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The Schrödinger equation with spatial white noise potential
Arnaud Debussche, Hendrik Weber
To cite this version:
Arnaud Debussche, Hendrik Weber. The Schrödinger equation with spatial white noise potential.
2016. �hal-01411205�
The Schr¨ odinger equation with spatial white noise potential
Arnaud Debussche
IRMAR, ENS Rennes, UBL, CNRS
Hendrik Weber University of Warwick
Abstract
We consider the linear and nonlinear Schr¨odinger equation with a spatial white noise as a potential in dimension 2. We prove existence and uniqueness of solutions thanks to a change of unknown originally used in [8] and conserved quantities.
2010 Mathematics Subject Classification AMS:
Key words: Nonlinear Schr¨ odinger, white noise.
1 Introduction
In this work we study a linear or nonlinear Schr¨ odinger equation on a periodic domain with a random potential given by a spatial white noise in dimension 2. This equation is important for various purposes. In the linear case, it is used to study Anderson localisation.
It is a complex version of the famous PAM model. In the nonlinear case, it describes the evolution of nonlinear dispersive waves in a totally disorder medium (see for instance [6], [7] and the references therein).
If u denotes the unknown, the equation is given by:
du
dt = ∆u + λ|u|
2u + uξ, x ∈ T
2, t ≥ 0,
where T
2denotes the two dimensional torus, identified with [0, 2π]
2, and ξ is a real-valued spatial white noise. Of course, λ = 0 for the linear equation. A positive λ corresponds to the focusing case and λ < 0 to the defocusing case. For simplicity, we take λ = ±1 in the nonlinear case. The qualitative properties of the solutions are completely different in these two cases.
We are here interested in the question of existence and uniqueness of solutions. This is a preliminary but important step before studying other phenomena: solitary waves, blow up, Anderson localisation... The main difficulty is of course due to the presence of the rough potential. Recall that in dimension 2, a white noise has a negative regularity which is strictly less than −1. Apparently, since the Schr¨ odinger equation has very few smoothing properties and since this smoothing is very difficult to use, it seems hopeless to regularize it.
However, this equation has many other properties. In particular, it is Hamiltonian and
preserves the mass. Using a transformation due to [8] in the context of PAM, we are able to
use these invariants and construct solutions which have regularity strictly less than 1. More precisely, we solve a transformed equation in almost H
2( T
2), the standard Sobolev space of functions with derivatives up to order 2 in L
2( T
2). This is rather surprising since the regularity is comparable to what is obtained in the parabolic case, when strong smoothing properties are available.
As in the parabolic case, a renormalization is necessary and at the level of the original equation, the renormalized equation rewrites formally:
du
dt = ∆u + λ|u|
2u + u(ξ − ∞), x ∈ T
2, t ≥ 0.
The transformation u → e
i∞tu transforms the original equation into the renormalized one.
Therefore, the renormalization amounts to renormalize only the phase. A similar remark was made in [3] in a related but different context.
For λ < 0, we obtain global solution for any initial data satisfying some smoothness assumptions. For λ > 0, as in the determinstic case, we need a smallness assumption on the initial data.
We could of course consider the equation with a more general nonlinearity: |u|
2σu with σ ≤ 1. For σ < 1, no restriction on the size of the initial data is required for λ > 0.
Another easy generalization is to consider a general bounded domain and Dirichlet boundary conditions, as long as they are sufficiently smooth and the properties of the Green function of the Laplace operator are sufficiently good so that Lemma 2.1 below holds.
The study of the linear equation is closely related to the understanding of the Schr¨ odinger operator with white noise potential. This is the subject of a recent very interesting article by Allez and Chouk ([1]) where the paracontrolled calculus is used to study the domain and spectrum of this operator. It is not clear how this can be used for the nonlinear equation.
We use the classical L
p= L
p( T
2) spaces for p ∈ [1, ∞], as well as the L
2based Sobolev spaces H
s= H
s( T
2) for s ∈ R and the Besov spaces B
p,qs= B
p,qs( T
2), for s ∈ R , p, q ∈ [1, ∞].
These are defined in terms of Fourier series and Littlewood-Paley theory (see [2]). Recall that H
s= B
2,2sand that, for k ∈ N , s ∈ (0, 1), B
k+s∞,∞coincide with the H¨older space C
k,s( T
2).
Throughout the article, c denotes a constant which may change from one line to the next. Also, we use a small parameter 0 < ε < e
−1and K
εis a random constant which can also change but such that E K
εpis uniformly bounded in ε for all p. Similarly, for 0 < ε
1, ε
2< e
−1, the random constant K
ε1,ε2may depend on ε
1, ε
2but E K
εp1,ε2is uniformly bounded in ε
1, ε
2for all p.
