rx f¯ + divv̄ (Ē
Ē(t, x) = mE(m1/2 t̄, x) =
rv¯0
¯
|S N
v)f¯
¯ = 0, f¯(0, .) = f¯0 ,
Z
x
⇤
⇢
¯
,
⇢
¯
=
f¯dv̄.
1 | |x|N
¯
v̄ f
This case corresponds to a small mass, large friction and large di↵usion asymptotic
limit, with a fixed total mass, or a strong interaction asymptotic limit.
Hence we have by Theorem B that
f¯(t̄, x, v̄) ! f¯s (x, v̄) ⌘
t̄!1
ūs = ¯
Denoting
⇣ ZZ
⇣ ZZ
we have
us =
R2N
¯
2
e (|v̄| /2+ū0 +ūs )/✓
R
¯ N/2 e (ū0 +ūs )/✓¯ ,
(2⇡ ✓)
fs = mN/2 f¯s (x, m1/2 v),
⇣ ZZ
and we obtain, for any t > 0, in L1 (RN
x )
|f (t, x, v)
⌘
⌘ e (ū0 +ūs )/✓¯
f¯0 R (ū +ū )/✓¯ .
e 0 s
us = ūs /m,
ZZ
f¯0
⌘ e (u0 +us )/✓
f0 R (u +u )/✓ ,
e 0 s
fs (x, v)| dx dv ! 0,
m!0
⇢(t, x) !
m!0
⇣ ZZ
⌘ e (ū0 +ūs )/✓¯
f¯0 R (ū +ū )/✓¯ .
e 0 s
F. BOUCHUT AND J. DOLBEAULT
506
III. Stationary solutions of the Vlasov-Poisson-Fokker-Planck system. The
Poisson-Boltzmann equation. We are now concerned with the Maxwellian solutions
f (t, x, v) = (2⇡✓)
N/2
|v|2 /2✓
⇢(t, x)e
0, x, v 2 RN ,
, t
(3.1)
of the Vlasov-Poisson-Fokker-Planck system
@t f + v · rx f
(rx u + rx u0 )rv f = divv ( vf + rv f ),
Z
u=
f (t, x, v) dv = ⇢(t, x).
RN
(VFP)(3.2)
(3.3)
Here > 0 corresponds to the coulombic case and < 0 corresponds to the newtonian
case. We assume that
N = 3,
RR
the mean velocity is zero:
vf (t, x, v) dx dv = 0,
the temperature is fixed by ✓ = / ,
the total mass is given by
ZZ
Z
M=
f (t, x, v) dx dv =
⇢(t, x) dx.
R3
Inserting (3.1) into (3.2), we can see that
@t ⇢(t, x) = 0,
1
(rx u + rx u0 )
✓
rx ln ⇢ =
for all (t, x) such that ⇢(t, x) 6= 0. Hence ⇢ does not depend on t and ⇢(x) = C ·
e (u+u0 )/✓ , where C is such that
Z
Z
M=
⇢(x) dx = C ·
e (u+u0 )/✓ dx.
R3
R3
The potential u is therefore a solution of the Poisson-Boltzmann equation
e
e
R3
u= MR
(u+u0 )/✓
(u+u0 )/✓
dx
.
(PB)(3.4)
We will refer to this equation with (PBc) in the coulombic case ( > 0) and with (PBn)
in the newtonian case ( < 0). The solutions are obviously defined up to an additive
constant, when boundary conditions are not specified. Let us assume that
lim u(x) = 0.
|x|!1
(3.5)
This “limit” of course depends on the functional space to which the solution of (PB)
belongs.
III.1. Coulombic case. Existence and uniqueness results. Let us define
⇢0 = exp (u0 /✓). The Poisson-Boltzmann equation now reads
⇢0 e
⇢ e
R3 0
u= MR
u/✓
u/✓ dx
,
lim u(x) = 0.
|x|!1
We have the following results (see [2], [15], [16, 17]).
(PBc)(3.6)
ON LONG TIME ASYMPTOTICS
507
Theorem 3.1. If ⇢0 is a nonnegative function of L1 (R3 ) such that ⇢0 is not identically
equal to 0, ✓ > 0, M
0, then there exists a unique solution u of (PBc) in the
Marcinkiewicz space L3,1 (R3 ), and ru belongs to L3/2,1 (R3 ). Moreover,
u(x) =
1
⇢0 e
⇤ MR
4⇡|x|
⇢ e
R3 0
u/✓
u/✓
dx
,
and the solution belongs to D1,2 (R3 ) = {⌫ 2 L6 (R3 ) : r⌫ 2 L2 (R3 )}.
