Stability of the Front under a
Vlasov-Fokker-Planck Dynamics
R. Esposito1 , Y. Guo2 and R. Marra3
1
2
3
Dipartimento di Matematica, Università di L’Aquila, Coppito, 67100 AQ, Italy
Division of Applied Mathematics, Brown University, Providence, RI 02812, U.S.A.
Dipartimento di Fisica and Unità INFN, Università di Roma Tor Vergata, 00133 Roma, Italy.
July 2, 2007
Abstract
We consider a kinetic model for a system of two species of particles interacting
through a long range repulsive potential and a reservoir at given temperature. The
model is described by a set of two coupled Vlasov-Fokker-Plank equations. The
important front solution, which represents the phase transition, is a one-dimensional
stationary solution on the real line with given asymptotic values at infinity. We prove
the asymptotic stability of the front for small symmetric perturbations.
Keywords: Vlasov, Fokker-Planck, stability, fronts
1
Introduction and Notation
The dynamical study of phase transitions has been tackled, among the others, with an
approach based on kinetic equations modeling short range and long range interactions
which are responsible of critical behaviors. An example of such models has been proposed
in [BL] where the authors study a system of two species of particles undergoing collisions
regardless of the species and interacting via long range repulsive forces between different
species. A simplification of such model has been considered in [MM] where a kinetic model
has been introduced for a system of two species of particles interacting through a long range
repulsive potential and with a reservoir at a given inverse temperature β. The interaction
with the reservoir is modeled by a Fokker-Plank operator and the interaction between the
two species by a Vlasov force. The system is described by the one-particle distribution
functions fi (x, v, t), i = 1, 2 solutions of the system of two coupled Vlasov-Fokker-Plank
(VFP) equations in a domain Ω
∂t fi + v · ∇x fi + Fi · ∇v fi = Lfi ,
1
(1.1)
where
fi
Lfi = ∇v · M ∇v
,
M
(1.2)
M is a Maxwellian with mean zero and variance β −1 ,
M=
β
2π
32
β 2
e− 2 v
and β −1 = T is the temperature of the reservoir. The self-consistent Vlasov force is
Z
Z
0
0
Fi = −∇x dx U (|x − x |)
dvfj (x0 , v, t) .
R3
Ω
The potential function U is smooth, monotone with compact support, and its integral over
the whole space is equal to one. There is a natural Liapunov functional, the free energy
functional, for this dynamics,
Z
Z
β
dxdv(f1 + f2 )v 2
G(f1 , f2 ) :=
dxdv [(f1 ln f1 )(x, v) + (f2 ln f2 )(x, v)] +
2
3
3
Ω×R
Z
Z
Z Ω×R
+β
dxdyU (|x − y|)
dvf1 (x, v)
dv 0 f2 (y, v 0 ) .
R3
Ω×Ω
R3
In fact, we have that
XZ
M2
fi
d
G(f1 , f2 ) = −
dxdv
[∇v ]2 ≤ 0
dt
fi
M
3
i=1,2 Ω×R
and the time derivative is zero if and only if fi are of the form fi = ρi M , where ρi are
functions only of the position. If we put these expressions back in the equations we get
that the stationary solutions of (1.1) have densities satisfying the equations
Z
ln ρi (x) + β dx0 U (|x − x0 |)ρj (x0 ) = Ci , x ∈ Ω, i, j = 1, 2, i 6= j
(1.3)
Ω
and Ci are arbitrary constants, related to the total masses of the components of the mixture.
Moreover, replacing fi by fi = ρi M in the functional G and integrating out the velocity
variable we get a functional on the densities ρi
Z
Z
F(ρ1 , ρ2 ) =
dx(ρ1 ln ρ1 + ρ2 ln ρ2 ) + β
dxdyU (|x − y|)ρ1 (x)ρ2 (y)
(1.4)
Ω
Ω×Ω
1
The Euler-Lagrange equations for this functional with the constraints
|Ω|
are exactly (1.3).
2
Z
dxρi (x) = ni
Ω
In [CCELM1] it is proved that for nβ ≤ 2, with n = n1 + n2 , the total mass density,
equations (1.3) in a torus have a unique homogeneous solution, while for nβ > 2 there are
non homogeneous solutions. To explain the physical meaning of these non homogeneous
solutions, we write the functional F(ρ1 , ρ2 ) in the following equivalent form
Z
Z
β
F(ρ1 , ρ2 ) =
dxf (ρ1 , ρ2 ) +
dxdyU (|x − y|)[ρ1 (x) − ρ1 (y)][ρ2 (y) − ρ2 (x)]
2 Ω×Ω
Ω
where f (ρ1 , ρ2 ) is the thermodynamic free energy made of the entropy and the internal
energy
ρ1 log ρ1 + ρ2 log ρ2 + βρ1 ρ2
The function f (ρ1 , ρ2 ) is not convex and has, for any given temperature
T , two symmetric
Z
(under the exchange 1 → 2) minimizers if the total mass
dx[ρ1 + ρ2 ] is larger than a
Ω
critical value 2T . In other words, this system undergoes a first order phase transition with
coexistence of two phases, one richer in the presence of species 1 and the other richer in the
presence of species 2. If we look for the minimizers of the functional under the constraints
on the total masses
Z
dxρi (x) = ni |Ω|
Ω
we get homogeneous minimizers if we fix (n1 , n2 ) equal to one of the two minimizers of
f (n1 , n2 ).
Otherwise, we get non homogeneous minimizers below the critical value. The structure
of the minimizing profiles of density will be as close as possible to one of the two minimizers of f : they will be close to one of the minimizing values in a region B, close to the
other minimizing value in the complement but for a separating interface along which the
minimizing profiles will interpolate smoothly between the two values.
We can conclude then that the minimizers of G in a torus will be Maxwellians times
densities ρi of the form discussed above. Since G is a Liapunov functional, we expect that
the minimizers are related to the stable solutions of the equations. In this paper we want
to study the stability of the non homogeneous stationary solutions of the equations (1.1),
which are minimizers of the functional. We restrict ourselves to the one-dimensional case
with
x = (0, 0, z)
−∞<z <∞ .
The reason for this is that in such a situation we know many properties of the minimizers.
In fact, consider the following variational problem. Define the excess free energy functional
in one dimension on the infinite line as
F̂(ρ1 , ρ2 ) := lim [FN (ρ1 , ρ2 ) − FN (M ρ+ , M ρ− )]
N →∞
(1.5)
where FN is the free nergy associated to the interval [−N, N ] and (ρ+ , ρ− ) is a homogeneous
minimizer of f . Note that FN (M ρ+ , M ρ− ) = FN (M ρ− , M ρ+ ). Look for the minimizers
of the excess free energy such that limx→±∞ f1 (x) = M ρ± , limx→±∞ f2 (x) = M ρ∓ . In
[CCELM2] it is proved that
3
THEOREM 1.1. There exists a unique C ∞ positive minimizer (front) w = [w1 (z), w2 (z)],
with w1 (z) = w2 (−z), for the one-dimensional excess free energy F̂, defined in (1.5), in
the class of functions ρ = (ρ1 , ρ2 ) such that
lim ρ1 = ρ± ,
lim ρ2 = ρ∓ .
z→±∞
z→±∞
ρ− < wi (z) < ρ+
for any z ∈ R. Moreover, the front w satisfies the Euler-Lagrange equations (1.3) and its
derivative w0 satisfies almost everywhere the equations
w10 (z)
+ β(U ∗ w20 )(z) = 0,
w1 (z)
w20 (z)
+ β(U ∗ w10 )(z) = 0
w2 (z)
(1.6)
The front w converges to its asymptotic values exponentially fast, in the sense that there
is α > 0 such that
|w1 (z) − ρ∓ |eα|z| → 0 as z → ∓∞,
|w2 (z) − ρ± |eα|z| → 0 as z → ∓∞.
All its derivatives have the same property.
Our main result is the stability of these fronts for the VFP dynamics, under suitable
assumptions on the initial data. To state the result, we write fi , solutions of (1.1), as
fi = wi M + hi .
Then, the perturbation hi satisfies
∂t hi + Gi hi = Lhi − Fi (h)∂vz hi ,
(1.7)
where the operators Gi are defined by
Gi hi = vz ∂z hi − U ∗
wj0 ∂vz hi
Z
+ βvz M wi U ∗ ∂z
dvhj ( · , v, t)
(1.8)
R3
while the force Fi (h) due to the perturbation is
Z
Z
0
0
Fi (h) = −∂z dz U (z − z )
dvhj (z 0 , v, t),
j 6= i .
(1.9)
R3
R
We define ( · , · ) as the L2 inner product for two scalar functions (on R or R×R3 depending
on the context), while h·, ·i denotes the L2 the inner product for vector-valued functions,
and we denote k · k as their corresponding L2 norms. Furthermore, we define the weighted
L2 norms as
Z
XZ
1
1
fi gi , hf, giM =
dzdv
f i gi ,
(fi , gi )M =
dzdv
wi M
wi M
3
R×R3
i=1,2 R×R
4
with corresponding weighted L2 norms by k · kM . We also define the dissipation rate as
kgk2D = k(I − P )gk2M + k∇v (I − P )gk2M ,
(1.10)
where P is the L2 projection on the null space of L = {cM, c ∈ R}, for any given t, z. We
also define the weighted norms as
kgkM,γ = kzγ gkM
kgkD,γ = kzγ gkD ,
with the notation
zγ = zγ .
In the following we will also denote by ∂t,z h the couple of derivatives (∂t h, ∂z h).
THEOREM 1.2. We assume that h = (h1 , h2 ) at time zero has the following symmetry
property in z, v
h1 (z, v, 0) = h2 (−z, Rv, 0), Rv = (vx , vy , −vz ).
(1.11)
There is δ0 small enough such that, if
(1)
kh(0)kM + k∂t,z h(0)kM ≤ δ0
then, there is a unique global solution to (1.7) such that for some K > 0
d
{K[kh(t)k2M + k∂t,z h(t)k2M ]} + kh(t)k2D + k∂t,z h(t)k2D ≤ 0.
dt
(1.12)
(2) For γ > 0 sufficiently small, if
kh(0)kM,γ + k∂t,z h(0)kM, 1 +γ ≤ δ0
2
then there is constant C > 0,
sup kh(t)kM,γ + sup k∂t,z h(t)kM, 1 +γ ≤ C(kh(0)kM,γ + k∂t,z h(0)kM, 1 +γ .