2 Preliminaries
We consider the following nonlinear Schr¨ odinger equation in dimension 2 on the torus, that is periodic boundary conditions are assumed, for the complex-valued unknown u = u(x, t):
i du
dt = ∆u + λ|u|
2u + uξ, x ∈ T
2, t ≥ 0. (2.1)
It is supplemented with initial data
u(x, 0) = u
0(x), x ∈ T
2.
We need some smoothness on the initial data, this will be made precise below. In the focusing case λ > 0 we need an extra assumption on the size of ku
0k
L2which has be small enough (see (4.3) below).
The potential ξ is random and is a real valued spatial white noise on T
2. For simplicity, we assume that it has a zero spatial average. The general case could be recovered by adding an additional Gaussian random potential which is constant in space. This would not change the analysis below.
Formally equation (2.1) has two invariant quantities. Given a solution u of (2.1), the mass:
N (u(t)) = Z
T2
|u(x, t)|
2dx is constant in time as well as the energy:
H(u(t)) = Z
T2
1
2 |∇u(x, t)|
2− λ
4 |u(x, t)|
4− 1
2 ξ(x)|u(x, t)|
2dx.
This is formal because the noise ξ is very rough. In dimension 1, the noise has regularity
−1/2
−and belongs to B
∞,∞αfor any α < −1/2, therefore the product ξ|u|
2can be defined rigorously for u ∈ H
1and this provides a bound in H
1. Existence and uniqueness through regularization of the noise and a compactness argument can then be obtained.
In dimension 2, the noise lives in any space with regularity −1
−, that is any regularity strictly less than −1, and the solution is not expected to be sufficiently smooth to compen- sate this. In fact, the product is almost well defined for u ∈ H
1and we are in a situation similar to the two dimensional nonlinear heat equation with space time white noise. We expect that a renormalization is necessary.
Inspired by [8], we introduce:
Y = ∆
−1ξ
(note that this is well defined since we consider a zero average noise, we choose Y also with a zero average) and v = ue
Y. Then the equation for v reads
i dv
dt = ∆v − 2∇v · ∇Y + v|∇Y |
2+ λ|v|
2ve
−2Y. (2.2)
Now Y has regularity 1
−and ∇Y is 0
−. Thus this transformation has lowered the roughness of the most irregular term on the right hand side. At this point It is easier to see why we need a renormalization: the term |∇Y |
2is not well defined since ∇Y is not a function. However, the roughness is mild here and it has been known for long that up to renormalization by a log divergent constant this square term can be defined in the second order Wiener chaos based on ξ.
Let us be more precise. Let ρ
ε= ε
−2ρ(
ε·) be a compactly supported smooth mollifier
and consider the smooth noise ξ
ε= ρ
ε∗ ξ. We denote by Y
ε= ∆
−1ξ
ε. Then it is proved in
[8] that for every κ > 0, ξ belongs almost surely to B
∞,∞−1−κand, as ε → 0, ξ
εconverges in probability to ξ in B
∞,∞−1−κ.
Also, denoting by
C
ε= E |∇Y
ε|
2the quantity : |∇Y
ε|
2:= |∇Y
ε|
2− C
εconverges in L
p(Ω; B
∞,∞−κ) for any p ≥ 1, κ > 0 to a random variable : |∇Y
ε|
2: in the second Wiener chaos associated to ξ. It is easy to see that C
εgoes to ∞ as | ln ε| as ε → ∞:
E |∇Y
ε|
2∼ K
0| ln ε|
for some K
0> 0. By stationarity this quantity does not depend on x ∈ T
2.
The precise result, whose proof can be found in [8] in the more difficult case of the space variable in R
2(see the proofs of Lemma 1.1 and Proposition 1.3 in this work), is the following.
Lemma 2.1. 1 ≥ κ
′> κ > 0 and any p ≥ 1, there exist a constant c independent of ε such that:
E
kY
ε− Y k
pB1∞−,κ∞′
1p≤ cε
κ−2pand
E
k : |∇Y
ε|
2: − : |∇Y |
2|k
pB∞−κ,∞′
1p≤ cε
κ−p2.
Remark 2.2. Using the monotonicity of stochastic L
pnorms in p, one can drop the expo- nent −
2pin the right hand side. We state the result in this way because this is the bound that one actually proves. Below, we use this bound with κ −
2p=
κ2′(and κ instead of κ
′).
Note that for s < ˜ s, p, r ≥ 1, we have B
∞,∞˜s⊂ B
p,rs. Thus, bounds in the latter Besov spaces follow.