This last result is proven using the minimization of a C 1 convex functional defined on
D (R3 ). The general case is a consequence of the Hardy-Littlewood-Sobolev inequality
and of a weak maximum principle.
1,2
III.2. Newtonian case. The situation is much more complicated than in the
coulombic case. First, we state results when M is small. For large masses, we will
exhibit two examples showing very di↵erent behaviors. Like before, let us define ⇢0 =
exp (u0 /✓), and V = u/✓, µ = | |M/✓. Then
⇢0 eV
⇢ eV dx
R3 0
V = µR
(PBn)(3.7)
and V satisfies the boundary condition
lim V (x) = 0.
|x|!1
(3.8)
The results are very similar to those obtained by [23] in the case of a bounded domain.
III.2.1. Small mass results.
Theorem 3.2. Let us assume that ⇢0 is a nonnegative function of L1 (R3 ) \ L1 (R3 )
such that ⇢0 is not identically equal to 0. Then there exists a constant µ0 > 0 such
that for any µ 2 [0, µ0 [ there exists a bounded continuous solution V of (PBn) satisfying
(3.8).
Proof. The proof relies on a fixed point argument like in [16, 17] (written for the
coulombic case) or as in [23] (for a bounded domain). Let us consider
X = {v 2 L1 (R3 ) : 0  v  1}, T : L1 (R3 ) ! L1 (R3 ),
1
⇢0 ev
v ! Tv = µ
⇤R
;
4⇡|x|
⇢ ev dx
R3 0
X is stable under the action of T if
µ
1
4⇡|x|
⇤ ⇢0
k⇢0 k1
1
e  1.
It is easy to see that X is a closed convex subset of L1 (R3 ).The point is to prove
that T (X) is precompact for µ small enough. Let us state the following lemma, due to
Dressler (see [16, 17]).
F. BOUCHUT AND J. DOLBEAULT
508
Lemma 3.3. Let d 3, G(x) = (|S d 1 |·|x|d 2 ) 1 and B(t) = {f 2 L1 (Rd )\L1 (Rd ) :
kf k1  1, kf k1  t}. Then for all f 2 B(t), G ⇤ f is di↵erentiable,
@
@
(G ⇤ F ) = (
G) ⇤ f
@xi
@xi
and there exists a constant m(d, t) such that
kG ⇤ f k1  m(d, t)
and
k
@
(G ⇤ f )k1  m(d, t).
@xi
Consequently, {krG⇤f k1 : f 2 B(t)} is bounded, which implies that the family (G⇤f ),
f 2 B(t) is equicontinuous.
If (vn )n2N is a sequence of functions of X, then
1
⇢0 evn
⇤R
,
4⇡|x|
⇢ evn dx
R3 0
n 2 N,
is equicontinuous, and according to Ascoli’s Theorem, there exists a continuous function
f 0 such that, up to extraction of a subsequence,
1
⇢0 evn
⇤R
! f.
4⇡|x|
⇢ evn dx
R3 0
Therefore, µf = lim T vn (in the sense of the uniform convergence) is such that
kµf k1  µm 3,
ek⇢0 k1
.
k⇢0 k1
Obviously, for µ small enough,
kµf k1  1
and µf belongs to X.
Proposition 3.4. For all A > 1, for µ small enough, there is at most one solution V
of (PBn) such that kV k1  A, and this solution satisfies kV k1  1.
Proof. The proof of this uniqueness result relies on a contraction argument in L1 . Let
us choose A > 1, U and V belonging to L1 , and such that
max(kU k1 , kV k1 )  A.
We have
kT U
µ
µ
T V k1 = µ
1
⇢0 eU
⇤ R
4⇡|x|
⇢0 eU
1
eU eV
⇤ ⇢0 R
4⇡|x|
⇢0 eU dx
1
4⇡|x|
⇤ ⇢0
k⇢0 k1
1 A
1
+µ
e (1 + eA )kU
⇢ eV
R 0
⇢0 eV
1
⇤ ⇢0
4⇡|x|
V k1  kU
1
1
eA R
1
⇢0 eU
V k1 ,
R
1
⇢0 eV
1
ON LONG TIME ASYMPTOTICS
for
µ
509
k⇢0 k1
.