0≤t≤∞
0≤t≤∞
2
2
(1.13)
Moreover, we have the decay estimate
i
t −2γ h
kh(0)k2M,γ + k∂t,z h(0)k2M, 1 +γ .
kh(t)k2M + k∂t,z h(t)k2M ≤ C 1 +
2
2γ
(1.14)
Since the equation preserves the symmetry property (1.11) we have that hi (z, v, t) have
the same simmetry property (1.11) at any time. The proof of the theorem is based on
energy estimates and takes advantage of the fact that at time zero the perturbation is
small in a norm involving also the spatial and the time derivatives. To close the energy
estimates, we use the spectral gap for the Fokker-Planck operator L to control (I − P )h,
the part of h orthogonal to the null space of L, and the conservation laws to control P h,
the component of h in the null space of L, in terms of (I − P )h, like the method used in
[Guo].
5
The key difficulty in our paper is the control of the hydrodynamic part P h, in the
presence of of the Vlasov force with large amplitude. Because of the presence of the Vlasov
force, the hydrodynamic equations do not give directly the control of P h but instead of a
norm involving the operator A, the second variation of the free energy F̂ at the front w,
which is given by
2 Z
X
d2
dzgi (z)(Ag)i (z) = 2 F̂(w + sg) s=0 .
hg, Agi :=
ds
i=1 R
The operator A acts on g = (g1 , g2 ) as
(Ag)1 =
g1
+ βU ∗ g2 ,
w1
(Ag)2 =
g2
+ βU ∗ g1 .
w2
(1.15)
Since w is a minimizer of F̂ the quadratic form on the left hand side is positive and
the first variation gives the Euler-Lagrange equations
δ F̂
(w) = log wi + βU ∗ wj − Ci = 0,
δρi
i 6= j ,
Differentiating with respect to z and using the prime to denote the derivative with respect
to the z variable
w0
(Aw0 )i = 1 + βU ∗ wj0 = 0, i 6= j ,
w1
0
which shows that w is in the null space of A. Indeed, one can show (see Section 2) that w0
spans the null space of A and that there exists a constant λ > 0 such that (spectral gap)
hg, Agi ≥ λ
2 Z
X
i=1
dz
R
1
|(I − P)gi |2
wi
where P is the projector on the null space of A.
Hence, by using the spectral gap for A one can control the component of P h on the
orthogonal to the null space of A, but not the component on the null space of A. Let us
write P h =R M a and a = αw0 + (I − P)a. What is missing at this stage is the control
of α(t) = R×R3 dzdvP h(z, v, t)w0 (z) for large time. We would like to show that α(t)
vanishes asymptotically in time, which amounts to prove that the solution of the VlasovFokker-Plank equations (VFP) converges to the initial front. The existence of a Liapunov
functional for this dynamics forces the system to relax to one of the stationary points for
the functional, which are of the form M wx , with wx any translate by x of the symmetric
front w. Then, it is the conservation law, in the form
Z
dzdv[f (z, v, t) − M (v)w(z)] = 0
R×R3
which should select the front the solution has to converge to. But this is a condition on the
L1 norm of the solution while the energy estimates control some L2 norm. In the approach
6
in [Guo] the conservation law is used in problems in finite domains or in infinite domains
but in dimension greater or equal than 3. The problem we are facing here is analogous
to the one in [CCO] and we refer to it for an exhaustive discussion. One can realize the
connections between the problem discussed here and the one in [CCO] by looking at the
hydrodynamic limit of the model. In [MM] it is proved that the diffusive limit of the VFP
dynamics is
!
δF
0
−1 ρ1
(1.16)
, M=β
∂t ρ̄ = ∇ · M∇
0 ρ2
δ ρ̄
δF
denotes the functional derivative of F with respect to ρ̄ = (ρ1 , ρ2 )
δ ρ̄
and M is the 2 × 2 mobility matrix. These equations are in the form of a gradient flow for
the free energy functional as the equation considered in [CCO], which is an equation for a
bounded magnetization m(x, t) ∈ [−1, 1]:
δF
∂t m = ∇ · σ(m)
δm
where ρ̄ = (ρ1 , ρ2 ),
where σ(m) = β(1 − m2 ) and F is a suitable non local free energy functional. In [CCO] the
stability result is obtained by using suitable weighted L2 norms, with a weight x, which
allow to control the tails of the distribution and hence a control of the L1 norm. This is
possible essentially because the equation is of diffusive type.
Unfortunately, we cannot use directly the approach in [CCO] since the dissipation in
the kinetic model is given by the Fokker-Plank operator and does not produces directly
diffusion on the space variable. In fact, we are able to use, as explained above, weighted
norms (in space) with a weight xα , α < 1, which are not enough to control the L1 norm.
Hence, to overcome the difficulty, we consider a special initial datum. We assume, as
explained before, that h is symmetric at initial time. It is easy to see that this property
is conserved by the dynamics so that h is symmetric at any later time. We note that also
wM
symmetric
while w0 is antisymmetric in the z variable. This implies the vanishing
R
Pis
2
of i=1 R×R3 dzdvhi (z, v, t)M (v)wi0 (z), the component of a on the null space of A, which
consequently is zero at any later time.
Even with such a symmetry assumption (1.11), the estimate for the hydrodynamic part
P h is delicate. Based on the precise spectral information of A, we need to further study
∂
Ag = (Ag)0 .
∂z
To this end, we employ a crucial decompostion (2.6) for each component of g and a con∂
tradiction argument to establish the important lower bound for ∂z
Ag (Theorem 2.4). Furthermore, in order to get the decay rate, we use polynomial additional weight function in
z and a trick of interpolation to carefully derive the corresponding energy estimate in a
bootstrap fashion. Once again, Theorem 2.4 is crucial to control local L2 norm of P h in
terms of its z-derivative.
7
It is worth to stress that our result does not rely on a smallness assumption on the
potential, like for example in [Vi], where it is proved the stability in L1 of the constant
stationary state for a one component VFP equation, on a torus, for general initial data.
The assumption of small U in [Vi] guarantees the uniqueness of the stationary state, namely
it means not to be in the phase transition region. On the contrary, we are working with
values of the parameters (temperature and asymptotic values of densities, ρ± ) in the phase
transition region. For values of the parameters ρ+ = ρ− , 2βρ+ ≤ 2 the minimizer is unique
and we can prove that the constant solution is stable, by a simplified version of the proof
given here. The critical value βρ+ = 2 is selected by the fact that the analogous of the
operator A, that comes out from the linearization around the constant solution, is positive
and has spectral gap for βρ+ < 2 (it coincides with the operator called L0 in Theorem 2.2).
We expect also that the constant solution will become unstable above this critical value.
Finally, we want to return to the kinetic model by [BL], mentioned at the beginning of
this section and studied in a series of papers [BELM], in which the Fokker-Planck term is
replaced by a Boltzmann kernel to model species blind collisions between the particles. The
dynamics is described by a set of two Vlasov-Boltzmann equations, coupled through the
Boltzmann collisions and the Vlasov terms and conserve not only the total masses but also
energy and momentum. The stationary solutions are the same as in the previous model,
Maxwellians times densities ρi satisfying (1.3), so that one could study the stability of these
solutions with respect to the Vlasov-Boltzmann dynamics. This result is more difficult to
get due to the non linearity of the Boltzmann terms. The first results on the stability of
the Maxwellian are due to [Uk], [Ma]. Recently, it has been proved by energy methods
in a finite domain or in R3 by [Guo] ([SG] for soft potentials) who has also extended
the method to cover other models involving self-consistent forces and singular potentials
[Guo1] and in Rd by [LY], who also proved the stability of a 1 − d shock. The stability of
the non homogeneous solution for a Boltzmann equation with a given small potential force
has been proved in [UYZ]. We are not aware of analogous results for non small force, but
a very recent one [As]. Our method is in principle suited to prove stability under VlasovBoltzmann dynamics on a finite interval, but what is still lacking is a detailed study of
stationary solutions in a bounded domain. We plan to report on that in the future.
The paper is organized as follows. In Section 2 we collect the properties of the operators
L and A and the properties of the fronts. In Section 3 we prove some Lemmas that allow
to control some z-derivative of P h in terms of (I − P )h. In Section 4 we give the energy
estimates for the function, the time derivative and the z-derivative.
2
Spectral Gaps of L and A
In this section we collect all the relevant properties of the operators L and A and also the
properties of the fronts.
LEMMA 2.1. There is a ν0 > 0 such that for all g = (g1 , g2 ),
hg, LgiM ≤ −ν0 k(I − P )gk2D .
8
(2.1)
Proof. Since wi is bounded from below for i = 1, 2, we only need consider the case
when g is a scalar.
Recall (1.2), the null space of L is clearly made of constants (in v) times M . Moreover,
Lg is orthogonal to the null space of L in the inner product ( · , · )M . We denote by P the
projector on the null space of L. Finally, the spectral gap property holds [LB]: for any g
(g, Lg)M ≤ −ν((I − P )g, (I − P )g)M
On the other hand, a direct computation yields
Z
Z
2
−1
dv M
|∇v (I − P )g| + 3β
(g, Lg)M = −
R3
dv M −1 |(I − P )g|2 .
R3
We thus conclude our lemma by splitting (Lg, g) = (1 − )(Lg, g) + (Lg, g) and applying
above two estimates respectively, for sufficiently small.
By (1.15), it is immediate to check that
F(w + u) − F(w) = 2 hAu, ui + o(2 ).
THEOREM 2.2. There exist ν > 0 such that
hu, Aui ≥ νh(I − P)u, (I − P)ui,
where P is the projector on (Null A):
Null A = {u ∈ L2 (R) × L2 (R) | u = cw0 , c ∈ R}.
Proof. We first characterize Null A. We note that(1.6) imply
u21
=−
w1
u1
w10
2
βw10 U
∗
u22
=−
w2
w20 ,
u2
w20
2
βw20 U ∗ w10 .