Instead of solving equation (2.2) for v
ε, we consider:
i dv
εdt = ∆v
ε− 2∇v
ε· ∇Y
ε+ v
ε: |∇Y
ε|
2: +λ|v
ε|
2v
εe
−2Yε(2.3) and setting u
ε= v
εe
−Yε:
i du
εdt = ∆u
ε+ λ|u
ε|
2u
ε+ (u
ε− C
ε)ξ
ε, x ∈ T
2, t ≥ 0.
Since Y
εis smooth, it is classical to prove that these equations have a unique solution in C([0, T ]; H
k) for an initial data in H
k, k = 1, 2, provided the L
2norm is small for λ > 0 (see for instance [5], Section 3.6). More details are given in Section 4.
The mass and energy are transformed into the two following quantities which are invari- ant under the dynamic for v
ε:
N ˜
ε(v
ε(t)) = Z
T2
|v
ε(x, t)|
2e
−2Yε(x)dx
and
H ˜
ε(v
ε(t)) = Z
T2
1
2 |∇v
ε(x, t)|
2+ 1
2 v
2ε: |∇Y
ε|
2: − λ
4 |v
ε(x, t)|
4e
−2Yε(x)e
−2Yε(x)dx.
Since the most irregular term : |∇Y
ε|
2: here is not as rough as ξ, this transformed energy is a much better quantity than the original one. It is possible to give a meaning to it for ε = 0 and use it to get bounds in H
1.
Below, we use the following simple results.
Lemma 2.3. For any κ ∈ (0, 1) and any p ≥ 1, there exist a constant independent on ε such that:
E
ke
−2Yε− e
−2Yk
pB∞1−,κ∞
≤ cε
−2p+κ.
Proof: Since B
1−κ∞,∞is equal to the H¨older space C
1−κ( T
2) we have:
ke
−2Yε− e
−2Yk
B1−κ∞,∞
= ke
−2Y(e
−2(Yε−Y)− 1)k
B1−κ∞,∞
≤ ke
−2Yk
B1−κ∞,∞
ke
−2(Yε−Y)− 1k
B1−κ∞,∞
. Then we write:
ke
−2Yk
B1−κ∞,∞
≤ 2ke
−2Yk
L∞kY k
B1−κ∞,∞
, ke
−2(Yε−Y)− 1k
B1−κ∞,∞
≤ 2ke
−2Yk
L∞ke
−2Yεk
L∞kY
ε− Y k
B1−κ∞,∞
The result follows by H¨older inequality, Lemma 2.1 and Gaussianity to bound exponential
moments of Y and Y
ε.
Lemma 2.4. There exists a contant c independent of ε such that:
E k∇Y
εk
4L4≤ c| ln ε|
2and
E k : |∇Y
ε|
2: k
4L4≤ c(| ln ε|)
4. Proof: It suffices to write:
E Z
T2
|∇Y
ε(x)|
4dx
= Z
T2
E |∇Y
ε(x)|
4dx = 12π
2C
ε2. Similarly:
E Z
T2
| : |∇Y
ε(x)|
2: |
4dx
= E Z
T2
(|∇Y
ε(x)|
2− C
ε)
4dx
= 51πC
ε4.
3 The linear case
In this section, we start with the linear case: λ = 0. Then the equation for v
εreads i dv
εdt = ∆v
ε− 2∇v
ε· ∇Y
ε+ v
ε: |∇Y
ε|
2: . (3.1)
There exists a unique solution in C([0, T ]; H
2) if v
ε(0) ∈ H
2. We take the initial data v
ε(0) = v
0= u
0e
Yand assume below that it belongs to H
2. The mass and energy of a solution are now:
N ˜
ε(v
ε(t)) = Z
T2
|v
ε(x, t)|
2e
−2Yε(x)dx and
H ˜
ε(v
ε(t)) = Z
T2
1
2 |∇v
ε(x, t)|
2+ 1
2 v
ε2: |∇Y
ε|
2:
e
−2Yε(x)dx.
They are constant in time under the evolution (3.1).
Since Y
εconverges in B
1−κ∞,∞for any κ > 0 as ε tends to zero, we see that the mass gives a uniform bound in L
2on v
ε. More precisely:
kv
ε(t)k
2L2≤ ke
2Yεk
L∞ke
−2Yεk
L∞kv
0k
2L2= K
εkv
0k
2L2(3.2) with
K
ε= ke
2Yεk
L∞ke
−2Yεk
L∞. (3.3)
The energy enables us to get a bound on the gradient.
Proposition 3.1. Let κ ∈ (0, 1/2), there exists a random constant K
εbounded in L
p(Ω) with respect to ε for any p ≥ 1 such that if v
0∈ H
1:
Z
T2
|∇v
ε(x, t)|
2dx ≤ K
εH ˜
ε(v
0) + kv
0k
2L2.