1
eA (1 + eA )k 4⇡|x|
⇤ ⇢0 k1
Using again the theorem, we can also state the following remark.
Remark. If there exists a solution V of (PBn) such that kV k1
is not unique.
1, then this solution
III.2.2. Large mass results. The following examples (with N=3) show that the
situation for large masses is strongly correlated with the local behavior of ⇢0 . In the first
example, solutions exist for arbitrarily large µ. On the opposite, in the second example,
there exists a critical parameter µ0 such that there are no solutions for µ larger than
µ0 .
Example 1. Let us consider the case where
⇢0 is nonnegative and radially symmetric,
⇢0 2 L1 , 8r 2 (0, 1), ⇢0 (r) = e
1/r2
/r5 .
(3.9)
Proposition 3.5. If ⇢0 satisfies conditions (3.9), then for all µ0 > 0, there exists
µ µ0 such that (PBn) has at least one solution.
Proof. Let us consider the equation (for all ⌘ 2]0, 1[)
⌘
1 d 2 d ⌘
(r
u ) = ⇢0 eu ,
2
r dr
dr
with the boundary conditions
d ⌘
u (0) = 0,
dr
u⌘ (0) =
1
⌘2
ln
4(⌘ 3 + 2)
.
⌘
The term ⇢0 exp u⌘ belongs to L1 (R3 ), because u⌘ is decreasing, and we shall prove that
(⌘) =
Z
R3
⌘
⇢0 eu dx
1
⌘
1 + O(⌘) as ⌘ ! 0+ ,
and that u⌘ is a solution of
⇢0 exp u⌘
.
⇢ exp u⌘ dx
R3 0
u⌘ = (⌘) R
Then to conclude, one has to take ⌘ small enough in order that µ0 < (⌘) = µ. Let us
give some details of the proof. We have
0
0
u⌘ (0)
d ⌘
u (r)
dr
d ⌘
u (⌘)
dr
u⌘ (⌘)
2
1
e 1/r exp u⌘ (0), for 0 < r  ⌘,
2
2r
2
1
e 1/⌘ exp u⌘ (0),
2
2⌘
⌘ 1/⌘2
⌘
u (0)
e
exp u⌘ (0),
4
F. BOUCHUT AND J. DOLBEAULT
510
and for r > ⌘,
1
k⇢0 exp u⌘ k1
4⇡⌘
There exists a limit
u⌘ (r)
u⌘ (⌘)
u⌘1 = lim u⌘ (r)
r!1
u⌘ (0)
u⌘ (0)
⌘
⌘
(1 + 2/⌘ 3 )eu (0)
4
⌘
⌘
(1 + 2/⌘ 3 )eu (0)
4
1/⌘ 2
1/⌘ 2
1
k⇢0 exp u⌘ k1 .
4⇡⌘
1
k⇢0 exp u⌘ k1
4⇡⌘
because u⌘ is decreasing, and
⌘
(⌘) = k⇢0 eu k1
⌘
k⇢0 k1 eu1
k⇢0 k1 exp(u⌘ (0)
⌘
⌘
(1 + 2/⌘ 3 )eu (0)
4
1/⌘ 2
1
(⌘)),
4⇡⌘
and therefore
(⌘)
⌘ ln k⇢0 k1
1
ln (⌘) +
1
⌘
+ ⌘ ln(
).
3
⌘
4(⌘ + 2)
Example 2. Let us consider the case where
⇢0 is nonnegative, radially symmetric, continuous and strictly decreasing
(1 + |x|3 )⇢0 (x) 2 L1 (R3 ),
8r 2 (0, 1), ⇢0 (r) = 1.
(3.10)
Proposition 3.6. If ⇢0 satisfies (3.10), then there exists µ0 > 0 such that for all µ > µ0
, (PBn) has no solution in L1 (R3 ).