From (1.15), (Au, u) takes the from
Z Z
Z 2
u22 (z)
u1 (z)
+
dz + 2β
u1 (z)u2 (z 0 )U (z − z 0 )dzdz 0 =
w
(z)
w
(z)
1
2
R R
R
2
Z Z u1 (z)
u2 (z 0 )
−β
− 0 0
U (z − z 0 )w10 (z)w20 (z 0 )dzdz 0 .
0
w2 (z )
R R w1 (z)
(2.2)
But, by the monotonicity properties of wi it follows that −w10 (z)w20 (z 0 )dzdz 0 is a positive
measure on R × R. Therefore the quadratic form is non negative and vanishes if and only
if h is parallel to w0 . In particular, this identifies the null space of the operator A.
9
To establish the spectral gap of A, it is sufficient to prove the lower bound for the
normalized operator Ã: L2 (R) × L2 (R) → L2 (R) × L2 (R) such that
√
√
(Ãu)i = wi (A(u w))i ,
√
√
√
with the abuse of notation u w = (u1 w1 , u2 w2 ). The explicit form is
√
√
(Ãu)1 = u1 + β w1 U ∗ ( w2 u2 ),
√
√
(Ãu)2 = u2 + β w2 U ∗ ( w1 u1 ).
The corresponding associated quadratic form is
Z
Z
√
√
2
2
w1 u1 U ∗ (u2 w2 ).
hu, Ãui = (u1 + u2 ) + 2β
R
R
The operator à is a bounded symmetric operator on H = L2 (R) × L2 (R). From the
previous considerations it is also non negative and positive on the orthogonal complement
of its null space. The spectral gap for à is established in [CCELM2]. For completeness,
we give a sketch of the proof below.
We decompose the operator as à = Ã0 + K where
p
p
(Ã0 u)1 = u1 + β ρ+ ρ− U ∗ u2 , (Ã0 u)2 = u2 + β ρ+ ρ− U ∗ u1
p
√
√
(Ku)1 = β w1 U ∗ ( w2 u2 ) − β ρ+ ρ− U ∗ u2
Z
h
i
p
√
0 √
0
+
−
= β dzdz
w1 (z) w2 (z ) − ρ ρ U (|z − z 0 |)u2 (z 0 ),
p
√
√
(Ku)2 = β w2 U ∗ ( w1 u1 ) − β ρ+ ρ− U ∗ u1
Z
h√
i
p
√
= β dzdz 0
w2 (z) w1 (z 0 ) − ρ+ ρ− U (|z − z 0 |)u1 (z 0 ).
(2.3)
(2.4)
The operator Ã0 has the spectral gap property. Consider the equation
Ã0 u = λu + f.
(2.5)
Denote by ũ(ξ), f˜(ξ) and Ũ (ξ) the Fourier transforms of u, f and U . For λ in the resolvent
set of Ã0 we can find a solution to (2.5) if the determinant of the matrix
√
1−λ
β Ũ ρ+ ρ−
√
β Ũ ρ+ ρ−
1−λ
is different from zero for any ξ ∈ R. This happens if λ is such that for all ξ ∈ R
(1 − λ)2 − β 2 (Ũ (ξ))2 ρ+ ρ− 6= 0.
10
Moreover, by the positivity of U , Ũ (ξ)| ≤ Ũ (0) = 1. As a consequence, the spectrum of
L0 is in the interval
p
p
[1 − β ρ+ ρ− , 1 + β ρ+ ρ− ].
√
Now, for β > βc it is immediate to check that β ρ+ ρ− < 1 and hence the spectrum is
contained in (k, +∞) for some positive k.
We claim that K is compact on H. Indeed, uniformly for kukL2 ≤ 1, K satisfies
(1) ∀ > 0 ∃Z > 0:
Z
|Ku|2 dz < ,
Z > Zε
|z|>Z
(2) ∀ > 0 ∃` > 0:
Z
|Ku(z + `) − Ku(z)|2 dz < ε,
` > `ε .
|z|>Z
These proofs follow trivially from the regularity
p of the convolution,
√ + −the fact that U has
compact support and the fact that limx,y→±∞ w1 (x)w2 (y) = ρ̄ ρ̄ . For the property
(2) the boundedness of wi0 and the regularity of U are used. Hence, by Weyl’s theorem we
have that the spectral gap holds also for Ã.
We are also interested into a lower bound on the norm of (Au)0 . To this purpose,
consider u = (u1 , u2 ) ∈ L2 (R) × L2 (R) with derivative u0 ∈ L2 (R) × L2 (R). Assume u
orthogonal to w0 = (w10 , w20 ): hu, w0 i = 0.
We now make orthogonal decomposition of each component of u with respect to the
corresponding w0 = (w10 , w20 ) in the scalar L2 inner product. In terms of the vector inner
product, by a direct computation, such a process leads to
u = αw̃0 + ũ
where ũ is such that
Z
dz ũ1 w10
Z
=0=
(2.6)
dz ũ2 w20
while w̃0 = (w10 , −w20 ) is orthogonal to w0 in the inner product h· , · i (note that w20 (z) =
−w10 (−z)) with the coefficient α computed as
α=
hu , w̃0 i
,
N
R
R
N = hw̃0 , w̃0 i = 2 dz(w10 )2 = 2 dz(w20 )2 . We first prove a Lemma for ũ.
LEMMA 2.3. There is a constant C such that
k(Aũ)0 k2 ≥ CkQũ0 k2 .
where Q is the orthogonal projection on the orthogonal complement of w00 .
11
(2.7)
Proof. We follow the proof in [CCO]. We have
d ũi
ũ0
w0
w0
[ + U ∗ uj ] = [ i + U ∗ ũ0j ] − 2i ui = (Aũ0 )i − 2i ũi .
dz wi
wi
wi
wi
(Aũ)0i =
By integrating over z̄ after multiplication by wi0 the equation
Z z
ũi (z) = ũi (z̄) +
ũ0i (s)ds
z̄
we get
(−1)i+1
ũi (z) = +
(ρ − ρ− )
R
Z
+∞
dz̄wi0 (z̄)
Z
−∞
z
ũ0i (s)ds .
z̄
dz ũi wi0
We have used
= 0.
From above, we can write (Aũ)0i in terms of an operator A + K acting on L2 (R) × L2 (R)
such that, if h = ũ0 then
(Aũ)0i = (Ah)i + (Kh)i ,
Z z
Z
(−1)i wi0 +∞ 0 0 0
(Kh)i (z) := +
hi (s)ds .
dz wi (z )
(ρ − ρ− ) wi2 −∞
z0
We prove first that
The operator K is compact on L2 . Indeed, we show that
• ∀ε > 0 ∃Zε > 0:
Z
|Kz̄ h|2 dz < ε,
Z > Zε ,
|z|>Z
• ∀ε > 0 ∃`ε > 0:
|Kh(z + `) − Kh(z)|2 < ε,
` < `ε .
The second is true because of the continuity of the integral. To prove the first, note that
Z +∞
Z z
Z +∞
p
0
0 0
(2.8)
dz|wi0 (z 0 )| |z − z 0 | ≤ C(1 + |z|)khk
dz wi (z )
hi (s)ds ≤ khk
z0
−∞
so that
Z
wi0
(z)
2
|z|>Z wi
2
Z
−∞
+∞
0
0
0
Z
2
z
dz wi (z )
hi (s)ds
z0
−∞
Z
dz ≤ CkhkL2
|z|>Z
Then, by the rapid decay property of wi0 ,
Z
|(Kh)i |2 dz → 0,
|z|>Z
12
Z → +∞,
2
wi0
(z) |(1 + |z|)2 .
wi2
which proves (2). Now,
Z
Z
0 2
|(Aũ) | dz =
dz ũ0 A2 + K ∗ A + AK ∗ + K ∗ K ũ0 dx .
R
R
The operator K ∗ A + AK ∗ + K ∗ K is compact because A is bounded and K compact and
its null space is spanned by w00 , because by definition of A + K
0 = (Aw0 )0 = (A + K)w00 .
But, A2 has a strictly positive essential spectrum, hence the result follows from Weyl’s
theorem. Moreover
Z
|(Aw̃0 )0 |2 = δ > 0,
R
0
because w̃ is orthogonal to the null space of A.
THEOREM 2.4. For any u ∈ L2 (R) × L2 (R), u0 ∈ L2 (R) × L2 (R) such that hu, w0 i = 0,
there exists a positive constant B such that
k(Au)0 k2 ≥ B(|α|2 + kQũ0 k2 ).
(2.9)
where Q is the projection on the orthogonal complement of w00 . Furthermore, if u0 = Qu0 ,
then there is a constant k > 0 such that
k(Au)0 k2 ≥ k(|α|2 + kũ0 k2 )2 .
(2.10)
Proof. First, we prove that there is a constant C such that, if u = (1 − P)u 6= 0 and
hu, w0 i = 0,
k(Au)0 k2 ≥ C(δα2 + k(Aũ)0 k2 ) .
(2.11)
We introduce the normalized vector ω and its decomposition along w0 and the orthogonal
complement by setting:
ω=
δα2
u
;
+ k(Aũ)0 k2
ω = η w̃ + ω̃ ,
so that equation (2.11) reads as
k(Aω)0 k2 ≥ C.
By the decomposition of ω we have
k(Aω)0 k2 = k(Aω̃)0 k2 + δη 2 + 2 ((Aω̃)0 , η(Aw̃0 )0 ) .
By definition, ω is such that
k(Aω̃)0 k2 + δη 2 = 1,
13
(2.12)
hence
k(Aω)0 k2 = 1 + 2h(Aω̃)0 , η(Aw̃0 )0 i .
Suppose now that the inequality (2.12) is not true. Then, for any n we can find ω̃n and
ηn such that
1
k(A[ω̃n + ηn w̃0 ])0 k2 = 1 + 2h(Aω̃)0k , ηn (Aw̃0 )0 i < .
n
By weak compactness, up to subsequences, there are ω̃0 and η0 such that ω̃n converges
weakly to ω̃0 ., ηn → η0 . By weak convergence,
h(Aω̃n )0 , ηn (Aw̃0 )0 i → h(Aω̃0 )0 , η0 (Aw̃0 )0 i
and
liminf[k(Aω̃n )0 k2 + δηn2 ] + 2h(Aω̃0 )0 , η0 (Aw̃0 )0 i = 0
By lower semicontinuity,
k(Aω̃0 )0 k2 + δη02 ≤ liminf k(Aω̃n )0 k2 + δηn2 = 1
Hence,
0 ≤ k(Aω0 )0 k2 = k(Aω̃0 )0 k2 + δη02 + 2h(Aω̃0 )0 , η0 (Aw̃0 )0 i ≤ 1 + 2h(Aω̃0 )0 , η0 (Aw̃0 )0 i ≤ 0 .