Proof: Since B
∞,2−κis in duality with B
1,2κwe deduce by the standard multiplication rule in Besov spaces (see e.g. [2], Section 2.8.1)
Z
T2
v
ε2: |∇Y
ε|
2: dx
≤ kv
ε2k
Bκ1,2k : |∇Y
ε|
2: k
B−κ∞,2
≤ K
εkv
2εk
B1,2κ≤ K
εkv
εk
2Bκ 2,2. Then we note that kv
εk
2Bκ2,2
= kv
εk
2Hκso that by interpolation
Z
T2
v
2ε: |∇Y
ε|
2: dx
≤ K
εkv
εk
2(1−κ)L2kv
εk
2κH1.
It follows Z
T2
|∇v
ε(x, t)|
2dx ≤ K
εH(v ˜
ε(t)) + K
εkv
ε(t)k
2(1−κ)L2kv
ε(t)k
2κH1≤ K
εH ˜
ε(v
0) + K
εkv
ε(t)k
2L2+ K
εkv
ε(t)k
2(1−κ)L2k∇v
εk
2κL2≤ K
εH ˜
ε(v
0) + K
εkv
0k
2L2+ 1
2 k∇v
εk
2L2and hence, by absorbing the last term in the left hand side, Z
T2
|∇v
ε(x, t)|
2dx ≤ K
εH ˜
ε(v
0) + kv
0k
2L2.
Since ˜ H
ε(v
0) is bounded for v
0∈ H
1, we obtain a (random) bound on v
εin H
1using similar arguments as above.
Corollary 3.2. There exists a random constant K
εbounded in L
p(Ω) with respect to ε for any p ≥ 1 such for any v
0∈ H
1:
kv
ε(t)k
H1≤ K
εkv
0k
H1, t ≥ 0.
Unfortunately, this regularity is not sufficient to control the product ∇v
ε· ∇Y
εon the right hand side of (3.1).
The next observation is that, if v
0is smooth enough, v
εis time differentiable and setting w
ε=
dvdtεit is a solution of:
i dw
εdt = ∆w
ε− 2∇w
ε· ∇Y
ε+ w
ε: |∇Y
ε|
2: . (3.4)
Since w
εsatisfies the same equation as v
ε, it has the same invariant quantities. We use in particular the mass:
N ˜ (w
ε(t)) = ˜ N (w
ε(0)).
Hence:
kw
ε(t)k
L2≤ K
εN ˜ (w
ε(0)) ≤ K
εkw
ε(0)k
L2.
This is still true under the assumption that w
ε(0) ∈ L
2, which is equivalent to v
0∈ H
2. Proposition 3.3. There exist a random constant K
εbounded in L
p(Ω) with respect to ε for any p ≥ 1 such that if v
0∈ H
2:
kv
εk
H2≤ cK
εkv
0k
H2+ kv
0k
L2| ln ε|
2. Proof: From (3.1), we have:
w
ε(0) = −i(∆v
0− 2∇v
0· ∇Y
ε+ v
0: |∇Y
ε|
2:), so that, thanks to the embedding H
1/2⊂ L
4,
kw
ε(0)k
L2≤ c kv
0k
H2+ kv
0k
H3/2k∇Y
εk
L4+ kv
0k
H1/2k : |∇Y
ε|
2: k
L4.
By interpolation we deduce:
kw
ε(0)k
L2≤ c
kv
0k
H2+ kv
0k
3/4H2kv
0k
1/4L2k∇Y
εk
L4+ kv
0k
1/4H2kv
0k
3/4L2k : |∇Y
ε|
2: k
L4≤ c
kv
0k
H2+ kv
0k
L2k∇Y
εk
4L4+ kv
0k
L2k : |∇Y
ε|
2: k
4/3L4≤ c kv
0k
H2+ K
εkv
0k
L2| ln ε|
2,
(3.5) where:
K
ε= k∇Y
εk
4L4| ln ε|
−2+ k : |∇Y
ε|
2: k
4/3L4| ln ε|
−2.
By Lemma 2.4 and gaussianity, we know that the moments of this random variable are bounded with respect to ε.
It follows
kw
ε(t)k
L2≤ K
εkv
0k
H2+ kv
0k
L2| ln ε|
2. This in turn allows us to control kv
εk
H2. Indeed, from (3.1),
k∆v
εk
L2≤ kw
ε(t)k
L2+ 2k∇v
ε· ∇Y
εk
L2+ kv
ε: |∇Y
ε|
2: k
L2and by similar arguments as above
k∆v
εk
L2≤ kw
ε(t)k
L2+ 1
2 k∆v
εk
L2+ cK
εkv
ε(t)k
L2| ln ε|
2and finally
k∆v
εk
L2≤ K
εkv
0k
H2+ kv
0k
L2| ln ε|
2.