Proof. First of all, the symmetry result stated in Appendix 2 proves that if V is a
solution of (PBn), then V is radial and decreasing along any radius. Any solution of
(PBn) with the boundary condition (3.8) belongs to L1 (R3 ). Let us consider
exp v
⌃ = max{ > 0 : 9v 2 H01 (B) solution of
v= R
},
exp v dx
B
where B is the unit ball of R3 with center at 0. According to [23], ⌃ exists and belongs
to ]0, 1[. Let us assume now that (PBn) has a solution V for
Z
1
µ > ⌃(1 +
⇢0 (x) dx).
|B| B c
Then v = V
We have
v=V
V (1) is solution in H01 (B) of
exp v
µ
v= R
, R
=R
.
exp
v
dx
exp
v
dx
⇢
exp
v dx
B
B
R3 0
V (1) > 0 for |x| < 1,
and
Z
exp v dx > |B|,
Z
Z
v = V V (1) < 0 for |x| > 1, and
⇢0 exp v dx <
⇢0 dx,
Bc
Bc
R
1
1 + |B|
⇢ (x) dx
µ
Bc 0
R
R
R
R
=
>⌃
> ⌃,
⇢ exp v dx/ B exp v dx
1 + B c ⇢0 exp v dx/ B exp v dx
R3 0
B
which is in contradiction with the definition of ⌃.
Remark. For µ small enough, the solution in a ball is unique (see [23]) so that the
solution of (PBn) in R3 , obtained through the resolution of an ordinary di↵erential
equation, is also unique in L3,1 (R3 ).
Appendix 1.A. Compactness Lemma. We prove the following.
ON LONG TIME ASYMPTOTICS
511
Lemma. Let > 0,
0, T > 0, 1  p < 1, f0 2 Lp (R2N ), h 2 L1 (]0, T [, Lp (R2N )),
and consider the solution f 2 C([0, T ], Lp (R2N )) of
L0 f ⌘ @t f + v · rx f
divv (vf )
= h in ]0, T [⇥R2N , f (0, .) = f0 .
vf
Assume that f0 2 F a bounded subset of Lp (R2N ), h 2 H a bounded subset of Lq (]0, T [,
Lp (R2N )), with 1 < q  1. Then for any ⌘ > 0 and ! bounded open subset of R2N , f
is compact in C([⌘, T ], Lp (!)).
Notice that the compactness lemma of R.J. DiPerna and P.L. Lions [11] applies to a
general di↵erential operator, but only gives compactness in L1 (]0, T [, Lp (!)).
Proof. f is given by
f (t, x, v) =
ZZ
G(t, x, v, ⇠, ⌫)f0 (⇠, ⌫) d⇠ d⌫
⇠,⌫
+
ZZZ
G(t
s, x, v, ⇠, ⌫)h(s, ⇠, ⌫) ds d⇠ d⌫ ⌘ f (1) + f (2) ,
0<s<t, ⇠,⌫
with G the fundamental solution of L0 . We have
kf
(1)
N T /p0
(t, .)kp  e
kf0 kp ,
kf
(2)
N T /p0
(t, .)kp  e
Z
t
0
kh(s, .)kp ds.
Following Ascoli’s and Frechet-Kolmogorov’s Theorems, we have to prove
(i) 8" > 0 9 > 0 8t  t0 2 [⌘, T ], t0 t  ) kf (t0 , .) f (t, .)kLp (!)  ",
(ii) 8t 2 [⌘, T ] 8" > 0 9 > 0 8z 2 R2N , |z|  ) kf (t, . + z) f (t, .)kLp (!)  ".
We just have to treat the case when q = 1, and f0 and h have compact support in v
(if not, just consider (v)f , 2 Cc1 (RN ) and use bounds on rv f in L1 (]⌘, T [, Lp )).
Since
ZZ
ZZ
ZZ
ZZ
G(t, x, v, ⇠, ⌫) dx dv = 1,
ZZ
|rx G| dx dv  C/t3/2 ,
1/2
|rv G| d⇠ d⌫  C/t
ZZ
,
G d⇠ d⌫ = eN t ,
|rv G| dx dv  C/t1/2 ,
|@t G| dx dv  C
1 + |⌫|
,
t3/2
we have for example
ZZ
ZZ
|G(t0 , .)
G(t, .)| dx dv  C
|G(t, x + zx , v + zv , ⇠, ⌫)
(t0
t)1/3
t1/2
(1 + |⌫|)1/3 ,
G(t, x, v, ⇠, ⌫)| dx dv  C
0 < t  t0  T,
|z|1/3
,
t1/2
0 < t  T.