(2.13)
As a consequence,
k(Aω0 )0 k2 = 0
which implies ω0 = 0: indeed hω0 , w0 i = limn→∞ hωn , w0 i = 0 because ωn is a sequence of
vectors orthogonal to w0 . Furthermore, since ωn → ω0 = 0 weakly, ηn → η0 = 0. Then,
h(Aω̃0 )0 , η0 (A + w̃0 )0 i = 0 in contradiction with last inequality in (2.13). Therefore (2.12)
is true and, together with (2.7), implies (2.9).
Finally, to prove (2.9), we notice that if u0 = Qu0 , then by (2.6),
αw̃00 + ũ0 = u0 = Qu0 = αQw̃00 + Qũ0 .
This implies that ũ0 is bounded by α and Qũ0 , which completes the proof.
3
Estimates of the hydrodynamic part P f .
Decompose the solution of (1.7) in the component in the null space of L and in the one
orthogonal to the null space:
hi = P hi + (I − P )hi . Denote by ai M the components in the
R
null space of L: P hi = R3 dvhi M = ai M , so that
hi = ai M + (I − P )hi .
14
By using this decomposition in (1.7) we have
M ∂t ai + vz ∂z ai − ai U ∗ wj0 M −1 ∂vz M + βvz wi U ∗ ∂z aj
= −∂t (I − P )hi − Gi (I − P )hi − Fi (h)∂vz hi + L(I − P )hi
Define
µi =
so that
∂z µi =
(3.1)
ai
+ βU ∗ aj := (Aa)i
wi
1
w0
∂z ai − ai 2i + βU ∗ ∂z aj
wi
wi
By using the equation for the front (1.6) we can write the equation (3.1) as
M [∂t ai + vz wi ∂z µi ]
= −∂t (I − P )hi − Gi (I − P )hi − Fi (h)∂vz hi + L(I − P )hi .
R
By integrating (3.2) over the velocity, since − R3 dv∂t (I − P )hi = 0, we have
Z
∂ t ai = −
dvGi (I − P )hi
(3.2)
R3
and, by the definition of Gi ,
Z
∂t ai = −∂z
dvvz (I − P )hi
(3.3)
R3
By integrating (3.2) over the velocity after multiplication by vz
Z
Z
Z
Z
dvvz Fi (h)∂vz hi +
dvvz Gi (I−P )hi −
T wi ∂z µi = − dvvz ∂t (I−P )hi −
Moreover, by integrating by parts,
Z
Z
Z
2
0
dvvz Gi (I − P )hi =
dvvz ∂z (I − P )hi + U ∗ wj
R3
R3
dvvz L(I−P )hi .
R3
R3
R3
Z
dv(I − P )hi = ∂z
R3
R3
dvvz2 (I − P )hi .
Hence
Z
T wi ∂z µi = −
Z
dvvz ∂t (I − P )hi − ∂z
R3
R3
dvvz2 (I
Z
− P )hi +
dvvz L(I − P )hi + F (hi )ai .
R3
(3.4)
Define
`ai
Z
dvvz (I − P )hi ,
=
R3
`bi
Z
=
R3
dvvz2 (I
Z
− P )hi ,
dvvz L(I − P )hi .
mi =
R3
By integrating twice by parts we get the identity:
Z
Z
(I − P )hi
mi = −
dvM ∂vz
=β
dvvz (I − P )hi = β`ai .
M
3
3
R
R
15
The following estimates are an easy consequence of (3.3) and (3.4).
k∂t ai k = k∂z `ai k
k∂z µi k ≤ k∂t `ai k + k∂z `bi k + kmi k + k∂z ai k kai k
From the definition, we have
`ai
√
Z
dvvz
=
R3
1
M √ (I − P )hi ≤ C
M
Z
|(I − P )hi |2
dv
M
21
.
This and the fact that w is bounded from above and below give
Z
1
a 2
+
dz |`ai |2 .
k`i k ≤ ρ
wi
R
Hence,
2
X
k`ai k ≤ Ck(I − P )hkM .
i=1
We will often use the notation ġ instead of ∂t g for the t-derivative of a function g(t, z, v)
and g 0 instead of ∂z g for its z-derivative. Set
Z
Z
1
(1)
dvvz (I − P )ḣi −
dvvz + L(I − P )hi − F (hi )ai
(3.5)
∂z µi = −
T wi R3
R3
Z
1
+∂z
dv vz2 (I − P )hi ,
T wi
R3
Z
1
(2)
2
∂z µi = −∂z
dv vz (I − P )hi
(3.6)
T wi R3
(1)
(2)
(1)
(2)
so that µi + µi = µi . Define a(j) by setting µi = (Aa(1) )i , µi = (Aa(2) )i . Since the
null space of A is given by αw0 = (αw10 , αw20 ) for α ∈ R, the equation
µi = Agi
has solutions iff
2 Z
X
i=1
dzwi0 µi = 0
R
and they are of the form
gi = (A−1 µ)i + αwi0
where (A−1 µ) is the unique solution orthogonal to the null space of A. Therefore, we need
to show that µ(j) are orthogonal to the null space of A. We shall prove it at the end of this
section. Moreover, we can always choose α = 0 since a = a(1) + a(2) and a does not have
component on the null space of A. In fact, a has by assumption at time zero the same
16
symmetry property of w and it is preserved in time. This implies that at any time a is
orthogonal to w0 and hence has no component in the null space of A. This is one of the
crucial points where we use the symmetry assumption on the initial perturbation.
(1)
(1)
We now estimate the L2 norm k∂z a(1) k. To this end, we first prove that Q∂z ai = ∂z ai
P R
(1)
which is equivalent to show that 2i=1 ∂z ai wi00 = 0.
LEMMA 3.1. Assume h1 (z, v, t) = h2 (−z, Rv, t) is valid. Then h∂z a, w00 i = 0.
Proof: We notice that this property is true for ∂z a because of the simmetry properties of
the solution. In fact, ∂z a1 (z) = −∂z a2 (−z) and w100 (z) = w200 (−z). We are left with proving
that the same symmetry property hold for each a(j) . It is enough to prove that for a(2) .
We have that
Z
1
(2)
−1
2
a = −A
dv vz (I − P )h
T wi R3
and since A does not change the symmetry properties we are done if we prove that
Z
Z
1
1
2
2
dv vz (I − P )h1 (z) =
dv vz (I − P )h2 (−z) .
w1 R3
w2 R3
By using the properties of h we have that the left hand side is equal to
Z
1
dv vz2 (I − P )[h1 (v, z) + h2 (v, −z)],
w1 (z) R3+
with R3+ the set of velocities with vz ≥ 0, and the right hand side to
Z
1
dv vz2 (I − P )[h1 (v, z) + h2 (v, −z)] .
w2 (−z) R3+
The simmetry properties of wi imply the result. The same argument also shows that µ(2)
is orthogonal to the kernel of A and hence µ(1) has the same property since µ = Aa is
orthogonal to the kernel of A by definition.
THEOREM 3.2. We have
2
X
(2)
kai k ≤ Ck(I − P )hkM ≤ khkD
i=1
Moreover, if khkM ≤ δ0
2
X
(1)
k∂z ai k ≤ C k(I − P )hkM + k(I − P )∂t hkM
i=1
17
Proof: From (3.6), by integration over z, since µi → 0 as z → ±∞,
Z
1
1 b
(2)
µi = −
dvvz2 (I − P )hi =
`
T wi R3
T wi i
which implies
2
X
(2)
kµi k ≤ Ck(I − P )hkM
i=1
(2)
Moreover, µi = (Aa(2) )i so that we have also, by Theorem 2.2,
2
X
(2)
kai k ≤ Ck(I − P )hkM .
i=1
From (3.5) we get
2
X
(1)
k∂z µi k
≤ Ck(I − P )∂t hkM + Ck(I − P )hkM + sup kF (hi )kL∞
i
i=1
2
X
kai k .
i=1
Now,
(1)
(2)
(1)
(2)
kF (hi )kL∞ = kU ∗ ∂z (aj + aj )kL∞ ≤ Ck∂z aj k + Ckaj k .
To apply Theorem 2.4 we need to show that a is orthogonal to w0 . We notice that the front
w is symmetric under the exchange 1 → 2 while the derivatives wi0 are antisymmetric. On
the other hand, as already observed, a has at time zero the same symmetry properties as
w and this implies that the component of a on the null space of A is zero at any time. In
addition, by Lemma 3.1, we can apply Lemma 2.4 to get
k∂z a(1) k ≤ C k(I − P )∂t hkM + k(I − P )hkM + kP hkM [k∂z a(1) k + ka(2) k
h
i
(1)
≤ C k(I − P )ḣkM + k(I − P )hkM + khkM k∂z a k + k(I − P )hkM .
To conclude the proof, using that property and the hypothesis we have that for δ0 small
enough
k∂z a(1) k ≤ C[k(I − P )∂t hkM + Ck(I − P )hkM + δ0 k(I − P )hkM
which proves Theorem 3.2.
As a consequence of the proof we have also
2
X
kF (hi )kL∞
h
i
≤ C k(I − P )∂t hkM + k(I − P )hkM .
(3.7)
i=1
From now on we use the more explicit notation ah M = P h. Moreover we use the previous
(1)
(2)
decomposition: ah = ah + ah .
18
LEMMA 3.3. Let 0 ≤ γ ≤
1
8
and khkM be sufficiently small, then we have
(2)
kah kγ ≤ CkhkD,γ ,
(1)
k∂z ah kγ
Z
(3.8)
≤ C{khkD,γ + k∂t hkD,γ },
(3.9)
z
|ah |2 dz ≤ C{khk2D,γ + k∂t hk2D,γ }.