The result follows thanks to (3.2).
This bound does not seem to be very useful since it explodes as ε → 0. To use it, we consider the difference of two solutions.
Proposition 3.4. Let ε
2> ε
1> 0 then for κ ∈ (0, 1], p ≥ 1 there exist a random constant K
ε1,ε2bounded in L
p(Ω) with respect to ε
1, ε
2for any p ≥ 1 such that if v
0∈ H
2sup
t∈[0,T]
kv
ε1(t) − v
ε2(t)k
2L2≤ K
ε1,ε2ε
κ/22| ln ε
2|
1+2κkv
0k
2H2Proof: We set r = v
ε1− v
ε2and write:
i dr
dt = ∆r − 2r · ∇Y
ε1+ r : |∇Y
ε1|
2: −2∇v
ε2· ∇(Y
ε1− Y
ε2) + v
ε2(: |∇Y
ε1|
2: − : |∇Y
ε2|
2:).
By standard computations, we deduce:
1 2
d dt
Z
T2
|r(x, t)|
2e
−2Yε1(x)dx
= Im Z
T2
−2∇v
ε2· ∇(Y
ε1− Y
ε2) + v
ε2(: |∇Y
ε1|
2: − : |∇Y
ε1|
2:)
¯
re
−2Yε1(x)dx
≤ 2k∇v
ε2re ¯
−2Yε1k
Bκ1,2k∇(Y
ε1− Y
ε2)k
B−κ∞,2
+ kv
ε2re ¯
−2Yε1k
Bκ1,2k : |∇Y
ε1|
2: − : |∇Y
ε2|
2: k
B−κ∞,2
the first term of the right hand side is bounded thanks to interpolation and paraproduct inequalities (see [2]) and we have
k∇v
ε2re ¯
−2Yε1k
B1,2κ≤ ckv
ε2k
H1+κ(kv
ε1k
Hκ+ kv
ε2k
Hκ) ke
−2Yε1k
B∞κ,2. Then, by Proposition 3.3 and interpolation:
kv
ε2k
H1+κ≤ ckv
ε2k
(1+κ)/2H2kv
ε2k
(1−κ)/2L2≤ K
ε2(kv
0k
H2+ kv
0k
L2| ln ǫ
2|
2)
(1+κ)/2kv
0k
(1−κ)/2L2and, by Corollary 3.2, for i = 1, 2
kv
εik
Hκ≤ kv
εik
κH1kv
εik
1−κL2≤ K
εikv
0k
κH1kv
0k
1−κL2≤ K
εikv
0k
κ/2H2kv
0k
1−κ/2L2. It follows
k∇v
ε2¯ re
−2Yε1k
B1,2κ≤ K
ǫ2ke
−2Yε1k
B∞κ,2kv
0k
L32−κ2(kv
0k
H2+ kv
0k
L2| ln ǫ
2|
2)
12+κThe second term is bounded by the same quantity and we deduce:
1 2
d dt
Z
T2
|r(x, t)|
2e
−2Yε1(x)dx
≤ K
ǫ2ke
−2Yε1k
B∞κ,2kv
0k
3 2−κ
L2
(kv
0k
H2+ kv
0k
L2| ln ǫ
2|
2)
12+κ×
k∇(Y
ε1− Y
ε2)k
B−κ∞,2
+ k : |∇Y
ε1|
2: − : |∇Y
ε2|
2: k
B−κ∞,2
.
The result follows thanks to integration in time and Lemma 2.1.
By interpolation, we deduce from Proposition 3.3 and 3.4 the following result.
Corollary 3.5. Let ε
2> ε
1> 0 then for κ ∈ (0, 1], γ ∈ [0, 2), p ≥ 1 there exist a random constant K
ε1,ε2bounded in L
p(Ω) with respect to ε
1, ε
2for any p ≥ 1 such that if v
0∈ H
2:
sup
t∈[0,T]
kv
ε1(t) − v
ε2(t)k
2Hγ≤ K
ε1,ε2ε
κ 2(1−γ2)
2
| ln ε
1|
4kv
0k
2H2We are now ready to state and prove the main result of this section.
Theorem 3.6. Assume that v
0= u
0e
Y∈ L
p(Ω; H
2) for some p > 1. For any T ≥ 0, q < p, γ ∈ (1, 2), when ε → 0, the solution v
ǫof (3.1) satisfying v
ε(0) = v
0converges in L
q(Ω; C([0, T ]; H
γ)) to v which is the unique solution to
i dv
dt = ∆v − 2∇v · ∇Y + v : |∇Y |
2: (3.6)
in C([0, T ]; H
γ) such that v(0) = v
0.