F. BOUCHUT AND J. DOLBEAULT
512
Let us examine condition (ii) for f (1) . We have
|f (1) (t, . + z) f (1) (t, .)|
ZZ

|G(t, x + zx , v + zv , ⇠, ⌫)
⇠,⌫

ZZ
|G( )
⇠,⌫
 2eN
0
T /p
G( )| d⇠ d⌫
ZZ
1/p0
G(t, x, v, ⇠, ⌫)| |f0 (⇠, ⌫)| d⇠ d⌫
ZZ
|G( )
G( )| |f0 (⇠, ⌫)|p d⇠ d⌫
1/p
⇠,⌫
|G(t, x + zx , v + zv , ⇠, ⌫)
G(t, x, v, ⇠, ⌫)| |f0 (⇠, ⌫)|p d⇠ d⌫
1/p
,
⇠,⌫
hence
kf (1) (t, . + z)
f (1) (t, .)kp  C
|z|1/3p
kf0 kp ,
t1/2p
which concludes for this first term. The condition (i) for f (1) , and (i), (ii) for f (2) are
treated in exactly the same way.
Appendix 2.A. Symmetry result à la Gidas, Ni and Nirenberg. Let us state
the following theorem, which is a straightforward adaptation of a theorem of Gidas, Ni
and Nirenberg (see [18, 19], and also [24]).
Theorem. Let N
3, u > 0 be a solution of
u = f (|x|, u),
8x 2 RN ,
where r 7! f (r, u) is strictly decreasing for all u > 0, (r, u) 7! f (r, u) and (r, u) 7!
@u f (r, u) are continuous for u
0, x 7! (1 + |x|3 )f (|x|, u(x)) belongs to L1 (RN ) and
1
N
@u f (|x|, u(x)) belongs to L (R ). Then u is radially symmetric and du /dr < 0 for all
r > 0.
Corollary. Let ⇢0 be a nonnegative radially symmetric strictly decreasing in r continuous function such that
(1 + |x|3 )⇢0 (x) 2 L1 (RN ).
If u 2 L1 (RN ) is a solution of
u = ⇢0 exp u
for some
> 0, then u is radial and du /dr < 0 for all r > 0.
Let us give the sketch of the theorem.
1. Let us define
Z
x0 =
x⇢0 exp u dx
RN
Z
RN
⇢0 exp u dx,
and (if x0 6= 0), e1 = |xx00 | . If x0 = 0, e1 is any unit vector of S N 1 . Replacing x by
x x0 , one can prove that (we use the convention of summation over repeated indices)
u(x) =
1
|x|N
2
(a0 +
aij xi xj
1
+ o( 2 )),
|x|4
|x|
@u
(x) =
@xi
N 2
1
(a0 xi + O( )),
|x|N
|x|
ON LONG TIME ASYMPTOTICS
for some a0 > 0, and thus @x1 u(x) < 0 for x1 C0 /|x|, |x|
and R1 (here the x1 -axis is defined by the unit vector e1 ).
2. If > 0 is such that @x1 u(x) < 0 for all x1 > , then
u(x) > u(x ),
x1 , x0 ),
x = (2
and for all x = ( , x0 ), x0 2 RN
1
513
R1 for some constants C0
(x1 , x0 ) 2 R ⇥ RN
1
,
x1 < ,
,
@u
(x) < 0.
@x1
3. There exists
4. v(x) = u(x )
0
1 such that for all
u(x) satisfies
Lv
Lv =
c(x) =
0,
u(x) > u(x ) if x1 < .
0 on {x = (x1 , x0 ) 2 R ⇥ RN
v + c(x) · v,
f (|x|, u(x ))
u(x )
1
: x1 < },
f (|x|, u(x))
,
u(x)
and Hopf’s lemma gives a contradiction if
inf{ > 0 : u(x) > u(x ), 8x = (x1 , x0 ) 2 R ⇥ RN
1
, x1 < } > 0.
5. {x = (0, x0 ) 2 R ⇥ RN 1 } is a symmetry plane: x0 = 0 because r 7! f (r, u) is
strictly decreasing. One can now apply the same argument to any e1 2 S N 1 ; there
is no specific direction in the equation, so that any plane containing 0 is a symmetry
plane. It is then easy to check that du /dr < 0 for all r > 0.
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