(1 + z 2 )2−2γ
(3.10)
2
Proof: We introduce the commutator:
(2)
(2)
(2)
[zγ , A]ah = zγ Aah − A(zγ ah ),
Notice that
(2)
zγ Aah
1
= −zγ
T wi
Z
dvvz2 (I − P )hi .
which implies
(2)
kzγ Aah k ≤ CkhkD,γ .
The commutator can be estimated by taking into account the property of the convolution
and the fact that U is of finite range. We have, for a suitable z∗ ,
Z
(2)
(2)
[zγ , A]ah (z) =
dz 0 U (|z − z 0 |)(zγ − zγ0 )ah (z 0 )
Z
(2)
=
dz 0 U (|z − z 0 |)2γz∗ {1 + z∗2 }γ−1 (z − z 0 )ah (z 0 )
Z
(2)
≤ C dz 0 U (|z − z 0 |)|z − z 0 |zγ− 1 ah (z 0 )
2
It follows, for 0 ≤ γ ≤ 12 ,
(2)
(2)
k[zγ , A]ah kL2 ≤ Ckzγ− 1 ah kL2 ≤ CkhkD .
2
The last two estimates together imply
(2)
kA(zγ ah )k ≤ CkhkD,γ .
(2)
(2)
Since zγ ah has the same symmetry properties of ah , it is orthogonal to w0 as well and
we can use Theorem 2.2 to deduce (3.8).
(1)
To estimate ∂z ah , using the decomposition (2.6) we write
(1)
(1)
ah = αh w̃0 + ãh
Then the argument leading to (2.8) provides the estimate
Z +∞
Z z
(1)
(1)
(1)
0
|ãh (z)| ≤
dz̄w (z̄)
dy|ãh (y)| ≤ (1 + |z|)k∂z ãh k.
−∞
z̄
19
By Theorem 2.4,
(1)
(1)
|αh | ≤ k(Aah )0 k ≤ Ck∂z ah k
and therefore
(1)
(1)
|ah (z)| ≤ C(1 + |z|)k∂z ah k.
Hence, by Theorem 3.2, we obtain
(1)
|ah (z)| ≤ (1 + |z|) (khkD + k∂t hkD ) .
(1)
(1)
But ∂z (Aah )i = A{∂z ah }i −
wi0
(1)
(ah )i .
wi2
Therefore
(1)
(1)
zγ (A∂z ah )i = zγ ∂z (Aah )i +
zγ wi0 (1)
(a )i .
wi2 h
(3.11)
Clearly, since wi0 decays exponentially, we have the following estimate for the second term
in (3.11)
zγ w0 (1)
k 2 i (ah )i k ≤ C{khkD + k∂t hkD }
wi
We examine now the first term in (3.11). We deduce from equation (3.5)
(1)
kzγ ∂z (Aah )k ≤ C (khkD,γ + Ck∂t hkD,γ + kzγ F (ah )kL∞ kah k) .
We further split kzγ F (ah )kL∞ in the last term as
(2)
(1)
(2)
(2)
kzγ F (ah )kL∞ + kzγ U ∗ (∂z ah )kL∞ ≤ kF (zγ (ah ))kL∞ + k[zγ , F ]ah kL∞
(1)
(1)
+kU ∗ (zγ ∂z ah )kL∞ + k[zγ , U ]∂z ah kL∞
(2)
(1)
(1)
≤ C kzγ ah kL2 + kzγ ∂z ah kL2 ≤ C khkD,γ + kzγ ∂z ah kL2 .
The commutators are estimated as before, by using 0 ≤ γ ≤ 12 , and we have used (3.8) in
the last inequality. We hence conclude that, for kak small,
(1)
(1)
(1)
kA(zγ ∂z ah )kL2 ≤ kzγ A{∂z ah }kL2 + k[zγ , A]∂z ah kL2
≤ C khkD,γ + k∂t hkD,γ + kzγ ∂z a1h kL2 kakL2 .
Therefore, by decomposing along the null space of A, we have by the spectral gap for the
operator A: Denote by τ = w0 kw0 kL−12 the unit vector in the direction w0 . We have
(1)
(1)
(1)
(1)
kzγ ∂z ah kL2 ≤ kzγ ∂z ah − hzγ ∂z ah , τ iτ kL2 + khzγ ∂z ah , τ iτ kL2
(1)
≤ CkA(zγ ∂z ah )kL2 + C{khkD + k∂t hkD }
(1)
≤ C{khkD,γ + k∂t hkD,γ } + Ckzγ ∂z ah kL2 kakL2
In conclusion, for kakL2 small, we deduce (3.9).
20
(1)
(2)
To prove (3.10), since k∂z ah kL2 ≤ C{khkD + k∂t hkD } as well as kah kL2 ≤ CkhkD ,
(1)
zah
the key is to estimate
for z large. In fact, let χ(z) be a smooth cutoff function
(1 + z 2 )1−γ
with χ(z) ≡ 1 for |z| ≥ k, for k large and χ(z) ≡ 0 for |z| ≤ k − 1. We have for the
contribution due to |z| ≤ k,
Z
Z
Z
z2
(1) 2
(1) 2
(1)
{1 − χ}|ah | dz ≤
|ah | ≤ Ck
|∂z ah |2 ≤ Ck {khk2D + k∂t hk2D }.
(1 + z 2 )2−γ
|z|≤k
|z|≤k
We now consider the contribution for |z| large. Since
Z
(1)
χz 2 |ah |2
dz = −
(1 + z 2 )2−2γ
Z
=
1
(1)
zχ|ah |2 dz
2
1−2γ
2(1 − 2γ)(1 + z )
Z
1
1
(1)
0 (1) 2
zχ |ah | +
χ|ah |2
2
1−2γ
2
1−2γ
2(1 − γ)(1 + z )
2(1 − 2γ)(1 + z )
Z
1
(1)
(1)
+
χah ∂z ah .
2
1−2γ
(1 − 2γ)(1 + z )
Z
d
dz
Therefore,
1
z2
(1)
−
χ|ah |2 dz
2
2−2γ
2
1−2γ
(1 + z )
2(1 − 2γ)(1 + z )
Z
Z
1
1
(1)
(1)
0 (1) 2
zχ |ah | +
χah ∂z ah .
=
2
1−2γ
2
1−2γ
2(1 − 2γ)(1 + z )
(1 − 2γ)(1 + z )
Z For γ ≤
1
8
and |z| > k,
z2
1
z2
−
≥
.
(1 + z 2 )2−2γ
2(1 − 2γ)(1 + z 2 )1−2γ
4(1 + z 2 )2−2γ
We thus have (χ0 ≡ 0 for |z| ≥ k)
Z
Z
Z
z2
1
z2
(1) 2
(1) 2
(1)
χ|a
|
dz
≤
C
|a
|
+
χ|ah |2 dz
k
h
h
2
2−2γ
2
2−2γ
4(1 + z )
8
(1 + z )
|z|≤k
(1)
+Ckzγ ∂z ah k2 .
We thus deduce from (3.9):
Z
z2
2
2
2
χa
dz
≤
C
k∂
hk
+
khk
t
D,γ
D,γ .
(1 + z 2 )2−2γ h
It is important to control ∂z [zγ ∂z a(1) ]. To this end, we have
21
(3)
LEMMA 3.4. Define hah , w0 i = 0 and
(3)
(Aah )i
Then, for 0 ≤ γ ≤
1
2
1
≡−
T wi
Z
vz ∂t (I − P )hi .
and kah kL2 + k∂z ah kL2 ≤ δ0 sufficiently small,
(2)
(3.12)
(3)
(3.13)
kzγ ∂z ah k ≤ C (k∂z hkD,γ + khkD ) ,
kzγ ah k ≤ Ck∂t hkD,γ ,
(1)
(3) 1
k∂z zγ ∂z ah − zγ ah kL2 ≤ C k∂t,z hkD,γ + khkD,γ− .
2
(3.14)
Proof. For notational simplicity, we denote
ag = zγ ∂z ah .
(3)
(2)
(1)
We need to estimate kag kL2 , kag kL2 and k∂z ag kL2 . First of all, we prove (3.13).
(3)
(3)
(3)
kAzγ ah ]kL2 ≤ kzγ Aah kL2 + k[A, zγ ]ah kL2
(3)
≤ k∂t hkD,γ + Ckzγ− 1 ah kL2
2
(3)
≤ k∂t hkD,γ + CkAah kL2
≤ Ck∂t hkD,γ .
To get the third inequality we have used γ < 1/2 and Theorem 2.2 while in the last
(3)
inequality comes from the definition of Aah . We thus deduce, again by Theorem 2.2,
(3)
kzγ ah kL2 ≤ Ck∂t hkD,γ .
and hence (3.13) is proved.
We now turn to (3.12). Note that
(2)
Aa(2)
= A(zγ ∂z ah )
g
(2)
(2)
= zγ A{∂z ah } − [zγ , A]∂z ah
w0 (2)
(2)
(2)
= zγ ∂z A{ah } + zγ 2i ah − [zγ , U ]∂z ah .
wi
w0
(2)
Clearly, kzγ w2i ah kL2 ≤ CkhkD . Since
i
(2)
zγ ∂z A{ah }
1
= −zγ ∂z {
T wi
Z
dvvz2 (I − P )hi }
it follows that
(2)
kzγ ∂z A{ah }kL2 ≤ Ck∂z hkD,γ .
22
(2)
And an integration by part in [zγ , U ]∂z ah yields:
Z
(2)
(2)
k U (|z − z 0 |)[zγ − zγ0 ]∂z ah (z 0 )kL2 ≤ Ckzγ− 1 ah k.
2
We therefore can decompose (remind that τ = w0 kw0 k−1 )
(2)
(2)
(2)
a(2)
g = hag , τ iτ + {ag − hag , τ iτ }.
Clearly,
(2)
kha(2)
g , τ iτ kL2 ≤ kah kL2 ≤ CkhkD .
(2)
(2)
But k{ag − hag τ iτ kL2 is bounded by using the spectral gap of A :
CkAa(2)
g kL2 ≤ C{khkD + k∂z hkD,γ },
Collecting terms, we deduce (3.12).