Proof: Pathwise uniqueness is clear. Indeed if v ∈ C([0, T ]; H
γ) is a solution of (3.6), it is not difficult to use a regularization argument to justify
d dt
Z
T2
|v(x, t)|
2e
−2Y(x)dx = 0.
Now let ε
k= 2
−k, Corollary 3.5 implies that (v
εk) is Cauchy in L
q(Ω; C([0, T ]; H
γ)). It is not difficult to prove that the limit v is a solution of (3.6) and
E sup
t∈[0,T]
kv
εk(t)k
qHγ!
≤ c E kv
0k
pH2 q/p,
E sup
t∈[0,T]
kv(t)k
qHγ!
≤ c E kv
0k
pH2 q/p. Then, by interpolation with γ < ˜ γ < 2:
sup
t∈[0,T]
kv
ε(t) − v
εk(t)k
2Hγ≤ c sup
t∈[0,T]
kv
ε(t) − v
εk(t)k
2(1−γ/˜L2 γ)sup
t∈[0,T]
kv
ε(t) − v
εk(t)k
2γ/˜H˜γγ≤ c sup
t∈[0,T]
kv
ε(t) − v
εk(t)k
2(1−γ/˜L2 γ)sup
t∈[0,T]
(kv
ε(t)k
H2+ kv
εk(t)k
H˜γ)
2γ/˜γ. By Proposition 3.4, Proposition 3.3 and the above inequality, we deduce:
E sup
t∈[0,T]
kv
ε(t) − v
εk(t)k
qHγ!
≤ cε
κ2(2−γ/˜γ))| ln ε|
4E kv
0k
pH2 q/p.
Letting k → ∞, we deduce that the whole family (v
ε)
ε>0converges to v in L
q(Ω; C([0, T ]; H
γ)).
4 The nonlinear equation
We now study the nonlinear equation (2.2) and consider its approximation (2.3) with initial condition
v
ε(0) = v
0= u
0e
Yand assume below that it belongs to H
2.
Again the mass gives a uniform bound in L
2on v
ε:
kv
ε(t)k
2L2≤ K
εkv
0k
2L2. (4.1)
The estimate on the H
1norm using the energy is similar to the linear case. Recall that the energy is given by:
H ˜
ε(v
ε(t)) = Z
T2
1
2 |∇v
ε(x, t)|
2+ 1
2 v
2ε: |∇Y
ε|
2: − λ
4 |v
ε(x, t)|
4e
−2Yε(x)e
−2Yε(x)dx
and it can be checked that for all t ≥ 0 we have ˜ H
ε(v
ε(t)) = ˜ H
ε(v
0).
Proposition 4.1. There exists a constant K
εbounded in L
p(Ω) for any p ≥ 1 such that if v
0∈ H
1and
ke
−2Yεk
3L∞ke
2Yεk
L∞kv
0k
L2≤ 1 if λ = 1 (4.2) then
kv
ε(t)k
2H1≤ K
εkv
0k
2H1+ kv
0k
2L2kv
0k
2H1. Moreover, the sequence (K
k) = (K
2−k) is bounded almost surely.
Remark 4.2. It is classical that the equation for u
εis locally well posed in H
1, see [5]
Section 3.6. The bound obtained here gives a global bound in H
1for u
εso that global existence and uniqueness hold for u
εand thus for v
ε.
Proof: We proceed as in the proof of Proposition 3.1. We first have
Z
T2
v
ε2: |∇Y
ε|
2: dx
≤ K
εkv
εk
2(1−κ)L2kv
εk
2κH1and Z
T2
|∇v
ε(x, t)|
2dx
≤ K
εH(v ˜
ε(t)) + K
εkv
ε(t)k
2(1−κ)L2kv
ε(t)k
2κH1+ λ 4
Z
T2
|v
ε(x, t)|
4e
−2Yε(x)e
−2Yε(x)dx
≤ K
εH ˜
ε(v
0) + K
εkv
0k
2L2+ 1
2 k∇v
εk
2L2+ λ 4
Z
T2
|v
ε(x, t)|
4e
−2Yε(x)e
−2Yε(x)dx.
For λ = −1, the result follows after dropping the last term and using Z
T2
|v
0(x)|
4e
−4Yε(x)dx ≤ K
εkv
0k
4L4≤ K
εkv
0k
4H1/2≤ K
εkv
0k
2L2kv
0k
2H1thanks to the Sobolev embedding H
1/2⊂ L
4and interpolation.