(1)
(3)
Finally, to estimate ∂z (ag − zγ ah ), we use the commutation relation
0
w
(A∂z ah )i = ∂z (Aah )i + 2i ah to get
wi
(1)
(Aa(1)
)
=
A(z
∂
a
)
i
γ
z
g
h
i
(1)
(1)
zγ ∂z (Aah )i
=
(1)
([zγ , A]∂z ah )i
−
zγ wi0 ah
+
wi2
(3)
By equation (3.5) and the definition of Aah
(1)
zγ ∂z (Aah )i =
(3)
zγ (Aah )i
+∂z
+ zγ
Z
1
T wi
R3
1
T wi
Z
dvvz L(I − P )hi + Fi (h)ai
2
dv vz (I − P )hi .
R3
Therefore,
(Aa(1)
g )i
+∂z
−
1
T wi
(3)
zγ (Aah )i
Z
dv
R3
= zγ
vz2 (I
Z
1
T wi
dvvz L(I − P )hi + F (hj )ai
R3
− P )hi −
23
(1)
(1)
([zγ , A]∂z ah )i
zγ wi0 (ah )i
+
.
wi2
Using again the commutation relation (A∂z a)i = ∂z (Aa)i +
wi0
ai , we find
wi2
(3) A∂z (a(1)
g − zγ ah ) i
wi0 (1)
(3) (3)
−
(A(z
a
= ∂z Aa(1)
)
−
{a − zγ ah }i
γ h
g
i
wi2 g
h
i w0
(3)
(3)
(3)
i
(a(1) − zγ ah )i
= ∂z (Aa(1)
)
−
z
(Aa
)
−
([z
,
A]a
)
γ
γ
g i
h i
h i −
wi2 g
Z
Z
1
1
2
= ∂z zγ
dvvz L(I − P )hi + Fi (h)(ah )i + ∂z
dv vz (I − P )hi
T wi
T wi
R3
R3
(1)
zγ wi0 (ah )i
(1)
− ([zγ , A]∂z ah )i
wi2
o w0
(3)
(3)
−([A, zγ ]ah )i + 2i (a(1)
− zγ ah )i
wi g
(1)
−([zγ , A]∂z ah )i −
The terms involving the commutator can be estimated by putting
the z-derivative
on the
zγ
potential U in the convolution. We only need to estimate ∂z
Fi (h)(ah )i . We expand
T wi
it as
zγ 2
zγ
× Fi (h)(ah )i −
(∂z U ∗ ∂z (ah )j )(ah )i − (∂z U ∗ ∂z (ah )j )∂z (ah )i
∂z
T wi
T wi
The first term is bounded by
(2)
(1)
kzγ− 1 ah kL2 + kzγ− 1 ∂z ah kL2 kah kL2
2
2
We modify the second term above (up to the factor (T wi )−1 ) as follows:
(∂z2 U ∗ zγ ∂z (ah )j )(ah )i − (∂z U ∗ zγ ∂z (ah )j )∂z (ah )i
+[∂z2 U, zγ ]∂z (ah )j )(ah )i + [∂z U, zγ ]∂z (ah )j )∂z (ah )i
The L2 norms of the last two terms are bounded by
(2)
(1)
kzγ− 1 ah kL2 + kzγ− 1 ∂z ah kL2 (kah kL2 + k∂z ah kL2 )
2
2
≤ δ0 (khD,γ− 1 k + k∂t hD,γ− 1 k)
2
2
The last inequality follows from Lemma 3.2 and the assumption. We write the first two
terms as
(3)
(3)
(2)
(1)
2
∂z U ∗ ∂z (ag )j − zγ (ah )j + ∂z U ∗ zγ (ah )j + ∂z U ∗ zγ ∂z (ah )j ∂z (ah )i
(3)
(3)
(2)
+ U ∗ ∂z (a(1)
)
−
z
(a
)
+
∂
U
∗
z
(a
)
+
U
∗
z
∂
(a
)
(ah )i
j
γ
j
z
γ
j
γ
z
j
g
h
h
h
24
Finally, we get
zγ
k∂z
Fi (h)(ah )i kL2 ≤ δ0 (khD,γ− 1 k + k∂t hkD,γ− 1 k
2
2
T wi
(3)
(3)
(2)
+ k∂z (a(1)
g − zγ ah )kL2 + kzγ ah k + kzγ ∂z ah k
× (kah kL2 + k∂z ah kL2 ) .
We use (3.12) and (3.13) to get
(3)
(2)
kzγ ah k + kzγ ∂z ah k ≤ khkD + k∂t,z hkD,γ .
We therefore conclude
(3)
kA(∂z a(1)
g − zγ ah )kL2 ≤ C k∂t,z hkD,γ
(3)
+khkD,γ− 1 + k∂z (a(1)
g − zγ ah kL2 )(kah kL2 + k∂z ah kL2 ).
2
(1)
(3.15)
(3)
We then split k∂z (ag − zγ ah )kL2 into
(3)
(3)
(3)
(1)
(1)
kh∂z {a(1)
g − zγ ah }, τ iτ kL2 + k∂z {ag − zγ ah } − h∂z (ag − zγ ah ), τ iτ kL2
The first term is bounded by khkD , while the second can be absorbed in the left hand side
for {kakkL2 + k∂z akL2 } small, by using the spectral gap for A. This concludes the proof of
(3.14).
4
Energy Estimates and Decay
LEMMA 4.1. Let
∂t gi + Gi gi − Lgi = Γi ,
(4.1)
then
1d
2 dt
Z
dzag Aag +
R
+
2 Z
X
i=1
2 Z
X
i=1
Z
1
dz
dv
|(I − P )gi |2
M
w
3
i
R
R
dzdv
R×R3
)
1
(I − P )gi L(I − P )gi
wi M
= hAag M, ΓiM + h(I − P )g, ΓiM .
Proof: Decompose the solution of (4.1) in the component orthogonal to the null space
of L and in the one in the nullR space: gi = P gi + (I − P )gi . Denote by ag M the component
in the null space of L: P g = R3 dvgM = ag M , so that
g = ag M + (I − P )g .
25
We shall denote a = ag in the proof of this lemma. Repeating the same computation as in
Section 3, we have
M [∂t ai + vz wi ∂z {Aa}i ] = −∂t (I − P )gi − Gi (I − P )gi + L(I − P )gi + Γi .
Take the scalar product ( · , · )M of (4.1) with M (Aa)i + wi−1 (I − P )gi to get:
Z
Z
2
1X d
1
2
dzdvM ai (Aa)i +
dzdv
|(I − P )gi |
2 i=1 dt R×R3
M wi
R×R3
Z
2 Z
X
= −
dzdvM wi (Aa)i vz ∂z (Aa)i −
dzdv(I − P )gi vz ∂z (Aa)i
i=1
R×R3
R×R3
Z
dzdv(Aa)i Gi (I − P )gi −
−
R×R3
dzdv
1
(I − P )gi Gi (I − P )gi
M wi
2
dzdv
R×R3
R×R3
i=1
Z
+
2 Z
X
X 1
1
(I − P )gi L(I − P )gi + hΓ, Aai +
(I − P )gi , Γi ).
(
M wi
M
w
i
i=1
The first term on the right hand side vanishes since (Aa)i ∂z (Aa)i are functions of z, t only.
By recalling the definition of Gi , (1.8),
Gi (I − P )gi = vz ∂z (I − P )gi + U ∗ wj0 ∂vz (I − P )gi ,
we have for the third term
Z
Z
−
dzdv(Aa)i Gi (I − P )gi = −
dzdv(Aa)i vz ∂z (I − P )gi
R×R3
R×R3
Z
=
dzdv∂z Aai vz (I − P )gi
R×R3
R
which exactly cancels with the second term (− R×R3 dzdv(I − P )gi vz ∂z (Aa)i ) in the right
hand side. By using the definition of Gi we get for the fourth term
Z
1
−
dzdv
(I − P )gi Gi (I − P )gi
M wi
R×R3
Z
1 1
= −
dzdv
vz ∂z ((I − P )gi )2 + U ∗ wj0 ∂vz ((I − P )gi )2
M wi 2
3
ZR×R
vz wi0
[ + βU ∗ wj0 ]((I − P )gi )2 = 0
= −
dzdv
2M
w
3
i wi
R×R
by using the equation for the front.
LEMMA 4.2. Let γ ≥ 0 be sufficiently small. Then if kh(t)kM,γ ≤ δ0
1d
kh(t)k2M,γ + ν0 kh(t)k2D,γ ≤ C{γ + δ0 }{k∂t h(t)k2D,γ + kh(t)k2D,γ },
2 dt
with ν0 given in Lemma 2.1.
26
(4.2)
Proof. Note that g = zγ h satisfies
∂t gi + Gi gi − Lgi =
where
Z
Ĝhi = vz M wi β
2zvz γgi
+ Ĝhi + Fi (h)∂vz gi ≡ Γi .
1 + z2
U 0 (z − z 0 ){zγ − zγ0 }hj ( z 0 , v, t)dz 0 dv.
(4.3)
We now apply Lemma 4.1. We first treat Fi (h)∂vz gi . Notice that (Fi (h)∂vz gi , M (Ag)i )M =
0, and
2
X
(Fi (h)∂vz gi ,
i=1
1
(I − P )gi )
M wi
≤ C{kF (h)kL∞ kag kL2 + kF (h)kL∞ k∂vz (I − P )gkM }k(I − P )gkM
≤ C{k∂t hkD + khkD }kgkM kgkD + khkM kgk2D
≤ Cδ0 {khkD + k∂t hkD }2 + Cδ0 kgk2D .
Next we estimate Ĝhi . Note hĜh, Aag i = 0. Since
|zγ − zγ0 | ≤ Cγ|z − z 0 |
(1)
(2)
for γ small, recalling ah = ah + ah , we deduce that
1
(I − P )gi )
(Ĝhi ,
M wi
Z
Z
0
0
0
0
0
= β vz
U (z − z )[zγ − zγ ]ahj ( z , v, t)dz dv (I − P )gi
Z
Z
0
0
0 (2)
0
0
≤ β vz
U (z − z )[zγ − zγ ]ahj ( z , v, t)]dz dv (I − P )gi
Z
Z
(1)
0
0
0
0
+β vz
U (z − z )∂z0 [(zγ − zγ )ahj ( z , v, t)]dz dv (I − P )gi
(1)
≤
(2)
Cγ{kah k
+
(1)
k∂z ah k
zah
}k}k(I − P )gkM
+ kU ∗ {
(1 + z 2 )1−γ
≤ Cγkgk2D .
We have used (3.10) and 3.8 in Lemma 3.3. For the third term
2zvz γgi
, we use again
1 + z2
estimate (3.10) to get
2
X
2zvz γgi 1
,
(I − P )gi )
(
1 + z 2 M wi
i=1
2
2
X
X
2zvz γ(I − P )gi 1
2zvz γP gi 1
=
(
,
(I − P )gi ) +
(
,
(I − P )gi )
2
2
1
+
z
M
w
1
+
z
M
w
i
i
i=1
i=1
≤ Cγ{k∂t gk2D + kgk2D }.
27
On the other hand,
h
2zvz γ(I − P )g
, Aag i
1 + z2
≤ Cγk(I − P )gkM {kA(
z
z
ag )k + k[A,
]ag k.
2
1 + |z|
1 + |z|2
Since A is bounded in L2
z
z
kA(
ag )k ≤ Ck
ag k ≤ C(k∂t hkD,γ + khkD,γ )
2
1 + |z|
1 + |z|2
where the last inequality is true because of (3.10). The commutator term
k[A,
z
z,
z
]ag k = kU 0 ∗ (
−
)ag k
2
2
1 + |z|
1 + |z|
1 + |z, |2
can be estimated as in the previous computation as C(k∂t hkD,γ + khkD,γ ). Therefore
2zvγ(I − P )g
, Aag i is bounded by
h
1 + z2
CγkgkD · (k∂t hkD,γ + khkD,γ ).
This concludes the proof of the lemma.
LEMMA 4.3. Recall ν0 in Lemma 2.1. If k∂t,z h(t)kM,γ + kh(t)kM,γ ≤ δ0 , then
1d
k∂t h(t)k2M,γ + ν0 k∂t h(t)k2D,γ ≤ C{γ + δ0 }k∂t,z h(t)k|2D,γ− 1 + Ckh(t)k2D,γ− 1 .
2
2
2 dt
(4.4)
Proof. Let g = {1 + z 2 }∂t h, we have
[∂t + Gi − L]gi =
2zvz γ gi
+ Ĝ∂t hi + Fi (h)∂vz gi + Fi (∂t h)∂vz gi := Γi .
1 + z2
By Lemma 4.1 we need to estimate
hAag M, ΓiM + h(I − P )g, ΓiM
We first estimate
h
2γzvz ∂t gi
. Notice that g = ag M + (I − P )g,
1 + z2
2γzvz g
2γzvz ag M
2γzvz (I − P )g
,
Aa
i
=
h
,
Aa
i
+
h
, Aag i.
g
g
1 + z2
1 + z2
1 + z2
The first term above vanishes. For the second term above, by an integration by part with
the kernel, we have
Z
1
ag
U (z − y){|z| − |y|}
k
Aag k ≤ kA{
}k + k
ag (y)dyk
1 + |z|
1 + |z|
{1 + |z|}{1 + |y|}
1
≤ C{k{1 + |z|2 }γ− 2 ∂t ah k ≤ Ck(I − P )∂z hkM,γ− 1 .
2
28
The last inequality is due to (3.3). We therefore have
h
2γzvz g
, Aag i ≤ εk∂t hk2D,γ + Cε γk∂z hk2D,γ− 1 .
2
1 + z2
Now by an integration by part in the v−variable, we turn to
2
X
≤
i=1
2
X
|(
1
zvz γgi
(I − P )gi ,
)|
M wi
1 + z2
|(
zvz γ(I − P )gi
1
(I − P )gi ,
)|
M wi
1 + z2
i=1
+
2
X
i=1
|(
zvz γ{1 + |z|2 }γ P ∂t hi
1
(I − P )gi ,
)
M wi
1 + z2
1
(I − P )g · ∇vz (I − P )gkM
1 + |z|
+Cγk(I − P )gk2M + Cε k{1 + |z|2 }γ−1/2 ∂t ah k2
≤ Cγk∂t,z hk2D,γ− 1 .
≤ Cγk
2
Now turn to the third term Ĝ∂t hi in Γ. Since
P2
i=1 (v Ĝ∂t hi , (Aag )i )
= 0,
2
2
X
X
1
1
|
(
(I − P )gi , Ĝ∂t hi v)| = |
(
(I − P )gi , Ĝ{∂t ah })|
M wi
M wi
i=1
i=1
≤ εkgk2D + Cε γk∂z hk2D,γ− 1 .
2
Now for the fifth term Fi (h)∂vz gi , we note that
2
X
(Fi (h)∂vz gi ,
i=1
2
X
P2
i=1 (Fi (h)∂vz gi , (Aag )i )
= 0, and
1
(I − P )gi )
M wi
2
X
1
1
(I − P )gi ) +
(Fi (h)∂vz (I − P )gi ,
(I − P )gi )
M
w
M
w
i
i
i=1
i=1
≤ CkFi (h)k∞ ag k · k(I − P )gi kM + k∂vz (I − P )gk2M
≤ Ck∂z ah k kag k · kgkD + kgk2D .
=
(Fi (h)∂vz P gi ,
Now for the sixth term zγ Fi (∂t ah )∂vz hi , we note
Z
zγ Fi (∂t ah )∂vz hi Aag dv = 0.
Since by (3.3),
kzγ Fi (∂t ah )k = kFi (zγ ∂t ah )k + k[Fi , zγ ]∂t ah k ≤ Ck(I − P )∂z hkD,γ ,
29
we have, by using the assumption and integrating by part on v,
2
X
(zγ Fi (∂t ah )∂vz gi ,
i=1
LEMMA 4.4. Let 0 ≤ γ ≤
1
2
1
(I − P )gi ) ≤ Cδ0 k∂z hkD,γ × k∂t hkD,γ .
M wi
+ 18 . If k∂t,z h(t)kM,γ + kh(t)kM,γ ≤ δ0 , then
1d
k∂z h(t)k2M,γ + ν0 k∂z h(t)k|2D,γ
2 dt 2
2
2
2
2
≤ Cγ khkD,γ− 1 + k∂t,z hkD,γ− 1 + khkD + k∂t hkD + δ0 k∂t hkD,γ .
2
(4.5)
2
whith ν0 given in Lemma 2.1. .
Proof. We define g = zγ ∂z h to get
(1 + 2γ)zvz gi
+ zγ ∂z U ∗ wj0 ∂vz hi
1 + z2
+Ĝ∂z hi + ∂z U ∗ wj0 ∂vz (zγ hi ) − Fi (h)∂vz gi − zγ Fi (∂z ah )∂vz hi ≡ Γi .
∂t gi + Gi gi − Lgi =
2
X
1
(I − P )gi , Γi ).
where Ĝ is defined in (4.3). By Lemma 4.1 we need to estimate hΓ, Aag i+ (
M wi
i=1
2γzvz gi
We first estimate the first term
. Since that g = ag M + (I − P )g,
1 + z2
h
2γzvz g
2γzvz ag M
2γzvz (I − P )g
,
Aa
i
=
h
,
Aa
i
+
h
, Aag i.
g
g
1 + z2
1 + z2
1 + z2
The first term above vanishes. For the second term above, we notice that by the splittng
(1)
(2)
(2)
ah = ah +ah , and by an integration by part with the kernel for ah , we have from Lemma
3.3,
1
Aag k
1 + |z|
Z
U (z − y){|z| − |y|}
ag
≤ kA(
)k + k
ag (y)dyk
1 + |z|
{1 + |z|}{1 + |y|}
!
(1)
a
g
(2)
≤ C kzγ− 1 ah k + k
k
2
1 + |z|
(1)
≤ C khk|D,γ− 1 + Ckzγ− 1 ∂z ah k
2
2
≤ C khkD,γ− 1 + k∂t hkD,γ− 1 .
k
2
2
30
Now by an integration by part in the v−variable, we turn to
2
X
≤
i=1
2
X
|(
1
zvz γgi
(I − P )gi ,
)|
M wi
1 + z2
|(
zvz γ(I − P )gi
1
(I − P )gi ,
)|
M wi
1 + z2
i=1
+
2
X
|(
i=1
zvz γzγ P ∂z hi
1
(I − P )gi ,
)|
M wi
1 + z2
1
(I − P )g · ∇v (I − P )gkM
1 + |z|
+εk(I − P )gk2M + Cε γkzγ ∂z ah k2
≤ εk(I − P )gk2M
+Cε γ{khk2D,γ− 1 + k∂z,t hk2D,γ− 1 + khk2D + k∂z,t hk2D }.
≤ Cγk
2
2
In the last inequality we have used Lemma 3.3 and Lemma 3.4.
We now estimate the second term zγ ∂z U ∗ wj0 ∂vz hi . Notice that
2
X
(zγ ∂z U ∗ wj0 ∂vz P hi , (Aag )i ) = 0,
i=1
(1)
(2)
and zγ ∂z U ∗ wj0 decays fast as |z| → ∞, by (3.10). We thus split ag = ag + ag and use
Lemma 3.3 and Lemma 3.4 to get
2
X
|(zγ ∂z U ∗ wj0 ∂vz (1 − P )hi , (Aag )i )|
i=1
(1)
(2)
≤ Ckzγ− 1 ∂vz (I − P )hi k2M + kzγ− 1 ∂z ah k2 + kzγ− 1 ah k2 }
2
2
2
≤ C{khkD + k∂t hkD + khkD,γ− 1 + k∂t hkD,γ− 1 }.
2
2
Notice that we are applying Lemma 3.3 for a bound on a norm with index γ − 21 . For that
we need γ ≤ 12 + 81 . On the other hand,
2
X
=
i=1
2
X
(zγ ∂z U ∗ wj0 ∂vz hi ,
1
(I − P )gi )
M wi
(zγ ∂z U ∗ wj0 ∂vz (I − P )hi ,
i=1
+
2
X
(zγ ∂z U ∗ wj0 ∂vz P hi ,
i=1
31
1
(I − P )gi )
M wi
1
(I − P )gi ).