For λ = 1, Gagliardo-Nirenberg inequality (see for instance [4] for a simple proof with the constant 1/2 used below):
Z
T2
|v
ε(x, t)|
4e
−4Yε(x)dx ≤ ke
−2Yεk
2L∞Z
T2
|v
ε(x, t)|
4dx
≤ 1
2 ke
−2Yεk
2L∞Z
T2
|v
ε(x, t)|
2dx Z
T2
|∇v
ε(x, t)|
2dx
≤ 1
2 ke
−2Yεk
3L∞ke
2Yεk
L∞kv
0k
2L2Z
T2
|∇v
ε(x, t)|
2dx,
where we have made use of kv
ε(t)k
2L2≤ ke
2Yεk
L∞ke
−2Yεk
L∞kv
0k
2L2according to (3.2). The result follows easily under assumption (4.2).
The constant K
εis a polynomial in k : |∇Y
ε|
2: k
B−κ∞,2
, ke
−2Yεk
L∞and ke
2Yεk
L∞. By Lemma 2.1 and Lemma 2.3 and Borel-Cantelli, we know that : |∇Y
2−k|
2: and Y
2−kconverge almost surely in B
∞,∞−κand B
∞,∞1−κso that K
kis indeed bounded almost surely.
We now proceed with the H
2bound.
Proposition 4.3. There exist a random constants K
εbounded in L
p(Ω) with respect to ε for any p ≥ 1 such that if v
0= u
0e
−Y∈ H
2and (4.2) holds:
kv
ε(t)k
H2≤ cK
ε1 + kv
0k
H2+ kv
0k
L2| ln ε|
4+ kv
0k
3H1+ kv
0k
3L2kv
0k
3H1 exp(Kε
kv0k2H1+kv0k2L2kv0k2H1 t)
. Moreover K
k= K
2−kis bounded almost surely.
Remark 4.4. We know that if u
ε(0) ∈ H
2, the solution u
εlives in H
2(see [4]). Since u
ε(0) = v
0e
−Yε, we can apply this result.
Proof: As in Section 3, we set w
ε=
dvdtεwhich now satisfies:
i dw
εdt = ∆w
ε− 2∇w
ε· ∇Y
ε+ w
ε: |∇Y |
2: +λ |v
ε|
2w
ε+ 2Re(v
εw ¯
ε)v
εe
−2Yε. From (2.3), we have:
w
ε(0) = −i(∆v
0− 2∇v
0· ∇Y
ε+ v
0: |∇Y
ε|
2:) + λ|v
0|
2v
0e
−2Yε, and as in Proposition 3.3 and using the embedding H
1⊂ L
6:
kw
ε(0)k
L2≤ ckv
0k
H2+ kv
0k
L2k∇Y
εk
4L4+ k : |∇Y
ε|
2: k
4/3L4+ ke
−2Yεk
L∞kv
0k
3H1. By Lemma 2.4:
P (k∇Y
εk
4L4+ k : |∇Y
ε|
2: k
4/3L4≥ | ln ε|
4) ≤ c| ln ε|
−2it follows that
kw
ε(0)k
L2≤ ckv
0k
H2+ K
εkv
0k
L2| ln ε|
4+ K
εkv
0k
3H1with K
εhaving all moments finite and such that K
2−kis almost finite by Borel-Cantelli.We have taken | ln ε|
4instead of | ln ε|
2in the estimate above in order to have this latter property.
Recall that Y
2−kconverges a.s. in L
∞.
We do not have preservation of the L
2norm but:
1 2
d dt
Z
T2
|w
ε(x, t)|
2e
−2Yε(x)dx
= 2λ Z
T2
Re(v
ε(x, t) ¯ w
ε(x, t))Im(v
ε(x, t) ¯ w
ε(x, t))e
−4Yε(x)dx
≤ K
εkv
ε(t)k
2L∞Z
T2
|w
ε(x, t)|
2e
−2Yε(x)dx
≤ K
εkv
ε(t)k
2H1(1 + ln(1 + kv
ε(t)k
H2)) Z
T2
|w
ε(x, t)|
2e
−2Yε(x)dx
≤ K
εkv
0k
2H1+ kv
0k
2L2kv
0k
2H1(1 + ln(1 + kv
ε(t)k
H2)) Z
T2
|w
ε(x, t)|
2e
−2Yε(x)dx
thanks to the Brezis-Gallouet inequality ([4]) and to Proposition 4.1. Then as above we have:
k∆v
ε(t)k
L2≤ 2kw
ε(t)k
L2+ cK
εkv
ε(t)k
L2| ln ε|
4+ λk|v
ε(t)|
2v
εe
−2Yεk
L2≤ 2kw
ε(t)k
L2+ K
εkv
ε(t)k
L2| ln ε|
4+ K
εkv
ε(t)k
3L6. By the embedding H
1⊂ L
6and Proposition 4.1, we deduce
kv
ε(t)k
H2≤ ckw
ε(t)k
L2+ K
εkv
0k
L2| ln ε|
4+ kv
0k
3H1+ kv
0k
3L2kv
0k
3H1.