M wi
The first term is clearly bounded by εk(I − P )gk2M + Cε khk2D . For the second term,
2
X
(zγ ∂z U ∗ wj0 ∂vz P hi ,
1
(I − P )gi )
M wi
i=1
ah
≤ εk(I − P )gk2M + Cε k
k2
1 + |z|N
≤ εkgk2D + Cε {khk2D + k∂t hk2D },
by (3.10) for some large N .
Now turn to the third term Ĝhi in Γ. Since hĜhi , Aag i = 0,
2
X
1
(I − P )gi , Ĝ∂z ah )|
|
(
M wi
i=1
≤ εk(I − P )gk2M + Cε k∂t hk2D,γ− 1 + Cε khk2D,γ− 1 .
2
2
We now estimate the fourth term ∂z U ∗ wj0 ∂vz (zγ hi ). Since
0
i=1 (∂z U ∗ wj ∂vz (zγ hi ), (Aag )i ) = 0, we have
P2
2
X
1
(
(I − P )gi , ∂z U ∗ wj0 ∂vz (zγ hi ))
M
w
i
i=1
=
2
X
i=1
(
1
(I − P )gi , ∂z U ∗ wj0 ∂vz P (zγ hi ))
M wi
2
X
1
(I − P )gi , ∂z U ∗ wj0 ∂vz (I − P )(zγ hi ))
M wi
i=1
(1) 2
(2) 2
2
2
≤ εk(I − P )gkM + Cε k∂z ah k + kah k + khkD .
+
(
To get the last inequality we have used an argument similar to the one in the proof of
(3.10) in Lemma 3.3. Since w0 decays exponentially fast, we can divide the integration
(1)
over z of the term involving ah in two pieces: for z small we can use Poincare’ inequality
to bound in terms of the z derivative. For z large we use the decay of w0 . Then, by using
Theorem 3.2 we get the final bound
(1)
(2)
εk(I − P )gk2M + Cε k∂z ah k2 + kah k2 + khk2D ≤ εk(I − P )gk2M + Cε k∂t hk2D + khk2D
32
Now for the fifth term Fi (h)∂vz gi , we note that
2
X
(Fi (h)∂vz gi ,
i=1
2
X
P2
i=1 (Fi (h)∂vz gi , Aag )
= 0 and
1
(I − P )gi )
M wi
2
X
1
1
=
(Fi (h)∂vz P gi ,
(I − P )gi ) +
(Fi (h)∂vz (I − P )gi ,
(I − P )gi )
M wi
M wi
i=1
i=1
≤ CkFi (h)k∞ {kag k · k(I − P )gi kM + k∇v (I − P )gk2M }
≤ Ck∂z ah k{kag k · k(I − P )gi kM + k∇v (I − P )gk2M }
≤ Cδ0 {kgk2D + khk2D + k∂t hk2D }.
Now for the sixth term zγ Fi (∂z ah )∂vz hi , we note
Z
zγ Fi (∂z ah )∂vz hi Aag dv = 0.
We need to employ Lemma 3.4 to treat the last term as
2
X
(zγ Fi (∂z ah )∂vz hi ,
i=1
2
X
1
(I − P )gi )
M wi
2
X
1
1
=
(Fi (ag )∂vz hi ,
(I − P )gi ) +
([zγ , Fi ](∂z ah )∂vz hi ,
(I − P )gi )
M wi
M wi
i=1
i=1
≤ ka(2)
g k · kah k + kag k · khkD k(I − P )gkM
(3)
(3)
+ (k∂z {a(1)
−
z
a
}k
+
kz
a
k)
·
ka
k
+
ka
k
·
khk
k(I − P )gkM
γ
γ
h
g
D
g
h
h
(1)
+ kzγ− 1 ∂z ah k · kah k k(I − P )gkM
2
≤ Cδ0 {kgk2D + khk2D + k∂t,z hk2D,γ + khk2D,γ− 1 + k∂t hk2D,γ− 1 }.
2
2
We deduce our lemma by letting ε small and using δ0 small.
Proof of Theorem 1.2: To prove the first part, we start with γ = 0 in all three
Lemmas 4.2, 4.3 and 4.4. We multiply by a positive number K (4.2) and (4.4) and add
both to (4.5):
1 d
2
2
2
2
2
k∂z h(t)kM + K(kh(t)kM + k∂t h(t)kM ) + Kν0 kh(t)kD + k∂t h(t)kD + ν0 k∂z h(t)k2D
2 dt
≤ KC δ0 {k∂t h(t)k2D + kh(t)k2D } + δ0 k∂t,z h(t)k|2D + kh(t)k2D + Ckhk2D + Ck∂t hk2D
By choosing K >
C
ν0
, and δ0 <
, we obtain that
4ν0
4CK
1d
ν0
{k∂z hk2M + K k∂t hk2M + khk2M } + {k∂z hk2D + K k∂t hk2D + khk2D } ≤ 0.
2 dt
2
33
(4.6)
A standard continuity argument completes the first part.
To prove the second part, we first prove an inequality like (??) for the weighted norms
with weight zγ . Once again, we multiply (4.2) and (4.3) by K and add them to (4.4)
1 d
K(kh(t)k2M,γ0 +k∂t h(t)k2M,γ0 )+k∂z h(t)k2M,γ0 +Kν0 kh(t)k2D,γ0 +∂t h(t)k2D,γ0 +ν0 k∂z h(t)k|2D,γ0
2 dt
≤ KC{γ0 + δ0 }{k∂t h(t)k2D,γ0 + kh(t)k2D,γ0 + k∂t,z h(t)k|2D,γ0 − 1 } + KCkh(t)k2D,γ0 − 1 ,
2
2
+Cγ0 khk2D,γ0 − 1 + k∂t,z hk2D,γ0 − 1 + khk2D + k∂t hk2D + δ0 k∂t hk2D,γ0 .
2
2
For γ0 small enough and also γ0 < 1/2, for δ0 small enough and K sufficiently large we get
1 d
K(kh(t)k2M,γ0 +k∂t h(t)k2M,γ0 )+k∂z h(t)k2M,γ0 +Kν0 kh(t)k2D,γ0 +∂t h(t)k2D,γ0 +ν0 k∂z h(t)k|2D,γ0
2 dt
≤ KC(k∂t,z h(t)k|2D } + kh(t)k2D ),
Finally, we let γ = γ0 sufficiently small in Lemma 4.2, while let γ =
Lemmas 4.3 and 4.4, while multiplying the first two by K. We get
1
2
+ γ0 in both
1d
Kkh(t)k2M,γ0 + Kν0 kh(t)k2D,γ0 ≤ KC{γ0 + δ0 }{k∂t h(t)k2D,γ0 + kh(t)k2D,γ0 },
2 dt
1
1d
Kk∂t h(t)k2M,γ0 + 1 +Kν0 k∂t h(t)k2D,γ0 + 1 ≤ KC{γ0 + +δ0 }k∂t,z h(t)k|2D,γ0 +KCkh(t)k2D,γ0 .
2
2
2 dt
2
1d
k∂z h(t)k2M,γ0 + 1 + ν0 k∂z h(t)k|2D,γ0 + 1
2
2
2 dt
≤ Cγ khk2D,γ0 + k∂t,z hk2D,γ0 + khk2D + k∂t hk2D + δ0 k∂t hk2D,γ0 + 1 .
2
Then, there is a large constant K such that
d
{k∂z hkM, 1 +γ0 +Kk∂t, hk2M, 1 +γ0 +Kkhk2M,γ0 }+ν0 k∂z hkM, 1 +γ0 +ν0 K{k∂t hk2D, 1 +γ0 +khk2D,γ0 }
2
2
2
2
dt
≤ KC{k∂t,z hk2D + khk2D } + K{kh(0)k2M,γ0 + k∂t,z h(0)k2M,γ0 }.
Using the first part and a standard continuity argument, we obtain:
sup {kh(t)kM,γ0 + k∂t,z h(t)kM, 1 +γ0 } ≤ C{kh(0)kM,γ0 + k∂t,z h(0)kM, 1 +γ0 }.
0≤t≤∞
2
2
34
(4.7)
We now turn back to (4.6). We want to control khkM + k∂t,z hkM but up to now we
only have a uniform bound on khkD + k∂t,z hkD . What is missing is a bound on kah k. But
from (4.7) and an interpolation,
(1)
1
(1)
1
(1)
(1)
2γ0
kah k ≤ k(1 + |z|2 )1/2 ∂z ah k ≤ Ck(1 + |z|2 ) 2 +γ0 ∂z ah k 1+2γ0 × k∂z ah k 1+2γ0
2γ0
1
≤ C{kh(0)kγ0 + k∂t,z h(0)k 1 +γ0 } 1+2γ0 khkD1+2γ0 .
2
(2)
As for ah , and ∂t,z ah , by Lemma 3.2, we conclude that they satisfy the same inequality
above with γ0 = 0. Therefore, let Eγ0 = {kh(0)k2γ0 + k∂t,z h(0)k21 +γ0 }
2
− 2γ1
{khk2D + k∂t,z hk2D } ≥ CE0
0
{khk2 + k∂t,z hk2 }
1+2γ0
2γ0
.
We thus conclude that:
− 1
d
1+ 1
{k∂z hk2M + K(khk2M + k∂t hk2M )} + CE0 2γ0 {k∂z hk2M + K(khk2M + k∂t hk2M )} 2γ0 ≤ 0.
dt
Denoting y(t) ≡ k∂t,z hk2M + Kkhk2M , we have
y0y
−1− 2γ1
0
− 2γ1
≤ −CE0
0
.
Integrating over 0 and t, we deduce
− 1
1
1
− 1
− 1
{y(0)} 2γ0 −
{y(t)} 2γ0 ≤ −CE0 2γ0 t.
2γ0
2γ0
Hence from y(0) ≤ E0
1
− 1
{y(t)} 2γ0
2γ0
C − 2γ10
− 1
+ {y(0)} 2γ0
E0
2γ0
− 1
C
≥ {t
+ 1}E0 2γ0 .
2γ0
≥ t
Acknowledgments. The authors thank their institutions for support of the collaborations
in this project. R.M. and R. E. are supported by in part by MIUR, INDAM-GNFM, and
Y. G. is supported in part by NSF grant 0603615.
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