Again, the almost sure boundedness of the different constant K
kis obtained thanks to Lemma 2.3, Lemma 2.4 and Borel-Cantelli.
To lighten the following computation, we use the temporary notations:
˜
w
ε= kw
εe
−Yεk
2L2, α
ε= K
εkv
0k
2H1+ kv
0k
2L2kv
0k
2H1, β
ε= K
εkv
0k
L2| ln ε|
4+ kv
0k
3H1+ kv
0k
3L2kv
0k
3H1. Then we have:
d
dt w ˜
ε≤ α
ε(1 + ln(1 + kv
εk
H2)) ˜ w
ε≤ α
ε(1 + ln(1 + ckw
εk
L2+ β
ε)) ˜ w
ε≤ α
ε(1 + ln(1 + K
εw ˜
ε+ β
ε)) ˜ w
ε. Hence
d
dt (1 + ln(1 + K
εw ˜
ε(t) + β
ε)) ≤ α
ε(1 + ln(1 + K
εw ˜
ε(t) + β
ε)).
By Gronwall’s Lemma we deduce:
1 + ln(1 + K
εw ˜
ε(t) + β
ε) ≤ (1 + ln(1 + K
εw ˜
ε(0) + β
ε)) exp(α
εt) and taking the exponential
kw
ε(t)k
L2≤ K
εw ˜
ε(t) ≤ cK
ε(1 + K
εw ˜
ε(0) + β
ε)
exp(αεt)≤ K
ε(1 + K
εkw
ε(0)k
L2+ β
ε)
exp(αεt)≤ K
ε(1 + K
εkv
0k
H2+ β
ε)
exp(αεt).
The result follows.
We see that this H
2bound is not as good as in the linear case. Due to the double exponential, we do not have moments here. However, this is sufficient to prove existence and uniqueness. We now state the main result of this section.
Theorem 4.5. Assume that v
0= u
0e
Y∈ H
2and
ke
−2Yk
3L∞ke
2Yk
L∞kv
0k
L2< 1 if λ = 1. (4.3)
For any T ≥ 0, p ≥ 1, γ ∈ (1, 2), when ε → 0, the solution v
ǫof (3.1) satisfying v
ε(0) = v
0converges in probability in C([0, T ]; H
γ)) to v which is the unique solution to i dv
dt = ∆v − 2∇v · ∇Y + v : |∇Y
2: +λ|v|
2ve
−2Y(4.4)
with paths in C([0, T ]; H
γ) such that v(0) = v
0. Proof: Again, pathwise uniqueness is easy.
Under assumption (4.3), we know that (4.2) holds for ε small enough. Thus we may use Propositions 4.1 and 4.3.
We take ε
2> ε
1> 0, set r = v
ε1− v
ε2and write:
i dr
dt = ∆r − 2r · ∇Y
ε1+ r : |∇Y
ε1|
2: −2∇v
ε2· ∇(Y
ε1− Y
ε2) + v
ε2(: |∇Y
ε1|
2: − : |∇Y
ε2|
2:) +λ|v
ε1|
2re
−2Yε1− λ(|v
ε2|
2− |v
ε1|
2)v
ε2e
−2Yε1+ λ|v
ε2|
2v
ε2(e
−2Yε1− e
−2Yε2).
By the same arguments as in Section 3 and standard estimates we obtain;
1 2
d dt
Z
T2
|r(x, t)|
2e
−2Yε1(x)dx
≤ cke
−2Yε1k
B∞κ,2kv
ε2(t)k
H1+κ(kv
ε1(t)k
Hκ+ kv
ε1(t)k
Hκ)
×
k∇(Y
ε1− Y
ε2)k
B−κ∞,2
+ k : |∇Y
ε1|
2: − : |∇Y
ε2|
2: k
B−κ∞,2
+K
ε1(kv
ε1k
L∞+ kv
ε2k
L∞)kv
ε2k
L∞Z
T2
|r(x, t)|
2e
−2Yε1(x)dx
+K
ε1kv
ε2k
3L∞(kv
ε1k
L2+ kv
ε2k
L2)ke
−2Yε1− e
−2Yε2k
L2≤ K
ε1,ε2kv
ε2(t)k
κH2(kv
ε1(t)k
H1+ kv
ε1(t)k
H1)
2−κε
κ/22+K
ε1(kv
ε1k
H1+ kv
ε2k
H1)
2(1 + ln(1 + kv
ε1k
H2) + ln(1 + kv
ε2k
H2))
× Z
T2